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  <row Id="1" PostTypeId="1" AcceptedAnswerId="879" CreationDate="2013-09-24T18:51:54.593" Score="36" ViewCount="1097" Body="&lt;p&gt;Suppose I would like to calculate the inclination of a satellite from the &lt;strong&gt;ecliptic&lt;/strong&gt;. Would it be possible to do this with an amateur telescope? How would I go about doing so?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note: A good answer should tell what kind of telescope an amateur would need, what measurements they would need to make, then what calculations they would have to perform to get the inclination (or the &quot;instantaneous angular measurement from the ecliptic at the time of measurement&quot;).&lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorUserId="6" LastEditDate="2013-11-01T11:23:02.907" LastActivityDate="2014-02-03T16:49:55.200" Title="How do I calculate the inclination of an object with an amateur telescope?" Tags="&lt;inclination&gt;&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2" PostTypeId="1" AcceptedAnswerId="45" CreationDate="2013-09-24T18:53:25.647" Score="24" ViewCount="265" Body="&lt;p&gt;How are exoplanetary atmosphere compositional spectra distinguished from those of the parent star(s), from the composition of the planetary surface or any other factor?  Is it actually possible to determine the atmospheric composition precisely using this method?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any specific examples of such an analysis having been performed on an exoplanet?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2020-12-04T17:02:08.540" Title="How are the compositional components of exoplanet atmospheres differentiated?" Tags="&lt;exoplanet&gt;&lt;planetary-atmosphere&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3" PostTypeId="1" CreationDate="2013-09-24T18:54:53.473" Score="28" ViewCount="945" Body="&lt;p&gt;Are there any easily resolvable, binary star, observing targets visible from ~N40°?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to be able to show, in one observing session, Mizar and Alcor (naked eye resolvable), another with binoculars, and finally one in a small (say, 4 to 6&quot; reflector) amateur scope.&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="7" LastEditDate="2013-09-24T19:00:42.437" LastActivityDate="2014-01-14T15:45:53.260" Title="Amateur observing targets for binary star systems?" Tags="&lt;amateur-observing&gt;&lt;binary-star&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4" PostTypeId="1" AcceptedAnswerId="5" CreationDate="2013-09-24T18:57:01.680" Score="28" ViewCount="22485" Body="&lt;p&gt;Sunspots, such as this one, appear dark:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VM9qym.jpg&quot; alt=&quot;sunspot&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="9527" LastEditDate="2020-06-02T10:41:14.250" LastActivityDate="2021-05-26T21:24:02.137" Title="Why do sunspots appear dark?" Tags="&lt;the-sun&gt;&lt;sunspots&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5" PostTypeId="2" ParentId="4" CreationDate="2013-09-24T18:57:01.680" Score="27" Body="&lt;p&gt;Typical sunspots have a dark region (umbra) surrounded by a lighter region, the penumbra. While sunspots have a temperature of about 6300 °F (3482.2 °C), the surface of the sun which surrounds it has a temperature of 10,000 °F (5537.8 °C).&lt;/p&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;https://image.gsfc.nasa.gov/poetry/workbook/sunspot.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this NASA resource&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Sunspots are actually regions of the solar surface where the magnetic field of the Sun becomes concentrated over 1000-fold. Scientists do not yet know how this happens. Magnetic fields produce pressure, and this pressure can cause gas inside the sunspot to be in balance with the gas outside the sunspot...but at a lower temperature. Sunspots are actually several thousand degrees cooler than the 5,770 K (5496.8 °C) surface of the Sun, and contain gases at temperature of 3000 to 4000 K (2726.9 - 3726.8 °C). They are dark only by contrast with the much hotter solar surface. &lt;strong&gt;If you were to put a sunspot in the night sky, it would glow brighter than the Full Moon with a crimson-orange color!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Sunspots are areas of intense magnetic activity, as is apparent in this image:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://image.gsfc.nasa.gov/poetry/workbook/sunspot1.jpg&quot; alt=&quot;sunspot&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;You can see the material kind of getting stretched into strands.&lt;/p&gt;&#xA;&lt;p&gt;As for the &lt;em&gt;reason&lt;/em&gt; it is cooler than the rest of the surface:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Although the details of sunspot generation are still a matter of research, it appears that sunspots are the visible counterparts of &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetic_flux_tube&quot; rel=&quot;nofollow noreferrer&quot;&gt;magnetic flux tubes&lt;/a&gt; in the Sun's &lt;a href=&quot;http://en.wikipedia.org/wiki/Convection_zone&quot; rel=&quot;nofollow noreferrer&quot;&gt;convective zone&lt;/a&gt; that get &amp;quot;wound up&amp;quot; by &lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_rotation&quot; rel=&quot;nofollow noreferrer&quot;&gt;differential rotation&lt;/a&gt;. If the stress on the tubes reaches a certain limit, they curl up like a rubber band and puncture the Sun's surface. &lt;strong&gt;Convection is inhibited at the puncture points; the energy flux from the Sun's interior decreases; and with it surface temperature.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;All in all, the sunspots appear dark because the are darker &lt;em&gt;than the surrounding surface&lt;/em&gt;. They're darker because they are cooler, and they're cooler because of the intense magnetic fields in them.&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="31410" LastEditDate="2021-05-26T21:24:02.137" LastActivityDate="2021-05-26T21:24:02.137" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="7" PostTypeId="1" AcceptedAnswerId="8" CreationDate="2013-09-24T19:14:05.603" Score="25" ViewCount="5192" Body="&lt;p&gt;I've heard that light can't escape from a &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_hole&quot;&gt;black hole&lt;/a&gt;. Can it? If not, why?&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2019-02-16T12:47:55.777" Title="Why can't light escape from a black hole?" Tags="&lt;black-hole&gt;&lt;light&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8" PostTypeId="2" ParentId="7" CreationDate="2013-09-24T19:14:05.603" Score="12" Body="&lt;p&gt;A black hole has an &lt;a href=&quot;http://en.wikipedia.org/wiki/Event_horizon&quot; rel=&quot;nofollow noreferrer&quot;&gt;event horizon&lt;/a&gt; which 'marks the point of no return'. So yes, light cannot escape from a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? Well, think of a 'spacetime fabric'. It's the easiest way to understand the physics at work here, in my opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, the fabric would look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7HKzL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7HKzL.jpg&quot; alt=&quot;fabric&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://whyfiles.org/wp-content/uploads/2011/05/gravity_probespacetime.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;whyfiles.org&lt;/a&gt;)&lt;/sub&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a black hole has so much gravity that one could say it 'rips' the spacetime fabric:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/z4Pni.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z4Pni.gif&quot; alt=&quot;black hole fabric&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;https://static.ddmcdn.com/gif/black-hole.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;ddmcdn.com&lt;/a&gt;)&lt;/sub&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the light hits this area of amazingly intense gravity, it simply cannot get out - the light travels 'along' the fabric, and since there is a rip in the fabric, one could say it simply goes away - it becomes part of the singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a simplification, of course, but it's enough to understand at least part of the physics behind this phenonenom.&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="7472" LastEditDate="2019-02-16T12:47:55.777" LastActivityDate="2019-02-16T12:47:55.777" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="9" PostTypeId="1" AcceptedAnswerId="35" CreationDate="2013-09-24T19:15:40.250" Score="20" ViewCount="9015" Body="&lt;p&gt;The equatorial coordinate system isn't too complicated. However, for people new to this, it seems rather intimidating at first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an easy way to explain it?&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="48" LastEditDate="2013-09-24T20:19:47.967" LastActivityDate="2017-08-17T18:39:56.977" Title="Could someone explain RA/dec in simple terms?" Tags="&lt;coordinate&gt;&lt;ascension&gt;&lt;declination&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10" PostTypeId="5" CreationDate="2013-09-24T19:17:46.940" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-24T19:17:46.940" LastActivityDate="2013-09-24T19:17:46.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11" PostTypeId="4" CreationDate="2013-09-24T19:17:46.940" Score="0" Body="Questions regarding planets that lie outside the Solar System." OwnerUserId="32" LastEditorUserId="2153" LastEditDate="2015-04-12T21:59:08.210" LastActivityDate="2015-04-12T21:59:08.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14" PostTypeId="1" AcceptedAnswerId="21" CreationDate="2013-09-24T19:28:29.673" Score="15" ViewCount="205" Body="&lt;p&gt;In many movies and in the popular culture wormholes are often referenced, as well as &quot;faster than light travel&quot;, which seems almost to be the same thing, is possible. &#xA;Are astronomers seriously considering and researching this phenomenon? What could be the signature of wormholes? Would next generations telescopes as ELT able to detect such signatures?&lt;/p&gt;&#xA;" OwnerUserId="36" LastEditorUserId="88" LastEditDate="2013-10-01T07:59:20.440" LastActivityDate="2013-10-01T07:59:20.750" Title="Are astronomers researching or trying to find signs of worm holes?" Tags="&lt;general-relativity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15" PostTypeId="1" AcceptedAnswerId="43" CreationDate="2013-09-24T19:30:09.823" Score="17" ViewCount="1282" Body="&lt;p&gt;On Venus, there is really inhospitable weather, as well as within the gas giants in our solar system. Are there examples of even more extreme weather on planets found in other solar systems than ours?&lt;/p&gt;&#xA;" OwnerUserId="36" LastEditorDisplayName="user8" LastEditDate="2013-09-24T19:51:27.420" LastActivityDate="2014-03-28T19:41:02.037" Title="What is the most extreme weather found on another planet?" Tags="&lt;exoplanet&gt;&lt;weather&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16" PostTypeId="1" AcceptedAnswerId="22" CreationDate="2013-09-24T19:31:17.000" Score="69" ViewCount="11219" Body="&lt;p&gt;Why do we only ever see the same side of the moon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is to do with gravity are there any variables which mean we might one day see more than we have before?&lt;/p&gt;&#xA;" OwnerUserId="41" LastEditorUserId="48" LastEditDate="2013-09-24T19:54:41.377" LastActivityDate="2020-12-19T11:56:36.173" Title="Why is only one side of the Moon visible from Earth?" Tags="&lt;the-moon&gt;&lt;orbit&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="17" PostTypeId="5" CreationDate="2013-09-24T19:32:28.077" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-24T19:32:28.077" LastActivityDate="2013-09-24T19:32:28.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18" PostTypeId="4" CreationDate="2013-09-24T19:32:28.077" Score="0" Body="Questions regarding Earth's moon and only natural satellite." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-24T19:50:59.997" LastActivityDate="2013-09-24T19:50:59.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19" PostTypeId="5" CreationDate="2013-09-24T19:34:17.320" Score="0" Body="&lt;p&gt;Questions regarding points of extremely high mass density, which creates an extremely strong gravitational field from which light cannot escape. Black holes are formed when a massive star (&lt;span class=&quot;math-container&quot;&gt;$&amp;gt;10M_\odot$&lt;/span&gt;) goes supernova and collapses into one.&lt;/p&gt;&#xA;" OwnerUserId="31410" LastEditorUserId="31410" LastEditDate="2020-12-13T22:57:43.007" LastActivityDate="2020-12-13T22:57:43.007" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="20" PostTypeId="4" CreationDate="2013-09-24T19:34:17.320" Score="0" Body="Questions regarding points of extremely high mass density, which creates an extremely strong gravitational field from which light cannot escape." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-24T19:53:01.993" LastActivityDate="2013-09-24T19:53:01.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21" PostTypeId="2" ParentId="14" CreationDate="2013-09-24T19:36:35.913" Score="11" Body="&lt;p&gt;This has been considered long ago (Here's &lt;a href=&quot;http://physics.aps.org/story/v2/st7&quot;&gt;a paper&lt;/a&gt; talking about this). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wormholes are not forbidden by physics, but the creation of wormholes is iffy ground. THere are two possible paths one can take to create a wormhole:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Choose a pre-existing wormhole in the quantum foam and &quot;expand&quot; it by feeding it exotic matter.&lt;/li&gt;&#xA;&lt;li&gt;&quot;Tear and sew up&quot; space &amp;mdash; we're not sure if this is allowed by physics, as it ventures into the area of physics that we don't have an adequate explanation for.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Whatever it is, the creation and sustenance of a wormhole requires us to have control over &lt;a href=&quot;http://en.wikipedia.org/wiki/Exotic_matter&quot;&gt;exotic matter&lt;/a&gt; (in this case, particles/waves with negative mass/energy density). &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_fluctuation&quot;&gt;Vacuum fluctuations&lt;/a&gt; already have regions of negative energy density, but they're an uncontrollable quantum phenomenon by current technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Faster than light travel&quot; is a different matter, though. While wormholes let one jump to another point in space, one does not attain a speed greater than c whilst in them.&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2013-09-24T19:36:35.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22" PostTypeId="2" ParentId="16" CreationDate="2013-09-24T19:42:15.507" Score="74" Body="&lt;p&gt;The reason for this is what we call &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_locking&quot; rel=&quot;nofollow noreferrer&quot;&gt;tidal locking&lt;/a&gt;&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Tidal locking (or captured rotation) occurs when the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_gradient&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravitational&#xA;gradient&lt;/a&gt; makes one side of an &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_body&quot; rel=&quot;nofollow noreferrer&quot;&gt;astronomical body&lt;/a&gt; always face another,&#xA;an effect known as synchronous rotation. For example, the same side of&#xA;the Earth's Moon always faces the Earth. A tidally locked body takes&#xA;just as long to rotate around its own axis as it does to revolve&#xA;around its partner. This causes one hemisphere constantly to face the&#xA;partner body. Usually, at any given time only the satellite is tidally&#xA;locked around the larger body, but if the difference in mass between&#xA;the two bodies and their physical separation is small, each may be&#xA;tidally locked to the other, as is the case between &lt;a href=&quot;https://en.wikipedia.org/wiki/Pluto&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pluto&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Charon_(moon)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Charon&lt;/a&gt;.&#xA;This effect is employed to stabilize some artificial satellites.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;         &lt;img src=&quot;https://i.stack.imgur.com/kUqtd.gif&quot; alt=&quot;Tidal locking of the Moon with the Earth&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;    &lt;sup&gt;&lt;strong&gt;Fig. 1&lt;/strong&gt;: Tidal locking results in the Moon rotating about its axis in about the same time it takes to orbit the Earth. (Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Tidal_locking_of_the_Moon_with_the_Earth.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Fig. 1, cont.&lt;/strong&gt;: Except for &lt;a href=&quot;https://en.wikipedia.org/wiki/Libration&quot; rel=&quot;nofollow noreferrer&quot;&gt;libration&lt;/a&gt; effects, this results in the Moon keeping the same face turned towards the Earth, as seen in the figure&#xA;on the left. (The Moon is shown in polar view, and is not drawn to&#xA;scale.) If the Moon were not spinning at all, it would alternately&#xA;show its near and far sides to the Earth while moving around our&#xA;planet in orbit, as shown in the figure on the right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;                         &lt;a href=&quot;https://web.archive.org/web/20201007012740/https://upload.wikimedia.org/wikipedia/commons/b/ba/Lunar_libration_with_phase_Oct_2007_450px.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://web.archive.org/web/20201007012740/https://upload.wikimedia.org/wikipedia/commons/b/ba/Lunar_libration_with_phase_Oct_2007_450px.gif&quot; alt=&quot;Tidal locking of the Moon with the Earth&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;                         &lt;sup&gt;&lt;strong&gt;Fig. 2&lt;/strong&gt;: Lunar librations in latitude and longitude over a period of one month (Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Lunar_libration_with_phase_Oct_2007_450px.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Libration is manifested as a slow rocking back and forth of the Moon&#xA;as viewed from Earth, permitting an observer to see slightly different&#xA;halves of the surface at different times.&lt;/p&gt;&#xA;&lt;p&gt;There are three types of lunar libration:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Libration in longitude&lt;/strong&gt; results from the eccentricity&#xA;of the Moon's orbit around Earth; the Moon's rotation sometimes leads&#xA;and sometimes lags its orbital position.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Libration in latitude&lt;/strong&gt; results&#xA;from a slight inclination between the Moon's axis of rotation and the&#xA;normal to the plane of its orbit around Earth. Its origin is analogous&#xA;to how the seasons arise from Earth's revolution about the Sun.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Diurnal_motion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Diurnal&lt;/a&gt; libration&lt;/strong&gt; is a small daily oscillation due to the Earth's&#xA;rotation, which carries an observer first to one side and then to the&#xA;other side of the straight line joining Earth's and the Moon's&#xA;centers, allowing the observer to look first around one side of the&#xA;Moon and then around the other—because the observer is on the surface&#xA;of the Earth, not at its center.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;All quotes and images from &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_locking&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia on Tidal locking&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Libration&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia on Libration&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="7472" LastEditDate="2020-12-19T11:56:36.173" LastActivityDate="2020-12-19T11:56:36.173" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23" PostTypeId="2" ParentId="16" CreationDate="2013-09-24T19:44:17.080" Score="25" Body="&lt;p&gt;The period of rotation of the moon is ~&lt;a href=&quot;http://www.wolframalpha.com/input/?i=period%20of%20rotation%20of%20moon&amp;amp;dataset=&quot;&gt;27.322 days&lt;/a&gt;, and the period of revolution is also ~&lt;a href=&quot;http://www.wolframalpha.com/input/?i=period%20of%20revolution%20of%20moon&quot;&gt;27.322&lt;/a&gt;. This means that for every degree it turns around the Earth, it turns a degree around itself, so the same side always faces us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is due to tidal forces coupling the various oscillators in the system (revolution of moon, orbit of moon, revolution of Earth). When oscillators are coupled, they have a tendency to settle to a state that is either in phase or 180 degrees out of phase*. Both cases give rise to tidal locking here.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This is an experiment you can try out suspend  a pendulum from each end of a ruler, and give them a small phase difference. Over time, the phases will match.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="28" LastActivityDate="2013-09-24T19:44:17.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24" PostTypeId="1" AcceptedAnswerId="40" CreationDate="2013-09-24T19:46:14.980" Score="40" ViewCount="2561" Body="&lt;p&gt;Black holes have so much gravity that &lt;a href=&quot;https://astronomy.stackexchange.com/questions/7/why-cant-light-escape-from-a-black-hole/8#8&quot;&gt;even light can't escape from them&lt;/a&gt;. If we can't see them, and the suck up all  electromagnetic radiation, then how can we find them? &lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-01-31T01:00:47.090" Title="How are black holes found?" Tags="&lt;black-hole&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="25" PostTypeId="2" ParentId="16" CreationDate="2013-09-24T19:48:54.653" Score="31" Body="&lt;p&gt;Primarily because of how the moon spins - it is spinning at &lt;em&gt;just&lt;/em&gt; the right rate to keep us from seeing one side of it. Here's a handmade diagram to show what I mean:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MDbbB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2013-09-24T19:48:54.653" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26" PostTypeId="1" AcceptedAnswerId="57" CreationDate="2013-09-24T19:50:44.980" Score="29" ViewCount="2650" Body="&lt;p&gt;According to &lt;a href=&quot;http://solarsystem.nasa.gov/planets/profile.cfm?Object=Venus&amp;amp;Display=Facts&amp;amp;System=Metric&quot;&gt;this NASA overview&lt;/a&gt;, the planet Venus is unique (amongst the major planets), Venus has a slow retrograde axial rotation, taking 243 Earth days to make one rotation (which is longer than its orbital revolution). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the current accepted theory as to why (and how) Venus developed this anomalous slow retrograde axial rotation?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-09-28T05:49:16.060" LastActivityDate="2019-08-29T06:49:35.013" Title="What is the current accepted theory as to why Venus has a slow retrograde rotation?" Tags="&lt;venus&gt;&lt;rotation&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27" PostTypeId="1" AcceptedAnswerId="32" CreationDate="2013-09-24T19:52:05.020" Score="16" ViewCount="550" Body="&lt;p&gt;Referring to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/7/why-cant-light-escape-from-a-black-hole&quot;&gt;this question&lt;/a&gt;, is there any information that can leave black holes? Are they causing a permanent information loss in our Universe?&lt;/p&gt;&#xA;" OwnerUserId="44" LastEditorUserId="6" LastEditDate="2020-09-18T12:38:19.000" LastActivityDate="2020-09-18T12:38:19.000" Title="Can any information leave a black hole?" Tags="&lt;black-hole&gt;&lt;universe&gt;&lt;hawking-radiation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28" PostTypeId="2" ParentId="15" CreationDate="2013-09-24T19:52:20.450" Score="18" Body="&lt;p&gt;So far, we don't have anywhere near enough detail to identify weather on planets in other star systems. We even have difficulty working out the size of some of them, so it may be a while till we can get a good idea as to weather.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is going to be in our solar system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for extremes, we have a couple of options:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;WIND&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn and Neptune have the highest wind speeds, and as Neptune has the most extreme difference between East-West and West-East winds, I would be tempted to say it has the most extreme weather.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SMw3x.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taken from &lt;a href=&quot;http://www.springerreference.com/docs/html/chapterdbid/366754.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.springerreference.com/docs/html/chapterdbid/366754.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TEMPERATURE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pluto is incredibly cold, but it doesn't vary much, so should we call it an extreme? Mercury has a much more extreme range of temperatures. From sunside (over 400C) to coldside (almost -200C) gives a difference of 600 degrees Celsius!&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2013-09-24T19:52:20.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29" PostTypeId="2" ParentId="24" CreationDate="2013-09-24T19:52:21.910" Score="18" Body="&lt;p&gt;One way is by following &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma-ray_burst&quot;&gt;Gamma Ray Bursts&lt;/a&gt;. When a black hole feeds on surrounding gas or swallows a star that got too close, they often emit gamma ray bursts which are very energetic and easy to spot (although they don't last long).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of &lt;a href=&quot;http://en.wikipedia.org/wiki/Supermassive_black_hole&quot;&gt;super massive blackholes&lt;/a&gt;, they are seemingly at the center of every medium and large galaxy. It makes &lt;em&gt;where&lt;/em&gt; to look rather easy.&lt;/p&gt;&#xA;" OwnerUserId="31" LastActivityDate="2013-09-24T19:52:21.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30" PostTypeId="5" CreationDate="2013-09-24T19:57:19.310" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-24T19:57:19.310" LastActivityDate="2013-09-24T19:57:19.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31" PostTypeId="4" CreationDate="2013-09-24T19:57:19.310" Score="0" Body="Questions regarding the planet Venus, the second planet from the Sun." OwnerUserId="19" LastEditorUserId="9527" LastEditDate="2020-01-22T14:53:56.503" LastActivityDate="2020-01-22T14:53:56.503" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32" PostTypeId="2" ParentId="27" CreationDate="2013-09-24T19:59:45.683" Score="15" Body="&lt;p&gt;Information cannot &quot;leave&quot; the black hole. There is no way (by our current framework of physics) that an entity inside the event horizon can send a signal out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, entropy can leave. The black hole has entropy proportional to its surface area, and this roughly corresponds to &quot;the number of ways that black hole could have been created&quot;. The surface area decreases due to &lt;a href=&quot;http://enwp.org/Hawking_radiation&quot;&gt;Hawking radiation&lt;/a&gt;, so it &lt;em&gt;can&lt;/em&gt; release entropy back to the universe. Of course, the growth rate due to infalling mas is usually many orders of magnitude more than the shrinkage.&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2013-09-24T19:59:45.683" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33" PostTypeId="1" AcceptedAnswerId="36" CreationDate="2013-09-24T20:00:00.203" Score="26" ViewCount="10688" Body="&lt;p&gt;From what I understand, the mass of a black hole should be nearly infinite, how much more &lt;em&gt;massive&lt;/em&gt; can something get? &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is the name to be literally interpreted such that a &lt;em&gt;Supermassive&lt;/em&gt; Black Hole just has more &lt;em&gt;mass&lt;/em&gt;?&lt;/li&gt;&#xA;&lt;li&gt;Or rather, is a Supermassive Black Hole just a regular Black Hole with nearly infinite mass that is larger in diameter?&lt;/li&gt;&#xA;&lt;li&gt;If the difference is in fact a change in diameter, how are the changes in size with the retention of the immense mass reflected in the Supermassive Black Hole's gravitational field?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="39" LastActivityDate="2014-07-11T21:06:39.213" Title="What are the differences between a Black Hole and a Supermassive Black Hole" Tags="&lt;black-hole&gt;&lt;supermassive-black-hole&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34" PostTypeId="1" CreationDate="2013-09-24T20:02:14.600" Score="14" ViewCount="1566" Body="&lt;p&gt;How to navigate with possible maximal precision using moon phases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon is the brightest celestial body seen in the night sky, and it is possible to find even through moderate clouds. So it's a good natural object to use in navigation when a compass is not available. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A full Moon is quite an easy case - it's directly on the opposite site from the Sun. However, in the other phases, are there tables or mnemotechnical methods allowing to determine the north based on moon shape, assuming the time is known? Are they different depending on geographical height? Is it possible to estimate both north and the time of the night, when only the geographical height is known?&lt;/p&gt;&#xA;" OwnerUserId="42" LastEditorUserId="19" LastEditDate="2013-09-24T20:12:01.720" LastActivityDate="2017-11-13T14:10:32.480" Title="Navigation using moon phases" Tags="&lt;the-moon&gt;&lt;navigation&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35" PostTypeId="2" ParentId="9" CreationDate="2013-09-24T20:02:18.120" Score="15" Body="&lt;p&gt;&lt;strong&gt;A system projected into the sky&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretend all the stars are painted on the inside of a large ball, and you are at the center of the ball. The imagined ball is called the Celestial Sphere. If the Earth wasn't blocking the lower half of your view, and the Sun wasn't making the blue sky so bright, you would be able to look at the stars in any direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the rotation of the Earth causes the imagined ball of stars to appear to move around the sky. If you lay on your back, (in the norther hemisphere,) with your head to the north, the stars slowly slide by from left to right. There are two points however, where the stars don't appear to move: These two points are directly above the Earth's north and south poles. These two points, projected out into the sky are labeled the North, and South, Celestial Poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, the Earth's equator is projected out into the sky and labeled the Celestial Equator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, sorting out the east-west positioning, (ie left-right if you lay on your back, head to the north), requires an arbitrary selection: There are an infinite number of circles one can imagine drawing on the sky. (For example, small circles the size of the moon, and also many much bigger circles.) The largest circle you can draw is called a &quot;great circle&quot;, from geometry. Point at the North Celestial Pole, (way above and behind your head if you face south in the northern hemisphere,) and draw a line straight south to the horizon, then down below your feet through the South celestial pole and back up to the North Celestial Pole. You have just drawn one Great Circle, through the Celestial Poles, all the way around the Celestial Sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that Great Circle is just one of an infinite number of similar ones. One could start at the North Celestial Pole and draw a Great Circle a little to the left or the right. So astronomers have chosen one specific Great Circle and labeled one side of that circle as &quot;zero degrees&quot; -- or, &quot;we will start measuring angles from here.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So how do we use this system?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;right ascension&quot; is measured rightward, (from your left to your right if you face south in the northern hemisphere,) from the zero-degrees half of that chosen great circle. So given a right ascension, one can find the corresponding half of a great circle. For a given right ascension, you have half of circle selected from the North Celestial Pole, to the South Celestial Pole. (This is called a line of equal longitude.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;declination&quot; is measured from the Celestial Equator. The Celestial Equator cuts through the middle of your line of longitude. From the Celestial Equator it is 90° northward(+) to the North Celestial Pole, and 90° southward(-) to the South Celestial Pole. So we can measure from +90° (northward) to -90° (southward) along that line of longitude, from the Celestial Equator. (Don't be confused by &quot;positive declinations&quot;; 45° of declination is northward from the Celestial Equator.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the system does not depend on where you are standing on Earth, nor on the time of day. If you have a Right Ascension and a Declination, you have an unambiguous spot specified on the Celestial Sphere. The system is actually very simple, but the rotation (time of day) of the Earth, and where you are geographically located needs to also be figured in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What's over my head right now?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Face south, (if you're in the northern hemisphere.) What is your latitude -- geographically, how far north are you from the Earth's equator? If you are 40° north, then Celestial declination of 40° north is directly over your head. Declination of 50° is 10° further northward from directly over your head, etc. But for each declination, you have a circle of points in the sky that all have that same declination.  The circle is centered around the celestial poles.  As the earth turns it appears to slowly rotate overhead. Note: If you are at the equator, this circle is just the celestial equator, and it is a great circle.  As you approach either pole, the circle will get smaller.  The poles, this circle collapses to a point.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right ascension you simply have to look up. That arbitrarily chosen, zero-degree line, appears to spin around the sky every day at the Earth rotates. Also, over the course of a year, the Earth's orbit about the sun makes an extra apparent turn of the sky. (What is overhead mid-winter, is directly under your feet mid-summer!) So there's an offset -- what right ascension is overhead &lt;strong&gt;on this date&lt;/strong&gt; at midnight, and how far (in hours) are we from midnight right now? Combining those two you can figure out what Right Ascension is over head at any time, on any date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(potential editors: I've intentionally not used acronyms for NCP, RA, etc as that makes it more complicated for people learning this material.)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="14232" LastEditDate="2016-09-13T11:35:34.127" LastActivityDate="2016-09-13T11:35:34.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36" PostTypeId="2" ParentId="33" CreationDate="2013-09-24T20:02:37.830" Score="16" Body="&lt;p&gt;Stellar mass black holes form from the collapse of massive stars at the end of their lives. You can then find them scattered throughout galaxies, just like you find massive stars. They typically have a mass a few times the mass of the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supermassive black holes are found at the centers of galaxies. They typically have mass of millions of Suns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently they have started to discover &lt;a href=&quot;http://en.wikipedia.org/wiki/Intermediate-mass_black_hole&quot;&gt;Intermediate Mass Black Holes&lt;/a&gt; which blur the lines between a stellar black hole and supermassive black hole. The typically have a mass in the range 100 to one million solar masses.&lt;/p&gt;&#xA;" OwnerUserId="31" LastEditorUserId="38" LastEditDate="2013-11-07T15:03:43.320" LastActivityDate="2013-11-07T15:03:43.320" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="37" PostTypeId="2" ParentId="33" CreationDate="2013-09-24T20:03:10.657" Score="9" Body="&lt;p&gt;The mass of a black hole is not infinite. In fact, if a black hole is created that is big enough to survive evaporation, its mass will be its starting mass, plus any mass swallowed up, minus the radiation that leaves it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is why you hear phrases such as &quot;A black hole with ten times the mass of our sun&quot; or in the case of a supermassive black hole, &quot;...of millions of suns&quot;&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2013-09-24T20:03:10.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="38" PostTypeId="1" AcceptedAnswerId="65" CreationDate="2013-09-24T20:06:13.763" Score="26" ViewCount="1435" Body="&lt;p&gt;The planet Uranus is another solar system anomaly, where according to the &lt;a href=&quot;http://solarsystem.nasa.gov/planets/profile.cfm?Object=Uranus&amp;amp;Display=Facts&amp;amp;System=Metric&quot; rel=&quot;noreferrer&quot;&gt;NASA profile&lt;/a&gt; has an axial tilt of 97.8 degrees, also considered to be retrograde.  This NASA summary &lt;a href=&quot;http://solarsystem.nasa.gov/planets/profile.cfm?Object=Uranus&amp;amp;Display=OverviewLong&quot; rel=&quot;noreferrer&quot;&gt;&quot;Uranus&quot;&lt;/a&gt; suggests the current theory of a large planet-sized impact earlier in its history.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the planet-impact theory still hold true or have new accepted theories come to light?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of all, are there any results from any simulations available?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A note, this is posted as a separate question to my other question &lt;a href=&quot;https://astronomy.stackexchange.com/questions/26/what-is-the-current-accepted-theory-as-to-why-venus-has-a-slow-retrograde-rotati&quot;&gt;&quot;What is the current accepted theory as to why Venus has a slow retrograde rotation?&quot;&lt;/a&gt; as the axial tilt is significantly different.&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorUserId="9527" LastEditorDisplayName="user8" LastEditDate="2020-07-22T09:06:29.287" LastActivityDate="2020-07-23T18:37:57.957" Title="What is the accepted theory as to why Uranus' axis is tilted so severely?" Tags="&lt;rotation&gt;&lt;uranus&gt;&lt;obliquity&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="39" PostTypeId="1" AcceptedAnswerId="234" CreationDate="2013-09-24T20:10:36.683" Score="10" ViewCount="575" Body="&lt;p&gt;Assume object A is moving through the space and is passing near the other object (B). Assume the gravitational influence of other objects can be ignored. How to find the equation describing the movement of the object B? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 2 cases, object A is moving straightforward or it's moving on orbit (around other object). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the problem is quite elementary, but I couldn't find anything that could help solve that problem using the physics on the level of basic university course (I've studied computer science, so I've got only 1 semester of physics, and basic mathematical knowledge - integrals, algebra etc.). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the problem can be solved numerically, but I'm interested in finding the equation describing the movement. &lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2014-04-02T11:57:33.483" Title="How to calculate the movement of the object passing near other object in space?" Tags="&lt;orbit&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="40" PostTypeId="2" ParentId="24" CreationDate="2013-09-24T20:13:56.917" Score="24" Body="&lt;p&gt;To add to John Conde's answer.  According to the NASA web page &lt;a href=&quot;http://science.nasa.gov/astrophysics/focus-areas/black-holes/&quot;&gt;&quot;Black Holes&quot;&lt;/a&gt;, detection of black holes can obviously not be performed by detection any form of electromagnetic radiation coming directly from it (hence, can not be 'seen').  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black hole is inferred by observing the interaction with surrounding matter, from the webpage:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We can, however, infer the presence of black holes and study them by detecting their effect on other matter nearby.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This also includes detection of x-ray radiation that radiates from matter accelerating towards the black hole. Although this seems contradictory to my first paragraph - it needs to be noted that this is not directly from the black hole, rather from the interaction with matter accelerating towards it.&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2013-09-24T20:13:56.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="41" PostTypeId="2" ParentId="9" CreationDate="2013-09-24T20:16:20.030" Score="12" Body="&lt;p&gt;The easiest way to think of equatorial coordinates is by extending lines of latitude/longitude (the geographic coordinate system we usually learn in school) out into space. The earth's equator becomes the celestial equator, and the north and south poles become the celestial north and south poles, respectively. By doing this you have a fairly simple way to describe the location of an object in space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each location is described by a pair of coordinates: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Declination (dec), which is measured in degrees (-90 for the south celestial pole, 0 for the celestial equator, and +90 for the north celestial pole). Partial degrees are usually described in terms of minutes (') and seconds (''). Think of declination as latitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Right Ascension (RA), which is usually measured in hours, minutes, and seconds. Think of this as terrestrial longitude extended out into space, then measuring it in hours makes a bit more sense (time zones). You could just as easily measure it in degrees, too, where 1 hour is 15 degrees. RA is measured from the point where the sun crosses the celestial equator at the March equinox (located in Pisces).  &lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2013-09-24T20:16:20.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="42" PostTypeId="1" AcceptedAnswerId="44" CreationDate="2013-09-24T20:19:24.813" Score="28" ViewCount="1841" Body="&lt;p&gt;What processes does a star undergo to become a pulsar? Does it take a very specific star with a certain set of qualities such as &quot;Just the right mass, diameter, and composition,&quot; or is it a freak accident that certain stars live out their remaining life as a pulsar?&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorDisplayName="user10" LastEditDate="2013-09-24T20:21:54.523" LastActivityDate="2015-07-26T10:48:41.157" Title="What causes a star to become a pulsar?" Tags="&lt;star&gt;&lt;pulsar&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="43" PostTypeId="2" ParentId="15" CreationDate="2013-09-24T20:21:31.950" Score="21" Body="&lt;p&gt;I'd say that &lt;a href=&quot;http://en.wikipedia.org/wiki/HD_189733b&quot;&gt;HD 189733b&lt;/a&gt; is a good candidate for the most extreme &lt;em&gt;known&lt;/em&gt; weather on another planet (outside our Solar System).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to some recent &lt;a href=&quot;http://www.space.com/22614-blue-alien-planet-glass-rain.html&quot;&gt;news accounts&lt;/a&gt;, the atmospheric temperature is believed to be over 1000° C, with 7000 kph winds. (For comparison to the data in Rory Alsop's answer, that's about 1900 meters per second.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it rains molten glass. Sideways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE :&lt;/strong&gt; As Guillochon points out in a comment, &lt;a href=&quot;http://en.wikipedia.org/wiki/HD_80606_b&quot;&gt;HD 80606 b&lt;/a&gt; likely has even higher winds, though they're not continuous. It's a Jovian with an extremely eccentric orbit. Quoting the Wikipedia article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Computer models predict the planet heats up 555 °C (1,000 °F) in just&#xA;  a matter of hours triggering &quot;shock wave storms&quot; with winds that move&#xA;  faster than the speed of sound, at 3 miles per second.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which, in civilized units, is about 4800 meters/second. Probably no molten glass rain, though, so it's not clear that it's more &quot;extreme&quot;.&lt;/p&gt;&#xA;" OwnerUserId="53" LastEditorUserId="53" LastEditDate="2013-11-10T05:29:52.387" LastActivityDate="2013-11-10T05:29:52.387" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="44" PostTypeId="2" ParentId="42" CreationDate="2013-09-24T20:25:36.830" Score="14" Body="&lt;p&gt;It's generally dictated by how massive the star is. Remember what a pulsar is, it's a very rapidly rotating, highly magnetized neutron star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://imagine.gsfc.nasa.gov/Images/basic/xray/pulsar.gif&quot; alt=&quot;Pulsar&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutron stars are a category of objects which have masses between 1.4 and 3.2 solar masses. This is the end stage of stars which are not massive enough to form black holes (they're held up by neutron degeneracy pressure), but are massive enough to overcome electron degeneracy pressure (which is what prevents white dwarves from further gravitational collapse). &lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:46.983" LastActivityDate="2013-10-21T13:53:29.723" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="45" PostTypeId="2" ParentId="2" CreationDate="2013-09-24T20:44:49.320" Score="14" Body="&lt;p&gt;In theory one should just be able to determine the difference between the spectra during the star's eclipse of the exoplanet (starlight alone) and the spectra of the star and exoplanet together, but in reality equipment is not precise enough for this. To remedy this problem, the analysis is integrated over many eclipses. Some other calibrations are made as well which you can read more about in the paper &lt;a href=&quot;http://arxiv.org/pdf/astro-ph/0702494.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;A &lt;em&gt;Spitzer&lt;/em&gt; Spectrum of the Exoplanet HD 189733b&lt;/a&gt;. This is called occultation spectroscopy.&lt;/p&gt;&#xA;&lt;p&gt;There's another method used as well called transmission spectroscopy, which &lt;a href=&quot;https://exoplanets.nasa.gov/resources/297/a-planets-transmission-spectrum/&quot; rel=&quot;nofollow noreferrer&quot;&gt;detects the change in light as the exoplanet passes in front of its star&lt;/a&gt;. You can read more about that in this paper: &lt;a href=&quot;https://iopscience.iop.org/article/10.1088/0004-637X/774/2/95/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;Infrared transmission spectroscopy of the exoplanets HD 209458b and XO-1b using the wide field Camera-3 on the Hubble Space Telescope&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorUserId="6" LastEditDate="2020-12-04T17:02:08.540" LastActivityDate="2020-12-04T17:02:08.540" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="46" PostTypeId="2" ParentId="9" CreationDate="2013-09-24T20:56:16.930" Score="14" Body="&lt;p&gt;The equatorial coordinate system is very similar to the system used on a globe or on maps. To specify a point on a sphere or a globe you need only two numbers. These are the longitudes and latitudes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a globe of the earth you have longitudes from -180º to +180º and latitudes from -90º to +90º. The south pole lies at -90º latitude, the north pole at +90º latitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sky uses the same system, projected onto the imaginary celestial sphere. The sky's equator is at the same position as the earth's equator. Only projected onto the (infinitely large) sky sphere. You can find lots of &lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/E/Equatorial+Coordinate+System&quot;&gt;very nice pictures&lt;/a&gt; of this on the web.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since the earth rotates, the sky's coordinate system would rotate also, if it only were a synchronized projection. So every star would have a coordinate that depended on the time of day! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve this, the equatorial coordinate system is fixed on the equinox points. Since the earth is tilted, the eclipitc (i.e. the earth's orbit around the sun) has two intersections with the sky's equator. These are the equinox points. The sun will be located there during the days of spring and autumn when day and night are of equal length. These points on the sky sphere are relatively fixed and well suited for pinning down the coordinate system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sky's coordinate system is not using the degree notation for both longitude and latitude. Instead we use the terms Right Acension (i.e. longitude), which goes from 0h to 24h, similar to a day on earth. And the other is Declination (i.e. latitude), which goes from -90º to +90º. This is similar to the earth's traditional coordinate system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So every fixed star can be assigned a coordinate in the sky which is independent of the time of day or year. For example, the star Sirius (α CMa) has the following coordinates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RA 06h 45m 08.9173s Dec −16° 42′ 58.017″&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These can be programmed into computerized telescope mounts, which can then find and track the star as it moves with the earth's rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a rather simplified explanation. One should read up on equatorial mounts for telescopes, which utilize this coordinate system in a very elegant manner.&lt;/p&gt;&#xA;" OwnerUserId="50" LastEditorUserId="50" LastEditDate="2013-09-24T21:04:43.613" LastActivityDate="2013-09-24T21:04:43.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="47" PostTypeId="1" CreationDate="2013-09-24T21:10:06.683" Score="15" ViewCount="152" Body="&lt;p&gt;In a museum in Lviv, I saw a pocket cellular clock. I don't have a photo, but it was a small disc that had 2 or 3 filaments in it which were pointed at the stars (one of them was Andromeda, I think). Through analyzing the position of the stars it was possible to determine the time of the night. I'm not sure if knowing the day of the year was promised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't find the information about the way the clock was used. Does anyone know how such device was used?&lt;/p&gt;&#xA;" OwnerUserId="42" LastEditorUserId="2153" LastEditDate="2015-01-19T14:16:59.647" LastActivityDate="2015-01-19T14:16:59.647" Title="How would the pocket cellular clock work?" Tags="&lt;star&gt;&lt;constellations&gt;&lt;clock&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="48" PostTypeId="5" CreationDate="2013-09-24T21:37:04.510" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-24T21:37:04.510" LastActivityDate="2013-09-24T21:37:04.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="49" PostTypeId="4" CreationDate="2013-09-24T21:37:04.510" Score="0" Body="Questions regarding a layer of gasses surrounding a celestial object." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-24T22:06:49.677" LastActivityDate="2013-09-24T22:06:49.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="50" PostTypeId="5" CreationDate="2013-09-24T21:37:51.497" Score="0" Body="&lt;p&gt;An orbit is the trajectory a body follows due to the gravitational field of another nearby object. &#xA;Most often, an orbit follows a closed path and is periodical, but the term can also refer to non-repeating trajectories. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of orbits include the path of the Moon around the Earth, the path of the Earth around the Sun, or the path of two stars in a binary pair around their common center of mass.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="9527" LastEditDate="2020-02-26T15:17:45.093" LastActivityDate="2020-02-26T15:17:45.093" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="51" PostTypeId="4" CreationDate="2013-09-24T21:37:51.497" Score="0" Body="Questions regarding an object 'falling around' another object, due to a combination of gravity and momentum.&#xD;&#xA;" OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-24T22:06:46.803" LastActivityDate="2013-09-24T22:06:46.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="52" PostTypeId="1" CreationDate="2013-09-24T21:56:20.273" Score="19" ViewCount="488" Body="&lt;p&gt;Given that some exoplanets, particular &lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2013/17aug_hotjupiters/&quot;&gt;&quot;Hot Jupiters&quot;&lt;/a&gt;, orbit very closely to their parent star, how hot can these planets become?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the hottest exoplanet discovered so far?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2016-12-26T19:23:03.970" Title="How hot can a planet be?" Tags="&lt;exoplanet&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="53" PostTypeId="2" ParentId="52" CreationDate="2013-09-24T21:56:20.273" Score="12" Body="&lt;p&gt;The temperature of an exoplanet is dependent on a number of factors, including:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The distance between the exoplanet and the parent star.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The type (energy output) of the parent star.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The ability of the exoplanet's atmosphere to distribute the heat energy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Recent discoveries, such as reported in the article &lt;a href=&quot;http://www.space.com/3781-sizzling-planet-stars-cool.html&quot;&gt;&quot;Sizzling Planet Makes Some Stars Look Cool&quot;&lt;/a&gt; (Than, 2007), have found that some exoplanets are hotter than some types of stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the exoplanet &lt;a href=&quot;http://www.nasa.gov/mission_pages/spitzer/multimedia/F-Black-artist.html&quot;&gt; HD 149026b &lt;/a&gt; has a star-facing side of over 2000C (3700F), this planet reflects so little light that it would appear like a &quot;glowing piece of charcoal&quot;. It is believed (from the article) that the planet is tidally locked and the 'dark side' of the planet is believed to be substantially cooler.&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2013-09-24T21:56:20.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="54" PostTypeId="2" ParentId="47" CreationDate="2013-09-24T21:59:32.217" Score="10" Body="&lt;p&gt;The device that you saw is called a nocturnal. It calculates the local time based on the month of the year and the position of Polaris and one or more other stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since stars change position throughout the night, they can be used to determine time; but the positions of stars change throughout the year, so the input of the month of the year is needed. Polaris is used as a relatively fixed reference point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;More information&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nocturnal_%28instrument%29&quot;&gt;Nocturnal (instrument) - Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6" LastActivityDate="2013-09-24T21:59:32.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="55" PostTypeId="5" CreationDate="2013-09-24T22:18:37.873" Score="0" Body="&lt;p&gt;Questions regarding devices used to track and display the current time.&lt;/p&gt;&#xA;" OwnerUserId="31410" LastEditorUserId="31410" LastEditDate="2020-12-14T16:37:54.573" LastActivityDate="2020-12-14T16:37:54.573" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="56" PostTypeId="4" CreationDate="2013-09-24T22:18:37.873" Score="0" Body="Questions regarding devices used to track and display the current time." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-25T04:05:23.360" LastActivityDate="2013-09-25T04:05:23.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="57" PostTypeId="2" ParentId="26" CreationDate="2013-09-24T22:22:59.173" Score="18" Body="&lt;p&gt;There seem to be a few, and none are accepted by the whole scientific community. The main ones:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Venus was struck by a large body during its early formation&lt;/li&gt;&#xA;&lt;li&gt;The spin axis flipped, as can happen with a gyroscope&lt;/li&gt;&#xA;&lt;li&gt;The spin slowed to a standstill and then reversed, caused by the sun's gravity, the dense atmosphere and friction between core and mantle&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That final one seems to be the most recent, being proposed by Alexandre Correira and Jacques Laskar in 2001. Their research seems to imply that the conditions on Venus and its distance for the sun make a retrograde spin slightly more likely than a forward one.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2013-09-24T22:22:59.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="58" PostTypeId="1" AcceptedAnswerId="60" CreationDate="2013-09-24T22:29:45.553" Score="19" ViewCount="679" Body="&lt;p&gt;Is it possible for two stars to exist close to each other? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Close&quot; is relative; let's assume that two stars are close to each other if they are at the center of the same solar system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's possible for this to happen with 3 stars. But what about more? How many stars can be the center of exactly one solar system without collapsing because of gravity?&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorDisplayName="user8" LastEditDate="2013-09-25T04:05:57.707" LastActivityDate="2018-01-03T15:40:31.283" Title="How many stars can stay close to each other without collapsing?" Tags="&lt;star&gt;&lt;binary-star&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="60" PostTypeId="2" ParentId="58" CreationDate="2013-09-24T22:55:19.680" Score="11" Body="&lt;p&gt;Well, if they have an angular momentum with respect to their center of mass, or in other words , if they are in orbit, gravitational collapse is not the issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=XG6ZaVWztzc#t=100&quot; rel=&quot;noreferrer&quot;&gt;This&lt;/a&gt; is what happens when three bodies are in orbit around each other. Don't get disappointed so soon. There is indeed a way to avoid this. If the three bodies are of comparable mass, and one of the bodies is situated at L4 or L5, tuning the orbital parameters may work just fine, but then you gotta worry about tidal forces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qn5Mn.png&quot; alt=&quot;Lagrange points&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if the size of our third star exceeds the area given to it at L4/L5, it will experience tidal forces and will start ejecting mass to either of the two other stars, or both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as Zolan commented above, astronomy with an infinite universe has a possibility for everything. What are listed above are the most common problems which arise, that is the reasons which are mainly responsible for making such systems highly unlikely of existence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here's a system which works the way you described : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A binary system exists, and the third star is far enough that it experiences the gravity as that of a single star. And the nearest planet(as you want it to be a &quot;solar system&quot;), is so far that it experiences the gravity of the three stars as that of one. Now, since &quot;close&quot; is a relative term, for the planet, the three stars are appreciably close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the answer to your question, &lt;strong&gt;YES&lt;/strong&gt; and &lt;strong&gt;NO&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2013-09-24T22:55:19.680" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="61" PostTypeId="2" ParentId="3" CreationDate="2013-09-24T23:13:54.810" Score="7" Body="&lt;p&gt;Okay! &lt;a href=&quot;http://pastebin.com/LAk0jdR5&quot;&gt;Here&lt;/a&gt; are some of the popular targets, search for the season of visibility yourself please.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above list describes the angular seperation between the stars and also the magnification required to resolve it, both explicitly. So, this list is not only for amateurs, but there are some intermediate ones which you can try if you have a good enough telescope.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2013-09-24T23:13:54.810" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="62" PostTypeId="2" ParentId="34" CreationDate="2013-09-24T23:20:11.593" Score="8" Body="&lt;p&gt;In other phases, the angular distance of Sun from Moon is equal to the phase of the moon, in the direction of the lit side. One of the methods could be to estimate the position of Sun based on this, and then you can, assuming time is known, estimate North.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't know time, use constellations (which can form a separate question) to estimate North and the reverse of above method for time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if there are clouds and you cannot see any constellations, well, you need to be able to calculate cosines and sines and their inverse in your heads. This method, which I will not describe depends on the rate of change of moon's position, which you'll have to estimate using your hands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will have to search for my notes for actual calculation of time, let's see if someone else answers it. But let me tell you how to find direction just using Moon, though this is less accurate than if you already know the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the center of the Moon, draw a line in the sky and follow it towards the lit side of the Moon, the spot where it cuts on the Horizon is either East or West. It is East if you are observing this in Waning Phase i.e. after a Full Moon, else it is West (if after the days of a New Moon). You can get the rest of the directions this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how it works. The line you just drew was approximately the ecliptic, the apparent path of Sun in sky. It is at an angle of 23.5° from the equator, so that is the maximum inaccuracy that you might have for your cardinal directions. This would actually lead you off in a wrong direction initially, but the ecliptic keeps on moving, so keep on doing this for the whole night, the positions of your North will be varying but the averaged out value will be better than your initial value.&lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorDisplayName="user5434" LastEditDate="2017-11-13T14:10:32.480" LastActivityDate="2017-11-13T14:10:32.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="63" PostTypeId="1" AcceptedAnswerId="68" CreationDate="2013-09-24T23:39:50.153" Score="24" ViewCount="6370" Body="&lt;p&gt;Do all the objects in the universe exert force on all other objects? Like a type of gravity; also, how much does it decrease as it gets farther away?&lt;/p&gt;&#xA;" OwnerUserId="69" LastActivityDate="2016-03-05T04:08:38.763" Title="Do all the objects in the universe exert force on all other objects?" Tags="&lt;gravity&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="65" PostTypeId="2" ParentId="38" CreationDate="2013-09-24T23:47:16.007" Score="17" Body="&lt;p&gt;&lt;a href=&quot;http://solarsystem.nasa.gov/faq/index.cfm?Category=Uranus&quot; rel=&quot;noreferrer&quot;&gt;Nasa.gov&lt;/a&gt; speculates that&#xA;the most likely possibility is that an object with a mass close to that of Earth's collided with Uranus, causing it to rotate on its side from then on.&lt;/p&gt;&#xA;&lt;p&gt;A recent test by &lt;a href=&quot;http://www.space.com/13231-planet-uranus-knocked-sideways-impacts.html&quot; rel=&quot;noreferrer&quot;&gt;Space.com&lt;/a&gt; suggests that &amp;quot;Planet Uranus Got Sideways Tilt From Multiple Impacts&amp;quot;. These findings suggest that two or more smaller collisions probably occurred, asserting that the early solar system was probably more turbulent than most would think.&lt;/p&gt;&#xA;" OwnerUserId="69" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-11-02T21:42:29.397" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="66" PostTypeId="1" AcceptedAnswerId="18998" CreationDate="2013-09-24T23:50:13.957" Score="17" ViewCount="2556" Body="&lt;p&gt;Do planets sometimes wobble and get off their paths? What if an asteroid were to hit it?&lt;/p&gt;&#xA;" OwnerUserId="69" LastEditorUserId="6" LastEditDate="2016-11-08T14:17:50.317" LastActivityDate="2016-11-08T16:37:10.297" Title="Do the orbits of planets change sometimes?" Tags="&lt;planet&gt;&lt;orbit&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="68" PostTypeId="2" ParentId="63" CreationDate="2013-09-24T23:50:42.790" Score="16" Body="&lt;p&gt;Yes - this is the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = G\frac{m_1m_2}{d^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this equation, we can say that all atoms in the universe exert force upon eachother. One &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbon-12&quot; rel=&quot;noreferrer&quot;&gt;carbon-12&lt;/a&gt; atom has a mass of $1.660538921(73)\times10^{-27} kg$. That's a crazy small mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's say that these two atoms are 100,000,000 light years apart. That's $9.461\times10^{23} m$, which is a very long distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if we plug these values into our equation, we get that the force is: $1.709191430132 \times 10^{-59} N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a very, very small amount of force. But it's still force.&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="19" LastEditDate="2014-12-20T18:19:01.867" LastActivityDate="2014-12-20T18:19:01.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="70" PostTypeId="1" CreationDate="2013-09-25T00:04:26.417" Score="15" ViewCount="1026" Body="&lt;p&gt;Is it because it has super massive gravity and energy? Or is it some sort of an anti-light component in it because it used to be a star?&lt;/p&gt;&#xA;" OwnerUserId="69" LastActivityDate="2016-11-09T16:45:51.750" Title="If nothing travels at the speed of light, except light, how can a black hole also pull light into itself?" Tags="&lt;black-hole&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="71" PostTypeId="1" CreationDate="2013-09-25T00:08:29.413" Score="16" ViewCount="764" Body="&lt;p&gt;I know someone or something sucked into a black hole could not survive, but is the bending of light so immense that an observer falling into a black hole will be able to witness any future events in the universe? That would definitely be cool :D I have a feeling this won't work though... so could you explain in detail why? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="69" LastEditorUserId="6" LastEditDate="2016-11-08T14:18:22.697" LastActivityDate="2016-11-08T14:18:22.697" Title="Will an observer falling into a black hole be able to witness any future events in the universe outside the black hole?" Tags="&lt;black-hole&gt;&lt;horizon&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ClosedDate="2016-11-09T08:08:15.593" ContentLicense="CC BY-SA 3.0" />
  <row Id="73" PostTypeId="2" ParentId="70" CreationDate="2013-09-25T00:09:14.533" Score="10" Body="&lt;p&gt;What happens is that the gravity produced by a black hole is simply so great that it warps the spacetime fabric in such a way that any light which crosses the &lt;a href=&quot;http://en.wikipedia.org/wiki/Event_horizon&quot; rel=&quot;nofollow noreferrer&quot;&gt;event horizon&lt;/a&gt; is funneled in by the fabric itself - it's a crazy notion, and one hard to get your head around. This diagram might help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DIswU.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2013-09-25T00:09:14.533" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="74" PostTypeId="2" ParentId="39" CreationDate="2013-09-25T00:20:25.563" Score="4" Body="&lt;p&gt;It isn't clear to me if you are merely interested in the equation so that you can plug numbers in and figure it out, or instead would like to understand how the equation is derived. Either way, is called the two-body problem and with that as a search key, it should not be hard to find online references explaining the derivation and illustrating the equation, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_two-body_problem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Gravitational_two-body_problem&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Two-body_problem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Two-body_problem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You refer to there being two cases, one where A is moving &quot;straightforward&quot; and one where A is orbiting B. This is the same problem, as your first one where 'A is moving straight forward' implies no influence, no gravitational effect from B at all, i.e. it isn't there. If it is there, there is a two-body gravitational effect, which makes it the second case.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2013-09-25T00:20:25.563" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="75" PostTypeId="1" AcceptedAnswerId="242" CreationDate="2013-09-25T00:26:24.657" Score="17" ViewCount="3707" Body="&lt;p&gt;What is the effect known as gravitational lensing? How does it work? What objects would be able to cause this effect?&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2014-03-20T01:59:16.583" Title="What is Gravitational Lensing?" Tags="&lt;light&gt;&lt;gravity&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="76" PostTypeId="2" ParentId="75" CreationDate="2013-09-25T00:34:45.893" Score="11" Body="&lt;p&gt;According to this &lt;a href=&quot;http://imagine.gsfc.nasa.gov/docs/features/news/grav_lens.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA tutorial&lt;/a&gt;, gravitational lensing is when the light emitted from a distant object will be bent by the gravitational field of a large object, such as a galaxy, that is in between us and the distant object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A diagram summarising this effect is below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/icb0w.jpg&quot; alt=&quot;gravity lensing&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.astro.caltech.edu/~george/qsolens/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Discovery of an Unusual Gravitational Lens&quot;&lt;/a&gt; (Courbin et al.)&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2013-09-25T00:34:45.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="77" PostTypeId="2" ParentId="75" CreationDate="2013-09-25T00:37:20.033" Score="6" Body="&lt;p&gt;One star passes in front of another from our perspective. Instead of just blocking the light from the distant star, the light from the distant star that was previously directed to one side, the other side, above, below, - in fact all around the other star - is now bent by the influence of gravity from the nearby star on the photons. As a result, more photons end up heading our way than there were before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What objects would cause this effect? A massive object passing in front of a star.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2013-09-25T00:37:20.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="78" PostTypeId="2" ParentId="75" CreationDate="2013-09-25T00:37:25.163" Score="15" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_lensing&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gravitational lensing&lt;/a&gt; is the effect that large amounts of gravity have on the path of light. Here's an example of the effect:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dh9Af.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The animation shows a black hole passing in front of a galaxy (simulated). In theory any object with a surface mass density larger than the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_lensing_formalism&quot; rel=&quot;nofollow noreferrer&quot;&gt;critical surface mass density&lt;/a&gt; would be able to produce similar such effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's happening is that light goes in a straight line along the spacetime fabric. When a very massive object is in the fabric, it bends the fabric - since the light follows a straight line &lt;em&gt;on the fabric&lt;/em&gt;, the line is also bent. This means that the light goes in a not-straight line, and thus a lensing effect.&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="45" LastEditDate="2013-09-26T15:18:32.143" LastActivityDate="2013-09-26T15:18:32.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="79" PostTypeId="1" AcceptedAnswerId="91" CreationDate="2013-09-25T00:48:51.853" Score="19" ViewCount="1465" Body="&lt;p&gt;What are the most extreme temperatures (both hot and cold) stars have been detected  at?  Is there an upper and lower limit for the detected temperature of stars?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorUserId="19" LastEditDate="2013-09-25T02:20:37.913" LastActivityDate="2019-05-28T19:47:17.963" Title="What is the upper and lower limit of temperatures found on stars?" Tags="&lt;star&gt;&lt;temperature&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="80" PostTypeId="1" AcceptedAnswerId="124" CreationDate="2013-09-25T01:09:44.843" Score="17" ViewCount="3437" Body="&lt;p&gt;According to &lt;a href=&quot;http://www.space.com/41-neptune-the-other-blue-planet-in-our-solar-system.html&quot;&gt;&quot;Neptune: The Other Blue Planet in Our Solar System&quot;&lt;/a&gt; (Choi, 2010), Neptune, despite its great distance from the sun, is the windiest planet in the solar system, with winds reaching 2000km/h (1500 mph) as detected by the Voyager 2 probe in 1989.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that Neptune is 30 AU from the Sun, what are the currently accepted theories for where the planet gets energy for these immense wind speeds?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorUserId="19" LastEditDate="2013-09-25T02:20:21.863" LastActivityDate="2013-09-25T13:13:33.133" Title="What is the current accepted theory for Neptune's immense wind speed?" Tags="&lt;neptune&gt;&lt;climate&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="81" PostTypeId="1" AcceptedAnswerId="86" CreationDate="2013-09-25T01:59:59.087" Score="17" ViewCount="278" Body="&lt;p&gt;Often, a newly discovered star is identified by a seemingly random string of letters and numbers. I'm sure that there is some order to it, though. What is the naming convention for newly discovered astronomical objects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure NASA doesn't ask the public, as something like this would happen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hYkwm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-25T04:32:31.390" LastActivityDate="2013-09-25T04:32:31.390" Title="What is the naming convention for newly discovered objects?" Tags="&lt;naming&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="82" PostTypeId="5" CreationDate="2013-09-25T02:00:34.713" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-25T02:00:34.713" LastActivityDate="2013-09-25T02:00:34.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="83" PostTypeId="4" CreationDate="2013-09-25T02:00:34.713" Score="0" Body="Questions regarding the process of assigning an identifying set of characters to an astronomical object." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-25T04:05:33.160" LastActivityDate="2013-09-25T04:05:33.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="84" PostTypeId="5" CreationDate="2013-09-25T02:16:18.047" Score="0" Body="&lt;p&gt;For questions specifically about the star Earth is orbiting around, use the &lt;a href=&quot;/questions/tagged/the-sun&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;the-sun&amp;#39;&quot; rel=&quot;tag&quot;&gt;the-sun&lt;/a&gt; tag.&lt;/p&gt;&#xA;" OwnerUserId="9527" LastEditorUserId="31410" LastEditDate="2020-12-02T20:42:35.453" LastActivityDate="2020-12-02T20:42:35.453" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="85" PostTypeId="4" CreationDate="2013-09-25T02:16:18.047" Score="0" Body="Questions regarding large spheres of plasma undergoing fusion." OwnerUserId="19" LastEditorUserId="9301" LastEditDate="2016-03-15T04:41:04.633" LastActivityDate="2016-03-15T04:41:04.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="86" PostTypeId="2" ParentId="81" CreationDate="2013-09-25T02:19:55.517" Score="12" Body="&lt;p&gt;The &lt;a href=&quot;https://www.iau.org/public/themes/naming/&quot;&gt;International Astronomical Union's Naming Astronomical Objects&lt;/a&gt; webpage is a comprehensive guide to how astronomical objects are named.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for &lt;a href=&quot;http://cdsweb.u-strasbg.fr/Dic/iau-spec.html&quot;&gt;&quot;Specifications concerning designations for astronomical radiation sources outside the solar system&quot;&lt;/a&gt;, the IAU states that the designation (or sequence of letters and numbers) follow a general form:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Acronym  ^ Sequence ^ (Specifier)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where the &lt;strong&gt;Acronym&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is a code (i.e., alphanumerical string of characters) that specifies the catalog or collection of sources.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;sequence&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The sequence (or numbering) is an alphanumeric string of characters, normally only numerical, that uniquely determines the source within a catalog or collection. It may be a sequence number within a catalog (e. g., HD^224801), a combination of fields, or it may be based on coordinates.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and finally, the &lt;strong&gt;Specifier&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The specifier is optional and allows one to indicate other source parameters. However, they are not required syntax and are enclosed in parentheses. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2013-09-25T02:19:55.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="87" PostTypeId="1" AcceptedAnswerId="88" CreationDate="2013-09-25T02:38:17.540" Score="21" ViewCount="1730" Body="&lt;p&gt;Specifically, what causes them, what sort of damage can they cause, and what are the differences in their composition? &lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2016-08-02T18:18:31.203" Title="What are the main differences between solar flares and coronal mass ejections?" Tags="&lt;solar-flare&gt;&lt;coronal-mass-ejection&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="88" PostTypeId="2" ParentId="87" CreationDate="2013-09-25T03:16:58.277" Score="11" Body="&lt;p&gt;The following definitions are according to NASA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://solarscience.msfc.nasa.gov/flares.shtml&quot;&gt;&quot;Solar Flares&quot;&lt;/a&gt; are gigantic explosions from the surface of the Sun, occurring near sunspots along the dividing/neutral line between opposite magnetic field areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://solarscience.msfc.nasa.gov/CMEs.shtml&quot;&gt;&quot;Coronal Mass Ejections&quot;&lt;/a&gt; (CME's) are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;huge bubbles of gas threaded with magnetic field lines that are ejected from the Sun over the course of several hours.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although the Earth's magnetic field protects the biosphere from the worst of these forms of 'space weather', according to the website &lt;a href=&quot;http://hesperia.gsfc.nasa.gov/sftheory/spaceweather.htm&quot;&gt;&quot;Space Weather: What impact do solar flares have on human activities?&quot;&lt;/a&gt;, the effects that do affect us are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Disturbance of the ionosphere, disrupting radio communications.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;heating, hence expansion of the atmosphere would cause drag on satellites, degrading their orbits, also affecting the accuracy of GPS.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From the Solar Weather site, an important distinction between CME's and solar flares in how they affect the Earth:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Coronal mass ejections are more likely to have a significant effect on our activities than flares because they carry more material into a larger volume of interplanetary space, increasing the likelihood that they will interact with the Earth. While a flare alone produces high-energy particles near the Sun, some of which escape into interplanetary space, a CME drives a shock wave which can continuously produce energetic particles as it propagates through interplanetary space. When a CME reaches the Earth, its impact disturbs the Earth's magnetosphere, setting off a geomagnetic storm. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Very severe CME's can also potentially disrupt power grids and communications.&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2013-09-25T03:16:58.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="89" PostTypeId="1" AcceptedAnswerId="278" CreationDate="2013-09-25T04:15:11.797" Score="11" ViewCount="441" Body="&lt;p&gt;Eventually, the Sun will run out of fuel, go dark, and other bad things will happen. About how much long is it predicted that we will have a sun functioning as it is currently?&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2013-09-27T01:40:42.267" Title="What is the current estimation for how much time the Sun will function properly?" Tags="&lt;the-sun&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="90" PostTypeId="2" ParentId="89" CreationDate="2013-09-25T04:15:11.797" Score="5" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Sun&quot;&gt;Wikipedia:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Sun does not have enough mass to explode as a supernova. Instead it will exit the main sequence in approximately &lt;strong&gt;5.4 billion years&lt;/strong&gt; and start to turn into a red giant. It is calculated that the Sun will become sufficiently large to engulf the current orbits of the solar system's inner planets, possibly including Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, the Sun has more nasty tricks up its sleeves:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Even &lt;strong&gt;before&lt;/strong&gt; it becomes a red giant, the luminosity of the Sun will have &lt;strong&gt;nearly doubled&lt;/strong&gt;, and the Earth will be &lt;strong&gt;hotter than Venus is today&lt;/strong&gt;. Once the core hydrogen is exhausted in 5.4 billion years, the Sun will expand into a subgiant phase and slowly double in size over about half a billion years. It will then expand more rapidly over about half a billion years until it is over &lt;strong&gt;two hundred times larger than today and a couple of thousand times more luminous.&lt;/strong&gt; This then starts the red giant branch (RGB) phase where the Sun will spend around a billion years and lose around a third of its mass.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, it's going to take long enough that we don't really have to worry about it. &lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2013-09-25T04:15:11.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="91" PostTypeId="2" ParentId="79" CreationDate="2013-09-25T04:26:03.407" Score="17" Body="&lt;p&gt;The answer depends on what you'd want to consider as a &quot;star.&quot; If you're just thinking about &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/6/6b/HRDiagram.png&quot;&gt;stars on the main sequence&lt;/a&gt;, then you can just refer to the classical stellar type letters, &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_classification&quot;&gt;OBAFGKM&lt;/a&gt;&quot; (which has relatively recently been extended to accommodate the coolest &lt;a href=&quot;http://en.wikipedia.org/wiki/Brown_dwarf&quot;&gt;brown dwarfs&lt;/a&gt; with the letters &quot;LTY&quot;), where &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_classification#Class_O&quot;&gt;O-stars&lt;/a&gt; are the hottest stars (~30,000 K) and &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_classification#Class_Y&quot;&gt;Y-stars&lt;/a&gt; are the coldest, so-called &quot;room-temperature&quot; stars (~300 K).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Self-gravitating, gaseous objects are incapable of fusing deuterium below about 13 Jupiter masses, and thus simply collapse and cool perpetually (as is the case for all the giant planets in our solar system). These objects can be colder than 300 K but are not technically stars as they do not undergo nuclear fusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For stars that leave the main sequence, two possible outcomes are a &lt;a href=&quot;http://en.wikipedia.org/wiki/White_dwarf&quot;&gt;white dwarf star&lt;/a&gt; or a &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutron_star&quot;&gt;neutron star&lt;/a&gt;, both of which are born extremely hot: White dwarfs are born with surface temperatures of ~10^9 K, whereas neutron stars are born with surface temperatures of ~10^12 K. However, both white dwarfs and neutron stars cool as they age, with the &lt;a href=&quot;http://adsabs.harvard.edu/abs/1997ApJ...489L.157H&quot;&gt;coldest known white dwarfs&lt;/a&gt; being ~3,000 K, and neutron stars cooling to ~10^6 K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer the first part of your question: The coldest known stars are &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_classification#Class_Y&quot;&gt;Y-stars&lt;/a&gt; (i.e. brown dwarfs) and the hottest known stars are either O-stars or young neutron stars, depending on whether you consider objects that have left the main sequence or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as for strict lower and upper limits, the coldest stars possible are likely &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_dwarf&quot;&gt;black dwarfs&lt;/a&gt;, which are what white dwarfs become after cooling for a very long time (&gt;10^15 years). The hottest stars are likely the newly-born neutron stars I previously mentioned, it is very difficult to get much hotter than 10^12 K because any excess energy is carried away via neutrinos.&lt;/p&gt;&#xA;" OwnerUserId="2" LastEditorUserId="2" LastEditDate="2013-09-25T14:17:02.157" LastActivityDate="2013-09-25T14:17:02.157" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="92" PostTypeId="1" CreationDate="2013-09-25T04:35:43.950" Score="24" ViewCount="1637" Body="&lt;p&gt;Distances in our universe can be calculated in several ways: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Stellar Parallax&lt;/li&gt;&#xA;&lt;li&gt;Standard Candles&lt;/li&gt;&#xA;&lt;li&gt;Redshift&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How are these methods linked, and how they are actually used to calculate   distances are measured in the Solar System, in the Galaxy, and in the Universe?&lt;/p&gt;&#xA;" OwnerUserId="12" LastActivityDate="2014-07-26T00:45:49.640" Title="How distances are measured in the Solar System, in the Galaxy, and in the Universe?" Tags="&lt;distances&gt;&lt;parallax&gt;&lt;redshift&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="93" PostTypeId="2" ParentId="70" CreationDate="2013-09-25T04:49:01.333" Score="11" Body="&lt;p&gt;Gravity is a force, and it need not have a &quot;speed&quot;&lt;sup&gt;1&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A gravitational body sets up a gravitational field around itself. Note that by the time the light wave approaches the body, &lt;em&gt;this field has already been set up&lt;/em&gt;. The gravitational force does not need to &quot;reach out&quot; and &quot;catch up to the light&quot; &amp;mdash; it's already there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, it's not even really a force (though we can treat it as one to an approximation). Gravity bends the fabric of spacetime around it, messing with the meaning of a &quot;straight line&quot;. From different reference frames, different lines appear straight. To the light wave, the inspiraling path seems &quot;straight&quot; and thus it follows that path. To an external observer, the light wave is not going straight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1. It does, but that's the speed at which changes to the field propagate, not the speed at which it &quot;catches up&quot; with other things. Changes in the gravitational field propagate at lightspeed.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2013-09-25T04:54:50.397" LastActivityDate="2013-09-25T04:54:50.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="94" PostTypeId="1" AcceptedAnswerId="180" CreationDate="2013-09-25T05:05:19.773" Score="28" ViewCount="3351" Body="&lt;p&gt;Astronomical units of measurement are mostly pretty straight-forward:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Astronomical Units&lt;/strong&gt; refer to the mean Earth-Sun distance (~150 million km or 93 million miles)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Light years&lt;/strong&gt; are the distance light travels in a year (~9.46 × 10^12 km)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Another astronomical unit of measurement is the &lt;strong&gt;parsec&lt;/strong&gt;.  What is a &lt;strong&gt;parsec&lt;/strong&gt; and how is it measured?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-09-28T05:46:58.207" LastActivityDate="2013-11-01T03:47:16.197" Title="What is a parsec and how is it measured?" Tags="&lt;distances&gt;&lt;parsec&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="95" PostTypeId="2" ParentId="94" CreationDate="2013-09-25T05:05:19.773" Score="28" Body="&lt;p&gt;A parsec is approximately 3.26 times a light year or 206,265 Astronomical Units (AU).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://coolcosmos.ipac.caltech.edu/cosmic_classroom/cosmic_reference/distance.html&quot; rel=&quot;noreferrer&quot;&gt;&quot;Cosmic Reference Guide&quot;&lt;/a&gt; (Caltech), the word 'parsec' stands for &quot;parallax of one arc second&quot;, and is (according to the website):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the length of the long leg of a right triangle, whose short leg is one astronomical unit when the angle between the Sun and the Earth, as seen from an object in space (a star for example), is one arcsecond&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Illustrated (from the link above):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/L1RaK.gif&quot; alt=&quot;parsec&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2013-09-25T05:05:19.773" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="96" PostTypeId="1" AcceptedAnswerId="97" CreationDate="2013-09-25T06:24:51.817" Score="16" ViewCount="1868" Body="&lt;p&gt;I know Andromeda is our nearest galaxy. But how many known galaxies have been discovered and what are they?&lt;/p&gt;&#xA;" OwnerUserId="80" LastActivityDate="2015-09-13T19:35:38.810" Title="How many galaxies have been discovered?" Tags="&lt;galaxy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="97" PostTypeId="2" ParentId="96" CreationDate="2013-09-25T06:58:41.663" Score="13" Body="&lt;p&gt;First, the nearest galaxy to ours is not Andromeda, according to the NASA based page &lt;a href=&quot;http://heasarc.nasa.gov/docs/cosmic/nearest_galaxy_info.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;The Nearest Galaxies&quot;&lt;/a&gt;, it was until recently considered to be the &lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/C/Canis+Major+Dwarf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Canis Major Dwarf Galaxy&lt;/a&gt; at approximately 42,000 light years away.  However, recently, there is suggestion that the closer &lt;a href=&quot;http://mcdonaldobservatory.org/news/releases/2008/0402.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Omega Centauri Globular Cluster&lt;/a&gt; may be a disrupted dwarf galaxy (&lt;a href=&quot;http://blackholes.stardate.org/directory/factsheet.php?p=omega-centauri&quot; rel=&quot;nofollow noreferrer&quot;&gt;Another source&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in answer to your other query of how many galaxies there are - there is a global project called &lt;a href=&quot;http://www.galaxyzoo.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Galaxy Zoo&lt;/a&gt; which is attempting to catalogue and classify all observed galaxies, so far, according to the &lt;a href=&quot;http://www.galaxyzoo.org/#/astronomers&quot; rel=&quot;nofollow noreferrer&quot;&gt;Galaxy Zoo for astronomers&lt;/a&gt; page, thus far they have&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;the entire Sloan Digital Sky Survey spectroscopic sample and all existing Hubble Space Telescope surveys (around 1.5 million galaxies in total).&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Pictured: The Hubble Ultra-Deep Field which contains over 10,000 objects, the majority of which are galaxies. The image is only a bit over 3 arcminutes across--a tiny sliver of the sky.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/0/0d/Hubble_ultra_deep_field_high_rez_edit1.jpg/600px-Hubble_ultra_deep_field_high_rez_edit1.jpg&quot; alt=&quot;Hubble Ultra Deep Field&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:46.993" LastActivityDate="2013-09-26T17:41:08.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="98" PostTypeId="1" AcceptedAnswerId="132" CreationDate="2013-09-25T07:04:36.490" Score="15" ViewCount="6623" Body="&lt;p&gt;Leading on from &lt;a href=&quot;https://astronomy.stackexchange.com/questions/42/what-causes-a-star-to-become-a-pulsar/44#44&quot;&gt;this pulsar question&lt;/a&gt;, what criteria/processes make neutron stars become pulsars, do all neutron stars become pulsars?&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2013-09-25T17:59:53.543" Title="What causes a neutron star to become a pulsar?" Tags="&lt;pulsar&gt;&lt;neutron-star&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="101" PostTypeId="1" AcceptedAnswerId="17933" CreationDate="2013-09-25T07:24:43.410" Score="11" ViewCount="186" Body="&lt;p&gt;How do professional astronomers measure and describe (ambient) observation conditions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know smog, light contamination from the city lights, fog, clouds, rain/snow, humidity etc all contribute negatively to quality of the image, obscuring the celestial objects more or less, but I'm fairly sure there is a more precise measurements of viewing capability for any given location or time than &quot;good weather and far from cities&quot; or &quot;Kinda can see M42 in Orion if you squint.&quot;&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2016-08-15T08:44:57.217" Title="What are the metrics of observation conditions?" Tags="&lt;observatory&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="102" PostTypeId="2" ParentId="98" CreationDate="2013-09-25T07:28:15.903" Score="4" Body="&lt;p&gt;I asked Google for this, and the first link that I followed took me to &lt;a href=&quot;http://en.wikipedia.org/wiki/Pulsar&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, where I found the basics as: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A pulsar (portmanteau of pulsating star) is a highly magnetized, rotating neutron star that emits a beam of electromagnetic radiation. This radiation can only be observed when the beam of emission is pointing toward the Earth, much the way a lighthouse can only be seen when the light is pointed in the direction of an observer, and is responsible for the pulsed appearance of emission. Neutron stars are very dense, and have short, regular rotational periods. This produces a very precise interval between pulses that range from roughly milliseconds to seconds for an individual pulsar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As defined, a neutron star &lt;strong&gt;only becomes a pulsar&lt;/strong&gt; once it starts pulsing; which is because of the rotation! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/43Px7.gif&quot; alt=&quot;pulsar&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So your answer would be:&lt;/strong&gt; Any neutron star becomes pulsar, when it starts to &lt;strong&gt;rotate&lt;/strong&gt;, due to rotation it starts to pulsar; which forms the name. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more about them, read this doc of &lt;strong&gt;NASA&lt;/strong&gt;: &lt;a href=&quot;http://imagine.gsfc.nasa.gov/docs/science/know_l1/pulsars.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://imagine.gsfc.nasa.gov/docs/science/know_l1/pulsars.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="79" LastEditorUserId="48" LastEditDate="2013-09-25T16:40:47.760" LastActivityDate="2013-09-25T16:40:47.760" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="104" PostTypeId="1" AcceptedAnswerId="109" CreationDate="2013-09-25T07:55:54.067" Score="20" ViewCount="724" Body="&lt;p&gt;According to the theory (or my understanding of it), the universe is not only expanding, but speeding up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the galaxies are moving apart, are the solar systems within them also moving apart from one another (&amp;amp; the planets within them), if they are, won't the Sun be moving away from the Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the expansion of the universe occur at all scales or just large (inter-galactic) scales?&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2021-01-31T21:09:59.530" Title="On what scale does the universe expand?" Tags="&lt;universe&gt;&lt;expansion&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="105" PostTypeId="2" ParentId="96" CreationDate="2013-09-25T08:00:36.880" Score="12" Body="&lt;p&gt;If you want to know the number of &lt;code&gt;spectroscopically confirmed&lt;/code&gt; galaxies then the number given by UV-D is roughly correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the actual number of galaxies that have been observed is order of magnitudes larger. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case I know is the Sloan Digital Sky Survey 3 (SDSS3):&#xA;they observed about a third of the sky using five different filters between optical and near infrared. Of all the objects detected &lt;code&gt;208,478,448&lt;/code&gt; are galaxies. And of those 1.5 to 2 millions will have the spectrum measured by mid next year. &#xA;&lt;a href=&quot;http://www.sdss3.org/dr8/scope.php&quot;&gt;More info here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To these number you have to add the galaxies observed by surveys in different areas of the sky, deeper that SDSS (which means that they can seen fainter and/or more fare away objects). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, if memory does not fail me, estimates gives about 10^10 galaxies in the observed universe&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2013-09-25T08:00:36.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="107" PostTypeId="1" AcceptedAnswerId="111" CreationDate="2013-09-25T08:08:51.417" Score="6" ViewCount="664" Body="&lt;p&gt;According to the NASA website &lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2013/17aug_hotjupiters/&quot;&gt;&quot;The strange attraction of Hot Jupiters&quot;&lt;/a&gt;, one of the main types of exoplanets that have been detected are Hot Jupiters, which are &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;These are behemoth worlds that orbit close to their parent stars, blocking a fraction of the star’s light when it transits in front.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Planets that are from as large as Saturn through to far larger than Jupiter, orbiting in a matter of days very close to their parent star.  Something our solar system does not have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What proportion of star systems have Hot Jupiters within their systems?  Is it a case that Hot-Jupiter-less systems such as our own are unusual?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2016-04-19T19:17:10.587" Title="What proportion of planetary systems have been found with 'Hot Jupiters'" Tags="&lt;exoplanet&gt;&lt;hot-jupiter&gt;&lt;solar-system&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="108" PostTypeId="2" ParentId="101" CreationDate="2013-09-25T08:09:46.547" Score="8" Body="&lt;p&gt;I'm not an observer, but there are a couple of quantities I know:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;code&gt;seeing&lt;/code&gt;: is a measure of atmospheric turbulence and airmass. Its units are &lt;code&gt;arcseconds&lt;/code&gt; and is the measured size of a point source, typically a star, as it appears in the image. In short: atmospheric turbulences shift slightly the position of the star, so if you observed it for long enought, the star will appear blurred. Good places for telescopes have seeing of few arcsecond; Excellent ones reach about 0.6 arcsec. &lt;a href=&quot;http://www.eso.org/gen-fac/pubs/astclim/paranal/seeing/&quot;&gt;This link&lt;/a&gt; can give some more info&lt;/li&gt;&#xA;&lt;li&gt;fog, cloud, rain, snow: you don't even bother opening the telescope domes as you won't be able to observe much&lt;/li&gt;&#xA;&lt;li&gt;humidity: I think that part enter in the &lt;code&gt;seeing&lt;/code&gt;. If you do just imaging what you get is some extra blurring of the image. If you do spectroscopy you will see a lot of atmospheric lines (due atoms and molecules, e.g. water, in the atmosphere) that you want to remove from the measured spectrum.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2013-09-25T09:12:49.883" LastActivityDate="2013-09-25T09:12:49.883" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="109" PostTypeId="2" ParentId="104" CreationDate="2013-09-25T08:22:07.337" Score="14" Body="&lt;p&gt;The short answer is: gravitationally bound system will not be ripped by the accelerated expansion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A longer answer: The current standard model (&lt;a href=&quot;http://en.wikipedia.org/wiki/Lambda-CDM_model&quot;&gt;$\Lambda$CDM model&lt;/a&gt;) says that everything started with the big bang. It released a lot of energy that pushed the Universe and is since then expanding. While it's expanding, gravitational attraction has been working to create the structures that we see today (planets, stars, galaxies, cluster of galaxies). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the Universe expands, the density of matter and radiation within the Universe decreases. At about 5 billions years after the big bang, these densities became smaller than the density of Dark Energy (&lt;a href=&quot;http://en.wikipedia.org/wiki/Dark-energy-dominated_era&quot;&gt;Dark-energy-dominated era&lt;/a&gt;). The strange things about Dark Energy are that, 1) its  density remains a constant regardless of the expansion of the universe, and 2) it has anti-gravitational properties - it pushes rather than pulls. Once Dark Energy came to dominate the density of the universe it began to dominate its overall dynamics, and has caused the expansion of the universe to accelerate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in structures which have already formed, gravity is much stronger than dark energy and so they don't feel it. &lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2013-10-28T14:38:22.667" LastActivityDate="2013-10-28T14:38:22.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="111" PostTypeId="2" ParentId="107" CreationDate="2013-09-25T08:33:36.823" Score="7" Body="&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1012.5676&quot; rel=&quot;nofollow&quot;&gt;In 2011, about 20% of the exoplanets found were hot Jupiters&lt;/a&gt;. That is a lot, but it is strongly biased simply because there are the easiest planets to detect. You can detect exoplanets by transit (that means that the planet passes in front of its host star which decreases the luminosity of the observed star during the transit), hot Jupiters being closer to their host star and bigger than smaller planets will therefore be easier to detect. You can detect exoplanets by radial velocity, which means that you detect variation of the position of the host star due to planet's gravity. The closer and the heavier the planet is, the larger the variation will be, and therefore the easier to detect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why it is important to say &quot;In 2011&quot;; as our detection techniques (and telescopes) improve, the proportion of hot Jupiters should decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;To put that in perspective, you can read &lt;a href=&quot;http://www.personal.psu.edu/jtw13/blogs/astrowright/2012/07/how-many-hot-jupiters.html&quot; rel=&quot;nofollow&quot;&gt;this article from Jason Wright&lt;/a&gt; in which he tries to estimate the &quot;real&quot; ratio of hot Jupiters orbiting around &quot;normal&quot; stars (F, G and K type stars). This ratio is actually much smaller than the currently observed ratio, about 1.2%.&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2013-09-25T09:20:20.667" LastActivityDate="2013-09-25T09:20:20.667" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="113" PostTypeId="2" ParentId="92" CreationDate="2013-09-25T08:58:13.880" Score="13" Body="&lt;h3&gt;Stellar Parallax&lt;/h3&gt;&#xA;&lt;p&gt;Stellar parallax uses differences in perspective to determine the distance from an object. When the earth goes around the sun, our perspective of the star, galaxy etc. changes and so the angle from us to the object changes. Because we know how the earth moves around the sun, we know the distance between the points that we take the measurements. From this we can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_sines&quot; rel=&quot;noreferrer&quot;&gt;Sine Rule&lt;/a&gt; to determine the distance from the object.&lt;/p&gt;&#xA;&lt;p&gt;For Example, if the distance between the two points of measurement were 2 units, and the two inside angles (&lt;em&gt;A&lt;/em&gt; and &lt;em&gt;B&lt;/em&gt;) were (&lt;em&gt;A&lt;/em&gt;)80° and (&lt;em&gt;B&lt;/em&gt;)75° (and therefore the other angle (&lt;em&gt;C&lt;/em&gt;) is 25°) we can say that:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$a = \frac{2\sin80}{\sin25}\ $&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &amp;quot;&lt;em&gt;a&lt;/em&gt;&amp;quot; is the opposite side to angle &amp;quot;&lt;em&gt;A&lt;/em&gt;&amp;quot;. This means that the distance between the object and the &lt;em&gt;&lt;strong&gt;second&lt;/strong&gt;&lt;/em&gt; (&lt;em&gt;B&lt;/em&gt;) point of measurement is approximately 4.66 units.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/5ObPo.png&quot; rel=&quot;noreferrer&quot;&gt;Diagram here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h3&gt;Standard Candles&lt;/h3&gt;&#xA;&lt;blockquote&gt; A standard candle is an astronomical object that has a known absolute magnitude. (&lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/S/Standard+Candle&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;) &lt;/blockquote&gt; &#xA;&lt;p&gt;By using the known absolute and the measured &lt;em&gt;apparent magnitude&lt;/em&gt; (what is seen by instruments etc.) it is possible to determine the distance of the object by using the formula:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;m-M = 5 log d – 5&lt;br /&gt;&#xA;where &lt;em&gt;m&lt;/em&gt; is the apparent magnitude of the object, &lt;em&gt;M&lt;/em&gt; is the absolute magnitude of the object, and &lt;em&gt;d&lt;/em&gt; is the distance to the object in parsecs.(&lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/S/Standard+Candle&quot; rel=&quot;noreferrer&quot;&gt;same source&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;One of the most widely used examples of these standard candles is the supernovae type Ia, which was used to discover that the universe is accelerating in its expansion. This led to the awarding of the 2011 Nobel Prize in Physics to &lt;a href=&quot;http://www.nobelprize.org/nobel_prizes/physics/laureates/2011/perlmutter-facts.html&quot; rel=&quot;noreferrer&quot;&gt;Saul Perlmutter&lt;/a&gt;, &lt;a href=&quot;http://www.nobelprize.org/nobel_prizes/physics/laureates/2011/schmidt-facts.html&quot; rel=&quot;noreferrer&quot;&gt;Brian Schmidt&lt;/a&gt; , and &lt;a href=&quot;http://www.nobelprize.org/nobel_prizes/physics/laureates/2011/riess-facts.html&quot; rel=&quot;noreferrer&quot;&gt;Adam Riess&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Additionally, &lt;a href=&quot;http://en.wikipedia.org/wiki/RR_Lyrae_variable&quot; rel=&quot;noreferrer&quot;&gt;RR Lyrae&lt;/a&gt; are another form of standard candle used to measure galactic distances (SN Ia are much more luminous and therefore can be used to much higher distances). RR Lyrae are a type of variable star with a relationship between pulsation period and absolute magnitude, which is what makes them good for use as standard candles.&lt;/p&gt;&#xA;&lt;p&gt;Other information &lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/S/Standard+Candle&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h3&gt;Redshift&lt;/h3&gt;&#xA;&lt;p&gt;Redshift is the effect where light from an object is &amp;quot;stretched&amp;quot; out so that the wavelength is moved towards the &amp;quot;red&amp;quot; end of the spectrum. Distances are calculated by using Hubble's Law, but I'll let others explain it: &lt;a href=&quot;http://www.astro.cornell.edu/academics/courses/astro201/hubbles_law.htm&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;, &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/astro/hubble.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://csep10.phys.utk.edu/astr162/lect/cosmology/hubble_constant.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2014-07-26T00:45:49.640" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="114" PostTypeId="1" AcceptedAnswerId="118" CreationDate="2013-09-25T10:57:55.510" Score="9" ViewCount="3152" Body="&lt;p&gt;I know that minimal light pollution is a must for stargazing, and a place which is away from civilization is better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does altitude of a place matter for light pollution? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it affect the quality of star gazing?&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="6" LastEditDate="2016-05-25T15:42:45.423" LastActivityDate="2016-05-25T15:42:45.423" Title="What qualifies as a good place for stargazing, i.e. with least light pollution?" Tags="&lt;star-gazing&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="115" PostTypeId="1" AcceptedAnswerId="236" CreationDate="2013-09-25T11:02:30.763" Score="15" ViewCount="1551" Body="&lt;p&gt;How hot must a star get before it actually becomes a star? Why does it need to get so hot? Please find an official site to quote from, if you can.&lt;/p&gt;&#xA;" OwnerUserId="69" LastEditorDisplayName="user96" LastEditDate="2013-09-25T13:27:06.540" LastActivityDate="2015-10-16T16:54:02.067" Title="How hot must a star get before it is considered to be a star?" Tags="&lt;star&gt;&lt;heat&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="117" PostTypeId="2" ParentId="114" CreationDate="2013-09-25T11:22:23.200" Score="3" Body="&lt;p&gt;Yes, altitude does play a role. The higher you go, less photons from civilization will reach you, as most of it will be scattered by dust in the air. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as to answer &quot;what qualifies as a good place&quot;, as if you are asking for a criterion, a popular one is : &quot;If you can see all the stars of Little Dipper, well that's a 'good' place to observe&quot;.&lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="63" LastEditDate="2013-09-26T01:53:41.130" LastActivityDate="2013-09-26T01:53:41.130" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="118" PostTypeId="2" ParentId="114" CreationDate="2013-09-25T11:43:02.043" Score="8" Body="&lt;p&gt;A good question, and in the early 2000s John Bortle published a categorization of a variety of conditions, with descriptions for each category. It is the commonly used scale to describe to others the sort of conditions at a location. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably one of the more significant factors provided by a dark sky site is: how faint do stars have to be for you not to see them anymore (overwhelmed by light pollution). There are specific stellar regions, each a triangle, and the idea is you could how many stars you can see in the triangle, and you look the number up in a table, and it will tell you the magnitude limit you are perceiving at that site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, high altitudes also help - less air between you and the stars means better seeing conditions/less atmospheric distortion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being able to see more stars IS better for star gazing. Living in a city, I am continually frustrated at the poor skies, and love being in dark sky locations like the International Dark Skies Reserve at Tekapo, New Zealand.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2013-09-25T11:43:02.043" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="120" PostTypeId="2" ParentId="66" CreationDate="2013-09-25T12:23:41.723" Score="5" Body="&lt;p&gt;The orbits of the planets and other bodies are changing all the time, because of the influence of other bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, planets can change orbit significantly because of the influence of other planets. There is a theory that has Neptune and Uranus swapping orbital positions a long time ago.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2013-09-25T12:23:41.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="122" PostTypeId="1" AcceptedAnswerId="123" CreationDate="2013-09-25T12:44:04.737" Score="15" ViewCount="669" Body="&lt;p&gt;Do other universes like ours exist?&#xA;If they exist, &lt;strong&gt;how do we know that they exist when we have even not seen the ends of our own universe?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="19" LastEditDate="2013-12-05T19:01:59.800" LastActivityDate="2017-08-14T10:02:14.483" Title="Do parallel universes exist?" Tags="&lt;universe&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="123" PostTypeId="2" ParentId="122" CreationDate="2013-09-25T12:59:54.703" Score="13" Body="&lt;p&gt;The simple answer is (as with so much in astronomy):&lt;br/&gt;&#xA;&lt;strong&gt;We Don't Know&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Parallel universes may or may not exist. There is no definitive way to prove that these universes do or don't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A parallel universe is a separate existence to ours. The Theories that suggest that there may be parallel universes are classified as theories of &lt;strong&gt;multiverses&lt;/strong&gt;. There are many theories of multiverse, all of which propose different ideas about what could exist beyond the limits of our universe. There are also theories that suggest that the multiverse doesn't exist, although the theories with most support are by far the multiverse theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a nice reference in book form, see Steinhardt and Turok's &quot;Endless Universe: Beyond the Big Bang&quot;. Also, see Max Tegmark's work on multiverses levels I-IV (&lt;a href=&quot;http://space.mit.edu/home/tegmark/crazy.html&quot;&gt;Max Tegmark -&gt; See his Scientific American article entitled Parallel Universes&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="45" LastEditDate="2013-09-26T01:34:24.047" LastActivityDate="2013-09-26T01:34:24.047" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="124" PostTypeId="2" ParentId="80" CreationDate="2013-09-25T13:13:33.133" Score="11" Body="&lt;p&gt;&lt;em&gt;Disclaimer: Most of what is written here is based highly on theory and speculation. Voyager 2 is the one and only trip we have taken to Neptune and most of these theories are built upon the data it has sent back. Leave large room for error&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is Neptune windy?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before we start answering why Neptune is windy we should first specify if Neptune is even windy in the first place. What proof do we have?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The Great Dark Spot&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;One observation was of a large abhorrence on the surface of Neptune aptly titled &quot;The Great Dark Spot&quot;. Repeated viewings placed the movement of the GDS to be about 700mph. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/T1StK.jpg&quot; alt=&quot;The Great Dark Spot!&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sourced from: &lt;a href=&quot;http://nineplanets.org/neptune.html&quot; rel=&quot;noreferrer&quot;&gt;http://nineplanets.org/neptune.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Keck Telescope&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A lot of observations and images of Neptune have come from the Keck II Telescope, which is able to observe other markings moving at similar to faster speeds on Neptune. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sourced From: &lt;a href=&quot;http://www.keckobservatory.org/recent/entry/keck_images_of_neptune_best_ever_captured1&quot; rel=&quot;noreferrer&quot;&gt;http://www.keckobservatory.org/recent/entry/keck_images_of_neptune_best_ever_captured1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What causes the winds on Neptune?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's theorised that there are two potential causes to Neptune's winds. Shallow processes in the outer atmosphere or much deeper atmospheric changes that extend into the interior. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.space.com/21157-uranus-neptune-winds-revealed.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.space.com/21157-uranus-neptune-winds-revealed.html&lt;/a&gt; (paraphrased for simplicity) Claims that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The researchers found the windy layers of Neptune occupy the outermost 0.2 percent of It's mass, suggesting that shallow processes drive those winds, such as swirling caused by moisture condensing and evaporating in the atmosphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the main source of energy for these winds?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now you are probably wondering where Neptune get's this energy from as it is so much further from the Sun than we are. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://nineplanets.org/neptune.html&quot; rel=&quot;noreferrer&quot;&gt;http://nineplanets.org/neptune.html&lt;/a&gt; Claims that this is due to Neptune's own internal heat source radiating twice as much energy as it receives from the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why does Neptune have stronger winds than Earth then?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So theoretically, if it was down to energy, it would make sense that Earth, being so much closer to the Sun, would have much more energy required to reach these speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes and No. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun is the main provider for our energy but it also creates turbulence in the atmosphere. It's cooling and heating of the layers keeps air constantly rising and falling, disturbing natural air flows and stemming the size of our storms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On top of this &lt;a href=&quot;http://www.space.com/18922-neptune-atmosphere.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.space.com/18922-neptune-atmosphere.html&lt;/a&gt; claims that 80% of Neptune's atmosphere is molecular Hydrogen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen being the smallest and lightest element requires less energy to reach high speeds than our larger, nitrogen based atmosphere. This may not be a main factor to consider but it is certainly contributing in no small amount. &lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastActivityDate="2013-09-25T13:13:33.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="125" PostTypeId="1" AcceptedAnswerId="163" CreationDate="2013-09-25T13:49:59.317" Score="12" ViewCount="722" Body="&lt;p&gt;&lt;a href=&quot;http://www.nasa.gov/mission_pages/cassini/whycassini/titan-clouds.html&quot;&gt;This NASA site&lt;/a&gt; says that hidden beneath the smog of Titan are Cirrus like clouds and &lt;a href=&quot;http://www.space.com/14247-saturn-moon-titan-atmosphere-earth.html&quot;&gt;This space article&lt;/a&gt; describes how Titan's atmosphere is similar to Earth's too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What other features does Titan boast that make scientists so ready to boast that &quot;Titan may well be the most Earth like planet we know of besides Earth itself&quot;? (paraphrased)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answers should focus on the features of Titan itself, obviously a similarity is that it orbits the same Sun, but this type of thing is not what I'm after. I am mainly searching for geological or atmospheric similarities that Titan shares all the way down to potential similarities in the composition and flow of the planetary core. &lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastActivityDate="2013-09-26T00:14:11.867" Title="What are the Earth-like features of Titan?" Tags="&lt;earth-like-planet&gt;&lt;titan&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="127" PostTypeId="2" ParentId="115" CreationDate="2013-09-25T14:23:26.527" Score="7" Body="&lt;p&gt;&lt;strong&gt;From a physics perspective&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a physics perspective an object is a star when it is undergoing nuclear fusion, generally of hydrogen atoms at its core, this is regardless of its temperature!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A star is not determined by its temperature, it is instead determined by it's internal processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does mean that if Jupiter began nuclear fusion it would be considered a star, albeit a minuscule one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case it is a yes/no distinction of if an object is a star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From an observational point of view once something is classified as a star there are 7 groups it can fall in to determined by its features.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sourced From: &lt;a href=&quot;http://en.wikipedia.org/wiki/Star#Classification&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Star#Classification&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Class &lt;strong&gt;Temperature&lt;/strong&gt; &lt;br/&gt; O:    &lt;strong&gt;33,000 K+&lt;/strong&gt;&lt;br/&gt;&#xA;  B:    &lt;strong&gt;10,500–30,000 K&lt;/strong&gt;  &lt;br/&gt; A:   &lt;strong&gt;7,500–10,000 K&lt;/strong&gt; &lt;br/&gt;&#xA;  F:    &lt;strong&gt;6,000–7,200 K&lt;/strong&gt;    &lt;br/&gt; G:   &lt;strong&gt;5,500–6,000 K&lt;/strong&gt; &lt;br/&gt;&#xA;  K:    &lt;strong&gt;4,000–5,250 K&lt;/strong&gt;    &lt;br/&gt; M:   &lt;strong&gt;2,600–3,850 K&lt;/strong&gt;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: Three more classifications L T and Y have been added to the colder end of this list, but I am unsure of the cut off points so omitted them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But strangely they are not classified by temperature but by their spectrum, it just so happens that their spectrum correlates to their temperature! The temperature spoken of here is of the photosphere of the star (where the photons begin free streaming), not its core (where photons are created from ongoing fusion reactions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dwarf stars have their own classification system prefixed by the letter D though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quote from Wiki article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;White dwarf stars have their own class that begins with the letter D. This is further sub-divided into the classes DA, DB, DC, DO, DZ, and DQ, depending on the types of prominent lines found in the spectrum. This is followed by a numerical value that indicates the temperature index.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user96" LastEditorDisplayName="user96" LastEditDate="2013-09-26T14:51:25.247" LastActivityDate="2013-09-26T14:51:25.247" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="130" PostTypeId="1" AcceptedAnswerId="137" CreationDate="2013-09-25T14:48:33.620" Score="34" ViewCount="8098" Body="&lt;p&gt;(Yes I'm excluding Pluto from this the same way it was excluded for not being a planet)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observing the planets orbit of the Sun they all seem relatively planar and roughly all orbit along the same plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this due to the way that our Solar system was formed or is this a physical phenomena observed in other systems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bjntY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastEditorUserId="6" LastEditDate="2020-09-17T14:13:22.657" LastActivityDate="2020-09-17T14:13:22.657" Title="Why do the planets in our solar system orbit in the same plane?" Tags="&lt;orbit&gt;&lt;planet&gt;&lt;solar-system&gt;&lt;celestial-mechanics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="131" PostTypeId="1" AcceptedAnswerId="135" CreationDate="2013-09-25T14:55:33.303" Score="21" ViewCount="13374" Body="&lt;p&gt;According to &lt;a href=&quot;http://lro.gsfc.nasa.gov/moonfacts.html&quot;&gt;The NASA Moon Facts&lt;/a&gt; page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The moon is actually moving away from earth at a rate of 1.5 inches per year.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why is the moon moving further away from the Earth? Is this a result of the moons formation that set it in motion to spiral away from us? Or is this a resultant force of the gravity from the Sun and other large bodies?&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastActivityDate="2017-09-09T15:10:08.333" Title="Is the moon moving further away from Earth and closer to the Sun? Why?" Tags="&lt;the-moon&gt;&lt;gravity&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="132" PostTypeId="2" ParentId="98" CreationDate="2013-09-25T14:56:39.043" Score="8" Body="&lt;p&gt;It is thought that &lt;strong&gt;all neutron stars are pulsars when they are first born&lt;/strong&gt;, as all spin extremely rapidly and have strong magnetic fields shortly after birth, arising from their dramatic collapse from a few Earth-sized massive star core into a city-sized neutron star. Conservation of angular momentum means that an initially slow-rotating core becomes a rapidly spinning neutron star, which also drives a dynamo action that generates a strong magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The emission from a neutron star comes as a result of pair-production in its magnetosphere, which has an approximately dipolar shape and thus the strongest field strength at its poles. This field is rarely aligned with the spin axis, thus giving the &quot;lighthouse&quot; effect where the beam sweeps over an observer at twice the spin frequency. These pair-produced particles radiate energy via synchrotron emission (primarily in the radio), sapping energy from the magnetic field. As the neutron star ages, the magnetic field eventually decays below a critical point where electrons/positrons cannot be pair-produced. This is called the &quot;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1993ApJ...402..264C&quot; rel=&quot;noreferrer&quot;&gt;death line&lt;/a&gt;,&quot; illustrated here and denoted by the word &quot;graveyard&quot; on a P-P_dot diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UvrDY.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After crossing this threshold, the pulsar loses its ability to emit radio waves. Pulsars only remain pulsars for the first ~10^7 years of their lives, and thus about 99% of neutron stars are no longer pulsars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, we can only see a pulsar if its beam crosses over the Earth while the pulsar rotates. As the beam angle is usually &lt;a href=&quot;http://adsabs.harvard.edu/abs/2010ApJ...715..230O&quot; rel=&quot;noreferrer&quot;&gt;tens of degrees&lt;/a&gt;, the total fraction of pulsars whose beams cross the Earth is approximately 10% the total that exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the lower-left of the diagram you can see very short period pulsars, these are &lt;a href=&quot;http://en.wikipedia.org/wiki/Millisecond_pulsar&quot; rel=&quot;noreferrer&quot;&gt;millisecond pulsars&lt;/a&gt; and have been spun up by accreting mass from a companion. These pulsars have very long lifetimes, but only represent a small subset of the neutron star population.&lt;/p&gt;&#xA;" OwnerUserId="2" LastEditorUserId="2" LastEditDate="2013-09-25T17:59:53.543" LastActivityDate="2013-09-25T17:59:53.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="133" PostTypeId="1" AcceptedAnswerId="141" CreationDate="2013-09-25T15:08:24.590" Score="17" ViewCount="335" Body="&lt;p&gt;Every time I hear about a black hole, it's always in conjunction with the collapse of a star. Have any other processes been observed to create a black hole?&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2014-12-17T16:15:50.700" Title="Are there ways other than the collapse of a star which have been observed to form black holes?" Tags="&lt;black-hole&gt;&lt;star&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="134" PostTypeId="1" AcceptedAnswerId="136" CreationDate="2013-09-25T15:09:58.843" Score="16" ViewCount="500" Body="&lt;p&gt;White holes are purely hypothetical at this point and are considered only because it is assumed something like it &lt;em&gt;could&lt;/em&gt; exist as the balance to black-holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reading &lt;a href=&quot;http://io9.com/5805202/mysterious-cosmic-explosion-might-be-first-ever-proof-of-white-holes&quot; rel=&quot;noreferrer&quot;&gt;this article&lt;/a&gt; seems mostly writers taking liberties with the information but poses an interesting question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a White Hole?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JBtvW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastActivityDate="2017-06-01T15:24:40.177" Title="What is a White Hole?" Tags="&lt;white-hole&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="135" PostTypeId="2" ParentId="131" CreationDate="2013-09-25T15:11:46.327" Score="21" Body="&lt;p&gt;Yes, the moon is moving away from Earth at around 1.48&amp;quot; per year. According to &lt;a href=&quot;http://www.bbc.co.uk/news/science-environment-12311119&quot; rel=&quot;noreferrer&quot;&gt;the BBC&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The Moon is kept in orbit by the gravitational force that the Earth exerts on it, but the Moon also exerts a gravitational force on our planet and this causes the movement of the Earth's oceans to form a tidal bulge.&lt;/p&gt;&#xA;&lt;p&gt;Due to the rotation of the Earth, this tidal bulge actually sits slightly ahead of the Moon. Some of the energy of the spinning Earth gets transferred to the tidal bulge via friction.&lt;/p&gt;&#xA;&lt;p&gt;This drives the bulge forward, keeping it ahead of the Moon. The tidal bulge feeds a small amount of energy into the Moon, pushing it into a higher orbit like the faster, outside lanes of a test track.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So, tidal forces are ultimately what causes this to happen.&lt;/p&gt;&#xA;&lt;p&gt;Also, there is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Tidal_acceleration&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article on tidal forces&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Tidal acceleration is an effect of the tidal forces between an orbiting natural satellite (e.g. the Moon), and the primary planet that it orbits (e.g. the Earth). The acceleration causes a gradual recession of a satellite in a prograde orbit away from the primary, and a corresponding slowdown of the primary's rotation. The process eventually leads to tidal locking of first the smaller, and later the larger body. The Earth–Moon system is the best studied case.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2014-11-27T20:00:27.187" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="136" PostTypeId="2" ParentId="134" CreationDate="2013-09-25T15:28:06.260" Score="8" Body="&lt;p&gt;Well according to the article &lt;a href=&quot;http://earthsky.org/space/have-we-seen-a-white-hole&quot;&gt;&quot;Have we seen a white hole?&quot;&lt;/a&gt; (Byrd, 2011) there are some claims that scientists (specifically &lt;a href=&quot;http://arxiv.org/vc/arxiv/papers/1105/1105.2776v1.pdf&quot;&gt;&quot;The Revival of White Holes as Small Bangs&quot;&lt;/a&gt; (Retter and Heller)) may have detected a white hole; however, this claim seems to be shrouded in controversy, as this observation may actually have a more stellar explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the article from Retter and Heller offer a hypothesis as to what white holes are, from the article linked before:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We thus suggest that the emergence of a white hole, which we name a ‘Small Bang’, is spontaneous – all the matter is ejected at a single pulse. Unlike black holes, white holes cannot be continuously observed rather their effect can only be detected around the event itself. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which, if the authors are correct, would explain why we don't see/detect them and white holes (unlike black holes) are unique because they supposedly spew matter, rather than extract.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One major caveat is that it is not clear as to where this matter that spews out comes from.&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-09-25T15:40:29.330" LastActivityDate="2013-09-25T15:40:29.330" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="137" PostTypeId="2" ParentId="130" CreationDate="2013-09-25T15:45:40.130" Score="25" Body="&lt;p&gt;In the protostar stage of the Sun, it was surrounded by a (spinning) gas cloud. This cloud behaved like a fluid (well, a gas &lt;em&gt;is&lt;/em&gt; a fluid), so it flattened out into an accretion disk due to conservation of angular momentum. The planets eventually formed from the dust/gas in the disk from compression of the dust in the disk. This process won't end up moving the dust out of the plane (all the vertical force of gravity is &lt;em&gt;toward&lt;/em&gt; the disk), so the final planet is in the plane too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does an accretion disk have to be flat? Well, firstly, let's imagine the protostar and gas cloud before the accretion disk formed. Usually such a setup will have particles spinning in mostly one direction. The ones spinning in retrograde orbits will end up reversing themselves due to collisions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this gas sphere, there will be an equal number of particles with positive and negative vertical velocities (at a given point in time; due to rotation the velocity signs will flip). From collisions, eventually these will all become zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A particle revolving around a planet will &lt;em&gt;always&lt;/em&gt; revolve such that the projection on the planet is a great circle. Thus we cannot have a particle with vertical velocity zero but with vertical position nonzero (as that would imply an orbit that isn't a great circle). So as the vertical velocity decreases, the orbit inclination decreases too. Eventually leading to an accretion disk with very little vertical spread. &lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="45" LastEditDate="2013-09-26T01:12:04.680" LastActivityDate="2013-09-26T01:12:04.680" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="138" PostTypeId="1" AcceptedAnswerId="140" CreationDate="2013-09-25T15:46:34.720" Score="86" ViewCount="7386" Body="&lt;p&gt;The planets rotate as an after effect of their creation, the dust clouds that compressed span as they did so and the inertia has kept it rotating ever since. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's fairly easy to prove that planetary bodies are rotating just by watching their features move across their respective horizons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems less easy to discern for amateur astronomers  in the case of the sun though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the Sun also rotate as a by-product of it's creation? What evidence is there to support this? Does the sun have any discerning features that make it evident it is rotating?&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastEditorUserId="9186" LastEditDate="2015-10-16T14:48:10.793" LastActivityDate="2018-04-10T15:20:16.850" Title="Does the Sun rotate?" Tags="&lt;the-sun&gt;&lt;rotation&gt;&lt;star-formation&gt;&lt;plasma-physics&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="139" PostTypeId="2" ParentId="133" CreationDate="2013-09-25T15:59:30.427" Score="10" Body="&lt;p&gt;Well, when two black holes collide a new black hole is created.  You could call that cheating, but this process is very different from that of a star collapsing. Also, &lt;a href=&quot;http://www.ligo.caltech.edu/&quot;&gt;these guys&lt;/a&gt; would be after your skin if you did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When two black holes collide, there are three stages. Firstly, there's the &quot;inspiral&quot;, which is exactly what it sounds like &amp;mdash; the two black holes slowly spiral in to each other. Cue waltz music. Then there's the &quot;Coalescence&quot;, where the two black holes merge. Note that prior to the merge, their absolute horizons actually &lt;em&gt;grow&lt;/em&gt; in size, as if in anticipation of the merge. The exact behavior of the coalescence stage is not well understood yet. Finally, there's a &quot;ringdown&quot;, where the new black hole (currently shaped like an eight) shakes down into an ellipsoid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has not been observed, but is something we're pretty sure occurs.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There also are primoridial black holes, from the time of the creation of the universe. They are extremely tiny (and unstable), though I doubt they can be formed at this current era.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2013-09-25T16:19:32.237" LastActivityDate="2013-09-25T16:19:32.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="140" PostTypeId="2" ParentId="138" CreationDate="2013-09-25T16:04:10.447" Score="89" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_rotation&quot;&gt;Yes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not rotate uniformly though, different portions have a different angular velocity (as a body made of plasma, it can get away with this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measuring this in theory is pretty easy, we just need to track the motion of the sunspots. This isn't as simple as calculating the changes in relative positions of the sunspots, though, as the Earth is rotating and revolving, which makes the calculations harder. This measurement can be done using the celestial sphere (the field of stars that we see) as a &quot;fixed&quot; reference point and seeing how the Earth and the sunspots move relative to that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost everything in the universe rotates/revolves, at least a little bit, because angular momentum is hard to get rid of. It can be transferred from body to body, but for a body to end up with zero angular momentum, it needs to meet another body with the exact same angular momentum and collide with it in a particular way. Given that this is pretty rare, all celestial bodies rotate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to that, a non rotating body that is revolving will eventually start spinning due to tidal forces.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2013-09-25T17:07:10.630" LastActivityDate="2013-09-25T17:07:10.630" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="141" PostTypeId="2" ParentId="133" CreationDate="2013-09-25T16:06:15.777" Score="11" Body="&lt;p&gt;There are currently three classifications of black hole and these are &lt;a href=&quot;http://hubblesite.org/reference_desk/faq/answer.php.id=62&amp;amp;cat=exotic&quot;&gt;stellar, super massive and miniature.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stellar&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stellar black holes are formed by a massive star collapsing and its mass being great enough to form a black hole of moderate size&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Super-massive&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These black holes can have the mass equivalent of a billion suns, these most likely exist at the centre of galaxies and are probably as old as time itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Miniature&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Miniature black holes are theoretical and are thought to have been created at the early stages of the universe. these require masses less than that of our sun and are thus very unlikely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;But Rhys&quot; I hear you cry, &quot;these are all from stars you fool!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, yes they are, this is because black holes are formed from mass, and lots of mass (except miniatures), coincidentally large amounts of mass tends to gravitate towards each other and form stars long before it forms black holes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore theoretically, a black hole does not even need stars, it just needs lots and lots of mass, but having that much mass without forming a star is no easy task. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short answer, you can make a black hole from anything as long as it has enough mass, but good luck stopping this from forming a star beforehand!&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastActivityDate="2013-09-25T16:06:15.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="142" PostTypeId="5" CreationDate="2013-09-25T16:11:13.457" Score="0" Body="&lt;p&gt;The Universe is all the contents of space and time — in essence, it is everything that exists, has existed, and will exist, everywhere.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2015-11-30T21:16:14.107" LastActivityDate="2015-11-30T21:16:14.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="143" PostTypeId="4" CreationDate="2013-09-25T16:11:13.457" Score="0" Body="Questions regarding all of time and space and its contents." OwnerUserId="19" LastEditorDisplayName="user14291" LastEditDate="2017-09-01T00:35:24.780" LastActivityDate="2017-09-01T00:35:24.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="144" PostTypeId="2" ParentId="24" CreationDate="2013-09-25T16:21:36.963" Score="20" Body="&lt;p&gt;A black hole can also be detected by how it bends light as various bodies move behind it. This phenomenon is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_lens&quot; rel=&quot;noreferrer&quot;&gt;gravitational lensing&lt;/a&gt;, and is the most visually stunning prediction of Einstein's theory of General Relativity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image portrays the geometry of gravitational lensing. Light from luminous background objects are bent due to the warping of space-time in the presence of mass (here, the red dot could conceivably be the black hole in question):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AJHEK.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomers have discovered the existence of a super-massive black hole at the center of our very own Milky Way Galaxy, and has been dubbed &lt;a href=&quot;http://en.wikipedia.org/wiki/Sagittarius_A*&quot; rel=&quot;noreferrer&quot;&gt;Sagittarius A*&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over a period of ten years, the trajectories of a small group of stars have been tracked, and the only explanation for their rapid movement is the existence of a highly compact object with the mass of about 4 million suns. Given the mass and distance scales involved, the conclusion is that it must be a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bYThB.png&quot; alt=&quot;Trajectories of stars around the Sagittarius A* within the Milky Way Galaxy.&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastEditorUserId="45" LastEditDate="2013-09-26T07:41:59.910" LastActivityDate="2013-09-26T07:41:59.910" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="145" PostTypeId="1" AcceptedAnswerId="146" CreationDate="2013-09-25T17:22:09.057" Score="25" ViewCount="4095" Body="&lt;p&gt;I've seen a lot of beautiful pictures of the Milky Way from Earth such as this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/W1Jgf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...but I can't understand what the cloudy ribbon at the bottom the horizon is. Is it a super-large nebula? Or is it due to the Gegenschein? What happens in the ribbon?&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="104" LastEditDate="2013-12-06T11:56:04.783" LastActivityDate="2013-12-06T11:56:04.783" Title="Milky Way Galaxy from Earth" Tags="&lt;galaxy&gt;&lt;milky-way&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="146" PostTypeId="2" ParentId="145" CreationDate="2013-09-25T17:28:34.007" Score="21" Body="&lt;p&gt;The Milky Way Galaxy is a large spiral galaxy with some characteristic features worth mentioning: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The bulge - This refers to the collection of tightly packed stars located in the central region of the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The spiral arms or the disk - This region extends from the inner region of the galaxy (where it meets the bulge) to outskirts of the galaxy, and contains stars, dust, and lots of gas. It's also very thin. The reason why a disk forms is largely thought to be due to slight asymmetries in the accretion of material early on in the galaxy's history. Over time, and due to conservation of angular momentum, the material collapses down into a disk. Incidentally, this is where the sun lives. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QxNbg.jpg&quot; alt=&quot;Diagram of galactic components.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That ribbon is the disk of the Milky Way. It looks cloudy because of the dust and gas which scatters light from the rest of the galaxy. The reason why gas accumulates here is because as things begin to collapse into a disk, the gas collides with itself, and sort of 'sticks' together. In other words, where a bunch of stars may very happily pass through a bunch of other stars without many of them colliding or being disturbed, gas has a much harder time doing this. If you want to learn more about what happens when galaxies and clusters collide, take a look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bullet_Cluster&quot; rel=&quot;noreferrer&quot;&gt;Bullet Cluster&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uJbpS.gif&quot; alt=&quot;Milky Way cartoon.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The majority of the other stars you you see which are not part of that structure are much more local. The reason why the rest of the sky looks rather transparent in comparison is because you are looking out of the plane of the galaxy. There is simply much less stuff (namely gas and dust which obscures your sight in optical wavelengths) that you are looking through. Because of this, most optical telescopes look in these directions when studying the rest of the universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more about the classification and structures of galaxies, see the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble_sequence&quot; rel=&quot;noreferrer&quot;&gt;Hubble sequence&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditDate="2013-09-26T03:22:39.577" LastActivityDate="2013-09-26T03:22:39.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="147" PostTypeId="2" ParentId="133" CreationDate="2013-09-25T17:29:08.130" Score="6" Body="&lt;p&gt;It is &lt;em&gt;theoretically&lt;/em&gt; possible to create black holes with radiation, it is called a  &lt;a href=&quot;http://en.wikipedia.org/wiki/Kugelblitz_%28astrophysics%29&quot;&gt;&quot;Kugelblitz&quot;&lt;/a&gt;. The technology needed to create bright enougth pulses of gamma radiation is beyond our current technology, but may be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possibilities for the usage of such miniature black holes are enormous, it would solve all energy problems because they can convert between 20% to 50% of the introduced mass directly to energy. If the hawking radiation exists its possible to get 100% but enougth scifi.&lt;/p&gt;&#xA;" OwnerUserId="102" LastEditorUserId="19" LastEditDate="2013-09-25T17:30:31.980" LastActivityDate="2013-09-25T17:30:31.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="148" PostTypeId="1" CreationDate="2013-09-25T17:50:54.560" Score="12" ViewCount="2640" Body="&lt;p&gt;I'm envisioning an analogy of the universe as a plane with marbles in that plane to represent different planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if this plane expands how does this work? I assume that it is not like 'nothing' becomes 'something' then but rather expanding/stretching the universe; but doesn't this imply that everything around us is actually stretching/becoming bigger (like every atom/quarks) becomes a little bit bigger? Or is it just the space between those subatomic particles and combinations of these become bigger? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example could be the structure of carbon dioxide becoming larger, but only because there is more space between the carbon and the oxygen atoms.&lt;/p&gt;&#xA;" OwnerUserId="105" LastEditorDisplayName="user8" LastEditDate="2013-09-26T21:35:42.713" LastActivityDate="2020-10-17T15:03:40.157" Title="Do all objects get larger in an &quot;expanding universe&quot;?" Tags="&lt;expansion&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="149" PostTypeId="2" ParentId="148" CreationDate="2013-09-25T18:04:56.717" Score="11" Body="&lt;p&gt;It would, if there were no interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, everyday matter is held together by other kinds of forces (electromagnetic, mainly). These are quite strong, and any &quot;stretching&quot; of space will be counteracted by these forces &quot;pulling&quot; the atom/molecule/object back to its original size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the expansion of the universe only matters on the larger, cosmic scale, between objects like galaxies which are separated by a large enough distance that they don't affect each other too much, gravitationally speaking.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2013-09-26T13:11:27.510" LastActivityDate="2013-09-26T13:11:27.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="150" PostTypeId="1" AcceptedAnswerId="281" CreationDate="2013-09-25T18:22:54.840" Score="22" ViewCount="13149" Body="&lt;p&gt;I've had a dobsonian telescope for a while now, but I've never been able to collimate  it. I thought I had it collimated once, but when I looked through it the star didn't look right, it looked kind of like a round disc with a hole in the middle. I know the basic idea behind doing it with a laser collimator, you need to get the laser to line up with the center of the primary mirror as well as line up in the center of the target on the laser itself. However, I always have a lot of difficult accomplishing this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For starters, I always have a lot of trouble getting the secondary mirror lined up. I'll get really close, but once I get close I have a lot of trouble doing the fine tuning to get the laser in the middle. I'll get it in, but when I got to tighten the other screws it will no longer be lined up. I'll try to account for the tightening as well, but it's always a guessing game for me when it comes to knowing which screws to loosen and tighten so that the laser winds up where I want.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have that many problems with the primary mirror, and can normally get that lined up fairly easily. However, usually after lining up the primary mirror the secondary mirror no longer seemed to be lined up and I need to do the secondary mirror all over again. I'm assuming this is to be expected, but considering the difficulty I have lining up the secondary mirror it's very frustrating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried looking around at resources online, but none of them go into enough detail into aligning the secondary mirror. They just say &quot;adjust the screws until it's lined up&quot;, which doesn't solve my problem, since I have trouble figuring out exactly which screws to adjust and when. I'm also confused as to what I should see through my telescope when it is properly aligned. Most of the sites just use a diagram, but I'm never really sure what to make of the diagram. The one time I did line it up it looked like a round disc with a hole in the middle, which I didn't think looked right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to summarize, the main problems I have are:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I have trouble fine tuning the secondary mirror, since I have trouble accounting for the movement of the mirror when tightening the two screws I didn't loosen   &lt;/li&gt;&#xA;&lt;li&gt;I'm not sure what a star should look like through my telescope when it is properly collimated &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Can anyone offer some detailed instructions, and possibly a video, of how to collimate a dobsonian telescope with a laser collimator? &lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="1378" LastEditDate="2014-09-17T00:11:32.553" LastActivityDate="2014-09-17T00:11:32.553" Title="How can I collimate a dobsonian telescope with a laser collimator?" Tags="&lt;dobsonian-telescope&gt;&lt;collimation&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="151" PostTypeId="2" ParentId="125" CreationDate="2013-09-25T18:48:37.547" Score="5" Body="&lt;p&gt;A lot - from &lt;a href=&quot;http://www.space.com/7103-titan-world-earth.html&quot;&gt;this space.com article&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Saturn's moon Titan may be worlds away from Earth, but the two bodies have some characteristics in common: &lt;strong&gt;Wind, rain, volcanoes, tectonics&lt;/strong&gt; and other Earth-like processes all sculpt features on Titan, but act in an environment &lt;strong&gt;more &#xA;  frigid than Antarctica&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The article goes on to say that while Titan has these earth like features, it only receives about 1% of the sunlight that Earth receives. Because of this, the average temperature on the moon has an average surface temperature of -180 C (-292 degrees F), and water can only exist there in a super-frozen form as hard as rock.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On Titan, &lt;strong&gt;methane takes water's place in the hydrological cycle&lt;/strong&gt; of evaporation and precipitation (rain or snow) and can appear as a gas, a liquid and a solid. Methane rain cuts channels and forms lakes on the surface and causes erosion, helping to erase the meteorite impact craters that pockmark most other rocky worlds, such as our own moon and the planet Mercury.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for the planet's core, &lt;a href=&quot;http://news.stanford.edu/pr/2012/pr-titan-saturn-moon-120312.html&quot;&gt;this Stanford paper&lt;/a&gt; has some info:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The picture of Titan that we get has an icy, rocky core with a radius of a little over 2,000 kilometers, an ocean somewhere in the range of 225 to 300 kilometers thick and an ice layer that is 200 kilometers thick,&quot; he said.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not going to quote the whole paper here, but it's a good read on the subject of Titan's core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://en.wikipedia.org/wiki/Atmosphere_of_Titan&quot;&gt;Wikipedia article&lt;/a&gt; has some information about Titan's atmosphere:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Observations from the Voyager space probes have shown that the Titanian atmosphere is denser than Earth's, with a surface pressure about 1.45 times that of Earth's. Titan's atmosphere is about 1.19 times as massive as Earth's overall, or about 7.3 times more massive on a per surface area basis. It supports opaque haze layers that block most visible light from the Sun and other sources and renders Titan's surface features obscure. The atmosphere is so thick and the gravity so low that humans could fly through it by flapping &quot;wings&quot; attached to their arms. Titan's lower gravity means that its atmosphere is far more extended than Earth's; even at a distance of 975 km, the Cassini spacecraft had to make adjustments to maintain a stable orbit against atmospheric drag. The atmosphere of Titan is opaque at many wavelengths and a complete reflectance spectrum of the surface is impossible to acquire from the outside. It was not until the arrival of the Cassini–Huygens mission in 2004 that the first direct images of Titan's surface were obtained. The Huygens probe was unable to detect the direction of the Sun during its descent, and although it was able to take images from the surface, the Huygens team likened the process to &quot;taking pictures of an asphalt parking lot at dusk&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19" LastActivityDate="2013-09-25T18:48:37.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="152" PostTypeId="2" ParentId="148" CreationDate="2013-09-25T19:55:09.920" Score="-1" Body="&lt;p&gt;Ummm this question made me think about a lecture from my teacher. When he was teaching us about the Big Bang theory!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;From Wikipedia:&lt;/strong&gt; The &lt;strong&gt;big bang&lt;/strong&gt; states: The Big Bang theory is the prevailing cosmological model that describes the early development of the Universe.&lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Bang&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt; According to the theory, the Big Bang occurred approximately 13.798 ± 0.037 billion years ago, which is thus considered the age of the universe. At this time, the Universe was in an extremely hot and dense state and began expanding rapidly. After the initial expansion, the Universe cooled sufficiently to allow energy to be converted into various subatomic particles, including protons, neutrons, and electrons. As stated &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Bang&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example of balloon:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what you can imagine, there is a baloon with some dots on it. When you are blowing it, the balloon gets bigger, its size expands. But the dots stay where they were. There do comes a change in their size too. Galaxies expand too, but that doesn't mean that there is amount being added, but that means there is some space some extra space being added to it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Objects expand:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objects around you aren't expanding. Because this change would occur in years, so its almost ~0 this size as the changes occur inside the atom, atom contains magnetic field by the movement of electrons that would lesser down expansion and bla bla which is Chemistry so skip this. However, galaxies, stars, and on whole universe is expanding! And its been years it was expanding, and will expand! :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your plane example:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And your plane? Imagine that the plane is expanding. Hey wait! try this one (if possible, as this one would explain it fairly), place some magnets over a magnetic field, make sure north faces the north so they keep floating. Now when the surface expands, the magnets would stay there! They won't move backward or forwards just as the universe (plane) is moving. Similarly, as universe expands, planets don't expand! As they are not stitched to the universe or clamped to the universe. They are free moving. &lt;/p&gt;&#xA;" OwnerUserId="79" LastActivityDate="2013-09-25T19:55:09.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="153" PostTypeId="2" ParentId="148" CreationDate="2013-09-25T20:32:22.313" Score="7" Body="&lt;p&gt;No - Distances between objects get larger, but galaxies and clusters are dominated locally by gravity. The scale at which the expansion of the universe is significant is larger than the largest bound structures in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you thought objects in the universe grew and stretched with the expansion, you would be forced to predict things which we do not actually observe, like the tearing apart of galaxies/clusters of galaxies at very high redshifts.&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2013-09-25T20:32:22.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="156" PostTypeId="1" CreationDate="2013-09-25T21:15:21.110" Score="9" ViewCount="15168" Body="&lt;p&gt;I wonder, why stars take a really long time to become stars? Is it because it needs to gain mass? Or heat up? Something like that? And could it be sped up at all?&lt;/p&gt;&#xA;" OwnerUserId="69" LastEditorUserId="13" LastEditDate="2013-09-28T06:06:22.353" LastActivityDate="2020-08-17T15:29:24.990" Title="How long does it take to produce a star? Why does it take that long?" Tags="&lt;star&gt;&lt;formation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="158" PostTypeId="1" AcceptedAnswerId="160" CreationDate="2013-09-25T21:18:32.467" Score="6" ViewCount="699" Body="&lt;p&gt;I have learned that there are white holes and black holes in the universe, but I am confused as to what a wormhole is. Is it a tunnel that connects black and white holes?&lt;/p&gt;&#xA;" OwnerUserId="69" LastEditorUserId="19" LastEditDate="2013-09-25T22:18:08.290" LastActivityDate="2013-09-26T00:00:18.580" Title="What is a wormhole?" Tags="&lt;black-hole&gt;&lt;white-hole&gt;&lt;wormhole&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="159" PostTypeId="2" ParentId="158" CreationDate="2013-09-25T23:09:42.300" Score="4" Body="&lt;p&gt;A wormhole is a theorised phenomenon that would, if they exist, connect two different parts of space-time. A common analogy for the wormhole is to represent space-time as a two-dimensional sheet. If the two-dimensional sheet is bended over so that two points touch each other, than the point of contact becomes a wormhole. This has been suggested as a way to effectively travel faster than the speed of light (without actually travelling faster than the speed of light). As with most things, there are different theories about how a wormhole could exist and what its properties would be.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2013-09-25T23:09:42.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="160" PostTypeId="2" ParentId="158" CreationDate="2013-09-26T00:00:18.580" Score="5" Body="&lt;p&gt;A wormhole is, in theory, a 'tunnel' which would be a shortcut to another point in spacetime. There's a &lt;a href=&quot;http://en.wikipedia.org/wiki/Wormholes&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; on the subject:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A wormhole, also known as an Einstein–Rosen bridge, is a hypothetical topological feature of spacetime that would be, fundamentally, a &amp;quot;shortcut&amp;quot; through spacetime. For a simple visual explanation of a wormhole, consider spacetime visualized as a two-dimensional (2D) surface. If this surface is folded along a third dimension, it allows one to picture a wormhole &amp;quot;bridge&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If you're the &lt;a href=&quot;https://astronomy.stackexchange.com/users/28/manishearth&quot;&gt;physicsy type&lt;/a&gt;, this might make some sense to you:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If a Minkowski spacetime contains a compact region &lt;em&gt;Ω&lt;/em&gt;, and if the topology of &lt;em&gt;Ω&lt;/em&gt; is of the form&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Ω ~ R x Σ&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;em&gt;Σ&lt;/em&gt; is a three-manifold of the nontrivial topology, whose boundary has topology of the form&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;∂Σ ~ S2&lt;/em&gt;,&lt;/p&gt;&#xA;&lt;p&gt;and if, furthermore, the hypersurfaces &lt;em&gt;Σ&lt;/em&gt; are all spacelike, then the region Ω contains a quasipermanent intra-universe wormhole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If this makes absolutely no sense to you, then you are just like me. Congratulations! For those that don't speak Greek math (such as me), here's an English explanation from &lt;a href=&quot;http://www.space.com/20881-wormholes.html&quot; rel=&quot;noreferrer&quot;&gt;space.com&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A wormhole is a theoretical passage through space-time that could create shortcuts for long journeys across the universe. Wormholes are predicted by the theory of general relativity. But be wary: wormholes bring with them the dangers of sudden collapse, high radiation and dangerous contact with exotic matter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And here's a pretty picture of the effect:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WkLcI.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;A model of 'folded' space-time illustrates how a wormhole bridge might form with at least two mouths that are connected to a single throat or tube.&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;In plain English, it's a shortcut through spacetime, and the primary conceivable way time travel could happen.&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2013-09-26T00:00:18.580" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="161" PostTypeId="5" CreationDate="2013-09-26T00:02:13.573" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-26T00:02:13.573" LastActivityDate="2013-09-26T00:02:13.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="162" PostTypeId="4" CreationDate="2013-09-26T00:02:13.573" Score="0" Body="Questions regarding a theoretical 'tunnel' through spacetime which could potentially allow time travel." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-26T01:12:13.760" LastActivityDate="2013-09-26T01:12:13.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="163" PostTypeId="2" ParentId="125" CreationDate="2013-09-26T00:14:11.867" Score="6" Body="&lt;p&gt;To expand a bit on Undo's answer, there are some geomorphological features on the surface of Titan that resemble similar features on Earth, &lt;em&gt;suggesting&lt;/em&gt; similar processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, according to &lt;a href=&quot;http://www.nasa.gov/mission_pages/cassini/multimedia/titan20120419-image.html&quot; rel=&quot;noreferrer&quot;&gt;this NASA image comparison&lt;/a&gt;, some lake features have features that are similar to groundwater based salt pans on Earth. Image below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lUuAf.jpg&quot; alt=&quot;NASA imagery&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The feature on the left is Ontario Lacus on Titan compared to the Etosha Pan of Earth on the right.  It should be noted, that the Titan feature is considerably larger than the Earth analogue.&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2013-09-26T00:14:11.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="164" PostTypeId="1" CreationDate="2013-09-26T00:53:58.033" Score="16" ViewCount="1562" Body="&lt;p&gt;Asides from the main Asteroid Belt, there are Trojan Asteroids - what are these and where are they located?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2013-09-26T00:53:58.033" Title="What are Trojan Asteroids and where are they located?" Tags="&lt;trojan-asteroids&gt;&lt;asteroids&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="165" PostTypeId="2" ParentId="164" CreationDate="2013-09-26T00:53:58.033" Score="10" Body="&lt;p&gt;A &lt;strong&gt;Trojan Asteroid&lt;/strong&gt; is defined by the Swinburne University Astro page &lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/T/Trojan+Asteroids&quot; rel=&quot;noreferrer&quot;&gt;&quot;Trojan Asteroids&quot;&lt;/a&gt; as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Asteroids sharing an orbit with a planet, but which are located at the leading (L4) and trailing (L5) Lagrangian points. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These are often divided into 'leading' and 'trailing', as per the diagram below (from the source linked above):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7h8MN.jpg&quot; alt=&quot;Trojan orbits&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though the term &lt;strong&gt;Trojan asteroid&lt;/strong&gt; is often associated with the 4800 or so that lead and follow Jupiter, it actually applies to any asteroid that displays this behaviour, associated with any planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are numerous Trojans co-orbiting with &lt;a href=&quot;http://www.minorplanetcenter.org/iau/lists/MarsTrojans.html&quot; rel=&quot;noreferrer&quot;&gt;Mars&lt;/a&gt; and &lt;a href=&quot;http://www.minorplanetcenter.org/iau/lists/NeptuneTrojans.html&quot; rel=&quot;noreferrer&quot;&gt;Neptune&lt;/a&gt;.  Recently, single Trojans have been found co-orbiting &lt;a href=&quot;http://rt.com/news/uranus-trojan-asteroid--252/&quot; rel=&quot;noreferrer&quot;&gt;Uranus&lt;/a&gt; and even &lt;a href=&quot;http://www.space.com/12443-earth-asteroid-companion-discovered-2010-tk7.html&quot; rel=&quot;noreferrer&quot;&gt;Earth&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2013-09-26T00:53:58.033" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="166" PostTypeId="1" AcceptedAnswerId="167" CreationDate="2013-09-26T02:04:09.627" Score="15" ViewCount="381" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Star number 12644769 from the Kepler Input Catalog was identified as&#xA;  an eclipsing binary with a 41-day period, from the detection of its&#xA;  mutual eclipses (9). &lt;strong&gt;Eclipses occur because the orbital plane of the&#xA;  stars is oriented nearly edge- on as viewed from Earth&lt;/strong&gt;. During primary&#xA;  eclipses, the larger star, denoted “A,” is partially eclipsed by the&#xA;  smaller star “B,” and the system flux de- clines by about 13%&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.sciencemag.org/content/333/6049/1602&quot;&gt;http://www.sciencemag.org/content/333/6049/1602&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the thing though: out of all possible edge-on configurations, there are far more configurations where a planet can never be in the position to be edge-on, rather than configurations where a planet could potentially be in the position to be edge on. (I'd guess that it happens in fewer than one in several hundred cases)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why are we able to observe so many transits? &lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="7982" LastEditDate="2021-04-16T23:20:30.080" LastActivityDate="2021-04-16T23:20:30.080" Title="What percent of planets are in the position that they could be viewed edge-on from Earth? (and thus able to undergo transits)" Tags="&lt;observational-astronomy&gt;&lt;exoplanet&gt;&lt;eclipsing-star-systems&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="167" PostTypeId="2" ParentId="166" CreationDate="2013-09-26T02:04:38.347" Score="11" Body="&lt;p&gt;Because there are so many planets out there!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://certificate.ulo.ucl.ac.uk/modules/year_one/NASA_Kepler/character.html&quot; rel=&quot;noreferrer&quot;&gt;There just happens to be an entire web page dedicated to calculating that answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Transits can only be detected if the planetary orbit is near the line-of-sight (LOS) between the observer and the star. This requires that the planet's orbital pole be within an angle of $d_*/a$ (part 1 of the figure below) measured from the center of the star and perpendicular to the LOS, where $d_{*}$ is the stellar diameter (= 0.0093 AU for the Sun) and $a$ is the planet's orbital radius.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This is possible for all $2\pi$ angles about the LOS, i.e., for a total of $4\pi d_*/2a$ steradians of pole positions on the celestial sphere (part 2 of figure).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Thus the geometric probability for seeing a transit for any random planetary orbit is simply $d_*/2a$ (part 3 of figure) (&lt;a href=&quot;http://dx.doi.org/10.1016/0019-1035%2884%2990102-7&quot; rel=&quot;noreferrer&quot;&gt;Borucki and Summers, 1984&lt;/a&gt;, &lt;a href=&quot;http://kepler.nasa.gov/files/mws/koch.circumstellar1.pdf&quot; rel=&quot;noreferrer&quot;&gt;Koch and Borucki, 1996&lt;/a&gt;).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2KYOW.gif&quot; alt=&quot;Diagram&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For the Earth and Venus this is 0.47% and 0.65% respectively (see above Table). Because grazing transits are not easily detected, those with a duration less than half of a central transit are ignored. Since a chord equal to half the diameter is at a distance of 0.866 of the radius from the center of a circle, the usable transits account for 86.6% of the total. If other planetary systems are similar to our solar system in that they also contain two Earth-size planets in inner orbits, and since the orbits are not co-planar to within $2d_*/D$, the probabilities can be added. Thus, approximately $0.011 \times 0.866$ $= 1\%$ of the solar-like stars with planets should show Earth-size transits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's pretty freaking amazing! Kepler has been up there for a short while, and has a possible list of nearly 2000 planets just looking at about 150,000 stars for only a couple of years! So if only 1% statistically transit, that would mean that just randomly 1500 systems would have the correct orientation (&lt;a href=&quot;http://en.wikipedia.org/wiki/Kepler_%28spacecraft%29#Mission_results_to_date&quot; rel=&quot;noreferrer&quot;&gt;given the results to date, that makes sense&lt;/a&gt;). And given that about 7500 stars were eliminated from consideration due to being variable of one sort or another... I think it would be pretty safe to say that pretty much every star out there has at least some sort of planetary body around it.&lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="2531" LastEditDate="2015-05-03T20:15:25.530" LastActivityDate="2015-05-03T20:15:25.530" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="168" PostTypeId="1" AcceptedAnswerId="169" CreationDate="2013-09-26T02:09:04.033" Score="15" ViewCount="295" Body="&lt;p&gt;I was watching the &lt;a href=&quot;http://en.wikipedia.org/wiki/TED_%28conference%29&quot;&gt;TED&lt;/a&gt; lecture &lt;em&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=rjECbQ1r-k0&amp;amp;feature=uploademail&quot;&gt;How to defend Earth from asteroids &lt;/a&gt;&lt;/em&gt;, and the speaker &lt;a href=&quot;http://en.wikipedia.org/wiki/Phil_Plait&quot;&gt;Phil Plait&lt;/a&gt; spoke of the possibility of this asteroid hitting Earth. Is this really a concern?&lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="9582" LastEditDate="2021-04-30T18:42:12.437" LastActivityDate="2021-04-30T18:42:12.437" Title="Is the Apophis asteroid a concern for planet Earth?" Tags="&lt;orbit&gt;&lt;asteroids&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="169" PostTypeId="2" ParentId="168" CreationDate="2013-09-26T02:09:35.223" Score="7" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/99942_Apophis&quot;&gt;Apophis&lt;/a&gt; is of no real concern to us as far as we can tell.  That said, it doesn't mean another asteroid doesn't have our name on it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.psi.edu/epo/faqs.htm&quot;&gt;According to the most reliable data we have regarding this particular asteroid&lt;/a&gt; it only has a 1 chance in about 250,000 of hitting the earth.  &lt;a href=&quot;http://www.ifa.hawaii.edu/info/press-releases/Apophis-TholenOct09/&quot;&gt;University of Hawaii states&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;“Our new orbit solution shows that Apophis will miss Earth’s surface in 2036 by a scant 20,270 miles, give or take 125 miles,” Tholen said. “That's slightly closer to Earth than most of our communications and weather satellites.” He credits the large telescopes and superb atmospheric conditions on Mauna Kea for being able to make these determinations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Considering how big space is, this is pretty much a hit in cosmic terms.  Keep in mind that &lt;a href=&quot;http://www.nasa.gov/home/hqnews/2009/oct/HQ_09-232_Apophis_Update.html&quot;&gt;this information has been readily available since 2009 at the NASA website as well&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="114" LastActivityDate="2013-09-26T02:09:35.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="170" PostTypeId="1" AcceptedAnswerId="171" CreationDate="2013-09-26T02:16:10.907" Score="17" ViewCount="2211" Body="&lt;p&gt;I've read various ideas about why the moon looks larger on the horizon. The most reasonable one in my opinion is that it is due to how our brain calculates (perceives) distance, with objects high above the horizon being generally further away than objects closer to the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But every once in a while, the moon looks absolutely huge and has a orange red color to it. Booth the size and color diminish as it moves further above the horizon. This does not seem to fit in with the regular perceived size changes that I already mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the name of this giant orange red effect and what causes it?&lt;/p&gt;&#xA;" OwnerUserId="114" LastActivityDate="2017-10-15T12:33:44.460" Title="Why does the moon sometimes appear giant and a orange red color near the horizon?" Tags="&lt;the-moon&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="171" PostTypeId="2" ParentId="170" CreationDate="2013-09-26T02:16:21.673" Score="14" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nEMol.jpg&quot; alt=&quot;Harvest Moon&quot;&gt; (Source, Wikipedia Commons)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon is generally called a &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Harvest_moon&quot; rel=&quot;noreferrer&quot;&gt;Harvest Moon&lt;/a&gt;&quot; when it appears that way (i.e. large and red) in autumn, amongst a few other names.  There are other names that are associated with specific timeframes as well.  The colour is due to &lt;a href=&quot;http://curious.astro.cornell.edu/question.php?number=280&quot; rel=&quot;noreferrer&quot;&gt;atmospheric scattering&lt;/a&gt; (Also known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Rayleigh_scattering&quot; rel=&quot;noreferrer&quot;&gt;Rayleigh scattering&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;may have noticed that they always occur when the Sun or Moon is close to the horizon. If you think about it, sunlight or moonlight must travel through the maximum amount of atmosphere to get to your eyes when the Sun or Moon is on the horizon (remember that that atmosphere is a sphere around the Earth). So, you expect &lt;em&gt;more&lt;/em&gt; blue light to be scattered from Sunlight or Moonlight when the Sun or Moon is on the horizon than when it is, say, overhead; this makes the object look redder.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As to the size, that is commonly referred to as the &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Moon_illusion&quot; rel=&quot;noreferrer&quot;&gt;Moon Illusion&lt;/a&gt;&quot;, which may be a combination of many factors.  The most common explanation is that the frame of reference just tricks our brains.  Also, if you look straight up, the perceived distance is much smaller to our brains than the distance to the horizon.  We don't perceive the sky to be a hemispherical bowl over us, but rather a much more shallow bowl.  Just ask anyone to point to the halfway point between the horizon and zenith, and you will see that the angle tends to be closer to 30 degrees as opposed to the 45 it should be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://facstaff.uww.edu/mccreadd/&quot; rel=&quot;noreferrer&quot;&gt;University of Wisconsin discussion on the Moon Illusion&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2005/20jun_moonillusion/&quot; rel=&quot;noreferrer&quot;&gt;NASA discussion on moon illusion&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.moillusions.com/2008/12/moon-optical-illusion.html&quot; rel=&quot;noreferrer&quot;&gt;A graphical representation of this&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uSP93.jpg&quot; alt=&quot;Optical Illusion illustrated&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://blogs.discovermagazine.com/badastronomy/2010/05/13/why-does-the-moon-look-so-huge-on-the-horizon/&quot; rel=&quot;noreferrer&quot;&gt;Dr. Phil Plait discusses the illusion in detail&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="114" LastActivityDate="2013-09-26T02:16:21.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="172" PostTypeId="1" AcceptedAnswerId="181" CreationDate="2013-09-26T02:18:01.870" Score="11" ViewCount="572" Body="&lt;p&gt;According to the &lt;a href=&quot;http://solarsystem.nasa.gov/planets/profile.cfm?Object=Nep_Triton&quot;&gt;NASA overview about Triton&lt;/a&gt;, this, the largest satellite of Neptune is unique in that it has a retrograde orbit around Neptune.  The page also suggests that Triton is a captured Kuiper Belt object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the current accepted hypothesis as to what caused Triton's retrograde orbit?  Also, is this orbit sustainable - meaning, is it degrading?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2013-09-28T05:31:31.810" Title="What is the current accepted hypothesis of what caused Triton's retrograde orbit around Neptune?" Tags="&lt;orbit&gt;&lt;neptune&gt;&lt;triton&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="173" PostTypeId="1" AcceptedAnswerId="174" CreationDate="2013-09-26T02:19:07.780" Score="22" ViewCount="651" Body="&lt;p&gt;There are 51 stars within 17 light years of the Earth (&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_nearest_stars&quot;&gt;source&lt;/a&gt;). If one of these stars was to become a supernova, how would they effect the Earth?&lt;/p&gt;&#xA;" OwnerUserId="114" LastActivityDate="2013-09-27T10:36:05.040" Title="Near-Earth supernova" Tags="&lt;supernova&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="174" PostTypeId="2" ParentId="173" CreationDate="2013-09-26T02:19:43.550" Score="19" Body="&lt;p&gt;None of those stars &lt;em&gt;can&lt;/em&gt; go supernova, so the question is rather moot.  If you look at the classifications, the most luminous is &lt;a href=&quot;http://en.wikipedia.org/wiki/Sirius#Sirius_A&quot;&gt;Sirius A&lt;/a&gt; (an &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_classification#Class_A&quot;&gt;A sequence&lt;/a&gt; star even) you can get an idea of its mass. If you look at your source page, and link to the explanation you see that &lt;a href=&quot;http://en.wikipedia.org/wiki/A_V_star&quot;&gt;A stars range from 1.4 to 2.1 stellar masses&lt;/a&gt;. In order to go supernova though, &lt;a href=&quot;http://en.wikipedia.org/wiki/Type_II_supernova&quot;&gt;you need about nine solar masses&lt;/a&gt;.  The nearest supernova candidate is &lt;a href=&quot;http://en.wikipedia.org/wiki/Supernova#Milky_Way_candidates&quot;&gt;IK Pegasi (HR 8210), located at a distance of 150 light-years&lt;/a&gt; (and that's a &lt;a href=&quot;http://en.wikipedia.org/wiki/Type_Ia_supernova&quot;&gt;Type Ia&lt;/a&gt;). &lt;a href=&quot;http://en.wikipedia.org/wiki/Spica&quot;&gt;The closest Type II candidate that I can think of is Spica&lt;/a&gt;, which is 260 light years away. &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_supernova_candidates&quot;&gt;Although this list doesn't include Spica&lt;/a&gt;, and has the closest &lt;a href=&quot;http://en.wikipedia.org/wiki/Type_II_supernova&quot;&gt;Type II&lt;/a&gt; as &lt;a href=&quot;http://en.wikipedia.org/wiki/Betelgeuse&quot;&gt;Betelgeuse&lt;/a&gt;, at 640 light years away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In his book, &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0670019976&quot;&gt;Death From The Skies&lt;/a&gt;&lt;/em&gt;, &lt;a href=&quot;http://blogs.discovermagazine.com/badastronomy/&quot;&gt;Dr. Phil Plait&lt;/a&gt; (a professional astronomer, writer, lecturer, etc.) covers exactly what would happen to earth if a star nearby &lt;em&gt;did&lt;/em&gt; go supernova, and it isn't pretty. Basically, it would strip us of our &lt;a href=&quot;http://en.wikipedia.org/wiki/Ozone_layer&quot;&gt;ozone layer&lt;/a&gt; and turn the upper layer of our atmosphere into a dirty brown smog layer, let in a lot of UV radiation that would devastate plankton and plant life, and be very bad for all life down on the surface. However, for a Type II, they need to be around 25 light years or closer to affect us.  Dr Plait's writing style is very much geared for a layman reader, and explains things in a very conversational tone, so I highly suggest that you get the book and read it. It will tell you exactly what to expect, and also tell you exactly why it's nothing to worry about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that said, there are two things in his book that you may want to consider as more likely than a supernova. A &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma-ray_burst&quot;&gt;GRB&lt;/a&gt; from &lt;a href=&quot;http://en.wikipedia.org/wiki/Eta_Carinae&quot;&gt;Eta Carinae&lt;/a&gt;, or an asteroid impact. The latter we may actually have the technology to eliminate as a concern though, so we're not doing too badly!&lt;/p&gt;&#xA;" OwnerUserId="114" LastActivityDate="2013-09-26T02:19:43.550" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="175" PostTypeId="2" ParentId="114" CreationDate="2013-09-26T02:33:15.220" Score="4" Body="&lt;p&gt;The other answers have some good points:  Lack of light pollution does make for much better skies, and higher elevation does help by eliminating some of the obscuring atmosphere (of course, you need to be careful how high you go and how quickly, as lack of oxygen can cause problems of several sorts).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to add a couple things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Obviously, obstructions to viewing (mountains, buildings, trees, etc.) can make viewing certain objects difficult, so a lack of obstructions is good.  This can be especially true for viewing something near the horizon (such as Mercury).&lt;/li&gt;&#xA;&lt;li&gt;An easily reached location can be useful.  While all the other issues make for good viewing, the fact is a perfect observing site is no good to you if you cannot get to it.  I live in a badly light polluted area; the closest pristine skies are a couple hours away in driving time.  While I do occasionally drive a few hours to get to nice skies, the majority of my observing has been from an unlit sports field on a hill that I can walk to in less than 15 minutes (drive time is about 3 minutes).  Yes, the limiting magnitude is usually about 3, but I can still indulge myself for a little while and get to work the next day, something I can't do when I drive to better skies.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="25" LastEditorUserId="25" LastEditDate="2013-09-26T14:21:55.937" LastActivityDate="2013-09-26T14:21:55.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="176" PostTypeId="5" CreationDate="2013-09-26T02:46:50.907" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-26T02:46:50.907" LastActivityDate="2013-09-26T02:46:50.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="177" PostTypeId="4" CreationDate="2013-09-26T02:46:50.907" Score="0" Body="Questions regarding Uranus, the 7th planet from Earth's sun." OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-09-26T04:16:19.207" LastActivityDate="2013-09-26T04:16:19.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="178" PostTypeId="5" CreationDate="2013-09-26T02:48:37.697" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-26T02:48:37.697" LastActivityDate="2013-09-26T02:48:37.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="179" PostTypeId="4" CreationDate="2013-09-26T02:48:37.697" Score="0" Body="Questions regarding Neptune, the 8th and furthest major planet orbiting Earth's sun." OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-09-26T05:47:13.997" LastActivityDate="2013-09-26T05:47:13.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="180" PostTypeId="2" ParentId="94" CreationDate="2013-09-26T02:54:02.650" Score="13" Body="&lt;p&gt;A parsec (abbreviated pc) is a unit of distance used by astronomers, cosmologists, and astrophysicists. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1 parsec is equal to $3.08567758 \times10^{16}$ meters, or $3.26163344$ light years (ly).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few typical scales to keep in mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The disks of galaxies like the Milky Way are a few 10's of kpc (that's pronounced kiloparsecs, which are 1000's of parsecs) in size. The dark matter halos surrounding them extend out to almost an order of magnitude further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FjHMd.jpg&quot; alt=&quot;Large spiral galaxy&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Galaxy clusters come in at around 1 Mpc (that's pronounced megaparsecs, or millions of parsecs), and are the largest bound objects in the universe. Pictured below is the galaxy cluster &lt;a href=&quot;http://apod.nasa.gov/apod/ap080210.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Abell 2218&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9PSJ2.jpg&quot; alt=&quot;Abell 2218 - Galaxy Cluster&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The closest star to the planet Earth (not counting the sun, of course) is the star system Alpha Centauri, at a distance of just about 1.3 parsecs. While you may think that this is incredibly close (and it is by cosmological standards), it would nonetheless still take us 4.24 years to travel to it if we traveled at the speed of light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AmUzD.jpg&quot; alt=&quot;Alpha Centauri&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditDate="2013-11-01T03:47:16.197" LastActivityDate="2013-11-01T03:47:16.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="181" PostTypeId="2" ParentId="172" CreationDate="2013-09-26T03:03:32.727" Score="5" Body="&lt;p&gt;According to &lt;a href=&quot;http://adsabs.harvard.edu/full/1989A&amp;amp;A...219L..23C&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; source from SOA/NASA Astrophysics Data System on Tidal evolution in the Neptune-Triton system (Chyba, C. F., Jankowski, D. G., &amp;amp; Nicholson, P. D.),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gXen1.png&quot; alt=&quot;Tidal evolution in the Neptune-Triton system, Chyba, C. F., Jankowski, D. G., &amp;amp; Nicholson, P. D.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Transcript: Many investigators have speculated on the reason for&#xA;  Triton's retrograde orbit. Lyttleton (1936) suggested that both Pluto&#xA;  and Triton originated as prograde satellites of Neptune, only to&#xA;  experience a catastrophic gravitational interaction. McCord (1966)&#xA;  showed that tidal evolution could have brought Triton inward from a&#xA;  near-parabolic orbit, lending plausibility to a capture origin for the&#xA;  satellite. Jarrington and Van Flandern (1979) suggested Triton's&#xA;  peculiar orbit and Pluto's supposed &quot;escape&quot; were caused by ab&#xA;  encounter with a massive &quot;rogue&quot; body, a conclusion contested ny&#xA;  Farinella et al. (1980). Finally, McKinnon (1984) combined momentum&#xA;  and energy constraints with an improved knowledge of Pluto's mass to&#xA;  show that all Pluto-Triton interaction scenarios for the origin of&#xA;  Triton's retrograde orbit are untenable, leaving the capture&#xA;  hypothesis as the most likely possibility.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, that summarizes all the theories one might have about it, at least all I could find.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as evolution of the system is concerned, the same document follows up to it and demonstrates how tidal effects are responsible for the evolution of the system. &quot;Degrading&quot; is a strange term. Let us go with evolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the possiblities stated is that Triton will reach a perfectly retrograde state and stay in it for ever! For a detailed analysis, you may read up section 6 in the link provided, &quot;&lt;strong&gt;The future of Triton&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="13" LastEditDate="2013-09-28T05:31:31.810" LastActivityDate="2013-09-28T05:31:31.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="182" PostTypeId="1" AcceptedAnswerId="211" CreationDate="2013-09-26T04:47:51.060" Score="16" ViewCount="408" Body="&lt;p&gt;My Astronomy book claims that scientists have discovered that Callisto, a moon of Jupiter, has no hot inner core. In fact, it says, Callisto has a core much like the nucleus of a comet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this still accepted theory, and how was this information discovered?&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2013-10-21T18:42:18.093" Title="How is it known that Callisto has no core?" Tags="&lt;callisto&gt;&lt;core&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="183" PostTypeId="5" CreationDate="2013-09-26T04:48:51.130" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-26T04:48:51.130" LastActivityDate="2013-09-26T04:48:51.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="184" PostTypeId="4" CreationDate="2013-09-26T04:48:51.130" Score="0" Body="Questions regarding Callisto, an icy moon of Jupiter." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-26T08:55:05.470" LastActivityDate="2013-09-26T08:55:05.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="185" PostTypeId="5" CreationDate="2013-09-26T04:49:26.803" Score="0" Body="&lt;p&gt;Questions regarding the Milky Way Galaxy, the galaxy that Earth's solar system is in. The Milky Way is a barred spiral galaxy that stretches across half of the night sky. The Sun's nearest stars and the brightest stars in our night sky are all in this galaxy.&lt;/p&gt;&#xA;" OwnerUserId="31410" LastEditorUserId="31410" LastEditDate="2021-01-08T00:27:23.940" LastActivityDate="2021-01-08T00:27:23.940" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="186" PostTypeId="4" CreationDate="2013-09-26T04:49:26.803" Score="0" Body="Questions regarding the Milky Way Galaxy, the galaxy that Earth's solar system is in." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-10-09T14:55:16.060" LastActivityDate="2013-10-09T14:55:16.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="187" PostTypeId="5" CreationDate="2013-09-26T04:50:27.293" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-26T04:50:27.293" LastActivityDate="2013-09-26T04:50:27.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="188" PostTypeId="4" CreationDate="2013-09-26T04:50:27.293" Score="0" Body="Questions regarding relatively small rocky bodies in an orbit around the Sun." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-26T08:55:55.240" LastActivityDate="2013-09-26T08:55:55.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="189" PostTypeId="5" CreationDate="2013-09-26T04:50:57.970" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-26T04:50:57.970" LastActivityDate="2013-09-26T04:50:57.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="190" PostTypeId="4" CreationDate="2013-09-26T04:50:57.970" Score="0" Body="Questions regarding Triton, the largest moon of Neptune." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-26T08:55:32.473" LastActivityDate="2013-09-26T08:55:32.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="191" PostTypeId="5" CreationDate="2013-09-26T04:51:37.050" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-26T04:51:37.050" LastActivityDate="2013-09-26T04:51:37.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="192" PostTypeId="4" CreationDate="2013-09-26T04:51:37.050" Score="0" Body="Questions regarding thermal energy and its effect on astronomical objects." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-26T08:55:44.880" LastActivityDate="2013-09-26T08:55:44.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="193" PostTypeId="5" CreationDate="2013-09-26T04:52:41.370" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-26T04:52:41.370" LastActivityDate="2013-09-26T04:52:41.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="194" PostTypeId="4" CreationDate="2013-09-26T04:52:41.370" Score="0" Body="Questions regarding the amount of space between two objects, the units, or measurement thereof." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-26T08:54:44.780" LastActivityDate="2013-09-26T08:54:44.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="195" PostTypeId="5" CreationDate="2013-09-26T04:53:15.840" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-26T04:53:15.840" LastActivityDate="2013-09-26T04:53:15.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="196" PostTypeId="4" CreationDate="2013-09-26T04:53:15.840" Score="0" Body="Questions regarding electromagnetic radiation in the visible spectrum." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-26T08:55:09.330" LastActivityDate="2013-09-26T08:55:09.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="197" PostTypeId="5" CreationDate="2013-09-26T04:54:00.237" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-26T04:54:00.237" LastActivityDate="2013-09-26T04:54:00.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="198" PostTypeId="4" CreationDate="2013-09-26T04:54:00.237" Score="0" Body="Questions regarding the attractive force which exists between any two bodies of matter." OwnerUserId="19" LastEditorUserId="1469" LastEditDate="2014-07-07T14:40:10.887" LastActivityDate="2014-07-07T14:40:10.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="199" PostTypeId="5" CreationDate="2013-09-26T04:55:04.020" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-26T04:55:04.020" LastActivityDate="2013-09-26T04:55:04.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="200" PostTypeId="4" CreationDate="2013-09-26T04:55:04.020" Score="0" Body="Questions regarding systems of large numbers of stars held together by gravity." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-26T08:55:13.610" LastActivityDate="2013-09-26T08:55:13.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="201" PostTypeId="5" CreationDate="2013-09-26T04:55:51.460" Score="0" Body="&lt;p&gt;Questions regarding the Sun and objects orbiting it. The 9 major bodies of the Solar System are (sorted by distance from the sun) the Sun, Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune. There are also hundreds of smaller minor planets, such as Ceres and Pluto, as well as millions of asteroids and comets in the Asteroid and Kuiper Belts and the Oort Cloud.&lt;/p&gt;&#xA;" OwnerUserId="31410" LastEditorUserId="31410" LastEditDate="2020-12-15T08:15:11.700" LastActivityDate="2020-12-15T08:15:11.700" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="202" PostTypeId="4" CreationDate="2013-09-26T04:55:51.460" Score="0" Body="Questions regarding the Sun and objects orbiting it." OwnerUserId="19" LastEditorUserId="9301" LastEditDate="2016-04-04T20:39:49.293" LastActivityDate="2016-04-04T20:39:49.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="203" PostTypeId="5" CreationDate="2013-09-26T04:56:40.667" Score="0" Body="&lt;p&gt;The Sun is the star that resides at the center of the Solar System and is closest to Earth, which orbits the Sun. It is a G2V star, with an apparent magnitude of -26.74 and an absolute magnitude of +4.83.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="31410" LastEditDate="2020-12-15T08:14:53.123" LastActivityDate="2020-12-15T08:14:53.123" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="204" PostTypeId="4" CreationDate="2013-09-26T04:56:40.667" Score="0" Body="Questions regarding the closest star to Earth, at the centre of the Solar System." OwnerUserId="19" LastEditorUserId="2153" LastEditDate="2015-04-12T21:59:33.153" LastActivityDate="2015-04-12T21:59:33.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="205" PostTypeId="5" CreationDate="2013-09-26T04:57:32.780" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-26T04:57:32.780" LastActivityDate="2013-09-26T04:57:32.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="206" PostTypeId="4" CreationDate="2013-09-26T04:57:32.780" Score="0" Body="Questions regarding structures designed to aid in the observation of astronomical objects." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-26T08:55:40.303" LastActivityDate="2013-09-26T08:55:40.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="207" PostTypeId="5" CreationDate="2013-09-26T04:58:22.173" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-26T04:58:22.173" LastActivityDate="2013-09-26T04:58:22.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="208" PostTypeId="4" CreationDate="2013-09-26T04:58:22.173" Score="0" Body="Questions regarding Earth's only natural satellite. For questions about the natural satellites of other planets, use the tag [natural-satellite]." OwnerUserId="19" LastEditorUserId="9527" LastEditorDisplayName="user14291" LastEditDate="2019-12-19T17:52:39.130" LastActivityDate="2019-12-19T17:52:39.130" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="209" PostTypeId="5" CreationDate="2013-09-26T04:59:46.070" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-26T04:59:46.070" LastActivityDate="2013-09-26T04:59:46.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="210" PostTypeId="4" CreationDate="2013-09-26T04:59:46.070" Score="0" Body="Questions regarding neutron stars which emit regular pulses of electromagnetic radiation." OwnerUserId="19" LastEditorUserId="2153" LastEditDate="2015-07-26T02:05:02.433" LastActivityDate="2015-07-26T02:05:02.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="211" PostTypeId="2" ParentId="182" CreationDate="2013-09-26T07:20:45.653" Score="7" Body="&lt;p&gt;It &lt;em&gt;may&lt;/em&gt; have a (silicate) core, but it's very small. There also is the possibility of &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0019103505001806&quot; rel=&quot;nofollow&quot;&gt;an internal liquid water &quot;ocean&quot;&lt;/a&gt;. Currently the most probable composition is that of rock and ice, &lt;a href=&quot;http://www.sciencemag.org/content/280/5369/1573&quot; rel=&quot;nofollow&quot;&gt;with the density of rock steadily increasing as one digs deeper&lt;/a&gt; (This is in stark contrast with planets like Earth, where there are discontinuities in the density between layers). &lt;a href=&quot;http://en.wikipedia.org/wiki/Planetary_differentiation&quot; rel=&quot;nofollow&quot;&gt;Planetary differentiation&lt;/a&gt; has not occurred appreciably because Callisto is not &lt;a href=&quot;http://en.wikipedia.org/wiki/Tidal_acceleration#Tidal_heating&quot; rel=&quot;nofollow&quot;&gt;tidally heated&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The composition can never be completely determined &amp;mdash; the composition of a spherical body cannot be fully discerned from just rotation and translation&lt;sup&gt;1&lt;/sup&gt;. However, we can make reasonable guesses. Basically, we can calculate its total mass and moment of inertia by observing its rotation and revolution. Then, assuming that the density ρ varies radially, we get the following two equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M = \int \rho \pi r^2dr$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = \int (\rho \pi r^2dr)r^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These do not completely determine ρ, however with other reasonable assumptions on the composition and the amount of planetary differentiation it can have, we can get a good estimate for ρ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1. It &lt;em&gt;can&lt;/em&gt; be determined if we take a larger gravitational field and pivot it, but this ventures into the Aristotlean &quot;Give me a place to stand on, and I will move the Earth.&quot; zone of thought experiments.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="6" LastEditDate="2013-10-21T18:42:18.093" LastActivityDate="2013-10-21T18:42:18.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="212" PostTypeId="1" AcceptedAnswerId="218" CreationDate="2013-09-26T07:33:33.467" Score="20" ViewCount="1274" Body="&lt;p&gt;Like many, when I was growing up, we always were taught, hence always learned that there were 9 planets.  However, recently, decisions were made and all of a sudden there were 8 major planets and a series of dwarf and minor planets.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What physical features delineate a body being referred to as a &lt;strong&gt;major planet&lt;/strong&gt; as opposed to &lt;strong&gt;dwarf planet&lt;/strong&gt; or &lt;strong&gt;minor planet&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorDisplayName="user96" LastEditDate="2013-09-26T08:18:30.217" LastActivityDate="2016-12-06T18:17:46.607" Title="What physical features determine if a planet is a major, minor or dwarf planet?" Tags="&lt;planet&gt;&lt;classification&gt;&lt;dwarf-planets&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="213" PostTypeId="1" AcceptedAnswerId="217" CreationDate="2013-09-26T07:46:36.753" Score="18" ViewCount="3779" Body="&lt;p&gt;Some celestial objects seem to be referred to as asteroids, some as comets, and some as meteors or meteorites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the distinction between all of these different objects? Are any of them the same?&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastActivityDate="2021-05-09T03:29:22.327" Title="What is the difference between asteroids, comets and meteors?" Tags="&lt;asteroids&gt;&lt;comets&gt;&lt;meteor&gt;&lt;meteorite&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="214" PostTypeId="1" AcceptedAnswerId="264" CreationDate="2013-09-26T07:46:45.707" Score="16" ViewCount="196" Body="&lt;p&gt;I am curious, if anybody knows of any stellar dynamical systems/environments, where relativistic effects could play a dynamical role on the motion of these stellar systems? As a subquestion - are there any known important weak, but cumulatively strong effects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, when can relativistic effects invalidate the applicability of N-Body/Collisionless Boltzman/Gas/.. models based on newtonian gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From these systems I would like to exclude the simplest well known case of compact binaries.&lt;/p&gt;&#xA;" OwnerUserId="115" LastEditorUserId="92" LastEditDate="2013-10-08T10:06:39.397" LastActivityDate="2013-10-25T18:22:07.127" Title="Relativistic effects in stellar dynamical systems" Tags="&lt;general-relativity&gt;&lt;stellar-dynamics&gt;&lt;newtonian-gravity&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="215" PostTypeId="2" ParentId="212" CreationDate="2013-09-26T07:53:45.783" Score="8" Body="&lt;p&gt;In &lt;a href=&quot;http://www.iau.org/static/archives/releases/doc/iau0603.doc&quot;&gt;2006 the Internaional Astronomical Union (IAU)&lt;/a&gt; (that's a download link, 85KB) came to the resolution that a &lt;strong&gt;planet&lt;/strong&gt; is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a celestial body that (a) is in orbit around the Sun, (b) has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape, and (c) has cleared the neighbourhood around its orbit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;Dwarf Planet&lt;/strong&gt; is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a celestial body that (a) is in orbit around the Sun, (b) has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape1, (c) has not cleared the neighbourhood around its orbit, and (d) is not a satellite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The same resolution also said that anything else, except satellites, is classified as a &quot;Small Solar-System Body&quot;.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2013-10-09T06:32:47.443" LastActivityDate="2013-10-09T06:32:47.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="216" PostTypeId="2" ParentId="213" CreationDate="2013-09-26T08:17:58.097" Score="6" Body="&lt;p&gt;&lt;strong&gt;Asteroids&lt;/strong&gt; are what are called minor planets - small inner solar system rocky bodies. Especially between Mars and Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Comets&lt;/strong&gt; typically have highly elliptical orbits but they don't have to. The key defining feature of a comet is that it has a gaseous atmosphere and a tail when it passes close to the Sun. Comets are generally loose aggregations of ice, dust, rocks etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Meteoroids&lt;/strong&gt; are very small rocky or metal bodies, from a single particle up to about 1m diameter. If they hit the atmosphere, the visible glowing trail of particles is called a &lt;strong&gt;meteor&lt;/strong&gt;. And if it hits the ground it is called a &lt;strong&gt;meteorite&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2013-09-26T08:17:58.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="217" PostTypeId="2" ParentId="213" CreationDate="2013-09-26T08:23:19.013" Score="13" Body="&lt;p&gt;The objects you are refering to are actually two different objects: asteroids and comets. Meteor and meteorite are other names for an asteroid, at a given time of its interaction with our planet. We'll get to that.&lt;/p&gt;&#xA;&lt;p&gt;So first, what is the difference between an asteroid and a comet?&lt;/p&gt;&#xA;&lt;p&gt;A &lt;strong&gt;comet&lt;/strong&gt; is a small solar system body that display a &amp;quot;coma&amp;quot; (an atmosphere of a sort) and sometimes a tail passing close to the Sun. They are mostly made of ice and dust, as well as some small rocky particles. We distinguish two kind of comets, with short or long orbital period. The short orbital period ones originated from the Kuiper Belt, a region composed of small bodies beyond the orbit of Neptune. The long orbital period ones originated from the Oort cloud, a scattered disk of icy planetesimals and small bodies laying around our solar system.&lt;/p&gt;&#xA;&lt;p&gt;An &lt;strong&gt;asteroid&lt;/strong&gt; is a small body, composed mostly of rocks and metals. In our solar system, they can be originated from the asteroid belt, laying between Mars and Jupiter, or from the orbit of Jupiter (the Jupiter Trojans), or from actually almost everywhere in the Solar System.&lt;/p&gt;&#xA;&lt;p&gt;A small asteroid ('meteoroid') that enters the Earth's atmosphere becomes a &lt;strong&gt;meteor&lt;/strong&gt;, what we also call a &amp;quot;shooting star&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;Eventually, a meteor that was massive enough not to be completely distroyed entering the Earth's atmosphere and hitting the ground is a &lt;strong&gt;meteorite&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="27993" LastEditDate="2020-09-15T16:08:53.360" LastActivityDate="2020-09-15T16:08:53.360" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="218" PostTypeId="2" ParentId="212" CreationDate="2013-09-26T08:42:06.133" Score="10" Body="&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Planet#Physical_characteristics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planets&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a body to be classified as a planet it must have a few physical characteristics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mass&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It must have enough mass to have a strong enough gravity to overcome electrostatic forces to bring it to a state of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrostatic_equilibrium&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrostatic equilibrium&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hydrostatic equilibrium is important because early in a planets life it is nearly entirely fluid, crust and all&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Internal Differentiation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The life cycle of a planet essentially leads to denser heavier metals being at the centre of a planet, surrounded by a mantle which must at some point have been fluid (it can still be solid and called a planet as long as it &lt;em&gt;used&lt;/em&gt; to be fluid.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Atmosphere&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is usually driven by its mass but a planet should have an atmosphere. This means it should be massive enough to have a strong enough gravity to hold some gasses to its surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More massive planets are capable of keeping lighter gasses, such as hydrogen, bound to them too. I.e Jupiter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Magnetosphere&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A magnetosphere suggests that the body is still geologically active. This means they have flows of elements that conduct electricity in their interiors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.iau.org/public/themes/pluto/&quot; rel=&quot;nofollow noreferrer&quot;&gt;As specified by the IAU&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A celestial body that (a) is in orbit around the Sun, (b) has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape, and (c) has cleared the neighbourhood around its orbit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Minor_planet#Physical_properties_of_comets_and_minor_planets&quot; rel=&quot;nofollow noreferrer&quot;&gt;Minor Planets&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Minor planets  are things such as asteroids, they are usually small mass, rocky bodies. This is different from small mass bodies that might be predominantly ice or water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dwarf_Planet#Characteristics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dwarf Planets&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Size and Mass&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/International_Astronomical_Union&quot; rel=&quot;nofollow noreferrer&quot;&gt;IAU&lt;/a&gt; have not specified an upper or lower limit for mass to be considered a Dwarf Planet, therefore it is predominantly determined by one other feature called:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Orbital Dominance&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbital dominance is achieved when a body has cleared it's orbit of all other bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example planets are able to remove small bodies out of their area through impact, capture, or gravitational disturbance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any body that is incapable of doing so is therefore classified as a dwarf planet. So if Jupiter was incapable of clearing its neighbourhood of bodies then it too would be a Dwarf planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/foQlV.jpg&quot; alt=&quot;planets and dwarf planets in our solar system&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastActivityDate="2013-09-26T08:42:06.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="219" PostTypeId="1" AcceptedAnswerId="227" CreationDate="2013-09-26T09:01:50.140" Score="13" ViewCount="341" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rz5eV.jpg&quot; alt=&quot;tropics and polar circles&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The middle line is the equator, the second and fourth line are the tropics, the first and fifth are the polar circles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the relevance of these bands on Earth? Do these exist on other planets? What purpose do they serve?&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastActivityDate="2013-09-26T11:09:30.440" Title="What is the relevance of the Tropics and the Polar Circles? Do they Exist on other planets?" Tags="&lt;planet&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="220" PostTypeId="1" AcceptedAnswerId="235" CreationDate="2013-09-26T09:04:21.743" Score="19" ViewCount="2679" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RwcKo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here on Earth we get to experience lunar and solar eclipses. Is this pure coincidence that everything happens to line up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do other planets in the solar system experience this phenomenon or is this something unique on Earth? &lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastActivityDate="2021-05-12T16:23:32.970" Title="Do other planets in our solar system experience eclipses or is this unique to Earth?" Tags="&lt;the-moon&gt;&lt;the-sun&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="221" PostTypeId="2" ParentId="156" CreationDate="2013-09-26T09:08:52.823" Score="10" Body="&lt;p&gt;The time to &amp;quot;produce&amp;quot; a star actually depends on the mass of the star. Let's start with a solar type star.&lt;/p&gt;&#xA;&lt;p&gt;For a typical solar type star, the formation time is about 10 million years (you can see &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Formation_stellaire_Andre_1994.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;this image&lt;/a&gt; on Wikimedia that gives an overview of the star formation process, adapted from Philippe André's figure in &lt;em&gt;Observations of protostars and protostellar stages&lt;/em&gt; in The cold Universe, 1994). You can distinguish different epochs in the star formation process, that are the signature of different dominant physical processes. The very first stage of star formation is a gravitational collapse that leads to the formation of the protostar itself. The timescale for this collapse is the so-called &lt;strong&gt;free-fall time&lt;/strong&gt; which depends only on the density of the object. When you get a central object in a hydrostatic equilibrium, things become more subtle: the core will contract adiabatically (without heat transfer) and when a temperature of about 2000 K is reached, dihydrogen dissociates (which is a highly endothermic reaction) which leads to a second phase of collapse, leading to the formation of the protostar itself. It takes about 1000 years to get to this stage, from the hydrostatic core.&lt;/p&gt;&#xA;&lt;p&gt;The next stage, the protostellar phase, is mostly an accretion phase. It means that the timescale for this epoch is given by an &lt;strong&gt;accretion time&lt;/strong&gt;, that varies with the accreted mass (which is quite low for a solar type star). It takes about 200 000 years to accrete 90% of the final mass of the star.&lt;/p&gt;&#xA;&lt;p&gt;Then, since the star is still contracting, the temperature at its center is increasing; when a temperature of 1 million Kelvin is reached, the protostar starts to burn its deuterium. At this stage, the Kelvin-Helmholtz mechanism allows the protostar to contract and to radiate away its gravitational energy. The meaningful timescale is then the &lt;strong&gt;Kelvin-Helmholtz time&lt;/strong&gt; (which varies as the square of the mass and the inverse of the radius and the luminosity), that is much longer than the previous timescales. Temperature continues to rise, up to 10 million Kelvin, when hydrogen eventually starts to burn, which is the birth certificate of a star. It takes around 10 million years to get to this point.&lt;/p&gt;&#xA;&lt;p&gt;But, as I said, this scenario depends on the mass of the star. It is valid for Sun-like stars, but not quite for massive stars. It is much faster for massive stars, and the star formation process is quite different. In particular, the accretion rate is much higher, the radiation pressure of the protostar is insanely higher, and their interplay is not completely understood. However, there are some theroretical works that gives an estimate of about 100 000 years to form a massive stars (see for example &lt;a href=&quot;http://www.astro.ufl.edu/%7Ejt/massivesf/massivesf.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;works from McKee and Tan&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="16685" LastEditDate="2020-08-17T15:29:24.990" LastActivityDate="2020-08-17T15:29:24.990" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="222" PostTypeId="1" AcceptedAnswerId="223" CreationDate="2013-09-26T10:08:58.087" Score="19" ViewCount="2848" Body="&lt;p&gt;Sometimes at night you will look up to the stars and they will appear to twinkle, getting brighter and darker in bursts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/V3plu.gif&quot; alt=&quot;twinkling stars&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this happen? Is this because of our atmosphere? Would they twinkle if you were just in space? How about looking from another planet in our solar system?&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastActivityDate="2014-06-12T01:26:07.320" Title="Why do stars appear to twinkle?" Tags="&lt;star&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="223" PostTypeId="2" ParentId="222" CreationDate="2013-09-26T10:24:31.897" Score="9" Body="&lt;p&gt;Stars tend to twinkle for two main reasons: first, stars are very far away (the closest star is about 4 light-year from the Sun) and are therefore seen as point sources. Second, Earth has an atmosphere. Earth's atmosphere is turbulent, and therefore all images view through it tends to &quot;swim&quot;. Therefore, sometimes a single point in &quot;object space&quot; is mapped to several points in &quot;image space&quot;, and sometimes it is not mapped at all. Since stars are seen as single points, they sometimes seems brighter, sometimes even seems to disapear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at it in another planet of our solar system, it will depends on the planet's own atmosphere. If you look at stars on Mars, the atmosphere being very thin, the stars won't twinkle that much. Same for Mercury. On Venus, the atmosphere is so thick than you won't see anything apart from the atmosphere itself (if you are not crunched by the atmosphere pressure, by the way...).&lt;/p&gt;&#xA;" OwnerUserId="88" LastActivityDate="2013-09-26T10:24:31.897" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="224" PostTypeId="2" ParentId="24" CreationDate="2013-09-26T10:48:28.680" Score="37" Body="&lt;p&gt;There are many, many ways of doing this.&lt;/p&gt;&#xA;&lt;h2&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_lens&quot; rel=&quot;noreferrer&quot;&gt;Gravitational lensing&lt;/a&gt;&lt;/h2&gt;&#xA;&lt;p&gt;This is by far the most well known. It has been mentioned by the others, but I'll touch on it.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LPNNP.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Light coming from distant bodies can be bent by gravity, creating a lens-like effect. This can lead to multiple or distorted images of the object (Multiple images give rise to &lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein_ring&quot; rel=&quot;noreferrer&quot;&gt;Einstein rings and crosses&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;p&gt;So, if we observe a lensing effect in a region where there isn't any visible massive body, there's probably a black hole there. The alternative is that we are peering through the dark matter 'halo' which surrounds (and extends passed) the luminous components of every galaxy and galaxy cluster (&lt;a href=&quot;http://apod.nasa.gov/apod/ap060824.html&quot; rel=&quot;noreferrer&quot;&gt;See: Bullet Cluster&lt;/a&gt;). On small enough scales (i.e. - the central regions of galaxies), this is not really an issue.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/q1dP2.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;(This is an artist's impression of a galaxy passing behind a BH)&lt;/p&gt;&#xA;&lt;h2&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_waves&quot; rel=&quot;noreferrer&quot;&gt;Gravitational waves&lt;/a&gt;&lt;/h2&gt;&#xA;&lt;p&gt;Spinning black holes and other dynamical systems involving black holes emit gravitational waves. Projects like &lt;a href=&quot;http://en.wikipedia.org/wiki/LIGO&quot; rel=&quot;noreferrer&quot;&gt;LIGO&lt;/a&gt; (and eventually, &lt;a href=&quot;http://en.wikipedia.org/wiki/Laser_Interferometer_Space_Antenna&quot; rel=&quot;noreferrer&quot;&gt;LISA&lt;/a&gt;) are able to detect these waves. One major candidate of interest for LIGO/VIRGO/LISA is the eventual collision of a binary black hole system.&lt;/p&gt;&#xA;&lt;h2&gt;Redshift&lt;/h2&gt;&#xA;&lt;p&gt;Sometimes we have a black hole in a binary system with a star. In such a case, the star will orbit the common barycenter.&lt;/p&gt;&#xA;&lt;p&gt;If we observe the star carefully, its light will be &lt;a href=&quot;http://en.wikipedia.org/wiki/Redshift&quot; rel=&quot;noreferrer&quot;&gt;redshifted&lt;/a&gt; when it is moving away from us, and blueshifted when it is coming towards us. The variation in redshift suggests rotation, and in the absence of a visible second body, we can usually conclude that there's a black hole or neutron star there.&lt;/p&gt;&#xA;&lt;h2&gt;Salpeter-Zel'dovitch / Zel'dovitch-Novikov proposals&lt;/h2&gt;&#xA;&lt;p&gt;Going in to a bit of history here, Salpeter and Zel'dovitch independently proposed that we can identify black holes from shock waves in gas clouds. If a black hole passes a gas cloud, the gases in the cloud will be forced to accelerate. This will emit radiation (X-rays, mostly), which we can measure.&lt;/p&gt;&#xA;&lt;p&gt;An improvement on this is the Zel'dovitch-Novikov proposal, which looks at black holes in a binary system with a star. Part of the solar winds from the star will be sucked in to the black hole. This abnormal acceleration of the winds will, again, lead to X-ray shock waves.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JNTmh.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;This method (more or less) led to the discovery of &lt;a href=&quot;http://en.wikipedia.org/wiki/Cygnus_X-1&quot; rel=&quot;noreferrer&quot;&gt;Cyg X-1&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h2&gt;Cosmic gyroscopes&lt;/h2&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cyg_A&quot; rel=&quot;noreferrer&quot;&gt;Cyg A&lt;/a&gt; is an example of this. Spinning black holes act like cosmic gyroscopes — they do not easily change their orientation.&lt;/p&gt;&#xA;&lt;p&gt;In the following radio image of Cyg A, we see these faint gas jets emanating from the central spot:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Oj75s.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;These jets are hundreds of thousands of light years long — yet they are very straight. Discontinuous, but straight. Whatever object lies at the center, it must be able to maintain its orientation for very long.&lt;/p&gt;&#xA;&lt;p&gt;That object is a spinning black hole.&lt;/p&gt;&#xA;&lt;h2&gt;Quasars&lt;/h2&gt;&#xA;&lt;p&gt;Most quasars are thought to be powered by black holes. Many (if not all) of the candidate explanations for their behavior involve black holes with accretion disks, e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Blandford%E2%80%93Znajek_process&quot; rel=&quot;noreferrer&quot;&gt;the Blandford-Znajek process&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2013-09-26T13:25:02.317" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="225" PostTypeId="1" AcceptedAnswerId="250" CreationDate="2013-09-26T10:51:51.540" Score="7" ViewCount="1269" Body="&lt;p&gt;I think I'd like to become an astronomer. What degrees and training/experience do I need to become one? I've heard I should take science and mathematics, but I don't know which specific courses I should take and which degrees I would need to become an astronomer.&lt;/p&gt;&#xA;" OwnerUserId="69" LastEditorUserId="6" LastEditDate="2016-05-25T15:42:35.110" LastActivityDate="2016-05-25T15:42:35.110" Title="I'd like to become an astronomer. What experience do I need?" Tags="&lt;observational-astronomy&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ClosedDate="2013-09-27T19:21:23.063" ContentLicense="CC BY-SA 3.0" />
  <row Id="226" PostTypeId="1" AcceptedAnswerId="228" CreationDate="2013-09-26T10:55:40.947" Score="16" ViewCount="1292" Body="&lt;p&gt;I've noticed that when I look into the sky at night, sometimes the moon appears very large near the horizon. Sometimes it also looks like it has a yellow tint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ulPd2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the light rays bouncing off the moon differently, or the atmosphere of the earth, or some type of sunspot blocking part of the light to the moon?&lt;/p&gt;&#xA;" OwnerUserId="69" LastEditorUserId="25" LastEditDate="2013-09-26T15:17:55.033" LastActivityDate="2013-09-27T23:46:32.727" Title="Why does the Moon seem larger when it is close to the horizon?" Tags="&lt;the-moon&gt;&lt;horizon&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2013-09-30T07:51:56.320" ContentLicense="CC BY-SA 3.0" />
  <row Id="227" PostTypeId="2" ParentId="219" CreationDate="2013-09-26T11:04:23.963" Score="6" Body="&lt;p&gt;The Tropics (Capricorn in the South and Cancer in the North) are the furthest point from the equator where the sun reaches a point directly overhead at any time during the year. The Arctic and Antarctic circles are the points at which the sun remains above the horizon for a full rotation of the Earth (i.e. 24 hours), commonly called &quot;Midnight Sun&quot;. Any planet that has axial tilt while have similar lines.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2013-09-26T11:09:30.440" LastActivityDate="2013-09-26T11:09:30.440" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="228" PostTypeId="2" ParentId="226" CreationDate="2013-09-26T11:11:44.967" Score="12" Body="&lt;p&gt;Interesting fact. &lt;strong&gt;It Isn't bigger!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a well known optical illusion that dates back hundreds of years. It's all in your head, it seems bigger because your brain perceives size as something that is relative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an effect called &lt;a href=&quot;http://en.wikipedia.org/wiki/Perspective_distortion_%28photography%29#Photograph_viewing_distance&quot; rel=&quot;nofollow noreferrer&quot;&gt;perspective distortion&lt;/a&gt; and is will understood and used in photography. &lt;a href=&quot;http://blogs.discovermagazine.com/badastronomy/2010/05/13/why-does-the-moon-look-so-huge-on-the-horizon/#.UkQ3Y99hKa4&quot; rel=&quot;nofollow noreferrer&quot;&gt;Phil Plait's Blog (A.K.A - The Bad Astronomer)&lt;/a&gt; also writes on this matter and would be worth reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the below picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3WgtW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the focal length of a lens gets longer the difference in relative size of foreground and background objects becomes more pronounced. Your brain then interprets this and respectively the moon seems huge!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But why does it look yellowish?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears yellow sometimes because our atmosphere blocks and stretches certain wavelengths. So when the moon is low in the sky, it has to pass through the most atmosphere to get to you, stretching and distorting the light through the yellows and oranges.&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastEditorUserId="45" LastEditDate="2013-09-27T23:46:32.727" LastActivityDate="2013-09-27T23:46:32.727" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="229" PostTypeId="2" ParentId="225" CreationDate="2013-09-26T11:13:17.753" Score="2" Body="&lt;p&gt;There are degrees in astronomy but many astronomers have physics degrees. But there aren't a lot of positions going for professional astronomers (and those that are, you'd better be doing it for love, not money) so besides the degree, you're going to need to stand out from the crowd. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you could become what I prefer to call an 'independently funded astronomer' which doesn't require a degree at all, and you still get to do the stuff you love.&lt;/p&gt;&#xA;" OwnerUserId="67" LastEditorUserId="67" LastEditDate="2014-03-18T10:20:31.413" LastActivityDate="2014-03-18T10:20:31.413" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="230" PostTypeId="2" ParentId="225" CreationDate="2013-09-26T12:02:55.247" Score="5" Body="&lt;p&gt;Well, if you want to become a professional astronomer, there is no obvious answer, since education systems can be quite different depending on your country. But astronomy &amp;amp; astrophysics is physics in the end, so a good training and strong knownledge in physics are definitely required (that involves math for sure). And in countries/universities where you can find an astronomy-oriented degree (at master level most of the time, before that it does not really make sense), you won't be surprised that most of the lectures are actually just physics or math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, to get a PhD in the field is the way to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the positions, you can find some statistics &lt;a href=&quot;https://www.google.fr/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=1&amp;amp;ved=0CCUQFjAA&amp;amp;url=https%3A%2F%2Fwww.eso.org%2Fwiki%2Fpub%2FScience%2FScienceNewsLetter%2FScienceNews-May-Jun2011.pdf&amp;amp;ei=ECtbUN_AFJSRhQfEk4GYDw&amp;amp;usg=AFQjCNHe2s7LfA-Gss8fX_aiMWyIsIJg6A&amp;amp;cad=rja&quot; rel=&quot;noreferrer&quot;&gt;in this paper&lt;/a&gt; released by ESO (the European Southern Observatory) for jobs in astronomy in the 2009/2010 hiring season. Bottom line is there were 65 opened positions worldwilde, that cannot be consider as &quot;a lot&quot;.&lt;/p&gt;&#xA;" OwnerUserId="88" LastActivityDate="2013-09-26T12:02:55.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="231" PostTypeId="1" CreationDate="2013-09-26T12:14:32.493" Score="24" ViewCount="46156" Body="&lt;p&gt;A number of ancient civilizations had devised methods to predict exact dates and times of such eclipses, marking them as important events.&#xA;Hence I assume the predictions were based on calculations, which should be quite easy to do now.&#xA;So what is the exact formula to predict the exact date and time(this is optional, but desirable) of lunar and solar eclipses?&#xA;Also how to calculate if the solar eclipse will be visible from a particular location or not?&lt;/p&gt;&#xA;" OwnerUserId="15" LastEditorUserId="15" LastEditDate="2013-09-27T07:40:42.100" LastActivityDate="2016-08-31T16:22:27.087" Title="What is the formula to predict lunar and solar eclipses accurately?" Tags="&lt;solar-eclipse&gt;&lt;lunar-eclipse&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="232" PostTypeId="2" ParentId="231" CreationDate="2013-09-26T12:50:30.057" Score="11" Body="&lt;p&gt;The NASA sites have some very useful resources for this I will list them below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lunar Eclipses&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://eclipse.gsfc.nasa.gov/LEcat5/LEcatalog.html&quot; rel=&quot;noreferrer&quot;&gt;This Link&lt;/a&gt; has an index for all lunar eclipses from -1999 to +3000, predominantly a statistics page but also has &lt;a href=&quot;http://eclipse.gsfc.nasa.gov/LEcat5/deltatpoly.html&quot; rel=&quot;noreferrer&quot;&gt;this page&lt;/a&gt; that contains how to calculate when lunar eclipses are. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is more than one formula depending on which time frame you are trying to look in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the formula for eclipses between the year 2005 and 2050:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\Delta T = 62.92 + 0.32217 * t + 0.005589 * t^2$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Where: &lt;br/&gt;  $$y = year + (month - 0.5)/12$$ &lt;br/&gt; $$t = y - 2000$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solar Eclipses&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://eclipse.gsfc.nasa.gov/SEcat5/SEcatalog.html&quot; rel=&quot;noreferrer&quot;&gt;This Link&lt;/a&gt; has an index like above but for all of the solar eclipses from -1999 to +3000. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://eclipse.gsfc.nasa.gov/SEcat5/deltatpoly.html&quot; rel=&quot;noreferrer&quot;&gt;This link&lt;/a&gt; has the formula for calculating solar eclipses. This is the formula for between 2005 and 2050:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\Delta T = 62.92 + 0.32217 * t + 0.005589 * t^2$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Where: &lt;br/&gt; $$y = year + (month - 0.5)/12$$ &lt;br/&gt; $$t = y - 2000$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user96" LastEditorUserId="6" LastEditDate="2013-10-21T18:57:59.920" LastActivityDate="2013-10-21T18:57:59.920" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="233" PostTypeId="2" ParentId="231" CreationDate="2013-09-26T12:58:28.137" Score="9" Body="&lt;p&gt;Calculation of solar eclipses can be done using &lt;strong&gt;Besselian elements&lt;/strong&gt;. The basic idea is to compute the motion of the Moon's shadow on a plane that crosses the Earth's center. Then, the shadow cone of the Moon can be projected on the Earth surface. The Besselian elements are the following: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;X and Y: the coordinates of the center of the shadow in the fundamental plane&lt;/li&gt;&#xA;&lt;li&gt;D: the direction of the shadow axis on the celestial sphere&lt;/li&gt;&#xA;&lt;li&gt;L1 and L2: the radii of the penumbral and umbral cone in the fundamental plane&lt;/li&gt;&#xA;&lt;li&gt;F1 and F2: the angles which the penumbral and umbral shadow cones make with the shadow axis&lt;/li&gt;&#xA;&lt;li&gt;$\mu$: the ephemeris hour angle&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;what you have to do now is to compute the variation of these parameters, which are time-dependent. It happens that it can be done using polynomial expensions for a given reference time $t_0$. The polynomial expension is of the form, for a Besselian element a:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a = a_0 + a_1\times t + a_2\times t^2 + a_3\times t^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a third order expension is enough in general), with $t = t_1 - t_0$, $t_0$ being the Terrestrial Dynamical Time (TDT) to the nearest hour of the instant of greatest eclipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sources:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://eclipse.gsfc.nasa.gov/SEmono/reference/explain.html&quot; rel=&quot;nofollow&quot;&gt;Nasa webpage about solar eclipse&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://eclipse.gsfc.nasa.gov/SEcat5/beselm.html&quot; rel=&quot;nofollow&quot;&gt;Besselian elements of solar eclipse&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="88" LastEditorUserId="9301" LastEditDate="2016-08-31T16:22:27.087" LastActivityDate="2016-08-31T16:22:27.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="234" PostTypeId="2" ParentId="39" CreationDate="2013-09-26T13:11:23.363" Score="7" Body="&lt;p&gt;Since you're interested in the equations of motion, I would solve this problem by using Lagrangian mechanics. Essentially, find the kinetic and potential energies for these two bodies, A and B. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Construct the Lagrangian:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L = T - V$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where T is the kinetic energy, and V is the potential energy. Then use the &lt;a href=&quot;http://mathworld.wolfram.com/Euler-LagrangeDifferentialEquation.html&quot; rel=&quot;nofollow&quot;&gt;Euler-Lagrange Equation&lt;/a&gt; to achieve the equations of motion (I would add it here, but I'm not sure of the specifics of your problem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two-body motion can always be constrained to a plane, so you may have to throw a constraint in there through the use of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_multiplier&quot; rel=&quot;nofollow&quot;&gt;Lagrange multiplier&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditDate="2014-04-02T11:57:33.483" LastActivityDate="2014-04-02T11:57:33.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="235" PostTypeId="2" ParentId="220" CreationDate="2013-09-26T13:11:58.027" Score="11" Body="&lt;p&gt;Pretty much every planet with a moon can have eclipses - you'll have seen photographs of Saturn with shadows cast from its moons, like this one from &lt;a href=&quot;http://wodumedia.com/checking-in-on-saturn/the-shadow-of-saturns-largest-moon-darkens-a-huge-portion-of-the-gas-giant-planet-titan-5150-km-or-3200-mi-across-is-not-pictured-here-but-its-shadow-is-elongated-across-saturns-upper-atmos/&quot; rel=&quot;nofollow noreferrer&quot;&gt;wodumedia.com:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SvZ2f.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were in a balloon where that shadow is, you would see a solar eclipse. It wouldn't be as exciting as from Earth, as the Sun would appear so much smaller from this distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the planet to eclipse the moons, this happens far more often than on Earth, as the moons are so much smaller than Saturn. They often are completely shadowed.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2013-09-26T13:11:58.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="236" PostTypeId="2" ParentId="115" CreationDate="2013-09-26T13:21:25.700" Score="5" Body="&lt;p&gt;Star temperature is an interesting question since temperature varies a lot in a star. I think that the more relevant temperature to this question is the core temperature of the star: a star is born when it starts to burn hydrodgen in its core.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Finally, hydrogen begins to fuse in the core of the star, and the rest&#xA;  of the enveloping material is cleared away. This ends the protostellar&#xA;  phase and begins the star's main sequence phase on the H–R diagram.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(See this &lt;a href=&quot;http://en.wikipedia.org/wiki/Star_formation#Protostar&quot;&gt;Wikipedia page&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature needed for hydrodgen burning is &lt;strong&gt;10 million Kelvin&lt;/strong&gt;, so that's how hot a star must be to be considered as a star. It needs to get so hot, because else it will fail to burn hydrodgen and will become a &quot;failed star&quot;: a &lt;a href=&quot;http://en.wikipedia.org/wiki/Brown_dwarf&quot;&gt;brown dwarf&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surface temperature can be misleading, since the temperature ranges in which lay stars are not populated only by stars, but also by other objects such as hot Jupiters, &lt;a href=&quot;http://www.ras.org.uk/news-and-press/224-news-2013/2309-hst-reveals-variation-between-hot-jupiters&quot;&gt;with surface temperature ranging from 1000 to 3000 K&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2013-09-26T13:49:26.290" LastActivityDate="2013-09-26T13:49:26.290" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="237" PostTypeId="1" AcceptedAnswerId="243" CreationDate="2013-09-26T13:37:48.340" Score="15" ViewCount="210" Body="&lt;p&gt;For a time, I thought Azimuth is always direction in degrees, 0 for local, geographic north, 90 for east etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, trying to observe moonrise I downloaded an app that gave azimuth in values between (if I recall correctly) -180 and 180, and I think &quot;0&quot; corresponded to south. I tried to read up on that and found very little fragmented information. I found other resources that frustrated me with numbers greater than 360, or provided values commonly associated with &quot;West&quot; for azimuth of moonrise and generally it took a bit of searching to find something with plain &quot;degrees from north&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone either post or provide a link to a comprehensive guide to various azimuth angle systems, when and where they are in use, and how to recognize which one is used?&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2013-10-14T22:02:27.207" Title="What azimuth description systems are in use?" Tags="&lt;navigation&gt;&lt;azimuth&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="238" PostTypeId="5" CreationDate="2013-09-26T13:44:36.790" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-26T13:44:36.790" LastActivityDate="2013-09-26T13:44:36.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="239" PostTypeId="4" CreationDate="2013-09-26T13:44:36.790" Score="0" Body="Questions regarding the action or process of observing a single object carefully to glean information." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-26T13:50:38.103" LastActivityDate="2013-09-26T13:50:38.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="240" PostTypeId="1" AcceptedAnswerId="715" CreationDate="2013-09-26T13:52:27.527" Score="14" ViewCount="7744" Body="&lt;p&gt;In my tenacious attempts to observe moonrise (surprisingly difficult in a poor weather heavily urbanized hilly area with frequent bad smog, with busy work schedule and a bicycle) I was frequently trying to find good observation spots that would provide view to open distant horizon at specific azimuth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While generally similar moon phases correlated with similar moonrise times I found the azimuth to vary in a way I have a difficult time to put my finger on. Sure there's a &lt;em&gt;roughly&lt;/em&gt; monthly cycle of &quot;wobble&quot; between deviating about +/- 30 degrees north or south from &quot;pure east&quot; but while the mechanics of correlation between hour when the Moon is visible from behind the horizon and the degree it's lit by the sun are quite clear to me, the way its - or is it Earth's? - orbital plane is &quot;off&quot; relative to the ecliptic plane and Earth rotation plane eludes me. Could someone provide a sort of guide - preferably something that acts more as help in understanding the mechanics, as opposed to providing raw mathematical solutions, what does this deviation depend on, and how it changes with time?&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2018-06-04T08:29:19.743" Title="How does moonrise/moonset azimuth vary with time?" Tags="&lt;the-moon&gt;&lt;azimuth&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="241" PostTypeId="1" AcceptedAnswerId="245" CreationDate="2013-09-26T14:05:24.540" Score="11" ViewCount="212" Body="&lt;p&gt;With about all visible stars catalogued, measured and photographed it seems that chapter is closed. I realize currently while amateur astronomy is big on cataloguing all asteroids in the Solar System, the &quot;Big Boys&quot; are now primarily busy with discovery of extrasolar planets. But that's certainly not all. Sure there's a lot of unique research, new, unexpected discoveries, assorted &quot;odd&quot; research like SETI, and so on, but I'd like to know what is the &quot;daily bread&quot; of every astronomer nowadays?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the current &lt;em&gt;typical&lt;/em&gt;, &lt;em&gt;routine&lt;/em&gt; tasks of an average professional astronomer? What is being observed/measured/searched for? What challenges occupy most effort and time? Or am I wrong assuming there even is such a thing, and every observatory is a unique snowflake?&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2013-09-27T12:02:34.780" Title="What is the current routine of modern astronomy?" Tags="&lt;observatory&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2013-09-27T18:35:37.530" ContentLicense="CC BY-SA 3.0" />
  <row Id="242" PostTypeId="2" ParentId="75" CreationDate="2013-09-26T14:18:49.667" Score="13" Body="&lt;p&gt;Gravitational lensing is the bending of light by massive objects in between the observer (us), and a background source of light. It is a direct prediction of Einstein's theory of General Relativity, and was tested and confirmed by Sir Aurther Eddington during the famous &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_eclipse_of_May_29,_1919&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solar exlipse of May 29, 1919&lt;/a&gt;, where the apparent position of a star very close to the sun was observed at a &lt;em&gt;different&lt;/em&gt; location - the exact position was successfully predicted by GR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/I8GDI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many situations which can give rise to gravitational lensing. These regimes are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Strong Lensing&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strong lensing is the most visually stunning form of gravitational lensing, and as its name suggests, requires an extremely massive object, and a good deal of alignment between the lens and the source. Galaxy clusters are the most common cause of strong gravitational lensing. Partial arcs, full arcs (&lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein_ring&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein rings&lt;/a&gt;), and multiple images are all strong gravitational lensing features one can observe. Some of the most commonly studied objects producing strong gravitational lensing features are the &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_Abell_clusters&quot; rel=&quot;nofollow noreferrer&quot;&gt;Abell clusters&lt;/a&gt;, the most famous of which is &lt;a href=&quot;http://en.wikipedia.org/wiki/Abell_1689&quot; rel=&quot;nofollow noreferrer&quot;&gt;Abell 1689&lt;/a&gt; (image below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lbIvw.jpg&quot; alt=&quot;Abell 1689&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strong lensing features like arcs and rings are typically due to extended objects (like background galaxies which are not part of the cluster itself), and multiple-image (mostly quad-image systems) are typically objects like background quasars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Weak Lensing&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weak gravitational lensing occurs much more frequently than strong lensing. Lenses can be clusters (in their outer regions), individual galaxies, or even large scale structure in the universe. Weak lensing is not a noticeable effect by eye, rather, it has to be done statistically. Ellipticities of a &lt;em&gt;field&lt;/em&gt; of background galaxies are observed on a grid, and are statistically averaged together to create the weak lensing signal. Shape distortions of these background galaxies due to lensing are on the percent scale. One important assumption is made, however, and that is that galaxies' isophotes (lines of constant light) are elliptical, and their orientations are completely random. With that, any &lt;em&gt;net&lt;/em&gt; tangential shearing produced is due to gravitational lensing. In the image below (from upper left to bottom right), the upper left frame shows an unlensed field of circular galaxies, and to its immediate right shows the effect of lensing. The image in the bottom right has added shape noise (a 'realistic' field of background galaxies), and to its right is how the field would lens.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jXnUj.png&quot; alt=&quot;WL Field.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, higher order shape distortion effects, most notably &lt;a href=&quot;http://www.physics.drexel.edu/~goldberg/flexion/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Flexion&lt;/a&gt;, can produce the extended galaxy sources to not only shear, but to flex. This is currently a tough thing to measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Microlensing&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Microlensing is a regime of lensing which is most common on the scale of the Milky Way galaxy. This can occur when background stars pass behind foreground stars. Microlensing is actually strong enough to produce multiple images of the background star, but since the image separations are so small (micro-arcsecond scale - hence the name), what we observe (since an angular resolution of a micro-arcsecond is tough to achieve)  is a change in the flux as the object moves into and out of alignment with the intermediate massive object. Interestingly, microlensing has actually proven useful in detecting planets around stellar lensing systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/B75do.jpg&quot; alt=&quot;Microlensing Planet&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2013-09-26T14:18:49.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="243" PostTypeId="2" ParentId="237" CreationDate="2013-09-26T14:24:00.253" Score="6" Body="&lt;p&gt;There are many indeed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be specified from South westwards(0° to 360°), or from North eastwards and westwards(180° to -180°).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most popular one, as used by Roy and Clarke in &quot;Astronomy : Principles and Practices&quot; is to measure the azimuth from North eastwards 0° to 360°. For a southern hemisphere observer, this system changes to from South eastwards 0° to 360°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With so much confusion, how do professional astronomers go about it? They simply specify the system they are using in a given document. Doesn't cause much trouble since the differences in all of the systems is by an angle of 180° or 360°. It just causes sign changes in all formula.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2013-09-26T14:24:00.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="244" PostTypeId="2" ParentId="237" CreationDate="2013-09-26T14:30:39.233" Score="5" Body="&lt;p&gt;Azimuth is conventionally calculated as a degree value between 0 (inclusive) and 360 (exclusive). &lt;strong&gt;North&lt;/strong&gt; is assigned 0 (also by convention), which puts &lt;strong&gt;East&lt;/strong&gt; at 90, &lt;strong&gt;South&lt;/strong&gt; at 180, and &lt;strong&gt;West&lt;/strong&gt; at 270.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes 0 is set to &lt;strong&gt;South&lt;/strong&gt; intead of &lt;strong&gt;North&lt;/strong&gt;. This is the traditional means of measuring solar azimuth, even though &lt;strong&gt;North&lt;/strong&gt; is the most accepted convention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Positive azimuths are noted when measuring east of 0, whether 0 is &lt;strong&gt;South&lt;/strong&gt; or &lt;strong&gt;North&lt;/strong&gt;. Negative azimuths are noted when measuring west of 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sources:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.susdesign.com/popups/sunangle/azimuth.php&quot; rel=&quot;nofollow&quot;&gt;Azimuth Angle - Sustainable by Design&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_azimuth_angle&quot; rel=&quot;nofollow&quot;&gt;Solar azimuth angle - Wikipedia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.astro.cornell.edu/academics/courses/astro201/alt_az.htm&quot; rel=&quot;nofollow&quot;&gt;Altitude and Azimuth - Martha Haynes, Cornell&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6" LastEditorUserId="6" LastEditDate="2013-10-14T22:02:27.207" LastActivityDate="2013-10-14T22:02:27.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="245" PostTypeId="2" ParentId="241" CreationDate="2013-09-26T14:39:28.353" Score="14" Body="&lt;p&gt;What you say is not quite true: the search for exoplanets is clearly intensive, but it is far from the only things astronomers are looking at. Most of the time, in two words, the situation is: &lt;strong&gt;resolution&lt;/strong&gt; &amp;amp; &lt;strong&gt;wavelength&lt;/strong&gt;. Whatever the field (if you are interested in galaxies, interstellar medium, stars and so on) you want &lt;strong&gt;more resolution&lt;/strong&gt;, to resolve smaller scales (most stars are still points even with our best telescopes, or we are still far from resolving individual stars in galaxies!), to have more informations, to understand better the underlying physics. You want more wavelength, because &lt;strong&gt;spectroscopy&lt;/strong&gt; gives you much more phsyical information than a single wavelength observation, for example. And to combine both is sometimes challenging: to have high resolution observations in infrared is not that easy, and it can be crucial for some fields (if you ever want to see a star forming, you better observe it in infrared, since this baby is embedded in its gas cloud that shields very efficiently its radiation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, the routine tasks of an astronomer would be&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;extract information from the current data. It involves a lot of coding, with Python, IDL, or more specific astronomy-oriented languages as IRAF or MIDAS. Data reduction is an important part of the job, because it is in general challenging to extract data from the raw signal you will get.&lt;/li&gt;&#xA;&lt;li&gt;write papers about these data and the inferred informations&lt;/li&gt;&#xA;&lt;li&gt;read a lot of papers to stay tune with the latest discoveries of other teams&lt;/li&gt;&#xA;&lt;li&gt;write proposals to ask for more observation time/better observations/bigger telescopes&lt;/li&gt;&#xA;&lt;li&gt;drink a lot of coffee&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The three first points probably takes an almost equal amount of time for any astronomer; point 4 takes even more time for older astronomers; point 5 is also crucial for all the good things that come out of discussions over a good ol' bowl of coffeine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Complements:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer to your comment and to give you an overview of the current research, I can think of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://sci.esa.int/herschel/&quot;&gt;Hershel data&lt;/a&gt; in infrared. People try to understand better the interstellar medium and the star formation processes in our galaxy, the formation of early galaxies, and the chemical composition and evolution of the Universe with these data.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://sci.esa.int/planck/30968-objectives/&quot;&gt;Planck data&lt;/a&gt; in longer wavelength. These data are useful to understand the first age of the Universe (searching for anistropy in the CMB), but also to have an other view of the galaxy and the interstellar medium in these wavelengths.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.eso.org/public/teles-instr/vlt/&quot;&gt;Very Large Telescope data&lt;/a&gt;. There are plenty of different kind of data out of these telescope, mostly in the visible and infrared ranges, and mostly in spectroscopy. Almost everything is studied with these data, from galaxies evolution to stars in the nearby galaxies.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.eso.org/public/teles-instr/alma/&quot;&gt;ALMA data&lt;/a&gt; in millimeter/submillimeter ranges. The same kind of objects are studied with ALMA and Herschel: early galaxies, interstellar medium and molecular clouds. How galaxies form and evolve? How stars form? In which environment? What are the dominent processes in star formation?&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.mpi-hd.mpg.de/hfm/HESS/&quot;&gt;HESS data&lt;/a&gt;, in the gamma ray range. Gamma rays offer a window on the non-thermal Universe, i.e. all the extreme events occuring in the Universe. It can give precious informations about gamma-ray bursts, supernovæ, AGN (the active galactic nuclei) etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That's for the big projects (with a strong European bias, sorry folks I know better what's done on this side of the ocean). You can add to that all the missions to study exoplanets (like Kepler), missions to study our solar system planets (Cassini, Huygens, Messenger, Juno, all the Mars missions etc.), plus all the other facilities around the world to study anything and everything, from stellar dynamic to planet's composition. The main problem is always to understand how such structure (from the large scale structures in the Universe to small scale structures in the galaxy), object (from galaxies to satellites), phenomenon occurs, forms, appears. To understand what are the dominant physical processes at play.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomy is still craving for data; the more data you have, the better your statistic will be, and hopefully the better your understanding will also be.&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2013-09-27T12:02:34.780" LastActivityDate="2013-09-27T12:02:34.780" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="246" PostTypeId="1" AcceptedAnswerId="13122" CreationDate="2013-09-26T14:53:45.853" Score="15" ViewCount="342" Body="&lt;p&gt;This was somewhat inspired by &lt;a href=&quot;https://astronomy.stackexchange.com/questions/220/do-other-planets-in-our-solar-system-experience-eclipses-or-is-this-unique-to-ea&quot;&gt;this question&lt;/a&gt; but is a bit different: I am interested in if there are any known systems in which the angular size of the eclipsing object is &lt;strong&gt;almost the same&lt;/strong&gt; as the angular size of the object being eclipsed from the surface of a planet, as is the case for the Moon and the Sun as viewed from Earth. This is the only way that one can get &quot;annular&quot; eclipses, or eclipses in which the host star's corona is clearly visible. This could be the case for some moons in the solar system (which requires that they eclipse at all), and at least Phobos in particular seems to have a similar angular size to the Sun as viewed from Mars,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/09ccx.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but it could also be true for some known (or candidate) multiple-planet exoplanetary systems in which the periods of the two planets are not too different. Probably a perusal of the public &lt;a href=&quot;http://archive.stsci.edu/kepler/planet_candidates.html&quot; rel=&quot;noreferrer&quot;&gt;Kepler candidate sample&lt;/a&gt; would be helpful here.&lt;/p&gt;&#xA;" OwnerUserId="2" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-01-07T20:26:13.937" Title="Do any known exoplanetary/solar bodies have &quot;annular&quot; eclipses similar to Earth's?" Tags="&lt;exoplanet&gt;&lt;solar-system&gt;&lt;solar-eclipse&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="247" PostTypeId="1" AcceptedAnswerId="248" CreationDate="2013-09-26T15:37:42.820" Score="25" ViewCount="964" Body="&lt;p&gt;The visible surface of the Sun is 6000 degrees Celsius but the gas in the solar corona is at least 1 million degrees. What heats the solar corona?&lt;/p&gt;&#xA;" OwnerUserId="117" LastActivityDate="2013-10-08T13:20:46.260" Title="Why is the Sun's atmosphere (the corona) so hot?" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="248" PostTypeId="2" ParentId="247" CreationDate="2013-09-26T16:05:34.500" Score="18" Body="&lt;p&gt;It is still an open question, even though it is clear that it is linked to the magnetic field of the Sun. The following hypothesis are the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Heating waves mechanism:&lt;/strong&gt; propagation of magnetohydrodynamic waves can heat significantly the corona. These waves can be produced in the solar photosphere and propagate carrying energy through the solar atmosphere; they turn in the end into shock waves that dissipate energy as heat in the corona.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Magnetic reconnection mechanism:&lt;/strong&gt; magnetic field lines are tied to the surface of the Sun and even when the plasma moves, they stay tied to their original position (their &quot;foot points&quot;); therefore, plasma movements can drag the magnetic field lines and entangled them in geometries such as loops, that can eventually &quot;reconnect&quot;. &quot;Reconnection&quot; corresponds to a modification of the field line topology (see the image that pictures it better than my words). During this process, heat and energy are released quite efficiently.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Nanoflares mechanism:&lt;/strong&gt; the dissipation of a lot of small scale structures (slow convective motions of magnetic foot points could create current sheets, coutinually dissipated and reformed, providing heating through Ohmic dissipation). You can see it as a kind of a lot of magnetic loops reconnecting together, instead of one big loop reconnecting alone.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/S7LEY.jpg&quot; alt=&quot;Magnetic reconnection&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black lines represent the magnetic field lines. They are tied to the surface of the Sun (the foot points). One of the magnetic field lines is looping, and it is &quot;crossing itself&quot;. When it reconnects, it forms, on the top, a magnetic loop and, on the bottom, an magnetic arch. As pointed out by this diagram, magnetic reconnection mechanism is also associated to Coronal Mass Ejection (CME), since the energy released by magnetic reconnection is powerful enough to eject a significant amount of matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt; &lt;a href=&quot;http://arxiv.org/abs/1206.6097&quot; rel=&quot;noreferrer&quot;&gt; A Contemporary View of Coronal Heating&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2013-10-08T13:20:46.260" LastActivityDate="2013-10-08T13:20:46.260" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="249" PostTypeId="1" AcceptedAnswerId="295" CreationDate="2013-09-26T16:40:43.187" Score="14" ViewCount="960" Body="&lt;p&gt;The temperature of the surface of the Sun (photosphere) is between 4500° - 6000° Kelvin. Inside the core, it's around 15.7 million degrees Kelvin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other types of stars (neutron stars, white dwarves, etc.), what is the temperature of these areas (though many do not have these same layers) and how to they compare to the temperatures of the sun?  &lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="37191" LastEditDate="2021-02-12T20:59:04.700" LastActivityDate="2021-02-12T20:59:04.700" Title="How do stellar temperatures vary?" Tags="&lt;star&gt;&lt;neutron-star&gt;&lt;temperature&gt;&lt;white-dwarf&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="250" PostTypeId="2" ParentId="225" CreationDate="2013-09-26T17:08:38.093" Score="5" Body="&lt;p&gt;To get into scientific fields as a professional pretty much requires a doctorate these days.  This can be in astronomy or your chosen specialty (see below), but many are also in physics; what the subject is called may depend on the institution you attend to get the Ph.D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Realize that what you and the general public thinks of as &quot;Astronomy&quot; is likely not the same as what the professional scientific world calls &quot;Astronomy&quot;.  An &quot;Astronomer&quot; studies stars and groupings of stars.  Someone who studies our Sun is usually called a &quot;Solar Physicist&quot; or something similar.  Someone who studies other planets is a &quot;Planetary Scientist&quot;; this can be further subdivided into &quot;Atmospheric Scientist&quot; (which can be even further divided into &quot;Atmospheric Chemist&quot;, &quot;Atmospheric Dynamicist&quot;, etc.) and &quot;Planetary Geologist&quot;, and so on.  A scientist studying the origins of the universe is a &quot;Cosmologist&quot;.  If you have a particular area you are interested in, research this to find out what you need to learn to move into that field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless of the degree you pursue, you will need a lot of mathematics and science.  During your undergraduate degree, I would recommend the following minimums: 3 semesters of Calculus, 1 semester of Ordinary Differential Equations, 1 semester of Linear Algebra (some schools combine ODE and LA), 3 semesters of Introductory Physics.  I'd recommend a semester or two of introductory astronomy, although some schools don't consider that as absolutely necessary, surprisingly. Other useful undergraduate courses will depend on what you think you will want to specialize in, but would include more physics, more astronomy, chemistry, geology, meteorology, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you aren't yet in college, take all the mathematics and physical science courses you can in high school... unless you're interested in becoming a astrobiologist (someone who studies living organisms on other planets - a speculative field at this point, but it does exist), in which case you should get some biology courses too!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Learning computer programming would be good also, as these fields often require programming for data processing and simulations.  Amongst general purpose languages, the &lt;a href=&quot;http://www.python.org/&quot; rel=&quot;nofollow&quot;&gt;Python&lt;/a&gt; programming language may be the most popular in scientific environments.  Others used include &lt;a href=&quot;http://www.exelisvis.com/ProductsServices/IDL.aspx&quot; rel=&quot;nofollow&quot;&gt;IDL&lt;/a&gt; and &lt;a href=&quot;http://www.mathworks.com/products/matlab/&quot; rel=&quot;nofollow&quot;&gt;MatLab&lt;/a&gt;.  Linux and Mac computers are used more in the science world than in the broader populace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted in Jeremy's answer and its comments, job prospects aren't great for astronomers and related fields.  That said, I don't want to discourage you, just prepare you for reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, as another answer noted, if you just want to observe the skies, no degree is required, just time and patience.&lt;/p&gt;&#xA;" OwnerUserId="25" LastEditorUserId="25" LastEditDate="2014-11-03T02:31:54.333" LastActivityDate="2014-11-03T02:31:54.333" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="251" PostTypeId="1" AcceptedAnswerId="269" CreationDate="2013-09-26T17:15:53.743" Score="13" ViewCount="446" Body="&lt;p&gt;This morning on the radio I heard about the discovery of a pulsar that, due to its proximity to a companion star, was flipping between a state where it emitted radio waves and one where it emitted x-rays. The flipping between states was said to be due to material from the companion star falling onto the pulsar, causing it to rotate more quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the significance of this discovery?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.independent.co.uk/news/science/astronomers-discover-pulsar-undergoing-radical-transformation-8841113.html&quot;&gt;This newspaper article from the UK on the discovery&lt;/a&gt; contains more information, but doesn't go into detail about &lt;em&gt;why&lt;/em&gt; this discovery is important.&lt;/p&gt;&#xA;" OwnerUserId="118" LastActivityDate="2013-09-26T23:23:25.370" Title="What is the significance of the discovery of a pulsar flipping between radio and x-ray emissions?" Tags="&lt;pulsar&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="252" PostTypeId="1" AcceptedAnswerId="259" CreationDate="2013-09-26T17:17:01.853" Score="16" ViewCount="623" Body="&lt;p&gt;How do I observe the Sun safely, with tools I could build myself and without breaking the bank? If I wanted to build a Sun funnel on my own and use it on my enthusiast grade telescope, how would I do that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What supplies and tools would I need, how do I attach it to my telescope so it can be applied to it at will, and are there any other, simpler yet safe ways to observe the Sun and its activity with my telescope? This is approximately what I had in mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/jbtte.png&quot; alt=&quot;Sun funnel&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I build something like that? I'm primarily interested in observing sunspots, but would also like to know, if there is any way for me to safely observe Sun's coronal activity to any appreciable detail with an enthusiast grade newtonian/dobsonian telescope? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, could I project the telescope's image directly on some screen behind it, without having to use any intermediary digital equipment, like a recording camera and a digital light projector? &lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2013-09-26T20:20:36.090" Title="How do I build a stargazing aficionado's Sun funnel to observe our nearest star safely?" Tags="&lt;the-sun&gt;&lt;observational-astronomy&gt;&lt;amateur-observing&gt;&lt;dobsonian-telescope&gt;&lt;newtonian-telescope&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="253" PostTypeId="1" AcceptedAnswerId="277" CreationDate="2013-09-26T18:04:25.677" Score="19" ViewCount="13782" Body="&lt;p&gt;It was estimated that the heat inside the core of the Sun inside around 15 000 000 °C - this value is extremely enormous. How did scientists estimate this value?&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="37191" LastEditDate="2021-02-17T22:25:53.253" LastActivityDate="2021-02-17T22:25:53.253" Title="How was the core temperature of the Sun estimated?" Tags="&lt;the-sun&gt;&lt;temperature&gt;&lt;core&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="254" PostTypeId="1" AcceptedAnswerId="8685" CreationDate="2013-09-26T18:15:26.707" Score="28" ViewCount="2591" Body="&lt;p&gt;The Oort cloud is a hypothetical structure based on our observation of long-period comets. There are currently proposals to design probes to confirm the existence of the Oort cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/0/03/Kuiper_oort.jpg&quot; alt=&quot;Oort cloud&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, sending a probe would have other benefits, but why can't we observe the Oort cloud with a telescope?&lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:47.007" LastActivityDate="2015-04-02T00:29:39.477" Title="Why can't we observe the Oort cloud with a telescope?" Tags="&lt;observational-astronomy&gt;&lt;oort-cloud&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="256" PostTypeId="1" AcceptedAnswerId="265" CreationDate="2013-09-26T19:32:54.407" Score="16" ViewCount="210" Body="&lt;p&gt;According to &lt;a href=&quot;http://www.bbc.co.uk/news/science-environment-24287207&quot;&gt;data from Curiosity, Mars' dust holds about 2% water by weight&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This wasn't previously detected, so the impression we have had of Mars being incredibly dry may need to be altered. Okay it is still very dry, but there is potentially extractable water in amounts usable by colonists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this sort of thing would be good to discover &lt;strong&gt;before&lt;/strong&gt; going to another planet, so how can we detect this sort of thing without actually going to the surface and heating up the local dirt?&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2019-09-12T03:53:53.367" Title="How can we detect water on Mars-like exoplanets?" Tags="&lt;water&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="257" PostTypeId="2" ParentId="253" CreationDate="2013-09-26T20:04:05.897" Score="15" Body="&lt;p&gt;Hydrodynamic models of the Sun allow one method of estimating its internal properties. To do this, the Mass, radius, surface temperature, and total luminosity (radiative energy emitted)/s of the Sun must be known (determined observationally). Making several assumptions, e.g., that the Sun behaves as a fluid and that local thermodynamic equilibrium applies, the stellar equations of state can be used. Numerical methods are applied to these equations to determine the internal properties of the Sun, such as its central temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A great example for how to work this problem your self can be found in the undergraduate text, 'An Introduction to Modern Astrophysics' by Carroll and Ostlie (Section 10.5). The FORTRAN code to run your own stellar model is included in Appendix H. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A comprehensive review paper on how stars of different masses evolve internally (e.g., with respect to T, P, etc.) that is worth reading is: &#xA;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1967ARA%26A...5..571I&quot;&gt;http://adsabs.harvard.edu/abs/1967ARA%26A...5..571I&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very interesting historical overview of the development of the Standard Solar Model:&#xA;&lt;a href=&quot;http://arxiv.org/abs/astro-ph/0209080&quot;&gt;http://arxiv.org/abs/astro-ph/0209080&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This (admittedly dry) paper gives you a good idea of how well the 'standard' solar models estimate the internal properties of the Sun using helioseismology and neutrino measurements to help tie down their boundary conditions:&#xA;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1997PhRvL..78..171B&quot;&gt;http://adsabs.harvard.edu/abs/1997PhRvL..78..171B&lt;/a&gt;&#xA;The answer is that they match incredibly well (&gt;0.2% error)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These were the least technical (but still academically published) references I could find.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a comprehensive page on the state-of-the-art in solar modelling and measuring the internal Sun using Helioseismology:&#xA;&lt;a href=&quot;http://www.sns.ias.edu/~jnb/Papers/Preprints/solarmodels.html&quot;&gt;http://www.sns.ias.edu/~jnb/Papers/Preprints/solarmodels.html&lt;/a&gt;&#xA;(highly technical)&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditDate="2013-09-30T21:16:33.860" LastActivityDate="2013-09-30T21:16:33.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="258" PostTypeId="2" ParentId="246" CreationDate="2013-09-26T20:19:48.173" Score="4" Body="&lt;p&gt;This will depend the position of the observer and (obviously) the relative sizes of the star and the eclipsing body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an intelligent observer standing on the surface of a planet the most obvious and likely candidate for an eclipsing body would be a moon of that planet. The nearer the planet is to the sun the larger that moon has to be, and the further the planet is away from the star the smaller the moon can be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding planets around stars is hard enough, finding moons that will be orbiting quite closely around those planets across interstellar distances is even harder (I'm not going to say impossible, but it's pretty close) with current technology, so at the moment the answer is &quot;None, that we know of&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also don't forget that, according to current theories, the moon was formed when a large Mars sized body collided with the nascent Earth, the chances of another Earth/Moon type system is probably remote. This means that most natural satellites are likely to be Phobos sized bodies. Therefore to get an eclipse the planet would have to be further out from the star (so the star looks smaller) and that would tend to put it on the edge of the &quot;Goldilocks&quot; zone, making the likelihood of there being an intelligent observer somewhat rare.&lt;/p&gt;&#xA;" OwnerUserId="111" LastEditorUserId="111" LastEditDate="2013-09-26T20:25:52.517" LastActivityDate="2013-09-26T20:25:52.517" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="259" PostTypeId="2" ParentId="252" CreationDate="2013-09-26T20:20:36.090" Score="12" Body="&lt;p&gt;As for projecting the Sun onto a screen at a low cost, I would recommend starting with a ~50-200$ sunspotter box, which is basically a lens mounted on a wooden box, that projects the sun onto a white piece of card. The advantage of using a telescope is that it can be programmed to track the Sun, so that if you want to trace sunspots, for instance, you can do so at leisure, instead of having to race to finish before the Sun moves off of the page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for observing coronal features- this is a very difficult task. A H-alpha filter (expensive) can be purchased for a home telescope outfitted for solar observing. This would allow you to see filaments/prominences, which is low T (10^4 K) chromospheric material that extends into the corona, suspended by magnetic field and buoyancy forces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To observe coronal material from Earth, you would need a coronagraph, which blocks out light from the central disk of the Sun. This is because coronal features are extremely dim, as compared with the photosphere. From Earth, coronal features are best observed with during a total eclipse. Using a telescope mounted coronagraph at home is not ideal (and probably not feasible) because a lot of light from the Sun would leak into your field-of-view, due to both diffraction, as well as sky brightness. The University of Hawaii takes these sorts of observations, but even with all of their expensive equipment, the space-based LASCO instrument is able to do far better. &lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2013-09-26T20:20:36.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="261" PostTypeId="1" AcceptedAnswerId="309" CreationDate="2013-09-26T20:55:04.263" Score="8" ViewCount="1973" Body="&lt;p&gt;What gases do you need to start the creation of a star, and why do you need these gases? What are their functions?&lt;/p&gt;&#xA;" OwnerUserId="69" LastActivityDate="2020-11-30T16:25:23.753" Title="What gases are needed to make a star?" Tags="&lt;star&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="264" PostTypeId="2" ParentId="214" CreationDate="2013-09-26T21:44:37.527" Score="8" Body="&lt;p&gt;Stellar clusters around supermassive black holes are systems in which relativity likely plays a role. Currently, only bright stars can be seen in our own galactic center because there is a ton of neutral gas between us and the galactic center that obscures it. As a result, we only have a few &quot;test particles&quot; out of the many stars that actually orbit the black hole at close distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, measuring &lt;a href=&quot;http://en.wikipedia.org/wiki/Relativistic_Precession&quot;&gt;relativistic precession&lt;/a&gt; may be possible for a star with one of the closest-known pericenter distances to Sagittarius A* (the central black hole in our galaxy), &lt;a href=&quot;http://en.wikipedia.org/wiki/S2_%28star%29&quot;&gt;S2&lt;/a&gt;, potentially within the next few years once enough data has been collected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to how relativistic effects can affect dynamics of the cluster, the precession induced by general relativity can suppress resonant interactions, including three-body resonances such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kozai_resonance&quot;&gt;Kozai&lt;/a&gt;. Depending on if these sorts of resonances are important compared to other relaxation processes, the relaxation time can increase significantly, resulting in the cluster evolving more-slowly over time. This can affect things such as the rate of &lt;a href=&quot;http://en.wikipedia.org/wiki/Mass_segregation&quot;&gt;mass segregation&lt;/a&gt;, &lt;a href=&quot;http://adsabs.harvard.edu/abs/2009ApJ...697L.149C&quot;&gt;tidal disruptions&lt;/a&gt;, and production of &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_kinematics#Hypervelocity_stars&quot;&gt;hypervelocity stars&lt;/a&gt;/&lt;a href=&quot;http://adsabs.harvard.edu/abs/2008ApJ...683L.151L&quot;&gt;S-stars&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2" LastEditorUserId="2" LastEditDate="2013-09-26T22:49:01.260" LastActivityDate="2013-09-26T22:49:01.260" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="265" PostTypeId="2" ParentId="256" CreationDate="2013-09-26T22:02:15.930" Score="9" Body="&lt;p&gt;It would be difficult to detect the water-by-weight in the dust of another planet to the precision that has been done for the Martian soil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, having said that, there are a couple of ways to detect the presence of water, and that would be, according to &lt;a href=&quot;http://www.space.com/21800-alien-planets-60-billion-habitable-exoplanets.html&quot;&gt;&quot;60 Billion Alien Planets Could Support Life, Study Suggests&quot;&lt;/a&gt; (Gannon, 2013), to detect the presence of clouds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To confirm that the clouds are water clouds, not something the iron-vapour etc, according to web article &lt;a href=&quot;http://www.astrobio.net/pressrelease/5541/hunting-for-water-on-exoplanets&quot;&gt;&quot;Hunting for Water on ExoPlanets&quot;&lt;/a&gt; (bolding mine),&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Using ESO’s Very Large Telescope (VLT), &lt;strong&gt;a team of astronomers have been able to detect the telltale spectral fingerprint of water molecules in the atmosphere of a planet in orbit around another star&lt;/strong&gt;. The discovery endorses a new technique that will let astronomers efficiently search for water on hundreds of worlds without the need for space-based telescopes. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2013-09-26T22:02:15.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="266" PostTypeId="1" AcceptedAnswerId="419" CreationDate="2013-09-26T22:56:19.733" Score="6" ViewCount="2796" Body="&lt;p&gt;What are shooting stars? How are they formed and how often do they occur during the night?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, why are there more shooting stars on some nights than on others?&lt;/p&gt;&#xA;" OwnerUserId="69" LastEditorUserId="325" LastEditDate="2014-05-27T10:40:52.980" LastActivityDate="2014-05-27T10:40:52.980" Title="What are shooting stars and how are they formed?" Tags="&lt;meteor&gt;" AnswerCount="2" CommentCount="5" ClosedDate="2013-10-01T14:10:10.360" ContentLicense="CC BY-SA 3.0" />
  <row Id="268" PostTypeId="2" ParentId="266" CreationDate="2013-09-26T23:07:07.290" Score="7" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Shooting_Star&quot; rel=&quot;nofollow noreferrer&quot;&gt;shooting star&lt;/a&gt; is simply the visible path a meteriod takes burning up in the atmosphere. As it flies through the atmosphere, it leaves a trail of fire, creating the streak of light you see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before they enter the atmosphere, they are simply chunks of rock. They could form in one of many ways - be broken off of a planet, etc. After they do this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TL5iK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;they look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RnUCl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the reasons for meteor frequency varying, it's because of where the Earth is in it's orbit around the sun. Since almost all meteors are in orbit around the sun, and there tend to be clumps of them, this makes a 'donut' of meteors in orbit around the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the earth intersects one of these donuts, there is a spike in meteor activity. Ultimately, it's about how many space-rocks happen to hit Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One annual 'high', for example, is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Perseids&quot; rel=&quot;nofollow noreferrer&quot;&gt;Perseid&lt;/a&gt; peak, which tends to occur mid-late summer each year. Sometimes, this even &lt;a href=&quot;https://space.stackexchange.com/questions/1199/does-the-perseid-peak-affect-mission-launches&quot;&gt;delays launches of space vehicles&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2013-09-26T23:12:39.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="269" PostTypeId="2" ParentId="251" CreationDate="2013-09-26T23:23:25.370" Score="4" Body="&lt;p&gt;The significance of this discovery, according to the European Space Agencies web article &lt;a href=&quot;http://sci.esa.int/integral/52866-volatile-pulsar-reveals-millisecond-missing-link/&quot; rel=&quot;nofollow&quot;&gt;&quot;Volatile pulsar reveals millisecond missing link&quot;&lt;/a&gt; is the first time a pulsar in a crucial transitional phase between has been observed, and that this explains the origin of mysterious millisecond pulsars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The significance is as stated by the ESA (bolding mine):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This bouncing behaviour is caused by a &lt;strong&gt;rhythmical interplay between the pulsar's magnetic field and the pressure of accreted matter.&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What this tells us about pulsar dynamics is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When accretion is more intense, the high density of accreted matter inhibits the acceleration of particles that cause radio emission, so the pulsar is not visible in radio waves but only through the X-rays radiated by the accreted matter. When the accretion rate decreases, the magnetosphere expands and pushes matter away from the pulsar: &lt;strong&gt;as a consequence, the X-ray emission becomes weaker and weaker, while the radio emission intensifies.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The article also contains contact details of scientists involved&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2013-09-26T23:23:25.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="270" PostTypeId="1" AcceptedAnswerId="275" CreationDate="2013-09-27T00:00:17.283" Score="22" ViewCount="33331" Body="&lt;p&gt;Each new star we find is generally considered to be part of the constellation it is nearest to. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our Sun is obviously a star, just much closer. Is our Sun part of any constellation? If so, which constellation is it a part of?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oqVJl.jpg&quot; alt=&quot;nearby contellations&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastEditorUserId="2153" LastEditDate="2015-01-19T14:16:22.557" LastActivityDate="2016-02-08T03:26:59.500" Title="Does the Sun belong to a constellation?" Tags="&lt;the-sun&gt;&lt;constellations&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="271" PostTypeId="1" AcceptedAnswerId="283" CreationDate="2013-09-27T00:14:09.570" Score="19" ViewCount="6264" Body="&lt;p&gt;What does it mean for a star when people say it goes 'nova' or super nova, what are the differences?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More importantly, can I safely observe these with an amateur telescope? I imagine they would be big enough to spot should one happen!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oOE5V.jpg&quot; alt=&quot;supernova&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastEditorUserId="141" LastEditDate="2013-11-15T15:44:27.397" LastActivityDate="2013-11-15T15:44:27.397" Title="What does it mean for a star to go nova or supernova? Can I safely observe these?" Tags="&lt;star&gt;&lt;supernova&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="272" PostTypeId="1" AcceptedAnswerId="282" CreationDate="2013-09-27T00:21:14.797" Score="11" ViewCount="1286" Body="&lt;p&gt;Inspired by the question &lt;a href=&quot;https://astronomy.stackexchange.com/questions/271/what-does-it-mean-for-a-star-to-go-nova-or-supernova-can-i-safely-observe-these&quot;&gt;&quot;What does it mean for a star to go nova or supernova? Can I safely observe these?&quot;&lt;/a&gt;, I am curious about what &lt;strong&gt;hypernovas&lt;/strong&gt; are?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have we observed any occurring, or the at least the after-effects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an upper limit of how large a stellar explosion can be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/T3c8J.jpg&quot; alt=&quot;hypernova&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2020-02-28T21:24:50.340" Title="What is a hypernova and have we observed any?" Tags="&lt;star&gt;&lt;supernova&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="273" PostTypeId="1" AcceptedAnswerId="284" CreationDate="2013-09-27T00:29:11.640" Score="22" ViewCount="781" Body="&lt;p&gt;Why is the Hubble Telescope in space? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we get enhanced clarity and range by placing it in space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can it achieve from space that it could not achieve from Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uy8rz.jpg&quot; alt=&quot;hubble telscope in space&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastActivityDate="2013-10-31T19:32:11.267" Title="Why is the Hubble Telescope in space?" Tags="&lt;hubble-telescope&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="274" PostTypeId="2" ParentId="270" CreationDate="2013-09-27T00:31:19.823" Score="16" Body="&lt;p&gt;No, it does not. The constellations are fixed (on time scales long enough for humans to consider as fixed, at least) patterns of stars which exist on the celestial sphere. This celestial sphere is a coordinate system which has the Earth at its center. From the Earth's perspective, the sun rises and sets at the same rate as the constellations, but as the Earth revolves around the sun, the constellation the sun will appear to be in will change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the constellation the sun appears to be in (if projected onto the celestial sphere), is exactly how the constellations of the Zodiac are determined. &lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditDate="2013-11-01T03:44:39.497" LastActivityDate="2013-11-01T03:44:39.497" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="275" PostTypeId="2" ParentId="270" CreationDate="2013-09-27T00:36:38.000" Score="37" Body="&lt;p&gt;Constellations are human constructs to make sense of the night sky.  When you are trying to find your way around, it helps to &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Chunking_(psychology)&quot; rel=&quot;noreferrer&quot;&gt;chunk&lt;/a&gt;&quot; stars into patterns and assign those groupings names.  When I want to point out a particular object in the sky (say Polaris, the North Star), I start by pointing out a familiar constellation (say &lt;em&gt;Ursa Major&lt;/em&gt;, the Big Dipper).  From there, I can tell my friend to &lt;a href=&quot;http://www.physics.ucla.edu/~huffman/finddip.html&quot; rel=&quot;noreferrer&quot;&gt;follow this or that line&lt;/a&gt; to get them to look where I'm looking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oWCZr.gif&quot; alt=&quot;Finding the North Star&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the advent of computerized telescopes and large data sets, constellations are less important for professional astronomers.  However, many stellar databases use &lt;a href=&quot;http://en.wikipedia.org/wiki/Flamsteed_designation&quot; rel=&quot;noreferrer&quot;&gt;Flamsteed&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Bayer_designation&quot; rel=&quot;noreferrer&quot;&gt;Bayer&lt;/a&gt; designations, which assign stars to constellations.  In order to include all stars, the sky is divided into irregular regions that encompass the familiar constellations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, which constellations is the Sun assigned to?  Well, from the perspective of someone on the Earth, the Sun moves through the constellations throughout the course of the year.  Or rather, Sol moves through the region of the sky where &lt;a href=&quot;http://en.wikipedia.org/wiki/Western_astrology#The_zodiac&quot; rel=&quot;noreferrer&quot;&gt;some of the constellations&lt;/a&gt; would be seen if its light did not drown out distant stars.  Our moon and the rest of the planets move through those same constellations.  (The Greek phrase which gives us the word &lt;a href=&quot;http://en.wikipedia.org/wiki/Planet&quot; rel=&quot;noreferrer&quot;&gt;&quot;planet&quot;&lt;/a&gt; means &quot;wandering star&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current position of the sun against the background of distant stars changes over the course of the year.  (This is important for &lt;a href=&quot;http://spaceplace.nasa.gov/starfinder2/&quot; rel=&quot;noreferrer&quot;&gt;astrology&lt;/a&gt;.)  It's a little easier to make sense of with a diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vku1F.gif&quot; alt=&quot;How the sun is aligned with a particular constellation.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So perhaps a better question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://spaceplace.nasa.gov/starfinder3/&quot; rel=&quot;noreferrer&quot;&gt;What constellation does the Sun belong to &lt;em&gt;today&lt;/em&gt;?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Presumably &lt;a href=&quot;http://www.projectrho.com/public_html/starmaps/alienskies.php&quot; rel=&quot;noreferrer&quot;&gt;an observer on an exoplanet&lt;/a&gt; would assign Sol to some constellation that is convenient from her perspective.  But from our perspective within the Solar system our sun, moon, and planets are not part of any constellation.&lt;/p&gt;&#xA;" OwnerUserId="48" LastEditorUserId="48" LastEditDate="2013-09-27T20:06:48.880" LastActivityDate="2013-09-27T20:06:48.880" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="276" PostTypeId="2" ParentId="273" CreationDate="2013-09-27T00:41:53.653" Score="12" Body="&lt;p&gt;There are a number of things which are important for a telescope. The first is the light collecting power of the telescope, which is simply a measure of how many photons the telescope can collect from an object. This, not surprisingly, is affected by the size of the primary mirror/lens of the telescope. The second issue is the telescope's angular resolution, which is a measure of the level of detail which can be seen within an image. Telescopes on the surface of the Earth have to worry about the atmosphere, which due to its non-uniformity causes blurring and smearing of images (this contributes significantly to what's known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Point_spread_function&quot;&gt;point spread function&lt;/a&gt;). The HST is in space to avoid atmospheric effects, and thus, it achieves higher angular resolution by doing so. &lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditDate="2013-10-28T16:01:14.660" LastActivityDate="2013-10-28T16:01:14.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="277" PostTypeId="2" ParentId="253" CreationDate="2013-09-27T01:04:35.140" Score="6" Body="&lt;p&gt;The composition can be determined by taking spectra. Additionally, the mass can be determined through dynamics. If you combine these two, under the assumption that the star is in a state of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrostatic_equilibrium&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrostatic equilibrium&lt;/a&gt; (which means that the outward thermal pressure of the star due to fusion of hydrogen into helium is in balance with the inward tug of gravity), you can make statements about what the temperature and density &lt;em&gt;must&lt;/em&gt; be in the core. You need high densities and high temperatures in order to fuse hydrogen into helium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember what's happening: Temperatures are hot enough for hydrogen in the core to be completely ionized, meaning that in order to fuse these protons into helium nuclei, you need to overcome the electromagnetic repulsion as two protons come close (like charges repel). Below is a diagram of the process of one particular type of fusion (&lt;a href=&quot;http://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Proton-proton chain reaction&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4WpuY.png&quot; alt=&quot;proton-proton&quot;&gt;     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other fusion reaction which occurs at the cores of stars is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/CNO_cycle&quot; rel=&quot;nofollow noreferrer&quot;&gt;carbon-nitrogen-oxygen&lt;/a&gt; (CNO) cycle, and is the dominant source of energy for stars more massive than about 1.3 solar masses. Below shows this process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5UMrk.png&quot; alt=&quot;CNO&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;br&gt;&#xA;Somebody pointed out that this doesn't actually answer the question at hand - which is true. Forgetting how to do some of the basic back of the envelope calculations myself (I admit, stellar astrophysics is definitely &lt;em&gt;not&lt;/em&gt; my specialty), I've stumbled across a &lt;em&gt;very crude&lt;/em&gt; and simple estimation of how to calculate the central pressure and temperature of the sun from. The &lt;a href=&quot;http://burro.astr.cwru.edu/Academics/Astr221/StarPhys/estcent.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;calculation&lt;/a&gt; does however point out the correct values and what one would need to know in order to get the details correct.   &lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="1437" LastEditDate="2014-07-16T16:38:18.537" LastActivityDate="2014-07-16T16:38:18.537" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="278" PostTypeId="2" ParentId="89" CreationDate="2013-09-27T01:21:42.783" Score="7" Body="&lt;p&gt;Just to add to Undo's answer, after the expansion to a red giant, the sun will become a &lt;a href=&quot;http://www.noao.edu/jacoby/&quot; rel=&quot;noreferrer&quot;&gt;planetary nebula&lt;/a&gt;, where (according to the link) the fusion reactions inside the star are 'overtaken' by the gravitational collapse, in turn causing the inner layers to condense and heat up causing the outer layers to be blown away.  After which, the hot core forms the remnant, a &lt;a href=&quot;http://imagine.gsfc.nasa.gov/docs/science/know_l2/dwarfs.html&quot; rel=&quot;noreferrer&quot;&gt;White Dwarf star&lt;/a&gt; (NASA source), which is likely to last for several more billions of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image below depicts the current ideas of the expected lifecycle and timeline of the life of the sun:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zob6i.png&quot; alt=&quot;Solar lifecycle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ifa.hawaii.edu/~barnes/ast110_06/tooe.html&quot; rel=&quot;noreferrer&quot;&gt;Image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How do we know what will happen to the sun?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, the main method to determine the solar lifecycle is described by the article &lt;a href=&quot;http://spaceref.com/solar-physics/the-future-of-the-sun.html&quot; rel=&quot;noreferrer&quot;&gt;&quot;The Future of the Sun&quot;&lt;/a&gt; (Cowing, 2013) is to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Studying stars with the same mass and composition as the Sun, the so-called &quot;solar twins,&quot; can give us more information about our own Sun; solar twins of various ages offer snapshots of the Sun's evolution at different phases &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where the mass and chemical composition of a star provide the information needed to determine its lifecycle.&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-09-27T01:40:42.267" LastActivityDate="2013-09-27T01:40:42.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="279" PostTypeId="1" CreationDate="2013-09-27T02:07:51.733" Score="14" ViewCount="18833" Body="&lt;p&gt;Given that the Earth and Moon are often said in the literature to have very similar geochemistry, what is the current theory as to how the Earth's Moon formed?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorDisplayName="user15217" LastEditDate="2017-01-18T13:43:58.843" LastActivityDate="2017-01-18T13:43:58.843" Title="What is the current theory for the formation of the Earth's Moon?" Tags="&lt;planet&gt;&lt;the-moon&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="280" PostTypeId="2" ParentId="279" CreationDate="2013-09-27T02:07:51.733" Score="10" Body="&lt;p&gt;The current accepted theory is known as the &lt;strong&gt;Giant Impact Hypothesis&lt;/strong&gt;, where according to this NASA webpage &lt;a href=&quot;http://solarsystem.nasa.gov/scitech/display.cfm?ST_ID=446&quot;&gt;&quot;Origin of the Earth and Moon&quot;&lt;/a&gt; (Taylor) a Mars sized object collided into the early Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This theory allows explanations of (from the link above):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chemical makeup:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The giant impact hypothesis is consistent with our ideas for how planets were assembled and explains some important features of the Earth-Moon system, such as why the Moon has only a tiny metallic core.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and in terms of orbits:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To account for the amount of angular momentum in the Earth-Moon system, Cameron estimated that the object would need to be about 10% the mass of Earth, about the size of Mars. (Angular momentum is the measure of motion of objects in curved paths, including both rotation and orbital motion. For the Earth and Moon this means the spin of each planet plus the orbital motion of the Moon around the Earth.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dykV0.gif&quot; alt=&quot;impact&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Giant_impact_hypothesis&quot;&gt;Image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, recently, there has been some revisions to this theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to NASA's page &lt;a href=&quot;http://www.nasa.gov/topics/solarsystem/features/moon_formation.html&quot;&gt;&quot;NASA Lunar Scientists Develop New Theory on Earth and Moon Formation&quot;&lt;/a&gt;, based on concerns that the Mars-sized would likely to have had a different composition to the Earth (inconsistent with the current similarities in geochemistry).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The new hypothesis is (from the link above):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;After colliding, the two similar-sized bodies then re-collided, forming an early Earth surrounded by a disk of material that combined to form the moon. The re-collision and subsequent merger left the two bodies with the similar chemical compositions seen today. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is also discussed in &lt;a href=&quot;http://www.space.com/18106-moon-formation-earth-giant-impact.html&quot;&gt;&quot;Huge Moon-Forming Collision Theory Gets New Spin&quot;&lt;/a&gt; (Wall, 2012), basing the revised hypothesis on the rotation rate, which is theorised to have been very fast, from the article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Earth's day had been just two to three hours long at the time of the impact, Cuk and Stewart calculate, the planet could well have thrown off enough material to form the moon (which is 1.2 percent as massive as Earth).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Further discussion, based on the geochemical makeup is discussed in &lt;a href=&quot;http://www.lpi.usra.edu/books/AsteroidsIII/4020.pdf&quot;&gt;&quot;Geochemical Constraints on the&#xA;Origin of the Earth and Moon&quot;&lt;/a&gt; (Jones and Palme) conclude that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Although none of these observations actually disproves&#xA;  the giant impact hypothesis, we find it disquieting that the&#xA;  obvious consequences expected of a giant impact are not&#xA;  observed to be fulfilled.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, there is still some question as to precisely how the Moon was formed.&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2013-09-27T02:07:51.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="281" PostTypeId="2" ParentId="150" CreationDate="2013-09-27T02:09:31.453" Score="12" Body="&lt;p&gt;&lt;sup&gt;&lt;strong&gt;Note to future answerers&lt;/strong&gt;: We have been working in the &lt;a href=&quot;http://chat.stackexchange.com/?tab=site&amp;amp;host=astronomy.stackexchange.com&quot;&gt;Astronomy.SE main chat room&lt;/a&gt; with OP to solve this problem and established it's actually a matter of focusing that resulted in problems described (telescope was either too intrafocal or extrafocal), not collimation itself. This is a &lt;a href=&quot;https://www.youtube.com/watch?v=T-9FIW3AzX0&quot; rel=&quot;noreferrer&quot;&gt;video on YouTube&lt;/a&gt; that OP said was exactly how the problems looked like. Granted, focusing can be technically called collimation too, and indeed that demonstration video calls it as such (albeit misspelled), since it's still about adjusting positions of primary and secondary mirror in &lt;a href=&quot;https://en.wikipedia.org/wiki/Dobsonian_telescope&quot; rel=&quot;noreferrer&quot;&gt;dobsonian&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/Newtonian_telescope&quot; rel=&quot;noreferrer&quot;&gt;newtonian telescopes&lt;/a&gt; (dobsonian is a type of newtonian), but the former is a lot simpler to solve by turning focusing knobs and achieve optimal illumination.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Now, let's answer the actual question.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Preparing our laser collimation setup&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First thing to know is that any laser collimator will likely have the beam strong enough to cause damage to your eyes if pointed directly in them, or damage strong light sensitive equipment, like for example digital cameras, so be careful with it and take precautions. Carefully read instructions on how to handle and use it, before you do so. This is how a laser collimator might look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/gyJob.jpg&quot; alt=&quot;laser collimator&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next thing to do is make it functional by inserting batteries or plugging its power adapter and testing it actually works by turning it on with the switch, if it has it (probably does). Once you'll establish your laser collimator is turned on and functioning by observing (usually ruby red) light beam spot it makes on any neutral surface, turn it back off and attach any adapters on it you might need to help you attach it to your telescope's focuser. This is how your telescope's OTA (Optical Tube Assembly) might look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/RyUiC.png&quot; alt=&quot;OTA (Optical Tube Assembly)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now remove the barlov and eyepiece from your telescope's focuser if you're using a direct focuser adapter for your telescope model, tighten the laser collimator's compression ring by rotating its rubber ring and attach the collimator into the focuser. Rotate collimator's compression rings to tighten it in place. If you're not using an adapter, your laser collimator might attach directly to your eyepiece. Consult your collimator's user manual, if unsure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotate your collimator's targeting faceplate towards your telescope's primary mirror (on the telescope's bottom) and turn on the laser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Angular alignment of the secondary mirror&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By looking into the telescope &lt;em&gt;top down&lt;/em&gt; from the secondary (top) mirror into the telescope, we will adjust the angular alignment of the secondary mirror by referencing the laser dot projecting on the primary (bottom) mirror. The secondary mirror's support has a center ring, the collimating cap, with three bolts on the top of the secondary mirror holder. Use Allen wrench or Phillip's screwdriver to rotate them individually to bring the laser dot into the center of the &lt;em&gt;doughnut&lt;/em&gt; ring of the primary mirror, like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/E79kX.png&quot; alt=&quot;primary mirror doughnut&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to bring the dot more directly into the &lt;em&gt;doughnut&lt;/em&gt; center than on the photograph above, though. I intentionally didn't select a photograph from when it was in its dead center, because of the light reflection covering the &lt;em&gt;doughnut&lt;/em&gt; ring completely, rendering it nearly invisible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Primary mirror collimation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlock the primary mirror plate locking screws, position yourself towards the panel side of the primary mirror tube so you can observe the laser's center spot on the targeting faceplate on the collimator and start adjusting collimation screws by turning them gently until the laser dot is in the center of the targeting faceplate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/fVorO.png&quot; alt=&quot;Primary mirror collimation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to check your collimation progress when adjusting collimation screws is by checking the projection laser dots on the secondary (top) mirror and making sure both laser dots are overlapping into one single spot, but that might make your adjustments of the primary mirror somewhat harder, so using the targeting faceplate of the laser collimator would be my suggestion and perhaps check alignment of the secondary mirrors laser dots projection for the last finishing touches to collimate your mirrors as precise as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you're done collimating your dobsonian or newtonian telescope, so it's time to remove the laser collimator by first switching it off, then untightening its compression ring and taking it out of the focuser. Once you've removed the laser collimator, insert in the focuser barlov and the eyepiece, and tighten them back in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, and as requested, here is the &lt;a href=&quot;https://www.youtube.com/watch?v=ME-CBtemo7Q&quot; rel=&quot;noreferrer&quot;&gt;Youtube video of the Newtonian Collimation using SCA Laser Collimator&lt;/a&gt; that are also the source of the last two example photographs and inspiration to write the collimation procedure, courtesy of &lt;a href=&quot;https://www.youtube.com/user/HoTechUSA?feature=watch&quot; rel=&quot;noreferrer&quot;&gt;HoTechUSA YouTube channel&lt;/a&gt;. Make sure to browse through other similar videos on telescope collimation in this same channel.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Collimation of a dobsonian without a laser collimator&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard collimation technique for the exact dobsonian telescope make OP mentioned has can be read in this &lt;a href=&quot;http://www.ursa.fi/fileadmin/ursa2010/kuvat/kirjakauppa/Dob2002.pdf&quot; rel=&quot;noreferrer&quot;&gt;Instruction Manual for Sky-Watcher dobsonians&lt;/a&gt; (PDF):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Collimation is the process of aligning the mirrors of your  telescope&#xA;  so that they work in concert with each other to  deliver properly&#xA;  focused light to your eyepiece. By observing  out-of-focus star&#xA;  images, you can test whether your  telescope's optics are aligned.&#xA;  Place a star in the centre of  the field of view and move the focuser&#xA;  so that the image is  slightly out of focus. If the seeing conditions&#xA;  are good, you  will see a central circle of light (the Airy disc)&#xA;  surrounded by a  number of diffraction rings. If the rings are&#xA;  symmetrical about  the Airy disc, the telescope's optics are correctly&#xA;  collimated  (Fig.g).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If you do not have a collimating tool, we suggest that you  make a&#xA;  &quot;collimating cap&quot; out of a plastic 35mm film canister  (black with&#xA;  gray lid). Drill or punch a small pinhole in the exact  center of the&#xA;  lid and cut off the bottom of the canister. This  device will keep&#xA;  your eye centered of the focuser tube. Insert  the collimating cap&#xA;  into the focuser in place of a regular  eyepiece. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Collimation is a painless process and works like this:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Pull off the lens cap which covers the front of the telescope  and&#xA;  look down the optical tube. At the bottom you will see the  primary&#xA;  mirror held in place by three clips 120º apart, and at  the top the&#xA;  small oval secondary mirror held in a support and  tilted 45º toward&#xA;  the focuser outside the tube wall (Fig.h). The secondary mirror is&#xA;  aligned by adjusting the central bolt  behind it, (which moves the&#xA;  mirror up and down the tube),  and the three smaller screws&#xA;  surrounding the bolt, (which  adjust the angle of the mirror). The&#xA;  primary mirror is adjusted  by the three adjusting screws at the back&#xA;  of your scope. The  three locking screws beside them serve to hold the&#xA;  mirror in  place after collimation. (Fig.i)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Aligning the Secondary Mirror&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Point the telescope at a lit wall and insert the  collimating cap into&#xA;  the focuser in place of a regular  eyepiece. Look into the focuser&#xA;  through your collimating  cap. You may have to twist the focus knob a&#xA;  few turns  until the reflected image of the focuser is out of your &#xA;  view. Note: keep your eye against the back of the focus  tube if&#xA;  collimating without a collimating cap. Ignore the  reflected image of&#xA;  the collimating cap or your eye for  now, instead look for the three&#xA;  clips holding the primary  mirror in place. If you can't see them&#xA;  (Fig.j), it means  that you will have to adjust the three bolts on the&#xA;  top of  the secondary mirror holder, with possibly an Allen  wrench or&#xA;  Phillip's screwdriver. You will have to  alternately loosen one and&#xA;  then compensate for the  slack by tightening the other two. Stop when&#xA;  you see all  three mirror clips (Fig.k). Make sure that all three&#xA;  small  alignment screws are tightened to secure the  secondary mirror&#xA;  in place.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Aligning the Primary Mirror&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;There are 3 hex bolts and 3 Phillip's head screws  at the back of your&#xA;  telescope, the hex bolts are the  locking screws and the&#xA;  Phillip's-head screws are  the adjusting screws (Fig.l). Use an Allen&#xA;  wrench  to loosen the hex bolts by a few turns. Now run  your hand&#xA;  around the front of your telescope  keeping your eye to the focuser,&#xA;  you will see the  reflected image of your hand. The idea here being &#xA;  to see which way the primary mirror is defected,  you do this by&#xA;  stopping at the point where the  reflected image of the secondary&#xA;  mirror is closest  to the primary mirrors' edge (Fig.m).  When you get&#xA;  to that point, stop and keep your  hand there while looking at the&#xA;  back end of your  telescope, is there a adjusting screw there? If&#xA;  there  is you will want to loosen it (turn the screw to the  left) to&#xA;  bring the mirror away from that point. If  there isn't a adjusting&#xA;  screw there, then go across  to the other side and tighten the&#xA;  adjusting screw on  the other side. This will gradually bring the&#xA;  mirror  into line until it looks like Fig.n. (It helps to have a &#xA;  friend to help for primary mirror collimation. Have  your partner&#xA;  adjust the adjusting screws according  to your directions while you&#xA;  look in the focuser.) After dark go out and point your telescope at &#xA;  Polaris, the North Star. With an eyepiece in the  focuser, take the&#xA;  image out of focus. You will see  the same image only now, it will be&#xA;  illuminated by  starlight. If necessary, repeat the collimating &#xA;  process only keep the star centered while tweaking  the mirror.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8Ju4o.png&quot; alt=&quot;collimation of a dobsonian&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditDate="2013-09-27T14:50:33.963" LastActivityDate="2013-09-27T14:50:33.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="282" PostTypeId="2" ParentId="272" CreationDate="2013-09-27T02:39:29.797" Score="4" Body="&lt;p&gt;A hypernova is just a really, really big supernova.  &lt;a href=&quot;http://www.astro.umass.edu/~wqd/hnr/faq.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;UMass has a (tacky designed) web page that explains it&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A hypernova explosion typically has a mechanical energy output of ~ 10^53 ergs, or about a factor of 100 greater than a supernova.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Regarding aftereffect, the page says this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The age of the hypernova remnant NGC5471B is about 30 thousand years, while the age of MF83 is about 1 million years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As best as I am able to determine, we have not had a chance to observe any hypernova candidates in the modern era, although &lt;a href=&quot;http://www.phy.mtu.edu/apod/ap990420.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;APOD featured the same one as mentioned by UMass&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KaYyK.jpg&quot; alt=&quot;APOD Picture of Hypernoava remnant NGC5471B&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.newscientist.com/article/dn7115-mysterious-limit-on-star-size-revealed.html#.UkTuToash8E&quot; rel=&quot;nofollow noreferrer&quot;&gt;According to New Scientist, research &lt;em&gt;suggest&lt;/em&gt; the upper limit to a star is about 150 solar masses&lt;/a&gt;. These are the stars that create the infamous supernovas.his is as derived from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Eddington_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eddington Limit&lt;/a&gt;.  Although &lt;a href=&quot;http://en.wikipedia.org/wiki/R136a1&quot; rel=&quot;nofollow noreferrer&quot;&gt;R136a1&lt;/a&gt; is an example of a star at nearly double that.  It is believed that these monsters are what are responsible for hypernovas. There is much to be learned from these stellar monsters!&lt;/p&gt;&#xA;" OwnerUserId="114" LastActivityDate="2013-09-27T02:39:29.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="283" PostTypeId="2" ParentId="271" CreationDate="2013-09-27T02:57:34.707" Score="13" Body="&lt;p&gt;It's a matter of size and stellar evolution.  There are many, many types of stellar explosions.  &lt;a href=&quot;http://ircamera.as.arizona.edu/NatSci102/NatSci102/lectures/supernovae.htm&quot; rel=&quot;noreferrer&quot;&gt;The University of Arizona has one page that describes these types&lt;/a&gt;.  generally, a Novs is not what we think of (i.e. a star exploding). That's actually a Type II Supernova.&lt;/p&gt;&#xA;&lt;p&gt;According to that site:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Novae are frequently (perhaps always) members of binary systems where the more massive star evolves more quickly and becomes a white dwarf. The less massive star loses some of its hydrogen-rich outer atmosphere to the white dwarf where it collects.  The high gravity and intense heat of the white dwarf make conditions suitable for hydrogen fusion into helium. When enough hydrogen collects, it explodes in a fusion reaction, causing a nova. The gas ejected in the explosion moves outwards at 1000-2000 km/sec.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The Supernova has two distinct types:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The &amp;quot;Type I&amp;quot; supernovae arise when the white dwarf star member of a binary system accretes so much matter from its companion that it is tipped over the 1.4M Chandrasekar limit. The white dwarf collapses very rapidly (just a few seconds) until the infalling layers hit the very hard surface of the neutron star. The resulting shock goes back through the collapsing layers, and they explode by hydrogen fusion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And the one we are more familiar with from popular media:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The &amp;quot;Type II&amp;quot; supernovae are the result of a massive star consuming all of its nuclear fuel and then exploding. The black line traces the evolution of a massive star schematically, until it reaches the supernova stage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://spot.pcc.edu/%7Eaodman/GS%20107%20web/end%20of%20main%20sequence/supernovae.htm&quot; rel=&quot;noreferrer&quot;&gt;Portland Community College actually lists a lot of the variations that happen in the different types of supernovas out there&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Type I: supernovae WITHOUT hydrogen absorption lines in their spectrum&#xA;Type II: supernovae WITH hydrogen absorption lines in their spectrum.&lt;/p&gt;&#xA;&lt;p&gt;Type II&#xA;The type II supernovae are massive stars whose iron cores collapse and then rebound, shock heating the outer layers of the star, which then explode outward.&lt;/p&gt;&#xA;&lt;p&gt;Type I&#xA;The type I supernovae are subdivided into three subclasses, called (boringly enough), Type Ia, Type Ib, and Type Ic.&#xA;Type Ia: no hydrogen lines, no helium lines, strong silicon lines&#xA;Type Ib: no hydrogen lines, strong helium lines&#xA;Type Ic: no hydrogen lines, no helium lines, no silicon lines&lt;/p&gt;&#xA;&lt;p&gt;Type Ib and type Ic supernovae are massive stars which lost their outer layers in a stellar wind before core collapse.&#xA;Type Ib supernovae lost their hydrogen-rich outer layer, revealing the helium-rich layer immediately below.&#xA;Type Ic supernovae suffered more mass loss as supergiants, losing both the hydrogen-rich layer and the helium-rich layer (revealing the carbon-rich layer below).&#xA;Type Ib and type Ic supernovae are essentially the same as type II supernovae.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;As for observing them, since none of them are expected to be close to us (i.e. no stars near us able to go supernova), it should be safe. The problem is predicting when one will happen.  For instance, Betelgeuse is expected to go supernova &amp;quot;soon&amp;quot;, but in stellar terms, that could be well over 100,000 years from now. &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_supernova_candidates&quot; rel=&quot;noreferrer&quot;&gt;The list of candidates in the Milky Way doesn't bode well for one in your lifetime&lt;/a&gt;. Although astronomers are often seeing them in other galaxies, and some are observable by amateur astronomers.  &lt;a href=&quot;http://www.space.com/22389-nova-star-explosion-delphinus-discovery.html&quot; rel=&quot;noreferrer&quot;&gt;In fact, just this past August (2013) an amateur discovered a nova&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2013-09-27T02:57:34.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="284" PostTypeId="2" ParentId="273" CreationDate="2013-09-27T03:48:08.737" Score="29" Body="&lt;p&gt;As @astromax observed, one of the primary factors that makes a space based telescope better than an equivalent telescope of the same size is &lt;a href=&quot;http://en.wikipedia.org/wiki/Atmosphere_of_Earth#Scattering&quot;&gt;scattering&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Along with scattering, there is also &lt;a href=&quot;http://en.wikipedia.org/wiki/Air#Refractive_index&quot;&gt;refraction&lt;/a&gt; which can be especially problematic when combined with atmospheric turbulence. In the modern era, this problem can be remedied to a certain extent using &lt;a href=&quot;http://en.wikipedia.org/wiki/Adaptive_optics&quot;&gt;adaptive optics&lt;/a&gt;, but since the Hubble was designed, built, and launched before AO became practical in the 1990's, a space telescope represented the pinnacle of optical clarity for that time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is another important &lt;a href=&quot;http://en.wikipedia.org/wiki/Atmosphere_of_Earth#Optical_properties&quot;&gt;optical property of the atmosphere&lt;/a&gt;, absorption. Although the Hubble is primarily a visible light telescope, it does have instruments that cover both the near UV and near IR, both of which are absorbed by the atmosphere more than visible light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, there are practical advantages of space based telescopes. There is neither weather, nor light pollution in space.&lt;/p&gt;&#xA;" OwnerUserId="38" LastEditorUserId="38" LastEditDate="2013-10-31T19:32:11.267" LastActivityDate="2013-10-31T19:32:11.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="285" PostTypeId="5" CreationDate="2013-09-27T04:21:38.633" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-27T04:21:38.633" LastActivityDate="2013-09-27T04:21:38.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="286" PostTypeId="4" CreationDate="2013-09-27T04:21:38.633" Score="0" Body="Questions regarding the planet on which humans live; the third planet from the Sun." OwnerUserId="19" LastEditorUserId="9301" LastEditDate="2016-08-16T07:46:54.650" LastActivityDate="2016-08-16T07:46:54.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="287" PostTypeId="5" CreationDate="2013-09-27T04:22:50.763" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-27T04:22:50.763" LastActivityDate="2013-09-27T04:22:50.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="288" PostTypeId="4" CreationDate="2013-09-27T04:22:50.763" Score="0" Body="Questions regarding stars which increase suddenly in energy output due to an explosion which ejected much of its mass." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-27T14:13:48.470" LastActivityDate="2013-09-27T14:13:48.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="289" PostTypeId="5" CreationDate="2013-09-27T04:23:31.407" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-27T04:23:31.407" LastActivityDate="2013-09-27T04:23:31.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="290" PostTypeId="4" CreationDate="2013-09-27T04:23:31.407" Score="0" Body="Questions regarding the Hubble Space Telescope, a visible light telescope orbiting Earth." OwnerUserId="19" LastEditorUserId="2153" LastEditDate="2015-05-25T14:33:13.647" LastActivityDate="2015-05-25T14:33:13.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="291" PostTypeId="5" CreationDate="2013-09-27T04:24:52.557" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-27T04:24:52.557" LastActivityDate="2013-09-27T04:24:52.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="292" PostTypeId="4" CreationDate="2013-09-27T04:24:52.557" Score="0" Body="Questions regarding the process of accurately finding your location, and getting to another place by planning and following a map." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-27T14:14:00.160" LastActivityDate="2013-09-27T14:14:00.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="293" PostTypeId="5" CreationDate="2013-09-27T04:25:44.810" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-27T04:25:44.810" LastActivityDate="2013-09-27T04:25:44.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="294" PostTypeId="4" CreationDate="2013-09-27T04:25:44.810" Score="0" Body="Questions regarding atmospheric conditions on either Earth or on other planets." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-27T14:14:22.753" LastActivityDate="2013-09-27T14:14:22.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="295" PostTypeId="2" ParentId="249" CreationDate="2013-09-27T06:47:33.487" Score="6" Body="&lt;p&gt;This question has two parts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Surface Temperatures&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very useful diagram which shows surface temperatures, and also gives you the temperature of any star you can observe is the Herzsprung-Russell Diagram, this one from &lt;a href=&quot;http://www.le.ac.uk/ph/faulkes/web/stars/o_st_overview.html&quot; rel=&quot;noreferrer&quot;&gt;le.ac.uk&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XyJyp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the yellow of our own sun places it in the 4.5 kKelvin to 6 kKelvin, as noted in the question. This temperature is down towards the lower end of average. The main sequence, where most stars are, tops out about 20 kKelvin, and there are some up towards the 40 kKelvin region - they aren't shown here as they are much rarer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;White dwarves are a little hotter than our sun - between 6 kKelvin and 10 kKelvin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutron stars are way off the main sequence - young ones can be over 1 MKelvin!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Core temperature:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Internally, the core temperatures are dependent on the mass of the star. In our sun, energy is delivered via the proton-proton chain mechanism, which occurs up to about 20 MKelvins, whereas more massive stars can use the Carbon-Nitrogen-Oxygen cycle - which happens from about 15 MKelvins upwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The differences are mainly down to convection and radiation differences - this extract from &lt;a href=&quot;http://en.wikipedia.org/wiki/Main_sequence&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia's Main Sequence page&lt;/a&gt; describes this in some detail:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because there is a temperature difference between the core and the surface, or photosphere, energy is transported outward via radiation and convection. A radiation&#xA;  zone, where energy is transported by radiation, is stable against&#xA;  convection and there is very little mixing of the plasma. By contrast,&#xA;  in a convection zone the energy is transported by bulk movement of&#xA;  plasma, with hotter material rising and cooler material descending.&#xA;  Convection is a more efficient mode for carrying energy than&#xA;  radiation, but it will only occur under conditions that create a steep&#xA;  temperature gradient. In massive stars (above 10 solar&#xA;  masses) the rate of energy generation by the CNO cycle is very&#xA;  sensitive to temperature, so the fusion is highly concentrated at the&#xA;  core. Consequently, there is a high temperature gradient in the core&#xA;  region, which results in a convection zone for more efficient energy&#xA;  transport. This mixing of material around the core removes the&#xA;  helium ash from the hydrogen-burning region, allowing more of the&#xA;  hydrogen in the star to be consumed during the main-sequence lifetime.&#xA;  The outer regions of a massive star transport energy by radiation,&#xA;  with little or no convection. Intermediate mass stars such as&#xA;  Sirius may transport energy primarily by radiation, with a small core&#xA;  convection region. Medium-sized, low mass stars like the Sun have&#xA;  a core region that is stable against convection, with a convection&#xA;  zone near the surface that mixes the outer layers. This results in a&#xA;  steady buildup of a helium-rich core, surrounded by a hydrogen-rich&#xA;  outer region. By contrast, cool, very low-mass stars (below 0.4 solar&#xA;  masses) are convective throughout. Thus the helium produced at the&#xA;  core is distributed across the star, producing a relatively uniform&#xA;  atmosphere and a proportionately longer main sequence lifespan.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="43" LastActivityDate="2013-09-27T06:47:33.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="296" PostTypeId="1" AcceptedAnswerId="297" CreationDate="2013-09-27T07:50:50.833" Score="23" ViewCount="5159" Body="&lt;p&gt;Some planets, specifically Jupiter, Saturn, Uranus, and Neptune in our solar system, have planetary rings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e3wDO.jpg&quot; alt=&quot;Jupiters rings, cross section&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do some planets have rings? How are they made and from what? Most importantly, will I be able to observe the rings on any planet with an amateur telescope?&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastActivityDate="2016-01-13T10:21:50.427" Title="Why do some planets have rings?" Tags="&lt;neptune&gt;&lt;uranus&gt;&lt;saturn&gt;&lt;jupiter&gt;&lt;planetary-ring&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="297" PostTypeId="2" ParentId="296" CreationDate="2013-09-27T07:59:03.663" Score="21" Body="&lt;p&gt;Rings are made up of tiny (and not so tiny) pieces of rock and ice that are in some way the bits &quot;left over&quot; from the formation of the planet. The theory involves the &lt;a href=&quot;http://blogstronomy.blogspot.com/2009/10/what-does-roche-limit-or-roche-radius.html&quot; rel=&quot;noreferrer&quot;&gt;Roche limit&lt;/a&gt; - and is that particles that are already within this limit can't accrete into a larger body because of the tidal forces involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another theory is that they are formed when a moon comes closer to a planet than the Roche limit, the tidal forces cause it to break up and form a ring. Though the presence of &quot;shepherd&quot; moons in the rings of Saturn does hint that this may not be major source of material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both explanations, to me, imply that you'd only get major ring systems around larger (gas giant) planets, though it doesn't preclude rings around smaller (rocky) planets. This seems to be borne out by our solar system where the gas giants have rings whereas the rocky planets don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://blogstronomy.blogspot.co.uk/2009/10/how-are-planetary-rings-formed.html&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to being able to see them, you should be able to see the rings of Saturn with an amateur telescope which has a 50 - 100 powers of magnification. With binoculars you'll probably see a misshapen blob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://astrobasics.blogspot.co.uk/2010/04/what-can-i-see-with-my-telescope-saturn.html&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="111" LastEditorUserId="111" LastEditDate="2016-01-13T10:21:50.427" LastActivityDate="2016-01-13T10:21:50.427" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="298" PostTypeId="1" AcceptedAnswerId="299" CreationDate="2013-09-27T09:09:22.557" Score="9" ViewCount="290" Body="&lt;p&gt;This question is inspired by &lt;a href=&quot;https://astronomy.stackexchange.com/questions/125/what-are-the-earth-like-features-of-titan&quot;&gt;&quot;What are the Earth-like features of Titan?&quot;&lt;/a&gt;.  According to NASA's &lt;a href=&quot;http://solarsystem.nasa.gov/planets/profile.cfm?Object=Jup_Europa&quot; rel=&quot;nofollow noreferrer&quot;&gt;Europa overview&lt;/a&gt;, Europa is believed to have an iron core, rocky mantle and an ocean under the frozen surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the tidal heating of Europa's interior by Jupiter make it a smaller analogue for Earth?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2013-09-27T10:00:00.053" Title="What evidence is there of Earth-Like internal features of Europa?" Tags="&lt;europa&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="299" PostTypeId="2" ParentId="298" CreationDate="2013-09-27T10:00:00.053" Score="6" Body="&lt;p&gt;This Image, Originally sourced from NASA shows the theorized interior of Europa:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iYvnR.jpg&quot; alt=&quot;Europa&amp;#39;s Interior&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://geology.com/nsta/earth-internal-structure.shtml&quot; rel=&quot;noreferrer&quot;&gt;this geology site&lt;/a&gt; Earth's core is predominantly iron and nickel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;a href=&quot;http://www.space.com/17777-what-is-earth-made-of.html&quot; rel=&quot;noreferrer&quot;&gt;this space site&lt;/a&gt; says that Earth's mantle is predominantly silicates&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the internal similarities so far is that they both possess iron cores and silicate rock mantles. However the core of Europa is mixed with sulphur rather than Earth's nickel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another similarity between Europa and Earth could be the composition of the oceans. &lt;a href=&quot;http://solarsystem.nasa.gov/europa/faq.cfm&quot; rel=&quot;noreferrer&quot;&gt;NASA&lt;/a&gt; speculates that Europa's oceans are also briny with salts and minerals dissolved in it. This is mostly theorised due to what we know about planet formation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this also comes with many differences, the Galileo spacecraft did not detect a magnetic field from Europa, which leads to speculation that, whilst Europa may have an Iron compound core, it is probable not molten any more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interior Similarities: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Iron compound core&lt;/li&gt;&#xA;&lt;li&gt;Silicate rock mantle&lt;/li&gt;&#xA;&lt;li&gt;Salty, briny, Oceans&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Differences: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Iron core mixed with sulphur not nickel&lt;/li&gt;&#xA;&lt;li&gt;Core likely to be solid, not molten&lt;/li&gt;&#xA;&lt;li&gt;Lack of internally generated magnetic field&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user96" LastActivityDate="2013-09-27T10:00:00.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="300" PostTypeId="1" AcceptedAnswerId="301" CreationDate="2013-09-27T10:07:08.227" Score="15" ViewCount="1658" Body="&lt;p&gt;Do all galaxies rotate?  If so, do they rotate in the same direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What observations have been made of galaxy rotation?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2013-09-27T11:37:28.730" Title="What is the evidence that galaxies rotate?" Tags="&lt;rotation&gt;&lt;galaxy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="301" PostTypeId="2" ParentId="300" CreationDate="2013-09-27T10:17:28.033" Score="15" Body="&lt;p&gt;&lt;strong&gt;Rotation:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the easiest way to show that galaxies rotates are Doppler observations of atomic or molecular lines. For example, if the observed galaxy is seen edge-on and rotates clockwise, the left side of the galaxy is going away from you, and is then Doppler-shifted in the red. The right side, going toward you, is Doppler-shifted in the blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://articles.adsabs.harvard.edu//full/1965SvA.....8..823A/0000823.000.html&quot;&gt;This kind of observations&lt;/a&gt; was realized already in the '60, in radio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spiral patern observed in spiral galaxies is also a clear signature of their rotation. Note that spiral are linked to density waves in the galactic disk (see for example &lt;a href=&quot;http://adsabs.harvard.edu/abs/1964ApJ...140..646L&quot;&gt;Lin &amp;amp; Shu 1964&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Direction:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for their directions, you can convinced yourself that there is no tendency, just looking at &lt;a href=&quot;http://hubblesite.org/gallery/album/galaxy/&quot;&gt;this Hubble telescope gallery&lt;/a&gt;. You can also think that galaxies probably rotate by angular momentum conservation during their formation history, and that it highly dependent on initial conditions, that are clearly random.&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2013-09-27T11:37:28.730" LastActivityDate="2013-09-27T11:37:28.730" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="302" PostTypeId="2" ParentId="300" CreationDate="2013-09-27T10:29:50.260" Score="2" Body="&lt;p&gt;We have not discovered every galaxy in existence, nor have we watched each to see how it is spinning. Most of this answer will heavily be speculation. You have been warned!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have &lt;a href=&quot;http://www.astro.cornell.edu/academics/courses/astro201/galaxies/types.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;three classifications&lt;/a&gt; of galaxy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spiral, Elliptical and Irregular. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Irregular galaxies are thought to be quite young, and predominantly gasses that are yet to combine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elliptical galaxies are thought to be like a disc, or elongated disk in shape. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spiral galaxies are theorised to be the older of the three, and it is heavily theorised that they are spiral in shape due to a &lt;a href=&quot;http://www.astro.ucla.edu/~ghezgroup/gc/journey/smbh.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;super massive black hole&lt;/a&gt; at their center. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see that it is thought elliptical galaxies will eventually become spiral galaxies&#xA;&lt;img src=&quot;https://i.stack.imgur.com/wVQ70.png&quot; alt=&quot;lifecycle of a galaxy&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E classifications are elliptical. S classifications are spiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore if we extrapolate the pattern and theories it could potentially be said that in the course of a galaxies life it will eventually become a spiral, orbiting a super massive black hole as more and more of its mass is drawn to the centre. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok so if it can be theorised that most galaxies will eventually spiral a black hole it leads us to believe that the rotation of a galaxy will then be because of the rotational direction of a black hole, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jpl.nasa.gov/news/news.php?release=2010-186&quot; rel=&quot;nofollow noreferrer&quot;&gt;This NASA theorist&lt;/a&gt; says there may be such a thing as a backwards black hole, one that spins in the opposite direction to its accretion disks, which in this case I assume would be the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at &lt;a href=&quot;http://hubblesite.org/newscenter/archive/releases/2002/03&quot; rel=&quot;nofollow noreferrer&quot;&gt;observations of galaxies&lt;/a&gt; leads to most of them, spinning with the arms trailing, but there are cases where the arms are facing the direction of the spin!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Yes, we have observed the spin of galaxies, no, not every galaxy rotates, irregular galaxies have no central point of mass of which to rotate about. As for if they all spin in the same direction, that is extremely hard to observe without more data points and further study.&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastActivityDate="2013-09-27T10:29:50.260" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="303" PostTypeId="2" ParentId="173" CreationDate="2013-09-27T10:36:05.040" Score="7" Body="&lt;p&gt;Just to add to the answer above, scientists can make assumptions as to how Earth would be affected by a nearby supernova, particularly by one of the main effects, as mentioned in Larian's answer - a gamma ray burst.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the article &lt;a href=&quot;http://paleo.ku.edu/geo/faculty/BSL/astrobiopaper.pdf&quot;&gt;&quot;Did a gamma-ray burst initiate the late Ordovician mass extinction?&quot;&lt;/a&gt; (Melott et al. 2004), a GRB is a probable contributor to the mass extinction that occurred around 440 million years ago.  The 2 possible mechanisms for the mass extinction discussed by the authors are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Increased UV radiation reaching the surface, due to the depletion of the ozone layer&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the increased production of nitrogen dioxide haze that would cause a global cooling (as was observed to also be a contributing factor in the Ordovician extinction).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2013-09-27T10:36:05.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="304" PostTypeId="1" AcceptedAnswerId="306" CreationDate="2013-09-27T11:23:54.283" Score="14" ViewCount="391" Body="&lt;p&gt;Could someone give the rough estimates of speeds of our travel through space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can easily find the two nearby ones:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Earth's own rotation - depending on latitude, up to 40,075 km/24h = 463 m/s&lt;/li&gt;&#xA;&lt;li&gt;Earth's orbital speed around the Sun - 940 000 000 km/year = 29,786 m/s&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now how do these fare?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Motion of Sun within the arm of the galaxy, around the center of the galaxy?&lt;/li&gt;&#xA;&lt;li&gt;Motion of the galaxy within the local supercluster?&lt;/li&gt;&#xA;&lt;li&gt;(if possible to measure at all) movement of the supercluster within the universe?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="83" LastActivityDate="2013-09-27T11:52:11.427" Title="How fast do we travel through space?" Tags="&lt;galaxy&gt;&lt;movement&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="305" PostTypeId="1" AcceptedAnswerId="308" CreationDate="2013-09-27T11:44:27.500" Score="21" ViewCount="777" Body="&lt;p&gt;So far I got to understand the expansion of space is not to be understood as stars drifting further apart through space. There's something more fundamental - e.g. you can't simply measure speed of it, nor tell where the center of the universe (the midpoint of expansion) is. Well, some smart people got me from knowing the answer that is &quot;Simple, obvious, neat and wrong&quot;  to &quot;I have no clue what it is&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a resource/method that would allow a layman to understand how to think about distortions of space like its expansion; to visualize and understand the concept correctly and without &lt;em&gt;erroneous&lt;/em&gt; simplifications; understand it before trying to delve into underlying details of physics?&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2013-10-02T14:41:24.487" Title="Space expansion in layman terms" Tags="&lt;expansion&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="306" PostTypeId="2" ParentId="304" CreationDate="2013-09-27T11:52:11.427" Score="10" Body="&lt;p&gt;According to the Astronomical Society of the Pacific's website &lt;a href=&quot;http://www.astrosociety.org/edu/publications/tnl/71/howfast.html&quot;&gt;&quot;How Fast Are You Moving When You Are Sitting Still?&quot;&lt;/a&gt;, we are moving:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotation of the Earth: 1600 km/h&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbital rotation of the Earth around the sun: 107,000 km/h&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun around the centre of the galaxy: 792,000 km/h&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going further, according to the article &lt;a href=&quot;http://www.space.com/19915-milky-way-galaxy.html&quot;&gt;&quot;Milky Way Galaxy: Facts About Our Galactic Home&quot;&lt;/a&gt; (Redd, 2013) quotes the sun's speed around the centre of the galaxy as 828,000 km/h&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Andromeda Galaxy and the Milky Way are headed for a collision at 112 km/sec&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.astronomy.com/news-observing/news/2007/06/milky%20way%20moving%20away%20from%20void&quot;&gt;&quot;Milky Way moving away from void&quot;&lt;/a&gt; (2007), the Milky Way galaxy is moving from the local void at 600,000 mph (960,000km/h)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the article &lt;a href=&quot;http://blogs.discovermagazine.com/badastronomy/2009/05/07/the-universe-is-expanding-at-742-kmsecmpc/&quot;&gt;&quot;The Universe is expanding at 74.2 km/sec/Mpc&quot;&lt;/a&gt; (Plait, 2009), the Universe is expanding at a rate of 74.2km/sec per megaparsec (3.26 million light years).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There seems to be some divergence of values in the articles, but these give a good estimate.&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2013-09-27T11:52:11.427" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="307" PostTypeId="1" AcceptedAnswerId="311" CreationDate="2013-09-27T11:55:34.173" Score="18" ViewCount="1578" Body="&lt;p&gt;What bodies can be encountered way outside galaxies - far beyond the farthest edges of galaxies, in the deep space between them? Are there single, galaxy-less stars, giant clouds of gas more dense than that the lower-than-elsewhere content of free hydrogen in space, or any other phenomena that span the void between the galaxies?&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2013-09-27T12:52:02.527" Title="What is there in the intergalactic space?" Tags="&lt;intergalactic-space&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="308" PostTypeId="2" ParentId="305" CreationDate="2013-09-27T12:10:03.853" Score="19" Body="&lt;p&gt;Basically, if two particles are placed with &lt;em&gt;no other interaction between them&lt;/em&gt;, the distance between them will increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine living on the surface of a balloon which is being blown up. Your size stays fixed, because you're more or less rigid, but items not attached to you will move further away. Your ruler, another rigid body, stays fixed in size (though it may bend to accommodate the new curvature &amp;mdash; this isn't so important). But &lt;em&gt;two&lt;/em&gt; rulers (which are not attached to each other) move further away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ULNK3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...also, if everything, including all our measurement devices expands at the same speed, how can we determine the fact it's expanding? :D &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the outset this seems true, however there are other forces at play here. Our measurement devices are held together by electromagnetic interactions, and the strength of these will not change. So the measurement device will hold itself together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine two faraway atoms. When space expands, the distance between the two atoms increases. However, the size&lt;sup&gt;1&lt;/sup&gt; of the atom does not &amp;mdash; this is determined by electrostatic equilibrium (and quantum mechanical considerations), and this remains unaffected. Even if the atom was stretched, it would rebound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This scales up to measurement devices, so they don't get distorted either. Indeed, the expansion of space only really makes sense when you look at galaxies &amp;mdash; these are pretty far away (when not in the same supercluster) and they don't have any interactions maintaining an equilibrium distance between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1. Whatever closest analog we have to &quot;size&quot; for atoms; eg the area which contains 99% of the charge density; or the nth Bohr radius.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2013-09-27T19:25:24.370" LastActivityDate="2013-09-27T19:25:24.370" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="309" PostTypeId="2" ParentId="261" CreationDate="2013-09-27T12:18:38.793" Score="10" Body="&lt;p&gt;It depends on the kind of star you want to make.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, what you really need is hydrodgen; as soon as enough hydrodgen is gathered somewhere and collapse, you end up with a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But&lt;/strong&gt;, if you just have hydrodgen, you will get a very massive star, and with a history different than the stars observed in the local Universe. Why so?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.cfa.harvard.edu/swas/science1.html&quot;&gt;hydrodgen is a poor coolant&lt;/a&gt;, therefore the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jeans_instability#Jeans_mass&quot;&gt;Jeans' mass&lt;/a&gt; (the minimum mass for a density structure to gravitationally collapse), which strongly depends on temperature, will be much higher if the temperature is higher. In practice, it means that the formed stars will be much more massive.&lt;/li&gt;&#xA;&lt;li&gt;you need at least a small fraction of carbon to launch the CNO cycle in stellar cores (CNO cycle is one of the two ways to burn hydrodgen and to turn it into helium); stellar cores are then hotter and denser, and stars become hotter and more luminous.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So if you want to make a star as we typically observe in our galaxy, you will need also some molecules (hopefully &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_interstellar_and_circumstellar_molecules&quot;&gt;there are plenty of them&lt;/a&gt;) and also some carbon. Else, you'll get big, massive, hot and luminous stars. Like the first generation of stars (called Population III stars, you can see &lt;a href=&quot;http://www.google.fr/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=1&amp;amp;ved=0CDAQFjAA&amp;amp;url=http%3A%2F%2Fwww.astro.psu.edu%2Fusers%2Frbc%2Fa534%2Fredman.pdf&amp;amp;ei=S0ZFUq3BGamH0AWAhYCQBA&amp;amp;usg=AFQjCNHgmWxmdBhD8QOM4VJYuLrWUhr66A&amp;amp;sig2=tsibya0AEYsbbx4iKJClDw&amp;amp;bvm=bv.53217764,d.bGE&quot;&gt;here&lt;/a&gt; if you want more dirty details) in the early Universe.&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2013-09-27T12:26:31.847" LastActivityDate="2013-09-27T12:26:31.847" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="310" PostTypeId="1" AcceptedAnswerId="327" CreationDate="2013-09-27T12:21:40.583" Score="13" ViewCount="654" Body="&lt;p&gt;I've read about and seen some rather sophisticated rigs for night sky photography that allow for ultra-long-exposure photos by counteracting the rotation of the sky by rotating the camera with it. The problem is mechanics for such rigs is rather complex, expensive and not entirely within reach of an amateur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was considering a different approach: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Set the camera on a firm tripod or whatever, leaving it motionlessly.&lt;/li&gt;&#xA;&lt;li&gt;Use intervalometer in the camera firmware to take multiple photos of moderately long exposure, short enough that the stars don't blur into lines.&lt;/li&gt;&#xA;&lt;li&gt;Save each as RAW so that the trace amounts of light captured in each of them don't get lost.&lt;/li&gt;&#xA;&lt;li&gt;Either using more bright stars as markers, or even calculating the misalignment of the photos basing on movement of the sky and lens geometry composite the images in software, to obtain good images of the night sky.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Obviously I wouldn't get very far with high zoom images, as the field of vision would shift away long before I'd get enough light in for a good photo, but obtaining a photo of a fair section of the sky from a relatively wide angle lens should be doable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is such approach used or even viable? Is there such software &quot;out there&quot; already or would I need to write my own from scratch? &lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="83" LastEditDate="2013-09-28T22:30:10.220" LastActivityDate="2013-09-29T21:12:02.753" Title="Sky photography through compositing images in software?" Tags="&lt;amateur-observing&gt;&lt;photography&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="311" PostTypeId="2" ParentId="307" CreationDate="2013-09-27T12:26:45.243" Score="12" Body="&lt;p&gt;There can be stars (and small star systems). Stars need not be found only in galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can be gas clouds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most nebulae are in intergalactic space. Indeed, for a while, both galaxies and nebulae were termed as &quot;nebulae&quot; until the differing nature of galaxies was discovered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost any (small) structure that can be present in a galaxy can be present in intergalactic space, for that matter. Black holes (neutron stars, white dwarfs, brown dwarfs, ... ), star clusters, star systems, etc can all be present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, there's a very rare (rare in the density sense) plasma, mainly of hydrogen, that makes up almost all of the intergalactic medium. Zero pressure is an unattainable ideal case.&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2013-09-27T12:26:45.243" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="312" PostTypeId="2" ParentId="261" CreationDate="2013-09-27T12:41:21.893" Score="6" Body="&lt;p&gt;Any gas that has a lower atomic mass than iron. The only condition on the gas is that nuclear fusion with it should be exothermic (the energy required to fuse the nuclei should be less than the energy obtained). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, usually a majority of hydrogen is the best. If the majority gas is of a higher element, we'll need a sufficient gravitational force to initiate the fusion, and that's not always easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it is possible for &lt;a href=&quot;http://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction&quot; rel=&quot;noreferrer&quot;&gt;just hydrogen to start fusion in a star&lt;/a&gt;, some carbon or nitrogen is a good thing to have to &lt;a href=&quot;http://en.wikipedia.org/wiki/CNO_cycle&quot; rel=&quot;noreferrer&quot;&gt;start fusion more efficiently&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different initial conditions lead to different kinds of stars.&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2013-09-27T12:41:21.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="313" PostTypeId="2" ParentId="307" CreationDate="2013-09-27T12:49:57.400" Score="8" Body="&lt;p&gt;Just to add to ManishEarth's answer, an article about what is in intergalactic space, &lt;a href=&quot;http://www.scientificamerican.com/article.cfm?id=wandering-stars&quot;&gt;&quot;Numerous &quot;Tramp&quot; Stars Adrift in Intergalactic Space Could Await Discovery&quot;&lt;/a&gt; (Matson, 2009) also describes what you would see if in the night sky, if Earth orbited a 'tramp star':&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;If you were on an Earth orbiting a sunlike star that was halfway between us and the Andromeda Galaxy, or halfway between us and the Virgo Cluster, you would look out in the night sky, and you would actually see with your naked eye zero stars,&quot; says study co-author Michael Shara, an astrophysicist at the American Museum of Natural History in New York City. &quot;You would see a few distant, blurry, fuzzy patches, and those would be the galaxies.&quot; Without a moon to reflect light, nighttime out there would be almost pitch-black.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So even though there are, as ManishEarth stated in his answer, everything you'd expect in a galaxy existing between galaxies, there would just be far less.&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2013-09-27T12:49:57.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="314" PostTypeId="2" ParentId="307" CreationDate="2013-09-27T12:52:02.527" Score="11" Body="&lt;p&gt;You can find all kind of objects, like &lt;a href=&quot;http://news.vanderbilt.edu/2012/04/rogue-stars-intergalactic-space/&quot;&gt;stars&lt;/a&gt;, &lt;a href=&quot;http://arxiv.org/abs/astro-ph/9609076&quot;&gt;planetary nebulæ&lt;/a&gt; etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it is mostly a &lt;a href=&quot;http://www.google.fr/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=6&amp;amp;ved=0CGgQFjAF&amp;amp;url=http%3A%2F%2Fwww.ast.cam.ac.uk%2F~pettini%2FPhysical%2520Cosmology%2Flecture09.pdf&amp;amp;ei=KXtFUqUIpbHQBZa-gfgB&amp;amp;usg=AFQjCNFhJOpTehKEIlaScFmOzOjzAEkQhQ&amp;amp;sig2=Bbt6c1308XUK_NGmOxQ95Q&amp;amp;bvm=bv.53217764,d.d2k&quot;&gt;hot, ionized void&lt;/a&gt;. How void? The density of the intergalactic medium is about 1 to 100 particles per cubic meter (you can compare it to the mean galactic density, of about a million particles per cubic meter, or that of Earth's atmosphere, of about 10^26 particles cubic meter). How hot? It can go from &lt;a href=&quot;http://adsabs.harvard.edu/abs/2010AAS...21631808G&quot;&gt;10^5 to 10^7 K&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fun thing is, even though it is &quot;very empty&quot;, is contains however most of the baryons (the &quot;classical matter&quot;) of the Universe, if you consider that only &lt;a href=&quot;http://www.google.fr/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=6&amp;amp;ved=0CGgQFjAF&amp;amp;url=http%3A%2F%2Fwww.ast.cam.ac.uk%2F~pettini%2FPhysical%2520Cosmology%2Flecture09.pdf&amp;amp;ei=KXtFUqUIpbHQBZa-gfgB&amp;amp;usg=AFQjCNFhJOpTehKEIlaScFmOzOjzAEkQhQ&amp;amp;sig2=Bbt6c1308XUK_NGmOxQ95Q&amp;amp;bvm=bv.53217764,d.d2k&quot;&gt;5-10% of the total mass of baryons is enclosed in stars and galaxies&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="88" LastActivityDate="2013-09-27T12:52:02.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="315" PostTypeId="5" CreationDate="2013-09-27T13:39:02.523" Score="0" Body="&lt;p&gt;Binary stars are double stars, though some double stars are actually far apart, but merely seem to be close together.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="2153" LastEditDate="2015-05-25T14:32:41.973" LastActivityDate="2015-05-25T14:32:41.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="316" PostTypeId="4" CreationDate="2013-09-27T13:39:02.523" Score="0" Body="Questions regarding a pair of stars that are gravitationally bound and orbit around their barycenter." OwnerUserId="88" LastEditorUserId="31410" LastEditDate="2020-12-12T22:29:39.483" LastActivityDate="2020-12-12T22:29:39.483" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="317" PostTypeId="5" CreationDate="2013-09-27T13:40:08.560" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-27T13:40:08.560" LastActivityDate="2013-09-27T13:40:08.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="318" PostTypeId="4" CreationDate="2013-09-27T13:40:08.560" Score="0" Body="Questions about the origin, history, evolution and fate of the Universe." OwnerUserId="88" LastEditorUserId="88" LastEditDate="2013-09-27T14:14:23.063" LastActivityDate="2013-09-27T14:14:23.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="319" PostTypeId="5" CreationDate="2013-09-27T13:43:25.260" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-27T13:43:25.260" LastActivityDate="2013-09-27T13:43:25.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="320" PostTypeId="4" CreationDate="2013-09-27T13:43:25.260" Score="0" Body="Questions about relativistic theory of gravity. General relativity describes gravity as a geometrical property of space and time." OwnerUserId="88" LastEditorUserId="88" LastEditDate="2013-09-27T14:14:02.453" LastActivityDate="2013-09-27T14:14:02.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="321" PostTypeId="1" AcceptedAnswerId="323" CreationDate="2013-09-27T14:01:12.247" Score="34" ViewCount="19518" Body="&lt;p&gt;Jupiter has a great many moons - in the hundreds, and they're still being discovered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the current theory for where all these moons came from? Are they rocks flying through space captured by Jupiter's gravity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dlQHH.jpg&quot; alt=&quot;Jupiters moons&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorDisplayName="user96" LastEditDate="2013-09-27T14:10:46.973" LastActivityDate="2016-09-29T17:40:57.847" Title="Why does Jupiter have so many moons?" Tags="&lt;jupiter&gt;&lt;natural-satellites&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="322" PostTypeId="1" CreationDate="2013-09-27T14:17:08.567" Score="22" ViewCount="412" Body="&lt;p&gt;Dark matter appears to dominate the matter component of the universe as compared to luminous, or baryonic, matter. Though it does not interact electromagnetically (it doesn't absorb, scatter, or emit photons), there is an ever-increasing mountain of evidence for its existence through it's gravitational interactions with stars, galaxies, and clusters, as well as it's influence on objects behind it through what's known as gravitational lensing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what are the most promising particle candidates of dark matter, and which experiments currently exist (or will exist) to attempt to answer this question?&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="37191" LastEditDate="2021-02-16T08:23:48.953" LastActivityDate="2021-02-16T08:23:48.953" Title="Dark Matter Particle Candidates" Tags="&lt;cosmology&gt;&lt;dark-matter&gt;&lt;matter&gt;&lt;particles&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="323" PostTypeId="2" ParentId="321" CreationDate="2013-09-27T14:22:37.710" Score="23" Body="&lt;p&gt;Mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more massive a body, the larger the gap between its lowest and highest orbit; the range of speeds at which a random body entering its gravity is likely to remain as its satellite. Sun has millions of satellites if you count all the asteroids; smaller planets tend to have one or two moons at most (Pluto with five being a notable and not fully explained exception)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To a lesser degree there's a matter of shape too. A regularly round body will have more regular and stable orbit than a potato-shaped one. Jupiter, being a gas giant is perfectly round. This doesn't play that much of a role though, especially with higher orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And last but not least, no destabilizing influence of other bodies. It's very hard to maintain a lunar orbit - artificial satellites around our Moon last only a couple years each, because relatively close neighborhood of Earth tends to destabilize orbit of anything orbiting the Moon. Jupiter being a single massive planet with relatively tiny (relatively to its mass) moons doesn't have them influence each other all that much. &lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2013-09-27T14:22:37.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="324" PostTypeId="2" ParentId="322" CreationDate="2013-09-27T14:58:05.690" Score="10" Body="&lt;p&gt;When you say &quot;particle&quot; candidates, I assume you're excluding &lt;a href=&quot;http://en.wikipedia.org/wiki/MACHO&quot; rel=&quot;noreferrer&quot;&gt;MACHOs&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Robust_Associations_of_Massive_Baryonic_Objects&quot; rel=&quot;noreferrer&quot;&gt;RAMBOs&lt;/a&gt;. MACHOs are &quot;dark&quot; objects at the stellar scale like black holes, neutron stars, brown dwarfs, etc. RAMBOs are clusters of similar dark objects. MACHOs and RAMBOs are made of primarily baryonic matter (everyday stuff like protons and neutrons &amp;mdash; electrons are not baryons but they can be a part of this too). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves us with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Weakly_interacting_massive_particle&quot; rel=&quot;noreferrer&quot;&gt;WIMPs&lt;/a&gt; and possible other particle candidates. I don't know the details of any experiments looking for these offhand (I'll see if I can find it out and update the answer in a few days), but the major particle candidates are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Neutrino&quot; rel=&quot;noreferrer&quot;&gt;Neutrinos&lt;/a&gt;: It's highly unlikely that these make up any significant chunk of the DM in our universe, but they deserve a mention.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sterile_neutrino&quot; rel=&quot;noreferrer&quot;&gt;Sterile neutrinos&lt;/a&gt; (I'm using the term in the specific sense here, not the general sense): This type of neutrino is right handed and only interacts gravitationally (note that there can be right handed active neutrinos, too). There is some preliminary evidence suggesting that these might exist.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Axion&quot; rel=&quot;noreferrer&quot;&gt;Axions&lt;/a&gt;: These are hypothetical, but if they exist (and make up the majority of DM), we kill two birds with one stone by both explaining DM and resolving the lack f symmetry breaking in the strong interaction.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitinos&quot; rel=&quot;noreferrer&quot;&gt;Gravitinos&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lightest_Supersymmetric_Particle&quot; rel=&quot;noreferrer&quot;&gt;Lightest Supersymmetric Partner&lt;/a&gt;: Both of these require supersymmetry to hold, but they are major DM candidates.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="28" LastEditorUserId="45" LastEditDate="2013-09-28T13:53:42.350" LastActivityDate="2013-09-28T13:53:42.350" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="325" PostTypeId="2" ParentId="321" CreationDate="2013-09-27T15:11:38.077" Score="18" Body="&lt;p&gt;Bigger is better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most moons, especially those of gas giants, are not &quot;formed&quot;, they are just &quot;captured&quot; (unlike our Moon, which &lt;em&gt;could&lt;/em&gt; have been captured, but &lt;a href=&quot;http://en.wikipedia.org/wiki/Origin_of_the_Moon&quot; rel=&quot;noreferrer&quot;&gt;probably was formed in a much more exciting way&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter is the most massive planet in the solar system. It stands to reason that it has a larger region of gravitational influence (where its influence outweighs the force due to the other planets and the sun). So, it's easy for it to capture rocky masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a look at the contours on the following image (Ignore the Lagrange points marked on it, I only want the contours)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bjrcc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the circular area around the Earth is more or less the area (there's a velocity dependence here which I'm not getting into) in which a moon-like body can form a reasonably stable orbit. The size of the small &quot;well&quot; will increase as the planet moves farther from the sun, and also when the planet is more massive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter is &lt;em&gt;both&lt;/em&gt; pretty far away from the Sun, and is very massive. This leads to a huge sphere of influence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Asteroid_belt&quot; rel=&quot;noreferrer&quot;&gt;asteroid belt&lt;/a&gt; may have something to do with this too, but I doubt it (it's pretty far away). However, if we assume the &quot;half-baked planet formation&quot; theory for the formation of the belt, Jupiter may have leeched off much of the mass that would have otherwise become part of that planet during the formative period.&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2013-09-27T15:11:38.077" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="327" PostTypeId="2" ParentId="310" CreationDate="2013-09-27T17:19:49.973" Score="7" Body="&lt;p&gt;What you are talking about is making a 'mosaic'. This technique is commonly used in astronomy to stitch together many images that are off-set from each other of some (apparently) large astronomical object (e.g., a star cluster or a nearby galaxy). In addition to covering a large field of view, making a mosaic generally also improves signal to noise where the images overlap. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to realign the images is to uses cross-correlation to find the optimal shift for consecutive images in order to align them. This can be done in common data analysis languages such as python (free), matlab or IDL (not free).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not used black-box software to do this, but here is a list of programs that might help: &lt;a href=&quot;http://astronomyonline.org/AstronomySoftware.asp#CCD_Control_and_Imaging_Software:_&quot;&gt;http://astronomyonline.org/AstronomySoftware.asp#CCD_Control_and_Imaging_Software:_&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully some astrophotographers here can recommend some software.&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2013-09-27T17:19:49.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="328" PostTypeId="1" AcceptedAnswerId="333" CreationDate="2013-09-27T17:26:31.793" Score="26" ViewCount="11489" Body="&lt;p&gt;Some stars are simply huge. Eventually, though, wouldn't there be simply too much pressure or mass for the star to sustain itself? Wouldn't it eventually collapse into a black hole?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a theoretical upper limit on the size of a star, and what is it based on?&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2018-04-02T19:05:56.187" Title="Is there a theoretical maximum size limit for a star?" Tags="&lt;star&gt;&lt;size&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="329" PostTypeId="5" CreationDate="2013-09-27T17:27:02.153" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-27T17:27:02.153" LastActivityDate="2013-09-27T17:27:02.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="330" PostTypeId="4" CreationDate="2013-09-27T17:27:02.153" Score="0" Body="Questions regarding the space or objects outside of galaxies." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-27T18:44:12.340" LastActivityDate="2013-09-27T18:44:12.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="331" PostTypeId="1" AcceptedAnswerId="414" CreationDate="2013-09-27T18:08:36.317" Score="33" ViewCount="1586" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_magnitude&quot; rel=&quot;noreferrer&quot;&gt;Apparent magnitude&lt;/a&gt; is a rather complex way to determine the brightness of a star. Quoting the introduction text from the linked to Wikipedia page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The apparent magnitude (m) of a celestial body is a measure of its&#xA;  brightness as seen by an observer on Earth, adjusted to the value it&#xA;  would have in the absence of the atmosphere. The brighter the object&#xA;  appears, the lower the value of its magnitude. Generally the visible&#xA;  spectrum (vmag) is used as a basis for the apparent magnitude, but&#xA;  other regions of the spectrum, such as the near-infrared J-band, are&#xA;  also used. In the visible spectrum Sirius is the brightest star in the&#xA;  night sky, whereas in the near-infrared J-band, Betelgeuse is the&#xA;  brightest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While it is of course a useful measure also for the slightly more casual, non-scientific observer to help determine the observed star from its neighboring cluster, or identification in general, I always wondered if there is a way to measure apparent magnitude with enthusiast-class equipment, and what would these procedures be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/BCsWU.jpg&quot; alt=&quot;apparent magnitude&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;sup&gt;Apparent magnitude scale and observational limits (Source: &lt;a href=&quot;http://sci.esa.int/education/35616-stellar-distances/?fbodylongid=1868&quot; rel=&quot;noreferrer&quot;&gt;ESA Science&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also interesting would be a description of what level of precision and enthusiast can get with such equipment while measuring apparent magnitude of a distant star. If you need specifics to answer the question, such as precise available equipment or subject of observations, please feel free to choose at will any that would broadly match the capabilities of enthusiast-grade equipment.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="7982" LastEditDate="2021-02-28T11:57:24.950" LastActivityDate="2021-02-28T11:57:24.950" Title="How could a hobbyist astronomer determine apparent magnitude of a star?" Tags="&lt;luminosity&gt;&lt;apparent-magnitude&gt;&lt;atmospheric-effects&gt;&lt;newtonian-telescope&gt;&lt;dobsonian-telescope&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="332" PostTypeId="1" AcceptedAnswerId="353" CreationDate="2013-09-27T18:35:43.900" Score="14" ViewCount="1885" Body="&lt;p&gt;We can often read in the scientific and also more casual reader literature and articles about the &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_resolution&quot; rel=&quot;noreferrer&quot;&gt;angular resolution&lt;/a&gt; of various telescopes and other optical equipment, be it ground based or onboard space probes. They would often list their angular resolution, or in other words, their ability to resolve or distinguish small, distant objects with today's digital era mostly on a per sensor pixel basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/hPakC.jpg&quot; alt=&quot;parallax&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Finding a star's distance from its parallax. Trigonometric Parallax method determines distance to star by measuring&lt;br /&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;its slight shift in apparent position as seen from opposite ends of Earth's orbit. (Source: &lt;a href=&quot;http://wwwhip.obspm.fr/heritage/hipparcos/SandT/hip-SandT.html&quot; rel=&quot;noreferrer&quot;&gt;Measuring the Universe&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm interested in is, is the precision in measurements of &lt;a href=&quot;https://en.wikipedia.org/wiki/Parallax&quot; rel=&quot;noreferrer&quot;&gt;parallax&lt;/a&gt; and with it our ability to determine distance of observed objects directly analogous to mentioned radial resolution and how could it be calculated using data on angular resolution of a telescope alone, if we assume both ground based and in space observatories have more or less the same &lt;a href=&quot;https://simple.wikipedia.org/wiki/Perihelion&quot; rel=&quot;noreferrer&quot;&gt;perihelion&lt;/a&gt; to &lt;a href=&quot;https://simple.wikipedia.org/wiki/Aphelion&quot; rel=&quot;noreferrer&quot;&gt;aphelion&lt;/a&gt; distance (i.e. the space observatory is in Earth's orbit).&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="6" LastEditDate="2014-01-07T15:43:30.707" LastActivityDate="2014-01-09T01:29:57.183" Title="How does angular resolution of a telescope translate to its parallax precision?" Tags="&lt;observational-astronomy&gt;&lt;distances&gt;&lt;parallax&gt;&lt;angular-resolution&gt;&lt;astrometry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="333" PostTypeId="2" ParentId="328" CreationDate="2013-09-27T19:18:55.347" Score="19" Body="&lt;p&gt;According to current knowledge, yes. If the gas cloud is too massive, the pressure of the radiation prevents the collapse and the star formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article &lt;a href=&quot;http://www.space.com/858-study-stars-size-limit.html&quot;&gt;Stars Have a Size Limit&lt;/a&gt; by Michael Schirber, it's about 150 Solar Masses. However, there's the Pistol Star, which is speculated to be 200 SM. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the article 'Das wechselhafte Leben der Sterne' by Ralf Launhard (Spektrum 8/2013) there's a diagram with information that when the mass is over 100 SM, the star can't form because of radiation pressure. The exact value of the limit is not speculated in the article. &lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2013-09-27T19:18:55.347" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="334" PostTypeId="1" AcceptedAnswerId="413" CreationDate="2013-09-27T19:25:53.543" Score="17" ViewCount="587" Body="&lt;p&gt;Moon is slowly receding from Earth, which means that after its formation it was much closer to the Earth than now. How would weather look like if it wasn't receding at it would be now as close as at the beginning?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely, the tides would be much higher, but how big would be the difference? I've also watched document stating that the hurricanes would be much stronger. Was there any simulations, how much stronger they would be?&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2014-12-19T16:39:05.583" Title="If the moon wasn't receding from Earth, what would be the impact on the weather and tides?" Tags="&lt;the-moon&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="335" PostTypeId="1" CreationDate="2013-09-27T20:41:48.637" Score="11" ViewCount="718" Body="&lt;p&gt;If you plot the movement of Mars across the Earth's sky it appears to move back on itself during its orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does it appear to have a loop shape from Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LfsrK.jpg&quot; alt=&quot;mars orbit loop&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastEditorUserId="14021" LastEditDate="2016-09-14T19:55:27.843" LastActivityDate="2016-09-15T02:07:02.337" Title="Why does Mars appear to retreat across the sky?" Tags="&lt;earth&gt;&lt;mars&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="336" PostTypeId="2" ParentId="335" CreationDate="2013-09-27T20:52:34.217" Score="11" Body="&lt;p&gt;The effect is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Apparent_retrograde_motion&quot; rel=&quot;nofollow noreferrer&quot;&gt;apparent retrograde motion&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens is that Mars has a 'direction opposite to that of other bodies within its system as observed from a particular vantage point' when this loop occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a bunch of words that don't mean a lot to me. A picture is worth a thousand clearer words:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/K8zUX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;(Imagine this turned sideways and you get the effect in your image)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, because Earth and Mars are orbiting the sun at different rates, our vantage point of Mars changes for each combination of points in the orbit of each planet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On this scale, the background of stars is pretty much stationary - any apparent movement of the stars due to this effect is going to be negligible. Thus, the stars are our point of reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As our vantage point of Mars changes, it appears to shift directions on the stellar background, creating the effect you describe.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2013-09-27T20:52:34.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="337" PostTypeId="5" CreationDate="2013-09-27T21:07:38.697" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-27T21:07:38.697" LastActivityDate="2013-09-27T21:07:38.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="338" PostTypeId="4" CreationDate="2013-09-27T21:07:38.697" Score="0" Body="Questions regarding Mars, the fourth planet from the Sun." OwnerUserId="19" LastEditorUserId="9301" LastEditDate="2016-10-04T19:44:29.077" LastActivityDate="2016-10-04T19:44:29.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="339" PostTypeId="1" CreationDate="2013-09-27T21:09:48.573" Score="10" ViewCount="4640" Body="&lt;p&gt;Once a body has been identified as a star it is classified into a subsection of either O,B,A,F,G,K or M (I think there are also 3 more letters recently added)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the origin of this classification system? Do each of the letters stand for anything? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JySO3.png&quot; alt=&quot;star classification&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastActivityDate="2020-12-15T00:10:01.673" Title="What is the origin of the OBAFGKM classification system?" Tags="&lt;star&gt;&lt;classification&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="340" PostTypeId="2" ParentId="339" CreationDate="2013-09-27T21:23:11.047" Score="9" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_classification&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; on stellar classification has the full list of letters:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;Class&lt;/th&gt;&#xA;&lt;th&gt;Surface Temperature&lt;/th&gt;&#xA;&lt;th&gt;Notes&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;O&lt;/td&gt;&#xA;&lt;td&gt;&amp;gt; 33,000 K&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;B&lt;/td&gt;&#xA;&lt;td&gt;10,000 - 33,000 K&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;A&lt;/td&gt;&#xA;&lt;td&gt;7500 - 10,000 K&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;F&lt;/td&gt;&#xA;&lt;td&gt;6000 - 7500 K&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;G&lt;/td&gt;&#xA;&lt;td&gt;5200 - 6000 K&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;K&lt;/td&gt;&#xA;&lt;td&gt;3700 - 5200 K&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;M&lt;/td&gt;&#xA;&lt;td&gt;2000 - 3700 K&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;L&lt;/td&gt;&#xA;&lt;td&gt;1300 - 2000 K&lt;/td&gt;&#xA;&lt;td&gt;L T and Y are those new letters&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;T&lt;/td&gt;&#xA;&lt;td&gt;700 - 1300 K&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Y&lt;/td&gt;&#xA;&lt;td&gt;&amp;lt; 700 K&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&lt;h2&gt;History&lt;/h2&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://curious.astro.cornell.edu/question.php?number=488&quot; rel=&quot;nofollow noreferrer&quot;&gt;This Cornell doc&lt;/a&gt; explains much of the history behind the system:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In the 1890s, many scientists were interested in developing a classification scheme for the stars. Edward C. Pickering at Harvard University, together with his assistant Williamina P. Fleming, assigned stars a letter according to how much Hydrogen could be observed in their spectra: stars labeled A had the most Hydrogen, B the next most, and so on through the alphabet. There were 22 types in all: this scheme was rather cumbersome, and it wasn't clear what its physical significance was.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In 1901, another of Pickering's assistants, Annie Jump Cannon, also begun to work on the classification sequence. Her meticulous observations led her to simplify the 22-type scheme into a simple sequence of &lt;em&gt;temperature&lt;/em&gt;: OBAFGKM. Not only was the scheme much simpler than the previous, but it related the amount of hydrogen observed to a physical property of the stars for the first time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19" LastEditorUserId="2365" LastEditDate="2020-12-15T00:10:01.673" LastActivityDate="2020-12-15T00:10:01.673" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="341" PostTypeId="5" CreationDate="2013-09-27T21:41:44.063" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-27T21:41:44.063" LastActivityDate="2013-09-27T21:41:44.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="342" PostTypeId="4" CreationDate="2013-09-27T21:41:44.063" Score="0" Body="Questions regarding the physical dimensions of a celestial object." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-28T06:44:34.447" LastActivityDate="2013-09-28T06:44:34.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="343" PostTypeId="2" ParentId="4" CreationDate="2013-09-27T23:04:33.573" Score="7" Body="&lt;p&gt;Sunspots are cooler because their magnetic fields inhibit the replenishing of heat from convective flows (due to the interaction between plasma and magnetic fields). This allows them to cool radiatively. The rest of the solar surface is constantly being replenished by convective cells that reheat it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solar plasma at the photosphere radiates roughly as a black-body, meaning that the energy (and wavelength) spectrum of radiation follows the Planck function. The Planck function scales up and down with temperature at all energies (and wavelengths). The brightness of a blackbody at a given energy (or wavelength) is determined by the value of the Planck function at that energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image you showed is taken in the &quot;photospheric continuum&quot;, which is a black-body dominated part of the radiation spectrum. So, because sunspots are cool (compared to their surroundings), this means that their Planck function is lower than their surroundings, and hence their brightness is lower, causing them to appear dark in the image.&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="31410" LastEditDate="2020-06-05T11:23:56.290" LastActivityDate="2020-06-05T11:23:56.290" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="344" PostTypeId="2" ParentId="148" CreationDate="2013-09-27T23:16:19.190" Score="10" Body="&lt;p&gt;The expansion of the universe it self does not cause atoms to increase in size (because of the electromagnetic forces holding them together), but it does cause galaxies to move apart from each other (e.g., observed galaxy redshifts and Hubble's Law).&lt;/p&gt;&#xA;&lt;p&gt;In reality, the expansion of the universe is known to be accelerating slightly (see Perlmutter et al., 1998). This acceleration puts a very small outward force on everything (including atoms). Although the electromagnetic forces in atoms hold everything together, it causes electrons to orbit protons at an (infinitesimally) larger distance from them.&lt;/p&gt;&#xA;&lt;p&gt;If the acceleration of the universe's expansion increased many times over, this could have the potential to tear matter apart. The case of extremely large expansion acceleration is called the 'Big Rip'.&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="22115" LastEditDate="2020-10-17T15:03:40.157" LastActivityDate="2020-10-17T15:03:40.157" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="345" PostTypeId="2" ParentId="138" CreationDate="2013-09-27T23:30:24.520" Score="36" Body="&lt;p&gt;Yes, the Sun rotates. This can be observed by tracking a variety of features on the Sun, such as sunspots, X-ray brightpoints, coronal holes, filaments, and small magnetic flux elements. Another way to determine the rotational speed of the Sun is to measure spectral lines at the edge of the Sun's disk and determine their redshift. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is thought that the rotation of the Sun is due to the way the primordial gas cloud collapsed in on itself to create the Sun. Also, it is likely that the Sun originally rotated much faster when it initially formed, than it does to day. This slow-down was probably caused by 'magnetic breaking' in which strong magnetic fields threading our primordial Sun out into the solar wind resisted the rotation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today, the reason that the Sun rotates at a different angular speed at different solar latitudes is due to hydromagnetic effects. One cause is thought to be the nature of convection within the outer third of the Sun. Interestingly, below a the convective envelope, beneath a boundary called the tachocline, the Sun rotates as a rigid body. You can find more about this in &lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/nph-data_query?bibcode=1998ApJ...505..390S&amp;amp;db_key=AST&amp;amp;link_type=ABSTRACT&amp;amp;high=524614d23814974&quot;&gt;Schou et al., 1998&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2013-09-27T23:30:24.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="346" PostTypeId="5" CreationDate="2013-09-28T00:18:00.687" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-28T00:18:00.687" LastActivityDate="2013-09-28T00:18:00.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="347" PostTypeId="4" CreationDate="2013-09-28T00:18:00.687" Score="0" Body="Questions regarding the rate of motion of an object, usually expressed as distance per unit time." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-28T06:44:45.587" LastActivityDate="2013-09-28T06:44:45.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="348" PostTypeId="1" AcceptedAnswerId="352" CreationDate="2013-09-28T00:24:01.500" Score="21" ViewCount="1787" Body="&lt;p&gt;Since the moon has no atmosphere, its surface has many more meteorite impacts than Earth's does. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://d1jqu7g1y74ds1.cloudfront.net/wp-content/uploads/2010/02/Earths-Moon.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.universetoday.com/62140/surface-of-the-moon/&quot; rel=&quot;noreferrer&quot;&gt;Image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have we ever observed a meteor strike the surface of the Moon?&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:47.013" LastActivityDate="2013-10-13T13:54:53.243" Title="Have we ever observed a large meteor hit the Moon?" Tags="&lt;the-moon&gt;&lt;meteor&gt;&lt;meteorite&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="349" PostTypeId="1" AcceptedAnswerId="350" CreationDate="2013-09-28T00:32:36.837" Score="18" ViewCount="19324" Body="&lt;p&gt;Since the moon has gravity, it's almost impossible that there aren't &lt;em&gt;some&lt;/em&gt; gasses trapped on the surface by the moon's gravity. Has any free-floating oxygen been found on the Moon? If so, in what concentration?&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-28T00:43:59.210" LastActivityDate="2015-12-01T23:06:44.200" Title="Does the Moon have any oxygen in its atmosphere?" Tags="&lt;the-moon&gt;&lt;planetary-atmosphere&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="350" PostTypeId="2" ParentId="349" CreationDate="2013-09-28T01:23:23.217" Score="22" Body="&lt;p&gt;The Moon's atmosphere is very thin compared to the Earth, so thin that it is usually said to have no atmosphere.  The Moon's gravity is not strong enough to retain lighter elements, so they escape into space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apollo 17 carried an instrument called the Lunar Atmosphere Composition Experiment (LACE).  Oxygen is not listed as one of the elements it found on &lt;a href=&quot;http://www.lpi.usra.edu/lunar/missions/apollo/apollo_17/experiments/lace/&quot;&gt;this NASA web page&lt;/a&gt;.  The principle gases found were neon, helium and hydrogen.  Others included methane, carbon dioxide, ammonia, and water.  Additionally, ground instruments have detected sodium and potassium in the lunar atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The recently launched &lt;a href=&quot;http://www.nasa.gov/mission_pages/ladee/main/index.html&quot;&gt;Lunar Atmosphere and Dust Environment Experiment&lt;/a&gt; (LADEE - pronounced laddie, not lady) should be able to tell us more soon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&#xA;According to &lt;a href=&quot;http://www.spaceflightnow.com/minotaur/ladee/140408eom/#.VX70Abyt1hE&quot;&gt;this article at Spaceflight Now&lt;/a&gt;, LADEE did detect oxygen in the Moon's atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="25" LastEditorUserId="25" LastEditDate="2015-06-15T15:55:10.877" LastActivityDate="2015-06-15T15:55:10.877" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="351" PostTypeId="1" CreationDate="2013-09-28T01:54:31.890" Score="9" ViewCount="387" Body="&lt;p&gt;At all places where I have read of Moving Cluster Method for measuring distance to a cluster and once applied myself on data, I have been told that the method has been successfully applied only to Hyades cluster in Taurus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rb1LP.png&quot; alt=&quot;hyades cluster moving cluster method&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question :&lt;/strong&gt; What factor made the method unsuccessful for other known clusters? &lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2014-02-05T17:24:40.403" Title="Why has Moving Cluster Method been successful only for Hyades?" Tags="&lt;distances&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="352" PostTypeId="2" ParentId="348" CreationDate="2013-09-28T02:02:11.713" Score="18" Body="&lt;p&gt;Yes, we have, and the impacts occur very frequently according to &lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2013/16may_lunarimpact/&quot; rel=&quot;noreferrer&quot;&gt;&quot;Bright explosion on the moon&quot;&lt;/a&gt; (NASA):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For the past 8 years, NASA astronomers have been monitoring the Moon for signs of explosions caused by meteoroids hitting the lunar surface. &quot;Lunar meteor showers&quot; have turned out to be more common than anyone expected, &lt;strong&gt;with hundreds of detectable impacts occurring every year&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On March 17, 2013 a significant impact occurred that could have been observed from the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Updated information about lunar impacts is found at the &lt;a href=&quot;http://www.nasa.gov/centers/marshall/news/lunar/&quot; rel=&quot;noreferrer&quot;&gt;Marshall Space Flight Centre's Lunar Impacts page.&lt;/a&gt;  Below is a map of the impacts that occurred on the lunar surface between 2005-2008, all of which, according to the article &lt;a href=&quot;http://science1.nasa.gov/science-news/science-at-nasa/2008/02sep_lunarperseids/&quot; rel=&quot;noreferrer&quot;&gt;&quot;Amateur Astronomers see Perseids hit the Moon&lt;/a&gt;, could have been seen from the Earth (with a telescope):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/meBXd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been instances throughout our history when, so is &lt;em&gt;believed&lt;/em&gt;, people on the ground have witnessed lunar impacts. One of the more famous ones was in 1178 AD, when monks witnessed fire coming from the moon; However, according to the article &lt;a href=&quot;http://news.bbc.co.uk/2/hi/science/nature/1304985.stm&quot; rel=&quot;noreferrer&quot;&gt;&quot;Historic lunar impact questioned&quot;&lt;/a&gt;, the accuracy of this historical account is questionable.&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorUserId="13" LastEditDate="2013-09-28T05:35:04.170" LastActivityDate="2013-09-28T05:35:04.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="353" PostTypeId="2" ParentId="332" CreationDate="2013-09-28T02:03:24.347" Score="5" Body="&lt;p&gt;Parallax measurement in practice is not as is explained above using the popular diagram you have used. The parallax causes the star to prescribe an ellipse in the sky, the semi-major axis of whose is equal to the parallactic angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;telescopes generally measure the shift in co-ordinates of star(RA and Dec)&lt;/strong&gt; and then translate the information to that of the &lt;a href=&quot;http://star-www.st-and.ac.uk/~fv/webnotes/chapt14.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;ellipse&lt;/a&gt; prescribed and determine the parallactic angle. Now, the &lt;strong&gt;ability of telescopes to note this change in RA and Dec is based on their least count, obviously which is dependent on their angular resolution.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2LA3K.png&quot; alt=&quot;parallactic ellipse&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, distance measurements using parallax method have their range based on the angular resolution of the telescope being used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image source:&lt;/em&gt; &lt;a href=&quot;http://documents.stsci.edu/hst/fgs/documents/handbooks/ihb_cycle16/c01_intro4.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://documents.stsci.edu/hst/fgs/documents/handbooks/ihb_cycle16/c01_intro4.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="63" LastEditDate="2014-01-09T01:29:57.183" LastActivityDate="2014-01-09T01:29:57.183" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="354" PostTypeId="5" CreationDate="2013-09-28T05:38:53.170" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-28T05:38:53.170" LastActivityDate="2013-09-28T05:38:53.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="355" PostTypeId="4" CreationDate="2013-09-28T05:38:53.170" Score="0" Body="Questions regarding the action of or an object rotating along an axis." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-28T06:44:39.557" LastActivityDate="2013-09-28T06:44:39.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="362" PostTypeId="1" AcceptedAnswerId="412" CreationDate="2013-09-28T10:51:07.643" Score="14" ViewCount="816" Body="&lt;p&gt;According to the NASA web page &lt;a href=&quot;http://solarsystem.nasa.gov/planets/profile.cfm?Object=Mercury&quot; rel=&quot;noreferrer&quot;&gt;overview about Mercury&lt;/a&gt;, despite the planet being just a bit larger than our moon, it's density is about 98.4% of Earth's.  This high density suggests a comparatively large iron core (compared to the rest of the planet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recent studies reported in &lt;a href=&quot;http://www.jpl.nasa.gov/news/news.php?release=2007-050&quot; rel=&quot;noreferrer&quot;&gt;&quot;NASA Antenna Cuts Mercury to Core, Solves 30 Year Mystery&quot;&lt;/a&gt; (NASA, 2007) have confirmed the presence of a molten core, surrounded by a silicate mantle (depicted in the illustration below (from the article):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KmXsP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is then, what is the current accepted theory as to Mercury has a still-molten core and a density like that of Earth?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2018-07-14T02:53:31.440" Title="What is the current accepted theory as to why Mercury, despite its size, has a similar density to Earth?" Tags="&lt;core&gt;&lt;mercury&gt;&lt;density&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="364" PostTypeId="5" CreationDate="2013-09-28T15:43:37.460" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-28T15:43:37.460" LastActivityDate="2013-09-28T15:43:37.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="365" PostTypeId="4" CreationDate="2013-09-28T15:43:37.460" Score="0" Body="Questions regarding celestial bodies in orbit around a star." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-28T20:17:14.523" LastActivityDate="2013-09-28T20:17:14.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="366" PostTypeId="1" AcceptedAnswerId="367" CreationDate="2013-09-28T16:22:23.880" Score="18" ViewCount="2946" Body="&lt;p&gt;I know Hawking Radiation has firm theoretical footing, but has a signal ever actually been observed? What observational research is being done to attempt to see this effect? Is it simply too tall an order (i.e. - too far away, too weak of a signal, etc..)? &lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2021-04-14T10:34:55.950" Title="Has Hawking Radiation Ever Been Observed?" Tags="&lt;black-hole&gt;&lt;observational-astronomy&gt;&lt;hawking-radiation&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="367" PostTypeId="2" ParentId="366" CreationDate="2013-09-28T21:00:23.273" Score="14" Body="&lt;p&gt;According to this MIT Technology Review &lt;a href=&quot;http://www.technologyreview.com/view/420940/first-observation-of-hawking-radiation/&quot;&gt;&quot;First Observation of Hawking Radiation&quot;&lt;/a&gt; and an associated article &lt;a href=&quot;http://arxiv.org/pdf/1009.4634v1.pdf&quot;&gt;&quot;Hawking radiation from ultrashort laser pulse filaments&quot;&lt;/a&gt; where the authors have simulated Hawking radiation in the laboratory, using ultrashort laser pulse filaments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in part answer to your question, yes Hawking Radiation is believed to have been observed, but in a laboratory environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://arxiv.org/pdf/1212.5392v1.pdf&quot;&gt;&quot;Observing Hawking radiation in Bose-Einstein condensates&#xA;via correlation measurements&quot;&lt;/a&gt; (Fabbri, 2012), direct observation of the Hawking radiation is (in their word): hopeless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, once again simulations are possible:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hawking effect, which depends only on kinematical properties of wave propagation&#xA;  in the presence of horizons, is present also in nongravitational contexts,&#xA;  for instance in stationary fluids undergoing supersonic flow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From these articles, it would seem that direct observations of Hawking radiation is not feasible currently; however, we are able to produce analogues of the phenomena in the laboratory.&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2013-09-28T21:00:23.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="368" PostTypeId="2" ParentId="305" CreationDate="2013-09-28T23:07:31.200" Score="12" Body="&lt;p&gt;A good analogy which shows that the expansion has no centre is one of a 2D sheet covered with uniform dots - representing the galaxies where the one in the centre is us:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vwILH.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the universe expands the dots (galaxies) get further away from each other:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sjgar.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you overlay the future and present images lining up on the central galaxy (us) you can see that all the other galaxies appear to be moving away from us, and that the further away they are the faster they appear to be receding:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zS3sP.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you line up on any other galaxy in the image you can see that all the other galaxies appear to be moving away from them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TcluC.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As no two points can be the centre of the universe we can only conclude that there is no centre to the the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://calgary.rasc.ca/universe_expansion.htm&quot; rel=&quot;noreferrer&quot;&gt;Image Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="111" LastEditorUserId="111" LastEditDate="2013-10-02T14:41:24.487" LastActivityDate="2013-10-02T14:41:24.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="369" PostTypeId="2" ParentId="349" CreationDate="2013-09-28T23:53:47.100" Score="11" Body="&lt;p&gt;Just to add to GreenMatt's answer, according to the article &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.21.9994&amp;amp;rep=rep1&amp;amp;type=pdf&quot;&gt;&quot;The Lunar Atmosphere: History, Status, Current Problems and Context&quot;&lt;/a&gt; (Stern, 1999), the lunar atmosphere is in fact a tenuous &lt;strong&gt;exosphere&lt;/strong&gt;, which the authors describe as being composed of &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;independent atmospheres&quot; occupying the same space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is further elaborated in &lt;a href=&quot;http://sservi.nasa.gov/wp-content/uploads/drupal/D&amp;amp;A_whitepaper_091409.pdf&quot;&gt;&quot;The Lunar Dusty Exosphere: The Extreme Case of an Inner Planetary Atmosphere&quot;&lt;/a&gt; (NASA), that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;in direct response to these intense and variable environmental drivers, the Moon releases a low density neutral gas forming a collisionless atmosphere. This ~100 tons of gas about the Moon is commonly called the &lt;strong&gt;lunar surface-bounded exosphere&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is also an ionosphere, due to (from the ANSA article):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ions are also created directly either by surface sputtering or subsequent neutral photoionization, forming a tenuous exo-ionosphere about the Moon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The authors also suggest that ionic oxygen may be present due to surface sputtering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to solar radiation and the solar wind, dust particles become charged as well, and can be subsequently listed from the lunar surface.  &lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-09-29T00:08:50.323" LastActivityDate="2013-09-29T00:08:50.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="370" PostTypeId="1" AcceptedAnswerId="371" CreationDate="2013-09-29T00:51:20.507" Score="32" ViewCount="7432" Body="&lt;p&gt;Most of the known gas planets (Jupiter, Saturn, etc.) are huge. They are even called &quot;gas &lt;em&gt;giants&lt;/em&gt;&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is it possible to exist a gas planet around the size of Earth? If yes, why; if no, why?&lt;/p&gt;&#xA;" OwnerUserId="62" LastActivityDate="2018-01-10T01:16:00.640" Title="Can small gas planets exist?" Tags="&lt;planet&gt;&lt;gas-giants&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="371" PostTypeId="2" ParentId="370" CreationDate="2013-09-29T01:50:50.433" Score="13" Body="&lt;p&gt;According to the article &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1989A&amp;amp;A...215..177S&amp;amp;db_key=AST&amp;amp;page_ind=0&amp;amp;data_type=GIF&amp;amp;type=SCREEN_VIEW&amp;amp;classic=YES&quot;&gt;&quot;Minimum planetary size for forming outer Jovian-type planets - Stability of an isothermal atmosphere surrounding a protoplanet&quot;&lt;/a&gt; (Sasaki, 1989), not only is this possible, but is suggested to have been an evolutionary stage for the development of Uranus and Neptune.  But there are a few conditions to allow this to happen, primarily due to distance from the sun, hence temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article's findings suggest that the critical mass for 'atmospheric collapse' is 0.2 of the current mass of the Earth for distances of 5 AU or greater for our solar system (this value of course would vary, depending on the parent star).  This critical mass increases the closer you are to the parent star - the article explicitly states that Mercury and Mars are too small for this to occur. &lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-09-29T01:57:01.543" LastActivityDate="2013-09-29T01:57:01.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="372" PostTypeId="2" ParentId="370" CreationDate="2013-09-29T02:01:59.057" Score="8" Body="&lt;p&gt;There are a few major problems with calling something a &lt;em&gt;small gas planet&lt;/em&gt; and of course depends on your definition of &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Planet&quot;&gt;a planet&lt;/a&gt;&lt;/em&gt;. By currently accepted classification (that excludes Pluto), a planet has to be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;massive enough to be rounded (i.e. held together) by its own mass, &lt;/li&gt;&#xA;&lt;li&gt;has to orbit a star, and &lt;/li&gt;&#xA;&lt;li&gt;has cleared its orbital neighborhood. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All of these points are largely argumentative, but two major problems that a &lt;em&gt;small gas planet&lt;/em&gt; would have to sustain itself for longer time during and after the formation within the &lt;a href=&quot;https://en.wikipedia.org/wiki/Protoplanetary_disk&quot;&gt;protoplanetary disk&lt;/a&gt; are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It will either be too close to the sun it orbits and the planet's own gravity holding it together wouldn't be sufficient to prevent it being effectively blown away layer by layer by sun's activity, so you might end up calling such &lt;em&gt;small gas planet&lt;/em&gt; a &lt;a href=&quot;https://en.wikipedia.org/wiki/Protoplanet&quot;&gt;protoplanet&lt;/a&gt; as it wouldn't persist, or &lt;/li&gt;&#xA;&lt;li&gt;it would likely be formed large distance from its sun and might not stay in orbit around it after it cleared with &lt;a href=&quot;https://en.wikipedia.org/wiki/Accretion_%28astrophysics%29&quot;&gt;accretion&lt;/a&gt; its path out of the protoplanetary disk and went rogue. So you might end up calling such &lt;em&gt;small gas planet&lt;/em&gt; a free-floating &lt;a href=&quot;https://en.wikipedia.org/wiki/Extrasolar_planet&quot;&gt;exoplanet&lt;/a&gt; or a &lt;a href=&quot;https://en.wikipedia.org/wiki/Rogue_planet&quot;&gt;rogue planet&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Another problem that we didn't yet discuss is that we have no observational proof that such &lt;em&gt;small gas planets&lt;/em&gt; actually exist, whatever you might end up calling them. Problem of course lies in the fact there simply aren't any smaller gas planets than &lt;a href=&quot;https://en.wikipedia.org/wiki/Uranus&quot;&gt;Uranus&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/Neptune&quot;&gt;Neptune&lt;/a&gt; (you pick your candidate for the &lt;em&gt;smallest&lt;/em&gt; one; former is least massive, and latter is least voluminous of the four gas giants) in our own Solar system, while &lt;a href=&quot;https://en.wikipedia.org/wiki/Extrasolar_planet&quot;&gt;extrasolar&lt;/a&gt; &lt;em&gt;small gas planets&lt;/em&gt; would likely be too small to be detectable over enormous distances to even our nearest stars by any &lt;a href=&quot;https://en.wikipedia.org/wiki/Methods_of_detecting_extrasolar_planets&quot;&gt;methods of detecting such extrasolar planets&lt;/a&gt;, or we wouldn't be observing enough into the system's past (i.e. distance) to see it still forming and it would have already lost grip to its former planets now exoplanets, or erode the smaller protoplanets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have this conundrum between whether they exist (likely they do), are in theory possible (I have no doubt), and how we'd end up calling them, because neither of the two possibilities I can think of for their existence would be considered &lt;em&gt;planets&lt;/em&gt; by mentioned current convention of what actually constitutes &lt;em&gt;a planet&lt;/em&gt;. Protoplanet wouldn't have cleared its neighborhood yet, and exoplanet wouldn't orbit a star.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2013-09-29T02:01:59.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="373" PostTypeId="1" AcceptedAnswerId="374" CreationDate="2013-09-29T02:26:11.650" Score="19" ViewCount="863" Body="&lt;p&gt;The James Webb space telescope is supposed to be located at the Earth-Sun L2 Lagrange point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we expect the region around that point to have a higher concentration of space debris, asteroids, dust, etc...? Would this be a matter of concern for bringing the telescope safely in position (i.e. require extra dust shield)? Do we expect however the exact Lagrange point to be free of any matter since it would take an acceleartion/deceleration for matter to actually reach that exact location? How large would be this 'debris-free' zone?&lt;/p&gt;&#xA;" OwnerUserId="148" LastActivityDate="2016-12-22T15:22:57.380" Title="Parking a telescope at a Lagrange point: is this a good idea from a debris point of view?" Tags="&lt;gravity&gt;&lt;space-telescope&gt;&lt;lagrange-point&gt;&lt;space-debris&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="374" PostTypeId="2" ParentId="373" CreationDate="2013-09-29T02:47:04.443" Score="19" Body="&lt;p&gt;Hmmm no, it wouldn't be cluttered with debris, and yes, it's a good idea to park the JWST (James Webb Space Telescope) at the Sun-Earth L&lt;sub&gt;2&lt;/sub&gt; point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The five &lt;a href=&quot;https://en.wikipedia.org/wiki/Lagrangian_point&quot; rel=&quot;noreferrer&quot;&gt;Lagrange points&lt;/a&gt; are unstable, for one because of the gravitational anomalies of the two massive bodies of the Lagrange system, eccentric orbits, and there are many other factors to their instability. At the same time, they are least gravitationally attractive points around two massive bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of L-points as parking your car on a flat space at the top of the hill. You'll have to approach with some control and then try to balance at your parking spot, if you don't plan using your handbrake and also remain there stationary:&#xA;&lt;br /&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/Mg2I3.jpg&quot; alt=&quot;Lagrange points&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;sup&gt;Visualisation of the relationship between two massive bodies and their five Lagrange points (Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Lagrangian_point&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There wouldn't be any debris clutter there, or any other matter like smaller particles, at least not any more likely than elsewhere around them, only transient in nature and possibly even less likely than elsewhere since all the other mass particles would gravitate towards the more massive bodies of two Lagrange system centers in their vicinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No body with &lt;a href=&quot;https://en.wikipedia.org/wiki/Invariant_mass&quot; rel=&quot;noreferrer&quot;&gt;rest mass&lt;/a&gt; would stay there on its own accord, not unless it has active attitude control to position itself there and constantly adjust for changes in gravitational attraction vector as the two mass bodies rotate around their axes, change distance while orbiting each other, or L points being influenced by other mass bodies of the same planetary system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same time, JWST will be shielded by the Earth from any Solar activities and also Sun's interference with JWST's sensitive equipment as it starts observing the Universe in the infrared spectrum. Most of the orbital debris from our own space exploration missions is cluttered in the LEO (Low Earth Orbit) belt at roughly an altitude of 500-1500 km above the Earth's surface: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/S2xEp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;sup&gt;Source: &lt;a href=&quot;http://ssi.org/2010/SM14-proceedings/Active-Debris-Removal-EDDE-ElectroDynamic-Debris-Eliminator-Pearson-Carroll-%20Levin.pdf&quot; rel=&quot;noreferrer&quot;&gt;Active Debris Removal: EDDE, the ElectroDynamic Debris Eliminator&lt;/a&gt;, Jerome Pearson et al. (PDF)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the JWST will be deployed to a very large 800,000 kilometres (500,000 mi) radius halo orbit around the Sun-Earth L&lt;sub&gt;2&lt;/sub&gt; point, that is 1,500,000 kilometers (930,000 mi) from the Earth, around 4 times farther than the distance between the Earth to the Moon. So not exactly in Earth's shadow and it will still use large deployable &lt;a href=&quot;http://webb.nasa.gov/sunshield.html&quot; rel=&quot;noreferrer&quot;&gt;sunshield&lt;/a&gt;, but that's a long way away from where we're littering our LEO or even GEO/GSO (Geostationary/Geosynchronous orbits) with GEO being the farthest of these frequently used orbits, at 35,786 kilometres (22,236 mi) above the Earth's equator (and its graveyard orbit a bit farther than that). JWST will be placed in halo orbit nearly 40 times as distant from the Earth as you'd expect any space debris cluttering our planet's orbits at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, it is a very good idea to place JWST around the Sun-Earth L&lt;sub&gt;2&lt;/sub&gt; point.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditDate="2014-06-22T11:22:34.347" LastActivityDate="2014-06-22T11:22:34.347" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="375" PostTypeId="5" CreationDate="2013-09-29T04:09:24.020" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-29T04:09:24.020" LastActivityDate="2013-09-29T04:09:24.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="376" PostTypeId="4" CreationDate="2013-09-29T04:09:24.020" Score="0" Body="Questions about Mercury, the first planet from Earth's sun.  Also the smallest major planet in the Solar System." OwnerDisplayName="user8" LastEditorUserId="13681" LastEditorDisplayName="user8" LastEditDate="2017-05-14T12:25:40.683" LastActivityDate="2017-05-14T12:25:40.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="377" PostTypeId="5" CreationDate="2013-09-29T04:12:08.940" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-29T04:12:08.940" LastActivityDate="2013-09-29T04:12:08.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="378" PostTypeId="4" CreationDate="2013-09-29T04:12:08.940" Score="0" Body="Questions about Europa, a moon of Jupiter with a planet-wide ocean under a thick ice crust." OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-09-29T06:27:24.313" LastActivityDate="2013-09-29T06:27:24.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="379" PostTypeId="5" CreationDate="2013-09-29T04:14:48.577" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-29T04:14:48.577" LastActivityDate="2013-09-29T04:14:48.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="380" PostTypeId="4" CreationDate="2013-09-29T04:14:48.577" Score="0" Body="Questions related to the unit of measurement, parsec, which is approximately 3.26 times the distance of a light year." OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-09-29T06:27:19.737" LastActivityDate="2013-09-29T06:27:19.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="381" PostTypeId="1" AcceptedAnswerId="384" CreationDate="2013-09-29T07:23:33.843" Score="26" ViewCount="598" Body="&lt;p&gt;After the big bang occurred what biased the formation of particles over anti-particles? Why are particles more common than anti-particles?&lt;/p&gt;&#xA;" OwnerUserId="150" LastEditorUserId="45" LastEditDate="2013-09-29T13:00:37.493" LastActivityDate="2013-09-29T13:09:23.997" Title="Why is there a matter/anti-matter asymmetry in the universe?" Tags="&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="382" PostTypeId="5" CreationDate="2013-09-29T07:29:38.060" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-29T07:29:38.060" LastActivityDate="2013-09-29T07:29:38.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="383" PostTypeId="4" CreationDate="2013-09-29T07:29:38.060" Score="0" Body="Questions regarding the big bang which is the prevailing cosmological model that describes the early development of the Universe." OwnerUserId="150" LastEditorUserId="150" LastEditDate="2013-09-29T13:22:22.767" LastActivityDate="2013-09-29T13:22:22.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="384" PostTypeId="2" ParentId="381" CreationDate="2013-09-29T08:00:19.183" Score="17" Body="&lt;p&gt;Currently, there's no single explanation for this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest explanation is that there isn't any imbalance &amp;mdash; there are large antimatter bodies in the universe; just separated from us. However, these would have to be pretty far away (otherwise the boundary annihilation would be detectable). Theoretically, this requires antimatter to have clumped early on, which is rather unlikely/impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A far more likely explanation is that &lt;a href=&quot;http://en.wikipedia.org/wiki/CP_violation&quot;&gt;charge-parity symmetry is violated&lt;/a&gt;. In other words, flipping the charges and parities in a reaction may not lead to the same dynamics &amp;mdash; the outcome (when the charge/parity is flipped back) or distributions of outcomes may not be the same. Usually, we expect that if we flip the charges on both sides of a reaction, the reaction proceeds in the same manner (with same probabilities, etc). There is evidence that suggests that this may not be true. If this is indeed the case, then the matter-antimatter imbalance is not that hard to explain.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2013-09-29T13:09:23.997" LastActivityDate="2013-09-29T13:09:23.997" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="385" PostTypeId="1" AcceptedAnswerId="386" CreationDate="2013-09-29T08:11:35.767" Score="15" ViewCount="715" Body="&lt;p&gt;Could Earth-like (suitable for Human habitation) planets exist in systems with 2 or 3 suns, have any been found?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With more than a single star, isn't it likely that the habitation zone would be too irregular, gravitational forces be irregular/strong and numerous other issues?&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2013-09-29T11:38:35.410" Title="Could there be Earth-like planets in binary or trinary systems?" Tags="&lt;earth-like-planet&gt;&lt;habitable-zone&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="386" PostTypeId="2" ParentId="385" CreationDate="2013-09-29T09:37:01.980" Score="11" Body="&lt;p&gt;Firstly, a few planets have been discovered around binary or trinary systems, mostly Neptune sized or larger.  An example is are the two planets described in the NASA article &lt;a href=&quot;http://kepler.nasa.gov/news/nasakeplernews/index.cfm?fuseaction=ShowNews&amp;amp;NewsID=228&quot; rel=&quot;noreferrer&quot;&gt;&quot;Kepler-47: Our First Binary Star 2-Planet System&quot;&lt;/a&gt;, where they are described as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now Kepler mission has discovered Kepler-47b and 47c, the first transiting circumbinary system — multiple planets orbiting two suns. To compound the excitement of the discovery, one of those planets is in the binary system's habitable zone (where liquid water may exist)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(A representation of the Kepler 47 system is pictured below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second planet discovered in the system, Kepler 47c, is within the binary stars' habitable zone, but is slightly larger than Neptune and is likely to be a gas giant - but that raises the potential for terrestrial satellites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is the Saturn sized Kepler 16b reported in the article &lt;a href=&quot;http://news.discovery.com/space/galaxies/kepler-16b-exoplanet-two-stars-tatooine-110915.htm&quot; rel=&quot;noreferrer&quot;&gt;&quot;Weird Exoplanet Discovered Orbiting Two Stars&quot;&lt;/a&gt; (Klotz, 2011), which orbits 2 stars that are smaller than our sun, thus has habitable zones closer to their parent stars, in this case, the equivalent of the Venus' orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of the habitable zones around multiple star systems, the article &lt;a href=&quot;http://www.astrobio.net/exclusive/5292/habitable-binary-star-systems&quot; rel=&quot;noreferrer&quot;&gt;&quot;Habitable Binary Star Systems&quot;&lt;/a&gt; suggests a major advantage of a binary (indeed a trinary) system is that the combined energy between the stars could indeed extend the habitable zones, particularly,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Low-mass twins could make the best hosts, because their combined energy extends the habitable region farther away than would exist around a single star. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, temperatures would vary significantly and the orbits would be irregular, and in simulations reported in &lt;a href=&quot;http://www.space.com/19141-alien-planets-two-stars-dangerous.html&quot; rel=&quot;noreferrer&quot;&gt;&quot;Double-Star Systems Can Be Dangerous for Exoplanets&quot;&lt;/a&gt; (Wall, 2013) results in such disruptions that one or more of the planets therein could be ejected and sent hurtling into interstellar space.  But such perturbations are likely to take a considerable amount of time (millions of years) to fully occur.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also discussed in &lt;a href=&quot;http://arxiv.org/pdf/1201.6582v1.pdf&quot; rel=&quot;noreferrer&quot;&gt;&quot;Exoplanets Bouncing Between Binary Stars&quot;&lt;/a&gt; (Moeckel and Veras, 2012), where the planet in such as system is described as 'bouncing' between the binary stars, then resuming a relatively regular orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The recent discovery of a terrestrial-like planet in the Alpha Centauri system, as reported in &lt;a href=&quot;http://www.space.com/18089-earth-size-alien-planet-alpha-centauri.html&quot; rel=&quot;noreferrer&quot;&gt;&quot;Discovery! Earth-Size Alien Planet at Alpha Centauri Is Closest Ever Seen&quot;&lt;/a&gt; (Wall, 2012) - despite this planet not being in the habitable zones of either star, this does raise the hopes for the possibility of Earth-like terrestrial planets existing in the habitable zones of other multiple star systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the possibility that an Earth sized moon could orbit the larger planets such as those in the Kepler systems mentioned earlier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MbPgR.jpg&quot; alt=&quot;kepler systems&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-09-29T11:38:35.410" LastActivityDate="2013-09-29T11:38:35.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="387" PostTypeId="1" AcceptedAnswerId="398" CreationDate="2013-09-29T10:35:02.707" Score="20" ViewCount="1590" Body="&lt;p&gt;Why do planets rotate and revolve in the universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/c1E8I.gif&quot; alt=&quot;rotating earth&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="150" LastEditorDisplayName="user96" LastEditDate="2013-09-29T11:10:04.353" LastActivityDate="2019-04-29T22:49:55.270" Title="Why does a planet rotate and revolve?" Tags="&lt;planet&gt;&lt;universe&gt;&lt;rotation&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="388" PostTypeId="5" CreationDate="2013-09-29T12:37:01.573" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-29T12:37:01.573" LastActivityDate="2013-09-29T12:37:01.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="389" PostTypeId="4" CreationDate="2013-09-29T12:37:01.573" Score="0" Body="Questions about solar system planets/moons and exoplanets that exhibit similar features to the Earth." OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-09-29T13:22:18.720" LastActivityDate="2013-09-29T13:22:18.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="390" PostTypeId="5" CreationDate="2013-09-29T12:38:39.600" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-29T12:38:39.600" LastActivityDate="2013-09-29T12:38:39.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="391" PostTypeId="4" CreationDate="2013-09-29T12:38:39.600" Score="0" Body="Questions regarding Jupiter, the 5th and largest planet in the Solar System." OwnerDisplayName="user8" LastEditorDisplayName="user14291" LastEditDate="2017-09-01T00:35:28.993" LastActivityDate="2017-09-01T00:35:28.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="392" PostTypeId="5" CreationDate="2013-09-29T12:40:52.127" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-29T12:40:52.127" LastActivityDate="2013-09-29T12:40:52.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="393" PostTypeId="4" CreationDate="2013-09-29T12:40:52.127" Score="0" Body="Questions about Titan, the largest moon of Saturn." OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-09-29T13:22:14.580" LastActivityDate="2013-09-29T13:22:14.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="394" PostTypeId="5" CreationDate="2013-09-29T12:45:24.480" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-29T12:45:24.480" LastActivityDate="2013-09-29T12:45:24.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="395" PostTypeId="4" CreationDate="2013-09-29T12:45:24.480" Score="0" Body="Questions about Saturn, the sixth planet from the Sun." OwnerDisplayName="user8" LastEditorUserId="9301" LastEditorDisplayName="user8" LastEditDate="2016-10-04T20:21:39.043" LastActivityDate="2016-10-04T20:21:39.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="396" PostTypeId="5" CreationDate="2013-09-29T12:47:10.880" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-29T12:47:10.880" LastActivityDate="2013-09-29T12:47:10.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="397" PostTypeId="4" CreationDate="2013-09-29T12:47:10.880" Score="0" Body="Questions about Trojan Asteroids, asteroids that co-orbit major planets." OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-09-29T13:22:09.673" LastActivityDate="2013-09-29T13:22:09.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="398" PostTypeId="2" ParentId="387" CreationDate="2013-09-29T13:48:23.480" Score="12" Body="&lt;p&gt;As the planets evolve during their protoplanetary stage and accrete materials from the protoplanetary disks, which are gravitationally collapsing interstellar dust and gases, these accreted particles retain some of the angular momentum from the materials they form from and being in constant motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/OLk0J.png&quot; alt=&quot;accretion period of the protoplanetary disk&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Generated image (virtual fly-by) from a simulation of the accretion period of the protoplanetary disk, showing preservation of&lt;br /&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;angular momentum in the orbit around a Jupiter-size planet, as it clears its neighborhood. (Source: &lt;a href=&quot;http://www.maths.qmul.ac.uk/~masset/moviesmpegs.html&quot;&gt;Frédéric Masset&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One nice description for this angular momentum preservation, and why the planets appear to rotate faster than their surrounding protoplanetary disk goes like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Conservation of angular momentum explains why an ice skater spins more&#xA;  rapidly as she pulls her arms in. As her arms come closer to her axis&#xA;  of rotation, her [rotation] speed increases and her angular momentum remains the&#xA;  same. Similarly, her rotation slows when she extends her arms at the&#xA;  conclusion of the spin.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;sup&gt;Source: &lt;a href=&quot;http://www.scientificamerican.com/article.cfm?id=why-and-how-do-planets-ro&quot;&gt;Scientific American article on Why and how do planets rotate?&lt;/a&gt; (George Spagna)&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So it could be described as this axial rotation of planets resulting in conservation of the angular momentum of the materials in the protoplanetary disk, forming during the accretion period of the planetary system as the protoplanets &lt;em&gt;gain in weight&lt;/em&gt;, and preserve this angular momentum due to inertia of their radial velocity.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2013-09-29T13:48:23.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="399" PostTypeId="2" ParentId="310" CreationDate="2013-09-29T21:12:02.753" Score="2" Body="&lt;p&gt;There are several ways to create mosaics. The most widely used by amateurs is probably Photoshop or &lt;a href=&quot;http://www.gimp.org/&quot; rel=&quot;nofollow&quot;&gt;GIMP&lt;/a&gt;. There are several tutorials on the web, and example of which is one on &lt;a href=&quot;http://www.astrophotography-tonight.com/mosaic-photoshop-astrophotography/&quot; rel=&quot;nofollow&quot;&gt;Astrophotography Tonight&lt;/a&gt;. Another good tutorial can be found on the &lt;a href=&quot;http://starizona.com/acb/ccd/advimagingmosaic.aspx&quot; rel=&quot;nofollow&quot;&gt;Starizona&lt;/a&gt; web page. There is also a software called &lt;a href=&quot;http://jaggedplanet.com/iMerge.html&quot; rel=&quot;nofollow&quot;&gt;iMerge&lt;/a&gt; and another called &lt;a href=&quot;http://montage.ipac.caltech.edu/&quot; rel=&quot;nofollow&quot;&gt;Montage&lt;/a&gt; that are worth looking at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also many panoramic image programs but they mostly work by distorting the images to make them fit. This may render nice photographs but they will of course not be geometrically correct.&lt;/p&gt;&#xA;" OwnerUserId="47" LastActivityDate="2013-09-29T21:12:02.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="400" PostTypeId="1" AcceptedAnswerId="495" CreationDate="2013-09-29T21:26:48.603" Score="16" ViewCount="665" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/TrES-2b&quot;&gt;The exoplanet TrES-2b is known as the darkest planet ever found, reflecting &amp;lt;1% of the light that hits it.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does it mean in respect to the surface? What would we see below the atmosphere of this planet? Is it a &quot;shadow planet&quot; with an atmosphere so thick and dense that it blocks all light that would reach the surface?&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="62" LastEditDate="2013-09-30T00:31:46.633" LastActivityDate="2013-10-06T07:44:23.250" Title="Is the surface of TrES-2b actually dark?" Tags="&lt;exoplanet&gt;&lt;light&gt;&lt;surface&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="401" PostTypeId="1" AcceptedAnswerId="415" CreationDate="2013-09-29T22:37:35.427" Score="18" ViewCount="853" Body="&lt;p&gt;It is well established that one main feature of many hot jupiters is their close proximity to their parent star, usually the equivalent of being within the orbit of Mercury.  So, these planets are gas giants and are very hot (hence their category).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a few discoveries have lead to questioning about what is the fate of these planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;example 1: HD 209458b  a.k.a. &quot;Osiris&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the NASA page &lt;a href=&quot;http://www.nasa.gov/vision/universe/newworlds/Osiris_leaks.html&quot;&gt;&quot;Dying Planet Leaks Carbon-Oxygen&quot;&lt;/a&gt;, Osiris is doing more than 'evaporating', it is leaking carbon, oxygen alongside hydrogen in an envelope behind the planet that has been detected from Earth.  The significance of carbon and oxygen is stated from the article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Although carbon and oxygen have been observed on Jupiter and Saturn, it is always in combined form as methane and water deep in the atmosphere. In HD 209458b the chemicals are broken down into the basic elements. But on Jupiter or Saturn, even as elements, they would still remain invisible low in the atmosphere. The fact that they are visible in the upper atmosphere of HD 209458b confirms that atmospheric 'blow off' is occurring.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is stated in the article that Osiris is likely to become a hypothesised class of exoplanet known as a &lt;strong&gt;Chthonian&lt;/strong&gt;, which is defined in &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0312384&quot;&gt;&quot;Evaporation rate of hot Jupiters and formation of Chthonian planets&quot;&lt;/a&gt; (Hebard et al. 2003) as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;new class of planets made of the residual central core of former hot&#xA;  Jupiters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These would be similar in size to Earth, but considerably denser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;example 2: CoRoT-7b&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the NASA article &lt;a href=&quot;http://www.nasa.gov/centers/goddard/news/topstory/2010/earthlike-exoplanet.html&quot;&gt;&quot;Most Earthlike Exoplanet Started out as Gas Giant&quot;&lt;/a&gt;, CoRoT-7b is an Earth-sized planet where a hot Jupiter usually is found, they describe it as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is almost 60 times closer to its star than Earth, so the star appears almost 360 times larger than the sun does in our sky,&quot; Jackson said. As a consequence, the planet's surface experiences extreme heating that may reach 3,600 degrees Fahrenheit on the daylight side. CoRoT-7b's size (70 percent larger than Earth) and mass (4.8 times Earth's) indicate that the world is probably made of rocky materials. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The high day time temperature means that the star-facing side of the planet is likely to be molten, any tenuous atmosphere is also blasted away.  Scientists estimate that many Earth-masses may have been boiled off.  It also seems that the decreasing mass is causing the planet to be drawn closer to the star  -cause more material to be boiled off, hence the mass to decrease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarise on of the scientists in the article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You could say that, one way or the other, this planet is disappearing before our eyes,&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As these are only 2 examples of a possible process, the question is, what is the current accepted theory as to the fate of hot Jupiter exoplanets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could this also be the reason a hot Jupiter does not exist in our solar system?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-09-30T00:38:38.947" LastActivityDate="2013-09-30T16:19:26.490" Title="What is the current accepted theory for the fate of hot Jupiters?" Tags="&lt;exoplanet&gt;&lt;hot-jupiter&gt;&lt;chthonian&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="402" PostTypeId="5" CreationDate="2013-09-29T22:46:40.250" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-29T22:46:40.250" LastActivityDate="2013-09-29T22:46:40.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="403" PostTypeId="4" CreationDate="2013-09-29T22:46:40.250" Score="0" Body="Questions regarding chthonians, a hypothesised class of exoplanets that are the remnant cores of hot Jupiters." OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-09-30T06:21:14.953" LastActivityDate="2013-09-30T06:21:14.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="404" PostTypeId="5" CreationDate="2013-09-30T01:43:19.770" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-30T01:43:19.770" LastActivityDate="2013-09-30T01:43:19.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="405" PostTypeId="4" CreationDate="2013-09-30T01:43:19.770" Score="0" Body="Questions about gas-giant planets in very close orbit to their parent star (closer than Mercury)." OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-09-30T06:21:29.297" LastActivityDate="2013-09-30T06:21:29.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="406" PostTypeId="5" CreationDate="2013-09-30T02:03:41.440" Score="0" Body="&lt;p&gt;The &lt;strong&gt;&lt;a href=&quot;https://simple.wikipedia.org/wiki/Density&quot; rel=&quot;nofollow noreferrer&quot;&gt;density&lt;/a&gt;&lt;/strong&gt; is usually abbreviated as &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt;, and with mass &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; and volume &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt;, it is defined as &lt;span class=&quot;math-container&quot;&gt;$\rho ={\frac {m}{V}}$&lt;/span&gt;. Common SI units for density are &lt;span class=&quot;math-container&quot;&gt;$g/cm^3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$kg/m^3$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;In general, density can be changed by changing either the pressure or the temperature.&lt;/p&gt;&#xA;" OwnerUserId="37191" LastEditorUserId="37191" LastEditDate="2021-02-15T23:20:34.880" LastActivityDate="2021-02-15T23:20:34.880" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="407" PostTypeId="4" CreationDate="2013-09-30T02:03:41.440" Score="0" Body="Questions on the ratio of an object's mass with its volume." OwnerDisplayName="user8" LastEditorUserId="37191" LastEditorDisplayName="user8" LastEditDate="2021-02-15T23:20:37.593" LastActivityDate="2021-02-15T23:20:37.593" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="408" PostTypeId="1" AcceptedAnswerId="411" CreationDate="2013-09-30T02:29:01.760" Score="14" ViewCount="431" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gmAeC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.universetoday.com/30308/sombrero-galaxy/&quot;&gt;The Sombrero Galaxy&lt;/a&gt; contains a rather peculiar ring of dust orbiting it (seen as the dark ring on the outer edge). What is the prevalent element in this dust? Carbon?&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2013-09-30T05:19:47.287" Title="What is the predominant element in the dust of the Sombrero Galaxy?" Tags="&lt;sombrero-galaxy&gt;&lt;dust&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="409" PostTypeId="5" CreationDate="2013-09-30T02:38:10.053" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-30T02:38:10.053" LastActivityDate="2013-09-30T02:38:10.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="410" PostTypeId="4" CreationDate="2013-09-30T02:38:10.053" Score="0" Body="Questions regarding the Sombrero Galaxy, also known as M104 or NGC 4594. It is an unbarred spiral galaxy, part of the Virgo constellation." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-09-30T06:21:41.487" LastActivityDate="2013-09-30T06:21:41.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="411" PostTypeId="2" ParentId="408" CreationDate="2013-09-30T05:19:47.287" Score="10" Body="&lt;p&gt;I think that first, we have to properly appreciate the size of the Sombrero Galaxy. It is roughly 50,000 light years (15 kilo parsecs) in diameter. That might be only half of the diameter of our own Milky Way galaxy, but still makes each and every pixel on the photograph you're attaching in your question stretch more than 100 light years in distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a huge distance, so we can probably agree that clouds made of nearly anything and at least as dense as what you'd on average expect the interstellar medium to be comprised of, would obscure even the strongests light sources within that galaxy. This is important to realize, so we somewhat lower your expectations; These areas aren't tightly packed with anything, we're just talking of such vast distances that the light has to travel through, that even almost nothing will be enough to stop it from shining through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to your question what it is comprised of. It is in essence matter from the interstellar medium that stops this light from reaching us, so nothing in particular that you wouldn't expect also in similar areas between stars within our Milky Way, or indeed any other galaxy that is roughly in the same age period of its development. Since the Sombrero Galaxy isn't even all that far away in cosmological sense, &lt;em&gt;merely&lt;/em&gt; around 29 million light years, it has an apparent magnitude of ~ 9 and is a rather good target for observations of even hobbyist astronomers with enthusiast telescopes. More importantly though, you aren't looking that far into the past. Again, in the cosmological sense. For comparison, our Sun is roughly 4.5 billion years old, and we're observing the Sombrero Galaxy as it was 29 million years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is this interstellar medium made of? On average, it is comprised of mostly gas clouds of hydrogen and helium forming during the primordial nucleosynthesis (read: Big Bang) and at most a few percent of heavier trace elements, some also in gas form and an itsy-bitsy percentage of these would combine into even heavier dust particles, mostly made of carbon, silicon, and oxygen (source for the interplanetary dust composition: &lt;a href=&quot;http://apod.nasa.gov/apod/ap030706.html&quot;&gt;NASA APOD&lt;/a&gt;). Remember, when I say &lt;em&gt;clouds&lt;/em&gt; and &lt;em&gt;dust&lt;/em&gt;, they are incredibly thin. So thin, you'd have to be traveling at a significant portion of C to even notice it much, unless you're looking for it with extremely sensitive equipment, like Voyager 1 and 2 do, for example. But even with such low particle density, some regions would be slightly denser, and some slightly thinner. And that is what creates that majestic dust lane of the Sombrero Galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are interstellar medium, mostly made of primordial hydrogen and helium, and dust made of carbon, silicon, oxygen and other trace elements, and all of it hundreds of light years in length and obscuring light from the galaxy's stars to penetrate through at varying levels, depending on background light strength, proximity, and local interstellar medium cloud density. The color of the cloud says nothing of its composition. It is mostly black with parts of it shifting towards red hues due to molecules of gas and dust in the interstellar medium scattering blue light from the source of light more than they scatter red light. I.e. the same reason why the Sun or the Moon appear more red when they're low on the horizon.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2013-09-30T05:19:47.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="412" PostTypeId="2" ParentId="362" CreationDate="2013-09-30T05:57:41.223" Score="8" Body="&lt;p&gt;The most-widely accepted hypothesis at the moment is that Mercury was struck by a large impactor that removed a significant fraction of its mantle (I believe this theory was originally proposed by &lt;a href=&quot;http://adsabs.harvard.edu/abs/1987LPI....18..151C&quot;&gt;Cameron &amp;amp; Benz in 1987&lt;/a&gt;, and the qualitative theory hasn't changed very much). For planets that are close to their parent stars (such as Mercury), the collision with the secondary body likely occurs at high velocity due to the rapid orbit speeds there (&gt; 40 km/s). As this is much larger than a typical rocky planet's escape velocity (~10 km/s), material is lost from the planet in these collisions. This collision is also more likely to be &quot;grazing,&quot; as &quot;direct hits&quot; are rare, and thus the outer parts of the planet are preferentially removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the mantles of rocky planets are primarily composed of silicates, which have a lower density than iron that resides in their cores, the collision results in a surviving planet with a higher average density. &lt;/p&gt;&#xA;" OwnerUserId="2" LastActivityDate="2013-09-30T05:57:41.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="413" PostTypeId="2" ParentId="334" CreationDate="2013-09-30T06:51:09.050" Score="11" Body="&lt;p&gt;Tidal amplitudes roughly scale inversely with distance to the fourth power (this comes from the tidal force scaling as $r^{-3}$ and the Earth's potential scaling as $r^{-1}$). At its formation, the Moon &lt;a href=&quot;http://adsabs.harvard.edu/abs/2004ARA&amp;amp;A..42..441C&quot; rel=&quot;nofollow&quot;&gt;was approximately seven Earth radii away from Earth&lt;/a&gt;, or about ten times closer than it is today. Thus, the amplitude of the tide on Earth would have been approximately $10^4$ times larger than it is today. As the typical ocean tide is 1&amp;nbsp;m, this means ocean tides would have been 10&amp;nbsp;km high on Earth! Of course, the collision that formed the moon turned the surface of the Earth molten, so really we're talking about a 10&amp;nbsp;km tall molten tide, which is far less scary...(kidding).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine all the noise the ocean makes when it crashes on a beach. That noise requires energy to produce, and is partially driven by the action of raising/lowering the ocean via tides. Now imagine how loud a 10 km wave crashing would be. This is why the Moon receded quite quickly from the Earth shortly after forming, the tides enable the dissipation of a &lt;strong&gt;lot&lt;/strong&gt; of orbital energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the hypothetical effect of this supertide on the present-day Earth's weather, the Earth's &lt;a href=&quot;http://en.wikipedia.org/wiki/Troposphere&quot; rel=&quot;nofollow&quot;&gt;troposphere&lt;/a&gt; (which contains pretty much all of the Earth's weather phenomena) extends 10&amp;ndash;20&amp;nbsp;km from the ground. Considering the height of the ground is changing by a similar amount due to the tide, I would imagine the effects on the weather would be quite dramatic.&lt;/p&gt;&#xA;" OwnerUserId="2" LastEditorUserId="1501" LastEditDate="2014-12-19T16:39:05.583" LastActivityDate="2014-12-19T16:39:05.583" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="414" PostTypeId="2" ParentId="331" CreationDate="2013-09-30T09:06:08.357" Score="19" Body="&lt;p&gt;The easiest way to determine the magnitude of a given star is probably to use the Pogson relation. The idea is to determine the magnitude of a star knowing the magnitude of a reference star; it is thus quite easy, using a well-known reference as Vega or Sirius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Pogson relation is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_1-m_2=-2.5\ log\ \left({\frac{E_1}{E_2}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m_1$ and $m_2$ are the magnitudes of star 1 and star 2 (your reference star), and $E_1$ and $E_2$ brightness (that could be in arbitrary units, which is a good news for you).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a CCD detector the brightness is easy to determine since it's just the flux of a pixel. You don't have to convert it in physical units (as Watt per square meter) since you have a ratio $\left({\frac{E_1}{E_2}}\right)$, so the raw intensity is enough to determine the magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for any other measurement with a CCD camera, you have of course to determine the bias, dark and flat-field and to reduce your data before estimating any magnitude. The precision will depend on the number of observations performed and, as always, it is better to perform several short exposure images than a single long exposure one.&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="13" LastEditDate="2013-12-03T06:53:50.777" LastActivityDate="2013-12-03T06:53:50.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="415" PostTypeId="2" ParentId="401" CreationDate="2013-09-30T14:33:10.630" Score="12" Body="&lt;p&gt;This is a fairly loaded question in that it depends heavily on what a &quot;hot Jupiter&quot; actually is defined to be. What is &quot;hot&quot;? What is a &quot;Jupiter&quot;? In reality, there's a continuum of planetary masses and distances from their parent star, and in the literature you'll commonly see references to &quot;hot Neptunes&quot;, &quot;hot Saturns,&quot; etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The predominant theory as to how giant planets form is that they first coalesce from rock and ice beyond the &lt;a href=&quot;http://en.wikipedia.org/wiki/Frost_line_%28astrophysics%29&quot;&gt;ice line&lt;/a&gt;, the distance from the parent star at which water becomes solid. This distance is approximately where Mars lies today in our solar system. What's surprising about &quot;hot gas planets&quot; is that they are found &lt;em&gt;within&lt;/em&gt; this ice line, significantly within. This implies that after they formed their cores, they migrated closer to their host stars via some currently undetermined process (for which there are several good candidates, but for now let's assume that the existence of hot planets shows that at least one of these processes operates quite regularly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what about the word &quot;hot&quot;? Well, for the planets that are closest to their parent stars, there is known to be a &lt;a href=&quot;http://oklo.org/2010/07/31/radius-anomalies-2/&quot;&gt;radius anomaly&lt;/a&gt;: The radii of these planets are significantly larger than models of giant planet structure irradiated by their host stars would predict. So I would define &quot;hot&quot; planets as gas giants whose radii are larger than what would be predicted by the standard models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we got some of the definitions out of the way, there's the question of survival. When giant planets are close to their parent stars, they become &lt;a href=&quot;http://en.wikipedia.org/wiki/Tidal_locking&quot;&gt;tidally locked&lt;/a&gt;. As a consequence, there is very little energy tidally dissipated on the surface of the giant planet, the shape of the planet is fixed and there are little internal motions. However, the giant planet &lt;em&gt;also&lt;/em&gt; raises a tide on its host star as well, and because it takes a lot of angular momentum to change the spin of an object with 1,000 times more mass, the host stars are almost never going to be tidally locked to their closest planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate at which energy is dissipated within the star is highly uncertain, and this uncertainty is typically swept into a fudge parameter &quot;Q,&quot; the quality factor, with lower quality factors reflecting more dissipation. &quot;Q&quot; is measured for certain bodies in our own solar system (i.e. Earth and Jupiter) and in some stellar binaries, but is highly variable from body to body, ranging from about 10 for the Earth to 10^8 for some stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether a planet survives to be observed today depends on how long the orbital decay time, which is determined by Q, compares to the age of the system. For some systems, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/WASP-12b&quot;&gt;WASP-12b&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/WASP-19b&quot;&gt;WASP-19b&lt;/a&gt;, which feature highly inflated hot Jupiters, Q is estimated to be small enough to cause them to fall into their host stars in a surprisingly short time (&amp;lt; 10^7 years).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One other possibility is that the gas surrounding the rock/ice core is blasted away by the tremendous amount of heat deposited into the planet. This leaves you with a relatively low-density planet that's somewhat devoid of iron, as the cores of giant planets form further from their host stars than the rocky planets. There are a few candidate close-in, Neptune-mass objects that may have been produced as a result of them losing the bulk of their atmospheres in this way (Example: &lt;a href=&quot;http://arxiv.org/abs/1301.6555&quot;&gt;GJ3470b&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for our own solar system, the formation of a hot Jupiter would have likely destroyed the inner solar system as it migrated close to the Sun, owing to the fact that it would violently perturb the inner planets' orbits. Additionally, the Sun would likely be enhanced in metals owing to the accretion of metal-rich material from this giant planet. While it's potentially possible that there was a hot Jupiter in our solar system before the other planets formed, it currently seems unlikely.&lt;/p&gt;&#xA;" OwnerUserId="2" LastEditorUserId="2" LastEditDate="2013-09-30T16:19:26.490" LastActivityDate="2013-09-30T16:19:26.490" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="416" PostTypeId="5" CreationDate="2013-09-30T16:05:37.313" Score="0" Body="&lt;p&gt;Supermassive black holes are a special type of black holes with masses in many millions of times that of the Sun. Their exact formation is unknown. They typically reside at the center of galaxies; our galaxy, the Milky Way, has a supermassive black hole at its center, Sagittarius A*.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="2153" LastEditDate="2015-04-12T22:03:04.333" LastActivityDate="2015-04-12T22:03:04.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="417" PostTypeId="4" CreationDate="2013-09-30T16:05:37.313" Score="0" Body="Questions regarding the most massive examples of black holes that typically reside at the centers of galaxies." OwnerUserId="2" LastEditorUserId="2153" LastEditDate="2015-04-12T22:03:04.333" LastActivityDate="2015-04-12T22:03:04.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="418" PostTypeId="1" AcceptedAnswerId="425" CreationDate="2013-09-30T17:05:48.710" Score="19" ViewCount="630" Body="&lt;p&gt;Obviously, I am not referring to actual viewing of the exoplanets themselves, but detecting their effects on the brightness of the light emitted from the parent star (as in the diagram below from &lt;a href=&quot;http://ifa.hawaii.edu/newsletters/article.cfm?a=407&amp;amp;n=34&quot; rel=&quot;noreferrer&quot;&gt;The Institute of Astronomy, University of Hawaii&lt;/a&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nX7hD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would imagine that a good quality telescope, would be about to detect and record the effects of the exoplanet transiting the parent star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What practical considerations in terms of equipment, software etc. would be required for 'backyard' amateur observations of the effects of transiting exoplanets? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If anyone has actually tried this, what has been your experience?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-09-30T17:21:50.107" LastActivityDate="2015-09-14T21:54:22.943" Title="What practical considerations are there for amateur observations of transiting exoplanets?" Tags="&lt;exoplanet&gt;&lt;amateur-observing&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="419" PostTypeId="2" ParentId="266" CreationDate="2013-09-30T18:25:38.650" Score="3" Body="&lt;p&gt;A shooting star is a rock from space that is entering the atmosphere at such a great speed that the air superheats it to a bright white-hot glow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When one of these rocks is floating around in space, it's called a &lt;strong&gt;meteoroid&lt;/strong&gt; (&lt;em&gt;-oid&lt;/em&gt; sounds like some sort of space-y future-y thing: think asteroid, android, humanoid). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it's found in the ground after it's already landed, it's called a &lt;strong&gt;meteorite&lt;/strong&gt; (&lt;em&gt;-ite&lt;/em&gt; sounds like a mineral-y rock-type thing you find in the ground: think pyrite, graphite, kryptonite).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the brief period of time between when they are meteoroids and meteorites, they are simply called &lt;strong&gt;meteors&lt;/strong&gt;, a.k.a. shooting stars. (Think: &lt;em&gt;Aaahh! ...look out for that METEOR!&lt;/em&gt;)&lt;/p&gt;&#xA;" OwnerUserId="158" LastActivityDate="2013-09-30T18:25:38.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="420" PostTypeId="1" AcceptedAnswerId="8621" CreationDate="2013-09-30T18:34:39.193" Score="18" ViewCount="4164" Body="&lt;p&gt;After reading the Q&amp;amp;A &lt;a href=&quot;https://astronomy.stackexchange.com/questions/131/is-the-moon-moving-further-away-from-earth-and-closer-to-the-sun-why&quot;&gt;Is the moon moving further away from Earth and closer to the Sun? Why?&lt;/a&gt; about the tides transferring energy to the Moon and pushing it from Earth, I have a question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is that energy actually being transferred to the Moon? The creation of tides requires energy, so I would expect that this should take energy from the Moon, slow it and cause it to eventually fall back to the Earth. Why isn't that happening?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if this is the general mechanism, would other moons that orbit around planets with a liquid surface and causing tides, be receding from their parent planets?&lt;/p&gt;&#xA;" OwnerUserId="42" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-01-20T21:08:25.463" Title="Why is the Moon receding from the Earth due to tides? Is this typical for other moons?" Tags="&lt;the-moon&gt;&lt;orbit&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="421" PostTypeId="1" CreationDate="2013-09-30T18:57:52.243" Score="12" ViewCount="255" Body="&lt;p&gt;In his answer to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/408/what-is-the-predominant-element-in-the-dust-of-the-sombrero-galaxy&quot;&gt;this question&lt;/a&gt;, TildalWave made this remark:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I think that first, we have to properly appreciate the size of the Sombrero Galaxy. It is roughly 50,000 light years (15 kilo parsecs) in diameter. That might be only half of the diameter of our own Milky Way galaxy, but still makes &lt;strong&gt;each and every pixel on the photograph you're attaching in your question stretch more than 100 light years in distance.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's the photograph:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gmAeC.jpg&quot; alt=&quot;sombrero galaxy&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it almost feels like I can see individual 'pieces' of dust in the lane, it would follow that the particles must be &lt;em&gt;very&lt;/em&gt; far apart. Just how far apart are they? Are they grouped in clumps, or relatively evenly space?&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-05-08T09:40:10.993" Title="How far apart is the dust in the Sombrero Galaxy's dust lane?" Tags="&lt;sombrero-galaxy&gt;&lt;dust&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="422" PostTypeId="2" ParentId="420" CreationDate="2013-09-30T19:07:51.813" Score="10" Body="&lt;p&gt;You have correctly identified that the tidal forces are transferring energy from the Earth to the Moon. This energy causes the Moon's orbit to get larger thus slowing it down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a bit counter intuitive, but if you think about it the Earth spins a rate of 1 spin per day The Moon is orbiting the Earth with a period of approximately 27.3 days. If it were to speed up it's orbit would actually decrease bringing it closer to the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your final point all other moons cause tides on their parent planets and are moving away from them, but the effects are much smaller due to the larger difference in sizes. The Earth/Moon system is unique in the Solar System as the ratio of the sizes of the bodies are relatively close to each other.&lt;/p&gt;&#xA;" OwnerUserId="111" LastEditorUserId="111" LastEditDate="2013-09-30T19:47:51.800" LastActivityDate="2013-09-30T19:47:51.800" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="423" PostTypeId="1" AcceptedAnswerId="427" CreationDate="2013-09-30T19:17:31.503" Score="8" ViewCount="648" Body="&lt;p&gt;In February, &lt;a href=&quot;http://www.youtube.com/watch?v=Rp5eg_lk8zA&quot; rel=&quot;noreferrer&quot;&gt;a meteor entered the Earth's atmosphere over Chelyabinsk, Russia&lt;/a&gt; and exploded, producing a shockwave blast that damaged hundreds of buildings around the explosion site in Russia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why did it explode? Why didn't it simply burn instead?  &lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="13" LastEditDate="2013-09-30T20:21:36.213" LastActivityDate="2013-10-01T06:02:19.583" Title="Why did the Chelyabinsk meteor explode?" Tags="&lt;planetary-atmosphere&gt;&lt;meteor&gt;&lt;explosion&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="424" PostTypeId="1" AcceptedAnswerId="426" CreationDate="2013-09-30T19:21:27.070" Score="8" ViewCount="496" Body="&lt;p&gt;Even though Pluto is not known as a planet anymore, theroetically it has/had a moon, called Charon. I've heard about something that their size are so close to each other that while Charon rotates around Pluto, it also rotates around a central point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does it work exactly?&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="2618" LastEditDate="2017-03-07T01:11:27.073" LastActivityDate="2017-03-07T01:11:27.073" Title="Effect of Charon on Pluto" Tags="&lt;solar-system&gt;&lt;natural-satellites&gt;&lt;pluto&gt;&lt;charon&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="425" PostTypeId="2" ParentId="418" CreationDate="2013-09-30T19:39:26.300" Score="12" Body="&lt;p&gt;This is actually quite straightforward with digital CCD's (it used to be quite tricky with film cameras as you'd have to carefully develop film that moved past the lens and assess the width of the trail)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Get yourself a good telescope - a 12&quot; Dobsonian or above if you want to give yourself a good chance of picking out the fluctuations against the noise background. Then select a decent CCD. Five hundered pounds gets a reasonable one, but expect to pay a couple of thousand pounds for a cooled CCD, which will also help to reduce noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Buying in US dollars?&lt;/em&gt; A reasonable CCD runs about $1000. A cooled CCD will cost you &lt;em&gt;at least&lt;/em&gt; $1500.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll want a good quality equatorial mount, with computer controlled servos to track the target smoothly over long periods of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally you will also slave a second telescope and CCD, pointed along the same path but slightly off - this will help you cancel out cloud and other fluctuations from our own atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, and get as far away from a city as possible - up into the mountains can be a good plan :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then arrange your viewing for a series of full nights. The more data points you can get, the better the noise reduction. Imagine the exoplanet is orbiting every 100 days, in order to get any useful data, you will need to track it over some multiple of 100 days. So assume you set up to track your target star for 2 years, plan for 3 or 4 star shots per night to give you a range of data points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These 600+ days of 4 data points per night gives you a reasonable stack of data - the challenge now is to work out whether there are any cyclic variations. Various data analysis tools can do this for you. As a first step, if you find a cycle around 365 days, it probably isn't the target, so try and normalise for this (of course this will make it very difficult to discover exoplanets with a period of exactly 1 year)&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="6" LastEditDate="2013-10-01T14:28:35.623" LastActivityDate="2013-10-01T14:28:35.623" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="426" PostTypeId="2" ParentId="424" CreationDate="2013-09-30T19:39:52.553" Score="9" Body="&lt;p&gt;Mathematically, the motion of the Pluto-Charon system can be decomposed into two parts: The motion of Pluto-Charon about the Sun, and the motion of Pluto and Charon about one another. If one sets the reference point to be the center of Pluto, the path the Pluto-Charon system would appear to follow about the Sun would be an &lt;a href=&quot;http://en.wikipedia.org/wiki/Deferent_and_epicycle&quot; rel=&quot;nofollow noreferrer&quot;&gt;epicycle&lt;/a&gt;, which is a far more complicated means of describing the pair's trajectory through space. By instead setting this reference point to be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Barycentric_coordinates_%28astronomy%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;barycenter&lt;/a&gt;, the trajectory followed by the Pluto-Charon system is conveniently a Keplerian ellipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Charon is 11.6% the mass of Pluto, and is (on average) 19,571 km away from Pluto (Source &lt;a href=&quot;http://en.wikipedia.org/wiki/Pluto&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Charon_%28moon%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;). The barycenter is the point at which the masses of the two bodies &quot;balance&quot;, and for the Pluto-Charon system it lies at a distance of $$\frac{m_{\text{Charon}}}{m_{\text{Pluto}}+m_{\text{Charon}}}\times\text{distance}=\frac{0.116}{1.116}\times\text{distance}=0.104\times 19,571\,\text{km}=2,034\,\text{km}$$ from the center of Pluto. As Pluto is only 1,153 km in radius, the barycenter lies ~900 km above its surface. The Pluto-Charon system is the only known (minor or major) planet for which this is the case, although the Earth-Moon system will likely satisfy this criteria billions of years in the future (see &lt;a href=&quot;https://physics.stackexchange.com/questions/26304/is-the-moon-a-planet&quot;&gt;Is the Moon a Planet?&lt;/a&gt; on Physics SE). There are some instances of &lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_asteroid&quot; rel=&quot;nofollow noreferrer&quot;&gt;binary asteroids&lt;/a&gt;, however, for which the barycenter lies outside the surface of both bodies.&lt;/p&gt;&#xA;" OwnerUserId="2" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2013-10-22T03:28:40.917" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="427" PostTypeId="2" ParentId="423" CreationDate="2013-09-30T19:47:36.920" Score="9" Body="&lt;p&gt;Same reason as the explosion at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tunguska_event&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tunguska event&lt;/a&gt; years ago: Pressure, due to boiling water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meteors usually cme into our atmosphere at incredible speeds. At these speeds, the meteor is highly affected by the viscosity of air and is heated by compression (and a little bit of friction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This starts boiling the ice present in the meteor, leading to a pressure buildup. If the conditions are right, it can just explode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also: &lt;a href=&quot;http://www.nature.com/nature/journal/v361/n6407/abs/361040a0.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.nature.com/nature/journal/v361/n6407/abs/361040a0.html&lt;/a&gt; , &lt;a href=&quot;https://physics.stackexchange.com/q/76045/7433&quot;&gt;https://physics.stackexchange.com/q/76045/7433&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2013-10-01T06:02:19.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="428" PostTypeId="2" ParentId="423" CreationDate="2013-09-30T19:49:56.360" Score="7" Body="&lt;p&gt;The meteorite decelerated from perhaps 30000 mph through friction and impact with the atmosphere, and this made it heat up to an incredible temperature. At this point the rock was boiling and burning, and liquids within it , such as water, became gaseous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some small explosions cast off burning pieces of the meteorite, and the major explosion was from the larger chunk being stressed beyond breaking point through pressure from the gases within.&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="43" LastEditDate="2013-09-30T19:59:59.467" LastActivityDate="2013-09-30T19:59:59.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="429" PostTypeId="1" CreationDate="2013-09-30T23:58:26.230" Score="15" ViewCount="607" Body="&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Dyson_sphere&quot; rel=&quot;noreferrer&quot;&gt;Dyson sphere/swarm/ring&lt;/a&gt; is a hypothetical structure an extra-terrestrial entity would construct to collect a large fraction of its host star's light, and would likely generate a fairly strong infrared signature that could potentially be detected by modern infrared surveys (such as &lt;a href=&quot;http://www.nasa.gov/mission_pages/WISE/main/index.html&quot; rel=&quot;noreferrer&quot;&gt;WISE&lt;/a&gt;). Do we have any meaningful constraints on the existence of these structures within the Milky Way (or beyond) from these surveys? What sorts of natural phenomena could be confused with such structures, if they existed?&lt;/p&gt;&#xA;" OwnerUserId="2" LastEditorUserId="2153" LastEditDate="2016-11-04T15:13:41.937" LastActivityDate="2016-11-04T15:13:41.937" Title="What are the current observational limits on the existence of Dyson spheres/swarms/rings?" Tags="&lt;observational-astronomy&gt;&lt;extra-terrestrial&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="430" PostTypeId="1" AcceptedAnswerId="1155" CreationDate="2013-10-01T00:06:31.483" Score="18" ViewCount="575" Body="&lt;p&gt;Nemesis is a hypothetical companion to the Sun on a very eccentric, long-period orbit. The star supposedly returns every few tens of millions years, driving comets into the inner solar system and causing extinction events. Given our very stringent observational limits from infrared surveys (such as WISE), is its existence definitively ruled out?&lt;/p&gt;&#xA;" OwnerUserId="2" LastEditorUserId="20348" LastEditDate="2019-09-26T09:22:43.443" LastActivityDate="2019-09-26T09:22:43.443" Title="What are the current observational constraints on the existence of Nemesis?" Tags="&lt;solar-system&gt;&lt;binary-star&gt;&lt;nemesis&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="431" PostTypeId="1" CreationDate="2013-10-01T00:09:34.760" Score="15" ViewCount="277" Body="&lt;p&gt;An object approximately the same size as Pluto, Eris, was discovered only 8 years ago (in 2005). Are there any Pluto-sized objects remaining to be discovered, and if so, how far away from the Sun would they have to be to not have been detected already?&lt;/p&gt;&#xA;" OwnerUserId="2" LastActivityDate="2020-11-14T09:39:37.867" Title="Are any Pluto-sized objects remaining to be discovered in the Kuiper Belt?" Tags="&lt;solar-system&gt;&lt;kuiper-belt&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="432" PostTypeId="1" CreationDate="2013-10-01T00:42:11.570" Score="13" ViewCount="146" Body="&lt;p&gt;The &lt;strong&gt;Late Heavy Bombardment&lt;/strong&gt; (LHB) or the &lt;strong&gt;Lunar Cataclysm&lt;/strong&gt; is when the inner solar system, including the Earth and the Moon, underwent multiple and sustained heavy impacts early in their history (around 3.7-4 billion years ago). This event formed many of the major impact scars we see on the Moon, (presumably as well as much of the impacts observed on Mars and Mercury). Two main theories are based on perturbations in either the Kuiper Belt, or in a primordial larger asteroid belt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What implications would a younger Earth and Moon, discussed in &lt;a href=&quot;https://www.llnl.gov/news/newsreleases/2011/Aug/NR-11-08-02.html&quot;&gt;&quot;Moon and Earth may be younger than originally thought&quot;&lt;/a&gt; (Lars Borg et al., 2011) have on hypotheses on what caused the LHB?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorUserId="13" LastEditDate="2013-10-01T05:40:53.813" LastActivityDate="2014-01-28T20:38:13.230" Title="What implications do younger Earth and Moon have on Late Heavy Bombardment genesis hypotheses?" Tags="&lt;solar-system&gt;&lt;terrestrial-planets&gt;&lt;lhb&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="433" PostTypeId="2" ParentId="421" CreationDate="2013-10-01T01:00:11.090" Score="1" Body="&lt;p&gt;Based on &lt;a href=&quot;http://articles.adsabs.harvard.edu/full/1995A&amp;amp;A...303..673E/0000673.000.html&quot; rel=&quot;nofollow&quot;&gt;1995A&amp;amp;A...303..673E&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How far apart they are?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No idea. Somebody else might help to provide a density distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are they grouped in clumps or relatively even spaced?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I really wanted to answer. They are in a concentric ring-like structure around the galaxy. The maximum optical depth is 0.8 with a mean value of 0.3. By the way, optical depth gives you a nice idea of how far apart they might be, not a good density distribution but you can get the physical picture in mind.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2013-10-01T01:00:11.090" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="435" PostTypeId="5" CreationDate="2013-10-01T05:59:55.997" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-01T05:59:55.997" LastActivityDate="2013-10-01T05:59:55.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="436" PostTypeId="4" CreationDate="2013-10-01T05:59:55.997" Score="0" Body="The near-vacuum extending between the planets and stars, containing small amounts of gas and dust. Also called outer space to refer to the physical universe beyond the Earth's atmosphere." OwnerUserId="150" LastEditorUserId="150" LastEditDate="2013-10-01T06:19:31.397" LastActivityDate="2013-10-01T06:19:31.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="437" PostTypeId="5" CreationDate="2013-10-01T06:01:19.553" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-01T06:01:19.553" LastActivityDate="2013-10-01T06:01:19.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="438" PostTypeId="4" CreationDate="2013-10-01T06:01:19.553" Score="0" Body="A region of space devoid of any fluid molecules resulting in a zero pressure in ideal cases. In practice, even space has a small but finite number of molecules. " OwnerUserId="150" LastEditorUserId="150" LastEditDate="2013-10-01T06:19:21.303" LastActivityDate="2013-10-01T06:19:21.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="439" PostTypeId="1" AcceptedAnswerId="441" CreationDate="2013-10-01T10:14:07.360" Score="11" ViewCount="107" Body="&lt;p&gt;This question is somewhat related to my earlier question &lt;a href=&quot;https://astronomy.stackexchange.com/questions/2/how-are-the-compositional-components-of-exoplanet-atmospheres-differentiated&quot;&gt;How are the compositional components of exoplanet atmospheres differentiated?&lt;/a&gt;, but this about a specific surface-atmospheric phenomena - volcanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using our solar system as a rough analogue (where asides from Earth, Venus, Io and Triton have active volcanism, probably more), volcanism should be not uncommon amongst exoplanets given the right conditions.  Obviously, we could probably only detect either the massive shield, plateau volcanism, or prolonged volcanism from multiple vents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What observational constraints are there in detecting the presence of volcanism on exoplanets?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2013-10-01T12:17:38.470" Title="What observational constraints are there in detecting the presence of volcanism on exoplanets?" Tags="&lt;exoplanet&gt;&lt;observational-astronomy&gt;&lt;surface&gt;&lt;volcanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="440" PostTypeId="2" ParentId="431" CreationDate="2013-10-01T12:16:54.017" Score="9" Body="&lt;p&gt;This is a part answer to your question, as it is difficult to answer without speculating, so here are some facts/observations related to your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asides from Pluto/Charon, Eris, Triton (&lt;a href=&quot;http://www.space.com/15216-dwarf-planets-facts-solar-system-sdcmp.html&quot;&gt;could be a captured Kuiper Belt object&lt;/a&gt;), &lt;a href=&quot;http://www.space.com/5606-fourth-dwarf-planet-named-polynesian-god.html&quot;&gt;Makemake&lt;/a&gt; and the football shaped &lt;a href=&quot;http://www.space.com/12414-dwarf-planet-haumea-water-ice-moons.html&quot;&gt;Haumea&lt;/a&gt;, most of the Kuiper Belt Objects (KBOs) are according to the article &lt;a href=&quot;http://www.space.com/16144-kuiper-belt-objects.html&quot;&gt;&quot;Kuiper Belt Objects: Facts about the Kuiper Belt &amp;amp; KBOs&quot;&lt;/a&gt; (Redd, 2012):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;thousands of bodies more than 62 miles (100 km) in diameter travel around the sun within this belt, along with trillions of smaller objects, many of which are short-period comets &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and is believed to have a total mass of only a tenth of the Earth, according to the article &lt;a href=&quot;http://www.gps.caltech.edu/classes/ge133/reading/levison.pdf&quot;&gt;&quot;Forming the Kuiper Belt by&#xA;the Outward Transport of Objects&#xA;During Neptune's Migration&quot;&lt;/a&gt; (Levison and Morbidelli).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;http://www.minorplanetcenter.net/iau/lists/TNOs.html&quot;&gt;list Of the many Transneptunian Objects&lt;/a&gt; that have been documented, detailing their absolute magnitudes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In regards to one of your main questions - according to Redd (2012), the challenge in their detection is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because of their small size and distant location, Kuiper Belt Objects are a challenge to spot from Earth. Infrared measurements from NASA's space-based telescope, Spitzer, have helped to nail down sizes for the largest objects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would add, their irregular elliptical orbits* and extreme (compared to the major planets) inclination to ecliptic make it that much more difficult to detect them.  Additionally, according to &lt;a href=&quot;http://burro.astr.cwru.edu/stu/solarsystem_edge.html&quot;&gt;&quot;The Edge of the Solar System&quot;&lt;/a&gt; website, further difficulties include low surface reflectivity.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;An example of a possible KBO with an extremely elliptical orbit is &lt;a href=&quot;http://www.gps.caltech.edu/~mbrown/sedna/&quot;&gt;Sedna&lt;/a&gt;, which is believed to take over 10,000 years to orbit the sun; is smaller than Pluto, but was observed at about 90AU (3 times further than Pluto).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, there could be many small Pluto-sized 'dark' worlds in highly elliptical irregular orbits in the Kuiper Belt and beyond.  However, beyond those listed, we have not seen that many and the total mass theorised does not support the idea of too many in existence, but that does not mean that they are not out there.&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-10-01T12:37:29.377" LastActivityDate="2013-10-01T12:37:29.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="441" PostTypeId="2" ParentId="439" CreationDate="2013-10-01T12:17:38.470" Score="4" Body="&lt;p&gt;Exoplanets are too far away to send satellites or to image them directly. So there is no way to go there and say: &lt;code&gt;there it is a volcano&lt;/code&gt;. My guess is that we have to guess from what we know working in the solar system. And we might just get a statistical probability that the planet is active.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say that there are two cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If the planet is rocky and near enough to the star or a giant planet, you might expect volcanism caused by tidal effects, similarly to what happen on Jupiter's &lt;code&gt;Io&lt;/code&gt; moon. In few words: the gravitational force of the nearby big planet/star deform the planet and this deformations are converted to heat due to friction. If the gravitational forces are large enough, you can end up having volcanism&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you are able to measure the spectral lines in the planet atmosphere you could try to identify signatures of gas or dust typically associated with volcanic activity. I don't know what you would want to look for, but my guess is that you want to search mostly in the infrared.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="86" LastActivityDate="2013-10-01T12:17:38.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="442" PostTypeId="1" AcceptedAnswerId="445" CreationDate="2013-10-01T13:20:01.323" Score="21" ViewCount="416" Body="&lt;p&gt;I had heard that comet ISON might not survive a close pass by our Sun, and I was curious about the odds of how other comets had fared. So, how often do other comets survive a close pass by our sun? Also, are there any connections between a comet's size or buildup(of material) that makes it more or less likely to not survive?&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorDisplayName="user8" LastEditDate="2013-10-01T13:22:27.797" LastActivityDate="2013-10-01T17:26:20.553" Title="How often do comets survive passage by the Sun?" Tags="&lt;the-sun&gt;&lt;solar-system&gt;&lt;observational-astronomy&gt;&lt;comets&gt;&lt;space&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="443" PostTypeId="1" CreationDate="2013-10-01T13:39:05.460" Score="22" ViewCount="509" Body="&lt;p&gt;I live in an urban area with a lot of light pollution and very little garden space. On a clear night I can see most of the sky and the brighter stars are visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't build an observatory - I don't have the space (let alone the money required) - so what can I do to minimise the effects of light pollution so I can see more of the fainter stars and planets?&lt;/p&gt;&#xA;" OwnerUserId="111" LastActivityDate="2013-10-01T15:53:47.700" Title="I live in an area with a lot of light pollution, how can I view the stars without building an observatory?" Tags="&lt;amateur-observing&gt;&lt;light-pollution&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="444" PostTypeId="1" CreationDate="2013-10-01T14:01:08.270" Score="7" ViewCount="95" Body="&lt;p&gt;The annual &lt;a href=&quot;http://en.wikipedia.org/wiki/Perseids&quot;&gt;Perseid&lt;/a&gt; peak occurs between August 9-14. During this time, about one meteor burns up in the atmosphere &lt;em&gt;per minute&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When these meteorites burn up, they generate light. When you have one of these per minute, does it produce enough light pollution to be a problem for those making astronomical observations? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8FPmR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2013-10-01T16:48:13.730" Title="Does the Perseid peak cause enough light pollution to be a problem?" Tags="&lt;light&gt;&lt;light-pollution&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="445" PostTypeId="2" ParentId="442" CreationDate="2013-10-01T14:17:11.540" Score="13" Body="&lt;p&gt;Key factors:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How close is perihelion? Too close and it may be destroyed on its first pass. We know Halley's Comet, which has a perihelion of about 0.6AU, has been orbiting for over 2000 years, passing the sun every 74-76 years and is still going strong.&lt;/li&gt;&#xA;&lt;li&gt;How big is it? Every pass loses material, so a bigger comet could last longer.&lt;/li&gt;&#xA;&lt;li&gt;What is its composition? Darker comets can absorb more energy when close to the sun, which may accelerate outgassing and breakup.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And of course some comets hit large objects - Shoemaker-Levy hit Jupiter and was destroyed.&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="43" LastEditDate="2013-10-01T17:26:20.553" LastActivityDate="2013-10-01T17:26:20.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="446" PostTypeId="2" ParentId="443" CreationDate="2013-10-01T14:48:15.053" Score="14" Body="&lt;p&gt;&lt;a href=&quot;http://www.telescope.com/Astronomy-Under-Light-Polluted-Skies-Whats-to-See/p/99825.uts&quot; rel=&quot;noreferrer&quot;&gt;telescope.com&lt;/a&gt; has a quick paragraph on astronomy. Below I summarise the important points with a few of my own suggestions thrown in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dark &amp;amp; dirty places&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set up on grass or dirt, pavements and buildings radiate the heat again at night and the air flow created by this can distort your image. If possible you might be able to make use of a public park.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Failing the above you might be better off going to the outskirts of the city and finding a nice dark hill and avoid as much of the light pollution as possible, but obviously this depends on your situation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Telescope&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't require a special telescope, you should be able to manage just fine with what your budget allows for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though a computerised mount can aid you plenty in locating objects, as you have fewer stars to use as way-points in light polluted areas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.buytelescopes.com/Products/30669-ioptron-zeq25-equatorial-mount.aspx&quot; rel=&quot;noreferrer&quot;&gt;Here is an example I found earlier&lt;/a&gt; - yes they are quite pricey. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uaa6R.jpg&quot; alt=&quot;goto mount&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: This picture and the mount in the link are NOT the same mount.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to set up out of direct view of a source of light, i.e street-light or porch light. This is because you will suffer from a lack of contrast, making it harder to view the objects. So dark spots are good!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Light Filters&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good, fairly cheap method*, for negating light pollution that I have found is to use a light filter, these usually just screw on over the telescope and filter out light of that wavelength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a very important point to note is that light filters will also filter out light from stars, galaxies and star clusters, making them appear duller. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light filters actually decrease 'sky glow' which is, in part, due to the upper atmosphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What these filters &lt;a href=&quot;http://www.knoxvilleobservers.org/dsonline/tips/lprfilters.html&quot; rel=&quot;noreferrer&quot;&gt;are best for&lt;/a&gt; is looking at nebulas&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nebulas emit light of a different wavelength that are not dulled by the filter and they pass straight through. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vqgXp.gif&quot; alt=&quot;light filters science&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are not hoping to look for nebulas then I do not suggest a light filter, as this will also dull the stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of which can be found &lt;a href=&quot;http://www.harrisontelescopes.co.uk/acatalog/Telescope_Light_Pollution_Filters.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/L2WwE.jpg&quot; alt=&quot;light filters&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*cheap relative to buying a go-to mount that is&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastEditorDisplayName="user96" LastEditDate="2013-10-01T15:53:47.700" LastActivityDate="2013-10-01T15:53:47.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="447" PostTypeId="1" CreationDate="2013-10-01T14:51:24.437" Score="23" ViewCount="659" Body="&lt;p&gt;The &lt;a href=&quot;http://www.esa.int/Our_Activities/Space_Science/Planck&quot;&gt;Planck satellite&lt;/a&gt; has been presented and awaited for a long time as the ultimate experiments for measuring temperature fluctuations in the cosmic microwave background (CMB) over the full sky. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the big questions that still need answer and that Planck might help clarify is about the dynamics and driving mechanisms in the first phases of the universe, in particular in the period called &lt;code&gt;inflation&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Thankfully&lt;/em&gt; there is room for improvements at small scales, i.e. small pieces of sky observed with extremely high resolution, and more importantly for experiments to measure the polarisation of CMB.&#xA;I know that for the next years a number of polarisation experiments, mostly from ground and balloons, are planned (I'm not sure about satellites).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For sure some of these result will rule out some of the possible inflationary scenarios, but to which level?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will we ever be able to say: &quot;inflation happened this way&quot;?&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="45" LastEditDate="2013-10-23T14:55:54.897" LastActivityDate="2013-10-31T21:40:52.710" Title="Future of CMB observations: How will our knowledge of the early universe change?" Tags="&lt;cosmology&gt;&lt;cmb&gt;&lt;early-universe&gt;&lt;cosmological-inflation&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="448" PostTypeId="5" CreationDate="2013-10-01T16:37:57.960" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-01T16:37:57.960" LastActivityDate="2013-10-01T16:37:57.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="449" PostTypeId="4" CreationDate="2013-10-01T16:37:57.960" Score="0" Body="Late Heavy Bombardment; occurring between 3.7-4 billion years ago when the inner solar system underwent heavy asteroidal impacts.  Also know as the Lunar Cataclysm." OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-10-01T16:38:45.340" LastActivityDate="2013-10-01T16:38:45.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="450" PostTypeId="2" ParentId="444" CreationDate="2013-10-01T16:48:13.730" Score="4" Body="&lt;p&gt;Tought to answer, because the amount of airborne dust due to meteor showers and its stability in the upper atmosphere will vary significantly enough. We had many meteor showers, Perseids included, in the past that were a bit of a letdown and didn't produce as many shooting stars as initially predicted, and also the other way around of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here's the thing tho, that discounting for any additional atmospheric glow during a few crucial hours of dusk and dawn, this airborne meteor dust might even improve the average atmospheric refraction with mostly extremely fine powder silicate dust settling onto the lower atmosphere and dispersing water molecules and binding to other solids (read: decrease humidity and pollutants), by being highly electromagnetically charged from the atmospheric reentry and electrostatically binding to larger atmospheric or pollution molecules, or simply for being hygroscopic and binding really good to water molecules, making them heavier enough to fall in a light spray back on the ground and remove the pollutants from the lower atmosphere in the process, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There might be other physical phenomena going on that I'm not aware of, but what I wanted to illustrate is, that there might even be positive side-effects in terms of light pollution vs increase in meteors, which is how we call meteoroids entering Earth's atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2013-10-01T16:48:13.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="451" PostTypeId="1" AcceptedAnswerId="454" CreationDate="2013-10-01T19:45:47.330" Score="11" ViewCount="304" Body="&lt;p&gt;In Robert Zubrin's &quot;The Case for Mars&quot; he outlines a plan to settle Mars by(among other things) finding water and using electrolysis to break up water into Hydrogen and Oxygen. This provides fuel for vehicles and air for the crew to breathe. Can a similar plan work to settle the moon? Are there enough resources on the moon for a settlement to be self-sufficent? Is there enough water on the moon for this to be viable?&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2013-10-02T16:26:47.960" Title="Does the Moon Have Enough Water for Robert Zubrin's Mars settlement Plan To Work?" Tags="&lt;the-moon&gt;&lt;solar-system&gt;&lt;water&gt;&lt;mars&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="452" PostTypeId="5" CreationDate="2013-10-01T21:57:43.807" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-01T21:57:43.807" LastActivityDate="2013-10-01T21:57:43.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="453" PostTypeId="4" CreationDate="2013-10-01T21:57:43.807" Score="0" Body="Questions regarding undesired brightening of the night sky inhibiting observation of astronomical objects." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-10-01T22:28:45.303" LastActivityDate="2013-10-01T22:28:45.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="454" PostTypeId="2" ParentId="451" CreationDate="2013-10-01T22:05:28.333" Score="8" Body="&lt;p&gt;Self-sufficiency is an incredibly broad term. We could argue that yes, there is water on the Moon, and that yes, there are viable ways to produce required electricity in self-sustainable ways, but the real question is, are there areas on the Moon that would be viable for both at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see, the most likely place where surface or near subsurface water could exist on the Moon and be suitable for mass extraction are its polar, permanently dark regions. Indeed, the ISRO (Indian Space Research Organisation) Chandrayaan-1 spacecraft has detected evidence for water locked in surface lunar regolith minerals in lunar south polar region, water that likely originates from asteroid and comet impacts embedding it deep within the lunar core and released as magmatic water closer to the surface. Any free-form water in other regions of the Moon that are exposed to sunlight and Solar radiation would sublimate to its gas form directly and with ionisation lose hydrogen atoms, so while hydrogen and oxygen atoms might still be present to some extent embedded into the surface layer minerals, extraction would likely be too elaborate there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, wherever you'd find your water source, you would still require a great deal of electricity to power your extraction plant, later use electrolysis to separate molecular water into its constituent atoms, and compress it in cryogenic conditions to their diatomic liquids that are suitable as propellant components, diatomic liquid oxygen (or LOX) as your oxidizer, and double as much in molecular quantity of diatomic liquid hydrogen (or LH2) as your rocket fuel. Problem with electricity is, unless you brought your own and a great deal of it on you to power your plants, you will likely want to use be it solar power, or tap into the in lunar regolith embedded helium-3 (or &lt;sup&gt;3&lt;/sup&gt;He) and power your third-generation Helium-3 fusion reactor. See for example &lt;a href=&quot;https://space.stackexchange.com/a/1845/49&quot;&gt;this answer&lt;/a&gt; of mine on &lt;a href=&quot;https://space.stackexchange.com/&quot;&gt;Space Exploration&lt;/a&gt; on how that could be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the main conundrum to exploiting Lunar resources, for the time being, remains finding sufficient and viably mineable resources of water where there is also self-sustainable ways of generating required electricity. One option that I can think of is staying on the most exposed to the Sun lunar equator and extracting deuterium and tritium hydrogen isotopes, as well as helium-3 from lunar regolith, all of them embedded there from Coronal Mass Ejections (CME). Required oxygen could be produced by crushing oxidized minerals and letting them &lt;em&gt;sweat&lt;/em&gt; with the presence of hydrogen isotopes into ionized water, and helium-3 could be used as previously mentioned to sustain a fusion reaction producing required electricity to later break water molecules into its constituent atoms of hydrogen and oxygen by electrolysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much of these hydrogen and helium isotopes are actually embedded in the lunar regolith, and how long these deposits persist in it, possibly staying there for at least some time due to the static charge of the regolith as it is bombarded by the Solar radiation, this is however a whole different question and one we can't currently yet answer. The study of the Lunar exosphere and dust environment is the sole purpose of the LADEE (Lunar Atmosphere and Dust Environment Explorer), that we barely just launched there. We will know in roughly one year, if it will be able to provide conclusive scientific evidence for these theories I've just mentioned.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2013-10-01T22:05:28.333" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="458" PostTypeId="2" ParentId="451" CreationDate="2013-10-02T01:18:39.087" Score="4" Body="&lt;p&gt;Additionally, Mars has a much more substantial atmosphere composed of ~95% CO2 (which is one of the major points Zubrin makes), whereas the atmosphere of the moon pales in comparison. Why is this important? Combined with the supply of Hydrogen which would be brought along, you could combine the CO2 with H2 to produce methane (CH4) which can be used as rocket fuel; water can also be produced. See the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sabatier_reaction&quot; rel=&quot;nofollow&quot;&gt;Sabatier reaction&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Page 60 in &quot;The case for mars&quot; also talks about the merits and drawbacks of CH4/O2 and CO/O2 propellant systems, the former is really the better alternative if hydrogen were available. Also, when talking about settlements, exploration is a crucial function. Fuel for vehicles can also be supplied through the use of the Martian atmospheric CO2.&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditDate="2013-10-02T16:26:47.960" LastActivityDate="2013-10-02T16:26:47.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="459" PostTypeId="1" AcceptedAnswerId="797" CreationDate="2013-10-02T03:21:10.763" Score="19" ViewCount="948" Body="&lt;p&gt;Evidently, direct observation of a black hole for an amateur astronomer, such as described for what professionals do in the question &lt;a href=&quot;https://astronomy.stackexchange.com/questions/24/how-are-black-holes-found&quot;&gt;&quot;How are black holes found?&quot;&lt;/a&gt; would be nigh on impossible, so the observations would be based on their effects of the surrounding matter (specifically, accretion disks and jets) and more likely detecting the radiation signals.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the locations of many black holes are known, what practical considerations would a backyard radio-astronomer need to consider to attempt to do detect black holes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amateur radio-astronomy detection of pulsars and gamma ray bursts are discussed practically in &lt;a href=&quot;http://www.radio-astronomy.org/pdf/sara-beginner-booklet.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Introduction to Radio Astronomy&lt;/a&gt; (Society of Radio Astronomers), this question is to see what practical considerations would be needed to extend those techniques to detecting black holes.&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorUserId="-1" LastEditorDisplayName="user8" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2013-10-30T07:57:31.027" Title="What are practical considerations for backyard radio-astronomy detection of black holes?" Tags="&lt;black-hole&gt;&lt;amateur-observing&gt;&lt;radio-astronomy&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="460" PostTypeId="1" CreationDate="2013-10-02T03:37:27.467" Score="12" ViewCount="457" Body="&lt;p&gt;Related to the question &lt;a href=&quot;https://astronomy.stackexchange.com/questions/431/are-any-pluto-sized-objects-remaining-to-be-discovered-in-the-kuiper-belt&quot;&gt;&quot;Are any Pluto-sized objects remaining to be discovered in the Kuiper Belt?&quot;&lt;/a&gt; and the fact that most of the Kuiper Belt objects have very elliptical orbits, the question begs, what observational techniques are used to confirm that the object being viewed is an orbiting Kuiper Belt Object, as opposed to the transit of a &lt;a href=&quot;http://www.space.com/18461-rogue-alien-planet-discovery.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;rogue or 'orphan' planet&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-12-10T11:48:15.667" Title="How can the orbit of a Kuiper Belt Object be differentiated from the transit of a rogue planet?" Tags="&lt;observational-astronomy&gt;&lt;orbit&gt;&lt;kuiper-belt&gt;&lt;kbo&gt;&lt;rogue-planet&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="461" PostTypeId="1" AcceptedAnswerId="474" CreationDate="2013-10-02T04:15:21.283" Score="9" ViewCount="111" Body="&lt;p&gt;I've been trying to wrap my head around capabilities of current radio-astronomical spectrometry technology to isolate not too distant tenuous sources, say, chemical composition and density of the the interstellar medium right in the neighborhood of our Solar system. In essence, what is our local interstellar cloud made out of and how much of it is there:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/LKxCg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've had much success so far in isolating distant radio wave sources and analyzing their hyperfine, i.e. molecular level composition. Same goes even for interstellar medium composition down to molecular ions of diazenylium (N&lt;sub&gt;2&lt;/sub&gt;H&lt;sup&gt;+&lt;/sup&gt;) that is surrogate to the presence of unobservable, nonpolar N&lt;sub&gt;2&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, P. Caselli et al. from the Harvard-Smithsonian Center for Astrophysics &lt;a href=&quot;http://iopscience.iop.org/1538-4357/455/1/L77/pdf/1538-4357_455_1_L77.pdf&quot;&gt;presented a paper&lt;/a&gt; (PDF) on J = 1 → 0 transition of N&lt;sub&gt;2&lt;/sub&gt;H&lt;sup&gt;+&lt;/sup&gt; at 93 GHz toward the quiescent low-mass cloud core L1512 in Taurus, back in 1995. That is 18 years ago!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand is, what's preventing us, with current technology, from isolating local low-mass formations in the radio wave spectra from its background noise and analyzing its hyperfine structure through radio-astronomical spectrometry? Is the signal to noise ratio simply unworkable and the density of the local cloud too thin to filter it out from its background? Or have such observations actually already been done, and we already have molecular data on the local cloud that I'm not aware of?&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2013-10-02T22:09:24.580" Title="Current limitations in radio-astronomical spectrometry analysis of local interstellar cloud's hyperfine structures" Tags="&lt;radio-astronomy&gt;&lt;spectrometry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="462" PostTypeId="5" CreationDate="2013-10-02T05:37:13.093" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-02T05:37:13.093" LastActivityDate="2013-10-02T05:37:13.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="463" PostTypeId="4" CreationDate="2013-10-02T05:37:13.093" Score="0" Body="Questions relating to the decisions made by the International Astronomical Union." OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-10-02T12:46:38.300" LastActivityDate="2013-10-02T12:46:38.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="464" PostTypeId="5" CreationDate="2013-10-02T05:38:18.737" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-02T05:38:18.737" LastActivityDate="2013-10-02T05:38:18.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="465" PostTypeId="4" CreationDate="2013-10-02T05:38:18.737" Score="0" Body="Questions related to the physical instruments for astronomical observations." OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-10-02T12:46:56.913" LastActivityDate="2013-10-02T12:46:56.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="466" PostTypeId="5" CreationDate="2013-10-02T05:40:50.370" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-02T05:40:50.370" LastActivityDate="2013-10-02T05:40:50.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="467" PostTypeId="4" CreationDate="2013-10-02T05:40:50.370" Score="0" Body="Questions about the &quot;Goldilocks Zone&quot; around a parent star where conditions are just right for the potential of life developing." OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-10-02T12:46:46.490" LastActivityDate="2013-10-02T12:46:46.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="468" PostTypeId="5" CreationDate="2013-10-02T05:43:33.093" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-02T05:43:33.093" LastActivityDate="2013-10-02T05:43:33.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="469" PostTypeId="4" CreationDate="2013-10-02T05:43:33.093" Score="0" Body="Questions related to Kuiper Belt Objects that exist in an annulus from 30AU to approximately 50AU from the Earth's Sun." OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-10-02T12:47:06.663" LastActivityDate="2013-10-02T12:47:06.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="470" PostTypeId="5" CreationDate="2013-10-02T05:47:09.520" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-02T05:47:09.520" LastActivityDate="2013-10-02T05:47:09.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="471" PostTypeId="4" CreationDate="2013-10-02T05:47:09.520" Score="0" Body="Questions regarding Charon, the largest satellite of the dwarf planet Pluto.  Often considered to be Pluto's binary." OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-10-02T12:46:52.960" LastActivityDate="2013-10-02T12:46:52.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="472" PostTypeId="1" CreationDate="2013-10-02T06:58:15.140" Score="0" ViewCount="2982" Body="&lt;p&gt;Pluto is no longer a member of the planets in the solar system.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why was it disqualified as a planet? &lt;/li&gt;&#xA;&lt;li&gt;What is the criteria for classifying an Astronomical Object as a Planet?&lt;/li&gt;&#xA;&lt;li&gt;Does this criteria apply to all Star Systems or just our Solar System?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="142" LastActivityDate="2019-10-18T09:45:48.653" Title="Reason for disqualifying Pluto as a Planet?" Tags="&lt;planet&gt;&lt;star-systems&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ClosedDate="2013-10-02T12:46:47.753" ContentLicense="CC BY-SA 3.0" />
  <row Id="473" PostTypeId="2" ParentId="472" CreationDate="2013-10-02T08:42:11.930" Score="2" Body="&lt;p&gt;&lt;strong&gt;Pluto:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was disqualified as a planet because orbital dominance was not achieved in the case of Pluto. Orbital dominance means that a the planet candidate should have remove all the small bodies from its orbit, by impact, capture of gravitational disturbance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Planet:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.iau.org/static/resolutions/Resolution_GA26-5-6.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;According to the IAU in its resolution B5&lt;/a&gt; (IAU is the &lt;a href=&quot;http://en.wikipedia.org/wiki/International_Astronomical_Union&quot; rel=&quot;nofollow noreferrer&quot;&gt;International Astronomical Union&lt;/a&gt;; in particular it is in charge of naming celestrial bodies), a planet is a celestial body:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;in orbit around the Sun&lt;/li&gt;&#xA;&lt;li&gt;with a sufficient mass so that self-gravity overcome rigid body forces (that assumes a hydrostatic equilibrium)&lt;/li&gt;&#xA;&lt;li&gt;has cleared its neighbourhood.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exoplanets:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you see in the resolution B5, this definition is specified for &quot;Planet in the Solar System&quot;. It is mainly because it does not really make sense now that it is already challenging to observe planets, so the distinction between a planet and a dwarf planet is not yet crucial for exoplanets. So &lt;a href=&quot;https://web.archive.org/web/20131209073011/http://www.dtm.ciw.edu/boss/definition.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;the official definition of an exoplanet by IAU&lt;/a&gt; makes mainly a difference between a planet and a brown dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I guess the IAU definition of a planet should be kept for the other planetary systems if we are able to detect objects such as dwarf planets.&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="23114" LastEditDate="2019-10-18T09:45:48.653" LastActivityDate="2019-10-18T09:45:48.653" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="474" PostTypeId="2" ParentId="461" CreationDate="2013-10-02T14:49:39.210" Score="4" Body="&lt;p&gt;The first thing to notice is that the Local Interstellar Cloud, in which the Sun is evolving right now, is a fairly diffuse region, with a typical density of about one to a few particles per cubic centimeters. Clouds with such low density are actually mostly atomic; as you can see on this plot (&lt;a href=&quot;http://www.annualreviews.org/doi/abs/10.1146/annurev.astro.43.072103.150624&quot; rel=&quot;nofollow noreferrer&quot;&gt;Snow &amp;amp; McCall 2006&lt;/a&gt;, adapted from &lt;a href=&quot;http://adsabs.harvard.edu/abs/2005ApJ...628..260N&quot; rel=&quot;nofollow noreferrer&quot;&gt;Neufeld et al. 2005&lt;/a&gt;):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Lw90e.png&quot; alt=&quot;Taken from Snow &amp;amp; McCall (2006)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is then quite challenging to observe molecules in this region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good news is, there is not only molecular lines to give informations about the physical properties (temperature, density, ionization, velocity structure and morphology) of the Local Interstellar Cloud. In particular, ultraviolet is a perfectly suited wavelength range to sudy this region in absorption, and there are some quite extensive studies of the subject (see for example &lt;a href=&quot;http://adsabs.harvard.edu/abs/2004AdSpR..34...41R&quot; rel=&quot;nofollow noreferrer&quot;&gt;Paper I&lt;/a&gt;, &lt;a href=&quot;http://adsabs.harvard.edu/abs/2004ApJ...602..776R&quot; rel=&quot;nofollow noreferrer&quot;&gt;Paper II&lt;/a&gt;, &lt;a href=&quot;http://adsabs.harvard.edu/abs/2004ApJ...613.1004R&quot; rel=&quot;nofollow noreferrer&quot;&gt;Paper III&lt;/a&gt; and &lt;a href=&quot;http://adsabs.harvard.edu/abs/2008ApJ...673..283R&quot; rel=&quot;nofollow noreferrer&quot;&gt;Paper IV&lt;/a&gt; of Redfield et al.).&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2013-10-02T22:09:24.580" LastActivityDate="2013-10-02T22:09:24.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="475" PostTypeId="1" AcceptedAnswerId="708" CreationDate="2013-10-02T18:52:26.703" Score="13" ViewCount="569" Body="&lt;p&gt;A lot of pop science articles (for example, &lt;a href=&quot;http://www.space.com/20330-cosmic-microwave-background-explained-infographic.html&quot; rel=&quot;noreferrer&quot;&gt;this Space.com article&lt;/a&gt;) discuss the cold spots on the WMAP data of Cosmic Microwave Background (CMB) radiation, but have any studies been done on the large mass of hot spots on the right of the image below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/KUwaF.png&quot; alt=&quot;CMB data from WMAP&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a side-effect of our observation methods or does it represent some actual feature of the early history of the Universe?&lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorUserId="6" LastEditDate="2013-10-18T12:29:01.820" LastActivityDate="2013-10-19T17:13:05.177" Title="What is the mass of hotspots in the Cosmic Microwave Background radiation?" Tags="&lt;cosmology&gt;&lt;cmb&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="476" PostTypeId="1" CreationDate="2013-10-02T20:10:06.110" Score="20" ViewCount="1078" Body="&lt;p&gt;In an academic study titled '&lt;a href=&quot;https://perso.imcce.fr/jacques-laskar/misc_files/PLS_amer.pdf&quot; rel=&quot;noreferrer&quot;&gt;The Moon and the Origin of Life on Earth&lt;/a&gt;'  it is said that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;'&lt;em&gt;If the Moon did not exist, the orientation of the Earth’s axis would not be stable, and would be subject to large chaotic variations over the ages. The resulting climatic changes very likely would have markedly disturbed the development of organized life&lt;/em&gt;'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since I don't have enough physics and mathematics knowledge in order to understand why the Moon stabilizes the Earth's axis, can anybody explain in simple words this phenomenon?&lt;/p&gt;&#xA;" OwnerUserId="159" LastEditorUserId="6841" LastEditDate="2018-01-10T04:02:46.413" LastActivityDate="2020-05-23T10:20:21.183" Title="Why does the Moon stablize Earth's axis?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="477" PostTypeId="2" ParentId="476" CreationDate="2013-10-02T22:41:30.543" Score="16" Body="&lt;p&gt;One way to look at this problem is to consider angular momentum. The Earth spins around its axis, and has therefore some angular momentum by itself. The angular momentum is proportional to the mass of the Earth, to its squared radius, and to its angular velocity. But the Earth is not alone; it has the Moon rotating around that adds angular momentum to the Earth-Moon system. And, even though the Moon is not as massive as the Earth (it is about 100 times less massive), nor rotates very fast around the Earth (and therefore a lesser angular velocity), it has a large orbit (about 300 000 km) and overall, it adds to the system an amount of angular momentum comparable to that of the Earth itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, think of a spinning-top: the faster it rotates (and thus the larger the angular momentum), the more stable it is. It is the same for the Earth-Moon system: without the Moon, the angular momentum of the Earth itself would be such that gravitational perturbations could be sufficient, in the long run, to perturbates significantly its axis (exactly like the spinning-top: if it does not rotate very fast, a small perturbation will rapidely increases and the spinning-top axis will start to oscillate more and more). But with the Moon, the global angular momentum of the system is larger, and it is therefore harder to pertubate sufficiently the system to get it oscillate strongly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sources:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those who wants the dirty details, you can have a look at &lt;a href=&quot;https://doi.org/10.1038/361615a0&quot; rel=&quot;nofollow noreferrer&quot;&gt;Laskar et al. 1993&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="-1" LastEditDate="2020-05-23T10:20:21.183" LastActivityDate="2020-05-23T10:20:21.183" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="480" PostTypeId="1" CreationDate="2013-10-03T07:51:33.793" Score="18" ViewCount="3688" Body="&lt;p&gt;I've noticed that the Pleiades look bright in the sky, but when I look directly at the constellation, it suddenly goes dim and it's more difficult to make out the individual stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this happen, and is there a way to &quot;beat the system&quot; by both being able to look directly at a star and see it clearly.&lt;/p&gt;&#xA;" OwnerUserId="158" LastActivityDate="2013-10-03T08:52:38.047" Title="Why do the Pleiades look clearer when viewed indirectly?" Tags="&lt;star-gazing&gt;&lt;constellations&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="481" PostTypeId="2" ParentId="480" CreationDate="2013-10-03T08:32:20.240" Score="5" Body="&lt;p&gt;The common explanation is linked to the eye structure: in the center of the eye, you have mostly cones, that are used to detect colors and that are efficient only in bright light (that's why we don't see colors so well during the night). The rest of the eye contains more rods, that are used to detect dim light and contrast. So the best way to observe some faint objects in the sky is to use averted vision, which uses peripherical vision and therefore takes profit of your rods more than your cones. That is exactly the same thing when you look at any nebula.&lt;/p&gt;&#xA;" OwnerUserId="88" LastActivityDate="2013-10-03T08:32:20.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="482" PostTypeId="2" ParentId="480" CreationDate="2013-10-03T08:36:36.663" Score="18" Body="&lt;p&gt;What you describe is a technique called &lt;a href=&quot;https://en.wikipedia.org/wiki/Averted_vision&quot;&gt;averted vision&lt;/a&gt; and takes advantage of the arrangement of &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/vision/rodcone.html&quot;&gt;cones and rods&lt;/a&gt;, two types of light sensitive elements within the eye:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The retina contains two types of photoreceptors, rods and cones. The&#xA;  rods are more numerous, some 120 million, and are more sensitive than&#xA;  the cones. However, they are not sensitive to color. The 6 to 7&#xA;  million cones provide the eye's color sensitivity and they are much&#xA;  more concentrated in the central yellow spot known as the macula. In&#xA;  the center of that region is the &quot; fovea centralis &quot;, a 0.3 mm&#xA;  diameter rod-free area with very thin, densely packed cones.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Shorter, the cones are sensitive to color and work best in bright light, while the rods are most effective in low light conditions. And since the rods are concentrated off center of retina, our night vision is best off center. This averted vision technique can be used with a naked eye, as well as with any optical equipment like binoculars or telescopes, although it can be a bit of a bother with eyeglasses and contact lenses. Especially with the latter, using this advert vision technique can cause additional irritation of the cornea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As per &lt;em&gt;&quot;beating the system&quot;&lt;/em&gt; part, with a bit of practice, you could train yourself to view dim objects better with the corner of your eyes at will, and not merely by chance or for only short periods of time. It might seem hard with first few tries, but it really isn't and you soon get used to not looking directly towards the observed object, countering what we tend to do naturally, direct our eyes towards the object of our observations. Practice makes perfect, but there is no long term and unwanted side effects to doing this, and you'll still be able to direct your eyes towards whatever you'll want to see clearer, just as you did before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one other thing to add to techniques to observing the night skies. Namely, our eyes need some time to adapt to darkness, and with our more frequent staring in computer monitors, television and other illuminated sources, our pupils need even longer to dilate and let the maximum of light to fall on the retina. So for best results, rest your eyes and do not expose them to anything except red light for 30 minutes prior to observing. This allows your pupils to expand to their maximum diameter and build up the levels of optical pigments, which are rapidly lost if exposed to bright light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also important to observe with both of your eyes open to avoid fatigue. If you find this difficult, you may try covering the other eye with your hand or an eye patch. It should also be mentioned that both smoking and drinking cause deterioration in your sight. Your blood-sugar levels are also important and a low level of blood sugar causes lowering of sensitivity of eyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, your eyes will tire using this technique, just as they would by focusing for longer periods of times as you normally would. Look away every now and then, preferably not towards too strong sources of light, but still at least slightly brighter than those you previously observed. Also stretch your legs and blink fast a few times. Stretch your back too, and relax your bottom. You might find it funny, but this technique of relaxing and contracting muscles of that part of our physique has been used by trained spies to cheat polygraph tests that also measure pupil dilation. How's that for &lt;em&gt;&quot;beating the system&quot;&lt;/em&gt;, huh? It is by far the fastest way of relaxing your eyes, at least that I know of.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditDate="2013-10-03T08:52:38.047" LastActivityDate="2013-10-03T08:52:38.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="483" PostTypeId="1" AcceptedAnswerId="608" CreationDate="2013-10-03T09:54:17.033" Score="18" ViewCount="5810" Body="&lt;p&gt;What modifications to a standard AM/FM or shortwave radio are needed in order to be able to detect radio-wave signals emitted from Jupiter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be possible to detect the transit of the major moons and even minor planets (maybe even Ceres and Mars) using this method?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-10-03T10:07:31.923" LastActivityDate="2019-04-11T03:30:56.493" Title="Jupiter FM - What are practical and inexpensive ways for the amateur detection of signals from Jupiter, especially of the transit of her moons?" Tags="&lt;amateur-observing&gt;&lt;jupiter&gt;&lt;radio-astronomy&gt;&lt;natural-satellites&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="484" PostTypeId="1" AcceptedAnswerId="485" CreationDate="2013-10-03T10:42:53.120" Score="10" ViewCount="674" Body="&lt;p&gt;The closest star to Earth (after the Sun) is well-known: Alpha Centauri, 4.3 light years away. But what about the closest exoplanet to Earth, outside of the Solar System?&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorDisplayName="user8" LastEditDate="2013-10-03T11:18:09.917" LastActivityDate="2016-08-25T15:39:05.097" Title="Which is the closest exoplanet to Earth?" Tags="&lt;exoplanet&gt;&lt;solar-system&gt;&lt;distances&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="485" PostTypeId="2" ParentId="484" CreationDate="2013-10-03T11:15:11.230" Score="12" Body="&lt;p&gt;Funnily enough, just until recently, the nearest exoplanet &lt;strong&gt;Alpha Centauri Bb&lt;/strong&gt; was believed to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a planet with a similar mass to Earth's just 4.3 light years away in the nearby Alpha Centauri system. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.newscientist.com/article/dn23609-nearest-exoplanet-to-earth-may-be-a-ghost-world.html#.Uk1Mdvh-8eE&quot;&gt;Nearest exoplanet to Earth may be a ghost world&lt;/a&gt; (Aron, 2013).  However, according to the article, recent observations have suggested that it may not actually be there.  So the answer to this question is not all that clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Paris Observatory has a list of all exoplanets in a sortable database at &lt;a href=&quot;http://exoplanet.eu/catalog/all_fields/&quot;&gt;exoplanet.eu&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the Paris Observatory catalogue, if Alpha Centauri Bb does not actually exist, the next closest is the just-larger-than Jupiter sized &lt;a href=&quot;http://exoplanet.eu/catalog/eps_eridani_b/&quot;&gt;Epsilon Eradani B&lt;/a&gt; (not a hot-Jupiter), at 10.5 light years away.  Which according to &lt;a href=&quot;http://www.space.com/2990-nearest-planet-solar-system-photogenic.html&quot;&gt;Nearest Planet Beyond Solar System Might Be Photogenic&lt;/a&gt; (Than, 2006):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;About 1.5 more massive than Jupiter, the planet takes 7-years to circle its star, Epsilon Eridani. The giant gas planet was originally detected in 2000, when astronomers noticed a rocking motion in the star which they attributed to the gravitational tug of an unseen planet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is speculation that more planets may exist in that system as with dust rings.&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2013-10-03T11:15:11.230" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="486" PostTypeId="1" AcceptedAnswerId="489" CreationDate="2013-10-03T22:43:55.840" Score="19" ViewCount="596" Body="&lt;p&gt;Some planets, such as Mars, have 2 or more moons. Giants, like Jupiter of Saturn have &lt;em&gt;a lot&lt;/em&gt; of moons!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How likely the orbit of one crosses another one's orbit? Is it possible for two moons to collide? If yes, what would exactly happen?&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="2618" LastEditDate="2017-03-07T01:11:08.930" LastActivityDate="2017-03-07T01:11:08.930" Title="Moon orbits crossing each other" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;natural-satellites&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="487" PostTypeId="2" ParentId="486" CreationDate="2013-10-04T01:02:39.083" Score="7" Body="&lt;p&gt;Though it's probably very unlikely in a solar system as old as ours (~5 billion years), I would not put it out of the realm of possibilities. Most of the collisions of this scale and magnitude have already happened - the early days of the formation of the solar system was violent and full of collisions between  protoplanets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, one hypothesis for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Giant_impact_hypothesis&quot;&gt;formation of the moon&lt;/a&gt; was that it formed from the debris of just such a collision. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Uranus#Axial_tilt&quot;&gt;axial tilt of Uranus&lt;/a&gt; could also be explained by the collision of the planet with a protoplanet billions of years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the likelihood zero? No. But I would bet money that collisions between planets and asteroids/comets are much more probable.  &lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditDate="2013-10-04T01:14:20.807" LastActivityDate="2013-10-04T01:14:20.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="488" PostTypeId="1" AcceptedAnswerId="491" CreationDate="2013-10-04T06:19:19.023" Score="18" ViewCount="881" Body="&lt;p&gt;How big can a black hole become and  how small can a black hole become?(minimum and maximum dimensions of a black hole)&lt;/p&gt;&#xA;" OwnerUserId="150" LastActivityDate="2016-09-24T04:12:18.030" Title="Dimensions of a black hole" Tags="&lt;black-hole&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="489" PostTypeId="2" ParentId="486" CreationDate="2013-10-04T09:13:04.040" Score="14" Body="&lt;p&gt;There is one known pair of moons in the solar system that seemingly swap orbits every couple of years. That's Saturn's &lt;a href=&quot;http://en.wikipedia.org/wiki/Epimetheus_%28moon%29#Orbital_relationship_between_Epimetheus_and_Janus&quot; rel=&quot;noreferrer&quot;&gt;Epimetheus and Janus&lt;/a&gt;. Their orbits are so close together that they interact gravitationally every couple of years (when the inner moon catches up to the outer moon), so that the outer moon is slowed down, and the inner moon is accelerated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each moon changed direction from prograde to retrograde and vice-versa in two co-orbital horseshoe orbits, never crossing paths as they exchange a bit of momentum when closer but Saturn's gravitational pull prevails and pulls each of these two moons back in the opposite direction where they came from, swapping from lower to higher orbit. It's a bit tricky to explain in words:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SdnjI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[From: &lt;a href=&quot;http://inspirehep.net/record/818547/plots&quot; rel=&quot;noreferrer&quot;&gt;The Rotation of Janus and Epimetheus - Tiscareno, Matthew S. et al. Icarus 204 (2009)&lt;/a&gt;.]&lt;/p&gt;&#xA;" OwnerUserId="50" LastEditorUserId="48" LastEditDate="2013-10-04T16:24:31.463" LastActivityDate="2013-10-04T16:24:31.463" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="490" PostTypeId="1" CreationDate="2013-10-04T12:16:07.193" Score="24" ViewCount="1439" Body="&lt;p&gt;In July 1994, the comet Shoemaker-Levy 9 (D/1993 F2) tidally fragmented and these fragments collided into Jupiter, as per the image below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pDJ3V.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.uni.edu/morgans/astro/course/Notes/section4/new22.html&quot; rel=&quot;noreferrer&quot;&gt;Image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, &lt;strong&gt;what mechanisms resulted in these small fragments causing such large explosions in the Jovian clouds?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2021-01-29T09:12:30.390" Title="Why did the Comet Shoemaker-Levy-9 fragments cause such large explosions on Jupiter?" Tags="&lt;jupiter&gt;&lt;comets&gt;&lt;impact&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="491" PostTypeId="2" ParentId="488" CreationDate="2013-10-04T12:49:48.117" Score="18" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Primordial_black_holes&quot;&gt;Primoridial black holes&lt;/a&gt; can be found (hypothetically; there is no experimental evidence yet) of any small size above the Planck mass. &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_black_hole&quot;&gt;Stellar black holes&lt;/a&gt;, however, cannot have a mass below the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tolman%E2%80%93Oppenheimer%E2%80%93Volkoff_limit&quot;&gt;TOV limit&lt;/a&gt; (1.5 to 3 solar masses)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There does seem to be an upper limit&lt;sup&gt;1&lt;/sup&gt; of 50 billion solar masses. However, I suspect&lt;sup&gt;2&lt;/sup&gt; that this takes into consideration formative constraints (i.e. the constraints posed on the &lt;em&gt;formation&lt;/em&gt; of such a BH); and does not prohibit such a black hole from existing. After all, the Schwarzschild metric certainly does not impose limits on the size of a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that talking about the limiting dimensions of a black hole is slightly meaningless as the dimensions change in different reference frames. It is far easier to talk about the &lt;em&gt;mass&lt;/em&gt; of a black hole; the radius can be calculated in various frames from that information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1. &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1111/j.1365-2966.2008.13864.x/full&quot;&gt;Natarajan, P. and Treister, E. (2009), Is there an upper limit to black hole masses?. Monthly Notices of the Royal Astronomical Society, 393: 838–845. doi: 10.1111/j.1365-2966.2008.13864.x&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2. but cannot confirm yet; I will have to read the paper more thoroughly&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2013-12-21T14:11:15.643" LastActivityDate="2013-12-21T14:11:15.643" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="492" PostTypeId="1" CreationDate="2013-10-05T02:49:17.953" Score="12" ViewCount="6257" Body="&lt;p&gt;In general, almost everyone describes Earth as an perfect sphere, but in reality it isn't a perfect sphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any specific reason why Earth is not perfect sphere ?&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="6" LastEditDate="2013-11-15T18:11:07.407" LastActivityDate="2015-12-22T17:14:23.160" Title="Why isn't Earth a perfect sphere?" Tags="&lt;earth&gt;&lt;rotation&gt;&lt;geology&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="5" ClosedDate="2013-12-12T14:44:32.880" ContentLicense="CC BY-SA 3.0" />
  <row Id="493" PostTypeId="2" ParentId="492" CreationDate="2013-10-05T07:36:50.397" Score="10" Body="&lt;p&gt;There is no reason it should be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are bound to be irregularities due to local geological differences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory as to why planets tend to be spherical is because any irregularity would cause a non-uniformity in the gravitational field at the surface, and the forces on the surface rocks would ultimately cause it to be spherical. Though, ignoring local irregularities, the Earth has a smaller radius (from its center of mass) at the poles, about 42 km less, than that at equator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be attributed to Earth's rotation. The distance of a point on the Earth's surface located at the equator is larger from the axis of rotation than at any other latitude. As it rotates, this causes a bulge at the equator. The technical term given to this shape is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Oblate_spheroid&quot; rel=&quot;noreferrer&quot;&gt;oblate spheroid&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read about the mathematical formula which describes the flattening and more about the formation &lt;a href=&quot;http://en.wikipedia.org/wiki/Equatorial_bulge&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="45" LastEditDate="2013-10-10T12:25:28.920" LastActivityDate="2013-10-10T12:25:28.920" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="494" PostTypeId="2" ParentId="492" CreationDate="2013-10-05T21:26:35.657" Score="5" Body="&lt;p&gt;Shape of Earth is geoid. It's because of the rotation of our planets around its axis. Because of the centrifugal force, the diameter of Earth is bigger at the Equator than at the (physical) North Pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Billions of years of rotating have deformed the perfect shape form of Earth - if it has ever had at all.&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="62" LastEditDate="2015-12-22T17:14:23.160" LastActivityDate="2015-12-22T17:14:23.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="495" PostTypeId="2" ParentId="400" CreationDate="2013-10-05T23:53:06.743" Score="5" Body="&lt;p&gt;I don't know, how it works there, but perhaps it is easier to imagine, how it could look. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;TrES-2b is a gas giant, so it doesn't have a well defined surface. When you are somewhere in the atmosphere then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-When looking up, you will see the light, as deep sea creatures do. The amount of light and its color will depend on how deep you are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-When looking down, it should be dark (black), comparing to how it looks up. Otherwise, the atmosphere would be reflecting light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-When looking to the sides, it should also be dark, for if the light were scattered at some angles, it would also be scattered upwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, one simplest possible model is: like in deep sea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, that at some other wavelengths, particularly in infrared, it all can look very differently.&lt;/p&gt;&#xA;" OwnerUserId="115" LastEditorUserId="115" LastEditDate="2013-10-06T07:44:23.250" LastActivityDate="2013-10-06T07:44:23.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="496" PostTypeId="1" AcceptedAnswerId="497" CreationDate="2013-10-06T11:35:07.050" Score="8" ViewCount="731" Body="&lt;p&gt;Galaxies are always in motion relative to the Milky Way. My question is, which galaxy is receding the fastest from our viewpoint?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the theorised mechanism that causes this?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorUserId="2531" LastEditDate="2014-12-23T17:59:21.810" LastActivityDate="2017-06-19T07:01:11.187" Title="Which galaxy is receding from the Milky Way the fastest? What is known of the mechanism behind its recession?" Tags="&lt;galaxy&gt;&lt;redshift&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="497" PostTypeId="2" ParentId="496" CreationDate="2013-10-06T15:12:12.570" Score="8" Body="&lt;p&gt;This question may be a duplicate to: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/104/on-what-scale-does-the-universe-expand/109#109&quot;&gt;On what scale does the universe expand?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism for the expansion of the universe is gravity, manifested through what's known as Dark Energy. Dark Energy is a component to the universe which today, composes &lt;a href=&quot;http://arxiv.org/abs/1303.5076&quot; rel=&quot;noreferrer&quot;&gt;68.3%&lt;/a&gt; of the energy density of the universe. You may be confused as to how gravity could be responsible for the expansion of the universe, but the equation of state of DE, &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;w = pressure/density&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;is negative (w_DE = -1), which means that it has anti-gravitational properties. Additionally, its density remains constant as the coordinates of the universe change, which is the reason why it &lt;a href=&quot;http://en.wikipedia.org/wiki/Dark-energy-dominated_era&quot; rel=&quot;noreferrer&quot;&gt;dominates&lt;/a&gt; later on in the universe's history (while all other components, matter (dark or otherwise) and radiation/relativistic matter, become diluted with the expansion of the universe).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question, because the expansion of the universe is happening at all points in space, it looks like the &quot;center&quot; of expansion is at every point in space (see: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/305/space-expansion-in-layman-terms&quot;&gt;Space expansion in layman terms&lt;/a&gt;), the galaxies furthest away (highest redshift), are the ones moving quickest away from the Milky Way Galaxy. &lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2013-10-06T15:12:12.570" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="498" PostTypeId="1" AcceptedAnswerId="2011" CreationDate="2013-10-06T20:40:44.357" Score="11" ViewCount="539" Body="&lt;p&gt;What barriers - of technology, physics and possibly economy (things that would be possible technologically but are simply too expensive) sets the upper bound on quality of telescopes for observation of the sky in visible spectrum - observing surface of distant objects at maximum resolution/detail?&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="83" LastEditDate="2013-10-07T06:52:52.663" LastActivityDate="2014-03-23T12:26:24.630" Title="What limits the usable focal length of telescopes currently?" Tags="&lt;telescope&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="499" PostTypeId="2" ParentId="492" CreationDate="2013-10-06T21:21:08.443" Score="1" Body="&lt;p&gt;Planet is not sphere because it is not in thermal equilibrium and is not closed. It internal core is still at higher temperature and it continues receives energy from the sun, as well as the gravity from other planet. This extra energy and external force could results in any shape of planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The appropriate question should be &quot;When can we treat the Earth as sphere?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is sphere when we look at it outside with at least few radius away from the Earth. The deviation from sphere is of order 100km, while compared with the radius of Earth 6400km, which is only 2%. In particular, the shape still look circle when your telescope resolution is low, and the gravity is very close to a spherical object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, you can also treat it as a point like object when you are very very far away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can treat it &quot;flat&quot; (or very rough mountain), when you look at it on the surface when a person stand there.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2013-10-06T21:21:08.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="500" PostTypeId="1" AcceptedAnswerId="504" CreationDate="2013-10-06T23:18:57.557" Score="15" ViewCount="467" Body="&lt;p&gt;Is it possible (for either a satellite or a planet) to orbit around a black hole? Do they attract everything around themselves into the center? Or they just affect gravitational force just like stars?&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="173" LastEditDate="2013-10-07T11:29:47.517" LastActivityDate="2014-01-14T11:20:01.487" Title="Orbiting around a black hole" Tags="&lt;black-hole&gt;&lt;orbit&gt;&lt;gravity&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="501" PostTypeId="1" AcceptedAnswerId="505" CreationDate="2013-10-07T02:40:03.707" Score="25" ViewCount="1395" Body="&lt;p&gt;I had heard that Mars once had a planetary magnetic field, but that it is now gone? What happened to it?&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="9527" LastEditDate="2020-09-24T07:29:25.847" LastActivityDate="2020-09-24T07:29:25.847" Title="Why did Mars lose its magnetic field?" Tags="&lt;mars&gt;&lt;terrestrial-planets&gt;&lt;magnetic-field&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="502" PostTypeId="2" ParentId="498" CreationDate="2013-10-07T05:30:07.827" Score="9" Body="&lt;p&gt;To answer the question reworded as: &quot;What limits the quality of telescopes currently?&quot; The answer is mostly: &lt;strong&gt;money&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It used to be: &lt;strong&gt;atmosphere&lt;/strong&gt;. But with advances in adaptive optics, ground-based telescopes are achieving what used to require a space telescope. Plus, we have the technology for space telescopes if we want, like the pending JWST. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So pretty much, it boils down to funding. Who is going to spend the money on the expensive technology to see better and further, when science is being squeezed for funding from all quarters, and astronomers can't put forward a business case for a return on investment like a biotech company can for developing a grass that will result in cows burping less methane.&lt;/p&gt;&#xA;" OwnerUserId="67" LastEditorUserId="67" LastEditDate="2014-03-18T10:35:47.287" LastActivityDate="2014-03-18T10:35:47.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="503" PostTypeId="2" ParentId="500" CreationDate="2013-10-07T05:32:25.077" Score="3" Body="&lt;p&gt;They don't attract gravitational force; they have mass, so they exert a gravitational influence upon other objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, it is possible for an object to indefinitely orbit around a black hole. Just because the mass it is orbiting is called a black hole does not mean the object is doomed to spiral in on the black hole.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2013-10-07T05:32:25.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="504" PostTypeId="2" ParentId="500" CreationDate="2013-10-07T10:02:03.730" Score="12" Body="&lt;p&gt;Absolutely possible. There's nothing magical about a black hole. The gravitational pull of a black hole reaches as far as gravity would for another object of the same mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you replace the Sun with a black hole of the same mass, everything would continue to orbit it just as it currently does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything with mass has a gravitational force itself, and a black hole will attract anything with mass. Again, it's the same as our star having an effect on the Earth, and the Earth having an effect on the Moon.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2013-10-07T10:02:03.730" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="505" PostTypeId="2" ParentId="501" CreationDate="2013-10-07T15:29:42.863" Score="18" Body="&lt;p&gt;Our own magnetic field is generated by convection currents in Earth's liquid outer core.&lt;/p&gt;&#xA;&lt;p&gt;A useful summary from Physics.org:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Differences in temperature, pressure and composition within the outer core cause convection currents in the molten metal as cool, dense matter sinks whilst warm, less dense matter rises.&lt;/p&gt;&#xA;&lt;p&gt;This flow of liquid iron generates electric currents, which in turn produce magnetic fields.&lt;/p&gt;&#xA;&lt;p&gt;The spiralling caused by the Coriolis force means that separate magnetic fields created are roughly aligned in the same direction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Mars used to have a liquid iron core, but it was never as extensive as Earth's and has long since solidified.&lt;/p&gt;&#xA;&lt;p&gt;Once our own core cools enough to solidify, we too will lose our magnetic field.&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2013-10-07T15:29:42.863" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="506" PostTypeId="1" AcceptedAnswerId="509" CreationDate="2013-10-07T17:31:13.553" Score="6" ViewCount="418" Body="&lt;p&gt;Earth's core is a giant liquid iron ball actually. If I know well, the magnetic field of our planet (that protects the surface from some particles comes from the Sun) can exist because as Earth rotates, the liquid iron core also rotates, and it generates a magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, ferromagnetism has strict conditions. How can the liquid and hot core still be ferromagnetic? How is it possible for Earth to have a magnetic field generated by a body that is not even solid?&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="6" LastEditDate="2013-11-15T18:11:46.970" LastActivityDate="2013-12-12T14:45:08.773" Title="Earth and ferromagnetism" Tags="&lt;earth&gt;&lt;rotation&gt;&lt;core&gt;&lt;magnetic-field&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2013-12-12T14:45:14.650" ContentLicense="CC BY-SA 3.0" />
  <row Id="507" PostTypeId="1" AcceptedAnswerId="521" CreationDate="2013-10-07T19:31:52.247" Score="8" ViewCount="398" Body="&lt;p&gt;I was looking over some of the data for the terrestrial planets, and noticed that Venus, despite its cloud covering, had been mapped by the Arecibo Observatory. Also that other planets such as mars, and some of the Jovian moons were mapped by the Voyagers. So, do any maps exist for Pluto or its moons?&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="19" LastEditDate="2013-10-07T20:36:33.267" LastActivityDate="2019-07-13T12:31:04.993" Title="Have Pluto or any of its moons been mapped?" Tags="&lt;solar-system&gt;&lt;kuiper-belt&gt;&lt;terrestrial-planets&gt;&lt;pluto&gt;&lt;topographic-map&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="508" PostTypeId="1" AcceptedAnswerId="510" CreationDate="2013-10-07T19:47:48.063" Score="6" ViewCount="529" Body="&lt;p&gt;According to Kepler's second law: A planet sweeps out equal areas in equal times (A simple definition). This means that if the orbit is somewhat eccentric, the planet will move faster when on the closest approach to it's parent star, and slowest when at aphelion. So why does this happen, is the curvature of space-time greater closer to the star?&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2014-01-01T23:19:03.003" Title="Why Do Planets Revolve Faster When they Are Closer to Their Parent Star?" Tags="&lt;star-systems&gt;&lt;space-time&gt;&lt;eccentric-orbit&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="509" PostTypeId="2" ParentId="506" CreationDate="2013-10-07T19:53:29.573" Score="7" Body="&lt;p&gt;Actually, the core has two parts. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Outer_core&quot;&gt;outer core&lt;/a&gt; is liquid, while the &lt;a href=&quot;http://en.wikipedia.org/wiki/Inner_core&quot;&gt;inner core&lt;/a&gt; is solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As explained in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Earth%27s_magnetic_field&quot;&gt;Wikipedia article about Earth's magnetic field&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Earth's magnetic field is mostly caused by electric currents in the liquid outer core, which is composed of highly conductive molten iron.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="38" LastActivityDate="2013-10-07T19:53:29.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="510" PostTypeId="2" ParentId="508" CreationDate="2013-10-07T19:58:51.387" Score="10" Body="&lt;p&gt;Velocity is a form of kinetic energy, while height within a gravity well is a form of potential energy. For an orbiting body, conservation of energy will keep the total energy constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as a planet moves away from the parent star, it loses velocity and gains potential energy. As it moves closer, it trades the potential energy back for velocity. The point of lowest potential has the highest velocity and the point of highest potential has the lowest velocity.&lt;/p&gt;&#xA;" OwnerUserId="38" LastActivityDate="2013-10-07T19:58:51.387" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="511" PostTypeId="5" CreationDate="2013-10-07T20:00:48.083" Score="0" Body="&lt;p&gt;A topographic map is a view of the surface features of something (usually from above). Should be used when talking about or referencing such maps.&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditDate="2013-10-07T21:32:41.477" LastActivityDate="2013-10-07T21:32:41.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="512" PostTypeId="4" CreationDate="2013-10-07T20:00:48.083" Score="0" Body="A map of the surface features of something." OwnerUserId="160" LastEditorUserId="160" LastEditDate="2013-10-07T21:32:45.900" LastActivityDate="2013-10-07T21:32:45.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="513" PostTypeId="5" CreationDate="2013-10-07T20:07:11.977" Score="0" Body="&lt;p&gt;Questions regarding magnetic fields.&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="2153" LastEditDate="2014-12-21T04:20:35.483" LastActivityDate="2014-12-21T04:20:35.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="514" PostTypeId="4" CreationDate="2013-10-07T20:07:11.977" Score="0" Body="Questions regarding magnetic fields." OwnerUserId="160" LastEditorUserId="2153" LastEditDate="2014-12-21T04:20:45.280" LastActivityDate="2014-12-21T04:20:45.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="515" PostTypeId="5" CreationDate="2013-10-07T20:10:27.730" Score="0" Body="&lt;p&gt;Icy bodies that usually consist of methane, water, and other materials that originate from beyond the jovian planets, and occasionally fall into the inner solar system giving off a massive tail as material is boiled off.&lt;/p&gt;&#xA;&lt;h3&gt;References&lt;/h3&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Comet&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Comet&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="160" LastEditorUserId="37191" LastEditDate="2021-02-16T12:28:48.473" LastActivityDate="2021-02-16T12:28:48.473" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="516" PostTypeId="4" CreationDate="2013-10-07T20:10:27.730" Score="0" Body="Icy bodies that usually consist of methane, water, and other materials that originate from beyond the jovian planets, and occasionally fall into the inner solar system giving off a massive tail as material is boiled off." OwnerUserId="160" LastEditorUserId="160" LastEditDate="2013-10-07T21:32:56.277" LastActivityDate="2013-10-07T21:32:56.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="517" PostTypeId="5" CreationDate="2013-10-07T20:12:29.680" Score="0" Body="&lt;p&gt;Questions regarding the former 9th (now dwarf) planet Pluto.&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditDate="2013-10-07T21:32:28.133" LastActivityDate="2013-10-07T21:32:28.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="518" PostTypeId="4" CreationDate="2013-10-07T20:12:29.680" Score="0" Body="Questions regarding the former 9th (now dwarf) planet Pluto." OwnerUserId="160" LastEditorUserId="160" LastEditDate="2013-10-07T21:32:48.697" LastActivityDate="2013-10-07T21:32:48.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="519" PostTypeId="2" ParentId="16" CreationDate="2013-10-07T20:14:44.190" Score="18" Body="&lt;p&gt;The other answers here are fantastic at explaining in a technical sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an everyday example, imagine taking something asymmetrical--like a marble with some clay stuck to it--and spinning it. The asymmetry eventually brings the object to spin in a certain way. The moon is like that, except way more complicated, since its interaction with Earth is a part of the equation.&lt;/p&gt;&#xA;" OwnerUserId="158" LastActivityDate="2013-10-07T20:14:44.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="520" PostTypeId="1" AcceptedAnswerId="522" CreationDate="2013-10-07T20:25:47.207" Score="10" ViewCount="428" Body="&lt;p&gt;So I, like most other astronomy enthusiasts, have, at some point, not been able to see something clearly, such as a star, and have had to squint to be able to see it better. When this happens, the object seems to become clearer, but the light seems to gain a cross, or sometimes a hexagon, or an irregular line shape when doing so. I've noticed a similar effect on different models of telescopes, especially the newtonian reflector series. More on this here: (&lt;a href=&quot;http://www.youtube.com/watch?v=ipe3NN1yPzM&quot;&gt;http://www.youtube.com/watch?v=ipe3NN1yPzM&lt;/a&gt;). Is this a similar effect on the eye? But more importantly how does squinting improve the quality of the picture?&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="19" LastEditDate="2013-10-08T00:09:09.423" LastActivityDate="2013-10-08T00:09:09.423" Title="Why does squinting make hard-to-see objects clearer?" Tags="&lt;star&gt;&lt;newtonian-telescope&gt;&lt;telescope&gt;&lt;space&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="521" PostTypeId="2" ParentId="507" CreationDate="2013-10-07T20:53:03.157" Score="14" Body="&lt;p&gt;Yes, &lt;strike&gt;but they're not very good&lt;/strike&gt; &lt;em&gt;and they're amazing&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Pluto&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article for Pluto&lt;/a&gt; shows a low-resolution map of the surface, generated from Hubble images:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y3C3c.gif&quot; alt=&quot;Spinning animated gif of Pluto&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the &lt;a href=&quot;https://en.wikipedia.org/wiki/Charon_(moon)&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article for Pluto's largest moon Charon&lt;/a&gt; shows a low-resolution map of the Pluto-facing side of Charon (not to scale):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NNcJv.png&quot; alt=&quot;Charon map&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/File:Charon.png&quot; rel=&quot;noreferrer&quot;&gt;Larger image here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only the Pluto-facing side is shown because the map was generated from brightness variations during Pluto–Charon occultations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maps with similar resolution would show no details of any of the other moons, which are much smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expect &lt;em&gt;much&lt;/em&gt; better images in &lt;a href=&quot;https://en.wikipedia.org/wiki/New_Horizons&quot; rel=&quot;noreferrer&quot;&gt;a couple of years&lt;/a&gt;, likely including moons that haven't yet been discovered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE :&lt;/strong&gt; Thanks to user HDE 226868 for reminding me: it's been a couple of years, and we have those &lt;em&gt;much&lt;/em&gt; better images now. Here's one from New Horizons via Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/2/2a/Nh-pluto-in-true-color_2x_JPEG-edit-frame.jpg&quot; alt=&quot;Pluto in true color&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the question was about a &lt;em&gt;map&lt;/em&gt; of Pluto, here you go. You can see the much lower resolution on the opposite hemisphere, which New Horizons saw only from a greater distance (it was a flyby mission).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://web.archive.org/web/20161223103605/http://pluto.jhuapl.edu/Multimedia/Science-Photos/pics/pmap_cyl_PS723_HR.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://web.archive.org/web/20161223103605/http://pluto.jhuapl.edu/Multimedia/Science-Photos/pics/pmap_cyl_PS723_HR.jpg&quot; alt=&quot;Map of Pluto&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also have some good images of Charon:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/2/2b/Charon_in_Color_(HQ).jpg&quot; alt=&quot;Charon&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here's a map of Charon:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://img.buzzfeed.com/buzzfeed-static/static/2015-07/27/17/enhanced/webdr03/longform-original-30076-1438032065-23.png&quot; alt=&quot;map of Charon&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and some lower resolution images of the other moons, &lt;a href=&quot;https://en.wikipedia.org/wiki/Styx_(moon)&quot; rel=&quot;noreferrer&quot;&gt;Styx&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Nix_(moon)&quot; rel=&quot;noreferrer&quot;&gt;Nix&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Kerberos_(moon)&quot; rel=&quot;noreferrer&quot;&gt;Kerberos&lt;/a&gt;, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydra_(moon)&quot; rel=&quot;noreferrer&quot;&gt;Hydra&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess that New Horizons would discover new moons turned out to be wrong.&lt;/p&gt;&#xA;" OwnerUserId="53" LastEditorUserId="7472" LastEditDate="2019-07-13T12:31:04.993" LastActivityDate="2019-07-13T12:31:04.993" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="522" PostTypeId="2" ParentId="520" CreationDate="2013-10-07T21:02:53.990" Score="13" Body="&lt;p&gt;Squinting works the same way as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Pinhole_camera&quot;&gt;pinhole camera&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally, light from a single point source entering your eye anywhere on your pupil will be focused on a single spot on your retina. But this works perfectly only if you have perfect vision; otherwise light entering near the top of your pupil may be directed to a slightly different spot on your retina than light entering near the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By squinting, you block out some of the light from the edges, effectively making your pupil narrower, creating a sharper but dimmer image on your retina. (You may find that it improves the vertical resolution more than the horizontal resolution.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you happen to be nearsighted (as I am), you can see a similar effect by looking through a small pinhole, or through a small aperture made with your fingers. If the light is bright enough, you'll see a dimmer but sharper image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The irregular shapes may be interference from your eyelashes.&lt;/p&gt;&#xA;" OwnerUserId="53" LastActivityDate="2013-10-07T21:02:53.990" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="523" PostTypeId="1" CreationDate="2013-10-08T09:16:20.597" Score="5" ViewCount="199" Body="&lt;p&gt;After reading the question and answer to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/424/effect-of-charon-on-pluto&quot;&gt;Effect of Charon on Pluto&lt;/a&gt;, a further question is this: how is the Pluto-Charon relationship explained?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, what theories are there in how did the Pluto-Charon system was formed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, does this explain the presence of other moons (Nix, Hydra etc)?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2013-12-13T14:06:49.097" Title="What theories are there in explaining the formation of the Pluto-Charon system?" Tags="&lt;pluto&gt;&lt;charon&gt;&lt;formation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="524" PostTypeId="1" AcceptedAnswerId="1134" CreationDate="2013-10-08T13:42:57.767" Score="15" ViewCount="433" Body="&lt;p&gt;The Sun exhibits a periodic cycle for solar activity as manifested by solar activity (e.g. sunspots, solar flares). This activity is related to magnetic fields which emerge from the interior of the Sun. The magnetic fields are created by the solar dynamo. The Earth has a dynamo as well but the Earth's magnetic field is (mostly) constant in time. What is some current theories (assuming there isn't one already accepted) that explains for why the solar dynamo generates magnetic fields periodically?&lt;/p&gt;&#xA;" OwnerUserId="117" LastEditorUserId="6" LastEditDate="2013-12-12T15:26:19.920" LastActivityDate="2019-06-15T04:55:09.047" Title="What are some theories that explain the 11 year solar cycle?" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="525" PostTypeId="1" AcceptedAnswerId="899" CreationDate="2013-10-08T13:43:41.017" Score="15" ViewCount="1010" Body="&lt;p&gt;I had heard that the large radiation belts around Jupiter may be formed by liquid metallic hydrogen in (or around) Jupiter's core (which Wikipedia says haven't been observed in labs yet due to the immense pressure required), but I don't understand how the liquid metallic hydrogen can form a planetary magnetic field. The article also says that the hydrogen can act as a conductor, is this somehow related to moving electric fields forming magnetic fields? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetosphere_of_Jupiter&quot;&gt;Magnetosphere of Jupiter - Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="6" LastEditDate="2021-02-22T17:04:07.630" LastActivityDate="2021-03-31T15:43:17.873" Title="What Causes the Large Radiation Fields Around Jupiter?" Tags="&lt;solar-system&gt;&lt;jupiter&gt;&lt;magnetic-field&gt;&lt;terrestrial-planets&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="526" PostTypeId="1" AcceptedAnswerId="527" CreationDate="2013-10-08T13:52:36.570" Score="16" ViewCount="384" Body="&lt;p&gt;This effect is frequently discussed in the context of solar time series. In particular, it a relationship between soft x-ray (thermal emission) and hard x-ray (nonthermal emission) from the Sun.&lt;/p&gt;&#xA;" OwnerUserId="117" LastActivityDate="2018-12-26T16:35:36.367" Title="What is the Neupert Effect?" Tags="&lt;the-sun&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="527" PostTypeId="2" ParentId="526" CreationDate="2013-10-08T13:58:50.323" Score="15" Body="&lt;p&gt;The Neupert Effect describes the commonly observed phenomenon that hard X-rays occur predominantly during the rise phase of soft X-rays during a solar flare, as illustrated by the graph below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4tw1y.gif&quot; alt=&quot;Neupert Effect during a flare observed by Yohkoh&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the graph, the first peak is that of the hard X-rays, which you can see occurs during the rise of the other line which represents the soft X-rays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://solar.physics.montana.edu/nuggets/1998/981016/981016.html&quot; rel=&quot;noreferrer&quot;&gt;The Neupert Effect - Solar Physics at Montana State University&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6" LastEditorUserId="7472" LastEditDate="2018-12-26T12:19:30.293" LastActivityDate="2018-12-26T12:19:30.293" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="528" PostTypeId="1" AcceptedAnswerId="591" CreationDate="2013-10-08T14:49:00.587" Score="8" ViewCount="102" Body="&lt;p&gt;X-ray binaries such as GRS-1915 feature a black hole and a regular star. The x-ray luminosity is likely related to the rate of mass flow into the black hole from the star so when there is a lot of mass flow the x-ray luminosity is high but this explanation seems to just beg the question as to why the mass flow would change over time.&lt;/p&gt;&#xA;" OwnerUserId="117" LastActivityDate="2013-10-09T15:18:13.703" Title="Why do X-ray binaries such as GRS-1915 have active and quiet states?" Tags="&lt;black-hole&gt;&lt;binary-star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="529" PostTypeId="1" AcceptedAnswerId="530" CreationDate="2013-10-08T20:06:04.260" Score="11" ViewCount="543" Body="&lt;p&gt;I've read once a popular science book, in which the author calculated how long would it take for the most massive black holes to evaporate because of Hawking radiation. He claimed that after that time the Universe will be the soup filled only with elementary particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what would happen with celestial bodies that aren't black holes? For example, a rogue planet of the size of Earth, drifting somewhere in deep space, in an extremely cold Universe? What known effects would cause that planet to evaporate? How long would it take?&lt;/p&gt;&#xA;" OwnerUserId="42" LastEditorUserId="2365" LastEditDate="2015-11-06T17:12:56.923" LastActivityDate="2016-04-18T20:20:10.140" Title="How long would it take for a rogue planet to evaporate in the late stages of the Universe?" Tags="&lt;cosmology&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="530" PostTypeId="2" ParentId="529" CreationDate="2013-10-08T20:22:36.780" Score="13" Body="&lt;p&gt;Before the evaporation of black holes that you mention, our universe will enter what is known as the &lt;strong&gt;Black Hole Era&lt;/strong&gt;. About $10^{40}$ years from now, all the protons and neutrons in the universe will have decayed into positrons and neutrinos. This is how the last of the planets will pass away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Black Hole Era is so named because at this point only black holes and elementary particles will remain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted that there is currently no evidence for proton decay, and so this phase of the Universe is strictly theoretical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.deepastronomy.com/the-decay-of-heaven.html&quot;&gt;The Decay of Heaven - DeepAstronomy.com&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6" LastEditorUserId="6" LastEditDate="2016-04-14T19:03:09.143" LastActivityDate="2016-04-14T19:03:09.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="531" PostTypeId="1" CreationDate="2013-10-08T20:39:08.893" Score="11" ViewCount="257" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Summa_Technologiae&quot; rel=&quot;noreferrer&quot;&gt;Summa Technologiae&lt;/a&gt;, a book by a Polish author &lt;a href=&quot;https://en.wikipedia.org/wiki/Stanis%C5%82aw_Lem&quot; rel=&quot;noreferrer&quot;&gt;Stanisław Lem&lt;/a&gt; that based his science fiction novels on scientific research of the era, the Earth crossed in the moment of forming of life the arms of the Galaxy, with high supernovae activity, which triggered life creation. Afterwards, it moved to regions with low supernovae activity, which enabled the preservation of existing life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to test the validity of this claim against current research. How often does the Earth cross regions with higher concentrations of stars (and therefore, much greater chance of gamma burst wipe)? Additionally, how many times the probability of nearby supernova explosion is higher in that regions?&lt;/p&gt;&#xA;" OwnerUserId="42" LastEditorUserId="13" LastEditDate="2013-10-12T07:26:50.137" LastActivityDate="2014-01-14T17:27:42.957" Title="How often does Earth cross the galaxy regions with higher supernovae activity?" Tags="&lt;earth&gt;&lt;galaxy&gt;&lt;milky-way&gt;&lt;supernova&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="533" PostTypeId="1" CreationDate="2013-10-08T21:18:50.483" Score="8" ViewCount="358" Body="&lt;p&gt;I was thinking about the impending collision between M31 and the milky way galaxy, and I've heard that since a galaxy is mostly space most of the contents of both galaxies would be just fine. Could a similar effect be responsible for the late heavy bombardment era? Just instead of galaxies, stellar systems?&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2014-09-23T12:08:26.293" Title="Did the Late Heavy Bombardment Period Happen Because of a Stellar System Collision?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;impact&gt;&lt;star-systems&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="534" PostTypeId="1" AcceptedAnswerId="582" CreationDate="2013-10-08T10:00:16.603" Score="21" ViewCount="15593" Body="&lt;p&gt;Let's assume there are three types of telescopes with the following specifications:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    Telescope                  Aperture              Focal Length&#xA;    --------------------------------------------------------------&#xA;    Example 1                     70 mm                    400 mm&#xA;    Example 2                     60 mm                    700 mm&#xA;    Example 3                     60 mm                    900 mm&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As a beginner, should I look for a higher value in the aperture or the focal length?&lt;/p&gt;&#xA;" OwnerUserId="210" OwnerDisplayName="user1754665" LastEditorUserId="19" LastEditDate="2013-10-08T22:28:00.960" LastActivityDate="2013-10-09T09:21:23.340" Title="Should I focus more on Aperture or Focal Length for a telescope?" Tags="&lt;telescope&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="535" PostTypeId="5" CreationDate="2013-10-08T22:06:06.837" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-08T22:06:06.837" LastActivityDate="2013-10-08T22:06:06.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="536" PostTypeId="4" CreationDate="2013-10-08T22:06:06.837" Score="0" Body="Questions regarding two objects colliding into each other with considerable force." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-10-09T14:55:07.480" LastActivityDate="2013-10-09T14:55:07.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="537" PostTypeId="1" AcceptedAnswerId="543" CreationDate="2013-10-08T22:18:17.050" Score="26" ViewCount="23034" Body="&lt;ul&gt;&#xA;&lt;li&gt;Is there a reason that the Earth has the tilt that it does (~23°)?  &lt;/li&gt;&#xA;&lt;li&gt;How do we know which way is supposed to be 0°? &lt;/li&gt;&#xA;&lt;li&gt;Does this tilt have major consequences on the planet? &lt;/li&gt;&#xA;&lt;li&gt;Has it changed and will it change in the future? (If so, would there be any consequences?)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="180" LastEditorUserId="181" LastEditDate="2013-10-09T02:24:41.730" LastActivityDate="2016-06-04T02:38:08.137" Title="Why does the Earth have a tilt of ~23°?" Tags="&lt;earth&gt;&lt;rotation&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="538" PostTypeId="2" ParentId="534" CreationDate="2013-10-08T22:38:07.933" Score="2" Body="&lt;p&gt;Right or wrong, myself as a beginner I started my decision based on aperture to get the appropriate light gathering ability for my area.  At that point focal length became pretty much a non-issue.  My choices were down to 2, and 1 of them would have been long enough that getting it outside easily would have been difficult.&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2013-10-08T22:38:07.933" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="539" PostTypeId="1" AcceptedAnswerId="544" CreationDate="2013-10-08T23:43:31.327" Score="58" ViewCount="36668" Body="&lt;p&gt;So, in school (that's a long time ago) they have been teaching us there are 9 planets in our solar system.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Mercury&lt;/li&gt;&#xA;&lt;li&gt;Venus&lt;/li&gt;&#xA;&lt;li&gt;Earth&lt;/li&gt;&#xA;&lt;li&gt;Mars&lt;/li&gt;&#xA;&lt;li&gt;Jupiter&lt;/li&gt;&#xA;&lt;li&gt;Saturn&lt;/li&gt;&#xA;&lt;li&gt;Uranus&lt;/li&gt;&#xA;&lt;li&gt;Neptune&lt;/li&gt;&#xA;&lt;li&gt;Pluto&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;But every now and then I keep reading stories about another &amp;quot;dwarf planet&amp;quot; (Eris, discovered in 2005) that - depending on what source tells the story - is another planet according to the astronomical definition, while other sources say that it isn't a planet. Some even say Pluto isn't a planet anymore either.&lt;/p&gt;&#xA;&lt;p&gt;The result: I'm confused due to the contradicting stories. Even &lt;a href=&quot;http://en.wikipedia.org/wiki/Eris_%28dwarf_planet%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia isn't clear about Eris&lt;/a&gt; and only writes (emphasis mine):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;NASA &lt;em&gt;&lt;strong&gt;initially&lt;/strong&gt;&lt;/em&gt; described it as the Solar System's tenth planet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Initially? So, is it a 10th planet or not? Fact is, there is another &amp;quot;something&amp;quot; out there and it surely seems to look like a planet. Yet, some people keep stating there are 9 planets in our solar system, while others say there are &lt;em&gt;more&lt;/em&gt; than 9 planets, and then again there are people stating that the latest definition of &amp;quot;planet&amp;quot; has kicked out Pluto too so there are actually &lt;em&gt;fewer&lt;/em&gt; than 9 planets in our solar system.&lt;/p&gt;&#xA;&lt;p&gt;Trying to get a definite, official, and astronomically correct answer I can actually rely on, I'm therefore asking: &lt;em&gt;&lt;strong&gt;How many planets are there in this solar system?&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Definition_of_planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;&amp;quot;Definition of planet&amp;quot; at Wikipedia&lt;/a&gt; doesn't really help either, as it states:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Many astronomers, claiming that the definition of planet was of little scientific importance, preferred to recognize Pluto's historical identity as a planet by &amp;quot;grandfathering&amp;quot; it into the planet list.*&lt;/p&gt;&#xA;&lt;p&gt;&lt;sub&gt;* Dr. Bonnie Buratti (2005), &amp;quot;Topic — First Mission to Pluto and the Kuiper Belt; &amp;quot;From Darkness to Light: The Exploration of the Planet Pluto&amp;quot;&amp;quot;, Jet Propulsion Laboratory. Retrieved 2007-02-22.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So, if you link somewhere to provide proof, it would be great if you could point me to a more trusted source than Wikipedia. Ideally, an astronomical trusted source and/or paper.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="19" LastEditDate="2021-01-26T01:37:02.710" LastActivityDate="2021-01-26T01:37:02.710" Title="How many planets are there in this solar system?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;dwarf-planets&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="17" ContentLicense="CC BY-SA 4.0" />
  <row Id="540" PostTypeId="5" CreationDate="2013-10-08T23:45:41.743" Score="0" Body="&lt;p&gt;A terrestrial planet is a planet that has a compact rocky surface like the Earth's; the four innermost planets in the solar system.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="181" LastEditDate="2013-10-09T14:51:46.810" LastActivityDate="2013-10-09T14:51:46.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="541" PostTypeId="4" CreationDate="2013-10-08T23:45:41.743" Score="0" Body="compact rocky planets, in our solar system these would be Mercury, Venus, Earth &amp; Mars " OwnerUserId="181" LastEditorUserId="7982" LastEditDate="2021-02-17T08:46:58.817" LastActivityDate="2021-02-17T08:46:58.817" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="542" PostTypeId="2" ParentId="539" CreationDate="2013-10-09T00:04:46.763" Score="12" Body="&lt;p&gt;The correct answer is 8 (Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pluto is not longer a planet since 2006 when the IAU adopted a &lt;a href=&quot;http://en.wikipedia.org/wiki/Definition_of_planet&quot; rel=&quot;noreferrer&quot;&gt;formal definition of planet&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="184" LastEditorUserId="-1" LastEditDate="2016-01-23T13:19:29.780" LastActivityDate="2016-01-23T13:19:29.780" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="543" PostTypeId="2" ParentId="537" CreationDate="2013-10-09T00:25:49.420" Score="25" Body="&lt;p&gt;First up, the tilt is exactly 23.45 degrees.&lt;/p&gt;&#xA;&lt;p&gt;The reason for Earth's tilt is still not yet really proven, but &lt;a href=&quot;https://www.princeton.edu/main/news/archive/S15/64/72A37/index.xml?section=newsreleases&quot; rel=&quot;noreferrer&quot;&gt;scientists at Princeton stated on August 25, 2006&lt;/a&gt; that planet Earth may have 'tilted' to keep its balance. Quote:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;By analyzing the magnetic composition of ancient sediments found in the remote Norwegian archipelago of Svalbard, Princeton University's Adam Maloof has lent credence to a 140-year-old theory regarding the way the Earth might restore its own balance if an unequal distribution of weight ever developed in its interior or on its surface.&lt;/p&gt;&#xA;&lt;p&gt;The theory, known as true polar wander, postulates that if an object of sufficient weight -- such as a supersized volcano -- ever formed far from the equator, the force of the planet's rotation would gradually pull the heavy object away from the axis the Earth spins around. If the volcanoes, land and other masses that exist within the spinning Earth ever became sufficiently imbalanced, the planet would tilt and rotate itself until this extra weight was relocated to a point along the equator.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Same goes for the lack of proof about the exact consequences on the planet or if it'll happen again. All that is still being researched and debated, but those Princeton scientists have thrown in some interesting perspectives which you'll discover when you visit the link I provided and completely read what they wrote.&lt;/p&gt;&#xA;&lt;p&gt;Besides &amp;quot;consequences&amp;quot; for Earth as a planet, it should be noted that the tilt of Earth is the reason why we have seasons. So even when we yet have to find out what the consequences for Earth (as a planet) have been and/or will be, we do know that the tilt surely has consequences for all beings that live on planet Earth… the seasons that influence us all: summer, fall, winter, and spring.&lt;/p&gt;&#xA;&lt;p&gt;And - last but not least - as for your question how they know where 0 degrees would be, you just have to look at the Earth's rotation around the sun. Take a 90° angle of the planet's orbit and you know where 0° would be. But I guess a picture explains more than a thousand words, so I quickly created the following graphic:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/K7p3s.png&quot; alt=&quot;tilt and axis explained&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2013-10-09T06:18:12.093" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="544" PostTypeId="2" ParentId="539" CreationDate="2013-10-09T00:34:45.137" Score="77" Body="&lt;p&gt;Since we're talking about terminology, we need to remember that none of this &lt;em&gt;really&lt;/em&gt; matters, outside of clarity when communicating. Still, some people tend to have rather strong opinions on it, thus confusion about how many planets are &lt;em&gt;really&lt;/em&gt; in the solar system arises.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The people&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The most trusted source in Astronomy would have to be the people that set the generally accepted rules. The &lt;a href=&quot;http://www.iau.org/about/&quot;&gt;IAU (International Astronomical Union)&lt;/a&gt; has been in existence since 1919 and is comprised of 10814 Individual Members in 93 different countries worldwide. Of those countries, 73 are National Members.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The key activity of the IAU is the organization of scientific meetings. Every year the IAU sponsors nine international IAU Symposia. The IAU Symposium Proceedings series is the flagship of the IAU publications. Every three years the IAU holds a General Assembly, which offers six IAU Symposia, some 25 Joint Discussions and Special Sessions, and individual business and scientific meetings of Divisions, Commissions, and Working Groups. The proceedings of Joint Discussions and Special Sessions are published in the Highlights of Astronomy series. The reports of the GA business meetings are published in the Transactions of the IAU - B series.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;The definition&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;At the 2006 IAU General Assembly in Prague, the accepted definition of a planet was debated vigorously. The outcome of the meeting was the currently accepted definition of a planet:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;A celestial body that (a) is in orbit around the Sun, (b) has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape, and (c) has cleared the neighbourhood around its orbit.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;What about Pluto?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;With this in mind, the group decided on Pluto's fate. From page 2 of &lt;a href=&quot;http://www.iau.org/static/resolutions/Resolution_GA26-5-6.pdf&quot;&gt;this official resolution document:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The IAU further resolves: &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Pluto is a &quot;dwarf planet&quot; by the above definition and is recognized as the &#xA;  prototype of a new category of Trans-Neptunian Objects. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason given in the FAQ on &lt;a href=&quot;http://www.iau.org/public/themes/pluto/&quot;&gt;this page&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Q: Why is Pluto now called a dwarf planet?&lt;/strong&gt; &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A: Pluto now falls into the dwarf planet category on account of its size and the fact that it resides within a zone of other similarly-sized objects known as the transneptunian region. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Basically, they decided that it isn't officially a planet anymore because it didn't match criteria &lt;em&gt;(c): has cleared the neighbourhood around its orbit.&lt;/em&gt; It hasn't done this, because it 'resides within a zone of &lt;strong&gt;other similarly-sized objects&lt;/strong&gt;'. Therefore, it hasn't cleared its neighborhood.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Soo... what about the number of planets?&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Q: Based on this new definition, how many planets are there in our Solar System?&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A: There are eight planets in our Solar System; Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune. Mnemonic: My Very Educated Mother Just Served Us Nachos.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But that's if you don't count Dwarf planets - if you &lt;em&gt;do&lt;/em&gt; count them, you end up with &lt;a href=&quot;http://solarsystem.nasa.gov/planets/profile.cfm?Object=Dwarf&amp;amp;Display=Sats&quot;&gt;five more&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ceres&lt;/li&gt;&#xA;&lt;li&gt;Pluto&lt;/li&gt;&#xA;&lt;li&gt;Eris&lt;/li&gt;&#xA;&lt;li&gt;Makemake&lt;/li&gt;&#xA;&lt;li&gt;Haumea&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So there are &lt;strong&gt;8&lt;/strong&gt; planets in the solar system if you don't count Dwarfs, &lt;strong&gt;13&lt;/strong&gt; if you do.&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-10-09T00:44:22.260" LastActivityDate="2013-10-09T00:44:22.260" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="545" PostTypeId="1" AcceptedAnswerId="574" CreationDate="2013-10-09T01:05:29.667" Score="18" ViewCount="8145" Body="&lt;p&gt;For many years scientists have studied our own solar system and modern technology allows them to look deeper and deeper into space. Knowing the Sun is only one of a few billion stars in the Milky Way galaxy, I can imagine scientists also started studying other planetary systems… at least enough to know how many planetary systems (like our solar system) reside in our galaxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any information available that would be able to (at least &quot;indicate&quot;) how many planetary systems reside in our galaxy?&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="6" LastEditDate="2014-03-05T13:51:17.070" LastActivityDate="2014-03-05T13:51:17.070" Title="How many planetary systems exist in our galaxy?" Tags="&lt;solar-system&gt;&lt;galaxy&gt;&lt;milky-way&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="546" PostTypeId="1" AcceptedAnswerId="547" CreationDate="2013-10-09T01:12:39.320" Score="40" ViewCount="7362" Body="&lt;p&gt;Our own solar system is &quot;tipped&quot; by about 63 degrees with respect to the plane of the galaxy. Has it ever been researched or is there any scientific theory which could explain the reason why our solar system and solar systems in general are not &quot;in-line&quot; to the galaxy's plane? What exactly determines the orientation of solar systems?&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="6" LastEditDate="2021-02-22T17:03:43.683" LastActivityDate="2021-02-22T17:03:43.683" Title="Why is our solar system &quot;tipped&quot; about 63° with respect to the plane of our galaxy?" Tags="&lt;galaxy&gt;&lt;solar-system&gt;&lt;inclination&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="547" PostTypeId="2" ParentId="546" CreationDate="2013-10-09T01:31:30.640" Score="18" Body="&lt;p&gt;The tilt of our solar system (or any star system) is determined by the net angular momentum of the gas cloud from which it formed. This might be a bit of a vague answer, but over time, the formation of stars and their respective planets is thought to look something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fMKYx.jpg&quot; alt=&quot;Formation of a star&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other influences (net forces: maybe nearby massive objects, or other components of the galaxy) on the tilt of the plane of a stellar system could certainly cause a precession of its axis with time. As an example of this see the precession of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Gyroscope&quot; rel=&quot;noreferrer&quot;&gt;gyroscope&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be really interesting to know the distribution of angles between the plane of the galaxy and the plane of stellar systems, though I'd bet that many of them are simply aligned with the disk itself.&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2013-10-09T01:31:30.640" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="548" PostTypeId="5" CreationDate="2013-10-09T02:05:45.500" Score="0" Body="" OwnerUserId="181" LastEditorUserId="4160" LastEditDate="2014-12-30T12:22:49.960" LastActivityDate="2014-12-30T12:22:49.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="549" PostTypeId="4" CreationDate="2013-10-09T02:05:45.500" Score="0" Body="Temperature is a measurement of the average kinetic energy of the molecules in an object or system. In simpler words: it's the degree or intensity of heat present in a substance or object that is commonly measured with a thermometer." OwnerUserId="4160" LastEditorUserId="4160" LastEditDate="2014-12-30T12:22:04.303" LastActivityDate="2014-12-30T12:22:04.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="550" PostTypeId="5" CreationDate="2013-10-09T02:08:37.293" Score="0" Body="&lt;p&gt;Constellations are arbitrary formations of stars perceived as a figure or design, especially one of 88 recognized groups, many of which were named by the ancient Greeks after animals, objects, or mythological persons.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="181" LastEditDate="2013-10-09T14:51:34.997" LastActivityDate="2013-10-09T14:51:34.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="551" PostTypeId="4" CreationDate="2013-10-09T02:08:37.293" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-09T02:08:37.293" LastActivityDate="2013-10-09T02:08:37.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="552" PostTypeId="5" CreationDate="2013-10-09T02:10:18.567" Score="0" Body="&lt;p&gt;Inclination defines the angle between the plane of the orbit of a planet or comet and another plane, usually that of the ecliptic.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="181" LastEditDate="2013-10-09T14:51:37.353" LastActivityDate="2013-10-09T14:51:37.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="553" PostTypeId="4" CreationDate="2013-10-09T02:10:18.567" Score="0" Body="Inclination defines the angle between the plane of the orbit of a planet or comet and another plane, usually that of the ecliptic." OwnerUserId="181" LastEditorUserId="181" LastEditDate="2013-10-09T14:51:23.337" LastActivityDate="2013-10-09T14:51:23.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="554" PostTypeId="5" CreationDate="2013-10-09T02:12:39.313" Score="0" Body="&lt;p&gt;Space debris (also known as &quot;orbital debris&quot;, &quot;space junk&quot;, or &quot;space waste&quot;) is the collection of defunct objects in orbit around Earth and includes everything from spent rocket stages, old satellites, fragments from disintegration, erosion, and collisions. &lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="181" LastEditDate="2013-10-09T14:55:03.247" LastActivityDate="2013-10-09T14:55:03.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="555" PostTypeId="4" CreationDate="2013-10-09T02:12:39.313" Score="0" Body="Space debris (also known as &quot;orbital debris&quot;, &quot;space junk&quot;, or &quot;space waste&quot;) is the collection of defunct objects in orbit around Earth and includes everything from spent rocket stages, old satellites, fragments from disintegration, erosion, and collisions. " OwnerUserId="181" LastEditorUserId="181" LastEditDate="2013-10-09T14:55:29.703" LastActivityDate="2013-10-09T14:55:29.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="556" PostTypeId="5" CreationDate="2013-10-09T02:16:18.957" Score="0" Body="&lt;p&gt;Naming astronomical objects using astronomical naming conventions, which includes assigning names to the most interesting objects — like in Planetary nomenclature and Meteorite Nomenclature — and (where relevant) naming important or interesting features of those objects. &lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="181" LastEditDate="2013-10-09T14:55:45.780" LastActivityDate="2013-10-09T14:55:45.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="557" PostTypeId="4" CreationDate="2013-10-09T02:16:18.957" Score="0" Body="Naming astronomical objects using astronomical naming conventions, which includes assigning names to the most interesting objects — like in Planetary nomenclature and Meteorite Nomenclature — and (where relevant) naming important or interesting features of those objects. " OwnerUserId="181" LastEditorUserId="181" LastEditDate="2013-10-09T14:56:35.597" LastActivityDate="2013-10-09T14:56:35.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="558" PostTypeId="5" CreationDate="2013-10-09T02:19:39.317" Score="0" Body="&lt;p&gt;Stellar dynamics describe the branch of astronomy that studies the regularities in the motions of stars in gravitational fields of a stellar systems, as well as the evolution of stellar systems.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="181" LastEditDate="2013-10-09T14:56:02.237" LastActivityDate="2013-10-09T14:56:02.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="559" PostTypeId="4" CreationDate="2013-10-09T02:19:39.317" Score="0" Body="Questions about the branch of astronomy that studies the regularities in the motions of stars in gravitational fields of stellar systems, as well as the evolution of stellar systems." OwnerUserId="181" LastEditorUserId="9301" LastEditDate="2016-10-04T20:45:06.750" LastActivityDate="2016-10-04T20:45:06.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="560" PostTypeId="5" CreationDate="2013-10-09T02:28:49.043" Score="0" Body="&lt;p&gt;A dobsonian telescope (popularized by John Dobson) is a relatively inexpensive Newtonian telescope, in which the tube assembly slips freely in the lower base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dobsonian telescopes can even build at home by amateurs as PopularMechanics shows in their related DIY project:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.popularmechanics.com/science/space/telescopes/4286700&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ex3zP.jpg&quot; alt=&quot;DIY Dobsonian telescope&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="181" LastEditDate="2013-10-09T14:55:35.140" LastActivityDate="2013-10-09T14:55:35.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="561" PostTypeId="4" CreationDate="2013-10-09T02:28:49.043" Score="0" Body="A Dobsonian telescope is a relatively inexpensive Newtonian telescope, in which the tube assembly slips freely in the lower base." OwnerUserId="181" LastEditorUserId="181" LastEditDate="2013-10-09T14:56:43.647" LastActivityDate="2013-10-09T14:56:43.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="562" PostTypeId="5" CreationDate="2013-10-09T02:31:35.953" Score="0" Body="&lt;p&gt;The Kuiper belt is a disk-shaped belt of icy debris orbiting in the outer solar system, lying beyond the orbit of Neptune and extending to a distance of about 50 astronomical units, containing thousands of small icy bodies, some of which are on highly elliptical orbits, periodically visiting the inner solar system as comets.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="181" LastEditDate="2013-10-09T14:55:32.203" LastActivityDate="2013-10-09T14:55:32.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="563" PostTypeId="4" CreationDate="2013-10-09T02:31:35.953" Score="0" Body="Questions about the circumstellar disk orbiting in the outer Solar System, beyond the orbit of Neptune." OwnerUserId="181" LastEditorUserId="9301" LastEditDate="2016-10-04T19:40:41.247" LastActivityDate="2016-10-04T19:40:41.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="564" PostTypeId="1" AcceptedAnswerId="636" CreationDate="2013-10-09T02:33:07.800" Score="15" ViewCount="425" Body="&lt;p&gt;From geological records in rocks and minerals we know that the magnetic field of Earth changed its polarity multiple times in the history. See &lt;a href=&quot;https://en.wikipedia.org/wiki/Geomagnetic_reversal&quot;&gt;Geomagnetic reversal&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was a similar process of a planetary magnetic field reversal observed on other planets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose that the record could be in rocks directly found on the planet (i.e. Mars), on its moons or in asteroids.&lt;/p&gt;&#xA;" OwnerUserId="188" LastEditorUserId="188" LastEditDate="2013-10-09T17:08:39.333" LastActivityDate="2021-01-05T00:50:48.133" Title="Were effects of a planetary magnetic field reversal observed on other planets than Earth?" Tags="&lt;earth&gt;&lt;planet&gt;&lt;core&gt;&lt;magnetic-field&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="565" PostTypeId="5" CreationDate="2013-10-09T02:34:05.767" Score="0" Body="&lt;p&gt;A meteor is a small body of matter from outer space that comes into the atmosphere of the Earth and that looks like a streak of light due to iridescence caused by friction on its body when entering Earth's atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="2674" LastEditDate="2014-11-09T14:20:57.873" LastActivityDate="2014-11-09T14:20:57.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="566" PostTypeId="4" CreationDate="2013-10-09T02:34:05.767" Score="0" Body="Questions on objects passing through the atmosphere which leave visible wakes of incandescent material." OwnerUserId="2674" LastEditorUserId="2674" LastEditDate="2014-11-09T14:20:16.127" LastActivityDate="2014-11-09T14:20:16.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="567" PostTypeId="5" CreationDate="2013-10-09T02:36:59.720" Score="0" Body="&lt;p&gt;An explosion describes the release of energy in a sudden and often violent manner caused by energy released from a very fast chemical reaction, a nuclear reaction, or the escape of gases under pressure.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="181" LastEditDate="2013-10-09T14:55:21.263" LastActivityDate="2013-10-09T14:55:21.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="568" PostTypeId="4" CreationDate="2013-10-09T02:36:59.720" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-09T02:36:59.720" LastActivityDate="2013-10-09T02:36:59.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="569" PostTypeId="5" CreationDate="2013-10-09T02:57:49.873" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-09T02:57:49.873" LastActivityDate="2013-10-09T02:57:49.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="570" PostTypeId="4" CreationDate="2013-10-09T02:57:49.873" Score="0" Body="Questions regarding the material at or near the center of an astronomical object." OwnerUserId="19" LastEditorUserId="19" LastEditDate="2013-10-09T14:55:59.080" LastActivityDate="2013-10-09T14:55:59.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="571" PostTypeId="2" ParentId="534" CreationDate="2013-10-09T03:49:38.017" Score="10" Body="&lt;p&gt;It depends on what you plan on using it for. For dimmer, deep sky objects you should be concerned with the balance of aperture and focal length (with a focus on aperture for light gathering power). For brighter objects, like the Moon or the solar system planets you probably don't need much light gathering power (aperture) so a long focal length for imaging finer details should be ok. Just a reminder though: never look at the Sun without proper filters, you'll go blind instantly, and looking at the Moon without filters will wreck your night vision and may hurt at first.&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2013-10-09T03:49:38.017" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="572" PostTypeId="2" ParentId="533" CreationDate="2013-10-09T06:06:01.020" Score="7" Body="&lt;p&gt;There are several potential explanation for the cause of the late heavy bombardment. One of them includes a passing star that disturbed the Oort cloud. A &quot;stellar system collision&quot; would have destroyed the solar system, though. While galaxies are mostly empty and a galactic collision won't involve that many actual collisions (if any), the gravitational interactions between the galaxies actually changes star orbits around the galactic centers. Similarly a &quot;stellar system collision&quot; would heavily change planetary orbits, even if there were no collisions at first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another hypothesis is a totally &quot;interior&quot; one: A 2:1 orbital resonance between Jupiter and Saturn changed the orbits of Neptune and Uranus, potentially even causing them to swap positions. Neptune's new orbit then affected many Kuiper belt objects, either sending them way out to form or join the Oort cloud, or sending them in as part of the late heavy bombardment.&lt;/p&gt;&#xA;" OwnerUserId="193" LastActivityDate="2013-10-09T06:06:01.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="573" PostTypeId="2" ParentId="545" CreationDate="2013-10-09T07:40:34.493" Score="4" Body="&lt;p&gt;According to observations by the Kepler space telescope and other ground based observations, it seems that about 5% of the stars in our galaxy have giant gas planets, similar to Jupiter (but often larger). Smaller planets are difficult to detect, but it is estimated that 40% of the stars have small planets orbiting them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all it is said that on average, 1.6 planets are orbiting each star. So almost all stars form a kind of solar system with planets. However, there is still a lot of uncertainty regarding the numbers, since small planets are very, very hard to detect. And other aspects of solar systems, like our asteroid belt, Kuiper belt and Oort cloud are hard to detect even in our own solar system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my own opinion is that it is reasonable to assume that we live in a pretty average solar system (because the odds are obviously suggesting this). So I think that almost every star will have a solar system with a couple of planets, some rocky, some gas giants, plus stuff like comets, asteroids and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Extrasolar_planet&amp;amp;oldid=576307814#Proportion_of_stars_with_planets&quot; rel=&quot;nofollow&quot;&gt;Wikipedia Extrasolar planets&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2013-10-09T07:40:34.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="574" PostTypeId="2" ParentId="545" CreationDate="2013-10-09T07:42:03.400" Score="8" Body="&lt;p&gt;A good, but hard question to answer - as you stated, only an indication could be provided.  But, there is some information available from the number of exoplanetary systems found 'locally', according to the NASA web page &lt;a href=&quot;http://spaceplace.nasa.gov/review/dr-marc-space/solar-systems-in-galaxy.html&quot;&gt;How many solar systems are in our galaxy?&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So far, astronomers have found more than 500 solar systems and are discovering new ones every year. Given how many they have found in our own neighborhood of the Milky Way galaxy, scientists estimate that there may be tens of billions of solar systems in our galaxy, perhaps even as many as 100 billion.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another estimate from the article &lt;a href=&quot;http://www.space.com/14200-160-billion-alien-planets-milky-galaxy.html&quot;&gt;160 Billion Alien Planets May Exist in Our Milky Way Galaxy&lt;/a&gt; (Hall, 2012), particularly with their quote from a scientist:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;This statistical study tells us that planets around stars are the rule, rather than the exception,&quot; said study lead author Arnaud Cassan of the Paris Institute of Astrophysics. &quot;From now on, we should see our galaxy populated not only with billions of bright stars, but imagine them surrounded by as many hidden extrasolar worlds.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An important caveat from the NASA site is that exoplanetary observations are still in its infancy - so over time, this estimate is likely to improve.&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2013-10-09T07:42:03.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="575" PostTypeId="1" AcceptedAnswerId="22611" CreationDate="2013-10-09T07:43:21.247" Score="17" ViewCount="618" Body="&lt;p&gt;As an amateur with limited budget, I'd be interested in taking photos of the night sky, trying to capture more detail than human eye armed with a lens of comparable parameters to what I have in my camera normally could see. I doubt I'd ever get down to details as fine as Jupiter's moons, but I'd hope to see detail of some nebulae I have a hard time seeing through my inexpensive telescope, stars too dim to notice in less-than-perfect conditions etc. I'm interested in taking full-sky images just as well as zooms on specific objects too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, I have a lower-end SLC camera, with two lenses - good sharpness though lower aperture with 50-120mm focal length, and a wide-angle, high-brightness one (about 12-50mm) currently. Firmware hacks allow me to take photos of arbitrarily long time, and I have the remote to start and stop it without touching the camera, and generally software-wise the camera is quite powerful. One of the lenses (the longer focal length) is of &quot;standard professional&quot; quality level too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this sufficient to get started? If so, what kind of settings should I use? If not, what other kind of entry-level equipment would I need to obtain/build on budget to get started with night sky photography?&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2017-09-10T17:38:38.597" Title="Affordable night sky photography" Tags="&lt;amateur-observing&gt;&lt;photography&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="576" PostTypeId="1" AcceptedAnswerId="587" CreationDate="2013-10-09T08:13:18.730" Score="11" ViewCount="7327" Body="&lt;p&gt;I'm using a 750mm (focal length) by 150mm (aperture) newtonian reflector and I've been curious about using Barlow lenses on deep sky objects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the astronomers at my local observatory say that if you get the lens selection correct so you don't lose any light through the eyepiece, you don't need a Barlow lens. While others say that the Barlow lens doubling the focal length of the telescope allows me to see deep sky objects better than without it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is a Barlow lens an effective tool for deep sky observing? If not, what should I use it for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I already have a 2x Barlow lens.&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="19" LastEditDate="2013-10-09T13:17:39.613" LastActivityDate="2021-01-18T09:00:43.087" Title="Are Barlow Lenses Good For Deep Sky Observing?" Tags="&lt;newtonian-telescope&gt;&lt;telescope-lens&gt;&lt;deep-sky-observing&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="577" PostTypeId="2" ParentId="575" CreationDate="2013-10-09T08:13:26.797" Score="7" Body="&lt;p&gt;For long exposure pictures you need to have a motorized mount for your camera. The earth's rotation will lead to streaks otherwise. An affordable way to do this is to use a standard tripod with a star tracker on top. There is a variety of products like:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ioptron.com/index.cfm?select=category&amp;amp;cid=91af533f-b0e2-4dd6-92eb-681025cbb317&quot;&gt;iOptron Sky Tracker&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.vixenoptics.com/mounts/polarie.html&quot;&gt;Vixen Polarie&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They cost around 400-500 USD and are fairly small. You need to align them to the sky's north pole (using the star Polaris in Ursa Minor), and then they will rotate your camera as the earth rotates during the night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you may want to start with your 50mm lens first. The greater the focal length, the higher the magnification, and also the more visible the tracking errors and misalignments of the sky tracker and camera system will be. Be prepared to spend quite some time trying to get good results!&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2013-10-09T08:13:26.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="578" PostTypeId="2" ParentId="546" CreationDate="2013-10-09T08:43:25.563" Score="21" Body="&lt;p&gt;What you could think at first, regarding the orientation of any planetary system, is that it should be roughly in the plane of the galaxy, simply by angular momentum conservation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, when &lt;a href=&quot;http://www.circumstellardisks.org/resolved.php?key=incl&quot; rel=&quot;noreferrer&quot;&gt;you take a look at observations&lt;/a&gt;, you see that protoplanetary disks orientation is not what you would expect, with no preferential orientation (protoplanetary disks are embryo of planetary systems, that makes them interesting). In the following figure, the orientation corresponds to the inclination between the line of sight and the rotation axis of the disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Zh8L.png&quot; alt=&quot;disk orientation (own work, CC-by-nc-sa)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why there is this distribution of orientation?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angular momentum scenario is nice but far to simple: star formation occurs in gas clouds in the interstellar medium, and &lt;a href=&quot;http://adsabs.harvard.edu/abs/1981MNRAS.194..809L&quot; rel=&quot;noreferrer&quot;&gt;these clouds are known to be turbulent (Larson, 1981)&lt;/a&gt;. Turbulence simply disrupts the gas, and is dominent over the global angular momentum of the cloud. Actually, you can even test that with numerical simulations of star formation: put an initial angular momentum &lt;a href=&quot;http://adsabs.harvard.edu/abs/1993ApJ...406..528G&quot; rel=&quot;noreferrer&quot;&gt;consistent with observations&lt;/a&gt;, and some turbulence (subsonic or slightly supersonic, &lt;a href=&quot;http://adsabs.harvard.edu/abs/2007prpl.conf...33W&quot; rel=&quot;noreferrer&quot;&gt;also in accordance to observations&lt;/a&gt;), and you will get &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013A%26A...554A..17J&quot; rel=&quot;noreferrer&quot;&gt;a misalignement of the rotation axis&lt;/a&gt;, due to turbulence.&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2013-10-21T14:33:30.103" LastActivityDate="2013-10-21T14:33:30.103" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="579" PostTypeId="1" AcceptedAnswerId="584" CreationDate="2013-10-09T08:43:31.667" Score="12" ViewCount="1642" Body="&lt;p&gt;I have &lt;a href=&quot;http://www.nbcnews.com/science/space-radiation-concerns-holding-back-female-astronauts-8C11043076&quot;&gt;recently been reading&lt;/a&gt; that space travel is strongly influenced by &quot;space radiation&quot; and how it poses a threat to human space exploration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this radiation originate from stars like our Sun, or is it an omnipresent — let's just call it — &lt;em&gt;&quot;force&quot;&lt;/em&gt; in space (such as cosmic noise) that doesn't have any specific source?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, can an amateur astronomer visualize this radiation in some way so as to be able to observe it?&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="181" LastEditDate="2013-10-09T11:35:34.480" LastActivityDate="2013-12-12T17:24:26.317" Title="Where does the radiation in space come from and can we observe it?" Tags="&lt;space&gt;&lt;radiation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="580" PostTypeId="5" CreationDate="2013-10-09T08:49:18.680" Score="0" Body="&lt;p&gt;Radiation is the emission of energy as electromagnetic waves or as moving subatomic particles, esp. high-energy particles that cause ionization.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="181" LastEditDate="2013-10-09T14:55:52.690" LastActivityDate="2013-10-09T14:55:52.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="581" PostTypeId="4" CreationDate="2013-10-09T08:49:18.680" Score="0" Body="Questions regarding the emission of high energy particles or waves through a vacuum or a partice medium." OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-02-25T08:06:47.573" LastActivityDate="2019-02-25T08:06:47.573" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="582" PostTypeId="2" ParentId="534" CreationDate="2013-10-09T09:21:23.340" Score="14" Body="&lt;p&gt;There is one rule that is generally true for all deep sky objects (nebulae, stars, galaxies,...): Aperture matters!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For solar system objects, aperture is not that important. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second most important thing is: What size are the objects you want to look at: Small objects need long focal lengths and high magnifications, large objects need short aperture for low magnifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 400mm you could watch objects like:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Andromeda galaxy core&lt;/li&gt;&#xA;&lt;li&gt;Orion nebula, other large emission or reflective nebulae (e.g. Pleiades)&lt;/li&gt;&#xA;&lt;li&gt;large star clusters&lt;/li&gt;&#xA;&lt;li&gt;low magnification lunar observations&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With 900mm you could watch objects like&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Planets (Saturn, Jupiter, Mars, ...)&lt;/li&gt;&#xA;&lt;li&gt;high magnification lunar observations&lt;/li&gt;&#xA;&lt;li&gt;planetary nebulae (e.g. ring nebula)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that 60 and 70mm aperture are still very small for telescopes! The aperture influences two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Light sensitvity: The larger the aperture, the more light you can collect. Very important if you live in a city!&lt;/li&gt;&#xA;&lt;li&gt;Maximum resolution: Rule of thumb is that you can do aperture in mm times two as maximum magnification. I.e. for 60mm a 120x magnification is the absolute maximum which still is feasible.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The magnification is created by the eyepiece. E.g. when you have a 400mm focal length telescope and use a 10mm eyepiece, you get 400mm/10mm = 40x magnification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: the shorter the eyepiece focal length, the more difficult it is to build. Good 5mm eyepieces can cost 100 USD and up. I personally started with a 750mm Newtonian with 150mm aperture and 25mm and 10mm eyepieces. That's a good allrounder, even though planets will appear rather small with the 10mm eyepiece. But you can later invest more money in good eyepieces, which you can re-use on better telescopes which you may buy later on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: One more thing -- the telescope mount is equally important as are the eyepieces and the telescope itself. A mount that fits the telescope easily is as expensive as the optical tube assembly itself. Hence many beginners start out with a Dobson telescope, which uses a very, very simple yet sturdy mount.&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2013-10-09T09:21:23.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="583" PostTypeId="2" ParentId="537" CreationDate="2013-10-09T10:32:56.957" Score="6" Body="&lt;p&gt;The tilt of the rotation axis of the Earth is probably twofold. First, it is probably due to &lt;strong&gt;Earth formation history&lt;/strong&gt;. Even though, by angular momentum conservation, planets should have a rotation axis aligned with that of the Sun, their formation history, characterized by accretion and collision of planetisimals, should have perturbated the initial axis. Second, it is due to &lt;strong&gt;gravitational perturbations&lt;/strong&gt;. And actually, regarding this effects, Earth axis should be much more tilted; hopefully, &lt;a href=&quot;https://astronomy.stackexchange.com/questions/476/why-does-the-moon-stablize-earths-axis/477#477&quot;&gt;the Moon is there to help&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2013-10-09T10:32:56.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="584" PostTypeId="2" ParentId="579" CreationDate="2013-10-09T11:20:16.903" Score="3" Body="&lt;p&gt;Cosmic rays consist of both electromagnetic radiation (i.e. photons) of different frequencies (radio waves, IR, light, UV light, x-rays, gamma rays), as well as charged particles (protons, electrons, maybe even ions of light elements), and other stuff like neutrinos. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vast majority of the radiation we encounter around earth will be from the sun, because it is so very close and basically a large radiating blob. Usually with isotropic (equally in all directions) radiating sources, the radiation intensity falls of with the square of the distance. That means radiation diminishes very, very fast. Go twice as far from the sun, and you only get a fourth of the radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The EM radiation from UV and up (X-rays and gamma rays) is probably the most harmful. The earth's magnetic field shields us from these rays, but interplanetary travel will not have this benefit. X-rays and gamma rays may also come from supernovae and other stellar objects, which are far away, but will probably be much too faint to have an effect on astronauts. However, it can be picked up by sensitive specialized telescopes and satellites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The charged particles may be a problem to spacecrafts and electronics onboard them, but can probably be dampened by shielding in the spacecraft, as to protect the astronauts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutrinos are I think of no concern, since they hardly interact with other matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an amateur, you will have problems detecting UV and above. Mainly because we are mostly shielded from this kind of radiation by the magnetosphere and the atmosphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could detect particle radiation, by taking photos of northern lights, though... :)&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2013-10-09T11:20:16.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="585" PostTypeId="2" ParentId="576" CreationDate="2013-10-09T12:02:22.097" Score="4" Body="&lt;p&gt;It depends. The Barlow is usually a cheap shortcut for avoiding to buy an expensive short focal length eyepiece. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you need Barlows and high magnification usually only for small objects. For example the ring nebula (M57) is pretty tiny, and might benefit. Your telescope is still rather wide field, so getting high magnifications might be neccessary to view M57. The Barlow may help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disadvantages of Barlow lenses:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Additional optical surfaces: Every glass surface produces light loss through reflection and also produces abberrations. Good Barlows will use multiple elements to reduce abberration.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Instabilities: Barlows are quite long. Putting this in the focuser, plus an eyepiece, and maybe attaching a camera to the eyepiece will be very fragile.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you cann afford it, buy a specialized eyepiece instead. A high quality 3.5mm or 5mm eyepiece will be much better with your Newtonian.&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2013-10-09T12:02:22.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="586" PostTypeId="2" ParentId="526" CreationDate="2013-10-09T12:32:15.450" Score="5" Body="&lt;p&gt;Following the previous answer stating that the HXR emission occurs during rise phase of the flare as evidenced by the full-Sun SXR flux observed i.e. by GOES in 1-8 Angstrom, I would add that the HXR maximum is typically observed during the strongest increase of the SXR flux, i.e., during d(flux_SXR)/dt = max.&lt;/p&gt;&#xA;" OwnerDisplayName="user203" LastActivityDate="2013-10-09T12:32:15.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="587" PostTypeId="2" ParentId="576" CreationDate="2013-10-09T13:29:41.950" Score="8" Body="&lt;p&gt;&lt;strong&gt;The Failures of High Magnification&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Higher magnification doesn't help you observe deep sky objects better. Deep sky objects unlike stars are extended objects. They subtend a finite solid angle on you. This ensures that the surface brightness(brightness per unit solid angle) of extended objects remains constant. Hence, a higher magnification would not make it brighter for you to see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worse for another reason. After a certain magnification, the apparent angular size of the extended object becomes comparable to the field of view, or sometimes even greater. Your eye won't be able to distinguish the object against the background since most of the background is the object itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Good Uses of Barlow&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the uses that you can put your Barlow:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Observe the Moon, Solar System objecs&lt;/li&gt;&#xA;&lt;li&gt;Some of the clusters which seem like point objects really &quot;explode&quot; when using a Barlow. The best example is Omega Centauri.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Experiences&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a 8&quot; Dobsonian Newtonian Reflector. The viewfinder is 20x80. I really get a better picture of deep sky objects like Ring Nebula, which have a small angular size, when using Barlow. But for cases like Lagoon and Trifid Nebula; they are visible to me from my viewfinder but not through the eyepiece(even in the lowest magnification).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Barlow really helps if you have a deep sky object with small angular size. For larger objects, a binocular may do better, especially for city conditions.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2013-10-09T13:29:41.950" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="588" PostTypeId="2" ParentId="579" CreationDate="2013-10-09T13:47:27.637" Score="2" Body="&lt;p&gt;All energized matter emits radiation. Radiation may consist of electromagnetic energy or particles, as covered in another answer. There are two types of radiation--ionizing and non-ionizing. Ionizing radiation is the type that we are predominantly concerned with danger of, because it can turn atoms that it passes through into ions--which is hazardous to human health. Non-ionizing radiation can still be dangerous if it generates enough heat to cause thermal ionization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ionizing radiation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ultraviolet (of 10 to 125 nm wavelength) - electromagnetic radiation that is absorbed by the Earth's atmosphere but is present in space&lt;/li&gt;&#xA;&lt;li&gt;X-ray - relatively harmless in the small doses that we get for medical work, but harmful in greater exposure&lt;/li&gt;&#xA;&lt;li&gt;Gamma radiation - extremely small wavelength electromagnetic radiation emitted during nuclear processes&lt;/li&gt;&#xA;&lt;li&gt;Alpha radiation - two protons and two neutrons bound as a single particle (helium-4 nucleus), cannot penetrate the skin at slow speeds, but high energy alpha particles can pose a danger to human health (cannot penetrate the atmosphere, but present in space)&lt;/li&gt;&#xA;&lt;li&gt;Beta radiation - may be electrons (Beta-minus) or positrons (Beta-plus), do not typically penetrate the atmosphere but can readily penetrate unshielded human tissue&lt;/li&gt;&#xA;&lt;li&gt;Neutron radiation - neutrons emitted by nuclear fission, highly dangerous, readily ionizes and can even make other materials radioactive&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Non-ionizing radiation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ultraviolet (lower part of the spectrum) - nonionizing but still high enough energy that it can have some dangerous effects on the human body&lt;/li&gt;&#xA;&lt;li&gt;Visible light - the electromagnetic energy that we see, about 380-750 nm wavelength&lt;/li&gt;&#xA;&lt;li&gt;Infrared - electromagnetic energy emitted by most objects at temperatures that we deal with on a daily basis, about 700 nm to 1 mm wavelength&lt;/li&gt;&#xA;&lt;li&gt;Microwave - electromagnetic energy of wavelengths from 1 mm to 1 meter&lt;/li&gt;&#xA;&lt;li&gt;Radio waves - electomagnetic energy of wavelengths greater than infrared&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt; Used &lt;a href=&quot;http://en.wikipedia.org/wiki/Radiation&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; as a reference to organize and bolster information &lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In space we have numerous sources of radiation, since all energized matter emits radiation. Stars are a big factor emitting most types of radiation. Supernovae and black holes also emit radiation. Finally, some radiation has been propagating through the universe since the Big Bang. The Cosmic Microwave Background (CMB) radiation gives us a glimpse of the early universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many ways to observe radiation. Traditional telescopes make use of our natural ability to take in visible light and amplify it with lenses. Radio telescopes are also relatively easy for an amateur to get a hold of. &lt;a href=&quot;http://www.radiotelescopebuilder.com/How_To_Build_Your_Own_Radio_Telescope_sample.pdf&quot; rel=&quot;nofollow&quot;&gt;Here's&lt;/a&gt; some instructions on how to build a simple radio telescope. Near-infrared light can be observed easily by an amateur with a regular telescope and infrared film, but this doesn't give us a lot more detail than visible light. Most infrared from space is absorbed by our atmosphere (&lt;a href=&quot;http://coolcosmos.ipac.caltech.edu/cosmic_classroom/ask_astronomer/faq/obs.shtml&quot; rel=&quot;nofollow&quot;&gt;more on infrared telescopes&lt;/a&gt;). UV and higher radiation would also be difficult to detect by an amateur since our atmosphere shields us from it, as well as particle radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As one clever answerer posted, we &lt;em&gt;can&lt;/em&gt; observe the stunning light effects that occur when particle radiation ionizes the upper atmosphere. Particle radiation is usually deflected by Earth's magnetic field, but sometimes travels along field lines toward the poles which is why the light effects of particle radiation are only observed in the arctic regions as the Northern and Southern lights.&lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorUserId="6" LastEditDate="2013-12-12T17:24:26.317" LastActivityDate="2013-12-12T17:24:26.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="591" PostTypeId="2" ParentId="528" CreationDate="2013-10-09T15:18:13.703" Score="4" Body="&lt;p&gt;Such objects are surrounded by an accretion disk, that is a disk of gas, the gas being accreted onto the central object. However, at this stage of its life, the accretion disk around the black hole of a binary system like GRS-1915+105 is strongly depleted. Therefore, the accretion rate is not sufficient to support a coutinuous burst. Matter fills the outer disk until a critical density is reached, that triggered an outburst. Mass from the companion star is lost by a &lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/R/roche-lobe+overflow&quot; rel=&quot;nofollow&quot;&gt;Roche lobe overflow mechanism&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;X-ray binaries in general: &lt;a href=&quot;http://cdsads.u-strasbg.fr/abs/2006ARA%26A..44...49R&quot; rel=&quot;nofollow&quot;&gt;Remillard &amp;amp; McClintock 2006&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;For the outburst X-ray transients in particular: &lt;a href=&quot;http://adsabs.harvard.edu/abs/2001A%26A...373..251D&quot; rel=&quot;nofollow&quot;&gt;Dubus et al. 2001&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="88" LastActivityDate="2013-10-09T15:18:13.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="592" PostTypeId="2" ParentId="539" CreationDate="2013-10-09T16:10:00.887" Score="86" Body="&lt;p&gt;In addition to Undo's fine answer, I would like to explain a bit about the motivation behind the definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Eris was discovered, it turned out to be really, &lt;em&gt;really&lt;/em&gt; similar to Pluto. This posed a bit of a quandary: should Eris be accepted as a new planet? Should it not? If not, then why keep Pluto? Most importantly, this pushed to the foreground the question&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;what, &lt;em&gt;exactly&lt;/em&gt;, is a planet, anyway?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This had been ignored until then because everyone &quot;knew&quot; which bodies were planets and which ones were not. However, with the discovery of Eris, and the newly-realized potential of more such bodies turning up, this was no longer really an option, and &lt;em&gt;some&lt;/em&gt; sort of hard definition had to be agreed upon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with coming up with a hard definition that decides what does make it to planethood and what doesn't is that nature very rarely presents us with clear, definite lines. Size, for example, is not a good discriminant, because solar system bodies come in a continuum of sizes from Jupiter down to meter-long asteroids. Where does one draw the line there? Any such size would be completely arbitrary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There &lt;em&gt;is&lt;/em&gt;, however, one characteristic that has a sharp distinction between some &quot;planets&quot; and some &quot;non-planets&quot;, and it is &lt;em&gt;the amount of other stuff in roughly the same orbit&lt;/em&gt;. This is still slightly arbitrary, because it's hard to put in numbers exactly what &quot;roughly&quot; means in this context, but it's more or less unambiguous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider, then a quantity called the &quot;planetary discriminant&quot; µ, equal to the ratio of the planet's mass to the total mass of other bodies that cross its orbital radius and have non-resonant periods (so e.g. Neptune doesn't count as sharing Pluto's orbit) up to a factor of 10 longer or shorter (to rule out comets, which has little effect in practice). This is still a bit arbitrary (why 10?) but it's otherwise quite an objective quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take this quantity and calculate it for the different bodies you might call planets, comparing it to both the objects' mass,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iKpkE.png&quot; alt=&quot;Planetary discriminant vs mass&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and their diameter,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MzTOZ.png&quot; alt=&quot;Planetary discriminant vs diameter&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or with an arbitrary horizontal axis, in order of decreasing discriminant,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yqfo1.png&quot; alt=&quot;Planetary discriminant on an arbitrary horizontal axis, in order of decreasing discriminant&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suddenly, &lt;strong&gt;a natural hard line emerges&lt;/strong&gt;. If you look only at the mass and the diameter of the objects (shown in the insets above the plots), then there is a pretty continuous spread of values, with bigger gaps between the gas giants and the terrestrial planets than between Mercury and Eris/Pluto. However, if you look at the planetary discriminant, on the vertical axis, you get a very clear grouping into two distinct populations, separated by over four orders of magnitude. There's a finite set of bodies that have &quot;cleared their orbits&quot;, and some other bodies which are well, well behind in that respect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the main reason that &quot;clearing its orbital zone&quot; was chosen as a criterion for planethood. It relies on a distinction that is actually there in the solar system, and very little on arbitrary human decisions. It's important to note that this criterion need not have worked: this parameter might also have come out as a continuum, with some bodies having emptier orbits and some others having slightly fuller ones, and no natural place to draw the line, in which case the definition would have been different. As it happens, this is indeed a good discriminant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further reading, I recommend &lt;a href=&quot;http://en.wikipedia.org/wiki/Clearing_the_neighbourhood&quot; rel=&quot;noreferrer&quot;&gt;the Wikipedia article on 'Clearing the neighbourhood'&lt;/a&gt;, as well as the original paper where this criterion was proposed,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is a planet? S Soter,  &lt;a href=&quot;http://dx.doi.org/10.1086/508861&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;The Astronomical Journal&lt;/em&gt; &lt;strong&gt;132&lt;/strong&gt; no.6 (2006), p. 2513&lt;/a&gt;. &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0608359&quot; rel=&quot;noreferrer&quot;&gt;arXiv:astro-ph/0608359&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is in general very readable (though there are some technical bits in the middle which are easy to spot and harmless to skip), and from which I took the discriminant data for the plots above.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I must apologize for having included, in previous versions of this post, an incorrect plot, caused by taking &lt;a href=&quot;http://web.archive.org/web/20131015071825/http://en.wikipedia.org/wiki/Clearing_the_neighbourhood&quot; rel=&quot;noreferrer&quot;&gt;data from Wikipedia&lt;/a&gt; without verifying it. In particular, the planetary discriminant for Mars was wrong (1.8×10&lt;sup&gt;5&lt;/sup&gt; instead of 5.1×10&lt;sup&gt;3&lt;/sup&gt;), which now puts it below Neptune's instead of just below Saturn's, but the overall conclusions are not affected. The Mathematica code for the graphics is available at &lt;code&gt;Import[&quot;http://goo.gl/NaH6rM&quot;][&quot;http://i.stack.imgur.com/CQA4T.png&quot;]&lt;/code&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;... and, as a final aside: Pluto is &lt;em&gt;awesome&lt;/em&gt;. It was visited in July 2015 by the &lt;a href=&quot;https://en.wikipedia.org/wiki/New_Horizons&quot; rel=&quot;noreferrer&quot;&gt;New Horizons&lt;/a&gt; probe, which found a world that was much more rich, dynamic, and active than anyone expected, including what appear to be churning lakes of solid nitrogen ringed by mountains of water ice, among other marvels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/File:Pluto-01_Stern_03_Pluto_Color_TXT.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gz9sM.jpg&quot; alt=&quot;Pluto as seen from the New Horizons flyby, in enhanced colour&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note the image has been colour-enhanced to bring out the variety of surface materials; the true-colour version of this image is &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Nh-pluto-in-true-color_2x_JPEG.jpg&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.) I, personally, don't feel it's at all necessary to 'grandfather' Pluto into the list of planets to really feel the awe at the amazing place it is - it's perfectly OK for it to be a cool place with cool science, that is also not a planet.&lt;/p&gt;&#xA;" OwnerUserId="207" LastEditorUserId="207" LastEditDate="2016-06-21T20:12:05.920" LastActivityDate="2016-06-21T20:12:05.920" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="595" PostTypeId="2" ParentId="490" CreationDate="2013-10-09T21:46:48.917" Score="10" Body="&lt;p&gt;Shoemaker Levy 9 was estimated to have released kinetic energy equivalent to 300 gigatons of TNT. That is &lt;span class=&quot;math-container&quot;&gt;$1.255 \times 10^{21} {\rm J}$&lt;/span&gt;. This release through friction and compression was sufficient to heat the atmosphere to 4000K at first. That in turn should probably be enough to break down the molecular hydrogen and methane, releasing further energy. But that's only my guess. The heating from the impact alone should make a huge fireball.&lt;/p&gt;&#xA;&lt;p&gt;Sources:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.lpi.usra.edu/meetings/lpsc97/pdf/1351.PDF&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fragment Size and Mass Estimates&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1995LPI....26.1483W&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Galileo observations&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="50" LastEditorUserId="37191" LastEditDate="2021-01-29T09:10:54.403" LastActivityDate="2021-01-29T09:10:54.403" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="597" PostTypeId="2" ParentId="483" CreationDate="2013-10-10T11:31:04.910" Score="-1" Body="&lt;p&gt;At a minimum, you'll need a highly directional antenna appropriate for the frequencies in question (I do not know what they are).  The radio would also have to able to detect the frequencies that you expect to hear signals on.&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2013-10-10T11:31:04.910" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="598" PostTypeId="2" ParentId="483" CreationDate="2013-10-10T13:38:46.663" Score="7" Body="&lt;p&gt;The StarGazers lounge &lt;a href=&quot;http://stargazerslounge.com/topic/75170-basic-jupitersolar-radio-astronomy/&quot; rel=&quot;noreferrer&quot;&gt;featured a radio kit article&lt;/a&gt; for Jupiter radio astronomy. The same article is also featured over at the &lt;a href=&quot;http://www.britastro.org/radio/projects/joveproj.html&quot; rel=&quot;noreferrer&quot;&gt;Radio Group of BritAstro&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that 20.1 MHz is the suitable frequency for amateurs observing Jupiter. I am far from being an expert for radio astronomy, but for a small source such as Jupiter, I would assume that you need a big parabolic dish (for antenna gain and directionality), and a proper feed antenna, suited for 20.1 MHz. You will probably also need to have a motorized mount for the antenna, to track the motion of Jupiter across the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will post more info, after I have asked a friend, who's an expert on antenna design, which configuration may be suitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Ok, the infos are in. A parabolic dish is out of the question -- 20 MHz corresponds to about 15m wavelength. And dishes need to be many multiples of a wavelength. So we could use a dipole antenna, but those are not very directional. Also, the dipole needs to be rather large, for good gain. To make it directional, you need an array of steerable dipole antennas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summary: I think it is not really feasible to do this on an amateur scale. Of course, I am gladly being proven wrong!&lt;/p&gt;&#xA;" OwnerUserId="50" LastEditorUserId="50" LastEditDate="2013-10-10T19:45:20.147" LastActivityDate="2013-10-10T19:45:20.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="599" PostTypeId="5" CreationDate="2013-10-10T18:08:46.147" Score="0" Body="&lt;p&gt;Volcanism describes any of various processes and phenomena associated with the surficial discharge of molten rock, pyroclastic fragments, or hot water and steam, including volcanoes, geysers, and fumaroles. &lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="181" LastEditDate="2013-10-10T20:11:10.783" LastActivityDate="2013-10-10T20:11:10.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="600" PostTypeId="4" CreationDate="2013-10-10T18:08:46.147" Score="0" Body="Questions about volcanic processes and observations." OwnerDisplayName="user2449" LastEditorDisplayName="user2449" LastEditDate="2015-02-08T14:16:59.103" LastActivityDate="2015-02-08T14:16:59.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="601" PostTypeId="1" AcceptedAnswerId="612" CreationDate="2013-10-10T18:20:12.637" Score="13" ViewCount="347" Body="&lt;p&gt;&lt;a href=&quot;http://www.ifa.hawaii.edu/info/press-releases/LonelyPlanet/&quot;&gt;Recent results&lt;/a&gt; from Institute for Astronomy at the University of Hawaii at Manoa claim that there is a Jupiter-sized gas giant planet that is independent of a star about 80 light-years from Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The press release refers to this planet as &quot;free-floating&quot; but I assume that it must be in some kind of predictable orbital path.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the orbital path of this newly discovered planet?&lt;/p&gt;&#xA;" OwnerUserId="219" LastEditorDisplayName="user8" LastEditDate="2013-10-10T19:13:52.357" LastActivityDate="2014-12-30T10:33:03.537" Title="What is the orbital path of the newly discovered star-less planet PSO J318.5-22?" Tags="&lt;exoplanet&gt;&lt;gas-giants&gt;&lt;rogue-planet&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="602" PostTypeId="1" AcceptedAnswerId="603" CreationDate="2013-10-10T18:42:55.477" Score="17" ViewCount="256" Body="&lt;p&gt;Sometimes, hobby-astronomers use rather professional means to observe the big voids of space. Every now and then (think in months, not days) even I can locate an NEO (near Earth object).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm assuming governments and institutions worldwide track such NEOs, have the best options to observe them, and most probably accumulated more data than we could wish for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For hobby-astronomers who spot an NEO, it could be of interest to verify the location of the spotted NEO, and maybe even learn more about a specific NEO from additional data like name, type, path, direction, speed, first observed, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any publicly available, NEO-related database out there? Or is there a specific institution a hobby-astronomer can/should turn to to be able to learn more about individual NEOs?&lt;/p&gt;&#xA;" OwnerUserId="181" LastActivityDate="2013-10-10T19:21:25.060" Title="How can an amateur astronomer verify the position of near Earth objects?" Tags="&lt;amateur-observing&gt;&lt;observational-astronomy&gt;&lt;near-earth-object&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="603" PostTypeId="2" ParentId="602" CreationDate="2013-10-10T19:15:46.170" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there any publicly available, NEO-related database out there? Or is&#xA;  there a specific institution a hobby-astronomer can/should turn to to&#xA;  be able to learn more about individual NEOs?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, there is NASA's &lt;a href=&quot;http://neo.jpl.nasa.gov/&quot;&gt;Near Earth Object Program&lt;/a&gt; that catalogues all  detected NEO's and had advanced reporting and seearch capabilities (a bit overwhelming number of options actually).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also ESA's own (actually DLR's, the German space agency) database under the &lt;a href=&quot;http://earn.dlr.de/&quot;&gt;EARN&lt;/a&gt; (European Asteroid Research Node) project within the &lt;a href=&quot;http://www.esa.int/Our_Activities/Operations/Space_Situational_Awareness/Near-Earth_Objects_-_NEO_Segment&quot;&gt;SSA-Programme&lt;/a&gt; (Space Situational Awareness) called &lt;a href=&quot;http://earn.dlr.de/nea/&quot;&gt;The Near-Earth Asteroids Data Base&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An on-line data-base of Physical and Dynamical properties NEOs, with&#xA;  the corresponding bibliographical references is available at the EARN&#xA;  site. This data-base, compiled and maintained at DLR by G&lt;a href=&quot;http://earn.dlr.de/ourgroup/gerhard.htm&quot;&gt;erhard Hahn&lt;/a&gt;,&#xA;  contains information on all known near-Earth objects&#xA;  (Atens-Apollo-Amor Asteroids), and is updated on a regular basis. It&#xA;  is also linked to, and reachable from the &lt;a href=&quot;http://newton.dm.unipi.it/neodys/&quot;&gt;NEO Dynamics Site NEODyS&lt;/a&gt; at&#xA;  the University of Pisa, which provides extensive information on all&#xA;  dynamical aspects of NEOs, including close approach predictions and&#xA;  collision probabilities.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Honestly though, I find the NEO Program one from NASA a lot more convenient to use, even if a bit overwhelming. For extensive research though, I'd suggest referring to both of these databases, with the latter probably more conveniently used through the already linked to &lt;a href=&quot;http://newton.dm.unipi.it/neodys/&quot;&gt;NEODyS Project&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditDate="2013-10-10T19:21:25.060" LastActivityDate="2013-10-10T19:21:25.060" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="604" PostTypeId="1" AcceptedAnswerId="606" CreationDate="2013-10-10T19:33:50.770" Score="11" ViewCount="3851" Body="&lt;p&gt;I wanted to ask a question; it's simple but I cannot find any possible and perfect solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth has poles, North and South. By which we can get directions using a compass or a needle compass, but that's not the concern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the Moon cause a change in the magnetic field of Earth when they are interpassing or colliding with each other, as the Moon does come between the magnetic path of the Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So would the Moon's magnetic field affect the Earth's magnetic field, just as its gravitational pull affects Earth's gravitational pull for oceans?&lt;/p&gt;&#xA;" OwnerUserId="79" LastEditorUserId="6" LastEditDate="2013-10-16T15:05:36.223" LastActivityDate="2015-05-16T18:21:10.703" Title="Does the Moon's magnetic field affect Earth's magnetic field?" Tags="&lt;the-moon&gt;&lt;earth&gt;&lt;magnetic-field&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="605" PostTypeId="1" CreationDate="2013-10-10T20:33:18.253" Score="1" ViewCount="142" Body="&lt;p&gt;I'm wondering if there's any good computer software for tracking the orbits of planets over time. I like to see the orbit's trails against the sky background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd prefer something that's free and of a good quality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've already tried Google Earth, but it doesn't seem to be able to do this.&lt;/p&gt;&#xA;" OwnerUserId="69" LastEditorUserId="50" LastEditDate="2013-10-11T08:10:36.697" LastActivityDate="2013-10-11T08:10:36.697" Title="How can I create tracks of the planet's orbits on my computer?" Tags="&lt;orbit&gt;&lt;planet&gt;&lt;software&gt;" AnswerCount="0" CommentCount="7" ClosedDate="2013-10-10T21:34:22.323" ContentLicense="CC BY-SA 3.0" />
  <row Id="606" PostTypeId="2" ParentId="604" CreationDate="2013-10-10T21:31:13.493" Score="13" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So would the Moon's magnetic field affect the Earth's magnetic field, just as its gravitational pull affects Earth's gravitational pull for oceans?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, but only slightly. Firstly, magnetic fields can superimpose, so the field at any point is the sum of the field due to the Earth and the field due to the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the moon is rather far away (and has a weak magnetic pole strength), so the magnetic field due to the moon on Earth's surface is nearly negligible (magnetic field also decreases as an inverse-square law)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, the magnetic field of the Moon may bolster or erode the Earth's field as magnets moving relative to each other tend to either lose magnetization or become stronger. But this process has a negligible effect when we take the Moon and Earth.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="6" LastEditDate="2013-10-16T15:07:23.763" LastActivityDate="2013-10-16T15:07:23.763" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="607" PostTypeId="1" AcceptedAnswerId="611" CreationDate="2013-10-10T22:37:07.967" Score="20" ViewCount="1530" Body="&lt;p&gt;I've learned that planets move in orbits around the Sun, but I really don't know how I would come to this conclusion myself. I've only seen planets in the sky a couple times (knowingly), and I am curious &lt;em&gt;how we know for certain today that planets really move in orbits around the Sun&lt;/em&gt; (i.e., rather than moving but not around the Sun or moving around the Sun in a shape that is not a regular path).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also heard that people in the past knew about orbits even when they thought that Earth was at the center of the solar system. &lt;em&gt;How did they figure this out in their times with their technology?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="180" LastActivityDate="2019-05-01T02:23:11.817" Title="How do/did we figure out that planets move in orbits?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;orbit&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="608" PostTypeId="2" ParentId="483" CreationDate="2013-10-10T22:55:24.073" Score="14" Body="&lt;p&gt;@Arne is right in &lt;a href=&quot;https://astronomy.stackexchange.com/a/598/13&quot;&gt;his answer&lt;/a&gt; about two things, that the most suitable frequency for Jovian amateur radio is 20.1 MHz, and that this is a 15 meter wavelength. However, the antenna can actually be half the wavelength, and amateur radio astronomers have had good results listening to all kinds of Jovian sounds, including detecting occultations of its many moons as they produce change in frequency due to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Doppler_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Doppler shift&lt;/a&gt; when passing in front of it, amplification due to echo of Jupiter's own radio wave signature when close beside it from the observer's vantage point, and other effects causing radio wave frequency and amplitude changes with a simple dual dipole array that might look something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/BEWgc.png&quot; alt=&quot;dual dipole array&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;The Jove dual dipole array. Dipoles are suspended between PVC masts. Signals from the dipoles go to a power combiner and&lt;br /&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;then to the receiver. Source: &lt;a href=&quot;http://www.radio-astronomy.org/pdf/qex/radio-jove-proof.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Amateur Radio Astronomy Projects — Radio Signals from Jupiter&lt;/a&gt; (PDF)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this dual dipole antenna is more or less a simple two-piece of roughly half the wavelength of coaxial cable, stripped of insulation at the dipole ends and run dipoles parallel to each other at roughly 6.1 meters (20 ft) apart suspended over PVC masts. That is perfectly acceptable for nearly any amateur radio astronomer in both space and the price of the parts needed. There are also ways to help oneself with a lot smaller antenna, which I'll mention a bit later. Let's first clear the frequency range discussion with this really informative quote from &lt;a href=&quot;http://receiverforjupiter.tripod.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Radio Receiver for Jupiter&lt;/a&gt; webpage (based on NASA's &lt;a href=&quot;http://radiojove.gsfc.nasa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Radio JOVE project&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The peak of Jupiter signals occur around 10 MHz. Still this frequency&#xA;  is not that suitable, as it is very near to Ionospheric cutoff. The&#xA;  best suitable frequencies are in between 18 to 22 MHz, as the chances&#xA;  of getting emissions are more. In practice 18.7 MHz, 20.1 MHz, 22.3&#xA;  MHz are common. The frequencies above 30 MHz are not suitable because&#xA;  of less strength. The frequency of 20.1 MHz is used for this project,&#xA;  as the probability of getting emission is high. So the entire receiver&#xA;  is designed considering 20.1 MHz as the operating frequency.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Antenna is of course merely one part of the story. Second thing that is required is the receiver. NASA is sponsoring &lt;a href=&quot;http://radiojove.gsfc.nasa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Radio JOVE project&lt;/a&gt; with an assembly manual for roughly 100 electronic components and pieces of hardware worth &lt;em&gt;RJ1.1 Receiver&lt;/em&gt; (&lt;em&gt;Radio Jove 1.1 Receiver&lt;/em&gt;) that can be almost completely assembled out of parts ordered from Radio Shack (it even includes RS part numbers). I'll add a photograph and a few other links, and you'll have to be on your own then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/EvNgT.jpg&quot; alt=&quot;Radio JOVE&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;sup&gt;Self-assembled Radio Jove Receiver's front plate with two turning knobs for volume and tuning. Source: &lt;a href=&quot;http://amateurradioastronomy.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;KB0LQJ Observatory&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some relevant links for building Radio Jove (or Jupiter FM, if you like, it will be your own receiver so name it as you wish), starting with already mentioned ones:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.radio-astronomy.org/pdf/qex/radio-jove-proof.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Amateur Radio Astronomy Projects — Radio Signals from Jupiter&lt;/a&gt; (PDF)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ten years ago a group of (mostly) &#xA;  University of Florida graduates working at &#xA;  NASA conceived an educational outreach &#xA;  program known as Radio Jove. The idea was &#xA;  to build an inexpensive radio telescope kit &#xA;  suitable for detecting signals from Jupiter. &#xA;  The Jove receiver (Figure 2) is a simple &#xA;  direct-conversion design operating over a &#xA;  few hundred kilohertz range centered at&#xA;  20.1 MHz.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://radiojove.gsfc.nasa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA's Radio JOVE project&lt;/a&gt; (PDF)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The site is currently down due to US government shutdown, so here's a &lt;a href=&quot;http://webcache.googleusercontent.com/search?q=cache:ppUt9qAWt_4J:radiojove.gsfc.nasa.gov/telescope/rcvr_manual.pdf%20&amp;amp;cd=2&amp;amp;hl=en&amp;amp;ct=clnk&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google cached version&lt;/a&gt; that sadly doesn't come with in-document images&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://amateurradioastronomy.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;KB0LQJ Observatory Amateur Radio Astronomy — Radio Jove observations&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For my home observatory, I started with the Radio Jove receiver from NASA’s Radio Jove Project. This was a fairly easy-to-build kit with great directions, not just for the receiver but also the antenna setup and installation.  Unfortunately, I don’t have enough space to put up a phased antenna array.  Also, I’m bordered by power lines on the north and south sides of my property.   Also, since I’m in an urban area, I knew that I’d pick up a lot of noise.   Not to worry.   Solar observations are also quite interesting, and since the Sun is such a good signal source (especially so far this year), I opted for an antenna in my attic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://receiverforjupiter.tripod.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Radio Receiver for Jupiter&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This website contains the Technical details of the antenna and receiver used for receiving natural radio emissions from Jupiter at 20.1MHz. The antenna and receiver discussed in this site is based on the design given by NASA Radio Jove Program. &#xA;  Natural radio emissions from Jupiter or from Sun are detected by using a dual dipole array as an antenna and with a sensitive receiver. The RF voltage developed at the antenna terminals is amplified with the RF amplifier and converted in to audio frequencies using a mixer. The audio signal thus generated is recorded on PC through a sound card in ‘wav’ format. Also a strip-chart recorder software is available to generate the strip-chart of data coming through sound card.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So there are indeed many ways to build your own antenna and receiver out of affordable and easily obtainable electronic parts, and some of these websites listed above will help you through the process of self-assembly, even providing some tricks how to do it easier, like for example assembling a smaller, in-house antenna array.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, one other thing that all these websites mention is also using various PC software that lets you analyze through computer's sound card received Jovian radio sounds, but since there's many different and free solutions out there, including NASA's own that runs on Windows PCs, I'll let you discover those on your own. &lt;a href=&quot;http://receiverforjupiter.tripod.com/softwares.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's one page&lt;/a&gt; listing many links, to get you started. More computer savvy of you might even write your own software for this purpose, this is Stack Exchange afterall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if anyone is wondering what such self-assembled radio receivers and antennas are capable of, here's a link to a collection of various &lt;a href=&quot;http://www.astrosurf.com/luxorion/audiofiles-jupiter.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jovian sounds in the radio wavelengths on Astrosurf.com&lt;/a&gt;, and another one as a &lt;a href=&quot;http://www.radiosky.com/rjcentral.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;collection of solely amateur radio astronomy sounds&lt;/a&gt; of Jupiter and its moons (scroll a bit down to the list of recordings). And here's a short description of various sound types one might hear:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Chorus&lt;/strong&gt; (quote source: &lt;a href=&quot;http://www-pw.physics.uiowa.edu/~dag/chorus.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The University of Iowa Department of Physics and Astronomy&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Chorus consists of brief, rising-frequency tones that sound like the chorus of birds singing at sunrise, hence the name &quot;chorus&quot; or &quot;dawn chorus.&quot; Chorus at Earth is generated by electrons in Earth's Van Allen radiation belts. Once generated, the chorus waves affect the motions of the electrons through a process called a wave-particle interaction. Wave-particle interactions disturb the trajectories of the radiation belt electrons and cause the electrons to hit the upper atmosphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Decametric noise storms&lt;/strong&gt; (quote source: &lt;a href=&quot;http://www.radiosky.com/rjcentral.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Radio-Jupiter Central on RadioSky.com&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The emissions we can hear are often referred to as decametric noise storms, because the waves are tens of meters long. Okay, it is possible to hear Jupiter from 15 to 38 MHz, but what are the optimal frequencies?  The consensus seems to be that 18 MHz up to about 28 MHz is a good place to listen.  A good rule would be to pick the lowest frequency in this range which was not being hindered by ionospheric refraction. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Whistlers&lt;/strong&gt; (quote and image source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Whistler_%28radio%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A whistler is a very low frequency or VLF electromagnetic (radio) wave generated by lightning.&lt;a href=&quot;https://en.wikipedia.org/wiki/Doppler_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt; Frequencies of terrestrial whistlers are 1 kHz to 30 kHz, with a maximum amplitude usually at 3 kHz to 5 kHz. Although they are electromagnetic waves, they occur at audio frequencies, and can be converted to audio using a suitable receiver. They are produced by lightning strokes (mostly intracloud and return-path) where the impulse travels along the Earth's magnetic field lines from one hemisphere to the other.&lt;br /&gt;&lt;br /&gt;&lt;img src=&quot;https://i.stack.imgur.com/cRdKF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Auroral Radio Emissions&lt;/strong&gt;  (quote source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Methods_of_detecting_extrasolar_planets#Auroral_radio_emissions&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Auroral radio emissions from giant planets with plasma sources such as Jupiter's volcanic moon Io could be detected with radio telescopes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And so on. The last two on the list are probably a bit of a stretch for amateur astronomers, with auroral radio emissions average frequency range at 100 to 500 kHz, and whistlers cutoff at usually around 30 kHz, and both requiring too big antennae, but I wouldn't bet against it being possible with smaller antennae either, although bow frequency cutoffs might prevent identification of actual &lt;em&gt;aural&lt;/em&gt; events. But there are many other sounds to listen to both from Jupiter, as well as it's many moons, mostly though the bigger and closer ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Good luck tuning in to your own Jupiter FM and happy hunting for rare radio frequency events!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-08-05T13:00:49.017" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="609" PostTypeId="1" AcceptedAnswerId="11625" CreationDate="2013-10-10T23:45:07.563" Score="8" ViewCount="4597" Body="&lt;p&gt;Speed, as we know, doesn't exist without first having a reference point. We then say that the reference point isn't moving at all, and speed is then measured in relation to the reference point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the standard reference point when determining the speed of objects in astronomy? Earth?&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="19" LastEditDate="2015-08-24T00:05:55.237" LastActivityDate="2015-08-24T08:15:29.533" Title="What is the standard reference point for measuring speed?" Tags="&lt;velocity&gt;&lt;speed&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="610" PostTypeId="2" ParentId="609" CreationDate="2013-10-11T00:24:58.987" Score="6" Body="&lt;p&gt;As Einstein realized and like you correctly state, you indeed can't measure the speed of an object by itself since it has to be measured relative to something else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a logic result, if you ask the question &lt;em&gt;&quot;How fast is X moving?&quot;&lt;/em&gt;, you will have to specify that you want the speed with respect to another object because motion cannot be measured without a reference point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you ask how fast Earth is moving with respect to its own axis, the center of the Earth will be your reference point. &lt;/li&gt;&#xA;&lt;li&gt;If you want to know how fast the Earth orbits the Sun, the Sun will be your reference point. &lt;/li&gt;&#xA;&lt;li&gt;If you want to know how fast the Moon's distance from Earth increases, Earth will be your reference point.&lt;/li&gt;&#xA;&lt;li&gt;If you want to know the speed of our solar system in the milky way galaxy, the center of the Milky Way will be your reference point.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;standard&quot; reference point&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no reference point is given, it could be assumed that the &lt;em&gt;&quot;standard reference point&quot;&lt;/em&gt; is the location of the observer. In the realms of astronomy, that would be the location of the astronomer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted that the astronomer could well be on an space mission outside Earth's atmosphere, or he/she could be using a telescope in Earth's orbit. Therefore, &quot;Earth&quot; can not generally be assumed to be a standard reference point for astronomy because, depending on the precision of the measurements and the location of (for example) the telescope, assuming &quot;Earth&quot; may result in inexact measurements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &lt;a href=&quot;https://astronomy.stackexchange.com/questions/609/what-is-the-standard-reference-point-for-measuring-speed/610#comment520_610&quot;&gt;@TidalWave correctly commented&lt;/a&gt;, there's also the &lt;a href=&quot;http://aa.usno.navy.mil/faq/docs/ICRS_doc.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;International Celestial Reference Frame (ICRS)&lt;/a&gt; which can help you find reference points, calculate distances, etc. according to a celestial reference system, which has been adopted by the International Astronomical Union (IAU) for high-precision positional astronomy. The origin of ICRS resides at the barycenter of the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Wrapping it up:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no reference point is defined and if we can assume that the astronomer works according to the rules and definitions of the International Astronomical Union (which would be the regular, if not ideal case), the International Celestial Reference Frame provides you with a &quot;standard reference point&quot; at it's center (which is the barycenter of the solar system). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In rare cases where IAU compliance can not be assumed (something that &lt;em&gt;&quot;might&quot;&lt;/em&gt; happen in amateur realms), it has to be assumed that the reference point is the point of observation.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2013-10-11T03:30:37.900" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="611" PostTypeId="2" ParentId="607" CreationDate="2013-10-11T05:27:41.707" Score="15" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;I've also heard that people in the past knew about orbits even when they thought that Earth was at the center of the solar system. &lt;em&gt;How did they figure this out in their times with their technology?&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The same celestial objects (stars, planets, the Moon) could be seen every year. So, people figured out there was a pattern to it.&lt;/p&gt;&#xA;&lt;p&gt;At first, geocentrism was prevalent due to prejudice and the  stars &amp;quot;orbiting&amp;quot; the Earth in a clean manner (no strange effects). If you stare up at night, stars move in a circle with the center approximately at the Pole Star. This can be mistaken for the Star moving around the Earth.&lt;/p&gt;&#xA;&lt;p&gt;Objects like visible planets (with retrograde motions) were relatively small in number — we have the inner solar system plus Jupiter and Saturn, so people gave plausible explanations for these. Ptolemy theorized that the planet orbits around a &amp;quot;ghost point&amp;quot;, for example:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6qIV8.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;(This is called the epicycle theory)&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;how we know for certain today that planets really move in orbits around the Sun&lt;/em&gt; (i.e., rather than moving but not around the Sun or moving around the Sun in a shape that is not a regular path).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The heliocentric model came up because of various smaller discoveries:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Jupiter has moons, so not all celestial bodies orbit the Earth&lt;/li&gt;&#xA;&lt;li&gt;Venus has full phases, but in the geocentric model only a small subset of the phases should be visible&lt;/li&gt;&#xA;&lt;li&gt;Theories like epicycles were not as accurate as they were expected to be, and people started realizing this over time.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;With all this in mind, the simple solution of heliocentrism explains everything in a much cleaner fashion, without leaving much doubt in mind. Kepler's laws make much cleaner predictions.&lt;/p&gt;&#xA;&lt;p&gt;Remember that people had catalogs of astronomical data to verify things with.&lt;/p&gt;&#xA;&lt;p&gt;Later on, once Newton came up with his theory of gravitation, Kepler's laws made even more sense (though I believe that Newton formulated his theory with some inspiration from Kepler's laws). I'm pretty sure that there were on-Earth experiments to verify the inverse-square as well.&lt;/p&gt;&#xA;&lt;p&gt;Nowadays, sending probes to space and getting a different perspective has confirmed this with 100% certainity for us. More or less.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2013-10-11T05:27:41.707" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="612" PostTypeId="2" ParentId="601" CreationDate="2013-10-11T08:36:17.243" Score="12" Body="&lt;p&gt;There is no &quot;orbital path&quot; detected, that's why it is a &quot;free-floating planet&quot;. There is no radial velocity mesured, but the informations given by its kinematic location show that it belongs to the beta Pictoris group, that is a stellar group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more dirty details, have a look at the submitted paper on PSO J318.5-22:&#xA;&lt;a href=&quot;http://arxiv.org/abs/1310.0457&quot;&gt;http://arxiv.org/abs/1310.0457&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Comment:&lt;/strong&gt;&#xA;Apart from that (the following reflects my personal opinion) the term &quot;free-floating planet&quot; is ill-chosen; it is clearly a very low-mass object, but since it is not orbiting around another larger object, it seems to me that the term &quot;planet&quot; is not pertinent. I think that it should be consider more than a &quot;very low-mass brown-dwarf&quot;. The problem arises from IAU definition of a brown-dwarf and a exoplanet, that is probably not well-fitted for this kind of objects (and that is not really physical). You will notice, by the way, that this object is moving in a stellar group, which kind of reinforce my point.&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2013-10-11T09:54:24.950" LastActivityDate="2013-10-11T09:54:24.950" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="613" PostTypeId="2" ParentId="601" CreationDate="2013-10-11T08:36:52.567" Score="7" Body="&lt;p&gt;We don't know the full story about the trajectory of the rogue planet PSO J318.5-22, more observations are needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though (according to the article you linked in the question) the planet is not orbiting a star, it is moving with a stellar group.  The &lt;a href=&quot;http://iopscience.iop.org/1538-4357/562/1/L87/fulltext/15754.text.html&quot;&gt;The β Pictoris Moving Group&lt;/a&gt; (Zuckerman et al. 2001), which is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;17 star systems, each with one or more characteristics indicative of extreme youth, that are moving through space together with β Pic. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user8" LastEditorUserId="6" LastEditDate="2013-10-21T19:52:33.847" LastActivityDate="2013-10-21T19:52:33.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="614" PostTypeId="5" CreationDate="2013-10-11T08:40:21.407" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-11T08:40:21.407" LastActivityDate="2013-10-11T08:40:21.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="615" PostTypeId="4" CreationDate="2013-10-11T08:40:21.407" Score="0" Body="Questions about the observation of celestial objects by non-professional astronomers." OwnerUserId="88" LastEditorUserId="15269" LastEditDate="2016-12-22T14:07:18.280" LastActivityDate="2016-12-22T14:07:18.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="616" PostTypeId="5" CreationDate="2013-10-11T08:46:23.763" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-11T08:46:23.763" LastActivityDate="2013-10-11T08:46:23.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="617" PostTypeId="4" CreationDate="2013-10-11T08:46:23.763" Score="0" Body="Questions about observations performed in the radio frequency range of the electromagnetic spectrum." OwnerUserId="88" LastEditorUserId="2365" LastEditDate="2020-04-29T06:45:59.867" LastActivityDate="2020-04-29T06:45:59.867" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="618" PostTypeId="5" CreationDate="2013-10-11T08:47:48.940" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-11T08:47:48.940" LastActivityDate="2013-10-11T08:47:48.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="619" PostTypeId="4" CreationDate="2013-10-11T08:47:48.940" Score="0" Body="Questions about photography of celestial objects by amateur or professional astronomers. Questions about photography of other objects are off-topic but might be asked on our sister site Photography Stack Exchange." OwnerUserId="88" LastEditorUserId="7472" LastEditDate="2018-04-11T23:04:51.073" LastActivityDate="2018-04-11T23:04:51.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="620" PostTypeId="5" CreationDate="2013-10-11T08:51:29.333" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-11T08:51:29.333" LastActivityDate="2013-10-11T08:51:29.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="621" PostTypeId="4" CreationDate="2013-10-11T08:51:29.333" Score="0" Body="Questions about patterns formed by group of prominent stars in the night sky." OwnerUserId="88" LastEditorUserId="88" LastEditDate="2013-10-11T15:59:58.623" LastActivityDate="2013-10-11T15:59:58.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="623" PostTypeId="1" AcceptedAnswerId="705" CreationDate="2013-10-11T14:32:38.377" Score="16" ViewCount="417" Body="&lt;p&gt;For the purposes of this question I would interpret the word &quot;recorded&quot; as loosely as possible meaning that historic records such as ice cores or tree rings would count. Also what effect did such a large flare have on life on Earth at the time?&lt;/p&gt;&#xA;" OwnerUserId="117" LastEditorUserId="6" LastEditDate="2013-10-15T22:16:33.673" LastActivityDate="2016-08-03T12:22:49.617" Title="What is the largest recorded solar flare?" Tags="&lt;the-sun&gt;&lt;solar-flare&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="624" PostTypeId="1" AcceptedAnswerId="626" CreationDate="2013-10-11T14:35:37.280" Score="16" ViewCount="410" Body="&lt;p&gt;Solar are violent releases of solar magnetic energy. Other stars are also known to have magnetic fields, in some cases much stronger than the Sun. How do the largest stellar flares compare to the largest solar flares in terms of energy released? It would also be interesting to normalize the answer by the strength of the magnetic field to be able to compare the efficiency of magnetic energy release on other stars compared to the Sun.&lt;/p&gt;&#xA;" OwnerUserId="117" LastActivityDate="2013-10-11T19:05:14.823" Title="How do large solar flares compare to flares on other stars?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;solar-flare&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="625" PostTypeId="1" AcceptedAnswerId="635" CreationDate="2013-10-11T16:38:54.387" Score="11" ViewCount="116" Body="&lt;p&gt;There are several articles out saying that a Warwick/Cambridge study of ultraviolet spectroscopy data from the Cosmic Origins Spectrograph on the Hubble Space Telescope found a large amount of oxygen in rocky debris around the white dwarf GD 61. They say that this amount of oxygen is indicative of a larger mass (asteroid), that was once composed of the rocky debris, which was 26% water.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4yYg7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4yYg7.jpg&quot; alt=&quot;Asteroid around GD 61&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;https://images.sciencedaily.com/2013/10/131010142739-large.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;sciencedaily.com&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;sub&gt;Image credit: copyright Mark A. Garlick, space-art.co.uk, University of Warwick and University of Cambridge&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;What I cannot seem to find is how much oxygen the spectroscopic examination found. Also, what methods did they use to determine how much water there would have been based on the level of oxygen?&lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-07-13T11:19:25.173" Title="How much oxygen did the Warwick/Cambridge study find in the rocky debris around white dwarf GD 61?" Tags="&lt;water&gt;&lt;spectrometry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="626" PostTypeId="2" ParentId="624" CreationDate="2013-10-11T18:52:20.830" Score="7" Body="&lt;p&gt;This is a part answer, a comparison of sun-like star superflare behaviour with our sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the article &lt;a href=&quot;http://arxiv.org/pdf/1308.1480v2.pdf&quot;&gt;Superflares on Solar Type Stars Observed with Kepler&#xA;I. Statistical Properties of Superflares&lt;/a&gt; (Shibayama et al. 2013), observations were made on solar-like stars (type G), over 500 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One key observation was that they&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;found 1547 superflares on 279 G-type dwarfs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Despite this seemingly massive amount, they deduced that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the occurrence frequency&#xA;  of superflares with energy of 10^34 − 10^35 erg is once in 800-5000 years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;in some G-type dwarfs the occurrence frequency of superflares was&#xA;  extremely high, ∼ 57 superflares in 500 days (i.e., once in 10 days). In the case of&#xA;  Sun-like stars, the most active stars show the frequency of one superflares (with&#xA;  10^34 erg) in 100 days.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These are associated with very large star-spots, far larger than those on our sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was an earlier theory that the presence of Hot Jupiters were a major contributor to super flares, hence the reason that our Sun did not exhibit this phenomenon often. However, there is some evidence of a possible past superflare occurred from our Sun:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;an occurrence of a energetic cosmic ray event in 8th century&#xA;  recorded in a tree ring of Japanese cedar trees. There is a possibility that this event was&#xA;  produced by a superflare (with energy of ∼ 10^35 erg) on our Sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and no hot Jupiters were detected around many of the stars observed either, so this theory is largely ruled out by the authors. Rather, they postulate that G-type stars 'store' magnetic energy.&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-10-11T19:05:14.823" LastActivityDate="2013-10-11T19:05:14.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="630" PostTypeId="1" AcceptedAnswerId="631" CreationDate="2013-10-13T14:37:21.977" Score="9" ViewCount="9730" Body="&lt;p&gt;Being able to see the Moon is dependant on various factors like sunset timing etc. Considering all these factors, which country or area of the world sees the new moon first? I am asking about the &lt;strong&gt;new&lt;/strong&gt; moon.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorDisplayName="user96" LastEditDate="2013-10-13T14:59:36.363" LastActivityDate="2018-08-12T15:25:36.903" Title="From which country or area is the new moon visible first?" Tags="&lt;the-moon&gt;&lt;moon-phases&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="631" PostTypeId="2" ParentId="630" CreationDate="2013-10-13T15:40:46.773" Score="10" Body="&lt;p&gt;The duration of one &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunation&quot; rel=&quot;noreferrer&quot;&gt;lunation&lt;/a&gt; (the period between one new moon to the next one) isn't neither constant as the Moon rotates around the Earth and it around the Sun (changes between 29.272 and 29.833 days due to the perturbing effects of the Sun's gravity on the Moon's eccentric orbit), nor integer divisible by 24 hours or one Earth's rotation around its axis. So this position of the Moon on the skies where the next &lt;a href=&quot;https://en.wikipedia.org/wiki/New_moon&quot; rel=&quot;noreferrer&quot;&gt;new moon&lt;/a&gt; as the first of its lunar phases will be first observable constantly changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/cmfTj.jpg&quot; alt=&quot;lunar phases&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Phases of the Moon, as seen looking southward from the Northern Hemisphere. The Southern Hemisphere will see each phase&lt;br /&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;rotated through 180°. (Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_phase&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia on Lunar phase&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saying it differently, by the time the Moon completes one lunation (or it's &lt;a href=&quot;https://en.wikipedia.org/wiki/Synodic_period&quot; rel=&quot;noreferrer&quot;&gt;synodic period&lt;/a&gt;) and starts the next one, it won't be positioned exactly above the same Earth's &lt;a href=&quot;https://en.wikipedia.org/wiki/Longitude&quot; rel=&quot;noreferrer&quot;&gt;longitude&lt;/a&gt; (East to West) as the one it started at.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2013-10-13T15:40:46.773" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="632" PostTypeId="1" AcceptedAnswerId="6239" CreationDate="2013-10-13T20:05:06.510" Score="14" ViewCount="533" Body="&lt;p&gt;Is there an analytical technique for determining the effect of a small variable transverse acceleration upon the rate of aspides precession (strictly not a precession but rotation of the line of aspides) of a planet orbitting around the Sun in a 2D plane according to Newtonian gravity law?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have modelled such effects in a reiterative computer model and would like to verify those measurements.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transverse acceleration formula is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$At = (K/c^2)*Vr*Vt * Ar.$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where:-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;c&lt;/em&gt; is speed of light, &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;K is a constant of magnitude between 0 and +/-3, such that $K/(c^2) &amp;lt;&amp;lt; 1$. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Ar is the acceleration of the planet towards the Sun due to Newtonian gravitational influence of the Sun, ($Ar = GM/r^2$).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Vr is radial component of planet velocity relative to the Sun (+ = motion away from the Sun)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Vt is transverse  component of planet velocity relative to the Sun (+ = direction of planet forward motion along its orbital path). Vectorially Vt = V - Vr where V is the total instantaneous velocity vector of the planet relative to the Sun. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Assume planet mass is small relative to the Sun&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;No other bodies are in the system&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;All motions and accelerations are confined to the two-dimensional plane of the orbit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;UPDATE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why this is interesting to me is that a value of K = +3 in my computer model produces anomalous (Non-Newtonian) periapse rotation rate values very close within about 1% of those predicted by General Relativity and within a few percent of those observed by astronomers (Le Verrier, updated by Newcomb).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formula (Einstein, 1915) for GR-derived periapse rotation (radians per orbit)&#xA;from &lt;a href=&quot;http://en.wikipedia.org/wiki/Apsidal_precession&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Apsidal_precession&lt;/a&gt;&#xA;$$&#xA;\boldsymbol{\omega}=24.\pi^3.a^2.T^{-2}.c^{-2}.(1-e^2)^{-1}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have accepted Walter's answer.  Not only did he answer the original question (Is there a technique...?) but also his analysis produces a formula which not only confirms the computer-simulated effects of the transverse acceleration formula (for K = 3) but which also (unexpectedly to me) is essentially equivalent to the Einstein 1915 formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from Walter's Summary (in Walter's answer below):-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;: (from first order peturbation analysis) semi-major axis and eccentricity are unchanged, but the direction of periapse rotates in the plane of the orbit at rate $$ \omega=\Omega \frac{v_c^2}{c^2} \frac{K}{1-e^2}, $$ where $\Omega$ is the orbital frequency and $v_c=\Omega a$ with $a$ the semi-major axis. Note that (for $K=3$) this agrees with the general relativity (GR) precession rate at order $v_c^2/c^2$ (given by Einstein 1915).&lt;/p&gt;&#xA;" OwnerUserId="233" LastEditorUserId="233" LastEditDate="2014-09-15T10:33:55.777" LastActivityDate="2014-09-15T10:33:55.777" Title="Determining effect of small variable force on planetary perihelion precession" Tags="&lt;gravity&gt;&lt;eccentric-orbit&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="635" PostTypeId="2" ParentId="625" CreationDate="2013-10-14T09:43:40.637" Score="5" Body="&lt;p&gt;In &lt;a href=&quot;http://www.sciencemag.org/content/342/6155/218.full&quot; rel=&quot;noreferrer&quot;&gt;Farihi et al. (2013)&lt;/a&gt; (it's a &lt;em&gt;Science&lt;/em&gt; paper, unfortunately I'm not sure its content is freely accessible), they actually measured metal excess in the white dwarf GD 61 (for an astronomer, everything that is nor hydrogen neither helium is a metal). Due to high surface gravity in white dwarfs, any heavy element should sink rapidely in its atmosphere; therefore, they infer that, if you find traces of metals, it means that it comes from accreted material, &quot;polluting&quot; its atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you find oxygen in a white dwarf spectrum, you know that it has to come from somewhere else, and if the white dwarf has a circumstellar disk, it has to come from planetesimals that are accreted from the disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what they did was to measure abundances (or upper limit for their abundance) for a bunch of chemical elements (O, Mg, Al, Si, Ca, Fe) and carbon. Then, they could infer different things to determine the total oxygen budget:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There is a carbon deficiency, that means that carbon abundance has no impact on the total oxygen budget;&lt;/li&gt;&#xA;&lt;li&gt;Elements Mg, Al, Si, Ca, are assumed to be carried as MgO, Al2O3, SiO2, and CaO at their maximum abundance, so once they had these abundances, they could substract it to the total oxygen budget;&lt;/li&gt;&#xA;&lt;li&gt;The remaining oxygen excess, found in FeO and debris, is interpreted as water-rich (if you want to form FeO, you need water somehow).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;That's how they were able to determine this water abundance in debris around the white dwarf GD 61.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure the oxygen abundance would be very meaningful here, but since you asked for it, it is of -5.95+/-0.13 dex.&lt;/p&gt;&#xA;" OwnerUserId="88" LastActivityDate="2013-10-14T09:43:40.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="636" PostTypeId="2" ParentId="564" CreationDate="2013-10-14T10:17:06.863" Score="5" Body="&lt;p&gt;There are such evidence in the case of Mars. Observations from &lt;a href=&quot;http://en.wikipedia.org/wiki/Mars_Global_Surveyor&quot;&gt;Mars Global Surveyor&lt;/a&gt; show evidences of crustal magnetization. In particular, this magnetization has extensive, east-west trending linear features in Terra Cimmeria and Terra Sirenum. These are probably reminiscent of magnetic features associated with a reversing dipole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no such evidence in the case of Venus: Venus has no intrinsic magnetic field today (contrary to the common belief, its inability to maintain a dynamo activity is either due to its chemical and physical conditions unable to form a solid core, either du to an early complete core solidification; it is not due to its slow rotation), and could not have maintained a remanent crustal magnetization, because its crust would have been to hot during its dynamo activity (above the Curie point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sources&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sciencemag.org/content/284/5415/794.abstract&quot;&gt;Acuna et al. (1999)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www-ssc.igpp.ucla.edu/personnel/russel/papers/venus_mag/&quot;&gt;Luhmann &amp;amp; Russell (1997)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="88" LastActivityDate="2013-10-14T10:17:06.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="637" PostTypeId="5" CreationDate="2013-10-14T11:49:20.617" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-14T11:49:20.617" LastActivityDate="2013-10-14T11:49:20.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="638" PostTypeId="4" CreationDate="2013-10-14T11:49:20.617" Score="0" Body="Questions about reflecting telescopes using a concave primary mirror and a flat diagonal secondary mirror." OwnerUserId="88" LastEditorUserId="88" LastEditDate="2013-10-14T13:05:54.080" LastActivityDate="2013-10-14T13:05:54.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="639" PostTypeId="1" AcceptedAnswerId="640" CreationDate="2013-10-14T14:30:22.200" Score="49" ViewCount="3604" Body="&lt;p&gt;This &lt;a href=&quot;https://space.stackexchange.com/a/635/58&quot;&gt;answer&lt;/a&gt; on Space Exploration to a question about Mars says that one reason Mars has such a thin atmosphere is because it lacks a magnetic field to protect it from the effect of double solar winds.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/636/6&quot;&gt;Here&lt;/a&gt; MBR explains that Venus does not have a magnetic field.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://sci.esa.int/documents/34571/35316/1567217260900-VenusSurface-410.jpg&quot; alt=&quot;Venus-atmopshere&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;sub&gt;Image credit: ESA&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;If this is the case, then why has Venus's atmosphere not been stripped away by solar wind like that of Mars?&lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorUserId="6" LastEditDate="2020-07-24T14:55:13.967" LastActivityDate="2021-01-05T00:42:31.437" Title="Why has Venus's atmosphere not been stripped away by solar wind?" Tags="&lt;planetary-atmosphere&gt;&lt;magnetic-field&gt;&lt;venus&gt;&lt;solar-wind&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="640" PostTypeId="2" ParentId="639" CreationDate="2013-10-14T14:56:26.140" Score="34" Body="&lt;p&gt;Venus has a strong ionosphere that protects it against violent solar winds. So, even though Venus has no intrisic magnetic field, it has an effective, induced magnetic field due to the interaction between the solar winds and the atmosphere, that protects it against solar winds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus atmosphere is thick enough to have a consequent ionosphere, that would be the difference between Mars and Venus (and Venus was able to keep a thicker atmosphere due to its greater mass, contrary to Mars).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sources:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www-ssc.igpp.ucla.edu/personnel/russell/papers/venus_mag/&quot;&gt;Luhmann &amp;amp; Russell (1997)&lt;/a&gt;,  on Venus&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www-ssc.igpp.ucla.edu/personnel/russell/papers/mars_mag/&quot;&gt;Luhmann &amp;amp; Russell (1997)&lt;/a&gt;,  on Mars&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="88" LastActivityDate="2013-10-14T14:56:26.140" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="641" PostTypeId="1" AcceptedAnswerId="644" CreationDate="2013-10-14T18:03:23.053" Score="12" ViewCount="3482" Body="&lt;p&gt;Watching &lt;a href=&quot;http://www.youtube.com/watch?v=a6kxVnPrbjs&quot; rel=&quot;nofollow&quot;&gt;this youtube video&lt;/a&gt; shows that our planet is nothing more than dust in the wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the &lt;a href=&quot;https://www.youtube.com/watch?v=a6kxVnPrbjs#t=2m40s&quot; rel=&quot;nofollow&quot;&gt;2:50&lt;/a&gt; mark in the video, you will see a view of the whole Milky Way. At the bottom you will see this text &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Light travel time from earth: 100,000 years&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which means if we travel at light speed we need 100,000 years to go far from the Milky Way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If nothing can travel faster than light, and if it takes 100,000 years to go far from the Milky Way at light speed &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How did they map our whole universe, which has a radius of 13.798±0.037 billion light years?&lt;/strong&gt;   &lt;/p&gt;&#xA;" OwnerUserId="239" LastEditorUserId="239" LastEditDate="2014-01-30T19:13:52.547" LastActivityDate="2019-04-30T13:39:23.293" Title="How exactly is our universe mapped?" Tags="&lt;universe&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="642" PostTypeId="2" ParentId="641" CreationDate="2013-10-14T19:18:15.190" Score="4" Body="&lt;p&gt;That's the part of the universe that is visible to us right now. In other words, we map it from light we receive from that zone right now, so the part being mapped is actually many, many years old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually this doesn't make a big difference because we're careful to analyze regions where the variation in the time of viewing is not large compared to the length scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, from the relativity point of view, it's almost as good as the faraway bodies being in the state they were billions of years ago as far as we are concerned. From our reference frame, they may have a &quot;future&quot; (really, the &quot;present&quot;, but calling it that gets a bit icky) but we just have no way of knowing it.&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2013-10-14T19:18:15.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="643" PostTypeId="1" CreationDate="2013-10-14T19:46:41.913" Score="4" ViewCount="211" Body="&lt;p&gt;I'm trying to understand the climatic effects of the far future scenario of an Earth-like planet with a reduced rotational speed caused by tidal locking with the moon (day-night period of 28 days, one hemisphere always facing the moon, the other never facing the moon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that observations of the moon's surface temperature have shown day temperatures of 120°C and freezing cold nights of -230°C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How would the day and night temperatures of tidal locked Earth vary?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also interested in climatic effects caused by the reduced tides, coriolis force, increased evaporation, but this may reasonably be outside the scope of this question and it's answers.&lt;/p&gt;&#xA;" OwnerUserId="241" LastEditorUserId="6" LastEditDate="2013-11-15T18:12:13.533" LastActivityDate="2013-12-12T14:45:28.493" Title="Day and night temperature on an earthlike planet with longer rotational period" Tags="&lt;the-moon&gt;&lt;earth&gt;&lt;temperature&gt;&lt;tidal-forces&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2013-12-12T14:45:35.557" ContentLicense="CC BY-SA 3.0" />
  <row Id="644" PostTypeId="2" ParentId="641" CreationDate="2013-10-14T21:30:39.797" Score="12" Body="&lt;p&gt;The simple answer is that light from the distant parts of the Milky Way Galaxy have &lt;strong&gt;already been traveling&lt;/strong&gt; for that amount of time. Because light does not travel instantaneously through space, we are seeing the galaxy not as it is now, but as it was 100,000 (or so) years ago. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, this is also how we are able to see back to earlier moments of the universe's history. In terms of distance, the further we look, the longer light would have had to leave in order to reach us today. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe did have a beginning (at least in the sense of having an earliest time). Because the universe is not infinitely old, there is actually a limit to how far we can see back into the universe's history (see: &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmological_horizon&quot; rel=&quot;noreferrer&quot;&gt;Cosmological horizon&lt;/a&gt;). We are able to make maps of the universe, like this one from &lt;a href=&quot;http://www.sdss.org/&quot; rel=&quot;noreferrer&quot;&gt;SDSS&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LHa3B.jpg&quot; alt=&quot;SDSS Map&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by running what's known as a survey. These are large projects which maps out the positions (and potentially other properties, depending on what their science goals are) of galaxies in the universe. How you should look at this picture is the following: 1) You're at the center of the image, 2) Each point in the picture represents an individual galaxy, 3) Distance from the observer, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Redshift&quot; rel=&quot;noreferrer&quot;&gt;redshift&lt;/a&gt; of a galaxy, increases with increasing distance from the center, and 4) as you travel around the circle in angle, you're sweeping through the &lt;a href=&quot;http://en.wikipedia.org/wiki/Right_ascension&quot; rel=&quot;noreferrer&quot;&gt;right ascension&lt;/a&gt; coordinate of the celestial coordinate system (think longitude). You're seeing a pizza slice instead of a sphere because this is a particular slice in &lt;a href=&quot;http://en.wikipedia.org/wiki/Declination&quot; rel=&quot;noreferrer&quot;&gt;declination&lt;/a&gt; (lattitude). As you look to higher redshifts, you are looking further back into time, and earlier into the universe's history. This is how astronomers/cosmologists can say something &lt;em&gt;statistically&lt;/em&gt; about models of structure formation of the universe (how objects like galaxies, and clusters of galaxies came to be).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with looking at galaxies at very high redshifts is that they become&#xA;very faint. At a certain point, you need more powerful telescopes to continue looking further back in time. Though, we can get around this somewhat by looking at the radiation component of the universe at microwave wavelengths (most surveys are in the optical region of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_spectrum&quot; rel=&quot;noreferrer&quot;&gt;electro-magnetic spectrum&lt;/a&gt;). This radiation is called the cosmic microwave background radiation (&lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;noreferrer&quot;&gt;CMB&lt;/a&gt;), and can also be mapped out (see: &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_Background_Explorer&quot; rel=&quot;noreferrer&quot;&gt;COBE&lt;/a&gt;,&lt;a href=&quot;http://en.wikipedia.org/wiki/Wmap&quot; rel=&quot;noreferrer&quot;&gt;WMAP&lt;/a&gt;, and most recently &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck_%28spacecraft%29&quot; rel=&quot;noreferrer&quot;&gt;Planck&lt;/a&gt; satellites).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YZEtL.jpg&quot; alt=&quot;Planck Map&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also a map of sorts. This map of low energy photons is a snapshot of the universe approximately ~380,000 (redshift of z=1100; meaning the universe was smaller by about a factor of 1101) years after the big bang, when the universe had very little of what we would recognize as structure. The job of the cosmologist is to connect the picture we see from the CMB to the picture we see from our surveys of galaxies through the use of physics.   &lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditDate="2013-10-14T21:37:23.543" LastActivityDate="2013-10-14T21:37:23.543" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="646" PostTypeId="1" AcceptedAnswerId="647" CreationDate="2013-10-15T04:25:01.693" Score="24" ViewCount="23340" Body="&lt;p&gt;Our nearest star &lt;a href=&quot;https://en.wikipedia.org/wiki/Proxima_Centauri&quot;&gt;Proxima Centauri&lt;/a&gt; is 4.243 light years away from Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that mean we are seeing light that is 4.243 years old everyday?  &lt;/p&gt;&#xA;" OwnerUserId="239" LastEditorUserId="254" LastEditDate="2013-10-15T11:12:22.593" LastActivityDate="2018-11-21T13:49:52.030" Title="Is the light we see from stars extremely old?" Tags="&lt;star&gt;&lt;distances&gt;&lt;light&gt;&lt;speed&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="647" PostTypeId="2" ParentId="646" CreationDate="2013-10-15T05:49:38.000" Score="21" Body="&lt;p&gt;Yes, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Speed_of_light&quot; rel=&quot;noreferrer&quot;&gt;speed of light&lt;/a&gt; in &lt;a href=&quot;https://en.wikipedia.org/wiki/Vacuum&quot; rel=&quot;noreferrer&quot;&gt;vacuum&lt;/a&gt; (or &lt;em&gt;c&lt;/em&gt;) is 299,792,458 m/s and one &lt;a href=&quot;https://en.wikipedia.org/wiki/Light-year&quot; rel=&quot;noreferrer&quot;&gt;light-year&lt;/a&gt; is the distance the light travels in one &lt;a href=&quot;https://en.wikipedia.org/wiki/Julian_year_%28astronomy%29&quot; rel=&quot;noreferrer&quot;&gt;Julian year&lt;/a&gt; (365.25 days), which comes out as 9.4605284 × 10&lt;sup&gt;15&lt;/sup&gt; meters. Since &lt;em&gt;c&lt;/em&gt; is the maximum speed at which all energy, matter, and information in the Universe can travel, it is the universal physical constant on which the light-year (&lt;em&gt;ly&lt;/em&gt;) as one of the astronomical units of length is based. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that &lt;a href=&quot;https://en.wikipedia.org/wiki/Light&quot; rel=&quot;noreferrer&quot;&gt;visible light&lt;/a&gt; as an &lt;a href=&quot;https://en.wikipedia.org/wiki/Electromagnetic_radiation&quot; rel=&quot;noreferrer&quot;&gt;electromagnetic radiation&lt;/a&gt; cannot travel faster than &lt;em&gt;c&lt;/em&gt; and in one Julian year it can traverse a maximum distance of &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;d = t * c&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;code&gt;d&lt;/code&gt;&lt;/em&gt; is distance in meters &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;code&gt;t&lt;/code&gt;&lt;/em&gt; time in seconds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;code&gt;c&lt;/code&gt;&lt;/em&gt; the speed of light in vacuum in meters per second&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we calculate this distance for a 4.243 &lt;em&gt;ly&lt;/em&gt; distant object, that comes out as &lt;em&gt;&lt;code&gt;4.243 * 365.25 * 86,400 s * 299,792,458 m * sˉ¹&lt;/code&gt;&lt;/em&gt; or exactly 40,141,879,395,160,334.4 meters (roughly 40 &lt;a href=&quot;https://en.wikipedia.org/wiki/Trillion&quot; rel=&quot;noreferrer&quot;&gt;trillion&lt;/a&gt; kilometers or 25 trillion miles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the distance the light traveled since it was last reflected of (or in our case emitted from, since Proxima Centauri is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Red_dwarf&quot; rel=&quot;noreferrer&quot;&gt;red dwarf&lt;/a&gt; star) the surface of a celestial object to be 4.243 Julian years later visible at our observation point, in this case our planet Earth from where the distance to Proxima Centauri you quoted was measured. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more powerful the telescope, the further into the past we can see because the light is much older! This goes the same regardless of the distance of the object you're observing, but astronomy is particularly neat in this regard and we can observe objects that are so distant that we see them from the time when they were still forming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further reading on other units used to measure far away objects you might be interested in reading &lt;a href=&quot;https://astronomy.stackexchange.com/questions/94/what-is-a-parsec-and-how-is-it-measured&quot;&gt;this question on the parsec.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2013-10-15T07:05:33.867" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="648" PostTypeId="1" AcceptedAnswerId="983" CreationDate="2013-10-15T08:26:44.220" Score="11" ViewCount="896" Body="&lt;p&gt;In slow-roll inflation models, the early inflation of the universe is driven by the flat non-zero part of the inflaton potential, and it ends as the ball rolls down the cliff and the potential energy released leads then to the formation of matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ylJYx.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as I have heard such slow-roll models are no longer deemed realistic today in inflationary cosmology, instead other models where the end of the inflation is explained by quantum tunneling of the universe from a state with higher vacuum energy density to another state with lower vacuum energy density are considered for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ljdEY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why (from a physical point of view) are slow-roll inflation models no longer considered to be realistic, what are their disadvantages? What can models that explain (the end of) the early inflation by a tunneling transition between to vacua explain or describe, that slow roll models can not do?&lt;/p&gt;&#xA;" OwnerUserId="92" LastEditorUserId="92" LastEditDate="2013-10-15T08:53:32.320" LastActivityDate="2013-11-21T06:59:41.517" Title="Inflation cosmology: slow-roll inflation versus tunneling between two vacua" Tags="&lt;cosmology&gt;&lt;cosmological-inflation&gt;&lt;early-universe&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="649" PostTypeId="1" CreationDate="2013-10-15T09:26:34.857" Score="8" ViewCount="182" Body="&lt;p&gt;Reading the article &lt;a href=&quot;http://orbit.psi.edu/~tricaric/pdf/skads.pdf&quot;&gt;On the asteroid belt’s orbital and size distribution&lt;/a&gt; (Gladman et al. 2009), there is an interest point made:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;asteroids a&#xA;  few kilometers and smaller in size are recent collisional shards of&#xA;  larger objects since their lifetime against collisional destruction is&#xA;  much less than the age of the Solar System. As such, these bodies preserve information about collisional breakup, the dominant&#xA;  physical process that has shaped the present asteroid belt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This suggests that there would have been a lot more asteroids etc in the main belt at the time prior to the Late Heavy Bombardment (~4 billion years ago) - which is after the major planets had formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is &lt;strong&gt;what was the greatest spatial extent of the Asteroid Belt in the period just before the Late Heavy Bombardment?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am after peer reviewed resources).&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-10-15T09:53:31.120" LastActivityDate="2014-09-23T12:39:22.163" Title="What was the greatest spatial extent of the Asteroid Belt prior to the Late Heavy Bombardment?" Tags="&lt;size&gt;&lt;lhb&gt;&lt;asteroid-belt&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="651" PostTypeId="1" CreationDate="2013-10-15T13:47:43.837" Score="3" ViewCount="187" Body="&lt;p&gt;The rotation axis of Uranus is 90 degrees away from vertical, &lt;a href=&quot;https://astronomy.stackexchange.com/questions/38/what-is-the-accepted-theory-as-to-why-uranus-axis-is-tilted-so-severely&quot;&gt;there are some theories about its reason.&lt;/a&gt;. But which are the consequences of this? Which are the unique features of Uranus that happened because of it?&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2013-10-15T13:59:11.443" Title="Which are the consequences of Uranus' high rotation axis angle?" Tags="&lt;rotation&gt;&lt;uranus&gt;" AnswerCount="0" CommentCount="6" ClosedDate="2013-10-15T22:09:08.857" ContentLicense="CC BY-SA 3.0" />
  <row Id="652" PostTypeId="2" ParentId="1" CreationDate="2013-10-15T14:33:22.527" Score="9" Body="&lt;p&gt;You just need a telescope with a &quot;wedge&quot; mount, (ie one for polar coordinates,) and not one which is simply a pan-and-tilt mount like commonly used for cameras. (BUT, it might be more fun to do it with a sextant -- see below.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a telescope that has a polar mount, you just need to set it up correctly. That means orienting the base of the mount in the correct north/south direction, and then setting the &quot;wedge&quot; angle to account for your geographic latitude. (Any amateur level scope with a polar mount will have instructions.) When you pan the scope it will sweep celestial lines of equal latitude, and when you tilt the scope it will sweep celestial lines of equal longitude. Point the scope at your target, and read the declination from the scope's mount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hardest part is going to be that the satellite will be moving pretty fast, and its orbit is going to have a continuously varying declination. (Unless you target a satellite in a circular, equatorial orbit. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fun with a sextant&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be easier to &quot;shoot&quot; the satellite with a simple (search for &quot;Davis Mark 15&quot; on ebay) sextant and something called an &quot;artificial horizon.&quot; You need any liquid reflecting surface... a swimming pool or kiddie pool might work for shooting a moving satellite. (You can buy a small &quot;artificial horizon&quot; that's a few inches square, but you'll never catch the satellite's reflection in that.) Using the sextant you measure the angle between the satellite and its reflection, also noting the compass direction. Then you crunch a bunch of trigonometry. But this is exactly how you practice astronomical navigation star sighting on land.&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2013-10-15T14:33:22.527" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="653" PostTypeId="1" CreationDate="2013-10-15T18:47:32.730" Score="16" ViewCount="1310" Body="&lt;p&gt;In case of Earth, we have many hints about the internal structure of our planet, from what I know, the most important of which is analyzing seismic waves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we have any instruments to detect if other solid planets, such as Mars or Venus, have solid or liquid core? Is it possible using remote sensing?&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2013-10-16T12:39:37.463" Title="How do we detect if a planet has a liquid core?" Tags="&lt;planet&gt;&lt;core&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="654" PostTypeId="1" CreationDate="2013-10-15T18:58:15.320" Score="10" ViewCount="228" Body="&lt;p&gt;Do we know (have we observed and cataloged) any rogue star, being not part of galaxy, but drifting somewhere in inter-galactic space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that determining if a star is a part of galaxy or not is a matter of definition, but let's take those, that inter-galactic space is where the gravitational influence of any galaxy isn't dominating. &lt;/p&gt;&#xA;" OwnerUserId="42" LastEditorDisplayName="user96" LastEditDate="2013-10-15T19:05:31.983" LastActivityDate="2013-10-16T09:17:44.770" Title="Have we observed any rogue/wandering stars?" Tags="&lt;star&gt;&lt;galaxy&gt;&lt;intergalactic-space&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="656" PostTypeId="1" AcceptedAnswerId="657" CreationDate="2013-10-15T22:33:02.420" Score="13" ViewCount="106" Body="&lt;p&gt;When an exocomet passes by our Sun in a trajectory that leads it to establish a Solar Orbit what eventually happens to it?&#xA;In addition to its original inertia and vector being altered what are the longer term affects that occur?&lt;/p&gt;&#xA;" OwnerUserId="257" LastActivityDate="2013-10-15T23:28:45.617" Title="Long term development of Comets" Tags="&lt;comets&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="657" PostTypeId="2" ParentId="656" CreationDate="2013-10-15T23:28:45.617" Score="6" Body="&lt;p&gt;Ice chunks would be melted off during closer approaches, potentially blasting non-ice bits off of the exocomet as it goes. This could affect the trajectory of the exocomet both by the inertial force of the separation and because the mass would be reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually it would run out of water, which means it would no longer be a comet--the streaming tail of gas wouldn't exist anymore, even on closer approaches to the sun. What would remain--assuming anything remained at all--would be a hard rocky core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there, the orbit would either stabilize or decay to a point that it crashed into the sun. Even a fairly stable orbit could eventually bring it close to other gravitational bodies in the solar system which could throw it into an unstable orbit, or alternatively, it could collide with a planet or asteroid.&lt;/p&gt;&#xA;" OwnerUserId="158" LastActivityDate="2013-10-15T23:28:45.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="658" PostTypeId="1" CreationDate="2013-10-16T01:00:02.003" Score="14" ViewCount="907" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Lyman-alpha_forest&quot; rel=&quot;noreferrer&quot;&gt;Lyman-alpha forest&lt;/a&gt; is an interesting feature found in the spectra of sufficiently distant objects. This series of absorption lines extends over a range of frequencies, and are a result of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lyman-alpha_forest&quot; rel=&quot;noreferrer&quot;&gt;Lyman-alpha electron transition&lt;/a&gt; of neutral hydrogen in between the source and the observer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ahImV.gif&quot; alt=&quot;Lyman-alpha forest&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &#xA;What information can be gleaned about the universe from this spectral feature?  In other words, what areas of research use the Lyman-alpha forest as a tool, instead of seeing it as noise? &lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorDisplayName="user96" LastEditDate="2013-10-16T11:30:37.300" LastActivityDate="2013-10-16T11:30:37.300" Title="What is the Lyman Alpha forest Used For?" Tags="&lt;spectra&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="659" PostTypeId="2" ParentId="658" CreationDate="2013-10-16T06:45:01.693" Score="9" Body="&lt;p&gt;According to UCLA's web-tutorial &lt;a href=&quot;http://www.astro.ucla.edu/~wright/Lyman-alpha-forest.html&quot; rel=&quot;noreferrer&quot;&gt;Lyman Alpha Forest&lt;/a&gt; (Wright, 2004), there are many 'clouds' of gas between us and a distant quasar (as an example) that absorb&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;ultraviolet light at the wavelength of the Lyman alpha line of hydrogen at a wavelength of 122 nm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, as the clouds of gas have less of a redshift than that of the distant quasar, then their absorption lines are less redshifted than the distant object - a cartoon example is shown below (from the UCLA website):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qo6sY.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the significance is of the forest is that it represents clouds that are smaller than the smallest galaxies, thus&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We can only see these very low mass clouds by the absorption they produce in the strongest line of the most abundant element: Lyman alpha. Thus by studying the Lyman alpha forest we can learn about the density fluctuations in the Universe on the smallest observable scales. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A similar explanation of the significance of the forest, is given by &lt;a href=&quot;http://astrobites.org/2011/10/20/quasars-the-lyman-alpha-forest-and-the-reionization-of-the-universe/&quot; rel=&quot;noreferrer&quot;&gt;Quasars, the Lyman Alpha Forest, and the Reionization of the Universe&lt;/a&gt; (Mortlock et al. 2011) is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;quasars like ULAS J1120+0641 are bright and at high redshift, lower redshift intervening material can absorb some of their light, leaving fingerprints on the final spectrum that we observe on Earth. Because hydrogen is the most abundant element in the universe, it should come as no surprise that it leaves the most prominent spectral signature, in the form of a forest of absorption lines.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Specifically, of major significance is that Mortlock et al. (2011) state that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Lyman alpha forest can be used to trace this re-ionization of the universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2013-10-16T06:45:01.693" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="660" PostTypeId="1" AcceptedAnswerId="662" CreationDate="2013-10-16T06:49:30.687" Score="5" ViewCount="282" Body="&lt;p&gt;I realize there are very few to none universities that offer direct specialization of astronomy. If I aspired to be make astronomy my career choice, I'd likely have to pick something like physics and specialize later - or something similar - but that's just my hunch, and I'd prefer to know for sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What educational and professional paths optimize one's education and skill set for the occupation of astronomer? How do I become one?&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="6" LastEditDate="2016-05-25T15:42:16.107" LastActivityDate="2016-05-27T06:49:48.560" Title="What study profiles could land me the job of astronomer?" Tags="&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="662" PostTypeId="2" ParentId="660" CreationDate="2013-10-16T08:42:58.783" Score="10" Body="&lt;p&gt;What you need would be at least the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Physics&lt;/strong&gt;: astronomy is physics in the end, and you'll need strong basis in physics;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Math&lt;/strong&gt;: as for any good education in physics, you'll need a reasonable amount of math to be able to become an astronomer;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Programming&lt;/strong&gt;: it's almost impossible nowadays to get away from it. Being an astronomer today implies lots of data processing (to reduce the data, to interpret it etc.), and it involves a lot of coding.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And, as far as I know, in lots of countries/universities, you can find at least few astronomy &amp;amp; astrophysics courses. In some countries, you can even find dedicated master degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With all of that, you'll be ready to start a PhD in astronomy to become a professional astronomer.&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2014-05-27T15:49:21.880" LastActivityDate="2014-05-27T15:49:21.880" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="663" PostTypeId="2" ParentId="654" CreationDate="2013-10-16T09:17:44.770" Score="3" Body="&lt;p&gt;Different types of observations have been made:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;hypervelocity stars&lt;/strong&gt;: there are stars in the halo of our Galaxy (it is kind of the suburbs of the Milky Way) with such velocity that they are escaping it, and aiming to the intergalactic medium (IGM). There not yet fully in the IGM, but the're on their way. (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2005ApJ...622L..33B&quot; rel=&quot;nofollow&quot;&gt;Reference&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;molecular clouds in the IGM&lt;/strong&gt;: there are molecular clouds detected in the IGM, that are forming stars. Stars are not directly observed, but these star forming molecular clouds are proof of their existence. (&lt;a href=&quot;http://arxiv.org/abs/1202.2862&quot; rel=&quot;nofollow&quot;&gt;Reference&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It's hard to do better, because it is hard to observe directly stars outside of the Milky Way.&lt;/p&gt;&#xA;" OwnerUserId="88" LastActivityDate="2013-10-16T09:17:44.770" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="664" PostTypeId="1" AcceptedAnswerId="670" CreationDate="2013-10-16T12:18:23.763" Score="12" ViewCount="1150" Body="&lt;p&gt;I've seen in documentaries that at the center of each galaxy is a super massive black hole which holds the galaxy together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since black holes have such a strong pull, are we slowly being pulled in closer and closer to this black hole?&lt;/p&gt;&#xA;" OwnerUserId="258" LastEditorDisplayName="user96" LastEditDate="2013-10-16T12:40:39.530" LastActivityDate="2020-10-06T13:49:55.407" Title="Are we moving ever closer to the center of our Galaxy due to a super massive black hole?" Tags="&lt;cosmology&gt;&lt;supermassive-black-hole&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="665" PostTypeId="2" ParentId="653" CreationDate="2013-10-16T12:39:37.463" Score="5" Body="&lt;p&gt;A good way to show how this is achieved is to show you a case study, below I will detail how we came to make the theory that Mercury has a liquid core. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is first important to note that we are never &lt;em&gt;definite&lt;/em&gt; of the properties but we are able to use the data to infer and create a theory that we believe to be an accurate representation of the truth.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The MESSENGER spacecraft was launched for the purpose of analyzing MErcuries Surface, Space ENvironment, GEochemistry and Ranging (hence MESSENGER).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Messenger had a number of pieces of equipment on board to allow it to conduct it's experiments, as detailed in the below image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QEgtn.jpg&quot; alt=&quot;Messengers Equipment&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://messenger.jhuapl.edu/instruments/MLA.html&quot; rel=&quot;noreferrer&quot;&gt;Mercury Laser Altimeter&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how does the MLA aide us in determining if Mercury has a liquid core? Well the purpose of the MLA is to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;track the planet’s slight, forced libration – a wobble about its spin axis – which will tell researchers about the state of Mercury’s core.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.geolas.com/Pages/laser.html&quot; rel=&quot;noreferrer&quot;&gt;How does a Laser Altimeter work?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A laser altimeter is operated from a plane, a helicopter or a satellite. It determines the distance to the [Planet's] surface by measuring the time-of-flight of a short flash of infrared laser radiation. The instrument emits laser pulses which travel to the surface, where they are reflected. Part of the reflected radiation returns to the laser altimeter, is detected, and stops a time counter which was started when the pulse was sent out. The distance is then easily calculated by taking the speed of light into consideration.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using this technology one can build up a 3D visualization of a planets surface&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What scientists &lt;a href=&quot;http://www.popsci.com/science/article/2012-03/mercury-smoother-moon-and-other-new-surprises-innermost-planet&quot; rel=&quot;noreferrer&quot;&gt;discovered from this mapping&lt;/a&gt; is that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The planet's crust is thicker in low latitudes and thinner at the poles, a distribution that suggests the planet could have a liquid outer core. Its core is also large relative to the planet, comprising 85 percent of the planet's radius, much more than Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How did they come to this conclusion? &lt;a href=&quot;http://www.sciencemag.org/content/336/6078/214.full&quot; rel=&quot;noreferrer&quot;&gt;This paper explains in depth.&lt;/a&gt; &lt;em&gt;(The full version is free, but requires signing up to the site, signing up is also free and has no costs and has the ability for you to turn off receiving emails and newsletters from them)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most notable part is this excerpt and image:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Resulting moment of inertia parameters for more than 1 million Monte Carlo models that include a silicate crust and mantle, as well as an Fe-rich core that may contain solid and liquid layers are shown in&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fiyy6.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(A) Outer radius of Mercury’s liquid core. (B) Average density of the solid shell that overlies the liquid core&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, specific readings taken by a Laser Altimeter allow us to map out the three dimensional shape of a planets surface. Analysis of the shape shows us that the crust is thinner at the poles which is likely due to a liquid core. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On top of this, Mercury is believed to have a small magnetic field (1% of our own) and this might be due to a liquid outer core circulating a solid inner core, acting like a dynamo. Though there is still much speculation given over what causes the magnetic field so this is unreliable to use for a solid footing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from that there is the logical thought process that larger planets are capable of creating huge amounts of heat and pressure at their cores which create a molten core, or at least a molten layer. This is why there is much skepticism over Mercury because it is such a small planet that its core &lt;em&gt;should&lt;/em&gt; have solidified long ago. &lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastActivityDate="2013-10-16T12:39:37.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="666" PostTypeId="1" AcceptedAnswerId="668" CreationDate="2013-10-16T13:37:49.987" Score="50" ViewCount="26872" Body="&lt;p&gt;This arose from a comment posted against &lt;a href=&quot;https://space.stackexchange.com/questions/2433/have-humans-ever-really-been-to-outer-space&quot;&gt;a question&lt;/a&gt; elsewhere on the stackexchange&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How powerful a telescope/binoculars would allow me to view the astronauts aboard the ISS do a space-walk? Aperture? Magnification? &lt;/p&gt;&#xA;" OwnerUserId="259" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2017-12-15T23:34:53.190" Title="How powerful a telescope would allow me to view the astronauts aboard ISS do a space-walk?" Tags="&lt;amateur-observing&gt;&lt;telescope&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="667" PostTypeId="2" ParentId="666" CreationDate="2013-10-16T15:15:40.597" Score="24" Body="&lt;p&gt;&lt;a href=&quot;http://ralfvandebergh.startje.be/&quot; rel=&quot;noreferrer&quot;&gt;Ralf Vandebergh&lt;/a&gt; is one of the best amateur astronomy photographers out there who does spacecraft photography. He is using a 10&quot; (25.4cm) &lt;a href=&quot;http://en.wikipedia.org/wiki/Newtonian_telescope&quot; rel=&quot;noreferrer&quot;&gt;Newtonian telescope&lt;/a&gt;, as far as I know, so this is pretty much an off the shelf telescope. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He supposedly has &lt;a href=&quot;http://www.startje.be/vieuw.php?qid=335219&quot; rel=&quot;noreferrer&quot;&gt;imaged spacewalkers&lt;/a&gt; on previous ISS and &lt;a href=&quot;http://en.wikipedia.org/wiki/Space_Transportation_System&quot; rel=&quot;noreferrer&quot;&gt;STS&lt;/a&gt; missions. Though they are only a few pixels in size, and you cannot make out any details. Nevertheless, his images are definitely, absolutely stunning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/V0IVN.jpg&quot; alt=&quot;STS-133 Astronaut during EVA-2 (by Ralf Vandebergh)&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="50" LastEditorUserId="259" LastEditDate="2013-10-16T16:18:48.330" LastActivityDate="2013-10-16T16:18:48.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="668" PostTypeId="2" ParentId="666" CreationDate="2013-10-16T15:16:45.020" Score="44" Body="&lt;p&gt;This has been done before, so I don't have to go through all the heavy calculations using Rayleigh criterion accounting for atmospheric diffraction and visible light wavelength. Ralf Vandebergh, a Dutch astronomer, professional photographer and veteran satellite spotter has been busy trying to do exactly this since the 2007 and has indeed succeeded on several occasions by now using a 10 inch (25.4 cm) Newtonian reflecting telescope that has a resolving power (angular resolution on the CCD sensor) of roughly one pixel per meter at the distance to the International Space Station (ISS) that is &lt;a href=&quot;https://encrypted.google.com/search?q=iss%20altitude&quot;&gt;currently in a 230 miles (370 km) orbit&lt;/a&gt; above the Earth:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/xwMEF.jpg&quot; alt=&quot;Ralf Vandebergh’s detail of an image he took on March 21, 2009 showing astronauts working outside the ISS. Credit: Ralf Vandebergh&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;sup&gt;Ralf Vandebergh’s detail of an image he took on Mar. 21, 2009 showing astronauts working outside the ISS. Credit: &lt;a href=&quot;http://ralfvandebergh.startje.be/&quot;&gt;R. Vandebergh&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ralfvandebergh.startje.be/&quot;&gt;Vandebergh's personal page&lt;/a&gt; also hosts all kinds of other successful observations of the ISS through his telescope and recorded in both photographs as well as some short videos. &lt;strong&gt;Why short? Because targeting the ISS as it moves at the speed of &lt;a href=&quot;https://encrypted.google.com/search?q=iss%20speed&quot;&gt;4.8 mi/s (7.7 km/s)&lt;/a&gt; is rather tricky, and the atmospheric conditions and times at which the ISS passes over some area on the surface of the Earth don't make it any easier either&lt;/strong&gt;. But perseverance and hard work have paid well for this individual astronomer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/Nkr7k.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Raw video of the ISS as seen through the air turbulence. Note the good visibility of the Lira antenna at the Russian&lt;br /&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Zvezda Module in the lower part of the image. Credit: &lt;a href=&quot;http://ralfvandebergh.startje.be/&quot;&gt;R. Vandebergh&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So again, skipping the math to calculate the required &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_resolution&quot;&gt;angular resolution&lt;/a&gt; of a telescope and apply that to some arbitrarily selected image sensor size and resolution, we can see that using a &lt;strong&gt;well collimated 10 inch &lt;a href=&quot;http://en.wikipedia.org/wiki/Newtonian_telescope&quot;&gt;Newtonian&lt;/a&gt; or a &lt;a href=&quot;http://en.wikipedia.org/wiki/Dobsonian_telescope&quot;&gt;Dobsonian telescope&lt;/a&gt; on a clear night&lt;/strong&gt; can, with some near perfect targeting, produce a direct proof of a spacewalker doing his or her job 230 miles (370 km) high during an EVA. More powerful telescopes would of course produce better resolution images, but the atmospheric effects limit their use and are of course much harder to target an object moving fast over the skies with.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorDisplayName="user96" LastEditDate="2013-10-28T13:41:08.670" LastActivityDate="2013-10-28T13:41:08.670" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="669" PostTypeId="1" CreationDate="2013-10-16T15:44:06.363" Score="44" ViewCount="6786" Body="&lt;p&gt;If the universe has formed &amp;amp; originated by a Big Bang Explosion, then there must be empty space left in the center of the explosion site, as all the matter is travelling at tremendous speeds away from the center, and there must be more matter, stars, galaxies and dust, etc near the present periphery or circumference or horizon of the present universe. As that big explosion has taken place about 13.7 billion years back, then the outer boundaries of our universe are 13.7 billion light years away from the centre of the explosion of Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have our astronomers discovered hollowness or emptiness anywhere in the centre of the universe or not? &lt;/p&gt;&#xA;" OwnerUserId="260" LastEditorUserId="14078" LastEditDate="2016-11-12T16:54:24.923" LastActivityDate="2018-06-20T04:38:28.713" Title="What is in the center of the universe?" Tags="&lt;universe&gt;&lt;big-bang-theory&gt;&lt;formation&gt;" AnswerCount="7" CommentCount="3" FavoriteCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="670" PostTypeId="2" ParentId="664" CreationDate="2013-10-16T15:53:50.237" Score="7" Body="&lt;p&gt;That's not really how gravity works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a galaxy forms, all the dust that falls into it retains angular momentum round the centre (and this pattern repeats locally round stars, and round planets round those stars etc) which leads to the relatively stable structures we see today with spiral arms etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Collisions and near collisions can change the direction of stuff orbiting a centre, but we are quite far out, so while we could change orbits we would be as likely to move wider than closer in the short term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over the longer term, it's a balance between the gradual loss of angular momentum of the Milky Way as a whole due to drag, gravity waves etc. and the slow growth of the black hole at the centre.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2013-10-16T15:53:50.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="673" PostTypeId="1" CreationDate="2013-10-16T16:43:12.473" Score="0" ViewCount="93" Body="&lt;p&gt;What shape does a three dimensional universe have ? and how did scientists determined that it is flat and not curved.&lt;/p&gt;&#xA;" OwnerUserId="261" LastActivityDate="2013-10-16T17:08:38.520" Title="What do shape of universe mean?" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2013-10-16T17:17:07.633" ContentLicense="CC BY-SA 3.0" />
  <row Id="674" PostTypeId="2" ParentId="673" CreationDate="2013-10-16T17:08:38.520" Score="1" Body="&lt;p&gt;I think it's useful to realize the meaning of &quot;shape&quot; is not the same as in &quot;what is the shape of this object I'm holding.&quot; Because we are a part of the universe, &quot;shape&quot; is about how things work (eg, how light moves through the universe), rather than &quot;is the universe a sphere, cube, hypersphere or some other 'shape'.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is probably too broad to be answered here. I suggest reading the &lt;a href=&quot;http://en.wikipedia.org/wiki/Shape_of_the_universe&quot; rel=&quot;nofollow&quot;&gt;WikiPedia article on the 'Shape of the Universe'&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2013-10-16T17:08:38.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="675" PostTypeId="1" AcceptedAnswerId="688" CreationDate="2013-10-17T02:00:54.913" Score="15" ViewCount="1417" Body="&lt;p&gt;I've never quite understood the theory behind why dark matter and dark energy exist. I know it has something to do with gravitational pull being stronger than what we calculate it SHOULD be, could someone explain this in more detail?&lt;/p&gt;&#xA;" OwnerUserId="258" LastEditorUserId="88" LastEditDate="2013-10-17T14:17:42.610" LastActivityDate="2014-01-29T15:46:08.050" Title="How do we know dark matter/dark energy exists?" Tags="&lt;cosmology&gt;&lt;dark-matter&gt;&lt;dark-energy&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="676" PostTypeId="1" AcceptedAnswerId="685" CreationDate="2013-10-17T02:21:56.480" Score="13" ViewCount="409" Body="&lt;p&gt;We know that the universe is approximately 13.7 billion years old. What methods did we use to discover this? How sure are we about its accuracy?&lt;/p&gt;&#xA;" OwnerUserId="258" LastActivityDate="2013-10-18T01:49:41.240" Title="Age of the universe" Tags="&lt;early-universe&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="677" PostTypeId="1" AcceptedAnswerId="680" CreationDate="2013-10-17T06:55:11.057" Score="15" ViewCount="1371" Body="&lt;p&gt;As I understand it, the difference between a reflector and a refractor is that a reflector uses mirrors on the back to reflect the light to the secondary mirror then the secondary mirror reflects the light to your eye focusing the image with each reflection. How then does a refractor use a lens to sharpen the image? Does it just sharpen the image and shoot it to the prism at the back, and into your eye?&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2013-10-17T08:23:02.500" Title="How Does a Refractor Telescope Work?" Tags="&lt;amateur-observing&gt;&lt;telescope&gt;&lt;refractor-telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="678" PostTypeId="1" AcceptedAnswerId="690" CreationDate="2013-10-17T07:00:51.557" Score="12" ViewCount="164" Body="&lt;p&gt;I own a newtonian reflector and occasionally use it (using the reflection method with paper) for looking at the sun. I happen to also own a refractor, but I'm not exactly sure if the reflection method will damage the equipment (notably the lens on the refractor) if I do.&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2013-10-17T17:10:28.283" Title="Does the Reflection Method for Looking at the Sun Work For Refractors?" Tags="&lt;the-sun&gt;&lt;amateur-observing&gt;&lt;telescope&gt;&lt;refractor-telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="679" PostTypeId="1" AcceptedAnswerId="681" CreationDate="2013-10-17T08:06:15.410" Score="16" ViewCount="356" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/314/83&quot;&gt;Answer&lt;/a&gt; to my question partially answers this one, about density of intergalactic matter and matter within galaxy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But it is mostly a hot, ionized void. How void? The density of the intergalactic medium is about 1 to 100 particles per cubic meter (you can compare it to the mean galactic density, of about a million particles per cubic meter, or that of Earth's atmosphere, of about 10^26 particles cubic meter). How hot? It can go from 10^5 to 10^7 K.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we skip the most dense concentrations of matter (stars, planets, generally everything solid, liquid or plasma, and border conditions like their atmosphere) how dense interstellar matter can we find? What is the concentration of matter in densest nebulae that still don't collapse into bodies like planets or stars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And conversely, how empty does the space get at its emptiest? I could imagine only very few particles in last their short moments of travel into the center exist under a black hole's event horizon, but other than that, how empty a space can be found in the universe, and where?&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2013-10-17T09:10:23.637" Title="How does interstellar matter density vary?" Tags="&lt;matter&gt;&lt;interstellar-medium&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="680" PostTypeId="2" ParentId="677" CreationDate="2013-10-17T08:23:02.500" Score="11" Body="&lt;p&gt;All telescopes have in common that they gather and focus light from far away objects. They use a primary opical element, such as a concave mirror or a (planar- or bi-)convex lense (or lense system), and they use an eyepiece with another lense system (for viewing) or a camera in their primary focus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A refractor telescope does not &lt;em&gt;sharpen&lt;/em&gt; the image per se. The convex lense concentrates the light rays, not unlike a magnifying glass. To actually focus the enlarged image on your retina, you need an eyepiece, which is another bi-convex lense (in its most simple form). This will re-align the light rays after they passed through the primary focus. See this image for a visual explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/crzqE.png&quot; alt=&quot;Light ray sketch for a refracting telescope&quot;&gt;&#xA;Source: &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Telescope-schematic-A.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above image also explains why the image of a refracting telescope appears upside down. You don't need (or want!) any prisms in this kind of setup. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, a reflector telescope uses a concave mirror plus an eyepiece. There are different configurations, but one of the most simple and most common ones is the Newtonian telescope:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eC2Db.png&quot; alt=&quot;Newton telescope ray diagram&quot;&gt;&#xA;Source: &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Newtontelescope.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So instead of refraction of light by a lense, we use the reflection of light on a mirror, to enlarge the image. Focusing on the retina is again done by an eyepiece in the same way as with the refracting telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantage of refracting telescopes is that there is no obstruction in the optical path inside the telescope. This is not the case with reflector telescopes. They usually have a secondary mirror in the middle of the optical paths, hence reducing light gathering performance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, reflector telescopes are often much lighter, and cheaper to assemble. Also, very compact models of reflector telescopes can be built. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, simple refractor telescopes will produce colourful fringes on object edges, called chromatic abberation, which is due to the glass used in the lenses. This can be compensated by multiple lenses, but this will make the refractor even heavier and more expensive.&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2013-10-17T08:23:02.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="681" PostTypeId="2" ParentId="679" CreationDate="2013-10-17T09:04:09.573" Score="10" Body="&lt;p&gt;The interstellar medium is a multiphase medium, and you can find (some references &lt;a href=&quot;http://www.maths.qmul.ac.uk/~wjs/MTH726U/chap3.pdf&quot;&gt;in this lecture&lt;/a&gt; and in this &lt;a href=&quot;http://www.lra.ens.fr/~levrier/Education/these.php&quot;&gt;thesis manuscript&lt;/a&gt; (this one is in French, but numbers are international)):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the &lt;strong&gt;hot ionized medium&lt;/strong&gt; (HIM) with density as low as 10^-3 cc (particles per cubic centimeter);&lt;/li&gt;&#xA;&lt;li&gt;the &lt;strong&gt;warm ionized medium&lt;/strong&gt; (WIM), with density of the order of 0.03 cc;&lt;/li&gt;&#xA;&lt;li&gt;the &lt;strong&gt;warm neutral medium&lt;/strong&gt; (WNM), with density of the order of 0.25 cc;&lt;/li&gt;&#xA;&lt;li&gt;the &lt;strong&gt;HII regions&lt;/strong&gt;, with densities ranging from 1 to 10^4 cc;&lt;/li&gt;&#xA;&lt;li&gt;the &lt;strong&gt;cold neutral medium&lt;/strong&gt; (CNM), with density of the order of 25 cc;&lt;/li&gt;&#xA;&lt;li&gt;the &lt;strong&gt;molecular clouds&lt;/strong&gt;, with densities above 10^3 cc, up to 10^6 cc roughly.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These different phases are due to the interplay of cooling and heating processes of its own components (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1995ApJ...443..152W&quot;&gt;Wolfire et al. 1995&lt;/a&gt;). The lowest density regions are hot and associated to &lt;strong&gt;supernova expansion bubbles&lt;/strong&gt;. The HII regions (HII for &quot;ionized hydrodgen&quot;) are associated with &lt;strong&gt;O stars&lt;/strong&gt; (massive stars that can efficiently ionized their environment).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The highest density you can get with no evidence of forming stars are of the order of 10^4 cc (see for example &lt;a href=&quot;http://www.google.fr/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=9&amp;amp;ved=0CGQQFjAI&amp;amp;url=http%3A%2F%2Fherschel.esac.esa.int%2FFirstResultsSymposium%2Fpresentations%2FA43_WardThompsonD_PolarisCores.pdf&amp;amp;ei=16JfUs3WGOWJ7Aba8IG4Cw&amp;amp;usg=AFQjCNEooNvjzWFwdxLEX7KD-3y6Bjz8HA&amp;amp;sig2=Fs6d6qQsiPeIk0w7qt76JQ&amp;amp;bvm=bv.54176721,d.Yms&quot;&gt;this talk by Ward-Thompson&lt;/a&gt; on starless cores). Dense, molecular clouds that are about to collapse and form a star have a density of the order of 10^6-10^7 cc.&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2013-10-17T09:10:23.637" LastActivityDate="2013-10-17T09:10:23.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="682" PostTypeId="5" CreationDate="2013-10-17T11:34:08.147" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-17T11:34:08.147" LastActivityDate="2013-10-17T11:34:08.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="683" PostTypeId="4" CreationDate="2013-10-17T11:34:08.147" Score="0" Body="Questions about the gas that lays between stars in galaxies." OwnerUserId="88" LastEditorUserId="88" LastEditDate="2013-10-17T12:25:54.563" LastActivityDate="2013-10-17T12:25:54.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="684" PostTypeId="2" ParentId="675" CreationDate="2013-10-17T12:16:28.983" Score="10" Body="&lt;p&gt;Dark matter and dark energy are two different things, accounting for different observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dark matter&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark matter is needed to explain, among other things, the rotation curve of galaxies. One could expect these rotation curves to decrease at large radii (because one should expect keplerian rotation for galaxies), and it is not the case, the rotation curves are flat, that accounts for a &quot;hidden mass&quot; in galaxies to sustain such a rotation pattern, that is the dark matter. Dark matter also accounts for some galaxy clusters properties, and CMB fluctuations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dark energy&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark energy is needed to explain the accelerated expansion of the Universe. This expansion was mostly measured using Type Ia supernovæ (thermonuclear supernovæ, that are the results of a white dwarf accreting matter and reaching Chandrasekhar mass) that are &quot;standard candles&quot;. The absolute magnitude of these objects are thought to be well-known, so if you observe them in the Universe, you can easily deduce their distance from their redshift. These observations show an accelerated expansion of the Universe (by expansion, we mean that the relative distance between two distant galaxies is increasing, distance in the local Universe are not affected). To account for this acceleration, one needs some energy uniformly distributed in the Universe, that is the dark energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Don't be confused:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark energy and dark matter cannot be mistaken one for the other, because dark energy is uniformly distributed in the Universe, whereas dark matter follows some distribution (it is generally more or less distributed with the matter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Final words:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, &quot;dark matter&quot; and &quot;dark energy&quot; are mostly words to characterize our ignorance on these observations that cannot be explained by standard laws of physics. That is why there is such intense work on these subjects, to try to characterize these dark things better and to understand what they could be.&lt;/p&gt;&#xA;" OwnerUserId="88" LastActivityDate="2013-10-17T12:16:28.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="685" PostTypeId="2" ParentId="676" CreationDate="2013-10-17T12:41:07.743" Score="15" Body="&lt;p&gt;The correct version of your first sentence would be something like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given our knowledge and the standard cosmological model, we estimate that the age of the universe is about 13.7 billion years old.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Every work quotes slightly different values for the age, depending on methods, observations used, assumptions, etc. As example, the &lt;a href=&quot;http://arxiv.org/pdf/1303.5076v1.pdf&quot;&gt;first results&lt;/a&gt; from the &lt;a href=&quot;http://www.rssd.esa.int/index.php?project=planck&quot;&gt;satellite Planck&lt;/a&gt; report an age of 13.813, with an uncertainty of 58 million years (0.5%, to my knowledge one of the best).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, first question is answered.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What methods did we use to discover this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;About 100 years ago a number of people realized that our Galaxy is not the only one in the Universe and that the &lt;strong&gt;all&lt;/strong&gt;&lt;sup&gt;1&lt;/sup&gt; other galaxies are receding from us with a speed increasing with distance (credit for this work is generally given to &lt;a href=&quot;http://en.wikipedia.org/wiki/Edwin_Hubble&quot;&gt;Edwin Hubble&lt;/a&gt; for his Redshift-Distance relation for galaxies formulated in 1929). This together with the theoretical results from general relativity, brought scientists to believe that our Universe is expanding. Which means that in the universe's past it was smaller, much smaller than it is now &lt;sup&gt;2&lt;/sup&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The age of the Universe depends on the rate of expansion at each time. To have this, you must estimate the mean composition of the universe (in various components like radiation, matter, dark energy, and curvature) and the present day expansion rate (also known as the &lt;a href=&quot;https://www.cfa.harvard.edu/~dfabricant/huchra/hubble/&quot;&gt;Hubble Constant&lt;/a&gt;). Once you have this, it's &lt;em&gt;easy&lt;/em&gt;&lt;sup&gt;3&lt;/sup&gt; to compute the age of the Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;footnotes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;except the ones in the local group that are gravitationally bound to us&lt;/li&gt;&#xA;&lt;li&gt;side note: the cosmic microwave background (a radiation emitted about 13 and a half billions of years ago) provided in the 1964 a striking confirmation that the universe once was indeed much hotter and smaller&lt;/li&gt;&#xA;&lt;li&gt;knowing what an integral is and with some basic of programming shouldn't be to difficult to solve equation 1 &lt;a href=&quot;http://www.astro.umd.edu/~miller/teaching/astr422/lecture14.pdf&quot;&gt;here&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="86" LastEditorUserId="45" LastEditDate="2013-10-18T01:49:41.240" LastActivityDate="2013-10-18T01:49:41.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="686" PostTypeId="1" AcceptedAnswerId="693" CreationDate="2013-10-17T15:53:35.890" Score="11" ViewCount="469" Body="&lt;p&gt;All the news-articles say that it is expected to &quot;eventually&quot; explode. Which doesn't really tell me much. When is &quot;eventually&quot;? Also, considering this is the largest star in the known universe and 16k light-years away, will this be visible? Will it be dangerous?&lt;/p&gt;&#xA;" OwnerUserId="273" LastActivityDate="2013-10-21T19:37:17.333" Title="In what timeframe is W26 expected to go supernova?" Tags="&lt;supernova&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="687" PostTypeId="1" AcceptedAnswerId="691" CreationDate="2013-10-17T16:58:32.890" Score="15" ViewCount="464" Body="&lt;p&gt;I have heard that the only planets able to be seen outside the Solar system are Jovian-sized planets with the occasional detection of planets three times the Earth's size. But, as far as I know, we haven't seen any Earth-sized (due to the range of our telescopes?) extrasolar planets. When will we be able to detect them?&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="6" LastEditDate="2016-05-23T17:57:23.737" LastActivityDate="2016-05-23T17:57:23.737" Title="When will we be able to detect Earth-sized exoplanets?" Tags="&lt;exoplanet&gt;&lt;earth-like-planet&gt;&lt;planetary-systems&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="688" PostTypeId="2" ParentId="675" CreationDate="2013-10-17T17:03:19.130" Score="18" Body="&lt;p&gt;Some additions to the answer of MBR:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, we do not know that dark matter and dark energy do exist, but we have indirect clues. You will often see claims that dark matter and dark energy are two of the major problems of cosmology today, including by professional astronomers, but this is an epistemological misconception: you cannot call a hypothesis &lt;em&gt;a problem&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently we have several observational anomalies (as defined per T. Kuhn): namely the rotation curve of galaxies, the luminosity of supernovae, the power spectrum of the CMB... (and some additional things like the difficulty to measure G or to detect directly gravitational waves).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the history of science, we have 3 ways to solve these anomalies:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;ontological approaches (dark matter and dark energy)&lt;/li&gt;&#xA;&lt;li&gt;legislatives approaches (extensions of General Relativity)&lt;/li&gt;&#xA;&lt;li&gt;paradigmatic approaches (inhomogeneous Universes)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Dark matter and dark energy are two ontological approaches that seems to work pretty well: by adding two simple components we can fit a wide range of observations. This is why astronomers like them: their modelization are relatively easy and allow to fit everything well. But for the moment we do not know their nature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some say that with &lt;a href=&quot;http://map.gsfc.nasa.gov/&quot;&gt;WMAP&lt;/a&gt; (Wilkinson Microwave Anisotropy Probe) and &lt;a href=&quot;http://www.esa.int/Our_Activities/Space_Science/Planck&quot;&gt;Planck space observatory&lt;/a&gt; we have entered in the era of precision cosmology: but this is only true if our paradigm of dark matter/dark energy approaches is correct. If this is not the case, we currently do not interpret correctly what we measure, but ultimately more accurate measures can lead to a better view of current cosmological anomalies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So dark matter and dark energy are two elegant and minimalist approaches to several fundamental problems. This is a very hot topic in cosmology, but we may realize, one day, that we only had problems because our current paradigm was erroneous.&lt;/p&gt;&#xA;" OwnerUserId="176" LastEditorUserId="13" LastEditDate="2013-12-08T06:33:23.407" LastActivityDate="2013-12-08T06:33:23.407" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="689" PostTypeId="1" CreationDate="2013-10-17T17:06:17.883" Score="7" ViewCount="132" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Arecibo_message&quot;&gt;Arecibo Message&lt;/a&gt; was aimed at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Great_Globular_Cluster_in_Hercules&quot;&gt;globular star cluster M13&lt;/a&gt;, but as far as I read, M13 will move out of the original position and thus not be able to receive it. So, who or what will be able to receive the message?&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="13" LastEditDate="2013-10-17T17:09:03.707" LastActivityDate="2013-10-18T08:39:11.827" Title="Who or what will the Arecibo message reach?" Tags="&lt;star-systems&gt;&lt;messier-catalog&gt;&lt;star-cluster&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="690" PostTypeId="2" ParentId="678" CreationDate="2013-10-17T17:10:28.283" Score="3" Body="&lt;p&gt;Actually, your Newtonian may run the risk of suffering from the projection method. The secondary, planar mirror in the Newtonian will be subject to heating. Although it is not in the prime focal point, the light rays are already pretty much bundled up, and this may lead to heating and ultimately to failure of the secondary mirror!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With refractors, you don't have anything in the optical path up until the eyepiece, so in a sense it is more safe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However:&lt;/strong&gt; I strongly suggest against using simple projection methods. One alternative is to use a &lt;a href=&quot;http://en.wikipedia.org/wiki/Herschel_wedge&quot; rel=&quot;nofollow&quot;&gt;Herschel-wedge&lt;/a&gt;. However, these are again not that useful on Newtonians, because they filter the light at the eyepiece end of the telescope. It is probably better to buy a solar filter which is rather inexpensive and can be fitted to any kind of telescope. Without wanting to promote any specific brand, the company &lt;a href=&quot;http://www.baader-planetarium.com/sofifolie/bauanleitung_e.htm&quot; rel=&quot;nofollow&quot;&gt;Baader has a good explanation&lt;/a&gt; on the topic on their website. The important thing is that the filter is securely fastened to the optical tube assembly, and that there are no scratches or cuts in the filter foil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another much more expensive method is to use a specialized solar telescope. These usually use very narrowband filters and coatings on the telescope lenses. Often they are specialized to one of the hydrogen or calcium spectral lines, showing great amount of structure in the suns atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2013-10-17T17:10:28.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="691" PostTypeId="2" ParentId="687" CreationDate="2013-10-17T17:18:32.340" Score="15" Body="&lt;p&gt;In 2013, the smallest detected exoplanet was Kepler-37-b, which is actually smaller in mass and size than Earth, so we already had a limited ability to detect these size exoplanets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worth noting that there wasn't any new technology that allowed this advance, &lt;a href=&quot;http://www.nature.com/nature/journal/v494/n7438/full/nature11914.html&quot; rel=&quot;nofollow&quot;&gt;the paper&lt;/a&gt; linked to by NASA indicates that the same methods were used as they would usually use. Making it remarkable that they were even able to discover something this small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, even more Earth-sized planets have been detected. One method used to aid in their detection is described as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We account for Kepler's imperfect detectability of such planets by injecting synthetic planet-caused dimmings into the Kepler brightness measurements and recording the fraction detected.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sources:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://kepler.nasa.gov/Mission/discoveries/kepler37/&quot; rel=&quot;nofollow&quot;&gt;Kepler-37 - NASA&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/1311.6806&quot; rel=&quot;nofollow&quot;&gt;Prevalence of Earth-size planets orbiting Sun-like stars - Petigura, Howard, and Marcy&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6" LastEditorUserId="6" LastEditDate="2015-06-18T12:11:59.803" LastActivityDate="2015-06-18T12:11:59.803" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="692" PostTypeId="2" ParentId="669" CreationDate="2013-10-17T17:21:51.723" Score="21" Body="&lt;p&gt;Your are misunderstanding the expansion of the Universe. The Big-Bang is not an explosion: this is the moment in time when the Universe had an (near) infinite density. So there is no center in the Universe as there is no center of the SURFACE of the earth (this is the most popular 2-dimensional analog). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this primordial ultra-high density state, the Universe is expanding, atoms have formed, stars and galaxies have formed and now, at very large scale, the distance between two clusters of galaxies continue to increase with time due to the expansion.&lt;/p&gt;&#xA;" OwnerUserId="176" LastEditorUserId="45" LastEditDate="2013-10-23T18:04:10.207" LastActivityDate="2013-10-23T18:04:10.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="693" PostTypeId="2" ParentId="686" CreationDate="2013-10-17T20:22:33.927" Score="5" Body="&lt;p&gt;There is little peer-reviewed information that gives a definitive time frame of when W26 would go supernova.  The reason for this is that we have models of stellar lifecycles, and have found candidates at each 'age'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that in mind, according to a very recent article: &lt;a href=&quot;http://arxiv.org/pdf/1309.4086v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Ionized Nebula surrounding the Red Supergiant W26&#xA;in Westerlund 1&lt;/a&gt; (Wright, 2013),  (RSG = Red Super Giant)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The presence of the nebula suggests extensive mass loss&#xA;  in the recent history of W26. Its late spectral type,&#xA;  very high luminosity and spectral variability all suggest the&#xA;  star to be highly evolved amongst the RSGs. Both the star and the nebula are comparable to the RSGs VY CMa and&#xA;  WOH G64, both of which are highly luminous late-type&#xA;  RSGs with evidence for circumstellar gas.&#xA;  W26 provides a rare opportunity to directly investigate&#xA;  an extreme mass-loss event from a highly evolved RSG.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Looking at the compared stars to W26 to see what theories, even timeframes are suggested:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the article &lt;a href=&quot;http://arxiv.org/pdf/1203.5194v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fundamental properties and atmospheric structure of the red&#xA;supergiant VY CMa based on VLTI/AMBER spectro-interferometry&lt;/a&gt; (Wittkowski et al. 2012), the supergiant star VY CMa is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;close to the Hayashi limit of recent evolutionary&#xA;  tracks of initial mass 25 M⊙ with rotation or 32 M⊙ without rotation, shortly before evolving blueward in the HR-diagram.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, according to Wittkowski et al. rather than being close to supernova, could well be close to entering the next phase of stellar evolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the article &lt;a href=&quot;http://arxiv.org/pdf/1302.0822v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rd Supergiants in the Local Group&lt;/a&gt; (Levesque 2013) and &lt;a href=&quot;http://arxiv.org/pdf/0803.3823.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Spatially resolved dusty torus toward the red supergiant WOH G64&#xA;in the Large Magellanic Cloud&lt;/a&gt; (Ohnaka et al. 2008), studies of WOH G64&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;implies that this object may be experiencing unstable, violent mass loss.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TL:DR&lt;/strong&gt; So, based on observations of W26 and comparable stars, there is no definitive timeframe, primarily due to these stars being close to the &lt;em&gt;Hayashi forbidden region&lt;/em&gt;, which, according to the article &lt;a href=&quot;http://arxiv.org/pdf/0705.3431v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Late-Type Red Supergiants: Too Cool for the Magellanic Clouds?&lt;/a&gt; (Levesque et al. 2007), results in them being&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;unstable hydrodynamically, which we&#xA;  expect to lead to this variability and behavior.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Hayashi track/forbidden zones in relation to stellar masses and main sequence is shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1SsgM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://onwardtotheedge.wordpress.com/author/astroisstellar/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to which Levesque et al. state&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Further monitoring of these stars, both&#xA;  photometrically and spectroscopically, may lead to an improved understanding of this phase&#xA;  of massive star evolution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suggesting that this behaviour may be a phase (albeit one of the final phases) in their evolution.  Also, at that distance, it is likely to put on a small light show, but not pose much danger to the Earth (except in the very unlikely event of a gamma ray burst).&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorUserId="6" LastEditDate="2013-10-21T19:37:17.333" LastActivityDate="2013-10-21T19:37:17.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="694" PostTypeId="1" AcceptedAnswerId="695" CreationDate="2013-10-17T22:23:16.317" Score="7" ViewCount="5090" Body="&lt;p&gt;I'm back to the refractor that I own, and I want to maximize the zoom capacity. I realized today that I own more than one barlow lens, and I was wondering if I could stack them together for looking at bright objects, like Jupiter. I'm aware that looking at anything else would probably be disastrous due to the stacked barlow lenses taking up more light. Is this an effective, or desirable way to approach the situation? What will happen if I do stack more than one barlow lens? Will it have the effect of an enhanced image, or will it do nothing but remove light?&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="13" LastEditDate="2013-10-17T23:47:14.027" LastActivityDate="2013-10-17T23:47:14.027" Title="Are Barlow lenses stackable for bright objects?" Tags="&lt;solar-system&gt;&lt;amateur-observing&gt;&lt;telescope&gt;&lt;jupiter&gt;&lt;refractor-telescope&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="695" PostTypeId="2" ParentId="694" CreationDate="2013-10-17T23:46:11.177" Score="7" Body="&lt;p&gt;Yes, stacking Barlow lenses is a common practice to effectively increase focal length by multiplying their individual focal lengths. When I say common, most advanced eyepieces actually have many glass elements and are a type of a Barlow lens themselves, so just by using a single Barlow lens in front of your eyepiece you'd already be, technically, stacking Barlows. It is a technique that is particularly useful for bright object like our Solar system's planets, with Jupiter that you mention perhaps even the best example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your image quality might vary substantially tho with the Barlow and eyepiece quality of glass elements and the manufacturing precision. You might also want to play with the order at which you stack them and find an optimal arrangement of all of the glass elements they add to the system. The more precisely they're made, the less the order will matter though. And you'd technically want the less precise ones at the end of the stack, so their deficiencies don't magnify with each added element. And yes, with any additional glass element you'll be losing a bit of the illumination so this is, like mentioned, most suitable for bright objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do not however, under any circumstances, do this when observing the Sun! You shouldn't need additional glass elements anyway, but should you try, you'll most certainly cause them to overheat, expand and permanently deform quite fast, regardless what material their tube and thread are made of, metal, alloy or plastic.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2013-10-17T23:46:11.177" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="696" PostTypeId="2" ParentId="689" CreationDate="2013-10-18T05:05:55.617" Score="6" Body="&lt;p&gt;The Arecibo Message was not broadcast with the aim that it would intercept with a notable astronomical object. Neither do we realistically expect that anyone will ever listen to the Voyager Golden Records.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such symbolic gestures are simply vectors for public relations and education. We send these &quot;time capsules&quot; into space &lt;em&gt;because we can&lt;/em&gt;, as a showcase of our technological achievements, and as a token of hope.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;As the choice of frequency, duration of message, and distance of the&#xA;  target clearly shows, the Arecibo message is very unlikely to produce&#xA;  interstellar discourse in the foreseeable future. Rather, it was&#xA;  intended as a concrete demonstration that terrestrial radio astronomy&#xA;  has now reached a level of advance entirely adequate for interstellar&#xA;  radio communication over immense distances.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;-- NAIC staff (1975), &lt;a href=&quot;http://adsabs.harvard.edu/abs/1975Icar...26..462&quot;&gt;http://adsabs.harvard.edu/abs/1975Icar...26..462&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The spacecraft will be encountered and the record played only if there are advanced spacefaring civilizations in interstellar space. But the launching of this bottle into the cosmic ocean says something very hopeful about life on this planet.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;-- Carl Sagan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="276" LastEditorUserId="276" LastEditDate="2013-10-18T08:39:11.827" LastActivityDate="2013-10-18T08:39:11.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="697" PostTypeId="1" AcceptedAnswerId="831" CreationDate="2013-10-18T06:54:33.457" Score="17" ViewCount="1176" Body="&lt;p&gt;Please note, I do &lt;strong&gt;not&lt;/strong&gt; necessarily believe in its existence, just after a scientific (hence, non-Wikipedia) reasoning to doubt the planet's existence, other than the &quot;we would have seen it&quot; argument.  I am, nor are the articles below, referring to the alleged 'Nemesis' planet, as in the question &lt;a href=&quot;https://astronomy.stackexchange.com/questions/430/what-are-the-current-observational-constraints-on-the-existence-of-nemesis&quot;&gt;What are the current observational constraints on the existence of Nemesis?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the article &lt;a href=&quot;http://www.livescience.com/33032-mystery-planet-tyche-probably-not-there.html&quot; rel=&quot;noreferrer&quot;&gt;Astronomers Doubt Giant Planet 'Tyche' Exists in Our Solar System&lt;/a&gt; (Wolchover, 2011), the planet Tyche is a &lt;strong&gt;hypothesised&lt;/strong&gt; massive planet &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;by John Matese and Daniel Whitmire of the University of Lousiana-Lafayette, is not new: They have been making a case for Tyche since 1999, suggesting that the giant planet's presence in a far-flung region of the solar system called the Oort cloud would explain the unusual orbital paths of some comets that originate there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Further information is found in the scientists' research paper &lt;a href=&quot;http://arxiv.org/pdf/1004.4584v1.pdf&quot; rel=&quot;noreferrer&quot;&gt;Persistent Evidence of a Jovian Mass Solar Companion in the Oort Cloud&lt;/a&gt; (Matese and Whitmire, 2010), describe the hypothesised planet to be &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;of mass  1 − 4 M(Jupiter) orbiting in the&#xA;  innermost region of the outer Oort cloud. Our most restrictive prediction is that the orientation&#xA;  angles of the orbit normal in galactic coordinates are centered on &#xA;  , the galactic longitude of&#xA;  the ascending node = 319 and i, the galactic inclination = 103 (or the opposite direction)&#xA;  with an uncertainty in the normal direction subtending  2% of the sky.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They posit that the elongated orbit of Sedna is a result of the presence of this planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So, the question, what reason is there to doubt the existence of Tyche?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2018-04-11T19:19:11.070" Title="What reason is there to doubt the existence of the hypothesised planet Tyche in the far distant solar system?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;oort-cloud&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="698" PostTypeId="1" CreationDate="2013-10-18T10:11:47.867" Score="1" ViewCount="58" Body="&lt;p&gt;Moon always faces Earth with one side, its spin period equal its orbital period - this is what we learn at school.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I know Nature rarely operates in absolutes. I'd find it hard to believe Moon wouldn't spin relative to Earth at least very slowly. I searched but found no data other than &quot;doesn't&quot;, so I'd like to ask - does the Moon stay turned the same side precisely - or do, over many years, new features show at one edge? If it does, how long would it take to make a full turn?&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2013-10-18T10:11:47.867" Title="Is the moon rotation at complete standstill?" Tags="&lt;the-moon&gt;" AnswerCount="0" CommentCount="13" ClosedDate="2013-10-18T12:37:45.780" ContentLicense="CC BY-SA 3.0" />
  <row Id="699" PostTypeId="1" AcceptedAnswerId="701" CreationDate="2013-10-18T10:54:10.537" Score="16" ViewCount="8765" Body="&lt;p&gt;If Titan has a methane atmosphere and seas of methane, then why doesn't some meteoroid / chemical reaction ignite and blow the whole thing?&lt;/p&gt;&#xA;" OwnerUserId="277" LastEditorUserId="2618" LastEditDate="2017-03-07T01:10:38.503" LastActivityDate="2020-03-10T17:16:03.300" Title="Why does Titan's atmosphere not start to burn?" Tags="&lt;titan&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="700" PostTypeId="1" AcceptedAnswerId="714" CreationDate="2013-10-18T11:32:21.377" Score="11" ViewCount="3024" Body="&lt;p&gt;Trying to compare density of Black Holes and Neutron Stars I came up with the following:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A typical neutron star has a mass between about 1.4 and 3.2 solar masses&lt;a href=&quot;http://curious.astro.cornell.edu/question.php?number=219&quot; rel=&quot;noreferrer&quot;&gt;1&lt;/a&gt;[3] (see Chandrasekhar Limit), with a corresponding radius of about 12 km. (...) Neutron stars have overall densities of 3.7×10^17 to 5.9×10^17 kg/m^3 &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutron_star&quot; rel=&quot;noreferrer&quot;&gt;[1]&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You can use the Schwarzschild radius to calculate the &amp;quot;density&amp;quot; of the black hole - i.e., the mass divided by the volume enclosed within the Schwarzschild radius. This is roughly equal to (1.8x10^16 g/cm^3) x (Msun / M)^2 (...)&lt;/p&gt;&#xA;&lt;p&gt;The value of the Schwarzschild radius works out to be about (3x10^5 cm) x (M / Msun) &lt;a href=&quot;http://curious.astro.cornell.edu/question.php?number=219&quot; rel=&quot;noreferrer&quot;&gt;[2]&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Let's take a neutron star from the top of the spectrum (3.2 Msun) and same mass black hole.&lt;/p&gt;&#xA;&lt;p&gt;Converting units:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Neutron star: 5.9×10^17 kg/m^3 = 5.9 × 10^14 g/cm^3&lt;/li&gt;&#xA;&lt;li&gt;Black hole: 1.8x10^16 g/cm^3 x (1/5.9)^2 = 5.2 x10^14 g/cm^3&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The radius of the black hole would be (3x10^5 cm) x ( 5.2 ) = 15.6km&lt;/p&gt;&#xA;&lt;p&gt;The 3.2Msun Neutron Star of this density would have volume of 1.08 x 10^13 m^3 which gives radius of 13.7 kilometers&lt;/p&gt;&#xA;&lt;p&gt;According to Shell Theorem, spherical objects' gravity field strength at given distance is the same for spheres as for point masses so at the same distance from center of same mass (point - black hole, sphere - neutron star) the gravity will be the same.&lt;/p&gt;&#xA;&lt;p&gt;That would put the surface of the neutron star below the surface of event horizon of equivalent black hole. Yet I never heard about even horizon of neutron stars.&lt;/p&gt;&#xA;&lt;p&gt;Either I made a mistake in my calculations (and if I did, could you point it out?) or... well, why?&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-04-01T06:24:32.020" Title="Why don't Neutron Stars form event horizon?" Tags="&lt;black-hole&gt;&lt;neutron-star&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="701" PostTypeId="2" ParentId="699" CreationDate="2013-10-18T11:40:39.473" Score="16" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Atmosphere_of_Titan&quot; rel=&quot;nofollow noreferrer&quot;&gt;Titan&lt;/a&gt; is one of Saturn's moons. Titan has a dense atmosphere, at about 1.5 bars. It also seems to have lakes of liquid methane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a conventional combustion, you would need a good Methane-Oxygen mix. Every combustion is basically an oxidization. Apart from very energetic events, such as &lt;a href=&quot;https://astronomy.stackexchange.com/questions/490/why-did-the-comet-shoemaker-levy-9-fragments-cause-such-large-explosions-on-jupi&quot;&gt;SL9 hitting Jupiter&lt;/a&gt;, you will need quite a bit of oxygen to get a nice, explosive fireball in Titan's atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Methane#Combustion&quot; rel=&quot;nofollow noreferrer&quot;&gt;According to Wikipedia&lt;/a&gt;, you need two Oxygen molecules for every Methane molecule you want to burn.&lt;/p&gt;&#xA;" OwnerUserId="50" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2013-10-21T17:01:02.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="702" PostTypeId="1" AcceptedAnswerId="787" CreationDate="2013-10-18T13:26:42.543" Score="7" ViewCount="509" Body="&lt;p&gt;Some astronomers in Ukraine have estimated that there is a small likelihood (1 in 63,000) that the asteroid &lt;strong&gt;2013 TV&lt;sub&gt;135&lt;/sub&gt;&lt;/strong&gt; will strike Earth in 2032. (&lt;a href=&quot;http://www.spaceanswers.com/news/3673/asteroid-may-impact-in-2032/&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don Yeoman of NASA JPL says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;with more observations, I fully expect we will be able to significantly reduce, or rule out entirely, any impact probability for the foreseeable future.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can we improve our estimation of the potential threat posed by 2013 TV&lt;sub&gt;135&lt;/sub&gt;?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, what process and techniques are used to calculate and increase the accuracy of our collision threat estimation?&lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorUserId="6" LastEditDate="2016-05-23T19:15:50.533" LastActivityDate="2016-05-23T19:15:50.533" Title="How can we tell if the asteroid 2013 TV135 will hit Earth in 2032?" Tags="&lt;asteroids&gt;&lt;impact&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="703" PostTypeId="1" CreationDate="2013-10-18T23:36:34.833" Score="7" ViewCount="102249" Body="&lt;p&gt;From some time now (few weeks as far as I can remember), there is a very bright star in the eastern sky. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I first thought it was Venus, but according to &lt;a href=&quot;http://earthsky.org/astronomy-essentials/visible-planets-tonight-mars-jupiter-venus-saturn-mercury&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt;, Venus is in the western sky and sets 2 hours after sunset. But this (very) bright star is still up even though it's midnight (6 hours after sunset).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know what is this star (or planet) ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please excuse my silly question as I know nothing about astronomy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. &lt;/p&gt;&#xA;" OwnerUserId="279" LastEditorUserId="2035" LastEditDate="2014-11-24T19:27:00.247" LastActivityDate="2016-12-15T17:53:56.627" Title="Very bright star in the east at northern hemisphere. What is it?" Tags="&lt;star&gt;" AnswerCount="6" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="704" PostTypeId="2" ParentId="703" CreationDate="2013-10-19T00:27:43.977" Score="5" Body="&lt;p&gt;It will be very difficult to tell you without knowing where you are located.  Particularly, if you are in the northern or southern hemisphere.  At this time of year, most of the brightest stars are in the sky.  Also where in the sky, the particular star that you are wondering about is.  And any other constellations that you can identify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For stars rising this time of year, particularly bright ones you might be thinking of Sirius.  Which would be near the constellation Orion below the hourglass shape of the it (&lt;a href=&quot;http://starchart.polaris.net/chart.php?constel=ori&amp;amp;layoutType=Key&amp;amp;layoutOri=Portrait&amp;amp;project=SANSONS&amp;amp;type=PNGC&quot; rel=&quot;nofollow&quot;&gt;http://starchart.polaris.net/chart.php?constel=ori&amp;amp;layoutType=Key&amp;amp;layoutOri=Portrait&amp;amp;project=SANSONS&amp;amp;type=PNGC&lt;/a&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility would be Aldebaran, it is located in a &quot;V&quot; shape of stars that make up the constellation Taurus.&#xA;&lt;a href=&quot;http://starchart.polaris.net/chart.php?constel=tau&amp;amp;layoutType=Key&amp;amp;layoutOri=Portrait&amp;amp;project=SANSONS&amp;amp;type=PNGC&quot; rel=&quot;nofollow&quot;&gt;http://starchart.polaris.net/chart.php?constel=tau&amp;amp;layoutType=Key&amp;amp;layoutOri=Portrait&amp;amp;project=SANSONS&amp;amp;type=PNGC&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find out I would take a look at some of the online star charts that are available for your time and location.  That will give you a good idea of what you are seeing.  Here is a list of some online: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.skyandtelescope.com/resources/internet/3305011.html?page=1&amp;amp;c=y&quot; rel=&quot;nofollow&quot;&gt;http://www.skyandtelescope.com/resources/internet/3305011.html?page=1&amp;amp;c=y&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="281" LastActivityDate="2013-10-19T00:27:43.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="705" PostTypeId="2" ParentId="623" CreationDate="2013-10-19T01:30:36.780" Score="13" Body="&lt;p&gt;A spike in Carbon-14 content was recorded from tree rings in Japan, &lt;em&gt;possibly&lt;/em&gt; due to a solar flare 20 times more energetic than the 1859 Carrington Event (the largest &lt;em&gt;definitively recorded&lt;/em&gt; solar storm). This event has been dated to 774 AD. I think this could be the most energetic (tentatively) recorded solar flare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a solar storm would be &lt;a href=&quot;http://news.nationalgeographic.com/news/2011/03/110302-solar-flares-sun-storms-earth-danger-carrington-event-science/&quot;&gt;far more devastating today&lt;/a&gt; than it was back then, in the sense that we rely so much on technology that the EM interference would cost trillions of dollars in damage to power grids and communications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A flare would need to be disastrously large, I think, to give us on Earth radiation poisoning. Certainly, animals that use Earth's magnetic field for navigation and would get rather confused. But our atmosphere and magnetic field are adequate protection when we're on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arstechnica.com/science/2012/12/mysterious-radiation-event-of-774-might-be-a-little-less-mysterious/&quot;&gt;Here is a good write-up&lt;/a&gt; of the situation, with a couple of Nature papers if you have access.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;If you're going to delve into considering the effects of solar storms on human health (i.e. the effects of magnetic field perturbations), put your tinfoil hat on first. There's a lot of pseudoscience and little substantive research.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="276" LastEditDate="2013-10-21T00:17:40.510" LastActivityDate="2013-10-21T00:17:40.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="706" PostTypeId="1" AcceptedAnswerId="707" CreationDate="2013-10-19T02:28:52.150" Score="11" ViewCount="506" Body="&lt;p&gt;Why didn't the center of the earth cool after millions or billions of years? What keeps it hot?&lt;/p&gt;&#xA;" OwnerUserId="282" LastEditorUserId="6" LastEditDate="2013-11-15T18:11:18.877" LastActivityDate="2013-12-12T14:44:44.990" Title="Why is the Earth's center still hot after millions of years?" Tags="&lt;earth&gt;&lt;geology&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ClosedDate="2013-12-12T14:44:50.757" ContentLicense="CC BY-SA 3.0" />
  <row Id="707" PostTypeId="2" ParentId="706" CreationDate="2013-10-19T04:05:27.320" Score="7" Body="&lt;p&gt;There is a nice article from &lt;a href=&quot;http://www.scientificamerican.com/article.cfm?id=why-is-the-earths-core-so&quot;&gt;Scientific American&lt;/a&gt;, but the main point is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are three main sources of heat in the deep earth: (1) heat from when the planet formed and accreted, which has not yet been lost; (2) frictional heating, caused by denser core material sinking to the center of the planet; and (3) heat from the decay of radioactive elements.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other planets may have more or less radioactive material and this will make a difference over time as to whether they continue to have a molten core, or become volcanically and tectonically inactive.&lt;/p&gt;&#xA;" OwnerUserId="38" LastActivityDate="2013-10-19T04:05:27.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="708" PostTypeId="2" ParentId="475" CreationDate="2013-10-19T17:13:05.177" Score="6" Body="&lt;p&gt;I don't remember reading any paper title specifically about the hot spots you refer to, so I guess that they are not as &quot;strange&quot; as the cold spot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the temperature fluctuations are well described by a &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_function&quot; rel=&quot;noreferrer&quot;&gt;Gaussian distribution&lt;/a&gt; with mean 2.72548 and variance 0.00057 &lt;a href=&quot;http://arxiv.org/abs/0911.1955&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;, the probability of having the cold spot of the observed size and temperature is very low. According to &lt;a href=&quot;http://arxiv.org/pdf/1303.5083v1.pdf&quot; rel=&quot;noreferrer&quot;&gt;a recent Planck paper&lt;/a&gt; this probability is of order of 1% or smaller. The probability of the hot spots is larger (1.5 to 5%), making them more &quot;normal&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This probably explains while the cold spot got more attention that the hot ones. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this a side-effect of our observation methods or does it represent some actual feature of the early history of the Universe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is tricky. A few scenarios:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The CMB map in the question shows fluctuations of order 10^-4 K. But extracting this signal from observations is very complicated by the fact that on top of the CMB radiation is &quot;polluted&quot; by foreground objects (mostly our own Galaxy, but also extragalactic point and extended sources) that create spurious fluctuations orders of magnitude larger than the CMB ones. &lt;a href=&quot;http://www.youtube.com/watch?v=5QvQp5j1hjs&quot; rel=&quot;noreferrer&quot;&gt;This video&lt;/a&gt; is great to have an idea of this. If any of the above foreground components is not well understood or some is not accounted for, we could be left with some residuals that we wrongly interpret as CMB anomalies.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The physics we know is the correct one and the cold spot is simply there. But we physicists don't like that in the Universe there are special places&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The cold spot is a signal of new/unknown physics or that our standard model is not correct. For instance could be that the assumption that the CMB is Gaussian could be just wrong and we have to re-think our theories about the early universe&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="86" LastActivityDate="2013-10-19T17:13:05.177" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="709" PostTypeId="2" ParentId="703" CreationDate="2013-10-19T18:26:46.830" Score="2" Body="&lt;p&gt;A good way to get to know what is visible at your location is to use a software that maps the sky. One such free software is &lt;a href=&quot;http://www.stellarium.org/&quot; rel=&quot;nofollow&quot;&gt;Stellarium&lt;/a&gt; available for all platforms. You can easily either find your place (or a nearby place) in its built in database of locations or alternatively provide the longitude/latitude for your location. With such software you will be able to see the sky map in real time and hence figure out what everything is. &lt;/p&gt;&#xA;" OwnerUserId="47" LastActivityDate="2013-10-19T18:26:46.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="710" PostTypeId="2" ParentId="706" CreationDate="2013-10-19T21:04:07.793" Score="4" Body="&lt;p&gt;From a simplistic point of view the rate of cooling depends on the ratio of surface area to volume.  It is complicated by the rate of heat transfer of various layers, and internal sources of &quot;new&quot; heat in addition to the heat generated when the planet formed. The main way heat is lost is radiation to space.  The atmosphere itself acts a a kind of &quot;blanket&quot; to sow the radiation loss(de Pater and Lissauer, Planetary Sciences)&lt;/p&gt;&#xA;" OwnerUserId="284" LastActivityDate="2013-10-19T21:04:07.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="711" PostTypeId="2" ParentId="220" CreationDate="2013-10-19T21:20:18.083" Score="13" Body="&lt;p&gt;Eclipses are common, but solar eclipses like the ones our moon causes would be unusual since it takes a combination of factors, the angular size of the moon and sun are nearly the same from the viewpoint of Earth so that the moon (depending on where it is in its slightly ellipitcal path around the Earth) can cover the sun nearly exactly.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;NASA has a nice picture of an eclipse as seen from the Mars Curiosity rover showing an eclipse by a moon with an angular size less than that of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nasa.gov/mission_pages/msl/multimedia/pia17356.html?cid=soc:103#.UmL2mfmsiM4&quot;&gt;Phobos eclipsing the Sun&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="284" LastActivityDate="2013-10-19T21:20:18.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="712" PostTypeId="1" AcceptedAnswerId="713" CreationDate="2013-10-19T21:37:06.343" Score="17" ViewCount="243" Body="&lt;p&gt;If the atmosphere on Mars was once much thicker, how was it likely lost?  Was it due to interaction with the solar wind, the small size of the planet, both, or something else, and approximately how long did it take to reach its current thickness?&lt;/p&gt;&#xA;" OwnerUserId="284" LastActivityDate="2013-10-28T12:41:45.033" Title="Loss of atmosphere on Mars" Tags="&lt;planetary-atmosphere&gt;&lt;mars&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="713" PostTypeId="2" ParentId="712" CreationDate="2013-10-20T01:35:32.220" Score="10" Body="&lt;p&gt;The loss of the Martian atmosphere can be mostly attributed to its mass. The reason why Earth still has an atmosphere made of lighter elements is because with larger mass comes larger &lt;a href=&quot;http://en.wikipedia.org/wiki/Escape_velocity&quot; rel=&quot;nofollow noreferrer&quot;&gt;escape velocity&lt;/a&gt;, which is the speed at which an atom's kinetic energy overcomes the gravitational potential energy of its planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distribution of speeds of most gasses can be described by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_distribution&quot; rel=&quot;nofollow noreferrer&quot;&gt;Maxwell–Boltzmann distribution&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7LPXz.png&quot; alt=&quot;Maxwell-Boltzmann&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This curve represents the probability of finding a particle with a given speed at a given temperature. Holding temperature constant, the above chart illustrates that lighter molecules will have higher probabilities of being found with higher speeds. &#xA;$$P(\epsilon) = \frac{2}{\sqrt{\pi}k_{B}T} \left( \frac{\epsilon}{k_{B}T}  \right)^{1/2} \exp{-\left(\frac{\epsilon}{k_{B}T}\right)} $$&#xA;where $\epsilon$ is the energy of the particle, $\frac{p^{2}}{2m}$ in the absence of a gravitational field (see &lt;a href=&quot;https://www.google.com/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=2&amp;amp;cad=rja&amp;amp;ved=0CC8QFjAB&amp;amp;url=http://highenergy.phys.ttu.edu/~slee/2402/2010_Fall/F10_2402_Lecture23.pdf&amp;amp;ei=bVluUob8E-ih4APN9IC4Cg&amp;amp;usg=AFQjCNFIhCzaBMVzBeTEHPu56JLMz_GZfw&quot; rel=&quot;nofollow noreferrer&quot;&gt;this lecture&lt;/a&gt; for information regarding how to include the gravitational term).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrated over time, the lighter gasses tend to exceed the escape velocity more often than their heavier counterparts. This is why the larger planets like Jupiter and Saturn still have atmospheres dominated by hydrogen and helium.   &lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditDate="2013-10-28T12:41:45.033" LastActivityDate="2013-10-28T12:41:45.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="714" PostTypeId="2" ParentId="700" CreationDate="2013-10-20T11:19:47.007" Score="15" Body="&lt;p&gt;As Francesco Montesano points out, using the wrong mass leads to the wrong answer.  Also, using the density here seems a complicated way to get to the answer; you could compute the Schwarzschild radius for the NS, and see whether it's smaller than its actual radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the density scales as &amp;rho; ~ M/R^3 and the Schwarzschild radius as R&lt;sub&gt;s&lt;/sub&gt;~M, the density of BHs scales as &amp;rho;~1/R^2; more massive BHs are less dense and simply testing whether a NS is denser than a BH alone is not sufficient - they must be of the same mass, which means that you are in fact comparing radii.&lt;/p&gt;&#xA;" OwnerUserId="286" LastActivityDate="2013-10-20T11:19:47.007" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="715" PostTypeId="2" ParentId="240" CreationDate="2013-10-20T12:00:52.737" Score="8" Body="&lt;p&gt;The monthly variations are caused by the change in declination of the Moon as it travels (roughly) along the ecliptic.  This is caused by the tilt of the Earth's rotation axis, which results in an angle of about 23&amp;deg; between the equator (declination = 0&amp;deg;) and the ecliptic.  Around the constellations of Taurus and Gemini the ecliptic lies at much higher declination than near Saggitarius or Ophiuchus.  The same effect causes the change in rise and set times of the Sun during a year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you're in the northern hemisphere, a higher declination means that the transit altitude of the Moon or Sun is higher.  In fact, their whole daily (apparent - caused by the Earth's rotation) paths lie closer to the North star.  Because of that, the points where a daily path intersects with your local horizon lie further north, which you see as a change in rise/set azimuth of the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this is true even without the 5&amp;deg; inclination of the Moon's orbit w.r.t. the ecliptic.  That inclination only mitigates the effect; the Moons declination does not exactly vary between -23 and +23&amp;deg; every month, but these extreme values themselves can lie between +/-18&amp;deg; or +/-28&amp;deg;, depending on whether the inclination effect increases (23+5&amp;deg;) or decreases (23-5&amp;deg;) the effect of the motion along the ecliptic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, the rise/set azimuths do not correlate to lunar phase, as do the rise/set times.  Both phases and rise/set times are defined based on the Sun's position; we roughly define &quot;noon&quot; as the moment when the Sun is transiting in the south, and call the Moon &quot;New&quot; when it's in the same direction as the Sun, so there's no wonder that the New Moon transits around noon.  This is not true for the effects described above, and hence the corresponding periods are different.  The mean lunar period between twice the same equinox point (e.g. Aries) is called the Tropical month and lasts ~27.3d (this is roughly the same as the siderial month, but influenced by precession).  The effect of the inclination of the Moon's orbit has a mean period of one Draconic month, with a period of ~27.2d.  Instead, the Synodic month, the mean period between two cases of New Moon, lasts 29.5d, about 2.2d longer.  Hence, the declination and rise/set azimuths of the Moon do not correlate with the phases. &lt;/p&gt;&#xA;" OwnerUserId="286" LastActivityDate="2013-10-20T12:00:52.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="716" PostTypeId="1" CreationDate="2013-10-20T18:33:24.970" Score="18" ViewCount="375" Body="&lt;p&gt;In other terms, did galaxies grow around black holes at their center? &lt;/p&gt;&#xA;" OwnerUserId="159" LastEditorUserId="37191" LastEditDate="2021-05-11T13:10:50.300" LastActivityDate="2021-05-11T16:57:23.040" Title="Which came first: black holes or galaxies?" Tags="&lt;black-hole&gt;&lt;galaxy&gt;&lt;cosmology&gt;&lt;galaxy-center&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="717" PostTypeId="1" AcceptedAnswerId="956" CreationDate="2013-10-21T11:46:34.350" Score="11" ViewCount="255" Body="&lt;p&gt;I've learned that the time period for a rotating neutron star, i.e., a pulsar is very accurate and can be used as an interstellar clock. But according to Walter Lewin's 8.01 Lec 19 on classical mechanics, he says that the rotational time period of neutron star in the Crab Nebula loses up to 364 ns &lt;em&gt;everyday&lt;/em&gt; in the form of neutron star emissions (this comes from rotational kinetic energy of the neutron star). So, how are pulsars used as precise clocks?&lt;/p&gt;&#xA;" OwnerUserId="104" LastActivityDate="2013-11-19T09:15:50.720" Title="Crab Nebula Time Period" Tags="&lt;pulsar&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="718" PostTypeId="1" AcceptedAnswerId="719" CreationDate="2013-10-21T13:18:01.660" Score="10" ViewCount="5902" Body="&lt;p&gt;Since all the galaxies are moving further away from each other at a faster and faster rate, eventually we won't be able to see stars from other galaxies. How long until we can't see stars from other galaxies?&lt;/p&gt;&#xA;" OwnerUserId="258" LastActivityDate="2015-08-14T08:51:00.750" Title="How long until we cannot see any stars from other galaxies?" Tags="&lt;galaxy&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="719" PostTypeId="2" ParentId="718" CreationDate="2013-10-21T14:43:03.737" Score="8" Body="&lt;p&gt;First of all, we can't see stars from other galaxies (with a few exceptions, Cepheid variable stars for example are regularly used to determine distances to nearby galaxies). As it currently stands, we can only see stars from our own Milky Way (in which I'm including the large and small Magellanic Clouds). Type Ia supernovae aside, it might be possible to see Cepheid variables in the Andromeda galaxy, but anything further is almost certainly not possible.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relevant number when it comes to the distance one could possibly see in the universe is the comoving distance to the cosmological horizon. This horizon defines the boundary between what can be seen and what cannot be seen, simply because the universe is not old enough for particles to have traveled that far (yes, even at the speed of light; in fact, if we're talking about what we can &lt;em&gt;observe&lt;/em&gt;, the photon is the particle we care about).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of the comoving distance is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\chi = \frac{c}{H_{0}} \int_{z=0}^{z=z_{Hor}} \frac{dz^{'}}{E(z^{'})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the function $$ E(z^{'}) $$ contains your choice of cosmology (the following is for flat universes only):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E(z^{'}) = \sqrt{ \Omega_{\gamma} (1+z)^{4} + \Omega_{m} (1+z)^{3} + \Omega_{\Lambda} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This distance can be extrapolated out into the future (given your choice of cosmology), and ultimately will tell you what the distance to the &lt;em&gt;cosmological event horizon&lt;/em&gt; will asymptotically reach. This is the distance at which no object beyond could ever come into causal contact with you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another &lt;a href=&quot;http://en.wikipedia.org/wiki/Particle_horizon&quot; rel=&quot;nofollow&quot;&gt;simpler&lt;/a&gt; way to calculate the particle horizon is to calculate the conformal time that has passed from the beginning of the universe, and multiply it by the speed of light $c$, where the conformal time is defined in the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta = \int_{0}^{t} \frac{dt^{'}}{a(t^{'})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a(t^{'})$ is the scale factor of the universe, and its relationship to time depends specifically on your choice of cosmology. Again, if the ultimate value of this horizon is desired, you would need to integrate this to infinity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SUMMARY&lt;/strong&gt;: I'm having a hard time finding the exact number of the comoving size of the cosmological event horizon, though I'll keep searching. If I can't find it I suppose I'll have to do the calculation when I have some free time. Almost certainly it is larger than the size of the local group of galaxies, making our night sky largely safe from the fate of the universe. Anything further than this horizon, however, will monotonically become both fainter and redder. I also think that it does asymptotically reach some value, rather than eventually increasing infinitely or decreasing passed some point in time. I'll have to get back to you on that.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/astroph/9905116&quot; rel=&quot;nofollow&quot;&gt;Here's&lt;/a&gt; a nice review of distance measurements in cosmology. &lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="609" LastEditDate="2015-08-14T08:51:00.750" LastActivityDate="2015-08-14T08:51:00.750" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="720" PostTypeId="1" AcceptedAnswerId="803" CreationDate="2013-10-21T14:46:01.750" Score="6" ViewCount="69" Body="&lt;p&gt;I would like to see the orionids sometime this week (October 20-26, 2013). What is the best way for me to observe them by naked eye?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, would I be benefitted by trying to look at them with a telescope or or are they too fast and spontaneous for most amateurs to capture?&lt;/p&gt;&#xA;" OwnerUserId="6" LastActivityDate="2013-10-30T19:41:35.620" Title="How can I observe the Orionid Meteor Shower?" Tags="&lt;amateur-observing&gt;&lt;meteor&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="721" PostTypeId="1" AcceptedAnswerId="723" CreationDate="2013-10-21T14:57:48.867" Score="13" ViewCount="511" Body="&lt;p&gt;What sources of turbulence exist within the &lt;a href=&quot;http://en.wikipedia.org/wiki/Interstellar_medium&quot;&gt;Interstellar medium (ISM)&lt;/a&gt;? Which ones are physically the most important for newly forming stellar systems?&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2013-10-22T10:27:37.747" Title="Sources of Turbulence in the ISM" Tags="&lt;star-systems&gt;&lt;interstellar-medium&gt;&lt;turbulence&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="722" PostTypeId="1" AcceptedAnswerId="744" CreationDate="2013-10-21T15:21:02.077" Score="30" ViewCount="6198" Body="&lt;p&gt;I assume that the A ring is the densest, but I might be wrong. Nevertheless, I failed to find any explicit information over the net about the minimum and maximum of densities for the different rings. As a less practical question: what would one see upon entering one of the denser rings? Would it be obvious that it is a dense but thin (few metres - 1 km) cloud of particles, continuously bombarding the spacecraft irregardless whether it orbits along within the ring, or one would not see more than just a distant haze (if lit by the Sun) that spans over the horizon?&lt;/p&gt;&#xA;" OwnerUserId="289" LastActivityDate="2014-11-04T13:52:37.040" Title="How dense are Saturn's rings?" Tags="&lt;saturn&gt;&lt;planetary-ring&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="723" PostTypeId="2" ParentId="721" CreationDate="2013-10-21T16:40:14.650" Score="9" Body="&lt;h2&gt;Turbulence sources:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There are numerous sources of turbulence in the interstellar medium, at all scales:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;at large scales, there is the &lt;strong&gt;shear from galactic rotation&lt;/strong&gt;. One way to sustain turbulence and to couple large and small scales would be the &lt;strong&gt;magnetorotational instability&lt;/strong&gt; (MRI).&lt;/li&gt;&#xA;&lt;li&gt;at large scales, &lt;strong&gt;gravitational instabilities&lt;/strong&gt; can also play a significant role, through spiral structures.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;outflows and jets&lt;/strong&gt; from forming stars play an important role, releasing a lot of energy in the ISM.&lt;/li&gt;&#xA;&lt;li&gt;in star forming regions, &lt;strong&gt;massive stars&lt;/strong&gt; are also important. Radiation and stellar winds from massive stars are an important input of energy in the ISM. And eventually, the most massive ones will blow up in supernovæ, releasing even more energy.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore, one could then regards separatly three processes related to massive stars:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;stellar winds&lt;/li&gt;&#xA;&lt;li&gt;ionizing radiation&lt;/li&gt;&#xA;&lt;li&gt;supernova explosion&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Importance for star formation:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;They are all relevant to star formation, one way or another. One key property of turbulence is to cascade from large to small scales; therefore, even if you inject turbulence at large scales (galactic scale) you'll get turbulent motions down to the scale of a molecular cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nice illustration of the tubulent cascade is the Larson's relation (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1981MNRAS.194..809L&quot; rel=&quot;noreferrer&quot;&gt;Larson 1981&lt;/a&gt;):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/BcSla.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Larson's relation shows the evolution of the velocity dispersion with the size of the structure you're looking at. Velocity dispersion is an indicator of turbulence. Indeed, these dispersions are non-thermal: knowing the typical temperature of the MIS (about 10 K), one can estimate the thermal velocity of, for example, the CO molecule ($v_th = \sqrt{2kT/\mu m_H}$, with $k$ the Boltzman constant, $T$ the temperature, $\mu$ the mean molecular wright and $m_H$ the hydrodgen atom mass) that is about 0.07 km s$^{-1}$. Measured velocity dispersions are of the order of 1 to 10 km s$^{-1}$, and these is interpreted as a turbulence signature (and estimate).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Details:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Energy rates:&lt;/strong&gt; values are given (roughly) for the Milky Way&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;MRI&lt;/strong&gt;: $ \small \dot e = 3 \times 10^{-29} {\rm erg\ cm}^{-3}\ {\rm s}^{-1} $;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Gravitational instabilities&lt;/strong&gt;: $\small \dot e = 4 \times 10^{-29} {\rm erg\ cm}^{-3}\ {\rm s}^{-1}$;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Outflows&lt;/strong&gt;: $\small \dot e = 2 \times 10^{-28} {\rm erg\ cm}^{-3}\ {\rm s}^{-1}$;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Ionizing radiation&lt;/strong&gt;: $\small \dot e = 5 \times 10^{-29} {\rm erg\ cm}^{-3}\ {\rm s}^{-1}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Supernovæ explosions&lt;/strong&gt;: $\small \dot e = 3 \times 10^{-26} {\rm erg\ cm}^{-3}\ {\rm s}^{-1}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Stellar winds&lt;/strong&gt;: it strongly depends on the type of star: it varies as the power of -6 of the star's luminosity. It therefore ranges from an energy comparable to a supernova explosion (or even more for Wolf-Rayet stars) to almost nothing.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sources:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2004RvMP...76..125M&quot; rel=&quot;noreferrer&quot;&gt;Mac Low &amp;amp; Klessen 2004&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2013-10-22T10:27:37.747" LastActivityDate="2013-10-22T10:27:37.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="724" PostTypeId="1" CreationDate="2013-10-21T17:23:16.240" Score="13" ViewCount="1510" Body="&lt;p&gt;Is there any handy module to detect emission lines in a spectrum like one we get from the Sloan Digital Sky Survey (SDSS)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see there are many emission lines like Ha,OI in the spectrum below. In fact the spectrum is a two column data set, wavelength and flux. It is a combination of lots of scatter points. The typical interval between two points nearyby is 1.5 angstroms. I need to find all the Gaussian bumps and their line center.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,first I need to find the the bumps and then I need to fit them to get their centers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/u0gKm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="2365" LastEditDate="2016-02-14T20:45:26.117" LastActivityDate="2017-06-20T04:49:06.127" Title="How to detect emission lines in optical spectra?" Tags="&lt;spectra&gt;&lt;sky-survey&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="725" PostTypeId="2" ParentId="716" CreationDate="2013-10-21T17:59:19.230" Score="8" Body="&lt;p&gt;&lt;sub&gt;Note: I am not an astrophysicist. If someone with the relevant expertise finds my analysis of the source paper to be invalid, I would appreciate your help with correcting my answer here.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;This is a somewhat complicated question to answer. Black holes may precede galaxies in the form of primordial black holes which according to some models arose in the early Universe from quantum fluctuations. Primordial black holes are currently hypothetical. Supermassive black holes (that you find at the center of galaxies) may not come from these, however. There are a few possible ways that supermassive black holes may have formed:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Molecular cloud / dwarf galaxies collapsing to form supermassive stars (SMS) which later further collapsed into &lt;a href=&quot;https://en.wikipedia.org/wiki/Supermassive_black_hole&quot; rel=&quot;nofollow noreferrer&quot;&gt;supermassive black holes&lt;/a&gt; (SMBH)&lt;/li&gt;&#xA;&lt;li&gt;Dwarf galaxies forming massive stars which collapse into massive black holes, which later coalesce into SMBHs&lt;/li&gt;&#xA;&lt;li&gt;Primordial black holes accreting sufficient matter to become massive black holes, which later coalesce into SMBHs&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Depending on how strictly you define &amp;quot;galaxy&amp;quot; it does seem most likely that the supermassive black holes formed inside early galaxies, but a type of black hole (primordial black holes) may have preceded galaxies.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://iopscience.iop.org/0004-637X/596/1/34/pdf/0004-637X_596_1_34.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Formation of the First Supermassive Black Holes - Bromm and Loeb&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE 10/21/2013&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;p&gt;This just in, recent observations have ruled out the possibility that SMBHs gain mass only through merging with other black holes. For more, see this &lt;a href=&quot;http://www.astronomy.com/news/2013/10/gravitational-waves-know-how-black-holes-grow&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomy.com article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorUserId="6" LastEditDate="2021-05-11T16:57:23.040" LastActivityDate="2021-05-11T16:57:23.040" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="726" PostTypeId="2" ParentId="460" CreationDate="2013-10-21T22:21:02.890" Score="3" Body="&lt;p&gt;I'm not speaking from an informed position here, but two things come to mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First is that a rogue planet is likely to be traveling very fast, relative to our solar system. An object from within our solar system (a Kuiper/Oort object) is going to have an orbital velocity. Something that isn't a part of the system at all could be traveling much, much faster. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, if a rogue planet somehow slowly drifted into our system at almost no speed, and then got pulled into a sort of orbital speed by the sun, the direction of its orbit might give it away. For example, it's orbital inclination could be way off--even perpendicular to the normal orbital plane. It also might be orbiting in the wrong direction--that is--clockwise, instead of ante-clockwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, there's no reason that some disturbed Kuiper belt object might not have an eccentric orbital inclination or even an anti-clockwise orbit. But the larger the object, the less likely this is, because it would need a larger and larger disturbance to throw it so out of whack.&lt;/p&gt;&#xA;" OwnerUserId="158" LastActivityDate="2013-10-21T22:21:02.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="727" PostTypeId="2" ParentId="724" CreationDate="2013-10-21T22:33:06.297" Score="5" Body="&lt;p&gt;Yes, there is are two python modules called &lt;a href=&quot;http://www.astropy.org/&quot; rel=&quot;noreferrer&quot;&gt;astropy&lt;/a&gt;, and &lt;a href=&quot;http://pythonhosted.org/Astropysics/&quot; rel=&quot;noreferrer&quot;&gt;astropysics&lt;/a&gt; that both claim to have spectral analysis tools. As a reference, here is a link for &lt;a href=&quot;http://www.astropython.org/resources&quot; rel=&quot;noreferrer&quot;&gt;resources&lt;/a&gt; for astronomers for the python programming language.  &lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2013-10-21T22:33:06.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="728" PostTypeId="2" ParentId="724" CreationDate="2013-10-22T09:07:13.330" Score="7" Body="&lt;p&gt;The programs I am aware of that you could use do require some programming expertise to operate. I would recommend looking at either &lt;a href=&quot;http://iraf.noao.edu&quot;&gt;IRAF&lt;/a&gt; (or &lt;a href=&quot;http://www.stsci.edu/institute/software_hardware/pyraf&quot;&gt;PyRAF&lt;/a&gt;, which uses Python to interface with IRAF), or &lt;a href=&quot;http://www.as.utexas.edu/~chris/spectre.html&quot;&gt;SPECTRE&lt;/a&gt;. Unfortunately, I'm not aware of a &quot;black box&quot; solution where you can just press a button and go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SPECTRE is written in FORTRAN77 and is quite easy to use once you get the hang of it, though it has a somewhat limited feature set and the documentation is brief. It will help you to manually fit Gaussians to spectral lines, giving their central wavelengths and equivalent widths. A caveat for SPECTRE is that it requires the plotting package &lt;a href=&quot;http://www.astro.princeton.edu/~rhl/sm/&quot;&gt;SuperMongo&lt;/a&gt;, which is not free software. If you are affiliated with a university, most Physics and Astronomy departments should have a license.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IRAF is a quite fully-featured, free, cross-platform application purposefully built for astronomical image analysis. It can help you measure the spectral lines, and can analyse the spectrum to determine the object's radial velocity.&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2013-10-22T09:07:13.330" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="731" PostTypeId="2" ParentId="724" CreationDate="2013-10-22T13:35:42.963" Score="5" Body="&lt;p&gt;There are plenty of software and tools available to do what you want:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://iraf.noao.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;IRAF&lt;/a&gt;, by NOAO;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.eso.org/sci/software/esomidas/&quot; rel=&quot;nofollow noreferrer&quot;&gt;MIDAS&lt;/a&gt; which is basically very similar to IRAF but developed by ESO;&lt;/li&gt;&#xA;&lt;li&gt;in Python, either &lt;a href=&quot;http://www.astropy.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;astropy&lt;/a&gt; or &lt;a href=&quot;http://www.stsci.edu/institute/software_hardware/pyraf&quot; rel=&quot;nofollow noreferrer&quot;&gt;pyRAF&lt;/a&gt; (to use IRAF with a Python interface).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would go with some Python tools (Python is more versatile than IRAF or MIDAS that are much more &quot;single-purpose&quot; oriented), to benefit from the other features offered by Python.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to plot and process your spectrum is just the first part of the work, what you would probably want to do in the end is to use spectral synthesis, to analyze your spectrum. Spectral synthesis will allow you:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;to identify efficiently the chemical composition of the celestrial object you study;&lt;/li&gt;&#xA;&lt;li&gt;to determine some properties of the celestial object you study.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Among the existing tools for spectral synthesis (for stellar spectra for example), there are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.pages-perso-bertrand-plez.univ-montp2.fr/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Turbospectrum&lt;/a&gt; by Bertrand Plez;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.astro.mat.uc.pt/BaSeL2/&quot; rel=&quot;nofollow noreferrer&quot;&gt;BaSeL&lt;/a&gt; by Thibault Lejeune&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="88" LastEditorUserId="-1" LastEditDate="2017-06-20T04:49:06.127" LastActivityDate="2017-06-20T04:49:06.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="732" PostTypeId="5" CreationDate="2013-10-22T15:29:42.150" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-22T15:29:42.150" LastActivityDate="2013-10-22T15:29:42.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="733" PostTypeId="4" CreationDate="2013-10-22T15:29:42.150" Score="0" Body="Questions about flow regime in astrophysical gas characterized by chaotic properties." OwnerUserId="88" LastEditorUserId="88" LastEditDate="2013-10-22T15:37:36.123" LastActivityDate="2013-10-22T15:37:36.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="734" PostTypeId="5" CreationDate="2013-10-22T15:33:11.450" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-22T15:33:11.450" LastActivityDate="2013-10-22T15:33:11.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="735" PostTypeId="4" CreationDate="2013-10-22T15:33:11.450" Score="0" Body="Questions about the apparent displacement of a celestial body with respect to its background when observed along two different lines of sight." OwnerUserId="88" LastEditorUserId="6" LastEditDate="2013-10-22T15:37:27.420" LastActivityDate="2013-10-22T15:37:27.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="736" PostTypeId="5" CreationDate="2013-10-22T15:49:05.640" Score="0" Body="&lt;p&gt;A neutron star is a type of stellar remnant resulting from the gravitational collapse of a massive star during a supernova event. They are mainly composed of neutrons because, as the pressure increases during the collapse of the star, electrons get closer to nuclei and eventually turn protons into neutrons by a beta reaction.&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2013-10-22T16:10:52.850" LastActivityDate="2013-10-22T16:10:52.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="737" PostTypeId="4" CreationDate="2013-10-22T15:49:05.640" Score="0" Body="Questions regarding a degenerate star that is mainly composed of neutrons." OwnerUserId="88" LastEditorUserId="2153" LastEditDate="2015-05-25T14:31:22.903" LastActivityDate="2015-05-25T14:31:22.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="738" PostTypeId="5" CreationDate="2013-10-22T15:51:11.167" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-22T15:51:11.167" LastActivityDate="2013-10-22T15:51:11.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="739" PostTypeId="4" CreationDate="2013-10-22T15:51:11.167" Score="0" Body="Questions about the increase with time of distance between two distant points in the Universe." OwnerUserId="88" LastEditorUserId="88" LastEditDate="2013-10-22T16:10:58.507" LastActivityDate="2013-10-22T16:10:58.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="740" PostTypeId="5" CreationDate="2013-10-22T15:53:32.603" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-22T15:53:32.603" LastActivityDate="2013-10-22T15:53:32.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="741" PostTypeId="4" CreationDate="2013-10-22T15:53:32.603" Score="0" Body="Questions about the water molecule (composed of two atoms of hydrodgen and one atom of oxygen) and its detection in celestial objects." OwnerUserId="88" LastEditorUserId="6" LastEditDate="2013-10-22T16:11:39.870" LastActivityDate="2013-10-22T16:11:39.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="742" PostTypeId="5" CreationDate="2013-10-22T15:55:22.287" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-22T15:55:22.287" LastActivityDate="2013-10-22T15:55:22.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="743" PostTypeId="4" CreationDate="2013-10-22T15:55:22.287" Score="0" Body="Questions about the dynamical processes leading to the assembling of a celestial object." OwnerUserId="88" LastEditorUserId="6" LastEditDate="2013-10-22T16:11:57.480" LastActivityDate="2013-10-22T16:11:57.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="744" PostTypeId="2" ParentId="722" CreationDate="2013-10-22T16:54:11.917" Score="17" Body="&lt;p&gt;&lt;strong&gt;Elements of answer&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not an easy question, as we lack of data to constrain strongly the density and mass of Saturn's rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a first clue of their density is their &lt;strong&gt;optical depth&lt;/strong&gt; (that is a measure of the transparency of a medium). The densest rings are the one of the &lt;em&gt;main ring system&lt;/em&gt; (rings A, Cassini Division, B and C) plus ring F; they are characterized by optical depth larger than 0.1 (up to about 5 for the densest part of ring B). Rings in the &lt;em&gt;faint ring system&lt;/em&gt; (E and G) have very small optical depth (lower than $10^{-5}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this regard, B is indeed the densest ring of Saturn's rings. &lt;strong&gt;(edit)&lt;/strong&gt; Looking around (&lt;a href=&quot;http://caps.gsfc.nasa.gov/simpson/kingswood/rings/&quot; rel=&quot;noreferrer&quot;&gt;[1]&lt;/a&gt; and &lt;a href=&quot;http://saturn.jpl.nasa.gov/files/questions_answers.pdf&quot; rel=&quot;noreferrer&quot;&gt;[2]&lt;/a&gt;) you can find numbers that help to figure out the density of this ring. Knowing its mass (about $2.8\times10^9$ kg) and its extend (from 1.527 to 1.951 Saturn radii, with a typical thickness of 100 m), you can deduce a mean density of about 0.0167 g.m$^{-3}$ (about $10^5$ times less dense than the terrestrial atmosphere at sea level).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Just for fun:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives you an idea of what means the optical depth. In the densest parts of ring B, optical depth is between 2.5 and 5, so you'll definitely notice you're in the ring. It is probably more complicated than that (with ring particles ranging from dust size to house size), but you get the idea.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/qkFXO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Side note:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Optical depth is a measure of transparency of one medium; it measures how radiation is neither scattered nor absorbed by the medium. It expression is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{I}{I_0} = {\rm e}^{-\tau}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $I_0$ is the source intensity, $I$ the observed intensity, and $\tau$ the optical depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sources:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/0912.3017&quot; rel=&quot;noreferrer&quot;&gt;Charnoz 2009&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2014-11-04T13:52:37.040" LastActivityDate="2014-11-04T13:52:37.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="745" PostTypeId="1" CreationDate="2013-10-23T03:35:35.267" Score="9" ViewCount="1378" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Kepler-10b orbits its parent star once every 0.84 days, which means&#xA;  it is more than 20 times closer to its star than Mercury is to our&#xA;  sun&quot; &lt;a href=&quot;http://www.space.com/10592-smallest-alien-planet-discovered-nasa.html&quot; rel=&quot;noreferrer&quot;&gt;space.com&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Are there any accepted theories that successfully predicted that so many planets would be found so close to their star?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; This is especially a question, since it seems that planets are moving away from their star. &lt;a href=&quot;http://www.newscientist.com/article/dn17228-why-is-the-earth-moving-away-from-the-sun.html#.UmgRDhAwfi0&quot; rel=&quot;noreferrer&quot;&gt;newscientist.com&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="296" LastEditorUserId="296" LastEditDate="2013-10-23T18:22:34.723" LastActivityDate="2013-10-23T23:48:05.430" Title="Why are some planets so close to their star?" Tags="&lt;exoplanet&gt;&lt;orbit&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="746" PostTypeId="2" ParentId="745" CreationDate="2013-10-23T04:26:30.807" Score="6" Body="&lt;p&gt;Because more massive, closely-orbiting planet exert much greater gravitational forces on their host star than smaller, more distant planets, there is a &lt;em&gt;significant&lt;/em&gt; observational bias. We are much more likely to detect these planets (i.e. &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hot_Jupiter&quot; rel=&quot;nofollow noreferrer&quot;&gt;hot Jupiters&lt;/a&gt;&quot;) because their observable effects (Doppler wobble, gravitational lensing, etc) are more significant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The below image (&lt;a href=&quot;http://en.wikipedia.org/wiki/Exoplanet#Orbital_parameters&quot; rel=&quot;nofollow noreferrer&quot;&gt;from here&lt;/a&gt;) shows you that we're still mostly unable to detect planets like those in our own Solar system (the gray circles are our Solar planets). It's just too hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't study planetary formation (yet!), so I can't speak well for the theoretical side of things. All we've managed to collect so far is the very lowest-hanging fruit, so it is highly likely that there are is a &lt;a href=&quot;http://www.space.com/19659-alien-earth-exoplanets-red-dwarfs.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;large number&lt;/a&gt; of Earth-like planets out there too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors of &lt;a href=&quot;http://mnras.oxfordjournals.org/content/430/2/951&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; on the Kepler-10 system (you'll need MNRAS access) suggest planet-planet gravitational scattering or collision-merger events. The &lt;a href=&quot;http://www.scholarpedia.org/article/Planetary_formation_and_migration&quot; rel=&quot;nofollow noreferrer&quot;&gt;Scholarpedia&lt;/a&gt; article provides a bit more brief reading on these mechanisms. &lt;a href=&quot;http://www.eos.ubc.ca/~mjelline/453website/eosc453/E_prints/newfer010/chambers_planetarymigration_AR09.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is a review article&lt;/a&gt; from 2009,  and a more &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0510487&quot; rel=&quot;nofollow noreferrer&quot;&gt;in-depth review&lt;/a&gt; from 2006.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XWp3B.png&quot; alt=&quot;Exoplanets discovered up until 2010-10-03.&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="276" LastEditDate="2013-10-23T23:48:05.430" LastActivityDate="2013-10-23T23:48:05.430" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="747" PostTypeId="1" AcceptedAnswerId="750" CreationDate="2013-10-23T09:14:46.190" Score="12" ViewCount="248" Body="&lt;p&gt;Planetary migration is defined by Lubow and Ida (2010) in their article &lt;a href=&quot;http://arxiv.org/pdf/1004.4137v1.pdf&quot;&gt;Planet Migration&lt;/a&gt; as &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the process by which a planet’s orbital radius changes in time. The&#xA;  main agent for causing gas giant planet migration is the gravitational interaction of the young&#xA;  planet with the gaseous disk from which it forms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Relevant, it is believed by some that the gas giant planets in our solar system experienced orbital migration.  According to the article &lt;a href=&quot;http://www.psrd.hawaii.edu/Aug06/cataclysmDynamics.html&quot;&gt;Wandering Gas Giants and Lunar Bombardment&lt;/a&gt; (Taylor, 2006), as an example,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Outward migration of Saturn might have triggered a dramatic increase in the bombardment rate on the Moon 3.9 billion years ago, an idea testable with lunar samples.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So my question is &lt;strong&gt;is there any evidence (not opinions) to suggest any form of migration occurring with our solar system's gas giants?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2013-10-23T14:35:29.357" Title="Is there any evidence that the Gas Giant planets in our solar system are experiencing orbital migration?" Tags="&lt;gas-giants&gt;&lt;orbital-migration&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="748" PostTypeId="1" AcceptedAnswerId="763" CreationDate="2013-10-23T09:19:16.200" Score="42" ViewCount="42399" Body="&lt;p&gt;We know the spectacular explosions of supernovae, that when heavy enough, form black holes. The explosive emission of both electromagnetic radiation and massive amounts of matter is clearly observable and studied quite thoroughly. If the star was massive enough, the remnant will be a black hole. If it wasn't massive enough, it will be a neutron star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there's another mode of creation of black holes: the neutron star captures enough matter, or two neutron stars collide, and their combined mass creates enough gravity force to cause another collapse - into a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What effects are associated with this? Is there an explosive release of some kind of radiation or particles? Is it observable? What physical processes occur in the neutrons as they are subjected to the critical increase of pressure? What is the mass of the new black hole, comparing to its neutron star of origin?&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2020-06-30T08:22:10.643" Title="How does neutron star collapse into black hole?" Tags="&lt;black-hole&gt;&lt;neutron-star&gt;&lt;astrophysics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="749" PostTypeId="2" ParentId="745" CreationDate="2013-10-23T09:36:26.037" Score="11" Body="&lt;h2&gt;Big picture:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The reason why there are planets so close to their host star is probably &lt;strong&gt;planetary migration&lt;/strong&gt;. These planets are difficult to explain by &lt;em&gt;in situ&lt;/em&gt; formation models (first of all there is not enough material to form that kind of massive objects in these regions). There are different theories to explain planetary migration:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;disk-planet interactions&lt;/strong&gt;: planets form in a protoplanetary disk that still contains lots of gas after planet formation, and that can interact with the formed planets. There are different interactions that can explain migration, mostly depending on the mass of the planet you consider. We distinguish 3 main types of migration processes: &lt;em&gt;Type I&lt;/em&gt; (migration of low-mass embedded planets), &lt;em&gt;Type II&lt;/em&gt; (migration and gap formation by massive planets) and &lt;em&gt;Type III&lt;/em&gt; migration (rapid migration of intermediate mass planet in relatively massive disk).&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;planet-planet interactions&lt;/strong&gt;: gravitational scattering can be also a very efficient process to modify planet orbits.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Kozai mechanism&lt;/strong&gt;: this mechanism happens in a three body problem. One object can be perturbated by an other object orbiting farther out; the object then starts to oscillate around its orbit, and can therefore migrates.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Interaction with planetesimals&lt;/strong&gt; (also refers as the &lt;em&gt;Nice model&lt;/em&gt; for the Solar System): interaction with a remnant planetesimal disk (as the Kuiper disk for the Solar System) can cause migration. Planets can exchange angular momentum with planetesimals in the outer edge of the disk by gravitational interaction, and thus migrating.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Disk-planet interaction:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Some more details about the &lt;em&gt;Type I, II and III&lt;/em&gt; migration mechanisms:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Type I&lt;/strong&gt;: planets are generating density waves propagating in the disk. Due to Keplerian rotation, this density waves become spiral waves, generating a &lt;em&gt;Lindblad torque&lt;/em&gt;. The inner spiral arm is pulling the planet forward, the planet gaining angular momentum and migrating outward, whereas the outer spiral arm is is pulling the planet back and drives inward migration. In most cases, the outer Lindblad resonances take over the inner Lindblad resonances because they lie closer to the planet, and therefore drive an inward migration.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Type II&lt;/strong&gt;: massive planets open gaps in the disk in which they form. The gravitational interactions become stronger and thus the angular momentum transfer to the disk. If angular momentum is deposit locally in the disk, material inside the planet loses angular momentum and material outside gains angular momentum, and therefore it recedes from the planet, opening a gap in the disk. This gap reduces the importance of Lindblad torques, and the planet is then coupled to the viscous evolution of the disk. Since material from the disk is accreted on the star and angular momentum transported away, massive planets are migrating inward.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Type III&lt;/strong&gt;: if the surface density distribution in the co-orbital region of the planet is asymmetric, it can lead to a large torque that can cause the planet to fall inwards on a short timescale.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sources:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2012ARA%26A..50..211K&quot;&gt;Kley &amp;amp; Nelson 2012&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="88" LastActivityDate="2013-10-23T09:36:26.037" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="750" PostTypeId="2" ParentId="747" CreationDate="2013-10-23T14:35:29.357" Score="8" Body="&lt;p&gt;There is no direct evidence of change of orbits of the gaseous planets of the Solar System. But one could wonder about the stability of the Solar System, and if such events could happen again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theory:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The Solar System is a chaotic system&lt;/em&gt;, as most of gravitational systems involving $N$ bodies. The &lt;em&gt;KAM theorems&lt;/em&gt; (for Kolmogorov, Arnold and Moser, the three authors of the theory) show that, for non-degenerate Hamiltonian systems, quasiperiodic trajectories exist, that could ensure the stability of the system. But unfortunatly the KAM theorems cannot be applied to the Solar System, because the planets are too massive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Simulations:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It leaves us with numerical simulations, that show that the Solar System is indeed chaotic, and that &lt;strong&gt;planetary collisions, ejection and migration can happen&lt;/strong&gt;. But if it is true for the telluric planets, &lt;strong&gt;it is not the case for the gaseous planets&lt;/strong&gt;: one reason could be that they are not much perturbated by the inner planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sources:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1994A%26A...287L...9L&quot; rel=&quot;noreferrer&quot;&gt;Laskar (1994)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/1209.5996&quot; rel=&quot;noreferrer&quot;&gt;Laskar (2012)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="88" LastActivityDate="2013-10-23T14:35:29.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="751" PostTypeId="1" CreationDate="2013-10-23T15:13:11.333" Score="15" ViewCount="232" Body="&lt;p&gt;I've asked this &lt;a href=&quot;https://physics.stackexchange.com/questions/81911/how-does-cosmology-constrain-the-number-of-neutrino-species&quot;&gt;question&lt;/a&gt; already on the Physics SE, but I felt it would be worth posting here as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that based upon theories of structure formation cosmologists can constrain the &lt;strong&gt;sum&lt;/strong&gt; of the masses of neutrinos - if neutrinos were too light or too heavy it would simply change the power spectrum of the universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how exactly is the number of neutrino species constrained by cosmological measurements (i.e. - $N_{eff}$ as measured by WMAP or Planck)? The recent results from the Planck Satellite give us $3.3\pm0.3$, which is consistent with three neutrino species. How is this measured?&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2014-01-06T15:24:43.827" Title="How Does Cosmology Constrain the Number of Neutrino Species?" Tags="&lt;cosmology&gt;&lt;neutrinos&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="752" PostTypeId="1" CreationDate="2013-10-23T16:58:09.063" Score="5" ViewCount="289" Body="&lt;p&gt;My question is about furthest object in the observable universe except &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot;&gt;Сosmic microwave background&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2013-10-25T21:34:14.137" Title="What is the furthest object in the observable universe?" Tags="&lt;cosmology&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="753" PostTypeId="1" AcceptedAnswerId="794" CreationDate="2013-10-23T18:11:34.513" Score="11" ViewCount="650" Body="&lt;p&gt;Why does Pluto's Orbit underlap inside of Neptunes Orbit. Clearly Neptunes Orbit is not overlapping on Pluto's. Yet, they fail to maintain the symmetric banding appearance all the other Planets maintain, similar to the quarter mile track you often see around football fields. &lt;/p&gt;&#xA;" OwnerUserId="257" LastActivityDate="2013-10-27T02:33:17.130" Title="Why does Pluto's Orbit underlap Neptune's Orbit" Tags="&lt;orbit&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="754" PostTypeId="2" ParentId="669" CreationDate="2013-10-23T18:31:14.327" Score="39" Body="&lt;p&gt;I think your question is on topic, but @RhysW has linked a very helpful post in understanding why your question is a common misconception about the Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;No Center&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no 'center' to the universe. At any point, a local observer will claim that they are at the center of the universe by the way galaxies move away from them. How can we possibly know this? The universe appears to be both homogeneous (has the same structure everywhere) and isotropic (there is no preferred direction). If these are indeed properties of the universe, then the expansion of the universe must be the same at all other locations (See: The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmological_principle&quot;&gt;cosmological principle&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How the Big Bang and Explosions Differ&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, the Big Bang is different from an explosion in the following ways:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Particles involved in an explosion slow down eventually due to frictional forces. Think of fireworks (&lt;a href=&quot;http://www.youtube.com/watch?v=qn_tkJDFG3s&quot;&gt;http://www.youtube.com/watch?v=qn_tkJDFG3s&lt;/a&gt;). Particles are moving the fastest at the instant of explosion, and slow monotonically with time. The expansion of the early universe does not follow this trend, though sometimes people use the word 'explosion' to describe the enormous volumetric increase (an increase by a factor of $\sim10^{76}$) which occurred between $10^{-36}- 10^{-32}$ seconds after the Big Bang, which is aptly named &lt;a href=&quot;http://en.wikipedia.org/wiki/Inflation_%28cosmology%29&quot;&gt;inflation&lt;/a&gt;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) An explosion implies the existence of space. For an explosion to take place, particles (whether we're talking about matter or light) must have space to explode into. Strictly speaking the inflation of the universe is an expansion of space-time coordinates, and so the word explosion cannot really apply since there was nothing for space-time to explode into.   &lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2013-10-23T18:31:14.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="755" PostTypeId="1" CreationDate="2013-10-23T18:48:29.917" Score="15" ViewCount="3252" Body="&lt;p&gt;A year is measured as the amount of time it takes Earth to orbit the Sun once, a Galactic year is the time it takes our sun to make one full orbit of the center of the Galaxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Galactic years, how old is our Sun?&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastEditorUserId="2531" LastEditDate="2014-12-22T23:49:15.743" LastActivityDate="2016-03-25T11:38:02.350" Title="How old is our Sun in Galactic years?" Tags="&lt;the-sun&gt;&lt;orbit&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="756" PostTypeId="2" ParentId="755" CreationDate="2013-10-23T19:33:54.477" Score="15" Body="&lt;p&gt;According to the blog &lt;a href=&quot;http://scienceblogs.com/startswithabang/2008/02/27/how-old-is-the-sun-in-galactic-years/&quot;&gt;How old is the Sun in Galactic years?&lt;/a&gt; (Siegel, 2008), a Galactic year is about 223 million Earth years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The age of the Sun, according to the Stanford Solar Centre's page &lt;a href=&quot;http://solar-center.stanford.edu/FAQ/Qage.html&quot;&gt;How old is the Sun?&lt;/a&gt; is approximately 4.57 billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would put the sun at &lt;strong&gt;approximately 20.5 galactic years old&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorUserId="38" LastEditDate="2013-10-23T19:36:34.317" LastActivityDate="2013-10-23T19:36:34.317" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="757" PostTypeId="1" CreationDate="2013-10-23T20:23:33.997" Score="11" ViewCount="3062" Body="&lt;p&gt;The gravitational pull of the moon is enough to create tidal forces of large bodies of liquid, i.e the sea. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was having a conversation the other day about how to terraform Mars, and someone suggested that if Mars was to have a moon artificially placed there then its gravitational effects would help keep a liquid core flowing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the moon therefore also have an effect on other large bodies of liquid on Earth? Does the moons gravitational pull affect in any way the flow of Earths liquid mantle?&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastEditorUserId="58" LastEditDate="2013-10-23T23:28:47.580" LastActivityDate="2017-11-21T20:58:10.630" Title="How much of an effect does the moon have on Earth's liquid mantle?" Tags="&lt;the-moon&gt;&lt;earth&gt;&lt;tidal-forces&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="758" PostTypeId="1" AcceptedAnswerId="759" CreationDate="2013-10-24T01:49:26.860" Score="0" ViewCount="173" Body="&lt;p&gt;Since Time is such a valuable variable in the calculation of Astronomic movement why does the Astronomy Community still rely on the the 6th Century Anno Domini system developed by Dionysius Exiguus to establish Time on Earth.&#xA;Obviously it is not 2013 on Earth, wouldn't it be more technically accurate to determine the actual time on Earth or get as close as possible to it than to use a time that is clearly wrong and developed by someone 15 centuries ago?&lt;/p&gt;&#xA;" OwnerUserId="257" LastActivityDate="2013-10-25T18:41:46.377" Title="Why does Astronomy still use the Anno Domini system for Time Synchronization" Tags="&lt;space-time&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="759" PostTypeId="2" ParentId="758" CreationDate="2013-10-24T02:18:25.083" Score="8" Body="&lt;p&gt;What makes you think that it is &lt;em&gt;&quot;obviously not 2013 on Earth&quot;&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In actual calculations, astronomers use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Julian_day&quot;&gt;Julian day&lt;/a&gt;, which is a decimal representation of time. A &lt;a href=&quot;http://en.wikipedia.org/wiki/Julian_year_%28astronomy%29&quot;&gt;Julian year&lt;/a&gt; is exactly 365.25 days of 86,400 seconds each. Astronomical coordinates are usually written in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Epoch_%28astronomy%29#Julian_years_and_J2000&quot;&gt;J2000 epoch&lt;/a&gt;, which allows us to compensate for Earth's &lt;a href=&quot;http://en.wikipedia.org/wiki/Precession_of_the_equinoxes&quot;&gt;axial precession&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our ability to measure and record time is limited only by the accuracy of our most precise clocks. There's nothing fundamentally wrong or imprecise about it.&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2013-10-24T02:18:25.083" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="760" PostTypeId="2" ParentId="757" CreationDate="2013-10-24T02:20:14.913" Score="8" Body="&lt;p&gt;Interesting question. I would say from an energy standpoint, it almost certainly it has no effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the extreme case is &lt;a href=&quot;http://en.wikipedia.org/wiki/Io_%28moon%29&quot; rel=&quot;noreferrer&quot;&gt;Io&lt;/a&gt;, one of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Galilean_moons&quot; rel=&quot;noreferrer&quot;&gt;Galilean moons&lt;/a&gt; whose heat source comes from the gravitational &lt;a href=&quot;http://www.planetaryexploration.net/jupiter/io/tidal_heating.html&quot; rel=&quot;noreferrer&quot;&gt;tidal stretching&lt;/a&gt; as it orbits very closely to the planet Jupiter. The heat that sustains the core of the Earth, however, is left over from its formation and also comes from radioactive decay of heavy elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The differential potential energy (and hence the tidal force) over the planet Io due to Jupiter, which is approximately 1300 times more massive than the Earth, is much larger than that of the Earth due to the moon. The relationship between force and differential potential energy is:&#xA;$$ F = - \nabla U $$&#xA;At a given location on the potential energy curve, the strength of the force is determined by its steepness (derivative) at that same location. Below is a quick plot I've generated for the Earth-Moon system, where the vertical red line represents the average Earth-Moon distance over a one-year period. As you can see it doesn't appear to be very 'steep', though keep the scales of the x- and y-axes in mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wIWKJ.png&quot; alt=&quot;Earth-Moon&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Admittedly this isn't that exciting of a plot. But, for comparison one could be made for the Jupiter-Io system, and numerical derivatives could be taken for both to calculate the magnitude of the tidal force in each situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To answer the question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the difference in the gravitational potential energy of object A on B over the scale of B is comparable to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_binding_energy&quot; rel=&quot;noreferrer&quot;&gt;self-gravitational energy&lt;/a&gt; of object B, then tidal forces will become important. This self-gravitational energy is the amount required to completely pull apart all massive particles infinitely far away. Formally, this limit is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Roche_limit&quot; rel=&quot;noreferrer&quot;&gt;Roche Limit&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2013-10-24T02:20:14.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="761" PostTypeId="2" ParentId="758" CreationDate="2013-10-24T07:13:31.127" Score="4" Body="&lt;p&gt;To consider this question we must consider four things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Humans work on a much smaller scale than what you are suggesting (i.e. seconds, days, years, centuries and maybe millennia) meaning that it is harE for us to comprehend the time scale of the existence of earth (billions of years). We also cannot get an accurate enough time for the age of the earth, creating the problem of when did it start. Would this year be 4 500 000 000 or would this year be 4 500 000 013 or would it be 4 487 575 864? With the current system we have defined a point in time from which to start our calendar (about four years after the real person called Jesus was born).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Most things that we do now operate on the BCE/CE system (BC/AD: new names, same system), meaning that &lt;strong&gt;everything&lt;/strong&gt; would need to be updated to the new system. All past observations, all past reports, all computer systems, all systems that control telescopes, &lt;strong&gt;EVERTHING&lt;/strong&gt;. It would cost hundreds of millions, if not billions or even trillions of dollars.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What do you consider as the point at which Earth became Earth? The formation of a planet spans over millions of years and we cannot simply define a point at which Earth was Earth.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When writing the long version of a date instead of simply writing eight characters, you would potentially have to write 14. it would become a lot harder to read and write and become impractical and produce more room for transcription errors and create confusion.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So all in all it becomes expensive and impractical to change the system and the system we would change to has no benefit worth the extra expense or effort.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2013-10-24T07:13:31.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="762" PostTypeId="1" AcceptedAnswerId="767" CreationDate="2013-10-24T08:50:42.803" Score="11" ViewCount="1831" Body="&lt;p&gt;For instance, Jupiter is the biggest planet in our solar system. Due to it's gravitational mass it can deflect meteorites and protect Earth from the Oort cloud.  While these are positives, the negative effects are that Jupiter can draw comets in from the Kuiper belt which may have catastrophic consequences.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do other planets have a similar effect on the solar system?&lt;/p&gt;&#xA;" OwnerUserId="301" LastEditorUserId="38" LastEditDate="2013-10-24T15:46:48.280" LastActivityDate="2013-10-24T15:46:48.280" Title="What effects do other planets have on the solar system?" Tags="&lt;solar-system&gt;&lt;planet&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="763" PostTypeId="2" ParentId="748" CreationDate="2013-10-24T09:22:33.430" Score="24" Body="&lt;p&gt;A neutron star must have a minimum mass of at least 1.4x solar masses (that is, 1.4x mass of our Sun) in order to become a neutron star in the first place.&#xA;See &lt;a href=&quot;https://en.wikipedia.org/wiki/Chandrasekhar_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chandrasekhar limit&lt;/a&gt; on wikipedia for details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A neutron star is formed during a &lt;a href=&quot;http://en.wikipedia.org/wiki/Supernova&quot; rel=&quot;nofollow noreferrer&quot;&gt;supernova&lt;/a&gt;, an explosion of a star that is at least 8 solar masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum mass of a neutron star is 3 solar masses. If it gets more massive than that, then it will collapse into a &lt;a href=&quot;https://en.wikipedia.org/wiki/Quark_star&quot; rel=&quot;nofollow noreferrer&quot;&gt;quark star&lt;/a&gt;, and then into a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that 1 electron + 1 proton = 1 neutron;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 neutron = 3 quarks = up quark + down quark + down quark;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 proton = 3 quarks = up quark + up quark + down quark;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A supernova results in either a neutron star (between 1.4 and 3 solar masses), a quark star(about 3 solar masses), or a black hole(greater than 3 solar masses), which is the remaining collapsed core of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During a supernova, most of the stellar mass is blown off into space, forming elements heavier than iron which cannot be generated through stellar nucleosynthesis, because beyond iron, the star requires more energy to fuse the atoms than it gets back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the supernova collapse, the atoms in the core break up into electrons, protons and neutrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case that the supernova results in a neutron star core, the electrons and protons in the core are merged to become neutrons, so the newly born 20-km-diameter neutron star containing between 1.4 and 3 solar masses is like a giant atomic nucleus containing only neutrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the neutron star's mass is then increased, neutrons become degenerate, breaking up into their constituent quarks, thus the star becomes a quark star; a further increase in mass results in a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The upper/lower mass limit for a quark star is not known (or at least I couldn't find it), in any case, it is a narrow band around 3 solar masses, which is the minimum stable mass of a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you talk about a black hole with a stable mass (at least 3 solar masses), it is good to consider that they come in 4 flavors: rotating-charged, &lt;a href=&quot;http://en.wikipedia.org/wiki/Kerr_metric&quot; rel=&quot;nofollow noreferrer&quot;&gt;rotating-uncharged&lt;/a&gt;, non-rotating-charged, non-rotating-uncharged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we would see visually during the transformation would be a hard radiation flash.&#xA;This is because during the collapse, the particles on/near the surface have time to emit hard radiation as they break up before going into the event horizon; so this could be one of the causes of gamma ray bursts (GRBs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that atoms break up into protons, neutrons, electrons under pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under more pressure, protons and electrons combine into neutrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under even more pressure, neutrons break down into quarks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under still more pressure, perhaps quarks break down into still smaller particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately the smallest particle is a &lt;a href=&quot;http://en.wikipedia.org/wiki/String_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;string&lt;/a&gt;: open or closed loop, and has a Planck length, which is many orders of magnitude smaller than a quark. if a string is magnified so it is 1 millimeter in length, then a proton would have a diameter that would fit snugly between the Sun and Epsilon Eridani, 10.5 light years away; that's how big a proton is compared to a string, so you can imagine there are perhaps quite a few intermediate things between quarks and strings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently it looks like several more decades will be needed to figure out all the math in string theory, and if there is anything smaller than strings then a new theory will be required, but so far string theory looks good; see the book &lt;strong&gt;Elegant Universe&lt;/strong&gt; by Brian Greene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A string is pure energy and Einstein said mass is just a form of energy, so the collapse into a black hole really breaks down the structure of energy that gives the appearance of mass/matter/baryonic particles, and leaves the mass in its most simple form, open or closed strings, that is, pure energy bound by gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that black holes (which are not really holes or singularities, as they do have mass, radius, rotation, charge and hence density, which varies with radius) can &lt;a href=&quot;http://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;evaporate&lt;/a&gt;, giving up their entire mass in the form of radiation, thus proving they are actually energy. Evaporation of a black hole occurs if its mass is below the minimum mass of a stable black hole, which is 3 solar masses; the &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schwarzschild radius&lt;/a&gt; equation even tells you what the radius of a black hole is given its mass, and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you could transform anything you want, such as your pencil, into a black hole if you wanted to, and could compress it into the required size for it to become a black hole; it is just that it would immediately transform itself (evaporate) completely into a flash of hard radiation, because a pencil is less than the stable black hole mass (3 solar masses).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why the CERN experiment could never have created a black hole to swallow the Earth - a subatomic black hole, even one with the mass of the entire Earth, or the Sun, would evaporate before swallowing anything; there is not enough mass in our solar system to make a stable (3 solar mass) black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple way for a neutron star to become more massive in order to be able to turn into a black hole is to be part of a binary system, where it is close enough to another star that the neutron star and its binary pair orbit each other, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Cataclysmic_variable&quot; rel=&quot;nofollow noreferrer&quot;&gt;the neutron star siphons off gas from the other star&lt;/a&gt;, thus gaining mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VCcfj.jpg&quot; alt=&quot;Cataclysmic variable binary&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a nice drawing showing exactly that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matter falling into a black hole is accelerated toward light speed. As it is accelerated, the matter breaks down into subatomic particles and hard radiation, that is, X-rays and gamma rays. A black hole itself is not visible, but the light from infalling matter that is accelerated and broken up into particles is visible. Black holes can also cause a gravitational lens effect on the light of background stars/galaxies.&lt;/p&gt;&#xA;" OwnerUserId="302" LastEditorUserId="1437" LastEditDate="2017-06-29T14:29:28.223" LastActivityDate="2017-06-29T14:29:28.223" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="764" PostTypeId="1" AcceptedAnswerId="799" CreationDate="2013-10-24T10:16:54.200" Score="14" ViewCount="996" Body="&lt;p&gt;What is the timescale of start of nuclear fusion as T Tauri type star transforms into a Main Sequence star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/T_Tauri_type_stars&quot;&gt;T Tauri type stars&lt;/a&gt; mentions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Their central temperatures are too low for hydrogen fusion. Instead, they are powered by gravitational energy released as the stars contract, while moving towards the main sequence, which they reach after about 100 million years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The mentioned 100 million years is the period of the star being in its stable (well, as stable as the turbulent T Tauri type stars get) state without nuclear fusion. Then, once the fusion starts, we get &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aRepresentative_lifetimes_of_stars_as_a_function_of_their_masses.jpg&quot;&gt;between 3 million and hundreds of billions years&lt;/a&gt; of main sequence, depending on mass of the resulting star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm interested is how long is the period of transition between the two - ignition of the nuclear reaction - time between &quot;all energy produced by gravitational contraction&quot; and &quot;most of energy produced by nuclear fusion&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine this period could be quite short, and the effect quite rapid and turbulent as the initial fusion increases local temperature (and as result pressure) drastically, leading to conditions conductive for fusion spreading rapidly over the volume which is already on brink of entering the fusion everywhere within the protostar, essentially a nuclear wildfire encompassing the gathered gas, a chain reaction starting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right in my guess this process is rather rapid? Was it ever observed? Or contrary, does the intensity of the fusion reaction rise gradually and slowly from zero over many millions years of star formation? &lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2013-10-27T12:45:59.967" Title="Timescale of ignition of a protostar?" Tags="&lt;astrophysics&gt;&lt;main-sequence&gt;&lt;protostar&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="765" PostTypeId="2" ParentId="752" CreationDate="2013-10-24T11:13:15.080" Score="4" Body="&lt;p&gt;About 13.1 billion light years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The z8 GND 5296 Galaxy mentioned in the &lt;a href=&quot;http://www.universetoday.com/105694/taking-measure-a-new-most-distant-galaxy/&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; linked by TildalWave, mere 700 millions years old (after Big Bang) would be the mentioned distance from us for the light to reach us at indicated speed, considering the &lt;a href=&quot;http://en.wikipedia.org/wiki/Age_of_the_universe&quot; rel=&quot;nofollow&quot;&gt;13.798&lt;/a&gt; bln years old universe. As we push technological barriers, we keep finding more distant objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The age of the universe is the hard limit in light years of distance of any observable object - there was nothing to emit light more than 13.798 bln years ago, or respectively, &lt;strike&gt;visible from more than 13.798 bln light years away.&lt;/strike&gt; visible from distance that what was 13.798 bln light years 13.798 bln years ago - or, after adjusting for universe expansion over that time, currently &lt;a href=&quot;http://en.wikipedia.org/wiki/Observable_universe&quot; rel=&quot;nofollow&quot;&gt;46.6 billion light years away&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, as RhysW mentions, this limit changes over time as the universe ages and grows. The exact rate of growth is &lt;a href=&quot;http://blogs.discovermagazine.com/badastronomy/2009/05/07/the-universe-is-expanding-at-742-kmsecmpc/&quot; rel=&quot;nofollow&quot;&gt;74.2km/sec per megaparsec&lt;/a&gt;, or about 0.000000007% per year. I'd say, comparing to our life and current technological progress, an insignificant factor.&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="83" LastEditDate="2013-10-24T22:01:45.560" LastActivityDate="2013-10-24T22:01:45.560" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="766" PostTypeId="2" ParentId="24" CreationDate="2013-10-24T11:56:48.147" Score="4" Body="&lt;p&gt;All 4 answers given prior to this one are very good and complete each other;&#xA;finding an object orbiting your target object enables you to also calculate the mass of your target object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matter falling into a black hole is accelerated toward light speed.&#xA;As it is accelerated, the matter breaks down into subatomic particles and hard radiation, that is, X-rays and gamma rays. A black hole itself is not visible, but the light (mostly X-rays, gamma rays) from infalling matter that is accelerated and broken up into particles is visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By looking toward the center of our galaxy, the Chandra X-ray space telescope has observed several black holes besides Sgr A*, indirectly, by catching the hard radiation of infalling matter flaring up as they swallow something; afterward, the black holes go dark again if there is nothing more to assimilate nearby;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chandra.harvard.edu/press/05_releases/press_011005.html&quot; rel=&quot;nofollow&quot;&gt;http://chandra.harvard.edu/press/05_releases/press_011005.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you can see some of this flaring in the swarm of black holes near the center of our galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methods to detect black holes (which are not really holes or singularities, as they do have mass, radius, rotation, charge and hence density, which varies with radius, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Schwarzschild_radius&lt;/a&gt; ).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;to passively detect a (stellar or supermassive) black hole, look/wait for hard radiation flares, which occur sporadically, then follow up with observations to see if you caught a grb (gamma ray burst) from an actual black hole or just a white dwarf or neutron star doing a periodic nova;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;to actively detect a black hole look for gravitational lensing, which is a continuous effect, or stars orbiting at a high speed around a seemingly empty point in space, such as S2 at 5000+km/sec, around Sgr A*&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/S2_%28star%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/S2_(star)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there will be nothing left to see what caused it; better have some observations of that spot in the sky before it happens.&lt;/p&gt;&#xA;" OwnerUserId="302" LastEditorUserId="2153" LastEditDate="2015-01-31T01:00:47.090" LastActivityDate="2015-01-31T01:00:47.090" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="767" PostTypeId="2" ParentId="762" CreationDate="2013-10-24T13:28:06.123" Score="5" Body="&lt;p&gt;The gravitational pull of all the planets and the sun, and the rest of the galaxy and the universe, all play a part, but gravitational effects fall off with distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Earth's orbit, the Sun is far and away the single biggest influence. Jupiter perturbs our orbit slightly, but with it or without it we have a simple elliptical orbit round the sun. We can measure the perturbation in our orbit from all the planets, but the effects are minor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to gravitational effects on small bodies passing one of the planets, however, the effects can be dramatic - an asteroid passing close to Jupiter will be moved far from its original path. The same would be true when passing close to Saturn or in fact any planet - just to a lesser degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while you can't say that Jupiter protects Earth from anything, especially not the Oort Cloud, you also can't say that it draws objects in. Anything out in the Kuiper belt or Oort cloud will be orbiting the barycentre of the solar system (which is within the Sun) and will keep doing that unless knocked out by something  like a collision with another object, or the pull of other Kuiper belt objects like Pluto, which can give a much greater tug.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2013-10-24T13:28:06.123" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="768" PostTypeId="1" CreationDate="2013-10-24T21:25:03.080" Score="11" ViewCount="342" Body="&lt;p&gt;Most of the planets in our solar system all orbit along the same plane, supposedly because of the almost flat dust clouds that formed the planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But some planets, such as the one featured below, have a significant tilt in their orbits that separates them from the normal plane of orbit for that star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AlH5r.jpg&quot; alt=&quot;Tilted orbit&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;a href=&quot;http://www.sciencedaily.com/releases/2013/10/131017144410.htm&quot; rel=&quot;noreferrer&quot;&gt;Image courtesy of Science Daily&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What might cause a planet to have such a tilt? Is this the star capturing a wandering body into orbit or is this a phenomena caused by the formation of these systems?&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastActivityDate="2013-10-27T03:24:46.903" Title="What might cause a planet to have a significant tilt in their orbit?" Tags="&lt;orbit&gt;&lt;planet&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="769" PostTypeId="1" CreationDate="2013-10-24T21:28:56.130" Score="14" ViewCount="2361" Body="&lt;p&gt;Solar Flares obviously release extreme amounts of energy and extend thousands of miles out into space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because they are so big I would like to be able to observe some of these events through a telescope but I am wary of damaging both my eyes or the sensitive telescopic equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How might I safely observe a solar flare through an amateur telescope? Might this be solved by using light filters to dim the amount of light entering the telescope or is a more specialist piece of equipment required?&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastActivityDate="2020-12-15T00:25:12.827" Title="How can I safely observe a Solar Flare?" Tags="&lt;amateur-observing&gt;&lt;observational-astronomy&gt;&lt;solar-flare&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="770" PostTypeId="1" CreationDate="2013-10-24T21:36:57.003" Score="11" ViewCount="501" Body="&lt;p&gt;Sometimes you can be trying to observe very faint or very small objects in the sky and it wouldn't pay to have these obscured or marred by a scratch or dust on the lenses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I keep to good practices and make sure to replace the cap on the lenses when finished in order to stop dust buildup but inevitably it will get a little dusty and dirty, taking this into consideration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What steps do I need to take to ensure that my lenses are kept in top condition?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I simply use a damp cloth or sponge and dab the lenses or should I use a softer material, like someone with glasses might have a softer lens cleaner?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I require a special cleaning fluid or can a good quality clean be done without any additional cleaning agents?&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastEditorDisplayName="user96" LastEditDate="2013-10-24T21:43:49.997" LastActivityDate="2013-10-25T03:22:19.953" Title="How can I clean my lenses without causing scratching or damages to the lens surface?" Tags="&lt;telescope-lens&gt;&lt;telescope-care&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="771" PostTypeId="1" CreationDate="2013-10-24T21:43:25.027" Score="11" ViewCount="214" Body="&lt;p&gt;Sometimes when using my telescope late at night or early in the morning, especially in the more humid of seasons, I notice i sometimes get dew buildup on the lenses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can I minimize damp buildup whilst using my telescope?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply dabbing away the liquid with a dry cloth doesn't seem to do the trick as I am still left with some water tracks and residue that I just can't soak up, this then means that all of the light is refracted providing a very fuzzy and blurry image. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What preventative measures can i take to stop damp build up in the first place?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastActivityDate="2013-10-25T03:57:44.797" Title="Dealing with damp and dew whilst stargazing at night or in the early morning" Tags="&lt;telescope-lens&gt;&lt;telescope-care&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="772" PostTypeId="1" AcceptedAnswerId="783" CreationDate="2013-10-24T22:17:07.163" Score="8" ViewCount="59" Body="&lt;p&gt;I remember reading in 2011 that 2013/2014 would be the peak of the current solar maximus, but I want to confirm whether that is still ongoing.  Will the upcoming winter season be a good time to see the Northern Lights?&lt;/p&gt;&#xA;" OwnerUserId="306" LastEditorUserId="306" LastEditDate="2013-10-25T18:38:49.650" LastActivityDate="2013-10-25T18:38:49.650" Title="Are we still in the Solar Maximus stage? Will this winter 2013/14 be a good aurora-viewing season?" Tags="&lt;light&gt;&lt;solar-flare&gt;&lt;coronal-mass-ejection&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="773" PostTypeId="2" ParentId="770" CreationDate="2013-10-25T03:17:03.127" Score="1" Body="&lt;p&gt;Clean your eyepieces in the same way you would clean a camera lens. An &lt;a href=&quot;http://www.photo-dictionary.com/photofiles/list/4436/6655air_blower.jpg&quot; rel=&quot;nofollow&quot;&gt;air blower&lt;/a&gt;, &lt;em&gt;soft&lt;/em&gt; (i.e. camel hair) brush, lens tissue or microfibre cloth, and a solvent or alcohol will do the trick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I was an undergraduate I did some work with lasers and very expensive dielectric coated mirrors. &lt;a href=&quot;http://www.newport.com/How-to-Clean-Optics/141176/1033/content.aspx&quot; rel=&quot;nofollow&quot;&gt;The techniques here&lt;/a&gt; will show you how to carefully clean optics. Never completely wet your lens tissue or cloth. One drop on a lens tissue is usually enough, or a couple of drops on a microfibre cloth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, you need to:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Use a &lt;a href=&quot;http://www.photo-dictionary.com/photofiles/list/4436/6655air_blower.jpg&quot; rel=&quot;nofollow&quot;&gt;blower&lt;/a&gt; to blow off as much dust as you can.&lt;/li&gt;&#xA;&lt;li&gt;Then use a soft brush.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Dampen&lt;/em&gt; a small square of lens tissue with pure alcohol or acetone (see warning below). Gently draw or wipe the tissue across the surface &lt;em&gt;once&lt;/em&gt;, and then discard. Repeat if necessary.&lt;/li&gt;&#xA;&lt;li&gt;You may need to fog the lens with your breath to help remove water-soluble gunk.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Be &lt;strong&gt;&lt;em&gt;insanely&lt;/em&gt; careful&lt;/strong&gt; if you are using a solvent such as acetone. Although a good for cleaning optics, it's not very nice to paint and plastic so it should be a last resort for very stubborn marks well away from the edges of the lens (so you'll never touch the paint with it). Methyl or isopropyl alcohol are safer but less effective, or a generic lens cleaner from a camera store.&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="276" LastEditDate="2013-10-25T03:22:19.953" LastActivityDate="2013-10-25T03:22:19.953" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="774" PostTypeId="2" ParentId="771" CreationDate="2013-10-25T03:57:44.797" Score="3" Body="&lt;p&gt;Condensation will build up because the temperature of your optics has fallen below the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dew_point&quot; rel=&quot;nofollow&quot;&gt;dew point&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You simply have to keep the temperature your telescope above the dew point, which can be quite difficult on humid nights. Dew shields can help, as can dew heaters (which literally just gently heat your telescope and/or eyepiece).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hypothetical: &quot;I'm looking at the graph of dew point vs air temperature on the link above. I don't understand how my optics can get colder than the air temperature!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At night, our source of heat (the Sun) has disappeared. The unlit side of the Earth (and everything on it) begins radiating its latent heat away into space (&lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow&quot;&gt;which has a temperature of ~3 Kelvin&lt;/a&gt;). Your telescope is trying to get into thermal equilibrium with outer space.&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2013-10-25T03:57:44.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="775" PostTypeId="5" CreationDate="2013-10-25T08:16:54.473" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-10-25T08:16:54.473" LastActivityDate="2013-10-25T08:16:54.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="776" PostTypeId="4" CreationDate="2013-10-25T08:16:54.473" Score="0" Body="Questions about the early phases of the star formation process. A protostar is a large mass of gas that results of the contraction of a molecular cloud." OwnerUserId="88" LastEditorUserId="88" LastEditDate="2013-10-25T09:25:57.303" LastActivityDate="2013-10-25T09:25:57.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="777" PostTypeId="5" CreationDate="2013-10-25T08:20:34.463" Score="0" Body="&lt;p&gt;Questions about the &amp;quot;adulthood&amp;quot; of stars. The main sequence is a distinctive band of stars that appears in a color-magnitude plot (known as Hertzsprung-Russell diagram). Stars appear in the main sequence when they are burning hydrogen in their core. When a star runs out of hydrogen, it evolves into a &lt;a href=&quot;/questions/tagged/red-giant&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;red-giant&amp;#39;&quot; rel=&quot;tag&quot;&gt;red-giant&lt;/a&gt; or &lt;a href=&quot;/questions/tagged/red-supergiant&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;red-supergiant&amp;#39;&quot; rel=&quot;tag&quot;&gt;red-supergiant&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="31410" LastEditorUserId="31410" LastEditDate="2021-01-16T22:54:12.603" LastActivityDate="2021-01-16T22:54:12.603" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="778" PostTypeId="4" CreationDate="2013-10-25T08:20:34.463" Score="0" Body="Questions about the &quot;adulthood&quot; of stars. The main sequence is a distinctive band of stars that appears in a color-magnitude plot (known as Hertzsprung-Russell diagram). Stars appear in the main sequence when they are burning hydrogen in their core." OwnerUserId="88" LastEditorUserId="88" LastEditDate="2013-10-25T09:26:05.833" LastActivityDate="2013-10-25T09:26:05.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="779" PostTypeId="5" CreationDate="2013-10-25T08:26:48.747" Score="0" Body="&lt;p&gt;Questions about the observed frequency range of the electromagnetic radiation of a celestial object.&lt;/p&gt;&#xA;" OwnerUserId="31410" LastEditorUserId="31410" LastEditDate="2020-12-15T11:35:21.997" LastActivityDate="2020-12-15T11:35:21.997" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="780" PostTypeId="4" CreationDate="2013-10-25T08:26:48.747" Score="0" Body="Questions about the observed frequency range of the electromagnetic radiation of a celestial object." OwnerUserId="88" LastEditorUserId="31410" LastEditDate="2020-12-15T04:48:08.653" LastActivityDate="2020-12-15T04:48:08.653" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="781" PostTypeId="2" ParentId="488" CreationDate="2013-10-25T15:10:35.287" Score="3" Body="&lt;p&gt;theoretically there should be no maximum size/mass of a black hole, or you could say the maximum mass would be if it contained all the mass in the universe;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;although the minimum dimension of a black hole would be a Planck length, the minimum mass of a &lt;strong&gt;stable&lt;/strong&gt; black hole is 3 solar masses;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a black hole that has a mass less than 3 solar masses will evaporate, transforming itself into radiation (energy); the smaller it is, the faster it evaporates; if it is small enough, it will instantly turn into a flash of hard radiation;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hawking_radiation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you know the mass of a black hole you can calculate its radius and if you know its radius you can calculate its mass;&#xA;note that a curious thing about this equation is that it shows a stellar mass black hole having a high density, whereas a supermassive black hole having a density that decreases as the black hole increases in mass;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Schwarzschild_radius&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so you could transform anything you want, such as your pencil, into a black hole if you wanted to, and could compress it into the required size for it to become a black hole; it is just that it would immediately transform itself (evaporate) completely into a flash of hard radiation, because a pencil is less than the stable black hole mass (3 solar masses);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this is why the CERN experiment could never have created a black hole to swallow the Earth - a subatomic black hole, even one with the mass of the entire Earth, or the Sun, would evaporate before swallowing anything;&#xA;there is not enough mass in our solar system to make a stable (3 solar mass) black hole;&lt;/p&gt;&#xA;" OwnerUserId="302" LastActivityDate="2013-10-25T15:10:35.287" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="782" PostTypeId="1" AcceptedAnswerId="796" CreationDate="2013-10-25T15:45:17.243" Score="48" ViewCount="1625" Body="&lt;p&gt;At this point in time, evidence for the existence of dark matter has accumulated in many ways: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;it affects galactic rotation curves&lt;/li&gt;&#xA;&lt;li&gt;plays a major role in cosmology, and the evolution of structure in the universe&lt;/li&gt;&#xA;&lt;li&gt;is predicted in copious amounts by gravitational lensing on a wide range of scales&lt;/li&gt;&#xA;&lt;li&gt;influences the dynamics of galaxy clusters&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;to name a few.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many known candidates for dark matter particles: &lt;a href=&quot;http://en.wikipedia.org/wiki/Weakly_interacting_massive_particles&quot;&gt;WIMPs&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Axion&quot;&gt;axions&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Wisp&quot;&gt;WISPs&lt;/a&gt;, neutrinos, etc (in fact, even bricks, though some other considerations would exclude them).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The question then is&lt;/strong&gt;: &#xA;Why do we expect that only one type of dark matter particles is responsible for phenomenological dark matter? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, $\Lambda$CDM cosmology, the standard cosmological model, requires dark matter to be cold (slow, non-relativistic), which is used to constrain the possible properties of dark matter particles. However, this doesn't actually imply, that dark matter is cold for all the astrophysical systems. For example, galactic halos could be made of warm dark matter, and halos of dwarf galaxies could be made of cold dark matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One might of course say that one-species model is the simplest one. The counter-argument would be that in reality there well may be many species. This in turn might have profound implications for astrophysical models. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize the question: &lt;strong&gt;Is there any good reason, preferably supported by observations, to think that only one species of dark matter is present in all the models currently used?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="115" LastEditorUserId="37191" LastEditDate="2021-02-16T08:55:19.810" LastActivityDate="2021-02-16T08:55:19.810" Title="Two species of dark matter?" Tags="&lt;cosmology&gt;&lt;dark-matter&gt;&lt;galactic-dynamics&gt;&lt;gravitational-lensing&gt;&lt;axion&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="783" PostTypeId="2" ParentId="772" CreationDate="2013-10-25T17:50:14.507" Score="3" Body="&lt;p&gt;From this data here: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.swpc.noaa.gov/SolarCycle/&quot; rel=&quot;nofollow&quot;&gt;http://www.swpc.noaa.gov/SolarCycle/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://solarscience.msfc.nasa.gov/SunspotCycle.shtml&quot; rel=&quot;nofollow&quot;&gt;http://solarscience.msfc.nasa.gov/SunspotCycle.shtml&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are in a local maximum but the actual values have been lower than predicted for this time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for getting an idea for the aurora, I would keep an eye on the sunspot numbers and the space weather.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.swpc.noaa.gov/today.html&quot; rel=&quot;nofollow&quot;&gt;http://www.swpc.noaa.gov/today.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.gi.alaska.edu/AuroraForecast&quot; rel=&quot;nofollow&quot;&gt;http://www.gi.alaska.edu/AuroraForecast&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="281" LastActivityDate="2013-10-25T17:50:14.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="784" PostTypeId="2" ParentId="214" CreationDate="2013-10-25T18:22:07.127" Score="3" Body="&lt;p&gt;Adding to @Guillochon's answer, there are even a number of &lt;a href=&quot;http://en.wikipedia.org/wiki/Tests_of_general_relativity&quot; rel=&quot;nofollow&quot;&gt;general relativistic tests&lt;/a&gt; in our solar system, the most famous being the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tests_of_general_relativity#Perihelion_precession_of_Mercury&quot; rel=&quot;nofollow&quot;&gt;precession of the perihelion of Mercury&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, the location of the point of closest approach to the Sun (perihelion) for the planet Mercury is a changing quantity. Essentially, given one full revolution, it doesn't trace out a closed shape. The distance this point moves per Julian year is &lt;strong&gt;not&lt;/strong&gt; well predicted by simply assuming a simple 2-body system evolving under Newtonian mechanics (the Sun and Mercury being these two bodies). Other things which are taken into account are the gravitational influences of other planets (most importantly Jupiter) on this 2-body system, and the fact that the sun is not perfectly spherical in shape (it's an &lt;a href=&quot;http://en.wikipedia.org/wiki/Oblate_spheroid&quot; rel=&quot;nofollow&quot;&gt;Oblate Spheroid&lt;/a&gt;).&#xA;It turns out that if you include a correction due to GR, its precession can be completely accounted for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other notable GR test was the deflection of light from a star by the Sun in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Tests_of_general_relativity#Deflection_of_light_by_the_Sun&quot; rel=&quot;nofollow&quot;&gt;1919 solar eclipse&lt;/a&gt;, proving only a few years its formulation that GR was a viable theory.&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2013-10-25T18:22:07.127" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="785" PostTypeId="2" ParentId="758" CreationDate="2013-10-25T18:41:46.377" Score="2" Body="&lt;p&gt;Hopefully this answer isn't redundant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomy is done over huge orders of magnitude. To make things manageable, we try to keep numbers under a thousand - it's both more practical and more intuitive. This is reason #1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second reason why we don't set our zero point to the beginning of the Earth is because it's a measured quantity with an error. If you set your clocks to the 'beginning' of the Earth (which by the way is not a well-defined statement since the time of the formation of the Earth is actually an inclusive range of times), you would have to report the year as:&#xA;$$ 4,540,002,013 \pm 50,000  \text{  years} $$&#xA;Though the zero-point is certainly arbitrary, it at least defines a point in time where we can use clocks to measure from. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, why do you think the age of the Earth is the relevant zero-point? Why not use the beginning of the Universe? &lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2013-10-25T18:41:46.377" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="787" PostTypeId="2" ParentId="702" CreationDate="2013-10-25T20:53:00.023" Score="7" Body="&lt;p&gt;The answer to the first question is: more observations.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;That is, what process and techniques are used to calculate and increase the accuracy of our collision threat estimation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You just have to know the orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the orbit of asteroids is very complicated to predict, because they are influenced, sometimes dramatically, by the planets and by smaller bodies, like other asteroids, that we might not be able to see. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On top of this we cannot measure directly the orbital speed of an asteroid. We can only observe changes in the angular position. &lt;strong&gt;If&lt;/strong&gt; we know the distance (which might not be easily measurable) we can at least put a lower limit to the speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic steps to study the orbit of an asteroids are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;find it and try to get a first guess about the position and velocity (this requires multiple observations far enough in time to be able to measure an angular movement)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;plug the position and velocity info (and likely their errors) into a computer simulation that simulate the solar system dynamics. Fast forward the simulation to predict the future orbit.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;observe again the asteroid and compare its position with the simulation. Re-estimate position and velocity. Go to point 2.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As the orbit can change significantly a continuous interplay between 2. and 3. is necessary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have an orbit you can estimate the probability for the asteroid to pass near enough to the Moon or Earth to be deflected and hit us. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But thankfully the solar system is huge, we are tiny and moving between the much much larger Sun and Jupiter :D&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2014-10-28T14:31:37.193" LastActivityDate="2014-10-28T14:31:37.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="789" PostTypeId="2" ParentId="752" CreationDate="2013-10-25T21:34:14.137" Score="4" Body="&lt;p&gt;There are few objects that might have even higher redshift. Googling I've found three accepted papers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;two are about a $z\approx10$ galaxy. One published in &lt;a href=&quot;http://arxiv.org/abs/0912.4263&quot; rel=&quot;nofollow&quot;&gt;Nature&lt;/a&gt; and the other one in &lt;a href=&quot;http://arxiv.org/abs/1211.3105&quot; rel=&quot;nofollow&quot;&gt;Astrophysics Journal Letters&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;is about a strong lensed candidate at $z\approx11$ published on &lt;a href=&quot;http://arxiv.org/abs/1211.3663&quot; rel=&quot;nofollow&quot;&gt;Astrophysical Journal&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are also few gamma ray burst associated with objects at large redshift. &lt;a href=&quot;http://www.mpe.mpg.de/~jcg/grbgen.html&quot; rel=&quot;nofollow&quot;&gt;This database&lt;/a&gt; containts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;z=8.26 &lt;a href=&quot;http://www.mpe.mpg.de/~jcg/grb090423.html&quot; rel=&quot;nofollow&quot;&gt;GRB090423&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;z=9.2 &lt;a href=&quot;http://www.mpe.mpg.de/~jcg/grb090429B.html&quot; rel=&quot;nofollow&quot;&gt;GRB090429B&lt;/a&gt;. The redshift if photometric, which means that the error on the determination is likely large &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There might be few other around&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2013-10-25T21:34:14.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="790" PostTypeId="2" ParentId="769" CreationDate="2013-10-26T06:28:49.263" Score="9" Body="&lt;p&gt;One truly safe way of solar flare observations is to not directly observe them (with your eyes or imaging technology) at all, such is the case with radio-astronomy.  There are a number of examples of how this could be done, for example, the &lt;a href=&quot;http://www.backyardastronomy.net/sid_receiver.html&quot;&gt;Sudden Ionosphere Disturbance Monitor&lt;/a&gt;, this device measures disturbances in the ionosphere that includes the effects of solar flares.  This operates in the VLF (3-30kHz) range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the &lt;a href=&quot;http://radio-astronomy.org/node/117&quot;&gt;Society of Amateur Radio Astronomers&lt;/a&gt;, the advice is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For VLF solar flare observations you will need a strip chart recorder and a radio receiver capable of operating in the noisy 20 to 100 kilohertz radio band. These receivers are quite simple and may be home constructed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Plans for these can be obtained from that group)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A repository of further links (far too many to list here) can be found on the website &lt;a href=&quot;http://www.starkenburg-sternwarte.de/radio/links.htm&quot;&gt;Amateur Radio Astronomy on the Internet&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorUserId="38" LastEditDate="2013-10-26T16:05:04.237" LastActivityDate="2013-10-26T16:05:04.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="791" PostTypeId="5" CreationDate="2013-10-26T09:25:14.640" Score="0" Body="" OwnerDisplayName="user96" LastEditorDisplayName="user96" LastEditDate="2013-10-26T09:25:14.640" LastActivityDate="2013-10-26T09:25:14.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="792" PostTypeId="4" CreationDate="2013-10-26T09:25:14.640" Score="0" Body="Questions regarding the universe shortly after its formation, up to the CMB or even as far as the formation of the first galaxies" OwnerDisplayName="user96" LastEditorUserId="6" LastEditDate="2020-08-18T18:06:27.160" LastActivityDate="2020-08-18T18:06:27.160" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="793" PostTypeId="1" CreationDate="2013-10-26T10:14:36.990" Score="9" ViewCount="124" Body="&lt;p&gt;Currently, the solar system has 8 major planets, a handful of dwarf planets and a plethora of asteroids, comets and natural debris, and of course, a rather locally dominant star (the Sun).  There are many references that describe the early solar system as a chaotic environment with many more protoplanets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on observational evidence, how many protoplanets are believed to have existed in the early solar system?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorUserId="38" LastEditDate="2013-10-26T16:03:28.247" LastActivityDate="2014-01-06T16:59:13.560" Title="What does observational evidence say about the number of protoplanets believed to exist in the early solar system?" Tags="&lt;solar-system&gt;&lt;formation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="794" PostTypeId="2" ParentId="753" CreationDate="2013-10-27T02:33:17.130" Score="4" Body="&lt;p&gt;There are a few peculiarities of Pluto's orbit. These are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Its high orbital &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_eccentricity&quot; rel=&quot;nofollow&quot;&gt;eccentricity&lt;/a&gt; (e = 0.25) causes Pluto's perihelion to be ever so slightly smaller than Neptune's perihelion.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Pluto has an orbital resonance with Neptune. Its orbital period is exactly 3/2 of Neptune's. This orbital resonance is the cause of its orbital eccentricity that you ask about.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Pluto has an orbital inclination of 17 degrees to the ecliptic. No one &lt;em&gt;knows&lt;/em&gt; why, we can only guess.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The origin of Pluto's eccentricity explained:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;During the later stages of planet formation, in this early Solar system, the planets were still forming. Neptune exchanged angular momentum with the remaining planetesimals, and its orbit expanded outwards. If Pluto were in a near-circular orbit larger than Neptune's (at about 33 &lt;a href=&quot;http://en.wikipedia.org/wiki/Astronomical_unit&quot; rel=&quot;nofollow&quot;&gt;AU&lt;/a&gt;), there is a high chance that Neptune could capture Pluto and lock it an orbital resonance (this probably happened when Neptune was at about 25 AU). As Neptune's orbit continued to expand outwards, this expansion drew out the eccentricity of Pluto's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source and further reading: &lt;a href=&quot;http://adsabs.harvard.edu/abs/1995AJ....110..420M&quot; rel=&quot;nofollow&quot;&gt;The Origin of Pluto's Orbit: Implications for the Solar System Beyond Neptune&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2013-10-27T02:33:17.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="795" PostTypeId="2" ParentId="768" CreationDate="2013-10-27T03:24:46.903" Score="8" Body="&lt;p&gt;Orbital tilts are almost certainly an effect of the formation and evolution of the protoplanetary disk (at least in the majority of cases). &lt;a href=&quot;http://arxiv.org/abs/1202.2362&quot; rel=&quot;nofollow noreferrer&quot;&gt;They could also be captured rogue planets&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the eccentricity of Pluto. Like Pluto's &lt;a href=&quot;https://astronomy.stackexchange.com/questions/753/why-does-plutos-orbit-underlap-neptunes-orbit&quot;&gt;eccentric orbit&lt;/a&gt;, it's probable that its current inclination is the result of orbital resonances with Neptune. &lt;a href=&quot;http://pages.physics.cornell.edu/~cmuhlberger/documents/astr399-paper.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This paper&lt;/a&gt; (unpublished grad student research) describes some such scenarios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is likely these resonances are between the Jupiter-induced precession frequencies of the ascending nodes of Pluto and Neptune. Pluto's inclination could either be relatively stable now, or it may still oscillate on the timescale of several million years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish I could give a more definitive and detailed explanation - it's just one of these cases where &lt;em&gt;no one knows&lt;/em&gt; for sure. &lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2013-10-27T03:24:46.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="796" PostTypeId="2" ParentId="782" CreationDate="2013-10-27T03:50:43.473" Score="23" Body="&lt;p&gt;Hot dark matter would be made from very light, fast moving particles. Such particles could not possibly be gravitationally bound to any structure, but rather would be dispersed all across the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But dark matter is always &quot;found&quot; (or &quot;inferred&quot;) either gravitationally bound to some visible structure (e.g. weak lensing detection of dark matter associated with &lt;a href=&quot;http://en.wikipedia.org/wiki/Bullet_Cluster&quot;&gt;colliding galaxy clusters&lt;/a&gt; / flat &lt;a href=&quot;http://en.wikipedia.org/wiki/Galaxy_rotation_curve&quot;&gt;rotation curves of spiral galaxies&lt;/a&gt; / abnormal &lt;a href=&quot;http://en.wikipedia.org/wiki/Virial_theorem#Galaxies_and_cosmology_.28virial_mass_and_radius.29&quot;&gt;velocity dispersion in galaxy clusters&lt;/a&gt;) or not associated to anything visible but nevertheless forming clumps (weak lensing detection of &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0104094&quot;&gt;galaxy clusters previously unseen&lt;/a&gt;). That is why dark matter is thought to be &lt;em&gt;cold&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, there is a clear distinction between both types: there is not such thing as dark matter that is &quot;not too cold but not too hot either&quot; (see footnote as well). Dark matter is &lt;a href=&quot;http://prl.aps.org/abstract/PRL/v29/i10/p669_1&quot;&gt;either made of particles with less than ~10 eV&lt;/a&gt; (hot dark matter, made of light particles, mostly dispersed everywhere) or &lt;a href=&quot;http://prl.aps.org/abstract/PRL/v39/i4/p165_1&quot;&gt;particles with more than ~2 GeV&lt;/a&gt; (heavier, slower particles gravitationally bound to some structure). Both limits are found when imposing the maximum amount in which the candidate particles (neutrinos or something more exotic) can possibly contribute to &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambda-CDM_model&quot;&gt;the actual value of the density parameter&lt;/a&gt; due to matter in our expaning Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, either DM appears gravitationally bound (cold DM) or dispersed (hot DM), and both types are clearly distinct (10 ev vs 2 Gev). Observations favour the first case. However, Cold Dark Matter is not the ultimate solution, and still faces some problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the possibility of mixed solutions, many of them have been already ruled out. Microlensing has ruled out the possibility of unseen compact objects (brown dwarfs, stars, stellar black holes) in galactic halos, &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0001272&quot;&gt;in our galactic neighbourhood&lt;/a&gt; as well as &lt;a href=&quot;http://arxiv.org/abs/0910.3645&quot;&gt;in the extragalactic domain&lt;/a&gt;. Ordinary matter (stones, bricks, dust) cannot possibly be, otherwise they would become hot and re-radiate. Any exotic mix of known particles doesn't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All we think we know is that DM must be made of some heavy particles yet to be discovered. In order to introduce a more complex model (e.g. different types of particles depending on the structure they appear attached to) one needs a justification (i.e. some predictions that better agree with reality) and nobody has been able to do that yet.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remark&lt;/strong&gt; Note that Dark Matter particles, either from the hot or the cold type, cannot possibly &quot;slow down&quot; and clump too much (e.g. forming planets) because they don't interact electromagnetically like ordinary matter, that is why DM is said to be &lt;em&gt;&lt;a href=&quot;http://arxiv.org/abs/astro-ph/0002308&quot;&gt;collisionless&lt;/a&gt;&lt;/em&gt;. Wherever infalling ordinary matter forms any structure (e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Protostar&quot;&gt;protostars&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Accretion_disc&quot;&gt;accretion disks&lt;/a&gt;), a very important part of the process is &lt;em&gt;thermalisation&lt;/em&gt;, i.e. the redistribution of energy of the infalling particles by means of numerous collisions. This cannot happen with Dark Matter.&lt;/p&gt;&#xA;" OwnerUserId="316" LastEditorUserId="316" LastEditDate="2013-11-09T12:29:01.997" LastActivityDate="2013-11-09T12:29:01.997" CommentCount="24" ContentLicense="CC BY-SA 3.0" />
  <row Id="797" PostTypeId="2" ParentId="459" CreationDate="2013-10-27T08:47:16.893" Score="4" Body="&lt;p&gt;The resolution/error box. Radioastronomy has always been hindered by the resolution, because it is inversely proportional to the size of the telescope and making larger telescopes (even with interferometry) is not always easy. No amount of modern technology can substitute for a large effective diameter.  (When I say &lt;em&gt;effective&lt;/em&gt; I'm including interferometry here; either way one needs a large area to build on).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take a look at Reber's initial paper&lt;sup&gt;&lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1944ApJ...100..279R&amp;amp;classic=YES&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;&lt;/sup&gt; where he first mapped the sky:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sP9tg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the diagram on the left, from top down, the three peaks in the contour map are &lt;a href=&quot;http://en.wikipedia.org/wiki/Cassiopeia_A&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cas A&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cygnus_A&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cyg A&lt;/a&gt;, and finally &lt;a href=&quot;http://en.wikipedia.org/wiki/Sagittarius_A&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sgr A&lt;/a&gt;. The latter two have black hole origins, the former is a supernova remnant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resolving power of Reber's telescope here seems to have been 6 degrees, and it was 31.4 feet in diameter (and he focused on a 1.9m wavelength).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, by the Rayleigh criterion, the angular resolution is proportional to the wavelength divided by the diameter. As mentioned before, this is the major limiting factor for radioastronomers, and will be what holds amateur radioastronomers from making great telescopes &amp;mdash; amateurs usually don't have acres or land in which to build a good interferometer (let alone the precision), and single-dish telescopes can't be made too large by an amateur. One may notice that I'm citing rather old observations here, on old telescopes; however given that radioastronomy &lt;em&gt;technology&lt;/em&gt; has not changed nearly as much as &lt;em&gt;size&lt;/em&gt; has, it should be OK to compare amateur telescopes with the smaller telescopes of the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, Cyg A was the first to be identified as a black hole, even though the radio-brightness of Sgr A was discovered at the same time. I'm focusing the rest of my analysis on Cyg A for this reason, as it stands to reason that the first confirmed BH out of the brighter radio sources would have more prominent indicators that it is a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's have a look at Cyg A with a better resolution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/O9QGB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(From this paper&lt;sup&gt;&lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1969MNRAS.146..221M&amp;amp;classic=YES&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;&lt;/sup&gt;, using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ryle_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;5km array&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the black blob in the center is the actual galaxy (probably an optical photograph superimposed on the contour map).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see that the lobes are less than a minute wide. (The actual galaxy is around &lt;a href=&quot;http://ned.ipac.caltech.edu/cgi-bin/datasearch?objname=Cyg%20A&amp;amp;search_type=Diameters&amp;amp;of=table&quot; rel=&quot;nofollow noreferrer&quot;&gt;50 arcseconds wide&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, the most interesting thing that one would like to see here are the gas jets coming from the central galaxy. As mentioned in my answer &lt;a href=&quot;https://astronomy.stackexchange.com/a/224/28&quot;&gt;here&lt;/a&gt;, these radio-emitting gas jets are in a steady line over thousands of light years, indicating that they come from some sort of cosmic gyroscope which has been steady for very long. However, even with the Ryle telescope the 1969 people could not get a picture of them; just a slight hint of their existence from the shape of the lobes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alright, so no gas jets. What else can indicate a black hole? They could try to look at the lobes themselves. They do not directly indicate the existence of a black hole, but their shape hints towards them being formed from jets (this is pretty much in retrospect).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, with lobe dimensions of less than an arc minute, there's not much an amateur can get here, either. It's possible that a really good amateur telescope would just manage to notice that there are two lobes, but not much else as far as I can tell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other interesting bits would be the central galaxy itself, but it's too small. In the optical region one may have a chance of seeing Baade's &quot;colliding galaxies&quot; (it only &lt;em&gt;looks&lt;/em&gt; like a pair of colliding galaxies). The gravitational effects (lensing, etc) are really only visible in the optical and beyond, for it to be visible in the radio we would need to be very lucky and have a huge radio source pass behind Cyg A &amp;mdash; not happening anytime soon.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure that a similar analysis would work for Sgr A or any other black hole candidate; the gas jets would be too small for an amateur radiofrequency resolution, and the gravitational effects of the black hole would only work well in the optical and X ray frequencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1. &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1944ApJ...100..279R&amp;amp;defaultprint=YES&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reber, G. (1944). Cosmic Static. &lt;em&gt;The Astrophysical Journal&lt;/em&gt;, 100, 279.&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2. &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1984MNRAS.209..851A&amp;amp;defaultprint=YES&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mitton, S., &amp;amp; Ryle, M. (1969). High resolution observations of Cygnus A at 2. 7 GHz and 5 GHz. &lt;em&gt;Monthly Notices of the Royal Astronomical Society&lt;/em&gt;, 146, 221.&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2013-10-27T08:47:16.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="799" PostTypeId="2" ParentId="764" CreationDate="2013-10-27T12:45:59.967" Score="9" Body="&lt;p&gt;I have mulled over this a couple of times (it's a really interesting question!), and hopefully come up with a somewhat enlightening answer. I haven't been able to find a good, modern reference for these details (perhaps I just suck at literature searches...) so there's a little mucking around in the history books&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total timescale of evolution onto the main sequence for a protostar in the T Tauri mass range ( &amp;lt; 3 solar masses) is on the order (of magnitude) of several tens of millions of years. The ignition of fusion is not precisely a &quot;runaway&quot; reaction: however it occurs relatively quickly and once it starts, gravitational contraction quickly ceases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The evolution of a 1 solar mass protostar follows these basic steps. Things are a little different for different masses - too complicated to explain here but the references should provide ample further reading!&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Jeans_instability&quot; rel=&quot;noreferrer&quot;&gt;Jeans-unstable&lt;/a&gt; cloud of gas and dust begins to contract, exchanging gravitational potential energy for kinetic energy, and thus heat. The luminosity of the protostellar cloud increases as it collapses. It takes around 100,000 years for the initial period of rapid collapse to finish, at this point the cloud is very luminous (perhaps 20 solar luminosities and 8000K).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Over the next 1 million years, the protostellar cloud slowly contracts and cools to around 4500K. The protostar then travels down the Hayashi track, contracting further but changing little in temperature - its luminosity continues to fall. This is the stage where T Tauri stars are at. Most T Tauri stars are younger than 3 million years old.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The star then follows the Henyey track, where the luminosity begins to slowly increase again as a radiative zone develops in the star's core and it continues to slowly contract. This can take some tens of millions of years.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, the conditions in the core are extreme enough for fusion to begin. The timescale from all the energy being provided by gravitational contraction to all the energy being provided by fusion, is on the order of 1 million years. The star's luminosity (counter-intuitively) decreases again when this happens, as the energy from fusion doesn't quite offset that from gravitational contraction, which ceases when fusion begins.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Figure: the The L&lt;sub&gt;g&lt;/sub&gt;/L curve describes the amount of energy gained from gravitational contraction over the total luminosity of the star. The logarithmic time axis is in seconds (reproduced from Iben (1965), Figure 3).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rBYVO.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;References:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1965ApJ...141..993I&quot; rel=&quot;noreferrer&quot;&gt;I. Iben (1965), Stellar Evolution. I. The Approach to the Main Sequence&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1972MNRAS.157..121L&quot; rel=&quot;noreferrer&quot;&gt;R. Larson (1972), The evolution of spherical protostars with masses 0.25 M_solar to 10 M_solar&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1955PASP...67..154H&quot; rel=&quot;noreferrer&quot;&gt;L. Henyey et al (1955), The Early Phases of Stellar Evolution&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Interesting reading I came across for somewhat higher mass protostellar formation:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1991ApJ...375..288P&quot; rel=&quot;noreferrer&quot;&gt;F. Palla and S. Stahler (1991), The evolution of intermediate-mass protostars. I - Basic results&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="276" LastActivityDate="2013-10-27T12:45:59.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="800" PostTypeId="1" CreationDate="2013-10-27T13:17:53.890" Score="11" ViewCount="6841" Body="&lt;p&gt;Please could someone advise the best method of finding the solar noon based on location?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pveducation.org/pvcdrom/properties-of-sunlight/suns-position&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://pveducation.org/pvcdrom/properties-of-sunlight/suns-position&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stuff on here has been helpful but I can't figure out what I should be finding out!&#xA;LST sounds about right, but I'm not 100% sure.&lt;/p&gt;&#xA;" OwnerUserId="319" LastEditorUserId="7472" LastEditDate="2017-07-19T10:42:24.063" LastActivityDate="2017-07-19T10:42:24.063" Title="Finding the solar noon based on longitude and latitude?" Tags="&lt;the-moon&gt;&lt;solar-system&gt;&lt;the-sun&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="801" PostTypeId="1" AcceptedAnswerId="805" CreationDate="2013-10-27T17:09:48.963" Score="-2" ViewCount="2114" Body="&lt;p&gt;For almost 3 centuries now the Nebular Hypothesis and Planetesimal Theory have become the preeminent explanations for how the Solar System and Planets evolved. Yet there is still no explanation for how Comets and Moons occur in the Solar System, and the Planetary explanation is still considered Theory. Is it possible that they are not correct? Yet since they hold such wide acceptance are actually preventing the discovery of how Comets, Moons, and Planets originate in a Solar System in an all inclusive explanation.&lt;/p&gt;&#xA;" OwnerUserId="257" LastActivityDate="2013-10-28T02:12:26.463" Title="Is it possible the Nebular Hypothesis and Planetesimal Theory are not correct?" Tags="&lt;the-moon&gt;&lt;planet&gt;&lt;comets&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ClosedDate="2013-10-28T11:45:43.863" ContentLicense="CC BY-SA 3.0" />
  <row Id="802" PostTypeId="1" CreationDate="2013-10-13T14:04:57.787" Score="6" ViewCount="1471" Body="&lt;p&gt;I am starting a study on automatic telescopes and the correction of their periodic error.&#xA;I found two ways of correcting it : with a CCD camera and manually by recording the error. I was wondering if it could be possible to model the error by a periodic signal in order to add it to the entry signal then correct it with a low-pass filter or an other filter ?&lt;/p&gt;&#xA;" OwnerDisplayName="user30394" LastEditorDisplayName="user96" LastEditDate="2013-10-28T16:20:10.843" LastActivityDate="2013-10-28T16:20:10.843" Title="Periodic error correction in automatic telescopes" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="803" PostTypeId="2" ParentId="720" CreationDate="2013-10-27T19:26:42.353" Score="2" Body="&lt;p&gt;While this shower is now largely over, the way to best observe meteors is the same for all showers in principle (including coming showers like the Leonids around November 17 and Geminids around December 14).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The name of most showers reference the name of the constellation their radiant (point in the sky the meteors seem to be coming from) lies in. In the case of the Orionids, this is the constellation of Orion, so look in the general direction of (and around) Orion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you suggest, meteors appear unexpectedly and move too fast to be caught with a telescope.  The best way to observe them is to lie down comfortably and use the naked eye.  The key word is &quot;comfortably&quot;: find a comfortable reclining garden chair or lilo and lie down.  That's one half of the comfort; the other is to stay warm.  Even when watching the Perseids in summer you're lying still under an open sky, and for the fall/winter showers mentioned here you'll really want to bundle up and perhaps take a thermos with a hot drink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The comfort is important - there may be several minutes between two meteors and hence lying down is a good way to relax while watching, which will allow you to stay out longer and see more.  It is definitely more relaxed than standing up with your head in your neck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One other tip for &quot;comfort&quot;: many people will enjoy this more in the company of other enthousiasts, and if you're one of them you should consider going out with a few other people.  It will allow you to discuss the meteors just sighted to pass the time between events, or even to help each other to see one (if they're not too fast).&lt;/p&gt;&#xA;" OwnerUserId="286" LastEditorUserId="6" LastEditDate="2013-10-30T19:41:35.620" LastActivityDate="2013-10-30T19:41:35.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="804" PostTypeId="2" ParentId="802" CreationDate="2013-10-28T00:22:25.703" Score="7" Body="&lt;p&gt;Quoting the most relevant parts from the &lt;a href=&quot;http://www.themcdonalds.net/richard/index.php?title=Astrophotography_Mounts:_Periodic_Error_Correction#Reducing_Periodic_Error&quot; rel=&quot;noreferrer&quot;&gt;Richard McDonald's Wiki on Astrophotography Mounts: Periodic Error Correction&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Periodic Error can be reduced to an acceptable level using a variety&#xA;  of techniques, only some of which are in the range of a beginner or&#xA;  mid-level astrophotographer.&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;&lt;strong&gt;Throw Money at the Problem&lt;/strong&gt;&lt;br /&gt;&lt;br /&gt;Very high-end mounts for astrophotography have very small periodic error because of the time&#xA;  and money spent on manufacturing high-precision gears. You can also&#xA;  buy higher-precision gear upgrades for some mounts. For the beginner,&#xA;  let's call this impractical. Buying a \$1000 replacement worm gear for&#xA;  your \$1000 mount is probably not a good balance.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;&lt;strong&gt;Don't Use Gears&lt;/strong&gt;&lt;br /&gt;&lt;br /&gt;This is really an extreme case of the &quot;throw money&quot; solution, mentioned just for fun. There are some&#xA;  very-high-end experimental drive systems showing up on the market that&#xA;  don't use worm gears, and that don't exhibit periodic error. Examples&#xA;  include &lt;a href=&quot;http://www.astrosysteme.at/&quot; rel=&quot;noreferrer&quot;&gt;direct drive&lt;/a&gt; systems and &lt;a href=&quot;http://www.chronosmount.com/&quot; rel=&quot;noreferrer&quot;&gt;harmonic drives&lt;/a&gt;. 'Way out of our&#xA;  price range.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;&lt;strong&gt;Autoguiding&lt;/strong&gt;&lt;br /&gt;&lt;br /&gt;A second guide camera and computer can be used to make frequent small corrections to the mount's pointing,&#xA;  and this can reduce or eliminate Periodic Error if the onset of the&#xA;  error is not too sudden. This is the subject of a &lt;a href=&quot;http://www.themcdonalds.net/richard/index.php?title=Astrophotography_Equipment:_Autoguiding&quot; rel=&quot;noreferrer&quot;&gt;separate article&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;&lt;strong&gt;Periodic Error Correction Feature&lt;/strong&gt;&lt;br /&gt;&lt;br /&gt;Training PEC with hand-control Most mounts intended for astrophotography include a&#xA;  feature called Periodic Error Correction (PEC), which can be used&#xA;  alone, or in conjunction with Autoguiding. PEC is used in two phases:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;strong&gt;Training&lt;/strong&gt;. During this phase you use the control panel or menus to say &quot;Hey! Pay attention to this!&quot; to the mount, then you manually&#xA;  keep a star perfectly centred for several worm periods. You do this by&#xA;  centering a star with a high-magnification eyepiece that includes a&#xA;  cross-hair reticle. You then stare at the star for 10 to 15 minutes&#xA;  and use the mount's hand controller to manually make the small&#xA;  adjustments necessary to keep it perfectly centred on the cross hair.&#xA;  The mount records the error corrections you supply, remembering where&#xA;  in the worm position each one was needed. By training through more&#xA;  than one worm cycle, the mount can record an average correction, in&#xA;  case you reacted slowly or over-reacted.&lt;/li&gt;&#xA;  &lt;li&gt;&lt;strong&gt;Playback&lt;/strong&gt;. Once you have trained the mount, you can turn on PEC. The mount will &quot;play back&quot; the recorded error correction information&#xA;  by slightly changing the speed of the Right Ascension drive to move&#xA;  ahead or back each time your manual error corrections did the same.&#xA;  This will cancel the periodic error, resulting in a smoother track.&lt;/li&gt;&#xA;  &lt;/ol&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/R69iL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;sup&gt;Training PEC with&#xA;  hand-control: Periodic Error Correction (PEC) button on the control&#xA;  panel of an equatorial mount.&lt;/sup&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The manual training phase is quite tedious and a more modern&#xA;  alternative is to use a camera and computer - usually the same one you&#xA;  will use for auto-guiding - to track a star while the mount records&#xA;  the training information.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;There is even specialized software available to help collect Periodic&#xA;  Error data, analyze and smooth it, and upload it to the mount. You do&#xA;  not need such software, as the manual techniques mentioned above will&#xA;  work just fine. However, it makes a tedious job simple and pleasant,&#xA;  and you may find it a worthwhile investment. I use &lt;a href=&quot;http://www.ccdware.com/products/pempro/&quot; rel=&quot;noreferrer&quot;&gt;PEMPRO&lt;/a&gt; and am&#xA;  very impressed with it. It's not free, but it is inexpensive and works&#xA;  very well (and includes another feature to help achieve perfect polar&#xA;  alignment).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Quoted text and photograph source, copyright and courtesy of &lt;a href=&quot;http://www.themcdonalds.net/richard/index.php?title=Richard_McDonald:_Main_Page&quot; rel=&quot;noreferrer&quot;&gt;Richard McDonald&lt;/a&gt;, no copyright infringement intended. I would also highly recommend reading same author's &lt;a href=&quot;http://www.themcdonalds.net/richard/index.php?title=Astrophotography_Equipment:_Autoguiding&quot; rel=&quot;noreferrer&quot;&gt;extensive article on Autoguiding&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2013-10-28T00:22:25.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="805" PostTypeId="2" ParentId="801" CreationDate="2013-10-28T02:00:55.307" Score="5" Body="&lt;p&gt;TidalWave says it perfectly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;&quot;What is the question? So far, all I can think of as an answer is explaining the meaning of scientific theory as an integral part of the scientific method. Of course it is possible it is not correct, science doesn't function on dogmatism. Theories are frequently corrected, expanded, or even completely dismissed. Having a status of the most accepted theory often means it's just the second best thing to the one that'll withstand the trial of reality better. Theory != Axiom.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The nebular hypothesis is simply the best model that fits our observations - that's how the scientific method works. In the last decade, we have been discovering and analysing increasing numbers of planetary and protoplanetary systems. Our models of the formation of these systems are continually refined as both our observations improve and our computing power increases.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;&quot;When I ask Senior Astronomers in the U.S. who specialize in Planetary Evolution to review the concept, they personally insult me for even asking rather than employ any scientific theory or scientific methods to evaluate it.&quot;&lt;/em&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're insulting &lt;em&gt;them&lt;/em&gt;, by asking them to justify why the work they base their livelihood upon isn't a load of rubbish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;You cannot justifiably criticize an accepted scientific theory without coming up with an equally plausible idea&lt;/strong&gt; (unless you manage to &lt;em&gt;prove&lt;/em&gt; that something is wrong). And unless someone does come up with a better idea, the nebular theory will continue to expand and evolve to better match our observations. &lt;em&gt;That&lt;/em&gt; is the scientific process.&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="276" LastEditDate="2013-10-28T02:12:26.463" LastActivityDate="2013-10-28T02:12:26.463" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="806" PostTypeId="1" AcceptedAnswerId="823" CreationDate="2013-10-28T10:24:36.300" Score="6" ViewCount="160" Body="&lt;p&gt;When was the nature and size of asteroids discovered? I was under the impression that it was not until the 1980s or 1990s with deeper space probes and Hubble imagery that we learned the nature of asteroids. However, today the wonderful &lt;code&gt;fortune&lt;/code&gt; UNIX script gave this quote from a man who died in 1910:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Noise proves nothing. Often a hen who has merely laid an egg cackles&#xA;  as if she laid an asteroid.  -- Mark Twain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How would Samuel Langhorne Clemens have known enough about asteroids to make such a statement in the early 20th century?&lt;/p&gt;&#xA;" OwnerUserId="209" LastEditorDisplayName="user96" LastEditDate="2013-10-28T10:41:40.610" LastActivityDate="2013-10-30T13:53:11.100" Title="How much did we know about asteroids at the beginning of the 20th century?" Tags="&lt;observational-astronomy&gt;&lt;asteroids&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="807" PostTypeId="1" CreationDate="2013-10-28T16:16:43.790" Score="16" ViewCount="1157" Body="&lt;p&gt;This is a follow up to: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/328/is-there-a-theoretical-maximum-size-limit-for-a-star&quot;&gt;Is there a theoretical maximum size limit for a star?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer there talks about the radiation pressure preventing a star from forming. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/a172j.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What reaction is causing this radiation pressure and why does this potentially prevent a star from forming?&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2013-10-31T21:41:32.650" Title="What is radiation pressure and how does it prevent a star from forming?" Tags="&lt;star&gt;&lt;formation&gt;&lt;radiation&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="809" PostTypeId="1" AcceptedAnswerId="813" CreationDate="2013-10-29T03:01:08.883" Score="11" ViewCount="1496" Body="&lt;p&gt;If supermassive black holes possess the gravity required to prevent the escape of photons and other mass particles, why do jets and accretion disks form? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They seem to indicate that something is coming &lt;em&gt;out&lt;/em&gt; of the black hole, while mathematics and physics seem to tell us that this is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would seem that supermassive black holes only exert gravity towards themselves. &lt;/p&gt;&#xA;" OwnerUserId="328" LastEditorUserId="37191" LastEditDate="2021-02-17T21:13:13.377" LastActivityDate="2021-02-17T21:13:13.377" Title="Why do black holes have jets and accretion disks?" Tags="&lt;black-hole&gt;&lt;supermassive-black-hole&gt;&lt;astronomical-jet&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="810" PostTypeId="2" ParentId="807" CreationDate="2013-10-29T07:31:53.120" Score="10" Body="&lt;p&gt;Radiation pressure is nothing but electromagnetic interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a hydrogen atom hit by a stream of photons coming from the same direction. Although the atom as a whole is neutral, the electron and the proton are physically displaced, forming a dipole, i.e. a positive-negative charge couple. Some of the photons therefore scatter against the dipole transferring to it some momentum. So the atom start moving in the same direction as the photons. If the photons are in the ultraviolet, the electron can be exited to higher orbitals and possibly stripped from the atom. In this case the scattering is even more efficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine a star surrounded by a layer of hydrogen. Gravity attracts the layer towards the star. The photons emitted by the star try to push the hydrogen atoms away from it, through the electromagnetic force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very massive stars are very luminous and hot, which means that they emit a lot of ultraviolet photons. When the pressure transferred from the photons to the layer is larger than the gravitational attraction, then the layer begins expanding, effectively stopping growth of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the figure posted by the OP there is also dust. I don't know the details of photons-dust-gas interactions (we need a stellar atmosphere expert, I guess), but the basic principle is nonetheless the same.&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="45" LastEditDate="2013-10-31T12:29:24.070" LastActivityDate="2013-10-31T12:29:24.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="811" PostTypeId="2" ParentId="500" CreationDate="2013-10-29T08:00:26.650" Score="2" Body="&lt;p&gt;proof exists that something can orbit around a black hole.&#xA;The star S2 orbits around the central 4 million solar mass black hole of our galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/S2_%28star%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/S2_(star)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if you find an object orbiting around your target object, then you can &#xA;&lt;a href=&quot;http://spacemath.gsfc.nasa.gov/weekly/5Page68.pdf&quot; rel=&quot;nofollow&quot;&gt;calculate the mass of your target object.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, for something orbiting around a black hole, it is very difficult to fall in. This is for the same reason why it is easier to send a probe from Earth to Mars (outward from Sun), than from Earth to Mercury (inward toward Sun), and also the same reason why it is impractical to get rid of toxic waste by throwing it into the Sun. It would require an enormous amount of energy to reach the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www2.jpl.nasa.gov/basics/index.php&quot; rel=&quot;nofollow&quot;&gt;See basics of space flight for details.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="302" LastActivityDate="2013-10-29T08:00:26.650" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="812" PostTypeId="2" ParentId="500" CreationDate="2013-10-29T08:17:03.133" Score="0" Body="&lt;p&gt;I think blackholes are dead star? Is not it? And i believe our nearest star is sun , and there is difference between star and blackholes(dead star) . And i strongly believe that blackholes ( dead star ) they&#xA; attract everything around themselves into the center.  And after that they must change the property of that object(which is at center), and then spread it's anu parts into the universe. :)&lt;/p&gt;&#xA;" OwnerUserId="330" LastActivityDate="2013-10-29T08:17:03.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="813" PostTypeId="2" ParentId="809" CreationDate="2013-10-29T08:23:29.660" Score="14" Body="&lt;p&gt;When it comes to accretion disks, nothing is coming &lt;em&gt;out&lt;/em&gt; of the black hole. That's just orbiting matter, though it is swirled around a bit by frame dragging. Even at high gravity, the ability to orbit around a massive body still exists. The gravitational force is already being &quot;used up&quot; to cause the orbiting (it accounts for the centripetal force), so there is no need for the gas to fall in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for jets, as far as I can tell there is no single explanation (I am not sure of this). One candidate explanation is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Blandford%E2%80%93Znajek_process&quot; rel=&quot;nofollow noreferrer&quot;&gt;Blandford-Znajek process&lt;/a&gt;&lt;sup&gt;1&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following image is from &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0393312763&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Black Holes and Time Warps: Einstein's Outrageous Legacy&lt;/em&gt;, by Kip S. Thorne&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jwVJ8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, most black holes rotate, and sometimes the intense rotation can cause forces that overcome gravity, even by a few orders of magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a black hole spins, magnetic field lines anchored to it&lt;sup&gt;2&lt;/sup&gt; spin along with it. Plasma (from the accretion disk) is then flung out along these lines, similar to what happens when you put a marble in a conical cup and rotate it. This is depicted in the first image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second image, current passes through the field lines (I do not understand this one as well as the first, however &lt;a href=&quot;https://physics.stackexchange.com/questions/20871/blandford-znajek-process-why-how-does-the-current-flow-along-the-magnetic-field&quot;&gt;this post has a reasonable explanation&lt;/a&gt;), accelerates plasma with a mechanism similar to an &lt;a href=&quot;http://en.wikipedia.org/wiki/Railgun&quot; rel=&quot;nofollow noreferrer&quot;&gt;electromagnetic railgun&lt;/a&gt;. This is another way of creating jets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the energy here comes from the rotational energy of the BH, &lt;em&gt;not&lt;/em&gt; the mass-energy of the &quot;contents&quot; of the BH (which is lost to the universe unless we consider Hawking radiation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I shall have a closer look at the paper when I have time and update the answer accordingly. Comments appreciated)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1. &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?db_key=AST&amp;amp;bibcode=1977MNRAS.179..433B&amp;amp;letter=.&amp;amp;classic=YES&amp;amp;defaultprint=YES&amp;amp;whole_paper=YES&amp;amp;page=433&amp;amp;epage=433&amp;amp;send=Send%20PDF&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Blandford, R. D., &amp;amp; Znajek, R. L. (1977). Electromagnetic extraction of energy from Kerr black holes. &lt;em&gt;Monthly Notices of the Royal Astronomical Society, 179, 433-456.&lt;/em&gt;&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2. While the no-hair theorem forbids a naked black hole from posessing magnetic field lines, one with an accretion disk may have them as the field lines cannot &quot;escape&quot; through the disk.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2013-11-15T08:49:20.690" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="814" PostTypeId="2" ParentId="79" CreationDate="2013-10-29T08:49:08.163" Score="4" Body="&lt;p&gt;This question already has a very good answer, I would just like to add a few details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astro.ucla.edu/~wright/BBhistory.html&quot; rel=&quot;nofollow&quot;&gt;http://www.astro.ucla.edu/~wright/BBhistory.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Says here that when the universe was 10^-33cm in diameter, its temperature was 10^32K. Therefore that should be the absolute max temperature reachable in this universe, and so the max temperature of a star should be below that; very interesting what Guillochon said above, that neutrinos carry away excess energy above 10^12K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The color of a star gives away its temperature. It is interesting to note that the corona of a star including our Sun can be well over a million K even though the surface temperature of our star is around 6000 K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Corona&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Corona&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in stellar cores, hydrogen fusion into helium starts at 3 million K, while carbon fusion starts at above 500 million K, and silicon fusion starts at over 2700 million K for comparison.&lt;/p&gt;&#xA;" OwnerUserId="302" LastActivityDate="2013-10-29T08:49:08.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="815" PostTypeId="1" AcceptedAnswerId="816" CreationDate="2013-10-29T09:04:37.960" Score="7" ViewCount="429" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/F479M.png&quot; alt=&quot;gravitational lensing around a black-hole&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light gets bent around high-gravity sources, such as a black-hole. A source of light, say &lt;strong&gt;A&lt;/strong&gt;, is directly observed in its 'true' position and some of its light is bent around the black-hole, &lt;strong&gt;X&lt;/strong&gt;, which gives a second observation of A, called &lt;strong&gt;A'&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does gravitational lensing cause the amount of emissions (detected as coming from the (point/position/direction) of &lt;strong&gt;X&lt;/strong&gt;) to far exceed anything else in that vicinity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this then make the 'black-hole' by far the brightest, noisiest thing to be observed? (kind of like waving &amp;amp; shouting in a quiet room)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[edit - based on astromax's answer]&lt;br/&gt;&#xA;I follow what is meant regarding super massive black-holes.&lt;br/&gt;&#xA;Consider the following scenario: A suitable massive sun on dying contracts into a (small) black-hole. This black-hole will now cause a lensing-effect, but since it is small the amount of emissions so 'lensed' around it is very small and maybe too small to detect? or, it is detected but erroneously ascribed to be a type of sun or other type of emission source?&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditDate="2013-10-30T12:08:49.477" LastActivityDate="2013-10-30T12:08:49.477" Title="Does gravitational lensing cause a black-hole to be the main 'source' of light in a given area?" Tags="&lt;black-hole&gt;&lt;gravitational-lensing&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="816" PostTypeId="2" ParentId="815" CreationDate="2013-10-29T12:12:07.113" Score="4" Body="&lt;p&gt;So the question is, Can black holes be really be extraordinarily bright due to lensing of background objects? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's first also specify that these black holes must certainly be super massive. If not, we're talking about stellar mass black holes (or intermediate mass black holes), and the lensing signal would be much weaker. More mass = more lensing generally speaking (there are some exceptions, e.g. - microlensing, where the lensing is done by stars in our own Milky Way galaxy and the signal is enhanced because the source and lens are aligned nearly perfectly). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an important statement to make because virtually &lt;a href=&quot;http://science1.nasa.gov/astrophysics/focus-areas/black-holes/&quot; rel=&quot;nofollow&quot;&gt;every galaxy has been identified with a super massive black hole at its center&lt;/a&gt;. The reverse is also true - every super massive black hole is associated with a galaxy. It's also important to note that, realistically speaking, galaxies have total masses of about $\sim10^{11} - 10^{12} M_{\odot}$, whereas super massive black holes typically have masses of millions to hundreds of millions (maybe even billions) of solar masses. This is a smallish fraction of the total mass. This means that most of the lensing will be done by the galactic halo and not the super massive black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we modeled the center of the lens galaxy as a point-lens, and the galactic halo as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Singular_isothermal_sphere_profile&quot; rel=&quot;nofollow&quot;&gt;singular isothermal sphere&lt;/a&gt;, the relevant question I would ask is what are their Einstein radii (which is a measure of how effective or efficient they are at lensing) individually, and what are they in combination? Essentially, how much does the existence of a central super massive black hole matter to the system as a whole. Strong lensing features (arcs, rings, or multiple images) generally occur at around the Einstein radius of the object. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My best guess:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite honestly I don't see the central super massive black hole mattering all that much when it comes to lensing. Many of these mass profile models for the lens galaxy halo are singular, or rise very rapidly to a central core. Furthermore, I've never really heard of a situation where a lone super massive black hole (not associated with a galaxy - call it a 'rogue' smbh if you will) has been found floating around in space to do this sort of lensing. They generally hide at the centers of galaxies, or show themselves only if they're actively accreting material. Correct me if one has been found (maybe it would come from a merging of two smbh's where one is kicked out of the system).    &lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditDate="2013-10-29T21:44:13.217" LastActivityDate="2013-10-29T21:44:13.217" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="817" PostTypeId="2" ParentId="7" CreationDate="2013-10-29T16:56:03.000" Score="2" Body="&lt;p&gt;Don't forget that if a black hole has less than the current stable mass of a black hole (3 solar masses) then it evaporates - transforming its mass in radiation, in which case it would give off light, mostly X rays and gamma, at an increasing rate as its mass decreases, until the entire black hole is turned into a flash of hard radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hawking_radiation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this light is the mass of the black hole escaping in the form of the most basic form of energy.&lt;/p&gt;&#xA;" OwnerUserId="302" LastEditorUserId="19" LastEditDate="2013-10-29T16:57:29.377" LastActivityDate="2013-10-29T16:57:29.377" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="818" PostTypeId="2" ParentId="7" CreationDate="2013-10-29T17:29:44.253" Score="11" Body="&lt;p&gt;I like to think of this in terms of &lt;a href=&quot;http://en.wikipedia.org/wiki/Escape_velocity&quot; rel=&quot;noreferrer&quot;&gt;escape velocity&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Escape velocity is the speed needed to escape the gravitational pull of a given object. For the Earth, that speed is 11.2 km/second (Mach 34!). When rockets blast off from Earth, they aren't trying to achieve a certain height or altitude, they're trying to reach a certain speed, the escape velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once a rocket reaches 11.2 kips*, it has attained &lt;a href=&quot;http://en.wikipedia.org/wiki/Escape_velocity#List_of_escape_velocities&quot; rel=&quot;noreferrer&quot;&gt;the speed needed&lt;/a&gt; to leave the Earth completely. If a rocket fails to attain that speed, regardless of its height, it will fall back to the Earth. (You can imagine a magical balloon that slowly lifts you up into space, up past the ISS and most satellites, and then you let go: since you aren't going fast enough, you will fall back down, past all the satellites, and crash into the earth.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smaller gravitational bodies, like the moon, have smaller escape velocities. That's why the lunar landers were able to leave the moon with such a &lt;a href=&quot;http://en.wikipedia.org/wiki/Apollo_Lunar_Module#Ascent_stage&quot; rel=&quot;noreferrer&quot;&gt;small ascent stage&lt;/a&gt;, compared to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Saturn_v&quot; rel=&quot;noreferrer&quot;&gt;massive Saturn V&lt;/a&gt; it took to leave Earth: they only had to go 2.4 km/second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HEMM8.gif&quot; alt=&quot;Saturn V vs Lunar Ascent Module &quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O2jvq.jpg&quot; alt=&quot;Lunar Ascent Module&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To escape the Sun, you'd have to go 617.5 km/second!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately for us, light goes faster than 617.5 kips, so we're able to see the light created on the Sun. However, as you increase the mass of an object, eventually the escape velocity would meet or exceed 299,792km/s, the speed of light. At that point not even light itself can go fast enough to escape the gravity well, and will always be pulled back down into the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Short for &quot;&lt;strong&gt;ki&lt;/strong&gt;&amp;#8203;lometers &lt;strong&gt;p&lt;/strong&gt;&amp;#8203;er &lt;strong&gt;s&lt;/strong&gt;&amp;#8203;econd&quot;&lt;/p&gt;&#xA;" OwnerUserId="158" LastEditorUserId="158" LastEditDate="2013-10-29T18:10:26.287" LastActivityDate="2013-10-29T18:10:26.287" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="819" PostTypeId="2" ParentId="459" CreationDate="2013-10-30T07:57:31.027" Score="1" Body="&lt;p&gt;another problem for an amateur looking for black holes is that the telltale signature of a black hole, coming in the form of continuous or bursts of X-rays and gamma rays, cannot be observed from the surface of the Earth, because the atmosphere (thankfully) protects us from that sort of stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ak1TG.gif&quot; alt=&quot;Transparency of Earth atmosphere to EM spectrum&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that is why you need to go in space, or on the surface of a planet without atmosphere (the Moon would be suitable, Mars should do as well, since the atmospheric pressure is 6 millibars there and space rays come all the way to the surface, compared to 700-1000 millibars on Earth at sealevel), or send up your own space telescope, such as the likes of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://xmm.esac.esa.int/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://xmm.esac.esa.int/ XMM Newton&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chandra.harvard.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://chandra.harvard.edu/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and especially&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nasa.gov/mission_pages/nustar/main/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.nasa.gov/mission_pages/nustar/main/index.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nustar - delivering the X-ray goods.&lt;/p&gt;&#xA;" OwnerUserId="302" LastActivityDate="2013-10-30T07:57:31.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="820" PostTypeId="1" AcceptedAnswerId="829" CreationDate="2013-10-30T10:03:43.313" Score="8" ViewCount="556" Body="&lt;p&gt;we know that stars fuse hydrogen into helium starting at 3 MK; 13 MK in the Sun's core; &#xA;carbon fusion starts at above 500 million K, and silicon fusion starts at over 2700 million K for comparison;&#xA;we know fusion stops at iron, because a star has to use more energy to fuse that than it gets back;&#xA;so heavier elements are created mostly in a supernova (but also possible in small quantities by special processes like neutron capture);&#xA;finally sun-like stars end up as white dwarfs, bigger stars as neutron stars, quark stars, black holes;&#xA;and black holes ultimately convert themselves into radiation, in the far distant future when the stable black hole mass limit goes up high enough that even the most massive black holes evaporate;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Vacuum_Diagrams&quot;&gt;http://en.wikipedia.org/wiki/Vacuum_Diagrams&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so my question is, will it be like Stephen Baxter said, that in the future only radiation will be left in the universe?&#xA;Specifically, is there a natural process out there by which hydrogen is spewed into the cosmos, converted back from heavier elements, to regenerate the fuel for stars so that they may shine in the far distant future as well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course we don't need to worry about this for the time being. This is only considering our concern with what will be 10^70 years from now.&lt;/p&gt;&#xA;" OwnerUserId="302" LastActivityDate="2013-11-05T11:04:00.480" Title="Is there a natural process by which hydrogen is generated from heavier elements in the cosmos?" Tags="&lt;cosmology&gt;&lt;universe&gt;&lt;radiation&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="821" PostTypeId="2" ParentId="820" CreationDate="2013-10-30T12:56:25.733" Score="2" Body="&lt;p&gt;It seems that &lt;a href=&quot;http://arxiv.org/pdf/astro-ph/0112486v2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;primordial black holes&lt;/a&gt; produce anti protons, and it is implied in the linked article that they are capable of producing all kinds of other particles. So maybe even protons.&lt;/p&gt;&#xA;&lt;p&gt;Also, I guess that during natural fission or nucleues collision reactions, there may be fragments being produced that are also single protons.&lt;/p&gt;&#xA;&lt;p&gt;Cosmic rays seem to &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_ray#Types&quot; rel=&quot;nofollow noreferrer&quot;&gt;consist primarily of protons&lt;/a&gt;. The question is, whether these protons were produced in the big bang, or if they stem from other sources. The article states that lots of cosmic rays stem from supernovae. However, this does not answer the question if the protons were produced in the supernova from heavier elements.&lt;/p&gt;&#xA;&lt;p&gt;Since I am not an astrophysicist, I am gladly waiting for comments or other answers!&lt;/p&gt;&#xA;&lt;p&gt;Edit: I read about another mechanism on how to create electrons and protons: &lt;a href=&quot;http://en.wikipedia.org/wiki/Matter_creation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Two-Photon interaction&lt;/a&gt;. I cite the Wikipedia article:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The law of conservation of energy sets a minimum photon energy required for creation of a pair of fermions: this threshold energy must be greater than the total rest energy of the fermions created. To create an electron-positron pair the total energy of the photons must be at least 2mec2 = 2 × 0.511 MeV = 1.022 MeV (me is the mass of one electron and c is the speed of light in vacuum), an energy value that corresponds to soft gamma ray photons. The creation of a much more massive pair, like a proton and antiproton, requires photons with energy of more than 1.88 GeV (hard gamma ray photons).&lt;/p&gt;&#xA;&lt;p&gt;First calculations of rate of e+–e− pair production in photon-photon collision was done by Lev Landau in 1934.&lt;a href=&quot;http://arxiv.org/pdf/astro-ph/0112486v2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt; It was predicted that the process of e+–e− pair creation (via collisions of photons) dominates in collision of ultra-relativistic charged particles—because those photons are radiated in narrow cones along the direction of motion of original particle greatly increasing photon flux.&lt;/p&gt;&#xA;&lt;p&gt;In high-energy particle colliders, matter creation events have yielded a wide variety of exotic heavy particles precipitating out of colliding photon jets (see two-photon physics). Currently, two-photon physics studies creation of various fermion pairs both theoretically and experimentally (using particle accelerators, air showers, radioactive isotopes, etc.).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So, in small amounts electron-positron pairs and proton anti-proton pairs should be created by soft and hard gamma radiation respectively (or other Fermion particles). The problem here again is that this event will happen only very rarely, not significantly producing new matter. The article goes on to say that this was the method in which matter was created during the Big Bang. But only one in &lt;span class=&quot;math-container&quot;&gt;$10^{10}$&lt;/span&gt; Fermions would have survived to form the current matter in the universe.&lt;/p&gt;&#xA;&lt;p&gt;All in all these processes will probably be not enough to form new stars.&lt;/p&gt;&#xA;" OwnerUserId="50" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2013-11-05T11:04:00.480" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="822" PostTypeId="1" AcceptedAnswerId="826" CreationDate="2013-10-30T13:20:40.497" Score="26" ViewCount="12763" Body="&lt;p&gt;How far is the Earth/Sun above/below the galactic plane, and is it heading toward/away from it?&#xA;We know the Sun is orbiting the galactic center every 225 million years.&#xA;But how far are we from the galactic plane, and is the inclination of the plane of the Sun's orbit around the galactic center, as compared to the galactic plane, known?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PERqj.jpg&quot; alt=&quot;Direction of Sun&amp;#39;s motion&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to this drawing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8ODdl.gif&quot; alt=&quot;Direction of Sun&amp;#39;s motion&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the Sun is traveling approximately in the ecliptic plane.&#xA;However, the ecliptic plane is steeply inclined to the galactic plane, so if this drawing is correct (from voyager.jpl.nasa.gov) then this would mean the plane of the Sun's orbit around the galactic center is also quite inclined.&#xA;And thus my question, what is the distance to the galactic plane, are we heading toward/away from it, and what is the inclination of the orbital plane of our Sun/solar system to the galactic plane? 3 questions I guess.&lt;/p&gt;&#xA;" OwnerUserId="302" LastActivityDate="2016-12-25T20:20:24.043" Title="How far is the Earth/Sun above/below the galactic plane, and is it heading toward/away from it?" Tags="&lt;the-sun&gt;&lt;galaxy&gt;&lt;milky-way&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="823" PostTypeId="2" ParentId="806" CreationDate="2013-10-30T13:53:11.100" Score="3" Body="&lt;p&gt;Well as late as 1942, with a reference to book &lt;em&gt;The Little Prince,&lt;/em&gt; by Antoine de Saint-Exupéry, we didn't even know the correct shape of asteroids, believing them to be round like planets. It was in 1971 that we got a look at the moons of Mars that we got a look at what an asteroid would really look like, although we are now thinking Phobos and Deimos were formed with an impacted. The best information for what your asking can come from XKCD's &lt;em&gt;What If&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Before the 1970s, it was common for science fiction to assume small&#xA;  asteroids would be round, like planets.[4] The Little Prince took this&#xA;  a step further, imagining an asteroid as a tiny planet with gravity,&#xA;  air, and a rose. There's no point in trying to critique the science&#xA;  here, because (1) it's not a story about asteroids, and (2) it opens&#xA;  with a parable about how foolish adults are for looking at everything&#xA;  too literally.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://what-if.xkcd.com/68/&quot; rel=&quot;nofollow&quot;&gt;http://what-if.xkcd.com/68/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we had we had fifteen asteroids discovered by 1851, and until 1891, all asteroids had been discovered by telescope and eye alone, at which point Max Wolf pioneered the use of astrophotography to detect asteroids. Max Wolf himself ended up credited with the discovery of 248 asteroids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while I don't know when Mark Twain made that quote but as far as refering to asteroid, I'll quote wikipedia for the term 'asteroid' &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Traditionally, small bodies orbiting the Sun were classified as&#xA;  asteroids, comets or meteoroids, with anything smaller than ten metres&#xA;  across being called a meteoroid.[21][22] The term &quot;asteroid&quot; is&#xA;  ill-defined. It never had a formal definition, with the broader term&#xA;  minor planet being preferred by the International Astronomical Union&#xA;  from 1853 on. In 2006, the term &quot;small Solar System body&quot; was&#xA;  introduced to cover both most minor planets and comets.[23]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The minor planets beyond Jupiter's orbit are sometimes also called&#xA;  &quot;asteroids&quot;, especially in popular presentations.[30] However, it is&#xA;  becoming increasingly common for the term &quot;asteroid&quot; to be restricted&#xA;  to minor planets of the inner Solar System.[26]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Asteroid&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Asteroid&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mark Twain could have been referring to quite a few objects with the unclear definition of asteroids.&lt;/p&gt;&#xA;" OwnerUserId="340" LastActivityDate="2013-10-30T13:53:11.100" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="824" PostTypeId="2" ParentId="809" CreationDate="2013-10-30T17:55:49.613" Score="3" Body="&lt;p&gt;An accretion disc is stuff orbiting around the black hole as it is falling into it.  Or it would be possible that the particles are in some sort of orbit around the black hole.  This disc would be outside of the event horizon so it isn't really &quot;in&quot; the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The jets are formed in a similar fashion.  Particles spinning into the black hole generate radiation which gets appears to be &quot;emitted&quot; by the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things only appear to be coming out of the black hole but they are really happening outside of the event horizon.  The disc and jets are formed by the interaction of particles as they are moving around the black hole and entering the event horizon.&lt;/p&gt;&#xA;" OwnerUserId="281" LastActivityDate="2013-10-30T17:55:49.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="825" PostTypeId="1" CreationDate="2013-10-30T18:44:43.610" Score="3" ViewCount="54" Body="&lt;p&gt;It's difficult for most people to grasp how distances between the Earth and distant celestial bodies are calculated. For example, when someone tells me that a given star is 100 light years away, how would I determine if that were accurate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that my question boils down to three general use cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How can I calculate the distance between Earth and a local body in our solar system such as the sun, Jupiter, our moon, or Neptune?&lt;/li&gt;&#xA;&lt;li&gt;How can I calculate the distance between Earth and the nearest star outside of our own solar system, Alpha Centauri?&lt;/li&gt;&#xA;&lt;li&gt;How can I calculate the distance between Earth and an extremely distant object, such as a star 100-1,000 light-years away?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I've heard of using red shift to calculate large distances, but what exactly is this and how does it work?&lt;/p&gt;&#xA;" OwnerUserId="328" LastActivityDate="2013-10-30T18:44:43.610" Title="How is the distance from Earth to distant celestial bodies calculated?" Tags="&lt;distances&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2013-10-30T20:53:18.243" ContentLicense="CC BY-SA 3.0" />
  <row Id="826" PostTypeId="2" ParentId="822" CreationDate="2013-10-30T20:21:11.410" Score="8" Body="&lt;p&gt;According to this article, we are somewhere between 75 and 101 light years above the plane and getting closer.  So we won't be crossing it for quite some time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://earthsky.org/astronomy-essentials/will-earth-pass-through-galactic-plane-in-2012&quot;&gt;http://earthsky.org/astronomy-essentials/will-earth-pass-through-galactic-plane-in-2012&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This page has references to the research in regards to the calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nature.com/nature/journal/v316/n6030/abs/316706a0.html&quot;&gt;http://www.nature.com/nature/journal/v316/n6030/abs/316706a0.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="281" LastActivityDate="2013-10-30T20:21:11.410" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="827" PostTypeId="2" ParentId="820" CreationDate="2013-10-30T20:21:55.080" Score="6" Body="&lt;p&gt;There are a couple of relevant questions one would want to ask:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Do protons decay, and if so, what do they decay into? The answer &lt;a href=&quot;http://en.wikipedia.org/wiki/Proton_decay&quot; rel=&quot;noreferrer&quot;&gt;appears to be no&lt;/a&gt;, or at least the theoretical lifetime of the proton must increase as a results of these experiments. If they do, eventually the universe could end up in a state of radiation (and dark energy, and dark matter, unless they also decay). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Is Hydrogen a bi-product of any natural decay process? Below is a table of all known nuclides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/budpw.png&quot; alt=&quot;isotopes&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the majority of the elements (not necessarily by number or mass in the universe) do decay through some type of process. There exists a 'stable' ridge (called the island of stability, surrounded by the sea of instability) of elements which will happily exist forever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, which modes of decay produce protons (Hydrogen nuclei)? Well, there is proton decay (not the proton itself decaying), which is colored in red, though I have to admit that I don't know exactly what this refers to. Fission bi-products are gamma rays (high energy photons), neutrons, and daughter nuclei (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Decay_chain&quot; rel=&quot;noreferrer&quot;&gt;Decay chain&lt;/a&gt;). Though, I should mention that free neutrons produced from this type of radioactive decay are not long lived, decaying into a &lt;strong&gt;proton&lt;/strong&gt; and an electron (this process takes on average approximately 11 minutes). By this logic, also isotopes which decay by emitting neutrons, colored purple, would also eventually produce protons. $\beta^{-}$ and $\beta^{+}$ refer to the beta decay process, where the minus sign refers to the emission of an electron and the plus sign refers to the emission of a positron (the anti-particle of the electron). $\alpha$ decay is the emission of a Helium nucleus, &lt;a href=&quot;http://en.wikipedia.org/wiki/Isotopes_of_helium&quot; rel=&quot;noreferrer&quot;&gt;which is stable&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, given that there are ways for heavy elements to naturally produce protons, the question I would ask is what is the rate of these processes in the universe compared to fusion processes occurring at the centers of stars. I'm not sure that I could give you an answer to this question (or even point you to the appropriate material), but in principle these rates are known. I'd imagine that it'd be quite a lot of bookkeeping to get it correct. &lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2013-10-30T20:21:55.080" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="828" PostTypeId="1" CreationDate="2013-10-30T23:20:08.360" Score="5" ViewCount="104" Body="&lt;p&gt;Why do some nebulae look like the billowing clouds? Wouldn't that require something like a viscosity in empty space?&lt;/p&gt;&#xA;" OwnerUserId="342" LastEditorUserId="19" LastEditDate="2013-10-31T00:14:54.267" LastActivityDate="2013-11-21T05:25:00.780" Title="Why is there billowing dust in nebulae?" Tags="&lt;star&gt;&lt;nebula&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="829" PostTypeId="2" ParentId="820" CreationDate="2013-10-31T03:41:00.173" Score="5" Body="&lt;p&gt;It is not possible to split a larger nucleus into hydrogen nuclei without &lt;em&gt;expending a greater amount of energy that you receive back&lt;/em&gt;. This is because Hydrogen has (by far) the lowest &lt;a href=&quot;http://csep10.phys.utk.edu/astr162/lect/energy/bindingE.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;nuclear binding energy&lt;/a&gt; per nucleon (protium has zero nuclear binding energy, though deuterium and tritium do have some). Therefore, such a process would &lt;strong&gt;decrease the entropy of the universe&lt;/strong&gt; - a violation of the laws of thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/T4JAA.png&quot; alt=&quot;Nuclear binding energy curve&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could not speak for if these laws would still hold true were there a &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Big_Crunch&quot; rel=&quot;nofollow noreferrer&quot;&gt;big crunch&lt;/a&gt;&quot; (though current observations support an expanding universe).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a scenario called &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_death_of_the_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;heat death&lt;/a&gt;, where the universe simply has no energy left to do anything - that is, everything is completely uniform. There would be no gradients or anisotropies in the distribution of energy or matter.&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="302" LastEditDate="2013-10-31T09:00:16.157" LastActivityDate="2013-10-31T09:00:16.157" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="830" PostTypeId="2" ParentId="807" CreationDate="2013-10-31T12:47:56.257" Score="3" Body="&lt;p&gt;To add to @Frencesco's answer, if you're talking about why temperature prevents a star from forming in the first place, the same principles apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a star to form you need a cloud of gas to cool off significantly so that it can condense, much further along it can begin fusing hydrogen at its core. For it to cool off and begin collapsing it needs radiates away energy (this is partly why there are thought to be no dark matter stars). This process can take time. However, once the star has formed, ionizing radiation works to act against gravitational collapse, transferring energy/momentum back into the gas.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summary: You first need to radiate away energy before you can create a star. For the most massive stars, the rate of energy/momentum transferred into the outer layers of the star is simply to great (see above).&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditDate="2013-10-31T21:41:32.650" LastActivityDate="2013-10-31T21:41:32.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="831" PostTypeId="2" ParentId="697" CreationDate="2013-10-31T13:41:10.493" Score="10" Body="&lt;p&gt;Generally it would be that peers of Matese and Whitmire looking at the statistics and viewing that the results are not conclusive enough to indicate the existence of Tyche.  There can also be alternative reasons to explain the results mentioned in the paper other than a Jupiter sized planet in the Oort Cloud. As was mentioned in the comments, it is almost impossible to prove the non-existence of something.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this time, the results from WISE are the best bet for locating Tyche if it exists.  As it will not be visible in the visible light spectrum but more likely in the infrared and the WISE mission scanned the entire sky which means that it won't have been missed due to looking in the wrong place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jpl.nasa.gov/news/news.php?release=2011-060&quot;&gt;http://www.jpl.nasa.gov/news/news.php?release=2011-060&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nasa.gov/mission_pages/WISE/main/index.html#.UnJb3xCJ5fs&quot;&gt;http://www.nasa.gov/mission_pages/WISE/main/index.html#.UnJb3xCJ5fs&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There isn't any hard science to show that Tyche doesn't exist.  It is skepticism that this is the explanation for observations to date.  These are the same reasons that Nemesis isn't thought to exist, the data isn't strong enough to support its existence.&lt;/p&gt;&#xA;" OwnerUserId="281" LastEditorUserId="281" LastEditDate="2013-10-31T21:17:17.083" LastActivityDate="2013-10-31T21:17:17.083" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="832" PostTypeId="2" ParentId="447" CreationDate="2013-10-31T21:40:52.710" Score="13" Body="&lt;p&gt;This is a great question. I know of a couple of really big things about inflation people want to be able to nail down by using the cosmic microwave background. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is measuring what are known as E- and B- modes, which are the curl-free and divergence-free components to the modes of cmb radiation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lmoh0.png&quot; alt=&quot;E/Bmodes&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, measuring &lt;a href=&quot;http://darkuniverse.uni-hd.de/pub/Main/WinterSchool12Slides/Stud_Crowe.pdf&quot; rel=&quot;noreferrer&quot;&gt;large scale Gaussian B-modes from primordial gravitational waves&lt;/a&gt; will help constrain the energy scale of inflation. It may also be able to rule out most &lt;a href=&quot;http://en.wikipedia.org/wiki/Ekpyrotic_universe&quot; rel=&quot;noreferrer&quot;&gt;ekpyrotic&lt;/a&gt; and pure curvaton/inhomogeneous reheating models (same source).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thing people are looking at is this idea of primordial non-Gaussianity, which are second order corrections to the Gaussian fluctuations present in the cmb (&lt;a href=&quot;http://arxiv.org/abs/1002.3173&quot; rel=&quot;noreferrer&quot;&gt;review article&lt;/a&gt;; &lt;a href=&quot;http://arxiv.org/abs/1303.5084&quot; rel=&quot;noreferrer&quot;&gt;early planck results&lt;/a&gt;). Measuring a parameter called &lt;a href=&quot;http://arxiv.org/pdf/0804.0136v2.pdf&quot; rel=&quot;noreferrer&quot;&gt;$f_{nl}$&lt;/a&gt; (deviation from Gaussianity) has been a fairly crucial part of current and future studies and will also help rule out various inflationary models. This $f_{nl}$ parameter is defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;In this case the multipole coeﬃcients $a_{lm}$ of the CMB&#xA;    temperature map can be written as&#xA;    $$a_{lm} = a_{lm}^{(G)} + f_{nl} a_{lm}^{(NG)} $$&#xA;    where $a_{lm}^{(G)}$ is the Gaussian contribution and $a_{lm}^{(NG)}$ is the non-Gaussian contribution.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="45" LastActivityDate="2013-10-31T21:40:52.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="834" PostTypeId="1" CreationDate="2013-11-01T00:12:56.400" Score="13" ViewCount="390" Body="&lt;p&gt;Any time a spacecraft comes in close proximity to a planet and if the spacecraft has the right angle then it is able to use the planet's velocity to move itself further into space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Newton's 3rd law: every action has an equal reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case when the spacecraft uses for example the Earth's gravity to speed up, Earth will move towards the spacecraft. Earth's orbital change will be very small because the spacecraft's mass is small compared to Earth's mass but what if a big asteroid comes in close proximity or what if we use Earth's gravity to catapult our spacecrafts and keep doing it for an extended period of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could happen in this case? Could that have a dramatic impact to Earth's orbit?&lt;/p&gt;&#xA;" OwnerUserId="146" LastEditorUserId="146" LastEditDate="2013-11-02T21:33:59.287" LastActivityDate="2013-11-02T21:33:59.287" Title="Changes to Earth's orbit" Tags="&lt;orbit&gt;&lt;planet&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="835" PostTypeId="2" ParentId="834" CreationDate="2013-11-01T07:10:13.600" Score="12" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gravity_assist&quot; rel=&quot;noreferrer&quot;&gt;Gravity assists&lt;/a&gt; such as this are a form of elastic collision. There's a bit of number crunching here (hopefully no mistakes!), so you'll want to be familiar with the basics of momentum, kinetic energy, and the conservation thereof.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question: If Ceres (the largest known asteroid and nearly 500 km in diameter) used Earth to perform a gravity assist to increase its own velocity, by how much would this slow the Earth down, and how much larger would Earth's orbit become?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.wolframalpha.com/input/?i=earth%20orbital%20speed&quot; rel=&quot;noreferrer&quot;&gt;orbital speed of Earth&lt;/a&gt; around the sun is $U = 29.8~\mathrm{km~s}^{-1}$. So at a mass of $$M = 5.97\times 10^{24}~\mathrm{kg},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it has a kinetic energy of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K = 2.65\times 10^{33}~\mathrm{J}$$ and momentum $$P = 1.78\times 10^{29}~\mathrm{kg~m~s^{-1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's say Ceres is performing a gravitational slingshot as in the simple diagram below. Ceres has a mass  $m = 9.47 \times 10^{20}~\mathrm{kg}$. It approaches Earth at velocity $v$, and after the slingshot its final velocity is (up to, for a low-mass object) a velocity of $2\times U+v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fQw5I.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total momentum of the system &lt;em&gt;must be conserved&lt;/em&gt;. Ceres has changed direction and thus gained a significant amount of momentum in the leftwards direction: the same momentum that Earth must then lose. Kinetic energy is also conserved. So, we have a system of equations, where the subscripts &lt;em&gt;i&lt;/em&gt; and &lt;em&gt;f&lt;/em&gt; are initial and final momenta and velocities. M and U are the mass and velocity of Earth, m and v are that of Ceres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$MU_i^2 + mv_i^2 = MU_f^2 + mv_f^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which says that the sum of the initial kinetic energies of the two objects must equal the sum of the final kinetic energy. We also have conservation of momentum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$MU_i + m\vec{v}_i = MU_f + m\vec{v}_f $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving these equations, the solution is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_f = \frac{(1-m/M)v_i + 2U_i}{1-m/M} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Ceres approached Earth at $v_i = 30~\mathrm{km~s}^{-1}$, I get a solution of $v_f = 89.6~\mathrm{km~s}^{-1}$ - even for such a massive object, the $v_f \approx 2U+v$ approximation is extremely good. This means that Ceres' velocity has nearly been tripled by the gravity assist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the final momentum of Earth is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$MU_f = MU_i - mv_i - mv_f = 1.78 \times 10^{29}~ \mathrm{kg~m~s^{-1}} $$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In fact, Earth's linear momentum will only decrease by $mv_i + mv_f =&#xA; 1.13 \times 10^{23} ~\mathrm{kg~m~s^{-1}}$. From this change in momentum and Earth's mass, we find its orbital velocity decreases by&#xA;  $0.019~\mathrm{m~s}^{-1}$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Approximating a circular orbit (using $r=GM_{sun} / v^2$), Earth's&#xA;  orbit widens by 190 km. Sounds like a lot, but bear in mind that's 190&#xA;  km out of 150 million!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ceres is many orders of magnitude larger than any satellite that we could launch. So we could never practically use spacecraft to change our orbit significantly, and even an enormous near-miss asteroid would be of little consequence. But, &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0102126&quot; rel=&quot;noreferrer&quot;&gt;it hasn't stopped some from trying&lt;/a&gt;!&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="276" LastEditDate="2013-11-01T23:20:09.857" LastActivityDate="2013-11-01T23:20:09.857" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="837" PostTypeId="2" ParentId="800" CreationDate="2013-11-01T12:25:05.663" Score="5" Body="&lt;p&gt;Taking solar noon as being the time when the sun is at its highest elevation (altitude) in the sky for a given day, then your assumption of the LST (local solar time) is correct, as it is stated in the website you linked as being:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Twelve noon local solar time (LST) is defined as when &lt;strong&gt;the sun is highest in the sky&lt;/strong&gt;. Local time (LT) usually varies from LST because of the eccentricity of the Earth's orbit, and because of human adjustments such as time zones and daylight saving.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is something I have, in part, programmed into an app for a separate project (quite unrelated to yours).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would have to equate the precedent quantities first (Local Standard Time Meridian, Equation of Time and Time Correction factors) as they have done on the website you linked (as these are reliant on the observer's position and time of observation), but to determine at what time solar noon occurred, the final step needs to have local solar time (LST) set to 12:00 (solar noon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A worked exampled is given within the first 7 pages/slides of &lt;a href=&quot;http://www.egr.unlv.edu/~eebag/Properties%20of%20Sunlight%20-%20Part%20II.pdf&quot;&gt;Sunlight and its Properties II&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorUserId="45" LastEditDate="2013-11-01T12:31:00.013" LastActivityDate="2013-11-01T12:31:00.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="838" PostTypeId="1" AcceptedAnswerId="839" CreationDate="2013-11-01T16:00:22.640" Score="16" ViewCount="395" Body="&lt;p&gt;Are there any bounds on where a planetary ring can form and maintain orbit relative to the planet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do they have to be in a circular equatorial orbit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a min or max altitude, say with respect to the Roche limit, geosynchronous orbit, or other points in space?&lt;/p&gt;&#xA;" OwnerUserId="349" LastActivityDate="2013-11-01T18:09:14.027" Title="Do planetary rings have geometric bounds?" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;planetary-ring&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="839" PostTypeId="2" ParentId="838" CreationDate="2013-11-01T17:24:48.237" Score="8" Body="&lt;p&gt;The Saturn's &lt;a href=&quot;https://space.stackexchange.com/a/2252/49&quot;&gt;Phoebe ring&lt;/a&gt; that has an orbital inclination 173° to the ecliptic so it is actually in a retrograde orbit and is tilted 27° to Saturn's inner rings, shows that there clearly isn't any limit to orbital inclination. It is feeding off the Saturn's moon Phoebe (possibly due to micrometeorite impacts) indicating that as long as the planetary ring has a source of its materials with already inclined orbit, the ring will follow it (preserve the angular momentum of its originating body). That of course doesn't exclude polar orbits, but we have yet to see any rings like that. Theoretically though, their orbit is just as stable as any other as long as it doesn't intersect other gravitational disturbances, like Lagrange points, or paths of other celestials that would stop it forming a complete ring with accretion (something that is only partially done by Iapetus intersecting the Phoebe ring). Majority of planetary rings might be made of planet's own materials though, and would then naturally follow its own rotation, more or less (depending on their formation) and again preserving angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phebe ring is also extremely huge, stretching from calculated 59 to 300 Saturn's radii (observed by NASA's Spitzer Space Telescope in the infrared range between 128 and 207 Saturn's radii, more can be read in &lt;a href=&quot;http://planetary.org/blogs/emily-lakdawalla/2009/2165.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Emily Lakdawalla's blog post&lt;/a&gt;), attesting that as long as the ring's materials would not reach escape velocity or be gravitationally attracted more by another celestial body (beyond the L&lt;sub&gt;1&lt;/sub&gt; point) there isn't much of a limit in their size either. In theory, if we for example imagine some rogue brown dwarf that isn't gravitationally bound to any solar system and floats freely in the interstellar medium not close to any stars, the ring size would only be limited by the interstellar medium's own pressure. At least until the brown dwarf in questions roams closer to some stronger gravitational influence and loses its ring, that is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the maximum size of the ring would in theory indeed be limited only to the L&lt;sub&gt;1&lt;/sub&gt; point (where gravitational attractions of two massive bodies cancel each other). As for the minimum distance, that depends on the size of particles that the ring is made of, stretch of planet's atmosphere, and strength of its &lt;a href=&quot;https://astronomy.stackexchange.com/q/807/13&quot;&gt;radiation pressure&lt;/a&gt; that would be negating gravitational attraction. So it is difficult to give any fair minimum value. If we take a fast spinning planet (e.g. &lt;a href=&quot;http://www.space.com/23091-haumea.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;dwarf planet Haumea&lt;/a&gt;) as an example, one that would spin so fast that it would actually lose some of its materials due to it reaching escape velocity at the planet's equator, and if we assume this process can be sustained for long enough period, then there isn't any minimum as the planetary disk would be essentially touching the planet's surface. Eventually, the planet would lose much of its own rotation, similar like a fast spinning figure skater would by stretching her arms, by shifting some of its mass into larger radius, and the planet would cease replenishing disk's materials.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2013-11-01T18:09:14.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="841" PostTypeId="1" AcceptedAnswerId="857" CreationDate="2013-11-02T08:20:19.973" Score="6" ViewCount="699" Body="&lt;p&gt;If I am right, we can see only those stars that lie within our cosmological horizon, and there may or may not be any stars beyond that. Given last 150 years of using telescopes, and since then our cosmological horizon must have increased by about 150 light years, have we seen any new galaxies, stars? If given enough time, say 500 years, new stars are found, can we be sure they have become visible against not being formed, say, 150 years after big bang?&lt;/p&gt;&#xA;" OwnerUserId="359" OwnerDisplayName="Ratna" LastEditorUserId="13" LastEditDate="2013-11-02T09:42:42.480" LastActivityDate="2013-11-04T13:22:53.797" Title="Can it be inferred that our cosmological horizon has increased over time?" Tags="&lt;big-bang-theory&gt;&lt;cmb&gt;&lt;cosmological-inflation&gt;&lt;space-time&gt;&lt;cosmological-horizon&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="842" PostTypeId="1" AcceptedAnswerId="844" CreationDate="2013-11-02T15:34:58.050" Score="14" ViewCount="2954" Body="&lt;p&gt;Is the Milky Way Visible from Nebraska?  If so, where is the best place to view it, and also what would be the best time of night to see it?  I know this is probably a very novice question, just trying to get a better understanding of what I can see.  If it helps, I'm south of Omaha.&lt;/p&gt;&#xA;" OwnerUserId="360" LastActivityDate="2015-06-23T13:04:35.270" Title="Is the Milky Way Visible from Nebraska?" Tags="&lt;amateur-observing&gt;&lt;milky-way&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="843" PostTypeId="2" ParentId="841" CreationDate="2013-11-02T16:35:03.067" Score="2" Body="&lt;p&gt;There are a couple of different horizons you should care about. The first is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmological_horizon&quot; rel=&quot;nofollow&quot;&gt;cosmological horizon&lt;/a&gt;, which is the furthest you could possibly see given that photons travel at a finite speed and the universe is not infinitely old. Since nothing travels faster than the speed of light, this is quite literally the furthest we could ever hope to see - everything outside is causally disconnected from us (though this horizon does increase with time, it asymptotically reaches some ultimate distance due to the accelerated expansion of space-time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Practically speaking, the horizon which we really care about is the cosmic microwave background. This is the point at which the universe became cool enough such the atoms could remain neutral. Why is this the important horizon? Before this, photons simply could not travel very far before interacting with charged particles. Before the surface of last scattering the universe was essentially opaque (it would be like looking through a cloud). Afterwards, however, it was transparent. Everything from the surface of last scattering and later (redshift of $z\sim 1100$ and lower) is said to be part of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Observable_universe&quot; rel=&quot;nofollow&quot;&gt;observable universe&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though the universe may still be infinite in size, we will never be able to see these objects (if they exist) if they are passed this horizon. The other thing to remember is that the further you look, the further back in time you're seeing. Galaxies did not always exist. It took time to form larger overdensities like stars, galaxies, and clusters from small perturbations in density during the early universe. After the surface of last scatter, there was a period of time in which the universe was very dark and potentially very boring from an astronomer's point of view. This is the time after atoms became largely neutral yet &lt;a href=&quot;http://en.wikipedia.org/wiki/Chronology_of_the_universe&quot; rel=&quot;nofollow&quot;&gt;before the first stars and galaxies &quot;turned on&quot;&lt;/a&gt;.     &lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2013-11-02T16:35:03.067" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="844" PostTypeId="2" ParentId="842" CreationDate="2013-11-02T17:09:15.017" Score="8" Body="&lt;p&gt;Well, because the axis of the rotation of the Earth is not the same as the axis of rotation of the disk of the Milky Way (and also because we're transforming a 2-dimensional spherical map into a 2-dimensional cartesian map), the path of the disk of the Milky Way galaxy looks something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8E4em.gif&quot; alt=&quot;milkywaydisk&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there is actually a wide range in declination that the Milky Way can be seen at. The range of declination you can see depends on your latitude (for a review of RA and declination, coordinates used in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Celestial_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;celestial coordinate system&lt;/a&gt;, see this &lt;a href=&quot;https://astronomy.stackexchange.com/questions/9/could-someone-explain-ra-dec-in-simple-terms/41#41&quot;&gt;post&lt;/a&gt;). For example, here in Philadelphia (just about $+40^{\circ}$ declination), I'd be able to see from from $-50^{\circ}$ to $+90^{\circ}$ in declination. For Nebraska, find the latitude of your location. To calculate the lower limit, add your latitude (which will be positive since you're in the northern hemisphere) to $-90^{\circ}$ (mine was $+40^{\circ}$, so: $-90^{\circ} + 40^{\circ} = -50^{\circ}$). To find the upper limit in declination, it's even easier. Since you're in the northern hemisphere, you can actually see the north celestial pole. This means that the upper limit is simply the maximum it can possibly be, which is $+90^{\circ}$. The larger the latitude the more &lt;a href=&quot;http://en.wikipedia.org/wiki/Circumpolar_star&quot; rel=&quot;nofollow noreferrer&quot;&gt;circumpolar&lt;/a&gt; your night sky gets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The good news is that you should definitely be able to see it in Nebraska. The only question during what season will you see it at its best? Take a look at the constellations which contain parts of the Milky Way and find the season you can see it in - now you know where to look. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last thing I want to add is that you need nice dark conditions. Local weather conditions or sources of light pollution may very easily hide the Milky Way. From Philadelphia, even on perfectly clear nights we have no chance of seeing it. If you've got a city on one of your horizons, try to plan around that - either go to a darker location or try looking opposite in the sky from any sources of light pollution.&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-06-23T13:04:35.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="845" PostTypeId="2" ParentId="841" CreationDate="2013-11-02T17:49:55.757" Score="1" Body="&lt;p&gt;Over the past 150 years, telescope technology has improved immensely, so we can now see much further than in the past. Hubble's 'Deep Field' images would not have been possible 30 years ago, and are among the first I know of that show objects near the cosmological horizon.&lt;br&gt;&#xA;Also, while the cosmological horizon increases over time, objects near the horizon have a huge redshift (speed) away from us. I don't have the data to prove it, but I suspect objects near the cosmological horizon would disappear past the horizon over time.  &lt;/p&gt;&#xA;" OwnerUserId="361" LastActivityDate="2013-11-02T17:49:55.757" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="846" PostTypeId="2" ParentId="769" CreationDate="2013-11-02T21:35:54.987" Score="8" Body="&lt;p&gt;Yes, there are &lt;a href=&quot;http://www.telescope.com/Accessories/Telescope-Eyepiece-Filters/Orion-Full-Aperture-Glass-Telescope-Solar-Filters/pc/-1/c/3/sc/48/e/4.uts&quot; rel=&quot;noreferrer&quot;&gt;filters&lt;/a&gt; which do block out the vast majority of light from the sun. I think it's actually only a very small (~1 angstrom) wavelength band of light which gets through. You can see some pretty amazing features, including sunspots, and solar flares. Here's a composite image as an example (taken through a &lt;a href=&quot;http://www.meade.com/product_pages/coronado/coronado.php&quot; rel=&quot;noreferrer&quot;&gt;Hydrogen alpha filter&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/v4tCc.jpg&quot; alt=&quot;suncomposite_halphafilter&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those smallish looking edge features are &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_prominence&quot; rel=&quot;noreferrer&quot;&gt;solar prominences&lt;/a&gt; which are large enough to swallow up multiple Earth's. This is not too far off from what you could observe with a solar scope.&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="137" LastEditDate="2013-11-05T20:37:11.280" LastActivityDate="2013-11-05T20:37:11.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="847" PostTypeId="1" AcceptedAnswerId="1231" CreationDate="2013-11-02T22:54:20.940" Score="8" ViewCount="4721" Body="&lt;p&gt;The background is that I'm trying to write a computer program to show at what angles each planet would be able to be seen transiting in front of the sun from an observer outside our solar system.  My endgame is to calculate the relative percentages of the sky that could see any group of planets transiting together.  For instance in the direction the Earth is from the sun in June and December, an alien observer that could see Earth transit could also see Venus transit, but from the angle where the Earth is in September or March they would not ever be able to see Venus transit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote &lt;a href=&quot;http://ideone.com/neujND&quot;&gt;a program&lt;/a&gt; using basic stats and circular orbits to test out the idea, but I would like to make it more robust using actual elliptical orbits and I'm having trouble.  For circular orbits with the sun at the center I was able to just take the inclination of Earth's orbit times the &lt;code&gt;sin&lt;/code&gt; of the viewing longitude relative to the longitude of ascention.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;double viewingLongitude = longitudeRadians - LongitudeOfAscent;&#xA;double planetAngle = Inclination * Math.Sin(viewingLongitude);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For instance Earth's orbit has an inclination of 1.57 degrees and longitude of ascent at 348.74 degrees.  At 348.74 degrees and 168.74 degrees Earth will have an actual inclination of 0 degrees.  At 78.74 degrees and 158.74 degrees Earth will have an actual inclination of 1.57 and -1.57 degrees from the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a similar easy way I could calculate that inclination for the elliptical orbit of the Earth or will I have to solve Kepler's equations?&lt;/p&gt;&#xA;" OwnerUserId="343" LastActivityDate="2013-12-22T14:43:04.537" Title="Calculate angle of inclination for a certain angle in a planet's orbit?" Tags="&lt;orbit&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="848" PostTypeId="1" AcceptedAnswerId="849" CreationDate="2013-11-02T23:01:05.447" Score="7" ViewCount="2080" Body="&lt;p&gt;As mentioned on the &lt;a href=&quot;http://eclipse.gsfc.nasa.gov/OH/OH2013.html#SE2013Nov03H&quot; rel=&quot;noreferrer&quot;&gt;NASA eclipse website&lt;/a&gt; the eclipse on the 3rd of November 2013 will be a &quot;Hybrid&quot; eclipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked at the diagram on that page but can't see what is different about the eclipse that would explain the word hybrid. This terminology also confuses me:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is one of the rare hybrid or annular/total eclipses in which some sections of the path are annular while other parts are total.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8AEiY.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2013-11-03T03:43:55.173" Title="What is a 'hybrid' eclipse?" Tags="&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="849" PostTypeId="2" ParentId="848" CreationDate="2013-11-03T01:55:01.937" Score="7" Body="&lt;p&gt;A hybrid eclipse is one in which it is a total eclipse in some places and an annular eclipse is seen in other places. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you may know, a total eclipse is one in which the moon completely covers the disc of the Sun; this allows the solar corona to be seen.  Usually the central disc of the Sun is so bright that the corona cannot be seen.  An annular eclipse is one in which the Moon is not quite &quot;large&quot; enough to completely block out the Sun - instead, a ring of the Sun's central disc can be seen around the Moon. This ring is too bright for the corona to be seen.  Photos of both eclipse types appear below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Annular Eclipse (credit: Fred Espenak at www.mreclipse.com):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/nkKRn.jpg&quot; alt=&quot;Annular eclipse photo&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total Eclipse (credit: Fred Espenak at www.mreclipse.com):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/U0tLY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The type of eclipse is determined by the geometry of the Sun, Earth, and Moon.  Since the Sun and Moon are not constantly at the same distance from the Earth, their apparent sizes will differ. When the Moon is closer to the Earth it appears larger, and when the Sun is farther away it will appear smaller.  For this particular eclipse, the geometry is such that the western part of the eclipse's center line will see an annular eclipse, but the rest will have a total eclipse.&lt;/p&gt;&#xA;" OwnerUserId="25" LastEditorUserId="25" LastEditDate="2013-11-03T03:43:55.173" LastActivityDate="2013-11-03T03:43:55.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="850" PostTypeId="1" AcceptedAnswerId="851" CreationDate="2013-11-03T03:28:50.637" Score="17" ViewCount="722" Body="&lt;p&gt;We are inside the Milky Way, right? So, how are we able to take photos of the Milky Way?&lt;/p&gt;&#xA;" OwnerUserId="150" LastEditorUserId="2153" LastEditDate="2015-03-09T20:59:40.853" LastActivityDate="2015-03-09T20:59:40.853" Title="How are the photos of the Milky Way taken?" Tags="&lt;milky-way&gt;&lt;photography&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="851" PostTypeId="2" ParentId="850" CreationDate="2013-11-03T03:47:59.643" Score="22" Body="&lt;p&gt;We are situated inside the galactic disk. The first image below, showing our approximate location, is a fabrication rather than a photograph. Such a photo would require a space probe to be tens of thousands of light years above the plane of the disk, which is patently infeasible with current technology. We don't even know exactly how many spiral arms our galaxy has!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But from where we are, we can still see a great density of gas and dust stretching 360 degrees across our sky, as you look along the galactic plane. These &quot;inside views&quot; are indeed real photographs - much the same as if you go outside on a dark night, well away from the city lights, you yourself will see its faint, dusty outline much like in the second image (which I can assure you is real).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gFwNY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GTYQt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2013-11-03T03:47:59.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="852" PostTypeId="2" ParentId="841" CreationDate="2013-11-03T23:55:23.753" Score="1" Body="&lt;p&gt;As far as observation goes, distance equates to age, so as our technology improves we effectively see farther into the past. The actual increase in distance visible to us is much much greater than 150 light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can actually detect the effects of the big bang itself, in the microwave background radiation of the universe, but sars and galaxies formed much later than 150 years after themig bang (try millions of years :-)&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2013-11-03T23:55:23.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="853" PostTypeId="1" AcceptedAnswerId="854" CreationDate="2013-11-04T05:09:38.503" Score="10" ViewCount="14669" Body="&lt;p&gt;As the title states, what would the effect on Earth's climate be if it's axis were tilted approximately 90 degrees like Uranus?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm specifically wondering about the effects on seasons, temperature, and life on Earth.  However, I'm happy to hear about other effects as well.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="6" LastEditDate="2013-11-15T18:10:29.420" LastActivityDate="2013-12-12T14:43:52.783" Title="How would Earth's climate differ if it's axis were tilted around 90 degrees like Uranus?" Tags="&lt;earth&gt;&lt;rotation&gt;&lt;climate&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ClosedDate="2013-12-12T14:43:59.393" ContentLicense="CC BY-SA 3.0" />
  <row Id="854" PostTypeId="2" ParentId="853" CreationDate="2013-11-04T08:54:07.070" Score="8" Body="&lt;p&gt;This is a good question, and has been somewhat a topic of study by a few.  So below is some consequences of if Earth had a tilt like that of Uranus (for this explanation, I am disregarding the wobbles in the Earth's current axis, just focussing on the 90 degree axis):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the article &lt;a href=&quot;http://www.spacedaily.com/news/extrasolar-99m.html&quot; rel=&quot;noreferrer&quot;&gt;Not All Habitable Zones Are Created Equal&lt;/a&gt; (Moomaw, 1999), each day and night would be 6 months each (including an epically long dusk and dawn), with daytime temperatures reaching up to 80C, but the night time side may not reach freezing - owing to the time to take to cool down from the 6 month day.  Parts of the equatorial regions would be, however, permanently encased in ice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretty much like this diagram of Uranus' rotation (from &lt;a href=&quot;http://www.ifa.hawaii.edu/~barnes/ast110_06/&quot; rel=&quot;noreferrer&quot;&gt;Source: University of Hawaii&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zCF1H.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A major consequence of this according to the author is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In such an environment, life could almost certainly still appear, but it would have much more difficulty evolving into forms that could survive such grotesque temperature extremes -- which would greatly slow down its evolution into more complex forms, maybe by billions of years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A great consequence of this, particularly the 6 month night is that photosynthesis would have been stilted, if it could have started at all.  This has a great consequence on oxygen levels of the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, the authors claim that if the Earth had an axial tilt of 90 degrees, but at 210 million km from the sun, then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;its climate would be positively balmy -- the equator would be 11 deg C (52 deg F), and the poles would never rise above 46 deg C (115 deg F) or fall below 3 deg C (37 deg F). Earth would have no ice anywhere on its surface, except on some of its highest mountains.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to the article &lt;a href=&quot;http://www.astrobio.net/exclusive/4543/high-planetary-tilt-lowers-odds-for-life&quot; rel=&quot;noreferrer&quot;&gt;High Planetary Tilt Lowers Odds for Life?&lt;/a&gt; (Hadhazy, 2012), has a great way of putting it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Your northern pole will be boiled during part of the year while the equator gets little sunlight,&quot; said Heller. Meanwhile, &quot;the southern pole freezes in total darkness.&quot; Essentially, the conventional notion of a scorching hell dominates one side of the planet, while an ultra-cold hell like that of Dante's Ninth Circle prevails on the other. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, to make matters worse, the hells reverse half a year later. &quot;The hemispheres are cyclically sterilized, either by too strong irradiation or by freezing,&quot; Heller said. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They also describe that if life were to evolve, extremophiles (specifically, thermophiles) would be dominant - seasonally.&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-11-04T10:38:38.243" LastActivityDate="2013-11-04T10:38:38.243" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="856" PostTypeId="1" AcceptedAnswerId="861" CreationDate="2013-11-04T12:53:42.537" Score="34" ViewCount="2872" Body="&lt;p&gt;Have we discovered any natural satellites of natural satellites of planets or dwarf planets? Even very small, or relatively short-lived - e.g. ringlets around Saturn's moons, some meteorites orbiting Jupiter moons, or something to orbit Charon? Or is the Star-Planets-Moons the deepest naturally ocurring orbital recursion level?&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2020-05-07T04:22:19.527" Title="Do moons have moons?" Tags="&lt;natural-satellites&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="857" PostTypeId="2" ParentId="841" CreationDate="2013-11-04T13:22:53.797" Score="3" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;No. Space expansion doesn't slow down, unlike the expansion of cosmic horizon. That means the cosmic horizon actually swallows galaxies as they travel away from us due to space expansion, reducing their number instead of increasing it. Nevertheless, we still haven't gathered all cosmic radiation from Big Bang that occurred within our current event horizon, so it's still a good while before it begins restricting distance of observable universe for us - currently the speed of light (+space expansion rate) vs age of the universe limits it to a considerably smaller radius.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes, we discovered more stars and galaxies, but that is due to improvement of technology, which is still quite far (roughly a third of the way) from observing stars near the Hubble Sphere as the universe formed (ones whose light would be observable today; as opposed to these at the space horizon, whose light &lt;em&gt;will&lt;/em&gt; reach us &lt;em&gt;eventually&lt;/em&gt;, in infinitely distant future.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Currently we're still quite a bit away from leaning against the cosmological limits of observability. Currently our limit of observability is the budget for building better devices.&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2013-11-04T13:22:53.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="858" PostTypeId="1" AcceptedAnswerId="900" CreationDate="2013-11-04T13:44:36.243" Score="14" ViewCount="732" Body="&lt;p&gt;We're fairly aware clusters of galaxies drift apart due to space expansion, which will drive them out of each other's cosmic event horizon eventually, leaving them separate, alone, each a single &quot;body&quot; within its local observable universe with no means to communicate or influence other clusters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clusters themselves are held together by gravitational binding that keeps them from being internally torn apart by space expansion, at least assuming density of dark energy doesn't increase drastically eventually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What will happen to the cluster though? At first I assumed its black holes will swallow all its available matter, and connect into one single, lonely black hole. But then, neither are black holes ultimately eternal, due to Hawking Radiation, nor do they swallow &lt;em&gt;all&lt;/em&gt; the stars, supposedly most stars escaping their gravity well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well then, what current theories say a local cluster of galaxies will transform into as t→∞ ?&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="1862" LastEditDate="2014-09-03T04:07:32.237" LastActivityDate="2014-09-18T12:47:04.970" Title="What is the ultimate fate of a cluster of galaxies?" Tags="&lt;cosmology&gt;&lt;galaxy-cluster&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="859" PostTypeId="1" CreationDate="2013-11-04T16:26:02.663" Score="8" ViewCount="2730" Body="&lt;p&gt;I recently created a &lt;a href=&quot;http://www.youtube.com/watch?v=HlO1qTTO8rU&quot; rel=&quot;nofollow&quot;&gt;visualisation of comet ISON's orbit&lt;/a&gt;. However, I received a critique that the dust tail of the comet should be pointing away from the Sun after perihelion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, the ion tail of the comet definitely points away from the Sun as it is blown away by the solar wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know what is the direction of the dust-tail of the comet, before and after the perihelion. Is it always trailing the path of the comet (to a fair degree of approximation) or are there other significant effects?&lt;/p&gt;&#xA;" OwnerUserId="337" LastEditorUserId="337" LastEditDate="2013-11-05T03:03:27.083" LastActivityDate="2019-03-04T18:05:04.497" Title="What is the direction of a comet's dust tail before and after perihelion?" Tags="&lt;comets&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="860" PostTypeId="2" ParentId="859" CreationDate="2013-11-04T17:25:27.860" Score="4" Body="&lt;p&gt;Comet tail always aims away from the sun, as it's torn from the comet's gas cloud by solar wind. It trails slightly behind, as the comet moves along its orbit while the gas travels directly away, but that's relatively minor - the speed at which the gas and particles are pulled away forming the tail is much higher than the comet's orbital velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your visualisation shows it trailing behind the comet as if it was moving in some gaseous/liquid medium that stops the lighter particles while letting the heavy comet head travel ahead. This is not the case - as the comet is far away from the sun, the gas cloud just forms its atmosphere and travels with the comet. As it approaches though, the intensity of solar wind increases and it pulls away the atmosphere which can't be protected against it by magnetic field or strong gravity, as the comet has neither.&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="83" LastEditDate="2013-11-04T17:40:05.190" LastActivityDate="2013-11-04T17:40:05.190" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="861" PostTypeId="2" ParentId="856" CreationDate="2013-11-04T18:52:38.130" Score="27" Body="&lt;p&gt;I don't think there are &lt;a href=&quot;http://en.wikipedia.org/wiki/Natural_satellite&quot; rel=&quot;nofollow noreferrer&quot;&gt;any&lt;/a&gt; in the Solar system. We do have around 250 &lt;a href=&quot;http://www.johnstonsarchive.net/astro/asteroidmoons.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;asteroids with moons&lt;/a&gt;. Rhea's &lt;a href=&quot;http://www.sciencemag.org/content/319/5868/1380.abstract?maxtoshow=&amp;amp;HITS=10&amp;amp;hits=10&amp;amp;RESULTFORMAT=&amp;amp;fulltext=rhea&amp;amp;searchid=1&amp;amp;FIRSTINDEX=0&amp;amp;resourcetype=HWCIT&quot; rel=&quot;nofollow noreferrer&quot;&gt;ring&lt;/a&gt; seems to be the only exception.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; Originally I said &quot;a moon with a moon would be an &lt;a href=&quot;http://www.popsci.com/scitech/article/2008-04/could-moon-have-moons&quot; rel=&quot;nofollow noreferrer&quot;&gt;unstable system&lt;/a&gt;, due to the gravitational influence of the planet.&quot; @Florian disagrees with this. However, the answer is more complex than the Hill sphere alone.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first approximation, the Hill sphere gives a radius in which orbits around a moon could be stable. Our Moon's Hill radius is 64000 km.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For our own Moon, we know that most low orbits are unstable due to &lt;a href=&quot;http://www.jpl.nasa.gov/news/news.php?release=2013-184&quot; rel=&quot;nofollow noreferrer&quot;&gt;mascons&lt;/a&gt;: mass concentrations below the surface which make the Moon's gravitational field noticeably uneven. There are &lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2006/06nov_loworbit/&quot; rel=&quot;nofollow noreferrer&quot;&gt;only four inclinations&lt;/a&gt; where an object orbiting the Moon avoids all mascons and would be stable: 27º, 50º, 76º, and 86º.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2006/30nov_highorbit/&quot; rel=&quot;nofollow noreferrer&quot;&gt;High orbits above the Moon&lt;/a&gt; aren't all safe either: above 1200 km and inclinations of more than 39.6º, Earth's gravity disrupts the satellite's orbit. Note that these orbits are comfortably within the Moon's Hill sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are stable orbits at high inclinations and high eccentricity:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/0Tqz1.jpg&quot; alt=&quot;Stable moon orbit&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for other moons in the solar system: most of them are smaller and orbit around larger planets, so their Hill spheres are small, and the planet's gravity will disrupt much of the volume inside the Hill sphere too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moons below the limit where their gravity is strong enough to make them spherical, will have problems with uneven gravitational fields. Mascons may also be present. &lt;/p&gt;&#xA;" OwnerUserId="361" LastEditorUserId="1987" LastEditDate="2019-05-21T16:51:47.157" LastActivityDate="2019-05-21T16:51:47.157" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="864" PostTypeId="2" ParentId="858" CreationDate="2013-11-04T23:38:11.057" Score="7" Body="&lt;p&gt;This depends where the cluster is. Large scale features of the universe (both observed and simulated) consist of filaments and voids. Below is a map produced by the Sloan Digital Sky Survey (SDSS) showing these features.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rLYOr.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each dot here is an individual galaxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simulations show remarkably the same thing. Here is a video of the &lt;a href=&quot;http://www.mpa-garching.mpg.de/galform/data_vis/millennium_sim_1024x768.avi&quot; rel=&quot;noreferrer&quot;&gt;Millennium Simulation&lt;/a&gt;. It zooms into smaller and smaller structures, and just at about the smallest scale it zooms to (~4Mpc) is about the size of an individual cluster. It turns out that collections of clusters which form at the intersections of filaments, super clusters, are even more massive, yet are still forming today. If the cluster belongs to these highly overdense regions, it will merge with other clusters to eventually form super clusters. If they live at the boundaries of these regions it is not guaranteed that they will fall into more massive regions. I know this answer is a bit lacking in detail, but structure formation theory is complicating stuff. Hopefully this gives you some idea of where clusters live and what could potentially happen to them as $t \rightarrow \infty$.&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2013-11-04T23:38:11.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="868" PostTypeId="1" AcceptedAnswerId="896" CreationDate="2013-11-05T16:58:22.873" Score="19" ViewCount="11964" Body="&lt;p&gt;After yesterday's announcement of the Kepler telescope finding a huge amount of newly observed exoplanets, i saw a &lt;a href=&quot;http://www.universetoday.com/106121/22-of-sun-like-stars-have-earth-sized-planets-in-the-habitable-zone/&quot;&gt;headline&lt;/a&gt; claiming that as much as 22% of sun like stars in the universe have planets in their habitable zone. There are loads of stars in the universe, so the number of planets in the habitable zone has to be enourmous. But how many of the stars in the universe are about the same size as our sun?&lt;/p&gt;&#xA;" OwnerUserId="36" LastEditorUserId="36" LastEditDate="2013-11-05T18:08:31.460" LastActivityDate="2016-02-14T20:58:54.333" Title="How many sun-like stars are there in the universe?" Tags="&lt;the-sun&gt;&lt;exoplanet&gt;&lt;universe&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="869" PostTypeId="1" AcceptedAnswerId="870" CreationDate="2013-11-05T17:01:21.747" Score="12" ViewCount="146" Body="&lt;p&gt;We can observe exoplanets, but is it possible with today's technology to observe if there is life as we know it on an exoplanet? What do you look for when doing this kind of observation? &lt;/p&gt;&#xA;" OwnerUserId="36" LastEditorUserId="204" LastEditDate="2013-12-05T19:26:54.807" LastActivityDate="2013-12-05T19:26:54.807" Title="Is it possible to observe if there is traces of life on an exoplanet?" Tags="&lt;observational-astronomy&gt;&lt;extra-terrestrial&gt;&lt;life&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="870" PostTypeId="2" ParentId="869" CreationDate="2013-11-05T17:30:20.730" Score="8" Body="&lt;p&gt;&lt;strong&gt;Direct&lt;/strong&gt; observation of exoplanets (and then of traces of life on them) is still a big challenge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up to now, there are several  &lt;a href=&quot;http://en.wikipedia.org/wiki/Methods_of_detecting_extrasolar_planets&quot;&gt;indirect methods&lt;/a&gt; (effects on the parent star or on other stars), and direct detection is performed by using devices such as &lt;strong&gt;coronagraphs&lt;/strong&gt; on faint stars, where very bright planets are imaged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A way to &lt;strong&gt;image&lt;/strong&gt; things such as as &quot;forests&quot; is very far in the future, and could be achieved by using extreme coronagraphy in &lt;strong&gt;space interferometers&lt;/strong&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way is to use &lt;strong&gt;spectroscopy&lt;/strong&gt; to find traces of biological activity in the athmosphere. This solution is very tricky due to the faint light reflected by extrasolar planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is anyway a field where progresses are very quick, so stay tuned!&lt;/p&gt;&#xA;" OwnerUserId="12" LastActivityDate="2013-11-05T17:30:20.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="871" PostTypeId="1" AcceptedAnswerId="873" CreationDate="2013-11-05T18:17:46.953" Score="12" ViewCount="7609" Body="&lt;p&gt;When our sun reaches the end of its lifetime , it will turn into a red giant. How long will it be a red giant, and what happens after a star has been a red giant? Does it turn into a planet after a long enough time spent, when it has used up all its helium?&lt;/p&gt;&#xA;" OwnerUserId="36" LastEditorUserId="6" LastEditDate="2013-11-07T12:47:45.787" LastActivityDate="2013-11-07T12:47:45.787" Title="What happens to a star after it has become a red giant?" Tags="&lt;star&gt;&lt;red-giant&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="872" PostTypeId="1" AcceptedAnswerId="890" CreationDate="2013-11-05T18:30:49.010" Score="24" ViewCount="3643" Body="&lt;p&gt;Our sun has 8 planets orbiting as well as a number of dwarf planets. Are there any calculations that hint as to whether this number is close to some theoretical maximum value or are we simply an average solar system in this particular way? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could imagine that if you have many planets, they will likely interact with each other. Can you calculate any theoretical value for the maximum number of planets which have long-term stable orbits around their own star? &lt;/p&gt;&#xA;" OwnerUserId="36" LastEditorUserId="45" LastEditDate="2013-11-05T22:29:19.473" LastActivityDate="2017-05-25T08:29:53.360" Title="Is there a upper limit to the number of planets orbiting a star?" Tags="&lt;orbit&gt;&lt;planet&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="873" PostTypeId="2" ParentId="871" CreationDate="2013-11-05T18:33:17.067" Score="11" Body="&lt;p&gt;It will become a &lt;a href=&quot;https://en.wikipedia.org/wiki/Planetary_nebula&quot; rel=&quot;noreferrer&quot;&gt;planetary nebula&lt;/a&gt; like e.g. the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cat%27s_Eye_Nebula&quot; rel=&quot;noreferrer&quot;&gt;Cat's Eye nebula&lt;/a&gt; that was formed by the death of a star with about &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_mass&quot; rel=&quot;noreferrer&quot;&gt;the same mass as the Sun&lt;/a&gt; ~ $1\ M_\odot $:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/t1qgY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;sup&gt;Composite image using optical images from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble_Space_Telescope&quot; rel=&quot;noreferrer&quot;&gt;HST&lt;/a&gt; and X-ray data from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Chandra_X-ray_Observatory&quot; rel=&quot;noreferrer&quot;&gt;Chandra X-ray Observatory&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun, and any &lt;a href=&quot;https://en.wikipedia.org/wiki/Red_dwarf&quot; rel=&quot;noreferrer&quot;&gt;red dwarfs&lt;/a&gt; above about 0.25 solar masses, will expand into what's called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Red_giant&quot; rel=&quot;noreferrer&quot;&gt;red giant&lt;/a&gt;, a late stage of stellar evolution. At this stage, the star starts to fuse different elements, and eventually throws off its layers as a planetary nebula, leaving behind a white dwarf made of carbon and oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A red dwarf that is too small to become a red giant will not turn into a planetary nebula: its fusion processes will eventually cease and it will probably produce a white &lt;a href=&quot;http://en.wikipedia.org/wiki/White_dwarf&quot; rel=&quot;noreferrer&quot;&gt;dwarf made&lt;/a&gt; mostly of helium. But the main-sequence lifetime of these very small stars is longer than the age of the universe, so this has never actually happened yet. The helium white dwarfs that do exist were formed in binary star systems (and were formed during their complex dynamics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Red dwarfs are small main sequence stars - the smallest and dimmest that are still able to fuse hydrogen in their core. &lt;a href=&quot;https://en.wikipedia.org/wiki/Brown_dwarf&quot; rel=&quot;noreferrer&quot;&gt;Brown dwarfs&lt;/a&gt; are even smaller (in terms of mass), but cannot even fuse hydrogen - they are thought to be able to fuse deuterium and lithium. Stars smaller than a brown dwarf turn out to be gas giants like our Jupiter, Saturn, Uranus and Neptune.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="276" LastEditDate="2013-11-05T21:30:25.907" LastActivityDate="2013-11-05T21:30:25.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="874" PostTypeId="1" AcceptedAnswerId="882" CreationDate="2013-11-05T19:04:49.320" Score="22" ViewCount="3608" Body="&lt;p&gt;If you rewind the universe back 14,7 billion years, all matter were in one spot, and then started expanding. Do we know where this is in reference to our own solar system? And is there anything there? Or are all matter in the universe just blown inn all different directions and none is left at the &quot;original&quot; spot where it originated&lt;/p&gt;&#xA;" OwnerUserId="36" LastActivityDate="2020-10-12T23:44:29.193" Title="Do we know the exact spot where big bang took place?" Tags="&lt;big-bang-theory&gt;&lt;matter&gt;" AnswerCount="7" CommentCount="3" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="875" PostTypeId="2" ParentId="769" CreationDate="2013-11-05T20:05:31.207" Score="3" Body="&lt;p&gt;Solar flares are explosive events that last not too long, so actually, your bigger problem will be to be able to see them when they are happening.&lt;/p&gt;&#xA;&lt;p&gt;Normally they are classified by their X-ray flux, and they can be either X, M, C, B or A class which corresponds to the logarithmic scale of &lt;span class=&quot;math-container&quot;&gt;$Watts/m^2$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$10^{-4}, 10^{-5}, 10^{-6}, 10^{-7}, 10^{-8}$&lt;/span&gt; respectively).&#xA;You can understand better what I mean by looking at the &lt;a href=&quot;http://www.swpc.noaa.gov/rt_plots/Xray.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;real time lightcurves obtained from GOES satellite&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;These peaks have different duration depending of the flare, there are some records of flares lasting few hours where the shortest can be shorter than a minute.&lt;/p&gt;&#xA;&lt;p&gt;All the above applies to X-ray flares, which we cannot seen from Earth, we need to put some instrument on board spacecraft or rocket so we go above the part of the atmosphere that absorbs them.&lt;/p&gt;&#xA;&lt;p&gt;Now, if you want to see them by using an optical telescope you have two options, either using the whole visible range by projection or through an Hydrogen alpha filter.  In this case the flare classification is different than in the X-ray case.  Where the flare class in X-ray is measured by the enhanced of flux produced, the optical class is based by the area it covers. The Solar Influence Data Analysis Center offers a table with their properties and the corresponding X-ray flare type:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;Area&lt;br&gt;(sq deg)&lt;/th&gt;&#xA;&lt;th&gt;Area&lt;br&gt; (&lt;span class=&quot;math-container&quot;&gt;$10^{-6}$&lt;/span&gt; solar A)&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Class&lt;/th&gt;&#xA;&lt;th&gt;Typical corresponding SXR Class&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;&amp;lt;= 2.0&lt;/td&gt;&#xA;&lt;td&gt;&amp;lt;= 200&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;S&lt;/td&gt;&#xA;&lt;td&gt;C2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;2.1-5.1&lt;/td&gt;&#xA;&lt;td&gt;200-500&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td&gt;M3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;5.2-12.4&lt;/td&gt;&#xA;&lt;td&gt;500-1200&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td&gt;X1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;12.5-24.7&lt;/td&gt;&#xA;&lt;td&gt;1200-2400&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td&gt;X5&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;&amp;gt;24.7&lt;/td&gt;&#xA;&lt;td&gt;&amp;gt;  2400&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td&gt;X9&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;For example, the famous Carrington's event (Sept. 1, 1859) was observed and hand-drawn from Carrington by projection.  But, as &lt;a href=&quot;http://www.noaa.gov/features/02_monitoring/1859solarstorm.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article says&lt;/a&gt;: &amp;quot;He was lucky enough to be in the right place at the right time...&amp;quot; you need too to be quite lucky to observe one of these.&lt;/p&gt;&#xA;" OwnerUserId="137" LastEditorUserId="2365" LastEditDate="2020-12-15T00:25:12.827" LastActivityDate="2020-12-15T00:25:12.827" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="876" PostTypeId="2" ParentId="874" CreationDate="2013-11-05T20:06:55.823" Score="6" Body="&lt;p&gt;My understanding is that the Big Bang took place &lt;em&gt;everywhere&lt;/em&gt;. That &quot;one spot&quot; expanded into the Universe as we see it now. There is no distinguished spot within the current Universe corresponding to the Big Bang.&lt;/p&gt;&#xA;" OwnerUserId="53" LastActivityDate="2013-11-05T20:06:55.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="877" PostTypeId="2" ParentId="874" CreationDate="2013-11-05T21:26:24.127" Score="9" Body="&lt;p&gt;It's all around us.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I explain it is that we are in the centre of the universe (I don't mean in the geocentric way as thought back in the middle ages where all is moving around us).  Everywhere we look we see the past: our Sun how it was 8 minutes ago; the closer star, Alpha Centauri, as it was 4.2 years ago, the Andromeda galaxy's light took 2.5 million years to reach us, and further on we go... but as you can imagine the universe goes in all the directions...so there's not a single point in space but in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this drawing helps to understand it, Earth is in the centre, the Sun could be at certain radius and each object in our universe will be placed somewhere between us and the beginning of the universe: the big bang at &lt;a href=&quot;http://map.gsfc.nasa.gov/universe/uni_age.html&quot; rel=&quot;noreferrer&quot;&gt;13.77 ± 0.059 billion years ago&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/66nbZ.png&quot; alt=&quot;Time diagram from Now to the past.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having this in mind you can then understand why we see the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_background_radiation&quot; rel=&quot;noreferrer&quot;&gt;cosmic background radiation&lt;/a&gt; coming from all the directions.&lt;/p&gt;&#xA;" OwnerUserId="137" LastEditorUserId="45" LastEditDate="2013-11-05T22:22:57.737" LastActivityDate="2013-11-05T22:22:57.737" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="878" PostTypeId="1" AcceptedAnswerId="880" CreationDate="2013-11-05T22:41:20.300" Score="6" ViewCount="6204" Body="&lt;p&gt;I was taking an astronomy test, the following question came up:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The most important function of the astronomical telescope is:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Resolving power&lt;/li&gt;&#xA;  &lt;li&gt;Light gathering power&lt;/li&gt;&#xA;  &lt;li&gt;magnifying power&lt;/li&gt;&#xA;  &lt;li&gt;focal length&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The focal length obviously means nothing here, as well as the magnification (that can always be done later). As far as resolving power and light gathering power go though, they both seem to be &quot;equally&quot; important, in that without the first, there is no detail, and without the second, there is no image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does seem that exposure time can compensate for a lack of light gathering power (what matters is the product between the two), but nothing can compensate for resolving power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My professor seems to think the answer is resolving power. Why is this? Is this some well known idea in astronomy?&lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2013-11-06T00:57:11.113" Title="Most important feature of a telescope" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="879" PostTypeId="2" ParentId="1" CreationDate="2013-11-05T23:46:37.250" Score="16" Body="&lt;p&gt;Any telescope can be made to give you the information that you are looking for.  The first thing that you will need to know is the location of the ecliptic which varies throughout the year.  Or are you looking to find in relation to the the celestial equator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Celestial_equator&quot;&gt;https://en.wikipedia.org/wiki/Celestial_equator&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, you would start of the same by find the Declination (D) and Right Ascension (R.A) of the object in question.  You do not need a telescope to find this, unless you are unable to see the object with the naked eye.  If the telescope has a polar mount and is properly set up.  You can read the R.A. and D off of the mount.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise it will require some trigonometry and knowing where you are (at least your latitude)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will need to determine the Altitude and Azimuth of the object.  Which is really just the direction from North and the angle from horizontal that your telescope is pointing.  You could do this with a compass and protractor or even an &lt;a href=&quot;https://en.wikipedia.org/wiki/Astrolab&quot;&gt;astrolabe&lt;/a&gt;.  Knowing this you can then convert to R.A and D with the following formulas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$RA = \arctan(\frac{- sin(Az) \times \cos(Alt)}{\cos(Lat) \times \sin(Alt) - \sin(Lat) \times \cos(Az) \times \cos(Alt)})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Dec = \arcsin(\sin(Lat) \times \sin(Alt) + \cos(Lat) \times \cos(Alt) \times \cos(Az))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are looking for the inclination to the ecliptic, so you are mostly concerned with the declination.  The ecliptic changes in declination throughout the year from 0 at the equinox to +/- 23.5 at the solstice.  So your inclination from the ecliptic would be the Declination of your object +/- the Declination of the ecliptic. &lt;/p&gt;&#xA;" OwnerUserId="281" LastEditorUserId="6" LastEditDate="2014-02-03T16:49:55.200" LastActivityDate="2014-02-03T16:49:55.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="880" PostTypeId="2" ParentId="878" CreationDate="2013-11-06T00:57:11.113" Score="4" Body="&lt;p&gt;Well the &lt;a href=&quot;http://en.wikipedia.org/wiki/Angular_resolution&quot; rel=&quot;nofollow&quot;&gt;angular resolution&lt;/a&gt; of a telescope depends on two things: 1) The wavelength of light you're looking in, and 2) the diameter of the primary lens/mirror. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \theta = 1.22 \frac{\lambda}{D}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\lambda$ is the wavelength of light and $D$ is the diameter of the lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light collecting power is simply how many photons you can collect in a particular wavelength in a given amount of time. Essentially, the bigger the bucket (or lens/mirror), the more rain you can catch. You can sort of get around light collecting power by increasing your exposure time. But, if the size of the object you're trying to observe is smaller than your telescope's resolving power, I don't believe there is really much you can do about it. Being limited by the resolution of your telescope is called being &lt;a href=&quot;http://en.wikipedia.org/wiki/Diffraction-limited_system&quot; rel=&quot;nofollow&quot;&gt;diffraction limited&lt;/a&gt;, but most telescopes on Earth are usually seeing limited due to the atmosphere's smearing effects.&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2013-11-06T00:57:11.113" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="881" PostTypeId="2" ParentId="874" CreationDate="2013-11-06T06:46:38.743" Score="19" Body="&lt;p&gt;&lt;strong&gt;Yes&lt;/strong&gt;, here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As has already been mentioned, the big bang happened everywhere due to the fact that all of time and space was in the same spot at the point of the Big Bang. And so the big bang happened here, there and everywhere. This is also similar to saying that the universe has no centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen this phenomenon explained using two pieces of clear plastic marked with the same pattern of dots, except that one has been scaled up (i.e. to represent an expanded universe). If any pair of corresponding points is selected, than it appears as though the selected point is at the centre of the universe because all other points have become further away. Points that were of equal distance from the selected points will be the same distance as each other from the selected point. This works for any corresponding pair of points.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2013-11-06T06:46:38.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="882" PostTypeId="2" ParentId="874" CreationDate="2013-11-06T18:01:41.017" Score="23" Body="&lt;p&gt;You're envisioning the Big Bang as a cosmological &quot;explosion&quot; in space where the universe is the resulting material expanding in outward all directions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, the universe doesn't work like that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe isn't a region of space that is expanding outward into another &lt;em&gt;thing&lt;/em&gt; where you can (even hypothetically) fly out to the border and say, &quot;Yup, that's about all there is. &lt;strong&gt;Let's use these borders to find the center of this expanding sphere.&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Universe is literally everything there is: space, time, and everything it contains and ever will contain. And the word &lt;em&gt;contains&lt;/em&gt; is even a bit of a misnomer in that it suggests a &lt;em&gt;container&lt;/em&gt; with boundaries. But the universe is simultaneously both unbounded and finite at the same time. &quot;Finite&quot; refers to the fixed amount of &lt;em&gt;stuff&lt;/em&gt; this universe contains (matter, energy, etc)&amp;hellip; and &lt;em&gt;unbounded&lt;/em&gt; because this thing we know as &quot;the universe&quot; has no borders in a sense that we point to or even experience. So how would you even &lt;em&gt;define&lt;/em&gt; a &quot;center&quot;? There isn't one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a famous way to help visualize an expanding universe that is both finite and unbounded without a center or borders; it's called &lt;strong&gt;&quot;The Balloon Analogy.&quot;&lt;/strong&gt; Imagine our friends in &lt;a href=&quot;http://en.wikipedia.org/wiki/Flatland&quot; rel=&quot;noreferrer&quot;&gt;Flatland&lt;/a&gt; living in two-dimensional space&amp;hellip; where everything they know (their entire universe) exists on the surface of a balloon. If you start to inflate that balloon (the expanding universe),  all the little astronomers on the surface will observe that all the surrounding galaxies are moving away from them &amp;mdash; and the farther away those galaxies are, the faster they seem to be moving away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's pretty much what most cosmologists believe is happening to us. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going back to that balloon analogy, picture yourself existing on the surface of that balloon-universe and ask these questions again &amp;mdash; In my expanding universe (the surface of the balloon), where is it's center? Everything is expanding outward in all directions, so if we rewind back in time, certainly everything should converge &lt;em&gt;somewhere,&lt;/em&gt; right? Can we just find the borders of this outward expansion and calculate the exact center? So basdically&amp;hellip; &lt;em&gt;&quot;Do we know the exact spot where big bang took place?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be careful taking the balloon analogy too far, because it starts to break down in many ways. But the answer to where the expansion of the universe emanates &lt;em&gt;from&lt;/em&gt; really &lt;strong&gt;&lt;em&gt;is&lt;/em&gt;&lt;/strong&gt; that it emanates from everywhere; everything we know is contained within the origin of our own existence, and there &lt;em&gt;is&lt;/em&gt; no center and there are no borders by any definition of existence that we can experience.&lt;/p&gt;&#xA;" OwnerUserId="58" LastEditorUserId="58" LastEditDate="2016-01-21T17:59:20.160" LastActivityDate="2016-01-21T17:59:20.160" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="884" PostTypeId="1" AcceptedAnswerId="887" CreationDate="2013-11-06T19:59:26.777" Score="11" ViewCount="335" Body="&lt;p&gt;We've had hubble for two decades. Its become the most famous space telescope, and lately &lt;a href=&quot;http://kepler.nasa.gov&quot;&gt;Kepler&lt;/a&gt; seems to be running for the second place. Are there any more powerful, better telescopes planned for launch in the near future? &lt;/p&gt;&#xA;" OwnerUserId="36" LastActivityDate="2013-11-07T02:48:28.220" Title="What are the next planned space telescopes?" Tags="&lt;telescope&gt;&lt;space&gt;&lt;space-telescope&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="887" PostTypeId="2" ParentId="884" CreationDate="2013-11-07T02:41:24.563" Score="8" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/James_Webb_Telescope&quot; rel=&quot;noreferrer&quot;&gt;The James Webb Telescope is the next one on the launchpad&lt;/a&gt; that you might be familiar with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although there are a few differences that one ought to be aware of.  NASA has an entire program of telescopes to observe the universe, and many of them are designed for different wavelengths of light.  The James Webb is primarily designed for the infrared part of the spectrum.  While many people view the Webb as the successor to the Hubble, it would be more fair to say that it's really a successor to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Spitzer_Space_Telescope&quot; rel=&quot;noreferrer&quot;&gt;Spitzer Telescope&lt;/a&gt;.  Hubble and Spitzer were both part of the &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Great_Observatories_program&quot; rel=&quot;noreferrer&quot;&gt;Great Observatories&lt;/a&gt;&quot; program.  The Webb telescope is barely hanging on to funding, but so far the US hasn't managed to entirely abandon the program, science, or reality. &lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="114" LastEditDate="2013-11-07T02:48:28.220" LastActivityDate="2013-11-07T02:48:28.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="888" PostTypeId="1" AcceptedAnswerId="936" CreationDate="2013-11-07T08:19:46.417" Score="10" ViewCount="806" Body="&lt;p&gt;The &lt;a href=&quot;http://chandra.harvard.edu/photo/2005/tycho/&quot;&gt;SN 1572&lt;/a&gt; remnant, also called Tycho's supernova remnant, is beautiful in X-ray images. It seems to be rather dim in visible light. Are there any amateur photographs of this object? How long are the exposure times needed to get a nice picture?&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2015-02-13T11:41:12.580" Title="Can the supernova remnant SN 1572 be observed by amateur astronomers?" Tags="&lt;amateur-observing&gt;&lt;supernova&gt;&lt;photography&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="889" PostTypeId="2" ParentId="872" CreationDate="2013-11-07T16:25:40.273" Score="3" Body="&lt;p&gt;The limit would depend on the size of the central star as well as the location and sizes of the planets in the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really the limit would be the number of planets that you can fit within the area of which the orbital velocity is &gt;0.  Once you reach that distance, you can't orbit anymore.  Though adding a planet would move this further out due to the added mass itself.  So in theory you could keep pushing this limit and stick more planets in for forever (depending on what you consider to be a planet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem comes more with having stable orbits.  Each planet that you add to the system would affect the rest of the system and could cause the orbits to not be stable anymore.  Also adding planets would allow more planets further out due to the additional mass but it does make figuring out if you have a stable orbit more complicated (&lt;a href=&quot;https://en.wikipedia.org/wiki/N-body_problem&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/N-body_problem&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="281" LastActivityDate="2013-11-07T16:25:40.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="890" PostTypeId="2" ParentId="872" CreationDate="2013-11-07T18:16:26.573" Score="12" Body="&lt;p&gt;There do exist somewhat trivial configurations, which are stable in the long term and which include arbitrarily many bodies. Consider, for example, a set of $N$ circularly moving bodies of the same mass $m$, which obeys the constraint $mN\ll M$, where $M$ is the mass of the star. So long as $mN\ll M$, the bodies move dominantly in the gravitational field of the star and are hence moving stably over long term period. However, as $N$ is arbitrary, one concludes that there is no upper limit on the number of planets, provided that their total mass is small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more physical example would be a protoplanetary disc, or an accretion disc, which is a limit $N\rightarrow \infty$ of an arbitrary planetary system (not not necessarily circular) of a given mass. A yet more physical example is an asteroid belt, consisting of a large number of bodies on, roughly, stable orbits. Finally, during planet formation process the star goes through stages, when it is surrounded by sets of pebbles and asteroids, which keep their structure constant over a large number of orbits (roughly, of order $10^5$). And these all are real physical examples of planetary-like systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to your question would start to alter, though, if you start imposing additional conditions apart from $N\rightarrow \infty$. For example, if you require that bodies do not collide in the long term, some of the above named systems would not work (for example, accretion disc model), but some other would (sets of concentric particles). If you additionally require that the object should obey the definition of a planet, that is have some range of masses, then interesting things will start happening when the total mass of the planets will start being comparable to the mass of the star. So the limit would certainly exist. Finally, you might be more strict about what do you really mean by stability here, and that could also have a bearing on the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To summarize, unless you impose any constraints, there do exist N-body systems orbiting a star in a stable fashion and having arbitrarily large $N$.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2013-11-07T18:16:26.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="891" PostTypeId="1" AcceptedAnswerId="895" CreationDate="2013-11-07T18:29:42.743" Score="34" ViewCount="22742" Body="&lt;p&gt;We know that nothing can have proper velocities larger than the speed of light in vacuum. But are there any objects in space that get close to it? Any comets, or other objects thrown by gravity or supernova explosions that were hurled to incredible speeds? &lt;/p&gt;&#xA;" OwnerUserId="36" LastEditorUserId="36" LastEditDate="2013-11-22T13:23:49.640" LastActivityDate="2021-05-24T20:27:35.883" Title="What's the fastest moving object in the universe?" Tags="&lt;gravity&gt;&lt;universe&gt;&lt;comets&gt;&lt;speed&gt;" AnswerCount="7" CommentCount="6" FavoriteCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="893" PostTypeId="2" ParentId="891" CreationDate="2013-11-07T20:08:53.023" Score="14" Body="&lt;p&gt;There are plenty of rapidly moving objects in astrophysics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good place where one can get moving relativistically is near an event horizon of a black hole. A simple Newtonian estimate illustrates the point. Black hole has all its mass $M$ hidden under an event horizon of the radius of order $r_{g}=\dfrac{2GM}{c^2}$. An object moving circularly in the gravitational field of a black hole at the radius $\alpha r_{g}$, where $\alpha&amp;gt;1$, would have Newtonian orbital velocity $v$ equal to $v=\sqrt{\dfrac{GM}{\alpha r_g}}=\dfrac{c}{\sqrt{2\alpha}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a quilitative estimate of the velocity scale. From general relativity there are no stable circular orbits at $\alpha&amp;lt;3$, but any body will have additional acceleration when inspiralling into the black hole. To add a bit of complexity, when one starts thinking in terms of general relativity, one has to wonder, what do we really mean by objects' velocity and about such kinds of questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, the above conclusion is correct: &lt;strong&gt;In the field of black holes intact objects can obtain relativistic velocities, which are comparable to the speed of light&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many physical examples of such systems: binary megring black holes, black holes merging with neutron stars, supermassive black holes and white dwarfs, etc. While all these systems are driven to eventual merger at relativistic velocities, it is hard for any of their components to get ejected and become free floating. To my knowledge, there are no known free floating relativistic astrophysical bodies, but some of them are indeed probably produced from the pieces of material ejected away at mildly relativistic speeds during mergers involving black holes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One other rare possibility is to have a compact binary system in the field of a supermassive black hole, which is being disrupted due to interaction with it. However, the probability for such a disruption happening when the compact binary is just about to merge is vanishingly low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another ubiquitous class of objects are relativistic jets, which are ultrarelativistic streams of plasma, produced mainly when some accretion onto a black hole is taking place. Particles in such jets move at highly relativictic velocities, though the exact nature of jet formation is yet not completely understood. Finally, there is plenty of relativistic particles present in the background, such as cosmic ray particles and neutrinos. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One last thing to mention would be plasmas which are at relativistic temperatures (of order $10^9 K$), and which hence contain particles (mainly, electrons) moving relativistically. It is rare that plasmas get temperatures that high, but it is definitely possible during core-collapse supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, at sufficiently early stages of the Big Bang, absolutely everything in the Universe was moving relativistically!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: A few more things which came to my mind afterwards: 1) Man-made particle beams in particle accelerators are relativistic, macroscopic, but not astrophysical objects. 2) If there exists intelligent life in the Universe, it might have also produced relativistic objects of macroscopic, but again probably not astrophysical, scale (like spacecrafts).&lt;/p&gt;&#xA;" OwnerUserId="115" LastEditorUserId="115" LastEditDate="2013-11-08T12:35:16.477" LastActivityDate="2013-11-08T12:35:16.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="894" PostTypeId="2" ParentId="891" CreationDate="2013-11-07T20:39:52.550" Score="17" Body="&lt;p&gt;There is also another mediator particle which moves at the speed of light other than the photon. This is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gluon&quot; rel=&quot;nofollow&quot;&gt;gluon&lt;/a&gt;, which is the exchange particle for the strong force. The odd thing about the gluon is that it's never seen by itself (that is, outside of collections of other gluons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, though neutrinos &lt;strong&gt;do&lt;/strong&gt; in fact have mass, they are neutral particles. Why I'm bringing this up is because in &lt;a href=&quot;http://en.wikipedia.org/wiki/Supernova&quot; rel=&quot;nofollow&quot;&gt;supernovae&lt;/a&gt; explosions neutrinos can arrive before the photons in some circumstances - they do not interact with charged particles. Also, because they are weakly interacting particles, they pass through considerable amounts of mass (namely dust and gas) before an interaction might occur. What this means is if you could detect the neutrinos coming from a supernova it could potentially give you early warning that the photons would soon follow. This would give you time to measure its light curve (see: &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0406214&quot; rel=&quot;nofollow&quot;&gt;SNEWS: The SuperNova Early Warning System&lt;/a&gt;). &lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditDate="2015-06-20T21:13:09.660" LastActivityDate="2015-06-20T21:13:09.660" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="895" PostTypeId="2" ParentId="891" CreationDate="2013-11-07T23:04:54.860" Score="37" Body="&lt;p&gt;The answer to this is surprising:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And many (if not all) other galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And they move faster than light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See, the universe is expanding, &lt;a href=&quot;http://en.wikipedia.org/wiki/Accelerating_universe&quot; rel=&quot;noreferrer&quot;&gt;at an accelerating rate&lt;/a&gt;. The fabric of spacetime itself stretches out, so that galaxies seem to move away from each other. The interesting thing is that relativity does not forbid these from moving away faster than light. While &lt;em&gt;local&lt;/em&gt; space is flat and the local speed of light must be upheld, this need not hold at a global scale, so it is possible to have frames which move away from each other faster than $c$. Indeed, &lt;a href=&quot;http://curious.astro.cornell.edu/question.php?number=575&quot; rel=&quot;noreferrer&quot;&gt;there are some galaxies that are moving away from us faster than light&lt;/a&gt; (&lt;a href=&quot;http://curious.astro.cornell.edu/question.php?number=56&quot; rel=&quot;noreferrer&quot;&gt;the only reason we see them is that they used to be closer and moving at a slower speed&lt;/a&gt;). Any pair of galaxies that are 4200 Mpc away from each other (that is, with a redshift of 1.4), are moving away from each other faster than light in each other's frames (numbers stolen from the linked page).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the only consistent way to talk about motion is relative, one can say that we are moving away from other galaxies faster than light, since the reverse is also true. This can put galaxies in the bucket of the fastest moving objects in the universe. As for &lt;em&gt;which&lt;/em&gt; is the fastest, I don't know, we would have to find a pair of galaxies which are the farthest apart (distance measured in the frame of the galaxy, of course), but since the universe is probably more than what we observe&lt;sup&gt;1&lt;/sup&gt;, we can't pinpoint the pair of galaxies for which this is true.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For those who think that it is cheating&lt;sup&gt;2&lt;/sup&gt; to short-circuit the question with space expansion, there are other objects that go faster than light (they are not the fastest objects in the universe though), and these can be found on good 'ol Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Electrons&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In nuclear reactor cooling pools&lt;sup&gt;3&lt;/sup&gt;, we have a phenomenon known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Cherenkov_radiation&quot; rel=&quot;noreferrer&quot;&gt;Cerenkov radiation&lt;/a&gt;. Basically, emitted beta particles move faster than the speed of light in water. This creates an effect of similar origin as the sonic boom, where strong light emanates from the medium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WQtF3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Saywhat? You think I'm cheating again&lt;sup&gt;2&lt;/sup&gt; by putting everything relative to the speed of light in a medium?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alright, fine. Here are some fast objects that &lt;em&gt;don't&lt;/em&gt; require space expansion to be fast, nor do they involve any trickery of semantics where the medium in which they are being measured is not mentioned. Many have already been mentioned by astromax.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tachyons&quot; rel=&quot;noreferrer&quot;&gt;Tachyons&lt;/a&gt;&lt;/strong&gt;:These are particles which go faster than $c$ &amp;mdash; this does not violate relativity as long as they never decelerate to subluminal speeds. However, there isn't much (any?) experimental evidence for these. A lot of BSM models do predict their existence, though. So there's still some cheating here, on to bradyonic matter:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gluon&quot; rel=&quot;noreferrer&quot;&gt;Gluons&lt;/a&gt;&lt;/strong&gt;: These are massless, and though they don't occur freely (except possibly in &lt;a href=&quot;http://en.wikipedia.org/wiki/Glueball&quot; rel=&quot;noreferrer&quot;&gt;glueballs&lt;/a&gt;, though these most probably have mass) they do travel &lt;em&gt;at&lt;/em&gt; $c$. But these can't move at any other speed, so again, this is slightly cheating. On to fermionic matter:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Neutrinos&quot; rel=&quot;noreferrer&quot;&gt;Neutrinos&lt;/a&gt;&lt;/strong&gt;: Now these are viable candidates. The electron neutrino is known to have very, very little mass (we have an upper bound for it, which gives ), and as a result it can easily attain very high speeds. Put it in a gravitational field, and it goes even faster. If you want macroscopic objects, however:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Stuff spiraling around spinning black holes&lt;/strong&gt;: Black holes have a strong gravitational field, and when rotating, they can impart angular momentum (lots of it) to nearby objects like accretion disks. Objects close to a black hole are accelerated to pretty high speeds. In fact, if an object is within the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ergosphere&quot; rel=&quot;noreferrer&quot;&gt;ergosphere&lt;/a&gt;, it moves faster than light from the point of view of certain frames of reference.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Stuff falling into black holes&lt;/strong&gt;: From the faraway frame, an object speeds up and approaches the speed of light as it approaches the horizon of a black hole. Arbitrarily large speeds bounded by $c$ can be attained here.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Black hole plasma jets&lt;/strong&gt;: Jets being flung out of black holes can get pretty fast relative to each other.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1. Due to cosmic expansion, there can be galaxies that are no longer visible to us. Some galaxies may never have been visible to us, if we start watching from when galaxies started forming.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2. I, for one, agree with you.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;3. And other places where you have massive particles being emitted really fast into  a medium&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="88" LastEditDate="2017-12-04T16:54:33.490" LastActivityDate="2017-12-04T16:54:33.490" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="896" PostTypeId="2" ParentId="868" CreationDate="2013-11-08T09:57:09.180" Score="14" Body="&lt;p&gt;This is a question that concerns the &lt;a href=&quot;http://en.wikipedia.org/wiki/Initial_mass_function&quot; rel=&quot;noreferrer&quot;&gt;initial mass function&lt;/a&gt; (IMF) - an empirical (that is, defined by observations rather than theory) function that describes the statistical distribution of stellar masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1955ApJ...121..161S&quot; rel=&quot;noreferrer&quot;&gt;Edwin Salpeter (1955)&lt;/a&gt; was the first to describe the IMF, though if you read &lt;a href=&quot;http://adsabs.harvard.edu//abs/2003PASP..115..763C&quot; rel=&quot;noreferrer&quot;&gt;Chabrier (2003)&lt;/a&gt; there are some reasonably comprehensive explanations of the theory and history. However, &lt;a href=&quot;http://es.ucsc.edu/~glatz/astr_112/lectures/notes19.pdf&quot; rel=&quot;noreferrer&quot;&gt;these lecture notes&lt;/a&gt; are a fair bit more accessible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the approximations in the UCSC lecture notes I linked above, I get that around 4% of stars are between 0.7 and 1.3 solar masses (92% are between 0.1 and 0.7 solar masses!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are perhaps 100 billion stars in a galaxy and 100 billion galaxies in the observable universe, giving something on the order of $4\times 10^{20}$ (400 billion billion) stars that are about ($\pm 30\%$) one solar mass.&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="2529" LastEditDate="2015-03-14T22:09:45.993" LastActivityDate="2015-03-14T22:09:45.993" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="897" PostTypeId="1" CreationDate="2013-11-08T12:33:12.823" Score="16" ViewCount="263" Body="&lt;p&gt;Thinking about the question and answers for &lt;a href=&quot;https://astronomy.stackexchange.com/questions/349/does-the-moon-have-any-oxygen-in-its-atmosphere&quot;&gt;Does the Moon have any oxygen in its atmosphere?&lt;/a&gt; begs the question, despite being considerably smaller than the Moon and further away from the sun, does Ceres have a similar atmosphere as the Moon?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2018-07-28T07:38:01.833" Title="Does Ceres have an exosphere similar to the Moon?" Tags="&lt;planetary-atmosphere&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="898" PostTypeId="1" CreationDate="2013-11-08T18:03:40.050" Score="12" ViewCount="3289" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Initial_mass_function&quot;&gt;Initial Mass Function (IMF)&lt;/a&gt; is the empirical function which describes the initial masses of a population of stars. My questions are,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What are the various IMF's which are used?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) For each, what type of population do they describe? (e.g. - galaxy, dwarf galaxy, globular cluster, etc..)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) And how are they actually calculated? (meaning, do they come from simulations/observations, and what assumptions are made about each?) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Entire answers and pieces of answers are all welcome. Formulae (in latex, please) are encouraged.&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2013-12-11T15:43:45.263" Title="How exactly is the Initial Mass Function (IMF) calculated?" Tags="&lt;stellar-astrophysics&gt;&lt;initial-mass-function&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="899" PostTypeId="2" ParentId="525" CreationDate="2013-11-09T10:20:24.917" Score="11" Body="&lt;p&gt;As you stated, we have not been able to simulate the pressure and temperatures required to generate those that are believed to exist in Jupiter's interior other than in short-lived shockwave experiments, according to the NASA webpage &lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2011/09aug_juno3/&quot; rel=&quot;noreferrer&quot;&gt;A Freaky Fluid inside Jupiter?&lt;/a&gt;, observing that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Liquid metallic hydrogen has low viscosity, like water, and it's a good electrical and thermal conductor,&quot; says Caltech's David Stevenson, an expert in planet formation, evolution, and structure. &quot;Like a mirror, it reflects light, so if you were immersed in it [here's hoping you never are], you wouldn't be able to see anything.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Going further, according to the article &lt;a href=&quot;https://www.llnl.gov/str/Nellis.html&quot; rel=&quot;noreferrer&quot;&gt;Jumpin' Jupiter! Metallic Hydrogen&lt;/a&gt; (Lawrence Livermore National Laboratory), discuss the shock wave results, finding the level at which hydrogen metallises as being&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;from 0.9 to 1.4 Mbar, resistivity in the shocked fluid decreases almost four orders of magnitude (i.e., conductivity increases); from 1.4 to 1.8 Mbar, resistivity is essentially constant at a value typical of that of liquid metals. Our data indicate a continuous transition from a semiconducting to metallic diatomic fluid at 1.4 Mbar, nine-fold compression of initial liquid density, and 3,000 K.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The findings from the researchers above are summarised in the diagram below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/heCDU.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The source is the Jumping Jupiter link above.&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2013-11-09T10:20:24.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="900" PostTypeId="2" ParentId="858" CreationDate="2013-11-10T05:07:14.433" Score="8" Body="&lt;p&gt;As you point out, in an accelerating Universe, large scale structures will become more and more isolated. So at a certain point you will have gravitationally bound superclusters separated by very large voids and less and less filamentary structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once isolated, we can then study the dynamics of these independent superclusters. On very large time scales, galaxies will collide and merge. After collisions you will tend to form elliptical galaxies. So I think that you will end up with a big single elliptical galaxy.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/xX0kp.jpg&quot; alt=&quot;Elliptical galaxy&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we can be interested in the future of the stars of these galaxies. First we currently see that the star formation rate has already peaked some billion years ago. So as the number of galaxies collision that usually targets star formation, the star formation rate will slowly continue to decrease. Moreover, as heavy elements (all elements apart from hydrogen and helium) are formed in stars, the future generation of stars will have more and more heavy elements. From the nuclear point of view, the most stable element is iron, so on very very large time scales light elements will be converted into iron, whereas heavy elements will decay into iron.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/TEJJf.png&quot; alt=&quot;Evolution of the star formation rate&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a little speculative, but I think that on large time scales, more and more interstellar gas and stars will fall at the center of the gravitational potential of super elliptical galaxy. So as the density will increase at the center, you will mechanically form a heavier and heavier supermassive black hole. Another interesting point is that we do not currently know whether proton are stable or not. So on time scales larger than $10^{30}$ years (see more details &lt;a href=&quot;http://en.wikipedia.org/wiki/Proton_decay&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;) protons may naturally decay into lighter subatomic particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So maybe as $t\rightarrow\infty$ we will end up with supermassive black holes and light particles. But as you mentioned, black holes will themselves slowly lose mass by Hawking radiation. In the same time, the expansion rate may have increased significantly leading ultimately to isolated particles in an expanding Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: this is an hypothetical scenario and there are a lot of unknowns&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="176" LastEditorUserId="411" LastEditDate="2013-11-12T04:25:09.257" LastActivityDate="2013-11-12T04:25:09.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="901" PostTypeId="2" ParentId="858" CreationDate="2013-11-10T07:54:47.517" Score="0" Body="&lt;p&gt;The galaxies could either be pulled away or collide with each other. This comes from the two possible theories: The Big Crunch and The Big freeze. But its true if we only consider the two galaxies of interest.      &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Voids and Black Holes should not ignored as there presence should escalate or demolish any of the existing forces. These two are equally responsible for the fate of galaxies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The Big Crunch and The Big Freeze are two theoretical possibilities that pose a future to our universe. Each of the theories conclude the end, where the universe either starts shrinking as it had once expanded and thus ends in a singularity as it once started, or the universe keeps expanding relentlessly making it colder and colder.&#xA;Voids and Black Holes should not directly relate to the above fates, but will be the eventual &lt;strong&gt;result&lt;/strong&gt; of them. &lt;/p&gt;&#xA;" OwnerUserId="400" LastEditorUserId="400" LastEditDate="2014-09-18T12:47:04.970" LastActivityDate="2014-09-18T12:47:04.970" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="902" PostTypeId="1" AcceptedAnswerId="905" CreationDate="2013-11-10T09:23:14.237" Score="6" ViewCount="202" Body="&lt;p&gt;I remember watching a talk by someone who designed a telescope to view a single star at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of the telescope was to detect waste-heat which would presumably be emitted by any civilization utilizing a large percentage of their planet's solar energy, due to the second law of thermodynamics. By focusing on a single star, it would be able to detect when the planet crossed the path of star-light, and the additional energy emitted by the planet - or something like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone know who this might be?&lt;/p&gt;&#xA;" OwnerUserId="304" LastActivityDate="2013-11-10T15:50:57.650" Title="Planned telescope to detect alien waste-heat" Tags="&lt;exoplanet&gt;&lt;observational-astronomy&gt;&lt;telescope&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="905" PostTypeId="2" ParentId="902" CreationDate="2013-11-10T15:50:57.650" Score="6" Body="&lt;p&gt;This would be the &lt;a href=&quot;http://www.the-colossus.com/&quot; rel=&quot;noreferrer&quot;&gt;SETI's Colossus telescope project&lt;/a&gt;, that aims to build a high-resolution, multiple-mirror instrument with ability to directly image the heat generated by other civilizations on planets orbiting stars near us:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/h9X98.jpg&quot; alt=&quot;SETI&amp;#39;s Colossus telescope&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;sup&gt;Artist's impression of the proposed SETI's Colossus Telescope (Credit: Innovative Optics/Colossus Corporation)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As Earth-like civilizations evolve, they use more power. For example,&#xA;  in Roman times, we estimate Ω [the ratio of a civilization’s power production to the amount of solar power it receives] was about 1/1000 what it is today.&#xA;  Humans’ global power consumption is growing by about 2.5 percent per&#xA;  year, even though the world’s population is growing at less than half&#xA;  this rate. In contrast, our knowledge base (the combined total of all&#xA;  recorded information) doubles in just two years. As cultures advance,&#xA;  their information content also must grow, and the power required to&#xA;  manipulate this knowledge eventually dominates a civilization’s total&#xA;  power use.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Using a sensitive coronagraph to remove scattered light that would&#xA;  obscure an exoplanet, Colossus would be able to find hundreds of&#xA;  Earth-sized or larger planets in the habitable zone including any&#xA;  civilizations on their surfaces.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;sup&gt;Source: &lt;a href=&quot;http://www.centauri-dreams.org/?p=27867&quot; rel=&quot;noreferrer&quot;&gt;Paul Gilster for Centauri Dreams&lt;/a&gt; quoting Jeff Kuhn (University of Hawaii), Svetlana V. Berdyugina (University of Freiburg), David Halliday (Dynamic Structures, Ltd., in British Columbia), and Caisey Harlingten (Searchlight Observatory Network, Norwich, England)&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is a Jeff Kuhn's, Institute for Astronomy, University of Hawaii, talk on &lt;a href=&quot;http://www.seti.org/weeky-lecture/colossus-project&quot; rel=&quot;noreferrer&quot;&gt;The Colossus Telescope and the search for ET&lt;/a&gt;, and a quite descriptive blog post by Brian Wang on &lt;a href=&quot;http://nextbigfuture.com/2013/06/technology-for-77-meter-primary-mirror.html&quot; rel=&quot;noreferrer&quot;&gt;Technology for 77 meter primary mirror ground based telescope exists today and it could survey earth sized exoplanets out to 60 light years&lt;/a&gt; advocating the project from a technology perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit more can also be read on the &lt;a href=&quot;http://www.innovativeoptics.ca/&quot; rel=&quot;noreferrer&quot;&gt;Innovative Optics Ltd.&lt;/a&gt; web pages, that are sadly under construction, but they do give a hint that an article about it was written in &lt;a href=&quot;http://www.astronomy.com/issues/2013/june-2013&quot; rel=&quot;noreferrer&quot;&gt;June 2013 edition of the Astronomy Magazine&lt;/a&gt;, titled &lt;em&gt;&lt;a href=&quot;http://www.astronomy.com/~/media/Files/PDF/Magazine%20articles/ET-with-infrared-light.pdf&quot; rel=&quot;noreferrer&quot;&gt;How to find ET with infrared light&lt;/a&gt;&lt;/em&gt; (PDF), by Jeff R Kuhn, Svetlana V Berdyugina, David Halliday, Caisey Harlingten. A nice digest is also available in &lt;a href=&quot;http://www.centauri-dreams.org/?p=27867&quot; rel=&quot;noreferrer&quot;&gt;this Centauri Dreams blog post&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2013-11-10T15:50:57.650" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="906" PostTypeId="1" AcceptedAnswerId="913" CreationDate="2013-11-11T15:00:21.700" Score="9" ViewCount="324" Body="&lt;p&gt;Is there any method to measure the mass of &lt;a href=&quot;https://en.wikipedia.org/wiki/Binary_star#Eclipsing_binaries&quot;&gt;eclipsing binaries&lt;/a&gt; without using Doppler shift effect?&lt;/p&gt;&#xA;" OwnerUserId="407" LastEditorUserId="7982" LastEditDate="2021-04-16T23:23:33.020" LastActivityDate="2021-04-16T23:23:33.020" Title="How can I measure the mass of eclipsing binary stars?" Tags="&lt;mass&gt;&lt;binary-star&gt;&lt;eclipsing-star-systems&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="907" PostTypeId="1" CreationDate="2013-11-11T15:06:05.680" Score="15" ViewCount="292" Body="&lt;p&gt;Accretion discs are ubiquitous in astrophisics. As a direct corollary, they are important for the following question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following model, representing one of the most simple models for accretion discs. A central object is a star (pre-MS, WD or NS, but not a BH) of mass $M$, surrounded by a thin flat disc of material, which continuosly feeds the star at a rate $\dot{M}$, such that $M/\dot{M}$ is much larger than thermal and dynamical timescale of the star (i.e. accretion rate is slow).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everywhere in the accretion disc its local motion is nearly circular and nearly Keplerian. Therefore, at the interface of the star and the disc the disc will always tend to make the star rotate at nearly-Keplerian velocities. From the other hand, if the stellar outer parts were to rotate at nearly-Keplerian velocities, these parts would become gravitationaly detached from the star, which would have significant consequences for the stellar shape and structure. Surely, though, the process is going to be slow and the acquired angular momentum will be redistributed within the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the question: &lt;strong&gt;What will be happening to the star if it approaches nearly break-up velocities due to such a spin-up?&lt;/strong&gt; This involves a few subquestions: How close the rotation rate can actually get to the critical one? If it can get close enough, how would the whole process look like? That is, what would happen in the short term to the star when the effects of rotation will start to affect its structure? What would happen to the star in the long term?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to keep this problem as a purely hydrodynamical one. That is, assume, that the only laws involved are hydrodynamical and gravitational ones, with some constant accretion rate supported. In reality magnetic fields would also play an important role for some stars, and stellar winds could also possibly be important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of the decribed systems are numerous. It might concern cataclysmic variables, millisecond pulsars, pre-main sequence star in a protoplanetary disc, and many more.&lt;/p&gt;&#xA;" OwnerUserId="115" LastEditorUserId="115" LastEditDate="2013-11-11T15:53:08.110" LastActivityDate="2013-12-10T16:37:38.967" Title="Stars at near break-up rotation rates" Tags="&lt;accretion-discs&gt;&lt;stellar-structure&gt;&lt;plasma-physics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="908" PostTypeId="1" AcceptedAnswerId="911" CreationDate="2013-11-12T01:12:37.857" Score="5" ViewCount="517" Body="&lt;p&gt;I was listening to &lt;a href=&quot;http://www.youtube.com/watch?v=QpGDs7cyRMM&quot; rel=&quot;nofollow&quot;&gt;Jeff Kuhn's talk&lt;/a&gt;&#xA;on &lt;a href=&quot;http://www.the-colossus.com/&quot; rel=&quot;nofollow&quot;&gt;SETI's Colossus telescope project&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He explains his theory that a civilization living somewhere the galaxy would want to hide, and thus would shortly learn to avoid broadcasting radio signals all over the galaxy, so as to avoid attracting attention (presumably to avoid inter-stellar predators).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then he comes up with a great idea; due to the second law of thermodynamics, they would have to dissipate heat-energy, and thus, one can detect them &lt;em&gt;anyway&lt;/em&gt;, because they have no way to hide this heat-energy. This, particularly when they are a &lt;a href=&quot;https://en.wikipedia.org/wiki/Type_II_civilization#Type_I&quot; rel=&quot;nofollow&quot;&gt;Type I&lt;/a&gt; civilization, and nearing &lt;a href=&quot;https://en.wikipedia.org/wiki/Type_II_civilization#Type_II&quot; rel=&quot;nofollow&quot;&gt;Type II&lt;/a&gt;; in this case, to avoid global warming, they would capture and use almost all solar energy, and emit an equivalent amount heat-energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus he proposes a telescope to focus on single stars, which would be capable of detecting this heat-energy (by the distinctive infrared light emmited).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a civilization was smart enough to hide all of its radio signals, wouldn't they &lt;em&gt;know&lt;/em&gt; that they cannot avoid emitting heat-energy? Thus their predators would be able to find them &lt;em&gt;anyway&lt;/em&gt;, and thus, it is pointless to hide the radio signals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we should already be seeing their radio signals, and this telescope is being built for a theory with some holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: I am not suggesting that the telescope shouldn't be built; I think it could have other very valuable uses, and it would also be interesting to eliminate the theory if it is indeed incorrect (or prove it correct ... somehow). I am just asking if there might be justifications for the theory, in light of my point-in-question, which I am missing.&lt;/p&gt;&#xA;" OwnerUserId="304" LastEditorUserId="13" LastEditDate="2013-11-12T01:52:13.220" LastActivityDate="2013-11-12T13:10:56.370" Title="Colossus telescope, trying to outsmart aliens?" Tags="&lt;exoplanet&gt;&lt;observational-astronomy&gt;&lt;telescope&gt;&lt;heat&gt;&lt;fermi-paradox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="909" PostTypeId="5" CreationDate="2013-11-12T01:52:55.453" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-11-12T01:52:55.453" LastActivityDate="2013-11-12T01:52:55.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="910" PostTypeId="4" CreationDate="2013-11-12T01:52:55.453" Score="0" Body="Fermi's Paradox is about the lack of evidence for alien life, despite the growing evidence that suitable planets are reasonably plentiful." OwnerUserId="304" LastEditorUserId="38" LastEditDate="2013-11-12T04:28:46.567" LastActivityDate="2013-11-12T04:28:46.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="911" PostTypeId="2" ParentId="908" CreationDate="2013-11-12T05:40:12.447" Score="10" Body="&lt;p&gt;The real reasoning has nothing to do with some civilization &quot;deliberately&quot; hiding its radio emissions. Rather, the problem is that we can not expect some other civilization to do something we would not do ourselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It makes no sense whatsoever to radiate large amounts of energy into space when there exist other, more economical alternatives. Radio transmitters circa 1911 required whole dedicated dynamo machines to run at abysmally low efficiency - these days we are using tiny battery powered devices with directed radio beam steering and adaptive power minimization as to not transmit a single joule more than absolutely necessary to get the data through to intended destination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem arises from the fact that most of the &quot;radio SETI&quot; effort to date was concentrated around the &quot;cosmic waterhole&quot; (the frequency of interstellar hydrogen emission). This is a reserved frequency here on Earth, so nobody broadcasts on it. Common sense suggests that aliens would probably do the same - after all, we expect them to have developed radio astronomy as well. Unfortunately, listening on any other frequency will probably pick up locally generated broadcasts first and completely mask the weak alien signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the above, what choices are we left with? One would be to construct a large radio telescope somewhere Earth radio emissions can not reach - the far side of the moon or even a trans-solar location on the Earth orbit come to mind. Unfortunately, such project, while doable, is hugely expensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other approach is to develop &quot;distributed&quot; telescopes which can be built gradually on low budget and to develop techniques to cancel the local noise. For low frequency radio telescopy such project is already ongoing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/LOFAR&quot;&gt;http://en.wikipedia.org/wiki/LOFAR&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For infrared, we've got this &quot;Collosus&quot; proposal, but place our hopes on JWST (&lt;a href=&quot;http://www.jwst.nasa.gov/&quot;&gt;http://www.jwst.nasa.gov/&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we don't expect to see much in visible light for all the obvious reasons.&lt;/p&gt;&#xA;" OwnerUserId="412" LastEditorUserId="38" LastEditDate="2013-11-12T13:10:56.370" LastActivityDate="2013-11-12T13:10:56.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="912" PostTypeId="1" AcceptedAnswerId="919" CreationDate="2013-11-12T13:45:04.650" Score="9" ViewCount="118" Body="&lt;p&gt;I have heard a certain statement, which can be summarized succinctly into &quot;we live in a local bubble&quot;, implying that solar system is located in an intestrellar dust underdensity region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This region is allegedly produced by a supernova explosion, which happened near the Sun in the distant past. The bubble size was claimed to be of order of few tens of parsec, and the fact of its existence allows one to easier deal with dust extinction for nearby stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another claim I have heard about this bubble, is that this concept is obsolete nowadays, for if even if there were a supernova taking place near the Sun, gas diffusion timescales are short enough, so that no vestige would have remained from this past event by now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to ask you, therefore, to comment on the text above and in particular indicate, whether such an underdensity really exists, and what do we know about it.&lt;/p&gt;&#xA;" OwnerUserId="115" LastEditorUserId="115" LastEditDate="2013-11-13T12:36:33.263" LastActivityDate="2013-11-13T12:36:33.263" Title="Do we live in a galactic bubble?" Tags="&lt;dust&gt;&lt;interstellar-medium&gt;&lt;galactic-dynamics&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="913" PostTypeId="2" ParentId="906" CreationDate="2013-11-12T15:36:17.797" Score="5" Body="&lt;p&gt;It depends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For many categories of stars, it is possible to determine the approximate mass of the star from its non-redshifted spectrum. Stars on the main sequence: this table of &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_classification#Harvard_spectral_classification&quot; rel=&quot;noreferrer&quot;&gt;spectral classifications&lt;/a&gt; lists the approximate mass for the various classifications. A more precise mass estimation should be possible with a more precise classification. Sol, for example, is a G2, near the high end of G, putting it near the top end of the range (0.80 - 1.04 stellar masses).&lt;/p&gt;&#xA;" OwnerUserId="38" LastActivityDate="2013-11-12T15:36:17.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="914" PostTypeId="1" CreationDate="2013-11-12T15:00:33.827" Score="14" ViewCount="7977" Body="&lt;p&gt;How could I know what phase the moon is in when it's high in the sky during sunrise? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could I tell in general what its height in the sky will be at any given time? &lt;/p&gt;&#xA;" OwnerUserId="414" OwnerDisplayName="JekylHyde" LastEditorUserId="6" LastEditDate="2013-11-13T13:29:40.610" LastActivityDate="2013-11-13T13:29:40.610" Title="How can I calculate moon phase and height?" Tags="&lt;the-moon&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="915" PostTypeId="2" ParentId="914" CreationDate="2013-11-12T23:26:40.420" Score="4" Body="&lt;p&gt;The first question is quite simple if you imagine how the Earth is looked from above. To help you with that I've added a little drawing below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably you know that the Sun rises on the East and sets on the West.  Imagine now how the sun illuminates the Earth: Europe is in the afternoon when the East cost of US is waking up. Imaging this you can see that the Earth is turning anti-clockwise if looked from the north.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, imagine you are somewhere in US (so we can use the drawing from below), the sun is rising (on the East) and the moon is above you (something close to the zenith).  This means the moon the configuration of the Sun-Earth-Moon system is something like in the drawing.  As you can see in the drawing I've kept in white what's illuminated from the Sun.  In Spanish we use to say that the moon is a bit of a liar, this comes because it shows a C shape when is waxing (decreciente) and a D shape when is crescent (creciente), so just the opposite letter of its action (don't know if there's a similar expression in other languages, and notice that the moon is not a liar for the Spanish speakers in the Souther hemisphere).  So, coming back to the drawing, you can imagine that from the red-star position you will see the moon with a C shape, therefore is waxing... which means in 7 days or so would be new moon so is going to move between the Sun and Earth.. and we can see how the moon rotates around the Earth in the same direction than the earth rotates around itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/I8HBE.png&quot; alt=&quot;Earth Moon Sun system - not in scale&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really understand your second question, so I cannot help much with that.  But, if you mean to get a roughly calculation as on where the moon is at a certain time knowing its phase you can calculate it by knowing that the Earth spins 15°/hour and that the moon positions for each phase.  Therefore, new moon the moon will be closer to the zenith at noon (local solar time), 6am for waxing moon, 12 am for full moon and 6pm for crescent moon. However, this is just a roughly approximation as the moon movement happens on a plane which is 5.14° inclined from the ecliptic plane which corresponds to positions between 18.29° and 28.58° latitude on Earth.  So the same phase moon can be at different heights at different times of the year at the same hour.&lt;/p&gt;&#xA;" OwnerUserId="137" LastActivityDate="2013-11-12T23:26:40.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="916" PostTypeId="1" AcceptedAnswerId="953" CreationDate="2013-11-13T06:31:58.680" Score="10" ViewCount="248" Body="&lt;p&gt;What are the newly discovered ring systems of Saturn, and the circumstances relating to the discovery? Is there something that makes them different from the old well-known rings, like their formation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ring in question is the &lt;a href=&quot;http://photojournal.jpl.nasa.gov/catalog/PIA08322&quot; rel=&quot;nofollow noreferrer&quot;&gt;Janus/Epimetheus ring system&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/f5yrE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="416" LastEditorUserId="6" LastEditDate="2013-11-14T16:33:25.633" LastActivityDate="2013-11-19T08:30:41.570" Title="How are the newly discovered Janus/Epimetheus rings different from the other rings of Saturn?" Tags="&lt;saturn&gt;&lt;planetary-ring&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="917" PostTypeId="1" CreationDate="2013-11-13T06:44:01.813" Score="19" ViewCount="3425" Body="&lt;p&gt;According to this &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_mountains_on_Mars_by_height&quot;&gt;Wikipedia page&lt;/a&gt;, the top five highest mountains on Mars (and the highest on Venus) are all taller than Mount Everest (and Mauna Kea as measured from the ocean floor).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does a planet's mass or gravity affect the height of it's mountains?  Is there an upper limit that a mountain can reach given the mass or gravity of a planet?  &lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2020-10-22T14:01:28.870" Title="Does a planet's mass or gravity affect the height of it's mountains?" Tags="&lt;planet&gt;&lt;gravity&gt;&lt;mass&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="918" PostTypeId="1" AcceptedAnswerId="8240" CreationDate="2013-11-13T11:38:27.183" Score="15" ViewCount="1721" Body="&lt;p&gt;I was just watching a lecture from Carl Sagan. He talked about figuring out the distance to the stars; it got me interested in learning more about the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, the Inverse square law and parallax can be used. Can anyone expand on these? Specifically with regards to what I could do to measure the distance from Earth to Proxima  Centauri.&lt;/p&gt;&#xA;" OwnerUserId="417" LastActivityDate="2015-04-05T10:53:34.603" Title="Calculate Distance To Stars" Tags="&lt;star&gt;&lt;distances&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="919" PostTypeId="2" ParentId="912" CreationDate="2013-11-13T12:35:50.873" Score="3" Body="&lt;p&gt;As I have discovered afterwards, the described bubble exists and is in fact a known phenomenon. For those, who want to read more, here is a wiki link &lt;a href=&quot;http://en.wikipedia.org/wiki/Local_Bubble&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Local_Bubble&lt;/a&gt; and the links therein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, it is an underdensity in ISM (not only dust) in the region of Orion Arm (which is a minor spiral arm). The underdensity is up to ten times comparing to normal density in the Milky way, depending on the component of ISM in consideration. The shape of the cavity is complex, typical sizes being of order $100 \textrm{pc}$, but when it comes to practical observations, one should be wary when neglecting dust extinction at distances over $50 \textrm{pc}$.&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2013-11-13T12:35:50.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="920" PostTypeId="2" ParentId="918" CreationDate="2013-11-13T12:58:54.983" Score="5" Body="&lt;p&gt;One way to find distance to a collection of stars is to hope for an &lt;a href=&quot;http://en.wikipedia.org/wiki/RR_Lyrae&quot; rel=&quot;noreferrer&quot;&gt;RRLyrae&lt;/a&gt;&#xA; in the bunch. Since RRLyrae are &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_candle#Standard_candles&quot; rel=&quot;noreferrer&quot;&gt;standard candles&lt;/a&gt;, you can use the inverse square law to extract the distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/369Q6.jpg&quot; alt=&quot;rrlyraePeriod&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2013-11-13T12:58:54.983" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="922" PostTypeId="1" AcceptedAnswerId="923" CreationDate="2013-11-13T16:46:41.953" Score="13" ViewCount="397" Body="&lt;p&gt;Considering their distance from their parent stars, might Oort cloud object such as comets be exchanged between passing stars (assuming that other stars have similar Oort clouds)?&lt;/p&gt;&#xA;" OwnerUserId="209" OwnerDisplayName="dotancohen" LastActivityDate="2014-10-19T15:28:18.197" Title="Might Oort cloud comets be exchanged between solar systems?" Tags="&lt;oort-cloud&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="923" PostTypeId="2" ParentId="922" CreationDate="2013-11-13T22:12:09.467" Score="4" Body="&lt;p&gt;You can exclude the 'considering the distance' piece - of course Oort Cloud objects could transfer between different gravitational fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However what is it you think will make this transfer? Without some sort of gravitational impetus why would one of these objects leave the solar system? And if you do manage to slingshot one out of the solar system at a high enough speed to exit the Sun's gravity, remember that most directions end up very far away from any other solar systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tl;Dr sure, but not very likely&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2013-11-13T22:12:09.467" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="924" PostTypeId="2" ParentId="917" CreationDate="2013-11-14T13:03:27.020" Score="5" Body="&lt;p&gt;In addition to the answer quoted by @UV-D, gravity affects the height of mountains composed of loose material (e.g. sand, volcanic ash). In a pile of loose material, the height is determined by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Angle_of_repose&quot;&gt;angle of repose&lt;/a&gt;, i.e. the steepest angle at which material will stay in place rather than roll down the sides of the mountain. This angle depends on gravity. &lt;/p&gt;&#xA;" OwnerUserId="361" LastActivityDate="2013-11-14T13:03:27.020" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="925" PostTypeId="1" AcceptedAnswerId="927" CreationDate="2013-11-14T18:16:15.383" Score="11" ViewCount="4123" Body="&lt;p&gt;Some asteroids and comets are non-spherical. But is the nature of big things and gravity so that large things in the universe are always spherical? What is the biggest astronomical object in terms of volume out there that's not spherical?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note: Astronomical object is a defined term. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Astronomical_object&quot; rel=&quot;nofollow&quot;&gt;Astronomical object - Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="36" LastEditorUserId="6" LastEditDate="2013-12-13T15:19:06.770" LastActivityDate="2019-06-06T11:30:03.393" Title="What's the largest non-spherical astronomical object in the universe?" Tags="&lt;universe&gt;" AnswerCount="10" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="927" PostTypeId="2" ParentId="925" CreationDate="2013-11-14T21:15:32.230" Score="13" Body="&lt;p&gt;The largest sub-galactic astronomical object (in volume) that we know of is the &lt;a href=&quot;http://astropixels.com/diffusenebulae/NGC3372-03.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carina Nebula&lt;/a&gt;, which is a non-spherical &lt;a href=&quot;http://en.wikipedia.org/wiki/Diffuse_nebula#Diffuse_nebulae&quot; rel=&quot;nofollow noreferrer&quot;&gt;diffuse nebula&lt;/a&gt;. The Carina Nebula has a radius of about 100 parsecs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/e/ea/Carina_Nebula.jpg/800px-Carina_Nebula.jpg&quot; alt=&quot;Carina Nebula&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Image credit: ESA&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider astronomical objects of the galactic scale, then the galaxy &lt;a href=&quot;http://en.wikipedia.org/wiki/IC_1101&quot; rel=&quot;nofollow noreferrer&quot;&gt;IC 1101&lt;/a&gt; is the largest astronomical object (in volume) that we know of. From Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The galaxy has a diameter of approximately 6 million light years, which makes it currently (as of 2013) the largest known galaxy in terms of breadth. It is the central galaxy of a massive cluster containing a mass (mostly dark matter) of roughly 100 trillion stars. Being more than 50 times the size of the Milky Way and 2000 times as massive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;IC 1101 is on the left in the image below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/c/cf/Abell_2029.png/800px-Abell_2029.png&quot; alt=&quot;IC 1101&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an image from the Sloan Digital Sky Survey (SDSS) of IC 1101:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oXRpD.jpg&quot; alt=&quot;SDSS IC 1101&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll leave it to you to decide if an elliptical galaxy counts as &quot;spherical&quot; or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't count elliptical galaxies, the largest spiral galaxy is &lt;a href=&quot;http://en.wikipedia.org/wiki/NGC_262&quot; rel=&quot;nofollow noreferrer&quot;&gt;NGC 262&lt;/a&gt;, pictured below. It has a diameter of 1.3 million light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dlkuc.jpg&quot; alt=&quot;NGC 262&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:47.053" LastActivityDate="2014-01-09T20:02:04.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="928" PostTypeId="2" ParentId="925" CreationDate="2013-11-15T08:50:44.383" Score="5" Body="&lt;p&gt;If you count conglomerates of galaxy style objects, the largest structure in the universe so far is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Huge-LQG&quot; rel=&quot;nofollow noreferrer&quot;&gt;Huge Large Quasar Group&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to be four billion ly in size, and contains dozens of Quasars, which in turn are very active galactic nuclei.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DJv9R.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2013-11-15T08:50:44.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="929" PostTypeId="2" ParentId="925" CreationDate="2013-11-15T10:27:15.720" Score="9" Body="&lt;p&gt;In the same vein as @arne's answer, this just in from Slashdot: &lt;a href=&quot;http://science.slashdot.org/story/13/11/15/003222/astronomers-discover-largest-structure-in-the-universe&quot; rel=&quot;noreferrer&quot;&gt;Astronomers Discover Largest Structure In the Universe&lt;/a&gt;, &lt;a href=&quot;https://medium.com/the-physics-arxiv-blog/267ddcb8057b&quot; rel=&quot;noreferrer&quot;&gt;linked article&lt;/a&gt; and &lt;a href=&quot;http://arxiv.org/abs/1311.1104&quot; rel=&quot;noreferrer&quot;&gt;the original&lt;/a&gt; - &lt;strong&gt;10 billion light years across&lt;/strong&gt;, based on data from gamma-ray bursts.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;So this thing, presumably another wall of even more distant galaxies, is the new largest structure in the universe.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;One of the fundamental tenets of cosmology is the Cosmological principle—this holds that the distribution of matter in the universe will appear uniform if viewed from a large enough scale.&lt;/p&gt;&#xA;&lt;p&gt;At whatever scale they look, large scale structures always seem to emerge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;(I haven't seen a name associated with it yet.)&lt;/p&gt;&#xA;" OwnerUserId="425" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2014-09-11T17:42:29.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="930" PostTypeId="2" ParentId="925" CreationDate="2013-11-15T13:39:36.700" Score="7" Body="&lt;p&gt;The largest non-spherical objects which are virialized (meaning essentially dynamically relaxed; not in the process of forming or collapsing) are super clusters of galaxies (&lt;a href=&quot;http://en.wikipedia.org/wiki/Supercluster#Nearby_superclusters&quot; rel=&quot;noreferrer&quot;&gt;here is a list of nearby super clusters&lt;/a&gt;). They're the most massive objects before you must begin to consider large scale structures like filaments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These objects are largely non-spherical. Most of them happen to be &lt;a href=&quot;http://en.wikipedia.org/wiki/Ellipsoid&quot; rel=&quot;noreferrer&quot;&gt;tri-axial ellipsoids&lt;/a&gt; (either footballs, frisbees, or somewhere in between) due to the nature of large scale structure formation occurring along two axes at different times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think tri-axial ellipsoids are just spheres squashed in certain ways (and that therefore clusters don't really count as &quot;non-spherical&quot;), then the answer would be filamentary structures (both the most massive, and the largest in scale). Here's an image from the Millennium Simulation to show what a super cluster would look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JdmmE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2013-11-15T13:39:36.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="931" PostTypeId="1" AcceptedAnswerId="972" CreationDate="2013-11-15T15:36:42.100" Score="10" ViewCount="212" Body="&lt;p&gt;Is there some ranking / statistics that shows how common given type of celestial objects is? A table that would allow the answers to questions like: What type of star is most common per volume of our galaxy? Are white dwarfs more common than red giants? What percentage in the number of celestial objects located are the black holes? Pulsars? How many red giants have been catalogued?&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="25" LastEditDate="2014-11-23T15:01:49.327" LastActivityDate="2014-11-23T15:01:49.327" Title="Appearance (frequency) ranking of stars?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="932" PostTypeId="1" AcceptedAnswerId="940" CreationDate="2013-11-15T16:01:11.913" Score="5" ViewCount="234" Body="&lt;p&gt;From reading &lt;a href=&quot;https://astronomy.stackexchange.com/q/918/43&quot;&gt;this question on calculating distance to stars&lt;/a&gt; and from a bit of background reading on the standard candle theory I still don't see how we can confirm that a star we see at one distance and a star we see at another distance (redshifted and at a different apparent luminosity) are the same type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond a certain distance, parallax is not possible, so how can we confirm star X and star Y are the same type therefore the difference in redshift and luminosity means Y must be at distance Z. How do we know it isn't a different type of star at a different distance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or are we just saying they look similar, so probably are?&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2013-11-17T08:49:10.667" Title="How can we be sure that we have identified very distant stars correctly?" Tags="&lt;star&gt;&lt;distances&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="933" PostTypeId="1" CreationDate="2013-11-15T16:47:23.047" Score="3" ViewCount="1619" Body="&lt;p&gt;I've been reading up on plate tectonics and found that there are several different driving forces involved. One of the main categories of driving force mentioned on Wikipedia is forces related to the Earth's rotation. The section on driving forces related to Earth rotation presents the information as a sort of historical dialogue of the scientific process, which is somewhat helpful but has left me confused as to exactly what role the Earth's rotation &lt;em&gt;does&lt;/em&gt; play in plate tectonics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So how does rotation affect the movement of plates?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="6" LastActivityDate="2013-12-12T14:45:43.010" Title="What effect does the Earth's rotation have on plate tectonics?" Tags="&lt;earth&gt;" AnswerCount="0" CommentCount="8" ClosedDate="2013-12-12T14:45:49.933" ContentLicense="CC BY-SA 3.0" />
  <row Id="935" PostTypeId="2" ParentId="888" CreationDate="2013-11-15T18:33:53.383" Score="5" Body="&lt;p&gt;It does not appear to be practical to photograph with amateur equipment.  According to the Wikipedia article the remnant was viewed visually with Palomar telescope.  Links to studies of the remnants were done using 2m + telescopes.  So trying to get a visible light photo would require an extremely large telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This table does not list a magnitude for the remnant (B CAS) &lt;a href=&quot;http://maps.seds.org/Const/Data/cassiopeia_tab.html&quot;&gt;http://maps.seds.org/Const/Data/cassiopeia_tab.html&lt;/a&gt;  So it appears that it might be extremely faint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this list of deep sky objects does not list the remnant either.  &lt;a href=&quot;http://www.dibonsmith.com/cas_con.htm&quot;&gt;http://www.dibonsmith.com/cas_con.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="281" LastActivityDate="2013-11-15T18:33:53.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="936" PostTypeId="2" ParentId="888" CreationDate="2013-11-15T18:34:47.577" Score="9" Body="&lt;p&gt;As you say, SN 1572 is not very bright in the optical.  There are some H&amp;alpha; regions that have been observed with world-class optical telescopes, but they do not look like the X-ray and infrared images that you normally see.  In fact, images from the Palomar Optical Sky Survey 2 (with a limiting magnitude of ~22) do not reveal any nebular emission from this object at all:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8l8bN.png&quot; alt=&quot;POSS2 of SN 1572&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the optical emission must be fainter than 22nd magnitude.  I do not think there are amateur photographs of an object this faint and I believe a CCD on an amateur telescope would saturate from noise before revealing any of the nebular emission.&lt;/p&gt;&#xA;" OwnerUserId="141" LastEditorUserId="141" LastEditDate="2014-02-25T12:29:15.160" LastActivityDate="2014-02-25T12:29:15.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="937" PostTypeId="1" AcceptedAnswerId="963" CreationDate="2013-11-15T19:08:59.477" Score="10" ViewCount="496" Body="&lt;p&gt;I'm in the northern hemisphere, facing north, and notice that the moon &#xA;is due north (azimuth 0) and is still above the horizon, albeit barely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean the moon is necessarily circumpolar (doesn't set), at &#xA;least for the next few hours? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the moon moves away from due north, it's altitude/elevation &#xA;normally increases. However, it's possible that the moon's declination &#xA;is decreasing so rapidly that the moon's elevation actually decreases, &#xA;and the moon sets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this possible? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is tangentially related to my question &lt;a href=&quot;https://stackoverflow.com/questions/19862226/&quot;&gt;&quot;libnova odd behavior for 89.5 degrees north latitude&quot;&lt;/a&gt; on Stack Overflow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT to answer questions: I am facing true north. I am referring to the Earth's true moon, but not necessarily in any given time frame. In other words, if something like this happens 10,000 years from now, that's fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: OK, I believe the moon CAN set even if it's up when due north. Example: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You are at latitude 89.5N &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The moon's declination when due north is +0.5N. This means the &#xA;moon grazes the horizon when due north. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the moon's declination were constant, it's highest elevation &#xA;for the day would be 1 degree. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since the moon's declination can change more than one degree per &#xA;day (average of 1.5 degrees/day), it's quite possible for the moon's &#xA;declination to go below -0.5N, at which point it would have set (it &#xA;will actually set before this point unless it happens to reach due &#xA;south). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In theory, this could happen with any object whose declination &#xA;changes, provide you're close enough to the North Pole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll leave this update open for critiques. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Actually, this whole question is just asking whether the midnight sun/moon can set at any position other than due north, which is fairly obviously true. Reasoning: once the moon/sun finally sets not due north, it was obviously still up the last time it was due north. &lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorUserId="-1" LastEditDate="2017-05-23T12:37:21.700" LastActivityDate="2013-11-25T19:18:31.560" Title="Can moon set after being up when due north? (pseudo circumpolarity)" Tags="&lt;horizon&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="938" PostTypeId="2" ParentId="937" CreationDate="2013-11-16T05:53:05.840" Score="6" Body="&lt;p&gt;Yes - &lt;a href=&quot;http://www.xefer.com/2011/10/lunar-arctic-circle&quot; rel=&quot;nofollow&quot;&gt;here is a good explanation&lt;/a&gt; of the situation. Just as it's possible to have continuous daylight for days or weeks on end, you can have continuous moonlight too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon's orbit is tilted at (a maximum of) just over 5 degrees to the ecliptic, however this tilt has an oscillatory &lt;a href=&quot;http://www.ifa.hawaii.edu/users/lin/ast110-6/applets/precession_of_moon_orbit.htm&quot; rel=&quot;nofollow&quot;&gt;precession&lt;/a&gt; with a period of 18.6 degrees. This precession means that our &quot;lunar arctic circle&quot; will vary in latitude over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the moon is sitting exactly on the ecliptic (latitudinally speaking), it would remain above the horizon until it is at a sufficiently low phase of illumination that its current ecliptic longitude is below the horizon. Of course in reality the moon only crosses the ecliptic twice during each lunar cycle (it's these two points that precess). &lt;a href=&quot;http://www.stellarium.org&quot; rel=&quot;nofollow&quot;&gt;Stellarium&lt;/a&gt; is a cool program to visualize this on if you turn on the ecliptic grids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One important thing to note is that this phenomenon is dependent on both the time of year &lt;em&gt;and&lt;/em&gt; the phase of the moon. In the summer months, new and near-new moons are the ones that can remain visible for days or weeks on end. Conversely, at the North pole, you might not see a full moon at all for a few months!&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="276" LastEditDate="2013-11-16T06:25:51.203" LastActivityDate="2013-11-16T06:25:51.203" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="940" PostTypeId="2" ParentId="932" CreationDate="2013-11-17T08:49:10.667" Score="7" Body="&lt;p&gt;We determine the spectral type (i.e. temperature) of a star using multicolour photometry, or (ideally) spectroscopy. By guesstimating the temperature, mass and radius of a star, we can say that two stars that have pretty similar observational properties probably are closely related to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cepheid_variables&quot;&gt;Cepheid variables&lt;/a&gt;, for example, display periodic pulsations that depend quite strongly on their intrinsic luminosity -- this is why they're reasonably good standard candles. Their characteristic variability makes them clearly identifiable as a Cepheid, and observations of their pulsation are backed up by data that place them in the same region of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hertzsprung%E2%80%93Russell_diagram&quot;&gt;HR diagram&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it looks like a duck, swims like a duck, and quacks like a duck, then chances are it's a duck. If it's as hot as a Cepheid and pulsates like a Cepheid, then it probably is a Cepheid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer your final paragraph: yep. Though with no spatial resolution and often (in the absence of parallax measurements) only tenuous distance data, classifying stars is a rather messy business.&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2013-11-17T08:49:10.667" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="941" PostTypeId="1" CreationDate="2013-11-17T08:50:52.540" Score="1" ViewCount="796" Body="&lt;p&gt;What are solar maximum and solar minimum?&lt;br&gt;&#xA;Is there a relation between sun spots and solar maximum and minimum?&lt;br&gt;&#xA;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="150" LastActivityDate="2013-11-17T08:55:10.217" Title="Solar maximum and minimum" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="942" PostTypeId="1" AcceptedAnswerId="952" CreationDate="2013-11-17T08:52:00.143" Score="15" ViewCount="6444" Body="&lt;p&gt;I was Googling about the oldest galaxies in the Universe. Everywhere is written that their age is known by the light. Referring to this line &lt;em&gt;&quot;Since light travels at a set speed, if you look at a star ten light years away, you're really seeing what it looked like ten years ago&quot;&lt;/em&gt; on &lt;a href=&quot;http://motherboard.vice.com/blog/astronomers-have-found-the-furthest-oldest-galaxy&quot; rel=&quot;noreferrer&quot;&gt;this site&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is, as we know light is continuous, how do scientists know that a particular ray has started its journey some 1000 light years back? How is the age of stars determined?&lt;/p&gt;&#xA;" OwnerUserId="379" LastEditorUserId="13" LastEditDate="2013-11-23T17:00:38.050" LastActivityDate="2017-05-12T17:06:03.297" Title="How do scientists determine the age of stars?" Tags="&lt;star&gt;&lt;light&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="943" PostTypeId="2" ParentId="941" CreationDate="2013-11-17T08:55:10.217" Score="2" Body="&lt;p&gt;&lt;strong&gt;Solar maximum&lt;/strong&gt;&#xA;&lt;em&gt;Solar maximum or solar max is a normal period of greatest solar activity in the 11 year solar cycle of the Sun. During solar maximum, large numbers of sunspots appear and the sun's irradiance output grows by about 0.07%.&lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_maximum&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt; The increased energy output of solar maxima can impact Earth's global climate and recent studies have shown some correlation with regional weather patterns.&lt;/em&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_maximum&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solar minimum&lt;/strong&gt;&#xA;&lt;em&gt;Solar minimum is the period of least solar activity in the 11 year solar cycle of the sun. During this time, sunspot and solar flare activity diminishes, and often does not occur for days at a time. The date of the minimum is described by a smoothed average over 12 months of sunspot activity, so identifying the date of the solar minimum usually can only happen 6 months after the minimum takes place. Solar minima are not generally correlated with changes in climate but recent studies have shown a correlation with regional weather patterns.&lt;/em&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_minimum&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="379" LastActivityDate="2013-11-17T08:55:10.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="945" PostTypeId="1" AcceptedAnswerId="946" CreationDate="2013-11-17T18:29:44.150" Score="4" ViewCount="958" Body="&lt;p&gt;Is there a site / service where I can get the data for the positions of satellites (I mean &lt;em&gt;artificial&lt;/em&gt; satellite, ISS, etc.)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that some sites provide a visualization of these positions, but I'm only interested in the data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit, for clarifications:&lt;/em&gt; What I want is &lt;strong&gt;only&lt;/strong&gt; the data. My final goal is to create my own site that displays the position of the satellite (I do that just for fun). That's why I just need the data, and not a service that displays the position on a map.&lt;/p&gt;&#xA;" OwnerUserId="431" LastEditorUserId="431" LastEditDate="2013-11-18T12:44:46.053" LastActivityDate="2013-11-18T12:44:46.053" Title="Open data for satellites positions" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="946" PostTypeId="2" ParentId="945" CreationDate="2013-11-18T12:24:14.443" Score="5" Body="&lt;p&gt;There are many such online services, most of them free and a Google search away, so I won't make this a list answer, and since you didn't specify what data in particular you're hoping to find, here's one of my favorite satellite trackers with fairly complete position data, ground track display, map options and so on: &lt;a href=&quot;http://science.nasa.gov/iSat/&quot;&gt;NASA iSAT (Interactive Satellite Viewer)&lt;/a&gt;. You might want to allow it to use your location for more convenient display, but it isn't necessary to enable that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's for a desktop computer, but if you're hoping to spot satellites (a.k.a. &lt;em&gt;tracking&lt;/em&gt;) on the go, then I'd personally suggest installing &lt;a href=&quot;https://play.google.com/store/apps/details?id=com.google.android.stardroid&quot;&gt;Sky Map +&lt;/a&gt; on your smartphone / tablet (it is free, but there are of course others), because it includes a lot more useful information and has both compass / gyroscope / GPS positioning for easier identification of spotted satellites (it pans automatically) as well as manual panning mode. You can tap on screen to display the toolbar and select objects that you want displayed, or switch to ruby &amp;amp; black monochromatic night mode, if default display is too bright. Satellite information can be updated in settings via an additional download, and all of it is free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For downloadable data alone (to perhaps include in your own software or import into your database), one of such sites is &lt;a href=&quot;http://www.celestrak.com/&quot;&gt;CelesTrak&lt;/a&gt; that includes &lt;a href=&quot;http://www.celestrak.com/NORAD/elements/&quot;&gt;TLE (Two-Line Element) sets&lt;/a&gt; and a &lt;a href=&quot;http://www.celestrak.com/SpaceTrack/TLERetriever3Help.asp&quot;&gt;Space Track TLE Retriever 3&lt;/a&gt; tool to download Space Track data sets.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditDate="2013-11-18T12:31:22.360" LastActivityDate="2013-11-18T12:31:22.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="947" PostTypeId="1" AcceptedAnswerId="980" CreationDate="2013-11-18T13:10:38.717" Score="6" ViewCount="519" Body="&lt;p&gt;In &lt;a href=&quot;http://motls.blogspot.com/2013/11/the-expansion-is-accelerating-due-to.html?m=1&quot;&gt;this&lt;/a&gt; article which nicely explains why it is really the quantity $\rho + 3p$ which is relevant to determine if the expansion of the universe is accelerating or decelerating by making use of the for this question relevant second Friedmann equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{\ddot{a}}{a} = -\frac{4\pi G}{3}(\rho + 3p)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is mentioned that for cosmic strings &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;p = -\frac{\rho}{3}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which has the effect that they dont contribute to the &quot;non-inertial&quot; expansion of the universe, and for cosmic domain walls we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;p = -\frac{2 \rho}{3} &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which leads to an accelerated expansion of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas I understand the derivations of such equations of state for radiation, &quot;ordinary&quot; matter, and a constant source of dark energy, I have not yet seen analog calculations for cosmic strings and domain walls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how can the equation of states for topological defects such as cosmic strings and cosmic domain walls be derived?&lt;/p&gt;&#xA;" OwnerUserId="92" LastEditorUserId="92" LastEditDate="2013-11-18T13:53:49.830" LastActivityDate="2013-11-21T06:05:02.307" Title="How can the equation of state for cosmic strings and domain walls be derived?" Tags="&lt;cosmology&gt;&lt;cosmological-inflation&gt;&lt;general-relativity&gt;&lt;dark-energy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="948" PostTypeId="1" AcceptedAnswerId="949" CreationDate="2013-11-18T14:33:01.907" Score="5" ViewCount="1126" Body="&lt;p&gt;For focusing the image of a telescope, one can use a Bahtinov mask. How does this mask work, and how did Mr Bahtinov get the idea to cut such a peculiar shape? Is it possible for me to design my own focusing masks?&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2013-11-19T14:24:36.000" Title="How does a Bahtinov mask work?" Tags="&lt;amateur-observing&gt;&lt;telescope&gt;&lt;optics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="949" PostTypeId="2" ParentId="948" CreationDate="2013-11-18T15:48:33.880" Score="4" Body="&lt;p&gt;The idea is to create a specific diffraction pattern so that you know if you image is in focus.  The pattern is so that you end up with 3 lines crossing the source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mask consists of three separate grids, positioned in such a way that the grids produce three angled diffraction spikes at the focal plane of the instrument for each bright image element (star). As the instrument's focus is changed the central spike appears to move from one side of the star to the other. In reality, all three spikes move but the central spike moves in the opposite direction to the two spikes forming the 'X'. Optimum focus is achieved when the middle spike is centered on the star and symmetrically positioned between the other two spikes. Small deviations from optimal focus are easily visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Bahtinov_mask&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Bahtinov_mask&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can construct a mask with some materials:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.deepskywatch.com/Articles/make-bahtinov-mask.html&quot; rel=&quot;nofollow&quot;&gt;http://www.deepskywatch.com/Articles/make-bahtinov-mask.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a Hartmann Mask that splits the object into multiple when not in focus.  This seems a little easier to construct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cloudynights.com/item.php?item_id=518&quot; rel=&quot;nofollow&quot;&gt;http://www.cloudynights.com/item.php?item_id=518&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="281" LastEditorUserId="281" LastEditDate="2013-11-19T14:24:36.000" LastActivityDate="2013-11-19T14:24:36.000" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="950" PostTypeId="1" CreationDate="2013-11-19T03:32:45.920" Score="6" ViewCount="262" Body="&lt;p&gt;I live in western Saudi Arabia, I am trying to see the comet ISON, reading here and there tells that I can see it by naked eye somewhere next to the constellation leo, I used an app to tells me where constellation leo is but I could not see the comet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it visible from Saudi Arabia? if so, in which direction (west, east, etc.) and what time is the best to see it?&lt;/p&gt;&#xA;" OwnerDisplayName="user444" LastActivityDate="2013-11-21T08:24:34.553" Title="Can I see comet ISON from Saudi Arabia?" Tags="&lt;amateur-observing&gt;&lt;comets&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="951" PostTypeId="2" ParentId="942" CreationDate="2013-11-19T06:38:19.437" Score="3" Body="&lt;p&gt;The most straightforward approach is to use a so-called Hertzsprung-Russell diagram: stars of a given mass &quot;burn&quot; in a fairly predictable way, so by looking at a star's spectrum and luminosity one can infer its possible age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hertzsprung%E2%80%93Russell_diagram&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hertzsprung%E2%80%93Russell_diagram&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia also has an overview of some additional methods:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_age_estimation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Stellar_age_estimation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="412" LastActivityDate="2013-11-19T06:38:19.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="952" PostTypeId="2" ParentId="942" CreationDate="2013-11-19T08:20:21.607" Score="3" Body="&lt;p&gt;Scientists look at groups of stars to determine the ages. As another poster already said, the HR diagram is the tool used to determine ages. The colors emitted from the stars is also used to determine ages because color is indicative of where the star is on its life cycle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cfa.harvard.edu/news/2011-15&quot; rel=&quot;nofollow&quot;&gt;How to Learn a Star's True Age&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the article says, stars are assessed by clusters. The stars formed around the same time and from there scientists determine the age. We have the following going on: brightness, color, and clusters. Size can be used as a determinant, however there are white dwarf stars which are very small, are dim as observed from Earth, yet are very old stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also to clarify-- just because a star is far away and therefore can take thousands of light years for the light to reach is does not mean it's old. The speed of light is simply a measurement of time of travel. &lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="417" LastEditDate="2013-11-23T16:58:19.620" LastActivityDate="2013-11-23T16:58:19.620" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="953" PostTypeId="2" ParentId="916" CreationDate="2013-11-19T08:30:41.570" Score="4" Body="&lt;p&gt;&lt;a href=&quot;http://saturn.jpl.nasa.gov/science/moons/epimetheus/&quot; rel=&quot;nofollow&quot;&gt;http://saturn.jpl.nasa.gov/science/moons/epimetheus/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scientists believe the ring formation was due to the break up of one moon. Cassini has taken photographs of the moons and Voyager was the satellite to allow scientists to discover the moons and the rings which follow their orbit. What is seemingly special about these rings is the structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://saturn.jpl.nasa.gov/photos/imagedetails/index.cfm?imageId=2277&quot; rel=&quot;nofollow&quot;&gt;http://saturn.jpl.nasa.gov/photos/imagedetails/index.cfm?imageId=2277&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2013-11-19T08:30:41.570" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="956" PostTypeId="2" ParentId="717" CreationDate="2013-11-19T09:15:50.720" Score="8" Body="&lt;p&gt;All pulsars are observed to slow down gradually,albeit with a wide range of rates.The following diagram,known as the $\dot{P}-P$ diagram for pulsars illustrates this.The crab pulsar yes indeed would be a bad choice as a clock due to it's high $\dot{P}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are pulsars to the bottom left of the diagram which have periods of the order of milliseconds(hence called millisecond) pulsars and their $\dot{P}$ is also very low.These are the pulsars which can be used as atomic clocks,for example:MSP J0437− 4715.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/f4j4G.gif&quot; alt=&quot;$\dot{P}-P$&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LFb9h.jpg&quot; alt=&quot;Parameters of MSP J0437− 4715. &quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We can see that the both the period and the period derivative of this&#xA;  particular pulsar are very low providing great precision to the period&#xA;  of the system.So it can be used as a precise clock.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&lt;a href=&quot;http://www.cv.nrao.edu/course/astr534/PulsarTiming.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.cv.nrao.edu/course/astr534/PulsarTiming.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.&lt;a href=&quot;http://www.cv.nrao.edu/course/astr534/Pulsars.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.cv.nrao.edu/course/astr534/Pulsars.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.&lt;a href=&quot;http://en.wikipedia.org/wiki/PSR_J0437-4715&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/PSR_J0437-4715&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="450" LastActivityDate="2013-11-19T09:15:50.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="961" PostTypeId="2" ParentId="917" CreationDate="2013-11-20T00:27:10.873" Score="-1" Body="&lt;p&gt;The mountains actually form as a result of the tectonic plates' movement at the earth mantle. Suggesting that a new mountain would be formed tomorrow, during the tectonic activity rock from the mantle and above moves upwards, only the tip of the newly formed mountain would consist of sediments like sand or soil. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I don't think that the gravity or mass of a planet affects the mountains' height. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plus, Earth is the only known planet to be affected by plate tectonics. So, the 'birth' mechanisms must be a lot different and cannot be compared with the ones' acting on our planet.&lt;/p&gt;&#xA;" OwnerUserId="454" LastActivityDate="2013-11-20T00:27:10.873" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="962" PostTypeId="1" CreationDate="2013-11-20T02:45:19.133" Score="6" ViewCount="521" Body="&lt;p&gt;Is lunar elevation at a given location for a given day unimodal: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Unimodal#Unimodal_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;Unimodal function&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, once the moon's elevation reaches a minimum (which may &#xA;be above or below the horizon depending on if the moon is &#xA;circumpolar), does it increase continuously to its maximum, or can it &#xA;decrease briefly and then increase again (resulting in a local minimum &#xA;that's not a global minimum) before reaching the maximum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also similar to the above as it goes from the maximum elevation &#xA;(which may also be above or below the horizon depending on whether the &#xA;moon rises at all on a given day). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why this is important: I think I found a bug in libnova that assumes &#xA;the moon is circumpolar if it is above the horizon when it's due &#xA;north. &lt;a href=&quot;https://astronomy.stackexchange.com/questions/937/&quot;&gt;This is untrue&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/libnova-fix.pl&quot; rel=&quot;nofollow noreferrer&quot;&gt;My fix&lt;/a&gt;&#xA;does not assume this, but does assume that lunar elevation is &#xA;unimodal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's untrue, I'll need to further fix my fix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Here's how far I've gotten: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The moon's declination and hour angle (at a given location) are &#xA;both functions of time. Thus, we can implicitly consider the moon's &#xA;declination to be a function of its hour angle. This only works &#xA;because the moon never experiences retrograde motion. Bodies that &#xA;experience retrograde motion could have two declinations for a given &#xA;hour angle, so the declination would no longer be a function of the &#xA;hour angle. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Using the standard formula, we see the moon's elevation is: &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$ &#xA;   \sin ^{-1}(\sin (\text{lat}) \sin (\text{dec}(\text{ha}))+\cos (\text{ha}) &#xA;    \cos (\text{lat}) \cos (\text{dec}(\text{ha}))) &#xA;$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where dec(ha) is the declination at hour angle, ha is the hour angle, &#xA;and lat is the latitude. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Since the moon's elevation is a differentiable function (except &#xA;potentially at the poles and equator), we find the derivative with &#xA;respect to the hour angle: &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;   \frac{\text{dec}'(\text{ha}) (\sin (\text{lat}) \cos &#xA;    (\text{dec}(\text{ha}))-\cos (\text{ha}) \cos (\text{lat}) \sin &#xA;    (\text{dec}(\text{ha})))-\sin (\text{ha}) \cos (\text{lat}) \cos &#xA;    (\text{dec}(\text{ha}))}{\sqrt{1-(\sin (\text{lat}) \sin &#xA;    (\text{dec}(\text{ha}))+\cos (\text{ha}) \cos (\text{lat}) \cos &#xA;    (\text{dec}(\text{ha})))^2}} &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The moon's elevation will reach a min/max (in the sense of &#xA;unimodality) when this derivative is 0. That occurs when: &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$ &#xA;   \sin (\text{ha})\text{ = } \text{dec}'(\text{ha}) (\tan (\text{lat})-\cos &#xA;    (\text{ha}) \tan (\text{dec}(\text{ha}))) &#xA;$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question becomes: can this equation have more than two &#xA;solutions for 0 &amp;lt; ha &amp;lt; 2*Pi &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which I hope someone answers at: &lt;a href=&quot;https://math.stackexchange.com/questions/587136/&quot;&gt;https://math.stackexchange.com/questions/587136/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2013-11-30T20:56:06.527" Title="Is lunar elevation at a given location for a given day unimodal?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="963" PostTypeId="2" ParentId="937" CreationDate="2013-11-20T04:30:55.917" Score="1" Body="&lt;p&gt;Answering my own question: yes, the moon &lt;em&gt;can&lt;/em&gt; set after being up &#xA;while due north, and this situation occurs quite frequently at far &#xA;northern/southern latitudes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: On 2013-08-03 21:45:21 at 72N, 0E, the moon is due north and up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rXNzy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Less than 24h later, at 2013-08-04 at 20:51:08, the moon sets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BQLXe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This situation is typical. If the moon has been up for more than 24 hours: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;It will almost always set at a location other than due north. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the 24 hours prior to moonset, it will have been due north and &#xA;above the horizon. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note: I used stellarium for dramatic effect, even though it tends to &#xA;give inaccurate results compared to other libraries ( &#xA;&lt;a href=&quot;https://stackoverflow.com/questions/16293146/&quot;&gt;https://stackoverflow.com/questions/16293146/&lt;/a&gt; ). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other libraries give similar, but nonidentical results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general case is the same, however: if the moon's been up for over &#xA;24 hour and sets, it will have been up and due north some time in the &#xA;last 24 hours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am downvoting and not accepting Moriarty's answer because: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;He originally answered the wrong question. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;He later answered the right question incorrectly. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;He later claimed that this is a &quot;rare borderline case&quot;. In fact, &#xA;this happens almost anywhere/anytime the moon is up for more than 24 &#xA;hours.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;EDIT: it turns out this is known. All planets (incl the Sun and Moon) do NOT reach maximum elevation when they culminate: &lt;a href=&quot;http://www.geoastro.de/TransitCulm/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.geoastro.de/TransitCulm/index.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorUserId="-1" LastEditDate="2017-05-23T12:37:21.700" LastActivityDate="2013-11-25T19:18:31.560" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="964" PostTypeId="1" AcceptedAnswerId="965" CreationDate="2013-11-20T04:55:00.393" Score="16" ViewCount="2260" Body="&lt;p&gt;On this &lt;a href=&quot;http://spaceinimages.esa.int/Images/2013/11/Hot_gas_sloshing_in_a_galactic_cauldron&quot;&gt;link&lt;/a&gt;, it states the following: &quot;large assemblies of galaxies that are permeated by even larger amounts of diffuse gas. With temperatures of 10 million degrees or more&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are these diffused (ionized) gas able to become so hot when they are in large distances from one another and have very little density? &lt;/p&gt;&#xA;" OwnerUserId="458" LastEditorUserId="6" LastEditDate="2013-11-20T16:23:36.533" LastActivityDate="2014-02-27T08:11:45.203" Title="Why is the interstellar medium so hot?" Tags="&lt;interstellar-medium&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="965" PostTypeId="2" ParentId="964" CreationDate="2013-11-20T06:31:02.240" Score="11" Body="&lt;p&gt;What that ESA (European Space Agency) page titled &lt;a href=&quot;http://spaceinimages.esa.int/Images/2013/11/Hot_gas_sloshing_in_a_galactic_cauldron&quot;&gt;Hot gas sloshing in a galactic cauldron&lt;/a&gt; that you link to describes are called &lt;a href=&quot;https://en.wikipedia.org/wiki/Warm%E2%80%93hot_intergalactic_medium&quot;&gt;WHIM&lt;/a&gt; (Warm–Hot Intergalactic Medium). They are not interstellar medium, but intergalactic medium gas. The difference in density is huge, with interstellar medium density at an average of $\rho ∼ 1\ ppcm$ (one proton per cubic centimeter), but the density of these WHIM being even a few orders of magnitude lower at $\rho ∼ 10^{−6}−10^{−5}\ ppcm$, or roughly 1 to 10 protons per cubic meter (&lt;a href=&quot;http://www.nasa.gov/mission_pages/chandra/news/10-048.html&quot;&gt;NASA's Chandra X-ray Observatory&lt;/a&gt; quotes average density of 6 protons per cubic meter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is interesting about WHIM is that they are absolutely huge. We're talking of distances that extend across clusters of galaxies (so stretching multiple millions of light years), which means that even as tenuous as they are, account for a large portion of the baryonic matter of the Universe:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Such matter is predicted to account for a sizable fraction ($∼ 50\%$)&#xA;  of all the baryons in the local&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; ($z &amp;lt; 1$ &lt;code&gt;[redshift in the infrared&#xA;  spectrum]&lt;/code&gt;) Universe, and it is thus considered the best candidate to&#xA;  host the baryons seen at high redshift and missing from the low&#xA;  redshift census.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So now about their heat emissions, and why are they detected in the X-ray range in the first place (the ESA's article mentions the photograph featured there was taken by the ESA’s XMM-Newton X-ray observatory):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Electrons and baryons in the WHIM are shock-heated during their infall&#xA;  in the dark matter LSS &lt;code&gt;[Large–Scale Structures]&lt;/code&gt; potential well, and settle in&#xA;  filamentary/sheet-like structures surrounding LSSs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've added a few clarifications in quotes encapsulated in square brackets, but what this means is that parts of these WHIM interact with &lt;a href=&quot;https://en.wikipedia.org/wiki/Active_galactic_nucleus&quot;&gt;AGN&lt;/a&gt; (Active Galactic Nucleus) as the galaxies pass by, and the X-ray emissions of AGN excite &lt;a href=&quot;https://en.wikipedia.org/wiki/Baryon&quot;&gt;Baryonic matter&lt;/a&gt; to a temperature $T ∼ 10^5−10^7 K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quote sources:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://iopscience.iop.org/0004-637X/503/2/518/&quot;&gt;The Cosmic Baryon Budget&lt;/a&gt;, Fukugita et al. 1998&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/bib_query?2004ApJ...616..643F&quot;&gt;The Cosmic Energy Inventory&lt;/a&gt;, Fukugita &amp;amp; Peebles 2004&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Additional reading:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nasa.gov/mission_pages/chandra/news/10-048.html&quot;&gt;X-ray Discovery Points to Location of Missing Matter&lt;/a&gt;, NASA&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.skyandtelescope.com/news/93797364.html&quot;&gt;Last “Missing” Normal Matter Is Found&lt;/a&gt;, Sky &amp;amp; Telescope&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/1004/1004.5359v1.pdf&quot;&gt;Studying the WHIM Content of the Galaxy Large–Scale Structures along the Line of Sight to H 2356-309&lt;/a&gt; (PDF), L. Zappacosta et al., Harvard-Smithsonian Center for Astrophysics&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditDate="2013-11-20T06:45:20.137" LastActivityDate="2013-11-20T06:45:20.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="966" PostTypeId="1" AcceptedAnswerId="967" CreationDate="2013-11-20T06:37:21.087" Score="20" ViewCount="32856" Body="&lt;p&gt;Also, does it always take the same amount of time, or does it fractionally differ on each revolution?&lt;/p&gt;&#xA;" OwnerUserId="460" LastEditorUserId="66" LastEditDate="2014-11-05T09:11:25.753" LastActivityDate="2018-03-03T12:46:51.990" Title="What is the distance that the Moon travels during one orbit around the Earth?" Tags="&lt;the-moon&gt;&lt;orbit&gt;&lt;earth&gt;&lt;distances&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="967" PostTypeId="2" ParentId="966" CreationDate="2013-11-20T07:08:34.320" Score="15" Body="&lt;p&gt;The Moon has an orbital eccentricity of 0.0549, so its path around the Earth is not perfectly circular and the distance between the Earth and the Moon will vary from the Earth's frame of reference (Perigee at 363,295 km and apogee at 405,503 km), see for example &lt;a href=&quot;https://astronomy.stackexchange.com/a/22/13&quot;&gt;second animation explaining Lunar librations in this answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But its orbit can be said, in an oversimplified manner, to be periodic, with no significant &lt;a href=&quot;https://en.wikipedia.org/wiki/Precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;apsidal precession&lt;/a&gt; (not really true, but somewhat irrelevant for my following musings here to be still close enough), so we can calculate its orbital length based on its quoted average orbital speed of 1.022 km/s and orbital period of 27.321582 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, plugging our numbers in a calculator, $l = v * t$, we get the Moon's orbital length of 2,412,517.5 km (or 1,499,070 miles). Should be close enough. Source of all orbital elements of the Moon is &lt;a href=&quot;https://en.wikipedia.org/wiki/Moon&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia on Moon&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2013-11-20T07:08:34.320" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="968" PostTypeId="2" ParentId="966" CreationDate="2013-11-20T07:48:36.053" Score="10" Body="&lt;p&gt;Concerning your first question, a simple estimation can be done assuming the distance Earth-Moon ≅ 4·10⁵km, and the orbit circular. So you can calculate the distance as a circumference (C=2πr) like that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2π·4·10⁵km&#xA;=8π·10⁵km &#xA;≅ &lt;strong&gt;2.4 millions of kilometers&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course you can do more precise calculations, but sometimes is good to have at first an idea of the orders of magnitude.&lt;/p&gt;&#xA;" OwnerUserId="12" LastActivityDate="2013-11-20T07:48:36.053" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="969" PostTypeId="1" AcceptedAnswerId="971" CreationDate="2013-11-20T12:26:06.027" Score="17" ViewCount="1550" Body="&lt;p&gt;Assume that I am standing on one of the many satellites of Neptune. Will I be able to see the planet with my naked eyes (without any sort of visual aid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I was to stand on Moon I would be able to see the Earth because the light from Sun is reflected off Earth into my eyes and I am able to see Earth. In case of Neptune, the distance between Neptune and the Sun is so great that the sun would look like any other star (a bit brighter). In that case if I am looking at Neptune would I be able to see it ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the other gas giants ?&lt;/p&gt;&#xA;" OwnerUserId="461" LastActivityDate="2015-12-29T09:59:40.770" Title="Will Neptune be visible with the naked eye if I am standing on its satellite" Tags="&lt;the-sun&gt;&lt;neptune&gt;&lt;uranus&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="970" PostTypeId="1" AcceptedAnswerId="974" CreationDate="2013-11-20T14:13:38.147" Score="15" ViewCount="640" Body="&lt;p&gt;How would a person measure that he is at a planetary pole? My first inclination is to use a sextant to ensure that the Sun remains at a constant inclination. However, due to the orbit around the Sun and a possible planetary axial tilt, the Sun's inclination may vary over the course of the year and that could affect measurements if the length of the day is a significant portion of the length of the year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second inclination is to try to identify a pole star or at least the point about which distant stars rotate during the course of a planetary rotation. The observer could then (at least in theory) measure the inclination of this point assuming that planetary precession takes a long enough time (as on Earth). However, in the most likely chance that there is no bright star near the pole, I don't see how a sextant could measure reliably the inclination of a blank point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the first method could be used, but with other objects in the sky (stars) in place of the Sun. Thus, if a given star's inclination does not change as it (apparently) revolves around the horizon, we are at the pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any better methods, assuming a completely self-contained system (i.e. no external satellites)?&lt;/p&gt;&#xA;" OwnerUserId="209" LastActivityDate="2014-06-29T03:56:16.413" Title="How would I measure that I'm at a pole?" Tags="&lt;planet&gt;&lt;observational-astronomy&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="971" PostTypeId="2" ParentId="969" CreationDate="2013-11-20T14:36:07.780" Score="15" Body="&lt;p&gt;&lt;a href=&quot;http://nssdc.gsfc.nasa.gov/planetary/factsheet/neptunefact.html&quot; rel=&quot;nofollow&quot;&gt;NASA's Neptune fact sheet&lt;/a&gt; states that Neptune's irradiance is $1.51 W/m^2$. This is roughly three orders of magnitude less than at Earth (at $1367.6 W/m^2$). This sounds like a lot, and it is quite an attenuation. So much that for example solar panels for interplanetary probes are not worthwile at these distances from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you compare this with the &lt;a href=&quot;http://www.ecse.rpi.edu/~schubert/Light-Emitting-Diodes-dot-org/Sample-Chapter.pdf&quot; rel=&quot;nofollow&quot;&gt;sensitivity of the human vision system&lt;/a&gt;, this seems still pretty bright. A sunny day outside will be approximately $10^5 cd/m^2$. So equally on Neptune a sunny day will probably be $10^2 cd/m^2$. This is comparable to an &lt;a href=&quot;http://www.wolframalpha.com/input/?i=100cd/m%5E2&quot; rel=&quot;nofollow&quot;&gt;overcast sky&lt;/a&gt;, and still well within human &lt;a href=&quot;https://en.wikipedia.org/wiki/Photopic_vision&quot; rel=&quot;nofollow&quot;&gt;photopic vision&lt;/a&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The perceived brightness of Neptune, as viewed from one of its moons, will not change, so yes, it will appear 1000 times less bright than earth does, but since you don't have any comparisons to brighter light sources there, you will perceive it as pretty bright!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the other gas giants, their surface irradiance will be equally higher, so they will in turn appear much brighter than Neptune.&lt;/p&gt;&#xA;" OwnerUserId="50" LastEditorDisplayName="user10250" LastEditDate="2015-12-29T09:59:40.770" LastActivityDate="2015-12-29T09:59:40.770" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="972" PostTypeId="2" ParentId="931" CreationDate="2013-11-20T14:43:54.123" Score="11" Body="&lt;p&gt;In general smaller stars are more common than larger, more massive stars.&#xA;This is typically described by the Initial Mass Function which describes the number of stars of a given mass which form initially. Moreover, larger, more massive stars burn through hydrogen quickly and live shorter lives, making them even less common as time passes. Thus, to determine the distribution of objects of various types of equal masses, you need to start with an initial mass function, and then use a stellar evolution model to evolve the system to the time of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, a Universal Mass Function, is an attempt to answer the question you pose in terms of mass. That is, are more massive or less massive objects more common in the universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Universal Mass Function provides a description the density of occurrence of objects of various masses per cubic parsec, shown in Figure 3 of &lt;a href=&quot;http://arxiv.org/abs/1112.3340&quot; rel=&quot;nofollow&quot;&gt;Kroupa et al. 2011&lt;/a&gt;. As you can see, the more massive something is, the less common its occurrence. Thus, black holes which form from massive stars will be far less common that small low mass stars which become white dwarfs. There are still uncertainties the exact distribution, shown at least partially by the shaded portions of the figure, and it will vary from one region of space to the next. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more info check out &lt;a href=&quot;http://arxiv.org/abs/1112.3340&quot; rel=&quot;nofollow&quot;&gt;Kroupa et al.&lt;/a&gt; and &lt;a href=&quot;http://arxiv.org/abs/0705.1599&quot; rel=&quot;nofollow&quot;&gt;Binggeli and Hascher. 2007.&lt;/a&gt; “Is There a Universal Mass Function?”.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kroupa, Pavel, Carsten Weidner, Jan Pflamm-Altenburg, Ingo Thies, Joerg Dabringhausen, Michael Marks, and Thomas Maschberger. 2011. “The Stellar and Sub-Stellar IMF of Simple and Composite Populations”. ArXiv e-print 1112.3340. &lt;/p&gt;&#xA;" OwnerUserId="457" LastEditorUserId="457" LastEditDate="2014-11-22T22:55:08.177" LastActivityDate="2014-11-22T22:55:08.177" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="973" PostTypeId="1" AcceptedAnswerId="982" CreationDate="2013-11-20T14:53:03.680" Score="12" ViewCount="259" Body="&lt;p&gt;I read &lt;a href=&quot;http://online.wsj.com/news/articles/SB10001424052702304672404579183940409194498&quot;&gt;some news stories&lt;/a&gt; about the current Solar maximum and the flipping of the magnetic poles. They say that one of the magnetic poles has switched polarity and the other will switch in about a month. I don't really understand magnetism, but I had thought you had to have 2 opposite polarity magnetic poles. How can the Sun's magnetic poles not switch in sync with each other?&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="38" LastEditDate="2013-11-20T21:54:37.387" LastActivityDate="2013-11-21T10:19:04.323" Title="How can only 1 magnetic pole of the Sun change at a time?" Tags="&lt;the-sun&gt;&lt;magnetic-field&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="974" PostTypeId="2" ParentId="970" CreationDate="2013-11-20T14:59:24.167" Score="13" Body="&lt;p&gt;You could confirm with taking an image of the star trails.  They would form a circle with the apparent center at the zenith of the location.  You do not need a pole star at all.  Just a night of viewing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would also be able to tell based on the height above the horizon that the sun is.  On the equinox, the sun would be on the horizon at noon (when it is at its maximum height).  On the summer solstice, its angle will be above the horizon equal to the angle of precession.  And below by the same amount on the winter solstice.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.pbs.org/weta/roughscience/series1/challenges/latlong/page4.html&quot; rel=&quot;noreferrer&quot;&gt;Measuring latitude using the sun&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any other day in the year, you will need to determine the max angle based where you are between the equinox and solstice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.pbs.org/weta/roughscience/series1/challenges/latlong/page5.html&quot; rel=&quot;noreferrer&quot;&gt;Compensating for the Earth's tilt&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, you would determine if you were at a pole the same way you determine any other latitude.  You are just looking for a result of 90 degrees.&lt;/p&gt;&#xA;" OwnerUserId="281" LastEditorUserId="417" LastEditDate="2013-11-23T16:56:02.737" LastActivityDate="2013-11-23T16:56:02.737" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="975" PostTypeId="2" ParentId="918" CreationDate="2013-11-20T15:50:10.057" Score="3" Body="&lt;p&gt;For close objects, the parallax method works perfectly. Though for higher distances, the Standard candles, as mentioned before, are used. The brightness of RR Lyrae, Supernovae type Ia, could be calculated, therefore, with the amount of light we get from these objects, we can estimate the distance. For even farther objects, the redshift method is used to calculate the distance, where a given line transition with a given frequency (iron emission for instance) is measured, and the shift in frequency, caused by the expansion of the universe (a phenomenon described mathematically) gives us a hint for the distance of the object.&lt;/p&gt;&#xA;" OwnerUserId="463" LastActivityDate="2013-11-20T15:50:10.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="976" PostTypeId="2" ParentId="964" CreationDate="2013-11-20T16:59:21.793" Score="10" Body="&lt;p&gt;Just to add to TidalWave's answer - something that is easier to simply imagine, the trivial &quot;why&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we call Temperature on thermodynamic level is Speed on atomic level. Saying the medium has a high temperature is equivalent to saying particles of that medium move awfully fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, they must be moving fast. They must be moving faster than the escape speed of the galaxies, or the ones ejected from galaxies wouldn't escape them and the primordial would get captured by the galaxies instead. Being so sparse they collide extremely rarely too - so whatever slowdowns resulting from collisions (expending energy e.g. as photons) simply almost never happens. In short, you're getting particles that were fast (hot) enough to get (and remain) there at all, and had no opportunity to cool down.&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="83" LastEditDate="2014-02-27T08:11:45.203" LastActivityDate="2014-02-27T08:11:45.203" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="977" PostTypeId="2" ParentId="970" CreationDate="2013-11-20T21:06:13.170" Score="4" Body="&lt;p&gt;To determine if you're at a pole:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Be there during that pole's half-year &quot;night&quot;.  Better around the solstice, when the sun will be well below the horizon&lt;/li&gt;&#xA;&lt;li&gt;Pick some of the brighter stars visible, scattered at different azimuths. &lt;/li&gt;&#xA;&lt;li&gt;Measure the altitude above the horizon of each. Tricky if you don't have a sea level horizon.&lt;/li&gt;&#xA;&lt;li&gt;Stay warm for a few hours.&lt;/li&gt;&#xA;&lt;li&gt;Measure all the altitudes again.&lt;/li&gt;&#xA;&lt;li&gt;Repeat Steps 3 through 5 for at least 12 hours.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If all the altitudes remain unchanged, you're at the pole.  Congratulations!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, all the stars will wiggle up and down the same amount.  A $2^\circ$ wiggle (diameter of 4 full moons) means you're $60$ nautical miles from the pole.  Pick a star that has just reached its highest point and move the appropriate distance &lt;strong&gt;&lt;em&gt;away&lt;/em&gt;&lt;/strong&gt; from that star.  Go back to Step 1 above and repeat...&lt;/p&gt;&#xA;" OwnerUserId="420" LastEditorUserId="38" LastEditDate="2013-11-20T21:50:44.363" LastActivityDate="2013-11-20T21:50:44.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="978" PostTypeId="1" AcceptedAnswerId="981" CreationDate="2013-11-21T04:48:01.427" Score="20" ViewCount="2592" Body="&lt;p&gt;If the formula to calculate the force of gravity between two objects is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = GM_1M_2/r^2,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;why do planets stay in orbit? Or is there another formula at work? &lt;/p&gt;&#xA;" OwnerUserId="468" LastEditorUserId="1577" LastEditDate="2015-09-24T09:34:09.103" LastActivityDate="2015-09-24T09:34:09.103" Title="Doesn't gravity attract objects in space until they collide?" Tags="&lt;planet&gt;&lt;gravity&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="979" PostTypeId="2" ParentId="828" CreationDate="2013-11-21T05:25:00.780" Score="3" Body="&lt;p&gt;This is a very complex answer. It has to do a lot with gravity and electromagnetism. Nebulae are made of particles that act according to those forces and give you the illusion of viscosity. Those particles are actually moving. Take a look at this simulation it might give you an idea: &lt;a href=&quot;https://www.youtube.com/watch?v=YbdwTwB8jtc&quot; rel=&quot;nofollow&quot;&gt;Simulation&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="468" LastActivityDate="2013-11-21T05:25:00.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="980" PostTypeId="2" ParentId="947" CreationDate="2013-11-21T05:35:16.923" Score="6" Body="&lt;p&gt;My topological defect cosmology is a little rusty, but I'm pretty sure this is how it goes.&#xA;Start with the fluid equation,&#xA;$$&#xA;\dot{\rho} + 3 {\dot{a} \over a} \left( \rho + p \right) = 0,&#xA;$$&#xA;and the equation of state,&#xA;$$&#xA;p = w \rho.&#xA;$$&#xA;Plug the equation of state into the fluid equation, assume a constant $w$, and you'll find&#xA;$$&#xA;\rho \propto a^{-3(1 + w)}.&#xA;$$&#xA;Now we'll find $\rho(a)$ for strings and sheets, and read $w$ off of them.&#xA;For cosmic strings, $\rho$ is&#xA;$$&#xA;\rho_{\rm string} = \sum^N_i {\lambda L_i \over V},&#xA;$$ &#xA;where $N$ is the number of strings in our cosmic horizon, $\lambda$ is the linear density of the strings, and $L_i$ is the length of each string.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the key: we've got to &lt;strong&gt;assume the length of any cosmic string scales with the expansion of the universe&lt;/strong&gt;, since they're topological defects. Given that assumption, we can get the dependence of $\rho_{\rm string}$ on $a$:&#xA;$$&#xA;\rho_{\rm string}(a) \propto {a \over a^3} = a^{-2}.&#xA;$$&#xA;Thus,&#xA;$$&#xA;2 = 3(1 + w_{\rm string}),&#xA;$$&#xA;and $w_{\rm string} = -{1 \over 3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, for domain walls, $\rho$ is&#xA;$$&#xA;\rho_{\rm wall} = \sum^N_i {\sigma A_i \over V},&#xA;$$&#xA;and since $A_i \propto a^2$, we get $w_{\rm wall} = -{2 \over 3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps!&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditDate="2013-11-21T06:05:02.307" LastActivityDate="2013-11-21T06:05:02.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="981" PostTypeId="2" ParentId="978" CreationDate="2013-11-21T06:52:40.260" Score="17" Body="&lt;p&gt;When an object is in orbit, there are &lt;em&gt;two&lt;/em&gt; factors at play, not just one. The first, as you mention, is the force of gravity pulling the objects together. However, each object also has a momentum component which is generally (in the case of circular orbits) perpendicular to the direction of the gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at the common situation of a small-mass object orbiting a large (massive) object, then we can ignore the perpendicular velocity (momentum) component of the larger object and reach a simplification: The smaller object is continually pulled towards the primary but perpetually 'misses' due to its own perpendicular momentum.&lt;/p&gt;&#xA;" OwnerUserId="209" LastEditorUserId="209" LastEditDate="2013-12-16T09:52:26.067" LastActivityDate="2013-12-16T09:52:26.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="982" PostTypeId="2" ParentId="973" CreationDate="2013-11-21T06:55:58.807" Score="8" Body="&lt;p&gt;The Sun is not a point object, in fact it occupies a huge swath of space. Nor is it solid, rather the Sun is mostly fluid. Therefore, there is not one single &quot;Sun's magnetic field&quot; but rather the aggregate of hundreds, thousands, or perhaps trillions of magnetic fields, each of which changes in complex ways and are affected by (and affect themselves) nearby fields. Thus, the aggregate field does not always resemble the simplification to one single large &quot;Sun's magnetic field&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the page the OP links to has a &lt;a href=&quot;http://s.wsj.net/public/resources/images/NA-BY856_FLIP_G_20131110184816.jpg&quot;&gt;nice diagram&lt;/a&gt; which explains the &quot;two south poles&quot; phenomenon. During the process of the Sun's poles swap, for a short time the magnetic positive 'pole' shifts to the Sun's equator and both geographic poles have magnetic negative poles. Then the once-magnetically-negative geographic pole becomes positive and the equator no longer acts as a magnetic pole. I've never seen this mentioned anywhere else, and I'm sure that it is an oversimplification, but it does explain the &quot;two south poles&quot; statement.&lt;/p&gt;&#xA;" OwnerUserId="209" LastEditorUserId="209" LastEditDate="2013-11-21T10:19:04.323" LastActivityDate="2013-11-21T10:19:04.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="983" PostTypeId="2" ParentId="648" CreationDate="2013-11-21T06:59:41.517" Score="8" Body="&lt;p&gt;I...think you've got it backwards? Slow-roll inflation is alive and well, unlike models which involve tunneling between two vacua. The model of inflation first proposed by Alan Guth back in 1980 was a tunneling model. But it had a serious problem: it didn't &lt;em&gt;reheat&lt;/em&gt;. Tunneling from a false vacuum to a true one (or a lower-energy false one) wouldn't release any radiation. And we need that radiation, because we &lt;em&gt;are&lt;/em&gt; that radiation, cooled down and condensed after nearly 14 billion years of expansion. The solution was slow-roll inflation: as you mentioned in your question, rolling down the potential well releases a great deal of energy, which in turn leads to the wide variety of matter and radiation in the universe today. Slow-roll inflation also does a nice job of predicting the tiny density fluctuations in the early universe that seeded the formation of all the structures in the universe. Wikipedia has &lt;a href=&quot;http://en.wikipedia.org/wiki/Inflation_%28cosmology%29#Early_inflationary_models&quot;&gt;more on all of this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small disclaimer: there's a lot of inflationary models out there in the literature (there's a hilarious partial list in footnote 3 of &lt;a href=&quot;http://arxiv.org/abs/1006.0275&quot;&gt;this excellent paper by Yadav and Wandelt&lt;/a&gt;), and for all I know, one or more of them involve vacuum decay. But the dominant paradigm for inflation today is definitely slow-roll.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2013-11-21T06:59:41.517" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="984" PostTypeId="2" ParentId="950" CreationDate="2013-11-21T07:09:41.570" Score="1" Body="&lt;p&gt;I'm not far from you (Israel) and I am also waiting to see comet ISON.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comets are very unpredictable, and comet ISON has not yet reached the critical part of its journey which will determine visibility for us. Comet visibility is usually due to reflection of the Sun's rays on the coma and tails of the comet, as the nucleus is to small to see directly. So ISON's visibility (and all comets for that matter) depend on how much gas, water, and other materials that it outgasses. This in turn depends on how much volatiles it has, how much the Sun melts them, how much the slingshot around the sun and tidal forces fracture the nucleus, and other factors. So the actual visibility that will occur is very difficult to determine in advance. That said, ISON has all the markings of this being its first trip around the Sun. Thus, we assume that it has many volatiles to release and we assume that ISON will be exceptionally bright.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, ISON will be most visible when it is closest to Earth, near the end of December. It should be most visible in the early morning sky, I'm guessing at about 4:00 AM or so. This is because ISON will be close to the Sun (thus visible close to the time that the Sun becomes visible). Like all celestial bodies, it will rise in the East. Each day the comet will move farther from the Sun and thus will be visible earlier and earlier in the night, until it disappears from view probably sometime in mid January.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are welcome to visit Beersheba and have a peak through my binoculars!&lt;/p&gt;&#xA;" OwnerUserId="209" LastEditorUserId="209" LastEditDate="2013-11-21T08:24:34.553" LastActivityDate="2013-11-21T08:24:34.553" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="985" PostTypeId="1" CreationDate="2013-11-21T10:14:43.240" Score="18" ViewCount="799" Body="&lt;p&gt;I know that the Milky Way will collide with Andromeda in the distant future but my knowledge about the lifecycle of a galaxy is very lacking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was surprised that Wikipedia is so sparse on information in this topic. So how can I visualize its lifecycle? What will happen in the future?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;FYI&lt;/strong&gt;: &lt;em&gt;I just found &lt;a href=&quot;https://www.youtube.com/watch?v=Qg4vb-KH5F4&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; very informative video on the topic.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="470" LastEditorUserId="470" LastEditDate="2019-07-04T09:30:01.863" LastActivityDate="2019-07-04T18:03:41.183" Title="What do we know about the lifecycle of the Milky Way (or any other spiral galaxy)?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;kinematics&gt;&lt;m31&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="986" PostTypeId="1" CreationDate="2013-11-21T19:17:05.723" Score="7" ViewCount="648" Body="&lt;p&gt;I've heard that our universe may be open or closed.  If it's closed it might have a toroid shape.  If this is the case, would that imply that our 3 spatial dimensions have to be embedded in a higher dimensional (spatially) space?  For example, the old asteroids video game was apparently a torus mapped to a 2-dim video screen, and it stayed in 2 dimensions, but with a specific behavior of moving objects.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2013-11-22T17:12:00.953" Title="Does our universe have to be embedded in a higher dimensional space?" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="987" PostTypeId="1" AcceptedAnswerId="6023" CreationDate="2013-11-21T19:52:42.050" Score="5" ViewCount="335" Body="&lt;p&gt;There is somewhat of an abstract way of generalising the notion of planets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard definition of planets is, obviously: &quot;&lt;strong&gt;planets are the objects formed from the residual material surrounding a newly formed main sequence star through a complex formation process&lt;/strong&gt;&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me introduce a bit more general, by no means commonly known, definition of &quot;&lt;strong&gt;generalised planets as bound macroscopic objects formed from some former accretion process taking place near some gravitating body over many orbital periods&lt;/strong&gt;&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, the latter definition is more general, as it doesn't specify the type of accretion event and the type of the central body. Now, for example, the accretion mechanism can be very different from that in protoplanetary discs: consider, for example, accretion on a compact object (a neutron star or a black hole), accrection on supermassive black hole, accretion from remnants of binary neutron stars merger, or any other possible accretion process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted in some answers, some stars also form in accretion-involving processes in the vicinity of other stars. To avoid including stars into 'generalised planet' category, the definition specifies that the objects should be formed over many orbital periods, implying that they should be sufficiently bound with respect to the primary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question: &lt;strong&gt;Are there generalised planets other than planets/exoplanets?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="115" LastEditorUserId="115" LastEditDate="2013-12-12T19:11:21.253" LastActivityDate="2014-08-03T20:16:31.123" Title="Generalised planets?" Tags="&lt;accretion-discs&gt;&lt;planetary-formation&gt;" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="988" PostTypeId="2" ParentId="985" CreationDate="2013-11-22T08:54:59.193" Score="11" Body="&lt;p&gt;If we look at Newtonian physics, and how galaxies will interact, a central black hole should just be considered as a massive, dense object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Milky Way does not fall into its own central black hole, it orbits the centre of mass - as physics tells us it should.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we get closer to Andromeda, the gravitational influence of Andromeda will act more strongly on us, and when we get really close, individual masses within each galaxy will have dramatic effects on each other, but as with any such system, the two black holes will not suck everything in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the two galaxies end up coalescing - which is not a given - the orbits of the stars and the black holes will be very complex. For billions of years the 2 black holes will orbit each other, getting closer as they shed energy, but during that time the stars around them will suffer many effects, including:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;some will be expelled&lt;/li&gt;&#xA;&lt;li&gt;some will hit the black holes&lt;/li&gt;&#xA;&lt;li&gt;some will go nova&lt;/li&gt;&#xA;&lt;li&gt;and so on&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The best way to understand what will happen is to avoid thinking too deeply about black holes being weird, and treat them as dense masses. For most purposes this will help you model galactic collisions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at &lt;a href=&quot;http://www.youtube.com/watch?v=4disyKG7XtU&quot; rel=&quot;nofollow noreferrer&quot;&gt;this NASA simulation&lt;/a&gt; of the collision between the Milky Way and Andromeda:&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="43" LastEditDate="2019-07-04T18:03:41.183" LastActivityDate="2019-07-04T18:03:41.183" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="989" PostTypeId="2" ParentId="970" CreationDate="2013-11-22T16:41:59.600" Score="5" Body="&lt;p&gt;The answers provided to this question so far seem to be good. The easiest way I'd say is to take a long exposure picture in one of two places:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) At the zenith (&lt;strong&gt;directly overhead&lt;/strong&gt;), if you see that your picture looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uHoZe.jpg&quot; alt=&quot;circumpolar&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you are at either one of the poles. If the center of rotation of the stars is off by some angle (in degrees), you are that same amount of degrees south (or north) of the one of the poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The alternative is to take a long exposure picture due East or West. The angle between the direction of the star trails to that of your horizon gives you your latitude in degrees. The picture below is at some mid-latitude location on the Earth's surface. What you'd be looking for on the North or South pole are for the star trails to be &lt;strong&gt;parallel&lt;/strong&gt; to your horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cOCUq.jpg&quot; alt=&quot;startrails&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To break the degeneracy about which pole you're at (if you're an alien and crash-landed on the planet Earth (or any planet for that matter)), ask a local if they're friendly. They may not define the North and South poles the same as we do. If you're a human being on planet Earth, you could use constellations to figure this out. Constellations like Draco, Ursa Major/Minor and Cassiopeia are constellations which should not be visible from the South Pole. One could also use GPS to figure this out. &lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2013-11-22T16:41:59.600" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="990" PostTypeId="2" ParentId="986" CreationDate="2013-11-22T17:12:00.953" Score="4" Body="&lt;p&gt;The universe being open, closed, or flat only determines the type of geometry one must use to describe distances (and time). For open and closed geometries, Euclidean Geometry is &lt;strong&gt;not&lt;/strong&gt; what one should use. I would also agree that our universe being open, closed, or flat has nothing to do with the number of dimensions it contains. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At present, there is &lt;strong&gt;no direct evidence&lt;/strong&gt; that there are any additional dimensions over 3+1 dimensions (this just means three spatial dimensions and one time dimension). However, many &lt;a href=&quot;http://en.wikipedia.org/wiki/Grand_Unified_Theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;GUT theories&lt;/a&gt; do include additional spatial dimensions in an attempt to unify all of the forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also at present, we have very good evidence that the universe is 'flat'. What this means is that the angles of a triangle have to add up to $180^{\circ}$ and distances are measured in the standard Euclidean way. When we talk about the universe being flat, this is a purely global statement. Locally, however, it is completely possible to live in curved space. We actually &lt;strong&gt;do&lt;/strong&gt; live in curved space. The mass of the Earth is curving space and time in a way that General Relativity predicts, and therefore &lt;a href=&quot;http://www.nist.gov/public_affairs/releases/aluminum-atomic-clock_092310.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;clocks run very slightly differently&lt;/a&gt; depending on where you are on the Earth's surface, and distances are very closely approximated by Euclidean distances, though they're not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we actually determine that the universe is globally flat, you may ask? We use what's known as a standard ruler. Much like a standard candle, if we think we understand the physics, any deviation from what we would predict gives us new information about the universe (in the case of supernovae, it's that they can be used to measure distances, and therefore map out the expansion of the universe). We use the angular size of fluctuations in the &lt;a href=&quot;http://www.nist.gov/public_affairs/releases/aluminum-atomic-clock_092310.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cosmic Microwave Background Radiation&lt;/a&gt; (we think we understand the physics behind the fluctuations fairly well) to test how much, if at all, the universe deviates from flatness. The &lt;a href=&quot;http://arxiv.org/abs/1303.5086&quot; rel=&quot;nofollow noreferrer&quot;&gt;latest results from Planck&lt;/a&gt; show very good agreement with the standard picture provided to us by the standard LCDM cosmological model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is the power spectrum of the temperature fluctuations of the CMB. The location of the first peak is what cosmologists use to measure flatness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NivGK.png&quot; alt=&quot;CMBPS&quot;&gt;          &lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2013-11-22T17:12:00.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="991" PostTypeId="2" ParentId="987" CreationDate="2013-11-22T17:17:09.617" Score="2" Body="&lt;p&gt;As per your definition, a star formed in a binary system (hence near some gravitating body) is a generalised planet.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-11-22T17:17:09.617" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="992" PostTypeId="2" ParentId="978" CreationDate="2013-11-22T17:27:57.770" Score="9" Body="&lt;p&gt;There are other formulas at work, but not any other forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to take into account ont only the force, thus the acceleration, but also the current velocity of a body orbiting another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it simply: if you move a ball sticked to a rope around your head, the only forces are the tension of the rope and gravity towards the floor. Ignoring gravity, the only force is the tension of the rope, but it anyway does not make the ball to orbit your head, it in fact makes the ball to orbit it, due to the speed you put on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravity for an orbit, like the rope, causes the already moving object to curve its otherwise stright trajectory into an ellipse/circumference, not to fall to center.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-11-22T17:27:57.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="993" PostTypeId="2" ParentId="964" CreationDate="2013-11-22T17:32:09.460" Score="3" Body="&lt;p&gt;They are hot in the sense of particle speed, but if you were out there, you will get freezed, since there is such a low density that any of these particles will likely impact you (and transfer you its energy, which is what you would notice as heat) while you will cold due to radiation.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-11-22T17:32:09.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="994" PostTypeId="2" ParentId="962" CreationDate="2013-11-22T17:38:57.287" Score="1" Body="&lt;p&gt;Yes, it is unimodal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth's movement is a simple rotation, thus if the Moon were fixed on sky it will describe a circumference like the fixed stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From that basic movement we need to introduce Moon's own movement on the fixed stars sphere, which makes Moon not to complete the circumference in 1 day. Instead, Moon performs an arc similar to that of the fixed stars but shorter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This circumference-like movement does not wobble, so it has just one maximun and one minimum in elevation.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-11-22T17:38:57.287" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="995" PostTypeId="2" ParentId="917" CreationDate="2013-11-22T17:48:25.513" Score="6" Body="&lt;p&gt;Yes, gravity definitely affects the maximum heigh of mountains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think on a solid bar of steel. It sticks straight because of the electronic forces. But when you make it larger and larger gravity makes it bend: gravity starts being considerable, but still smaller than electronic forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you make the bar larger, there will be a moment in which the weight of the whole bar will be larger than the short-range electronic force: your bar will break purely due to gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exactly the same happens to mountains made of solid rock (as opposed to sedimentary ones quoted by Hobbes). There is a point,k depending on the strenght of planetary gravity, where it takes over short-range electronic forces, making the mountain collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is exactly the force that &quot;rounds up&quot; the planets, as opposed to the non-spherical asteroids.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-11-22T17:48:25.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="997" PostTypeId="1" CreationDate="2013-11-23T16:44:37.567" Score="8" ViewCount="454" Body="&lt;p&gt;This is a question about the basics of astronomy, which I have never happened to see a good discussion for. It is about how well would we be able to measure effective temperature of a star, if we had any arbitrarily perfect measurement devices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is some context. Canonical definition of $T_{\textrm{eff}}$ of a star is based on its bolometric luminosity $L$ (total electromagnetic energy radiated by the star per unit time) and its photospheric radius R (radius, at which the optical depth at a given wavelength is equal to unity). This way, the definition specifies  $T_{\textrm{eff}}$ through $L=4\pi \sigma R^2 T_{\textrm{eff}}^4$, where $\sigma$ is Stefan-Boltzmann constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition clearly alludes to black-body law. Many stars, including our own Sun, have a spectrum that does not follow it. For this reason, one often talks about another effective temperature, which is the temperature of stellar material at photospheric radius, and which can be determined by examining stellar spectrum. There are a few more complications to that, but let's put them aside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determining  $T_{\textrm{eff}}$ is extremely important in characterising stars, therefore there exists a variety of methods of measuring it, and naturally researchers strive for obtaining the best possible precision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the question: &lt;strong&gt;How well can one in principle measure $T_{\textrm{eff}}$, if one could have arbitrarily perfect instruments?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &lt;strong&gt;I would like to see a quantitative estimate in your answer. Is the best possible precision for $T_{\textrm{eff}}$ of order $10\textrm{K}$, or is it $1\textrm{K}$, or some $10^{-4}\textrm{K}$, or can we measure it arbitrarily well?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are just a few sources of uncertainty/arbitrariness: convection in stars, dependence of photospheric radius on wavelength, limb darkening, stellar variability, to name a few.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would encourage the answers to be in the format &quot;Source of uncertainty&quot; - &quot;Simple derivation&quot; - &quot;Estimate of the effect&quot;. If there are more than a few estimates, I will add a summary of them in the question or in a separate reply. Please, also feel free to edit the question if you might like to.&lt;/p&gt;&#xA;" OwnerUserId="115" LastEditorUserId="616" LastEditDate="2021-03-26T09:28:09.350" LastActivityDate="2021-03-26T09:28:09.350" Title="How well can we in principle determine $T_{\textrm{eff}}$ of a star?" Tags="&lt;spectroscopy&gt;&lt;stellar-atmospheres&gt;&lt;radiative-transfer&gt;&lt;stellar-structure&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="998" PostTypeId="2" ParentId="942" CreationDate="2013-11-23T19:52:40.847" Score="3" Body="&lt;p&gt;fractal21 and user16653 are more or less right, but not completely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you use with a cluster of stars is an isocrone, a line theoretically calculated that gives you where in the HR diagram the stars of a given age are depending on their mass. This can be done because bigger stars start moving on the HR diagram earlier than smaller ones when they became supergiants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what we astrophysicists do is to compare a set of isocrones with the HR diagram of a cluster, and we assign to that cluster the age of the best fitting isocrone.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-11-23T19:52:40.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="999" PostTypeId="1" CreationDate="2013-11-23T22:00:45.603" Score="9" ViewCount="221" Body="&lt;p&gt;Analogies between (typically) ideal gas and stellar systems are not only intuitively valid to some extent, but have been established and used in the studies of stellar clusters and galactic systems, most often as a simplification to collisionless Boltzmann equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea behind the analogy is that if a stellar system can be represented as a set of point masses, and if the number of point masses is large, then we can consider them from the point of view of kinetic theory of gases. One thing to remember here, though, is that stellar gas system is neither relaxed, nor can be relaxed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curious here: &lt;strong&gt;How far can the described analogy be pushed?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, there is a range of gas-specific phenomena (or we could be talking about plasma, if you may prefer), which would be fascinating to imagine for stellar systems, such as shocks, turbulence or viscosity. Can such, or some other, characteristic phenomena exist in stellar systems and are there actual systems exhibiting such a behavior? (of the named ones, viscosity analogue exists and is rather common)&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2013-12-28T21:15:23.943" Title="On analogies between gas and stellar systems" Tags="&lt;stellar-dynamics&gt;&lt;plasma-physics&gt;&lt;n-body-simulations&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1000" PostTypeId="2" ParentId="997" CreationDate="2013-11-24T13:37:41.127" Score="1" Body="&lt;p&gt;It is quite easy. In fact you do not need a bolometer. You just need to perform Intensity measurements in several parts of the spectrum, and then fit these to a theoretical black body spectrum. Three uses to be enough if it does not happen that you are measuring on a spike or valley in the spectrum caused by an emission or absorption line. The black body spectrum that best fits your measurements will give you &lt;span class=&quot;math-container&quot;&gt;$T_{eff}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="473" LastEditorUserId="31410" LastEditDate="2021-03-25T15:54:08.310" LastActivityDate="2021-03-25T15:54:08.310" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="1001" PostTypeId="1" AcceptedAnswerId="1002" CreationDate="2013-11-24T15:48:08.387" Score="26" ViewCount="7242" Body="&lt;p&gt;The age of the Universe is estimated at 13.8 billion years, and current theory states nothing can exceed the speed of light, which can lead to the incorrect conclusion that the universe can't have a radius of more than 13.8 billion light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Observable_universe&quot;&gt;Wikipedia&lt;/a&gt; deals with this misconception as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This reasoning would only make sense if the flat, static &lt;a href=&quot;https://en.wikipedia.org/wiki/Minkowski_space&quot;&gt;Minkowski spacetime&lt;/a&gt; conception under special relativity were correct. In the real Universe, &lt;a href=&quot;https://en.wikipedia.org/wiki/Spacetime&quot;&gt;spacetime&lt;/a&gt; is curved in a way that corresponds to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Metric_expansion_of_space&quot;&gt;expansion of space&lt;/a&gt;, as evidenced by &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble%27s_law&quot;&gt;Hubble's law&lt;/a&gt;. Distances obtained as the speed of light multiplied by a cosmological time interval have no direct physical significance. → &lt;em&gt;Ned Wright, &lt;a href=&quot;http://www.astro.ucla.edu/~wright/Dltt_is_Dumb.html&quot;&gt;&quot;Why the Light Travel Time Distance should not be used in Press Releases&quot;&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That doesn't clear the matter up for me, and having no science or maths background beyond high school, further reading into Hubble's law isn't helping much either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One layman's explanation I've seen offers explanation that the Universe itself isn't bound by the same laws as things &lt;em&gt;within&lt;/em&gt; it. That would make sense – insofar as these things can – but the above quote (&lt;em&gt;&quot;Distances obtained as the speed of light multiplied by a cosmological time interval have no direct physical significance&quot;&lt;/em&gt;) seems more general than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone offer (or direct me to) a good layman's explanation?&lt;/p&gt;&#xA;" OwnerUserId="479" LastEditorUserId="37191" LastEditDate="2021-02-12T21:03:45.967" LastActivityDate="2021-02-12T21:03:45.967" Title="Why is the observable Universe larger than its age would suggest?" Tags="&lt;universe&gt;&lt;distances&gt;&lt;space-time&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1002" PostTypeId="2" ParentId="1001" CreationDate="2013-11-24T17:19:24.797" Score="22" Body="&lt;p&gt;The easiest explanation for why the maximum distance one can see is not simply the product of the speed of light with the age of the universe is because the universe is non-static. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different things (i.e. matter vs. dark energy) have different effects on the coordinates of the universe, and their influence can change with time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good starting point in all of this is to analyze the Hubble parameter, which gives us the Hubble constant at any point in the past or in the future given that we can measure what the universe is &lt;em&gt;currently&lt;/em&gt; made of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H(a) = H_{0} \sqrt{\frac{\Omega_{m,0}}{a^{3}} + \frac{\Omega_{\gamma,0}}{a^{4}} + \frac{\Omega_{k,0}}{a^{2}} + \Omega_{\Lambda,0}} $$&#xA;where the subscripts $m$, $\gamma$, $k$, and $\Lambda$ on $\Omega$ refer to the density parameters of matter (dark and baryonic), radiation (photons, and other relativistic particles), curvature (this only comes into play if the universe globally deviates from being spatially flat; evidence indicates that it is consistent with being flat), and lastly dark energy (which as you'll notice remains a &lt;strong&gt;constant&lt;/strong&gt; regardless of how the dynamics of the universe play out). I should also point out that the $0$ subscript notation means as measured &lt;strong&gt;today&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $a$ in the above Hubble parameter is called the scale factor, which is equal to 1 today and zero at the beginning of the universe. Why do the various components scale differently with $a$? Well, it all depends upon what happens when you increase the size of a box containing the stuff inside. If you have a kilogram of matter inside of a cube 1 meter on a side, and you increase each side to 2 meters, what happens to the density of matter inside of this new cube? It decreases by a factor of 8 (or $2^{3}$). For radiation, you get a similar decrease of $a^{3}$ in number density of particles within it, and also an additional factor of $a$ because of the stretching of its wavelength with the size of the box, giving us $a^{4}$. The density of dark energy remains constant in this same type of thought experiment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because different components act differently as the coordinates of the universe change, there are corresponding eras in the universe's history where each component dominates the overall dynamics. It's quite simple to figure out, too. At small scale factor (very early on), the most important component was radiation. The Hubble parameter early on could be very closely approximated by the following expression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H(a) = H_{0} \frac{\sqrt{\Omega_{\gamma,0}}}{a^{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At around:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\Omega_{m,0}}{a^{3}} = \frac{\Omega_{\gamma,0}}{a^{4}} $$&#xA;$$ a = \frac{\Omega_{\gamma,0}}{\Omega_{m,0}}  $$&#xA;we have matter-radiation equality, and from this point onward we now have matter dominating the dynamics of the universe. This can be done once more for matter-dark energy, in which one would find that we are now living in the dark energy dominated phase of the universe. One prediction of living in a phase like this is an &lt;strong&gt;acceleration&lt;/strong&gt; of the coordinates of universe - something which has been confirmed (see: &lt;a href=&quot;http://www.nobelprize.org/nobel_prizes/physics/laureates/2011/&quot; rel=&quot;noreferrer&quot;&gt;2011 Nobel Prize in Physics&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you see, it would a bit more complicating to find the distance to the cosmological horizon than just multiplying the speed of light by the age of the universe. In fact, if you'd like to find this distance (formally known as the comoving distance to the cosmic horizon), you would have to perform the following integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ D_{h} = \frac{c}{H_{0}} \int_{0}^{z_{e}} \frac{\mathrm{d}z}{\sqrt{\Omega_{m,0}(1+z)^{3} + \Omega_{\Lambda}}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the emission redshift $z_{e}$ is usually taken to be $\sim 1100$, the surface of last scatter. It turns out this is the true horizon we have as observers. Curvature is usually set to zero since our most successful model indicates a flat (or very nearly flat) universe, and radiation is unimportant here since it dominates at a higher redshift. I would also like to point out that this relationship is derived from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Friedmann%E2%80%93Lema%C3%AEtre%E2%80%93Robertson%E2%80%93Walker_metric&quot; rel=&quot;noreferrer&quot;&gt;Friedmann–Lemaître–Robertson–Walker metric&lt;/a&gt;, a metric which includes curvature and expansion. This is something that the Minkowski metric lacks.&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="9301" LastEditDate="2016-10-29T04:22:19.237" LastActivityDate="2016-10-29T04:22:19.237" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1003" PostTypeId="1" AcceptedAnswerId="1015" CreationDate="2013-11-24T17:56:50.880" Score="16" ViewCount="1006" Body="&lt;p&gt;What object is in this photo?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;View coordinates: $47.25103 \ \ 38.81697$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time: $2013$-$11$-$23 ~21$:$00 +4 ~UTC$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sony $A580, \ 50mm f/1.4 15&quot;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KMF67.jpg&quot; alt=&quot;M31 Andromeda Galaxy&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="481" LastEditorUserId="6" LastEditDate="2013-11-27T20:35:08.877" LastActivityDate="2013-11-27T20:35:08.877" Title="What is the object in this photo?" Tags="&lt;amateur-observing&gt;&lt;observational-astronomy&gt;&lt;photography&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1004" PostTypeId="2" ParentId="1001" CreationDate="2013-11-24T19:59:26.340" Score="12" Body="&lt;p&gt;In short: things can not move faster that light by theirselves, but they can move faster than light due to universal expansion. The more far away, the faster they go away.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-11-24T19:59:26.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1005" PostTypeId="1" CreationDate="2013-11-24T20:30:47.817" Score="13" ViewCount="316" Body="&lt;p&gt;As has been pointed out by @Envite in the context of a more general discussion (see &lt;a href=&quot;https://astronomy.stackexchange.com/questions/987/generalised-planets&quot;&gt;Generalised planets?&lt;/a&gt;), there seems to be a moderate possibility for protoplanetary discs forming main-sequence stars (MS stars).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Main argument here is that Jupiter has mass of about $10^{-3}M_\odot$, while in principle for an object to be a MS star, it would need the mass of at least about $0.07M_\odot$. Also, there is a handful of systems, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/HD_29587&quot; rel=&quot;nofollow noreferrer&quot;&gt;HD 29587&lt;/a&gt;, which contain a star and a brown dwarf, possibly forming from a protoplanetary disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, a few questions. Can low-mass MS stars form in protoplanetary discs? If no, why? If yes, how often does it happen? If yes, which implications would it have for the subsequent dynamics of the planetary system, if one of the objects became a MS star, as opposed to it being a brown dwarf?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, thanks to @Envite for the idea.&lt;/p&gt;&#xA;" OwnerUserId="115" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2013-12-12T08:50:45.913" Title="Can protoplanetary disks form main-sequence stars?" Tags="&lt;planetary-formation&gt;&lt;star-formation&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1007" PostTypeId="1" AcceptedAnswerId="1230" CreationDate="2013-11-25T04:21:29.460" Score="21" ViewCount="2584" Body="&lt;p&gt;I often hear that nothing can escape a black hole because its &amp;quot;escape velocity&amp;quot; is greater than &lt;em&gt;c&lt;/em&gt;. If that is accurate, what about the following? I know that the following has a lot of most likely impossible assumptions, but I'm trying to understand whether the escape velocity explanation makes sense.&lt;/p&gt;&#xA;&lt;p&gt;Suppose you've got some sort of space ship that's inside the event horizon, and it hasn't been destroyed by the gravity or tidal forces. Also suppose that you've got a whole lot of reaction mass and a fantastic power supply.&lt;/p&gt;&#xA;&lt;p&gt;Could this space ship, through applying enough constant force escape the event horizon?&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="7472" LastEditDate="2020-12-07T19:57:16.687" LastActivityDate="2020-12-07T19:57:16.687" Title="Escaping a black hole" Tags="&lt;black-hole&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="1008" PostTypeId="2" ParentId="999" CreationDate="2013-11-25T05:20:06.707" Score="1" Body="&lt;p&gt;There is an &lt;a href=&quot;http://adsabs.harvard.edu/abs/1996MNRAS.280.1089M&quot; rel=&quot;nofollow&quot;&gt;interesting paper by Jes Madsen&lt;/a&gt;, which has some success modelling globular clusters as isothermal spheres.&lt;/p&gt;&#xA;" OwnerUserId="364" LastActivityDate="2013-11-25T05:20:06.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1009" PostTypeId="2" ParentId="1007" CreationDate="2013-11-25T05:27:02.417" Score="4" Body="&lt;p&gt;This is more of a physics question than it is astronomy, but nevertheless, think of it this way (most of what I say requires verification as I am not qualified to give a 100% correct in details answer): Light can't escape a black hole, and no propulsion system can accelerate an object to the speed of light. The faster you go, the more you weigh, the more force you have to input to maintain acceleration, the more energy you waste over time. If drawn on a graph, the energy required to increase speed by dX when at speed X, the graph has an asymptote at the speed of light. Meaning that it won't reach it no matter how much energy is input. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at LHC, where hadrons are accelerated to such a point when not their speed is measured anymore, because it hardly changes, but rather the energy of the particles, which increases a lot when approaching the speed of light. This will give you perspective about how unreachable it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you can't accelerate to the speed of light then you, like light, are trapped inside the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, AFAIK, light doesn't exit not because photons have weight, but because time stops. So light doesn't travel, nothing happens because there is no time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, I'm not qualified. I just like to read about this stuff.&lt;/p&gt;&#xA;" OwnerUserId="484" LastActivityDate="2013-11-25T05:27:02.417" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1010" PostTypeId="2" ParentId="978" CreationDate="2013-11-25T05:31:59.863" Score="5" Body="&lt;p&gt;Well, Kepler has explained that 2 randomly moving objects, attracted towards each other, will always form elliptical orbits. The Aphelion and Perihelion depend on that initial motion, position, force of attraction. The only case when 2 objects collide is when the perihelion is closer to the orbit edge than the sum of the radii of the 2 objects.&lt;/p&gt;&#xA;" OwnerUserId="484" LastActivityDate="2013-11-25T05:31:59.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1011" PostTypeId="1" CreationDate="2013-11-25T09:11:06.047" Score="3" ViewCount="4100" Body="&lt;p&gt;This &lt;a href=&quot;http://www.cometison2013.co.uk/ison-fact-sheet/&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt; may be nebulous (no pun intended!) for some. It writes to say&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Starting at around 95,000mph at the start of this month, Comet ISON will eventually reach an astonishing 845,000mph on November 28th when it catapults round the Sun at perihelion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm no astronomer; the only comets I even know the names of are &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Shoemaker Levy&lt;/li&gt;&#xA;&lt;li&gt;Swift Tuttle&lt;/li&gt;&#xA;&lt;li&gt;Wild&lt;/li&gt;&#xA;&lt;li&gt;Halley Bopp&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To the best of my knowledge (feel free to poke me in the rib if i'm wrong!) none of the above list achieved a 3-digit velocity at perihelion. In contrast, going by the figure quoted above, S1 is expected to achieve a mind-boggling 375 km/s!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is C/2012 S1 the fastest comet on record to-date?&lt;/p&gt;&#xA;" OwnerUserId="259" LastActivityDate="2016-08-07T23:59:50.980" Title="Is C/2012 S1 (aka ISON) the fastest comet on record to-date?" Tags="&lt;comets&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1012" PostTypeId="2" ParentId="1011" CreationDate="2013-11-25T11:12:23.583" Score="1" Body="&lt;p&gt;ISON is a sungrazer. But there were legendary comets like the Kreutz Family of Sungrazing comets. The Great comet of 1843 is a member of the kreutz family and it passed even more close to the sun at perihelion. It was a time before SOHO was deployed and there are not accurate observations of the speed of the Great Comet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since, it was travelling closer to the sun, considering the angular momentum created by the gravity of the sun, The great comet might have been a bit faster than the ISON. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot just term, speed of the comet. Comets travel with higher speeds due to the sun's gravitational pull and can be accelerated to extreme speeds when they catapult around the sun. However, when the comet is moving away from the sun, it'll be travelling slower because of the gravitational pull of the sun. &lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2013-11-25T11:12:23.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1013" PostTypeId="1" CreationDate="2013-11-25T19:32:04.043" Score="12" ViewCount="5206" Body="&lt;p&gt;Determining effective temperature of a star is in general a non-trivial task. Simple reason for this is that we can only study the electromagnetic radiation from a star, but not the temperature directly. The complexity is due to the fact that the radiation is produced in stratified stellar atmospheres, which are only partially characterised by stellar temperature, but also by many other factors, such as stellar mass, elemental abundances, stellar rotation, etc. What is more, the temperature of atmospheres varies with depth, whereas effective temperature is just a number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the other hand, temperatures and magnitudes are the most important quantities, characterising stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So, the question&lt;/strong&gt;: How exactly does one use the spectrum to extract the information about the temperature of a star? By temperature here I mean effective temperature, or even the temperature profile of the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: This a rather textbook question. I created it because I have encountered a good existing answer by @Carl, previously posted in a bit less textbook discussion &lt;a href=&quot;https://astronomy.stackexchange.com/questions/997/how-well-can-we-in-principle-determine-t-textrmeff-of-a-star&quot;&gt;How well can we in principle determine $T_{\textrm{eff}}$ of a star?&lt;/a&gt; . This question seems to be a much better place for the answer.&lt;/p&gt;&#xA;" OwnerUserId="115" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2018-11-29T00:40:27.810" Title="How does one determine the effective temperature of a star from its spectrum?" Tags="&lt;spectroscopy&gt;&lt;stellar-atmospheres&gt;&lt;radiative-transfer&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1015" PostTypeId="2" ParentId="1003" CreationDate="2013-11-25T22:14:34.847" Score="11" Body="&lt;p&gt;Your guess was correct. It is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Andromeda_Galaxy&quot;&gt;Andromeda Galaxy, M31&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a map of the part of the sky near zenith at the place and time you provided: &lt;a href=&quot;http://www.skymaponline.net/view.aspx?r=4556&amp;amp;x=4454&amp;amp;y=4666&amp;amp;lat=47.2492935&amp;amp;long=38.8201646&amp;amp;time=20131123170000&amp;amp;rotation=90&amp;amp;w=1300&amp;amp;h=800&amp;amp;gmtoffset=60&amp;amp;loc=Taganrog,%20Russia&amp;amp;id=7e9537b31b&quot;&gt;Sky map for Taganrog, Russia on 11/23/2013 5:00:00 PM UTC&lt;/a&gt;. Even the rotation is small. The sky map is rotated approximately 30° counter clockwise relative to the photo. You were approximately facing south when taking it.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/la0PF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On your picture there is a part of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Andromeda_%28constellation%29&quot;&gt;Andromeda constellation&lt;/a&gt;. I will use only a small bottom part of your picture (below the imaginary horizontal line going through M31) to identify the constellation.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The bright star to the left of M31, very close to the edge is &lt;a href=&quot;https://en.wikipedia.org/wiki/Mu_Andromedae&quot;&gt;μ Andromedae&lt;/a&gt; (marked &lt;code&gt;37 μ And&lt;/code&gt; on the sky map). It is the star between &lt;a href=&quot;https://en.wikipedia.org/wiki/Beta_Andromedae&quot;&gt;Mirach&lt;/a&gt; (out of photo) and M31.&lt;/li&gt;&#xA;&lt;li&gt;On the bottom of the right part of the photograph you see three stars on a imaginary vertical line: &lt;a href=&quot;https://en.wikipedia.org/wiki/Psi_Andromedae&quot;&gt;ψ&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Kappa_Andromedae&quot;&gt;κ&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Iota_Andromedae&quot;&gt;ι&lt;/a&gt; Andromedae.&lt;/li&gt;&#xA;&lt;li&gt;Between ψ and κ, to the right there is &lt;a href=&quot;https://en.wikipedia.org/wiki/Lambda_Andromedae&quot;&gt;λ Andromedae&lt;/a&gt; (marked &lt;code&gt;16 λ And&lt;/code&gt; and connected with μ on the sky map).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The bottom of the three stars in the upper-left corner of your photo is &lt;a href=&quot;https://en.wikipedia.org/wiki/Phi_Andromedae&quot;&gt;φ Andromedae&lt;/a&gt; (connected by the vertical line with &lt;code&gt;37 μ And&lt;/code&gt; on the sky map).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;code&gt;1 ο And&lt;/code&gt; on the sky map is not on your photo.&lt;/p&gt;&#xA;" OwnerUserId="188" LastActivityDate="2013-11-25T22:14:34.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1016" PostTypeId="2" ParentId="1013" CreationDate="2013-11-25T23:44:52.150" Score="4" Body="&lt;p&gt;Temperature (&lt;span class=&quot;math-container&quot;&gt;$T_{eff}$&lt;/span&gt;) can be quite tricky to determine accurately as it interrelates to a number of other fundamental measurements.&lt;/p&gt;&#xA;&lt;p&gt;Firstly, remember that the spectrum we observe from stars are pin-point, they give us the entire overall result and not a specific location or part of the star. We need to dissect the various parts to arrive at the fundamental parameters. We arrive at our results by iterating the values of the fundamental parameters until a model spectrum matches the true spectrum we observe. The issue is, like you say, the existence of a whole lot of uncertainties.&lt;/p&gt;&#xA;&lt;p&gt;The first of these (although it doesn't have a large effect) is the Uncertainty Principle itself. This creates &lt;em&gt;natural&lt;/em&gt; line broadening due to the emitted photon having a range of frequencies. The width of the line is determined by;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta E \approx \frac{h}{T_{\text{decay}}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\Delta E$&lt;/span&gt; is the uncertainty in the energy,&#xA;&lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; is the Planck constant, and&#xA;&lt;span class=&quot;math-container&quot;&gt;$T_{\text{decay}}$&lt;/span&gt; is the amount of time the electron stays in a high energy state before decaying.&lt;/p&gt;&#xA;&lt;h2&gt;Fundamental parameters&lt;/h2&gt;&#xA;&lt;p&gt;The &lt;strong&gt;rotation&lt;/strong&gt; of the star causes a Doppler shift effect on the line spectra making it broaden. The faster the rotation, the broader (yet smaller) the line. Like the Uncertainty Principle, this is &lt;em&gt;natural broadening&lt;/em&gt; as it doesn't impact the abundance of any particular element in the star.&lt;/p&gt;&#xA;&lt;p&gt;Measuring the rotational velocity (&lt;span class=&quot;math-container&quot;&gt;$V_{\text{proj}}$&lt;/span&gt;) depends on both its axis of rotation and our line of sight to the star. Therefore, we use a combination of both velocity about the equator (&lt;span class=&quot;math-container&quot;&gt;$v_e$&lt;/span&gt;) and the star’s polar inclination (&lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt;) to determine the projected radial velocity;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V_{\text{proj}} = v_e \sin i$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Temperature&lt;/strong&gt; (&lt;span class=&quot;math-container&quot;&gt;$T_{eff}$&lt;/span&gt;)impacts wavelength in such a way that higher temperatures impart higher random motions on the atoms. When these photons collide with an atom, they can cause the atom to become ionised, i.e. lose an electron. Different energy levels (and therefore temperature) will create different abundances at the various ionisation stages of atoms.&lt;/p&gt;&#xA;&lt;p&gt;The temperature of the stellar photosphere decreases as we move away from the core. Therefore the line profile represents a range of temperatures. The wings of the line arise from deeper, hotter gas which displays a larger range of wavelengths due to increased motion. The higher the temperature, the broader the wings of the line profile ([Robinson 2007, pg 58][1]).&lt;/p&gt;&#xA;&lt;p&gt;Here you can see the effect of various temperature values on the synthetic spectral line of FE I 6593 A. Red: &lt;span class=&quot;math-container&quot;&gt;$T_{eff}$&lt;/span&gt; = 4000K; Black: &lt;span class=&quot;math-container&quot;&gt;$T_{eff}$&lt;/span&gt; = 5217K; Blue: &lt;span class=&quot;math-container&quot;&gt;$T_{eff}$&lt;/span&gt; = 6000K;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/W5rA2.png&quot; alt=&quot;Effect of &lt;span class=&quot;math-container&quot;&gt;$T_{eff}$&lt;/span&gt; on spectral lines&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Microturbulence&lt;/strong&gt; (&lt;span class=&quot;math-container&quot;&gt;$v_{\text{mic}}$&lt;/span&gt;) is the non-thermal localised random motion of the stellar atmosphere. It works in a similar way to temperature - an increase in the motion of atoms creates a wider range of wavelengths observed and therefore broader line profiles.&lt;/p&gt;&#xA;&lt;p&gt;In strong lines, saturation can occur when there aren’t any more photons to be absorbed. As microturbulence in these areas increases, it presents more opportunities for photons to be absorbed. This broadens the wings of the line profile increasing the overall strength of the line. We can use this fact to determine &lt;span class=&quot;math-container&quot;&gt;$v_{\text{mic}}$&lt;/span&gt;, by ensuring that the strength of the lines (equivalent width) have no correlation with their abundances.&lt;/p&gt;&#xA;&lt;p&gt;Finally, &lt;strong&gt;surface gravity&lt;/strong&gt; which is a function of the star’s mass and size:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\log g = \log M - 2 \log R + 4.437$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;with &lt;span class=&quot;math-container&quot;&gt;$M, R$&lt;/span&gt; being in solar units and &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; in cgs.&lt;/p&gt;&#xA;&lt;p&gt;A star with a higher mass but smaller radius will invariably be denser and under greater pressure. By definition, denser gas has a higher number of atoms per unit of area (abundance), leading to stronger spectral lines.&lt;/p&gt;&#xA;&lt;p&gt;A gas under pressure provides more opportunities for free electrons to recombine with ionised atoms. For a given temperature, ionisation is expected to decrease with an increase of surface gravity, in turn increasing the abundance of atoms in the neutral or low ionisation states.&lt;/p&gt;&#xA;&lt;h2&gt;The measuring of &lt;span class=&quot;math-container&quot;&gt;$T_{eff}$&lt;/span&gt;&lt;/h2&gt;&#xA;&lt;p&gt;As we've seen, there are a number of ways in which a spectrum of a star can be altered. The one you're interested in is temperature. As temperature is interrelated to all the other fundamental parameters, we need to treat them together as a whole and tease out the value of &lt;span class=&quot;math-container&quot;&gt;$T_{eff}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;We begin with a synthetic spectrum and modify its properties iteratively until it matches the shape of the star’s spectrum. Adjustments of one parameter will invariably affect the others. The spectra will match when the temperature, surface gravity, and microturbulence values (amongst others) are correct. This is obviously very time consuming although programs exist to help.&lt;/p&gt;&#xA;&lt;p&gt;Atmospheric properties can also be determined by other less time consuming means. Photometric colours can be used as a proxy for temperature, and absolute magnitudes for surface gravity. However, these determinations can suffer from inaccuracies due to interstellar extinction and are at best a close approximation.&lt;/p&gt;&#xA;&lt;p&gt;[1] Robinson, K. 2007, Spectroscopy: The Key to the Stars (Springer)&lt;/p&gt;&#xA;" OwnerUserId="173" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-11-28T18:59:51.780" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="1017" PostTypeId="2" ParentId="1007" CreationDate="2013-11-26T06:01:32.877" Score="-5" Body="&lt;p&gt;Although I partially agree with @Alexander here are a few more things. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Einstein's &lt;strong&gt;special theory of relativity&lt;/strong&gt;, particles cannot travel at the speed of light (c). It would take infinite energy as stated above. However, the special theory of relativity states that an object can travel &lt;strong&gt;slower or faster&lt;/strong&gt; than the speed of light but not at the speed of c. We may refer to the objects that travel faster than the speed of light as &lt;strong&gt;tachyons&lt;/strong&gt;. There are celestial objects which are accelerated faster than the speed of light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common thing that everyone says is that nothing can escape a blackhole. But it is totally &lt;a href=&quot;http://www.space.com/16867-black-holes-quantum-mechanics-theory.html&quot; rel=&quot;nofollow&quot;&gt;wrong&lt;/a&gt;. Blackholes do emit particles and radiation called as &lt;strong&gt;Hawking radiation&lt;/strong&gt;. Even stephen hawking has changed  his mind about blackholes as the general concept that blackholes absorb everything can be disproved by quantum mechanics theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So, If there exists a possibility that we can indeed escape a blackhole.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At &lt;strong&gt;CERN,&lt;/strong&gt; &lt;a href=&quot;http://news.nationalgeographic.co.in/news/2011/09/110923-neutrinos-speed-of-light-particles-cern-physics-einstein-science/&quot; rel=&quot;nofollow&quot;&gt;reports&lt;/a&gt; says that &lt;strong&gt;ghostly subatomic  particles&lt;/strong&gt; called &lt;strong&gt;neutrinos&lt;/strong&gt; were observed to travel faster than light. Researches are still trying to confirm it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also another possible theory existing around the scientific community is the concept of wormholes. If a ship were trying to escape a blackhole, opening a wormhole to escape the blackhole and reaching another spot is a proposed theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also researches that say that objects are &lt;a href=&quot;http://science.time.com/2012/10/04/how-to-escape-from-a-black-hole/&quot; rel=&quot;nofollow&quot;&gt;emanated from blackhole&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2013-11-26T06:01:32.877" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1018" PostTypeId="1" AcceptedAnswerId="1019" CreationDate="2013-11-26T13:55:43.463" Score="27" ViewCount="23259" Body="&lt;p&gt;How many of the luminous dots that we see naked are galaxies and not stars from our galaxy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine that the majority of the luminous points that we see naked eye during the night, are actually stars from our galaxy. But how many of them are other objects (other galaxies, nebula, etc.), excluding planets from our Solar System?&lt;/p&gt;&#xA;" OwnerUserId="490" OwnerDisplayName="Mario Stefanutti" LastEditorUserId="2153" LastEditDate="2015-04-12T14:40:52.190" LastActivityDate="2020-12-16T13:37:24.080" Title="How many stars and galaxies can be seen by the naked eye?" Tags="&lt;star&gt;&lt;galaxy&gt;&lt;nebula&gt;&lt;naked-eye&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1019" PostTypeId="2" ParentId="1018" CreationDate="2013-11-26T16:02:12.950" Score="14" Body="&lt;p&gt;In the best sky conditions, the naked eye (with effort) can see objects with an apparent magnitude of 8.0. This reveals about &lt;strong&gt;43,197 objects&lt;/strong&gt; in the sky.&lt;/p&gt;&#xA;&lt;p&gt;There are &lt;strong&gt;9 galaxies&lt;/strong&gt; visible to the naked eye that you might see when observing the sky, and there are about &lt;strong&gt;13 nebulae&lt;/strong&gt; that you might see.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Sources:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://skyandtelescope.org/astronomy-resources/light-pollution-and-astronomy-the-bortle-dark-sky-scale&quot; rel=&quot;noreferrer&quot;&gt;The Bortle Dark-Sky Scale - John E. Bortle&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://spacemath.gsfc.nasa.gov/weekly/6Page103.pdf&quot; rel=&quot;noreferrer&quot;&gt;How many stars are in the sky? - NASA&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_galaxies#Naked-eye_galaxies&quot; rel=&quot;noreferrer&quot;&gt;Naked-eye galaxies - Wikipedia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_planetary_nebulae&quot; rel=&quot;noreferrer&quot;&gt;List of planetary nebulae - Wikipedia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_diffuse_nebulae&quot; rel=&quot;noreferrer&quot;&gt;List of diffuse nebulae - Wikipedia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6" LastEditorUserId="37001" LastEditDate="2020-11-22T21:20:23.310" LastActivityDate="2020-11-22T21:20:23.310" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="1020" PostTypeId="2" ParentId="1005" CreationDate="2013-11-26T17:50:12.657" Score="9" Body="&lt;p&gt;&lt;strong&gt;Yes&lt;/strong&gt;, unless you want to get really particular with the &quot;protoplanetary&quot; part. For example, there are stars forming in the circumstellar disks around Wolf-Rayet stars [&lt;a href=&quot;http://arxiv.org/abs/1203.4659&quot;&gt;reference&lt;/a&gt;]. If we were picky, we might not call the circumstellar disk around this Wolf-Rayet star a protoplanetary disk (and instead refer to the planet-forming circumstellar disks of the newly formed stars as such), but there exists no clear line between the disks of gas within and the larger disk of gas around the Wolf-Rayet star of which they are a part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/1203.4659&quot;&gt;Triggered star formation surrounding Wolf-Rayet star HD 211853 - Liu, Wu, Zhang, and Qin&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6" LastActivityDate="2013-11-26T17:50:12.657" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1021" PostTypeId="1" AcceptedAnswerId="1022" CreationDate="2013-11-26T18:20:48.700" Score="9" ViewCount="304" Body="&lt;p&gt;&lt;a href=&quot;http://www.astro.ex.ac.uk/exoclimes/2010/talks/Forget_exoclimes10.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;These notes from an exoplanet talk&lt;/a&gt; discuss modeling the climate of exoplanets. How would we do that with our current technology? Would data be primarily gathered by spectrum analysis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://www.drroyspencer.com/wp-content/uploads/climate-model-2.jpg&quot; alt=&quot;Climate model&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:47.067" LastActivityDate="2014-11-06T14:08:40.647" Title="Climate modeling of exoplanets" Tags="&lt;exoplanet&gt;&lt;climate&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1022" PostTypeId="2" ParentId="1021" CreationDate="2013-11-27T04:34:07.660" Score="6" Body="&lt;p&gt;The presentation linked to seems to primarily  discuss General Circulation Models, solutions to the equations of fluid dynamics which predict the behavior of a planet's atmosphere, a component of more complete Global Climate Models (confusingly, both are GCMs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spectral data on exoplanets is extremely limited to date and poorly constrains the composition, but a lot can be inferred from a planet's orbit and eccentricity (how far from circular the orbit is). At the simplest level, planet temperature depends on how much light it receives from its parent star, and this will vary as a planet goes around a non-circular orbit. Starting from there and making reasonable assumptions about the composition of the planet (usually starting with a solar system analogy) you can test out a GCM model in the various extreme cases we observe (hot Jupiters etc.), to be subjected to future observational constraints. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One exciting new development is looking at a planet's phase curve (the observed brightness of the planet as it orbits its parent star) and using that to infer properties of the atmosphere. One team recently found that Kepler 7b's phase curve  may best be explained by the presence of clouds: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Demory, et al. 2013. “Inference of Inhomogeneous Clouds in an Exoplanet Atmosphere.” arXiv:1309.7894 [astro-Ph] (September 30). &lt;a href=&quot;http://arxiv.org/abs/1309.7894&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1309.7894&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="457" LastEditorUserId="457" LastEditDate="2014-11-06T14:08:40.647" LastActivityDate="2014-11-06T14:08:40.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1023" PostTypeId="1" CreationDate="2013-11-27T06:39:44.730" Score="11" ViewCount="603" Body="&lt;p&gt;If gravitation (attraction of mass) is the cause of the formation of all celestial bodies then how come the numerous small bodies found in asteroid belts spread over an orbit &lt;strong&gt;instead&lt;/strong&gt; of clustering together ?!&lt;/p&gt;&#xA;" OwnerUserId="493" LastActivityDate="2016-05-18T00:36:44.310" Title="Why haven't asteroid belts turned into new large bodies?" Tags="&lt;planetary-formation&gt;&lt;asteroid-belt&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1025" PostTypeId="1" AcceptedAnswerId="1026" CreationDate="2013-11-27T10:41:16.097" Score="12" ViewCount="1638" Body="&lt;p&gt;I am trying to understand something I read on wikipedia about M31.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia says that M31 appears more than six times as wide as the full Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I remember that when I watched it naked eye it seamed a lot smaller, even considering that I was able to see only the central region. For this reason I also used the software Stellarium to check it and setting the same magnification for the two objects (see pictures), the wikipedia information still seems wrong ... or most likely I am wrong!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is that I fail? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;At an apparent magnitude of 3.4, the Andromeda Galaxy is one of the brightest Messier objects,[15] making it visible to the naked eye on moonless nights even when viewed from areas with moderate light pollution. Although &lt;strong&gt;it appears more than six times as wide as the full Moon&lt;/strong&gt; when photographed through a larger telescope, only the brighter central region is visible to the naked eye or when viewed using binoculars or a small telescope&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://4coloring.files.wordpress.com/2013/11/m31.jpg&quot; alt=&quot;M31&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://4coloring.files.wordpress.com/2013/11/moon.jpg&quot; alt=&quot;Moon&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="490" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:47.077" LastActivityDate="2013-11-27T12:17:27.293" Title="Apparent size of M31" Tags="&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1026" PostTypeId="2" ParentId="1025" CreationDate="2013-11-27T12:17:27.293" Score="9" Body="&lt;p&gt;The answer is right there in your screenshots. The 'dimensione' or angular diameter on the sky is 3 degrees for M31, whereas the moon has an angular diameter of only 0.5 degrees. In other words, M31 is about 6 times as wide as the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As wikipedia correctly states, you need a large telescope to see the full extend of M31. With the naked eye of even a small telescope you simply cannot see the spiral arms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this image shows it very neatly:&#xA;&lt;a href=&quot;http://apod.nasa.gov/apod/ap061228.html&quot;&gt;http://apod.nasa.gov/apod/ap061228.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="494" LastActivityDate="2013-11-27T12:17:27.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1027" PostTypeId="2" ParentId="1023" CreationDate="2013-11-27T14:10:43.080" Score="5" Body="&lt;p&gt;From Wikipedia:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Asteroid_belt#Formation&quot;&gt;http://en.wikipedia.org/wiki/Asteroid_belt#Formation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planetesimals within the region which would become the asteroid belt were too strongly perturbed by Jupiter's gravity to form a planet. Instead they continued to orbit the Sun as before, and occasionally colliding.[27] In regions where the average velocity of the collisions was too high, the shattering of planetesimals tended to dominate over accretion,[28] preventing the formation of planet-sized bodies. Orbital resonances occurred where the orbital period of an object in the belt formed an integer fraction of the orbital period of Jupiter, perturbing the object into a different orbit; the region lying between the orbits of Mars and Jupiter contains many such orbital resonances. As Jupiter migrated inward following its formation, these resonances would have swept across the asteroid belt, dynamically exciting the region's population and increasing their velocities relative to each other.[29]&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-11-27T14:10:43.080" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1028" PostTypeId="1" AcceptedAnswerId="1061" CreationDate="2013-11-27T16:19:55.220" Score="10" ViewCount="15759" Body="&lt;p&gt;I am confused about this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I see an object from Mount Teide (longitude is 16&quot;30'E and latitude is 28&quot;18'N) that passes the meridian (azimuth=0) at 5h (am) UTC, and I also know that the elevation of the star is 43&quot;40', the stellar time at Greenwich at 0h UTC is 22h20min.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate the right ascension and declination?&lt;/p&gt;&#xA;" OwnerUserId="496" LastEditorUserId="47" LastEditDate="2013-11-29T21:55:32.243" LastActivityDate="2013-12-07T10:35:40.550" Title="Calculation of right ascension and declination" Tags="&lt;fundamental-astronomy&gt;&lt;ascension&gt;&lt;declination&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1029" PostTypeId="1" AcceptedAnswerId="1038" CreationDate="2013-11-27T20:49:04.123" Score="12" ViewCount="1084" Body="&lt;p&gt;I have seen pictures of clusters of galaxies, usually used in regards to theories of dark matter and galaxy formations. One of the most famous ones has the perceived shape of a stick-figure. If I am not mistaken some of these clusters seem to be bigger than our observable Universe horizon of $14$ billion light-years, which is as far as we can see because of the age of the Universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My assumption is that these are just coordinates plotted into a computer simulation. However, are these clusters the size of our observable Universe horizon or are these more theoretical abstractions?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: Interesting new find on the subject. &lt;a href=&quot;http://en.wikipedia.org/wiki/Laniakea_Supercluster&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Laniakea Supercluster&lt;/strong&gt;&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eBu2T.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PeF26.gif&quot; alt=&quot;Stick Figure&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="417" LastEditorUserId="417" LastEditDate="2014-09-14T20:13:59.093" LastActivityDate="2014-09-14T20:13:59.093" Title="How are Galaxy Super Clusters Generated" Tags="&lt;galaxy&gt;&lt;universe&gt;&lt;galaxy-cluster&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1030" PostTypeId="2" ParentId="1029" CreationDate="2013-11-27T21:24:30.120" Score="4" Body="&lt;p&gt;No, the structures are not of the size of the observable Universe. In fact modern cosmolgical models rely on the Universe being homogeneous over scales above $100 \textrm{Mpc}$. This is an observational fact, which serves as a base for the so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmological_Principle&quot; rel=&quot;nofollow&quot;&gt;cosmological principle&lt;/a&gt;. See also a wiki article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Large-scale_structure_of_the_cosmos#Large-scale_structure&quot; rel=&quot;nofollow&quot;&gt;large-scale structure&lt;/a&gt;, where the scale is mentioned in the section &quot;End of Greatness&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scale corresponds to a few hundred million light years. The largest structures, therefore, are clearly way smaller than the size of the observable Universe, which is of order of ten billion light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, that because it takes time for the light to reach us, the current size of the Universe we observe is larger than its apparent size due to cosmological expansion.&lt;/p&gt;&#xA;" OwnerUserId="115" LastEditorUserId="115" LastEditDate="2013-11-28T12:40:52.213" LastActivityDate="2013-11-28T12:40:52.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1031" PostTypeId="1" AcceptedAnswerId="1040" CreationDate="2013-11-28T20:23:04.347" Score="8" ViewCount="6009" Body="&lt;p&gt;When you emit the radio signal it starts moving at the speed of light. Radio beam is diffusing with each kilometer the signal has traveled. To the nearby receiver the signal is strong. But if the receiver is far away, the signal will become weaker and weaker until it becomes a noise. So the question is does the same happen in deep space and what distance radio signal (or any electromagnetic wave frequency) could travel until it becomes noise?&lt;/p&gt;&#xA;" OwnerUserId="499" LastActivityDate="2013-11-30T03:23:31.637" Title="Does the radio signal decay when it travels through the intergalactic space?" Tags="&lt;radio-astronomy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1032" PostTypeId="1" CreationDate="2013-11-28T20:31:53.367" Score="7" ViewCount="11570" Body="&lt;p&gt;I’ll put this question in a form of a physics problem:&#xA;Transmitter at a point A sends a signal to a receiver at a point B. What is the longest distance in light years from points A to B for which receiver can still understand the signal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Signal type: electromagnetic wave.&lt;/li&gt;&#xA;&lt;li&gt;Signal frequency: frequency for which signal fades the least - probably radio.&lt;/li&gt;&#xA;&lt;li&gt;Signal form:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Variant 1: beam of highest concentration possible.&lt;/li&gt;&#xA;&lt;li&gt;Variant 2: full sphere.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Transmitter power: power of the most powerful transmitter known to be built in the Earth.&lt;/li&gt;&#xA;&lt;li&gt;Receiver: sensitivity of the most sensitive receiver known to be built in the Earth.&lt;/li&gt;&#xA;&lt;li&gt;Transmission medium: interstellar space.&lt;/li&gt;&#xA;&lt;li&gt;Understanding of a signal:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Variant A: Can detect that the signal is not a noise.&lt;/li&gt;&#xA;&lt;li&gt;Variant B: Can decode the information in the signal.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="499" LastActivityDate="2019-07-25T20:15:36.230" Title="What is the maximum transmission distance of the radio signal in the outer space which could still be understood?" Tags="&lt;radio-astronomy&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1033" PostTypeId="2" ParentId="1031" CreationDate="2013-11-28T22:14:03.407" Score="8" Body="&lt;p&gt;All electromagnetic radiation from a point source - which a normal radio transmitter is - propagates according to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Inverse-square_law&quot;&gt;inverse square law&lt;/a&gt; which means that the intensity of the signal is inversely proportional to the square of the distance. This happens on earth and in deep space equally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this will mean that for any signal there will be a distance at which it's becomes indistinguishable from the background noise of the universe. However, that distance will depend on the initial strength of the signal.&lt;/p&gt;&#xA;" OwnerUserId="111" LastEditorUserId="111" LastEditDate="2013-11-29T08:47:16.293" LastActivityDate="2013-11-29T08:47:16.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1034" PostTypeId="2" ParentId="1032" CreationDate="2013-11-29T12:13:28.470" Score="2" Body="&lt;p&gt;It cannot be said correctly, since we humans have hardly traveled to the moon and sent space probes to explore other planets in our solar system. So, theoretically anything might be possible. I'm trying to be a bit practical here. The only man made object that has gone really far is &lt;strong&gt;Voyager 1&lt;/strong&gt;, which is at a distance of 18.7 billion kilometers (125.3 AU) from the sun. Although launched in 1977, it is the only live transmitter and receiver which is that far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Voyager_1&quot; rel=&quot;nofollow&quot;&gt;radio communication system of Voyager 1&lt;/a&gt;&lt;/strong&gt; was designed to be used up to and beyond the limits of the Solar System. The communication system includes a 3*&lt;em&gt;.7 meters (12 ft) diameter parabolic dish high-gain antenna&lt;/em&gt;* to send and receive radio waves via the three Deep Space Network stations on the Earth. Voyager 1 normally transmits data to Earth over Deep Space Network Channel 18, using a frequency of either 2296.481481 MHz or 8420.432097 MHz, while signals from Earth to Voyager are broadcast at 2114.676697 MHz.&#xA;As of 2013, signals from Voyager 1 take over 17 hours to reach Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree that there are powerful transmitters in the world than what is present in the Voyager 1, but, most of them still remain &lt;strong&gt;untested&lt;/strong&gt;. So, we can be exact with the measurements.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2013-12-04T13:35:57.197" LastActivityDate="2013-12-04T13:35:57.197" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1035" PostTypeId="1" AcceptedAnswerId="1036" CreationDate="2013-11-29T17:59:27.930" Score="5" ViewCount="614" Body="&lt;p&gt;I visited: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://aa.usno.navy.mil/imagery/moon&quot; rel=&quot;noreferrer&quot;&gt;http://aa.usno.navy.mil/imagery/moon&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.usno.navy.mil/USNO/time/moon-phase-images&quot; rel=&quot;noreferrer&quot;&gt;http://www.usno.navy.mil/USNO/time/moon-phase-images&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;within seconds of each other, and saw two very different &quot;current moon images&quot;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EcmtF.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/97cNe.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is one of these images wrong? The larger image seems to display a much &#xA;bigger crescent than the smaller image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've noticed this earlier as well.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2020-12-19T11:58:41.233" Title="USNO moon images look quite different?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1036" PostTypeId="2" ParentId="1035" CreationDate="2013-11-29T19:29:45.040" Score="5" Body="&lt;p&gt;This is an unofficial explanation, but I doubt you'll get an answer from official sources here, unless you ask them directly (e.g. their Twitter account is &lt;a href=&quot;https://twitter.com/NavyOceans&quot; rel=&quot;nofollow noreferrer&quot;&gt;@NavyOceans&lt;/a&gt;), so I'll give it a shot. The small thumbs appear to be merely approximation, representation icons that aren't resized from the real-time image, but merely prepared in advance images, displayed depending on the current Moon's phase, calculated by the server.&lt;/p&gt;&#xA;&lt;p&gt;If you try their Moon phase image generator (at the bottom of &lt;a href=&quot;https://www.usno.navy.mil/USNO/time/moon-phase-images&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;), you'll see why they did it like that. The prepared thumbs are numbered from &lt;a href=&quot;https://web.archive.org/web/20030111224231/http://tycho.usno.navy.mil:80/gif/moons/m180.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;m180.gif&lt;/a&gt; for Full Moon to &lt;a href=&quot;https://web.archive.org/web/20030111221525/http://tycho.usno.navy.mil:80/gif/moons/m360.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;m360.gif&lt;/a&gt; for New Moon, with a 2° step, and apparently the server calculates based on input parameters the Moon phase and displays prepared thumbnail with a minimum approximation precision of 2° (roughly 8 hour step). The same page also offers an explanation for how this thumbnail generation works:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;These lunar phase images were created by R. Schmidt from ray-traced&#xA;images of the Moon. A Clementine spacecraft mosaic of the lunar&#xA;surface was mapped onto a sphere, and scenes were rendered as a&#xA;virtual Sun &amp;quot;orbited&amp;quot; the Moon. The depiction of lunar surface&#xA;features suffers geometric distortion but the terminator is correct&#xA;with respect to the spherical Moon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The lunar phase calculator offers date selection that goes forward in time or before the time any images of actual observations were made and stored, so it's unreasonable to expect actual images displayed for, say, some date in 22nd or 19th century. Some examples off U.S. Naval Observatory's server:&lt;/p&gt;&#xA;&lt;p&gt;    &lt;a href=&quot;https://i.stack.imgur.com/GRyT5.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GRyT5.gif&quot; alt=&quot;180&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;https://web.archive.org/web/20030111224231/http://tycho.usno.navy.mil:80/gif/moons/m180.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;navy.mil&lt;/a&gt;)&lt;/sub&gt;         &#xA;&lt;a href=&quot;https://i.stack.imgur.com/RQFKE.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RQFKE.gif&quot; alt=&quot;246&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;https://web.archive.org/web/20030111215708/http://tycho.usno.navy.mil:80/gif/moons/m246.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;navy.mil&lt;/a&gt;)&lt;/sub&gt;         &#xA;&lt;a href=&quot;https://i.stack.imgur.com/3j36N.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3j36N.gif&quot; alt=&quot;270&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;https://web.archive.org/web/20091027195130/http://tycho.usno.navy.mil/gif/moons/m270.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;navy.mil&lt;/a&gt;)&lt;/sub&gt;         &#xA;&lt;a href=&quot;https://i.stack.imgur.com/F8gAc.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F8gAc.gif&quot; alt=&quot;294&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://tycho.usno.navy.mil/gif/moons/m294.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;navy.mil&lt;/a&gt;)&lt;/sub&gt;         &#xA;&lt;a href=&quot;https://i.stack.imgur.com/Fl97q.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fl97q.gif&quot; alt=&quot;360&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;https://web.archive.org/web/20030111221525/http://tycho.usno.navy.mil:80/gif/moons/m360.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;navy.mil&lt;/a&gt;)&lt;/sub&gt;  &lt;br /&gt;&#xA;    &lt;sup&gt;        Full Moon                        Waning Gibbons                     Last Quarter                    Waning Crescent                     New Moon&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;The larger image of &lt;a href=&quot;https://web.archive.org/web/20190213000838/https://aa.usno.navy.mil/imagery/moon&quot; rel=&quot;nofollow noreferrer&quot;&gt;how the Moon looks like now&lt;/a&gt; seems to be generated by a lot more precise renderer (image is refreshed every minute by a JavaScript call) that takes into account not merely lunar phase, but also its &lt;a href=&quot;https://astronomy.stackexchange.com/questions/16/why-is-only-one-side-of-the-moon-visible-from-earth/22#22&quot;&gt;libration&lt;/a&gt; in respect to the surface on the Earth. It would look more like any frame off this animation:&lt;/p&gt;&#xA;&lt;p&gt;                         &lt;a href=&quot;https://web.archive.org/web/20201007012740/https://upload.wikimedia.org/wikipedia/commons/b/ba/Lunar_libration_with_phase_Oct_2007_450px.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://web.archive.org/web/20201007012740/https://upload.wikimedia.org/wikipedia/commons/b/ba/Lunar_libration_with_phase_Oct_2007_450px.gif&quot; alt=&quot;Tidal locking of the Moon with the Earth&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;                         &lt;sup&gt;&lt;strong&gt;Fig. 2&lt;/strong&gt;: Lunar librations in latitude and longitude over a period of one month (Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Lunar_libration_with_phase_Oct_2007_450px.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;And inclination is likely calculated for geolocation of the &lt;a href=&quot;https://web.archive.org/web/20191011175433/https://aa.usno.navy.mil/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomical Applications Department of the U.S. Naval Observatory&lt;/a&gt; in Washington, DC. The page however doesn't explain if this assumption of mine is correct, but it would certainly appear so.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="7472" LastEditDate="2020-12-19T11:58:41.233" LastActivityDate="2020-12-19T11:58:41.233" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="1037" PostTypeId="1" AcceptedAnswerId="1112" CreationDate="2013-11-29T21:30:45.270" Score="16" ViewCount="805" Body="&lt;p&gt;With our current technology, or technology available in the near future (up to 2025), how would we detect a planet exactly like our own, and how close would it have to be to be detectable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which methods thus far have proven the most effective for detecting Earth-like planets and what would they be able to reveal regarding our planet's atmosphere, orbital properties, and our species?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This question is interesting food for thought on both how easily an extraterrestrial civilization with comparable technology might find us, and for how easily we might find them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the purpose of the question we will assume that there is a planet with a civilisation with equivalent technological advancements that could exist anywhere.&lt;/p&gt;&#xA;" OwnerUserId="500" LastEditorUserId="500" LastEditDate="2013-12-07T06:15:46.250" LastActivityDate="2015-05-03T15:09:19.783" Title="How would we detect an Earth doppelganger planet?" Tags="&lt;exoplanet&gt;&lt;extra-terrestrial&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1038" PostTypeId="2" ParentId="1029" CreationDate="2013-11-29T22:17:55.383" Score="4" Body="&lt;p&gt;@AlexeyBobrick is correct. To add to his answer: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scale of galaxy clusters are on the order of ~1Mpc (~3.14 Million light years in size), and are therefore much smaller than the cosmological horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cosmological principle is extraordinarily important for cosmologists, and makes the assumption about the universe's global properties (which turn out to be really good assumptions, so far): &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Homogeneity on the scale of about 100 Mpc. This means that one would see very little change in density within a bubble of radiua 100 Mpc as you move said bubble around the universe.&lt;/li&gt;&#xA;&lt;li&gt;Isotropy means that there is no preferred directions to look. No matter where you look, you should see roughly the same picture of the universe. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As for how these clusters are generated (addressing the title of your question), large N-body simulations are run over the age of the universe from different sets of initial conditions (this is where different models and assumptions come into it). These are dark matter only simulations and involve only the force of gravity. Some people are including electromagnetic interactions in their code, but it's far from being the norm, and are really only important on small scales, i.e. - inner regions of galaxies and clusters. These simulations can contain upwards of ~10 billion dark matter particles.&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2013-11-29T22:17:55.383" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1039" PostTypeId="2" ParentId="897" CreationDate="2013-11-29T22:44:44.980" Score="3" Body="&lt;p&gt;In general, asteroids experience the same fundamental processes as the moon: solar wind, micrometeorites  and photons from the sun are striking them and sputtering material off to form an exosphere.  So there must be some sort of exosphere, but its density and composition will vary with distance from the sun, as well as the gravity and composition of the asteroid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 1992 &lt;a href=&quot;http://adsabs.harvard.edu/abs/1992Icar...98...54A&quot; rel=&quot;nofollow noreferrer&quot;&gt;ultraviolet study of Ceres&lt;/a&gt; found $OH$ emission, consistent with escaping water, and suggesting a ice cap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a more &lt;a href=&quot;http://adsabs.harvard.edu/abs/2011AJ....142..125R&quot; rel=&quot;nofollow noreferrer&quot;&gt;recent study&lt;/a&gt; with the more sensitive &lt;a href=&quot;https://en.wikipedia.org/wiki/Very_Large_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;VLT&lt;/a&gt;  did not detect $OH$, limiting the maximum amount of water produced  per second to $7 \times 10^{25}$ molecules per second at the time of the observation. &lt;/p&gt;&#xA;" OwnerUserId="457" LastEditorDisplayName="user15217" LastEditDate="2016-12-21T12:36:38.293" LastActivityDate="2016-12-21T12:36:38.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1040" PostTypeId="2" ParentId="1031" CreationDate="2013-11-30T03:23:31.637" Score="8" Body="&lt;p&gt;The first thing to consider is that the area of a beam will, over long distances, diffuse. The best situation we can hope for is a diffraction-limited system, where this diffusion is minimized thus maximizing our received signal. That is, in theory we have a perfectly collimated transmission beam that neither diverges nor converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, we are still limited by diffraction. A diffraction limited system is described by the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sin\theta = \frac{1.22\lambda}{D}, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which prescribes an angular resolution $\theta$ in terms of wavelength $\lambda$ and a circular aperture of diameter $D$. This is called the &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/raylei.html&quot; rel=&quot;noreferrer&quot;&gt;Rayleigh criterion&lt;/a&gt;. The definition of angular resolution, in this case, is when two point sources are just discernable from each other where the principal maximum of the airy disk pattern of one source coincides with the first minimum of another. It's this definition that results in the apparently arbitrary constant of $1.22$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We usually think of diffraction being applicable in terms of receiving a signal - for example, a space telescope will usually have a diffraction-limited optical system. However, the exact same laws hold true whether we are receiving or sending a signal. The optical path is the same. Everything is just in reverse!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Side note: if we instead &lt;a href=&quot;http://i380.photobucket.com/albums/oo247/Vuldacon/BatmanSignalLightWonder.png&quot; rel=&quot;noreferrer&quot;&gt;projected an image into space&lt;/a&gt;, in order to acceptably resolve the image a receiver would need to have an angular resolution equal to, or greater than, than the projection. This includes a spatial resolution criterion in addition to the signal-to-noise performance discussed below.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To make a real-life example, let's consider a radio signal. Since a distant receiver will be getting a frequency-modulated signal not unlike FM radio, we are not concerned with angular resolution. We don't care if the &quot;image&quot; is blurred, or even if some areas of the originally transmitted beam entirely miss our receiver. All we are concerned with is the modulation of frequency over time - it's a one-dimesional signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, a receiver is a noise-limited system. &lt;a href=&quot;http://ipnpr.jpl.nasa.gov/progress_report2/42-45/45R.PDF&quot; rel=&quot;noreferrer&quot;&gt;This NASA report&lt;/a&gt; outlines some of the limitations that a realistic implementation of interstellar communication must deal with. Even in the case of a quantum noise-limited system, we can still &lt;a href=&quot;http://www.nature.com/news/squeezed-light-mutes-quantum-noise-1.13510&quot; rel=&quot;noreferrer&quot;&gt;make the best&lt;/a&gt; of the limitations dealt to us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the signal-to-noise ratio is above an acceptable threshold then the signal will be received well. There are so many factors to consider that really only an order-of-magnitude estimate is feasible. I don't know enough about this to come up with a good estimate myself of the noise levels of a particular system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://seti.berkeley.edu/sites/default/files/19730010095_1973010095.pdf&quot; rel=&quot;noreferrer&quot;&gt;Project Cyclops&lt;/a&gt; (1971) was the initial investigation into the feasibility of a search for extraterrestrial intelligence. For example, on page 41, we can see that the minimum &lt;a href=&quot;http://www.cv.nrao.edu/~rfisher/Papers/rcvr_basics.pdf&quot; rel=&quot;noreferrer&quot;&gt;noise temperature&lt;/a&gt; of a receiver receiving the 2.4 GHz &lt;a href=&quot;http://en.wikipedia.org/wiki/Arecibo_message&quot; rel=&quot;noreferrer&quot;&gt;Arecibo message&lt;/a&gt; is about 4K - the major contributor to noise here is the CMB. Frequencies of this order of magnitude will usually provide the best possible noise performance - too high and quantum noise and atmospheric effects become significant. Too low, and galactic noise takes over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This noise temperature provides a noise floor for the signal. The receiver usually introduces a significant noise temperature to the degree of some tens or hundreds of Kelvin, so any practical limitations on interstellar communication tend to become a function of our equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the Arecibo message was broadcast at a good frequency, for very long-distance communication amplitude modulation is superior to frequency modulation as it's easy to increase the pulse duration and interval to compensate for a weaker signal strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This table from page 50 of the well-worth-reading Cyclops report shows that a single $100~ \mathrm{m}$ transmitter/receiver combination, with a transmission power of $10^5~\mathrm{W}$, could function at a distance of 500 light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VWc5B.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Building bigger transmitters and receivers will increase the maximum distance of communications. So will increasing the transmission power, pulse duration, and pulse interval. Current technology could let us communicate over tens or hundreds of light years. To communicate further, just &lt;em&gt;build something bigger&lt;/em&gt;. The laws of physics place few limits on the distance we can communicate.&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2013-11-30T03:23:31.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1041" PostTypeId="1" CreationDate="2013-11-30T06:56:08.547" Score="10" ViewCount="567" Body="&lt;p&gt;Usual methods of estimating stellar ages involve &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_isochrone&quot;&gt;isochrone&lt;/a&gt; approximations. It can also help to estimate a star's radius by correlating its absolute magnitude with effective temperature and apparent magnitude. In the absence of these measurements or observations of tell-tale variability, how might you guess a star's age and evolutionary status?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a single, high-resolution $(R\gtrsim 50000)$ spectrum as your only data point, how easy is it to accurately infer the age and evolutionary status of a star? For example, how would the spectrum differ between a red dwarf and red giant, both with of $T_{\mathrm{eff}} = 4000~\mathrm{K}$? Or between two red dwarfs of ages $2~\mathrm{Gyr}$ and $8~\mathrm{Gyr}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good answer &lt;em&gt;could&lt;/em&gt; describe how surface gravity $(\log g)$ affects spectral lines (and how this relates to stellar mass and radius), what elements we might observe more strongly at different stages of evolution, and some observational results of &lt;a href=&quot;http://adsabs.harvard.edu/abs/2003ApJ...586..464B&quot;&gt;gyrochronology&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="276" LastEditDate="2013-11-30T07:28:06.197" LastActivityDate="2020-02-20T11:21:39.963" Title="Without using absolute magnitudes or isochrones, how might we tell a star's age and evolutionary status?" Tags="&lt;star&gt;&lt;stellar-evolution&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1042" PostTypeId="5" CreationDate="2013-11-30T07:04:44.723" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-11-30T07:04:44.723" LastActivityDate="2013-11-30T07:04:44.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1043" PostTypeId="4" CreationDate="2013-11-30T07:04:44.723" Score="0" Body="Questions related to the evolution of stars." OwnerUserId="276" LastEditorUserId="276" LastEditDate="2013-11-30T14:10:21.370" LastActivityDate="2013-11-30T14:10:21.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1044" PostTypeId="1" AcceptedAnswerId="1045" CreationDate="2013-12-01T11:16:59.027" Score="6" ViewCount="85" Body="&lt;p&gt;Iapetus's hemisphere facing Saturn is dark, whereas the opposite one is bright.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could a cryo-volcano be the reason behind this colour difference in Iapetus's hemispheres?&lt;/p&gt;&#xA;" OwnerUserId="502" LastEditorUserId="181" LastEditDate="2013-12-01T20:23:03.513" LastActivityDate="2013-12-01T20:23:03.513" Title="Could a cryo-volcano be the reason behind this colour difference in Iapetus's hemispheres?" Tags="&lt;natural-satellites&gt;&lt;volcanism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1045" PostTypeId="2" ParentId="1044" CreationDate="2013-12-01T12:46:57.673" Score="3" Body="&lt;p&gt;Reposting &lt;a href=&quot;https://space.stackexchange.com/a/2252/49&quot;&gt;my own answer&lt;/a&gt; from &lt;a href=&quot;http://chat.stackexchange.com/&quot;&gt;Space Exploration&lt;/a&gt; on &lt;a href=&quot;https://space.stackexchange.com/questions/2244/what-is-responsible-for-the-different-colors-of-iapetus&quot;&gt;nearly identical question&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The color dichotomy of Iapetus is due to the darker half, the &lt;em&gt;Cassini Regio&lt;/em&gt;, being a result of the moon's accumulation of the dust in the Saturn's largest, yet extremely tenuous, diffuse dust ring called the Phoebe ring depositing onto the moon's leading hemisphere, while the trailing hemisphere's surface remains uninfluenced by this source of dust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phoebe ring was only observed first in late 2009 by the Spitzer telescope in the infrared range, and is believed to have been caused by bombardment of micrometeorites of Phoebe, another Saturn's moon that the prevailing theories suggest is a captured Kuiper belt originating planetesimal orbiting Saturn in a retrograde, 173° to the ecliptic inclined orbit. It is the largest discovered ring of the Saturn, stretching from calculated 59 to 300 Saturn's radii (observed between 128 and 207 Saturn's radii by the Spitzer telescope), is tilted 27° to Saturn's inner rings and with an orbital inclination of 175°, suggesting its origins and again retrograde orbit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/t8MJU.jpg&quot; alt=&quot;Spitzer&amp;#39;s infrared Phoebe ring observations&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;sup&gt;The Phoebe ring was discovered by NASA's Spitzer Space Telescope, detecting it in the infrared spectrum. (Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Infrared_Ring_Around_Saturn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikimedia&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iapetus, being in a prograde orbit with a 17.28° inclination to the ecliptic, is directly in the path of this retrograde dust ring at an average distance from Saturn of 215 Saturn radii. Since it's &lt;a href=&quot;https://astronomy.stackexchange.com/a/22/13&quot;&gt;tidally locked&lt;/a&gt; to Saturn, it would always face same side towards any debris and dust in its path. Joseph A. Burns, Cornell's Irving Porter Church Professor of Engineering and professor of astronomy &lt;a href=&quot;http://www.news.cornell.edu/stories/2009/12/cassini-unveils-history-saturns-moon-iapetus&quot; rel=&quot;nofollow noreferrer&quot;&gt;said in a press release for Cornell Chronicle&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;&quot;The ring of collisional debris that has come off Phoebe and its&#xA;  companion moons is out there, and now we understand the process&#xA;  whereby the stuff is coming in. When you see the coating pattern on&#xA;  Iapetus, you know you've got the right mechanism for producing it.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So this is the theory. Iapetus is in a prograde orbit intersecting at inclination Phoebe dust ring in a retrograde orbit around Saturn, and since Iapetus is tidally locked and always hitting this dusty path &lt;em&gt;head on&lt;/em&gt;, only one hemisphere (the darker one, called &lt;em&gt;Cassini Regio&lt;/em&gt;) is covered in Phoebe's dust, while the other hemisphere (the lighter one, called &lt;em&gt;Roncevaux Terra&lt;/em&gt;) remains free from it, unveiling this Saturn moon's true surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/0YTGe.jpg&quot; alt=&quot;Saturn&amp;#39;s Phoebe ring&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;sup&gt;Artist's conception showing nearly invisible and largest Saturn's Phoebe ring (Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Saturn_largest_ring_Spitzer_telescope_20091006.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikimedia&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.news.cornell.edu/stories/2009/12/cassini-unveils-history-saturns-moon-iapetus&quot; rel=&quot;nofollow noreferrer&quot;&gt;Same mentioned Cornell Chronicle press release&lt;/a&gt; explains why the transition from the dark hemisphere to the whiter parts of the moon isn't seamless, but rather a mottled, patchy array of bright and dark spots:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Small, white craters that dot Iapetus' darker half indicate a veneer&#xA;  of dark dust, only meters deep, covering a white, icy surface that&#xA;  matches the rest of the satellite. The imaging data also revealed that&#xA;  all the materials on the leading side are much redder than the&#xA;  shielded and brighter trailing side -- another indication that the&#xA;  leading side's dust came from elsewhere.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The pattern, the scientists say, supports a theory described in a&#xA;  companion paper in Science that the darker parts of the moon tend to&#xA;  heat up when struck by sunlight, encouraging the ice to evaporate&#xA;  underneath.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This causes any dark spots to get even darker, creating the mottled&#xA;  look.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/9xgnz.jpg&quot; alt=&quot;Iapetus as seen by the Cassini probe&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mosaic of Iapetus taken by Cassini probe during NASA's Cassini Solstice mission, showing the bright trailing &lt;em&gt;Roncevaux&lt;br /&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Terra&lt;/em&gt; hemisphere with part of the dark area &lt;em&gt;Cassini Regio&lt;/em&gt; appearing on the right. (Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Iapetus_as_seen_by_the_Cassini_probe_-_20071008.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikimedia&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2013-12-01T12:46:57.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1046" PostTypeId="1" CreationDate="2013-12-01T21:55:00.127" Score="12" ViewCount="585" Body="&lt;p&gt;I've watched documentaries about the solar system, where it is suggested that Venus once had oceans of liquid water similar to those that cover most of Earth today.  Venus is now in a period of runaway greenhouse effect and has lost its oceans.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've always wondered whether the apparent overabundance of water on Earth could be partially explained by our planet sweeping up the water, pushed into Earth's path by the solar wind, after it had been liberated from Venus.  I know one theory on where our water came from is a bombardment of comets - but I've always wondered whether some could also have arrived after Venus heated up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How stupid is this as an idea?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;edits:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10/12/2014 - article from the bbc - &lt;a href=&quot;http://www.bbc.co.uk/news/science-environment-30414519&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rosetta results: Comets 'did not bring water to Earth'&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;20/09/2017 - article from www.spacetelescope.org - &lt;a href=&quot;http://www.spacetelescope.org/news/heic1715/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble discovers a unique type of object in the Solar System&lt;/a&gt; which includes an interesting note - &quot;[2] Current research indicates that water came to Earth not via comets, as long thought, but via icy asteroids.&quot;&lt;/p&gt;&#xA;" OwnerUserId="389" LastEditorUserId="389" LastEditDate="2017-09-21T19:46:09.200" LastActivityDate="2017-09-21T19:46:09.200" Title="Could Venus be a source of Earth's apparent overabundance of water?" Tags="&lt;earth&gt;&lt;water&gt;&lt;venus&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1047" PostTypeId="2" ParentId="1046" CreationDate="2013-12-02T06:16:08.120" Score="5" Body="&lt;p&gt;No idea is stupid &lt;em&gt;per se&lt;/em&gt;.  But in order to place an answer to your question, please consider Venus as a comet. It once had, in that image, a coma (a tail) pointing outwards from the sun, made of evaporated water pushed away by solar wind. How frequently is the Earth exactly inside that coma?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a comparison, when the Earth strikes a real comet coma we see meteor showers that last no more than one day. So if we consider that Earth may go through Venus imaginary coma once a year, we have an upper limit of 1/365 of Venus' water captured by Earth. This, without taking into effect that Venus had not have a real coma, that evaporation and expelletion of water from Venus was not a so directional process, and surely other factors I can not think about just now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides, do not consider comets the main source of water on Earth. You need to count also with the water vapour generated on volcanoes and that generated on acid-base reactions among Earth's rocks.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-12-02T06:16:08.120" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1048" PostTypeId="2" ParentId="1046" CreationDate="2013-12-03T02:56:45.337" Score="5" Body="&lt;p&gt;First, one should observe that water is very abundant in the Solar system and most of it comes from before the planetary formation stage. Even Mercury has got its deal of water, albeit a small one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think, it will be safer to assume that earth had had it water from the very beginning of its formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional thing to consider are the temperatures, routinely reachable on various planetary bodies. Mars and everything beyond are way too cool for liquid water to exist, but solid water is present there in considerable quantities. Similarly, Venus and Mercury  ended up with temperatures above point of water vaporization or even chemical breakdown in the presence of other compounds. Only the Earth happens to occupy a nice orbital zone where water can exist in all 3 phases (solid, liquid and vapor). But it's not something Venus has any effect on.&lt;/p&gt;&#xA;" OwnerUserId="412" LastActivityDate="2013-12-03T02:56:45.337" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1049" PostTypeId="1" AcceptedAnswerId="1050" CreationDate="2013-12-03T05:09:32.093" Score="12" ViewCount="595" Body="&lt;p&gt;How do you figure out the star magnitude of a star? Do they just give it a guess or what?&#xA;If not, how do they figure it out accurately? &lt;/p&gt;&#xA;" OwnerUserId="503" LastActivityDate="2013-12-09T19:56:12.497" Title="How do you figure out the magnitude of stars?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1050" PostTypeId="2" ParentId="1049" CreationDate="2013-12-03T06:22:12.237" Score="5" Body="&lt;p&gt;You must remember that there are &lt;a href=&quot;http://en.wikipedia.org/wiki/Absolute_magnitude&quot; rel=&quot;nofollow&quot;&gt;Absolute magnitude&lt;/a&gt; (that measured from a fixed distance of 10pc) and &lt;a href=&quot;http://en.wikipedia.org/wiki/Apparent_magnitude&quot; rel=&quot;nofollow&quot;&gt;Apparent magnitude&lt;/a&gt; (which is measured from Earth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Historically magnitudes ranged from First to Sixth and were assigned by guess by Ptolomaeus. His First Magnitude was the equivalent of saying first size, so it was the brightest, and Sixth Magnitude was the assignment he made for the dimmest seen with the naked eye. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays, we use roughly the same scale but with real numbers, ranging from -26.74 (Sun) up. We defined the star Vega to be the exact 0.0 apparent magnitude (and 0.0 colour in all filters, for the case), and then we use the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_1-m_{ref}=-2.5log{I\over I_{ref}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to define all other apparent magnitudes from that one by measuring Intensities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer you, yes, we measure apparent magnitudes rather accurately. For stars other than Sun, they range from -1.46 for Sirius up to +8 in a perfect night for a trained naked eye (typically +6 for the normal eye on a normal, far from city lights, night).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnitude_(astronomy)&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Magnitude_(astronomy)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="473" LastEditorUserId="473" LastEditDate="2013-12-09T19:56:12.497" LastActivityDate="2013-12-09T19:56:12.497" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1051" PostTypeId="1" CreationDate="2013-12-03T19:46:17.480" Score="17" ViewCount="449" Body="&lt;p&gt;Jupiter's moon Io is arguably one of the most volcanically active bodies in the Solar System.  The reason, according to NASA's page &lt;a href=&quot;http://www.nasa.gov/topics/solarsystem/features/io-volcanoes-displaced.html&quot;&gt;Scientists to Io: Your Volcanoes Are in the Wrong Place&lt;/a&gt; is believed to be caused by Io being&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;caught in a tug-of-war between Jupiter's massive gravity and the smaller but precisely timed pulls from two neighboring moons that orbit further from Jupiter – Europa and Ganymede. Io orbits faster than these other moons, completing two orbits every time Europa finishes one, and four orbits for each one Ganymede makes. This regular timing means that Io feels the strongest gravitational pull from its neighboring moons in the same orbital location, which distorts Io's orbit into an oval shape. This in turn causes Io to flex as it moves around Jupiter. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, how did Io form in the first place, given the tidal stresses acting upon it? Does this suggest (and what evidence is there) that Io 'migrated' into its present orbit?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2018-02-14T10:12:23.050" Title="How was Io not torn apart by tidal forces during its formation?" Tags="&lt;orbit&gt;&lt;jupiter&gt;&lt;volcanism&gt;&lt;tidal-forces&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1052" PostTypeId="2" ParentId="1051" CreationDate="2013-12-03T20:01:27.160" Score="6" Body="&lt;p&gt;No, it is not just a matter of migration. You need to take into account two facts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is that (as experience shows) Io's own gravity is enough to avoid it breaking by tidal forces. It has been like that through all its history: Io could not have been formed if it started aggregating today, but it was formed at the same time Europa and Ganymede did: they three were growing in parallel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other is that of the orbital resonances, which makes precisely that orbit with so simple integer number relations with those of Europa and Ganymede a stable one. Io could not have been formed in another place.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-12-03T20:01:27.160" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1053" PostTypeId="2" ParentId="1041" CreationDate="2013-12-03T23:29:44.120" Score="2" Body="&lt;p&gt;In short: you can not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In length: best you can do is to match up your spectrum with a library of known spectra, and find the best match. But for these spectra to be useful you need to have determined their ages, masses, Y's (contents of Helium) and Z's (contents of metals, that is, evrything beyond Helium). And their ages come from... yes, isochrones, so you would be using isochrones indirectly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in short again, yes, you can determine the mass, age and Y and Z of a star with its spectrum and without its own isochrone, maybe up to 5% of its main-sequence lifetime during main sequence status (e.g. 0.5 Gyr for a 10 Gyr main-secuence lifetime star like our Sun).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes again, this match-up of spectra gives additional info like surface gravity, which is not useful on its own but needs previous knowledge of mass and radius.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-12-03T23:29:44.120" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1054" PostTypeId="1" CreationDate="2013-12-04T00:06:27.963" Score="4" ViewCount="249" Body="&lt;p&gt;I've just watched a &lt;a href=&quot;http://en.wikipedia.org/wiki/Europa_Report&quot; rel=&quot;nofollow&quot;&gt;movie&lt;/a&gt; that made me curious as to whether life has really been discovered on another world such as Europa??&lt;/p&gt;&#xA;" OwnerUserId="510" LastEditorUserId="204" LastEditDate="2013-12-05T18:33:34.267" LastActivityDate="2013-12-05T18:33:34.267" Title="Has life been discovered outside of Earth?" Tags="&lt;universe&gt;&lt;life&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1055" PostTypeId="2" ParentId="1054" CreationDate="2013-12-04T01:25:00.317" Score="4" Body="&lt;p&gt;No life has been discovered out of the Earth. It is still discussed that bacteria once existed at Mars.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-12-04T01:25:00.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1056" PostTypeId="1" AcceptedAnswerId="1057" CreationDate="2013-12-04T07:46:33.977" Score="10" ViewCount="912" Body="&lt;p&gt;How do you find the distance from a star/planet/black hole to another? I know people can calculate the distance from Earth to a star, but what about from one to another? &lt;/p&gt;&#xA;" OwnerUserId="511" LastActivityDate="2015-02-15T02:20:47.117" Title="Find distance from star to star?" Tags="&lt;star&gt;&lt;planet&gt;&lt;black-hole&gt;&lt;distances&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1057" PostTypeId="2" ParentId="1056" CreationDate="2013-12-04T09:05:54.247" Score="6" Body="&lt;p&gt;If you know distance from Earth to both objects and the angle between them viewed from Earth, it is just a matter of trigonometry.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-12-04T09:05:54.247" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1058" PostTypeId="2" ParentId="1054" CreationDate="2013-12-04T12:16:22.437" Score="2" Body="&lt;p&gt;Quotation source : &lt;a href=&quot;http://en.wikipedia.org/wiki/Life_on_Mars&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Life_on_Mars&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The &lt;strong&gt;ALH84001 meteorite&lt;/strong&gt; was found in December 1984 in Antarctica, by members of the ANSMET project; the meteorite weighs 1.93 kilograms (4.3 lb). The sample was &lt;strong&gt;ejected from Mars&lt;/strong&gt; about 17 million years ago and spent 11,000 years in or on the Antarctic ice sheets. Composition analysis by NASA revealed a kind of magnetite that on Earth, is only found in association with certain microorganisms. Then, in August 2002, another NASA team led by Thomas-Keptra published a study indicating that 25% of the magnetite in ALH 84001 occurs as small, uniform-sized crystals that, on Earth, is &lt;strong&gt;associated only with biologic activity&lt;/strong&gt;, and that the remainder of the material appears to be normal inorganic magnetite. The extraction technique did not permit determination as to whether the possibly biological magnetite was organized into chains as would be expected.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In November 2009, NASA scientists reported after more detailed analyses, that a biogenic explanation is a more viable hypothesis for the origin of the magnetites in the meteorite.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The Nakhla meteorite&lt;/strong&gt; fell on Earth on June 28, 1911 on the locality of Nakhla, Alexandria, Egypt.&#xA;  In 1998, a team from NASA's Johnson Space Center obtained a small sample for analysis. Researchers found &lt;strong&gt;preterrestrial aqueous alteration phases and objects of the size and shape consistent with Earthly fossilized nanobacteria&lt;/strong&gt;, but the existence of nanobacteria itself is controversial. Analysis with gas chromatography and mass spectrometry (GC-MS) studied its high molecular weight polycyclic aromatic hydrocarbons in 2000, and NASA scientists concluded that as much as 75% of the organic matter in Nakhla &quot;may &lt;strong&gt;not&lt;/strong&gt; be recent terrestrial contamination&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may also refer this &lt;a href=&quot;http://en.wikipedia.org/wiki/Extraterrestrial_life&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; to get a detailed explanation of documented evidences of life outside earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a popular hypothesis exists that even life on earth started by a meteor or asteroid which landed on earth. Bacteria and viruses are known to survive extreme temperatures and vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;kepler probe&lt;/strong&gt; data and various other data exists that there are &lt;strong&gt;planets like earth.&lt;/strong&gt; There hasn't been a solid proof that we are not alone. But there are various signs and much more like above. There are many solar systems and many other universes, which CAN support life. Let's wait if anyone visits us.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2013-12-04T13:47:51.580" LastActivityDate="2013-12-04T13:47:51.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1061" PostTypeId="2" ParentId="1028" CreationDate="2013-12-04T15:27:02.477" Score="5" Body="&lt;p&gt;The answers to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/9/could-someone-explain-ra-dec-in-simple-terms&quot;&gt;this question&lt;/a&gt; are very clear and much better than I could have written.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what you mean by &lt;code&gt;&quot;the stellar time at Greenwich at 0h UTC is 22h20min&quot;&lt;/code&gt;, but I'm assuming that you mean that that is the amount of time since the culmination of Aries over the Prime Meridian (Greenwich). That being the case, the Right Ascension (RA) of your body can be easily calculated. All we need to do is figure out what RA is passing your meridian at the time in question. This is probably a good time to point out that the longitude of Mt Teide is 16° 30'W not E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We already know that RA 22H 20m is passing over Greenwich at the time of interest (or, at least that is what I am assuming as your explanation is not at all clear.), as we are to the West of Greenwich, then we know that any RA on our meridian is earlier and we can calculate how much earlier simply by dividing our longitude by 15 (as the earth spins at a nice steady 15°/hour):-&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;diff in RA = Long/15 &#xA;           = 16° 30'/15 &#xA;           = 1H 06m&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then:-&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;RA = RA at Greenwich - diff in RA &#xA;   = 22H 20m - 1H 06m &#xA;   = 21H 14m&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now for Declination:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first stage is to take a piece of ruled or graph paper and near the middle mark a short, horizontal line. This is your horizon, and is marked 'H'. Now draw upwards 9 lines and mark the top ,'Z' and downwards 9 lines and mark the bottom 'N'. You should end up with something that looks like this:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DtOhK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is, basically a picture of your meridian as you sit and gaze south at it. 'Z' is your zenith, the point above you in the sky and 'N' is the nadir, it's opposite out on the celestial sphere through the earth beneath your feet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need to add a little more detail. Imagine that the celestial sphere is actually a glass sphere surrounding the earth with the stars and stuff stuck to it that we can project anything we want on to (we'll ignore porn and coke adverts for now), where would the celestial equator show and where would it go on our picture of the meridian? Well, it would be a distance above the horizon equal to our latitude (lets round it off to 30°), so we now have:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6A532.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we put in the body we are looking at. We know it's elevation, just over 43°, we can just call it x as you didn't specify a body. We can also label some of the distances we know, to end up with:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TMkp7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We already knew that Lat was H -&gt; Q, that's how we place Q, we know that elevation, or altitude is between the horizon and the body and that declination is measured from the celestial equator. We want to find dec and the diagram shows us that:-&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;dec = Alt - Lat &#xA;    = 43° 40' - 28° 18' &#xA;    = 15° 18'N&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As 'x' is to the north of 'Q', then Declination is north.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we know that we are looking at a body with RA 21H 14m and Dec N 15° 18'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is much more complex to explain here than it is to do, or to teach in person. With a bit of practice you'll be doing it in your head. I hope this has helped a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should point out that I have ignored observation errors, such as height of eye, which may be significant from the place you specify and parallax but I wanted to keep it as simple as possible.&lt;/p&gt;&#xA;" OwnerUserId="509" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2013-12-07T10:35:40.550" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="1062" PostTypeId="1" AcceptedAnswerId="1063" CreationDate="2013-12-04T11:53:49.823" Score="22" ViewCount="2874" Body="&lt;p&gt;When I stand outside looking at the night sky, to my untrained eye, everything except the moon looks like a star.  I know intellectually that some are planets circling our sun, and some are entire galaxies far away, but they all look basically the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How far out do you have be be from our sun for it to appear the same as any other 'star' in the sky? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit to clarify&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we migrate out through the solar system, when we look up to the sky the sun will get dimmer the farther away we get. On earth there is no doubt which star is our sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we occupy bodies in the solar system farther from the sun, where will we be when the sun appears to be the same &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_brightest_stars&quot;&gt;brightness&lt;/a&gt; as any other star in the sky? &lt;/p&gt;&#xA;" OwnerUserId="513" OwnerDisplayName="James Jenkins" LastEditorUserId="6" LastEditDate="2013-12-04T16:59:34.287" LastActivityDate="2016-08-28T18:54:06.067" Title="At what distance from Earth would our Sun be the same apparent magnitude as the next brightest star in the sky?" Tags="&lt;the-sun&gt;&lt;observational-astronomy&gt;&lt;apparent-magnitude&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1063" PostTypeId="2" ParentId="1062" CreationDate="2013-12-04T16:38:39.687" Score="26" Body="&lt;p&gt;One way to answer would be to consider the brightest star in our sky (other than the Sun), which is Sirius.  Then determine how far you would have to be from our Sun for it to be as bright as Sirius is from here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That turns out to be 1.8 light years.  That's not even halfway to the nearest star, so if you're in any other star system, then our Sun is just another star.  If you're anywhere in our solar system, even way out in the Oort cloud, then our Sun is way brighter than anything else.&lt;/p&gt;&#xA;" OwnerUserId="512" OwnerDisplayName="Mark Adler" LastEditorUserId="512" LastEditDate="2013-12-04T17:48:56.210" LastActivityDate="2013-12-04T17:48:56.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1064" PostTypeId="2" ParentId="1062" CreationDate="2013-12-04T20:48:16.427" Score="16" Body="&lt;p&gt;As Mark Adler mentioned, the best way is to compare the brightness to other nearby stars. I'm going to assume that you have instantaneous travel time, and also take into account that you are actually getting closer to stars depending on the direction you go. I'm using &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_nearest_stars&quot; rel=&quot;nofollow&quot;&gt;this table&lt;/a&gt; from Wikipedia. I'm going to go no further on the list than Sirius, and assume in each instance we are heading strait towards the Star. The formula for calculating the apperant magnitude given absolute magnitude, which is provided, is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = M - 5 (1-\log_{10}{d})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting up for our situation, the problem becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4.85 - 5(1-\log_{10}({d_{☉})})= M - 5 (1-\log_{10}({d-d_{☉}}))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{M-4.85}{5}=\log_{10}{\frac{d_{☉}}{d-d_{☉}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing to solve for $d_{☉}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_{☉}=\frac{d\cdot10^{\frac{M-4.85}{5}}}{10^{\frac{M-4.85}{5}}+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging that into a spreadsheet gives the following distances where the two stars are equally bright (Only including the strongest contenders)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;α Centauri A- 1.94 Light Years&lt;/li&gt;&#xA;&lt;li&gt;α Centauri B- 2.61 Light Years&lt;/li&gt;&#xA;&lt;li&gt;Sirius A- 1.46 light years&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Bottom line, heading 1.46 light years towards Sirius, you'll see both Sirius and the Sun as equally bright. This is approximately the edge of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Oort_cloud&quot; rel=&quot;nofollow&quot;&gt;Oort Cloud&lt;/a&gt;, and still within the gravitational influence of the Sun, but is well on our way to another star system.&lt;/p&gt;&#xA;" OwnerUserId="182" LastEditorUserId="9301" LastEditDate="2016-08-28T18:54:06.067" LastActivityDate="2016-08-28T18:54:06.067" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1065" PostTypeId="1" AcceptedAnswerId="1067" CreationDate="2013-12-05T09:10:38.903" Score="12" ViewCount="913" Body="&lt;p&gt;That is, it's twice the radius where the radius is from the centre of the sun to some edge. But what is that edge?&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditDate="2013-12-05T10:10:47.773" LastActivityDate="2014-12-19T09:55:46.420" Title="How do you define the diameter of the Sun" Tags="&lt;the-sun&gt;" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1066" PostTypeId="1" CreationDate="2013-12-05T13:13:42.343" Score="2" ViewCount="84" Body="&lt;p&gt;I am sure I get parts of the terminology wrong but if anyone can shed some light in the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that for a given right ascension (RA) and declination (DEC), one has defined a ray (half-line) in the sky starting from the center of the earth towards infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if I also provide a angle, say &lt;code&gt;d&lt;/code&gt; degrees around that ray, I have essentially defined a cone in the sky. My question is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which steps (&lt;code&gt;RA_step&lt;/code&gt; and &lt;code&gt;DEC_step&lt;/code&gt;) should I use in the following loop, as a function of &lt;code&gt;d&lt;/code&gt; to ensure that I cover the entire sky and don't leave any patches anywhere?.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;for (RA = 0 ; RA &amp;lt;= 360 ; RA += RA_step)&#xA;    for (DEC = -90 ; DEC &amp;lt;= 90 ; DEC += DEC_step)&#xA;         examine-cone(RA, DEC, d)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="522" LastEditorUserId="522" LastEditDate="2013-12-05T13:55:16.270" LastActivityDate="2013-12-05T20:54:53.813" Title="iteration to cover the whole sky with right ascension, declination, angle" Tags="&lt;ascension&gt;&lt;inclination&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1067" PostTypeId="2" ParentId="1065" CreationDate="2013-12-05T13:40:11.527" Score="6" Body="&lt;p&gt;Fusion reactions taking place inside the core of the star produce a huge amount of energy, most of which becomes heat. These reactions are not evenly distributed through the star and so there are phenomena such as sun spots and solar flares, however the total amount of energy produced tends to be reasonably constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say that the edge is defined by the average point where the gravity reaches equilibrium with the pressure of the star's super-heated gases (as a result of internal fusion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the picture of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sun&quot; rel=&quot;noreferrer&quot;&gt;Sun on Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That edge/balance will change when the sun begins to run low on hydrogen. At this time, the reactions inside the star will change causing it to become become a &lt;a href=&quot;http://en.wikipedia.org/wiki/Sun#After_core_hydrogen_exhaustion&quot; rel=&quot;noreferrer&quot;&gt;giant red star&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess you could compare it with the surface of sea water on Earth. It's technically not still and stable, but we can calculate an average value of the sea level. And it is because it's an average value that we can rely on that to determine altitude and earth radius as well.&lt;/p&gt;&#xA;" OwnerUserId="204" LastEditorUserId="38" LastEditDate="2013-12-05T15:39:08.623" LastActivityDate="2013-12-05T15:39:08.623" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1068" PostTypeId="1" AcceptedAnswerId="1079" CreationDate="2013-12-05T14:19:22.300" Score="32" ViewCount="2635" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/1054/has-life-been-discovered-outside-of-earth&quot;&gt;No life has been discovered outside of Earth&lt;/a&gt; (yet?), but do we know if anything that would be considered &quot;living&quot; on Earth could conquer &lt;a href=&quot;http://en.wikipedia.org/wiki/Mars&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mars&lt;/a&gt;? (or maybe &lt;a href=&quot;http://en.wikipedia.org/wiki/Venus&quot; rel=&quot;nofollow noreferrer&quot;&gt;Venus&lt;/a&gt;?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mars_One&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mars One&lt;/a&gt; project on the way, I was wondering if it was possible to transplant some of Earth's life to Mars, and if something could survive there (naturally).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many kinds of life are based on air and sun. Animals have to breathe oxygen, plants need the sun, and both need water. But there are many other kinds of life on Earth, like:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Very deep seawater life&lt;/strong&gt; that lives under high pressure and without sunlight,&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Extremophile&quot; rel=&quot;nofollow noreferrer&quot;&gt;extremophile organisms&lt;/a&gt;&lt;/strong&gt; that live in very hot water,&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;weird &quot;arsenic life&quot;&lt;/strong&gt; which was debunked as being fully arsenic-based, but turned out to be something different...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Have there been any theories or studies about unusual forms of life on Earth that might survive on Mars?&lt;/p&gt;&#xA;" OwnerUserId="204" LastEditorUserId="25994" LastEditDate="2019-03-07T23:33:40.750" LastActivityDate="2020-10-12T13:59:10.160" Title="Could any known, living organisms on Earth survive on Mars?" Tags="&lt;earth&gt;&lt;mars&gt;&lt;life&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="1069" PostTypeId="2" ParentId="122" CreationDate="2013-12-05T20:39:51.000" Score="3" Body="&lt;p&gt;If they exist, they are out of anything we can ever reach. They are (by definition) out of our Universe, so we can not affect them and they can not affect us in any way. That's the same as saying that, for any experimental definition, they do not exist.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-12-05T20:39:51.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1070" PostTypeId="2" ParentId="1037" CreationDate="2013-12-05T20:42:24.747" Score="1" Body="&lt;p&gt;If there were a planet with same civilazion &quot;exactly like us&quot;, probability that it were near enough to be detectable are extremely remote. Not zero, but as low as those of trespassing a wall by means of quantum mechanics.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-12-05T20:42:24.747" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1071" PostTypeId="2" ParentId="1066" CreationDate="2013-12-05T20:54:53.813" Score="2" Body="&lt;p&gt;Imagine you are patching a plane with circles, instead of patching the sky sphere with circles (which is the same as filling the space with cones). Now, for every circle of diameter D define a square of diagonal D (D=2d). It is easier to patch the plane with squares, isn't it? So you have squares of side D/2*sqrt(2), and that is exactly your DEC_step, for each vertical sweep, and your RA_step, from sweep to sweep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will lead to a lot of duplicated examination (in fact you will examinate exactly the same place at the Poles once for each sweep) but you can be sure you'll not leave anything out of your squares.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-12-05T20:54:53.813" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1072" PostTypeId="2" ParentId="703" CreationDate="2013-12-05T20:57:17.820" Score="1" Body="&lt;p&gt;There's &lt;a href=&quot;http://earthsky.org/astronomy-essentials/visible-planets-tonight-mars-jupiter-venus-saturn-mercury#jupiter&quot; rel=&quot;nofollow&quot;&gt;a link&lt;/a&gt; on the very page you link to for other planets that shows Jupiter apparently rising about 7 or 8 pm so that may be it.  The article says &quot;Watch for the moon to swing near Jupiter on December 18 and December 19.&quot; so if that bright object is near the moon on those dates that is probably what you are looking at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.stellarium.org/&quot; rel=&quot;nofollow&quot;&gt;Stellarium&lt;/a&gt;  is a great free program for viewing positions of planets and stars.&lt;/p&gt;&#xA;" OwnerUserId="343" LastActivityDate="2013-12-05T20:57:17.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1073" PostTypeId="2" ParentId="1056" CreationDate="2013-12-05T21:12:32.480" Score="5" Body="&lt;p&gt;You only need two-dimensional trigonometry if you know the distances to the two stars and their angular separation. Any two-dimensional plane can be defined by three points that lie on it so we just use the plane containing the two stars and the Earth.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use Earth as the origin and the closest star as a point on the $x$-axis ($x_1$,$y_1$) where $x_1$ is the distance and $y_1$ is zero. You then can use the distance of the second star and its angular separation from the first star (which is on the $x$-axis) to plot a point ($x_2 = \text{distance} \times \cos(\text{angle})$, $y_2 = \text{distance} \times \sin(\text{angle})$). The distance between those two points is $$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$&lt;/p&gt;&#xA;" OwnerUserId="343" LastEditorUserId="2153" LastEditDate="2015-02-15T02:20:47.117" LastActivityDate="2015-02-15T02:20:47.117" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1074" PostTypeId="2" ParentId="700" CreationDate="2013-12-05T22:17:12.133" Score="8" Body="&lt;p&gt;Using density is invalid.  As the radius of the event horizon for a given mass increases linearly, the volume of that radius increases as the cube and the density therefore decreases.  Looking at it the other way, the density increases as the event horizon decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can calculate the size of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Event_horizon&quot; rel=&quot;noreferrer&quot;&gt;event horizon&lt;/a&gt; for any given mass.  You just need to find the point at which the &lt;a href=&quot;https://en.wikipedia.org/wiki/Escape_velocity&quot; rel=&quot;noreferrer&quot;&gt;escape velocity&lt;/a&gt; exceeds the speed of light. We can use the speed of light in the formula for escape velocity and solve for the radius&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Escape velocity formula &lt;img src=&quot;https://i.stack.imgur.com/xbdbH.png&quot; alt=&quot;enter image description here&quot;&gt; solving for r gives &#xA;&lt;img src=&quot;https://i.stack.imgur.com/QsssV.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I put together a &lt;a href=&quot;https://docs.google.com/spreadsheet/ccc?key=0Akd7u7y3k3jVdEZVQ0F1emliSGZreG1hekdmVWlfUmc&amp;amp;usp=sharing&quot; rel=&quot;noreferrer&quot;&gt;spreadsheet&lt;/a&gt; with the numbers. I calculate that a 3.2 solar mass black hole would have a radius of 4.752km, meaning that a neutron star of 3.2 solar masses were to become a black hole it would have to shrink to 9.504km and have a density of 7.13E18 kg/m^3.    Conversely the super-massive black hole at the center of our galaxy has an event horizon radius of about 6 billion km and a density of only 4.34E6 kg/m^3.  A black hole the size of a proton would need 350 million metric tons and have a density of 1.5E56 kg/m^3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you are probably off on some of your numbers.  Specifically you are using ranges of numbers at the top end of a spectrum and an 'about' figure for the radius of a neutron star as though 12km is a single constant radius for all neutron stars.  In fact a 1.4 solar mass neutron star would nave a radius somewhere between 10.4 and 12.9 km (&lt;a href=&quot;http://arxiv.org/abs/1205.6871&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://heasarc.gsfc.nasa.gov/docs/nicer/nicer_about.html&quot; rel=&quot;noreferrer&quot;&gt;https://heasarc.gsfc.nasa.gov/docs/nicer/nicer_about.html&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/lYH3Z.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="343" LastEditorUserId="343" LastEditDate="2013-12-05T22:48:49.463" LastActivityDate="2013-12-05T22:48:49.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1075" PostTypeId="1" CreationDate="2013-12-06T04:51:52.863" Score="6" ViewCount="273" Body="&lt;p&gt;What particularly causes the T Tauri star to eject the strong bipolar wind which clears the gases around the star?&lt;/p&gt;&#xA;" OwnerUserId="231" LastActivityDate="2013-12-10T21:16:06.163" Title="Source of T Tauri wind?" Tags="&lt;solar-system&gt;&lt;solar-wind&gt;&lt;protostar&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1076" PostTypeId="2" ParentId="1037" CreationDate="2013-12-06T08:51:26.647" Score="3" Body="&lt;p&gt;&quot;probably not&quot;. Unless it were so close that television or radio signals like we've been inadvertently sending out can be detected above background radiation levels we can't know about the civilisation. &lt;br/&gt;&#xA;The signals are extremely weak, and not on any band scientifically interesting (deliberately of course, because we don't want our televisions to pick up natural signals that would interfere with our artificial signals) unless your scientific interest is specifically to snoop on communications sent out by potential aliens (SETI comes to mind, a nice hobby and something to do with radio telescopes for the times there's nothing more interesting to listen to).&lt;br/&gt;&#xA;&lt;br/&gt;&#xA;As to planets the size of earth, we've never yet been able to find one, let alone a rocky one in the habitable zone of a star. The mechanisms used to detect exoplanets simply aren't sensitive enough.&lt;br/&gt;&#xA;&lt;br/&gt;&#xA;Leaves one option: their equivalent of Voyager 1 enters our solar system and we actually spot and recognise it for what it is, and find a way to recover it. But If they're at our level of technology it will be thousands of years yet before that happens.&lt;/p&gt;&#xA;" OwnerUserId="534" LastActivityDate="2013-12-06T08:51:26.647" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="1079" PostTypeId="2" ParentId="1068" CreationDate="2013-12-06T09:54:27.983" Score="17" Body="&lt;p&gt;The most likely candidate would be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tardigrade&quot;&gt;Tardigrade&lt;/a&gt;. These little guys handle vacuum and radiation just fine. So long as water is provided, according to tests done in LEO the Tardigrade would survive on Mars. Even if they do dehydrate, they spring back to life once water is provided again.&lt;/p&gt;&#xA;" OwnerUserId="209" LastActivityDate="2013-12-06T09:54:27.983" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1080" PostTypeId="1" AcceptedAnswerId="1560" CreationDate="2013-12-06T16:43:40.367" Score="12" ViewCount="885" Body="&lt;p&gt;Three of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Galilean_moons&quot;&gt;first four moons&lt;/a&gt; ever discovered outside of our own planetary sphere of gravitational influence orbit in very close to perfect resonance. Europa's orbit is almost exactly twice as long as Io's, and Ganymede's is almost exactly twice as long as Europa's. Perhaps it is just coincidence. If you look at enough random things, you start to see patterns that aren't necessarily there. But it would be a pretty big coincidence. When I see something like this in nature, I have to wonder if there is an underlying cause.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratios aren't quite exactly 2, so if there is a gravitational influence or something that pulls them into this resonance, the influence must not be overpowering enough to force exact resonance. Or maybe it is, given enough time, but &quot;recent&quot; collisions or the pull of other satellites in the system may have driven them a little bit off of their resonance. Has anyone studied this and found a probable reason for this phenomenon?&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2021-02-04T16:43:12.313" Title="Does anyone know why three of Jupiter's largest moons orbit in 1:2:4 resonance?" Tags="&lt;orbit&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1081" PostTypeId="1" AcceptedAnswerId="1082" CreationDate="2013-12-06T17:04:43.907" Score="19" ViewCount="562" Body="&lt;p&gt;It always seemed counter-intuitive to me that we observed an measured a gravitational force and since the universe has not enough mass to account for it, the conclusion was to say there's a different kind of matter wich doesn't interact with ordinary matter nor can be seen o measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if the force was applied by other adjacent or overlapsed universes or by something completely different? Are there any other theories propossed? or would we just call whatever we find out to be the source of this force &quot;Dark matter&quot; even if it´s nothing alike matter?&lt;/p&gt;&#xA;" OwnerUserId="537" LastActivityDate="2013-12-06T21:15:33.150" Title="Are there any alternative explanations for the gravitational force we attribute to dark matter?" Tags="&lt;universe&gt;&lt;gravity&gt;&lt;matter&gt;&lt;dark-matter&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1082" PostTypeId="2" ParentId="1081" CreationDate="2013-12-06T17:36:03.460" Score="12" Body="&lt;p&gt;There are certainly people who study alternative (non-General Relativistic) theories of gravity. The most popular theories have so far been:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Modified_Newtonian_dynamics&quot; rel=&quot;noreferrer&quot;&gt;Modified Newtonian Dynamics (MOND)&lt;/a&gt; - which essentially postulates that Newtonian Mechanics break down on some scale, leading to the rotation curves we see in galaxies. &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tensor%E2%80%93vector%E2%80%93scalar_gravity&quot; rel=&quot;noreferrer&quot;&gt;Tensor–vector–scalar gravity&lt;/a&gt; (TeVeS) - this is a relativistic generalization of MOND.&lt;/li&gt;&#xA;&lt;li&gt;Falling into the MOND category is &lt;a href=&quot;http://en.wikipedia.org/wiki/F%28R%29_gravity&quot; rel=&quot;noreferrer&quot;&gt;f(R) gravity&lt;/a&gt;, which alters the general relativistic expression relating the curvature of space-time to the stress-energy tensor (the 'source' of gravity - like mass and pressure). These theories relate different functions of the Ricci scalar to this stress-energy tensor.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;TeVes seems to be the most promising alternative to General Relativity (it may be able to explain things like rotation curves, predicts gravitational lensing, and some other things), but there are still many problems it doesn't do well with. So far it seems like modified theories of gravity are creating more problems than they are fixing - though I must admit, the existence of things like dark matter and dark energy are non-trivial issues cosmologists must come to terms with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to things like alternative universes and the like as the cause of our current mysteries, I'm sure these things have been proposed. The problem here is coming up with an experiment to test them. Dark matter has been named as such because we know that it doesn't emit or absorb any radiation. As far as detecting such a particle, many real experiments have been conducted and will be conducted, and is an exciting field of research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below shows a few of these experiments and the range of cross-section/mass parameter space it has/will probe:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UwBFc.jpg&quot; alt=&quot;dmdetectionexperiments&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditDate="2013-12-06T21:15:33.150" LastActivityDate="2013-12-06T21:15:33.150" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1083" PostTypeId="1" CreationDate="2013-12-06T21:26:36.937" Score="13" ViewCount="3967" Body="&lt;p&gt;Just to think about this is mind boggling. But how do scientist get these numbers? What technology/system/theory do they they use?&lt;/p&gt;&#xA;" OwnerUserId="540" LastEditorUserId="540" LastEditDate="2013-12-06T23:01:50.213" LastActivityDate="2019-02-14T01:54:51.577" Title="How do scientists know there are about 300 billion stars in a galaxy and there are about 100 billion galaxies?" Tags="&lt;galaxy&gt;&lt;star-systems&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1084" PostTypeId="1" AcceptedAnswerId="1091" CreationDate="2013-12-07T04:02:32.407" Score="1" ViewCount="1073" Body="&lt;p&gt;I have tried to search for this but it won't show me the formula or &quot;how to&quot; figure out the formula. So what is the formula, and how do you figure it out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The formula for stars within 400 light years.)&lt;/p&gt;&#xA;" OwnerUserId="503" LastEditorUserId="204" LastEditDate="2013-12-10T14:51:04.753" LastActivityDate="2013-12-10T14:51:04.753" Title="What is the formula to figure out how far away a star is?" Tags="&lt;distances&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ClosedDate="2013-12-10T15:13:29.520" ContentLicense="CC BY-SA 3.0" />
  <row Id="1085" PostTypeId="1" CreationDate="2013-12-07T12:40:13.593" Score="2" ViewCount="3852" Body="&lt;p&gt;Simple, I fell in love with the sky when my parents were constructing our new house I was a kid then and one day I ventured out on the terrace, to my amazement it was a full moon, ever since I've felt spiritually connected with the Universe. Its been around 20 years since that day, out of my busy schedule I looked at the stars yesterday, I found love again! Wanted to buy the best telescope out there, but its out of my pocket. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Want to build a home made powerful telescope? Any help fellow sky lovers and star chasers? &lt;/p&gt;&#xA;" OwnerUserId="543" LastActivityDate="2013-12-07T13:31:05.853" Title="How to build a powerful home made telescope" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ClosedDate="2013-12-08T01:58:35.890" ContentLicense="CC BY-SA 3.0" />
  <row Id="1086" PostTypeId="2" ParentId="1085" CreationDate="2013-12-07T13:31:05.853" Score="2" Body="&lt;p&gt;You can buy remarkably cheap reflecting telescopes these days on a mount called the Dobsonian mount. They don't (usually) have fancy GoTo automated pointing and tracking, but give fantastic bang for buck as the majority of their cost is invested into the biggest primary mirror possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Building a telescope would be a great hobby and learning experience, and there are &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0943396557&quot; rel=&quot;nofollow&quot;&gt;many books&lt;/a&gt; which will show you how. Just don't expect to save a significant sum of money over a commercially built scope - &lt;a href=&quot;http://www.garyseronik.com/?q=node/211&quot; rel=&quot;nofollow&quot;&gt;it could even be more expensive to build your own&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fancy GoTo tracking mounts are not usually worth the expense and hassle (in setting them up) for a first-timer, though the Orion Intelliscope series is an intriguing compromise. For an extra $100 or so, the telescope has no guiding motors, but tracks where you push the telescope and a hand controller will tell you how much further you have to go till you reach your target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My recommendation? Get the biggest Dobsonian mount scope you can afford. With a mirror greater than 10&quot; things begin to get large and unwieldy, so 10 inches (mirror diameter, not tube length) is my recommended maximum size for a first telescope. Just go for the biggest aperture you can buy and easily carry. Don't invest much in GoTo systems, as although they can be quite useful they are also a pain to set up if you don't have the scope permanently mounted.&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2013-12-07T13:31:05.853" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1087" PostTypeId="2" ParentId="1065" CreationDate="2013-12-07T14:02:00.747" Score="2" Body="&lt;p&gt;Look at the Sun.  You shouldn't do this directly with the naked eye, but you can do so through a very dark filter, or project a suitably dark image through a pinhole.  You can even find &lt;a href=&quot;http://apod.nasa.gov/apod/ap120605.html&quot; rel=&quot;nofollow&quot;&gt;photos of the Sun on the internet&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you see is a disk, uniformly bright and with a sharp boundary, surrounded by a comparatively much darker sky.  The bright region is the part we consider the Sun, and that's how we get the radius.  &lt;/p&gt;&#xA;" OwnerUserId="544" LastEditorUserId="544" LastEditDate="2014-12-18T11:57:42.557" LastActivityDate="2014-12-18T11:57:42.557" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1089" PostTypeId="1" CreationDate="2013-12-07T20:20:57.020" Score="13" ViewCount="390" Body="&lt;p&gt;Titan (moon of Saturn) is unique in that it possesses a very thick atmosphere.  However, Titan is certainly is not the largest of the moons - Ganymede being larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the current accepted theory as to why Titan has retained its thick atmosphere?&lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastActivityDate="2014-10-27T04:26:22.367" Title="What is the current accepted theory as to why Titan has retained its atmosphere?" Tags="&lt;planetary-atmosphere&gt;&lt;titan&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1090" PostTypeId="2" ParentId="1084" CreationDate="2013-12-07T20:26:09.570" Score="2" Body="&lt;p&gt;There is no formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a star is near enough so you can see its spectrum, you can deduce its Absolute magnitude, and you can deduce the distance from that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no spectrum, you can not guess the distance, best you can do is find some nearby star that you can use as a standard candle.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-12-07T20:26:09.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1091" PostTypeId="2" ParentId="1084" CreationDate="2013-12-07T23:39:02.903" Score="4" Body="&lt;p&gt;There is no single formula that tells you how far away a star is. In general, it is difficult to find the exact distance to a star just using its brightness since the brightness of stars vary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For close stars &lt;a href=&quot;http://en.wikipedia.org/wiki/Parallax&quot; rel=&quot;nofollow&quot; title=&quot;parallax&quot;&gt;parallax&lt;/a&gt; is used to accurately tell the distance to the star. For stars farther away more complex ways of finding the distance to the star must be used based on how far away the star is, these methods make up the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_distance_ladder&quot; rel=&quot;nofollow&quot; title=&quot;Cosmic Distance Ladder&quot;&gt;cosmic distance ladder&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="546" LastActivityDate="2013-12-07T23:39:02.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1092" PostTypeId="2" ParentId="675" CreationDate="2013-12-08T06:01:19.010" Score="4" Body="&lt;p&gt;Another thing about dark energy/matter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People have a pretty good idea that dark energy exist because, when you chart the expansion of two objects in the universe over time, from its origin, there is a bell curve. Basically, the speed of the universe's expansion started out faster, than slowed down, and recently (or at least relatively), the expansion has sped back up again. This implies that when the Universe began, it was rapidly expanding, and as this momentum began to flag, gravity began to pull things back together. However, to explain the recent acceleration of expansion, there has to be a force that has increased in strength over time and has been able to cause acceleration in expansion (all of this occurs between galaxies, in areas like solar systems and galaxies, gravity is much to strong to be affected). It is believed that as the universe expands, dark energy increases uniformly to occupy the new space. This means that dark energy increases, thus accelerating expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the universe is expanding because of red shift. Basically, as light travels through the universe, the universe is expanding. Over enough distance, the light will perceptibly lose energy, so that violet light will fall towards red (hence the name red shift). This is because, viewed from our expanding universe, the light appears to be losing energy, when really it is having to travel further as space expands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for dark matter, in short, we know that there is something that has gravity that we cannot detect otherwise. Dark matter, unlike dark energy, is not uniform. It forms strand like formations, which correspond to the strand like formations of galaxies.&lt;/p&gt;&#xA;" OwnerUserId="548" LastActivityDate="2013-12-08T06:01:19.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1093" PostTypeId="1" AcceptedAnswerId="1094" CreationDate="2013-12-08T07:29:52.590" Score="12" ViewCount="3152" Body="&lt;p&gt;How do people measure the distance between the Earth and The Moon? &#xA;What methods were used? Where can I find the official data?&lt;/p&gt;&#xA;" OwnerUserId="549" LastEditorUserId="14021" LastEditDate="2016-09-14T20:28:27.937" LastActivityDate="2017-10-15T17:16:14.817" Title="How do people measure the distance between the Earth and The Moon?" Tags="&lt;the-moon&gt;&lt;earth&gt;&lt;distances&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1094" PostTypeId="2" ParentId="1093" CreationDate="2013-12-08T08:13:06.597" Score="10" Body="&lt;p&gt;There are many ways and I'm not entirely sure who you mean with &lt;em&gt;&quot;how do people measure the distance&quot;&lt;/em&gt; (does this exclude space observatories like e.g. &lt;a href=&quot;https://en.wikipedia.org/wiki/Clementine_%28spacecraft%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Clementine&lt;/a&gt;, probes currently in lunar orbit like e.g. &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_Atmosphere_and_Dust_Environment_Explorer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lunar Atmosphere and Dust Environment Explorer&lt;/a&gt; a.k.a. LADEE, or any other currently available technology?), but one interesting and extremely precise way is by laser ranging, pointing them towards one or more of the retroreflectors that were left on the surface of the Moon by the Apollo landers, in what is known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lunar Laser Ranging Experiment&lt;/a&gt;. Since they are laid on the Moon's near side that &lt;a href=&quot;https://astronomy.stackexchange.com/a/22/13&quot;&gt;is always pointed towards the Earth&lt;/a&gt;, these retroreflectors are available for measurements to any properly equipped researcher, no matter which country they're from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By measuring the time it takes for the light to reflect back, we can infer distance by simply multiplying the &lt;em&gt;c&lt;/em&gt; (&lt;a href=&quot;https://en.wikipedia.org/wiki/Speed_of_light&quot; rel=&quot;nofollow noreferrer&quot;&gt;speed of light in vacuum&lt;/a&gt;) with the time taken between the light signal being transmitted and received and then dividing all of it by two to get a single leg of the distance. I'll let you read the linked to &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page on LLRE&lt;/a&gt; for more information, but if you find that too boring, here's a &lt;a href=&quot;https://www.youtube.com/watch?v=VmVxSFnjYCA&quot; rel=&quot;nofollow noreferrer&quot;&gt;YouTube video on Mythbusters: Moon Hoax Retroreflectors&lt;/a&gt; that explains it from the practical perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for results, again quoting that same Wikipedia page on LLRE:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some of the findings of this long-term experiment are: &lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;The Moon is&#xA;  spiraling away from Earth at a rate of 3.8 cm (about 1.5 inches) per year. This rate&#xA;  has been described as anomalously high.&lt;/li&gt;&#xA;  &lt;li&gt;The Moon probably has a&#xA;  liquid core of about 20% of the Moon's radius.&lt;/li&gt;&#xA;  &lt;li&gt;The universal force&#xA;  of gravity is very stable. The experiments have put an upper limit on&#xA;  the change in Newton's gravitational constant G of less than 1 part in&#xA;  $10^{11}$ since 1969. The likelihood of any &quot;Nordtvedt effect&quot; (a&#xA;  composition-dependent differential acceleration of the Moon and Earth&#xA;  towards the Sun) has been ruled out to high precision, strongly&#xA;  supporting the validity of the Strong Equivalence Principle.&lt;/li&gt;&#xA;  &lt;li&gt;Einstein's theory of gravity (the general theory of relativity)&#xA;  predicts the Moon's orbit to within the accuracy of the laser ranging&#xA;  measurements.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So these distance measurements are so precise they're a lot more interesting and reliable than merely telling us how far away the Moon is. For one, it also blows any Moon landing hoax theories right out of the water. Links to many more official results can be found in &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment#References&quot; rel=&quot;nofollow noreferrer&quot;&gt;references and external links sections of Wiki on LLRE&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a plethora of other ways to measure the distance to the Moon, with different precision and equipment requirements, but I will let others add some other methods in their answers. If however you had some specific &lt;em&gt;&quot;people measuring the distance&quot;&lt;/em&gt; equipped with specific measuring devices in mind, don't forget to mention that in your question. ;)&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="1891" LastEditDate="2017-10-15T17:16:14.817" LastActivityDate="2017-10-15T17:16:14.817" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1095" PostTypeId="1" AcceptedAnswerId="1152" CreationDate="2013-12-08T10:42:52.113" Score="9" ViewCount="946" Body="&lt;p&gt;An earlier question addressed why all planets formed in the same orbital plane, but how is this angle maintained? What prevents the planets from taking on a different orbital plane?&lt;/p&gt;&#xA;" OwnerUserId="551" LastEditorUserId="6" LastEditDate="2013-12-12T14:50:46.077" LastActivityDate="2013-12-13T09:50:36.873" Title="Why do the planets in the Solar system stay in the same orbital plane?" Tags="&lt;solar-system&gt;&lt;planet&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1097" PostTypeId="2" ParentId="1083" CreationDate="2013-12-08T23:27:31.367" Score="5" Body="&lt;p&gt;It's a matter of statistics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scientists take a small amount of the space (let's say 1 &lt;a href=&quot;http://en.wikipedia.org/wiki/Minute_of_arc&quot; rel=&quot;nofollow&quot;&gt;second of arc&lt;/a&gt;). They look at it carefully with strong telescopes, and count all the stars and galaxies they see. Then, they extrapolate that number at the total visible space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course they can compute several spots of the space and make an average count.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the number is extrapolated, that's why it doesn't really matter if it is 100 billion or 300 billion of stars. The goal is to have an order of magnitude as pointed by Moriarty.&lt;/p&gt;&#xA;" OwnerUserId="204" LastEditorUserId="204" LastEditDate="2013-12-09T08:16:20.943" LastActivityDate="2013-12-09T08:16:20.943" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1098" PostTypeId="1" CreationDate="2013-12-09T14:23:08.600" Score="12" ViewCount="132" Body="&lt;p&gt;I was looking at Wikipedia's &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_system#Structure_and_composition&quot;&gt;Solar system page&lt;/a&gt;, and it says that Sun represents 99.86% of the whole solar system mass. I found that pretty huge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So i calculated ratio of masses : Earth / (&lt;a href=&quot;http://en.wikipedia.org/wiki/Earth&quot;&gt;Earth&lt;/a&gt; + &lt;a href=&quot;http://en.wikipedia.org/wiki/Moon&quot;&gt;Moon&lt;/a&gt;) and it's about 98.78%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did the same with Jupiter : Jupiter / (&lt;a href=&quot;http://en.wikipedia.org/wiki/Jupiter&quot;&gt;Jupiter&lt;/a&gt; + &lt;a href=&quot;http://en.wikipedia.org/wiki/Io_%28moon%29&quot;&gt;Io&lt;/a&gt; + &lt;a href=&quot;http://en.wikipedia.org/wiki/Europa_%28moon%29&quot;&gt;Europa&lt;/a&gt; + &lt;a href=&quot;http://en.wikipedia.org/wiki/Ganymede_%28moon%29&quot;&gt;Ganymede&lt;/a&gt; + &lt;a href=&quot;http://en.wikipedia.org/wiki/Callisto_%28moon%29&quot;&gt;Callisto&lt;/a&gt;) and it's about 99.97% (I ignored small satellites).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why is it such a high ratio in this 3 examples?&lt;/li&gt;&#xA;&lt;li&gt;Is 1% a regular number for satellites' mass?&lt;/li&gt;&#xA;&lt;li&gt;It is simple gravitation maths ?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I understand that one body has to be much more massive in order to be a planet-satellite system, otherwise it's a &lt;a href=&quot;http://en.wikipedia.org/wiki/Double_planet&quot;&gt;double planet&lt;/a&gt; system. But I would have thought 90% would be enough.&lt;/p&gt;&#xA;" OwnerUserId="204" LastActivityDate="2013-12-09T14:34:18.253" Title="Is there a pattern between the mass of a body and the mass of orbiting objects around it?" Tags="&lt;the-moon&gt;&lt;solar-system&gt;&lt;earth&gt;&lt;the-sun&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1099" PostTypeId="2" ParentId="1098" CreationDate="2013-12-09T14:34:18.253" Score="7" Body="&lt;p&gt;No, there is no pattern on 1%. There are cases of &lt;a href=&quot;http://en.wikipedia.org/wiki/Double_star&quot;&gt;double stars&lt;/a&gt; and (allegedly) of &lt;a href=&quot;http://en.wikipedia.org/wiki/Double_planets&quot;&gt;double planets&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our Solar System, Charon has a mass of 11.6% of that of Pluto.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-12-09T14:34:18.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1100" PostTypeId="1" AcceptedAnswerId="1104" CreationDate="2013-12-09T18:03:21.887" Score="6" ViewCount="6298" Body="&lt;p&gt;Given that the star is crossing the local meridian line in a certain location, I've tried calculating the altitude of a star by finding the difference between the declination of the star and the latitude of the position from where it is observed (which I believe is not correct). I'm so confused about that and it would be of great help if someone could explain the altitude calculation. Also, I'm wondering how to calculate the position of a place (latitude and longitude) while the declination and the right ascension of a star at the zenith is known.&lt;/p&gt;&#xA;" OwnerUserId="557" LastEditorUserId="557" LastEditDate="2013-12-09T19:48:44.770" LastActivityDate="2013-12-09T20:08:18.470" Title="How to measure the altitude and azimuth of a star?" Tags="&lt;amateur-observing&gt;&lt;observational-astronomy&gt;&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1101" PostTypeId="1" AcceptedAnswerId="1113" CreationDate="2013-12-09T18:34:20.593" Score="15" ViewCount="736" Body="&lt;p&gt;While looking at sunspot information in connection with Amateur Radio, I found that the Sun rotates on a period from 27 days to 31 days.  Its rotation is differential, at the equator it spins at a rate of about 27 days, while at the poles it's closer to 31 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth's moon also rotates at a rate of 27.3 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect this is nothing more than coincidence, but was wondering if there is more than a coincidental connection between these two rates of rotation, similar to how tidal locking forces the moon's rotation and orbit around earth to be the same rate.&lt;/p&gt;&#xA;" OwnerUserId="559" OwnerDisplayName="Adam Davis" LastEditorUserId="6" LastEditDate="2013-12-09T18:56:28.433" LastActivityDate="2014-06-17T20:58:25.053" Title="Is the rotation of the Sun and the rotation/orbit of the Moon around the Earth a coincidence?" Tags="&lt;the-moon&gt;&lt;the-sun&gt;&lt;orbit&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1102" PostTypeId="1" CreationDate="2013-12-09T19:19:34.810" Score="10" ViewCount="273" Body="&lt;p&gt;Miranda, a moon of Uranus is unique in that it has a very fractured surface&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BNsYt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.uoregon.edu/~jimbrau/astr121/Notes/Chapter13.html&quot; rel=&quot;noreferrer&quot;&gt;Source: University of Oregon&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface is said to be jagged and fractured, with comparatively large disjointed cliffs and great faultlines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the current accepted theories for the formation of such a strange world?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Please include authoritive links in any reply&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerDisplayName="user8" LastEditorDisplayName="user8" LastEditDate="2013-12-10T10:36:33.143" LastActivityDate="2013-12-10T10:36:33.143" Title="What are the current accepted theories of the formation of the Uranian moon Miranda?" Tags="&lt;formation&gt;&lt;uranus&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1103" PostTypeId="2" ParentId="1101" CreationDate="2013-12-09T19:58:03.147" Score="4" Body="&lt;p&gt;Pure coincidence. Keep in mind that the Moon is going far from the Earth (and thus slowing its motion), and that Earth is no particularly important planet for the Sun.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-12-09T19:58:03.147" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1104" PostTypeId="2" ParentId="1100" CreationDate="2013-12-09T20:08:18.470" Score="2" Body="&lt;p&gt;For the second question: If you know the coordinates of the Zenith, your latitude is exactly Zenith's declination. For your longitude you can not rely on the Zenith: the same star will be at the Zenith at the same sidereous time for every place in the same latitude, so you need a clock besides the telescope. &lt;em&gt;cf&lt;/em&gt;. &lt;a href=&quot;http://en.wikipedia.org/wiki/History_of_longitude#Problem_of_longitude&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/History_of_longitude#Problem_of_longitude&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first question: The altitude of a star that is crossing the local meridian is also quite easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Northern Hemisphere, a star that is crossing the North local meridian (that is, between the Zenith and the North Horizon) the latitude is&#xA;$90-alt=dec-lat$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Northern Hemisphere, a star that is crossing the South local meridian (that is, between the Zenith and the South Horizon) the latitude is&#xA;$90-alt=lat-dec$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same goes for Southern Hemisphere if you change also North Horizon and South Horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You did not asked specifically for the Azimuth, but again you need a clock for that.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-12-09T20:08:18.470" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1105" PostTypeId="1" AcceptedAnswerId="1109" CreationDate="2013-12-09T21:10:48.710" Score="15" ViewCount="1991" Body="&lt;p&gt;I was wondering if it would be possible to measure red shift from distant galaxies using consumer grade telescope equipment and spectroscopy filters. (&lt;a href=&quot;http://www.rspec-astro.com/star-analyser/&quot;&gt;like this one&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine it would require a tracking mount, and enough aperture to gather data on distant faint objects, but I wonder if it would require more sensitive equipment then a consumer grade tools available for under 5-6 thousand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A long term goal would be to replicate Hubble's work, and directly measure the expansion rate of the universe from backyard equipment.&lt;/p&gt;&#xA;" OwnerUserId="539" LastActivityDate="2014-01-31T00:28:31.207" Title="Is it possible to measure galactic red shift using consumer telescope equipment?" Tags="&lt;spectroscopy&gt;&lt;redshift&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1106" PostTypeId="2" ParentId="1089" CreationDate="2013-12-09T22:52:27.190" Score="5" Body="&lt;p&gt;It's not really known why only Titan has a thick atmosphere while moons like Ganymede don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thought is that temperatures may have been too high (well above ~40K) in the Jovian subnebula of our solar system due to the greater gravitational potential energy release, mass, and proximity to the Sun, greatly reducing the $NH_{3}$-hydrate inventory accreted by both Callisto and Ganymede. The resulting $N_{2}$ atmospheres might have been too thin to survive the atmospheric erosion effects that Titan has withstood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: some links:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://perso.utinam.cnrs.fr/~mousis/papier3.pdf&quot;&gt;http://perso.utinam.cnrs.fr/~mousis/papier3.pdf&lt;/a&gt; (though it's from 2002).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2012ttt..work...25N&quot;&gt;http://adsabs.harvard.edu/abs/2012ttt..work...25N&lt;/a&gt; (this paper critiques the previous idea of a $CH_{4}$ supported atmosphere)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also think the NASA/JPL Cassini pages would have any up-to-date info in regards to Titan.&lt;/p&gt;&#xA;" OwnerUserId="555" LastEditorUserId="555" LastEditDate="2013-12-09T23:19:14.027" LastActivityDate="2013-12-09T23:19:14.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1107" PostTypeId="1" AcceptedAnswerId="1111" CreationDate="2013-12-10T01:16:49.263" Score="5" ViewCount="317" Body="&lt;p&gt;Suppose there are two observational frames of reference with origins $O$ and $O'$, respectively, separated by some constant distance. A body located at point $P$ has Cartesian coordinates $\left(x,y,z\right)$ and $\left(x',y',z',\right)$ in $O$ and $O'$, respectively; and similarly, the spherical coordinates of $P$ in the unprimed and primed frames are $\left(r,\theta,\phi\right)$ and $\left(r',\theta',\phi'\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say $O$ determines the Cartesian coordinates of $O'$ to be $\left(a,b,c\right)$. Then the Cartesian coordinates of $P$ in $O$ are simply related to the primed Cartesian coordinates: $$\left(x,y,z\right)=\left(x'+a,y'+b,z'+c\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question, then, is how does one compute the affect of translation on spherical coordinates. That is, given $\left(a,b,c\right)$, how can one write $\left(r,\theta,\phi\right)$ as a function of $\left(r',\theta',\phi'\right)?$&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="13" LastEditDate="2013-12-10T06:05:36.957" LastActivityDate="2013-12-10T10:04:01.490" Title="Coordinate transformations between reference frames in spherical astronomy" Tags="&lt;fundamental-astronomy&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1108" PostTypeId="2" ParentId="1102" CreationDate="2013-12-10T01:19:49.123" Score="7" Body="&lt;p&gt;As Miranda is a rather small object (only 235km diameter) it may simply have not enough mass to make it matter settle into hydrostatic equilibrium under the influence of its gravity, thus it basically retains the irregular shape most asteroids have. This nice arxiv paper discusses the mass limits necessary for a stellar body to settle into a nice round shape:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/ftp/arxiv/papers/1004/1004.1091.pdf&quot;&gt;http://arxiv.org/ftp/arxiv/papers/1004/1004.1091.pdf&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="412" LastActivityDate="2013-12-10T01:19:49.123" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1109" PostTypeId="2" ParentId="1105" CreationDate="2013-12-10T01:40:05.550" Score="9" Body="&lt;p&gt;Looks like you can measure the cosmological redshift of quasars using that equipment and an 14&quot; reflector:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.rspec-astro.com/sample-projects/&quot;&gt;http://www.rspec-astro.com/sample-projects/&lt;/a&gt;&#xA;(halfway down the page)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer appears to be yes, you can do it, assuming that page isn't a total fabrication. Seems plausible to me: the redshift of that quasar is sizable at 0.15 (though that's small for a quasar), which means that the spectral lines will be shifted by something like 15%. So you  don't need to have a very fine diffraction grating to see the shift. And &lt;a href=&quot;https://en.wikipedia.org/wiki/3C_273&quot;&gt;that quasar&lt;/a&gt; is pretty bright for a quasar, at magnitude 13, so you can easily spot it with a good scope. Sounds like a fun project!&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditDate="2013-12-10T01:47:45.900" LastActivityDate="2013-12-10T01:47:45.900" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1110" PostTypeId="2" ParentId="1084" CreationDate="2013-12-10T04:42:28.697" Score="3" Body="&lt;p&gt;One method involves finding a &lt;a href=&quot;http://outreach.atnf.csiro.au/education/senior/astrophysics/variable_cepheids.html&quot; rel=&quot;nofollow&quot;&gt;cepheid variable&lt;/a&gt; in the vicinity of the star, or even in the galaxy you are investigating. These stars vary in brightness with a period closely related to their absolute magnitude. By measuring the period, you can know how bright it would appear at a distance of, say, one light year. Using a photometer, you can measure it's apparent magnitude, and figure out how far away it must be for you to see it as you do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the referenced web page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 1924 Edwin Hubble detected Cepheids in the Andromeda nebula, M31&#xA;  and the Triangulum nebula M33. Using these he determined that their&#xA;  distances were 900,000 and 850,000 light years respectively. He thus&#xA;  established conclusively that these &quot;spiral nebulae&quot; were in fact&#xA;  other galaxies and not part of our Milky Way. This was a momentous&#xA;  discovery and dramatically expanded the scale of he known Universe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For another Method, imagine this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look carefully at Sirius with your naked eye tonight. Tomorrow, take a piece of brass with small holes a varying size, and see which one, when held up to the sun, seems as bright as Sirius. Knowing the area of that hole, knowing what fraction of the area of the sun's disk you could see through that hole, &lt;em&gt;and assuming the sun is a star just like Sirius&lt;/em&gt;, you now know how bright the sun would be if it were as far away as Sirius, and you can make an educated guess how far Sirius is from the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's around 1650, you're Christiaan Huygens, and you assume all stars are the same, then you've made a massive contribution to our understanding of the universe. However, since the Sirius puts out something like 25 times as much light as the sun, Huygens answer was only some 30,000AU, an epic underestimate. Still, the first scientific evidence hinting at the true size of our universe.&lt;/p&gt;&#xA;" OwnerUserId="562" LastEditorUserId="562" LastEditDate="2013-12-10T05:09:24.210" LastActivityDate="2013-12-10T05:09:24.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1111" PostTypeId="2" ParentId="1107" CreationDate="2013-12-10T10:04:01.490" Score="3" Body="&lt;p&gt;There is kind of an answer &lt;a href=&quot;https://math.stackexchange.com/questions/92301/transforming-from-one-spherical-coordinate-system-to-another&quot;&gt;over at Math&lt;/a&gt;. All you can do in spherical coordinates is to change the position of your &quot;pole&quot;, i.e. you have $(1,0,0)$ in your first coordinate system, which is mapped to some $(r', \vartheta', \varphi')$ in the second coordinate system. The two angles represent a rotation, and the $r$ represents a scaling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think since we still deal with a vector space (for the angular part), you can simply add the origin of coordinate system two to vectors of the first coordinate system. The radius is a scaling, and needs to be multiplied. Hence for a point $p=(r, \vartheta, \varphi)$ in coordinate system 1 you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;p'=(r \cdot r', \vartheta+\vartheta', \varphi+\varphi')&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="50" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:35.350" LastActivityDate="2013-12-10T10:04:01.490" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1112" PostTypeId="2" ParentId="1037" CreationDate="2013-12-10T14:29:37.677" Score="5" Body="&lt;p&gt;Avi Loeb and Edwin Turner have written a paper about the possibility of detecting a civilization similar to our own on another world by looking for city lights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their proposed method suggests observing the dark sides of planets when they transit in front of their parent star. Though they said that this method will require future generations of telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.cfa.harvard.edu/news/2011-30&quot; rel=&quot;noreferrer&quot;&gt;City Lights Could Reveal E.T. Civilization - Harvard-Smithsonian Center for Astrophysics&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6" LastActivityDate="2013-12-10T14:29:37.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1113" PostTypeId="2" ParentId="1101" CreationDate="2013-12-10T14:43:43.743" Score="13" Body="&lt;p&gt;This is purely coincidental. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why?&lt;/strong&gt; Because the moon used to rotate faster and is continually &lt;a href=&quot;http://curious.astro.cornell.edu/question.php?number=124&quot;&gt;slowing down&lt;/a&gt;. Long story short, you just happen to be at a point in history where they two numbers roughly coincide. A few million years ago it wasn't true, and a few million years from now it wont be either. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And just to cover all my bases, there's nothing to suggest that the slowing orbit of the moon can have any reciprocal effect on the sun's rotational period.&lt;/p&gt;&#xA;" OwnerUserId="58" LastActivityDate="2013-12-10T14:43:43.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1114" PostTypeId="1" AcceptedAnswerId="1119" CreationDate="2013-12-10T15:22:18.090" Score="14" ViewCount="904" Body="&lt;p&gt;Many sources indicate that there is &lt;a href=&quot;http://en.wikipedia.org/wiki/Martian_polar_ice_caps&quot; rel=&quot;nofollow noreferrer&quot;&gt;ice water&lt;/a&gt; on &lt;a href=&quot;http://en.wikipedia.org/wiki/Mars&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mars&lt;/a&gt; today, but it is mainly &lt;a href=&quot;http://www.esa.int/Our_Activities/Space_Science/Mars_Express/Water_ice_in_crater_at_Martian_north_pole&quot; rel=&quot;nofollow noreferrer&quot;&gt;traces&lt;/a&gt;, here and there, but not that much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, solid hypothesis state that there were oceans on Mars, see Wikipedia's &lt;a href=&quot;http://en.wikipedia.org/wiki/Mars_ocean_hypothesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mars ocean hypothesis&lt;/a&gt; or at least &lt;a href=&quot;http://www.esa.int/Our_Activities/Space_Science/Results_from_Mars_Express_and_Huygens/Mars_Express_evidence_for_large_aquifers_on_early_Mars&quot; rel=&quot;nofollow noreferrer&quot;&gt;large aquifers on early Mars&lt;/a&gt;. They all concern early Mars so around 3.5 to 4 billions years ago. These theories suggest that Mars oceans could have formed Mesa landscapes (just like&lt;a href=&quot;http://en.wikipedia.org/wiki/Grand_Canyon&quot; rel=&quot;nofollow noreferrer&quot;&gt;Grand Canyon&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Monument_Valley&quot; rel=&quot;nofollow noreferrer&quot;&gt;Monument Valley&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is simple, if all these hypotheses are true, where did that water go?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible that so much water could have been evacuated from the planet? I guess it would have needed A LOT of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it may have been transformed on the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible that the water has oxidized iron, creating Iron oxide (ie rust) and that it is the main reason for Mars red color?&#xA;The Mars ocean hypothesis page suggests some leads about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mars_ocean_hypothesis#Fate_of_the_ocean&quot; rel=&quot;nofollow noreferrer&quot;&gt;fate of that ocean&lt;/a&gt;, but none suggests the oxidation of iron consuming the entire oceans. Moreover, oxidation would imply the presence of water and dioxygen...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PKeNw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PKeNw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;http://en.yibada.com/articles/18182/20150309/red-planet-once-held-large-shallow-ocean-mean-positive-signs.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.yibada.com/articles/18182/20150309/red-planet-once-held-large-shallow-ocean-mean-positive-signs.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="204" LastEditorUserId="9527" LastEditDate="2020-08-04T09:35:46.320" LastActivityDate="2020-08-04T09:35:46.320" Title="Where did Mars' water go?" Tags="&lt;mars&gt;&lt;water&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="1115" PostTypeId="2" ParentId="907" CreationDate="2013-12-10T16:37:38.967" Score="4" Body="&lt;p&gt;I have not the qualification to answer the question in its whole but the question is interesting (I worked on Be Stars which are episodically surrounded by an decretion disk and which rotates at nearly critical velocities. The phenomenon in Be stars is different from accreting stars. The only consequences of subcritical velocity is a flattened envelope and modification of its inner structure and of the oscillations modes found in these stars&#xA;(if you have time and curiosity, a good example of flattened star with Keplerian rotating decretion disk is Achernar, a Be star observed using interferometry --&gt; Have a look at Meilland et al. 2007: www.aanda.org/articles/aa/pdf/2007/10/aa4848-06.pdf)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this paper about critically-rotating accretors. May be you will find answers to your questions here or in its references (use nasa ads site for your query: &lt;a href=&quot;http://adsabs.harvard.edu/&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/&lt;/a&gt;).&#xA;http: //arxiv.org/pdf/1306.1348v2.pdf&#xA;It seems that in the introduction, there is some answers to your questions about reaching the critical velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The accreted mass can increase the rotation rate until the star reaches the critical velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is said : &quot;For a typical 6 + 3.6 M⊙ system, with initial period&#xA;Pinit = 2.5 days, in the absence of spin-down mechanisms, only&#xA;3 per cent (0.12 M⊙) of the total amount of matter transferred&#xA;by RLOF (more than 5 M⊙) is enough to spin the gainer up&#xA;to the critical rotation.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But&lt;/strong&gt; we still don't know if the gainer can really reach the critical velocity. Some papers are dealing with break-up mechanims which does not allow the gainer to reach critical velocity: spin down by tides, magnetic breaking, limitation of the accretion angular momentum through interaction with the accretion disk, stopping of the accretion mechanism...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sure that you will find many papers on nasa ads that will give you answers to your questions.&lt;/p&gt;&#xA;" OwnerUserId="567" LastActivityDate="2013-12-10T16:37:38.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1116" PostTypeId="1" AcceptedAnswerId="1122" CreationDate="2013-12-10T16:38:05.943" Score="7" ViewCount="166" Body="&lt;p&gt;I'm sure I'm just doing something dumb, but: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Yildun has a declination of approximately 86.5. This puts it 3.5 &#xA;degrees from the celestial north pole. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Per the snapshot below, Yildun's azimuth can be as high as 4.17 &#xA;degrees. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since the celestial north pole always has an azimuth of 0, &#xA;wouldn't this put Yildun 4.17 degrees from the pole, contradicting &#xA;the distance calculated from declination? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What am I missing? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e3aj6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I thought I'd figured this out:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;On Earth, one degree of latitude is not the same as one degree of longitude (except at the equator). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Analogously, in the sky, one declination degree is not the same as one right ascension degree. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I thought the situation for right ascension applied to azimuth, but it doesn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both azimuth and declination are measured on &quot;great circles&quot; (declination is measured on a half circle, but same general idea), so degrees in azimuth should equal degrees in declination.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2013-12-10T20:33:25.453" LastActivityDate="2013-12-11T09:55:37.303" Title="How can absolute value of azimuth exceed distance from pole?" Tags="&lt;azimuth&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1117" PostTypeId="1" AcceptedAnswerId="1118" CreationDate="2013-12-10T19:38:52.020" Score="19" ViewCount="3682" Body="&lt;p&gt;If I have the right numbers, it seems to me that even Hubble telescope might barely be able to make out a carcass of a blue whale on the surface of the moon, which puts objects as small as the lunar rovers or the American flag left there by Apollo out of range. Hubble is often lauded as the best telescope we have built. I know a large part of that is because it is in space where it's free from all of the interference of the atmosphere, but is there any ground-based observatory with better resolution? Is there any place on Earth where I could point a scope at the Sea of Tranquility and see what we left there over 40 years ago?&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2013-12-19T22:11:25.663" Title="Is there any telescope on Earth that can see the lunar rovers on the moon?" Tags="&lt;the-moon&gt;&lt;telescope&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1118" PostTypeId="2" ParentId="1117" CreationDate="2013-12-10T19:50:31.833" Score="12" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The largest optical wavelength telescope that we have now is the Keck&#xA;  Telscope in Hawaii which is 10 meters in diameter. The Hubble Space&#xA;  Telescope is only 2.4 meters in diameter.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Resolving the larger lunar rover (which has a length of 3.1 meters)&#xA;  would require a telescope 75 meters in diameter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Information extracted from &lt;a href=&quot;http://curious.astro.cornell.edu/question.php?number=134&quot; rel=&quot;nofollow&quot;&gt;The Curious Team answers: Are there telescopes that can see the flag and lunar rover on the Moon?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: In a comment to my answer, @Envite mentions &lt;a href=&quot;http://en.wikipedia.org/wiki/Astronomical_interferometer&quot; rel=&quot;nofollow&quot;&gt;Astronomical Interferometers&lt;/a&gt; which are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...an array of telescopes or mirror segments acting together to probe&#xA;  structures with higher resolution by means of interferometry. The&#xA;  benefit of the interferometer is that the angular resolution of the&#xA;  instrument is nearly that of a telescope with the same aperture as a&#xA;  single large instrument encompassing all of the individual&#xA;  photon-collecting sub-components. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It also must be noted that probably this wouldn't be the best idea for watching the moon rovers since:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The drawback is that it does not&#xA;  collect as many photons as a large instrument of that size. Thus it is&#xA;  mainly useful for fine resolution of the more luminous astronomical&#xA;  objects, such as close binary stars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="537" LastEditorUserId="537" LastEditDate="2013-12-19T22:11:25.663" LastActivityDate="2013-12-19T22:11:25.663" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1119" PostTypeId="2" ParentId="1114" CreationDate="2013-12-10T20:02:59.793" Score="9" Body="&lt;p&gt;Some of the water evaporated, some of it froze and another good part of it went into the surface and is now subterranean. For more info read &lt;a href=&quot;http://www.dailygalaxy.com/my_weblog/2011/08/-mars-vanished-water-atmosphere-where-did-it-go.html&quot; rel=&quot;noreferrer&quot;&gt;Mars' Vanished Water &amp;amp; Atmosphere --Where Did It Go?&lt;/a&gt; and &lt;a href=&quot;http://www.universetoday.com/106297/mavens-quest-where-did-mars-water-go/&quot; rel=&quot;noreferrer&quot;&gt;MAVEN’s Quest – Where Did Mars Water Go?&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's much still to learn and experiments in the future will help us to better understand the evolution of Mars' atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="537" LastEditorUserId="22872" LastEditDate="2018-05-01T12:17:18.497" LastActivityDate="2018-05-01T12:17:18.497" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1120" PostTypeId="1" AcceptedAnswerId="10903" CreationDate="2013-12-10T20:43:13.060" Score="5" ViewCount="716" Body="&lt;p&gt;I've visited several &quot;how sundials work&quot; sites and can't seem to get a &#xA;clear answer to this: is &quot;sundial time&quot; just a linear function of &#xA;solar azimuth? More specifically: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;When the sun is due south (northern hemisphere), it is sundial &#xA;noon. All sites I've visited agree on this. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When the sun is due west (azimuth 270 degrees), I say the sundial &#xA;time is 6pm, a quarter turn/day from noon. However, I can't find a &#xA;site that actually says this, and some sites seem to disagree with &#xA;this. Same for it being 6am sundial time when the sun is due east. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I know there are different types of sundials, but had always assumed &#xA;they would give the same sundial time. Is that not true? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a note, the sundial I describe above can be fairly inaccurate at &#xA;times, which makes me question whether I'm correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT (to clarify question): Forgetting entirely about clock time for a &#xA;moment, suppose I build a sundial. When the sun is due west, my &#xA;sundial reads 6pm. When the sun is due east, my sundial reads &#xA;6am. When the sun is due south, my sundial reads noon. My question: &#xA;have I built my sundial correctly? The type of answer I'm hoping to &#xA;get: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;No. When you build a sundial, it should read (something else) when &#xA;the sun is due west. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are many different types of sundials. Depending on type, &#xA;your sundial may or may not read 6pm when the sun is due west. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It depends on your latitude: sundials are &#xA;latitude-specific. There's no such thing as a global sundial. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes, your understanding of a sundial is correct. Sundials can be &#xA;off by as much as (some number/formula) from mean solar time, &#xA;depending on your latitude. They can be as much as (some &#xA;number/formula) from clock time, depending on your longitude and &#xA;time zone.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user21" LastEditorUserId="6980" LastEditDate="2015-06-02T19:50:00.810" LastActivityDate="2016-08-26T05:26:29.100" Title="Is sundial time entirely dependent on solar azimuth?" Tags="&lt;time&gt;&lt;azimuth&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1121" PostTypeId="2" ParentId="1075" CreationDate="2013-12-10T21:16:06.163" Score="3" Body="&lt;p&gt;If you want to trigger outflows, there are two main processes that can help:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;magnetic field;&lt;/li&gt;&#xA;&lt;li&gt;radiation pressure.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Magnetic fields&lt;/strong&gt; is well-suited to launch such outflows, through magneto-centrifugal effects. Basically, if there is magnetic fields tighed to the disk surrounding the protostar and the angle between the field lines and the disk is right, a particle from the disk that follows the magnetic field lines can be ejected. This is the well-known &lt;a href=&quot;http://adsabs.harvard.edu/abs/1982MNRAS.199..883B&quot; rel=&quot;nofollow&quot;&gt;Blandford &amp;amp; Payne (1982)&lt;/a&gt; mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, at earlier stages of star formation, magnetic field could also trigger outflows by a &quot;magnetic field tower&quot; growth. Basically, during the collapse of the prestellar core into a protostar, the field lines are dragged and twisted (due to the rotation of the core), the twist propagating from the inner to the outer region of the core, that triggers outflows. This kind of thing is discussedd in &lt;a href=&quot;http://adsabs.harvard.edu/abs/2008A%26A...477....9H&quot; rel=&quot;nofollow&quot;&gt;Hennebelle &amp;amp; Fromang (2008)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, these two mechanisms are complementary, because slow outflows (of the order of 10 km s$^{-1}$) are observed around young protostar, whereas &quot;fast outflows&quot; -- jets, of the order of 100 km s$^{-1}$ -- are observed around late protostars, and these two mechanisms explained well the former and the later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more massive stars, &lt;strong&gt;radiation pressure&lt;/strong&gt; will plays a role in outflows triggering. The protostar radiates away a part of its accreting energy, and strong a radiation pressure can trigger the growth of a kind of &quot;bubble&quot;, that is an outflow (you can see &lt;a href=&quot;http://adsabs.harvard.edu/abs/2012arXiv1209.1786H&quot; rel=&quot;nofollow&quot;&gt;Hennebelle &amp;amp; Commerçon (2012)&lt;/a&gt; for a brief review of the question).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your comment on your own question, it is mostly radiation pressure that will clear up the gas around the new-born star, that is clearly not an outflow or anything alike.&lt;/p&gt;&#xA;" OwnerUserId="88" LastActivityDate="2013-12-10T21:16:06.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1122" PostTypeId="2" ParentId="1116" CreationDate="2013-12-10T21:47:22.243" Score="2" Body="&lt;p&gt;It turns out I was right when I thought I was wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the azimuth circle for 85 degrees elevation. It's very small. For example, 85 degrees elevation and 0 azimuth is only 10 degrees away from 85 degrees elevation and 180 azimuth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the azimuth circle for a given elevation is smaller than a great circle. It's only cosine(elevation) the size of a great circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 35 degrees, the azimuth circle is cosine(35 degrees) = 0.819152 the size of a great circle, so 4.17 degrees in azimuth = 4.17 degrees * 0.819152 = right around 3.5 degrees on a great circle.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2013-12-10T21:47:22.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1123" PostTypeId="1" AcceptedAnswerId="1129" CreationDate="2013-12-10T21:49:54.383" Score="13" ViewCount="93" Body="&lt;p&gt;What's the definitive source for names of geographic objects on the moon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reason: As a computer graphics project I've rendered a scene based on LROC terrain data. I want to know what I'm looking at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most detailed atlas I can find online is this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://planetarynames.wr.usgs.gov/Page/Moon1to1MAtlas&quot;&gt;http://planetarynames.wr.usgs.gov/Page/Moon1to1MAtlas&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...but it's not brilliant and there are lots of small but prominent craters and ridges which aren't labelled. It may be the some of these simply don't have names, but it's also plausible that there are simply too many to list in that atlas. How can I find out for sure?&lt;/p&gt;&#xA;" OwnerUserId="572" LastActivityDate="2013-12-11T14:12:04.293" Title="Official registry of lunar placenames?" Tags="&lt;the-moon&gt;&lt;naming&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1124" PostTypeId="1" CreationDate="2013-12-10T22:54:11.213" Score="6" ViewCount="202" Body="&lt;p&gt;I would have assumed that the brightest radio source in the Universe is a quasar of some kind (perhaps &lt;a href=&quot;http://en.wikipedia.org/wiki/3C_273&quot; rel=&quot;nofollow&quot;&gt;3C 273&lt;/a&gt;) given that the average pulsar has luminosity of $~10^{40} \text{Watts}$ and this is the brightest (optically) and is radio loud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My professor seems to be of the opinion that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Arecibo_message&quot; rel=&quot;nofollow&quot;&gt;Arecibo message&lt;/a&gt; is louder, but that does not seem to pass the straight face test, given that it was only transmitted at a power of only $1\text{MW}$, and only a very small fraction of the pulsars output (a mere $1/10^{34}$th) would be sufficient to outshine the message.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is one of these the brightest radio source in the universe (excluding the Big Bang if it matters), or is it something else entirely?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, is Watts even the right unit to be measuring this in?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; Here is the text of the question (it was a true/false, and the answer is true):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1975, Frank Drake and Carl Sagan sent a message in to the Universe, directed to extraterrestrials, from the radio telescope at Arecibo in Puerto Rico that, for a brief time, made the Earth the brightest radio object in the Universe.&lt;/p&gt;&#xA;" OwnerUserId="383" LastEditorUserId="383" LastEditDate="2013-12-12T04:50:50.657" LastActivityDate="2013-12-12T04:50:50.657" Title="Brightest Radio Source In the Universe" Tags="&lt;radio-astronomy&gt;&lt;pulsar&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1126" PostTypeId="1" AcceptedAnswerId="1127" CreationDate="2013-12-11T10:55:03.567" Score="11" ViewCount="1217" Body="&lt;p&gt;It is known as a fact that Earth's rotation is slowing every year, that's why we sometimes have 61 seconds long minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Java documentation of Time : &lt;a href=&quot;http://docs.oracle.com/javase/7/docs/api/java/util/Date.html&quot; rel=&quot;noreferrer&quot;&gt;http://docs.oracle.com/javase/7/docs/api/java/util/Date.html&lt;/a&gt; , it is said :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some computer standards are defined in terms of Greenwich mean time&#xA;  (GMT), which is equivalent to universal time (UT). GMT is the &quot;civil&quot;&#xA;  name for the standard; UT is the &quot;scientific&quot; name for the same&#xA;  standard. The distinction between UTC and UT is that UTC is based on&#xA;  an atomic clock and UT is based on astronomical observations, which&#xA;  for all practical purposes is an invisibly fine hair to split. Because&#xA;  &lt;strong&gt;the earth's rotation is not uniform (it slows down and speeds up in&#xA;  complicated ways)&lt;/strong&gt;, UT does not always flow uniformly. Leap seconds are&#xA;  introduced as needed into UTC so as to keep UTC within 0.9 seconds of&#xA;  UT1, which is a version of UT with certain corrections applied. There&#xA;  are other time and date systems as well; for example, the time scale&#xA;  used by the satellite-based global positioning system (GPS) is&#xA;  synchronized to UTC but is not adjusted for leap seconds.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why is that so? How can this occur?&lt;/p&gt;&#xA;" OwnerUserId="579" LastEditorUserId="204" LastEditDate="2013-12-12T09:10:27.350" LastActivityDate="2013-12-12T09:10:27.350" Title="How can the Earth's rotation speed up?" Tags="&lt;earth&gt;&lt;rotation&gt;" AnswerCount="3" CommentCount="2" ClosedDate="2013-12-12T14:55:11.453" ContentLicense="CC BY-SA 3.0" />
  <row Id="1127" PostTypeId="2" ParentId="1126" CreationDate="2013-12-11T11:05:51.050" Score="3" Body="&lt;p&gt;Any body with an irregular mass distribution will experience those wobblings. And more if it is not just irregular but fluid.&lt;/p&gt;&#xA;" OwnerUserId="473" LastEditorUserId="473" LastEditDate="2013-12-11T11:19:09.450" LastActivityDate="2013-12-11T11:19:09.450" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1128" PostTypeId="2" ParentId="1123" CreationDate="2013-12-11T11:22:12.367" Score="6" Body="&lt;p&gt;&lt;a href=&quot;http://planetarynames.wr.usgs.gov/Page/MOON/target&quot;&gt;http://planetarynames.wr.usgs.gov/Page/MOON/target&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Official list endorsed by IAU&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-12-11T11:22:12.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1129" PostTypeId="2" ParentId="1123" CreationDate="2013-12-11T11:29:50.877" Score="8" Body="&lt;p&gt;The &lt;a href=&quot;http://planetarynames.wr.usgs.gov/Page/Moon1to1MAtlas&quot;&gt;resource you cite&lt;/a&gt;, in addition to this &lt;a href=&quot;http://planetarynames.wr.usgs.gov/SearchResults?target=MOON&amp;amp;featureType=Crater,%20craters&quot;&gt;full list of crater's names&lt;/a&gt; are the official, definitive lists of named craters. Note that you have the coordinates of each crater in the complete list; I guess you could reuse these informations for your project. As for the atlas, it is specified in the &quot;introduction&quot; of the atlas that the list is up-to-date with the approved nomenclature of lunar craters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The IAU (the International Astronomical Union) is the official organisation in charge of naming celestrial objects et al. You can rely on it.&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2013-12-11T14:12:04.293" LastActivityDate="2013-12-11T14:12:04.293" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1130" PostTypeId="2" ParentId="1126" CreationDate="2013-12-11T11:32:26.840" Score="2" Body="&lt;p&gt;To complete Envite's answer, I would point that you can easily experience that irregularity yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a fresh egg. Rotate it on a table. Stop it briefly and release it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The egg will continue to rotate irregularly because of the fluids moving inside. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the egg is not a mathematic ball, and the fluids inside have several phases : yellow part, white part and gaz, the forces inside are complex, so the rotation is irregular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover I think the Moon is not helping with the regularity of fluids moving (cf sea tides).&lt;/p&gt;&#xA;" OwnerUserId="204" LastActivityDate="2013-12-11T11:32:26.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1131" PostTypeId="2" ParentId="1126" CreationDate="2013-12-11T14:50:20.377" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Earth%27s_rotation#Changes_in_rotation&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's article on Earth's Rotation&lt;/a&gt; proposes an example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;By affecting Earths moment of inertia, some large scale events, such&#xA;  as the 2004 Indian Ocean earthquake, have caused the rotation to speed&#xA;  up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another one is seen in &lt;a href=&quot;http://io9.com/5884476/shifting-ocean-currents-can-actually-speed-up-earths-rotation&quot; rel=&quot;nofollow&quot;&gt;Shifting ocean currents can (and do) actually speed up Earth's rotation&lt;/a&gt; where it goes on to say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Back in November 2009, something strange happened the Southern Ocean.&#xA;  The Antarctic Circumpolar Current suddenly slowed down. The change in&#xA;  current was extreme enough to throw off the planet's angular momentum,&#xA;  and so to compensate, the Earth sped up its rotation. For about two&#xA;  weeks, the days were a tenth of a millisecond shorter than they should&#xA;  be. Then, with just as little warning, the current returned to its&#xA;  normal speed, and the Earth slowed back down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is understandable since currents affect the amount of water on any given place and thus the distribution of mass on earth. &lt;/p&gt;&#xA;" OwnerUserId="537" LastActivityDate="2013-12-11T14:50:20.377" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1132" PostTypeId="2" ParentId="898" CreationDate="2013-12-11T15:43:45.263" Score="6" Body="&lt;h2&gt;What is it?&lt;/h2&gt;&#xA;&lt;p&gt;An IMF, &lt;span class=&quot;math-container&quot;&gt;$\Phi(m)$&lt;/span&gt;, is defined such as &lt;span class=&quot;math-container&quot;&gt;$\Phi(m){\rm d}m$&lt;/span&gt; gives the fractions of stars with a mass between &lt;span class=&quot;math-container&quot;&gt;$m - {\rm d}m/2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$m + {\rm d}m/2$&lt;/span&gt;, and with a normalized distribution&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\int_{m_{\rm min}}^{m_{\rm max}}m\Phi(m){\rm d}m = 1\ M_{\odot}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Note that these boundaries (&lt;span class=&quot;math-container&quot;&gt;$m_{\rm min}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$m_{\rm max}$&lt;/span&gt;) are ill-defined, but typically of the order of 0.1 &lt;span class=&quot;math-container&quot;&gt;$M_{\odot}$&lt;/span&gt; and 100 &lt;span class=&quot;math-container&quot;&gt;$M_{\odot}$&lt;/span&gt;, respectively.&lt;/p&gt;&#xA;&lt;h2&gt;IMFs&lt;/h2&gt;&#xA;&lt;p&gt;The various IMF used are the following, with their main characteristics:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the &lt;a href=&quot;http://adsabs.harvard.edu/abs/1955ApJ...121..161S&quot; rel=&quot;noreferrer&quot;&gt;Salpeter's IMF&lt;/a&gt;, that is a parametrization of the IMF by a simple power-law, of the form&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\Phi(m){\rm d}m \propto m^{-\alpha}{\rm d}m;$$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;the &lt;a href=&quot;http://adsabs.harvard.edu/abs/1979ApJS...41..513M&quot; rel=&quot;noreferrer&quot;&gt;Miller &amp;amp; Scalo's IMF&lt;/a&gt;, that is a parametrization of the IMF by a log-normal distribution of the form&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\xi\left(\log(m)\right) = A_0 + A_1 \log(m) + A_2\left(\log(m)\right)^2;$$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;the &lt;a href=&quot;http://adsabs.harvard.edu/abs/2001MNRAS.322..231K&quot; rel=&quot;noreferrer&quot;&gt;Kroupa's IMF&lt;/a&gt;, that is a parametrization of the IMF by a broken power-law;&lt;/li&gt;&#xA;&lt;li&gt;the &lt;a href=&quot;http://adsabs.harvard.edu/abs/2003PASP..115..763C&quot; rel=&quot;noreferrer&quot;&gt;Chabrier's IMF&lt;/a&gt; and &lt;a href=&quot;http://adsabs.harvard.edu/abs/2005ASSL..327...41C&quot; rel=&quot;noreferrer&quot;&gt;Chabrier's system IMF&lt;/a&gt;, that is a combination of log-normal distribution (for low mass stars with masses less than 1 &lt;span class=&quot;math-container&quot;&gt;$M_{\odot}$&lt;/span&gt;) and and a power-law distribution (for larger masses). The difference between the IMF and the system IMF is to merge resolved objects into multiple systems to compute the magnitude of systems instead of individual stars.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h2&gt;Determination&lt;/h2&gt;&#xA;&lt;p&gt;As you see, all these IMF are parametrization, deduced from observations. In general, the observations used to infer these mass functions come from star clusters in our galaxy. All you need to do it is to find a mass-magnitude relationship to deduce, from an observed luminosity, a mass function. In general, the number density distribution per mess interval, &lt;span class=&quot;math-container&quot;&gt;${\rm d}n/{\rm d}m$&lt;/span&gt;, takes the following form&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{{\rm d}n}{{\rm d}m}\left(m\right)_\tau = \left(\frac{{\rm d}n}{{\rm d}M_\lambda(m)}\right) \times \left(\frac{{\rm d}m}{{\rm d}M_\lambda(m)}\right)^{-1}_\tau,$$&lt;/span&gt;&#xA;for a given age &lt;span class=&quot;math-container&quot;&gt;$\tau$&lt;/span&gt; and an observed magnitude &lt;span class=&quot;math-container&quot;&gt;$M_\lambda$&lt;/span&gt;. Then, it is just a matter of parametrization, but also of how well it can arise from a proper theory.&lt;/p&gt;&#xA;&lt;p&gt;For this matter, the Chabrier's IMF is probably the one that is best back up by theoretical arguments. It relies on a gravo-turbulent theory, taking into account all the possible supports (thermal support, turbulent support and magnetic support) plus the dual nature of turbulence, that both favors star formation by compressing the gas, and impedes star formation, by dispersing the fluid. All the dirty details are given in &lt;a href=&quot;http://adsabs.harvard.edu/abs/2008ApJ...684..395H&quot; rel=&quot;noreferrer&quot;&gt;Hennebelle &amp;amp; Chabrier (2008)&lt;/a&gt; and &lt;a href=&quot;http://adsabs.harvard.edu/abs/2009ApJ...702.1428H&quot; rel=&quot;noreferrer&quot;&gt;Hennebelle &amp;amp; Chabrier (2009)&lt;/a&gt;, showing how you can analytically deduce an IMF from these theoretical considerations.&lt;/p&gt;&#xA;&lt;h2&gt;Applications&lt;/h2&gt;&#xA;&lt;p&gt;As far as I know, these IMF are more or less used for every type of population. However, you won't favor Salpeter's IMF if you have enough resolution to resolve low-mass objects, that are not at all well-taken into account with this IMF. You should also favor the Chabrier's &lt;em&gt;system&lt;/em&gt; IMF in case of unresolved objects.&lt;/p&gt;&#xA;&lt;p&gt;To know if all these IMFs are really well-suited to any kind of population is an open and difficult question (the so-called question of the universality of the IMF), in particular because you need to resolve individual stars in clearly identified clusters to deduce an IMF. There are some papers investigating the question (for example, you could have a look at &lt;a href=&quot;http://adsabs.harvard.edu/abs/2012Natur.484..485C&quot; rel=&quot;noreferrer&quot;&gt;Cappellari et al. (2012)&lt;/a&gt; for a recent discussion of the problem).&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2013-12-11T15:43:45.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1133" PostTypeId="1" AcceptedAnswerId="1135" CreationDate="2013-12-11T16:16:54.367" Score="14" ViewCount="4212" Body="&lt;p&gt;Since Jupiter is very massive, it is the only planet (in our solar system) that has a center of mass with the Sun that lies outside the volume of the Sun. (&lt;a href=&quot;http://en.wikipedia.org/wiki/Jupiter_%28planet%29#Orbit_and_rotation&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Jupiter was a star, they would form a « binary star ».&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Sun was a planet, they would form a « double planet ».&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Sun is a star and Jupiter is a planet, does this have a particular name?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does Jupiter have a special status or a particular effect in our solar system because of its heavy mass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Jupiter-Sun's center of mass lies outside the volume of the Sun, that means that the Sun moves around that center of mass. Does this have an effect on Mercury, Venus, Earth and Mars orbits?&lt;/p&gt;&#xA;" OwnerUserId="204" LastEditorUserId="6" LastEditDate="2013-12-11T20:39:57.487" LastActivityDate="2017-06-23T09:57:35.863" Title="Is the Jupiter-Sun system considered a binary system of some type?" Tags="&lt;the-sun&gt;&lt;solar-system&gt;&lt;jupiter&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1134" PostTypeId="2" ParentId="524" CreationDate="2013-12-11T16:55:02.503" Score="6" Body="&lt;p&gt;It is linked to the solar dynamo and the dynamical behavior of the so-called $\alpha$-effect. One way to understand easily the solar dynamo and the the $\alpha$-effect with the hand is the following: you can generate a magnetic field by differential rotation (the $\Omega$-effect, $\Omega$ being the classical notation for the angular velocity). But this sole effect, twisting the magnetic field, would fold it so much that it will reconnect and dissipate. If you don't want to lose the magnetic field, you need a process to &quot;regenerate&quot; it, and a good way to do it is to create a poloidal component of the magnetic field. This poloidal component can be generated by convection in the solar interior; the magnetic field is, in the same time, twisted and pulled and that explains the solar dynamo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you can show that dynamo waves will propagate from the equator to the poles, and when the poles are reached, a reversal is observed. This propagation of dynamo waves is well observed by the sunspots patterns during the solar cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1955ApJ...122..293P&quot; rel=&quot;noreferrer&quot;&gt;Parker (1955)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2006EAS....21..241P&quot; rel=&quot;noreferrer&quot;&gt;Proctor (2006)&lt;/a&gt;; you can find it freely available here: &lt;a href=&quot;http://damtp.cam.ac.uk/user/mrep/solcyc/paper.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://damtp.cam.ac.uk/user/mrep/solcyc/paper.pdf&lt;/a&gt;‎&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="88" LastActivityDate="2013-12-11T16:55:02.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1135" PostTypeId="2" ParentId="1133" CreationDate="2013-12-11T19:26:47.703" Score="10" Body="&lt;p&gt;I'm not sure I understand your question entirely, but i'll do my best to offer a decent answer. It's true that the composition of Jupiter is very similar to that of the Sun (very similar approx. $H$ and $He$ abundance and pretty similar in density). The problem is that Jupiter is not nearly massive enough to have the internal pressure and temperature to undergo nuclear fusion. Jupiter doesn't have any particular special status aside from being the King of Planets in our solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the last part of your question, all objects orbit around a center of mass. Though because the Sun is much more massive, the center of mass lies very close to the center of the star (except in the case of Jupiter, where the CoM lies outside of the sun and is approx the length of its radius). This is why all planets in our solar system orbit around the sun. This will indeed cause a slight perturbation of orbital alignment, but I don't believe it's significant.&lt;/p&gt;&#xA;" OwnerUserId="555" LastEditorUserId="555" LastEditDate="2013-12-11T21:18:22.010" LastActivityDate="2013-12-11T21:18:22.010" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1137" PostTypeId="1" AcceptedAnswerId="1153" CreationDate="2013-12-12T00:59:56.273" Score="3" ViewCount="352" Body="&lt;p&gt;How would one go about determining the apsides of an orbit, knowing only the current distance between the 2 bodies, the orbital velocity, and the period? Every formula I've seen requires eccentricity, but conveniently, every formula for eccentricity requires the apsides. Surely there must be some way to find one without knowing the other?&lt;/p&gt;&#xA;" OwnerUserId="585" LastEditorUserId="13" LastEditDate="2013-12-12T03:44:28.810" LastActivityDate="2015-11-30T22:24:55.530" Title="Calculate apsides without knowing eccentricity" Tags="&lt;orbit&gt;&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1138" PostTypeId="2" ParentId="1124" CreationDate="2013-12-12T02:25:58.530" Score="2" Body="&lt;p&gt;1MW of transmitted power is very modest by RADAR terms, to give one example. Thus, you probably need to establish what exactly your professor means by &quot;brightest&quot; (may it be a specific wavelength or point in time, or some other definition). Otherwise, there's no reason to believe that Arecibo transmission was any special or outstanding from radio engineering stand point, while there were such nice toys operating at the very same time: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Russian_Woodpecker&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Russian_Woodpecker&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="412" LastActivityDate="2013-12-12T02:25:58.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1139" PostTypeId="2" ParentId="987" CreationDate="2013-12-12T08:22:43.423" Score="3" Body="&lt;p&gt;The main flaw with your definition of a &quot;generalised planet&quot; would be the boundary between stars and planets. And, in particular, the status of brown dwarfs, that are kind of &quot;failed stars&quot;, that could not start to burn their hydrodgen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this regard, the definition of a planet as it is now is probably the best one (or the &lt;em&gt;less worst one&lt;/em&gt;), because it is then possible to draw a line between planets and stars (and in particular brown dwarfs), because a planet is then an object formed by accretion in the debris material of a forming/newly formed star, whereas a star is, in general, an independent object formed by collapse and accretion (there are now observations of forming brown dwarfs; see for example the review by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2012ARA%26A..50...65L&quot; rel=&quot;nofollow&quot;&gt;Luhman (2102)&lt;/a&gt; or the recent paper by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2012Sci...337...69A&quot; rel=&quot;nofollow&quot;&gt;André et al. (2012)&lt;/a&gt;). So you have two types of celestrial objects, well defined by two types of formation mechanisms (core accretion for the planets vs. gravo-turbulent collapse for the stars (brown dwarfs included). Then you can also clearly distinguish between &quot;real&quot; free-floating planets (that are then planets expelled from their orbit and floating around in the galaxy) and brown dwarfs (that form independently, where they lay).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bottom line&lt;/strong&gt;: if your object, in any environment (including any type of accretion disk) is form by core accretion: it is a planet. If your object is form by gravitational collapse (with a later, unavoidable phase of accretion onto the central object): it is a star.&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2013-12-12T15:36:07.813" LastActivityDate="2013-12-12T15:36:07.813" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1140" PostTypeId="1" CreationDate="2013-12-12T08:38:03.203" Score="4" ViewCount="7311" Body="&lt;p&gt;Considering that a star of certain declination is crossing the local meridian at the observer's zenith at an unknown location on the earth. Here, the declination of the star is equal to the observer's latitude. Knowing the Right Ascension of the star, how could the observer calculate his longitude? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: The star Miaplacidus (Dec: -69° 42' and RA: 9h 13m) crossing at the observer's zenith!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Latitude&lt;/strong&gt; = 69° 42' S &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[ &lt;strong&gt;&lt;em&gt;Longitude&lt;/strong&gt;= RA(Decimal Conversion) X 15°&lt;/em&gt; ... &lt;em&gt;I'm not sure how to proceed from here ...&lt;/em&gt; ]&lt;/p&gt;&#xA;" OwnerUserId="557" LastEditorUserId="557" LastEditDate="2013-12-13T17:36:11.933" LastActivityDate="2014-05-29T14:54:45.777" Title="How to calculate Longitude from Right Ascension?" Tags="&lt;star&gt;&lt;amateur-observing&gt;&lt;observational-astronomy&gt;&lt;fundamental-astronomy&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1141" PostTypeId="2" ParentId="1005" CreationDate="2013-12-12T08:50:45.913" Score="8" Body="&lt;p&gt;No, not really.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars can form in circumstellar disks, that are, in general, disks surrounding forming stars, but not in protoplanetary disks. Protoplanetary disks are, by definition, flat, rotationing disks composed of gas and dust, found around newly born low-mass stars (see the review by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2011ARA&amp;amp;A..49...67W&quot;&gt;Williams &amp;amp; Cieza (2011)&lt;/a&gt;). There are two important point in this definition: protoplanetary disks orbit around &lt;strong&gt;low-mass&lt;/strong&gt; stars, and, in particular, &lt;strong&gt;newly-born&lt;/strong&gt; low-mass stars. These two points are important, because at this stage of the life of the disk and for this range of stellar mass, its mass it clearly too small to form any star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, it is not impossible to form stars in the protostellar disk of a forming star (that is one of the possibility to form multiple systems, that are still hard to explain), as discussed by various authors (see, for example  &lt;a href=&quot;http://adsabs.harvard.edu/abs/2009MNRAS.392..413S&quot;&gt;Stamatellos et al. (2009)&lt;/a&gt;, &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013MNRAS.433.3256V&quot;&gt;Vorobyov et al. (2013)&lt;/a&gt;, &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013A%26A...554A..17J&quot;&gt;Joos et al. (2013)&lt;/a&gt; and many others), but it is more likely to form brown dwarfs of very low-mass stars in these disks than main sequence stars (their mass is simply too small, once again).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said (&lt;em&gt;bis repetita&lt;/em&gt;), you could, as pointed out by called2voyage, you could form a main sequence star in the circumstellar disk of a high-mass star (as a Wolf-Rayet type star).&lt;/p&gt;&#xA;" OwnerUserId="88" LastActivityDate="2013-12-12T08:50:45.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1142" PostTypeId="2" ParentId="1095" CreationDate="2013-12-12T15:33:37.733" Score="8" Body="&lt;p&gt;Answer to the NEW question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the Angular Momentum Conservation Law states that, for any moving body, its angular momentum does not change unless you exercise an external force different from the central force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an orbiting body like a planet, this means that Sun's gravity, being the central force, does not modify Angular Momentum, but any other external force will do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of external forces are collisions or the forces made by Jupiter on another planet, or by Neptune on Pluto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the Solar System was formed, these external forces are quite small, and thus does not change greatly the Angular Momentum of any major body. But you can see how passing near a body can alter a comet's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, the external forces made by bodies that are in the same plane as an orbiting body does modify the &lt;em&gt;value&lt;/em&gt; of its Angular Momentum, but not the &lt;em&gt;direction&lt;/em&gt;. This causes that  the orbiting body changes its orbit but can not not change planes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you add small forces from objects in the same plane, you end up with no changes to planes.&lt;/p&gt;&#xA;" OwnerUserId="473" LastEditorUserId="473" LastEditDate="2013-12-12T15:43:19.677" LastActivityDate="2013-12-12T15:43:19.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1143" PostTypeId="1" AcceptedAnswerId="1144" CreationDate="2013-12-12T18:50:03.147" Score="6" ViewCount="92" Body="&lt;p&gt;Was comet ISON in a Kepler orbit prior to disintegrating? If so, it is possible over time that Kepler orbits can be altered by increases or decreases in the mass of of celestial bodies?&#xA;For instance if ISON was in a Kepler orbit and it lost mass did it maintain its orbit and disintegrate in its orbital track or does it alter its track due to the loss of mass?&lt;/p&gt;&#xA;" OwnerUserId="257" LastEditorUserId="204" LastEditDate="2013-12-12T23:29:31.560" LastActivityDate="2013-12-18T13:50:55.130" Title="Was comet ISON in a Kepler orbit prior to disintegrating?" Tags="&lt;comets&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1144" PostTypeId="2" ParentId="1143" CreationDate="2013-12-12T20:20:54.393" Score="3" Body="&lt;p&gt;To your first question, yes, it was in a Kepler orbit, (this interactive model shows just how it &lt;a href=&quot;http://www.solarsystemscope.com/ison/&quot; rel=&quot;nofollow&quot;&gt;appeared&lt;/a&gt;). ISON never completed a full circuit around the Sun since this was it's first time coming out of the &lt;strong&gt;Oort Cloud&lt;/strong&gt; and it disintegrated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further answering, the orbit of a body around another in space is not affected by it's mass. Remember the experiment that proves two object fall at the same speed regardless of it's mass &lt;a href=&quot;http://www.youtube.com/watch?v=KDp1tiUsZw8&quot; rel=&quot;nofollow&quot;&gt;(here the experiment is made on the moon with a feather and a hammer)&lt;/a&gt; and remember that &lt;strong&gt;orbiting is just falling at a certain speed and angle in respect to another body&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="537" LastEditorUserId="259" LastEditDate="2013-12-18T13:50:55.130" LastActivityDate="2013-12-18T13:50:55.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1145" PostTypeId="1" CreationDate="2013-12-12T20:41:46.693" Score="19" ViewCount="9424" Body="&lt;p&gt;I always see galaxies in &quot;disk&quot; shape in pictures. It is like, there is a rotational movement on a plane and the system is balanced by the pulling-in gravitational force which prevents the stars to leap out of the galaxy. What is the physical reason behind this? I have never seen a star in disk shape. A star is nothing but a much much smaller non-solid mass revolving around an axis, just like the galaxies. Does size matter? What makes the difference? Why doesn't a typical galaxy form in a spherical shape rather than a disk shape?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qyKPA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2020-05-19T05:08:53.380" Title="Why are galaxies disk shaped?" Tags="&lt;galaxy&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1146" PostTypeId="1" AcceptedAnswerId="1159" CreationDate="2013-12-12T20:50:44.937" Score="5" ViewCount="140" Body="&lt;p&gt;&lt;em&gt;[some spoilers]&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were in the ISS and orbiting the Earth roughly every 90 minutes, would it be possible for debris to be orbiting such that it passes you every 90 minutes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While Neil DeGrass Tyson pointed out scientific &lt;a href=&quot;http://gawker.com/neil-degrasse-tyson-thinks-gravity-is-scientifically-i-1441842497&quot;&gt;flaws&lt;/a&gt; in the movie on &lt;a href=&quot;https://twitter.com/neiltyson&quot;&gt;twitter&lt;/a&gt;, he &lt;a href=&quot;http://www.theverge.com/2013/10/10/4823078/neil-degrasse-tyson-gravity-movie-high-quality&quot;&gt;praised&lt;/a&gt; many things the movie got right in interviews, including &quot;The 90-minute orbital time for objects at that altitude.&quot;  I think I recall an interview on a podcast that I can't remember where he specifically said that referring to the debris recurring every 90 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality wouldn't the debris &lt;em&gt;have&lt;/em&gt; to be in a different orbit?  If the ISS orbits every 90 minutes (92.87 really but I'm saying 90) then after 90 minutes they would be in the same relative position.  The Earth would have rotated in the mean time so they wouldn't be over the same point on Earth, but they would be in the same position in orbit about the Earth.  Is there an orbit that would allow the same debris to cross the same spot in their orbit every 90 minutes?  Would it be nearly the same orbit, just inclined slightly?  Would that explain why the debris was moving so slowly relative to the ISS as well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;note:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are problems with the orbits and location in the movie, I simplified my question to ask about the particular orbit of the ISS.&lt;/p&gt;&#xA;" OwnerUserId="343" LastEditorUserId="343" LastEditDate="2013-12-13T00:29:29.520" LastActivityDate="2013-12-13T17:07:45.213" Title="Is there an orbit that could cause regular debris showers like in the movie Gravity?" Tags="&lt;orbit&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1147" PostTypeId="2" ParentId="1145" CreationDate="2013-12-12T21:14:07.917" Score="10" Body="&lt;p&gt;Spiral galaxies may be the most commonly depicted probably because we find ourselves in one, however, galaxies come in many sizes and shapes. &lt;a href=&quot;http://en.wikipedia.org/wiki/Galaxy_morphological_classification&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia's page on Galaxy morphological classification&lt;/a&gt; explains the most common classifications. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answering the question on the title:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The basic answer to your question is Conservation of Angular Momentum.&#xA;  Astronomers have been interested in this question for a long time and&#xA;  are still writing papers on the subject.   Even if you assume that at&#xA;  the time of the big bang there was no net angular momentum in the&#xA;  universe, one would expect that there were local fluctuations.  When&#xA;  galaxies initially formed by gravitational attraction these net local&#xA;  amounts remained.  Now angular momentum is conserved so when the&#xA;  initial collection of stars and gas in one galactic volume start to&#xA;  collapse the net angular momentum has to stay the same.  Since angular&#xA;  momentum $\vec{L} =m\vec{v}\times\vec{r}$,  for a given $v$ its easier to conserve $L$ if the $r$ is&#xA;  large.   In the orthogonal direction (along the axis of the accidental&#xA;  angular momentum) there is no such hindrance to collapse and so the&#xA;  collection turns out to be disk shaped.   By the way there has been&#xA;  several studies on the net angular momentum of a large collection of&#xA;  galaxies coming from the Sloan Digital Sky Survey and the answer is&#xA;  pretty close to but not quite zero.  Stay tuned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Extracted from an &lt;a href=&quot;http://van.physics.illinois.edu/qa/listing.php?id=18318&quot; rel=&quot;noreferrer&quot;&gt;University of Phoenix Department of Physics article&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="537" LastEditorUserId="1038" LastEditDate="2015-03-04T09:23:26.683" LastActivityDate="2015-03-04T09:23:26.683" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1148" PostTypeId="2" ParentId="1133" CreationDate="2013-12-12T23:02:55.253" Score="5" Body="&lt;p&gt;The Sun burns hidrogen, so it is a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter does not, so it is not a star. Not even a brown dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since a non-star body orbits a star, it is a planet, a dwarf planet or an asteroid/comet. Since Jupiter has cleared his orbit, it is a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do I say that Jupiter orbits a star? This is where I go to the center of your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbit of the Sun around the galactic center is (more or less) a circle. If you draw Jupiter's orbit around the galactic center it has some retrogradations and is not always convex.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-12-12T23:02:55.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1149" PostTypeId="2" ParentId="1146" CreationDate="2013-12-12T23:14:14.190" Score="1" Body="&lt;p&gt;There are orbits with 90min period. The difficult point is that if you are not in these orbits, the debris will not impact you, and if you are, you have their same speed, so they will not impact you but float around you.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-12-12T23:14:14.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1150" PostTypeId="1" CreationDate="2013-12-13T01:56:52.013" Score="8" ViewCount="195" Body="&lt;p&gt;What is the smallest scale at which we have detected &quot;dark matter&quot; (more kinetic energy than would be predicted)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't mind indulging a follow-up: Would its detection on smaller scales be beneficial at all to advancing our understanding?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(aside: I base this question purely on the notion that our instruments have surely improved a bit since the 1930s.)&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2013-12-13T14:04:14.987" Title="What's the smallest scale at which dark matter has been measured?" Tags="&lt;observational-astronomy&gt;&lt;gravity&gt;&lt;dark-matter&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1152" PostTypeId="2" ParentId="1095" CreationDate="2013-12-13T09:50:36.873" Score="7" Body="&lt;h2&gt;Angular momentum conservation&lt;/h2&gt;&#xA;&lt;p&gt;To put it in more mathematical terms, you can play with the energy and the angular momentum of a bunch of particles orbinting a central mass &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt;, given by&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$E = \sum_i m_i \left(\frac{1}{2}v_i^2 - \frac{GM}{r_i}\right),$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;for the energy and&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$${\bf I} = \sum_i m_i {\bf r}_i \times {\bf v_i},$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;for the angular momentum. Now, let's try to extremize the energy for a given angular momentum, keeping in mind that the system has to conserve angular momentum, and that collisions between the particles can reduce the energy. One good way to do it is to use &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_multiplier&quot; rel=&quot;noreferrer&quot;&gt;Lagrange multiplier&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\delta E - \lambda\cdot\delta {\bf I} = \sum_i\left[\delta {\bf v}_i \cdot \left({\bf v}_i - \lambda \cdot {\bf r}_i \right) + \delta {\bf r}_i \cdot \left( \frac{GM}{r_i^3} + \lambda \times {\bf v}_i\right)\right],$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;that requires&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\lambda\cdot{\bf r}_i = 0, \qquad {\bf v}_i = \lambda \times {\bf r}_i, \qquad \lambda^2 = \frac{GM}{r_i^3},$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;that means that all orbits are coplanar and circular.&lt;/p&gt;&#xA;&lt;h2&gt;Is this true in general?&lt;/h2&gt;&#xA;&lt;p&gt;That's the principle. Note, however, that all the planetary systems &lt;a href=&quot;http://exoplanet.eu/diagrams/?t=s&amp;amp;f=&amp;amp;x=inclination&amp;amp;xmin=&amp;amp;xmax=&amp;amp;y=mass&amp;amp;ymin=min&amp;amp;ymax=max&quot; rel=&quot;noreferrer&quot;&gt;do not always stay in an orbital plane&lt;/a&gt;. Such systems can be explained by &lt;a href=&quot;http://en.wikipedia.org/wiki/Kozai_mechanism&quot; rel=&quot;noreferrer&quot;&gt;Lidov-Kozai oscillations&lt;/a&gt;, typically trigger by &amp;quot;high-excentricity migration&amp;quot; of hot Jupiters (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2010AAS...21536708F&quot; rel=&quot;noreferrer&quot;&gt;Fabrycky, 2012&lt;/a&gt;). As far as we know now, we can say that:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;our Solar System is flat!&lt;/li&gt;&#xA;&lt;li&gt;planetary systems observed by Kepler are mostly flat (there is kind of an observational bias, due to the transit method);&lt;/li&gt;&#xA;&lt;li&gt;planetary systems observed by radial-velocity method are more or less flat (with an mean angle between 10 and 20°);&lt;/li&gt;&#xA;&lt;li&gt;planetary systems with hot Jupiters are not flat in general.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;strong&gt;More dirty details&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;p&gt;There is an excellent talk by &lt;a href=&quot;http://www.eso.org/sci/meetings/garching/abs13.html&quot; rel=&quot;noreferrer&quot;&gt;Scott Tremaine, given at ESO last year&lt;/a&gt; you could watch online.&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2013-12-13T09:50:36.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1153" PostTypeId="2" ParentId="1137" CreationDate="2013-12-13T10:13:34.197" Score="3" Body="&lt;p&gt;You mention this is a system in which a vessel is orbiting a planet. In that case it's fair to assume the mass of the vessel is negligible compared with the planet mass and&#xA;the planet will sit in one of the focal points of the ellipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then use Kepler's &lt;a href=&quot;http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Second_law&quot; rel=&quot;nofollow&quot;&gt;second law&lt;/a&gt; to find the semi-minor axis as&#xA;$$ \frac{1}{2} r v = \frac{\pi a b}{P} $$&#xA;where $r$ the current distance and $v$ the current velocity, $P$ is the period and $a$ and $b$ give the respective semi-major and semi-minor axes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From $b$ you find the eccentricity as&#xA;$$ b=a \sqrt{1-e^2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the apsides follow as&#xA;$$ \mathrm{periapsis} = a ( 1 - e ) $$&#xA;$$ \mathrm{apoapsis}  = a ( 1 + e ) $$&lt;/p&gt;&#xA;" OwnerUserId="494" LastEditorUserId="2365" LastEditDate="2015-11-30T22:24:55.530" LastActivityDate="2015-11-30T22:24:55.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1154" PostTypeId="2" ParentId="987" CreationDate="2013-12-13T10:16:18.367" Score="1" Body="&lt;p&gt;Will be adding some relevant information here as I find it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This article speculates that there might be planets forming from the remnants of common envelope phase in binaries: &lt;a href=&quot;http://arxiv.org/abs/1312.3479&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1312.3479&lt;/a&gt; . These are not normal planets, as the material is not coming from star formation. &lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2013-12-13T10:16:18.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1155" PostTypeId="2" ParentId="430" CreationDate="2013-12-13T13:20:08.623" Score="12" Body="&lt;p&gt;Few important points about WISE:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;it was able to detect anything with a temperature above 70-100 K, whereas the coolest brown dwarfs are in the 500-600 K range (the coolest was discovered by WISE itself, see &lt;a href=&quot;http://adsabs.harvard.edu/abs/2011ApJ...726...30M&quot;&gt;Mainzner et al., 2011&lt;/a&gt;);&lt;/li&gt;&#xA;&lt;li&gt;it was able to detect objects larger than 1km up to 3 AU from the Sun, or objects of 2-3 Jupiter masses in a distance up to 7-10 light-years (see &lt;a href=&quot;http://wise.ssl.berkeley.edu/gallery_images/WISE-mission_overview.ppt&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/2009/2070.html&quot;&gt;here&lt;/a&gt;);&lt;/li&gt;&#xA;&lt;li&gt;the closest brown dwarf detected by WISE is a &lt;a href=&quot;http://en.wikipedia.org/wiki/WISE_1049-5319&quot;&gt;brown dwarfs binary system&lt;/a&gt;, located at 6.6 light-years from the Sun.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So I would say we can be pretty confident that Nemesis existence has been ruled out.&lt;/p&gt;&#xA;" OwnerUserId="88" LastActivityDate="2013-12-13T13:20:08.623" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1156" PostTypeId="2" ParentId="1150" CreationDate="2013-12-13T14:04:14.987" Score="1" Body="&lt;p&gt;It looks like &lt;a href=&quot;http://www.keckobservatory.org/recent/entry/uci_scientists_size_up_universes_most_lightweight_dwarf_galaxy_with_keck_ob&quot; rel=&quot;nofollow&quot;&gt;Segue 2&lt;/a&gt;, found by the Keck Observatory, is a candidate for the &lt;em&gt;lightest&lt;/em&gt; galaxy, but I don't know if its the &lt;em&gt;smallest&lt;/em&gt; one.&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2013-12-13T14:04:14.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1157" PostTypeId="2" ParentId="523" CreationDate="2013-12-13T14:06:49.097" Score="4" Body="&lt;p&gt;The favored theory is a giant impact, at the end of the final stage of planetary formation (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2005Sci...307..546C&quot; rel=&quot;nofollow&quot;&gt;Canup, 2005&lt;/a&gt;). A &quot;giant impact&quot; is a collision with an object of a comparable size of the planet. That is also the most probable scenario for lunar formation (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2004ARA%26A..42..441C&quot; rel=&quot;nofollow&quot;&gt;Canup, 2004&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="88" LastActivityDate="2013-12-13T14:06:49.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1158" PostTypeId="1" AcceptedAnswerId="1175" CreationDate="2013-12-13T15:10:55.913" Score="5" ViewCount="342" Body="&lt;p&gt;I bought &lt;a href=&quot;http://www.opticsplanet.com/celestron-binoculars-upclose-71144.html&quot;&gt;this binocular&lt;/a&gt; because I've read countless times that one should start with a pair of binoculars before diving into telescopes. I've seen details of the moon and I can notice Venus round shape on some nights. I also saw Jupiter and it's two larger moons, as faint as the faintest stars, but this was on a beach trip so the sky was clearer than what I'm used to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I live on a place with a &lt;strong&gt;7&lt;/strong&gt; or &lt;strong&gt;6&lt;/strong&gt; on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bortle_scale&quot;&gt;&lt;strong&gt;Bortle Scale&lt;/strong&gt;&lt;/a&gt;, is there something else I should try to see?&lt;/p&gt;&#xA;" OwnerUserId="537" LastEditorUserId="537" LastEditDate="2013-12-13T15:17:05.567" LastActivityDate="2016-03-19T23:53:43.683" Title="What are some night sky objects I could see with my Celestron UpClose 20x50 Porro Binocular?" Tags="&lt;amateur-observing&gt;&lt;observational-astronomy&gt;&lt;bortle-scale&gt;&lt;binoculars&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1159" PostTypeId="2" ParentId="1146" CreationDate="2013-12-13T17:07:45.213" Score="5" Body="&lt;p&gt;If I understand you correctly (and I haven't seen the movie ^_^), you want to know if it is possible for two different objects in orbit to come close to each other periodically, with roughly the same period as one of the object's orbital period. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, that is possible, but not probable.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Obviously, if the objects are in the &lt;em&gt;same&lt;/em&gt; orbit, they are always close to each other &lt;/li&gt;&#xA;&lt;li&gt;If they are in &lt;em&gt;slightly&lt;/em&gt; different orbits, one of the objects will seem to move about the other object in regular or irregular patterns, depending on orbit specifics&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If they are in completely different orbits, the necessary conditions are &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the orbital period of both is equal&lt;/li&gt;&#xA;&lt;li&gt;they have at most 1 point in common (or at least very closely approach each other)&lt;/li&gt;&#xA;&lt;li&gt;the phasing of both objects is such that the objects are in the close approach region at the same time.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In ordinary two-body Keplerian celestial mechanics, an orbit's semi-major axis is the only orbital element that determines the orbital period. Therefore, one orbit may be circular and the other elliptic and inclined in all sorts of ways, but as long as the semi major axis of both is the same, they will have the same period. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this whole scenario is completely unstable for any real-world celestial body. The (relatively small) asphericity of the Earth, as well as the presence of the Moon for example, causes all orbits around it to drift in some way or another. If you put two objects in orbits like you describe, they will not likely meet up very often before never seeing each other again for centuries. &lt;/p&gt;&#xA;" OwnerUserId="599" LastActivityDate="2013-12-13T17:07:45.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1160" PostTypeId="1" AcceptedAnswerId="1162" CreationDate="2013-12-13T18:10:31.763" Score="12" ViewCount="511" Body="&lt;p&gt;How do we calculate the escape velocity of galaxies?&#xA;Do we have to know the total mass?&#xA;What is the escape velocity for Milky Way?&lt;/p&gt;&#xA;" OwnerUserId="465" LastEditorUserId="6" LastEditDate="2013-12-13T18:19:04.843" LastActivityDate="2013-12-13T21:29:12.287" Title="How do we calculate the escape velocity of galaxies?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;velocity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1161" PostTypeId="2" ParentId="1140" CreationDate="2013-12-13T18:22:45.710" Score="2" Body="&lt;p&gt;Let's think it is september 21 midnight AT GREENWHICH (you didn't specify midnight where).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's go to &lt;a href=&quot;http://www.csgnetwork.com/siderealjuliantimecalc.html&quot; rel=&quot;nofollow&quot;&gt;http://www.csgnetwork.com/siderealjuliantimecalc.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There input september 21, midnight at Greenwich. It says Sidereal time (ST) is 00:00:6.62 or, in hours, 0.001839h&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This number is the difference between any star RA and its Azimuth at Greenwich.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Miaplacidus has a RA of 9h13min (9,216667h) so its Azimuth at Greenwich is 9,218506 at that moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the star is at Azimuth 0 for you, you are -9,218506h away from greenwich, that is (multiplying by 15) -138,27759º away from Greenwich. This is your Longitude.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-12-13T18:22:45.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1162" PostTypeId="2" ParentId="1160" CreationDate="2013-12-13T21:29:12.287" Score="5" Body="&lt;p&gt;In short, the escape velocity from the region we find ourselves in the milky way is &lt;strong&gt;544 Km/s +- 10%&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This number, and an extensive explanation of how it was calculated can be found on a paper called &lt;a href=&quot;http://arxiv.org/pdf/astro-ph/0611671v2.pdf&quot;&gt;The RAVE Survey: Constraining the Local Galactic Escape Speed&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also important to consider is the fact that we already are at aproximately &lt;strong&gt;220 Km/s&lt;/strong&gt; because of the &lt;strong&gt;sun's orbital velocity&lt;/strong&gt;, so the actual value needed to get there from earth is much lower.&lt;/p&gt;&#xA;" OwnerUserId="537" LastActivityDate="2013-12-13T21:29:12.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1163" PostTypeId="5" CreationDate="2013-12-13T21:35:51.913" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-12-13T21:35:51.913" LastActivityDate="2013-12-13T21:35:51.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1164" PostTypeId="4" CreationDate="2013-12-13T21:35:51.913" Score="0" Body="A scale that quantifies the interference caused by light pollution for the observation of astronomical objects in the night sky. " OwnerUserId="9527" LastEditorUserId="9527" LastEditDate="2020-01-28T23:37:40.270" LastActivityDate="2020-01-28T23:37:40.270" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="1165" PostTypeId="1" AcceptedAnswerId="1187" CreationDate="2013-12-14T00:30:52.867" Score="7" ViewCount="1302" Body="&lt;p&gt;From &lt;a href=&quot;http://www.youtube.com/watch?v=KDp1tiUsZw8&quot; rel=&quot;noreferrer&quot;&gt;this exhibition by an astronaut on the moon&lt;/a&gt; dropping a hammer and a feather and showing they move at the same speed, why do the particles of the comets tail move away from the comet, slow down and fall away as it disintegrates? Shouldn't they move at the same speed of the comet? &#xA;For example if the feather was attached to the hammer and it was falling and broke off wouldn't it fall at the same speed, not start to trail behind and eventually fall away?&lt;/p&gt;&#xA;" OwnerUserId="257" LastEditorUserId="537" LastEditDate="2013-12-16T14:17:59.050" LastActivityDate="2013-12-16T21:23:25.483" Title="If the hammer and feather move at the same speed why do comet and the tail particles move at different speeds?" Tags="&lt;gravity&gt;&lt;comets&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1166" PostTypeId="2" ParentId="1165" CreationDate="2013-12-14T01:27:21.890" Score="2" Body="&lt;p&gt;The tail particles from a comet are ejected, so they have a distinct velocity than that of the comet. Besides that, they orbit the Sun in an orbit similar to that of the comet when they were ejected, they don't slow down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the hammer and the feather, it is not the same to just break off than to be ejected. If the hammer blows the feather up while falling, the feather will delay a lot more than the hammer to fall.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-12-14T01:27:21.890" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1169" PostTypeId="2" ParentId="1158" CreationDate="2013-12-14T01:58:46.870" Score="2" Body="&lt;p&gt;Try galaxies like Andromeda, and clouds like Orion's sword, as a start.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-12-14T01:58:46.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1170" PostTypeId="2" ParentId="1165" CreationDate="2013-12-14T11:24:04.790" Score="-6" Body="&lt;p&gt;There are a few more elements which you might consider. Like, when dropping a hammer along with a feather in earth, the feather descends slower than the hammer. Reason: although the force of gravity is same, their mass, and the resistance from external forces faced by them is an entirely different story. The air between the object and the ground must also be taken into consideration. Whereas, moon lacks atmosphere and hence they fall at relatively same speed (with only low gravity acting upon them).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in &lt;a href=&quot;http://en.wikipedia.org/wiki/Comet_tail&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Comet_tail&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As a comet approaches the inner Solar System, solar radiation causes&#xA;  the volatile materials within the comet to vaporize and stream out of&#xA;  the nucleus, carrying dust away with them. The streams of dust and gas&#xA;  thus released form a huge, extremely tenuous atmosphere around the&#xA;  comet called the coma, and the force exerted on the coma by the Sun's&#xA;  radiation pressure and solar wind cause an enormous tail to form,&#xA;  which points away from the Sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Similar to the air acting on the detached feather (as said above), the solar radiation pressure and solar winds act on the tail debris of the comet. For instance, you drop an article from a moving vehicle, it falls to the ground rather than flying in the air at the same speed. Deceleration and opposite forces like solar winds cause it to slow down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt; The reason why mass is taken into consideration is due to the fact that mass is directly proportional to gravity. So, increasing the mass means increased gravitational pull. That is why heavier objects like hammer fall faster than lighter objects like feather. Explaining this would be a matter of physics and not astronomy which is irrelevant to this website and should be discussed in the physics stack exchange website.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2013-12-14T13:38:07.757" LastActivityDate="2013-12-14T13:38:07.757" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1171" PostTypeId="2" ParentId="1120" CreationDate="2013-12-14T11:36:34.747" Score="2" Body="&lt;p&gt;The sundial translates the position of the sun to the time of day, so it depends on the path the sun takes across the sky, this is called the ecliptic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the earth's rotation is tilted with respect to its orbit around the sun, the ecliptic shifts across the sky during the year, which is also the cause of seasonal change on earth. Yet the highest altitude of the ecliptic will always be due south, which means noon is well defined every day of the year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to answer the question; if the day length (from sun-rise to sun-down) is exactly 12 hours,  the sun will rise at an azimuth of 90 degrees and set at 270 degrees. However, this happens only twice a year, namely at the &lt;em&gt;vernal equinox&lt;/em&gt; (first day of spring) and the &lt;em&gt;autumnal equinox&lt;/em&gt; (first day of fall). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the longest day, &lt;em&gt;the summer solstice&lt;/em&gt;, both sun-rise and sun-down azimuths will be at their highest shift northward. On the other hand, during the shortest day, the &lt;em&gt;winter solstice&lt;/em&gt;, the azimuth-shift will be maximally southward. This last bit applies to the northern hemisphere, for the southern hemisphere the seasons and therefore day lengths are reversed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in short, you have to account for what place on earth your sundial is at and what time of year it is. &lt;/p&gt;&#xA;" OwnerUserId="494" LastActivityDate="2013-12-14T11:36:34.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1172" PostTypeId="2" ParentId="1120" CreationDate="2013-12-14T11:57:59.023" Score="1" Body="&lt;p&gt;The sun is not on the same Azimuth for the same hour (not even at noon!). You need to take the analema figure (&lt;a href=&quot;http://en.wikipedia.org/wiki/Equation_of_time&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Equation_of_time&lt;/a&gt;) into your accounts. If you do not, your clock will be exact only 4 times a year.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-12-14T11:57:59.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1173" PostTypeId="2" ParentId="1133" CreationDate="2013-12-14T19:22:45.587" Score="5" Body="&lt;p&gt;No, it is not since Jupiter is not a star. We call the system of a star with its planets surrounding the center of mass a simply &quot;solar System&quot;.&lt;/p&gt;&#xA;" OwnerUserId="465" LastActivityDate="2013-12-14T19:22:45.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1175" PostTypeId="2" ParentId="1158" CreationDate="2013-12-15T10:48:26.493" Score="4" Body="&lt;p&gt;Depending on the season you can try different objects. Try large star clusters and nebulae:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Pleiades&lt;/li&gt;&#xA;&lt;li&gt;Orion nebula&lt;/li&gt;&#xA;&lt;li&gt;Andromeda galaxy&lt;/li&gt;&#xA;&lt;li&gt;M13 globular cluster&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Also double stars like epsilon Lyrae. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It helps to have a foto tripod and an adaptor for the binoculars. Gives much better view when the binoculars are resting on the tripod. &lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2013-12-15T10:48:26.493" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1176" PostTypeId="1" AcceptedAnswerId="1180" CreationDate="2013-12-16T00:09:07.880" Score="6" ViewCount="510" Body="&lt;p&gt;I was watching a video of the &lt;a href=&quot;http://digg.com/video/video-of-the-chinese-lunar-landing-released&quot; rel=&quot;nofollow&quot;&gt;chinese lunar landing&lt;/a&gt; and saw numerous bumps in the soil. What are they? Is there any theory that explains how they came into existence?&lt;/p&gt;&#xA;" OwnerUserId="279" LastEditorUserId="6593" LastEditDate="2015-06-11T00:41:28.957" LastActivityDate="2015-06-11T00:41:28.957" Title="Lunar Soil: what are those &quot;bumps&quot; and how were they formed?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1177" PostTypeId="1" AcceptedAnswerId="1178" CreationDate="2013-12-16T01:03:55.783" Score="7" ViewCount="480" Body="&lt;p&gt;As I understand it, beyond parallax the next way of estimating the distance to astronomical objects is by using a relationship between the period and intrinsic luminosity of Cepheid variables.  How do we know that relationship?  Are there some close enough that we used parallax to validate it, or is there a theoretical reason, or something else?&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorDisplayName="user1569" LastEditDate="2016-08-18T15:05:57.513" LastActivityDate="2017-01-01T15:16:37.783" Title="How do we know that the intrinsic brightness of Cepheid variables corresponds to their period?" Tags="&lt;distances&gt;&lt;cepheids&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1178" PostTypeId="2" ParentId="1177" CreationDate="2013-12-16T02:15:14.887" Score="10" Body="&lt;p&gt;This relationship was discovered empirically by &lt;a href=&quot;https://en.wikiquote.org/wiki/Henrietta_Swan_Leavitt#Periods_of_25_Variable_Stars_in_the_Small_Magellanic_Cloud_.281912.29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Henrietta Leavitt&lt;/a&gt; by comparing the apparent magnitudes of stars in the Magellanic clouds. Since these clouds are far away (200,000 ly) and relatively small (7,000 ly), the difference in distances to the 47 different Cephid variables she observed could account for only a small difference in brightness.  By plotting the brightness against the period she noticed that there was a relationship, cephids with longer periods were brighter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you guessed, modern measurements are made via parallax for close stars. The Hipparcos satellite compiled a vast list of parallaxes, and by measuring a large number of periods, we can calculate the &lt;a href=&quot;http://adsabs.harvard.edu/abs/1999A&amp;amp;A...352..547P&quot; rel=&quot;nofollow noreferrer&quot;&gt;relation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="608" LastEditorUserId="608" LastEditDate="2017-01-01T15:16:37.783" LastActivityDate="2017-01-01T15:16:37.783" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1179" PostTypeId="1" AcceptedAnswerId="13358" CreationDate="2013-12-16T04:21:08.577" Score="20" ViewCount="688" Body="&lt;p&gt;The planet HD 106906 b was found at 650 AU from its home star, a star not so different from the Sun. Fomalhaut b is also a young planet that is orbiting very far away from its star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two planets were discovered mainly because they are young enough to still be hot enough to be seen in the infrared. I think that old planets in very high orbits would have temperatures below 10K, and thus be black and invisible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people agree with the hypothesis that, in our Solar System, somewhere out beyond Neptune, some non-dwarf planets might be orbiting the Sun. The existence of HD 106906 b and Fomalhaut b suggests that this is possible. What are the odds that we could find at least one of these planets in our Solar System? If they do exist, how could we find them?&lt;/p&gt;&#xA;" OwnerUserId="609" LastEditorUserId="13591" LastEditDate="2016-08-26T21:51:28.203" LastActivityDate="2016-08-26T21:51:28.203" Title="Hypothetical trans-Neptunian planets orbiting the Sun" Tags="&lt;solar-system&gt;&lt;exoplanet&gt;&lt;planet&gt;&lt;9th-planet&gt;&lt;tno&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1180" PostTypeId="2" ParentId="1176" CreationDate="2013-12-16T06:08:29.260" Score="10" Body="&lt;p&gt;It's a bit difficult to understand what exactly you're asking about, but I will presume you meant the topological features in images like this one, taken from &lt;a href=&quot;http://digg.com/video/video-of-the-chinese-lunar-landing-released&quot; rel=&quot;noreferrer&quot;&gt;the video&lt;/a&gt; of which you provided a link to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/LHix3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the most part, they are meteorite and micrometeorite impact craters, with some regolith boulders and impact ejecta scattered along the edge of the larger impact crater at the bottom right of the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These impact craters appear &lt;em&gt;strange&lt;/em&gt; to an untrained eye when they're the only visible feature on images, because there is not atmospheric scattering of light on the Moon that would somewhat illuminate the shadow side too and reduce contrast to what we're used to seeing on Earth, so the high contrast can appear as bumps or dents interchangingly and at will, similar to how you can see either either a vase / candlestick in the middle (yellow) or two faces facing each other (black) on this image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/KfjP6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A trained eye will have less problems recognising them for what they are, though. One nifty trick is to search the image for different looking topological features, such as already mentioned larger crater ejecta along its edge at the bottom right of the image, or perhaps prominent features within the features themselves, like some larger micrometeorite impact craters show shockwave formed uplifted centers, and so on. It also helps to understand how these features were likely made, but once you orient yourself, it will prove difficult to convince yourself into seeing what you saw initially, if you saw indentations where there are perturbations, or vice versa.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2013-12-16T06:08:29.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1183" PostTypeId="2" ParentId="1179" CreationDate="2013-12-16T13:38:48.790" Score="11" Body="&lt;p&gt;If such a planet exists, WISE should observe it. &lt;a href=&quot;http://wise.ssl.berkeley.edu/&quot; rel=&quot;noreferrer&quot;&gt;WISE&lt;/a&gt; is an infrared satellite that imaged the entire sky. In particular:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;it was able to detect anything with a temperature above 70-100 K, whereas the coolest known exoplanets are in the 100 K range (see histogram below, taken from &lt;a href=&quot;http://www.openexoplanetcatalogue.com/histogram.html&quot; rel=&quot;noreferrer&quot;&gt;the Exoplanets Catalogue&lt;/a&gt;);&lt;/li&gt;&#xA;&lt;li&gt;it was able to detect objects larger than 1km up to 3 AU from the Sun, or objects of 2-3 Jupiter masses in a distance up to 7-10 light-years (so roughly up to 440,000 to 630,000 AU, see &lt;a href=&quot;http://wise.ssl.berkeley.edu/gallery_images/WISE-mission_overview.ppt&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/2009/2070.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;);&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;so I would say that its existence is quite ruled out now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ec5uX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="88" LastActivityDate="2013-12-16T13:38:48.790" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1184" PostTypeId="1" AcceptedAnswerId="1189" CreationDate="2013-12-16T13:46:03.487" Score="16" ViewCount="6014" Body="&lt;p&gt;Due to Earth's elliptical orbit, its distance from Sun varies by almost &lt;strong&gt;5 million Kilometers&lt;/strong&gt; (147 million Kilometers at closest point &amp;amp; 152 million Kilometers at farthest point, i.e. almost 3% of the average distance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As evident from the fact that that Venus has hotter environment than Mars due to their respective distances from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why then Earth does not observe two winters (at farthest points) and two summers (at closest points)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Additional Note&lt;/strong&gt;:&#xA;I know that Earth's seasonal climate change is caused by its 23 degrees tilt that causes the sunlight density variations for the hemispheres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to me this &lt;strong&gt;5 million Km&lt;/strong&gt; distance seems more relevant than the 23 degrees tilt.&lt;/p&gt;&#xA;" OwnerUserId="614" LastActivityDate="2016-06-22T16:32:24.407" Title="Why don't we have 2 Summers and 2 Winters?" Tags="&lt;orbit&gt;&lt;earth&gt;&lt;climate&gt;" AnswerCount="6" CommentCount="6" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1185" PostTypeId="2" ParentId="1184" CreationDate="2013-12-16T14:02:00.397" Score="2" Body="&lt;p&gt;There are two factors here. One is that the 23 degrees tilt is way more important than the tiny 5 million Km (remember that that's only 0.033 AU). Proof for that is that when it is Summer in the north it is winter in the south, regardless of the distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other is that if there were no tilt, we would have one very light summer and one very light winter (very very light, both) because the Sun is not at the center of the ellipse (two nearest points and two farthest points) but on one focus (one nearest point and one farthest point).&lt;/p&gt;&#xA;" OwnerUserId="473" LastEditorUserId="13" LastEditDate="2014-01-01T08:20:11.777" LastActivityDate="2014-01-01T08:20:11.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1186" PostTypeId="2" ParentId="1184" CreationDate="2013-12-16T14:47:57.680" Score="1" Body="&lt;p&gt;It is because the 23 degrees tilt of the earth and the ellipse orbit of the earth around the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you make an equation with all the variables we have here&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;23 degrees tilt of earth&lt;/li&gt;&#xA;&lt;li&gt;Ellipse orbit of the earth around the sun&lt;/li&gt;&#xA;&lt;li&gt;Earth speed around itself&lt;/li&gt;&#xA;&lt;li&gt;Earth orbital speed around the sun&lt;/li&gt;&#xA;&lt;li&gt;Distance between the earth and the sun&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So you will note a variation in the weather conditions regarding this variables, so we have 2 winters and 2 summers, but 2 of them are just a transition state.&lt;br&gt;&#xA;A proof of that, at the earth's poles, we can only have summer and winter because the limitation of variable (1).&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2013-12-16T14:47:57.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1187" PostTypeId="2" ParentId="1165" CreationDate="2013-12-16T14:48:48.277" Score="11" Body="&lt;p&gt;The tail of a comet is &lt;strong&gt;&lt;em&gt;not&lt;/em&gt;&lt;/strong&gt; actually &quot;slowing down and falling away&quot; from the comet, like you might expect to see when smoke streams out from behind a moving object on earth. The tail of a comet is actually being &lt;strong&gt;&lt;em&gt;pushed&lt;/em&gt; away from the sun by the solar winds&lt;/strong&gt; and radiation. That's why the tail of a comment always points &lt;em&gt;away&lt;/em&gt; from the sun, and doesn't stream out behind it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If there were no other forces at work, the coma &lt;em&gt;would&lt;/em&gt; actually travel at about the same speed at the comet.&lt;/strong&gt; As the material surrounding the core of the comet vaporized, it would form a coma (&quot;atmosphere&quot;) made up of sublimated gas mixed with dust, moving along  at the same speed (more or less&lt;sup&gt;(1)&lt;/sup&gt;) as the comet&amp;hellip; like you observed in the video.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://spaceplace.nasa.gov/comet-quest/en/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://spaceplace.nasa.gov/review/comet-quest/comet_orbit-lrg.en.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;(1)&lt;/sup&gt; &lt;em&gt;In reality, without the solar winds, the coma of the comet would continue to move outward, as its expulsion velocity would easily overcome the comet's weak gravitational pull.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="58" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:47.090" LastActivityDate="2013-12-16T21:23:25.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1188" PostTypeId="1" AcceptedAnswerId="1325" CreationDate="2013-12-16T15:50:15.717" Score="9" ViewCount="3366" Body="&lt;p&gt;With the known latitude coordinate of the observing position, how to find the altitude of the moon when it is high, i.e when it crosses the local meridian? &lt;/p&gt;&#xA;" OwnerUserId="557" LastActivityDate="2014-01-06T16:22:58.767" Title="How to calculate the altitude of the Moon?" Tags="&lt;the-moon&gt;&lt;amateur-observing&gt;&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1189" PostTypeId="2" ParentId="1184" CreationDate="2013-12-16T17:44:51.417" Score="38" Body="&lt;p&gt;There are a few incorrect assumptions in your post, so it is difficult to answer as asked. But I can address the misconceptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. The seasons are not caused by our distance from the sun&lt;/strong&gt;&lt;br&gt;&#xA;The seasons are caused by the 23.5° tilt in Earth's axis. When the Northern Hemisphere is tilted towards the sun (summer), the Southern Hemisphere is simultaneously tilted &lt;em&gt;away&lt;/em&gt; from the sun (winter). So the seasonal temperature difference has little to do with the Earth's position in its elliptical orbit. Without this tilt, there would be no seasons and the temperature day to day across the globe would be relatively uniform. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Even the GLOBAL temperature is NOT consistent with our change in distance&lt;/strong&gt;&lt;br&gt;&#xA;As a matter of fact, the &lt;em&gt;average&lt;/em&gt; temperature of the Earth globally is &lt;strong&gt;&lt;em&gt;hottest&lt;/em&gt; when it is the &lt;em&gt;furthest&lt;/em&gt;&lt;/strong&gt; from the sun &amp;mdash; hotter by about 2.3°C (&lt;a href=&quot;http://www.todayifoundout.com/index.php/2011/12/the-earth-is-hottest-when-it-is-furthest-from-the-sun-on-its-orbit-not-when-it-is-closest/&quot; rel=&quot;noreferrer&quot;&gt;ref&lt;/a&gt;). That's because there is a lot more landmass in the Northern Hemisphere facing the sun (when Earth is &lt;em&gt;farthest&lt;/em&gt; away in its orbit). So even though there is less intensity of sunlight, the land is able to be heated up much faster than the vast oceans which have to be heated at perihelion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This distance-temperature inconsistency isn't unique to the Earth. Look at the average temperature of the &lt;em&gt;other&lt;/em&gt; inner planets as we move away from the sun:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Mercury (167°C)&lt;/li&gt;&#xA;&lt;li&gt;Venus (460°C)  &lt;strong&gt;&amp;larr; &lt;em&gt;farther, but hotter than Mercury?&lt;/em&gt;&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Earth (14.0°C)&lt;/li&gt;&#xA;&lt;li&gt;Mars (-60°C)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Venus is actually warmer than Mercury because of the thick carbon dioxide atmosphere causing runaway global warming. So it isn't simply the distance from the sun that determines the average temperature of a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3. There's only ONE aphelion/perihelion&lt;/strong&gt;&lt;br&gt;&#xA;The closest point of the Earth's orbit (perihelion) and the farthest (aphelion) only happens &lt;strong&gt;once&lt;/strong&gt; per year; not twice. That is because the elliptical orbit of the Earth is such so the sun is at one of the foci, not the center (as illustrated below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SzyXD.png&quot; alt=&quot;Earth&amp;#39;s Elliptical Orbit](http://i.imgur.com/hxAbz8y.png)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;em&gt;Note that the size of the bodies and the eccentricity of the orbits are greatly exagerated here.&lt;/sup&gt;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="58" LastEditorUserId="58" LastEditDate="2016-06-22T16:32:24.407" LastActivityDate="2016-06-22T16:32:24.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1190" PostTypeId="2" ParentId="1184" CreationDate="2013-12-16T21:46:26.803" Score="3" Body="&lt;p&gt;I suspect you're thinking that we'd have a summer when the northern hemisphere, for example, is tilted toward the Sun, and a second summer during the perihelion, when the Earth is closest to the Sun. For one thing, the timing doesn't work; the perihelion takes place in early January, close to the northern midwinter. That probably moderates the effects of axial tilt for the northern hemisphere (and amplifies them for the southern hemisphere), but it's not enough to override them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other answers have said that the axial tilt is a more significant&#xA;factor than the variation in distance from the Sun, but they haven't&#xA;explained &lt;em&gt;why&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is a rough back-of-the-envelope guesstimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference in illumination caused by the varying distance from&#xA;the Sun can be computed from the ratio between the perihelion and&#xA;aphelion distance, which is about a factor of 0.967.  Applying the&#xA;inverse square law indicates that amount of sunlight at aphelion is&#xA;about 93.5% of what it is at perihelion.  Reference:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Perihelion#Planetary_perihelion_and_aphelion&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Perihelion#Planetary_perihelion_and_aphelion&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At my current location (about 33° north latitude), at this time of&#xA;year (close to the northern winter solstice), we're getting about 10&#xA;hours of sunlight and 14 hours of darkness each day.  (Reference: the weather app on my phone.) That's about 83% of what we'd get with 12 hours of daylight during&#xA;either equinox, and about 71% of what we'd get with 14 hours of daylight and 10 hours of darkness&#xA;per day during the summer solstice. The effect is greater at higher latitudes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to that, the sun is lower in the sky during the winter &#xA;than it is during the summer, meaning that a given amount of sunlight&#xA;is spread over a larger area of the Earth's surface, which makes&#xA;the ratio even larger.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have the numbers for that, but it's enough to show that the &#xA;effect of the axial tilt is substantially greater than the effect of&#xA;the varying distance between the Earth and the Sun.&lt;/p&gt;&#xA;" OwnerUserId="53" LastActivityDate="2013-12-16T21:46:26.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1191" PostTypeId="1" CreationDate="2013-12-17T00:12:52.577" Score="11" ViewCount="995" Body="&lt;p&gt;I read on &lt;a href=&quot;http://en.wikipedia.org/wiki/Milky_Way#Appearance&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;depending on the time of night and the year, the arc of Milky Way can&#xA;  appear relatively low or relatively high in the sky. For observers&#xA;  from about 65 degrees north to 65 degrees south on the Earth's surface&#xA;  the Milky Way passes directly overhead twice a day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What happens and how does the Milky Way look like out of that range?&#xA;If it matters, consider also the all-summer light and the dark winter.&lt;/p&gt;&#xA;" OwnerUserId="623" LastActivityDate="2013-12-19T20:45:28.707" Title="How does the Milky Way look like above 66° North and below 66° South?" Tags="&lt;milky-way&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1192" PostTypeId="2" ParentId="1188" CreationDate="2013-12-17T02:50:29.557" Score="3" Body="&lt;p&gt;With a lot of difficulty if you want to do it numerically by yourself. The Moon's orbit is very complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Far and away the easiest method would be to use NASA's &lt;a href=&quot;http://ssd.jpl.nasa.gov/?horizons&quot; rel=&quot;nofollow&quot;&gt;HORIZONS&lt;/a&gt; service for calculating &lt;a href=&quot;http://ssd.jpl.nasa.gov/?glossary&amp;amp;term=ephemeris&quot; rel=&quot;nofollow&quot;&gt;ephemerides&lt;/a&gt;, which will get you very comprehensive and accurate information on objects at your specified times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think that HORIZONS can directly return the time that the Moon is at its maximum altitude. However, if you can write some basic computer code then it shouldn't be difficult to query the ephemerides with a precise (say 1 minute) time interval and find the maximum altitude from there.&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="276" LastEditDate="2013-12-17T02:58:51.857" LastActivityDate="2013-12-17T02:58:51.857" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1194" PostTypeId="2" ParentId="925" CreationDate="2013-12-17T09:49:35.710" Score="3" Body="&lt;p&gt;Let me see, some objects are listed because of their records. First, nebulae, which are huge clouds of dust and gas. The largest in the Milky Way's neighborhood is the Tarantula Nebula (30 Doradus, NGC 2070 or Caldwell 103).:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://2.bp.blogspot.com/_viputHRpZ2A/SRL1YHQxmKI/AAAAAAAABSY/IQN4-D4FClY/s400/NGC+2070+by+Joe+Brimacombe.jpg&quot; alt=&quot;Tarantula Nebula&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;This is the largest nebula in the Milky Way's neighborhood, at 300 light-years (almost 100 pc). But if we are in the universal record, that would be NGC 604, which is 740 light-years (240 pc) across.:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RdfBm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RdfBm.jpg&quot; alt=&quot;NGC 604&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://www.wolaver.org/Space/ngc604.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;wolaver.org&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;Next, galaxies, which come in different shapes. Spirals first, and what like called2voyage has said, is NGC 262. Because he can't provide a pic, I will!:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/msj9f.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/msj9f.jpg&quot; alt=&quot;NGC 262&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://cseligman.com/text/atlas/ngc262.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;cseligman.com&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;NGC 262 is a giant Seyfert spiral, and is 1.3 million light-years across, more than two times larger than the mistakenly clamed largest spiral last January 2013, NGC 6872.:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://www.nasa.gov/sites/default/files/images/718359main_composite_no_labels.jpg&quot; alt=&quot;NGC 6872&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;I don't know why they claimed this one as largest, at 522,000 light-years across &lt;em&gt;only&lt;/em&gt; is less than half as that of NGC 262.&lt;/p&gt;&#xA;&lt;p&gt;But the hands-down king of all galaxies is IC 1101, at 5.6 million light-years across, and is the record holder since 1989.:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://4.bp.blogspot.com/-kFoYO-Y4AW4/T3_zoxObL8I/AAAAAAAAfH8/02vUYCn5q44/s400/ic_1101.jpg&quot; alt=&quot;IC 1101&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;A size comparison of this monster:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hVh0B.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hVh0B.jpg&quot; alt=&quot;IC 1101 vs. Milky Way and other galaxies&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;https://astrobob.areavoices.com/files/2013/07/IC-1101-compared-to-Milky-Way1-1024x563.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;areavoices.com&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;Next, structures. aneroid has the right answer. It's name is Hercules-Corona Borealis Great Wall, some 10 billion light-years across.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-02-16T12:50:16.720" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="1195" PostTypeId="1" CreationDate="2013-12-17T14:04:22.330" Score="16" ViewCount="171" Body="&lt;p&gt;It is believed that dark matter is made of particles, which interact with matter only weakly and gravitationally. One common candidate for dark matter are so called &lt;a href=&quot;http://en.wikipedia.org/wiki/Weakly_interacting_massive_particle&quot;&gt;WIMPs&lt;/a&gt;. WIMPs, specifically, are heavy and may be their own antiparticles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as any other particles dark matter particles can be produced at sufficiently high energies. The mass of dark matter particles is unknown, but is estimated to be of order $1$-$100 \textrm{GeV}$, which corresponds to temperatures of $T_{DM}\approx 10^{13}$-$10^{15}\textrm{K}$, at which these particles may be expected to be produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such enormous temperatures are barely attainable in any reasonable astrophysical processes, but say in core-collapse supernovae newly formed core has temperatures of $T_{SN,after}\approx 10^{11}\textrm{K}$, and probably more during the collapse phase. Then a crude estimate would suggest that the amount of dark matter produced is $M_{DM}\approx e^{-T_{DM}/T_{SN,max}}M_\odot$. Or, in number form $\log_{10}(M_{DM}/\textrm{kg})=30.3-0.43(T_{DM}/T_{SN})$. This means that at $T_{SN}=1.4\cdot 10^{-2}T_{DM}$ the amount of dark matter produced during a supernova will be around one kilogram. Such temperatures are fairly reachable for $1 \textrm{GeV}$ DM particles. So one can optimistically expect few kilograms of dark matter produced per supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the question. &lt;strong&gt;What is a typical dark matter production in core-collapse supernovae?&lt;/strong&gt; A good answer, I imagine, would be a more robust expansion on the existing estimate. Any constructive comments are welcome.&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2014-09-24T21:55:01.553" Title="On (minuscule) dark matter production in supernovae" Tags="&lt;dark-matter&gt;&lt;high-energy-astrophysics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1197" PostTypeId="1" AcceptedAnswerId="1232" CreationDate="2013-12-17T20:40:52.657" Score="4" ViewCount="5111" Body="&lt;p&gt;Is there a difference between solar wind and solar radiation or are they just different ways of describing the same thing?&lt;/p&gt;&#xA;" OwnerUserId="257" LastActivityDate="2015-06-16T01:49:41.790" Title="What is the difference between solar wind and solar radiation?" Tags="&lt;solar-wind&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1198" PostTypeId="5" CreationDate="2013-12-17T21:52:31.903" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-12-17T21:52:31.903" LastActivityDate="2013-12-17T21:52:31.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1199" PostTypeId="4" CreationDate="2013-12-17T21:52:31.903" Score="0" Body="Questions related to the physical processes involved in or to the observations of the process of forming a star." OwnerUserId="88" LastEditorUserId="88" LastEditDate="2013-12-18T16:58:14.513" LastActivityDate="2013-12-18T16:58:14.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1200" PostTypeId="1" CreationDate="2013-12-17T23:24:00.253" Score="12" ViewCount="307" Body="&lt;p&gt;For some &quot;average sized&quot; black hole, how long would it take for a distant (spacesuit-wearing!) astronaut to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;be able to see the gravitational lensing with the naked eye &lt;/li&gt;&#xA;&lt;li&gt;clearly feel the gravity gradient&lt;/li&gt;&#xA;&lt;li&gt;reach a lethal gravity gradient&lt;/li&gt;&#xA;&lt;li&gt;cross the event horizon&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Ultimately I'm interested in the subjective experience of a human falling into a black hole. Would it be drawn out and unpleasant? Or is it that by the time the gradient would be painful, there'd be barely any time at all before annihilation?&lt;/p&gt;&#xA;" OwnerUserId="628" LastEditorUserId="628" LastEditDate="2013-12-18T02:03:58.500" LastActivityDate="2013-12-18T19:27:11.963" Title="Timeline of falling into a blackhole" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="1201" PostTypeId="1" AcceptedAnswerId="1205" CreationDate="2013-12-18T06:12:29.327" Score="12" ViewCount="1443" Body="&lt;p&gt;We know that the universe is gradually expanding and this indirectly means that the gravitational force between sun, earth, planets and other stars (roughly anything in the universe) is gradually decreasing as gravitational force is indirectly proportional to square of distance between the objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think this also effects to length of the year. If yes then is it possible to know how days does 1 year had 1 million year back?&lt;/p&gt;&#xA;" OwnerUserId="379" LastEditorUserId="204" LastEditDate="2014-01-04T12:20:16.813" LastActivityDate="2014-04-11T14:05:58.900" Title="What was the length of year 1 million years back?" Tags="&lt;solar-system&gt;&lt;gravity&gt;&lt;newtonian-gravity&gt;" AnswerCount="2" CommentCount="10" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1202" PostTypeId="2" ParentId="1201" CreationDate="2013-12-18T11:18:17.783" Score="9" Body="&lt;p&gt;&lt;em&gt;(Disclaimer: As I already pointed out in a comment to the question above, I never did a calculation with $H_0$ before and I might be utterly, horrible wrong with my interpretation.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you completely ignore the slowly changing orbit of earth and only take expansion of space into account and assume the Hubble-parameter to be pretty constant in the timeframe of 1 My, we can calculate the difference of the orbital period of earth using Keppler's third law [3]:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T = 2\pi\sqrt(a^3/GM)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a = 1.4959789*10^{11} m$ (semi-major axis of earth today) [1]&lt;br&gt;&#xA;$G = 6.67*10^{-11} Nm^2/kg^2$ (gravitational constant)&lt;br&gt;&#xA;$M = 1.988435*10^{30} kg$ (mass sun) [1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also assume: $H_0 = 2.3*10^{-18} s^-1$ [2] (Hubble parameter then and today in SI-units) which basically means &quot;in every second a meter get $2.3*10^{-18} m$ longer&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of taking the length of an (siderial) orbital period of earth from some source, let's calculate it manually first and take it as a reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_{today} = 2 \pi \sqrt((1.4959789*10^{11}m)^3/(6.67*10^{-11} Nm^2/kg^2 * 1.988435*10^{30} kg))$ = 365 days 8 hours 56 minutes 13.45 seconds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretty close and a good reference for more calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what was earth's semi-major axis 1 million years ago, only taking into account a constant $H_0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x - (2.3*10^{-18} s^-1 * 1 My * x) = 1.4959789*10^{11} m$&lt;br&gt;&#xA;Solving for $x$ leads to $x = 1.49598*10^{11} m$.&#xA;&lt;br&gt;(Sorry for the lousy precision; I only have Wolfram Alpha at my hands right now.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The old semi-major axis is a little smaller. Using Keppler's law again we can calculate the orbital period again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_{old} = 2 \pi \sqrt((1.496*10^{11} m)^3/(6.67*10^{-11} Nm^2/kg^2 * 1.988435*10^{30} kg))$ = 365 days 8 hours 56 minutes 48.26 seconds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, subtracting both times from another we can say that &lt;em&gt;1 My ago the year was indeed 34.81 seconds shorter&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However. This probably doesn't mean much; the orbit changes slightly over time anyway; the Hubble-parameter is not considered a constant any more, it changes slightly over time; and while this was an interesting question I don't trust my interpretation much and hope that someone else who's more qualified than me could enlighten the question better than I ever could.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I hope I didn't botch anything somewhere. I need more coffee.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Source: Wolfram Alpha&lt;br&gt;&#xA;[2] Source for Hubble-parameter in SI-units taken from the German Wikipedia: &lt;a href=&quot;http://de.wikipedia.org/wiki/Hubble-Konstante#Definition&quot; rel=&quot;nofollow&quot;&gt;http://de.wikipedia.org/wiki/Hubble-Konstante#Definition&lt;/a&gt;&lt;br&gt;&#xA;[3] &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_period#Small_body_orbiting_a_central_body&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Orbital_period#Small_body_orbiting_a_central_body&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="6" LastEditDate="2014-04-11T14:05:58.900" LastActivityDate="2014-04-11T14:05:58.900" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1203" PostTypeId="1" CreationDate="2013-12-18T15:03:08.570" Score="8" ViewCount="167" Body="&lt;p&gt;What are commonly important astrophysical systems/models, where optical refraction is important or necessary to account for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would kindly ask you not to consider refraction in Earth's atmosphere or inside instrumentation in this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comment: Gravitational lensing is different from optical refraction even though it affects light paths. I would like to ask the authors to avoid mentioning it in the following.&lt;/p&gt;&#xA;" OwnerUserId="115" LastEditorUserId="115" LastEditDate="2013-12-18T18:53:31.037" LastActivityDate="2019-02-22T11:25:26.063" Title="When is optical refraction important in astronomy?" Tags="&lt;radiative-transfer&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1204" PostTypeId="2" ParentId="1203" CreationDate="2013-12-18T17:07:46.723" Score="8" Body="&lt;p&gt;Optical refraction is related to the change in direction of a light ray when the refractive index changes. Excluding Earth's atmospheres and instruments, I think that refraction has little/no impact in astronomy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only cases that comes to my mind where we can (probably) have some important refraction is in eclipsing star binaries or near edge on planetary systems. &#xA;Let's imagine a planet transiting behind his star. Some of it's light passes through the stellar atmosphere and gets refracted. As the atmosphere is curved and likely changes the refractive index with height, it acts like a lens dispersing (intuition says so) the planet light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; A similar description holds in general for any object passing behind an other one that have atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;And there is &lt;strong&gt;Gravitational Lensing&lt;/strong&gt; (if you allow me), which has much larger impact on observations. This is caused by gravity bending light rays when passing near galaxies/cluster of galaxies(/stars/...). One of the differences of gravitational lensing with respect to standard lenses is that there is no change in refractive index, so it's achromatic (all wavelengths get bent by the same angle). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effective index of refraction can be described as ( source: &lt;a href=&quot;http://www.ita.uni-heidelberg.de/research/bartelmann/Publications/Proceedings/JeruLect.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Narayan and Bartelmann&lt;/a&gt;(pdf) ):&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ n = 1 + \frac{2}{c^{2}} |\Phi|$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\Phi$&lt;/span&gt; is the gravitational potential and is generally a function of position of the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational lensing is canonically divided in three groups:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Strong lensing, usually observed in galaxy clusters or around massive galaxies. The gravitational potential is so strong that the the image of a background galaxy is heavily distorted into arcs and rings, like in this striking image of Abell 2218 from HST:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VFkpu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VFkpu.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://imgsrc.hubblesite.org/hu/db/images/hs-2000-07-b-print.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;hubblesite.org&lt;/a&gt;)&lt;/sub&gt;  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Weak lensing. The light of a galaxy encounters matter (and a lot of dark matter) travelling to us and gets refracted. This doesn't have a dramatic effect as in strong lensing, but distorts the shape of the galaxy. And this distortion can be used to study, e.g., the dark matter distribution around some object or the content of the universe.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Micro lensing. Imagine to observe a star and somehow know that a blob of dark matter is going to pass in front of the star. The blob is not big enough to distort the star shape, but for sure it will increase by a small amount the luminosity of the star.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="86" LastEditorUserId="7472" LastEditDate="2019-02-18T07:50:35.060" LastActivityDate="2019-02-18T07:50:35.060" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="1205" PostTypeId="2" ParentId="1201" CreationDate="2013-12-18T17:10:27.770" Score="15" Body="&lt;p&gt;The Hubble expansion has no bearing whatsoever on the length of the year. This is because the whole Milky Way galaxy (and in fact most galaxies, if not all, and even local groups) has decoupled from the Hubble flow long ago. In fact, it could only form after it decoupled. Note that M31, our sister galaxy, is in fact falling onto the Milky Way rather than receding (as the Hubble flow would imply), demonstrating that the whole of the Local Group (of galaxies) is decoupled from the Hubble flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens is that any over-density expands at less than the Hubble rate and thereby grows. Galaxies (and larger structures) form from small relative over-densities that eventually grow large enough to withstand the overall expansion and instead collapse under their own gravity to form bound objects, such as galaxy clusters, galaxies, star clusters, and stars. This implies that the Hubble flow has no bearing on the inner dynamics of such systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the number of days in a year was higher in the past than today, but that is only because the Earth is spinning down (due to tidal friction with the Moon), so that the days become longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anything has had an effect on the semi-major axis of the Earth orbit (and hence on its period), then that is gravitational interactions with the other planets. However, weak interactions (secular perturbations) can only alter the orbital eccentricity and leave the semi-major axis unaltered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, there is a tiny effect from the Sun loosing mass (to the Solar wind). The period of any orbiting bodies is proportional to $M_\odot^{-1/2}$.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2013-12-25T13:44:11.220" LastActivityDate="2013-12-25T13:44:11.220" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="1206" PostTypeId="1" AcceptedAnswerId="1263" CreationDate="2013-12-18T18:39:10.683" Score="8" ViewCount="641" Body="&lt;p&gt;From recent &lt;a href=&quot;http://www.mpifr-bonn.mpg.de/announcements/2013/8&quot; rel=&quot;nofollow noreferrer&quot;&gt;news from the Max Planck Institute for Radio Astronomy:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The European Research Council (ERC) has awarded 14 Million Euros to a&#xA;  team of European astrophysicists to construct the first accurate image&#xA;  of a black hole. The team will test the predictions of current&#xA;  theories of gravity, including Einstein's theory of General&#xA;  Relativity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've seen computer depictions of black holes in the past, are they similiar to what we can expect from this new endeavour? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the next image we can see a computer model of a black hole of ten solar masses as seen from a distance of 600km with the Milky Way in the background with a horizontal camera opening angle of 90 degrees:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xF5EP.jpg&quot; alt=&quot;A Black Hole of ten solar masses as seen from a distance of 600km with the Milky Way in the background with a horizontal camera opening angle of 90 degrees&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.tempolimit-lichtgeschwindigkeit.de/galerie/galerie.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="537" LastEditorUserId="2365" LastEditDate="2016-04-03T18:38:01.680" LastActivityDate="2016-04-03T18:38:01.680" Title="What can we expect from the first accurate image of a black hole?" Tags="&lt;black-hole&gt;&lt;milky-way&gt;&lt;radio-astronomy&gt;&lt;general-relativity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1207" PostTypeId="2" ParentId="1200" CreationDate="2013-12-18T19:27:11.963" Score="2" Body="&lt;p&gt;Interestingly enough, Neil deGrasse Tyson, &lt;a href=&quot;http://en.wikipedia.org/wiki/Neil_deGrasse_Tyson&quot; rel=&quot;nofollow&quot;&gt;an american astrophysicist&lt;/a&gt;, wrote a book titled &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0393330168&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Death by black hole&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://fora.tv/2008/02/19/Neil_DeGrasse_Tyson_Death_by_Black_Hole&quot; rel=&quot;nofollow&quot;&gt;this video&lt;/a&gt; he goes on to explain the details of what you asked, the subjective experience of the astronaut being sucked into the black hole. &lt;strong&gt;It is definitely painful!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="537" LastActivityDate="2013-12-18T19:27:11.963" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1208" PostTypeId="1" AcceptedAnswerId="1211" CreationDate="2013-12-18T23:18:38.867" Score="10" ViewCount="2302" Body="&lt;p&gt;I just got a 4.5&quot; Newtonian reflector.  The skies haven't been clear where I live lately so I've only been able to use it to see the moon and Jupiter (I was also able to see the Galilean moons).  A friend of mine with a similar telescope says he can see Saturn with its rings and the phases of Venus with his telescope.  I am wondering what else I can see with a telescope my size, specifically which Messier objects I can see with my telescope.  I live in a suburban area near a small urban area and so light pollution is a concern but I have access to darker skies close by.&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2013-12-19T10:43:32.430" Title="What can be seen with a 4.5&quot; telescope" Tags="&lt;amateur-observing&gt;&lt;newtonian-telescope&gt;&lt;messier-catalog&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1209" PostTypeId="2" ParentId="1191" CreationDate="2013-12-19T05:29:04.670" Score="1" Body="&lt;p&gt;Outside of this range you are close to the poles of our planet.  The milky way will be low on the horizon and hard to see.&lt;/p&gt;&#xA;" OwnerUserId="639" LastActivityDate="2013-12-19T05:29:04.670" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1210" PostTypeId="2" ParentId="1191" CreationDate="2013-12-19T09:02:29.267" Score="5" Body="&lt;p&gt;On midnight, right around this time of year, the Milky Way will be in the zenith. Here is an XEphem rendering for the north of finland (65th latitude) for yesterday midnight (the brown outline marks the Milky Way):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/X81db.png&quot; alt=&quot;Sky map for 65th latitude including milky way&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also see this on photographs by finnish photographer &lt;a href=&quot;http://www.tommyeliassen.com/#!/index/15&quot; rel=&quot;noreferrer&quot;&gt;Tommy Eliassen&lt;/a&gt;. He has many examples of this on his website. I won't put any in here, because I guess they are copyrighted, and not free to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an animated 24 hour version of the above sky map, which shows the rotation of the milkyway around the zenith:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/K4s5U.gif&quot; alt=&quot;Animated sky map including the milky way, 24 hours.&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="50" LastEditorUserId="50" LastEditDate="2013-12-19T20:45:28.707" LastActivityDate="2013-12-19T20:45:28.707" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1211" PostTypeId="2" ParentId="1208" CreationDate="2013-12-19T10:43:32.430" Score="5" Body="&lt;p&gt;The aperture of your 4.5&quot; telescope is one thing, it's also important what focal length you have. Is it a f/5 or rather a f/8? The f/8 would be suitable for viewing the Moon, Jupiter, Saturn, maybe even Mars and Venus. You can also buy a good solar filter, attach it in the front of the optical tube assembly, and view the sun. But be careful with that, and inform yourself beforehand! Your eyesight might be in danger otherwise. For the Moon it is advisable to buy a neutral grey filter, because it is also very, very bright.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the Messier objects: you can probably view the larger ones. The Ring nebula (M57) will be tough, because it is very small and faint. But open clusters, globular clusters and Andromeda's core will be possible. A few bright nebulas will also be possible, like M42 (Orion Nebula).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a good list of all Messier objects over at &lt;a href=&quot;http://astropixels.com/messier/messiercat.html&quot;&gt;Astropixels&lt;/a&gt;. You should try objects of magnitude 4 or better, and that have a size of more than 5 arcminutes. Otherwise they may be too dim or small to find.&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2013-12-19T10:43:32.430" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1212" PostTypeId="2" ParentId="1206" CreationDate="2013-12-19T11:12:09.543" Score="3" Body="&lt;p&gt;I think the image you posted is not quite reallistic. On it, objects are just inverted from some radius on, while what you can expect from a real black hole seen from near enough is a combination of these:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) an accretion disc&#xA;b) a companion being sucked&#xA;c) Hawking's radiation&#xA;d) X-Ray burst from the poles (really starting out of the event horizon)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will not see an actual black circle, as most probably there will be a lot of matter out of the event horizon radiating.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-12-19T11:12:09.543" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1213" PostTypeId="1" AcceptedAnswerId="1215" CreationDate="2013-12-19T13:27:59.530" Score="8" ViewCount="1716" Body="&lt;p&gt;According to the first paragraph of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Noon&quot;&gt;Noon&lt;/a&gt; Wikipedia page, &lt;strong&gt;solar noon&lt;/strong&gt; is the moment when the Sun crosses the meridian and is at its highest elevation in the sky. The wording there suggests that there is an exact identity between the solar meridian crossing time and the solar maximum elevation time on a given day. However, later in same article in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Noon#Solar_noon&quot;&gt;Solar Noon&lt;/a&gt; section it states that, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Solar noon is the moment when the Sun transits the celestial meridian – &lt;strong&gt;&lt;em&gt;roughly&lt;/em&gt;&lt;/strong&gt; the time when it is highest above the horizon on that day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Emphasis mine)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The use of the word &quot;roughly&quot; really surprised me. I was strongly under the impression that these times were necessarily equal for sub-polar regions. Am I wrong and missing something, or is the wiki article just poorly worded? &lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2013-12-19T17:58:37.897" Title="Solar Noon: meridian crossing time versus time of maximum elevation" Tags="&lt;the-sun&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1215" PostTypeId="2" ParentId="1213" CreationDate="2013-12-19T17:51:52.460" Score="5" Body="&lt;p&gt;There are a couple of things you have to think about in order to understand the problem at hand. The first is that there are three coordinate systems at play. The first is the celestial coordinate system, which is a coordinate system based upon the latitude and longitude of the Earth (the celestial equator is the Earth's equator projected out into space). See this &lt;a href=&quot;https://astronomy.stackexchange.com/questions/9/could-someone-explain-ra-dec-in-simple-terms/41#41&quot;&gt;post&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second coordinate system is that of the solar system: The plane of the solar system (which contains the sun and planets, and traces out a line in the sky called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ecliptic&quot; rel=&quot;nofollow noreferrer&quot;&gt;ecliptic&lt;/a&gt;). Now because of the tilt of the Earth's axis with respect to the normal vector of the plane of the solar system, the celestial equator is not the same as the ecliptic. What's more, is that the relative position of the ecliptic and the celestial equator change throughout the course of the year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, there is the altitude/azimuth coordinate system which locked to the observer. The zenith (the point at which all horizons are equidistant from the point you are looking at) is $+90^{\circ}$ in altitude and is a degenerate point in azimuth, and all horizons are $0^{\circ}$ altitude but lay on a unique azimuthal arc stemming from the zenith and intersecting the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a diagram showing the relationship between the first two coordinate systems, the celestial coordinate system and the coordinate system of the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/H1gLU.png&quot; alt=&quot;eclipticdiagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does all of this matter for the question you've asked? It's because as you look due East and due West, you're looking at $0^{\circ}$ declination (the celestial equator intersects the horizon at exactly due East and due West) whereas the ecliptic intersects the horizon at different points. These points change throughout the year as the relative position of the ecliptic changes with respect to the celestial equator. Now, the amount the sun travels on the ecliptic over a 24 hour period is $0.9856^{\circ}$ ($360^{\circ}/365.25 \text{ days}$; actually I may want to use a &lt;a href=&quot;http://en.wikipedia.org/wiki/Sidereal_time#Sidereal_time_and_solar_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;sidereal day&lt;/a&gt; here but nonetheless it's about a degree). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to this, there is potentially a little bit extra in altitude that the sun may travel throughout the course of a day (since the path of the ecliptic is neither aligned with the celestial coordinate system, nor aligned with the local alt/az coordinate system of the observer; unless you're at the north or south poles in which case your local alt/az coordinate system is the same as the celestial coordinate system, however both are still mis-aligned with the coordinate system of the solar system). I think the word &quot;approximate&quot; needs to be there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All told, I don't personally know how to quantify the extra bit of altitude the sun may gain from the misalignment of the ecliptic with lines of constant altitude and from the sun's motion along the ecliptic, but I would welcome references and calculations from other people. This little bit extra may not be a really small number. In other words, you might want to see how the rate of change of altitude of the sun at that point changes simply due to the Earth's rotation and how it may compare to the rate of change of altitude due to the motion of the sun along the ecliptic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &#xA;I also just looked to see what the ecliptic looks like on the sky (as seen from Philadelphia), and how it compares to the local alt/az coordinate system. Below is a picture with all coordinate systems present to try to illustrate the point I was trying to make.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1EtMA.png&quot; alt=&quot;coordinatesystems&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The green line is the meridian, the orange(ish) lines are alt/az coordiantes, the blue line is the celestial equator, and the red line is the path of the ecliptic. Hopefully you can make this out sufficiently well.&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2013-12-19T17:58:37.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1216" PostTypeId="1" AcceptedAnswerId="21571" CreationDate="2013-12-19T20:29:40.223" Score="13" ViewCount="530" Body="&lt;p&gt;A friend showed me an &lt;a href=&quot;http://www.livescience.com/42048-halleys-comet-linked-to-ancient-famine.html&quot;&gt;article&lt;/a&gt; saying that there's evidence that a piece of Halley's comet may have struck the earth about 1,500 years ago. Is anyone here aware of such evidence?&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="2365" LastEditDate="2015-11-10T22:02:42.507" LastActivityDate="2017-07-08T16:34:52.897" Title="Did a piece of Halley's comet strike the Earth 1,500 years ago?" Tags="&lt;comets&gt;&lt;impact&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1217" PostTypeId="2" ParentId="1203" CreationDate="2013-12-20T01:29:49.170" Score="3" Body="&lt;p&gt;Here's a &lt;a href=&quot;http://arxiv.org/abs/0904.1612&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper&lt;/a&gt; I found which talks about the index of refraction of dark matter (different from gravitational lensing) and how a signal might attenuate. The paper is entitled, &lt;strong&gt;&amp;quot;Dark Matter Constraints from a Cosmic Index of Refraction&amp;quot;&lt;/strong&gt;, and here is the abstract:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The dark-matter candidates of particle physics invariably possess electromagnetic interactions, if only via quantum fluctuations. Taken en masse, dark matter can thus engender an index of refraction which deviates from its vacuum value. Its presence is signaled through frequency-dependent effects in the propagation and attenuation of light. We discuss theoretical constraints on the expansion of the index of refraction with frequency, the physical interpretation of the terms, and the particular observations needed to isolate its coefficients. This, with the advent of new opportunities to view gamma-ray bursts at cosmological distance scales, gives us a new probe of dark matter and a new possibility for its direct detection. As a first application we use the time delay determined from radio afterglow observations of distant gamma-ray bursts to realize a direct limit on the electric-charge-to-mass ratio of dark matter of |varepsilon|/M &amp;lt; 1 x 10^{-5} eV^{-1} at 95% CL.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;To be honest I don't really understand how quantum fluctuations of dark matter could produce such an effect. Aside from instrumental or atmospheric effects (and theoretical considerations), I honestly don't recall refraction ever coming up as an important effect.&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2013-12-20T01:29:49.170" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1218" PostTypeId="2" ParentId="1216" CreationDate="2013-12-20T08:13:09.603" Score="1" Body="&lt;p&gt;I have not read the complete scientific article, but I'm not prone to trust this. On that year the Chinese civilization were quite well established and annotating about comets, and I have not read about them annotating about a comet that approached so much to Earth.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2013-12-20T08:13:09.603" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1219" PostTypeId="1" CreationDate="2013-12-20T08:46:17.877" Score="21" ViewCount="604" Body="&lt;p&gt;Now that the Gaia Space Telescope is on it's way to the Sun-Earth L2 Lagrangian point (SEL2), I start wondering about the stability of Gaia's orbit there. The Planck Telescope is already there, as was Wilkinson Microwave Anisotropy Probe (WMAP) and other probes, and from &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrangian_point&quot;&gt;Wikipedia&lt;/a&gt; I learned that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In practice, any orbit around Lagrangian points L1, L2, or L3 is&#xA;  dynamically unstable, meaning small departures from equilibrium grow&#xA;  exponentially over time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gaia has some kind of Orbital Maneuvering System (to borrow a Space Shuttle term) and some propellant on board, so has Planck, however I wonder how deterministic these orbits are and if both Planck and Gaia have automatic corrections and collision detection in their flight computers; L2 is &quot;only&quot; 1.5 million km (or about 5 light-seconds) away so surely there's time for manual correction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know a source which tells how different Gaia's and Planck's orbits are, if there are intersections between their orbital planes or even how likely the need for an unplanned orbital correction is? I know Lissajou-shapes from maths classes and I know how much the projected trail can differ depending on the precision of data types used in calculations (e.g. float vs. double). How does ESA/NASA handle this, now that SEL2 seems it will become a crowded place?&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="13" LastEditDate="2013-12-20T09:27:54.020" LastActivityDate="2013-12-21T21:42:14.123" Title="How stable are Lissajous orbits?" Tags="&lt;orbit&gt;&lt;the-sun&gt;&lt;earth&gt;&lt;lagrange-point&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1220" PostTypeId="2" ParentId="1197" CreationDate="2013-12-20T09:14:20.443" Score="2" Body="&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_wind&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's article on solar winds&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The solar wind is a stream of charged particles (a plasma) released&#xA;  from the upper atmosphere of the Sun. It mostly consists of electrons&#xA;  and protons with energies usually between 1.5 and 10 keV.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solar radiation would be everything in general, including solar wind.  Everything else is electromagnetic radiation (Radio, Microwave, Infra-red, Visible, Ultraviolet, X-ray, Gamma Ray)&lt;/p&gt;&#xA;" OwnerUserId="639" LastEditorUserId="537" LastEditDate="2013-12-22T04:34:24.790" LastActivityDate="2013-12-22T04:34:24.790" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1221" PostTypeId="1" AcceptedAnswerId="1222" CreationDate="2013-12-20T10:05:16.893" Score="6" ViewCount="351" Body="&lt;p&gt;In the Christian tradition the first Christians are believed to be three wise men who followed the star of Bethlehem to the stable where Jesus was supposed to be born. Is it with some accuracy known if this is an actual star and if so will I be able to see that star during Christmas in North-western Europe?&lt;/p&gt;&#xA;" OwnerUserId="653" LastActivityDate="2013-12-23T06:00:46.657" Title="Can I see the star of Bethlehem at Christmas?" Tags="&lt;star&gt;&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2013-12-20T20:39:21.643" ContentLicense="CC BY-SA 3.0" />
  <row Id="1222" PostTypeId="2" ParentId="1221" CreationDate="2013-12-20T10:48:15.347" Score="5" Body="&lt;p&gt;The most recent suggestions are that it was a nova or supernova, however no one really knows what the object was. The verbatim records describe it as a star, though it is not certain whether or not it was a &quot;true&quot; star (in those times, a star being any object that had a fixed right ascension and declination). Perhaps it was just a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case I suspect it is unlikely you will be able to see it with your naked eye today. Even if you could, its true identity has certainly been lost in history.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some references for further reading (click on the &quot;&lt;em&gt;Full Printable Article (PDF/Postscript)&lt;/em&gt;&quot; link for a PDF of the articles):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1918PA.....26..150H&quot; rel=&quot;nofollow&quot;&gt;What was the star of Bethlehem?&lt;/a&gt;, S. Hagar (1918)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://adsabs.harvard.edu/full/1968ASPL...10..185L&quot; rel=&quot;nofollow&quot;&gt;Interpretations of the Star of Bethlehem&lt;/a&gt;, E. Little (1968)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1979QJRAS..20..153C&quot; rel=&quot;nofollow&quot;&gt;Can we find the Star of Bethlehem in Far Eastern Records?&lt;/a&gt;, C. Cullen (1979)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://adsabs.harvard.edu//abs/2005Obs...125..168T&quot; rel=&quot;nofollow&quot;&gt;The Star of Bethlehem: a type Ia/Ic supernova in the Andromeda galaxy?&lt;/a&gt;, F. Tipler (2005)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="276" LastEditorUserId="276" LastEditDate="2013-12-23T06:00:46.657" LastActivityDate="2013-12-23T06:00:46.657" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1223" PostTypeId="2" ParentId="1184" CreationDate="2013-12-20T11:38:24.650" Score="9" Body="&lt;p&gt;The 5 million kilometer variation in distance to the Sun may seem like a lot, 5 million km is a large distance, but it doesn't affect Earth's weather as much as the axial tilt. A 3% variation in distance leads to an ~7% variation in the intensity of solar radiation reaching Earth, give or take.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But consider the variations from Earth's axial tilt, and here I will use the example of a location at 45 deg. latitude, similar to the latitudes of cities like New York, London, Berlin, Paris, etc. The first factor is day length. Axial tilt means that during Summer the day will be longer and during Winter the day will be shorter. At 45 deg. latitude this translates into the Sun being above the horizon for a full 15.5 hours on the Summer solstice, and only 8.7 hours during the Winter solstice. That's nearly a 2:1 difference in the number of daylight hours, which totally swamps a small 7% variation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, the angle of the Sun in the winter is very low, which means that the amount of Sun falling on the ground is lower per area than if it were at a higher angle. On the Winter solstice at 45 deg. latitude the Sun's rays are diminished to 37% of their strength per square meter of ground compared to how strong they would shine directly above some location nearer the equator. But during the Summer solstice they are 97% as strong as they would be if they were directly overhead. Which is a 2.5:1 difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here you have a combined total of roughly a 400% difference in the amount of light and heat shining down on a given patch of ground every day between the Winter and the Summer.&lt;/p&gt;&#xA;" OwnerUserId="654" LastEditorUserId="654" LastEditDate="2013-12-20T22:31:46.973" LastActivityDate="2013-12-20T22:31:46.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1225" PostTypeId="1" CreationDate="2013-12-21T10:23:56.430" Score="33" ViewCount="11946" Body="&lt;p&gt;From Earth, we can distinguish the type of another galaxy very easily simply by observing the shape, colour, and structure of the galaxy when we image it. But since we are located &lt;em&gt;inside&lt;/em&gt; the Milky Way, we cannot get an &quot;outside view&quot;. Thus, how can we tell if we are located in either a spiral or an elliptical galaxy?&lt;/p&gt;&#xA;" OwnerUserId="662" LastEditorUserId="276" LastEditDate="2013-12-21T12:02:09.050" LastActivityDate="2016-02-16T05:01:16.917" Title="How can we tell that the Milky Way is a spiral galaxy?" Tags="&lt;milky-way&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1226" PostTypeId="2" ParentId="1225" CreationDate="2013-12-21T10:47:33.737" Score="36" Body="&lt;p&gt;There are several clues.&lt;/p&gt;&#xA;&lt;h2&gt;The Milky Way is a flat disk&lt;/h2&gt;&#xA;&lt;p&gt;The first one, and the simplest one, is that we live in a disk. As you can see on images like the ones from the 2MASS survey in the infrared range:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/qsTp9.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;We clearly see that we are inside a disk, since what we see when we look around us is a disk seen edge-on. It does not look like an ellipsoid or any other shape; it looks like a disk.&lt;/p&gt;&#xA;&lt;h2&gt;The Milky Way has arms&lt;/h2&gt;&#xA;&lt;p&gt;The other strong clue we have is the velocity of stars of the Milky Way, we can measure. First, we see an overall rotational motions, and, when you look closely to the patterns (like in the following image, that is observations in the radio range of the Milky Way at Green Bank Observatory), you can deduce that stars are even ordered into spirals:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/LkhXt.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2013-12-21T10:47:33.737" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1228" PostTypeId="2" ParentId="1219" CreationDate="2013-12-21T21:42:14.123" Score="9" Body="&lt;p&gt;Gaia's original &lt;a href=&quot;http://www.rssd.esa.int/SYS/docs/ll_transfers/project=PUBDB&amp;amp;id=359232.pdf&quot;&gt;science and technology report&lt;/a&gt; (see page 221, see also the &lt;a href=&quot;http://arxiv.org/pdf/astro-ph/0101235v1.pdf&quot;&gt;summary&lt;/a&gt;) gives an analysis of the Lissajous orbit. From what I understand Gaia will be placed in a small amplitude Lissajous orbit, giving it an orbital radius of $\sim400000$ km away from $\sim100000$ km along the Sun-Earth axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to the fact that this orbit is intrinsically unstable, variable radiation pressure from the sun causes &lt;em&gt;stochastic&lt;/em&gt; destabilization of the orbit. The prediction is that once in orbit, a small velocity correction of $\sim1$ m/s is needed about once a month in order to maintain the orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a &lt;a href=&quot;http://sci.esa.int/planck/34728-orbit-navigation/&quot;&gt;summary&lt;/a&gt; on Planck's orbit it seems that this applies to Planck as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That the SEL2 point is being 'crowded' doesn't seem like such a big problem to me. Once the satellites are turned off (both Planck and Herschel are no longer active) their orbit will rapidly destabilize, effectively removing then from the region. But more importantly, the orbital radius is comparable to the distance to the moon, this really is a tremendously large amount of space. Since these satellites are only a few metres across, the probability of a collision is likely negligible.  &lt;/p&gt;&#xA;" OwnerUserId="494" LastActivityDate="2013-12-21T21:42:14.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1230" PostTypeId="2" ParentId="1007" CreationDate="2013-12-22T02:38:47.537" Score="21" Body="&lt;p&gt;One of the clearest way to see what's going on is to look at a Penrose diagram of a Schwarzschild black hole. A Penrose diagram is like a map of the spacetime drawn in such a way as to preserve angles and put the light rays diagonally at $45^\circ$ angles, forming the light cones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we're mapping all of an infinite space(time) into a finite drawing, distances are necessarily distorted, but that's a small price to pay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UwOxF.gif&quot; alt=&quot;Schwarzschild causal diagram, original by Andrew Hamilton&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time goes up on the diagram, and a typical infalling trajectory is in blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because every massive object must be locally slower than light, it must stay within those light cones. Hence, no matter how you accelerate, at every point of your trajectory you must go in a direction that stays within those $45^\circ$ diagonals at that point. But once you're inside the horizon, every direction that stays within the light cones leads to the singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accelerating this way or that just means that you get to choose where you hit the singularity--a little on the left on the diagram or a little to the right. Trying to escape using high acceleration brings you closer to the light's $45^\circ$ lines, which because of time dilation will actually &lt;em&gt;shorten&lt;/em&gt; your lifespan. You'll actually get to the singularity &lt;em&gt;sooner&lt;/em&gt; if you struggle in that manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This particular image came from Prof. &lt;a href=&quot;http://casa.colorado.edu/~ajsh/home.html&quot; rel=&quot;noreferrer&quot;&gt;Andrew Hamilton&lt;/a&gt;. Note that it pictures an &lt;em&gt;eternal&lt;/em&gt; Schwarzschild black hole, i.e. one that has always and will exist. An actual black hole is formed by stellar collapse and will eventually evaporate, so its diagram will be slightly different (in particular, there will be no &quot;antihorizon&quot;). However, in all respects relevant here, it's the same situation.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2013-12-22T02:38:47.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1231" PostTypeId="2" ParentId="847" CreationDate="2013-12-22T14:43:04.537" Score="3" Body="&lt;p&gt;In short, no, you don't have to solve Kepler's equation. If you define a viewing longitude and you know the full set of Keplerian orbital elements, you can calculate if a planet will show a transit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me first clear up some of the jargon, though. The term inclination is traditionally reserved for the angle between the orbital plane and the reference plane. So unless the entire orbit is wobbling (for instance due to precession) the inclination will be constant for all positions of the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle that varies as the planet moves through the reference plane is the latitude. However, in this particular problem calculating the latitude isn't particularly useful. The more practical coordinate to look at is the height of the planet away from the plane, I'll call this $r_z$. If along your observers line of sight you find $r_z$ to be smaller than the solar radius, the planet will show a transit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working out a fully general elliptical orbit in 3 dimensions is somewhat tricky, so I'll start out with a simple Cartesian coordinate system aligned with the orbital plane. In subsequent steps I work my way back using a number of coordinate transformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First a list of the Kepler parameters and their meaning:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\nu$ - true anomaly - The anglular position with respect to perigee in the orbital plane.&lt;/li&gt;&#xA;&lt;li&gt;$\omega$ - argument of periapsis - The angular position of perigee with respect to the ascending node in the orbital plane.&lt;/li&gt;&#xA;&lt;li&gt;$\lambda_{asc}$ - the longitude of ascending node in the invariable plane ($\Omega$ in the picture).&lt;/li&gt;&#xA;&lt;li&gt;$i$ - inclination - the angle between the orbital plane and the invariable plane, such that $\lambda_{asc}$ gives the rotation axis. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/t8Zsq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in the orbital plane, the radius of the planet orbit is given as&#xA;$$&#xA;r(\nu) = \frac{a(1-e^2)}{1+e\cos(\nu)}&#xA;$$&#xA;If I define the x-axis to align with the ascending node, I can write the position vector as&#xA;$$&#xA;\vec{r}_{orb} = r(\nu) &#xA;\begin{bmatrix} &#xA;\cos(\omega+\nu) \\ &#xA;\sin(\omega+\nu) \\ &#xA;0 &#xA;\end{bmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To move to a Cartesian coordinate system aligned with the invariable plane I have to apply a &lt;a href=&quot;http://en.wikipedia.org/wiki/Rotation_matrix#Basic_rotations&quot; rel=&quot;nofollow noreferrer&quot;&gt;rotation&lt;/a&gt; of angle $i$ about the x-axis, so&#xA;$$&#xA;\vec{r} = R_x(i)\vec{r}_{orb}= r(\nu)&#xA;\begin{bmatrix} &#xA;\cos(\omega+\nu) \\ &#xA;\sin(\omega+\nu)\cos(i) \\ &#xA;\sin(\omega+\nu)\sin(i)&#xA;\end{bmatrix}&#xA;$$&#xA;So now I have an expression for $r_z$, I just need to know how $\nu$ relates to the planet longitude in the invariable plane, $\lambda_{planet}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can find the longitude of the planet relative to $\lambda_{asc}$ as&#xA;$$&#xA;\lambda'_{planet} = \arctan( \frac{r_y}{r_x} ) &#xA;$$ &#xA;such that the actual longitude is&#xA;$$&#xA;\lambda_{planet} = \lambda_{asc} + \lambda'_{planet}&#xA;$$&#xA;Inverting this last expression gives&#xA;$$&#xA;\nu(\lambda) = \arctan( \frac{\tan(\lambda - \lambda_{asc})}{\cos(i)} ) - \omega&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have all the equations to answer the question:&#xA;If the planet is along the observer's line of sight, then is has a true anomaly &#xA;of $\nu(\lambda_{obs})$, which gives a height above the plane&#xA;$$&#xA;r_z(\nu) = r(\nu)\sin(\omega+\nu)\sin(i)&#xA;$$&#xA;If $|r_{z}| &amp;lt; R_{sun}$ the planet can be seen in transit.&lt;/p&gt;&#xA;" OwnerUserId="494" LastActivityDate="2013-12-22T14:43:04.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1232" PostTypeId="2" ParentId="1197" CreationDate="2013-12-23T10:02:56.957" Score="5" Body="&lt;p&gt;Okay we can state it as follow:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Solar Radiation:&lt;br&gt;&#xA;It is the radiation from the sun which includes all wavelengths of electromagnetic radiation coming from the sun.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_wind&quot; rel=&quot;nofollow&quot;&gt;Solar Wind&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The solar wind is a stream of charged particles (a plasma) released from the upper atmosphere of the Sun. It mostly consists of electrons and protons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I can simplify it as like it is something like when you stand in front of an operating fan and this fan is rotating to spread the air all over the place, so this will make an air &lt;strong&gt;wind&lt;/strong&gt; around it, this is the wind you can sense. Back to sun, Solar wind is just like that it is the sun rotating particles produced out of the sun energy around the sun shaped by the sun's rotating magnetic field.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_flare&quot; rel=&quot;nofollow&quot;&gt;Solar Flare&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A solar flare is a sudden brightening observed over the Sun's surface or the solar limb, which is interpreted as a large energy release of up to 6 × 1025 joules of energy.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And for more clarification, it is an &lt;strong&gt;event&lt;/strong&gt; happen from the sun due too a high charged accelerated particles, this event could have some damages on the earth like the big flare happens every 11 years.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="615" LastEditorUserId="2153" LastEditDate="2015-06-16T01:49:41.790" LastActivityDate="2015-06-16T01:49:41.790" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1233" PostTypeId="1" CreationDate="2013-12-23T15:03:03.713" Score="3" ViewCount="57" Body="&lt;p&gt;Why do  the Earth or any other planet continue to revolve around the sun or a star in general and that also with constant time period? What is the reason behind it? Other than gravitational force is there any other guiding force?&lt;/p&gt;&#xA;" OwnerUserId="662" LastEditorUserId="14021" LastEditDate="2016-09-14T20:28:31.523" LastActivityDate="2016-09-14T20:28:31.523" Title="Why the Earth revolves?" Tags="&lt;earth&gt;&lt;rotation&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="2" ClosedDate="2013-12-23T19:36:56.353" ContentLicense="CC BY-SA 3.0" />
  <row Id="1234" PostTypeId="1" AcceptedAnswerId="1239" CreationDate="2013-12-23T15:10:05.213" Score="21" ViewCount="2843" Body="&lt;p&gt;Why do planets rotate around a star in a specific elliptical orbit with the star at one of it's foci? Why isn't the orbit a circle?&lt;/p&gt;&#xA;" OwnerUserId="662" LastEditorUserId="537" LastEditDate="2013-12-23T16:45:25.420" LastActivityDate="2020-11-18T20:53:31.587" Title="Why are orbits elliptical instead of circular?" Tags="&lt;star&gt;&lt;orbit&gt;&lt;planet&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1235" PostTypeId="2" ParentId="1234" CreationDate="2013-12-23T16:24:48.317" Score="9" Body="&lt;p&gt;It is posible for a planet to have a circular orbit, a circle, after all, is an ellipse where both foci are in the same place; this is known as having an &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_eccentricity&quot; rel=&quot;noreferrer&quot;&gt;eccentricity&lt;/a&gt; of 0&lt;/strong&gt;. Eccentricity is defined in the following way:&#xA;$$ e = \frac{r_{a} - r_{p}}{r_{a} + r_{p}} $$&#xA;where $r_{a}$ is the apoapsis (farthest point in the orbit from the center of mass), and $r_{p}$ is the periapsis (the closest distance). Just to build some intuition here, if the apoapsis is twice the distance of the periapsis, the eccentricity will be $e=0.333$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From all the planets of the solar system, &lt;strong&gt;Venus&lt;/strong&gt;, with an eccentricity of &lt;strong&gt;0.007&lt;/strong&gt; has the most circular orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to why all orbits aren't round, it comes down to &lt;strong&gt;kinetic energy&lt;/strong&gt;. The kinetic energy is proportional to the square of the speed. In the orbital plane and in polar coordinates about the star, we can decompose this is into a combination of radial velocity $\dot{r}$ and angular velocity $\dot{\phi}$:&#xA;$$v^2 = \dot{r}^2 + r^2\dot{\phi}^2\text{.}$$&#xA;Since circles have constant radii, for the orbit to be circular around the star, the planet's radial velocity must be exactly zero. Additionally, the angular speed must be such that the &lt;strong&gt;centrifugal force&lt;/strong&gt; in the corotating frame exactly balances the gravitational force--a little more or a little less, the imbalance will change the radial velocity, spoiling the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the fact that velocities vary for a large number of reasons, it's no wonder that only a few orbits end up being circular, and considering that actual orbits &lt;strong&gt;change with time&lt;/strong&gt;, we know they can't stay this way for long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're looking for a mathematical proof, &lt;a href=&quot;http://www.askamathematician.com/2010/08/q-why-are-orbits-elliptical-why-is-the-sun-in-one-focus-and-whats-in-the-other/&quot; rel=&quot;noreferrer&quot;&gt;this link shares some details about it&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an image showing eccentricity of some bodies in the solar system extracted from &lt;a href=&quot;http://www.spiralwishingwells.com/guide/planets.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OrcTm.gif&quot; alt=&quot;Some solar system bodies and their excentricities&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="537" LastEditorUserId="661" LastEditDate="2013-12-25T00:12:59.330" LastActivityDate="2013-12-25T00:12:59.330" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1236" PostTypeId="1" AcceptedAnswerId="1237" CreationDate="2013-12-23T17:38:13.063" Score="15" ViewCount="462" Body="&lt;p&gt;&lt;a href=&quot;http://news.yahoo.com/first-exomoon-possibly-glimpsed-110000768.html&quot;&gt;This article&lt;/a&gt; got me thinking, can a planet hold a moon in orbit if it is just floating in the galaxy by itself not as part of a star system? Can a celestial body even qualify as a planet if it is floating around by itself?&lt;/p&gt;&#xA;" OwnerUserId="257" LastEditorUserId="2618" LastEditDate="2017-03-07T01:10:28.107" LastActivityDate="2017-03-07T01:10:28.107" Title="Is it possible for a moon to orbit a planet floating free in the galaxy rather than orbiting a star" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;exoplanet&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1237" PostTypeId="2" ParentId="1236" CreationDate="2013-12-23T18:31:40.203" Score="19" Body="&lt;p&gt;The answer is &lt;strong&gt;Yes&lt;/strong&gt;.. Planets that don't orbit around a star are known as &lt;strong&gt;Rogue Planets&lt;/strong&gt;. There is nothing preventing a rogue planet from having one or many moons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not so long ago, the first &lt;strong&gt;candidate for a free-floating exoplanet-exomoon system&lt;/strong&gt; was presented in &lt;a href=&quot;http://arxiv.org/abs/1312.3951&quot;&gt;this paper&lt;/a&gt;. It looks like a gas giant several times larger than Jupiter with a sub-Earth mass moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/0709.0945&quot;&gt;Another study&lt;/a&gt; calculated and simulated scenarios where planets where ejected from their orbits around a star and concluded that around &lt;strong&gt;five percent&lt;/strong&gt; of Earth-sized planets who are accompanied by Moon-sized natural satellites would retain them after the event.&lt;/p&gt;&#xA;" OwnerUserId="537" LastActivityDate="2013-12-23T18:31:40.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1238" PostTypeId="1" CreationDate="2013-12-23T22:07:19.707" Score="15" ViewCount="255" Body="&lt;p&gt;As far as I can tell, we do not yet have the precision to even put reasonable bounds on an exoplanets obliquity, but wikipedia seems to indicate that this may be possible in the &quot;near future.&quot; It seems like this would have to be accomplished by direct imaging, either by directly observing rotational flattening of an exoplanets, or by looking for moons and assuming that the planet is tidally locked to the same plane as its satellite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How close would you estimate we are to this kind of precision? Are there other approaches to measuring planetary obliquity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, I'm not expecting a definite answer. Just wondering if anyone knows of any research in this area or has any thoughts about it.&lt;/p&gt;&#xA;" OwnerUserId="674" LastActivityDate="2017-01-08T16:09:37.207" Title="How close are we to having the technology to measure planetary obliquity for exoplanets?" Tags="&lt;exoplanet&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1239" PostTypeId="2" ParentId="1234" CreationDate="2013-12-24T08:35:57.890" Score="16" Body="&lt;p&gt;Assume the planet has a negligible mass compared to the star, that both are spherically symmetric (so Newton's law of gravitation holds, but this normally happens to a very good approximation anyway), and that there aren't any forces besides the gravity between them. If the first condition does not hold, then the acceleration of each is going to be towards the &lt;a href=&quot;http://en.wikipedia.org/wiki/Barycentric_coordinates_%28astronomy%29&quot;&gt;barycenter&lt;/a&gt; of the system, as if barycenter was attracting them a gravitational force with a certain reduced mass, so the problem is mathematically equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the star to be at the origin. By Newton's law of gravitation, the force is $\mathbf{F} = -\frac{m\mu}{r^3}\mathbf{r}$, where $\mathbf{r}$ is the vector to the planet, $m$ is its mass, and $\mu = GM$ is the standard gravitational parameter of the star.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Conservation Laws&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Because the force is purely radial $(\mathbf{F}\parallel\mathbf{r})$, angular momentum $\mathbf{L} = \mathbf{r}\times\mathbf{p}$ is conserved:&#xA;$$\dot{\mathbf{L}} = \frac{\mathrm{d}}{\mathrm{d}t}\left(\mathbf{r}\times\mathbf{p}\right) = m(\dot{\mathbf{r}}\times \dot{\mathbf{r}}) + \mathbf{r}\times\mathbf{F} = \mathbf{0}\text{.}$$&#xA;If the initial velocity is nonzero and the star is at the origin, then in terms of the initial position and velocity, the orbit must be confined to the plane of all points with vectors $\mathbf{x}$ from the origin that satisify $\mathbf{L}\cdot\mathbf{x} = 0$. If the initial velocity is zero, then the motion is purely radial, and we can take any one of infinitely many planes that contain the barycenter and initial position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total orbital energy is given by&#xA;$$\mathcal{E} = \frac{p^2}{2m} - \frac{m\mu}{r}\text{,}$$&#xA;where the first term part is the kinetic energy and the second term is the gravitational potential energy of the planet. Its conservation, as well as the fact that it invokes the correct potential energy, can be proven by the fundamental theorem of calculus for line integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the Laplace-Runge-Lenz vector to be&#xA;$$\mathbf{A} = \mathbf{p}\times\mathbf{L} - \frac{m^2\mu}{r}\mathbf{r}\text{.}$$&#xA;It is also conserved:&#xA;$$\begin{eqnarray*}&#xA;\dot{\mathbf{A}} &amp;amp;=&amp;amp; \mathbf{F}\times\mathbf{L} + \mathbf{p}\times\dot{\mathbf{L}} - \frac{m\mu}{r}\mathbf{p} + \frac{m\mu}{r^3}(\mathbf{p}\cdot\mathbf{r})\mathbf{r}\\&#xA;&amp;amp;=&amp;amp; -\frac{m\mu}{r^3}\underbrace{\left(\mathbf{r}\times(\mathbf{r}\times\mathbf{p})\right)}_{(\mathbf{r}\cdot\mathbf{p})\mathbf{r} - r^2\mathbf{p}} - \frac{m\mu}{r}\mathbf{p} + \frac{m\mu}{r^3}(\mathbf{p}\cdot\mathbf{r})\mathbf{r}\\&#xA;&amp;amp;=&amp;amp; \mathbf{0}\text{.}&#xA;\end{eqnarray*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, let's also take $\mathbf{f} = \mathbf{A}/(m\mathcal{E})$, which has the same units as $\mathbf{r}$, and since $\mathbf{L}\cdot\mathbf{f} = 0$, it lies along the orbital plane. As it's a conserved vector scaled by a conserved scalar, it's easy to show that $\mathbf{f}$ is conserved as well, as long as $\mathcal{E}\neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Simplifying&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;By employing the vector triple product, we can write&#xA;$$\begin{eqnarray*}&#xA;\frac{1}{m}\mathbf{A} &amp;amp;=&amp;amp; \frac{1}{m}\left[p^2\mathbf{r}-(\mathbf{p}\cdot\mathbf{r})\mathbf{p}\right] -\frac{m\mu}{r}\mathbf{r}\\&#xA;&amp;amp;=&amp;amp; \left(\mathcal{E}+\frac{p^2}{2m}\right)\mathbf{r} - \frac{1}{m}\left(\mathbf{p}\cdot\mathbf{r}\right)\mathbf{p}\\&#xA;\mathcal{E}(\mathbf{f}-\mathbf{r}) &amp;amp;=&amp;amp; \left(\frac{p^2}{2m}\right)\mathbf{r} - \frac{1}{m}\left(\mathbf{p}\cdot\mathbf{r}\right)\mathbf{p}\text{,}&#xA;\end{eqnarray*}$$&#xA;the norm-squared of which is easy to crank out:&#xA;$$\mathcal{E}^2|\mathbf{f}-\mathbf{r}|^2 = \left(\mathcal{E} + \frac{m\mu}{r}\right)^2r^2\text{,}$$&#xA;where $\mathcal{E}$ was used throughout to switch between kinetic and potential terms.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Why Ellipses?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Since $\mathcal{E}$ is energy relative to infinity, to have a bound orbit we need $\mathcal{E}&amp;lt;0$. Thus, from the previous section, $|\mathbf{f}-\mathbf{r}| = -\mathcal{E}^{-1}\left(\mathcal{E}r + m\mu\right)$ and therefore&#xA;$$|\mathbf{f}-\mathbf{r}| + |\mathbf{r}| = -\frac{m\mu}{\mathcal{E}}\text{,}$$&#xA;which defines an ellipse with foci $\mathbf{0},\,\mathbf{f}$ and major axis $2a=-m\mu/\mathcal{E}$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Why Not Circles?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The circle is a special case where the foci are the same point, $\mathbf{f} = \mathbf{0}$, which can be restated as&#xA;$$\mathcal{E} = -\frac{1}{2}\frac{m\mu}{r} = -\frac{p^2}{2m}\text{.}$$&#xA;In other words, circular orbits require the orbital energy to be the negative of the kinetic energy. This is possible, but almost certain not to hold exactly. Since any values of $\mathcal{E}&amp;lt;0$ are allowed for bound orbits, there are many more ways to have elliptic orbits. (Although some of them would actually crash because the star and planet have positive size.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that hyperbolic orbits have $\mathcal{E}&amp;gt;0$, and we can still find the foci using the above method, though being careful with the signs. For $\mathcal{E}=0$, the second focus $\mathbf{f}$ is undefined because this is a parabolic orbit, and parabolas only have one focus within a finite distance from the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, the &lt;em&gt;eccentricity vector&lt;/em&gt; $\mathbf{e} = \mathbf{A}/(m^2\mu)$ is an alternative choice for the LRL vector; as the name suggests, its magnitude is the orbital eccentricity.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="332" LastEditDate="2014-01-01T13:56:13.333" LastActivityDate="2014-01-01T13:56:13.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1240" PostTypeId="1" AcceptedAnswerId="1329" CreationDate="2013-12-24T15:10:51.093" Score="9" ViewCount="947" Body="&lt;p&gt;When we see a star it looks much bigger in diameter to us than it really is, this picture (&lt;a href=&quot;http://www.gemini.edu/sciops/future-instrumentation&quot; rel=&quot;noreferrer&quot;&gt;extracted from here&lt;/a&gt;) explains what I mean:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/K6GSj.jpg&quot; alt=&quot;Star as we see it and as it really is&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Notice that the point we see in the skynight, represented by the yellow outer circle, would actually encompass even planets orbiting the actual star, represented by the black dot in the center. In this case we see one planet inside the white circle.&lt;/p&gt;&#xA;&lt;p&gt;My question is, can the effect of distance and brightness be measured as to know how far would we must be to watch a star that encompasses neptune?, the earth? or does it depend on different factors?&lt;/p&gt;&#xA;" OwnerUserId="537" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2014-01-06T17:57:52.060" Title="If we were to see the Sun with our naked eyes from the Orion belt, would all planets be encompassed inside the star? Is this calculable?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;observational-astronomy&gt;&lt;distances&gt;&lt;size&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1241" PostTypeId="1" CreationDate="2013-12-24T17:53:35.807" Score="23" ViewCount="3633" Body="&lt;p&gt;When the weather is clear, we can look at the stars. And we normally would see several thousands of them, they all being more than a $\textrm{pc}$ away from us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there are &lt;a href=&quot;http://en.wikipedia.org/wiki/Globular_cluster&quot; rel=&quot;noreferrer&quot;&gt;globular clusters&lt;/a&gt;, which consist of some $10^5$ stars concentrated is a few $\textrm{pc}$ area. From oustside they look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nbyj1.jpg&quot; alt=&quot;A globular cluster&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, how do they (and the whole sky) look from inside? Imagine that solar system is inside such a cluster. Would it be a big change? Would there be a significant difference between night and day? Would it be easier or harder to study astronomy there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. The credits for the question go to Ross Church.&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2017-01-19T05:15:06.580" Title="How would the night sky look from inside a globular cluster?" Tags="&lt;amateur-observing&gt;&lt;fundamental-astronomy&gt;&lt;globular-clusters&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1242" PostTypeId="2" ParentId="1241" CreationDate="2013-12-26T04:11:43.330" Score="6" Body="&lt;p&gt;In a typical position in a globular cluster (maybe halfway between center and edge), there'd be many more bright stars in the sky due to the star density. These would be distributed unevenly in the sky, with more light coming from the center of the globular cluster. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the globular cluster's orbit, we might be able to see the Milky Way face-on. This would be beautiful and would take up a large fraction of the sky. However, it wouldn't be particularly bright--the apparent magnitude would be $\approx -3$, the brightness of Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If our alien eyes were good at discerning wavelengths, the sky would seem &lt;em&gt;redder&lt;/em&gt;. Globular clusters have many old stars (more red), and very few massive stars (fewer blue) since they are not sites of active star formation.&lt;/p&gt;&#xA;" OwnerUserId="608" LastActivityDate="2013-12-26T04:11:43.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1243" PostTypeId="1" CreationDate="2013-12-26T04:54:43.590" Score="2" ViewCount="142" Body="&lt;p&gt;I have a more mathematical background but have always been interested in astrophysics, cosmology, astronomy, etc. There are a lot of popular books on the subject, but I am getting more interested in looking at a more rigorous approach to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the &lt;strong&gt;best introductory textbooks&lt;/strong&gt; on the subject? (Calculus+ and assuming introductory courses in Mechanics and Heat and Electricity and Magnetism, etc.)&lt;/p&gt;&#xA;" OwnerUserId="684" LastEditorUserId="6" LastEditDate="2016-05-25T15:41:10.383" LastActivityDate="2016-05-25T15:41:10.383" Title="Good Introductory Textbooks" Tags="&lt;astrophysics&gt;" AnswerCount="0" CommentCount="10" ClosedDate="2013-12-27T12:07:37.517" ContentLicense="CC BY-SA 3.0" />
  <row Id="1244" PostTypeId="1" CreationDate="2013-12-26T11:14:56.110" Score="8" ViewCount="236" Body="&lt;p&gt;So curious to know why comet ISON disintegrated? Well, it went near the Sun but was the Sun the only reason it &lt;em&gt;died&lt;/em&gt;? Or is there any other reason for it's fading away?&lt;/p&gt;&#xA;" OwnerUserId="662" LastEditorUserId="13" LastEditDate="2013-12-26T13:19:07.023" LastActivityDate="2013-12-30T19:36:32.177" Title="What caused the comet ISON to disintegrate?" Tags="&lt;the-sun&gt;&lt;comets&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1245" PostTypeId="2" ParentId="1244" CreationDate="2013-12-26T13:16:47.857" Score="10" Body="&lt;p&gt;ISON slingshot past the Sun close enough (around one Solar diameter) for its surface temperature pointed towards it to reach 6,000 °C and it wasn't a fast spinner either to help this temperature average out along its total surface. In a nutshell, it melted to its nucleus into a hot gaseous plasma of materials it was composed of long before it reached perihelion. These materials still followed more or less same ballistic trajectory prior to perihelion, it merely elongated substantially due to strong gravitational influence of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it was approaching perihelion, the Sun's tidal forces broke this plasma into a wave-like pattern, increasing its surface area and exposing its materials to solar winds (and possibly hit by one CME, Coronal Mass Ejection that was detected near comet ISON on the perihelion day), changing orbit of different mass and size particles differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was apparent past perihelion when the comet failed to create tails pointing away from the direction of solar winds like comets with a nucleus do, but instead started splitting into two  masses inclined roughly 90° to each other, suggesting different mass particles separated and the comet was indeed torn apart by Sun's gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since most of its materials thus gained angular momentum different enough to negate its own gravity, and they weren't bound to each other as they would if it remained solid, it entered a spin of death, slowly evaporated and dissipated entirely. No nucleus was observed to survive the perihelion, end even Hubble wasn't successful in finding any of its remains, further proving this fate of the comet ISON.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also additionally described with the help of ISON perihelion videos from various spacecraft in &lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/2013/12021624-multiple-views-of-comet-ison.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Emily Lakdawalla's blog post&lt;/a&gt;, and since a picture tells a thousand words, here's my capture of a frame from &lt;a href=&quot;http://sdo.gsfc.nasa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;SDO&lt;/a&gt; (Solar Dynamics Observatory) that matches the time, estimated size and distance, suggesting it could indeed be comet ISON being torn apart by tidal forces on its perihelion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;a href=&quot;https://i.stack.imgur.com/23zTN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2owp3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;sup&gt;A frame from SDO/AIA - 4500 203/11/28 18:00:07 (ISON pre-perihelion time). &lt;a href=&quot;https://i.stack.imgur.com/23zTN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Click on image to download full frame JPeG&lt;/a&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This SDO frame (&lt;code&gt;SDO/AIA - 4500 203/11/28 18:00:07&lt;/code&gt;) seems to have been missed by all other astronomers that were following the demise of ISON, likely because SDO was just positioning itself for ISON's perihelion at the time and this isn't a complete frame. So, in a sense, this is &lt;a href=&quot;https://astronomy.stackexchange.com/&quot;&gt;Astronomy Stack Exchange&lt;/a&gt; exclusive. :)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: Turns out it was a &lt;em&gt;reflection / filter defect&lt;/em&gt; after all. I would like to point out that I never claimed this SDO/AIA frame shows comet ISON, merely that it seemed to me that it might. Position did seem odd, but not being equipped with all required SDO data, I had no way of confirming that. Clearly, I'm not experienced in interpreting SDO data, let alone a qualified expert, but my perseverance did pay off regardless and &lt;a href=&quot;http://sdo.gsfc.nasa.gov/mission/team/team_pesnell.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dean Pesnell of the SDO team&lt;/a&gt; kindly looked into this and wrote an eye-opening explanation for this SDO/AIA frame in his SDO blog post &lt;a href=&quot;http://sdoisgo.blogspot.com/2013/12/was-comet-ison-seen-in-aia-4500.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Was Comet ISON Seen in AIA 4500?&lt;/a&gt; A polite debunking of my suspicions, an interesting read, and another win for science over one layman's speculation. Kudos to Dean Pesnell and the &lt;a href=&quot;https://twitter.com/NASA_SDO&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA SDO on Twitter&lt;/a&gt;!&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2013-12-30T19:36:32.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1246" PostTypeId="2" ParentId="1241" CreationDate="2013-12-26T14:20:37.493" Score="4" Body="&lt;p&gt;Let us assume the data for a globular cluster to be equivalent to that of &lt;a href=&quot;http://en.wikipedia.org/wiki/Messier_13&quot; rel=&quot;nofollow&quot;&gt;M13&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given 300,000 stars and a radius of 1 ly, let us assume uniform density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another assumption is to consider all stars to be Sun-like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number density can be calculated as $$\rho = \frac{300000*3}{4\pi(1\ ly)^3} \approx 9 \times10^{-44}\ m^{-3} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, using the formula for mean free path $$\lambda = \frac{1}{\sqrt{2}\pi D^2 \rho}$$, and using $D = 14 \times 10^{8}\ m$,&#xA;we have $$\lambda \approx 10^{24}\ m \equiv 10^{8}\ ly$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you don't have to be a genius to see how the sky will look if the nearest stars to you are this far. It will be almost like our own sky but just a lot of stars in every direction. There will be no special hike in flux received.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number density is much higher at the centre. But even if you assume the density at centre to be orders higher than the average value, the uninteresting view exists! As Zack pointed out, we'll have a lot of long wavelength light due to abundance of old stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is mentioned that the optical view won't be too attractive, but there is a certain thrill in being at the centre of a globular cluster. It is very difficult to stay there for a long time avoiding collisions or surviving the radiation and stellar winds from collisions and novae which are frequent for globular cluster.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2013-12-26T14:20:37.493" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1247" PostTypeId="2" ParentId="1241" CreationDate="2013-12-26T17:50:42.123" Score="14" Body="&lt;p&gt;Globular clusters occupy an interesting place in the spectrum of composite stellar systems. As you point out, they are highly concentrated populations of stars, and seem to lack any dark matter component, unlike more massive dwarf galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Binary interactions become very important in simulating globular clusters, and interestingly enough (maybe unsurprisingly), the one example of a discovery of a planet found in a globular cluster has been around a binary star system (see: &lt;a href=&quot;http://en.wikipedia.org/wiki/PSR_B1620-26_b&quot;&gt;PSR B1620-26 b&lt;/a&gt;; this circumbinary planet was found orbiting a pulsar and a white dwarf.). This is not to say there are not other examples, however, this was the easiest for me to come across. I would be interested to know how common this situation is, and in addition, how stable it is given the potentially highly chaotic environment it lives in. These speculations don't answer your question, but I thought it interesting enough to bring up as evidence in favor of your question not being an unreasonable one to ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the wiki page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Globular clusters can contain a high density of stars; on average about 0.4 stars per cubic parsec, increasing to 100 or 1000 stars per cubic parsec in the core of the cluster.&lt;a href=&quot;http://www.astro.keele.ac.uk/workx/globulars/globulars.html&quot;&gt;[26]&lt;/a&gt; The typical distance between stars in a globular cluster is about 1 light year,&lt;a href=&quot;http://www.dur.ac.uk/ian.smail/gcCm/gcCm_intro.html&quot;&gt;[27]&lt;/a&gt; but at its core, the separation is comparable to the size of the Solar System (100 to 1000 times closer than stars near the Solar System).&lt;a href=&quot;http://www.eso.org/public/news/eso0107/&quot;&gt;[28]&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This seems to indicate to me that location within the globular cluster would matter quite a bit. If at the core the average distance between stars is about three thousand times closer than our nearest neighbor is to our sun (my estimate to give some perspective: a few lightyears to Proxima Centauri divided by 100 is about 3000AU (about 100 times further than Pluto from the sun)), then stable orbits may be shifted inward, or simply may not exist due to two-body interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if life were to exist (an assumption we're going to make for the purposes of your question), one would see a very different night sky. According to this &lt;a href=&quot;http://arxiv.org/abs/1302.1714&quot;&gt;paper&lt;/a&gt;, the number density profile of stars within the globular cluster M92 follows a Wilson Profile fairly well, which has the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f_{W} = A\{ e^{-aE} - e^{-aE_{0}} [1 - a(E-E_{0}) ] \} $$&#xA;where $E \le E_{0}$. E is the specific energy of the star:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E = v^{2}/2 + \Phi(r) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and where $\Phi(r)$ is the mean-field gravitational potential, determined from the Poisson equation. For each family of models, the constants A, $E_{0}$, and a in the above distribution function define two dimensional scales (a typical radius and a typical mass or velocity) and one dimensionless parameter, the central depth of the potential well (related to the concentration parameter) (all of the information is taken from the paper I've linked).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to be the case that globular clusters are not &quot;simple stellar populations&quot;, in that they are usually made of multiple generations (sources: &lt;a href=&quot;http://arxiv.org/pdf/1201.6526v1.pdf&quot;&gt;1&lt;/a&gt;,&lt;a href=&quot;http://arxiv.org/pdf/1106.3174v1.pdf&quot;&gt;2&lt;/a&gt;). However, globular clusters generally consist of &lt;a href=&quot;http://en.wikipedia.org/wiki/Metallicity&quot;&gt;population II&lt;/a&gt; stars and are older stellar systems when compared to other star clusters. I bring all of this up because in addition to the number density of stars, the stellar distribution of stellar types would certainly be an important factor in how the night sky would look. If you lived a thousandth of a lightyear from a blue supergiant, you could imagine that that would make a huge difference in what you would see on a day to day basis. At the same distance, a supergiant star is on the order of $10^{5}$ times the luminosity of our sun (and therefore is $10^{5}$ the flux, since $L \propto f$ holding $D_{L}$ constant). At the same distance as our sun, the magnitude of a star with $10^{5}$ times the flux would have an apparent magnitude of about -38 (I used Rigel as my test case; this produces a star in our sky which is 12 magnitudes brighter than our sun). Moving this to the average distance between stars at the center of a globular cluster we would get an apparent magnitude of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$M = -6.43 $ from $m=-38$ at distance of $d=1$AU.&lt;/li&gt;&#xA;&lt;li&gt;$m=-26.43$ from $M=-6.43$ at a distance of $d = .00326 ly$ (the new average distance between stars at the center of a globular cluster)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In other words, a blue supergiant at an average distance between stars within a globular cluster would appear to be as bright as our sun is to us! This is absolutely crazy. Depending on where it is in relation to the sun, it could effective cause two days, or potentially one day which is greater than half the time it takes your planet to rotate once. I would image that this would certainly interfere with observing in optical (and shorter wavelengths).&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2013-12-26T17:50:42.123" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1249" PostTypeId="1" CreationDate="2013-12-27T07:26:27.440" Score="3" ViewCount="23114" Body="&lt;p&gt;I read that our day is not exactly 24 hours long. If it is not then why weather seasons come at the same time (Summer, Winter, Autumn Spring) ? how it is equalized ?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; Sorry I could not make it clear. My meaning was if 3.8 seconds are shorter of everyday the how is the calculation made that summer starts in May and winter start december everytime why it does not move further since. How time is neutralized ? I hope you guys understood this time.&lt;/p&gt;&#xA;" OwnerUserId="460" LastEditorUserId="1891" LastEditDate="2019-04-02T11:52:21.440" LastActivityDate="2019-04-02T11:52:21.440" Title="What is the exact time Earth takes to revolve around its axis?" Tags="&lt;earth&gt;&lt;space-time&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ClosedDate="2013-12-27T20:05:55.233" ContentLicense="CC BY-SA 4.0" />
  <row Id="1250" PostTypeId="2" ParentId="1249" CreationDate="2013-12-27T11:38:31.483" Score="7" Body="&lt;p&gt;Our day is 23 hours and 56 minutes long, and slowing by an infinitesimal (but measurable) amount each year due to tidal losses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our day has a connection with the weather, in that the sun drives all our weather systems, so heating over each part of the globe happens every day, but aside from that, your question doesn't make much sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weather changes may come at the same time where you live (on a 24 hour cycle) but here in Scotland, we still have low reliability on even a 3 day weather forecast, because the weather systems that impact the UK are so complex as heating from the sun drives various air flows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;After your update&lt;/strong&gt;, I still cannot understand what you mean. There is slippage in accuracy, but this is counteracted by leap seconds and leap years. It has nothing to do with weather. The shortest and longest days happen when they happen and are measurable. They help us define the year.&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="43" LastEditDate="2013-12-30T23:41:59.863" LastActivityDate="2013-12-30T23:41:59.863" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1252" PostTypeId="5" CreationDate="2013-12-27T13:57:41.620" Score="0" Body="&lt;p&gt;Analysis of data is a process of inspecting, cleaning, transforming, and modeling data with the goal of discovering useful information, suggesting conclusions, and supporting decision making. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Data analysis has multiple facets and approaches, encompassing diverse techniques under a variety of names, in different business, science, and social science domains. In astronomy, data in the form of images, calculations and all kind of waves is abundant, it's analysis is a priority for the study of space.&lt;/p&gt;&#xA;" OwnerUserId="537" LastEditorUserId="537" LastEditDate="2013-12-27T14:13:31.913" LastActivityDate="2013-12-27T14:13:31.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1253" PostTypeId="4" CreationDate="2013-12-27T13:57:41.620" Score="0" Body="Analysis of data is a process of inspecting, cleaning, transforming, and modeling data with the goal of discovering useful information, suggesting conclusions, and supporting decision making. " OwnerUserId="537" LastEditorUserId="537" LastEditDate="2013-12-27T14:13:37.447" LastActivityDate="2013-12-27T14:13:37.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1255" PostTypeId="2" ParentId="1249" CreationDate="2013-12-27T19:42:36.633" Score="6" Body="&lt;p&gt;The length of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_day&quot; rel=&quot;noreferrer&quot;&gt;solar day&lt;/a&gt; is very close to 24 hours. A solar day is the time it takes the Sun to return to the same position in the sky. The exact length varies slightly over the course of the year, because the Earth's orbit around the Sun is not perfectly circular; the tilt of the Earth's axis with respect to its orbit also has an effect. See the linked article for details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Sidereal_day&quot; rel=&quot;noreferrer&quot;&gt;sidereal day&lt;/a&gt; is slightly shorter. It's the time it takes for a distant star to return to the same position in the sky, and it differs from the solar day because the Earth revolves around the Sun as it rotates on its own axis; there is one more sidereal day than solar day each year. A sidereal day is approximately 23 hours, 56 minutes, 4.0916 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weather can actually have a minuscule effect on the Earth's rotation, as the mass of the atmosphere is distributed differently. This effect is barely measurable. I don't think that's the kind of &quot;weather changes&quot; you were asking about; can you clarify? Obviously it gets colder at night, and that's controlled by the length of the solar day.&lt;/p&gt;&#xA;" OwnerUserId="53" LastActivityDate="2013-12-27T19:42:36.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1257" PostTypeId="1" AcceptedAnswerId="1312" CreationDate="2013-12-28T00:13:28.147" Score="8" ViewCount="163" Body="&lt;p&gt;&lt;a href=&quot;http://www.space.com/23985-alien-planet-nearby-brown-dwarfs.html&quot; rel=&quot;noreferrer&quot;&gt;This article&lt;/a&gt; left me wondering what do they mean by the term alien planet? Aren't all planets simply alien by default, or do they mean planets outside our solar system are alien? Do they mean there is alien life there? Is this a a new keyword the astronomy community is using to generate more buzz and get more attention to their articles? &lt;/p&gt;&#xA;" OwnerUserId="257" LastEditorUserId="257" LastEditDate="2013-12-29T19:21:15.023" LastActivityDate="2014-01-05T16:05:50.187" Title="What does the term alien planet actually refer to?" Tags="&lt;planet&gt;&lt;exoplanet&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1258" PostTypeId="2" ParentId="1257" CreationDate="2013-12-28T02:33:42.963" Score="6" Body="&lt;p&gt;Very interesting article!. The term &lt;strong&gt;alien planet&lt;/strong&gt; is not a scientific one, so I think you already answered your own question by saying that it &lt;em&gt;was used in the article to generate more buzz&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the English language the word is used to describe something that's not familiar, so you could argue that this is in fact an alien planet, but because of the subject in question being an exoplanet I would bet they did this as a technique to get more attention.&lt;/p&gt;&#xA;" OwnerUserId="537" LastActivityDate="2013-12-28T02:33:42.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1259" PostTypeId="1" AcceptedAnswerId="1260" CreationDate="2013-12-28T03:07:39.797" Score="10" ViewCount="5387" Body="&lt;p&gt;In my area, from December 17th - 26th, you can see the space station every day, twice&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4G7WA.png&quot; alt=&quot;Screenshot 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But from December 27th - January 6th, you can only see it twice&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3aU2S.png&quot; alt=&quot;Screenshot 2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, why does it appear for almost the whole month of December, and then not appear for a few days? Is it based on my location? I live in Alberta, Canada if that helps. &lt;/p&gt;&#xA;" OwnerUserId="701" LastEditorUserId="537" LastEditDate="2013-12-28T16:43:50.850" LastActivityDate="2014-11-07T22:40:27.437" Title="Why can you see the space station on some days but not on others?" Tags="&lt;observational-astronomy&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1260" PostTypeId="2" ParentId="1259" CreationDate="2013-12-28T09:34:01.260" Score="10" Body="&lt;p&gt;Two basic conditions need to be met for anyone on the globe to observe the International Space Station (ISS) from any location:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;ISS needs to pass roughly overhead of your location, and&lt;/li&gt;&#xA;&lt;li&gt;it needs to do that during night so it's visible to the naked eye&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, obviously there are other requirements, like e.g. weather (if it's overcast we won't be able to see it through the clouds), but those screens you attach don't account for that, since it would be nigh impossible to predict weather so far in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's discuss the two basic conditions mentioned a bit more. ISS has an orbital inclination of 51.65°. Why is that is explained in &lt;a href=&quot;https://space.stackexchange.com/a/2536/49&quot;&gt;this answer on Space.SE&lt;/a&gt;, but to quickly recap, this orbital inclination was chosen because it's near ideal for launches from the Baikonur Cosmodrome, where most launches towards the ISS are made from, it is still achievable from the US launch facilities and most US built launchers, and it allows the ISS to go over majority of the populated areas on Earth. So it was decided both for economic reasons, as well as to allow ISS to do as much science as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what this inclination means is that the ISS will make a ground track (path along the surface of the Earth directly below it) that moves towards the West with each new orbit, even though the ISS itself moves West to East (or saying it otherwise, it's in a prograde Low-Earth orbit). Its ground track could look something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/tkXR0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Ground track of the International Space Station with an area where it is observable from the Earth's surface&lt;br /&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;shown in the white circle (Image source: &lt;a href=&quot;http://www.isstracker.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ISS Tracker&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ISS orbital period, the time it takes it to circumnavigate the globe and complete one orbit, is roughly 93 minutes. As you can see in the screen grab from the &lt;a href=&quot;http://www.isstracker.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ISS Tracker&lt;/a&gt; above, its ground track moves westward on each new orbit, which is due to Earth's rotation on its axis, so this shift towards the West will be equal to how much the Earth rotates in the meantime, which in those 93 minutes comes out at roughly 22,5° or a good 2,505 km (1,556 mi) of the Earth's total equatorial circumference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But ISS's orbit isn't Sun-synchronous, meaning it won't be over any given Earth latitude at the same local mean solar time, so when it will be observable depends also on its orbital position relative to the position of the Sun. We require that it is dark, remember? ISS isn't bright enough (doesn't reflect enough light with its truss and solar panels) during the day to be brighter than the day skies are. So we have another parameter, which is local time. As you probably anticipated by now, the length of the day isn't really equal at any day of the year, and moving further away from the equator only increases this difference of the length of the day in winter, or the length of it during summer. This is another range or values that your software you're attaching print screens from has to account for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when your software collects data from you regarding your geolocation and current date and time (might do that automatically if you're using it on a device that has GPS enabled), it will establish the times (timetable) of when the ISS ground track is over or around your area, and if that happens during night. The ISS doesn't have to be exactly overhead, at zenith, so it will also calculate angle to the zenith for any area that the ISS goes over, at which you would still be able to see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Understanding this, even if we don't look at your timetable, we could establish that if you're at latitude to the equator that enables you to see the ISS, you'd be able to do that roughly a few times in a row separated by around 93 minutes, then wait for the ISS ground track to come back towards your longitude (16 orbits or about one day and an hour), and do that during night when you'd be able to observe it. Nights are long now during December at Northern latitudes, so you see a long row of couples of such chances during each night. As long as the weather holds, of course.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2014-11-07T20:10:09.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1261" PostTypeId="1" CreationDate="2013-12-28T15:02:22.933" Score="4" ViewCount="234" Body="&lt;p&gt;Could you explain to me the life course for a massive star (30-40 solar masses) from birth to death using the HR Diagram (by showing key events in its life)?&lt;/p&gt;&#xA;" OwnerUserId="414" LastEditorUserId="31410" LastEditDate="2021-04-28T22:16:22.697" LastActivityDate="2021-04-28T22:16:22.697" Title="The life course for a massive star from birth to death using the HR Diagram" Tags="&lt;star&gt;&lt;formation&gt;&lt;hr-diagram&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="1263" PostTypeId="2" ParentId="1206" CreationDate="2013-12-28T15:53:04.067" Score="6" Body="&lt;p&gt;Since the astronomers are using radio telescopes and not optical telescopes, I'd like to point out why they are doing so - The centre of the Milky Way is a &lt;em&gt;very&lt;/em&gt; dusty place. Wavelengths from the millimeter to optical get easily absorbed by all this dust, so it's very difficult to see the centre of the galaxy in the optical spectrum. But radio waves do not get absorbed, and the centre of our galaxy is a very strong source of radio waves. Therefore looking in radio will give us the clearest picture of what's happening over there.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;They mention in the article that they use a technique called VLBI (Very Long Baseline Interferometry) to image the blackhole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The images you get from &lt;a href=&quot;https://en.wikipedia.org/wiki/VLBI&quot;&gt;VLBI&lt;/a&gt; aren't images in the traditional sense, like you would get from an optical telescope. VLBI measures the phase difference of waves arriving at two different antennas (on two different continents, perhaps) and use this phase difference to infer the size of the source in the sky. Contradictorily, even though they can resolve sources of less than an arcsecond in size, it's very difficult to localize the &lt;em&gt;absolute&lt;/em&gt; position in the sky to a very high degree. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So hopefully we will be able to resolve the event horizon of the black hole, but we will (almost definitely) not be able to see a picture like the one you posted.&lt;/p&gt;&#xA;" OwnerUserId="694" LastActivityDate="2013-12-28T15:53:04.067" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1264" PostTypeId="1" AcceptedAnswerId="1265" CreationDate="2013-12-28T16:44:07.070" Score="9" ViewCount="2450" Body="&lt;p&gt;I am so sorry for being ambiguous. I'm not an astronomer and have have never actually even bothered to look at the sky before. I don't have a telescope nor camera. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet, I'm really really curious about what it is I saw in the sky tonight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Towards, the east I first noticed a &quot;constellation&quot; the looked much like what is described to be the big dipper (It is possible that it could have been Andromeda and Pegasus). It was squarish and had somewhat of a ladle towards it's north eastern corner. The most distinguishable feature of the structure was that it had 3 stars in a straight line. It was the straightest line in the entire night sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About a hand's length to the west was another &quot;constellation&quot;, my guts said it's name was Orion. Again, I have never seen, let alone distinguished, a constellation properly in my life. It was like a triangle with it's nose pointing towards the east. The sides that converged to the nose &lt;em&gt;seemed&lt;/em&gt; to extend a bit backwards like antennas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And to the west (a few inches north to the Orion was this strange glow. &#xA;At first I thought it was my eyes playing tricks on me as sometimes whenever you stare into light bulb or some other light source, a faint glow in that area of your vision often persists. And as I looked directly at it, it slowly faded away into darkness and all that was distinguishable was an extremely tiny star or two. But whenever I looked to something near it (like a tree or constellation) , the patch reappeared. Then, I noticed that there was cluster of tiny little stars. I couldn't make out much. &lt;strong&gt;What caused that light patch?&lt;/strong&gt; It seemed to glow in and out like it was breathing. I am sure that my eyes weren't playing tricks on me because even as I turned around and tilted my head and jumped up and down in the process, the patch remained in the same spot in the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The location which I observed this is from Ernakulam, Kerala, India. The nearest current sky map I could find is the following from Chennai (which is really far away):&lt;img src=&quot;https://i.stack.imgur.com/L7AGh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't make head or tail of it. Please help and no I'm not crazy. I know it sounds like one of those crazy UFO stories but I assure you it's not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything like what I've said observed in the night sky before?&lt;/p&gt;&#xA;" OwnerUserId="704" LastEditorUserId="704" LastEditDate="2013-12-29T04:19:51.877" LastActivityDate="2013-12-29T04:19:51.877" Title="Light patch in the night sky" Tags="&lt;amateur-observing&gt;&lt;supernova&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2013-12-29T02:46:01.967" ContentLicense="CC BY-SA 3.0" />
  <row Id="1265" PostTypeId="2" ParentId="1264" CreationDate="2013-12-28T17:06:18.143" Score="6" Body="&lt;p&gt;Given the description, that &lt;strong&gt;light patch&lt;/strong&gt; was probably &lt;a href=&quot;http://en.wikipedia.org/wiki/Pleiades&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pleiades&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4VGBv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a star cluster in the constellation of Taurus which is inspired from a popular Greek mythology of Seven Sisters. You can read more about it &lt;a href=&quot;http://www.naic.edu/~gibson/pleiades/pleiades_myth.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; One of the great softwares, free and open source, for amateur level observational astronomy, though professionals use it as frequently is &lt;a href=&quot;http://www.stellarium.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellarium&lt;/a&gt;. It's easy to use and should clear all your doubts.&lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="63" LastEditDate="2013-12-28T17:19:23.663" LastActivityDate="2013-12-28T17:19:23.663" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1266" PostTypeId="1" AcceptedAnswerId="1268" CreationDate="2013-12-28T20:04:41.440" Score="4" ViewCount="136" Body="&lt;p&gt;I read in a scientific magazine there was new gamma bursts that last longer than usual. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One was detected during december 2010, also known as Christmas Burst: &lt;a href=&quot;http://en.wikipedia.org/wiki/GRB_101225A&quot; rel=&quot;nofollow&quot;&gt;GRB 101225A&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;and another one in 9th of December 2011, &lt;a href=&quot;http://en.wikipedia.org/wiki/GRB_111209A&quot; rel=&quot;nofollow&quot;&gt;GRB 111209A&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They were both detected by the &quot;Swift&quot; program and &lt;a href=&quot;http://www.dailygalaxy.com/my_weblog/2013/04/christmas-burst-of-2011-source-discovered-most-powerful-known-gamma-ray-blast-to-hit-earth-.html&quot; rel=&quot;nofollow&quot;&gt;this article goes into detail&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know if there is another programs for GRB detection and I would be curious to find a resource on Internet which registers/records all Ultra-long Gamma-Ray Bursts on one single page/site (across all research/programs).&lt;/p&gt;&#xA;" OwnerUserId="703" LastEditorUserId="537" LastEditDate="2014-02-01T16:53:25.950" LastActivityDate="2014-02-01T16:53:25.950" Title="Internet Resource which register all Ultra-long Gamma-Ray Bursts" Tags="&lt;black-hole&gt;&lt;explosion&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1267" PostTypeId="2" ParentId="999" CreationDate="2013-12-28T21:15:23.943" Score="3" Body="&lt;p&gt;The analogy is rather weak and not really useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So-called collisionless stellar systems (those for which relaxation by stellar encounters has no appreciable effect over their lifetime), such as galaxies, can be described by the collisionless Boltzman equation, but never settle into thermodynamic equilibrium (only into some dynamical or virial equilibrium). Thus, the only other systems with somewhat similar behaviour are collisionless plasmas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sound, turbulence, viscosity etc are all effected by close-range collisions (not mere encounters) between the molecules. These also maintain thermodynamic equilibrium and a Maxwell-Boltzmann velocity distribution. Stellar systems have none of these processes and their velocities are in general anisotropically distributed and don't follow a Maxwell distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gases are in some sense simpler to understand, because their dynamics is driven by local processes and because statistical methods are very useful. Stellar systems are driven by gravity, i.e. long-range non-local processes, and intuition from the physics of gases is often very misleading (for example, a self-gravitating system has negative heat capacity -- this also applies to gas spheres, such as stars).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that the number of particles in a gas is &lt;em&gt;much much larger&lt;/em&gt; ($\sim10^{26}$) than the number of stars in a galaxy ($\sim10^{11}$), though the number of dark-matter particles may be much higher.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2013-12-28T21:15:23.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1268" PostTypeId="2" ParentId="1266" CreationDate="2013-12-28T22:30:28.390" Score="4" Body="&lt;p&gt;I guess one of them you're referring to was &lt;a href=&quot;https://en.wikipedia.org/wiki/GRB_130427A&quot; rel=&quot;nofollow&quot;&gt;GRB 130427A&lt;/a&gt; (not in December of 2010 or 2011 tho, but in April 2013)? But yes, there are several such catalogs, as are different designators and phenomenological classification methods for these gamma-ray bursts (GBR). For example, &lt;em&gt;GRB 130427A&lt;/em&gt; is a simple nomenclature that consists of a date of discovery (&lt;em&gt;130427&lt;/em&gt; denotes 27th of April, 2013 and suffix A is just the first letter from the alphabet, increased when there's more than a single such event in a day), but it doesn't indicate e.g. its type or origin. The same GRB event is so cataloged under designator &lt;em&gt;CSS130502:113233+274156&lt;/em&gt; in the &lt;a href=&quot;http://crts.caltech.edu/&quot; rel=&quot;nofollow&quot;&gt;Catalina Real-time Transient Survey catalog&lt;/a&gt;, &lt;em&gt;SDSS J113232.84+274155.4&lt;/em&gt; in &lt;a href=&quot;http://cas.sdss.org/dr7/en/&quot; rel=&quot;nofollow&quot;&gt;Sloan Digital Sky Survey catalog&lt;/a&gt;, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as you can see, it would help if you could clarify which designators you saw, so we can identify the catalog easier. But not wanting to leave you empty-handed, one such catalog that is frequently used and is relatively easily searchable is the Goddard Space Flight Center's &lt;a href=&quot;http://heasarc.gsfc.nasa.gov/grbcat/&quot; rel=&quot;nofollow&quot;&gt;GRBCAT (Gamma Ray Burst CATalog)&lt;/a&gt; (on top of those already mentioned).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Post update to the question, I've managed to identify SWIFT pages for the two GRB events in question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://swift.gsfc.nasa.gov/archive/grb_table/fullview/101225A/&quot; rel=&quot;nofollow&quot;&gt;101225A&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://swift.gsfc.nasa.gov/archive/grb_table/fullview/111209A/&quot; rel=&quot;nofollow&quot;&gt;111209A&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Catalog numbers, references and data are available on linked pages. And here is a &lt;a href=&quot;http://swift.gsfc.nasa.gov/archive/grb_table/fullview/&quot; rel=&quot;nofollow&quot;&gt;complete SWIFT GRB catalog&lt;/a&gt; (note, it's rather long, listing all GRB in a single page).&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditDate="2013-12-28T23:00:15.320" LastActivityDate="2013-12-28T23:00:15.320" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1269" PostTypeId="1" AcceptedAnswerId="1275" CreationDate="2013-12-29T00:20:11.773" Score="19" ViewCount="2801" Body="&lt;p&gt;In reference to the question, &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1225/how-can-we-tell-that-the-milky-way-is-a-spiral-galaxy&quot;&gt;&quot;How can we tell that the milky way is a spiral galaxy?&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers there clearly sum up the question asked. But Milky Way is not just a spiral galaxy. It is further classified as a &lt;strong&gt;barred&lt;/strong&gt; spiral galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Which particular feature in the distribution of stars, or in general a feature in observations led us to believe that it is a barred galaxy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; The edge-on picture is not sufficient to establish it, since a dense distribution at smaller radii could also arise from a non-uniform density on the spiral disk if modeled to fit the observations. We don't have any data from other angles or orientations.&lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2020-01-29T08:22:37.010" Title="How do we know Milky Way is a 'barred' spiral galaxy?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1270" PostTypeId="1" AcceptedAnswerId="1277" CreationDate="2013-12-29T07:06:46.027" Score="10" ViewCount="695" Body="&lt;p&gt;As you know, Doppler effect can change the wavelength of the light.&lt;/p&gt;&#xA;&lt;p&gt;Suppose there are two stars at same distance and same temperature. If one star is receding and the other star is approaching, their colors would look different.&lt;/p&gt;&#xA;&lt;p&gt;What I want to know is, how probable is it that two stars show different colors because of Doppler shift? In other words, if two stars at same distance have different colors, say blue and red, then is it safe to assume that two stars have different surface temperature?&lt;/p&gt;&#xA;&lt;p&gt;Actually, this was the question from the final exam for Wave &amp;amp; Heat.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Q. Stars emit radiation whose spectrum is very similar to that of an ideal blackbody. Two stars, identical in size, are at the same distance from us. One of the stars appears reddish in color while the other one looks very blue. Choose the correct statement.&lt;/p&gt;&#xA;&lt;p&gt;(a) The red star is hotter.&lt;br&gt;&#xA;(b) The blue star is hotter.&lt;br&gt;&#xA;(C) They could both have the same surface temperature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I read that some stars near the galactic center moves very fast. &lt;a href=&quot;http://en.wikipedia.org/wiki/S2_(star)&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/S2_(star)&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;If we only consider Planck distribution, only (b) is possible. I'm wondering whether (c) is also a realistic case, or it is only conceptually possible.&lt;/p&gt;&#xA;" OwnerUserId="707" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2013-12-29T19:39:55.470" Title="Can Doppler shift make colors of two stars at same distance different?" Tags="&lt;light&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1271" PostTypeId="2" ParentId="782" CreationDate="2013-12-29T12:01:50.863" Score="2" Body="&lt;p&gt;Essentially, the answer is &lt;strong&gt;Occam's razor&lt;/strong&gt;: look for the simplest solution and &lt;strong&gt;avoid complicated&lt;/strong&gt; and contrived &lt;strong&gt;solutions, unless&lt;/strong&gt; (observational) &lt;strong&gt;evidence requires them&lt;/strong&gt;. Yes, it is possible that two or more types of dark-matter particles exist. But any solutions where not one species dominates require fine-tuning and hence are unfavourable. So, unless there is a theory that would &lt;em&gt;naturally&lt;/em&gt; come with a mix of dark-matter particles (with different properties regarding their astrophysical implications, i.e. hot and cold etc, &lt;em&gt;when Occam's razor does not apply&lt;/em&gt;), we should expect only one species to dominate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If such a theory fails to explain the evidence, only then does it make sense to go to a more complicated model with more than one type of dark-matter particle.&#xA;Currently, we are not at that stage.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2015-10-15T07:18:19.347" LastActivityDate="2015-10-15T07:18:19.347" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="1272" PostTypeId="1" CreationDate="2013-12-29T14:35:58.503" Score="4" ViewCount="132" Body="&lt;p&gt;Now that we are making it to Mars, continuing onward, in an effort for humanity to civilize other planets, not just setting up residential colonies, but to mine minerals and carry on geological and scientific research, which planet or moon has all resources that can be used to sustain life in a controlled biosphere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And how can the available resources on that body be converted to suit human requirement?&lt;/p&gt;&#xA;" OwnerUserId="538" LastEditorUserId="2618" LastEditDate="2017-03-07T01:10:14.100" LastActivityDate="2017-03-07T01:10:14.100" Title="Which planet or moon has all resources that can be used to sustain life in a controlled biosphere?" Tags="&lt;planet&gt;&lt;natural-satellites&gt;&lt;terrestrial-planets&gt;&lt;space-debris&gt;" AnswerCount="1" CommentCount="7" ClosedDate="2013-12-30T16:48:06.373" ContentLicense="CC BY-SA 3.0" />
  <row Id="1273" PostTypeId="1" CreationDate="2013-12-29T14:42:34.590" Score="5" ViewCount="2198" Body="&lt;p&gt;I am located in the southern part of India and am looking forward to buy a telescope to gaze up to look ay farther planets, moons in our solar system and take pictures. How should I go about getting a telescope as I know nothing about telescope and things I need to know?&lt;/p&gt;&#xA;" OwnerUserId="538" LastEditorUserId="2618" LastEditDate="2017-03-07T01:09:57.053" LastActivityDate="2017-03-07T01:09:57.053" Title="Telescope buying guide for a beginner in India" Tags="&lt;planet&gt;&lt;telescope&gt;&lt;asteroids&gt;&lt;natural-satellites&gt;&lt;space-debris&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1274" PostTypeId="2" ParentId="1272" CreationDate="2013-12-29T16:14:46.297" Score="2" Body="&lt;p&gt;No known planet except Earth can be colonized by a human civilization. There are at least three serious issues: temperatures at around 300K, an atmosphere of appropirate pressure, and damaging cosmic radiation (low gravity is also a worry for long-term human presence). Minerals are less of a problem (and water can be synthesized). Mars and the Moon are close enough to the Sun for pressurized greenhouses to provide an environment where humans could stay without space suit. However, the problem of cosmic radiation remains and such &quot;biospheres&quot; must forever dependent on supportive technology. Venus's inhospitable atmosphere makes it completely unfeasible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using robots to mine planets and moons in the Solar system is certainly a realistic and interesting option. Sending humans there is very expensive and not necessarily sensible. Moreover, it is not clear whether humanity should send any live (including microorganisms) to such places, as this would inevitably forever change their nature (like sending rabbits/cats to Australia); all ongoing space missions are sterilized before being sent on their way.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2013-12-30T11:26:45.693" LastActivityDate="2013-12-30T11:26:45.693" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1275" PostTypeId="2" ParentId="1269" CreationDate="2013-12-29T16:25:31.440" Score="16" Body="&lt;p&gt;There are several different lines of evidence which together form a coherent picture: that of a barred galaxy. Moreover, as most disc galaxies are barred, we should expect the same from the Milky Way. The various evidences are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The observed light distribution (2MASS) shows a left-right asymmetry in brightness and the vertical height. This is explained by the near end of the bar being located on that side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The observed gas velocities show velocities which are &quot;forbidden&quot; in an axisymmetric or near-axisymmetric (spiral arms only) galaxy. These velocities occur naturally from the orbits of gas in a barred potential&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The velocity distribution of stars in the Solar neighbourhood shows some asymmetries and clumping which is most naturally explained by orbital resonance with the bar rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extent, pattern speed, and orientation of the bar is consistent between all three of these.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="7411" LastEditDate="2019-04-16T11:06:26.630" LastActivityDate="2019-04-16T11:06:26.630" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="1276" PostTypeId="1" CreationDate="2013-12-29T16:50:29.353" Score="6" ViewCount="895" Body="&lt;p&gt;I am in the field of software programming. What skills should I brush up and gear up to get a job in space exploration and the observatory industry? What are various job profiles in terms of software development in this field?&lt;/p&gt;&#xA;" OwnerUserId="538" LastEditorUserId="332" LastEditDate="2014-01-01T13:56:23.460" LastActivityDate="2014-01-01T13:56:23.460" Title="What skills would a software developer need to get a job in space research or exploration organization?" Tags="&lt;space&gt;&lt;software&gt;" AnswerCount="0" CommentCount="4" FavoriteCount="1" ClosedDate="2013-12-30T16:48:41.360" ContentLicense="CC BY-SA 3.0" />
  <row Id="1277" PostTypeId="2" ParentId="1270" CreationDate="2013-12-29T19:39:55.470" Score="6" Body="&lt;p&gt;As you correctly mention, there are rare stars which move particularly fast. Check also hypervelocity stars section &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_kinematics&quot;&gt;here&lt;/a&gt;, where a stellar population moving at around $10^3 \textrm{km/s}$ is described (by the way, these velocities were most likely produced in the galactic center). Note also, that these stars are too fast to stay in the disk for long times. As another sidenote, galactic center in not accessible in visual due to dust extinction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Redshift&quot;&gt;Redshifts&lt;/a&gt; are calculated particularly easy: &#xA;$$z=\dfrac{v}{c}$$, &#xA;where $v$ is the velocity along the line of sight, $c$ is the speed of light, $z=\dfrac{\delta\lambda}{\lambda}$ and $(v/c)^2\approx 0$ (which holds even for hypervelocity stars).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because hypervelocity stars are rare, they most likely can be found only close to regular stars. Yet, being very optimistic, take $v=3\cdot 10^3 \textrm{km/s}$, and you obtain that $\dfrac{\delta\lambda}{\lambda}=10^{-2}$ under very favourable conditions. In general it is not small, but when it comes to colors, the change is negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say, green light at $\lambda=555 \textrm{nm}$ would now have $\lambda=550 \textrm{nm}$ or $560 \textrm{nm}$ in wavelegth, i.e. would still be very green. So Doppler effect has no bearing on your problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What has possible bearing, however, is &lt;a href=&quot;http://www.chm.bris.ac.uk/webprojects2002/murphy/colour_and_interstellar_clouds.htm&quot;&gt;dust reddening&lt;/a&gt;. One of the two stars might be obscured by the dust, unlike the other one, and look reddish. So, both b) and c) are possible answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. And actually, if a star &quot;appears reddish&quot;, it is most likely dust extinction. &lt;a href=&quot;http://sustainabilityworkshop.autodesk.com/sites/default/files/styles/600px/public/core-page-inserted-images/light_colors.jpg&quot;&gt;Black body temperature colors&lt;/a&gt; are either blue, bluish, white, yelowish, yellow, orange or red.&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2013-12-29T19:39:55.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1278" PostTypeId="1" AcceptedAnswerId="1290" CreationDate="2013-12-30T19:21:27.407" Score="19" ViewCount="2200" Body="&lt;p&gt;Are there different classifications or just the one? How does our galaxy group compare with others?&lt;/p&gt;&#xA;" OwnerUserId="537" LastEditorUserId="537" LastEditDate="2013-12-31T13:06:30.813" LastActivityDate="2014-01-02T20:43:51.150" Title="Our galaxy, the Milky Way, forms part of a galaxy group called The Local Group. What are the parameters that dictate what a galaxy group is?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;classification&gt;&lt;local-group&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1279" PostTypeId="5" CreationDate="2013-12-30T19:22:51.867" Score="0" Body="&lt;p&gt;The term, &quot;The Local Group&quot; was introduced by Edwin Hubble in Chapter VI of his book The Realm of the Nebulae (Hubble 1936, pp. 124–151). There, he described it as &quot;a typical small group of nebulae which is isolated in the general field.&quot; &lt;a href=&quot;http://en.wikipedia.org/wiki/Local_Group&quot; rel=&quot;nofollow&quot;&gt;From Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="537" LastEditorUserId="537" LastEditDate="2013-12-30T23:28:10.813" LastActivityDate="2013-12-30T23:28:10.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1280" PostTypeId="4" CreationDate="2013-12-30T19:22:51.867" Score="0" Body="The Local Group is the group of galaxies that includes the Milky Way among others." OwnerUserId="537" LastEditorUserId="537" LastEditDate="2013-12-30T23:28:15.050" LastActivityDate="2013-12-30T23:28:15.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1281" PostTypeId="2" ParentId="1259" CreationDate="2013-12-31T21:17:15.530" Score="2" Body="&lt;p&gt;Best viewing will be shortly before sunrise or shortly after sunset so your sky is dark, but not so late that everything above you will be in the Earth's shadow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will be brightest when it's near the opposite horizon from the sunrise/sunset, because the solar cells will be perpendicular to the sun, and nearly perpendicular to you.&lt;/p&gt;&#xA;" OwnerUserId="562" LastActivityDate="2013-12-31T21:17:15.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1282" PostTypeId="2" ParentId="1273" CreationDate="2014-01-01T17:42:47.697" Score="3" Body="&lt;p&gt;If you are a beginning astronomer, there is not much point worrying about all the bells and whistles you can get with a good telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, the key points to look for are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;light-gathering diameter - this gives an indication of how dim an object you will be able to see&lt;/li&gt;&#xA;&lt;li&gt;supports/stand/mount - a basic tripod will be fine if you want to see no further than Jupiter, but it will wobble too much for longer distance viewing, so a solid mount will be essential&lt;/li&gt;&#xA;&lt;li&gt;motor/computer drive - at high magnification, objects will pass the eyepiece very quickly, so you will want to look at tracking drives that can follow objects smoothly&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But to be honest, you could start with a $100 telescope and really enjoy yourself learning the sky.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2014-01-01T17:42:47.697" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1285" PostTypeId="2" ParentId="508" CreationDate="2014-01-01T23:19:03.003" Score="5" Body="&lt;p&gt;The reason is that gravity is a radial force, and so conserves angular momentum $\mathbf{L} = \mathbf{r}\times m\mathbf{v}$, where $\mathbf{r}$ is the vector to the planet from the star and $\mathbf{v}$ is its velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, remember that the cross product of two vectors $\mathbf{a}$ and $\mathbf{b}$ has a magnitude that's equal to the area of a parallelogram with those sides, and hence is twice the area of the &lt;em&gt;triangle&lt;/em&gt; with those sides:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vVqSG.png&quot; alt=&quot;Vector Cross Product (Wikipedia)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the planet is at position $\mathbf{r}$ and you wait a small amount of time $\mathrm{d}t$, in that time it will be displaced by $\mathrm{d}\mathbf{r} = \mathbf{v}\,\mathrm{d}t$. Think of the area $\mathrm{d}A$ it sweeps out in that time as the area of the triangle defined by its old position $\mathbf{r}$ and its displacement $\mathrm{d}\mathbf{r}$, and therefore its rate of change of the swept area is:&#xA;$$\frac{\mathrm{d}A}{\mathrm{d}t} = \frac{1}{2}\left|\mathbf{r}\times\mathbf{v}\right| = \frac{1}{2m}|\mathbf{L}|\text{,}$$&#xA;which is constant by conservation of angular momentum. Since the rate of increase swept area does not change, a line joining the star and planet sweeps out equal areas in equal times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, Kepler's second law is exactly equivalent to the statement that the magnitude of the angular momentum of the planet is conserved. Of course, if you know that the orbit is planar, the direction of the angular momentum must be conserved as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, conservation of angular momentum implies that the orbit is planar and that Kepler's second law holds.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-01-01T23:19:03.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1286" PostTypeId="1" AcceptedAnswerId="1288" CreationDate="2014-01-02T01:37:59.990" Score="9" ViewCount="261" Body="&lt;p&gt;I am looking to buy a Celestron 21023 FirstScope 76 Telescope, which is quite inexpensive, and it has good reviews on Amazon.  One thing that many reviewers have said is that the two lenses that it comes with are of low quality, and the suggestion is to toss them and buy a Barlow lens, and perhaps some other more quality lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what a Barlow lens is.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a brand name?  The name of a type of lens?  What distinguishes it from a non-Barlow lens?  &lt;/p&gt;&#xA;" OwnerUserId="718" LastActivityDate="2014-01-02T20:49:38.400" Title="What is a Barlow lens?" Tags="&lt;telescope-lens&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1287" PostTypeId="1" CreationDate="2014-01-02T01:47:40.543" Score="7" ViewCount="250" Body="&lt;p&gt;I've heard that there is something called the Great Attractor which seems to have some kind of gravitational influence on objects in the Universe.  Is this some kind of theoretical notion posited in order to help answer a question in Astronomy, or is there an actual object which is greatly attracting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is this an obsolete notion?&lt;/p&gt;&#xA;" OwnerUserId="718" LastActivityDate="2014-01-02T20:25:16.333" Title="What Is The Great Attractor?" Tags="&lt;gravity&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1288" PostTypeId="2" ParentId="1286" CreationDate="2014-01-02T19:55:39.247" Score="10" Body="&lt;p&gt;I have been an amateur astronomer for over 30 years so I can give you some practical advice on this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;Barlow lens&lt;/strong&gt; is an attachment which fits into the eyepiece holder. It's main purpose is to give increased magnification with your existing eyepieces. They are usually marked as 2x or 3x which indicates how many times they will multiply the magnifying power of an eyepiece. They can be useful for large telescopes which can comfortably handle high powers but for a smaller scope they can be of limited use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two problems you would encounter if you used one are, first, that the more you magnify an image the fainter it gets, with a smallish diameter scope like yours this means that you will find it very difficult to see faint objects. The other one is the fact that locating and tracking objects as they move (due to the rotation of the earth) becomes much harder as you increase magnification, the field of view of the eyepiece gets much smaller as you increase the power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would personally suggest that you get used to using your new scope for a while as it is, then, see whether you might need one in the future. If your main area of interest turns out to be the moon and planets then a Barlow might be useful at some point, but if you enjoy wide field views of nebulae and star clusters then that is less likely.&#xA;Hope this is of some help.&lt;/p&gt;&#xA;" OwnerUserId="720" LastEditorUserId="13" LastEditDate="2014-01-02T20:49:38.400" LastActivityDate="2014-01-02T20:49:38.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1289" PostTypeId="2" ParentId="1287" CreationDate="2014-01-02T20:17:05.520" Score="3" Body="&lt;p&gt;No; it's real - observations of the movement of galaxies indicate that there is an unusual concentration of mass (probably not a massive &quot;object&quot; of course, but totalling as much as tens of thousands of galaxies) in a place around 200 million lightyears from our galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's hard to study since the view in that direction happens to be blocked by our own galaxy, but given that it has been studied for decades, it seems unlikely that the phenomenon turns out to be based on some sort of methodical error.&lt;/p&gt;&#xA;" OwnerUserId="668" LastEditorUserId="668" LastEditDate="2014-01-02T20:25:16.333" LastActivityDate="2014-01-02T20:25:16.333" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1290" PostTypeId="2" ParentId="1278" CreationDate="2014-01-02T20:43:51.150" Score="9" Body="&lt;p&gt;A galaxy group is a set of galaxies that are close together and gravitationally bound, i.e. barring outside influences they will stay together indefinitely. That means a galaxy that is in the space occupied by the group but &quot;passing through&quot; it at a high velocity would not be considered part of the group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The upper limit for calling it a group is roughly 50, bigger accumulations of galaxies are called clusters (which may consist of many groups). The borders between adjacent groups may be uncertain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These aren't mathematical definitions and actual usage is often based on tradition more than on measurements. It's just like geography: you &lt;em&gt;could&lt;/em&gt; debate endlessly whether a given piece of high ground should be called a hill or a mountain, and whether something is a separate mountain or a subpeak, but sensible people recognize it as a waste of time, make up a clear but arbitrary numeric distinction without strongly insisting on its usage.&lt;/p&gt;&#xA;" OwnerUserId="668" LastActivityDate="2014-01-02T20:43:51.150" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1291" PostTypeId="1" AcceptedAnswerId="1292" CreationDate="2014-01-03T02:46:59.027" Score="10" ViewCount="14825" Body="&lt;p&gt;What is the fastest spinning rotation of a Neutron star? I have heard that Neutron stars have a specially fast spinning rotation. What is the fastest?&lt;/p&gt;&#xA;" OwnerUserId="503" LastEditorUserId="537" LastEditDate="2014-01-03T14:01:01.970" LastActivityDate="2019-01-28T20:04:57.733" Title="What is the fastest spinning rotation of a Neutron star?" Tags="&lt;rotation&gt;&lt;neutron-star&gt;&lt;speed&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1292" PostTypeId="2" ParentId="1291" CreationDate="2014-01-03T13:25:02.627" Score="12" Body="&lt;p&gt;As it turns out, the fastest spinning neutron star found yet is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Pulsar&quot;&gt;pulsar&lt;/a&gt; 18000 light years away in the constelation of Sagittarius which scientist catalogued as &lt;a href=&quot;http://en.wikipedia.org/wiki/PSR_J1748-2446ad&quot;&gt;&lt;strong&gt;PSR J1748-2446ad&lt;/strong&gt;&lt;/a&gt;. Pulsars are neutron stars that rotate, are highly magnetic and emit a strong perpendicular beam of electromagnetic radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This pulsar's speed is such that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At its equator it is spinning at approximately 24% of the speed of light, or over 70,000 km per second.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PSR J1748-2446ad&lt;/strong&gt; rotates a little over 700 times a second, and scientists have this to say on the theoretical limits of the rotation speed of a pulsar (&lt;a href=&quot;http://en.wikipedia.org/wiki/Millisecond_pulsar&quot;&gt;from here&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Current theories of neutron star structure and evolution predict that pulsars would break apart if they spun at a rate of ~1500 rotations per second or more, and that at a rate of above about 1000 rotations per second they would lose energy by gravitational radiation faster than the accretion process would speed them up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="537" LastEditorUserId="537" LastEditDate="2014-01-03T13:48:47.870" LastActivityDate="2014-01-03T13:48:47.870" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1293" PostTypeId="1" AcceptedAnswerId="1303" CreationDate="2014-01-03T17:26:59.120" Score="6" ViewCount="527" Body="&lt;p&gt;Would it be possible for something to cause the Moon to start rotating at a different rate, breaking its tidal lock with Earth, without ripping it apart?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, would the Moon's tidal lock eventually stabilize?&lt;/p&gt;&#xA;" OwnerUserId="730" LastEditorUserId="730" LastEditDate="2014-01-05T05:33:41.090" LastActivityDate="2014-01-07T23:20:44.703" Title="Could something temporarily break the Moon's tidal lock?" Tags="&lt;the-moon&gt;&lt;earth&gt;&lt;tidal-forces&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1295" PostTypeId="1" AcceptedAnswerId="1302" CreationDate="2014-01-03T17:50:34.377" Score="5" ViewCount="842" Body="&lt;p&gt;Everybody knows that the shortest and longest days of the year occur on the solstices.  However, examination of sunrise/sunset tables shows that the maxima and minima of sunrise/sunset times are offset from the solstices by about 10 days. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.timeanddate.com/worldclock/astronomy.html?n=202&amp;amp;month=12&amp;amp;year=2013&amp;amp;obj=sun&amp;amp;afl=-11&amp;amp;day=1&quot;&gt;timeanddate.com&lt;/a&gt; for my location (Portland, OR), for the winter solstice, the earliest sunset is about December 10 (16:27), and the latest sunrise is around January 1 (07:51).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is counterintuitive... what's the explanation?&lt;/p&gt;&#xA;" OwnerUserId="731" LastActivityDate="2014-01-04T00:29:01.460" Title="Latest sunrise/earliest sunset out of phase with solstice?" Tags="&lt;earth&gt;&lt;the-sun&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1296" PostTypeId="1" AcceptedAnswerId="1298" CreationDate="2014-01-03T18:50:56.613" Score="5" ViewCount="157" Body="&lt;p&gt;It says in &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_singularity#Entropy&quot;&gt;this Wikipedia's article&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Entropy, however, implies heat and therefore temperature. The loss of energy also suggests that black holes do not last forever, but rather &quot;evaporate&quot; slowly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Would a black hole, given enough time, dissapear? Has this happened before?&lt;/p&gt;&#xA;" OwnerUserId="537" LastActivityDate="2014-01-03T19:58:34.957" Title="What does it mean for a black hole to &quot;evaporate&quot;?" Tags="&lt;black-hole&gt;&lt;temperature&gt;&lt;heat&gt;&lt;supermassive-black-hole&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1298" PostTypeId="2" ParentId="1296" CreationDate="2014-01-03T19:58:34.957" Score="4" Body="&lt;p&gt;It's due to &lt;a href=&quot;https://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow&quot;&gt;Hawking radiation&lt;/a&gt;, a black body radiation due to quantum effects near the event horizon of black holes and named after Stephen Hawking that predicted this phenomena. From the same Wikipedia article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hawking radiation reduces the mass and the energy of the black hole&#xA;  and is therefore also known as &lt;em&gt;black hole evaporation&lt;/em&gt;. Because of&#xA;  this, black holes that lose more mass than they gain through other&#xA;  means are expected to shrink and ultimately vanish. &lt;a href=&quot;https://en.wikipedia.org/wiki/Micro_black_hole&quot; rel=&quot;nofollow&quot;&gt;Micro black holes&lt;/a&gt;&#xA;  (MBHs) are predicted to be larger net emitters of radiation than&#xA;  larger black holes and should shrink and dissipate faster.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't think it was ever observed to actually lower mass of black holes that we observe as astronomy phenomena, the process would simply take too long for any measurable effects to manifest, but scientists did try to &lt;a href=&quot;http://www.technologyreview.com/view/420940/first-observation-of-hawking-radiation/&quot; rel=&quot;nofollow&quot;&gt;observe it experimentally&lt;/a&gt;. In September 2010, Belgiorno et al. claimed they've:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... produced Hawking radiation by firing an intense laser pulse through&#xA;  a so-called nonlinear material, that is one in which the light itself&#xA;  changes the refractive index of the medium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But this is yet to be reproduced and confirmed by experiments of others.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2014-01-03T19:58:34.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1299" PostTypeId="1" AcceptedAnswerId="1342" CreationDate="2014-01-03T22:51:58.677" Score="9" ViewCount="2722" Body="&lt;p&gt;I'm not asking about theoretical ball, vs saddle, vs flat surface which is just a metaphor with 2D space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's hard to say as we see very little of it, and we see them in the past because light travels for so long. But what we do know is that it is inflating (not exploding as one might thing from the &quot;big bang&quot; naming). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would the universe look like, if we were to freeze it in a moment, is it likely to be a ball, or rugby ball, a cone or a sort of an irregular shape?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it filled throughout with galaxies, dust, black-holes, or does it live on the edges of its 3D shape, and the central part is &quot;empty&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it have a giant black hole or a star in the middle around which everything revolves? &lt;/p&gt;&#xA;" OwnerUserId="692" LastActivityDate="2016-07-29T07:35:56.683" Title="What is the physical, geometric shape of the universe?" Tags="&lt;universe&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1301" PostTypeId="2" ParentId="1299" CreationDate="2014-01-04T00:22:17.603" Score="6" Body="&lt;p&gt;Ok, maybe you have some misconceptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Universe has no center at all. It looks the same from avaery point, to wherever you look to. It is approximatedly as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tiPsH.gif&quot; alt=&quot;Density filaments&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image represents a very large scale box on our universe on current time (not on perceived time, based on received light). Of course, it is just a computer simulation. Each dot represents a cluster of galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need to imagine an infinite tridimensional space filled with filament-like structures like these. And infinite means it has no bounds, so it has no &quot;external&quot; shape. No ball, rugby-ball nor cone there. Also not irregular external shape, just infinite. Any of these shapes have a 2D boundary on a 3D space, but the universe has no boundary.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-01-04T00:22:17.603" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="1302" PostTypeId="2" ParentId="1295" CreationDate="2014-01-04T00:29:01.460" Score="2" Body="&lt;p&gt;The point here is that sunrise and sunset does not only move towards or away from noon, but noon itself moves around 12:00. There is a difference between &quot;civil time&quot; (the one counted by wall clocks) and &quot;solar time&quot; (the one marked by sun clocks). This difference is called the &quot;Equation of Time&quot; and its geometrical shape is called the Analema.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in order to have the earliest sunset you need a combination of both a short day and a noon moved before 12:00. Same reasoning for latest sunrise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see more about the Equation of Time on Wikipedia: &lt;a href=&quot;http://en.wikipedia.org/wiki/Equation_of_time&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Equation_of_time&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-01-04T00:29:01.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1303" PostTypeId="2" ParentId="1293" CreationDate="2014-01-04T00:35:39.097" Score="4" Body="&lt;p&gt;Yes, theoretically it is possible to impact the Moon with enough moment and low enough energy to have it change its rotation periode, which actually is the same as its orbital periode (&lt;strong&gt;this&lt;/strong&gt; is the tidal lock). Best candidate will be a high-mass, low-speed asteriod impacting almost tangentially and on the rotation plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this really possible? I would bet that it will never happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supossing that it happens, will Moon get tidal locked again? Yes. As long as there are water on Earth suffering tides from Moon, these same water will attract one side of the Moon differentially from the other, so Moon will get locked again with the same face towards us.&lt;/p&gt;&#xA;" OwnerUserId="473" LastEditorUserId="58" LastEditDate="2014-01-07T23:20:44.703" LastActivityDate="2014-01-07T23:20:44.703" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1304" PostTypeId="1" AcceptedAnswerId="1306" CreationDate="2014-01-04T10:18:37.580" Score="13" ViewCount="10131" Body="&lt;p&gt;When we look at the sky, and see the &lt;a href=&quot;http://en.wikipedia.org/wiki/Milky_Way&quot; rel=&quot;noreferrer&quot;&gt;Milky Way&lt;/a&gt;, we mainly see a band, with no real center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a more accurate picture we can guess the center of the galaxy (see &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aMilky_Way_Arch.jpg&quot; rel=&quot;noreferrer&quot;&gt;this picture&lt;/a&gt;), but we still don't see the center as some kind of big bright &quot;star&quot;, like on &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aNGC_4414_%28NASA-med%29.jpg&quot; rel=&quot;noreferrer&quot;&gt;this picture&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there is a lot of objects (stars, planets and dust) between us and the center of the galaxy. But stars are bright and they shouldn't &quot;hide&quot; the galaxy center's light since they shine too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why don't we see the galaxy center as a &quot;night sun&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much dust is there so it's blocking the galaxy center's light?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because of all &lt;a href=&quot;http://en.wikipedia.org/wiki/Oort_cloud&quot; rel=&quot;noreferrer&quot;&gt;Oort clouds&lt;/a&gt; of all solar systems that we don't see the galaxy center clearly?&lt;/p&gt;&#xA;" OwnerUserId="204" LastEditorUserId="633" LastEditDate="2014-01-06T19:02:54.783" LastActivityDate="2015-07-12T00:10:28.407" Title="Why don't we see the galaxy center?" Tags="&lt;milky-way&gt;&lt;galaxy-center&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1305" PostTypeId="1" AcceptedAnswerId="1307" CreationDate="2014-01-04T14:20:27.647" Score="8" ViewCount="202" Body="&lt;p&gt;I always wondered how those tiny dots representing moving stars or whatever forming an interesting event (supernova explosions, stars being sucked into black holes etc.) get caught in the huge solid angle of $4\pi$? It seems absolutely improbable to find any such event in that big solid angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do astronomers just appear lucky looking at the right place at the right time to find an interesting event, or do they use some systematic methods of finding them? Do they maybe use special software which analyzes tons of data and identifies such interesting events?&lt;/p&gt;&#xA;" OwnerUserId="735" OwnerDisplayName="Ruslan" LastEditorDisplayName="user96" LastEditDate="2014-01-04T15:39:47.450" LastActivityDate="2014-01-04T16:22:47.447" Title="How do astronomers find interesting events?" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1306" PostTypeId="2" ParentId="1304" CreationDate="2014-01-04T15:08:59.273" Score="18" Body="&lt;p&gt;The main reason we don't see the bright center of our galaxy, which is composed of millions of stars, is &lt;strong&gt;dust&lt;/strong&gt;. Visible light is absorbed and scattered by interstellar dust, but that doesn't mean we can't see it on other waves of the spectrum, for example, infrared light doesn't suffer as much because of the dust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice on this image how bright the galaxy center looks on the infrared and the near-infrared pictures!:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FrTOc.jpg&quot; alt=&quot;Milky Way galaxy at different wavelengths&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image taken from the &lt;a href=&quot;http://mwmw.gsfc.nasa.gov/mmw_product.html#slides&quot; rel=&quot;noreferrer&quot;&gt;MultiwaveLength Milky Way website&lt;/a&gt; of NASA's Goddard Space Flight Center.&lt;/p&gt;&#xA;" OwnerUserId="537" LastEditorUserId="45" LastEditDate="2014-01-04T16:29:01.047" LastActivityDate="2014-01-04T16:29:01.047" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1307" PostTypeId="2" ParentId="1305" CreationDate="2014-01-04T16:22:47.447" Score="6" Body="&lt;p&gt;The answer to your question is yes. Many times discoveries in astronomy are serendipitous in nature - some of the most well-known discoveries fall into this category: the discovery of the cosmic microwave background, the discovery of some of the planets in our solar system, are but two examples of this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, many times they are not serendipitous. Astronomers plan to discover things - look at the &lt;a href=&quot;http://kepler.nasa.gov/&quot;&gt;Kepler satellite&lt;/a&gt; which finds nearby exosolar planets by detecting planetary &lt;a href=&quot;http://en.wikipedia.org/wiki/Transit_%28astronomy%29&quot;&gt;transits&lt;/a&gt;. The &lt;a href=&quot;http://www.sdss.org/&quot;&gt;Sloan Digital Sky Survey&lt;/a&gt; was a planned galactic survey which gave us new insights about the properties of galaxies as well as structure formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One also has to consider time scales. Many things astronomers care about have timescales which are millions or billions of years (merging clusters, the evolution of main sequence stars, etc..), however, some phenomena are on very short time scales, namely supernovae. Microlensing events can also be somewhat serendipitous in their discovery, though researchers still devise experiments to look for them (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Optical_Gravitational_Lensing_Experiment&quot;&gt;OGLE&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is certainly a combination of luck and planning. This answer was intended to be a bit vague, since the question doesn't specifically ask about one phenomenon. Software does play a big role in identifying and classifying objects and events in the universe. I can talk a little bit more about this in the context of gravitational lensing if people care to know about it.&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2014-01-04T16:22:47.447" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1309" PostTypeId="1" AcceptedAnswerId="1321" CreationDate="2014-01-04T18:37:20.490" Score="15" ViewCount="1824" Body="&lt;p&gt;Knowing it has exited the solar system, is it traveling in the direction of the center of the galaxy, away from it, orbiting it, or maybe even going perpendicular from the disk-shaped galaxy?&lt;/p&gt;&#xA;" OwnerUserId="692" LastEditorUserId="13" LastEditDate="2014-02-04T07:27:01.870" LastActivityDate="2015-05-09T17:29:54.630" Title="Where is Voyager 1 (or 2) going? In which direction?" Tags="&lt;galaxy&gt;&lt;artificial-satellite&gt;&lt;navigation&gt;&lt;interstellar-travel&gt;&lt;space-probe&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1312" PostTypeId="2" ParentId="1257" CreationDate="2014-01-05T16:05:50.187" Score="0" Body="&lt;p&gt;Apparently the term is being used as a more common way to say exoplanet. Here is an &lt;a href=&quot;http://news.yahoo.com/39-super-bowl-astronomy-39-conference-kicks-off-162254438.html&quot; rel=&quot;nofollow&quot;&gt;Article&lt;/a&gt; with a hyperlink to the term Alien Planet. The &lt;a href=&quot;http://www.space.com/17738-exoplanets.html&quot; rel=&quot;nofollow&quot;&gt;hyperlink&lt;/a&gt; then references an article on exoplanets: worlds beyond our solar system.&lt;/p&gt;&#xA;" OwnerUserId="257" LastActivityDate="2014-01-05T16:05:50.187" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1313" PostTypeId="1" AcceptedAnswerId="1322" CreationDate="2014-01-05T19:35:17.857" Score="3" ViewCount="60" Body="&lt;p&gt;Hey I want to describe an event A happening roughtly 5-7 after event B. I'd like to describe it as the 'A happened after the sound of B echo'd off X', where X is recognisable space thingy... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm talking mildly poetically, so I'm aware there is no sound in space, but I'm interested in what X might be, if such a thing was possible - if it makes you feel better you can use light and I'll describe 'A happened before X saw B') :) &lt;/p&gt;&#xA;" OwnerUserId="740" LastActivityDate="2014-01-06T06:10:39.253" Title="(Sound) Echo from X takes 5-7 days," Tags="&lt;space&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2014-01-06T12:28:31.913" ContentLicense="CC BY-SA 3.0" />
  <row Id="1314" PostTypeId="1" CreationDate="2014-01-05T21:56:49.750" Score="11" ViewCount="3584" Body="&lt;p&gt;I know how stars' distances are measured by using parallax, but I want to know how the actual telescope measures the parallax angle. Any helpful links/ explanations are needed.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2014-01-06T06:20:12.200" Title="How does a telescope measure parallax angle?" Tags="&lt;telescope&gt;&lt;parallax&gt;&lt;angular-resolution&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1315" PostTypeId="2" ParentId="1314" CreationDate="2014-01-06T00:03:56.600" Score="7" Body="&lt;p&gt;No, the telescope doesn't measure the parallax. A sextant or any other angle measuring device fit on the telescope does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, we don't(can't) directly measure the parallax angle. Instead, we just track the position of the star/object throughout the year. A little bit of spherical astronomy math shows us that the path of a star in the celestial sphere defined by a fixed reference through the year is an ellipse around it's mean postion, called the &lt;em&gt;parallactic ellipse&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vlESX.jpg&quot; alt=&quot;parallactic ellipse&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://star-www.st-and.ac.uk/~fv/webnotes/chapt14.htm&quot; rel=&quot;noreferrer&quot;&gt;Here&lt;/a&gt; is a link which does the rough math involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The semi-major axis of this ellipse is equal to the parllax angle P, while the semi-minor axis is equal to P*sin(b), where b is the stars ecliptic latitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EXTRA NOTE:&lt;/strong&gt; These parallactic ellipses are often rotated and parameters modified as the abberational ellipse is superposed on it. In practice, things get nigh complicated.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2014-01-06T00:03:56.600" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1317" PostTypeId="2" ParentId="1304" CreationDate="2014-01-06T03:39:10.800" Score="1" Body="&lt;p&gt;You may like this youtube video: &lt;a href=&quot;http://www.youtube.com/watch?v=duoHtJpo4GY&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=duoHtJpo4GY&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A detailed explanation can be found in Wikipedia: &lt;a href=&quot;http://en.wikipedia.org/wiki/Sagittarius_A&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sagittarius_A&lt;/a&gt;*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not the Oort clouds, neither stars nor planets, that make the center of the Milky Way invisible. It's 25 magnitudes of extinction by interstellar dust.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-01-06T03:39:10.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1318" PostTypeId="2" ParentId="1195" CreationDate="2014-01-06T04:12:43.317" Score="4" Body="&lt;p&gt;The most favoured WIMPS at the moment are probably neutralinos, see&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Neutralino&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Neutralino&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These particles are purely hypothetical at the moment.&#xA;The mass estimates in the above Wikipedia article for the lightest neutralino range between &#xA;10 and 10,000 GeV, meaning that the production rates in SNs will be much lower than with an assumed 1 GeV.&#xA;Higher production rates should have already been detected at LHC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence from the non-detection (in the form of energy loss) of WIMPS at LHC an estimate of an upper bound of the production rates in SNs should be possible.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-01-06T04:12:43.317" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="1320" PostTypeId="2" ParentId="751" CreationDate="2014-01-06T05:14:12.253" Score="4" Body="&lt;p&gt;The original research &lt;a href=&quot;http://arxiv.org/pdf/1303.5076v1.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt;, p.43, section 6.3.2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I leave it to someone else to find comprehensible words to describe the connection between CMB, redshift and effective neutrino number.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="537" LastEditDate="2014-01-06T15:24:43.827" LastActivityDate="2014-01-06T15:24:43.827" CommentCount="2" CommunityOwnedDate="2016-10-26T18:05:31.837" ContentLicense="CC BY-SA 3.0" />
  <row Id="1321" PostTypeId="2" ParentId="1309" CreationDate="2014-01-06T05:30:39.790" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Stars and gases at a wide range of distances from the Galactic center orbit at approximately 220 kilometers per second.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Milky_Way&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's much faster than the Voyager probes relative to the Sun (Voyager 1: 17 km/s). Hence the probes will orbit the galactic center roughly the same way as our Solar System, even after occasional hyperbolic encounters with other stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Voyager_1&quot; rel=&quot;nofollow&quot;&gt;Voyager 1&lt;/a&gt; is travalling in the direction of &lt;a href=&quot;http://en.wikipedia.org/wiki/Ophiuchus&quot; rel=&quot;nofollow&quot;&gt;Ophiuchus&lt;/a&gt;, seen from Earth. She's at or beyond the border of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heliosphere&quot; rel=&quot;nofollow&quot;&gt;heliosphere&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="2153" LastEditDate="2015-05-09T17:29:54.630" LastActivityDate="2015-05-09T17:29:54.630" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1322" PostTypeId="2" ParentId="1313" CreationDate="2014-01-06T06:10:39.253" Score="2" Body="&lt;p&gt;If there would be air at room temperature between Earth and Moon, sound would take about 28 days from Earth to Moon and echoed back, with helium instead of air 9 days, and with hydrogen about 7 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculated from &lt;a href=&quot;http://www.phy.mtu.edu/~suits/SpeedofSoundOther.html&quot; rel=&quot;nofollow&quot;&gt;http://www.phy.mtu.edu/~suits/SpeedofSoundOther.html&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Moon&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Moon&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also reflect Earth light from Pluto back to Earth, 11 times this double Earth - Pluto distance to get about 5 days of travel of light, much depending of the orbital position of Pluto.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-01-06T06:10:39.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1323" PostTypeId="2" ParentId="1314" CreationDate="2014-01-06T06:20:12.200" Score="2" Body="&lt;p&gt;In addition: It's often better to measure the angular distance to reference stars, instead of absolute angles, because those angles can be measured with much higher accuracy.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-01-06T06:20:12.200" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1324" PostTypeId="1" CreationDate="2014-01-06T15:45:57.840" Score="5" ViewCount="800" Body="&lt;p&gt;I've read that even light cannot escape from black holes.&lt;br&gt;&#xA;The speed of light is theoretically predicted to be the escape velocity of a black hole.&lt;br&gt;&#xA;I've also read that the general notion that nothing can escape from a black hole is wrong.  Even black holes emit particles and radiation known as the Hawking radiation.&lt;br&gt;&#xA;My question is: How can black holes emit &lt;em&gt;any&lt;/em&gt; kind of radiation?&#xA;If they even do emit, wouldn't the speed of the radiation be less than or equal to velocity of light?&lt;br&gt;&#xA;If the escape velocity of a black hole is &lt;em&gt;c&lt;/em&gt;, then how can the Hawking radiation even escape the black hole? Wouldn't it be pulled back into the black hole? Then, how is it even &#xA;&lt;em&gt;emitting&lt;/em&gt; anything?&lt;/p&gt;&#xA;" OwnerUserId="748" LastActivityDate="2020-05-26T23:32:46.073" Title="Black Holes emitting Hawking radiation" Tags="&lt;black-hole&gt;&lt;hawking-radiation&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1325" PostTypeId="2" ParentId="1188" CreationDate="2014-01-06T16:22:58.767" Score="2" Body="&lt;p&gt;The transit altitude of an object does not only depend on your latitude ($b$), but also on the declination of the object ($\delta$):&#xA;$$h_\mathrm{tr} = \arcsin\left(\sin b \sin \delta + \cos b \cos\delta\right).$$&lt;/p&gt;&#xA;" OwnerUserId="286" LastActivityDate="2014-01-06T16:22:58.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1326" PostTypeId="2" ParentId="1324" CreationDate="2014-01-06T16:31:25.357" Score="4" Body="&lt;p&gt;The escape velocity is c at the event horizon of a black hole. Above the event horizon the escape velocity is below c. Escaping particles of the Hawking radiation form above the event horizon; that's why they can escape, if they are pointing towards a sufficiently narrow angle to vertical upward, and if they are sufficiently energetic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Escaping particles form as virtual particle-antiparticle pairs in the &quot;infalling&quot; coordinate system: One of the two particles forms outside the event horizon; the counterpart forms below the event horizon. Thus the originally virtual particles cannot annihilate, and therefore become real particles; one particle can escape; the counterpart falls toward the singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy needed to form the escaping particle, and its remaining kinetic energy after escape, is subtracted from the mass of the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The described mechanism probably works also for both particles forming very close above the event horizon if the tidal forces are high enough to separate the virtual particle pair, before it can annihilate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formation of virtual particles is due to the Heisenberg uncertainty applied to time and energy: Very short time intervals require energy uncertainty, leading to short-lived particle-antiparticle pairs.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-01-06T16:31:25.357" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1327" PostTypeId="2" ParentId="1140" CreationDate="2014-01-06T16:44:06.370" Score="4" Body="&lt;p&gt;In other &quot;words&quot;, the connection between the time of transit $t_\mathrm{tr}$ of an object, its right ascension $\alpha$ and the geographical longitude of an observer $l$ is the (apparent) siderial time at Greenwich $\theta_0$ (if you know your local siderial time $\theta$, you don't need $\theta_0$ or $l$):&#xA;$$t_\mathrm{tr} = \alpha - \theta_0 - l = \alpha - \theta.$$&lt;/p&gt;&#xA;" OwnerUserId="286" LastActivityDate="2014-01-06T16:44:06.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1328" PostTypeId="2" ParentId="793" CreationDate="2014-01-06T16:59:13.560" Score="3" Body="&lt;p&gt;Earth and moon are thought to be formed by the collision of two planets. So we have at least these two protoplants. The cores probably ended mainly in the core of Earth; that's inferred by the much higher mean density of Earth in relation to Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron-nickel meteorites strongly indicate at least one pre-existing planet large enough to form an iron-nickel core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comets of the Kuiper belt are hypothized to be a collision product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you go further back in time the building blocks of planets get ever smaller and more numerous till the interstellar dust and supernova remnants our solar system is formed of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the early material will most likely have been ejected from the solar system; this may include protoplanets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the chaotic nature of multi-body systems like our solar system it's getting more and more difficult to track single bodies back into the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Nice_model&quot; rel=&quot;nofollow&quot;&gt;NICE models&lt;/a&gt; try to describe the early solar system.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-01-06T16:59:13.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1329" PostTypeId="2" ParentId="1240" CreationDate="2014-01-06T17:32:13.067" Score="2" Body="&lt;p&gt;Such observations are possible, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Fomalhaut_b&quot; rel=&quot;nofollow&quot;&gt;Fomalhaut b&lt;/a&gt;.&#xA;It depends mainly on the diameter of the objective of the telescope (&lt;a href=&quot;http://en.wikipedia.org/wiki/Diffraction&quot; rel=&quot;nofollow&quot;&gt;diffraction&lt;/a&gt; limit) and contrast. Telescopes can (at least in theory) be combined to an effective telescope of larger diameter (&lt;a href=&quot;http://en.wikipedia.org/wiki/Aperture&quot; rel=&quot;nofollow&quot;&gt;aperture&lt;/a&gt;) by &lt;a href=&quot;http://en.wikipedia.org/wiki/Interferometry&quot; rel=&quot;nofollow&quot;&gt;interferomentry&lt;/a&gt;.&#xA;Masks can help to occult bright stars to overcome the contrast limitatios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are indirect methods to &quot;see&quot; faint planets, as e.g. applied in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kepler_%28spacecraft%29&quot; rel=&quot;nofollow&quot;&gt;Kepler mission&lt;/a&gt;, or by &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaia_%28spacecraft%29&quot; rel=&quot;nofollow&quot;&gt;Gaia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, it is calculable. With naked eye we wouldn't be able to distinguish the sun from the planet orbits from a distance of about 1000 lightyears, as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Orion_%28constellation%29&quot; rel=&quot;nofollow&quot;&gt;orion&lt;/a&gt; belt stars: 1000 lightyears are about 300 parsec. This means, that the earth orbit would be about 1/300 arc second. The resolution of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Naked_eye&quot; rel=&quot;nofollow&quot;&gt;human eye&lt;/a&gt; is about 4 arc minutes, the 72,000-fold. For Pluto it's still more than 1000-fold.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-01-06T17:57:52.060" LastActivityDate="2014-01-06T17:57:52.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1330" PostTypeId="1" AcceptedAnswerId="1331" CreationDate="2014-01-06T18:04:33.853" Score="10" ViewCount="1041" Body="&lt;p&gt;If, hypothetically, me and my rocket powered flashlight were falling straight toward the center of a black hole. The flashlight is a few kilometers behind me in our travels toward the center of the black hole, but since it is rocket powered, it manages to maintain the exact distance to me for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is; The distance between me and my flashlight is constant as long as I am observing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The photons coming from the flashlight would obviously not be rocket powered - and they would be affected by the black holes gravitation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the light I see from the flashlight be shifted towards red or blue, even though the distance between me and my dear flashlight is maintained?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so; switching the positions of me and my flashlight, would it change the color I'd observe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we turn off the rocket on the flashlight, I assume it would be redshifted regardless of which were closer to the singularity, and the magnitude of redshift would appear to accelerate?&lt;/p&gt;&#xA;" OwnerUserId="749" LastEditorUserId="45" LastEditDate="2014-01-07T05:13:01.163" LastActivityDate="2014-01-07T22:13:50.537" Title="How does gravity affect the wavelength of light?" Tags="&lt;gravity&gt;&lt;redshift&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1331" PostTypeId="2" ParentId="1330" CreationDate="2014-01-06T18:51:48.937" Score="3" Body="&lt;p&gt;A partial answer: The lower the gravitational potential the slower the clocks, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_time_dilation&quot; rel=&quot;nofollow&quot;&gt;Gravitational time dilation&lt;/a&gt;. This means, as long as your flashlight is behind you in constant distance, it appears blue-shifted to you. With roles exchanged (light below you in constant distance) it appears red-shifted.&#xA;By the heterogeneity of the gravitational field the shift grows with the time, while falling in constant distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is called the Einstein shift, which is not the same as the doppler effect. The doppler effect is caused by an objects velocity relative to the observer.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="749" LastEditDate="2014-01-07T22:13:50.537" LastActivityDate="2014-01-07T22:13:50.537" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1332" PostTypeId="2" ParentId="1261" CreationDate="2014-01-07T02:22:30.963" Score="2" Body="&lt;p&gt;I won't primarily explain the H-R diagrams, because I think focussing on some of the underlying physics is essential for understanding, especially &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_nucleosynthesis&quot; rel=&quot;nofollow&quot;&gt;stellar nucleosytheses&lt;/a&gt;.&#xA;For simplification let's assume the star initially consists of nothing else than ordinary hydrogen and traces of carbon and nitrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/CNO_cycle&quot; rel=&quot;nofollow&quot;&gt;Nuclear fusion of hydrogen forms helium&lt;/a&gt;; &lt;a href=&quot;http://en.wikipedia.org/wiki/Triple-alpha_process&quot; rel=&quot;nofollow&quot;&gt;fusion of helium forms carbon&lt;/a&gt;; &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbon_burning_process&quot; rel=&quot;nofollow&quot;&gt;fusion of carbon&lt;/a&gt; leads to heavier elements like neon, also oxygen, sodium, magnesium; &lt;a href=&quot;http://en.wikipedia.org/wiki/Neon-burning_process&quot; rel=&quot;nofollow&quot;&gt;neon decays to oxygen&lt;/a&gt;; &lt;a href=&quot;http://en.wikipedia.org/wiki/Oxygen-burning_process&quot; rel=&quot;nofollow&quot;&gt;oxygen fuses to silicon and others&lt;/a&gt;; &lt;a href=&quot;http://en.wikipedia.org/wiki/Silicon_burning_process&quot; rel=&quot;nofollow&quot;&gt;silicon fuses stepwise with helium to iron&lt;/a&gt;. Each of these phases of burning needs higher temperature, and it releases energy. The phases can be subdivided in first core burning, then shell burning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earlier phases last long enough to be be visible from outside, resulting in motion within the HR diagram. Heating from phase to phase generally results in an overall expansion of the star.&#xA;The last phases last only a short time, too short to propagate effects to outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fusion of iron consumes energy. Therefore the star gets a problem after the core consists of iron. Outer pressure cannot be stopped by fusion and further heating: The core collapses to a neutron star or a black hole, while the outer part is blown off by a &lt;a href=&quot;http://en.wikipedia.org/wiki/Type_II_supernova&quot; rel=&quot;nofollow&quot;&gt;supernova&lt;/a&gt;.&#xA;Neutron stars usually rotate rapidly, and cool down over very long periods of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this overall framework you may go deeper into details of the phases.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-01-07T02:22:30.963" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1333" PostTypeId="1" AcceptedAnswerId="13531" CreationDate="2014-01-07T10:31:30.340" Score="11" ViewCount="1216" Body="&lt;p&gt;When light is emitted by for example a star, that star loses energy - which causes it to reduce its gravity. Then that energy begins a journey for potentially billions of years, until it reaches some other object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When that light reaches a surface, such as another star or galaxy, it will give that energy to the destination star in the form of heat. This causes the receiver to increase its energy, in turn restoring a sort of balance. It also causes the receiver to emit a minute amount of more light again, almost like a reflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will also excert pressure on the receiving surface once it reaches its destination, be it a star, a rock or anything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But while that light is travelling through space, its energy is &quot;unavailable&quot; to the rest of the universe. Naturally I ask the following question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Will light cause gravity, while it is traveling?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every single star emits light in every direction, and will eventually reach every other star in the universe. At any single point in the universe, there must be a continous ray of light coming from every single other star in the universe, that has a direct path to that point. Given that all stars on the sky is sending photons that reaches every square centimeter of the earth surface, the amount of pressure should sum up to be quite large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is the amount of pressure really neglible, given that every single atom on any surface is receiving light from every single lightsource on the sky?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on a calculation found at &lt;a href=&quot;http://solar-center.stanford.edu/FAQ/Qshrink.html&quot; rel=&quot;noreferrer&quot;&gt;http://solar-center.stanford.edu/FAQ/Qshrink.html&lt;/a&gt; the sun will during its lifetime emit 0.034 % of its total mass as energy. Assuming the sun is average, and that there are about 10^24 stars in the universe, and all of these stars on average are half way through their lifetime, there should be energy amounting to the gravity of about 1.7*10^22 suns distributed throughout the universe.&lt;/p&gt;&#xA;" OwnerUserId="749" LastEditorUserId="774" LastEditDate="2014-01-11T22:34:08.810" LastActivityDate="2016-02-04T14:51:26.970" Title="How does light affect the universe?" Tags="&lt;gravity&gt;&lt;universe&gt;&lt;light&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1334" PostTypeId="2" ParentId="1333" CreationDate="2014-01-07T12:48:25.377" Score="4" Body="&lt;p&gt;Light causes gravity while travelling, a clear yes, by &lt;a href=&quot;http://en.wikipedia.org/wiki/Mass%E2%80%93energy_equivalence&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein's famous mass-energy equvalence&lt;/a&gt;. (Compare &lt;a href=&quot;https://physics.stackexchange.com/questions/6197/do-two-beams-of-light-attract-each-other-in-general-theory-of-relativity&quot;&gt;this discussion on StackExchange&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational pull of light is negligible to other mass in large scale.&#xA;Only a small fraction of mass of a star is transformed into light during its lifetime, and only a small part of the ordinary matter has ever been a star. A fraction of the ordinary (standard model particles) matter consists of neutrinos (neutrinos and electrons are leptons). The baryonic matter consists mainly of hydrogen and some helium (nuclei) formed shortly after the big bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small fraction of mass of a star consists of photons, travalling out of the star. &lt;a href=&quot;http://www.astro.uni-bonn.de/~nlanger/siu_web/ssescript/new/chapter4-5.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This travel can take millions of years&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect of light on asteroids is not negligible, but it's not the gravitational pull. It' mainly the &lt;a href=&quot;http://en.wikipedia.org/wiki/Yarkovsky%E2%80%93O%27Keefe%E2%80%93Radzievskii%E2%80%93Paddack_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;YORP&lt;/a&gt; effect.&#xA;Dust is also affected by light.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2014-01-10T23:05:18.273" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="1335" PostTypeId="1" AcceptedAnswerId="1336" CreationDate="2014-01-07T15:54:47.670" Score="1" ViewCount="82" Body="&lt;p&gt;According to some documentaries History and Discovery, fossils have been found in a meteorite that fell to earth. This meteorite came from Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that the Martians always try to be bacteria or unicellular. It is possible that single-celled organisms that evolved into multicellular beings and then to beings we know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So possibly we are Martians (acestral have a Martian, a bacterium).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, does on Mars, before the earth existed humans?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that were to be existed underground, or another planet were extinct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there were other humans on Mars, then we live in parallel ignorance of each other. So we may be extinct before meeting our Martian (or other planets now) brother?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have we discovered but scare our discoverers by our violent attitudes and our boundless ambition?&lt;/p&gt;&#xA;" OwnerUserId="753" LastActivityDate="2014-01-07T16:28:13.040" Title="Martian Brother" Tags="&lt;mars&gt;&lt;life&gt;" AnswerCount="1" CommentCount="15" ClosedDate="2014-01-07T16:34:16.927" ContentLicense="CC BY-SA 3.0" />
  <row Id="1336" PostTypeId="2" ParentId="1335" CreationDate="2014-01-07T16:28:13.040" Score="2" Body="&lt;p&gt;It's exceedingly unlikely, that human-like species ever existed on Mars, from all we know thus far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very likely, that matter has been exchanged between Earth and Mars by asteroids, especially from Mars to Earth, and especially in the early solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very likely, that Mars was habitable for microorganisms in its early history, although there is no definitve evidence yet, that there ever existed life on Mars. We only know, that it could have survived for some time (millions of years), if it ever existed. But there are reasonable arguments, that there could have existed microbial life on Mars. A lot of effort is on the way to find evidence for this hypotheses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence it cannot be ruled out, that we have Martian ancestors. No observational evidence at all exits yet to verify this hyptheses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are reasonable arguments, that there could be underground extant life on Mars, although no experimental evidence at all has been found until now. But probes to investigate this are planned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We didn't discover life on other planets, we are searching, and don't scare.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-01-07T16:28:13.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1337" PostTypeId="2" ParentId="332" CreationDate="2014-01-07T19:59:04.547" Score="5" Body="&lt;p&gt;The probably most advanced system for determination of parallaxes is &lt;a href=&quot;http://www.astro.lu.se/~david/posters/UweLammers.pdf&quot; rel=&quot;noreferrer&quot;&gt;AGIS&lt;/a&gt; as used for &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaia_%28spacecraft%29&quot; rel=&quot;noreferrer&quot;&gt;Gaia&lt;/a&gt;. It's able to go far beyond the angular resolution of the telescopes. Angular resolution is just one parameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually it's just necessary to determine the &lt;a href=&quot;http://en.wikipedia.org/wiki/Luminosity&quot; rel=&quot;noreferrer&quot;&gt;luminosity&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Centroid&quot; rel=&quot;noreferrer&quot;&gt;centroids&lt;/a&gt; of the stars, almost independent of the resolution of the telescopes. That's mainly a statistical challenge, and it is especially dependent of the image noise, the brightness of the star and the number of observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calibration of the telescope and the determination of the luminosity centroids can be done in one solution algorithm.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-01-07T19:59:04.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1338" PostTypeId="2" ParentId="63" CreationDate="2014-01-08T01:10:26.133" Score="3" Body="&lt;p&gt;No. It's impossible for &lt;em&gt;every&lt;/em&gt; object to interact with every other object, due to the assertion by general relativity, that the universe can, and is, expanding faster than the speed of light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then assume that the universe initially was expanding at, or close to the speed of light, and that it immediately after the big bang was expanding faster than the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the particles/forms of energy that would have reached us are also bound to have been &quot;held back or deflected&quot;, even in the young stages of the big bang, and are now at a distance at which they can never reach us. They could have been held back by for example a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Potentially, if the expansion of the universe at one point was so slow that gravity from every particle had time to propagate to every other particle, then yes - every particle and energy in the universe affects every other particle.&lt;/p&gt;&#xA;" OwnerUserId="749" LastActivityDate="2014-01-08T01:10:26.133" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="1339" PostTypeId="1" CreationDate="2014-01-08T01:33:24.677" Score="5" ViewCount="156" Body="&lt;p&gt;Is the discovery of Alien or Extra Terrestrial Life a function of the science of Astronomy? When i read astronomy articles on the search for new Planets or discussions on Mars a lot of the topic is on the planets occurring in the habitable zone and the possibility of life on them or stories on the possibility of previous life on Mars. Is this actually part of the science objectives for the Astronomy community, or is there another branch of science who is tasked with this? Is Astronomy only focused on the mechanics of celestial bodies and the discovery of Alien or Extra Terrestrial life is just and interesting sidebar?  &lt;/p&gt;&#xA;" OwnerUserId="257" LastEditorUserId="746" LastEditDate="2014-01-08T03:03:22.447" LastActivityDate="2014-01-08T21:10:43.507" Title="Is the discovery of Alien or Extra Terrestrial Life a function of the science of Astronomy?" Tags="&lt;planet&gt;&lt;extra-terrestrial&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1340" PostTypeId="2" ParentId="1339" CreationDate="2014-01-08T02:38:04.303" Score="4" Body="&lt;p&gt;The search for extraterrestrial life is usually covered by &quot;exobiology&quot;. But astronomy can provide clues of whether a planet or an exoplanet provides some of the conditions, which allow life or indicate life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The habitable zone - in its various versions - defines, in which distance from a star a planet needs to be to allow for e.g. liquid surface water, a condition regarded as necessary for most life-forms we know from Earth. With the discovery of extremophiles the habitable zone for that kind of life has been extended.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomy can provide information about the central star and the distance of a planet from the central star. From these data a decision is possible, whether the planet fulfills the surface temperature condition for habitability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomy can provide information about the stability of a star, necessary for life to evolve on its planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomy can - in principle - provide information about the chemical composition of a potential atmosphere of a planet. Oxygen and methane at the same time would be regarded as an indicator of possible life on the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomy can determine size and density information about the planet. That way it can be decided, whether it's a rocky planet or a gas planet, and infer the gravitational acceleration at the surface. The gravitational acceleration is relevant, e.g. for constraints of the atmospheric composition, or for possible properties of life, if extant.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-01-08T02:43:54.013" LastActivityDate="2014-01-08T02:43:54.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1341" PostTypeId="2" ParentId="1339" CreationDate="2014-01-08T03:23:51.103" Score="4" Body="&lt;p&gt;Certainly it's a function of astronomy - chances are, if you're doing science outside the Earth then by definition it's astronomy in some form or another. There are many sub-fields of astronomy that are useful in the search for extra-terrestrial life:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Astrophysics is used to discover exoplanets, and to make crude approximations as to their habitability. There are also programs like SETI, which try to find radio transmissions from extrasolar sources.&lt;/li&gt;&#xA;&lt;li&gt;It's also a matter of &lt;a href=&quot;http://en.wikipedia.org/wiki/Astrobiology&quot; rel=&quot;nofollow&quot;&gt;astrobiology&lt;/a&gt; - given certain conditions, could &quot;life as we know it&quot; survive there? What about the possibility of life forms &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypothetical_types_of_biochemistry&quot; rel=&quot;nofollow&quot;&gt;based on other elements or molecules&lt;/a&gt;, such as arsenic-based life? (the possibility of arsenic-based life is disputed, however)&lt;/li&gt;&#xA;&lt;li&gt;There's also &lt;a href=&quot;http://en.wikipedia.org/wiki/Astrochemistry&quot; rel=&quot;nofollow&quot;&gt;astrochemistry&lt;/a&gt;. What is the composition of a planet's atmosphere? Is the planet's environment conducive to the formation of amino acids, one of the building blocks of life on Earth?&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Astrogeology&quot; rel=&quot;nofollow&quot;&gt;Astrogeology&lt;/a&gt; and the planetary sciences look at how the planets formed and evolved. Even if life doesn't exist now, what was the planet like in the past? Could life have existed there at some stage?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All of these sub-fields contribute very interesting and valuable information towards what is by no means a simple question to answer.&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="276" LastEditDate="2014-01-08T21:10:43.507" LastActivityDate="2014-01-08T21:10:43.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1342" PostTypeId="2" ParentId="1299" CreationDate="2014-01-08T03:29:08.580" Score="4" Body="&lt;p&gt;The overall geometry and topology of the universe has been investigated by the Planck mission. Some results are described in &lt;a href=&quot;http://arxiv.org/pdf/1303.5086v2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;. Final results are not yet available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An excerpt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We have calculated the&#xA;  Bayesian likelihood for speciﬁc topological models in universes&#xA;  with locally ﬂat, hyperbolic and spherical geometries, all of&#xA;  which ﬁnd no evidence for a multiply-connected topology with&#xA;  a fundamental domain within the last scattering surface. After&#xA;  calibration on simulations, direct searches for matching circles&#xA;  resulting from the intersection of the fundamental topological&#xA;  domain with the surface of last scattering also give a null result&#xA;  at high conﬁdence&#xA;  ...&#xA;  Future Planck&#xA;  measurement of CMB polarization will allow us to further test&#xA;  models of anisotropic geometries and non-trivial topologies and&#xA;  may provide more deﬁnitive conclusions, for example allowing&#xA;  us to moderately extend the sensitivity to large-scale topology.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The amount of anisotropy of the universe is going to be inferred from the cosmic microwave background (CMB).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kaSsv.jpg&quot; alt=&quot;Cosmic microwave background, as inferred from the Planck data, Image Credit: European Space Agency, Planck Collaboration&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image Credit: European Space Agency, Planck Collaboration&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Higher resolved images of the CMB can be found &lt;a href=&quot;http://apod.nasa.gov/apod/ap130325.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe is roughly a 4-dimensional spacetime with the big bang as a singularity. It has no edges in the 3d space when travelling. When looking to the past the border, if you like to call it like this, is the big bang. The big bang looks to us on Earth like being in a distance of 13.81 billion (13.81e9) light years in any direction. Or being 13.81 billion years in the past as light needed that time to travel to us. But we cannot travel to that boundary, because the universe expands faster than we (or light) can travel. We had to travel into the past or faster than light to get there, no matter in which spacial direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no black hole in the center of the universe, but the big bang, if you like to call it the center of a 4-d spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe, when looking to a fixed age of say 13.81 billion years is filled almost homogeneously with galaxies on the very large scale. Locally galaxies are grouped to clusters and superclusters. Superclusters form kind of a 3d-net. But there aren't totally void regions. There is always some gas or some dust or some plasma or some fast-travelling cosmic rays, neutrinos, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could stop the expansion of the universe at a given &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;cosmic time&lt;/a&gt;, you would see yourself in either direction in approximately the same distance, and in approximately the same past. (Such a structure is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/3-sphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;3-sphere&lt;/a&gt;. The surface of a &lt;a href=&quot;http://en.wikipedia.org/wiki/N-ball&quot; rel=&quot;nofollow noreferrer&quot;&gt;4-ball&lt;/a&gt; is an example of a 3-sphere. &lt;a href=&quot;http://www.youtube.com/watch?v=BqfwPQvb7KA&quot; rel=&quot;nofollow noreferrer&quot;&gt;This youtube video&lt;/a&gt; tries to visualize a rotating 3-sphere.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the fast expansion of spacetime, light cannot travel fast enough around the universe to make this possible. Therefore we can at best look back to the big bang, no matter which direction we look. The light needs more time to travel around the universe as the universe is existing after the big bang.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-01-09T15:51:52.510" LastActivityDate="2014-01-09T15:51:52.510" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="1344" PostTypeId="1" AcceptedAnswerId="1345" CreationDate="2014-01-08T13:46:50.040" Score="5" ViewCount="319" Body="&lt;p&gt;How  can artificial satellites be safe in the atmosphere when there are hundreds of tons of dust and asteroids coming into earth's atmosphere everyday?&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is an orbit of an artificial satellite?&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;At what approximate distance (or which orbit) from artificial satellites do these asteroids, dust, rocks annihilate?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="760" LastEditorUserId="6" LastEditDate="2014-01-08T15:23:35.050" LastActivityDate="2014-01-08T17:10:01.430" Title="How can satellites be safe in the atmosphere when there are so many asteroids coming in?" Tags="&lt;planetary-atmosphere&gt;&lt;asteroids&gt;&lt;artificial-satellite&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1345" PostTypeId="2" ParentId="1344" CreationDate="2014-01-08T14:05:51.543" Score="7" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Space_debris&quot; rel=&quot;nofollow noreferrer&quot;&gt;Space debris&lt;/a&gt; is a big problem for space industry. Natural bodies like dust or asteroids are not the main challenge, although probes are frequently hit by fine dust. The dust causes tiny craters on the surface of the probes or can change rotation or orbit slightly. This can mostly be corrected, e.g. by wheels fueled from solar panels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most concerning objects are remnants of human-made objects circling around the earth. With each collition of two such objects clouds of fragments form, leading to ever more collitions. Those fragments are tracked by radar, and probes try to avoid collitions by adjusting their orbits. But sometimes they are hit and destroyed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An orbit is the path along which a satellite travels around e.g. Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A satellite must remain outside the lower parts of the atmosphere, otherwise it will be slowed down and fall to the ground. The lower limit is somewhere near 200 km above ground. A satellite travels with about 8 kilometers per second around Earth, when in a lower Earth orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many satellites are in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Geostationary_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;geo-stationary&lt;/a&gt; orbit, about 36,000 km above ground. There Earth's atmosphere is neglectably thin, only some minor magnetic field can be found there; that part of the atmosphere is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;magnetosphere&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The further away a satellite orbits the Earth the slower it travels to compensate Earth's gravity, because Earth's gravity grows weaker with distance from the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At present the probability of collisions, which destroy satellites, is very low for a single satellite. Actually there are only a few documented cases: &quot;only one major incident has occurred: the 2009 satellite collision between Iridium 33 and Cosmos 2251&quot;, of the above wikipedia article. And &quot;China was widely condemned after their 2007 anti-satellite missile test, both for the military implications as well as the huge amount of debris it created&quot; of the same Wikipedia article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The central problem is the growing number of fragments to be expected for the future, as a consequence of more satellites in the orbit, and the expected chain reaction of fragmentation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Details of surface degradation of satellites due to impacts of dust particles can be read e.g. &lt;a href=&quot;http://www.intechopen.com/download/get/type/pdfs/id/12221&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. Most of that dust ist human-made. Surface degradation has consequences, e.g. for solar panels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;https://space.stackexchange.com/questions/1703/does-cosmic-dust-pose-a-problem-for-long-term-satellites-telescopes-and-probes&quot;&gt;a related question at StackExchange&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the second part of the question:&#xA;The dust particles collide with the satellites and evaporate from the impact heat. So the annihilation distance is more or less zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When entering the earth's atmosphere &quot;they usually disintegrate at altitudes of 50 to 95 km &quot; (&lt;a href=&quot;http://en.wikipedia.org/wiki/Meteoroid&quot; rel=&quot;nofollow noreferrer&quot;&gt;Meteorid&lt;/a&gt;). That's at least 100 km below the lowest satellite orbits.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2014-01-08T17:10:01.430" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1346" PostTypeId="5" CreationDate="2014-01-08T15:34:05.293" Score="0" Body="&lt;p&gt;A satellite is an object that orbits another object. In space, satellites may be made by man, or they may be natural. Most man-made satellites also orbit the Earth, but some orbit other planets, such as Saturn, Venus or Mars, or the moon.&lt;/p&gt;&#xA;" OwnerUserId="537" LastEditorUserId="537" LastEditDate="2014-01-08T15:37:29.633" LastActivityDate="2014-01-08T15:37:29.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1347" PostTypeId="4" CreationDate="2014-01-08T15:34:05.293" Score="0" Body="Questions about man-made objects orbiting astronomical objects." OwnerUserId="537" LastEditorUserId="9301" LastEditDate="2016-10-04T20:22:27.180" LastActivityDate="2016-10-04T20:22:27.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1348" PostTypeId="1" AcceptedAnswerId="1420" CreationDate="2014-01-08T22:49:24.967" Score="7" ViewCount="1443" Body="&lt;p&gt;I have a set of dates, position angles ($\theta$) and angular separations ($\rho$) for visual binary star. For example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1994.3132  57.0    0.097&#xA;1996.1783  73.4    0.051&#xA;2006.1938 182.0    0.126&#xA;2009.2597 192.9    0.152&#xA;etc.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What are algorithms to calculate orbital elements and to plot the orbit of the binary star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there available codes (or programs) for this calculations?&lt;/p&gt;&#xA;" OwnerUserId="761" LastActivityDate="2015-05-08T19:35:24.850" Title="How to plot orbit of binary star and calculate its orbital elements?" Tags="&lt;orbit&gt;&lt;fundamental-astronomy&gt;&lt;binary-star&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1349" PostTypeId="1" AcceptedAnswerId="1351" CreationDate="2014-01-09T00:27:30.437" Score="4" ViewCount="429" Body="&lt;p&gt;This article: &lt;a href=&quot;http://physicsworld.com/cws/article/news/2011/jul/25/was-the-universe-born-spinning&quot; rel=&quot;nofollow&quot;&gt;http://physicsworld.com/cws/article/news/2011/jul/25/was-the-universe-born-spinning&lt;/a&gt; asserts that the universe was &quot;born spinning&quot; and that there is a preferred rotation direction for galaxies in the universe.  My question is whether there is a preferred galaxy orientation in the universe, that is, whether there is a preferred plane for the galaxy disk?&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2014-01-09T01:33:45.917" Title="Do galaxies have a preferred orientation?" Tags="&lt;galaxy&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1350" PostTypeId="1" AcceptedAnswerId="1352" CreationDate="2014-01-09T01:24:35.220" Score="8" ViewCount="730" Body="&lt;p&gt;There's plenty of data about the density of nebulae, and obviously they're pretty thin.  This is space, after all, and these things are really, really big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm wondering about the practical experience of that density.  Are the colorful patterns of gas only visible when you're far away and looking through them, or if you were inside a nebula would you see colors all around you instead of the black of space?  Are any dense enough to obscure the view between planets of the same solar system?  Between a planet and its moon?&lt;/p&gt;&#xA;" OwnerUserId="529" LastActivityDate="2016-10-22T19:28:53.447" Title="Are nebulae only visible from a distance?" Tags="&lt;nebula&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1351" PostTypeId="2" ParentId="1349" CreationDate="2014-01-09T01:33:45.917" Score="3" Body="&lt;p&gt;There is no evidence for that. That's either a systematic observation error, or a random result within usual statistical fluctuations, compare &lt;a href=&quot;http://van.physics.illinois.edu/qa/listing.php?id=18609&quot; rel=&quot;nofollow&quot;&gt;this Q&amp;amp;A site&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-01-09T01:33:45.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1352" PostTypeId="2" ParentId="1350" CreationDate="2014-01-09T02:15:24.610" Score="6" Body="&lt;p&gt;The colorful patterns of gas are caused by excitation of gas atoms, e.g. oxygen, from radiation of a nearby star. You would see the same colors as from a distance, as the light is emitted in almost random directions. But you wouldn't see the same region of nebula shining at the same time as seen from Earth, when excitation is caused by a flash of radiation from a central star, because the radiation travels concentrical away from the star, and the distances the light has to travel from the excited part of the nebula to the observer is dependent of the location of the observer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light may be scattered non-uniformly by dust-particles, so looking different from differnt positions, e.g. forming a &lt;a href=&quot;http://en.wikipedia.org/wiki/Halo_%28optical_phenomenon%29&quot;&gt;halo&lt;/a&gt; around stars, depending of the composition and shape of the particles. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Interplanetary_dust_cloud&quot;&gt;interplanetary dust&lt;/a&gt; in our solar system causes &lt;a href=&quot;http://en.wikipedia.org/wiki/Zodiacal_light&quot;&gt;zodiacal light&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our solar system is in an interstellar dust cloud right now, but it's not very dense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very dense clouds can form, but would become instable and collapse to a star or a &lt;a href=&quot;http://en.wikipedia.org/wiki/Protoplanetary_disk&quot;&gt;protoplanetary disk&lt;/a&gt;, compare &lt;a href=&quot;http://abyss.uoregon.edu/~js/glossary/dark_nebula.html&quot;&gt;this site about dark nebula&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-01-09T02:15:24.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1353" PostTypeId="2" ParentId="1348" CreationDate="2014-01-09T03:00:28.070" Score="3" Body="&lt;p&gt;If it's a binary, it's fairly simple (in comparison with a ternary system), because the stars of the binary orbit about their common &lt;a href=&quot;http://en.wikipedia.org/wiki/Barycentric_coordinates_%28astronomy%29&quot; rel=&quot;nofollow&quot;&gt;barycenter&lt;/a&gt; in &lt;a href=&quot;http://en.wikipedia.org/wiki/Kepler_orbit&quot; rel=&quot;nofollow&quot;&gt;Kepler ellipses&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An orbit simulator, see &lt;a href=&quot;http://astro.unl.edu/classaction/animations/renaissance/kepler.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;; replace the central star by the barycenter of the binaries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://science.larouchepac.com/gauss/ceres/InterimII/Astronomy/KeplerProblem.html&quot; rel=&quot;nofollow&quot;&gt;Calculation of an Orbit from Three Observations&lt;/a&gt;. &lt;a href=&quot;http://en.wikipedia.org/wiki/Kepler_problem&quot; rel=&quot;nofollow&quot;&gt;Kepler problem on Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't find a ready-to-use software, try to solve it numerically by simulating the orbit of the binaries according to Kepler's law, and vary mass, distance, excentricity assumptions until they match to the observations. Use optimization methods, e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Hill_climbing&quot; rel=&quot;nofollow&quot;&gt;hill climbing algorithms&lt;/a&gt;, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Gradient_method&quot; rel=&quot;nofollow&quot;&gt;gradient methods&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-01-09T03:00:28.070" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1354" PostTypeId="2" ParentId="1350" CreationDate="2014-01-09T07:23:02.350" Score="3" Body="&lt;p&gt;You might be interested in these two answers from the Physics.SE site:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/26326/how-dense-are-nebulae&quot;&gt;How dense are nebulae?&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Excerpt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you were within a nebula, it is hard to say what it would look like. But nebulae are so large that the optical depth of the cloud would actually probably be quite high, and I would guess that it would look like you were surrounded by glowing green and red gas in the far distance - instead of space looking black and dark, it would be colored all over. But this would only be an effect caused by the fact that you are looking through so much gas - even if your spaceship were a thousand kilometers away, it probably wouldn't look much different if you were inside a nebula versus outside of it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/26562/if-we-were-to-travel-through-space-sci-fi-style-how-close-to-the-false-color/26563#26563&quot;&gt;If we were to travel through space, how close to the false-color images would the galaxies we see be?&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Excerpt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Unfortunately, even from a near distance, most of these objects won't look much better. They are, after all, faint, rarefied clouds of dust and gas. They are just not bright enough for the human eye to see color.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is another question that I cannot find which asks what a nebula would look like from a nearby spaceship. It basically states that the nebula are too disperse and too large to see from up close. I'm mentioning this in an answer instead of a question with the hope of finding the relevant link and providing more details.&lt;/p&gt;&#xA;" OwnerUserId="209" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-10-22T17:31:08.963" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1356" PostTypeId="1" CreationDate="2014-01-09T16:12:57.137" Score="5" ViewCount="253" Body="&lt;p&gt;Gravity is sometimes described as a curvature in space-time. Due to relativity, doesn't this imply that gravity doesn't propagate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A thought experiment. If you are moving toward a black hole, this could also be considered as if the black hole is moving toward you. If a black hole was moving toward you at the speed of light, and gravity propagates at the speed of light, you would experience something similar to a supersonic blast, except its a gravity blast, and it would allow you to remain unaffected until you arrived at the very centre of the black hole?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were able to turn around, just before you arrive at the center, and travel away from the black hole at the speed of light you would never notice it, even though you would be travelling at the same velocity as the black hole in the same direction?&lt;/p&gt;&#xA;" OwnerUserId="749" LastActivityDate="2014-01-10T02:49:02.610" Title="Does gravity propagate?" Tags="&lt;gravity&gt;&lt;general-relativity&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1357" PostTypeId="1" AcceptedAnswerId="1366" CreationDate="2014-01-09T16:35:23.733" Score="6" ViewCount="482" Body="&lt;p&gt;A heavy star should look redshifted, due to gravitational time dilation. How is that incorporated into the calculations of distances to the stars, or is it negligible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about an entire region of space that is denser or more energetic than our region? Would that not appear further away, than another region that is less dense than our region?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were in a region of low density and energy, and the density of space increases as we look further away, is that something that could theoretically explain Einsteins &quot;blunder&quot;? Can one be sure that this is not the case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the universe were collapsing, and gravity does indeed propagate. Wouldn't all distant regions seem to be affected more by gravity than the region of the observer, and thus ever more redshifted? (this due to less gravity in &quot;front&quot; of an approaching region of space than &quot;behind&quot; it, relative to us)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious of how these things are found to be important or unimportant on a cosmic scale. &lt;/p&gt;&#xA;" OwnerUserId="749" LastEditorUserId="6" LastEditDate="2014-01-09T18:07:11.630" LastActivityDate="2014-01-09T21:30:29.983" Title="Heavy star and redshift" Tags="&lt;general-relativity&gt;&lt;redshift&gt;&lt;expansion&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1358" PostTypeId="2" ParentId="1356" CreationDate="2014-01-09T16:54:32.503" Score="4" Body="&lt;p&gt;Black holes distort the geometry of spacetime. So you need to take care about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changes of gravity propagate with the speed of light. That's thought to occur for accelerated mass, e.g. a binary of black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second fundamental point: Speeds don't add up in an additive way, but in a subadditive way: speed of light + speed of light = speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you or the black hole travel with the speed of light, you resp. the black hole needs to be massless at rest, otherwise your relativistic mass, or that of the black hole would be infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photons are massless at rest, so you're actually talking of light, or something which behaves rather similar to light in many aspects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at the end you're talking of light directed to a black hole and reflected by a mirror inside the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The light can get inside the black hole, but it cannot leave the hole after being reflected from the mirror due to the high curvature of spacetime below the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are travelling towards a binary of black holes, you get blue-shifted changes of gravity.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-01-09T16:54:32.503" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1359" PostTypeId="1" AcceptedAnswerId="1360" CreationDate="2014-01-09T17:16:17.643" Score="13" ViewCount="607" Body="&lt;p&gt;I am a complete novice when it comes to astronomy, but someone asked me the above question casually. It seems as though planets are all roughly spherical, whereas other celestial bodies aren't necessarily. Is there are particular reason for this?&lt;/p&gt;&#xA;" OwnerUserId="294" LastActivityDate="2014-01-10T15:14:32.940" Title="Why are planets spherical?" Tags="&lt;planet&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1360" PostTypeId="2" ParentId="1359" CreationDate="2014-01-09T18:22:52.720" Score="5" Body="&lt;p&gt;That's mainly because of the gravity, and of size. Small objects, e.g. a stone, create only a very weak gravitation. The stone is much harder than necessary to withstand this force caused by its own gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth is much larger. The gravity at the surface causes the weight we feel. A huge mountain on Earth would be flattened under its own weight, because the matter it's made of isn't hard enough to withstand this weight.&#xA;By hills filling up valleys the planet as a whole becomes spherical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fast-rotating planet gets a slightly flattened ellipsoid shape.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-01-09T18:22:52.720" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1361" PostTypeId="1" CreationDate="2014-01-09T18:24:58.447" Score="4" ViewCount="829" Body="&lt;p&gt;How do you determine the optimal length of a guide scope for a given imaging telescope?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine it would have to take into consideration the pixel size of the guide scope, but would the aperture matter as well?&lt;/p&gt;&#xA;" OwnerUserId="539" LastActivityDate="2014-03-17T23:28:39.537" Title="What is the relationship between the focal length and f number of a telescope compared to a guide scope?" Tags="&lt;photography&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1362" PostTypeId="1" CreationDate="2013-12-10T16:07:01.070" Score="5" ViewCount="606" Body="&lt;p&gt;I'm planning to create a sundial in Ghana (10° 28.852', 0° 26.032') in July 2014.&#xA;I will stay there for about 3 weeks.&#xA;Do I need to adjust the numbers anywhere in these 3 weeks?&#xA;I know the shadow's path changes through the year, but how much does it change?&#xA;Is the change significant for a sundial?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(btw, is this the correct site to ask or should I use the astronomy site?)&lt;/p&gt;&#xA;" OwnerDisplayName="user737" LastActivityDate="2014-01-19T11:03:44.593" Title="How many minutes does a sundial change every day?" Tags="&lt;the-sun&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1363" PostTypeId="2" ParentId="1362" CreationDate="2013-12-10T16:55:58.600" Score="1" Body="&lt;p&gt;According to the wiki entry, the error runs up to roughly 15 minutes.  One can use the &#xA;&quot;equation of time&quot; , for which there appear to be published tables, to generate a correction curve. (The main contributors to the error are eccentricity and obliquity of Earth's orbit.&lt;/p&gt;&#xA;" OwnerUserId="907" OwnerDisplayName="Carl Witthoft" LastActivityDate="2013-12-10T16:55:58.600" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1364" PostTypeId="2" ParentId="925" CreationDate="2014-01-09T18:56:52.167" Score="2" Body="&lt;p&gt;All these answers are, of course, correct. I only want to make a simple observation. According to the definition linked in the OP, an &lt;em&gt;astronomical object&lt;/em&gt; is gravitationally bound. So we are looking for the largest gravitationally bound object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the universe were closed, it's bound and hence the largest astronomical object ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it appears the universe is expanding forever and hence not bound. The largest bound objects are galaxy clusters and galaxy super clusters. These objects have just formed or are still forming: the dynamical (orbital) time scale at their outskirts is comparable to the age of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the largest bound objects will be larger if you wait a few giga years ...&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-01-09T18:56:52.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1365" PostTypeId="2" ParentId="1357" CreationDate="2014-01-09T19:19:08.147" Score="4" Body="&lt;p&gt;You have many questions. I only answer the first. It's doesn't only matter how heavy a star is, but also how big. For ordinary stars, the effect is neglible (work it out by yourself -- it's a useful exercise). Even for compact stars, such as white dwarves or neutron stars, the effect is small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what astronomers commonly refer to as (stellar-mass) black holes, may actually be &lt;em&gt;strange stars&lt;/em&gt; consisting of some quark-gluon plasma (a white dwarf is like a big crystal, a neutron star like a big atomic nucleus, a strange star like a big neutron). These stars would have a high gravitational redshift (1000 or more) at their surface, such that the surface is effectively invisible. This makes them very hard/impossible to distinguish form &quot;real&quot; black holes.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2014-01-09T19:28:26.157" LastActivityDate="2014-01-09T19:28:26.157" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1366" PostTypeId="2" ParentId="1357" CreationDate="2014-01-09T19:22:55.627" Score="4" Body="&lt;p&gt;That's a lot of not quite trivial questions! I'll try to answer part of it. First, the red-shift can be composed of relativistic Doppler-effect and gravitational red-shift. When neglecting the gravitationl part, we get a higher radial velocity. The radial velocity can be used to calculate a distance estimate via the Hubble &quot;constant&quot;. So for low velocities the gravity part is not negligible in respect to relative errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars are moving more or less randomly. Therefore it's necessary to look at a sufficiently large population of stars or galaxies to get a distance estimate that way. For low red-shifts this doesn't work reliably. For larger red-shifts, the gravity part plays a minor relative role, as long as one looks at usual stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A region of high gravity can be detected by gravitational lensing effects. So this source of error can be avoided by working properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Einstein's &quot;blunder&quot; was the assumption, that the universe needs to be static on large scales. Therefore he induced a non-zero cosmological constant to avoid the universe to expand or collapse. He could have predicted the big bang by assuming the constant to be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking further away means looking into the past, when the universe has been denser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hollow-world scenario with a dense shell, heavy enough to cause the observed redshift would probably collapse rapidly to the shell. If the shell as a whole doesn't collapse, kind of anti-gravity provided by a cosmological constant or function would have to be provided. But this would probably also annihilate the red-shift, thus not consistent with observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a collapsing universe objects would look blue-shifted instead of red-shifted. The degree of blue-shift would be dependent of the way the collapse would take place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These things are relevant to be considered to exclude an option of an alternate spacetime which could explain observations. I could recommend reading more about the Planck results, that many options are actually considered, e.g. starting with &lt;a href=&quot;http://scienceblogs.com/startswithabang/2013/03/21/what-the-entire-universe-is-made-of-thanks-to-planck/&quot; rel=&quot;nofollow&quot;&gt;this blog&lt;/a&gt;, then continuing with &lt;a href=&quot;http://www.sciops.esa.int/index.php?project=PLANCK&amp;amp;page=Planck_Published_Papers&quot; rel=&quot;nofollow&quot;&gt;the original Planck papers&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-01-09T19:22:55.627" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="1367" PostTypeId="1" AcceptedAnswerId="1370" CreationDate="2014-01-09T20:45:06.780" Score="7" ViewCount="995" Body="&lt;p&gt;When we see a distant galaxy, the light we are watching started it's journey millions of years ago. In all that time, space has been expanding, so if the initial distance was A, right now the distance would be A+B and, surely, light would take more time to travel A+B than it originally did A.&lt;/p&gt;&#xA;" OwnerUserId="537" LastActivityDate="2014-01-09T23:42:10.350" Title="When we say a galaxy is 200 million light years away, does this account for the expansion of space in the time it took it's light to reach us?" Tags="&lt;distances&gt;&lt;light&gt;&lt;space-time&gt;&lt;intergalactic-space&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1368" PostTypeId="1" CreationDate="2014-01-09T21:07:36.150" Score="7" ViewCount="418" Body="&lt;p&gt;Since the cosmological constant is not required to explain that the universe seem to be expanding, why do we have it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What &lt;em&gt;other factors&lt;/em&gt; cause us to have that constant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Background:&#xA;Without the cosmological constant, distant stars should be affected by a great redshift. The amount of redshift is a function of their distance from us. This is due to gravitational time dilation. We are looking 13 billion years into the past, where the universe was very dense. Those stars should be experiencing extreme gravity, causing Einstein shift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we DO have the cosmological constant, we are now looking for other explanations for the redshift.&lt;/p&gt;&#xA;" OwnerUserId="749" LastActivityDate="2014-05-14T14:47:11.777" Title="Why do we have the cosmological constant?" Tags="&lt;gravity&gt;&lt;space-time&gt;&lt;redshift&gt;&lt;cosmological-inflation&gt;" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="1369" PostTypeId="2" ParentId="1357" CreationDate="2014-01-09T21:30:29.983" Score="8" Body="&lt;p&gt;I will make a small calculation here, but please proceed to the results if you may like to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Calculation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars are spherical and static, so metric near their surface (photosphere) and outside on is Schwarzschild. Hence time-time metric component on the surface is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g_{44}=1-\dfrac{R_{grav,*}}{R_*}$$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R_*$ is the radius of the star and $R_{grav,*}$ is its gravitational radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, if the velocity of the star is much smaller than the speed of light, gravitational redshift in the lowest order does not depend on this velocity. Therefore, the emitting star can be assumed at rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The light from the star propagates along isotropic geodesic in Schwarzschild metric. The geodesic is described by lagrangian:&#xA;$$ \mathcal{L} = \dfrac{1}{2}g_{\mu\nu}k^{\mu}k^{\nu}$$,&#xA;where $k^\mu=(\vec{k},\omega/c)$ is the 4-vector of the light wave and $\omega$ is the frequency of light. Since the metric is static $\dfrac{d\mathcal{L}}{dk^4}=g_{\mu 4} k^4=g_{44}k^4=\textrm{const}$. Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1-\dfrac{R_{grav,*}}{R_*})\omega = \textrm{const}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the light as it travels towards us. So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega_{obs}=\omega_{emitted}(1-\dfrac{R_{grav,*}}{R_*})\Longleftrightarrow \lambda_{obs}=\dfrac{\lambda_{emitted}}{(1-\dfrac{R_{grav,*}}{R_*})}$$,&#xA;where $\lambda$ is the wavelength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Redshift is simply $z=\dfrac{\lambda_{obs}-\lambda_{emitted}}{\lambda_{emitted}}$. Assuming $z\ll 1$ one has a simple formula:&#xA;$$&#xA;z_0=\dfrac{R_{grav,*}}{R_*}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $z_0$ turns out to be comparable to unity, one should compute&#xA;$$&#xA;z=\dfrac{1}{1-z_0}-1,&#xA;$$&#xA;which then gives the correct value of redshift. Note that redshift does not depend on $\lambda$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nice numerical forms for this would come from $R_{grav,*}=2.95\textrm{km}\dfrac{M_*}{M_\odot}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z_0=0.295\dfrac{10\textrm{km}}{R_*}\dfrac{M_*}{M_\odot}\Longleftrightarrow z_0=4.24\cdot 10^{-6}\dfrac{R_\odot}{R_*}\dfrac{M_*}{M_\odot}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also nice to express the redshift in $\textrm{km}/\textrm{s}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;z_0=8.84\cdot 10^4 \dfrac{10\textrm{km}}{R_*}\dfrac{M_*}{M_\odot} \textrm{km/s}\Longleftrightarrow z_0=1.27 \dfrac{R_\odot}{R_*}\dfrac{M_*}{M_\odot}\textrm{km/s}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary and discussion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, when redshift $z$ is small, it is approximated by $z_0$, the numerical expressions for which are given just above. If $z_0$ turns out to be not small, one can calculate $z=\dfrac{1}{1-z_0}-1$, which then gives correct redshift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Stars&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can see that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For normal stars like the Sun ($R_*\sim R_\odot, M_*\sim M_\odot$) the redshift is of order $1\textrm{km/s}$. It is almost important since stars in solar neighbourhood move normaly at the velocity of a few tens of $\textrm{km/s}$&lt;/li&gt;&#xA;&lt;li&gt;For white dwarfs ($R_*\sim 10^4 \textrm{km}, M_*\sim M_\odot$) redshift is a few times $100 \textrm{km/s}$ and gets very important when doing proper spectroscopy. Therefore one typically accounts for it.&lt;/li&gt;&#xA;&lt;li&gt;For neutron stars ($R_*\sim 10 \textrm{km}, M_*\sim M_\odot$) redshift is very important $z\sim 0.4$, but neutron stars are general relativistic objects anyway, so it would be expected beforehand.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, in summary, when measuring light from individual stars one does have to account for gravitational redshifts to get accurate results, and particularly so when studying white dwarfs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Groups of objects&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the same formulae are correct to an order of magnitude when applied for larger volumes of space, with $R_*$ and $M_*$ now meaning the size of the volume and the mass inside it. However, as typical interstellar distances are of order of parsec and $\textrm{pc}=3\cdot10^{13}\textrm{km}$, resulting $z$ will be very small even for such dense groups as globular clusters ($z$ is of order $10^{-8}$ in this case). So, groups of objects do not affect the redshift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Cosmological overdensities&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, cosmological scale underdensities of order few $100\textrm{Mpc}$ in size may affect the apparent redshift of distant objects, as we would be inside the underdensity. However, such an underdensity would have to be significantly symmetric around us in order to explain the lack of corresponding anisotropy in cosmic microwave background. Therefore, it is considered unlikely.&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2014-01-09T21:30:29.983" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1370" PostTypeId="2" ParentId="1367" CreationDate="2014-01-09T22:14:16.087" Score="6" Body="&lt;p&gt;Usually the distance data doesn't take the expansion of the universe fully into account. That's not important for a distance of 200 million lightyears, but as we get close to the border of the visible universe, it makes a relevant difference. Therefore astronomers prefer to talk of redshift instead of distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be a little more precise: The light travelling from a galaxy to Earth takes only that part of the expansion of the space between the galaxy and the Earth into account, which it still has to traverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a proper distance measurement in the everyday's sense, we would have to send light to the galaxy and reflect it back to Earth, measure the time, divide it by the speed of light, and take the half of the resulting distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this experiment would be performed, some of the now visible objects would never reflect light back to Earth, because they are already too far away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more detailed discussion about the various notions of distance, read &lt;a href=&quot;http://en.wikipedia.org/wiki/Distance_measures_%28cosmology%29&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's article on distance measures in cosmology&lt;/a&gt;.&#xA;The usual distance data is provided in the (one-way) light travel distance.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-01-09T23:42:10.350" LastActivityDate="2014-01-09T23:42:10.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1373" PostTypeId="1" CreationDate="2014-01-10T00:35:25.417" Score="4" ViewCount="84" Body="&lt;p&gt;When a planet is spinning around its own axis, it has an effect on the trajectory of its satellites. I believe it is called frame dragging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spin increases the kinetic energy of an object, contributing to its mass-energy, but some other effect must come into play as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one differentiate between spin and orbit velocity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the &lt;em&gt;effect&lt;/em&gt; of the spin? Lets say that Earth suddenly started spinning faster, what would happen to the moons trajectory?&lt;/p&gt;&#xA;" OwnerUserId="749" LastActivityDate="2014-03-13T21:15:47.920" Title="Rotation and relativity" Tags="&lt;rotation&gt;&lt;general-relativity&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1374" PostTypeId="2" ParentId="925" CreationDate="2014-01-10T01:02:28.013" Score="10" Body="&lt;p&gt;I think, by his mention of asteroids, that the OP means an object in the scale of asteroids or dwarf planets, that is, almost-solid objects. In this case, asteroids the size of 2 Pallas or 4 Vesta are ellipsoids, and smaller than that, we have 511 Davida which is the biggest clearly non-spherical asteroid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9jPuo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-01-10T01:02:28.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1375" PostTypeId="1" AcceptedAnswerId="1396" CreationDate="2014-01-10T01:11:12.637" Score="9" ViewCount="1511" Body="&lt;p&gt;The speed of light in a vacuum is presumably the fastest speed possible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;if gravity bends the course of light, does this imply that gravity the retards light so it is moving at a slower speed?  If it affects its course, why cannot gravity affect its speed -- or does it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if gravity does affect the speed of light, what does that say about our measurements of the distance to the farthest observable object?  Could we assume that all the gravity effects across 15 billion light-years even themselves out?  Or is the actual distance across the observable universe subject to unknowable variations due to gravity effects?&lt;/p&gt;&#xA;" OwnerUserId="718" LastEditorUserId="774" LastEditDate="2014-01-11T23:49:52.677" LastActivityDate="2017-09-07T23:58:25.540" Title="Does gravity slow or speed light up?" Tags="&lt;gravity&gt;&lt;light&gt;&lt;general-relativity&gt;&lt;speed&gt;&lt;special-relativity&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1376" PostTypeId="1" AcceptedAnswerId="1384" CreationDate="2014-01-10T01:56:22.027" Score="14" ViewCount="14785" Body="&lt;p&gt;At some points in the year, the moon at my location can be clearly visible during the day. Why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is any help, I live in Alberta, Canada. &lt;/p&gt;&#xA;" OwnerUserId="701" LastActivityDate="2014-07-23T20:58:30.493" Title="Why can we sometimes see the moon during the day?" Tags="&lt;the-moon&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1377" PostTypeId="2" ParentId="1375" CreationDate="2014-01-10T02:34:44.563" Score="2" Body="&lt;p&gt;This is a tough one, especially as I'm not used to giving explanations in non technical terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting at the top:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conditionally yes. In the emptiest space possible - not that between stars, not that between galaxies, not that between families of galaxies and so on.....  in the very emptiest space between the super-clusters of galaxies, there it's at its fastest, where gravity is at its weakest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had the time to do so, and a nice clear target black hole and shot a blue laser just at the event horizon on one side (say it's transmitting the whole of the works of Shakespeare followed by the rest of project Gutenberg) - in such a way that it skimmed all the way round and then exited back in your direction, like a slingshot of the moon the first orbit of the moon did, what would happen? Would the light colour change?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closer the beam got to the event horizon the more stretched space is - think of it that way, then the light has to travel further, and the same all the way round the black hole - the closer to the event horizon the deeper the well, the more the space is stretched and the longer the light takes to get around. From your point of view, the black hole is X distance away, the path the light took is Y in apparent length. Using your handy slide rule you calculate the Taming of the shrew should arrive at time Z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't turn up on time. Why? Remember the light had to take a very long path because of the gravity field density making the journey longer. When it does turn up finally What colour is it? Still blue - this does not depend on if the black hole is moving away or closer - there is no red or blue shift.&#xA;(I am being slightly disingenuous here as the wavelength would have shifted a minute amount to the red - it does this as it travels, the further it does the more it shifts, partly from collisions with free floating atoms that absorb then re emit at a lower frequency eg. the big bang (Very hot) - the light from this is very long wavelength indeed, (red shifted to the extreme) but space is expanding remember. To put it in a nutshell, entropy cannot be reversed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The odd thing is the distance the light travels from the point of view of the observer who shot the laser, he would extrapolate that the light waves containing The Shrew, since they arrived so late must have not only slowed down but gotten closer together (blue shifted) - but when it comes back to the observer it's just the same colour as before. (Space lengthened apparently, that would explain this wouldn't it?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To say that gravity slows light is the same as saying a watched kettle never boils, it has a kind of truth to it from a particular point of view - a perceptual point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the whole universe, there are visible hot spots and cold spots, places with more and less matter - this can be observed. The trouble we're having at the moment is with dark matter, and dark energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We started with observations in our own solar system. Distant objects are all measured relative to each other. A large number of observations are made of many objects, their luminosity, their aggregate luminosity, their red or blue shift - and interestingly their change in Doppler shift. Various different type of star, pulsating ones, stars that emit hard radiation, co-orbiting stars of all sorts, the accretion disks at the center of galaxies and their temperatures, This accumulation of data since Copernicus, or at least since the Renaissance has all been put together, adjusting along the way taking into account  world changing paradigm shifts such as relativity, and huge advances in the resolution of our observations of the universe, from land and space based platforms has (we think!) made our estimations of the error margin of our calculations of the age of the universe smaller and smaller. &lt;/p&gt;&#xA;" OwnerUserId="768" LastEditorUserId="111" LastEditDate="2014-01-10T23:46:18.077" LastActivityDate="2014-01-10T23:46:18.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1378" PostTypeId="2" ParentId="1356" CreationDate="2014-01-10T02:49:02.610" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Gravity is sometimes described as a curvature in space-time. Due to relativity, doesn't this imply that gravity doesn't propagate?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's a fairly precise sense in which gravity propagates: if you have a spacetime and you perturb it a bit, then you can think of the new spacetime as the old &quot;background&quot; spacetime with a small change on top of it. Then it makes sense to discuss the speed at which this change propagates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the general case, the speed of propagation of gravity has no particular meaning. This is sensible: to talk about speed, you need some standard to compare it to, and hence a background spacetime. But gravity is nonlinear, so to have an objective answer to that question, the changes to that background need to be either small or tightly constrained.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If a black hole was moving toward you at the speed of light ... [would you] experience something similar to a supersonic blast, except its a gravity blast, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not analogously to a supersonic (so here, superluminal?) one, no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an electric charge, the ultra-relativistic limit of its electromagnetic field is a plane wave that's impulsive, an infinitely-thin Dirac delta profile. This travels at the speed of light. The gravitational analogue of this (for a Schwarzschild black hole) is the Aichelburg-Sexl ultraboost, which is a spacetime that's axially symmetric and everywhere flat except for an impulsive plane wave.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... and it would allow you to remain unaffected until you arrived at the very centre of the black hole?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you were able to turn around, just before you arrive at the center, and travel away from the black hole at the speed of light you would never notice it, even though you would be travelling at the same velocity as the black hole in the same direction?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's the same thing as saying you're stationary above the black hole. Hence, you would notice it because you'd need to be accelerated to stay stationary. Plus, if you're turning around after passing the horizon, it's far too late.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-01-10T02:49:02.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1379" PostTypeId="2" ParentId="1376" CreationDate="2014-01-10T02:59:00.943" Score="11" Body="&lt;p&gt;Well, the Moon reflects quite a bit of the Sun's radiation. In fact, &lt;a href=&quot;http://web.archive.org/web/20080523151225/http://jeff.medkeff.com/astro/lunar/obs_tech/albedo.htm&quot; rel=&quot;noreferrer&quot;&gt;this person&lt;/a&gt; states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The visual geometric &lt;a href=&quot;http://en.wikipedia.org/wiki/Albedo&quot; rel=&quot;noreferrer&quot;&gt;albedo&lt;/a&gt; of the full moon is 12.5%, but much less at other phases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In combination with this, the Moon is also the closest astronomical object to us, and since the intensity of light falls off as $\propto \frac{1}{r^{2}}$ (&lt;a href=&quot;https://en.wikipedia.org/wiki/Inverse-square_law&quot; rel=&quot;noreferrer&quot;&gt;inverse-square law&lt;/a&gt;), this plays arguably the biggest role in being able to see the Moon during the day. In fact, so much light from the Moon gets reflected to Earth that its apparent magnitude must be larger than the sky during the day (otherwise it would get washed out by sky noise).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to this &lt;a href=&quot;http://newswatch.nationalgeographic.com/2011/07/10/how-bright-is-the-daytime-sky/&quot; rel=&quot;noreferrer&quot;&gt;article&lt;/a&gt;, the apparent magnitude of the daytime sky is somewhere between 1.5 and 3. When you compare this to the full moon (–12.74; though you really can't see this during the day due to the geometry of the Earth/Sun/Moon system), or even a new moon at its brightest (-2.50), the Moon can almost always be seen in the sky.&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="13" LastEditDate="2014-07-23T01:42:24.213" LastActivityDate="2014-07-23T01:42:24.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1380" PostTypeId="2" ParentId="1368" CreationDate="2014-01-10T04:17:55.497" Score="9" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Reason 1:&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let's look at the Friedmann equations without the cosmological constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  \frac{\dot{a}^2 }{a^2} = \frac{8 \pi G \rho}{3}-\frac{kc^2}{a^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term on the LHS is just the Hubble constant squared $H^2$ which can be measured the direct measurement of recession velocity of galaxies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The density term can be said to be a combination of $\rho_{matter}+\rho_{dark- matter}$ both of which can be measured directly;$p_{matter}$ by observation of matter in our galaxy and other galaxies while $\rho_{dark- matter}$ by rotation curves of galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The curvature constant $k$ can be estimated today by the anisotropy measurements in the CMBR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it turns out the parameters don't fit and we need more mass-energy in the universe(almost 2-3 times of that we had estimated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So comes along &lt;strong&gt;Dark energy&lt;/strong&gt; or basically the cosmological constant. Cosmological constant or the dark energy are just two ways of looking at the equation,either as just a constant or a form of mass-energy(though we have solid reasons to believe the latter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is our picture of the universe today:&lt;img src=&quot;https://i.stack.imgur.com/PBW4z.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Reason 2:&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now historically the cosmological constant was necessary for an altogether different reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second Friedmann equation without the cosmological constant looks:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$   \frac{\ddot{a}}{a} = -\frac{4 \pi G}{3}\left(\rho+\frac{3p}{c^2}\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this predicts for normal type of matter,the universe must decelerate.($\ddot{a}&amp;lt;0$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now,people measured the redshift of the type-1a supernovae and found out the quite paradoxical result that the universe was being accelerated in its expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TuP1e.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since normal matter can't explain this type or behaviour,we again have to look at Dark Energy(or the cosmological constant).And so with the cosmological constant the equation becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\ddot{a}}{a} = -\frac{4 \pi G}{3}\left(\rho+\frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\ddot{a}&amp;gt;0$ is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Therefore the cosmological constant is necessary to both explain the current rate of expansion and the accelerated expansion.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So finally the accelerated expansion can be explained and today we have the $ΛCDM$ model of the universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1:&lt;a href=&quot;http://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Friedmann_equations&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2:&lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/astro/univacc.html&quot; rel=&quot;noreferrer&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/hbase/astro/univacc.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3:&lt;a href=&quot;http://en.wikipedia.org/wiki/Lambda-CDM_model&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Lambda-CDM_model&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="450" LastActivityDate="2014-01-10T04:17:55.497" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1382" PostTypeId="2" ParentId="1359" CreationDate="2014-01-10T06:03:52.270" Score="1" Body="&lt;p&gt;If the planet was made of gas or liquid, and was not spinning it would be a perfect sphere, because with each atom settling as close to the center as possible, that's the shape you get. We can see this on Earth whenever we take a ride in a boat and look out at the ocean. We see the surface of a sphere. (If the planet is spinning, it bulges out around the equator a bit.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rocky planet like Earth is a bit more complicated than a gas or water world (but only a bit).  Continents are lighter than the mantle, so they float in it, but they're rigid, so they can &quot;stick up&quot; a bit like a boat in water, rather than form a smooth layer over the mantle like oil on water.&lt;/p&gt;&#xA;" OwnerUserId="562" LastActivityDate="2014-01-10T06:03:52.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1384" PostTypeId="2" ParentId="1376" CreationDate="2014-01-10T08:21:44.040" Score="14" Body="&lt;p&gt;The Moon makes a full orbit around the Earth once per month. From full moon to full moon, it is 29.53058886 days (synodic month).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Full moon happens when the Moon is at one side of the Earth (right side in the image), and the Sun is at the other side. This way, Moon is fully iluminated by Sun (except in the cases where it is exactly in the opposite side of the same line, where you have a Moon eclipse). In this phase, Moon is at its highest altitude at midnight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, as the Moon orbits, it comes closer to the Sun in the sky. This means both that it shows us less iluminated percentage and that its rise and set get delayed. At Last Quarter, Moon is at its highest altitude at sunrise. This means that you will be able to see the half moon figure on the sky during the morning (and last half of the night).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bew2Q.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then it comes New Moon, where the Moon is nearest to the Sun in the sky, and almost invisible, since it shows us the dark side during the day. Its highest altitude is at noon, but you will not be able to see it. You can, if you are lucky, see a Sun Eclipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally it comes First Quarter. To put it short, highest altitude is at sunset and you are able to see Moon during all the afternoon (and first half of the night).&lt;/p&gt;&#xA;" OwnerUserId="473" LastEditorUserId="473" LastEditDate="2014-07-23T20:58:30.493" LastActivityDate="2014-07-23T20:58:30.493" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1385" PostTypeId="1" AcceptedAnswerId="1386" CreationDate="2014-01-10T08:21:49.813" Score="5" ViewCount="139" Body="&lt;p&gt;What's the name of the measure of apparent (visible) size of a body? e.g. the Sun and the Moon appear to be roughly the same size, despite the Sun being vastly bigger - because the Moon is much closer. What's this apparent size called?&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2017-03-27T14:20:02.137" Title="apparent size of an astronomical body" Tags="&lt;units&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1386" PostTypeId="2" ParentId="1385" CreationDate="2014-01-10T09:21:51.870" Score="6" Body="&lt;p&gt;Angular size. That is, the vieving angle that the body spans on sky.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-01-10T09:21:51.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1387" PostTypeId="1" AcceptedAnswerId="1404" CreationDate="2014-01-10T10:46:04.420" Score="10" ViewCount="1008" Body="&lt;p&gt;Orrery is a clockwork model of the solar system. I am trying to emulate one in 2D. Now, to emulate, I need to know what goes on inside. Can someone please explain the basic principle behind the clockwork? Or direct me to a resource that will explain all the machinery inside a simple Orrery.&lt;/p&gt;&#xA;" OwnerUserId="774" LastEditorUserId="774" LastEditDate="2014-01-11T20:22:06.550" LastActivityDate="2014-01-12T14:21:46.913" Title="Emulation of an Orrery" Tags="&lt;solar-system&gt;&lt;n-body-simulations&gt;&lt;orrery&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1388" PostTypeId="1" AcceptedAnswerId="1395" CreationDate="2014-01-10T10:59:54.160" Score="1" ViewCount="136" Body="&lt;p&gt;This is a question, where I start with some assertions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to consider the universe as a four-coordinate system, x,y,z,t where t is time and where we view a change in t as a change in position, and as a velocity in the same way as any other change in either of the other components of the coordinate. Everything is then moving at the speed of c. The difference between a photon and another particle would be that a photon &quot;spends&quot; its velocity in any of the x, y or z dimensions - but not in the time dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that all photons are on the same coordinate in the t-dimension - only the x,y,z values change while t remains 0. It also implies that time would appear to stop for something travelling in only the other dimensions, while it would pass fastest for something completely stationary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then also consider space-time as a function of acceleration. A physical objects' coordinate could be derived from its velocity, which would have to be derived from its acceleration. Acceleration would be a function of entropy, or &quot;age&quot; I believe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the basis of general relativity? Is it at least an initial starting point for general relativity, or is it completely wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would spin have to be part of the coordinate system?&lt;/p&gt;&#xA;" OwnerUserId="749" LastEditorUserId="749" LastEditDate="2014-01-10T11:27:14.573" LastActivityDate="2014-01-11T00:11:06.057" Title="Age and relativity" Tags="&lt;general-relativity&gt;" AnswerCount="1" CommentCount="16" ClosedDate="2014-01-13T16:10:53.913" ContentLicense="CC BY-SA 3.0" />
  <row Id="1389" PostTypeId="2" ParentId="1333" CreationDate="2014-01-10T12:20:22.903" Score="6" Body="&lt;p&gt;Yes, light gravitates. The gravitational charge is energy. Well, gravity is a spin-2 force, so you really have momentum and stress as well, but they are analogous to a generalization of electric current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, anything that contributes to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stress-energy_tensor&quot; rel=&quot;noreferrer&quot;&gt;stress-energy tensor&lt;/a&gt; will have some gravitational effect, and light does that, having both an energy density and putting a pressure in the direction of propagation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But while that light is travelling through space, its energy is &quot;unavailable&quot; to the rest of the universe. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not quite. It still gravitates. However, the radiation-dominated era was before about 50k years after the Big Bang, but it is long past. Today the gravitational effect of radiation is cosmologically negligible. We live in a transition between matter-dominated and dark-energy-dominated eras.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given that all stars on the sky is sending photons that reaches every square centimeter of the earth surface, the amount of pressure should sum up to be quite large.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The light pressure on any surface is proportional to the light energy density incident on it. Thus we can check this line of reasoning directly by observing that the sky is dark at night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Why&lt;/em&gt; it is dark at night is probably deserves its own question (cf. also &lt;a href=&quot;http://en.wikipedia.org/wiki/Olbers%27_paradox&quot; rel=&quot;noreferrer&quot;&gt;Olbers' paradox&lt;/a&gt;), but it is pretty clear that it is in fact quite small. To be fair, we should check more than the visible range, but even so the sky is pretty dark. Thus on average, light pressure is very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have the privilege of being close to a star, but even during the day, the light pressure due to the Sun is on the order of micropascals.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... there should be energy amounting to the gravity of about 1.7*10^22 suns distributed throughout the universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And this is a tiny amount. As you just said, this is the equivalent of about 0.034% of the total mass of stars in the universe, which is in turn constitute but a fraction of the matter in the universe. So why you are surprised that its effect is negligible? It's literally thousands of times less than the uncertainty in the measurements of the amount of matter in the universe.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-01-10T12:20:22.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1390" PostTypeId="2" ParentId="1385" CreationDate="2014-01-10T13:26:14.757" Score="12" Body="&lt;p&gt;It's &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_diameter&quot; rel=&quot;nofollow noreferrer&quot;&gt;angular diameter&lt;/a&gt;&lt;/strong&gt; (a.k.a &lt;strong&gt;angular size&lt;/strong&gt; or &lt;strong&gt;apparent size&lt;/strong&gt;). This &lt;a href=&quot;http://xkcd.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;XKCD&lt;/a&gt; comic strip shows in an interesting way the angular size of some of the most remarkable objects in the sky:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EaP5q.png&quot; alt=&quot;Comic about apparent size of celestial bodies&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="537" LastEditorUserId="16433" LastEditDate="2017-03-27T14:20:02.137" LastActivityDate="2017-03-27T14:20:02.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1391" PostTypeId="2" ParentId="1359" CreationDate="2014-01-10T15:14:32.940" Score="3" Body="&lt;p&gt;When a planetary system is born, most material to build bodies is in fine dust(gas state/plasma) because it has termo-dynamic high , thus is easy that gravity force create regular spherical forms width very weight materials as iron (earth center) after that system planetary is yougest (lower termo-dynamic),it has planets and smal bodies with regular forms but with trajectory collision, after has collisions occurred , it has regular and irregular bodies, when it reach state like as solar system, it has regular forms(spherical) semi-regular(with flat surface but irregular form as phobos) and irregular (asteroids),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other razons:   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;because the main axis force from Centripetal force of solar system, has a direction to west-east axis (earth)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some bodies(planets and satellites) has liquid state matter and this dragg solid matter in her surface&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tectonic plate changes create deformation in surface bodies( Monts etc.)&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2014-01-10T15:14:32.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1392" PostTypeId="1" AcceptedAnswerId="1393" CreationDate="2014-01-10T17:38:07.320" Score="9" ViewCount="130" Body="&lt;p&gt;As for example explained at the beginning of &lt;a href=&quot;http://motls.blogspot.de/2014/01/trinary-star-wont-disprove-equivalence.html?m=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; blog post, the trinary system consists of a millisecond pulsar ($1.438$ times the mass of the sun) orbited by two white dwarfs. One of the white dwarfs ( $0.198$ solar masses) is very close to the pulsar and has an orbit period of $1.6$ d, whereas the other one ($0.410$ solar masses) is farther away and and needs about a year ($327$ d) to orbit the central pulsar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a three-body system is in principle expected to show a chaotic behavior sooner or later, which means that collisions between these three celestial bodies can be expected and a finit life time of the system can be assumed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making some in my opinion way too hand-waving arguments, the blog post further explains that the collisions can not be expected too soon however, by taking into account that the distant white dwarf &quot;sees&quot; the inner white dwarf and the pulsar as a single central body and the relative motion of the inner white dwarf around the pulsar is rather stable and eliptic too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking about such multiple star systems as chaotic dynamical systems, another approach to estimate the liftime could be to make use of some chaos-theoretic methods which could for example involve the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lyapunov_exponent&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lyapunov Exponent&lt;/a&gt; of the system, such that a large exponent would mean that the collisions happen soon and the stellar system has a rather short life time, whereas the converse would be true if the Lyapunov exponent is small (which is what I would expect for the system in my question).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in short my question is: how can the liftime of a multiple star system be calculated in a not just hand-waving way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/q/58/92&quot;&gt;This&lt;/a&gt; question is interestingly related to my issue, but it does not yet answer it...&lt;/p&gt;&#xA;" OwnerUserId="92" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2014-01-10T21:46:17.103" Title="How can the life time of a multiple star system, such as for example the trinary system PSR J0337+1715 be derived?" Tags="&lt;stellar-dynamics&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1393" PostTypeId="2" ParentId="1392" CreationDate="2014-01-10T20:46:53.307" Score="6" Body="&lt;p&gt;&lt;em&gt;Such a three-body system is in principle expected to show a chaotic behavior sooner or later.&lt;/em&gt; &lt;strong&gt;Nope&lt;/strong&gt;. &lt;em&gt;Hierchical&lt;/em&gt; multiple systems (such as this), where the semi-major axes differ by a factor ten or larger may well be stable for ever (never become chaotic), in particular if the eccentricities are low and if the most massive object is in a tight binary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An unstable three-particle system will eventually result in (typically) the two most massive objects in a tight binary and the third particle ejected (unbound). The time scale for this to happen is of the order of several (10-100) dynamical times and is indeed a highly chaotic process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept of the Lyapunov time scale is not too useful here. One issue is that as soon as one object is ejected (unbound), the system is no longer bounded, when the concept of Lyapunov becomes problematic. Another issue is that the Lyapunov time is defined in the limit of infinite time and does not necessary reflect the behaviour of the system over any finite time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, to &lt;strong&gt;answer your question&lt;/strong&gt;. I think there is no stringent way. What one can do is to integrate numerically many realisations of the system, each equally in agreement with the data (and their uncertainties). Then one can see whether there are any stable configurations and how frequent they occur. Given that the system didn't form yesterday, it appears likely that it is indeed stable.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2014-01-10T21:46:17.103" LastActivityDate="2014-01-10T21:46:17.103" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1394" PostTypeId="2" ParentId="1375" CreationDate="2014-01-10T21:53:27.693" Score="14" Body="&lt;p&gt;Gravity doesn't affect the &lt;em&gt;speed&lt;/em&gt; of light. It affects the space-time geometry and hence the &lt;em&gt;paths&lt;/em&gt; of light. However, this can have a similar effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light emitted at source $S$ to pass a massive object $M$ that is very close on the otherwise (if &lt;em&gt;M&lt;/em&gt; weren't there) straight path to an observer $O$ has to &quot;go around&quot; $M$, which takes longer than following the straight path in the absence of $M$. The light that reaches $O$ from $S$ is not emitted from $S$ in the &quot;straight&quot; (in the absence of $M$) direction to $O$, but slightly off that direction, such that the &quot;bending&quot; of its path by the gravity of $M$ &quot;deflects&quot; it onto $O$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, light is never bent, but always follows a straight path. What is bent is the space-time as compared to Euclidean space-time in the absence of distorting masses (see: &lt;a href=&quot;http://en.wikipedia.org/wiki/Geodesics_in_general_relativity&quot;&gt;geodesic&lt;/a&gt;). This distortion in the fabric of space time is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_lens&quot;&gt;gravitational lensing&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="45" LastEditDate="2014-01-12T19:49:01.547" LastActivityDate="2014-01-12T19:49:01.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1395" PostTypeId="2" ParentId="1388" CreationDate="2014-01-11T00:11:06.057" Score="2" Body="&lt;p&gt;I think, you are trying to describe a &lt;a href=&quot;http://en.wikipedia.org/wiki/World_line&quot; rel=&quot;nofollow&quot;&gt;world line&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-01-11T00:11:06.057" CommentCount="6" CommunityOwnedDate="2016-10-26T18:08:53.133" ContentLicense="CC BY-SA 3.0" />
  <row Id="1396" PostTypeId="2" ParentId="1375" CreationDate="2014-01-11T02:40:57.950" Score="13" Body="&lt;p&gt;As Walter says, gravity doesn't bend light. Light travels along null geodesics, a particular type of straight path. Since (affine) geodesics don't change direction by definition, geometrically light trajectories are straight. Moreover, the speed of light in vacuum is $c$ in every inertial frame, regardless of whether or not spacetime is curved, although a curved spacetime inertial frames can only ever be local.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can change, however, is the &lt;em&gt;coordinate speed&lt;/em&gt; of light. Since coordinates are just labels for spacetime events, this is true even in completely flat spacetime. For example, in the Rindler coordinate chart, the Minkowski metric of flat spacetime takes the form&#xA;$$\mathrm{d}s^2 = -\frac{g^2x^2}{c^2}\,\mathrm{d}t^2 + \underbrace{\mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2}_{\mathrm{d}S^2_\text{Euclid}}\text{,}$$&#xA;where $g$ has units of acceleration. Since light travels along null ($\mathrm{d}s^2 = 0$) wordlines, the coordinate speed of light is&#xA;$$\frac{\mathrm{d}S}{\mathrm{d}t} = \frac{|gx|}{c}\text{,}$$&#xA;which is position-dependent and can even be $0$, as there is an apparent event horizon. An observer that's stationary in Rindler coordinates actually has proper acceleration $g$, so the Rindler chart of flat spacetime is a natural analogue of a &quot;uniform gravitational field&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;if gravity bends the course of light, does this imply that gravity the retards light so it is moving at a slower speed?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, but what we can say is this. For weak, slowly-changing gravitational fields, the following  metric is appropriate to describe spacetime in terms of the Newtonian gravitational potential $\Phi$:&#xA;$$\mathrm{d}s^2 = -\left(1+2\frac{\Phi}{c^2}\right)c^2\,\mathrm{d}t^2 + \left(1-2\frac{\Phi}{c^2}\right)\,\mathrm{d}S^2\text{,}$$&#xA;as we can readily calculate the coordinate speed of light (again $\mathrm{d}s^2 = 0$):&#xA;$$\frac{\mathrm{d}S}{\mathrm{d}t} = c\sqrt{\frac{1+2\Phi/c^2}{1-2\Phi/c^2}}\text{,}$$&#xA;and so by expanding its reciprocal in a Taylor-MacLaurin series, we find that light travels &quot;&lt;em&gt;as if&lt;/em&gt;&quot; we had an &lt;a href=&quot;http://en.wikipedia.org/wiki/Refractive_index&quot;&gt;index of refraction&lt;/a&gt;&#xA;$$n = c \frac{\mathrm{d}t}{\mathrm{d}S} \approx 1 - 2\frac{\Phi}{c^2} + \mathcal{O}\left(\frac{\Phi^2}{c^4}\right)\text{.}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we keep in mind that we are dealing with just the coordinate speed of light, then yes, we could say that gravity (rather, gravitational potential) retards light. Another way of thinking of this is like so: if we pretend that we're dealing with ordinary flat Minwkoski spacetime in the usual inertial coordinates, then we need a medium with the above refractive index to reproduce the trajectories of light. But of course taking this literally is not legitimate, since (1) the metric affects more than the propagation of light, and (2) such an interpretation would fail to explain gravitational redshift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter approach is morally similar to what's described in Walter's answer, since it depends on a hypothetical comparison with flat spacetime. The difference is that by limiting ourselves to talk about what happens far from the gravitating bodies, Walter can sidestep the gravitational redshift issue, but then cannot ascribe any local refractive index (on the plus side, his approach is not limited to weak, slowly-changing gravity).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And if gravity does affect the speed of light, what does that say about our measurements of the distance to the farthest observable object? Could we assume that all the gravity effects across 15 billion light-years even themselves out?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our cosmological models &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmological_principle&quot;&gt;assume&lt;/a&gt; that the universe is on the large scale homogeneous and isotropic, an assumption which is backed by &lt;a href=&quot;http://sci.esa.int/planck/51551-simple-but-challenging-the-universe-according-to-planck/&quot;&gt;observations&lt;/a&gt; of the parts of it we can see. In a homogeneous and isotropic universe, it's fairly easy to account for how light behaves when traversing it. So no, we don't need to assume that the effects gravity even themselves out--on the contrary, we use such gravitational effects on light to fit the parameters of our models.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditDate="2014-01-14T01:14:13.793" LastActivityDate="2014-01-14T01:14:13.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1397" PostTypeId="1" CreationDate="2014-01-11T11:50:28.673" Score="13" ViewCount="623" Body="&lt;p&gt;The universe supposedly started off with equal parts matter and antimatter and they are said to annihilate each other. Also, nature is famous for it's balance between everything. So all we see is matter and no antimatter (I guess that's a good thing, we would all have been annihilated otherwise), so where did all the antimatter go?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, does antimatter exist in the same space as matter? What are the leading theories about it?&lt;/p&gt;&#xA;" OwnerUserId="774" LastEditorUserId="774" LastEditDate="2014-01-13T16:30:32.870" LastActivityDate="2014-02-22T13:44:55.030" Title="Where is all the antimatter?" Tags="&lt;universe&gt;&lt;formation&gt;&lt;matter&gt;&lt;early-universe&gt;&lt;antimatter&gt;" AnswerCount="1" CommentCount="12" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1398" PostTypeId="1" CreationDate="2014-01-11T16:45:29.157" Score="9" ViewCount="639" Body="&lt;p&gt;I am involved in the fight between palaeontologists and aerodynamicists on the theoretical flight capabilities of the largest pterosaurs (who reached their climax in the Cretaceous and then disappeared at the K/T extinction event).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is it possible that the earth's atmosphere was reduced at this time by a passing rogue planet or other sufficiently large body?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, the atmosphere would have been thicker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the principle of Occam's razor, this one single hypothesis solves the problem of flight in the Mesozoic; the problem of how the dinosaurs could carry such weight; why they all had tails that would be much more use in a thick medium such as water...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The K/T extinction has attracted too many theories about what was 'added' to life on earth such as the comet/asteroid/Deccan Traps theories etc. No-one has looked at what might have been taken away from earth to cause the same effect!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the earth's atmosphere had been much thicker (far higher pressure/density at the surface) the dinosaurs would have had a much easier time. Losing a huge amount of their atmosphere would kill them off because their 'effective weight' would have been doubled almost overnight.&lt;/p&gt;&#xA;" OwnerUserId="782" LastEditorUserId="6" LastEditDate="2014-01-15T15:51:50.657" LastActivityDate="2014-01-15T15:51:50.657" Title="Could the earth's atmosphere have been partially stripped by a passing rogue planet?" Tags="&lt;planetary-atmosphere&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1399" PostTypeId="2" ParentId="1105" CreationDate="2014-01-11T17:30:30.040" Score="5" Body="&lt;p&gt;As freelanceastro noted in his answer: &lt;strong&gt;Yes&lt;/strong&gt;, you can measure the cosmological redshift of quasars with simple equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The page he referred to is not a fabrication. (See the article on amateur astronomical spectroscopy in the August 2011 issue of &lt;em&gt;Sky &amp;amp; Telescope&lt;/em&gt;.) Many amateurs have made this measurement with a variety of instruments, including a modified security camera on an 8&quot; SCT. (That one took just 15 minutes total integration time, which is about what most most amateurs find using typical equipment.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you noted, asawyer, quasars are good targets for red shift measurements because their radial velocity is so high.  The Doppler shift of objects that are closer than 3C 273 is more difficult to measure because their radial velocity is so much lower. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As astromax notes, capturing the spectra of extended objects generally requires a slit, but not always!  For example, a lot of amateurs captured spectra of comet ISON, which certainly isn't a compact object! Check out this one, done with just a DSLR and an 80 mm refractor:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9kWVF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're interested in 3C 273, in case you missed it, there's a wonderful transcript of an interview with its discoverer, Maarten Schmidt here: &lt;a href=&quot;http://www.aip.org/history/ohilist/4861.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.aip.org/history/ohilist/4861.html&lt;/a&gt;. It's rather long. Search the text for the text &quot;December 1962&quot; for the relevant section.  It's a fun read about how he made the discovery and his personal experience of the process.&lt;/p&gt;&#xA;" OwnerUserId="783" LastEditorUserId="783" LastEditDate="2014-01-12T02:29:53.330" LastActivityDate="2014-01-12T02:29:53.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1400" PostTypeId="5" CreationDate="2014-01-11T20:23:55.933" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-01-11T20:23:55.933" LastActivityDate="2014-01-11T20:23:55.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1401" PostTypeId="4" CreationDate="2014-01-11T20:23:55.933" Score="0" Body="A clockwork device that produces a small scale model of the orbits of objects in the solar system" OwnerUserId="774" LastEditorUserId="6" LastEditDate="2014-01-13T17:24:13.157" LastActivityDate="2014-01-13T17:24:13.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1402" PostTypeId="5" CreationDate="2014-01-11T20:24:32.203" Score="0" Body="&lt;p&gt;For every &lt;a href=&quot;/questions/tagged/particle&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;particle&amp;#39;&quot; rel=&quot;tag&quot;&gt;particle&lt;/a&gt; there is an antiparticle, e.g. the positron as antiparticle to the electron. There are also antiatoms, for example antihydrogen which consists from an antiproton and a positron.&lt;/p&gt;&#xA;&lt;p&gt;Particles and antiparticles differ only in that their charges (for example the electric charge) or more precisely their charge-like quantum numbers are opposite. While the electron is electrically negatively charged, the positron is correspondingly positively charged. Other properties such as mass are exactly the same (as far as known so far).&lt;/p&gt;&#xA;&lt;p&gt;In 1928 the physicist Paul Dirac predicted for the first time the existence of antimatter from theoretical considerations, positrons were proved as first antiparticles then in 1932.&lt;/p&gt;&#xA;&lt;h3&gt;Further reading&lt;/h3&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Antimatter&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Antimatter&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Antiparticle&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Antiparticle&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://home.cern/science/physics/antimatter&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://home.cern/science/physics/antimatter&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="37191" LastEditorUserId="37191" LastEditDate="2021-02-17T15:00:00.123" LastActivityDate="2021-02-17T15:00:00.123" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="1403" PostTypeId="4" CreationDate="2014-01-11T20:24:32.203" Score="0" Body="Questions about material composed of antiparticles, which have the same mass as particles of ordinary matter but opposite charge and properties, such as lepton and baryon number. " OwnerUserId="774" LastEditorUserId="37191" LastEditDate="2021-02-17T15:01:13.257" LastActivityDate="2021-02-17T15:01:13.257" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="1404" PostTypeId="2" ParentId="1387" CreationDate="2014-01-12T14:21:46.913" Score="8" Body="&lt;p&gt;The principle is almost exactly the same as a watch or clock, but instead of three concentric axles, you need 9 for the planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a google for Orrery kit - there are loads available. It is really all simple maths - you just need to know relative orbital periods in order to calculate cog sizes.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/wHjjr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;(picture from &lt;a href=&quot;http://www.curiousminds.co.uk/large-fergusons-orrery&quot; rel=&quot;noreferrer&quot;&gt;curiousminds.co.uk&lt;/a&gt;)&#xA;For moons, you do add a little complexity in the form of a transmission to support the Earth, so that the difference in movement can move the Moon. &lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2014-01-12T14:21:46.913" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1405" PostTypeId="1" CreationDate="2014-01-12T16:56:06.117" Score="24" ViewCount="1355" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;[T]his coincidence enabled stars in the late stages of their lives to turn helium into carbon, oxygen and most of the other atoms that you and I are made of. [...] [W]e're made of star stuff.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Max Tegmark, &lt;a href=&quot;https://www.goodreads.com/book/show/17797249&quot;&gt;&lt;em&gt;Our Mathematical Universe: My Quest for the Ultimate Nature of Reality&lt;/em&gt;&lt;/a&gt;, p. 64&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the Sun is not in the late stages of its life, I read that as that we are made mostly of star dust, but &lt;em&gt;not&lt;/em&gt; Sun dust. Rather, we apparently are mostly made of atoms produced by earlier (now dead and buried) stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is whether it is likely that there was &lt;em&gt;one&lt;/em&gt; immidiate (o.a. carbon-producing) predecessor star to the Solar System, or whether the Solar system likely has more than one (perhaps many) (o.a. carbon-producing) &quot;parent stars&quot;? (And, there seems to be also also a third &quot;possibility&quot;: that the build-up of, say, carbon takes more than one &lt;em&gt;generation&lt;/em&gt; of stars.) What's the (hi)story here?&lt;/p&gt;&#xA;" OwnerUserId="789" LastActivityDate="2014-01-15T16:51:01.673" Title="Are we made of the stuff of one star or more stars?" Tags="&lt;star&gt;&lt;earth&gt;&lt;solar-system&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1406" PostTypeId="2" ParentId="1405" CreationDate="2014-01-12T19:44:42.813" Score="17" Body="&lt;p&gt;The precise presolar history isn't known. I'll try to tell a likely story backward in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our sun probably has brothers and sisters scattered throughout the Milky Way. Further back in time our sun likely was part of on open star cluster like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyades_%28star_cluster%29&quot;&gt;Hyades&lt;/a&gt;. Open star clusters are not stable and eject stars over hundreds of millions of years. (&lt;a href=&quot;http://news.nationalgeographic.com/news/2012/01/120117-sun-solar-system-born-stars-m67-space-science/&quot;&gt;More about attempts to find Sun's siblings&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such star clusters typically form in nebulae like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Orion_Nebula&quot;&gt;Orion Nebula&lt;/a&gt;. Dense nebulae typically contain short-lived stars with multiple solar masses, which typically end as &lt;a href=&quot;http://en.wikipedia.org/wiki/Supernova&quot;&gt;supernova&lt;/a&gt;, adding heavy atoms to the nebula and surrounding &lt;a href=&quot;http://en.wikipedia.org/wiki/Interstellar_medium&quot;&gt;interstellar medium&lt;/a&gt;. Our sun is rich in heavy atoms, easy to detect in spectral lines of sunlight or in solar wind; those atoms cannot have formed in the sun, because its core is not yet hot enough to form these elements; some will never form in the sun. Earth mainly consists of chemical elements heavier than helium. Stars like our sun, rich in heavy elements, are called &lt;a href=&quot;http://en.wikipedia.org/wiki/Metallicity&quot;&gt;population I stars&lt;/a&gt; (astronomers call all chemical elements besides hydrogen and helium metals). Population I stars (and Earth) can only have formed, at least in parts, from supernova remnants. Old stars, born in a time, when the immediate predecessor(s) of our sun were born, can still be found in the Milky Way. They are called population II stars. They contain less material made of heavy elements than our sun, but still more than provided by the big bang. Therefore there must have existed even older stars, called population III stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supernovae also can destabilize the interstellar medium due to shock fronts, eventually leading to local collapse especially of nebulae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Presolar_grains&quot;&gt;Presolar grains&lt;/a&gt; are still raining down on Earth; &quot;the elements in these particles were made at different times (and places) in the early Milky Way&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By this, the likely answer to the question is, that we probably consist of remnants of several supernovae, and of several generations of supernovae.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-01-12T23:36:15.677" LastActivityDate="2014-01-12T23:36:15.677" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1407" PostTypeId="2" ParentId="1398" CreationDate="2014-01-12T20:38:14.403" Score="4" Body="&lt;p&gt;Very interesting question. I'm okay with this question being here since you specifically ask for the effect of rogue planets on Earth's atmosphere. Thinking &quot;astronomically&quot; could be useful in coming to conclusions about whether or not your idea is even a possibility.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I saw this I instantly thought of ice cores, since trapped gas bubbles could tell you about the composition of the atmosphere. Unfortunately ice cores go nowhere near that far back to be able to tell you about the composition of the atmosphere during the Mesozoic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now for the answer:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't really speak to the title of your question effectively, since I am unaware of any research involving rogue planets and atmospheric stripping, however, from an astronomy point of view, here is one &lt;em&gt;other&lt;/em&gt; thing I would like to point out that may work to reduce the thickness of the atmosphere with time.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A small fraction of the gasses of the atmosphere exceed the escape velocity of Earth at any given time. Over time, many of these gasses get bombarded by high energy photons (as well as other particles) and fly out of Earth's gravitational potential well. If there are no other sources of these lighter gasses then there is a net loss.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9MU7f.gif&quot; alt=&quot;maxwell-boltzmann&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Temperature does play a key role here (as the above plot suggests), but regardless, this process works to reduce the amount of atoms and molecules in the atmosphere (and it affects smaller mass atoms/molecules more so than heavier ones - this is why the Martian atmosphere is primarily carbon dioxide. Essentially it wasn't massive enough to be able to hold on to its lighter gasses, and over time the vast majority of them escaped).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had temperature information going back as far as the Mesozoic era, you could in theory run the clock backwards by running a simulation of the evolution of the atmosphere. I don't know how helpful this would really be (or how accurate), since there are potentially other things like volcanic outgassing and the bi-products of living things to complicate the situation. Quite honestly, I know little about these topics and so I couldn't tell you whether there is a net loss or gain of atmospheric gasses.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;As far as rogue planets are concerned, I would find it rather unlikely that such a massive object would still be lurking around the solar system between about 252 to 66 million years ago. The early solar system was certainly more chaotic than it is now (for example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Giant_impact_hypothesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Giant impact hypothesis&lt;/a&gt;: $\sim 4.5$ Gya; &lt;a href=&quot;http://news.nationalgeographic.com/news/2011/10/111010-uranus-planets-tilted-impact-double-blows-moons-space-science/&quot; rel=&quot;nofollow noreferrer&quot;&gt;the collision which caused Uranus' tilt&lt;/a&gt;: a few billion years ago), however it is not out of the realm of possibility. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="45" LastActivityDate="2014-01-12T20:38:14.403" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1410" PostTypeId="2" ParentId="1405" CreationDate="2014-01-13T10:47:48.410" Score="3" Body="&lt;p&gt;The short answer is: we are made from the &quot;stuff&quot; contributed by many stars. The interstellar medium from which the Solar system formed was enriched by all sorts of elements synthesised in previous generations of stars. Most of the heavy elements are formed and injected into the ISM in a supernova explosion of a massive star.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-01-13T10:47:48.410" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1411" PostTypeId="2" ParentId="1405" CreationDate="2014-01-13T13:33:30.703" Score="4" Body="&lt;p&gt;It is highly unlikely that we're made out of stuff from only one star. The simplest reason for this, is that we have gold on earth. Gold is (we believe) created by the collision of two massive stars (neutron stars probably).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there was only one predecessor to the sun, it would be extremely likely that all of its mass would still be nearby. The total mass of the entire solar system is much to small to have formed for example nickel and copper in any noticable amount. You would have to have a star of about 5 times the mass of the solar system. That mass doesn't go away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could, of course speculate that a couple of neutron stars at 2-3 times the solar system initially created everything, then split up into around 2-3 different solar systems. But we have quite a lot of chemical elements such as hydrogen and helium in the sun, which there would not be a lot of in neutron stars (due to it having to have fused into heavier elements).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All stars go through a few stages from their inception to their death: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially, the theory is that hydrogen was the prevalent chemical element. It's the &quot;simplest&quot; chemical element. When helium is compressed, due to the gravity of a star it's temperature will rise up. Once it reaches around 10 million degrees, fusion starts. Fusion of hydrogen creates helium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Helium, being heavier than hydrogen will sink to the center of the star. Later, you will have helium fusion, if the star is massive enough to compress helium enough. Helium will generate oxygen and carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very massive star, multiple suns in mass, is able to fuse oxygen and carbon - which creates a lot of different chemical elements; neon, sodium, magnesium, sulfur and silicon. Later reactions transform these elements into calcium, iron, nickel, chromium, copper and others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, many stars will explode into a supernova. In that process, a lot of these materials will be ejected into space, where they will gather due to gravity again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun was probably created in a messy universe, where hydrogen and perhaps helium came together in a region where there was a lot of these other chemical elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://curious.astro.cornell.edu/supernovae.php&quot; rel=&quot;nofollow&quot;&gt;http://curious.astro.cornell.edu/supernovae.php&lt;/a&gt;, &lt;a href=&quot;http://www.enchantedlearning.com/subjects/astronomy/stars/fusion.shtml&quot; rel=&quot;nofollow&quot;&gt;http://www.enchantedlearning.com/subjects/astronomy/stars/fusion.shtml&lt;/a&gt; - stars must be able to fuse progressivly heavier elements.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.cfa.harvard.edu/news/2013-19&quot; rel=&quot;nofollow&quot;&gt;http://www.cfa.harvard.edu/news/2013-19&lt;/a&gt; - Gold created by colliding neutron stars.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="749" LastEditorUserId="749" LastEditDate="2014-01-15T16:51:01.673" LastActivityDate="2014-01-15T16:51:01.673" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="1412" PostTypeId="1" CreationDate="2013-12-13T19:08:03.317" Score="2" ViewCount="54" Body="&lt;p&gt;As of the time of writing this question, the meteors should have appeared, but so far I haven't seen any. What time would be the best to view them?&lt;/p&gt;&#xA;" OwnerDisplayName="Sumit Singh" LastActivityDate="2014-01-13T14:29:24.440" Title="In India, what is the best time to view the geminid meteor shower in 2013" Tags="&lt;space&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1413" PostTypeId="2" ParentId="1412" CreationDate="2013-12-13T21:38:09.377" Score="2" Body="&lt;p&gt;They should be visible any time at night (note that longitude has little to do with observing meteors), but there are many things that &lt;em&gt;could&lt;/em&gt; thwart your observing.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Bright lights:&lt;/strong&gt; In a big city you will see far fewer meteors. In a big enough city, you won't see any. On that note, if you keep checking a mobile device with a backlit screen, you probably won't see anything.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;The Moon:&lt;/strong&gt; The Moon will unfortunately be up until something like 4 AM local time (assuming I plugged numbers in correctly to the US Naval Observatory's &lt;a href=&quot;http://aa.usno.navy.mil/data/docs/RS_OneYear.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;table&lt;/a&gt;). Just as with cities, this will cover up smaller meteors. Just try not to stare directly at the Moon lest you lose night vision entirely.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Limited view of the sky:&lt;/strong&gt; The key to watching meteors is to see a large amount of the sky at once. Never use binoculars or a telescope. The idea is to get comfortable by laying down (yes, this is important - people who crane their necks get tired after 20 seconds and then stop looking up!) and just look at as much sky as possible. Your eyes naturally catch and follow movement, so you'll instantly focus on a meteor once it comes into view.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As for &lt;em&gt;where&lt;/em&gt; to look, the answer is &quot;everywhere.&quot; Meteor showers are named after the &lt;em&gt;radiant&lt;/em&gt; - the point in the sky from which the trails seem to originate &lt;em&gt;if you trace them backward far enough&lt;/em&gt;. This shower originates in the constellation Gemini, which isn't terribly recognizable, but can be found by locating &lt;a href=&quot;http://en.wikipedia.org/wiki/File:Orion_IAU.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Orion&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/f/ff/Orion_IAU.svg&quot; alt=&quot;Orion skychart&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Move from his belt to his right arm (the one indicated by the very red star Betelgeuse) and keep going until you come across two reasonably bright stars. That is Gemini. It is important to keep in mind, though, that meteors streak across other parts of the sky. Just because they seem to originate from Gemini doesn't mean they're actually visible there. Knowing the radiant just helps with knowing what direction to expect the streak to be going.&lt;/p&gt;&#xA;" OwnerDisplayName="Chris White" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:47.180" LastActivityDate="2013-12-13T21:38:09.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1414" PostTypeId="1" AcceptedAnswerId="1422" CreationDate="2014-01-13T16:06:28.103" Score="19" ViewCount="960" Body="&lt;p&gt;New stars keep forming in the universe thanks to all the nebulae. Now, we need Hydrogen to form stars and there would a time when all the hydrogen will get exhausted, and no more star formation will take place, theoretically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will there practically be such a point of time? I guess there is no place where hydrogen is replenished after star formation. Or is there a feedback cycle?&lt;/p&gt;&#xA;" OwnerUserId="774" LastActivityDate="2014-01-13T22:20:30.003" Title="Will new stars stop forming at some point of time?" Tags="&lt;star&gt;&lt;nebula&gt;&lt;star-formation&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1415" PostTypeId="5" CreationDate="2014-01-13T16:07:13.700" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-01-13T16:07:13.700" LastActivityDate="2014-01-13T16:07:13.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1416" PostTypeId="4" CreationDate="2014-01-13T16:07:13.700" Score="0" Body="A collection of ionized gases and dust in interstellar space" OwnerUserId="774" LastEditorUserId="6" LastEditDate="2014-01-13T17:22:30.790" LastActivityDate="2014-01-13T17:22:30.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1417" PostTypeId="5" CreationDate="2014-01-13T16:44:12.567" Score="0" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Redshift&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia entry on redshift&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In physics, redshift happens when light or other electromagnetic radiation from an object moving away from the observer is increased in wavelength, or shifted to the red end of the spectrum. In general, whether or not the radiation is within the visible spectrum, &quot;redder&quot; means an increase in wavelength – equivalent to a lower frequency and a lower photon energy, in accordance with, respectively, the wave and quantum theories of light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Redshift also occurs when light leaves a gravitational potential (gravitational redshift) or due to the expansion of the universe (cosmological redshift). In the latter case, it indicates the change in the scale factor between the lights emission and observation.&lt;/p&gt;&#xA;" OwnerUserId="774" LastEditorUserId="13681" LastEditDate="2016-08-13T14:48:30.880" LastActivityDate="2016-08-13T14:48:30.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1418" PostTypeId="4" CreationDate="2014-01-13T16:44:12.567" Score="0" Body="Questions related to the phenomenon whereby electromagnetic radiation (such as visible light) generated by an object moving away from an observer will have increase in wavelength (i.e. shifted toward the red end of the spectrum) once it reaches the observer." OwnerUserId="774" LastEditorUserId="9527" LastEditDate="2020-02-26T15:07:25.513" LastActivityDate="2020-02-26T15:07:25.513" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="1419" PostTypeId="2" ParentId="1414" CreationDate="2014-01-13T17:04:32.943" Score="2" Body="&lt;p&gt;Given what we know of entropy and of the expansion of the Universe, the answer to your question is of course &quot;Yes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will take a really really long time, however. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited to add:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;...a really really long time...&quot; is an intentionally vague description.  Whether it is 10&lt;sup&gt;10&lt;/sup&gt; or 10&lt;sup&gt;10&lt;sup&gt;10&lt;/sup&gt;&lt;/sup&gt; Hubble times is open to question.  &lt;/p&gt;&#xA;" OwnerUserId="718" LastEditorUserId="718" LastEditDate="2014-01-13T22:20:30.003" LastActivityDate="2014-01-13T22:20:30.003" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1420" PostTypeId="2" ParentId="1348" CreationDate="2014-01-13T17:16:44.963" Score="1" Body="&lt;p&gt;The good software for orbit calculation is &quot;Binary&quot; (Binary Star Combined Solution Package) by CHARA team:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astro.gsu.edu/~gudehus/binary.html&quot; rel=&quot;nofollow&quot;&gt;http://www.astro.gsu.edu/~gudehus/binary.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="761" LastEditorUserId="1288" LastEditDate="2015-05-08T19:35:24.850" LastActivityDate="2015-05-08T19:35:24.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1422" PostTypeId="2" ParentId="1414" CreationDate="2014-01-13T20:21:19.280" Score="15" Body="&lt;p&gt;&lt;a href=&quot;http://phys.org/news/2012-11-cosmic-gdp-star-formation-slumps.html&quot; rel=&quot;noreferrer&quot;&gt;Cosmic GDP has already crashed&lt;/a&gt;, as Peak Star was ~11 billion years ago.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/rDhA4.jpg&quot; alt=&quot;Cosmic GDP&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://arxiv.org/abs/1202.3436&quot; rel=&quot;noreferrer&quot;&gt;Sobral et al&lt;/a&gt;'s prediction, the future star production by mass will give only 5% of the stars in the universe today, &quot;even if we wait forever.&quot; More theoretical predictions, such as &lt;a href=&quot;http://arxiv.org/abs/astro-ph/9701131&quot; rel=&quot;noreferrer&quot;&gt;this one&lt;/a&gt;, suggest that nebulae will run out of hydrogen on the order of $10^{13}$ years, while star formation will occasionally happen due to collision of brown dwarfs until somewhere on the order of $10^{14}$ years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, hydrogen itself may have a finite lifetime. The half-life of a proton is experimentally known to be longer than $10^{34}$ years, but it may still be quite finite.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-01-13T20:21:19.280" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="1423" PostTypeId="1" AcceptedAnswerId="1424" CreationDate="2014-01-13T23:50:49.733" Score="4" ViewCount="102" Body="&lt;p&gt;The Andromeda Galaxy is tens of thousands of light years in diameter. That suggests that the most distant stars in the Andromeda galaxy is up to tens of thousands of years &quot;behind&quot; in their orbit around the galaxy, compared to the closest stars. Have anybody generated pictures of the galaxy, which shows the approximate real position of stars? Is this effect visible when looking at the Andromeda galaxy through an amateur telescope?&lt;/p&gt;&#xA;" OwnerUserId="749" LastActivityDate="2014-01-14T21:59:23.830" Title="Pictures of large galaxies" Tags="&lt;telescope&gt;&lt;light&gt;&lt;rotation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1424" PostTypeId="2" ParentId="1423" CreationDate="2014-01-14T01:40:28.043" Score="5" Body="&lt;p&gt;M31 is actually something like 141,000 light-years in diameter (there has been some variation in our understanding of its size, but 141K is good compromise).  M31 is 2.54 million light-years from us.  The rotational speed of stars near the edge of the galaxy's disk is in the vicinity of 200 km/sec. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given all that, in one year a star in the vicinity of the outer edge of the galaxy's disk is going to traverse 6.3 x 10&lt;sup&gt;9&lt;/sup&gt; km.  At 2.54 x 10&lt;sup&gt;6&lt;/sup&gt; light-years away, the apparent movement of such a star in the course of a year is going to be about 1.5 x 10&lt;sup&gt;-8&lt;/sup&gt; degrees, or 0.000054 arcseconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Keck Telescopes in Hawaii, one of the most powerful telescopes in the world, has an angular resolution of 0.04 arcseconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the best telescope in the world cannot distinguish any change of position of any star at that distance over the course of a year.  It would take about 740 years before such a star would move far enough for a telescope with the angular resolution of the Keck to distinguish any movement at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it isn't visible through an amateur telescope either.&lt;/p&gt;&#xA;" OwnerUserId="718" LastEditorUserId="718" LastEditDate="2014-01-14T21:59:23.830" LastActivityDate="2014-01-14T21:59:23.830" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="1426" PostTypeId="1" AcceptedAnswerId="1725" CreationDate="2014-01-14T09:30:18.450" Score="2" ViewCount="97" Body="&lt;p&gt;Are there pictures that contain high enough dynamic width, so that amateurs can use fourier spectrum analysis to observe the redshift themselves? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know pictures usually are encoded using RGB and/or CMYK and therefore lose a lot of color information from a physical perspective, but I'm wondering if there exists a picture format that could store that much information, and if so, could amateurs then use fourier spectrum analysis to actually see the red-/blueshift?&lt;/p&gt;&#xA;" OwnerUserId="749" LastActivityDate="2014-02-17T14:18:09.360" Title="Can redshift be measured using fourier?" Tags="&lt;photography&gt;&lt;redshift&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1427" PostTypeId="1" AcceptedAnswerId="1430" CreationDate="2014-01-14T10:39:20.787" Score="9" ViewCount="743" Body="&lt;p&gt;13 billion years ago, the universe was about 600 million years old according to many scientists. At that time, all matter in the universe would have to be closer, or in other words denser. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did time pass more slowly, at that time, relative to now? I ask, because increased gravity causes time to pass more slowly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we'd been there, and measured the speed of light as we measure it today, would the speed of light be different?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, does distant stars look magnified, given that a &quot;tiny&quot; universe is stretched on our entire sky?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume then that we'd measure time based on caesium 133 and distance not as a function of the speed of light, but as a theoretical physical &quot;device&quot; that would be the same physical length then and now.&lt;/p&gt;&#xA;" OwnerUserId="749" LastEditorUserId="749" LastEditDate="2014-01-14T11:03:32.170" LastActivityDate="2014-01-15T00:40:20.143" Title="Did time pass more slowly in the past?" Tags="&lt;light&gt;&lt;general-relativity&gt;&lt;special-relativity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1428" PostTypeId="1" CreationDate="2014-01-14T10:44:56.263" Score="4" ViewCount="231" Body="&lt;p&gt;Where can I find good places to watch stars near me? I live in Leicestershire, England and am looking for good places to watch the sky nearby. &lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2014-01-14T14:02:35.263" Title="Where can I find good places to watch stars?" Tags="&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2014-01-14T15:59:57.493" ContentLicense="CC BY-SA 3.0" />
  <row Id="1429" PostTypeId="2" ParentId="500" CreationDate="2014-01-14T11:20:01.487" Score="4" Body="&lt;p&gt;For the purposes of comparison, here's flat, Minkowski spacetime in spherical coordinates:&#xA;$$\mathrm{d}s^2 = -\mathrm{d}t^2 + \underbrace{\mathrm{d}r^2 + r^2(\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\phi^2)}_\text{Euclidean 3-space}\text{.}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The soure of misunderstanding was that I wasn't clear enough, how black holes work. I always imagined them as &quot;suckholes&quot; like whirlpools in the water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is not entirely incorrect. The Schwarzschild spacetime of an uncharged, nonrotating black hole in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates&quot; rel=&quot;nofollow&quot;&gt;Gullstrand-Painlevé coordinates&lt;/a&gt; is&#xA;$$\mathrm{d}s^2 = -\mathrm{d}t^2 + \underbrace{\left(\mathrm{d}r + \sqrt{\frac{2M}{r}}\,\mathrm{d}t\right)^2}_\text{suckhole} + r^2(\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\phi^2)\text{.}$$&#xA;Where it deviates from ordinary, flat Minkowski spacetime is entirely in the middle square term. Here, the time coordinate $t$ is not the Schwarzschild time, but rather the time measured by an observer free-falling from rest at infinity. The last bit, if adjoined with the $\mathrm{d}r^2$ term one would get by multiplying out the middle part, is ordinary Euclidean $3$-space written in spherical coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you recognize from Newtonian gravity the quantity $\sqrt{2M/r}$, or $\sqrt{2GM/r}$ in ordinary units, as the escape velocity, then the picture is very peculiar indeed: according to an observer free-falling from rest at infinity, Euclidean space is sucked into the singularity at the local escape velocity. The event horizon is the surface at which the speed at which space is &quot;falling&quot; at the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an additional reason why &lt;a href=&quot;http://en.wikipedia.org/wiki/Sonic_black_hole&quot; rel=&quot;nofollow&quot;&gt;sonic black holes&lt;/a&gt; are good analogues to their gravitational counterparts. In a sonic black hole, there can be an actual &quot;suckhole&quot; that drains a low-viscosity fluid at an increasing velocity, up to and faster than the speed of sound in that fluid. This forms an &lt;em&gt;acoustic event horizon&lt;/em&gt; that is one-way to sound and is expected to have an analogue of Hawking radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding structure for charged black holes is similar, and for a rotating one more complicated, although can still be described as &quot;sucking&quot; with a certain additional twist that rotates the free-falling observers.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-01-14T11:20:01.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1430" PostTypeId="2" ParentId="1427" CreationDate="2014-01-14T11:44:02.640" Score="3" Body="&lt;p&gt;Without some reference to compare to, time passing more slowly makes no sense at all. But every clock has its own &lt;em&gt;proper time&lt;/em&gt; that measures time along its own worldline. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In cosmological models, the &lt;em&gt;cosmological time&lt;/em&gt; is the proper time of a certain kind of ideal observer: one comoving with the Hubble flow. In other words, imagine space filled with observers for whom the universe looks as close to the same in every direction as possible (zero &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_microwave_background#CMBR_dipole_anisotropy&quot; rel=&quot;nofollow&quot;&gt;CMBR dipole anisotropy&lt;/a&gt;). The cosmological time is what's measured by their clocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, there is no time dilation... because we &lt;em&gt;defined&lt;/em&gt; our time coordinate as that which is measured by the comoving clocks no matter how small or dense the universe is. You can see this in the standard form of the FLRW metric:&#xA;$$\mathrm{d}s^2 = -\mathrm{d}t^2 + a^2(t)\left[\frac{\mathrm{d}r^2}{1-kr^2} + r^2\,\mathrm{d}\Omega^2\right]\text{,}$$&#xA;where nothing happens to the time component no matter what the scale factor $a(t)$ is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could we compare the rate of clocks &lt;em&gt;now&lt;/em&gt; with rate of clocks &lt;em&gt;then&lt;/em&gt;? We could have the past clock emit some pattern, say a light at a specific frequency, and then later catch and measure it. If we find a different frequency, we could interpret this as evidence of time dilation. But we could also fix the clocks as our time standard and interpret the frequency shift as the result of a changing scale factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could also arbitrarily rescale the time coordinate any way we like, but there's no point in doing so because it wouldn't change the time experienced by comoving clocks.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;are you saying that we've decided to look at it as an increasing scale factor where clocks are fixed, instead of time dilation - but in fact both are just different point of views?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That was phrased too poorly, so let me clarify. If you're watching a far-away supernova, the collapse and explode process will take a different amount of time in the supernova's own local rest frame than is seen by you. Taken alone, it's up to you whether you interpret this fact as time dilation of your clock relative to the supernova or a stretching of wavelengths due to cosmic expansion, or any combination of both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what we have decided is to use a frame in which the homogeneity and isotropy of the universe is directly manifest, in which case we're not &quot;looking&quot; at a single supernova but all around us. For a general FRW spacetime, we cannot re-interpret cosmological redshift as due entirely to time dilation without spoiling those conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is that homogeneity and isotropy picks out a sequence of spatial hypersurfaces, i.e. a sequence of &quot;nows&quot;, snapshots of the universe at instants of time. Then generally the distances between different galaxies are different at different times.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditDate="2014-01-15T00:40:20.143" LastActivityDate="2014-01-15T00:40:20.143" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1432" PostTypeId="2" ParentId="531" CreationDate="2014-01-14T12:09:17.223" Score="1" Body="&lt;p&gt;Actually i read some where that in the entire time of humankind on earth, the earth has just covered 1/10th of a percent around the milky way galaxy, so assuming earth would survive that long and humankind would survive to witness it, i would say it would take billions-billions of year, before earth would move to closer to giant stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the supernova activity, humankind hasn't seen a supernova (excluding some theories), but our best bet to witness on would be &lt;code&gt;Betelgeuse&lt;/code&gt;, it is already old for its size class and is expected to explode relatively soon (it may explode tomorrow or after a million years nobody knows exactly when) compared to its age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, i would say, we would be lucky if we witness a supernova in our lifetime. &lt;/p&gt;&#xA;" OwnerUserId="769" LastActivityDate="2014-01-14T12:09:17.223" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1433" PostTypeId="1" CreationDate="2014-01-14T13:31:37.613" Score="4" ViewCount="89" Body="&lt;p&gt;I was having a discussing about big scientific projects with colleagues, and I postulated that astronomy is still one of the scientific fields with biggest projects per amount of data dealt with, after the high-energy physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know what is the highest throughput astronomy project as defined by the number of pixels in the images analysed in the project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something like &quot;project XYZ&quot; with an estimated &quot;1.2 trillion pixels&quot; worth of pictures taked and analysed over 5 years.&lt;/p&gt;&#xA;" OwnerUserId="798" LastActivityDate="2014-01-15T00:39:58.567" Title="what is the highest throughput astronomy project in pixels?" Tags="&lt;galaxy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1434" PostTypeId="2" ParentId="1433" CreationDate="2014-01-14T13:35:59.493" Score="3" Body="&lt;p&gt;If you refer to &quot;data volume&quot; instead of real image pixels, it would be the Millenium Run (25TB of data):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mpa-garching.mpg.de/galform/virgo/millennium/&quot; rel=&quot;nofollow&quot;&gt;http://www.mpa-garching.mpg.de/galform/virgo/millennium/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Millennium Run used more than 10 billion particles to trace the&#xA;  evolution of the matter distribution in a cubic region of the Universe&#xA;  over 2 billion light-years on a side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-01-14T13:35:59.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1435" PostTypeId="1" CreationDate="2014-01-14T13:40:44.357" Score="3" ViewCount="258" Body="&lt;p&gt;Is it possible that dark matter particles are unstable and the existing dark matter will decay in the long run and turn into hydrogen, for example? What would be possible mechanisms for such a decay?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could some of the hydrogen we observe be &quot;new&quot; hydrogen, i.e. be the product of dark matter decay?&lt;/p&gt;&#xA;" OwnerUserId="749" LastEditorUserId="115" LastEditDate="2014-01-14T19:36:42.903" LastActivityDate="2014-01-15T15:47:17.730" Title="Could dark matter particles be unstable?" Tags="&lt;dark-matter&gt;&lt;dark-energy&gt;&lt;hydrogen&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1436" PostTypeId="2" ParentId="1435" CreationDate="2014-01-14T13:58:16.677" Score="4" Body="&lt;p&gt;Yes, as mentioned elsewhere, it is possibly possible. Dark matter particles may be intrinsically unstable (though having long lifetimes, which are at least significantly longer than Hubble time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check for more info here:&#xA;&lt;a href=&quot;http://arxiv.org/abs/1307.6434&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1307.6434&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="115" LastEditorUserId="115" LastEditDate="2014-01-15T15:47:17.730" LastActivityDate="2014-01-15T15:47:17.730" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="1437" PostTypeId="2" ParentId="1435" CreationDate="2014-01-14T13:59:30.687" Score="1" Body="&lt;p&gt;Dark matter is actually matter. Real matter, just that we can not observe it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of it is actually cold hydrogen, and even cold dust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other parts of it are actually different particles that may decay. But that does not mean formation of new Hydrogen. To form Hydrogen you need both a proton and an electron to get bound together, and this implies they meeting, which is very difficult in low density regions like those of extragalactic dark matter.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-01-14T13:59:30.687" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="1438" PostTypeId="1" CreationDate="2014-01-14T14:01:34.877" Score="15" ViewCount="380" Body="&lt;p&gt;Do &lt;a href=&quot;https://en.wikipedia.org/wiki/Mercury-manganese_star&quot;&gt;HgMn&lt;/a&gt; (peculiar A type) stars really possess global magnetic fields? See, for example, &lt;a href=&quot;http://arxiv.org/abs/1208.2910&quot;&gt;this paper&lt;/a&gt; by Hubrig et al. from 2012.&lt;/p&gt;&#xA;" OwnerUserId="761" LastEditorUserId="2153" LastEditDate="2014-12-21T15:05:53.390" LastActivityDate="2016-04-09T20:30:41.057" Title="Magnetic fields of peculiar HgMn A type stars" Tags="&lt;star&gt;&lt;magnetic-field&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1439" PostTypeId="2" ParentId="1428" CreationDate="2014-01-14T14:02:35.263" Score="2" Body="&lt;p&gt;It would have been better to ask about &quot;which conditions must have a good starsighting place&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It must be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Dark&lt;/li&gt;&#xA;&lt;li&gt;Far from cities&lt;/li&gt;&#xA;&lt;li&gt;Not under main flight paths (this depends on which altitude is the object you want to observe)&lt;/li&gt;&#xA;&lt;li&gt;High altitude (or any special place with low cloud formation and low turbulence)&lt;/li&gt;&#xA;&lt;li&gt;Far from big water bodies that can cause local humidity&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-01-14T14:02:35.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1440" PostTypeId="1" CreationDate="2014-01-14T15:25:08.547" Score="14" ViewCount="191" Body="&lt;p&gt;What are mechanisms of binary/multiple star formation in different mass ranges (low, intermediate and high stellar masses)?&lt;/p&gt;&#xA;" OwnerUserId="761" LastEditorUserId="761" LastEditDate="2014-03-27T17:46:08.930" LastActivityDate="2014-05-19T08:46:02.967" Title="Mechanisms of binary/multiple star formation" Tags="&lt;formation&gt;&lt;binary-star&gt;&lt;mass&gt;&lt;star-formation&gt;&lt;initial-mass-function&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1441" PostTypeId="2" ParentId="3" CreationDate="2014-01-14T15:45:53.260" Score="3" Body="&lt;p&gt;I advise you to use this form (appropriate for professional astronomers also) from Washington Double Star Catalog (WDS): &lt;a href=&quot;http://ad.usno.navy.mil/wds/request.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ad.usno.navy.mil/wds/request.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Observing List Request&lt;/p&gt;&#xA;&lt;p&gt;If you would like to have an observing list custom made to your specifications please enter that data in the box provided. Pertinent data include what portion of the sky you can observe (RA and Dec limits), what separation you can observe (lower and upper limits), and what magnitudes are observable by you (both primary, secondary, and delta-mag). In addition, if no magnitude information is provided should we include the selected object? Also, do you wish to observe infrequently observed (N = ?) objects, or those which have not been observed for a long time (DATE = ?)? Please make the subject line &amp;quot;Data Request&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="761" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2014-01-14T15:45:53.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1442" PostTypeId="2" ParentId="531" CreationDate="2014-01-14T17:27:42.957" Score="4" Body="&lt;p&gt;The paper &quot;Frequency of nearby supernovae and climatic and biological catastrophes&quot; by Clark, McCrea, and Stephenson published in &lt;em&gt;Nature&lt;/em&gt; estimates (at 50% probability) that the Solar System passes within 10 parsecs of a supernova every 100 million years. This supernova would be part of a 20-parsec strip in which an estimated 50 supernovae occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They do speculate that a connection between the Solar System passing through these regions and climatic and biological events (such as ice ages) is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read the paper yourself for more details. It's only two pages and you can purchase 48 hour access for $5 from Readcube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nature.com/nature/journal/v265/n5592/abs/265318a0.html&quot; rel=&quot;nofollow&quot;&gt;&quot;Frequency of nearby supernovae and climatic and biological catastrophes&quot; - Clark, McCrea, and Stephenson&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6" LastActivityDate="2014-01-14T17:27:42.957" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1443" PostTypeId="1" AcceptedAnswerId="1446" CreationDate="2014-01-14T20:21:30.870" Score="19" ViewCount="21547" Body="&lt;p&gt;I only ask this because of how fast light travels. The question remains in the title. Why, or why not, would this work? &lt;/p&gt;&#xA;" OwnerUserId="802" LastEditorUserId="537" LastEditDate="2014-01-14T20:37:37.477" LastActivityDate="2019-12-25T14:42:27.027" Title="By putting a mirror in space, would we be able to see into the past?" Tags="&lt;light&gt;&lt;space-time&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1444" PostTypeId="2" ParentId="1443" CreationDate="2014-01-14T20:31:24.893" Score="10" Body="&lt;p&gt;Yes, we always look into the past, when looking somewhere. There is for instance a mirror on the moon. When sending a laser beam to that mirror, we can detect the reflected light about 2.5 seconds later. This could be interpreted as looking 2.5 seconds into the past, when the laser has been fired. Details &lt;a href=&quot;http://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="45" LastEditDate="2014-01-15T00:49:52.813" LastActivityDate="2014-01-15T00:49:52.813" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1445" PostTypeId="2" ParentId="1426" CreationDate="2014-01-14T20:47:16.430" Score="0" Body="&lt;p&gt;You'll probably need &lt;a href=&quot;http://en.wikipedia.org/wiki/Astronomical_spectroscopy&quot; rel=&quot;nofollow noreferrer&quot;&gt;spectrometer&lt;/a&gt; data to distinguish and identify spectral lines. That kind of spectrographic images look e.g. like this:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Op3AB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="1050" LastEditDate="2014-02-17T14:18:09.360" LastActivityDate="2014-02-17T14:18:09.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1446" PostTypeId="2" ParentId="1443" CreationDate="2014-01-14T22:09:21.853" Score="15" Body="&lt;p&gt;I think the question is referring to situating a very large mirror in space facing earth.  If we were to put it several light minutes away, then events occurring opposite the mirror could be reviewed &lt;em&gt;de novo&lt;/em&gt; with more preparation upon the warning we received upon the first light of the event arriving at earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a supernova going off in M31 might not be under observation at the moment its light first arrives, and so the initial observations might be lost.  However, with a mirror facing M31, we would be able to observe that mirror as the event unfolded, having been warned that there was something worth watching, in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nice idea!  But it would likely be much less expensive to simply have multiple telescopes always watching &quot;prime&quot; starscape for unexpected events.&lt;/p&gt;&#xA;" OwnerUserId="718" LastActivityDate="2014-01-14T22:09:21.853" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1447" PostTypeId="2" ParentId="1433" CreationDate="2014-01-15T00:39:58.567" Score="3" Body="&lt;p&gt;Another &quot;mega project&quot; currently starting to produce data is the &lt;a href=&quot;http://www.esa.int/Our_Activities/Space_Science/Gaia/Gaia_factsheet&quot; rel=&quot;nofollow&quot;&gt;Gaia project&lt;/a&gt;: &quot;By the end of the mission, the data archive will exceed 1 Petabyte (1 million Gigabytes)&quot;. They intend to &quot;measure the positions and velocity of approximately one billion stars in our Galaxy&quot;. &quot;Gaia will carry the largest digital camera into space with nearly one billion pixels.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Petabyte&quot; rel=&quot;nofollow&quot;&gt;Petabyte&lt;/a&gt; is about 1e15 bytes (1000 Terabytes).&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-01-15T00:39:58.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1448" PostTypeId="1" AcceptedAnswerId="1450" CreationDate="2014-01-15T02:10:00.010" Score="11" ViewCount="332" Body="&lt;p&gt;I read about the 1995 Lamb-Paczynski debate on whether GRBs are galactic or extragalactic in origin.  With our current experimental evidence, have we found whether GRBs are galactic or extragalactic?&lt;/p&gt;&#xA;" OwnerUserId="806" LastEditorUserId="718" LastEditDate="2014-01-17T22:11:32.227" LastActivityDate="2014-05-25T01:29:59.407" Title="Are Gamma Ray Bursts of galactic or extragalactic origin?" Tags="&lt;galaxy&gt;&lt;gamma-ray-bursts&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1449" PostTypeId="1" CreationDate="2014-01-15T03:29:04.210" Score="15" ViewCount="18170" Body="&lt;p&gt;I am trying to understand the phenomenon of space-time. But, everything on internet seems to be too complicated for me since I do not have a background in physics. Can anyone give me simple explanations for the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the difference between time and space-time?&lt;/li&gt;&#xA;&lt;li&gt;How does gravity affect the passage of time? &lt;/li&gt;&#xA;&lt;li&gt;What is the speed of light and how does it relate to time? &lt;/li&gt;&#xA;&lt;li&gt;How do scientists deal with timescales on the order of billions of years if time is not constant for all observers in the universe?&lt;/li&gt;&#xA;&lt;li&gt;How is time, or for example the age of the universe, actually measured experimentally? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="460" LastEditorUserId="1469" LastEditDate="2014-04-23T13:01:54.387" LastActivityDate="2015-07-07T22:47:01.903" Title="What is the difference between our time and space time?" Tags="&lt;gravity&gt;&lt;light&gt;&lt;space-time&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1450" PostTypeId="2" ParentId="1448" CreationDate="2014-01-15T05:42:56.677" Score="6" Body="&lt;p&gt;Based on our understanding of what might be possible causes for GRB, there's nothing to suggest that this phenomena is isolated to other galaxies alone, but we have so far only detected GRB originating from outside of the Milky Way. Good that we did too, because if any of them happened in our cosmic neighborhood and was pointing towards us, it would be the last thing we'd ever see and would effectively sterilize our planet. More information is available on &lt;a href=&quot;https://en.wikipedia.org/wiki/Gamma-ray_burst&quot;&gt;Wikipedia on Gamma-ray burst&lt;/a&gt;, it's actually a pretty good article presenting extensive information on GRB (I guess that's why it's featured there).&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2014-01-15T05:42:56.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1451" PostTypeId="1" AcceptedAnswerId="1453" CreationDate="2014-01-15T10:04:36.763" Score="5" ViewCount="737" Body="&lt;p&gt;When a particle travels through spacetime, this is sometimes illustrated moving in flat grid with various depths. How accurate is this way of imagining gravity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the gravity well be infinitely deep?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://blakemaybank.com/wp-content/uploads/2012/12/Gravity_well.gif&quot; rel=&quot;noreferrer&quot;&gt;Gravity well http://blakemaybank.com/wp-content/uploads/2012/12/Gravity_well.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at this illustration, it suggests that there is an amount of distance going down the well. Is movement &lt;em&gt;down the well&lt;/em&gt; limited by the speed of light? Remember, I'm aware that it is an illustration, and the question is about the accuracy of the illustration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is all gravitational wells infinitely deep at the exact center? So that if a particle could be made small enough to fit within the center, it would become a black hole.&lt;/p&gt;&#xA;" OwnerUserId="749" LastActivityDate="2018-03-05T07:26:09.067" Title="Spacetime curvature illustration accuracy" Tags="&lt;gravity&gt;&lt;general-relativity&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1452" PostTypeId="1" AcceptedAnswerId="1454" CreationDate="2014-01-15T12:21:06.143" Score="7" ViewCount="191" Body="&lt;p&gt;I have one problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On picture below we can see Erigone family asteroids in plane semimajor axis - absolute magnitude. Based on V-type as a consequence of Yarkovsky effect, estimate age of this family bearing in mind that members of this family have spectral of C-type. For average value of change of semimajor axis which belongs to this family take $da=5 \times 10^{-4} \text{AU}$ (for object with diameter $D=1 \,\text{km}$ and time $t=1$ million years). Ignore initial size of family.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rQ6vh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really don't know how to solve this problem (I am studying math, not astronomy). Professor won't tell us how to do it, and I tried to find the answer in his textbook and in two more books in my language, but there is no similar problem at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you point me in the right direction?&lt;/p&gt;&#xA;" OwnerUserId="808" LastEditorUserId="6" LastEditDate="2014-01-15T17:08:33.417" LastActivityDate="2014-01-15T17:08:33.417" Title="How to estimate age of asteroid family Erigon" Tags="&lt;asteroids&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1453" PostTypeId="2" ParentId="1451" CreationDate="2014-01-15T12:43:07.740" Score="8" Body="&lt;p&gt;Newtonian gravity of a point-source can described by a potential $\Phi = -\mu/r$. If we suppress one spatial dimension and use it to graph the value of this potential instead, we get something that looks very close to this illustration, and is indeed infinitely deep at the center--at least, in the idealization of a point-mass. And farther away from the center, it goes flat, just as many illustrations like this have it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Illustrations like this are fairly common, and I'm guessing that they're ultimately inspired by the Newtonian potential, because they have almost nothing to do with with the spacetime curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an isometric embedding of the Schwarzschild geometry at an instant of Schwarzschild time, again with one dimension supressed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lNYcH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above the horizon (red circle), the surface is a piece of a paraboloid (the &lt;em&gt;Flamm paraboloid&lt;/em&gt;). Unlike the potential, it does not go flat at large distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being isometric means that it correctly represents the spatial distances above the horizon at an instant of Schwarzschild time. Below the horizon, the embedding isn't technically accurate because the Schwarzschild radial coordinate does not represent space there, but time. Although if we pretend it's spacelike below the horizon, that would be the correct embedding. Picture the below-horizon part as having one-directional flow into the singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we've only represented space and not time, the embedding is not enough to reconstruct the trajectories of particles in this spacetime. Still, it is a more accurate representation of a &lt;em&gt;part&lt;/em&gt; of the spacetime curvature of point-source--specifically the spatial part.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The velocity of the object from this perspective, would seem to increase, until a point - where the velocity in x,y coordinates starts to decrease due to most of the motion happening &quot;down&quot; the time dimension. Is this also correct? Would a photon seem to slow down when moving down the well, if seen from above?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The above is an embedding of a slice of spatial geometry, and is not a gravity well. The mathematical form of the paraboloid above the horizon is best described in &lt;a href=&quot;http://en.wikipedia.org/wiki/Cylindrical_coordinates&quot; rel=&quot;noreferrer&quot;&gt;cylindrical coordinates&lt;/a&gt; as&#xA;$$r = 2M + \frac{z^2}{8M}\text{.}$$&#xA;Here the vertical $z$ coordinate doesn't mean anything physically. It's purely an artifact of creating a surface of the same intrinsic curvature in Euclidean $3$-space as the $2$-dimensional spatial slice of Schwarzschild geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Schwarzschild spacetime, radial freefall is actually exactly Newtonian in the Schwarzschild radial coordinate and proper time, i.e. time experienced by the freefalling object, rather than Schwarzschild time. So the Newtonian gravity well isn't actually a bad picture for the physics--it's just not the geometry and so is not a good representation of how any part of spacetime is curved. For non-radial orbits, the effective potential is somewhat different that than the Newtonian one, but ignoring the effects of angular momentum gets us the Newtonian form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Schwarzschild time, yes, a photon (or anything else) does slow down as gets near the horizon. In fact, in Schwarzschild time it never reaches the horizon, which is one indication that the Schwarzschild coordinates are badly behaved at the horizon. The coordinate acceleration actually becomes repulsive close to the horizon--and for a fast enough infalling object, is &lt;em&gt;always&lt;/em&gt; repulsive. This can be understood as the particle moving to places with more and more gravitational time dilation. In proper time of any infalling observer, however, close to the horizon the acceleration is always attractive.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditDate="2014-01-15T22:22:16.047" LastActivityDate="2014-01-15T22:22:16.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1454" PostTypeId="2" ParentId="1452" CreationDate="2014-01-15T13:48:41.917" Score="2" Body="&lt;p&gt;I'm not a professional astronomer, but I'm going to take a stab at this. If anyone sees any problems, feel free to correct them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Yarkovsky effect basically means that spinning asteroids over time will be pushed into a further orbit (more precisely, it increases the semimajor axis of the orbit) due to uneven heating and cooling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently to estimate the age of an asteroid family, using only the Yarkovsky effect, we would start out with this equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0.2H = \log(\Delta a / C)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation assumes a fixed geometric albedo for the whole family. $H$ here is the absolute magnitude. $\Delta a$ is equal to $a - a_c$ where $a$ is the semimajor axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C = \sqrt{p_v} (da/dt)_0 T \cos \epsilon$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_v$ is the geometric albedo.  $(da/dt)_0$ is the maximum Yarkovsky drift rate for a body of size $D_0$, where $D_0$ is an arbitrary reference size (we could use your provided $D$ value above). $T$ is the age of the family that we are trying to calculate. $\epsilon$ is the spin axis obliquity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: I've had to remove my example solution because I could not figure out how to correct my errors from accidentally putting $C$ outside the logarithm.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can check out more details in my source below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www-n.oca.eu/morby/papers/YarkoChrono.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Yarkovsky/YORP chronology of asteroid families&quot; - Vokrouhlickýa, Broža, Bottke, Nesvorný, Morbidelli&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorUserId="6" LastEditDate="2014-01-15T14:24:17.327" LastActivityDate="2014-01-15T14:24:17.327" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1455" PostTypeId="2" ParentId="1449" CreationDate="2014-01-15T16:50:06.573" Score="0" Body="&lt;p&gt;Particle like, molecule or atoms or any smallest particle behaves differently in different environment. For example, these particle behaves different on earth or on moon or on earth's orbit or near to black hole horizon. &lt;br&gt; &#xA;This was actually proved by the theory of relativity given by &lt;a href=&quot;http://en.wikipedia.org/wiki/Albert_Einstein&quot; rel=&quot;nofollow&quot;&gt;Albert Einstein&lt;/a&gt;. We (Human) have to adjust artificial satellite's clock everyday 0.0000000001 second. Otherwise we will get wrong data from satellite on earth computers and all. &lt;br&gt;&#xA;Now IMO, Any object in Gravitational environment or  the object travelling at a speed of light must be getting some effect at atomic level and all electromagnetic level also. Because of that clock become slow ( because electromagnetic/atomic movement which requires to produce output in clock becomes slow ).&lt;br&gt;&#xA;In the same way, IMO (not validated ), when we travel at the speed of light or roaming near to the black-holes horizon then our biological activity or any electromagnetism  which is required to &lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_theories_of_consciousness&quot; rel=&quot;nofollow&quot;&gt;THINK&lt;/a&gt; becomes slow in such a way that difference produced by this effect is dramatically big. It is said that, a person on earth become old but the person travelling at a speed of light will think &lt;strong&gt;'I traveled only for a week!'&lt;/strong&gt;(see, this person's biological/atomic activity become slow, even about thinking)  but actually from earth's person's point of view he/she (traveller) traveled for at least 75 year at a speed of light which is real fact. Normally after 75 years anybody will become old on earth. This is all IMO.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;What is the speed of light and how does it relate to time?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;speed of light is 300000 Km/sec.&lt;/p&gt;&#xA;" OwnerUserId="760" LastActivityDate="2014-01-15T16:50:06.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1456" PostTypeId="1" AcceptedAnswerId="1458" CreationDate="2014-01-15T19:43:21.020" Score="8" ViewCount="159" Body="&lt;p&gt;After a couple months of watching a very bright Venus appear at dusk and set soon after the Sun, I've noticed that it seems to have temporarily disappeared to the naked eye. I used a virtual sky simulator (Stellarium) to confirm that the Evening Star has indeed transitioned to being the Morning Star (though due to obstacles blocking my view of the horizon in my backyard, Venus is still too close to the Sun for me to witness it at dawn before sunrise renders it invisible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the period of Venus's alternation between morning and evening visibility? I gather this isn't a simple seasonal period as its visibility should depend on the Sun-Earth-Venus angle rather than just Earth's yearly orbit.&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2014-01-15T20:44:36.760" Title="Periodic behavior of Venus" Tags="&lt;amateur-observing&gt;&lt;venus&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1458" PostTypeId="2" ParentId="1456" CreationDate="2014-01-15T20:36:29.577" Score="4" Body="&lt;p&gt;Assumed a &lt;a href=&quot;http://en.wikipedia.org/wiki/Venus&quot; rel=&quot;nofollow&quot;&gt;Venus&lt;/a&gt; year is 0.615 198 Earth years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.615 198 yr / (1 - 0.615 198) = 1.598739 yr = 583 days 11 hours 24 minutes for one period in mean, or roughly 19.5 months.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's 1.598739 periods for &lt;a href=&quot;http://en.wikipedia.org/wiki/Earth%27s_orbit&quot; rel=&quot;nofollow&quot;&gt;Earth&lt;/a&gt; and 2.598739 periods for &lt;a href=&quot;http://en.wikipedia.org/wiki/Venus&quot; rel=&quot;nofollow&quot;&gt;Venus&lt;/a&gt;:&#xA;1.598739 * 365.256 days = 583.95 days = 2.598739 * 224.705 days&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relative accuracy should be somewhere near 1e-5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since both orbits are elliptical the actual period oscillates around that mean value.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-01-15T20:44:36.760" LastActivityDate="2014-01-15T20:44:36.760" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1460" PostTypeId="1" AcceptedAnswerId="1462" CreationDate="2014-01-15T22:15:07.253" Score="6" ViewCount="2397" Body="&lt;p&gt;Is there a graph of known black holes, with their estimated mass in the X axis and their estimated radius in the Y axis? If so, where can we find it? I would like to know if a black whole with the whole estimated mass of our universe would have the estimated radius of our universe (which means our universe could be a black hole, that's why light can't escape it and it looks &quot;finite&quot;).&lt;/p&gt;&#xA;" OwnerUserId="813" LastActivityDate="2016-07-13T20:18:24.313" Title="Relation between black hole mass and radius, and our universe's" Tags="&lt;black-hole&gt;&lt;universe&gt;&lt;mass&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="2" ClosedDate="2018-03-14T21:33:26.027" ContentLicense="CC BY-SA 3.0" />
  <row Id="1461" PostTypeId="2" ParentId="1449" CreationDate="2014-01-15T22:16:46.590" Score="13" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the difference between time and space-time?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Space-time is time plus space.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How does gravity affect the passage of time?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The higher the gravity of a planet or star and the closer to that body the slower the time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the speed of light and how does it relate to time?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The speed of light is 299,792.4580 km/s in vacuum, the speed at which light propagates, roughly 1.3 seconds from Earth to Moon. Velocity is distance divided by time; this applies also to the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do scientists deal with timescales on the order of billions of years if time is not constant for all observers in the universe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They treat time dependent of the observer. For different observers on Earth variations are tiny, in many cases neglectable in comparison to measurement errors, although not for precision measurements.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How is time, or for example the age of the universe, actually measured experimentally?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are many ways to measure or estimate ages. The age of the earth can be estimated by ratios of certain radionuclides in the oldest rocks. Estimates of the age of the universe are obtained by simulations based on observations (redshift and distance) of distant galaxies, and on observations of the cosmic microwave background. The redshift is used for a velocity estimate; together with a distance estimate (obtained by stars of known brightness) one can calculate back, when all objects of the universe should have been close together.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-01-15T22:16:46.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1462" PostTypeId="2" ParentId="1460" CreationDate="2014-01-15T22:57:38.137" Score="6" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_radius&quot;&gt;Schwarzschild radius&lt;/a&gt; of a black hole is probably the closest we can get to your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;r_s = (2G/c^2) \cdot m \mbox{, with }\ 2G/c^2 = 2.95\ \mbox{km}/\mbox{solar mass}.&#xA;$$&#xA;This means, that the Schwarzschild radius for a given mass is proportional to that mass.&#xA;The radius shouldn't be taken too literal in the physical sense, because space is highly non-euclidean close to a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Present (light-travel) radius of the visible universe, as seen from the earth:&#xA;$$13.81 \cdot 10^9\ \mbox{lightyears} = 13.81 \cdot 10^9 * 9.4607 * 10^{12}\ \mbox{km} = 1.3065\cdot 10^{23}\ \mbox{km}.$$&#xA; So we need $$1.3065\cdot 10^{23}\ \mbox{km} / 2.95\ \mbox{km} =4.429 \cdot 10^{22}$$&#xA;solar masses to get a black hole of the light-travel Schwarzschild radius of the visible universe, pretty close (by order of magnitude) to the number of stars estimated for the visible universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia author(s) gets a similar result: &quot;The observable universe's mass has a Schwarzschild radius of approximately 10 billion light years&quot;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-01-15T23:59:18.743" LastActivityDate="2014-01-15T23:59:18.743" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1463" PostTypeId="2" ParentId="1460" CreationDate="2014-01-16T00:52:41.753" Score="7" Body="&lt;p&gt;According to the standard &lt;a href=&quot;http://en.wikipedia.org/wiki/%CE%9BCDM&quot;&gt;ΛCDM&lt;/a&gt; cosmological model, the observable universe has a density of about $\rho = 2.5\!\times\!10^{-27}\;\mathrm{kg/m^3}$, with a cosmological consant of about $\Lambda = 1.3\!\times\!10^{-52}\;\mathrm{m^{-2}}$, is very close to spatially flat, and has a current proper radius of about $r =  14.3\,\mathrm{Gpc}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, we can conclude that the total mass of the observable universe is about&#xA;$$M = \frac{4}{3}\pi r^3\rho \sim 9.1\!\times\!10^{53}\,\mathrm{kg}\text{.}$$ &#xA;Sine the universe at large is nonrotating and uncharged, it's natural to compare this to a Schwarzschild black hole. The Schwarzschild radius of such a black hole is&#xA;$$R_s = \frac{2GM}{c^2}\sim 44\,\mathrm{Gpc}.$$&#xA;Well! Larger that the observable universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the Schwarzschild spacetime has zero cosmological constant, whereas ours is positive, so we should instead compare this to a Schwarzschild-de Sitter black hole. The SdS metric is related to the Schwarzchild one by&#xA;$$1-\frac{R_s}{r}\quad\mapsto\quad1 - \frac{R_s}{r} - \frac{1}{3}\Lambda r^2,$$&#xA;and for our values we have $9\Lambda(GM/c^2)^2 \sim 520$. This quantity is important because the black hole event horizon and the cosmological horizon become close in $r$-coordinate when it is close to $1$, a condition that creates a maximum possible mass for an SdS black hole for a given positive cosmological constant. For our $\Lambda$, that extremal limit gives $M_\text{Nariai} \sim 4\!\times\!10^{52}\,\mathrm{kg}$, smaller than the mass of the observable universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, the mass of the observable universe cannot make a black hole.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Well, we don't fully comprehend black matter, do we? And it was just &quot;yesterday&quot; that we discovered the &quot;black energy&quot;, wasn't it? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If GTR with cosmological constant is right, we don't need to &quot;fully comprehend&quot; it to know its gravitational effect, which is what the calculation is based on. If GTR is wrong, which is of course quite possible, then we could be living in some analogue of a black hole. But then it's rather unclear what theory of gravity you wish for us to use to try to answer the question. There's no remotely competitive theory that's even approaching general acceptance.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From the perspective of our huge ignorance, I think that 14.3Gpc and 44Gpc are not even one order of magnitude apart, which I consider a good approximation. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Actually, the point of that calculation was to show that it's at least &lt;em&gt;prima facie&lt;/em&gt; plausible. The Schwarzschild radius calculation doesn't rule out the black hole--quite the opposite. However, it's also not appropriate for reasons I explained above. The more relevant one actually does have mass more than one order of magnitude apart, and shows inconsistency. So if GTR with Λ is correct, it's unlikely because the ΛCDM error bars aren't that bad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, even if we still treat it as &quot;close enough&quot;, that does not by itself imply what you want. The question of what kind of black hole all the mass of the observable universe would make, if any, is quite different from whether or not we're living in one. The black hypothetical needs to be larger still.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest point of uncertainly, though, is the cosmological constant, even if GTR is otherwise correct. If we're allowed to have very different conditions outside our hypothetical black hole, then we could still have one, but then we get into very speculative physics at best, and just complete guesswork at worst.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So treat the above answer as conditional on the mainstream physics; if that's not what you want, then there can be no general answer besides &quot;we don't know&quot;. And that's always a possibility, although not a very interesting one.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditDate="2014-01-16T02:56:12.160" LastActivityDate="2014-01-16T02:56:12.160" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1464" PostTypeId="2" ParentId="1449" CreationDate="2014-01-16T01:16:39.477" Score="0" Body="&lt;p&gt;What is the difference between time and space-time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;time:  is a change of position and/or energy of matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;space-time: is a configuration between change of energy and/or position storage into a space , ie. an atom from (Fe) Iron vs foton, gallaxy andromeda vs via lactea, black hole vs star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does gravity affect the passage of time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a set from space-time ( as black hole or star)  can change other sets space- time, in this change acelerate position and energy (time). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the speed of light and how does it relate to time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the speed of light is a specific set space-time (configuration) that have a position/energy with a limit into space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do scientists deal with timescales on the order of billions of years if time is not constant for all observers in the universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is a ilussion because a speed light or energy emmision as gamma radiation or infrared (space-time sets) has different configurations and interactions, it is only a reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is time, or for example the age of the universe, actually measured experimentally?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;they shall create a relation parent set space-time between children sets space-time (Im not sure if they do it). &lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2014-01-16T01:16:39.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1465" PostTypeId="1" CreationDate="2014-01-16T01:19:52.183" Score="4" ViewCount="611" Body="&lt;p&gt;Following the Big Bang the Universe continues to expand, presumably and roughly equally in all directions. It is understood that the Big Bang occurred 13.798 ± 0.037 billion years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way for us to know how far we are away from the nearest edge of the expansion front of the Big Bang?&lt;/p&gt;&#xA;" OwnerUserId="718" LastEditorUserId="1891" LastEditDate="2017-12-11T13:36:47.537" LastActivityDate="2017-12-11T13:36:47.537" Title="How far are we from the edge of the Universe?" Tags="&lt;universe&gt;&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ClosedDate="2014-02-06T08:45:20.630" ContentLicense="CC BY-SA 3.0" />
  <row Id="1466" PostTypeId="2" ParentId="1465" CreationDate="2014-01-16T03:06:24.777" Score="9" Body="&lt;p&gt;In the standard models of the Big Bang, there is no such thing as the &quot;edge of the expansion front&quot;. The universe, as far as we can see and as far as standard cosmology assumes, is homogeneous and isotropic on the large scale, so there is no edge or anything analogous to an the shock front of an explosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the seven-year WMAP results [&lt;a href=&quot;http://lambda.gsfc.nasa.gov/product/map/dr4/pub_papers/sevenyear/basic_results/wmap_7yr_basic_results.pdf&quot;&gt;pdf&lt;/a&gt;], the proper distance to the surface of last scattering that emitted the cosmic microwave backgrounds is approximately $46.0\,\mathrm{Gly}$, which is basically as far as we can actually see. The true cosmological horizon slightly more distant that this, closer to $46.6\,\mathrm{Gly}$. What's beyond the horizon is not known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another cosmologically significant distance scale is the Hubble radius, at which the galaxies comoving with the Hubble flow recede from us at the speed of light. It is $c/H_0 = 13.9\pm0.3\,\mathrm{Gly}$.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-01-16T03:06:24.777" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1468" PostTypeId="1" CreationDate="2014-01-16T15:41:44.723" Score="2" ViewCount="113" Body="&lt;p&gt;A lot of you are probably already aware of our &lt;a href=&quot;https://astronomy.meta.stackexchange.com/&quot;&gt;meta site&lt;/a&gt;, but I wanted to take the time to remind you that it only requires 5 reputation points to participate--and this is a critical part of our site. Meta is where we decide what kinds of questions we like in the community, it is where we pick our topic of the week, it is where we can deal with different issues of site growth. Did you know that we can suggest revisions to the help center? These are the types of things that meta allows us to discuss. In fact, this post would be best suited on meta because it is not about the field of astronomy, but about this site. I have posted it here so that you (especially new users) can see it. Maintaining activity on our meta site is important to the long term health of our site. If we don't have activity there, we could wither and ultimately get shut down. I appreciate the recent surge of questions, but I hope that many of you will stick around and participate in making this site work!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more on Meta, read this &lt;a href=&quot;http://blog.stackoverflow.com/2010/07/new-per-site-metas/&quot;&gt;blog post&lt;/a&gt;. Though, note that the link to meta is now located under the Stack Exchange dropdown on the black top bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also check out &lt;a href=&quot;http://blog.stackoverflow.com/2010/07/the-7-essential-meta-questions-of-every-beta/&quot;&gt;this blog post&lt;/a&gt; on the important questions to ask on Meta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your continued contribution to Astronomy Stack Exchange!&lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorUserId="6" LastEditDate="2017-09-01T16:28:45.240" LastActivityDate="2017-09-01T16:28:45.240" Title="Please take the time to visit Meta!" Tags="&lt;fundamental-astronomy&gt;" AnswerCount="0" CommentCount="10" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1469" PostTypeId="1" CreationDate="2014-01-16T17:22:57.637" Score="3" ViewCount="212" Body="&lt;p&gt;Imagine that hypothetically the black hole in the center of the milky way gradually increased in mass by for example 50% every year. That is exponential increase in mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which visual effects would we see on for example Alpha Centauri, stars more distant and on M31? Red or blueshift? Lensing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about linear increase in mass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there software I can use to visualize it as seen from earth?&lt;/p&gt;&#xA;" OwnerUserId="749" LastActivityDate="2014-01-18T10:42:17.707" Title="What if the black hole in the center of the galaxy grew faster?" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1470" PostTypeId="1" AcceptedAnswerId="1487" CreationDate="2014-01-16T18:08:05.497" Score="7" ViewCount="127" Body="&lt;p&gt;A gamma ray burst produces somewhere around 10&lt;sup&gt;44&lt;/sup&gt; J of energy, according to Wikipedia. Only a small portion of that energy will be absorbed by other objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest of that energy will go on to travel the universe indefinitely I presume?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we know how much of the universes total mass-energy is in &quot;in transit&quot; in the form of gamma rays and other forms of radiation?&lt;/p&gt;&#xA;" OwnerUserId="749" LastEditorUserId="718" LastEditDate="2014-01-17T22:11:49.307" LastActivityDate="2014-01-18T10:27:44.890" Title="What happens to the energy from a GRB?" Tags="&lt;radiation&gt;&lt;gamma-ray-bursts&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1471" PostTypeId="2" ParentId="1470" CreationDate="2014-01-16T20:28:08.770" Score="2" Body="&lt;p&gt;No energy is in transit as GRBs, just like Gamma rays that started in a GRB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you remember that all that energy is in the form of radiation, it obeys the same laws as ordinary light: the farther, the dimmer (inverse of square of the distance law).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you see stars dimmer as they are farther from you, same happens to the RGBs. Radiation from the stars expands infinitely (becoming dimmer and dimmer) until absorbed by clouds or just becoming dimmer than the background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same happens to gamma rays from GRBs. It is just that since rays are more energetic, they are less easily absorbed, and since GRBs themselves are more energetic, the becoming dimmer than background distance is bigger.&lt;/p&gt;&#xA;" OwnerUserId="473" LastEditorUserId="718" LastEditDate="2014-01-17T22:11:16.117" LastActivityDate="2014-01-17T22:11:16.117" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1472" PostTypeId="1" AcceptedAnswerId="1491" CreationDate="2014-01-16T21:57:05.267" Score="6" ViewCount="93" Body="&lt;p&gt;I am highly interested in how the moon originated. Theories suggest that the moon is a chunk of the earth, or rather, both were at one point one celestial body. Is there any literature out there with regard to this topic? I mean papers and simulation applets and algorithms. &lt;/p&gt;&#xA;" OwnerUserId="401" LastActivityDate="2014-01-18T13:57:18.413" Title="Recommended papers" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1473" PostTypeId="1" AcceptedAnswerId="1475" CreationDate="2014-01-17T09:04:45.153" Score="2" ViewCount="562" Body="&lt;ol&gt;&#xA;&lt;li&gt;A few days ago I just read &lt;a href=&quot;https://astronomy.stackexchange.com/questions/891/whats-the-fastest-moving-object-in-the-universe&quot;&gt;here&lt;/a&gt; that our galaxies (one of which we are part of) move or expand faster than light,&lt;/li&gt;&#xA;&lt;li&gt;And read &lt;a href=&quot;https://astronomy.stackexchange.com/questions/646/is-the-light-we-see-from-stars-extremely-old/647#647&quot;&gt;here&lt;/a&gt; that light we see from stars is extremely old, maybe more than several years.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If galaxies' expansion is faster than the light of stars, then is it possible that we see our own Sun as a star somewhere because light may reach us later than expansion? &lt;/p&gt;&#xA;" OwnerUserId="460" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-01-17T16:05:56.950" Title="Is it possible that we see our solar system as a star?" Tags="&lt;star&gt;&lt;galaxy&gt;&lt;light&gt;&lt;speed&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1475" PostTypeId="2" ParentId="1473" CreationDate="2014-01-17T10:11:29.510" Score="8" Body="&lt;p&gt;I think the original answer has confused you somewhat - it is spacetime that is expanding faster than light - &lt;em&gt;when viewed from sufficiently far away&lt;/em&gt; - we certainly are not moving away from the Sun at anything even remotely close to the speed of light (in fact the expansion of spacetime is not a motion at all in the true sense, but a change in the underlying metric).&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2014-01-17T10:11:29.510" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1476" PostTypeId="1" CreationDate="2014-01-17T13:09:44.787" Score="9" ViewCount="134" Body="&lt;p&gt;Would dense molecular structures on heavier planets (crystalline or other compounds which are generally unknown to us) allow neutrinos to pass through as easily as on Earth?&lt;/p&gt;&#xA;" OwnerUserId="825" LastEditorUserId="13" LastEditDate="2014-01-22T13:37:53.887" LastActivityDate="2014-01-22T13:37:53.887" Title="Neutrino interaction with dense molecular structures" Tags="&lt;matter&gt;&lt;neutrinos&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1477" PostTypeId="2" ParentId="1476" CreationDate="2014-01-17T13:50:46.240" Score="3" Body="&lt;p&gt;One impressive thing about neutrinos is that you would need about one light year of lead to stop a neutrino about &lt;a href=&quot;http://www.sciencedaily.com/articles/n/neutrino.htm&quot; rel=&quot;nofollow&quot;&gt;50% of the time&lt;/a&gt;. Which is due to neutrinos interacting only weakly with matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to answer your question, it would make $X$ time difference in neutrino opacity if you take you planet $X$ times larger in radius or $X$ times denser. However, the net effect will be still vanishingly small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, molecular-level structures do not affect the interaction with neutrinos, but rather the nuclear structure.&lt;/p&gt;&#xA;" OwnerUserId="115" LastEditorUserId="45" LastEditDate="2014-01-19T16:05:09.140" LastActivityDate="2014-01-19T16:05:09.140" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1478" PostTypeId="1" CreationDate="2014-01-17T16:55:18.970" Score="7" ViewCount="429" Body="&lt;p&gt;Are there any simulations that can be obtained that use smooth particle hydrodynamics and can be configured to include different initial conditions? I wish to simulate planetary collisions and their impact. I wish to &lt;em&gt;exactly replicate&lt;/em&gt; the following link.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=Fwl_JBQtH9o&quot;&gt;https://www.youtube.com/watch?v=Fwl_JBQtH9o&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="401" LastEditorUserId="401" LastEditDate="2014-01-18T03:10:23.430" LastActivityDate="2017-05-24T06:22:27.260" Title="SPH simulations" Tags="&lt;software&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1479" PostTypeId="1" AcceptedAnswerId="1480" CreationDate="2014-01-17T17:14:57.007" Score="19" ViewCount="1845" Body="&lt;p&gt;If I were floating in a void (far away from any galaxy or cluster), would the light from these objects at such distances be enough to see for example, my hand?&lt;/p&gt;&#xA;" OwnerUserId="828" LastEditorUserId="45" LastEditDate="2014-01-17T18:00:51.180" LastActivityDate="2014-01-17T18:00:51.180" Title="Can I see &quot;my own hand&quot; in the void space?" Tags="&lt;space&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1480" PostTypeId="2" ParentId="1479" CreationDate="2014-01-17T18:00:23.683" Score="11" Body="&lt;p&gt;Yes, and no. You will be able to see your hand's silhouette as it blocks incoming light from other sources, but you won't be able to see any features on it. For example, illuminance of starlight overcast moonless night sky is &lt;a href=&quot;https://en.wikipedia.org/wiki/Lux&quot;&gt;quoted&lt;/a&gt; at $0.0001\ lux$ which means that in absence of a brighter light source, human eye simply isn't that sensitive to such low illuminance, even if your hand reflected all of the incoming light (it doesn't). What is the minimum lux rating that a human eye can resolve is not as straightforward to answer though, as it varies greatly by specific wavelengths. This is explained in more detail for example in this &lt;a href=&quot;http://www.ecse.rpi.edu/~schubert/Light-Emitting-Diodes-dot-org/Sample-Chapter.pdf&quot;&gt;Human eye sensitivity and photometric quantities (PDF)&lt;/a&gt; document, or perhaps &lt;a href=&quot;https://en.wikipedia.org/wiki/Scotopic_vision&quot;&gt;Wikipedia on Scotopic vision&lt;/a&gt;. But in a nutshell, no, you wouldn't be able to see any features on your hand in outer space, unless closer to a brighter visible wavelength light source. You'll still be able to infer that it's still there and all fingers on it accounted for, though.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2014-01-17T18:00:23.683" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1481" PostTypeId="2" ParentId="1478" CreationDate="2014-01-17T18:04:20.430" Score="6" Body="&lt;p&gt;I would recommend MPA Garching's &lt;a href=&quot;http://www.mpa-garching.mpg.de/gadget/&quot;&gt;Gadget&lt;/a&gt; code for cosmological simulations of structure formation. It's primarily gravitational, but I do believe you can include gas effects as well:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;GADGET computes gravitational forces with a hierarchical tree algorithm (optionally in combination with a particle-mesh scheme for long-range gravitational forces) and represents fluids by means of smoothed particle hydrodynamics (SPH). The code can be used for studies of isolated systems, or for simulations that include the cosmological expansion of space, both with or without periodic boundary conditions. In all these types of simulations, GADGET follows the evolution of a self-gravitating collisionless N-body system, and allows gas dynamics to be optionally included. Both the force computation and the time stepping of GADGET are fully adaptive, with a dynamic range which is, in principle, unlimited. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know if this is what is commonly used for tasks such as yours, but it might be a place to start looking.&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2014-01-17T18:04:20.430" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1482" PostTypeId="1" AcceptedAnswerId="1483" CreationDate="2014-01-18T01:57:45.250" Score="5" ViewCount="488" Body="&lt;p&gt;Do all moons orbit their planets on the ecliptic? Do they all follow this same process as the moon that orbits Earth, or is their some random distribution with some even having extreme polar orbits.&lt;/p&gt;&#xA;" OwnerUserId="257" LastEditorUserId="2618" LastEditDate="2017-03-07T01:09:21.687" LastActivityDate="2017-03-07T01:09:21.687" Title="Do all moons orbit their planets on the ecliptic?" Tags="&lt;orbit&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1483" PostTypeId="2" ParentId="1482" CreationDate="2014-01-18T03:18:03.880" Score="5" Body="&lt;p&gt;Of course not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Neptune's moons. Triton's orbit has a tilt of 157º from Neptune's rotation plane, which is in itself tilted 30º from Neptune's orbit, just 2º away from the Ecliptic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Moons_of_Uranus&quot;&gt;http://en.wikipedia.org/wiki/Moons_of_Uranus&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-01-18T03:18:03.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1484" PostTypeId="1" CreationDate="2014-01-18T03:51:49.317" Score="18" ViewCount="1145" Body="&lt;p&gt;I understand the evidence for the big bang (expansion, background radiation, etc), but how do we know that it was the start of the universe? Why couldn't it have occurred in an existing, but very sparsely populated universe?&lt;/p&gt;&#xA;" OwnerUserId="829" LastActivityDate="2014-01-27T21:49:32.857" Title="How do we know the big bang didn't happen in an existing universe?" Tags="&lt;early-universe&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1485" PostTypeId="2" ParentId="1484" CreationDate="2014-01-18T05:35:24.337" Score="16" Body="&lt;p&gt;We don't know what's on the other side of the Big Bang, if anything at all. The Big Bang theory as well as the rest of physics is agnostic on this possibility. During the time period after the Big Bang, the entire universe was basically in a hot, dense plasma state; any physical signals propagating from a period before the Big Bang that could potentially influence the present or provide us with information of a pre-Bang past would be eradicated by the extreme conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the singularity of the  Big Bang is effectively the start of our universe because it is effectively an impenetrable cosmic firewall disconnecting us from anything that came before. Maybe there's another room on the other side of the wall, or maybe it's just more wall. Until someone comes up with a more refined theory or an experiment finds a way to detect a tiny signal that somehow made it through, it's pointless to speculate.&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2014-01-18T05:35:24.337" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1487" PostTypeId="2" ParentId="1470" CreationDate="2014-01-18T10:27:44.890" Score="2" Body="&lt;p&gt;The energy density of the universe is quite well known. Most seems in the form of what is called &quot;dark energy&quot;, which accelerates cosmic expansion. Next comes dark matter, accounting for some 20%, then baryonic matter, about 4%. Electromagnetic radiation (photons) contributes so little that I forgot how much. Most of the energy density in photons is in the cosmic background (CMB), the remnant of an ealier epoch when electromagnetic radiation dominated the energy budget of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cosmic expansion makes the energy density of photons decay faster than that of matter, because (in addition to the decrease in spatial density $\propto(1+z)^{-3}$) photons also get red-shifted, adding another factor $(1+z)^{-1}$.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-01-18T10:27:44.890" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1488" PostTypeId="2" ParentId="1469" CreationDate="2014-01-18T10:42:17.707" Score="1" Body="&lt;p&gt;this is not really an answer to your question, but to the assumption you made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The BH in the galactic center &lt;strong&gt;cannot grow that much&lt;/strong&gt;. There are two problems. First, there is not enough food close by. Any potential food (gas clouds, stars, and dark matter) will have some non-zero angular momentum preventing it from coming close enough. Loosing this angular momentum is difficult (it cannot be radiated away like energy), the only way is to exchange it with other objects either via impact (of gas clouds) or gravitational interactions with other objects (not the BH).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second problem is that a feeding BH is surrounded by an accretion disc of hot gas. In that disc, angular momentum is slowly transported outwards and mass inwards via viscosity (that viscosity most likely originates from turbulent magnetic fields that become unstable -- the magneto-rotational instability). This process inevitably heats the accretion disc to very high temperatures ($10^{6-9}$K) such that it emits a wind of raditation and particles (similar to the Solar wind, but much much stronger). If the BH feeds too much, this wind becomes so strong that it pushes away any further infalling material (potential food).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This second process limits the growth of any BH to double in mass in no less than about $10^6$ years, I think (I'm not too sure--if you want a precise value, consult the literature), even if the first problem was no issue.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-01-18T10:42:17.707" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1489" PostTypeId="2" ParentId="1449" CreationDate="2014-01-18T10:51:21.717" Score="5" Body="&lt;p&gt;&lt;strong&gt;Space&lt;/strong&gt;, as we experience it, is simply three-dimensional Euklidean (flat) space. A flat space is one in which parallel lines never intersect. Consider the two-dimensional space of your kitchen table: (if your table is new) this space is flat. But the surface of a sphere, also a two-dimensional space, is not flat. In the same way a three-dimensional space can be curved (though this is hard/impossible to imagine).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Time&lt;/strong&gt; as we experience it is simply linearly progressing. You can combine space and time to a four-dimensional &lt;strong&gt;space-time&lt;/strong&gt;, one dimension of which is time-like. The theory of relativity now says that this four-dimensional space-time is not always exactly flat and that any curvature of space-time is equivalent to a force. Any form of energy (and rest-mass energy is the most compact such form) curves space-time and hence exerts a force -- known as gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This curving of space time affects both space and time, hence gravity also affects clocks etc. For weak forces (any we experience as humans), the curvature effects are extremely minute.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-01-18T10:51:21.717" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1490" PostTypeId="2" ParentId="646" CreationDate="2014-01-18T12:25:01.743" Score="11" Body="&lt;p&gt;A deeper answer is &quot;yes and no&quot;. In the frame of reference of the light itself the journey from Proxima to here is instantaneous. In our frame of reference it takes four years - this is all bound up in relativity and the nature of spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the everyday sense we are indeed looking back in time at light from the stars.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2014-01-18T12:25:01.743" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1491" PostTypeId="2" ParentId="1472" CreationDate="2014-01-18T13:57:18.413" Score="4" Body="&lt;p&gt;I'd recommend taking a look at this &lt;a href=&quot;http://www.annualreviews.org/doi/abs/10.1146/annurev.astro.41.082201.113457?journalCode=astro&quot; rel=&quot;nofollow&quot;&gt;review paper&lt;/a&gt;. The author is well known for doing SPH simulations of lunar formation models, so it should provide a good starting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a side note, the most popular moon formation theory is probably the &lt;a href=&quot;http://en.wikipedia.org/wiki/Giant_impact_hypothesis&quot; rel=&quot;nofollow&quot;&gt;giant impact hypothesis&lt;/a&gt; which suggests that the moon formed after a large body collided with the proto-earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you should be aware that there are severe problems with this model as it cannot properly explain the moon composition and/or the the angular momentum in the earth-moon system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://arxiv.org/abs/1001.4243&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; gives a rather critical (and technical) discussion of the giant impact model and might also provide an interesting read.  &lt;/p&gt;&#xA;" OwnerUserId="494" LastActivityDate="2014-01-18T13:57:18.413" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1492" PostTypeId="2" ParentId="1484" CreationDate="2014-01-18T17:51:11.210" Score="4" Body="&lt;p&gt;We don't is the simple answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it is difficult to test any such theory and so it veers in the direction of metaphysics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one theory is that the physics in our neighbourhood changed (a state change, much like a change in the quantum state of a molecule migh lead to a release of energy as light) and that the energy released caused the early rapid expansion of the universe.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2014-01-18T17:51:11.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1493" PostTypeId="1" AcceptedAnswerId="1495" CreationDate="2014-01-18T18:14:45.903" Score="1" ViewCount="274" Body="&lt;p&gt;Something moving toward or away from us will be contracted in length according to special relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the M31 galaxy is moving toward us at great speed it's &quot;depth&quot; should appear slightly flattened for us. A sphere moving toward us at the speed of light will appear &quot;pancaked&quot; in lack of a better word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was predicted by Lorentz and has been proven in experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this also mean that electromagnetic radiation (light) from a flattened star is focused in the direction of motion relative to the observer?&lt;/p&gt;&#xA;" OwnerUserId="749" LastEditorUserId="749" LastEditDate="2014-01-19T02:15:09.360" LastActivityDate="2014-01-19T02:15:09.360" Title="Length contraction of a star" Tags="&lt;special-relativity&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1494" PostTypeId="2" ParentId="1493" CreationDate="2014-01-18T19:13:48.430" Score="0" Body="&lt;p&gt;Yes, this is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Relativistic_beaming&quot; rel=&quot;nofollow&quot;&gt;relativistic beaming&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="494" LastActivityDate="2014-01-18T19:13:48.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1495" PostTypeId="2" ParentId="1493" CreationDate="2014-01-18T19:30:48.093" Score="6" Body="&lt;p&gt;Note that for cosmic expansion, special relativity only applies when the spacetime region in question is approximately flat, which in our case happens when it is small.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As the M31 galaxy is moving toward us at great speed it's &quot;depth&quot; should appear slightly flattened for us. A sphere moving toward us at the speed of light will appear &quot;pancaked&quot; in lack of a better word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, that's not correct. You have a worldline in spacetime, and its direction at an event is your four-velocity. The orthogonal complement of your four-velocity, i.e., all directions perpendicular to it, is your space. What Lorentz-FitzGerald length contraction says is that if you project a moving sphere onto your space, the result is contracted along the direction of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between this and what you say is that there is no mention of how the sphere will &lt;em&gt;appear&lt;/em&gt;. The sphere &lt;em&gt;is&lt;/em&gt; contracted in the inertial frame comoving with you at that instant. It doesn't &lt;em&gt;appear&lt;/em&gt; contracted--in fact, if you took a picture of it with a hypothetical super-fast camera, the relativistically moving sphere will appear... still completely spherical. But not quite the same as a non-moving sphere if it has features on the surface, as &lt;a href=&quot;http://en.wikipedia.org/wiki/Terrell_rotation&quot;&gt;Penrose-Terrell rotation&lt;/a&gt; makes some of its &quot;back parts&quot; visible.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does this also mean that electromagnetic radiation (light) from a flattened star is focused in the direction of motion relative to the observer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. Just as a vertically falling rain comes at an angle when you're riding in a car, the angles at which any signal comes are different in a moving frame compared to a stationary frame. This is described by &lt;a href=&quot;http://en.wikipedia.org/wiki/Relativistic_aberration&quot;&gt;relativistic aberration&lt;/a&gt;. Additionally, Doppler shift changes the intensities of a radiating source along different directions, making the overall effect &lt;a href=&quot;http://en.wikipedia.org/wiki/Relativistic_beaming&quot;&gt;relativistic beaming&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-01-18T19:30:48.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1496" PostTypeId="2" ParentId="1362" CreationDate="2014-01-19T11:03:44.593" Score="2" Body="&lt;p&gt;You need to look up &quot;the equation of time&quot; - this is a correction to the time reported by the sun on any given day - the apparent solar time - (and measured by your sundial) and the standard civil time - the mean solar time - it varies throughout the year and the maximum your sundial will be out is about 15 minutes - but the difference and the rate of change varies. &lt;a href=&quot;http://en.wikipedia.org/wiki/Equation_of_time&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Equation_of_time&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your sundial is merely for educative/demonstration purposes, this might not matter so much of course - as the sundial will surely demonstrate that the sun's shadow can be used to tell the time.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2014-01-19T11:03:44.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1497" PostTypeId="1" CreationDate="2014-01-19T20:45:27.050" Score="8" ViewCount="2688" Body="&lt;p&gt;I can't seem to find this algorithm, if it exists. Can you calculate the longitude and latitude of where you are standing based on the date and the angle of the sun? Also, would the algorithm work for any date past or future?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help, AD&lt;/p&gt;&#xA;" OwnerUserId="837" LastActivityDate="2014-06-24T06:52:33.173" Title="Calculate latitude and longitude based on date and sun" Tags="&lt;the-sun&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1498" PostTypeId="1" AcceptedAnswerId="1500" CreationDate="2014-01-20T22:04:35.963" Score="8" ViewCount="360" Body="&lt;p&gt;Is it sensible to ask for the overall increase in potential energy when looking at the baryonic mass in the universe moving away from each other, that is, &lt;em&gt;moving against the direction of the force&lt;/em&gt; towards a higher energy state in each others however weak gravitation field? I can't imagine that nature has the same take on this matter like a numerical simulation (a distance which it deems &quot;large enough to make effects negligible so let's forget about it completely&quot;), i.e. the energy can't be zero... where does it come from?&lt;/p&gt;&#xA;" OwnerUserId="844" LastEditorUserId="11264" LastEditDate="2017-06-01T08:24:43.687" LastActivityDate="2017-06-01T08:24:43.687" Title="Expansion again: where does the energy come from?" Tags="&lt;expansion&gt;&lt;cosmological-inflation&gt;&lt;matter&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1499" PostTypeId="2" ParentId="1498" CreationDate="2014-01-21T00:17:19.510" Score="1" Body="&lt;p&gt;The totally unsatisfying answer is: &lt;a href=&quot;http://en.wikipedia.org/wiki/Dark_energy&quot; rel=&quot;nofollow&quot;&gt;Dark energy&lt;/a&gt;. It's formally quantified by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmological_constant&quot; rel=&quot;nofollow&quot;&gt;cosmological constant&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of hypotheses about the nature of this energy. Quantum theoretical explanations are considered as likely candidates; the &lt;a href=&quot;http://en.wikipedia.org/wiki/Casimir_effect&quot; rel=&quot;nofollow&quot;&gt;Casimir effect&lt;/a&gt; is at least an experimentally accessible way to show the existence of &lt;a href=&quot;http://en.wikipedia.org/wiki/Vacuum_energy&quot; rel=&quot;nofollow&quot;&gt;vacuum energy&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-01-21T00:17:19.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1500" PostTypeId="2" ParentId="1498" CreationDate="2014-01-21T00:29:43.120" Score="6" Body="&lt;p&gt;Conceptually there are several things going on here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does energy conservation come from? In modern understanding, energy is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Noether%27s_theorem&quot; rel=&quot;noreferrer&quot;&gt;Noether charge&lt;/a&gt; of time translation symmetry, as found by Noether's first theorem.&#xA; But in general relativity, the metric is dynamical, and so in general we simply don't have any time translation symmetry. Static spacetimes do, and there is also a form of energy conservation for spacetimes that regain time translation symmetric far away from the gravitating system (e.g. ADM energy of asymptotically flat spacetimes). But those are the exceptions, not the rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, in general relativity we don't have any scalar notion of &quot;energy&quot; that's applicable globally. Vacuously, it is neither conserved nor violated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about locally? In a local inertial frame, energy is exactly conserved, but the gravitational forces exactly vanish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing you can do in the context of cosmology is look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;noreferrer&quot;&gt;Friedmann equations&lt;/a&gt; as some sort of analogue of energy conservation, by making a balance between the terms describing cosmic expansion and the energy density, pressure, and cosmological constant. The Friedmann equations come from the components of the Einstein field equation connecting the Einstein curvature tensor and the stress-energy tensor: $G_{\mu\nu} = 8\pi T_{\mu\nu}$. According to this interpretation, Einstein curvature always exactly balances the stress-energy of the matter in spacetime. But this is just a restatement of a &lt;em&gt;dynamical&lt;/em&gt; law, so it's not really a &quot;conservation&quot; law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Einstein field equation itself can be found from the Einstein-Hilbert action, and trying to apply Noether's &lt;em&gt;second&lt;/em&gt; theorem simply shows that the covariant derivative of the stress-energy tensor vanishes: $\nabla_\nu T^{\mu\nu} = 0$, which is analogous to $\nabla\cdot\mathbf{B} = 0$ of electromagnetism: &quot;there are no local sources or sinks of [stress-energy/magnetic field] anywhere.&quot; This is actually trivial, because the covariant derivative of Einstein curvature always vanishes (a theorem of geometry lacking any physics), so Noether's second theorem didn't tell us much more than we would have known otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the derivative is covariant rather than partial, many people don't consider this to be a true conservation law either. Certainly it gives no information as to &quot;how much&quot; energy there in spacetime--that's still undefined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have the following issues:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There is no global energy conservation is general relativity, except for very special spacetimes, and the FRW family used for Big Bang models don't qualify.&lt;/li&gt;&#xA;&lt;li&gt;In a local inertial frame, energy is exactly conserved, but there are no gravitational forces. (Local inertial frames only exist as first-order approximations anyway.)&lt;/li&gt;&#xA;&lt;li&gt;One can interpret the Einstein field equation as Einstein curvature exactly balancing the stress-energy of matter, which is also motivated by interpreting the Friedmann equations of cosmology as a balance between cosmic expansion and local energy, pressure, and cosmological constant. However, this is actually a dynamical law.&lt;/li&gt;&#xA;&lt;li&gt;The vanishing of the covariant derivative of stress-energy can be construed as an analogue of local energy conservation, although doing so is conceptually misleading. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum&lt;/strong&gt;: It's notable that there is yet another sense in which the total energy of the a spatially finite universe is exactly zero. Intuitively, one can try to measure the content inside some closed surface, and then expand that surface to try to enclose everything in the universe. However, for a closed universe, that surface will contract to a point, thus enclosing nothing (picture a circle around the north pole of the Earth, and expand it to try to enclose all of Earth's surface--it just contracts to a point at the south pole).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More formally, one can find a sequence of asymptotically flat universes (for which, again, the energy actually &lt;em&gt;is&lt;/em&gt; defined) that approximate a spatially finite universe. In the limit in which the approximating universes &quot;pinch off&quot; and separate from the asymptotically flat region (thus becoming actually finite), the ADM energy goes to $0$.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditDate="2014-01-22T14:01:27.680" LastActivityDate="2014-01-22T14:01:27.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1502" PostTypeId="2" ParentId="1476" CreationDate="2014-01-21T21:23:12.550" Score="4" Body="&lt;p&gt;Neutrinos wouldn't pass through &lt;a href=&quot;http://en.wikipedia.org/wiki/White_dwarf&quot; rel=&quot;nofollow&quot;&gt;white dwarfs&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutron_star&quot; rel=&quot;nofollow&quot;&gt;neutron stars&lt;/a&gt; as easily as through Earth, measured by the scattering rate per distance, since these types of stars are very dense: white dwarfs about one million times denser than water, neutron stars about as dense as atomic nuclei.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To scatter a neutrino with a chance of &gt; 50%, you &quot;just&quot; need the inner of a neutron star along 10 times the diameter of Earth's orbit around the sun:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The neutrino-nucleon scattering cross section is about $10^{-38}cm^2=10^{-42}m^2$ for neutrinos with energies above about 1 GeV, according to &lt;a href=&quot;http://pdg.lbl.gov/2013/reviews/rpp2012-rev-nu-cross-sections.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;. A &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_nucleus&quot; rel=&quot;nofollow&quot;&gt;proton&lt;/a&gt; has a &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutron&quot; rel=&quot;nofollow&quot;&gt;diameter&lt;/a&gt; of about $1.75\cdot 10^{-15}m$.&#xA;With $$1.75\cdot 10^{-15}m/10^{-42}m^2=1.75 \cdot 10^{27}$$ we need about $1.75\cdot 10^{27}$ nucleons in a row to scatter a neutrino. That's about $$1.75\cdot 10^{27}\cdot 1.75\cdot 10^{-15}m=3\cdot 10^{12}m=20\mbox{ a.u.}$$ as a rough estimate. (For simplicity I've neglected the exponentially decreasing intensity of the neutrino beam in the scattering medium.) Neutrinos will probably scatter a little more easily in a neutron star than in the same number of single traversed neutrons, but that's a guess based on the scattering cross section for heavier nuclei.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-01-21T21:35:48.363" LastActivityDate="2014-01-21T21:35:48.363" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1503" PostTypeId="1" AcceptedAnswerId="1504" CreationDate="2014-01-21T21:30:54.017" Score="7" ViewCount="1942" Body="&lt;p&gt;For every year, the moon is orbiting further away from earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What distance has the moon been from earth at the closest?&lt;/p&gt;&#xA;" OwnerUserId="16629" LastEditorUserId="6" LastEditDate="2014-01-22T16:44:37.857" LastActivityDate="2014-01-22T16:44:37.857" Title="What distance has the moon been from earth at the closest?" Tags="&lt;the-moon&gt;&lt;earth&gt;&lt;near-earth-object&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1504" PostTypeId="2" ParentId="1503" CreationDate="2014-01-21T21:36:55.240" Score="6" Body="&lt;p&gt;If we accept &lt;a href=&quot;http://en.wikipedia.org/wiki/Giant_impact_hypothesis&quot; rel=&quot;noreferrer&quot;&gt;the Giant impact hypothesis&lt;/a&gt;, which says that the Moon was once a part of the Earth pushed away by a big impact, the answer is obviously 0.&lt;/p&gt;&#xA;" OwnerUserId="473" LastEditorUserId="537" LastEditDate="2014-01-22T14:15:30.023" LastActivityDate="2014-01-22T14:15:30.023" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1505" PostTypeId="2" ParentId="1497" CreationDate="2014-01-22T00:54:07.407" Score="5" Body="&lt;p&gt;What you are looking for is the navigation method used by ships and aircraft before the advent of GPS.  It requires not only an instrument for measuring the angle between the sun or a star (such as Polaris) and the horizon, but also an accurate time measurement, and charts that can be used to interpret the numbers.  And of course an accurate chronometer -- you'll need to know the time at Greenwich Obervatory, in London.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The traditional instrument used for determining this angle is the sextant.  Here is an article describing the use of a sextant for deterining one's position:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ehow.com/how_7562747_use-nautical-sextant.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ehow.com/how_7562747_use-nautical-sextant.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose that an algorithm for doing what you are asking is available where good sextants are sold - as to the charts, well, perhaps there are downloadable charts somewhere.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, &lt;a href=&quot;https://wiki.brown.edu/confluence/download/attachments/29391/latitude+longitude.pdf?version=1&quot; rel=&quot;nofollow&quot;&gt;here is a good article&lt;/a&gt; on determining latitude and longitude by the stars.  Latitude is what a sextant will tell you, longitude is determined differently.  For that you need two clocks: one set to GMT; one set to your local UTC time (NOT your timezone time).  The method is described here: &lt;a href=&quot;http://www.ehow.com/how_6225015_calculate-longitude.html&quot; rel=&quot;nofollow&quot;&gt;How to Calculate Longitude&lt;/a&gt;.  Which brings us to the question of how to calculate your local solar time, which is described in this article &lt;a href=&quot;http://www.ehow.com/how_8612288_calculate-solar-time.html&quot; rel=&quot;nofollow&quot;&gt;How to Calculate Solar Time&lt;/a&gt;.  Clearly you can't use this for navigation - it's convenient for human economics, but not finding your location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to make it clear why you can't use local civil time for determining longitude is that the solar time is nearly the same on both Hawaii and Kodiak islands, but Hawaii's civil time is one hour after Kodiak's.  And China stretches some 5,026 kilometers across the East Asian landmass, which is about 4 hours of solar time, but it all has the same civil time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question might better be asked in the Sailing SE.  Oh, wait, there isn't one, yet.&lt;/p&gt;&#xA;" OwnerUserId="718" LastEditorUserId="718" LastEditDate="2014-01-22T17:50:28.097" LastActivityDate="2014-01-22T17:50:28.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1506" PostTypeId="2" ParentId="646" CreationDate="2014-01-22T01:07:27.410" Score="7" Body="&lt;p&gt;Actually, the light hitting us from Proxima Centauri is not necessarily 4.243 years old.  Perhaps some of the photons arriving here were created in the photosphere of Proxima.  But some of them will have been created in the center of the star, and these photons may take many years to arrive at the photosphere, where they are then &quot;emitted&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For our sun, it is written (in Wikipedia's article about &lt;a href=&quot;https://en.wikipedia.org/wiki/Sun#Core&quot;&gt;our Sun&lt;/a&gt;): &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The gamma rays (high-energy photons) released in fusion reactions are&#xA;  absorbed in only a few millimeters of solar plasma and then re-emitted&#xA;  again in a random direction and at slightly lower energy. Therefore it&#xA;  takes a long time for radiation to reach the Sun's surface.&#xA;  &lt;strong&gt;Estimates of the photon travel time range between 10,000 and 170,000 years.&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Similarly, many of the photons arriving from Proxima may be many tens of thousands of years old.  Their travel time from Proxima's photosphere is only a small part of their journey to Earth.&lt;/p&gt;&#xA;" OwnerUserId="718" LastActivityDate="2014-01-22T01:07:27.410" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1507" PostTypeId="1" AcceptedAnswerId="1817" CreationDate="2014-01-22T16:12:52.687" Score="5" ViewCount="674" Body="&lt;p&gt;If Venus was entirely inside the habitable zone, or life zone, would its proximity to the Earth provoke any remarkable changes to the Earth? &lt;/p&gt;&#xA;" OwnerUserId="828" LastEditorUserId="863" LastEditDate="2014-02-22T13:23:06.673" LastActivityDate="2014-02-22T19:49:36.030" Title="Would Venus have any significant effects on Earth if its orbit were entirely within the habitable zone?" Tags="&lt;planet&gt;&lt;venus&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1508" PostTypeId="2" ParentId="1507" CreationDate="2014-01-22T19:40:34.113" Score="1" Body="&lt;p&gt;Changes on Earth:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Except for possible tides and possible eclipses, if it were near enough, no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changes on Earth's orbit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus would establish some resonance with Earth's orbit, if it were near enough or in one of the exact possible spots. This will not have a major impact except on year's lenght and Sun illumination on Earth.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-01-22T19:40:34.113" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1509" PostTypeId="1" AcceptedAnswerId="1511" CreationDate="2014-01-22T20:28:49.457" Score="21" ViewCount="1552" Body="&lt;p&gt;Planets form from disks of matter orbiting around a star; some moons form from disks of matter orbiting planets. If this were going to happen around Saturn, approximately how much time would it take?&lt;/p&gt;&#xA;" OwnerUserId="537" LastEditorUserId="9186" LastEditDate="2015-10-02T20:35:52.850" LastActivityDate="2015-10-02T20:35:52.850" Title="Will Saturn's rings become a moon?" Tags="&lt;gravity&gt;&lt;formation&gt;&lt;natural-satellites&gt;&lt;saturn&gt;&lt;planetary-ring&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1510" PostTypeId="2" ParentId="1509" CreationDate="2014-01-23T00:13:44.360" Score="2" Body="&lt;p&gt;It will not happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Existing moons have orbits that avoid these matter fragments to become bigger, by posing tidal forces into them.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-01-23T00:13:44.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1511" PostTypeId="2" ParentId="1509" CreationDate="2014-01-23T00:24:08.153" Score="24" Body="&lt;p&gt;The answer to the headline question is: No. Most of Saturn's rings are below the &lt;a href=&quot;http://en.wikipedia.org/wiki/Roche_limit&quot;&gt;Roche limit&lt;/a&gt; of about 2.5 Saturn radii. Hence tidal forces will prevent that part of the rings to form a (large) moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, part of the rings may be caused by loss of material from some of Saturn's moons, as &lt;a href=&quot;http://saturn.jpl.nasa.gov/science/moons/enceladus/enceladusfeedring/&quot;&gt;suspected from observations of Enceladus&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.lpi.usra.edu/meetings/metsoc99/pdf/5060.pdf&quot;&gt;Accretion of Earth&lt;/a&gt; is still going on. So any number between millions of years and billions of years of accretion time for a planet can be justified.&#xA;Half of Earth's mass should have accreted within 10 million years, see &lt;a href=&quot;http://www.pnas.org/content/early/2011/10/10/1108544108.full.pdf&quot;&gt;this paper&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-01-23T00:24:08.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1512" PostTypeId="1" AcceptedAnswerId="1513" CreationDate="2014-01-23T09:04:10.547" Score="6" ViewCount="1290" Body="&lt;p&gt;It is my understanding that one can measure the temperature of the universe by measuring cosmic microwave background radiation. Could we also use this method for calculating the average temperature of the heliosphere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average temperature of the universe has supposedly been measured in at 2.735 degrees above absolute zero. Presumably this measurement takes into account the massive amounts of 'empty' space which I imagine has a huge effect on the number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question: Can we measure the average temperature of the heliosphere? If so, then what is it?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might be quite hard to define as how do you determine when our solar system ends? So for the purpose of this question the following assumptions should be made:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The distance from the Sun to the edge of the heliosphere is &lt;a href=&quot;http://www.solarviews.com/eng/solarsys.htm&quot; rel=&quot;nofollow&quot;&gt;approximately 100AU&lt;/a&gt; (Roughly the size of the heliosphere as measured by &lt;em&gt;Voyager 1&lt;/em&gt;).&lt;/li&gt;&#xA;&lt;li&gt;Assume the heliosphere is a perfect sphere.&lt;/li&gt;&#xA;&lt;li&gt;Assume we are measuring the black body equivalent temperature (as this is what was measured to calculate the temperature of the universe).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If we don't know how to measure this average temperature (which is likely if we can't differentiate between microwaves from in and outside the heliosphere) then how could we theoretically measure this value assuming a perfect environment?&lt;/p&gt;&#xA;" OwnerDisplayName="user96" LastEditorUserId="2153" LastEditDate="2016-05-02T00:03:15.713" LastActivityDate="2016-05-02T00:03:15.713" Title="Can we calculate the average temperature of the heliosphere?" Tags="&lt;solar-system&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1513" PostTypeId="2" ParentId="1512" CreationDate="2014-01-23T15:37:35.870" Score="6" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Heliosphere&quot;&gt;heliosphere&lt;/a&gt; is mainly defined by the region dominated by solar wind against the &lt;a href=&quot;http://en.wikipedia.org/wiki/Interstellar_medium&quot;&gt;interstellar medium&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The solar wind is divided into two components, respectively termed the slow solar wind and the fast solar wind. The slow &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_wind&quot;&gt;solar wind&lt;/a&gt; has a velocity of about 400 km/s, a temperature of $1.4–1.6×10^6 K$ and a composition that is a close match to the corona. By contrast, the fast solar wind has a typical velocity of 750 km/s, a temperature of $8×10^5 K$ and it nearly matches the composition of the Sun's photosphere.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence with 1 million Kelvin we get a good estimate of the average temperature of the heliosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature is the freeze-in temperature derived from carbon ion charge states &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1029/1999JA900384/abstract&quot;&gt;measured by SOHO&lt;/a&gt;. Hence it's not the black-body temperature, but a reasonable definition of temperature for solar wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Coronal temperatures are inferred from density ratios of adjacent charge states. The freeze-in temperature derived from a given density ratio is the electron temperature that reproduces this ratio in a static situation. For the analyzed iron charge states we use the ionization and recombination rates of Arnaud and Raymond [1992]&quot;. More details about SOHO's temperature measurement &lt;a href=&quot;http://www.space.unibe.ch/~wurz/c_fep_51.pdf&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the heliosphere would be in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_equilibrium&quot;&gt;thermodynamic equilibrium&lt;/a&gt;, according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck%27s_law&quot;&gt;Planck's law&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Wien%27s_displacement_law&quot;&gt;Wien's law&lt;/a&gt;, you would get the peak at about 3nm, in the soft &lt;a href=&quot;http://en.wikipedia.org/wiki/X-ray&quot;&gt;x-ray&lt;/a&gt; range. But the equilibrium assumption probably doesn't hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the hot interstellar medium extreme ultraviolet radiation measurements have been tried (&lt;a href=&quot;http://en.wikipedia.org/wiki/CHIPSat&quot;&gt;CHIPSat&lt;/a&gt;), some successful (&lt;a href=&quot;http://www.ssl.berkeley.edu/euve/sci/Resources_high_960301.html&quot;&gt;EUVE&lt;/a&gt;).&#xA;Hence, in an ideal environment we would measure the radiation intensity across the electromagnetic spectrum, look for the peak intensity, apply Wien's law, and retrieve the  &lt;a href=&quot;http://en.wikipedia.org/wiki/Effective_temperature&quot;&gt;effective temperature&lt;/a&gt;, which would be identical to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_body&quot;&gt;black body&lt;/a&gt; temperature in an ideal world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plasma of the heliosphere is very rare, therefore almost transparent. That's why we don't feel that heat of about 1 milion degrees. Microwaves play a minor role in that temparature range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A temperature profile of Sun's corona and heliosphere is shown e.g. on pages 42 and 47 of &lt;a href=&quot;http://www.solar-system-school.de/germerode/Solar-Corona-Heliosphere.pdf&quot;&gt;this survey&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-01-24T17:45:38.710" LastActivityDate="2014-01-24T17:45:38.710" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="1514" PostTypeId="1" AcceptedAnswerId="1516" CreationDate="2014-01-23T19:12:34.553" Score="14" ViewCount="220" Body="&lt;p&gt;A supernova has been recently spotted on M82, also known as the cigar galaxy. Being the closest to earth in 27 years and given the advances of technology &lt;a href=&quot;http://www.newscientist.com/article/dn24923-closest-supernova-in-27-years-may-reveal-fate-of-cosmos.html#.UuFWdxAo7IU&quot;&gt;this article&lt;/a&gt; proposes that: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...as the Cigar Galaxy has been studied in detail thanks to images from the Hubble Space Telescope. &quot;It is likely that the star that blew up has been directly imaged&quot;...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If astronomers do find the dying star in past images and are able to make accurate readings of its brightness before and after the explosion, we might be able to reduce the uncertainties in our measurements of dark energy by a factor of two.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How will it help reduce the uncertainty?. Furthermore, how would this help us better understand the way the universe is going to end?.&lt;/p&gt;&#xA;" OwnerUserId="537" LastEditorUserId="537" LastEditDate="2014-03-14T17:36:25.157" LastActivityDate="2014-03-14T17:36:25.157" Title="How will the recent appearance of a supernova, the closest in 27 years, help reduce the uncertainties in our measurements of dark energy?" Tags="&lt;supernova&gt;&lt;dark-matter&gt;&lt;dark-energy&gt;&lt;high-energy-astrophysics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1516" PostTypeId="2" ParentId="1514" CreationDate="2014-01-23T19:53:32.263" Score="7" Body="&lt;p&gt;The current supernova is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Type_Ia_supernova&quot;&gt;supernova of type Ia&lt;/a&gt;. Supernovae of type Ia are used as &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_distance_ladder&quot;&gt;standard candles&lt;/a&gt; for distance estimates, especially used to determine the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble%27s_law&quot;&gt;Hubble constant&lt;/a&gt;. Hence by a better calibration of this kind of supernovae, more about the reliability and accuracy of distance estimates can be learned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expansion rate (in relation to the distance) of the universe, expressed by the Hubble constant, is an important incredient for the estimation of the fate of the universe. &lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-01-23T19:53:32.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1518" PostTypeId="1" AcceptedAnswerId="1521" CreationDate="2014-01-24T06:54:44.470" Score="30" ViewCount="5231" Body="&lt;p&gt;We know that the orbits of &lt;strong&gt;Pluto&lt;/strong&gt; and &lt;strong&gt;Neptune&lt;/strong&gt; overlap.&#xA;This means that pluto sometimes crosses the orbit of Neptune; will Pluto hit Neptune in any circumstance?&lt;/p&gt;&#xA;" OwnerUserId="863" LastEditorUserId="537" LastEditDate="2014-01-24T13:11:22.190" LastActivityDate="2018-03-28T14:14:27.203" Title="Pluto's orbit overlaps Neptune's, does this mean Pluto will hit Neptune sometime?" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;neptune&gt;&lt;pluto&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1519" PostTypeId="2" ParentId="1518" CreationDate="2014-01-24T08:42:25.633" Score="8" Body="&lt;p&gt;No, their orbits do not cross. They are just tangled, but in no point they coincide.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-01-24T08:42:25.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1520" PostTypeId="1" CreationDate="2014-01-24T09:35:19.030" Score="9" ViewCount="537" Body="&lt;p&gt;Is the cluster of asteroids between Mars and Jupiter two dimensional as seen in every model of solar system shown in books or television? Or is it three dimensional as i think it should be?&lt;/p&gt;&#xA;" OwnerUserId="662" LastActivityDate="2014-01-24T13:14:02.893" Title="Asteroids between Mars and Jupiter" Tags="&lt;solar-system&gt;&lt;asteroids&gt;&lt;jupiter&gt;&lt;mars&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1521" PostTypeId="2" ParentId="1518" CreationDate="2014-01-24T11:14:42.007" Score="26" Body="&lt;p&gt;No, Pluto is a so called resonant trans-neptunian object; the orbital period of Pluto is almost exactly 3:2 (1.5) times that of Neptune. This means that every time Pluto nears perihelion and is therefore closest to the Sun and also closest to the orbit of Neptune, Neptune is always at a specific angle (50° according to Wikipedia) in front or behind Pluto. &#xA;(See for instance Figure 3 and surrounding text in Jewitt, D., Morbidelli, A., &amp;amp; Rauer, H. (2007). Trans-Neptunian Objects and Comets. Springer.)&lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2014-01-24T11:14:42.007" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1522" PostTypeId="2" ParentId="1520" CreationDate="2014-01-24T11:18:31.497" Score="12" Body="&lt;p&gt;Many models shown in books or television show a very populated asteroid belt but in fact the belt is mostly empty. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question, the inclination of the asteroids vary a lot going from 0° to 40° although most off them are in between 0° and 30°; See &lt;a href=&quot;http://www.nature.com/nature/journal/v460/n7253/fig_tab/nature08094_F1.html&quot;&gt;The orbital element distributions of real and modelled asteroids&lt;/a&gt;. So yes it would be 3 dimensional.&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="537" LastEditDate="2014-01-24T13:14:02.893" LastActivityDate="2014-01-24T13:14:02.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1523" PostTypeId="1" CreationDate="2014-01-25T07:23:05.990" Score="9" ViewCount="765" Body="&lt;p&gt;I am trying to calculate the Sérsic profile of various galaxies from the SDSS based on the images provided by the galaxy zoo site. I am doing this as part of a &lt;a href=&quot;http://www.kaggle.com/c/galaxy-zoo-the-galaxy-challenge/leaderboard&quot; rel=&quot;noreferrer&quot;&gt;kaggle competition&lt;/a&gt; on using machine learning to predict galaxy morphology. I have no chance of a high rank in this competition so I'm not hesitant to ask for help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used the R contourLines function to identify the isophotes of the galaxy and then fit ellipses to each isophote. This seemed to work well, the isophotes are almost always well fit by the ellipses and the ellipses are nearly concentric. Then letting &lt;code&gt;I&lt;/code&gt; be the pixel intensity of an isophote and &lt;code&gt;R&lt;/code&gt; be the length of the semi-major axis of the corresponding ellipse, I need to fit an equation of the form&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;log I(R) = log I_0 - k * R^(1/n)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The simple approach seemed to be to take the log of both sides and use OLS regression, so I fit a linear model in R of the form&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;log(log(I)) ~ log(R)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The resulting graphs showed a good fit but the resulting Sérsic indices n are almost always less than one and never as big as two. This doesn't seem right since indices of 4 or higher seem common in my reading. I don't get anywhere near 4 for an image of M87.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possibly taking log log flattens things out too much and the index is not responsive enough. I tried using nls to work with just the log but it didn't move the indices much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any standard software or algorithm for computing the Sérsic index from an image? Are there reference images I can work from that would let me check if my algorithm is reasonable? Any recommendations on how to proceed would be welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: I have found the programs GALFIT and GIM2D which look like they might be useful. Any other software that is commonly used for this?&lt;/p&gt;&#xA;" OwnerUserId="868" LastEditorUserId="537" LastEditDate="2014-01-25T16:30:28.580" LastActivityDate="2015-06-05T11:45:33.177" Title="Computing the Sérsic profile of a galaxy from jpg images" Tags="&lt;galaxy&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1524" PostTypeId="2" ParentId="1523" CreationDate="2014-01-25T11:17:40.203" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there any standard software or algorithm for computing the Sérsic index from an image?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't think it is standard but &lt;a href=&quot;http://arxiv.org/pdf/1307.4996v1.pdf&quot; rel=&quot;nofollow&quot;&gt;Vika et al (2013)&lt;/a&gt; have used a modified version of GALFIT to extract Sérsic profiles. EDIT&gt; But I see you already found it ;-)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there reference images I can work from that would let me check if my algorithm is reasonable?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are references in literature that provide images with Sérsic profiles, but I haven't been able to find a database which I think is what you are looking for.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Any recommendations on how to proceed would be welcome.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For your purposes it is not really necessary to calculate the Sérsic index so that you can compare it with literature. If your Sérsic index, which is just one feature for your classification algorithm, is distinctive enough to be able to distinguish between different morphological classes than that is fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine that it might be difficult to calculate reliable Sérsic indices from the images provided by GalaxyZoo. The images I remember are often low resolution images of galaxies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would proceed by calculating your Sérsic index for your training and test sets and see how well this feature performs in a classification task.&lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2014-01-25T11:17:40.203" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1525" PostTypeId="1" AcceptedAnswerId="1534" CreationDate="2014-01-25T18:42:30.900" Score="6" ViewCount="166" Body="&lt;p&gt;How would a receiver on Earth hear a radio transmission from an object circling earth at 99% of the speed of light for 24 hours. The transmission from the object circulating would be non-stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because time dilation would occur, would the transmission be slowed to the receiver? The object would travel so near earth that the signal would not have any delay reaching earth.&lt;/p&gt;&#xA;" OwnerUserId="736" LastEditorUserId="537" LastEditDate="2014-01-25T18:44:19.437" LastActivityDate="2014-01-27T12:25:36.257" Title="Time dilation on an object circling earth" Tags="&lt;light&gt;&lt;general-relativity&gt;&lt;space-time&gt;&lt;speed&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1528" PostTypeId="1" CreationDate="2014-01-26T11:15:21.970" Score="3" ViewCount="1802" Body="&lt;p&gt;How to measure distances to stars by means of &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectroscopic_parallax&quot; rel=&quot;nofollow&quot;&gt;spectroscopic parallaxes&lt;/a&gt; on practice?&#xA;What is the accuracy of measuring distances using this method compared with distances based on HIPPARCOS trigonometric parallaxes? My sample of objects consists of A-B type stars.&lt;/p&gt;&#xA;" OwnerUserId="761" LastActivityDate="2014-02-01T01:28:52.937" Title="How to measure distances to stars by means of spectroscopic parallaxes?" Tags="&lt;star&gt;&lt;distances&gt;&lt;spectroscopy&gt;&lt;spectra&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1529" PostTypeId="2" ParentId="1528" CreationDate="2014-01-26T14:05:23.720" Score="4" Body="&lt;p&gt;Actually the method described on Wikipedia is not the method that is meant by Spectroscopic Parallax. To determine the spectroscopic parallax, you'll need a spectrum of the star and measure the widths of the spectral lines. Compact stars have higher surface gravity, which means that their spectral lines are broader. This means that the radius of the star can be calculated from line widening. Using the observed effective temperature $T_{eff}$ and the radius R, you can get the Luminosity from Stefan's Law, which gives an absolute magnitude and together with the apparent magnitude, a distance modulus and therefore a distance.&#xA;This gives an accuracy of 10% in luminosity and 5% in distance for main sequence stars and 25-30% in distance for giant stars. (see &lt;a href=&quot;http://books.google.nl/books?id=f2r0v98HQeQC&amp;amp;pg=PA283&amp;amp;dq=distance%20ladder&amp;amp;hl=&amp;amp;cd=7&amp;amp;source=gbs_api&quot; rel=&quot;nofollow&quot;&gt;de Grijs, 2011&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wilson and Bappu (1957) proposed to use the width (W_0) of the emission core of the CaIII absorption line at 3933 Angstrom. Using data from Hipparcos the relation between the absolute magnitude in V and the width of the absorption line was calibrated to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M_V = 33.2-18.0 \log(W_0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by Pace er al. (2003).&#xA;So to use this method you would need to be able to measure the width of the CaIII absorption line, but other calibrations using the MgIIk line also exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As these distance measurements are calibrated using Hipparcos, they will never be able to be more accurate until better calibration data is available (i.e. with Gaia).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The much more simpler method as described on Wikipedia might also be used but that is even less accurate. NB. The relation between the colour (colour index B-V) and the spectral class is not linear. Many Hertzsprung-Russel diagrams show both colour and spectral type on the X axis but this is not really correct. A K0Ia supergiant has a different colour index than a K0V star on the main sequence.&#xA;On page 50 in &lt;a href=&quot;http://books.google.nl/books?id=f2r0v98HQeQC&amp;amp;pg=PA283&amp;amp;dq=distance%20ladder&amp;amp;hl=&amp;amp;cd=7&amp;amp;source=gbs_api&quot; rel=&quot;nofollow&quot;&gt;de Grijs, 2011&lt;/a&gt;, this relation is shown in Fig 2.11. It should also be somewhere on Wikimedia (to which it is attributed) but I cannot find it.&#xA;EDIT: Here it is: &lt;a href=&quot;http://en.wikipedia.org/wiki/File:H-R_diagram.svg&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/File:H-R_diagram.svg&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="864" LastEditorUserId="864" LastEditDate="2014-02-01T00:04:28.123" LastActivityDate="2014-02-01T00:04:28.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1530" PostTypeId="1" CreationDate="2014-01-26T14:14:45.517" Score="1" ViewCount="199" Body="&lt;p&gt;Lets say that we observe in a distant galaxy cluster, which in all the galaxies move far away from us. &#xA;and here we see in the same cluster a galaxy that move away from us in a much lower velocity than the other galaxies. &#xA;is this contradict the expansion of the universe?&lt;/p&gt;&#xA;" OwnerUserId="414" LastEditorUserId="1862" LastEditDate="2014-09-03T04:07:24.510" LastActivityDate="2014-09-06T10:32:53.980" Title="Contradict the expansion of the universe?" Tags="&lt;universe&gt;&lt;expansion&gt;&lt;galaxy-cluster&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1531" PostTypeId="1" AcceptedAnswerId="1532" CreationDate="2014-01-26T16:36:15.483" Score="5" ViewCount="33706" Body="&lt;p&gt;I want to know why don't we see a change in the position of stars during the night while we do see a change in the position of the moon. I've checked other online sources and some answers say that it is because they are too far away, is this true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean for example if i'm laying down on the earth since sunset looking at the stars, and let's say I'm close to the equator, no matter the distance, I'll be looking at the stars and earth will turn about 180° until sunrise, so my angle of view will be about 180°.&lt;/p&gt;&#xA;" OwnerUserId="871" LastEditorUserId="537" LastEditDate="2014-01-26T23:07:41.577" LastActivityDate="2017-03-08T06:46:33.877" Title="Why don't stars move in the night-sky as the moon does?" Tags="&lt;the-moon&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1532" PostTypeId="2" ParentId="1531" CreationDate="2014-01-26T17:56:12.877" Score="13" Body="&lt;p&gt;This is referred to as &lt;a href=&quot;https://en.wikipedia.org/wiki/Diurnal_motion&quot; rel=&quot;noreferrer&quot;&gt;diurnal motion&lt;/a&gt;, due to Earth's rotation on its axis, and it affects apparent motion of stars differently depending on their position on the skies relative to the axis of Earth's rotation. For example, on northern hemisphere, the star that appears not to move at all is positioned so that the earth's axis of rotation points directly towards it, and is called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Polaris&quot; rel=&quot;noreferrer&quot;&gt;Polaris&lt;/a&gt; due to its role as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Pole_Star&quot; rel=&quot;noreferrer&quot;&gt;pole star&lt;/a&gt;. Stars further away from true north (not to be confused with magnetic north), will appear to prescribe a circle around it as the Earth completes one rotation on its axis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/Xhgdk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;sup&gt;A long exposure photograph of the night skies, showing their apparent motion&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what you assert, that the stars don't seem to move isn't exactly true. Stars closer to celestial equator will move at equal apparent radial velocity as any object &lt;em&gt;fixed&lt;/em&gt; on the night skies would (so at the speed of Earth's rotation), while the stars closer to celestial poles, while maintaining this radial velocity, will prescribe a smaller circle and appear to be more stationary. And the stars exactly aligned with the Earth's axis (like Polaris, but there isn't any such exact counterpart on the southern hemisphere), would appear not to move at all.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2014-01-26T17:56:12.877" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1533" PostTypeId="2" ParentId="1530" CreationDate="2014-01-26T18:19:01.277" Score="5" Body="&lt;p&gt;No, it doesn't contradict the paradigm of the universal inflation. It would merely mean that the example galactic cluster that you mention has its proper (own) motion at a smaller velocity relative to the proper motion of the Milky Way than the speed of the universal inflation at the distance between them. I.e. it would move towards us, if the Universe wasn't expanding. But since it does, galactic cluster's proper motion only managed to decrease relative velocity at which it moves away from us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it in analogy, say you're on a speeding train that moves at 120 km/h relative to the station it just passed by. You then lean out of its window and throw an apple towards the station that the train is now moving away from, but you can only manage to throw the apple at a velocity of 40 km/h. That apple, since it came from the train moving at 120 km/h and shared its momentum, is now travelling away from the station's platform at 80 km/h, even though you threw it towards the station, because you only managed to reduce its velocity relative to the station for 40 km/h. So it would still move away from it, but for 40 km/h slower than the train does.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditDate="2014-01-28T20:15:01.077" LastActivityDate="2014-01-28T20:15:01.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1534" PostTypeId="2" ParentId="1525" CreationDate="2014-01-27T02:54:09.320" Score="3" Body="&lt;p&gt;At 99% the speed of light the behaviour would be almost completely determined by special relativity. The scenario is well-investigated for &lt;a href=&quot;http://en.wikipedia.org/wiki/Synchrotron&quot; rel=&quot;nofollow&quot;&gt;synchrotrons&lt;/a&gt;. In principle a synchrotron or a &lt;a href=&quot;http://en.wikipedia.org/wiki/Storage_ring&quot; rel=&quot;nofollow&quot;&gt;storage ring&lt;/a&gt;, e.g. around the equator of Earth, could be built.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 99% the speed of light the frequency $f_s$ of the circling object should occur red-shifted by a factor of a little more than 7 to an observer in the center of the circle due to the transverse &lt;a href=&quot;http://en.wikipedia.org/wiki/Relativistic_Doppler_effect&quot; rel=&quot;nofollow&quot;&gt;relativistic Doppler effect&lt;/a&gt;:&#xA;$$f_o = f_s/\gamma=f_s\cdot\sqrt{1-v^2/c^2}=f_s\cdot\sqrt{1-0.99^2}=f_s\cdot\sqrt{0.0199}=0.141067 f_s.$$&#xA;For an observer immediately near the ring, the frequency $f_o$ is the same as for the observer at the center for the signal emitted, when the particle was diametral on the other side of the ring. When approaching the observer along the lign of sight, the signal of the particle is blue-shifted to&#xA;$$f_o=f_s\cdot\sqrt{(1+v/c)/(1-v/c)}=f_s\cdot\sqrt{1.99/0.01}=f_s\cdot\sqrt{199}=14.1067\cdot f_s.$$&#xA;When leaving the observer along the lign of sight, the signal of the particle is red-shifted to&#xA;$$f_o=f_s\cdot\sqrt{(1+v/c)/(1-v/c)}=f_s\cdot\sqrt{0.01/1.99}=0.070888\cdot f_s.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By this we now have calculated the observed frequencies for three positions of the circling particle/radio to give some idea about the oscillation of the observed frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More details about the relativistic transverse Doppler shift, see e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Ives%E2%80%93Stilwell_experiment&quot; rel=&quot;nofollow&quot;&gt;Ives-Stilwell experiment&lt;/a&gt;. Close to the experiment with the observer in the center of the circling particles are &lt;a href=&quot;http://en.wikipedia.org/wiki/Ives%E2%80%93Stilwell_experiment#M.C3.B6ssbauer_rotor_experiments&quot; rel=&quot;nofollow&quot;&gt;Mössbauer rotor experiments&lt;/a&gt;. In this cases ions or atomic nuclei emitting or absorbing at known wavelengths are used as &quot;radios&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/vc/physics/papers/0502/0502007v2.pdf&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; describes a slower version of the transverse Doppler shift, as observed by using the GPS satellites moving with just 4 km/s. In this slow case the gravitational frequency shift, as predicted by general relativity to be induced by the gravitational field of Earth, plays a relevant role, relative to the (in this case) small transverse Doppler shift. Here the GPS satellites are the moving radios.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-01-27T12:25:36.257" LastActivityDate="2014-01-27T12:25:36.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1535" PostTypeId="1" CreationDate="2014-01-27T03:59:34.730" Score="-1" ViewCount="245" Body="&lt;p&gt;Following question &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1476&quot;&gt;enter link description here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After watching the Thunderbolt Project on Youtube I have a very very very fresh perspective on the universe - the electric universe. From the electric comet theory that pretty much proves water forms at the comet, and comets are not frozen water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the solar wind all that hydrogen missing an electron and/or free electrons, fuse/react with oxygen-silicate elements of long orbit bodies that have been long long long holding opposite --- charges +++ inside and at the outside surfaces. When the rocky body nears a sun it gains plasma discharged from the sun, it arch's like a plasma torch, and water is formed, stealing the oxygen long held in the rock formation.  A comet is an asteroid that has just had a long long long time to build up a charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So with that and deep space showing dancing archs of electric plasma, ie space lightning, across the universe being as, or more influential as a force/energy, than gravity even, and with Einstein's admission that he was missing something in a unified theory, my question is: while neutrinos interact weekly with quote unquote solid matter, in a magnetic electrified universe are neutrinos subject to capture/interaction in Electro-magnetic currents or where voltage potential and capacitance is built up ? A?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand particle accelerators speed up sub atomic particles using electromagnetism but does it also translate that due to speed of atomic particles they are more likely to react with neutrinos than as a solid in out environment? B?&lt;/p&gt;&#xA;" OwnerUserId="825" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-01-27T16:19:16.597" Title="Nuetrino interaction with plasma and electromagnetism" Tags="&lt;neutrinos&gt;&lt;plasma-physics&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="0" ClosedDate="2014-01-27T22:02:08.193" ContentLicense="CC BY-SA 3.0" />
  <row Id="1536" PostTypeId="2" ParentId="1535" CreationDate="2014-01-27T06:50:07.287" Score="7" Body="&lt;p&gt;I can tell you just from &lt;a href=&quot;http://www.thunderbolts.info/wp/about/our-team/&quot;&gt;this page alone&lt;/a&gt; that it's a bunch of pseudoscience and quackery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These folks are conspiracy theorists and &lt;a href=&quot;http://en.wikipedia.org/wiki/Catastrophism&quot;&gt;catastrophists&lt;/a&gt;. Please correct me if there is, but I cannot find a single article on this that is published in a &lt;em&gt;reputable&lt;/em&gt;, peer-reviewed journal.&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2014-01-27T06:50:07.287" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1537" PostTypeId="2" ParentId="1535" CreationDate="2014-01-27T13:42:16.383" Score="5" Body="&lt;p&gt;Silicates are rather stable, and there are only small amounts of protons in the interplanetary space near a comet, hence only tiny amounts of hydroxyl or water can form within a short period of time by photochemical processes. Thus the hypotheses of the Thunderbolt Project can easily be discarded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface of comets and asteroids will be charged by ultraviolet light due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Photoelectric_effect&quot;&gt;photoelectricity&lt;/a&gt;. This can lead to &lt;a href=&quot;http://www.lpi.usra.edu/meetings/lpsc2013/pdf/2205.pdf&quot;&gt;dust levitation&lt;/a&gt;. But electric charge cannot be accumulated arbitrarily. At some point the comet would desintegrate due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Coulomb%27s_law&quot;&gt;electrostatic repulsion&lt;/a&gt;. Even before that point charge is mostly neutralized by (electrons of) the surrounding interplanetary medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Neutrino&quot;&gt;Neutrinos&lt;/a&gt; are electrically neutral and don't have any known inner structure of possibly charged electric particles. Therefore no interaction with electromagnetic fields is to be expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At high energies (more than 100 GeV) electromagnetism and weak force are thought to fuse to one &lt;a href=&quot;http://en.wikipedia.org/wiki/Electroweak_interaction&quot;&gt;electroweak interaction&lt;/a&gt;. You get neutrino reactions via $Z^0$ bosons at that energy level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These extreme conditions will be found in or near &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutron_star&quot;&gt;neutron stars&lt;/a&gt;, or near &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_hole&quot;&gt;black holes&lt;/a&gt;. They will also be caused by collisions with &lt;a href=&quot;http://en.wikipedia.org/wiki/Ultra-high-energy_cosmic_ray&quot;&gt;high energy cosmic rays&lt;/a&gt;, and will have been prevailing early after the big bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electroweak energy region is accessible to &lt;a href=&quot;http://en.wikipedia.org/wiki/Large_Hadron_Collider&quot;&gt;large colliders&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.itp.phys.ethz.ch/education/fs11/ppp2/PPP2_11.pdf&quot;&gt;In-depth paper about the electroweak interaction&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;High-energy collisions take a tiny amount of space and time; therefore it's still unikely, that some lurching neutrino gets involved. With the exception of the very dense and hot very early universe.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-01-27T16:19:16.597" LastActivityDate="2014-01-27T16:19:16.597" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1538" PostTypeId="2" ParentId="1484" CreationDate="2014-01-27T21:49:32.857" Score="4" Body="&lt;p&gt;The current model of cosmology starting with the Big Bang states that all spatial and temporal dimensions (Length, width, height, and time) as well as the four fundamental forces originated from a single point.  With time starting with the big bang, there was no 'before', just like there was no up or down, no gravity or electromagnetism.  There was no existing universe when the big bang occurred.  It happened, and then the universe existed (causality). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to say that it started in another universe would require a different hypothesis than the current cosmological model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many challenges to formulating a hypothesis that could explain what you are asking.  Some of these challenges are how to account for the laws of conservation of mass and energy.  For example - if there was a 'sparsely' populated pre-existing universe, where did the mass and energy for our universe come from?  Was it always there?  If so it must have been in the form of some type of black hole.  This much should be self evident - we can have black holes that exist in our universe with only the tiniest fraction of the total mass.  If ALL of the mass were concentrated in one place, then it too would exist within it's own Schwarzschild radius.  As a side note, it's a common exercise in college astronomy classes to calculate what the observable universe's Schwarzschild radius is.  Surprisingly enough, our observable universe can be computed as existing within an event horizon, but these calculations require that there be a lot of assumptions of total mass and maximum extent of the observable universe (which usually limits the diameter to being smaller than what the Schwarzschild radius is prior to starting the experiment).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another difficulty might be trying explain a black hole suddenly going into reverse and ejecting all matter/energy, spatial dimensions, fundamental forces, and time into an existing universe.  How does that explanation then exclude all of the black holes that are thought to exist in the present epoch of our universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short:  We don't know for a fact that the universe we see today didn't already start in a universe which was in an existing state.  However,  to the best of my knowledge, there have been no attempts made to explain how a sparsely populated universe could suddenly grow a densely populated mini-universe inside of it.  Any attempts that are out there may not have gained wide enough audience yet to include me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, remember that for a hypothesis to become theory, it must be testable.  There is no way to test events that occurred before the big bang.  Therefore there is no possible way to test for the sparsely populated universe you asked about in your question.&lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2014-01-27T21:49:32.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1539" PostTypeId="1" AcceptedAnswerId="1540" CreationDate="2014-01-27T22:24:05.803" Score="8" ViewCount="9227" Body="&lt;p&gt;Even if we see a crescent moon, always the lower circumference is visible. Why we never see this? &lt;img src=&quot;https://i.stack.imgur.com/bD52c.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: this image is vertically inverted.&lt;/p&gt;&#xA;" OwnerUserId="878" LastActivityDate="2015-02-01T23:00:59.613" Title="When we see half moon, why is it always the lower half?" Tags="&lt;the-moon&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1540" PostTypeId="2" ParentId="1539" CreationDate="2014-01-27T22:43:19.497" Score="17" Body="&lt;p&gt;Actually, you sometimes can see the Moon illuminated by the Sun from the top, and therein lies the answer to your question - during day. When looking at the Moon during night time, the side of the Earth where you're standing and looking at it is pointed away from the Sun, while this is reversed during the day. The illuminated side of the Moon is also facing the Sun, of course. So we have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Night time on Earth: Side illuminated by the Sun is &quot;below us&quot; and that's where the Moon is illuminated from too.&lt;/li&gt;&#xA;&lt;li&gt;Day time on Earth: Side illuminated by the Sun is &quot;above us&quot; and that's where the Moon is illuminated from too:&lt;br /&gt;&lt;br /&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/EkzCL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;br /&gt;&#xA;&lt;sup&gt;&lt;strong&gt;Blue Moon in the daytime sky&lt;/strong&gt; (Photograph source: &lt;a href=&quot;http://www.lpod.org/?p=1406&quot; rel=&quot;noreferrer&quot;&gt;LPOD&lt;/a&gt;)&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, the direction from which the Moon is illuminated from is the same as the direction the Earth is, both illuminated by the same celestial body - our Sun. So, when you see the Sun high above the skies, and if the Moon is visible during daytime, it's going to be illuminated from the same direction from which you are, and during night time, when you'd expect the Sun &quot;below where you're standing&quot;, it's illuminated from that direction.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditDate="2014-01-27T23:36:04.320" LastActivityDate="2014-01-27T23:36:04.320" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1541" PostTypeId="2" ParentId="1539" CreationDate="2014-01-28T00:00:36.460" Score="6" Body="&lt;p&gt;To add to TidalWave's answer, here's another way to think about it. Here are some facts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The &quot;horns&quot; of the crescent always point away from the sun. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The sun and the moon follow &lt;em&gt;roughly&lt;/em&gt; the same path through the sky on any given day; they rise in the east and set in the west.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The crescent is very thin only when the moon appears close to the sun in the sky.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A thin crescent is very dim compared to a full moon, and extraordinarily dim compared to the sun.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you don't understand why any of these facts are true then think about the relative positions of the earth, moon and sun in space until you do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how then can we possibly have a downward-opening thin crescent? These facts imply that this only happens when the moon appears to be very close to the sun and farther west than the sun. Which means that the moon is going to set before the sun does. So therefore the only time when such a moon is visible is &lt;em&gt;during the day&lt;/em&gt;, and during the day the sun is much, much brighter than the thin crescent moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, the upward-opening crescent happens when the moon is close to the sun &lt;em&gt;and farther east&lt;/em&gt;, which means that there is time &lt;em&gt;after&lt;/em&gt; the sun has set but &lt;em&gt;before&lt;/em&gt; the moon has set when it is still in the sky and bright enough to be seen in the dusk.&lt;/p&gt;&#xA;" OwnerUserId="879" LastActivityDate="2014-01-28T00:00:36.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1542" PostTypeId="1" AcceptedAnswerId="1544" CreationDate="2014-01-28T09:11:51.807" Score="5" ViewCount="282" Body="&lt;p&gt;We know, everything is in existence due to big bang from singularity. Time also started from that big bang point.&lt;br&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, did big bang happen only once or it might happen again somewhere million time away from our observable universe ? &lt;/p&gt;&#xA;" OwnerUserId="760" LastEditorUserId="760" LastEditDate="2014-01-28T16:52:03.247" LastActivityDate="2014-01-28T16:52:03.247" Title="Did big bang happen only one time or it might happen again somewhere million time away from our observable universe?" Tags="&lt;big-bang-theory&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1543" PostTypeId="2" ParentId="1542" CreationDate="2014-01-28T09:23:30.310" Score="3" Body="&lt;p&gt;Time, as we know it, started with Big Bang, so there is no &quot;before&quot;. And if there is &quot;another&quot; Big Bang (which I doubt as the universe is likely to expand forever) it will not be &quot;another&quot; Big Bang, but &quot;the&quot; Big Bang, as time will start anew with it.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-01-28T09:23:30.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1544" PostTypeId="2" ParentId="1542" CreationDate="2014-01-28T13:11:15.880" Score="2" Body="&lt;p&gt;There is a fundamental difficulty in answering this question because, while theories exist about multiple repeating events similar to the Big Bang it is difficult, if not impossible, to design experiments that would (dis)prove them - in other words they are potentially metaphysical theories. Roger Penrose, an eminent cosmologist/mathematician, recently proposed such a theory - which he sought to popularise in his book &quot;Cycles of Time&quot; - so certainly it has been suggested.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2014-01-28T13:11:15.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1545" PostTypeId="2" ParentId="1542" CreationDate="2014-01-28T13:51:45.703" Score="6" Body="&lt;p&gt;We don't know it, because the question isn't feasible thus far to experiments or observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's possible to develop theories far beyond our observable universe, even far beyond &lt;a href=&quot;http://en.wikipedia.org/wiki/Multiverse&quot;&gt;the multiverse theories described here in Wikipedia&lt;/a&gt;. There exist theories about a tranfinite hierachy of meta-levels of theories in &lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematical_physics&quot;&gt;mathematical physics&lt;/a&gt;, each level comprising all possible theories on the level below (beyond any notion of &lt;a href=&quot;http://en.wikipedia.org/wiki/Aleph_number&quot;&gt;infinity&lt;/a&gt;) and all their interconnections (including hierarchies of interconnections between interconnections). To explain this in reasonable detail would need hundreds of mind-boggling pages, at least. So I won't even try to explain more details. At the end it's not even clear, whether there is a difference between mathematics (including a transfinite number of meta-levels) and physics (including a transfinite number meta-levels).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Anthropic_principle&quot;&gt;anthropic principle&lt;/a&gt; tries to answer the follow-up question &quot;Why do we live in exactly &lt;em&gt;this&lt;/em&gt; universe with exactly &lt;em&gt;these&lt;/em&gt; laws of nature?&quot;. But this kind of answer isn't satisfying from a scientific point of view, when looking for cause and effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're looking for a final answer comprising everything: There is no, because the notion &quot;everything&quot; leads to paradoxa, when used in an absolute sense. (See &lt;a href=&quot;http://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems&quot;&gt;Gödel's incompleteness theorems&lt;/a&gt;, if you need more details.)&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-01-28T13:51:45.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1547" PostTypeId="2" ParentId="1225" CreationDate="2014-01-28T16:09:12.027" Score="7" Body="&lt;p&gt;One additional source of evidence comes from other spiral galaxies. We can see what they look like, how they are composed and we can see that the evidence we have from observations of our own galaxy fits with those galaxies. It's not as strong as the observational evidence shown elsewhere in the answers here, but it does support them.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2014-01-28T16:09:12.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1548" PostTypeId="1" AcceptedAnswerId="1549" CreationDate="2014-01-28T16:59:01.850" Score="19" ViewCount="1532" Body="&lt;p&gt;With my little knowledge, I know this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dark Matter&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The center of a galaxy controls/attracts its objects (stars, planets, comets etc.) towards itself because of gravity. But the mass of the center of this galaxy appears to be less than what it should take to attract all objects of the galaxy. Hence Dark Matter is there, which makes this attraction or controlling possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dark Energy&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the force which is pulling galaxies away from each other, and hence making the universe expand. If we don't have this hypothetical force, we cannot explain why distances between galaxies are increasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Question&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please let me know how far I am from the actual meaning/definition of these terms? And how can I explain these to a layman, who has no or very little knowledge about astronomy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: I have read &lt;a href=&quot;https://astronomy.stackexchange.com/questions/675/how-do-we-know-dark-matter-dark-energy-exists&quot;&gt;this question here&lt;/a&gt;, which is excellent but I could not get an idea of how to explain to a layman.&lt;/p&gt;&#xA;" OwnerUserId="698" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-08-29T04:55:09.783" Title="Explaining Dark Matter and Dark Energy to layman" Tags="&lt;galaxy&gt;&lt;universe&gt;&lt;dark-matter&gt;&lt;dark-energy&gt;&lt;galaxy-center&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1549" PostTypeId="2" ParentId="1548" CreationDate="2014-01-28T20:19:48.493" Score="19" Body="&lt;p&gt;&lt;strong&gt;Dark Matter&lt;/strong&gt;&#xA;Your understanding of dark matter isn't bad, but here's a few clarifying details.&#xA;Orbits: The speed of an object's orbit is related to 2 things: the radius of its orbit and the mass inside of it. In the solar system, over 99% of the mass is concentrated at the centre, so radius is the dominant effect on orbital speed. As we look at planets further away from the sun, their orbital speed decreases. In the galaxy, it's a similar story, but with increasing orbital radius, you're also getting more and more mass inside of your orbit since the galaxy is full of other stars. Even so, we would expect orbital speeds to drop as you look at objects toward the edge of the galaxy, since the stars get more diffuse as you move out and not enough mass is inside the orbit to compensate for the increasing radius. Instead, we see that after a certain point, the orbital speeds stay the same, suggesting that there is more mass there than we can account for with stars and other visible objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example of a so-called rotation curve of the &lt;a href=&quot;http://adsabs.harvard.edu/abs/1985ApJ...295..305V&quot; rel=&quot;noreferrer&quot;&gt;spiral galaxy NGC 3198&lt;/a&gt;:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/2jsX9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without dark matter, we would expect the speed of objects as a function of radius (from the center of mass) to follow the curve labeled 'disk'. However, what we see appears to be the sum of the two contributions (the disk and the 'dark matter halo' which surrounds this disk), which is overplotted with actual data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also have evidence of this extra mass from gravitational lenses. As light passes by a large object like a galaxy cluster, its path can get bent by that object's gravity. We can see this effect when a galaxy cluster is in front of an even more distant galaxy, the light from the background object gets magnified and distorted. We can calculate the path the light must have taken to appear as we see it, and the mass needed to bend it that way can't be accounted for with just the stars and gas we see in the lensing cluster. This again suggests that there is extra mass that we can't see. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gaAuX.jpg&quot; alt=&quot;Here&amp;#39;s a great picture from CFHT showing how gravity lensing works.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dark Energy&lt;/strong&gt;&#xA;You're not too far off with this one, with one major distinction: Dark energy is the 'force' that's causing the universe to &lt;em&gt;accelerate&lt;/em&gt; its expansion. We expect the universe to be expanding from the Big Bang cosmology model, but we'd expect it to be slowing down as the mutual gravity of everything in the universe acted upon each other. Dark energy appears to be a &quot;pressure&quot; on every part of space to expand, but it's very weak and has been dominated by gravity and other factors throughout most of the universe's history. It's only in the past few billion years it's started to take over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It works something like this. Imagine you have a 1-D region of spacetime 1 unit long (just for convenience's sake). Here it is: |--------|&#xA;Now this region has this dark energy 'expansion pressure' acting on it. Let's say it's 0.07 units per year per unit. This means that every year, this region of spacetime gets bigger by 7%. In 10 years, the length will be double:&#xA;|--------|--------|&#xA;Now the thing is, each of these regions has the same expansion pressure as the original one! So both will double in 10 years, and then those will double, and so on. So what happens is you get a small expansion locally, but the further away something gets the faster it's accelerating away from you. The real effect of dark energy is orders of magnitude smaller, something like 67.15 ± 1.2 (km/s)/Mpc (&lt;a href=&quot;http://en.wikipedia.org/wiki/Metric_expansion_of_space#How_is_the_expansion_of_the_universe_measured_and_how_does_the_rate_of_expansion_change.3F&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;), but it still means that any galaxies more than 4.5 Gigaparsecs away from us are currently being carried away from us faster than the speed of light. (We can still see them right now because the light we see was emitted long before the expansion rate got that high.) The expansion adds up over the huge distances we see in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark energy doesn't affect things like planets, solar systems, and galaxies because the effect is so weak on small scales that gravity more than counteracts it. It's effect can mostly be seen over the vast empty stretches of space between galaxy clusters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps!&lt;/p&gt;&#xA;" OwnerUserId="889" LastEditorUserId="45" LastEditDate="2014-01-29T21:14:40.713" LastActivityDate="2014-01-29T21:14:40.713" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1550" PostTypeId="2" ParentId="432" CreationDate="2014-01-28T20:38:13.230" Score="2" Body="&lt;p&gt;From the article: &quot;In fact, its age may be 4.36 billion years old.&quot; This is still older than the estimated age of the Late Heavy Bombardment, so it wouldn't affect that hypothesis much. In fact, it may actually help the hypothesis, as it means the Moon would have more of its interior molten (though probably not by much). It's suggested that the magma from underneath the crust filled the lunar mare after the giant impacts through fissures in the crust. (&lt;a href=&quot;http://en.wikipedia.org/wiki/Lunar_mare&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="889" LastActivityDate="2014-01-28T20:38:13.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1551" PostTypeId="1" AcceptedAnswerId="1552" CreationDate="2014-01-29T01:26:46.003" Score="14" ViewCount="8926" Body="&lt;p&gt;When one planet passes near another during its trip around the sun, does their gravitational pull is strong enough to disrupt noticeably each other's orbit ?&lt;/p&gt;&#xA;" OwnerUserId="202" LastEditorUserId="537" LastEditDate="2014-01-29T13:27:49.037" LastActivityDate="2018-10-11T17:10:14.220" Title="Does the gravity of the planets affect the orbit of other planets in our solar system?" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;gravity&gt;&lt;planet&gt;" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1552" PostTypeId="2" ParentId="1551" CreationDate="2014-01-29T05:50:47.943" Score="10" Body="&lt;p&gt;It does - although the term 'disrupt' may be a bit too strong to describe the effect; personally, I think 'influence' would fit better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting consequence of such iterations is something called &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_resonance&quot; rel=&quot;nofollow noreferrer&quot;&gt;orbital resonance&lt;/a&gt;; after long periods of time - and remember that the current estimate for our planet's existence is 4.54 billion years - the ebb and flow of tiny gravitational pulls cause nearby celestial bodies to develop an interlocked behavior. It's a double-edged sword, though; it may de-estabilize a system, or lock it into stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quoting the Wikipedia entry, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Orbital resonances greatly enhance the mutual gravitational influence&#xA;  of the bodies, i.e., their ability to alter or constrain each other's&#xA;  orbits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another gravity-related effect (although, as pointed out by &lt;a href=&quot;https://astronomy.stackexchange.com/users/864/dieudonne&quot;&gt;Dieudonné&lt;/a&gt;, present only on our solar system between bodies that have very close orbits like the Earth-Moon and Sun-Mercury systems) is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Tidal_locking#Planets&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tidal locking&lt;/a&gt;, or captured rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More about orbital resonance on this ASP Conference Series paper: &lt;a href=&quot;http://www.lpl.arizona.edu/~renu/malhotra_preprints/rio97.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Renu Malhotra, Orbital Resonances and Chaos in the Solar System&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="893" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2014-01-29T21:00:05.303" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1553" PostTypeId="2" ParentId="669" CreationDate="2014-01-29T09:55:09.933" Score="14" Body="&lt;p&gt;In one sense any point you choose is at &quot;the centre&quot; of the universe and at any point in the universe, at a large scale, the universe looks the same as at any other point. This is &lt;em&gt;not&lt;/em&gt; the same as saying the universe is infinite, though (but it could be).&#xA;The analogy with an explosion is a poor one, as explosions expand into existing space. With the Big Bang space itself expands. But by definition space does not have an edge (if it did then there would be a &quot;meta space&quot; which would be the real space and so on) and so everywhere is the centre and/or nowhere is.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2014-01-29T09:55:09.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1555" PostTypeId="2" ParentId="1551" CreationDate="2014-01-29T13:45:59.470" Score="7" Body="&lt;p&gt;Absolutely yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the planet &lt;a href=&quot;https://en.wikipedia.org/wiki/Discovery_of_Neptune&quot; rel=&quot;noreferrer&quot;&gt;Neptune was discovered&lt;/a&gt; only after differences between the observed and calculated orbit of Uranus became apparent, and astronomers were able to make predictions about the position of the 8th planet that were eventually confirmed via telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar process led to the discovery of the now ex-planet Pluto.&lt;/p&gt;&#xA;" OwnerUserId="668" LastActivityDate="2014-01-29T13:45:59.470" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1556" PostTypeId="1" AcceptedAnswerId="1562" CreationDate="2014-01-29T15:52:56.023" Score="6" ViewCount="2488" Body="&lt;p&gt;I've imported the HYG Database from &lt;a href=&quot;http://www.astronexus.com/node/34&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.astronexus.com/node/34&lt;/a&gt; and now I need to convert all RA/dec stars positions to fit in a WGS 84 coordinates (SRID=4326) PostGIS map (-180 to 180, 90 to -90).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;BayerFlamsteed  ProperName  RA            DEC&#xA;21Alp And       Alpheratz   0.13976888    29.09082805&#xA;11Bet Cas       Caph        0.15280269    59.15021814&#xA;88Gam Peg       Algenib     0.22059721    15.18361593&#xA;Alp Phe         Ankaa       0.43801871    -42.30512197&#xA;18Alp Cas       Shedir      0.67510756    56.53740928&#xA;16Bet Cet       Diphda      0.7264523     -17.98668410&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Related to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://gis.stackexchange.com/questions/84389/spatial-database-containing-coordinates-for-stars-in-the-sky&quot;&gt;https://gis.stackexchange.com/questions/84389/spatial-database-containing-coordinates-for-stars-in-the-sky&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://gis.stackexchange.com/questions/2459/what-coordinate-system-should-be-used-to-store-geography-data-for-celestial-coor&quot;&gt;https://gis.stackexchange.com/questions/2459/what-coordinate-system-should-be-used-to-store-geography-data-for-celestial-coor&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT: Ok, I think MerseyViking gives the answer in &lt;a href=&quot;https://gis.stackexchange.com/questions/2459/what-coordinate-system-should-be-used-to-store-geography-data-for-celestial-coor&quot;&gt;https://gis.stackexchange.com/questions/2459/what-coordinate-system-should-be-used-to-store-geography-data-for-celestial-coor&lt;/a&gt; telling to create a new coordinate system. Can someone give a look and some opinion?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 2 (After answered): Fantastic!! I need to show the results of whuber and Dieudonné's answer. The first picture I show the query in QGIS. In second picture you can see the real positions in a map. The scale is different - Find Shaula, Nunki and Antares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the conversion method ('H' is for Hyparcus in my database):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;insert into stars &#xA;    select &#xA;        id, &#xA;        proper_name, &#xA;        ST_SetSRID(ST_MakePoint( (-ra * 15), dec),4326),&#xA;        mag,&#xA;        'H'&#xA;    from hygn&#xA;    where (proper_name is not null)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HgQrA.png&quot; alt=&quot;My qgis query&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fT5SP.png&quot; alt=&quot;A real map&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="-1" LastEditDate="2017-04-13T12:34:15.910" LastActivityDate="2014-02-03T13:54:22.837" Title="Convert coordinates between RA/dec and WGS 84 (SRID=4326)" Tags="&lt;coordinate&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1558" PostTypeId="2" ParentId="1551" CreationDate="2014-01-29T17:49:18.290" Score="5" Body="&lt;p&gt;Depends on what you would call noticeable. The perturbations between the planets are quite small and you will only notice them if you either measure the positions of the planets very accurately or over a very long period. So don't expect two planets to suddenly change direction and move towards each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These effect are so small because planets don't actually get very near to each other. If they did then their orbits would be very unstable. If any such planet existed then they would have collided or been ejected from the solar system long ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you want to calculate accurate positions then you will have to take these effects into account.&lt;/p&gt;&#xA;" OwnerUserId="864" LastEditorUserId="864" LastEditDate="2014-01-29T17:56:56.713" LastActivityDate="2014-01-29T17:56:56.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1559" PostTypeId="2" ParentId="1397" CreationDate="2014-01-29T19:59:04.117" Score="4" Body="&lt;p&gt;To answer your second question first: Yes, antimatter does exist in the same space as matter. In fact, the universe creates antimatter (and an equal amount of matter) every day as a matter of course in events like &lt;a href=&quot;http://www.nasa.gov/mission_pages/GLAST/news/fermi-thunderstorms.html&quot; rel=&quot;nofollow&quot;&gt;lightning strikes&lt;/a&gt; and supernovae, and even in certain nuclear decays. Humans create it in particle accelerators for research and for commercial/medical applications such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Positron_emission_tomography&quot; rel=&quot;nofollow&quot;&gt;Positron Emission Tomography&lt;/a&gt;. The thing is, when we create antimatter, we also create an equal amount of matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the hot flash of energy after the Big Bang, particle-antiparticle pairs were popping into existence and annihilating each other constantly. There were almost exactly equal amounts. For some reason, though, for every 100 trillion (10^11) particles of antimatter, there were 100 trillion and one particles of matter. In the ensuing few minutes, all the antimatter and all but that tiny fraction of matter annihilated each other and turned back into energy. Everything we can see today, all the galaxies, stars, and planets, are made up of that tiny amount of matter that was left over. Particle physicists still aren't sure why there was this tiny imbalance in the amount of matter and antimatter, because all interactions we've seen so far produce equal amounts of both. This is one question particle colliders like the Large Hadron Collider are attempting to answer.&lt;/p&gt;&#xA;" OwnerUserId="889" LastActivityDate="2014-01-29T19:59:04.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1560" PostTypeId="2" ParentId="1080" CreationDate="2014-01-29T20:15:44.143" Score="8" Body="&lt;p&gt;When the Galilean moons formed, they weren't in resonance with each other. All of them were in slightly smaller orbits than they are now. Over time after their formation, Io's orbit slowly moved outward due to tides from Jupiter. This is the same effect that is causing our moon to slowly move away from the Earth (at about the same rate your fingernails grow). It goes like this. The Moon's gravity causes tides to form in Earth's oceans. This bulge of water gets carried forward with Earth's rotation, because the Earth is spinning faster than the Moon is orbiting. Because the moon is still attracting the bulge, it causes a drag on the Earth, slowing its spin. At the same time, the bulge is attracting the moon, causing it to go faster in its orbit. As the moon speeds up, its orbit gets bigger. So essentially, the Earth's spin energy is getting transferred into the Moon's orbital energy. The same thing happens with Jupiter and Io, with a bulge in Jupiter's atmosphere causing Io's orbit to get bigger.&lt;/p&gt;&#xA;&lt;p&gt;As Io's orbit expanded, its 'year' got longer, until it approached a 2/1 resonance with Europa. Once they reached resonance, they got 'locked in', their mutual gravity acting on each other reinforced it. Io was still raising tides on Jupiter, though, and its orbit was still trying to expand. As Io's orbit kept expanding, it gave a gravitational kick to Callisto on each pass, expanding both of their orbits until Callisto reached a 2/1 resonance with Ganymede. This is where the inner 3 Galilean moons got their resonance. The orbits are still expanding, but much more slowly because with each one you add on, it gets harder to transfer the energy. Given enough time, all 4 Galilean moons would probably reach resonance, although the sun will die before that happens.&lt;/p&gt;&#xA;&lt;p&gt;I may have some of the minor details wrong here, but this is the story as I understand it.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;  Seven years later, &lt;a href=&quot;https://astronomy.stackexchange.com/questions/40701/when-will-callisto-be-in-orbital-resonance-with-the-rest-of-jupiters-big-moons&quot;&gt;When will Callisto be in orbital resonance with the rest of Jupiter&amp;#39;s big moons?&lt;/a&gt; was asked.  The newest answer for when the first four moons will be in orbital resonance is much sooner than the Sun will die.  Specifically, about 1.5 billion years from now.&lt;/p&gt;&#xA;" OwnerUserId="889" LastEditorUserId="36644" LastEditDate="2021-02-04T16:43:12.313" LastActivityDate="2021-02-04T16:43:12.313" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="1561" PostTypeId="2" ParentId="1551" CreationDate="2014-01-29T20:19:52.913" Score="4" Body="&lt;p&gt;Indeed it was the perturbations of Uranus's orbit that led to the discovery of Neptune - which was itself a major triumph of the scientific process and the mathematician Le Verrier is often credited with the discovery even though he did not actually make the observational discovery.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2014-01-29T20:19:52.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1562" PostTypeId="2" ParentId="1556" CreationDate="2014-01-29T22:18:46.373" Score="3" Body="&lt;p&gt;Assuming negative longitude in WGS84 is west of Greenwich than you would get LONGITUDE = RA*15 and LATITUDE = DEC. The celestial longitude RA=0 through the vernal equinox would then be projected on the meridian of Greenwich.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you subtract 180 from the LONGITUDE then you would just rotate the celestial sphere by 180°. The meridian of Greenwich would then equate to RA=12h.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the stars to move with the rotation of the Earth, then you would use LONGITUDE = RA*15 - THETA*15, where THETA is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sidereal_time&quot; rel=&quot;nofollow&quot;&gt;sidereal time&lt;/a&gt; at Greenwich in decimal hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be aware that for the celestial sphere you look from the centre towards the inside of the surface of the sphere, while for geographical purposes you look from the outside (above) towards the spheroid (down). So if you project the stars on a globe like this, you will notice that the constellations will look inverted from what you're used to in star maps. If you find one of those old celestial globes, you will also see the inverted constellations. For instance on &lt;a href=&quot;http://www.mhs.ox.ac.uk/wp-content/uploads/17.-Mercator-globe1.jpg&quot; rel=&quot;nofollow&quot;&gt;this image&lt;/a&gt;, you'll see Leo to the left of Virgo and Hercules, while on most star maps, Leo will be to the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the constellations to look 'right' then you would need to use LONGITUDE = -RA*15 (mind the minus sign). But then you would also need to invert the rotation of the Earth as well. &lt;/p&gt;&#xA;" OwnerUserId="864" LastEditorUserId="864" LastEditDate="2014-01-29T22:42:39.893" LastActivityDate="2014-01-29T22:42:39.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1563" PostTypeId="1" AcceptedAnswerId="1565" CreationDate="2014-01-29T22:44:38.983" Score="3" ViewCount="113" Body="&lt;p&gt;&lt;a href=&quot;http://news.yahoo.com/asteroid-belt-reveals-drama-early-solar-system-evolution-180746410.html&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt; got me thinking if kepler orbits explain how a planet moves in a solar system how can a planet migrate as explained in the article. At some point don't the kepler orbits must need some other model to explain their positioning?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Jupiter would have moved right through the primordial asteroid belt, emptying it and then repopulating it with scrambled material from both the inner and outer solar system as Jupiter then reversed course and headed back towards the outer solar system,&quot; DeMeo and Carry write, stressing that further observations should nail down how faithfully this scenario represents reality.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="257" LastActivityDate="2014-01-30T02:35:38.857" Title="How do kepler orbits account for planetary migration?" Tags="&lt;orbit&gt;&lt;planet&gt;&lt;orbital-migration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1564" PostTypeId="1" AcceptedAnswerId="1566" CreationDate="2014-01-29T22:47:53.710" Score="4" ViewCount="976" Body="&lt;p&gt;&lt;strong&gt;Two Questions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;&lt;strong&gt;1)&lt;/strong&gt; Are rogue planets in motion, i.e. are they just wandering freely in space or stationary?&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1)&lt;/strong&gt; Do rogue planets have defined path or a one which can be anticipated? I understand that they cannot be at absolute rest as they are moving along with the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2)&lt;/strong&gt; Do rogue planets rotate around their axes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they are in motion, and they outnumber the stars in our galaxy, why our solar system never had a guest?&lt;/p&gt;&#xA;" OwnerUserId="698" LastEditorUserId="698" LastEditDate="2014-05-20T18:42:57.150" LastActivityDate="2017-10-27T08:23:12.883" Title="Motion of rogue planets" Tags="&lt;exoplanet&gt;&lt;rotation&gt;&lt;rogue-planet&gt;&lt;movement&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1565" PostTypeId="2" ParentId="1563" CreationDate="2014-01-29T22:57:52.520" Score="2" Body="&lt;p&gt;Kepler orbits cannot explain planetary migration as they assume constant orbital parameters such as semi-major and -minor axes and constant eccentricity. To explain planetary migration you would need at least Newtonian mechanics and numerical models, i.e. you would need the concept of gravity and the means to calculate the effects of gravity. &lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2014-01-29T22:57:52.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1566" PostTypeId="2" ParentId="1564" CreationDate="2014-01-29T23:09:47.477" Score="4" Body="&lt;p&gt;1) this question has no real answer as it depends on the reference frame being used. It is very unlikely that they will be stationary except in their own reference frame. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;rephrased 1) In theory their paths can be calculated precisely if you know their speed and the positions and motions of all other bodies (also gas and dust clouds) that might influence their motion. Practically, you will not have that knowledge, as most non-stellar bodies that the rogue planet may encounter will be very dim and you will not be able to detect them. On the other hand space is for the most part empty, so you might reasonably expect their paths to be quite predictable (within a limited time frame). They'll probably be in orbit (either an elliptical, parabolic, or hyperbolic) around the centre of the milky way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) yes, they will probably have some angular momentum left from where they formed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space is very large so it is not very likely that one will appear in our solar system in our lifetime. And of course we do not know if none ever moved through the solar system. &lt;/p&gt;&#xA;" OwnerUserId="864" LastEditorUserId="864" LastEditDate="2014-01-30T18:59:33.097" LastActivityDate="2014-01-30T18:59:33.097" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1567" PostTypeId="2" ParentId="1563" CreationDate="2014-01-30T02:28:15.817" Score="3" Body="&lt;p&gt;Kepler orbits apply only to &lt;a href=&quot;http://en.wikipedia.org/wiki/Two-body_problem&quot; rel=&quot;nofollow&quot;&gt;2-body systems&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They apply only, if the masses aren't too large and the orbits aren't too close. Otherwise &lt;a href=&quot;http://en.wikipedia.org/wiki/Apsidal_precession&quot; rel=&quot;nofollow&quot;&gt;relativistic effects&lt;/a&gt; occur, as for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Perihelion_precession_of_Mercury#Perihelion_precession_of_Mercury&quot; rel=&quot;nofollow&quot;&gt;Sun - Mercury system&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kepler orbits only apply, if the bodies are sufficiently spherical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As soon as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Three-body_problem&quot; rel=&quot;nofollow&quot;&gt;third body&lt;/a&gt; comes into play, the system can become chaotic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Too many conditions, especially for our early solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Co-orbital_configuration&quot; rel=&quot;nofollow&quot;&gt;Strange orbits&lt;/a&gt; like those around &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrangian_point&quot; rel=&quot;nofollow&quot;&gt;Lagrangian points&lt;/a&gt;, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Horseshoe_orbit&quot; rel=&quot;nofollow&quot;&gt;horseshoe orbits&lt;/a&gt; can occur. &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_resonance&quot; rel=&quot;nofollow&quot;&gt;Orbital resonance&lt;/a&gt; can stabilize as well as destablize orbits.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-01-30T02:35:38.857" LastActivityDate="2014-01-30T02:35:38.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1568" PostTypeId="1" AcceptedAnswerId="1569" CreationDate="2014-01-30T06:04:12.190" Score="5" ViewCount="149" Body="&lt;p&gt;If the planets move out of their current orbits into other orbits over long periods of time could this cause a planet to crash into the sun? Maybe the term crash is a little dramatic, could the orbit shrink to the point that the planet intersects with or is vectored into the sun?&lt;/p&gt;&#xA;" OwnerUserId="257" LastActivityDate="2014-01-30T06:49:09.347" Title="Can orbital migration cause a planet to crash into the sun?" Tags="&lt;orbit&gt;&lt;planet&gt;&lt;orbital-migration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1569" PostTypeId="2" ParentId="1568" CreationDate="2014-01-30T06:49:09.347" Score="5" Body="&lt;p&gt;Definitely. In the early solar system, there were probably several more planets than there are today. Gravitational interactions between the planets caused some to be flung into the sun, while others were sent spiralling into the sun until it settled out into the more or less stable configuration we have today. With the Kepler spacecraft and other exoplanet finding programs, we've come to realize just how big a factor migration can be. The best example of this is what are called 'hot Jupiters'. These are gas giant planets up to several times the mass of Jupiter that are in orbits that would be well within Mercury's orbit in our own solar system. These planets could not have formed where they are now, as when their parent stars ignited fusion they would have blown away all the gas in their immediate vicinity. This means that these planets must have migrated all the way in towards their sun. For each one we see in a stable orbit tight around its star or spiralling its way in, there were probably several more that collided with their parents. We just see the 'lucky' ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to assuage any fears: This sort of thing mostly happens on such large scales in the first few million years of a solar system's history. A major migration isn't going to happen in our system. Our planets' orbits aren't entirely stable, but they'll last longer than the Sun will.&lt;/p&gt;&#xA;" OwnerUserId="889" LastActivityDate="2014-01-30T06:49:09.347" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1571" PostTypeId="1" CreationDate="2014-01-30T12:00:52.697" Score="9" ViewCount="208" Body="&lt;ol&gt;&#xA;&lt;li&gt;What is the expected maximum apparent magnitude of supernova SN 2014J?&lt;/li&gt;&#xA;&lt;li&gt;When will it attain peak brightness?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="905" LastEditorUserId="537" LastEditDate="2014-01-30T14:46:26.467" LastActivityDate="2016-04-20T13:21:47.587" Title="What will be Supernova SN 2014J's peak brightness?" Tags="&lt;amateur-observing&gt;&lt;supernova&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1573" PostTypeId="1" AcceptedAnswerId="1578" CreationDate="2014-01-30T12:31:43.313" Score="4" ViewCount="1783" Body="&lt;p&gt;Related to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1556/convert-coordinates-between-ra-dec-and-wgs-84-srid-4326&quot;&gt;Convert coordinates between RA/dec and WGS 84 (SRID=4326)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WARNING: I'm a noob in astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After import all stars from HYG Database (&lt;a href=&quot;http://www.astronexus.com/node/34&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.astronexus.com/node/34&lt;/a&gt;) I noticed some missings:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) NGC catalog seems to have other names ( maybe Gliese )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Some stars have catalog names when could have simple (commom) names ('Mimosa' is 'Bet Cru').&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, I think the HYG Database is too professional (&#xA;&lt;code&gt;select count(*) from hygn where proper_name is not null&lt;/code&gt; gives me only 87 stars) and keep the stars names more scientific.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there another database like &lt;code&gt;HYG Database&lt;/code&gt; I can try?&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-01-30T16:22:52.793" Title="Star names and the HYG Database" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1574" PostTypeId="2" ParentId="370" CreationDate="2014-01-30T13:28:36.403" Score="6" Body="&lt;p&gt;&lt;a href=&quot;http://www.cfa.harvard.edu/news/2014-01&quot; rel=&quot;noreferrer&quot;&gt;This may help you&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;KOI-314c is small even when compared to Uranus or Neptune.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, look for &quot;Mini-Neptune&quot; on Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it seems those &quot;dwarf-giant&quot; planets are possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I do not have the required knowledge, I wonder if it is possible to exist a planet &lt;em&gt;smaller than&lt;/em&gt; the earth that still could be considered a gas planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, just think of something mars-sized or between mars and the earth in size and mass (by size I mean not only the &quot;solid&quot; part, but including the atmosphere outer layers, just as one would see at the eyepiece - Venus, Titan). Now, think of it as a heftier version of Titan with a much extended and denser atmosphere... could that be called a Gas planet? Besides, nature don't have to fit exactly to our classifications/expectations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know Hydrogen and Helium are very volatile and have a small atomic mass, but I guess in a very cold environment, for example few AUs around a red/brown dwarf, a small planet with a small mass (that means low escape velocity) could retain all of its original atmosphere for aeons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I dare say that even the most creative sci-fi authors coudn't predict the variety of planets that are being discovered nowadays!&lt;/p&gt;&#xA;" OwnerUserId="910" LastEditorUserId="537" LastEditDate="2014-01-30T14:43:59.300" LastActivityDate="2014-01-30T14:43:59.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1575" PostTypeId="2" ParentId="601" CreationDate="2014-01-30T13:41:39.333" Score="4" Body="&lt;p&gt;This rogue planet orbits the center of the galaxy Milk Way, (or rather, the center of mass of the system PSO J318 - rest of MW, wherever it is localised) just like the Sun and the other stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think the details of this orbit, that is eccentricity, period, inclination, etc are not well known.&#xA;If PSO J318.5-22 be confirmed as a Beta Pictoris association member, I guess it will have an orbit similar to the other members of this association/moving group.&lt;/p&gt;&#xA;" OwnerUserId="910" LastActivityDate="2014-01-30T13:41:39.333" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1576" PostTypeId="2" ParentId="1571" CreationDate="2014-01-30T13:41:51.143" Score="5" Body="&lt;p&gt;According to &lt;a href=&quot;http://www.skyandtelescope.com/observing/home/Bright-Supernova-in-M82-241477661.html&quot;&gt;this website&lt;/a&gt; the peak visible magnitude will be about 10.5 around February 2nd.&#xA;Earlier estimates had been a little brighter around the same date.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-01-30T15:16:37.707" LastActivityDate="2014-01-30T15:16:37.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1577" PostTypeId="2" ParentId="15" CreationDate="2014-01-30T13:54:11.887" Score="5" Body="&lt;p&gt;Molten Iron and hot sand - &lt;a href=&quot;http://www.nasa.gov/jpl/spitzer/brown-dwarf-20140107/&quot; rel=&quot;noreferrer&quot;&gt;it's not easy to think of something more extreme than this&lt;/a&gt;.&#xA;Add that to the high gravity of a small, dense Brown Dwarf. Theory predicts they can be as small as 0.75 RJup (if old enough) and much more massive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know, BDs are not planets. Howewer at some places, the boundaries are not well defined.&#xA;Then, I imagine that a high mass planet close to or even above what the astronomers define as the minimum mass for a BD could have a similar weather. Or could it not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This mass could be anywhere between 10 and 17 MJ depending on metallicity and other parameters. I saw this numbers in a paper at arxiv.org - sorry for not remembering it's name now. And it seems there are planets heavier than that in the net lists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Jupiter's gravity, Io have a molten core and plenty of volcanoes.&#xA;Now, what would be the climate of a planet that is subject to the much stronger pull of a pulsar or even a black hole? or the most extreme 'hot jovians' orbiting around an A/F star in few days (or just hours) and with a temperature above 3000 K...&#xA;Or going to another opposite, an ancient rogue world as cold as the circundant space or even colder (if immersed in the 'boomerang' nebula for instance) with enough atmosphere pressure to have pools of liquid Helium.&lt;/p&gt;&#xA;" OwnerUserId="910" LastEditorUserId="537" LastEditDate="2014-01-30T14:40:56.587" LastActivityDate="2014-01-30T14:40:56.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1578" PostTypeId="2" ParentId="1573" CreationDate="2014-01-30T13:57:08.087" Score="3" Body="&lt;p&gt;What about &lt;a href=&quot;http://heasarc.gsfc.nasa.gov/db-perl/W3Browse/w3table.pl?tablehead=name%3Dbsc5p&amp;amp;Action=More+Options&quot; rel=&quot;nofollow&quot;&gt;BSC5p&lt;/a&gt; for stars brighter than magnitude 6.5? The &quot;&lt;a href=&quot;http://heasarc.gsfc.nasa.gov/W3Browse/star-catalog/bsc5p.html#alt_name&quot; rel=&quot;nofollow&quot;&gt;alt_name&lt;/a&gt;&quot; &quot;is usually the Flamsteed and/or Bayer constellation-based name for the star&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://tdc-www.harvard.edu/catalogs/index.html&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; a list of more advanced catalogs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.rssd.esa.int/index.php?page=common&amp;amp;project=HIPPARCOS&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; a list of some common star names for Hipparcos numbers. Probably not much different from your query result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_proper_names_of_stars_in_alphabetical_order&quot; rel=&quot;nofollow&quot;&gt;On Wikipedia&lt;/a&gt; a probably more complete list of more or less wide-spread common star names.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those common names often are not unique, and not all common names are widely accepted. That's probably why they aren't provided in the star catalogs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way I could imagine to get a solution is to convert the Wikipedia table to a csv file, and to import it to a database table used as a relation from the various catalog names to the proper names. That will probably need quite a bit of manual work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of the naming systems is complete, not even necessarily unique. Stars frequently turn out to be not one, but several stars, or no star at all. At some point it's easier to use 'scientifics'.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-01-30T16:22:52.793" LastActivityDate="2014-01-30T16:22:52.793" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="1579" PostTypeId="2" ParentId="1361" CreationDate="2014-01-30T20:10:13.077" Score="2" Body="&lt;p&gt;The aperture of the guide scope matters because it affects how faint the stars are which are available for guiding - the larger the aperture the fainter the stars it will pick up which gives you more choice when picking a good one as a guide star. This is especially important if you are imaging from an urban area where light pollution may otherwise limit the number of visible stars. &#xA;The length of the guide scope is not directly important, although generally larger diameter objective means longer anyway. The weight of the guide scope does however have to be also considered as your mount has to be able to deal with the extra load you are placing on it, lighter is better here of course.&#xA;When you talk about pixel size I presume you mean pixel size of the CCD camera you are attaching to the guide scope itself? I am not expert enough to go into great details there, if you need more in depth advice on that side of astrophotography I can thoroughly recommend the Cloudy Nights forum, there are some very experienced astrophotographers on there who are very willing to share their knowledge and experience with all no matter how much of a beginner you may be.&lt;/p&gt;&#xA;" OwnerUserId="720" LastActivityDate="2014-01-30T20:10:13.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1580" PostTypeId="2" ParentId="1539" CreationDate="2014-01-30T23:25:18.787" Score="2" Body="&lt;p&gt;It very much depends on where on the globe you are in terms of your latitude. From mid-northern or mid-southern latitudes the terminator (line between dark and light parts of the moon) will appear roughly vertical around midnight local time at first and third quarter ('half full'). If you observed the same phase on the same night at the same local time from the equator the terminator would be roughly horizontal. Same goes for the constellations, it took some getting used to when I visited the southern hemisphere to see Orion (and all the other constellations I was familiar with) 'upside down'. &lt;/p&gt;&#xA;" OwnerUserId="720" LastActivityDate="2014-01-30T23:25:18.787" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1581" PostTypeId="2" ParentId="1105" CreationDate="2014-01-31T00:28:31.207" Score="2" Body="&lt;p&gt;Also see this: &lt;a href=&quot;http://www.threehillsobservatory.co.uk/astro/spectra_21.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.threehillsobservatory.co.uk/astro/spectra_21.htm&lt;/a&gt;. The designer of the Star Analyser grating has observed the red shift of compact galaxies with a video camera!&lt;/p&gt;&#xA;" OwnerUserId="783" LastActivityDate="2014-01-31T00:28:31.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1582" PostTypeId="2" ParentId="1518" CreationDate="2014-01-31T13:20:29.483" Score="12" Body="&lt;p&gt;No. From 1979 to 1999, Pluto was the eighth planet from the sun. In 1999, it slipped beyond Neptune to become the ninth. But Pluto's 248-year orbit around the sun takes it 17 degrees above and below the plane in which Neptune and the other planets travel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So their paths don't actually cross as they swap positions. Imagine you are the sun in the middle of your back yard. The fence is Neptune's orbit. You toss a boomerang way out over the neighbor's houses and it comes back, being on both sides of your fence during its travels without hitting the fence. Of course, activity like that can be frowned upon, and in Pluto's case helped lead to its demotion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: &lt;a href=&quot;http://www.livescience.com/32505-will-pluto-ever-hit-neptune.html&quot;&gt;Will Pluto Neptune Hit&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="903" LastEditorUserId="903" LastEditDate="2014-02-01T04:51:42.783" LastActivityDate="2014-02-01T04:51:42.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1583" PostTypeId="2" ParentId="249" CreationDate="2014-01-31T14:39:35.480" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Neutron_star&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; you read: &quot;The temperature inside a newly formed neutron star is from around 10&lt;sup&gt;11&lt;/sup&gt; to 10&lt;sup&gt;12&lt;/sup&gt; kelvin.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to McCook and Sion &lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR-3?-source=III/235B&amp;amp;-out.max=50&amp;amp;-out.form=HTML%20Table&amp;amp;-out.add=_r&amp;amp;-out.add=_RAJ,_DEJ&amp;amp;-sort=_r&amp;amp;-oc.form=sexa&quot; rel=&quot;nofollow&quot;&gt;Spectroscopically Identified White Dwarfs&lt;/a&gt; catalog, the hottest White Dwarf is&#xA;&lt;strong&gt;RE J150208+661224&lt;/strong&gt; with 170 kK.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read somewhere that the coldest WDs have Teffs between 3000 and 4000 K.&#xA;If the universe were old enough, the first WDs would be now Black Dwarfs as cold as the space around them, 3 K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For non-degenerate stars, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possibly, the hottest known main sequence star is &lt;strong&gt;HD 93129 A&lt;/strong&gt; with 52 kK.&#xA;The hypothetical &lt;a href=&quot;http://en.wikipedia.org/wiki/Population_III_stars#Population_III_stars&quot; rel=&quot;nofollow&quot;&gt;Population III stars&lt;/a&gt; could be hotter than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparison, Sun's temperature is 5778 K (wikipedia).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://arxiv.org/abs/1312.1736&quot; rel=&quot;nofollow&quot;&gt;coldest known main sequence star&lt;/a&gt; is possibly &#xA;&lt;strong&gt;2MASS J0523-1403&lt;/strong&gt; with only 2075 K. Dieterich's paper suggests that the coldest possible star could not be much colder than that, or else it would be not a star, but a Brown Dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For fusors (objects that fuse Hydrogen - stars - plus objects that fuse Deuterium - Brown Dwarfs), models predict that at the present age of the universe a BD would have cooled down to ~260 K (sorry for not remembering the reference now). Like WDs, BDs could be as cold as the space if the universe were old enough, I guess.&#xA;Then, black dwarfs aside, it seems that it is secure to consider objects colder than 260 K as planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that all temperatures listed here excepting those of the Neutron stars are &lt;em&gt;temperatures measured at the surface of these stars&lt;/em&gt;. Their centers are much hotter than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I forgot about another hypothetical objects like Quark stars, Q-stars, etc.&#xA;I would not be surprised if (they really exist outside theory) that their central temperatures would be higher than 10&lt;sup&gt;12&lt;/sup&gt; kelvin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the temperature of a supermassive Black Hole?&lt;/p&gt;&#xA;" OwnerUserId="910" LastEditorUserId="910" LastEditDate="2014-01-31T15:05:07.187" LastActivityDate="2014-01-31T15:05:07.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1584" PostTypeId="1" AcceptedAnswerId="1585" CreationDate="2014-01-31T15:33:42.933" Score="10" ViewCount="318" Body="&lt;p&gt;...what would be the immediate consequences?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or putting it better, if (from big-bang or whathever), the universe were always made of what we call Anti-matter what would it be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scientists of this universe would call 'Matter' 'Anti-Matter'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the universe looks like exactly the same? By &quot;looks&quot;, I mean the laws of Physics, the constitution of the universe with its superclusters of galaxies containing clusters containing galaxies and so on...&lt;/p&gt;&#xA;" OwnerUserId="910" LastEditorUserId="37191" LastEditDate="2021-02-17T22:26:21.937" LastActivityDate="2021-02-17T22:26:21.937" Title="If we &quot;turned universe upside-down&quot; and changed all matter to anti-matter instantly" Tags="&lt;cosmology&gt;&lt;antimatter&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1585" PostTypeId="2" ParentId="1584" CreationDate="2014-01-31T15:48:22.697" Score="5" Body="&lt;p&gt;It would look very similar, but tiny differences are suspected, see e.g. &lt;a href=&quot;http://home.web.cern.ch/about/updates/2013/04/lhcb-experiment-observes-new-matter-antimatter-difference&quot; rel=&quot;nofollow&quot;&gt;this LHC experiment&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://home.web.cern.ch/topics/antimatter&quot; rel=&quot;nofollow&quot;&gt;Here a list of more antimatter experiments at CERN&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://en.wikipedia.org/wiki/CP_violation&quot; rel=&quot;nofollow&quot;&gt;CP violation on Wikipedia&lt;/a&gt; hinting towards tiny differences between matter and antimatter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Immediate consequence: The CP violation parameter of the &lt;a href=&quot;http://pdg.lbl.gov/2008/reviews/kmmixrpp.pdf&quot; rel=&quot;nofollow&quot;&gt;CKM matrix&lt;/a&gt; in &lt;a href=&quot;http://en.wikipedia.org/wiki/Cabibbo%E2%80%93Kobayashi%E2%80%93Maskawa_matrix&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia article&lt;/a&gt; would be different, because of different quark generation probabilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you also reverse time, you'll get to the &lt;a href=&quot;http://en.wikipedia.org/wiki/CPT_symmetry&quot; rel=&quot;nofollow&quot;&gt;CPT theorem&lt;/a&gt; (Charge, Parity, Time). Matter and antimatter are thought to be symmetric in this sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or to state it with simpler words: Antimatter behaves like matter travelling backward in time.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-01-31T16:53:36.793" LastActivityDate="2014-01-31T16:53:36.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1586" PostTypeId="1" AcceptedAnswerId="1589" CreationDate="2014-01-31T22:33:47.083" Score="1" ViewCount="338" Body="&lt;p&gt;What are diapasons of galactic coordinates for &lt;a href=&quot;http://en.wikipedia.org/wiki/Thin_disk&quot; rel=&quot;nofollow&quot;&gt;thin&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Thick_disk&quot; rel=&quot;nofollow&quot;&gt;thick&lt;/a&gt; disks of our Galaxy?&lt;/p&gt;&#xA;" OwnerUserId="761" LastActivityDate="2014-02-01T14:37:35.350" Title="What are angular sizes of thin and thick disks in our Galaxy?" Tags="&lt;milky-way&gt;&lt;coordinate&gt;&lt;size&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1587" PostTypeId="2" ParentId="1571" CreationDate="2014-02-01T00:48:59.597" Score="2" Body="&lt;p&gt;The &lt;a href=&quot;http://www.aavso.org/lcg/plot?auid=000-BLG-310&amp;amp;starname=PSN+J09554214%2B6940260&amp;amp;lastdays=20&amp;amp;start=2456674&amp;amp;obscode_symbol=2&amp;amp;obstotals=yes&amp;amp;calendar=calendar&amp;amp;grid=on&amp;amp;visual=on&amp;amp;r=on&amp;amp;fainterthan=on&amp;amp;bband=on&amp;amp;v=on&amp;amp;pointsize=1&amp;amp;width=800&amp;amp;height=450&quot; rel=&quot;nofollow&quot;&gt;AAVSO data&lt;/a&gt; seems to indicate that it might have already peaked, at around 10.5 (visual).  The infrared is quite a bit brighter, partly because it penetrates dust better.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sky and Telescope says it peaked on Jan 31.&lt;/p&gt;&#xA;" OwnerUserId="924" LastEditorUserId="537" LastEditDate="2014-02-01T15:08:53.147" LastActivityDate="2014-02-01T15:08:53.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1588" PostTypeId="2" ParentId="1528" CreationDate="2014-02-01T01:28:52.937" Score="2" Body="&lt;p&gt;If you really want to use the method outlined in Wikipedia than you might use the Hipparcos catalogue to determine the mean absolute magnitude for each spectral type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried this and I got the following values for A and B type main sequence stars:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Spec        N    M_V           &amp;lt;d_err&amp;gt;  max(d_err)&#xA;B0V       [23]  -1.8 ± 2.3     130 %    1014 %&#xA;B1V       [95]  -2.1 ± 1.8      81 %     833 %&#xA;B2V      [223]  -1.6 ± 1.5      69 %    1562 %&#xA;B3V      [279]  -1.3 ± 1.4      59 %     739 %&#xA;B4V       [91]  -1.1 ± 1.1      44 %     232 %&#xA;B5V      [251]  -0.8 ± 1.2      44 %     465 %&#xA;B6V      [145]  -0.6 ± 1.3      53 %     881 %&#xA;B7V      [122]  -0.5 ± 1.4      70 %    1405 %&#xA;B8V      [522]  -0.2 ± 1.3      58 %    1988 %&#xA;B9V      [887]   0.4 ± 1.2      57 %    4613 %&#xA;A0V     [1206]   0.8 ± 1.2      54 %    2990 %&#xA;A1V      [735]   0.9 ± 1.1      50 %    2509 %&#xA;A2V      [491]   1.0 ± 1.0      42 %    1076 %&#xA;A3V      [451]   1.1 ± 1.1      51 %    3526 %&#xA;A4V      [178]   1.3 ± 1.1      49 %    1106 %&#xA;A5V      [214]   1.3 ± 1.1      45 %     762 %&#xA;A6V       [99]   1.4 ± 1.1      54 %    1924 %&#xA;A7V      [188]   1.6 ± 1.2      65 %    5371 %&#xA;A8V      [114]   1.7 ± 1.0      42 %     392 %&#xA;A9V      [329]   2.0 ± 1.2      57 %    2788 %&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where Spec is the spectral type, N is the number of stars in the Hipparcos catalogue with that spectral type (nb. many stars do not have a spectral type, or a spectral type with the star type (e.g. main sequence or supergiant) missing), the average and standard deviation of the absolute magnitude (M_V), the average error in distance in percentages() and the maximum error in distance (max(d_err)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB. I did not take extinction into account!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can use these values to calculate the distance to a main sequence A or B type star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D = 10^{(m-M_V)/5+1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as you can see the errors are quite large.&#xA;For A/B type supergiants, even the average error is more than 100%. The lowest average errors I found (25%) are for F type sub giant stars (type IV).&lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2014-02-01T01:28:52.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1589" PostTypeId="2" ParentId="1586" CreationDate="2014-02-01T02:34:13.190" Score="2" Body="&lt;p&gt;The sun is usually taken as the center of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Galactic_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;galactic coordinate systems&lt;/a&gt;. Our solar system is well within the thin disk of our galaxy (population I stars). So taking the headline question literally, we see the thin disk as well as the surrounding thick disk in almost every direction, meaning any galactic latitude and longitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radius interval, where we find mostly stars of the thin or thick disk, varies with angles. Along the &lt;a href=&quot;http://en.wikipedia.org/wiki/Galactic_plane&quot; rel=&quot;nofollow&quot;&gt;galactic plane&lt;/a&gt; the radius interval is across the galaxy (up to about 30 kilo parsecs, varying a bit with direction, because we are not in the center of the galaxy) for the thin disk, containing the sun as origin; perpendicular to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Galactic_plane&quot; rel=&quot;nofollow&quot;&gt;galactic plane&lt;/a&gt; it's about 1 kilo parsec to both directions for the thin disk, followed by a respective radius interval from 1 to 3 kilo parsecs for the thick disk, if I follow the Wikipedia articles referenced in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The region dominated by the thick disk is the farther away from us the closer we look along the galactic plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are certainly no sharp boundaries between the star populations. Actually the thick disk penetrates the thin disk, but it's much sparser populated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaia_%28spacecraft%29&quot; rel=&quot;nofollow&quot;&gt;Gaia&lt;/a&gt; project may give a much more detailed answer about the distribution of stars within our galaxy in the next 5 to 10 years.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-02-01T14:37:35.350" LastActivityDate="2014-02-01T14:37:35.350" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1590" PostTypeId="1" AcceptedAnswerId="1591" CreationDate="2014-02-01T13:17:45.387" Score="10" ViewCount="305" Body="&lt;p&gt;It's a fact that our planet's location from our parent star (among other factors) is of crucial importance to the existence of life. But can our position in the galaxy be considered a factor when considering potential planets to carry life?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any given position in the galaxy for instance there are distinct radiation levels, distinct gravitational effects, space-time, etc. These factors could contribute to the existence of life on earth or even its continuity, wouldn't they?&lt;/p&gt;&#xA;" OwnerUserId="927" LastEditorUserId="537" LastEditDate="2014-02-01T15:07:48.360" LastActivityDate="2014-02-03T13:08:08.003" Title="Is there anything special about our position in the galaxy thats necessary for life?" Tags="&lt;planet&gt;&lt;space-time&gt;&lt;life&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1591" PostTypeId="2" ParentId="1590" CreationDate="2014-02-01T15:10:52.650" Score="5" Body="&lt;p&gt;Certain regions of the galaxy are more favourable for life, others less. Life takes some time to develop, and it needs chemical elements heavier than helium, e.g. carbon or silicon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get sufficient time to develop, the solar system must not be disturbed too much by neigbouring stars. Therefore inner densely populated parts of the galaxy are less suitable than the periphery we are located. (The central star of the planetary system must not be too young, as well, to allow evolution.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heavier chemical chemical elements are sufficiently abundant mainly in parts of the galaxy with relatively young stars (population I). This kind of stars is found mainly in the thin disk close to the galactic plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very old stars tend to expand, and may eventually destroy potential life on their planets, if they are surrounded by rocky planets at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Galactic_habitable_zone&quot;&gt;More discussion on a potential galactic habitable zone on Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-02-01T15:17:09.407" LastActivityDate="2014-02-01T15:17:09.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1592" PostTypeId="1" AcceptedAnswerId="1595" CreationDate="2014-02-01T16:33:21.397" Score="6" ViewCount="1809" Body="&lt;p&gt;I know that Halley's comet is seen from Earth every 76 years or so since a couple centuries. As it travels, and particularly when it comes near the sun, it loses some of its material (ice). Someday, all of this material will be exhausted and the comet will disappear (or explode I think). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an estimation of how long the comet still have to live ?&lt;/p&gt;&#xA;" OwnerUserId="202" LastActivityDate="2017-05-16T02:01:56.047" Title="What's the estimated remaining life of Halley's Comet?" Tags="&lt;comets&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1593" PostTypeId="1" AcceptedAnswerId="1598" CreationDate="2014-02-01T16:53:46.337" Score="5" ViewCount="297" Body="&lt;p&gt;How to find acronym of specific catalog in &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/&quot; rel=&quot;noreferrer&quot;&gt;SIMBAD&lt;/a&gt; database? For example, I want to select &#xA;stellar names, object types and all binary stars from &quot;Stellar effective magnetic fields. II.&quot; catalog (Bychkov+, 2009; J/MNRAS/394/1338) via &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-fscript&quot; rel=&quot;noreferrer&quot;&gt;SIMBAD script page&lt;/a&gt;. How to do this?&lt;/p&gt;&#xA;" OwnerUserId="761" LastEditorUserId="761" LastEditDate="2014-02-01T17:12:03.843" LastActivityDate="2014-02-01T20:59:36.053" Title="Acronyms of catalogs in SIMBAD database" Tags="&lt;fundamental-astronomy&gt;&lt;data-analysis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1594" PostTypeId="1" AcceptedAnswerId="1597" CreationDate="2014-02-01T17:18:06.833" Score="1" ViewCount="3076" Body="&lt;p&gt;My birthday : &lt;strong&gt;4 December 1982&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have such astronomical calculator &lt;strong&gt;on your computer&lt;/strong&gt; please send me the result !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively if you know an &lt;strong&gt;online&lt;/strong&gt; astronomical service for my purpose please link it here ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am so curious to discover geometrical proportions between distance and angle of planets at that &lt;strong&gt;date&lt;/strong&gt; ...&lt;/p&gt;&#xA;" OwnerUserId="862" LastActivityDate="2016-03-17T08:00:47.150" Title="the most accurate system that can show the positions of planets at my birthday" Tags="&lt;planet&gt;&lt;solar-system&gt;" AnswerCount="2" CommentCount="5" ClosedDate="2014-02-03T16:21:12.047" ContentLicense="CC BY-SA 3.0" />
  <row Id="1595" PostTypeId="2" ParentId="1592" CreationDate="2014-02-01T17:38:30.100" Score="9" Body="&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/full/1985MNRAS.213..103H&quot;&gt;This paper&lt;/a&gt; estimates &quot;another 2300 close perihelion passages&quot;. Multiplied with around 76 years per period, we get roughly 175,000 years. That's probably only a very rough estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take the estimated mass of $2.2\cdot 10^{17}g$ and divide it naively by the estimated 1910 mass loss of $2.8\cdot 10^{14}g$, we get just 786 periods or roughly 60,000 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, after about 200,000 years the remnant of Halley's comet will probably ressemble an asteroid, if any.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-01T17:38:30.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1596" PostTypeId="2" ParentId="1593" CreationDate="2014-02-01T20:08:48.847" Score="2" Body="&lt;p&gt;Click on the link &quot;SIMBAD Objects (1224)&quot; shown &lt;a href=&quot;http://adsabs.harvard.edu/abs/2009MNRAS.394.1338B&quot; rel=&quot;nofollow&quot;&gt;in this page&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems Simbad doesn't have an specific acronym for the stars studied by Bychkov et al, because I was unable to find such acronym using Simbad's &lt;a href=&quot;http://cds.u-strasbg.fr/cgi-bin/Dic-Simbad&quot; rel=&quot;nofollow&quot;&gt;Dictionary of Nomenclature of Celestial Objects&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I knew that ADS shows a Simbad link for some articles, I searched for articles written by Bychkov in 2009 using &lt;a href=&quot;http://adsabs.harvard.edu/abstract_service.html&quot; rel=&quot;nofollow&quot;&gt;The SAO/NASA Astrophysics Data System Astronomy Query Form&lt;/a&gt;. That's how I find the first link in this answer.&lt;/p&gt;&#xA;" OwnerUserId="910" LastActivityDate="2014-02-01T20:08:48.847" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1597" PostTypeId="2" ParentId="1594" CreationDate="2014-02-01T20:13:08.023" Score="4" Body="&lt;p&gt;The most accurate system is the ephemeris service from Jet Propulsion Laboratory (JPL): &lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow&quot;&gt;http://ssd.jpl.nasa.gov/horizons.cgi&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2014-02-01T20:13:08.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1598" PostTypeId="2" ParentId="1593" CreationDate="2014-02-01T20:19:09.773" Score="4" Body="&lt;p&gt;It's roughly like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;set limit 100&#xA;format object form1 &quot;%IDLIST(1) : %OTYPE(S)&quot;&#xA;query sample (otypes = '*i*') &amp;amp; (bibcode='2009MNRAS.394.1338B')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To get the list of the bibcodes choose&#xA;&quot;Other&quot; / &quot;Catalogue collection&quot;&#xA;from the menue bar.&#xA;The screen should look like&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ajO7q.jpg&quot; alt=&quot;SIMBA1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then click the icon &quot;Click to display the menu&quot; at the left border:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Je1iR.jpg&quot; alt=&quot;SIMBA2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you find the bibcode:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/EEVTk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-02-01T20:59:36.053" LastActivityDate="2014-02-01T20:59:36.053" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1599" PostTypeId="1" AcceptedAnswerId="1600" CreationDate="2014-02-02T01:12:36.523" Score="2" ViewCount="4220" Body="&lt;p&gt;I wanted to know if the tilt of the earth were changed to 30 degrees would the tropics of cancer and Capricorn also both change to 30 degrees?&lt;/p&gt;&#xA;" OwnerUserId="931" LastEditorUserId="202" LastEditDate="2014-02-02T23:14:10.140" LastActivityDate="2014-02-02T23:14:10.140" Title="If the tilt of the Earth changed, would the tropics latitude change too?" Tags="&lt;earth&gt;&lt;inclination&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2014-02-03T16:24:57.733" ContentLicense="CC BY-SA 3.0" />
  <row Id="1600" PostTypeId="2" ParentId="1599" CreationDate="2014-02-02T06:57:45.163" Score="4" Body="&lt;p&gt;Yes, the Tropics of Cancer and Capricorn are defined as the highest and lowest latitudes where the sun is directly overhead at least one day a year. The sun's highest position appears to move up and down on the sky over the year because of the Earth's tilt, so if the Earth's tilt were greater or lesser, so would be the extent of the Tropics. (Side note: I may be wrong, but I think the Tropics were named after the Zodiac sign the sun is in when it reaches its highest or lowest point.)&lt;/p&gt;&#xA;" OwnerUserId="889" LastActivityDate="2014-02-02T06:57:45.163" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1602" PostTypeId="2" ParentId="1590" CreationDate="2014-02-02T14:22:22.160" Score="1" Body="&lt;p&gt;Yes, in the sense that there are some places that it would not be great to be - for instance if we were near the galactic centre we would risk being gobbled up by the super-massive black hole that is likely to exist there and so on. But spatially most of the galaxy is much like where we are, so it is not as though we are in a particularly special place.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2014-02-02T14:22:22.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1603" PostTypeId="2" ParentId="1594" CreationDate="2014-02-02T22:21:10.260" Score="2" Body="&lt;p&gt;If you have a smartphone(android, iPhone, windows phone) there are lots of apps with features like &quot;time travelling&quot; . Location can be set up automatically or manually.&#xA;They are all very accurate. One example is google's &quot;Sky Map&quot; , but check other apps too. Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="938" LastEditorUserId="938" LastEditDate="2016-03-17T08:00:47.150" LastActivityDate="2016-03-17T08:00:47.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1604" PostTypeId="2" ParentId="1590" CreationDate="2014-02-03T10:41:04.930" Score="2" Body="&lt;p&gt;The chance of the host star being part of a Binary is also of significance, since the majority of stars in our galaxy form one member of a binary or higher order stellar system. &lt;a href=&quot;http://arxiv.org/abs/0705.3444&quot; rel=&quot;nofollow&quot;&gt;Research has hinted&lt;/a&gt; that the number of planets in binary systems is larger due to gravitational perturbations collecting the dust disc quicker which could in turn lead to a higher chance of a planet landing in the habitable zone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are still unsure of the effect of a stellar companion on the chances for life on a planet. It could be detrimental by perturbing the orbit of any planets around either star, leading to it moving in and out of the habitable zone (not good for creating steady conditions for life).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chance of a star being in a binary system increases as the number density of stars increases; areas such as the Bulge at the centre of a disc galaxy, or in elliptical galaxies. The disc would have a lower density of binaries.&lt;/p&gt;&#xA;" OwnerUserId="941" LastEditorUserId="537" LastEditDate="2014-02-03T13:08:08.003" LastActivityDate="2014-02-03T13:08:08.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1605" PostTypeId="1" AcceptedAnswerId="1608" CreationDate="2014-02-03T11:23:18.097" Score="16" ViewCount="339" Body="&lt;p&gt;We all focused on our moon, to find water or life, or any thing related to life. But my question is, is there any chance for life in moons of other planets in our solar system, others have too many moons....&lt;/p&gt;&#xA;" OwnerUserId="863" LastActivityDate="2014-02-03T11:47:50.930" Title="Any possibility for life on moons of planets in our Solar System" Tags="&lt;the-moon&gt;&lt;solar-system&gt;&lt;planet&gt;&lt;life&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1606" PostTypeId="1" AcceptedAnswerId="1607" CreationDate="2014-02-03T11:25:59.923" Score="1" ViewCount="133" Body="&lt;p&gt;According to the theory of relativity (If I understand correctly), nothing is supposed to be able to travel faster then the speed of light. I believe, to my limited knowledge and grasp of the concept, that quantum theorie speaks against, i.e. extend, the theory of relativity, hence my question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we find (or imagine) a place in the universe where there is no obstruction for this thought experiment to take place. We take a spherical object and attach a spire at both sides so you get an object shaped like this: -O-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We start turning the object so the ends of the spire travel in an orbital mather according to the spherical object. What if the sphere starts to orbit near the spead of light, wouldn't the ends of the spires travel faster than the speed of light?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if this is the wrong place for this question.&lt;/p&gt;&#xA;" OwnerUserId="942" LastActivityDate="2014-02-03T11:42:48.860" Title="traveling past the speed of light" Tags="&lt;light&gt;&lt;space-time&gt;&lt;pulsar&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2014-02-03T16:35:36.113" ContentLicense="CC BY-SA 3.0" />
  <row Id="1607" PostTypeId="2" ParentId="1606" CreationDate="2014-02-03T11:42:48.860" Score="3" Body="&lt;p&gt;This would be more appropiate at Physics.s.c but here is the answer nonetheless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your object may be perfectly rigid or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the object is perfectly rigid, you would not be able to accelerate it to the point at which the extreme points of the spires reach the speed of light, because when they are approaching it, their mass will increment towards infinity and the energy you will need to continue accelerating will increment to infinity as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you will simply not be able to accelerate the whole thing to a rotation speed where the extreme points will reach the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the object is not rigid, well, it will simply bend, plus the previous explanation.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-02-03T11:42:48.860" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1608" PostTypeId="2" ParentId="1605" CreationDate="2014-02-03T11:47:50.930" Score="10" Body="&lt;p&gt;If you check 2001 A Space Odyssey saga, Europa is actually a satellite with Life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coming back to real space, the &quot;comfort zone&quot; for life is approximately from Venus' orbit to Mars' orbit. There, only Moon, Phobos and Deimos are satellites, and all of them are too dry to have any kind of life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond the comfort zone you need some other way for water not to frozen (a commonly referred prerequisite of Life), so you'll need a volcanic or tidal-heatened satellite. There are a few of these. Europa and Io come to my mind just now. Life there could be possible.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-02-03T11:47:50.930" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1609" PostTypeId="1" AcceptedAnswerId="1615" CreationDate="2014-02-03T11:50:23.463" Score="3" ViewCount="807" Body="&lt;p&gt;Assuming that the Universe is and has been expanding ever since its creation (&lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Bang&quot; rel=&quot;nofollow&quot;&gt;Big Bang theory&lt;/a&gt;), the galaxies that are now at the 'edge' (not visible theoretically) must have been (at some point in time) at a place around where the Earth is at now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is assumed to be true, then the photons emitted by these galaxies from this point onwards should be around for us to see these galaxies (and even the first light emerged from the Big Bang must be visible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then why is the observable so small? It should depend upon the direction in which we look. That is, if we look in the direction of the center of the expanding sphere, it should be a small observable universe. If we train our telesopes in 'exactly' opposite direction of the center, it should be 'fully' observable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it work this way? Or are my basics wrong?&lt;/p&gt;&#xA;" OwnerUserId="943" LastEditorUserId="332" LastEditDate="2014-04-21T19:55:38.857" LastActivityDate="2014-04-21T19:55:38.857" Title="Why are distant galaxies not visible in the observable Universe?" Tags="&lt;amateur-observing&gt;&lt;universe&gt;&lt;light&gt;&lt;big-bang-theory&gt;&lt;early-universe&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1610" PostTypeId="2" ParentId="1609" CreationDate="2014-02-03T15:55:11.243" Score="2" Body="&lt;p&gt;The space we observe behaves roughly like the surface of an expanding sphere, and we can only look along its surface. The interior of the sphere has existed in the past, the outside will exist in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface of a sphere is two-dimensional. Our space has one more dimension; it may be thought of being embedded in a 4-dimensional space (&lt;a href=&quot;http://en.wikipedia.org/wiki/N-dimensional_space&quot; rel=&quot;nofollow&quot;&gt;hyperspace&lt;/a&gt;), although this embedding hyperspace doesn't need to exist to describe our universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like the surface of an expanding sphere has finite area and no boundary, our three-dimensional space can be finite for a fixed time of expansion, with no boundary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe (the radius of the sphere, if you want) expands faster than the speed of light, therefore we cannot look completely around the universe to see our own past, but just to the point it started its expansion. The starting point (big bang) seems to be opaque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see the afterglow of the big bang in any direction as &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow&quot;&gt;cosmic microwave background&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-03T15:55:11.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1611" PostTypeId="1" AcceptedAnswerId="1667" CreationDate="2014-02-03T17:33:46.310" Score="16" ViewCount="1770" Body="&lt;p&gt;I`ve heard about different extremofiles living on earth. Is water a requirement for life because we have not seen life survive on earth without it, is it a chemical  reason which makes life without water seem impossible? Could life exist with water, but in gas form? What makes the liquid part so important? Can life be based on other  liquids on potential exoplanets?&lt;/p&gt;&#xA;" OwnerUserId="36" LastEditorUserId="36" LastEditDate="2014-02-03T17:39:57.207" LastActivityDate="2015-01-13T15:21:12.493" Title="Why is liquid water considered a requirement for life?" Tags="&lt;life&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1612" PostTypeId="1" AcceptedAnswerId="1616" CreationDate="2014-02-03T17:49:04.243" Score="2" ViewCount="152" Body="&lt;p&gt;How big does a satelite we send have to be to be able to chrush through the ice layer and dive into the ocean on Europa? If a mission should head there, is this a better way of examining the ocean than landing on the surface and drilling through?&lt;/p&gt;&#xA;" OwnerUserId="36" LastActivityDate="2014-02-03T20:17:07.873" Title="What is needed to break through the surface of Europa?" Tags="&lt;europa&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2014-02-03T21:03:07.253" ContentLicense="CC BY-SA 3.0" />
  <row Id="1613" PostTypeId="5" CreationDate="2014-02-03T17:56:51.760" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-02-03T17:56:51.760" LastActivityDate="2014-02-03T17:56:51.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1614" PostTypeId="4" CreationDate="2014-02-03T17:56:51.760" Score="0" Body="Questions regarding life in the universe, especially the possibility of life outside our planet, or the influence of astronomical forces on life on Earth." OwnerUserId="2674" LastEditorUserId="2153" LastEditDate="2015-04-12T22:00:12.900" LastActivityDate="2015-04-12T22:00:12.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1615" PostTypeId="2" ParentId="1609" CreationDate="2014-02-03T18:08:54.430" Score="8" Body="&lt;p&gt;Quick answer: Because they didn't entered our event horizon. Some never will. And some will move out of our event horizon - their last photons that'll be received here being sent right now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's do some fact checks first:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[...]the galaxies that are now at the 'edge' (not visible&#xA;  theoretically) must have been (at some point in time) at place around&#xA;  where the Earth is at now[...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.space.com/11386-galaxies-formation-big-bang-hubble-telescope.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;According to current calculations&lt;/a&gt;&lt;/strong&gt;, the first galaxies may have formed around 200 million years after the Big Bang - older estimates went with the 400-500 MY range. For a long, long while, there were no stars to be seen. So if you go back in time you won't be seeing the same structures we see today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, and that may be an awkward mental experiment, nothing else was occupying our place other than ourselves. I'll ask you to excuse the cliché, but the old balloon example is very apt to explain this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://astrobioloblog.files.wordpress.com/2011/08/hubble_expansion.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the universe expands, the distance between celestial bodies increases. Now, here's a way to put it: &lt;em&gt;space is being generated in between the objects&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And not only that - while there's a limit on how fast you can move on the balloon's surface, &lt;em&gt;it doesn't apply to the amount of space being generated&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a direct consequence there's a bubble around us that basically works in the very same way as a black hole's event horizon does:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In a black hole, gravity is so strong that its pull on photons exceeds &lt;em&gt;C&lt;/em&gt; (the speed of light, or 299,792,458 m/s);&lt;/li&gt;&#xA;&lt;li&gt;On a fringe object, the amount of space generated by the expansion, per second, may exceed &lt;em&gt;C&lt;/em&gt;; its photons will never reach Earth.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And it seems that we live in a &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Accelerating_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;accelerated universe&lt;/a&gt;&lt;/strong&gt; - that is, the spatial acceleration ratio is actually going up. If that's correct, some fringe objects will  slip away form our event horizon, disappearing (from our point of view) into a sample of the &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_death_of_the_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;heat death of the universe&lt;/a&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scary, huh?&lt;/p&gt;&#xA;" OwnerUserId="893" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:47.197" LastActivityDate="2014-02-03T18:08:54.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1616" PostTypeId="2" ParentId="1612" CreationDate="2014-02-03T20:10:48.203" Score="3" Body="&lt;p&gt;There is no feasible way for a human-made probe to crush through Europa's ice layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;It is predicted that the outer crust of solid ice is approximately 10–30 km (6–19 mi) thick.&quot; (&lt;a href=&quot;http://en.wikipedia.org/wiki/Europa_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not even known, whether there is a liqud ocean at all under Europa's ice layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of drilling one could try to slowly melt trough the ice.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-02-03T20:17:07.873" LastActivityDate="2014-02-03T20:17:07.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1620" PostTypeId="1" AcceptedAnswerId="1629" CreationDate="2014-02-04T08:19:15.427" Score="4" ViewCount="156" Body="&lt;p&gt;It is somewhat imaginable for all the matter or energy to be condensed in sub-atomic size. But what about space? The space is neither 'matter' nor 'Energy'.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there were no space at singularity where did it come from?&lt;/p&gt;&#xA;" OwnerUserId="943" LastActivityDate="2014-02-04T16:19:35.833" Title="In what form was the space at singularity?" Tags="&lt;space&gt;&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1621" PostTypeId="1" AcceptedAnswerId="1630" CreationDate="2014-02-04T09:58:43.600" Score="4" ViewCount="4000" Body="&lt;p&gt;Since Neptune/Uranus have high percentages of methane, wouldn't it be highly likely that there will be combustion (triggered by the lightning storms or any other factor) if oxygen was abundant? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the hydrogen combust as well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gi3qc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gi3qc.jpg&quot; alt=&quot;Neptune&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="927" LastEditorUserId="-1" LastEditDate="2015-12-08T21:51:02.097" LastActivityDate="2018-09-14T00:07:41.983" Title="If oxygen was abundant in Neptune, would there be combustion?" Tags="&lt;planet&gt;&lt;exoplanet&gt;&lt;neptune&gt;&lt;gas-giants&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1622" PostTypeId="1" CreationDate="2014-02-04T10:48:53.410" Score="5" ViewCount="1736" Body="&lt;p&gt;Some quick googling has given me the answer to whether seeing part of the moon that receives no sunlight is an optical illusion or not: it isn't. Earthshine lights the moon enough for the human eye to notice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it seems clear to me that the outline of the unlit part of the crescent moon is brighter than the surface. I was riding my bike home yesterday on a clear winter night, and became confused on why that could be. What mechanism is responsible for making the edge brighter than the dark part of the moon? Or is it my brain trying to make a circle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://31.media.tumblr.com/1bb65c125d6ceda03e76b5b787dd82ef/tumblr_mrqi7hcZ0b1qmvxavo2_1280.jpg&quot; rel=&quot;nofollow&quot;&gt;Here's an example of what I mean.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="948" LastEditorUserId="948" LastEditDate="2014-02-04T15:52:32.987" LastActivityDate="2014-02-06T17:55:49.707" Title="Why is the outline of a crescent moon brighter?" Tags="&lt;the-moon&gt;&lt;light&gt;" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1623" PostTypeId="1" CreationDate="2014-02-04T10:50:50.920" Score="2" ViewCount="71" Body="&lt;p&gt;We know that planets form as a result of gaseous material revolving around a center of mass that froze to form planets. However shouldn't the angular speed of that matter increase as they go closer to the center to maintain the angular momentum? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why then the average rotational period of the external planets are faster than that for the terrestrial planets?&lt;/p&gt;&#xA;" OwnerUserId="927" LastEditorUserId="927" LastEditDate="2014-02-04T20:44:32.400" LastActivityDate="2014-02-04T21:09:27.750" Title="Rotational speeds of the material forming planets should increase, not decrease" Tags="&lt;orbit&gt;&lt;planet&gt;&lt;rotation&gt;&lt;planetary-formation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1624" PostTypeId="1" AcceptedAnswerId="1626" CreationDate="2014-02-04T13:15:07.070" Score="6" ViewCount="269" Body="&lt;p&gt;I just read that according to the BBT the universe is expanding. This is evident from the fact that distant stars (and galaxies) are moving away from us. But it could also happen that the universe is in fact shrinking to a very small point and we being far away from the point are experiencing its pull less than those that are near it. It may be like a race where a very fast (in 1st place) , a moderately fast (in 2nd place) and a slow car (at last) are racing . To the moderately fast car it would seem that the other two are moving away from it,but actually they all are moving towards the same point -the end. So it may happen that earth is like the moderately fast car moving towards the collapse point at a different relative speed. Is this proposition possible? If not,why?&lt;/p&gt;&#xA;" OwnerUserId="949" LastActivityDate="2015-01-24T23:32:19.283" Title="Can there be a 2nd possibility deduced from the fact that galaxies are moving away from us?" Tags="&lt;universe&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1625" PostTypeId="2" ParentId="1624" CreationDate="2014-02-04T13:27:36.480" Score="2" Body="&lt;p&gt;Simple answe would be No.&lt;br&gt;&#xA;Why? Because these galaxies are also moving away from each other.&lt;br&gt;&#xA;I.e. if these galaxies are moving towards the same point, then distance between them should decrease. Which is not the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nice question though.&lt;/p&gt;&#xA;" OwnerUserId="943" LastActivityDate="2014-02-04T13:27:36.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1626" PostTypeId="2" ParentId="1624" CreationDate="2014-02-04T15:19:14.733" Score="6" Body="&lt;p&gt;Galaxies are moving away from us in every direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were like your race, galaxies at our same distance from the central point will be slowly approximating. That is, we will note that galaxies nearer to the center than us will move away from us (since they are faster), galaxies in the opposite side of the sky will move away from us (since they are slower), and galaxies in the plane that lies between these two point will be slowly approximating to us (since we converge with them to the same point at the same speed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is no, that is not a possibility.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-02-04T15:19:14.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1627" PostTypeId="2" ParentId="1623" CreationDate="2014-02-04T15:22:28.497" Score="1" Body="&lt;p&gt;You need to consider which material the planets are formed of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Terrestrial planets are formed from less material (way less), and with less momentum (since they were orbiting near from the Sun). Gaseous planets (which is another name for external planets, due to their material), also known as gaseous giants, are formed with a lot more of material with a lot more of angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Angular momentum formula is $\vec L=\vec r\times m\vec v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbital speed formula is $\vec v=\vec\omega\times\vec r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus: $\vec L=\vec r\times m\vec\omega\times\vec r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since all vectors are perpendicular:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L=r^2 m \omega$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus a higher mass $m$ makes a higher momentum $L$, and a higher distance to the Sun $r$ makes a quadratically higher momentum $L$.&lt;/p&gt;&#xA;" OwnerUserId="473" LastEditorUserId="473" LastEditDate="2014-02-04T21:09:27.750" LastActivityDate="2014-02-04T21:09:27.750" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="1628" PostTypeId="2" ParentId="1611" CreationDate="2014-02-04T15:26:51.607" Score="5" Body="&lt;p&gt;The reason for which water is needed for Life as we know it (and we have not been able to successfuly imagine a different model from that of DNA that has ability to self-replicate, react to the ambient, and evolve), is that water is the &quot;universal solvent&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that most chemical reactions needed for life happen on a liquid environment, more specifically a water-based environment for most of them.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-02-04T15:26:51.607" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1629" PostTypeId="2" ParentId="1620" CreationDate="2014-02-04T15:33:04.980" Score="2" Body="&lt;p&gt;A singularity in the strict sense is not assumed in the beginning of the big bang, but instead a &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck_epoch&quot; rel=&quot;nofollow&quot;&gt;Planck epoch&lt;/a&gt; governed by a yet to confirm &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_gravity&quot; rel=&quot;nofollow&quot;&gt;quantum gravity&lt;/a&gt;. General relativity and quantum theory, are not consistent in this phase. This means, that space and time, if they existed at all in some sense, didn't follow the laws of physics we inferred from experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This epoch is thought to have been followed by a &lt;a href=&quot;http://en.wikipedia.org/wiki/Grand_unification_epoch&quot; rel=&quot;nofollow&quot;&gt;grand unification epoch&lt;/a&gt;, where gravity is already separated from a unified strong, weak and electromagnetic interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then &lt;a href=&quot;http://en.wikipedia.org/wiki/Inflation_%28cosmology%29&quot; rel=&quot;nofollow&quot;&gt;inflation&lt;/a&gt; is thought to have followed, &quot;driven by a negative-pressure vacuum energy density&quot;. That's when spacetime expansion is thought to have accelerated rapidly, weak and electromagnetic interaction still unified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this is thought to have happened within the first $10^{-32}$ seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the physics in those early phases is not (yet) accessible to experiments. A few &lt;a href=&quot;http://www.mpi-hd.mpg.de/hfm/CosmicRay/CosmicRaySites.html&quot; rel=&quot;nofollow&quot;&gt;cosmic ray experiments&lt;/a&gt; may try to look into the energy ranges, thought to have been present in the early epochs of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Chronology_of_the_universe&quot; rel=&quot;nofollow&quot;&gt;More on the chronology of the universe&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-02-04T16:19:35.833" LastActivityDate="2014-02-04T16:19:35.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1630" PostTypeId="2" ParentId="1621" CreationDate="2014-02-04T16:01:02.887" Score="6" Body="&lt;p&gt;&quot;At high altitudes, Neptune's atmosphere is 80% hydrogen and 19% helium&quot; (&lt;a href=&quot;http://en.wikipedia.org/wiki/Neptune&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;). No significant abundancy of free oxygen to react with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A source of oxygen could easily made burn on Neptune, like a source of hydrogen on Earth.&#xA;Or take a sample of Neptun's atmosphere. It would easily burn in Earth's atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Oxyhydrogen&quot; rel=&quot;nofollow&quot;&gt;Hydrogen oxygen combustion&lt;/a&gt; is sufficiently &lt;a href=&quot;http://en.wikipedia.org/wiki/Exothermic_reaction&quot; rel=&quot;nofollow&quot;&gt;exothermic&lt;/a&gt; (483.6 kJ/mol of O2) to sustain burning even at the low temperatures on Neptune, provided the atmospheric pressure is high enough: With a &lt;a href=&quot;http://en.wikipedia.org/wiki/Gas_constant&quot; rel=&quot;nofollow&quot;&gt;gas constant&lt;/a&gt; of about 8.3 J / mol K, the gas can easily be heated up to the point of combustion by the energy released by the combustion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming oxygen would be provided, just sufficiently to burn up hydrogen and trace hydrocarbons:&#xA;After the explosion and condensation Neptune would get a huge ocean of water, later freezing to water ice, with an atmosphere of helium, and some less abundant gases like CO2, nitrogen, and argon, traces of water vapor. Traces of hydrogen and methane could survive the presence of oxygen. Hence some oxygen would remain in the atmosphere. Most of the CO2 would be resolved in the water ice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average density of Neptune would increase, it would shrink a bit, the core would be compressed and heated by &lt;a href=&quot;http://en.wikipedia.org/wiki/Adiabatic_process&quot; rel=&quot;nofollow&quot;&gt;adiabatic compression&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before combustion Neptune's gravity is strong enough to hold an atmosphere of hydrogen, the easiest gas to escape. Hence the denser planet after combustion wouldn't improve its capability to keep an atmosphere significantly: The &lt;a href=&quot;http://nssdc.gsfc.nasa.gov/planetary/factsheet/neptunefact.html&quot; rel=&quot;nofollow&quot;&gt;escape velocity of Neptune&lt;/a&gt; is 23.5 km/s, its surface &lt;a href=&quot;http://en.wikipedia.org/wiki/Temperature&quot; rel=&quot;nofollow&quot;&gt;temperature&lt;/a&gt; is below 100K. With &lt;a href=&quot;http://en.wikipedia.org/wiki/Boltzmann%27s_constant&quot; rel=&quot;nofollow&quot;&gt;Boltzmann constant&lt;/a&gt; $k=1.38\cdot 10^{-23}J/K$ and $E=3kT/2$ we get $$E=3\cdot 1.38\cdot 10^{-23}J/K\mbox{ or }4.14\cdot 10^{-21}J\mbox{ at }100K$$ for an average $H_2$ molecule with a mass of about $3.35\cdot 10^{-27}kg$. Hence for Neptun's escape velocity the kinetic energy of a hydrogen molecule is $$E=0.5\ mv^2=0.5\cdot 3.35\cdot 10^{-27}kg \cdot (23.5\cdot 10^3 m/s)^2 = 9.25\cdot 10^{-19} J.$$ Hence the energy needed to escape is more than 200-times the kinetic energy of the hydrogen molecules; no way to escape, 2000K or more would be needed to let escape some hydrogen over time (to get below a factor of about ten between the two energies). Photolyic dissociated atomic hydrogen would still need more than 1000K to escape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability to catch asteroids or comets would be reduced due to the reduced radius, if the total mass is adjusted to the mass before combustion, since asteroids would miss instead of hit the planet.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-02-04T22:41:39.757" LastActivityDate="2014-02-04T22:41:39.757" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="1632" PostTypeId="1" AcceptedAnswerId="1673" CreationDate="2014-02-05T03:44:10.177" Score="6" ViewCount="252" Body="&lt;p&gt;Obviously, bodies with low gravity force orbit around another one with higher gravity force, if they are close enough to each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true also for objects like black holes? Are they even able to be moved by another force?&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="62" LastEditDate="2014-02-05T10:16:07.933" LastActivityDate="2014-02-08T08:27:47.433" Title="Are black holes orbiting around each other?" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;supermassive-black-hole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1637" PostTypeId="2" ParentId="1624" CreationDate="2014-02-05T13:43:24.953" Score="1" Body="&lt;p&gt;As has been pointed out the expansion of the galaxies is seen in all directions. But there is other evidence to support theory of cosmic expansion - such as the red shift in the cosmic microwave background (CMB). The CMB is the radiation left over from a specific event - the moment when the primeval fireball's temperature dropped sufficiently to allow the stable recombination of electrons and protons into hydrogen atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That happened when the universe's temperature (and remember this was a relatively smooth plasma so this happened more or less everywhere at the same time) fell to about 4000K&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the CMB we see today has been so redshifted that the CMB looks like the signature of an event that took place at a temperature of about 3K - the reason is that the universe has expanded by so much (in all directions) that the radiation from the CMB has been &quot;stretched&quot; into these much longer wavelengths on its journey to us.&lt;/p&gt;&#xA;" OwnerUserId="823" LastEditorUserId="823" LastEditDate="2014-02-05T14:21:31.580" LastActivityDate="2014-02-05T14:21:31.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1638" PostTypeId="1" AcceptedAnswerId="1639" CreationDate="2014-02-05T15:16:09.867" Score="12" ViewCount="938" Body="&lt;p&gt;Complete astronomy noob over here who would be happy if he get a simple answer (and who is also aware that this may be not possible)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've learned from a tv documentary that the stars at the edge of the galaxy are not traveling more slowely than the ones closer to the center. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this also mean that all stars in a galaxy do not change their positions relatively to each other? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it simply: If I note the relative position of a star compared to our sun, and then do it again 10 (100 / 1000) years later: Are the coordinates of that star identical or will the star be in a different relative position?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would assume that the star is in a completely different (relative) place given the speed star systems are travelling because the galaxy is rotating AND the &quot;fact&quot; (read: I don't know if this is really a fact) that the solar system is oscillating through the galactic plane. &lt;/p&gt;&#xA;" OwnerUserId="956" LastEditorUserId="956" LastEditDate="2014-02-05T17:59:59.763" LastActivityDate="2014-02-05T17:59:59.763" Title="Do stars of a galaxy change their positions relatively to each other?" Tags="&lt;star&gt;&lt;orbit&gt;&lt;galaxy&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1639" PostTypeId="2" ParentId="1638" CreationDate="2014-02-05T16:30:13.530" Score="5" Body="&lt;p&gt;Stars do in fact move relative to one another within galaxies of all types. The orbital period of stars in a typical spiral galaxy (at around the same distance as the Sun is from the center of the Milky Way) is on the order of hundreds of millions of years. For the Sun it's something like 230 million years (&lt;a href=&quot;http://hypertextbook.com/facts/2002/StacyLeong.shtml&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;). A particularly old person (~100 years old), will have been around long enough to see something like one-tenthousandth of 1% of the orbit of stars in this region of the galaxy! Most of these stars simply do not move much relative to us in that amount of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exception to this view of the motions of the stars is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sagittarius_A*&quot; rel=&quot;noreferrer&quot;&gt;Sagittarius A*&lt;/a&gt; (which is not a star at all, but is in fact a supermassive black hole at the center of the Milky Way). The period of stars traveling around it are on the order of 10's of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cfA3v.png&quot; alt=&quot;SagA*&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, to give you a rough idea of how galaxies look dynamically, here's a link to a particularly nice &lt;a href=&quot;http://viz.adrian.pw/galaxy/&quot; rel=&quot;noreferrer&quot;&gt;galaxy simulator&lt;/a&gt; implemented in the browser (created by Adrian Price-Whelan).&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2014-02-05T16:30:13.530" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1640" PostTypeId="2" ParentId="351" CreationDate="2014-02-05T17:24:40.403" Score="3" Body="&lt;p&gt;Do you know this paper? &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0507416&quot; rel=&quot;nofollow&quot;&gt;Mamajek (2005). &quot;A Moving Cluster Distance to the Exoplanet 2M1207b in the TW Hydrae Association&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll risk an opinion:&#xA;There is about a thousand clusters with kinematic measurements, but besides no more than half a dozen, all pm components are under 20 mas (milli-arcseconds) with errors more or less in the range 0.3-6 mas. Such errors are too high in relation to the pm values and as a result, the vectors will not converge.&lt;/p&gt;&#xA;" OwnerUserId="910" LastActivityDate="2014-02-05T17:24:40.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1641" PostTypeId="2" ParentId="1622" CreationDate="2014-02-05T20:30:40.467" Score="3" Body="&lt;p&gt;In &lt;a href=&quot;http://31.media.tumblr.com/1bb65c125d6ceda03e76b5b787dd82ef/tumblr_mrqi7hcZ0b1qmvxavo2_1280.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;the image you are now refering to&lt;/a&gt; the only significant brightness difference is between the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lunar_mare&quot; rel=&quot;nofollow noreferrer&quot;&gt;mare&lt;/a&gt; and the &quot;highlands&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Retina&quot; rel=&quot;nofollow noreferrer&quot;&gt;retina&lt;/a&gt; &lt;a href=&quot;http://www.colorado.edu/physics/phys1230/phys1230_fa08/Illusions.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;enhances edges&lt;/a&gt;. So you'll see edges enhanced, even if they aren't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &lt;a href=&quot;http://en.wikipedia.org/wiki/Edge_enhancement&quot; rel=&quot;nofollow noreferrer&quot;&gt;image processing&lt;/a&gt; edges can also be enhanced to make contours clearer. This is not the case for the referred image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To better assess, which part of the perception is real, and which is illusion, here the illusion part:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/WPhal.jpg&quot; alt=&quot;Moon illusion part&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-02-06T05:31:32.913" LastActivityDate="2014-02-06T05:31:32.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1642" PostTypeId="2" ParentId="1622" CreationDate="2014-02-05T23:08:37.073" Score="0" Body="&lt;p&gt;Taking the two answers, the comments, and the lack of a reason why it would actually happen I'll attribute my misperception to the cognitive illusion illustrated by Ernst Mach's &lt;a href=&quot;http://en.wikipedia.org/wiki/Mach_bands&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mach Bands&lt;/a&gt;. See also the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Lateral_inhibition&quot; rel=&quot;nofollow noreferrer&quot;&gt;lateral inhibition&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a shame... I was hoping for some physics effect for bending light around the earth into the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cKLHs.png&quot; alt=&quot;Mach bands illusion&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="948" LastActivityDate="2014-02-05T23:08:37.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1643" PostTypeId="1" AcceptedAnswerId="1658" CreationDate="2014-02-06T02:57:49.503" Score="4" ViewCount="208" Body="&lt;p&gt;I was watching a Nottingham Science video &lt;a href=&quot;http://www.youtube.com/watch?v=4ESW_NTIhBM&quot; rel=&quot;nofollow&quot;&gt;linked here (periodic videos)&lt;/a&gt; and they mention modern reflectors are made of aluminum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can understand why mirrors are preferred over lenses, but what inhibits the aluminum oxide coating that would naturally form on the surface of the mirror? Why do they work so well?&lt;/p&gt;&#xA;" OwnerUserId="960" LastActivityDate="2014-02-07T03:30:47.340" Title="Is there an oxide layer on aluminum reflectors for telescopes?" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1645" PostTypeId="1" CreationDate="2014-02-06T09:23:11.510" Score="0" ViewCount="127" Body="&lt;p&gt;Here are some &lt;a href=&quot;http://www.google.com.au/search?client=safari&amp;amp;rls=en&amp;amp;q=Solar%20storm%202014&amp;amp;ie=UTF-8&amp;amp;oe=UTF-8&amp;amp;gfe_rd=ctrl&amp;amp;ei=N1PzUqHkDayN8Qev1ID4Cw&amp;amp;gws_rd=cr&quot; rel=&quot;nofollow&quot;&gt;Articles&lt;/a&gt; I've found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true? If it is, How big could it be?, what might be possible or will be possible?, what will happen?&lt;/p&gt;&#xA;" OwnerUserId="503" LastEditorUserId="537" LastEditDate="2014-02-06T13:13:46.807" LastActivityDate="2014-02-06T13:13:46.807" Title="Is there going to be a really bad Solar Storm in 2014?" Tags="&lt;solar-storm&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1646" PostTypeId="1" CreationDate="2014-02-06T09:29:48.873" Score="6" ViewCount="724" Body="&lt;p&gt;I own a Nikon OceanPro 7x50 pair of binoculars, but I find that when I look at the moon, I would like more magnification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My requirements are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;costs less than $400&lt;/li&gt;&#xA;&lt;li&gt;weighs less than 10kg&lt;/li&gt;&#xA;&lt;li&gt;the length is less than 1-2 feet&lt;/li&gt;&#xA;&lt;li&gt;much higher maginification than 7x (I already have a pair of 7x binoculars, as I said)&lt;/li&gt;&#xA;&lt;li&gt;at least an 100mm objective I (I already have a device with a 50mm objective)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What kind of device should I go for? A spotting scope? Binoculars? A telescope of some kind?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally, I'd go for binoculars, so that I can see with both eyes, but I'm told binoculars don't come with objectives larger than 100mm, for my budget.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could buy a spotting scope, like &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B0002862UW&quot;&gt;this&lt;/a&gt;, but Wikipedia says that spotting scopes are not designed for astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I live in an apartment, so I can't use a big, heavy telescope, like a 4-feet 20kg thing. I don't have room in my apartment for such a big telescope, and I can't carry it over the threshold into the balcony.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what kind of device should I consider, and what are the tradeoffs? What kind of astronomical objects will I be able to see?&lt;/p&gt;&#xA;" OwnerUserId="962" LastEditorUserId="67" LastEditDate="2014-02-17T05:37:27.070" LastActivityDate="2016-12-06T06:55:29.107" Title="What's a good step up from 7x50 binoculars?" Tags="&lt;telescope&gt;&lt;binoculars&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1647" PostTypeId="2" ParentId="1645" CreationDate="2014-02-06T09:32:50.873" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it true?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solar storms are like Earth storms: unpredictable until they are quite near.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What will happen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If there is a massive solar storm, we will experience power outages in long lines (e.g. U.S.A. mainland, Russia), communications disruption on satellite links (and again in long copper lines), and very nice auroras.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-02-06T09:32:50.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1648" PostTypeId="2" ParentId="1225" CreationDate="2014-02-06T12:10:09.937" Score="13" Body="&lt;p&gt;In addition to @MBR's answer; there are other (less confusing) ways to represent the data. &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?db_key=AST&amp;amp;bibcode=1958MNRAS.118..379O&amp;amp;letter=.&amp;amp;classic=YES&amp;amp;defaultprint=YES&amp;amp;whole_paper=YES&amp;amp;page=379&amp;amp;epage=379&amp;amp;send=Send%20PDF&amp;amp;filetype=.pdf&quot; rel=&quot;noreferrer&quot;&gt;This paper&lt;/a&gt; describes the method used; basically they plot the position and motion of &lt;a href=&quot;http://www.natureworldnews.com/articles/5334/20131217/study-massive-stars-confirms-milky-way-four-spiral-arms.htm&quot; rel=&quot;noreferrer&quot;&gt;the stars&lt;/a&gt;, or of &lt;a href=&quot;http://www.atlasoftheuniverse.com/milkyway.html&quot; rel=&quot;noreferrer&quot;&gt;hydrogen clouds&lt;/a&gt;. The hydrogen distribution of our galaxy looks like this:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/EzvT1.gif&quot; alt=&quot;hydrogen map of the Galaxy&quot;&gt;&lt;br&gt;&#xA;Now there's a lot of noise in these measurements, but generally you can see that the hydrogen clumps together in spirals. &lt;/p&gt;&#xA;" OwnerUserId="361" LastActivityDate="2014-02-06T12:10:09.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1649" PostTypeId="1" CreationDate="2014-02-06T13:59:45.310" Score="7" ViewCount="381" Body="&lt;p&gt;How can I adjust the axis labels in ds9?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to have something like:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/P3txS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I only managed to get floating points (tl h%f3):&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/EYWEj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or the leading hours are missing (tl h%f1 m):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EsPkK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Edit:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I tried &quot;tlhm.1&quot; as kindly suggested by &lt;a href=&quot;https://astronomy.stackexchange.com/a/1958/950&quot;&gt;Jeremy&lt;/a&gt;, but unfortunatly it still does not work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jl7Vk.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="950" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-04-04T16:20:03.070" Title="Adjust axis labels in ds9" Tags="&lt;data-analysis&gt;&lt;ds9&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1650" PostTypeId="1" CreationDate="2014-02-06T17:12:36.963" Score="5" ViewCount="243" Body="&lt;p&gt;I'm not in the industry myself, but as an interested member of the public the terminology of astronomy seems a bit funny. Astronomers who today talk publicly about the interstellar medium say that the solar system is in the &quot;local fluff&quot; inside the &quot;local bubble&quot; near a &quot;local chimney&quot;. Cosmology uses a term like &quot;the dark ages&quot; to describe an era in the history of the phenomena they study. But geologists instead prefer to use terms like &quot;cenozoic&quot; to name a period. Early on, astronomers also used pompous words and names from antique languages. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Was &quot;Big Bang&quot; (coined in 1949, I think) the term which started off this fluffy bubble blob bang type of terminology? Or were there precursors?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there a connection to sci-fi literature?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Are the &quot;blobby bang fluff&quot; words just a mass media (&quot;reach out&quot;) construction which professional astronomers don't use at all even as quick references in conversations with colleagues?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1030" OwnerDisplayName="LocalFluff" LastEditorUserId="6" LastEditDate="2014-02-06T17:18:48.827" LastActivityDate="2014-02-17T01:56:50.270" Title="What's the origin and culture of funny astronomical terminology?" Tags="&lt;terminology&gt;&lt;history&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1651" PostTypeId="1" AcceptedAnswerId="1728" CreationDate="2014-02-06T17:30:24.577" Score="22" ViewCount="10096" Body="&lt;p&gt;Is there any evidence that the Milky Way could have been a quasar in it's early history? Is it thought that most galaxies come from quasars? &lt;/p&gt;&#xA;" OwnerUserId="966" LastActivityDate="2019-12-07T16:15:59.317" Title="Was the Milky Way ever a quasar?" Tags="&lt;galaxy&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1652" PostTypeId="2" ParentId="1622" CreationDate="2014-02-06T17:34:26.687" Score="3" Body="&lt;p&gt;It appears that way because there is more land reflecting sun light to your retina at the edges. This is because the Moon is a sphere. As you move your eye from the center of the moon towards the edges, you are scanning over increasingly more amounts of reflective rock. Hence, the edges appear very bright relative to the center. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's all about surface area.&lt;/p&gt;&#xA;" OwnerUserId="966" LastEditorUserId="966" LastEditDate="2014-02-06T17:55:49.707" LastActivityDate="2014-02-06T17:55:49.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1653" PostTypeId="2" ParentId="1651" CreationDate="2014-02-06T17:48:52.207" Score="6" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Quasar&quot; rel=&quot;nofollow&quot;&gt;Quasars&lt;/a&gt; are thought to be &lt;a href=&quot;http://en.wikipedia.org/wiki/Active_galactic_nucleus&quot; rel=&quot;nofollow&quot;&gt;active galactic nuclei&lt;/a&gt;. The Milky Way contains a very dense region in its center, probably a &lt;a href=&quot;http://en.wikipedia.org/wiki/Supermassive_black_hole&quot; rel=&quot;nofollow&quot;&gt;supermassive black hole&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a dense region or black hole is thought to form by &lt;a href=&quot;http://en.wikipedia.org/wiki/Accretion_%28astrophysics%29&quot; rel=&quot;nofollow&quot;&gt;accretion&lt;/a&gt; of dust, gas and stars. This accretion process of a supermassive black hole releases huge amounts of energy emitted perpendicular to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Accretion_disc&quot; rel=&quot;nofollow&quot;&gt;accretion disk&lt;/a&gt;. If such a &lt;a href=&quot;http://en.wikipedia.org/wiki/Polar_jet&quot; rel=&quot;nofollow&quot;&gt;jet&lt;/a&gt; at the center of a distant galaxy points towards Earth we may see it as quasar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the answer is most likely yes, the Milky Way or some of its predecessor galaxies will probably have had quasars at their centers, at some period when consuming lot of material, and seen from appropriate direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most galaxies are thought to contain a supermassive black hole at their center. Therefore the same applies to most galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here two nice simulations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=FMPuujaldZk&quot; rel=&quot;nofollow&quot;&gt;A simulation of a merger of two galaxies forming active galactic nuclei&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=-feWoMSTiDY&quot; rel=&quot;nofollow&quot;&gt;Computer simulation of the formation of a disk galaxy&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nasa.gov/mission_pages/chandra/multimedia/photo10-009.html&quot; rel=&quot;nofollow&quot;&gt;A rare ongoing merger of two quasars observed by Chandra&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-02-07T01:15:00.530" LastActivityDate="2014-02-07T01:15:00.530" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1654" PostTypeId="1" CreationDate="2014-02-06T21:22:42.203" Score="1" ViewCount="499" Body="&lt;p&gt;The Hubble law states that the recessional velocity of a galaxy is proportional to its distance from us.&lt;br&gt;&#xA;It takes redshift and angular sizes (for distance) of the galaxy to determine velocity, and thus rate of expantion.&lt;br&gt;&#xA;But I guess the factor of direction in which the galaxy is moving may substantially affect the 'expantion rate' calculations. i.e. If the galaxy A is moving in same direction as ours, will have less relative speed and the galaxy B moving in opposite direction will have more relative speed. The actual local speed may be the same.&lt;br&gt;&#xA;I wonder whether this fact can be considered when we determine the expansion rate?&lt;/p&gt;&#xA;" OwnerUserId="943" LastEditorUserId="943" LastEditDate="2014-02-06T23:17:12.363" LastActivityDate="2014-02-07T10:15:52.167" Title="Can we know whether any distant galaxy is moving away in same direction or in opposite direction w.r.t. our galaxy?" Tags="&lt;galaxy&gt;&lt;universe&gt;&lt;milky-way&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1655" PostTypeId="2" ParentId="1654" CreationDate="2014-02-06T22:41:06.073" Score="1" Body="&lt;p&gt;Yes, it can. For instance the Andromeda Galaxy is moving towards us, due to local gravitational attraction but that does not affect the validity of the theory of the expansion of spacetime because one result based on local factors is perfectly compatible with a cosmological theory which, by definition, includes every galaxy.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2014-02-06T22:41:06.073" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1656" PostTypeId="2" ParentId="1654" CreationDate="2014-02-06T22:41:29.143" Score="1" Body="&lt;p&gt;For galaxies it's difficult to determine their &lt;a href=&quot;http://en.wikipedia.org/wiki/Proper_motion&quot; rel=&quot;nofollow&quot;&gt;proper motions&lt;/a&gt;, but possible with the &lt;a href=&quot;http://www3.mpifr-bonn.mpg.de/staff/abrunthaler/proper.shtml&quot; rel=&quot;nofollow&quot;&gt;Very Long Baseline Array&lt;/a&gt; for nearby galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more distant galaxies (more than several megaparsecs), relevant for the determination of the Hubble constant, no direct measurements are available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An overall parallel motion relative to the majority of distant galaxies would be detected via redshift as radial motion when looking in any direction other than perpendicular to the motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proper motion of the majority of distant galaxies circling us as observer without or with the same radial velocity would lead over time to at least one compactification and one thinning, according to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hairy_ball_theorem&quot; rel=&quot;nofollow&quot;&gt;hairy ball theorem&lt;/a&gt; (&quot;Any smooth vector field on a sphere has a singular point.&quot;) This would be possible to be observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Large-scale isotropy and homogeneity is in parts conjectural, but e.g. supported by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow&quot;&gt;cosmic microwave background&lt;/a&gt;, although not quite undisputed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmological_principle&quot; rel=&quot;nofollow&quot;&gt;More about the cosmological principle&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-02-07T10:15:52.167" LastActivityDate="2014-02-07T10:15:52.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1657" PostTypeId="1" AcceptedAnswerId="1659" CreationDate="2014-02-07T01:53:26.377" Score="4" ViewCount="2509" Body="&lt;p&gt;Which planet is also the name of an element?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came across this question on cosmos quest and need and answer&lt;/p&gt;&#xA;" OwnerUserId="969" LastActivityDate="2014-02-09T11:16:27.403" Title="Which planet is also the name of an element?" Tags="&lt;planet&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" ClosedDate="2014-02-07T12:48:25.503" ContentLicense="CC BY-SA 3.0" />
  <row Id="1658" PostTypeId="2" ParentId="1643" CreationDate="2014-02-07T03:30:47.340" Score="2" Body="&lt;p&gt;The aluminum oxide layer forms as soon as oxygen comes in contact with the mirror surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aluminum oxide is tough, firmly attached and transparent;  it serves to protect the aluminum surface.  Rubies and sapphires are slightly impure aluminum oxide&lt;/p&gt;&#xA;" OwnerUserId="420" LastActivityDate="2014-02-07T03:30:47.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1659" PostTypeId="2" ParentId="1657" CreationDate="2014-02-07T05:14:21.073" Score="11" Body="&lt;p&gt;The answer is &lt;strong&gt;Mercury&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Plutonium&lt;/strong&gt; was made after &lt;strong&gt;Pluto&lt;/strong&gt; was discovered.&lt;/p&gt;&#xA;" OwnerUserId="863" LastActivityDate="2014-02-07T05:14:21.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1660" PostTypeId="2" ParentId="1657" CreationDate="2014-02-07T05:19:06.113" Score="2" Body="&lt;pre&gt;&lt;code&gt;Mercury -  Mercury(hg-80)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Mercury&lt;/code&gt; will be the Answer, Because &lt;code&gt;Mercury&lt;/code&gt; is Very clear name with the 80th Element so planet is(However, Mercury was named after the &lt;code&gt;Roman God&lt;/code&gt; which was related to what they called &lt;code&gt;quicksilver&lt;/code&gt;).&lt;/p&gt;&#xA;" OwnerUserId="903" LastEditorUserId="903" LastEditDate="2014-02-09T11:16:27.403" LastActivityDate="2014-02-09T11:16:27.403" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1661" PostTypeId="1" AcceptedAnswerId="1666" CreationDate="2014-02-07T09:08:43.483" Score="3" ViewCount="257" Body="&lt;p&gt;How does one measure the brightness of a star?, I did not find anything on the internet.&lt;/p&gt;&#xA;" OwnerUserId="756" LastEditorUserId="537" LastEditDate="2014-02-07T14:53:07.943" LastActivityDate="2014-02-07T20:41:03.647" Title="How is the magnitude of a star measured?" Tags="&lt;star&gt;&lt;light&gt;&lt;apparent-magnitude&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1662" PostTypeId="2" ParentId="1225" CreationDate="2014-02-07T09:12:03.950" Score="4" Body="&lt;p&gt;The clues we have to the shape of the Milky Way are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) When you look toward the galactic center with your eye, you see a long, thin strip. This suggests a disk seen edge-on, rather than a ellipsoid or another shape. We can also detect the bulge at the center. Since we see spiral galaxies which are disks with central bulges, this is a bit of a tipoff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) When we measure velocities of stars and gas in our galaxy, we see an overall rotational motion greater than random motions. This is another characteristic of a spiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The gas fraction, color, and dust content of our galaxy are spiral-like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, overall, it's a pretty convincing argument. Of course, we have to assume our galaxy is not completely unlike the other galaxies we see--but I suppose once a civilization has accepted that it does not occupy any special place in the universe, arguments about similarity seem sensible.&lt;/p&gt;&#xA;" OwnerUserId="756" LastActivityDate="2014-02-07T09:12:03.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1663" PostTypeId="2" ParentId="1657" CreationDate="2014-02-07T10:19:08.083" Score="5" Body="&lt;p&gt;As per ReNiSh A R's post, the answer is Mercury.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick Google search took me here:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Timeline_of_chemical_element_discoveries&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Timeline_of_chemical_element_discoveries&lt;/a&gt; &amp;lt;= second result!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you control-F search the page, you'll find the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury - discovered before 2000BC, so the planet will have been named after the element&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plutonium - About 1940/41, meaning it would have been named after Pluto (before it was declassified as a planet)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uranus - Uranium discovered around 1789&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neptune - Neptunium discovered 1940&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, the only one of those having exactly the same name for both the planet and the element is Mercury.&lt;/p&gt;&#xA;" OwnerUserId="972" LastActivityDate="2014-02-07T10:19:08.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1664" PostTypeId="1" AcceptedAnswerId="1766" CreationDate="2014-02-07T13:30:28.917" Score="6" ViewCount="98" Body="&lt;p&gt;I hear conflicting claims about our understanding of the structure formation in our universe. I heard that &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambda-CDM_model&quot;&gt;ΛCDM&lt;/a&gt; explaines well the large scale structure of the universe i.e.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Comparison of the model with observations is very successful on large&#xA;  scales (larger than galaxies, up to the observable horizon)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and then I read the opposite on &lt;a href=&quot;http://imagine.gsfc.nasa.gov/docs/science/mysteries_l1/structures.html&quot;&gt;nasa.gov&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, we still do not understand how these structures condensed&#xA;  out of the 'primordial soup'. The formation of structure appears much&#xA;  quicker than the current theories would have it!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So what is Lambda-CDM actually good at and what are its potential weaknesses, if there is any?&lt;/p&gt;&#xA;" OwnerUserId="959" LastActivityDate="2014-02-17T06:42:35.227" Title="To what extent are structure formations unexplained?" Tags="&lt;cosmology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1665" PostTypeId="1" AcceptedAnswerId="1672" CreationDate="2014-02-07T14:35:23.690" Score="10" ViewCount="3384" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Andromeda_Galaxy#Future_collision_with_the_Milky_Way&quot; rel=&quot;noreferrer&quot;&gt;Andromeda Galaxy&lt;/a&gt; is approaching the Milky Way at about 684000 mi/hours, making it one of the few &lt;a href=&quot;http://en.wikipedia.org/wiki/Blueshift&quot; rel=&quot;noreferrer&quot;&gt;blueshifted&lt;/a&gt; galaxies. The Andromeda Galaxy and the Milky Way are thus expected to collide in about 3.75 or 4.5 billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are some galaxies moving away and why are our galaxy and Andromeda coming towards each other?&lt;/p&gt;&#xA;" OwnerUserId="760" LastEditorUserId="13426" LastEditDate="2017-12-14T06:45:42.947" LastActivityDate="2017-12-14T06:45:42.947" Title="If galaxies are moving away from each other then why are the Milky Way and Andromeda galaxy coming towards each other?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;formation&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1666" PostTypeId="2" ParentId="1661" CreationDate="2014-02-07T20:16:11.967" Score="5" Body="&lt;p&gt;A device called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Photomultiplier&quot; rel=&quot;nofollow&quot;&gt;photomultiplier&lt;/a&gt; is attached to a telescope. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The telescope is pointed at the target star and an aperture is&#xA;closed down to avoid as much spurious light as possible, and a&#xA;reading is taken. The photomultiplier literally count the photons&#xA;hitting it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Next the telescope is pointed at a well-known reference star in the&#xA;same part of the sky, the aperture is closed, and a reading is&#xA;taken.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, the telescope is pointed at a dark part of the sky, the&#xA;aperture is again closed, and a reading taken.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Computer software compares the data from the target star, the reference star, and the dark sky to remove as much noise from light scattered by the atmosphere as possible. It's important to know how much of the measured light is coming from the sky and subtract it from the light measured from the target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often, this same procedure is repeated with a series of filters to get the luminosity of the target star at many wavelengths. Searching for a variable star's period requires switching from target star to reference star to dark sky time after time, hour after hour, night after night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I paid my way through college doing this.&lt;/p&gt;&#xA;" OwnerUserId="562" LastEditorUserId="562" LastEditDate="2014-02-07T20:41:03.647" LastActivityDate="2014-02-07T20:41:03.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1667" PostTypeId="2" ParentId="1611" CreationDate="2014-02-07T20:57:52.347" Score="5" Body="&lt;p&gt;&quot;Life&quot; seems to be an &lt;a href=&quot;http://en.wikipedia.org/wiki/Emergence&quot; rel=&quot;nofollow&quot;&gt;emergent phenomenon&lt;/a&gt;, a complex result of a very large number of simple parts or interactions. The whole is more than the sum of it's parts, unpredicted by the properties of the individual parts. If those individual parts are atoms and molecules, they have to move to be able to interact, to build larger, more complex structures. To move, they need to be suspended in something that permits movement and allows chemical reactions. On Earth, that would be liquid water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other liquids might work, but they'd need to allow for the movement and reactions life requires. It's hard to imagine something other than a liquid doing the job.&lt;/p&gt;&#xA;" OwnerUserId="562" LastEditorUserId="562" LastEditDate="2014-02-07T23:11:44.377" LastActivityDate="2014-02-07T23:11:44.377" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1668" PostTypeId="1" CreationDate="2014-02-07T22:22:19.550" Score="5" ViewCount="539" Body="&lt;p&gt;The usual explanation is that escape velocity of a black hole is above speed of light, therefore nothing can escape because nothing can move faster than light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is however a fallacy, because escape velocity only pertains to bodies &quot;without propulsion&quot;, or in other words without any force other than gravity acting on them. As an example I could theoretically escape Earth's gravity at a walking speed, just give me a staircase and I could walk all the way to the Moon. No need to accelerate to 11 km/s at any point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this affect black holes? Why cannot I have a staircase and climb out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the staircase idea must be wrong, but I would like to know why. My guess is that I am applying Newtonian mechanics to a problem which requires general relativity approach, because Newtonian mechanics approximation fails badly around such massive gravity fields?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, according to Newtonian mechanics one can easily escape a black hole, even if we assume that speed of light cannot be exceeded.&lt;/p&gt;&#xA;" OwnerUserId="978" LastActivityDate="2014-02-07T23:46:59.257" Title="Why one cannot escape from a black hole" Tags="&lt;black-hole&gt;" AnswerCount="2" CommentCount="0" ClosedDate="2014-06-18T19:51:53.423" ContentLicense="CC BY-SA 3.0" />
  <row Id="1669" PostTypeId="2" ParentId="1668" CreationDate="2014-02-07T23:19:07.750" Score="3" Body="&lt;p&gt;A great explanation; all credit goes to: &lt;a href=&quot;https://groups.google.com/forum/#!original/sci.physics.research/PZh0yawL2nA/YKbv-EpWJPsJ&quot; rel=&quot;nofollow&quot;&gt;https://groups.google.com/forum/#!original/sci.physics.research/PZh0yawL2nA/YKbv-EpWJPsJ&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The explaination of a black hole in terms of escape velocity is not a&#xA;  good way to think of a black hole. The escape velocity is merely the&#xA;  intial speed at which a projectile must be fired from a large body in&#xA;  order so that, given that only gravitational forces act on it (which&#xA;  is a scenario very far from reality) it will never return to that&#xA;  body.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;However, anything that passes through the event horizon of a black&#xA;  hole will never be on the other side. Even if it were to constantly&#xA;  accelerate away from the black hole, it would still be going inward.&#xA;  This is not true of Newtonian gravity. In fact, in Newtonian gravity,&#xA;  there are no black holes, because there is no universal speed limit.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The best way to think of an event horizon is where spacetime is so&#xA;  bent that all paths 'curve inwards', IE, the lightcones all point&#xA;  inward, so there is no chance of escape.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;BTW, in GR, it's not good to explain the effects of gravity as a&#xA;  FORCE. It is as much a force as the Centrifugal Force or the Coriolis&#xA;  Force. There is an amount of force needed to stay the same distance&#xA;  from an object, IE, to overcome gravity, however. And as you approach&#xA;  the black hole, the amount of force required approaches infinity. But&#xA;  to say that there is an infinite amount of force PULLING on someone is&#xA;  not very accurate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="818" LastActivityDate="2014-02-07T23:19:07.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1670" PostTypeId="2" ParentId="1668" CreationDate="2014-02-07T23:39:42.063" Score="2" Body="&lt;p&gt;That subject is highly controverse and paradoxical, see e.g. &lt;a href=&quot;http://www.nature.com/news/stephen-hawking-there-are-no-black-holes-1.14583&quot; rel=&quot;nofollow&quot;&gt;Stephen Hawking's recent paper&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_metric&quot; rel=&quot;nofollow&quot;&gt;Schwarzschild black hole&lt;/a&gt; as the simplest case, &lt;a href=&quot;http://jila.colorado.edu/~ajsh/insidebh/waterfall.html&quot; rel=&quot;nofollow&quot;&gt;space is falling into the black hole&lt;/a&gt;. At the event horizon this speed is the speed of light. Since nothing can travel faster than light, you cannot escape. Actually you'll be slower than light, therefore dragged into the black hole together with the space.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-02-07T23:46:59.257" LastActivityDate="2014-02-07T23:46:59.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1671" PostTypeId="2" ParentId="1665" CreationDate="2014-02-08T00:00:49.450" Score="8" Body="&lt;p&gt;The farther away, the faster the galaxies move away from us. But that's only the overall expansion of the universe. Locally velocities can differ and are to be added to the overall expansion. The Andromeda Galaxy is close enough to Milky Way to be able overcome the overall expansion of space. The Andromeda Galaxy and the Milky Way attract each other by gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble%27s_law&quot;&gt;Hubble's law&lt;/a&gt; applied to &lt;a href=&quot;http://en.wikipedia.org/wiki/Andromeda_Galaxy&quot;&gt;Andromeda Galaxy&lt;/a&gt;/Milky Way returns an expansion of space of less than 70 km/s, since Hubble's constant is about 70 km/s/Mpc, and the distance is about 2.5 million light-years, less than 1 Mpc (= 3.26 light-years, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Parsec&quot;&gt;Parsec&lt;/a&gt;). That speed is slower than the about -300 km/s helio-radial velocity (space expansion already subtracted) of the Andromeda galaxy, hence feasible to overcome.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-02-08T00:16:44.390" LastActivityDate="2014-02-08T00:16:44.390" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1672" PostTypeId="2" ParentId="1665" CreationDate="2014-02-08T04:19:14.847" Score="8" Body="&lt;p&gt;Here is my answer to a &lt;a href=&quot;https://physics.stackexchange.com/questions/96679/expanding-universe-and-the-peculiar-velocity/96682#96682&quot;&gt;similar question&lt;/a&gt; posted on the physics stack exchange website.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hubble's law (the law that deals with the expansion of the universe) applies to the expansion of space itself, i.e., if two objects stationary to each other that had no force between them were left alone the distance between would increase with time because space itself is expanding. This is what Hubble's law addresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the Milky Way and Andromeda galaxies (and all galaxies for that matter) there is a force between them: gravity. The gravitational force between the Milky Way and Andromeda galaxies has produced an acceleration that is causing the two galaxies to be moving towards each other faster than the space between them is expanding as calculated by Hubble's law. However, the vast majority of galaxies lie far enough away from the Milky Way that the gravitational force between us and them is small compared to the Hubble expansion and Hubble's law dominates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, Hubble's law applies throughout the universe, but localized systems may have enough gravitational attraction between them that the gravitational effects dominate&lt;/p&gt;&#xA;" OwnerUserId="128" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2014-02-08T04:19:14.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1673" PostTypeId="2" ParentId="1632" CreationDate="2014-02-08T08:02:55.040" Score="4" Body="&lt;p&gt;A black hole (briefly), is nothing more than a dead star whose mass was more than 3 solar masses. When that star's life ended it collapsed under its own gravity and went through a supernova, losing a portion of its original gas in the process, thus losing part of its mass, and thus its total gravitational pull is diminished compared to the original star (contrary to common believe for a black hole having the &quot;ultimate gravity&quot;). So first of all, black holes vary in size and are not all the same (thing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we say the original star was orbiting around another star (A Binary system), (putting other factors aside) the effect on the other star from the black hole's gravitational forces would be &lt;em&gt;less&lt;/em&gt; compared to its original star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However that gravitational pull increases drastically as you go closer to the blackhole, since all the remaining matter is concentrated into a single point (singularity). Gravity becomes too severe, close to a  blackhole that even light can not escape, an area called &quot;the event horizon&quot; of the black hole. But that's too close to the black hole. Nothing is special about its gravitational effect on the further objects as I pointed out with the orbiting star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your statement: &quot;Is it true also for objects like black holes? Are they even able to be moved by another force?&quot; I don't know what you mean to be moved by another force, but black holes do move as any other object in space, they are nothing more than the remainder of their original star, and they inherited its motion relative to the other objects around. If say the original star was rotating around the center of it's galaxy, it will continue to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your statement: &quot;bodies with low gravity force orbit around another one with higher gravity force&quot; is not very accurate, since (bodies with higher gravity force - the more massive object), and  (bodies with low gravity force - lesser mass) both orbit one another. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, as the Moon orbits the Earth, the Earth also orbits the Moon, they both orbit around a point between the two of them called the &quot;Barycenter&quot;. This point is always closer to the more massive object. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the Earth-Moon system it's inside the Earth's volume. Therefore it's common believe that only the Moon's orbiting the Earth, without the other way around as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Topics you must check out besides the link provided in the previous answer:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_mechanics&quot; rel=&quot;nofollow&quot;&gt;Orbital Mechanics&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation&quot; rel=&quot;nofollow&quot;&gt;Newton's Universal Law of Gravitation&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Black_hole&quot; rel=&quot;nofollow&quot;&gt;Black Holes&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="927" LastEditorUserId="927" LastEditDate="2014-02-08T08:27:47.433" LastActivityDate="2014-02-08T08:27:47.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1674" PostTypeId="2" ParentId="1611" CreationDate="2014-02-08T08:35:26.573" Score="0" Body="&lt;p&gt;Water contains oxygen and hydrogen which are suitable for growth of life so yes it is necessary for life...at least now... We may find something else on different parts of universe which is necessary for growth of life.&lt;/p&gt;&#xA;" OwnerUserId="979" LastActivityDate="2014-02-08T08:35:26.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1675" PostTypeId="1" CreationDate="2014-02-08T08:38:33.333" Score="3" ViewCount="478" Body="&lt;p&gt;Was there space before big bang? Or who knows there might be different heavenly bodies that we might not perceive?&lt;/p&gt;&#xA;" OwnerUserId="979" LastActivityDate="2014-02-08T19:02:37.127" Title="What happened before the big bang?" Tags="&lt;space&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2014-02-18T22:40:41.520" ContentLicense="CC BY-SA 3.0" />
  <row Id="1676" PostTypeId="1" AcceptedAnswerId="1678" CreationDate="2014-02-08T10:12:02.823" Score="10" ViewCount="2502" Body="&lt;p&gt;We know neutron stars as a very massive object with extremely strong gravitational forces that composes mostly of neutrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't help but wonder, what would happen if an object fell into a neutron star, what would become of it? Will it turn it into neutrons as well? And will be any sort of accompanying radiation emission?&lt;/p&gt;&#xA;" OwnerUserId="927" LastEditorUserId="927" LastEditDate="2014-02-09T14:22:11.490" LastActivityDate="2015-04-06T15:23:16.917" Title="What would happen if a body were to fall into a neutron star?" Tags="&lt;star&gt;&lt;gravity&gt;&lt;neutron-star&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1677" PostTypeId="2" ParentId="1675" CreationDate="2014-02-08T14:12:16.797" Score="0" Body="&lt;p&gt;The answer to this questions turns out to be a &lt;strong&gt;&quot;yes&quot;&lt;/strong&gt;. The big bang is a sudden rapid expansion of space, so yes, there was space before the big bang. To understand more, check this vulgarization &lt;a href=&quot;http://www.youtube.com/watch?v=q3MWRvLndzs&quot; rel=&quot;nofollow&quot;&gt;video&lt;/a&gt; by MinutePhysics who corrects some misconceptions about the BB, since you seem to not know the Friedmann-Walker-Robertson-Lemaître metric.&lt;/p&gt;&#xA;" OwnerDisplayName="user569" LastActivityDate="2014-02-08T14:12:16.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1678" PostTypeId="2" ParentId="1676" CreationDate="2014-02-08T14:51:03.933" Score="5" Body="&lt;p&gt;No detailed calculations, but a qualitative answer: Depending on the trajectory of the impactor, the results will vary a bit, but it's clear, that the potential energy of the impactor will be transformed into a high amount of kinetic energy before the impact happens. The kinetic energy will then be transformed mainly into heat during the impact, transforming a substantial part of the mass of the impactor into x-rays and gamma rays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remnants of the impactor will be transformed into a plasma, with most of the electrons moving independently of their former nuclei, and dispersed mainly into the atmosphere (a thin layer of a few millimeters) of the neutron star. The energies will be high enough to trigger nuclear fusion as well as fission, together with other high-energy particle reactions. Part of the energy will be transformed into magnetic fields, which can also be very strong on neutron stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No much intermixing with the interior of the neutron star is to be expected in the first instant for small impactors due to the high inertia and density of the inner parts of the neutron star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases the impact could trigger the collapse of the neutron star into a black hole, depending of the mass of the neutron star, and the mass of the impactor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Neutron_star&quot; rel=&quot;noreferrer&quot;&gt;More on the inner structure of neutron stars on Wikipedia&lt;/a&gt;.&#xA;(&quot;Matter falling onto the surface of a neutron star would be accelerated to tremendous speed by the star's gravity. The force of impact would likely destroy the object's component atoms, rendering all its matter identical, in most respects, to the rest of the star.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/010607a.html&quot; rel=&quot;noreferrer&quot;&gt;More about the Chandrasekhar limit of neutron stars&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-02-08T15:08:18.580" LastActivityDate="2014-02-08T15:08:18.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1679" PostTypeId="1" AcceptedAnswerId="1680" CreationDate="2014-02-08T17:27:38.117" Score="3" ViewCount="155" Body="&lt;p&gt;I've always wondered how someone looking at the sky into the enormous amount of stars and other things out there figures out that whether it is an already identified object or is he looking at something new? Is there some sort of tools like latitude and longitude to pin point something in the sky or some catalog where you can see if this is an identified object and what it is called, how it looks like and where it is located and all? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you might have guessed, I don't have any science background, just asking out of a common man's curiosity, so please be gentle with the jargon you use.  &lt;/p&gt;&#xA;" OwnerUserId="980" LastActivityDate="2014-02-08T20:22:49.673" Title="How do one identify something they see in the sky?" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1680" PostTypeId="2" ParentId="1679" CreationDate="2014-02-08T17:55:09.320" Score="5" Body="&lt;p&gt;There exist many &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_astronomical_catalogues&quot; rel=&quot;nofollow&quot;&gt;astronomical catalogs&lt;/a&gt;, reaching from rather simple catalogs like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bayer_designation&quot; rel=&quot;nofollow&quot;&gt;Bayer catalog&lt;/a&gt; for stars, or the &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_Messier_objects&quot; rel=&quot;nofollow&quot;&gt;Messier catalog&lt;/a&gt; for galaxies, nebulae, etc., to huge catalogs with millions of entries. &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaia_%28spacecraft%29&quot; rel=&quot;nofollow&quot;&gt;A catalog with about a billion objects&lt;/a&gt; is scheduled for 2022.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exist also &lt;a href=&quot;http://www.deepskywatch.com/Deep-Sky-atlas/comparison.html&quot; rel=&quot;nofollow&quot;&gt;printed editions&lt;/a&gt; of some catalogs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objects in the sky have several properties, by which they can be described and mostly identified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the most frequently used properties is the position in the sky. From the local geographic position and the time, the observed position of the object can be translated into a celestial coordinate system, e.g. the &lt;a href=&quot;http://en.wikipedia.org/wiki/Equatorial_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;equatorial coordinate system&lt;/a&gt;. The position of an object is described in this system by &lt;a href=&quot;http://en.wikipedia.org/wiki/Declination&quot; rel=&quot;nofollow&quot;&gt;declination&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Right_ascension&quot; rel=&quot;nofollow&quot;&gt;right ascension&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two other frequently used properties are &lt;a href=&quot;http://en.wikipedia.org/wiki/Apparent_magnitude&quot; rel=&quot;nofollow&quot;&gt;magnitude&lt;/a&gt; and &lt;a href=&quot;http://www.physics.uc.edu/~hanson/ASTRO/LECTURENOTES/W07/Stars/Page3.html&quot; rel=&quot;nofollow&quot;&gt;color&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also frequently used are patterns of stars, meaning their relative position to each other, beginning with &lt;a href=&quot;http://en.wikipedia.org/wiki/Lists_of_stars_by_constellation&quot; rel=&quot;nofollow&quot;&gt;star constellations&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analysis of &lt;a href=&quot;http://www.physics.sjsu.edu/tomley/ObjectSpectra/Rigel.html&quot; rel=&quot;nofollow&quot;&gt;steller spectra&lt;/a&gt; allows a more detailed identification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longer term observations may uncover &lt;a href=&quot;http://en.wikipedia.org/wiki/Variable_star&quot; rel=&quot;nofollow&quot;&gt;variability&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Proper_motion&quot; rel=&quot;nofollow&quot;&gt;proper motion&lt;/a&gt; as additional properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://heasarc.gsfc.nasa.gov/W3Browse/star-catalog/bsc5p.html&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a description of parameters (including properties) provided by the BSC5p catalog.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-02-08T20:22:49.673" LastActivityDate="2014-02-08T20:22:49.673" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1681" PostTypeId="2" ParentId="1675" CreationDate="2014-02-08T19:02:37.127" Score="2" Body="&lt;p&gt;&quot;What happened before the Big Bang&quot; is like asking what Alexander the Great's phone number was. It's a nonsensical question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space and time began with the Big Bang. &quot;When&quot; and &quot;where&quot; began with the Big Bang.&lt;/p&gt;&#xA;" OwnerUserId="562" LastActivityDate="2014-02-08T19:02:37.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1682" PostTypeId="1" CreationDate="2014-02-09T08:27:11.517" Score="3" ViewCount="1554" Body="&lt;p&gt;I heard this on Prof. Neil Tyson's StarTalk podcast (&lt;a href=&quot;https://www.nerdist.com/2011/11/nerdist-podcast-139-neil-degrasse-tyson/&quot; rel=&quot;nofollow&quot;&gt;this episode&lt;/a&gt;) and had a question about it. In a Q&amp;amp;A session, a listener had asked what would happen if Superman really did stop the earth from rotating and rotated it in the other direction. Prof. Tyson's response was that if the earth, which is rotating at approx. 1000mph about its axis, is stopped, everything that is not anchored on earth will move the opposite direction at 1000mph and we'd all go crashing into things at that speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose that is just simple inertia: if a body moving at 1000mph is stopped abruptly, everything that was moving with it but not attached to it would indeed attempt to continue in the same direction at 1000mph. However, the earth is &lt;em&gt;rotating&lt;/em&gt; rather than moving in a straight line. Essentially, gravity is strong enough to ensure that we are not thrown off into space by the centrifugal force from the earth's rotation (we aren't holding on to our dear lives as the earth turns at 1000mph). So in a sense, aren't we &lt;em&gt;anchored&lt;/em&gt; to earth by gravity? And therefore, doesn't it mean that if the earth stopped abruptly, we'd just stop with it? (I'm thinking of say a centrifuge with test tubes attached to it. Though as I think about this, I feel that we'd sway at the point of attachment instead of stand still, were the earth to stop).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone help resolve this for me?&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2014-02-10T11:38:55.653" Title="If the earth stopped rotating, would we really fly inertially at 1000mph?" Tags="&lt;earth&gt;&lt;rotation&gt;" AnswerCount="2" CommentCount="5" ClosedDate="2014-02-10T19:03:12.973" ContentLicense="CC BY-SA 3.0" />
  <row Id="1684" PostTypeId="2" ParentId="1682" CreationDate="2014-02-09T11:45:58.183" Score="6" Body="&lt;p&gt;You would be sliding, because the friction between you and Earth would be too weak, or else you wouldn't survive the acceleration, if the rotation of Earth would be changed abruptly. That's like crashing into a wall with several hundred mph. This is with the exception of close to the poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your weight wouldn't change much. On a smooth, plane, and horizontal surface you would slide (due to your inertia). Your velocity you would be sliding depends on your geographic latitude. At the equator it would be at the maximum. At the poles it would be zero. More precisely the velocity would be proportional to the cosine of the latitude (latitude 0 at the equator, and (+/-) 90 degrees, resp. $\pi/2$ at the poles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even more precisely, the speed you would be sliding would be $$2\cdot 465.1~ m/s \cdot cos ~\varphi,$$ with $\varphi$ the &lt;a href=&quot;http://en.wikipedia.org/wiki/Latitude&quot; rel=&quot;noreferrer&quot;&gt;latitude&lt;/a&gt;.&#xA;Factor two, because superman accelerates to the opposite direction. &lt;a href=&quot;http://en.wikipedia.org/wiki/Earth&quot; rel=&quot;noreferrer&quot;&gt;Earth&lt;/a&gt;'s equatorial rotation velocity is $465.1~m/s$.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-02-10T11:38:55.653" LastActivityDate="2014-02-10T11:38:55.653" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1685" PostTypeId="1" AcceptedAnswerId="1689" CreationDate="2014-02-09T16:15:17.943" Score="3" ViewCount="6654" Body="&lt;p&gt;Planets (at least, some of them like Earth) aren't exactly spherical - but what about stars? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is Sun perfectly spherical, for example? What may be the reasons if it isn't? What about other stars? &lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="6" LastEditDate="2014-02-10T19:05:00.113" LastActivityDate="2016-02-14T07:49:35.967" Title="Do stars have exactly sphere shape?" Tags="&lt;star&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1686" PostTypeId="2" ParentId="1685" CreationDate="2014-02-09T18:01:09.737" Score="2" Body="&lt;p&gt;&lt;strong&gt;No&lt;/strong&gt; (1. All stars rotate to some degree and hence are slightly oblate. 2. stars have no surface like a rocky planet, what we see is the photosphere, i.e. the surface from which light escapes without being (much) absorbed or scattered. This surface, however, is constantly moving due to convection. 3. close binary stars are affected by their partner's tide.)&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2014-02-09T18:16:49.557" LastActivityDate="2014-02-09T18:16:49.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1687" PostTypeId="2" ParentId="1682" CreationDate="2014-02-09T18:14:26.623" Score="3" Body="&lt;p&gt;Depends on how the angular momentum (that manifests itself as Earth's rotation) would be absorbed. If, magically, it's absorbed smoothly and evenly, then not much dramatic happens initially (a small change in your weight and some earthquakes as response to the missing centrifugal forces), but the Earth climate and weather would change dramatically within about 100 hours (mind you that the day==year), the day side heating and drying up and the night side freezing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, on the other hand, the angular momentum is absorbed only at certain points, for example if some cosmic giant stops rotation with its finger, then all other structures would need to decelerate from the rotation, likely destroying most human constructions, causing massive flooding of the continents, crumbling of mountains, huge earthquakes, instant death to most non-microbic life forms.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-02-09T18:14:26.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1689" PostTypeId="2" ParentId="1685" CreationDate="2014-02-09T21:21:04.090" Score="10" Body="&lt;p&gt;No, no stars have an exactly spherical shape. The reason for this is that the centrifugal force of the star's rotation is much greater at the equator of the star than it is at the poles, for the simple reason that the rotational velocity is greater. This greater centrifugal force pushes the equator outwards, stretching the star into an oblate shape. This is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravity_darkening&quot; rel=&quot;noreferrer&quot;&gt;gravity darkening&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because we have only visually resolved the surfaces of a few other stars, it's not something that is commonly &lt;em&gt;directly&lt;/em&gt; observed (though the effect can also be observed from stellar spectra). &lt;a href=&quot;http://en.wikipedia.org/wiki/Regulus&quot; rel=&quot;noreferrer&quot;&gt;Regulus&lt;/a&gt; is one star that has been &lt;a href=&quot;http://adsabs.harvard.edu/abs/2007ASPC..361..425G&quot; rel=&quot;noreferrer&quot;&gt;observed as an oblate spheroid&lt;/a&gt; spectrographically. The image below shows the star &lt;a href=&quot;http://en.wikipedia.org/wiki/Altair&quot; rel=&quot;noreferrer&quot;&gt;Altair&lt;/a&gt;, directly imaged using the CHANDRA space telescope. &lt;a href=&quot;http://dept.astro.lsa.umich.edu/~monnier/Altair2007/altair2007.html&quot; rel=&quot;noreferrer&quot;&gt;Go here&lt;/a&gt; for an animation!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5RaSB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some stars are also non-spherical due to the effects of a nearby star in a close binary orbit. Much like how the Moon causes tides on Earth, two stars can stretch each other's surfaces. If they're very close to one another, as in the picture below, there can even be mass transfer between them (&lt;a href=&quot;http://en.wikipedia.org/wiki/Common_envelope&quot; rel=&quot;noreferrer&quot;&gt;ref common envelope&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fZVSy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2014-02-09T21:21:04.090" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1690" PostTypeId="2" ParentId="1685" CreationDate="2014-02-09T21:21:50.587" Score="4" Body="&lt;p&gt;As well as being oblate, like most large spinning bodies, there is a high degree of turbulence in any heated gaseous body, whether that be large gas planets, like Jupiter, or stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at this NASA photo of the sun as an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FoPBO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see a wide range of non-spherical details.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2014-02-09T21:21:50.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1691" PostTypeId="2" ParentId="1650" CreationDate="2014-02-09T23:48:48.777" Score="1" Body="&lt;p&gt;&quot;Big Bang&quot; was coined by supporters of the alternative 'steady state' cosmological theory, essentially as a form of ridicule, but the name stuck, even after the advocates of 'steady state' were forced to concede that their theory (essentially that, at a grand scale, the universe 'looks' the same at all times and places) was flawed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discovery of the Cosmic Microwave Background (CMB) was essentially the nail in the coffin of steady state. CMB is well explained by the &quot;big bang&quot; but not by steady state.&#xA;Of course, what we think of as the &quot;big bang&quot; has also changed over the years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Fred Hoyle, the principal advocate of the steady state theory wrote some interesting 'hard' science fiction, but I am not aware of any link between this and the term 'big bang'.)&lt;/p&gt;&#xA;" OwnerUserId="823" LastEditorUserId="562" LastEditDate="2014-02-10T05:21:03.490" LastActivityDate="2014-02-10T05:21:03.490" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1692" PostTypeId="1" CreationDate="2014-02-10T03:32:54.600" Score="4" ViewCount="6342" Body="&lt;p&gt;I've sometimes heard of galaxies that are millions or billions of light years away but if we could instantly travel 786 trillion light years in any direction, what are we likely to find? And what if we continued on for another 786 trillion light years?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I'm not a physicist.&lt;/p&gt;&#xA;" OwnerUserId="989" LastActivityDate="2015-04-12T13:41:17.103" Title="What would we find if we could travel 786 trillion light years in any direction" Tags="&lt;universe&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1694" PostTypeId="1" CreationDate="2014-02-10T09:01:30.603" Score="5" ViewCount="494" Body="&lt;p&gt;When a clock travels close to the speed of light as observed by us, it should tick slower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that, when using this clock to measure a second - it could travel much more than 300 000 km during a second? In fact, this clock could travel infinitely fast when viewed only from its own perspective?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this imply that if I traveled on a space ship, at the speed of 99 % of light, I would age 4 years, but could reach a destination perhaps 400 light years away? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My friends and family on earth would have perished, but the people I saw on the destination planet &lt;em&gt;as I started my journey&lt;/em&gt; will have aged only 4 years as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the last assumption is correct; it means that we could watch a launching space ship at alpha centauri, it could arrive here in just 3-4 days - appearing to travel much faster than the speed of light, without breaking the laws of relativity. We would just say that it was relativistic blue shift that made it seem to travel faster than the speed of light.&lt;/p&gt;&#xA;" OwnerUserId="749" LastEditorUserId="749" LastEditDate="2014-02-10T15:25:08.230" LastActivityDate="2014-02-15T04:37:40.230" Title="A clock travelling &quot;faster than the speed of light&quot;" Tags="&lt;general-relativity&gt;&lt;speed&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" ClosedDate="2014-02-15T10:49:09.740" ContentLicense="CC BY-SA 3.0" />
  <row Id="1695" PostTypeId="2" ParentId="1694" CreationDate="2014-02-10T12:55:51.863" Score="4" Body="&lt;p&gt;Let's restrict to &lt;a href=&quot;http://en.wikipedia.org/wiki/Introduction_to_special_relativity&quot; rel=&quot;nofollow&quot;&gt;special relativity&lt;/a&gt;, meaning two &lt;a href=&quot;http://en.wikipedia.org/wiki/Inertial_frame_of_reference&quot; rel=&quot;nofollow&quot;&gt;inertial frames&lt;/a&gt; moving in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Minkowski_space&quot; rel=&quot;nofollow&quot;&gt;Minkowski spacetime&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A clock in the first inertial frame ticks slower, when seen from the second. A clock in the second inertial frames ticks slower, when seen from the first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume, that you are fixed to one of the two inertial frames. Usually you measure velocities within your own inertial frame, meaning measuring distance and time in your inertial frame to calculate the velocity of the other frame. In this case you'll get a velocity below the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you measure the distance in your inertial frame, and divide this distance by the time you observe on the clock of the moving inertial frame, you'll get higher velocities, which may exceed the speed of light. But this result has only &lt;em&gt;the dimension&lt;/em&gt; of a velocity; it is not a velocity with respect of any of the two inertial frames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moving observer would observe a slow-down of external events, not a speed-up.&#xA;The observer would observe a speed-up only in the sense of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Relativistic_Doppler_effect&quot; rel=&quot;nofollow&quot;&gt;relativistic Doppler blue-shift&lt;/a&gt; when moving towards the observed object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The people on the destination planet age the same as on the Earth, provided they don't move relative to the Earth. It's just you, since you travel that fast, the distance gets shorter, and therefore the time for the travel also gets shorter. By the Doppler shift and space contraction you'll see them age faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Alpha Centauri, it's again the space contraction for the traveller, making the distance shorter. Again we have a combined space contraction/Doppler effect looking the people on Earth aging rapidly for about 4.5 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For people on Earth observing the travel takes 4.5 years. For the traveller it may take just a few days. With an acceleration of just 1g you could traverse the Milky Way in about 20 years, seen from the spaceship. The same travel seen from Earth would take about 100,000 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Twin_paradox&quot; rel=&quot;nofollow&quot;&gt;More on the twin paradox on Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-02-10T17:53:26.607" LastActivityDate="2014-02-10T17:53:26.607" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="1696" PostTypeId="2" ParentId="1692" CreationDate="2014-02-10T13:53:52.407" Score="6" Body="&lt;p&gt;We would likely find galaxies. The universe, as far as we can observe, is very uniform on the large scale. Depending on the geometry of the universe, may be the place we left for travalling. For a fixed &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_time&quot;&gt;cosmic time&lt;/a&gt; the spacetime seems to be curved in a way that the universe looks finite when travelling along a &lt;a href=&quot;http://en.wikipedia.org/wiki/Geodesic&quot;&gt;straight (geodesic) line&lt;/a&gt;, although without boundary.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-10T13:53:52.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1697" PostTypeId="2" ParentId="1692" CreationDate="2014-02-10T14:30:49.827" Score="3" Body="&lt;p&gt;There is &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmological_principle&quot; rel=&quot;nofollow&quot;&gt;one principle&lt;/a&gt; that says &quot;The universe is isotropic&quot;. It means it looks the same (on large scales) from any point you look from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means, in turn, that if you travel 786 trillion light years instantaneously in some direction, you'll find a very differnt sky but with exactly the same general structure: stars, galaxies, clusters of galaxies, and attractors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Altough you can NOT travel so far so fast, you can make this mind experiment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Send a spaceship so far at (almost) speed of light with a photo camera, take a 360º photo, come back, arrive 1572 triilion years afterwards.&#xA;While awaiting for the spaceship return, wait 786 trillion light years, take a 360º photo, wait the othe 768 trillion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare the two photos. They were taken at the same &quot;Cosmic time&quot;, one of them 768 trillion years ago, here, and the other one maybe two or three years of spaceship time ago, 768 trillion light years afar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The photos will look the the same on the rough scale.&lt;/p&gt;&#xA;" OwnerUserId="473" LastEditorUserId="45" LastEditDate="2014-02-11T14:36:29.883" LastActivityDate="2014-02-11T14:36:29.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1698" PostTypeId="1" AcceptedAnswerId="1702" CreationDate="2014-02-10T15:41:25.580" Score="2" ViewCount="223" Body="&lt;p&gt;At the tremendous forces of gravity inside a black hole, can &quot;matter&quot; be transformed into a more  degenerate form; &quot;dark matter&quot; (if that truly what dark matter is)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the inside of a black hole isn't fully understood, but can this be an acceptable possibility? Why?&lt;/p&gt;&#xA;" OwnerUserId="927" LastEditorUserId="927" LastEditDate="2014-02-10T17:48:40.693" LastActivityDate="2016-12-09T18:54:26.183" Title="Could black holes be creators of dark matter?" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;dark-matter&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1700" PostTypeId="1" AcceptedAnswerId="1701" CreationDate="2014-02-10T18:04:12.233" Score="5" ViewCount="244" Body="&lt;p&gt;Well, the title says everything...&#xA;How many earth-like planets have we discovered out there?&#xA;Discard too big (made of gas), too cold, too hot, etc.&#xA;Consider earth-like sun distance, size, temperature, the rocky ones...&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2015-08-06T20:47:54.630" Title="How many earth-like planets do we know at this moment?" Tags="&lt;exoplanet&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1701" PostTypeId="2" ParentId="1700" CreationDate="2014-02-10T18:09:25.850" Score="6" Body="&lt;p&gt;&lt;a href=&quot;http://phl.upr.edu/projects/habitable-exoplanets-catalog&quot;&gt;Twelve plus Mars are considered as potentially habitable&lt;/a&gt; (as of February 10, 2014).&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-10T18:09:25.850" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1702" PostTypeId="2" ParentId="1698" CreationDate="2014-02-10T20:07:18.760" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ultra-high-energy_cosmic_ray&quot; rel=&quot;nofollow&quot;&gt;Ultra-high energetic cosmic rays&lt;/a&gt; are thought to be caused by black holes as one option. Those energies should be sufficient for the formation of (hypothetical) heavy &lt;a href=&quot;http://en.wikipedia.org/wiki/Minimal_Supersymmetric_Standard_Model&quot; rel=&quot;nofollow&quot;&gt;supersymmetric particles&lt;/a&gt;, which should decay to stable (hypothetical) neutralinos, candidates for (hypothetical) WIMPS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming this theoretical framework, black holes (or similar dense objects) may cause the generation of dark matter. But the amount of dark matter formed that way won't explain the amount of dark matter needed for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambda-CDM_model&quot; rel=&quot;nofollow&quot;&gt;Lambda-CDM model&lt;/a&gt;. Also dark matter forming this way should be hot (fast moving). Lambda-CDM needs cold dark matter (CDM).&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-10T20:07:18.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1703" PostTypeId="1" CreationDate="2014-02-10T20:19:23.140" Score="2" ViewCount="114" Body="&lt;p&gt;I am trying to determine the time of day of some images I have.  I know the lat/lon, the day and year of image.  I also know the 3 solar angles (zenith, azimuth, and elevation).  Is there MATLAB or any other code that can do this calculation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="995" LastActivityDate="2014-02-14T08:18:12.433" Title="Is it possible to calculate the time of day given latitude, longitude and solar elevation angles (zenith, azimuth, elevation)?" Tags="&lt;fundamental-astronomy&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1707" PostTypeId="1" AcceptedAnswerId="1710" CreationDate="2014-02-11T16:30:01.983" Score="1" ViewCount="467" Body="&lt;p&gt;It is said that before big bang there was nothing. Not even space/time. So, now where does space exists?&lt;/p&gt;&#xA;" OwnerUserId="760" LastActivityDate="2014-02-11T18:00:22.153" Title="Matter/mass exist into the space. Where does space exist?" Tags="&lt;space-time&gt;&lt;space&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2014-02-13T15:15:01.897" ContentLicense="CC BY-SA 3.0" />
  <row Id="1708" PostTypeId="1" AcceptedAnswerId="1711" CreationDate="2014-02-11T17:16:54.663" Score="6" ViewCount="697" Body="&lt;p&gt;Is it possible for a solar system like ours to exist and evolve apart from all other star systems and galaxies. Sort of like a stand alone solar system or rogue solar system for lack of a better name. Or is there some requirement that a solar system capable of managing and growing planets especially ones that will occur in the habitable zone originate in a galaxy? &lt;/p&gt;&#xA;" OwnerUserId="257" LastActivityDate="2014-04-11T07:57:05.547" Title="Do solar systems have to evolve in a galaxy?" Tags="&lt;star&gt;&lt;solar-system&gt;&lt;galaxy&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1709" PostTypeId="1" CreationDate="2014-02-11T17:33:04.580" Score="6" ViewCount="2501" Body="&lt;p&gt;The phase of the Moon (the bright limb) looks different depending on where on Earth the observer is located and the time of the observation. I've located &lt;a href=&quot;https://www.youtube.com/watch?v=zdTpIGHIBSU&quot; rel=&quot;nofollow&quot;&gt;this video&lt;/a&gt; and &lt;a href=&quot;http://dropcanvas.com/0hn26&quot; rel=&quot;nofollow&quot;&gt;the accompanying spreadsheet&lt;/a&gt; that supposedly handles this sort of calculation, however I am somewhat confused on what formulas are actually used. I would greatly appreciate if someone with a bit more experience could decipher the spreadsheet a bit and perhaps write down or link me to the formulas being used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm most interested in determining the illuminated phase (i.e the size of the bright limb) and the apparent orientation of the illuminated portion relative to the horizon. My goal is to understand how this type of calculation is done, in order to re-implement it in code for an application to use - displaying the typical crescent moon icon for night-time, but at the correct angle for a specific location and time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS! First post here, so I hope I have used the correct terminology for things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Eventhough the author of the spreadsheet left an answer to the topic, it doesn't feel like it really answers the question. Not to sound clueless and needy, but the question talks about formulas, which I know can be derived from the spreadsheet, however as I have no idea how to approach this, then the answer I was hoping for would help a bit more.&lt;/p&gt;&#xA;" OwnerUserId="999" LastEditorUserId="999" LastEditDate="2014-02-19T20:50:04.777" LastActivityDate="2020-12-09T01:22:33.630" Title="Formulas to determine the illuminated phase and orientation of the Moon" Tags="&lt;the-moon&gt;&lt;moon-phases&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1710" PostTypeId="2" ParentId="1707" CreationDate="2014-02-11T17:54:13.653" Score="3" Body="&lt;p&gt;There is still no way to solve this question for all people.&#xA;You also need to keep in mind that time is directly linked to space. So before the universe there was no space and no time. However, if you want describe processes or even states you should also be able to talk about time.&#xA;Time is the perception of a sequence of events. So before the big bang no events could happen. To be more detailed time is what physicsists call the increase of entropy. Entropy itself is the randomness of a closed physical system. Each physical systems wants to have maximum entropy or randomness. The big bang at point of time -zero- can be seen as a point with the lowest possible value of entropy, because any matter/energy was located at a single point - at a singularity. There was ne place for particles to get at a random place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the expansion of this point space and time expanded into &quot;nothing&quot;. If you think about it: how could it expand if there was something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next: If there is something outside, we still have a neverending question: What's outside the outside? Is it infinite?&#xA;Our physical laws don't work outside - so how should we talk about things nobody could ever describe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there are theories that try to explain why or how the big bang happened - the ethereal expansion by Alan H. Guth and Alexander Vilenkin, for example. Universes are &quot;created&quot; like holes in a cheese by massive discharges of energy (like lightnings during a storm), but those are very hypothetical theories without any prove or even evidence. And all those theories need parallel universes / multiverses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for my opinion, I keep at this point: There was &lt;em&gt;nothing&lt;/em&gt;. As human beings we cannot deal with &lt;em&gt;nothing&lt;/em&gt;. For us there is always something. If we try to think about nothing, we still think on how much time has passed not thinking about anything - for example. Nothing is mathematically described as 0. But also 0 is a number. It's more than nothing. It still consumes place in a formula. So don't think about nothing as &quot;nothing&quot;. Think about nothing as &lt;em&gt;everything&lt;/em&gt;. Because or universe was created out of nothing, this nothing is also everything for any or every time.&lt;/p&gt;&#xA;" OwnerUserId="1000" LastEditorUserId="1000" LastEditDate="2014-02-11T18:00:22.153" LastActivityDate="2014-02-11T18:00:22.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1711" PostTypeId="2" ParentId="1708" CreationDate="2014-02-11T21:11:18.153" Score="6" Body="&lt;p&gt;Probably not. You'd need a could of gas dense enough to contract under its own gravity. In the space between galaxies, matter is spread too thin for a solar-system sized cloud to exist. A much larger cloud could be more diffuse and still collapse, but being much larger, it would collapse into a galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even within a galaxy, stars seem to form when shock waves from nova or some other event initiates the gravitational collapse.&lt;/p&gt;&#xA;" OwnerUserId="562" LastActivityDate="2014-02-11T21:11:18.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1712" PostTypeId="2" ParentId="1708" CreationDate="2014-02-12T20:40:33.570" Score="4" Body="&lt;p&gt;It may be possible for a solar system like ours to exist outside of a galaxy. This topic was discussed briefly in &lt;a href=&quot;http://www.cbc.ca/player/Radio/Quirks+and+Quarks/ID/2434218308/&quot; rel=&quot;nofollow&quot;&gt;this episode of CBC's Quirks and Quarks&lt;/a&gt; where they were talking about extremely fast moving stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These stars are believed to have originally come from the centre of the galaxy but have achieved such huge speeds that they were flung out, and some that have been observed are going in such bizarre angles that they may actually have come from other galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conclusion was basically, if such a solar system does exist, the view from the planet is probably pretty damn good. &lt;/p&gt;&#xA;" OwnerUserId="1009" LastEditorUserId="537" LastEditDate="2014-02-12T21:23:43.490" LastActivityDate="2014-02-12T21:23:43.490" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1714" PostTypeId="1" AcceptedAnswerId="1716" CreationDate="2014-02-13T06:53:37.757" Score="10" ViewCount="5394" Body="&lt;p&gt;Ok, I am majoring in physics (4th year) and I never understood this fundamental (kinda) question. Maybe I haven't explored it enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, why does it take 8min20sec for the light from the sun to get to us?&#xA;I know the answer to this question on a 'surface' scale. The sun is 1AU away, c=3E8 m/s, and d=v/t to get approx 8 min 20 sec. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;my question is on a deeper level. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you could &quot;ride a photon&quot; I KNOW THIS IS IMPOSSIBLE, but just say you could. Or a better question: what does a photon experience? The photon, as per my understanding, would leave the sun and (if it is on the right trajectory) hit the Earth instantaneously. A photon leaving Alpha Centauri would see the universe all at once, in a infinitesimal small unit of time (if directed out to space). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a photon sees everything all at once, why do we perceive it to have a speed? I am sure this has something to do with frames of reference, special relatively, Lorentz transforms? but just seems strange. why is the speed of light finite to us... if it was infinite would this be problematic? &lt;/p&gt;&#xA;" OwnerUserId="1015" LastEditorUserId="58" LastEditDate="2014-02-13T16:18:30.943" LastActivityDate="2016-12-02T06:50:44.600" Title="Why is the speed of light 299,792,458 meters/sec?" Tags="&lt;light&gt;&lt;speed&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1715" PostTypeId="1" AcceptedAnswerId="1717" CreationDate="2014-02-13T07:02:50.463" Score="9" ViewCount="987" Body="&lt;p&gt;Planets are usually found by observing a star and waiting for the light level to drop when a planet passes in front of it, but what about rogue planets that don't have host stars?&lt;/p&gt;&#xA;" OwnerUserId="1016" LastEditorUserId="2153" LastEditDate="2014-12-29T14:20:55.910" LastActivityDate="2016-05-07T06:10:26.047" Title="How are rogue planets discovered?" Tags="&lt;planet&gt;&lt;observational-astronomy&gt;&lt;rogue-planet&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1716" PostTypeId="2" ParentId="1714" CreationDate="2014-02-13T08:50:19.147" Score="10" Body="&lt;p&gt;Speed of light being finite is one of the fundamentals of our Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were infinite, this would have a major implication in causality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides, in non-quantum physics, light is just an electromagnetic wave. Eelectromagnetic field is described by Maxwell's equations, which predict that the speed $c$ of electromagnetic waves propagating through the vacuum depends on the dielectric permittivity $ε_0$ and the magnetic permeability $μ_0$ by the equation $c = {1\over\sqrt{ε_{0}μ_{0}}}$ so you can not have an infinite speed of light unless electric permittivity or magnetic permeability were zero, which in turn would cause all sorts of odd things to electromagnetical attraction (and thus, to matter existence beyond elemental particles).&lt;/p&gt;&#xA;" OwnerUserId="473" LastEditorUserId="473" LastEditDate="2014-02-19T16:28:05.517" LastActivityDate="2014-02-19T16:28:05.517" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1717" PostTypeId="2" ParentId="1715" CreationDate="2014-02-13T11:04:32.557" Score="1" Body="&lt;p&gt;The only way really is through the transit method you describe in your question, however It's pretty much a statistical improbability that a rogue planet will pass through the line of sight between us and another star of which it is not a planetary member.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Transiting_Exoplanet_Survey_Satellite&quot; rel=&quot;nofollow&quot;&gt;Transiting Exoplanet Survey Satellite&lt;/a&gt; would provide a glimmer of hope of identifying some of these events. It would require this kind of constant observation since the transit will only occur once and not regularly as an orbiting planet transit would.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A distant stars light could be gravitationally microlensed by the rogue planet, however the planet would have to be very large to produce a noticeable effect (more of a brown dwarf than a rogue planet) and even then the effect would be fleeting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Direct imaging would be pretty much impossible since the rogue planet would not be close enough to a star to reflect a substantial amount of its light.&lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2014-02-13T11:04:32.557" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1719" PostTypeId="2" ParentId="1715" CreationDate="2014-02-13T12:43:25.620" Score="10" Body="&lt;p&gt;The most successful methods are by &lt;a href=&quot;http://arxiv.org/abs/1105.3544&quot;&gt;gravitational microlensing&lt;/a&gt; and &lt;a href=&quot;http://www.ifa.hawaii.edu/info/press-releases/LonelyPlanet/&quot;&gt;direct observation in the infrared&lt;/a&gt; or far infrared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Methods_of_detecting_exoplanets&quot;&gt;transit method&lt;/a&gt; doesn't work well for rogue planets, because usually at least three transits are needed to confirm a planet.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-13T12:43:25.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1720" PostTypeId="2" ParentId="1709" CreationDate="2014-02-13T18:41:13.343" Score="3" Body="&lt;p&gt;I created the spreadsheet.  It's a bit complicated, to say the least.  Initially I was inputting the formulas by hand from the book I cited in the video, but the process was rather error-prone so I ditched those tables and simply scanned the pages electronically and used the numbers from the scans to construct the tables of the Fourier series used to compute the moon's position.  Some of the tables are therefore redundant and aren't used.  You may be able to use a much simpler method to compute the position of the moon though; for my purposes I needed arcsecond accuracy.  If all you're interested in is computing the approximate phase angle and apparent orientation you can get away with a much more simplified calculation for the moon's position that is accurate to within a few tenths of a degree.  In that case the main thing you need from my spreadsheet are the formulas for computing the phase angle and apparent orientation.  You'll need to calculate the approximate location of the sun as well in order to determine these things.  The spreadsheet does this as well, using a simplified method described by Jean Meeus in &quot;Astronomical Formulae for Calculators.&quot;  He also gives a simplified method for computing the position of the moon.  Either &quot;Astronomical Formulae for Calculators&quot; or &quot;Astronomical Algorithms&quot; (both by Jean Meeus) would be excellent references for your project.  I can walk you through the formulas you need if you need help with that part.&lt;/p&gt;&#xA;" OwnerUserId="1022" LastActivityDate="2014-02-13T18:41:13.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1721" PostTypeId="1" CreationDate="2014-02-13T21:29:34.147" Score="6" ViewCount="142" Body="&lt;p&gt;Can the expected frequency of all sizes of impact craters on Earth be calculated based on the preservation of impact craters seen on the Moon? Taking into account the difference in gravity and atmosphere of the Earth and Moon, and assuming a similar age for both, it should be possible to come up with a reasonable estimate of how many meteorites, etc have hit the earth and their size distribution.     &lt;/p&gt;&#xA;" OwnerUserId="1023" LastEditorUserId="1023" LastEditDate="2014-02-15T02:26:56.213" LastActivityDate="2014-03-18T10:44:44.807" Title="Would it be possible to calculate the expected frequency of impact craters of all sizes on Earth" Tags="&lt;earth&gt;&lt;geology&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1722" PostTypeId="1" CreationDate="2014-02-13T22:23:57.643" Score="4" ViewCount="239" Body="&lt;p&gt;I have a pair of Solar Viewing Eclipse Shades that say &quot;Safe for Direct Solar Viewing&quot; and would like to know if my eyes would be harmed if I looked at the sun everyday for about a minute wearing them.&lt;/p&gt;&#xA;" OwnerUserId="1025" LastActivityDate="2014-02-24T16:33:41.677" Title="Can looking at the sun through solar viewing eclipse shades everyday harm your eyes?" Tags="&lt;the-sun&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1723" PostTypeId="1" CreationDate="2014-02-13T23:14:15.153" Score="5" ViewCount="245" Body="&lt;p&gt;In the same way, as a black hole emits Hawking radiation corresponding to its temperature&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;T = \frac{\hbar}{r_{BH}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the accelerating universe can be described as all of the stuff falling towards the cosmic horizon, such that it is analogue to a black hole where the cosmic horizon takes the role of the event horizon with the black hole &quot;interior&quot; being outside. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding temperature of the accelerating universe is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;T = \hbar H = \frac{\hbar}{R_{H}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where H is the Hubble constant. Therefore, the cosmic horizon emits something similar to Hawking radiation too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fate of a black hole is to finally evaporate. My question is: What does black hole evaporation correspond to in the accelerating universe when making use of this analogy?&lt;/p&gt;&#xA;" OwnerUserId="92" LastActivityDate="2021-02-21T07:32:43.603" Title="What does black hole evaporation correspond to in the accelerating universe / black hole analogy?" Tags="&lt;black-hole&gt;&lt;cosmology&gt;&lt;hawking-radiation&gt;&lt;cosmological-horizon&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1724" PostTypeId="1" CreationDate="2014-02-14T00:16:39.873" Score="3" ViewCount="3654" Body="&lt;p&gt;&lt;em&gt;If anyone would at least be willing to point me in the direction of finding the answer, or help me solve it I'd appreciate it. I don't know how to go about this, and need to know how to solve this on my own (plus I need the answer for a homework assignment, but just an answer isn't gonna help).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Question:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 10 months. The lines of both stars shift by equal amounts, and the amount of the Doppler shift indicates that each star has an orbital speed of $~8.0\cdot 10^4~ \mbox{m}/\mbox{s}.$ &#xA;What are the masses of the two stars? Assume that each of the two stars traces a circular orbit around their center of mass.&lt;/p&gt;&#xA;" OwnerUserId="1026" LastEditorUserId="746" LastEditDate="2014-02-14T12:04:53.400" LastActivityDate="2016-04-05T16:06:36.800" Title="What are the masses of the two stars (given the information provided)?" Tags="&lt;mass&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1725" PostTypeId="2" ParentId="1426" CreationDate="2014-02-14T02:57:22.873" Score="5" Body="&lt;p&gt;No, amateur astronomy equipment is not capable of doing such an analysis. Most color CCDs use something called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Bayer_filter&quot; rel=&quot;nofollow&quot;&gt;Bayer filter&lt;/a&gt; to detect color. Essentially, certain pixels detect only red photons, other pixels detect only green, and still others detect only blue. Each filter has a high transmission over some band of wavelengths (say, for red, 600 - 700 nm). Photons outside of this band are absorbed. The CCD pixels behind the red Bayer filter then count how many photons they see (they actually measure charge, but that's a detail). The point is, the CCD can't measure a photon's precise wavelength; it can only count how many it sees over some wavelength band.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this is NOT necessarily the case when it comes to professional astronomy. To my knowledge, there do not currently exist optical CCDs that can do spectroscopy without a diffraction grating. For one thing, optical diffraction gratings work really well, so there is no reason. The other reasons are that optical photons don't generate a lot of charge in a CCD, so it would be hard to measure, and there are generally a LOT of optical photons, so you'd get a lot of &lt;a href=&quot;http://xmm.esac.esa.int/sas/current/doc/epileupmask/node4.html&quot; rel=&quot;nofollow&quot;&gt;pile up&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, X-ray CCDs often do spectroscopy, albeit not with Fourier analysis (for example, the &lt;a href=&quot;http://cxc.harvard.edu/cal/Acis/&quot; rel=&quot;nofollow&quot;&gt;ACIS&lt;/a&gt; CCDs on Chandra). There are a lot fewer X-rays than optical photons, and they deposit a lot more energy on a CCD. Unfortunately, you can't do X-ray astronomy from the ground, so even if an amateur astronomer could somehow afford one, it wouldn't do them any good. Even in the case of X-ray astronomy, you generally use an x-ray diffraction grating to get a spectrum if that is what you are interested in.&lt;/p&gt;&#xA;" OwnerUserId="1027" LastEditorUserId="1027" LastEditDate="2014-02-14T21:36:35.250" LastActivityDate="2014-02-14T21:36:35.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1726" PostTypeId="2" ParentId="1724" CreationDate="2014-02-14T03:06:21.707" Score="1" Body="&lt;p&gt;You need to use the generalized (a la Newton) form of Kepler's third law; see &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/orbv.html#bo&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1027" LastActivityDate="2014-02-14T03:06:21.707" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1727" PostTypeId="1" AcceptedAnswerId="1734" CreationDate="2014-02-14T03:18:30.783" Score="8" ViewCount="6733" Body="&lt;p&gt;So, I am learning about quantum entanglement in my quantum course and had a thought. If the Big Bang is the creation of everything - time included, and this came from a single &quot;source&quot;. Would it not follow that all particles in the universe are entangled to each other? And perhaps, we cannot observe this entanglement because we are &quot;inside the system&quot; sort to speak. Are we not all connected? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the last part is cheesy, but the point is serious.  &lt;/p&gt;&#xA;" OwnerUserId="1015" LastEditorUserId="37191" LastEditDate="2021-02-17T22:26:49.603" LastActivityDate="2021-02-17T22:26:49.603" Title="Is everything quantum entangled?" Tags="&lt;cosmology&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1728" PostTypeId="2" ParentId="1651" CreationDate="2014-02-14T03:34:31.590" Score="18" Body="&lt;p&gt;A quasar is simply an active galactic nucleus (AGN) that is viewed from a particular angle; see the picture below, in which quasars are labeled &quot;QSO&quot;. This is really a remarkable figure because historically all of the names in the figure were thought to correspond to different types of objects, when really they all refer to the same thing! &lt;img src=&quot;https://ned.ipac.caltech.edu/level5/March04/Torres/Figures/figure5.jpg&quot; alt=&quot;AGN&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question really shouldn't be &quot;Was there ever a quasar in the Milky Way?&quot;, since the dotted line in the figure would correspond to the Galactic plane and we would not see Sagittarius A* (the Milky Way's super-massive black hole) from the correct angle. A better question might be, &quot;Has Sagittarius (Sgr) A* ever been active?&quot; The answer to that question is yes; according to &lt;a href=&quot;http://science1.nasa.gov/science-news/science-at-nasa/2002/21feb_mwbh/&quot; rel=&quot;noreferrer&quot;&gt;this page&lt;/a&gt; it was probably active (very bright with a jet) about 10,000 years ago. However, at the moment, it isn't really doing anything, since it isn't currently accreting anything (to put it plainly, it isn't eating anything, so it doesn't have enough energy to be active). However, many astronomers (myself included!) are anxiously waiting for &lt;a href=&quot;http://en.wikipedia.org/wiki/Sagittarius_A*#Discovery_of_G2_gas_cloud_on_an_accretion_course&quot; rel=&quot;noreferrer&quot;&gt;a cloud of gas called G2 to fall into Sgr A*&lt;/a&gt;. We are hoping that Sgr A* will burp or do something interesting.&lt;/p&gt;&#xA;" OwnerUserId="1027" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:47.210" LastActivityDate="2014-02-14T21:26:06.360" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1729" PostTypeId="2" ParentId="1703" CreationDate="2014-02-14T03:42:26.447" Score="1" Body="&lt;p&gt;Of course it is! How do you think &lt;a href=&quot;http://en.wikipedia.org/wiki/Sundial&quot; rel=&quot;nofollow noreferrer&quot;&gt;sundials&lt;/a&gt; work? That said, there are better, more precise ways to keep time. I don't know how to do it for an image; that is really a computational problem rather than an astronomy question. You might be better off asking it at &lt;a href=&quot;https://stackoverflow.com/&quot;&gt;Stack Overflow&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1027" LastEditorUserId="-1" LastEditDate="2017-05-23T12:37:21.700" LastActivityDate="2014-02-14T03:42:26.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1730" PostTypeId="2" ParentId="1703" CreationDate="2014-02-14T08:18:12.433" Score="1" Body="&lt;p&gt;Yes, you can, and it is easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You only need a Nautical Almaniac.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll find there both the tables you need for the Equation of Time and the formulas to use.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-02-14T08:18:12.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1731" PostTypeId="2" ParentId="1714" CreationDate="2014-02-14T10:03:25.893" Score="4" Body="&lt;p&gt;Why is the speed of light 299,792,458 m/s, and not (for instance) 3,1 or 4,3 x 10^44 m/s?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that all those numbers are consequences of &lt;em&gt;arbitrary&lt;/em&gt; choices of the units of measure. In fact there is nothing special about the number &lt;em&gt;299,792,458&lt;/em&gt;, so much so that one can correctly write that the speed of light is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1079252848.8 [km/h]&lt;/li&gt;&#xA;&lt;li&gt;0.3 [pc/year]&lt;/li&gt;&#xA;&lt;li&gt;1 [lightyear/year]&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So much so, that in particle physics it is common to use the speed of light as the unit of measure of speed, so that it is effectively equal to unity. You would write that the speed of light is 1 [c], and that the speed of that particle was 0.99998 [c], while I typically get to work at a speed of 1. x 10^{-8} [c].&lt;/p&gt;&#xA;" OwnerUserId="198" LastActivityDate="2014-02-14T10:03:25.893" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1732" PostTypeId="1" AcceptedAnswerId="1739" CreationDate="2014-02-14T10:53:12.060" Score="3" ViewCount="126" Body="&lt;p&gt;In &lt;a href=&quot;http://www.youtube.com/watch?v=50-WpB14OMo#t=18&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; talk, the not yet settled down idea that the wave function of the universe could potentially be written as  the partition function of a scale invariant statistical field theory is mentioned:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Psi[g] = Z[g] = \int D(\text{Fields}) e^{-S[g,\text{Fields}]}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If our universe were AdS, this relation could already be well enough explained by the AdS/CFT correspondence, but as our expanding universe correspond to a dS geometry, things are less clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the concrete technical arguments, ideas, or hints that this relationship should hold for our dS universe too? What work has already been done on this?&lt;/p&gt;&#xA;" OwnerUserId="92" LastEditorUserId="9301" LastEditDate="2016-10-22T19:36:02.833" LastActivityDate="2016-10-22T19:36:02.833" Title="What are the concrete technical arguments supporting the idea that the wave function of the universe can be written as partition function?" Tags="&lt;cosmology&gt;&lt;cosmological-inflation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1733" PostTypeId="1" AcceptedAnswerId="1735" CreationDate="2014-02-14T12:10:05.960" Score="17" ViewCount="1594" Body="&lt;p&gt;Why haven't some meteoroids gotten caught in Earth's or Venus' orbit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AFAIK most meteors are tiny fragments from comets. Shouldn't some comet tail sometime have passed Earth orbit at velocities suitable for our planet to capture such fragements? And 100 000s of asteroids have been detected. Why haven't the inner planets gotten asteroid like moons like the outer planets have? The asteroid belt isn't too far away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth's large Luna might clean them away, but that won't explain Venus' lack of tiny moons.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="2618" LastEditDate="2017-03-07T01:09:04.613" LastActivityDate="2017-03-07T01:09:04.613" Title="Why haven't Earth and Venus got any tiny moons? Or have they?" Tags="&lt;gravity&gt;&lt;earth&gt;&lt;asteroids&gt;&lt;natural-satellites&gt;&lt;venus&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1734" PostTypeId="2" ParentId="1727" CreationDate="2014-02-14T13:44:40.053" Score="6" Body="&lt;p&gt;By far the most particles in the visible universe aren't quantum entangled. That's obvious from observation, since if all electron spins e.g. would be entangled, all electrons would flip their spin at the same time, and we couldn't observe varying statistics of electron spins in a sample, resulting in varying degrees of magnetism, e.g. by applying gradually changing exterior magnetic fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second example: All (bosonic) atoms of spin 0 would form a &lt;a href=&quot;http://en.wikipedia.org/wiki/Bose%E2%80%93Einstein_condensate&quot;&gt;Bose-Einstein condensate&lt;/a&gt; not just near absolute zero, since they would be in the same quantum state, hence in the lowest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;single source&quot; isn't a single quantum state. In the early phase of the big bang temperatures were very high, allowing for a huge amount of possible quantum states within a small volume. Without warranty you may think of the initial state of the universe as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_superposition&quot;&gt;superposition&lt;/a&gt; of all possible quantum states, the wave functions of which collapse relative to an observer, regarding the observer as a particular quantum state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That way we get to &lt;a href=&quot;http://en.wikipedia.org/wiki/Many-worlds_interpretation&quot;&gt;the many world interpretation of quantum theory&lt;/a&gt; and a &lt;a href=&quot;http://en.wikipedia.org/wiki/Universal_wavefunction&quot;&gt;universal wave function&lt;/a&gt;. See also a proposed &lt;a href=&quot;http://en.wikipedia.org/wiki/Hartle%E2%80%93Hawking_state&quot;&gt;Hartle-Hawking state&lt;/a&gt; as a pre-Planck epoch of the universe.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-14T13:44:40.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1735" PostTypeId="2" ParentId="1733" CreationDate="2014-02-14T15:33:12.897" Score="13" Body="&lt;p&gt;The strength of the Earth's gravitational field compared to the Moon and the Sun is not enough to capture and hold satellites - there are too many disruptive forces that would rip them away over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there are some objects at the Lagrangian points - the points where the gravitational fields of the Earth and other objects are equal and so it is possible to have a (likely-meta) stable orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives some details about what might be found at the various Lagrangian points: &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_objects_at_Lagrangian_points&quot;&gt;http://en.wikipedia.org/wiki/List_of_objects_at_Lagrangian_points&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2014-02-14T15:33:12.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1736" PostTypeId="2" ParentId="1723" CreationDate="2014-02-14T17:38:12.433" Score="4" Body="&lt;p&gt;An analogy in the settings of the question would be a deceleration down to an expansion rate, such that eventually light from anywhere in the universe can reach us, meaning the cosmic horizon can vanish. A constant expansion rate of the universe above the speed of light would be the analogy of a black hole of fixed mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an infinite universe this would mean, that its expansion needs to slow down to an asymptotic speed for infinite distance below the speed of light. For a finite &lt;a href=&quot;http://en.wikipedia.org/wiki/N-sphere&quot; rel=&quot;nofollow&quot;&gt;3-spherical universe&lt;/a&gt; for a fixed cosmic time a slow-down to below the speed of light would be sufficient. This would correspond to a massive follow-up object of the black hole without event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an analogy to a completely evaporating black hole the expansion rate needs to slow down to zero. The slow down between speed of light and zero corresponds to the very last phase of black hole evaporation, when its mass drops below the &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck_particle&quot; rel=&quot;nofollow&quot;&gt;Planck mass&lt;/a&gt;, then without event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analogy to Hawking radiation would be radiation or matter emerging from behind the former cosmic horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking in &lt;a href=&quot;http://en.wikipedia.org/wiki/Stereographic_projection&quot; rel=&quot;nofollow&quot;&gt;stereographic projections&lt;/a&gt; may help for a better understanding: Project infinity of a plane expanding universe to the singularity of a black hole at the north pole of an expanding sphere, as an initial point of further reasoning.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-14T17:38:12.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1737" PostTypeId="1" AcceptedAnswerId="1738" CreationDate="2014-02-14T17:57:47.253" Score="5" ViewCount="721" Body="&lt;p&gt;I would know what would happen if we put a small part (let me assume 1 cm$^3$) of &lt;code&gt;White Dwarf&lt;/code&gt; or &lt;code&gt;Neutron Star&lt;/code&gt; on &lt;code&gt;Earth&lt;/code&gt;. Will it explode or maybe it collapses to the Earth's core?&lt;/p&gt;&#xA;" OwnerUserId="1033" LastEditorUserId="1033" LastEditDate="2014-02-14T18:42:14.253" LastActivityDate="2014-12-13T17:58:14.993" Title="What would happen if we put a small part of White Dwarf or Neutron Star on Earth?" Tags="&lt;earth&gt;&lt;neutron-star&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1738" PostTypeId="2" ParentId="1737" CreationDate="2014-02-14T18:31:28.510" Score="6" Body="&lt;p&gt;A piece of a white dwarf would explode, because atoms would expand to their normal size. This would hold for outer layers of a neutron star, too. For inner parts of the neutron star, that's not quite clear, but that kind of matter would probably decay rapidly, also resulting in an explosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the material of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutron_star&quot;&gt;neutron star&lt;/a&gt; could be kept stable, it would fall down to Earth's core, up to the other side of the Earth, and oscillate a while that way, before being slowed down by friction, and stopped close to Earth's center. (1 cm³ neutron star matter has a mass of about 400 million tons.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One cm³ &lt;a href=&quot;http://en.wikipedia.org/wiki/White_dwarf&quot;&gt;white dwarf&lt;/a&gt; matter would have a mass of about 1 ton. Assumed it could be kept stable, solid rock could withstand this load.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-14T18:31:28.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1739" PostTypeId="2" ParentId="1732" CreationDate="2014-02-14T22:24:25.653" Score="2" Body="&lt;p&gt;Similar approaches have been generalized to more general spacetimes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technical details for dS space and further references in &lt;a href=&quot;http://arxiv.org/abs/hep-th/0203101&quot; rel=&quot;nofollow&quot;&gt;this paper about the holographic principle&lt;/a&gt;, p.43 ff., may provide some idea.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-14T22:24:25.653" CommentCount="4" CommunityOwnedDate="2016-10-26T18:07:04.177" ContentLicense="CC BY-SA 3.0" />
  <row Id="1741" PostTypeId="1" CreationDate="2014-02-14T23:22:15.703" Score="3" ViewCount="95" Body="&lt;p&gt;So I was reading some interesting stuff about universe and Big Bang and wanted to make sure one thing. As far as I understand everything we can see about universe is the past, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean all the galaxies and known stars are many light years away so we have no way to knowing if they still exist, so pretty much we don't know anything about present state of the universe, galaxies and stars. Am I right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, all of the facts are said in present tense like &quot;There are billions of galaxies&quot; or &quot;there are black holes&quot;, but in fact, they WERE, because we don't know if they exist now.&lt;/p&gt;&#xA;" OwnerUserId="1035" LastEditorUserId="1035" LastEditDate="2014-02-14T23:27:41.553" LastActivityDate="2014-02-14T23:27:41.553" Title="Can we only see the past of the universe?" Tags="&lt;universe&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2014-02-15T08:35:14.510" ContentLicense="CC BY-SA 3.0" />
  <row Id="1742" PostTypeId="1" CreationDate="2014-02-15T00:03:07.923" Score="0" ViewCount="570" Body="&lt;p&gt;Assuming one could create a staircase and walk to the moon, would escape velocity still be a factor in leaving the Earth's gravitational pull? I'm not a physicist but am just curious to know.&lt;/p&gt;&#xA;" OwnerUserId="1036" LastActivityDate="2014-02-15T03:48:16.727" Title="Could I create a staircase to the moon and walk to it?" Tags="&lt;gravity&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2014-02-17T09:08:16.437" ContentLicense="CC BY-SA 3.0" />
  <row Id="1743" PostTypeId="2" ParentId="1727" CreationDate="2014-02-15T01:14:59.287" Score="6" Body="&lt;p&gt;If you try to consider some hypothetical state of all particles, that state is almost certainly entangled. However, that does not mean that you can't (say) treat two particles at random as very probably independent of each other--effectively unentangled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard definition of an &lt;em&gt;entangled state&lt;/em&gt; is a state that is not fully separable, i.e. one that is not probability mixture of product states. This is just the quantum version of the notion of &quot;independent random variables&quot; in more ordinary probability theory. Thus, for any compound system, almost all states are entangled, as the non-entangled ones are vanishly small (measure zero) subset of all possible states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, any time you measure a particle with apparatus, after measurement the apparatus indicates something about the measured system. Thus, the joint state of the compound &quot;apparatus + particle&quot; system is not separable because its parts are not independent of one another. Measurement produces a particular case of entanglement between the measuring and the measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a macroscopic system interacts with its environment, information about it diffuses into the environment, producing entanglement between parts of the system and parts of its environment. Overall, this leakage of information is responsible for &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_decoherence&quot; rel=&quot;noreferrer&quot;&gt;quantum decoherence&lt;/a&gt; and the second law of thermodynamics. But its chaotic nature means that for all practical purposes, that information is lost, and we can think of entropy increase as 'missing information'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you just pick particles at random, chances overwhelming are that you can treat them as essentially independent--unentangled. Intuitively, one can think information about something becoming so mixed up in its environement in the large that any particular microscopic parts of it and its environment are going to tell you next to nothing about each other. In other words, they'd be almost entirely independent from each other--unentangled.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-02-15T01:14:59.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1744" PostTypeId="2" ParentId="1742" CreationDate="2014-02-15T03:48:16.727" Score="2" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Space_elevator&quot; rel=&quot;nofollow&quot;&gt;space elevator&lt;/a&gt; has been proposed. With currently available technology it would be very expensive to build one, but not necessarily impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the staircase or elevator being fixed at some point on Earth near the equator, the Earth's gravitational pull is compensated by centrifugal force at a height of about 36,000 km, in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Geostationary_orbit&quot; rel=&quot;nofollow&quot;&gt;geostationary orbit&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the height you would have to climb to, before you could slide &quot;down&quot; to the Moon along an even more expensive construction stepwise adjusting your velocity to that of the Moon. The total trip would be at least about 400,000 km, that's ten times around the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your weight would decrease, the higher you climb from ground to the 36,000 km height. Assuming you make 3 km of height each day, it would take you about 34 years to reach geostationary orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then sliding down with 10 m/s would take at least one more year to reach the Moon. Hence, it's possible with a one-way trip of about 35 years, assumed the technical infrastructure is provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it looks easier to take an appropriate vehicle.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-15T03:48:16.727" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1745" PostTypeId="2" ParentId="1694" CreationDate="2014-02-15T04:37:40.230" Score="1" Body="&lt;p&gt;One is always traveling. Even if traveling at the speed of light, one is still affected by the forces. If the clock depends on gravity, then inertia will have an affect on it. If the clock uses a battery, then maybe the clock will only affect itself if nothing else is nearby.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not an expert, but even at the speed of light, forces will still have an effect.&lt;/p&gt;&#xA;" OwnerUserId="1037" LastActivityDate="2014-02-15T04:37:40.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1746" PostTypeId="1" AcceptedAnswerId="1778" CreationDate="2014-02-15T13:38:12.350" Score="11" ViewCount="791" Body="&lt;p&gt;I know of stars actually dying (exploding as a supernova) but I have never heard of what happen to black holes. Do they continue to live forever, waiting there for more matter to absorb? Do they explode at some point ?&lt;/p&gt;&#xA;" OwnerUserId="202" LastActivityDate="2016-10-05T02:07:56.963" Title="Do black holes die?" Tags="&lt;black-hole&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1748" PostTypeId="2" ParentId="1733" CreationDate="2014-02-15T15:01:35.513" Score="5" Body="&lt;p&gt;@adrianmcmenamin Has the right idea here. These objects he is referring to are called &lt;a href=&quot;http://en.wikipedia.org/wiki/Trojan_%28astronomy%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Trojans&lt;/a&gt;, and are defined to be: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a minor planet or natural satellite (moon) that shares an orbit with a planet or larger moon, but does not collide with it because it orbits around one of the two Lagrangian points of stability (trojan points), L4 and L5, which lie approximately 60° ahead of and behind the larger body, respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Near to each massive body in the solar system (Earth included) are minima, maxima, and saddle points in the potential energy landscape due to the combination of said object and the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FNvGv.jpg&quot; alt=&quot;lagrangepoints&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible for objects to essentially get caught in the more stable of these extrema. Jupiter, being the most massive, has quite a number of these types of objects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Aw6eU.png&quot; alt=&quot;JupiterTrojans&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, one such object has been discovered orbiting around a stable Lagrange point in 2010, called &lt;a href=&quot;http://en.wikipedia.org/wiki/2010_TK7&quot; rel=&quot;nofollow noreferrer&quot;&gt;2010 TK7&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2014-02-15T15:01:35.513" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1749" PostTypeId="1" CreationDate="2014-02-15T15:55:58.613" Score="1" ViewCount="208" Body="&lt;p&gt;during a visit to New Zealand we observed (on 3.2 2014) a strange moonset in Te Anau. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Around 10 pm, shortly after sunset, the moon appeared, but then quickly began to go down, and about 1 hour or less it &quot;touched&quot; the mountains on horizon and shortly after it completely disappeared and then we did not see it anymore. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did not bring a camera or camcorder, so I only have this bad photo from a mobile phone:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gj9gU.jpg&quot; alt=&quot;moonset in Te Anau, NZ&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole process was pretty fast, so I only have one picture. I've never seen anything like it - usually (in Europe) the moon comes out after sunset rather than going down :-) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was also one guy living there more than 10 years, but he was apparently surprised too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody explain this? Is it something normal in this location? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2014-02-15T16:48:00.937" Title="strange moonset around 10 pm (3.2 2014, NZ) - explanation?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1750" PostTypeId="2" ParentId="1749" CreationDate="2014-02-15T16:48:00.937" Score="1" Body="&lt;p&gt;February 3, 2014 has been short (three days) after new moon. The moon is then about 40 degrees or about 3 hours behind the sun. Hence, what you observed is quite normal, in Europe, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only difference is, that the moon looks to the other side when seen from the southern hemisphere. &lt;a href=&quot;http://lossofthenight.blogspot.de/2013/11/moon-phase-calendar-for-2014.html&quot; rel=&quot;nofollow&quot;&gt;Here, for 2014, a moon phase calendar of the northern hemisphere&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-15T16:48:00.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1751" PostTypeId="1" AcceptedAnswerId="1753" CreationDate="2014-02-15T17:38:36.100" Score="3" ViewCount="1632" Body="&lt;p&gt;According to Wikipedia (&lt;a href=&quot;http://en.wikipedia.org/wiki/Geostationary_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;),  a geostationary orbit, geostationary Earth orbit or geosynchronous equatorial orbit (GEO), is a circular orbit 35,786 kilometres (22,236 mi) above the Earth's equator and following the direction of the Earth's rotation. An object in such an orbit has an orbital period equal to the Earth's rotational period (one sidereal day), and thus appears motionless, at a fixed position in the sky, to ground observers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is then couldn't any orbit be a GEO? I mean, what about a satellite in an orbit that's say 70,000km, if the angular speed of the satellite was the same as the earth's rotation such that it always stayed in a fixed position when observed from the Earth, then isn't that a GEO orbit as well?&lt;/p&gt;&#xA;" OwnerUserId="1036" LastEditorUserId="2365" LastEditDate="2017-03-31T13:47:36.093" LastActivityDate="2017-03-31T13:47:36.093" Title="Can someone explain a Geostationary Orbit in a simpler way?" Tags="&lt;orbit&gt;" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1752" PostTypeId="2" ParentId="1751" CreationDate="2014-02-15T20:01:41.920" Score="1" Body="&lt;p&gt;You have yourself provided us with the answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any satellite revolving over the &lt;strong&gt;Earth's Equator&lt;/strong&gt; would be considered as a GeoStationary object. Provided, it should move with the rotation of the Earth and it should be &lt;strong&gt;35,786 kilometres&lt;/strong&gt; above of the sea so that the observer, that is on the ground would think that the satellite is &lt;strong&gt;stationary&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If to say for simple: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Geostationary object is an object, which is fully synchronized to the movement of the earth. Its position is as it is fixed to the Equator of the earth at zero degree latitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything that would stay synchronized with the Earth's rotation would be a Geostationary object. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from the Wikipedia article you can read more here: &lt;a href=&quot;http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Geostationary_orbit.html&quot; rel=&quot;nofollow&quot;&gt;http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Geostationary_orbit.html&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has clear and more simple information, and has described that the object must be this_many kms above the sea level and should be at the 0 degree latitude. &lt;/p&gt;&#xA;" OwnerUserId="79" LastActivityDate="2014-02-15T20:01:41.920" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1753" PostTypeId="2" ParentId="1751" CreationDate="2014-02-15T20:04:16.267" Score="6" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Centrifugal_force&quot; rel=&quot;noreferrer&quot;&gt;centrifugal force&lt;/a&gt; increases with the radius of the orbit for a fixed angular speed. The gravitational pull decreases with the radius of the orbit around Earth. The geostationary orbit is, where the two forces annihilate for the geosynchronous angular speed. The gravitational pull takes the role of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Centripetal_force&quot; rel=&quot;noreferrer&quot;&gt;centripetal force&lt;/a&gt; to keep the satellite on the circular orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the principle, which keeps satellites, moons, planets on their respective orbit. If there is some additional force, they leave their former orbit.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-15T20:04:16.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1754" PostTypeId="1" CreationDate="2014-02-15T22:51:01.220" Score="-1" ViewCount="129" Body="&lt;p&gt;&lt;a href=&quot;http://www.space.com/20606-nasa-asteroid-capture-mission-images.html&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt; got me thinking If NASA is going to use our primitive space exploration technology to capture an asteroid wouldn't an Extra Terrestrial species have used UFO's to maneuver celestial bodies?&#xA;They clearly would have come across this problem along time ago and the videos of UFO's levitating and vectoring themselves demonstrates they operate on similar applied electromagnetic forces that the celestial bodies use to maneuver themselves around in solar systems.&#xA;Couldn't they maneuver meteors, asteroids, possibly even speed the advance of collisions developing protoplanets and move moons into orbits with planets?&#xA;Maybe they build solar systems to live in they way earthlings build housing developments?&lt;/p&gt;&#xA;" OwnerUserId="257" LastActivityDate="2014-02-16T12:08:49.767" Title="If NASA is going to capture an asteroid with a spacecraft, wouldn't a UFO be able to manuever a celestial body?" Tags="&lt;solar-system&gt;&lt;asteroids&gt;&lt;nasa&gt;&lt;identify-this-object&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2014-02-16T17:31:43.817" ContentLicense="CC BY-SA 3.0" />
  <row Id="1755" PostTypeId="2" ParentId="1721" CreationDate="2014-02-16T03:20:34.240" Score="1" Body="&lt;p&gt;Yes. By looking at a nearby object that hasn't been subject to erosion and has been subject to the same position around the sun, you could estimate the distribution of impacts on the Earth in time and in size. The moon is excellent for this purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an article touching on the subject: &lt;a href=&quot;http://adsabs.harvard.edu/full/1975LPSC....6.2597N&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/full/1975LPSC....6.2597N&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="562" LastActivityDate="2014-02-16T03:20:34.240" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1756" PostTypeId="2" ParentId="1754" CreationDate="2014-02-16T03:32:00.427" Score="3" Body="&lt;p&gt;You are assuming&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;there are extraterrestrial space travelers&lt;/li&gt;&#xA;&lt;li&gt;those travelers have technologies we know nothing of&lt;/li&gt;&#xA;&lt;li&gt;those technologies we know nothing of allow &quot;levitation&quot; and&#xA;&quot;vectoring themselves&quot; with &quot;electromagnetic forces.&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Asking what they would do if they existed and if they had capabilities is really the realm of science fiction. There can be no good answer to your question at this time. Maybe later we'll know, maybe never.&lt;/p&gt;&#xA;" OwnerUserId="562" LastActivityDate="2014-02-16T03:32:00.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1757" PostTypeId="2" ParentId="1751" CreationDate="2014-02-16T07:09:01.580" Score="0" Body="&lt;p&gt;In a simpler way, if that satellite is above 35,786 kilometres above you country place, for example, Inidia-Dehli, then satellite will remain at the same position throughout its life. That is why it is called as Geostationary satellite. It revolves with the speed of earth, but circumference of earth and circumference of geostationary satellite's orbit is different. Circumference of geostationary satellite is much greater than the earths circumference. So speed of geostationary satellite is greater than speed of earth because it has to stick to you country place above earth so it have to move fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also in this orbit, gravitational force of earth which attracts satellite towards earth and centrifugal force which tries to escape satellite from that orbit (due to the velocity of satellite), works in synchronous way, and it becomes neutral. So that satellite keeps revolving around at fixed position. So, such orbit is known as &lt;strong&gt;Geostationary orbit&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="760" LastEditorUserId="760" LastEditDate="2014-02-17T08:43:16.737" LastActivityDate="2014-02-17T08:43:16.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1758" PostTypeId="2" ParentId="1721" CreationDate="2014-02-16T09:48:09.917" Score="0" Body="&lt;p&gt;The simple answer to your question is, simply, yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All objects in the inner solar system is generally assumed to have been impacted by the same population of impactors which is mostly comprised of asteroids, and perhaps up to 10% comets.  The outer solar system likely has a much larger percentage of cometary impactors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the five main inner solar system terrestrial bodies, the Moon and Mercury have the most preserved crater record, idealized by a death of volcanism early in their histories, no atmosphere to speak of, and therefore craters are best preserved there.  Mars is second-best (some atmosphere, localized volcanism over the last few billion years, and water in the first ~billion years).  Venus is a mess for cratering (entire planet resurfaced ~400-800 million years ago plus a thick atmosphere that prevents craters &amp;lt;3-5 km from forming), while Earth has the most modified crater record.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the moon next door, it is our best historical record for what also likely hit Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The factors that will alter the lunar &lt;em&gt;crater&lt;/em&gt; population from what would have formed on Earth are primarily gravity, surface type, and atmosphere.  &lt;strong&gt;Atmosphere&lt;/strong&gt; will not only screen out the smallest impactors (and hence not make craters), but it will also fragment less competent objects, changing what could have made a single large impact on the moon into something that will more likely make numerous smaller, clustered craters on Earth (from the surviving fragments).  A larger &lt;strong&gt;surface gravity&lt;/strong&gt; will tend to decrease the size of the final crater caused by a given impactor, but the dependence is small (difference of around 35% between Earth and the Moon despite the factor of 6 difference in gravity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Surface type&lt;/strong&gt; is much less well understood in the cratering community.  Earth has stronger, denser crust than the Moon, but it's also ~70% covered by water -- the surface of Venus is like being under 1 km of water on Earth, so any ocean &gt;1 km deep is going to prevent craters $\lesssim 3-5$ km from forming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These all combine for the much longer and more complicated answer of, &quot;yes, but it's hard to figure out.&quot;  There are lots of knobs in the models that we still don't know how to correctly turn, but we can, to first-order, use the lunar crater population and rate to estimate the population of craters that should have formed on Earth.  At the time of writing this, there are &lt;a href=&quot;http://www.passc.net/EarthImpactDatabase/&quot; rel=&quot;nofollow&quot;&gt;184 confirmed terrestrial impact craters&lt;/a&gt;, which is certainly a tiny fraction of a percent the number that have formed over Earth's history.&lt;/p&gt;&#xA;" OwnerUserId="934" LastActivityDate="2014-02-16T09:48:09.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1760" PostTypeId="2" ParentId="1754" CreationDate="2014-02-16T12:08:49.767" Score="3" Body="&lt;p&gt;One of a million possible answers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An advanced civilization wouldn't do this without need.&#xA;If necessary, they would use &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitation&quot; rel=&quot;nofollow&quot;&gt;gravity&lt;/a&gt; instead of &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetism&quot; rel=&quot;nofollow&quot;&gt;magnetism&lt;/a&gt;, since magnetism acts differently on different kind of matter, gravity is much better suited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They would exploit the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chaos_theory&quot; rel=&quot;nofollow&quot;&gt;chaotic&lt;/a&gt; behaviour of &lt;a href=&quot;http://en.wikipedia.org/wiki/N-body_problem&quot; rel=&quot;nofollow&quot;&gt;n-body systems&lt;/a&gt; like planetary systems mainly governed by gravity, and find out which distortion of minimal intervention would return the desired result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They would know the need long time in advance, hence be able to act, when there is still enough time to get the desired result with a reasonably small effort.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-16T12:08:49.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1762" PostTypeId="2" ParentId="1722" CreationDate="2014-02-17T00:48:02.903" Score="1" Body="&lt;p&gt;Assuming that the glasses are safe for direct viewing of the sun as claimed, and assuming you don't damage them, then yes, they should be safe for viewing the sun daily in the manner you suggest. There would not be an accumulative harm effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, assuming the glasses are safe for direct viewing, you'll probably be at more risk for the rest of the day when you are walking around with naked eyes.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-02-17T00:48:02.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1764" PostTypeId="2" ParentId="1646" CreationDate="2014-02-17T01:27:59.007" Score="2" Body="&lt;p&gt;For higher magnifications, you are going to get into the territory where a tripod becomes a must. I mention this because you didn't say anything about a tripod, so the implication may not have crossed your mind. Unfortunately, the cost of a tripod will eat into your budget. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am inclined to think that your best option would be a telescope. (Unless you also want to view terrestrial objects, then a cheap spotting scope might be what you want.) Maybe your best bet, to get a good one, would be to ask around on forums like cloudynights or iceinspace that have some support from astronomers local to you, let them know you are looking to buy, and maybe you can get one second-hand off someone who is trading up. You might get a better telescope for your money than trying to buy new.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a smaller length telescope, you may be put off something like a newtonian reflector or another simple 2-mirror reflector. Still, try one out (maybe you can attend a local star-party and see some up close, even if there isn't a telescope store handy). A 114mm newtonian on a tripod would meet almost all your requirements including budget, even purchased new. It will be a bit less than 2' long; slightly bigger newtonians like 130mm start getting longer than 2'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, what might fit the bill is a small cassegrain-style reflector - these usually also have a corrector lens, making them a catadioptric type telescope - such as a Maksutov-Cassegrain. The extra manufacturing cost of making the lenses pushes up the price of these well past the simple reflecting telescopes, however. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refracting telescopes with an objective of 100+mm are probably in your &quot;too big&quot; basket, and you're also not likely to find one with that budget. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You use different eyepieces to provide different magnifications using the same telescope. If you buy a telescope as a kit, it will come with at least one eyepiece, and it's likely the magnification provided will be more than 10x (divide the focal length of the telescope by the focal length of the eyepiece to find out).&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-02-17T01:27:59.007" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1765" PostTypeId="2" ParentId="1650" CreationDate="2014-02-17T01:56:50.270" Score="2" Body="&lt;p&gt;Any causal claims or associations relating to language use are never going to be able to be proven, so the best answer to your overall question is: &quot;nobody knows&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use Google's ngram viewer to visualise the appearance of a specific term or terms over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps astronomers generally have been better as a field in making their research more approachable by not pitching their discoveries in superior-sounding high-falutin sesquipedal pomposity. But it isn't universal; for every approachable 'Hanny's Voorwerp' there is an ancient-language-originating label like 'heliosphere'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geologists also use non-latin terms like &quot;slip&quot; and &quot;fault&quot; and &quot;hanging wall&quot;. Is that to make geology more approachable for mass-media public reach-out PR?&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-02-17T01:56:50.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1766" PostTypeId="2" ParentId="1664" CreationDate="2014-02-17T06:42:35.227" Score="2" Body="&lt;p&gt;The info you reported are not opposite.&#xA;They refers to different scales on structure formation, and are both true.&#xA;As you said, ΛCDM model is very useful to explain &lt;em&gt;large&lt;/em&gt; scale structures, as galaxies, clusters, and larger.&#xA;This means that the model fits really well to the observables, first of all the CMB map and its anisotropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the model can NOT explain is the original formation of the structures, at the observed &lt;em&gt;rate&lt;/em&gt; (&lt;a href=&quot;http://burro.astr.cwru.edu/Academics/Astr222/Cosmo/Structure/primordial.html&quot; rel=&quot;nofollow&quot;&gt;take a look at this&lt;/a&gt;).&#xA;We know the age of universe (within this model), and we know the age of the observed galaxies. Then it results that galaxies have formed much faster then in a &quot;simple&quot; ΛCDM scenario. We can not explain how the primordial structures started to collapse and to grow in what now we see as galaxies.&#xA;To account for this observation we need to invoke &lt;a href=&quot;http://en.wikipedia.org/wiki/Primordial_fluctuations&quot; rel=&quot;nofollow&quot;&gt;dark matter primordial fluctuations&lt;/a&gt;. But this is an assumption, and does not give us the original spectrum of the fluctuations, neither it explain how it is formed. And this is a parameter that the ΛCDM model can not account for, at least at this stage.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastActivityDate="2014-02-17T06:42:35.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1767" PostTypeId="1" AcceptedAnswerId="1777" CreationDate="2014-02-17T07:12:11.813" Score="20" ViewCount="35862" Body="&lt;p&gt;I think I am not alone who saw videos about that we (humans) are made of same atoms which someday were in stars. In other words, some atoms in our bodies are from stars which exploded billions of years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if it is indeed true. I mean human's life begins when sperm cell fertilize egg cell. Now does that sperm cell or egg cell indeed contains some of the exact atoms from those stars? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know little bit strange question, but would be interesting to hear if it is indeed true, that atoms in our body are same which someday were in stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you are wondering that's the video I am talking about: &lt;a href=&quot;http://www.youtube.com/watch?v=9D05ej8u-gU&quot;&gt;http://www.youtube.com/watch?v=9D05ej8u-gU&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1035" LastEditorUserId="1035" LastEditDate="2014-02-17T10:38:02.560" LastActivityDate="2018-12-16T21:26:27.727" Title="Did atoms in human body indeed come from stars?" Tags="&lt;star&gt;" AnswerCount="9" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1768" PostTypeId="2" ParentId="1746" CreationDate="2014-02-17T08:51:10.810" Score="2" Body="&lt;p&gt;Yes, everything that is in this universe has to say Good bye to its existence. That's what Black hole also has to follow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stephen Hawking gave this concept and it is known as Hawking Radiation, in which the Black Hole radiates energy because of the quantum mechanical process. But they donot stay there forever to consume more matter nor do they burst up. They just simply evaporate slowly; a concept of the Hawking's theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hawking_radiation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://curious.astro.cornell.edu/question.php?number=163&quot; rel=&quot;nofollow&quot;&gt;http://curious.astro.cornell.edu/question.php?number=163&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="79" LastActivityDate="2014-02-17T08:51:10.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1769" PostTypeId="1" AcceptedAnswerId="1771" CreationDate="2014-02-17T09:33:55.217" Score="3" ViewCount="146" Body="&lt;p&gt;I know it is possible to infer the mass of a supermassive black hole (SMBH) by many methods, i.e., stellar orbits for out Galaxy, Iron line profile from the accretion disk, and probably other methods (perhaps from the spectrum of the radiation disk itself, that can be related with the central mass, if supposed thermal in origin).&#xA;What I don't know is: how can we infer the mass of SMBH in galaxies that are not active anymore?&lt;/p&gt;&#xA;" OwnerUserId="1038" LastActivityDate="2014-02-17T12:50:02.390" Title="How can we infer the mass of SMBH in galaxies that are not active anymore?" Tags="&lt;supermassive-black-hole&gt;&lt;galaxy-center&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1770" PostTypeId="2" ParentId="1767" CreationDate="2014-02-17T10:05:09.093" Score="-3" Body="&lt;p&gt;I might not be that much conceptual to you, but do you believe that stars are made up of Hydrogen (and Helium too). Hydrogen is a part of all the organic compounds. We humans are made up of billions of organic compounds, which in turn do contain Carbon, Hydrogen and Oxygen as their basic element and other compounds such as Sulphur, Phosphorus and other fats. And I guess a biology student would tell you more clearly what a sperm contains and what and how do these elements ressembles to those who are in Stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a star collapses, it liberates out elements such as Carbon, Iron etc. Which maybe you're talking about. So that sperm consists of the same elements that the Star had at the time of its death (well not death you might know a better word for the star's ending period). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this doesn't make sense that these atoms are really the exact same or the actuall ones that were in the Stars many years ago. No such theory has been provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does Star contain and what it liberates: &lt;a href=&quot;http://answers.yahoo.com/question/index?qid=20080908195830AA5Iheb&quot; rel=&quot;nofollow&quot;&gt;http://answers.yahoo.com/question/index?qid=20080908195830AA5Iheb&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="79" LastActivityDate="2014-02-17T10:05:09.093" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1771" PostTypeId="2" ParentId="1769" CreationDate="2014-02-17T10:09:39.850" Score="1" Body="&lt;p&gt;I can think of two methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both rely on the dynamics of material surrounding the SMBH, which is affected up to a distance of the order of the &quot;sphere of influence&quot;. This is the region where the BH dominates the dynamics as compared to the enclosed mass of the galaxy. The sphere of influence is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ R_{\mathrm{infl}} \equiv \frac{G M_{\mathrm{SMBH}}}{\sigma_{\mathrm{bulge}}^2} &amp;gt;&amp;gt; R_{\mathrm{horizon}} \approx R_{\mathrm{Schwartzschild}} \equiv \frac{G M_{\mathrm{SMBH}}}{c^2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While typically $\sigma_{\mathrm{bulge}} \approx 250 km s^{-1}$, it is well known that $c \approx 300000 km s^{-1}$. This means that the influence of a BH can be felt much further away than its event horizon, which is where the accretion takes place. In fact the ratio between the two distances is about 1000000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exploiting this fact, astronomers have been using two methods to probe extragalactic*, quiescent SMBH:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The first method is to observe CO lines (radio astronomy) to trace gas circling the BH. The gas does not need to be near the event horizon, which is much smaller than the sphere of influence. In fact CO observations rely on the gas being relatively dense but cold. Essentially the speed at which the CO gas will rotate is a (quadrature) sum of the declining component due to the stellar mass, plus the Keplerian component due to SMBH mass.&lt;/li&gt;&#xA;&lt;li&gt;The second method is completely analogous, but relies on measuring the &lt;em&gt;unresolved&lt;/em&gt; kinematics of the stars surrounding the SMBH. This can be done in various bands, but most authors use visible line absorptions to measure the velocity and velocity dispersion (and other moments) of the stars in the regions surrounding the BH. If the kinematics cannot be explained without including a point mass in the middle, then you are done.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.google.com/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=4&amp;amp;cad=rja&amp;amp;ved=0CD8QFjAD&amp;amp;url=http://mnras.oxfordjournals.org/content/370/2/559.full.pdf&amp;amp;ei=XgQCU73mDsmX0AW1toGwAg&amp;amp;usg=AFQjCNFQMjYGMM9Gbz4XTZ6cF8sTiGob7g&amp;amp;bvm=bv.61535280,d.d2k&quot; rel=&quot;nofollow&quot;&gt;this work&lt;/a&gt;, for a comparison of the two methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*(extragalactic means outside of &lt;em&gt;our own&lt;/em&gt; galaxy)&lt;/p&gt;&#xA;" OwnerUserId="198" LastEditorUserId="198" LastEditDate="2014-02-17T12:50:02.390" LastActivityDate="2014-02-17T12:50:02.390" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1772" PostTypeId="2" ParentId="1751" CreationDate="2014-02-17T10:20:51.427" Score="4" Body="&lt;p&gt;The speed of a satellite in orbit is directly linked to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_speed&quot; rel=&quot;nofollow&quot;&gt;radius of that orbit&lt;/a&gt;:&lt;br&gt;&#xA;$V=\frac{2*\pi*a}{T}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;V is the orbital speed, a is the radius and T is the orbital period. This means that for any radius, there is only one possible orbital speed and vice versa. So there is only one radius where the period is 24 hours.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really want a satellite to orbit at 70,000 km and still be geostationary, you'll have to use thrusters constantly to provide an extra force towards the Earth. This uses far too much fuel to be economical. &lt;/p&gt;&#xA;" OwnerUserId="361" LastActivityDate="2014-02-17T10:20:51.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1773" PostTypeId="2" ParentId="1751" CreationDate="2014-02-17T10:41:53.327" Score="1" Body="&lt;p&gt;@Hobbes put the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V={2\pi a\over T}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you substitute on it $T$ for 1 sidereal day you'll have a one-to-one relation between $V$ and $a$. This seems to imply that you can choose any $a$ and use the corresponding $V$, but this is not true since it does not take into account which force are you using.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula above is valid of any object in a circular motion, like a stone on a sling, or a car in a loop. It is also valid for satellites around planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is another formula stating a relation among $a$ and $T$ when the force is that of Gravity, so you are not free to chose $a$ anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It happens that centripetal acceleration (that of the sling, the Normal force on the loop, or gravity for the satellite) is the Universal Gravity in our case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_c={GM\over r^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but it being the centripetal acceleration means&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_c=\omega^2 r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega^2 r={GM\over r^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega^2 r^3=GM$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega$ is $v\over r$ so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v^2 r=GM$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives you a fixed relation between $v$ and $r$ for any planet with mass $M$.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-02-17T10:41:53.327" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1774" PostTypeId="2" ParentId="1767" CreationDate="2014-02-17T10:49:22.983" Score="-2" Body="&lt;p&gt;Think about it this way, there are two options about how life began on earth: abiogenesis(life began on earth) and panspermia (life began somewhere else and from something like a meteor strike it continued to evolve on earth),at least those are the ones the are more dominant than other theories. Either way if life began to evolve on earth from matter that was on earth than it is suitable that whatever life evolved here would contain the same materials as earth, and earth originated from other starts, dust and debris that was around 4.5 billion years ago same thing for the second theory I mentioned. So I would suggest reading more on abiogenesis and panspermia it may satisfy your curiousity and answer your question. &lt;/p&gt;&#xA;" OwnerUserId="1049" LastActivityDate="2014-02-17T10:49:22.983" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1775" PostTypeId="2" ParentId="1767" CreationDate="2014-02-17T10:54:36.990" Score="0" Body="&lt;p&gt;Your body has hydrogen and heavier elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most hydrogen in your body (really most of it) is primitive hydrogen from the origins of the Universe. Same would happen for Helium but we don't have it (almost none) on our bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all other elements, yes, they come (really most of them) from a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sequence is approximately as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the Universe was young, our Galaxy was young: it was a cloud made of Hydrogen and Helium. Then, some stars (called Population III stars) were born and started burning hydrogen into helium, and on the latter stages of their lives they burned up helium into weighter elements like carbon, nitrogen, up to very small amounts of iron (the most stable element), and beyond that up to uranium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on their mass, of course. The smaller of these stars may be still around us, and the bigger ones exploded, sending these new elements to the galactic (a.k.a. interstellar) medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, from the now enriched medium, new stars were born (called Pupulation II stars). These ones had of course a lot of hydrogen and helium, but they had also some of the heavier elements. In turn, some of them are still observable (smaller ones, which last longer) and some of them exploded (bigger ones, which burn faster).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This second wave of stellar explosions enriched more the interstellar medium (gallactic medium) so a new generation of stars could be born. These are known as Pupullation I stars. Our Sun is one of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But not all the mass of the cloud which formed our Sun went into the Sun. Some if it made up the planets, and thus, ourselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the atoms on our bodies come from the pre-planetary cloud, which consisted in original hydrogen enriched by Popullation III and Popullation II explosions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note: due to chemical reactions, electrons in the atoms do not need to be the same they were while they were ejected by the stars, but the nucleus are.&lt;/p&gt;&#xA;" OwnerUserId="473" LastEditorUserId="473" LastEditDate="2014-02-17T11:00:10.357" LastActivityDate="2014-02-17T11:00:10.357" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1776" PostTypeId="2" ParentId="1767" CreationDate="2014-02-17T11:09:32.010" Score="4" Body="&lt;p&gt;Almost all hydrogen nuclei (protons), some helium atomic nuclei and traces of lithium nuclei are thought to have formed early in the universe, after the big bang. Almost all other atomic nuclei are thought to have formed in stars or have decayed from atomic nuclei, which have formed in stars. A minor fraction forms by high-energy collisions with cosmic rays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electrons of the hull of atoms in parts formed during the big bang, part of them come into existence, when neutrons decay to protons. These neutrons may have been free neutrons, or neutrons bound in instable atomic nuclei.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence our body doesn't contain the exact same atoms as they formed in stars. But without stars most of the atoms besides hydrogen in our body wouldn't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our body contains many of the exact same atomic nuclei, as they formed in stars, not the exact same atoms/ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be a little more precise: Our body doesn't contain many free atoms, but mainly molecules and ions.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-02-17T11:26:09.157" LastActivityDate="2014-02-17T11:26:09.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1777" PostTypeId="2" ParentId="1767" CreationDate="2014-02-17T11:16:27.703" Score="7" Body="&lt;p&gt;When the first atoms came into being in the early universe they were mainly hydrogen (the smallest atoms there are) and some helium. All over the universe those atoms lumped together under gravity until the pressure and temperature became so high that the hydrogen atoms fused together to form heavier elements. The reaction is nuclear fusion, and it's the engine of all stars. First hydrogen fuses to form helium, and then in a cascade helium atoms fuse to form heavier elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many stars die as a supernova, without doubt the most violent explosions in the universe. The supernova which was just a single star becomes as bright as the complete galaxy it's part of. Remember that such a galaxy typically consists of 100 billion stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the supernova explosion all the elements from helium to the most heavy elements are thrown into space. Later they will coalesce to form planets around new stars. So indeed, everything the earth consists of comes from such an exploding star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the next step is life. A single cell consists mainly of carbon, hydrogen, oxygen and nitrogen, all ultimately coming from the earth. For instance a plant will take these elements from the soil and the air, and we animals get it from plants. So the elements from the soil, which came from stars, ultimately ends up in each of our cells.&lt;/p&gt;&#xA;" OwnerUserId="1050" LastActivityDate="2014-02-17T11:16:27.703" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="1778" PostTypeId="2" ParentId="1746" CreationDate="2014-02-17T11:51:13.527" Score="8" Body="&lt;p&gt;Like &lt;em&gt;Afzaal Ahmad&lt;/em&gt; said black holes &lt;em&gt;evaporate&lt;/em&gt; through &lt;strong&gt;Hawking radiation&lt;/strong&gt;, proposed by Stephen Hawking in 1974. Black holes have a temperature just a little bit above absolute zero; for a black hole with the weight of the Earth it's about 0.02 K. (Such a black hole would be about 2.5 cm in diameter.) And black bodies radiate energy mainly in the form of photons as their temperature is larger, so the radiation for a black hole isn't very high: for the Earth-weigth black hole it's about $10^{-17}$ W. Larger black holes are even colder, and will radiate even less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how can the photon escape the black hole's gravity when we know light can't escape it? It happens through quantum &lt;strong&gt;tunneling&lt;/strong&gt;. A photon is not just a particle, it's also a wave, and quantum mechanics shows a &lt;strong&gt;probability distribution&lt;/strong&gt; for the location of a particle. The black hole's event horizon is a barrier for the photon, but it could exist beyond that barrier. Under these conditions the photon can tunnel through the event horizon and suddenly find itself outside of the black hole. (Actually tunneling &lt;em&gt;through&lt;/em&gt; is not a good choice of words; the photon doesn't move through the barrier, it just suddenly is at the other side.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So energy &lt;em&gt;can&lt;/em&gt; and &lt;em&gt;does&lt;/em&gt; radiate away from the black hole, but this happens so slowly that it may take a trillion times the age of the universe before the whole hole :-) has evaporated.&lt;/p&gt;&#xA;" OwnerUserId="1050" LastActivityDate="2014-02-17T11:51:13.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1779" PostTypeId="1" AcceptedAnswerId="1781" CreationDate="2014-02-17T12:07:56.687" Score="4" ViewCount="34785" Body="&lt;p&gt;Why do all objects burn when they enter our atmosphere? is this because of our atmosphere composition? and does this happen on other planets as well?&lt;/p&gt;&#xA;" OwnerUserId="1049" LastEditorUserId="1049" LastEditDate="2014-02-17T12:45:05.280" LastActivityDate="2017-05-02T21:13:05.900" Title="Why do objects burn when they enter earth's atmosphere?" Tags="&lt;planetary-atmosphere&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1780" PostTypeId="1" CreationDate="2014-02-17T12:46:30.020" Score="2" ViewCount="239" Body="&lt;p&gt;If the big bang expanded the universe from a singularity point, that means that there was matter before, so where and how did matter form for the first time if it's even possible to know such a thing?&lt;/p&gt;&#xA;" OwnerUserId="1049" LastActivityDate="2014-02-17T13:03:21.930" Title="if the big bang only expanded the universe when and how did it originate?" Tags="&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2014-02-18T21:11:01.603" ContentLicense="CC BY-SA 3.0" />
  <row Id="1781" PostTypeId="2" ParentId="1779" CreationDate="2014-02-17T12:54:01.220" Score="8" Body="&lt;p&gt;You'll often hear that it's because of friction, but that's often not the main factor. For larger objects it's more likely the pressure they create. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases the reason is the enormous speed, often tens of kilometers per second. When a larger object enters the atmosphere at these speed the air in front of it gets compressed so much that it becomes extremely hot. (Think of pumping up a tire; you're also compressing air and you can feel the valve becoming hot.) The compressed air will often disintegrate the object in the air, and then the debris may burn because of the heat. This is exactly what happened to the asteroid above Russia last year: it exploded with an enormous flash in the air, and left little traces on the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This happens on other planets as well, if they have a sufficiently dense atmosphere. In 1994 the comet Shoemaker-Levy crashed into Jupiter. It disintegrated before entering Jupiter's atmosphere due to the strong gravitation, but when the fragments entered the atmosphere they could easily be seen lighting up as they burned up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt;&lt;br&gt;&#xA;Remember the Space Shuttle? It had heat resistant tiles on the bottom of the craft to protect it from burning when it entered the atmosphere, even though its speed is only a fraction of a meteorite's speed when that enters the atmosphere.&lt;br&gt;&#xA;During the last launch of the Space Shuttle Columbia some material from the external fuel tank damaged this heat shield, and upon re-entry the heat and the highly pressurized air under the craft could enter it, causing the craft to disintegrate and kill all crew.&lt;/p&gt;&#xA;" OwnerUserId="1050" LastEditorUserId="1050" LastEditDate="2014-02-17T16:34:36.143" LastActivityDate="2014-02-17T16:34:36.143" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1782" PostTypeId="2" ParentId="1780" CreationDate="2014-02-17T13:03:21.930" Score="3" Body="&lt;p&gt;&quot;before&quot; is not the right word. It suggests that there was an arrow of time which allowed to trace events. But time was created only with the big bang, there was no before. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't know where the matter came from, or what caused the Big Bang. One hypothesis suggests that it was caused by the collision of &lt;em&gt;branes&lt;/em&gt; in other universes, but there's no way to verify this, so in a strict sense it isn't science. Many researchers think we'll never have an answer this.&lt;/p&gt;&#xA;" OwnerUserId="1050" LastActivityDate="2014-02-17T13:03:21.930" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1783" PostTypeId="2" ParentId="646" CreationDate="2014-02-17T15:26:07.100" Score="5" Body="&lt;p&gt;All light we see is from the past. The light from a light bulb at 3 meter distance arrives 10 ns after it left the bulb in your eye. For short distances this delay is negligible (10 ns is 10 billionths of a second), but at astronomical scale it becomes significant. Light from the Sun takes 8 minutes and 20 seconds to reach the Earth, so when we see the Sun it's the Sun like it was 8 minutes ago. If the Sun would suddenly die we wouldn't notice for 8 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same goes for other stars in our Galaxy. The light from a star at 4 light years takes 4 years to reach us; it's the definition of a light year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could make the following comparison: suppose there's a town at 100 &lt;em&gt;car years&lt;/em&gt; from where you live. That means it takes a car 100 years to reach you. When a car from that town reaches you today it left in 1914. It won't be a 2010 sedan, but a Ford T. As the car arrives you're looking 100 years in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looking into history is very convenient for cosmologists. You want to know what galaxies looked like 13.5 billion years ago, when the Universe was still young? Well, look for light that's been underway for that time. It left the galaxy under study 13.5 billion years ago and shows you what that galaxy looked like at the time. It doesn't tell you anything about the current state of it. It may have collided with another galaxy or absorbed by a black hole. There's no way of knowing other than waiting another 13.5 billion years, until the light emitted now will reach us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another interesting thing to observe from that far past is the Cosmic Microwave Background Radiation (CMB). It's the radiation from the Big Bang, which has been underway for 13.8 billion years. Of course today the Big Bang is history, but thanks to the &quot;limited&quot; light speed this history is continuously underway to us.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt;&lt;br&gt;&#xA;Enter relativity. So we say the light from Proxima Centauri has been underway for 4.2 years, but only &lt;em&gt;from our point of view&lt;/em&gt;. As objects go nearer the speed of light their time slows down, and ultimately when you would reach light speed time would stop completely. Now photons travel at the speed of light, so for them time is at a standstill. &lt;em&gt;From the photon's point of view&lt;/em&gt; it travels the whole distance from Proxima Centauri to Earth &lt;strong&gt;instantaneously&lt;/strong&gt;: it arrives at Earth at the same time it leaves Proxima Centauri! (You can't do this with objects which have mass.)&lt;/p&gt;&#xA;" OwnerUserId="1050" LastEditorUserId="1050" LastEditDate="2014-02-24T18:16:14.557" LastActivityDate="2014-02-24T18:16:14.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1784" PostTypeId="2" ParentId="1779" CreationDate="2014-02-17T17:15:03.707" Score="0" Body="&lt;p&gt;An object hitting the Earth at high speed has a lot of &lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic_energy&quot; rel=&quot;nofollow&quot;&gt;kinetic energy&lt;/a&gt;, \begin{align}&#xA;  \frac{1}{2}mv^2&#xA;\end{align}&#xA;where &lt;em&gt;m&lt;/em&gt; is the object's mass and &lt;em&gt;v&lt;/em&gt; is it's velocity. If the velocity goes from lots to nothing relative to the Earth, that kinetic energy has to go somewhere. That somewhere ends up as heat.&lt;/p&gt;&#xA;" OwnerUserId="562" LastActivityDate="2014-02-17T17:15:03.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1786" PostTypeId="1" AcceptedAnswerId="1795" CreationDate="2014-02-17T22:45:47.827" Score="6" ViewCount="233" Body="&lt;p&gt;What sort of algorithm is used to calculate the relative quality of an image, such as is performed by PIPP, Registax, or Avistack? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such applications take planetary or solar images and prior to stacking and sharpening, they can optionally select the best images. When looking at images to see which have been selected by the algorithms, I sometimes see images that I wouldn't have selected. This has piqued my curiosity as to what methodology is employed in ranking or grading the images. With a bit of a primer to understand the concepts, I would like to put some thought into how such algorithms might be improved...&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-02-19T01:05:26.020" Title="What method is used to calculate the 'quality' of a solar/planetary image?" Tags="&lt;the-moon&gt;&lt;planet&gt;&lt;the-sun&gt;&lt;photography&gt;&lt;algorithm&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1787" PostTypeId="1" CreationDate="2014-02-18T05:36:18.583" Score="2" ViewCount="219" Body="&lt;p&gt;I have a doubt in Earth's curvature. This question may be dumb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much we need to travel on the Earth to feel the curvature of Earth? Hypothetical case: for example, we are building a vehicle. How much is the length of that vehicle to feel the curvature of the Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more question, what is about the speed of the Earth. Again a hypothetical case. If we are travelling in aeroplane from one country to other, whether we can take Earth's rotational speed (rotating around it's own axis), and aeroplane speeds as comparables and find the resultant speed?&lt;/p&gt;&#xA;" OwnerUserId="1054" LastEditorUserId="1891" LastEditDate="2019-03-16T16:49:13.550" LastActivityDate="2019-03-16T16:49:13.550" Title="Regarding the curvature of the Earth" Tags="&lt;earth&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2014-02-19T15:15:14.793" ContentLicense="CC BY-SA 4.0" />
  <row Id="1788" PostTypeId="2" ParentId="1787" CreationDate="2014-02-18T08:22:50.160" Score="6" Body="&lt;p&gt;About curvature, you can see it. Just look at a ship some hundred meters away: you'll notice if you are careful that you can not see where the ship touches water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see at &lt;a href=&quot;http://mathcentral.uregina.ca/qq/database/QQ.09.97/dyck2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://mathcentral.uregina.ca/qq/database/QQ.09.97/dyck2.html&lt;/a&gt; that the curvature is 8 inches per mile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, about rotation, planes do actually take advantage of earth rotation, but not directly. This was answered in &lt;a href=&quot;https://physics.stackexchange.com/questions/16390/does-the-rotation-of-the-earth-dramatically-affect-airplane-flight-time&quot;&gt;https://physics.stackexchange.com/questions/16390/does-the-rotation-of-the-earth-dramatically-affect-airplane-flight-time&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="473" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2014-02-18T08:22:50.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1789" PostTypeId="2" ParentId="1787" CreationDate="2014-02-18T13:14:47.047" Score="2" Body="&lt;p&gt;Earth's rotation doesn't help airplanes much: the air is almost stationary with respect to the earth and airplanes move relative to the air. Suppose a hot air balloon. Air is dragged along with the rotation of earth, so the balloon will almost be stationary. (Ok, it moves, but only slowly.) So whether the plane travels east or west for 1000 km, it has to travel 1000 km against air which is 900 km/h slower.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a meteorological effect that planes can use, though. The &lt;strong&gt;jet stream&lt;/strong&gt; is a channel of very high speed winds at high altitude.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mpwh5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planes traveling east, for instance from the US west coast to the east coast, or from America to Europe, take advantage of this jet stream to save fuel. Note that the jet stream isn't anywhere as fast as the earth's rotation. At 50 degrees north the earth's rotation is about 1000 km/h, while the jet stream is about 200 km/h.&lt;/p&gt;&#xA;" OwnerUserId="1050" LastActivityDate="2014-02-18T13:14:47.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1792" PostTypeId="1" AcceptedAnswerId="1793" CreationDate="2014-02-18T20:54:05.073" Score="5" ViewCount="939" Body="&lt;p&gt;First off, I apologize if I am breaking any site-etiquitte. Secondly, this is probably extremely elementary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a lab, I was given the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When we are sitting in a planetarium, where are we located on(or in)&#xA;  the &quot;celestial sphere&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To me, this seems like a silly question - because the answer is really dependent on where the operator places you. Is this assumption wrong? Is there a particular place in the celestial sphere that you would be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also thought that it might just be the very center of the celestial sphere.&lt;/p&gt;&#xA;" OwnerUserId="1059" LastActivityDate="2014-02-19T00:02:45.480" Title="Location on (or in) the celestial sphere" Tags="&lt;amateur-observing&gt;&lt;fundamental-astronomy&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1793" PostTypeId="2" ParentId="1792" CreationDate="2014-02-19T00:00:20.840" Score="4" Body="&lt;p&gt;Well, at least you did some thinking and proposed a couple of personal thoughts instead of just asking for the answer to a homework question. Both observations you make are pertinent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, a celestial sphere doesn't actually exist. It is an imaginary concept, a simplification based upon &lt;em&gt;our perspective&lt;/em&gt; of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A planetarium, whether it is an old-fashioned one that depicts an earth-bound perspective of the universe, or a modern digital projection system that can &quot;fly&quot; you through space to view the universe from a different place, is projecting a perspective of the view of the universe from a particular place, that is, you see the &quot;celestial sphere&quot; as it would appear from that place. In the case of a view from elsewhere in space, it would be a different celestial sphere from &lt;em&gt;our&lt;/em&gt; celestial sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, you would be at &quot;the centre&quot; of the celestial sphere. Maybe not the &quot;very&quot; centre in the case of old fashioned projectors with fixed star fields, as they wouldn't account for the 150 million kilometre imprecision caused by the movement of the earth around the sun. But, as this is an insignificant distance in astronomical terms, that doesn't matter a great deal, so &quot;the centre&quot; is probably a sufficient description.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-02-19T00:00:20.840" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1794" PostTypeId="2" ParentId="1792" CreationDate="2014-02-19T00:02:45.480" Score="1" Body="&lt;p&gt;If we follow &lt;a href=&quot;http://en.wikipedia.org/wiki/Celestial_sphere&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia definition of &quot;celestial sphere&quot;&lt;/a&gt;, &quot;the celestial sphere can be considered to be infinite in radius. This means any point within it, including that occupied by the observer, can be considered the center.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the question could be answered, as you presume, by &lt;strong&gt;in the center of the celestial sphere&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although, of course, if we aren't located in the center of the planetarium, the celestial sphere will look distorted to us.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-19T00:02:45.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1795" PostTypeId="2" ParentId="1786" CreationDate="2014-02-19T01:05:26.020" Score="3" Body="&lt;p&gt;A little further research, and I found out that the basic principle, from which several variations derive, is to take the sum of the squares of the difference between adjacent pixels, to get a score. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principle is that a higher quality picture has a higher probability that there will be significant differences in adjacent pixel values, i.e. that there can be a significant variation from pixel to pixel, whereas a lower quality picture will have similar valued adjacent pixels, as they have &quot;blurred&quot; together. Therefore, when taking the square of the difference of adjacent pixels and summing them together, higher quality pictures will show a significantly higher score, or summation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to demonstrate this with a simplified example below, but I'm interpreting the meaning of &quot;adjacent pixels&quot; in a particular way, and I am hoping that if I am incorrect, someone will tell me and I can modify this answer appropriately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine we are imaging an object that exactly fills the frame of our small 16-pixel camera. In the middle of this object is a very dark square, that coincidentally happens to exactly fill the central 4 pixels of our camera. If the seeing for this object is perfect, then the camera will be exposed with a value of 255 for each of the light pixels, and 0 for each of the dark pixels.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ldinT.png&quot; alt=&quot;A perfect image of our target object&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how the differences between adjacent pixels would be calculated (I'm assuming only horizontal and vertical differences, not diagonal):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Enf42.png&quot; alt=&quot;Differences between adjacent pixels&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, movement of the camera or poor seeing resulted in the light being spread more between the pixels, the calculations of the differences between adjacent pixels would be different:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/6luQG.png&quot; alt=&quot;Differences between adjacent pixels when the image is blurred&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The score, that is the sum of the squares of the differences, for the first (perfect) image is &lt;strong&gt;520200&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The score for the blurred image is &lt;strong&gt;304200&lt;/strong&gt;. Therefore the sum of the squares of the differences between adjacent pixels suggests that the first picture is the better one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This strategy isn't perfect, obviously. The presumption is that better images will exhibit a better score due to larger differences between adjacent pixels, but this is an assumption. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The squaring of the difference is intended to provide greater weight to larger differences, but I came across a comment on the PIPP support website that says that PIPP has modified the basic algorithm to give still greater weight to sharper contrast changes over lower contrast changes - so maybe squaring the differences isn't enough sometimes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently this quality measurement strategy doesn't work when the object is small in size on the sensor, and may be bright or even over-exposed. The above methods may select blurred and smeared frames over frames that are actually better. In this case, a strategy that looks at the histogram of values for the frame and selects the one with the higher peak, may produce a better selection of appropriate 'best quality' frames. It assumes that the histogram will have a higher count of frames sharing a pixel value across a smaller range, rather than spreading the light out across a wider range of pixel values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative quality measurement strategy simply adds up the values of all the pixels in an image. A higher score will indicate more light got through to the sensor, which implies that less cloud (for instance) got in the way.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-02-19T01:05:26.020" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1796" PostTypeId="1" CreationDate="2014-02-19T12:09:50.697" Score="2" ViewCount="780" Body="&lt;p&gt;From what I know the sun (or any other star) nuclear fission keeps it from collapsing under its own gravity until the fuel runs out, if that's so why doesn't the earth surface like the tectonic plates collapse to the earth's core? is it because of similar processes that happen in the earth's outer core (nuclear fission)? and if so what would happen if they would once stop?&lt;/p&gt;&#xA;" OwnerUserId="1049" LastEditorUserId="1062" LastEditDate="2014-02-19T12:20:17.623" LastActivityDate="2014-02-19T15:13:09.757" Title="Why doesn't the earth's surface collapse onto itself?" Tags="&lt;earth&gt;&lt;gravity&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-02-19T15:10:54.857" ContentLicense="CC BY-SA 3.0" />
  <row Id="1797" PostTypeId="1" AcceptedAnswerId="1805" CreationDate="2014-02-19T13:03:15.477" Score="6" ViewCount="265" Body="&lt;p&gt;The only tidally locked planet in the Solar system is Mercury. But it is synchronously tidally locked 3:2, because of the relatively high eccentricity of its orbit, so doesn't turn the same side towards the Sun. I wonder if this is common for exoplanets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many discovered exoplanets are close to their star and must be tidally locked. Has any of them been determined to be synchronously tidally locked, like Mercury? Is Mercury a rare freak in this respect, or a representant of a common phenomenon?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1891" LastEditDate="2019-07-29T17:21:47.833" LastActivityDate="2019-07-29T17:21:47.833" Title="Are many exoplanets synchronously tidally locked like Mercury?" Tags="&lt;exoplanet&gt;&lt;tidal-forces&gt;&lt;mercury&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="1798" PostTypeId="2" ParentId="1796" CreationDate="2014-02-19T15:03:20.170" Score="2" Body="&lt;p&gt;For the same (an more prosaic) reason milk skin floats over boiling milk - because it is lighter than the layer below. During the formation of the planet it went through a process called &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Iron_catastrophe&quot; rel=&quot;nofollow noreferrer&quot;&gt;Iron Catastrophe&lt;/a&gt;&lt;/em&gt;, were a large parcel of the denser materials like iron and nickel sunk to the core, leaving the lighter materials close to the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nice graph with relative densities follow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vESUp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fission contributes to keep our internal engine super-hot and well-oiled, allowing for our metallic core to spin and generate our magnetosphere. Eventually we'll run out of heat, the outer core will solidify and stop spinning.&lt;/p&gt;&#xA;" OwnerUserId="893" LastEditorUserId="893" LastEditDate="2014-02-19T15:13:09.757" LastActivityDate="2014-02-19T15:13:09.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1799" PostTypeId="1" CreationDate="2014-02-19T22:58:57.233" Score="3" ViewCount="731" Body="&lt;p&gt;Where to find comprehensive list of absolute magnitudes for stars of different spectral types?  I need published paper or web page with data and reference to the source.&lt;/p&gt;&#xA;" OwnerUserId="761" LastEditorUserId="761" LastEditDate="2014-02-19T23:15:54.767" LastActivityDate="2014-12-10T23:24:06.343" Title="Absolute magnitudes of stars" Tags="&lt;star&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1800" PostTypeId="1" AcceptedAnswerId="1804" CreationDate="2014-02-20T01:34:47.650" Score="3" ViewCount="174" Body="&lt;p&gt;My understanding is that quasars are the early phase of formation of galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the time quasars were forming, the universe would have been smaller and therefore things closer to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About how close would quasars have been to each other?&lt;/p&gt;&#xA;" OwnerUserId="1068" LastActivityDate="2014-02-26T15:40:29.467" Title="How close would quasars have been to each other?" Tags="&lt;distances&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1801" PostTypeId="1" AcceptedAnswerId="1806" CreationDate="2014-02-20T06:36:30.733" Score="12" ViewCount="3514" Body="&lt;p&gt;In anticipation of the upcoming &lt;a href=&quot;http://en.wikipedia.org/wiki/April_2014_lunar_eclipse&quot;&gt;April 2014 lunar eclipse&lt;/a&gt;, I decided to use the 3D virtual planetarium software &lt;a href=&quot;http://www.stellarium.org/&quot;&gt;Stellarium&lt;/a&gt; to simulate what it is supposed look like where I live, since I've never actually witnessed an actual eclipse before. I set the clock to begin at the moonrise before the beginning of the event and to tick at a rate of a few virtual-minutes per real-second, pressed play, and waited for the eclipse to begin. When it did, I saw something I did not expect. To be honest, I realized my expectations were embarrassingly naïve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I watched the Moon rise in the east and begin its upward left-to-right path across the night sky, towards its zenith in the south. My expectation was that the moment the Moon enters Earth's shadow which marks the beginning of the eclipse would be visually apparent by the darkening of the full moon's outer limb on the right side, and the darkening would progress from right to left from there. When I saw that the darkening actually progresses in the opposite direction, I immediately realized that my understanding of the geometry and kinematics of the Earth-Moon-Sun system must be much shallower than I would have thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My main question is must lunar eclipses happen this way every time, and if so why? My gut is now telling me that it does, so I'm mostly interested in an answer of reasonable depth to the 'why' part of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, it occurs to me that the observation of left-to-right progression of darkening of the Moon's face during a lunar eclipse is especially counter-intuitive from the point of views of ancient geocentric cosmologies based on a stationary Earth that doesn't rotate. Can this observation be taken as empirical evidence that the Earth rotates? (I could be completely off base here. If so, please correct me.)&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2015-10-01T16:24:46.433" Title="Why does the darkening of the Moon go from left-to-right during an eclipse?" Tags="&lt;the-moon&gt;&lt;solar-system&gt;&lt;lunar-eclipse&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1802" PostTypeId="1" CreationDate="2014-02-20T06:56:45.380" Score="3" ViewCount="72" Body="&lt;p&gt;We understand that this is the universe and it is essentially full of space and time, but what is on the outside of the universe, and isn't that nothing, something, and how far does that reach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. isn't that another aspect of the universe? For it to go into it, it must be something for it to go into it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I throw a ball through the air, the ball travels through air. If I throw a universe through the air so to speak, doesn't that have a characteristic, frame and limitation?&lt;/p&gt;&#xA;" OwnerUserId="1070" LastActivityDate="2014-02-20T06:56:45.380" Title="What is that void that the universe is growing into?" Tags="&lt;universe&gt;" AnswerCount="0" CommentCount="7" ClosedDate="2014-02-20T16:58:06.923" ContentLicense="CC BY-SA 3.0" />
  <row Id="1804" PostTypeId="2" ParentId="1800" CreationDate="2014-02-20T16:15:24.337" Score="4" Body="&lt;p&gt;&lt;a href=&quot;http://www.nasa.gov/mission_pages/chandra/multimedia/photo10-009.html&quot; rel=&quot;nofollow&quot;&gt;That's an image of two quasars, very close to each other&lt;/a&gt;, &quot;separated on the sky by only about 70 thousand light years&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0602007&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; the peak density should have been about $10^{-6}$ quasars per Mpc, resulting in a peak mean distance between two quasars of about 100 Mpc. (Density of $10^{-6}$ quasars per Mpc results in one quasar per cube of $10^6 Mpc^3$, hence on average one quasar each $\sqrt[3]{10^6 \mbox{Mpc}^3}=100 \mbox{ Mpc}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With &lt;a href=&quot;http://www.nasa.gov/mission_pages/chandra/multimedia/photo10-009.html&quot; rel=&quot;nofollow&quot;&gt;about 1 million known quasars&lt;/a&gt; in the observable universe I get a mean distance of about half that distance. (With a considered light-travel distance of about $4 \mbox{ Gpc} = 4\cdot 3.26 \mbox{ Gly} = 13.04\mbox{ Gly},~~$  a little less than the 13.81 Gly light travel radius of the observable universe, the volume of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Sphere&quot; rel=&quot;nofollow&quot;&gt;sphere&lt;/a&gt; with this radius is about $\frac{4}{3}\pi(4\cdot 10^3 \mbox{Mpc})^3=2.68\cdot 10^{11}\mbox{Mpc}^3$. With 1 million quasars, it's about one quasar per $268000 \mbox{ Mpc}^3$ or one quasar per $\sqrt[3]{268000 \mbox{ Mpc}^3}=64\mbox{ Mpc}$ in one dimension. Due to non-constant quasar frequency over cosmic time, the minimum mean distance should have been below this overall average distance. Using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Distance_measures_%28cosmology%29&quot; rel=&quot;nofollow&quot;&gt;light-travel distance&lt;/a&gt; considers the higher density of the earlier universe better than the comoving distance of 13.8 billion years after big bang. 4 Gpc are used as light travel radius to restrict roughly to redshift z &amp;lt; 7, the range where quasars have been observed. A redshift calculator can be found e.g. &lt;a href=&quot;http://www.kempner.net/cosmic.php&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, take $H_0=70.4$, $\Omega_m = 0.265$, $\Omega_{\Lambda}=0.728$, and $z=7$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The peak number density seems to be somewhere between a redshift of 2.0 and 2.5 (interpreting &lt;a href=&quot;http://en.wikipedia.org/wiki/Distance_measures_%28cosmology%29&quot; rel=&quot;nofollow&quot;&gt;the paper&lt;/a&gt;). If it will turn out by future observations, that the peak number density has been earlier in the universe, we get much shorter peak mean distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have at least an order of magnitude: Somewhere between 150 and 300 million lightyears (between 50 and 100 Mpc, 1 &lt;a href=&quot;http://en.wikipedia.org/wiki/Parsec&quot; rel=&quot;nofollow&quot;&gt;parsec&lt;/a&gt; equals 3.26 lightyears) should have been the peak average distance between two quasars. Might be this estimate will shrink over time, as more (distant) quasars get known.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-02-26T15:40:29.467" LastActivityDate="2014-02-26T15:40:29.467" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1805" PostTypeId="2" ParentId="1797" CreationDate="2014-02-20T17:35:01.297" Score="5" Body="&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1208.0814&quot; rel=&quot;noreferrer&quot;&gt;GJ 581d&lt;/a&gt; and &lt;a href=&quot;http://arxiv.org/abs/1311.4831&quot; rel=&quot;noreferrer&quot;&gt;GJ 667c&lt;/a&gt; are candidates for showing spin-orbit resonance.&#xA;&lt;a href=&quot;http://gravitation.web.ua.pt/sites/gravitation.web.ua.pt/files/Colluci08102012.pdf&quot; rel=&quot;noreferrer&quot;&gt;This paper&lt;/a&gt; mentions GJ 3634b and 55 Cnc b as two further candidates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spin-orbit resonances different from 1:1 are expected to be rather common among rocky planets orbiting close to a star. Direct observational evidence is difficult to obtain; results are based on model calculations.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-20T17:35:01.297" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1806" PostTypeId="2" ParentId="1801" CreationDate="2014-02-20T18:00:29.060" Score="16" Body="&lt;p&gt;On any given day, you see the moon rise in the east and (apparently) travel across the sky from your left to the right; therefore, you would assume (incorrectly) that the leading edge of the moon's movement must be to the right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is not correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The apparent motion you are seeing is predominantly from the &lt;em&gt;Earth's&lt;/em&gt; ~24 hour rotation. &lt;strong&gt;But the orbit of the moon is ACTUALLY from &lt;em&gt;west to east&lt;/em&gt;&lt;/strong&gt; as it revolves around the Earth every 29.5 days. Look a the diagram below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were looking down at the earth-moon-sun system far above the ecliptic over the north pole, you would see the Earth spinning in a counter-clockwise direction… and the moon is also revolving around the Earth in a counter-clockwise direction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So forget about the Earth's rotation for a moment; it has nothing to do with the moon passing into the Earth's shadow (i.e. the eclipse.) Look at the diagram below, and place yourself on that little ball labeled &quot;Earth&quot; and look into the night sky. Which side of the moon is covered first as it passes behind the shadow cast by the Earth? What do you see?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Left to right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5qypj.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="58" LastEditorUserId="58" LastEditDate="2014-02-24T14:31:53.650" LastActivityDate="2014-02-24T14:31:53.650" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1807" PostTypeId="1" CreationDate="2014-02-21T08:52:08.590" Score="2" ViewCount="68" Body="&lt;p&gt;If I stood on a theoretical earth at the time that galaxies were just starting to form, what would the sky have looked like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the smaller size of the universe would it have been densely packed with objects? If yes, what might I have seen?&lt;/p&gt;&#xA;" OwnerUserId="1068" LastActivityDate="2014-02-21T15:07:53.290" Title="What would the sky have looked like when galaxies first started to form?" Tags="&lt;early-universe&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1808" PostTypeId="1" CreationDate="2014-02-21T12:38:30.340" Score="6" ViewCount="10335" Body="&lt;p&gt;I have two stars, with known masses and known orbital radius. How do I calculate the orbital periods of both stars?&lt;/p&gt;&#xA;" OwnerUserId="327" LastActivityDate="2017-12-07T15:29:46.717" Title="How can I calculate the orbital periods in a binary star system?" Tags="&lt;orbit&gt;&lt;gravity&gt;&lt;binary-star&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1809" PostTypeId="2" ParentId="1807" CreationDate="2014-02-21T15:07:53.290" Score="3" Body="&lt;p&gt;In early times of galaxy formation you would have seen a sky similar to &lt;a href=&quot;http://hubblesite.org/newscenter/archive/releases/2012/37/image/a/&quot; rel=&quot;nofollow&quot;&gt;this Hubble Extreme Deep Field image&lt;/a&gt;. A sky filled with small, irregularly shaped, young galaxies and galaxy predecessor fragments (sub-galaxies). &lt;a href=&quot;https://www.e-education.psu.edu/astro801/content/l9_p7.html&quot; rel=&quot;nofollow&quot;&gt;Galaxies probably formed bottom-up from smaller fragments&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further back in time it would have been dark. Hydrogen clouds would begin to contract, as predecessors of first stars (&lt;a href=&quot;http://www2.astro.psu.edu/users/rbc/a534/redman.pdf&quot; rel=&quot;nofollow&quot;&gt;population III&lt;/a&gt;) and &lt;a href=&quot;http://jwst.nasa.gov/firstlight.html&quot; rel=&quot;nofollow&quot;&gt;clusters of stars&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even further in the past (redshift between 300 and 400) the sky would have been glowing red by what we now call &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow&quot;&gt;cosmic microwave background&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-21T15:07:53.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1810" PostTypeId="2" ParentId="1808" CreationDate="2014-02-21T17:21:47.373" Score="5" Body="&lt;p&gt;If you need a rough estimation, you can sample the positions of the stellar objects from their acceleration, using Newton's law. A full picture is drawn &lt;a href=&quot;http://en.wikipedia.org/wiki/N-body_problem&quot; rel=&quot;nofollow&quot;&gt;on this Wikipedia page&lt;/a&gt;, but basically, given N stellar objects at position P(i), with their respective mass M(i):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;M_i.\vec{acc_i} = -G\sum_{n \neq i}^N \frac {M_i.M_n.(\vec{pos_i}-\vec{pos_n})}{\begin{vmatrix} \vec{pos_i}-\vec{pos_n}\end{vmatrix}^3 }&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then derivate acceleration into speed, then into position using a small enough time delta. With initial positions and speed (which are the trickiest and also funniest to choose), you can simulate a rough N-body system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the fun and simulation side, this is what &lt;a href=&quot;http://store.steampowered.com/app/72200/&quot; rel=&quot;nofollow&quot;&gt;Universe Sandbox&lt;/a&gt; aims to present (in advance, sorry for linking to a commercial store, I'm not related to this studio).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lagrangian points are also interesting to look at when simulating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, to simplify things, you may like to think nearly all stellar objects &quot;close&quot; to the binary system would have been eaten by the massive couple over time, and then consider only the barycenter of the pair of stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; although OP was clear, my answer is for N objects. The simplification to N=2, with A and B being the positions of the two objects, results in:&#xA;$$&#xA;\vec{acc_A} = \frac{G.M_B}{AB^3}\vec{AB}&#xA;$$&#xA;$$&#xA;\vec{acc_B} = \frac{G.M_A}{BA^3}\vec{BA}&#xA;$$&#xA;And with an iterative process, once acceleration is computed for step k, using known initial conditions for speed and position:&#xA;$$&#xA;\vec{spd_{k+1}} = \vec{acc_k}.\Delta{t} + \vec{spd_k}&#xA;$$&#xA;$$&#xA;\vec{pos_{k+1}} = \vec{spd_k}.\Delta{t} + \vec{pos_k}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT2:&lt;/strong&gt; well, another rough estimation, for the period duration (which I misunderstood as &quot;trajectory&quot; from the question). I'd use cylindrical coordinates as reference, for their clean spatial representation, and for the single-angle conversion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left( \begin{array}{c} R_k \\ \theta_k \\ h_k \end{array} \right) = \left( \begin{array}{c} \sqrt{pos_{x,k}^2+pos_{y,k}^2} \\ \arctan \left( \frac{pos_{y,k}}{pos_{x,k}} \right) \\ pos_{z,k} \end{array} \right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...with adequate precaution. Introduce part of that conversion at each iteration, and wait for theta to complete a rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(in advance, sorry for the weird notation, which I just tried to make consistent from my previous posts)&lt;/p&gt;&#xA;" OwnerUserId="1080" LastEditorUserId="1080" LastEditDate="2014-02-26T09:46:16.503" LastActivityDate="2014-02-26T09:46:16.503" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1811" PostTypeId="2" ParentId="1799" CreationDate="2014-02-22T00:35:38.370" Score="1" Body="&lt;p&gt;I have found next list based on HIPPARCOS data:&#xA;&lt;a href=&quot;http://www.astro.lu.se/~lennart/MVstars.html&quot; rel=&quot;nofollow&quot;&gt;http://www.astro.lu.se/~lennart/MVstars.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="761" LastActivityDate="2014-02-22T00:35:38.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1812" PostTypeId="1" AcceptedAnswerId="1828" CreationDate="2014-02-22T02:50:00.600" Score="17" ViewCount="14241" Body="&lt;p&gt;This is the supposition:&#xA;Venus is too hot, Mars is too cold. If they switched places then Solar energy would change to make both more Earth like. We might as well have had three instead of just one &quot;Earth&quot; in the Solar system, if the roulette of early planet formation had played out just a little bit different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's of course too simple. But wouldn't it help alot if it were the case from the beginning that Venus had had Mars' mass, and Mars had had Venus' mass? Or do other factors dominate mass and distance from the Sun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would Venus still have had a thick, but not too thick, atmosphere if it were at 1.5 AU, because it has mass enough to keep one, and wouldn't that have kept it warmer than Mars is today? Including flowing water on its surface under its atmospheric pressure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would a Mars at 0.7 AU have been warmer and maybe have had a passing atmosphere created from melting volatiles during a longer era in its history?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2014-02-22T09:48:22.710" LastActivityDate="2017-06-13T11:23:19.557" Title="If Venus and Mars changed places, would we then have 3 habitable planets?" Tags="&lt;mars&gt;&lt;venus&gt;&lt;earth-like-planet&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1813" PostTypeId="2" ParentId="1812" CreationDate="2014-02-22T04:22:16.423" Score="2" Body="&lt;p&gt;Simply &quot;swapping&quot; position would not in and of itself be cause for a habitable planet.&#xA;The key is &quot;changing&quot; position. If Venus was to undergo orbital migration it could conceivably move into Earths current position and cool down, continue to develop a more stable atmosphere, and produce oceans. Then further orbital migration could cause the atmosphere to collapse and marine succession. The question then is would it become like Ceres and Jupiter.&#xA;Simulating a vector reversal, where Mars undergoes orbital migration and moves into the position currently taken by Earth, would warming it up cause it to further develop an atmosphere, and a marine environment. Then, when its vector moves it to the spot currently taken by Venus would marine succession and atmospheric reduction appear as Venus is to day.&#xA;From physical appearance alone I would say there is a natural progression where the planets evolve as the orbital migration causes their vector to move away from the Sun.&lt;/p&gt;&#xA;" OwnerUserId="257" LastActivityDate="2014-02-22T04:22:16.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1814" PostTypeId="2" ParentId="1799" CreationDate="2014-02-22T05:54:10.253" Score="6" Body="&lt;p&gt;&lt;a href=&quot;http://vizier.u-strasbg.fr/&quot;&gt;VizieR&lt;/a&gt; is an online source for all sorts of astronomical data published in scientific papers. As you mentioned, The &lt;a href=&quot;http://adsabs.harvard.edu/abs/1997yCat.1239....0E&quot;&gt;HIPPARCOS catalogue&lt;/a&gt; contains visual magnitude data.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Open the query page for the &lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR-3?-source=I/239/hip_main&quot;&gt;main HIPPARCOS catalogue&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;Select the fields you want (defaults are ok for you)&lt;/li&gt;&#xA;&lt;li&gt;Hit submit to see the results&lt;/li&gt;&#xA;&lt;li&gt;You can limit the number of results and format under &lt;em&gt;Preferences&lt;/em&gt; on the left&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This table gives you the measured visual magnitude, i.e. the &lt;strong&gt;Apparent Magnitude&lt;/strong&gt; ($m_V$, &lt;code&gt;V&lt;/code&gt; column). To convert that into &lt;strong&gt;Absolute Magnitude&lt;/strong&gt; ($M_V$) you need to know the distance to the star. This can be calculated using the &lt;strong&gt;Parallax&lt;/strong&gt; field (&lt;code&gt;Plx&lt;/code&gt; column).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the formula for you:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M_V = m_V + 5 * log_{10}( Plx / 100 )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can easily dump the data into Excel or something, put a formula into an extra column and calculate the &lt;strong&gt;Absolute Magnitude&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2014-02-22T05:54:10.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1816" PostTypeId="1" CreationDate="2014-02-22T16:00:02.977" Score="2" ViewCount="86" Body="&lt;p&gt;I was always taught and lead to believe that at the beginning of our solar system, there was a period of heavy bombardment, which is conceivable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why has it just stopped and we are at peace from this, and, there is no such things happening today? &lt;/p&gt;&#xA;" OwnerUserId="1084" LastActivityDate="2014-02-23T20:22:52.490" Title="Heavy bombardment - Why has things cooled?" Tags="&lt;solar-system&gt;&lt;asteroids&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1817" PostTypeId="2" ParentId="1507" CreationDate="2014-02-22T19:49:36.030" Score="1" Body="&lt;p&gt;There are several scenarios: An Earth-Venus &lt;a href=&quot;http://en.wikipedia.org/wiki/Double_planet&quot; rel=&quot;nofollow&quot;&gt;binary&lt;/a&gt; would be possible, or &lt;a href=&quot;http://arxiv.org/abs/1207.5250&quot; rel=&quot;nofollow&quot;&gt;coplanar&lt;/a&gt; prograde or &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_inclination&quot; rel=&quot;nofollow&quot;&gt;inclined&lt;/a&gt; up to &lt;a href=&quot;http://en.wikipedia.org/wiki/Retrograde_motion&quot; rel=&quot;nofollow&quot;&gt;retrograde&lt;/a&gt; on &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_resonance&quot; rel=&quot;nofollow&quot;&gt;resonant&lt;/a&gt; or &lt;a href=&quot;http://www.univie.ac.at/adg/Conferences/hcm8/talks/hadjidemetriou.pdf&quot; rel=&quot;nofollow&quot;&gt;non-resonant&lt;/a&gt; orbits, &lt;a href=&quot;http://en.wikipedia.org/wiki/Elliptic_orbit&quot; rel=&quot;nofollow&quot;&gt;elliptic&lt;/a&gt; or almost &lt;a href=&quot;http://en.wikipedia.org/wiki/Circular_orbit&quot; rel=&quot;nofollow&quot;&gt;circular&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One option would be a &lt;a href=&quot;http://www.ualberta.ca/~urban/Samples/Stars/Coorbitals/images/strobe.mpg&quot; rel=&quot;nofollow&quot;&gt;co-orbital scenario&lt;/a&gt;, which is described in more detail &lt;a href=&quot;http://www.ualberta.ca/~urban/Samples/Stars/Coorbitals/Coorbitals.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. The latter would allow Earth and Venus to stay both in the habitable zone. But the effects for Earth could be very remarkable in some cases: Seasons would change over time, since &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_eccentricity&quot; rel=&quot;nofollow&quot;&gt;eccentricity&lt;/a&gt; of Earth's orbit would change. If Earth and Venus would follow a &lt;a href=&quot;http://en.wikipedia.org/wiki/Horseshoe_orbit&quot; rel=&quot;nofollow&quot;&gt;horseshoe orbit&lt;/a&gt;, we would get variations in the length of a year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://link.springer.com/article/10.1007/s10569-010-9288-0&quot; rel=&quot;nofollow&quot;&gt;Co-orbital systems&lt;/a&gt; can become instable by the gravity of other planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the long run &lt;a href=&quot;http://arxiv.org/abs/1306.4293&quot; rel=&quot;nofollow&quot;&gt;tidal forces may lead to a collision of the two co-orbiting planets&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-22T19:49:36.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1818" PostTypeId="1" CreationDate="2014-02-22T20:07:02.240" Score="3" ViewCount="125" Body="&lt;p&gt;I was hoping that astronomy enthusiasts/scientists would help me out with a simple question about light. Hold your outstretched hand up to any bright light source (the sun, or even a light bulb) so that the light is completely blocked by at least one finger, and then slowly move your finger to the side until the light just peeks around. &lt;strong&gt;There seems to be a glow when you do this, and I was wondering what causes this.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the light bending around my finger from some sort of diffraction? What's going on?&lt;/p&gt;&#xA;" OwnerUserId="1085" LastActivityDate="2014-02-22T20:07:02.240" Title="Is light bending around my finger?" Tags="&lt;light&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2014-02-23T19:15:53.943" ContentLicense="CC BY-SA 3.0" />
  <row Id="1819" PostTypeId="1" AcceptedAnswerId="1820" CreationDate="2014-02-22T22:23:04.857" Score="1" ViewCount="970" Body="&lt;p&gt;We live in the Milky Way Galaxy, but we don't just call it &amp;quot;The Galaxy&amp;quot;, because we know there are multiple different galaxies. So if we lived in a multiverse, it wouldn't make much sense to keep calling our universe &amp;quot;The Universe&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;How are names for things like this created, and what would a likely name for our universe be?&lt;/p&gt;&#xA;" OwnerUserId="1059" LastEditorUserId="37191" LastEditDate="2021-02-01T22:34:24.277" LastActivityDate="2021-02-01T22:34:24.277" Title="If we lived in a multiverse, what would our universe most likely then be named?" Tags="&lt;universe&gt;&lt;naming&gt;&lt;multiverse&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2014-02-25T16:27:13.237" ContentLicense="CC BY-SA 4.0" />
  <row Id="1820" PostTypeId="2" ParentId="1819" CreationDate="2014-02-23T11:25:04.707" Score="3" Body="&lt;p&gt;New objects usually get some preliminary name or number according to one or more naming or numbering conventions (e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Bright_Star_Catalogue&quot; rel=&quot;nofollow&quot;&gt;HR numbers for stars&lt;/a&gt;). If there aren't too many objects of interest, they are named by the discoverer(s) (e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble_volume&quot; rel=&quot;nofollow&quot;&gt;Hubble volume&lt;/a&gt;), an institution (e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/C/2011_L4&quot; rel=&quot;nofollow&quot;&gt;PANSTARRS&lt;/a&gt;), an occasional nickname liked by the public (e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_hole&quot; rel=&quot;nofollow&quot;&gt;black hole&lt;/a&gt;), or by a majority in a commitee or an election (e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydra_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Hydra&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For our universe I can only speculate; probably a notion of the respective multiverse theory would become part of the name, e.g. &quot;Herman brane&quot;, if we take a &lt;a href=&quot;http://en.wikipedia.org/wiki/Brane_cosmology&quot; rel=&quot;nofollow&quot;&gt;brane cosmology&lt;/a&gt; as an example.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-23T11:25:04.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1821" PostTypeId="1" CreationDate="2014-01-28T20:43:29.037" Score="6" ViewCount="140" Body="&lt;p&gt;Black holes release &lt;a href=&quot;http://en.wikipedia.org/wiki/Hawking_radiation&quot;&gt;Hawking radiation&lt;/a&gt;. Now assume that the universe after all collapses back again, bringing all matter to a single black hole. I assume that this would bend the whole space-time of the universe as well, wrapping it around the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what happens to the shrinking universe in the presence of the pressure of Hawking radiation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could there be a stable point where pressure and gravity are balanced?&lt;/p&gt;&#xA;" OwnerUserId="218" OwnerDisplayName="draks ..." LastActivityDate="2014-02-23T22:08:43.937" Title="What happens to the shrinking universe in the presence of the pressure of Hawking radiation?" Tags="&lt;cosmology&gt;&lt;hawking-radiation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1822" PostTypeId="1" AcceptedAnswerId="1831" CreationDate="2014-02-23T13:50:24.707" Score="5" ViewCount="281" Body="&lt;p&gt;A Galactic year is estimated to be around 250 million &quot;terrestrial&quot; years. That means there will be about 4 rotations per billion years. And age of the universe is estimated to be near 13.8 billion &quot;terrestrial&quot; years. It looks improbable to think that our galaxy has rotated 50 times since the beginning of time. Could someone explain this please?&lt;/p&gt;&#xA;" OwnerUserId="1087" OwnerDisplayName="Colombo" LastEditorUserId="92" LastEditDate="2014-02-26T12:41:14.180" LastActivityDate="2014-02-26T12:41:14.180" Title="Age of the universe" Tags="&lt;galaxy&gt;&lt;universe&gt;&lt;galactic-dynamics&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1823" PostTypeId="1" CreationDate="2014-02-23T16:33:13.743" Score="1" ViewCount="276" Body="&lt;p&gt;Some say that a black hole is a rip in space-time, a portal linking two points in the universe as one. How sound is this? I've read that when a giant star collapses, what is left behind (a massive neutron star?) is so dense that light cannot even escape its gravitational pull. Is this phenomenon caused by an actual object? If one were to travel into a black hole, should not one find only really dense matter?&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2014-02-24T16:45:29.497" Title="Black Hole, Object or Portal?" Tags="&lt;black-hole&gt;&lt;supermassive-black-hole&gt;&lt;matter&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ClosedDate="2014-03-04T22:18:42.473" ContentLicense="CC BY-SA 3.0" />
  <row Id="1824" PostTypeId="2" ParentId="1822" CreationDate="2014-02-23T17:51:24.820" Score="-3" Body="&lt;p&gt;Currently Astronomy relates to Time in a manner similar to the Geocentric / Heliocentric syndrome that plagued early Astronomers that did not have enough information available to accurately describe how the Solar System maneuvered.&lt;br&gt;&#xA;Earthlings use Time as it relates to Earth to measure the Age of Celestial objects because we don't currently understand how Time works. More specifically we use the J2000 epoch to tell time on Earth because it was designed hundreds of years ago, appears to work for the time being and is commonly agreed upon.&lt;br&gt;&#xA;The problem is that we believe our planet is approximately 4.5 billion years old and a year is an approximate amount of time that is consistent over long periods of time. So we extrapolate out and use that to measure objects outside our system since that is the only reference point we know and understand and we do not have access to a Time system that explains other Systems and is capable of measuring them.&lt;br&gt;&#xA;The Universe if it is the largest encapsulating structure needs its own time system that starts from its origin and is consistent to measure against other systems. For example using a second instead of a year (presuming a second takes the same duration in all systems).&lt;br&gt;&#xA;A Galaxy would need its own Time reference from its origin with in the Universe using a similar consistent standard like a second so you could accurately compare their ages.&#xA;If you just look at the Earth year we don't know how long a year was a billion years ago, it may have only been 100 days and we don't know how long one will be in a billion years, it may be 1000 days, just to make an example. So an Earth year is a poor way to measure the age of other Celestial objects but it is the best we have at the present time so we continue to use it. Technically we shouldn't be able to measure anything older than approximate 4.5 billions years because theoretically that time doesn't exist because Earth didn't exist. You can see how this resembles the Geocentric / Heliocentric problem.&lt;br&gt;&#xA;In the future we will realize the J2000 epoch is based on the birth of Jesus Christ and that we are not accurately measuring time on our own planet based on its beginning but rather basing it on an individuals life. At some point we will agree on a common starting point for the planet and develop technology to determine that moment similar to how we developed carbon dating.&lt;br&gt;&#xA;You can see how this will even effect the term light year as we realize the length of a Earth year changes over long periods of time and is not accurate to measure distances that light travels relating them to orbital period of our planet. Note also light appears to travel a linear fashion whereas our planetary orbit is circular. The key would be to have a time standard that is linear in fashion as well.&lt;br&gt;&#xA;Your point about how many times the galaxy has rotated is similar to the age of an earth year problem, no one from Earth really knows how fast or how long a galaxy evolves so they are just estimating based on the best information available and that is commonly agreed upon.&lt;/p&gt;&#xA;" OwnerUserId="257" LastEditorUserId="257" LastEditDate="2014-02-23T18:01:56.653" LastActivityDate="2014-02-23T18:01:56.653" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1825" PostTypeId="2" ParentId="1816" CreationDate="2014-02-23T20:22:52.490" Score="4" Body="&lt;p&gt;According to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nice_model&quot; rel=&quot;nofollow&quot;&gt;Nice model&lt;/a&gt; -- that's Nice, France (rhymes with Reese or geese), not nice (rhymes with mice) -- the solar system formed with the giant planets (the outer four) in the following order: Jupiter, Saturn, Neptune, Uranus. Outside of these planets existed a large field of asteroids (also called planetesimals), the remnants of which form our Kuiper belt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Nice model hypothesizes that Neptune and Uranus actually swapped places due to an &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_resonance&quot; rel=&quot;nofollow&quot;&gt;orbital resonance&lt;/a&gt; between Jupiter and Saturn. This had the effect of moving Jupiter inward (towards the Sun) and kicking Neptune farther out. This orbital disruption also caused a thinning of the Kuiper belt, where some of the planetesimals gained kinetic energy and were knocked into the Oort cloud, and some of them slowed down and fell into the inner solar system, causing a period of heavy bombardment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is thought that the period of heavy bombardment ended once the orbits of the outer four planets stabilized and an equilibrium was reached. The result was the current order of the outer planets we see today (Jupiter, Saturn, Uranus, Neptune), a thinner Kuiper belt, and an outer Oort cloud. Of course, this is just a model; the Nice model itself has already undergone some revisions, even though it is relatively recent. A nice video illustrating the Nice model can be found &lt;a href=&quot;http://www.skyandtelescope.com/skytel/beyondthepage/123632899.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this answers your question.&lt;/p&gt;&#xA;" OwnerUserId="1027" LastActivityDate="2014-02-23T20:22:52.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1827" PostTypeId="2" ParentId="1800" CreationDate="2014-02-23T20:46:48.723" Score="1" Body="&lt;p&gt;Quasars are simply Active Galactic Nuclei (AGN) viewed from the correct inclination angle from the earth. Quasars are not &quot;the early phase of formation of galaxies.&quot; It is currently thought that most galaxies (if not all) contain a supermassive black hole at their center -- for example, at the center of the Milky Way resides Sagittarius A* (Sgr A*), which is thought to be a black hole with mass of $\sim \!4.6 \times 10^6$ solar masses! However, Sgr A* is currently not active.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to your question is that AGN are generally only as close as galaxies are to each other. It is true when the universe was younger, galaxies were closer together. However, as Gerald's link shows, galaxies are still colliding today, so occasionally AGN can become very close and can even merge to form an even larger supermassive black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note according to our current understanding of the history of the universe, galaxies and AGN did not form right away. In order for &lt;a href=&quot;http://en.wikipedia.org/wiki/Galaxy_formation_and_evolution&quot; rel=&quot;nofollow&quot;&gt;galaxies to form&lt;/a&gt;, perturbations of dark matter had to develop which accreted baryonic matter, which then cooled and condensed to form galaxies. The formation of supermassive black holes also takes time. First, you have to create a black hole, which are formed from core-collapse supernovae of stars. Then that black hole has to accrete mass to become large, and smaller black holes are actually less efficient at accreting mass from surrounding gas clouds than bigger ones are. Therefore, there was never a period where most supermassive black holes (and thus quasars) were ever right on top of each other.&lt;/p&gt;&#xA;" OwnerUserId="1027" LastActivityDate="2014-02-23T20:46:48.723" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1828" PostTypeId="2" ParentId="1812" CreationDate="2014-02-23T21:11:25.020" Score="14" Body="&lt;p&gt;This is a very interesting question. Of course, as you noted, you have simplified things quite a bit; there are other factors besides temperature that affect habitability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding Venus, you probably know that Venus is extremely hot at its surface not just because it is closer to the Sun, but because it has a thick CO2 atmosphere and is warmed by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Greenhouse_effect&quot;&gt;greenhouse effect&lt;/a&gt;. There are, in fact, two things about Venus that would prevent it from being habitable wherever you put it. One is the lack of a magnetosphere, which is necessary to prevent ionizing radiation (particularly from the solar wind) from reaching the planet's surface. The lack of a magnetosphere appears to be due to the lack of a geodynamo on Venus, which has to do with the structure of its core. Second, Venus appears to lack tectonic plates, which you may know are responsible for earthquakes here on Earth. Interestingly, tectonic plates play a major role in controlling the buildup of CO2 in the atmosphere (see &lt;a href=&quot;http://rainbow.ldeo.columbia.edu/courses/v1001/6.html&quot;&gt;here&lt;/a&gt; for details). Lacking tectonic plates, Venus is doomed to have a large CO2 atmosphere wherever you put it, which would not make it a nice place to live.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars, on the other hand, is a very different matter. It has both a magnetosphere (albeit it is very weak) and it likely has tectonic plates (although last I heard it is thought to only have two). One of the reasons that NASA has sent so many probes to Mars is that it was thought to be habitable at one point. It is thought that Mars's growth was stunted because of gravitation effects from Jupiter and Saturn. So, in another universe, Mars could have ended up very much like another Earth.&lt;/p&gt;&#xA;" OwnerUserId="1027" LastActivityDate="2014-02-23T21:11:25.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1829" PostTypeId="2" ParentId="1821" CreationDate="2014-02-23T22:08:43.937" Score="4" Body="&lt;p&gt;In a shrinking, roughly 3-spherical universe with only a black hole, Hawking radiation should follow a &lt;a href=&quot;http://en.wikipedia.org/wiki/Geodesic&quot; rel=&quot;nofollow&quot;&gt;geodesic line&lt;/a&gt; and return to the black hole, without excerting radiation pressure to the universe as a whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore it's hard to see, how Hawking radiation should establish an equilibrium with gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More feasible seems, that the shrinking universe can prevent the black hole from further evaporation at some point, since evaporated particles and radiation are directed back into the black hole by the overall curvature of spacetime.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-23T22:08:43.937" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1830" PostTypeId="1" CreationDate="2014-02-23T22:26:32.543" Score="6" ViewCount="1263" Body="&lt;p&gt;At present, it seems like we are alone in our solar system. How many earth-like planets would fit into the habitable zone of the solar system? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbits should be stable for quite a time, i.e. for life to evolve, let's say &lt;a href=&quot;http://en.wikipedia.org/wiki/Age_of_the_Earth&quot;&gt;4.6 billion years&lt;/a&gt;.&#xA;The habitable zone may be estimated from 0.725 to 3.0 astronomical units, according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Habitable_zone#Solar_System_estimates&quot;&gt;Wiki&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could one use perturbation theory to answer this?&lt;/p&gt;&#xA;" OwnerUserId="218" LastEditorUserId="1062" LastEditDate="2014-02-24T09:24:58.913" LastActivityDate="2014-08-03T20:18:52.503" Title="What is the maximum number of planets in the habitable zone?" Tags="&lt;solar-system&gt;&lt;exoplanet&gt;&lt;earth-like-planet&gt;&lt;habitable-zone&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1831" PostTypeId="2" ParentId="1822" CreationDate="2014-02-23T22:54:21.340" Score="12" Body="&lt;p&gt;Your calculation sounds correct. It is however based on assumptions that are non-trivial. An analogy with the Earth would give that the Earth rotated around the Sun 13.8 billion times since the Big Bang. Which is meaningless since the Earth was created only a few billion years ago. Our galaxy, the Milky-Way, may have had a long and quiet history since 10 billion years ago (&lt;a href=&quot;http://arxiv.org/abs/astro-ph/0702585&quot;&gt;http://arxiv.org/abs/astro-ph/0702585&lt;/a&gt;), which is not too far from the age of the Universe. Our Galaxy, however, evolved, and its rotation speed too. As for the solar system, the duration of a &quot;year&quot; (ie one rotation over the Sun) on Earth is very different from the duration of a &quot;year&quot; on other planets. It is the same in the Milky Way where the revolution period is different depending on the distance from the center of the galaxy. As an order of magnitude, and considering in first approximation that our Galaxy is old, and that its properties did not change too much, an order a magnitude of a few tenths revolutions since our Galaxy is a stable disk sounds plausible. As a first approximation. I hope this will help you to find that such a value is not improbable.&lt;/p&gt;&#xA;" OwnerUserId="1092" LastActivityDate="2014-02-23T22:54:21.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1832" PostTypeId="2" ParentId="1722" CreationDate="2014-02-24T16:33:41.677" Score="2" Body="&lt;p&gt;It &lt;strong&gt;depends on the quality of glasses&lt;/strong&gt; and &lt;strong&gt;intensity, nature and degree of harmfulness of sunrays&lt;/strong&gt; which you are watching to. Every eclipse has different intensity so as it can harm you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When there was a solar eclipse around 3-4 years ago, it was a full solar eclipse. That time I watched it from my school with total naked eyes. My eyes were not harmed from that, but my school principle did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I still recommend you to &lt;strong&gt;not to watch it directly&lt;/strong&gt;, &lt;strong&gt;use projectors&lt;/strong&gt; which would allow you to get the view without directly watching the sun. Only &lt;strong&gt;use original products&lt;/strong&gt; which are being certified.&lt;/p&gt;&#xA;" OwnerUserId="1096" LastActivityDate="2014-02-24T16:33:41.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1833" PostTypeId="2" ParentId="1708" CreationDate="2014-02-24T16:37:12.343" Score="1" Body="&lt;p&gt;Evidence for stars evolving alone outside galaxies is very hard to get. And for a good reason: those stars are very hard to detect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxies are quite easy to observe: one of the reasons for that, is that the surface brightness of a galaxy does not change with the distance (a good explanation here: &lt;a href=&quot;http://mysite.verizon.net/vze55p46/id18.html&quot; rel=&quot;nofollow&quot;&gt;http://mysite.verizon.net/vze55p46/id18.html&lt;/a&gt;, 5th paragraph) (Note: this is not true anymore for cosmological distances because the expansion plays a role, but this demonstration is usefull for the nearby universe in which we focus now).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, observing stars is very hard as they do not have a surface brightness: the light we receive evolves in 1/radius2. Do stars outside galaxies exist? Yes: we have evidence for intergalactic stars. See here for example: &lt;a href=&quot;http://hubblesite.org/newscenter/archive/releases/1997/02/text/&quot; rel=&quot;nofollow&quot;&gt;http://hubblesite.org/newscenter/archive/releases/1997/02/text/&lt;/a&gt; which I think was the first discovery (anyone can back me up?). Or see this more recent paper about &quot;rogue stars&quot; evaded from our Galaxy: &lt;a href=&quot;http://lanl.arxiv.org/abs/1202.2152&quot; rel=&quot;nofollow&quot;&gt;http://lanl.arxiv.org/abs/1202.2152&lt;/a&gt;. Those findings are not surprising as we were expecting that some stars are sometimes ejected from galaxies by gravitational forces. Can these stars carry on their planetary system with them? On this question I am not an expert, so I won't give an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That was for stars created inside galaxies. What about star formation outside galaxies? This is a recent subject of discussion but we have growing evidence that this could be possible. Here is a paper about the discovery of intergalactic HII regions, which are regions of star formation outside galaxies: &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0310674&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/astro-ph/0310674&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore we showed that:&#xA;1/ there are stars outside galaxies&#xA;2/ there is evidence for star formation outside galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first case, it may be possible that these stars carry their planetary systems with them. In the second case, it sounds plausible that the forming stars will build their torque of material which might give birth to planetary systems. There is however no evidence yet for this (to my mind).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope my answer clarifies your mind!&lt;/p&gt;&#xA;" OwnerUserId="1092" LastActivityDate="2014-02-24T16:37:12.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1834" PostTypeId="2" ParentId="1823" CreationDate="2014-02-24T16:45:29.497" Score="3" Body="&lt;p&gt;There should be nothing special about crossing the &quot;event horizon&quot; - the line beyond which light cannot escape and so, in theory at least, one might &quot;travel into a black hole&quot; and not be aware of any physical changes at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I suspect what you mean is what would happen if one travelled to the centre of a black hole. Here we may speculate but our current theories of physics breakdown at this &quot;singularity&quot; and so we cannot know - nor is it easy to think of how we might design an experiment, given our currently available technologies, to test any theory we have.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2014-02-24T16:45:29.497" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1835" PostTypeId="2" ParentId="664" CreationDate="2014-02-24T17:36:13.973" Score="3" Body="&lt;p&gt;There is an incorrect assumption in this question. There is no &quot;super massive black hole which holds the galaxy together&quot;. There is just a super massive black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is often believed that black holes have a huge gravitational force. Actually, they simply have the gravitational force of their mass. A black hole of 50 solar masses has no more &quot;gravitational power&quot; than a star of 50 solar masses. And if the sun would instantly change into a black hole (of equal mass), the Earth trajectory would not change at all (but yes, you would have to turn on the heater!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the question is: considering that the black hole at the center of our galaxy is super massive, does it plays an important role for stars like the Sun, at our distance? To answer this, you simply have to ask yourself if the mass of the black hole is important or negligible in front of the mass of the other components that are inside the orbit of our Sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the black hole at the center of our galaxy is about 10^6 (1 million) solar masses (from this paper: &lt;a href=&quot;http://arxiv.org/abs/0810.4674&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0810.4674&lt;/a&gt;). The total mass of the Galaxy is about 10^12 (a thousand billion) solar masses (&lt;a href=&quot;http://arxiv.org/abs/1102.4340&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1102.4340&lt;/a&gt;). The mass inside the radius of the Sun is way lower (you could calculate it from the speed of the Sun and its distance from the center), but still, the mass of the black hole is very small compared to the rest of the mass inside the Sun's radius. So for the effect of the black hole to become dominant, you would need to be very close from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would happen if instead of this super massive black hole there would be many stars, totalizing the mass of the black hole? Nothing. If would not change anything on the Sun trajectory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To change the Sun's trajectory and make the Sun &quot;fall&quot;, what you would need would be:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;More &quot;pulling&quot; mass, but there is no reason this mass suddenly appears.&lt;/li&gt;&#xA;&lt;li&gt;Something to decrease the speed of the Sun around its orbit. It is probably the case as the Sun is slowed down by the gas of the interstellar medium, but this effect is very small.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1092" LastActivityDate="2014-02-24T17:36:13.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1836" PostTypeId="1" AcceptedAnswerId="1843" CreationDate="2014-02-24T18:59:23.883" Score="7" ViewCount="189" Body="&lt;p&gt;How would things look if we saw them through the light of a blue or red star?&#xA;Would there be any changes?&lt;/p&gt;&#xA;" OwnerUserId="828" LastEditorUserId="6" LastEditDate="2014-02-25T17:30:47.170" LastActivityDate="2014-02-25T17:30:47.170" Title="How would light from a blue or red star affect the way we see?" Tags="&lt;star&gt;&lt;light&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1837" PostTypeId="1" AcceptedAnswerId="1857" CreationDate="2014-02-24T19:33:53.293" Score="5" ViewCount="135" Body="&lt;p&gt;Are there any known correlations between the changes in planetary surface temperatures in a solar system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, do the farthest planets have smaller albeit correlated changes?&lt;/p&gt;&#xA;" OwnerDisplayName="user1098" LastActivityDate="2014-02-27T04:38:37.987" Title="Do planetary surface temperatures change in unison in a solar system?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;temperature&gt;&lt;surface&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1838" PostTypeId="1" CreationDate="2014-02-24T20:59:12.600" Score="7" ViewCount="1179" Body="&lt;p&gt;The SOHO observatory has an opaque object in the light path that blocks the direct light from the sun making it possible to observe the corona.&#xA;Hubble and many other telescopes have a big primary mirror in the bottom of a tube which bounces light on to a smaller secondary mirror at the opening of the tube. Why doesn't that other mirror cause a circular area of &quot;no data&quot; in the middle of the resulting images?&lt;/p&gt;&#xA;" OwnerDisplayName="Joakim Rosqvist" LastEditorDisplayName="user1569" LastEditDate="2018-09-30T16:54:39.497" LastActivityDate="2018-09-30T16:54:39.497" Title="Why does e.g. Hubble's secondary mirror not block part of the picture?" Tags="&lt;hubble-telescope&gt;&lt;optics&gt;&lt;soho&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1839" PostTypeId="2" ParentId="1838" CreationDate="2014-02-24T22:01:26.623" Score="2" Body="&lt;p&gt;A particular point in the image is formed by the convergence of all the rays of light that leave from that point in the object, and bounce off the mirror.  As long as there are some rays that can do that, then the only effect of the secondary is to block &lt;strong&gt;&lt;em&gt;some&lt;/em&gt;&lt;/strong&gt; of the rays, dimming those parts of the image slightly.&lt;/p&gt;&#xA;" OwnerUserId="420" OwnerDisplayName="User58220" LastActivityDate="2014-02-24T22:01:26.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1840" PostTypeId="1" AcceptedAnswerId="1841" CreationDate="2014-02-24T23:02:50.247" Score="8" ViewCount="1996" Body="&lt;p&gt;Obviously there's a limited part of earths surface where the moon is visible at any given time. I imagine the area of this surface would be a lot smaller than the area where the sun is visible at any given time, due to the fact that the moon is a lot closer to earth. But how big is that area exactly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, to put it differently: Imagine you are standing outside your house and looking at the moon. At the exact same moment someone else does the same in another location. But how far apart can you (theoretically) be?&lt;/p&gt;&#xA;" OwnerUserId="1099" LastActivityDate="2014-02-26T10:15:29.547" Title="How far apart can two people watch the moon simultaneously?" Tags="&lt;the-moon&gt;&lt;distances&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1841" PostTypeId="2" ParentId="1840" CreationDate="2014-02-25T01:00:54.940" Score="4" Body="&lt;p&gt;If we take 6371.0 km as the mean radius of &lt;a href=&quot;http://en.wikipedia.org/wiki/Earth&quot; rel=&quot;nofollow&quot;&gt;Earth&lt;/a&gt;, an apogee of &lt;a href=&quot;http://en.wikipedia.org/wiki/Moon&quot; rel=&quot;nofollow&quot;&gt;Moon&lt;/a&gt; of 405503 km, and a perigee of 363295 km, we get ratios of 6371.0 km / 405503 km = 0.01571 = sin 0.9002° resp. 6371.0 km / 363295 km = 0.01754 = sin 1.005°. So on both sides of Earth between $$40030\mbox{ km}\cdot 0.9002°/360°=100\mbox{ km}$$ and $$40030\mbox{ km}\cdot 1.005°/360° =112\mbox{ km},$$ with an average circumference of $2\pi\cdot 6371\mbox{ km}=40030\mbox{ km},~$ get lost of the 20015 km of the half circumference of Earth as zone of visibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculation is simplified to the simultaneous visibility of a point at the distance of the center of moon. More precisely one part of the moon can be visible from one observer, while an other part of the moon is visible from a second observer. This adds between 29.3 and 34.1 arc minutes or between 54.3 (apogee) and 63.2 km (perigee) to the diameter of the zone of visibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The visibility of the Moon is influenced by additional factors like air temperature, height of the observer above sea level or the geographic position by Earth's flattening. But this applies also for the visibility of the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding calculations for the sun: With aphelion of Earth of 152098232 km and perihelion of 147098290 km, we get ratios of 6371.0 km / 152098232 km = 0.000041887 = sin 0.0024000° resp. 6371.0 km / 147098290 km = 0.000043311 = sin 0.0024815°.&#xA;So on both sides of Earth between $$40030\mbox{ km}\cdot 0.0024000°/360°=0.26686\mbox{ km}=266.86\mbox{ m}$$&#xA;and&#xA;$$40030\mbox{ km}\cdot 0.0024815/360°=0.27780\mbox{ km}=277.80\mbox{ m}$$&#xA;get lost of the 20015 km of the half circumference of Earth as zone of visibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For a rough estimate you may take the 100 km loss of the moon visibility, and multiply it by the quotient of the distances Earth-Moon and Earth-Sun roughly equal to 400,000 km / 150,000,000 km = 0.002667, to get 266.7 m visibility loss of the sun.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is again simplified, and applies to the center of the sun. The appearent diameter of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sun&quot; rel=&quot;nofollow&quot;&gt;sun&lt;/a&gt; varies between 31.6 and 32.7 arc minutes, adding between 58.6 (aphelion) and 60.6 km (perihelion) to the diameter of the zone of visibility for observers looking at different parts of the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on which scenarios to be compared, the lost area varies. As an example, a loss of a radius of the visible zone of the moon in comparison to the sun of 100 km corresponds to a roughly cylindrical area of $$100\mbox{ km}\cdot \pi\cdot 40030\mbox{ km}=12.6 \mbox{ million square kilometers}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How far can two observers be apart at maximum, to see two differnt parts of the moon at the same time? $$20015 \mbox{ km} - 2\cdot 100 \mbox{ km} + 54.3 \mbox{ km} = 19869.3 \mbox{ km}$$ (apogee) for the average Earth radius. For the same part of the moon it's $$20015 \mbox{ km} - 2\cdot 100 \mbox{ km} = 19815\mbox{ km}.$$&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-02-26T10:15:29.547" LastActivityDate="2014-02-26T10:15:29.547" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1842" PostTypeId="1" CreationDate="2014-02-25T04:51:51.770" Score="3" ViewCount="328" Body="&lt;p&gt;The boundary of a black hole is said to be surrounded by event horizon - the point of no return! What is its significance in terms of general relativity? &lt;/p&gt;&#xA;" OwnerUserId="1100" LastActivityDate="2014-02-27T01:43:52.833" Title="What is Event Horizon of a Black Hole?" Tags="&lt;black-hole&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2014-03-10T18:38:21.217" ContentLicense="CC BY-SA 3.0" />
  <row Id="1843" PostTypeId="2" ParentId="1836" CreationDate="2014-02-25T08:43:57.920" Score="4" Body="&lt;p&gt;I guess it would be influenced. Right now, our vision is already influenced by that &lt;a href=&quot;http://patarnott.com/atms749/pdf/blueSkyHumanResponse.pdf&quot; rel=&quot;nofollow&quot;&gt;massive blue sky above us&lt;/a&gt;. The human visual system corresponds to the large influx of blue light that gets scattered in the atmosphere. Having a red or a blue star would probably also have some influence, as would have a different kind of atmosphere (which is not mainly nitrogen and oxygen). The brain would probably compensate, but the rods in the eye would have a different composition.&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2014-02-25T08:43:57.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1844" PostTypeId="2" ParentId="1842" CreationDate="2014-02-25T08:59:25.613" Score="3" Body="&lt;p&gt;It is exactly the point of no return for light. In other words, it is the point (actually more like a sphere) in which the escape velocity from the Black Hole gravity reaches $c$.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-02-25T08:59:25.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1845" PostTypeId="1" AcceptedAnswerId="1917" CreationDate="2014-02-25T07:47:38.917" Score="5" ViewCount="874" Body="&lt;p&gt;Recently, I visited the NASA website which provides sighting locations to see the &lt;a href=&quot;https://en.wikipedia.org/wiki/International_Space_Station&quot; rel=&quot;nofollow&quot;&gt;International Space Station&lt;/a&gt; (ISS). As we all know, we can even see ISS with the naked eye in a clear sky as a passing star without any blinking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But unfortunately, I don't know how to measure the angle on the sky to see it. On the official NASA website there's given height (altitude) and direction as 21° and 11 above N respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the link: &lt;a href=&quot;http://spotthestation.nasa.gov/sightings/view.cfm?country=India&amp;amp;region=None&amp;amp;city=Hyderabad#.UwxIRs6haTs&quot; rel=&quot;nofollow&quot;&gt;NASA - Spot The Station&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I trace ISS?&lt;/p&gt;&#xA;" OwnerUserId="1103" OwnerDisplayName="Praveen Kadambari" LastEditorUserId="332" LastEditDate="2014-03-10T12:19:07.380" LastActivityDate="2014-09-22T07:45:07.370" Title="Spotting the International Space Station (ISS)" Tags="&lt;orbit&gt;&lt;amateur-observing&gt;&lt;nasa&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1847" PostTypeId="2" ParentId="1838" CreationDate="2014-02-25T19:34:04.580" Score="2" Body="&lt;p&gt;The function of a telescope or your eye is to translates angles into distances on a detector. You can see in the figure below (from Hyperphysics' excellent telescope page: &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/teles.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/teles.html&lt;/a&gt;) that rays of light coming from a single object (such an unresolved point of light from a star in space) enter the telescope from a single direction (angle on the sky), in parallel. They pass through the telescope and an image is formed on the back of your eye (Or in Hubble's case, on an electronic detector, which you can think of being were the little red arrow is). The image is a representation of those angles as positions separated by some distance on a 2-D surface.  On the detector all of the light from that single angle angle is light at a single point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BdoYq.gif&quot; alt=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/imggo/astel.gif&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If your pupils constrict, you don't see less of the world, the angles entering your eye stay the same but the number of rays from each angle decreases and less light from every object (light from every direction) gets dimmer.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The opening at the front of Hubble is like your eye's pupil, if it were smaller, less light would come in, but just as many directions could be seen. The secondary mirror being in the way just means Hubble different shaped pupil from yours, a few rays from the object don't make it to the detector, but the image doesn't change, just the brightness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the secondary mirror were not there (as in a harder to manufacturer refracting telescope) Hubble would collect a little bit more light from each direction and could observe things a little quicker, but the view of the sky on the detector would be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(SOHO blocks the sun by effectively placing a block in an image formed inside the telescope, for example where the arrow showing the image is in the simple telescope above, then SOHO's &quot;eye&quot; only sees the rays from around the edges of the sun, by blocking them at an &quot;image plane&quot;, this is a &quot;coronagraph&quot;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Coronagraph&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Coronagraph&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="457" LastEditorUserId="457" LastEditDate="2014-02-26T00:43:55.153" LastActivityDate="2014-02-26T00:43:55.153" CommentCount="2" CommunityOwnedDate="2014-02-25T19:34:04.580" ContentLicense="CC BY-SA 3.0" />
  <row Id="1848" PostTypeId="2" ParentId="1830" CreationDate="2014-02-25T22:22:32.270" Score="2" Body="&lt;p&gt;This question not only makes you think about the temperature, pressure but also the &lt;strong&gt;impact of gravity&lt;/strong&gt;. Not moving away from topic, I must say that the region you are saying about is &lt;strong&gt;quite large in the view of metrics but quite short again if we really sense gravity.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance of a planet from the sun is maintained by gravity, but we must also not forget that distance between two planets is also controlled by the same. You must appreciate gravity because most of the features on earth are because of the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you want to calculate it more precisely you should know each and every postulates, theorems and properties shown by the space in nourishing our Earth for habitability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perturbation may give you an idea of adjustments, that how our earth is adjusting day by day for its decreasing distance from sun or increasing size of the sun. Talking according to theory of perturbation there &lt;strong&gt;may be more than 2-3 planets in the habitable zone&lt;/strong&gt;. But still you &lt;strong&gt;cannot annoy the word &quot;more&quot;&lt;/strong&gt; in that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It shall depend mainly on the following factors : -&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Size of Planets&lt;/li&gt;&#xA;&lt;li&gt;Distance between each of them&lt;/li&gt;&#xA;&lt;li&gt;Distance from sun accordingly&lt;/li&gt;&#xA;&lt;li&gt;Their mass (priorly)&lt;/li&gt;&#xA;&lt;li&gt;Speed of revolution around sun&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I know this information is not much for your question but this is enough according to recent knowledge about space. We even don't know the drastic changes happening in the space....the dimensions of space or how vast it is....whether it is measurable or not.....and if yes then are the measurements precise. Each answer tackles you more to have questions....But don't stop, just keep asking!&lt;/p&gt;&#xA;" OwnerUserId="1096" LastActivityDate="2014-02-25T22:22:32.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1850" PostTypeId="1" AcceptedAnswerId="1861" CreationDate="2014-02-26T11:24:32.597" Score="14" ViewCount="2286" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/E1IuE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any reason why the Earth side of the Moon has enormous plains and mountains, while the far side contains just few small ones inside largest craters?&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2014-03-12T01:10:56.027" Title="Why are most lunar maria on the visible side?" Tags="&lt;the-moon&gt;&lt;geology&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1851" PostTypeId="1" AcceptedAnswerId="1853" CreationDate="2014-02-26T17:42:46.487" Score="8" ViewCount="337" Body="&lt;p&gt;You can watch a video of a meteorite impacting the moon &lt;a href=&quot;http://www.npr.org/blogs/thetwo-way/2014/02/24/282201355/meteorite-impact-on-moon-sets-record-as-brightest-ever-seen&quot; rel=&quot;noreferrer&quot;&gt;at the npr.org website&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the impact produce white light?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the impact is like an explosion, but there is no air on the moon to burn material. If it was the result of superheating rock from the impact, then why does it fade so fast. White light emitted by heated rocks should last longer as it takes time to cool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see meteorites breaking up in Earth's atmosphere as a streak of light, but there is no atmosphere on the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why isn't it just a big black puff of smoke?&lt;/p&gt;&#xA;" OwnerUserId="695" LastEditorUserId="181" LastEditDate="2014-03-02T13:24:15.453" LastActivityDate="2014-03-02T13:24:15.453" Title="Why is a meteorite impact on the moon visible as white light?" Tags="&lt;the-moon&gt;&lt;meteorite&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1852" PostTypeId="1" AcceptedAnswerId="1854" CreationDate="2014-02-26T17:49:05.663" Score="10" ViewCount="9159" Body="&lt;p&gt;In the TV show &quot;Big Bang Theory&quot; episode &quot;&lt;a href=&quot;http://bigbangtheory.wikia.com/wiki/The_Lunar_Excitation&quot; rel=&quot;noreferrer&quot;&gt;The Lunar Excitation&lt;/a&gt;&quot;, the gang fires a laser from their rooftop, bounces it off mirrors on the moon, and measures the laser coming back on a computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this really possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know scientists have successfully done this, because it's why NASA put the mirrors on the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wouldn't this require very precise targeting by the laser to hit the mirrors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, is it possible for an amateur to successfully perform this experiment?&lt;/p&gt;&#xA;" OwnerUserId="695" LastEditorUserId="202" LastEditDate="2014-02-27T15:30:06.880" LastActivityDate="2014-02-27T15:30:06.880" Title="Can an amateur astronomer bounce a laser off the moon?" Tags="&lt;the-moon&gt;&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1853" PostTypeId="2" ParentId="1851" CreationDate="2014-02-26T20:02:34.013" Score="10" Body="&lt;p&gt;The velocity of the impactor of tens of kilometers per second provides enough energy to heat the impactor and parts of the target to several thousands of Kelvins, so that parts are converted to plasma or to vapor, at least.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck%27s_law&quot;&gt;Planck's law&lt;/a&gt; the color at these temperatures is white or bluish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law&quot;&gt;Stefan Boltzman law&lt;/a&gt; the total emitted energy is proportional to the forth power of the absolute temperature. Hence material which isn't heated to thousands of degrees, and may glow yellowish or reddish is too dim to be noticible in comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since most of the heated ejecta are finely distributed, they cool down rapidly by the emitted radiation, and also after contact with the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After cooling there will certainly be some fine dust, which still has to fall back to Moon's surface, but you won't see this on a video from a distance, since it's cooled down and distributed over a wider area.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-26T20:02:34.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1854" PostTypeId="2" ParentId="1852" CreationDate="2014-02-26T20:32:22.657" Score="5" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment&quot; rel=&quot;noreferrer&quot;&gt;This experiment&lt;/a&gt; is still done frequently, but you need a powerful pulsed laser, precise pointing (error below about 20 arc seconds) and a sensitive detector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amateurs will fail most likely, because they won't get access to an appropriate equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=LpjbdH1y_ds&quot; rel=&quot;noreferrer&quot;&gt;Here is a youtube video&lt;/a&gt;, which may give an impression of how difficult it is to perform the experiment.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-26T20:32:22.657" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1855" PostTypeId="1" CreationDate="2014-02-26T20:49:42.103" Score="5" ViewCount="5337" Body="&lt;p&gt;How do I get the longitude/latitude from zenith/azimuth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or how can I position my sun in a 3d application? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;code extracted from &lt;a href=&quot;https://github.com/KlausBrunner/solarpositioning&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;public static AzimuthZenithAngle getSolarAzimuthZenithFromLatLon(float lat, float lon) {&#xA;   GregorianCalendar time = new GregorianCalendar(TimeZone.getDefault());&#xA;   return PSA.calculateSolarPosition(time, lat, lon);&#xA;}&#xA;&#xA;public static Vector3 getSolarPosition(float radius, AzimuthZenithAngle az) {&#xA;    double a = az.getAzimuth();&#xA;    double z = az.getZenithAngle();&#xA;&#xA;    float X  = (float) (radius*Math.sin(z)*Math.cos(a));&#xA;    float Y  = (float) (radius*Math.sin(z)*Math.sin(a));&#xA;    float Z  = (float) (radius*Math.cos(z));&#xA;&#xA;    return new Vector3(X, Y, Z);&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;this is my code, but it's not working very well. &lt;/p&gt;&#xA;" OwnerUserId="1114" LastEditorUserId="537" LastEditDate="2014-02-26T23:11:54.737" LastActivityDate="2017-08-30T15:52:39.317" Title="How to get the longitude/latitude from solar zenith/azimuth?" Tags="&lt;the-sun&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1856" PostTypeId="2" ParentId="1842" CreationDate="2014-02-27T01:43:52.833" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What I mean to say is, how are the characteristics of event horizon related to Space-Time?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Formally, an event horizon is the boundary of a region of spacetime that's not in the causal past of future null infinity. In other words, the boundary of a region from which even idealized light rays cannot escape to infinity. Whenever the event horizon is smooth, it is also null hypersurface--i.e., the direction perpendicular to it is a light ray.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the event horizon is defined in terms of the infinite future, the definition is very non-local, and one would need to know the entire future history to be sure where the event horizon is. As such, it is only one of a half-dozen different types of horizons used to study black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though the event-horizon itself is a three-dimensional hypersurface in spacetime, it can also be viewed as an evolving two-dimensional membrane made of a viscuous electrically conducting fluid with finite temperature and entropy but zero thermal conductivity.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-02-27T01:43:52.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1857" PostTypeId="2" ParentId="1837" CreationDate="2014-02-27T04:38:37.987" Score="1" Body="&lt;p&gt;The simple answer to your question is yes. Taking a simplified equation from Carroll &amp;amp; Ostlie, An Introduction to Modern Astrophysics Second Edition, the temperature of a planet can be estimated as:&#xA;$$&#xA;T_{p} = T_{\odot}(1-a)^{\frac{1}{4}}\sqrt{\frac{R_\odot}{2D}}&#xA;$$&#xA;Where $T_p$ is the predicted temperature of a planet in a circular orbit of radius $D$ with an albedo of $a$ around a star with a temperature of $T_\odot$ and a radius of $R_\odot$. If the energy output of the star were to increase, raising $T_\odot$, then there would be a corresponding increase in the temperature of all planets orbiting said star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice there are factors which can make this correlation difficult to measure. The albedo  of a planet during the course of a day can vary greatly and the distance of a planet from the host star changes throughout the year. This equation also assumes that the planet is a perfect black body which most are not which can also change a planets temperature and obscure any changes caused by the host star.  &lt;/p&gt;&#xA;" OwnerUserId="1118" LastActivityDate="2014-02-27T04:38:37.987" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1859" PostTypeId="1" AcceptedAnswerId="1862" CreationDate="2014-02-27T08:01:17.637" Score="7" ViewCount="6120" Body="&lt;p&gt;According to various theories the Moon was created around 4.5 billion years ago. About all of these theories suggest that it was rotating around its axis at that time though. Currently, Moon is at tidal lock with Earth, despite some monthly &quot;wiggling&quot; a flat zero on the long-term rotation speed relative to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to ask when did that stop occur - relative to Moon's age, how long was the period of rotating Moon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer would shed some light on my other question - &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1850/why-are-most-lunar-maria-on-the-visible-side&quot;&gt;Why are most lunar maria on the visible side?&lt;/a&gt; as Earth tends to catch or deflect many bodies heading for Moon surface from &quot;our&quot; direction - still, there is no erosion on the Moon, so craters once formed are extremely slow to vanish - if that period was relatively long, Earth's &quot;protection&quot; wouldn't explain the maria, as rotating Moon would get 'cratered' uniformly all over its surface.&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="43" LastEditDate="2020-04-27T09:47:14.620" LastActivityDate="2020-04-27T09:47:14.620" Title="When did the Moon become tidally locked to Earth?" Tags="&lt;the-moon&gt;&lt;tidal-forces&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="1860" PostTypeId="1" AcceptedAnswerId="1863" CreationDate="2014-02-27T13:02:20.537" Score="5" ViewCount="361" Body="&lt;p&gt;I've hear several theories stating that the universe is expanding faster than the speed of light, others claim that the universe expands faster the further away you measure it. Which of this is correct and how do you prove it (mathematically)? Furthermore, does this correctly imply, then, that eventually galaxies will be so far away, and moving so fast, that we will never see them again?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance!&lt;/p&gt;&#xA;" OwnerUserId="1111" LastActivityDate="2017-03-15T10:10:36.617" Title="How fast is the universe expanding?" Tags="&lt;galaxy&gt;&lt;space&gt;&lt;astrophysics&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1861" PostTypeId="2" ParentId="1850" CreationDate="2014-02-27T13:07:13.660" Score="10" Body="&lt;p&gt;The two hemispheres have distinctly different appearances, with the near side covered in multiple, large maria (Latin for 'seas,' since the earliest astronomers incorrectly thought that these plains were seas of lunar water). The far side has a battered, densely cratered appearance with few maria. Only 1% of the surface of the far side is covered by maria, compared to 31.2% on the near side. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One commonly accepted explanation for this difference is related to a higher concentration of heat-producing elements on the near-side hemisphere, as has been demonstrated by geochemical maps obtained from the Lunar Prospector gamma-ray spectrometer. While other factors such as surface elevation and crustal thickness could also affect where basalts erupt, these do not explain why the farside South Pole-Aitken basin (which contains the lowest elevations of the Moon and possesses a thin crust) was not as volcanically active as Oceanus Procellarum on the near side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has also been proposed that the differences between the two hemispheres may have been caused by a collision with a smaller companion moon that also originated from the Theia collision. In this model the impact led to an accretionary pile rather than a crater, contributing a hemispheric layer of extent and thickness that may be consistent with the dimensions of the farside highlands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though the far side has more visible craters, this is due to the effects of lunar lava flows, which cover and obscure craters, rather than a shielding effect from the Earth. NASA calculates that the Earth obscures only about 4 square degrees out of 41,000 square degrees of the sky as seen from the moon. &quot;This makes the Earth negligible as a shield for the Moon. [...] It is likely that each side of the Moon has received equal numbers of impacts, but the resurfacing by lava results in fewer craters visible on the near side than the far side, even though the both sides have received the same number of impacts.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Far_side_of_the_Moon&quot; rel=&quot;noreferrer&quot;&gt;wikipedia page&lt;/a&gt; and &lt;a href=&quot;http://lunarscience.nasa.gov/?question=3318&quot; rel=&quot;noreferrer&quot;&gt;an other source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: to complete the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/F2KgF.jpg&quot; alt=&quot;The moon’s centre of mass is not its geometric centre.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From space craft orbiting the moon, it has been found that the centre of mass of the moon is displaced toward Earth. An explanation of seas distribution states that lunar core, aesthenosphere and the mantle is off-centre, whereas the crust is not — so the crust is thicker on the far side (see figure). Thus the basalt would reach the surface more easily on the near side due to thinner crust (to 60 km near side, 100 km far) than on the far side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://creation.com/lunar-maria&quot; rel=&quot;noreferrer&quot;&gt;this source&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastEditorUserId="1038" LastEditDate="2014-03-10T06:58:12.560" LastActivityDate="2014-03-10T06:58:12.560" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1862" PostTypeId="2" ParentId="1859" CreationDate="2014-02-27T14:16:59.483" Score="14" Body="&lt;p&gt;&quot;Protection&quot; isn't the only effect of Earth. &lt;a href=&quot;http://www.scientificpsychic.com/etc/timeline/moon-asteroid-impacts.html&quot; rel=&quot;noreferrer&quot;&gt;Here is a different POV&lt;/a&gt;: Earth may have accelerated impactors by &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravity_assist&quot; rel=&quot;noreferrer&quot;&gt;gravity assist&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A different approch is the thinner crust, as suggested for the near side, which may have allowed asteroids to penetrate Moon's crust, such that lava could flow into the basins, or which may have favoured volcanism on the near side (see &quot;Lunar interior&quot; on &lt;a href=&quot;http://ase.tufts.edu/cosmos/print_images.asp?id=5&quot; rel=&quot;noreferrer&quot;&gt;this site&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A third approach is the protective property of Earth preventing the near side to be covered with many new craters, hence leave the maria visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Tidal_locking&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; the time to lock tidally is about&#xA;$$t_{\mbox{lock}}=\frac{wa^6IQ}{3G{m_p}^2k_2R^5},$$&#xA;with $$I=0.4m_sR^2.$$&#xA;For Moon $k_2/Q = 0.0011$, hence&#xA;$$t_{\mbox{lock,Moon}}=121\frac{wa^6m_s}{G{m_p}^2R^3}.$$&#xA;With &lt;a href=&quot;http://en.wikipedia.org/wiki/Earth&quot; rel=&quot;noreferrer&quot;&gt;Earth&lt;/a&gt;'s mass $m_p=5.97219\cdot 10^{24}\mbox{ kg}$, &lt;a href=&quot;http://en.wikipedia.org/wiki/Moon&quot; rel=&quot;noreferrer&quot;&gt;Moon&lt;/a&gt;'s mass $m_s=7.3477\cdot 10^{22}\mbox{ kg}$, Moon's mean radius of $R=1737.10\mbox{ km}$, $G=6.672\cdot 10^{-11}\frac{\mbox{Nm}^2}{\mbox{kg}^2}$we get&#xA;$$t_{\mbox{lock,Moon}}=121\frac{wa^67.3477\cdot 10^{22}\mbox{ kg}}{6.672\cdot 10^{-11}\frac{\mbox{Nm}^2}{\mbox{kg}^2}\cdot{(5.97219\cdot 10^{24}\mbox{ kg})}^2(1737.10\mbox{ km})^3},$$&#xA;or&#xA;$$t_{\mbox{lock,Moon}}=7.12753\cdot 10^{-25}wa^6 \frac{\mbox{kg}}{\mbox{Nm}^2 \mbox{km}^3}.$$&#xA;Parameters are $w$ the spin rate in radians per second, and $a$ the semi-major axis of the moon orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take the the current simi-major axis of the moon orbit of 384399 km&#xA;and a maximum possible spin rate of&#xA;$$w=v/(2\pi R)=\frac{2.38 \mbox{ km}/\mbox{s}}{2\pi\cdot 1737.10\mbox{ km}}=\frac{1}{4586 \mbox{ s}},$$&#xA;with $v=2.38 \mbox{ km}/\mbox{s}$, Moon's escape velocity, 1737.1 km Moon's radius,&#xA;we get&#xA;$$t_{\mbox{lock,Moon}}=7.12753\cdot 10^{-25}\cdot \frac{1}{4586 \mbox{ s}}\cdot (384399\mbox{ km})^6 \frac{\mbox{kg}}{\mbox{Nm}^2 \mbox{km}^3}\\&#xA;=501416\mbox{ s}^{-1}\cdot \mbox{ km}^6 \frac{\mbox{kg}}{\mbox{Nm}^2 \mbox{km}^3}=&#xA;5.01416\cdot 10^{14} \mbox{ s}.$$&#xA;That's about 16 million years, as an upper bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume a higher &lt;a href=&quot;http://en.wikipedia.org/wiki/Love_number&quot; rel=&quot;noreferrer&quot;&gt;Love number&lt;/a&gt; for the early moon, or slower initial rotation, the time may have been shorter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time for getting locked is very sensitive to the distance Earth-Moon (6th power). Hence if tidal locking occurred closer to Earth, the time will have been shorter, too. That's likely, because Moon is spiraling away from Earth.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-03-01T12:03:25.103" LastActivityDate="2014-03-01T12:03:25.103" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1863" PostTypeId="2" ParentId="1860" CreationDate="2014-02-27T14:46:35.673" Score="1" Body="&lt;p&gt;This was supposed to be more a comment than an answer, but since I can not comment due to reputation lack I will spend few words here.&#xA;First of all, the &quot;theories&quot; you mentioned are not inconsistent each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know the simple &lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble%27s_law&quot; rel=&quot;nofollow&quot;&gt;Hubble law&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v = H D$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $v$ is the receding velocity of a galaxy, $H$ is the Hubble constant, $D$ is the distance of the considered galaxy. This means that the further is the galaxy you observe, the faster this galaxy is receding. At some point it will become faster than light (or &lt;a href=&quot;http://en.wikipedia.org/wiki/Faster-than-light&quot; rel=&quot;nofollow&quot;&gt;superluminal&lt;/a&gt;). At some point, the space between us and the light emitter will grow so fast that the light can never reach us, and this will make those objects invisible. Indeed, all we can observe is by definition our &lt;a href=&quot;http://en.wikipedia.org/wiki/Observable_universe&quot; rel=&quot;nofollow&quot;&gt;observable universe&lt;/a&gt;. This is growing with time, but still some objects will stay invisible forever. The very first thing you mention, I suppose you should put it more correctly, since it should be better to talk about &lt;em&gt;expansion rate&lt;/em&gt; of the universe (instead of velocity), and this is given itself by the Hubble constant, around $70 km/s/Mpc$. Take care of the units of this &quot;constant&quot;, and you will grasp why this argument is not so intuitive.&#xA;Please, wait for more experienced people, since this was just a very rough summary of cosmology concepts.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastActivityDate="2014-02-27T14:46:35.673" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1864" PostTypeId="2" ParentId="1860" CreationDate="2014-02-27T14:47:17.047" Score="3" Body="&lt;p&gt;Both are correct, although the first can be further explained a bit. I won't give you a mathematical proof, though; Instead I'll play with characters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume, for the sake of gedankenexperiment, that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The speed of light is 5 characters/sec;&lt;/li&gt;&#xA;&lt;li&gt;our universe is expanding at 1 character per 5 characters per sec.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is our current universe, and we launch a photon from body A aiming at body E.&#xA;(Space generated each second is marked with a &lt;code&gt;#&lt;/code&gt; symbol.)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T=0s A----B----C----D----E           Bodies&#xA;     *                               Photon     - 19 chars to E&#xA;T=1s A--#--B--#--C--#--D--#--E &#xA;      --#--*                                      17 chars to E&#xA;T=2s A--#---B-#----C#-----#D----#-E &#xA;      --#-----#-*                                 17 chars to E&#xA;T=3s A--#----B#-----#-C---#----D#-----#-E &#xA;      --#-----#-----#*                            18 chars to E&#xA;T=4s A--#-----#B----#----C#-----#---D-#-----#--E &#xA;      --#-----#-----#-----#*                      19 chars to E&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I know, the graph isn't too granular, and the space generation isn't evenly distribute. I apologize for that, but it's for the sake of demonstration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that at &lt;code&gt;T=2&lt;/code&gt; some space is already generated between A and the photon. But that's irrelevant: E is sitting at the event horizon, and will never be reached by the photon, because the amount of space being generated between photon &lt;code&gt;*&lt;/code&gt; and body &lt;code&gt;E&lt;/code&gt; is equal, or superior, to the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any positive expansion rate, there will be an event horizon - a point where the accumulated dilation of space is more than the amount of space a particle moving at the speed of light can travel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A galaxy sitting initially at say, 1000 characters from A at T=0, will be at staggering 1200C at T=1 - that's 40 times our speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At T=16s, B (that was passed by the original photon at T=1) will be sitting exactly where E was relative to A, and at T=17 will fall out of our event horizon. A new photon emitted from A will never reach it.&lt;/p&gt;&#xA;" OwnerUserId="893" LastEditorUserId="893" LastEditDate="2014-02-27T18:48:47.167" LastActivityDate="2014-02-27T18:48:47.167" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1865" PostTypeId="2" ParentId="1860" CreationDate="2014-02-27T14:49:06.687" Score="1" Body="&lt;p&gt;The universe expands with about 70 km/s per Mega parsec, due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble%27s_law&quot; rel=&quot;nofollow&quot;&gt;Hubble's law&lt;/a&gt;. This means, that the velocity two objects move away from each other is proportional to their distance. At one Mega parsec it's 70 km/s. That's an average value, which doesn't need to hold for each single object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By dividing the speed of light of about 300,000 km/s by the Hubble constant, you get, that objects further away than about 4300 Mega parsecs move faster away from each other than the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expansion is measured by the redshift of spectra, meaning absorption and emission lines are shifted. Together with distance estimates, based on several methods, the expansion per distance i.e. the Hubble constant, can be estimated.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-27T14:49:06.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1866" PostTypeId="1" AcceptedAnswerId="1867" CreationDate="2014-02-27T16:12:36.577" Score="5" ViewCount="6812" Body="&lt;p&gt;Once while camping with my family as a child we all watched a meteor shower in the summer (in the month of August).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the meteorites streaked across the sky in the same direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, this was the result of the Earth passing thru a group of meteors but I've always wondered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do meteorites always streak in the same direction? While I know it's not likely to be exactly the same direction, but generally speaking. Would one say meteorites always travel for example: &lt;code&gt;east to west&lt;/code&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought this was a plausible fact since the Earth is travelling around the Sun in a constant direction, and additionally it's always rotating in the same direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While meteorites can travel in all directions. Does the Earth's gravity pull them straight down towards the surface resulting in the streak to be always in the same direction.&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2020-08-03T06:36:50.377" Title="Do meteorites streak across the sky in the same direction?" Tags="&lt;meteor&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1867" PostTypeId="2" ParentId="1866" CreationDate="2014-02-27T17:12:26.983" Score="12" Body="&lt;p&gt;No, meteorites of the same shower look as coming from the same point in the sky. They are usually named according to the point they seen to come from, e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Leonids&quot;&gt;the leonids&lt;/a&gt;, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Perseids&quot;&gt;the perseids&lt;/a&gt;. The point, they seem to originate is a result of the relative velocity of the meteorites to the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meteor showers are thought to be remnants of comets, hence the similar orbits of the meteorites within a shower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_meteor_showers&quot;&gt;Here a more complete list&lt;/a&gt; of &lt;a href=&quot;http://en.wikipedia.org/wiki/Meteor_shower&quot;&gt;meteor showers&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-27T17:12:26.983" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1868" PostTypeId="1" AcceptedAnswerId="1881" CreationDate="2014-02-27T17:47:22.780" Score="1" ViewCount="82" Body="&lt;p&gt;This question may sound silly but I'm really not pretty sure whether the stars are rotating or stationary.If they are not rotating what makes them to be stable?&lt;/p&gt;&#xA;" OwnerUserId="1103" LastActivityDate="2014-03-02T03:14:43.867" Title="State of the stars" Tags="&lt;star&gt;&lt;orbit&gt;&lt;astrophysics&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1869" PostTypeId="2" ParentId="1868" CreationDate="2014-02-27T18:00:14.713" Score="2" Body="&lt;p&gt;Stars are rotating, but that's not the cause for their stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stability is caused by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrostatic_equilibrium&quot; rel=&quot;nofollow&quot;&gt;hydrostatic equilibrium&lt;/a&gt; between gas/plasma pressure and gravity. The pressure is mainly caused by the heat of several millions degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More details &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_structure&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-02-27T18:00:14.713" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1870" PostTypeId="1" AcceptedAnswerId="1874" CreationDate="2014-02-27T18:17:48.457" Score="0" ViewCount="109" Body="&lt;p&gt;What causes an asteroid to loose control over it's orbit and enter in to another objects atmosphere?&lt;/p&gt;&#xA;" OwnerUserId="1103" LastActivityDate="2014-03-10T18:37:21.387" Title="Asteroids turning out to be Meteorites" Tags="&lt;orbit&gt;&lt;asteroids&gt;&lt;meteorite&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2014-03-10T18:37:26.590" ContentLicense="CC BY-SA 3.0" />
  <row Id="1871" PostTypeId="2" ParentId="1870" CreationDate="2014-02-27T18:58:55.200" Score="1" Body="&lt;p&gt;Asteroids don't have control over their orbit. The orbits are a consequence of the laws of physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This way asteroids can just happen to cross the orbit of an other celestial body. Products of such a collision take new orbits, which may or may not result in new collisions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbits, which had been free of intersections with orbits of other bodies, may get moved away from &lt;a href=&quot;http://en.wikipedia.org/wiki/Kepler%27s_equation&quot; rel=&quot;nofollow&quot;&gt;Kepler ellipses&lt;/a&gt; over time by the gravitational pull of other celestial objects, e.g. planets like Jupiter or even by stars lightyears away. An orbit can also leave its ellipsis within a short time due to a close encounter with a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbits around a planet can &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_decay&quot; rel=&quot;nofollow&quot;&gt;decay&lt;/a&gt; e.g. due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Mass_concentration_%28astronomy%29&quot; rel=&quot;nofollow&quot;&gt;uneven mass distribution&lt;/a&gt; on the planet.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-02-28T08:54:46.437" LastActivityDate="2014-02-28T08:54:46.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1872" PostTypeId="1" AcceptedAnswerId="1873" CreationDate="2014-02-27T19:35:11.740" Score="3" ViewCount="45" Body="&lt;p&gt;I was interested in tracking the orbit of 1986 DA. I couldn't find any resource when I looked. Just for fun, so any help appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1125" LastActivityDate="2014-02-27T22:10:23.653" Title="Locate asteroid in asteroid belt" Tags="&lt;orbit&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1873" PostTypeId="2" ParentId="1872" CreationDate="2014-02-27T20:06:40.727" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://www.minorplanetcenter.net/mpec/J94/J94T11.html&quot; rel=&quot;nofollow&quot;&gt;It has been observed 1977, and in 1994&lt;/a&gt;.&#xA;It should be near its perihelion this year (2014) around august 7 (using &lt;a href=&quot;http://en.wikipedia.org/wiki/%286178%29_1986_DA&quot; rel=&quot;nofollow&quot;&gt;wikipedia data&lt;/a&gt;, perihelion at March 6, 2005 and period 4.71 years), give or take a few days.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-02-27T20:27:30.993" LastActivityDate="2014-02-27T20:27:30.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1874" PostTypeId="2" ParentId="1870" CreationDate="2014-02-27T22:03:47.457" Score="3" Body="&lt;p&gt;Asteroids don't have control of their orbit. They are influenced by gravity; they obey the laws of physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There aren't any traffic lights for them in space to obey so they know to slow down and stop to avoid a collision with a planet. They'll just carry right on going like someone driving through a red light. If it goes right in front of a planet at the wrong moment then it gets t-boned. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or it might miss. But here is where the analogy falls apart. Because if someone drives through a red light and misses everyone, they carry on without any subsequent repercussions (except maybe a ticket). An asteroid passing close by a planet will be influenced by gravity between the planet and the asteroid, and it's path will change. That might mean it is flung far away, or it may mean it is captured, putting it on a looping path that eventually draws it in to crash onto the planet.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-02-27T22:03:47.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1875" PostTypeId="2" ParentId="1872" CreationDate="2014-02-27T22:10:23.653" Score="0" Body="&lt;p&gt;The IAU's &lt;a href=&quot;http://www.minorplanetcenter.net/&quot; rel=&quot;nofollow&quot;&gt;Minor Planet Center&lt;/a&gt; is the authoritative place to go to for information on any asteroid.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-02-27T22:10:23.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1876" PostTypeId="1" AcceptedAnswerId="1878" CreationDate="2014-02-28T06:22:38.427" Score="2" ViewCount="591" Body="&lt;p&gt;From what I understand the universe is expanding faster than the speed of light and accelerating so that it's possible that a given galaxy: $A$ will no longer be visible to another given galaxy: $B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if a light wave is emitted from galaxy $A$ toward galaxy $B$ after it has passed the point where neither galaxy is visible to each other, how can the speed of the light wave be measured if there is no fixed point in space for it to be measure against (as the light wave will never be visible to either galaxy again)?&lt;/p&gt;&#xA;" OwnerUserId="953" LastEditorUserId="181" LastEditDate="2014-03-03T01:25:39.070" LastActivityDate="2016-10-08T02:06:27.280" Title="What is the speed of light relative to (in space)?" Tags="&lt;galaxy&gt;&lt;universe&gt;&lt;expansion&gt;" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1878" PostTypeId="2" ParentId="1876" CreationDate="2014-02-28T11:06:26.730" Score="3" Body="&lt;p&gt;To any observer between 'A' and 'B' where the light wave passes, it passes with &lt;a href=&quot;http://en.wikipedia.org/wiki/Speed_of_light&quot; rel=&quot;nofollow&quot;&gt;the speed of light&lt;/a&gt;. Adding a velocity $v$ below the speed of light $c$ to the speed of light results in the speed of light, simplified $c+v=c$, where the '$+$' means &lt;a href=&quot;http://en.wikipedia.org/wiki/Velocity-addition_formula&quot; rel=&quot;nofollow&quot;&gt;relativistic addition of velocities&lt;/a&gt;, not the '$+$' of arithmetics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the relative velocity of two observers at the (almost) same point they will observe different &lt;a href=&quot;http://en.wikipedia.org/wiki/Relativistic_Doppler_effect&quot; rel=&quot;nofollow&quot;&gt;Doppler shifts&lt;/a&gt; of the light, more precisely of spectral properties of the light. By analysing the &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectral_line&quot; rel=&quot;nofollow&quot;&gt;spectrum of the light&lt;/a&gt; the observers may be able to determine the Doppler shift, meaning relative velocity, to the object which caused the emission or absorption line in the spectrum.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-02-28T12:05:55.083" LastActivityDate="2014-02-28T12:05:55.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1879" PostTypeId="2" ParentId="1876" CreationDate="2014-02-28T11:19:53.843" Score="-1" Body="&lt;p&gt;There is nothing special about the point in space where the galaxies move out of view of each other. This is just the point where their respective Hubble spheres (the sphere at which the expansion speed is equal to the speed of light for the given observer) overlap. The speed of the light will still be c from the perspective of an observer between the Hubble spheres of A and B&lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2014-02-28T11:19:53.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1880" PostTypeId="2" ParentId="1876" CreationDate="2014-02-28T13:37:29.540" Score="0" Body="&lt;p&gt;To answer your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed of light is relative to the metric of the space itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, if you imagine that there is a network of rules and clocks at rest with respect to each other floating in space, the speed of light is relative to that group of rules and clocks if the group is inertial.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-02-28T13:37:29.540" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1881" PostTypeId="2" ParentId="1868" CreationDate="2014-02-28T13:45:25.853" Score="3" Body="&lt;p&gt;Stars rotate due to the angular momentum of the gas they formed from. This angular momentum must be conserved, and remains as the rotation of the star and it's satellites. If a star collapsed from a completely static gas cloud it would not rotate, but would still be stable. The stability is provided by the hydrostatic equilibrium between radiation and thermal pressure with gravitation collapsing the star - not the stars rotation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know any statistics but I expect all stars rotate to some degree; It is usually the turbulence of gas clouds that leads to overdensities that then collapse in to stars. &lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2014-02-28T13:45:25.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1882" PostTypeId="1" AcceptedAnswerId="1883" CreationDate="2014-02-28T17:16:05.017" Score="4" ViewCount="551" Body="&lt;p&gt;I don't know whether we can observe a sun through a telescope. But if we can then will it harm our eyes and if such telescopes exists what is their configuration?&lt;/p&gt;&#xA;" OwnerUserId="1096" LastEditorUserId="1135" LastEditDate="2014-03-01T17:15:58.767" LastActivityDate="2014-03-01T17:15:58.767" Title="Do telescopes harm while observing sun through them?" Tags="&lt;the-sun&gt;&lt;telescope&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1883" PostTypeId="2" ParentId="1882" CreationDate="2014-02-28T17:38:31.193" Score="8" Body="&lt;p&gt;Observing the Sun through a telescope is very dangerous, whatever the telescope you use, if you don't use the appropriate tools.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A telescope a basically a light collector: its purpose is to collect all the light that is arriving on his primary mirror and focus it on a point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may have already tried to make the Sun light converge through a little magnifying glass (which works quite similarly to a telescope). If you make it converge on something like paper, it will simply burn it. This can eventually be the cause for big fires (example, a glass bottle in a dry forest, concentrating the Sun's light on leaves). Here is an example to convince you: &lt;a href=&quot;http://www.youtube.com/watch?v=N7nJ3wIxt3o&quot;&gt;http://www.youtube.com/watch?v=N7nJ3wIxt3o&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A telescope works the same, except that the collecting surface is usually bigger, so it will be more effective at burning ... your eye for example (astronomers make this joke that you can observe the Sun with a telescope twice in your life: once with your left eye, once with your right eye). Or, for bigger telescope, it could even start a fire, or burn instruments...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To observe the Sun, you will need a proper filter that you will put usually &quot;before&quot; the telescope. You may also observe it indirectly on small telescope by projecting the image of the Sun.You can see examples here: &lt;a href=&quot;http://www.skyandtelescope.com/observing/objects/sun/Observing_the_Sun.html&quot;&gt;http://www.skyandtelescope.com/observing/objects/sun/Observing_the_Sun.html&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1092" LastActivityDate="2014-02-28T17:38:31.193" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1884" PostTypeId="1" AcceptedAnswerId="1885" CreationDate="2014-03-01T11:17:16.030" Score="4" ViewCount="201" Body="&lt;p&gt;What is the maximum distance for a satellite to orbit the earth? Does earth's gravity has the impact on satellite? I do know that earth's gravity will never be zero and it's gravity is inversely proportional to the distance.I heard that satellites escape earth's gravitational pull when they achieve a speed of 11kmph.How far it is true?&lt;/p&gt;&#xA;" OwnerUserId="1103" LastEditorUserId="2365" LastEditDate="2016-04-16T19:04:24.180" LastActivityDate="2016-04-16T19:04:24.180" Title="Satellite's orbit" Tags="&lt;orbit&gt;&lt;gravity&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1885" PostTypeId="2" ParentId="1884" CreationDate="2014-03-01T12:27:38.773" Score="3" Body="&lt;p&gt;Orbits around Earth need to be within its &lt;a href=&quot;http://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;nofollow&quot;&gt;Hill sphere&lt;/a&gt;. Earth's Hill sphere is the region where Earth's gravity field dominates over the gravity fields of Moon and Sun. For Earth the radius of the Hill sphere is about 1.5 million kilometers. But for long-term stability the satellite needs to be well within the Hill sphere, less than half the maximum radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Escape_velocity&quot; rel=&quot;nofollow&quot;&gt;escape velocity&lt;/a&gt; from Earth's surface is about 11.2 km per second. That's the velocity an objects needs at the surface of Earth, relative to a non-rotating coordinate system, to escape the gravity field of Earth, no atmospheric drag assumed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The escape velocity is dependent of the distance from Earth's center of gravity, the further away from Earth the slower the escape velocity. Its roughly proporional to the inverse of the square root of the distance from Earth's center, as long as points above Earth's surface are considered.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-03-01T12:40:49.413" LastActivityDate="2014-03-01T12:40:49.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1886" PostTypeId="1" AcceptedAnswerId="1887" CreationDate="2014-03-01T14:52:11.163" Score="2" ViewCount="52" Body="&lt;p&gt;I would assume that, with light from more distant areas in space reaching the earth, we should be able to see more distant (and older) objects.&lt;br&gt;&#xA;However there's lot of &lt;a href=&quot;https://astronomy.stackexchange.com/q/1860/1135&quot;&gt;weird stuff regarding the space expansion&lt;/a&gt; that could eventually make the objects leaving us so fast the we'd never see them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, did we reach any boundary of what we can possibly see, or are we still working on more sensitive exploration devices?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;I'll be grateful for some useful re-tag&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1135" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2014-03-01T21:28:29.207" Title="Do more objects appear on sky in time?" Tags="&lt;expansion&gt;&lt;dark-energy&gt;&lt;cosmological-horizon&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1887" PostTypeId="2" ParentId="1886" CreationDate="2014-03-01T15:50:02.450" Score="2" Body="&lt;p&gt;No, the expansion of the universe appears to be accelerating. Hence distant objects get more and more red-shifted. Our horizon is shrinking instead of expanding. At the end, only galaxies which are sufficiently bound by gravity remain visible; this could turn out to be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Local_supercluster&quot; rel=&quot;nofollow&quot;&gt;Virgo supercluster&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More reading about the accelerating universe on &lt;a href=&quot;http://en.wikipedia.org/wiki/Accelerating_universe&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, and on &lt;a href=&quot;http://news.nationalgeographic.com/news/2011/10/111004-nobel-prize-physics-universe-expansion-what-is-dark-energy-science/&quot; rel=&quot;nofollow&quot;&gt;National Geographic&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nobelprize.org/nobel_prizes/physics/laureates/2011/advanced-physicsprize2011.pdf&quot; rel=&quot;nofollow&quot;&gt;Scientific background to the according Nobel Prize in physics 2011&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-03-01T19:01:43.820" LastActivityDate="2014-03-01T19:01:43.820" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1888" PostTypeId="1" AcceptedAnswerId="1907" CreationDate="2014-03-01T18:29:40.043" Score="1" ViewCount="752" Body="&lt;p&gt;How do they make International space station to orbit the earth beyond Earth's gravity acting on it? We all know that ISS is rotating at an altitude of just 350km away. How could the ISS escape Earth's gravitational pull?&lt;/p&gt;&#xA;" OwnerUserId="1103" LastEditorUserId="37191" LastEditDate="2021-01-18T21:15:22.717" LastActivityDate="2021-01-18T21:15:22.717" Title="Earth's gravitational pull on ISS" Tags="&lt;orbit&gt;&lt;gravity&gt;&lt;astrophysics&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="1889" PostTypeId="1" AcceptedAnswerId="1890" CreationDate="2014-03-01T19:05:54.467" Score="3" ViewCount="545" Body="&lt;p&gt;I am quite curious about black holes. Once I just thought whether it is possible to create an artificial Black Hole, i.e. an human made black hole. Also explain there formation if it is possible.&lt;/p&gt;&#xA;" OwnerUserId="1096" LastActivityDate="2014-03-06T08:38:49.713" Title="Is it possible to create an artificial Black Hole?" Tags="&lt;black-hole&gt;&lt;universe&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1890" PostTypeId="2" ParentId="1889" CreationDate="2014-03-01T19:45:17.947" Score="3" Body="&lt;p&gt;There has been a lot of speculation, whether it's possible with particle accelerators to create &lt;a href=&quot;http://en.wikipedia.org/wiki/Micro_black_hole&quot; rel=&quot;nofollow&quot;&gt;micro black holes&lt;/a&gt;. &lt;a href=&quot;http://arxiv.org/abs/hep-ph/0106295&quot; rel=&quot;nofollow&quot;&gt;This is only feasible, if space is higher dimensional on very short distances&lt;/a&gt;. This kind of black holes would be very short-lived due to decay by &lt;a href=&quot;http://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow&quot;&gt;Hawking radiation&lt;/a&gt;. This kind of black holes could be detected by their decay products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.bbc.com/future/story/20130716-will-we-ever-create-a-black-hole&quot; rel=&quot;nofollow&quot;&gt;There seems to be no evidence, that black holes have formed in a particle accelerator&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In standard 3d space the smallest black hole needs the energy-equivalent of more than $10^{19}\mbox{GeV}$ of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck_mass&quot; rel=&quot;nofollow&quot;&gt;Planck mass&lt;/a&gt; of about $20\mu\mbox{g}$ to form. That's far beyond the energy a &lt;a href=&quot;http://en.wikipedia.org/wiki/Large_Hadron_Collider&quot; rel=&quot;nofollow&quot;&gt;particle collider&lt;/a&gt; can focus onto a single particle collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ultra-high-energy_cosmic_ray&quot; rel=&quot;nofollow&quot;&gt;most energetic cosmic rays&lt;/a&gt; are several orders of magnitude below the needed energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LHC provided up to 8 TeV (8,000 GeV) for proton-proton collisions, scheduled are 13 TeV for the upgrade. That's 15 orders of magnitude below a smallest 3d black hole, but might allow short-lived black holes for higher-dimensional space. &lt;a href=&quot;http://pprc.qmul.ac.uk/~rizvi/Talks/QuantumBlackHoles.pdf&quot; rel=&quot;nofollow&quot;&gt;This way LHC can constrain the structure of space on short distances&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-03-01T20:43:51.780" LastActivityDate="2014-03-01T20:43:51.780" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1891" PostTypeId="2" ParentId="1888" CreationDate="2014-03-01T21:01:32.637" Score="0" Body="&lt;p&gt;ISS is travelling with about 8 km/s around Earth. At that velocity the centrifugal force on the circular orbit is strong enough to cancel out gravity. The parts ISS is made of, needed to be accelerated to that speed by rockets.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-03-03T20:28:33.320" LastActivityDate="2014-03-03T20:28:33.320" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1892" PostTypeId="2" ParentId="1889" CreationDate="2014-03-01T22:22:20.447" Score="2" Body="&lt;p&gt;It is thought likely that black holes would produce Hawking radiation and evaporate.  We can't yet describe the end products of an evaporating black hole, and space is too warm to detect Hawking radiation anywhere.  There are some recent new ideas regarding complementarity that are difficult to reach a decision on as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hawking seems to be pulling away from the black hole concept recently.  I don't know what to make of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding particle colliders: every few years some team gets into the general news cycle with an announcement about detections that resemble particular models of black hole behavior.  Miscommunication tends to produce journalism about real black holes being created.  Now there may be developing interest in the possibility of the galactic core lighting up.  Maybe that would lead to better journalism.&lt;/p&gt;&#xA;" OwnerUserId="1136" LastActivityDate="2014-03-01T22:22:20.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1893" PostTypeId="1" CreationDate="2014-03-02T03:05:09.987" Score="0" ViewCount="122" Body="&lt;p&gt;If we would take the average number of supernovae per year and multiply that by the number of stars, could we figure out how many years it will be before all stars are gone?&lt;/p&gt;&#xA;" OwnerUserId="1138" LastEditorUserId="181" LastEditDate="2014-03-02T18:18:28.590" LastActivityDate="2014-03-02T18:18:28.590" Title="Average rate of supernova x number of stars" Tags="&lt;star&gt;&lt;supernova&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1894" PostTypeId="2" ParentId="1868" CreationDate="2014-03-02T03:14:43.867" Score="0" Body="&lt;p&gt;Like Chris said above, stars rotate to conserve their angular momentum. They have an initial angular momentum from their formation. In a stellar system both energy and momentum, must be conserved for the system to be stable. Throughout the evolution, the system may radiate some angular momentum away in the form of gravity waves (ie. double degenerate binaries). During this process the separation of two stars will be reduced and the rotation rate of the star will spike to conserve angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More on this here: &lt;a href=&quot;http://www.selmademink.com/science/on-the-origin-of-stellar-rotation-rates-and-the-role-binary-interaction/&quot; rel=&quot;nofollow&quot;&gt;http://www.selmademink.com/science/on-the-origin-of-stellar-rotation-rates-and-the-role-binary-interaction/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1137" LastActivityDate="2014-03-02T03:14:43.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1895" PostTypeId="1" AcceptedAnswerId="1898" CreationDate="2014-03-02T05:35:04.607" Score="3" ViewCount="858" Body="&lt;p&gt;What is the length of the apogalacticon and perigalacticon of the Sun and Milky Way?  The general terms seem to be apoapsis and periapsis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My greatest efforts at Googling have failed miserably.  If you can provide references as well, please do!&lt;/p&gt;&#xA;" OwnerUserId="1139" LastEditorUserId="181" LastEditDate="2014-03-02T16:56:26.193" LastActivityDate="2017-08-30T17:08:23.173" Title="Apogalacticon and Perigalacticon" Tags="&lt;orbit&gt;&lt;the-sun&gt;&lt;galaxy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1896" PostTypeId="1" AcceptedAnswerId="1899" CreationDate="2014-03-02T08:18:40.983" Score="2" ViewCount="5561" Body="&lt;p&gt;I just bought my first telescope - a Celestron CDC 1100.  It came with a 40mm lens so with a primary focal length of 2800mm the power should be 70x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is that I've looked at Sirius and Mars and they look clearer to my naked eye than with my telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am able to focus from way out so I can see the 'blind spot' of the secondary mirror, go through 'focused' and then continue moving the mirror in the same direction until I again see the blind spot.  So I know I'm focusing through when it should be 'good'.  But, like I said, what I see is a relatively unfocused spot of light that is clearer and steadier with my naked eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Admittedly, my 'seeing conditions' aren't very good since I am looking through my window in a suburban neighborhood where light pollution is pretty bad.  That being said, I still see Mars and Sirius quite brightly - it's just that they never come into clear focus - nor are they any larger than when I look using my eyes.  And when I move my eye around the eyepiece I see streaks of light in different directions originating from the target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many stars to be seen, though they are just pinpricks as I would expect given the seeing conditions.  What I mean to say is that I cannot see all those stars with my naked eye but can see tens or hundreds more looking through the scope so I know it is gathering plenty of light as an 11&quot; scope should.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried both with and without the diagonal with no discernable difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not collimated but supposedly it was collimated before leaving the factory and I would hope that even if it were off a bit from shipping I would still be able to see a sizable though fuzzy image of Mars instead of just a small speck of orangish light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thoughts as to what I may have done wrong or what may be causing this lack of clarity?&lt;/p&gt;&#xA;" OwnerUserId="1142" LastActivityDate="2015-01-24T03:43:56.180" Title="Why can't I see Mars clearly?" Tags="&lt;solar-system&gt;&lt;amateur-observing&gt;&lt;telescope&gt;" AnswerCount="5" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1897" PostTypeId="1" CreationDate="2014-03-02T09:28:34.740" Score="2" ViewCount="1099" Body="&lt;p&gt;I've written a quick and dirty n-body simulator, but currently it requires initial positions and velocities. I've been looking for a way to convert Keplerian orbital elements to positions and velocities. I've searched the web, but I only seem to get conversion from orbital elements to positions and velocities in ECI coordinates, or to a system that orbits the Earth. Ideally, I am hoping that there is an easy conversion between orbital elements and position and velocities for an observer on Earth looking at this n-body system somewhere out in space. I'd appreciate any help in determining this conversion.&lt;/p&gt;&#xA;" OwnerUserId="1143" LastActivityDate="2014-03-06T19:37:53.610" Title="Convert orbital elements to positions and velocities" Tags="&lt;velocity&gt;&lt;n-body-simulations&gt;&lt;orbital-elements&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1898" PostTypeId="2" ParentId="1895" CreationDate="2014-03-02T11:56:21.510" Score="4" Body="&lt;p&gt;According to &lt;a href=&quot;http://ottawa-rasc.ca/astronotes/1997/an9701p3.html&quot; rel=&quot;nofollow&quot;&gt;this website&lt;/a&gt; we need another 15 million years to the perigalacticon.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Recall the Sun's motion ... 30 degrees toward the galactic centre from circular motion, and 23 degrees upward out of the galactic plane. We will be at perigalacticon (closest to the nucleus) in 15 million years. At apogalacticon the Sun is seven percent farther out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;During the &lt;a href=&quot;http://en.wikipedia.org/wiki/Galactic_year&quot; rel=&quot;nofollow&quot;&gt;galactic year&lt;/a&gt; of about 250 million years our solar system isn't orbiting on a &lt;a href=&quot;http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion&quot; rel=&quot;nofollow&quot;&gt;Kepler ellipse&lt;/a&gt; around the galactic center, but instead oscillating up and down the &lt;a href=&quot;http://en.wikipedia.org/wiki/Galactic_plane&quot; rel=&quot;nofollow&quot;&gt;galactic plane&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This vertical oscillation cycles at 3.5 times per galactic year.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are now about 27,000 lightyears away from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Galactic_Center&quot; rel=&quot;nofollow&quot;&gt;galactic center&lt;/a&gt;, close to the perigalacticon. The difference between apogalacticon and perigalacticon is a little more than 4,000 lightyears (15% of 27,000 lightyears). The 15% number is calculated from the 7% eccentricity of the orbit by $1.07/0.93=1.15~~$ according to the &lt;a href=&quot;http://mathworld.wolfram.com/Ellipse.html&quot; rel=&quot;nofollow&quot;&gt;definition of eccentricity&lt;/a&gt;, and Kepler's 1st law.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-03-02T12:28:03.337" LastActivityDate="2014-03-02T12:28:03.337" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1899" PostTypeId="2" ParentId="1896" CreationDate="2014-03-02T13:25:40.513" Score="2" Body="&lt;p&gt;Light pollution doesn't disturb the focussing, just the contrast.&#xA;But non-uniform &lt;a href=&quot;http://en.wikipedia.org/wiki/Refraction&quot; rel=&quot;nofollow&quot;&gt;refraction&lt;/a&gt; of the window can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So try to open the window, if you can, or take a try outside.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-03-02T13:25:40.513" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1900" PostTypeId="2" ParentId="1893" CreationDate="2014-03-02T14:10:47.500" Score="3" Body="&lt;p&gt;Not that easy way. Only stars heavy enough will undergo a &lt;a href=&quot;http://en.wikipedia.org/wiki/Supernova&quot; rel=&quot;nofollow&quot;&gt;supernova&lt;/a&gt; explosion. The majority of stars is too light. The lifetime of a star is mostly determined by its &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_mass&quot; rel=&quot;nofollow&quot;&gt;mass&lt;/a&gt;. In some cases (supernova type Ia) a companion star provides mass to white dwarf, which originally has been too light to explode as a supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence your technique can only work, if it is applied to stars of a certain mass and composition separately, and then summed or integrated up over all these restricted classes of stars. All together you'll get results for stars which eventually will end up as supernova, i.e. the heavy ones with more than about 8 solar masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... And in the meanwhile &lt;a href=&quot;http://en.wikipedia.org/wiki/Star_formation&quot; rel=&quot;nofollow&quot;&gt;new stars will form&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-03-02T14:10:47.500" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="1901" PostTypeId="1" AcceptedAnswerId="1905" CreationDate="2014-03-02T16:39:50.533" Score="4" ViewCount="1348" Body="&lt;p&gt;Is it true that when a star 'runs out' of its fuel, it hasn't really used up all its hydrogen, it still has hydrogen left, but the hydrogen is not at its center where the fusion happens. If something crashed into it rearranging its content a bit, could its life be extended?&lt;/p&gt;&#xA;" OwnerUserId="36" LastActivityDate="2014-08-19T09:28:55.173" Title="Is it possible to extend the life of a star?" Tags="&lt;star&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1902" PostTypeId="1" AcceptedAnswerId="1903" CreationDate="2014-03-02T19:22:20.637" Score="2" ViewCount="128" Body="&lt;p&gt;So, elements such as iron, gold, etc. are formed from different types of supernovas, whether it's type 1A or 2. My question is, is it possible that there are elements that are formed that we never received during the formation of Earth and that we are unaware of? Or have we discovered every non man-made element?&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2014-03-02T21:25:30.680" Title="Question about the formation of elements" Tags="&lt;supernova&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1903" PostTypeId="2" ParentId="1902" CreationDate="2014-03-02T21:25:30.680" Score="3" Body="&lt;p&gt;It's very likely, that we don't have discovered every non man-made element. For some elements there exist only very short-lived isotopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Plutonium&quot; rel=&quot;nofollow&quot;&gt;Plutonium&lt;/a&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is the heaviest primordial element by virtue of its most stable isotope, plutonium-244, whose half-life of about 80 million years is just long enough for the element to be found in trace quantities in nature&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Primordial_nuclide&quot; rel=&quot;nofollow&quot;&gt;More about naturally occurring nuclides&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://nopr.niscair.res.in/bitstream/123456789/9595/1/IJPAP%2042%287%29%20479-484.pdf&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; mentions the detection of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Curium&quot; rel=&quot;nofollow&quot;&gt;curium&lt;/a&gt; ion in cosmic rays.&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Neutron_capture_nucleosynthesis&quot; rel=&quot;nofollow&quot;&gt;The process which forms curium&lt;/a&gt; can also form elements beyond curium, hence it's likely that there exist short-lived isotopes of elements in nature, which have not yet been detected. &lt;a href=&quot;http://www.ift.uam.es/paginaspersonales/bellido/gravitacion/exp/CR_en.html&quot; rel=&quot;nofollow&quot;&gt;Research in this field is going on&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-03-02T21:25:30.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1904" PostTypeId="1" AcceptedAnswerId="1906" CreationDate="2014-03-02T21:48:45.487" Score="1" ViewCount="461" Body="&lt;p&gt;It always amuse us to research about things around us and we always start our research with a question that - what must be the reason for happening of that phenomenon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we know that the formation of universe is being postulated with theory of Big Bang. &#xA;But now, I want to ask that &lt;strong&gt;what was the reason for Big Bang&lt;/strong&gt;. I mean &lt;strong&gt;why it happened, what must be the circumstances or what made it happened?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1096" LastActivityDate="2014-03-02T22:23:27.127" Title="The reason behind Big Bang" Tags="&lt;universe&gt;&lt;formation&gt;&lt;big-bang-theory&gt;&lt;early-universe&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1905" PostTypeId="2" ParentId="1901" CreationDate="2014-03-02T21:58:12.267" Score="4" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_nucleosynthesis&quot; rel=&quot;nofollow&quot;&gt;The outer shell of the star&lt;/a&gt; contains still a lot of hydrogen, after running out of fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elements other than hydrogen are denser than hydrogen. This makes it difficult to get hydrogen to the core of a star, or to the region of &lt;a href=&quot;http://en.wikipedia.org/wiki/CNO_cycle&quot; rel=&quot;nofollow&quot;&gt;hydrogen-burning&lt;/a&gt; at least, after the core has been burnt to helium or heavier elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the life of a star can be extended in some way, e.g. by adding hydrogen to a white dwarf. This can trigger &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbon_burning_process&quot; rel=&quot;nofollow&quot;&gt;burning of carbon&lt;/a&gt; with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Type_Ia_supernova&quot; rel=&quot;nofollow&quot;&gt;supernova Ia&lt;/a&gt; as result.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-03-02T21:58:12.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1906" PostTypeId="2" ParentId="1904" CreationDate="2014-03-02T22:23:27.127" Score="4" Body="&lt;p&gt;&lt;strong&gt;WHY&lt;/strong&gt; implies a &lt;strong&gt;reason&lt;/strong&gt;, which is venturing in to the religious realm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The circumstances under which an event such as the Big Bang occurs are guess work at best. Since the Big Bang is a singularity from which time itself started there is no real &lt;em&gt;before&lt;/em&gt; the big bang. That's the first problem. The second is that because of the spatial singular nature of the big bang no information from 'before' it could ever pass through to our current universe and provide us with any hints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of oddball theories that explore the idea of physics before the big bang. &lt;a href=&quot;http://arxiv.org/pdf/hep-th/9802057v2.pdf&quot; rel=&quot;nofollow&quot;&gt;Some&lt;/a&gt; investigate what occurs on scales of time shorter than that of the Planck time, where quantum mechanical processes may give us hints in to the cause or mechanics of a big bang event. Proving or, indeed, disproving these theories is currently impossible experimentally.&lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2014-03-02T22:23:27.127" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1907" PostTypeId="2" ParentId="1888" CreationDate="2014-03-02T22:58:49.790" Score="11" Body="&lt;p&gt;Your question presumes that the ISS is beyond Earth's gravity, that it has escaped earth's gravitational pull. This is not correct. All objects with mass in the universe affect all other things with mass in the universe, the effect just gets weaker with distance. So the ISS is feeling the effect of gravity from Earth significantly more than the moon is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason the ISS doesn't just fall to the earth, either directly or gradually spiralling towards the earth, is that it is travelling fast enough around the earth that it is continually &quot;missing&quot; earth. It is sometimes described as 'falling' constantly around earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am to be properly correct though, the reality is that ISS &lt;strong&gt;is&lt;/strong&gt; in fact falling towards the earth, &lt;a href=&quot;http://www.heavens-above.com/IssHeight.aspx&quot;&gt;getting closer and closer to Earth all the time&lt;/a&gt;. It needs occasional boosts to push it further back out in it's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to blow your mind a little bit: The ISS is pulling on the Earth with the same force that the Earth is pulling on ISS.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-03-02T22:58:49.790" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1908" PostTypeId="1" CreationDate="2014-03-03T07:09:18.163" Score="1" ViewCount="619" Body="&lt;p&gt;Is there any star atlas someone recommends?&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then for some kind of star with a definite spectral type, we can know its absolute magnitude.&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2014-03-04T02:06:32.717" Title="star atlas with absolute magnitude and spectral type" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1909" PostTypeId="2" ParentId="1897" CreationDate="2014-03-03T09:45:56.390" Score="2" Body="&lt;p&gt;Is &lt;a href=&quot;https://www.google.co.uk/url?sa=t&amp;amp;source=web&amp;amp;rct=j&amp;amp;ei=RE0UU8yVIMOd7QbJ-YHgDw&amp;amp;url=https://downloads.rene-schwarz.com/download/M001-Keplerian_Orbit_Elements_to_Cartesian_State_Vectors.pdf&amp;amp;cd=9&amp;amp;ved=0CD0QFjAI&amp;amp;usg=AFQjCNHxGtCwuWay1mR1xLkxyKCbm_WnpQ&amp;amp;sig2=wjAztcsoPcC8g10mpctHjg&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; what you're looking for? When you say you want something defined for an observer on the earth looking at the system, do you mean you wish to set the origin of the system at the earth? This can be done, but I don't really see why you would do it unless the earth was intrinsic to the system. &lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2014-03-03T09:45:56.390" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1910" PostTypeId="2" ParentId="1908" CreationDate="2014-03-03T11:08:50.660" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://heasarc.gsfc.nasa.gov/db-perl/W3Browse/w3table.pl?tablehead=name=bsc5p&amp;amp;Action=More%20Options&quot; rel=&quot;nofollow&quot;&gt;BSC5p&lt;/a&gt; may be a nice database to start with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parallax (check according box) and vmag will provide a basis for the calculation of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Absolute_magnitude&quot; rel=&quot;nofollow&quot;&gt;absolute visual magnitude&lt;/a&gt;. The absolute visual magnitude may be lower (star may be absolutely brighter), if the &lt;a href=&quot;http://en.wikipedia.org/wiki/Extinction_%28astronomy%29&quot; rel=&quot;nofollow&quot;&gt;extinction&lt;/a&gt; isn't negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/parallax is a good estimate for the distance. Take this distance as an approximate for the luminosity distance, if you can assume negligible extinction. Now calculate the visual magnitude of the star at 10 parsec distance to get the absolute visual magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: Alp1Cen (HR 5459) has vmag -0.01, and parallax 0.751 arc seconds. Therefore the distance is about 1/0.751 = 1.33 parsec. We get absolute visual magnitude&#xA;$$ M=m-5\cdot (\mbox{log}_{10}D_L - 1)=-0.01-5\cdot (\mbox{log}_{10}1.33 - 1)=4.37.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.uni.edu/morgans/astro/course/Notes/section2/spectraltemps.html&quot; rel=&quot;nofollow&quot;&gt;An overview of absolute magnitude and spectral types&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More detailed discussion, and tables can be found in &lt;a href=&quot;http://mnras.oxfordjournals.org/content/374/4/1549.full.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://xoomer.virgilio.it/hrtrace/Sk.htm&quot; rel=&quot;nofollow&quot;&gt;More tables&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://press.princeton.edu/titles/8963.html&quot; rel=&quot;nofollow&quot;&gt;This book&lt;/a&gt; seems to contain a table with values as of 2009, which is cited &lt;a href=&quot;http://www.handprint.com/ASTRO/specclass.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-03-04T02:06:32.717" LastActivityDate="2014-03-04T02:06:32.717" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1911" PostTypeId="1" AcceptedAnswerId="1916" CreationDate="2014-03-04T07:44:29.917" Score="23" ViewCount="30259" Body="&lt;p&gt;It is said that the Andromeda and Milky Way galaxies are coming close to each other with a speed of approximately 400000 km/hour. They will be together in the next 4 billion years.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What will happen to life on Earth or human beings on Earth?&lt;/li&gt;&#xA;&lt;li&gt;If we are about to collide in the next 4 billion years then how long before we should take action for interstellar voyage?&lt;/li&gt;&#xA;&lt;li&gt;Are scientists working on such projects for an interstellar voyage?&lt;/li&gt;&#xA;&lt;li&gt;Will we be able to get the something like Earth to go away from this galaxy/Earth/ solar system and, considering speed of human instruments/spaceships, how long will it take to go to a safe place?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="760" LastEditorUserId="2153" LastEditDate="2015-04-12T17:05:48.557" LastActivityDate="2015-10-03T07:56:10.130" Title="What will happen to life on Earth when the Andromeda and Milky Way galaxies collide?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;life&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1912" PostTypeId="2" ParentId="1911" CreationDate="2014-03-04T08:59:04.500" Score="10" Body="&lt;p&gt;Normally when two galaxies collide, it is the gas that interacts with each other. The odds of stars impacting each other are nearly zero due to the huge distances between the stars. The same goes for planets hitting each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The timescales on this happening are so large that it is difficult for our mind to understand these distances (and the timescales at which these processes take place), and by that time, many other things might have happened that change the situation of our solar system making it difficult to predict what will happen to humans. For example a meteorite impact like the one that killed the dinosaurs may kill all life on earth one billion years from now, so that we are not even there to see what happens to human life when andromeda interacts with the milky way. &lt;/p&gt;&#xA;" OwnerUserId="1062" LastActivityDate="2014-03-04T08:59:04.500" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1913" PostTypeId="2" ParentId="1911" CreationDate="2014-03-04T11:30:31.353" Score="19" Body="&lt;p&gt;4 billion years is the same timeframe of the life remaining to our Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we have not yet invented interstellar voyages, we're screwed, with or without Andromeda.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides, stars do not interact directly with each other in a galactic collision. What we will notice from the several stars we are on is that star orbits around the galactic centre will be altered by the massive gravitational perturbation made by the other galaxy. Almost all stars will change orbits from that of orbiting around the galactic centre to orbiting around the center of mass of both galaxies. Some stars will be ejected out of the new bigger galaxy. It is quite safe to think that planets will continue orbiting their stars, but not that they will not have some alterations in their orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, from the gas to gas interaction, there will be a lot of new pressure waves in the interstellar medium, which will account for the formation of billions of new stars in new nebulae.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-03-04T11:30:31.353" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1914" PostTypeId="2" ParentId="1911" CreationDate="2014-03-04T12:00:04.247" Score="12" Body="&lt;p&gt;Direct collisions between stars and planets is highly unlikely, due to the relatively low density of objects in the Milky Way and Andromeda. For instance, the stellar density in the solar neighborhood is only 0.004 stars per cubic light year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that gravitational interactions between objects is not low. Stars that eventually pass too close to other systems might disrupt the orbits of planets, asteroids and comets, and this can be problematic if we are still around. I would speculate that planets and other objects would be flung out everywhere, and the system may become a shooting gallery.&lt;/p&gt;&#xA;" OwnerUserId="1151" LastActivityDate="2014-03-04T12:00:04.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1916" PostTypeId="2" ParentId="1911" CreationDate="2014-03-04T16:05:26.310" Score="23" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What will happen to life on earth or human beings on earth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming that human beings, or life, still exists on Earth at that time, they will have survived so much due to the ongoing death of the sun, that the gravitational pertubations due to the galactic collision will be nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that in about 1-2 billion years, the sun will be so hot and large that all the water will have boiled off the earth into space.  About 3 billion years from now, the surface of the Earth will be so hot that metals will be melting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any life that has survived those events and still lives on Earth will surely take a galactic collision in stride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine, though, that most humans will have fled Earth - if not for distant star systems, then at least for planets in our own system that are going to be warming up enough for human habitation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If we are about to collide in next 4 billion year then how long before we should take action for inter-stellar voyage? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As soon as possible. When interstellar voyage becomes possible, we should start sending out ships to colonize other planets and star systems.  This will likely take a long time, but if we are to survive more than a billion years, it is necessary. Keep in mind that the Sun and Andromeda are events we can predict.  We don't know, and can't predict, the next cataclysmic asteroid strike, which is likely to happen in a shorter time than a billion years.  There are lots of reasons to exit the planet, we should be worried about the ones we can't predict or see, not the ones we can predict.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are scientists working on such projects for inter-stellar voyage?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, but in small steps.  Manned missions to space, to the moon, and living aboard the ISS have provided significantly valuable information that will be used in such interstellar missions.  As we continue to push the boundaries of our ability to survive in space we eventually will be able to live in space, perhaps whole lifetimes will be spent in space.  As engine technology progresses beyond simply lifting people out of Earth's gravitational well, we will eventually be sending people on long voyages outside our solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a very, very long way off, but each advance takes us closer to that eventual goal.&lt;/p&gt;&#xA;" OwnerUserId="559" LastEditorUserId="559" LastEditDate="2014-03-04T16:16:56.603" LastActivityDate="2014-03-04T16:16:56.603" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1917" PostTypeId="2" ParentId="1845" CreationDate="2014-03-04T23:20:53.750" Score="3" Body="&lt;p&gt;The official site you point to has 3 pieces of location information:&#xA;1) the max height angle&#xA;2) the point it appears&#xA;3) the point it disappears&#xA;The only one of these that is especially accurate is the max height angle, and that isn't the number you want to use to try and see it. However, passes with a bigger number for the max height angle will be easier to see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the ISS, get the reference to the point it &quot;Appears&quot; - that is the &quot;11 above N&quot; you have described. Now, notice that the reference to the compass direction is simply N or NNE or SW or whatever. This is not a precise directional reference, but it doesn't need to be, as if you're looking towards that direction, you shouldn't have any problem seeing ISS appear. It isn't clear to me if the website is using True North or Magnetic North for the direction; Magnetic North might be easier for people with a simple compass to find, but usually these directions will be relative to True North, which is not where a compass points. From a quick check using my own location, it appears that they have used True North, which is maybe a bit unfortunate. If you don't know where True North is for your location, &lt;a href=&quot;http://www.ngdc.noaa.gov/geomag-web/#declination&quot; rel=&quot;nofollow&quot;&gt;you can look it up&lt;/a&gt;, and then you have to remember to adjust the reading off your compass by the offset for your location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the description is &quot;11 above N&quot;. This means 11 degrees above the N direction. The reason for it appearing some distance above the horizon is because it is coming up out of the shadow of the Earth, which is being cast some distance above the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To quickly and easily estimate angular distances, &lt;a href=&quot;http://oneminuteastronomer.com/860/measuring-sky/&quot; rel=&quot;nofollow&quot;&gt;use the trick of observing your hand at arms length&lt;/a&gt;. Different parts give different angular dimensions. It is a pretty fair estimate, because no matter how big or small you are, the relative size of your hands and the length of your arms is probably about the same as everyone else... not good enough to use as a survey instrument, but good enough to guesstimate locations on the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second position description is the &quot;Disappears&quot; location. That is where the ISS will again drop into the shadow of the Earth. Similarly, it is described as degrees above a True North compass direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can figure out (approx) the path the ISS will take by drawing an imaginary line between these two points. However, don't draw that imaginary line so it goes over your head, note the &quot;Max Height&quot; angle again: this is the highest point on the line that you draw between the &quot;Appears&quot; and &quot;Disappears&quot; spots. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just give it a go, especially one with a pass that starts around 30mins or so after sunset, and that has a high &quot;Max Height&quot; angle. It won't be hard to see.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-03-04T23:20:53.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1918" PostTypeId="1" AcceptedAnswerId="1957" CreationDate="2014-03-05T01:13:21.583" Score="11" ViewCount="9283" Body="&lt;p&gt;Right now, I think the winter solstice is around December 21. Has it always been around this date constantly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the earth's axis's 23.45° tilt will be on the exact opposite side in 13000 years (or was 13000 years ago) due to precession, it'd seem like the seasons would flip every 13000 years. So, +/- 13000 years from now, it would've turned summer around December 21 in the Northern Hemisphere, right?&lt;/p&gt;&#xA;" OwnerUserId="1166" LastActivityDate="2019-12-01T14:37:39.930" Title="Do solstices and equinoxes shift over time?" Tags="&lt;rotation&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1919" PostTypeId="2" ParentId="1876" CreationDate="2014-03-05T04:27:58.660" Score="1" Body="&lt;p&gt;The light will eventually reach galaxy B even though the space between them is expanding. It is important to remember that it is the space that is expanding not galaxy B moving away from galaxy A. Imagine putting two dots on a balloon and blowing it up. It is the rubber between the dots that is stretching, not one dot moving about the balloon's surface away from the other dot.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a photon of light leaving galaxy A. After a period of time it will have travelled some distance towards galaxy B. Let's say it has travelled 1% of the distance so it has 99% to go. Space is expanding equally in front of and behind the photon, so even if it were to stop moving it would stay at 1% of the total distance. But it doesn't stop, it moves nearer. It will eventually get half way and again the space is expanding both behind and in front of the photon. Now there is the same amount of space in between the photon and galaxy A as there is between the photon and galaxy B. No matter how much expansion there is the photon is still in the middle. Eventually the photon will get all the way to galaxy B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No matter how fast the universe is expanding and how slow the photon is moving it will eventually reach galaxy B although it may take a very very long time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the space the photon is in is expanding the wavelength of the photon will slowly change. This is red shift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the problem with this answer. I believe this answer is only valid if the speed of the expansion and the speed of the photon is constant, but the expansion of the universe appears to be accelerating. &lt;/p&gt;&#xA;" OwnerUserId="1168" LastActivityDate="2014-03-05T04:27:58.660" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1920" PostTypeId="2" ParentId="1896" CreationDate="2014-03-05T07:00:54.097" Score="8" Body="&lt;p&gt;You're experiencing a few issues all of which have one solution. Take the Scope outside. Let it acclimate to the ambient air tempurature, and then try looking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 1st issue is the mirror needs to be the same temp as the air you're observing in. Since you are indoors you have a very warm optical path this causes distortion and trouble focusing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: Because you're looking through a piece of window glass (or more if it's a multipane window) you're introducing distortions that way. It's no wonder you can't reach focus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light pollution won't really affect Mars or the other planets too much, so that's not your worry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are you observing from inside anyway?&lt;/p&gt;&#xA;" OwnerUserId="912" LastActivityDate="2014-03-05T07:00:54.097" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1921" PostTypeId="2" ParentId="1911" CreationDate="2014-03-05T08:59:25.837" Score="18" Body="&lt;p&gt;First, note that by the time Andromeda is close enough for collisions with wandering stars to become a concern, Earth's average temperature will have changed significantly, and the planet will be unrecognizable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Sol is 8.5 billion years old, it will still have hydrogen available for fusion, but as it fuses it contracts and expands differentially.  The contraction causes hydrogen fusion to become more favorable, so that Sol will have 50% greater power output ($6 \times {10}^{26}\ \mathrm{W}$) and 3% greater effective temperature ($6000\ \mathrm{K}$).  Fusion also causes Sol to lose mass at a prodigious rate (currently $4 \times {10}^9 \mathrm{kg/s}$); it will release $6 \times {10}^{43} \mathrm{J}$ from fusion, which corresponds to $7 \times {10}^{26}\ \mathrm{kg}$.  That is about one hundred Earth masses of sunlight but only $1 \over 3000$ the mass of Sol.  Gravitation with Earth decreases proportionally, so Earth's orbit might on average expand $3000\ \mathrm{km}$ per billion years.  Other gravitational effects might change Earth's average distance by as much as $6 \times {10}^5\ \mathrm{km}$, 4‰ of an astronomical unit.  Expansion of Sol's outer layers due to reduced gravitation will increase its radius by 20%, $3 \times {10}^5\ \mathrm{km}$.  Thus Earth will receive nearly 50% more power as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy balance of Earth wrt Sol gives the expected surface temperature:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\bar{a} = &amp;amp; 0.7 &amp;amp; \small\text{(Average absorption)} \\&#xA;P_p = &amp;amp; 1366\ \mathrm{W/m^2} &amp;amp; \small\text{(Average solar flux incident on Earth at present)} \\&#xA;P_f = &amp;amp; P_p \cdot 1.5 \approx 2000\ \mathrm{W/m^2} &amp;amp; \small\text{(In future)} \\&#xA;\sigma = &amp;amp; 5.670373 \times {10}^{-8}\ \mathrm{W/m^2/K^4} &amp;amp; \small\text{(Stefan-Boltzmann constant)} \\&#xA;\\&#xA;T_p^4 = &amp;amp; \frac{\bar{a} P_p}{4 \sigma} \\&#xA;\approx &amp;amp; \frac{0.7 \cdot 1366\ \mathrm{W/m^2}}{2.268149 \times {10}^{-7}\ \mathrm{W/m^2/K^4}} \\&#xA;\approx &amp;amp; 4.2 \times {10}^9\ \mathrm{K^4} \\&#xA;T_p \approx &amp;amp; 250\ \mathrm{K} \\&#xA;\\&#xA;T_f \approx &amp;amp; T_p \cdot {1.5}^{1/4} \approx T_p \cdot 1.11 \\&#xA;\approx &amp;amp; 280\ \mathrm{K}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the average surface temperature on Earth is not $-20\ \mathrm{°C}$ — it is $+15\ \mathrm{°C}$ and already around $8\ \mathrm{K}$ warmer than in an airless future — we can see the atmosphere has a significant role in retaining heat.  Assuming increasing cooling needs &lt;em&gt;do not&lt;/em&gt; lead to the atmosphere retaining more heat, the average surface temperature can be expected to rise to $+50\ \mathrm{°C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average temperature of Antarctica is now $240\ \mathrm{K}$ in winter and $270\ \mathrm{K}$ in summer.  These can be expected to rise to $270\ \mathrm{K}$ (just below freezing) and $300\ \mathrm{K}$ (well above freezing) respectively, and this is a best-case scenario.  Antarctica will melt.  That will produce the largest component (60%) of sea level increase, in total around $100\ \mathrm{m}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If Earth were still inhabited four billion years from now, it is extremely unlikely that Earth would fall into a star from Andromeda.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Space is big.  Really big.  You just won't believe how vastly, hugely, mindbogglingly big it is.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;— Douglas Adams, &lt;em&gt;The Hitchhiker's Guide to the Galaxy&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Milky Way is about 100,000 light years in diameter and contains about 400 billion stars.  Andromeda is bigger and denser; it may have one trillion stars and a diameter of 140,000 light years.  It is 2.5 million light years away but appears six times larger than Sol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;d_M \approx &amp;amp; \frac{4 \times {10}^{11}\ \mathrm{stars}}{{10}^{10} \pi/4\ \mathrm{{ly}^2}} \\&#xA;\approx &amp;amp; 50\ \mathrm{stars/{ly}^2} \\&#xA;\\&#xA;d_A \approx &amp;amp; \frac{{10}^{12}\ \mathrm{stars}}{2 \times {10}^{10} \pi/4\ \mathrm{{ly}^2}} \\&#xA;\approx &amp;amp; 60\ \mathrm{stars/{ly}^2} \\&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the two galaxies were simply superposed, there would be about one hundred stars per square light year, viewed from infinitely far along the rotation axis.  However, the Milky Way is a 2:1 ellipse as seen from Andromeda, while we see Andromeda as a 3:1 ellipse.  Projecting both onto a plane between them, perpendicular to a line between their central black holes, would give a region of overlap with dimensions between $50 \times 50\ \mathrm{{kly}^2}$ and $50 \times 100\ \mathrm{{kly}^2}$, with at most half the Milky Way outside it.  Sol is likely to be involved in the collision, since it is about 27,200 light years from the galactic center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That doesn't mean, though, that Earth will come close to another star, that Sol might collide, or that the solar system will be disrupted.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering probabilistically the worst-case scenario (the entire Milky Way falls through Andromeda on their first pass), there is a mean free path for stars.  The actual stellar density of the colliding galaxies is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\rho \approx 1.4 \times {10}^{12}\ \mathrm{stars}\ /\ V_{A \cup M}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the union of the two galaxies' volumes would be a very complicated expression.  &lt;em&gt;Very roughly&lt;/em&gt;, their volumes can be described as joined cones, ignoring their spheroidal dark matter halos (which are mostly harmless).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\rho \approx &amp;amp; \frac{1.4 \times {10}^{12}\ \mathrm{stars}}&#xA;{\left(&#xA;\frac{1}{2} \cdot&#xA;\left(&#xA;{10}^3\ \mathrm{ly} \cdot {10}^{10} \pi/4\ \mathrm{{ly}^2} +&#xA;1.4 \times {10}^3\ \mathrm{ly} \cdot 2 \times {10}^{10} \pi/4\ \mathrm{{ly}^2}&#xA;\right)&#xA;\cdot \frac{1}{3}&#xA;\right)} \\&#xA;\approx &amp;amp; 0.28\ \mathrm{stars/{ly}^3} \\&#xA;\\&#xA;V_\star \approx &amp;amp; 3.6\ \mathrm{{ly}^3} \\&#xA;\\&#xA;r_\star \approx &amp;amp; {\left( V_\star \cdot \frac{3}{4 \pi} \right)}^{1/3} \\&#xA;\approx &amp;amp; 0.95\ \mathrm{ly}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a distance of 1.9 light years, Betelgeuse would look a lot like Mars.  If we assume disaster results from a star closer than the diameter of the heliosphere (about 200 AU), then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;m = &amp;amp; \frac{1\ \mathrm{star}}{\rho \cdot \pi \cdot 4 \times {10}^4\ \mathrm{{AU}^2}} \\&#xA;\approx &amp;amp; 1.1 \times {10}^{21}\ \mathrm{m} \\&#xA;\approx &amp;amp; 7.2 \times {10}^9\ \mathrm{AU} \\&#xA;\approx &amp;amp; 1.1 \times {10}^5\ \mathrm{ly}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On average, a star can travel 110 thousand light years before it grazes past another, slightly less than the diameter of Andromeda.  The proportion of stars from the Milky Way that do not approach within 200 AU of stars in Andromeda is at least $1/e^{1.4 / 1.1} \approx 100/400\ \mathrm{billion\ stars}$.  For Earth to approach within 4 AU of another star (one Betelgeuse radius), it can be expected to travel at least 2500 times farther, which at a relative velocity of 300 km/s would take $9 \times {10}^{18}\ \mathrm{s} \approx 300\ \mathrm{billion\ years}$.&lt;/p&gt;&#xA;" OwnerUserId="1136" LastEditorUserId="1136" LastEditDate="2014-03-05T22:47:29.647" LastActivityDate="2014-03-05T22:47:29.647" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1922" PostTypeId="1" CreationDate="2014-03-05T14:44:11.257" Score="3" ViewCount="2854" Body="&lt;p&gt;I couldn't find on the web how to get the velocity dispersion and velocity maps of galaxies from the 3D data cube that I get from the integral field spectrograph.&lt;/p&gt;&#xA;" OwnerUserId="1172" LastEditorUserId="537" LastEditDate="2014-03-05T21:13:26.450" LastActivityDate="2014-03-06T14:16:09.117" Title="How do I calculate the velocity dispersion of a galaxy given an integral field spectrograph?" Tags="&lt;spectroscopy&gt;&lt;data-analysis&gt;&lt;observatory&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1924" PostTypeId="2" ParentId="1896" CreationDate="2014-03-05T21:59:37.060" Score="1" Body="&lt;p&gt;Regarding the fact of using the scope indoors: it's understandable, but silly. Unless you live inside a clean room (like those ones in labs), your scope will get dirty anyway, because an average person's room is not free of dust and particles. In fact, it might be even worse, because you won't move the scope frequently and the dust is definitely going to build up. Besides, cleaning a scope is not that difficult, as long as you have the patience to put everything back together later, and then doing the re-collimation. Also, the dust grains on the mirrors are not big enough to depreciate the image, so you might as well just not bother with them. I can guarantee to you that the only dust that cause Astronomers trouble is the dust up there, in space, and rarely this one down here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other things to check: collimation and quality of the optics. Things can get bumpy during transport, so it is possible that the mirrors got misaligned or maybe you got damaged optics (or just poor quality ones). A collimation problem can be easily identified: you'll get comet-like distortions (tails) in stars and planets. Poor quality optics is problematic, because it is generally the item that buyers do not check, so the manufacturer may shove the cheapest optics elements in the package to get a lower price tag. So, if that's the problem, you might want a replacement, or new (better) elements.&lt;/p&gt;&#xA;" OwnerUserId="1151" LastActivityDate="2014-03-05T21:59:37.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1925" PostTypeId="2" ParentId="1896" CreationDate="2014-03-06T02:13:12.920" Score="2" Body="&lt;p&gt;If you are asking 'Why is Mars so small in my Field of View, I thought it would be larger' then you need to consider how big you can expect to see it for various eyepieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now (2014 03 06) Mars is approx 12.6 arcsec in apparent size. It can be as big as 25 arcsec, and as small as 3 arcsec. (Why the variation? We orbit the sun at a different rate to Mars.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is your field of view? Well, for a 2800mm focal length telescope and if your 40mm focal length eyepiece is a Celestron Plossl, then your true field of view will be about 3/4 of a degree (0.74 deg). That is 2664 arcsecs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars will therefore cover 0.0047 of the field of view. That isn't big. 0.5% of the diameter of what you see in the eyepiece. So, it will be barely more than a pin-prick. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had a Celestron 7.5mm Plossl, for a magnification of 373x, then the true field of view for that combination with your telescope would be 0.14 degrees, or 504 arcsec. Then Mars will cover 0.025 of the field of view, or 2.5% of the diameter of the view. So if you mentally divide the field into 10 equally sized stripes or rings, Mars will cover about a quarter of one of those imaginary divisions. Still not big, but a lot bigger than what you are currently seeing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could try going bigger using a shorter focal length eyepiece, but 400x is about the usable limit. Sorry, you aren't going to get an eye-piece filling view of Mars, not unless you're riding on the Mars Global Surveyor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use the equipment to get a picture of Mars that will impress your friends, you might try getting a camera with a small chip with a nice fine pixel size to record video frames. &lt;a href=&quot;http://www.skyandtelescope.com/howto/astrophotography/How_to_Process_Planetary_Images.html&quot; rel=&quot;nofollow&quot;&gt;These you then 'stack' using software&lt;/a&gt; to remove the noise that comes from various sources including atmospheric refraction variations, and you end up with an image that is often quite impressive, and orders of magnitude better than what you can perceive with the naked eye.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-03-06T02:13:12.920" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1929" PostTypeId="2" ParentId="1922" CreationDate="2014-03-06T12:43:40.133" Score="2" Body="&lt;p&gt;If the 3D data cube is sufficiently highly resolved, try to find &lt;a href=&quot;http://en.wikipedia.org/wiki/Relativistic_Doppler_effect&quot; rel=&quot;nofollow&quot;&gt;Doppler shifts&lt;/a&gt; in the &lt;a href=&quot;http://astronomy.nmsu.edu/nicole/teaching/ASTR505/lectures/lecture26/slide01.html&quot; rel=&quot;nofollow&quot;&gt;spectra&lt;/a&gt; by looking for &lt;a href=&quot;http://en.wikipedia.org/wiki/H-alpha&quot; rel=&quot;nofollow&quot;&gt;typical absorption/emission lines&lt;/a&gt;. Doppler shifts correspond to &lt;a href=&quot;http://en.wikipedia.org/wiki/Radial_velocity&quot; rel=&quot;nofollow&quot;&gt;radial velocities&lt;/a&gt;. Calculate the mean radial velocity, and e.g. the &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_deviation&quot; rel=&quot;nofollow&quot;&gt;standard deviation&lt;/a&gt; of the radial velocities, as velocity dispersion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The velocity dispersion of galaxies leads to a widening of spectral lines in comparison to a reference star. A couple of methods are discussed in &lt;a href=&quot;http://mnras.oxfordjournals.org/content/339/1/215.full.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;, mostly some kind of &lt;a href=&quot;http://cas.sdss.org/stripe82/en/help/docs/algorithm.asp?key=veldisp&quot; rel=&quot;nofollow&quot;&gt;Fourier fitting, but direct fitting is used, too&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Redshift&quot; rel=&quot;nofollow&quot;&gt;Redshifts&lt;/a&gt; may contain a &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_redshift&quot; rel=&quot;nofollow&quot;&gt;gravitational&lt;/a&gt; component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://p3d.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;There exist data reduction software tools&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.eso.org/~hkuntsch/papers/MNRAS_373_906.pdf&quot; rel=&quot;nofollow&quot;&gt;A paper&lt;/a&gt; describing data reduction and velocity analysis using the &lt;a href=&quot;http://cral.univ-lyon1.fr/labo/oasis/download/&quot; rel=&quot;nofollow&quot;&gt;XOASIS tool&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-03-06T14:16:09.117" LastActivityDate="2014-03-06T14:16:09.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1930" PostTypeId="1" CreationDate="2014-03-06T17:20:45.380" Score="6" ViewCount="3852" Body="&lt;p&gt;The surface gravities of the Sun’s planets are all close to 1g, 0.38-2.53 (about a factor of e, by chance). The cloud-top gravity of the gas giants is not too interesting here, but I suppose that the gravity on their solid surfaces (if any) does not tend to be higher. So called super-Earth exoplanets seem likely to generally accumulate light gas atmospheres, which decreases their expected solid surface gravity relative to their size as measured during for example transits (I suppose so anyway).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there good reasons to expect that in the zoo of planets in the universe, our Earth scores high in the surface gravity top-list? That it is unusual with solid real estate to stand on which has substantially (as in a factor of a few tenths) greater gravity than we have here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surface gravities from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mercury  0.38&lt;/li&gt;&#xA;&lt;li&gt;Venus    0.90&lt;/li&gt;&#xA;&lt;li&gt;Earth    1.00&lt;/li&gt;&#xA;&lt;li&gt;Mars     0.38&lt;/li&gt;&#xA;&lt;li&gt;Jupiter  2.53&lt;/li&gt;&#xA;&lt;li&gt;Saturn   1.07&lt;/li&gt;&#xA;&lt;li&gt;Uranus   0.89&lt;/li&gt;&#xA;&lt;li&gt;Neptune  1.14&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2014-03-06T17:33:57.263" LastActivityDate="2015-04-14T10:58:15.537" Title="Is Earth's 1g solid surface gravity unusually high for exoplanets?" Tags="&lt;gravity&gt;&lt;exoplanet&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1931" PostTypeId="2" ParentId="1930" CreationDate="2014-03-06T18:08:20.947" Score="4" Body="&lt;p&gt;Surface gravity is a function of two things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Mass&lt;/li&gt;&#xA;&lt;li&gt;Radius&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Mass is proportional to the cube of the radius times the mean density. Gravity is proportional to the square of the radius. As a result, is you have two planets with the same density, the larger planet will have the higher gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But mean density can be a big factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mean density 5.515 g/cm3&lt;/li&gt;&#xA;&lt;li&gt;Mean radius 6371.0 km&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Venus:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mean density 5.243 g/cm3&lt;/li&gt;&#xA;&lt;li&gt;Mean radius 6051.8 km&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Jupiter:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mean density 1.326 g/cm3&lt;/li&gt;&#xA;&lt;li&gt;Mean radius  69911±6 km&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(source: Wikipedia pages for &lt;a href=&quot;http://en.wikipedia.org/wiki/Earth&quot; rel=&quot;nofollow&quot;&gt;Earth&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Venus&quot; rel=&quot;nofollow&quot;&gt;Venus&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Jupiter&quot; rel=&quot;nofollow&quot;&gt;Jupiter&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Venus is both slightly smaller and slightly less dense than Earth, giving it 90% of Earth gravity. Jupiter is much larger (more than 10 times) but much less dense (about one fourth) giving it 10/4 or about 250% of Earth gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that the Earth is fairly small, but unusually dense. Of planets that we can detect with current technology, I would expect that more of them would be larger and/or heavier, as those properties would make them easier for us to detect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of all the planets in the universe, I expect that we will find a distribution not too unlike our own Solar System and that Earth will fall just on the high side of average.&lt;/p&gt;&#xA;" OwnerUserId="38" LastActivityDate="2014-03-06T18:08:20.947" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1932" PostTypeId="2" ParentId="1897" CreationDate="2014-03-06T19:37:53.610" Score="0" Body="&lt;p&gt;I wrote a similar N-body orbital program a few years ago.&#xA;To obtain the initial velocities and positions for Solar System bodies I used a free program called PLANEPH.&#xA;Download available at:- &lt;a href=&quot;ftp://cdsarc.u-strasbg.fr/pub/cats/VI/87/&quot; rel=&quot;nofollow&quot;&gt;ftp://cdsarc.u-strasbg.fr/pub/cats/VI/87/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can give you 3D xyz position and velocity of solar system bodies at a particular point in time with resolution of a few seconds of time as I remember.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The user interface is a bit old-fashioned (DOS-style) but it seems to work OK.  I ran it on Windows 2000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used PLANEPH to extract Heliocentric Coordinates then I had to convert to Solar System Barycentre frame using data on planetary masses obtainable elsewhere (e.g. NASA).&lt;/p&gt;&#xA;" OwnerUserId="233" LastActivityDate="2014-03-06T19:37:53.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1934" PostTypeId="2" ParentId="1930" CreationDate="2014-03-06T22:33:02.093" Score="9" Body="&lt;p&gt;According to &lt;a href=&quot;http://iopscience.iop.org/0004-637X/669/2/1279/pdf/71144.web.pdf&quot; rel=&quot;noreferrer&quot;&gt;this paper&lt;/a&gt;, fig. 4, solid planets of arbitrary mass (up to 3,000 times Earth's mass) don't grow much about three- or four-times the Earth in diameter. That's because interior parts of the planet become compressed by the high pressure. Planets heavier than 3,000 times Earth's mass are in the transition zone to stars. In the case of rocky planets they would become substellar objects, similar in their physical properties (&lt;a href=&quot;http://en.wikipedia.org/wiki/Degenerate_matter&quot; rel=&quot;noreferrer&quot;&gt;electron degeneracy&lt;/a&gt; in the core) to white or &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_dwarf&quot; rel=&quot;noreferrer&quot;&gt;black dwarfs&lt;/a&gt;, not capable to start nuclear fusion due to lack of light elements in their core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Surface_gravity&quot; rel=&quot;noreferrer&quot;&gt;Surface gravity&lt;/a&gt; is proportioal to the mean density $\rho$, and to the radius $r$ of the planet: $g=\frac{4\pi}{3}G\rho r$, with $G$ the gravitational constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A planet with roughly the composition of Earth, and 3,000 times the mass of Earth would have about 3-times the Earth's radius, hence the 27-fold volume, and the 3000/27 = 111-fold density. Its surface gravity would hence be 333 g. That's close to the upper bound of surface gravity of what would be called a planet. With a pure iron planet we could go a little beyond. Any surface gravity below this upper bound is possible for a solid planet, at least in theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to page 1284 of &lt;a href=&quot;http://iopscience.iop.org/0004-637X/669/2/1279/pdf/71144.web.pdf&quot; rel=&quot;noreferrer&quot;&gt;the paper&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Massive solid exoplanets of&#xA;  hundreds to thousands of Earth masses may be able to form around&#xA;  massive stars (B and O stars; 5-120 solar masses) where the protoplanetary&#xA;  disk would contain enough heavy elements.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_classification&quot; rel=&quot;noreferrer&quot;&gt;B and O type stars&lt;/a&gt; are rare (0.13% of main sequence stars) and &lt;a href=&quot;http://en.wikipedia.org/wiki/Main_sequence&quot; rel=&quot;noreferrer&quot;&gt;short-lived&lt;/a&gt; (less than 100 million years; more accurately less than $10^{10}\cdot 18^{-2.5}$ years, for mass &gt; 18 solar masses). Hence large solid planets of the described type will also be rare, although a priori not impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The planet with the highest estimated surface gravity discovered thus far (March 8, 2014) is&#xA;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2008A&amp;amp;A...491..889D&quot; rel=&quot;noreferrer&quot;&gt;CoRoT-Exo-3b&lt;/a&gt; (add &quot;surface gravity&quot; column to the &lt;a href=&quot;http://exoplanets.org/table&quot; rel=&quot;noreferrer&quot;&gt;exoplanets table&lt;/a&gt;, and sort by that column):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;CoRoT-Exo-3b has a radius of 1.01 ± 0.07 R_Jup and transits around its F3-type primary every 4.26 days in a synchronous orbit. Its mass of 21.66 ± 1.0 M_Jup, density of 26.4 ± 5.6 g cm-3, and surface gravity of logg = 4.72 clearly distinguish it from the regular close-in planet population, making it the most intriguing transiting substellar object discovered so far.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A surface gravity of logg = 4.72 means surface gravity equals $10^{4.72}\mbox{ cm}/\mbox{s}^2=52,480 \mbox{ cm}/\mbox{s}^2 = 53.5~g.$ &#xA;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2011PASP..123..412W&quot; rel=&quot;noreferrer&quot;&gt;This research has made use of the Exoplanet Orbit Database&#xA;and the Exoplanet Data Explorer at exoplanets.org.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/COROT-3b&quot; rel=&quot;noreferrer&quot;&gt;CoRoT-3b on Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-03-08T15:03:36.317" LastActivityDate="2014-03-08T15:03:36.317" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1935" PostTypeId="1" AcceptedAnswerId="1945" CreationDate="2014-03-06T22:53:08.813" Score="4" ViewCount="4533" Body="&lt;p&gt;So I was reading &lt;a href=&quot;https://astronomy.stackexchange.com/a/895/1016&quot;&gt;this&lt;/a&gt; answer about how galaxies are the fastest moving objects in the universe because space is expanding faster than the speed of light. This got me wondering, would it be possible to be 'still' inside of space so that galaxies are moving away from you and others towards you? If so, wouldn't this allow us to travel from place to place in the universe much quicker? I know that this wouldn't be something that's even close to possible today, but I was just curious if such a thing is possible. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2020-08-10T00:03:39.550" Title="Is it possible to be 'still' in space?" Tags="&lt;universe&gt;&lt;speed&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1936" PostTypeId="2" ParentId="1935" CreationDate="2014-03-06T23:09:12.277" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow&quot;&gt;We are moving&lt;/a&gt; &quot;... relative to the comoving cosmic rest frame ... at some 371 km/s towards the constellation Leo&quot;. That's slow in comparison to the speed of light.&#xA;...And it would be possible to be &quot;still&quot; relative to the &quot;comoving cosmic rest frame&quot;: Just travel with 371 km/s to the opposite direction relative to the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distant galaxies are moving fast relative to us due the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble%27s_law&quot; rel=&quot;nofollow&quot;&gt;expansion of space&lt;/a&gt;. They are moving slow relative to the cosmic microwave background seen from their location. Slowing down relative to one of those distant galaxies would mean to accelerate to a high velocity relative to our local environment, and relative to the cosmic microwave background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect would be as you described: Some galaxies would approach, some would move away, according to the direction we would travel.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-03-06T23:22:32.257" LastActivityDate="2014-03-06T23:22:32.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1937" PostTypeId="2" ParentId="1935" CreationDate="2014-03-07T00:09:01.687" Score="5" Body="&lt;p&gt;There is no 'still' that is not relative to some other object. So yes, you can be still with respect to one object, but you'll be moving with respect to every other object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for using it as a form of travel, letting your destination come to you, well, there's no difference between that and simply moving toward your destination other than the way you phrase it.&lt;/p&gt;&#xA;" OwnerUserId="562" LastEditorUserId="562" LastEditDate="2014-03-07T02:37:50.843" LastActivityDate="2014-03-07T02:37:50.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1940" PostTypeId="1" AcceptedAnswerId="1943" CreationDate="2014-03-07T05:43:10.523" Score="6" ViewCount="930" Body="&lt;p&gt;I'm working on a planetary motion simulator. I've been working through the equations anomaly, eccentricity, etc. The one thing I'm curious about is if all the ellipses are oriented the exact same way (the direction of the periapsis of all planets are pointing the same way), or is it different for each planet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would this be called? I've only seen eccentricity and inclination.&lt;/p&gt;&#xA;" OwnerUserId="1190" LastActivityDate="2014-03-07T06:19:54.153" Title="What is the orientation of planetary orbits?" Tags="&lt;orbit&gt;&lt;planet&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1942" PostTypeId="2" ParentId="1935" CreationDate="2014-03-07T06:13:39.520" Score="2" Body="&lt;p&gt;The best definition of &quot;still in space&quot; you can have is that you make a weighted average of galaxy velocities around you and their mean movement is zero. This means all of them recedes from you at the same speed when at the same distance, so you are &quot;still&quot; regading what we can call &quot;the cosmic expansion frame&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you must discard nearby galaxies, where their proper motion is big. And note also that you will not be still in reference to some other viewer that perform the same experiment. In fact, you both will be receding from the other at the cosmic expansion rate.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-03-07T06:13:39.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1943" PostTypeId="2" ParentId="1940" CreationDate="2014-03-07T06:19:54.153" Score="5" Body="&lt;p&gt;No, the planet orbits do not have their major axis aligned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_elements&quot;&gt;http://en.wikipedia.org/wiki/Orbital_elements&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have &lt;strong&gt;Inclination&lt;/strong&gt; and &lt;strong&gt;Longitude of the ascending node&lt;/strong&gt; defining the plane, and &lt;strong&gt;Argument of periapsis&lt;/strong&gt; defining the ellipse orientation in that plane.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-03-07T06:19:54.153" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1945" PostTypeId="2" ParentId="1935" CreationDate="2014-03-07T18:19:03.910" Score="12" Body="&lt;p&gt;The so called expansion of the universe is not as trivial as most people think. What is happening, in fact, is that the distance between two points in space (note that I'm not talking about objects with velocities, but just coordinates in space) increases with time in a manner proportional to a given factor (in this case, the Hubble constant - which is actually not a constant in time, but let's ignore that for now).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that galaxies are not exactly moving away from a specific point with a particular velocity (i.e., a particular rest frame), but that the space between them is expanding. It is somewhat hard to wrap your head around the difference between these two situations, but this has to do with the geometry of space, and not with what's inside it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is widely accepted that the expansion of the universe is &lt;a href=&quot;http://physics.ucr.edu/~wudka/Physics7/Notes_www/node124.html&quot;&gt;homogeneous and isotropic&lt;/a&gt; (the Cosmological Principle), which means that there is no &quot;special&quot; position (rest frame) on the universe, and whatever your velocity and your location, the universe will seem to expand the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way to imagine this is to reduce to 2 dimensions expanding in a third dimension. For example, suppose that you live in a galaxy embedded on the surface (a 2D universe) of a balloon (completely unaware of the third dimension), with other galaxies distributed homogeneously. If, for some reason, the balloon is expanding isotropically and, every point of the surface will be moving away from every other point, so will every galaxy from each other. Think about this: is it possible to find a special, &quot;still&quot;, point (without resorting to the elusive third dimension) where you will see galaxies moving from you in one side and galaxies moving towards you on the other side?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our universe works similarly, but we're embedded on a three-dimensional space instead of a two-dimensional one.&lt;/p&gt;&#xA;" OwnerUserId="1151" LastEditorUserId="1151" LastEditDate="2014-03-07T18:35:01.063" LastActivityDate="2014-03-07T18:35:01.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1947" PostTypeId="2" ParentId="718" CreationDate="2014-03-08T00:47:11.813" Score="0" Body="&lt;p&gt;In most cases clusters of galaxies are moving away from each other. The gravitational attraction between galaxies is usually enough in concentrated clusters to overcome the cosmological expansion. The Milky Way and Andromeda, both the largest galaxies in the local group, the cluster of which we are a part of, are close and massive enough to overcome this expansion. Therefore we will never lose sight of the stars in the Andromeda galaxy until the stars within it die and/or fade out, in approximately 10 billion years. However, if you are talking about individual stars, not the collective starlight of stars in a given galaxy, then the answer is we still cant!&lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2014-03-08T00:47:11.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1948" PostTypeId="1" CreationDate="2014-03-08T06:59:08.443" Score="6" ViewCount="401" Body="&lt;p&gt;The celestial bodies like stars and galaxies are moving away from each other. How fast are they moving apart? Is that speed more than the speed of light?&lt;/p&gt;&#xA;" OwnerUserId="1100" LastActivityDate="2014-03-12T08:17:35.613" Title="Is the universe expanding faster than speed of light?" Tags="&lt;universe&gt;&lt;expansion&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1949" PostTypeId="2" ParentId="1948" CreationDate="2014-03-08T11:09:34.223" Score="5" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble%27s_law&quot;&gt;Hubble's law&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For [proper] distances $D$ larger than the radius of the Hubble sphere $r_{\mbox{HS}}$ , objects recede at a rate faster than the speed of light ... $$r_{\mbox{HS}}=\frac{c}{H_0}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With $H_0=67.8\mbox{ }(\mbox{km}/{\mbox{s}})/\mbox{Mpc}$ the Hubble constant according to the &lt;a href=&quot;http://www.nasa.gov/mission_pages/planck/news/planck20130321.html&quot;&gt;Planck mission&lt;/a&gt;, $c=299,792,458\mbox{ m}/\mbox{s}$ the &lt;a href=&quot;http://en.wikipedia.org/wiki/Speed_of_light&quot;&gt;speed of light&lt;/a&gt;, and one Mega&lt;a href=&quot;http://en.wikipedia.org/wiki/Parsec&quot;&gt;parsec&lt;/a&gt; $1\mbox{ Mpc}=3.0857\cdot 10^{22}\mbox{ m}$, we get&#xA;$$r_{\mbox{HS}}=\frac{c}{H_0}=\frac{299,792,458\mbox{ m}/\mbox{s}}{67.8\cdot 10^3\mbox{ }(\mbox{m}/{\mbox{s}})/3.0857\cdot 10^{22}\mbox{ m}}=1.3644\cdot 10^{26}\mbox{ m}.$$&#xA;That's $$1.3644\cdot 10^{26}\mbox{ m}/1.3644\cdot 10^{26}\mbox{ m}= 4,421 \mbox{ Mpc}$$&#xA;or $r_{\mbox{HS}}=14.422$ billion lightyears (1 pc = 3.26156 ly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence objects further away than this proper distance (what you would measure with a chain of rulers) recede faster than the speed of light. The distance is also called &lt;em&gt;Hubble length&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may find 13.8 billion lightyears elsewhere. That's calculated with $H_0=70.4\mbox{ }(\mbox{km}/{\mbox{s}})/\mbox{Mpc},$ according to the estimates of 2010, based on WMAP data, or with even older data. The exact value is not known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Hubble constant isn't really constant over time, the universe is thought to be a little younger (13.8 billion years) than the 14.422 billion years you get by just dividing the Hubble distance by the speed of light.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-03-08T11:09:34.223" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1950" PostTypeId="1" CreationDate="2014-03-09T16:07:24.267" Score="3" ViewCount="579" Body="&lt;p&gt;It's believed the main reason that the interior of the Earth hasn't cooled down in 4 billion years is because of the ongoing decay of radioactive elements that keeps adding more heat than is lost to the various cooling processes. Is this a proven fact or are there other possibilities?  &lt;/p&gt;&#xA;" OwnerUserId="1023" LastActivityDate="2014-03-09T19:23:30.370" Title="Radioactive decay is the main source of the earth's interior heat: fact or theory?" Tags="&lt;earth&gt;&lt;geology&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-03-10T14:37:48.110" ContentLicense="CC BY-SA 3.0" />
  <row Id="1951" PostTypeId="2" ParentId="1950" CreationDate="2014-03-09T19:23:30.370" Score="0" Body="&lt;p&gt;There are really only six ways you could heat the interior of a planet. Gravitational collapse and compression, fission, fusion, crystallization, a very large impactor, and tidal flexing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fusion is out because Earth would be a rock-star. (Been waiting to use that one.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the remaining possibilities, nuclear fission explains the majority of Earth's current temperature gradient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Geothermal_gradient&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Geothermal_gradient&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="562" LastActivityDate="2014-03-09T19:23:30.370" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1952" PostTypeId="1" AcceptedAnswerId="1953" CreationDate="2014-03-10T22:59:17.907" Score="10" ViewCount="404" Body="&lt;p&gt;We probably all know the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wow!_signal&quot; rel=&quot;noreferrer&quot;&gt;Wow!-Signal&lt;/a&gt;, happened on August 15, 1977. Assume that it wasn't meant for us. What was it pointing at then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The signal at that time came from RA= 19h25m31s ± 10s or 19h28m22s ± 10s and the declination becomes −26°57′ ± 20′. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which stars could it point at and did anybody investigate this?&lt;/p&gt;&#xA;" OwnerUserId="218" LastEditorUserId="525" LastEditDate="2014-04-01T14:52:58.193" LastActivityDate="2014-11-11T04:26:26.183" Title="If the Wow! signal was not directed at us, then where?" Tags="&lt;extra-terrestrial&gt;&lt;astrometry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1953" PostTypeId="2" ParentId="1952" CreationDate="2014-03-11T14:24:35.533" Score="3" Body="&lt;p&gt;In the opposite direction (RA= 07h 25m 31s ± 10s, +26° 57′ ± 20′, J2000.0, seen from Earth) of the positive horn of the &lt;a href=&quot;http://www.youtube.com/watch?v=ZAKy_08klrg&quot; rel=&quot;nofollow&quot;&gt;Wow!-signal&lt;/a&gt; (RA= 19h 25m 31s ± 10s, −26° 57′ ± 20′, in J2000.0 coordinates), &lt;a href=&quot;http://en.wikipedia.org/wiki/Iota_Geminorum&quot; rel=&quot;nofollow&quot;&gt;Iota Geminorum&lt;/a&gt;, aka Propus, (RA= 07h 25m 43.6s, +27° 47' 53&quot;) is within a 3-sigma error ellipse (1.26 sigma in RA, 2.54 sigma in Dec, 2.84 sigma total error).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this doesn't reveal, whether the signal pointed to anywhere.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-03-11T14:24:35.533" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1954" PostTypeId="2" ParentId="1918" CreationDate="2014-03-11T20:42:07.557" Score="1" Body="&lt;p&gt;Dates are set by progress through the currently used calendar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Season, solstices and equinoxes are set by the direction of tilt of the earth towards/away from the sun, based on the earth's orbital position and the absolute direction of that tilt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The date of seasons, solstices and equinoxes will shift as long as the calendar system does not match the reality of the earth's orbital motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dates of seasons, solstices and equinoxes had shifted by about 14 days by the time the Gregorian calendar began to replace the Julian calendar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole purpose of a particular pattern of leap-years is to &lt;strong&gt;&lt;em&gt;keep&lt;/em&gt;&lt;/strong&gt; the seasons/solstices/equinoxes in long-term step with the calendar. Over four years, there will be small wiggles back and forth...&lt;/p&gt;&#xA;" OwnerUserId="420" LastActivityDate="2014-03-11T20:42:07.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1955" PostTypeId="2" ParentId="1850" CreationDate="2014-03-12T01:10:56.027" Score="1" Body="&lt;p&gt;This is largely speculation on my part; take it with a Moon-sized grain of salt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For whatever reason, the Moon's mass is distributed unevenly; see &lt;a href=&quot;https://astronomy.stackexchange.com/a/1861/53&quot;&gt;Py-ser's answer&lt;/a&gt; for some plausible explanations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon's rotation is tidally locked, so one side always faces the Earth. Given this fact, in principle the Moon's orientation could be anything -- but the actual orientation is controlled by the uneven pull of the Earth's gravity on the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's not the case that the near side has more maria because it's the near side. Rather, it's the near side &lt;em&gt;because it has more maria&lt;/em&gt; (and because of the mass concentrations associated with the maria). If the orientation were off by, say, 90° from what it is now, that would be an unstable configuration, and it would gradually rotate (perhaps wobbling back and forth for a while) until it returned to something like the current more stable configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wild speculation: My guess is that an orientation 180° away from the current one, with the maria facing &lt;em&gt;away&lt;/em&gt; from Earth, would also be stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This answer would be better if it included some actual numbers, but I don't have them.)&lt;/p&gt;&#xA;" OwnerUserId="53" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2014-03-12T01:10:56.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1956" PostTypeId="2" ParentId="1948" CreationDate="2014-03-12T08:17:35.613" Score="1" Body="&lt;p&gt;The comoving radial distance, which is also the proper distance at current epoch, at redshift $z$ is given by&#xA;$$D_\mathrm{C} = \frac{c}{H_0}\int_0^z\frac{\mathrm{d}x}{\sqrt{\Omega_\mathrm{M}(1+x)^3 + \Omega_k(1+x)^2+\Omega_\Lambda}}\text{.}$$&#xA;For a spatially flat cosmology, $\Omega_k = 0$, and so matter and dark energy density parameters are related through $\Omega_\mathrm{M} = 1-\Omega_\Lambda$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Wilkinson_Microwave_Anisotropy_Probe&quot; rel=&quot;nofollow&quot;&gt;WMAP&lt;/a&gt;, $\Omega_\Lambda = 0.721\pm 0.025$ (and various refinements for measurements of dark energy density), from which it follows that every object comoving with the Hubble flow at redshift $z\gtrsim 1.4$ has superluminal recession velocity from us. This corresponds to a light travel time of $c^{-1}D_\mathrm{ltt}\gtrsim 9.1\,\mathrm{Gyr}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most distant objects found by the Hubble telescope have $z\approx 12$, corresponding to a light travel time of approximately $13.4\,\mathrm{Gyr}$. Therefore, we do routinely see objects that have superluminal recession velocity from us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the light travel times with the help of &lt;a href=&quot;http://www.astro.ucla.edu/~wright/CosmoCalc.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; cosmology calculator.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Clarification:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... however, the fact still remains that [Gerald] has quoted that the distance at which objects are moving faster than c is larger than the distance traveled by light since the big bang.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a conflation of distances. The light travel time distance $D_\mathrm{ltt}$ of objects with recession velocity of $c$ is only $\sim 9.1\,\mathrm{Gly}$. Intuitively, since the light we see now has been emitted in the distant past, the universe has expanded a lot in the time it took for light to get here, and therefore the object's recession velocity &lt;em&gt;now&lt;/em&gt; is much higher. Assuming the object still exists.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-03-12T08:17:35.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1957" PostTypeId="2" ParentId="1918" CreationDate="2014-03-12T23:15:47.633" Score="6" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Gregorian_calendar&quot; rel=&quot;nofollow&quot;&gt;Gregorian Calendar&lt;/a&gt; was created so that annual astronomical events, specifically the vernal equinox (used to determine when Easter is), would on average keep their places in the calendar year over time.  It is the best official approximation to the definition of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tropical_year&quot; rel=&quot;nofollow&quot;&gt;tropical year&lt;/a&gt;, which is defined as &quot;the length of time that the Sun takes to return to the same position in the cycle of seasons&quot;.  Because this calendar describes 97 leap years out of every 400 years, it defines the average year as exactly 365.2425 solar days, or exactly 365 days, 5 hours, 49 minutes, and 12 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the mean tropical year is in reality about 365 days, 5 hours, 48 minutes, and 45 seconds, or 27 seconds shorter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the Gregorian Calendar is based on the tropical year, the calendar dates of the year will keep up with the solstices and equinoxes, and thus the seasons.  If this calendar were exactly the length of the tropical year, then the calendar would keep the vernal (northward) equinox around March 20th for all time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But because of the slight inaccuracy, it will take about 3,200 years (60 s/min * 60 min/hr * 24 hr/day / 27 s/year) for these 27 seconds to add up every year to be 1 full day, and that will result in the solstices and equinoxes marching backwards in the calendar by 1 day every 3,200 years or so, depending on the accuracy of the 27 seconds difference.  This very slow shift is due to the slight inaccuracy in the Gregorian calendar in, on average, matching the tropical year, not because of the precession of the equinoxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3,200 years from now, if the Gregorian Calendar is still used, the date of the vernal (northward) equinox will be on average one day earlier in March.  The precession of the equinoxes will still occur, so the Earth's axis tilt will be significantly different from today.  The Earth will be at a noticeably different position with respect to the Sun on the vernal (northward) equinox from where it is today, in 2014, on the vernal (northward) equinox, but it will still be in March.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This inaccuracy may very slowly increase over time, because according to the same Wikipedia page for the tropical year, the tropical year is very slowly getting shorter, and the mean solar day is even more slowly getting longer.  But for 10,000 years to come, the Gregorian Calendar will keep the vernal (northward) equinox in March, even if it slowly shifts earlier in the month.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is in contrast to the scenario that you imply, where the calendar date would correspond to the relative position of the Earth in its orbit around the Sun.  That describes the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sidereal_year&quot; rel=&quot;nofollow&quot;&gt;sidereal year&lt;/a&gt;, the time taken for the Sun to reach the same spot in the sky relative to the stars, which is 365 days, 6 hours, 9 minutes, and 10 seconds.  A sidereal calendar would explain why you might think that precession would cause the dates of equinoxes and solstices to change in the calendar year.  That would result in a shift in the calendar of one full month in 1/12th the cycle length of the precession of the equinoxes, or about 1 full month about every 2,000 years.&lt;/p&gt;&#xA;" OwnerUserId="1082" LastEditorUserId="1082" LastEditDate="2014-03-13T20:54:40.047" LastActivityDate="2014-03-13T20:54:40.047" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1958" PostTypeId="2" ParentId="1649" CreationDate="2014-03-13T11:05:12.880" Score="2" Body="&lt;p&gt;The DS9 Reference Manual, p42, 'Numeric Formats'. &lt;a href=&quot;http://ds9.si.edu/doc/ref/ref.pdf#page48&quot; rel=&quot;nofollow&quot;&gt;http://ds9.si.edu/doc/ref/ref.pdf#page48&lt;/a&gt; has a bunch of information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on that, I expected that &quot;tlhm.1&quot; would achieve what you proposed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it seems that SAOImage DS9 is very keen on dropping leading hours it sees as unnecessary, even if you have specified them in a format specifier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't been able to figure out a format specifier combination that retains the hours for every gridline while also showing the minutes or seconds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have searched elsewhere for advice on this and haven't found it either. In fact, I haven't been able to find any comment on the hours being suppressed, nothing that explains why it does this, nor even anyone else remarking or complaining about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest sending an email to the DS9 Help Desk address found in the menu under Help &gt; Help Desk.&lt;/p&gt;&#xA;" OwnerUserId="67" LastEditorUserId="67" LastEditDate="2014-04-04T10:59:45.490" LastActivityDate="2014-04-04T10:59:45.490" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1959" PostTypeId="2" ParentId="1918" CreationDate="2014-03-13T20:07:37.627" Score="1" Body="&lt;p&gt;Yes, and no. yes, the equinox will drift over time, but the seasons will not invert every 13000 years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is the calendar we use is already adjusting for most of the drift due to the precession of the equinoxes. In 10,000 years the drift in seasons will be only around 10 days. &lt;a href=&quot;http://en.wikipedia.org/wiki/Gregorian_calendar#Accuracy&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Gregorian_calendar#Accuracy&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-03-13T20:07:37.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1960" PostTypeId="2" ParentId="1373" CreationDate="2014-03-13T21:15:47.920" Score="1" Body="&lt;p&gt;The moon's rotation plane would precess (more) in the same direction as Earth's rotation, if the Earth spun faster.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-03-13T21:15:47.920" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1961" PostTypeId="1" AcceptedAnswerId="11828" CreationDate="2014-03-14T04:35:19.563" Score="17" ViewCount="8613" Body="&lt;p&gt;&lt;strong&gt;Disclaimer:&lt;/strong&gt; I’m not a career astronomer.  I don’t own a telescope.  I have no professional credentials.  But I do find this stuff fascinating, and I consume all the astronomy documentaries I can.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, I’ve watched lots of documentaries describing stellar evolution.  I understand that below a certain threshold, stellar death does not involve supernovae.  I understand that above that threshold, supernovae may create neutron stars, magnetars, or (if the supernova qualifies as a hypernova) black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for a long time, I was curious about &lt;strong&gt;why&lt;/strong&gt; stars below the supernova threshold—like our own Sun—become Red Giants.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From documentaries, I have been instructed that (for stars below the supernova threshold), when the star’s core’s fusion cannot continue…fusion ceases, and the star begins to collapse under gravity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As gravity crushes the star, I understand that the star heats up as gravity crushes it.  As a result, although the stellar core remains “dead” (no fusion occurs), a “shell” of gas around the stellar core becomes hot enough to begin fusing helium.  Since the fusion occurs as a “shell” around the stellar core, the outward-push from the fusion is what pushes the star’s outer layers further.  The result is that the star grows into a Red Giant.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question is this:&lt;/strong&gt;  Why does fusion cease in the core?!  It seems to me that as gravity crushes the star, stellar fusion would reignite in the core itself—not in a sphere around the core.  Why does the stellar core remain “dead” while its “shell” begins fusion???&lt;/p&gt;&#xA;" OwnerUserId="1216" LastEditorUserId="4085" LastEditDate="2015-09-17T07:09:01.573" LastActivityDate="2019-06-06T18:06:50.757" Title="Why do stars become red giants?" Tags="&lt;the-sun&gt;&lt;red-giant&gt;&lt;stellar-evolution&gt;" AnswerCount="6" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1963" PostTypeId="2" ParentId="1961" CreationDate="2014-03-14T07:29:38.417" Score="3" Body="&lt;p&gt;The destiny of a star basically depends upon its mass.&#xA;All its activities variety depends upon its mass.&#xA;If a star's core has a mass that is below the Chandraseckhar limit ($M\sim1.4M_{sun}$), then is destined to die as a white dwarf (or, actually, as a black dwarf in the end).&#xA;The &lt;em&gt;composition&lt;/em&gt; of the white dwarf, also depends upon the original mass of the star. Different masses will lead to different compositions.&#xA;More precisely, the more massive is the star, the heavier are the elements composing the final object.&#xA;This is because more mass means more &lt;a href=&quot;http://burro.astr.cwru.edu/Academics/Astr221/StarPhys/gravity.html&quot; rel=&quot;nofollow&quot;&gt;gravitational potential energy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dU = - \frac{GM(r)dm}{r}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that in turns can be converted into heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydrogen nuclear fusion starts, for the &lt;a href=&quot;https://en.wikipedia.org/wiki/Proton-proton_chain&quot; rel=&quot;nofollow&quot;&gt;proton-proton reaction&lt;/a&gt;(that is the dominant process for Sun-like stars) at around $10^7 K$. This is the value that allows the particles to overcome their coulombian barrier (i.e., to &lt;em&gt;fuse&lt;/em&gt;).&#xA;After the hydrogen fusion, when the most of the core is composed by helium, then of course the hydrogen fusion can't happen anymore. The core starts to collapse, and heats itself. For a Sun-like star, there is enough mass to compress up to a level that heats the core enough to start the He burning. But that is all. When also the Helium is converted into Carbon, the star has not enough mass to compress again up to a level that starts another nuclear fusion reaction. &lt;em&gt;This is why the core nuclear reactions stop&lt;/em&gt;. For the shell burning question, it is important to understand two things: $(1)$ the shell structure of a star is only an approximation, and $(2)$ there is a gradient of temperature within Sun-like stars, that means that (besides the corona) the temperature increases when you go from the outside to the core. Now, if the core is compressed and became so hot to burn helium, the shell &quot;outside&quot; the core (that in a onion-like schema was within the radius of the previous hydrogen burning core), is still hot enough to burn hydrogen. The &lt;em&gt;size&lt;/em&gt; of the helium-burning core is smaller than the hydrogen-burning core (this is compression &lt;em&gt;by definition&lt;/em&gt;). The shell has still enough hydrogen, and contemporary is deep enough inside the star (that means high temperature), to allow nuclear fusion of hydrogen.&#xA;If the star was more massive, more things could happen, like heavier elements core fusion, and more and more burning shells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at these:&#xA;&lt;a href=&quot;http://abyss.uoregon.edu/~js/ast122/lectures/lec16.html&quot; rel=&quot;nofollow&quot;&gt;Ref 1&lt;/a&gt;, &lt;a href=&quot;http://www.astro.virginia.edu/class/whittle/astr1220/17_Star_Evolution/star_evolution.pdf&quot; rel=&quot;nofollow&quot;&gt;Ref 2&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astro.princeton.edu/~burrows/classes/403/shell.burning.pdf&quot; rel=&quot;nofollow&quot;&gt;Ref 3&lt;/a&gt; for some numbers too.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastEditorUserId="1038" LastEditDate="2014-03-14T07:59:37.267" LastActivityDate="2014-03-14T07:59:37.267" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1965" PostTypeId="1" CreationDate="2014-03-14T08:27:40.197" Score="3" ViewCount="483" Body="&lt;p&gt;I know black dwarf stars do not exist yet, and that they are what white dwarfs end up after a long time. The question is really simple: would it be possible to have a binary system in which one of the two is a black dwarf, and the other companion star is dumping mass on the black dwarf, causing the black dwarf to supernova? I guess based on the timescales, the maximal initial mass of the second object could be calculated to see if it is possible that one object is at the black dwarf stage and the other still a giant. Or is the timescale on which the star goes from white dwarf to black dwarf big enough that it will never reach the black dwarf stage while the other one is still an AGB star? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second part: If not possible that one is black dwarf and the other AGB star, is it possible if the companion is a brown dwarf, that they are close binary, and that a brown dwarf would dump mass on a black dwarf? This would need the binary system to be a very close binary system at this stage of the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third part of the question: would the spectrum of the supernova be any different from the normal one with a white dwarf and a companion AGB star? I guess the spectral lines would be different, since the companion would have to be a very low-mass AGB star, since that companion would be about 20billion years or so, when the black dwarf star becomes a black dwarf.&lt;/p&gt;&#xA;" OwnerUserId="1062" LastActivityDate="2020-12-11T00:25:43.360" Title="Black dwarf binary systems" Tags="&lt;supernova&gt;&lt;binary-star&gt;&lt;stellar-evolution&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1966" PostTypeId="1" AcceptedAnswerId="2038" CreationDate="2014-03-14T15:57:23.503" Score="8" ViewCount="370" Body="&lt;p&gt;Do galaxies have habitable zones the same as stars do? Say in a galaxy with a very active nucleus producing a lot of heat and radiation, would there be a point at which no star's planets could harbor life do to the effects of the black hole? Also would there possibly be a habitable zone for galaxy clusters? If there were many galaxies with extremely active galactic nuclei condensed closely together would this possibly also hinder the evolution of life? &lt;/p&gt;&#xA;" OwnerUserId="1206" LastEditorUserId="1206" LastEditDate="2014-03-17T18:51:21.590" LastActivityDate="2015-09-14T17:35:43.967" Title="Galactic Habitable Zone" Tags="&lt;solar-system&gt;&lt;exoplanet&gt;&lt;planet&gt;&lt;earth-like-planet&gt;&lt;habitable-zone&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1968" PostTypeId="2" ParentId="1966" CreationDate="2014-03-14T20:18:45.173" Score="0" Body="&lt;p&gt;The zone, you mean, in galaxies would be very unstable - huge changes in electromagnetic flux (something like explosions) because of more dynamic changes of the environment there (than in the outer rim of galaxy: black hole, jets, etc...) - so there is considered lower probability of life in the center of galaxies (or Galaxy). But in fact I think we have to few information about galaxies to seriously think about probability of life on free planets close to the center of the Galaxy (the more other galaxies).&lt;/p&gt;&#xA;" OwnerUserId="1221" LastActivityDate="2014-03-14T20:18:45.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1969" PostTypeId="1" AcceptedAnswerId="1987" CreationDate="2014-03-14T20:31:20.517" Score="1" ViewCount="2098" Body="&lt;p&gt;consider two stars of the spectral type and subtype, but not necessarily both on the main sequence. Describe whether this information is sufficient to tell you how the stars differ in surface temperature, size, and distance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so if they have they same spectral type and subtype i think they're supposed to have the same surface temperature. however this is not enough information to tell us about their sizes and distances. is this correct? any insight would be great, thanks!&lt;/p&gt;&#xA;" OwnerUserId="1222" LastActivityDate="2014-07-10T03:38:38.923" Title="regarding spectral type of a star" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1970" PostTypeId="1" AcceptedAnswerId="1977" CreationDate="2014-03-15T11:26:24.953" Score="3" ViewCount="595" Body="&lt;p&gt;Within a year now it has been decided to launch TESS in 2017 and PLATO in 2024, space telecopes of NASA and ESA. They seem to have very similar science goals, both searching for transiting Earth sized planets in the HZ of the brighest (nearest) stars across all or most of the sky. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these missions as overlapping as they seem on first sight, or are there important complementarities? Is it perhaps the intention for PLATO to continue the observations of the planets discovered by TESS in order to get a longer time series?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="37191" LastEditDate="2021-02-19T10:18:29.813" LastActivityDate="2021-02-19T10:18:29.813" Title="What are the differences between TESS and PLATO exoplanet telescopes?" Tags="&lt;exoplanet&gt;&lt;space-telescope&gt;&lt;nasa&gt;&lt;esa&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1972" PostTypeId="1" AcceptedAnswerId="2040" CreationDate="2014-03-15T14:35:54.023" Score="5" ViewCount="331" Body="&lt;p&gt;As explained in &lt;a href=&quot;http://motls.blogspot.de/2014/03/rumor-inflation-related-primordial-b.html?m=1&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; article, as the CMB is created from a vaiable (matter?) density $\kappa$ its gradient $\nabla \kappa = \vec{u}$ can be separated into a gradient and curl part, called the $E$ and $B$ modes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\nabla^2 \kappa^E = \nabla \cdot \vec{u} &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\nabla^2 \kappa^B = \nabla \times \vec{u}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking about electromagnetism, these definitions look slightly (but not exactly) familiar, if the vector $\vec{u}$ would take the role of the electromagnetic potential $\vec{A}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the similarity to the terminology of electromagnetism just a coincidence, or do these $E$ and $B$ modes have indeed something to do with electric and magnetic fields ?&lt;/p&gt;&#xA;" OwnerUserId="92" LastEditorUserId="92" LastEditDate="2014-03-16T12:37:25.037" LastActivityDate="2014-03-24T15:02:04.697" Title="Do the E and B modes of the CMB polarization have anything to do with electric and magnetic fields?" Tags="&lt;cosmology&gt;&lt;cmb&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1973" PostTypeId="1" CreationDate="2014-03-15T15:56:08.537" Score="5" ViewCount="262" Body="&lt;p&gt;On episode 10 of the original Cosmos TV series, Carl Sagan mentions that some scientist don't agree that the redshift observed in distant galaxies is evidence of the big bang, stating that probably the doppler effect is not responsible for this effect. What other explanations are there?&lt;/p&gt;&#xA;" OwnerUserId="537" LastActivityDate="2019-04-09T18:27:27.300" Title="Is there another explanation, apart from the doppler effect, for the redshift observed in distant galaxies?" Tags="&lt;galaxy&gt;&lt;big-bang-theory&gt;&lt;radio-astronomy&gt;&lt;astrophysics&gt;&lt;redshift&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1974" PostTypeId="1" CreationDate="2014-03-15T17:29:04.270" Score="2" ViewCount="57" Body="&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: &lt;em&gt;Which is the average or maximum speed of a telematically sent message that a space probe can send&lt;/em&gt;? Is as simple as radio waves travel at the speed of light in a vacuum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Context&lt;/strong&gt;: I have been asked to make a first estimation about how much time is needed to send a space probe and then receive the signal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am supposing that the space probe is sent at $15 \ km/s$ but I have not idea of the telematic signal speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance! &lt;/p&gt;&#xA;" OwnerUserId="1226" LastEditorUserId="1226" LastEditDate="2014-03-15T17:44:52.303" LastActivityDate="2014-03-15T17:44:52.303" Title="Max. speed that a space probe can communite with the Earth?" Tags="&lt;exoplanet&gt;&lt;space&gt;&lt;extra-terrestrial&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2014-03-17T15:08:04.757" ContentLicense="CC BY-SA 3.0" />
  <row Id="1975" PostTypeId="1" CreationDate="2014-03-15T23:42:09.500" Score="3" ViewCount="75" Body="&lt;p&gt;I've been playing around with Near Star Map, and I think I kind of want something more sophisticated, so for all stars within 100ly/30pc, is the following known?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;velocity (it would be nice to have a vector field showing how fast in which direction stars are moving)&lt;/li&gt;&#xA;&lt;li&gt;magnetic field intensity&lt;/li&gt;&#xA;&lt;li&gt;estimated age&lt;/li&gt;&#xA;&lt;li&gt;local ecliptic plane (and as much as that's meaningful for three-star systems with Poincare-befuddling orbital dynamics)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2014-10-05T18:06:02.517" Title="local stellar mapping data" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1976" PostTypeId="1" CreationDate="2014-03-16T01:20:33.040" Score="3" ViewCount="131" Body="&lt;p&gt;I'm designing a sensor system to perform specialized [astronomy and space-sciences] experiments, and need a &quot;reality check&quot; to support or adjust my theoretical calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I need is the &quot;counts per second&quot; produced by any modern APD (avalanche photo-diode) sensor through a telescope of any specific &quot;aperture&quot; of a star of any &quot;visual magnitude&quot;.  I also need the number of &quot;counts per second&quot; of &quot;nothing&quot; (the &quot;dark count&quot;) to subtract that from the &quot;counts per second&quot; when illuminated by the star (to determine the &quot;counts per second&quot; generated by the star alone).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &quot;reality check&quot; will help me assure various &quot;inefficiency allowances&quot; I made are realistic.  Examples:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#1: overall detector QE over relevant visual [and near IR] wavelengths.&#xA;#2: loss of light in atmosphere before entering telescope.&#xA;#3: loss of light in telescope optics.&#xA;#4: loss of light in fiber (if any).&#xA;#5: anything/everything else.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As implied, I am only interested in the APD operating in &quot;photon counting mode&quot; (not analog).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read about 5 dozen articles that discuss APDs for astronomy, but none give a straightforward value.  The closest I found was a vague statement that the limited magnitude was 22nd magnitude on a 6-meter telescope based upon observations of the crab nebula pulsar.  But this is not specific and the object is highly variable (on a short time frame).  They did not say, for example, whether they consider their &quot;limiting magnitude&quot; is where the count rate rises from 200 per second (dark count) to 220 per second (measurement), or 200cps to 400cps, or over what time period, or any other indication of their definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I need is ONE clear statement of cps for any aperture and visual magnitude star.  You'd think I could find that in dozens if not hundreds of articles, but... no.  Probably a clear statement like I need exists in some article somewhere, but I haven't seen one.  Have you?  Or better yet, have you made such an observation yourself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following detail is not very important (but just to be complete), my primary applications perform fairly high time-resolution measures on fairly bright stars.  In other words, the experiments generally need to measure in the range of &quot;counts per microsecond&quot; to &quot;counts per millisecond&quot;.  Typically APDs max out at around 15 to 50 million counts per second, and most of my experiments will be working at 10K to 10M counts per second to observe the short time-period phenomenon I need to measure.&lt;/p&gt;&#xA;" OwnerUserId="1229" LastEditorUserId="1229" LastEditDate="2014-03-16T01:34:46.500" LastActivityDate="2014-03-16T01:34:46.500" Title="need data-point: actual counts per second with APD (avalanche photodiode) detector" Tags="&lt;star&gt;&lt;telescope&gt;&lt;apparent-magnitude&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1977" PostTypeId="2" ParentId="1970" CreationDate="2014-03-16T06:31:40.837" Score="3" Body="&lt;p&gt;Here is what one commentator has said about TESS in comparison to PLATO:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The Transiting Exoplanet Survey Satellite (TESS), to launch in 2017, seems superficially to be a similar mission to Plato. It will potentially discover hundreds of planets before Plato even gets off the ground in 2024. However, the limited sensitivity of its cameras mean it is completely blind to Earth-like worlds around sun-like stars. Astroseismology is also off-limits for TESS, meaning the size of any worlds it does discover will be highly uncertain. Unlike Plato, it will also move between patches of sky every 30 days, allowing only hot, short-period planets to be found.&#xA;With all other new telescopes, both in space and on the ground, limited to finding super-Earths around small stars, Plato is the only mission on the table truly capable of discovering an Earth-like world around a star like our Sun. And by targeting bright stars that allow atmospheric follow-up, it is not impossible to think that, as well as the first truly habitable planet, Plato could find the first inhabited one too.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Credit: Hugh Osborn, PhD student, University of Warwick&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://lostintransits.wordpress.com/2014/01/30/what-can-plato-do-for-exoplanet-astronomy/&quot; rel=&quot;nofollow&quot;&gt;http://lostintransits.wordpress.com/2014/01/30/what-can-plato-do-for-exoplanet-astronomy/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-03-16T06:31:40.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1979" PostTypeId="1" AcceptedAnswerId="1982" CreationDate="2014-03-16T14:07:16.187" Score="10" ViewCount="5171" Body="&lt;p&gt;In our solar system, with the exception of Pluto all planets follow a relatively circular orbit around the Sun, at the same inclination. They also all rotate in the same direction, none are 'retrograde'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bY8GT.gif&quot; alt=&quot;Solar System&quot;&gt;&lt;br&gt;&#xA;An image of the orbits of our solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How and why have they all formed into these orbits? Is this reflected in most orbital systems? What factors affect the orbits of newly formed systems?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="480" OwnerDisplayName="Vedant Chandra" LastActivityDate="2020-02-02T01:32:10.520" Title="Why are most planetary orbits nearly circular" Tags="&lt;orbit&gt;&lt;solar-system&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1980" PostTypeId="1" AcceptedAnswerId="1989" CreationDate="2014-03-16T18:52:22.267" Score="2" ViewCount="1006" Body="&lt;p&gt;Does outer space contain any gases? Recently i watched a video on youtube,in which NASA installs a camera on a spacecraft.The footage shows some dust particles or some matter floating over the space craft's surface and as well from the thrust.What exactly is that? If there is nothing in vacuum,how come newton third law works out to make it move in turn? Here is the link&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=HSnOTaCeSuk&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=HSnOTaCeSuk&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1103" LastActivityDate="2014-03-17T17:08:12.153" Title="What exactly does outer space contain?" Tags="&lt;space&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1981" PostTypeId="1" AcceptedAnswerId="1983" CreationDate="2014-03-16T19:01:32.313" Score="2" ViewCount="603" Body="&lt;p&gt;Why Earth is like a magnet? how can earth attract any object irrespective of what it is made of? What is the reason behind Celestial Body's gravity? i.e Sun,Jupiter,Moon etc.&lt;/p&gt;&#xA;" OwnerUserId="1103" LastActivityDate="2016-03-14T17:32:30.027" Title="What causes Earth's gravity?" Tags="&lt;gravity&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2014-03-28T13:38:18.097" ContentLicense="CC BY-SA 3.0" />
  <row Id="1982" PostTypeId="2" ParentId="1979" CreationDate="2014-03-16T19:14:28.477" Score="9" Body="&lt;p&gt;The combination of the conservation of angular momentum and gravity give you the inclination and direction.  Gravity will condense material in the axis direction of the angular momentum, and collisions and gravitational interactions will damp oscillations in that direction, forming a disk of material.  The angular momentum resists gravity in the plane perpendicular to the axis, maintaining the extent of the disk.  Over time, the direction of rotation of the angular momentum is the direction of the vast majority of the material, due to collisions and interactions with that smaller population going the wrong way on a one-way street.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The circularity of the orbits is the result of a more dynamical process.  Hopefully someone else here can explain it better than the following.  My simplistic understanding is that a bunch of orbits crossing each other is not stable.  Planets and planetesimals have their orbits changed continuously until such time as they settle into orbits that have fewer interactions with other bodies.  Sort of a natural selection.  In the long haul, this results in a relatively stable configuration with circular zones that tend to not interfere with the other circular zones.  In fact, this is now part of the definition of the word &quot;planet&quot;, in that to be called such, a planet needs to clear the neighborhood of its orbit.  Though its not clear how it could occur, a bunch of elliptical zones that happen to be co-aligned would also not be stable since orbits precess due to various influences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the material was less dense than where the planets are currently must certainly have many objects in eccentric, retrograde, and/or highly inclined orbits.  Pluto is the first hint at that, with a relatively high inclination and eccentricity compared to the planets.  Sedna is far more eccentric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for other systems, I'm sure there must be some oddballs out there.  However the &lt;a href=&quot;http://arxiv.org/abs/1310.7942&quot; rel=&quot;noreferrer&quot;&gt;Kepler mission has observed mostly circular orbits for planets around other stars&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="512" LastActivityDate="2014-03-16T19:14:28.477" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1983" PostTypeId="2" ParentId="1981" CreationDate="2014-03-16T22:25:00.327" Score="6" Body="&lt;p&gt;Any object's mass warps the space around it towards the center of the object which causes other objects to 'fall' in towards that object or towards each other. Bigger objects have more mass therefore they warp space to higher degree causing more gravity. It doesn't matter what a substance is made of as long as it has mass. If it has mass it causes gravity. &lt;/p&gt;&#xA;" OwnerUserId="1206" LastActivityDate="2014-03-16T22:25:00.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1984" PostTypeId="1" CreationDate="2014-03-16T21:35:46.383" Score="3" ViewCount="193" Body="&lt;p&gt;I am hoping to find axial tilts for asteroids and also their spring and fall equinox. Some of the asteroids I'm interested in are:&#xA;4 Vesta, 1 Ceres, 24 Themis, 65 Cybele, 153 Hilda, 624 Hektor&lt;/p&gt;&#xA;" OwnerUserId="1232" OwnerDisplayName="HopDavid" LastEditorUserId="67" LastEditDate="2014-03-17T21:43:55.580" LastActivityDate="2014-03-17T21:43:55.580" Title="Is there online data on asteroid axial tilts?" Tags="&lt;orbit&gt;&lt;rotation&gt;&lt;asteroids&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1985" PostTypeId="2" ParentId="1984" CreationDate="2014-03-16T22:20:27.570" Score="1" Body="&lt;p&gt;From this article, &lt;a href=&quot;http://hubblesite.org/pubinfo/pdf/2010/33/pdf.pdf&quot; rel=&quot;nofollow&quot;&gt;http://hubblesite.org/pubinfo/pdf/2010/33/pdf.pdf&lt;/a&gt;, here's the rotation axis for 4 Vesta: RA=307.5°±3.1°, Dec=43.1°±1.2°. Update, per TildalWave's comment: &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0019103510003726&quot; rel=&quot;nofollow&quot;&gt;RA=305.8°±3.1°, Dec=41.4°±1.5°&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://yly-mac.gps.caltech.edu/A_DPS/dps%202011%20/a_dps%202011%20program%20+%20abstracts/pdf/EPSC-DPS2011-829.pdf&quot; rel=&quot;nofollow&quot;&gt;Rotational motion of Ceres&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An old article (1986): Magnusson, P., &lt;em&gt;Distribution of spin axes and senses of rotation for 20 large asteroids&lt;/em&gt;, Icarus 68:1 1-39 &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0019103586900722&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencedirect.com/science/article/pii/0019103586900722&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2618" OwnerDisplayName="David Hammen" LastEditorUserId="13" LastEditDate="2014-03-16T23:11:21.023" LastActivityDate="2014-03-16T23:11:21.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1987" PostTypeId="2" ParentId="1969" CreationDate="2014-03-17T03:51:29.193" Score="2" Body="&lt;p&gt;The key for stellar classification is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hertzsprung%E2%80%93Russell_diagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hertzsprung-Russell diagram&lt;/a&gt; (or &lt;em&gt;HRD&lt;/em&gt; for brevity), also know as the &lt;em&gt;color-magnitude diagram&lt;/em&gt;.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/DQ8fd.png&quot; alt=&quot;H-R diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows the relation between the surface temperature (on the x-axis) of a star and its luminosity (on the y-axis). The temperature is an index of the &lt;a href=&quot;http://www.uni.edu/morgans/astro/course/Notes/section2/spectraltemps.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Spectral type&lt;/a&gt;. For distant stars also spectroscopy is taken into account, to classify spectral types according to their absorption lines &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/starlog/staspe.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here some examples&lt;/a&gt;. The usual classification is called MKK (from the astronomers that had developed it - Morgan, Keenan and Kellman in 1943), and divides the stars in 7 stellar types O, B, A, F, G, K, M. Please note that from the 1943 other stellar types have added, but these are the most used. This sequence goes from the hottest (O) to coolest (M) stars. Each letter class is then subdivided using a numeric digit with 0 being hottest and 9 being coolest. Our Sun is a G2 type star. This classification also uses an additional notation to avoid the &quot;degeneracy&quot; you mentioned: a roman numeral between &quot;I&quot; (one) and &quot;V&quot; (five). The higher the number, the wider the lines in the observed spectrum. However, despite of the quantity these roman numbers indicate, they are an index of &lt;strong&gt;luminosity class&lt;/strong&gt;. This is based on the width of certain absorption lines in the star's spectrum which vary with the density of the atmosphere and so distinguish giant stars from dwarfs. Luminosity class I stars are supergiants, class III regular giants, and class V dwarfs or main-sequence stars, with II for bright giants, IV for sub-giants, and VI for sub-dwarfs (&lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_classification&quot; rel=&quot;nofollow noreferrer&quot;&gt;from here&lt;/a&gt;). Then, our Sun is &quot;completely&quot; identified by the type G2V. You can take a look at the figure below, and check out by yourself where it belongs:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/BBYuw.png&quot; alt=&quot;H-R diagram 2&quot;&gt;.&#xA;Also note that other indexes are used to characterize the stars, as for the width of the lines in the luminosity class, or for some &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_classification#Spectral_peculiarities&quot; rel=&quot;nofollow noreferrer&quot;&gt;peculiarities&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if you are interested in the study of stars, you should always keep in mind the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law#Temperature_of_stars&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stefan-Boltzmann law&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L=4\pi\sigma R^2 T^4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By this you can identify isoradii lines in the HR diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iN7tT.png&quot; alt=&quot;The red cross is our Sun&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there is a proportionality between luminosity and mass:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L\sim M^{3.5}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can identify &quot;isomasses&quot; lines in the HR diagram, but I can't find such a figure to show.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastEditorUserId="1038" LastEditDate="2014-07-10T03:38:38.923" LastActivityDate="2014-07-10T03:38:38.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1989" PostTypeId="2" ParentId="1980" CreationDate="2014-03-17T17:08:12.153" Score="4" Body="&lt;p&gt;Any gas or particles you see in the video are not matter of outer space, but part of the propulsion system or the spacecraft. The high velocity of the gas is a consequence of the propulsion. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Rocket_engine&quot; rel=&quot;nofollow&quot;&gt;exhausted gas&lt;/a&gt; (water vapor, aluminium, carbon dioxide, soot, many other compounds, depending on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rocket_fuel&quot; rel=&quot;nofollow&quot;&gt;propellant&lt;/a&gt;) continues to move roughly with constant velocity relative to the rocket as soon as it has left the thrusters (&lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion&quot; rel=&quot;nofollow&quot;&gt;Newton's first law&lt;/a&gt;); the rocket accelerates as long as the thrusters are active (Newton's third law), but that's not noticible in the video relative to the exhaust fumes. While the rocket is moving through Earth's atmosphere there may be some interaction of the exhaust fumes with Earth's atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no interaction of gas or fragments with matter of outer space visible  in the video.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although &lt;a href=&quot;http://en.wikipedia.org/wiki/Interplanetary_medium&quot; rel=&quot;nofollow&quot;&gt;outer space&lt;/a&gt; isn't quite empty. It contains e.g.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_dust&quot; rel=&quot;nofollow&quot;&gt;interplanetary and interstellar dust&lt;/a&gt;, mainly consisting of silicates, but also of other compounds like oxides, water ice, traces of organic compounds,&lt;/li&gt;&#xA;&lt;li&gt;hot plasma, mainly consisting of protons and electrons emitted by the sun (&lt;a href=&quot;http://en.wikipedia.org/wiki/Heliosphere&quot; rel=&quot;nofollow&quot;&gt;solar wind&lt;/a&gt;),&lt;/li&gt;&#xA;&lt;li&gt;galactic &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_ray&quot; rel=&quot;nofollow&quot;&gt;cosmic rays&lt;/a&gt;, mainly consisting of fast-moving atomic nuclei,&lt;/li&gt;&#xA;&lt;li&gt;electromagnetic radiation across the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_spectrum&quot; rel=&quot;nofollow&quot;&gt;electromagnetic spectrum&lt;/a&gt;, from &lt;a href=&quot;http://en.wikipedia.org/wiki/Radio_galaxy&quot; rel=&quot;nofollow&quot;&gt;radio waves&lt;/a&gt; to &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma-ray_burst&quot; rel=&quot;nofollow&quot;&gt;gamma rays&lt;/a&gt;,&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetosphere&quot; rel=&quot;nofollow&quot;&gt;magnetic fields&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-03-17T17:08:12.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1990" PostTypeId="2" ParentId="1981" CreationDate="2014-03-17T18:41:55.813" Score="5" Body="&lt;p&gt;Since you're wanting to know why mass causes gravity (from your question marked as duplicate &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1988/how-does-an-astronomical-objects-mass-make-it-attract-the-things-around-it&quot;&gt;https://astronomy.stackexchange.com/questions/1988/how-does-an-astronomical-objects-mass-make-it-attract-the-things-around-it&lt;/a&gt;) I'll give a little better of an explanation. Since the discovery of Higgs particle last year--which I'm sure you've heard about--if not here's a link &lt;a href=&quot;http://en.wikipedia.org/wiki/Higgs_boson&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Higgs_boson&lt;/a&gt;, we've essentially confirmed that mass is caused by a fundamental particle called the &lt;em&gt;Higgs boson&lt;/em&gt;. I think this article may actually answer your question &lt;a href=&quot;http://physics.aps.org/articles/v6/111&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://physics.aps.org/articles/v6/111&lt;/a&gt;. I'm a mathematician not a physicist so I'm not an expert in this area but hopefully the other article can help answer your question better. &lt;/p&gt;&#xA;" OwnerUserId="1206" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-03-14T17:32:30.027" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1991" PostTypeId="1" CreationDate="2014-03-17T18:44:47.303" Score="5" ViewCount="2201" Body="&lt;p&gt;Why exactly does time get slow around a black hole? And does it provide a way to go into the future?&lt;/p&gt;&#xA;" OwnerUserId="1091" LastEditorUserId="67" LastEditDate="2014-03-17T20:12:14.690" LastActivityDate="2014-03-25T22:25:46.720" Title="Why does time get slow near a black hole?" Tags="&lt;black-hole&gt;&lt;space-time&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2014-03-28T13:41:36.920" ContentLicense="CC BY-SA 3.0" />
  <row Id="1992" PostTypeId="1" CreationDate="2014-03-17T18:56:21.290" Score="1" ViewCount="89" Body="&lt;p&gt;I have some image data that may be interesting to someone, somewhere, but I don't know who to contact or notify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would imagine it would depend on the type of data, but maybe there is a central place people can contact?&lt;/p&gt;&#xA;" OwnerUserId="539" LastActivityDate="2014-03-17T20:00:19.207" Title="If I have amateur astrophotography image data that might be of scientific value, what should I do?" Tags="&lt;photography&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1993" PostTypeId="2" ParentId="1992" CreationDate="2014-03-17T20:00:19.207" Score="3" Body="&lt;p&gt;It depends, as you say, on the subject matter. For instance, images useful in variable star study can be provided to the AAVSO &lt;a href=&quot;http://www.aavso.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.aavso.org/&lt;/a&gt; and images relating to asteroids can be provided to the IAU Minor Planet Center. &lt;a href=&quot;http://www.minorplanetcenter.net/&quot; rel=&quot;nofollow&quot;&gt;http://www.minorplanetcenter.net/&lt;/a&gt; Other subject matter will be accepted elsewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for the images to be useful in scientific fields, you will need to comply with the submission instructions, which a casual snap probably will not comply with. On the MPC guide for beginners there is a list of 44 technical suggestions for submitting scientific data: &lt;a href=&quot;http://www.minorplanetcenter.net/iau/info/Astrometry.html&quot; rel=&quot;nofollow&quot;&gt;http://www.minorplanetcenter.net/iau/info/Astrometry.html&lt;/a&gt; The AAVSO has a tutorial in six chapters on photometry using a DSLR camera (and another for astro CCD cameras) &lt;a href=&quot;http://www.citizensky.org/content/dslr-documentation-and-reduction&quot; rel=&quot;nofollow&quot;&gt;http://www.citizensky.org/content/dslr-documentation-and-reduction&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just want to get it out there and have people comment on what is in it, post it to Google+.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-03-17T20:00:19.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1994" PostTypeId="1" CreationDate="2014-03-17T21:21:16.140" Score="7" ViewCount="105" Body="&lt;p&gt;Why can primordial tensor perturbations of the CMB be ascribed to gravitational waves? Is this attribution unique, or are there other mechanisms that could lead to the excitation of tensor modes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an explanation I have read recently that gravitational waves turn the E-modes into the now observed B-modes somehow, but I don't understand this. So can somebody give a more detailed explanation of this interaction of gravitational waves with the CMB?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LaTeX and equations are welcome and appreciated :-)&lt;/p&gt;&#xA;" OwnerUserId="92" LastEditorUserId="7472" LastEditDate="2017-07-15T20:42:27.090" LastActivityDate="2017-07-15T20:42:27.090" Title="Why can primordial tensor perturbations of the CMB be ascribed to gravitational waves?" Tags="&lt;cosmology&gt;&lt;cmb&gt;&lt;gravitational-waves&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1995" PostTypeId="2" ParentId="1991" CreationDate="2014-03-17T22:04:02.080" Score="3" Body="&lt;p&gt;The simplified answer is: Due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_time_dilation&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravitational time dilation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a consequence of &lt;a href=&quot;http://en.wikipedia.org/wiki/General_relativity&quot; rel=&quot;nofollow noreferrer&quot;&gt;general relativity&lt;/a&gt;. One fundamental idea of general relativity is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Equivalence_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;equivalence of inertial and gravitational mass&lt;/a&gt;, meaning forces induced by acceleration aren't distinguishable from gravity.&#xA;Persuing this principle together with the constant speed of light lead to &lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein_field_equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein's field equations&lt;/a&gt;. One of the solutions of these equations is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_metric&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schwarzschild solution&lt;/a&gt; for non-rotating, electrically neutral black holes.&#xA;This solution can be used to derive gravitational time dilation, as part of &lt;a href=&quot;https://physics.stackexchange.com/questions/3009/how-exactly-does-curved-space-time-describe-the-force-of-gravity&quot;&gt;spacetime warp&lt;/a&gt; near mass/energy.&#xA;(&lt;a href=&quot;http://www.youtube.com/watch?v=foRPKAKZWx8&quot; rel=&quot;nofollow noreferrer&quot;&gt;Introductory video to general relativity&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A black hole provides a way to go faster into the future of a region of higher gravitational potential away from the black hole.&#xA;But the body near the black hole needs to withstand the high gravity. Circular orbits around a black hole are only possible down to 3/2 times, stable only down to 3 times, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_geodesics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schwarzschild radius&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2014-03-17T22:04:02.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1996" PostTypeId="2" ParentId="1361" CreationDate="2014-03-17T23:28:39.537" Score="1" Body="&lt;p&gt;Don't worry about the main telescope. There doesn't have to be a relationship between the focal length and f ratio of the telescope and the guidescope. I think this is proven by the use of off-axis-guiders, which use the same focal length as the telescope. They just come with the headache of finding guide stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is important is: can the guidescope provide data for the controlling software to accurately automate the telescope?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want a wide, fast field of view, so that you are guaranteed to be able to find a bright star to guide by.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want a camera on the guidescope that has sufficiently small pixels that the guide star that gets picked will at least cover a few pixels - the more the better. To some extent, this conflicts with the wish for a wide, fast view, as that will result in smaller stars on the sensor. This can be mitigated by ever so slightly de-focusing the guidescope so the stars form fuzzier, larger circles on the sensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is more important that the focal length of your telescope will be: (a) what the limit of your seeing tends to be, and (b) the accuracy with which your mount can be controlled, and (c) how accurately the guiding software can place the guide star (usually a fraction of a pixel - use maybe 0.2 pixels as a worst case, if unknown).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is not much point in being more accurate than the seeing permits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula for figuring out how much each pixel sees is:&#xA;$$&#xA;px_{fov}=206*px_{microns} / f_l&#xA;$$&#xA;where f_l is the focal length of a telescope and px_microns is how large each pixel is on the sensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can therefore estimate a suitable focal length guide scope this way:&#xA;$$&#xA;f_l = px_{microns} / (res_{seeing}/G_{acc}) * 206&#xA;$$&#xA;where px_microns is the size of the sensor pixels,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;res_seeing is how accurate you need the guiding to be,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G_acc is the guider software's centroid accuracy (use 0.2 as a worst case if unknown)&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-03-17T23:28:39.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1997" PostTypeId="1" AcceptedAnswerId="2012" CreationDate="2014-03-18T01:57:15.347" Score="12" ViewCount="4428" Body="&lt;p&gt;In a &lt;a href=&quot;http://arxiv.org/abs/1403.0576&quot; rel=&quot;noreferrer&quot;&gt;recent paper&lt;/a&gt; (&lt;a href=&quot;http://www.nature.com/news/did-dark-matter-kill-the-dinosaurs-1.14839&quot; rel=&quot;noreferrer&quot;&gt;news release here&lt;/a&gt;) Lisa Randall and Matthew Reece propose that a dark matter disk coinciding with the galactic plane together with the solar system's oscillations through the galactic plane could explain the 35 million year periodicity in mass extinctions.  They propose that the solar system passes through this dark matter disk in the galactic plane every 35 million years, disrupting bodies in the Oort cloud and causing some of them to collide with Earth.  An image of the proposed cycles from the news release is below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: What mechanism is responsible for this 35 million year oscillation about the galactic plane?  Is there a companion body to the sun? Are we orbiting around one of the arms of our galaxy?  Is this a well known phenomenon or are they proposing the 35 million year oscillation as well as the dark matter disk?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DfAo0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="821" LastEditorUserId="821" LastEditDate="2014-03-18T12:55:57.773" LastActivityDate="2019-10-18T07:07:08.113" Title="What mechanism causes oscillations of the solar system's orbit about the galactic plane?" Tags="&lt;orbit&gt;&lt;earth&gt;&lt;dark-matter&gt;&lt;galactic-dynamics&gt;&lt;near-earth-object&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1998" PostTypeId="1" AcceptedAnswerId="2000" CreationDate="2014-03-18T05:29:03.573" Score="5" ViewCount="3776" Body="&lt;p&gt;When we move our vehicle, we &quot;feel&quot; the speed of it. However, when stationary, we still are moving,but moving by quiet a speed. We know that earth moves with &lt;code&gt;X&lt;/code&gt; speed, our own solar system moves at &lt;code&gt;Y&lt;/code&gt; speed. etc.&lt;br&gt;&#xA;Why do we not feel this speed? Even sitting at my home, my mind must be gyrating at an incredible speed.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="-1" LastEditDate="2014-04-08T12:51:18.717" LastActivityDate="2018-10-15T03:23:48.803" Title="Why don't we feel the speed of earth moving" Tags="&lt;earth&gt;&lt;gravity&gt;&lt;speed&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1999" PostTypeId="2" ParentId="1998" CreationDate="2014-03-18T07:52:53.390" Score="3" Body="&lt;p&gt;When you sit in a car, and you suddenly brake, you feel the change in velocity.&#xA;On earth, the change in velocity of the earth is too small for you to be noticed.&lt;/p&gt;&#xA;" OwnerUserId="1175" LastActivityDate="2014-03-18T07:52:53.390" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2000" PostTypeId="2" ParentId="1998" CreationDate="2014-03-18T10:05:00.087" Score="13" Body="&lt;p&gt;In a car, you have a perception of speed because of (a) the &quot;wind&quot; passing by as you rush through the air which is not moving at the same speed as the vehicle, and (b) you perceive the stationary objects nearby as &quot;moving&quot; off into the distance behind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the earth moves in its orbit, you don't notice any &quot;wind&quot; from the planet rushing through space, as the atmosphere of the earth is moving along with the planet. There isn't any substance to the space around the planet to create an impression of &quot;wind&quot;. The car is surrounded by air, so when you move through the air, you have an impression of moving because of the air rushing past. Because the planet isn't moving through an especially gaseous medium, there is no impression of movement, no &quot;wind&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also when it comes to our planet, there is nothing &quot;nearby&quot; to create a reference point that would make it appear as if we are moving relative to that reference. Even the closest of objects, such as the moon, are sufficiently far off that the only obvious impression is that these move and we are still. Hence the reason why the geocentric view of our world was dominant for so long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even sitting still at home, your entire body is moving at the same velocity as the rest of the planet, so your brain isn't being jostled or 'gyrated' at all. If someone shook you, then your velocity would be changing and your body would feel the effects of acceleration, and &lt;strong&gt;then&lt;/strong&gt; you would be jostled and gyrated. But that is because you are perceiving sudden and rapid changes in velocity, accelerations. &lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-03-18T10:05:00.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2001" PostTypeId="1" AcceptedAnswerId="2004" CreationDate="2014-03-18T11:29:25.070" Score="2" ViewCount="316" Body="&lt;p&gt;I want to know whether the temperature in Outer space is higher than earth's or cold.Is the temperature in outer space vary as it does on earth?&lt;/p&gt;&#xA;" OwnerUserId="1103" LastActivityDate="2014-03-19T10:13:55.240" Title="What is the temperature of outerspace?" Tags="&lt;space&gt;&lt;temperature&gt;&lt;weather&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2003" PostTypeId="2" ParentId="2001" CreationDate="2014-03-18T13:48:25.043" Score="0" Body="&lt;p&gt;Than answer to whether it varies or is definitely yes but not in the same way as on Earth. The temperature in completely empty space with no stars for large distances would approach absolute zero ( −459.67°F or −273.15°C depending on the system to which you are accustomed) the further you go away from any heat producing body such as a star or active black hole. If you get close to stars or active black holes or many other things, the temperature could be thousands of degrees higher than on Earth. It just depends on distance. &lt;/p&gt;&#xA;" OwnerUserId="1206" LastActivityDate="2014-03-18T13:48:25.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2004" PostTypeId="2" ParentId="2001" CreationDate="2014-03-18T13:48:39.380" Score="4" Body="&lt;p&gt;The temperature in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heliosphere&quot; rel=&quot;nofollow&quot;&gt;heliosphere&lt;/a&gt; is about one million Kelvin, lower (but still very hot) in the outer regions of the heliosphere, higher &lt;a href=&quot;http://en.wikipedia.org/wiki/Heliosphere&quot; rel=&quot;nofollow&quot;&gt;close to the sun&lt;/a&gt;.&#xA;But the density of the &lt;a href=&quot;http://de.wikipedia.org/wiki/Plasma_%28Physik%29&quot; rel=&quot;nofollow&quot;&gt;plasma&lt;/a&gt; is very low, so you wouldn't feel the heat (except close to the sun).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Within &lt;a href=&quot;http://airs.jpl.nasa.gov/maps/satellite_feed/atmosphere_layers/&quot; rel=&quot;nofollow&quot;&gt;Earth's exosphere&lt;/a&gt;, reaching up to 10,000 km, the temperature is lower, roughly 1,000 to 1,500 C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://en.wikipedia.org/wiki/Interstellar_medium&quot; rel=&quot;nofollow&quot;&gt;interstellar space&lt;/a&gt; there exist &lt;a href=&quot;http://en.wikipedia.org/wiki/H_II_region&quot; rel=&quot;nofollow&quot;&gt;hot&lt;/a&gt; (e.g. 10,000 K) and &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_cloud&quot; rel=&quot;nofollow&quot;&gt;cold gas clouds&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the very low density, hot and cooler fractions can penetrate each other, such that there doesn't need to be a uniquely defined temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The large-scale average temperature of &lt;a href=&quot;http://en.wikipedia.org/wiki/Outer_space&quot; rel=&quot;nofollow&quot;&gt;outer space&lt;/a&gt; should be 2.7 K (about -270.5 C), close to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow&quot;&gt;CMB&lt;/a&gt; black body temperature.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-03-19T10:13:55.240" LastActivityDate="2014-03-19T10:13:55.240" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2005" PostTypeId="1" AcceptedAnswerId="2039" CreationDate="2014-03-18T18:55:58.697" Score="8" ViewCount="2342" Body="&lt;p&gt;For instance, If a large globular cluster were somehow ejected from its galaxy, and was observed by astronomers, would they call it a galaxy, or would there be characteristics of the star cluster that make it obviously not a galaxy by definition? And why are the Magellanic Clouds dwarf galaxies, and not large star clusters? Could we remove stars until the Magellanic clouds became star clusters, or is there something intrinsically galaxy like about these objects?  &lt;/p&gt;&#xA;" OwnerUserId="1244" LastActivityDate="2014-03-25T10:04:13.733" Title="Is there a fundamental difference between a small galaxy, and a large star cluster?" Tags="&lt;galaxy&gt;&lt;star-cluster&gt;&lt;globular-clusters&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2006" PostTypeId="1" AcceptedAnswerId="2010" CreationDate="2014-03-18T20:18:28.900" Score="2" ViewCount="354" Body="&lt;p&gt;From a mathematical point of view,  what is the probability that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) There is organic life on other planets &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) There are advanced species like us or even greater than us on other planets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a formula or equation or a way we can calculate this probability?&lt;/p&gt;&#xA;" OwnerUserId="1246" LastEditorUserId="11264" LastEditDate="2017-11-06T20:56:15.337" LastActivityDate="2017-11-06T20:56:15.337" Title="What is the probability that there is life on other planets?" Tags="&lt;galaxy&gt;&lt;life&gt;&lt;mathematics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2007" PostTypeId="2" ParentId="2005" CreationDate="2014-03-18T20:33:07.957" Score="1" Body="&lt;p&gt;An overly simplistic distinction might be that globular clusters do not have any cloudy/gaseous regions, with no current star forming activity occurring. Globulars tend to have very similar stars, which together with their lack of gas regions, indicates that they formed early on in a single star forming period - though actually some globulars show evidence of additional period(s) of star forming. A galaxy will tend to have ongoing star forming activity and have evidence of stars of various ages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that is over-simplifying; there isn't always that clear-cut a difference between globulars and galaxies. &lt;a href=&quot;http://arxiv.org/pdf/0711.4795v1.pdf&quot; rel=&quot;nofollow&quot;&gt;At least one recent paper&lt;/a&gt; suggests that globular clusters and dwarf spheroidal galaxies may not be clearly distinct groups. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the Magellanic Clouds do make themselves quite distinct from globular clusters in being rich in dust and gas. There is a lot of current star-formation in progress, and there is a lot of variety in the stars in these galaxies.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-03-18T20:33:07.957" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2009" PostTypeId="1" CreationDate="2014-03-19T01:51:05.340" Score="1" ViewCount="357" Body="&lt;p&gt;I was poking around &lt;a href=&quot;http://lroc.sese.asu.edu/gigapan/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Gigapan image of the moon's north polar region&lt;/a&gt;, when I noticed this area:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7E1ba.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the source of these light streaks?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For position reference, here's a zoomed-out shot (I couldn't easily get coordinates from the Gigapan thing):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6rZtb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1249" LastEditorUserId="1249" LastEditDate="2014-03-20T13:53:44.947" LastActivityDate="2014-03-23T12:02:59.920" Title="What are these faint streaks on the moon?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2010" PostTypeId="2" ParentId="2006" CreationDate="2014-03-19T12:09:16.570" Score="7" Body="&lt;p&gt;The probabilites are unknown at the moment (March 2014), since there is only one known planet (Earth) harboring life. This doesn't allow any meaningful probability estimates for the occurence of life, based on empirical data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall formation of life is too complex to allow simulations based on current technology.&#xA;Although some intermediate steps can be &lt;a href=&quot;http://en.wikipedia.org/wiki/Evolutionary_algorithm&quot;&gt;simulated&lt;/a&gt; or performed by experiment, e.g. by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Miller%E2%80%93Urey_experiment&quot;&gt;Miller-Urey experiment&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Drake_equation&quot;&gt;Drake equation&lt;/a&gt; is a simplified attempt to decompose the probability into factors. Some of the factors aren't known yet. But there has been much progress in estimating the &lt;a href=&quot;http://www.nasa.gov/ames/kepler/nasas-kepler-mission-announces-a-planet-bonanza/#.Uyl8TKh5Oqi&quot;&gt;frequency of exoplanets&lt;/a&gt;, and the &lt;a href=&quot;http://arxiv.org/ftp/arxiv/papers/1212/1212.0113.pdf&quot;&gt;probability of planets to be habitable&lt;/a&gt;. From &lt;a href=&quot;http://planetquest.jpl.nasa.gov/news/139&quot;&gt;this news release&lt;/a&gt; one could estimate (order of magnitude), that there is one habitable planet around a Sun-like star per about 1,700 (about 12³) cubic light years in our region of the Milky Way (&quot;... the nearest sun-like star with an Earth-size planet in its habitable zone is probably only 12 light years away...&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't know the probability for habitable planets to develop life, nor to be &lt;a href=&quot;http://en.wikipedia.org/wiki/Panspermia&quot;&gt;colonized&lt;/a&gt;. Probabilities of complex life forms are even more difficult to estimate (yet). &lt;a href=&quot;http://en.wikipedia.org/wiki/Planetary_habitability&quot;&gt;More discussion about habitability of exoplanets on Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-03-19T12:09:16.570" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2011" PostTypeId="2" ParentId="498" CreationDate="2014-03-19T13:49:33.177" Score="7" Body="&lt;p&gt;Visual resolution of a telescope is directly proportional to the aperture of the telescope. The focal length, and hence the magnification that can be achieved, is then just following on the visual resolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The telescopes today are usually so well build that they are &lt;a href=&quot;http://en.wikipedia.org/wiki/Diffraction-limited_system&quot;&gt;diffraction limited&lt;/a&gt;, which means optical resolution due to diffraction is the limiting factor. If you want to have &quot;higher magnification&quot; in a telescope, you &lt;em&gt;always&lt;/em&gt; want to have a larger aperture. The longer focal length may help, but is not quite necessary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, as Jeremy said, the limiting resource in this is money. There are some engineering problems with building extremely large telescopes, but most of these can be solved, given enough money, time and resources.&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2014-03-19T13:49:33.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2012" PostTypeId="2" ParentId="1997" CreationDate="2014-03-19T14:24:46.813" Score="7" Body="&lt;p&gt;The cause for the oscillations perpendicular to the galactic plane is the gravity of the non-spherical mass distribution (needed for a plane &lt;a href=&quot;http://en.wikipedia.org/wiki/Kepler_orbit&quot;&gt;Kepler ellipse&lt;/a&gt;) in the Milky Way. Simplified, there is a dense galactic plane. The density is not exactly known; therefore there is some uncertainty (a few million years) about the precise oscillation period. Details see &lt;a href=&quot;http://www.phys.huji.ac.il/~shaviv/articles/ShavivChapter.pdf&quot;&gt;this article&lt;/a&gt;, subsection 3.3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea of a &lt;a href=&quot;http://www.nature.com/nature/journal/v308/n5961/abs/308712a0.html&quot;&gt;correlation of mass extinctions with this oscillation&lt;/a&gt; isn't new, it originates back probably to about 1970, or earlier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The outer solar system probably does not contain a large gas giant planet, or a small, companion star&quot;, see &lt;a href=&quot;http://www.nasa.gov/press/2014/march/nasas-wise-survey-finds-thousands-of-new-stars-but-no-planet-x/#.UymjCqh5Oqg&quot;&gt;this press release&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are probably not orbiting around a spiral arm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dark matter disk is a hypotheses, an idea to investigate. Usually only a small fraction of this kind of hypotheses can be confirmed definitively later, most of them can be ruled out after some time, some remain unsolved, some can be refined to match observations.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-03-19T14:24:46.813" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2013" PostTypeId="1" AcceptedAnswerId="2019" CreationDate="2014-03-19T16:42:19.017" Score="0" ViewCount="95" Body="&lt;p&gt;If there were an asteroid orbiting its star within the habitable zone and this asteroid contained liquid water, would it be possible for simple life such as bacteria to form? &lt;/p&gt;&#xA;" OwnerUserId="1206" LastActivityDate="2014-03-20T18:09:06.257" Title="Simple life on asteroids" Tags="&lt;life&gt;&lt;habitable-zone&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2014" PostTypeId="1" CreationDate="2014-03-19T17:37:41.890" Score="4" ViewCount="234" Body="&lt;p&gt;I've read that the further away a Galaxy is from us the faster it moves away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By this logic how can scientists predict that there will eventually be a big crunch when every piece of matter is seemingly getting further and further apart?&lt;/p&gt;&#xA;" OwnerUserId="1156" LastEditorDisplayName="user96" LastEditDate="2014-03-21T13:14:50.043" LastActivityDate="2014-03-21T13:14:50.043" Title="How can we predict a Big Crunch when all galaxies are moving further apart?" Tags="&lt;galaxy&gt;&lt;universe&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2015" PostTypeId="2" ParentId="2014" CreationDate="2014-03-19T19:11:12.253" Score="1" Body="&lt;p&gt;Well, the original idea of the Big Crunch came about when it was thought that gravity was slowing down the expansion of the universe and that one day it will stop expanding (a finite universe). At which point the gravitational pull of all the objects in the universe would make it contract, growing smaller into itself until eventually it is just a singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, today we see that the universe is expanding and accelerating due to dark energy's effect on gravity making it an infinite universe. This makes the original idea of the Big Crunch not possible (since the universe would be infinite), but a big crunch can still occur in an alternate way if dark energy becomes too weak to counteract gravity which would set in motion the Big Crunch. However there is still so much unknown about dark energy that it is also possible that it will never weaken and the universe will just keep expanding, which opens different end of the universe theories.  &lt;/p&gt;&#xA;" OwnerUserId="1253" LastEditorUserId="1178" LastEditDate="2014-03-20T16:47:51.207" LastActivityDate="2014-03-20T16:47:51.207" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2017" PostTypeId="1" CreationDate="2014-03-20T09:01:49.947" Score="2" ViewCount="108" Body="&lt;p&gt;Academic papers often make claims about observational data and results from computations. The papers only refer to names of data sources (such as an observatory) and of mathematical methods without any specifics, other than some of it illustrated in a few graphs and tables. References lead almost only to other papers of similar format.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to replicate (or even peer review) a paper, obviously one must have access to specifics such as the actual calculations, the raw data, the software, the blue prints of the measurement equipment and logs of how it has been operated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What data do peer reviewers and replicating scientists (and the public) have access to beyond a paper making some claims?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two examples: The results of the infamous cold fusion press conference of 1990 could never be satisfactorly replicated. But were the attempts based only on the paper they then published, or did they publish on the side also a wealth of data about their procedures? And the more recent &quot;neutrinos travel faster than light&quot; media debacle was, as far as I know, all about the details of equipment, was all relevant information about it made public (for the science community)?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="6" LastEditDate="2016-05-25T15:40:41.847" LastActivityDate="2016-05-25T15:40:41.847" Title="What must/do astronomers reveal beyond their academic papers?" Tags="&lt;data-analysis&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2018" PostTypeId="2" ParentId="1973" CreationDate="2014-03-20T11:07:20.990" Score="5" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tired_light&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Tired light&lt;/em&gt;&lt;/a&gt; has been used as an explanation before, whereby light loses energy whilst travelling through space---a sort of drag effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think anyone actually supports it nowadays, though.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="537" LastEditDate="2014-03-20T13:07:24.247" LastActivityDate="2014-03-20T13:07:24.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2019" PostTypeId="2" ParentId="2013" CreationDate="2014-03-20T16:16:43.100" Score="3" Body="&lt;p&gt;If you take a slowly rotating asteroid with a large basin of trapped water, some solved minerals, and may be some gases, close to the asteroid's surface, to get an environment similar to a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrothermal_vent#Black_smokers_and_white_smokers&quot; rel=&quot;nofollow&quot;&gt;black smoker&lt;/a&gt; by the heat of the sun, I wouldn't call it totally impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formation of such an asteroid is close to the limits of my imagination, might be as a remnant of a comet with a crust of dust baked together to a water-proof surface layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that's pure speculation at the moment. May be we can learn more this year with &lt;a href=&quot;http://en.wikipedia.org/wiki/Rosetta_%28spacecraft%29&quot; rel=&quot;nofollow&quot;&gt;Rosetta&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some biochemistry details: By the rotating asteroid part of the water undergoes a heating/cooling cycle. Provided there are enough prebiotic molecules, by this heating/cooling cycle kind of &lt;a href=&quot;http://en.wikipedia.org/wiki/Polymerase_chain_reaction&quot; rel=&quot;nofollow&quot;&gt;polymerase chain reaction&lt;/a&gt; (PCR) could multiply short occasional RNA or DNA fragments (&lt;a href=&quot;http://en.wikipedia.org/wiki/Nucleotide&quot; rel=&quot;nofollow&quot;&gt;nucleotide&lt;/a&gt; sequences). The difficult part is providing &lt;a href=&quot;http://en.wikipedia.org/wiki/Taq_polymerase&quot; rel=&quot;nofollow&quot;&gt;polymerase&lt;/a&gt;. It has originally been isolated from a &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermus_aquaticus&quot; rel=&quot;nofollow&quot;&gt;bacterium&lt;/a&gt; living in hot springs, not quite dissimilar to black smokers. Information storage and replication (e.g. in RNA/DNA) is one important ingredient for life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/DNA_polymerase_III&quot; rel=&quot;nofollow&quot;&gt;Polymerase&lt;/a&gt; is an enzyme, mainly consisting of amino acids. &lt;a href=&quot;http://science.gsfc.nasa.gov/691/analytical/PDF/Elsila2009.pdf&quot; rel=&quot;nofollow&quot;&gt;Glycine is the first amino acid found in cometary dust in 2009&lt;/a&gt;. That's a first step, but there is a big gap. As an enzyme polymerase is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Protein&quot; rel=&quot;nofollow&quot;&gt;protein&lt;/a&gt;, hence can be encoded by RNA or DNA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To decode RNA to a proteine, kind of &lt;a href=&quot;http://en.wikipedia.org/wiki/Ribosome&quot; rel=&quot;nofollow&quot;&gt;ribosome&lt;/a&gt; is needed. A ribosome itself consists of &lt;a href=&quot;http://en.wikipedia.org/wiki/Ribonucleoprotein&quot; rel=&quot;nofollow&quot;&gt;RNA and protein&lt;/a&gt; (chain of amino acids). That's the basics of the replication mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need a containment.&#xA;The simplest form of a cell-like structure is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Micelle&quot; rel=&quot;nofollow&quot;&gt;micelle&lt;/a&gt;. Organic molecules with a polar group can form micelles.&#xA;&lt;a href=&quot;http://www.gb.nrao.edu/astrobiology/gbtModel/astroBioARAA.38.1.427.pdf&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; contains a list of organic molecules found in interstellar medium.&#xA;Some of them could be used to synthesize appropriate molecules, but it's also quite a gap in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A long way from what we know of extraterrestrial organic molecules to a bacterium.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-03-20T18:09:06.257" LastActivityDate="2014-03-20T18:09:06.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2020" PostTypeId="2" ParentId="2017" CreationDate="2014-03-20T17:09:39.330" Score="5" Body="&lt;p&gt;Most catalogs have a public release somewhere on the web. Here is the best listing of astronomical catalogs that I know of:&#xA;&lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR&quot; rel=&quot;nofollow&quot;&gt;http://vizier.u-strasbg.fr/viz-bin/VizieR&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a NASA Extragalactic Database (NED) that can be helpful: &lt;a href=&quot;http://ned.ipac.caltech.edu/&quot; rel=&quot;nofollow&quot;&gt;http://ned.ipac.caltech.edu/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, astronomical observations are available to the public to be analyzed. Different telescopes/space missions maintain their own archives. Some give an observer a proprietary period (6-7 months is the time frame I have heard given for most observatories) to give the person who requested the observations time to analyze the data before it is released to the general public.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In publications the author(s) will list they catalogs the used to come to their conclusions so replicating scientist can use the same dataset. &lt;/p&gt;&#xA;" OwnerUserId="1118" LastEditorUserId="1118" LastEditDate="2014-03-23T15:36:58.143" LastActivityDate="2014-03-23T15:36:58.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2021" PostTypeId="2" ParentId="2014" CreationDate="2014-03-21T03:45:13.840" Score="3" Body="&lt;p&gt;In a homogeneous and isotropic Universe (even if recent observations challenge this hypothesis), you can derive the &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Friedmann equations&lt;/a&gt;, which describe the evolution of the Hubble constant with time:&#xA;$\frac{\dot{a}}{a} = H(t) = \frac{8 \pi G}{3}\rho - \frac{k}{a^2} + \frac{\Lambda}{3}$   (with $c=1$) (Equation $1$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a=a(t)$ is the scale factor, $\dot{a}$ its derivative, $G$ the gravitational constant, $\rho$ the matter density, $\frac{k}{a^2}$ the spatial curvature (a parameter that describes the metric of the Universe), and $\Lambda$ the cosmological constant (an integration constant added by Einstein).&#xA;It could be useful to rewrite the equation as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H^2 = \frac{8 \pi G}{3}(\rho + \rho_{\Lambda}) - \frac{k}{a^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\rho_{\Lambda} = \frac{\Lambda}{8 \pi G}$ is the &quot;density of cosmological constant&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also expand the matter density as $\rho = \rho_{matter} + \rho_{radiation}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have a &quot;total&quot; density $\rho_{tot} = \rho_{matter} + \rho_{radiation} + \rho_{\Lambda}$. The destiny of the Universe depends on this amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of $\rho_{tot} &amp;gt; \rho_{crit}$, or equivalently a &lt;em&gt;closed&lt;/em&gt; Universe ($k=+1$), the equation $(1)$ becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot a^2 = \frac{8 \pi G}{3}\rho a^2 -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which points out that the scale factor must have an upper limit $a_{max}$ ($\dot a^2$ must be positive).&#xA;This in turn means that the second derivative $\ddot a$ of the scale factor must be negative, when approaching $a_{max}$, that is &lt;em&gt;the scale factor function inverts its behavior&lt;/em&gt;:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/LI2GJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at &lt;a href=&quot;http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll8.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.ita.uni-heidelberg.de/~dullemond/lectures/cosmology_2011/Chapter_4.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; if you want to go deeper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Bardathehobo This figure shows what I mean when I say that a currently accelerating Universe can still crunch. This is because we are basically ignorant upon the dark energy issue.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastActivityDate="2014-03-21T03:45:13.840" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2025" PostTypeId="1" CreationDate="2014-03-21T18:00:13.967" Score="1" ViewCount="139" Body="&lt;p&gt;There are numerous theories about the shape of the Universe. I've read somewhere most probably in 'A Brief History of Time' that there is a possibility that the Universe is spherical. So is it possible that all galaxies are moving towards each other? Imagine the Universe as a sphere. Now, imagine a diameter AB. The Big Bang took place at A and the galaxies started drifting away and as they moved away further they stared attracting each other on the other side i.e.Gravitational attraction. So we feel that the farther galaxies are the faster they are moving away(Hubble's Law) but actually they are moving towards each other and will eventually meet at point B. Also, space-time is spherical and so light travels along the sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So, my question is: Is it possible for the universe to be spherical with galaxies apparently moving away but actually moving closer?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1156" LastActivityDate="2014-03-22T05:39:17.927" Title="Are we moving away or towards each other?" Tags="&lt;universe&gt;&lt;expansion&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2026" PostTypeId="2" ParentId="2025" CreationDate="2014-03-21T18:17:13.167" Score="2" Body="&lt;p&gt;At the present time this doesn't hold for the observable universe, since the observable universe is only a tiny fraction of the surface of the 3-sphere (assumed it's a sphere at all). &lt;a href=&quot;http://wmap.gsfc.nasa.gov/universe/uni_shape.html&quot; rel=&quot;nofollow&quot;&gt;The observable universe is flat within a 0.4% error&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on this, we are moving away from each other. And the universe is expanding. So even if it would be curved, we would move away from each other. Only in a future Big Crunch scenario, we would eventually move towards each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://terpconnect.umd.edu/~wmg/GoldmanKamishima.pdf&quot; rel=&quot;nofollow&quot;&gt;Finite flat cyclic spaces&lt;/a&gt; can be constructed, but there is &lt;a href=&quot;http://arxiv.org/pdf/1303.5086v2.pdf&quot; rel=&quot;nofollow&quot;&gt;no significant hint&lt;/a&gt;, that the universe is of such a shape, or any other non-trivial topology.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-03-21T18:52:40.207" LastActivityDate="2014-03-21T18:52:40.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2030" PostTypeId="1" AcceptedAnswerId="2031" CreationDate="2014-03-22T14:24:14.733" Score="2" ViewCount="615" Body="&lt;p&gt;whether its in movies,television,real-life or some planetarium different galaxies are shown in different color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For ex&lt;/strong&gt;:- Milky way is always seen in Blue where as Andromeda gives out a bit reddish color. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because different galaxies contain different kinds of stars?&#xA;Does this effect the life the galaxy harbors...   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;^_^&lt;/p&gt;&#xA;" OwnerUserId="1274" LastActivityDate="2016-09-15T15:02:33.353" Title="why are galaxies different in color?" Tags="&lt;star&gt;&lt;galaxy&gt;&lt;universe&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2031" PostTypeId="2" ParentId="2030" CreationDate="2014-03-22T14:49:48.187" Score="2" Body="&lt;p&gt;See &lt;a href=&quot;https://web.archive.org/web/20150117214502/http://curious.astro.cornell.edu/question.php?number=566&quot; rel=&quot;nofollow&quot;&gt;What do a galaxy's colors mean? Are they its true colors?&lt;/a&gt; This article should help answer your question. There's really nothing I can add to the article. It lists all the reasons I could think of. &lt;/p&gt;&#xA;" OwnerUserId="1206" LastEditorUserId="-1" LastEditDate="2016-09-15T15:02:33.353" LastActivityDate="2016-09-15T15:02:33.353" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2032" PostTypeId="1" AcceptedAnswerId="2033" CreationDate="2014-03-23T04:33:26.010" Score="2" ViewCount="182" Body="&lt;p&gt;When we find an exoplanet by observing the dimming of its parent star, this means that the planet passes in front of the star from our view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if our view of a star system is from its &quot;top&quot; or &quot;bottom&quot;? Then from our view, none of its planets would pass in front of the star so we could observe a dimming. How would we know, then, whether such a star has or doesn't have planets?&lt;/p&gt;&#xA;" OwnerUserId="1166" LastActivityDate="2014-12-13T21:57:12.610" Title="How would an exoplanet be found from earth if our view of its star system is &quot;from the top&quot;?" Tags="&lt;exoplanet&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2033" PostTypeId="2" ParentId="2032" CreationDate="2014-03-23T07:20:38.950" Score="9" Body="&lt;p&gt;First of all, it is not that you can always find the planet (if it is there).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, if this is the case, when we observe the system &lt;em&gt;not totally face-on&lt;/em&gt;, we resort to &lt;a href=&quot;http://en.wikipedia.org/wiki/Radial_velocity&quot; rel=&quot;noreferrer&quot;&gt;radial velocity&lt;/a&gt; measurements. The principle is simple, the actual much less.&#xA;The orbiting of the planet around the star causes the star to &quot;oscillate&quot; around the center of mass.&#xA;Here you can see a good example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/S4Rsm.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when the system is &lt;em&gt;exactly face-on&lt;/em&gt;, the Doppler shift is useless, but still the gravitational effect of the planet onto the star is useful and used.&#xA;Here is what happens in the latter case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JGwsd.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that the star undergoes a change in position.&#xA;Then usually, the radial measurements go along with astrometrical measurements, to identify the position of the star along the planet's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you need some conditions are verified to allow this measurement.&#xA;First of all, you planet must be massive enough to bring a sensible change in the position of the star. Then your instrument must be sensible enough to resolve the tiny effect originated by this change.&#xA;Possibly, also other conditions play a role, but I am not an expert (you can take a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Doppler_spectroscopy#Problems&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a general comprehension, also a wiki for &lt;a href=&quot;http://en.wikipedia.org/wiki/Methods_of_detecting_exoplanets&quot; rel=&quot;noreferrer&quot;&gt;Methods of detecting exoplanets&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastEditorUserId="1038" LastEditDate="2014-03-23T13:02:28.373" LastActivityDate="2014-03-23T13:02:28.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2034" PostTypeId="2" ParentId="2009" CreationDate="2014-03-23T12:02:59.920" Score="1" Body="&lt;p&gt;i would rather call the streaks you are referring to as ridges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the surface of the moon is ofcourse not a perfect sphere and the same way the earth had ups and downs, there are ridges on the moon too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;you say that most of them are diagonal - top-left to bottom-right but on the right hand corner of the image, there are ones vertical ones. So, there is no specific orientation for these &quot;streaks&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;on the other hand, if there were streaks around a crater such that concave part of the streaks point towards the crater, they are due to shock waves travelling outwards from the point of impact.&lt;/p&gt;&#xA;" OwnerUserId="1278" LastActivityDate="2014-03-23T12:02:59.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2035" PostTypeId="2" ParentId="498" CreationDate="2014-03-23T12:26:24.630" Score="6" Body="&lt;p&gt;Comparing telescopes that observe the visible spectrum to the radio spectrum, radio astronomers have been able to create telescopes with apertures of the order of kms, thanks to &lt;a href=&quot;https://science.nrao.edu/science/meetings/presentation/jdf.webinar.2.pdf&quot;&gt;aperture synthesis&lt;/a&gt;. This is extremely hard in optical telescopes and the only telescope (afaik) that does so is the &lt;a href=&quot;http://www.lbto.org/index.htm&quot;&gt;Large Binocular Telescope&lt;/a&gt;. The reason this is possible in radio astronomy is because we can measure the phase of the incoming wave using radio telescopes where as information about the phase is not captured by optical telescopes. Maybe in the future, technology will help us make optical detectors that can measure the phase of the wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;coming to the size of the aperture itself, larger and larger sizes doesn't help as long as we don't account for the atmospheric seeing. the reason stars twinkle is because of atmospheric seeing. effects seeing can be negated using adaptive and active optics and advancement of these technologies will help astronomy move ahead. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;coming to the actual detectors, the intrinsic noise from radio detectors (eg. bolometers) is much smaller than that in optical detectors (eg. CCDs). so again, maybe in the future, we have better detectors with extremely low noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(sorry coulnd't add more links. needed more rep :D)&lt;/p&gt;&#xA;" OwnerUserId="1278" LastActivityDate="2014-03-23T12:26:24.630" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2036" PostTypeId="1" AcceptedAnswerId="2041" CreationDate="2014-03-23T12:35:25.043" Score="2" ViewCount="235" Body="&lt;p&gt;I'm not sure how scientists determine that a planet is really an Earth-like planet. How do they do it?&lt;/p&gt;&#xA;" OwnerUserId="1141" LastEditorUserId="50" LastEditDate="2014-03-24T08:35:25.953" LastActivityDate="2014-03-24T08:35:25.953" Title="How do scientists know if an Earth-like planet is really Earth-like?" Tags="&lt;exoplanet&gt;&lt;earth-like-planet&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2037" PostTypeId="2" ParentId="2036" CreationDate="2014-03-23T16:06:51.243" Score="0" Body="&lt;p&gt;the composition of an exoplanet is what is being referred to when someone calls it earth-like. one a planet is found, scientists try to calculate the distance of the planet from the star - by studying the rotational period of the planet, wobble of the parent star and so on. Once we know the distance to the star and the type of star, we can infer if the planet is in the relevant &lt;a href=&quot;http://en.wikipedia.org/wiki/Goldilocks_zone&quot; rel=&quot;nofollow&quot;&gt;Goldilocks zone&lt;/a&gt; i.e if the planet can have liquid water, tolerable temperatures and other such factors which are necessary for the sustenance of life. &lt;/p&gt;&#xA;" OwnerUserId="1278" LastActivityDate="2014-03-23T16:06:51.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2038" PostTypeId="2" ParentId="1966" CreationDate="2014-03-23T16:19:45.083" Score="1" Body="&lt;p&gt;stars themselves, like our sun does, have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Heliosphere&quot; rel=&quot;nofollow&quot;&gt;heliosphere&lt;/a&gt;. the size of this heliosphere would vary depending on the size/strength of the star and it's position in the galaxy. and what happens outside the heliosphere has usually little effect inside, in terms of radiation. so, the presence of a black hole or an agn near the star doesn't matter as long as the star itself is strong to create a large heliosphere (in which the planets reside). we can talk about the problem case by case but i don't know of a general galactic version of the goldilocks zone.&lt;/p&gt;&#xA;" OwnerUserId="1278" LastActivityDate="2014-03-23T16:19:45.083" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2039" PostTypeId="2" ParentId="2005" CreationDate="2014-03-23T21:06:09.643" Score="5" Body="&lt;p&gt;There are a couple of important distinctions between the two types of objects. Galaxies are objects which range in mass from about $10^{9}-10^{12} M_{\odot}$, and contain 'halos' of dark matter which represent the majority of the mass of the object. Now, there are things called dwarf galaxies (which are less massive than regular galaxies; I'd imagine that there really &lt;em&gt;could&lt;/em&gt; be a bit of a blurred distinction between these two kinds of objects), however dwarf galaxies still seem to have dark matter components to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Globular clusters are tightly bound collections of about $10^{6}$ stars, and do not have significant (if any) dark matter contained within or around them. So for these reasons in practice it is not hard to confuse the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some References:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;a href=&quot;http://arxiv.org/abs/1310.0828&quot;&gt;Evidence that dwarf galaxies are dominated by a dark-matter component.&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;a href=&quot;http://arxiv.org/abs/1210.7787&quot;&gt;Dark matter present in globular clusters are either unlikely, or incredibly highly concentrated and approximately the same mass as the stellar component.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditDate="2014-03-23T21:13:44.623" LastActivityDate="2014-03-23T21:13:44.623" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2040" PostTypeId="2" ParentId="1972" CreationDate="2014-03-23T21:34:58.603" Score="5" Body="&lt;p&gt;In this context, the $\kappa$ you are referring to is called the dimensionless matter density field. It is gravitational lensing jargon, and is usually just referred to as the 'convergence' field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you have written there (that a field can be split up into divergence-less and curl-less components) is generally speaking true for any field which can be expressed as the curl of another field $A$. @chris is correct in that they refer not to the electromagnetism you are probably used to hearing these types of things discussed in, but refer to the polarization states of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow noreferrer&quot;&gt;cosmic microwave background&lt;/a&gt; due to quantum fluctuations in the early universe. They are called 'E' and 'B' only because they are reminiscent of electric and magnetic field lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big find was the measurement of B-modes. As a reminder, here is what these E- and B-mode patterns in the CMB would look like:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/XuEUk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big thing here is that E-modes are symmetric upon reflection, whereas B-modes are clearly not ($B&amp;gt;0$ maps into $B&amp;lt;0$). In order to produce B-mode polarization, you need two things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;a href=&quot;http://en.wikipedia.org/wiki/Thomson_scattering&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thompson scattering&lt;/a&gt; of photons by free electrons (present in the &lt;a href=&quot;http://ned.ipac.caltech.edu/level5/March03/Lineweaver/Lineweaver7_2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;surface of last scattering&lt;/a&gt; as well as &lt;a href=&quot;http://en.wikipedia.org/wiki/Reionization&quot; rel=&quot;nofollow noreferrer&quot;&gt;reionization&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) A temperature quadrupole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exciting thing about the second item is that there are only a few things which can produce just such a quadrupole moment, and one of them is gravity waves! Gravitational waves physically warp the coordinates of space-time as it travels through space, which of course have existing particles, causing these types of distortions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gKD3u.gif&quot; alt=&quot;gravmode1&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BdkVf.gif&quot; alt=&quot;gravmode2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the source of the temperature quadrupole moments. Sides which are squeezed are hotter and sides which are stretched are cooler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational lensing can also produce small amounts of B-mode like features in the cmb polarization, but I believe they occur on much smaller scales and can therefore be ruled out as the cause of the BICEP2 findings. Lensing tends to stretch things out perpendicular to the radial vector (in the $\hat{\theta}$ direction):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/amRZX.png&quot; alt=&quot;weaklensing&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why this matters is because lensing provides a mechanism for turning E-mode polarization (take the $E&amp;lt;0$ mode for example) into a B-mode type polarization. Most plots of the BICEP2 results show the effect of lensing at much higher multipole moment (i.e. - smaller physical scales), so I believe they have a good way of differentiating between the two types of secondary cmb anisotropies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aUuxa.png&quot; alt=&quot;bicep2&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditDate="2014-03-24T15:02:04.697" LastActivityDate="2014-03-24T15:02:04.697" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2041" PostTypeId="2" ParentId="2036" CreationDate="2014-03-23T22:25:32.400" Score="2" Body="&lt;p&gt;The earth similarity index (ESI) is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Weighted_geometric_mean&quot; rel=&quot;nofollow&quot;&gt;weighted geometric mean&lt;/a&gt; of four similarities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula documented on&#xA;&lt;a href=&quot;http://phl.upr.edu/projects/earth-similarity-index-esi&quot; rel=&quot;nofollow&quot;&gt;http://phl.upr.edu/projects/earth-similarity-index-esi&lt;/a&gt;&#xA;(as of march 23, 2014) should be adjusted a bit, since&#xA;$n$ should be the weight sum instead of the number of planetary properties.&#xA;With the weights provided on the site we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ESI =\left( \left(1-\left | \frac{r_E - r_P}{r_E + r_P} \right |\right)^{0.57}\cdot&#xA; \left(1-\left | \frac{\rho_E - \rho_P}{\rho_E + \rho_P} \right |\right)^{1.07}\cdot\\&#xA; \left(1-\left | \frac{v_E - v_P}{v_E + v_P} \right |\right)^{0.70}\cdot&#xA; \left(1-\left | \frac{288\mbox{K} - \vartheta_P}{288\mbox{K} + \vartheta_P} \right |\right)^{5.58}\right)^{\frac{1}{7.92}},$$&#xA;with $r_E=6,371 \mbox{ km}~~$  &lt;a href=&quot;http://en.wikipedia.org/wiki/Earth&quot; rel=&quot;nofollow&quot;&gt;Earth&lt;/a&gt;'s radius, $r_P$  the radius of the planet,&#xA;$\rho_E=5.515\mbox{ g}/\mbox{cm}^3$ Earth's bulk density, $\rho_P$ the bulk density of the planet,&#xA;$v_E=11.2 \mbox{ km}/\mbox{s}~~$ the escape velocity on the surface of Earth, $v_P$ the escape velocity on the surface of  the planet, and&#xA;$\vartheta_P$ the surface temperature of the planet; the weight sum is $0.57+1.07+0.70+5.58=7.92$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars as an &lt;strong&gt;example&lt;/strong&gt;:&#xA;With $r_P=0.53 r_E$, $\rho_P=0.71 \rho_E$, $v_P=0.45v_E$, $\vartheta_P=227\mbox{ K}$ we get&#xA;$$ESI_s =\left( \left(1-\left | \frac{0.47}{1.53} \right |\right)^{0.57}\cdot&#xA; \left(1-\left | \frac{0.29}{1.71} \right |\right)^{1.07}\cdot\\&#xA; \left(1-\left | \frac{0.55}{1.45} \right |\right)^{0.70}\cdot&#xA; \left(1-\left | \frac{61\mbox{ K}}{515\mbox{ K}} \right |\right)^{5.58}\right)^{\frac{1}{7.92}}=\\&#xA;(0.811241627\cdot 0.819676889\cdot 0.716163454\cdot 0.494865663)^{\frac{1}{7.92}}=&#xA;0.833189885$$ as surface similarity. (Some of the data used from &lt;a href=&quot;https://docs.google.com/document/d/1bwIPT9C6EhHkYM5667kbPtZeLoE61fZLjfc1XiazODQ/edit?hl=en&amp;amp;authkey=CJTO5v4E&amp;amp;pli=1&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.)&#xA;Global similarity combines surface similarity with interior similarity.&#xA;Global similarity of Mars with Earth is about 0.7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surface gravity can be calculated from radius and bulk density of a planet.&#xA;The radius of an expoplanet can be estimated by the transit method, relating the estimated diameter of the star to the brightness change of the &#xA;star during planet transit. The mass of the planet can be estimated by the wobble of the radial velocity of the star (using Doppler shift).&#xA;By mass estimate of the star and the orbital period of the planet the distance of the planet to the star can be estimated. An estimate of the &#xA;absolute brightness of the star can then be used to estimate the surface temperture of the planet. There exist more methods.&#xA;The accuracy of these estimates vary with the quality of the observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ESI values between 0.8 and 1.0 are considered as Earth-like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Details of the planet's atmosphere, amount of surface water, and other details are not considered in the formula. So it's just a very rough prioritization. With future spectroscopic data further refinement could be possible.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-03-23T22:40:50.600" LastActivityDate="2014-03-23T22:40:50.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2042" PostTypeId="1" CreationDate="2014-03-24T07:42:22.883" Score="3" ViewCount="257" Body="&lt;p&gt;AS the articles on the web suggest, Dark energy is the reason behind the expansion of universe. If so, why isn't it getting used up in doing so. And consequently, the rate of expansion should reduce! &lt;/p&gt;&#xA;" OwnerUserId="1100" LastActivityDate="2014-03-24T10:39:50.123" Title="Why isn't the dark energy getting decreased?" Tags="&lt;dark-energy&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2043" PostTypeId="2" ParentId="2042" CreationDate="2014-03-24T10:39:50.123" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;AS the articles on the web suggest, Dark energy is the reason behind the expansion of universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If some articles on the web suggest that, they are mistaken. Dark energy affects the acceleration of the cosmic expansion, but it is not necessary for the universe to be expanding, so it cannot be &quot;the reason behind the expansion&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How the dark energy density changes as the universe expands depends on its current energy density and its equation of state--the relationship between its energy density and pressure. For cosmological constant provides the simplest dark energy model in cosmological models--it's the $\Lambda$ in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambda-CDM_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;ΛDCM model&lt;/a&gt;--for which the pressure is exactly the negative of its energy density: $p = -\rho$. In general relativity, both energy density and pressure gravitate, and in three spatial dimensions expansion the acceleration is proportional to $\rho + 3p$ (cf. also &lt;a href=&quot;http://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Friedmann equations&lt;/a&gt;). As a result, a cosmological constant with &lt;em&gt;positive&lt;/em&gt; energy density $\rho&amp;gt;0$ leads to &lt;em&gt;accelerated expansion&lt;/em&gt;, because the contribution of its negative pressure means the overall effect is repulsive.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If so, why isn't it getting used up in doing so.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Somewhat ironically the answer to your question for cosmological constant can be interpreted like so: it doesn't &quot;get used up&quot; &lt;em&gt;because&lt;/em&gt; of a kind of energy conservation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a given small volume expands, it keeps a constant energy density $\rho$ and a constant pressure $p$. Is this situation sensible? Suppose its volume increased by $\Delta V$, so its energy content increased by $\rho\Delta V$. If the pressure is constant, this is some kind of &lt;a href=&quot;http://en.wikipedia.org/wiki/Isobaric_process&quot; rel=&quot;nofollow noreferrer&quot;&gt;isobaric process&lt;/a&gt;, so the work done is $W = p\Delta V$. But if $p = -\rho$, then everything is consistent because the dark energy does negative work as it expands to exactly balance the increase in its energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, you shouldn't take the above reasoning too literally. There are many issues in interpreting anything like &quot;how much energy&quot; there is in general relativity and whether or not it's conserved, some of which I've covered in &lt;a href=&quot;https://astronomy.stackexchange.com/a/1500/661&quot;&gt;this&lt;/a&gt; answer, though some of which can be alleviated by the stressing of the &quot;&lt;em&gt;small&lt;/em&gt; volume&quot; qualifier. Still, &quot;it doesn't get used up because the equations of the theory say it shouldn't be getting used up&quot;, while perfectly true, would not be a very interesting answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, it should be clear from the above that the cosmological constant case of $p = -\rho$ is very itself special. The reason for its simplicity is that it's completely Lorentz-invariant: in different local inertial frames, the energy density and pressure of a perfect fluid get intermixed and transformed into different values... except in the case where the pressure is exactly the negative of energy density. Every observer would agree about the cosmological constant, so one can interpret it as an intrinsic energy density of the vacuum. In fact that's how it was originally introduced; other forms of dark energy are a generalization of $\Lambda$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And consequently, the rate of expansion should reduce! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a slightly more general equation of state $p = w\rho$, the cosmological constant corresponds to $w = -1$. Other cases give a perfect fluid, which might indeed &quot;get used up&quot; in the sense of getting more dilute, and for positive energy density, $w&amp;lt;-1/3$ is necessary to outwardly accelerate cosmic expansion. But something curious happens if $w&amp;lt;-1$: we should expect from the work done argument for the energy density to &lt;em&gt;accumulate&lt;/em&gt; instead of remaining constant. This kind of &quot;phantom energy&quot; would, if positive, produce a &lt;a href=&quot;http://en.wikipedia.org/wiki/Big_Rip&quot; rel=&quot;nofollow noreferrer&quot;&gt;Big Rip&lt;/a&gt; scenario, in which the cosmic expansion accelerates to an infinite value in finite cosmological time, ending the universe by tearing everything apart.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2014-03-24T10:39:50.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2044" PostTypeId="1" CreationDate="2014-03-24T11:59:10.457" Score="16" ViewCount="419" Body="&lt;p&gt;In the course of &lt;a href=&quot;https://www.youtube.com/watch?v=Iasqtm1prlI&quot;&gt;this&lt;/a&gt; very enjoyable press announcement, it is mentioned that inflation can create gravity waves by amplifying gravity fluctuations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not properly understand this statement. I always thought that quantum fluctuations (fluctuation of the metric in this case?) should occur pretty randomly whereas waves are rather coherent motions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I understand somehow that inflation can amplify quantum fluctuations, but I don't see how it can convert these random &quot;microscopic&quot; processes into &quot;macroscopically&quot; observable coherent phenomena called gravitational waves.&#xA;For example, how are the sources of these gravitational waves distributed? Does each point in space-time behave as some kind of a point source? And what is observed at a specific point as the superposition of all these excitations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, in the same video it is said that not all inflation models produce gravitational waves, the ones that have them are favoured by the data now, etc.&lt;/p&gt;&#xA;" OwnerUserId="92" LastEditorUserId="1702" LastEditDate="2014-05-24T17:08:07.213" LastActivityDate="2014-05-26T20:29:35.397" Title="How exactly does inflation convert random gravity fluctuations into coherent gravitational waves?" Tags="&lt;cosmology&gt;&lt;general-relativity&gt;&lt;cosmological-inflation&gt;&lt;gravitational-waves&gt;&lt;quantum-field-theory&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2045" PostTypeId="1" AcceptedAnswerId="2046" CreationDate="2014-03-24T21:11:20.930" Score="6" ViewCount="1136" Body="&lt;p&gt;Earlier today I was exploring the astronomy features on WolframAlpha, when I stumbled upon an intriguing numerical relationship involving the values for the lunar synodic and sidereal months. Using the values&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{syn}=1\,\text{synodic month}=29.530588\,\text{days},~~P_{sid}=1\,\text{sidereal month}=27.321661\,\text{days},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I noticed that this algebraic combination is very close to one year:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{P_{sid}^{-1}-P_{syn}^{-1}}=365.256396\,\text{days}\\&#xA;\frac{1}{P_{sid}^{-1}-P_{syn}^{-1}}-1\,\text{tropical year}=0.0142053\,\text{days}\approx20\,\text{minutes}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really doubt this this is a coincidence. Is there a simple way to understand what makes this approximate equality hold?&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2018-09-04T08:41:51.070" Title="A curious relationship between lunar periods and the solar year" Tags="&lt;the-moon&gt;&lt;fundamental-astronomy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2046" PostTypeId="2" ParentId="2045" CreationDate="2014-03-24T22:26:25.643" Score="6" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Month#Synodic_month&quot; rel=&quot;nofollow&quot;&gt;synodic month&lt;/a&gt; is the &quot;average period of the Moon's revolution with respect to the line joining the Sun and Earth&quot;.  However, the Earth also moves in its orbit around the Sun during this month.  From our vantage point, the Sun has appeared to move in the sky with respect to the background stars, in the same direction as the Moon moves in the sky with respect to the background stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your calculation deals with the sidereal month, so your result is the calculation of the sidereal year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the Sun returns to the same spot in the sky, that is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sidereal_year&quot; rel=&quot;nofollow&quot;&gt;sidereal year&lt;/a&gt;, whose length is 365.256363004 days, very close to your calculation.  So why is that off from the tropical year by 20 minutes?  Because the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tropical_year&quot; rel=&quot;nofollow&quot;&gt;tropical year&lt;/a&gt; (the year that keeps the seasons in place throughout the calendar year) is slightly shorter than the sidereal year.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The tropical year is about 20 minutes shorter than the time it takes Earth to complete one full orbit around the Sun as measured with respect to the fixed stars (the sidereal year).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You just found the difference between the sidereal year and the tropical year, and that was no coincidence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addition&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of sidereal months in a sidereal year is one more than the number of synodic months in a sidereal year.  That is because the Earth goes around the Sun once of year (of course), leading to one less synodic month than sidereal month.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $P_{syn}$ is the synodic period and $P_{sid}$ is the sidereal period, and $Y_{sid}$ is the sidereal year, all in days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{Y_{sid}}{P_{sid}} = \frac{Y_{sid}}{P_{syn}} + 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing both sides by $Y_{sid}$ yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{P_{sid}} = \frac{1}{P_{syn}} + \frac{1}{Y_{sid}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $Y_{sid}$...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{P_{sid}} - \frac{1}{P_{syn}} = \frac{1}{Y_{sid}}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{sid}^{-1} - P_{syn}^{-1} = \frac{1}{Y_{sid}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying both sides by $Y_{sid}$ and dividing both sides by $P_{sid}^{-1} - P_{syn}^{-1}$ yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Y_{sid} = \frac{1}{P_{sid}^{-1} - P_{syn}^{-1}}$$&lt;/p&gt;&#xA;" OwnerUserId="1082" LastEditorUserId="1082" LastEditDate="2014-03-25T17:29:41.703" LastActivityDate="2014-03-25T17:29:41.703" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2048" PostTypeId="2" ParentId="2005" CreationDate="2014-03-25T10:04:13.733" Score="0" Body="&lt;p&gt;Main difference (beyond mass, which we can extrapolate to make giant clusters and small galaxies touch) is star age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a cluster, all stars were formed in a single burst. Moreover this uses to happen when a pressure wave travels through the proto-cluster cloud inside a galaxy (This gives us a secondary difference: if it is inside another galaxy or not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a galaxy, pressure waves move forward and backward through the gas, causing constand bursts of star formation, so you get stars of varied ages (This gives us another difference: if there is gas or not, as mentioned by Jeremy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another difference is how tightly gravitationally bound are the stars of the body, as mentioned by astromax.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-03-25T10:04:13.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2050" PostTypeId="2" ParentId="1991" CreationDate="2014-03-25T22:25:46.720" Score="3" Body="&lt;p&gt;Time does &lt;b&gt;not&lt;/b&gt; pass slower near a black hole.  Stay-Home-Sally far from a black hole observing Astro-Bob descending into a black hole would see Bob's time passing slower as Bob neared the event horizon, but Bob would not experience this. On the contrary, Bob, looking back at Sally, would see time pass faster for her as he neared the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who is right, or is time relative to the motion of the observers? &lt;/p&gt;&#xA;" OwnerUserId="562" LastActivityDate="2014-03-25T22:25:46.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2051" PostTypeId="1" CreationDate="2014-03-25T23:28:15.393" Score="5" ViewCount="194" Body="&lt;p&gt;I recently saw the International Space Station without a telescope. Are there any other man-made space objects that can be seen using bare eyes?&lt;/p&gt;&#xA;" OwnerUserId="1292" LastEditorUserId="63" LastEditDate="2014-03-26T10:02:08.703" LastActivityDate="2014-03-26T13:57:04.910" Title="Space objects visible from earth" Tags="&lt;space&gt;&lt;deep-sky-observing&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2053" PostTypeId="2" ParentId="874" CreationDate="2014-03-26T09:33:16.770" Score="-4" Body="&lt;p&gt;I have pondered this issue for nearly 35 years. If the universe came into being by way of a Big Bang process we will probably never find the exact center at which it commenced:&lt;/p&gt;&#xA;&lt;h2&gt;REASONS BB &lt;em&gt;CENTER&lt;/em&gt; MAY NEVER BE FOUND&lt;/h2&gt;&#xA;&lt;p&gt;&lt;strong&gt;1.&lt;/strong&gt; We should always keep this in mind as our starting point: We did not see the beginning of our universe take place. There are &lt;em&gt;no eyewitnesses&lt;/em&gt; who can tell us what actually happened. Therefore, it is a 13.7+ &lt;em&gt;billions&lt;/em&gt; of years old &lt;strong&gt;COLD&lt;/strong&gt; case. In other words everything we put forward regarding the beginning of the universe will always be only speculation.&lt;/p&gt;&#xA;&lt;p&gt;We have no way to prove anything that we hypothesize regarding how our universe commenced (no matter how many theories are proposed regarding the beginning of the universe nor how good they and their corresponding math equations appear to be, there is no way to fully test them to prove out anything they make claims for).&lt;/p&gt;&#xA;&lt;p&gt;In other words, even if the numbers do not add up perfectly you can always come up with another &lt;em&gt;constant&lt;/em&gt; or sub-theory that makes it look &lt;em&gt;more&lt;/em&gt; correct, none of which is provable in our real world predicament. This means it may be impossible to determine if the BB took place or if a &lt;em&gt;center&lt;/em&gt; ever existed at all.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;2.&lt;/strong&gt; That said, even if the Big Bang is responsible for how the universe expanded to its current state, problems or barriers may prevent us from locating the exact location of the center from which it all started.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;3.&lt;/strong&gt; The starting point, regarding the BB, was supposedly a state of singularity. What singularity actually would be is not really known. However, its state would be a point about the size of a marble in which presumably all the laws of physics as we know them are broken down, a kind of state of total annihilation of all atomic structure and its various particles. If the supposed early &amp;quot;super&amp;quot; inflation phase occurred it would be opaque because photons would not be present nor released during this phase. Therefore, our view of events up to this point would be obscured.&lt;/p&gt;&#xA;&lt;p&gt;In other words, we can never see through this phase nor would there be any way to see this phase of inflation because we would need to see photons. Only where photons are released are we able to see even part of what was there (the so-called &lt;em&gt;visible universe&lt;/em&gt; limit).&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;4.&lt;/strong&gt; Because we can not see beyond the visible universe (which would include the parts of the electromagnetic spectrum that can not be seen by the naked eye, that can only be seen through equipment that can detect invisible wavelengths) we would have almost no chance of ever finding the exact center of the BB.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;5.&lt;/strong&gt; According to one theory space exists and inflates separate from matter and energy. That is we are located within and are a part of this inflation. According to this proposal space (fabric of spacetime) is inflating in all directions all around us, that we are within the &lt;strong&gt;spot&lt;/strong&gt;. The result being that we can not detect a direction in which the BB started from. Supposedly, because we are located within the BB inflation, this precludes us from having a frame of reference that would allow us to have the ability to trace back and locate the exact center or even the general location of the BB starting point.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;6.&lt;/strong&gt; The real world frame of reference for our universe is three dimensional plus time. The visible universe &lt;em&gt;matter&lt;/em&gt; is not uniform in all directions. That is all the galaxies, etc. are not uniformly spaced along its x, y &amp;amp; z axis (unlike the two dimensional demonstration models with evenly spaced matter). Since, supposedly the Big Bang started with a very tiny singularity blob (at least for this theory), smaller than a marble, then there ought to be various vector trails (even if they are fragmented or somewhat askew, blurred or dissipated) of some sort going back to the marble size singularity state or at least into the outward edge area of the 'super' iflation phase (especially since matter is not uniformly spaced and would have to change direction and consistancy in order for large areas or gaps at irregular intervals to be present now). However, so far we have not been able to locate any evidence that would suggest that the BB came from a particular direction which has also precluded us from locating a BB &lt;em&gt;center&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;7.&lt;/strong&gt; Unfortunately, two dimensional demonstation models lack ability to show true three dimensional effect regarding starting at a tiny point through the present state of inflation or expansion of 13.7 billion light years. In other words, if we look across &lt;em&gt;through&lt;/em&gt; our universe &lt;strong&gt;sphere&lt;/strong&gt; (matter does not exist evenly spaced on two dimensional planes here) somewhere at least near part of the edge of our visible universe we should see at least some large area where galaxies are much closer together compared to say an eight or ten billion light years area around where our galaxy is located now. So far we have not found any evidence. Perhaps this means the BB did not occur.&lt;/p&gt;&#xA;&lt;p&gt;Certainly if the universe had a tiny starting point there should be a visible change somewhere in an inflation or expansion that covers such a huge area of more than 13.7 billion light years (there would be some indication in which direction we should concentrate our effort to find something regarding at least the general direction of the BB's original location). So far such indication has not been found.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;8.&lt;/strong&gt; Also, there may be a force(s) outside our universe that is the driving force for the inflation or expansion of our universe, in which case it would most likely make it impossible to ever find evidence of a center if one ever existed.&lt;/p&gt;&#xA;" OwnerUserId="982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2014-03-31T18:06:56.357" CommentCount="2" CommunityOwnedDate="2014-03-28T20:18:48.500" ContentLicense="CC BY-SA 3.0" />
  <row Id="2054" PostTypeId="1" CreationDate="2014-03-26T13:52:53.470" Score="2" ViewCount="332" Body="&lt;p&gt;After Bradley discovered &lt;em&gt;stellar aberration&lt;/em&gt; and the corresponding &lt;em&gt;constant of aberration&lt;/em&gt; he was able to calculate the speed of light, since he knew the speed of the Earth around the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I think I understand the math behind his calculations. Yet, I believe that Bradley (and anyone else until Relativity) didn't consider the velocity of the Solar System itself. It seems (to me, that is) that Bradley assumed a Solar System at rest. If the Solar System is moving, this certainly affects the result of the calculation (especially so since at that time the speed of light was not considered constant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd appreciate if someone pointed out to me where my thinking is wrong.&lt;/p&gt;&#xA;" OwnerUserId="1295" LastActivityDate="2017-06-28T20:12:32.333" Title="How did Bradley arrive at the +/- correct speed of light when he calculated it?" Tags="&lt;star&gt;&lt;solar-system&gt;&lt;light&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2055" PostTypeId="2" ParentId="2051" CreationDate="2014-03-26T13:57:04.910" Score="2" Body="&lt;p&gt;Yes, there are &lt;a href=&quot;http://www.n2yo.com/satellites/?c=1&amp;amp;srt=4&amp;amp;dir=1&quot; rel=&quot;nofollow&quot;&gt;quite a lot of them&lt;/a&gt;, but ISS is by far the brightest. The other satellites can be visible with magnitude fainter than 3. &quot;Can be&quot; means in dusk/dawn, when the satellite is in the sunlight.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-03-26T13:57:04.910" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2056" PostTypeId="1" CreationDate="2014-03-26T14:14:37.550" Score="3" ViewCount="288" Body="&lt;p&gt;As it is for example explained &lt;a href=&quot;http://motls.blogspot.com/2014/03/bicep2-some-winners-and-losers.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, Andrei Linde's chaotic inflation is among the potential winners or survivors of the spring cleaning induced by the potential detection of primoridal gravitational waves. As this type of inflation is eternal, the new data might also lend some updraft to the multiverse business etc ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I always thought that because of the holographic principle, which means when applied to the exponentially expanding universe that stuff moving away can be thought of as being attracted to the cosmic horizon which takes to role of a black hole horizon, it does not make sense to talk about regions or any other universes beyond the cosmic horizon anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, does the holographic principle contradict eternal inflation as it arises for example from Linde's chaotic inflation? What does the detaction of primordial gravitational waves then mean when taking the holographic principle into account?&lt;/p&gt;&#xA;" OwnerUserId="92" LastEditorUserId="37191" LastEditDate="2021-02-01T21:50:15.790" LastActivityDate="2021-02-01T21:50:15.790" Title="Does the holographic principle contradict eternal inflation?" Tags="&lt;cosmology&gt;&lt;universe&gt;&lt;cosmological-inflation&gt;&lt;multiverse&gt;&lt;holographic-principle&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2057" PostTypeId="2" ParentId="2054" CreationDate="2014-03-26T14:32:50.443" Score="3" Body="&lt;p&gt;He didn't need to know the absolute position of the stars. The change of the appearent position  has been sufficient to get a good estimate.&#xA;For velocites small in comparison to the speed of light, the shift of the angle is still small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For small angles the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sine&quot; rel=&quot;nofollow&quot;&gt;sine&lt;/a&gt; is proportional to the angle with 2nd order precision.&#xA;The first &lt;a href=&quot;http://en.wikipedia.org/wiki/Taylor_series&quot; rel=&quot;nofollow&quot;&gt;Taylor summands&lt;/a&gt; are&#xA;$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - + ... $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 2nd order approximation $$\sin x\approx x$$ can be applied on stars perpendicular, or at least apearently perpendicular, to the plane of the Earth orbit, where the method is most accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the underlying reason, for which it works, is the smoothness of sine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This argument can be adjusted to less optimal conditions.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-03-26T14:51:24.373" LastActivityDate="2014-03-26T14:51:24.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2058" PostTypeId="2" ParentId="2056" CreationDate="2014-03-26T15:46:43.530" Score="2" Body="&lt;p&gt;There is no obvious implication to re-assess the large-scale topology of the universe.&#xA;You can still &lt;a href=&quot;http://en.wikipedia.org/wiki/Stereographic_projection&quot; rel=&quot;nofollow&quot;&gt;stereographically&lt;/a&gt; project infinity to a point on a sphere, if you need.&#xA;Implications to the holographic principle aren't immediate, at least.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When talking about objects beyond the cosmic horizon, it's a difference, whether we talk about observation or mathematical models. When talking about mathemetical models of the universe, it makes sense to talk about space and objects beyond the cosmic horizon, because we can get a higher symmetry of the theory. We can use mathematically simple structures like &lt;a href=&quot;http://en.wikipedia.org/wiki/De_Sitter_universe&quot; rel=&quot;nofollow&quot;&gt;de Sitter spaces&lt;/a&gt; as simplified models of the universe, with no need to add an observer to the structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an evidence-based science it's of course an extrapolation far beyond immediate observation, hence questionable. But there is also no evidence-based argument to assume, that the universe looks fundamentally different from a distant observer with a different cosmic horizon overlapping with our cosmic horizon, but reaching beyond.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-03-26T15:46:43.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2061" PostTypeId="1" AcceptedAnswerId="2062" CreationDate="2014-03-27T23:37:02.153" Score="17" ViewCount="3032" Body="&lt;p&gt;Is it possible that all dark matter is made of rogue planets (free-floating planet)? (and other stuff like asteroids or meteoroids)&lt;/p&gt;&#xA;" OwnerUserId="827" LastActivityDate="2020-11-18T08:10:51.767" Title="Is it possible that all dark matter is made of rogue planets (free-floating planet)?" Tags="&lt;dark-matter&gt;&lt;rogue-planet&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2062" PostTypeId="2" ParentId="2061" CreationDate="2014-03-28T07:35:31.770" Score="18" Body="&lt;p&gt;First of all I'll start with a few ideas:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;Baryonic Matter&lt;/em&gt;: Baryons are elementary particles made up of 3 quarks. This includes protons and neutrons, and the term &lt;em&gt;baryonic matter&lt;/em&gt; refers to matter made of baryons, such as atoms. Examples of non-baryonic matter includes neutrinos, free electrons and other exotic matter.&lt;/li&gt;&#xA;&lt;li&gt;Things like planets, stars, dust, etc. are all made of atoms, and so are classified as baryonic matter.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, &lt;em&gt;how do we know that dark matter is present in the universe?&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomers measure the gravitational pull of galaxies and galaxy groups/clusters based on how objects behave when interacting with these objects. Some examples of this include tidal gas/dust stripping, the orbit of stars in a galaxy and gravitational lensing of distant light from a large cluster. Using this they determine the mass of the galaxy (or galaxy group).&#xA;We can also determine the mass of a galaxy or group by looking at it and adding up the mass of all the objects (like stars, dust, gas, black holes, and other baryonic matter). While these methods both give us approximations, it is clear that the gravitational mass of galaxies and groups exceeds the baryonic mass by a factor of 10-100.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When astrophysicists first found this phenomenon they had to come up with a plausible explanation, so they suggested that there is some new, invisible matter called dark matter. (Aside: some astrophysicists also came up with other explanations like modified gravity, but so far dark matter does the best job at explaining observations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, so now &lt;em&gt;how do we know dark matter is not any sort of baryonic matter?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few reasons astrophysicists know that it is extremely unlikely that dark matter is baryonic. First of all if all the stars in a galaxy shine on an object it heats up, this heat causes the release of radiation, called &lt;em&gt;thermal radiation&lt;/em&gt;, and every (baryonic) object above zero kelvin (or -273.14 deg celcius) emits this radiation. However, dark matter does not emit any radiation at all (hence the name dark!) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If dark matter were baryonic it would also mean that it could become light emitting. If we got a clump of baryonic matter* and put it in space it would gravitationally contract, and would eventually form a star or black hole** - both of which we would be able to see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, because of these reasons the dark matter in galaxies and in galaxy groups/clusters &lt;em&gt;cannot&lt;/em&gt; be baryonic, and so cannot be planets, dead stars, asteroids, etc. It would definetely not be planets as there is &lt;strong&gt;no way&lt;/strong&gt; 10-100 times the mass of the stars in a galaxy would be planets, as the mechanism for making planets relies on supernovae, and the number of supernovae needed for the that many planets would be far too high to match our observations. I hope that this answered your question!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*provided the clump of baryonic matter was large, and the amount there is in galaxies definitely is!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;** we don't observe black holes directly, but can see radiation from their accretion disks.&lt;/p&gt;&#xA;" OwnerUserId="1263" LastEditorUserId="1263" LastEditDate="2014-03-29T02:14:17.510" LastActivityDate="2014-03-29T02:14:17.510" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2063" PostTypeId="1" AcceptedAnswerId="2064" CreationDate="2014-03-28T15:59:57.440" Score="2" ViewCount="567" Body="&lt;p&gt;I was watching this &lt;a href=&quot;http://www.space.com/25177-asteroid-between-saturn-and-uranus-has-rings-animation.html&quot; rel=&quot;nofollow&quot;&gt;video&lt;/a&gt; and they said that the rock probably has a moon. Is the reason they assume it has a moon is because there is a gap between the rings created by the moon's orbit around the rock? Or the rings created by the moon breaking down in its orbit and leaving debris behind in its orbit?  &lt;/p&gt;&#xA;" OwnerUserId="1305" LastActivityDate="2014-03-30T13:32:25.233" Title="What is the connection between moons, rings, and gaps?" Tags="&lt;the-moon&gt;&lt;asteroids&gt;&lt;planetary-ring&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2064" PostTypeId="2" ParentId="2063" CreationDate="2014-03-28T17:36:12.050" Score="3" Body="&lt;p&gt;It's mainly by the clear definition of the rings, and their mere existence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without a moon the rings would be short-lived, hence unlikely to be detected just in time shortly after they've formed. And they would tend to wash out to broader rings. The space between the rings seems to be empty. That's easier to explain by one or more moons keeping the gap clean, and keeping the rings on their track (&lt;a href=&quot;http://en.wikipedia.org/wiki/Planetary_ring&quot; rel=&quot;nofollow&quot;&gt;shepherd moons&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A way of thinking of the shepherd moons:&#xA;Our Earth's moon slows down Earth's rotation by tidal forces and friction.&#xA;Think of a ring a little inside the orbit of a small moon as an analogon of Earth inside the orbit of Moon: It's slowed down by tidal forces and friction (non-elastic collisions between ring particles). Slowing down ring particles results in an orbit closer to the planet/asteroid. This way a small moon can act repulsive to a nearby ring, confining it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More details about the dynamics of planetary rings, including confinement by shepherd moons, see &lt;a href=&quot;http://adsabs.harvard.edu/full/1982ARA&amp;amp;A..20..249G&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;, p. 491 ff.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-03-30T13:32:25.233" LastActivityDate="2014-03-30T13:32:25.233" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2065" PostTypeId="2" ParentId="270" CreationDate="2014-03-28T19:18:25.613" Score="-1" Body="&lt;p&gt;Yes, but it does not stay in the same constellation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun, Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus and Neptune and all the other solar system objects move across the constellations of the Zodiac: Aries, Taurus, Gemini, Cancer, Leo, Virgo, Libra, Scorpius, Sagittarius, Capricorn, Aquarius, Pices and Ophiuchus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As of March 28, 2014, the Solar System is most likely traveling through Cancer, Virgo or Libra and will move on to Scorpius.&lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2014-03-28T19:18:25.613" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2066" PostTypeId="1" CreationDate="2014-03-28T19:31:13.807" Score="5" ViewCount="1209" Body="&lt;p&gt;I am interested in building a high quality portable telescope from scratch in an attempt to better understand how telescopes work. What kind of materials do I need? How would I go about putting the telescope together? Are there any online guides that you can recommend?&lt;/p&gt;&#xA;" OwnerUserId="1308" LastActivityDate="2014-03-29T00:02:06.997" Title="Build a high quality telescope from scratch" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-05-05T06:39:14.533" ContentLicense="CC BY-SA 3.0" />
  <row Id="2067" PostTypeId="2" ParentId="1551" CreationDate="2014-03-28T19:32:18.790" Score="-1" Body="&lt;p&gt;Yes it does, especially with the gas giants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter's magnetic field is so strong it yanks on almost everything in the Solar System. Life on Earth also relies on the Jovian magnetic field, because if Jupiter's magnetic field was not there, the gravity would feel much weaker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the discovery of Uranus, astronomers predicted another gas giant beyond Uranus due to Uranus being yanked on by something large, which led to the discovery of Neptune by Urbain Le Verrier and John Galle in September 1846.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Saturnian magnetic field primarily affects Uranus and Neptune.&lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2014-03-28T19:32:18.790" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2068" PostTypeId="2" ParentId="15" CreationDate="2014-03-28T19:41:02.037" Score="2" Body="&lt;p&gt;I would say on the gas giants: Jupiter, Saturn, Uranus and Neptune.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Storms on Jupiter, such as the Great Red Spot or Oval BA, are well known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Storms on Saturn, such as the Dragon Storm or the Great White Spot, are well known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Storms on Uranus are not common, but there is one: the Uranus Dark Spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Storms on Neptune are very common: the former Great Dark Spot, the former Scooter and much more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Storms on the inner planets also happen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Storms on Mars are pretty rare, but one type is not, a dust devil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Storms on Venus are common, cyclones similar to Earth's storms appear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Storms on Mercury are extremely rare, mostly due to the lack of an atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Storms on Earth, you all know.&lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2014-03-28T19:41:02.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2069" PostTypeId="2" ParentId="2066" CreationDate="2014-03-28T19:45:33.167" Score="4" Body="&lt;p&gt;Amateur telescope making (&lt;a href=&quot;https://www.google.co.nz/search?q=atm%20telescopes&quot; rel=&quot;nofollow&quot;&gt;ATM&lt;/a&gt;) is a popular past-time for astronomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can grind &lt;a href=&quot;http://www.telescopemirrorblanks.com/&quot; rel=&quot;nofollow&quot;&gt;your own&lt;/a&gt; &lt;a href=&quot;http://www.newportglass.com/apxblank.htm&quot; rel=&quot;nofollow&quot;&gt;mirror&lt;/a&gt;, or you &lt;a href=&quot;http://www.rfroyce.com/stand_mirrors.htm&quot; rel=&quot;nofollow&quot;&gt;can&lt;/a&gt; &lt;a href=&quot;http://www.gs-telescope.com/content.asp?id=142&quot; rel=&quot;nofollow&quot;&gt;buy&lt;/a&gt; &lt;a href=&quot;http://www.hubble-optics.com/index.html&quot; rel=&quot;nofollow&quot;&gt;ready&lt;/a&gt;-&lt;a href=&quot;http://www.orionoptics.co.uk/OPTICS/telescopemirrors.html&quot; rel=&quot;nofollow&quot;&gt;made&lt;/a&gt; &lt;a href=&quot;http://zambutomirrors.com/mirrors.html&quot; rel=&quot;nofollow&quot;&gt;mirrors&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can coat your mirrors with silver using chemical solutions yourself, or have a coating of aluminium applied using a vacuum chamber by an outfit offering such a service.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main materials are a thick round slab of glass, usually Pyrex, and a smaller elliptical flat mirror. Using this you can make a Newtonian reflector. The support tube or structure for the mirrors can be one of a huge variety of designs, or you can be inventive yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are websites to guide you in this, such as &lt;a href=&quot;http://www.bbastrodesigns.com/tm.html&quot; rel=&quot;nofollow&quot;&gt;Mel Bartel's&lt;/a&gt;, &lt;a href=&quot;https://stellafane.org/tm/&quot; rel=&quot;nofollow&quot;&gt;Stellafane&lt;/a&gt;, and &lt;a href=&quot;http://www.amateurtelescopemaker.com/&quot; rel=&quot;nofollow&quot;&gt;AmateurTelescopeMaker.com&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="67" LastEditorUserId="67" LastEditDate="2014-03-29T00:02:06.997" LastActivityDate="2014-03-29T00:02:06.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2071" PostTypeId="2" ParentId="16" CreationDate="2014-03-29T17:49:54.823" Score="22" Body="&lt;p&gt;As an complement to the other answers, let me address the question of &lt;em&gt;why&lt;/em&gt; planets tend towards tidal locking. In short, the torque applied by the differential gravitational force between both sides of the surface of the planet induces friction, which in turn dissipates aways the excess spin of the (proto) moon when it is not tidally locked. When locking occurs the dissipation is minimized.&lt;br&gt;&#xA;Another artifact is that the moon is also moving away from the earth (as it looses angular momentum). See e.g. &lt;a href=&quot;http://curious.astro.cornell.edu/question.php?number=124&quot;&gt;http://curious.astro.cornell.edu/question.php?number=124&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1227" LastEditorUserId="1227" LastEditDate="2014-03-29T17:57:05.910" LastActivityDate="2014-03-29T17:57:05.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2072" PostTypeId="2" ParentId="978" CreationDate="2014-03-29T18:32:03.403" Score="2" Body="&lt;p&gt;This very good question (I wondered myself the same 30 years ago! :-) has an important yet simple  answer: because of inertia, in most cases they miss collision. In short, e.g. planets's trajectories are a compromise between their tendency to move on straight lines (inertia) and the gravitational pull applied by other objets. When the gravitational  pull becomes stronger, velocity increases hence inertia increases, which typically allows the planet to zip next to the source of pull (it has gained so much velocity by then that it just overshoots). &#xA;So, in practice, only a tiny set of initial conditions lead to actual collision. Those which hit have zero angular momentum to start with (so they are on a purely radial collisional orbit). &lt;/p&gt;&#xA;" OwnerUserId="1227" LastEditorUserId="1227" LastEditDate="2014-03-29T20:22:53.340" LastActivityDate="2014-03-29T20:22:53.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2074" PostTypeId="1" AcceptedAnswerId="2076" CreationDate="2014-03-30T17:51:40.027" Score="6" ViewCount="642" Body="&lt;p&gt;If I select satellite imagery in Google maps and zoom out, the Earth is shown in space. It shows light/dark regions correctly updated, but there’s an object between the Earth and the sun. It looks too small to be The Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JZA6j.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it real? What is it?&lt;/p&gt;&#xA;" OwnerUserId="1314" LastEditorUserId="14021" LastEditDate="2016-09-14T20:59:07.300" LastActivityDate="2016-09-14T20:59:07.300" Title="What’s the object between the Earth and The Sun currently showing in Google maps?" Tags="&lt;near-earth-object&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2076" PostTypeId="2" ParentId="2074" CreationDate="2014-03-30T19:13:38.303" Score="7" Body="&lt;p&gt;That's the Moon alright, and it's definitely real and definitely there. If you go outside and look at the Sun right now, the Moon will be almost but not quite on top of it, though it's impossible to make out due to the Sun's glare. If it were any closer we would have had a solar eclipse around the time of the New Moon that took place 20 minutes ago. Two weeks from now at Full Moon, it'll be in even better alignment, though on the opposite side of the Earth from the Sun this time, which will result in a total lunar eclipse. Mark your calendar!&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2014-03-30T19:13:38.303" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2078" PostTypeId="1" AcceptedAnswerId="2084" CreationDate="2014-03-31T02:08:13.040" Score="5" ViewCount="399" Body="&lt;p&gt;I have come across a number of star catalogs that list stars by right ascension and declination, along with other data such as magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a star catalog that lists the 3-D position of stars (e.g. includes distance from the sun, or otherwise specifies a 3-D position)?&lt;/p&gt;&#xA;" OwnerUserId="1315" LastActivityDate="2014-04-01T02:05:04.563" Title="3D Positions of Nearby Stars" Tags="&lt;star&gt;&lt;distances&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2080" PostTypeId="1" AcceptedAnswerId="2081" CreationDate="2014-03-31T12:10:31.027" Score="4" ViewCount="631" Body="&lt;p&gt;I was just watching a Geoff Marcy talk on YouTube showing how they infer the presence of planets transiting distant stars. The supposed periodicity wasn't always quite regular from what I could see, and so I wondered if there might be something tugging at the planets as they orbit their star. Then I thought of the earth: Surely there are some years when we are aligned with Jupiter so much so that it would exert some sort of perceptible influence on the planet, thus lengthening the year, even if only by some hours or days, considering that the sun is much more massive. And what would happen every few thousand (?) million (?) years when all the planets and even some of the larger asteroids, Kuiper Belt Objects - everything but the kitchen sink - come into alignment? Wouldn't the earth be drawn outwards to some extent, thus making a year perhaps 370 days long (or whatever)? And maybe temperatures would drop with the planet a fraction of an AU farther from the sun? I know that this is a website for the already enlightened, but I would appreciate an answer that even Josephine Bloggs would understand.&lt;/p&gt;&#xA;" OwnerUserId="564" LastActivityDate="2020-10-11T10:28:50.490" Title="Why is every year the same number of days despite the gravity in the solar system?" Tags="&lt;earth&gt;&lt;solar-system&gt;&lt;gravity&gt;&lt;jupiter&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2081" PostTypeId="2" ParentId="2080" CreationDate="2014-03-31T14:24:11.717" Score="7" Body="&lt;p&gt;Derived from &lt;a href=&quot;http://adsabs.harvard.edu/abs/1992JBAA..102...40M&quot;&gt;this paper&lt;/a&gt; the interval between March equinoxes for 1989/1990 has been 4 minutes 28 seconds longer than the corresponding interval for 1987/1988.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the length of a year can vary several minutes, e.g. by gravitational effects of other planets. For calendars a mean year is used. Atomic clocks run far more precise than the more or less periodic motion of Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All planets aligned don't cause anything grave. A similar kind of &lt;a href=&quot;http://en.wikipedia.org/wiki/Syzygy_%28astronomy%29&quot;&gt;sysygy&lt;/a&gt; has been 1982.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-03-31T14:24:11.717" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2082" PostTypeId="1" AcceptedAnswerId="2083" CreationDate="2014-03-31T17:12:36.063" Score="10" ViewCount="368" Body="&lt;p&gt;Can a black hole reach a limit where it can no longer attract more matter? If so, what happens with that black hole? Does the black hole die? Does the black hole decrease its size?&lt;/p&gt;&#xA;" OwnerUserId="827" LastEditorUserId="6" LastEditDate="2014-03-31T18:30:26.480" LastActivityDate="2014-04-01T15:16:04.640" Title="Can a black hole reach a limit where it can no longer attract more matter?" Tags="&lt;black-hole&gt;&lt;matter&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2083" PostTypeId="2" ParentId="2082" CreationDate="2014-03-31T20:15:00.077" Score="4" Body="&lt;p&gt;There is no known upper limit for the mass of a black hole. &#xA;A black hole stops attracting matter, when there is no matter within reach.&#xA;In this case the (macroscopic) black hole is assumed to evaporate by &lt;a href=&quot;http://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow&quot;&gt;Hawking radiation&lt;/a&gt; over very long time scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://chandra.harvard.edu/press/12_releases/press_121812.html&quot; rel=&quot;nofollow&quot;&gt;estimates for the currently (as of early 2014) known most massive black holes&lt;/a&gt; (ultramassive black holes) are below about 40 billion ($4\cdot 10^{10}$) solar masses. A black hole of that mass would have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;nofollow&quot;&gt;Schwarzschild radius&lt;/a&gt; of about 780 &lt;a href=&quot;http://en.wikipedia.org/wiki/Astronomical_unit&quot; rel=&quot;nofollow&quot;&gt;times the distance Earth-Sun&lt;/a&gt;, or about 4.5 light-days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Hypothetical) &lt;a href=&quot;http://en.wikipedia.org/wiki/Micro_black_hole&quot; rel=&quot;nofollow&quot;&gt;micro black holes&lt;/a&gt; are assumed to be able to evaporate faster than they can accrete matter.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-04-01T15:16:04.640" LastActivityDate="2014-04-01T15:16:04.640" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2084" PostTypeId="2" ParentId="2078" CreationDate="2014-04-01T02:05:04.563" Score="3" Body="&lt;p&gt;One of the most popular ones known as &lt;a href=&quot;http://heasarc.gsfc.nasa.gov/W3Browse/all/hipparcos.html&quot; rel=&quot;nofollow&quot;&gt;HIPPARCOS&lt;/a&gt; lists parallax which is indicative of distance.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2014-04-01T02:05:04.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2085" PostTypeId="1" AcceptedAnswerId="2088" CreationDate="2014-04-01T05:16:38.577" Score="6" ViewCount="4612" Body="&lt;p&gt;From &lt;a href=&quot;https://www.xkcd.com/1342/&quot; rel=&quot;noreferrer&quot;&gt;XKCD&lt;/a&gt;:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/8DjPX.png&quot; alt=&quot;http://imgs.xkcd.com/comics/ancient_stars.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Caption: &lt;em&gt;'The light from those millions of stars you see is probably many thousands of years old' is a rare example of laypeople substantially OVERestimating astronomical numbers.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what is the furthest object visible to the naked eye in the night sky from the surface of the earth? Assume a person with good eyes, on a clear, dry night at sea level, away from bright lights.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2015-01-08T13:38:17.350" Title="What's the furthest object observable by the naked eye from earth?" Tags="&lt;distances&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2088" PostTypeId="2" ParentId="2085" CreationDate="2014-04-01T09:32:43.127" Score="11" Body="&lt;p&gt;This is partly a question of 'limiting magnitude' - the faintest magnitude that is visible using a particular instrument, or in this case, just the eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;+6 - maybe +6.5  - magnitude is sometimes used as a baseline expectation of what can be seen in dark skies sites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most distant star &lt;a href=&quot;http://www.armaghplanet.com/blog/how-far-away-is-the-farthest-star.html&quot;&gt;that can easily be seen&lt;/a&gt; is Deneb. It is actually quite bright, at brighter than +2 Magnitude, but is (probably) 1425ly away. So though it is at quite some distance, it's luminosity means that actually, it isn't limiting magnitude that is at play.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are stars much further away that are at the barely perceivable limit, and seeing them will depend on conditions. The +4 mag mu Cephei is probably fairly achievable, at 5900ly, and &lt;a href=&quot;http://astrobob.areavoices.com/2010/10/02/how-to-see-the-farthest-thing-you-can-see/&quot;&gt;V762 in Cassiopeia&lt;/a&gt; is about +6 mag and maybe 16000ly, which would probably be the record if you can actually see it, but you'll need special conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to galaxies, Andromeda is 2500000ly away, and +3.4, very visible and much further away than any individual star. If you have good conditions, the +5.7 mag M33 is slightly further away at 2900000ly, but it would be much harder to see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the generally accepted limit is maybe +6.5, some people claim to have seen to +8. If it is possible, M83 is 14700000ly away at +8.2, &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_galaxies#Naked-eye_galaxies&quot;&gt;with naked-eye claims reported&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="67" LastEditorUserId="67" LastEditDate="2014-04-01T09:41:32.407" LastActivityDate="2014-04-01T09:41:32.407" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2089" PostTypeId="1" AcceptedAnswerId="2094" CreationDate="2014-04-01T12:05:35.947" Score="11" ViewCount="361" Body="&lt;p&gt;It's &quot;commonly known&quot; how distant are our solar system planets from Sun. But we can't easily say that about planets, which distances can differ greatly, without some observations (or simulations, knowing their state in some moment in time). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I can check 'actual' relative distances or positions of planets?&lt;/p&gt;&#xA;" OwnerUserId="1326" LastEditorUserId="1326" LastEditDate="2014-04-01T13:49:13.510" LastActivityDate="2014-04-01T20:44:07.307" Title="Present distances between planet. How can I find them?" Tags="&lt;planet&gt;&lt;observational-astronomy&gt;&lt;distances&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2090" PostTypeId="2" ParentId="2089" CreationDate="2014-04-01T15:43:19.167" Score="4" Body="&lt;p&gt;&lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi#top&quot; rel=&quot;nofollow&quot;&gt;JPL's HORIZONS&lt;/a&gt; system should be a fairly good starting point to calculate &lt;a href=&quot;http://ssd.jpl.nasa.gov/?ephemerides&quot; rel=&quot;nofollow&quot;&gt;ephemerides&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-04-01T15:43:19.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2091" PostTypeId="1" AcceptedAnswerId="2108" CreationDate="2014-04-01T16:54:11.393" Score="3" ViewCount="133" Body="&lt;p&gt;I recently came across an article about the Murchison Wide Field Array which said directivity is achieved by Electronic Beam Steering and not by Mechanical methods. It would be helpful if someone can tell me: how do they achieve it?&lt;/p&gt;&#xA;" OwnerUserId="1328" LastEditorUserId="67" LastEditDate="2014-04-01T20:57:38.137" LastActivityDate="2014-04-04T02:23:05.237" Title="What is meant by Electronic Beam Steering and how is it achieved?" Tags="&lt;telescope&gt;&lt;radio-astronomy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2092" PostTypeId="1" AcceptedAnswerId="2093" CreationDate="2014-04-01T19:17:38.360" Score="12" ViewCount="13376" Body="&lt;p&gt;Is this minimum mass known? or maybe, is it given in terms of density? If so, how much density is the minimum to have an spherical object due to its own gravity?&lt;/p&gt;&#xA;" OwnerUserId="827" LastActivityDate="2020-03-25T14:14:29.537" Title="What is the minimum mass required so that objects become spherical due to its own gravity?" Tags="&lt;gravity&gt;&lt;mass&gt;&lt;density&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2093" PostTypeId="2" ParentId="2092" CreationDate="2014-04-01T20:10:34.637" Score="5" Body="&lt;p&gt;This question is more complicated than it seems like it should be!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no threshold mass or density beyond which an object becomes perfectly spherical; even supermassive stars are slightly oblong. The only exception is black holes, which are perfectly round up until you reach the quantum level. If we want a simple answer, most guesses are somewhere around $\frac{1}{10000}$ the mass of earth, or $6\cdot10^{20}$ kg, but that is very approximate and depends on the composition of the object.&lt;/p&gt;&#xA;" OwnerUserId="1311" LastEditorDisplayName="user2764" LastEditDate="2015-09-27T06:28:25.917" LastActivityDate="2015-09-27T06:28:25.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2094" PostTypeId="2" ParentId="2089" CreationDate="2014-04-01T20:44:07.307" Score="10" Body="&lt;p&gt;Ask Wolfram Alpha. &lt;a href=&quot;https://www.wolframalpha.com/input/?i=How%20far%20is%20Mars%20from%20Jupiter%20right%20now?&quot;&gt;Here is an example&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-04-01T20:44:07.307" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2095" PostTypeId="2" ParentId="2091" CreationDate="2014-04-01T20:46:14.573" Score="1" Body="&lt;p&gt;Check out the &lt;a href=&quot;http://arxiv.org/ftp/arxiv/papers/0903/0903.1828.pdf&quot; rel=&quot;nofollow&quot;&gt;Design Overview&lt;/a&gt; and the &lt;a href=&quot;http://www.mwatelescope.org/index.php/telescope/antenna-beamformer&quot; rel=&quot;nofollow&quot;&gt;Antenna/Beamformer page&lt;/a&gt; on the MWA website.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic idea is that the signal arriving at an antenna is delayed in the circuitry, which makes it seem to the processing engine like the tile is being tilted in a particular direction.&lt;/p&gt;&#xA;" OwnerUserId="67" LastEditorUserId="67" LastEditDate="2014-04-01T20:51:43.077" LastActivityDate="2014-04-01T20:51:43.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2097" PostTypeId="1" CreationDate="2014-04-02T03:20:18.433" Score="5" ViewCount="933" Body="&lt;p&gt;Would an &quot;indoor&quot; pool of water on the Moon or on Mars behave differently than on Earth? Would the waves caused by a splash be higher or lower? Would they propagate faster or slower? Would boyance be different? And what would it be like to shower in slowly falling water?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2017-01-04T22:14:27.597" Title="How would water waves behave in partial gravity?" Tags="&lt;gravity&gt;&lt;water&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2098" PostTypeId="2" ParentId="2097" CreationDate="2014-04-02T03:25:05.513" Score="2" Body="&lt;p&gt;Would an &quot;indoor&quot; pool of water on the Moon or on Mars behave differently than on Earth?&#xA;Depends on what you are observing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the waves caused by a splash be higher or lower? The waves would be higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would they propagate faster or slower? They would propagate at the same rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would boyance be different? No, because buoyancy depends on the displacement of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what would it be like to shower in slowly falling water? How can anyone here answer this since only a few people (&amp;lt;0.001%) have experienced this.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-04-02T03:25:05.513" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2099" PostTypeId="1" CreationDate="2014-04-02T09:35:22.777" Score="13" ViewCount="1320" Body="&lt;p&gt;I've been working on a space game in my spare time, and lately I've been thinking on how to lay out the universe. Though I've searched around and found it hard to get a good view of what the universe looks like. I did find a lot of good sources, though I'm a very visual learner and found it hard to get a good picture of the universe with what I found. I also don't know what the good, reliable, accurate and up-to-date sources were.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, are there any good visual sources to get a picture of the universe? Preferably a TV-series or a very well illustrated book, or perhaps even computer program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I'm looking for a more general view of the universe. Like what do galaxies look like, and what is their scale. As I understand they contain a combination of planets, solar systems, star clusters, ... Though I couldn't get a good picture of how dense they are. Is there a lot of space (relatively) between these solar systems and star-clusters. I found that our solar system is about 4 light-hours big, though are the neighbouring ones average light years or thousands of light-years away? And in the space between them, is it mostly empty or is there space dust?&#xA;The same question for whatever our galaxy circles around (if it does) and what is between different galaxies. (And perhaps what else is there of galaxy-size out there?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got a vague view of all of this from reading texts, though I feel like I'm missing a whole lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2: I'm not looking for properties of different planets, that's on a too small scale. In the game I'm planning on making a fictional universe (perhaps randomly generated, though not yet decided), but I want the broad layout to be realistic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 3: I just found this: &lt;a href=&quot;http://htwins.net/scale2/&quot; rel=&quot;noreferrer&quot;&gt;http://htwins.net/scale2/&lt;/a&gt; This helped me A LOT in getting a sense of the relative size of things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="1333" LastEditorUserId="1333" LastEditDate="2015-01-27T15:18:57.993" LastActivityDate="2015-01-27T15:18:57.993" Title="Layout of the universe" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2102" PostTypeId="2" ParentId="2099" CreationDate="2014-04-02T14:19:31.303" Score="22" Body="&lt;p&gt;&lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/0/0f/Earth%27s_Location_in_the_Universe_SMALLER_%28JPEG%29.jpg&quot;&gt;Here&lt;/a&gt; is an overview of different scales we may look at the universe. On scales beyond it's almost homogeneous, and we get at the border of the visible universe. Many detailed images are available, e.g. from the &lt;a href=&quot;http://www.nasa.gov/mission_pages/hubble/multimedia/index.html#.UzwUl6h_uqh&quot;&gt;Hubble Space Telescope&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need a three-dimensional description of the universe, your program will probably need to read portions of star, cloud, star cluster or galaxy catalogs, similar to a planetarium software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general structure of the universe, and its contents may be summarized the following way: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Observable_universe&quot;&gt;observable universe&lt;/a&gt; has a diameter in the order of 10s of billions of lightyears; it depends on which of the several &lt;a href=&quot;http://en.wikipedia.org/wiki/Distance_measures_%28cosmology%29&quot;&gt;notions of distance&lt;/a&gt; is chosen. For distances below a few hundred million lightyears the notions get similar.&lt;/li&gt;&#xA;&lt;li&gt;The universe is about 13.8 billion years old, started as a hot and dense state (Big Bang), expands, and is almost 3-dimensional Euclidean plus one time dimension on large scales; that kind of space plus time is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Minkowski_space&quot;&gt;Minkowski space-time&lt;/a&gt;. Its average temperature is now near 2.7 K; we can detect a background of this temperature as &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_microwave_background&quot;&gt;cosmic microwave background&lt;/a&gt; (CMB); the CMB is thought to be the redshifted afterglow of its state as hot plasma at 380,000 years after the Big Bang.&lt;/li&gt;&#xA;&lt;li&gt;The universe mainly consists (according to current models) of &lt;a href=&quot;http://en.wikipedia.org/wiki/Dark_energy&quot;&gt;dark energy&lt;/a&gt; (about 70%); that's invisible; dark energy is thought to be equally distributed, although this is going to be investigated in more detail.&lt;/li&gt;&#xA;&lt;li&gt;Most (about 63%) of the matter is cold (slow) and &quot;dark&quot;, but not equally distributed. &lt;a href=&quot;http://en.wikipedia.org/wiki/Dark_matter&quot;&gt;Dark matter&lt;/a&gt; is thought to form large halos around galaxies and galaxy clusters. It's invisible and interacts via gravity with ordinary matter. It can be inferred from &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_lens&quot;&gt;gravitational lensing&lt;/a&gt; and the velocity distribution of stars within galaxies.&lt;/li&gt;&#xA;&lt;li&gt;Much of the remaining matter consists of hot (fast) &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutrino&quot;&gt;neutrinos&lt;/a&gt;; they are very hard to detect, but they can be detected, and they can be inferred from nuclear decay measurements.&lt;/li&gt;&#xA;&lt;li&gt;Part of the energy of the universe consists of photons (e.g. radiowaves, light, x-rays, gamma rays, magnetic and electric fields).&lt;/li&gt;&#xA;&lt;li&gt;A minor part (about 5%) of the universe consists of atoms and ions (simplified called &lt;a href=&quot;http://en.wikipedia.org/wiki/Baryon#Baryonic_matter&quot;&gt;baryonic matter&lt;/a&gt;); that's the kind of matter we know from our immediate experience.&lt;/li&gt;&#xA;&lt;li&gt;About half of the baryonic matter in the universe is thought to be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Warm%E2%80%93hot_intergalactic_medium&quot;&gt;intergalactic medium&lt;/a&gt; made of a rarified plasma, consisting of hydrogen atoms, protons, electrons, helium ions. It's thought to form huge filaments around &lt;a href=&quot;http://en.wikipedia.org/wiki/Void_%28astronomy%29&quot;&gt;voids&lt;/a&gt; of roughly 100 million light years diameter.&lt;/li&gt;&#xA;&lt;li&gt;The filaments connect and embed galaxy &lt;a href=&quot;http://en.wikipedia.org/wiki/Supercluster&quot;&gt;superclusters&lt;/a&gt;. Our local galaxy group is part of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Virgo_Supercluster&quot;&gt;Virgo supercluster&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Galaxy superclusters are made of &lt;a href=&quot;http://en.wikipedia.org/wiki/Galaxy_cluster&quot;&gt;galaxy clusters&lt;/a&gt; (diameter up to about 30 million lightyears), like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Virgo_Cluster&quot;&gt;Virgo cluster&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Galaxy_group&quot;&gt;galaxy groups&lt;/a&gt; (diameter up to about 10 million lightyears), like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Local_Group&quot;&gt;Local Group&lt;/a&gt;. Sometimes an intermediate grouping (between supercluster and cluster) like a galaxy cloud is defined.&lt;/li&gt;&#xA;&lt;li&gt;Galaxy clusters contain up to about 1000 galaxies, galaxy groups less than 50. Galaxies may have sizes about 100,000 lightyears across, and consist (besides a halo of dark matter) mainly of &lt;a href=&quot;http://en.wikipedia.org/wiki/Interstellar_medium&quot;&gt;interstellar gas and dust&lt;/a&gt;. After galaxy collisions this gas may be consumed by star formation or removed from the galaxy. A galaxy typically contains billions of stars. Interstellar gas of different temperature is distinguished.&lt;/li&gt;&#xA;&lt;li&gt;Galaxies can contain stars across the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hertzsprung%E2%80%93Russell_diagram&quot;&gt;Hertzsprung-Russel diagram&lt;/a&gt;, neutron stars, black holes, sometimes supernovae, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Rogue_planet&quot;&gt;rougue planets&lt;/a&gt;, probably also smaller rocks ejected from planetary systems. Stars can be grouped to binary, ternary or larger systems of stars up to open and &lt;a href=&quot;http://en.wikipedia.org/wiki/Globular_cluster&quot;&gt;globular star clusters&lt;/a&gt;. Globular clusters are usually old, located in the halo of a galaxy, and contain up to about a million, mostly old stars, may be with a black hole (&lt;a href=&quot;http://en.wikipedia.org/wiki/Intermediate-mass_black_hole&quot;&gt;IMBH&lt;/a&gt;) in their center. &lt;a href=&quot;http://en.wikipedia.org/wiki/Open_cluster&quot;&gt;Open clusters&lt;/a&gt; are mostly young and in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Disc_%28galaxy%29&quot;&gt;galactic disk&lt;/a&gt;. A couple of different galaxy types are distinguished, e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Elliptical_galaxy&quot;&gt;elliptical&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Spiral_galaxy&quot;&gt;spiral galaxies&lt;/a&gt;. Dense &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_cloud&quot;&gt;molecular clouds&lt;/a&gt; may be the location where &lt;a href=&quot;http://en.wikipedia.org/wiki/Star_formation&quot;&gt;stars form&lt;/a&gt; and start as open clusters.&lt;/li&gt;&#xA;&lt;li&gt;The distance between stars is usually a few lightyears; in globular clusters it's well below a lightyear. In the galactic center (&lt;a href=&quot;http://en.wikipedia.org/wiki/Bulge_%28astronomy%29&quot;&gt;bulge&lt;/a&gt;) stars are also closer to each other. The diameter of a star can vary from about 10 kilometers (&lt;a href=&quot;http://en.wikipedia.org/wiki/Neutron_star&quot;&gt;neutron stars&lt;/a&gt;) to several astronomical units (&lt;a href=&quot;http://en.wikipedia.org/wiki/Red_supergiants&quot;&gt;red supergiants&lt;/a&gt;), the mass from &lt;a href=&quot;http://en.wikipedia.org/wiki/Brown_dwarf&quot;&gt;about 13 Jupiter masses for brown dwarfs&lt;/a&gt; (or by a stricter definition &lt;a href=&quot;http://en.wikipedia.org/wiki/Red_dwarf&quot;&gt;from 0.075 solar masses for red dwarfs&lt;/a&gt;) to  &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_most_massive_known_stars&quot;&gt;several 100 solar masses&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;A galaxy center is mostly populated by a &lt;a href=&quot;http://en.wikipedia.org/wiki/Supermassive_black_hole&quot;&gt;supermassive black hole&lt;/a&gt; (SMBH), with about 0.1% the mass of the galaxy, the mass of the SMBH can vary from galaxy to galaxy.&lt;/li&gt;&#xA;&lt;li&gt;Galaxies can collide and temporarily form &lt;a href=&quot;http://en.wikipedia.org/wiki/Irregular_galaxy&quot;&gt;irregular galaxies&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Planetary_system&quot;&gt;Planetary systems&lt;/a&gt; around stars seem to be common; they are thought to form from &lt;a href=&quot;http://en.wikipedia.org/wiki/Protoplanetary_disk&quot;&gt;protoplanetary disks&lt;/a&gt; shorty after the central star has formed. The outermost part of planetary systems may reach out half a lightyear, to the limits of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hill_sphere&quot;&gt;Hill sphere&lt;/a&gt; of the central star; it depends on the distance and mass of neighbouring stars. Planets can form outside the &lt;a href=&quot;http://en.wikipedia.org/wiki/Roche_limit&quot;&gt;Roche limit&lt;/a&gt; of a star; small asteroids can survive closer to the central star for some time. A planetary system can contain planets, dwarf planets, moons, asteroids, trojans, comets, dust, magnetic fields, stellar wind.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That's certainly still incomplete, and most of the notions can be subdivided to much more detail.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-04-02T21:41:55.217" LastActivityDate="2014-04-02T21:41:55.217" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2103" PostTypeId="1" CreationDate="2014-04-02T18:44:47.877" Score="8" ViewCount="286" Body="&lt;p&gt;I'm looking for any exoplanet nomenclature, official, semi-official, proposed, or just a good idea, beyond what can already be found on Wikipedia at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Exoplanet#Other_naming_systems&quot;&gt;http://en.wikipedia.org/wiki/Exoplanet#Other_naming_systems&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or via the obvious Google searches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a vaguely sci-fi related request.  Lowercase Latin alphabet isn't going to do it once we get out there, and I like the proposal for Greece-roman mythology but there are simply too many planets now, and will be more over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side issue I'm already familiar with Ian Banks Culture addressing nomenclature, so I accept that humanity being what it is, we can expect about a hundred or so &quot;Vulcan&quot; orbiting various different stars, much as there are 37 &quot;Greenville&quot;s in the USA, currently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pointer to a really great fictional idea or naming scheme aside from the schemes mentioned about might be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interesting technical (non literary) naming schemes beyond the most obvious and trivial ordinal systems could be interesting.&lt;/p&gt;&#xA;" OwnerUserId="1334" LastActivityDate="2014-10-05T17:48:27.117" Title="Exoplanet naming conventions" Tags="&lt;exoplanet&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2104" PostTypeId="2" ParentId="666" CreationDate="2014-04-02T18:52:17.673" Score="2" Body="&lt;p&gt;Note that its more or less symmetric, so hardware of a caliber sufficient to observe human activity on the ground from space, would, if mounted on the ground, be more or less of the same size and technology required to observe human sized activity in space.  So you can cross check both theoretical and experimental results vs what parameters are known about military surveillance satellites.  So you &quot;know&quot; it doesn't have to be 200 feet in diameter, and has to be way larger than binoculars, in fact about the size of a spy satellite seems to be adequate for military levels of reliability and illumination.&lt;/p&gt;&#xA;" OwnerUserId="1334" LastActivityDate="2014-04-02T18:52:17.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2105" PostTypeId="2" ParentId="2097" CreationDate="2014-04-02T23:16:11.177" Score="2" Body="&lt;p&gt;With regard to buoyancy:  The buoyant force (in $\text{newtons}$) on a partially or totally submerged body depends on the volume of fluid displaced (in $\text{meters}^3$) , the density of the fluid (in $\text{kilograms}/\text{meter}^3$) , and the &lt;strong&gt;&lt;em&gt;local acceleration of gravity&lt;/em&gt;&lt;/strong&gt; (in $\text{meter}/\text{second}^2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the force of gravity on the object itself will depend on the local acceleration of gravity.  So:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A substance that floats in a particular liquid on Earth will float in the same liquid on the moon, or Mars;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A substance that sinks in a particular liquid on Earth will sink in the same liquid on the moon, or Mars;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A body of some substance that floats immersed to a certain depth in some liquid on Earth will float at the same depth in that liquid on the moon or Mars.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BUT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The external force needed to &lt;strong&gt;make&lt;/strong&gt; an object submerge or &lt;strong&gt;keep&lt;/strong&gt; it submerged will be less on the moon than on earth&lt;/p&gt;&#xA;" OwnerUserId="420" LastEditorUserId="9301" LastEditDate="2017-01-04T22:14:27.597" LastActivityDate="2017-01-04T22:14:27.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2106" PostTypeId="2" ParentId="1065" CreationDate="2014-04-03T10:24:18.403" Score="7" Body="&lt;p&gt;Most literature will define the diameter of the Sun up to the photosphere, the layer of the solar atmosphere you would see if you were to observe the Sun in white light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The base of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Photosphere&quot; rel=&quot;noreferrer&quot;&gt;Photosphere&lt;/a&gt; is defined as the region where the &lt;a href=&quot;https://en.wikipedia.org/wiki/Optical_depth&quot; rel=&quot;noreferrer&quot;&gt;optical depth&lt;/a&gt; is around 2/3, or the region where the plasma becomes transparent to most optical light wavelengths. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the true edge of the solar atmosphere could be considered as the heliopause, where the direct influence of the Sun's magnetic field and solar wind end and interstellar space begins.&lt;/p&gt;&#xA;" OwnerUserId="93" LastEditorUserId="1206" LastEditDate="2014-04-03T19:09:32.660" LastActivityDate="2014-04-03T19:09:32.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2107" PostTypeId="2" ParentId="1065" CreationDate="2014-04-03T15:29:26.573" Score="3" Body="&lt;p&gt;The radius of the Sun depends highly on which wavelength you are looking (well, taking a photo). In each of them, you'll have a well defined sharp boundary as explained by Zsbán Ambrus in his answer, but it is not the same: it varies with the wavelenght.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-04-03T15:29:26.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2108" PostTypeId="2" ParentId="2091" CreationDate="2014-04-04T02:23:05.237" Score="2" Body="&lt;p&gt;This can actually mean a couple of different things, but in the case of steering an MWA tile, Jeremy is quite correct. Each tile has a 'beamformer' box that all of the individual antenna cables plug into. This box physically delays each signal appropriately (by having a longer path to go through for more delay) based on the signals sent to it for each pointing. Because of the quantised path lengths there are what are called 'sweet spots' that are used, in other words arbitrary pointings don't normally happen you just dial in the closest 'sweet spot' to where you'd like to point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also version that do this in software (and in fact in a special phased mode the MWA does this between tiles sometimes), where 'turns of phase' are used instead but it is essentially the same idea.&lt;/p&gt;&#xA;" OwnerUserId="1340" LastActivityDate="2014-04-04T02:23:05.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2110" PostTypeId="2" ParentId="1649" CreationDate="2014-04-04T16:20:03.070" Score="0" Body="&lt;p&gt;If there is anybody out there how is able to answer this question, please feel free to do so ... in the meanwhile I do not what to post an answer but an (annoyingly timeconsuming and unautomated) workaround:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Workaround&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Print your ds9 plot to a .ps file&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fbk3V.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;convert this .ps file to .pdf&lt;/li&gt;&#xA;&lt;li&gt;adjust the axis labels with some vector graphic editing tool like &lt;code&gt;Inkscape&lt;/code&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/v1Hv8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="950" LastActivityDate="2014-04-04T16:20:03.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2111" PostTypeId="1" AcceptedAnswerId="2112" CreationDate="2014-04-05T10:00:28.607" Score="8" ViewCount="458" Body="&lt;p&gt;There is a large number of visible &lt;a href=&quot;https://en.wikipedia.org/wiki/Supernova&quot;&gt;Supernova&lt;/a&gt;s. Each week - about a 20 Supernovas Type 1a around the Universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there collisions of Black Holes? What is this collision called?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. 2 black holes come close to each other. There are event horizons for both black holes, will this made such collision super long in Earth's time?&lt;/p&gt;&#xA;" OwnerUserId="1346" LastEditorUserId="417" LastEditDate="2014-04-13T09:55:21.690" LastActivityDate="2014-04-13T09:55:21.690" Title="Collision of 2 black holes" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2112" PostTypeId="2" ParentId="2111" CreationDate="2014-04-05T12:48:40.997" Score="6" Body="&lt;p&gt;It's called a black hole merger, or coalescence. &lt;a href=&quot;http://www.youtube.com/watch?v=GVds0q0y5RM&quot;&gt;Here&lt;/a&gt; a simulation video.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even the formation of the event horizons of the two initial black holes takes &quot;super long&quot; in Earth's time. That's similar with the merger. On the other hand we are very close to the completed merger within a short &quot;Earth's&quot; time seen from a distance, as soon as the merger starts. General relativity as well as quantum theory are incomplete with what will happen very close to a presumed singularity or at the presumed event horizon; this will remain disputed until a satisfying theory of &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_gravity&quot;&gt;quantum gravity&lt;/a&gt; is found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mergers of black holes are likely to occur, e.g. when two galaxies collide, the momentum of the central supermassive black holes (SMBH) is slowed down by consumption of gas, dust and stars, until the SMBHs merge to the central SMBH of the merged galaxy.&#xA;&lt;a href=&quot;http://www.youtube.com/watch?v=_FlGiqPdbx8&quot;&gt;Here&lt;/a&gt; a galaxy merger simulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=eTlF_0rJF8Y&quot;&gt;Here&lt;/a&gt; a simulation of the coalescence of two black holes within a collapsing star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_black_hole&quot;&gt;More on black hole binaries on Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-04-05T13:02:53.243" LastActivityDate="2014-04-05T13:02:53.243" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2115" PostTypeId="1" CreationDate="2014-04-05T17:07:41.213" Score="13" ViewCount="1543" Body="&lt;p&gt;As you go up the periodic table (more protons), the ratio of neutrons to protons steadily increases as well. Are we sure there are absolutely no protons and electrons in a neutron star, or could there be so much more neutrons that we cannot measure any protons and electrons? Perhaps then a neutron star is a nucleus of some huge element with a neutron:proton ratio higher than we can distinguish.&lt;/p&gt;&#xA;" OwnerUserId="1350" LastActivityDate="2017-10-19T15:06:28.497" Title="Is it possible neutron stars are actual elements?" Tags="&lt;neutron-star&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2117" PostTypeId="1" CreationDate="2014-04-05T17:43:13.437" Score="2" ViewCount="354" Body="&lt;p&gt;I watch this movie:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How big is the universe &#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=b34gwCKIM50#t=50m00s&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=b34gwCKIM50#t=50m00s&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't understand where is there triangle with 180 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exactly is this red spot on the cosmic microwave background? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a base of triangle? How does it appear that this spot is smooth surface but not a concavity?&lt;/p&gt;&#xA;" OwnerUserId="1346" LastEditorUserId="1346" LastEditDate="2014-04-06T08:25:44.583" LastActivityDate="2014-06-06T22:16:18.030" Title="How big is the universe. How does it works?" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2118" PostTypeId="1" CreationDate="2014-04-05T20:37:11.613" Score="7" ViewCount="123" Body="&lt;p&gt;From watching a video on youtube, it states that Dark Matter is 6 times more abundant than known matter. I was wondering if it was possible to have hydrogen clouds between the &lt;strike&gt;universes&lt;/strike&gt; galaxies. The density would need to be less than inside &lt;strike&gt;universes&lt;/strike&gt; galaxies (since star formation would be possible). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible that hydrogen clouds be a major component of Dark Matter?&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="1322" LastEditDate="2014-04-06T04:03:00.457" LastActivityDate="2014-04-06T04:03:00.457" Title="Density of hydrogen between galaxies" Tags="&lt;dark-matter&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2119" PostTypeId="1" AcceptedAnswerId="2120" CreationDate="2014-04-05T21:36:27.330" Score="3" ViewCount="1260" Body="&lt;p&gt;&lt;s&gt;The documentary/mini-series &lt;a href=&quot;https://en.wikipedia.org/wiki/How_the_Universe_Works&quot; rel=&quot;nofollow&quot;&gt;How the Universe Works&lt;/a&gt; is based on scientific evidence and according to what is known nowadays, or sheer bunk?&lt;/s&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Throughout the series, are there any particular topics which are still disputed among the scientific community and that were portrayed in a different light than they should've had been?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I've only watched a few episodes but they always talk in a way that gives you the feeling everything is well-established and accepted but that might not be the case. I figured that an astronomer that watched the series would have remember if this happened or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the subjects were researched and there is footage of known scientists in the episodes, but that doesn't mean everyone agrees with them. (By everyone I mean the most part of scientists working on those subjects, and not literally every single one of them.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give you an example, on the second episode they state that there are massive black holes at the center of galaxies. From my perspective (i.e. of someone which is not an astronomer) if they say so it's because it's probably true. However, in the real world &lt;em&gt;maybe&lt;/em&gt; this is a highly disputed theory and a lot of astronomers disagree with this stance, and they're note telling the audience about it. Why would they do that? I don't know. I don't know if they did, I'm just asking if someone knows whether it happened, or if anyone saw something in the series with which they disagree.&lt;/p&gt;&#xA;" OwnerUserId="1351" LastEditorUserId="1351" LastEditDate="2014-04-07T16:13:02.477" LastActivityDate="2014-04-07T16:13:02.477" Title="&quot;How the Universe Works&quot;: is it scientifically accurate?" Tags="&lt;universe&gt;&lt;cosmology&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ClosedDate="2014-04-07T14:10:16.213" ContentLicense="CC BY-SA 3.0" />
  <row Id="2120" PostTypeId="2" ParentId="2119" CreationDate="2014-04-05T21:43:00.260" Score="0" Body="&lt;p&gt;I don't think that the Discovery Channel would produce science fiction and call it a documentary. The episodes are bases on factual information at the time of production.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-04-05T21:43:00.260" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2121" PostTypeId="2" ParentId="2118" CreationDate="2014-04-05T22:21:05.573" Score="1" Body="&lt;p&gt;Hydrogen clouds don't even make up a small component of dark matter, because hydrogen is not dark. The image below depicts the emission spectrum of hydrogen in the visible regime (a.k.a., the &lt;a href=&quot;http://en.wikipedia.org/wiki/Balmer_series&quot; rel=&quot;nofollow noreferrer&quot;&gt;Balmer series&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/79N24.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The emission spectrum of dark matter on the other hand would be completely black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark matter is gets its name from the fact that it doesn't absorb or emit light, i.e., electromagnetic radiation. From an elementary particle standpoint, this means dark matter particles can't interact with photons via the electromagnetic force at all, so dark matter particles must be electrically neutral, which electrons and protons certainly are not. No electrons are protons means no hydrogen atoms.&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2014-04-05T22:21:05.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2122" PostTypeId="1" AcceptedAnswerId="2123" CreationDate="2014-04-06T09:27:16.970" Score="7" ViewCount="131" Body="&lt;p&gt;A telescope located in the gravitational focus of the Sun can use the Sun as a magnifying lens. The focus begins 550 AU away, but maybe a 700 or 1000 AU distance is needed to get rid of disturbances from the Corona, and the focus extends practically indefinately. Here are some slides by Dr. Maccone who has promoted this idea he calls FOCAL:&#xA;&lt;a href=&quot;http://www.spaceroutes.com/astrocon/AstroconVTalks/Maccone-AstroconV.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://www.spaceroutes.com/astrocon/AstroconVTalks/Maccone-AstroconV.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I intend to ask about the technical design and feasibility of such a project in the Space Exploration SE. Here I rather ask about the scientific value and challenges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;POINTING: The magnification would occur only in the exact direction of the Sun. But since the Sun moves and the background objects magnified move, I suppose that the observed targets would change continuously. Would it even be practically possible to give the telescope a trajectory which keeps it aiming at for example Alpha Centauri? Would there most of the time be nothing in the right direction as the line between the telescope and the Sun sweeps across space, or would there alway be some star or galaxy in sight? Like CMB if nothing else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GAIN: In the slides linked above, Maccone has calculated the expected gain to 114 dB for infrared wavelengths. How many times &quot;magnification&quot; does this mean? I don't think I understand the units here, I get a ridiculously large number. Can it be explained somewhat intuitively? Would a FOCAL mission be a unique revolution in astronomy, or could similar results be achieved by building an interferometer with interplanetary sized baselines here nearer to the Sun? How does the science value of a gravity lens compare to that of a wide baseline? Are they good for different tasks?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DISTORTIONS: Could the lensed signals be reconstructed thanks to our knowledge of the Sun and measurements of corona activity? If pointed towards a central part of the Milky Way, wouldn't signals come from multiple objects at the same time, some much further away than others? Would the gravity lens of the Sun have bigger problems with distortion than the intergalactic gravity lenses we know of today?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, can any natural strong lensing inside the Milky Way be used today, for example using a globular cluster as a lens?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2014-04-06T15:03:41.197" Title="Telescope in Sun's gravity lens focus - pointing, gain, distortions" Tags="&lt;radio-astronomy&gt;&lt;space-telescope&gt;&lt;gravitational-lensing&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2123" PostTypeId="2" ParentId="2122" CreationDate="2014-04-06T14:21:49.113" Score="3" Body="&lt;p&gt;The &lt;strong&gt;pointing&lt;/strong&gt; is not a fundamental problem with the suggested design: The suggested trajectory is designed to include a Sun flyby as the last flyby. This ensures an asymptotically radial trajectory away from the sun after the flyby, hence maintaining the pointing relative to the sun. The proper motion of the observed object may be some challenge, but the trajectory could be adjusted appropriately by an additional burn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;114 dB are an &lt;strong&gt;amplification&lt;/strong&gt; of a factor of about $2.51\cdot 10^{11}$.&#xA;It doesn't refer just to the magnification, but to the intensity of the signal.&#xA;Therefore interferometry with a long baseline isn't the same; the latter provides a high resolution.&#xA;Whether those numbers are achievable in practice is a different question; a factor of 1000 for naturally occuring gravitational lenses would mostly be regarded as excellent.&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_microlensing&quot; rel=&quot;nofollow&quot;&gt;Theory&lt;/a&gt; allows arbitrary amplification for perfect alignment of observer, lense and observed object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the &lt;strong&gt;distortions&lt;/strong&gt; can be &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0612094&quot; rel=&quot;nofollow&quot;&gt;deconvoluted&lt;/a&gt;, revealing the field of gravity of the sun, and the shape of the observed object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational lensing is used today, &lt;a href=&quot;http://www.astro.uni-bonn.de/~spp/files/school10/Bad_Honnef_final.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; some lecture notes, and &lt;a href=&quot;http://www.nasa.gov/mission_pages/hubble/science/quasar-lens.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; quasars as an example.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-04-06T15:03:41.197" LastActivityDate="2014-04-06T15:03:41.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2124" PostTypeId="2" ParentId="2117" CreationDate="2014-04-07T12:33:38.843" Score="1" Body="&lt;p&gt;There are features in the CMB, the distance and size of which can be derived from cosmological models (6-parameter &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambda-CDM_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lambda CDM model&lt;/a&gt;). The expected size of the features can be predicted in terms of angles in the sky, depending on how &lt;a href=&quot;http://en.wikipedia.org/wiki/Curvature&quot; rel=&quot;nofollow noreferrer&quot;&gt;curved&lt;/a&gt; large-scale space-time is. (The features are a statistical property of tiny temperature fluctuations in the CMB indicating the largest scale of causal connection - &lt;a href=&quot;http://en.wikipedia.org/wiki/Baryon_acoustic_oscillations&quot; rel=&quot;nofollow noreferrer&quot;&gt;kind of sound waves&lt;/a&gt; - in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Photon_epoch&quot; rel=&quot;nofollow noreferrer&quot;&gt;baryon-photon plasma&lt;/a&gt; before &lt;a href=&quot;http://en.wikipedia.org/wiki/Plasma_recombination&quot; rel=&quot;nofollow noreferrer&quot;&gt;recombination&lt;/a&gt; 380,000 years after the Big Bang, called first &lt;a href=&quot;http://ned.ipac.caltech.edu/level5/Sept02/Reid/Reid5_2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;acoustic peak&lt;/a&gt; of the &lt;a href=&quot;http://zuserver2.star.ucl.ac.uk/~hiranya/PHAS3136/PHAS3136/PHAS3136_files/Cosmo6_101112_cmbpower.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;CMB power spectrum&lt;/a&gt;. The angle should be about 1 degree for a flat space-time, in accordance with observation.) &lt;a href=&quot;http://ned.ipac.caltech.edu/level5/March02/Sahni/Sahni4_5.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Things aren't quite straightforward&lt;/a&gt;, e.g. due to the way the universe has been expanding, therefore it's done with simulations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Angles look larger in a positively curved space-time (e.g. a sphere) than in a flat space-time (e.g. a plane) than in a negatively curved space-time (e.g. a saddle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The curvature of space-time follows the same principles as curvature of 2-dimensional surfaces. A triangle in an &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_geometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Euclidean&lt;/a&gt; plane has an angle sum of 180 degrees, while a &lt;a href=&quot;http://en.wikipedia.org/wiki/Spherical_geometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;triangle on a sphere&lt;/a&gt; has an angle sum of more than 180 degrees. Hence an object of a fixed size and a fixed distance looks larger (takes a larger angle) on a sphere than on a plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here a sketch of a triangle on a sphere (may be interpreted as an observer looking to an object) and a triangle of the same size - meaning the same lengths of the sides in this case - in a plane:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FSgqx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://en.wikipedia.org/wiki/Non-Euclidean_geometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;More on non-Euclidean geometry&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the observable universe is positively curved, it could be a fragment of the surface of a 3-sphere resp. 3+1-de Sitter space-time. A 3-sphere is finite. If the observable universe is flat (curvature 0), it could be a fragment of an infinite Euclidean space resp. 4-dimensional Minkowski space-time. That's plausible, but not the only possible extrapolation beyond the observable universe. An almost flat observable universe could e.g. be a fragment of a huge 3-spherical universe or of a huge &lt;a href=&quot;http://en.wikipedia.org/wiki/Torus&quot; rel=&quot;nofollow noreferrer&quot;&gt;3-torus&lt;/a&gt;. Planck data don't indicate (in a significant way) that kind of structure outside the observable universe, but extrapolations become less valid the further we go beyond the observable universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the current (Lambda-CDM) model of the universe together with the best-fit parameters, the universe is flat within the limits of measurement. Hence it's plausible (by extrapolation), but not necessary, that the universe is either huge or even infinite.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-04-07T21:48:59.257" LastActivityDate="2014-04-07T21:48:59.257" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2125" PostTypeId="1" AcceptedAnswerId="2126" CreationDate="2014-04-07T22:30:43.200" Score="9" ViewCount="6001" Body="&lt;p&gt;If Spock came with a super machine capable of moving planets, and placed Earth so close to Jupiter as the Earth's moon, the gravity of Jupiter would make us fall into it?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/MdYvq.jpg&quot; alt=&quot;Jupiter as close as the Moon&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="827" LastActivityDate="2019-05-28T16:12:39.147" Title="What would happen if Jupiter and Earth were at the same distance as the Moon is from Earth?" Tags="&lt;the-moon&gt;&lt;gravity&gt;&lt;jupiter&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2126" PostTypeId="2" ParentId="2125" CreationDate="2014-04-07T23:07:17.060" Score="16" Body="&lt;p&gt;That's no good idea. Earth wouldn't necessarily fall into Jupiter in the short run, provided it orbits Jupiter fast enough (within about 1.7 days), and on a circular orbit, but we would risk to collide with &lt;a href=&quot;http://en.wikipedia.org/wiki/Io_%28moon%29&quot;&gt;Io&lt;/a&gt;, destroy it by tidal forces, or change its orbit heavily.&#xA;The other &lt;a href=&quot;http://en.wikipedia.org/wiki/Galilean_moons&quot;&gt;Galilean moons&lt;/a&gt; would get out of sync and change their orbits over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tides would be severe on Earth, not just limited to oceans, but also for &quot;solid&quot; ground, as long as Earth isn't tidally locked. This would result in severe earth quakes and volcanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our days would be dim due to the distance to the sun. After tidal locking of Earth and ejection/destruction of Io the tides as heat source would be lost, oceans would freeze, temperature would fall to about -160°C mean temperature. During the polar night &lt;a href=&quot;http://en.wikipedia.org/wiki/Oxygen&quot;&gt;oxygen&lt;/a&gt; would probably condense from the atmosphere and form lakes, may be even &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrogen&quot;&gt;nitrogen&lt;/a&gt;. By this atmospheric pressure would drop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Spock is smart enough to know these consequences in advance, he wouldn't do it.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-04-07T23:07:17.060" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2127" PostTypeId="1" CreationDate="2014-04-08T04:19:01.950" Score="4" ViewCount="115" Body="&lt;p&gt;As Einstein's General relativity suggests that the space &#xA;time curves in the presence of matter, which is experimentally&#xA;proven too, this phenomenon also explains orbits. My question&#xA;is if the presence of mass curves space-time, in which direction &#xA;does it bend, if we assume that it bends in one of all possible&#xA;directions then what makes the body to choose that particular &#xA;direction. Does all the bodies bend the space-time in the same&#xA;direction ?&lt;/p&gt;&#xA;" OwnerUserId="1361" LastActivityDate="2014-04-08T05:18:53.620" Title="Einstein General relativity and bending of space" Tags="&lt;general-relativity&gt;&lt;space-time&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-04-09T14:54:51.577" ContentLicense="CC BY-SA 3.0" />
  <row Id="2128" PostTypeId="2" ParentId="2127" CreationDate="2014-04-08T05:18:53.620" Score="-1" Body="&lt;p&gt;We cannot say that it bends in a particular direction because space-time in relativity is 4-dimensional. The space-time diagram which we see frequently is just a simple way of explaining it in two dimensions because 4-dimensional space is hard to visualize.&lt;/p&gt;&#xA;" OwnerUserId="1156" LastActivityDate="2014-04-08T05:18:53.620" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2129" PostTypeId="2" ParentId="1998" CreationDate="2014-04-08T06:07:30.037" Score="8" Body="&lt;p&gt;From a (more) physics standpoint our acceleration on earth is basically zero from what we can feel. Just like the others that posted about cars traveling at a certain &lt;em&gt;constant&lt;/em&gt; velocity, you won't feel a change. If a car is traveling at a &lt;em&gt;constant&lt;/em&gt; velocity there is essentially no feelable force acting on your body. Therefore you you don't feel any effects of the speed in which you are traveling. But on the other hand if you slam on the breaks or accelerate rapidly, you would feel the respected motion on your body (so will the loose items in your car).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a plane, it's the same thing: you feel the lift off and the altitude rise due to an increase in acceleration to meet a certain altitude in which it's safe to fly. Same with the descent of an airplane: you decelerate to a point in which it's safe to let friction and the planes mechanisms to land timely and in a matter to reach your gate for off boarding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example would be elevators vs. Free falls at a theme park. Most of the time elevators decent without much acceleration but at a somewhat constat velocity so that you don't feel like you're free falling due to gravity (this isn't ALWAYS the feeling we get in elevators though). But if you were to go to a theme park which suspends you in the air and drops you close to the rate of gravity, you know you are plummeting and have that feeling in your stomach. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The earth is the same way as the examples previously described. Same as the others but hopefully in a more explanatory physics way.&lt;/p&gt;&#xA;" OwnerUserId="1356" LastEditorUserId="1356" LastEditDate="2014-04-16T08:03:52.183" LastActivityDate="2014-04-16T08:03:52.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2130" PostTypeId="1" CreationDate="2014-04-08T06:31:34.943" Score="6" ViewCount="222" Body="&lt;p&gt;if a single source of light quite small, was placed somewhere inside a completely reflective sphere, quite large, what would the reflections look like from the light source's point of view within that sphere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;:)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;extra-credit: what would it look like over the course of time with movement starting from the moment the light turned on?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://mathoverflow.net/questions/36221/light-rays-bouncing-around-inside-a-sphere-in-d-dimensions&quot;&gt;Found this question has been asked over in MathOverflow, sum clue?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:47.677" LastActivityDate="2014-04-09T12:39:45.653" Title="Might the universe be really small yet seem infinite from spherical reflections?" Tags="&lt;cosmology&gt;&lt;big-bang-theory&gt;&lt;redshift&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2014-05-02T16:10:13.723" ContentLicense="CC BY-SA 3.0" />
  <row Id="2131" PostTypeId="2" ParentId="2130" CreationDate="2014-04-08T09:04:57.947" Score="3" Body="&lt;p&gt;The simple answer to your headline question is: No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That kind of correlation between different angles has been investigated, and no significant correlation beyond 1 degree (first acoustic peak), e.g. large-scale circular structures in the CMB, has been found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Details about the investigation of non-trivial topologies within the limits of recombination, see &lt;a href=&quot;http://arxiv.org/pdf/1303.5086v2.pdf&quot; rel=&quot;nofollow&quot;&gt;this Planck paper&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Closer-by structures (like galaxy clusters) visible by telescopic observations are correlated to gravitational lensing of the CMB, as described in &lt;a href=&quot;http://arxiv.org/abs/1303.5077&quot; rel=&quot;nofollow&quot;&gt;this Planck paper&lt;/a&gt;, or by the integrated Sachs-Wolfe effect, as described in &lt;a href=&quot;http://planck.caltech.edu/pub/2013results/Planck_2013_results_19.pdf&quot; rel=&quot;nofollow&quot;&gt;this Planck paper&lt;/a&gt;, since galaxy clusters tend to be correlated with dark matter halos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeating pattens (from mirroring) would be evident from 3d maps (like &lt;a href=&quot;http://www.youtube.com/watch?v=NrMubFvV-VU&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt; for the local 300 Mpc environment) of the mass distribution or telescopic observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiple images of distant objects are well-known, e.g. as &lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein%27s_Cross&quot; rel=&quot;nofollow&quot;&gt;Einstein crosses&lt;/a&gt;, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein_ring&quot; rel=&quot;nofollow&quot;&gt;Einstein rings&lt;/a&gt;, caused by &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_lens&quot; rel=&quot;nofollow&quot;&gt;gravitational lensing&lt;/a&gt;.&#xA;This can include &lt;a href=&quot;http://cosmo.fis.fc.ul.pt/users/crawford/papers/Stuckey.pdf&quot; rel=&quot;nofollow&quot;&gt;mirroring in extreme cases&lt;/a&gt;; &lt;a href=&quot;http://hubblesite.org/explore_astronomy/black_holes/encyc_mod3_q11.html&quot; rel=&quot;nofollow&quot;&gt;it's well-known&lt;/a&gt; how this would look like.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-04-09T12:39:45.653" LastActivityDate="2014-04-09T12:39:45.653" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2132" PostTypeId="1" AcceptedAnswerId="2134" CreationDate="2014-04-09T13:31:49.317" Score="10" ViewCount="1113" Body="&lt;p&gt;Nowadays we come across with these two terms - parallel universes and multiverse. I wonder if this is the same thing or its two entirely different principles because the names are a bit confusing to me. Can anybody please shed some light on this?&lt;/p&gt;&#xA;" OwnerUserId="1369" LastEditorUserId="37191" LastEditDate="2021-02-01T21:48:44.167" LastActivityDate="2021-02-01T21:48:44.167" Title="Is there any difference between parallel universes and multiverse?" Tags="&lt;universe&gt;&lt;multiverse&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2133" PostTypeId="1" CreationDate="2014-04-09T15:24:04.213" Score="10" ViewCount="246" Body="&lt;p&gt;If we want to check a QSO's lyman alpha narrow band image, could a ground based telescope do that?&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean under the circumstance that its lyman alpha shifts to 3300A.&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="417" LastEditDate="2014-04-13T01:23:54.393" LastActivityDate="2016-02-10T03:10:07.297" Title="What is the bluest narrow band filter mounted on ground based telescopes?" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2134" PostTypeId="2" ParentId="2132" CreationDate="2014-04-09T18:53:24.590" Score="4" Body="&lt;p&gt;I'd like to see the notion &quot;parallel universe&quot; restricted to universes similar to each other in some sense. This applies best to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Many-worlds_interpretation&quot; rel=&quot;nofollow&quot;&gt;many-worlds interpretation&lt;/a&gt; of quantum theory. Two universes in this sense can be almost identical, just different in a couple of quantum states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two such parallel universes could be interpreted (in the simplest case) as two elements of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert_space&quot; rel=&quot;nofollow&quot;&gt;Hilbert space&lt;/a&gt;. A Hilbert space defines a &lt;a href=&quot;http://en.wikipedia.org/wiki/Metric_space&quot; rel=&quot;nofollow&quot;&gt;metric&lt;/a&gt; as a measure of distance between two elements (here applied to universes).&#xA;This may be seen in some analogy to points (universes as analog) in an Euclidean plane (many-worlds multiverse as analog). Here the notion &quot;parallel&quot; can make some sense. At least, it's possible to compare the items in some way (by the distance between two points/universes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Talking of universes as parallel in the very general case of items in a multiverse, when there is no similarity at all between these items, would be a bit too lax in my eyes.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-04-09T18:53:24.590" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2135" PostTypeId="2" ParentId="1998" CreationDate="2014-04-10T03:34:52.050" Score="2" Body="&lt;p&gt;If you're big enough and fluid enough, like an atmosphere, you would feel acceleration do to the Earth's motion. &lt;a href=&quot;http://en.wikipedia.org/wiki/Hadley_cell&quot; rel=&quot;nofollow&quot;&gt;Hadley cells&lt;/a&gt; are patterns of warm air rising in the tropics, flowing north, and sinking at higher latitudes. The Earth's rotation causes this north-south flow to curve, causing storms to have that distinctive spin you see in in hurricanes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if you are an exceptionally long pendulum, you'd try to oscillate in the same plane. However, the Earth's rotation would make that plane appearto rotate with respect to the surface of the Earth. Called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Foucault_pendulum&quot; rel=&quot;nofollow&quot;&gt;Foucault pendulum&lt;/a&gt;, it was the first proof the Earth rotates visible from the surface. &lt;/p&gt;&#xA;" OwnerUserId="562" LastActivityDate="2014-04-10T03:34:52.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2136" PostTypeId="1" AcceptedAnswerId="2143" CreationDate="2014-04-10T06:36:06.640" Score="4" ViewCount="133" Body="&lt;p&gt;When reading &lt;a href=&quot;http://www.lawrencehallofscience.org/pass/passv07/jupmoons.html&quot; rel=&quot;nofollow&quot;&gt;an article on using Jovian moons to calculate longitude&lt;/a&gt;, I came across this passage:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Periodically, Jupiter eclipses each of the four large moons as they pass into the planet's mammoth shadow. In a telescope, a moon's brightness takes several minutes to fade to black as it enters the shadow. Reappearances are just as gradual.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How can the moons take this long to fade in and out? Is it simply the shadow creeping over the surface of the moons? What does this look like?&lt;/p&gt;&#xA;" OwnerUserId="1374" LastActivityDate="2014-04-10T17:27:30.423" Title="Why do Jovian moons fade when transiting Jupiter's shadow?" Tags="&lt;observational-astronomy&gt;&lt;jupiter&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2137" PostTypeId="1" AcceptedAnswerId="2138" CreationDate="2014-04-09T22:34:31.330" Score="13" ViewCount="5658" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Universe&quot;&gt;&quot;&lt;em&gt;The Universe is commonly defined as the totality of existence&lt;/em&gt;&quot;&lt;/a&gt; but that is more of a philosophical position. Empirically we know (&lt;em&gt;or believe&lt;/em&gt;) that it is of finite size some believe it is &lt;a href=&quot;https://en.wikipedia.org/wiki/Expanding_universe&quot;&gt;expanding&lt;/a&gt; others that is (&lt;em&gt;or will be&lt;/em&gt;) &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Crunch&quot;&gt;contracting&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we gain knowledge about distant places, our scope of area has expanded from Planet, to Solar system, to &lt;a href=&quot;https://en.wikipedia.org/wiki/Galaxy&quot;&gt;Galaxy&lt;/a&gt;, to Universe. In the same way we continue to find smaller and smaller building blocks, from &lt;a href=&quot;https://en.wikipedia.org/wiki/Classical_element&quot;&gt;Elements&lt;/a&gt;, to &lt;a href=&quot;https://en.wikipedia.org/wiki/Atom&quot;&gt;Atoms&lt;/a&gt;, to &lt;a href=&quot;https://en.wikipedia.org/wiki/Quark&quot;&gt;Quarks&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is reasonable to assume, that the same gains in knowledge the removed the burden from &lt;a href=&quot;https://en.wikipedia.org/wiki/Atlas_%28mythology%29&quot;&gt;Atlas&lt;/a&gt;, will find the edge of the Universe, and something that is beyond that edge.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the name of that which exists beyond the Universe? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naysayers, who may claim that there is &quot;nothing&quot; beyond universe, are reminded that their beliefs are no less strong than, those who claimed that &quot;nothing&quot; was supporting the &lt;a href=&quot;https://en.wikipedia.org/wiki/World_Turtle&quot;&gt;Turtle&lt;/a&gt; or the &lt;a href=&quot;https://en.wikipedia.org/wiki/World_Elephant&quot;&gt;Elephant&lt;/a&gt; who like Atlas also supported the world.  Lack of knowledge does not mean lack of existence. &lt;/p&gt;&#xA;" OwnerUserId="513" OwnerDisplayName="James Jenkins" LastActivityDate="2018-02-10T11:48:37.130" Title="What is the name of that which exists beyond the Universe?" Tags="&lt;terminology&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2138" PostTypeId="2" ParentId="2137" CreationDate="2014-04-09T23:37:52.127" Score="14" Body="&lt;p&gt;By definition, the capitalized word Universe denotes &lt;em&gt;everything there is&lt;/em&gt;, so even if we one day discovered we're just a part of a Multiverse, all the parallel universes of it would still be parts of the Universe as a whole, where Multiverse would just describe its nature. Or, if some yet undiscovered regions of it would defy our current understanding of its physical laws and constants as we know them, all of it would still be a part of the whole Universe. So it really doesn't matter, if beyond the known universe, there are regions of honey, milk and chocolate biscuits and all of it is carried by a giant tortoise. All of it would be the Universe, the physical universe as we can observe, the honey, milk, chocolate biscuits and the tortoise. The lot. All of it. The whole shebang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that I use capitalisation here, i.e. you can have more than one universe, but they're all a part of the Universe. Without capitalisation, it's just any domain, a particular sphere in physical or metaphysical sense, and only a part of the whole Universe. Sadly, this capitalisation is often neglected or used inconsistently, as is often the case with earth vs the Earth (the top soil vs the planet), sun vs the Sun (any star with planets vs our Sol), moon vs the Moon (any natural satellite vs our Luna), even galaxy vs the Galaxy (any galaxy vs our Milky Way). For example, observable universe is a sphere, a region of space, within the Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The beauty of this naming convention is, that we already know the name for (but not necessarily of) everything there is, even if we don't know or can agree on what all that encompasses, or what laws govern some regions of space, time, or some other, yet unknown existence of it. It is universally true regardless of anyone's beliefs, even if they choose to call all of it by other names or attribute this existence to a sentient being, super being, or God. We're all a part of everything there is - the Universe.&lt;/p&gt;&#xA;" OwnerUserId="13" OwnerDisplayName="TildalWave" LastActivityDate="2014-04-09T23:37:52.127" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2139" PostTypeId="2" ParentId="2137" CreationDate="2014-04-10T13:11:07.617" Score="8" Body="&lt;p&gt;It is far from empirically known that the universe is finite in size. The observable universe is certainly of finite size, but that is just a product of the constraints of our observation. We don't know for sure whether the universe is finite or infinite, bounded or unbounded. &lt;sup&gt;&lt;a href=&quot;http://www.public.iastate.edu/~astro/courses/a120/workb32.pdf&quot; rel=&quot;nofollow&quot;&gt;[Source]&lt;/a&gt;&lt;/sup&gt; What we do know is that right now the universe looks &lt;em&gt;very&lt;/em&gt; flat as far as we can measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned in other answers, the Universe has traditionally been defined as &quot;all that is&quot;, though in practice in astronomy it is &lt;em&gt;this&lt;/em&gt; particular &quot;field&quot; of space-time that we exist in. New theories in physics have brought with them the possibility that there may be other universes in this sense. The word we use to encompass all possible universes is &quot;the Multiverse&quot;. There are different types of universes proposed, and you can read a good summary of them in the Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Multiverse&quot; rel=&quot;nofollow&quot;&gt;Multiverse&lt;/a&gt; article. From there you might read of Brian Greene and others that you might want to pursue further. For example, Brian Greene has a book titled &lt;em&gt;The Hidden Reality: Parallel Universes and the Deep Laws of the Cosmos&lt;/em&gt; which might be of interest to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Note: There is a sense in which we can say the universe is unbounded in 3D as mentioned in the linked source, but there are other senses in which the bounded/unbounded nature of the universe is questioned but which cannot be adequately discussed in this space.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorUserId="6" LastEditDate="2014-04-11T15:02:26.777" LastActivityDate="2014-04-11T15:02:26.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2141" PostTypeId="1" AcceptedAnswerId="2144" CreationDate="2014-04-10T15:53:24.373" Score="9" ViewCount="2276" Body="&lt;p&gt;Comets have tails. We all know that, and a lucky few have seen the better known. So why don't the planets (e.g. Earth, Mars) have tails? Why don't asteroids don't have tails?&lt;/p&gt;&#xA;" OwnerUserId="1377" LastActivityDate="2018-01-05T20:44:43.063" Title="Why don't planets and asteroids have (cometary) tails?" Tags="&lt;planet&gt;&lt;asteroids&gt;&lt;comets&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2142" PostTypeId="2" ParentId="2141" CreationDate="2014-04-10T16:59:22.447" Score="1" Body="&lt;p&gt;In order for an object to have a tail like a comet it must first possess a number of qualities:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The object must have a lot of ice or gas that is near the surface and able to boil off.&lt;/li&gt;&#xA;&lt;li&gt;The object must pass close enough to the sun for its solar wind to vaporize said ice/gas.&lt;/li&gt;&#xA;&lt;li&gt;The object must not have any sort of magnetosphere protecting it from solar wind.&lt;/li&gt;&#xA;&lt;li&gt;The object must not have already burned off all its ice and gas from being previously close to the sun.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Most planets and moons won't have a tail because of &lt;code&gt;2&lt;/code&gt; or &lt;code&gt;3&lt;/code&gt;. Most asteroids won't have a tail because of &lt;code&gt;1&lt;/code&gt;, &lt;code&gt;2&lt;/code&gt;, or &lt;code&gt;4&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Item &lt;code&gt;4&lt;/code&gt; alone eliminates a lot of potential comets. The best shot something has at becoming a comet is if it has a &lt;strong&gt;huge&lt;/strong&gt; orbit, taking it way out into the oort cloud, so that it almost never comes in close to get melted away, or, alternatively, something that has been in a relatively stable orbit far beyond Pluto, but for some reason (gravitational perturbation of some kind) is now coming in close to the sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: This answer is 100% speculation and based on no research or sources.&lt;/p&gt;&#xA;" OwnerUserId="158" LastEditorUserId="158" LastEditDate="2014-04-11T15:49:15.250" LastActivityDate="2014-04-11T15:49:15.250" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2143" PostTypeId="2" ParentId="2136" CreationDate="2014-04-10T17:27:30.423" Score="3" Body="&lt;p&gt;It's pretty simple, really. The shadow of Jupiter, just like the shadow of the Earth, is not razor-sharp. There is an intermediate region, called penumbra, where illumination is partial. This is due mostly to the fact that the Sun is not a mere dot, but has a measurable size. The fact that the planet itself does not have a sharp edge also contributes a small amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Umbra&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Umbra&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Penumbra can be clearly seen during Moon eclipses on Earth. The next one is on April 15 - a few days from today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://eclipse.gsfc.nasa.gov/LEplot/LEplot2001/LE2014Apr15T.pdf&quot; rel=&quot;nofollow&quot;&gt;http://eclipse.gsfc.nasa.gov/LEplot/LEplot2001/LE2014Apr15T.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.skyandtelescope.com/astronomy-news/observing-news/aprils-total-eclipse-of-the-moon/&quot; rel=&quot;nofollow&quot;&gt;http://www.skyandtelescope.com/astronomy-news/observing-news/aprils-total-eclipse-of-the-moon/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2014-04-10T17:27:30.423" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2144" PostTypeId="2" ParentId="2141" CreationDate="2014-04-11T00:47:33.130" Score="8" Body="&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Atmospheric_escape&quot;&gt;Jeans Escape&lt;/a&gt;. If the average velocity of the volatile molecules is above escape velocity, volatiles will escape. And the with the shallow gravity wells of comets, escape velocity is very low. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The earth and Mars have lots volatile gases and ice. But with their deeper gravity wells, sublimated volatile ices aren't hurled into space as they are with comets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of asteroids have high enough temperatures that any volatiles would exceed their tiny escape velocity. Which is why a lot of asteroids in our neighborhood have little or no volatile ices. If they ever had ice, it would have long since boiled off. For example, after &lt;a href=&quot;http://en.wikipedia.org/wiki/107P/Wilson-Harrington&quot;&gt;Comet Wilson Harrington&lt;/a&gt; outgassed much of its volatiles, it'd no longer have a tail when coming near the sun. They thought they had lost the comet when they stopped sighting it. Then later they discovered asteroid 1979 VA. Then they noticed they were the same object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's speculated there are more &quot;dead&quot; comets like Asteroid 1979 VA. Former comets who've outgassed most their volatiles. &lt;/p&gt;&#xA;" OwnerUserId="1232" LastActivityDate="2014-04-11T00:47:33.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2145" PostTypeId="2" ParentId="1708" CreationDate="2014-04-11T06:38:40.720" Score="2" Body="&lt;p&gt;Stars form in star-formation regions, which itself develop from molecular clouds. The paper &quot;&lt;em&gt;The Minimum Mass of Molecular Cloud Cores&lt;/em&gt;&quot; [1] claims that there is a minimum mass and density for a molecular cloud to turn into a star-formation region and they claim it's $10 \times M_{sol}$ with the cloud having a core diameter of $0.1\; pc$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if there were regions satisfying these properties elsewhere you could say that star formations is possible, though I can't quote a source which has found such a region yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Apparently such regions were found [2] between M81 and M82, although their mass is quite higher that the theoretical low limit; the mass is supposed to be $10^{4.3} - 10^{6.5}\;M_{sol}$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Kamaya, Hideyuki. &quot;The Minimum Mass of Molecular Cloud Cores.&quot; The Astrophysical Journal Letters 466.2 (1996): L99.; &lt;a href=&quot;http://iopscience.iop.org/1538-4357/466/2/L99/fulltext/5258.text.html&quot; rel=&quot;nofollow&quot;&gt;http://iopscience.iop.org/1538-4357/466/2/L99/fulltext/5258.text.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;de Mello, D. F. &quot;Blue Blobs: Star-Forming Regions Outside Galaxies.&quot; Revista Mexicana de Astronomía y Astrofísica 35 (2009): 211-212.; &lt;a href=&quot;http://www.astroscu.unam.mx/rmaa/RMxAC..35/PDF/RMxAC..35_ddemello.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.astroscu.unam.mx/rmaa/RMxAC..35/PDF/RMxAC..35_ddemello.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="627" LastEditorUserId="627" LastEditDate="2014-04-11T07:57:05.547" LastActivityDate="2014-04-11T07:57:05.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2146" PostTypeId="1" AcceptedAnswerId="2147" CreationDate="2014-04-12T02:11:57.820" Score="3" ViewCount="128" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;All planetary orbits contain 5 unusually stable points. These points are particularly important because they allow man-made satellites to orbit the Sun with a period equal to that of Earth’s. 3 of these points are collinear. Suppose that   is the distance between the centers of mass of Earth and the Sun. &lt;strike&gt;Find the distance from Earth’s center of mass to either one of the other stable points in the Earth-Sun system in terms of...&lt;/strike&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(I'm not looking for a full solution; I just want to know what these points are &lt;em&gt;called&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are these points it talks about, &lt;strike&gt;and what's their mathematical relation to Earth's orbit&lt;/strike&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This still isn't homework; I'm just looking for the name of these points.&lt;/p&gt;&#xA;" OwnerUserId="1385" LastEditorUserId="1385" LastEditDate="2014-06-17T20:36:06.127" LastActivityDate="2014-06-17T20:36:06.127" Title="Satellite/Planetary Orbits" Tags="&lt;orbit&gt;&lt;artificial-satellite&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2147" PostTypeId="2" ParentId="2146" CreationDate="2014-04-12T04:10:58.053" Score="2" Body="&lt;p&gt;These are the Langrangian Points &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrangian_point&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lagrangian_point&lt;/a&gt; -you'll find an explanation of the maths at a variety of sources if you search using that.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-04-12T04:10:58.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2148" PostTypeId="1" AcceptedAnswerId="2149" CreationDate="2014-04-12T06:25:26.313" Score="3" ViewCount="795" Body="&lt;p&gt;One of my favorite tools for modeling the solar system is the &lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow&quot;&gt;Horizons Ephemeris generator&lt;/a&gt; where I can get position and velocity vectors of various bodies. Is there something similar for neighboring stars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest I've come so far is &lt;a href=&quot;http://www.astrostudio.org/500nearest.html&quot; rel=&quot;nofollow&quot;&gt;Ashland Studio's 500 nearest stars in the Hipparcos Catalog&lt;/a&gt;. It gives distance of each star. Clicking on a star in this list will often give the star's relative velocity wrt our solar system. But it doesn't the direction of it's motion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the time being I'd like the model the stars within about 20 lightyears of Sol.&lt;/p&gt;&#xA;" OwnerUserId="1232" LastActivityDate="2014-04-12T08:43:16.790" Title="Position and velocity vectors of nearby stars?" Tags="&lt;star&gt;&lt;solar-system&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2149" PostTypeId="2" ParentId="2148" CreationDate="2014-04-12T08:43:16.790" Score="2" Body="&lt;p&gt;You may use the &lt;a href=&quot;http://cdsarc.u-strasbg.fr/viz-bin/Cat?V/137D&quot; rel=&quot;nofollow&quot;&gt;XHIP&lt;/a&gt; star catalog, using the fields X,Y,Z for the heliocentric position and U,V,W for the heliocentric velocity vector.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-04-12T08:43:16.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2150" PostTypeId="1" AcceptedAnswerId="2154" CreationDate="2014-04-13T10:00:57.100" Score="3" ViewCount="488" Body="&lt;p&gt;I was recently reading about a galaxy or quasar (not sure which, so lets just say quasar) where the estimated distance to this thing was some very significant fraction of the age of the universe.  What I mean is this:  I can't recall the exact particulars, but assuming for argument's sake that the age of the universe is 13.8 billion years old, this thing was like 13 billion light years away.  (Again, some detail here might be slightly off; the thing might have been estimated at 12 billion light years away - so focus on the general principle I am layout out here, not any particulars).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand is this.  If the thing is 13 billion light years away, then the light reaching my eyes as I look at it now took 13 billion years to get to me.  Yet the age of the universe is 13.8 billion years old (again if I'm slightly wrong in some of these numbers, just accept them for now for argument's sake).  It means the light must have left this object when the universe was much &quot;smaller&quot; or more compact, and things were much closer together.  Which means the light should have passed my position a long time ago if I have been either &quot;moving&quot; or &quot;expanding&quot; (or some combination) away from the object at anything except some rather large percentage of the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To try to boil this down, or restate it in simpler terms:  1) either it seems I am moving/expanding/red-shifting away from this object at some significant fraction of the speed of light (haven't done the numbers, but might be 80% or 90% or so) and have been for a while, or 2) something is &quot;off&quot; (the age of the universe, or the distance of this object).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or 3) there is something I'm not understanding.  Which is why I came here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have never seen it stated that we are &quot;expanding&quot; or &quot;redshifting&quot; or [insert whatever term you like] at some significant fraction of the speed of light; I have only ever seen it stated that for some brief period of time after the big bang the universe could have (or did) expand faster than light, but again that was for like less than a second if memory serves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, anyone care to explain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1392" LastActivityDate="2014-09-18T12:30:03.213" Title="Conundrum involving distance to object, universal expansion, age of universe, etc" Tags="&lt;distances&gt;&lt;expansion&gt;&lt;redshift&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2151" PostTypeId="1" AcceptedAnswerId="2153" CreationDate="2014-04-13T10:56:36.980" Score="5" ViewCount="144" Body="&lt;p&gt;I live in the UK and am aged 15. I have become extremely interested in astronomy, astrophysics, the lot. Basically I have never seen a tetrad and believe this will be my first time - at least understanding what it is. I'm hoping it will be possible and if so where will it be seen in the sky? As in North, South etc. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1393" LastEditorUserId="1393" LastEditDate="2014-05-04T23:30:57.530" LastActivityDate="2014-05-04T23:30:57.530" Title="Will the UK be able to witness the tetrad in April 2014?" Tags="&lt;lunar-eclipse&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2152" PostTypeId="1" CreationDate="2014-04-13T17:14:56.003" Score="5" ViewCount="131" Body="&lt;p&gt;Is the movement of the satellites (moons) of a planet coplanar, like the planets being coplanar around the local Sun?&lt;/p&gt;&#xA;" OwnerUserId="1394" LastActivityDate="2014-04-14T23:42:53.700" Title="Movement of the satellites of the planets . . ." Tags="&lt;planet&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2153" PostTypeId="2" ParentId="2151" CreationDate="2014-04-13T20:12:12.880" Score="6" Body="&lt;p&gt;It is great that you have an interest in astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the UK will be on the wrong side of the Earth to see the lunar eclipse on April 15. If you you want to experiment with various places on earth that will see it, and what they will see, get a planetarium application like Stellarium or SkySafari or TheSkyX. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lunar eclipse is always easy to spot - it is where the full moon is. Also, because it is the result of the Earth getting in the way of the Sun shining on the Moon, it is always after the sun sets and after the full moon rises in the East. That is, if you're going to be in a place to see the whole sequence, not just a part of it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just so you know, the tetrad is not one but a series of lunar eclipses over 2 years, this is just &lt;a href=&quot;http://eclipse.gsfc.nasa.gov/OH/OH2014.html#LE2014Apr15T&quot; rel=&quot;nofollow&quot;&gt;the first lunar eclipse in the series&lt;/a&gt;. The last tetrad was 2003-2004, so you were around, but probably not paying as much attention at that time :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, Cambridge UK will not see any of the first 3 eclipses in this tetrad. However the 28 Sept 2015 eclipse will be visible from that location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, one of the better places to see this lunar eclipse will be Easter Island (Rapa Nui) which will see the whole eclipse, and with the totality occuring high in the sky, near the meridian. Also, not much light pollution.&lt;/p&gt;&#xA;" OwnerUserId="67" LastEditorUserId="67" LastEditDate="2014-04-15T01:41:00.450" LastActivityDate="2014-04-15T01:41:00.450" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2154" PostTypeId="2" ParentId="2150" CreationDate="2014-04-14T00:54:27.520" Score="2" Body="&lt;p&gt;The 13 billion lightyears distance of the quasar mean 13 billion lightyears &lt;a href=&quot;http://en.wikipedia.org/wiki/Distance_measures_%28cosmology%29&quot; rel=&quot;nofollow&quot;&gt;light-travel distance&lt;/a&gt;. In other words, the light took 13 billion years to reach us, independent of the distance the quasar is now away from us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current proper distance (a chain of rulers would measure) is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Comoving_distance&quot; rel=&quot;nofollow&quot;&gt;comoving distance&lt;/a&gt;; it is much larger than the light-travel distance for large redshifts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quasar is moving away from us in an accelerated fashion, and is now moving away faster than light. This means, light which is emitted now (in the sense of &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_time&quot; rel=&quot;nofollow&quot;&gt;cosmic time&lt;/a&gt;) by the galaxy, the quasar may have evolved into in the meanwhile, will never reach us.&#xA;The light, the quasar emitted 13 billion years ago, left the quasar just in time to leave the regions of the universe which later have been receding faster than light from us, to eventually reach us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a thought experiment, imagine walking with 5 km/h (our playground speed of light) on a rubber band which is expanding with a constant rate of 1 km/h per meter actual length of the band. (This leads to an exponential acceleration of the distance between the two marks on the band). If you start closer than 5 meters away from your goal, e.g. from a start mark (our quasar) 4.50 m away, you'll finally reach the goal mark. The start mark will soon be further away from the goal mark than 5 meters, therefore receding with more than 5 km/h from the goal shortly after you left the start mark.&#xA;At the moment you arrive at the goal mark, the start mark will be much further away (comoving distance) than the distance you needed to walk (light travel distance). And you've been walking a longer distance than the (proper) distance between the marks was at the time you started walking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Btw.: Acceleration is only felt as a force, when the velocity to the local rubber band (mataphoric space-time) is changed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt; calculations with a protogalaxy of redshift $z=11.9$:&#xA;Based on the &lt;a href=&quot;http://adsabs.harvard.edu/abs/2006PASP..118.1711W&quot; rel=&quot;nofollow&quot;&gt;Cosmology Calculator&lt;/a&gt; on &lt;a href=&quot;http://www.astro.ucla.edu/~wright/CosmoCalc.html&quot; rel=&quot;nofollow&quot;&gt;this website&lt;/a&gt;, the cosmological parameters $H_0 = 67.11$ km/s/Mpc,&#xA;$\Omega_{\Lambda} = 0.6825$ provided by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck_%28spacecraft%29&quot; rel=&quot;nofollow&quot;&gt;Planck&lt;/a&gt; project, and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Scale_factor_%28Universe%29&quot; rel=&quot;nofollow&quot;&gt;scale factor&lt;/a&gt; $d(t) = d_0 / (1+z)$, setting $\Omega_M = 1- \Omega_{\Lambda} = 0.3175$, the&#xA;age of the universe is $13.820$ Gyr, and the comoving distance of the protogalaxy is $d_0 = 32.644$ Gly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The age of the universe, we see the protogalaxy (at redshift 11.9), was 0.370 Gyr, light-travel distance has been 13.450 Gly, proper distance was 2.531 Gly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the protogalaxy has been emitting light 0.370 Gyr after the big bang, the light travelled towards us through space of redshift beginning with 11.9 shrinking to 0; the light arrived at us 13.820 Gyr after the big bang. The comoving distance (to us) of the space traversed by the light started with 32.644 Gly shrinking to 0. The remaining distance, the light needed to travel, started with 13.450 Gly shrinking to 0. The proper distance between the protogalaxy and us started with 2.531 Gly increasing to 32.644 Gly due to the expansion of spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here some intermediate states described by&#xA;a couple of tuples, consisting of&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;redshift $z$,&lt;/li&gt;&#xA;&lt;li&gt;according age $t$ of the universe (Gyr),&lt;/li&gt;&#xA;&lt;li&gt;comoving radial distance (at age $t$) of the emitted light, we can now detect from the protogalaxy (Gly),&lt;/li&gt;&#xA;&lt;li&gt;remaining light travel distance of that emitted light (Gly),&lt;/li&gt;&#xA;&lt;li&gt;proper distance of the protogalaxy at age $t$, according to $d(t) = d_0 / (1+z)$:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$(11.9,  0.370, 32.644, 13.450,  2.531),$$&#xA;$$(11.0,  0.413, 32.115, 13.407,  2.720),$$&#xA;$$(10.0,  0.470, 31.453, 13.349,  2.968),$$&#xA;$$( 9.0,  0.543, 30.693, 13.277,  3.264),$$&#xA;$$( 8.0,  0.636, 29.811, 13.184,  3.627),$$&#xA;$$( 7.0,  0.759, 28.769, 13.061,  4.081),$$&#xA;$$( 6.0,  0.927, 27.511, 12.892,  4.663),$$&#xA;$$( 5.0,  1.168, 25.952, 12.651,  5.441),$$&#xA;$$( 4.0,  1.534, 23.952, 12.285,  6.529),$$&#xA;$$( 3.0,  2.139, 21.257, 11.680,  8.161),$$&#xA;$$( 2.0,  3.271, 17.362, 10.549, 10.881),$$&#xA;$$( 1.0,  5.845, 11.124,  7.974, 16.322),$$&#xA;$$( 0.0, 13.820,  0.0  ,  0.0  , 32.644).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hubble parameter, meaning the expansion rate of space per fixed proper distance, is decreasing with time. This allowed the protogalaxy to recede almost with the speed of light, although it was just about 2.5 Gly away from us (proper distance) in the time, when it emitted the light we detect now.&#xA;Nevertheless distant objects in this space accelerate away from us, since their increasing distance is multiplied with the expansion rate of space.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-04-15T23:49:35.430" LastActivityDate="2014-04-15T23:49:35.430" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2155" PostTypeId="1" AcceptedAnswerId="2181" CreationDate="2014-04-14T07:12:05.070" Score="5" ViewCount="1534" Body="&lt;p&gt;The heliocentric model marked the beginning of modern astronomy. But what were the reasons which led Copernicus to think that the earth orbited the sun. I've read that his model was based on observations made by some other astronomer. Can anyone describe how he got this idea?&lt;/p&gt;&#xA;" OwnerUserId="1156" LastActivityDate="2017-09-28T16:56:58.563" Title="How did Copernicus get the idea of heliocentric model?" Tags="&lt;fundamental-astronomy&gt;&lt;history&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2156" PostTypeId="1" AcceptedAnswerId="2218" CreationDate="2014-04-14T14:51:06.797" Score="2" ViewCount="13277" Body="&lt;p&gt;Could someone give a lay description of why the Solar System's depiction of motion as a vortex is wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ref: Youtube video &lt;a href=&quot;http://youtu.be/0jHsq36%5fNTU&quot; rel=&quot;nofollow&quot;&gt;The helical model - our solar system is a vortex&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please don't take for granted that because of my previous question about why planets or asteroids don't have cometary tails, that I believe the 'tails' shown in the video are real. I do not. I know that the tails as shown are simply a graphic representation of the path 'flown' through space by the planets. It seems that the solar system we inhabit has to be moving through interstellar space in some fashion. If the paths of the planets could be shown in a video, it seems to me (a lay person) that there would have to be motion 'forward' in addition to the eliptical path around Sol.&lt;/p&gt;&#xA;" OwnerUserId="1377" LastActivityDate="2015-09-29T11:28:31.263" Title="Why is the Solar Helical (Vortex) model wrong?" Tags="&lt;milky-way&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2157" PostTypeId="1" CreationDate="2014-04-14T18:30:06.220" Score="2" ViewCount="78" Body="&lt;p&gt;If the big bang is true, after the emission of light from the hydrogen plasma, the universe was still expanding. Why would we expect to see uniform radiation if earth very well could have formed outside of this hydrogen plasma. Then we would expect to see bits of background radiation coming from one direction. People sometimes answer this question by saying that the big bang happened everywhere and that where the earth is right now was a haze of hydrogen plasma 13.7 billion years ago. I do not see how one can assume this as since the universe expands we could have formed outside of the plasma and therefore should not expect to detect uniform background radiation. &lt;/p&gt;&#xA;" OwnerUserId="1397" LastEditorUserId="37191" LastEditDate="2021-02-12T21:20:14.013" LastActivityDate="2021-02-12T21:20:14.013" Title="Cosmic microwave background radiation" Tags="&lt;cosmology&gt;&lt;cosmic-microwave-background&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2158" PostTypeId="2" ParentId="2157" CreationDate="2014-04-14T19:51:34.927" Score="2" Body="&lt;p&gt;We could not have formed &quot;outside&quot; the expanding universe, as there would not be any substance to form our galaxy, and solar system -including our Sun and planet.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-04-14T19:51:34.927" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2159" PostTypeId="1" CreationDate="2014-04-14T19:49:32.333" Score="2" ViewCount="185" Body="&lt;p&gt;I understand that Kerr Black Holes are purely theoretical and have not been found anywhere in the Universe. However, they are theorized for possible means of time travel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am attempting to prove in a research paper that controlled time travel is statistically improbable for the human species. Therefore if scientists have evidence that there does not exist a black hole within a distance of X from Earth, and it is statistically improbable for the human species to ever be able to travel a distance of X, I can rule out time travel by means of kerr black holes with a significant degree of certainty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I cannot find this &quot;X&quot; threshold, which is why I am asking this question. Have scientists verified that there are no kerr black holes within a distance of X from earth for some X?&lt;/p&gt;&#xA;" OwnerUserId="1408" OwnerDisplayName="user1380792" LastActivityDate="2014-04-14T20:28:16.443" Title="Finding a Kerr Black Holes" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2160" PostTypeId="2" ParentId="2159" CreationDate="2014-04-14T20:28:16.443" Score="5" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Rotating_black_hole&quot;&gt;1) Nearly all black holes are Kerr black holes.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hubblesite.org/explore_astronomy/black_holes/encyc_mod3_q7.html&quot;&gt;2) There are roughly 100 million black holes estimated in our galaxy.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.space.com/7678-black-hole-closer-earth-thought.html&quot;&gt;3) There is a confirmed black hole about 7800 light years away.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://io9.com/5949616/is-earths-closest-black-hole-hiding-inside-this-nebula&quot;&gt;4) There is a potential black hole about 1500 light years away.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://io9.com/5949616/is-earths-closest-black-hole-hiding-inside-this-nebula&quot;&gt;5) It is very likely that there are black holes even closer than that, but they are difficult to detect.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of your assumptions are wrong, but for now the closest known potential black hole is still a pretty good distance away. Traveling that far with our current technology is impossible, and overcoming that hurdle is not trivial. Regardless, using a Kerr black hole for time travel &lt;a href=&quot;http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/041130a.html&quot;&gt;may not be feasible anyway&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6" LastActivityDate="2014-04-14T20:28:16.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2161" PostTypeId="2" ParentId="2152" CreationDate="2014-04-14T23:42:53.700" Score="1" Body="&lt;p&gt;Inner moon orbits tend to low inclinations with regard to planet's equatorial plane. Charon is an exception to this general rule. Nineplanets.org has a &lt;a href=&quot;http://nineplanets.org/data.html&quot; rel=&quot;nofollow&quot;&gt;nice data base&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1232" LastActivityDate="2014-04-14T23:42:53.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2164" PostTypeId="1" AcceptedAnswerId="2165" CreationDate="2014-04-15T11:58:28.070" Score="4" ViewCount="86" Body="&lt;p&gt;So supposedly the Lyrids meteor shower will take place between early morning April 16th to 22nd - correct me if I am wrong. I am aged 15 and am interested in skywatching, I just wondered will it be visible from the northern Hemisphere? Or will it be like the lunar eclipse and only be visible in the Western Hemisphere? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1393" LastEditorUserId="1393" LastEditDate="2014-05-04T23:32:55.503" LastActivityDate="2014-05-04T23:32:55.503" Title="Where will I be able to see the Lyrid meteor shower?" Tags="&lt;meteor&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2165" PostTypeId="2" ParentId="2164" CreationDate="2014-04-15T13:36:37.433" Score="2" Body="&lt;p&gt;You should definitely be able to see this meteor shower. It will be in the north east from your viewpoint near the star vega. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astronomy.co.uk/skymap&quot; rel=&quot;nofollow&quot;&gt;http://www.astronomy.co.uk/skymap&lt;/a&gt;           &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://earthsky.org/astronomy-essentials/earthskys-meteor-shower-guide&quot; rel=&quot;nofollow&quot;&gt;http://earthsky.org/astronomy-essentials/earthskys-meteor-shower-guide&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may find these websites useful. If you're interested in amateur astronomy, there is free software for interactive star maps. I use one called stellarium. I'm also an amateur astronomer and I've learned so much with it. You may enjoy it as well. &lt;/p&gt;&#xA;" OwnerUserId="1206" LastActivityDate="2014-04-15T13:36:37.433" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2166" PostTypeId="1" AcceptedAnswerId="2192" CreationDate="2014-04-15T19:37:31.723" Score="10" ViewCount="888" Body="&lt;p&gt;I understand that the redness of the moon during the total lunar eclipse last night/this morning (April 14-15 2014) is a reflection of the red light caused in earth's atmosphere due to Rayleigh scattering (basically, a reflection of all the sunsets and sunrises happening around the earth).  Because of this, it seems to me that the moon will appear red during every total lunar eclipse that happens.  Is this the case, or are there some lunar eclipses where the moon does not turn red?  If it is the case that all total lunar eclipses produce a red moon, why was the media so hyped up about last night's eclipse producing a red moon?  Was it especially red?&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2014-04-24T04:36:35.517" Title="Red Moon a Characteristic of all Total Lunar Eclipses?" Tags="&lt;the-moon&gt;&lt;lunar-eclipse&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2168" PostTypeId="2" ParentId="2166" CreationDate="2014-04-15T20:35:26.543" Score="2" Body="&lt;p&gt;Pretty most all lunar eclipses will turn the moon red like that. The amount of redness does vary, and sometimes so little light gets to the sun it is almost completely dark. However the redness is so typical of a lunar eclipse that NASA describes it as a &quot;&lt;a href=&quot;http://eclipse.gsfc.nasa.gov/LEcat5/appearance.html&quot; rel=&quot;nofollow&quot;&gt;characteristic orange-red color&lt;/a&gt;&quot;. That link has a neat table with a categorization of the colour ranges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably what triggered the media's &quot;do a story on that!&quot; bells was lots of chatter they noticed online, mostly from the loony fringe making more of it than it deserves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for the extra attention was probably because this lunar eclipse (I can't bring myself to write 'blood moon' ick I've done it) is coincident with the Jewish Passover - not a surprising event, as Passover is always when the moon is full after the spring equinox. It happens that the Passover that was celebrated the evening before the day Jesus was executed was followed by a lunar eclipse, and the Bible makes mention of 'the moon being turned to blood.' Simple and straightforward, but then some people start going off on their own weird supposes and imagining that Jesus' second coming will occur on another lunar eclipse, as if a lunar eclipse after the spring equinox was the only time it is possible for him to visit. So you get lots of internet chatter, the media notices what is trending on twitter and does a half-baked story.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lunar eclipse will occur in other months too, but without this association, people don't seem to care as much.&lt;/p&gt;&#xA;" OwnerUserId="67" LastEditorUserId="67" LastEditDate="2014-04-17T20:25:28.597" LastActivityDate="2014-04-17T20:25:28.597" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2169" PostTypeId="1" AcceptedAnswerId="2170" CreationDate="2014-04-15T22:19:14.113" Score="2" ViewCount="34" Body="&lt;p&gt;So I am a newcomer to Astronomy - only aged 15 - and am currently looking upon the sky in hopes of witnessing several meteors. This is my first time ever - maybe - seeing them. I just went inside for food and drink, come out and i saw a trail (this trail was not here before I went in). I do have a few doubts, one it was seen below the moon, when it was supposedly meant to be near Vega - however I read about them having a tendency to be different to predictions. Also, considering it is so close to the Moon and I can't currently see Vega from this position I highly doubt it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;please view the below image in the hopes of identifying the object (please excuse the camera quality, I am currently setting up the DSLR and this was photographed on my Nexus.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FAlWL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1393" LastActivityDate="2014-04-16T00:08:52.243" Title="First meteor(s) of Lyrid meteor shower?" Tags="&lt;meteor&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2170" PostTypeId="2" ParentId="2169" CreationDate="2014-04-16T00:08:52.243" Score="4" Body="&lt;p&gt;The feature in the photograph ressembles a &lt;a href=&quot;http://en.wikipedia.org/wiki/Contrail&quot; rel=&quot;nofollow&quot;&gt;condensation trail&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meteors usually move rather rapidly, mostly perish within a few seconds or less.&#xA;In rare cases they may leave long trails; the meteor would have been very bright (brighter than the moon) in those cases, moving much faster than an aeroplane.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-04-16T00:08:52.243" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2171" PostTypeId="1" CreationDate="2014-04-16T00:28:52.503" Score="6" ViewCount="3691" Body="&lt;p&gt;The cosmic microwave background that we observe uniformly around us is usually explained by assuming that our universe is the surface of a four dimensional sphere. That way the uniformity makes sense since there is no center. My question is if this is true then what is the explanation that describes the fact that the farther we look into space, the further we look back in time. I can't perfectly picture this and see how it would coexist. Help me out.&lt;/p&gt;&#xA;" OwnerUserId="1397" LastActivityDate="2014-09-11T00:53:13.457" Title="Our universe the surface of a 4-dimensional sphere?" Tags="&lt;universe&gt;&lt;space-time&gt;&lt;cosmological-inflation&gt;&lt;astrophysics&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2172" PostTypeId="2" ParentId="2171" CreationDate="2014-04-16T02:03:53.867" Score="5" Body="&lt;p&gt;The surface of the 4-dimensional &lt;a href=&quot;http://en.wikipedia.org/wiki/N-ball&quot; rel=&quot;noreferrer&quot;&gt;ball&lt;/a&gt; (called &lt;a href=&quot;http://en.wikipedia.org/wiki/3-sphere&quot; rel=&quot;noreferrer&quot;&gt;3-sphere&lt;/a&gt;) is a slice through the universe as a whole for a fixed &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_time&quot; rel=&quot;noreferrer&quot;&gt;cosmic time&lt;/a&gt;. This slice describes just three spatial dimensions. The observable universe is a tiny part of this 3-sphere; hence it looks flat (3-dimensional &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_space&quot; rel=&quot;noreferrer&quot;&gt;Euclidean space&lt;/a&gt;) up to measurement precision (&lt;a href=&quot;http://map.gsfc.nasa.gov/universe/uni_shape.html&quot; rel=&quot;noreferrer&quot;&gt;about 0.4% at the moment&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding time makes the universe 4-dimensional. The observable part is similar to a 3+1-dimensional &lt;a href=&quot;http://en.wikipedia.org/wiki/Minkowski_space&quot; rel=&quot;noreferrer&quot;&gt;Minkowski space-time&lt;/a&gt;. The universe as a whole may be a de Sitter space-time. A de Sitter space-time is the analogon of a sphere ( = surface of a ball), embedded in a Minkowski space, instead of an Euclidean space, but it's not literally a sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If time would be taken as an additional spatial dimension, the &lt;a href=&quot;http://en.wikipedia.org/wiki/De_Sitter_space&quot; rel=&quot;noreferrer&quot;&gt;de Sitter space&lt;/a&gt; would ressemble a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperboloid&quot; rel=&quot;noreferrer&quot;&gt;hyperboloid of revolution&lt;/a&gt;, if embedded in a 5-dimensional hyperspace.&#xA;The difference to an Euclidean space is due to the different definition of the distance: In a 4-dimensional Euclidean space the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_distance&quot; rel=&quot;noreferrer&quot;&gt;distance&lt;/a&gt; between two points is defined by&#xA;$l = \sqrt{\Delta x^2+\Delta y^2+\Delta z^2+\Delta t^2};$ for a &lt;a href=&quot;http://en.wikipedia.org/wiki/Pseudo-Euclidean_space&quot; rel=&quot;noreferrer&quot;&gt;3+1-dimensional Minkowski space&lt;/a&gt; it's $l = \sqrt{\Delta x^2+\Delta y^2+\Delta z^2-\Delta t^2}.$ For simplicity the speed of light has been set to $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This model of the universe as a whole can hold, if it actually originated (almost) as a (0-dimensional) singularity (a point); but our horizon is restricted to the observable part, so everything beyond is a theoretical model; other theoretical models could be defined in a way to be similar in the observable part of the universe, but different far beyond, see e.g. &lt;a href=&quot;http://arxiv.org/abs/1303.5086&quot; rel=&quot;noreferrer&quot;&gt;this Planck paper&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-04-16T22:44:54.337" LastActivityDate="2014-04-16T22:44:54.337" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="2173" PostTypeId="2" ParentId="2146" CreationDate="2014-04-16T02:27:51.240" Score="1" Body="&lt;p&gt;As Jeremy explained, these are the Lagrangian points (see his link to the Wikipedia article).  At these points, earth's gravity and the sun's gravity partially cancel each other to cause the orbital period of an object at that point to match the orbital period of the earth-sun system.&lt;/p&gt;&#xA;" OwnerUserId="128" LastEditorUserId="128" LastEditDate="2014-04-21T05:15:34.073" LastActivityDate="2014-04-21T05:15:34.073" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2174" PostTypeId="1" CreationDate="2014-04-16T02:52:57.287" Score="2" ViewCount="2986" Body="&lt;p&gt;It is said that absorption lines originate from regions higher in the photosphere where the gas is cooler. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gas should be absorbing photons, then re-emitting them... absorb, re-emit, absorb, re-emit...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is where are the photons produced by re-emission? Why are there absorption lines? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although photons could be scattered, photons at other places could be scattered to our line our sight. Surely the total number of photons does not drop the further out you go? So while photons originally headed in the direction of an observer are possibly absorbed and re-emitted in a direction away from the observer, surely this will be made up for by a photon being emitted from another location in the direction of the observer? Otherwise wouldn't there be a location somewhere else where a bunch of extra scattered photons can be observed? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we integrate all the photons at the lower surface of the photosphere A and integrated all the photons at the upper surface of the chromosphere C, surely there are the same number of photons? Otherwise, where have they gone? Where is the lost energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there less photons emitted from the surface C because some have been emitted back down into the Sun? But that has to reach an equilibrium, or there would be a cascading increase in the number of photons inside the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/R5Ev0.png&quot; alt=&quot;Diagram showing photosphere, chromosphere, and photons&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="67" LastEditDate="2014-04-21T08:21:25.543" LastActivityDate="2014-04-21T08:21:25.543" Title="The formation of absorption lines in solar spectrum" Tags="&lt;the-sun&gt;&lt;radiation&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2175" PostTypeId="2" ParentId="2174" CreationDate="2014-04-16T03:15:22.340" Score="2" Body="&lt;p&gt;In the upper photosphere, the electrons join with the ions to neutralize the atom. When a photon strikes this atom (the atom absorbs the photon), it promotes a electron from a low energy shell to an higher energy shell. The electron then falls back to the lower energy shell emitting a photon in a different direction. When you break the sunlight into wavelengths (by a prism), you will see dark bands where the light was absorbed by the atoms and emitted in a different direction. These bands correspond to the energy absorbed and released when an electron changes energy shells. &lt;a href=&quot;http://www.pbs.org/wgbh/nova/education/activities/3113_origins_01.html&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is the sun's light spread out to show the absorption bands. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: &lt;a href=&quot;http://www.solarnavigator.net/the_sun.htm&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a high resolution of the wavelengths from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apr 20: You have the wrong interpretation of Figure 16.9. At level A, there is about the same number of photons at all wavelengths coming at the observer on Earth. As the photons pass through the Chromosphere, the photons with the wavelengths of the atom's electron transition are absorbed and redirected away from the observer. The photons with different wavelengths of the atom's electron transition eventually arrive at the observer. This is why there are dark lines in the spectrum. It is because most of the photons are absorbed and redirected in a different direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The absorption lines are formed at a higher region since the photons are removed from the spectrum.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="1322" LastEditDate="2014-04-21T04:11:14.477" LastActivityDate="2014-04-21T04:11:14.477" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="2176" PostTypeId="1" CreationDate="2014-04-16T06:10:05.623" Score="15" ViewCount="1535" Body="&lt;p&gt;What will a lunar eclipse look like from moon? Will earth become a completely dark circle?&lt;/p&gt;&#xA;" OwnerUserId="1411" LastEditorUserId="67" LastEditDate="2014-04-20T23:15:47.867" LastActivityDate="2015-09-26T15:31:28.000" Title="What will a lunar eclipse look like from moon?" Tags="&lt;the-moon&gt;&lt;earth&gt;&lt;the-sun&gt;&lt;lunar-eclipse&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2177" PostTypeId="1" AcceptedAnswerId="2180" CreationDate="2014-04-16T06:11:26.877" Score="4" ViewCount="259" Body="&lt;p&gt;The Hubble space telescope performs incredibly detailed and stunning imaging of celestial bodies that are many, many light years away, and yet when it attempts to image objects within our own Solar System, such as of Pluto and Charon, the results appear out of focus and smeared. Why is this so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can only guess that the objects within the Solar System have a higher relative intrinsic motion than objects light years away, and thus Hubble can't track them so precisely? Or maybe Hubble's optics are &quot;fixed&quot; at an &quot;infinity point&quot; beyond &quot;nearby&quot; objects? Or...?&lt;/p&gt;&#xA;" OwnerUserId="1410" LastActivityDate="2016-05-15T18:29:34.527" Title="Inability of Hubble to clearly resolve nearby celestial objects" Tags="&lt;hubble-telescope&gt;&lt;angular-resolution&gt;&lt;photography&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2178" PostTypeId="2" ParentId="2176" CreationDate="2014-04-16T07:00:56.007" Score="10" Body="&lt;p&gt;It would look like a solar eclipse! What happens is that Earth gets in the way between the Sun and the Moon. From Earth we see the Moon disappear, from the Moon one would see the Sun disappear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth is fix on the Moon's sky, because the Moon always turns the same side towards Earth. Earth as seen from the Moon has phases, like the Moon has when seen from Earth. You would see the Earth crescent waning and becoming completely dark as the Sun moves towards it and disappears behind the then invisable Earth. While as seen from Earth, the Moon and the Sun have the same size on the sky, when seen from the Moon Earth will look almost 4 times larger (in diameter) than the Sun, so solar eclipses will last that much longer and be more common than on Earth.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2014-04-16T07:00:56.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2179" PostTypeId="2" ParentId="2176" CreationDate="2014-04-16T07:01:36.360" Score="3" Body="&lt;p&gt;Yeah, you'll see earth as all dark, if it is total lunar eclipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a Lunar eclipse &lt;strong&gt;earth comes in between the sun and the moon&lt;/strong&gt;. Because of this, the sun rays don't reach the surface of moon.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now answer to your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As earth's sunward side is blocking light the other side will be darker. As you are standing towards shadowed side of earth, you cannot see earth at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All above explanation is accurate in the case of a &lt;strong&gt;total lunar eclipse&lt;/strong&gt;.    &lt;/p&gt;&#xA;" OwnerUserId="1414" LastEditorUserId="67" LastEditDate="2014-04-20T23:10:29.257" LastActivityDate="2014-04-20T23:10:29.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2180" PostTypeId="2" ParentId="2177" CreationDate="2014-04-16T08:25:06.253" Score="10" Body="&lt;p&gt;This has to do with the angular resolution of the Hubble telescope and the ratio between the distance of an object in space and its size in space. The galaxies that the Hubble telescope can see are bigger in size than they are far in light years away compared to pluto from earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the galaxy NGC 5584 for example:&lt;br/&gt;&#xA;It spans 50,000 light-years and it's 72 million light-years away which gives us a ratio of 0.00069&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then for pluto:&lt;br/&gt;&#xA;Pluto spans 2400 km in size and 4675 million km away which gives the ratio of 0.00000051&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take for example a clear view at a mountain with trees on it. You would see the mountain fairly clearly: you can see it has trees on it, perhaps snow, and other aspects. But if you look at any given tree on that mountain you can't make anything other than it's observable color and that it's most likely a tree. The relationship of distance in size is what gives you a good view of the mountain but not a tree on the mountain; the trees size to distance ratio is much smaller than the mountain's size per distance ratio.&lt;/p&gt;&#xA;" OwnerUserId="1356" LastEditorUserId="11525" LastEditDate="2016-05-15T18:29:34.527" LastActivityDate="2016-05-15T18:29:34.527" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2181" PostTypeId="2" ParentId="2155" CreationDate="2014-04-16T08:41:10.140" Score="3" Body="&lt;p&gt;The heliocentric model is historically thought first by Aristarchus. He eventually gave up on the idea due to not being able to give evidence for a concept called stellar parallax: If earth orbits the sun then we should see the closer stars shift in relation to the other stars (strong evidence for this didn't come until Galileo).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Copernicus became to accept the heliocentric model given by Aristarchus due to his study of the Ptolemaic model. His studies showed that planetary motion in a geocentric model was inaccurate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;His model, though, was still based on the fact that &quot;heavenly objects&quot; moved in complete circles. He had to add much complexity to account for this fact which is why the heliocentric model was still not widely accepted. Side note, it was Johannes Kepler, over 100 years later, that claimed that planets move in ellipses rather than circles.&lt;/p&gt;&#xA;" OwnerUserId="1356" LastEditorUserId="1356" LastEditDate="2014-04-16T17:27:03.787" LastActivityDate="2014-04-16T17:27:03.787" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2183" PostTypeId="1" CreationDate="2014-04-16T13:31:06.440" Score="4" ViewCount="757" Body="&lt;p&gt;I am trying to build an astronomy app that will use the user's latitude and longitude, along with the current date and time, to create a 3D view of the celestial sphere as seen based on that data.  However, the information I've found so far has been very general and less-than-practical, and has included virtually nothing about how to take date and time into account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know of a more concrete source of information on sky mapping in this way?&lt;/p&gt;&#xA;" OwnerUserId="1420" LastActivityDate="2014-09-21T17:36:14.407" Title="Are there standard algorithms and procedures for creating unique sky maps based on latitude/longitude/date/time?" Tags="&lt;star-gazing&gt;&lt;algorithm&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2184" PostTypeId="2" ParentId="2132" CreationDate="2014-04-16T16:24:46.177" Score="-1" Body="&lt;p&gt;In my own mind the multiverse could consist of either multiple universes occupying the same &quot;space&quot; (as in different dimensions) and/or different areas of space where what we consider to be our own universe is only one &quot;bubble&quot;, outside of which are an infinite number of other bubbles, possibly created by their own Big Bangs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first variant above is what I would refer to as parallel universes, i.e. not the &quot;bubbles&quot;. Or to put it another way, The different universe instances in the multiverse &lt;em&gt;could&lt;/em&gt; be parallel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that my understanding of current hypothesis and theories in this area is very limited. These are just my own thoughts.&lt;/p&gt;&#xA;" OwnerUserId="1421" LastActivityDate="2014-04-16T16:24:46.177" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2185" PostTypeId="1" AcceptedAnswerId="2190" CreationDate="2014-04-17T03:20:37.953" Score="4" ViewCount="1787" Body="&lt;p&gt;Given the magnitudes (in the i-band) of certain galaxies, I would like to calculate their &lt;strong&gt;stellar mass (in terms of solar masses)&lt;/strong&gt;. So far, I have calculated their absolute magnitudes and gotten to working out the mass-light ratio $M/L$ for each galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. $M/L=0.563$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Values I have are the calculated $M/L$ for each galaxy, and the $i$-band apparent ($13.25$) and absolute ($-18.06$) magnitudes for the galaxy, as well as the distance ($18.44Mpc$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this I need to get the mass of the galaxy $M$ in terms of solar masses. Therefore I assume I first need to calculate the i-band luminosity for the galaxy in solar masses $L_i$. This is where I am stuck. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, once I have $L_i$  next step would be ...&#xA;$$M_g = 0.563 * L_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, given these values, how would I go about estimating the stellar mass of a galaxy in terms of solar masses?&lt;/p&gt;&#xA;" OwnerUserId="1428" LastEditorUserId="1428" LastEditDate="2014-04-18T14:32:50.480" LastActivityDate="2014-07-22T19:30:18.993" Title="Stellar mass of galaxies" Tags="&lt;mass&gt;&lt;luminosity&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2186" PostTypeId="2" ParentId="2174" CreationDate="2014-04-17T04:50:44.767" Score="2" Body="&lt;p&gt;All the photons being emitted do not 'fill in the blank' of the absorption line, because when they are emitted, most do not come towards us. You suppose that there will be some emitted from a different hydrogen atom elsewhere, that will just happen to aim a photon at us, but where will it come from? Think about how much of the sky the Sun subtends from our perspective. Now think about how much of the sky the Earth subtends from the perspective of an atom of hydrogen in the Sun's photosphere that is thinking about popping a photon out and wondering if it will head in the direction of Earth. Here is a 2D diagram - bear in mind that the atom actually is in 3D space and instead of being limited to only 360 degrees, can pop that photon off in any direction at all. &#xA;&lt;img src=&quot;https://i.stack.imgur.com/lGr71.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;So it doesn't matter where ANY of the hydrogen atoms are in the photosphere, they have next to no chance of actually hitting Earth with a photon that they emit. Even though there are a lot of them, they are ALL firing off photons in almost every direction EXCEPT towards Earth.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-04-17T04:50:44.767" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2187" PostTypeId="1" CreationDate="2014-04-17T06:57:07.040" Score="2" ViewCount="58" Body="&lt;p&gt;It seems as if this is necessary to assume in order for the CMB to make an sense. Otherwise, an instantaneous infinite Euclidean space model would work. The issue is, with any visualization of the Big Bang, let's take a Big Bang visualization in 3 - dimensions, the radiation emitted from the creation of hydrogen atoms would not necessarily be all around us, as it is. Our position in space would dictate what we would observe in terms of background radiation. &lt;/p&gt;&#xA;" OwnerUserId="1397" LastActivityDate="2014-04-17T10:20:23.763" Title="Assumption of our universe being the surface of a 4-D sphere in order to describe relative uniformity in the cosmic microwave background?" Tags="&lt;light&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2188" PostTypeId="2" ParentId="2187" CreationDate="2014-04-17T10:20:23.763" Score="1" Body="&lt;p&gt;To a high degree of accuracy, the cosmic background radiation we observe is homogeneous and isotropic... except for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_microwave_background#CMBR_dipole_anisotropy&quot; rel=&quot;nofollow&quot;&gt;CMBR dipole anisotropy&lt;/a&gt;, which is there because we are moving at $369\pm 0.9\;\mathrm{km/s}$ relative to the CMB rest frame, in which this anisotropy would vanish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relevantly, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Friedmann%E2%80%93Lema%C3%AEtre%E2%80%93Robertson%E2%80%93Walker_metric&quot; rel=&quot;nofollow&quot;&gt;Friedmann-Robertson-Walker family&lt;/a&gt; of solutions of general relativity used to model the Big Bang cosmology assume that the universe is, on the large scale, homogeneous and isotropic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It seems as if this is necessary to assume in order for the CMB to make an sense. ... Our position in space would dictate what we would observe in terms of background radiation. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's not the case. There are four distinct homogeneous and isotropic spatial geometries: the Euclidean space $\mathbf{E}^3$, the hyperbolic space $\mathbf{H}^3$, the real projective space $\mathbf{RP}^3$, and the sphere $S^3$. In any of those cases, the universe would look the same all around us and be independent of our position. Thus, the sphere is possible but not necessary. The first two of those cases correspond to the &quot;flat&quot; and &quot;open&quot; Big Bang cosmologies, respectively, and the last two are variations of the &quot;closed&quot; cosmology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if we don't require the universe to be completely homogeneous and isotropic, but merely look like it in the part of it we observe, then much more exotic geometries are also possible. With &lt;a href=&quot;http://en.wikipedia.org/wiki/Inflation_%28cosmology%29&quot; rel=&quot;nofollow&quot;&gt;cosmological inflation&lt;/a&gt;, virtually any large-scale geometry can be &quot;blown up&quot; to look flat, homogenous, and isotropic within our observable portion--even though it might not be s beyond the our cosmological horizon.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Otherwise, an instantaneous infinite Euclidean space model would work.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It does work, in the sense of being consistent with observations. In fact, in the standard &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambda-CDM_model&quot; rel=&quot;nofollow&quot;&gt;ΛCDM cosmological model&lt;/a&gt;, space is infinite and Euclidean.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-04-17T10:20:23.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2189" PostTypeId="2" ParentId="2176" CreationDate="2014-04-17T15:03:06.773" Score="3" Body="&lt;p&gt;During a lunar eclipse on Earth, the Sun would be eclipsed by the Earth as seen from the Moon:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{3cm}$&lt;img src=&quot;https://i.stack.imgur.com/F25tE.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this animation does not show is the corona around the Sun (which would be partially visible when the sun has been occulted) and the reddish ring of light that would encircle the Earth due to the sunlight refracted by the atmosphere of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike a solar eclipse as seen on Earth, the stars are always visible since there is virtually no atmosphere on the Moon. Of course, until the Sun is totally occulted, it would be hard to see stars close to the Sun and the solar corona due to the brightness of the Sun. Furthermore, the Earth is 4 times as large as the Moon and so would cover most of the corona (or possibly all of the corona if the eclipse is central).&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2014-04-17T15:03:06.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2190" PostTypeId="2" ParentId="2185" CreationDate="2014-04-17T15:05:35.707" Score="1" Body="&lt;p&gt;The relation between &lt;a href=&quot;http://en.wikipedia.org/wiki/Absolute_magnitude&quot; rel=&quot;nofollow&quot;&gt;absolute magnitude&lt;/a&gt; $M$ and luminosity $L$ for stars&#xA;$$\frac{L_{Star}}{L_{Sun}}=10^{(M_{Sun}-M_{Star})/2.5}$$&#xA;should also be appliable to i-band luminosities of galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ucolick.org/~cnaw/sun.html&quot; rel=&quot;nofollow&quot;&gt;Taking the absolute magnitude 4.08 of the sun on the I-band&lt;/a&gt; the luminosity of a galaxy with absolute magnitude −18.06 would be&#xA;$$\frac{L_{Galaxy}}{L_{Sun}}=10^{(4.08−(-18.06))/2.5}=0.7178\cdot 10^9.$$&#xA;An order of magnitude estimate for the mass of the galaxy would be&#xA;$0.563\cdot 0.7178\cdot 10^9=0.404\cdot 10^9$ solar masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $M/L$ isn't necessarily biased the same way for the sun to $M_i/L_i$ as for the galaxy.&#xA;Therefore you'll probably need to compare the spectrum of the sun with the spectrum of the galaxy to find out the ratios of the i-band fraction of the total emission. Absorption and extinction at different wavelength may be different, therefore determining kind of mean temperature of stars in the galaxy could help finding a more realistic estimate for the stellar mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similiar approach has been used in &lt;a href=&quot;http://www.ifa.hawaii.edu/reu/2007research/whiting.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-04-18T19:28:06.380" LastActivityDate="2014-04-18T19:28:06.380" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2191" PostTypeId="2" ParentId="2176" CreationDate="2014-04-17T17:14:29.370" Score="15" Body="&lt;p&gt;It would look like this (actual picture of the Earth, seen from the Moon, during a lunar eclipse):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pVTHh.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Link to full page: &lt;a href=&quot;http://apod.nasa.gov/apod/ap140407.html&quot; rel=&quot;noreferrer&quot;&gt;http://apod.nasa.gov/apod/ap140407.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth would appear surrounded by a bright ring, even though the Sun is completely hidden behind it. The ring is sunlight refracted through the atmosphere. It's basically all the sunrises and sunsets, seen at once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth would never appear fully dark. The bright ring around it will always be there. This is due to the atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2014-04-17T17:14:29.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2192" PostTypeId="2" ParentId="2166" CreationDate="2014-04-17T17:29:31.857" Score="8" Body="&lt;p&gt;No, not all total lunar eclipses will turn the Moon deep red. Most of them do, but not all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were standing on the Moon during the eclipse, you'd see the Earth passing in front of, and obscuring, the Sun. But the Earth will never become fully dark, even when the Sun is fully covered. A bright ring will always surround the Earth. Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That ring is sunlight refracted by the atmosphere. It's there because the Earth has an atmosphere. You could say it's all the sunrises and sunsets of the Earth, all seen at once. It's this light that continues to illumine the Moon during the eclipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why is the Moon red, instead of some other color? This is because the blue end of the spectrum is scattered more easily in all directions (same mechanism that explains why the sky is blue on Earth), whereas the red part of the spectrum is scattered less easily and moves on a straighter path along the refraction lines. The bright ring around the Earth, as seen from the Moon, is probably red, because most of the blue in it has been scattered away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the Earth's atmosphere is full of dust particles from huge volcanic eruptions, the bright ring is a lot weaker. That makes the Moon during the eclipse a much darker shade of red. Sometimes the Moon is a very dark, dull grey during the eclipse, no red hue at all - so dark in fact that it's hard to see in the sky while the eclipse is full. This has happened some decades ago after the Pinatubo eruption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lunar_eclipse#Appearance&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Lunar_eclipse#Appearance&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://eclipse.gsfc.nasa.gov/LEcat5/appearance.html&quot; rel=&quot;noreferrer&quot;&gt;http://eclipse.gsfc.nasa.gov/LEcat5/appearance.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A measure of the brightness and color of a lunar eclipse is the Danjon scale. Eclipses are rated between 0 (almost invisible, black or very dark grey) and 4 (bright orange with bluish rim).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Danjon_scale&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Danjon_scale&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://eclipse.gsfc.nasa.gov/OH/Danjon.html&quot; rel=&quot;noreferrer&quot;&gt;http://eclipse.gsfc.nasa.gov/OH/Danjon.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My estimate is that the last eclipse was a 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't find a list of recent eclipses rated on a Danjon scale, but here's a list of 20th century eclipses, with a measure of the magnitude of the umbra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://eclipse.gsfc.nasa.gov/LEcat5/LE-1999--1900.html&quot; rel=&quot;noreferrer&quot;&gt;http://eclipse.gsfc.nasa.gov/LEcat5/LE-1999--1900.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the magnitude varies quite a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bottom line&lt;/strong&gt;: each eclipse is a bit different. A majority would have some kind of orange, copper or red tint. A minority are too dark to see any color. Unusual colors (outside of the red-yellow interval) are very rarely possible. Brightness and hue varies from one instance to another, since it depends on Earth's atmosphere, which is a system that changes greatly over time.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2014-04-18T20:04:21.213" LastActivityDate="2014-04-18T20:04:21.213" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2193" PostTypeId="1" AcceptedAnswerId="2198" CreationDate="2014-04-17T23:25:19.037" Score="15" ViewCount="19914" Body="&lt;p&gt;So, I am a 15 year old interested in astronomy. My father used to have a reflector telescope, but was never so serious about it that he could recommend me a good telescope. I am looking to buy a telescope up to £400 (possibly £500) and would like one that is mainly great for looking at nebulae. Stars and Planets would be great also, but are not necessary. Light Pollution isn't too bad. I have been recommended a Newtonian Reflector, but am not too sure if this would be suitable for my conditions. Also, does anyone recommend a good, British website where I could buy such a scope from? Finally, is there much point in me buying a computerized scope, will this allow me to align the scope better, or is it just going to be a waste of money.&lt;/p&gt;&#xA;" OwnerUserId="1393" LastEditorUserId="2365" LastEditDate="2017-05-01T14:31:14.680" LastActivityDate="2018-02-09T10:18:17.310" Title="A good telescope for the viewing of Nebulae, Stars and Planets" Tags="&lt;telescope&gt;&lt;newtonian-telescope&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="2194" PostTypeId="1" AcceptedAnswerId="2195" CreationDate="2014-04-18T11:52:35.153" Score="6" ViewCount="136" Body="&lt;p&gt;How long did it take after the Big Bang for the average density of the universe to reach one Earth atmosphere at sea level? How about the density of water? Is there a chart of densities that relate to everyday objects (so easier to understand)?&lt;/p&gt;&#xA;" OwnerUserId="1435" LastActivityDate="2014-04-18T21:13:49.010" Title="What was the age of the universe when the average density was one atmosphere?" Tags="&lt;cosmological-inflation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2195" PostTypeId="2" ParentId="2194" CreationDate="2014-04-18T21:13:49.010" Score="5" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Critical_density_%28cosmology%29#Density_parameter&quot;&gt;critical density of the universe&lt;/a&gt; is &#xA;$$\rho_c=\frac{3H_0^2}{8\pi G},$$&#xA;with $H_0=67.8 \frac{\mbox{km}/\mbox{s}}{\mbox{Mpc}}$ the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble%27s_law&quot;&gt;Hubble constant&lt;/a&gt; and $G=6.673848\cdot 10^{-11}\frac{\mbox{m}^3}{\mbox{kg }\mbox{s}^2}$ Newton's &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_constant&quot;&gt;gravitational constant&lt;/a&gt;.&#xA;Hence with one &lt;a href=&quot;http://en.wikipedia.org/wiki/Parsec&quot;&gt;parsec&lt;/a&gt; $1\mbox{pc}=3.0857×10^{16} \mbox{m}$,&#xA;$$\rho_c=\frac{3\cdot (67.8\cdot 10^3\cdot \frac{\mbox{m}/\mbox{s}}{3.0857×10^{22} \mbox{m}})^2}{8\pi\cdot 6.673848\cdot 10^{-11}\frac{\mbox{m}^3}{\mbox{kg}\mbox{s}^2}}=8.635 \cdot 10^{-27}\frac{\mbox{kg}}{\mbox{m}^3}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ordinary matter content is $4.9\%$, the universe has total density close to critical density, hence ordinary matter density is&#xA;$$0.049\cdot 8.635 \cdot 10^{-27}\frac{\mbox{kg}}{\mbox{m}^3}=4.23 \cdot 10^{-28}\frac{\mbox{kg}}{\mbox{m}^3}.$$&#xA;(The Hubble constant is not precisely known, so the calculated density depends a bit on the precise value.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Density of water:&lt;/strong&gt;&#xA;The density of water is about $10^3 \mbox{kg}/\mbox{m}^3$.&#xA;That's a factor $4.23 \cdot 10^{31}$ denser than the average universe.&#xA;The space needs to have been reduced by a factor $\sqrt[3]{4.23 \cdot 10^{31}}=3.48\cdot 10^{10}$ in each if the three spatial dimensions. That's the inverse of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Scale_factor_%28cosmology%29&quot;&gt;scale factor&lt;/a&gt;, hence corresponds to a redshift of $z=3.48\cdot 10^{10}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the &lt;a href=&quot;http://adsabs.harvard.edu/abs/2006PASP..118.1711W&quot;&gt;Cosmology Calculator&lt;/a&gt; on &lt;a href=&quot;http://www.astro.ucla.edu/~wright/CosmoCalc.html&quot;&gt;this website&lt;/a&gt;, the cosmological parameters $H_0 = 67.11$ km/s/Mpc (slightly different to the above value),&#xA;$\Omega_{\Lambda} = 0.6825$ provided by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck_%28spacecraft%29&quot;&gt;Planck&lt;/a&gt; project, and setting $\Omega_M = 1- \Omega_{\Lambda} = 0.3175$ the age of the universe was &lt;strong&gt;0.022 seconds&lt;/strong&gt; at the redshift $z=3.48\cdot 10^{10}$, hence when it was of the average density of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Density of air at sea level:&lt;/strong&gt;&#xA;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Density_of_air&quot;&gt;density of air&lt;/a&gt; at sea level is $1.225 \frac{\mbox{kg}}{\mbox{m}^3}$. The corresponding redshift is about $z=1.423\cdot 10^9$ yielding an age of the universe of &lt;strong&gt;12.68 seconds&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-04-18T21:13:49.010" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2196" PostTypeId="1" AcceptedAnswerId="2199" CreationDate="2014-04-18T22:44:06.317" Score="4" ViewCount="5342" Body="&lt;p&gt;So I recently got into astronomy and I am using my fathers old Meade EQ1-B reflector telescope. This is my first night using it and I have come across a few questions. To begin, the telescope was left alone for several years (the scope is a fairly old model) and may have become damaged over time, yet nothing too severe. We were cleaning it and got to the mirror to find that over the years the &quot;mirror&quot; cover on top of the glass had chipped away all over, giving the result of a patchy look. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We were testing the scope out during the day and the image was fine - the mirror didn't affect it. Now, I am looking upon stars and have noticed a couple of things. When viewing these stars the image sort of moves, not the scope, the image. As if it were an animation, the star isn't just a white blob, it looks like the sun, where the blob is moving - as if it were a liquid. No I am not stupid enough to point the scope at the sun... during night time. The stars just move like a liquid in the center, is this an effect of the mirror? I can't see why it would be as the image was fine during daylight. Is this how all stars work? I haven't had a chance to point it at the moon or any other planets so I can't test it on an object that isn't a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as the scope is a reflector it has the &quot;three lines&quot; on the front, however this interferes with the object I am looking at as i can see the lines on the star. Is there any way to stop this, or is it just the result of using a reflector?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image below:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/EG8re.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1393" LastEditorUserId="1437" LastEditDate="2014-04-20T21:32:35.257" LastActivityDate="2014-04-20T21:32:35.257" Title="First night on a telescope questions" Tags="&lt;star&gt;&lt;telescope&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2197" PostTypeId="2" ParentId="2196" CreationDate="2014-04-18T22:59:30.973" Score="1" Body="&lt;p&gt;When you say 'move ' do you mean it travels across the view, this will be the earth rotating. Or is it sparkling, this will be the interference from earths atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="953" LastActivityDate="2014-04-18T22:59:30.973" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2198" PostTypeId="2" ParentId="2193" CreationDate="2014-04-19T00:28:25.587" Score="20" Body="&lt;p&gt;This is a very common question, yet very hard to answer if you prefer a clear, concise, uncontroversial answer that applies to all situations. So I'm not going to do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, I'm going to describe your main options, and let you choose. Be aware that you'll make the choice while still not knowing much about optics. So, in a sense, it will be just the beginning of a learning journey, and you'll have to keep learning and trying things out. Hope you like that sort of thing, because it's kind of mandatory in this hobby.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there are two kinds of people in this field: those who enjoy &lt;em&gt;using&lt;/em&gt; the instruments, and those who enjoy &lt;em&gt;making&lt;/em&gt; the instruments. I'm 25% of the former, 75% of the latter kind (I make telescopes and mirrors, and I do some optics design). So that's the perspective you'll get.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;First off, a few words about the objects you'll observe. There are two kinds of objects:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HIGH RESOLUTION, HIGH BRIGHTNESS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the planets, the Moon, many double stars, and the like. These are typically pretty bright, so the amount of light you capture is not essential. But these objects show high-resolution detail, so the resolving power of your scope is crucial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You get more resolving power if the aperture (the diameter) of the scope is bigger. But there's a problem - seeing (the inverse of turbulence).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Astronomical_seeing&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Astronomical_seeing&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When air is turbulent, it will reduce the resolving power of your instrument. If your scope is 200 mm (8&quot;) or bigger, this will happen A LOT. If your instrument is 100 mm (4&quot;) or smaller, it's not so noticeable because you don't have that much resolving power to begin with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are ways to forecast seeing, because it depends on weather:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cleardarksky.com/csk/&quot; rel=&quot;noreferrer&quot;&gt;http://cleardarksky.com/csk/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for high resolution targets, the point of having a large instrument (like a huge dobsonian) is because once in a while seeing is so good (turbulence is so low) that you can actually use all its huge power - but most of the time you'll be using it at much less than its theoretical power, as if it's a 4&quot; instrument or whatever the seeing is at the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These objects are not affected by light pollution. You could observe the Moon from a brightly illumined parking lot in the middle of the city. You don't need to be in deep darkness while observing them - in fact, it's beneficial if you're not adapted to very deep darkness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LOW BRIGHTNESS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are most of the nebulae, star clusters, and galaxies. These are distant, faint objects. Their light is very faint, so it's essential you capture as much of it as possible. Again, bigger aperture is better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the kind of objects that huge dobsonians (half-meter aperture, or more) are made for. In a very large scope, there's galaxies everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But again, there's a catch. These objects are affected by light pollution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jshine.net/astronomy/dark_sky/&quot; rel=&quot;noreferrer&quot;&gt;http://www.jshine.net/astronomy/dark_sky/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to take your scope far and away from any city, town and industrial complex (at least a 1 hour trip at highway speed). This is pretty hard to do in Europe; in the US, population density is low and this is somewhat easier. Anyway, just stay away from city lights and the view will improve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few of these objects will be visible from the city: the Andromeda galaxy, the M13 cluster, the Orion nebula, etc. But many of them look much, much better away from light polution. In any case, you should be able to observe the whole Messier catalogue (over 100 objects) from the city - but out in the boondocks these things look better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_Messier_objects&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/List_of_Messier_objects&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important that your eyes are adapted to very deep darkness when observing these things, so stay away from any source of light. Even the screen of your phone is far too bright. Street lights 500 meters away are too bright. And turn off that flashlight and keep it off. One second of exposure to a bright light destroys a dark adaptation that took 20 minutes to build.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since these objects are so faint, the eye doesn't see a whole lot of high-resolution detail, so they are not affected by seeing (as far as the human eye is concerned).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Okay, now here are the main types of instruments that you're most likely to encounter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NEWTONIAN REFLECTORS IN GENERAL (DOBSONIANS IN PARTICULAR)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many people will say &quot;just get an 8 inch dob as your first scope&quot;, and that's usually correct. This design gives the biggest aperture per price. Biggest bang for the buck. It scales up easily to very, very large sizes. It's the ideal instrument for observing the &quot;faint fuzzies&quot; (nebulae, galaxies). It's great for learning, because you can easily observe its guts. When seeing is good, a large dob will open up the sky for you - you can crank up the magnification a lot (depending on aperture).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the flip side, due to the low price, you will see many dobs out there that are shoddily made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to remember two things with a dob: collimation, and thermal management.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Collimation&lt;/strong&gt; means keeping all the optics aligned. The mirrors will move a little, due to the way a dob is built, and they'll lose collimation. So don't forget to re-adjust it once in a while. In time, you'll get so good at it, you'll give it a two-minute quick adjustment every time you observe, and it'll be good to go; that's what I do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why collimation is important:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://legault.perso.sfr.fr/collim.html&quot; rel=&quot;noreferrer&quot;&gt;http://legault.perso.sfr.fr/collim.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a good introduction to newtonian collimation (which is a vast topic, you could write a whole book on it):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cloudynights.com/documents/primer.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://www.cloudynights.com/documents/primer.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thermal management&lt;/strong&gt; means keeping the scope at the same temperature as the ambient air. If you don't do this, internal convection will destroy the image. This is why:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-1/&quot; rel=&quot;noreferrer&quot;&gt;http://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-1/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-2/&quot; rel=&quot;noreferrer&quot;&gt;http://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-2/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small dobs (up to 6...8&quot;), just take it outside at least 1 hour before you observe - or better yet, 2 hours. Let it &quot;breathe&quot;. Dobs 10&quot; or larger, you MUST use a fan on the primary mirror to force cooling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REFRACTORS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These instruments are very convenient and very low maintenance. Take it out of the box, and it's good to go. It's collimated from factory, and it doesn't care about thermal issues. A refractor makes a great &quot;travelscope&quot;, and a great grab-and-go scope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the flip side, they are the most expensive, in terms of how much aperture you get for the money you spend. And they are typically made in small apertures; most commercial refractors are not bigger than 6&quot; or so; those that are bigger are extremely expensive. The small aperture limits the maximum performance of these instruments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is this myth out there that refractors are somehow magically better for planets (and other high-resolution targets), whereas dobs, even very large ones, are not so great. This is false; load up the designs in Zemax, do a basic performance analysis, and you'll see that aperture is the dominant factor, with everything else being secondary. But there are several reasons why this myth is around, and it's instructive to see what these reasons are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;refractors perform at 100% of their potential no matter what, right out of the box. Dobs require maintenance and fiddling to perform at 100%.&lt;/li&gt;&#xA;&lt;li&gt;many dob owners simply don't do the required maintenance - they don't collimate their scope, and don't give it time to reach thermal equilibrium. Then of course it will perform miserably on planets. There are 12&quot; dobs out there that perform worse than 4&quot; refractors because of how out of collimation they are and how hot the mirror is.&lt;/li&gt;&#xA;&lt;li&gt;many refractors out there are quite well made (they ought to be, given the higher price you're paying), whereas many dobs are cheaply made in a sweatshop. So you're not comparing apples to apples.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Bottom line: don't buy a refractor because you &quot;want to watch planets&quot;. But do buy a refractor if simplicity of use and the low maintenance are very important to you (and you don't care much about price).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A decent refractor on an excellent mount can do pretty good astrophotography. More on that below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;CATADIOPTRICS (CASSEGRAIN, MAKSUTOV, DALL-KIRKHAM, RITCHIE-CHRETIEN)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Broadly speaking, these are for astrophotography (not always, but usually). But can you do visual astronomy with them? Sure. It's just that it's easier to optimize them for photo, with this design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, they come installed on a motorized mount. All but the very cheapest ones have a computer on-board; you just punch the name of the object into the remote control, and the scope will turn around automatically to face the object. Sounds cool, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, I am going to say something a little controversial here: this is not a good place to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, astrophotography has an extremely steep learning curve. It's very easy to take a little blurry photo of Jupiter that will impress nobody. It's very hard to take a great, high resolution photo like these:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.acquerra.com.au/astro/&quot; rel=&quot;noreferrer&quot;&gt;http://www.acquerra.com.au/astro/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The skills and knowledge required to take that kind of photos are more easily acquired by doing purely visual astronomy, on a dob or whatever, for a couple years or more. Also, the equipment required for that sort of stunt costs many thousands of dollars - your cheap little Cassegrain cannot do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why buy a small Cass? Well, if you have an overwhelming interest in astrophoto, and are prepared to deal with the difficult learning, then go ahead. But realize that the first images are not going to be very impressive, and the motorized mount will not teach you some valuable skills that you would acquire more easily pushing a dob around and keeping your eye plugged into the ocular for a while. Also, money spent on the motorized mount is money NOT spent on aperture, and the aperture is so important for overall performance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For astrophotography, the mount is of huge importance; the scope per se is relatively less important (I'm generalizing). A so-so scope on a great mount is much better than the other way around. But a great GEM mount can be very expensive.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Well, that's it. Do some more research on your own, then make a decision. Don't forget that learning only begins with the moment you acquire the scope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above all, have fun, and keep looking up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clear skies to you!&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2017-06-08T18:18:38.557" LastActivityDate="2017-06-08T18:18:38.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2199" PostTypeId="2" ParentId="2196" CreationDate="2014-04-19T01:05:54.623" Score="13" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;the &quot;mirror&quot; cover on top of the glass had chipped away all over,&#xA;  giving the result of a patchy look.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Optics can take a huge amount of damage (chipping, scratching) before it really starts to affect performance. You'll be surprised at the amount of abuse a telescope can take like that. As long as it's not properly cracked in two, don't lose any sleep over it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The stars just move like a liquid in the center&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The moving around is normal. It is due to atmospheric turbulence, which happens all the time. Unless you're in a vacuum, the images will always shimmer a little. Sometimes they will shimmer a lot. Other times not so much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In astronomy we have a word for this - it's called seeing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Astronomical_seeing&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Astronomical_seeing&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, as the scope is a reflector it has the &quot;three lines&quot; on the&#xA;  front, however this interferes with the object I am looking at as i&#xA;  can see the lines on the star.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are way, WAY, &lt;strong&gt;WAY&lt;/strong&gt; out of focus. The scope might also be completely miscollimated. The star must not appear like a big blob, with the &quot;lines&quot; and the image of the secondary mirror visible in the middle, like in your photo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;lines&quot; are called spider vanes - they are parts of the assembly called spider, which holds the secondary mirror in place. The vanes, or the secondary mirror itself, must never be visible in the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, do a rough, purely visual collimation. Do this in a place with plenty of light (inside the house). Take out the eyepiece. Look into the focuser. You'll see the secondary mirror. Reflected in it, you'll see the primary mirror from the bottom of the tube. Reflected again in the primary, you'll see the focuser, with your eye in the middle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qqXkR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If things are more or less centered, like in the third diagram above, leave it alone. If they are way out of center, like in the first diagram, use the adjustment screws on the mirrors and center everything. Collimation is very important to performance. A miscollimated scope will perform miserably. This is FAR more important than all the scratches on the mirror.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, this is purely visual, no-tools collimation. This is all you need for now. You can get a laser collimator later, after all your problems are solved. This is the article where I copied that image from, if you need more info:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.skyandtelescope.com/astronomy-resources/how-to-collimate-your-newtonian-reflector/&quot; rel=&quot;noreferrer&quot;&gt;http://www.skyandtelescope.com/astronomy-resources/how-to-collimate-your-newtonian-reflector/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also see this document:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.etnaastros.com/documents/No_Tools_Collimation.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://www.etnaastros.com/documents/No_Tools_Collimation.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once collimation is done, you need to find true focus. Go outside at night. Point the scope at a medium-bright star. Polaris, &lt;a href=&quot;http://www.wikihow.com/Find-the-North-Star&quot; rel=&quot;noreferrer&quot;&gt;if you can find it&lt;/a&gt;, is great because it doesn't move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now moving the focuser back and forth, try and determine which way you can make the image of the star appear smaller. Now keep adjusting the focuser until the image of the star is super tiny, then it starts to grow again. Then back off until you make it as tiny as possible. That's when it's in perfect focus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now point the scope at a very bright star. If your scope has 3 spider vanes on the secondary mirror, you may see 6 spikes coming out of the star. Like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NktTr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your scope has 4 vanes, then you should see 4 spikes, like this (it's really 8 spikes, 2 times the number of vanes, but in this case they are superimposed 2 by 2):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1g8vB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diffraction spikes are easily visible on very bright stars. On medium-size and small stars, they are not easily visible, or not visible at all. This is normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless the stars are super-tiny dots like in the images above, your scope is not ready for use. Do your best to collimate it and find true focus. Make the stars super-tiny.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck and let me know if you need more advice. It would help a lot if you could find an experienced astronomer nearby to help you fix the scope.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2014-04-19T01:17:17.447" LastActivityDate="2014-04-19T01:17:17.447" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2200" PostTypeId="2" ParentId="2196" CreationDate="2014-04-19T09:27:57.843" Score="2" Body="&lt;p&gt;Florin's answer covers most of what I would have said, except one problem which I have faced and I think deserves mention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The telescope assembly might vibrate&lt;/strong&gt;. You don't need a gale for that to happen. Depending upon where you place it, even a slight breeze might make it vibrate. The solution is to &lt;strong&gt;fix your mount on a terrain which is flat for at least a 100 sq m area around it&lt;/strong&gt;. It does seem cool in pictures, and perhaps movies to fix it on a high mound but a height gradient makes wind currents(local, of course) to turn an twist which causes the telescope to shake. I am sorry I couldn't find an image to support it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. This information came in the manual of my telescope which is a Skywatcher 8&quot; Dobsonian&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2014-04-19T09:27:57.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2201" PostTypeId="1" CreationDate="2014-04-19T20:48:52.887" Score="1" ViewCount="92" Body="&lt;p&gt;Since we are spiraling around the black whole will we spiral into the black hole?&lt;/p&gt;&#xA;" OwnerUserId="1442" LastEditorUserId="1322" LastEditDate="2014-04-20T22:12:21.247" LastActivityDate="2014-04-20T22:12:21.247" Title="Will we ever end up in the black hole?" Tags="&lt;galactic-dynamics&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-04-29T13:48:39.677" ContentLicense="CC BY-SA 3.0" />
  <row Id="2202" PostTypeId="2" ParentId="2201" CreationDate="2014-04-19T20:55:11.797" Score="0" Body="&lt;p&gt;It is very difficult to predict since the Milky Way galaxy and the Andromeda galaxy are going to collide in about 5 billion years. Since we cannot predict the gravity influences so far in advance, it is impossible to say. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 3 outcomes of this collision. The solar system falls into the black hole, the solar system goes into orbit in the new galaxy, or the solar system get propelled away from the new galaxy.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-04-19T20:55:11.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2203" PostTypeId="1" AcceptedAnswerId="2207" CreationDate="2014-04-19T22:30:48.980" Score="2" ViewCount="1759" Body="&lt;p&gt;I noticed that East and West appear swapped on maps of the Sun, even though the Sun rotates around its polar axis in the same sense as Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See for example below Sun disk with a Venus transit diagram or &lt;a href=&quot;http://www.raben.com/maps&quot; rel=&quot;nofollow noreferrer&quot;&gt;also here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, at least for the Sun, East and West are defined from an Earth observer viewpoint. This makes me wonder: How are East and West defined on other bodies of our solar system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hfUHM.png&quot; alt=&quot;solar map&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1443" LastActivityDate="2014-04-20T05:08:08.027" Title="How are East and West defined on other bodies of our solar system?" Tags="&lt;solar-system&gt;&lt;the-sun&gt;&lt;rotation&gt;&lt;coordinate&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2204" PostTypeId="1" CreationDate="2014-04-19T22:48:36.727" Score="4" ViewCount="1686" Body="&lt;p&gt;I'm 13 - I'd like to know how we found/discovered important details about the Universe. I know we have found many different things about the Universe, but when did we find how amazingly large the Universe is?&lt;/p&gt;&#xA;" OwnerUserId="1442" LastEditorUserId="67" LastEditDate="2014-04-20T23:00:28.330" LastActivityDate="2014-04-20T23:00:28.330" Title="How and when did we learn about the universe?" Tags="&lt;universe&gt;&lt;space&gt;&lt;fundamental-astronomy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2205" PostTypeId="2" ParentId="2203" CreationDate="2014-04-20T01:18:34.593" Score="0" Body="&lt;p&gt;Since the direction of revolution of all the planets is the same, the direction of angular velocity vector on the planet is taken to be North, and the direction of revolution is said to be east.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like Earth.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2014-04-20T01:18:34.593" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2206" PostTypeId="2" ParentId="2174" CreationDate="2014-04-20T01:35:02.233" Score="1" Body="&lt;p&gt;Emission and Absorption lines are due to &lt;a href=&quot;http://astronomy.nju.edu.cn/~lixd/GA/AT4/AT404/HTML/AT40401.htm&quot; rel=&quot;nofollow&quot;&gt;Kirchhoff's Laws&lt;/a&gt; that explain what happens when light passes through a gas. This is due to &lt;a href=&quot;http://astronomy.nju.edu.cn/~lixd/GA/AT4/AT404/HTML/AT40402.htm&quot; rel=&quot;nofollow&quot;&gt;photons being absorbed by an atom or ion causing an electron to jump to a higher energy state, and photons being emitted when an electron returns to a lower energy state&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photons will only be absorbed by an atom or ion if they are of precisely the energy required to bump an electron up to an excited state. If they are not of the right energy, they pass unimpeded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photons of particular wavelengths that are not absorbed by ions in the photosphere come from deep within the photosphere where it is hotter, and travel straight through unimpeded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photons of the right energy to be absorbed by a gas present in the photosphere, say hydrogen seeing as you have mentioned h-alpha, have a high probability of being absorbed by an atom on the way through the photosphere, and then being re-emitted in a random direction (including back down!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photons that have a high probability of being absorbed and re-emitted are therefore more likely to be coming from the top of the photosphere, or the lower chromosphere, where it happens to be cooler. (During the travel from the lower to upper photosphere, they will have been absorbed and re-emitted).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the Stephan-Boltzmann Law for black bodies, the energy radiated per unit of surface area is related to the temperature of the black body. A black body that is cooler will be less bright than a black body that is hotter. So when photons are radiated from a cooler part of the photosphere, &lt;a href=&quot;http://astronomy.nju.edu.cn/~lixd/GA/AT4/AT416/HTML/AT41603.htm&quot; rel=&quot;nofollow&quot;&gt;they will appear as dips in the spectrum&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-04-20T01:35:02.233" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2207" PostTypeId="2" ParentId="2203" CreationDate="2014-04-20T02:06:09.933" Score="3" Body="&lt;p&gt;Actually, it is a bit contentious. The IAU defined the North Pole of a planet to be &lt;a href=&quot;http://link.springer.com/article/10.1007/BF00695716&quot; rel=&quot;nofollow&quot;&gt;that axis of rotation which lies north of the solar system's invariable plane&lt;/a&gt;. Which means that it is possible for bodies to rotate retrograde, and Venus and Uranus are considered to rotate retrograde with this definition. This definition is not universally accepted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From early periods when solar maps were drawn, they have been labelled such that &lt;a href=&quot;http://soho.nascom.nasa.gov/classroom/notsofaq.html#SPIN&quot; rel=&quot;nofollow&quot;&gt;features appear on the &quot;Eastern&quot; limb of the sun and move to the &quot;Western&quot; limb&lt;/a&gt;, though this means that they are moving from Earth's East to Earth's West. This convention hasn't changed, &lt;a href=&quot;https://the-moon.wikispaces.com/IAU+directions&quot; rel=&quot;nofollow&quot;&gt;though things have changed for the moon, just to confuse things.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="67" LastEditorUserId="67" LastEditDate="2014-04-20T05:08:08.027" LastActivityDate="2014-04-20T05:08:08.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2208" PostTypeId="2" ParentId="2204" CreationDate="2014-04-20T02:22:42.470" Score="6" Body="&lt;p&gt;Throughout man's history, we have been learning things about the universe we live in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably one of the most interesting discoveries that let us know how amazingly large the universe really is was not really very long ago, only in the 1920s when Edwin Hubble demonstrated that the Milky Way was not in fact the total sum of the universe, but just a galaxy, and there were other galaxies beyond our own, much further away than any of the stars that we see in the night sky with our naked eyes. This fundamentally changed our perspective of our place in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have subsequently made many more significant discoveries about the nature of the universe, and many of those are covered in other questions already. If there are specific details you would like to know about, raise a specific question. The whole topic cannot be broached in one answer.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-04-20T02:22:42.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2209" PostTypeId="2" ParentId="2133" CreationDate="2014-04-20T04:36:18.120" Score="2" Body="&lt;p&gt;You probably will have difficulty imaging Ly$\alpha$ from a z=1.7 quasar at 330nm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Says P. Veron: &lt;a href=&quot;http://adsabs.harvard.edu/full/1986A&amp;amp;A...170...37V&quot; rel=&quot;nofollow&quot;&gt;The lower limit [z=1.8] is set by the ultraviolet transmission of the atmosphere... between 3300 and 3600Å&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-04-20T04:36:18.120" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2210" PostTypeId="2" ParentId="2204" CreationDate="2014-04-20T05:20:31.877" Score="4" Body="&lt;p&gt;You're asking a very big question - basically, &quot;what is the history of astronomy?&quot; It's a very long story. Our knowledge about the size of the Universe grew slowly, bit by bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ancient Greeks knew that Earth is round, like a sphere, and measured its size with surprisingly good precision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Eratosthenes&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Eratosthenes&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 1600s, with the invention of the telescope, and the idea that the Earth revolves around the Sun, scientists (Johannes Kepler and others) figured out how measure the distance from Earth to Sun. After that, it was easy to measure distances within the Solar System.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.tbp.org/pubs/Features/Su04Bell.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.tbp.org/pubs/Features/Su04Bell.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 1800s, we made the next step: measure distances to other stars nearby. This is something called parallax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_parallax&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Stellar_parallax&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 1910s, we figured out how to measure not just the distance to nearby stars, but the whole size of our galaxy. This was done by measuring variable stars (stars that go brighter, then dimmer, then brighter again, and so on).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.aip.org/history/cosmology/ideas/island.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.aip.org/history/cosmology/ideas/island.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, in the 1920s, Hubble figured how to measure the distance to other galaxies - that has been answered already on this page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Edwin_Hubble#The_universe_goes_beyond_the_Milky_Way_galaxy&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Edwin_Hubble#The_universe_goes_beyond_the_Milky_Way_galaxy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Different methods are used to measure different distances. Each method measures up to a certain distance; after that, a different method must be used. This is called the Cosmic Distance Ladder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_distance_ladder&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Cosmic_distance_ladder&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, there's a lot to talk about on this topic. You may want to just grab a book on the history of astronomy and read. What you're asking is basically what's the whole history of astronomy, after all. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/History_of_astronomy&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/History_of_astronomy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a book that, even though pretty big, should be quite accessible to someone your age:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1554075378&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/The-History-Astronomy-Heather-Couper/dp/1554075378/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2014-04-20T05:25:53.977" LastActivityDate="2014-04-20T05:25:53.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2211" PostTypeId="1" AcceptedAnswerId="2322" CreationDate="2014-04-20T11:29:47.053" Score="3" ViewCount="1389" Body="&lt;p&gt;I am a beginner interested in astronomy. I bought a Celestron AstroMaster 130EQ telescope. It's a Newtonian Reflector and it's not computerised. I did manage to see the Moon in some magnifications but my real question is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I want to see Proxima Centauri and I know it's coordinates, like Right ascension is 14h 29m 43.0s and Declination is -62° 40' 46&quot;. Having three analogue dials on my telescope (&lt;a href=&quot;https://www.dropbox.com/s/6bk213cggel5x25/20140420_111105000_iOS.jpg&quot; rel=&quot;nofollow&quot;&gt;picture from the front&lt;/a&gt;) (&lt;a href=&quot;https://www.dropbox.com/s/nrrgik4pa1ia3x8/20140420_111123000_iOS.jpg&quot; rel=&quot;nofollow&quot;&gt;zoomed in&lt;/a&gt;), how do I adjust my telescope with the help of the dials with the coordinates so I can view them from my telescope? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am wrong that these are the dials I have to use to adjust my telescope to the coordinates then please explain how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this question will also be useful to other amateur astronomers. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1449" LastEditorUserId="67" LastEditDate="2014-04-20T23:06:27.143" LastActivityDate="2014-04-30T22:25:47.327" Title="How to use a telescope to find a specific celestial body?" Tags="&lt;newtonian-telescope&gt;&lt;star-gazing&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2212" PostTypeId="1" AcceptedAnswerId="2216" CreationDate="2014-04-20T16:28:52.830" Score="8" ViewCount="886" Body="&lt;p&gt;I am not very much interested in observational physics but I just like to look at the night sky. It gives me a lot of inspiration. So, i was thinking about making a telescope for basic observations. I googled about the design and found two models. One consisted of two convex lenses of different focal lengths and the other one consisted of a concave lens as the objective lens and a convex lens as the eye-piece(that's called a Galilean one I guess). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am confused about which one should I make. Can anyone give me some insight into both these types?&lt;/p&gt;&#xA;" OwnerUserId="1156" LastEditorUserId="37191" LastEditDate="2021-05-28T09:13:34.443" LastActivityDate="2021-05-28T09:13:34.443" Title="Which type of telescope should I make?" Tags="&lt;telescope&gt;&lt;telescope-lens&gt;&lt;diy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2213" PostTypeId="1" AcceptedAnswerId="2214" CreationDate="2014-04-20T18:31:44.397" Score="5" ViewCount="261" Body="&lt;p&gt;To be honest my knowledge about astronomy and related physics is close to zero. However, since I studied math a bit, I would like to ask a couple maybe stupid questions, which I am curious about and I don't know who and where to ask.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;A) Projection of life&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Premises:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Carl Sagan explained in his &quot;flatlanders&quot; story, we can imagine high-dimension space by making projections into space with lower dimension. &#xA;As we learned at elementary school time is the fourth dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it right that what we see/feel in every moment is projection of some 4D reality into our 3D world noticed by our senses. &#xA;For example, If my eyes were able to &quot;somehow&quot; see in 4D and I pointed my eyes to a person, would I see that person in all stages of his/her life from birth to death as a single &quot;entity&quot;? In every second we are only a projection of our 4D representation to our 3D world? &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;B) Parallel realities&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Premises:&lt;/strong&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we imagine 1D space as a line without ends, 2D space as an unbounded plane and 3D space as an unbounded space, then I could put infinity number of parallel lines into 2D space and infinity number of parallel planes to the 3D space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read that there are at least 10 dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I put infinite number of 3D spaces into a single 4D space? If yes, then if we for example chose geographic coordinates of Earth-Italy-Caesena as place and 49 B.C. as time coordinate, we would be probably close to event when Caesar screamed his famous: &quot;alea iacta est&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we considered that 5th dimension exists and that we could put infinite number of 4D spaces into 5D space, then we would end up with infinite number of parallel realities, in which Caesar did not crossed Rubicon, or did not even exist. Is it right or complete nonsense? Can be somehow proved that this is right/nonsense?&lt;/p&gt;&#xA;" OwnerUserId="1450" LastActivityDate="2014-04-24T04:09:32.150" Title="Questions about time and space (from beginner)" Tags="&lt;space-time&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2214" PostTypeId="2" ParentId="2213" CreationDate="2014-04-20T21:17:58.413" Score="4" Body="&lt;p&gt;A)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it right that what we see/feel in every moment is projection of&#xA;  some 4D reality into our 3D world noticed by our senses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's more similar to a slice, hyperplane or hypermanifold, less to a projection.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If my eyes were able to &quot;somehow&quot; see in 4D and I pointed my eyes to a&#xA;  person, would I see that person in all stages of his/her life from&#xA;  birth to death as a single &quot;entity&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In a classical world, yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can I put infinite number of 3D spaces into a single 4D space?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, a possible mathematical description of such a space-time is a 3+1-&lt;a href=&quot;http://en.wikipedia.org/wiki/Minkowski_space&quot; rel=&quot;nofollow&quot;&gt;Minkowski space-time&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If we considered that 5th dimension exists and that we could put&#xA;  infinite number of 4D spaces into 5D space, then we would end up with&#xA;  infinite number of parallel realities, in which Caesar did not crossed&#xA;  Rubicon, or did not even exist. Is it right or complete nonsense? Can&#xA;  be somehow proved that this is right/nonsense?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's close to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Many-worlds_interpretation&quot; rel=&quot;nofollow&quot;&gt;many-world interpretation&lt;/a&gt; of quantum theory.&#xA;But it's thought of as an infinite-dimensional &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert_space&quot; rel=&quot;nofollow&quot;&gt;Hilbert space&lt;/a&gt;, as one of the simpler ways of thinking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking of a 5D space-time as a collection of all possible 4D space-times is less suitable, since it would be a mere set of 4D space-times without meaningful structure. Take Ceasar crossing Rubikon a day earlier or day later or a meter to the left or a meter to the right with or without a leaf falling from a tree, or a huge number of other versions. Which would be the correct order to pile up these versions to a 5D world? The infinite-dimensional Hilbert space model solves this ambiguity, and provides a meaningful &lt;a href=&quot;http://en.wikipedia.org/wiki/Metric_space&quot; rel=&quot;nofollow&quot;&gt;metric&lt;/a&gt; for the set of all possible worlds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 10-dimensional models of some &lt;a href=&quot;http://en.wikipedia.org/wiki/String_theory&quot; rel=&quot;nofollow&quot;&gt;string theories&lt;/a&gt; are something else: The 6 extra-dimensions are thought to be curled up, and tiny.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have no access to parallel worlds, so it cannot be proven by observational evidence. It's a mathematical model. It's possible to prove theorems within the mathematical theory, but that's usually based on axioms, or assumptions. Mathematical theories may, but don't need to match with observation, depending on how the axioms match to physical &quot;reality&quot;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-04-21T11:57:10.740" LastActivityDate="2014-04-21T11:57:10.740" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2215" PostTypeId="1" AcceptedAnswerId="2217" CreationDate="2014-04-20T22:18:28.410" Score="3" ViewCount="361" Body="&lt;p&gt;I am interested in doing some solar spectroscopy on my own for fun. How do I go about choosing the appropriate diffraction grating?&lt;/p&gt;&#xA;" OwnerUserId="1308" LastActivityDate="2014-04-20T23:12:14.133" Title="Choosing a diffraction grating" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2216" PostTypeId="2" ParentId="2212" CreationDate="2014-04-20T22:59:51.210" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;One consisted of two convex lenses of different focal lengths and the&#xA;  other one consisted of a concave lens as the objective lens and a&#xA;  convex lens as the eye-piece(that's called a Galilean one I guess).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's the ABC of refractor design - the most basic types of refractor. They are only interesting from a historic perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Refracting_telescope&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Refracting_telescope&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can certainly purchase a large convergent lens (bi-convex, or plan-convex) and use that as an objective. Then, for the eyepiece (ocular), you can use either a small convergent lens (bi-convex, or plan-convex), or small divergent (bi-concave, or plan-concave). Probably only the first one will work well: the image will be inverted, but the field of view will be nice and wide. The second design (divergent lens for eyepiece, Galilean refractor) provides a non-inverted image, but the field of view is extremely narrow and hard to use; avoid it, it's frustrating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this sort of bare-bones design is only interesting if you want to reconstruct the history of astronomy. At least for the eyepiece you could just &quot;steal&quot; any cheap 1-1/4&quot; eyepiece from a telescope or microscope. Even the cheapest one will work much better than your single-lens &quot;eyepiece&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.telescope.com/Accessories/Telescope-Eyepieces/pc/3/47.uts?&amp;amp;refinementValueIds=4469&quot; rel=&quot;nofollow&quot;&gt;http://www.telescope.com/Accessories/Telescope-Eyepieces/pc/3/47.uts?&amp;amp;refinementValueIds=4469&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the objective, again, I suggest you don't use a single convergent lens. It will work, but it will be full of aberrations (image distortions, color fringes, etc). Instead, go on Surplus Shed, click on the Lens Finder. For lens type, choose Achromat. For diameter, anything over 45 mm. For focal length, anything over 190 mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.surplusshed.com/lens.cfm&quot; rel=&quot;nofollow&quot;&gt;http://www.surplusshed.com/lens.cfm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An achromat is a combination of two lenses, glued together, that are designed to reduce aberrations. The image will be much better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their stock varies greatly. Right now, I've found these two:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.surplusshed.com/pages/item/pa1038.html&quot; rel=&quot;nofollow&quot;&gt;http://www.surplusshed.com/pages/item/pa1038.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.surplusshed.com/pages/item/pa1039.html&quot; rel=&quot;nofollow&quot;&gt;http://www.surplusshed.com/pages/item/pa1039.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 50 mm diameter, 300 mm focal length achromat, combined with a 15 mm focal length eyepiece, will provide 300 / 15 = 20x magnification. With a 5 mm eyepiece instead, it will give 300 / 5 = 60x magnification. Don't go much higher than this (more magnification is not always better).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The focal length of the objective dictates the length of the telescope tube. E.g. with the 300 mm achromat, the telescope length will be 300 mm plus some extra at the back for the eyepiece.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Be mindful of the f-ratio. Basically, divide the focal length by the diameter, and that gives you the focal ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lens with a diameter of 50 mm and focal length of 500 mm has an f-ratio of f/10. A lens with a diameter of 50 mm and focal length of 300 mm has an f-ratio of f/6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's best to use an achromat with an f-ratio between f/8 and f/20. If you go below f/8 it will be very hard to collimate (fine tune) the scope. If you go above f/20, it will be too long.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Sometimes you find super-cheap eyepieces there, too, like this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.surplusshed.com/pages/item/l2019.html&quot; rel=&quot;nofollow&quot;&gt;http://www.surplusshed.com/pages/item/l2019.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also keep digging around that site, maybe you'll find various other parts necessary for building the refractor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enormous care must be given to keeping the objective and the eyepiece collimated (aligned and centered). Even a tiny collimation error will greatly affect performance. This is not a trivial engineering problem.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There's also an easier way. You could purchase a complete kit, like the Galileoscope:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://galileoscope.org/&quot; rel=&quot;nofollow&quot;&gt;http://galileoscope.org/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a 50 mm (2&quot;) f/10 refractor that has about the same performance like good binoculars. The price is very affordable (and very cheap for what it does). As a telescope and mirror maker, I was impressed by the quality of the achromat doublet objective that they provide (50 mm diameter, 500 mm focal length, f/10). For the price and the size, it is well made and well corrected, and punches well above its price category.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kit comes with everything you need: all the lenses, the tube, various small parts, etc., including components for the ocular (eyepiece). Another nice thing is that you can swap its eyepiece for a standard 1-1/4&quot; eyepiece used in a regular telescope (like a cheapie on Surplus Shed). Whoever designed this thing put some good thought in it. There are several different instruction documents on the site, that you can choose from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My 9 and 12 year old sons put it together in an afternoon. It's that easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; What can you do with the Galileoscope?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, you can watch some of the planets. You'll see the two big equatorial belts on Jupiter, and the 4 galilean moons of Jupiter. Saturn's rings will be quite obvious. Venus will show a crescent - it will look just like the Moon, only much smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can watch the Moon, too. The big craters are quite visible in an instrument this size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can watch some of the deep space objects too. The Orion Nebula is quite visible actually. The Andromeda galaxy is somewhat visible from a place outside the city.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Double stars such as Albireo, which appear as single star to the naked eye, are easily split by a 50 mm aperture. Albireo is very pretty, and easy to see in a Galileoscope - an orange/blue double, easy and wide, very beautiful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mizar, one of the main stars in the Big Dipper, is also a double. This one is a lot more tight. I think you should be able to split it in a G-scope anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.synapses.co.uk/astro/bearings.html&quot; rel=&quot;nofollow&quot;&gt;http://www.synapses.co.uk/astro/bearings.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the planets and the Moon, you'll probably want high magnification. 50x or more, up to 100x (the maximum supported by this instrument).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure the instrument is sitting on a stable, firm support. A flimsy, shaky tripod is the enemy of high magnification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look on Surplus Shed and other online stores for 1-1/4&quot; eyepieces. A 5 mm eyepiece will give you 100x magnification - the maximum. A 10 mm eyepiece will give 50x. A 20 mm eyepiece will give 25x - nice and wide, good for big objects like the Orion Nebula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look for eyepieces with a wide field of view (at least 50 degrees, up to 80 degrees, or more, if you can find them). A wide field makes everything easier.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It is also possible to build everything from scratch, including making the lenses and/or mirrors yourself from a piece of glass. But that is a MUCH bigger commitment, and a much more long-term project (a few months to a year).&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2014-04-22T08:00:56.737" LastActivityDate="2014-04-22T08:00:56.737" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2217" PostTypeId="2" ParentId="2215" CreationDate="2014-04-20T23:12:14.133" Score="4" Body="&lt;p&gt;There isn't a lot of engineering here. Just use a grating dense enough to give you a comfortable angle. Less dense grating (fewer lines per mm) means a more shallow angle and narrow spectrum, and that can be harder to use. Super-dense grating might be expensive and fragile. Stay somewhere in the middle - you'll have to do a few trial runs on your own, see what works for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made this spectrograph for my kids, works well enough:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://sci-toys.com/scitoys/scitoys/light/spectrograph/spectrograph.html&quot; rel=&quot;nofollow&quot;&gt;http://sci-toys.com/scitoys/scitoys/light/spectrograph/spectrograph.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used the cheap grating that they provide on their own store, 1000 lines / mm, not bad:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.scitoyscatalog.com/product/DIFFRACTION.html&quot; rel=&quot;nofollow&quot;&gt;http://www.scitoyscatalog.com/product/DIFFRACTION.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2014-04-20T23:12:14.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2218" PostTypeId="2" ParentId="2156" CreationDate="2014-04-21T00:11:32.610" Score="6" Body="&lt;p&gt;It isn't correct, because a vortex is not a helix, and so while the planets do trace a helical path as they move through the galaxy, this is not evidence of a vortex. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, the sun actually is moving through space, as it traces a path around the centre of the galaxy. The whole mass of the solar system moves with it, so the planets are not left behind as the sun moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.universetoday.com/107322/is-the-solar-system-really-a-vortex/&quot; rel=&quot;noreferrer&quot;&gt;Rhys Taylor&lt;/a&gt; and &lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2013/03/04/vortex_motion_viral_video_showing_sun_s_motion_through_galaxy_is_wrong.html&quot; rel=&quot;noreferrer&quot;&gt;Phil Plait&lt;/a&gt; have comprehensive smackdowns debunking this vortex idea and other misunderstandings/delusions by the author.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-04-21T00:11:32.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2219" PostTypeId="1" CreationDate="2014-04-21T01:30:16.963" Score="11" ViewCount="1936" Body="&lt;p&gt;I was just lying under the sky trying to possibly see some meteorites, unfortunately never seeing any I might add, but I saw three objects all moving at about the same speed( all at different times). They were far too faint and small to be aircraft or anything inside of the atmosphere I would think and there were no blinking lights. They were also moving fairly quickly it seemed. Could these have been satellites in low earth orbit? It's the only other thing I would think that they could be.&lt;/p&gt;&#xA;" OwnerUserId="1206" LastEditorUserId="1206" LastEditDate="2014-04-21T16:47:19.380" LastActivityDate="2014-04-21T16:47:19.380" Title="Are satellites orbiting around earth visible to the naked eye?" Tags="&lt;amateur-observing&gt;&lt;observational-astronomy&gt;&lt;satellite&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2220" PostTypeId="2" ParentId="2219" CreationDate="2014-04-21T01:43:57.443" Score="16" Body="&lt;p&gt;A lot of satellites are visible under the right conditions. Usually up to 2 hours after sunset and 2 hours before sunrise. This allows the sun to strike the satellite when you are on the dark side. Depending on the orbit, it will take between 1 and 5 minutes to traverse most of the sky. Usually, they will enter the shadow and you lose sight of them.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="1461" LastEditDate="2014-04-21T16:12:43.417" LastActivityDate="2014-04-21T16:12:43.417" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2221" PostTypeId="1" CreationDate="2014-04-21T13:06:40.633" Score="19" ViewCount="6224" Body="&lt;p&gt;I may have some specifics wrong here.  If so, don't focus on those.  Just focus on the general thrust of my question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &quot;understand&quot; (cough) that particle/anti-particle pairs form spontaneously in space.  I understand that they can form near the event horizon of a black hole, and that one particle can fall in, where as the other particle can barely escape.  I understand that an anti-particle will annihilate with a particle.  What I don't understand is why only the anti-particles of these virtual particle pairs fall into the black hole, while the other ones just manage to escape.  Shouldn't both particle and anti-particle have equal chance to be the one to fall in, or just manage to escape?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that there should be an equal chance of either the particle, or the anti-particle, would be captured while the other &quot;ejected.&quot;  So it seems that the black hole should be somewhat steady-state as far as mass change with respect to virtual particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explain?&lt;/p&gt;&#xA;" OwnerUserId="1392" LastActivityDate="2019-10-11T11:04:05.643" Title="Black Hole / Hawking Radiation: Why only capture anti-particle?" Tags="&lt;black-hole&gt;&lt;hawking-radiation&gt;&lt;quantum-field-theory&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2222" PostTypeId="1" CreationDate="2014-04-21T13:18:35.063" Score="4" ViewCount="304" Body="&lt;p&gt;A star &lt;em&gt;A&lt;/em&gt; goes through the body of another star &lt;em&gt;B&lt;/em&gt;, or is swallowed by &lt;em&gt;B&lt;/em&gt;. There is no tidal disruption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sounds like &lt;em&gt;kerzan&lt;/em&gt;. Does anybody recognise this pronunciation?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="13" LastEditDate="2014-04-21T23:23:13.537" LastActivityDate="2014-04-22T04:34:40.230" Title="What is the term for a star swallowing another star?" Tags="&lt;star&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2223" PostTypeId="1" AcceptedAnswerId="2227" CreationDate="2014-04-21T14:26:25.740" Score="3" ViewCount="556" Body="&lt;p&gt;I am a beginner and I've planned to buy a telescope(Newtonian reflector)for viewing planets and deep sky objects. As far as I know the most important factor to get a brighter view is that the telescope need to have a good aperture. I need a telescope that would be good both for viewing planets and deep sky objects.So my question is here as a &lt;strong&gt;beginner&lt;/strong&gt; how much aperture is necessary to get amazed with the space. And I don't know much regarding telescope so please let me know the other factors for getting good view of celestial objects.&lt;/p&gt;&#xA;" OwnerUserId="1441" LastActivityDate="2014-04-21T23:14:34.503" Title="Please Guide me to buy my first Telescope" Tags="&lt;amateur-observing&gt;&lt;observational-astronomy&gt;&lt;telescope&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" ClosedDate="2014-05-04T13:48:42.617" ContentLicense="CC BY-SA 3.0" />
  <row Id="2224" PostTypeId="1" CreationDate="2014-04-21T14:54:53.743" Score="16" ViewCount="3977" Body="&lt;p&gt;Whenever I have learned about the solar system I always see the orbits displayed as a virtually flat plane. &#xA;&lt;img src=&quot;https://i.stack.imgur.com/gaBPU.jpg&quot; alt=&quot;2D Solar System&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are all of the orbits in the solar system really like this? If so, why? It seems like a rather large coincidence for all of our planets + the asteroid belt to be on virtually the same plane. &lt;/p&gt;&#xA;" OwnerUserId="1549" OwnerDisplayName="Dgrin91" LastEditorUserId="1469" LastEditDate="2014-04-24T09:44:27.977" LastActivityDate="2014-04-24T09:44:27.977" Title="Why is the solar system often shown as a 2D plane?" Tags="&lt;orbit&gt;&lt;solar-system&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2225" PostTypeId="2" ParentId="2224" CreationDate="2014-04-21T14:59:41.693" Score="16" Body="&lt;p&gt;This is not a coincidence at all, but a direct consequence of &lt;a href=&quot;https://en.wikipedia.org/wiki/Formation_and_evolution_of_the_Solar_System&quot;&gt;the way the solar system was formed&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The generally accepted model is that solar systems (including our own) form out of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Protoplanetary_disc&quot;&gt;Protoplanetary disc&lt;/a&gt;.  Gravitation causes mass to collapse around a protostar, which always has &lt;em&gt;some&lt;/em&gt; angular momentum (as does everything).  Wikipedia explains it better than I can:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Protostars typically form from molecular clouds consisting primarily of molecular hydrogen. When a portion of a molecular cloud reaches a critical size, mass, or density, it begins to collapse under its own gravity. As this collapsing cloud, called a solar nebula, becomes denser, random gas motions originally present in the cloud average out in favor of the direction of the nebula's net angular momentum. Conservation of angular momentum causes the rotation to increase as the nebula radius decreases. This rotation causes the cloud to flatten out—much like forming a flat pizza out of dough—and take the form of a disk.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And then, from this protoplanetary disk, planets form.  Consequently, they're all in the same plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, the inclinations of each planet's orbit are pretty close to that of Earth's:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Planet  Orbital Inclination&#xA;Mercury 7°&#xA;Venus   3.39°&#xA;Earth   0°&#xA;Mars    1.85°&#xA;Jupiter 1.3°&#xA;Saturn  2.49°&#xA;Uranus  0.77°&#xA;Neptune 1.77°&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1462" OwnerDisplayName="gerrit" LastActivityDate="2014-04-21T15:14:49.573" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="2226" PostTypeId="2" ParentId="2221" CreationDate="2014-04-21T17:14:41.047" Score="14" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I thought that the anti-particle was annihilating with &quot;normal&quot; mass inside the black hole? No?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. First, both particles and anti-particles have &quot;normal&quot; mass (should they have mass in the first place) and &quot;normal&quot; (positive) energy. The distinction between them is either a matter of convention or a question of which type is more common in the universe. Furthermore, for typical-massed blacked holes, the bulk of Hawking radiation would be made of photons, which properly speaking do not even have anti-particles, though one could also say that they are their own anti-particle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Shouldn't both particle and anti-particle have equal chance to be the one to fall in, or just manage to escape?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, and uncharged ones do. A smaller black hole would radiate both neutrinos and anti-neutrinos, assuming all neutrinos are massive (otherwise, all black holes would do it already), and a sufficiently small (and thus sufficiently hot) one would radiate both electrons and positrons. Very roughly, a black hole will radiate non-negligible amounts of massive particles when the temperature of the black hole is on the order of the particle mass or greater, in natural units.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It seems that there should be an equal chance of either the particle, or the anti-particle, would be captured while the other &quot;ejected.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Correct, with a minor exception that if a hot black hole has electric charge, it is more likely to radiate particles of the same sign of charge.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So it seems that the black hole should be somewhat steady-state as far as mass change with respect to virtual particles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If either a particle or an anti-particle falls into a black hole, its mass will go up. &lt;em&gt;It doesn't matter.&lt;/em&gt; Fundamentally, the &quot;reason&quot; for Hawking radiation is that the vacuum state in quantum field theory is a state of lowest energy, but different observers can disagree about &lt;em&gt;which&lt;/em&gt; state is the vacuum. Thus, since particles are fluctuation on top of the vacuum, they can disagree about whether or not there are particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think there is a good way to repair the &quot;antiparticle falls in&quot; story except some roundabout appeal to energy conservation: if the escaping particle is real and have positive energy, the one that fell in must have negative energy, and would therefore decrease the mass of the black hole. Unfortunately, that only shows what must happen for the situation to be consistent, not that it does actually happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although with some knowledge of general relativity, one can motivate this slightly further--e.g., for the Schwarzschild black hole, there is energy conservation given by a Killing vector field, which goes from timelike to spacelike at the horizon--so what an external observer considers time/energy would be space/momentum inside the black hole, and momentum &lt;em&gt;is&lt;/em&gt; allowed to be negative.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-04-21T17:14:41.047" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2227" PostTypeId="2" ParentId="2223" CreationDate="2014-04-21T19:57:34.200" Score="5" Body="&lt;p&gt;It's a broad question, but I'll take a stab trying to provide the essentials for a good start. Also see this post which contains important additional information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/2193/best-telescope-for-the-viewing-of-nebulae-stars-and-planets/2198#2198&quot;&gt;Best telescope for the viewing of Nebulae, Stars and Planets&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many, many factors involved in choosing a telescope. You seem settled on a dobsonian. That's not the only possible choice, but a dob is a good scope to start with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are correct that aperture (diameter of the primary lens or mirror) is the primary driver of telescope performance. Bigger aperture = better performance. However, performance is not the only thing you should worry about. Think of a car - do you only care about speed when buying a new car?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any size scope will show you something. I've a 50 mm (2&quot;) achromat refractor that does double duty as finderscope. I can push it up to 100x magnification, and it will show lots of craters and mountains on the Moon, will show Jupiter as a disk with two equatorial belts, will show the rings of Saturn, will show the M13 globular cluster (barely) and the M31 Andromeda galaxy (again, barely). Not bad for a little scope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small 150 mm ... 250 mm (6&quot; ... 10&quot;) dob will show you more. When seeing is good, you see several belts on Jupiter, and even individual curls on the big belts. Saturn's rings start to show the divisions, such as Cassini. These days Mars is at opposition - such a scope will show you the polar ice cap and the major land features. M13 fills the eyepiece with a dust of stars. The Orion Nebula looks big and clear even in the city. The Ring Nebula is visible, etc. Under a very dark sky, you can see (barely) the 3C 273 quasar, at 2.4 billion light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A mid-size dob (12&quot; ... 18&quot;) will show you even more. More details on the planets, when seeing permits. More faint fuzzies (galaxies, nebulae, star clusters) especially out and far away from the city.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A large dob (20&quot; and over) is an amazing instrument. On those very rare occasions when seeing cooperates, Jupiter at 1000x in such a scope is unforgettable. Also, under a very dark sky out in the desert, point it anywhere and the sky is full of faint fuzzies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, which size should you choose? It depends on price, maintenance, and ease of use (size, etc). Let's talk about maintenance.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As a dob owner, there are two things you must never forget: collimation, and cooling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Collimation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It means &quot;keeping all the optics aligned and centered&quot;. The mirrors will move around a bit, and in time will lose collimation. As a dob owner, you have to collimate the scope once in a while (and every time before you observe high-resolution targets such as planets). Once you learn how to do it, it takes about 2 minutes, it's super-quick and easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why collimation is important:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://legault.perso.sfr.fr/collim.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://legault.perso.sfr.fr/collim.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an introduction to dobsonian collimation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cloudynights.com/documents/primer.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.cloudynights.com/documents/primer.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some simpler collimation techniques, which you can use until you acquire all the tools:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.etnaastros.com/documents/No_Tools_Collimation.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.etnaastros.com/documents/No_Tools_Collimation.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.skyandtelescope.com/astronomy-resources/how-to-collimate-your-newtonian-reflector/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.skyandtelescope.com/astronomy-resources/how-to-collimate-your-newtonian-reflector/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also see the manual that comes with the scope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cooling&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you keep the scope in the house or garage, and then take it out for observation, the primary mirror will be much hotter than the night air. As a result, air convection will appear on the mirror and distort the image - just like the shimmering you see in the summer along the hot walls of a house under sunlight. The mirror is too &quot;hot&quot; for the ambient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what happens:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.garyseronik.com/?q=node/55&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.garyseronik.com/?q=node/55&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.garyseronik.com/?q=node/69&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.garyseronik.com/?q=node/69&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To combat convection, take the scope outside at least 1 hour prior to observation, and let it cool off. Very small dobs (up to 6&quot; or so) will probably work fine just with passive cooling. Small-medium and above (10&quot; or above) basically require a fan on the primary mirror (forced cooling).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Always collimate and cool off the scope before you do an observation requiring high resolution (when observing planets, the Moon, double stars). If you don't need a lot of high resolution (like when observing galaxies, nebulae, star clusters) then collimation and cooling are not so important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you keep your dob collimated and cooled, it will provide great views of the planets. There is this myth out there that dobs &quot;are not great for planets&quot;, and refractors are somehow magically &quot;so much better for planets&quot;. This is not true. The myth exists only because of the large number of people out there who keep their dobs in terrible shape; those poor scopes underperform by a very large margin. Don't be one of those folks.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let me just draw a line in the sand for you: An 8&quot; f/6 dob is a good start overall. Why is that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 8&quot;, it is big enough that it will be seeing-limited most of the time. In other words, it is so big, atmospheric turbulence will limit the resolution more often than the size of the scope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is big enough to show you some faint fuzzies even from the city. All the Messier objects should be visible no matter where you are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not too big to fit on the back seat of a sedan car, if you want to drive to a place with a darker sky. It is not too big to move around in the backyard easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mirror is not so big and thick as to mandate active cooling (fan on the mirror); you can pretty much get by with passive cooling only - take it out at least 1 hour before observation. However, a fan may improve things a bit anyway (you need to experiment with that, but it's not a priority at this size).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At f/6 it's not too hard on the eyepieces, and it's not too hard to collimate. A shorter scope (f/5 and below) produces a steeper light cone; you need higher quality (and price) eyepieces to use it, and collimation requires much better precision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the price is right - a few hundred dollars, which is affordable for quite a few people. Take care of it and it will last you a lifetime.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, if you can afford it and think you can deal with the steeper learning curve, could you buy a bigger instrument? Sure. 16&quot;, 18&quot; - the wallet is the limit. Such big dobs are made differently - they are called &quot;truss dobs&quot;. They don't have a big tube where everything is contained. Instead, the tube is replaced by a truss made of a bunch of poles - looks like the arm of a crane machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sRw1Y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you're done observing, you disassemble it. It all fits in a sedan car. It's not too heavy for transport, because you handle each part separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, such an instrument is usually f/5 or less. Collimation is much more difficult. You need a coma corrector (like a Paracorr) to fix coma at the edge, and that's additional money. You need better eyepieces. Cooling doesn't work unless you use active cooling (fan). And it's just more expensive overall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you start on a large dob like that? It's not impossible, but it requires more money and a much more significant commitment. This is not a toy that you can discard after two week-ends.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-04-21T23:14:34.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2228" PostTypeId="2" ParentId="2222" CreationDate="2014-04-21T23:05:18.557" Score="6" Body="&lt;p&gt;The term you're searching for is probably &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Accretion_%28astrophysics%29&quot; rel=&quot;noreferrer&quot;&gt;accretion&lt;/a&gt;&lt;/em&gt; (I guess it might sound similar to &lt;em&gt;a kerzan&lt;/em&gt; if the speaker had a cold or the listener couldn't hear it well enough for other reasons). This term is not limited to close binary stars (accreting binaries) cannibalizing each other because they orbit each other at such proximity that the L&lt;sub&gt;1&lt;/sub&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Lagrangian_point&quot; rel=&quot;noreferrer&quot;&gt;Lagrange point&lt;/a&gt; between them extends into their own &lt;a href=&quot;https://en.wikipedia.org/wiki/Roche_lobe&quot; rel=&quot;noreferrer&quot;&gt;Roche lobe&lt;/a&gt;, they form accretion disks connected through an accretion stream, and they feed off each other (thus why this is often referred to as cannibalism):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/uTv0Z.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;sup&gt;Artist's conception shows an accreting binary at the center of the globular cluster NGC 6624 (Source: &lt;a href=&quot;http://csep10.phys.utk.edu/astr162/lect/binaries/accreting.html&quot; rel=&quot;noreferrer&quot;&gt;U. of Tennessee AstroWiki&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some examples of accreting binaries are &lt;a href=&quot;https://en.wikipedia.org/wiki/Symbiotic_variable_star&quot; rel=&quot;noreferrer&quot;&gt;symbiotic novae&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/X-ray_binary&quot; rel=&quot;noreferrer&quot;&gt;X-ray binaries&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/X-ray_binary#Microquasar&quot; rel=&quot;noreferrer&quot;&gt;microquasars&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/X-ray_pulsar&quot; rel=&quot;noreferrer&quot;&gt;X-ray pulsars&lt;/a&gt;, and so on. See links for more examples and explanations. Other examples of accretion include &lt;a href=&quot;http://www.nasa.gov/mission_pages/hubble/science/planet-eater.html&quot; rel=&quot;noreferrer&quot;&gt;stellar accretion of planetary materials of a planet in a low altitude orbit around its parent star&lt;/a&gt;, formation of planets via accretion of materials of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Protoplanetary_disk&quot; rel=&quot;noreferrer&quot;&gt;protoplanetary disk&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Active_galactic_nucleus&quot; rel=&quot;noreferrer&quot;&gt;active galactic nucleus&lt;/a&gt; (AGN) like &lt;a href=&quot;https://en.wikipedia.org/wiki/Quasar&quot; rel=&quot;noreferrer&quot;&gt;quasars&lt;/a&gt; (could also sound similar), &lt;a href=&quot;https://en.wikipedia.org/wiki/Nova&quot; rel=&quot;noreferrer&quot;&gt;novae&lt;/a&gt;, and e.g. the &lt;a href=&quot;http://www.sciencemag.org/content/262/5133/550.abstract&quot; rel=&quot;noreferrer&quot;&gt;Earth still accretes on average ~ 110 metric tons of new materials per day&lt;/a&gt;, too.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditDate="2014-04-21T23:24:03.427" LastActivityDate="2014-04-21T23:24:03.427" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2229" PostTypeId="2" ParentId="2222" CreationDate="2014-04-22T01:17:54.173" Score="4" Body="&lt;p&gt;Is it possible you refer to a &lt;a href=&quot;https://en.wikipedia.org/wiki/Thorne%E2%80%93%C5%BBytkow_object&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thorne–Żytkow object&lt;/a&gt; ? &#xA;This is literally a star within another star, more specifically a compact object (Neutron star) inside a giant star (red giant or supergiant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WmKSM.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, this is an only theoretical object, it has never been observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &lt;a href=&quot;http://adsabs.harvard.edu/doi/10.1086/155109&quot; rel=&quot;nofollow noreferrer&quot;&gt;original paper&lt;/a&gt; from the authors, they provide some observable characteristics of these objects, like some peculiar emission lines. You can imagine the chemistry of such an object would be really unique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here an article for a possible identification &lt;a href=&quot;http://phys.org/news/2014-01-discovery-thorne-zytkow.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;link 1&lt;/a&gt;, very recent work (2014).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here some calculations &lt;a href=&quot;http://cow.physics.wisc.edu/~ogelman/guide/tz/&quot; rel=&quot;nofollow noreferrer&quot;&gt;link2&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastEditorUserId="1038" LastEditDate="2014-04-22T04:34:40.230" LastActivityDate="2014-04-22T04:34:40.230" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2230" PostTypeId="1" CreationDate="2014-04-22T04:10:56.617" Score="6" ViewCount="212" Body="&lt;p&gt;I stumbled on The &lt;a href=&quot;https://en.wikipedia.org/wiki/Lagrangian_point#L4_and_L5&quot; rel=&quot;nofollow&quot;&gt;Lagrangian points&lt;/a&gt; during some wikipedia reading.  After looking at the gravity contours, I naturally come to the conclusion that the L4 &amp;amp; L5 should have a wave pattern and then found the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lissajous_orbit&quot; rel=&quot;nofollow&quot;&gt;Lissajous orbit&lt;/a&gt; page.  It states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Orbits about Lagrangian points L4 and L5 are dynamically stable in theory so long as the ratio of the masses of the two main objects is greater than about 25, meaning the natural dynamics keep the [third object] in the vicinity of the Lagrangian point even when slightly perturbed from equilibrium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After reading it I started to wonder if there is a maximum possible amplitude (height of the peaks and troughs relative to the orbital plane of the second object) of the pattern?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if there theoretically is none, for cases where it is extremely large, say larger than the 2 times the radius of the second orbiting object, what would the respective object weight ratios have to be to keep such a perturbed orbit stable for any realistic period of time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FYI, I'm a computer scientist who loves reading about physical cosmology but can be kinda a noob sometimes. Please forgive me if I'm asking for the wrong parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: &lt;a href=&quot;https://www.youtube.com/watch?v=VGmdSbtkFS0&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is an animation of 2010 TK7, Earth's first trojan asteroid, showing the wave pattern I'm referring to. Recall that my question is referring to the height of the peaks and troughs relative to Earth's orbital plane. Since the video is a top-down view, the peaks and troughs are going into and out of the screen.&lt;/p&gt;&#xA;" OwnerUserId="1463" LastEditorUserId="1463" LastEditDate="2014-12-02T20:17:29.247" LastActivityDate="2014-12-02T20:17:29.247" Title="Maximum Amplitude of a Lissajous Orbiting Object in a L4 or L5 Position" Tags="&lt;gravity&gt;&lt;lagrange-point&gt;&lt;satellite&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2231" PostTypeId="1" AcceptedAnswerId="2236" CreationDate="2014-04-22T10:21:30.570" Score="5" ViewCount="9658" Body="&lt;p&gt;Why is the element iron responsible for supernova? Can any star create more element than iron within the span of its life?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that when star dies due to supernova, other elements are created (Gold and other 92 elements) due to more heat generated than required not only for fusion of Iron element but also for the fusion of higher element with each other.&lt;/p&gt;&#xA;" OwnerUserId="760" LastEditorUserId="760" LastEditDate="2014-04-23T10:17:07.190" LastActivityDate="2020-09-12T11:35:55.177" Title="Why is iron responsible for causing a supernova?" Tags="&lt;supernova&gt;&lt;explosion&gt;" AnswerCount="4" CommentCount="9" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2232" PostTypeId="2" ParentId="2231" CreationDate="2014-04-22T11:22:52.070" Score="0" Body="&lt;p&gt;When a star has exhausted its supply of elements below iron the fusion process is no longer able to prevent the outer layers from collapsing in towards the core. If the star is sufficiently massive it will then eject the outer layers explosively as a supernova. So iron is not the cause, just the point in the fusion process at which this occurs.&lt;/p&gt;&#xA;" OwnerUserId="1464" LastActivityDate="2014-04-22T11:22:52.070" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2233" PostTypeId="2" ParentId="2231" CreationDate="2014-04-22T13:23:26.633" Score="5" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aBinding_energy_curve_-_common_isotopes.svg&quot; rel=&quot;noreferrer&quot;&gt;binding energy per nucleon&lt;/a&gt; is among the highest for iron-56. Therefore &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_fusion&quot; rel=&quot;noreferrer&quot;&gt;nuclear fusion&lt;/a&gt; as well as &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_fission&quot; rel=&quot;noreferrer&quot;&gt;fission&lt;/a&gt;/&lt;a href=&quot;http://en.wikipedia.org/wiki/Photodisintegration&quot; rel=&quot;noreferrer&quot;&gt;photodisintegration&lt;/a&gt; of iron-56 consumes energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat production is needed to prevent a star from collapsing to a much denser state. Iron-56 provides no way to produce heat by nuclear reactions. Hence &lt;a href=&quot;http://en.wikipedia.org/wiki/Supernova#Core_collapse&quot; rel=&quot;noreferrer&quot;&gt;core collapse&lt;/a&gt; is unavoidable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the star isn't large enough, nuclear fusion may stop, before the core of the star is fused to iron, since very high temperature (more than 2 billion Kelvin) and pressure are needed to &lt;a href=&quot;http://en.wikipedia.org/wiki/Silicon_burning_process&quot; rel=&quot;noreferrer&quot;&gt;fuse silicon to iron&lt;/a&gt;. This way the core may collapse earlier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The collapse releases energy responsible for much of the luminosity of the supernova. Some additional (exothermic) fusion in outer layers of the exploding star may be triggered by the collapse during the explosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike most supernovae, type Ia supernovae explode by &lt;a href=&quot;http://en.wikipedia.org/wiki/Supernova#Thermal_runaway&quot; rel=&quot;noreferrer&quot;&gt;runaway nuclear fusion&lt;/a&gt;, not mainly by core collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypernova&quot; rel=&quot;noreferrer&quot;&gt;hypernova&lt;/a&gt; may form either due to lack of nuclear fuel, the core of the star first collapsing to a neutron star, followed by a collapse to a black hole, or by &lt;a href=&quot;http://en.wikipedia.org/wiki/Pair-instability_supernova&quot; rel=&quot;noreferrer&quot;&gt;pair instability&lt;/a&gt; leaving no dense remnant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cause for the pair instability supernovae isn't lack of nuclear fuel, but the heat of the core. Due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck%27s_law&quot; rel=&quot;noreferrer&quot;&gt;Planck's law&lt;/a&gt; the energy of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_radiation&quot; rel=&quot;noreferrer&quot;&gt;radiated photons&lt;/a&gt; increases with temperature. As soon as they reach the energy needed to &lt;a href=&quot;http://en.wikipedia.org/wiki/Pair_production&quot; rel=&quot;noreferrer&quot;&gt;form electron-positron pairs&lt;/a&gt;, at least 1.02 &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronvolt&quot; rel=&quot;noreferrer&quot;&gt;MeV&lt;/a&gt;, heat &lt;a href=&quot;http://en.wikipedia.org/wiki/Mass%E2%80%93energy_equivalence&quot; rel=&quot;noreferrer&quot;&gt;energy is transformed to mass&lt;/a&gt;. This way pressure needed to keep the core stable, gets lost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pair-instability supernovae and -hypernovae are indicated by high amounts of released nickel-56, which will decay to iron-56.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-04-22T13:23:26.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2234" PostTypeId="1" AcceptedAnswerId="2239" CreationDate="2014-04-22T13:47:25.673" Score="5" ViewCount="2967" Body="&lt;p&gt;If so by how much does it &quot;spoil&quot; the view of stars and galaxies etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this error noticable, when compared to pictures taken from outer space.&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="7982" LastEditDate="2021-02-28T11:56:29.087" LastActivityDate="2021-02-28T11:56:29.087" Title="Does the earth's atmosphere act as a spherical lens and refract light from space?" Tags="&lt;star&gt;&lt;space&gt;&lt;atmospheric-effects&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2235" PostTypeId="2" ParentId="2234" CreationDate="2014-04-22T13:52:43.897" Score="0" Body="&lt;p&gt;It does not act as a solid lens, because it does not have a limit surface. Instead, atmosphere's density varies slowly with altitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since its refractive index is very close to 1 it is hardly noticeable any refraction. You will note it mostly on the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three main effects of atmosphere are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Reddening, caused by differential scattering on the wavelenghts&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Absorption&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Seeing (Airy's disc)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-04-22T13:52:43.897" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2236" PostTypeId="2" ParentId="2231" CreationDate="2014-04-22T16:17:02.000" Score="13" Body="&lt;p&gt;Your question is a bit oversimplified because there are many types of supernovae based on the size and configuration of the star. But I can answer your question about &lt;em&gt;&quot;why iron&quot;&lt;/em&gt; by considering what keeps a star from exploding in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the simplest terms of star formation, when material from an interstellar nebula starts to collapse under its own gravity, the pressure and temperatures involved will become great enough to eventually start fusing hydrogen into helium (it's a bit more complicated than that, but I'm speaking in generalities). If you were to consider the helium atoms created by that process, you'll notice that each helium atom weighs just a bit &lt;em&gt;less&lt;/em&gt; than the two hydrogen atoms that formed it. That bit of extra mass is given off as energy which is produced in great quantities as the hydrogen continues to be fused into helium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the star's &quot;main sequence&quot;, the release of energy by the hydrogen-helium fusion helps counteract the weight of the star's gasses pushing inward. Material presses in; energy pushes out in perfect balance. This balance of gravity and energy output continues until the star uses up most of its hydrogen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's at this this point (when there is no hydrogen left in the &lt;em&gt;core&lt;/em&gt; of the star to fuse into helium) that the fusion reaction stops and gravity will resume to collapse the star further. As this star collapses, it will quickly become denser and hotter until the temperature and pressures of the interior are great enough to start fusing helium into heavier elements&amp;hellip; and the process continues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;That is, until the star starts fusing elements into iron&amp;hellip;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fusion into iron is the first element that does &lt;strong&gt;&lt;em&gt;not&lt;/em&gt;&lt;/strong&gt; create more energy than it takes to produce. The effect is that there is no net energy being produced to counteract the gravity pushing inward. So the outer layers will quickly collapse into a much denser and smaller ball causing the remaining star material to undergo fusion all at once, causing the supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in that sense, &lt;strong&gt;iron is not the &lt;em&gt;cause&lt;/em&gt; of the supernova,&lt;/strong&gt; but its presence marks the inevitable end of this star's life cycle&amp;hellip; in this particular scenario. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But understand that this is an oversimplification to illustrate the process you asked about. There are many other sequences of a star's life cycle. Our sun, for example, does not have sufficient mass to keep collapsing down with sufficient pressures to fuse heavier elements into iron. Without getting into &lt;em&gt;other&lt;/em&gt; pathways for the production of heavy elements (even in smaller stars like our sun) &amp;mdash; once our sun starts creating carbon and oxygen, the fuel starts to run out and the core will simply collapse and rebound as it swells up into a red giant, before losing its outer layers as a planetary nebula while the core shrinks to become a white dwarf (and eventually cooling into a black dwarf). &lt;/p&gt;&#xA;" OwnerUserId="58" LastEditorUserId="58" LastEditDate="2019-01-22T19:35:44.237" LastActivityDate="2019-01-22T19:35:44.237" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="2237" PostTypeId="1" AcceptedAnswerId="2238" CreationDate="2014-04-22T16:33:11.837" Score="10" ViewCount="698" Body="&lt;p&gt;In both books and documentaries I often see the Goldilocks Zone as described between Venus and Mars with the Earth &quot;just right.&quot; This seems sort of ill thought out to me, because it assumes that Venus' atmosphere comes from its distance from the sun and the same with Mars. I'm not too familiar with current ideas about how Venus' atmosphere formed, but with Mars I'm fairly sure it's widely understood the reason the atmosphere is so small is because the gravity is so weak. Even some discussions about terra forming Mars mention that within a few million years, perhaps, the atmosphere would drift off into space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how can we be so certain that's the limits of the zones, especially with Mars? If Mars was as large as Earth/Venus and could hold on to a thicker atmosphere and thus retain more heat, then wouldn't that automatically push out the Goldilocks Zone? It seems like the edges have been places arbitrarily without consideration as to &lt;em&gt;why&lt;/em&gt; they're the edges instead of &quot;they just are.&quot; This comes up especially because even with its weak atmosphere, at the equator, it can reach up to 20 degrees which is well within the realm of liquid water.&lt;/p&gt;&#xA;" OwnerUserId="1768" LastActivityDate="2018-12-27T19:11:17.650" Title="Why is Mars considered the outer edge of the &quot;goldilocks zone&quot;?" Tags="&lt;mars&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2238" PostTypeId="2" ParentId="2237" CreationDate="2014-04-22T18:14:08.543" Score="10" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Circumstellar_habitable_zone&quot;&gt;circumstellar habitalbe zone&lt;/a&gt; can be defined as the distance range around a star, where the mean temperature of a rotating planet would be between 0 and 100 centigrades, if radiation (heat) received from the star and thermal radiation emitted by the planet form an equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that's only a rule of thumb. It has been redefined, and is still &lt;a href=&quot;http://www.astro.umd.edu/~miller/teaching/astr380f09/lecture14.pdf&quot;&gt;disputed&lt;/a&gt;.&#xA;Hence your criticism is justified to some degree.&#xA;There may exist exceptions of habitable planets outside the presumed habitable zone; and there exist certainy many planets within the habitable zone which aren't habitable.&#xA;But the probability for a planet similar to Earth to be habitable on its surface is thought to be highest in the habitable zone. When looking for habitable exoplanets, it's therefore more efficient to focus on the habitable zone.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-04-22T18:14:08.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2239" PostTypeId="2" ParentId="2234" CreationDate="2014-04-22T19:17:44.203" Score="12" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If so by how much does it &quot;spoil&quot; the view of stars and galaxies etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are several very different issues related to your question. Let's tackle them one by one.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Atmospheric refraction&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, the Earth's atmosphere refracts light. One notable effect is that objects near horizon appear higher than they should be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pCJOn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the Sun (or any other object) rises earlier than it should, and sets later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Atmospheric_refraction&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Atmospheric_refraction&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It affects astrometry, because it messes up the coordinates of objects near the horizon. Other than that, it does not cause great harm to astronomy.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;And then there are several effects harmful to astronomy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. Seeing&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we call in astronomy &quot;seeing&quot; is just air turbulence. Because air moves around due to heat, it distorts the image continuously. This is how the Moon looks like in a telescope - notice the shifting around due to seeing (animated GIF, reload if it's not moving, or open in new tab):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZxTIF.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we lived in a vacuum, that image would be static, and perfectly clear. But turbulence causes the shifting. Also, it causes a blur - the image is less sharp than it could be. It's also what causes the stars to &quot;twinkle&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's basically the same phenomenon that you see in summer along the hot walls of a house under sunlight - the same shimmering effect of the image seen through the hot rising air along the walls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For amateur astronomers, seeing affects high-resolution targets such as planets, the Moon, double stars. It does not affect targets that are perceived to be &quot;low-resolution&quot; in an amateur telescope: galaxies, nebulae, star clusters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More reading:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Astronomical_seeing&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Astronomical_seeing&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since seeing depends on weather, it can be predicted to some extent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cleardarksky.com/csk/&quot; rel=&quot;noreferrer&quot;&gt;http://cleardarksky.com/csk/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Light pollution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atmosphere also scatters and reflects light back to Earth. All the city lights around you are scattered by haze and dust in the air, and some of that light goes back down into your eye, creating sort of a &quot;light fog&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, objects that are not too bright are masked by light pollution and become less visible, or invisible altogether.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yngVX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bigger telescope will help you pierce through the haze of light pollution, but only to some extent. The best way to see faint objects is to put the scope in the car and drive far away from the city.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a light pollution map:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jshine.net/astronomy/dark_sky/&quot; rel=&quot;noreferrer&quot;&gt;http://www.jshine.net/astronomy/dark_sky/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For amateur astronomers, light pollution affects the &quot;faint fuzzies&quot;: galaxies, nebulae, star clusters. It has no effect whatsoever on bright objects such as: planets, the Moon, many double stars.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Other ways the atmosphere can negatively impact observations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;clouds - this is an obvious one&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;transparency - different from clouds, it includes haze, dust, particulate matter, water vapor. It affects low contrast objects such as galaxies.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2014-04-22T19:17:44.203" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2240" PostTypeId="1" AcceptedAnswerId="2254" CreationDate="2014-04-23T01:10:08.957" Score="23" ViewCount="3566" Body="&lt;p&gt;So because I can only really think of space-time in 2-dimensions like a sheet of something, my assumptions might be wrong to begin with. I was watching a YouTube video on black-holes and there was a quip regarding black holes and their warping of space time, to the point where the narrator said something along the lines of &lt;em&gt;“there is nothing at the center of the black hole but gravity has warped space time so much that it's this ’pinhole’ like affect that causes the blackhole behavior”&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;When a star dies and collapses into a black hole, what is at its center? The star's mass compacted into the size of the plank length, or something similarly small? &lt;strong&gt;Is there really nothing at the center of a black hole? Surely, the core collapsed into something really small right?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;What is meant by singularity? Is it just the warping of space-time that makes it this way?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1468" LastEditorUserId="14078" LastEditDate="2016-11-11T19:40:35.097" LastActivityDate="2020-01-05T19:19:33.487" Title="What is a singularity? What is at the center of a black hole? Specifically regarding space-time" Tags="&lt;black-hole&gt;&lt;space-time&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="2241" PostTypeId="1" AcceptedAnswerId="2245" CreationDate="2014-04-23T01:50:51.147" Score="14" ViewCount="8889" Body="&lt;p&gt;I've heard from renowned astrophysicists that we don't yet know whether or not the Universe is infinite. How is that possible regarding the big bang theory is accepted (as they all do)? Are they referring to the existence of other Universes when they say it could be infinite, or what?&lt;/p&gt;&#xA;" OwnerUserId="1469" LastEditorUserId="67" LastEditDate="2014-04-23T10:02:43.567" LastActivityDate="2019-08-04T07:34:14.903" Title="How can the universe be infinite?" Tags="&lt;universe&gt;&lt;big-bang-theory&gt;&lt;size&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2242" PostTypeId="2" ParentId="2241" CreationDate="2014-04-23T01:55:38.880" Score="5" Body="&lt;p&gt;It is known that the universe that we can see in our telescopes is less than the total universe. Since we cannot see what is beyond the visual edge, we cannot determine if the universe is infinite or finite.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-04-23T01:55:38.880" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2245" PostTypeId="2" ParentId="2241" CreationDate="2014-04-23T02:28:46.453" Score="14" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What I mean is, even if it is capable of expanding, if everything originated at the big bang how can that space turn to be infinite in a finite time - the age of the universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the standard &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambda-CDM_model&quot; rel=&quot;noreferrer&quot;&gt;ΛCDM model&lt;/a&gt; of the Big Bang, the universe is infinite and has &lt;em&gt;always&lt;/em&gt; been such. The Big Bang singularity happened &lt;em&gt;everywhere&lt;/em&gt;, in the sense that far back enough in time, the density diverges to infinity at every place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is just a particular model--it assumes that the universe if spatially flat and is globally homogeneous and isotropic. There are extended models in which it is not exactly flat, and so could be finite even if it is still homogeneous and isotropic (if the curvature is even slightly positive). And of course we don't actually &lt;em&gt;know&lt;/em&gt; whether it is homogeneous and isotropic at scales much larger than we actually see. Some inflationary models imply that it isn't.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To clarify: the ΛCDM model uses the assumes a spatially flat &lt;a href=&quot;http://en.wikipedia.org/wiki/Friedmann%E2%80%93Lema%C3%AEtre%E2%80%93Robertson%E2%80%93Walker_metric&quot; rel=&quot;noreferrer&quot;&gt;FLRW solution&lt;/a&gt; of general relativity, in which space is the Euclidean $3$-space The Euclidean $3$-space is the only &lt;em&gt;flat&lt;/em&gt; homogeneous and isotropic $3$-manifold, so there no way to make it finite without violating at least one of those modeling assumption (e.g., a flat torus could have the same form for the metric, but would not be globally isotropic).&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditDate="2014-04-23T17:08:27.600" LastActivityDate="2014-04-23T17:08:27.600" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2246" PostTypeId="2" ParentId="2240" CreationDate="2014-04-23T03:24:00.207" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;2) What is a singularity? Is it just the warping of space time that makes it this way?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A singularity is a point in space-time where the current physical models we use to describe forces and how particles behave break down i.e. gives us no information (or give us &lt;em&gt;all&lt;/em&gt; the information) and thus no predictions can be made.&#xA;Also a singularity can be described from different perspectives:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A spacetime singularity is a breakdown in the geometrical structure of space and time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In general relativity, a singularity is a place that objects or light rays can reach in a finite time where the curvature becomes infinite, or space-time stops being a manifold.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But it all falls back to what I mention at the beginning.&lt;/p&gt;&#xA;" OwnerUserId="1469" LastActivityDate="2014-04-23T03:24:00.207" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2247" PostTypeId="1" AcceptedAnswerId="2253" CreationDate="2014-04-23T05:30:55.520" Score="22" ViewCount="1923" Body="&lt;p&gt;The Sun is huge, when compared to the Moon. Despite the huge difference in their size and distance from Earth, is it purely coincidental that they both look almost the same from Earth?&lt;/p&gt;&#xA;" OwnerUserId="1470" LastEditorUserId="1891" LastEditDate="2020-12-23T22:52:54.797" LastActivityDate="2020-12-23T22:52:54.797" Title="Is it a coincidence that both the Sun and the Moon look of same size from the Earth?" Tags="&lt;the-moon&gt;&lt;earth&gt;&lt;the-sun&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="2249" PostTypeId="1" CreationDate="2014-04-23T11:49:00.957" Score="1" ViewCount="470" Body="&lt;p&gt;How do the scientists calculate the correct time to launch the Mars mission so that the satellite travel time is less? How they are synchronizing the speed of satellite with respect to earth and mars ?&#xA;Refer this link for more understanding: &#xA;&lt;a href=&quot;http://www.sankara.net/mom.html&quot; rel=&quot;nofollow&quot;&gt;Animation of MOM and Maven launch&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1411" LastEditorUserId="67" LastEditDate="2014-04-23T20:43:13.383" LastActivityDate="2017-05-04T19:28:00.857" Title="How are the speed, distance and time of India's MOM and USA's Maven Mars missions calculated?" Tags="&lt;distances&gt;&lt;mars&gt;&lt;speed&gt;&lt;astrophysics&gt;&lt;artificial-satellite&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2250" PostTypeId="1" AcceptedAnswerId="2258" CreationDate="2014-04-23T12:28:30.593" Score="0" ViewCount="439" Body="&lt;p&gt;I think that the title is completely clear, but here's an expansion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was just reading about &lt;a href=&quot;http://en.wikipedia.org/wiki/Mass%E2%80%93luminosity_relation&quot; rel=&quot;nofollow&quot;&gt;Mass-luminosity relation&lt;/a&gt; that says &lt;strong&gt;massive stars are more luminous than tiny ones&lt;/strong&gt;. Well, let's talk about main-sequence stars for now. This relation becomes interesting when it's mixed with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hertzsprung%E2%80%93Russell_diagram&quot; rel=&quot;nofollow&quot;&gt;Hertzsprung-Russel diagram&lt;/a&gt; that says &lt;strong&gt;young stars are more luminous than older ones&lt;/strong&gt; (Originally says that &lt;em&gt;hotter&lt;/em&gt; stars are more luminous and we know that hotter stars are younger).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So mixing these two relations produces some kind of &lt;em&gt;Mass-age relation&lt;/em&gt; which says &lt;strong&gt;young stars are more massive&lt;/strong&gt;, that is a star loses its mass over time, right? (Please tell me if I'm going wrong)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this Mass-age relation is right, how do stars lose this extra mass during their evolution? And where does this stray matter go?&lt;/p&gt;&#xA;" OwnerUserId="1473" LastEditorUserId="354" LastEditDate="2015-05-26T23:25:06.613" LastActivityDate="2015-05-26T23:25:06.613" Title="Putting mass-luminosity relation and Hertzsprung-Russel diagrams together leads us to a mass-age relation; so how do stars lose their mass over time?" Tags="&lt;star&gt;&lt;mass&gt;&lt;stellar-evolution&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2251" PostTypeId="2" ParentId="2250" CreationDate="2014-04-23T12:33:01.837" Score="1" Body="&lt;p&gt;There are several causes for mass loss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is transformation of mass into energy by means of nuclear fusion, and the energy is radiated away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the big one is stellar winds: particles ejected from the star surface.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-04-23T12:33:01.837" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2253" PostTypeId="2" ParentId="2247" CreationDate="2014-04-23T15:24:51.953" Score="32" Body="&lt;p&gt;The coincidence isn't so much that they appear very similar sizes from Earth, but that we are alive to see them at the point in time in which they appear very similar sizes.  The moon is slowly moving away from the Earth, and at some point in the future the moon will be unable to totally eclipse the sun and conversely, if you could step far into prehistory, you would be able to see the moon with a much greater angular diameter than you see it now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most research I've found on the topic seem to be unavailable through my institute, however I did find one paper, &lt;a href=&quot;http://articles.adsabs.harvard.edu//full/1973ApJ...180..307C/0000313.000.html&quot; rel=&quot;noreferrer&quot;&gt;&quot;Outcomes of tidal evolution&quot;&lt;/a&gt;, which references results from Goldreich's research on the subject.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This qualitative description of the eventual disruption of the Earth-Moon system is confirmed by the results of Goldreich's numerical integration, which showed that the moon will recede to 75 Earth radii, when spin-orbit synchronism will be reached; then the Moon's orbit will decay steadily inward because of the influence of the Sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For reference, the Moon is currently at a distance of approximately 60.3 Earth radii.  As such, the moon will steadily move away until synchronism would be reached, and from that point begin to recede towards the Earth due to the tidal affects of the Sun on the Earth disturbing the synchronization. It would seem that at some eventual point in the far distant future, it will return to this coincidental position once again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Counselman III, Charles C. &quot;Outcomes of tidal evolution.&quot; The Astrophysical Journal 180 (1973): 307-316.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="1288" LastEditDate="2014-08-10T13:00:33.917" LastActivityDate="2014-08-10T13:00:33.917" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2254" PostTypeId="2" ParentId="2240" CreationDate="2014-04-23T17:18:41.103" Score="30" Body="&lt;p&gt;This is more of a question for the Physics stack, but I'll give it a shot, since it's fairly basic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to understand something before we begin. The theoretical framework we have to gauge and answer this sort of thing is called General Relativity, which was proposed by Einstein in 1915. It describes things such as gravity, black holes, or just about any phenomena where large densities of mass or energy are involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's another chapter in Physics called Quantum Mechanics. This describes, usually, what happens at very small scales - things that are super-tiny.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both GR and QM are fine in their own way. Both are tested against reality and work very well. &lt;em&gt;But they are not compatible with each other.&lt;/em&gt; Meaning: you cannot describe a phenomenon from a GR and a QM perspective, both at once. Or meaning: we don't have a coherent set of equations that we could write down, and then &quot;extract&quot; out of them either a GR-like view of reality, or a QM-like view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, the center of a black hole is both very high mass density and very high gravity (and therefore right in the field of GR), and very small (and therefore &quot;quantum-like&quot;). To properly deal with it, we'd have to reconcile GR and QM and work with both at once. &lt;em&gt;This is not possible with current physics.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We pretty much have to stick to GR only for now, when talking about black holes. This basically means that anything we say about the center of a black hole is probably incomplete, and subject to further revision.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A star dies, collapses into a black hole, what is at the center? The&#xA;  star's mass compacted into the size of the plank length or something&#xA;  similarly small? Is there really nothing at the center of a black&#xA;  hole?, surely the core collapsed into something, just really small&#xA;  right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to General Relativity, it collapses all the way down to nothing. Not just &quot;very small&quot;, but smaller and smaller until it's exactly zero in size. Density becomes infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't say &quot;Plank length&quot; because, remember, we can't combine GR and QM, we just don't know how. All we have here is GR, and GR says it goes all the way down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is quite possible that the singularity is not physical, but just mathematical - in other words, whatever is at the center is not actually zero size. Quantum mechanics in particular would be offended by zero-size things. But we can't say for sure because our knowledge here is incomplete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using words such as &quot;size&quot; (which implies space) and &quot;becomes&quot; (which implies time). But both space and time in the context of a black hole are very seriously warped. The &quot;becoming&quot; of a black hole all the way down to the zero-size dot is a reality only for the unlucky observer that gets caught in it. But for a distant, external observer, this process is slowed and extended all the way to plus infinity (it's only complete after an infinitely long time). Both observers are correct, BTW.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when we are saying &quot;density is infinite and size is zero at the singularity&quot;, this language applies to the unfortunate observer being dragged down in the middle of the initial collapse of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But from the perspective of the distant observer, a black hole is still a chunk of mass (the original star) in a non-zero volume (the event horizon of the BH). To this observer, the density of that object is finite, and its size is definitely not zero. From this perspective, anything falling into the BH never quite finishes falling, but just slows down more and more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both observers are correct. So, keep in mind, when I talk about &quot;infinite density&quot;, that's the inside observer point of view.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is a singularity? Is it just the warping of space time that makes&#xA;  it this way?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You get a singularity whenever there's a division by zero in the equations, or when the equations misbehave somehow at that point. There are many different kinds of singularities in science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematical_singularity&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Mathematical_singularity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of a black hole, the center is said to be a gravitational singularity, because density and gravity are suggested to become infinite, according to the GR equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GR says: when you have a lump of matter that's big enough, it starts to collapse into itself so hard, there's nothing to stop it. It keeps falling and falling into itself, with no limit whatsoever. Extrapolate this process, and it's easy to see that the size of it tends to zero, and density tends to an infinite value.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put another way - if density becomes large enough, gravity is so huge, no other force is strong enough to resist it. It just crushes all barriers that matter raises to oppose further crushing. That lump of matter simply crushes itself, its own gravity pulls it together smaller and smaller... and smaller... and so on. According to current theories, there's nothing to stop it (QM might stop it, but we cannot prove it, because we don't have the math). So it just spirals down in a vicious cycle of ever-increasing gravity that increases itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space and time are really pathologic inside the event horizon. If you are already inside, there's no way out. This is not because you can't move out fast enough, but because &lt;em&gt;there's really no way out&lt;/em&gt;. No matter which way you turn, you're looking towards the central singularity - in both space and time. There is no conceivable trajectory that you could draw, starting from the inside of the event horizon, that leads outside. All trajectories point at the singularity. All your possible futures, if you're inside the event horizon, end at the central singularity.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, why the center of a black hole is called a &quot;singularity&quot;? Because all sorts of discontinuities and divisions by zero jump out of the equations, when you push math to the limit, trying to describe the very center of a black hole, within a GR frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_singularity&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Gravitational_singularity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speaking in general, physicists don't like singularities. In most cases, this is an indication that the mathematical apparatus has broken down, and some other calculations are necessary at that point. Or it might indicate that new physics are taking place there, superseding the old physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One last thing: just because we don't have a combined GR/QM theory to fully describe the center of black holes, that doesn't mean a pure GR research in this area is &quot;wrong&quot; or &quot;useless&quot;. It doesn't mean one could imagine some arbitrary fantasy taking place inside a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomers these days are starting to &lt;a href=&quot;http://chandra.harvard.edu/photo/2013/4c2930/&quot; rel=&quot;noreferrer&quot;&gt;observe cosmic objects that are very much like black holes&lt;/a&gt;, and their observed properties are in very close accord with what GR predicts for such things. So research in this field must continue, because it's clearly on the right track, at least in the ways we can verify today in astronomy.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2019-01-30T08:10:07.073" LastActivityDate="2019-01-30T08:10:07.073" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="2256" PostTypeId="2" ParentId="2183" CreationDate="2014-04-23T21:32:07.640" Score="0" Body="&lt;p&gt;I think you should think of something in the area of apps that have a (public) API, and from there you could build up your own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;APIs I can think of are those of Google Earth. You can find a reference &lt;a href=&quot;https://developers.google.com/earth/documentation/reference/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1469" LastActivityDate="2014-04-23T21:32:07.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2257" PostTypeId="2" ParentId="2249" CreationDate="2014-04-23T21:48:03.747" Score="1" Body="&lt;p&gt;There is an extremely accurate model of the solar system and the positions of the planets and so on. So basically you have to lay Newton's equations for Mars, the probe, Earth and so on, you have an equation that models Mars's path, and you have equations for your probe, so you get some x's and y's that eventually will have to match if you want to reach Mars. From there you can then calculate the values for the initial conditions of the probe -such as velocity and direction- that will satisfy you equations.&lt;/p&gt;&#xA;" OwnerUserId="1469" LastEditorDisplayName="user5434" LastEditDate="2017-05-04T19:28:00.857" LastActivityDate="2017-05-04T19:28:00.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2258" PostTypeId="2" ParentId="2250" CreationDate="2014-04-23T23:04:28.460" Score="5" Body="&lt;p&gt;There is an alternative interpretation: Massive stars burn their fuel much faster than tiny ones. Hence massive stars are short-lived in comparison to tiny stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.astro.umd.edu/~ssm/ASTR100/lecture20.pdf&quot; rel=&quot;noreferrer&quot;&gt;lifetime of a star&lt;/a&gt; is proportional to about the inverse of the cube of its mass.&#xA;The Sun's lifetime is about $10^{10}$ years, hence for stars with 10 solar masses the lifetime is about $10^7$ years, and for stars with 0.1 solar masses it's about $10^{13}$ years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But stars also loose mass, e.g. by &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_wind&quot; rel=&quot;noreferrer&quot;&gt;stellar wind&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Coronal_mass_ejection&quot; rel=&quot;noreferrer&quot;&gt;coronal mass ejections&lt;/a&gt;. Heavy stars may &lt;a href=&quot;http://www.astro.ru.nl/~onnop/education/stev_utrecht_notes/chapter12-13.pdf&quot; rel=&quot;noreferrer&quot;&gt;loose mass&lt;/a&gt; (parts of their envelope) e.g. by &lt;a href=&quot;http://en.wikipedia.org/wiki/Variable_star#Eruptive_variable_stars&quot; rel=&quot;noreferrer&quot;&gt;pulsation&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Radiation_pressure&quot; rel=&quot;noreferrer&quot;&gt;radiation pressure&lt;/a&gt; during their &lt;a href=&quot;http://en.wikipedia.org/wiki/Red_giant&quot; rel=&quot;noreferrer&quot;&gt;red giant&lt;/a&gt; or their &lt;a href=&quot;http://en.wikipedia.org/wiki/Wolf%E2%80%93Rayet_star&quot; rel=&quot;noreferrer&quot;&gt;Wolf-Rayet&lt;/a&gt; phases, before they explode as a supernova.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-04-23T23:04:28.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2259" PostTypeId="1" CreationDate="2014-04-24T02:07:43.480" Score="-2" ViewCount="140" Body="&lt;p&gt;A theory has been said that the patterns and positions help identify where valuable gemstones are hidden. I am wondering if this has any truth in it.&lt;/p&gt;&#xA;" OwnerUserId="1480" LastActivityDate="2014-04-24T09:52:52.473" Title="Relation between position of stars and locations of gemstones?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2014-04-24T15:12:39.090" ContentLicense="CC BY-SA 3.0" />
  <row Id="2261" PostTypeId="2" ParentId="2213" CreationDate="2014-04-24T04:09:32.150" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it right that what we see/feel in every moment is projection of some 4D reality into our 3D world noticed by our senses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your and our world is not 3D, it is 4D, you just said it but then mysteriously you changed your approach. Our 'reality', our 'world' is 4D, there are 3 spacial dimensions and 1 time dimension. But you can live only in one point of space-time &lt;em&gt;at a time&lt;/em&gt;.&#xA;Furthermore what you can &lt;em&gt;see&lt;/em&gt; or experience at any given time is a slice of the space-time that we live in, the slice depends on your speed and point in space-time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If we considered that 5th dimension exists and that we could put&#xA;  infinite number of 4D spaces into 5D space, then we would end up with&#xA;  infinite number of parallel realities&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think part 2 of your question has been answered. My short answer is yes! That's what string theory predicts.&lt;/p&gt;&#xA;" OwnerUserId="1469" LastActivityDate="2014-04-24T04:09:32.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2262" PostTypeId="2" ParentId="2166" CreationDate="2014-04-24T04:36:35.517" Score="0" Body="&lt;p&gt;The redness of the moon is due to Earth's atmosphere refraction of Sun rays and pollution - dust particles etc. The intensity of the colouring will depend on the path the moon takes through the Earth's corona.&#xA;If it goes right through the middle  then we will see a darker moon, while if it travels close to the edges then the influence of refraction and pollution will be greater and thus a redder moon.&lt;/p&gt;&#xA;" OwnerUserId="1469" LastActivityDate="2014-04-24T04:36:35.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2263" PostTypeId="2" ParentId="2259" CreationDate="2014-04-24T09:52:52.473" Score="2" Body="&lt;p&gt;Short answer: No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer: No, there is no truth in that, at all. No body out of our Solar System could interfere in Earth's techtonics nor in Geology in any way.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-04-24T09:52:52.473" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2264" PostTypeId="2" ParentId="2237" CreationDate="2014-04-24T21:03:23.773" Score="5" Body="&lt;p&gt;We could err by being 'chauvinists' as Carl Sagan would have said, because we are reasoning taking in account the biology and chemistry of life of 1 planet only: a sample of 1...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before 1995, we thought that alien solar systems would be similar to ours, with small rocky planets closer to the star and giant planets further out.&#xA;Almost nobody thought about circumbinary planets, planets with periods under one day, scorched jupiters ridiculously close to their stars, ultra compact systems with four or more planets crowded in few tenths of Astronomical Units, free-floating cloudy dwarfs with precipitation of melted iron and hot sand (brown dwarfs? planets? planemos?) and the list goes on and on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enceladus, that is quite small for planetary standards and that are far from the most optimistic and inclusive Goldilocks boundary has a subsurface ocean, of water.&lt;/p&gt;&#xA;" OwnerUserId="910" LastActivityDate="2014-04-24T21:03:23.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2265" PostTypeId="1" AcceptedAnswerId="2270" CreationDate="2014-04-24T21:06:52.620" Score="9" ViewCount="301" Body="&lt;p&gt;I learned that gravity always pulls, it is always attractive. There is no antigravity, no &quot;negative&quot;  mass. I also learned that most matter is dark, invisible. We know that because we see visible matter moving towards invisible bundles of dark matter and other effects like accelaration or gravity lenses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I ask myself is: Couldn't there be &quot;negative&quot; bundles of mass just the other way that pushes matter away instead of invisible dark matter that pulls it? Has someone tried to find something like that? &#xA;What is the proof for the fact &quot;Gravity is always attractive&quot; beside that fact that it seems to be? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Granted, it may be a stupid question, but I lack a bit in physics education that is not very related to chemistry.&lt;/p&gt;&#xA;" OwnerUserId="1483" LastActivityDate="2020-04-24T20:59:54.510" Title="Has someone looked the other way?" Tags="&lt;gravity&gt;&lt;dark-matter&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2266" PostTypeId="2" ParentId="2265" CreationDate="2014-04-24T21:49:55.573" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Negative_mass&quot; rel=&quot;nofollow&quot;&gt;Negative energy&lt;/a&gt; should be gravitationally repulsive. But naturally occuring negative energy is too weak to be noticible on a large scale. Macroscopic known is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Casimir_effect&quot; rel=&quot;nofollow&quot;&gt;Casimir effect&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From chemistry you'll certainly know &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_force&quot; rel=&quot;nofollow&quot;&gt;van-der-Waals forces&lt;/a&gt;. There are several kinds of van-der-Waals forces. One of them are London-van der Waals forces, thought to be related to the Casimir effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But electromagnetism is too strong compared to gravity in theses cases for reasonable experiments to show gravitational repulsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dark_energy&quot; rel=&quot;nofollow&quot;&gt;Dark energy&lt;/a&gt; might be interpreted as &quot;gravitational repulsion&quot;; but this won't be effective for the gravitational force between two objects, if it's distributed equally, as &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambda-CDM_model&quot; rel=&quot;nofollow&quot;&gt;current models&lt;/a&gt; suggest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.earthtech.org/publications/davis_STAIF_conference_1.pdf&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; a paper about a couple of attempts to produce negative energy in the lab.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-04-25T13:59:48.347" LastActivityDate="2014-04-25T13:59:48.347" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2267" PostTypeId="1" AcceptedAnswerId="2273" CreationDate="2014-04-24T23:22:11.583" Score="3" ViewCount="437" Body="&lt;p&gt;Does the Sun spread hard radiation waves around the Solar System?  If so, why are we safe here (or are we even safe)?&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="1386" LastEditDate="2014-04-25T22:40:04.210" LastActivityDate="2014-07-02T06:29:13.077" Title="Does the Sun have hard radiation?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;radiation&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2268" PostTypeId="2" ParentId="2267" CreationDate="2014-04-25T02:29:57.987" Score="2" Body="&lt;p&gt;The sun sends out solar radiation, which is strong enough to be dangerous to living things. Much of it, however, is prevented from reaching us due to the Earth's electromagnetic field, which surrounds the Earth.  This sphere originates at the poles (flowing out of the negative pole and into the positive pole) much like one on a standard magnet. The grade results in it being more near the surface at the poles, which is why we can see aurora in the more northern and southern parts of the Earth.  The aurora is generated by a reaction between solar particles as they travel along the electromagnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From an &lt;a href=&quot;http://www.esa.int/Our_Activities/Space_Science/Cluster/Earth_s_magnetic_field_provides_vital_protection&quot; rel=&quot;nofollow&quot;&gt;article at Universe Today&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the magnetic field of the Earth protects the planet from space radiation. The biggest culprit is the Sun’s solar wind. These are highly charged particles blasted out from the Sun like a steady wind. The Earth’s magnetosphere channels the solar wind around the planet, so that it doesn’t impact us. Without the magnetic field, the solar wind would strip away our atmosphere – this is what probably happened to Mars. The Sun also releases enormous amounts of energy and material in coronal mass ejections. These CMEs send a hail of radioactive particles into space. Once again, the Earth’s magnetic field protects us, channeling the particles away from the planet, and sparing us from getting irradiated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Read more: &lt;a href=&quot;http://www.universetoday.com/27005/earths-magnetic-field/#ixzz2zrSVQnsX&quot; rel=&quot;nofollow&quot;&gt;http://www.universetoday.com/27005/earths-magnetic-field/#ixzz2zrSVQnsX&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="473" LastEditDate="2014-04-25T09:26:14.663" LastActivityDate="2014-04-25T09:26:14.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2270" PostTypeId="2" ParentId="2265" CreationDate="2014-04-25T04:46:45.510" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What I ask myself is: Couldn't there be &quot;negative&quot; bundles of mass just the other way that pushes matter away instead of invisible dark matter that pulls it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The galaxy rotation curve indicates a (positively massed) dark matter distribution that is close to spherically symmetric; cf. &lt;a href=&quot;http://en.wikipedia.org/wiki/Dark_matter_halo&quot;&gt;dark matter halo&lt;/a&gt;. I take it that you are asking whether instead of modeling it as a positive-mass distribution around the galaxy, we can model it instead as a negative-mass distribution surrounding the galaxy pushing the stars inward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, that's not possible. Since the effect is spherically symmetric, the hypothetical negative-mass distribution would need to be spherically symmetric as well. However, Newton's shell theorem guarantees that &lt;em&gt;everywhere&lt;/em&gt; inside a void surrounded by such a distribution with this symmetry, whatever the sign of the mass is, the gravitational force vanishes. So a negatively-massed dark matter distribution outside the galaxy would do nothing at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general relativity, the analogous statement is guaranteed by Birkhoff's theorem, and although GTR is nonlinear (and there are stars in the way), in our galaxy gravity is weak enough to be handled by the linearized theory.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the proof for the fact &quot;Gravity is always attractive&quot; beside that fact that it seems to be? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In general relativity, gravity doesn't have to be attractive. For example, inside a perfect fluid with density energy $\rho$ and pressure $p$, $\rho+3p&amp;lt;0$ would imply that gravity is locally repulsive in the sense that a ball of test particles initially at rest expands rather than contracts under gravitational freefall. More generally, the &lt;em&gt;strong energy condition&lt;/em&gt; characterizes whether or not gravity is attractive in this sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is no known material that violates the strong energy condition, besides possibly dark energy, which is a cosmological constant in the standard ΛCDM model, as Gerald says.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Has someone tried to find something like that?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are many studies and simulations that consider dark matter with a nonzero pressure component, sometimes even anisotropic pressure components. For example, &quot;hot&quot; dark matter would have $p\sim\rho/3$, while &quot;cold&quot; dark matter would have $p\sim 0$, &quot;warm&quot; dark matter somewhere inbetween. The CDM in the ΛCDM model stands for &quot;cold dark matter&quot;, naturally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are more exotic possibilities with &lt;em&gt;anisotropic&lt;/em&gt; pressure components, etc. But I don't know of any paper that looked specifically for $\rho&amp;lt;0$ dark matter, and for the above reasons I'd be surprised if one existed.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditDate="2014-04-25T08:35:01.050" LastActivityDate="2014-04-25T08:35:01.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2271" PostTypeId="1" AcceptedAnswerId="2272" CreationDate="2014-04-25T15:49:22.303" Score="1" ViewCount="920" Body="&lt;p&gt;I know not all stars can &quot;go nova&quot;, and that of those that do there are different types of novae, among which are some that aren't even explosions (at least not as we think of them), and that even the ones that &lt;em&gt;are&lt;/em&gt; explosions are really not like what we do with TNT or C4 anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, I can't shake the comparison from my head, and that got me thinking: Are there any circumstances under which one star going nova can trigger novae in other nearby stars, creating a chain reaction?&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2014-04-25T16:55:32.037" Title="&quot;Chain reaction&quot; supernovae?" Tags="&lt;star&gt;&lt;supernova&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2272" PostTypeId="2" ParentId="2271" CreationDate="2014-04-25T16:55:32.037" Score="2" Body="&lt;p&gt;Be careful when saying &quot;nearby stars&quot;. The scales involved in &quot;nearby&quot; are beyond what most people imagine. If our Sun was the size of a soccer ball located in California, the nearest star would be another soccer ball located in Greenland.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say the average distance between &quot;nearby&quot; stars is a few light years. Let's say an average supernova is like the recent one in the M82 galaxy. That supernova, at, say, 10 light years, would be less bright and less hot than the Sun as seen from Earth. To make that supernova appear as bright and as hot as the mid-day Sun, you'd have to be at 1/3 light years from it (the supernova would have to be in the middle of the Oort cloud).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interstellar distances are stupendous. Even a supernova cannot do much to compensate for distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if the stars are very close to each other? They would pretty much have to orbit each other, forming a double star. Moreover, both would have to be on the verge of going supernova. That's a pretty unlikely situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even then, there are many kinds of supernova. There's type Ia, where a dwarf orbiting a red giant leeches too much material from the giant and eventually explodes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/16ZQZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there's the core collapse supernova, where a star is simply too bloated and too old to sustain enough pressure in the core; the core collapses into itself due to gravity and re-ignites fusion, effectively exploding like a bomb and ejecting the outer layers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e2etE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is a nearby stellar explosion going to increase instability in its about-to-blow-up companion, or is it going to reduce its chances of exploding? The answer is not trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line: I wouldn't say it's mathematically impossible, but it's a combination of quite unlikely scenarios. In a &quot;normal&quot; galactic neighborhood like ours, it's never going to happen.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2014-04-25T16:55:32.037" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2273" PostTypeId="2" ParentId="2267" CreationDate="2014-04-25T17:13:54.927" Score="5" Body="&lt;p&gt;The Sun outputs several different kinds of things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Electromagnetic radiation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun is (partially) a &lt;a href=&quot;http://en.wikipedia.org/wiki/Black-body_radiation&quot; rel=&quot;noreferrer&quot;&gt;black-body radiator&lt;/a&gt; at a temperature of near 6000 K, and therefore emits all sorts of electromagnetic energy, including UV and X rays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CpCQh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UV is &lt;a href=&quot;http://en.wikipedia.org/wiki/Ozone#Importance_to_surface-dwelling_life_on_Earth&quot; rel=&quot;noreferrer&quot;&gt;stopped in the upper atmosphere&lt;/a&gt;. X rays are absorbed by the whole atmosphere, and are pretty weak anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun's EM activity in terms of harsh radiation increases greatly if there's a &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_flare&quot; rel=&quot;noreferrer&quot;&gt;solar flare&lt;/a&gt; blowing up. Even then, we're pretty safe on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;High energy particles&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_wind&quot; rel=&quot;noreferrer&quot;&gt;solar wind&lt;/a&gt; is basically a flux of particles (nuclei of hydrogen, etc) shooting out of the Sun at pretty high energies. These are charged particles. When approaching the Earth, the Earth's magnetic field for the most part deflects them. If you're on Earth, you're safe. Even in space, the flux is not very great; you'd be safe inside a spacecraft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/46Oe0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, during a solar flare, the flux of particles increases greatly. The Earth is safe, but an astronaut en route to Mars would be in a pretty unsafe situation.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2014-04-25T17:13:54.927" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2275" PostTypeId="1" AcceptedAnswerId="2277" CreationDate="2014-04-25T22:20:56.983" Score="5" ViewCount="252" Body="&lt;p&gt;I realize radiation and extreme temperatures would probably destroy life on the surface of most planets and moons, but could life exist beneath the surface (e.g. like earth worms on earth)?&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2014-05-14T08:24:50.650" Title="Could there be life beneath the surface of Mars or moons?" Tags="&lt;the-moon&gt;&lt;mars&gt;&lt;life&gt;&lt;surface&gt;" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2276" PostTypeId="1" CreationDate="2014-04-26T01:11:09.360" Score="3" ViewCount="473" Body="&lt;p&gt;The virgo cluster is currently red-shifted by 0.003 according to Wikipedia; this indicates that the cluster is expanding. Will the cluster's enormous gravity eventually &quot;win&quot; and pull the milky way/local group back toward itself, or will our current recessional velocity combined with dark energy cause us to recede off to infinity?&lt;/p&gt;&#xA;" OwnerUserId="1490" LastActivityDate="2014-04-26T10:59:59.867" Title="Is the local group bound to the Virgo cluster?" Tags="&lt;cosmology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2277" PostTypeId="2" ParentId="2275" CreationDate="2014-04-26T03:11:13.940" Score="7" Body="&lt;p&gt;That's one of the big questions. ESA scientists, at least, think it's worth looking for underground life. See &lt;a href=&quot;http://www.esa.int/esapub/bulletin/bulletin126/bul126c_vago.pdf&quot;&gt;ExoMars mission&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More likely than earth worms are microbes, since some microbes on Earth live under similar conditions as presumed for underground Mars.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-04-26T03:11:13.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2278" PostTypeId="1" AcceptedAnswerId="2297" CreationDate="2014-04-26T09:45:16.310" Score="4" ViewCount="6448" Body="&lt;p&gt;Could a (regional or global) fallout of radioactive material be a &quot;bonus&quot; disaster effect of an asteroid impact? My reasoning for how such a scenario maybe could be possible:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) Some asteroids maybe consist of lots of radioactive heavy metals because they were initially formed by ejecta from the cores of planets during planetary collisions. Maybe even an airburst of a rather small such an asteroid could cause widespread dangerous radiak fallout?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, Earth's initial uranium has fallen into the core. The uranium mines dig into concentrated deposits of uranium in the crust which came from uranium rich asteroids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) If any kind of asteroid large enough hits a uranium rich part of the Earth crust (making it the second one on the same spot), it could maybe eject much of Earth's own uranium and thus cause a dangerous fallout?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2014-04-26T10:12:49.073" LastActivityDate="2016-08-16T17:02:07.197" Title="Could an asteroid impact cause radioactive fallout?" Tags="&lt;asteroids&gt;&lt;impact&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2279" PostTypeId="2" ParentId="2276" CreationDate="2014-04-26T10:59:59.867" Score="2" Body="&lt;p&gt;It would seem so. The local group is part of the Virgo cluster and as such is considered to be gravitationally bound. Although the Virgo cluster and the local group are currently moving apart, the mass if the Virgo cluster will likely slow and reverse the recession over time, with the local group ultimately merging with the cluster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&#xA;&lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/V/Virgo+Cluster&quot; rel=&quot;nofollow&quot;&gt;http://astronomy.swin.edu.au/cosmos/V/Virgo+Cluster&lt;/a&gt;&#xA;&lt;a href=&quot;http://heasarc.nasa.gov/docs/cosmic/local_supercluster_info.html&quot; rel=&quot;nofollow&quot;&gt;http://heasarc.nasa.gov/docs/cosmic/local_supercluster_info.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1464" LastActivityDate="2014-04-26T10:59:59.867" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2280" PostTypeId="1" AcceptedAnswerId="2281" CreationDate="2014-04-26T16:11:55.247" Score="3" ViewCount="758" Body="&lt;p&gt;I was wondering, is the surface of Venus so hot that it would glow red in the dark (e.g. on the dark side of Venus)?  I am working on making a solar system simulator, and that would make for a great level if it is accurate.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2017-10-27T01:24:59.297" Title="Is the surface of Venus red hot?" Tags="&lt;venus&gt;&lt;planet&gt;&lt;surface&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2281" PostTypeId="2" ParentId="2280" CreationDate="2014-04-26T16:23:54.570" Score="4" Body="&lt;p&gt;The average surface temperature on Venus is 462 °C. You would need a temperature above 800 °C before you would notice any light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:&lt;br&gt;&#xA;Here are some calculated radiation outputs of the surface &lt;a href=&quot;https://drive.google.com/file/d/0B80IPeLPxBV3WnFCWWF6RTNuV28/edit?usp=sharing&quot; rel=&quot;nofollow&quot;&gt;temperature of Venus&lt;/a&gt;, &lt;a href=&quot;https://drive.google.com/file/d/0B80IPeLPxBV3a094ZkxveTdqdG8/edit?usp=sharing&quot; rel=&quot;nofollow&quot;&gt;a 1000 °C black body&lt;/a&gt; and &lt;a href=&quot;https://drive.google.com/file/d/0B80IPeLPxBV3VTk0UU5mZjc5dkk/edit?usp=sharing&quot; rel=&quot;nofollow&quot;&gt;the sun&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="1322" LastEditDate="2014-04-26T18:12:06.457" LastActivityDate="2014-04-26T18:12:06.457" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2283" PostTypeId="1" AcceptedAnswerId="2290" CreationDate="2014-04-26T23:05:28.680" Score="3" ViewCount="268" Body="&lt;p&gt;Let's assume the Moon had a considerable magnetic field to prevent the atmosphere dispersion caused by solar wind. Given the low gravity of the Moon, how many years would pass before all the gases (of an Earth like atmosphere) would escape its body? Would its gravity be sufficient to trap heavier gases like CO2?&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditDate="2014-04-27T10:37:48.397" LastActivityDate="2014-04-28T01:24:03.067" Title="If the moon still had a magnetic field how long would have been possible to keep an atmosphere?" Tags="&lt;the-moon&gt;&lt;gravity&gt;&lt;planetary-atmosphere&gt;&lt;magnetic-field&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2284" PostTypeId="1" AcceptedAnswerId="20886" CreationDate="2014-04-26T23:28:28.593" Score="1" ViewCount="454" Body="&lt;p&gt;Jupiter's magnetosphere extends &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetosphere_of_Jupiter&quot; rel=&quot;nofollow&quot;&gt;&quot;seven million kilometers in the Sun's direction and almost to the orbit of Saturn in the opposite direction&quot;&lt;/a&gt; and Saturn's magnetosphere extends towards the Sun on average &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetosphere_of_Saturn&quot; rel=&quot;nofollow&quot;&gt;22 Rs (Rs=60,330 km is the equatorial radius of Saturn).&lt;/a&gt;&#xA;My question is: do these magnetospheres actually touch each other when the planets are on the same line?&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditDate="2014-04-27T02:57:27.700" LastActivityDate="2017-06-03T13:42:24.310" Title="Jupiter's magnetosphere interacts with the one from Saturn?" Tags="&lt;jupiter&gt;&lt;magnetic-field&gt;&lt;saturn&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2285" PostTypeId="2" ParentId="2284" CreationDate="2014-04-27T02:11:59.997" Score="2" Body="&lt;p&gt;The simple answer is yes, because magnetic fields extend to infinity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question should ask if a significant interaction is present. My question would be, what is a significant interaction?&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-04-27T02:11:59.997" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2286" PostTypeId="2" ParentId="2283" CreationDate="2014-04-27T08:21:56.973" Score="3" Body="&lt;p&gt;Your question could be interpreted in two ways: Could a strong magnetic field be enough to hold an atmosphere in place, or could a magnetic field provide enough shielding for an atmosphere to be kept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, a magnetic field could never be enough to keep an atmosphere on a body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, a magnetic field would stop an atmosphere from being heavily irradiated and blown away, but the Moon's gravity is just too low.&lt;/p&gt;&#xA;" OwnerUserId="1494" LastActivityDate="2014-04-27T08:21:56.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2287" PostTypeId="1" AcceptedAnswerId="2288" CreationDate="2014-04-27T17:05:38.507" Score="2" ViewCount="713" Body="&lt;p&gt;Assuming you could be standing inside a black hole, would it be light or dark around you? &lt;/p&gt;&#xA;" OwnerUserId="1497" LastActivityDate="2014-04-27T18:10:09.203" Title="Is it dark inside a black hole?" Tags="&lt;black-hole&gt;&lt;light&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2288" PostTypeId="2" ParentId="2287" CreationDate="2014-04-27T17:27:11.690" Score="2" Body="&lt;p&gt;The only light you would see, would be from the stars outside the black hole. Any light that is generated inside the event horizon would be refracted towards the singularity. You wouldn't see the object in front of you since the light would not reach your eye.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-04-27T17:27:11.690" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2289" PostTypeId="2" ParentId="2287" CreationDate="2014-04-27T18:10:09.203" Score="1" Body="&lt;p&gt;As said above, you'd see the light coming into the black hole but you mught also see a dim light from Hawking radiation.&lt;/p&gt;&#xA;" OwnerUserId="1494" LastActivityDate="2014-04-27T18:10:09.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2290" PostTypeId="2" ParentId="2283" CreationDate="2014-04-28T01:24:03.067" Score="5" Body="&lt;p&gt;I'll give a slightly different answer: The escape velocity at the surface of our &lt;a href=&quot;http://en.wikipedia.org/wiki/Moon&quot;&gt;Moon&lt;/a&gt; is about 2.38 km/s. Derived from &lt;a href=&quot;http://www.astro.washington.edu/users/solontoi/150/atm.pdf&quot;&gt;this paper&lt;/a&gt;, as a rule of thumb, an atmosphere can survive 4.5 billion years, if it's average molecule velocity is below 1/6 of the escape velocity of the planet/moon. Applied to our Moon, it's 2380 m/s / 6 = 396.67 m/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon dioxide has a molecule mass of 44u.&#xA;The according temperature for an average molecule velocity for carbon dioxide is hence&#xA;$$\mbox{temperature}=(v_{\mbox{gas}}/157)^2 \cdot \mbox{molecule mass}\mbox=(396.67/157)^2\cdot 44\mbox{ K}=280.87\mbox{ K}.$$&#xA;Moon's mean surface temperature at the equator is about 220 K. That's well below the allowed 280.87 K. A carbon dioxide atmosphere could have survived 4.5 billion years according to these oversimplified assumptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When looking closer to the atmosphere, it turns out, that the higher layers of the &lt;a href=&quot;http://www.windows2universe.org/earth/Atmosphere/layers_activity_print.html&quot;&gt;thermosphere and the exosphere&lt;/a&gt; can reach more than 1000 K.&#xA;Even carbon dioxide can escape from Moon at this temperature over time:&#xA;The average molecule velocity for carbon dioxide at 1000 K is&#xA;$$157\cdot \sqrt{\frac{1000}{44}}\mbox{ m/s}= 748.5\mbox{ m/s}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This results in a &lt;a href=&quot;http://arxiv.org/ftp/arxiv/papers/1009/1009.5110.pdf&quot;&gt;Jeans escape parameter&lt;/a&gt; of&#xA;$$\lambda_0=\left(\frac{v_{\mbox{esc}}}{v_{\mbox{gas}}}\right)^2 =\left(\frac{2380}{748.5}\right)^2= 10.1,$$&#xA;resulting in an escape rate of about $10^{-4}$ relative to a free molecular flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this it isn't yet straightforward to calculate the period of time, Moon would loose a carbon dioxide atmosphere, since it's a function of the height-dependent temperature curve of the atmosphere. But it's at least clear, that Moon wouldn't loose a carbon dioxide atmosphere suddenly, since this is likely just for $\lambda_0 &amp;lt; 3$.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-04-28T01:24:03.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2291" PostTypeId="1" AcceptedAnswerId="2312" CreationDate="2014-04-28T04:00:56.207" Score="1" ViewCount="351" Body="&lt;p&gt;I know that tidal forces are pushing Jupiter farther from the Sun, but I couldn't find exactly the yearly amount. In a few billion years would this effect (and subsequent decrease in gravity pull) allow the formation of a new planet from the asteroid belt or would Jupiter take the asteroids with it farther away?&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditDate="2014-04-29T22:50:52.053" LastActivityDate="2014-04-29T23:59:24.420" Title="Will a new planet form if Jupiter's influence on asteroid belt will diminish in a few billion years?" Tags="&lt;gravity&gt;&lt;jupiter&gt;&lt;tidal-forces&gt;&lt;asteroid-belt&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2292" PostTypeId="2" ParentId="2291" CreationDate="2014-04-28T04:12:54.740" Score="2" Body="&lt;p&gt;Sometime after 4.5 billion years, the sun will change into a red giant star and expand to about 1 AU. Some people think it will be beyond Earth's orbit. (In any case, Earth's orbit would have changed and likely to have struck the surface of the expanded sun.) The asteroids would experience a stronger solar wind and their orbits would become more elliptical, so Jupiter's influence would be significant.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-04-28T04:12:54.740" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2293" PostTypeId="1" CreationDate="2014-04-28T06:58:53.537" Score="5" ViewCount="416" Body="&lt;p&gt;At least Atacama Desert of northern Chile is a good place to build telescopes.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;VLT and ALMA locate there.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about Sahara desert?&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Atacama_Large_Millimeter_Array&quot;&gt;http://en.wikipedia.org/wiki/Atacama_Large_Millimeter_Array&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Very_Large_Telescope&quot;&gt;http://en.wikipedia.org/wiki/Very_Large_Telescope&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2014-05-03T07:41:58.223" Title="Is Sahara desert a good place to build radio/optical telescopes?" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2294" PostTypeId="2" ParentId="2293" CreationDate="2014-04-28T07:19:40.210" Score="12" Body="&lt;p&gt;No, the Sahara isn't a good place to build telescopes.&lt;br&gt;&#xA;The Atacama desert is used because it is at high altitude, which means that there is less atmosphere to get in the way. Other telescopes are located on mountaintops for the same reason.&lt;br&gt;&#xA;The Sahara is mostly at sea level. It's also very hot, so you get lots of turbulence due to rising air, which distorts the image.&lt;br&gt;&#xA;The Sahara is also undesirable for other reasons: no infrastructure, unstable regimes, lots of erosion due to sand storms. &lt;/p&gt;&#xA;" OwnerUserId="361" LastEditorUserId="361" LastEditDate="2014-05-03T07:41:58.223" LastActivityDate="2014-05-03T07:41:58.223" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2295" PostTypeId="1" CreationDate="2014-04-28T08:43:41.097" Score="5" ViewCount="520" Body="&lt;p&gt;My simple understanding is that a parabolic dish focuses radio waves to a receiver. It is obvious that any parabola focuses visible light and heat, so I assume that it focuses all radio wavelengths too. So evidently the design of the receiver determines which wavelengths it is sensitive to.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;How is such a receiver designed in radio telescopes?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What properties of the receiver determine the wavelengths it is sensitive to?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is the sensitivity range electronically adjustable?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why aren't different receivers used so that the same parabola, for example the huge Arecibo dish, could be used for all wavelengths? (Or is it?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What importance does the size and material of the parabola have for the sensitivity to different radio wavelengths?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1501" LastEditDate="2014-04-28T14:32:21.453" LastActivityDate="2014-04-28T14:32:21.453" Title="What determines which frequency range a radio telescope has?" Tags="&lt;telescope&gt;&lt;radio-astronomy&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2296" PostTypeId="2" ParentId="2295" CreationDate="2014-04-28T11:28:16.380" Score="7" Body="&lt;p&gt;A parabola does indeed focus over a broad frequency range. The lower limit is determined by the dish diameter, the upper by the construction (mesh size, parabolic accuracy etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The collector placed at the focal point may be a simple dipole or other fixed frequency antenna, or more commonly, a waveguide that leads the collected signals to a low noise amplifier and receiver. The physical dimensions of the antenna or waveguide set the operating frequency range of the antenna system. More than one range may be tuned by means of multiple switched collectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The receiver is fundamentally no different to any other receiver; a tuning element selects the signals of interest which are then detected and amplified.&lt;/p&gt;&#xA;" OwnerUserId="1464" LastActivityDate="2014-04-28T11:28:16.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2297" PostTypeId="2" ParentId="2278" CreationDate="2014-04-28T12:48:10.880" Score="6" Body="&lt;p&gt;I'm not aware of any relevant &lt;a href=&quot;http://en.wikipedia.org/wiki/Uranium_ore&quot; rel=&quot;noreferrer&quot;&gt;uranium ore&lt;/a&gt; deposit, which is related to meteoritic material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dangerous fallout is by far the most caused by short-lived radioactive isotopes. Those isotopes are rare in meteorites as well as in rock on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Natural_nuclear_fission_reactor&quot; rel=&quot;noreferrer&quot;&gt;Natural nuclear reactors&lt;/a&gt;, which would produce short-lived radioactive isotopes, don't occur any more on Earth, since the natural U-235/U-238 isotope ratio is too low, now. This is due to the half-life of &lt;a href=&quot;http://en.wikipedia.org/wiki/Uranium&quot; rel=&quot;noreferrer&quot;&gt;U-235&lt;/a&gt; of about 713 million years in contrast to 4,468 million years for U-238.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an asteroid would impact into a uranium ore deposit, most of the molten ejected material would fall back to Earth as &lt;a href=&quot;http://en.wikipedia.org/wiki/Tektite&quot; rel=&quot;noreferrer&quot;&gt;tektites&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evaporated ejecta and aerosols (dust) would probably increase health risk for some time, until it's washed out from the atmosphere. But other, less-radioactive aerosols would be a health risk, too.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-04-28T12:53:21.310" LastActivityDate="2014-04-28T12:53:21.310" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2298" PostTypeId="1" AcceptedAnswerId="2301" CreationDate="2014-04-29T07:55:52.817" Score="1" ViewCount="1253" Body="&lt;p&gt;I know the colours of the stars relate to their heat, however if I were standing on a different planet identical to Earth, but orbiting a blue Bright Giant or a Red Giant, or perhaps even a Red Dwarf, what colours in visible light would I be able to see? Is it the same as with our star, i.e. would any be missing such as if I'm in a room with a black light?&lt;/p&gt;&#xA;" OwnerUserId="1768" LastActivityDate="2014-04-29T14:21:33.053" Title="What is visible light colour output of different stars?" Tags="&lt;star&gt;&lt;light&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2299" PostTypeId="1" CreationDate="2014-04-29T09:11:34.850" Score="0" ViewCount="331" Body="&lt;p&gt;Anybody knows that?&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data products are all images or star catalogs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the link:&#xA;&lt;a href=&quot;http://pan-starrs.ifa.hawaii.edu/public/&quot; rel=&quot;nofollow&quot;&gt;http://pan-starrs.ifa.hawaii.edu/public/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2014-04-29T12:51:02.877" Title="is pan-starrs' data available to public users?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2300" PostTypeId="2" ParentId="2299" CreationDate="2014-04-29T12:51:02.877" Score="1" Body="&lt;p&gt;Looks like you've to be member of the &lt;a href=&quot;http://www.ps1sc.org/&quot; rel=&quot;nofollow&quot;&gt;PS1 Science Consortium&lt;/a&gt; to be able to &lt;a href=&quot;http://web01.psps.ifa.hawaii.edu/PSI/login.php?referencePage=/PSI/index.php&quot; rel=&quot;nofollow&quot;&gt;log in&lt;/a&gt;.&#xA;And you'll need some &lt;a href=&quot;http://psps.ifa.hawaii.edu/psps-trac/attachment/wiki/PsiFigs/query_builder_step4.png&quot; rel=&quot;nofollow&quot;&gt;database query&lt;/a&gt; knowledge, at least by using the &lt;a href=&quot;http://psps.ifa.hawaii.edu/psps-trac/wiki/PSIQueryBuilder&quot; rel=&quot;nofollow&quot;&gt;PSPS Query Builder&lt;/a&gt;, or more advanced programming capabilities (XML, SOAP, Python, Perl). A short documentation of the database tables can be found &lt;a href=&quot;http://web01.psps.ifa.hawaii.edu/psps-trac/wiki/ODMTables&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&#xA;If you like to dig deeper into the database, &lt;a href=&quot;http://web01.psps.ifa.hawaii.edu/psps-trac/wiki/PSPSFAQ&quot; rel=&quot;nofollow&quot;&gt;this FAQ&lt;/a&gt; page may be a starting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image processing pipeline (IPP) is described in &lt;a href=&quot;http://www.amostech.com/TechnicalPapers/2006/Pan-STARRS/Magnier.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-04-29T12:51:02.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2301" PostTypeId="2" ParentId="2298" CreationDate="2014-04-29T14:01:27.183" Score="0" Body="&lt;p&gt;Compare the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_classification#Harvard_spectral_classification&quot; rel=&quot;nofollow&quot;&gt;effective temperature&lt;/a&gt; of the stars with &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck%27s_law&quot; rel=&quot;nofollow&quot;&gt;Planck's law&lt;/a&gt;, and with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Incandescent_light_bulb&quot; rel=&quot;nofollow&quot;&gt;light bulb&lt;/a&gt; (2000 to 3300 K): The color of a light bulb corresponds to spectral class L (red brown star) or class M (red star, including dwarfs and giants); violet, and some blue are missing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light of Brown Dwarfs (spectral class T) would look red from their black body radiation, but due to their chemical composition the actual color may vary a bit over different tones of red. They would mainly feel hot, and look dull red at the same time, similar to glowing iron. Colors can hardly be distinguished under these lighting conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the very hottest stars (spectral class O) may ressemble &quot;black&quot; light, looking bluish, and causing fluorescence due to to the shift of the spectrum into the ultraviolet, including x-rays.&#xA;Without protective ozone/atmosphere layer, this wouldn't be healthy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may apply to some degree also to spectral class B (blue white stars); plasma cores of &lt;a href=&quot;http://en.wikipedia.org/wiki/Lightning&quot; rel=&quot;nofollow&quot;&gt;lightnings&lt;/a&gt; are of a similar temperature (up to 30,000 K or more) as the effectve temperature of class B or class O stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any spectral class in-between (A, F, G, K) would look almost white, a little bluish for hot stars, yellowish for cooler stars, after some adaption of the eyes.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-04-29T14:21:33.053" LastActivityDate="2014-04-29T14:21:33.053" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2302" PostTypeId="1" AcceptedAnswerId="2303" CreationDate="2014-04-29T15:45:40.920" Score="11" ViewCount="3658" Body="&lt;p&gt;How does the Sun burn without oxygen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might not be burning but a big part of the society speaks of it as burning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how does it work?&lt;/p&gt;&#xA;" OwnerUserId="1505" LastEditorUserId="1437" LastEditDate="2019-03-22T15:52:12.697" LastActivityDate="2019-07-01T13:16:46.033" Title="How can the Sun burn without oxygen?" Tags="&lt;the-sun&gt;&lt;nucleosynthesis&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="2303" PostTypeId="2" ParentId="2302" CreationDate="2014-04-29T16:56:17.310" Score="16" Body="&lt;p&gt;As you are suspecting, the sun burns in a different sense, not by chemical reaction with oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Atom&quot;&gt;Atoms&lt;/a&gt; consist of a tiny, heavy &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_nucleus&quot;&gt;nucleus&lt;/a&gt;, surrounded by an almost empty space, populated by &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrons&quot;&gt;electrons&lt;/a&gt;. Burning by chemical reaction with oxygen doesn't change the nucleus of atoms, but takes place in the hull of atoms: Atoms may assemble to form &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecule&quot;&gt;molecules&lt;/a&gt;; electrons change their &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_orbital&quot;&gt;orbitals&lt;/a&gt; (the way they surround the nucleus), and release some energy as heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atomic nuclei are (positively) electrically charged, and repell each other.&#xA;But if small nuclei, like those of hydrogen atoms, come close together, they can &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_fusion&quot;&gt;fuse&lt;/a&gt; and form a larger nucleus. This nuclear fusion of hydrogen to helium (in this case) releases much energy, more even than &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_fission&quot;&gt;fission&lt;/a&gt; of uranium in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_power_plant&quot;&gt;nuclear power plant&lt;/a&gt;. The notion &quot;burning&quot; is used sometimes for reactions of atomic nuclei, too, if they release energy as heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To overcome the &lt;a href=&quot;http://en.wikipedia.org/wiki/Coulomb%27s_law&quot;&gt;electrostatic repulsion&lt;/a&gt; of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_atom&quot;&gt;hydrogen&lt;/a&gt; nuclei, high pressure and temperature are needed. &lt;a href=&quot;http://en.wikipedia.org/wiki/Proton-proton_chain&quot;&gt;These conditions&lt;/a&gt; occur in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_core&quot;&gt;core of the Sun&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-04-29T16:56:17.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2304" PostTypeId="1" AcceptedAnswerId="2306" CreationDate="2014-04-29T17:32:38.740" Score="5" ViewCount="626" Body="&lt;p&gt;We know that redshift and blueshift is the result of the frequency of light waves reflected or emitted by an object lengthening or shortening due to the relative velocity of the observer moving away or toward the object. This is the Doppler effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it was stated in response to &lt;a href=&quot;https://space.stackexchange.com/questions/4365/how-quickly-could-a-craft-find-its-location-using-only-pulsars&quot;&gt;a question I asked on a sister site&lt;/a&gt; that the Doppler effect would &lt;em&gt;also&lt;/em&gt; result in the apparent frequency of a pulsar's pulses being longer or shorter due to the relative velocity of the star. Given that the speed of light is constant regardless of the frame of reference (a concept I have a really hard time understanding by the way), is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thinking is that it is incorrect. Using the analogy of a launcher lobbing medicine balls at me at a fixed rate, I can &quot;see&quot; the rate of launch by the frequency of when I get hit. However, if the launcher is moving away from me, the velocity of the balls lobbed at me will be slower (relative to my frame of reference), and thus will hit me with less frequency. However, if the launcher were to adjust its firing speed (i.e. speed of the ball relative to itself; frequency remains constant in this analogy) so that the balls always hit me at the same speed (thereby simulating the concept of the always-fixed speed of light), the time between them would not change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, given the always-constant speed of light, is it correct that a pulsar's pulse rate would appear to be different at different relative velocities?&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2014-04-29T19:29:12.220" Title="Does Doppler shift affect apparent pulsar frequency?" Tags="&lt;pulsar&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2305" PostTypeId="1" CreationDate="2014-04-29T18:14:55.700" Score="4" ViewCount="872" Body="&lt;p&gt;i try to extend my analysis from planar restricted three body problem to the general one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anyone that can help me to find out the Hamiltonian for this prolem in cartesian coordinates? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need the hamiltonian of 3D case somethink like eq.(1) in this &lt;a href=&quot;http://iopscience.iop.org/1538-3881/145/3/63/pdf/1538-3881_145_3_63.pdf&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; or in this one &lt;a href=&quot;http://www.aanda.org/articles/aa/pdf/2009/02/aa10843-08.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; again eq.1 . Because my symplectic integrator can be used only if I have the fixed cartesian coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="1310" LastEditorUserId="1310" LastEditDate="2014-04-29T19:16:43.757" LastActivityDate="2014-12-19T18:16:57.110" Title="Hamiltonian of general three body problem" Tags="&lt;solar-system&gt;&lt;stellar-dynamics&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2306" PostTypeId="2" ParentId="2304" CreationDate="2014-04-29T19:17:20.380" Score="6" Body="&lt;p&gt;Yes, it is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the classic example of a (say) $1000$ Hz siren on an ambulance approaching you.  Once a single wave is produced, the source moves towards you before the next wave is produced.  The wavelength of the sound waves in the air is reduced, so the frequency you hear is increased.  Simple, basic Doppler shift...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But consider the case where the siren is being switched on and off with a five-second overall period.  So you have a $1$ kilohertz signal modulated at $0.2$ Hertz. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the same argument applies to this modulation frequency.  A chain of $1$ kilohertz waves is generated for $2.5$ seconds and sent on its way.  The ambulance approaches you for the next silent $2.5$ seconds, then starts the next chain on its way, &lt;strong&gt;closer to the last chain&lt;/strong&gt; because of the distance travelled in the silent phase.  The same Doppler shift factor applies to the carrier frequency and the modulation frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same physics almost derailed the Cassini-Huygens probe. The extreme velocity changes of the probe caused a Doppler shift in both the transmission frequency and the &lt;strong&gt;data rate&lt;/strong&gt; carried by this transmission. The first was handled;  the second was not!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.thespacereview.com/article/306/1&quot;&gt;http://www.thespacereview.com/article/306/1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://spectrum.ieee.org/aerospace/space-flight/titan-calling&quot;&gt;http://spectrum.ieee.org/aerospace/space-flight/titan-calling&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="420" LastEditorUserId="420" LastEditDate="2014-04-29T19:29:12.220" LastActivityDate="2014-04-29T19:29:12.220" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2307" PostTypeId="2" ParentId="2305" CreationDate="2014-04-29T20:12:18.443" Score="3" Body="&lt;p&gt;The condition $$\sum_{i=1}^{3}m_iq_i'=0$$&#xA;of &lt;a href=&quot;http://iopscience.iop.org/1538-3881/145/3/63/pdf/1538-3881_145_3_63.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; generalizes to&#xA;$$\sum_{i=1}^{n}m_iq_i'=0,$$&#xA;meaning the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic_energy&quot; rel=&quot;nofollow&quot;&gt;barycenter&lt;/a&gt; should be at the origin of the frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation (1) generalizes to&#xA;$$H=\frac{1}{2}\sum\frac{|p_i'|^2}{m_i}-\sum_{i=1}^n\sum_{j=1, j\neq i}^n\frac{m_im_j}{|q_i'-q_j'|}.$$&#xA;The first part is the total &lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic_energy&quot; rel=&quot;nofollow&quot;&gt;kinetic energy&lt;/a&gt; of the system relative to the barycenter, written in terms of &lt;a href=&quot;http://en.wikipedia.org/wiki/Momentum&quot; rel=&quot;nofollow&quot;&gt;momenta&lt;/a&gt;: $$E=\frac{1}{2}mv^2=\frac{1}{2}m\left(\frac{p}{m}\right)^2=\frac{1}{2}\frac{p^2}{m}.$$&#xA;The second part is the sum of all &lt;a href=&quot;http://en.wikipedia.org/wiki/Potential_energy&quot; rel=&quot;nofollow&quot;&gt;self-potential energy&lt;/a&gt; between each pair of (spherical) objects, with Newton's gravitational constant $G$ set to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://en.wikipedia.org/wiki/N-body_problem&quot; rel=&quot;nofollow&quot;&gt;n-body problem on Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 3-dimensional 3-body (or n-body) problem is like the planar version. Just take 3-dimensional vectors for $p_i'$ and $q_i'$.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-04-29T21:51:38.193" LastActivityDate="2014-04-29T21:51:38.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2308" PostTypeId="1" AcceptedAnswerId="2309" CreationDate="2014-04-29T20:21:59.103" Score="6" ViewCount="893" Body="&lt;p&gt;This is not a science fiction question, but I was considering the visual accuracy of ships in space depcicted in Battlestar Galactica and the like. They are always strongly lit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they are in between stars, would they be visible at all or bathed in light equally on all sides?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tAIDy.jpg&quot; alt=&quot;Sample image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pursuing it further, at what distance from a star does an object no longer reflect it's light as the dominant light source and after that is it dark?&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="417" LastEditDate="2014-05-14T12:30:07.933" LastActivityDate="2014-05-14T12:30:07.933" Title="Would space-faring vehicles in interstellar space be in pitch darkness, show reflections from the nearest star or be bathed in light completely?" Tags="&lt;interstellar-travel&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2309" PostTypeId="2" ParentId="2308" CreationDate="2014-04-29T21:15:01.230" Score="6" Body="&lt;p&gt;The brightness of the spaceship follows (almost) the &lt;a href=&quot;http://en.wikipedia.org/wiki/Inverse-square_law&quot;&gt;inverse square law&lt;/a&gt;, meaning twice the distance from the star, the brightness will be a quater.&#xA;In the middle of nowhere, but within a galaxy, it would look like in a moonless, and cloudless night, far away from any artificial light source.&#xA;It wouldn't be pitchblack, but much too dark to read a newspaper. You would see the stars in the background, and the fuzzy band of the Milky Way. The starship would occult the stars; that way you would see the black silhouette of the starship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 630.1-fold the distance of the Earth from the Sun (about 3.65 light-days), it would look like at full moon (derived from &lt;a href=&quot;http://en.wikipedia.org/wiki/Apparent_magnitude#Example%3a_Sun_and_Moon&quot;&gt;apparent magnitudes&lt;/a&gt; of full moon and Sun)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But maybe the starship is illuminated artificially.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-04-29T21:27:52.067" LastActivityDate="2014-04-29T21:27:52.067" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2310" PostTypeId="1" AcceptedAnswerId="2311" CreationDate="2014-04-29T21:45:02.123" Score="1" ViewCount="622" Body="&lt;p&gt;Is it because there isn't life to exhale enough oxygen and carbon dioxide? The size of Mars is about the same as Earth, so isn't there enough gravity to hold enough air?&lt;/p&gt;&#xA;" OwnerUserId="1510" LastEditorUserId="361" LastEditDate="2014-04-30T17:37:46.947" LastActivityDate="2014-04-30T17:37:46.947" Title="Why is Mars' atmosphere so thin?" Tags="&lt;mars&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2311" PostTypeId="2" ParentId="2310" CreationDate="2014-04-29T22:25:57.280" Score="3" Body="&lt;p&gt;The escape velocity of &lt;a href=&quot;http://en.wikipedia.org/wiki/Mars&quot; rel=&quot;nofollow&quot;&gt;Mars&lt;/a&gt; is about $5.03 \mbox{ km}/\mbox{s}$ , that of Earth about $11.2 \mbox{ km}/\mbox{s}$ , that of &lt;a href=&quot;http://en.wikipedia.org/wiki/Moon&quot; rel=&quot;nofollow&quot;&gt;Moon&lt;/a&gt; about $2.38 \mbox{ km}/\mbox{s}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The escape velocity is an important ingredient for the stability of a planetary atmosphere.&#xA;From this it's not surprising, that the density of the Martian atmosphere is between that of Moon and that of Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The loss of most of the magnetic field of Mars makes its atmosphere less resistant to erosive solar wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Actually things get more complicated, since &lt;a href=&quot;http://perso.utinam.cnrs.fr/~mousis/papier91.pdf&quot; rel=&quot;nofollow&quot;&gt;sequestration and outgassing&lt;/a&gt;, meaning an exchange of atmosphere with rock, as well as sublimation, and resublimation of water and carbon dioxide take place. Water vapor is decomposed into hydrogen and oxygen by &lt;a href=&quot;http://en.wikipedia.org/wiki/Photodissociation&quot; rel=&quot;nofollow&quot;&gt;photolysis&lt;/a&gt;, hydrogen escapes to space due to its low molecular weight, and oxygen tends to be bound chemically to iron-bearing rocks. On the other hand &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_energetic_particles&quot; rel=&quot;nofollow&quot;&gt;solar wind provides a steady supply of hydrogen&lt;/a&gt;, which is bound by rocks in the top few micrometers; &quot;&lt;a href=&quot;http://www.lanl.gov/newsroom/news-releases/2012/August/08.23-chemcam-results.php&quot; rel=&quot;nofollow&quot;&gt;ChemCam saw peaks of hydrogen and magnesium during the first shots that we didn’t see in subsequent firings&lt;/a&gt;.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The history of the Marsian atmosphere isn't yet known in full detail. NASA's &lt;a href=&quot;http://en.wikipedia.org/wiki/MAVEN&quot; rel=&quot;nofollow&quot;&gt;MAVEN&lt;/a&gt; mission is designed to find out more.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-04-30T13:17:31.103" LastActivityDate="2014-04-30T13:17:31.103" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2312" PostTypeId="2" ParentId="2291" CreationDate="2014-04-29T23:59:24.420" Score="2" Body="&lt;p&gt;The total mass of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Asteroid_belt&quot; rel=&quot;nofollow&quot;&gt;asteroid belt&lt;/a&gt; is just about 4% of the mass of our Moon.&#xA;Even if the asteroids don't collide with other planets in the meanwhile, the mass is too low to form a planet in the sense of the 2006 IAU definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if &lt;a href=&quot;http://en.wikipedia.org/wiki/Ceres_%28dwarf_planet%29&quot; rel=&quot;nofollow&quot;&gt;Ceres&lt;/a&gt; would accrete all asteroids in the asteroid belt, its radius wouldn't grow to the double of its present radius, hence stay smaller than &lt;a href=&quot;http://en.wikipedia.org/wiki/Mercury_%28planet%29&quot; rel=&quot;nofollow&quot;&gt;Mercury&lt;/a&gt;. But Ceres is already large enough to fit the present &lt;a href=&quot;http://en.wikipedia.org/wiki/2006_definition_of_planet&quot; rel=&quot;nofollow&quot;&gt;IAU definition&lt;/a&gt; of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Planet&quot; rel=&quot;nofollow&quot;&gt;planet&lt;/a&gt;, with the exception, that it hasn't &lt;a href=&quot;http://en.wikipedia.org/wiki/Clearing_the_neighbourhood&quot; rel=&quot;nofollow&quot;&gt;cleared its neighbourhood&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a Stern-Levison parameter of $8.32 \cdot 10^{−4},$ Ceres is too small to fulfill the third IAU criterion for a planet, and too small to clear its neighbourhood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even a dwarf planet of the total mass of the asteroid belt couldn't fulfill the criterion on a pure Stern-Levison parameter basis, hence wouldn't be able to keep its neighbourhood clear. Its Stern-Levison parameter would be about $7.5\cdot 10^{-3}$. That's well below 1, the estimated value necessary to fulfill the third IAU criterion, but well above the Stern-Levison parameter of $2.95 × 10^{−3}$ for Pluto.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-04-29T23:59:24.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2313" PostTypeId="1" CreationDate="2014-04-30T00:19:32.837" Score="13" ViewCount="15682" Body="&lt;p&gt;Suppose the Earth's rotation slowed for some reason. Would the lack of centrifugal force cause us to feel heavier than normal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, if Earth's rotation increased, would we feel lighter as centrifugal force lifts us from the ground?&lt;/p&gt;&#xA;" OwnerUserId="1512" LastActivityDate="2014-04-30T00:54:22.323" Title="If Earth didn't rotate, would we feel heavier?" Tags="&lt;earth&gt;&lt;rotation&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ClosedDate="2014-05-08T12:13:00.253" ContentLicense="CC BY-SA 3.0" />
  <row Id="2314" PostTypeId="2" ParentId="2313" CreationDate="2014-04-30T00:51:05.363" Score="7" Body="&lt;p&gt;The centrifugal force at the equator is only 1/289 of normal gravity, or 0.35%. That's too small to be perceived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/?id=mE4GAQAAIAAJ&amp;amp;pg=PA358&amp;amp;lpg=PA358&quot; rel=&quot;noreferrer&quot;&gt;http://books.google.com/?id=mE4GAQAAIAAJ&amp;amp;pg=PA358&amp;amp;lpg=PA358&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It gradually diminishes as you move away from the equator, and becomes zero at the poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, if Earth stopped rotating, the added weight at the equator would be the least of your worries. There would be major perturbations in terms of weather, etc, that would amount to global catastrophe.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Likewise, if Earth's rotation increased, would we feel lighter as&#xA;  centrifugal force lifts us from the ground?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In theory, yes. But the rotation increase would have to be huge. Again, in that case, there would be other, much bigger effects, that would be much more important than your loss of weight. E.g. the whole planet would bulge at the equator and become more flat at the poles, with incalculable effects on plate tectonics, weather, etc.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2014-04-30T00:51:05.363" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2315" PostTypeId="1" CreationDate="2014-04-30T00:52:53.497" Score="9" ViewCount="1929" Body="&lt;p&gt;A planet made from denser material than Earth might have equal gravity to Earth but a smaller radius. How small can a planet be and still have Earth gravity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally it should be habitable by humans, so not made of anything dangerous, radioactive or unstable.&lt;/p&gt;&#xA;" OwnerUserId="1512" LastActivityDate="2014-04-30T01:51:19.680" Title="How small can a planet be and still have Earth-like gravity?" Tags="&lt;planet&gt;&lt;earth-like-planet&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2316" PostTypeId="2" ParentId="2313" CreationDate="2014-04-30T00:54:22.323" Score="7" Body="&lt;p&gt;If the rotation would stop, we would feel an additional gravity of $0.034 \mbox{ m}/\mbox{s}^2$, or about 0.35%, at the equator, (incorrectly) assuming the shape of Earth isn't changed by the changing rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The centrifugal acceleration is $v^2/r$, with $v=465.1 \mbox{ m}/\mbox{s},$ and $r=6378100 \mbox{ m}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With shorter rotation period, the other way, as you say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polar regions wouldn't be affected, with the exception, that the ellipsoid of Earth would change due to changing rotation period. Respecting this, gravity would increase at the poles with faster rotation, since the poles would get closer to Earth's center.&#xA;In equatorial regions the change of surface gravity would be affected more than in the simplified calculation above. In contrast to the poles, faster rotation would reduce surface gravity more than just by the centrifugal force.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-04-30T00:54:22.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2317" PostTypeId="2" ParentId="2315" CreationDate="2014-04-30T01:51:19.680" Score="9" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Surface_gravity&quot;&gt;surface gravity&lt;/a&gt; of a planet is very close to &#xA;$$g=\frac{4\pi G}{3}\rho r.$$&#xA;With $g$ to be kept constant, and $\frac{4\pi G}{3}$ a constant, we need&#xA;$\rho_Pr_P=\rho_Er_E$, or&#xA;$$r_P=\frac{\rho_E}{\rho_P}r_E,$$&#xA;with $\rho_E=5.515 \mbox{ g}/\mbox{cm}^3$ the mean density of &lt;a href=&quot;http://en.wikipedia.org/wiki/Earth&quot;&gt;Earth&lt;/a&gt;, $r_E=6371.0 \mbox{ km}$ the mean radius of Earth, $\rho_P=22.59\mbox{ g}/\mbox{cm}^3$ the density of densest natural element &lt;a href=&quot;http://en.wikipedia.org/wiki/Osmium&quot;&gt;osmium&lt;/a&gt;, and $r_P$ the radius of the fictive osmium planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $$r_P=\frac{5.515}{22.59}r_E=0.2441~r_E=1555\mbox{ km}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some compression of the core of an osmium planet due to pressure is neglected.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-04-30T01:51:19.680" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2318" PostTypeId="1" AcceptedAnswerId="2319" CreationDate="2014-04-30T03:52:59.020" Score="5" ViewCount="920" Body="&lt;p&gt;Due to tidal deceleration, Phobos is &lt;a href=&quot;http://en.wikipedia.org/wiki/Phobos_%28moon%29#Predicted_destruction&quot; rel=&quot;noreferrer&quot;&gt;predicted&lt;/a&gt; to be destroyed by Mars in the next 30-50 million years either by collision or by breaking up in a planetary ring. If Phobos falls on Mars in one piece could the impact could be comparable with &lt;a href=&quot;http://en.wikipedia.org/wiki/Chicxulub_crater#Impact_specifics&quot; rel=&quot;noreferrer&quot;&gt;Chicxulub meteorite&lt;/a&gt;? Since Mars is smaller that Earth, would the effects of the impact cause a global restart of volcanic activity and temporary warm the planet? Would this be sufficient to make Mars more friendly to life for a few thousand years?&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2015-09-27T11:52:47.330" Title="Would a Phobos full impact make Mars temporary habitable?" Tags="&lt;mars&gt;&lt;life&gt;&lt;impact&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2319" PostTypeId="2" ParentId="2318" CreationDate="2014-04-30T12:44:19.737" Score="10" Body="&lt;p&gt;The impact of Phobos, even in one piece (less likely), would be different from Chicxulub. Phobos is probably a little larger than the Chicxulub impactor, but much slower, and comes in almost perfectly tangential. The impact energy would be less than a 10th of the Chicxulub impact, and the energy would be distributed over a large region around the Martian equator. It wouldn't start volcanism, and it wouldn't warm Mars in a relevant way. It wouldn't make Mars more friendly to life by warming. The only noteworthy contribution to habitability would be additional organic matter which Phobos is presumed to contain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phobos will (probably) desintegrate due to tidal forces, before its fragments will collide with Mars. Any loose rocks, dust, regolith on the surface will separate from Phobos below the &lt;a href=&quot;http://en.wikipedia.org/wiki/Roche_limit&quot;&gt;Roche limit&lt;/a&gt; of&#xA;$$d=1.26~R_M\cdot\left(\frac{\rho_M}{\rho_m}\right)^\frac{1}{3},$$&#xA;with $R_M=3389.5\mbox{ km}$ the radius of Mars, $\rho_M=3.9335 \mbox{ g/cm}^3$ the mean density of Mars, and $\rho_m=1.887\mbox{ g/cm}^3$ the mean density of Phobos. This results in a Roche limit of&#xA;$$d=1.26\cdot 3389.5\mbox{ km}\cdot\left(\frac{3.9335 \mbox{ g/cm}^3}{1.887\mbox{ g/cm}^3}\right)^\frac{1}{3}=5455.6\mbox{ km}.$$&#xA;That's for Phobos' center about 2059 km above the equatorial surface of Mars (taking 3396.2 km as equatorial radius.)&#xA;Therefore desintegration will start long before the impact. Loose rocks will collide with Phobos and unlock more material. Whether there will be a considerable solid remnant to finally cause a big (tangential) impact is difficult to predict.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_speed&quot;&gt;orbital speed&lt;/a&gt; of the fragments at the time of collision will be about&#xA;$$v_0=\frac{v_e}{\sqrt 2},$$&#xA;with $v_e=5.03\mbox{ km/s}$ the escape velocity of &lt;a href=&quot;http://en.wikipedia.org/wiki/Mars&quot;&gt;Mars&lt;/a&gt;.&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Phobos_%28moon%29&quot;&gt;Phobos&lt;/a&gt;' inclination of 1.093° (relative to Mars's equator) is almost neglectable for the calculation of the impact velocity.&#xA;Mars' equatorial rotation velocity is about 241.17 m/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The impact velocity will hence be about&#xA;$$\frac{5030\mbox{ m/s}}{\sqrt 2}-241.17 \mbox{ m/s}=3316\mbox{ m/s}.$$&#xA;Phobos' mass is about $1.0659\cdot 10^{16}\mbox{ kg}.$&#xA;This results in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic_energy&quot;&gt;kinetic impact energy&lt;/a&gt; of&#xA;$$\frac{1}{2}mv^2=\frac{1}{2}\cdot 1.0659\cdot 10^{16}\mbox{ kg}\cdot (3316\mbox{ m/s})^2=5.86\cdot 10^{22}\mbox{ J}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Enthalpy_of_fusion&quot;&gt;heat of fusion&lt;/a&gt; of water is about 334 kJ/kg, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Properties_of_water&quot;&gt;heat of vaporization of water&lt;/a&gt; at 0°C is about 2504 kJ/kg, together 2838 kJ/kg are needed to evaporate water ice of 0°C.&#xA;Hence the impact energy of Phobos would be sufficient to evaporate&#xA;$$\frac{5.86\cdot 10^{22}\mbox{ J}}{2.838 \cdot 10^{6}\mbox{ J/kg}}=2.06\cdot 10^{16}\mbox{ kg}$$ of water ice of 0°C, corresponding to $2.06\cdot 10^{13}\mbox{ m}^3$ of liquid water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars' surface area is about 144,798,500 km², or $1.447985\cdot 10^{14}\mbox{ m}^2$.&#xA;Equally distributed, we could get a layer of water ice corresponding to&#xA;$$\frac{2.06\cdot 10^{13}\mbox{ m}^3}{1.447985\cdot 10^{14}\mbox{ m}^2}=0.142\mbox{ m},$$&#xA;(correspoding to 14.2 kg/m² of liquid water) evaporated by the impact energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a surface gravity of 3.711 m/s², this would correspond to&#xA;a pressure of $14.2 \mbox{ kg/m}^2\cdot 3.711 \mbox{ m/s}^2= 52.8 \mbox{ N/m}^2=0.528\mbox{ hPa}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current atmospheric pressure on Mars is about 6.3 hPa, with diurnal, seasonal and regional changes.&#xA;Thus we could get a maximum of 10% short-term pressure increase in the Martian atmosphere by water vapor, in an ideal case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end, an impact of Phobos, or its fragments, would result in a regional reworking of the Martian surface in the equatorial region, but not in a long-term global change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just melting water ice would lead to a global ocean with a depth of about 1.2 m. But this ocean would boil due to the low atmospheric pressure, and freeze out again. Same with local lakes due to the uneven topography.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Model calculatios with a release of carbon dioxide would cause some temporary greenhouse effect, but in the equatorial region a release of carbon dioxide is even more unlikely than a release of water vapor.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-04-30T20:14:08.553" LastActivityDate="2014-04-30T20:14:08.553" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2320" PostTypeId="1" CreationDate="2014-04-30T15:12:41.393" Score="2" ViewCount="163" Body="&lt;p&gt;It is said that in Blackhole 100% part of the mass is getting converted into energy. Then how come Blackhole is having mass ?&lt;/p&gt;&#xA;" OwnerUserId="760" LastEditorUserId="1469" LastEditDate="2014-05-02T18:33:40.227" LastActivityDate="2014-05-02T18:33:40.227" Title="How can a Blackhole have mass?" Tags="&lt;black-hole&gt;&lt;supermassive-black-hole&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2321" PostTypeId="2" ParentId="2320" CreationDate="2014-04-30T17:50:43.637" Score="0" Body="&lt;p&gt;The truth is no one truly knows if 100% of the mass is being converted into energy. So without this assumption many other things are possible. Some think that since there is a constant flow of mass into the black hole, the conversion into energy is not keeping up with the rate the matter is being consumed, therefore the mass would remain substantially large.&lt;/p&gt;&#xA;" OwnerUserId="1397" LastActivityDate="2014-04-30T17:50:43.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2322" PostTypeId="2" ParentId="2211" CreationDate="2014-04-30T22:25:47.327" Score="1" Body="&lt;p&gt;You have to set your equatorial axis to tilt to match your latitude eg if you were at 5 degrees North then the axis needs to tilt to 5 degrees. Then you need to find an object of known RA and declination (at my latitude I always used Polaris as it didn't move) and then set the setting circles (the dials) to those, then direct the telescope to the RA and declination you seek.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2014-04-30T22:25:47.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2323" PostTypeId="1" AcceptedAnswerId="2326" CreationDate="2014-05-01T00:41:03.740" Score="1" ViewCount="85" Body="&lt;p&gt;I'm doing a project for a statistics class, and I thought it may be interesting work with a dataset associated with astronomy. I know absolutely nothing about astronomy, unfortunately, and I wouldn't even know where to look for any large sets of data. Preferably I would want some data that could conform to a regression model, ie explanatory variables plus a response variable. Does anyone know of a place to look for what I am describing? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the question is kind of general - I guess the important part is that the data is somewhat 'interesting' (something a layman understands), and that there are multiple variables that might possibly have a correlation.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastActivityDate="2014-05-01T03:23:38.057" Title="Sources of astronomy-related dataset?" Tags="&lt;universe&gt;&lt;early-universe&gt;&lt;earth-like-planet&gt;&lt;data-analysis&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2325" PostTypeId="1" CreationDate="2014-05-01T03:20:29.037" Score="6" ViewCount="974" Body="&lt;p&gt;I have always been curious about what black holes (or their event horizons) would look like if we could directly observe them. I notice that they tend to be depicted as flat holes in space but it seems to me that they would be spheres. Is there a scientific consensus on this. Or is it a vague question?&lt;/p&gt;&#xA;" OwnerUserId="1525" LastActivityDate="2014-05-01T18:10:05.673" Title="What shape are black holes theorized to be?" Tags="&lt;black-hole&gt;&lt;space-time&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2326" PostTypeId="2" ParentId="2323" CreationDate="2014-05-01T03:23:38.057" Score="2" Body="&lt;p&gt;You have plenty of choices!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am specialized in X-ray astronomy, then I can suggest you many facilities: they usually give at least some of their data publicly available. Tipically, you have data for light curves (photon count rates versus time), and/or spectra (that is flux versus energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest you to work on the light curves: they are easily available, already reduced (you don't have to play with the data to make them usable, you can directly use them), and they fit your request of conforming to a regression model. Also spectra are enjoying, still I know some theoreticians who are not familiar with spectra, plus they usually need some more knowledge, because of response files, spectral fitting packages, etc. It is up to you (and your available time)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here a list of some facilities with available data:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://xte.mit.edu/ASM_lc.html&quot; rel=&quot;nofollow&quot;&gt;Rossi X-ray Time Explorer&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://swift.gsfc.nasa.gov/results/transients/&quot; rel=&quot;nofollow&quot;&gt;Swift/BAT&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.isdc.unige.ch/integral/archive&quot; rel=&quot;nofollow&quot;&gt;INTEGRAL&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cxc.harvard.edu/cda/&quot; rel=&quot;nofollow&quot;&gt;Chandra&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://maxi.riken.jp/top/index.php?cid=000000000001&quot; rel=&quot;nofollow&quot;&gt;MAXI&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also &lt;a href=&quot;http://heasarc.gsfc.nasa.gov/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; you can find a list of many facilities with NASA participation, so the list is much larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course you can also use other kind of data (optical, infrared, radio) by the &lt;a href=&quot;http://archive.stsci.edu/&quot; rel=&quot;nofollow&quot;&gt;MAST project&lt;/a&gt;, but I have no experience on those. I guess, in the optical band, the &lt;a href=&quot;http://adass.org/adass/proceedings/adass94/bornek.html&quot; rel=&quot;nofollow&quot;&gt;Hubble Space Telescope&lt;/a&gt; will have a huge amount of data!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need a hand, beyond the SE vote up/vote down neurotic fashion, let me know. Good luck!&lt;/p&gt;&#xA;" OwnerUserId="1038" LastActivityDate="2014-05-01T03:23:38.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2327" PostTypeId="2" ParentId="2325" CreationDate="2014-05-01T04:13:30.187" Score="4" Body="&lt;p&gt;The event horizon would be spherical. What is usually depicted in drawings is the accretion disk, where all the debris gather before entering the event horizon.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-05-01T04:13:30.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2330" PostTypeId="1" AcceptedAnswerId="8488" CreationDate="2014-05-01T15:02:10.933" Score="1" ViewCount="90" Body="&lt;p&gt;If we consider a point on the central part of a contracting object as observed from earth, the point is continuously moving away from us as the object contracts. Moreover, light emerges from a deeper gravitational well as the object contracts. So, shouldn't contracting objects show a redshift? And if so it should be less towards the edges than towards the centre, right?&lt;/p&gt;&#xA;" OwnerUserId="1156" LastEditorUserId="1156" LastEditDate="2014-05-02T03:21:10.410" LastActivityDate="2015-01-06T23:06:42.690" Title="Do contracting objects show redshift?" Tags="&lt;light&gt;&lt;redshift&gt;" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="2332" PostTypeId="1" AcceptedAnswerId="2338" CreationDate="2014-05-01T16:14:12.030" Score="7" ViewCount="358" Body="&lt;p&gt;I'm trying to imagine what a hypothetical observer on the surface of Alpha Centauri Bb would see in the daytime sky. I know it would vary depending on the orbit of &lt;strong&gt;b&lt;/strong&gt; around &lt;strong&gt;B&lt;/strong&gt;, as well as the position of B and A in relation to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would the &quot;sun&quot; in the sky look like at different times on Bb? I know that at times B and A would overlap and appear as a single sun, but what about when they don't?&lt;/p&gt;&#xA;" OwnerUserId="6" LastActivityDate="2014-05-01T18:53:58.263" Title="How would Alpha Centauri A appear from the surface of Alpha Centauri Bb?" Tags="&lt;exoplanet&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2333" PostTypeId="2" ParentId="856" CreationDate="2014-05-01T17:09:23.913" Score="18" Body="&lt;p&gt;There is a prior answer here claiming that &quot;a moon with a moon would be an unstable system&quot;. That is incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively: Of course satellites can have satellites with long-term stable orbits. Think of the Earth orbiting the Sun, and the Moon orbiting the Earth. The orbit of the Moon (a satellite's satellite) is long-term stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More rigorously:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbit of a satellite's satellite will be stable if it's deep enough inside the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;noreferrer&quot;&gt;Hill sphere&lt;/a&gt;, within the so-called true region of stability. The limits are a bit fuzzy, but the true region of stability is typically the lower 1/3 to 1/2 of the Hill sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the gravitational potential, the Hill sphere is the area where contours become circular. Deep into that zone, orbits are long-term stable:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Oa9oX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line is: A moon can have its own moons if it's big enough, and far enough from the planet, and if the secondary moons are close enough to the primary moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to calculate the Hill sphere is given on the wiki page linked above. Some more math can be found here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jgiesen.de/astro/stars/roche.htm&quot; rel=&quot;noreferrer&quot;&gt;http://www.jgiesen.de/astro/stars/roche.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few extra articles about the issue of long-term stability of satellite orbits:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mnras.oxfordjournals.org/content/391/2/675.full&quot; rel=&quot;noreferrer&quot;&gt;http://mnras.oxfordjournals.org/content/391/2/675.full&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/001910359190039V&quot; rel=&quot;noreferrer&quot;&gt;http://www.sciencedirect.com/science/article/pii/001910359190039V&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mnras.oxfordjournals.org/content/373/3/1227.full&quot; rel=&quot;noreferrer&quot;&gt;http://mnras.oxfordjournals.org/content/373/3/1227.full&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2014-05-01T17:09:23.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2334" PostTypeId="2" ParentId="2325" CreationDate="2014-05-01T17:42:59.077" Score="4" Body="&lt;p&gt;It's complicated, and it depends on what you mean by &quot;black hole&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, for a distant observer, the &quot;black hole&quot; is the event horizon - the surface within which there is no escape to the outside. The more massive the black hole, the bigger the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Event_horizon#Event_horizon_of_a_black_hole&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Event_horizon#Event_horizon_of_a_black_hole&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If the black hole is not rotating&lt;/strong&gt;, the event horizon is a sphere, plain and simple. In a perfect vacuum, and if the black hole is big enough (is not microscopic), looking at it you'll see a perfectly black, round &quot;thing&quot;, with the stellar background behind it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KT8eg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the background image near the black hole would be distorted as if seen through a bad lens. This would become apparent as you move around it. This is because the powerful gravity of the BH bends the rays of light. See the first half of this video:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=8s-pWPqFQBE&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=8s-pWPqFQBE&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the black hole is not in a perfect vacuum, but is surrounded by various pieces of matter, it may capture that matter, and an accretion disk will grow around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JYTuV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If the black hole is rotating&lt;/strong&gt;, things are a bit more complicated. This is an object described by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kerr_metric&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kerr metric&lt;/a&gt;, and it has two important surfaces: the usual event horizon, and the outer skin of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ergosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;ergosphere&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5mVD3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing can escape from within the event horizon. It is possible to escape from within the ergosphere if you rotate with the black hole - this is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Penrose_process&quot; rel=&quot;nofollow noreferrer&quot;&gt;Penrose process&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was not able to find visualisations of a Kerr (rotating) black hole, but it should look somewhat similar to a non-rotating one, except it would appear &quot;flattened&quot; at the poles, and bulging at the equator.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2014-05-01T17:58:17.097" LastActivityDate="2014-05-01T17:58:17.097" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2335" PostTypeId="2" ParentId="2330" CreationDate="2014-05-01T18:06:11.757" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;shouldn't contracting objects show a redshift?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, they should. But do they, actually? That's the better question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To have any appreciable redshift, the speed of the object needs to be huge. That will be a very short implosion. Also, there aren't many mechanisms that can accelerate implosions that much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm saying is - it's theoretically possible, but in practice you won't see it that often, if ever.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And if so it should be less towards the edges than towards the centre,&#xA;  right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sure. If the object is transparent to its own radiation, then you'd see blueshift from the other side, too. Again, this is very theoretical.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2014-05-01T18:06:11.757" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2336" PostTypeId="2" ParentId="2325" CreationDate="2014-05-01T18:10:05.673" Score="4" Body="&lt;p&gt;It depends on whether the black hole is rotating or not. If it is, The &lt;a href=&quot;https://en.wikipedia.org/wiki/Ergosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;ergosphere&lt;/a&gt; and singularity are oblate spheroids. If not, both would be spherical. The event horizon is always spherical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accreting matter around a black hole would form an accretion disk, however this should not be mistaken for the black hole itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZcQAL.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2014-05-01T18:10:05.673" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2338" PostTypeId="2" ParentId="2332" CreationDate="2014-05-01T18:39:02.523" Score="10" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm trying to imagine what a hypothetical observer on the surface of&#xA;  Alpha Centauri Bb would see in the daytime sky. I know it would vary&#xA;  depending on the orbit of b around B, as well as the position of B and&#xA;  A in relation to each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My calculations are based on data from these wiki pages:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Alpha_Centauri&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Alpha_Centauri&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Alpha_Centauri_Bb&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Alpha_Centauri_Bb&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The planet Alpha Centauri Bb orbits the star Alpha Centauri B on an orbit with a semi-major axis of 0.04 AU. That's 25x closer to its star than Earth is to the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alpha Centauri B has 86% the diameter of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, ACB would appear in the sky of ACBb as a huge, extremely bright and hot disk, about 21x bigger than the Sun as seen from Earth (25 x 0.86). The heat would destroy and melt many materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ACB looks a bit more orange than the Sun, but the color difference is small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ACA and ACB orbit each other, being separated by as little as 11 AU, or as much as 35 AU. ACA is about 22% bigger in diameter than the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, ACA as seen from ACBb would appear as a very small &quot;sun&quot; (about 1/10 the size of the Sun as seen from Earth) when at the closest approach, or basically an extremely bright dot, with no apparent size, so shiny it would hurt the eye, when at the maximal distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ACA would provide essentially no heat, and very little light, when compared to ACB (or even when compared to what we get from the Sun here on Earth). But it would be very similar in color to the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the closest approach, ACA would look very similar to the Sun as seen from Saturn. At the maximal distance, it would look like the Sun seen from Neptune.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ACB is a huge fireball of doom, 20x bigger than how we see the Sun, a bit of an orange hue, scorching hot, melting and igniting everything. ACA is a tiny disk, or even just a dot, yellow-white, very bright, providing essentially no heat and much less light than ACB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ACA would grow bigger, then smaller again, following a cycle of 80 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ACA and ACB would get closer together in the sky, then drift further apart, on a cycle following ACBb's &quot;year&quot; - which is only 3.2 days in our terms. A very, very short &quot;year&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not unlikely, although this is speculative, that ACBb is tidally locked to ACB, because they are so close together. If that's the case, ACB would only be visible from one side of ACBb, and would appear to remain &quot;fixed&quot; in the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an artist's impression. Keep in mind that ACB is a giant fireball, so this is more like a fisheye lens image (horizon to horizon in one snapshot). The size of ACA is overestimated. The Sun as seen from Earth would be the size of the letter B in that image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9u3hb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I know that at times B and A would overlap and appear as a single sun,&#xA;  but what about when they don't?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, do they? Do we know that for sure? For the overlap to happen, the orbit of Bb would have to be relatively coplanar to the mutual orbits of A and B. This seems likely, but I'm not sure if it has been established as a fact.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2014-05-01T18:53:58.263" LastActivityDate="2014-05-01T18:53:58.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2339" PostTypeId="2" ParentId="2320" CreationDate="2014-05-02T03:14:30.387" Score="2" Body="&lt;p&gt;I believe when most scientists refer to the mass of a blackhole - i.e. a blackhole having the mass of a billion Suns, what they really mean is that the calculated gravitational pull of the blackhole is equivalent to that of such a mass would it exists at the location of the blackhole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Perhaps you could also bring some reference with respect to where exactly have you heard that 'It is said that in Blackhole 100% part of the mass is getting converted into energy'&lt;/p&gt;&#xA;" OwnerUserId="1469" LastActivityDate="2014-05-02T03:14:30.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2340" PostTypeId="2" ParentId="2115" CreationDate="2014-05-02T03:47:51.280" Score="8" Body="&lt;p&gt;Neutron stars are not to be considered as an homogeneous object, they have different properties at different &lt;em&gt;layers&lt;/em&gt; which are dependant on pressure and temperature for example. Thus in the core below some critical temperature you could have superconductant protons (or any charged baryon) thus meaning you would not find atoms, but more like a soup of &lt;em&gt;free&lt;/em&gt; particles.&#xA;In conclusion I don't see how could this fall into the definition of a chemical element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: &lt;a href=&quot;http://books.google.com.uy/books?id=fgj_TZ06niYC&quot;&gt;Neutron Stars 1: Equation of State and Structure.&lt;/a&gt; By P. Haensel, A.Y. Potekhin, D.G. &lt;/p&gt;&#xA;" OwnerUserId="1469" LastActivityDate="2014-05-02T03:47:51.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2341" PostTypeId="2" ParentId="1855" CreationDate="2014-05-02T05:27:09.557" Score="5" Body="&lt;p&gt;Well let's see:&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Local Standard Time Meridian (LSTM)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The Local Standard Time Meridian (LSTM) is a reference meridian used for a particular time zone and is similar to the Prime Meridian, which is used for Greenwich Mean Time. &#xA;The (LSTM) is calculated according to the equation:&#xA;$$&#xA;LSTM = 15^{o}.\Delta T_{GMT}&#xA;$$&#xA;where $\Delta T_{GMT}$ is the difference of the Local Time (LT) from Greenwich Mean Time (GMT) in hours.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Equation of Time (EoT)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The equation of time (EoT) (in minutes) is an empirical equation that corrects for the eccentricity of the Earth's orbit and the Earth's axial tilt.&#xA;$$&#xA;EoT = 9.87 sin\left(2B\right) - 7.53cos\left(B\right) - 1.5 sin\left(B\right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $B = \frac{360}{365}\left(d - 81\right) $ in degrees and d is the number of days since the start of the year.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Time Correction Factor (TC)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The net Time Correction Factor (in minutes) accounts for the variation of the Local Solar Time (LST) within a given time zone due to the longitude variations within the time zone and also incorporates the EoT above.&#xA;$$&#xA;TC = 4\left(Longitude - LSTM\right) + EoT&#xA;$$&#xA;The factor of 4 minutes comes from the fact that the Earth rotates 1° every 4 minutes.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Local Solar Time (LST)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The Local Solar Time (LST) can be found by using the previous two corrections to adjust the local time (LT).&#xA;$$&#xA;LST = LT + \frac{TC}{60}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Hour Angle (HRA)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The Hour Angle converts the local solar time (LST) into the number of degrees which the sun moves across the sky. By definition, the Hour Angle is 0° at solar noon. Since the Earth rotates 15° per hour, each hour away from solar noon corresponds to an angular motion of the sun in the sky of 15°. In the morning the hour angle is negative, in the afternoon the hour angle is positive.&#xA;$$&#xA;HRA = 15^{o}\left(LST - 12\right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Declination angle:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The declination angle denoted by $\delta$, varies seasonally due to the tilt of the Earth on its axis of rotation and the rotation of the Earth around the sun. If the Earth were not tilted on its axis of rotation, the declination would always be 0°. However, the Earth is tilted by 23.45° and the declination angle varies plus or minus this amount. Only at the spring and fall equinoxes is the declination angle equal to 0°.&#xA;$$&#xA;\delta = 23.45^{o} sin\left[\frac{360}{365}\left(d - 81\right)\right]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;strong&gt;d&lt;/strong&gt; is the day of the year with Jan 1 as d = 1.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Elevation angle:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The elevation angle is the angular height of the sun in the sky measured from the horizontal. &#xA;$$&#xA;α=sin^{−1}[sin\delta sin\phi + cos\delta cos\phi cos(HRA)]&#xA;$$&#xA;Where $\delta$ is the &lt;strong&gt;declination angle&lt;/strong&gt;, $\phi$ is the &lt;strong&gt;local latitude&lt;/strong&gt; and &lt;em&gt;HRA&lt;/em&gt; is the &lt;strong&gt;Hour angle&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Azimuth Angle:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The azimuth angle is the compass direction from which the sunlight is coming. At solar noon, the sun is always directly south in the northern hemisphere and directly north in the southern hemisphere.&#xA;$$&#xA;Azimuth = cos^{-1}\left[\frac{sin\delta cos\phi - cos\delta sen\phi cos(HRA)}{cos\alpha}\right]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\delta$ is the &lt;strong&gt;declination angle&lt;/strong&gt;, $\phi$ is the &lt;strong&gt;local latitude&lt;/strong&gt; and &lt;em&gt;HRA&lt;/em&gt; is the &lt;strong&gt;Hour angle&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Zenith Angle:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The zenith angle is the angle between the sun and the vertical. The zenith angle is similar to the elevation angle but it is measured from the vertical rather than from the horizontal, thus making the zenith&#xA;$$&#xA;Zenith = 90° - \alpha&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\alpha$ is the &lt;strong&gt;elevation angle&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note&lt;/em&gt; that your input parameters are going to be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Longitude&lt;/li&gt;&#xA;&lt;li&gt;$\Delta T_{GMT}$ is the difference of the Local Time (LT) from Greenwich Mean Time (GMT) in hours&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;LT&lt;/strong&gt; local military time in hours&lt;/li&gt;&#xA;&lt;li&gt;$\phi$ the &lt;strong&gt;local latitude&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;d&lt;/strong&gt; the day of the year&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1469" LastEditorUserId="1469" LastEditDate="2014-05-02T06:27:09.053" LastActivityDate="2014-05-02T06:27:09.053" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2342" PostTypeId="1" CreationDate="2014-05-02T08:41:31.957" Score="-2" ViewCount="251" Body="&lt;p&gt;Given the vast infinite nature of universe, billions of worlds, distances and masses that we cannot comprehend, do you feel Big Bang theory is inadequate and in some ways tells you that it was all there from beginning.&lt;/p&gt;&#xA;" OwnerUserId="1541" LastActivityDate="2014-05-02T12:54:31.030" Title="Big Bang Thoery, what triggered it and why" Tags="&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-05-03T05:17:06.717" ContentLicense="CC BY-SA 3.0" />
  <row Id="2343" PostTypeId="1" AcceptedAnswerId="2349" CreationDate="2014-05-02T09:00:22.137" Score="3" ViewCount="823" Body="&lt;p&gt;We know that, in stars, hydrogen is used by the stars and due to the fusion in two hydrogen elements we get helium -&gt; Carbon -&gt; Oxygen -&gt;Other elements -&gt; Iron -&gt; then Super/Hypernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If everywhere hydrogen is getting converted to other element then will we(universe) have hydrogen to create/retain stars ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or in the end what is the ultimate matter in the universe ?&lt;/p&gt;&#xA;" OwnerUserId="760" LastEditorUserId="760" LastEditDate="2014-05-02T09:47:59.913" LastActivityDate="2019-08-01T15:01:56.423" Title="In the end what is the ultimate matter/element in the universe, due to fusion process in stars?" Tags="&lt;star&gt;&lt;universe&gt;&lt;planetary-atmosphere&gt;&lt;formation&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2345" PostTypeId="2" ParentId="2342" CreationDate="2014-05-02T12:54:31.030" Score="3" Body="&lt;p&gt;Big Bang theory is rather adequate, with respect to observations. But it's incomplete. Your question is about the pre-planck epoch, a state of the universe, when space and time didn't exist in a way we are used to in standard physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing is really &lt;em&gt;known&lt;/em&gt; (by empirical evidence) about this state, although there are theoretical approches (e.g. &lt;a href=&quot;http://arxiv.org/pdf/gr-qc/9906072.pdf&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;), which try to describe even those highly non-local and non-causal structures, from which &quot;eventually&quot; the Planck epoch could emerge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck_epoch&quot; rel=&quot;nofollow&quot;&gt;Planck epoch&lt;/a&gt;, lasting from zero to about $10^{-43}\mbox{ s}$, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck_time&quot; rel=&quot;nofollow&quot;&gt;Planck time&lt;/a&gt;, is still far from being understood, since the energy needed to fully investigate this state of the universe is far beyond accessible to experiments. We just know, that neither &lt;a href=&quot;http://en.wikipedia.org/wiki/General_relativity&quot; rel=&quot;nofollow&quot;&gt;general relativity&lt;/a&gt; nor &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_mechanics&quot; rel=&quot;nofollow&quot;&gt;quantum theory&lt;/a&gt; describe this state correctly. Research about &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_gravity&quot; rel=&quot;nofollow&quot;&gt;quantum gravity&lt;/a&gt; tries to fill this gap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may think of the pre-Planck and the Planck epoch as a state of the universe, which is dominated by &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_fluctuation&quot; rel=&quot;nofollow&quot;&gt;quantum fluctuations&lt;/a&gt;. They don't need a reason or a cause to occur. They are a consequence of &lt;a href=&quot;http://en.wikipedia.org/wiki/Uncertainty_principle&quot; rel=&quot;nofollow&quot;&gt;Heisenberg's uncertainty principle&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, if you try to measure time with very high precision, you'll get an increasing uncertainty in the energy measurement of the observed particle (&lt;a href=&quot;http://ocw.usu.edu/physics/classical-mechanics/pdf_lectures/14.pdf&quot; rel=&quot;nofollow&quot;&gt;time-energy-uncertainty&lt;/a&gt;). Within the very short Planck epoch, high energy fluctuations needed to occur, since both values (time and energy) cannot be pinned down together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for the Big Bang is thought to be such a kind of quantum fluctuations, yet not understood in very detail.&#xA;Questions like where?, when?, and why? don't apply in the intuitive sense to quantum phenomena. It's more a matter of probabilities, &lt;a href=&quot;http://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation&quot; rel=&quot;nofollow&quot;&gt;waves&lt;/a&gt;, and relations between observable values, without the observable values themselves being defined independently.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-05-02T12:54:31.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2346" PostTypeId="1" CreationDate="2014-05-02T13:26:54.063" Score="5" ViewCount="478" Body="&lt;p&gt;I mean those kind of stars should be very stable and barely have activity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their light curves should be very very flat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our sun, as a G type star, are not so stable. At lest there are lots of floating solar spots.&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2015-04-22T16:15:18.713" Title="which or what kind of star has a very stable luminosity?" Tags="&lt;star&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2347" PostTypeId="1" AcceptedAnswerId="2369" CreationDate="2014-05-02T15:37:20.757" Score="2" ViewCount="265" Body="&lt;p&gt;Am asking if a neutrino can be pull by the gravity field of a quasar or a black hole, assuming the neutrino have momentum. It could be possible in the that a neutrino can orbit a black hole at an orbit equivalent to the Schwarzschild radius give or take. If it even possible for a neutrino to go through a black hole?&lt;/p&gt;&#xA;" OwnerUserId="1546" LastActivityDate="2014-05-05T19:30:25.227" Title="Are neutrinos affected by gravity fields?" Tags="&lt;black-hole&gt;&lt;neutrinos&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2348" PostTypeId="2" ParentId="2346" CreationDate="2014-05-02T20:47:20.053" Score="6" Body="&lt;p&gt;Hipparchos numbers of the 26 most stable stars known:&#xA;&lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR-S?HIP%202021&quot;&gt;2021&lt;/a&gt;, &lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR-S?HIP%202854&quot;&gt;2854&lt;/a&gt;, &lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR-S?HIP%205542&quot;&gt;5542&lt;/a&gt;, &lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR-S?HIP%2016611&quot;&gt;16611&lt;/a&gt;, 19747, 24927, 32537, 38414, 42913, 45556, 50191, 57363, 71053, 73555, 74666, 76440, 74946, 90139, 94648, 96052, 10239, 102488, 104732, 111169, 116631, 118322.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare Hpmax and Hpmin: They differ 0.01. Standard error e_Hpmag is 0.0003 mag.&#xA;There are 681 stars in the Hipparchos catalog with an amplitude of 0.01 mag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More detail in &lt;a href=&quot;http://www.aanda.org/articles/aa/pdf/2001/07/aa10185.pdf&quot;&gt;this paper&lt;/a&gt;, including a link to the appropriate VizieR query.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-05-02T20:47:20.053" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2349" PostTypeId="2" ParentId="2343" CreationDate="2014-05-02T22:37:09.440" Score="8" Body="&lt;p&gt;If we don't wait for too long, where we don't know for sure, whether &lt;a href=&quot;http://en.wikipedia.org/wiki/Proton_decay&quot;&gt;protons or atomic nuclei stay stable forever&lt;/a&gt;, there won't be just one single element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most abundant element in the universe will probably stay hydrogen, since it will stay in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Warm%E2%80%93hot_intergalactic_medium&quot;&gt;intergalactic medium&lt;/a&gt;, and thin out by &lt;a href=&quot;http://en.wikipedia.org/wiki/Accelerating_universe&quot;&gt;accelerating cosmic expansion&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second-most abundant element will probably stay helium, mostly for the same reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most frequent stars are probably &lt;a href=&quot;http://en.wikipedia.org/wiki/Red_dwarf&quot;&gt;red dwarfs&lt;/a&gt;.&#xA;They'll burn hydrogen and helium to oxygen and carbon, and end up as white dwarfs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only a small fraction of stars is large enough to burn oxygen and carbon to silicon and iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very heavy elements, like uranium and beyond, as formed in supernovae, will decay to lead, helium (alpha particles), some hydrogen (via neutrons decaying to hydrogen), and a couple of less abundant elements formed by &lt;a href=&quot;http://en.wikipedia.org/wiki/Spontaneous_fission&quot;&gt;spontaneous fission&lt;/a&gt;. Other stable (isotopes of) elements heavier than iron are also formed in, or a consequence of, supernova explosions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of the dead stars will collide, and merge to heavier stars, which will end as intergalactic dust (mixture of elements, e.g. iron and silicon), including planets, after supernova explosions, and as Black Holes. Others will be scattered out of reach of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Supermassive_black_hole&quot;&gt;supermassive black hole&lt;/a&gt; in the center of the galaxy (&lt;a href=&quot;http://en.wikipedia.org/wiki/Galaxy#Future_trends&quot;&gt;gravitational relaxation&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-05-02T22:37:09.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2350" PostTypeId="1" CreationDate="2014-05-03T09:46:48.497" Score="0" ViewCount="201" Body="&lt;p&gt;Science is trying to explain how the universe is made. The beginning of the universe explains the Big Bang theory. But what happened before Big Bang? Science cannot go before that. So, does this mean that always there be space for religion and philosophy to explain things that science cannot?&lt;/p&gt;&#xA;" OwnerUserId="1550" LastActivityDate="2014-05-04T12:28:39.717" Title="Does science need support from religion or philosophy to explain the creation?" Tags="&lt;universe&gt;&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2014-05-04T14:19:51.090" ContentLicense="CC BY-SA 3.0" />
  <row Id="2351" PostTypeId="1" CreationDate="2014-05-03T12:00:33.587" Score="3" ViewCount="74" Body="&lt;p&gt;In the introduction of &lt;a href=&quot;http://arxiv.org/abs/1302.0884&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; paper at the top of p11, it is mentioned that a hypothetical enhancement of the scale invariance of the CMB spectrum to conformal invariance could potentially be established by analyzing the higher n-point functions of the tensor modes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, why are the tensor modes needed for this? Does conformal invariance not appear in the scalar modes?&#xA;How does this work exactly (mathematically), how does conformal invariance (not just scale invariance...!) manifest itself in the n-point functions of the tensor modes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has this already been looked at or will it be done for the tensor modes in the BICEP2 data for example?&lt;/p&gt;&#xA;" OwnerUserId="92" LastActivityDate="2014-05-03T12:00:33.587" Title="How exactly can the hypothetical conformal invariance of the CMB spectrum be established by analyzing tensor modes?" Tags="&lt;cosmology&gt;&lt;cmb&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2353" PostTypeId="1" AcceptedAnswerId="2354" CreationDate="2014-05-03T22:46:00.133" Score="4" ViewCount="148" Body="&lt;p&gt;If all stars suddenly stopped producing heat (or heating up) which stars would cool to absolute zero first and how long would it take?&lt;/p&gt;&#xA;" OwnerUserId="1494" LastActivityDate="2014-05-04T19:26:53.533" Title="Cooling of stars" Tags="&lt;star&gt;&lt;the-sun&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2354" PostTypeId="2" ParentId="2353" CreationDate="2014-05-03T23:12:13.093" Score="4" Body="&lt;p&gt;In our current universe white dwarfs are the first ones that should cool, because they are already &quot;cold&quot; (not producing anything anymore, just radiating heat)  remnants of a former star. The time for this to happen is disputed (10^15 or 10^37 years), but is far bigger than the age of universe, so nobody expects to find one &quot;cooled star&quot; yet. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_dwarf&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; for details on &quot;black dwarfs&quot; as they are called.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your sudden stop hypothetical universe probably the first one to cool would be the smallest and coolest stars, a M9V red dwarf (with 7.5% solar mass, 8% solar radius, 2,300K temperature). Note that brown dwarfs are categorized as substellar objects, so they shouldn't be considered. I don't think the poorly understood &quot;Mpemba effect&quot; can be yet applied to stars.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditDate="2014-05-04T19:26:53.533" LastActivityDate="2014-05-04T19:26:53.533" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2355" PostTypeId="2" ParentId="2350" CreationDate="2014-05-04T11:44:13.087" Score="4" Body="&lt;p&gt;The question seems to be based on the notion &quot;before&quot;. This notion relies of the concept of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Arrow_of_time&quot; rel=&quot;nofollow&quot;&gt;arrow of time&lt;/a&gt;, meaning time &quot;flowing&quot; in some &quot;direction&quot;. The arrow of time itself is based, among others, on the notion of &quot;time&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck_epoch&quot; rel=&quot;nofollow&quot;&gt;Planck epoch&lt;/a&gt; - the first epoch of the current Big Bang model - lasted a little less than $10^{-43}\mbox{ s}$. That's the shortest unit of time. To ask within this short time interval for an arrow of time doesn't make any sense. Hence a before/after didn't exist within  the Planck epoch, and hence asking for a &quot;before&quot; isn't meaningful within the Planck epoch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Planck epoch to be able to exist needs the mere existence of space and time, not yet an arrow of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cause for the existence of space and time at all is a matter of the pre-Planck epoch. In this phase using any notion which is based on space or time isn't meaningful. Cause and effect in the temporal sense, which underly most &quot;why?&quot;-questions, are based on time, and are meaningless in this epoch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's possible to construct other &quot;pre-Big-Bang&quot; models. But it's hard to find observational evidence for any of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you suggest a creator of whatever kind, you just push the ultimate question a little bit further to the &quot;past&quot;, and making it much more complicated. How could a creator be created, able to create such a complex thing like the universe? You need a chain of ever more complex meta-creators without end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you argue, that the first creator doesn't need a cause to exist, this applies also for the universe, formed from a pre-Planck medium. Hence nothing would be won.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore it's rather unlikely, that religion or philosophy could be able to provide a more satisfying answer than science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://en.wikipedia.org/wiki/Multiverse&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; some metaphysical multiverse theories.)&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-05-04T12:28:39.717" LastActivityDate="2014-05-04T12:28:39.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2357" PostTypeId="1" AcceptedAnswerId="2363" CreationDate="2014-05-04T15:01:12.820" Score="51" ViewCount="14512" Body="&lt;p&gt;Recently I boarded a flight and noticed outside air temperature as -53°C at an altitude of 36860ft (11.23km). I don't know what causes such a freezing temperature in that altitude but was wondering higher altitudes (space) may have even freezing temperatures. Here I got a doubt i.e what happens if an ice cube is left in space? Would it be melting or stay as it is?&lt;/p&gt;&#xA;" OwnerUserId="1103" LastEditorUserId="5278" LastEditDate="2016-12-13T19:37:40.123" LastActivityDate="2016-12-13T19:37:40.123" Title="What would happen if an ice cube is left in space?" Tags="&lt;space&gt;&lt;temperature&gt;&lt;water&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2358" PostTypeId="2" ParentId="2357" CreationDate="2014-05-04T15:10:06.723" Score="12" Body="&lt;p&gt;It would sublimate. The frozen mass of water would decrease in size as the water converts from a solid to a gas (without becoming a liquid) and drifts away.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-05-04T15:10:06.723" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2359" PostTypeId="1" AcceptedAnswerId="2361" CreationDate="2014-05-04T21:27:18.940" Score="2" ViewCount="108" Body="&lt;p&gt;The Sun currently shines because H is being burned to He, and it has been doing this for about 4.57 Ga. But when we analyze the Sun's atmosphere, we do not find unusually large amounts of He. Why not?&lt;/p&gt;&#xA;" OwnerUserId="1555" LastActivityDate="2014-05-04T21:44:28.837" Title="Why do we not find larger quantities of helium in the Sun's atmosphere?" Tags="&lt;the-sun&gt;&lt;stellar-atmospheres&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2361" PostTypeId="2" ParentId="2359" CreationDate="2014-05-04T21:44:28.837" Score="2" Body="&lt;p&gt;Helium is burnt in Sun's core, and it is heavier (specific weight) than hydrogen.&#xA;Hence most of the helium stays in Sun's core.&#xA;(&lt;a href=&quot;http://www.uni.edu/morgans/astro/course/Notes/section2/new8.html&quot; rel=&quot;nofollow&quot;&gt;More details&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-05-04T21:44:28.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2362" PostTypeId="1" AcceptedAnswerId="2364" CreationDate="2014-05-04T22:32:09.740" Score="17" ViewCount="2660" Body="&lt;p&gt;Could large craters on the moon be used as reflective lenses for radio signals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acting like a large radio telescope reflecting radio waves to a satellite positioned over the crater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xAMMj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2018-12-17T10:39:06.723" Title="Can impact craters on the moon act like giant radio telescopes?" Tags="&lt;the-moon&gt;&lt;radio-astronomy&gt;" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2363" PostTypeId="2" ParentId="2357" CreationDate="2014-05-04T22:54:03.050" Score="87" Body="&lt;p&gt;It depends on where in outer space you are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you simply stick it in orbit around the Earth, it'll sublimate: the mean surface temperature of something at Earth's distance from the Sun is about 220K, which is solidly in the vapor phase for water in a vacuum, and the solid-vapor transition at that temperature doesn't pass through the liquid phase.  On the other hand, if you stick your ice cube out in the Oort Cloud, it'll grow: the mean surface temperature is 40K or below, well into the solid phase, so it'll pick up (or be picked up by) gas and other objects in space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A comet is a rough approximation to an ice cube.  If you think of what happens to a comet at various places, that's about what would happen to your ice cube.&lt;/p&gt;&#xA;" OwnerUserId="1559" LastActivityDate="2014-05-04T22:54:03.050" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2364" PostTypeId="2" ParentId="2362" CreationDate="2014-05-05T03:04:20.457" Score="13" Body="&lt;p&gt;Interesting idea. I think the answer is both yes and no -- yes with a manufactured dish but no in the crater's raw state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Arecibo telescope sits in a natural crater, but adds a dish which has a couple of important things required by a radio dish:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;a radio-reflective surface&lt;/li&gt;&#xA;&lt;li&gt;a specific curvature, classically parabolic, but also shaped&lt;/li&gt;&#xA;&lt;li&gt;low surface roughness typically in the order of mm to µm for radio frequencies we're interested in. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Ruze%27s_Equation&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Ruze%27s_Equation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;a receiver close to the reflective surface (focal length), a satellite would probably be beyond the focal point&lt;/li&gt;&#xA;&lt;li&gt;probably some some other things, but that's probably enough.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Edit: Added #4&lt;/p&gt;&#xA;" OwnerUserId="1563" LastEditorUserId="1563" LastEditDate="2014-05-09T00:08:08.773" LastActivityDate="2014-05-09T00:08:08.773" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2365" PostTypeId="2" ParentId="2362" CreationDate="2014-05-05T03:09:16.633" Score="1" Body="&lt;p&gt;I agree that the crater needs to be lined, but you also have the problem of maintaining a satellite in stationary orbit above the crater. Nearly impossible unless the crater is on the equatorial plane. Also, a stationary satellite around the moon would be influenced by the Earth, so you would need to burn fuel to keep the satellite in position.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-05-05T03:09:16.633" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2369" PostTypeId="2" ParentId="2347" CreationDate="2014-05-05T19:30:25.227" Score="4" Body="&lt;p&gt;If you buy into General Relativity (you need to throw out an awful lot of modern physics if you don't), gravity is a manifestation of the shape of spacetime around massive objects. It's not that they are being &quot;pulled,&quot; but rather that they are just following the straightest path through spacetime that they can. Neutrinos would be the same as any other particles in that respect, and following anything other than this path would require some extraordinary gymnastics and external forces.&lt;/p&gt;&#xA;" OwnerUserId="366" LastActivityDate="2014-05-05T19:30:25.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2370" PostTypeId="1" CreationDate="2014-05-05T20:46:20.493" Score="3" ViewCount="161" Body="&lt;p&gt;Considering the vast amount of insulating gas emitted by human activity on Earth without a coincident release on all other planets in the solar system as well as the phenomenon that planets in a solar system normally exhibit similar coincident temperature changes described &lt;a href=&quot;https://astronomy.stackexchange.com/a/1857/1098&quot;&gt;here&lt;/a&gt;, how far have temperature changes diverged from Earth's?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please include raw data, graphed if possible.&lt;/p&gt;&#xA;" OwnerDisplayName="user1098" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-05-05T20:46:20.493" Title="Intrasolar planetary surface temperature change divergence from Earth" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;temperature&gt;&lt;surface&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2371" PostTypeId="2" ParentId="2362" CreationDate="2014-05-06T00:13:57.220" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Could large craters on the moon be used as reflective lenses for radio&#xA;  signals?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'd have to line the surface with something reflective to microwaves, like a metallic mesh, or similar materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, the shape of the crater is probably not quite ideal, so it would have to be adjusted a little, carved up a bit in various places. But it's a good start, and definitely better than starting with a flat ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the question of stability - you need to make sure that whatever changes you make (carving a different shape, lining it with mesh) do not affect the stability of the crater, or else various parts may slide or collapse. This is an engineering problem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Acting like a large radio telescope reflecting radio waves to a&#xA;  satellite positioned over the crater.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not possible unless the crater is exactly on the equator, and even then it would be tricky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a crater like the one in your picture is so strongly curved, the focal length is about the same as the diameter. In other words, if the diameter of the hole is X, the altitude of the receiver is pretty close to X - give or take something like 50% or so, depending on the exact curvature. It might be easier to just build a giant arch over the crater. Again, this is a matter of engineering.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2014-05-06T00:13:57.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2372" PostTypeId="1" AcceptedAnswerId="2376" CreationDate="2014-05-06T00:49:18.813" Score="11" ViewCount="826" Body="&lt;p&gt;There are many documented cases of meteorites being reported as &lt;em&gt;from Mars&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Allan_Hills_84001&quot;&gt;http://en.wikipedia.org/wiki/Allan_Hills_84001&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I can understand that it's possible to know if a rock originated as a meteorite. What I don't understand is how scientists can say it came from Mars with certainty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can they verify such a claim, and differentiate between a rock that is &lt;em&gt;similar&lt;/em&gt; to one from Mars to one that &lt;em&gt;really&lt;/em&gt; is from Mars.&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2020-10-07T09:10:10.317" Title="How do scientists know that a rock found on Earth came from Mars?" Tags="&lt;mars&gt;&lt;meteorite&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2373" PostTypeId="1" AcceptedAnswerId="2378" CreationDate="2014-05-06T07:55:21.260" Score="9" ViewCount="309" Body="&lt;p&gt;It has been speculated that Venus billions of years ago could've had a much different atmosphere with liquid water on its surface and possibly life. Partly thanks to the cooler young Sun. But is it possible for science to find out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What traces would remain that our probes could measure? AFAIK Venus was completely resurfaced 300-600 million years ago so I suppose there won't be any geological signs of water of the kind found on Mars. And the atmosphere obviously has changed dramatically if Venus was once Earth-like. What should we look for to discover the evolution of Venus? What instruments should we land there, or fly in its atmosphere or orbit?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2014-05-06T10:02:20.797" LastActivityDate="2014-08-18T13:05:25.543" Title="Can we find out whether early Venus was Earth-like or not?" Tags="&lt;planetary-atmosphere&gt;&lt;earth-like-planet&gt;&lt;venus&gt;&lt;instruments&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2375" PostTypeId="1" CreationDate="2014-05-06T10:36:41.840" Score="4" ViewCount="17652" Body="&lt;p&gt;This is a very basic question, but I am a little confused. As far as I know, the temperature of a star is analyzed based on the color of the light it emits. So, if a star is moving away from us, then the light emitted by it will be redshifted(or if it is stationary with respect to us and the light undergoes gravitational redshift), then how do we know the exact temperature of that star or any other object because it is possible that we observe red light but actually the star might be emitting yellow light.&lt;/p&gt;&#xA;" OwnerUserId="1156" LastEditorUserId="1156" LastEditDate="2015-04-05T17:04:02.370" LastActivityDate="2016-12-13T00:44:03.273" Title="How do we find the exact temperature of a star?" Tags="&lt;star&gt;&lt;temperature&gt;&lt;redshift&gt;&lt;stellar-astrophysics&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2376" PostTypeId="2" ParentId="2372" CreationDate="2014-05-06T11:07:21.910" Score="5" Body="&lt;p&gt;Mars is the only planet providing rocks with a similar chemical composition and age. So the origin of the meteorite is evident.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't mean, that it's absolutely waterproof. There might have existed protoplanets similar to Mars 4 billion years ago, which since then have been swallowed by Jupiter, the Sun, or have been ejected out of the solar system. But this would be much of coincidence without leaving convincing evidence. (&lt;a href=&quot;http://arxiv.org/pdf/1306.0975.pdf&quot;&gt;Here a paper about a couple of simulations&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-05-06T11:07:21.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2377" PostTypeId="2" ParentId="2375" CreationDate="2014-05-06T11:17:01.513" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://skyserver.sdss.org/dr1/en/proj/basic/spectraltypes/lines.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;Spectral lines&lt;/a&gt; occur at defined wavelengths. By their redshift you can calculate the radial velocity (or gravitational redshift) of the star, or the absorbing medium, and hence the amount you've to shift the &lt;a href=&quot;https://www.google.de/#q=black+body+radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;black body&lt;/a&gt; radiation to obtain the surface temperature (and the radial velocity of a possibly absorbing medium between the star and Earth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Schematic &lt;strong&gt;example&lt;/strong&gt;:&#xA;Assume, you measure the following two stellar spectra, and you're able to identify the typical &lt;a href=&quot;http://en.wikipedia.org/wiki/Balmer_series&quot; rel=&quot;nofollow noreferrer&quot;&gt;H-alpha spectral emission line&lt;/a&gt;. This line should be at 565.3 nm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iChlw.jpg&quot; alt=&quot;spectrum schemes of stars of different redshift, but almost the same measured color&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second spectrum, H-alpha is at the correct position: no redshift. In the first spectrum, it's redshifted (towards longer wavelength).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the measured intensity may be the same elsewhere in the spectrum, you'll know, that the first spectrum is of a hotter star, since the maximum intensity (besides the H-alpha line) is left (towards blue) of the H-alpha wavelength, whereas the maximum intensity in the second spetrum is right (towards red) of the H-alpha line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both stars would look reddish, but the first one is the hotter one, and it's redshifted, either due to Doppler shift, due to gravity, or due to cosmic expansion.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2014-05-07T00:06:36.087" LastActivityDate="2014-05-07T00:06:36.087" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2378" PostTypeId="2" ParentId="2373" CreationDate="2014-05-06T12:32:34.057" Score="7" Body="&lt;p&gt;My ad-hoc opinion: This wouldn't be the first step of Venus exploration. Geologic in-situ investigation of the resurfacing hypotheses would already be a very challenging mission.&#xA;Might be, one could find some metamorphic remnants which have survived the last resurfacing, and one could determine the age of rocks. Might be there exist some old layers below the surface, which haven't been molten up completely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some kinds of crystals are more resistant to heat than others. They may also contain some records of more ancient epochs. Comparing their isotopic ratios might tells something about the ancient atmosphere, or at least about ancient geology. But conditions on the surface of Venus are very hard for probes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More feasible would be a detailed isotopic analysis of Venus' atmosphere, as e.g. proposed as &lt;a href=&quot;http://www.planetaryprobe.org/sessionfiles/session2/abstracts/a137.pdf&quot;&gt;VDAP&lt;/a&gt; mission concept, or by follow-up balloon missions. A submarine-like ballon diving into the hellish conditions of the lower atmosphere or even to the surface for a short period might be an approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A TLS (tunable laser spectrometer) has been suggested to analyse the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;May be an additional mass spectrograph would be useful. If short dives close to the surface become feasible, cameras would make sense. A LIBS (laser induced breakdown spectroscope) would make sense only on direct contact with the surface due to the atmospheric absorption. APXS wouldn't probably work due to the heat, unless samples could be taken for analysis higher into the (cooler) atmosphere. IR spectroscopy could possibly work, if cooling can be achieved somehow (e.g. by insulation and adiabatic decompression). If surface samples can be taken during short surface contact, more ways to analyse rocks would become possible. The more instruments, the more weight, the larger the ballon, the larger the probe, the more expensive. But I'm far from elaborating technical or scientific proposals here for space agencies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence VDAP with a small TLS and a QMS (quadrupole mass spectrometer) - &lt;a href=&quot;http://solarsystem.nasa.gov/docs/pr454.pdf&quot;&gt;as already proposed by hugely experienced scientists&lt;/a&gt; - would be a good start. &lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-05-06T12:32:34.057" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2379" PostTypeId="1" AcceptedAnswerId="2380" CreationDate="2014-05-06T16:55:45.190" Score="7" ViewCount="466" Body="&lt;p&gt;Yesterday night i witnessed something very strange when i looked outside the window. I saw the moon (crescent) but it was dull red and right on the horizon ,which is strange considering that it is usually on the upper right of the night sky and white in colour. On further inspection with my binoculars i noticed it was lowering down until it was hidden by the mountain range (5km away) next to my building, this all occurred within a few minutes (about 5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tonight i saw the moon (crescent) had again returned to its normal position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please explain the cause for this, i'm completely baffled!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry for the poor wording, i'm not familiar with all the astronomical terms!)&lt;/p&gt;&#xA;" OwnerUserId="1587" LastEditorUserId="1587" LastEditDate="2014-10-12T12:37:49.920" LastActivityDate="2014-10-12T12:37:49.920" Title="Red cresent moon" Tags="&lt;the-moon&gt;&lt;moon-phases&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2380" PostTypeId="2" ParentId="2379" CreationDate="2014-05-06T23:16:41.597" Score="5" Body="&lt;p&gt;The dull red color has been due to atmospheric causes, like the reddish sun close to sunset. There hasn't been an astronomical reason for the reddish color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few days after New Moon moonset occurs short after sunset, so you won't see the Moon high over the horizon at those evenings. With each day the Moon is a little higher above the horizon after sunset. It's hence less close to the horizon at the same time of the day. Less close to the horizon means less atomospheric absorption/scattering responsible for the dull red color, assuming the same weather conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At Full Moon the Moon is at the opposite side of the Sun relative to Earth. Moon is then rising shortly after sunset.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-05-06T23:16:41.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2381" PostTypeId="1" AcceptedAnswerId="2391" CreationDate="2014-05-07T00:34:46.340" Score="4" ViewCount="559" Body="&lt;p&gt;I know from Rayleigh scattering that in the atmosphere redder photons are scattered less than bluer photons by the particles that both make up and are floating around in the atmosphere.  This is why the sky is blue and sunsets are red.  My question is, is the physics behind Rayleigh scattering (which, according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Rayleigh_scattering&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;, is due to the electric polarizability of the molecules) the same physics behind dust extinction? Additionally, is it the only physics in play or is there additional physics that causes the phenomenon?&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2014-05-09T00:36:46.297" Title="Is dust extinction/reddening caused by Rayleigh scattering or some other physics phenomenon (or both)?" Tags="&lt;dust&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2382" PostTypeId="2" ParentId="2381" CreationDate="2014-05-07T01:41:57.777" Score="0" Body="&lt;p&gt;Rayleigh scattering is one kind of elastic scattering. Besides other kinds of &lt;a href=&quot;http://en.wikipedia.org/wiki/Scattering&quot; rel=&quot;nofollow&quot;&gt;scattering&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Absorption_%28electromagnetic_radiation%29&quot; rel=&quot;nofollow&quot;&gt;absorption&lt;/a&gt; contributes to &lt;a href=&quot;http://en.wikipedia.org/wiki/Extinction_%28astronomy%29&quot; rel=&quot;nofollow&quot;&gt;extinction&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Absorption is dependent of the chemical composition of the dust or gas. But generally blue light is easier absorbed than red light, hence the reddening (not to be confused with red-shift!)&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-05-07T01:41:57.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2383" PostTypeId="1" AcceptedAnswerId="2386" CreationDate="2014-05-07T02:53:46.057" Score="3" ViewCount="674" Body="&lt;p&gt;Mercury is the most unstable planet of the solar system, with an oscillating orbital eccentricity (between 0 to 0.45). &lt;a href=&quot;http://en.wikipedia.org/wiki/Mercury_%28planet%29#Spin.E2.80.93orbit_resonance&quot; rel=&quot;nofollow&quot;&gt;It seems&lt;/a&gt; there is a 1% probability in the next 5 billion years Jupiter and Mercury might enter in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Stability_of_the_Solar_System#Mercury.E2.80.93Jupiter_1:1_resonance&quot; rel=&quot;nofollow&quot;&gt;1:1 orbital resonance&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ending variants for this event are :&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;colliding with Venus&lt;/li&gt;&#xA;&lt;li&gt;colliding with the Sun&lt;/li&gt;&#xA;&lt;li&gt;being ejected from solar system.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Can it hit other targets on its way out of the solar system beside those enumerated, like Earth, Mars or the outer planets?&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="204" LastEditDate="2014-05-13T13:11:59.607" LastActivityDate="2014-05-13T13:11:59.607" Title="Can Mercury hit Earth or Mars in the next 5 billion years?" Tags="&lt;impact&gt;&lt;mercury&gt;" AnswerCount="1" CommentCount="16" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2384" PostTypeId="1" CreationDate="2014-05-07T07:40:04.433" Score="2" ViewCount="154" Body="&lt;p&gt;There are many more stars in the universe than the years that the universe has been in existence, so wouldn't this mean that stars were/are forming at an astounding rate?&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2014-05-07T10:30:32.437" Title="How often do stars form?" Tags="&lt;star&gt;&lt;universe&gt;&lt;space-time&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2385" PostTypeId="2" ParentId="2384" CreationDate="2014-05-07T10:30:32.437" Score="4" Body="&lt;p&gt;By an estimated order of magnitude there are about $10^{22}$ in our observable universe (&lt;a href=&quot;http://en.wikipedia.org/wiki/Observable_universe#Extrapolation_from_number_of_stars&quot; rel=&quot;nofollow&quot;&gt;hundred billion galaxies with hundred billion stars, each, as a rough estimate&lt;/a&gt;), and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Age_of_the_universe&quot; rel=&quot;nofollow&quot;&gt;universe exits since&lt;/a&gt; (order of magnitude) $10^{10}$ years. Hence the average formation rate of stars has been about $10^{12}$ per year, meaning more than 10,000 per second, give or take a factor of 100. This calculation ignores stars which don't exist any more, but for an order-of-magnitude estimate numbers should be ok.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the star formation rate has been much higher in the younger universe than today. (&lt;a href=&quot;http://www.nasa.gov/mission_pages/herschel/news/herschel20130417.html#.U2oKMYF_uqg&quot; rel=&quot;nofollow&quot;&gt;One of the press releases&lt;/a&gt; about star formation in the early universe)&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-05-07T10:30:32.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2386" PostTypeId="2" ParentId="2383" CreationDate="2014-05-07T11:01:20.117" Score="3" Body="&lt;p&gt;Yes, after a close encounter with other planets, almost anything can happen, including a split of Mercury into smaller bodies by tidal forces, or capturing as one or more moons of one or more planets, or a sequence of captures by, and escapes from planets, conversion to a ring of all or some of the fragments, collisions with other moons, fragments falling into the Sun, others ejected from the solar system.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-05-07T11:01:20.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2387" PostTypeId="1" AcceptedAnswerId="2388" CreationDate="2014-05-07T11:08:47.933" Score="3" ViewCount="283" Body="&lt;p&gt;Why planet's orbit is not perpendicular or random ? It always seems each planet is revolving on the same geometric-plane around the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lyIL6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="760" LastActivityDate="2014-05-07T11:17:56.493" Title="Why planet's orbit is not perpendicular or random ?" Tags="&lt;star&gt;&lt;orbit&gt;&lt;planet&gt;&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2388" PostTypeId="2" ParentId="2387" CreationDate="2014-05-07T11:17:56.493" Score="3" Body="&lt;p&gt;Short answer: conservation of angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The origin of almost any planetary system is a sparse cloud. That cloud starts to contract due (typically) to a pressure wave crossing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cloud fragments as it contracts, and each fragment is what we know as a pre-star cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since almost always there is some movement in the matter in each cloud, the cloud as a whole starts to rotate, very slowly. Contraction helps because, due to angular momentum conservation law, when the cloud contracts, the rotation accelerates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soon we get a protostar with the most contracted matter, surrounded by a protoplanetary disk composed with the less contracted matter. The rotation of the whole system is in the same plane, due to conservation of angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The protostar becomes a star, and the protoplanetary disk becomes a bunch of planets. Each planet, in turn, orbits the star and rotates on itself, all in the same direction, based on which point of the protoplanetary disk started accreting mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later on, interaction among massive bodies disturbs this shiny process by changing some orbits and by having some collisions, but that's minor.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-05-07T11:17:56.493" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2389" PostTypeId="1" CreationDate="2014-05-07T15:11:16.107" Score="2" ViewCount="180" Body="&lt;p&gt;As far as I know we stick to the surface of the Earth as its gravity is pulling us and Earth rotation is in a way reducing the gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At what speed should Earth be spinning so that we all get ejected / catapulted into space (if we all line up on the equator).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sub question: at what speed of spinning would we all “stick” a few meters above the ground?&lt;/p&gt;&#xA;" OwnerUserId="1594" LastActivityDate="2014-05-07T15:11:16.107" Title="Faster Earth spinning speed to overcome gravity, possible?" Tags="&lt;gravity&gt;&lt;rotation&gt;&lt;speed&gt;" AnswerCount="0" CommentCount="8" ClosedDate="2014-05-07T18:29:32.900" ContentLicense="CC BY-SA 3.0" />
  <row Id="2391" PostTypeId="2" ParentId="2381" CreationDate="2014-05-08T03:35:11.947" Score="5" Body="&lt;p&gt;In an astronomical context, Rayleigh scattering does not take a huge place.&#xA;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Extinction_%28astronomy%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;extinction&lt;/a&gt; is mostly due to absorption and scattering, from gas and dust. Its effect is to diminish the flux from the source (that is, a change in magnitude). Please, take care this depends on the observed wavelength!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity, we can assume the extinction factor $Q_{ext}$ as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q_{ext} = Q_{scattering} + Q_{absorption}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $Q_{scattering}$ is in turn composed of many elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the OP above said, blue light generally suffers of higher extinction than other frequencies, so causing &lt;em&gt;reddened&lt;/em&gt; observed light. This is due to the cross-section dependence on the frequency: if the light has a wavelength that is comparable to the size of the grain dusts (high frequency), it is more effectively absorbed, while in the opposite case it can pass through undisturbed.&#xA;This is the reason why obscured fields are observed in radio (in our galaxy, especially) or in infrared (extragalactic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the answer to you question, &lt;strong&gt;in space&lt;/strong&gt; is: Rayleigh scattering is not the most responsible of the extinction/reddening, you have to invoke absorption/scattering by dust and gas.&#xA;For comparison, &lt;a href=&quot;http://www.astro.ncu.edu.tw/~wchen/Courses/ISM/11.Extinction.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; some explanatory numbers about the &lt;em&gt;extinction efficiency&lt;/em&gt; in different contests:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Extinction \sim\lambda^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Thomson \sim\lambda^0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Rayleigh \sim\lambda^{-4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You basically would need impressive amount of gas to account for the extinction just by Rayleigh scattering.&#xA;Instead, &lt;a href=&quot;https://en.wikipedia.org/wiki/Mie_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mie scattering&lt;/a&gt; is more important in this context. Here a figure to quickly grab the concept (&lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/atmos/blusky.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;from here&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i29zG.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here a very popular plot about extinction versus frequency &lt;a href=&quot;http://ned.ipac.caltech.edu/level5/Sept07/Li2/Li2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;from here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sqCzh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you observe from ground, you should take care of Rayleigh scattering as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other references:&#xA;&lt;a href=&quot;http://astro.berkeley.edu/~ay216/08/NOTES/Lecture05-08.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lecture 1&lt;/a&gt;&#xA;&lt;a href=&quot;http://astro.berkeley.edu/~ay216/06/NOTES/ay216_2006_07_Dust1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lecture 2&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1038" LastEditorUserId="128" LastEditDate="2014-05-09T00:36:46.297" LastActivityDate="2014-05-09T00:36:46.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2392" PostTypeId="1" AcceptedAnswerId="2395" CreationDate="2014-05-08T05:57:07.870" Score="11" ViewCount="727" Body="&lt;p&gt;I recently found &lt;a href=&quot;http://messenger-education.org/Interactives/ANIMATIONS/Day_On_Mercury/day_on_mercury.php&quot;&gt;this&lt;/a&gt; animation which shows the motion of sun as observed from mercury. It seems as if the sun stops in between, retraces a bit and then continues towards the west. What is the reason for this strange motion? Is it because of variation in the distance between mercury and the sun? Or is it because a day on Mercury is longer than Mercury's year?&lt;/p&gt;&#xA;" OwnerUserId="1156" LastActivityDate="2015-08-19T20:43:20.200" Title="Motion of the sun as observed from mercury" Tags="&lt;the-sun&gt;&lt;mercury&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2395" PostTypeId="2" ParentId="2392" CreationDate="2014-05-08T09:05:41.453" Score="7" Body="&lt;p&gt;Both effects combined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Day being longer than year would just have retrograde motion of the Sun on sky, but no direction change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Variation of the distance alone happens on Earth, and we have no such effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the combination of both factors, in the precise amount they have on Mercury, makes this effect happen.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-05-08T09:05:41.453" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2396" PostTypeId="1" CreationDate="2014-05-08T20:12:45.130" Score="1" ViewCount="63" Body="&lt;p&gt;Using the SIMBAD database for Alnitak, Saiph, and Sirius -- i.e.: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alnitak: FK5 coord. (J2000): RA = 05 40 45.527 DEC = -01 56 33.26; &#xA;Saiph; FK5 coord. (J2000): RA = 05 47 45.389 DEC = -09 40 10.58; &#xA;Sirius; FK5 coord. (J2000): RA = 06 45 08.917 DEC = -16 42 58.02; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the trigonometric relationship in degrees -- length of sides and inclusive angles to four decimal places -- between these three stars when they are used to form a triangular aterism? &lt;/p&gt;&#xA;" OwnerUserId="1601" LastActivityDate="2014-05-08T20:12:45.130" Title="Triangular Asterism Trigonometry" Tags="&lt;star&gt;&lt;distances&gt;&lt;stellar-dynamics&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2398" PostTypeId="1" AcceptedAnswerId="2410" CreationDate="2014-05-09T15:14:16.523" Score="73" ViewCount="9701" Body="&lt;p&gt;In doing research on vision, I have learned that &quot;20/20&quot; vision corresponds to a visual acuity of being able to resolve details 1 arcminute in size, that &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1111/j.1600-0420.2005.00516.x/pdf&quot; rel=&quot;noreferrer&quot;&gt;most people have around 20/15 vision&lt;/a&gt;, and that due to the limits of physiology basically nobody has vision better than 20/10 vision. This is an upper limit of resolving details about 0.5 arcminutes in size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia the moon is around &lt;a href=&quot;http://en.wikipedia.org/wiki/Angular_diameter&quot; rel=&quot;noreferrer&quot;&gt;30 arcminutes wide&lt;/a&gt; when seen by the naked eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put these together, and it seems to say that when looking at the moon with the naked eye nobody can see more details than would be visible in a 60×60 image of the moon&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/VVppV.png&quot; alt=&quot;Moon on black background, 60×60 pixels&quot;&gt;&lt;br&gt;&#xA;and that the average person can't see any more details than in a 40×40 version&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/JGAOU.png&quot; alt=&quot;Moon on black background, 40×40 pixels&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those seem so small on my monitor. Can that really be all the detail that I can see on the moon with the naked eye?&lt;/p&gt;&#xA;" OwnerUserId="1607" LastEditorUserId="9186" LastEditDate="2015-10-16T14:47:59.810" LastActivityDate="2018-02-01T13:24:17.933" Title="Is the moon only 60 pixels?" Tags="&lt;the-moon&gt;&lt;naked-eye&gt;&lt;angular-resolution&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="2399" PostTypeId="2" ParentId="2398" CreationDate="2014-05-09T17:28:41.547" Score="68" Body="&lt;p&gt;Yes and no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, it's true that the apparent size of the Moon is 30 arcmin. It's true that the visual acuity of most people is 1 arcmin. So it's true that if you take the angular size of the smallest detail you can see on the Moon, and you put a bunch of those lined up straight in a row, you could span a Moon diameter with only a few dozen of them. In that sense, you are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when you try to reproduce the situation on a computer screen, the comparison breaks down. First off, the eye doesn't see in &quot;pixels&quot;. Like most optical systems, there's a &lt;a href=&quot;http://en.wikipedia.org/wiki/Point_spread_function&quot;&gt;point-spread function&lt;/a&gt;, that takes very tiny details and smears them up to a larger spot. The resolution of the eye is not the pixel size, but the size of the bell curve coming out of the point-spread function, and that has soft edges and is round, and it's everywhere and it's not fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/b9P6p.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You assimilate the size of that larger spot with the size of a pixel on a digital screen, in your comparison. But that's not the same. The pixel grid in those thumbnails is fixed, so whatever falls between pixels is lost forever. Aliasing intervenes and creates artifacts that are not there in the original image. The dynamic range of the monitor is not the same like the dynamic range of the eye (the eye is much better). Color and brightness levels on the monitor are discrete, whereas the eye sees them as a continuum. Finally, the visual center in your brain is like a powerful computer that applies intelligent correction algorithms to the live image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The list goes on and on. The bottom line is - all these effects combine and allow you to perceive a live image that is slightly more rich than those dead, frozen thumbnails that you posted. Not a whole lot better, but a little bit better. It's not like the eye can &quot;work around&quot; limitations, but it's more like you lose too much when you shrink a large image into a tiny fixed pixel grid on a computer screen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very hard to reproduce reality on a computer screen. A much better way would be to take a 2000px by 2000px image of the Moon, put it on a big super HD monitor, and move it back to the point where the apparent size of that image is 30 arcmin. I know that doesn't sound satisfactory in the context of your original query, but it's a much better simulation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Similar problems appear whenever you try to map the resolution of any continuous optical system (like a telescope) to a fixed digital grid (like a camera).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you're using a sensor with a pixel size of 4 microns. Let's say your telescope has a linear resolution in prime focus equal to 4 microns. You might be tempted to say - great, the sensor matches the telescope, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, not really. When that happens, you actually lose a bit of resolution. The image is good, but it's a little bit softer than it really should. See below an image of the Moon I took a while ago, with a system having exactly the parameters indicated above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can tell it's a bit soft, it's not really down to the pixel. Turbulence also plays a role, but part of the problem is that linear resolution is equal to pixel size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click the image below and open in new tab; if your browser shrinks it again to fit the window, left-click the big image to expand to full size - you must do this to see the full resolution image and notice the effects I'm talking about. The fuzziness is not visible on this small version here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/d9MYA.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/d9MYA.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way around that phenomenon, as an example, is to blow up the image in the telescope with a barlow until the linear resolution in prime focus is much greater than the camera pixel size, maybe 4x bigger. You do all your processing, and then you shrink it back, if you like, and you'll get a sharper image. Combine it with stacking multiple frames, and the overall quality can get pretty close to 100% the theoretical performance of the telescope.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TLDR:&lt;/strong&gt; Continuous optical systems, and discrete grids of pixels, are very different things and cannot be easily compared.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="220" LastEditDate="2014-05-12T09:13:17.197" LastActivityDate="2014-05-12T09:13:17.197" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2400" PostTypeId="1" AcceptedAnswerId="2412" CreationDate="2014-05-10T05:33:02.643" Score="3" ViewCount="240" Body="&lt;p&gt;I have read that our sun was created from older star(s) which had exploded in a supernova.&#xA;If all the matter is travelling away from the central point of explosion, how does it coalesce back into a nebula and then form a sun?  Wouldn't all the matter just keep moving further and further apart?  Wouldn't gravity be far too weak to bring it all back together into one place?&lt;/p&gt;&#xA;" OwnerUserId="1619" LastEditorUserId="1619" LastEditDate="2014-05-10T05:47:47.580" LastActivityDate="2016-06-04T18:04:09.107" Title="Sun from SuperNova" Tags="&lt;star&gt;&lt;supernova&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2402" PostTypeId="1" CreationDate="2014-05-10T10:05:47.677" Score="3" ViewCount="1330" Body="&lt;p&gt;Heavier elements like gold, uranium etc. are formed at the end of a star's life. As the star explodes into a supernova, it gives rise to nebula which is the birthplace of new stars. But as the star has already fused lighter elements into heavier elements, where does the new hydrogen required for the formation of new stars come from? Do heavier elements breakdown during supernova or is it some other process?&lt;/p&gt;&#xA;" OwnerUserId="1156" LastActivityDate="2015-06-05T01:43:18.833" Title="Do heavier elements breakdown during supernova?" Tags="&lt;supernova&gt;&lt;nebula&gt;&lt;star-formation&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2403" PostTypeId="2" ParentId="2398" CreationDate="2014-05-10T10:19:30.177" Score="22" Body="&lt;p&gt;When you gaze at the moon &quot;live&quot;, you are not seeing a still image. You're seeing a &quot;video&quot;: your retina is gathering multiple images over time. Those pixels have to be taken into account; they amount to extra pixels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that 60x60 pixel images are taken of a scene using a tripod-mounted camera which slightly jitters. From the multiple images, a higher-resolution image could be reconstructed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you ever noticed how a sharp-looking video can appear blurry when paused or stepped frame by frame?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, another thing to remember is, that a pixel is not a unit of information; not unless you specify how many bits encode a pixel. Suppose you sample 60x60 points, but with continuous amplitude resolution, and zero noise. The 60x60 pixel image then contains infinite information (though, of course, its ability to resolve adjacent details is still limited).&lt;/p&gt;&#xA;" OwnerUserId="1623" LastActivityDate="2014-05-10T10:19:30.177" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2404" PostTypeId="2" ParentId="2402" CreationDate="2014-05-10T12:55:50.930" Score="1" Body="&lt;p&gt;Only a very small fraction of the elements in the core the supernova get converted into heavier elements. Most of the matter remains unchanged.&lt;/p&gt;&#xA;" OwnerUserId="818" LastActivityDate="2014-05-10T12:55:50.930" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2406" PostTypeId="2" ParentId="2402" CreationDate="2014-05-11T06:13:29.773" Score="3" Body="&lt;p&gt;For a star to become a nova, it needs to have a mass at least 8 times greater than our sun. For a supernova, it needs to be larger. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first stage is the hydrogen fusion into heavier elements. The energy created pushes the hydrogen outwards. If there was no fusion, the star would be a low smaller. When the hydrogen fusion in the core slows down (due to a lack of hydrogen), the star shrinks and helium fusion takes over as the main process. At this time, it is believed that the core of the star is about 25% hydrogen and 75% helium. The outer layer of hydrogen doesn't change. Eventually, the star makes iron as a fusion product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fusion to make an element heavier than iron consumes more energy than what it produces, so when the &quot;iron fusion&quot; (creation of iron) process slows down significantly, the star collapses. This collapse causes a massive pressure and temperature increase in the core, which triggers the fusion of heavier elements; although it is a net loss of energy. This burst of energy blows the star apart. The difference between a nova and a supernova is the amount of fusion that happens before the core is blasted apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since fusion only takes place in the core, and the core is less 1/100 the volume of the star, when the star goes nova (or supernova), the outer layer of the star is blown outwards.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="1322" LastEditDate="2015-06-05T01:43:18.833" LastActivityDate="2015-06-05T01:43:18.833" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="2407" PostTypeId="1" AcceptedAnswerId="2409" CreationDate="2014-05-11T07:56:25.093" Score="2" ViewCount="5628" Body="&lt;p&gt;I am trying to make sense of the following formula that claims to calculate the day of the year given today's date represented by (day/month/year):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N1 = floor(275 * month / 9)&#xA;N2 = floor((month + 9) / 12)&#xA;N3 = (1 + floor((year - 4 * floor(year / 4) + 2) / 3))&#xA;N = N1 - (N2 * N3) + day - 30&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Finally N supposedly refers to the day of the year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do the magic numbers 275,9 etc represent in the equation ?&lt;/p&gt;&#xA;" OwnerUserId="1636" LastActivityDate="2019-09-12T17:02:18.153" Title="Calculate Day of the Year for a given date" Tags="&lt;earth&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2408" PostTypeId="1" AcceptedAnswerId="2411" CreationDate="2014-05-11T11:28:11.923" Score="12" ViewCount="7221" Body="&lt;p&gt;I recently decided to set my alarm clock to wake me up when it is &quot;dark&quot; out.  In the end, I decided to set my clock to the earliest time that nautical sunrise is in my state (Illinois) and stick with that all year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While doing some research for this, I noticed something that surprised me.  Check out this disparity between astronomical sunrise to civil sunrise for the solstices and equinox (the latter two adjusted for daylight saving time):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Date/Astronomical/Civil/Disparity&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Dec 20: 0533 - 0640 (67 minutes)&lt;/li&gt;&#xA;&lt;li&gt;Mar 20: 0526 - 0630 (64 minutes)&lt;/li&gt;&#xA;&lt;li&gt;Jun 20: 0320 - 0451 (91 minutes)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To be honest, these sets of ranges surprise me for multiple reasons.  I clearly don't know what I don't know, but here are some questions I can formulate:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Why would twilight be longer in summer than winter?&lt;/strong&gt;  Before seeing this data, I had assumed that since the sun makes a more perpendicular path through the horizon in summer that twilight would be shorter in summer than in winter.  After all, in winter the sun takes a &quot;slanted&quot; path across the horizon.  Wouldn't the summer's path be more direct and therefore quicker?&lt;/li&gt;&#xA;&lt;li&gt;OK: seeing this empirical evidence I conclude that something is wrong with my premise that forms my first question.  Summer twilight is longer in summer than winter.  However, I still would have assumed that March 20 would have had a twilight length in between the two solstices.  But it's not!  &lt;strong&gt;Why does the equinox have the shortest twilight?&lt;/strong&gt; &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Appended 5/13/2014:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't want to leave my original incorrect statement in here without flagging it. As Cheekhu points out below, the sun does &lt;strong&gt;not&lt;/strong&gt; follow a more perpendicular path in summer than winter, as I had erroneously assumed and stated above.  See his post for more details.&lt;/p&gt;&#xA;" OwnerUserId="1637" LastEditorUserId="1637" LastEditDate="2014-05-13T10:59:37.200" LastActivityDate="2017-06-18T21:24:28.997" Title="Why is twilight longer in summer than winter and shortest at the equinox" Tags="&lt;the-sun&gt;&lt;light&gt;&lt;rotation&gt;&lt;planetary-atmosphere&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2409" PostTypeId="2" ParentId="2407" CreationDate="2014-05-11T12:34:39.763" Score="2" Body="&lt;p&gt;This would really be better off in Stackoverflow, or another page.  Presumably, you assumed some of the calculations rely on concepts in Astronomy, which as far as I can tell having gone through them, they do not with the exception of the concept of leap years; which I think is general enough that it doesn't really mandate being in Astronomy.  Regardless:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation is a little abstract, so probably the easiest way to understand it would be through understanding each bit of data. Let's start with N2, I'll come back to N1:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N2 = floor((month + 9) / 12)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;N2 will equal 0 if month is less than 3, and will equal 1 if it is greater.  This formula is to determine whether February has passed.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N3 = (1 + floor((year - 4 * floor(year / 4) + 2) / 3))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;N3 will be equal to 2 if the year is not a leap year, and will be equal to 1 if it is a leap year. The math here is just to determine whether the current year happens to be a leap year.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N = N1 - (N2 * N3) + day - 30&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now we take N1 (which I haven't covered yet) and we subtract it by the product of N2 and N3, which will equal 0 if we haven't reached March, or if we have it will be 1 on a leap year, and 2 on a non-leap year. We then add the number of the current day of the month, and subtract 30.  So what does N1 do?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N1 = floor(275 * month / 9)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Well, lets look at the values returned compared to days in a non-leap year:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Month     N1         Day       Diff&#xA;1.        30         31          -1&#xA;2.        61         59          +2&#xA;3.        91         90          +1&#xA;4.        122        120         +2&#xA;5.        152        151         +1&#xA;6.        183        181         +2&#xA;7.        213        212         +1&#xA;8.        244        243         +1&#xA;9.        275        273         +2&#xA;10.       305        304         +1&#xA;11.       336        334         +2&#xA;12.       366        365         +1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A bit messy, right?  Now, remember that you're subtracting 30 from the total at the end to get N, and we're adding in the current date. This means that although we multiply our current month to get N1, we're actually using this to calculate the dates from the months &lt;em&gt;prior&lt;/em&gt; to our current month! Thus if we take the value of N1, subtract it by 30, and compare it to the preceding month, the chart will come out like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Month     N1         Day       Diff&#xA;1.        31         31           0&#xA;2.        61         59          +2&#xA;3.        92         90          +2&#xA;4.        122        120         +2&#xA;5.        153        151         +2&#xA;6.        183        181         +2&#xA;7.        214        212         +2&#xA;8.        245        243         +2&#xA;9.        275        273         +2&#xA;10.       306        304         +2&#xA;11.       336        334         +2&#xA;12.       ---        365         ---&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From this, you can see that the value of N1 will equal 2 greater than the actual date for any day in which it is March or later.  This is perfect, as N2 is already a formula determining this for catching leap days. Note, these would all equal +1 on a leap year, as in another day would have been added in February. Thus coming back to the final calculation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N = N1 - (N2 * N3) + day - 30&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We take &lt;code&gt;N1&lt;/code&gt; and subtract in by the value of &lt;code&gt;(N2 * N3)&lt;/code&gt;, which will be 2 on a non-leap year if February has passed, 1 on a leap year, or 0 if we haven't passed February.  We add in the days in the current month, and we subtract 30 to get the offset in &lt;code&gt;N1&lt;/code&gt; values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should consistently give you the day of the year.  As for special significance, they're being used only for mathematical properties. Dividing a number between 1 and 12 by 9, just so happens to equal less than 1 for numbers less than 3, and greater than or equal to 1 and less than 2 all the way up to 12, which is why it is we have started March. The &lt;code&gt;275 * (month / 9)&lt;/code&gt; formula I'm not sure where they came up with. AFAIK, they have no special meaning to time or the Sun-Earth system - rather, they were chosen because they calculate well in this particular usage.&lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="1288" LastEditDate="2014-05-12T01:36:44.607" LastActivityDate="2014-05-12T01:36:44.607" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2410" PostTypeId="2" ParentId="2398" CreationDate="2014-05-11T17:54:43.027" Score="54" Body="&lt;p&gt;It doesn't seem so far-fetched to me.  Sure, you might be off by a few pixels, due to differences between the human eye and a computer monitor, but the order of magnitude seems about right &amp;mdash; the detail in your images, viewed closely, more or less matches what I see when I look at the full moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you could fairly easily test it yourself: go outside on a dark night, when the moon is full, and see if you can spot with your naked eye any details that are not visible (even under magnification) in the image scaled to match your eyesight.  I suspect you might be able to see &lt;em&gt;some&lt;/em&gt; extra detail (especially near the terminator, if the moon is not perfectly full), but not very much.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For a more objective test, we could try to look for early maps or sketches of the moon made by astronomers before the invention of the telescope, which should presumably represent the limit of what the naked human eye could resolve.  (You &lt;em&gt;needed&lt;/em&gt; to have good eyesight to be an astronomer in those days.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alas, it turns out that, while the invention of the telescope in the early 1600s brought on a veritable flood of lunar drawings, with every astronomer starting from Galileo himself rushing to look at the moon through a telescope and sketch what they saw, &lt;em&gt;very&lt;/em&gt; few astronomical (as opposed to purely artistic) drawings of the moon are known from before that period.  Apparently, while those early astronomers were busy compiling remarkably accurate star charts and tracking planetary motions with the naked eye, nobody really though it important to draw an accurate picture of the moon &amp;mdash; after all, if you wanted to know what the moon looked like, all you had to do was look at it yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps this behavior may be partly explained by the prevailing philosophical opinions at the time, which, influenced by Aristotle, held the heavens to be the realm of order and perfection, as opposed to earthly corruption and imperfection.  The clearly visible &quot;spots&quot; on the face of the moon, therefore, were mainly regarded as something of a philosophical embarrassment &amp;mdash; not something to be studied or catalogued, but merely something to be explained away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the first and last known &quot;map of the moon&quot; drawn purely based on naked-eye observations was drawn by William Gilbert (1540&amp;ndash;1603) and included in his posthumously published work &lt;em&gt;De Mundo Nostro Sublunari&lt;/em&gt;.  It is quite remarkable how little detail his map actually includes, even compared to a tiny 40 by 40 pixel image as shown above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xcVXH.gif&quot; rel=&quot;noreferrer&quot; title=&quot;William Gilbert&amp;#39;s map of the moon&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xcVXHm.gif&quot; alt=&quot;William Gilbert&amp;#39;s map of the moon&quot;&gt;&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/i8bxF.jpg&quot; alt=&quot;The moon, scaled down to 40 px radius and back up to 320 px&quot; title=&quot;The moon, scaled down to 40 px radius and back up to 320 px&quot;&gt;&lt;br&gt;&#xA;&lt;sup&gt;Left: William Gilbert's map of the moon, from &lt;a href=&quot;http://galileo.rice.edu/sci/observations/moon.html&quot; rel=&quot;noreferrer&quot;&gt;The Galileo Project&lt;/a&gt;; Right: a photograph of the full moon, scaled down to 40 pixels across and back up to 320 px.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, even the sketches of the moon published by Galileo Galilei in his famous &lt;em&gt;Sidereus Nuncius&lt;/em&gt; in 1610, notable for being based on his telescopic observations, are not much better; they show little detail except near the terminator, and the few details there are appear to be inaccurate bordering on fanciful.  They are, perhaps, better regarded as &quot;artist's impressions&quot; than as accurate astronomical depictions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Galileo%27s_sketches_of_the_moon.png&quot; rel=&quot;noreferrer&quot; title=&quot;Galileo&amp;#39;s sketches of the moon from Sidereus Nuncius (1610)&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Af0qi.png&quot; alt=&quot;Galileo&amp;#39;s sketches of the moon from Sidereus Nuncius (1610)&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sup&gt;Galileo's sketches of the moon, based on early telescopic observations, from &lt;em&gt;Sidereus Nuncius&lt;/em&gt; (1610), via Wikimedia Commons.  Few, if any, of the depicted details can be confidently matched to actual lunar features.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much more accurate drawings of the moon, also based on early telescopic observations, were produced around the same time by Thomas Harriott (1560&amp;ndash;1621), but his work remained unpublished until long after his death.  Harriott's map actually starts to approach, and in some respects exceeds, the detail level of even the 60 pixel photograph above, showing e.g. the shapes of the maria relatively accurately.  It is, however, to be noted that it is presumably based on extensive observations using a telescope, over several lunar cycles (allowing e.g. craters the be more clearly seen when they're close to the terminator):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://academic.oup.com/view-large/figure/2042835/50-1-1.27-fig003.gif&quot; rel=&quot;noreferrer&quot; title=&quot;Thomas Harriott&amp;#39;s lunar map, c. 1610-1613&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2usRH.png&quot; alt=&quot;Thomas Harriott&amp;#39;s lunar map, c. 1609&quot;&gt;&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/YOw7B.jpg&quot; alt=&quot;The moon, scaled down to 60 px radius and back up to 320 px&quot; title=&quot;The moon, scaled down to 40 px radius and back up to 320 px&quot;&gt;&lt;br&gt;&#xA;&lt;sup&gt;Left: Thomas Harriott's lunar map, undated but probably drawn c. 1610-1613, based on early telescopic observations, quoted from Chapman, A. &lt;a href=&quot;https://academic.oup.com/astrogeo/article/50/1/1.27/201448&quot; rel=&quot;noreferrer&quot;&gt;&quot;A new perceived reality: Thomas Harriot's Moon maps&quot;&lt;/a&gt;, &lt;em&gt;Astronomy &amp;amp; Geophysics&lt;/em&gt; 50(1), 2009; Right: same photograph of the full moon as above, scaled down to 60 pixels across and back up to 320 px.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on this historical digression, we may thus conclude that the 40 pixel image of the moon, as shown in the question above, indeed &lt;em&gt;does&lt;/em&gt; fairly accurately represent the level of detail visible to an unaided observer, while the 60 pixel image even matches the detail level visible to an observer using a primitive telescope from the early 1600s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sources and further reading:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Kopal, Zdeněk (1969). &lt;a href=&quot;http://adsabs.harvard.edu/full/1969Moon....1...59K&quot; rel=&quot;noreferrer&quot;&gt;&quot;The Earliest Maps of the Moon&quot;&lt;/a&gt;. &lt;em&gt;The Moon&lt;/em&gt;, Volume 1, Issue 1, pp. 59&amp;ndash;66. Available courtesy of the SAO/NASA Astrophysics Data System (ADS).&lt;/li&gt;&#xA;&lt;li&gt;Van Helden, Al (1995). &lt;a href=&quot;http://galileo.rice.edu/sci/observations/moon.html&quot; rel=&quot;noreferrer&quot;&gt;&quot;The Moon&quot;&lt;/a&gt;. &lt;em&gt;The Galileo Project&lt;/em&gt; (web site).&lt;/li&gt;&#xA;&lt;li&gt;Wikipedia articles on the &lt;a href=&quot;//en.wikipedia.org/wiki/Moon&quot; rel=&quot;noreferrer&quot;&gt;Moon&lt;/a&gt; and &lt;a href=&quot;//en.wikipedia.org/wiki/Selenography&quot; rel=&quot;noreferrer&quot;&gt;Selenography&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="464" LastEditorUserId="464" LastEditDate="2018-02-01T13:24:17.933" LastActivityDate="2018-02-01T13:24:17.933" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2411" PostTypeId="2" ParentId="2408" CreationDate="2014-05-12T02:10:31.783" Score="8" Body="&lt;p&gt;See this diagram,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4FlyA.png&quot; alt=&quot;Celestial Sphere&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the &lt;em&gt;slant&lt;/em&gt; appears to be the same for June and December Solstices, doesn't it? First thing, it is a wrong conception that Sun travels more &quot;perpendicular&quot; to horizon in the summers. It just reaches a &lt;em&gt;higher&lt;/em&gt; point as you can see, not that the slant is more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does the disparity come in?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Civil Sunrise begins when the Sun is 6° below the horizon, Nautical Sunrise begins when it is 12° below the horizon and 18° for Astronomical Sunrise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the Civil/Astronomical Disparity is due to the extra 12° of altitude that the Sun has to cover. Now, the diagram I have posted is better to explain the sunset, so let it be that way. It is just the same for our purpose. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motion of the sun is uniform along the circles that you see. But these angles which decide our sunrise are measured perpendicular to the horizon, from the zenith. Are you aware of what happens when you project circular motion along a straight line?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LrtC7.jpg&quot; alt=&quot;circular motion projection on  line&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's right. You get a sinusoidal curve. The special thing about this curve is that it changes fast at some points and slow at others. &#xA;Transit is the position when the Sun is highest in the sky. In the &lt;em&gt;curve for equinox&lt;/em&gt; (as in Fig 1), if you assign 0° to that point, the Sun is at 90° at sunset and -90° at sunrise. Both positions give you the peak of a sine curve and the change is lowest at that point. That might explain why you get the least difference at Equinox. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you consider the curves for June/December solstice (as in Fig 1), the position for sunset/sunrise is a little off 90°, hence rate of change is faster there as compared to the equinox. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That should explain why the disparity is larger at solstices as compared to equinox. But what about the difference in June and Decemeber solstice?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, my dear, you are entering dangerous territories. Let me remind you that the line/plane along which we are making our degree measurements is perpendicular to horizon. But the plane of circular motion is still at an angle to it. So, the sinusoidal curve has to go through another set of projections before you get the exact thing. &#xA;On Equator, the disparity between the solstices vanish, yes, as the aforementioned two planes align.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I was hoping to give you a qualitative description just so you get the gist of it. Here are the equations if you need a &lt;em&gt;mathematical&lt;/em&gt; idea&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos a = \sin \delta \sin \phi + \cos \delta \cos \phi \cos H$$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a$ is (90+6)° for civil sunrise/set, and (90+18)° for astronomical sunrise/set,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta$ is declination of Sun on that date. Google it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi$ is latitude of the place&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the disparity, you need to find $H$ from the formula for two values of $a$ and get their difference. Remember to divide by 15 to convert from degrees to hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a graph showing disparity along declination(-23.5° to +23.5°)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WCXId.png&quot; alt=&quot;disparity curve&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="63" LastEditDate="2014-05-12T02:21:42.690" LastActivityDate="2014-05-12T02:21:42.690" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2412" PostTypeId="2" ParentId="2400" CreationDate="2014-05-12T06:14:26.037" Score="4" Body="&lt;p&gt;Actually yes, gravity is too weak to do the job by itself.&#xA;But as you mention, another force is acting on the system too, in a very strong way: it is the pressure that pushes outwards. In fact, the energy released by the explosion of the supernova, helps to compress the layers of released material (gas and clouds). You have shocks between the expelled gas (fast) and the ISM (slooow). Also, you need cold gas to produce a collapse event, and the ejected material from an explosion is not cold at all!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you need an efficient mechanism to induce collapse of the gas cloud, and this mechanism is furnished by compression instabilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You start from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Virial_theorem&quot; rel=&quot;nofollow&quot;&gt;Virial theorem&lt;/a&gt;: $2K + U = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with K kinetic energy and U potential energy. What you want here, for your collapse is that the gravitational potential energy is larger than the kinetic energy (otherwise particles energy will overcome the gravity and avoid collapse - not exactly true, but you can think as if high energy particles pushes outward).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can rewrite the Virial theorem as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3 N K T = \frac{3}{5}\frac{G M^2}{R}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where K boltzmann constant, N atoms number, M mass of the cloud, R its size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case (collapse case) the left term must be less than the right one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can transform $N = M/m$, with m mass of the single particles, and $R = \frac{3 M}{4\pi\rho}$, with $\rho$ cloud density (assumed constant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Than you can calculate the Jean mass as: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M_J = (\frac{5KT}{Gm})^{3/2}(\frac{3}{4\pi\rho})^{1/2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the limit after which your cloud can collapse.&#xA;Look at the dependence on the temperature, and the inverse dependence on the density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to give an idea of the typical values encountered, we can still rewrite the critical mass as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M_J = 2M_{sun}(\frac{n}{10^5 cm^{-3}})^{-1/2}(\frac{T}{10 K})^{3/2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://burro.astr.cwru.edu/Academics/Astr221/LifeCycle/jeans.html&quot; rel=&quot;nofollow&quot;&gt;this lesson&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: a similar mechanism is triggered by the rotating arms of spiral galaxies.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastEditorUserId="9301" LastEditDate="2016-06-04T18:04:09.107" LastActivityDate="2016-06-04T18:04:09.107" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2414" PostTypeId="1" AcceptedAnswerId="2415" CreationDate="2014-05-12T20:03:29.267" Score="2" ViewCount="2319" Body="&lt;p&gt;More exactly what would to happen to us/Earth in general if any of the gas giants is removed by magic from the solar system ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How long would it take after such change before we notice it ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How major and would life resist to it ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance if Saturn or Jupiter is removed, what would happen ?&lt;/p&gt;&#xA;" OwnerUserId="1630" LastEditorUserId="1630" LastEditDate="2014-05-12T23:57:34.130" LastActivityDate="2014-05-13T01:33:30.053" Title="What would happen if a planet is removed from the solar system?" Tags="&lt;solar-system&gt;&lt;planet&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ClosedDate="2014-05-13T13:11:46.483" ContentLicense="CC BY-SA 3.0" />
  <row Id="2415" PostTypeId="2" ParentId="2414" CreationDate="2014-05-12T22:28:04.873" Score="6" Body="&lt;p&gt;This question isn't really very clear or practical. You need to add quite a bit of information, eg which planet? &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If Sedna vanished, it would be detectable by instruments, but we would not notice. &lt;/li&gt;&#xA;&lt;li&gt;Jupiter or Mars might make a bigger difference. We could spot it straight away by the naked eye. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of course Earth life would be fine at first (unless it was Earth that was removed ...) but the orbits of asteroids and other bodies may be perturbed, which could lead to an influx of asteroid strikes - potentially - in the longer term.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2014-05-12T22:28:04.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2416" PostTypeId="1" AcceptedAnswerId="2419" CreationDate="2014-05-13T00:18:49.360" Score="9" ViewCount="2601" Body="&lt;p&gt;As the title suggests, I'm in need of a set of data of our solar system. Similar to this &lt;a href=&quot;http://bima.astro.umd.edu/nemo/archive/#iau25&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://bima.astro.umd.edu/nemo/archive/#iau25&lt;/a&gt;, but more specifically, the Dubinsky Milky Way-Andromeda data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm creating a n-body simulation for school and can't seem the find any initial particle conditions that I could use to simulate our solar system in the software I'm developing. I need initial positions, velocity and mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea where I could find this?&lt;/p&gt;&#xA;" OwnerUserId="1657" LastEditorUserId="14094" LastEditDate="2020-06-12T22:50:01.657" LastActivityDate="2020-06-12T22:50:01.657" Title="Where can I find a set of data of the initial conditions of our solar system?" Tags="&lt;solar-system&gt;&lt;software&gt;&lt;n-body-simulations&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2417" PostTypeId="2" ParentId="2416" CreationDate="2014-05-13T01:26:17.920" Score="0" Body="&lt;p&gt;What language are you writing this simulation in? Is it 2D or 3D? Do you only need positions and velocities for our solar system planets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done this exact thing (simulated the solar system in Fortran) and I didn't need exact initial positions, all I needed were initial radii (in AU from the Sun/centre of mass) and initial velocities. Use a random number generator to distribute the planets at random places along their orbits. In Fortran, this looked like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;CALL RANDOM_NUMBER(randNum)&#xA;degrees = 2*3.141592653&#xA;theta(1:15) = degrees*randNum(1:15)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And there I have an array of 15 random radial positions. You can obtain initial velocities of our solar system's planets from any reputable resource.&lt;/p&gt;&#xA;" OwnerUserId="1658" LastActivityDate="2014-05-13T01:26:17.920" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2418" PostTypeId="2" ParentId="2414" CreationDate="2014-05-13T01:33:30.053" Score="3" Body="&lt;p&gt;Short answer: the solar system would be less stable in the short run due to the massive gravitational influence of Jupiter vanishing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer: The current layout of the solar system is actually remarkably well balanced; without outside interference, it'll stay stable for a good long while. Anything that changes the current equilibrium will force the system to adjust until it finds a new equilibrium. Removing a huge mass will certainly affect this in &lt;em&gt;some&lt;/em&gt; way, but it's difficult to say exactly how without running a few hundred simulations. Jupiter's gravitational influence kept the asteroid belt from coalescing into a planet, so that might happen, for one. &lt;/p&gt;&#xA;" OwnerUserId="1658" LastActivityDate="2014-05-13T01:33:30.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2419" PostTypeId="2" ParentId="2416" CreationDate="2014-05-13T07:20:40.933" Score="4" Body="&lt;p&gt;&lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi#top&quot; rel=&quot;nofollow noreferrer&quot;&gt;Horizons Ephemeris generator&lt;/a&gt; can give you a planet's position and velocity vectors at a specified time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one set of possible options:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BFbfc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clicking &lt;strong&gt;Generate Ephemeris&lt;/strong&gt; on that page will give you position and velocity vectors:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/h6kAv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above the position and velocity vectors are the Julian date as well as the more conventional date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia can give the masses of the sun and planets.&lt;/p&gt;&#xA;" OwnerUserId="1232" LastActivityDate="2014-05-13T07:20:40.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2420" PostTypeId="2" ParentId="2398" CreationDate="2014-05-13T09:35:38.573" Score="4" Body="&lt;p&gt;After all these astronomic answers, I will add a computer one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pixels are not the same on all monitors.&#xA;Take a 1990's monitor and take the latest smartphone screen, the 60 pixels won't be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did you calculate the pixel size according to the vision accuracy ?&lt;/p&gt;&#xA;" OwnerUserId="204" LastActivityDate="2014-05-13T09:35:38.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2421" PostTypeId="2" ParentId="2275" CreationDate="2014-05-13T09:47:29.190" Score="1" Body="&lt;p&gt;I asked a question on the same topic few months ago : &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1068/could-any-known-living-organisms-on-earth-survive-on-mars&quot;&gt;Could any known, living organisms on Earth survive on Mars?&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer from &lt;code&gt;dotancohen&lt;/code&gt; pointed the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tardigrade&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tardigrade&lt;/a&gt; that can live in extreme conditions.&#xA;The other life forms I mentioned may also be resistant to extreme conditions.&lt;/p&gt;&#xA;" OwnerUserId="204" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-05-13T09:47:29.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2422" PostTypeId="1" AcceptedAnswerId="2423" CreationDate="2014-05-13T13:09:36.833" Score="5" ViewCount="7569" Body="&lt;p&gt;I was pondering our insignificance, when I wondered - how much smaller is our planet then the (observable) universe? And being as I don't know how to do the math, I'm asking it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how many of our planet (in space it occupies - i.e. ignoring space between the space between the spheres) can fit inside the known/observable universe?&lt;/p&gt;&#xA;" OwnerUserId="1418" LastEditorUserId="2365" LastEditDate="2017-05-09T16:08:28.150" LastActivityDate="2021-01-21T13:46:19.470" Title="How many earths fit in the observable universe?" Tags="&lt;earth&gt;&lt;universe&gt;&lt;size&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2423" PostTypeId="2" ParentId="2422" CreationDate="2014-05-13T16:09:38.560" Score="16" Body="&lt;p&gt;Without checking the numbers in detail, according to Wikipedia, the volume of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Observable_universe&quot;&gt;observable universe&lt;/a&gt; is about $3.5\cdot 10^{80} \mbox{ m}^3$, and the volume of &lt;a href=&quot;http://en.wikipedia.org/wiki/Earth&quot;&gt;Earth&lt;/a&gt; is about&#xA;$1.08321\cdot 10^{21} \mbox{ m}^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By dividing the two volumes we get a factor of $3.2\cdot 10^{59}$, or written as decimal number: The observable comoving volume of the universe is about&#xA;320,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000-times the volume of Earth.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-05-13T16:09:38.560" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2424" PostTypeId="1" CreationDate="2014-05-13T19:09:43.873" Score="6" ViewCount="251" Body="&lt;p&gt;How does inflation explain why the curvature in the early universe was flattened out and now results in a flat universe?&lt;/p&gt;&#xA;" OwnerUserId="1542" LastActivityDate="2014-05-15T04:08:08.423" Title="Flatness of the universe?" Tags="&lt;early-universe&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2425" PostTypeId="2" ParentId="2424" CreationDate="2014-05-13T19:57:04.190" Score="8" Body="&lt;p&gt;This is really simple once you get it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual analogy here is to look at the surface of a balloon (which is a surface in 3D space, rather than 3D space itself, but the analogy still works)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A partially inflated balloon has tight curvature, but as it is inflated, the curve decreases - the surface at any point becomes 'flatter'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From an article on this topic on &lt;a href=&quot;http://scienceblogs.com/startswithabang/2009/07/31/the-size-of-the-universe-a-har/&quot; rel=&quot;nofollow noreferrer&quot;&gt;scienceblogs.com:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mp40Y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe has expanded &lt;strong&gt;a lot&lt;/strong&gt; since the big bang - any curves, ripples or inconsitencies have largely been flattened, so it is difficult to detect them now. &lt;a href=&quot;http://www.space.com/25088-gravitational-waves.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article&lt;/a&gt; on space.com quotes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Harvard-Smithsonian study spotted gravitational waves as ripples in space-time possible left over from the rapid expansion of the universe (called inflation) right after the Big Bang nearly 13.8 billion years ago.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="43" LastEditorUserId="43" LastEditDate="2014-05-13T21:14:01.990" LastActivityDate="2014-05-13T21:14:01.990" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2426" PostTypeId="2" ParentId="2275" CreationDate="2014-05-13T20:27:39.677" Score="1" Body="&lt;p&gt;Some of our gas giant's icey moons are thought to have liquid water interiors. Tidal flexing is thought to be a source of heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth's smoker ecosystem demonstrate life can thrive sans sunlight. This &lt;a href=&quot;http://science1.nasa.gov/science-news/science-at-nasa/2001/ast13apr_1/&quot; rel=&quot;nofollow&quot;&gt;Nasa article&lt;/a&gt; talks about Smoker colonies and the possibility of life beneath the surface of icey moons like &lt;a href=&quot;http://en.wikipedia.org/wiki/Europa_%28moon%29#Potential_for_extraterrestrial_life&quot; rel=&quot;nofollow&quot;&gt;Europa&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1232" LastActivityDate="2014-05-13T20:27:39.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2427" PostTypeId="1" CreationDate="2014-05-13T21:57:15.107" Score="17" ViewCount="3107" Body="&lt;p&gt;&lt;em&gt;Ignoring expansion of the universe, entropy, decaying orbits, and interference from any bodies colliding with or otherwise interfering with their orbits&lt;/em&gt;, will the eight planets known planets in our solar system ever align?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the &quot;period&quot; of the planets; &lt;em&gt;how often&lt;/em&gt; would they align perfectly? And based on their current positions, how far off into the future is their next theoretical alignment?&lt;/p&gt;&#xA;" OwnerUserId="660" LastActivityDate="2020-12-24T01:44:28.727" Title="When will all eight planets in our solar system align?" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;planet&gt;" AnswerCount="5" CommentCount="10" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2428" PostTypeId="2" ParentId="2427" CreationDate="2014-05-14T00:24:45.157" Score="9" Body="&lt;h1&gt;This is a low accuracy - yet simple - answer&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;It allows you to calculate only &lt;em&gt;radial&lt;/em&gt; alignment configuration of the planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you would like an approximation, let's say, you approximate the position of the planets as hands in a clock, you could work the math out by something like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $\theta_i$ is the initial angle for planet $i$ at time $t_0$ - measured from an arbitrary but fixed position, and $l_i$ is the length of the year - in days - for planet $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then it resumes to solving this system of equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x \equiv \theta_i \left( \ mod\ l_i\right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here you would then simply apply the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chinese_remainder_theorem&quot;&gt;Chinese Remainder Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding the minimum x, will give you the angle that the planet that at $t_0$ had angle $\theta_i = 0$ would have travelled until an &lt;em&gt;alignment&lt;/em&gt; configuration was reached. Asuming you choose Earth as the mentioned planet, then divide that angle by a complete revolution ($360^{o}$) and you will get the number of years for that configuration to be reached - from the $t_0$ configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The different $\theta_i$ in degrees for all the planets on 01 Jan 2014 - you can use this as your $t_0$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;Mercury &amp;amp;\quad 285.55&#xA;\\Venus &amp;amp;\quad 94.13&#xA;\\Earth &amp;amp;\quad 100.46&#xA;\\Mars &amp;amp;\quad 155.60&#xA;\\Jupiter &amp;amp;\quad 104.92&#xA;\\Saturn &amp;amp;\quad 226.71&#xA;\\Uranus &amp;amp;\quad 11.93&#xA;\\Neptune &amp;amp;\quad 334.90&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://omniweb.gsfc.nasa.gov/coho/helios/planet.html&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The different $l_i$ in days for all the planets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;Mercury &amp;amp;\quad 88&#xA;\\Venus &amp;amp;\quad 224.7&#xA;\\Earth &amp;amp;\quad 365.26&#xA;\\Mars &amp;amp;\quad 687&#xA;\\Jupiter &amp;amp;\quad 4332.6	&#xA;\\Saturn &amp;amp;\quad 10759.2	&#xA;\\Uranus &amp;amp;\quad 30685.4	&#xA;\\Neptune &amp;amp;\quad 60189&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Finally&lt;/strong&gt; under an integer values aproximation and using &lt;a href=&quot;http://davidwees.com/chineseremaindertheorem/&quot;&gt;this&lt;/a&gt; online solver for the system of equations the answer is $x = 4.0384877779832565 \times 10^{26}$ which divided by $360^{o}$ gives you roughly $$1.1218 \times 10^{24} \quad\text{years}$$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Edit 1&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Just found &lt;a href=&quot;http://www.solarsystemscope.com/&quot;&gt;this site&lt;/a&gt; you may like to play around with. It's an interactive flash application with the accurate position of the planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know that the all the information can be obtained from &lt;a href=&quot;http://ssd.jpl.nasa.gov/&quot;&gt;this NASA page&lt;/a&gt; and that is as accurate you can get, but it is just incomprehensible for me now. I will try to revise it later when I find time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also &lt;a href=&quot;http://dzucconi.github.io/calendrical/docs/Jean-Meeus_Astronomical_Algorithms.pdf&quot;&gt;this book&lt;/a&gt; by Jean Meeus called Astronomical Algorithms covers all the fundamental euqations and formulas - it has nothing to do with programming algorithms though.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Edit 2&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Seeing that you are a programmer, it could be worth for you checking out the NASA site I mentioned above, the data for all the planets can even be accessed via $\tt{telnet}$.&#xA;Or &lt;a href=&quot;http://sourceforge.net/projects/astroalgorithms/&quot;&gt;this Sourceforge site&lt;/a&gt; where they have implementations for many of the equations described in the book also mentioned above.&lt;/p&gt;&#xA;" OwnerUserId="1469" LastEditorUserId="1469" LastEditDate="2014-05-14T05:02:07.060" LastActivityDate="2014-05-14T05:02:07.060" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2430" PostTypeId="2" ParentId="2275" CreationDate="2014-05-14T04:58:43.043" Score="1" Body="&lt;p&gt;While such locations might be &lt;em&gt;habitable&lt;/em&gt; to some (of Earth's) lifeforms, for life to actually exist there it must also be an &lt;em&gt;originable&lt;/em&gt; location for life (or somehow had life transported to it). The origin of life on Earth is a big unknown. For sure, Earth was a very different place when it happened and it could've happened underground.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2014-05-14T04:58:43.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2431" PostTypeId="2" ParentId="1368" CreationDate="2014-05-14T07:07:22.743" Score="-2" Body="&lt;p&gt;I propose the following model:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe consists already of a giant central black hole attracting our milky way and all other galaxies. The observed red shift can be explained by the 1/r² law of gravitation: galaxies that are nearer than our one to the central black hole have higher speeds towards it than we have. So we see them moving away from us. Galaxies that are further away from the central black hole than our one don't move so fast towards it as we do. So looking at them we see them escaping from us as well. The black hole and our distance to it are so big, that the field gradient is quite low, so we experience no tidal forces. Adjusting Newton's 1/r² law by Einstein's general relativity field equations does not make a big difference to this model. The general pattern of galaxy motion relative to each other remains the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can these claims be verified? Here is the principle algorithm. It can be implemented and executed even on a PC:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take all recorded quasar spectra from the known quasar catalogs. Normalize their spectra to zero red shift. Compare each spectral signature which each other. Try reasonable compensation for differences according to dust or other noise adding effects. If you find two identical or similar ones, check the respective quasar's positions. If their angular positions differ by several degrees, you have some evidence. Because you found the same quasar twice: Once seen in direct (curved) line, and once its light wrapped around by the central black hole's gravitational field.&#xA;Twin quasars are already known but turned out to be the result of &quot;small&quot; (means: a massive galaxy in our line of sight to them) gravitational lens effects. In these cases their seeming positions differed by some arc seconds or minutes. A number of twin quasars with significant (several degrees) angular position differences or even at opposite sites of the universe would be a proof for the above model&lt;/p&gt;&#xA;" OwnerUserId="1668" LastEditorUserId="1668" LastEditDate="2014-05-14T14:47:11.777" LastActivityDate="2014-05-14T14:47:11.777" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2432" PostTypeId="2" ParentId="2275" CreationDate="2014-05-14T08:24:50.650" Score="1" Body="&lt;p&gt;And define &quot;life&quot;. It's conceivable that lifeforms exist that are for example silicon based, don't need oxygen, and can survive in hard vacuum, even directly use gamma and X-rays the way plants on earth photosynthesise visible light.&lt;br/&gt;&#xA;Such could well survive there, even thrive. But would we even recognise them as alive?&lt;/p&gt;&#xA;" OwnerUserId="534" LastActivityDate="2014-05-14T08:24:50.650" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2433" PostTypeId="2" ParentId="254" CreationDate="2014-05-14T09:19:54.323" Score="15" Body="&lt;p&gt;I had a chat with a European PhD student who plans to make an attempt to find Oort cloud objects in data from the Gaia space telescope. This could be possible thanks to microlensing events when an Oort object transits (near) a background star and relativistically magnifies the star's light for a moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best case is that in a few years we will have a map of a statistically useful number of Oort cloud objects. Enough to claim that we have &quot;seen&quot; it.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2014-05-14T09:19:54.323" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2434" PostTypeId="1" AcceptedAnswerId="2440" CreationDate="2014-05-14T11:54:27.343" Score="5" ViewCount="238" Body="&lt;p&gt;How many black holes exist and what is their distrubtion over mass?&#xA;There is a &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_black_holes&quot; rel=&quot;noreferrer&quot;&gt;list of black holes&lt;/a&gt;, but I'm interested in an estimate of all black holes in the visible universe.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2014-05-14T14:12:45.583" Title="How many black holes exist?" Tags="&lt;black-hole&gt;&lt;supermassive-black-hole&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2435" PostTypeId="2" ParentId="2424" CreationDate="2014-05-14T12:18:26.670" Score="1" Body="&lt;p&gt;Curvature of spacetime is predicted by General Relativity and is a function of the density of the universe. Because inflation rapidly expands the volume of the universe it also rapidly decreases the density, so flattening space time.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2014-05-14T12:18:26.670" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2436" PostTypeId="1" AcceptedAnswerId="2437" CreationDate="2014-05-14T12:43:24.523" Score="0" ViewCount="944" Body="&lt;p&gt;As we know Earth rotates around Sun and Sun around Milky Way but then Sun must have some velocity/angular momentum (I don't know that much physics terms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as Earth revolves around Sun, it must adjust its orbit as Sun also revolves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it so? Or am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="1669" LastEditorUserId="204" LastEditDate="2014-05-14T20:12:22.193" LastActivityDate="2014-05-14T20:12:22.193" Title="Does Earth revolves around Milky Way?" Tags="&lt;orbit&gt;&lt;earth&gt;&lt;the-sun&gt;&lt;milky-way&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2437" PostTypeId="2" ParentId="2436" CreationDate="2014-05-14T13:13:22.083" Score="4" Body="&lt;p&gt;Earth is bound gravitationally to the Sun. So it is orbiting together with the Sun around the center of the Milky Way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This takes one &lt;a href=&quot;http://en.wikipedia.org/wiki/Galactic_year&quot; rel=&quot;nofollow&quot;&gt;galactic year&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence your presumption is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seen from a non-rotating system relative to the Milky Way barycenter, Earth's orbit isn't an ellipse, but ressembles a distorted &lt;a href=&quot;http://en.wikipedia.org/wiki/Helix&quot; rel=&quot;nofollow&quot;&gt;helical&lt;/a&gt; trajectory. Distorted, since the orbit around the Sun isn't perpendicular to the orbit around the center of the Milky Way.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-05-14T13:13:22.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2438" PostTypeId="2" ParentId="2436" CreationDate="2014-05-14T13:19:49.480" Score="3" Body="&lt;p&gt;It is a composite movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth does not need to adjust anything: Earth's reference frame for orbiting is Sun, almost alone. You can take into account Jupiter, Saturn, Moon and that's pretty all. If you want to do things really really precise you use General Relativity and take into account Venus, Mars, Mercury, Uranus and Neptune. If you want even further precision you put into Dwarf planets and asteroids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only once you have all of that into account, you need to think &quot;hey, there are other gravity forces out there&quot;, because Solar System's movement is almost rectilinear, 'cos the enormous radius of the orbit. So it affects almost exactly the same to Sun and Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In resume, you can simply think &quot;Earth orbits Sun, and idependently of that, the System moves towards the Apex&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About angular momentum, yes, there is a good amount of angular momentum in those movements. And also velocity: the sun moves through Milky Way at 72,000 kilometers per hour.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-05-14T13:19:49.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2439" PostTypeId="1" CreationDate="2014-05-14T13:44:14.603" Score="4" ViewCount="228" Body="&lt;p&gt;When matter crosses the event horizon, it is easy to imagine that matter is torn apart into their individual components by tidal wave effects. Atoms would be ripped apart eventually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At one time, the particles will fall so deep into the gravity well, that a single particle would be touching both sides of the gravity well at the same time. At that moment, I choose to assume that the particle particle core may be spaghettified, but externally it will appear to be shrinking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing that spaghettification is dependent on the size of the object, as well as the difference in gravity on both sides, a shrinking particle may eventually become so small that the steepness of the gravity well is no longer a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The particle continues to shrink forever. Over time, gravity between all the particles that have shrunk, will again start to take effect - and I imagine that that is what happened in the big bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the cosmic inflation, could just as well be particle shrinking, as a result of all particles in our universe being within the same black superhole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking this idea, we could look at our own universe and I guess we would expect to see certain effects such as accelerating cosmic expansion, perhaps a cosmic background radiation and we would expect &lt;em&gt;our&lt;/em&gt; universe to end as a cold, dead place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a known theory or idea, and where can I find papers on it?&lt;/p&gt;&#xA;" OwnerUserId="749" LastEditorUserId="749" LastEditDate="2014-05-15T10:20:11.150" LastActivityDate="2017-04-25T17:31:01.380" Title="A black &quot;superhole&quot; possibility?" Tags="&lt;cosmological-inflation&gt;&lt;supermassive-black-hole&gt;&lt;cosmological-horizon&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2440" PostTypeId="2" ParentId="2434" CreationDate="2014-05-14T14:12:45.583" Score="8" Body="&lt;p&gt;There exists probably about one &lt;a href=&quot;http://en.wikipedia.org/wiki/Supermassive_black_hole&quot;&gt;supermassive black hole&lt;/a&gt; per galaxy, hence about 100 billion ($10^{11}$) in the observable universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take 1% of the stars ending as a black hole as a very rough order of magnitude estimate, we'll get about 1 billion stellar black holes per galaxy, hence about $10^{20}$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_black_hole&quot;&gt;stellar black holes&lt;/a&gt; in the observable universe, give or take one order of magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tolman%E2%80%93Oppenheimer%E2%80%93Volkoff_limit&quot;&gt;Stars of more than about 15 stellar masses&lt;/a&gt; (&lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_classification#Class_O&quot;&gt;O class stars&lt;/a&gt;) will probably end up as black holes. &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_classification#Harvard_spectral_classification&quot;&gt;About 0.00003% of the stars&lt;/a&gt; are O-class. The &lt;a href=&quot;http://www.asc-csa.gc.ca/eng/educators/resources/astronomy/module2/calculator.asp&quot;&gt;lifespan&lt;/a&gt; of a star with 15 solar masses is about 11.5 million years. So, we need to multiply the 0.00003% at least by a factor 1000, since the age of the universe is more than 1000-times the life of an O-class star, resulting in 0.03% of all stars over time should have been O-class stars.&#xA;(Most stars are red dwarfs, and have a longer life-span than the current age of the universe.) Star formation, especially of heavy stars, in the early universe has probably been &lt;a href=&quot;http://www.caltech.edu/content/bursts-star-formation-early-universe&quot;&gt;higher than today&lt;/a&gt;.&#xA;Therefore I'm estimating the number of stellar black holes as about 1% the number of all stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Intermediate-mass_black_hole&quot;&gt;Intermediate-mass black holes&lt;/a&gt; as presumed centers of &lt;a href=&quot;http://en.wikipedia.org/wiki/Globular_cluster&quot;&gt;globular clusters&lt;/a&gt; are harder to estimate, since there isn't much observational evidence. If the hypotheses turns out to be correct, and we take 1000 globular clusters per galaxy, we get about $10^{14}$ IMBHs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no observational evidence for the existence of small black holes (smaller than a few solar masses).&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-05-14T14:12:45.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2441" PostTypeId="1" AcceptedAnswerId="14239" CreationDate="2014-05-14T17:27:41.100" Score="43" ViewCount="7904" Body="&lt;p&gt;My understanding is that time slows and approaches stopping when approaching the event horizon of a black hole.  I have seen this explained several places, including a brief explanation in the last paragraph under: &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_hole#General_relativity&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Black_hole#General_relativity&lt;/a&gt;, quoted below:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Oppenheimer and his co-authors interpreted the singularity at the boundary of the Schwarzschild radius as indicating that this was the boundary of a bubble in which time stopped. This is a valid point of view for external observers, but not for infalling observers. Because of this property, the collapsed stars were called &quot;frozen stars&quot;,[17] because an outside observer  would see the surface of the star frozen in time at the instant where its collapse takes it inside the Schwarzschild radius.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does this mean then that no matter actually falls into a black hole (except possibly what was there at its formation)?  Would this also mean matter is accumulating just outside its event horizon?  As I understand it, this would be the perspective from outside the black hole.  If this is the case, I wonder if we would observe a tremendous amount of matter surrounding the event horizon, but it would be extremely red shifted?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I noticed an answer to a different question, especially the end portion, provides some insight here as well:&#xA;&lt;a href=&quot;https://astronomy.stackexchange.com/a/1009/1386&quot;&gt;https://astronomy.stackexchange.com/a/1009/1386&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These YouTube videos someone put together explain the concept very well, and seem to indicate this idea is gaining traction!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=yZvgeAbrjgc&amp;amp;list=PL57CC037B74307650&amp;amp;index=118&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=yZvgeAbrjgc&amp;amp;list=PL57CC037B74307650&amp;amp;index=118&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=b1s7omTe1HI&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=b1s7omTe1HI&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This new YouTube video describes this idea very well, and describes it as the way black holes work!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/mquEWFutlbs&quot; rel=&quot;noreferrer&quot;&gt;https://youtu.be/mquEWFutlbs&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="1386" LastEditDate="2019-02-06T12:24:19.767" LastActivityDate="2020-11-26T07:22:35.783" Title="Does matter accumulate just outside the event horizon of a black hole?" Tags="&lt;black-hole&gt;&lt;space-time&gt;&lt;redshift&gt;&lt;matter&gt;&lt;horizon&gt;" AnswerCount="8" CommentCount="3" FavoriteCount="17" ContentLicense="CC BY-SA 4.0" />
  <row Id="2442" PostTypeId="1" AcceptedAnswerId="2444" CreationDate="2014-05-14T18:11:36.457" Score="1" ViewCount="173" Body="&lt;p&gt;I am not sure if this is the right place to post at, but I tried to find the best place to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my point, I wonder if it is posssible, to determine where north is, &lt;strong&gt;just by one shadow&lt;/strong&gt;, that means no Sun, no time, just a shadow in a picture or video...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you get my point? Like this picture, can I tell where north is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like this image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OczGZ.jpg&quot; alt=&quot;Shadow&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1671" LastEditorUserId="1671" LastEditDate="2014-05-15T17:32:13.727" LastActivityDate="2014-05-15T17:32:13.727" Title="Determine north just by one shadow" Tags="&lt;the-sun&gt;&lt;light&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2443" PostTypeId="2" ParentId="2442" CreationDate="2014-05-14T18:49:16.100" Score="1" Body="&lt;p&gt;With only the shadow in the picture, I do not think it would be possible.  Consider a stick on the equator with a shadow. In the winter, at noon, since the sun would be &quot;south&quot; of the equator, it would point the opposite direction as in the summer (when the sun is &quot;north&quot; of the equator).  If we assume it is summer, the calculations would show north is the opposite direction of if we assume it is winter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the &quot;fuzziness&quot; of the shadow could possibly be used to establish a time (at least help determine the season) and aid in calculations.  When we are closer to the sun, the shadow should be more fuzzy.  The length of the shadow could be used to determine the time if the location were known.  Still, I believe there are too many unknowns to determine the direction of North.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2014-05-14T18:49:16.100" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2444" PostTypeId="2" ParentId="2442" CreationDate="2014-05-14T20:29:06.160" Score="2" Body="&lt;p&gt;With only one shadow in a picture, no, you can not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to get North's position from a shadow you need to know where you are and which solar time and date is it there. These are two extra data and you can not disentangle them, even if the picture is perfect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. suppose a picture in which you see exact measurements of shadow and people, and they are both exactly of the same size. This means Sun is 45º above Horizon. But that may be local noon at Bourdeaux on March 20th (thus shadow pointing North), or local noon at Christchurch on same date (thus shadow pointing South), or local 3PM solar on Quito on March 20th again (thus shadow pointing East).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I chose March 20th because it is easier to visualize where the Sun would be (exactly over the Equator), and these cities due to their latitudes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bourdeaux: 44º 50' N (almost 45ºN)&lt;/li&gt;&#xA;&lt;li&gt;Christchurch: 43º 31' S (almost 45ºS)&lt;/li&gt;&#xA;&lt;li&gt;Quito: 0º 13' N (almost on Equator)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-05-14T20:29:06.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2445" PostTypeId="1" AcceptedAnswerId="8374" CreationDate="2014-05-15T03:40:33.343" Score="2" ViewCount="134" Body="&lt;p&gt;We have discovered planets in  Beta_Pictoris(A6V)&lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_Pictoris_b&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and Fomalhaut alpha(A3v)&lt;a href=&quot;http://en.wikipedia.org/wiki/Fomalhaut&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;br/&gt;&#xA;However they were discovered by direct images.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There should be an stellar type limit by RV? &lt;br/&gt; &#xA;What is the most early type ever found?&lt;br/&gt;&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="63" LastEditDate="2014-05-15T14:41:21.720" LastActivityDate="2014-12-27T11:33:27.270" Title="Which is the most early type star with a planet discovered by radial velocity method?" Tags="&lt;planet&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2446" PostTypeId="2" ParentId="2424" CreationDate="2014-05-15T04:08:08.423" Score="0" Body="&lt;p&gt;One of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;nofollow&quot;&gt;Friedmann equations&lt;/a&gt; describing the cosmological evolution of the universe is:&#xA;$$H^2 \equiv \frac{\dot{a}^2}{a^2} = \frac{8\pi G}{3}\rho - \frac{k}{a^2}\text{,}$$&#xA;where $a$ is the scale factor describing by how much the space expands (related to the Hubble parameter by $H = \dot{a}/a$), and $k\in\{-1,0,+1\}$ is a parameter corresponding to whether the spatial curvature is negative, zero, or positive, respectively. Note that this parameter doesn't tell us the magnitude of the spatial curvature, only its sign, e.g., a very slightly positively curved universe would still have $k=+1$. Also, the cosmological constant is considered to be part of the energy density $\rho$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want a spatially flat universe, we can put it $k = 0$ and infer a &lt;em&gt;critical density&lt;/em&gt; that the universe would have to have in order to be flat:&#xA;$$\rho_\text{c} = \frac{3H^2}{8\pi G}\text{.}$$&#xA;Using this value, we can re-write the first Friedmann equation as:&#xA;$$\rho_\text{c} = \rho - \frac{3k}{8\pi G a^2}\text{,}$$&#xA;or in terms of the density parameter $\Omega = \rho/\rho_\text{c}$:&#xA;$$(\Omega-1)\rho_\text{c}a^2 = \frac{3k}{8\pi G}\text{.}$$&#xA;What inflation does is exponentially increase scale factor $a$, which keeps $H$ and therefore $\rho_\text{c}$ constant. Since the right-hand side is always constant, that means $(\Omega-1)\rightarrow 0$, i.e., the universe is driven towards the density that makes it flat.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-05-15T04:08:08.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2447" PostTypeId="2" ParentId="2439" CreationDate="2014-05-15T11:17:42.760" Score="3" Body="&lt;p&gt;Let's say you have a rocket near a Schwarzschild black hole. Near the horizon, the thrust required to keep stationary goes to infinity; in this specific sense, the gravitational force becomes infinite at the horizon. If the rocket is hovering above the horizon, then the difference in gravitational forces becomes arbitrarily large near the horizon. This can tear it apart. &lt;em&gt;However&lt;/em&gt;, an object falling into the black hole will also gain a velocity arbitrarily close to the speed of light (as measured by a stationary observer as it passes) as it gets closer to the horizon, meaning it will be length-contracted. It &lt;em&gt;shrinks&lt;/em&gt; to zero size. As a result, the difference in 'gravitational force' across it is more complicated, and in fact remains finite.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Knowing that spaghettification is dependent on the size of the object, as well as the difference in gravity on both sides, a shrinking particle may eventually become so small that the steepness of the gravity well is no longer a problem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A shrinking object tames the infinite 'gravitational force' and keeps the tidal forces finite. But the curvature is &lt;em&gt;already&lt;/em&gt; defined in the limit of the size of the measurement probe tending to zero. (The static tidal forces correspond to six of the twenty independent curvature components, but let's not worry about that much.) So no, you can't 'get rid of curvature' by shrinking your particle/system/whatever. That's conceptually broken.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this a known theory or idea, and where can I find papers on it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You would have to have a much clearer idea before that can be answered reasonably. I've skipped all the way to the end of your question because I haven't understood anything more of what you're trying to say.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-05-15T11:17:42.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2448" PostTypeId="2" ParentId="2441" CreationDate="2014-05-15T11:33:38.697" Score="6" Body="&lt;p&gt;The logical consequence is, that an event horizon cannot form, since the first particle slows down asymptotically to zero, just before the event horizon forms (&lt;a href=&quot;http://en.wikipedia.org/wiki/Proof_by_infinite_descent&quot; rel=&quot;noreferrer&quot;&gt;Fermat's infinite descent&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The emergence of the event horizon therefore takes infinite time seen from outside. But due to Hawking radiation a black hole exists only a finite time. Hence an event horizon doesn't form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The frustrating thing about this is, that &lt;a href=&quot;http://www.nature.com/news/stephen-hawking-there-are-no-black-holes-1.14583&quot; rel=&quot;noreferrer&quot;&gt;you need to be at least Stephen Hawking&lt;/a&gt;, to not be called a geek.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current mainstream way to circumvent this paradoxon is to switch to a purely general relativistic geometry of infalling space-time, which doesn't experience the event horizon. That way you avoid the event horizon as a pole, but you get the singularity at the center of the black hole, governed by yet to investigate physical laws of quantum gravity.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-05-15T11:33:38.697" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2449" PostTypeId="2" ParentId="2441" CreationDate="2014-05-15T15:32:53.753" Score="28" Body="&lt;p&gt;What you're describing is basically the &quot;collapsed star&quot; (Eng) or &quot;frozen star&quot; (Rus) interpretation of black holes that was common prior to the late mid-1960s. It was a mistake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you are distant and stationary relative to the black hole. You will observe infalling matter asymptotically approaching the horizon, growing ever fainter as it redshifts. Does it mean that matter &quot;clumps&quot; around the horizon? To find out, suppose you throw yourself towards the black hole to try to catch the matter that you see. What you will find is that it fell into the black hole long ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the most sensible way to answer whether or not infalling matter clumps on the horizon is to look at the situation from the frame of that infalling matter. And there, it is clear: no, it does not clump, as it crosses the horizon in finite proper time. (As an aside, for a Schwarzschild black hole, falling from rest is exactly Newtonian in Schwarzschild radial coordinate and proper time.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;comoving viewpoint&quot; was recognized by Oppenheimer and Snyder in 1939, but it was not until the 1960s, with the work of Zel'dovich, Novikov, et al., that it was generally recognized as truly significant in the community. In 1965, Penrose introduced conformal diagrams based on the Eddington-Finkelstein coordinates (1924/1958) that showed quite clearly that the stellar collapse is not slowed, but instead continues to a singularity. For an overview of the history of this change of viewpoint, cf. Kip Thorne, et al., &lt;em&gt;The Memberane Paradigm&lt;/em&gt; (1986). These topics are commonly covered in many relativity textbooks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, but since it still takes an infinite amount time in the frame adapted to a stationary distant observer, does that mean that the horizon never forms in that frame? It does form: the underlying assumption in the argument that it does not would be either that the infalling matter needs to reach the center for the horizon to form or cross a pre-existing horizon to make it expand. But that assumption is simply not true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An event horizon is defined in terms of future lightlike infinity, roughly speaking in terms of whether or not light rays escape if one waits an infinite amount of time. That means the location of the horizon at any time depends on not just what has happened, but also what &lt;em&gt;will&lt;/em&gt; happen in the future. In the frame of the distant stationary observer, as matter falls towards the event horizon, it does slow down to asymptotically approach... but the horizon also expands to meet it. Similarly, the initial collapsing matter does not need to collapse all the way to the center for the event horizon to form.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can the finite life-time of the Black hole due to Hawking radiation be made consistant with the infinite amount of time (future) needed for the expansion of the event horizon (in the outer time-frame)? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's no need to: [edit]that a particular time coordinate doesn't cover the full manifold is a fault of the coordinate chart, not of spacetime[/edit]. From every event, send out an omnidirectional locus of idealized light rays. The event horizon is the boundary of the spacetime region from which none of these light rays escape to infinity. This question has an objective answer--for any given light ray, either it will escape or it won't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An external observer would need to wait infinitely long to &lt;em&gt;know for sure&lt;/em&gt; where the event horizon is exactly, but that's a completely different issue.  With Hawking radiation, the black hole shrinks, but it doesn't change the fact that light rays from some events will fail to escape, and thus that an event horizon will exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a Penrose diagram of a spherically collapsing star forming a black hole that subsequently evaporates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2i0ao.png&quot; alt=&quot;Penrose diagram of an evaporating black hole&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light rays run diagonally at ±45° on the diagram. Note that there is a region from which outgoing light rays (running diagonally lower-left to upper-right) don't escape and instead meet the $r = 0$ singularity (the bolded, undashed horizontal line). The horizon itself is the the $r = 2m$ line marked on the diagram &lt;em&gt;and&lt;/em&gt; its extension into the star: it should actually go from the (dashed, vertical) $r = 0$ line on the left, rather than extending from the star's collapsing surface. That's because some of the (idealized, noninteracting) light rays from inside the star will also fail to escape to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that on this diagram you draw timelike curves that stubbornly stay away from the horizon, and you insist on using a parameter along them as a time coordinate. Does the fact that you've chosen coordinates that exclude the horizon needs to be made consistent with whether or not the event horizon actually exist? The resolution is simple: if you want to talk about the horizon, stop using coordinates that exclude it.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditDate="2016-03-26T03:45:45.010" LastActivityDate="2016-03-26T03:45:45.010" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2450" PostTypeId="2" ParentId="2445" CreationDate="2014-05-15T22:18:03.963" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://exoplanet.eu/catalog/hd_103774_b/&quot; rel=&quot;nofollow&quot;&gt;HD 103774 b&lt;/a&gt;, detected by radial velocity, orbits a star of spectral type F5V according to &lt;a href=&quot;http://exoplanet.eu/catalog/all_fields/?f=%22radial%22%20IN%20detection%20OR%20%22astrometry%22%20IN%20detection&quot; rel=&quot;nofollow&quot;&gt;The Extrasolar Planet Encyclopeadia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://exoplanet.eu/catalog/hd_102956_b/&quot; rel=&quot;nofollow&quot;&gt;HD 102956 b&lt;/a&gt;, detected by radial velocity, &lt;a href=&quot;http://iopscience.iop.org/2041-8205/721/2/L153/pdf/2041-8205_721_2_L153.pdf&quot; rel=&quot;nofollow&quot;&gt;orbits a spectral type A subgiant&lt;/a&gt;, the star of the &quot;earliest&quot; type harboring a planet detected by radial velocity, I could find (as of May 2014).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The entry for&#xA;&lt;a href=&quot;http://exoplanet.eu/catalog/kepler-88_c/&quot; rel=&quot;nofollow&quot;&gt;Kepler-88 c&lt;/a&gt; around a spectral type B star (according to the catalog as of May 2014) appears to be incorrect, or encodes something else, &lt;a href=&quot;http://arxiv.org/pdf/1311.4335v2.pdf&quot; rel=&quot;nofollow&quot;&gt;the star should be classified as spectral type G6V&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-05-15T22:18:03.963" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2451" PostTypeId="1" AcceptedAnswerId="2453" CreationDate="2014-05-16T20:04:14.110" Score="2" ViewCount="725" Body="&lt;p&gt;By followig the laws of physics, is it possible, that in a place far far away that one day (at that place), is counted as 1000 years in the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am familiar with black holes having impact on time. But would it be possible without black holes and such,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think i've heard that the gravity has impact with the time too? &lt;/p&gt;&#xA;" OwnerUserId="1671" LastActivityDate="2014-05-16T20:48:14.067" Title="A day in earth, a thousand years somewhere else" Tags="&lt;gravity&gt;&lt;space-time&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2452" PostTypeId="1" AcceptedAnswerId="2456" CreationDate="2014-05-16T20:10:29.413" Score="4" ViewCount="262" Body="&lt;p&gt;Is the expansion of the universe proof of the Big Bang theory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or are there other proofs?&lt;/p&gt;&#xA;" OwnerUserId="1671" LastEditorUserId="67" LastEditDate="2014-05-18T19:52:46.077" LastActivityDate="2014-05-18T19:52:46.077" Title="Is the expansion of the universe proof of the big bang?" Tags="&lt;big-bang-theory&gt;&lt;expansion&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2453" PostTypeId="2" ParentId="2451" CreationDate="2014-05-16T20:30:37.947" Score="4" Body="&lt;p&gt;Assuming you mean a &lt;em&gt;day&lt;/em&gt; in terms of an Earth day in both cases, the phenomenon I believe you're referring to is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Time_dilation&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;time dilation&lt;/em&gt;&lt;/a&gt;, which affects the speed at which time is experienced by one body relative to another. This rate is affected by velocity and gravity, each of which can cause the bending of spacetime that results in the discrepancy between objects. Both gravity and velocity slow one's time as they increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation you describe is &lt;em&gt;probably&lt;/em&gt; not naturally occurring, however is not outside the realm of reality.  As referenced by Wiki from Calder's book &lt;a href=&quot;http://books.google.com/books?id=kbPmQX-OoBgC&amp;amp;pg=PA378&quot; rel=&quot;nofollow&quot;&gt;Magic Universe&lt;/a&gt;, Calder claims that an acceleration of a constant 1G (what we feel on Earth all the time) would result in the affects of time dilation allowing you to traverse the entirety of the known universe in a single human lifetime (for the traveler). Conversely, during your trip, the rest of the universe would be aging 'normally'. There are plenty of technical problems with the feasibility of achieving this, but is conceptually possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not particularly qualified to explain the equations behind it, so I'll leave that to our more capable members. Coincidentally, time dilation is the excuse I give when asked why I walk so quickly.&lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="1288" LastEditDate="2014-05-16T20:48:14.067" LastActivityDate="2014-05-16T20:48:14.067" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2454" PostTypeId="1" AcceptedAnswerId="2458" CreationDate="2014-05-17T04:14:50.257" Score="6" ViewCount="29088" Body="&lt;p&gt;Three or four weeks ago I was outside showing my neighbors how cool the moon was through a 'backyard' telescope in broad daylight (~8PM PDT).  It was 3/4 of the way across the sky (setting behind the houses behind us.)  Two weeks later the moon was not even up yet at 8PM.  The planets I've been watching (Jupiter, Saturn, Mars) are all rising and setting a little earlier each day as I would expect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why did the moon abruptly change its position in the sky a couple of weeks ago?&lt;/p&gt;&#xA;" OwnerUserId="1142" LastActivityDate="2016-05-11T02:42:08.337" Title="Why did the moon abruptly change positions in the sky?" Tags="&lt;the-moon&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2455" PostTypeId="2" ParentId="2454" CreationDate="2014-05-17T04:33:19.870" Score="5" Body="&lt;p&gt;The moon orbits the Earth in roughly 29 days. So in 14 days, it would be on the other side of the Earth.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-05-17T04:33:19.870" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2456" PostTypeId="2" ParentId="2452" CreationDate="2014-05-17T10:40:25.930" Score="3" Body="&lt;p&gt;In and of itself, no the expansion of the universe is not proof of the big bang. Other theories could be constructed that would also be explained by the expansion. For example Fred Hoyle and others proposed the &lt;a href=&quot;https://en.wikipedia.org/wiki/Steady_State_theory&quot; rel=&quot;nofollow&quot;&gt;Steady State Theory&lt;/a&gt; which proposed that new matter was constantly being created thus causing the expansion. A the time there was no other evidence for either theory so either could have been true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when you couple the expansion with other evidence such as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow&quot;&gt;cosmic microwave background radiation&lt;/a&gt; discovered in 1964 then it does strengthen the case for the big bang being the cause.&lt;/p&gt;&#xA;" OwnerUserId="111" LastEditorUserId="111" LastEditDate="2014-05-17T16:22:36.823" LastActivityDate="2014-05-17T16:22:36.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2457" PostTypeId="2" ParentId="2454" CreationDate="2014-05-17T11:18:14.423" Score="6" Body="&lt;p&gt;The moon didn't abruptly change position, it is gradually and continually changing position. You've obviously never really noticed what it is doing until now. You would really benefit from getting a planetarium application for your PC or device and adjusting the time base to see how everything moves. You'll get a much better understanding of the mechanics of it all, why the moon goes through phases, and why it rises and sets at different times.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-05-17T11:18:14.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2458" PostTypeId="2" ParentId="2454" CreationDate="2014-05-17T13:59:48.940" Score="12" Body="&lt;p&gt;Compared to the planets the moon changes it's rise and set times very quickly.  Using the calculator on &lt;a href=&quot;http://www.timeanddate.com/worldclock/astronomy.html?n=840&amp;amp;month=5&amp;amp;year=2014&amp;amp;obj=moon&amp;amp;afl=-11&amp;amp;day=1&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt; it can be seen that if one were looking at the sky on May 1st, 2014 from Irvine, California the moon set at 10:08PM (which at 8PM would have had the moon most of the way across the sky as noted in the question.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon rises and sets approximately one hour later each day.  So, by May 14th the moon did not even &lt;strong&gt;rise&lt;/strong&gt; until nearly 8PM - which also correlates with the observations noted above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contrasting that with the other context point made in the OP, using &lt;a href=&quot;http://www.almanac.com/astronomy/rise/CA/Irvine/2014-05-14&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt; we can see that on May 1st Saturn rose at 8:07PM and on May 14th rose at 7:11PM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while the moon changes its rise a set times by almost an hour &lt;strong&gt;each day&lt;/strong&gt;, the planets change much more slowly by comparison, almost an hour over &lt;strong&gt;two weeks&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one weren't continually watching the moon's progression it would appear to abruptly change positions when comparing it to the planets' movements.&lt;/p&gt;&#xA;" OwnerUserId="1142" LastEditorUserId="1142" LastEditDate="2015-11-09T19:38:09.087" LastActivityDate="2015-11-09T19:38:09.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2459" PostTypeId="1" AcceptedAnswerId="2484" CreationDate="2014-05-17T14:26:27.597" Score="2" ViewCount="250" Body="&lt;p&gt;I understand that astronomers use right ascension and declination for the position of stars. However both these values strike me as very anthropocentric and, more importantly, unreliable over time as I understand that the Earth wobbles a bit over time so the celestial equator shifts and I guess Sun's rotation around the galaxy and Milky Way's own trajectory should play a role in depreciating the accuracy of those values over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my questions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] how steady are right ascension and declination values for distant stars over time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] is there a more stable coordinate system or do we lack an absolutely fixed frame of reference?&lt;/p&gt;&#xA;" OwnerUserId="522" LastEditorUserId="522" LastEditDate="2014-05-17T16:21:19.490" LastActivityDate="2014-05-20T02:01:03.463" Title="how steady over time are right ascension and declination values?" Tags="&lt;coordinate&gt;&lt;declination&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2460" PostTypeId="1" AcceptedAnswerId="2463" CreationDate="2014-05-17T15:39:40.113" Score="8" ViewCount="365" Body="&lt;p&gt;I was reading a very old book (around 1850) on astronomy and the author explains Sun spots as a phenomena where you can see the black planetary like hard core of the Sun through two different atmospheres.  It got me interested that somewhere between 1850 and now it became understood that the Sun's core is a gas.  How was this determined conclusively and approximately when in history?&lt;/p&gt;&#xA;" OwnerUserId="1683" LastEditorUserId="9527" LastEditDate="2020-06-02T10:42:35.790" LastActivityDate="2020-06-02T10:42:35.790" Title="How did astronomers first come to understand that the Sun's core was a gas?" Tags="&lt;the-sun&gt;&lt;history&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2461" PostTypeId="1" CreationDate="2014-05-17T18:04:02.383" Score="4" ViewCount="720" Body="&lt;p&gt;Quasars are known to contain supermassive Black Holes at their cores. So does that mean that the number of quasars is equal to that of the SMBHs out there?&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2014-07-22T21:54:36.157" Title="Quasars and SMBH" Tags="&lt;galaxy&gt;&lt;supermassive-black-hole&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2462" PostTypeId="2" ParentId="2461" CreationDate="2014-05-17T20:20:09.617" Score="3" Body="&lt;p&gt;It means that the number of quasars must be less than or equal to the number of SMBH out there.  It's believed that supermassive black holes are found at the center of most galaxies and the &lt;a href=&quot;http://en.wikipedia.org/wiki/M%E2%80%93sigma_relation&quot; rel=&quot;nofollow&quot;&gt;SMBH mass correlates with the velocity distribution of stars in the galaxy&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, quasars only form in particularly large and active galaxies.  Or perhaps particularly massive galaxies can only form around and be supported by particularly massive black holes? Regardless, based on this, while every quasar must possess a supermassive black hole, not every supermassive black hole is part of a system that contains a quasar.  &lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="1862" LastEditDate="2014-07-22T21:54:36.157" LastActivityDate="2014-07-22T21:54:36.157" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2463" PostTypeId="2" ParentId="2460" CreationDate="2014-05-17T22:43:03.930" Score="7" Body="&lt;p&gt;Well simply said: All data fits the premise that it should be a gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what data?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First of all, in the late 1960's the &lt;a href=&quot;http://en.wikipedia.org/wiki/Helioseismology&quot;&gt;5-minute solar oscillations&lt;/a&gt; on the surface of the sun have beend discovered. Those have been recognized as surface features of internal pressure waves with penetration depth $d$, wavenumber $k$ and their relation roughly $d \sim  \frac{1}{\sqrt{k}}$.&lt;br&gt;&#xA;So from the measurment and simple existence of certain wave-ripples on top of the photospheric plasma-ocean we can learn about the interior of the sun. To relate the strength of certain wavemodes to the interior structure of this plasma-ball we use usually a reverse-engineering approach called modelling. We take what we know about gases, energy transport etc. and calculate from this how a solar-oscillations-spectrum on a modelled sun would look like. Then compare, scratch your head, change the modell and repeat until it fits. More mathematical details, if you are interested can be found in the excellent &lt;a href=&quot;http://users-phys.au.dk/jcd/oscilnotes/&quot;&gt;lecture notes&lt;/a&gt; on Helioseismology from Christensen-Daalsgard.&lt;br&gt;&#xA;So all this gave us a pretty good look into the sun's interior and it was quite clear, that the observed solar parameters like, size, energy output, surface temperature must lead to certain Pressure, Temperature and Density values for the core corresponding to a fusing plasma.&lt;/li&gt;&#xA;&lt;li&gt;But those theoretical predictions would imply the core-plasma to create neutrinos from nuclear reactions, which famously (wiki: Solar_neutrino_problem) were off by a factor of about 3. With the discovery of neutrino oscillations (wiki: Neutrino_oscillation) this problem was also solved in that the sun didn't produce too few neutrinos, but the produced neutrinos would simply change their types. After some distance from the sun those changes would redistribute the numbers from 3:0:0 for electron-myon-tau neutrinos to 1:1:1.&#xA;This problem persisted until approx. the year 2000.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2014-05-17T22:43:03.930" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2464" PostTypeId="1" CreationDate="2014-05-18T00:19:05.537" Score="10" ViewCount="162" Body="&lt;p&gt;I was scrolling idly through &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Orion_%28constellation%29&quot;&gt;the Wikipedia article on Orion&lt;/a&gt; when I read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Orion will still be recognizable long after most of the other constellations—composed of relatively nearby stars—have distorted into new configurations, with the exception of a few of its stars eventually exploding as supernovae, for example Betelgeuse, which is predicted to explode sometime in the next million years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given that Betelgeuse is only 640 light-years away from Earth, the question of whether we're talking about Betelgeuse going supernova in the next million years, or about &lt;em&gt;news of the supernova reaching us&lt;/em&gt; in the next million years, is irrelevant, since 640 years is hardly noticeable when you're talking about ten thousand centuries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about a star further away? Say, something in the Andromeda Galaxy, 2.5 million light-years from us? If I read an article stating that a star in the Andromeda Galaxy was going nova in a million years, would that mean:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) That we think the star will go nova one million years from now, and its light will reach us in 3.5 million years; or&lt;br&gt;&#xA;b) That we believe the star went nova 1.5 million years ago, and its light will reach us in one million years &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, are we making estimates in our time, or in celestial-body-local time?&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2014-06-26T15:21:10.833" Title="Do we make predictions in our time, or local time?" Tags="&lt;distances&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2465" PostTypeId="1" AcceptedAnswerId="2475" CreationDate="2014-05-18T04:05:52.370" Score="3" ViewCount="173" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UCidd.png&quot; alt=&quot;Broadband Spectrum of Sun&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Broadband spectrum is one which is inclusive of all the observed wavelength ranges. In the case of the BB spectra of our Sun, there is a sharp (non-differentiable) kink in the curve in the UV-range (as shown above). What is the significance of this kink?&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2020-11-16T11:18:59.713" Title="Broadband spectrum of Sun" Tags="&lt;the-sun&gt;&lt;spectra&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2466" PostTypeId="1" CreationDate="2014-05-18T04:11:48.553" Score="2" ViewCount="150" Body="&lt;p&gt;I've read that there are 4 distinct although related forms of AGs : Seyferts, Quasars, Radio and Blazars. Are there any other forms possible? Also do these forms ever intermix to give something new? &lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="35056" LastEditDate="2020-09-12T16:35:25.410" LastActivityDate="2020-09-12T16:35:25.410" Title="Classification of Active Galaxies" Tags="&lt;galaxy&gt;&lt;active-galaxy&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2467" PostTypeId="1" CreationDate="2014-05-18T05:48:09.373" Score="1" ViewCount="1664" Body="&lt;p&gt;If I know specific time, t0, how can I calculate RA of Greenwich?&lt;/p&gt;&#xA;" OwnerUserId="1686" LastEditorUserId="1322" LastEditDate="2014-05-18T19:50:25.917" LastActivityDate="2014-09-16T09:10:45.503" Title="How to calculate right ascension of Greenwich?" Tags="&lt;earth&gt;" AnswerCount="2" CommentCount="0" ClosedDate="2014-09-28T13:09:10.473" ContentLicense="CC BY-SA 3.0" />
  <row Id="2468" PostTypeId="1" CreationDate="2014-05-18T08:17:47.933" Score="6" ViewCount="206" Body="&lt;p&gt;The companion of a neutron star could be a main sequence star, white dwarf, neutron star or black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We detected a &lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_pulsar&quot; rel=&quot;nofollow noreferrer&quot;&gt;pulsar-pulsar binary&lt;/a&gt; several decades ago. But if one of the pulsar does not radiate towards us, we may be not able to tell it is a pulsar or a black hole. If the companion is not a main sequence star, how do we know the nature of the companion in a neutron star binary?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the binary systems with one single neutron star, is that possible there is another neutron star or a hidden black hole?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="2153" LastEditDate="2016-11-03T20:03:51.423" LastActivityDate="2016-11-03T20:03:51.423" Title="How can we know the nature of the companion in a neutron star binary?" Tags="&lt;star&gt;&lt;black-hole&gt;&lt;observational-astronomy&gt;&lt;neutron-star&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2469" PostTypeId="2" ParentId="2467" CreationDate="2014-05-18T10:02:57.097" Score="2" Body="&lt;p&gt;I think you mean to ask, &quot;What is the local sidereal time of Greenwhich?&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the GMST (Greenwich Mean Sidereal Time): &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To an accuracy within 0.1 second per century, Greenwich (Mean) Sidereal Time (in hours and decimal parts of an hour) can be calculated as GMST = 18.697 374 558 + 24.065 709 824 419 08 * D , where D is the interval, in UT1 days including any fraction of a day, since 2000 January 1, at 12h UT (interval counted positive if forwards to a later time than the 2000 reference instant), and the result is freed from any integer multiples of 24 hours to reduce it to a value in the range 0–24.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[&lt;a href=&quot;http://en.wikipedia.org/wiki/Sidereal_time&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;]&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2014-05-18T10:02:57.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2470" PostTypeId="1" CreationDate="2014-05-18T12:49:35.710" Score="2" ViewCount="370" Body="&lt;p&gt;Normally, when people say &lt;em&gt;&quot;the universe is infinite&quot;&lt;/em&gt; they generally mean something like &quot;the &lt;em&gt;observable universe is locally flat&lt;/em&gt;, it is easy to assume that it is flat way beyond the observable universe, I don't know what lies beyond that, I won't talk about that, that is enough for practical purposes&quot; (or any other proposition implying infinity in any direction and assumed for simplicity's sake while actually focussing on a small region of the space, the observable universe). So, &quot;the universe is infinite&quot; does not seriously mean that physical space goes on forever because this simply makes no sense at all, &quot;infinite&quot; is not a number, it is not a magnitude, it is not a property of real things (if you disagree with this premise please explain how the notion of infinity can be seriously applied to any direction of physical space, i.e. how you can seriously think to extend, sic et simpliciter, such a property of your mathematical model to the real world). (Personally I don't think this should be counted as a theory of the shape of the universe at all).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the contrary, &lt;em&gt;&quot;the universe is finite and unbounded, compact, closed in all directions&quot;&lt;/em&gt; is indeed a meaningful and in principle acceptable theory because it can be thought, described, discussed for what it actually implies. Of course, we don't know what lies beyond the observable universe but the idea of a compact universe makes sense and it seems to me this is the only meaningful explanation currently available of what &quot;could&quot; be (it is not necessarily the &lt;em&gt;right, true&lt;/em&gt; explanation, of course).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any other &lt;strong&gt;meaningful&lt;/strong&gt; alternative (which can be thought, described, discussed)? Is there any alternative to the dichotomy &quot;finite universe&quot; vs &quot;infinite universe&quot;? Is there anybody out there working on models that overcome this dichotomy?&lt;/p&gt;&#xA;" OwnerUserId="1676" LastEditorUserId="1676" LastEditDate="2014-05-21T13:22:25.980" LastActivityDate="2016-02-12T11:29:33.867" Title="Meaningful theories of the shape of the universe" Tags="&lt;universe&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2471" PostTypeId="2" ParentId="2470" CreationDate="2014-05-18T13:30:29.727" Score="3" Body="&lt;p&gt;I disagree with your premise. First, the universe being infinite does not imply anything at all about its spatial curvature without additional assumptions. Second, even with the usual cosmological assumptions of large-scale homogeneity and isotropy, being infinite would not imply spatial flatness, as negative curvature would still be possible.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On the contrary, &quot;&lt;em&gt;the universe is finite and unbounded, compact, closed in all directions&lt;/em&gt;&quot; is indeed a meaningful and acceptable theory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This this alternative still makes grand assumptions about the structure of the universe beyond what we can observe, I don't see why you attribute more 'meaning' or 'acceptability' to it. At the end of the day, all we really know is that the bit of the universe around us is very close to being flat. Any statement whatsoever about the larger structure of the universe is going to introduce some assumptions not backed by observations, and you haven't provided much guidance to what kind are 'meaningful and acceptable'.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there any other &lt;strong&gt;meaningful&lt;/strong&gt; alternative?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You seem to be using that word in an idiosyncratic sense I'm not familiar with. In any case, you're mistaken about what &quot;the universe is infinite&quot; implies.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-05-18T13:30:29.727" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2472" PostTypeId="1" CreationDate="2014-05-19T00:14:04.560" Score="4" ViewCount="175" Body="&lt;p&gt;When asteroids crash into the Moon they give of huge amounts of energy resulting in a &quot;bright&quot; flash, as can be seen in &lt;a href=&quot;http://www.theguardian.com/science/video/2013/may/18/meteor-crashes-moons-surface-flash-video&quot; rel=&quot;nofollow&quot;&gt;this video&lt;/a&gt;. At the end of the video its also says that anyone looking at the Moon at that moment would have been able to observe it without a telescope.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what kind of collision would be required to be able to be seen from Earth when it would involve more distant bodies? Would one of Jupiter's retrograde moons colliding with one of its prograde moons be visible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;br&gt;&#xA;A real example of a high energy collision would be the collisions of the fragments of comet Shoemaker-Levy-9 with Jupiter. However according to &lt;a href=&quot;http://www.esa.int/spaceinvideos/Videos/1994/06/Collision_Comet_Shoemaker_Levy_9&quot; rel=&quot;nofollow&quot;&gt;this video&lt;/a&gt; from ESA it happened on the opposite side of Jupiter, so not in our view. &lt;/p&gt;&#xA;" OwnerUserId="1689" OwnerDisplayName="fibonatic" LastEditorUserId="1689" LastEditDate="2014-05-19T01:02:12.840" LastActivityDate="2014-05-19T06:26:29.230" Title="What kind of collision of celestial bodies would be visible to the naked eye?" Tags="&lt;impact&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2475" PostTypeId="2" ParentId="2465" CreationDate="2014-05-19T01:34:25.663" Score="2" Body="&lt;p&gt;The point here is, that the kink you're pointing out is not belonging to the BB-spectrum of the sun.&#xA;This part of the solar spectrum is a classic example for a broad class of radiation, called nonthermal radiation.&#xA;The physical origin of this is in general that, every &lt;strong&gt;accelerated&lt;/strong&gt; particle will radiate part of its energy away, with the experienced acceleration a, radiated power P and their relation $P \sim a^2$ in a wavelength-range depending on the exact process.&lt;br&gt;&#xA;The origins of those accelerations are usually thought to be strong magnetic interactions on the solar surface plasma, creating the famous sunspots and &lt;a href=&quot;http://en.wikipedia.org/wiki/Corona&quot; rel=&quot;nofollow&quot;&gt;coronal loops&lt;/a&gt;. In those loops, electrons and protons are being shot into space, following the loop, and then return back onto the solar 'surface'. Upon return they release &lt;a href=&quot;http://en.wikipedia.org/wiki/Bremsstrahlung&quot; rel=&quot;nofollow&quot;&gt;Bremsstrahlung&lt;/a&gt;, as the interactions between loop-particles and surface plasma create strong braking accelerations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spectrae for such processes are derived in every standard textbook about theoretical astrophysics and are given by roughly $\frac{dP}{dt} \sim exp(-\frac{h \nu}{k_BT})$, which should fit the UV-part in the spectrum you gave pretty well.&#xA;This now also explains the variability of this part of the solar spectrum: As the number of loops and sunspots correlates with &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_cycle&quot; rel=&quot;nofollow&quot;&gt;solar activity&lt;/a&gt;, it is clear that the UV-bremsstrahlung-production should too.&lt;br&gt;&#xA;There is however much more to this, if you are interested. Given certain &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_flares&quot; rel=&quot;nofollow&quot;&gt;violent outbreaks&lt;/a&gt; that happen upon release of magnetic energy, there are modulations of the emitted radio flux (also indicated in your graph) up higher-energetic regions of x-rays.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2014-05-19T01:34:25.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2476" PostTypeId="2" ParentId="2472" CreationDate="2014-05-19T06:26:29.230" Score="2" Body="&lt;p&gt;OK, so... you said the collision with Shoemaker-Levy 9 and Jupiter happened on the non-visible side. What if we'd seen it? The energy released by the largest fragment collision was estimated at 6 million megatons of TNT, which is 2.52e22 joules. The Sun's luminosity is 3.846e26 watts, by comparison. So obviously the collision won't rival the Sun. Using the luminosity-distance relation and taking the 6 million megatons as pure luminosity (an overestimate, but we're not going for precision here) gives the result of 0.0033 watts/square metre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relating flux to apparent magnitude, we get an apparent magnitude of the collision of... approximately -12. That's about the brightness of a full moon. Bear in mind this is a very rough calculation and it makes the assumption that all the energy was released over a second and all of it was light, so probably the real value would be a few magnitudes down But, assuming I've made no terrible calculation errors, it would definitely have been visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that was a relatively tiny comet, your example of two of Jupiter's moons colliding would release FAR more energy, therefore would be clearly visible, I'd imagine. All depends on how much energy gets emitted as light.&lt;/p&gt;&#xA;" OwnerUserId="1658" LastActivityDate="2014-05-19T06:26:29.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2477" PostTypeId="2" ParentId="2467" CreationDate="2014-05-19T07:28:24.357" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Right_ascension&quot; rel=&quot;nofollow&quot;&gt;Right Ascension&lt;/a&gt; is a celestial coordinate, not a terrestrial one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RA is used to locate a star on sky, and is measured between a fixed point in the sky (the crossing of the Ecliptic and the Celestial Equator) and the star's meridian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It corresponds to terrestrial coordinate Latitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, Greenwich has no Right Ascension.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-05-19T07:28:24.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2478" PostTypeId="2" ParentId="1440" CreationDate="2014-05-19T08:46:02.967" Score="6" Body="&lt;p&gt;I don't have enough reputation to comment...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this might help you understand the formation or binary and more stars systems. This of course is not the only possible method but it might explain the systems with big mass differences. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the initial rotation speed increases (marked in the videos as beta) you will see how the protoplanetary disk breaks up and form more than one protostar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astro.ex.ac.uk/people/mbate/Animations/prestellar_discs.html&quot;&gt;http://www.astro.ex.ac.uk/people/mbate/Animations/prestellar_discs.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1678" LastActivityDate="2014-05-19T08:46:02.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2479" PostTypeId="2" ParentId="2468" CreationDate="2014-05-19T14:21:55.497" Score="1" Body="&lt;p&gt;For something to be possible in this universe, &lt;em&gt;doesn't even have to be consistent with the physics&lt;/em&gt; we know. Astronomical surprises have often led to revisions in our theories. Thus, all the rare situations you described are possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How do we detect such evasive systems?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no definite answer. Based on the data we have about the system, you can get creative while doing data analysis and come up with new methods. That is one of the reasons science continues to be fun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe someone who has had some experience with the concerned matter can expand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, such systems are rare due to a physical reason. Mass transfer between binary systems ensures that one of the stars get evolved earlier than the other. The other star either ends up being a low mass star and evolves accordingly in most cases. The remaining cases account for your rare description of systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Algol_paradox&quot; rel=&quot;nofollow&quot;&gt;Algol Paradox&lt;/a&gt; for an interesting read.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2014-05-19T14:21:55.497" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2480" PostTypeId="2" ParentId="2464" CreationDate="2014-05-19T17:29:32.187" Score="8" Body="&lt;p&gt;We express when events are going to happen (or have happened) in &lt;strong&gt;&lt;em&gt;our&lt;/em&gt; observed time frame.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, we say &lt;a href=&quot;http://en.wikipedia.org/wiki/SN_185&quot;&gt;SN 185&lt;/a&gt; went supernova in the year AD 185; but it didn't. The star actually &lt;em&gt;physically&lt;/em&gt; exploded about 9,100 years earlier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we say the sun will set tonight at 8:07 PM ET, that is when we observe the event. But our &lt;em&gt;actual&lt;/em&gt; position relative to the tangent from the sun will have already passed about 8 minutes 20 seconds earlier (plus several minutes due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Atmospheric_refraction&quot;&gt;atmospheric refraction&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when you say Betelgeuse is predicted to explode sometime in the next million years, that's how many more Earth years are expected to pass until we &lt;strong&gt;&lt;em&gt;see&lt;/em&gt;&lt;/strong&gt; it happen&amp;hellip; from &lt;em&gt;here.&lt;/em&gt; From Betelgeuse's point of view, it would have happened about 642 years earlier&amp;hellip; or it may have happened already.&lt;/p&gt;&#xA;" OwnerUserId="58" LastEditorUserId="58" LastEditDate="2014-06-26T15:21:10.833" LastActivityDate="2014-06-26T15:21:10.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2481" PostTypeId="1" CreationDate="2014-05-18T23:50:43.937" Score="6" ViewCount="836" Body="&lt;p&gt;I was scrolling through the list of &lt;a href=&quot;http://en.wikipedia.org/wiki/Moons_of_Jupiter&quot;&gt;moons of Jupiter&lt;/a&gt; and I wondered how astronomers where able to determine the masses of those objects. When I tried to google it I meanly found results of people deriving the mass of Jupiter from the masses of its moons using Kepler's laws. So the mass of celestial bodies with satellite(s) can be estimated this way. But what about the rest? Or are they all derived from the trajectories of man-made satellites passing near them?&lt;/p&gt;&#xA;" OwnerUserId="1689" OwnerDisplayName="fibonatic" LastActivityDate="2014-05-19T21:06:58.557" Title="How are the masses of celestial bodies determined?" Tags="&lt;mass&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2482" PostTypeId="2" ParentId="2481" CreationDate="2014-05-19T03:04:18.127" Score="4" Body="&lt;p&gt;Newton's analysis of Kepler's laws lets you determine the ratios of the masses of everything in the solar system with quite high precision, but it doesn't let you determine the actual mass.  The actual masses were one of the large unsolved problems of physics for almost a century, until 1798 when &lt;a href=&quot;https://en.wikipedia.org/wiki/Cavendish_experiment&quot; rel=&quot;nofollow&quot;&gt;Cavendish's work with a torsion balance&lt;/a&gt; (described by him as &quot;weighing the world&quot;) determined the density of Earth, from which the masses of everything could be computed.&lt;/p&gt;&#xA;" OwnerUserId="1559" OwnerDisplayName="Mark" LastActivityDate="2014-05-19T03:04:18.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2483" PostTypeId="1" AcceptedAnswerId="2485" CreationDate="2014-05-19T23:47:59.663" Score="0" ViewCount="621" Body="&lt;p&gt;Like this, but floating in space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aK6ds.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine that this would also be influenced by the distance of the object from the sun, and that there is some ideal distance for this object to be? (and obviously the distance of the Earth from the Sun as well)&lt;/p&gt;&#xA;" OwnerUserId="1697" LastActivityDate="2014-05-20T03:32:58.683" Title="How big would a disc have to be to blot out the sun covering 5% of the land mass of earth?" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2484" PostTypeId="2" ParentId="2459" CreationDate="2014-05-20T02:01:03.463" Score="3" Body="&lt;p&gt;The values are more steady for objects close to Ecliptic poles, but in genenral, they change by a few arcseconds every year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, we lack an absolutely fixed frame of reference.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2014-05-20T02:01:03.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2485" PostTypeId="2" ParentId="2483" CreationDate="2014-05-20T02:05:46.543" Score="3" Body="&lt;p&gt;The angular diameter of the sun is approximately 30 arcminutes (roughly .0087 radians), or half of a degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula for angular diameter of a round object in radians is: &#xA;$$&#xA;\delta = 2arctang\frac{diameter}{(2)distance}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this, we should also think about position. Assuming we want to continually block the same spot on Earth we need to the orbital period to one year. &lt;em&gt;(I'm not sure if there is a technical difference between orbiting the Earth once per year such that you're always between the Sun and Earth, and just plain orbiting the Sun)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.calctool.org/CALC/phys/astronomy/earth_orbit&quot; rel=&quot;nofollow&quot;&gt;this calculator&lt;/a&gt;, our sun blocker would need at a height of &lt;strong&gt;2,152,050km&lt;/strong&gt; to orbit Earth once per year. So, given that we can set up our formula with...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;.0087 = 2arctang\frac{diameter}{(2)2152050}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...and the needed diameter of an object should come out to be about &lt;strong&gt;18,722km in diameter&lt;/strong&gt;. Given that, Mr. Burns's terrestrial solution is far more practical, though certainly a nice orbital one is far more devious. Being evil isn't cheap.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Edit:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This answer is a bit rough as it doesn't calculate for the 5% of the Earth's land mass stipulation. The answer which would meet this would depend on whether only an &lt;a href=&quot;http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit2/eclipses.html&quot; rel=&quot;nofollow&quot;&gt;umbral shadow qualifies, or if a penumbral shadow&lt;/a&gt; would also suffice. I suspect probably the prior. The shadow &lt;em&gt;would&lt;/em&gt; certainly reach the Earth, but probably not cover 5% of it, which is a shame considering its diameter is actually &lt;em&gt;larger&lt;/em&gt; than Earth's. Regardless, it's not a very elegant solution.&lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="1288" LastEditDate="2014-05-20T03:32:58.683" LastActivityDate="2014-05-20T03:32:58.683" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2487" PostTypeId="1" AcceptedAnswerId="2491" CreationDate="2014-05-20T13:51:19.757" Score="1" ViewCount="82" Body="&lt;p&gt;What I was wondering is that, for example, toss a few balls into zero gravity space randomly, what would be the steady state motion?   Would not all the balls go on weird directions and undergo irregular orbits due to the interacting and irregular gravitations. I imagine the largest or more dense planets to begin to influence the smaller and eventually it would stabalize.  Is this what origin of the universe theories imagine, or am I totally missing something here? Would the smaller ones eventually all rotate around the largest? All in the same plane, more or less, and all in the same direction? It seems possible but I have no idea if this is mathematically derivable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are theories that explain chaotic motions into the present, same plane, same direction planetary orbits around the Sun?&lt;/p&gt;&#xA;" OwnerUserId="1683" LastEditorUserId="1322" LastEditDate="2014-05-21T07:57:02.773" LastActivityDate="2014-05-21T07:57:02.773" Title="Do origin theories imagine each planet to first orbit the Sun very irregularly before stabilizing?" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;formation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2488" PostTypeId="1" AcceptedAnswerId="2489" CreationDate="2014-05-20T15:06:29.860" Score="6" ViewCount="2391" Body="&lt;p&gt;When viewing star maps in Stellarium, I noticed that the meridians of the equatorial grid were denoted in hours and not degrees, and that always puzzled me. Why did astronomers choose this system? I understand it has something to do with being relative to the Earth's rotation which takes 24 hours, but I can't figure out exactly what.&lt;/p&gt;&#xA;" OwnerUserId="1703" LastActivityDate="2018-12-24T15:49:09.937" Title="Why is right ascension measured in hours?" Tags="&lt;coordinate&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2489" PostTypeId="2" ParentId="2488" CreationDate="2014-05-20T15:24:59.873" Score="5" Body="&lt;p&gt;In order to know when a star will be above horizon, you'll need an equation with times, not &quot;celestial latitudes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you end up adding RA and sidereal time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to find out the position of a star you need to find out how its Hour Angle varies with time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hour Angle is how far the star is from the observer's meridian:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1YAB9.gif&quot; alt=&quot;Diagram of angles&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this image, it is the angle between the site's celestial meridian (North Celestial &lt;strong&gt;P&lt;/strong&gt; ole-&lt;strong&gt;Z&lt;/strong&gt; enith-&lt;strong&gt;S&lt;/strong&gt; outh horizon) and the star's celestial meridian (&lt;strong&gt;P&lt;/strong&gt;-&lt;strong&gt;X&lt;/strong&gt;-&lt;strong&gt;Y&lt;/strong&gt;). It is the same angle as between where X culminates and &lt;strong&gt;X&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it comes than &lt;span class=&quot;math-container&quot;&gt;$HA=ST-RA$&lt;/span&gt; and since you may not have a &quot;local sidereal time&quot; clock at hand, you can use Greenwich's Sidereal Time: &lt;span class=&quot;math-container&quot;&gt;$HA=GST-Lat-RA$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$HA$&lt;/span&gt; (local) Hour Angle of a star&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$ST$&lt;/span&gt; Local Sidereal Time&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$GST$&lt;/span&gt; Greenwich Sidereal Time&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$RA$&lt;/span&gt; Right Ascension of a star&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$Lat$&lt;/span&gt; (local) Geographical Latitude&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="473" LastEditorUserId="7472" LastEditDate="2018-12-24T15:49:09.937" LastActivityDate="2018-12-24T15:49:09.937" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="2490" PostTypeId="2" ParentId="2487" CreationDate="2014-05-20T15:27:21.357" Score="0" Body="&lt;p&gt;That chaotic movements would make sens in a place, like a globullar cluster, with similarly-sized bodies. On the other hand, in a place like a planetary system where almost all mass belongs to a single body, you can be pretty certain taht you will have almost-independent orbits, like we have here.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-05-20T15:27:21.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2491" PostTypeId="2" ParentId="2487" CreationDate="2014-05-21T01:29:48.620" Score="2" Body="&lt;p&gt;Actually, the sun, planets and other small bodies are all created from the cloud of dust and gas that existed before the solar system. As the gas and dust collect towards the center, the mass rotates and soon an accretion disk forms (which is why the planets are nearly on the same plane). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your notion of the beginning of the universe is incorrect. It did not start with a bunch of planets (or stars) randomly thrown into space. Also all the stars in a globular cluster are rotating in the same direction; the stars are in orbit in the cluster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:&lt;br&gt;&#xA;Yes, they did have irregular movements. It started with a large cloud of gas and dust, some from the big bang, mostly from exploded stars systems. The particles all had different directions and speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When gravity influenced enough particles to group and start accumulating is when the mass began to spin (conservation of momentum). The spinning mass flattens out to an accretion disk and planets can start to form (along with the sun). &lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="1322" LastEditDate="2014-05-21T04:48:35.687" LastActivityDate="2014-05-21T04:48:35.687" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2492" PostTypeId="2" ParentId="2461" CreationDate="2014-05-21T01:43:38.040" Score="0" Body="&lt;p&gt;Not really. Quasars do indeed need a SMBH to be powered, and this is a necessary condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, Quasar phase only lasts for &lt;a href=&quot;http://www.phys.vt.edu/~jhs/faq/quasars.html#q5&quot; rel=&quot;nofollow&quot;&gt;tens millions to few billions&lt;/a&gt; years.&#xA;This means that, in a more recent universe, Quasars are turned off but the host galaxy... is still there. And the SMBH also do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is exactly as our Galaxy case (&lt;a href=&quot;http://en.wikipedia.org/wiki/Sagittarius_A%2a&quot; rel=&quot;nofollow&quot;&gt;$Sgr A^*$&lt;/a&gt;), where the evidences for the presence of a SMBH are among the strongest ones, but still no nuclear activity is present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are tons of study to infer the proper amount of AGNs in the whole Universe by using &lt;a href=&quot;http://www.astro.virginia.edu/class/whittle/astr553/Topic04/Lecture_4.html#sec4&quot; rel=&quot;nofollow&quot;&gt;luminosity functions&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1038" LastActivityDate="2014-05-21T01:43:38.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2493" PostTypeId="2" ParentId="2466" CreationDate="2014-05-21T05:31:17.653" Score="3" Body="&lt;p&gt;Yes of course, there are other forms apart from the four stated above. Basically you have just two kinds: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the radio-quiet, and&lt;/li&gt;&#xA;&lt;li&gt;the radio-loud.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But then again, there are two Seyferts&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Seyfert I, and&lt;/li&gt;&#xA;&lt;li&gt;Seyfert II&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Apart from Quasar, Blazar and Radio galaxy, we also have presently,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;BL Lac (Named after its prototype BL Lacarte (original member of Blazar type)), and&lt;/li&gt;&#xA;&lt;li&gt;OVV (Optically Variable Violent Quasar or OVV Quasar; subtype of Blazar)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Best source to begin with is wikipedia.org of course.&lt;/p&gt;&#xA;" OwnerUserId="1702" LastActivityDate="2014-05-21T05:31:17.653" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2495" PostTypeId="1" AcceptedAnswerId="2496" CreationDate="2014-05-21T08:13:27.603" Score="2" ViewCount="707" Body="&lt;p&gt;I'm writing a space game that I want to give a more &quot;realistic&quot; feel. When looking for a reference as to distribute elements in planets (for mining, etc) Instead of finding the more usual percentage tables, it seems that all elements are presented in a &quot;normalized to Si 10 to the power 6&quot;, and furthermore, it seems that the data is presented as number of atoms, instead of actual mass. I've been searching this for days, and even the &quot;parts per million/billion/trillion&quot; tables sometimes just jump for percentages, where is is again no clear if what is being referred to are masses or number of atoms. Could someone clarify this, or better, point me to a mass percentage table of elements in a solar system/galaxy for reference? Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="1707" LastActivityDate="2014-05-21T10:53:20.407" Title="What does &quot;Normalized to Si 10 to the power 6&quot; mean?" Tags="&lt;galaxy&gt;&lt;planetary-formation&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2496" PostTypeId="2" ParentId="2495" CreationDate="2014-05-21T09:03:46.697" Score="4" Body="&lt;p&gt;The table you referenced is normalized so that the abundance of Silicon (Si) is exactly $10^6$, and all the other numbers are related to that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Abundance&quot; in this table refers to the number of atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that this table is for the entire Solar System, taking into account the huge amounts of Hydrogen and Helium in the Sun. It will not be appropriate for creating Earth-like planets.&lt;/p&gt;&#xA;" OwnerUserId="473" LastEditorUserId="173" LastEditDate="2014-05-21T10:53:20.407" LastActivityDate="2014-05-21T10:53:20.407" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2497" PostTypeId="2" ParentId="2495" CreationDate="2014-05-21T09:07:44.780" Score="2" Body="&lt;p&gt;It simply means that the scale is such that the relative abundance of silicon is $10^6$. For example, if you see a statement on this scale that in atom number $\mathrm{C} = 1.17\times 10^7$, that would tell you that carbon atoms are $11.7$ times as numerous as silicon ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you're probably looking at is sometimes &lt;em&gt;cosmochemical scale&lt;/em&gt;, normalized to this way. A common alternative scale in astronomy is to measure things in $10^{12}$ hydrogen atoms and logarithmically (so that abundance of hydrogen is exactly $12$ and $11$ would mean one-tenth as abundant as hydrogen).&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-05-21T09:07:44.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2498" PostTypeId="1" CreationDate="2014-05-21T10:20:39.967" Score="4" ViewCount="870" Body="&lt;p&gt;I'm parsing FITS files for a project based on data from a telescope. These files include 'DEC', 'RA' and lat long values. I understand roughly the concept of celestial coordinates and I assume that these values are sufficient to calculate the direction that the telescope is facing and the angle that it's at but my trigonometry is not up to the task.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using astropy for the project so any answers which reference that would be great.&lt;/p&gt;&#xA;" OwnerUserId="1708" LastActivityDate="2014-12-11T04:11:15.890" Title="Calculate telescope orientation based on RA, DEC and Lat/Long" Tags="&lt;telescope&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2499" PostTypeId="1" AcceptedAnswerId="2506" CreationDate="2014-05-21T13:31:19.607" Score="3" ViewCount="4288" Body="&lt;p&gt;Let's imagine that the Earth's moon is hollow and the remaining mass is structurally strong enough to retain its spherical shape (like a really big tennis ball).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this scenario, where would its center of gravity be? What would the conditions be like inside? Also would one still be able to stand on the moon?&lt;/p&gt;&#xA;" OwnerUserId="1089" LastEditorUserId="1089" LastEditDate="2017-09-20T15:33:11.563" LastActivityDate="2017-09-20T15:33:11.563" Title="Where is a Hollow Object's Center of Gravity?" Tags="&lt;gravity&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ClosedDate="2014-05-24T17:03:08.683" ContentLicense="CC BY-SA 3.0" />
  <row Id="2500" PostTypeId="1" AcceptedAnswerId="2502" CreationDate="2014-05-21T13:40:29.173" Score="3" ViewCount="157" Body="&lt;p&gt;According to Einstein's theory of Special Relativity, one's speed in relation to a relatively stationary object would 'slow down' time on the moving object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind, would it be possible to travel in a straight line away from Earth (stationary object) in a space craft able to travel at +90% the speed of light for 2 years and return with significant time dilation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What technologies have we generated that could reach speeds for significant time dilation (+10 years)?&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2014-05-22T13:58:47.490" Title="Taking Advantage of Time Dilation" Tags="&lt;space-time&gt;&lt;speed&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2501" PostTypeId="2" ParentId="2499" CreationDate="2014-05-21T13:43:35.373" Score="2" Body="&lt;p&gt;I'm not an expert but I'll share a few ideas about it:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Since the space inside of the moon would be empty there would not be any kind of pressure so the moon would end up collapsing to it's center. For example thermal pressure due to particle interactions inside a molecular cloud prevents it from collapsing.&lt;/li&gt;&#xA;&lt;li&gt;In the case the moon doesn't collapse, at it's center there would be no gravity at all since all the forces applied from every single point of the sphere would cancel.&lt;/li&gt;&#xA;&lt;li&gt;It's a very weird situation. I think it would be possible but it would depend on how thick the sphere surface is. It the gravity at a point + the gravity at the far opposite is enough to pull you i guess yes, it's possible.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I hope this gives some clues and helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: from all the above statements I think the most plausible one is first one.&lt;/p&gt;&#xA;" OwnerUserId="1678" LastEditorUserId="1678" LastEditDate="2014-05-21T13:53:11.807" LastActivityDate="2014-05-21T13:53:11.807" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2502" PostTypeId="2" ParentId="2500" CreationDate="2014-05-21T14:00:35.407" Score="3" Body="&lt;p&gt;For the fist question it is theoretically possible, and it is explained in the popular &lt;a href=&quot;http://en.wikipedia.org/wiki/Twin_paradox&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Twin Paradox&quot;&gt;Twin Paradox&lt;/a&gt; example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second one I'm not sure if there are some advances but as far as I know there is an important problem about relativistic speeds: the mass of the object traveling at such speed increases dramatically &lt;strike&gt;to the point it would weight so much it would collapse&lt;/strike&gt;(&lt;a href=&quot;https://physics.stackexchange.com/questions/29632/reletivistic-mass-as-seen-by-different-observers/29633#29633&quot; title=&quot;Relativistic Mass&quot;&gt;It wouldn't collapse actually&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="1678" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2014-05-22T13:58:47.490" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2503" PostTypeId="1" AcceptedAnswerId="5895" CreationDate="2014-05-21T14:58:54.863" Score="4" ViewCount="492" Body="&lt;p&gt;There are various modes of Cepheids like&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;single modes: &#xA;   fundamental (F), &#xA;   first overtone (1O), and likewise.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And then we have &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;double modes: F/1O, 1O/2O etc.&lt;/li&gt;&#xA;&lt;li&gt;triple modes: 1O/2O/3O etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What are these &lt;code&gt;modes&lt;/code&gt; actually? What do they mean and which frequencies do they correspond to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Stellar pulsations&lt;/code&gt; are caused by &lt;code&gt;expansions&lt;/code&gt; and &lt;code&gt;contractions&lt;/code&gt; in the outer layers as a star seeks to maintain equilibrium. &lt;strong&gt;What kind of &lt;em&gt;equilibrium&lt;/em&gt; is being referred to here?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1702" LastEditorDisplayName="user1569" LastEditDate="2016-08-18T15:06:42.920" LastActivityDate="2016-08-18T15:06:42.920" Title="Pulsation Modes of Cepheids" Tags="&lt;galaxy&gt;&lt;stellar-evolution&gt;&lt;star-systems&gt;&lt;stellar-dynamics&gt;&lt;cepheids&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2505" PostTypeId="1" AcceptedAnswerId="2511" CreationDate="2014-05-21T17:44:11.813" Score="5" ViewCount="260" Body="&lt;p&gt;In literature about the instrumentation of radio astronomy, I frequently come across a type of mixer described as &quot;2SB&quot;.  I'm familiar with single-sideband (SSB) and double-sideband (DSB) mixers, but I have never studied a 2SB mixer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a 2SB mixer and how does it compare to a DSB mixer?  Ideally, I would also like to know of formal references that directly address the topic for further reading.&lt;/p&gt;&#xA;" OwnerUserId="1710" LastActivityDate="2020-08-20T01:48:34.240" Title="2SB mixer in radio astronomy?" Tags="&lt;radio-astronomy&gt;&lt;instruments&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2506" PostTypeId="2" ParentId="2499" CreationDate="2014-05-21T18:36:44.240" Score="7" Body="&lt;p&gt;If by &quot;a hollow object&quot; you mean a spherical shell -- a theoretically perfect sphere Moon, with a perfect sphere of material removed from the center: The center of mass, and therefore the center of gravity remains at the geometric center. Ref &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation#Bodies_with_spatial_extent&quot;&gt;Newton's Law; Bodies with spatial extent&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The body of the moon is plastic; Meaning the gravitational forces are strong enough to reshape the material... it basically flows into a sphere. So if you magically removed a small amount of the center, it would collapse into a smaller sphere. If you removed most of the center, then the &quot;thin&quot; shell may or may not be strong enough to support it's shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside a hollow spherical shell, there is no net gravitational force. So you would be in a zero-G environment everywhere within the shell. From outside the shell, the gravity is the same as if all the remaining mass was located at a point at the geometric center.&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2014-05-21T18:36:44.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2507" PostTypeId="1" CreationDate="2014-05-21T19:01:50.777" Score="7" ViewCount="559" Body="&lt;p&gt;Given a fixed coordinate location on Earth (i.e. a latitude and longitude), and a fixed equatorial coordinate location of a particular astronomical object in the sky (i.e. right ascension and declination), how can I calculate the horizontal coordinates as a function of date and time on Earth?&lt;/p&gt;&#xA;" OwnerUserId="1712" LastEditorUserId="1712" LastEditDate="2014-05-21T21:27:43.997" LastActivityDate="2014-05-21T21:27:43.997" Title="Calculation of Horizontal Coordinates" Tags="&lt;fundamental-astronomy&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2508" PostTypeId="2" ParentId="2507" CreationDate="2014-05-21T20:18:52.017" Score="4" Body="&lt;p&gt;You can't, not with that information. However if you also have the date and time, you'll be good to go: there are plenty of sites in the web with this information, such as: &lt;a href=&quot;http://www.stargazing.net/kepler/altaz.html&quot; rel=&quot;nofollow&quot;&gt;RA and DEC to ALT and AZ&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-05-21T20:18:52.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2509" PostTypeId="1" AcceptedAnswerId="2512" CreationDate="2014-05-22T08:49:16.307" Score="7" ViewCount="507" Body="&lt;p&gt;When looking at the diagram of the focal plane of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble_Space_Telescope&quot; rel=&quot;noreferrer&quot;&gt;Hubble Space Telescope&lt;/a&gt; I noticed that the aperture sizes of most of the scientific instruments are very small. They occupy a very tiny fraction of the focal plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that for a spectrograph (like the &lt;a href=&quot;http://www.spacetelescope.org/about/general/instruments/cos/&quot; rel=&quot;noreferrer&quot;&gt;COS&lt;/a&gt;) you need a very small aperture to look at point sources. But why the &lt;a href=&quot;http://www.spacetelescope.org/about/general/instruments/acs/&quot; rel=&quot;noreferrer&quot;&gt;Advanced Cmaera for Surveys&lt;/a&gt; or &lt;a href=&quot;http://www.spacetelescope.org/about/general/instruments/wfc3/&quot; rel=&quot;noreferrer&quot;&gt;Wide Field Camera 3&lt;/a&gt; also have so small apertures? Most of the focal plane is not used and it must be very tedious for the guidance system to move the target object exactly into the small aperture of the selected instrument. Even though the WFC3 is called &quot;wide field&quot;, its field of view is &lt;a href=&quot;http://www.stsci.edu/hst/wfc3/documents/handbooks/currentIHB/c03_optimum_instr4.html&quot; rel=&quot;noreferrer&quot;&gt;less than 3 arc minutes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What were the reasons behind selecting these aperture sizes? Is HST designed to observe details of small targets? Are the sizes optimized for majority of observations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand the HST is taking images of large objects like nebulae and usually it is necessary to stitch a number of individual frames to create the whole image of a bigger object. For example the &lt;a href=&quot;http://hubblesite.org/newscenter/archive/releases/2007/16/image/a/format/xlarge_web/&quot; rel=&quot;noreferrer&quot;&gt;Carina nebula&lt;/a&gt; (&lt;a href=&quot;http://hubblesite.org/newscenter/archive/releases/2007/16/image/a/&quot; rel=&quot;noreferrer&quot;&gt;stitched from 48 frames&lt;/a&gt;) or the Crab nebula (&lt;a href=&quot;http://hubblesite.org/newscenter/archive/releases/2005/37/image/a/&quot; rel=&quot;noreferrer&quot;&gt;24 frames&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nZW70.gif&quot; alt=&quot;Hubble telescope focal plane&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1714" LastActivityDate="2014-05-22T16:36:14.823" Title="Instrument aperture sizes on Hubble Telescope" Tags="&lt;hubble-telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2510" PostTypeId="1" CreationDate="2014-05-22T10:58:43.473" Score="4" ViewCount="84" Body="&lt;p&gt;As we know AGNs are considered to be one of the most luminous objects to be found in the cosmos.Based on the observer, for instance on Earth,  we categorize them on the basis of the angle between the observer and the object. Suppose for instance there are two observers, one on Earth and one on some distant planet (far away from our solar system,  let it be on the opposite side of the Milky Way relative to us). Would both the observers relate an AGN belonging to the same class? In other words, can these categories be relative to their viewing? Like an object can be a QSO to us but to the other observer can the same object be a Seyfert or a Blazar or an RG? &lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1702" LastEditDate="2014-05-23T03:15:42.883" LastActivityDate="2014-05-23T03:40:35.127" Title="Active Galactic Nuclei - categorisation on the basis of observer?" Tags="&lt;galaxy&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2511" PostTypeId="2" ParentId="2505" CreationDate="2014-05-22T11:08:19.390" Score="3" Body="&lt;p&gt;2SB is Dual Sideband, as opposed to DSB - Double Sideband. Here are a couple of papers you might find relevant and interesting:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nrao.edu/meetings/isstt/papers/2009/2009257260.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;a 2SB upgrade replacing a DSB&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.das.uchile.cl/lab_mwl/doc.pdf/dsss_rfinger_uchile.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;a paper that mentions the advantages of 2SB over DSB&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Answers to &lt;a href=&quot;https://dsp.stackexchange.com/q/69815/25659&quot;&gt;What is the functional difference between DSB and 2SB mixers, and why exactly would 2SB be better for radio astronomy?&lt;/a&gt; in Signal Processing SE go into further detail.&lt;/p&gt;&#xA;" OwnerUserId="67" LastEditorUserId="7982" LastEditDate="2020-08-20T01:48:34.240" LastActivityDate="2020-08-20T01:48:34.240" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="2512" PostTypeId="2" ParentId="2509" CreationDate="2014-05-22T16:29:58.927" Score="6" Body="&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; You can't have a very wide field, and simultaneously a very high resolving power, in an instrument with a large focal ratio (or any instrument, really), unless you use sensors with an unrealistic pixel count. Details below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The linear size of the focal plane is large - &lt;a href=&quot;http://hubblesite.org/the_telescope/nuts_.and._bolts/optics/optics2.php&quot; rel=&quot;noreferrer&quot;&gt;one document&lt;/a&gt; says it's the size of a &quot;dinner plate&quot;. But the instrument is an f/24, which is a large focal ratio. At that ratio, a big linear size in the focal plane doesn't mean much in terms of angular size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I've seen, the Wide Field Camera has a total field of 4.7 arcmin, or 280 arcsec according to this document:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://asd.gsfc.nasa.gov/archive/hubble/a_pdf/news/SM2-MediaGuide.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://asd.gsfc.nasa.gov/archive/hubble/a_pdf/news/SM2-MediaGuide.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, yes, that's a small field, if you compare it with ground-based amateur astrographs, which can have a field as wide as several degrees of arc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the Hubble telescope has a very high resolving power - 0.05 arcsec. Let's take your information and assume the WFC has a total field of 3 arcmin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=3%20arcmin%20/%200.05%20arcsec&quot; rel=&quot;noreferrer&quot;&gt;3 arcmin / 0.05 arcsec = 3600 pixels&lt;/a&gt; in theory, as a linear size of the WFC&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted the WFC to have a bigger total field, you'd have to either sacrifice resolving power, or use a sensor with an enormous pixel count, which is not feasible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The WFC uses several sensors, each one of them 800 x 800 pixels.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TLDR:&lt;/strong&gt; Wide field, or great resolving power - pick one. They chose resolving power, and they compensate the narrow field by doing multiple passes.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2014-05-22T16:36:14.823" LastActivityDate="2014-05-22T16:36:14.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2513" PostTypeId="1" AcceptedAnswerId="12051" CreationDate="2014-05-22T18:32:41.560" Score="20" ViewCount="1173" Body="&lt;p&gt;&lt;code&gt;Metallicity&lt;/code&gt; of objects refers to the amount of chemical elements present in it other than Hydrogen and Helium.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Note: The other elements may or may not be actual metals in the true sense of their defintion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But why did astronomers use such a term as &lt;code&gt;metallicity&lt;/code&gt;? What is the history and cause behind coining such a term that may be (or rather actually is being) confused with &lt;code&gt;metal-content&lt;/code&gt; of that celestial object ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any scientific objective or explanation to this? I do not believe it is just a random acceptance, but perhaps there is a not-so-solid motive behind doing this. If there is, what is it?&lt;/p&gt;&#xA;" OwnerUserId="1702" LastEditorUserId="9527" LastEditDate="2020-01-29T14:49:01.253" LastActivityDate="2020-01-29T14:49:01.253" Title="Metallicity of Celestial Objects: Why &quot;Metal = Non-metal&quot;?" Tags="&lt;star&gt;&lt;galaxy&gt;&lt;history&gt;&lt;definition&gt;&lt;stellar-structure&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2514" PostTypeId="2" ParentId="2513" CreationDate="2014-05-23T00:32:43.217" Score="6" Body="&lt;p&gt;There is no 'history' behind it. Stellar physics is less than a 100 years old. Thus, it is not a terminology that came to be due to some anecdotal reason. This is the way it has been since the start. Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because we don't care. Why, really?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen and Helium are more important and abundant, so we need something to measure the others, which can be grouped under one name due to their low significance to us. One fine day, one fine astronomer was sitting in his balcony, drinking coffee, and he thought, &quot;What do I call a group of elements most of which are metals?&quot; Surprise surprise, He decided to call them &lt;strong&gt;metals.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not kidding with you, although it may have sounded like that. But that is how it is with astronomical terminologies. I am assuming you are new to the field, because this is not the only crazy thing we've done to express how much crap we give in for terms and notations and units.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2014-05-23T00:32:43.217" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2515" PostTypeId="1" AcceptedAnswerId="3717" CreationDate="2014-05-23T00:59:07.163" Score="6" ViewCount="124" Body="&lt;p&gt;As a Sun-like star is born from a collapsed nebula and generates energy by nuclear fusion some of that energy is converted to magnetic fields which fuel solar activity. How does that activity evolve as the star itself evolves?&lt;/p&gt;&#xA;" OwnerUserId="117" LastActivityDate="2014-06-20T23:20:10.133" Title="How does solar activity (e.g. flares, coronal mass ejections) change over the life cycle of a Sun-like star?" Tags="&lt;the-sun&gt;&lt;solar-flare&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2516" PostTypeId="2" ParentId="2510" CreationDate="2014-05-23T03:40:35.127" Score="8" Body="&lt;p&gt;Yes, but most probably not in your mentioned example.&#xA;You need larger distances to make a difference, I guess.&#xA;However, it is by now largely accepted a unification model for AGNs, where the difference are mostly due to the angle of line of sight.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/YvoCI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First introduced by &lt;a href=&quot;http://arxiv.org/abs/astro-ph/9506063&quot; rel=&quot;nofollow noreferrer&quot;&gt;Urry &amp;amp; Padovani&lt;/a&gt;, then developed for further observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, this is only half of the truth, since the AGNs evolve, and in this evolution feedback on the host galaxy is exerted. Also, the amount of feeding/obscuring gas is varying with time (redshift) and this has to be taken into account for a correct classification.&#xA;For the jets too, it is also possible that they forms only in certain stages of the AGN life...&#xA;As usual in astronomy, things are much more complicated than simple catalogs/models !&lt;/p&gt;&#xA;" OwnerUserId="1038" LastActivityDate="2014-05-23T03:40:35.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2517" PostTypeId="2" ParentId="2515" CreationDate="2014-05-23T06:39:13.050" Score="2" Body="&lt;p&gt;As in the Earth and more than in the the giant gaseous planets the Sun's magnetic field is generated like in a dynamo. A few facts about it:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The rotating plasma inside the Sun acts like a dynamo creating it's&#xA;magnetic field.&lt;/li&gt;&#xA;&lt;li&gt;The plasma is ionized so all interact with the magnetic field and&#xA;vice-versa.&lt;/li&gt;&#xA;&lt;li&gt;The sun as a fluid body also has &lt;strong&gt;Differential rotation&lt;/strong&gt;, that&#xA;means the equatorial part of it rotates faster than the poles. This provokes what is known as the &lt;strong&gt;Solar cycle&lt;/strong&gt; every 11 years.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;At the begining of the &lt;strong&gt;Solar cycle&lt;/strong&gt; the Sun's magnetic field is pointed North-South as in the Earth. During the cycle the differential rotation makes the plasma interact with the magnetic field causing it to be displaced. This makes the first the &lt;strong&gt;Sun's spots&lt;/strong&gt; appear near the poles. As the cycle advances the magnetic field get more and more scrambled, it's orientation changes to west-east and the spots appear closet to the equator. Finally the magnetic field can't hold the &quot;pressure&quot; and &quot;breaks&quot; like a rubber band creating the biggest spots and flares. The magnetic field stabilizes again but with a south-north polarity and the cycles begins again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at this &lt;a href=&quot;http://ircamera.as.arizona.edu/NatSci102/NatSci102/movies/solarcycle.gif&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Solar Cycle&quot;&gt;gif image&lt;/a&gt; that shows how the cycle go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in this image you can see what is know as the butterfly diagram that shows how the Sun's spots are located closer to the equator as the cycle advances. And it lasts 11 years.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/WjzxJ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1678" LastActivityDate="2014-05-23T06:39:13.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2518" PostTypeId="2" ParentId="2503" CreationDate="2014-05-23T08:12:10.503" Score="7" Body="&lt;p&gt;Some classical Cepheids pulsate simultaneously in two or even three modes.&#xA;Their lightcurves can be explained as a overposition of fundamental plus overtone modes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/u7TjS.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The terminology you used is also found &lt;a href=&quot;http://www.google.co.jp/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=1&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0CCgQFjAA&amp;amp;url=http://mnras.oxfordjournals.org/content/286/3/L33.full.pdf&amp;amp;ei=GgJ_U7-YL8yA8gWZm4DoAQ&amp;amp;usg=AFQjCNFiZaxYDMa-kXosuP-7fEv4jWf6HA&amp;amp;sig2=ECz3B_Rh2BWLhqBpCvmyqQ&amp;amp;bvm=bv.67720277,d.dGc&quot; rel=&quot;nofollow noreferrer&quot;&gt;in this paper&lt;/a&gt; (where you have also other references):&#xA;&quot;the discovery of many double-mode Cepheids (DMCs) pulsating in both the fundamental and first-overtone modes (F/10) and the first and second-overtone modes (10/20)&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other reference: &lt;a href=&quot;http://ogle.astrouw.edu.pl/atlas/classical_Cepheids.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Classical cepheids&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1038" LastActivityDate="2014-05-23T08:12:10.503" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2519" PostTypeId="1" AcceptedAnswerId="3615" CreationDate="2014-05-23T10:53:01.480" Score="3" ViewCount="123" Body="&lt;p&gt;Following &lt;a href=&quot;https://physics.stackexchange.com/questions/73964/accretion-disk-physics-stellar-formation&quot;&gt;this line&lt;/a&gt;, I am wondering about the following question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accreting pulsars in binary systems are usually thought to accrete from a prograde disk, so increasing their spin in the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why is the disk supposed to be prograde?&lt;/p&gt;&#xA;" OwnerUserId="1038" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2014-06-04T20:50:07.990" Title="Pulsars with accreting disk in binary system" Tags="&lt;neutron-star&gt;&lt;pulsar&gt;&lt;accretion-discs&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2520" PostTypeId="2" ParentId="2519" CreationDate="2014-05-23T11:56:15.463" Score="2" Body="&lt;p&gt;As far as I know all accretion disks are prograde. When the pulsar starts accreting mass from it's companion this mass is pulled to follow the pulsar's rotation.&lt;/p&gt;&#xA;" OwnerUserId="1678" LastActivityDate="2014-05-23T11:56:15.463" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2521" PostTypeId="1" AcceptedAnswerId="2523" CreationDate="2014-05-23T12:26:52.150" Score="5" ViewCount="978" Body="&lt;p&gt;When I see pictures of the Milky Way they show a spiral shape. Why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QEQJj.jpg&quot; alt=&quot;artist depiction of Milky Way from above&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1683" LastEditorUserId="1683" LastEditDate="2014-05-24T02:02:34.650" LastActivityDate="2015-06-21T18:20:58.467" Title="Why do pictures of the Milky Way look like a spiral?" Tags="&lt;milky-way&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2522" PostTypeId="2" ParentId="2521" CreationDate="2014-05-23T12:37:03.187" Score="9" Body="&lt;p&gt;The objects in a galaxy orbit it's center as a planet does with it's star, they keep the same distance. They look like a spiral for something that is called &lt;strong&gt;density waves&lt;/strong&gt; that make some parts of the galaxy (the arms) more dense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to imagine a star system like ours orbiting the galaxy in your picture following those radial lines, so sometime the system will be inside of and arm and sometimes outside of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to this extra density in the arms is where most of the star formation takes place.&lt;/p&gt;&#xA;" OwnerUserId="1678" LastActivityDate="2014-05-23T12:37:03.187" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2523" PostTypeId="2" ParentId="2521" CreationDate="2014-05-23T13:46:52.323" Score="9" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Density_wave_theory&quot;&gt;Density waves&lt;/a&gt; cause the build up of galactic material in to spiral arms. An object does not have to stay in a spiral arm; it can join or leave it as the density wave passes through the galaxy. However, the likelihood of an object being in a spiral arm is greater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape of the arms does not relate to matter falling in; this is an illusion. Objects perform elliptical or near-circular orbits around the center of the galaxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The origin of these density waves is not known conclusively, though there are many simulations that re-create the effect through satellite galaxy perturbations.&lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2014-05-23T13:46:52.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2524" PostTypeId="1" AcceptedAnswerId="2525" CreationDate="2014-05-23T16:08:34.213" Score="14" ViewCount="11068" Body="&lt;p&gt;If you were to fall into a black hole, my understanding is that from your reference point, time would speed up (looking out to the rest of the universe), approaching infinity when approaching the event horizon.  If this is correct, would you see the whole universe's future &quot;life&quot; flash before your eyes as you fall in, assuming you could somehow withstand the tremendous forces, and assuming black holes don't evaporate?  If it is correct that black holes evaporate due to Hawking radiation, would you be &quot;transported&quot; forward in time to where the black hole fully evaporates?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is considering the &quot;alternate&quot; frame of reference from my question: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/2441/does-matter-accumulate-just-outside-the-event-horizon-of-a-black-hole&quot;&gt;Does matter accumulate just outside the event horizon of a black hole?&lt;/a&gt;  In that question, I thought about what happens to matter falling into a black hole from the perspective of someone observing from the outside (e.g. as seen from earth).  Here I am considering the perspective of the thing falling into a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also takes into account ideas discussed in: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1991/why-does-time-get-slow-near-a-black-hole/2050#2050&quot;&gt;Why does time get slow near a black hole?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: This answer to another question provides some insight here too (refer to the last part of the answer): &lt;a href=&quot;https://astronomy.stackexchange.com/a/3713/1386&quot;&gt;https://astronomy.stackexchange.com/a/3713/1386&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-03-19T16:24:32.740" Title="Would time go by infinitely fast when crossing the event horizon of a black hole?" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;space-time&gt;&lt;special-relativity&gt;&lt;horizon&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2525" PostTypeId="2" ParentId="2524" CreationDate="2014-05-23T22:16:39.933" Score="18" Body="&lt;p&gt;(I will assume a Schwarzschild black hole for simplicity, but much of the following is morally the same for other black holes.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you were to fall into a black hole, my understanding is that from your reference point, time would speed up (looking out to the rest of the universe), approaching infinity when approaching the event horizon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In Schwarzschild coordinates,&#xA;$$\mathrm{d}\tau^2 = \left(1-\frac{2m}{r}\right)\mathrm{d}t^2 - \left(1-\frac{2m}{r}\right)^{-1}\mathrm{d}r^2 - r^2\,\mathrm{d}\Omega^2\text{,}$$&#xA;the gravitational redshift $\sqrt{1-\frac{2m}{r}}$ describes the time dilation of a &lt;em&gt;stationary observer&lt;/em&gt; at a given Schwarzschild radial coordinate $r$, compared to a stationary observer at infinity. You can check this easily: plug in $\mathrm{d}r = \mathrm{d}\Omega = 0$, the condition that neither the radial nor the angular coordinates are changing (i.e. stationary observer), and solve for $\mathrm{d}\tau/\mathrm{d}t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conclusion is that if you have the rocket power to hover arbitrarily close to the horizon, you will be able to see arbitrarily far into the history of the universe over your lifetime. However, that doesn't actually cover what happens to an observer that crosses the horizon. In that case, $\mathrm{d}r\not=0$, and the coefficient of $\mathrm{d}r^2$ above becomes undefined at the horizon: as in the other question, the Schwarzschild coordinate chart simply fails to cover the horizon and so is ill-suited for talking about situations cross the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that's a fault of the coordinate chart, not of spacetime. There are other coordinate charts that are better adapted to questions like that. For example, the two Eddington-Finkelstein charts are best suited for incoming and outgoing light rays, respectively, and the Gullstrand-Painlevé chart is adapted to a freely falling observer starting from rest at infinity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If this is correct, would you see the whole universe's future &quot;life&quot; flash before your eyes as you fall in, assuming you could somehow withstand the tremendous forces, and assuming black holes don't evaporate?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. I think this is best seen from the Penrose diagram of Schwarzschild spacetime:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/grB5f.png&quot; alt=&quot;Penrose diagram of Schwarazschild spacetime, modified from one by A.Hamilton&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light rays run diagonally. In blue is an example infalling trajectory, not necessarily freely falling. Note the two events where it crosses the horizon and where it reaches the singularity. Shown in red are inward light rays that intersect those events. Thus, the events that the infalling observer can see of the external universe consist of the region between those light rays and the horizon. The events occurring after that won't be seen because the the observer will have already reached the singularity by then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose the observer tries a different trajectory after crossing the horizon, accelerating outward as much as possible in order to see more of the future history of the external universe. This will only work up to a point: the best the observer can do is hug the outgoing light ray (diagonally from lower-left to upper-right) as much as possible... but since the observer is not actually allowed to go at the speed of light, seeing &lt;em&gt;all&lt;/em&gt; of the future of history will be impossible. The best the observer can do is to meet the singularity a bit more on the right of the diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, since the light ray worldlines have zero proper time, trying to do that will actually shorten the the observer's lifespan. If you're in a Schwarzschild black hole, you would live longer if you don't struggle to get out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is for an eternal, non-evaporating black hole, as that's what you're asking about here. (The 'antihorizon' is there because the full Schwarzschild spacetime is actually an eternal black hole and its mirror image, a white hole in a mirror 'anti-verse', which not shown on this diagram. That's unphysical, but not relevant to the situation we're considering here.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If it is correct that black holes evaporate due to Hawking radiation, would you be &quot;transported&quot; forward in time to where the black hole fully evaporates? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An evaporating black hole is morally the same as above: only an ideal light ray can reach the point when the black hole fully evaporates; everyone else gets the singularity. (Since this ideal light ray right along the horizon would be infinitely redshifted, arguably not even that.) You can repeat the above reasoning on its Penrose diagram yourself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2i0ao.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Addendum&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have thought a bit about this, and does this solution take into account the relativistic time effects near the horizon of the black hole (e.g. is my understanding correct that the observer would observe time in the universe passing approaching infinitely fast when approaching the event horizon)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How much time dilation happens depends entirely on what coordinates we're talking about (more generally, which frame field). What a given observer will actually see, however, is completely independent of choice of coordinates. In particular, Penrose diagrams illustrate the light cone structure of the given spacetime, and what an observer can in principle see depends entirely on what light rays intersect the observer's wordline. So yes, it's taken into account by default.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're actually falling in it, no, your understanding is mistaken, for reasons explained above. For additional motivation, flip the question around: what does the very distant stationary observer see of the infalling object? On the above Penrose diagram, outwardly directed light rays are diagonal, from lower-left to upper-right. Draw some some outward light rays from the blue infalling worldline. You will see that no matter how far into the far future (&lt;em&gt;up&lt;/em&gt; on the diagram) you pick an event outside the black hole to be, you can connect that event with an outward light ray originating from the blue infalling worldline &lt;em&gt;before&lt;/em&gt; it crosses the horizon. The conclusion would be that an observer that stays outside the black hole would be able to see the infalling object arbitrarily far into the future. No matter how much time passes for someone who stays out of the black hole, the image of the infalling object would still be visible as it was before it crossed the horizon. (In principle at least; in practice it will get too faint to see after a while.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the usual result of &quot;infinite gravitational time dilation makes the image of the infalling object hover forever near the horizon&quot; is &lt;em&gt;also&lt;/em&gt; straightforwardly deducible from the diagram, and so is completely consistent with the infalling object being able to see a finite part into the future of the external universe. Perhaps it is best to emphasize that the situation is not actually symmetric: what the external observer sees of the infalling object is not some straightforward flip-around of what the infalling object sees of the external universe. The black hole itself breaks that symmetry.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditDate="2014-06-02T02:05:05.380" LastActivityDate="2014-06-02T02:05:05.380" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2526" PostTypeId="1" AcceptedAnswerId="2536" CreationDate="2014-05-24T04:28:32.117" Score="3" ViewCount="324" Body="&lt;p&gt;How can a pre-main sequence star radiate more energy by gravitational contraction than a main-sequence star can by hydrogen fusion?&lt;/p&gt;&#xA;" OwnerUserId="1728" LastActivityDate="2014-05-25T01:10:42.817" Title="How can pre-main sequence stars radiate more energy than main-sequence stars?" Tags="&lt;star&gt;&lt;star-formation&gt;&lt;protostar&gt;&lt;main-sequence&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2528" PostTypeId="1" CreationDate="2014-05-24T14:06:21.940" Score="6" ViewCount="219" Body="&lt;p&gt;Hypothetically, if space could be moved or scaled, what would happen to everything inside?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know if an element of momentum could be transferred to objects by contorting space-time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than gravity, what effects does space-time contortion have on mass?&lt;/p&gt;&#xA;" OwnerUserId="1089" LastEditorUserId="1702" LastEditDate="2014-05-24T16:24:01.850" LastActivityDate="2014-10-03T00:34:30.387" Title="Can Space-Time Itself Have Energy Qualities Like Momentum?" Tags="&lt;space-time&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2531" PostTypeId="1" AcceptedAnswerId="2554" CreationDate="2014-05-24T20:24:41.120" Score="19" ViewCount="7352" Body="&lt;p&gt;Does our Sun have a counterpart, i.e., is it a part of a binary-system? If so, how does the other star look like and where is it? &lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2015-02-16T23:53:52.803" Title="Is Sun a part of a binary system?" Tags="&lt;the-sun&gt;&lt;binary-star&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2532" PostTypeId="2" ParentId="2531" CreationDate="2014-05-24T20:29:24.553" Score="5" Body="&lt;p&gt;No, the sun is not part of a binary system. If it was, then there would be a wobble in the stars during the year, due to sun's orbit in a binary system.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-05-24T20:29:24.553" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2533" PostTypeId="1" CreationDate="2014-05-24T20:39:09.243" Score="27" ViewCount="2925" Body="&lt;p&gt;I can't imagine the forces involved in black holes' lives. So please, help me to find out, if it is possible or not to destroy black hole in this specific way.&lt;/p&gt;&#xA;" OwnerUserId="1734" LastEditorUserId="1702" LastEditDate="2014-05-26T03:16:14.510" LastActivityDate="2021-04-13T21:25:21.230" Title="Can we (theoretically) spin the black hole so strong that it will be broken apart by centrifugal force?" Tags="&lt;black-hole&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2534" PostTypeId="2" ParentId="2533" CreationDate="2014-05-24T20:41:05.597" Score="0" Body="&lt;p&gt;As far as we know there is nothing that can stop a black hole. For this notion to make sense you must first look at what is currently known about &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_hole&quot; rel=&quot;nofollow noreferrer&quot;&gt;black holes&lt;/a&gt;. Once you get a grasp of it, then you will see that due to our current understanding of the Cosmos there is nothing we can do to black holes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that Hawking Radiation can affect a black hole, but that is only for very very small black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, in Physics there is no centrifugal force -- this is actually a misconception many people have. However, there is &lt;a href=&quot;http://en.wikipedia.org/wiki/Centripetal_force&quot; rel=&quot;nofollow noreferrer&quot;&gt;centripetal force&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WlSJS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="417" LastEditorUserId="417" LastEditDate="2014-05-24T20:46:15.653" LastActivityDate="2014-05-24T20:46:15.653" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2535" PostTypeId="2" ParentId="2533" CreationDate="2014-05-24T20:46:56.833" Score="1" Body="&lt;p&gt;Black holes can evaporate through a quantum process known as &lt;a href=&quot;http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html&quot; rel=&quot;nofollow&quot;&gt;Hawking Radiation&lt;/a&gt; and that's it.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-05-24T20:46:56.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2536" PostTypeId="2" ParentId="2526" CreationDate="2014-05-25T01:10:42.817" Score="2" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lpZDP.jpg&quot; alt=&quot;pre main sequence evolution&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although pre-main sequence stars have lower temperature, they are essentially huge clouds of gas, often as large as 1 pc wide. The luminousity being proportional to square of the radius is essentially large for pre-main sequence stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plus, the problem with gravitational contraction is not the amount of energy that can be generated per second. That can actually be larger than that produced from fusion if you insist on really massive objects. The logic, when we ruled out gravitational contraction as a source of energy for Sun, was that gravitational contraction cannot sustain it for long. Geologists knew from their dating science that Sun had to be older than grav. contraction accounts for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, admit it. Gravitational contraction makes pre-main sequence stars radiate more energy than main sequence starts, but that's just luminousity and not really energy.&#xA;Luminousity is energy per second.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2014-05-25T01:10:42.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2537" PostTypeId="2" ParentId="1448" CreationDate="2014-05-25T01:29:59.407" Score="1" Body="&lt;p&gt;Shortly after their discovery, astronomers realized there were at least two classes of GRB: short events (&amp;lt;2 seconds) and long events (&gt;2 seconds). The long GRB are widely believed to by hypernova, the explosions of massive black holes in very distant galaxies. In fact, they are much further away than even Paczynski and his followers believed at the time of the debate. Consensus on short GRB has yet to be reached, though the neutron star-neutron star merger theory, no longer applicable to long GRB, is still in play for the short ones. Also, some small percentage of short GRB are certainly magnetar flares (such as the famous March 7, 1979 event) but in other nearby galaxies.&lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2014-05-25T01:29:59.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2538" PostTypeId="2" ParentId="2533" CreationDate="2014-05-25T02:12:49.340" Score="23" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can we (theoretically) spin the black hole so strong so it will be broken apart by centrifugal force?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a Kerr-Newman (rotating, charged, isolated) black hole of mass $M$, angular momentum $J$, and charge $Q$, the surface area of the event horizon is given by&#xA;$$A = 8M\left[M^2 + (M^2-a^2-Q^2)^{1/2} - Q^2/2\right]\text{,}$$&#xA;where $a = J/M$. An &lt;em&gt;extremal black hole&lt;/em&gt; occurs when $M^2 = a^2 + Q^2$. Beyond that, if the black hole is even more overspun or overcharged, is an &quot;overextremal&quot; Kerr-Newman spacetime, which wouldn't really be a black hole at all, but rather a naked singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, I interpret your question as asking whether or not a black hole can be be spun up to the extremal limit and beyond, so as to destroy the event horizon. It's very probable that it can't be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wald proved in 1974 that as one flings matter into a black hole to try to increase its angular momentum, the nearer to an extremal black hole it is, the harder it is to continue this process: a fast-spinning black hole will repel matter that would take it beyond the extremal limit. There are other schemes, and though I'm not aware of any &lt;em&gt;completely general&lt;/em&gt; proof within classical general relativity, the continual failure of schemes like this is well-motivated by the connection between black hole dynamics and thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the Hawking temperature of the black hole is $T_\text{H} = \kappa/2\pi$, where&#xA;$$\kappa = \frac{\sqrt{M^2-a^2-Q^2}}{2M\left(M+\sqrt{M^2-a^2-Q^2}\right)-Q^2}$$&#xA;is the black hole's surface gravity. Thus, even reaching the extremal limit is thermodynamically equivalent to cooling a system to absolute zero.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-05-25T02:12:49.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2539" PostTypeId="2" ParentId="2533" CreationDate="2014-05-25T04:38:54.573" Score="2" Body="&lt;p&gt;I don't have all of the math off the top of my head, but from my conceptual understanding, it is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes have a large enough gravitational attraction that even light cannot escape even from well beyond the &quot;surface&quot; (that is if the black hole has a low enough mass that it still has a surface and has not collapsed into a singularity).  That would mean that it would have to spin fast enough that the surface is moving significantly faster than the speed of light in order to have enough linear momentum (often colloquially called &quot;centrifugal force&quot; in a circular reference frame) to escape, which according to the theory of relativity is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hawking radiation is only possible because the electromagnetic radiation is moving very nearly to orthogonally to the &quot;surface&quot; of the black hole and light can only be &quot;bent&quot; by gravity, it can't be pulled to a stop.&lt;/p&gt;&#xA;" OwnerUserId="1738" LastActivityDate="2014-05-25T04:38:54.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2540" PostTypeId="1" AcceptedAnswerId="2541" CreationDate="2014-05-25T08:59:42.547" Score="3" ViewCount="1002" Body="&lt;p&gt;&lt;code&gt;Ternary&lt;/code&gt; (or &lt;code&gt;trinary&lt;/code&gt;) stars are not the same as &lt;code&gt;triple&lt;/code&gt; stars. So, what is the basic underlying difference between these two? What are some popular examples of these two varieties?&lt;/p&gt;&#xA;" OwnerUserId="1702" LastActivityDate="2014-05-25T15:54:48.070" Title="Triple Stars v/s Ternary Stars" Tags="&lt;star-systems&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2541" PostTypeId="2" ParentId="2540" CreationDate="2014-05-25T15:54:48.070" Score="6" Body="&lt;p&gt;Exactly the difference between &lt;strong&gt;binary&lt;/strong&gt; and &lt;strong&gt;double&lt;/strong&gt; stars.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Trinary System&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Three stars gravitationally bound to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Star_system#Trinary&quot;&gt;Here&lt;/a&gt; are some examples&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Triple Star System&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Three stars that appear to be together but any one of them is gravitationally unbound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Multiple_star#Triple_star_systems&quot;&gt;Here&lt;/a&gt; are some examples.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2014-05-25T15:54:48.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2542" PostTypeId="1" AcceptedAnswerId="2543" CreationDate="2014-05-25T21:47:02.910" Score="5" ViewCount="2106" Body="&lt;p&gt;Given the horizontal coordinates of two objects in the sky, how does one go about calculating the angular separation between them?&lt;/p&gt;&#xA;" OwnerUserId="1712" LastEditorUserId="616" LastEditDate="2019-11-03T20:45:45.637" LastActivityDate="2019-11-03T20:45:45.637" Title="Calculating Angular Distance" Tags="&lt;positional-astronomy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2543" PostTypeId="2" ParentId="2542" CreationDate="2014-05-25T23:21:15.070" Score="6" Body="&lt;p&gt;You use &lt;a href=&quot;http://en.wikipedia.org/wiki/Spherical_trigonometry&quot; rel=&quot;noreferrer&quot;&gt;Spherical Trigonometry&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $A_1$ and $A_2$ are the respective azimuthal coordinates of the two objects, and $a_1$, $a_2$ their respective altitudes,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the angular seperation $\theta$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos \theta = \sin a_1 \sin a_2 + \cos a_1 \cos a_2 \cos (A_1-A_2)$$&lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="45" LastEditDate="2014-05-26T13:33:08.340" LastActivityDate="2014-05-26T13:33:08.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2544" PostTypeId="1" CreationDate="2014-05-26T07:59:09.813" Score="4" ViewCount="178" Body="&lt;p&gt;Mappings of practically all potentially dangerous asteroids is well underway. But comets seem almost hopeless to protect against.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;How far away are approaching comets detected today? How long time before they would pass Earth orbit?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is it feasible in our time to map comets in their beyond Neptune orbits, or will we only detect them while approaching?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How much more difficult is it to reach a comet compared to an asteroid? I imagine that the high speed and potentially high inclination of a comet is a challenge for rocketry. Rosetta reaches a comet soon after 10 year trip, but I suppose they picked the easiest to reach comet known.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2017-12-08T15:25:52.380" Title="How would a planetary defence against comets work?" Tags="&lt;comets&gt;&lt;impact&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2545" PostTypeId="1" AcceptedAnswerId="2574" CreationDate="2014-05-26T14:11:32.490" Score="8" ViewCount="2885" Body="&lt;p&gt;Did we ever actually see the Earth revolving around the Sun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is the geocentric model completely disproved or was it sidelined because the heliocentric model made things easier to understand? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(BTW I know Earth revolves around the Sun and am just asking out of curiosity.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1416" LastEditorUserId="1702" LastEditDate="2014-05-28T10:35:26.440" LastActivityDate="2018-03-05T19:45:44.760" Title="Did we ever actually see the earth revolving around the sun? Is the geocentric model completely disproved?" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;rotation&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2546" PostTypeId="2" ParentId="2545" CreationDate="2014-05-26T15:38:02.040" Score="6" Body="&lt;p&gt;According to Newton, any binary system revolves around the center of mass. For the Sun-Earth system, the center of mass is inside the Sun. I believe most people will then say that the Earth revolves around the Sun and not the other way around.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-05-26T15:38:02.040" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2547" PostTypeId="2" ParentId="2545" CreationDate="2014-05-26T17:06:03.663" Score="7" Body="&lt;p&gt;Well, I'd argue with our satellites that are currently orbiting various planets in our solar system (Venus Express, Mars Reconnaisance Orbiter, Cassini, etc.).&lt;br&gt;&#xA;Those have directional antennas to transmit the big chunks of data (omnidirectional antennas for low-gain communication like wake-up calls in case of a safety-mode event). So if those would not take into account the revolution of Earth around the Sun, we'd not have any scientific data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there won't be any more 'direct' and 'visible' proof than this. Ofc if the person you want to convince about the truth of the Copernican system is more open to physical arguments (and doesn't run away at the sight of a bit math) you can start by laying out the gravitational potential for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Restricted_3_body_problem&quot;&gt;restricted 3-body-problem&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Horseshoe_orbit&quot;&gt;see also&lt;/a&gt;, explain how literally none of our Space missions would have succeeded if the system were indeed Ptolemeian (lack of centrifugal forces around the sun -&gt; everything shoots in a different direction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could go on here forever, all the dynamics which is explained nicely by Newtons law of Gravity (until a certain point..) would collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning your question whether the Ptolemeian model is completely disproved: It is as disproven as it gets. If your theory makes wrong predictions, it's disproven. You can add more and more virtual spheres of higher and higher order (modify the theory) in your imagined medieval sky, but at a certain point it becomes ridiculous.&#xA;Even before the space age there was so much evidence pointing towards a Heliocentric viewpoint (Phases of Venus, Comets around the sun, discovery of Neptune and Uranus under heliocentric assumption...)&#xA;Also here I could go on forever, but if you're interested in this, best read one of the books that are free on &lt;a href=&quot;http://www.gutenberg.org/&quot;&gt;Project Gutenberg&lt;/a&gt;. They're legaly there as their copyright expired long ago and I found it very funny to read how already in the 1850's so much evidence for the Heliocentric theory was piled up.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2014-05-26T17:06:03.663" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2548" PostTypeId="2" ParentId="2044" CreationDate="2014-05-26T20:29:35.397" Score="2" Body="&lt;p&gt;When inflation occurs the stretching of the gravitational field is dependent on the perturbations within that field originally. These perturbations are caused by quantum fluctuations. This leads to fluctuations in the magnitude of the inflation itself at a given point in space, causing gravitational waves due to the gravitational potential between neighboring points. The waves are fed by the inflationary energy, but are seeded by these quantum fluctuations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.preposterousuniverse.com/blog/2014/03/16/gravitational-waves-in-the-cosmic-microwave-background/&quot; rel=&quot;nofollow&quot;&gt;This site&lt;/a&gt; gives a good explanation, as well as this particular quote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Gravitational-wave perturbations are different. They are not modulated&#xA;  by some unknown potential; they are produced by inflation, and we&#xA;  observe them directly. In straightforward models of inflation, the&#xA;  amplitude of the gravitational waves is directly proportional to the&#xA;  inflationary energy scale.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="941" LastActivityDate="2014-05-26T20:29:35.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2549" PostTypeId="1" AcceptedAnswerId="6558" CreationDate="2014-05-26T20:41:09.910" Score="12" ViewCount="301" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dark_flow&quot;&gt;Dark Flow&lt;/a&gt; has been thrown around as a term a lot lately, especially with the latest Planck results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.newscientist.com/article/dn23340-blow-for-dark-flow-in-plancks-new-view-of-the-cosmos.html#.U5M1y_ldUgo_&quot;&gt;New Scientist&lt;/a&gt; reported last year that it was now statistically improbable, but the paper by &lt;a href=&quot;http://arxiv.org/pdf/1303.6614v2.pdf&quot;&gt;Atrio Barandela&lt;/a&gt; mentioned in the article places doubt on the statistical significance of the dipole measurement by the Planck team.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know what observational limits lie on it's existence as a physical phenomena (giant attractor/multiverse/new region of spacetime/other weirdness) and not just a statistical anomaly in the overall matter distribution, I.e. what future observations could prove or disprove this theory?&lt;/p&gt;&#xA;" OwnerUserId="941" LastEditorUserId="941" LastEditDate="2014-06-07T16:13:58.090" LastActivityDate="2014-10-06T22:09:06.080" Title="Dark Flow: statistical limits on existence" Tags="&lt;gravity&gt;&lt;galaxy&gt;&lt;cosmology&gt;&lt;galactic-dynamics&gt;&lt;intergalactic-space&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2551" PostTypeId="2" ParentId="2544" CreationDate="2014-05-26T23:20:55.057" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://www.esa.int/esatv/Videos/2013/12/Rosetta_s_Journey_B-Roll/Solar_system_animation_showing_Rosetta_trajectory&quot; rel=&quot;nofollow&quot;&gt;They used several Mars and earth gravity assists to raise Rosetta's aphelion to around 5 AU&lt;/a&gt;. The comet's aphelion is also in that ball park. The comets from Jupiter's neighborhood (about 5 AU) are moving about 40 km/s when they're in our neck of the woods (earth moves about 30 km/s). A comet moving 40 km/s wrt the sun at 1 AU can be moving anywhere from 70 to 10 km/s wrt to the earth, depending on inclination. A head on collision would be 30 + 40. If the comet was moving the same direction as the earth, it'd be 40 - 30.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A comet from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kuiper_belt&quot; rel=&quot;nofollow&quot;&gt;Kuiper Belt&lt;/a&gt; would be moving just a hair under escape velocity when in our neighborhood. Escape is sqrt(2) * circular velocity. So a comet from the outer system would be moving 42 km/s wrt to the sun.  The velocities wrt to the earth could range from 72 to 12 km/s. The Kuiper Belt is 30 to 50 AU from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A comet from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Oort_cloud&quot; rel=&quot;nofollow&quot;&gt;Oort Cloud&lt;/a&gt; would also be moving a hair under escape velocity, 42 km/s in our neighborhood. But Aphelion would be much higher, up to 50,000 AU. Presumably many of the &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_non-periodic_comets&quot; rel=&quot;nofollow&quot;&gt;non-periodic comets&lt;/a&gt; are from the Oort. Click the comet links in &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_non-periodic_comets#After_1910&quot; rel=&quot;nofollow&quot;&gt;this list&lt;/a&gt; and you can compare discovery date with last perihelion. For example &lt;a href=&quot;http://en.wikipedia.org/wiki/C/2011_W3_%28Lovejoy%29&quot; rel=&quot;nofollow&quot;&gt;Comet C/2011 W3&lt;/a&gt; (Lovejoy) was discovered November 27, 2011 and it's .55 AU perihelion occurred December 16, 2011. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In JPL's &lt;a href=&quot;http://neo.jpl.nasa.gov/missions/linear.html&quot; rel=&quot;nofollow&quot;&gt;description of the LINEAR NEO&lt;/a&gt; search program, they say &quot;… with most of the efforts going into searching &lt;strong&gt;along the ecliptic plane&lt;/strong&gt; where most NEOs would be expected. &quot; But objects from the Oort don't tend to the ecliptic plane. In my opinion a high inclination comet from the Oort with our name might escape undetected until it was way too late to do anything about it. But an Oort comet with our name on it is very unlikely.&lt;/p&gt;&#xA;" OwnerUserId="1232" LastActivityDate="2014-05-26T23:20:55.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2553" PostTypeId="1" CreationDate="2014-05-27T15:04:31.760" Score="12" ViewCount="1617" Body="&lt;p&gt;The 1800's was a century of fantastic scientific discoveries in chemistry, geology, biology, engineering and so on.  Is it correct to say that astronomy did not keep up with this development? And if so, why? If not, what might be some main reasons for me (and others) having this impression?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The late 1700's saw a couple of revolutions in astronomy such as the discovery of Uranus as the first non-antique planet and the measurement of solar system distances thanks to the transits of Venus. As I understand it, the next big leaps in astronomy came in the early 1900's with the HR-diagram and the &quot;discovery&quot; of galaxies and more. The view of the universe did not seem to change much between 1800 and 1900. Nowadays every decade has its revolution such as inflation, dark matter, dark energy, exoplanets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spectroscopy, optics, photography, Doppler effect, and electricity all developed greatly during the 1800's but don't seem to have brought any revolution to astronomy until later. Is relativity and quantum physics behind this new flourishing? Had the Newtonian physics reached its limit for potentially revolutionizing astronomy already about year 1800? Or was astronomy just out of fashion, maybe because of more profitable employment for the brightest brains in the age of industrialization?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1386" LastEditDate="2014-06-24T14:22:37.870" LastActivityDate="2014-06-24T20:36:32.443" Title="Was the progress of astronomy in the 1800s surprisingly slow, and if so why?" Tags="&lt;fundamental-astronomy&gt;&lt;history&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2014-07-07T12:45:59.093" ContentLicense="CC BY-SA 3.0" />
  <row Id="2554" PostTypeId="2" ParentId="2531" CreationDate="2014-05-27T17:12:23.627" Score="20" Body="&lt;p&gt;There is no observational evidence that the sun is a member of a binary (trinary, or more) star system, where &quot;star&quot; means an object that is at least ~80 times the mass of jupiter and emits energy/light via standard hydrogen fusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some evidence that people point to is that the majority of stars in the Galaxy (perhaps 60% or so) are binary.  However, that does not mean that we are a binary system, just that we are among those which are not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Nemesis_(hypothetical_star)&quot;&gt;Nemesis star&lt;/a&gt; idea which was a hypothetical binary companion to the sun to account for periodic mass extinctions roughy every 26 million years (as in, the two stars would therefore be on a 26 million year period, so the companion is a few light-years away).  There are two problems with that:  First, that a 2010 reanalysis of the data showed that it was &lt;em&gt;too&lt;/em&gt; periodic over the past 500 million years, during which time we've gone around the galaxy twice, and so our hypothetical binary orbits should have been perturbed and not be perfect (though others say that the precision of the geologic timescale isn't good enough to say this).  But, second, we can see stars out to a few light years -- we can see small, faint stars out to thousands of light years.  We have all-sky infrared surveys that should be able to pick up the faintest even star-like objects to at least a few light-years, and yet ... nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other problem is what LDC3 pointed out in their answer: We should see some systematic motion of our own star in orbit around the common center of mass of the hypothetical binary.  This would not be a yearly wobble, but rather it would be a slow motion of the entire sky on top of the other &gt;1 year motions that we see.  We now have very accurate astronomical records dating back at least a century of star positions, especially close stars.  Even if we were on a 26 million year orbit - and especially one much shorter as some people claim - we should see effectively that our star is making a small arc, a part of a circle as it orbits the center of mass of it and the binary.  We don't.  There does not seem to be any systematic signal in the motions of stars that require the binary star model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to summarize: A binary companion simply lacks any observational data; if it existed, we should not only have been able to see it by now, but we should also observe not only the stars in our sky show a systematic motion due to our orbit around it, but we should again see THAT binary star also move very quickly, relative to other stars, as it orbits the common center of mass.&lt;/p&gt;&#xA;" OwnerUserId="934" LastActivityDate="2014-05-27T17:12:23.627" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2555" PostTypeId="1" CreationDate="2014-05-27T17:47:35.397" Score="16" ViewCount="3755" Body="&lt;p&gt;I was camping recently (May 2014), and observed several shooting stars (very fast, short lived), a few satellites (very slow, long lived) and lots of aircraft (flashing lights) in the night sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the shooting stars appeared normal, as very fast streaks across the sky, which disappeared within a fraction of a second. All except one. One started fairly straight as normal, then took a completely unexpected wavy path (like releasing a balloon to fly around the room) towards the end of its flight. Then it winked out. What caused this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, I was completely sober at the time, the sky was clear, and I believe this could not have been an aircraft or satellite because of its tremendous speed. Apart from this 'meteor', there was nothing else in the sky which could be interpreted as a firework.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could this really have been a meteor, and what could have made it take this path? If not a meteor, what else could it have been?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: I have tagged this 'UFO' because I'm not sure I can identify this object, not because I think it's an alien space craft).&lt;/p&gt;&#xA;" OwnerUserId="1756" LastEditorUserId="1469" LastEditDate="2014-05-30T08:42:03.140" LastActivityDate="2020-12-24T02:59:18.747" Title="Is a wavy path possible for a shooting star?" Tags="&lt;meteor&gt;&lt;identify-this-object&gt;" AnswerCount="1" CommentCount="13" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2556" PostTypeId="2" ParentId="2408" CreationDate="2014-05-27T20:59:21.633" Score="1" Body="&lt;p&gt;I have a friend who is an astronomy professor.  I pointed him to this post and he shared an answer I'll post on his behalf in case it's helpful to others.  His comment here references the first diagram posted by Cheeku.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An easier way to see this is to look at the circles that the Sun makes&#xA;  at different declinations (angles north or south of the equator).  The&#xA;  biggest circle is when the Sun is at the celestial equator, which&#xA;  corresponds to the equinoxes.  Since the Sun needs to complete its&#xA;  circle in 24 hours regardless of how big the circle is, the Sun's&#xA;  daily motion will be fastest near the equinoxes and so twilight will&#xA;  be the shortest then.  Comparing the two solstices, the main&#xA;  difference is that in summer the Sun doesn't get that far below the&#xA;  horizon before it turns around and starts to rise.  Its vertical&#xA;  motion at night is thus very slow.  That makes the twilight longest in&#xA;  the summer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1637" LastActivityDate="2014-05-27T20:59:21.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2558" PostTypeId="1" CreationDate="2014-05-28T06:02:45.440" Score="5" ViewCount="106" Body="&lt;p&gt;How much do we know about the structure of the torus surrounding the &lt;a href=&quot;https://en.wikipedia.org/wiki/Active_galactic_nucleus&quot; rel=&quot;nofollow noreferrer&quot;&gt;Active Galactic Nucleus&lt;/a&gt; (AGN)? Is the shape defined or still under speculations? And are there different structures for this torus possible i.e., does it vary from galaxy to galaxy?&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="37191" LastEditDate="2021-04-16T08:37:42.147" LastActivityDate="2021-04-16T08:37:42.147" Title="Structure of the torus of the Active Galactic Nucleus" Tags="&lt;galaxy&gt;&lt;accretion-discs&gt;&lt;galaxy-center&gt;&lt;active-galaxy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="2559" PostTypeId="1" CreationDate="2014-05-28T07:04:43.543" Score="3" ViewCount="39" Body="&lt;p&gt;The Swift gamma ray burst space telescope has measured a GRB according to the link below. But why the oscillations in the measured gamma rate? Why wasn't just an increase detected, but also as big decreases? The average rate doesn't look to have increased during the burst. If averaged out over a minute the line would've been flat. Is there a fundamental astronomical explanation or has it to do with the technicalities of the instrument?&#xA;&lt;a href=&quot;http://www.universetoday.com/112194/possible-gamma-ray-burst-detected-in-andromeda-would-be-closest-ever-observed/&quot; rel=&quot;nofollow&quot;&gt;http://www.universetoday.com/112194/possible-gamma-ray-burst-detected-in-andromeda-would-be-closest-ever-observed/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2014-05-29T05:19:05.717" Title="Why the lows of the GRB 2014-04-27 in Andromeda?" Tags="&lt;gamma-ray-bursts&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2560" PostTypeId="1" CreationDate="2014-05-28T07:06:09.467" Score="3" ViewCount="522" Body="&lt;p&gt;It's mentioned that the Babylonians decreed that there were exactly 360 days in a year. This statement was based upon the utilization of 360 as a standard (based on the sexagesimal system) i.e., they divided the circle into 360 equal parts.&#xA;Although, there is a mention about the Sun making one full circle across the sky in one year (as estimated by them), but, how were they able to measure it? And did they believe in the Heliocentric version of the solar system?&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1702" LastEditDate="2014-05-28T10:35:22.970" LastActivityDate="2014-05-28T10:35:22.970" Title="Did the Babylonians believe in the Heliocentric version?" Tags="&lt;the-sun&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2561" PostTypeId="2" ParentId="2560" CreationDate="2014-05-28T07:21:47.660" Score="4" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Seleucus_of_Seleucia&quot; rel=&quot;nofollow noreferrer&quot;&gt;In 150 BC&lt;/a&gt; a babylonian heliocentric model is documented. Heliocentric models were not uncommon historically, but Ptolemaios geocentric model was actually better at predicting the movements of the planets until Kepler came up with the idea of elliptic rather than circular orbits in the early 1600s. In the 1500s Copernicus hesitated to publish his heliocentric ideas because they did not improve predictions.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2014-05-28T07:21:47.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2563" PostTypeId="2" ParentId="2559" CreationDate="2014-05-28T10:55:18.137" Score="2" Body="&lt;p&gt;The caption for the graph in the article linked says &quot;Raw Data&quot;. And as far as I am acquainted with GRB measurements, the scale of oscillations would fit quite well within error margins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Final Verdict:&lt;/strong&gt; Could be anything, background noise, instrumental noise/error. It is raw data. Without the characteristics and calibration of instruments, the only relevant thing that can be said is &quot;Look, a GRB!&quot;&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2014-05-28T10:55:18.137" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2564" PostTypeId="1" CreationDate="2014-05-28T12:16:04.960" Score="3" ViewCount="279" Body="&lt;p&gt;I've got a file which contains data about photons coming from some pulsar. For each photon I know: a) time, when it was registered, b) a probability that this photon came from pulsar but not from galactic background, c) energy of a photon. My task is to find a period of pulsar. &#xA;I suppose I have to use FFT to find a period but there are two issues. First of all, there are lots of photons with low probability (background photons), so I have to somehow select relevant photons. Second, I've got an unequally spaced data, so there is a problem with using FFT.&#xA;I am a novice in data processing. Could you please give me links to some lectures or books where this topic is discussed clearly?&lt;/p&gt;&#xA;" OwnerUserId="1762" LastActivityDate="2014-06-29T17:40:37.640" Title="How to determine period of pulsar?" Tags="&lt;data-analysis&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2566" PostTypeId="1" AcceptedAnswerId="3633" CreationDate="2014-05-28T17:10:06.130" Score="12" ViewCount="503" Body="&lt;p&gt;I am looking for recommendations for an introductory text (or texts) on cosmology on the advanced undergrad or beginning grad level. I am coming from primarily a physics undergrad background (doing some undergrad research on galaxy Lyman-alpha emission and attending some of my institution's astrophysics journal club meetings) and will beginning a graduate astrophysics program in the fall. I am looking especially for texts that also provide a good introduction into GR since I have never had a chance to learn GR formally.  I have had a lot of experience with Lyman-alpha radiative transfer and some experience with large-scale structure and baryon oscillation, if there are any books that can play off those strengths.&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2014-06-07T15:22:07.843" Title="Recommendation for introductory cosmology text" Tags="&lt;cosmology&gt;&lt;general-relativity&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2567" PostTypeId="2" ParentId="2559" CreationDate="2014-05-29T05:19:05.717" Score="3" Body="&lt;p&gt;It turned out to be not real GRB event, just a &lt;a href=&quot;http://www.universetoday.com/112197/update-possible-nearby-gamma-ray-burst-alert-was-false-alarm/&quot; rel=&quot;nofollow noreferrer&quot;&gt;false alarm&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that, since the source is actually a steady X-ray source, you have a high count rate level at the base:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y6edf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not expert of GRBs, but this high-oscillation behavior seems a characteristics of GRBs: you can see &lt;a href=&quot;http://www.mpe.mpg.de/~jcg/grb090709.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;few examples here&lt;/a&gt;.&#xA;In fact, the BAT time resolution is very high ($\sim100\mu s$), then you can resolve very tiny details of the source's timing event.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the source variability, this is something more theoretical (so I can not be of much help), and there is no accepted theory, but qualitatively, in some models the gamma rays are produced by clumpy clouds of relativistically ejected material, which could account for the high variability indeed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma-ray_burst#Classification&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia classification&lt;/a&gt; .&lt;/p&gt;&#xA;" OwnerUserId="1038" LastActivityDate="2014-05-29T05:19:05.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2568" PostTypeId="1" AcceptedAnswerId="2569" CreationDate="2014-05-29T10:27:40.173" Score="6" ViewCount="3777" Body="&lt;p&gt;I know this is such a thing as a binary star where for example two stars rotate around a central point of gravity but to the naked eye are just a star.  What I do not know is how common this is (approximately)?&lt;/p&gt;&#xA;" OwnerUserId="1683" LastEditorUserId="1683" LastEditDate="2014-05-29T15:26:14.210" LastActivityDate="2014-05-29T18:47:41.303" Title="What percentage of visual stars are actually binary stars?" Tags="&lt;binary-star&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2569" PostTypeId="2" ParentId="2568" CreationDate="2014-05-29T11:13:39.707" Score="10" Body="&lt;p&gt;This is a difficult question, It's been changing through the years and it's difficult to calculate accurately. Now it's know that this percentage changes depending on the star type and ranges from &lt;strong&gt;50% for Sun like stars&lt;/strong&gt; up to a &lt;strong&gt;80% for type O stars&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fonts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.space.com/1995-astronomers-wrong-stars-single.html&quot;&gt;http://www.space.com/1995-astronomers-wrong-stars-single.html&lt;/a&gt; (2006)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.space.com/22509-binary-stars.html&quot;&gt;http://www.space.com/22509-binary-stars.html&lt;/a&gt; (2013)&lt;/p&gt;&#xA;" OwnerUserId="1678" LastEditorUserId="1378" LastEditDate="2014-05-29T18:47:41.303" LastActivityDate="2014-05-29T18:47:41.303" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2570" PostTypeId="2" ParentId="2566" CreationDate="2014-05-29T11:18:50.267" Score="6" Body="&lt;p&gt;I don't know any specific text but I can recommend you a course I took last year at &lt;a href=&quot;https://www.coursera.org/course/introastro&quot;&gt;&lt;strong&gt;Coursera.org&lt;/strong&gt;&lt;/a&gt;. It says &quot;introduction&quot; but I can assure you it is really complete, 12 week that include a lot of mathematics and physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.coursera.org/course/introastro&quot;&gt;https://www.coursera.org/course/introastro&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1678" LastActivityDate="2014-05-29T11:18:50.267" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2572" PostTypeId="1" CreationDate="2014-05-29T15:56:00.167" Score="15" ViewCount="848" Body="&lt;p&gt;Often on television shows and in articles I see it mentioned that it'd always be bad to blow up an asteroid or comet because then the energy would just be spread out and cause even more damage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to some estimates I've seen around 100 tons (or more) of meteoroids hit the Earth each day. If all of this were combined together into a single asteroid, it could destroy an entire, large city.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that alone, it makes it seem to me that it be more logical to take the chance and blow up an asteroid and thereby trim its weight down, causing a lot more to be more easily burnt up on entry, so that when it did hit, it would cause less damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this logical at all? If my science/math/physics is incorrect, I want to understand why having more burn up by spreading out is worse compared to having it more concentrated and vastly more dangerous.&lt;/p&gt;&#xA;" OwnerUserId="1768" LastActivityDate="2015-10-01T18:16:13.177" Title="Blowing up an asteroid/comet really potentially worse?" Tags="&lt;meteorite&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2573" PostTypeId="1" AcceptedAnswerId="2576" CreationDate="2014-05-29T17:02:32.723" Score="3" ViewCount="514" Body="&lt;p&gt;There is a scientific journal article having a line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;In order to study the star formation scenario in the radial direction&#xA;  of the LMC....&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is meant by 'radial direction' of a galaxy ?&lt;/p&gt;&#xA;" OwnerUserId="1702" LastEditorUserId="1702" LastEditDate="2014-05-30T05:35:13.103" LastActivityDate="2014-05-30T05:35:13.103" Title="What is meant by &quot;Radial Direction&quot; of a galaxy?" Tags="&lt;galaxy&gt;&lt;star-formation&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2574" PostTypeId="2" ParentId="2545" CreationDate="2014-05-29T17:39:05.080" Score="26" Body="&lt;p&gt;What you're asking, basically, is whether there are any proofs for the heliocentric model of the Solar System.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A literal naked-eye observation of the Earth revolving around the Sun would be rather difficult, since human beings have never gone to another planet yet, and have only been to the Moon briefly, decades ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are several proofs; some of them are historically relevant also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Kepler's laws of planetary motion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This became one of the earliest proofs, as soon as Newton figured out the law of universal gravitation, and the &quot;fluxions&quot; (what we would call today differential equations). When you assume a heliocentric model, and the inverse square law for gravity, then Kepler's laws in a heliocentric model come out of the equations naturally, as soon as you do the math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is like saying: &quot;if it's heliocentric, and knowing that the law of gravity is correct, then Kepler's laws should be such-and-such&quot;. And then: &quot;oh, but the theoretical calculations for Kepler's laws match the actual observations with great precision. Therefore, our hypothesis (heliocentric, inverse square law) must be correct.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pOlgX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was the earliest strong indication that the heliocentric model is in natural accord with the basic laws of science, whereas the geocentric view was becoming more and more contrived as evidence accumulated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tycho Brahe in late 1500s provided the enormous mass of observations of planetary motion. Johannes Kepler in early 1600s used Brahe's observations to come up with his laws empyrically (and also arguing for the heliocentric model). Isaac Newton in late 1600s said &quot;yes, Kepler is right, because of mathematics and the law of gravity, and here's the proof from calculus&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stellar parallax&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An early argument &lt;em&gt;against&lt;/em&gt; heliocentrism was that, if the Earth was really revolving around the Sun, then very distant objects, such as the stars, would appear to be ever-so-slightly bobbing back and forth around their average positions. Since that's not the case, it was argued, therefore the Earth must be fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see this argument in historical archives, proposed by theologians in the late 1600s, in favor of the geocentric model, and against the growing consensus then of &quot;natural philosophers&quot; (what we would call today scientists) that the heliocentric model was correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/91QZA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, parallax does exist, it's just very small. It was measured experimentally in the 1800s, and was then quickly used to determine, for the first time, the distance to the nearest stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_parallax&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Stellar_parallax&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Aberration of starlight&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The direction where we see a distant star also changes when Earth's speed vector changes during its revolution around the Sun. This is different from parallax; it's more akin to the way raindrops on the side windows of a car leave diagonal traces when the car starts moving (even though the raindrops fall vertically).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CO54M.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's essentially a relativistic phenomenon (when applied to light), but it can be partially explained in a classic framework. It was actually observed before parallax in the late 1600s (the heyday of Newton), but it went unexplained until early 1700s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Aberration_of_light&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Aberration_of_light&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Orbital mechanics of interplanetary probes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Landing a probe on Mars or Venus would simply not work if you assumed a geocentric model. A geocentric description of the Solar System might remain valid in a purely kinematic perspective (just the geometry of motion) as long as you remain on Earth. But the illusion breaks down quickly as soon as you start to consider dynamics (see Kepler's laws), and/or when you try to actually leave Earth (space probes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me reinforce this point, since several answers and comments got it wrong: the geocentric and the heliocentric models are not completely interchangeable, or a matter of relativity. You could build an &quot;explanatory&quot; geocentric model, and it would be &quot;correct&quot;, purely in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Kinematics&quot; rel=&quot;noreferrer&quot;&gt;kinematic&lt;/a&gt; fashion (the geometry of motion), and only as seen from Earth. But the model breaks down as soon as you consider &lt;a href=&quot;http://en.wikipedia.org/wiki/Analytical_dynamics&quot; rel=&quot;noreferrer&quot;&gt;dynamics&lt;/a&gt; (forces and masses); it would also reveal itself as incorrect even from a kinematic perspective as soon as you leave Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not just an artifice to simplify calculations. The dynamic calculations are &lt;em&gt;wrong&lt;/em&gt; in a geocentric model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to compute the very high precision trajectory of the space vehicle carrying &lt;a href=&quot;http://en.wikipedia.org/wiki/Curiosity_%28rover%29&quot; rel=&quot;noreferrer&quot;&gt;the Curiosity rover&lt;/a&gt; and successfully place it on Mars, you must operate from a heliocentric perspective. The dynamics are all wrong otherwise. You would not miss the target only a little, in a geocentric approach, you would miss it by &lt;em&gt;a lot&lt;/em&gt; - the vehicle would not even go in the general direction of Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Miscellaneous&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When observed in a telescope, Venus has phases like the Moon, and also grows and then shrinks in size, synchronized with its phases (it's large as a thin crescent, it's small when it's gibbous). In a geocentric model, the size changes could be explained by an elliptic orbit of Venus around Earth, but the phase changes synchronized with that are harder to explain. Both phenomena become trivial to explain in a heliocentric model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted that this does not necessarily prove the heliocentric model, just the fact that Venus is orbiting the &lt;em&gt;Sun&lt;/em&gt;, not the &lt;em&gt;Earth&lt;/em&gt;. So it's an argument against pure (or strict) geocentrism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter, when observed in a telescope, clearly has its own satellites. This was an early blow against a strict geocentric model, which assumed that &lt;em&gt;everything&lt;/em&gt; must orbit the Earth. It opened the door to the idea that orbits could be centered on other celestial bodies too, and to the idea that things in orbit around larger objects could have their own smaller satellites (and therefore the Earth could orbit the Sun without losing the Moon).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The list could continue (and the full list is &lt;em&gt;very&lt;/em&gt; long) but these arguments should suffice. You don't necessarily have to see something with your own eyes in order to know with certainty that it's there. In the case of Earth's revolution around the Sun, it was simply a matter of an overwhelming amount of evidence piling up in favor of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geocentrism simply doesn't make any sense whatsoever in modern science and space exploration.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2018-03-05T19:45:44.760" LastActivityDate="2018-03-05T19:45:44.760" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2576" PostTypeId="2" ParentId="2573" CreationDate="2014-05-29T19:23:19.677" Score="8" Body="&lt;p&gt;'Radial direction' typically means from the center, moving outwards. Typically, these kinds of studies will break a galaxy up into several annuli, or rings, and examine the star formation activity within each ring. (Imagine drawing a series of concentric circles, each one a little bigger than the last.) Then, if the star formation rate or star formation history changes as a function of radius, those changes can be more easily quantified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is often difficult to do this well because galaxies are not perfectly symmetric. The LMC, for instance, is a dwarf irregular galaxy, meaning its structure is not nearly as orderly as some other galaxies (e.g. The Milky Way, which is a spiral). The LMC does contain more stars near its center, and it even has a bar, but it is difficult to discern much more structure than that. To define the center of such a galaxy, the total light profile is measured and the centroid calculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we are rarely fortunate enough to be looking at another galaxy perfectly face-on, we also have to consider the effects of inclination on our observations. (If you were to hold up a plate and angle it a little bit, it would begin to look like an oval, but that doesn't mean it isn't still actually round.) If they have taken inclination into account, they should reference the &quot;deprojected radius.&quot; That means that they know that the inclination to our line of sight makes some parts of the galaxy look closer to the center than they actually are, and they corrected for that effect.&lt;/p&gt;&#xA;" OwnerUserId="477" LastActivityDate="2014-05-29T19:23:19.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2577" PostTypeId="2" ParentId="2573" CreationDate="2014-05-29T22:58:08.107" Score="1" Body="&lt;p&gt;In &lt;a href=&quot;http://en.wikipedia.org/wiki/Spherical_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;spherical coordinates&lt;/a&gt; (galaxy viewed as a ball) it would mean any vector going through the center point, in &lt;a href=&quot;http://en.wikipedia.org/wiki/Cylindrical_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;zylindrical coordinates&lt;/a&gt; (galaxy viewed as a disc) it is any vector going through the rotation axis of the disc.&lt;/p&gt;&#xA;" OwnerUserId="1774" LastActivityDate="2014-05-29T22:58:08.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2579" PostTypeId="2" ParentId="2572" CreationDate="2014-05-30T03:44:19.603" Score="6" Body="&lt;p&gt;Well there are some things to consider. Initially if you could make sure that after you blow up an asteroid you will end up with numerous but small enough pieces so that they will either: one, burn up in the atmosphere or two, be headed away from Earth (and not hitting us five years later) then we are OK, and blowing up the asteroid with a missile would be a fair solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem here lies in the fact that we know little about the internal composition of asteroids in general, and presumably even less about a particular one, so it is very hard to predict exactly where the pieces of the asteroid generated by an impact are or aren't going to end up or, be headed towards or even its size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another scenario could be that if you effectively smashed the asteroid into small pieces that could then burn into the atmosphere, and if those pieces were coincidentally to end up being consumed by Earth's atmosphere, it would heat up provoking presumably an unpleasant day on Earth of course depending on the mass of the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is a much better solution than that Armageddon-Hollywood inspired one. It is call gravitational tethering. There is something we know, and we know very well about asteroids, and that would be their trajectory paths or orbits. Even when a new asteroid is discovered, its orbit can be computed pretty quick and with great accuracy (because we know the solar system's gravity very well). So if an asteroid is to impact Earth, it is likely that we will know with years, probably decades in advance. And so we can just send a space vehicle (called &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravity_tractor&quot;&gt;gravity tractor&lt;/a&gt;), with enough mass and time in advance, and place it just beside the asteroid, hence allowing us to tilt its orbit by just a tiny amount, due to the gravitational pull between the two objects. Now when you consider the effect of that tiny amount in the long run, it effectively deflects the path of the given asteroid from that of the Earth so that it won't hit us 20 or 30 years later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is something we have control over, and something we can predict with great accuracy. It is the (safe) way to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are still not happy with my answer, you can listen to Neil de Grasse Tyson himself explaining it in &lt;a href=&quot;https://www.youtube.com/watch?v=1-ReuLZ2quc&quot;&gt;this 5 min video&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also check out this talk from the American Museum of Natural History on &quot;Defending Earth from Asteroids&quot; &lt;a href=&quot;https://www.youtube.com/watch?v=Hi54HYX9pWc&quot;&gt;LINK&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further reference &lt;a href=&quot;http://en.wikipedia.org/wiki/Asteroid_impact_avoidance&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1469" LastEditorUserId="1469" LastEditDate="2014-05-30T07:42:22.513" LastActivityDate="2014-05-30T07:42:22.513" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2580" PostTypeId="1" CreationDate="2014-05-30T03:55:14.103" Score="5" ViewCount="361" Body="&lt;p&gt;We think that the mass boundary between neutron stars and stellar mass black holes is  around three solar masses. &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum mass of the neutron stars now is two solar masses and we may find a 2.6 solar mass neutron star in the future. That is based on the equation of state.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stellar mass black holes may have different origins. Even the smallest stellar mass black hole originates in SN explosion,  we do not know the details of the explosion.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do not we think a two solar mass BH, or even a smaller BH, could exist?&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do we think that the companion of Taylor-Hulse binary is a neutron star instead of a BH? &lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditDate="2014-05-30T06:51:12.593" LastActivityDate="2014-06-06T12:30:53.300" Title="Why do we think that there is no two-solar-mass black hole?" Tags="&lt;star&gt;&lt;black-hole&gt;&lt;neutron-star&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2581" PostTypeId="1" AcceptedAnswerId="2589" CreationDate="2014-05-30T06:10:30.203" Score="3" ViewCount="712" Body="&lt;p&gt;Apart from the obvious difference as suggested by their names that the Large Magellanic Cloud is 'larger' than the Small Magellanic Cloud, what are the other differences between them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do give all the details you can. Thank you!&lt;/p&gt;&#xA;" OwnerUserId="1702" LastEditorUserId="9527" LastEditDate="2020-09-25T08:10:28.660" LastActivityDate="2020-09-25T08:10:28.660" Title="What is the difference between LMC and SMC?" Tags="&lt;star&gt;&lt;galaxy&gt;&lt;star-formation&gt;&lt;magellanic-cloud&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2582" PostTypeId="2" ParentId="2581" CreationDate="2014-05-30T07:06:06.743" Score="1" Body="&lt;p&gt;I think wikipedia is enough.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are two different satellite galaxies of our Milk Way now, ALTHOUGH maybe they were gravitationally bound before and even now there is a bridge of gas connecting them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ferdinand Magellan brought them into common knowledge, although he was not the first people to mention them in writings.&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2014-05-30T07:06:06.743" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2584" PostTypeId="2" ParentId="2580" CreationDate="2014-05-30T09:36:33.957" Score="4" Body="&lt;p&gt;As you said in your question we only know about Black Holes creating in Super Novae, which lead to relatively big BH. The reason they can't be smaller in these cases it that they are created when the core of the star reaches a critical mass that overflows both the &lt;a href=&quot;http://en.wikipedia.org/wiki/Degenerate_matter&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;electron and the neutron degeneracy&lt;/strong&gt;&lt;/a&gt;. Lower masses can be hold by those degenerate mass states and won't collapse any further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically BH with masses less than the solar one could exists if there was some method capable to compress it's mass reaching radii smaller than their &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Schwarzschild radius&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1678" LastActivityDate="2014-05-30T09:36:33.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2585" PostTypeId="1" AcceptedAnswerId="8039" CreationDate="2014-05-30T11:36:25.920" Score="9" ViewCount="1359" Body="&lt;p&gt;In the National Geographic article: &lt;a href=&quot;http://news.nationalgeographic.com/news/2009/04/090403-gamma-ray-extinction_2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://news.nationalgeographic.com/news/2009/04/090403-gamma-ray-extinction_2.html&lt;/a&gt; it is suggested that a gamma ray burst likely caused a mass extinction in earth's history.  How severe is the threat of a gamma ray burst to life on earth?  It would be nice to see a chart that considers the threat within the next hundred years as follows (I have no idea what the actual numbers are, hence I am asking):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Severity - Likelihood in 100 years:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Global extinction - 0.0000001%&lt;/li&gt;&#xA;&lt;li&gt;Major impact - 0.001%&lt;/li&gt;&#xA;&lt;li&gt;Moderate impact - 0.1%&lt;/li&gt;&#xA;&lt;li&gt;Minor impact (e.g. deplete ozone layer by 0.01%) - 10%&lt;/li&gt;&#xA;&lt;li&gt;Observable - 100% (&quot;minor&quot; gamma ray bursts are frequently detected)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Follow up questions are: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Would there be any warning of an impending severe gamma ray burst?&lt;/li&gt;&#xA;&lt;li&gt;Could we survive in gamma ray burst shelters? (it would appear being underground may be sufficient according to: &lt;a href=&quot;http://en.wikipedia.org/wiki/Radiation_protection#Shielding_design&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Radiation_protection#Shielding_design&lt;/a&gt; and &lt;a href=&quot;http://2013.org/showthread.php/188-How-does-one-prepare-sufficiently-for-gamma-radiation-in-a-shelter-etc&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://2013.org/showthread.php/188-How-does-one-prepare-sufficiently-for-gamma-radiation-in-a-shelter-etc&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Something else I found to be shocking is that the Bible appears to predict such an event.  I would rate the predicted event on the major impact level.  I asked about this in the Christian Stack Exchange here:&#xA;&lt;a href=&quot;https://hermeneutics.stackexchange.com/questions/9324/interpretations-of-revelation-168-11/9335#9335&quot;&gt;https://hermeneutics.stackexchange.com/questions/9324/interpretations-of-revelation-168-11/9335#9335&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="-1" LastEditDate="2017-04-13T12:47:47.630" LastActivityDate="2014-11-26T20:31:57.780" Title="How likely and severe is the threat of a gamma ray burst to earth?" Tags="&lt;earth&gt;&lt;cosmology&gt;&lt;life&gt;&lt;surface&gt;&lt;gamma-ray-bursts&gt;" AnswerCount="1" CommentCount="12" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2587" PostTypeId="2" ParentId="2564" CreationDate="2014-05-30T16:57:16.163" Score="2" Body="&lt;p&gt;Naively, I would say that you're looking for a periodicity in the luminosity.  What about multiplying the probability with the photon energy, and binning those data?  That would allow you to use an FFT.&lt;/p&gt;&#xA;" OwnerUserId="286" LastActivityDate="2014-05-30T16:57:16.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2589" PostTypeId="2" ParentId="2581" CreationDate="2014-05-30T21:37:48.240" Score="7" Body="&lt;p&gt;The LMC has an apparent size of about 645x550 arc mins, the SMC 320x205. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both contain several hundred million stars each. The LMC is about 14000ly in size, and is about 10 billion solar masses; the SMC is 7000ly in size, and is about 7 billion solar masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The visual magnitude of the LMC is +0.28, the SMC is +2.23.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both feature a number of interesting clusters, nebulae, and supernova remnants. Notably, LMC is home to NGC 2070, the Tarantula Nebula, and Supernova 1987A.&#xA;L&#xA;The LMC is usually considered an irregular galaxy, though it has a prominent bar, somewhat warped, and a spiral arm. The SMC is a dwarf irregular galaxy, or maybe a barred disc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LMC is the fourth largest galaxy in our local group, and the third closest to us. SMC is the fifth? largest galaxy in the local group, and the fourth closest to us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{In a previous version of this response I said: A number of websites proclaim that the LMC and SMC are gravitationally bound to the Milky Way. This has been known to be incorrect since 2007 when Hubble observations showed they are travelling too fast to be orbiting the Milky Way. &lt;a href=&quot;http://www.sciencedaily.com/releases/2007/09/070926123359.htm&quot; rel=&quot;nofollow&quot;&gt;See also&lt;/a&gt;.}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A number of websites proclaim that the LMC and SMC ARE gravitationally bound to the Milky Way. This was based on old information. More &lt;a href=&quot;http://www.sciencedaily.com/releases/2007/09/070926123359.htm&quot; rel=&quot;nofollow&quot;&gt;recent measurements&lt;/a&gt; have challenged this assumption, though &lt;a href=&quot;http://arxiv.org/pdf/0808.0104v2.pdf&quot; rel=&quot;nofollow&quot;&gt;further analysis&lt;/a&gt; has left open the door for the possibility that the LMC and SMC are gravitationally bound to the Milky Way. We can, however, safely say that the LMC and SMC &lt;strong&gt;&lt;em&gt;may&lt;/em&gt;&lt;/strong&gt; be gravitationally bound to the MW.&lt;/p&gt;&#xA;" OwnerUserId="67" LastEditorUserId="67" LastEditDate="2015-08-12T23:04:56.913" LastActivityDate="2015-08-12T23:04:56.913" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2593" PostTypeId="1" AcceptedAnswerId="3603" CreationDate="2014-05-31T03:04:01.277" Score="6" ViewCount="278" Body="&lt;p&gt;There is a principle in Geography called &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tobler%27s_first_law_of_geography&quot; rel=&quot;noreferrer&quot;&gt;Tobler's First Law of Geography&lt;/a&gt;&lt;/strong&gt; which states that &quot;Everything is related to everything else, but near things are more related than distant things.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this principle apply to star composition?  Do stars that are closer together have similar compositions than those that are far apart?  I'm guessing that stars that are formed in the same stellar nursery would have similar compositions.  Have studies been done to compare the compositions and proximity of stars?&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2014-05-31T12:00:03.657" LastActivityDate="2014-12-14T22:33:18.973" Title="Does Tobler's First Law of Geography Apply to Star Composition?" Tags="&lt;star&gt;&lt;distances&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2594" PostTypeId="1" AcceptedAnswerId="12529" CreationDate="2014-05-31T05:25:50.120" Score="10" ViewCount="86" Body="&lt;p&gt;&lt;a href=&quot;http://www.nasa.gov/topics/solarsystem/features/dream-cme.html&quot;&gt;This article&lt;/a&gt; from 2011 mentions simulations about the sputtering effect caused by a solar flare hitting the Moon:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;We found that when this massive cloud of plasma strikes the moon, it&#xA;  acts like a sandblaster and easily removes volatile material from the&#xA;  surface,&quot; said William Farrell, DREAM team lead at NASA Goddard. &quot;The&#xA;  model predicts 100 to 200 tons of lunar material – the equivalent of&#xA;  10 dump truck loads – could be stripped off the lunar surface during&#xA;  the typical 2-day passage of a CME.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The researchers said they were waiting for &lt;a href=&quot;http://en.wikipedia.org/wiki/Lunar_Atmosphere_and_Dust_Environment_Explorer&quot;&gt;LADEE&lt;/a&gt; to confirm these simulations and add new data, but I didn't find any conclusions on this subject on the &lt;a href=&quot;http://www.nasa.gov/mission_pages/LADEE/news/index.html&quot;&gt;mission website&lt;/a&gt;. I was interested if this event can create a significant static electricity difference between the 2 sides of the Moon, causing some sort of lightnings on the edges or even a mini-version of a dust storm.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2015-11-17T17:23:47.977" Title="Any new info about solar flares hitting the Moon added by the LADEE mission?" Tags="&lt;the-moon&gt;&lt;solar-flare&gt;&lt;dust&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2595" PostTypeId="1" CreationDate="2014-05-31T15:53:02.947" Score="3" ViewCount="102" Body="&lt;p&gt;A friend of mine asked me this question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If we have a guess that there may be underground water on Mars, why do&#xA;  we not send rovers with diggers?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do have a few ideas, but I want alternative views by the community.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2014-06-01T22:23:07.827" Title="Why do we not send diggers to Mars if we think there's underground water?" Tags="&lt;planet&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3595" PostTypeId="1" CreationDate="2014-06-01T06:16:01.713" Score="8" ViewCount="474" Body="&lt;p&gt;L4 and L5, the Lagrange points 60 degrees leading and trailing an orbiting body, are famous for being stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A well known example are the Trojan aseroids at the Sun Jupiter L4 and L5. Nodding to these bodies I label the central mass S, orbiting mass J and the L4 mass T:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nG3JF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrangian_point#Stability&quot; rel=&quot;nofollow noreferrer&quot;&gt;It's said S/J needs to be equal to or greater than 24.96 for the system to be stable&lt;/a&gt;. For a stable system there's a ceiling on J, it can't be more than 4% of S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: Is there a ceiling on mass T? If T were as massive as J, could the system still be stable?&lt;/p&gt;&#xA;" OwnerUserId="1232" LastEditorUserId="1232" LastEditDate="2015-02-13T16:51:13.933" LastActivityDate="2015-02-14T16:25:13.367" Title="Is there a ceiling for stable L4 or L5 masses?" Tags="&lt;orbit&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3596" PostTypeId="1" CreationDate="2014-06-01T06:31:14.200" Score="3" ViewCount="44" Body="&lt;p&gt;Few days back, the Pleiades, the Sun and the Moon were forming an almost perfect equilateral triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a term that describes such momentary geometrical patterns? &lt;code&gt;Asterism&lt;/code&gt; and &lt;code&gt;conjunction&lt;/code&gt; are close, but not exactly applicable here.&lt;/p&gt;&#xA;" OwnerUserId="337" LastActivityDate="2014-06-01T08:46:17.540" Title="Term for a momentary geometric pattern formed by astronomical objects" Tags="&lt;terminology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3597" PostTypeId="2" ParentId="3596" CreationDate="2014-06-01T08:46:17.540" Score="4" Body="&lt;p&gt;Momentary events in general (comets, supernovae, solar flares, etc.) can be lumped under the general heading of &quot;transient phenomena&quot;. 'Transient asterisms' may be a candidate term for the concept you are describing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then again, we now know that on a large enough time scale &lt;em&gt;all&lt;/em&gt; asterisms are eventually transient in nature. So one might justifiably take the point of view that the adjective &quot;transient&quot; is redundant. So perhaps calling the patterns you describe simply as &quot;asterisms&quot; is not as bad of a candidate term as you might suspect.&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2014-06-01T08:46:17.540" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3598" PostTypeId="1" CreationDate="2014-06-01T15:40:05.780" Score="5" ViewCount="232" Body="&lt;p&gt;Consider a space ship that is being drawn closer to a black hole.  The crew begins to notice the effects, and discovers that they are nearing the black hole.  They then manage to halt their progression toward the black hole.  What would be the optimal escape trajectory, assuming they have not yet wandered too close?  I am thinking it would likely either be directly away from the black hole (figure 1), or parallel to the black hole (figure 2).  However, it is conceivable that there would be some other angle that would be optimal (e.g. 45 degrees in figure 3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure 1: Go directly away from the black hole&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/b8ib8.png&quot; alt=&quot;Black hole away&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure 2: Go &quot;parallel&quot; to the black hole&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lWDe3.png&quot; alt=&quot;Black hole parallel&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure 3: Go 45 degrees from the black hole&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VUUmY.png&quot; alt=&quot;Black Hole 45 degrees&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A follow up question is, would this answer be the same for other high gravity objects, such as neutron stars, or even regular stars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say for simplicity sake that they have already managed to halt their advance toward the black hole and they are now still (not moving) in relation to the black hole. If distance factors in, it would be interesting to see what the answer is at different distances (e.g. fairly close to the event horizon where I realize a tremendous amount of energy would be required, and for far away).&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="1386" LastEditDate="2014-06-01T17:08:34.187" LastActivityDate="2014-06-01T17:08:34.187" Title="What is the optimal escape trajectory from near a black hole?" Tags="&lt;orbit&gt;&lt;black-hole&gt;&lt;gravity&gt;&lt;velocity&gt;&lt;interstellar-travel&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3599" PostTypeId="2" ParentId="3598" CreationDate="2014-06-01T16:46:04.877" Score="6" Body="&lt;p&gt;I'd like to know a little more about the geometry of the ship's trajectory. I would be asking for clarification in comments but I don't know how to put images in comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good distance away the ship is moving nearly a straight line with regard to the large mass. As the ship gets closer the path gradually bends towards the large mass. If you're still a fair distance from the star, the path can be fairly well modeled with Newtonian mechanics and the curving path can be modeled as a hyperbola. The straight line the hyperbola is gradually deviating from is called an asymptote:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dbsug.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This illustration (page 36 of my coloring book) is a hyperbola about the earth, but it could also be a hyperbola about a larger mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Escape velocity climbs as you get closer to the mass. The hyperbola's speed is sqrt(Vescape^2 + Vinf^2). I use this right triangle and the Pythagorean theorem as a memory device:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/M7ymg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you did a burn in the opposite direction from your Vinf vector, it could reduce your Vinf and drop you even closer to the mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your burn vector is at right angles to your velocity vector, it would increase your vinf (and thus raise point of closest approach aka periapsis). It would also change the direction of the hyperbola's asymptote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To better answer your question, I'd need to know more about the geometry of this scenario, what Vesc and Vinf is and how much delta V the ship was capable of. It would be helpful to know at what distance our heroes discover they're in trouble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the ship has already fallen close enough that it's traveling an appreciable fraction of c, the above doesn't apply. Conics from Newtonian mechanics is a good approximation until you get too close to the black hole. Then you'd need general relativity to model the trajectory -- and that's above my pay grade.&lt;/p&gt;&#xA;" OwnerUserId="1232" LastActivityDate="2014-06-01T16:46:04.877" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3600" PostTypeId="2" ParentId="2595" CreationDate="2014-06-01T22:23:07.827" Score="2" Body="&lt;p&gt;Sending anything to Mars is expensive. Sending a machine capable of drilling to the required depths to find subsurface water would be prohibitively expensive with current rocket payload limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also far more accessible sources of water on the surface - lakes of ice in craters, water rich minerals, and even an &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Water_on_Mars&quot; rel=&quot;nofollow&quot;&gt;equatorial frozen sea&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also water at the surface of Mars on the poles, but the conditions here are much harsher than nearer the equator, and there is less sunlight, reducing the capability of any solar powered rover.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The subsurface water is not necessarily that far below the surface, but it is also not a torrent; more likely trapped as ice particles in between the rock or in secondary minerals.&lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2014-06-01T22:23:07.827" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3601" PostTypeId="1" AcceptedAnswerId="8248" CreationDate="2014-06-02T09:23:21.273" Score="21" ViewCount="1967" Body="&lt;p&gt;Interstellar exchange of massive objects is difficult across several light years. But as the stars orbit the galaxy the distances between them change. I don't find data for neighbor star distances which covers more than a couple of a hundred thousand years. I can imagine that it is difficult to trace their orbits further given todays uncertainties in distance measurements. But I'm interested in a purely statistical/geometrical estimation of how frequent nearby Sun-star passages have been.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How many times has a star come closer than, for example, 1 light year from the Sun since it formed?&lt;/li&gt;&#xA;&lt;li&gt;How well mixed have the nearest stars become over time? What fraction of them have followed the Sun since formation and made multiple passages?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I suppose that 1 light year's distance is enough to cause disturbances in the Oort cloud and likely exhange comets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2BYKf.png&quot; alt=&quot;From Wiki&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2014-06-02T09:31:10.970" LastActivityDate="2014-12-15T00:00:11.427" Title="How often do stars pass close (~1ly) to the Sun?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;distances&gt;&lt;movement&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3603" PostTypeId="2" ParentId="2593" CreationDate="2014-06-02T11:03:31.550" Score="4" Body="&lt;p&gt;As you say stars composition depend on the composition of the molecular cloud where they were created. You have a &lt;strong&gt;molecular cloud&lt;/strong&gt; where many stars are being created at the same time, all with the same composition and age, but some are bigger than others. As time goes the bigger stars will die earlier, adding new heavier materials to the cloud which will lead to a new generation of stars with a bigger &lt;a href=&quot;http://en.wikipedia.org/wiki/Metallicity&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;metalicity&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the star's composition not only depends on where it was created but also on &lt;a href=&quot;http://en.wikipedia.org/wiki/Metallicity#Stellar_populations&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;how old it is&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way this can be checked is by studying the stars within a &lt;strong&gt;star cluster&lt;/strong&gt;. You may find &lt;a href=&quot;http://www.astro.ex.ac.uk/people/mbate/Cluster/index.html&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt; interesting.&lt;/p&gt;&#xA;" OwnerUserId="1678" LastActivityDate="2014-06-02T11:03:31.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3605" PostTypeId="2" ParentId="3601" CreationDate="2014-06-02T14:39:47.427" Score="13" Body="&lt;p&gt;I was curious about the same things. I believe it was in the astronomy stack exchange I was referred to an online data base that gives position and velocity vectors for neighboring stars. From those I put together a spreadsheet. Here's a screen capture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rYyUq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only entered 48 of the closest stars so it's by no mean an exhaustive list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like your graphic matches my estimates which is reassuring. I don't know why the Ross Stars aren't in my list, possibly the omission is an error on my part when I was entering data to the spreadsheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looks like the closest approaches are around 3 lightyears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If each star has an Oort cloud, I believe the comets' velocity with regard to our sun would be pretty close to the star's relative velocity. Slowest star wrt our solar system seems to be Gliese 729 which is moving ~14 km/s wrt the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, for example, some of Van Maanen's Oort cloud came within a light year of our sun, they would have been moving 270 km/s. Those snowballs would have zoomed in and out of our neighborhood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these distances and relative velocities I don't see much opportunity for swapping comets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's speculated that our sun swapped comets with neighboring stars when our solar system was being formed. From &lt;a href=&quot;http://en.wikipedia.org/wiki/Oort_cloud#Origin&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Recent research has been cited by NASA hypothesizing that a large&#xA;  number of Oort cloud objects are the product of an exchange of&#xA;  materials between the Sun and its sibling stars as they formed and&#xA;  drifted apart, and it is suggested that many—possibly the majority—of&#xA;  Oort cloud objects were not formed in close proximity to the Sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1232" LastActivityDate="2014-06-02T14:39:47.427" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3606" PostTypeId="1" AcceptedAnswerId="3619" CreationDate="2014-06-03T13:47:56.587" Score="3" ViewCount="850" Body="&lt;p&gt;In this &lt;a href=&quot;http://arxiv.org/pdf/1207.3797v1.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; (&quot;&lt;em&gt;The Apache Point Observatory Galactic Evolution Experiment: First Detection of High Velocity Milky Way Bar Stars&lt;/em&gt;&quot;) they plot and calculate $v_{GSR}$. $GSR$ stands for &lt;em&gt;Galactic Standard of Rest&lt;/em&gt;, but I am not sure which velocity they mean. Is it the Radial Velocity, with respect to the GSR, or something else? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe to  clarify: suppose I have a database of stars with Cartesian location and velocities, how do I calculate the $v_{GSR}$ of a single star?&lt;/p&gt;&#xA;" OwnerUserId="1309" LastEditorUserId="1309" LastEditDate="2014-06-03T13:55:36.917" LastActivityDate="2014-06-05T20:37:19.897" Title="What is $v_{GSR}$" Tags="&lt;galactic-dynamics&gt;&lt;velocity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3607" PostTypeId="1" AcceptedAnswerId="3610" CreationDate="2014-06-03T14:07:58.470" Score="5" ViewCount="703" Body="&lt;p&gt;Does the Sun have the same composition and density everywhere within it? How does it vary in other stars?&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="37191" LastEditDate="2021-04-16T08:40:46.113" LastActivityDate="2021-04-16T08:40:46.113" Title="Is the Sun homogeneous?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;density&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3608" PostTypeId="1" CreationDate="2014-06-03T14:30:38.953" Score="6" ViewCount="172" Body="&lt;p&gt;I am puzzled by this question, because by the laws between the two object the relative gravitation force should be infinite. Anyways we can take two objects where the sum of speeds are equaling c (speed of light).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer should be not, but many theoretical physicists making examples to demonstrate what happens if we closing to the speed of light. But what happens if we reach 90% of the speed of light and an object flies on the opposite direction by 10% of speed of light? By rational thinking there should be an extreme gravitational force between the two items.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these are true if I understood correctly the the General Relativity Theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can somebody point out what is the problem with my thinking if there is?&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="1799" LastEditorUserId="1799" LastEditDate="2014-06-03T14:52:01.483" LastActivityDate="2014-06-03T14:52:01.483" Title="Is it possible to have two objects moving by sum speed of light (c) in the opposite direction?" Tags="&lt;light&gt;&lt;general-relativity&gt;&lt;speed&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" ClosedDate="2014-06-04T12:58:10.910" ContentLicense="CC BY-SA 3.0" />
  <row Id="3609" PostTypeId="2" ParentId="3607" CreationDate="2014-06-03T17:44:18.003" Score="5" Body="&lt;p&gt;No it does not have the same composition everywhere. In the core hydrogen is fused into helium, so the fraction of hydrogen (denoted by $X$, between 0 and 1) decreases while the fraction of helium ($Y$) increases as time goes by. There is not much exchange of matter between core and envelope so the envelope will essentially have the same constitution as when the Sun formed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other stars the convection zone extends into the core, and for these stars there will be more exchange of the different elements within the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evolved stars (e.g. red giant stars, horizontal branch stars) often have multiple shells where different nuclear fusion processes occur. An Asymptotic branch star, for instance, has a carbon-oxygen core surrounded by a helium burning shell, surrounded by a inert helium shell, surrounded by a hydrogen burning shell, surrounded by a very large envelope consisting mostly of hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For non-degenerate matter density depends on the pressure. The deeper you descend into the star, the higher the pressure and therefore the higher the density.&lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2014-06-03T17:44:18.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3610" PostTypeId="2" ParentId="3607" CreationDate="2014-06-03T21:54:29.217" Score="7" Body="&lt;p&gt;No, the sun and all other stars do not have the same composition and density within them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;composition&lt;/strong&gt; varies with depth. Most stars overall composition reflects the interstellar medium from which they form. Convection, settling and nuclear processes then lead to layers of different elements at differing depths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fusion occurs at the core of our sun (where pressure and temperature are greatest) therefore there is a greater abundance of Helium at this depth (through conversion of hydrogen). Other stars that have entered the red giant phase will also have many more heavier elements. Very late stage stars are thought to fuse these elements progressively, leading to layers of decreasing atomic number elements stacked on top of each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Convection can also lead to mixing of elements between layers. The diagram below shows theoretically how these convective layers differ over a range of stellar masses&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CjlRU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The density increases as you move &lt;a href=&quot;https://i.stack.imgur.com/taOAy.png&quot; rel=&quot;noreferrer&quot;&gt;closer to the center&lt;/a&gt; of the star. Note how the density is less than that of water for up to half the sun's radius!&lt;/p&gt;&#xA;" OwnerUserId="941" LastEditorUserId="941" LastEditDate="2014-06-04T06:11:57.890" LastActivityDate="2014-06-04T06:11:57.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3611" PostTypeId="1" AcceptedAnswerId="3612" CreationDate="2014-06-04T12:50:32.940" Score="8" ViewCount="345" Body="&lt;p&gt;I love seeing all the long-exposure photographs of the Milky Way, but I'm going to Cherry Springs State Park in Pennsylvania next month, and I'm wondering if there are any photos that approximate what our galaxy looks like to the unaided eye. I'm really just trying to set my expectations, because I know it's not going to look like what you generally see in long-exposure photos. (Naturally, people are generally showing off their carefully crafted and enhanced images, rather than trying to show what we actually see with our eyes.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I did find &lt;a href=&quot;http://amazingsky.net/2013/02/03/the-natural-naked-eye-milky-way/&quot;&gt;this post&lt;/a&gt;, which, according to the author, comes close to approximating the view with the unaided eye. For those that have seen the Milky Way from a Dark Sky site, though, I'd be interested to hear your opinion on the relative accuracy of this image.&lt;/p&gt;&#xA;" OwnerUserId="1803" LastEditorUserId="1803" LastEditDate="2014-06-04T14:33:12.967" LastActivityDate="2015-10-10T01:21:40.140" Title="Are there are any photographs online that approximate what the Milky Way looks like to the unaided eye?" Tags="&lt;milky-way&gt;&lt;photography&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3612" PostTypeId="2" ParentId="3611" CreationDate="2014-06-04T13:14:24.230" Score="2" Body="&lt;p&gt;It may be difficult to find proper photos that emulate this, both because given the tools and the subject, most people would opt for longer exposures, and also because of the general difficulty in emulating how our brain processes visual information into an image, especially in low-light or high contrast situations like this.  I think the photo at the bottom of &lt;a href=&quot;http://utahscanyoncountry.wordpress.com/2012/06/06/astronomy-program-at-natural-bridges-national-monument/&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; may be a decent recreation of it, though it's a bit fuzzy and not so wide. &lt;em&gt;Note: There are star trails forming in the image, which is indicative of it being a longer exposure. The image (unless cropped) doesn't appear to be from a wide-angle lens - I would guess somewhere in the 35-50mm range.  Perhaps a 10 secondish exposure, but my judgement is more based on result than technique.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can see will depend a lot upon the conditions of the night. You're probably not expecting a lot of light pollution, but you also have to keep in mind that your eyes will adjust to light as well.  A cell phone or a full moon can be enough to prevent you from being able to see well.  If conditions are perfect - little to no light pollution, no light from cars, flashlights, etc, a new moon, low humidity/no clouds, etc, and the milky way is in a good position, you will be able to pick out the galaxy quite well. You should be able to notice both the high concentration of stars as well as the 'milky' lighter shade caused by the many stars too dim or distant to appear bright.&lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="1288" LastEditDate="2014-06-04T19:57:36.637" LastActivityDate="2014-06-04T19:57:36.637" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3613" PostTypeId="1" CreationDate="2014-06-04T15:10:22.937" Score="2" ViewCount="568" Body="&lt;p&gt;I'm a third year undergrad, and for the project I'm doing this summer, I need to figure out the Earth's position and velocity from the moon's point of view. From the reading I've done, it seems like there are formulae to convert from the orbital elements of the earth-moon system to a state vector describing the moon's position and velocity in the earth's reference frame. So, I'm wondering if because of the symmetry of the problem (i.e. from the moon's POV, the earth orbits it in an ellipse), there is a way to convert the easy-to-find orbital elements of the earth-moon system to the orbital elements of the earth as seen from the moon? I could then just do the conversion to the state vectors that I'm looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any sources/help is much appreciated!&lt;/p&gt;&#xA;" OwnerUserId="1800" LastActivityDate="2014-06-04T19:33:56.053" Title="Orbital Elements Transformations" Tags="&lt;the-moon&gt;&lt;orbit&gt;&lt;earth&gt;&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3614" PostTypeId="2" ParentId="3613" CreationDate="2014-06-04T19:33:56.053" Score="2" Body="&lt;p&gt;You can get moon centered position and velocity vectors at the &lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;Horizons Ephemeris Generator&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a screen capture:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/MJ5z0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Code for moon is 301, Code for earth is 399&lt;/p&gt;&#xA;" OwnerUserId="1232" LastActivityDate="2014-06-04T19:33:56.053" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3615" PostTypeId="2" ParentId="2519" CreationDate="2014-06-04T20:50:07.990" Score="2" Body="&lt;p&gt;Consider a binary system with a neutron star and a companion star. It is fair to assume that the two stars have had tidal interaction, such that their angular momentum is aligned. This is just saying that both stars rotate in the same direction as their orbital motion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now say that the companion is filling its &lt;a href=&quot;http://en.wikipedia.org/wiki/Roche_lobe&quot; rel=&quot;nofollow&quot;&gt;Roche Lobe&lt;/a&gt; and steadily transferring mass into the gravitational potential of the neutron star. &#xA;At this point there are two things we need to realize&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The transferred matter has velocity and angular momentum which is set by the motion of the companion star.&lt;/li&gt;&#xA;&lt;li&gt;The matter is being transferred in a rotating system, so from the neutron star's reference frame it is subject to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Coriolis_effect&quot; rel=&quot;nofollow&quot;&gt;Coriolis effect&lt;/a&gt;. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Then, as the matter stream moves down the potential well, it will fall behind the neutron star (Coriolis effect), only to loop around and settle into an orbit (angular momentum). The disk then rotates in the same direction as the orbital motion, making it prograde with with neutron star rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this explanation is rather simple and by no means comprehensive. There may very well be peculiar situations in which a retrograde disk forms. I know that at least for black hole binaries there are some systems suggested to be retrograde. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details on mass transfer you can look at &lt;a href=&quot;http://eagle.phys.utk.edu/guidry/astro411/lectures/lecture_ch13.pdf&quot; rel=&quot;nofollow&quot;&gt;this file&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="494" LastActivityDate="2014-06-04T20:50:07.990" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3616" PostTypeId="2" ParentId="1234" CreationDate="2014-06-04T21:16:38.163" Score="2" Body="&lt;p&gt;I always prefer answers which try to avoid any formula and reply on argumentation instead.  Concerning the part of the question why not all orbits are circular, an argumentation would be like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a stationary star and a moving planet.  For each impulse the planet can have, a curve for its further movement can be predicted.  If this impulse is directed &lt;em&gt;exactly orthogonal&lt;/em&gt; to the line from the star to the planet, and if the velocity has the &lt;em&gt;exact amount&lt;/em&gt;, then this curve of movement can be an exact circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for every deviance of this one exact impulse, the resulting curve cannot be a circle:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the speed is too low, the planet will &lt;em&gt;fall&lt;/em&gt; towards the star (in the extreme case of an impulse of zero, this fall will be in a straight line).&lt;/li&gt;&#xA;&lt;li&gt;If the speed is too high, the planet will gain distance from the star (similar to a slingshot).&lt;/li&gt;&#xA;&lt;li&gt;If the impulse is not directly orthogonal to the line to the star, the first movement will move the towards or from the star, so again the curve will not be a circle.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, one can simply argue, a circle is a very special case for the curve a planet can take around a star.&lt;/p&gt;&#xA;" OwnerUserId="1808" LastActivityDate="2014-06-04T21:16:38.163" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3617" PostTypeId="2" ParentId="3595" CreationDate="2014-06-05T04:15:53.953" Score="7" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Giant_impact_hypothesis&quot; rel=&quot;nofollow&quot;&gt;Giant Impact Hypothesis&lt;/a&gt;, a theory of how the moon was created, says that once it exceeds 10% of the mass of your 'J', the orbit of an L4 or L5 (your 'T') destabilizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Possible origin of Theia&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 2004, Princeton University mathematician Edward Belbruno and astrophysicist J. Richard Gott III proposed that Theia coalesced at the L4 or L5 Lagrangian point relative to Earth (in about the same orbit and about 60° ahead or behind), similar to a trojan asteroid. Two-dimensional computer models suggest that the stability of Theia's proposed trojan orbit would have been affected when its growing mass exceeded a threshold of approximately 10% of the Earth's mass (the mass of Mars). In this scenario, gravitational perturbations by planetesimals caused Theia to depart from its stable Lagrangian location, and subsequent interactions with proto-Earth led to a collision between the two bodies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1810" LastEditorUserId="1810" LastEditDate="2015-02-13T22:43:40.207" LastActivityDate="2015-02-13T22:43:40.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3618" PostTypeId="1" AcceptedAnswerId="3620" CreationDate="2014-06-05T10:26:21.083" Score="4" ViewCount="384" Body="&lt;p&gt;If it were possible to replace the Moon with Mars, how distant would it have to be to essentially create the same oceanic tides as the Moon currently does? Mars seems to be roughly 3 times the mass of the Moon, so does that mean it should be 300% the distance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, would the increased distance cause any issues with Earth's orbit around the Sun or potentially cause some sort of situation like Charon and Pluto where the centre of mass is between them?&lt;/p&gt;&#xA;" OwnerUserId="1768" LastEditorUserId="1768" LastEditDate="2014-06-05T17:20:12.313" LastActivityDate="2014-06-06T06:37:03.017" Title="If Mars orbited the Earth how distant would it have to be to cause the same tides?" Tags="&lt;gravity&gt;&lt;mars&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3619" PostTypeId="2" ParentId="3606" CreationDate="2014-06-05T13:29:18.400" Score="3" Body="&lt;p&gt;The $v_{GSR}$ in the linked paper seems to be the radial velocity (RV, in $km.s^{-1}$) with respect to the GSR. This information is mentioned in the Abstract section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know exactly what kind of coordinates you have and how exactly are your velocities expressed (Cartesian vectors?), but what you probably need to do is to convert the velocity vector of a star into the radial component using the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r=\sqrt{x^2 + y^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to make sure where the origin of your coordinates is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.atnf.csiro.au/people/Tobias.Westmeier/tools_hihelpers.php#restframes&quot; rel=&quot;nofollow&quot;&gt;This link&lt;/a&gt; clarifies what GSR is. Basically there are several frames of reference that are used on the galactic scales. LSR (Local Standard of Rest) is co-rotating with the gas and dust of the galaxy (so an average star is at rest in LSR), while GSR is not rotating with the galaxy and can be calculated from LSR using a formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_{GSR}=v_{LSR}+220sin(l)cos(b)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where l and b are the Galactic longitude and latitude.&lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditDate="2014-06-05T20:37:19.897" LastActivityDate="2014-06-05T20:37:19.897" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3620" PostTypeId="2" ParentId="3618" CreationDate="2014-06-05T14:25:26.840" Score="5" Body="&lt;p&gt;The Tidal acceleration between 2 bodies is calculated with this formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;a_T = \frac{2GM_MR_1}{Dm^3}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $M_M$ is Mar's mass, $R_1$ Earth Radius and $D_m$ the distance to the Moon. If you equal this to Moon's Tidal acceleration you will get $D_M$ as distance to Mars to get the same Tidal acceleration having $\frac{M_M}{M_m}=8.73328184501$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;D_M = \sqrt[3]{8.73}D_m\simeq2.06Dm&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1678" LastEditorUserId="1678" LastEditDate="2014-06-06T06:37:03.017" LastActivityDate="2014-06-06T06:37:03.017" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3621" PostTypeId="1" AcceptedAnswerId="3642" CreationDate="2014-06-06T07:55:57.450" Score="5" ViewCount="1576" Body="&lt;p&gt;I've never fully understood why we can still see galaxies that are 10 billion light years away. The age of the universe is calculated to be 13.9 billion years old and we live on a planet that it is roughly 4 billion years old. The universe came into existence 13.9 billion years ago and expanded incredibly quickly but those galaxies that were created in those early years are still visible to us via the amazing Hubble telescope (&lt;a href=&quot;http://hubblesite.org/newscenter/archive/releases/2014/27/&quot; rel=&quot;noreferrer&quot;&gt;see the latest from Hubble&lt;/a&gt;). Those galaxies would surely have long gone so why can we still see the light from so long ago? How can we see further and further back to the beginning of the universe because surely the light from the moment of the big bang and subsequent eons would have past by us long ago?&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="1818" LastEditDate="2014-06-06T08:47:56.670" LastActivityDate="2016-12-18T18:07:28.790" Title="Why can we still see 10 billion year old galaxies?" Tags="&lt;big-bang-theory&gt;&lt;hubble-telescope&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3622" PostTypeId="2" ParentId="3621" CreationDate="2014-06-06T08:49:13.877" Score="3" Body="&lt;p&gt;The deepest we look into the space the further in the past we look. So if we take pictures of galaxies that are 10 billion ly away we see them as they were 10 billion years ago.&lt;/p&gt;&#xA;" OwnerUserId="1678" LastActivityDate="2014-06-06T08:49:13.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3623" PostTypeId="2" ParentId="3621" CreationDate="2014-06-06T09:02:31.127" Score="9" Body="&lt;p&gt;Those galaxies were 10 billion light years away from Earth. So light would take much more time to reach here and that light which is now 10 billion years older can be seen now. Even light from the Sun takes 8 minutes to reach to us. So if somehow sun disappears suddenly(very unlikely) we wouldn't know for 8 minutes.&lt;/p&gt;&#xA;" OwnerUserId="1669" LastEditorUserId="1669" LastEditDate="2014-12-10T06:22:43.580" LastActivityDate="2014-12-10T06:22:43.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3624" PostTypeId="2" ParentId="2580" CreationDate="2014-06-06T12:30:53.300" Score="1" Body="&lt;p&gt;The mass of a black hole is dependent on the formation mechanism, which as far as we are aware is predominantly as a supernovae remnant. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Tolman%E2%80%93Oppenheimer%E2%80%93Volkoff_limit&quot; rel=&quot;nofollow&quot;&gt;Tolman–Oppenheimer–Volkoff limit&lt;/a&gt; provides the theoretical lower mass bound of a black hole formed from a collapsing star; this is between 1.5-3.0 solar masses depending on the interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must also remember that the mass of a Black Hole is not static; due to Hawking radiation and absorbtion from the Cosmic Microwave Background it can grow or shrink &lt;a href=&quot;http://www.einstein-online.info/elementary/quantum/evaporating_bh/?set_language=en&quot; rel=&quot;nofollow&quot;&gt;depending on its size&lt;/a&gt;. The following wiki quote sums it up well&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A stellar black hole of one solar mass has a Hawking temperature of&#xA;  about 100 nanokelvins. This is far less than the 2.7 K temperature of&#xA;  the cosmic microwave background radiation. Stellar-mass or larger&#xA;  black holes receive more mass from the cosmic microwave background&#xA;  than they emit through Hawking radiation and thus will grow instead of&#xA;  shrink. To have a Hawking temperature larger than 2.7 K (and be able&#xA;  to evaporate), a black hole needs to have less mass than the Moon.&#xA;  Such a black hole would have a diameter of less than a tenth of a&#xA;  millimeter.[87]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If a single solar mass black hole did form through some exotic process it would grow through absorbtion of the CMB. &lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2014-06-06T12:30:53.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3625" PostTypeId="1" AcceptedAnswerId="3632" CreationDate="2014-06-06T21:53:28.113" Score="7" ViewCount="28569" Body="&lt;p&gt;If this is the wrong group please direct me to the correct one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems intuitively obvious that the amount of daylight per annum should be the same for any latitude on earth. For example, 12 hours per day at the equator. The poles have daylight for half the year and darkness for the other half (crudely).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to get an answer to this apparently simple question - is the annual amount of daylight the same at any point on earth?&lt;/p&gt;&#xA;" OwnerUserId="1821" LastActivityDate="2018-02-23T22:34:24.247" Title="Average amount of annual daylight at any place on earth" Tags="&lt;orbit&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3626" PostTypeId="1" CreationDate="2014-06-07T01:27:14.033" Score="5" ViewCount="754" Body="&lt;p&gt;Are there other fragments aside from the Moon that resulted from the collision of Earth and &quot;Theia&quot; (as it's called in recent main-stream press stories)?  And if not, why not?  And what happened to &quot;Theia&quot; (like were did it go?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has there been any consideration or hypothesizing that Earth may have had a larger orbit (say between Mars/Jupiter) until a large body (Theia) bashed into it, sending Earth closer to the sun along with its new satellite, while also creating the asteroid-belt?  Could something like Ceres be the missing &quot;Theia&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know it sounds a bit far-fetched, but then so does the whole concept of the Moon being born from a planet collision with Earth (with very little evidence to back it up, at least till recently).&lt;/p&gt;&#xA;" OwnerUserId="1826" LastEditorUserId="22074" LastEditDate="2019-09-26T06:54:04.433" LastActivityDate="2019-09-26T20:36:32.180" Title="Connection between Earth, Moon, Theia, and asteroid belt?" Tags="&lt;the-moon&gt;&lt;earth&gt;&lt;asteroid-belt&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3628" PostTypeId="1" CreationDate="2014-06-07T05:17:23.700" Score="23" ViewCount="1199" Body="&lt;p&gt;Please forgive: I am a layman when it comes to physics and cosmology, and have tried finding an answer to this that I can understand, with no luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, the solar system evolved from a massive molecular cloud. To me, this seems to break the second law of thermodynamics, as I think it suggests order from disorder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there must be something wrong with my logic, but am really stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain this one in layman's terms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Posting to both &quot;Astronomy&quot; and &quot;Physics&quot;, as it seems to overlap these subjects)&lt;/p&gt;&#xA;" OwnerUserId="1828" LastEditorUserId="6" LastEditDate="2021-02-22T17:03:08.073" LastActivityDate="2021-02-22T17:03:08.073" Title="How does the evolution of a solar system not break the second law of thermodynamics?" Tags="&lt;astrophysics&gt;&lt;cosmology&gt;&lt;stellar-evolution&gt;&lt;stellar-astrophysics&gt;&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3629" PostTypeId="2" ParentId="3628" CreationDate="2014-06-07T12:11:24.567" Score="4" Body="&lt;p&gt;This comes from a misunderstanding of local and absolute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is nothing to prevent a local increase in order - overall, order still decreases (or in common terminology, entropy increases)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;According to the second law of thermodynamics the entropy of an isolated system never decreases, because isolated systems spontaneously evolve towards thermodynamic equilibrium, the configuration with maximum entropy. Systems which are not isolated may decrease in entropy. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the universe is considered an isolated system, but our local solar system is not isolated, so our local decrease in entropy does not breach the 2nd law of thermodynamics as the overall entropy of the universe does not decrease.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2014-06-07T12:11:24.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3631" PostTypeId="2" ParentId="3628" CreationDate="2014-06-07T14:25:18.707" Score="18" Body="&lt;p&gt;The total entropy actually increases, as the molecular cloud shrinks under gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may seem that as the molecules are getting closer, they are more ordered, which means less entropy. That is however only one part of the process. The second (important) part is: when the molecules are closer, they also have higher kinetic energy (since they descended into lower gravitational potential). So the gas is getting hotter, as it shrinks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The increase in temperature of the gas is increasing its entropy, because the molecules occupy more momentum space. This increase in entropy via temperature is bigger than the decrease in entropy via the shrinking itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later on, the hot condensed gas (or hot planet) radiates the heat into the space and cools down. You end up with a cold planet which indeed has lower entropy than the original gas cloud, because it is no longer hot. But the entropy increase was carried away by the radiated photons. So in total - entropy of the universe increased (the radiated photons are out there somewhere).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find some more detailed discussion of this topic on the excellent &lt;a href=&quot;http://math.ucr.edu/home/baez/entropy.html&quot; rel=&quot;noreferrer&quot;&gt;web page by John Baez&lt;/a&gt; or &lt;a href=&quot;https://physics.stackexchange.com/questions/4546/please-clarify-how-entropy-increases-when-matter-gravitationally-coalesces&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2014-06-07T14:25:18.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3632" PostTypeId="2" ParentId="3625" CreationDate="2014-06-07T14:53:47.317" Score="11" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sunshine_duration&quot;&gt;Wikipedia&lt;/a&gt; strikes again:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The naive expectation is that, for every place on Earth, the Sun will&#xA;  appear to be above the horizon for exactly half the time. Thus, for a&#xA;  standard year consisting of 8760 hours, apparent maximal daytime&#xA;  duration would be 4380 hours. However, there are physical and&#xA;  astronomical effects which change that picture. Namely, atmospheric&#xA;  refraction allows the Sun to be still visible even when it physically&#xA;  sets below the horizon line. For that reason, average daytime&#xA;  (disregarding cloud effects) is longest in polar areas, where the&#xA;  apparent Sun spends the most time around the horizon. Places on the&#xA;  Arctic Circle have the longest total annual daytime of 4647 hours,&#xA;  while the North Pole receives 4575. Because of elliptic nature of the&#xA;  Earth's orbit, the Southern Hemisphere is not symmetrical: Antarctic&#xA;  Circle at 4530 hours receives 5 days less of sunshine than its&#xA;  antipodes. The Equator has the total daytime of 4422 hours per&#xA;  year.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Further details &lt;a href=&quot;http://en.wikipedia.org/wiki/Daytime_%28astronomy%29&quot;&gt;here&lt;/a&gt; on astronomical causes of average daytime variation, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Insolation&quot;&gt;here&lt;/a&gt; on Insolation, the solar radiation received at the top of the atmosphere and its effects on the energy received at ground level.&lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2014-06-07T14:53:47.317" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3633" PostTypeId="2" ParentId="2566" CreationDate="2014-06-07T15:22:07.843" Score="6" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0470848359&quot;&gt;This book&lt;/a&gt; by Andrew Liddle is fantastic for explaining the basic concepts and giving the reader a good mental picture of the ideas, but it may be a little more basic than what you're looking for (I still keep it as a good reference when reading more advanced texts). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're looking for something more advanced &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471489093&quot;&gt;Peter Coles &amp;amp; Francesco Lucchins&lt;/a&gt; guide may be of interest. Unfortunately some of the discussions have already dated a little (there is a large section on non-gaussianity if I remember rightly, whose proponents have been all but silenced by the latest Planck results). &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/052141072X&quot;&gt;This&lt;/a&gt; is another good resource I went to when completing my graduate cosmology course, but I couldn't recommend it on it's own; many of the derivations are piecemeal at best, so ideally read in conjunction with a more thorough text.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure if you came across &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691130272&quot;&gt;this&lt;/a&gt; text during your undergrad investigations but it is basically the bible of galactic physics along with &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691025657&quot;&gt;this&lt;/a&gt;. It goes in to some of the touch points of galaxy physics and cosmology which may prove useful for your understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521546230&quot;&gt;this&lt;/a&gt; may be of interest - I can't vouch for it but it popped up when looking for links to the books above, has good reviews and seems to be at the level you're looking for as well as discussing some topics you may already have exposure to. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all the above I'd recommend getting them out from a library first and working thoroughly through a few chapters to get a feel for the text before splashing out.&lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2014-06-07T15:22:07.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3634" PostTypeId="1" AcceptedAnswerId="3639" CreationDate="2014-06-08T09:29:45.737" Score="4" ViewCount="1488" Body="&lt;p&gt;If we take the convex hull of the 3D coordinates of the stars in Ursa Major and Ursa Minor: which hull has larger volume?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is meant semi-humorous but we might also learn some interesting points from good answers, e.g. which metric to use when mass distorts space or how to measure with respect to frames of reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: to those asking about which stars to consider, just choose e.g. the canonical 7 stars of each. But this question is less about one arbitrary number and more about which things to consider, e.g. to show beginners that a bright star is not always close (and the other way around), and that size can be different in 3d than our arbitrary 2d-perspective.&lt;/p&gt;&#xA;" OwnerUserId="1836" LastEditorUserId="1836" LastEditDate="2015-05-16T06:35:39.893" LastActivityDate="2015-05-16T06:35:39.893" Title="Which is really larger, Big Dipper or Small Dipper, in 3D" Tags="&lt;constellations&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3635" PostTypeId="1" CreationDate="2014-06-08T20:04:58.073" Score="18" ViewCount="3054" Body="&lt;p&gt;The universe is about 13.8 billion years old. Assume at the big-bang it starts from a small region and the maximum possible speed according Einstein is the speed of light how can the universe got a radius of 46 billion light years? According my understanding it could have a max. radius of 13.8 billion light years only. Can someone explain to a non-astrophysicists.&lt;/p&gt;&#xA;" OwnerUserId="1838" LastEditorUserId="14078" LastEditDate="2016-10-17T18:20:08.780" LastActivityDate="2016-10-17T18:20:08.780" Title="How can the 13.8 billion years old universe have a radius of 46 billion light years?" Tags="&lt;universe&gt;&lt;big-bang-theory&gt;&lt;size&gt;" AnswerCount="3" CommentCount="1" ClosedDate="2016-09-02T13:01:28.727" ContentLicense="CC BY-SA 3.0" />
  <row Id="3637" PostTypeId="2" ParentId="3635" CreationDate="2014-06-08T22:09:55.543" Score="8" Body="&lt;p&gt;You assumption is wrong. Universe can( and is ) expand faster than the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The photon emitted towards our planet in the early universe, had to traverse a universe that is expanding. That photon experienced redshift, that means his wavelenght increased and frequency decreased. We measure that photon's wavelength, compare it to a photon that was not redshifted and calculate the scale, which is then applied to the radius of the observable universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reading this Wikipedia page should give you more complete understanding of the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Observable_universe&quot;&gt;http://en.wikipedia.org/wiki/Observable_universe&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="818" LastActivityDate="2014-06-08T22:09:55.543" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3638" PostTypeId="2" ParentId="3634" CreationDate="2014-06-09T01:54:33.857" Score="1" Body="&lt;p&gt;It's hard to say without some research, but I don't think the answer really matters much. Both constellations are chance alignments that have little or nothing to do with how close the stars really are to each other. It's not hard to envision a constellation composed of a few nearby stars and a few galaxies millions of light years away.&lt;/p&gt;&#xA;" OwnerUserId="562" LastEditorUserId="562" LastEditDate="2014-06-09T15:09:50.470" LastActivityDate="2014-06-09T15:09:50.470" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3639" PostTypeId="2" ParentId="3634" CreationDate="2014-06-09T06:29:41.707" Score="7" Body="&lt;p&gt;To calculate that you will have to build a 3D model based upon the distance of each of their stars and then calculate the resulting 3D object's volume. I know this may not answer the question, but I give the way to answering it. It's not an easy job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making a quick guess based upon the distances of the stars of both constellations in my opinion I guess the &lt;strong&gt;Small Dipper&lt;/strong&gt; has a bigger volume since it's starts have bigger distance differences &lt;strong&gt;ranging from &amp;lt; 100 ly up to &gt; 500 ly&lt;/strong&gt;. In the case of the Big Dipper all distance are very similar since all stars come from the same open cluster. Their distance range between &lt;strong&gt;78 and 124 ly&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_stars_in_Ursa_Minor&quot; rel=&quot;nofollow&quot;&gt;Small Dipper stars&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Big_Dipper#Stars&quot; rel=&quot;nofollow&quot;&gt;Big Dipper stars&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Useful resources:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://astro.unl.edu/classaction/animations/coordsmotion/bigdipper.html&quot; rel=&quot;nofollow&quot;&gt;Big Dipper 3D model&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1678" LastEditorUserId="1678" LastEditDate="2015-05-15T08:11:30.160" LastActivityDate="2015-05-15T08:11:30.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3641" PostTypeId="1" CreationDate="2014-06-09T07:44:05.790" Score="3" ViewCount="1108" Body="&lt;p&gt;There's a limit to how well sunlight can be focused by a parabolic reflector or a lens because the Sun is not a point source. I wonder if the sunlight could be engineered to work as if it was (much more similar to) a point source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If almost all of the Solar disc was covered by a coronograph, the light which gets through a small hole would be (almost) a point source. Is that correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If then an array of such covered reflectors or lenses concentrated their light to one and the same point, would a better focus be achieved than without covers?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2014-06-09T13:28:30.637" Title="Can the Sun be used as a point source of light to achive better focus?" Tags="&lt;the-sun&gt;&lt;optics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3642" PostTypeId="2" ParentId="3621" CreationDate="2014-06-09T11:15:30.493" Score="4" Body="&lt;blockquote&gt;How can we see further and further back to the beginning of the universe because surely the light from the moment of the big bang and subsequent eons would have past by us long ago?&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A simple intuitive way to understand that part of your question:&#xA;Bigbang happened everywhere, in all of space. As time passes we see more distant parts of the bigbang, or rather, of the microwave background which is the furthest away phenomena we can see. The bigbang as it happened for example 1 billion years ago has passed by us long ago. Today we see the bigbang as it happened 13.8 billion years ago.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2014-06-09T11:15:30.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3643" PostTypeId="2" ParentId="3641" CreationDate="2014-06-09T13:28:30.637" Score="2" Body="&lt;p&gt;No, it won't work. After all, it is recommended to make a pinhole camera to watch a solar eclipse. What you suggested is similar to a pinhole camera. Why it won't work, is that the rays that pass through the pinhole are not parallel, but divergent. Light from a point source would have (nearly) parallel rays.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-06-09T13:28:30.637" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3644" PostTypeId="1" AcceptedAnswerId="3647" CreationDate="2014-06-10T03:43:06.343" Score="12" ViewCount="854" Body="&lt;p&gt;Can the Sun be used as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_lens&quot; rel=&quot;noreferrer&quot;&gt;gravitational lens&lt;/a&gt; to achieve better telescopic viewing? Can this effect be practically used to view celestial objects?&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="7991" LastEditDate="2016-01-25T16:36:56.840" LastActivityDate="2016-02-06T03:29:48.817" Title="Using the Sun as a Gravitational Lens" Tags="&lt;the-sun&gt;&lt;solar-system&gt;&lt;gravity&gt;&lt;telescope&gt;&lt;gravitational-lensing&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3645" PostTypeId="2" ParentId="3644" CreationDate="2014-06-10T04:04:26.863" Score="-1" Body="&lt;p&gt;In a word, no. The light path for a &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_lens&quot; rel=&quot;nofollow&quot;&gt;gravitational lens&lt;/a&gt; around the sun would be too close to the surface to see the star behind the sun. Usually, a gravitational lens is from a galaxy far away so that the light can converge after passing the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from the sun, do you see any stars during the day?&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-06-10T04:04:26.863" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3646" PostTypeId="2" ParentId="3635" CreationDate="2014-06-10T04:22:52.167" Score="5" Body="&lt;p&gt;To expand on self.'s answer - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Derek from Veritasium on Youtube explains the &quot;(and is)&quot; portion -&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...This doesn't violate Einstein's theory of relativity since nothing is moving through space faster than light, it's just that space itself is expanding such that far away objects are receding rapidly from each other. ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=XBr4GkRnY04&quot; rel=&quot;noreferrer&quot;&gt;Link to video on Youtube&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1849" LastActivityDate="2014-06-10T04:22:52.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3647" PostTypeId="2" ParentId="3644" CreationDate="2014-06-10T04:53:16.873" Score="10" Body="&lt;p&gt;Yes, it would be possible. There are two roads here:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Visible light&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;In case of detecting light in the range of visible wavelength, perhaps you would consider that rare occasion when there is a &lt;code&gt;solar eclipse&lt;/code&gt;.&#xA;And it may also be possible at times of &lt;code&gt;'early' dawn&lt;/code&gt; and &lt;code&gt;'late' dusk&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Invisible light&lt;/strong&gt; (outside visible range, beyond the IR and the UV)&lt;/p&gt;&#xA;&lt;p&gt;Now, other than that, we have telescopes for radio, infra-red, UV rays and X-rays which wouldn't require us to look into the 'eyepiece'. Astronomers have detectors for that and light received is stored in 'pixels'.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;You might also want to look at a similar question in &lt;code&gt;Physics-SE&lt;/code&gt; &lt;a href=&quot;https://physics.stackexchange.com/questions/25498/the-sun-as-a-gravitational-lens&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;As pointed out by Jeremy in a comment below and also pointed out &lt;a href=&quot;http://news.discovery.com/space/using-the-sun-as-a-magnifying-glass.htm&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;, it is actually a far fetched idea. You need to be &lt;strong&gt;really far&lt;/strong&gt; away from the sun to use it as a &lt;strong&gt;G-lens&lt;/strong&gt;. The telescope has to be around 50 billion miles away and to cover that distance within one generation, probably we haven't achieved that technologically advanced level yet.&lt;/p&gt;&#xA;" OwnerUserId="1702" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2014-06-12T04:39:38.473" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3648" PostTypeId="1" CreationDate="2014-06-10T17:38:11.937" Score="2" ViewCount="59" Body="&lt;p&gt;How does one compute the right ascension and declination of the Moon at a given time (or for that matter, the altitude and azimuth)?&lt;/p&gt;&#xA;" OwnerUserId="1712" LastActivityDate="2014-06-10T17:38:11.937" Title="Coordinates of the Moon" Tags="&lt;fundamental-astronomy&gt;" AnswerCount="0" CommentCount="5" ClosedDate="2016-12-30T23:04:13.857" ContentLicense="CC BY-SA 3.0" />
  <row Id="3649" PostTypeId="1" CreationDate="2014-06-10T18:32:42.730" Score="5" ViewCount="512" Body="&lt;p&gt;According to Hubble's Law, the farther a galaxy is, the farther it is moving away. But do we take into account the fact that we are actually looking in the past?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, there are two galaxies A and B at distance of 5 and 10 billion years respectively. Now, when we observe A we are looking at how it was moving 5 billion years ago. The same applies for B. So, now we conclude that 5 billion years ago space was expanding at a slower rate while it was expanding comparatively faster 10 billion years ago. What's wrong with this conclusion?&lt;/p&gt;&#xA;" OwnerUserId="1156" LastEditorUserId="1156" LastEditDate="2014-06-10T19:30:54.327" LastActivityDate="2014-06-11T01:43:03.277" Title="Is the expansion accelerating or Decelerating?" Tags="&lt;space&gt;&lt;redshift&gt;&lt;expansion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3650" PostTypeId="1" CreationDate="2014-06-10T18:45:59.723" Score="2" ViewCount="243" Body="&lt;p&gt;I know this question had often been asked on Physics.SE as well as this site but I don't get them. What do mean when we say that space is expanding? I mean that space is no physical entity that can expand. Space is just (apparently) &quot;nothingness&quot;. Please solve my confusion.&lt;/p&gt;&#xA;" OwnerUserId="1156" LastEditorUserId="1156" LastEditDate="2014-06-11T07:57:39.680" LastActivityDate="2014-06-12T02:38:33.447" Title="What do we mean by space is expanding?" Tags="&lt;space&gt;&lt;general-relativity&gt;&lt;expansion&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3651" PostTypeId="2" ParentId="3635" CreationDate="2014-06-10T21:24:50.207" Score="7" Body="&lt;p&gt;Far away galaxies are receding from us faster than $c_{0}$ = 299 792 458 m/s, but it does not mean that they are breaking the speed of light limit. These galaxies would still measure the speed of light locally to be $c_{0}$ (assuming the speed of light and the laws of physics are the same throughout the universe), and they would never catch up and move faster than photons moving past them: photons passing by these galaxies in the direction away from us still recede from us faster than these galaxies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We observe that on average the further away a galaxy is the faster it recedes from us. If we consider that our galaxy is not in a special place in the universe (just like our Sun is not in a special place in our galaxy and like Earth is not a planet around which the rest of the universe revolves), then we think any other galaxy should make the same observation than ours, that is that other galaxies recede away from them and that the further away the faster they are seen receding. We call that the expansion of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people will tell you that this expansion is due to space being &quot;stretched&quot; or &quot;created&quot; between the galaxies, as if space was an entity or substance actually being stretched or created. To me this is a poor explanation because no one has ever detected that entity or substance we call space, for now it is a theoretical abstraction rather than something having a tangible reality. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In case you're wondering, the story as it goes is that the furthest matter we see (the Cosmic Microwave Background or CMB for short) was 42 million light years away from us 380 000 years after the big bang, back when it emitted the light we are detecting now, and this matter is now 46 billion light years away (which we call the radius of our observable universe).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason we cannot see light from further back than 380 000 years after the big bang is that before that time, the universe is thought to have been too dense to allow for light to travel freely without being constantly reabsorbed by the surrounding matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason it took so long for this light to reach us is that 380 000 years after the big bang, the matter that was 42 million light years away (the matter which emitted the light that we call the CMB) was receding much faster than $c_{0}$ from the soon-to-be Milky Way, hence the light it emitted towards us was in fact getting further away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the reason this light reached us at all is that during the first 9 billion years following the big bang, &lt;a href=&quot;http://www.scientificamerican.com/article/expanding-universe-slows-then-speeds/&quot; rel=&quot;noreferrer&quot;&gt;the universe is thought to have been decelerating&lt;/a&gt; (as opposed to the last 5 billion years when it is thought to have been accelerating). What this means is that galaxies (or soon-to-be galaxies) were moving away from each other at a decelerating velocity during the first 9 billion years. The matter which emitted the CMB didn't slow down enough to recede at less than $c_{0}$ from us, but the light that it emitted 380 000 years after the big bang eventually reached a point where galaxies receded from us slower than $c_{0}$ , i.e. a point where that light started to get closer to us (in other words, the spherical region inside which galaxies recede from us slower than $c_{0}$ enlarged fast enough so that the light of the CMB eventually entered it), and 13.8 billion years later we receive it at last.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The values for the age of the universe and the radius of the observable universe are calculated by solving the equations of Einstein's theory of general relativity, assuming the universe is homogeneous and isotropic, and using cosmological observations to estimate the various parameters in the equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to learn more, this is a useful read about galaxies receding from us faster than the speed of light and the expansion of the universe:&#xA;&lt;a href=&quot;http://www.publish.csiro.au/?act=view_file&amp;amp;file_id=AS03040.pdf&quot; rel=&quot;noreferrer&quot;&gt;Expanding Confusion: Common Misconceptions of Cosmological Horizons and the Superluminal Expansion of the Universe&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1856" LastActivityDate="2014-06-10T21:24:50.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3652" PostTypeId="2" ParentId="3649" CreationDate="2014-06-11T01:43:03.277" Score="4" Body="&lt;p&gt;The light we see now is &lt;em&gt;not&lt;/em&gt; a direct indication of how the galaxy was moving at the time the light was emitted. In the cosmological frame, the galaxies aren't moving (on average, at least); rather, space between them is expanding. The rate at which it is expanding is the Hubble parameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If cosmic expansion is homogeneous and isotropic, then all distances on the cosmological scale should be affected in the same manner, proportionally, unless other forces are involved. Thus the distance between some galaxy and us is affected in the same proportion as the wavelength of the light:&#xA;$$\frac{D_\text{now}}{D_\text{then}} = \frac{\lambda_\text{now}}{\lambda_\text{then}}\text{.}$$&#xA;The light we see now is a direct indication of the &lt;em&gt;scale factor&lt;/em&gt;, i.e. by what factor the distance has grown since the time of emission. We can also consider distance $D$ as a function of cosmological time $t$ and define a &lt;em&gt;recession velocity&lt;/em&gt; as the rate at which it is changing:&#xA;$$v_\text{r} \equiv \frac{\mathrm{d}D}{\mathrm{d}t} = \underbrace{\left[{\dot{D}}/{D}\right]}_{H(t)}D\text{.}$$ &#xA;Therefore, the recession velocity &lt;em&gt;now&lt;/em&gt; is given by the Hubble parameter $H(t)$ &lt;em&gt;now&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-06-11T01:43:03.277" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="3653" PostTypeId="2" ParentId="3650" CreationDate="2014-06-11T11:06:42.433" Score="0" Body="&lt;p&gt;All we know is this : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Distant objects are receding from us at a rate proportional to their distance from us.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All versions of &quot;space is expanding&quot; are philosophical variants of this.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2014-06-11T11:06:42.433" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3654" PostTypeId="2" ParentId="3650" CreationDate="2014-06-11T13:47:38.917" Score="1" Body="&lt;p&gt;The expansion of space means that objects in cosmological distances are receding away from each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may ask: is it the same thing like if there were 2 rockets in empty space (stationary with respect to each other) and then one of them starts its engines and accelerates away?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is no. The above described case is not expansion of space. It is a simple movement with respect to an inertial frame. Notice that the accelerating rocket would be able to detect its motion with an accelerometer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real expasion of space would manifest itself in the following way: you have 2 rockets in empty space, stationary with respect to each other. None of them starts its engines, but yet, they start moving away from each other. As if dragged by some invisible force, but not feeling any acceleration (the accelerometers would indicate zero).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is very similar to the case of inhomogenous gravitational field. Tidal forces would also increase the separation between the 2 rockets without any readout on accelerometers. However in the case of inhomogenout gravitational field there is usually a nearby gravitating body and the 2 rockets would be in free fall towards the body. But in expanding space there is no such body towards which objects would gravitate. The space is expanding because the spacetime metric is changing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this means that in expanding universe you cannot find one global inertial frame. Distant objects initially at rest would start moving away from each other immediately. This is not how an inertial frame behaves.&lt;/p&gt;&#xA;" OwnerUserId="1714" LastActivityDate="2014-06-11T13:47:38.917" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="3655" PostTypeId="1" CreationDate="2014-06-12T00:19:59.123" Score="3" ViewCount="420" Body="&lt;p&gt;Does the sun have a protective layer that at least diminishes its radiation. I know the earth's atmosphere protects us from harmful rays. But does the sun have a sort of shield as well that if removed would radiate more heat. &lt;/p&gt;&#xA;" OwnerUserId="1861" LastActivityDate="2014-06-12T05:00:47.940" Title="Does the sun have a protective shield" Tags="&lt;the-sun&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3656" PostTypeId="2" ParentId="2513" CreationDate="2014-06-12T01:15:58.797" Score="6" Body="&lt;p&gt;In 1812 Fraunhofer measured a set of absorption lines in the sun. It wasn't until later that Kirchoff and Bunsen figured out that these absorption lines matched emission features from metals they were burning in the lab. (&lt;a href=&quot;http://www.chemteam.info/Chem-History/Kirchhoff-Bunsen-1860.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.chemteam.info/Chem-History/Kirchhoff-Bunsen-1860.html&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In chemistry, most things in the periodic table are known as &quot;metals&quot;(  &lt;a href=&quot;http://en.wikipedia.org/wiki/Periodic_table_%28metals_and_nonmetals%29&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Periodic_table_%28metals_and_nonmetals%29&lt;/a&gt;), so given the small fraction that &quot;metals&quot; make up of the overall star, it may not be that far off to call them metals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomy does tend to keep historical conventions around (see magnitudes), but using the name metallicity does have some use. In many cases objects, like stars, can be classified into &quot;metal rich&quot; or &quot;metal poor&quot; to describe differing behavior; the specific elemental abundances aren't necessarily important for some of these studies. Using a blanket name like &quot;metals&quot; makes categorization much simpler.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-06-12T01:15:58.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3657" PostTypeId="2" ParentId="222" CreationDate="2014-06-12T01:26:07.320" Score="4" Body="&lt;p&gt;Twinkling is caused by turbulence AND refraction in the atmosphere. The refraction bends the light from the star and the random motion from the turbulence causes the bending to change direction, making the star appear to twinkle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will happen with any object, but it is most noticeable in stars because they are so far away that they appear like a point of light. The effect is less obvious in planets since we can resolve their size, though on particularly turbulent times, planets may also appear to be twinkling.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-06-12T01:26:07.320" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3658" PostTypeId="2" ParentId="3650" CreationDate="2014-06-12T02:38:33.447" Score="3" Body="&lt;p&gt;Indeed space is not a physical entity (as far as we know). Saying space expands is another way to say that galaxies recede away from each other at a rate proportional to their distance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;expanding space&quot; picture is unfortunately a source of endless confusion and misconceptions, as can be seen on the page of this very question: mpv mentioned that &quot;&lt;em&gt;The real expansion of space would manifest itself in the following way: you have 2 rockets in empty space, stationary with respect to each other. None of them starts its engines, but yet, they start moving away from each other.&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is plain wrong, the &quot;expansion of space&quot; is not some magical force that pulls objects away from each other. If you mean empty space as in no dark energy, no dark matter and no visible matter then the two rockets would move towards each other due to their mutual gravitational attraction (if you neglect their mass then they would not move at all). In the general case what will dictate the motion of the rockets is their gravitational attraction, the matter nearby (visible and dark) and dark energy (whatever that might be), the combination of which might pull them closer or further away from each other. The expansion of space is not an additional force. Galaxies don't keep moving away from each other because some undetected substance is created between them, but because they have an initial velocity which they acquired at the time of the big bang: without anything to stop them they keep moving away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Joan.bdm said: &quot;&lt;em&gt;As I said in a large scale all objects recede from us, if it wasn't for the space expansion it would mean we are at the center of the universe.&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second part of the above sentence is wrong as well. As long as all galaxies recede from one point at a rate proportional to their distance, it follows that from ANY point all galaxies will be seen to recede at a rate proportional to their distance. This is easily seen with a little drawing (each * corresponds to a galaxy):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Before&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TuenC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;After&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G50ld.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all galaxies are seen to recede at a rate proportional to their distance from the point of view of the galaxy at the center, then from any galaxy all other galaxies will be also seen to recede at a rate proportional to their distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think we have any satisfying explanation as to why galaxies move away from each other at a rate proportional to their distance in the first place. The &quot;expansion of space&quot; is not an explanation, it is an analogy. A related and interesting read: &lt;a href=&quot;http://arxiv.org/pdf/0707.0380.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Expanding Space: the Root of all Evil?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1856" LastActivityDate="2014-06-12T02:38:33.447" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3659" PostTypeId="2" ParentId="3644" CreationDate="2014-06-12T03:30:36.170" Score="6" Body="&lt;p&gt;Actually, one of the first confirmation of GR, is by Sir A. Eddington et al., who measured the deviation of light trays of an unsaid-as-far-as-i-know stars, on May 29, 1919.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an original snapshot from their experiment:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/nrEBk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;They took advantage of a solar eclipse to measure the light bending from the expected position of the sources.&#xA;The experiment has been a bit contested, but successive experiments of the same type (measures of light bending of Galactic stars during solar eclipses) confirmed those results. And, he was Eddington in the end!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find the &lt;a href=&quot;http://adsabs.harvard.edu/abs/1919Obs....42..119E&quot; rel=&quot;nofollow noreferrer&quot;&gt;original paper here&lt;/a&gt;, and some other information on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tests_of_general_relativity#Deflection_of_light_by_the_Sun&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wiki page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastActivityDate="2014-06-12T03:30:36.170" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3660" PostTypeId="2" ParentId="3655" CreationDate="2014-06-12T05:00:47.940" Score="3" Body="&lt;p&gt;You seriously cannot expect the sun to have a layer that would contain some of its harmful radiation. Sun consists of a plasma and is not solid. Learn more about its structure &lt;a href=&quot;http://en.wikipedia.org/wiki/Sun&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Magnetic field is a kind of layer you might think of apart from all gases and energy it radiates. But that is harmful and not protective for us. It is highly unlike earth which has a solid crust. &lt;strong&gt;Any layer around sun wouldn't actually protect us as it would have to contain a huge amount of energy and that would make it more unstable&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth just happens to be at the right distance from the sun that was conducive enough to harbor a suitable atmosphere with just the right combination of the right elements and compounds (especially oxygen for breathing, ozone for protection and nitrogen for maintaining atmospheric balance). This is why life has been possible here.&#xA;Earth also happens to have a magnetic field strong enough to protect us from the solar wind, as described in details &lt;a href=&quot;http://en.wikipedia.org/wiki/Earth%27s_magnetic_field&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. It just diverts the harmful energy away from us. &lt;strong&gt;Containment of huge amount of energy is the major concern here and nobody is doing that&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So life has been possible only because the way earth has been created, well of course all energy comes from the sun. But the sun just provides energy and doesn't save us from itself.&lt;/p&gt;&#xA;" OwnerUserId="1702" LastActivityDate="2014-06-12T05:00:47.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3661" PostTypeId="1" AcceptedAnswerId="4818" CreationDate="2014-06-12T07:59:38.233" Score="3" ViewCount="547" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Thales&quot; rel=&quot;nofollow&quot;&gt;Thales of Miletus&lt;/a&gt; purportedly &lt;a href=&quot;http://en.wikipedia.org/wiki/Eclipse_of_Thales&quot; rel=&quot;nofollow&quot;&gt;predicted a solar eclipse&lt;/a&gt; (most likely the solar eclipse of May 28, 585 BCE). The historicity of Thales' prediction can't be verified with any certainty given the sparsity of written records documenting the pre-Socratic era that have survived to present day. So, putting the historical question of whether or not he did aside, what plausible answers are there to the question of how he &lt;em&gt;could/would&lt;/em&gt; have made such a prediction?&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2014-07-04T14:51:12.423" Title="How might Thales have predicted a solar eclipse?" Tags="&lt;fundamental-astronomy&gt;&lt;history&gt;&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3662" PostTypeId="1" AcceptedAnswerId="3665" CreationDate="2014-06-12T08:23:39.407" Score="8" ViewCount="545" Body="&lt;p&gt;Per the Big Bang Theory, the universe was formed from a dense singular point which existed 13 billion years ago and has been expanding ever since. This suggests that the point should have existed somewhere in space implying an outer universe. If we call this point the &quot;Universe&quot;, then what do we call the space that held that point? Also how did this single point come into existence in the first place?&lt;/p&gt;&#xA;" OwnerUserId="1863" LastActivityDate="2014-06-12T12:14:56.060" Title="Outside of a Universe and its origin" Tags="&lt;universe&gt;&lt;big-bang-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3663" PostTypeId="1" AcceptedAnswerId="3670" CreationDate="2014-06-12T10:57:00.957" Score="16" ViewCount="3195" Body="&lt;p&gt;Is there a strict difference between gas and dust?&#xA;In Earthly environment most things become gaseous if heated enough. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Interstellar_medium#Interstellar_matter&quot;&gt;temperature&lt;/a&gt; of interstellar medium seems to range mostly between 10 and 10 000 Kelvin. Is gas/dust an analog for hot/cold, or does the phase diagram of the element in question matter too? Can metals and molecules be gas in astronomical terms?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2018-06-13T07:59:05.503" Title="What is the difference between gas and dust in astronomy?" Tags="&lt;temperature&gt;&lt;dust&gt;&lt;interstellar-medium&gt;&lt;gas&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3664" PostTypeId="2" ParentId="3662" CreationDate="2014-06-12T11:43:25.307" Score="1" Body="&lt;p&gt;This has not been answered yet for sure I guess. For example string theory (M-Theory?) suggests that our Universe lies on a higher dimensional &quot;membrane&quot; that &quot;colided&quot; with another one and thus created the big bang.&#xA;This implies stuff like extra dimensions, strings, multiverses and other yet not proven theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even why the big bang took place and how it could be that the matter in this universe is so unevenly spread out from a point that once consisted of perfectly even distributed energy is not clarified. &quot;Who messed with it?&quot; )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Big Question for sure! And you see how many quotation marks I use just to be careful with the wording. Maybe someone here can give a more scientific answer with some maths :) I can't.&lt;/p&gt;&#xA;" OwnerUserId="1859" LastActivityDate="2014-06-12T11:43:25.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3665" PostTypeId="2" ParentId="3662" CreationDate="2014-06-12T12:14:56.060" Score="8" Body="&lt;p&gt;You are making the mistake of thinking that there was a single spot in space that everything expanded from. Rather, the expansion was not of things in space, it was space that expanded. It was an expansion OF space, not IN space. There wasn't &quot;space&quot; around the singularity.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-06-12T12:14:56.060" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3668" PostTypeId="1" CreationDate="2014-06-12T19:42:53.890" Score="12" ViewCount="7481" Body="&lt;p&gt;Is there a catalog of all known stars or brightest stars in our galaxy? Preferably with some sort of galactic coordinates instead of just night sky coordinates (right ascension, etc.) I'm trying to make a model of the milky way, explorable with an oculus rift.&lt;/p&gt;&#xA;" OwnerUserId="1190" LastActivityDate="2020-05-07T07:49:31.547" Title="Where can I find a catalog of all stars in the Milky Way?" Tags="&lt;star&gt;&lt;data-analysis&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3669" PostTypeId="2" ParentId="3668" CreationDate="2014-06-12T20:15:52.553" Score="2" Body="&lt;p&gt;My understanding is that any star catalog today represents only such a very tiny and local part of the Milky Way, that you would have very little use of it for your purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a year or two the Gaia space telescope will have mapped the one billion or 1% of the nearest and brightest stars in the Milky Way. Even then, to model a galaxy one needs other ideas than maps of individual stars.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2014-06-12T20:15:52.553" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3670" PostTypeId="2" ParentId="3663" CreationDate="2014-06-12T21:31:42.700" Score="19" Body="&lt;p&gt;Yes, metals and other elements and molecules can exist in gaseous form under the right conditions of temperature and pressure. A &quot;gas&quot; is simply one of the fundamental states of matter, as in &lt;em&gt;solid&lt;/em&gt;, &lt;em&gt;liquid&lt;/em&gt;, or &lt;em&gt;gas&lt;/em&gt; (and a few other states outside the scope of this question). But as a gas, these substances exist entirely as either &lt;strong&gt;&lt;em&gt;individual&lt;/em&gt;&lt;/strong&gt; atoms, &lt;em&gt;individual&lt;/em&gt; elemental molecules, or &lt;em&gt;individual&lt;/em&gt; compound molecules of multiple atoms (e.g. carbon dioxide).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dust, on the other hand, is comprised of tiny particulate matter that has undergone the stronger intermolecular bonds to create substances like ice, silicates, and carbon compounds that float around in varying densities between the stars and between the galaxies. Since these particles are still extremely small (typically a fraction of a micron across), they can &lt;em&gt;appear&lt;/em&gt; to be a gas, but these tiny, irregularly-shaped objects still exist individually in a solid or liquid state.&lt;/p&gt;&#xA;" OwnerUserId="58" LastEditorUserId="58" LastEditDate="2014-06-12T21:41:57.567" LastActivityDate="2014-06-12T21:41:57.567" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3671" PostTypeId="2" ParentId="3668" CreationDate="2014-06-13T08:39:21.277" Score="7" Body="&lt;p&gt;Hipparcos, the predecessor to Gaia, has a dataset (&lt;a href=&quot;http://cdsarc.u-strasbg.fr/viz-bin/Cat?I/239&quot; rel=&quot;noreferrer&quot;&gt;http://cdsarc.u-strasbg.fr/viz-bin/Cat?I/239&lt;/a&gt;) with 3D positions for 100,000 stars. While we have much larger datasets of galactic stars, such as SDSS, finding the distances to stars is much harder. Parallax is the most precise technique for finding distances. Even for Gaia, we will only be able to measure distances to 10% precisions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a &quot;galactic coordinate system&quot;, but that stil has the sun at the center and the line between the sun and the galactic center as its reference. There are calculators that can convert RA and Dec to galactic coordinates (e.g. &lt;a href=&quot;http://python4astronomers.github.io/astropy/coordinates.html&quot; rel=&quot;noreferrer&quot;&gt;http://python4astronomers.github.io/astropy/coordinates.html&lt;/a&gt;). You can use that and the distance to place them in space.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-06-13T08:39:21.277" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3672" PostTypeId="1" AcceptedAnswerId="7770" CreationDate="2014-06-13T13:28:07.650" Score="4" ViewCount="9192" Body="&lt;p&gt;I found that the solstices work on a 400 year cycle based on &lt;a href=&quot;http://www.rmg.co.uk/explore/astronomy-and-time/time-facts/equinoxes-and-solstices&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. I can't find anything similar for the lunar cycle. Is there a formula I can use to calculate the date of the full moon?&lt;/p&gt;&#xA;" OwnerUserId="1869" LastActivityDate="2016-06-20T14:44:59.717" Title="In 2016, the summer solstice will coincide with a full moon. How often does this happen?" Tags="&lt;the-moon&gt;&lt;the-sun&gt;&lt;orbit&gt;" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="3674" PostTypeId="2" ParentId="1101" CreationDate="2014-06-14T00:14:02.023" Score="2" Body="&lt;p&gt;Rotation of these object is caused by conservation of angular momentum. So while the rotation of the moon has changed as it became tidally locked with Earth, the total angular momentum of the Earth-moon system hasn't. So it could affect it but given the rotation periods of planets (&lt;a href=&quot;http://en.wikipedia.org/wiki/Rotation_period&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Rotation_period&lt;/a&gt;) this is just a coincidence.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-06-14T00:14:02.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3675" PostTypeId="2" ParentId="2498" CreationDate="2014-06-14T00:29:47.023" Score="1" Body="&lt;p&gt;PyEphem will get you Alt, Az from RA, Dec, Lat, Lng, Date, Time, and Elevation &#xA;&lt;a href=&quot;http://rhodesmill.org/pyephem/&quot; rel=&quot;nofollow&quot;&gt;http://rhodesmill.org/pyephem/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-06-14T00:29:47.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3676" PostTypeId="2" ParentId="3668" CreationDate="2014-06-15T17:10:05.993" Score="3" Body="&lt;p&gt;No, such a catalogues does not (yet) exist. There are two reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1&lt;/strong&gt; The Milky Way galaxy is about 20kpc (1pc ~= 3 lyr) across and only the very brightest stars are individually identifyable across such large a distance (such bright stars by their nature are very massive and hence young). Astronomers tend to cataloge stars by their &lt;em&gt;apparent&lt;/em&gt; brightness, which for stars of identical luminosity declines as $1/d^2$ ($d$=distance). As a consequence, most catalogues contain only stars in the immediate galactic neighbourhood of the Sun. The Hipparcos catalogue (mentioned in another answer), for example, has most stars within a mere 100pc of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2&lt;/strong&gt; Obtaining distances for individual stars is inherently difficult, in particular the more distant the star in question is. Accurate distances for stare several kpc away can currently only be obtained by indirect methods applicable only to certain types of stars (such as RR Lyrae variables). The classical trigonometric parallax measurement for such distances, however, is subject of ESA's ongoing Gaia mission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ESA's Gaia satellite launched last year aims at cataloguing about $10^9$ stars across the Milky Way, including their velocity. The first preliminary versions of resulting catalogue, however, will still take some time to appear.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-06-15T17:10:05.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3677" PostTypeId="2" ParentId="3663" CreationDate="2014-06-15T17:20:59.987" Score="4" Body="&lt;p&gt;I may just add to the excellent answer by Robert that interstellar dust particles, very much like cigarette smoke in air, hangs in the interstellar gas and interacts with it both kinematically (is dragged along with it depending on the particle size) and energetically (exchanges heat, which can result in significant cooling of the gas). Dust particles also interact with the (stellar) radiation and can be evaporated due to high-energy radiation, but can also grow by condensing from the surrounding gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All larger solid astronomical objects (planets, asteroids etc, but not stellar remnants) have formed from dust, which in turn has formed from the heavier elements in the inter stellar gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For many astronomical purposes, dust is annoying, as it blocks the light, in particular the shorter wave lengths (&lt;em&gt;reddening&lt;/em&gt; and &lt;em&gt;darking&lt;/em&gt; the light of stars), hiding stars, in particular in the mid-plane of the Milky Way. As a consequence, the Galactic centre, a place of great astronomical interest, is largely invisible and can only be studied by observing other wave lengths than visible light, in particular infrared which is hardly affected by dust absorption.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2014-06-15T21:53:43.147" LastActivityDate="2014-06-15T21:53:43.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3678" PostTypeId="1" AcceptedAnswerId="3682" CreationDate="2014-06-16T02:58:28.137" Score="12" ViewCount="735" Body="&lt;p&gt;I'm watching MIT OpenCourseWare lectures on general relativity and not too long into the first lecture the professor stated that the Kerr Black Hole solution allowed for travel between universes. How is that possible to know? How does one derive this/come to this conclusion?&lt;/p&gt;&#xA;" OwnerUserId="1206" LastEditorUserId="1702" LastEditDate="2014-06-16T13:22:31.883" LastActivityDate="2014-06-16T18:45:44.953" Title="How are black holes doors to other universes?" Tags="&lt;black-hole&gt;&lt;general-relativity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3681" PostTypeId="2" ParentId="1241" CreationDate="2014-06-16T09:43:39.380" Score="7" Body="&lt;p&gt;In fact, some people had looked at it more seriously recently and conducted a computer simulation to visualize the night sky as seen from within a globular cluster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article has appeared recently in Astronomy journal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just one example of a typical image inside a globular cluster:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FbyaO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some more discussion can be found here: &lt;a href=&quot;http://io9.com/what-the-night-sky-would-look-like-from-inside-a-globul-1589324556&quot;&gt;http://io9.com/what-the-night-sky-would-look-like-from-inside-a-globul-1589324556&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the journal issue containing the article by William Harris and Jeremy Webb can be found here: &lt;a href=&quot;http://www.astronomy.com/magazine/press-releases/2014/05/july-2014&quot;&gt;http://www.astronomy.com/magazine/press-releases/2014/05/july-2014&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2014-06-16T09:43:39.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3682" PostTypeId="2" ParentId="3678" CreationDate="2014-06-16T14:06:06.233" Score="10" Body="&lt;p&gt;It is correct that the Kerr black hole solution of GTR allows travel between universes. However, that does not mean that if you actually jump into any kind of black hole that you could go to another universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To motivate the resolution to this conundrum, let's start off very easy: suppose you stand on the ground with a ball in your hand, and you throw it with some initial velocity. For simplicity, let's ignore everything except a uniform gravity. Mathematics will then tell you that the ball follows a parabolic arc, and when and where the ball will hit the ground. And if you take the resulting equations too literally, then it will also tell you that the ball hits the ground &lt;em&gt;twice&lt;/em&gt;: once in the future, once in the past. But you know the past solution isn't right: you held the ball; it didn't actually continue its parabolic arc into the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A morally similar kind of thing occurs for, say, a Schwarzschild black hole. If you look at it in the usual Schwarzschild coordinates, there's a problem at the horizon. Mathematics will then tell you that the problem is just with the coordinate chart, and that there's actually an interior region to the black hole that becomes apparent in different coordinates. And if you do this generally enough, it will tell you that there's &lt;em&gt;more&lt;/em&gt; to it than even that: there's also white hole with a reverse horizon and its exterior region--another universe. This full &quot;maximally extended&quot; Schwarzschild spacetime has this other universe connect to ours via an &quot;Einstein-Rosen bridge&quot; and then &quot;pinch off&quot;, producing separate black and white holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, that too is an artifact of mathematical idealization: and actual black hole is not infinitely extended in the past and future; it was actually produced by something, a stellar collapse. (And the &quot;bridge&quot; isn't traversable anyway; one will be destroyed in the singularity if one tries.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, on to the Kerr solution, it's a bit better because formally the singularity is avoidable, unlike the Schwarzschild case. However, it's still physically unreasonable: in addition to the fact that actual black holes aren't eternal, the interior of the Kerr solution is unstable in regard to any infalling matter, which will perturb the solution into something else entirely. Therefore, it cannot be taken as a physically meaningful. Still, it is true that full Kerr spacetime contains a way into another universe--in fact, infinitely many of them, chained one after another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're interested in the details of its structure, you could look at some &lt;a href=&quot;http://jila.colorado.edu/~ajsh/insidebh/penrose.html&quot;&gt;Penrose diagrams&lt;/a&gt; of those black hole solutions.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-06-16T14:06:06.233" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3683" PostTypeId="1" AcceptedAnswerId="3688" CreationDate="2014-06-16T15:50:17.110" Score="15" ViewCount="699" Body="&lt;p&gt;If a star or black hole neared the solar system, which would do the least damage?  For the sake of a good comparison, lets assume the candidate star is an &quot;average&quot; 5 solar mass star, and the black hole is a 5 solar mass black hole.  My thoughts at first were that the star would be the least devastating, but now that I think about it more, since not even radiation, etc... is being emitted by the black hole, (except possibly from hawking radiation and from any in falling matter), the black hole might actually be safer!&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2014-06-16T18:34:59.670" Title="Is it safer to be near a star or a black hole?" Tags="&lt;star&gt;&lt;black-hole&gt;&lt;gravity&gt;&lt;radiation&gt;&lt;hawking-radiation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3684" PostTypeId="1" AcceptedAnswerId="3687" CreationDate="2014-06-16T16:38:45.300" Score="1" ViewCount="1135" Body="&lt;p&gt;This might be a weird question, but I'll ask it anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one year on earth is the time it takes for the earth to orbit around the sun, then (if it were possible for humans to survive on the sun), how long would be a year on the sun?&lt;/p&gt;&#xA;" OwnerUserId="1880" LastEditorUserId="53" LastEditDate="2014-06-17T00:19:12.273" LastActivityDate="2014-06-17T00:19:12.273" Title="One year on the sun" Tags="&lt;the-sun&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" ClosedDate="2014-07-03T20:05:05.703" ContentLicense="CC BY-SA 3.0" />
  <row Id="3685" PostTypeId="1" AcceptedAnswerId="3702" CreationDate="2014-06-16T16:58:11.343" Score="2" ViewCount="159" Body="&lt;p&gt;There have been many questions and answers about finding the distance of a star from the earth. But as I did some research on the net, I found that we have specific approaches for finding the distances of different types of stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is more specific now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How to determine how far a Classical Cepheid is?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What if the Cepheid is extra-galactic? Is there a good method to finding the solution?&lt;/strong&gt;&#xA;&lt;strong&gt;Can anybody suggest how to make a start at deriving a relation at finding the distance of a Cepheid from us?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is this &lt;strong&gt;Period-Luminosity-Color&lt;/strong&gt; relation. Can anyone at least enlist the journal articles where I can make a study to get closer to finding the answer to my question?&lt;/p&gt;&#xA;" OwnerUserId="1702" LastEditorDisplayName="user1569" LastEditDate="2016-08-18T15:05:40.887" LastActivityDate="2016-08-18T15:05:40.887" Title="Distance of extra-galactic Classical Cepheids" Tags="&lt;star&gt;&lt;galaxy&gt;&lt;distances&gt;&lt;star-formation&gt;&lt;cepheids&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3686" PostTypeId="2" ParentId="3684" CreationDate="2014-06-16T17:17:23.237" Score="2" Body="&lt;p&gt;I hope you mean if a &lt;strong&gt;sun&lt;/strong&gt; was 'rotating around' another &lt;strong&gt;sun&lt;/strong&gt;, how long it would take to complete a revolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is this thought process. Since they both have equal mass, they would definitely rotate around their center of mass. So now, many factors come into the picture.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Distance: How far they are from each other.&lt;/li&gt;&#xA;&lt;li&gt;Mass: Mass of sun, i.e. their individual masses (same in this case of course)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Using the mass and distance we can make some basic calculations using $F = G*\frac{M^2}{d^2}$, where $M$ is the mass of sun and $d$ is the distance of separation between the two 'suns'. Keeping in mind of the fact they must obviously be rotating about their center of mass which is half-way in this case, we can use $a = \frac{v^2}{r}$ accordingly and get to the solution you are seeking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, depending on the above factors (there may be few more), we may find the linear or angular velocity with which they are rotating around their center of mass, and accordingly we can calculate the period of rotation.&lt;/p&gt;&#xA;" OwnerUserId="1702" LastActivityDate="2014-06-16T17:17:23.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3687" PostTypeId="2" ParentId="3684" CreationDate="2014-06-16T18:31:47.090" Score="10" Body="&lt;p&gt;There are (at least) two possible answers.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The concept of a &quot;year&quot; has no meaning for the Sun. For Earth, a year is the time it takes to complete one orbit around the Sun. Since the Sun doesn't orbit around the Sun, it doesn't have a year.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;But the Sun &lt;em&gt;does&lt;/em&gt; orbit around the core of the Milky Way Galaxy. One orbit takes from 225 million to 250 million years. (Current estimates are uncertain, and the chaotic nature of star motions is likely to cause it to vary a bit.) This is called a &quot;galactic year&quot; or &quot;cosmic year&quot;. Reference: &lt;a href=&quot;http://en.wikipedia.org/wiki/Galactic_year&quot;&gt;http://en.wikipedia.org/wiki/Galactic_year&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(I suppose there could be a third answer. If humans were able to put a colony on the surface of the Sun, using some advanced technology that obviously doesn't exist yet, it's likely they'd still use an Earth-based calendar to measure time.)&lt;/p&gt;&#xA;" OwnerUserId="53" LastEditorUserId="53" LastEditDate="2014-06-16T20:59:26.033" LastActivityDate="2014-06-16T20:59:26.033" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3688" PostTypeId="2" ParentId="3683" CreationDate="2014-06-16T18:34:59.670" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;the candidate star is an &quot;average&quot; 5 solar mass star, and the black&#xA;  hole is a 5 solar mass black hole&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then their gravity is identical. Black holes don't have magic powers. A 5 M☉ star and a 5 M☉ black hole exert the same attraction from the same distance. The only difference is that the black hole would be much, MUCH smaller (about 30 km diameter in this case), so you could get a lot closer to the center, which is where the extreme gravity happens. But at cosmic distances they are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hawking radiation from a 5 M☉ black hole is negligible, likely too small to measure - its temperature is about 10^-8 Kelvin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it has an accretion disk, that might be a problem, but usually it only generates two relativistic jets at the poles - if you're not hit by the jets, you're fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall thermal glow from the accretion disk of a 5 M☉ black hole might be intense, but it can't last very long. It's not a continuous burn like a star, unless it's a much bigger black hole with a huge accretion disk and plenty of source material nearby to feed it - which is not the case here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your conclusion is technically correct. Gravitationally they are the same, but the star has the added problem of generating extra heat. Not that it would matter anyway, because both would completely disrupt orbits in the solar system, and the Earth is going to be flung out into space and would freeze anyway. It's just that the star would bake us first, and THEN we would freeze solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or we would settle into a highly elliptical orbit around either the Sun or the invader body, which will have us alternately baking to death and freezing to death.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the very tiny chance that, due to orbital disruptions, we would collide with something else, either with the invader, or with the Sun, or with another planet. This would mean a quicker death and might be overall &quot;preferable&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless, having a 5 M☉ body entering the system is not a good scenario for life on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://xaonon.dyndns.org/hawking/&quot; rel=&quot;noreferrer&quot;&gt;http://xaonon.dyndns.org/hawking/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2014-06-16T18:34:59.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3689" PostTypeId="2" ParentId="3678" CreationDate="2014-06-16T18:45:44.953" Score="5" Body="&lt;p&gt;&quot;Allowed for&quot; does not mean &quot;necessarily cause&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the professor implied is that the solutions look, from a mathematical p.o.v., just like what you would expect from a bridge between universes - IF multiple universes exist, and IF the bridge is passable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's all there is to it. A mathematical solution that looks like a bridge. But has it ever been verified experimentally? No. Do we have proof that other universes exist? No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have the math that describes what looks, for all intents and purposes, like a door. But is the door separating this room and another room, or is it just a fake door built into a solid brick wall, like in movie comedies? We don't know. Would the door open at all? We don't know. Has anyone actually seen such a door yet? No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That doesn't mean the professor was wrong. It only means this is just a hypothesis at this point. We don't know yet if reality matches it or not.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2014-06-16T18:45:44.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3692" PostTypeId="1" AcceptedAnswerId="3694" CreationDate="2014-06-17T03:54:58.170" Score="2" ViewCount="153" Body="&lt;p&gt;Is Mars visible at night sky of Michigan these days? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was trying to search for Mars from Michigan (at elevation of 30 deg and azimuth of 220 deg) around 11:45 PM with binoculars of 20x60 but didn't get successful.&lt;/p&gt;&#xA;" OwnerUserId="1842" LastEditorUserId="43" LastEditDate="2014-06-19T12:52:20.850" LastActivityDate="2014-06-19T12:52:20.850" Title="Is Mars visible from naked-eye from Michigan (47N, 88W) these days?" Tags="&lt;planet&gt;&lt;mars&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3693" PostTypeId="1" AcceptedAnswerId="3697" CreationDate="2014-06-17T05:52:38.727" Score="3" ViewCount="6318" Body="&lt;p&gt;In the &lt;a href=&quot;https://astronomy.stackexchange.com/a/1019/1884&quot;&gt;best sky conditions, we can see about 44,000 stars in the sky&lt;/a&gt;. But what if I am located around Equator area? How many stars I can see? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I know, the number of stars I can see should be far less ( I don't know why, though). Is it true? If yes, why?&lt;/p&gt;&#xA;" OwnerUserId="1884" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-06-17T19:18:22.307" Title="How many stars are visible to naked eyes around Equator area?" Tags="&lt;star&gt;&lt;galaxy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3694" PostTypeId="2" ParentId="3692" CreationDate="2014-06-17T11:20:13.000" Score="6" Body="&lt;p&gt;On 16 June 2014 at 11:45pm, it was up not far from 220az 30alt. It was actually 232az, 27alt, almost level and to the right of Spica. It would have been easily seen without binoculars being -0.2 visual magnitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can answer this sort of question yourself by getting a planetarium app like Stellarium, The Sky X, Sky Safari, or one of many others.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-06-17T11:20:13.000" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3695" PostTypeId="2" ParentId="3693" CreationDate="2014-06-17T11:29:19.617" Score="0" Body="&lt;p&gt;No, it isn't true. In fact over time you will see more stars at the equator than at any other latitude. The further you go towards the poles, the more stars are never seen because the planet gets in the way. &lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-06-17T11:29:19.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3697" PostTypeId="2" ParentId="3693" CreationDate="2014-06-17T19:03:08.087" Score="3" Body="&lt;p&gt;The numbers you're quoting are extreme, more like a theoretical limit, and meaningless in practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 44k number would be true if you assume a limiting magnitude of 8. There aren't many people alive today who have seen at least one object at mag +8. This is not so much due to vision problems (although that may be important for some people), as to the fact that a mag +8 sky is an incredibly rare occurrence today due to light pollution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you live on a continent with a low population density (such as North America, Australia, or the remote parts of Asia), then yes, maybe you have a chance of catching a mag +8 sky - if you drive far out into the desert, many hours away from any city or town. But if you live in Europe or the like, forget it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Light_pollution&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Light_pollution&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jshine.net/astronomy/dark_sky/&quot; rel=&quot;nofollow&quot;&gt;http://www.jshine.net/astronomy/dark_sky/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the 44k number is obtained by assuming the mag 8 conditions, and then counting all known objects in the sky whose brightness exceeds that limit. It is very unlikely that any one person would see them all in one lifetime. Seeing even a single star at mag 8 can be a difficult undertaking for many people, even in ideal conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very hard to quote a number, since it is so much dependent on light pollution, transparency, visual acuity of the observer (nearsighted people can't focus well enough to see very faint stars), the observer's age (acuity gets worse with age), the observer's experience (visual astronomers with years of experience will be able to see much fainter stars because &lt;em&gt;they know how to look&lt;/em&gt;), etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're on a farm, away from cities, in a place with reasonably low light pollution, and your eyes are good, and you've been sitting in perfect pitch black darkness for at least 30 minutes prior - when you look up you can reasonably expect to see a few thousand objects, mostly stars. Keep looking, and after a while you will distinguish one or two thousand more stars, very faint, that you could not see at first sight. Practice this steadily for a few years, and you'll add maybe another thousand; but you won't be able to see those all at once - only one at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now travel to the Cerro Tololo site in Chile, up in the mountaineous desert, zero light pollution, excellent transparency, and you'll multiply all those numbers by a factor of 2x ... 5x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the numbers are very flexible because there are so many factors involved. You can't just slap a 44k label on it and call it a day; that doesn't make any sense in reality. Astronomers know that the pure magnitude number doesn't mean much by itself, because it is just one factor among many.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a place with very high light pollution (like where I live, in the middle of a large, dense, sprawling urban area in California), you'd be lucky if you can see a hundred stars at night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, in a place with zero light pollution, shine a flashlight into someone's face, and you've temporarily blinded them. You've reduced the number of stars they could see by an order of magnitude for the next half hour (night vision gradually recovers, and it takes 30 minutes to fully recover, according to US military manuals and visual astronomers practice).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, Equator versus poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spin axis of the Earth has a fixed direction in space. The north - south axis always points at the same places in the sky, near Polaris in the north, and near the Southern Cross in the south.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this reason, at the North Pole, you only ever see half the sky - the northern half, centered on Polaris. At the South Pole, you only see the other half. You'll never see Polaris from Australia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the Equator, you pretty much see the whole sky - not all at once, of course, but as the Earth rotates, it gradually exposes all parts of the sky at different times. However, the parts near Polaris and near the Southern Cross are always low in the sky and a bit hard to see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, bottom line is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any place on Earth, in any given instant, you see exactly half of the sky. At the poles, that half is all you &lt;em&gt;ever&lt;/em&gt; see. At the Equator, if you're patient, you'll eventually see the whole sky in time, as the Earth rotates and gradually exposes different parts of the sky at different times.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2014-06-17T19:18:22.307" LastActivityDate="2014-06-17T19:18:22.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3700" PostTypeId="1" CreationDate="2014-06-17T21:20:10.997" Score="4" ViewCount="1265" Body="&lt;p&gt;I'm looking for the total mass of the moons in the solar system in Earth Mass&#xA;this would be include the 67's moons of jupiter, the 62's of saturn, the 27's Uranus and the 14's for Neptune.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it's a crazy question I know, if you have some list of the object with mass in same range i can make the addition ^^&lt;/p&gt;&#xA;" OwnerUserId="1865" LastEditorUserId="2618" LastEditDate="2017-03-07T01:07:17.723" LastActivityDate="2017-03-07T01:07:17.723" Title="Sum of the Masses of All Moons in the Solar System" Tags="&lt;solar-system&gt;&lt;mass&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3701" PostTypeId="2" ParentId="3700" CreationDate="2014-06-18T01:58:27.080" Score="6" Body="&lt;p&gt;Here you can find a list of &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_natural_satellites&quot;&gt;all the natural satellites in our Solar System&lt;/a&gt;. You can check one by one (good luck!) OR you can check &lt;a href=&quot;http://nineplanets.org/data1.html&quot;&gt;this webpage&lt;/a&gt;, and just add the terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please, keep in mind that the latter website is kind of unknown, so double-check at least some of the masses, before to trust it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps, you can cross check with &lt;a href=&quot;http://www.windows2universe.org/our_solar_system/moons_table.html&quot;&gt;this list&lt;/a&gt; as well, and see if other parameters (size, distance, etc.) agree.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastActivityDate="2014-06-18T01:58:27.080" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3702" PostTypeId="2" ParentId="3685" CreationDate="2014-06-18T06:14:59.493" Score="2" Body="&lt;p&gt;Since it seems you don't know the equations, I will try to keep it simple.&#xA;However, just keep in mind that, in principle, there is no difference between Galactic and extragalactic Cepheids, in this context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we know that a Period-Luminosity relation holds for the Cepheids:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P\sim L$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $P$ is the pulsation period observed from the Cepheid, and $L$ is the observed luminosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also know that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L\sim d$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $d$ is the distance of the source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then it is like to write $P\sim d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that, if we observe the pulsation period of a Cepheid, we know its distance, and this allows to know the distance of the host galaxy as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some basic sources:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cepheid_variable&quot; rel=&quot;nofollow&quot;&gt;Cepheid variable&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Classical_Cepheid_variable&quot; rel=&quot;nofollow&quot;&gt;Classical Cepheid variable with some numbers&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astronomynotes.com/ismnotes/s5.htm&quot; rel=&quot;nofollow&quot;&gt;And some diagrams too&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1038" LastActivityDate="2014-06-18T06:14:59.493" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3704" PostTypeId="1" AcceptedAnswerId="5892" CreationDate="2014-06-18T14:52:26.837" Score="2" ViewCount="220" Body="&lt;p&gt;Is there any catalog or any paper published in any journal that lists the stars discovered under whichever star cluster of the Small Magellanic Cloud (SMC) they belong to? There is one for Large Magellanic Cloud (LMC) in the paper Efremov 2003 &quot;Cepheids in LMC clusters and the period-age relation&quot;. But I can't find any for SMC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be better still if the stars happen to be Cepheids.&lt;/p&gt;&#xA;" OwnerUserId="1702" LastEditorUserId="9527" LastEditorDisplayName="user1569" LastEditDate="2020-09-25T08:11:06.570" LastActivityDate="2020-09-25T08:11:06.570" Title="Stars in star clusters in SMC and LMC" Tags="&lt;star&gt;&lt;star-cluster&gt;&lt;cepheids&gt;&lt;dwarf-galaxy&gt;&lt;magellanic-cloud&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3705" PostTypeId="2" ParentId="3692" CreationDate="2014-06-18T15:22:32.467" Score="3" Body="&lt;p&gt;Yes, it is. Mars is bright (-0.14 magnitude) and is in the constellation Virgo. It's the brightest star in this region of the sky (SW during the evening) and should appear as a very bright reddish star at a good altitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need a guide, try to find Leo (W). Virgo is at the left of the constellation.&lt;/p&gt;&#xA;" OwnerUserId="1890" LastActivityDate="2014-06-18T15:22:32.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3706" PostTypeId="1" CreationDate="2014-06-19T08:48:16.490" Score="4" ViewCount="652" Body="&lt;p&gt;My supposition is that within the distance of the so called &lt;a href=&quot;https://en.wikipedia.org/wiki/Frost_line_%28astrophysics%29&quot; rel=&quot;nofollow&quot;&gt;frost line&lt;/a&gt; around a star, water in empty space sublimates, evaporates. I further suppose that this means that vaporized water is pushed to outside the frost line by solar wind or some other mechanism. (Please correct me if this is wrong)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are some ways for water (not necessarily liquid) to exist within the frost line. I wonder under what conditions that is possible. What the origin or history such water can have. And how common or likely it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest some possibilities to be extended, elaborated or refuted:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Water under enough atmospheric pressure to keep it from escaping into space.&lt;/li&gt;&#xA;&lt;li&gt;Water in cold shadow, as in some craters on the lunar poles.&lt;/li&gt;&#xA;&lt;li&gt;Water inside rock, as in the mantle of the Earth (and other planets?)&lt;/li&gt;&#xA;&lt;li&gt;Water ice transported by comets across the frost line.&lt;/li&gt;&#xA;&lt;li&gt;Water vapor hanging around in empty inner inter-planetary space, for example after a planetary collision which melts most matter into a cloud. Or maybe permanently.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2014-06-20T11:56:29.700" Title="How does water exist within the frost line of a star?" Tags="&lt;solar-system&gt;&lt;planetary-atmosphere&gt;&lt;water&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3708" PostTypeId="1" CreationDate="2014-06-19T23:36:30.093" Score="4" ViewCount="87" Body="&lt;p&gt;recently I heard an astronomer on the radio claim that the supernova we have observed in our own galaxy lasted about 6 months, while in distant galaxies they last about 7-8 months, due to relativity and the expanding universe. He didn't give any references. Just to be clear, I don't mean to imply that he implied the actual duration was different, but that it only appeared different because of expansion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found records of supernova, but only 1 mentioned how long it lasted, the first recorded one was said to last 8 months according to wikipedia. &#xA;Can someone point me to a list that contains both the location and duration of supernova's .&lt;/p&gt;&#xA;" OwnerUserId="1901" LastEditorUserId="1901" LastEditDate="2014-06-20T02:15:58.857" LastActivityDate="2014-06-20T22:29:37.420" Title="What record do we have of the length of supernovas?" Tags="&lt;supernova&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3709" PostTypeId="2" ParentId="3706" CreationDate="2014-06-20T11:56:29.700" Score="4" Body="&lt;p&gt;When you talk about the frost line you are talking about a protostar, and its protoplanetary disk, not an actual star. So, there is no empty space nor strong solar winds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside the frost line, water does not condense into ice grains, but the individual molecules could as well become attracted by the planets-to-be.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-06-20T11:56:29.700" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3714" PostTypeId="1" CreationDate="2014-06-20T21:38:10.893" Score="4" ViewCount="316" Body="&lt;p&gt;The moon is not challenging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to observe the big planets and their rings.&lt;/p&gt;&#xA;" OwnerUserId="1905" LastEditorUserId="-1" LastEditDate="2014-06-21T01:26:13.520" LastActivityDate="2014-06-21T01:26:13.520" Title="Any cheap (less than 100 euro) telescope for observing Jupiter?" Tags="&lt;planet&gt;&lt;telescope&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ClosedDate="2014-08-12T13:39:04.290" ContentLicense="CC BY-SA 3.0" />
  <row Id="3715" PostTypeId="1" AcceptedAnswerId="3718" CreationDate="2014-06-20T21:59:37.350" Score="10" ViewCount="1348" Body="&lt;p&gt;&lt;strong&gt;Main question:&lt;/strong&gt; Could dark energy (the mysterious accelerating expansion of the universe) be explained by &quot;negative gravity&quot;?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;Spin off&quot; questions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Does antimatter have negative gravity?  &lt;/li&gt;&#xA;&lt;li&gt;If antimatter has negative gravity, and is present in the universe with equal amounts of matter and antimatter, would it produce the observed amount of dark energy in the universe?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My theory that I am trying to validate:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity is the flow of space into matter, and it flows out of antimatter.  If this is true, antimatter should have &quot;negative gravity&quot;, I would expect it to have a repelling effect, while matter has an attracting effect.  For example, if antimatter were on earth (somehow without annihilating), it would fall to the planet.  However, if there were an antimatter planet, objects would &quot;fall up&quot; off it.  In fact, I believe the antimatter planet would break apart because of this quality.  Furthermore, I don't believe antimatter planets or stars would form because of this quality (e.g. it would remain anti-hydrogen).  I also theorize that it would be in parts of the universe where matter isn't (e.g. in the voids outside the galactic clusters), and would be very diffuse (because it would repel itself).  I believe the theorized repelling effect of antimatter could be what we are observing as dark energy (hence why I am asking these questions).  I just searched for this online to see if anyone else has this theory, and yes, someone has something very similar here: &lt;a href=&quot;http://www.universetoday.com/84934/antigravity-could-replace-dark-energy-as-cause-of-universes-expansion/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.universetoday.com/84934/antigravity-could-replace-dark-energy-as-cause-of-universes-expansion/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;** Edit **&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems others have caught on and simulated the idea of negative mass as dark energy in the universe.  They have run simulations with equal mass and negative mass and ended up with a result that resembles the current universe!  They say negative mass might also explain dark matter!  Check it out here: &lt;a href=&quot;https://youtu.be/MZtS7cBMIc4&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/MZtS7cBMIc4&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This video also shows how negative mass might react: &lt;a href=&quot;https://youtu.be/uAJlg8MDAlU&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/uAJlg8MDAlU&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="1386" LastEditDate="2019-02-19T02:46:05.500" LastActivityDate="2019-02-19T02:46:05.500" Title="Could dark energy be negative gravity?" Tags="&lt;gravity&gt;&lt;dark-energy&gt;&lt;galactic-dynamics&gt;&lt;intergalactic-space&gt;&lt;antimatter&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="3716" PostTypeId="2" ParentId="3708" CreationDate="2014-06-20T22:29:37.420" Score="1" Body="&lt;p&gt;I would expect that the further away a supernova is, the &quot;longer&quot; we observe it will be active because of an increased red shift.  Objects in the universe that are further away are generally going away from us faster than objects that are close, hence the increased red shift.  This increase in time due to redshift is very similar to the Doppler effect, where a siren is high pitched (and faster) when approaching you (e.g. blue shift with light), and low pitched (and slower) when going away from you (e.g. red shift with light).  The faster the siren is moving away, the more the effect (both lower pitch and longer time) will be increased, likewise with light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some sources that may help with what you are looking for (although I haven't been able to find a list exactly like you are looking for):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=supernova&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=supernova&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Supernova#Light_curves&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Supernova#Light_curves&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_distance_ladder#Type_Ia_light_curves&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Cosmic_distance_ladder#Type_Ia_light_curves&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://curious.astro.cornell.edu/question.php?number=49&quot; rel=&quot;nofollow&quot;&gt;http://curious.astro.cornell.edu/question.php?number=49&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2014-06-20T22:29:37.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3717" PostTypeId="2" ParentId="2515" CreationDate="2014-06-20T23:20:10.133" Score="2" Body="&lt;p&gt;Magnetic activity depends on rotation due to the dynamo mechanism. Younger stars in general rotate faster. They (stars like the sun) spin down because their ionised winds interact with their magnetic fields and gradually remove angular momentum. A star like the Sun at 100 million years old may have had a rotation period of 0.5-5 days - much faster than the present-day Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly speaking, magnetic activity scales with the square of angular velocity, so young stars can be orders of magnitude more active than the Sun, manifested as much stronger coronal X-ray emission, flares and much greater coverage of starspots and strong magnetic fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, stars older than the Sun should be a bit less active, although the rate of spindown slows with rotation rate and so the changes beyond a few billion years are not so great and not so well calibrated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A useful review article would be Ribas et al. (2005)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2005ApJ...622..680R&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/2005ApJ...622..680R&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1907" LastActivityDate="2014-06-20T23:20:10.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3718" PostTypeId="2" ParentId="3715" CreationDate="2014-06-21T00:14:35.747" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Could dark energy (the mysterious accelerating expansion of the universe) be explained by &quot;negative gravity&quot;? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But it already is &quot;negative gravity&quot;. In general relativity, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor&quot; rel=&quot;noreferrer&quot;&gt;stress-energy tensor&lt;/a&gt; $T_{\mu\nu}$ describes the energy, momentum, and stress of matter in spacetime. Through the Einstein field equation, it is connected with Ricci curvature $R_{\mu\nu}$, which consists of half of the twenty independent degrees of freedom of spacetime curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intuitive geometrical meaning of Ricci curvature is as follows: suppose you have a small ball of test particles, initially all at the same velocity with respect to one another, and allow them to fall freely under gravity. Then the Ricci curvature (contracted with their intitial velocity) measures the initial acceleration of the volume of the ball of test particles divided by the initial volume. Thus, this is a good way to tell if something is locally &quot;attractive&quot; or &quot;repulsive&quot;, based on whether a ball of test particles around it starts shrinking or expanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Through the Einstein field equation, stress-energy is connected to Ricci curvature. As a result, the volume of the small ball of test particles shrinks proportionally to $\rho+3p$, where $\rho$ is the energy density and $p$ is the average of the principal stresses--for a perfect fluid, the pressure. Naturally, if $\rho+3p &amp;lt; 0$, this is &quot;negative gravity&quot; in the sense that the test ball will start expanding instead. This quantity is also what's seen in the cosmological Friedmann equations. Since the Friedmann-Robertson-Walker family of solutions assume that the universe is homogeneous and isotropic on the large scale, it's sensible that treat the large-scale the same as the above local condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark energy has $\rho+3p&amp;lt;0$, and in particular (for cosmological constant), $\rho = -p &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To recap: (1) in general relativity, gravity depends on the stress-energy tensor, which has more than just the mass-energy density, and numerically pressure is more important because there are three spatial dimensions while only one time dimension, (2) we can characterize locally &quot;negative gravity&quot; as $\rho+3p &amp;lt; 0$, or, more generally, as any the violation of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Energy_condition#Strong_energy_condition&quot; rel=&quot;noreferrer&quot;&gt;strong energy condition&lt;/a&gt;, because this characterizes whether a small ball of test particles is attracted or repulsed by the local stress-energy content.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does antimatter have negative gravity? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not in general relativity. Energy is the gravitational charge, and antimatter has positive energy; thus it should gravitate in the same way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Gravity is the flow of space into matter, and it flows out of antimatter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you're thinking of some analogy with electric charge, with electric field lines flowing out of positive charge and into negative charge, antimatter has positive mass, which is required by quantum field theory.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... I just searched for this online to see if anyone else has this theory, and yes, someone has something very similar here: ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some particles, such as photons, are their own anti-particle. This has immediate implication for Villata's claim that particles and antiparticles behave oppositely in a (normal) matter-produced gravitational field: it predicts that gravitational lensing cannot happen, since if light is bent one way by a gravitational field, Villata's stance predicts it's &lt;em&gt;also&lt;/em&gt; bent in the the reverse direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your stance is different from Villata's, but the corresponding statement for your theory is that light cannot produce a gravitational field. That's a lot harder to specifically test for, but it's clear that it's massively inconsistent with general relativity.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-06-21T00:14:35.747" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4718" PostTypeId="2" ParentId="3714" CreationDate="2014-06-21T01:16:35.470" Score="5" Body="&lt;p&gt;Jupiter is pretty easy. A little refractor with a 50 mm aperture (diameter of the primary lens) will show you the disk of Jupiter and the biggest two equatorial belts, and the four galilean satellites. Or Saturn and its rings - that's doable too. The image will be tiny, but you'll see those features - assuming the instrument is not junk and it can take a decent magnification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, that's all that you'll see at that level of investment. No small equatorial belts on Jupiter, or inter-belt swirls. With the Great Red Spot shrinking and becoming washed out each year, you may not even see that one. Also, Saturn will be disk + rings, that's it; no Cassini division, no ring shadow, or anything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's okay, then keep reading.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Availability of instruments depends on where you live. I'm more familiar with the market in North America, but EU can't be too different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surprisingly, a very decent cheap telescope is the Galileoscope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://galileoscope.org/&quot;&gt;http://galileoscope.org/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, it's a DIY kit that you assemble yourself. But the primary lens is a surprisingly good achromat doublet; the design makes sense for an instrument like this; it accepts standard 1.25&quot; eyepieces used for bigger telescopes; and the price is within your limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an amateur optician and telescope maker, I was pleasantly surprised by this instrument. It works well up to about 100x, which is the theoretical limit for its aperture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What will be difficult with this scope - and with any cheap refractor - is the mount. For astronomy you need a rock-solid mount, which is hard to do. The Galileoscope can be plugged into a standard photographic tripod - maybe you can find a good one second-hand. The site also gives suggestions for DIY mounts. Try different things and see what works. I've seen many commercial scopes, more expensive than this, that were basically unusable due to really poorly thought out (or poorly executed) mounts. Beware.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If the DIY route is not appealing, try some ready-made scopes. Here are a couple suggestions of vendors (I'm not affiliated with anyone):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.telescope.com/Telescopes/Best-Telescopes-for-Kids/pc/1/459.uts&quot;&gt;http://www.telescope.com/Telescopes/Best-Telescopes-for-Kids/pc/1/459.uts&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.optcorp.com/telescopes/all-telescopes.html&quot;&gt;http://www.optcorp.com/telescopes/all-telescopes.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sort by price and keep digging. The mount will be a weak point, always, in this price range. Those refractors with tall, thin tripods tend to be very wobbly. It's a true hell to push the scope a bit this way, only to have the tripod rebound the other way - very annoying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may have better success with one of those short, stubby 76 mm table-top reflectors. Tweak the mount until it moves relatively easy, but not too easy. Hold it with both hands and guide it to track the planet. It may actually work better than one of those tall refractors. OTOH, the focal ratio for this reflector is only f/3.9, which is awfully short (it's hard to control aberrations for such a short instrument, unless you invest in high-end optics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do get a refractor, get one of those with short legs. The nice thing about these cheap refractors is that they have a long focal length (large focal ratio), which helps keep aberrations low.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Frankly, there will be some limitations in this price range no matter what. It would help if you could buy from a store with a good return policy - just in case things don't work well for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you say &quot;the Moon is not challenging&quot;? It can be very challenging, once you get into exploring the tiny craters and so on. Just a suggestion. ;)&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2014-06-21T01:16:35.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4719" PostTypeId="1" AcceptedAnswerId="4721" CreationDate="2014-06-21T06:03:59.407" Score="8" ViewCount="247" Body="&lt;p&gt;The new Cosmos extra features mentions that at about 6,771,500,000 years ago the universe began an accelerated expansion. How do we know this? What evidence do we have for this renewed and accelerated expansion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the interactive cosmic calendar on the new Cosmos bluray, dates above taken from the heading&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Until [now], gravity has been slowing the expansion of the universe,&#xA;  but suddenly a mysterious, unknown force that we call dark energy&#xA;  takes over, the expansion of the universe begins to accelerate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How do we know it was dark energy? Given we can't define or detect dark energy, why do we believe this is the cause? How do we even know this new acceleration happened? Something in the red shift? Is there a gap in the red shift in things we can see that would imply a slow expansion and then a fast expansion? If there is a gap, why is dark energy the only explanation? Or are there other theories?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Yes that is a lot of questions, but the main question is &lt;strong&gt;how do we know an accelerated expansion happened at that time?&lt;/strong&gt; the rest of the questions stem from that, mainly in asking for data to back up claims, for or against&lt;/p&gt;&#xA;" OwnerUserId="1909" LastEditorUserId="1386" LastEditDate="2014-06-21T20:34:15.217" LastActivityDate="2014-06-21T20:34:15.217" Title="Dark Energy Expansion" Tags="&lt;gravity&gt;&lt;expansion&gt;&lt;redshift&gt;&lt;cosmological-inflation&gt;&lt;dark-energy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4720" PostTypeId="1" AcceptedAnswerId="4723" CreationDate="2014-06-21T12:12:36.003" Score="2" ViewCount="2104" Body="&lt;p&gt;Would an object with mass traveling the speed of light destroy the whole universe because it would have infinite energy / mass?  According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Speed_of_light#Upper_limit_on_speeds&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt; and other sources, it takes infinite energy to go the speed of light, which should be impossible.  However, for the sake of the question, let's assume it were somehow achieved e.g. potentially as described &lt;a href=&quot;http://www.popsci.com/technology/article/2012-03/pro-tip-flying-faster-speed-light-could-have-devastating-consequences&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My premise behind it possibly destroying the whole universe is that an object with infinite energy should also have infinite mass ($e=mc^2$, and infinity / $c^2$ is still infinity).  Infinite mass should have infinite gravity, which may well destroy the entire universe (although, would this destruction travel at the speed of light, so it would &quot;slowly&quot; destroy the universe vs. instantly?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thought is that it could just form a black hole instead, but as I understand it, a black hole is an infinitely small (singularity) of infinite density, with a &quot;net mass&quot; equal to the net mass that it has ingested.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="23378" LastEditDate="2018-06-15T16:38:17.943" LastActivityDate="2018-06-16T13:52:23.347" Title="If an object with mass were to somehow go the speed of light, would it destroy the whole universe?" Tags="&lt;universe&gt;&lt;light&gt;&lt;speed&gt;&lt;mass&gt;&lt;special-relativity&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2018-06-21T13:12:00.283" ContentLicense="CC BY-SA 4.0" />
  <row Id="4721" PostTypeId="2" ParentId="4719" CreationDate="2014-06-21T14:09:18.190" Score="6" Body="&lt;p&gt;Is Cosmos seriously using that exact number? Egads... if they are, don't take it too seriously, but otherwise they're probably conceptually correct.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do we know it was dark energy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In cosmology, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambda-CDM_model&quot; rel=&quot;noreferrer&quot;&gt;ΛCDM model&lt;/a&gt; fits numerous observations, most notably those observed by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wilkinson_Microwave_Anisotropy_Probe&quot; rel=&quot;noreferrer&quot;&gt;WMAP&lt;/a&gt; satellite, but also others, to the Friedmann-Robertson-Walker family of solutions of general relativity, which describe a homogeneous and isotropic universe. The $\Lambda$ refers to the cosmological constant, which is the simplest model of dark energy, and is equivalent to an intrinsic energy density of the vacuum--hence, it must be Lorentz-invariant, and since time and space mix in a Lorentz transformation, the only way for that to be so is for its pressure to be equal to exactly the negative of its energy density. With $\Lambda&amp;gt;0$, the pressure is negative, and this is what's responsible for the outward gravitational acceleration of the universe.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given we can't define or detect dark energy, why do we believe this is the cause?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The cosmological constant was defined eight decades before the discovery of cosmic acceleration, and we believe it to be the cause because it's theoretically very simple and it fits observations. There are other models for dark energy, though.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do we even know this new acceleration happened?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cosmological evolution is described by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;noreferrer&quot;&gt;Friedmann equations&lt;/a&gt;, which has the principal contribution of radiation, matter, and dark energy, in addition to spatial curvature of the universe. As the universe expands, both radiation and matter get diluted, but dark energy does not. Therefore, if the universe keeps expanding, the proportion of dark energy becomes dominant over both radiation and matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most likely, the article is simply an exaggerated reference to the transition between the &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Matter-dominated_era&quot; rel=&quot;noreferrer&quot;&gt;matter-dominated era&lt;/a&gt;&lt;/em&gt; and the &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dark-energy-dominated_era&quot; rel=&quot;noreferrer&quot;&gt;dark-energy-dominated era&lt;/a&gt;&lt;/em&gt;. The particular number $6,771,500,000$ years seems comically precise for much of anything in cosmology, but since the nine-year best-fit WMAP+eCMB+BAO+H₀ (see above WMAP link) for the age of the universe is $13.772\pm0.059\,\mathrm{Ga}$, I propose the following explanation:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The author hears that the age of the universe is $13.772$ billion years, and completely ignores any error bounds on that estimate.&lt;/li&gt;&#xA;&lt;li&gt;The author hears from somewhere else that the dark-energy-dominated era began when the universe was around $7$ billion years old, and again treats this number as exact.&lt;/li&gt;&#xA;&lt;li&gt;The author then subtracts the two.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This accounts for all but the last $0.5\,\mathrm{Ma}$, which about one-ten-thousandth. I conclude that my theory of the origin of the number agrees rather well with the empirical evidence of the author's words. Perhaps the $0.5$ as the next digit was rationalized as some sort of midway-point in the author's mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to where $7$ billion years comes from, I have two guesses.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In cosmology, the Hubble parameter is sometimes taken to be unitless, by scaling to $100\,\mathrm{km/s/Mpc}$. For example, we sometimes say that $h = 0.704$ when we mean that $H_0 = 70.4\,\mathrm{km/s/Mpc}$. Thus, $100\,\mathrm{km/s/Mpc}$ is kind of a de facto standard of comparison. It might be a coincidence, but it's rather suspicious that if one ignores the contribution of radiation (which is justifiable, since it's negligible except in the very early universe) and pretends that $H_0 = 100\,\mathrm{km/s/Mpc}$, then the age of the universe when the matter and dark energy densities are equal would be&#xA;$$\frac{2\ln(1+\sqrt{2})}{3\sqrt{1-\Omega_\text{M}}}\frac{1}{H_0}\bigg\vert_{\Omega_\text{M} = 0.2865} \approx 0.70H_0^{-1}\bigg\vert_{h = 1.0} \approx 6.8\,\mathrm{Ga}$$&#xA;I'm using the Ryden reference given in wiki for matter-dominated era above. Thus, I think someone did a guesstimate using a nice, round number for the Hubble parameter and rounded to $7$ billion years.&lt;/li&gt;&#xA;&lt;li&gt;The transition between matter-dominated and dark-energy-dominated is a somewhat arbitrary matter of convention. We &lt;em&gt;could&lt;/em&gt; take it mean when matter and dark energy densities are equal, as above, but on the cosmological scale, pressure is three times as important as energy density (cf. Friedmann equations). Thus, we could take it instead as the time when the matter density was &lt;strong&gt;twice&lt;/strong&gt; the dark energy density. This would be earlier than the $9.8\,\mathrm{Ga}$ given in the Ryden reference, but arguably more appropriate because the Friedmann equations would predict zero acceleration, i.e., it would be the transition between inward and outward acceleration. I haven't calculated when that would be, but this sort of conventional fuzziness is &lt;em&gt;might&lt;/em&gt; be responsible for the discrepancy between the wiki's &lt;a href=&quot;http://en.wikipedia.org/wiki/Matter-dominated_era&quot; rel=&quot;noreferrer&quot;&gt;matter-dominated era&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dark-energy-dominated_era&quot; rel=&quot;noreferrer&quot;&gt;dark-energy-dominated era&lt;/a&gt; pages. Although the latter claims $5\,\mathrm{Ga}$ instead, rather than $7\,\mathrm{Ga}$ (but it doesn't explain the source).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So Cosmos may or may not be right about dark-energy-dominated era beginning when the universe about $7$ billion years old, but concluding that it happened $6.7715\,\mathrm{Ga}$ ago is certainly taking some numbers as too exact to be justifiable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Something in the red shift?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think at this point you're better off making a separate question on specific issues with the ΛCDM model, if you have any. For the purposes of this question, I'm taking ΛCDM as a given, as almost certainly Cosmos is doing as well.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Yes that is a lot of questions, but the main question is &lt;strong&gt;how do we know an accelerated expansion happened at that time?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The standard cosmological model is that dark energy is a cosmological constant. Thus, its energy density does not change. It then becomes straightforward that at some point, cosmological expansion makes matter too dilute to compete with dark energy, which is when dark energy starts dominating. It was there all along, but before then, matter was dense enough to be dominant instead.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-06-21T14:09:18.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4722" PostTypeId="1" AcceptedAnswerId="4748" CreationDate="2014-06-21T15:39:46.753" Score="10" ViewCount="545" Body="&lt;p&gt;It seems a little counter intuitive that neutron stars possess such strong magnetic fields. Its electric charge is presumably zero, so however fast it spins, it shouldn't generate any magnetic field. Or is it due to the electric charges of quarks or their intrinsic spins?&lt;/p&gt;&#xA;" OwnerUserId="1510" LastActivityDate="2017-05-03T18:44:02.410" Title="Origin of the magnetic field of neutron stars" Tags="&lt;neutron-star&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4723" PostTypeId="2" ParentId="4720" CreationDate="2014-06-21T16:26:58.390" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Would an object with mass traveling the speed of light destroy the whole universe because it would have infinite energy / mass?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we understand the question as a limiting process, which is the only way it makes any sense, the answer is no. For illustrative simplicity, take a spherically symmetric isolated body, so that its exterior gravitational field is the Schwarzschild spacetime. Now boost this to ultrarelativistic speeds. In the speed of light limit, the result is a linearly polarized axisymmetric gravitational pp-wave, the &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Aichelburg%E2%80%93Sexl_ultraboost&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aichelburg-Sexl ultraboost&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its effect on test particles it passes is an &lt;em&gt;impulse&lt;/em&gt;: it instantaneously bends the worldlines, but does not destroy anything. You can think of this as infinite force acting for an infinitesimally small time, making the overall effect finite and nondestructive, somewhat analogously to a particle encountering a Dirac-delta potential in Newtonian physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is similar to the ultrarelativistic limit of a moving electric charge. The electric field lines are Lorentz-contracted along the direction of travel, squeezing them together (image credit: &lt;a href=&quot;http://en.wikipedia.org/wiki/Relativistic_electromagnetism&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;kbd&gt;&lt;img src=&quot;https://i.stack.imgur.com/wGyCk.png&quot; alt=&quot;Stationary charge electric field&quot;&gt;&lt;/kbd&gt;&lt;kbd&gt;&lt;img src=&quot;https://i.stack.imgur.com/NhPz1.png&quot; alt=&quot;Moving charge electric field&quot;&gt;&lt;/kbd&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the ultrarelativistic limit, the electric field in the transeverse direction becomes infinitely strong, but it is also Lorentz-contracted to be infinitely thin. Without ignoring the magnetic field, the result is an electromagnetic plane wave with a Dirac delta profile. The gravitational case is analogous, though quantitatively different. Some details on this can be found in &lt;a href=&quot;http://arxiv.org/abs/gr-qc/0110032/&quot; rel=&quot;nofollow noreferrer&quot;&gt;arXiv:gr-qc/0110032&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My premise behind it possibly destroying the whole universe is that an object with infinite energy should also have infinite mass (E=mc^2, and infinity / c^2 is still infinity). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The concept of 'relativistic mass' is largely deprecated in modern physics. Mass is more properly related to energy and momentum via $(mc^2)^2 = E^2 - (pc)^2$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One thought is that it could just form a black hole instead, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, that's completely wrong. It doesn't become a black hole for the same reason it doesn't destroy the universe: the gravitating body moving relativistically past a stationary observer is physically equivalent to a relativistic observer moving past a stationary gravitating body. But the observer moving relativistically obviously does nothing to the gravitating body!&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="16685" LastEditDate="2018-06-16T13:52:23.347" LastActivityDate="2018-06-16T13:52:23.347" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4724" PostTypeId="1" CreationDate="2014-06-21T16:07:40.063" Score="5" ViewCount="487" Body="&lt;p&gt;Each day light has more time to reach our eyes from distant galaxies. In one day, light travels 2.59×10^10 km. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So our observable universe (assuming my simple math skills apply here and there isn't something about relativity to take into account), expands that much every day in all directions. So at what rate are new galaxies being encompassed by the expanding observable universe?&lt;/p&gt;&#xA;" OwnerDisplayName="crclayton" LastEditorUserId="67" LastEditDate="2014-06-21T20:32:04.537" LastActivityDate="2014-11-07T16:57:33.693" Title="How many new galaxies enter the observable universe each day/year/decade?" Tags="&lt;expansion&gt;&lt;observable-universe&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4725" PostTypeId="1" CreationDate="2014-06-21T21:42:46.977" Score="14" ViewCount="1132" Body="&lt;p&gt;The closest confirmed black hole is several thousand light years away from earth. Our galaxy has about 100 billion stars. I didn't find any reliable information on the black hole count of ratio versus stars for our galaxy. Some sources say about one in a thousand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to estimate how close a black hole would be given the data we have. If I did the calculation myself, I would use the volume of the galaxy and the number of stars in it. This would give the average number of stars per some volume, and out of that, the average number of black holes per volume. The distribution of stars would also be have to taken into account, since our part of the galaxy isn't as densely populated. Out of this the statistically approximate distance of the closest black hole could be obtained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is anyone equipped with enough information to give me an estimate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically what is needed is the ratio of black holes to stars for our galaxy and then list of x nearest stars and their distances, where x is the ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;I'm asking this because black holes are object of interest, and will have to be visited in the future. Given that our space capabilities might be limited to a distance of a couple of dozens of light years in the near future, this number is interesting.&lt;/em&gt;)&lt;/p&gt;&#xA;" OwnerUserId="818" LastEditorUserId="818" LastEditDate="2014-06-22T22:41:56.623" LastActivityDate="2018-02-28T19:26:18.507" Title="Statistically, what would the average distance of the closest black hole be?" Tags="&lt;star&gt;&lt;galaxy&gt;&lt;black-hole&gt;&lt;milky-way&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4730" PostTypeId="1" AcceptedAnswerId="4734" CreationDate="2014-06-22T15:49:25.120" Score="4" ViewCount="96" Body="&lt;p&gt;How are stars formed? &#xA;I read somewhere that when dust, gas, rocks etc come close to each other due to gravity, slowly and gradually build up a lot of pressure and form stars. &#xA;If what I read is right, then from where did these materials come from?&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2014-06-23T06:24:02.940" Title="Formation of stars" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4731" PostTypeId="2" ParentId="4725" CreationDate="2014-06-22T23:42:39.493" Score="2" Body="&lt;p&gt;I have found a database of the nearest stars within 25 parsecs. The database contains 2608 stars, given the not very accurate estimate of 1 black hole per 1000 stars, would make 2.6 black holes within 81.5 ly( 1 parsec = 3.26 lightyears ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking only the closest 1000 stars from the database, then the maximum distance is 50.9 ly, so there is on average one black hole within that distance. The average distance of all 1000 stars is 35.8 ly, and that is the average distance to that probable black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more accurate ratio would make this much more interesting. Imagine a ration of 1 to 100. Then the average distance becomes only 14.3 ly. &lt;/p&gt;&#xA;" OwnerUserId="818" LastEditorUserId="818" LastEditDate="2014-06-23T02:14:30.487" LastActivityDate="2014-06-23T02:14:30.487" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4732" PostTypeId="1" CreationDate="2014-06-23T01:50:30.670" Score="3" ViewCount="616" Body="&lt;p&gt;What is gravity?  I want to know more than it being simply the &quot;mysterious force&quot; that attracts things to earth.  Is it a particle, a wave, or something else entirely?&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2020-11-10T12:18:53.550" Title="What is gravity really?" Tags="&lt;gravity&gt;&lt;space-time&gt;&lt;space&gt;&lt;newtonian-gravity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4733" PostTypeId="2" ParentId="4732" CreationDate="2014-06-23T01:52:22.777" Score="-3" Body="&lt;p&gt;This is one of the great remaining mysteries of the universe.  I have an idea on this that one can derive known equations for gravity, as described below (this has not been proven, but I believe it is very likely to be the case).  Other people have thought of this as well, but it is not yet the mainstream view.  In fact, someone wrote a paper on it, discussed here: &lt;a href=&quot;http://www.angelfire.com/pq/spaceflow/part2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.angelfire.com/pq/spaceflow/part2.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;My gravity idea (simplified):&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;This recent YouTube video by ScienceClic English actually illustrates my idea very well:&#xA;&lt;div class=&quot;youtube-embed&quot;&gt;&lt;div&gt;&#xD;&#xA;                &lt;iframe width=&quot;640px&quot; height=&quot;395px&quot; src=&quot;https://www.youtube.com/embed/wrwgIjBUYVc?start=0&quot;&gt;&lt;/iframe&gt;&#xD;&#xA;            &lt;/div&gt;&lt;/div&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;I believe gravity is the flow of space itself into mass.&lt;/strong&gt;  By space, I am referring to the 3 dimensions of space that we observe in the universe.  I believe all mass constantly &amp;quot;absorbs&amp;quot; the space around it, bringing everything around closer to itself.  To conceptualize this, imagine mass being a vacuum cleaner, sucking up the air, and the dust that is in the air all around it.  This has the greatest effect on the air that is nearest to the vacuum, and I believe mass behaves likewise with space.  Since the linked video is explaining Relativity, perhaps my idea is actually an explanation of Relativity.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Several things this explains:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;This explains why objects with more mass have more gravity (e.g. each bit of mass is absorbing a small amount of space, so has gravity).  The more mass you have, the more gravity you have.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This explains why gravity is strong when near an object (e.g. when close to earth), and decreases inversely proportional to square distance from the object.  If you imagine an expanding &amp;quot;spherical wave&amp;quot; of space moving inward, surface area of the sphere increases in relation to the square of the distance.  However, the effect is dispersed over this increased surface area, hence it decreases with the square of the distance from the object.  A fairly simple (but 2D vs. 3D) analogy is how when you throw a stone into water, the ripple will be strong at first, near the entry point, and then the ripple disperses as it expands into the surrounding pond.  A good 3D analogy is how sound (or light) expands spherically, and is more faint the further you are away from the source.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The orbit of &amp;quot;binary&amp;quot; systems, including the dynamics between the earth and moon.  The earth is more massive than the moon, and is absorbing more space than the moon.  Therefore, it is moving the moon toward it fairly quickly (but the moon has enough momentum that it remains in orbit vs. falling toward earth).  However, the moon is also pulling on the earth, so the earth is also &amp;quot;falling toward the moon&amp;quot; a little.  Again, the earth has enough momentum not to &amp;quot;fall into the moon&amp;quot;.  This results in mostly the moon orbiting around the earth, but the earth is also &amp;quot;slightly&amp;quot; orbiting around the moon, which would be observed as a slight wobble.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This explains why large mass objects and small mass objects are affected the same by gravity.  Since it is space itself that is moving, whatever is in that space will be moved accordingly.  If it is an anvil, it will be moved just the same as a feather.  Therefore (if it weren't for other effects such as air resistance), a feather would fall to earth the same speed as an anvil.  In practice, air resistance makes the feather small much slower (but repeat the experiment on the moon, and they should hit at the same time since it has no air).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This even explains why things without mass (e.g. light) are also affected by gravity.  Since space itself is moving, light moves along with it.  This would explain gravitational lensing, and why light cannot escape a black hole.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;strong&gt;Second part of my idea (more speculative):&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;I also believe antimatter somehow constantly ejects space back into the universe.  In the video, imagine the reverse happening (e.g. if you could play it in reverse).  I believe when matter and antimatter form, they are somehow &amp;quot;linked&amp;quot; so that space absorbed by matter is ejected by the corresponding antimatter particle.  This would give antimatter a &amp;quot;negative&amp;quot; gravity effect.  I believe there is equal amounts of matter and antimatter in the universe, and the antimatter is in a diffuse cloud spread throughout the universe, where matter is not present.  I believe this accounts for dark energy - the accelerating expansion of the universe.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="1386" LastEditDate="2020-11-10T12:18:53.550" LastActivityDate="2020-11-10T12:18:53.550" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="4734" PostTypeId="2" ParentId="4730" CreationDate="2014-06-23T06:24:02.940" Score="4" Body="&lt;p&gt;It all starts in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_cloud&quot; rel=&quot;nofollow&quot;&gt;molecular cloud&lt;/a&gt;, which are clouds in space with high densities. They can contain all kind of gases and &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_dust&quot; rel=&quot;nofollow&quot;&gt;dust&lt;/a&gt; and also due to their densities and relatively low temperatures they also contain molecules like H2, CO2 and H2O. These are in a thermodynamic equilibrium until something disturbs them (usually nearby supernovae). For some reason a zone of this cloud becomes more dense and gravity wins to and starts to collapse the could until it forms a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theses molecular clouds can be the remains of dead stars or simply a bunch of gas and dust collected and hold at the high density zones at the galaxies spiral arms or arround the core.&lt;/p&gt;&#xA;" OwnerUserId="1678" LastActivityDate="2014-06-23T06:24:02.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4736" PostTypeId="1" CreationDate="2014-06-23T12:31:32.887" Score="3" ViewCount="80" Body="&lt;p&gt;I have a question about Laplace angle-only method for orbit determination where the line of sight vectors are being interpolated. I read somewhere that the method fails (due to a matrix being singular) when the observation site is in the motion plane of the observed body, but failed to see proof for that. &#xA;Could anyone explain to me why is that?&lt;/p&gt;&#xA;" OwnerUserId="1922" LastActivityDate="2014-06-23T12:31:32.887" Title="Singularity in Laplace Method for Orbit Determination" Tags="&lt;orbit&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4737" PostTypeId="2" ParentId="2183" CreationDate="2014-06-23T13:48:46.623" Score="2" Body="&lt;p&gt;If you want to use the user's co-ordinates (longitude and latitude) to calculate the topocentric co-ordinates of an object to a not very high accuracy then you should be able to find a few software libraries (e.g. &lt;a href=&quot;http://libnova.sourceforge.net&quot; rel=&quot;nofollow&quot;&gt;libnova&lt;/a&gt;) that do just that. Many use the algorithms from Jean Meeus' book Astronomical Algorithms. You can use catalogues such as the &lt;a href=&quot;http://cdsarc.u-strasbg.fr/viz-bin/Cat?V/50&quot; rel=&quot;nofollow&quot;&gt;Bright Star Catalogue&lt;/a&gt; to find the equatorial co-ordinates of a star and then use the algorithms in this book to convert to topocentric co-ordinates (also called 'apparent place of a star').&#xA;Of course if you only want to calculate the 3D positions of stars, then it would be useless to use topocentric co-ordinates as the distances to the stars are so great that the error introduced by using geocentric (centre of the earth) instead of topocentric (position on earth) is very very small compared to the distance to the stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to calculate the position of solar system object (planets, asteroids and comets) than you would need to be able to calculate the position of these objects to a very high precision. Meeus' book and software libraries derived from his book contain these algorithms. For planets you could use &lt;a href=&quot;http://cdsarc.u-strasbg.fr/cgi-bin/myqcat3?VI/81/&quot; rel=&quot;nofollow&quot;&gt;VSOP&lt;/a&gt;, but many software libraries include these. Meeus' book also includes algorithms to calculate the positions of the Moon and Pluto and the moons of the major planets. For asteroids and comets you can get the data needed for the algorithms from the &lt;a href=&quot;http://www.minorplanetcenter.net/iau/Ephemerides/EphemOrbEls.html&quot; rel=&quot;nofollow&quot;&gt;IAU Minor Planet Center&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For high accuracy the algorithms provided in these libraries will not suffice and you may need to use professional software libraries such as &lt;a href=&quot;http://aa.usno.navy.mil/software/novas/novas_info.php&quot; rel=&quot;nofollow&quot;&gt;NOVAS&lt;/a&gt; or &lt;a href=&quot;http://www.iausofa.org&quot; rel=&quot;nofollow&quot;&gt;SOFA&lt;/a&gt;. But then you are in milliarcsecond accuracy.&lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2014-06-23T13:48:46.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4738" PostTypeId="1" CreationDate="2014-06-23T15:45:07.670" Score="3" ViewCount="132" Body="&lt;p&gt;I am now reading the book ''The story of Helium and the Birth of Astrophysics''.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really want to observe the lines (dark or bright) in the solar spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any easily available tools for an amateur? &lt;/p&gt;&#xA;" OwnerUserId="1905" LastActivityDate="2016-06-20T00:47:36.543" Title="Any cheap (&lt;100 euro) tools for observing the lines in solar spectrum?" Tags="&lt;spectroscopy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4741" PostTypeId="1" AcceptedAnswerId="4744" CreationDate="2014-06-23T17:11:45.777" Score="8" ViewCount="918" Body="&lt;p&gt;When I was younger, I was looking at HR diagrams, and began to experiment with the HR diagram. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was looking at plotting different groups of stars on the HR diagram when I found RV Tauri stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are stars between 0.4 and 9 solar masses; they have reached the end of their &quot;life&quot; and the fusion shell of such a swollen star is advancing towards its surface (when it hits, the planetary nebula starts to form), thus its heats up and swells further. This phase occurs after the Asymptotic Giant Branch/Red Clump phases (AGB for high-metallicity stars, RC for low-metallicity stars).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking into account the above information, RV Tauri stars are BOTH more luminous and hotter at the surface than red giants (and AGB/RC stars), they should appear to the upper-left of the latter of the aforementioned groups (given the inversion of the temperature axis). All seemed well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then I saw this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AsJVE.jpg&quot; alt=&quot;Additional correlations&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My knowledge of astrophysics tells me that due to stellar winds and the equation E=MC^2, a star's mass (after &quot;birth&quot;) will usually decrease over time (unless receiving matter from another star). But if RV Tauri stars are above and to the left of red giants, it means that according to the graph, an increase in mass occurs every time a star migrates to the RV Tauri stage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that all graphs have their limitations, but in my mind it was wrong to claim that these correlations were set in stone. Only axes could ever display data in a perfect way (they treat different units of data as separate simple measures, not &quot;more unstable&quot; compound measures). So I added a third axis of mass (as this is whole problem is to do with plotting the evolutionary &quot;tracks&quot; of stars); this is what is known to affect stellar evolution to the greatest extent. Here is what I came up with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GSrH1.png&quot; alt=&quot;3DHRD&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did this in order to show that the HR diagram shows general trends of stars (not the evolutionary ones that scientists claim).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was this a valid flaw with the HR diagram - and if so, is mass the best axis to add to it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--EDIT--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The HR diagram is not flawed, I respect that. I also know that there will always be outliers. But the extra correlation of mass seems to be too flawed for that (as detailed above). My new question: is the attribution of these extra correlations wrong?&lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="616" LastEditDate="2021-04-28T15:52:32.027" LastActivityDate="2021-04-28T15:52:32.027" Title="Is there a flaw with the newer purposes and correlations attributed to the HR diagram? (And would a third axis of mass correct the enclosed flaw?)" Tags="&lt;star&gt;&lt;mass&gt;&lt;hr-diagram&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="4742" PostTypeId="2" ParentId="4738" CreationDate="2014-06-23T18:18:24.110" Score="2" Body="&lt;p&gt;Yes, with a piece of 1000 lines / mm of diffraction grating that costs a few dollars, and some odd bits and ends, you could see the Fraunhofer lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a way to build it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://sci-toys.com/scitoys/scitoys/light/spectrograph/spectrograph.html&quot; rel=&quot;nofollow&quot;&gt;http://sci-toys.com/scitoys/scitoys/light/spectrograph/spectrograph.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the bottom there's a link to a place where you could purchase diffraction grating - but any grating around 1000 lines / mm is fine, and you could probably find it in other places too, if this store doesn't work for you for some reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could try to use a digital camera and photograph the spectrum, instead of looking at it directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beware, looking into the Sun poses a certain amount of danger. As long as you keep the slit pretty narrow, this spectrograph is safe. If the spectrum is too bright to look at comfortably, the slit is probably too wide and the instrument is not safe for direct viewing (but probably safe for a camera).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also use it to inspect other sources of light: LED lightbulbs, fluorescent lamps, open flames, etc. It's quite fascinating.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2014-06-23T18:18:24.110" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4744" PostTypeId="2" ParentId="4741" CreationDate="2014-06-24T07:46:32.667" Score="6" Body="&lt;p&gt;Yours is a very, very interesting work. If you publish it it may happen that 3D diagrams will be used in the future. Please publish it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this does not mean that there is a flaw in the H-R diagram. It is correct as it is: it displays Luminosity Vs. Color and is perfect at that. And then it happens that some extra relations appear: stars are not randomly scattered on the HR diagram, but they group in branches, and these branches are related to stages of their life. The position on each branch is related to each star's mass. This is not a general relation holding for all the diagram. Almost the same happens with size, but this is more tightly related to Luminosity, per definition.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-06-24T07:46:32.667" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4746" PostTypeId="1" CreationDate="2014-06-24T16:36:56.630" Score="4" ViewCount="182" Body="&lt;p&gt;There are many formulas for atmospheric pressure on earth, but how does gas behave in free space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am thinking about why stars form. I am guessing that the gas density will influence pressure, as well as gravitational force. So both factors must determine if a given gas mass in a given volume in space will collapse into a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a more well defined question would be: what is the relation between mass and volume in space (let's think about hydrogen only) so that this mass will collapse into a star (or at least forms a spherical body)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second question is: given a mass and volume so that the gas collapses into a spherical body, what is the pressure vs. distance from the center of the sphere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite interesting I think!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="1932" LastEditorUserId="1702" LastEditDate="2014-06-26T15:20:21.967" LastActivityDate="2014-06-26T15:20:21.967" Title="What is the physics of a gas mass subject to gravity in space?" Tags="&lt;star&gt;&lt;gravity&gt;&lt;space&gt;&lt;star-formation&gt;&lt;gas&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4747" PostTypeId="2" ParentId="2553" CreationDate="2014-06-24T20:29:52.983" Score="5" Body="&lt;p&gt;I feel this question is so broad, it might well be the topic of an interesting book on the history of astronomy, should someone be inclined to write it. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I think a few points could be made briefly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. Collecting data&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In astronomy, that means observing the cosmos. That means using an instrument of some kind, typically a telescope, and gathering information through it. Telescope performance is dictated by many factors, but the most important one is size (or aperture).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Telescope size grew rapidly through 1600s and 1700s, from Galileo's 1.5 cm refractor in early 1600s, surpassing the 1 meter aperture in early 1800s - &lt;a href=&quot;https://en.wikipedia.org/wiki/40-foot_telescope&quot; rel=&quot;noreferrer&quot;&gt;Herschel's 40 foot reflector&lt;/a&gt;. There was a steady stream of improvements regularly throughout that 200 year period. One could say that the first golden age of the telescope aperture race both culminated and ended with Herschel and his giant telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there was a lull, briefly interrupted by Lord Rosse's 1.83 meter telescope, &lt;a href=&quot;https://en.wikipedia.org/wiki/Leviathan_of_Parsonstown&quot; rel=&quot;noreferrer&quot;&gt;the Leviathan of Parsonstown&lt;/a&gt;, in mid-1800s. Then nothing again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aperture race was resumed only in early 1900s, with the 2.5 meter reflector on Mt. Wilson, &lt;a href=&quot;https://en.wikipedia.org/wiki/Mount_Wilson_Observatory&quot; rel=&quot;noreferrer&quot;&gt;the Hooker telescope&lt;/a&gt;. Afterwards, throughout the 20th century, and now in early 21st, the race is going strong, with &lt;a href=&quot;https://en.wikipedia.org/wiki/Gran_Telescopio_Canarias&quot; rel=&quot;noreferrer&quot;&gt;the 10.4 meter Gran Canarias segmented reflector&lt;/a&gt; currently in the lead, and &lt;a href=&quot;https://en.wikipedia.org/wiki/European_Extremely_Large_Telescope&quot; rel=&quot;noreferrer&quot;&gt;the 39 meter E-ELT reflector&lt;/a&gt; being under construction at Cerro Armazones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_largest_optical_telescopes_historically&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/List_of_largest_optical_telescopes_historically&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Interpreting data&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The year 1900 marks the boundary between classic physics and the new physics. After that year, relativity and quantum mechanics took off. This is what enabled the new cosmology to emerge in the 20th century.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, with the 1800s science, even with tons of data, there would have been no way to figure out, basically, everything. Supernovas? The expansion of the universe? Dark matter and the rotation of galaxies? This is all based on 20th century physics. 19th century physics would have been clueless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomy used classic physics to derive interpretations from data pretty quickly, and that process had achieved great success already well into the 1700s. That's when &lt;a href=&quot;https://en.wikipedia.org/wiki/Discovery_and_exploration_of_the_Solar_System&quot; rel=&quot;noreferrer&quot;&gt;the structure of the solar system was figured out&lt;/a&gt;, way back to Kepler in the 1600s. Herschel found Uranus in late 1700s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some exceptions here. &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_parallax&quot; rel=&quot;noreferrer&quot;&gt;Stellar parallax&lt;/a&gt; was detected in early 1800s, which enabled an estimate to nearest stars. Spectroscopy showed that distant stars are made of the same elements like the Earth in the 1850s. Around that same time, Neptune was discovered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the 1800s was not quite a completely dry period, in terms of theoretical progress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, a limit was reached anyway in late 1800s, because what was needed was new paradigms in physics to give new life to the interpretation process. That boost occurred after 1900, with relativity and quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cosmology is highly dependent on physics (and vice-versa).&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2014-06-24T20:36:32.443" LastActivityDate="2014-06-24T20:36:32.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4748" PostTypeId="2" ParentId="4722" CreationDate="2014-06-25T03:58:08.360" Score="6" Body="&lt;p&gt;The strong magnetic fields in neutron stars are supposed to come from &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetic_flux#Description&quot; rel=&quot;nofollow noreferrer&quot;&gt;magnetic flux&lt;/a&gt; conservation. If we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Phi_B = \int B\ \mathrm{d}S = \text{const}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Phi_B$ is the magnetic field flux, $B$ is the magnetic field strength, and $\mathrm{d}S$ is the elemental closed surface; then, this integral is constant through the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider the star surface over which take the integral, than &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S = 4\pi R^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R$ is the star radius. This can be translated, altogether with the magnetic flux conservation law,  as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_f = B_i \left(\frac{R_i}{R_f} \right)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $i$ and $f$ are the indices for &lt;em&gt;initial&lt;/em&gt; and &lt;em&gt;final&lt;/em&gt; stages.&#xA;We know that the star implodes from a whatever star size to $\sim10 \; \mathrm{km}$. So the radii ratio is huge. You just need a starting magnetic field of $10-100 \ \mathrm{G}$, to get a final magnetic field of the order of $10^{12}  \ \mathrm{G}$, that is typical in neutron stars.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastEditorUserId="9301" LastEditDate="2017-05-03T18:44:02.410" LastActivityDate="2017-05-03T18:44:02.410" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4749" PostTypeId="1" AcceptedAnswerId="6192" CreationDate="2014-06-26T03:15:32.967" Score="7" ViewCount="454" Body="&lt;p&gt;I know people bounced lasers off of Venus and Mercury in 1960s in order to test the &lt;a href=&quot;http://en.wikipedia.org/wiki/Shapiro_delay&quot; rel=&quot;nofollow&quot;&gt;Shapiro delay&lt;/a&gt;. Since then, we've developed higher power lasers and radio transmitters (&lt;a href=&quot;http://en.wikipedia.org/wiki/Bolshakovo_transmitter&quot; rel=&quot;nofollow&quot;&gt;Bolshakovo transmitter&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any farther objects where we've been able to detect the reflected wave?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="67" LastEditDate="2014-09-16T12:12:38.687" LastActivityDate="2014-09-16T12:12:38.687" Title="What is the farthest object we've been able to bounce signals off of to date?" Tags="&lt;distances&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4750" PostTypeId="1" CreationDate="2014-06-26T11:16:28.443" Score="2" ViewCount="335" Body="&lt;p&gt;What are the various values of metallicity of various known galaxies, like Milky Way, Large Magellanic Cloud, Small Magellanic Cloud, Andromeda and so on?&lt;/p&gt;&#xA;" OwnerUserId="1702" LastActivityDate="2014-07-02T16:37:57.067" Title="Metallicity: What are the metallicity values of various galaxies?" Tags="&lt;galaxy&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4751" PostTypeId="2" ParentId="4746" CreationDate="2014-06-26T13:03:52.923" Score="1" Body="&lt;p&gt;The formula for the change of pressure with altitude is the same-you need hydrostatic equilibrium.  This means that at constant temperature and over small height differences (so gravity can be considered constant) the pressure falls exponentially with the scale height.  You have $\frac {dp}{dz}=-{\rho g}$  You can compute the local $g$ by taking everything closer to the center as a point mass and ignoring all the mass outside the current radius.  You can consider the compression adiabatic, but need to use the appropriate value of $\gamma$ in $PV^\gamma=k$ for the local temperature.  Calculating the radiation is hard unless there is enough dust around, in which case the grains can be black bodies.  If you assume rotational symmetry, you have a 2D problem, which you can model on a grid and step through.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2014-06-26T13:03:52.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4752" PostTypeId="1" AcceptedAnswerId="4756" CreationDate="2014-06-26T18:10:35.157" Score="18" ViewCount="1022" Body="&lt;p&gt;The fact is, many, if not most large galaxies have a supermassive black hole at their center. My question is why? Is it because when these galaxies were first formed supermassive black holes were created out of super-dense clouds of matter, which then attracted the rest of the galaxy? Or is the black hole attracted to the center of the galaxy in such a way that due to its large mass and ability to perturb everything around it, the interactions with the galaxy tend to move the center to where the black hole is located?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems interesting that Sagittarius A* is at the precise center of the galaxy, when it has so much less mass than the rest of the galaxy put together. The sun being at the center of the solar system makes a lot of sense - it accounts for 99.8% of the mass of the solar system. But the mass of Sagittarius A* is on the order of one millionth the mass of the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to phrase it like a chicken and egg problem – what came first, the super-massive black holes, or the galaxies surrounding them?&lt;/p&gt;&#xA;" OwnerUserId="1943" LastEditorUserId="14006" LastEditDate="2016-09-13T08:31:46.330" LastActivityDate="2016-09-13T08:31:46.330" Title="Supermassive black holes at the center of galaxies" Tags="&lt;galaxy&gt;&lt;supermassive-black-hole&gt;&lt;galaxy-center&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4753" PostTypeId="1" AcceptedAnswerId="4754" CreationDate="2014-06-26T19:49:11.450" Score="11" ViewCount="14299" Body="&lt;p&gt;What's the chance that there might be undiscovered chemical elements in the Solar System - either on planets or around the Sun or on asteroids of the Oort-cloud?&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="1469" LastEditDate="2014-07-07T12:44:42.703" LastActivityDate="2019-07-22T20:17:39.177" Title="What's the likelihood of the existence of unknown elements in the Solar System?" Tags="&lt;solar-system&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4754" PostTypeId="2" ParentId="4753" CreationDate="2014-06-26T20:13:19.927" Score="18" Body="&lt;p&gt;As far as elements (e.g. on the periodic table) go, I would say the odds are very slim.  We already discovered or produced all the elements of the Periodic Table up to atomic number 112 at least.  As the number increases, the half lives of the elements generally decreases, and is very short for elements above 102. If this trend holds true as the number increases, practically all the &quot;undiscovered&quot; elements should have turned into the lower known atomic number elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is hope.  There is a theorized &quot;island of stability&quot; where a narrow range of yet to be discovered high atomic number elements may be stable: &lt;a href=&quot;http://en.wikipedia.org/wiki/Island_of_stability&quot;&gt;http://en.wikipedia.org/wiki/Island_of_stability&lt;/a&gt;&#xA;I would say there is a slight chance this element could be discovered in the solar system.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2014-06-26T20:13:19.927" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4755" PostTypeId="1" AcceptedAnswerId="4781" CreationDate="2014-06-26T23:24:09.790" Score="7" ViewCount="16142" Body="&lt;p&gt;What level of percent-illumination of moon (waxing crescent), given by Stellarium, is enough to make it visible with naked eye, in clear sky?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;" OwnerUserId="1842" LastActivityDate="2019-06-05T19:44:18.947" Title="Percent-illumination of crescent moon and its naked-eye visibility?" Tags="&lt;the-moon&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4756" PostTypeId="2" ParentId="4752" CreationDate="2014-06-27T02:08:05.083" Score="6" Body="&lt;p&gt;This is a very well posed question, thanks!&#xA;The problem is, we still don't know.&#xA;What we suppose, is that dark matter came first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Primordial fluctuations of dark matter made the right conditions to accrete enough mass to build SMBHs, and then the surrounding galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But still we don't know if, the SMBH comes from merging, of other SMBHs or &lt;a href=&quot;http://www.nature.com/nature/journal/v466/n7310/full/4661049a.html&quot; rel=&quot;noreferrer&quot;&gt;galaxies&lt;/a&gt;, or from high massive BH that are called &lt;a href=&quot;http://www.google.co.jp/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=6&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0CDwQFjAF&amp;amp;url=http%3A%2F%2Fwww.as.utexas.edu%2F~fsgcon%2Fabstracts_and_talks%2Fd_whalen_talk.pdf&amp;amp;ei=etGsU6LDNITt8AWrqoDQAw&amp;amp;usg=AFQjCNE5gsM8jX83nbTd33xkFDIwHIrZMA&amp;amp;sig2=kE-nLXy0dfGmQCpNSSM8LQ&amp;amp;bvm=bv.69837884,d.dGc&quot; rel=&quot;noreferrer&quot;&gt;seed (of the order of $100\,M_{Sun}$)&lt;/a&gt;.&#xA;Perhaps, there are very recent developments, which could be known by the experts, but until few years ago, this was the scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the mass issue, we could have a better reasoning if we think to the influence spheres. We discovered a relation between the &lt;a href=&quot;http://www.scirp.org/journal/PaperInformation.aspx?paperID=34804#.U6zOo_HMGhc&quot; rel=&quot;noreferrer&quot;&gt;SMBH mass and many bulge properties&lt;/a&gt; (i.e., the luminosity, the velocity dispersion, and others). This means that an object of the order of $\sim1$ pc, influences an object of the order of $\sim1$ kpc (maybe $\sim3$). Plus, &lt;a href=&quot;https://dept.astro.lsa.umich.edu/ugactivities/Labs/MWgalStruct/index.html&quot; rel=&quot;noreferrer&quot;&gt;the bulge structure is essential to the formation&lt;/a&gt; and characterization and classification of the host galaxy.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastActivityDate="2014-06-27T02:08:05.083" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4757" PostTypeId="1" AcceptedAnswerId="4780" CreationDate="2014-06-27T02:10:58.183" Score="9" ViewCount="2354" Body="&lt;p&gt;In the book &quot;A Really Short History Of Nearly Everything&quot;, I read that the larger the star is, the faster it burns itself. Whys that? Wouldn't there be more energy to burn if it's larger, and just be brighter? Our sun can burn for billions of years, but a planet bigger than our sun will only burn for millions of years. Why is that?&lt;/p&gt;&#xA;" OwnerUserId="1944" OwnerDisplayName="Bob the zealot" LastEditorUserId="1944" LastEditDate="2015-02-06T23:44:58.953" LastActivityDate="2015-02-06T23:44:58.953" Title="Burning Out Stars" Tags="&lt;star&gt;&lt;stellar-evolution&gt;" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4758" PostTypeId="2" ParentId="4757" CreationDate="2014-06-27T06:11:42.503" Score="4" Body="&lt;p&gt;It's easy: the biggest the star is the more matter it has to burn but also:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It has &lt;strong&gt;bigger burning zones&lt;/strong&gt;, so it burns at higher rates.&lt;/li&gt;&#xA;&lt;li&gt;As the star is bigger it reaches to &lt;strong&gt;higher pressures and temperatures&lt;/strong&gt; at it's core, so the star can start heavier elements (than H) burning earlier than smaller and cooler stars This creates an onion like structure where different elements are fused at each layer ending in a &lt;strong&gt;core of inert Ni and Fe&lt;/strong&gt; that can't be fused to create energy.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gUiOr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could compare a big star to a small one like a Formula 1 car to a simple car: the F1 is faster, but it runs out of fuel earlier.&lt;/p&gt;&#xA;" OwnerUserId="1678" LastEditorUserId="1678" LastEditDate="2014-07-03T06:34:16.953" LastActivityDate="2014-07-03T06:34:16.953" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4760" PostTypeId="2" ParentId="4755" CreationDate="2014-06-27T19:27:10.783" Score="5" Body="&lt;p&gt;Any percentage at all would be visible, if the sky is dark enough. The surface on the illumined part is as bright as the street in front of your house during the day. So even a very tiny sliver would be clearly visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you're running into a different problem here. As the illumined portion decreases, the apparent position of the Moon in the sky gets closer to the Sun. At some point it will be lost in the glare. It's a contrast issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your best bet is a coincidence: if the Moon's apparent position is very close to the Sun's, while the Sun is setting, and you have clear visibility to a flat western horizon. In that case, the Moon's crescent could become very thin indeed, and still remain visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen it like that, once. The Moon was very close to the Sun, several lunar diameters away. The Sun had just descended under the horizon. The Moon was very low in the sky, extremely thin, almost too thin to estimate the width of the illumined portion. The expression &quot;razor edge&quot; comes to mind. The thin curved line made less than half a circle (maybe 1/3?), with the two ends gradually tapering off in terms of brightness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the unusual shape, it took me a second or two to realize that &lt;em&gt;that was actually the Moon&lt;/em&gt;. I showed it to my kids and their reaction was: &quot;That's the Moon? NO! Wait... Really?&quot; It was the Moon, really, confirmed with SkySafari on my phone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was not drawing attention to itself by any means, being almost lost in the sunset glare - not bright, just a thin, discreet, whitish curve in a yellow-red sky. But once you saw it, it was clearly there. It would have been invisible if the Sun was not below horizon, I think, or maybe not. &lt;em&gt;Definitely&lt;/em&gt; invisible at mid-day, without any doubt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish I had access to a telescope during that rare sight. :( Although, with the Moon at such low elevation, &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_seeing&quot; rel=&quot;nofollow noreferrer&quot;&gt;seeing&lt;/a&gt; must have been atrocious.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2018-09-10T18:01:35.730" LastActivityDate="2018-09-10T18:01:35.730" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="4761" PostTypeId="1" CreationDate="2014-06-28T08:51:42.800" Score="2" ViewCount="816" Body="&lt;p&gt;Imagine two cubes each with the side 1 light year. One of the cubes is empty of any matter. The other cube contains a massive star.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Does the cube which contains the massive star have a larger volume than the empty cube? (Because the mass makes space curvy)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the star is replaced by a gas cloud of the same mass but which is evenly distributed inside the cube, does the volume of the cube change?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is it not enough with geometry to calculate volumes? Is knowledge of the mass (and its distribution?) also necessary?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Does mass (concentrations) create new space volume, or does it just change the properties of the space around it?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-05-03T15:04:33.107" Title="Does mass create space?" Tags="&lt;space&gt;&lt;general-relativity&gt;&lt;mass&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4762" PostTypeId="2" ParentId="4761" CreationDate="2014-06-28T12:54:21.400" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it not enough with geometry to calculate volumes? Is knowledge of the mass (and its distribution?) also necessary?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To answer this question. Mass distribution does make a difference. For example, if a star collapses into a black hole, the mass of the black hole is less than the original mass of the star due to some matter lost during the supernova. But now the star is compressed into a more compact space. So, now one can reach much more closer to its center(and experience more gravitational force) than one could have before, So, a BH is a deadlier object than normal stars.&lt;/p&gt;&#xA;" OwnerUserId="1156" LastActivityDate="2014-06-28T12:54:21.400" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4763" PostTypeId="2" ParentId="4753" CreationDate="2014-06-28T16:58:13.493" Score="9" Body="&lt;p&gt;Further to the answer of @Jonathan, the thing that distinguishes one chemical element from another is the number of protons in the nucleus, which in turn determines the number of orbital electrons in the uncharged atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we already know the element that corresponds to any given number of protons between 1 and 112;  that's the atomic number. And you can't have a fraction of a proton. The only room for possible new elements is on the end.&lt;/p&gt;&#xA;" OwnerUserId="420" LastActivityDate="2014-06-28T16:58:13.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4764" PostTypeId="1" AcceptedAnswerId="4772" CreationDate="2014-06-28T17:57:39.070" Score="5" ViewCount="137" Body="&lt;p&gt;I live in a light polluted city of the northern hemisphere and I would like to know what I can see with my Skywatcher BK 707AZ2. I have already seen Jupiter (and its moons), Saturn, Mars, M45, the Double cluster, M42 and, obviously, the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M31, M13 and maybe M33 are also probably visible, but what else?&lt;/p&gt;&#xA;" OwnerUserId="1890" LastEditorUserId="1890" LastEditDate="2014-06-28T18:04:00.260" LastActivityDate="2014-06-29T15:33:17.933" Title="What are the objects that are visible from a city?" Tags="&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4765" PostTypeId="2" ParentId="4764" CreationDate="2014-06-28T20:26:04.320" Score="5" Body="&lt;p&gt;We can't know this without knowing what the sky is like for your city. However, you can &lt;a href=&quot;http://www.imo.net/visual/major/observation/lm&quot;&gt;determine what the limiting magnitude&lt;/a&gt; is for your location, then you can set planetarium software to only display catalogued stars that are brighter than this limit. You can do this for naked eye objects, but you could also do this using your telescope to find the limiting magnitude for your equipment also.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-06-28T20:26:04.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4766" PostTypeId="1" AcceptedAnswerId="4782" CreationDate="2014-06-28T22:23:43.720" Score="5" ViewCount="365" Body="&lt;p&gt;(Hope this is the right Stack-exchange site for this question)&lt;br/&gt;&#xA;I'm working on a sci-fi RPG campaign, set on a very atypical location. Since this is a work of fiction, there's enough room for speculation and artistic license. But I'd still like to start from a more astronomically sound perspective.&lt;br/&gt;&#xA;&lt;br/&gt;&lt;/p&gt;&#xA;&lt;h3&gt;Background&lt;/h3&gt;&#xA;&lt;p&gt;Consider the following system:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A single &lt;strong&gt;moon&lt;/strong&gt;, with properties roughly comparable to Earth's so that it can support human life (1G gravity, Nitrogen-Oxygen-Carbon atmosphere, soil and liquid water etc.). This moon is tidally locked with it's planet.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A single gas-giant &lt;strong&gt;planet&lt;/strong&gt;, with a size roughly comparable to Saturn. Orbiting it's star so that it completes a cycle about every 12 earth years.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;strong&gt;Star&lt;/strong&gt;, larger than Sol and emitting enough energy to keep the moon in a comfortably warm temperature.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;The features above may be altered to ensure possibility of human survival on the moon, and if possible, to keep the moon tidally locked with a large planet.&lt;/p&gt;&#xA;&lt;p&gt;Now, the complications (inspired by a short story by Larry Niven):&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The whole system (star, planet and moon) are &lt;strong&gt;traveling at a relativistic speed&lt;/strong&gt; across the galaxy.&lt;/li&gt;&#xA;&lt;li&gt;(Based on the theory that some parts of the universe might be made from anti-matter) The whole system is &lt;strong&gt;made from anti-matter&lt;/strong&gt;, but is currently traveling through a normal-matter area of space.&#xA;Note: the &amp;quot;human life&amp;quot; on the moon are also based on anti-matter...&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Finally, the story will begin with a small community of primitive &amp;quot;anti-matter humans&amp;quot;, living on the moon in a place from which the planet fully visible, and at the zenith.&lt;/p&gt;&#xA;&lt;h3&gt;The Questions&lt;/h3&gt;&#xA;&lt;p&gt;I'd appreciate ideas, insights and comments on the following:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;From the moon, how will the sky look like?&lt;/li&gt;&#xA;&lt;li&gt;What will the equivalent of a &amp;quot;day and night cycle&amp;quot; (sun-up to sun-up) look like?&lt;/li&gt;&#xA;&lt;li&gt;Will the winds on the moon have any &amp;quot;non-earthlike&amp;quot; patterns?&lt;/li&gt;&#xA;&lt;li&gt;Considering that at relativistic speeds, the odds of the bodies in the system hitting hydrogen particles in the vacuum of interstellar space are significant (is that even correct?), will that create any special aura or energy streams from the sun or planet?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Thank you for your help.&lt;/p&gt;&#xA;" OwnerUserId="1957" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-05-25T15:40:04.173" Title="Help in determining the features of an unusual, fictional star system" Tags="&lt;tidal-forces&gt;&lt;earth-like-planet&gt;&lt;habitable-zone&gt;&lt;antimatter&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4767" PostTypeId="2" ParentId="970" CreationDate="2014-06-29T03:56:16.413" Score="2" Body="&lt;p&gt;You can actually figure out if you are at a pole without ever going outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set a pendulum swinging, the bigger the better. If it takes 23hours and 56 minutes (a sidereal day) to complete one full precession clockwise, you are at the north pole. If it takes that long and goes counterclockwise then you are at the south pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it doesn't precess at all, then break out your summer shorts because you are on the equator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can actually figure out what your latitude is this way in general, it will be the arcsin of how many degrees it precesses in one day.&lt;/p&gt;&#xA;" OwnerUserId="1958" LastActivityDate="2014-06-29T03:56:16.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4769" PostTypeId="1" AcceptedAnswerId="4779" CreationDate="2014-06-29T14:05:53.227" Score="1" ViewCount="367" Body="&lt;p&gt;&lt;a href=&quot;http://www.celestron.com/browse-shop/astronomy/telescopes/cpc-800-gps-%28xlt%29-computerized-telescope&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; is a Celestron CPC 800 XLT telescope. It doesn't seem to have charge-coupled device (CCD) attached. Can a CCD be attached separately to this 8&quot; Schmidt-Cassegrain telescope even if it didn't come along with the package?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For astro-phtography how cheaper and cost-effective would a DSLR camera be than CCD? I think it is better to suit yourself with a CCD than a DSLR camera for quality photography. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do excuse this novice who is here to learn.&lt;/p&gt;&#xA;" OwnerUserId="1702" LastEditorUserId="1702" LastEditDate="2014-07-02T16:36:13.650" LastActivityDate="2018-10-14T23:35:36.043" Title="CCD in telescopes: Observation and Astro-photography" Tags="&lt;telescope&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4770" PostTypeId="1" CreationDate="2014-06-29T14:12:47.910" Score="2" ViewCount="335" Body="&lt;p&gt;What is a Schmidt-Cassegrain (SC) telescope? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have Newtonian telescopes and also Galilean ones. How is SC telescope different from other optical telescopes?&#xA;What is its construction detail and what different mechanism does it work with?&lt;/p&gt;&#xA;" OwnerUserId="1702" LastActivityDate="2014-07-04T18:42:15.637" Title="Schmidt-Cassegrain telescope" Tags="&lt;telescope&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4771" PostTypeId="1" AcceptedAnswerId="6485" CreationDate="2014-06-29T14:41:45.330" Score="13" ViewCount="1274" Body="&lt;p&gt;Scientists talk about the possibility of life underground on Mars, and I assume they are talking about microscopic life that can live in small gaps in the soil or rock.  But is there any evidence that there might be large caverns under the surface of Mars relative to the human scale or even larger?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I wonder how could we even detect or map such subsurface features on Mars or other rocky bodies in the solar system such as Titan, Venus, or even the Moon.  Could we use radar or maybe sonar for bodies with substantial atmospheres from an orbiting spacecraft?  Are there other methods for detecting sub-surface cavities? &lt;/p&gt;&#xA;" OwnerUserId="1962" LastEditorUserId="37191" LastEditDate="2021-02-25T13:11:56.367" LastActivityDate="2021-02-25T13:11:56.367" Title="Are there large underground caverns on Mars?" Tags="&lt;planet&gt;&lt;mars&gt;&lt;volcanism&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4772" PostTypeId="2" ParentId="4764" CreationDate="2014-06-29T15:33:17.933" Score="5" Body="&lt;p&gt;M6, the Butterfly cluster, and M7, the Ptolemy cluster,  are open clusters in the Scorpius constellation. They are interesting amateur target to observe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M4, a globular cluster, is a very interesting object too. It is one of the brightest globular clusters. M4 is in the Scorpius cluster too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure about your location, but you should try observing M81 and M82. M82 is known as the Cigar galaxy. Both messier objects are galaxies. They could be tricky to find but they are worth the try. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Happy observations and clear skies :)&lt;/p&gt;&#xA;" OwnerUserId="1963" LastActivityDate="2014-06-29T15:33:17.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4773" PostTypeId="1" CreationDate="2014-06-29T16:34:15.533" Score="2" ViewCount="53" Body="&lt;p&gt;I'm reading many of the Wikipedia pages about exoplanets and the different methods they are using to detect them.  But I wonder, it seems that the emphasis is on detecting and finding new exoplanets, but are astronomers continuously monitoring the exoplanets that have already been discovered?  For instance, if there is a collision between a detected exoplanet and a previously unknown planet, this could cause a change in some of the orbital parameters, and could be potentially detected.  If this happens, would we learn of this change, or will it have to wait until astronomers again take another look at the system?&lt;/p&gt;&#xA;" OwnerUserId="1962" LastActivityDate="2014-07-02T11:14:01.523" Title="Are astronomers continuously monitoring exoplanetary systems?" Tags="&lt;exoplanet&gt;&lt;orbital-migration&gt;&lt;astrometry&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4774" PostTypeId="1" CreationDate="2014-06-29T17:32:11.923" Score="10" ViewCount="415" Body="&lt;p&gt;I'm reading on Wikipedia about Halo Stars that orbit the center of the galaxy at a high inclination away from the plane of the Milky Way.  It seems that at some point, these stars must dive back into the core of our galaxy at very high relative velocities.  This causes me to wonder...wouldn't this, combined with the greater density of stars in the core, cause a large number of stellar collisions in the galactic core?&lt;/p&gt;&#xA;" OwnerUserId="1962" LastActivityDate="2014-07-10T18:12:03.180" Title="Are there a lot of collisions between stars in the core of the galaxy?" Tags="&lt;star&gt;&lt;galaxy&gt;&lt;galactic-dynamics&gt;&lt;eccentric-orbit&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4775" PostTypeId="1" AcceptedAnswerId="4777" CreationDate="2014-06-30T09:51:03.753" Score="1" ViewCount="55" Body="&lt;p&gt;I have a HYG catalogue and I'd like to filter out stars visible at night and in Europe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doesn't have to be exact. I was assuming that it'll be all stars with positive declination, but I wasn't sure about RA values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, Michal&lt;/p&gt;&#xA;" OwnerUserId="1969" LastActivityDate="2014-06-30T17:17:06.197" Title="Naive star filter visible at night in certain country" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4776" PostTypeId="1" AcceptedAnswerId="6035" CreationDate="2014-06-30T15:07:58.010" Score="14" ViewCount="1945" Body="&lt;p&gt;The accelerating expansion of space means that the space between us and far away galaxies expands faster than light can travel through space. There is a horizon of possible observation beyond which galaxies disappear from us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At what rate do galaxies, or some other measurement of mass/energy, disappear beyond that horizon at present rate? Will future telescopes actually see galaxies red shift themselves into oblivion? I imagine there being issues about the definitions of &quot;when&quot; and &quot;where&quot; in this context. Answers should try to address how they deal with that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the radius of the observable universe expands at the speed of light, my intuition tells me that there should've been a point in time when the expansion of space balanced the arrival of light of more distant parts of space. if so, when did that happen?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-07-07T15:48:21.513" Title="How many galaxies disappear beyond the Hubble Bubble horizon every year now?" Tags="&lt;expansion&gt;&lt;cosmological-horizon&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ClosedDate="2016-07-07T16:10:25.353" ContentLicense="CC BY-SA 3.0" />
  <row Id="4777" PostTypeId="2" ParentId="4775" CreationDate="2014-06-30T17:17:06.197" Score="1" Body="&lt;p&gt;In the answer &lt;a href=&quot;https://physics.stackexchange.com/a/26787&quot;&gt;https://physics.stackexchange.com/a/26787&lt;/a&gt; you can find an example how to convert celestial coordinates to horizon coordinates an than check, if the altitude is above the horizon &lt;/p&gt;&#xA;" OwnerUserId="950" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2014-06-30T17:17:06.197" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4778" PostTypeId="2" ParentId="4770" CreationDate="2014-07-01T10:28:09.567" Score="5" Body="&lt;p&gt;There are many other variations on telescope designs others than the ones you have mentioned. They differ in the light path, the shape of the optics utilised, the use of mirrors and/or lenses, and the physical structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Schmidt-Cassegrain is a particular type of Cassegrain reflector. A Cassegrain reflector uses two mirrors, a primary and a secondary. Both are curved, and the secondary is placed in front of the primary, which blocks some of the light to the primary. The primary directs light onto the secondary, and the secondary directs light through a hole in the centre of the primary. This 'folds' the lightpath, resulting in a short telescope, even at quite wide apertures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Schmidt-Cassegrain is a Cassegrain with particular design details. The primary mirror has a spherical curve, and the secondary usually has a spherical curve, though some manufacturers modify the design with a parabolic secondary to improve the performance. A key characteristic is the addition of a lens to correct for the optical aberrations produced by the spherical mirror(s). This lens is also used to mount the secondary mirror, so no vanes are required to hold it in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Galilean telescope is a refractor, and so it uses lenses exclusively instead of mirrors. Lenses may be able to out-perform mirrors, but it is easier to manufacture high-quality large mirrors than large lenses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newtonian telescopes are also reflectors with two mirrors, but one is flat. It also blocks some of the light in the centre of the objective, but it is held in place by vanes instead of a lens. The flat mirror angles the light out the side of the telescope near the objective end of the telescope. While this may frequently place the eyepiece in a more convenient position to access it, especially when aiming near the zenith, it is also necessary to rotate the telescope in the mount to get the eyepiece in the right place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are pros and cons to all telescopes, and selecting a telescope requires knowing what you will be doing with it, and choosing something that complements that purpose well with its pros, and that has cons that are not relevant.&lt;/p&gt;&#xA;" OwnerUserId="67" LastEditorUserId="67" LastEditDate="2014-07-01T11:22:41.270" LastActivityDate="2014-07-01T11:22:41.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4779" PostTypeId="2" ParentId="4769" CreationDate="2014-07-01T10:51:32.380" Score="2" Body="&lt;p&gt;You will typically find that CCD cameras are sold separately to the telescopes. That way the telescope can be used visually or for astrophotography, by either fitting an eyepiece, or a camera. Most telescopes do ship with eyepieces, as they are much less expensive than a CCD and it is sort of expected, but most observers will typically replace this eyepiece with a better one anyway, so to all intents and purposes, when you buy a telescope you are just buying the Optical Tube Assembly and maybe a mount, and you add to that all the accessories you need to accomplish your objectives. Different CCDs will be used for different purposes, for instance planetary observations or deep space objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That telescope has a fork mount. Note that this constrains how much equipment you can put on the end of the telescope  without getting in the way of the mount. It also is an Alt-Az mount, which means it is not polar aligned; stars will rotate in the field of view, requiring a field de-rotator in addition to a CCD. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like a nice enough scope for visual observations, but if you are interested in astrophotography, you might want to hold off for a bit, maybe get a scope for visual use first, and maybe make use of well-set-up robotic astrographs that you can hire over the internet, like slooh.com and itelescope.net - get some experience using premium gear that you don't have to worry about setting up or maintaining, and while you do that, work out what gear you will be happy about getting for yourself. The fact you are asking this question about CCDs suggests you aren't yet ready to spend money on astrophotography gear...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting set up well for visual stuff can set you back hundreds... Getting set up for astrophotography can set you back many thousands, so you want to be well equipped to make good purchasing decisions. And you can rent a lot of time on some pretty good telescopes for that same $...&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-07-01T10:51:32.380" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4780" PostTypeId="2" ParentId="4757" CreationDate="2014-07-01T12:59:15.273" Score="8" Body="&lt;p&gt;Nuclear fusion rates in the core of a star have very non-linear and strong dependences on temperature, pressure and density (for temperature it's like T^40 for some processes - &lt;a href=&quot;http://www.astro.soton.ac.uk/~pac/PH112/notes/notes/node117.html&quot;&gt;http://www.astro.soton.ac.uk/~pac/PH112/notes/notes/node117.html&lt;/a&gt;). So, as the star gets bigger, denser and hotter at the centre, the rates of fusion rise much faster and the star loses energy quicker, hence living a shorter life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the star tries to balance itself to maintain stability, so the actual dependence is not that high, and ends up being related to the mass luminosity relationship, where the dependence is not that steep.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2014-07-01T12:59:15.273" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4781" PostTypeId="2" ParentId="4755" CreationDate="2014-07-01T13:09:41.307" Score="5" Body="&lt;p&gt;To add to Florin's great answer, even the dark part of the moon is visible due to illumination by the earth (has a magnitude of around -3, probably calculated for a new moon/solar eclipse configuration - &lt;a href=&quot;http://adsabs.harvard.edu/full/1943MNSSA...2...24J&quot; rel=&quot;noreferrer&quot;&gt;http://adsabs.harvard.edu/full/1943MNSSA...2...24J&lt;/a&gt;, first paragraph of 'brightness of earthshine on moon'). So, the moon is always visible if not in the 'glow' of the Sun.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2014-07-01T13:09:41.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4782" PostTypeId="2" ParentId="4766" CreationDate="2014-07-01T13:53:16.113" Score="5" Body="&lt;p&gt;First it will be important to consider the term 'relativistic speed'. If by that you mean something like 0.1c, it will only change the colour of the stars as you mentioned in the bounty description. However, if it means something with higher Lorentz Gammas (like 0.9c or 0.99c), all other relativistic effects come into play. There's relativistic beaming and aberration combined, which means that all the light that you receive from systems with respect to which you're moving at relativistic speeds will only be received in small beams towards the direction you're moving in and those will be strongly blueshifted and intensified. Watch this animated video for an illustrated view of what I mean - &lt;a href=&quot;https://www.youtube.com/watch?v=JQnHTKZBTI4&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=JQnHTKZBTI4&lt;/a&gt;. (It's not a high quality video, but it is very effective in demonstrating what I just wrote.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in effect, you will possibly see only three major sources of light in the sky:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The parent planet&lt;/li&gt;&#xA;&lt;li&gt;The parent star&lt;/li&gt;&#xA;&lt;li&gt;The bright patch of the sky in the direction of motion, highly blueshifted, with brightness dependent on your speed&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Their relative brightnesses will depend on:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;distance to parent star&lt;/li&gt;&#xA;&lt;li&gt;distance to parent planet&lt;/li&gt;&#xA;&lt;li&gt;albedo of the parent planet&lt;/li&gt;&#xA;&lt;li&gt;speed of motion through the ism&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For a day and night cycle:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The parent planet will obviously be at the zenith the whole time (or fixed location in the sky if you're at a different location).&lt;/li&gt;&#xA;&lt;li&gt;The parent star will appear to move around the sky with the same period as the revolution period of the moon - say a few days at the most (will be fairly small as compared to the 12 year revolution period of the planet, so you can ignore 'synodic' effects to the first order for a single day/night cycle).&lt;/li&gt;&#xA;&lt;li&gt;The sky will also move with roughly the same period as the sun, ignoring synodic effects.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Winds will be very complexly affected by the three sources (since heating due to radiation affects winds). There will also be tidal effects due to the parent planet which also play a role in the winds (similar to how tides work on the earth). So, you will have to take into account heating from the star, heating from the planet (which is constant due to tidal locking), tidal effects of the planet (again constant) and heating from the sky. Now the radiation from the sky can also act in a weird way, since blueshifting can give rise to strong emissions in UV, X-ray and Gamma ray photons, plus relativistic cosmic rays (everything you encounter will be in the form of cosmic rays due to the relativistic speed of your motion). These can ionize the atmosphere, heat it, cause there to be particle showers and aurorae, which can all affect wind patterns in complex ways, which I cannot exactly say much about, given my limited knowledge in atmospheric sciences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I mentioned above, yes, all interactions will be relativistic and will be similar to cosmic ray bombardment onto the moon you live on. Things will get nasty and you'll be having geomagnetic storms all the time. Satellite communications will probably never work. Expect these particles to arrive in the previously mentioned bright region of the sky due to aberration still applying to this case. You will see similar aurora like regions on the parent planet (&lt;a href=&quot;http://www.spacetelescope.org/static/archives/postcards/screen/hst_postcard_0008.jpg&quot; rel=&quot;noreferrer&quot;&gt;http://www.spacetelescope.org/static/archives/postcards/screen/hst_postcard_0008.jpg&lt;/a&gt;), and the star will probably not be affected that much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh and by the way, due to the antimatter nature of your system and the matter nature of the ism, the atmosphere would thin over time, and later these antimatter cosmic rays will start bombarding the moon. So if you stay in the ism for long enough, living on the moon will be difficult. Also, I am assuming that life evolved on your moon in an antimatter ISM and you have entered the matter ism only recently, as otherwise, the moon won't be stable for long enough for life (intelligent life) to evolve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will try to see if I can think of some other effects, but for now, I guess this will do?!&lt;/p&gt;&#xA;" OwnerUserId="1971" LastEditorUserId="1971" LastEditDate="2014-07-01T14:05:33.373" LastActivityDate="2014-07-01T14:05:33.373" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4783" PostTypeId="1" AcceptedAnswerId="4784" CreationDate="2014-07-01T14:17:48.957" Score="2" ViewCount="108" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Milankovitch_cycles&quot; rel=&quot;nofollow&quot;&gt;Milankovitch cycle&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;describes the collective effects of changes in the Earth's movements upon its climate[.]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By this theory, the Earth is now experiencing an interglacial.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the theory, when will the interglacial end and the next ice age begin?&lt;/p&gt;&#xA;" OwnerDisplayName="user1098" LastActivityDate="2014-07-01T14:28:14.887" Title="Can the next possible ice age be projected based upon the Milankovitch cycle?" Tags="&lt;earth&gt;&lt;climate&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4784" PostTypeId="2" ParentId="4783" CreationDate="2014-07-01T14:28:14.887" Score="1" Body="&lt;p&gt;This paper seems to be discussing what you've asked: &lt;a href=&quot;http://www.sciencemag.org/content/297/5585/1287.full&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencemag.org/content/297/5585/1287.full&lt;/a&gt;. The next ice age is expected to begin about 50,000 years from now. However, I think it's still debatable how accurate the predictions of Milankovitch cycles are, and how much of an effect humans are likely to have on this earth system.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2014-07-01T14:28:14.887" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4785" PostTypeId="2" ParentId="4766" CreationDate="2014-07-01T19:11:12.307" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;1) From the moon, how will the sky look like?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the moon is tidally locked in a circular orbit then the giant gas planet will always be in the same spot in the sky. It wouldn't move. It would always be there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if the orbit is eccentric, as is that of Mercury, then it would turn and expose all its sides to its mother planet, even though it is tidally locked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mother gas giant planet would change its phases and its atmospheric storms. Its hugeness and nearness would take up a really large fraction of the sky for anyone on the side of that moon which permanently faces the planet. You couldn't have a better GPS...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2) What will the equivalent of a &quot;day and night cycle&quot; (sun-up to&#xA;  sun-up) look like?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Day, as in dawn to dusk, for a moon would be its orbital period. For most large moons in the Solar system, their orbital periods are a few days to a week or so. Not extremely different from our Earth day rhythm, but likely a few times longer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3) Will the winds on the moon have any &quot;non-earthlike&quot; patterns?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Titan of Saturn is the only moon which has atmosphere, and thus winds. And its winds seem to be very calm. And the winds on Mars too move grains of sand quite slowly most of the time. It's more of a geological phenomenon than a weather phenomenon, if I've understood it correctly. But weather patterns in gas giant planets can be enormous. As in the atmospheres of stars. And then there's the Venusian atmosphere. There are four atmospheres of rocky/icy worlds in the Solar system, and they are all very different. Very different. So your fiction may be believable almost anyway you make it in that respect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a SETI talk about tidally locked exoplanetary insolation which I think might interest you:&#xA;&lt;a href=&quot;http://www.seti.org/weeky-lecture/patterns-sunlight-extra-solar-planets&quot; rel=&quot;nofollow&quot;&gt;http://www.seti.org/weeky-lecture/patterns-sunlight-extra-solar-planets&lt;/a&gt;&#xA;13 minutes into it maybe it gets more interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I think that the solar wind from the star would keep all the interstellar medium anti-anti-matter from interacting with its planets. ISM is about one atom per cubic centimeter. If its matter or antimatter I think it wouldn't change much as our Sun plows through it.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2014-07-01T19:54:17.703" LastActivityDate="2014-07-01T19:54:17.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4786" PostTypeId="1" AcceptedAnswerId="4790" CreationDate="2014-07-01T19:35:17.980" Score="2" ViewCount="119" Body="&lt;p&gt;This may be a stupid question, but I've been wondering this for a while and haven't yet found a solid answer. I'm worried the answer will be something along the lines of &quot;of course not, you dummy&quot; but it's bothering me and I haven't found the answer so I figured I'd take the risk.&#xA;Is it possible for matter to ONLY be visible in an area of the spectrum other than visible light? For instance, can any matter be invisible to us but visible in the UV or IR range?&lt;/p&gt;&#xA;" OwnerUserId="1980" LastActivityDate="2014-07-02T06:07:14.757" Title="Invisible in visible light but visible in non-visible light?" Tags="&lt;spectra&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4787" PostTypeId="1" CreationDate="2014-07-01T22:31:46.980" Score="1" ViewCount="59" Body="&lt;p&gt;These days, we have some very precise ways of making measurements, but I'm sure it wasn't so in Kepler's day. So I am wondering how astronomers of that time could make such accurate determinations of planetary orbits, given the likely limitations of the available instrumentation. To put the question another way: How does each added (or lost) digit of precision in the measurement affect the accuracy of the calculated orbital elements? Is there significant compounding of errors due to the nature of the calculations, or is it essentially one-for-one? How much accuracy (to how many digits of precision) must the &quot;three observations&quot; be in order to determine the orbital elements to a given precision? How does that precision degrade over time? If, for example, we wanted to prepare an almanac for Neptune that will be accurate to say, one second of arc in our sky after 100 years? &lt;/p&gt;&#xA;" OwnerUserId="1983" LastActivityDate="2014-08-31T06:20:46.240" Title="Sensitivity of calculated orbital elements to observational errors" Tags="&lt;observational-astronomy&gt;&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4789" PostTypeId="2" ParentId="4787" CreationDate="2014-07-02T05:54:14.740" Score="1" Body="&lt;p&gt;Kepler used data from catalogs made by Tycho Brahe which had a varying level of precision and accuracy down to about 0.5 arcminutes for a &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1978JHA.....9...42W&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow&quot;&gt;measurement&lt;/a&gt;. However, since Kepler thought all systems should operate in the same way, he could use a larger sample verify his laws. Once you know what the relation should be, you can then map the observations to what you would expect. For an individual case, the precision of measuring orbital parameters depeneds on how much of the orbit you have actually observed. &lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-07-02T05:54:14.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4790" PostTypeId="2" ParentId="4786" CreationDate="2014-07-02T06:07:14.757" Score="3" Body="&lt;p&gt;While individual atoms or molecules can emit energy at discrete or narrow ranges of wavelengths, once you have larger chunks of matter (and they are above absolute zero) they will emit, at least, a continuous &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_body&quot; rel=&quot;nofollow&quot;&gt;black body spectrum&lt;/a&gt;. While black body radiation can be observed at all wavelengths, far away from the peak wavelength it may be below our ability to detect its emission.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-07-02T06:07:14.757" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4791" PostTypeId="2" ParentId="2267" CreationDate="2014-07-02T06:29:13.077" Score="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_Allen_radiation_belt&quot; rel=&quot;nofollow&quot;&gt;Van Allen radiation belt&lt;/a&gt;, which is outside of the Earth's magnetic field is believed to get its radiation from solar wind.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-07-02T06:29:13.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4792" PostTypeId="2" ParentId="4757" CreationDate="2014-07-02T06:47:01.687" Score="5" Body="&lt;p&gt;The &lt;a href=&quot;http://csep10.phys.utk.edu/astr162/lect/binaries/masslum.html&quot;&gt;mass luminosity relation&lt;/a&gt; is non-linear (L = M^3.5 for the main sequence) so more massive stars burn at a much faster rate. This is a non-linear relationship because the fusion is caused by pressure in the core from the star trying to gravitationally collapse on itself.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-07-02T06:47:01.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4793" PostTypeId="1" AcceptedAnswerId="4795" CreationDate="2014-07-02T08:46:18.467" Score="6" ViewCount="382" Body="&lt;p&gt;If we make the assumption that the Universe is infinite, and has an infinite number of hydrogen atoms, then why is it not of infinite density - because, under Schrodinger's wave equation the probability of an electron being at any given point is non-zero and any non-zero number multiplied by infinity is itself infinity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the answer (a) I have made some basic error in physics, (b) the Universe is provably not infinite because of this - effectively a version of Obler's Paradox or (c) the Pauli exclusion principle means that electrons just cannot be anywhere?&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2019-12-16T19:03:13.617" Title="If the Universe is infinite, why isn't it of infinite density?" Tags="&lt;cosmology&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4794" PostTypeId="1" AcceptedAnswerId="11095" CreationDate="2014-07-02T08:54:53.120" Score="2" ViewCount="385" Body="&lt;p&gt;As I understand it, one theory of the multiverse is that there are an infinite number of universes separated by small distances in other (than our observable 3/4) dimensions and that gravity is weak because it permeates these other dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if there are an infinite number of these universes then surely the gravitational force we would feel would be infinite?&lt;/p&gt;&#xA;" OwnerUserId="823" LastEditorUserId="37191" LastEditDate="2021-02-01T21:49:01.653" LastActivityDate="2021-02-01T21:49:01.653" Title="Why doesn't this paradox disprove (some) multiverse quantum gravitational theories?" Tags="&lt;cosmology&gt;&lt;space-time&gt;&lt;multiverse&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4795" PostTypeId="2" ParentId="4793" CreationDate="2014-07-02T09:19:58.030" Score="7" Body="&lt;p&gt;I would like to point out that any information (even quantum mechanical) travels at most at the speed of light. (There are some problems with entangled systems etc, but we can assume most systems to not be entangled, which is a fair assumption, imo). So, if you consider the age of the universe to be finite, any point will have only finitely many overlapping wavefunctions (those within the range of visibility, and the probability can be very very small at most points). So that means that there is never an infinite amount of atoms in the visible region at any finite time, meaning that the density never reaches infinity (or does so only at t=inf).&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2014-07-02T09:19:58.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4796" PostTypeId="1" AcceptedAnswerId="4797" CreationDate="2014-07-02T09:22:36.577" Score="4" ViewCount="197" Body="&lt;p&gt;I came across &lt;a href=&quot;http://voyager.jpl.nasa.gov/where/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; page which provides (mostly extrapolated) speeds of both voyagers. Assuming they are fairly accurate, any reason why the speed of Voyager 2 (5-6 km/sec) is less than that of Voyager 1 (13 km/sec). &lt;/p&gt;&#xA;" OwnerUserId="1985" LastActivityDate="2014-07-02T09:29:37.210" Title="Speed comparison of both voyagers" Tags="&lt;speed&gt;&lt;interstellar-travel&gt;&lt;space-probe&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4797" PostTypeId="2" ParentId="4796" CreationDate="2014-07-02T09:29:37.210" Score="4" Body="&lt;p&gt;If you notice carefully, you'll see that their speeds away from the sun are fairly similar (7-9 km/s), and the difference is only in their speeds away from the earth. This is because of a small component of earth's motion around the sun (which is at about 30 km/s) adding to Voyager 1's motion and subtracting from Voyager 2's motion.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2014-07-02T09:29:37.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4798" PostTypeId="1" CreationDate="2014-07-02T10:45:44.520" Score="1" ViewCount="228" Body="&lt;p&gt;Images taken by landers on Titan and Venus and Mars show landscapes where rocks, to me at least, are surprisingly flat. Being used to walking around in forests with roundish meter sized boulders, I'd be very surprised to find myself in a landscape of flat rocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their atmospheric density range between about 100 times, half and 1/100th of Earth's atmosphere. So that can't be the main explanation. Actually, I think that the landscape of the Moon is more Earth like than that of the atmospheric worlds out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this my impression of &quot;flatness&quot; a real phenomenon, and if so, why and how?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="67" LastEditDate="2014-07-04T19:40:27.363" LastActivityDate="2016-11-08T06:14:39.943" Title="Why do rocks on other solar system bodies that have an atmosphere seem to be flat?" Tags="&lt;planetary-atmosphere&gt;&lt;venus&gt;&lt;surface&gt;&lt;geology&gt;&lt;titan&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4799" PostTypeId="2" ParentId="4773" CreationDate="2014-07-02T11:14:01.523" Score="3" Body="&lt;p&gt;Exoplanets is an almost over populated field of research now, so yes they are monitored. But maybe not too much beyond what is needed to confirm their existence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you have to humble yourself in the enormity of time. There are some transient events in astronomy, but mostly, watching the sky is like watching a rock. Nothing much changes. Phoebes is crashing into Mars. But not until several million years. It will be a very slow process of it breaking up and parts of it landing one by one. Even in the midst of that event human intuition would think of it as a frozen constant non-event. Like the rings of Saturn. Look at a mountain and see how fast it grows or shrinks. Astronomers cannot see exoplanets crashing into each other in the way I think you mean.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2014-07-02T11:14:01.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4800" PostTypeId="1" AcceptedAnswerId="4807" CreationDate="2014-07-02T11:56:05.587" Score="18" ViewCount="2393" Body="&lt;p&gt;Is it possible that light itself is dark matter?  I am speaking of photons (e.g. visible light, infrared, ultraviolet, etc...).  I realize light is understood to be massless, but it is obvious it at least contains energy because we can see with it (e.g. it energizes the cells in our retinas).  I wonder if light has a very tiny &quot;net mass&quot; (e.g. 0 mass * relativistically infinite speed).  I would think that light at least has a little mass, in proportion to its energy.  For example, take E=mc^2, then m = E/c^2 would describe how much mass it has.  If this is true, light should have a very little gravity too.  Although the effect would seem minimal, light is practically everywhere. Gravity from light would be more concentrated inside galaxies, and even more concentrated in the center of galaxies where there are many stars (like dark matter is).  It would be interesting to run the calculations, assuming light does have gravity, and see if this matches the gravitational observations of dark matter in the universe.  It would be funny and ironic if dark matter really is light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Note that it appears light does have gravity as per the discussions here: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1333/how-does-light-affect-the-universe?rq=1&quot;&gt;How does light affect the universe?&lt;/a&gt;  If that much is true, I wonder if this is significant enough to account for dark matter?&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2019-10-21T20:59:18.820" Title="Could light be dark matter?" Tags="&lt;galaxy&gt;&lt;gravity&gt;&lt;light&gt;&lt;dark-matter&gt;&lt;galaxy-center&gt;" AnswerCount="6" CommentCount="10" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4801" PostTypeId="1" AcceptedAnswerId="4802" CreationDate="2014-07-02T12:36:04.783" Score="1" ViewCount="1329" Body="&lt;p&gt;I having been studying some astronomy papers related to galaxy observations and I realized that every time they want to express the size of the Point Spread Function (PSF) of a system which can be approximated by Gaussian they use the Full Width Half Maximum (FWHM) and not the sigma. Is there a reason behind that? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="1986" LastActivityDate="2014-07-02T12:51:32.957" Title="Point Spread Function size: Full Width Half Maximum (FWHM) vs Sigma" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4802" PostTypeId="2" ParentId="4801" CreationDate="2014-07-02T12:51:32.957" Score="1" Body="&lt;p&gt;FWHM is the indicator of the width of a Gaussian that is easiest to measure, and is least error prone in terms of actual physical measurement (due to the slope of a Gaussian being the highest near half maximum - not exactly sure about this, but it does seem like that's the case - so error in y contributes least in the measurement of width, relatively. If this is not clear, look at a noisy Gaussian - something like this: &lt;a href=&quot;http://www.astronomie-amateur.fr/Documents_Supernovae/Mesure_Vexp_gauss.PNG&quot; rel=&quot;nofollow&quot;&gt;http://www.astronomie-amateur.fr/Documents_Supernovae/Mesure_Vexp_gauss.PNG&lt;/a&gt; - and you'll know what I mean. It seems most sensible to estimate width using FWHM). And it has a simple correlation to the sigma with a factor of 2.35 if the distribution is actually Gaussian (which is often the assumption for PSFs - correct me if I'm wrong). Measuring actual standard deviation of data with errors is much more complicated than just simply measuring FWHM, and often it is enough to know the FWHM estimate. Hope this is the answer you were looking for.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2014-07-02T12:51:32.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4803" PostTypeId="1" CreationDate="2014-07-02T16:24:21.067" Score="6" ViewCount="91" Body="&lt;p&gt;Another way of stating this question: Does the fact that a star system (or comparable concentrated mass) exists in a galaxy affect its long-term evolution, in terms of expansion, contraction, nebulation, etc.?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question was inspired by a view of the &lt;a href=&quot;http://apod.nasa.gov/apod/ap140702.html&quot;&gt;&quot;Umbrella Galaxy&quot;&lt;/a&gt; where previous interactions have torn streams of stars far from the galactic disk.&lt;/p&gt;&#xA;" OwnerUserId="1987" LastActivityDate="2014-07-29T07:17:03.947" Title="Do star systems stripped from galaxies have different expected behavior?" Tags="&lt;intergalactic-space&gt;&lt;stellar-astrophysics&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4804" PostTypeId="2" ParentId="4750" CreationDate="2014-07-02T16:37:57.067" Score="3" Body="&lt;p&gt;metallicity is a complicated issue, and giving just one number (usually Z, but also 12+log(O/H), ...) is very rough. But, here are the standard values for the Magellanic clouds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LMC: Z~1/2 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;SMC: Z~1/5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with Z=1 being the solar value. Practically, the values will vary across any given galaxy (the Milky Way also has a metallicity gradient). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Madden et al. 2014, PUBLICATIONS OF THE ASTRONOMICAL SOCIETY OF THE PACIFIC, 125:600–635, 2013 June&#xA;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2013PASP..125..600M&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/2013PASP..125..600M&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article is also &lt;a href=&quot;http://arxiv.org/abs/1305.2628&quot; rel=&quot;nofollow&quot;&gt;available on astroPH&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1988" LastActivityDate="2014-07-02T16:37:57.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4805" PostTypeId="2" ParentId="3644" CreationDate="2014-07-02T17:29:40.053" Score="1" Body="&lt;p&gt;Yes, it is possible to use the sun as a gravitational lens and to achieve better telescopic viewing. As you know space is curved by mass and so light is deflected by mass, it is possible to focus light using gravitational lenses and thus achieving greater telescopic viewing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the sun does have corona fluctuations around it. So, to better exploit the gravitational lensing effect of the Sun, one should try to focus a little bit further away from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, it was already being planned for a space mission to exploit the sun's gravitational lensing effect to communicate interstellarly. The mission is called FOCAL (For &quot;Fast Outgoing Cyclopean Astronomical Lens&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more further information, search for either &quot;Dr.Claudio Maccone&quot; or &quot;FOCAL Space Mission&quot;&lt;/p&gt;&#xA;" OwnerUserId="1931" LastActivityDate="2014-07-02T17:29:40.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4806" PostTypeId="2" ParentId="4800" CreationDate="2014-07-02T21:32:45.260" Score="3" Body="&lt;p&gt;As Yashbhatt said, we can detect light: with our eyes (visible light only) and with special machines. We can also see the effects of some type of lights. Dark matter, however, cannot be detected for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, light is energy, dark matter is matter. Why does your skin tan? It's because of the ultraviolet light. Why are you hot each summer? It's because of the infrared light. Light has an effect, so that's energy: your skin won't tan if there's no energy provided to make your skin tan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, light is not dark matter.&lt;/p&gt;&#xA;" OwnerUserId="1890" LastEditorUserId="1890" LastEditDate="2014-07-02T22:12:25.793" LastActivityDate="2014-07-02T22:12:25.793" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4807" PostTypeId="2" ParentId="4800" CreationDate="2014-07-02T22:37:55.907" Score="21" Body="&lt;p&gt;Dark matter, is just a name for something we know nothing of. It was named to account for an extra gravity source for which there have been indirect observations, but yet we cannot explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force of gravity exerted by light is negligibly small yet we have measured the gravitational pull of Dark Matter to be big enough to affect whole galaxies; it is what binds galaxies together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, we have included everything we can observe (all ordinary matter including photons) when we do the calculations for the amount of gravity there should be. So light is already there. 'Dark matter' is that extra gravity which we cannot account for.&lt;/p&gt;&#xA;" OwnerUserId="1469" LastEditorUserId="1469" LastEditDate="2014-07-03T20:40:16.407" LastActivityDate="2014-07-03T20:40:16.407" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4808" PostTypeId="2" ParentId="4753" CreationDate="2014-07-03T06:56:50.390" Score="5" Body="&lt;p&gt;Another way to look at this question is to consider how elements are produced. The elements with larger atomic numbers (i.e.: 26 (iron) or so) on the periodic table are primarily produced during supernovae explosions. Based on a lot of findings in stellar physics and nuclear physics in the past half century, it's unlikely that a &lt;a href=&quot;http://en.wikipedia.org/wiki/Transfermium_element&quot; rel=&quot;noreferrer&quot;&gt;transfermionic&lt;/a&gt; element (an element with 92 or more protons) can be produced in that process. Further, these elements tend to decay with half lives measured in hours or minutes (or less), so even if they were produced in a supernova, they are long since gone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @Jonathan pointed out, there is some potential for such elements due to the so-called island of stability, but they are still likely to be highly unstable, with very short decay times.&lt;/p&gt;&#xA;" OwnerUserId="1991" LastActivityDate="2014-07-03T06:56:50.390" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4809" PostTypeId="1" CreationDate="2014-07-03T07:58:23.887" Score="2" ViewCount="192" Body="&lt;p&gt;TMT will be put at Mauna Kea in Hawaii. It should be near Keck and Gemini.&lt;br/&gt;&#xA;However it seems there is no place left at that site.&lt;br/&gt;&#xA;What is the exact position? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pmxg2.jpg&quot; alt=&quot;Mauna Kea&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditDate="2014-07-03T08:38:33.183" LastActivityDate="2014-07-03T08:38:33.183" Title="what will be the exact position of the Thirty Meter Telescope" Tags="&lt;telescope&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2014-07-07T16:50:29.963" ContentLicense="CC BY-SA 3.0" />
  <row Id="4810" PostTypeId="2" ParentId="4793" CreationDate="2014-07-03T11:45:28.883" Score="5" Body="&lt;p&gt;Your assumption is not true. For example: if the probability density function of the atoms decreases quadratic exponentially (i.e. &lt;span class=&quot;math-container&quot;&gt;$e^{-r^2}$&lt;/span&gt;), then their sum will be finite even in an infinite universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of infinite many positive real numbers can be finite, although they need to decrease on a long-term.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="1891" LastEditDate="2019-12-16T19:03:13.617" LastActivityDate="2019-12-16T19:03:13.617" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="4811" PostTypeId="2" ParentId="4800" CreationDate="2014-07-03T11:53:29.083" Score="3" Body="&lt;p&gt;Light may account for a small portion of dark matter, but it is unlikely to account for most/all of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a Wikipedia article on dark matter: &lt;a href=&quot;http://en.wikipedia.org/wiki/Dark_matter&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Dark_matter&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the total mass–energy of the known universe contains 4.9% ordinary matter, 26.8% dark matter and 68.3% dark energy.[2][3] Thus, dark matter is estimated to constitute 84.5% of the total matter in the universe, while dark energy plus dark matter constitute 95.1% of the total content of the universe.[4][5]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now that I think more about it, this would mean most of the mass in the universe would be light instead of matter if dark matter is light.  I doubt that most matter (e.g. in stars) turned into light, even throughout the life of the universe.  I believe hydrogen looses only a tiny portion of its mass (as energy / light) when it is fuzed into Helium.  However, it is possible that the cosmic background radiation, passing through almost all parts of the universe could account for more of the light.  I still doubt that would be enough to account for dark matter, but would like to see what effect this does have.  This tells me that light probably does not account for all dark matter, but that it could still account for a portion of it.  I welcome further input on this (e.g. how much of dark matter is light, if any).&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2014-07-03T11:53:29.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4812" PostTypeId="1" AcceptedAnswerId="4814" CreationDate="2014-07-03T11:55:03.750" Score="10" ViewCount="758" Body="&lt;p&gt;I came across this video recently where artists depicted what it would look like if Earth had rings like Saturn. &lt;a href=&quot;http://www.wimp.com/earthrings/&quot;&gt;http://www.wimp.com/earthrings/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note: The video contains already some information on what I ask here, but I can't assess its reliability, therefore I'd appreciate any scientific thoughts on it :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Earth has roughly a distance of 1 AU to the sun, Saturn between 9 and 10. Sun's gravity would therefore have a much stronger influence on these rings. Would it already be enough to distort the rings (e.g. even leading to the sun &quot;stealing&quot; parts of it) or would the rings still have a ring shape?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) How well would the objects in such rings be visible from earth? The Saturn rings are mostly dust and ice as far as I know, but there are surely some larger objects. Depending on material and size some object would probably be visible to the naked eye, right? Is there (probably yes) a formula to calculate what materials and sizes are required to achieve this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Would those rings not be constantly raining particles down on earth? Would there be a much higher risk of dangerous impacts for the earth's population?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Are the colours used in the video precise? Don't they depend on the angle? What influence would our atmosphere have onto them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Would those rings have the same orbital inclination or would theirs differ from earth's to an extent where the difference becomes visible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) What shadow would these rings cause on earth? Would they be dense enough to &quot;shade out&quot; complete stretchs of land?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware that these are hypothetical questions. But I find it, especially as an author of sci-fi and fantasy, nonetheless very interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your interest!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pat&lt;/p&gt;&#xA;" OwnerUserId="1992" LastActivityDate="2014-07-03T14:13:25.447" Title="&quot;If earth had rings like Saturn&quot;: A few further thoughts on the consequences" Tags="&lt;earth&gt;&lt;saturn&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4813" PostTypeId="1" AcceptedAnswerId="39807" CreationDate="2014-07-03T12:03:02.990" Score="3" ViewCount="231" Body="&lt;p&gt;I would like to know the ratio of cosmic microwave background radiation to normal radiation in the universe.  I am considering cosmic microwave background radiation to include the microwave, and any other radiation that is being emanated from near the &quot;edge of the universe&quot;, while normal radiation is radiation emmitted by stars, nebulae, and other sources within the universe (excluding the cosmic microwave background).  Since we are in a galaxy, I know that the normal radiation here far exceeds the cosmic microwave background, but I am interested in the &quot;average&quot; value over the universe.  For example: total cosmic microwave background radation in the universe / total normal radiation in the universe.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="2153" LastEditDate="2015-07-08T21:27:00.630" LastActivityDate="2020-11-12T19:36:25.693" Title="What is the ratio of cosmic microwave background radiation to normal radiation?" Tags="&lt;light&gt;&lt;cmb&gt;&lt;radiation&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4814" PostTypeId="2" ParentId="4812" CreationDate="2014-07-03T14:13:25.447" Score="13" Body="&lt;p&gt;This is a big one, so I'll split it into parts based on the questions you asked:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) Earth has roughly a distance of 1 AU to the sun, Saturn between 9 and 10. Sun's gravity would therefore have a much stronger influence on these rings. Would it already be enough to distort the rings (e.g. even leading to the sun &quot;stealing&quot; parts of it) or would the rings still have a ring shape?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sun would not have too much of an influence because the rings are all individually gravitationally bound to the earth (similar to the moon, and you don't see moon being affected by the sun, gravitationally; the rings will be much closer to the earth and hence even less affected).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon, on the other hand, would distort a few rings due to orbital resonances (similar to Titan Ringlets: &lt;a href=&quot;https://en.wikipedia.org/wiki/Rings_of_Saturn#Colombo_Gap_and_Titan_Ringlet&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Rings_of_Saturn#Colombo_Gap_and_Titan_Ringlet&lt;/a&gt; and resonant gaps in asteroid belts due to Jovian planets: &lt;a href=&quot;https://en.wikipedia.org/wiki/Kirkwood_gap&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Kirkwood_gap&lt;/a&gt;), and its effect would be much more pronounced because it is closer to the earth than Titan is to Saturn, though limited to a few resonant regions. There will also be ripples if the plane of the moon is different from the plane of the rings (&lt;a href=&quot;http://blogs.discovermagazine.com/badastronomy/2009/06/11/saturns-rings-do-the-wave/#.U7VjpvmSw08&quot; rel=&quot;noreferrer&quot;&gt;http://blogs.discovermagazine.com/badastronomy/2009/06/11/saturns-rings-do-the-wave/#.U7VjpvmSw08&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2) How well would the objects in such rings be visible from earth? The&#xA;  Saturn rings are mostly dust and ice as far as I know, but there are&#xA;  surely some larger objects. Depending on material and size some object&#xA;  would probably be visible to the naked eye, right? Is there (probably&#xA;  yes) a formula to calculate what materials and sizes are required to&#xA;  achieve this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most matter in Saturn's ring is particles up to a few metres in size (&lt;a href=&quot;http://saturn.jpl.nasa.gov/faq/FAQSaturn/#q6&quot; rel=&quot;noreferrer&quot;&gt;http://saturn.jpl.nasa.gov/faq/FAQSaturn/#q6&lt;/a&gt;). Resolution of the human eye is about 1' (&lt;a href=&quot;https://astronomy.stackexchange.com/questions/2398/is-the-moon-only-60-pixels&quot;&gt;Is the moon only 60 pixels?&lt;/a&gt;), which means at a distance of around 1 to 1.5 Earth radii (given by Roche limit of the earth; this is where the rings will be), the eye can resolve things above a couple of kilometres in size. However, Saturn also has Shepherd moons, which are a few tens of kilometers in size. Such moons, if present, would be visible to naked eye. Also, the gaps would be visible clearly, since they are often big enough.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3) Would those rings not be constantly raining particles down on&#xA;  earth? Would there be a much higher risk of dangerous impacts for the&#xA;  earth's population?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There would be collisions/interactions in the rings (I'm guessing, correct me if I'm wrong) which would send particles earthwards. However, the particles which will get thrown will be the lighter, smaller ones (conservation of momentum etc.), and usually small meteoroids are fairly harmless. They will also be slower than the usual meteoroids that enter the earth at above 20 km/s, since the orbital motion of the earth doesn't add up in this case. I am a little unsure about the consequences of them being slower, but anyway, smaller meteoroids should technically be fairly harmless. (Also I couldn't find decent references other than: &lt;a href=&quot;http://science.howstuffworks.com/question486.htm&quot; rel=&quot;noreferrer&quot;&gt;http://science.howstuffworks.com/question486.htm&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;4) Are the colours used in the video precise? Don't they depend on the&#xA;  angle? What influence would our atmosphere have onto them?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe so. This section shows true colour images of the rings from two different angles, and there doesn't seem to be a significant change in the colouration: &lt;a href=&quot;https://en.wikipedia.org/wiki/Rings_of_Saturn#Subdivisions_and_structures_within_the_rings&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Rings_of_Saturn#Subdivisions_and_structures_within_the_rings&lt;/a&gt;. The atmosphere will probably affect the rings closer to the horizon, with redder colouration similar to sunrise/sunset and moonrise/moonset.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;5) Would those rings have the same orbital inclination or would theirs&#xA;  differ from earth's to an extent where the difference becomes visible?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Depends largely on the formation history. This FAQ question here discusses this fairly well: &lt;a href=&quot;http://saturn.jpl.nasa.gov/faq/FAQSaturn/#q14&quot; rel=&quot;noreferrer&quot;&gt;http://saturn.jpl.nasa.gov/faq/FAQSaturn/#q14&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;6) What shadow would these rings cause on earth? Would they be dense&#xA;  enough to &quot;shade out&quot; complete stretchs of land?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There will be shadow similar to that due to clouds, since the rings are not too thick or dense. In many of the Cassini photos, you can see through the rings, so it is likely that the sunlight would still pass through, but would be dimmed (it would be interesting to see by what factor; I cannot say for sure with my knowledge). Also, remember that the earth will still be rotating but the ring system will be dynamically static, so there will be extended periods of darkness during the day for some regions at different times of the year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will add more to this if something comes to mind. I hope this was helpful.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-07-03T14:13:25.447" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4815" PostTypeId="1" AcceptedAnswerId="4817" CreationDate="2014-07-04T00:17:53.893" Score="7" ViewCount="2116" Body="&lt;p&gt;Suppose I wake up from a coma on a desert island in the 19th century (i.e. we already use the Gregorian calendar but have no satellites yet). I have a clear view of the sky and a couple of days to make observations and calculations. Can I determine the current day of the week? Also, can I find the answer instantly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a related note, is it possible to define days of the week in a purely astronomical (e.g. planet-agnostic) manner such that for the Earth we get the traditional meaning?&lt;/p&gt;&#xA;" OwnerUserId="1996" LastActivityDate="2016-12-24T20:23:28.477" Title="Can I look at the sky and find the day of the week?" Tags="&lt;earth&gt;&lt;amateur-observing&gt;&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4816" PostTypeId="1" CreationDate="2014-07-04T01:43:49.083" Score="5" ViewCount="476" Body="&lt;p&gt;The massive elliptical galaxy M87 in the Virgo cluster is 53,490,000 light years away. It also contains one of the largest, heaviest supermassive black holes in the known Universe. But it's also my understanding that the stellar population of this galaxy is mostly old, red supergiants. Is there any evidence for new star formation in M87? If there isn't, what is the fate of this galaxy and what is it like in modern times? That may be impossible to know for sure but what I wonder is if the remaining stars and matter in the galaxy have been swallowed up by central black hole which may be thousands of lightyears across by now.&lt;/p&gt;&#xA;" OwnerUserId="1998" LastActivityDate="2016-11-08T20:22:52.047" Title="What is the galaxy M87 doing these days?" Tags="&lt;supermassive-black-hole&gt;&lt;messier-catalog&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4817" PostTypeId="2" ParentId="4815" CreationDate="2014-07-04T05:46:15.517" Score="6" Body="&lt;p&gt;It is possible to determine the day of the year using the distance of the sun from the vernal equinox (you have to account for the equation of time etc, but it is possible). It is also possible to determine the year if your measurements are precise enough, using the current position of the vernal equinox (which changes due to precession). The ambiguity of the factor of 26000 years can be removed by noting the positions of other stars (for example, Barnard's star). Knowing these two, you should be able to find the day of the week.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About your second question, Astronomers like to deal with Julian date, which is a continuous day counting system, and you can determine the day of the week from the Julian date, though only if given a reference, for example Jan 1, 2000, was a Saturday. As such, a week and days of a week are man made (&lt;a href=&quot;https://en.wikipedia.org/wiki/Week&quot;&gt;https://en.wikipedia.org/wiki/Week&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="1971" LastEditorUserId="1971" LastEditDate="2014-07-04T10:17:20.397" LastActivityDate="2014-07-04T10:17:20.397" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="4818" PostTypeId="2" ParentId="3661" CreationDate="2014-07-04T14:51:12.423" Score="2" Body="&lt;p&gt;I recently read the relevant parts of the Cambridge Concise History of Astronomy and I'd suggest it was partly good observation, partly luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good observation in that Thales would have noted that the movements of the Sun and Moon were such that eclipses were possible on certain dates, and luck in that the shadow of the eclipse he suggested was possible passed over him or those to whom he had suggested the eclipse took place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I very much doubt he had the observational data (about the size of the Earth for instance) to have been precise about the shadow path - but would have had the data to show when the Sun and Moon were in alignment.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2014-07-04T14:51:12.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4819" PostTypeId="2" ParentId="4770" CreationDate="2014-07-04T18:42:15.637" Score="5" Body="&lt;p&gt;The differences are in the designs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newtonian:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/EA3pC.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gallilean:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/tGNYH.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Schmidt-Cassegrain:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/TfwT1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Schmidt-Cassegrain telescopes are characterized by including 1 convex mirror, 1 concave mirror and 1 correction plate.&lt;/p&gt;&#xA;" OwnerUserId="1931" LastActivityDate="2014-07-04T18:42:15.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4820" PostTypeId="2" ParentId="4798" CreationDate="2014-07-04T20:12:44.597" Score="4" Body="&lt;p&gt;I would say that your initial observation is flawed, so the question is moot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.esa.int/var/esa/storage/images/esa_multimedia/images/2006/01/area_surrounding_the_huygens_landing_site/9704260-4-eng-GB/Area_surrounding_the_Huygens_landing_site_medium.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Huygens landing site, Titan&lt;/a&gt;:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/fKRbp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-07-04T20:12:44.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4821" PostTypeId="2" ParentId="4815" CreationDate="2014-07-04T20:26:07.343" Score="4" Body="&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; &lt;em&gt;(massive rewrite in response to comments)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's nothing &quot;out there&quot; that will tell you what day of the week it is.&#xA;The concept of a week is determined by religion or culture but there's no objective reason why a week should have seven days.  From 1793-1805 the French Republic had a calendar with a &lt;a href=&quot;http://en.wikipedia.org/wiki/French_Republican_Calendar#Ten_days_of_the_week&quot; rel=&quot;nofollow&quot;&gt;10-day week&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to try to extrapolate our idea of a week to other planets, be aware that the planets in our solar system have very different &lt;a href=&quot;http://www.bobthealien.co.uk/table.htm&quot; rel=&quot;nofollow&quot;&gt;day lengths&lt;/a&gt;.  One Venus day is 243 Earth days, while the planets Jupiter through Neptune all have days that are less than one Earth day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't happen in our own solar system, but it's theoretically possible for a planet to have one side always facing its star.  Then it would be impossible to define a day in the way that you've intended.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you wanted to know what day of the year it is, that's possible but quite hard.  In particular you would have to account for things like time zones, leap years, and daylight savings, which are not empirically observable phenomena.  A few hundred years ago the definition of noon was when the sun was at its highest point in the sky.  I think that definition is closer to what you're asking, but that's not the calendar system that we currently use.&lt;/p&gt;&#xA;" OwnerUserId="2001" LastEditorUserId="2001" LastEditDate="2014-07-05T21:10:01.067" LastActivityDate="2014-07-05T21:10:01.067" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4822" PostTypeId="1" AcceptedAnswerId="4825" CreationDate="2014-07-05T08:29:22.753" Score="1" ViewCount="96" Body="&lt;p&gt;I learnt that the spectrum of a black body can be explained using&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Planck distribution&lt;/li&gt;&#xA;&lt;li&gt;Rayleigh -Jeans equation and&lt;/li&gt;&#xA;&lt;li&gt;Weins displacement law&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Can anyone tell me why there are three different equations that explain the same physical phenomena and when should I consider which equation?&lt;/p&gt;&#xA;" OwnerUserId="1328" LastActivityDate="2014-07-05T12:53:29.703" Title="Why there are three equations governing blackbody emission?" Tags="&lt;the-sun&gt;&lt;radiation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4823" PostTypeId="1" AcceptedAnswerId="4824" CreationDate="2014-07-05T12:05:45.320" Score="1" ViewCount="983" Body="&lt;p&gt;Knowing the apoapsis, periapsis and therefore period of an orbit, how can I find the radius of an object in an orbit at a given angle or time, whichever is needed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I have an object m orbiting around object M with an apoapsis of 100m and a periapsis of 50m, what is the distance between m and M after t seconds, or at $\theta$ degrees. Which ever is needed. I say 'or' because I don't know what you would use to find the radius.&lt;/p&gt;&#xA;" OwnerUserId="2004" LastEditorUserId="37191" LastEditDate="2021-02-12T20:53:50.903" LastActivityDate="2021-02-12T20:53:50.903" Title="Finding the radius of an eccentric orbit at any point" Tags="&lt;orbit&gt;&lt;gravity&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4824" PostTypeId="2" ParentId="4823" CreationDate="2014-07-05T12:41:56.420" Score="2" Body="&lt;p&gt;If you know the apoapsis and periapsis, you can find the eccentricity, using $r_a = a (1+e), r_p = a (1-e)$ (&lt;a href=&quot;https://en.wikipedia.org/wiki/Apsis#Mathematical_formulae&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Apsis#Mathematical_formulae&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the eccentricity, you can find the radius at any angle $\theta$ from the periapsis (this angle is also called the true anomaly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r(\theta) = \frac{a(1-e^2)}{1+e.\cos \theta}$(&lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler_orbit#Johannes_Kepler&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Kepler_orbit#Johannes_Kepler&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With time, it is little more difficult. The mean anomaly (M) is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M = \frac{2\pi t}{T}$ where t is the time since last periapsis, and T is the time period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This mean anomaly relates to true anomaly through a variable called eccentric anomaly (E). From mean anomaly (which you get from the time), you can get eccentric anomaly using&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M=E-e.\sin E$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this equation isn't analytically solvable, but you can solve it graphically or numerically, or using a simple iteration which I would demonstrate here (technically included in the numerical methods, but it's easy enough to show here). It is easy enough to be done on a simple scientific calculator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_{n+1}=M+e.\sin E_{n}$ where $E_0=M$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this E, you can find the true anomaly $\theta$ using&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta=2 \tan^{-1}[ \sqrt{\frac{1+e}{1-e}} \tan \frac {E}{2}]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can find the radius as described above.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2014-07-05T12:41:56.420" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="4825" PostTypeId="2" ParentId="4822" CreationDate="2014-07-05T12:53:29.703" Score="4" Body="&lt;p&gt;Historically, two people (or groups of people) independently came up with different equations to model the blackbody equations in different parts of the spectrum. Rayleigh-Jeans law (classically derived) is valid for longer wavelengths and Wien's law (not Wien's displacement law) is valid for shorter wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Planck Distribution approaches the two laws in its limits, as in, for shorter wavelengths it is approximately equivalent to Wien's law and for longer wavelengths, it is approximately equivalent to Rayleigh-Jeans law. However, the quantum mechanically formulated &lt;strong&gt;Planck's law is accurate at all wavelengths, so it is the one that should always be used.&lt;/strong&gt; One can use the other laws, for example, for brightness temperature definitions in radio astronomy, they use the RJ law for convenience, but that's because the wavelengths are long enough, and the approximation of the Planck's law gives the same result. (&lt;a href=&quot;https://en.wikipedia.org/wiki/Planck%27s_law#Approximations&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Planck%27s_law#Approximations&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wien's displacement law only relates the peak wavelength to the temperature, which is a different law completely, though Planck's law (differentiating the function to find the maximum) also gives the same result. Though in this case, the displacement law is not approximate. It is accurate and can be used whenever you want.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2014-07-05T12:53:29.703" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4826" PostTypeId="1" AcceptedAnswerId="4827" CreationDate="2014-07-05T15:00:36.547" Score="5" ViewCount="1010" Body="&lt;p&gt;At places other than the poles, the sun is seen to &quot;rise&quot; on the Eastern horizon and to &quot;set&quot; on the Western horizon. After a period of night, the sun &quot;reappears&quot; on the Eastern horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the poles, during the short periods of 24-hour daylight, what does the sun path look like? Where does the sun move after it has reached its Western most position? Back Eastwards along the same path in reverse?&lt;/p&gt;&#xA;" OwnerUserId="2005" LastActivityDate="2017-02-09T08:23:04.730" Title="Sun path at poles" Tags="&lt;the-sun&gt;&lt;horizon&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4827" PostTypeId="2" ParentId="4826" CreationDate="2014-07-05T15:07:54.837" Score="8" Body="&lt;p&gt;It moves in circles. At the poles there is no direction defined (east, west, north or south). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how the sun moves at the north pole: &lt;a href=&quot;http://www.jaloxa.eu/resources/daylighting/docs/sunpath_90_north.pdf&quot;&gt;http://www.jaloxa.eu/resources/daylighting/docs/sunpath_90_north.pdf&lt;/a&gt; . You can compare it to the other latitudes to get an idea. (&lt;a href=&quot;http://www.jaloxa.eu/resources/daylighting/sunpath.shtml&quot;&gt;http://www.jaloxa.eu/resources/daylighting/sunpath.shtml&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2014-07-05T15:07:54.837" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4828" PostTypeId="1" AcceptedAnswerId="4832" CreationDate="2014-07-05T15:08:22.027" Score="3" ViewCount="841" Body="&lt;p&gt;My astronomy teachers never would answer this for me... Redshift obviously indicates an object (such as another galaxy) is moving away, but how do we know its acceleration from this? It's my understanding that the instantaneous redshift of an object is an indicator solely of its instantaneous velocity, and we have certainly not been observing redshift long enough to compare and determine a change in velocity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can, please avoid using complex math in your answer.&lt;/p&gt;&#xA;" OwnerUserId="2006" LastActivityDate="2014-07-06T10:14:20.493" Title="how does redshift prove expansion is accelerating?" Tags="&lt;observational-astronomy&gt;&lt;expansion&gt;&lt;redshift&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4829" PostTypeId="2" ParentId="4828" CreationDate="2014-07-05T15:21:15.580" Score="2" Body="&lt;p&gt;I don't know the exact details (so I'll be leaving the math out anyway), but this is my understanding and I hope it helps you get some idea about what's being done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you look at galaxies farther away, you're looking at them farther back in time (because speed of light is constant). So, to sample the velocities of galaxies over time, you don't need to observe them for a long time, but you can work with galaxies that are at different distances, and hence at different times in the history of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you notice that the recession velocities of galaxies that are farther away are &lt;strong&gt;slower&lt;/strong&gt; than what you would expect from a constant expansion scenario (Hubble's law, for example), that means the universe was expanding slower, back then, than it is, now. If there is a clear trend that this is happening, then you can say that the speed of expansion of the universe has been increasing over time. This means that the universe is accelerating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is a lot more complicated, since if the universe is accelerating, Hubble's law itself becomes shaky, if not invalid (as it assumes a direct proportionality between distance and recession velocity). Also, if you cannot find distance using Hubble's law, then you have to have an independent distance measurement that far away, which is also problematic, since afaik, there aren't such distance indicators that are valid that far away (since physics is different due to changes in stellar metallicities etc.; a lot of problems start showing up). So I don't know how this is done in practice, but in theory what I described above is how one would arrive at the conclusion of an expanding universe. I would like to know how this is really done in practice, as well.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2014-07-05T15:21:15.580" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4830" PostTypeId="1" CreationDate="2014-07-06T07:43:57.607" Score="2" ViewCount="56" Body="&lt;p&gt;Could a spectrometer on a probe or space telescope identify minerals, not only elements and molecules, from the spectral lines? Is there any technology for identifying minerals remotely?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2014-07-06T09:44:44.003" Title="Can spectroscopes identify minerals?" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4831" PostTypeId="2" ParentId="4830" CreationDate="2014-07-06T09:44:44.003" Score="2" Body="&lt;p&gt;Yes, minerals can be observed using spectroscopy on a telescope, typically from their thermal signature. We can determine what type of mineral based on the elements that make up the mineral as well as the type of crystal structure, which will give a different &lt;a href=&quot;http://www.boulder.swri.edu/~hamilton/VEH/TIR-SNCs3.html&quot; rel=&quot;nofollow&quot;&gt;spectral signature&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-07-06T09:44:44.003" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4832" PostTypeId="2" ParentId="4828" CreationDate="2014-07-06T10:14:20.493" Score="3" Body="&lt;p&gt;You're right, a redshift will give you a measure of a velocity, not an acceleration. I'm assuming when you say acceleration, you're referring to the &lt;a href=&quot;http://www.nytimes.com/2011/10/05/science/space/05nobel.html&quot; rel=&quot;nofollow&quot;&gt;accelerating expansion of the universe&lt;/a&gt; so you're not looking at the particular acceleration of an object (which you could get from examining the force of gravity imparted on that object). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in addition to a redshift, you need to find a way to independently measure the distance. To get out to an appreciable redshift (z~0.5), Type Ia supernova &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_distance_ladder#Type_Ia_light_curves&quot; rel=&quot;nofollow&quot;&gt;are used&lt;/a&gt; as &quot;standard candles&quot;. They are considered standard because they're believed to be from white dwarfs that explode because they have reached their maximum possible mass, the Chandrasekhar limit, of 1.4 solar masses. Therefore, when they explode, they should all look about the same and any difference in brightness would be from how far they are away from us. So they measured the distance and the redshift of the galaxy the supernova originated from to determine the expansion of the universe appears to be accelerating over time.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-07-06T10:14:20.493" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4833" PostTypeId="2" ParentId="33" CreationDate="2014-07-06T10:26:51.300" Score="1" Body="&lt;p&gt;Stellar mass black holes and supermassive black holes likely form from very different mechanisms. While we have a pretty good idea how stellar mass black holes form, we're still not sure about how supermassive black holes form because they occur &lt;a href=&quot;http://www.nasa.gov/mission_pages/chandra/news/H-11-183.html&quot; rel=&quot;nofollow&quot;&gt;very early&lt;/a&gt; on in the Universe.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-07-06T10:26:51.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4834" PostTypeId="1" CreationDate="2014-07-06T19:27:53.887" Score="8" ViewCount="3699" Body="&lt;p&gt;As I understand it, Kepler used the orbital period of Mars, along with observational data of Mars' and the sun's position in the sky to derive the orbits of Earth and Mars.  (As described, here: &lt;a href=&quot;https://faculty.uca.edu/njaustin/PHYS1401/Laboratory/kepler.pdf&quot;&gt;https://faculty.uca.edu/njaustin/PHYS1401/Laboratory/kepler.pdf&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how did he determine the orbital period of Mars without already knowing the orbit of the Earth?&lt;/p&gt;&#xA;" OwnerUserId="2009" LastActivityDate="2014-07-07T10:15:10.833" Title="How did Kepler determine the orbital period of Mars?" Tags="&lt;orbit&gt;&lt;observational-astronomy&gt;&lt;history&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4835" PostTypeId="2" ParentId="4757" CreationDate="2014-07-06T23:24:25.733" Score="3" Body="&lt;p&gt;The easiest way to think of it is terms of the cube square law coming into play. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed at which a star can radiate heat is mostly dependent on its surface area. but it's ability to generate heat is dependent on its volume. Surface area increases with the square of the increase in radius but volume increases with cube of the radius. As a star grows in a diameter it's inside grows faster than its outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if all other factors stayed the same, the star would grow hotter because it has less surface area to radiate from. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since gravity is dependent on volume, the gravitational compression (simplistically) scales with the cube as well. So, a larger star compresses its interior more per unit of mass than does a smaller star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Joan.bdm pointed out, stars have an interior volume, the burning zone down towards the core in which the pressures are high enough for fusion to take place. Probably should call it the burning volume&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The burning volume is the volume in which fusion can occur. Fusion occurs once a specific threshold of heat and pressure have been reached. Below that threshold, fuel consumption is zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cube square law means that the relative size of the burning volume to total volume increases faster than the diameter. E.g If a small star has ten percent of its mass inside the burning zone, a star with twice the diameter will have 20% of its total volume inside the burning zone. (actual ratios invented for demonstration purposes.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A higher percentage of the larger star's total mass is inside the burning zone compared to a smaller star so a higher percentage of the star's mass is subject to fusion at any given time, so a higher percentage of the star's total &quot;fuel&quot; is being &quot;burned&quot; at any one time than in a smaller star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The addition of more fusible isotopes also does not scale linearly. Fusing heavier elements does not release as much energy per unit of mass as lighter ones and they require higher temperatures to start i.e. heavier elements are a poorer fuel. The heavier elements are also produced by fusing lighter ones so the star has to burn fuel to make fuel. Since heavier elements require higher temperatures to fuse, the burning volume for them is smaller than that for lighter elements. Stars usually can't burn more than small percentage of the elements isotopes they produce so the notional increase in fuel supply doesn't compensate for the higher burn rate of lighter elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's like the bard said, &quot;The candle that burns twice as bright, burns half as long.&quot; &lt;/p&gt;&#xA;" OwnerUserId="1628" LastActivityDate="2014-07-06T23:24:25.733" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4836" PostTypeId="1" AcceptedAnswerId="4840" CreationDate="2014-07-06T23:57:43.030" Score="1" ViewCount="71" Body="&lt;p&gt;It is believed that our sun will become a red giant with a diameter of about 1 AU. When the fusion slows down, gravity will collapse the sun. Since the energy release from the fusion diminishes slowly (maybe hundreds of years for a significant change of 1%). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume that only a small fraction of the sun (10%) is needed to collapse (to 1/1000 the volume) to produce a nova. Would it still be possible for a star to collapse in a few hour (instead of decades)?&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-07-07T12:29:25.743" Title="Rate of star collpse" Tags="&lt;stellar-evolution&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4837" PostTypeId="2" ParentId="4834" CreationDate="2014-07-07T00:53:40.220" Score="5" Body="&lt;p&gt;Kepler did know the orbital period of the Earth, Mars, Venus, Mercury, Jupiter, and Saturn. He had access to observations going back centuries, including Brahe's most excellent data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What couldn't know was the absolute size of the Earth's orbit, so the best he could do was give the size of the plantets' orbits in terms of the Earth-Sun distance, 1 A.U. Nobody else could do better for centuries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the transit of Venus across the face of the sun, timing the instant the edge of Venus &quot;touches&quot; the edge of the Sun from two different locations on Earth a known distance apart makes a parallax measurable, and the absolute distance between the Earth and Venus a matter of trigonometry. Kepler's Law lets us convert all the other relative distances into absolute distances. I believe it was 1769 when this was first accomplished.&lt;/p&gt;&#xA;" OwnerUserId="562" LastActivityDate="2014-07-07T00:53:40.220" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4838" PostTypeId="2" ParentId="4836" CreationDate="2014-07-07T05:24:22.177" Score="2" Body="&lt;p&gt;Well, I'm not sure exactly what you mean by &quot;collapse&quot; (if it takes $10^9$ years to die, the last few hours satisfy what your asking about). However, maybe some explanation would help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're asking about the interaction between two timescales - the Dynamical Timescale, which is the time it takes for a star to free fall into a nova, and the Nuclear time scale (I think), which is the rate at which the sun exhausts it's fuel. The dynamical timescale for the Sun is like 1000 s, and the nuclear timescale is like a Gyr (usually what we think of when we think &quot;lifetime of the Sun&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any nuclear effects would totally dominate over dynamical effects (since they take much, much longer), and without some specific collusion of behavior (angular momentum, metalicity, what have you), collapse would occur with the burning of fuel and not with the dynamical timescale until the Sun exhausted all it's fuel, and would collapse in a matter of minutes.&lt;/p&gt;&#xA;" OwnerUserId="2010" LastActivityDate="2014-07-07T05:24:22.177" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4839" PostTypeId="2" ParentId="4834" CreationDate="2014-07-07T10:15:10.833" Score="8" Body="&lt;p&gt;He knew the orbital period of Earth. You can do this by observing the sun's motion in the sky over a year. (And then you know that the orbital period of the earth is a year.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now from earth, you can observe Mars from one Opposition (&lt;a href=&quot;https://en.wikipedia.org/wiki/Opposition_%28planets%29&quot;&gt;https://en.wikipedia.org/wiki/Opposition_(planets)&lt;/a&gt;) to another. You can find out the time ($T_\text{relative}$ it takes for Mars to do this, using which, you find the relative angular velocity between the orbital motions of earth and Mars ($\omega_\text{relative}=2\pi/T_\text{relative}$). This angular velocity is the difference between the angular velocities of earth and mars. Using this and knowing the orbital period of earth, you can find the orbital period of mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2\pi/T_\text{relative}=2\pi/T_\text{earth}-2\pi/T_\text{Mars}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this is clear.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2014-07-07T10:15:10.833" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4840" PostTypeId="2" ParentId="4836" CreationDate="2014-07-07T12:29:25.743" Score="2" Body="&lt;p&gt;Technically, collapsing of a star as a whole at any point of time happens dynamically, but due to the thermal timescale being much higher, the compression can be considered nearly adiabatic, which causes the star to become hotter and brighter and maintains hydrostatic equilibrium at a dynamical timescale. (There is little involvement of the core here since the photon diffusion timescales are high - ~ 100,000 years - and any effect of the core on the star will be seen much much later.) This is why you never see a 'star' collapse. You can, at best, see them pulsate. Typically the core is what collapses, which alters fusion rates and hence is NOT adiabatic. I will try to explain this process in a little detail; let me know if that answers you question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the star is almost out of fuel, it cannot burn fuel as effectively as before, so, the core compresses a little, increasing the temperature, which increases the efficiency and rate of fusion, which makes the 'running out of fuel' part faster leading to a positive feedback process (compression $\rightarrow$ faster exhaustion $\rightarrow$ more compression). This gradually keeps 'collapsing' and burning fuel faster till a point where it cannot counter the collapse using an increased fusion rate due to lack of sufficient fuel (this might still take a few million years, maybe, but it keeps collapsing faster in a runaway fashion till that point). Beyond this point, the collapse happens at the dynamical timescale till it hits degeneracy (or ignition temperature of the previously inert core), which leads to puffing, novae and supernovae based on the mass of the core and its composition.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2014-07-07T12:29:25.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4842" PostTypeId="1" AcceptedAnswerId="4859" CreationDate="2014-07-08T01:46:00.373" Score="8" ViewCount="7503" Body="&lt;p&gt;I have a &lt;a href=&quot;https://web.archive.org/web/20061216215456/http://plettstone.com/telescopes/products.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Plettstone 18&quot;&lt;/a&gt; Dobsonian telescope and use a &lt;a href=&quot;https://www.collimator.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Howie Glatter&lt;/a&gt; tuBlug, laser collimator, and Square Grid Projection Attachment to adjust the mirrors.  Proper collimation requires an accurate identification of the mirror's center. I would like to hear of good methods to very accurately place a white sticky donut concentric to the mirror's optical center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mirror already has a donut, but I wonder if it is accurately placed as to my eye it appears slightly off center, but perhaps the optical center is not the physical center?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen a &lt;a href=&quot;http://www.skyandtelescope.com/astronomy-resources/how-to-align-your-newtonian-reflector-telescope/&quot; rel=&quot;nofollow noreferrer&quot;&gt;recommendation in Sky &amp;amp; Telescope&lt;/a&gt; to get two clear plastic rulers that bend and place them at exact right angles to each other with the donut upside down lightly stuck to the bottom center cross point and then positioning the rulers evenly and pressing down to bend them toward the mirror and affix the donut.  However, I don't think this method is well suited for an 18&quot; mirror, perhaps I am mistaken?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also seen a &lt;a href=&quot;http://garyseronik.com/centre-dotting-your-scopes-primary-mirror/&quot; rel=&quot;nofollow noreferrer&quot;&gt;recommendation by Gary Seronik&lt;/a&gt; to make a paper cutout that matches the mirror shape and to find its center and make a small hole and then place that over the mirror and use a pen to mark the center and center the donut around the marker, however, again, I don't think this is well suited for an 18&quot; mirror.  This method is also &lt;a href=&quot;http://www.alcoat.net/pap_dot.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;mentioned here&lt;/a&gt; in some detail, but they specifically state not for use with a laser collimator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please suggest how I can first gauge the existing donut for accuracy and if it is found to deviate significantly how to remove it and the adhesive and place a new one more accurately.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2017-06-18T20:51:12.733" LastActivityDate="2017-06-18T20:51:12.733" Title="How can I accurately find &amp; apply the white donut to the center of my mirror?" Tags="&lt;dobsonian-telescope&gt;&lt;collimation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4843" PostTypeId="2" ParentId="4774" CreationDate="2014-07-08T14:01:38.057" Score="4" Body="&lt;p&gt;The answer is yes. Stellar collisions are events that occur mainly in Globular clusters due to their high stellar density. The likelihood of these events is highly dependent on the stellar density. As we know the galactic centers concentrate large amounts of matter and, of course, thousads of stars which are mainly in the old red main sequence. However, recent observations hane shown as the existence of young stellar clusters located in the Galactic center of Milky in contrast with our models and predictions about the galactic structures. It is currently believed that in these clusters occur a lot of stellar collisions. For more information there is a related paper &lt;a href=&quot;http://cds.cern.ch/record/478379/files/0011424.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1956" LastEditorUserId="1956" LastEditDate="2014-07-10T18:12:03.180" LastActivityDate="2014-07-10T18:12:03.180" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4845" PostTypeId="2" ParentId="4774" CreationDate="2014-07-08T21:19:46.323" Score="2" Body="&lt;p&gt;A high density of stars definitely increases the chances of collisions, however, the high velocity of a halo star orbiting wouldn't increase its chances. Since the halo star is traveling very quickly, the halo star would only spend a small amount of time near the galactic core. Additionally, it would have less time to get deflected, gravitationally, by other stars in the core.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-07-08T21:19:46.323" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4846" PostTypeId="1" CreationDate="2014-07-09T02:39:55.813" Score="4" ViewCount="264" Body="&lt;p&gt;I often hear about cosmic inflation occurring rapidly after the Big Bang, but have never came across any explanation for it. What supposedly caused this inflationary period in the early universe? Also, my assumption is that the laws of physics, i.e. light speed and relativity, did not exist during inflation, but were after effects. Is that a correct assumption? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2025" LastEditorUserId="473" LastEditDate="2014-07-09T06:50:39.920" LastActivityDate="2015-01-24T21:52:36.853" Title="What is the explanation for rapid inflation just after the big bang?" Tags="&lt;big-bang-theory&gt;&lt;expansion&gt;&lt;cosmological-inflation&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4847" PostTypeId="1" AcceptedAnswerId="8368" CreationDate="2014-07-09T03:04:34.907" Score="3" ViewCount="128" Body="&lt;p&gt;I'm having trouble understanding what exactly &quot;information&quot; is in the context of the holographic principle suggested by string theory. Can it be equated to a matrix of ones and zeros? Does this information have its own laws of physics or are all laws of physics in our universe a result of this information, as in, does information supersede everything? &lt;/p&gt;&#xA;" OwnerUserId="2025" LastEditorUserId="6" LastEditDate="2021-02-22T17:02:45.390" LastActivityDate="2021-02-22T17:02:45.390" Title="In regards to the holographic principle, what exactly is information?" Tags="&lt;black-hole&gt;&lt;supermassive-black-hole&gt;&lt;holographic-principle&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ClosedDate="2014-12-31T00:55:36.430" ContentLicense="CC BY-SA 3.0" />
  <row Id="4848" PostTypeId="1" CreationDate="2014-07-09T05:44:19.590" Score="2" ViewCount="77" Body="&lt;p&gt;The Andromeda galaxy is about 600,000 light years away and is doppler blue-shifted.  Are there objects closer to us than Andromeda which are red-shifted?  Are there objects farther than Andromeda that are blue-shifted?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious as to whether there is a more or less uniform radius from earth outside of which objects are generally all moving away from us, and inside of which objects are generally all moving towards us.&lt;/p&gt;&#xA;" OwnerUserId="2026" LastActivityDate="2014-07-09T13:35:37.757" Title="Is there a radius inside of which objects are (doppler) blue shifted?" Tags="&lt;gravity&gt;&lt;light&gt;&lt;redshift&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4849" PostTypeId="2" ParentId="4848" CreationDate="2014-07-09T13:35:37.757" Score="3" Body="&lt;p&gt;No, there is no spherical volume of space within which objects all tend to approach us and thereby show blue-shift. The fact that the Andromeda galaxy is blue-shifted is because at these relatively small distances the cosmological red-shift is much smaller than further away. Andromeda's own (non-cosmological speed) is larger then the cosmological speed away from us and it happens to move towards us and is therefore blue-shifted. It could also have been moving away from us, and be red-shifted as a result, but that red-shift would not be cosmological. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd expect that the chance that a nearby (Local Group) galaxy is either moving away from us or moving towards us is 50%. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the Andromeda galaxy, and all other galaxies belonging to the Local Group, are gravitationally bound and would therefore not show any cosmological red-shift anyhow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB. The Andromeda Galaxy is actually 2.5 million light years away from us.&lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2014-07-09T13:35:37.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4850" PostTypeId="1" AcceptedAnswerId="4854" CreationDate="2014-07-09T14:01:57.987" Score="5" ViewCount="2355" Body="&lt;p&gt;If I focus my torch into the night sky,what would be the farthest point the light can reach? Recently I looked at the night sky and noticed a light spark moving around the clouds for about 15 min.After observation I jumped into a conclusion that its probably because of the light house near by or any powerful light emitter.&lt;/p&gt;&#xA;" OwnerUserId="1103" LastEditorUserId="1103" LastEditDate="2014-07-10T06:45:14.537" LastActivityDate="2014-07-11T21:01:50.823" Title="What is the farthest point of light that is emitted by a torch?" Tags="&lt;light&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4851" PostTypeId="2" ParentId="2488" CreationDate="2014-07-09T14:27:05.267" Score="2" Body="&lt;p&gt;Astronomers use this system simply because it is (was) convenient. The sky shifts about 15° every hour or a full circle, 360° every 24 hours (actually 23h56m). So if you want to know where a star is in an hour, simply add an hour to the local hour angle. &lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2014-07-09T14:27:05.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4852" PostTypeId="1" AcceptedAnswerId="4855" CreationDate="2014-07-09T13:30:48.523" Score="15" ViewCount="1359" Body="&lt;p&gt;There are several sci-fi movies where planetary bodies are shown in the sky in order to indicate that the characters are on an alien planet. A good example is in Predators (2010):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/M5Vjr.png&quot; alt=&quot;Alien Planet&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it actually possible for such planetary bodies to exist so close to each other as to be visible in such detail to the naked eye? Wouldn't gravitational forces cause disruptions on them that inhibit the development of life (whether plant or animal)? In our Solar system there's a good example in &lt;a href=&quot;http://en.wikipedia.org/wiki/Io_(moon)&quot; rel=&quot;noreferrer&quot;&gt;Io&lt;/a&gt; which has severe volcanic activity resulting from tidal heating due to its proximity to Jupiter.&lt;/p&gt;&#xA;" OwnerUserId="2030" OwnerDisplayName="KalenGi" LastActivityDate="2014-07-09T15:26:34.640" Title="Is it Possible for Planetary Bodies To Exist In Close Proximity Without Adverse Effects?" Tags="&lt;planet&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4853" PostTypeId="2" ParentId="4852" CreationDate="2014-07-09T15:11:31.947" Score="3" Body="&lt;p&gt;Have a look at this question (&lt;a href=&quot;https://astronomy.stackexchange.com/questions/4766/help-in-determining-the-features-of-an-unusual-fictional-star-system&quot;&gt;Help in determining the features of an unusual, fictional star system&lt;/a&gt;) for one possible hypothetical scenario in which such a planet (or a satellite) could exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a satellite of a giant planet, there will be tidal forces, sure, but after a few hundred million years of formation of the system, it is likely that there will be tidal locking due to the dissipative forces within the satellite, which will ensure that the effects of tides are minimal after that (or, in other words, the effects are 'static'). On a related note, I do find it interesting that Io isn't tidally locked to Jupiter yet, but this is a different question altogether.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I believe it actually helps to have a Jupiter-sized body around as it drives meteoroids away from the moon you live on, making life a little safer (this might be wrong, but I think this is how it will work out; also, this does not help if 'panspermia hypothesis' turns out to be correct).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having multiple planets nearby, as shown in the picture, is a little more difficult due to many body systems not being very stable. You could possibly have a binary planetary systems and life in the moons at Lagrangian points, but it is very likely that such systems will lose some of their bodies due to perturbations from other planets or similar second order effects due to them being unstable. You can have a moon revolving around the binary planetary system some distance away as an alternative, which would be fairly stable, I guess. However, with the current planet formation hypotheses, it seems very unlikely that you will have large planets near each other, since they won't be formed in the protoplanetary discs in a way that they don't coalesce to form a single larger planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also try this: &lt;a href=&quot;http://www.stefanom.org/spc/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.stefanom.org/spc/&lt;/a&gt; . You'll soon realize that having many planets close by, in a stable system is difficult.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2014-07-09T15:11:31.947" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4854" PostTypeId="2" ParentId="4850" CreationDate="2014-07-09T15:22:06.927" Score="8" Body="&lt;p&gt;If you point a flashlight up, then the light will reach as far as the other ends of the universe, if one has a sensitive enough measuring device, so long as it is not absorbed. Of course, said device might require a lot of work, and a huge telescope. Lights can easily be seen by satellites, but it would take a huge telescope to see a single light from, say, the Moon, and even larger the further you go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of your moving light, I would suspect a laser of some kind. Lasers are focused enough where one could easily see a reflection from the clouds, and pretty focused as well. It could be a moving light source, such as a plane or balloon of some kind as well.&lt;/p&gt;&#xA;" OwnerUserId="182" LastActivityDate="2014-07-09T15:22:06.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4855" PostTypeId="2" ParentId="4852" CreationDate="2014-07-09T15:26:34.640" Score="7" Body="&lt;p&gt;Quick calculation of &lt;a href=&quot;http://en.wikipedia.org/wiki/Roche_limit&quot;&gt;Roche limit&lt;/a&gt; for two identical earthlike (fluid) bodies.&#xA;Roche distance is about equal to 2.44 X radius X(density1/density2)^1/3.&#xA;For identical bodies the density conveniently cancels, so we get&#xA;Rd = 2.44 X radius.&#xA;Earth radius is 6378 km.&#xA;So Rd = 2.44 X 6378 = 15500 km (center to center).&#xA;The distance surface to surface (15500- 6378) is 9184km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At that distance, by trigonometry, another earth would subtend about 44° of sky, pretty big!, and just at the bleeding edge of stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nearest planet in the provided picture looks like it might be at about that distance. Evacuation might be in order as the other three planets or moons will degrade the close binary's orbital stability.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2014-07-09T15:26:34.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4856" PostTypeId="2" ParentId="4850" CreationDate="2014-07-09T15:29:20.663" Score="4" Body="&lt;p&gt;Potentially photons emitted by your torch could travel to infinite distances, assuming the universe will keep on expanding. If the light gets further away, the photons of your torch will be spread out over an increasing area and the density (number of photons per square metre) will decrease rapidly. At some distance, only one photon will strike the area of your eye/detector. And you may not be able to observe a single photon.&#xA;It is also likely (but not certain) that the photons will be absorbed by atoms. &lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2014-07-09T15:29:20.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4857" PostTypeId="1" CreationDate="2014-07-09T16:29:55.320" Score="5" ViewCount="170" Body="&lt;p&gt;Does the &quot;amount&quot; of dark energy in the universe change over time, or is it constant?  The dark energy I am asking about is the rate of acceleration of the universe's expansion.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="1386" LastEditDate="2014-09-04T16:54:47.017" LastActivityDate="2014-09-04T16:54:47.017" Title="Is dark energy constant?" Tags="&lt;expansion&gt;&lt;cosmological-inflation&gt;&lt;dark-energy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4858" PostTypeId="1" CreationDate="2014-07-09T19:11:11.907" Score="4" ViewCount="765" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Segue_1&quot; rel=&quot;nofollow&quot;&gt;This Wikipedia article&lt;/a&gt; states that Segue 1 is either a dwarf spheroidal galaxy or globular cluster associated with our own Milky Way.  I am wondering what the difference is between dwarf spheroidal galaxies and globular clusters, since at first glance is seems that the morphological properties and stellar ages are similar between the two classes of objects.&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2021-04-16T00:07:45.570" Title="What is the difference between a dwarf spheroidal galaxy and a globular cluster?" Tags="&lt;galaxy&gt;&lt;globular-clusters&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4859" PostTypeId="2" ParentId="4842" CreationDate="2014-07-09T19:52:55.883" Score="8" Body="&lt;p&gt;I make my own mirrors and telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method recommended by Gary Seronik, that you linked above, is very good, and it's widely used. A few comments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't be afraid to touch the mirror with clean paper. The vast majority of mirrors these days are over-coated with SiO(x) and are quite sturdy. Maybe wash the mirror before you do anything else, this way you'll remove potentially abrasive dust particles. This is how: &lt;a href=&quot;https://www.youtube.com/watch?v=9Y8xFnXFVGQ&quot;&gt;https://www.youtube.com/watch?v=9Y8xFnXFVGQ&lt;/a&gt; - it's a good idea to wash the mirror anyway, like once a year or every other year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When making the big paper template, cut the disk from a piece of paper without wrinkles. Draw a good circle, and cut it carefully. Make sure it's flat and it doesn't want to roll up. When centering it, go over the edge again and again, and adjust it so that it's perfectly centered. You center it by working out the edge alignment with the mirror. Take all the time you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also cut the template out of transparent plastic, larger than the mirror, and just draw a circle on it, the same size like the mirror. This method is recommended by Jim Fly (of Catseye) even for quite large mirrors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both versions (paper or plastic) work about the same IMO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may take a bit of trial and error to put the central dot on the mirror without smearing it. The marker tends to smear up underneath the paper. Practice on a random piece of glass or polished metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why Bob Fies (of Alcoat) says it's not for lasers is that, the way he shows it, the paper dot is solid. Of course it wouldn't work with a bare laser - you need a donut shape, with a hollow center, to allow the laser to reflect (when using a bare laser, not with a tuBlug). Just use a little donut instead of a solid paper dot, for the final step - like Gary shows on his website.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you never use a bare laser to adjust the primary, a solid dot is fine. But a donut gives you more options.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done business with Bob directly, he's very experienced. Like Gary Seronik and Jim Fly, he should be considered an expert in this field. But take his advice in context and look at all the details - he says what he says for a good reason.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The way to gauge the existing donut is also by using Gary's method. Make a paper disk with a hole in the middle, center it on the mirror, and put a little dot with a marker on the glass. Compare that dot with your existing donut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wipe the marker dot off with alcohol and a q-tip when you're done. You can't damage the mirror unless you press very hard. The center is shadowed by the secondary mirror anyway, so it doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do everything very slowly and be as precise as humanly possible. The precision of the central donut is critical for the precision of your collimation. Since you only put the donut on once, it makes sense to do a very slow but very high quality job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the existing donut is off-center even a little bit, don't hesitate to remove it. If alcohol soaking doesn't remove it, try acetone. You may have to clean the center afterwards, but that's okay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After everything is finished, clean the inside of the donut with alcohol and a q-tip. You need that surface relatively clean, for the laser to reflect back into your collimator when not using the tuBlug (just bare laser). It doesn't have to be squeaky clean, just remove the marker smears, more or less, and make it nice and tidy.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Congratulations on taking the right steps. Many, many people don't care about collimation, and it's so important for overall performance. Keep that scope collimated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tuBlug is an excellent tool for collimation, it makes the process a snap. Great choice. Catseye is another great collimator, but if you have the tuBlug and a Glatter laser (one of the most precise lasers out there, BTW) you probably don't need anything else, unless you have some extremely low f/ratio on your scope (not your case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use the tuBlug every time I do an observation. It's taking maybe 2 minutes to do the adjustments. Totally worth it.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2014-07-09T23:26:38.920" LastActivityDate="2014-07-09T23:26:38.920" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4860" PostTypeId="2" ParentId="4857" CreationDate="2014-07-09T21:14:04.007" Score="3" Body="&lt;p&gt;As mentioned above the density of vacuum energy in empty space is absolute constant, even as the volume of a region of space grows as the universe expands. So the total energy, density times volume, goes up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More information &lt;a href=&quot;http://www.preposterousuniverse.com/blog/2010/02/22/energy-is-not-conserved/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2034" LastEditorUserId="1469" LastEditDate="2014-07-10T16:12:12.120" LastActivityDate="2014-07-10T16:12:12.120" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4861" PostTypeId="1" AcceptedAnswerId="4863" CreationDate="2014-07-09T21:28:11.813" Score="7" ViewCount="770" Body="&lt;p&gt;I am working on a (personal) programming project, part of which involves generating random stars and their parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to create a procedurally generated star, I want to randomly select as few parameter values as possible, and then extrapolate the rest using formulas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance by interpreting the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law&quot; rel=&quot;noreferrer&quot;&gt;Stefan-Boltzmann law&lt;/a&gt; I can derive the star's luminosity based on the surface temperature and the star's radius:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;L = 4 \pi R^2 \sigma T_{e}^4&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are however other parameters that I would like to generate, such as the mass of the star. What seems to be able to help me is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Mass%E2%80%93luminosity_relation&quot; rel=&quot;noreferrer&quot;&gt;mass-luminosity relation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{L}{L_{\odot}} = \left(\frac{M}{M_{\odot}}\right)^a&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By knowing $a$ (based on my randomly selected mass) it seems that I can derive the star's luminosity. By glancing at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hertzsprung%E2%80%93Russell_diagram&quot; rel=&quot;noreferrer&quot;&gt;Hertzsprung–Russell diagram&lt;/a&gt; (e.g. &lt;a href=&quot;https://en.wikipedia.org/wiki/Hertzsprung%E2%80%93Russell_diagram#mediaviewer/File%3aHertzsprung-Russel_StarData.png&quot; rel=&quot;noreferrer&quot;&gt;this one&lt;/a&gt;) I can see that I can perhaps somehow calculate the radius of the star, but I'm not entirely sure how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an existing star &quot;generator&quot; which can be found &lt;a href=&quot;http://web.archive.org/web/20030218125318/www.geocities.com/Area51/Corridor/8611/mseqstar.htm&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;, however this one interpolates existing data using logarithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My full question would be, is there a way to &quot;generate&quot; a &quot;regular&quot;, main sequence star, starting from one or two parameters and extrapolating the rest using formulas? Is there a sensible relation between the mass of the star and its radius that can be used to generate a &quot;realistic&quot; star? Or would most parameters need to be interpolated from existing data?&lt;/p&gt;&#xA;" OwnerUserId="2035" LastEditorUserId="2035" LastEditDate="2014-07-10T08:43:16.077" LastActivityDate="2014-07-14T03:17:39.443" Title="Accurate way to programmatically generate a main sequence star?" Tags="&lt;star&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4862" PostTypeId="2" ParentId="4858" CreationDate="2014-07-09T21:29:52.260" Score="2" Body="&lt;p&gt;There are many differences starting from size to the dark matter content... Please check the following link  it has given much information :&#xA; &lt;a href=&quot;http://www.answers.com/topic/dwarf-spheroidal-galaxy&quot; rel=&quot;nofollow&quot;&gt;http://www.answers.com/topic/dwarf-spheroidal-galaxy&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2034" LastActivityDate="2014-07-09T21:29:52.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4863" PostTypeId="2" ParentId="4861" CreationDate="2014-07-10T03:48:42.967" Score="4" Body="&lt;p&gt;This is a tricky question, not immediate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we talk about Main Sequence stars, yes, you can constrain your fictional star by two parameters. The crucial parameters in the diagram, as you said, are the mass and the radius. This is why the HR diagram well represents the stellar population with two axis. Of course, for your scope, we need to do some assumptions. For example, the metallicity, that for your purposes can be fixed as the Solar one. Or, that the initial mass is already in a condition to contract.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see it this way: the luminosity is connected with the radius (because it depends on the irradiating surface), while the temperature is connected with the mass (the more is the mass, the more is the pressure at the center of the star).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have $L \sim RT$ and $L \sim M$, then you also have $RT \sim M$. &#xA;The degeneracy from one parameter can be overcame if you have your reference star (the Sun), which allows you to trace isoradii lines on the HR diagram. See here for example:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/0sS3I.gif&quot; alt=&quot;enter image description here&quot;&gt;&#xA;Or, that is equivalent, you can express your first equation (Stefan-Boltzmann), in Solar units as well (&lt;a href=&quot;http://abyss.uoregon.edu/~js/ast122/lectures/lec11.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;as shown here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try it out just by your self: choose a mass, or observationally speaking, choose a luminosity. Then &lt;a href=&quot;http://www.kcvs.ca/martin/astro/au/unit4/85/chp8_5.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;trace your line on the HR diagram&lt;/a&gt;, and find your star. That star has a defined temperature (spectral class), Mass and Radius as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need the exact equations, feel free to ask please.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nice aim, anyway. Have fun!&lt;/p&gt;&#xA;" OwnerUserId="1038" LastEditorUserId="1038" LastEditDate="2014-07-11T09:31:21.807" LastActivityDate="2014-07-11T09:31:21.807" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5863" PostTypeId="2" ParentId="4858" CreationDate="2014-07-10T05:56:05.123" Score="0" Body="&lt;p&gt;The main difference is if it is gravitationally bound to another galaxy or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A globullar cluster is a group of old stars inside a galaxy. It is not independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A galaxy is a group of stars, and a dwarf sheroidal galaxy is a group of (mainly) old stars, gravitationally bound in itself, but not bound inside a larger body (even when it can be orbiting a local group center of mass).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Segue 1 is orbiting around Milky Way's center, it is a globullar cluster. If it is orbiting with Milky way about Local Group's center near Andromeda, then it is a dwarf spheroidal galaxy.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-07-10T05:56:05.123" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5864" PostTypeId="1" AcceptedAnswerId="5873" CreationDate="2014-07-10T11:25:43.490" Score="5" ViewCount="1582" Body="&lt;p&gt;If we were to put a highly reflective substance into Venus's upper atmosphere, could we cool it down?  I am envisioning adding a highly reflective fine dust or gas, possibly lighter than &quot;air&quot; to the atmosphere.  I believe to have the proper effect, it should also reflect infrared light, which causes heat.  Apparently Venus already has a high albedo of .75 (reflects 75% of light): &lt;a href=&quot;http://www.universetoday.com/36833/albedo-of-venus/&quot; rel=&quot;nofollow&quot;&gt;http://www.universetoday.com/36833/albedo-of-venus/&lt;/a&gt; .  However, if we were able to raise that higher, even up to 99%, would that cool it down, perhaps even making it habitable (temperature wise), or possibly even making it ice cold?&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="2365" LastEditDate="2018-12-07T17:55:04.003" LastActivityDate="2018-12-07T17:55:04.003" Title="Would a high albedo reflective substance cool down Venus?" Tags="&lt;planetary-atmosphere&gt;&lt;temperature&gt;&lt;venus&gt;&lt;albedo&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5865" PostTypeId="1" AcceptedAnswerId="5872" CreationDate="2014-07-10T11:34:43.100" Score="1" ViewCount="192" Body="&lt;p&gt;If Mars were coated in a dark substance that absorbs sunlight (including infrared light), would this heat up Mars?  I am envisioning a fine dust that is dispersed into the atmosphere, but possibly settles on Mars's surface all around.  If we were able to change the albedo, possibly down to as low as 1%, would that make Mars habitable (temperature wise)?  If the albedo were changed all the way to 1%, could this even make Mars too hot?&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="2365" LastEditDate="2018-12-07T17:54:29.440" LastActivityDate="2018-12-07T17:54:29.440" Title="Would a low albedo substance warm up Mars?" Tags="&lt;mars&gt;&lt;temperature&gt;&lt;albedo&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5866" PostTypeId="1" CreationDate="2014-07-10T12:08:44.557" Score="1" ViewCount="842" Body="&lt;p&gt;As part of small project i want to measure temperature of the distant star on basis of optical observations. so by what exact method i can do it? which instruments will be necessary? what is the precision of the experiment?&lt;/p&gt;&#xA;" OwnerUserId="2034" LastActivityDate="2014-07-10T14:23:37.233" Title="how to measure temperature of the distant star" Tags="&lt;star&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2014-07-10T14:41:34.670" ContentLicense="CC BY-SA 3.0" />
  <row Id="5867" PostTypeId="2" ParentId="4850" CreationDate="2014-07-10T13:06:30.720" Score="5" Body="&lt;p&gt;&lt;a href=&quot;http://physicsworld.com/cws/article/news/2013/jul/24/what-is-the-lifetime-of-a-photon&quot; rel=&quot;noreferrer&quot;&gt;Some theories allow for photon decay&lt;/a&gt;. In that case, the average photon would decay after travelling for 10&lt;sup&gt;18&lt;/sup&gt; years, which would give us around 9.461×10&lt;sup&gt;30&lt;/sup&gt; kilometers. That's about a million times the current estimate for the diameter of the observable universe.&lt;/p&gt;&#xA;" OwnerUserId="893" LastActivityDate="2014-07-10T13:06:30.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5868" PostTypeId="2" ParentId="5866" CreationDate="2014-07-10T14:23:37.233" Score="1" Body="&lt;p&gt;To measure the surface temperature of a star it is used is it's &lt;strong&gt;black body spectrum&lt;/strong&gt;. You would have to get the light curve of the light from tat star and then by checking it's peak you could guess its temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really don't know which instruments to use to get the light curve from a distant star but you could use a tool like this one for the final conversion: &lt;a href=&quot;http://astro.unl.edu/naap/blackbody/animations/blackbody.html&quot; rel=&quot;nofollow&quot;&gt;http://astro.unl.edu/naap/blackbody/animations/blackbody.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1678" LastActivityDate="2014-07-10T14:23:37.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5869" PostTypeId="1" AcceptedAnswerId="5871" CreationDate="2014-07-10T15:42:44.690" Score="29" ViewCount="6608" Body="&lt;p&gt;When I was in college, I posed to my astronomy professor a thought experiment that had been puzzling my mind for some time: &quot;If all the matter in the Sun magically disappeared instantly, how long would it take its gravity to stop having an effect on us?&quot; His response was that the force of gravity is instant, unlike the speed of light, which appears instant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My big question is: &lt;strong&gt;&quot;Do we know it's instant?&quot;&lt;/strong&gt; We can't possibly move an object large enough to have a noticeable gravitational influence fast enough to measure if it creates (or doesn't create) a doppler-like phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he was wrong, &lt;strong&gt;how do we know it's not?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2006" LastEditorUserId="13092" LastEditDate="2018-02-24T21:17:31.397" LastActivityDate="2018-05-27T09:35:09.340" Title="Is the influence of gravity instantaneous?" Tags="&lt;gravity&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5870" PostTypeId="2" ParentId="5869" CreationDate="2014-07-10T16:13:29.643" Score="8" Body="&lt;p&gt;Actually, &quot;instant&quot; gravity was part of Newton's theory of gravity.  It is now understood that the &quot;speed of gravity&quot; is equal to the speed of light according to general relativity.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Speed_of_gravity&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Speed_of_gravity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has apparently fairly recently been confirmed, and is described in greater detail here: &lt;a href=&quot;https://medium.com/starts-with-a-bang/what-is-the-speed-of-gravity-8ada2eb08430#.kgnvcvxo2&quot; rel=&quot;nofollow&quot;&gt;https://medium.com/starts-with-a-bang/what-is-the-speed-of-gravity-8ada2eb08430#.kgnvcvxo2&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="1386" LastEditDate="2016-09-16T22:10:12.060" LastActivityDate="2016-09-16T22:10:12.060" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5871" PostTypeId="2" ParentId="5869" CreationDate="2014-07-10T16:49:30.117" Score="31" Body="&lt;p&gt;The first question as stated has a rather trivial answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;If the sun magically disappeared, instantly, &lt;strong&gt;along with all its influences&lt;/strong&gt;, how long would it take its gravity to stop having an effect on us?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since the Sun's gravity is among its influences, it would instantly stop having an effect on us. That's just part of the magical situation, and doesn't even involve any physics. A bit more interesting is the question without the bolded part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general relativity, changes in the gravitational field propagate at the speed of light. Thus, one might expect that the magical and instant disappearance of the Sun would not affect earth for about eight minutes, since that's how long light from the Sun takes to reach Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is mistaken because the instant disappearance of the Sun itself violates general relativity, as the Einstein field equation enforces a kind of &lt;em&gt;local&lt;/em&gt; conservation law on the stress-energy tensor analogous to the non-divergence of the magnetic field in the electromagnetism: in any small neighborhood of spacetime, there are no local sources or sinks of stress-energy; it must come from somewhere and go somewhere. Since the magical instant disappearance of the Sun violates general relativity, it does not make sense to use that theory to predict what happens in such a situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the Sun's gravity instantly ceasing any effect on the Earth is just as consistent with general relativity as having any sort of time-delay. Or to be precise, it's no more inconsistent.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My big question, now, is: &quot;How do we know it's instant?&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not instant, but it can appear that way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We can't possibly move an object large enough to have a noticeable gravitational influence fast enough to measure if it creates (or doesn't create) a doppler-like phenomenon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We don't have to: solar system dynamics are quite fast enough. An simple calculation due to Laplace in the early nineteenth century concluded that if gravity aberrated, Earth's orbit would crash into the Sun on the time-scale of about four centuries. Thus gravity does not aberrate appreciably--more careful analyses concluded that in the Newtonian framework, the speed of gravity must be more than $2\times10^{10}$ the speed of light to be consistent with the observed lack of aberration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may seem quite a bit puzzling with how it fits with general relativity's claim that changes in the gravitational field propagate at the speed of light, but it's actually not that peculiar. As an analogy, the electric field of a uniformly moving electric charge is directed toward the instantaneous position of the charge--&lt;em&gt;not&lt;/em&gt; where the charge used to be, as one might expect from a speed of light delay. This doesn't mean that electromagnetism propagates instantaneously--if you wiggle the charge, that information will be limited by $c$, as the electromagnetic field changes in response to your action. Instead, it's just something that's true for &lt;em&gt;uniformly moving&lt;/em&gt; charges: the electric field &quot;anticipates&quot; where the change will be if no influence acts on it. If the charge velocity changes slowly enough, it will look like electromagnetism is instantaneous, even though it really isn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity does this even better: the gravitational field of a uniformly accelerating mass is toward its current position. Thus, gravity &quot;anticipates&quot; where the mass will be based on not just current velocity, but also acceleration. Thus, if conditions are such that the acceleration of gravitating bodies changes slowly (as is the case in the solar system), gravity will look instantaneous. But this is only approximately true if the acceleration changes slowly--it's just a very good approximation under the conditions of the solar system. After all, Newtonian gravity works well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A detailed analysis of this can be found in Steve Carlip's &lt;em&gt;Aberration and the Speed of Gravity&lt;/em&gt;, Phys.Lett.A &lt;strong&gt;267&lt;/strong&gt;:81-87 (2000) [arXiV:&lt;a href=&quot;http://arxiv.org/abs/gr-qc/9909087&quot;&gt;gr-qc/9909087&lt;/a&gt;].&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If he was wrong, &lt;strong&gt;how do we know it's not?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We have a lot of evidence for general relativity, but the best current evidence that gravitational radiation behaves as GTR says it does is &lt;a href=&quot;http://en.wikipedia.org/wiki/Hulse%E2%80%93Taylor_binary&quot;&gt;Hulse-Taylor binary&lt;/a&gt;. However, there is no direct observation of gravitational radiation yet. The connection between the degree of apparent cancellation of velocity-dependent effects in both electromagnetism and gravity, including its connection with the dipole nature of EM radiation and quadrupole nature of gravitational radiation, can also be found in Carlip's paper.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditDate="2014-07-10T16:58:48.573" LastActivityDate="2014-07-10T16:58:48.573" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5872" PostTypeId="2" ParentId="5865" CreationDate="2014-07-10T18:36:12.620" Score="2" Body="&lt;p&gt;Of course! Actually this is one of the possible ways we have in order to make Mars friendlier &#xA;to humans. This technique is called Terraformation and it inludes the altering of the Mars surface's albedo. However, the feasibility of this plan is yet questionable and there are maybe better ways to achieve this increase in the temperature like CO$_2$ emissions. &lt;a href=&quot;http://www.nexialquest.com/The%20Terraformation%20of%20Worlds.pdf&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a nice article discussing exactly this subject.&lt;/p&gt;&#xA;" OwnerUserId="1956" LastActivityDate="2014-07-10T18:36:12.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5873" PostTypeId="2" ParentId="5864" CreationDate="2014-07-10T20:00:37.107" Score="6" Body="&lt;p&gt;Eventually, yes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interesting information about Venus: Venus is hotter than Mercury, despite being nearly twice as far from the Sun. Earth, despite being further from the Sun, receives &lt;strong&gt;more&lt;/strong&gt;&#xA;energy from the Sun than Venus, due to Venus's very high albedo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you might guess by this information, the major factor that keeps Venus hot isn't how much energy it receives from the Sun, but rather its inability to rid itself of energy. While adding a cloud of some high albedo substance to the atmosphere may effectively prevent the Sun from heating Venus, it would also likely prevent infrared radiation from leaving Venus, thus keeping it warm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you know, Venus already has a very high albedo. It's thick atmosphere prevents energy from leaving the planet, which is what caused the temperature to rise far above the Earth-like temperatures we believe it once enjoyed. This is caused by IR radiation being emitted from the surface and being reflected back down towards the planet by the atmosphere &lt;a href=&quot;http://www.weatherquestions.com/What_is_infrared_radiation.htm&quot; rel=&quot;noreferrer&quot;&gt;more detail&lt;/a&gt;, which plays an essential part in the greenhouse effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, a more effectual method in terms of making it habitable would be to find a way to drastically &lt;em&gt;lower&lt;/em&gt; the albedo of Venus to allow all the excess energy to escape.&lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="1288" LastEditDate="2014-07-10T20:27:25.960" LastActivityDate="2014-07-10T20:27:25.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5875" PostTypeId="2" ParentId="5865" CreationDate="2014-07-10T23:55:40.783" Score="0" Body="&lt;p&gt;Asphalt roads do a great job at &lt;a href=&quot;http://www.asla.org/lamag/lam07/february/ecology.html&quot; rel=&quot;nofollow&quot;&gt;heating up cities&lt;/a&gt;! Mars can actually get up to 80F, so part of the challenge is preventing losing that heat in the low pressure &lt;a href=&quot;http://en.wikipedia.org/wiki/Atmosphere_of_Mars&quot; rel=&quot;nofollow&quot;&gt;atmosphere&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-07-10T23:55:40.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5876" PostTypeId="2" ParentId="5864" CreationDate="2014-07-11T09:39:34.633" Score="4" Body="&lt;p&gt;If you can add a material which can reflect optical light (high optical albedo) and &lt;em&gt;transmit&lt;/em&gt; IR light (i.e. it does not absorb or reflect too much in IR), it will reduce further heating of Venus and it will gradually cool down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot lower the IR albedo (by using an IR absorbing material, for example) of Venus since it already has a thick atmosphere with greenhouse gases and if the atmosphere absorbs more IR, it will heat up. You cannot increase IR albedo because it will reflect emitted energy back to Venus and it will still heat up. You need to have a layer which transmits energy and does not reflect or absorb. The transmission of the atmosphere is already low and it cannot be increased, but if you can reduce the amount of incoming light by increasing the optical albedo, and not reduce the transmission at IR wavelengths, Venus will cool down gradually.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2014-07-11T09:39:34.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5877" PostTypeId="1" AcceptedAnswerId="5888" CreationDate="2014-07-11T15:13:14.713" Score="3" ViewCount="142" Body="&lt;p&gt;I am trying to get some data from the &lt;a href=&quot;http://www.sdss3.org/dr10/data_access/&quot; rel=&quot;nofollow&quot;&gt;Sloan digital sky survey&lt;/a&gt; (SDSS). This task turned out to be challenging for me as someone with only little knowledge about astronomy and maybe someone here can help me. What I need is data for all galaxies of the SDSS with their optical spectroscopy and position. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose I need:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;redshifts&lt;/li&gt;&#xA;&lt;li&gt;RA (right ascension)&lt;/li&gt;&#xA;&lt;li&gt;dec (declination)&lt;/li&gt;&#xA;&lt;li&gt;classifications of all SDSS spectra&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;from the &lt;a href=&quot;http://www.sdss3.org/dr10/data_access/bulk.php&quot; rel=&quot;nofollow&quot;&gt;Optical Spectra Catalog Data&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a script to convert ra/dec/redshift to x/y/z coordinates.&#xA;Will I need &quot;the associated photometric position based matches&quot; and &quot;stellar parameter (SSPP) results&quot; to get RA and dec? I have no idea what these are.&lt;/p&gt;&#xA;" OwnerUserId="2042" LastEditorUserId="2365" LastEditDate="2016-02-14T20:44:43.950" LastActivityDate="2016-02-14T20:44:43.950" Title="Sloan digital sky survey galaxies" Tags="&lt;galaxy&gt;&lt;data-analysis&gt;&lt;deep-sky-observing&gt;&lt;sky-survey&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5879" PostTypeId="1" AcceptedAnswerId="8654" CreationDate="2014-07-11T18:12:08.073" Score="2" ViewCount="1911" Body="&lt;p&gt;I want to know if bringing the image of a DSO in eyepiece generally represents a problem for beginners in astronomy.&lt;/p&gt;&#xA;" OwnerUserId="2044" LastActivityDate="2020-08-13T19:28:32.667" Title="Is it difficult to see DSO in your eyepiece?" Tags="&lt;amateur-observing&gt;&lt;observational-astronomy&gt;&lt;deep-sky-observing&gt;" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5880" PostTypeId="2" ParentId="2357" CreationDate="2014-07-11T18:14:47.223" Score="5" Body="&lt;p&gt;In the vacuum of space the most important consideration is to consider how much radiation an ice cube would absorb from, for example, nearby stars and how fast the ice cube itself would radiate away energy (using &lt;a href=&quot;http://en.wikipedia.org/wiki/Wien%27s_displacement_law&quot; rel=&quot;noreferrer&quot;&gt;Wien's law&lt;/a&gt;), finding what ice cube temperature would produce an equilibrium (the temperature at which the ice cube radiate energy at the same rate it absorbed energy) and then determining if that temperature is above or below the melting point of the ice cube.  If it is above the melting point (of water in a vacuum), then as the other answers have said the ice cube would sublimate; if it is below the melting point then the ice cube would stay frozen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically for an ice cube that is a cube in orbit around the sun with one side facing the sun you would need to calculate how much energy the side facing the sun absorbs from the sun as well as how much energy radiates away from all six sides of the cube and then find the equilibrium temperature.&lt;/p&gt;&#xA;" OwnerUserId="128" LastEditorUserId="128" LastEditDate="2014-07-11T18:20:55.377" LastActivityDate="2014-07-11T18:20:55.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5881" PostTypeId="2" ParentId="4850" CreationDate="2014-07-11T21:01:50.823" Score="2" Body="&lt;p&gt;Potentially forever assuming nothing blocks it's path fully, however past a relatively short distance a torch would not be observable without an immense effort (large optics.)  It may be next to impossible to discern if there is competing light emission (for example, if you shine a torch to the sky in the middle of the day, then the Sun's reflection would overpower it to a potential observer.)&lt;/p&gt;&#xA;" OwnerUserId="2046" LastActivityDate="2014-07-11T21:01:50.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5882" PostTypeId="2" ParentId="33" CreationDate="2014-07-11T21:06:39.213" Score="2" Body="&lt;p&gt;Black Holes do not have infinite mass; they have presumably infinite &lt;em&gt;density&lt;/em&gt; at the point of singularity.  The mass of the black hole plays a large role in the diameter of the event horizon (the sphere where escape velocity required exceeds the speed of light.  The 'black' part.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A supermassive black hole, as you suggested, just has more mass; It has a lot more mass than a stellar mass black hole.  Supermassive black holes are usually at the center of galaxies and can have gravitation influences across the entire galaxy, while stellar mass black holes will have the gravitation influence of the size of the former large star they formed from.&lt;/p&gt;&#xA;" OwnerUserId="2046" LastActivityDate="2014-07-11T21:06:39.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5883" PostTypeId="1" CreationDate="2014-07-12T03:02:09.243" Score="2" ViewCount="1023" Body="&lt;p&gt;I am trying to program a sunrise/sunset calculator using &lt;a href=&quot;http://users.electromagnetic.net/bu/astro/sunrise-set.php&quot; rel=&quot;nofollow&quot;&gt;this formula tutorial&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you follow the steps, eventually you get to this formula for &quot;H&quot; (the &quot;hour angle&quot;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H = \arccos( [\sin(-0.83) - \sin(ln) * \sin(δ)] / [\cos(ln) * \cos(δ)] )$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which comes with a note attached:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Note: If H is undefined, then there is either no sunrise (in winter) or no sunset (in summer) for the supplied latitude.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, as it turns out, this happens quite often (because inverse cosine has a domain of -1 to 1). I ran a program and here are all of the dates where the H is defined from Jan 2008 through Dec 2014 (using a longitude west of -77.0368710 and latitude north of 38.9071920 [Washington D.C.]):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;January 1 2008&#xA;January 3 2008&#xA;January 4 2008&#xA;January 11 2008&#xA;January 30 2008&#xA;******************************&#xA;February 5 2008&#xA;February 24 2008&#xA;******************************&#xA;March 2 2008&#xA;March 24 2008&#xA;March 25 2008&#xA;March 29 2008&#xA;March 31 2008&#xA;******************************&#xA;April 3 2008&#xA;April 4 2008&#xA;April 6 2008&#xA;April 11 2008&#xA;April 12 2008&#xA;April 13 2008&#xA;April 14 2008&#xA;April 17 2008&#xA;April 19 2008&#xA;April 22 2008&#xA;April 23 2008&#xA;April 25 2008&#xA;April 26 2008&#xA;April 30 2008&#xA;******************************&#xA;May 2 2008&#xA;May 3 2008&#xA;May 5 2008&#xA;May 6 2008&#xA;May 8 2008&#xA;May 14 2008&#xA;May 15 2008&#xA;May 19 2008&#xA;May 21 2008&#xA;May 24 2008&#xA;May 25 2008&#xA;May 27 2008&#xA;******************************&#xA;June 1 2008&#xA;June 2 2008&#xA;June 3 2008&#xA;June 4 2008&#xA;June 7 2008&#xA;June 9 2008&#xA;June 12 2008&#xA;June 13 2008&#xA;June 15 2008&#xA;June 16 2008&#xA;June 20 2008&#xA;June 22 2008&#xA;June 23 2008&#xA;June 25 2008&#xA;June 26 2008&#xA;June 28 2008&#xA;******************************&#xA;July 4 2008&#xA;July 5 2008&#xA;July 9 2008&#xA;July 11 2008&#xA;July 14 2008&#xA;July 15 2008&#xA;July 17 2008&#xA;July 22 2008&#xA;July 23 2008&#xA;July 24 2008&#xA;July 25 2008&#xA;July 28 2008&#xA;July 30 2008&#xA;******************************&#xA;August 2 2008&#xA;August 3 2008&#xA;August 5 2008&#xA;August 6 2008&#xA;August 10 2008&#xA;August 12 2008&#xA;August 13 2008&#xA;August 15 2008&#xA;August 16 2008&#xA;August 18 2008&#xA;August 24 2008&#xA;August 25 2008&#xA;August 29 2008&#xA;August 31 2008&#xA;******************************&#xA;September 3 2008&#xA;September 4 2008&#xA;September 6 2008&#xA;September 11 2008&#xA;September 12 2008&#xA;September 13 2008&#xA;September 14 2008&#xA;September 17 2008&#xA;September 19 2008&#xA;September 22 2008&#xA;September 23 2008&#xA;September 25 2008&#xA;September 26 2008&#xA;September 30 2008&#xA;******************************&#xA;October 2 2008&#xA;October 3 2008&#xA;October 5 2008&#xA;October 6 2008&#xA;October 8 2008&#xA;October 14 2008&#xA;October 15 2008&#xA;October 19 2008&#xA;October 21 2008&#xA;October 24 2008&#xA;October 25 2008&#xA;October 27 2008&#xA;******************************&#xA;November 1 2008&#xA;November 2 2008&#xA;November 3 2008&#xA;November 4 2008&#xA;November 7 2008&#xA;November 9 2008&#xA;November 12 2008&#xA;November 13 2008&#xA;November 15 2008&#xA;November 16 2008&#xA;November 20 2008&#xA;November 22 2008&#xA;November 23 2008&#xA;November 25 2008&#xA;November 26 2008&#xA;November 28 2008&#xA;******************************&#xA;December 4 2008&#xA;December 5 2008&#xA;December 9 2008&#xA;December 11 2008&#xA;December 14 2008&#xA;December 15 2008&#xA;December 17 2008&#xA;December 22 2008&#xA;December 23 2008&#xA;December 24 2008&#xA;December 25 2008&#xA;December 28 2008&#xA;December 30 2008&#xA;******************************&#xA;January 2 2009&#xA;January 3 2009&#xA;January 16 2009&#xA;January 23 2009&#xA;******************************&#xA;February 11 2009&#xA;February 17 2009&#xA;******************************&#xA;March 8 2009&#xA;March 15 2009&#xA;March 23 2009&#xA;March 25 2009&#xA;March 26 2009&#xA;March 28 2009&#xA;March 29 2009&#xA;March 31 2009&#xA;******************************&#xA;April 6 2009&#xA;April 7 2009&#xA;April 11 2009&#xA;April 13 2009&#xA;April 16 2009&#xA;April 17 2009&#xA;April 19 2009&#xA;April 24 2009&#xA;April 25 2009&#xA;April 26 2009&#xA;April 27 2009&#xA;April 30 2009&#xA;******************************&#xA;May 2 2009&#xA;May 5 2009&#xA;May 6 2009&#xA;May 8 2009&#xA;May 9 2009&#xA;May 13 2009&#xA;May 15 2009&#xA;May 16 2009&#xA;May 18 2009&#xA;May 19 2009&#xA;May 21 2009&#xA;May 27 2009&#xA;May 28 2009&#xA;******************************&#xA;June 1 2009&#xA;June 3 2009&#xA;June 6 2009&#xA;June 7 2009&#xA;June 9 2009&#xA;June 14 2009&#xA;June 15 2009&#xA;June 16 2009&#xA;June 17 2009&#xA;June 20 2009&#xA;June 22 2009&#xA;June 25 2009&#xA;June 26 2009&#xA;June 28 2009&#xA;June 29 2009&#xA;******************************&#xA;July 3 2009&#xA;July 5 2009&#xA;July 6 2009&#xA;July 8 2009&#xA;July 9 2009&#xA;July 11 2009&#xA;July 17 2009&#xA;July 18 2009&#xA;July 22 2009&#xA;July 24 2009&#xA;July 27 2009&#xA;July 28 2009&#xA;July 30 2009&#xA;******************************&#xA;August 4 2009&#xA;August 5 2009&#xA;August 6 2009&#xA;August 7 2009&#xA;August 10 2009&#xA;August 12 2009&#xA;August 15 2009&#xA;August 16 2009&#xA;August 18 2009&#xA;August 19 2009&#xA;August 23 2009&#xA;August 25 2009&#xA;August 26 2009&#xA;August 28 2009&#xA;August 29 2009&#xA;August 31 2009&#xA;******************************&#xA;September 6 2009&#xA;September 7 2009&#xA;September 11 2009&#xA;September 13 2009&#xA;September 16 2009&#xA;September 17 2009&#xA;September 19 2009&#xA;September 24 2009&#xA;September 25 2009&#xA;September 26 2009&#xA;September 27 2009&#xA;September 30 2009&#xA;******************************&#xA;October 2 2009&#xA;October 5 2009&#xA;October 6 2009&#xA;October 8 2009&#xA;October 9 2009&#xA;October 13 2009&#xA;October 15 2009&#xA;October 16 2009&#xA;October 18 2009&#xA;October 19 2009&#xA;October 21 2009&#xA;October 27 2009&#xA;October 28 2009&#xA;******************************&#xA;November 1 2009&#xA;November 3 2009&#xA;November 6 2009&#xA;November 7 2009&#xA;November 9 2009&#xA;November 14 2009&#xA;November 15 2009&#xA;November 16 2009&#xA;November 17 2009&#xA;November 20 2009&#xA;November 22 2009&#xA;November 25 2009&#xA;November 26 2009&#xA;November 28 2009&#xA;November 29 2009&#xA;******************************&#xA;December 3 2009&#xA;December 5 2009&#xA;December 6 2009&#xA;December 8 2009&#xA;December 9 2009&#xA;December 11 2009&#xA;December 17 2009&#xA;December 18 2009&#xA;December 22 2009&#xA;December 24 2009&#xA;December 27 2009&#xA;December 28 2009&#xA;December 30 2009&#xA;******************************&#xA;January 4 2010&#xA;January 10 2010&#xA;January 17 2010&#xA;January 29 2010&#xA;******************************&#xA;February 5 2010&#xA;February 24 2010&#xA;******************************&#xA;March 2 2010&#xA;March 9 2010&#xA;March 21 2010&#xA;March 23 2010&#xA;March 25 2010&#xA;March 28 2010&#xA;March 29 2010&#xA;March 31 2010&#xA;******************************&#xA;April 1 2010&#xA;April 5 2010&#xA;April 7 2010&#xA;April 8 2010&#xA;April 10 2010&#xA;April 11 2010&#xA;April 13 2010&#xA;April 19 2010&#xA;April 20 2010&#xA;April 24 2010&#xA;April 26 2010&#xA;April 29 2010&#xA;April 30 2010&#xA;******************************&#xA;May 2 2010&#xA;May 7 2010&#xA;May 8 2010&#xA;May 9 2010&#xA;May 10 2010&#xA;May 13 2010&#xA;May 15 2010&#xA;May 18 2010&#xA;May 19 2010&#xA;May 21 2010&#xA;May 22 2010&#xA;May 26 2010&#xA;May 28 2010&#xA;May 29 2010&#xA;May 31 2010&#xA;******************************&#xA;June 1 2010&#xA;June 3 2010&#xA;June 9 2010&#xA;June 10 2010&#xA;June 14 2010&#xA;June 16 2010&#xA;June 19 2010&#xA;June 20 2010&#xA;June 22 2010&#xA;June 27 2010&#xA;June 28 2010&#xA;June 29 2010&#xA;June 30 2010&#xA;******************************&#xA;July 3 2010&#xA;July 5 2010&#xA;July 8 2010&#xA;July 9 2010&#xA;July 11 2010&#xA;July 12 2010&#xA;July 16 2010&#xA;July 18 2010&#xA;July 19 2010&#xA;July 21 2010&#xA;July 22 2010&#xA;July 24 2010&#xA;July 30 2010&#xA;July 31 2010&#xA;******************************&#xA;August 4 2010&#xA;August 6 2010&#xA;August 9 2010&#xA;August 10 2010&#xA;August 12 2010&#xA;August 17 2010&#xA;August 18 2010&#xA;August 19 2010&#xA;August 20 2010&#xA;August 23 2010&#xA;August 25 2010&#xA;August 28 2010&#xA;August 29 2010&#xA;August 31 2010&#xA;******************************&#xA;September 1 2010&#xA;September 5 2010&#xA;September 7 2010&#xA;September 8 2010&#xA;September 10 2010&#xA;September 11 2010&#xA;September 13 2010&#xA;September 19 2010&#xA;September 20 2010&#xA;September 24 2010&#xA;September 26 2010&#xA;September 29 2010&#xA;September 30 2010&#xA;******************************&#xA;October 2 2010&#xA;October 7 2010&#xA;October 8 2010&#xA;October 9 2010&#xA;October 10 2010&#xA;October 13 2010&#xA;October 15 2010&#xA;October 18 2010&#xA;October 19 2010&#xA;October 21 2010&#xA;October 22 2010&#xA;October 26 2010&#xA;October 28 2010&#xA;October 29 2010&#xA;October 31 2010&#xA;******************************&#xA;November 1 2010&#xA;November 3 2010&#xA;November 9 2010&#xA;November 10 2010&#xA;November 14 2010&#xA;November 16 2010&#xA;November 19 2010&#xA;November 20 2010&#xA;November 22 2010&#xA;November 27 2010&#xA;November 28 2010&#xA;November 29 2010&#xA;November 30 2010&#xA;******************************&#xA;December 3 2010&#xA;December 5 2010&#xA;December 8 2010&#xA;December 9 2010&#xA;December 11 2010&#xA;December 12 2010&#xA;December 16 2010&#xA;December 18 2010&#xA;December 19 2010&#xA;December 21 2010&#xA;December 22 2010&#xA;December 24 2010&#xA;December 30 2010&#xA;December 31 2010&#xA;******************************&#xA;January 17 2011&#xA;January 23 2011&#xA;January 30 2011&#xA;******************************&#xA;February 11 2011&#xA;February 18 2011&#xA;******************************&#xA;March 9 2011&#xA;March 15 2011&#xA;March 22 2011&#xA;March 23 2011&#xA;March 25 2011&#xA;March 30 2011&#xA;March 31 2011&#xA;******************************&#xA;April 1 2011&#xA;April 2 2011&#xA;April 5 2011&#xA;April 7 2011&#xA;April 10 2011&#xA;April 11 2011&#xA;April 13 2011&#xA;April 14 2011&#xA;April 18 2011&#xA;April 20 2011&#xA;April 21 2011&#xA;April 23 2011&#xA;April 24 2011&#xA;April 26 2011&#xA;******************************&#xA;May 2 2011&#xA;May 3 2011&#xA;May 7 2011&#xA;May 9 2011&#xA;May 12 2011&#xA;May 13 2011&#xA;May 15 2011&#xA;May 20 2011&#xA;May 21 2011&#xA;May 22 2011&#xA;May 23 2011&#xA;May 26 2011&#xA;May 28 2011&#xA;May 31 2011&#xA;******************************&#xA;June 1 2011&#xA;June 3 2011&#xA;June 4 2011&#xA;June 8 2011&#xA;June 10 2011&#xA;June 11 2011&#xA;June 13 2011&#xA;June 14 2011&#xA;June 16 2011&#xA;June 22 2011&#xA;June 23 2011&#xA;June 27 2011&#xA;June 29 2011&#xA;******************************&#xA;July 2 2011&#xA;July 3 2011&#xA;July 5 2011&#xA;July 10 2011&#xA;July 11 2011&#xA;July 12 2011&#xA;July 13 2011&#xA;July 16 2011&#xA;July 18 2011&#xA;July 21 2011&#xA;July 22 2011&#xA;July 24 2011&#xA;July 25 2011&#xA;July 29 2011&#xA;July 31 2011&#xA;******************************&#xA;August 1 2011&#xA;August 3 2011&#xA;August 4 2011&#xA;August 6 2011&#xA;August 12 2011&#xA;August 13 2011&#xA;August 17 2011&#xA;August 19 2011&#xA;August 22 2011&#xA;August 23 2011&#xA;August 25 2011&#xA;August 30 2011&#xA;August 31 2011&#xA;******************************&#xA;September 1 2011&#xA;September 2 2011&#xA;September 5 2011&#xA;September 7 2011&#xA;September 10 2011&#xA;September 11 2011&#xA;September 13 2011&#xA;September 14 2011&#xA;September 18 2011&#xA;September 20 2011&#xA;September 21 2011&#xA;September 23 2011&#xA;September 24 2011&#xA;September 26 2011&#xA;******************************&#xA;October 2 2011&#xA;October 3 2011&#xA;October 7 2011&#xA;October 9 2011&#xA;October 12 2011&#xA;October 13 2011&#xA;October 15 2011&#xA;October 20 2011&#xA;October 21 2011&#xA;October 22 2011&#xA;October 23 2011&#xA;October 26 2011&#xA;October 28 2011&#xA;October 31 2011&#xA;******************************&#xA;November 1 2011&#xA;November 3 2011&#xA;November 4 2011&#xA;November 8 2011&#xA;November 10 2011&#xA;November 11 2011&#xA;November 13 2011&#xA;November 14 2011&#xA;November 16 2011&#xA;November 22 2011&#xA;November 23 2011&#xA;November 27 2011&#xA;November 29 2011&#xA;******************************&#xA;December 2 2011&#xA;December 3 2011&#xA;December 5 2011&#xA;December 10 2011&#xA;December 11 2011&#xA;December 12 2011&#xA;December 13 2011&#xA;December 16 2011&#xA;December 18 2011&#xA;December 21 2011&#xA;December 22 2011&#xA;December 24 2011&#xA;December 25 2011&#xA;December 29 2011&#xA;December 31 2011&#xA;******************************&#xA;January 1 2012&#xA;January 3 2012&#xA;January 4 2012&#xA;January 11 2012&#xA;January 30 2012&#xA;******************************&#xA;February 5 2012&#xA;February 12 2012&#xA;February 24 2012&#xA;******************************&#xA;March 2 2012&#xA;March 24 2012&#xA;March 25 2012&#xA;March 26 2012&#xA;March 29 2012&#xA;March 31 2012&#xA;******************************&#xA;April 3 2012&#xA;April 4 2012&#xA;April 6 2012&#xA;April 11 2012&#xA;April 12 2012&#xA;April 13 2012&#xA;April 14 2012&#xA;April 17 2012&#xA;April 19 2012&#xA;April 22 2012&#xA;April 23 2012&#xA;April 25 2012&#xA;April 26 2012&#xA;April 30 2012&#xA;******************************&#xA;May 2 2012&#xA;May 3 2012&#xA;May 5 2012&#xA;May 6 2012&#xA;May 8 2012&#xA;May 14 2012&#xA;May 15 2012&#xA;May 16 2012&#xA;May 19 2012&#xA;May 21 2012&#xA;May 24 2012&#xA;May 25 2012&#xA;May 27 2012&#xA;******************************&#xA;June 1 2012&#xA;June 2 2012&#xA;June 3 2012&#xA;June 4 2012&#xA;June 7 2012&#xA;June 9 2012&#xA;June 12 2012&#xA;June 13 2012&#xA;June 15 2012&#xA;June 16 2012&#xA;June 20 2012&#xA;June 22 2012&#xA;June 23 2012&#xA;June 25 2012&#xA;June 26 2012&#xA;June 28 2012&#xA;******************************&#xA;July 4 2012&#xA;July 5 2012&#xA;July 6 2012&#xA;July 9 2012&#xA;July 11 2012&#xA;July 14 2012&#xA;July 15 2012&#xA;July 17 2012&#xA;July 22 2012&#xA;July 23 2012&#xA;July 24 2012&#xA;July 25 2012&#xA;July 28 2012&#xA;July 30 2012&#xA;******************************&#xA;August 2 2012&#xA;August 3 2012&#xA;August 5 2012&#xA;August 6 2012&#xA;August 10 2012&#xA;August 12 2012&#xA;August 13 2012&#xA;August 15 2012&#xA;August 16 2012&#xA;August 18 2012&#xA;August 24 2012&#xA;August 25 2012&#xA;August 26 2012&#xA;August 29 2012&#xA;August 31 2012&#xA;******************************&#xA;September 3 2012&#xA;September 4 2012&#xA;September 6 2012&#xA;September 11 2012&#xA;September 12 2012&#xA;September 13 2012&#xA;September 14 2012&#xA;September 17 2012&#xA;September 19 2012&#xA;September 22 2012&#xA;September 23 2012&#xA;September 25 2012&#xA;September 26 2012&#xA;September 30 2012&#xA;******************************&#xA;October 2 2012&#xA;October 3 2012&#xA;October 5 2012&#xA;October 6 2012&#xA;October 8 2012&#xA;October 14 2012&#xA;October 15 2012&#xA;October 16 2012&#xA;October 19 2012&#xA;October 21 2012&#xA;October 24 2012&#xA;October 25 2012&#xA;October 27 2012&#xA;******************************&#xA;November 1 2012&#xA;November 2 2012&#xA;November 3 2012&#xA;November 4 2012&#xA;November 7 2012&#xA;November 9 2012&#xA;November 12 2012&#xA;November 13 2012&#xA;November 15 2012&#xA;November 16 2012&#xA;November 20 2012&#xA;November 22 2012&#xA;November 23 2012&#xA;November 25 2012&#xA;November 26 2012&#xA;November 28 2012&#xA;******************************&#xA;December 4 2012&#xA;December 5 2012&#xA;December 6 2012&#xA;December 9 2012&#xA;December 11 2012&#xA;December 14 2012&#xA;December 15 2012&#xA;December 17 2012&#xA;December 22 2012&#xA;December 23 2012&#xA;December 24 2012&#xA;December 25 2012&#xA;December 28 2012&#xA;December 30 2012&#xA;******************************&#xA;January 2 2013&#xA;January 3 2013&#xA;January 4 2013&#xA;January 16 2013&#xA;January 23 2013&#xA;******************************&#xA;February 11 2013&#xA;February 17 2013&#xA;February 24 2013&#xA;******************************&#xA;March 8 2013&#xA;March 15 2013&#xA;March 23 2013&#xA;March 25 2013&#xA;March 26 2013&#xA;March 28 2013&#xA;March 29 2013&#xA;March 31 2013&#xA;******************************&#xA;April 6 2013&#xA;April 7 2013&#xA;April 8 2013&#xA;April 11 2013&#xA;April 13 2013&#xA;April 16 2013&#xA;April 17 2013&#xA;April 19 2013&#xA;April 24 2013&#xA;April 25 2013&#xA;April 26 2013&#xA;April 27 2013&#xA;April 30 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2013&#xA;August 25 2013&#xA;August 26 2013&#xA;August 28 2013&#xA;August 29 2013&#xA;August 31 2013&#xA;******************************&#xA;September 6 2013&#xA;September 7 2013&#xA;September 8 2013&#xA;September 11 2013&#xA;September 13 2013&#xA;September 16 2013&#xA;September 17 2013&#xA;September 19 2013&#xA;September 24 2013&#xA;September 25 2013&#xA;September 26 2013&#xA;September 27 2013&#xA;September 30 2013&#xA;******************************&#xA;October 2 2013&#xA;October 6 2013&#xA;October 8 2013&#xA;October 9 2013&#xA;October 13 2013&#xA;October 15 2013&#xA;October 16 2013&#xA;October 18 2013&#xA;October 19 2013&#xA;October 21 2013&#xA;October 27 2013&#xA;October 28 2013&#xA;October 29 2013&#xA;******************************&#xA;November 1 2013&#xA;November 3 2013&#xA;November 6 2013&#xA;November 7 2013&#xA;November 9 2013&#xA;November 14 2013&#xA;November 15 2013&#xA;November 16 2013&#xA;November 17 2013&#xA;November 20 2013&#xA;November 22 2013&#xA;November 26 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2014&#xA;******************************&#xA;May 2 2014&#xA;May 7 2014&#xA;May 8 2014&#xA;May 9 2014&#xA;May 10 2014&#xA;May 13 2014&#xA;May 15 2014&#xA;May 19 2014&#xA;May 21 2014&#xA;May 22 2014&#xA;May 26 2014&#xA;May 28 2014&#xA;May 29 2014&#xA;May 31 2014&#xA;******************************&#xA;June 1 2014&#xA;June 3 2014&#xA;June 9 2014&#xA;June 10 2014&#xA;June 11 2014&#xA;June 14 2014&#xA;June 16 2014&#xA;June 19 2014&#xA;June 20 2014&#xA;June 22 2014&#xA;June 27 2014&#xA;June 28 2014&#xA;June 29 2014&#xA;June 30 2014&#xA;******************************&#xA;July 3 2014&#xA;July 5 2014&#xA;July 9 2014&#xA;July 11 2014&#xA;July 12 2014&#xA;July 16 2014&#xA;July 18 2014&#xA;July 19 2014&#xA;July 21 2014&#xA;July 22 2014&#xA;July 24 2014&#xA;July 30 2014&#xA;July 31 2014&#xA;******************************&#xA;August 1 2014&#xA;August 4 2014&#xA;August 6 2014&#xA;August 9 2014&#xA;August 10 2014&#xA;August 12 2014&#xA;August 17 2014&#xA;August 18 2014&#xA;August 19 2014&#xA;August 20 2014&#xA;August 23 2014&#xA;August 25 2014&#xA;August 29 2014&#xA;August 31 2014&#xA;******************************&#xA;September 1 2014&#xA;September 5 2014&#xA;September 7 2014&#xA;September 8 2014&#xA;September 10 2014&#xA;September 11 2014&#xA;September 13 2014&#xA;September 19 2014&#xA;September 20 2014&#xA;September 21 2014&#xA;September 24 2014&#xA;September 26 2014&#xA;September 29 2014&#xA;September 30 2014&#xA;******************************&#xA;October 2 2014&#xA;October 7 2014&#xA;October 8 2014&#xA;October 9 2014&#xA;October 10 2014&#xA;October 13 2014&#xA;October 15 2014&#xA;October 19 2014&#xA;October 21 2014&#xA;October 22 2014&#xA;October 26 2014&#xA;October 28 2014&#xA;October 29 2014&#xA;October 31 2014&#xA;******************************&#xA;November 1 2014&#xA;November 3 2014&#xA;November 9 2014&#xA;November 10 2014&#xA;November 11 2014&#xA;November 14 2014&#xA;November 16 2014&#xA;November 19 2014&#xA;November 20 2014&#xA;November 22 2014&#xA;November 27 2014&#xA;November 28 2014&#xA;November 29 2014&#xA;November 30 2014&#xA;******************************&#xA;December 3 2014&#xA;December 5 2014&#xA;December 9 2014&#xA;December 11 2014&#xA;December 12 2014&#xA;December 16 2014&#xA;December 18 2014&#xA;December 19 2014&#xA;December 21 2014&#xA;December 22 2014&#xA;December 24 2014&#xA;December 30 2014&#xA;December 31 2014&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You'll notice there are quite a lot of dates where H (the hour angle) is undefined. I understand from a trigonometric standpoint why it is incalculable, but is there a scientific reason for this? Also, how do they calculate when the sun will set and when the sun will rise on those days? Is it an average of the last known date and the next known date?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the hour angle, exactly, and how do you calculate the sunrise/sunset if you can't calculate it?&lt;/p&gt;&#xA;" OwnerUserId="2043" LastActivityDate="2014-08-11T08:01:18.257" Title="Hour angle sunrise calculation problem" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;clock&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5884" PostTypeId="2" ParentId="5883" CreationDate="2014-07-12T07:54:45.087" Score="2" Body="&lt;p&gt;Did you consider the &lt;em&gt;degree to radian&lt;/em&gt; conversion and vice versa? There seems to be something like that happening. Note that -0.83 is in degrees - which is the correction for the size of the sun and refraction effects at horizon. (More details here: &lt;a href=&quot;https://en.wikipedia.org/wiki/Sunrise_equation&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Sunrise_equation&lt;/a&gt;) At latitude 38 N, there shouldn't be any day where sunrise/sunset is not defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, Hour Angle at any time is the time since culmination (culmination is the passing of a body through the meridian; meridian is the great circle passing through the North cardinal point, North Pole, Zenith and South cardinal point. In this case, Hour Angle gives the time between sunrise and noon (i.e. culmination of the sun), or noon and sunset.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2014-07-12T07:54:45.087" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5885" PostTypeId="1" AcceptedAnswerId="5886" CreationDate="2014-07-12T13:27:46.333" Score="4" ViewCount="2493" Body="&lt;p&gt;If you only read the question, you will answer: ''&lt;em&gt;That depends on the light pollution&lt;/em&gt;''. Yes, it does, but if you are in the darkest night sky (Bortle class 1), what magnitude would have the faintest star? In the Wikipedia article about the magnitude, it's written that +6.0 would be the typical limit, but in the article about the Bortle scale, I saw the fainter magnitude +8.0. So, what is the real limit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the question has been already asked, please give me the link.&lt;/p&gt;&#xA;" OwnerUserId="1890" LastActivityDate="2014-07-12T21:41:56.473" Title="What is the apparent magnitude limit for the naked eye?" Tags="&lt;magnitude&gt;&lt;naked-eye&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5886" PostTypeId="2" ParentId="5885" CreationDate="2014-07-12T21:41:56.473" Score="6" Body="&lt;p&gt;The apparent magnitude classification was done arbitrarily by Ptolemy. His thought was to set the 20 brightest stars to the first position, the less bright stars to the second position and so on up to the faintest stars which were given the sixth position. After the use of Pogson's law we were able to give to stars not only natural numbers, but also numbers with decimal digits. When we started to use the first telescopes, we realised that the numbers which are greater than 6 have meaning. So the natural limit of naked eye is the apparent magnitude of +6. &lt;a href=&quot;http://www.inquinamentoluminoso.it/dmsp/starvis.html&quot;&gt;Here&lt;/a&gt; you can find a nice work from the italian Light Pollution Science and Technology Institute which shows as the typical limit of naked eye compared to the light pollution of each area.&lt;/p&gt;&#xA;" OwnerUserId="1956" LastActivityDate="2014-07-12T21:41:56.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5887" PostTypeId="1" CreationDate="2014-07-13T00:36:16.943" Score="2" ViewCount="1740" Body="&lt;p&gt;I've been looking into the document &lt;a href=&quot;http://www.gps.gov/technical/icwg/IS-GPS-200G.pdf&quot; rel=&quot;nofollow&quot;&gt;IS-GPS-200H&lt;/a&gt; to understand how to calculate satellite location in the ECEF coordinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having problem understanding the formula to derive $\Omega$, the longitude of the ascending node (LAN) relative to Greenwich at given time $t$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Omega = \Omega_0 + \left( \dot{\Omega - w} \right)\times t_k - w \times t_{oe}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&#xA;$$&#xA;\Omega_0: \text{LAN relative to vernal equinox, at the beginning of the week}\\&#xA;\dot{\Omega}: \text{angular velocity for LAN, relative to vernal equinox.}\\&#xA;w: \text{angular velocity of earth, relative to vernal equinox.}\\&#xA;t_k: t - t_{oe}\\&#xA;t_{oe}: \text{ephemeris reference epoch}\\&#xA;$$&#xA;(and let us denote the beginning of the week as $t_0$ for brevity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if I try to work out this from scratch:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;At $t = t_0$, LAN was $\Omega_0$. But since what we really need is the difference of LAN and longitude of Greenwich, we also need to know $w_0$, the initial longitude of Greenwich at $t = t_0$.&#xA;$$&#xA;\Omega(t = t_0) = \Omega_0 - w_0&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;At the ephemeris reference epoch time $t = t_{oe}$, LAN and the earth both rotate with their respective angular momentum and hence:&#xA;$$&#xA;\Omega(t = toe) = \Omega_0 + w_0 + (\dot{\Omega} - w) \times t_{oe}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;As time varies from $toe$ to $t$, again LAN and the earth both rotate with their own respective angular momentum and hence&#xA;$$&#xA;\Omega(t) = \Omega_0 + w_0 + (\dot{\Omega} - w) \times t_{oe} + (\dot{\Omega} - w) \times t_k&#xA;$$&#xA;which obviously differs from the right formula by $w_0 + \dot{\Omega} \times t_{oe}$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My question is where am I making mistakes/misunderstanding the eqution? &#xA;Explain also why we don't need to know $w_0$ or equivalent input, that would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2051" LastEditorUserId="1469" LastEditDate="2014-07-13T17:52:24.073" LastActivityDate="2014-08-12T22:06:44.160" Title="Derivation of the formula for longitude of ascending node for a satellite" Tags="&lt;orbit&gt;&lt;orbital-elements&gt;&lt;satellite&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5888" PostTypeId="2" ParentId="5877" CreationDate="2014-07-13T09:32:54.653" Score="3" Body="&lt;p&gt;Use the &lt;a href=&quot;http://skyserver.sdss3.org/dr10/en/tools/search/SQS.aspx&quot; rel=&quot;nofollow&quot;&gt;Optical Spectra Query Form&lt;/a&gt;. You'll want to set PRIMTARGET to &quot;GALAXY&quot;. Here's the list of possible parameters it can return &lt;a href=&quot;http://skyserver.sdss3.org/dr10/en/help/docs/QS_UserGuide.aspx#Spectroscopy&quot; rel=&quot;nofollow&quot;&gt;http://skyserver.sdss3.org/dr10/en/help/docs/QS_UserGuide.aspx#Spectroscopy&lt;/a&gt;. Note, there is a query limit of 500,000 rows, so you'll have to break up the query since you'll have more spectra than that.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-07-13T09:32:54.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5889" PostTypeId="2" ParentId="4861" CreationDate="2014-07-14T03:17:39.443" Score="2" Body="&lt;p&gt;Cool aim. I've been intermittently working on a similar program, feel free to use any pieces you like. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's hosted at:&#xA;&lt;a href=&quot;http://thestarsbetween.com/maker/builder?stellar=o&amp;amp;rand=1233&quot; rel=&quot;nofollow&quot;&gt;http://thestarsbetween.com/maker/builder?stellar=o&amp;amp;rand=1233&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the code is shared open source at:&#xA;&lt;a href=&quot;https://github.com/jaycrossler/procyon&quot; rel=&quot;nofollow&quot;&gt;https://github.com/jaycrossler/procyon&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2053" LastActivityDate="2014-07-14T03:17:39.443" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5890" PostTypeId="1" CreationDate="2014-07-14T05:26:51.263" Score="5" ViewCount="125" Body="&lt;p&gt;A randomness beacon is a source of random data that is broadcast to multiple parties. Users listening to the beacon receive the same sequence of random strings and no one can predict the values in advance. This is a construct that has many applications in cryptography and distributed communications. Such random sequences have been generated using &lt;a href=&quot;http://www.random.org&quot;&gt;atmospheric noise&lt;/a&gt;, &lt;a href=&quot;http://www.nist.gov/itl/csd/ct/nist_beacon.cfm&quot;&gt;quantum measurements&lt;/a&gt;, and financial data (*).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using astronomical radio sources (such as meteor bursts, solar noise, planetary observations, CMBR, etc.), is it possible for multiple independent observers spread around the globe, using relatively inexpensive equipment, to detect the same unpredictable cosmic signals, and agree on the time and properties of those signals (to within some margin of error)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this were feasible, you could create a publicly-verifiable randomness beacon that didn't require trusting a central authority to generate the sequence in an unbiased way. Instead, observers can certify the random values independently using their own local measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) usenix dot org /legacy/events/evtwote10/tech/full_papers/Clark.pdf&lt;/p&gt;&#xA;" OwnerUserId="2054" LastEditorUserId="2054" LastEditDate="2014-07-14T05:34:44.153" LastActivityDate="2014-10-07T21:46:39.713" Title="Can astronomical radio sources be used as a verifiable randomness beacon?" Tags="&lt;radio-astronomy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5891" PostTypeId="1" AcceptedAnswerId="8666" CreationDate="2014-07-14T12:18:55.657" Score="3" ViewCount="224" Body="&lt;p&gt;I have a Meade wedge that I am to install on one of their LX200 telescopes. Will I be able to align in alt-az afterwards? Is polar easier or harder to make an alignment? Any other tips welcome. Thanks&lt;/p&gt;&#xA;" OwnerUserId="1873" LastActivityDate="2015-01-24T21:31:45.150" Title="Alt-Az or Polar?" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5892" PostTypeId="2" ParentId="3704" CreationDate="2014-07-14T12:53:51.210" Score="3" Body="&lt;p&gt;Finally, after searching quite a deal, I found &lt;a href=&quot;http://arxiv.org/abs/astro-ph/9912190&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; paper which would aptly answer the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper from 1999 published by Pietrzynski and Udalski in Acta Astronomica lists the Cepheids in the star clusters of Magellanic Clouds.&lt;/p&gt;&#xA;" OwnerUserId="1702" LastActivityDate="2014-07-14T12:53:51.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5893" PostTypeId="2" ParentId="2503" CreationDate="2014-07-14T13:03:56.090" Score="1" Body="&lt;p&gt;[In addition to the answer already provided by Py-ser]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the body of a Cepheid, there is a regular or periodical variation of difference between the gravitational pressure and radiation pressure. This radiation pressure is due to the consumption of fuel in it, lighter elements like hydrogen and helium undergoing nuclear fusion into forming heavier elements. The gravitational pressure occurs of course due to the elements and compounds in the star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One increases, then decreases, equals the other and then further decreases, and again increases and this happens periodically. This equilibrium between both the pressures is being referred to here.&lt;/p&gt;&#xA;" OwnerUserId="1702" LastActivityDate="2014-07-14T13:03:56.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5894" PostTypeId="1" CreationDate="2014-07-15T02:30:44.250" Score="19" ViewCount="11549" Body="&lt;p&gt;When I was teaching a class why the earth has seasons today, I mentioned how the poles of the earth experience months of daylight and darkness. Then one of my students asked whether the moon disappears as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using an orrery and a string, we looked at the orbit of the moon around the earth and tried to find a place where the moon was below the horizon for an entire orbit. In our attempts, we could find no such place, though some students speculated that if you were in the right place at the right time, the moon could make a full orbit while partially obscured by the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a place on the earth where the moon cannot be seen on for long periods of time, like the sun in northern latitudes? Or does the fact that it orbits around the earth make this impossible?&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2020-02-20T21:15:24.663" Title="Is there any point on earth where the moon stays below the horizon for an extended period of time?" Tags="&lt;the-moon&gt;&lt;orbit&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5895" PostTypeId="2" ParentId="2503" CreationDate="2014-07-15T05:48:02.197" Score="5" Body="&lt;p&gt;&lt;strong&gt;Cepheid pulsations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic description of the mechanism behind Cepheid pulsations is given &lt;a href=&quot;https://en.wikipedia.org/wiki/Cepheid_variable#Dynamics_of_the_pulsation&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The accepted explanation for the pulsation of Cepheids is called the Eddington valve,[38] or κ-mechanism, where the Greek letter κ (kappa) denotes gas opacity. Helium is the gas thought to be most active in the process. Doubly ionized helium (helium whose atoms are missing two electrons) is more opaque than singly ionized helium. The more helium is heated, the more ionized it becomes. At the dimmest part of a Cepheid's cycle, the ionized gas in the outer layers of the star is opaque, and so is heated by the star's radiation, and due to the increased temperature, begins to expand. As it expands, it cools, and so becomes less ionized and therefore more transparent, allowing the radiation to escape. Then the expansion stops, and reverses due to the star's gravitational attraction. The process then repeats.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't really describe it much better than this, but let me know if it needs to be made more clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this mechanism isn't restricted to helium II/III.  Other classes of variable star operate in the same way. e.g. RR Lyraes and $\beta$ Cepheids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The equilibrium state of stars&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all phases, stars are presumed to be in &lt;em&gt;hydrostatic equilibrium&lt;/em&gt;: the outward force of pressure in the star (from the gas, radiation and sometimes electron or neutron degeneracy) is precisely balanced by the inward force of gravity. In many phases, we can additional regard the star as being in &lt;em&gt;local thermodynamic equilibrium&lt;/em&gt;.  What we really mean here is that the star is not generating (or absorbing) energy through expansion and contraction.  This isn't true for, say, pre-main-sequence stars, which are only generating energy by their contraction toward the main sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cepheid pulsations are oscillations about the equilibrium state.  They aren't linearly stable but the non-linear components mean that the oscillations aren't big enough to tear the star apart.  The calculation of the frequencies is either a tricky sixth-order eigenvalue problem or a simulation of the oscillation from which the frequencies can be determined by analysing the output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which modes are excited is a similarly complicated problem.  There are many things we haven't actually solved!  But you can think of the system as a driven, damped oscillator. The driving is given by the opacity mechanism described above, and the damping is related to the equilibrium structure of the star. So, the excited modes are usually around where these match best, which may happen to be the fundamental mode (where the entire star expands or contracts as a whole) or the overtones (where different layers are expanding or contracting).&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2014-07-15T05:48:02.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5896" PostTypeId="1" CreationDate="2014-07-15T07:27:47.377" Score="8" ViewCount="2283" Body="&lt;p&gt;It seems our universe is neutral in large scale.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is CP violation&lt;a href=&quot;http://en.wikipedia.org/wiki/CP_violation&quot;&gt;wiki&lt;/a&gt; problem about matter and dark matter. &lt;br/&gt;&#xA;Similarly is there any theory about &lt;strong&gt;whether our universe is electrically neutral&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cosmic rays are not neutral. However the atomic nucleus and electrons are just separated. There should be no charge excess in total.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the largest scale that is severely non-neutral? &lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="2365" LastEditDate="2018-07-27T22:34:33.467" LastActivityDate="2018-07-27T22:34:33.467" Title="is there any theory or observational evidence that our universe is electrically neutral or not?" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5897" PostTypeId="1" AcceptedAnswerId="5912" CreationDate="2014-07-15T09:05:37.527" Score="5" ViewCount="59" Body="&lt;p&gt;I used to have a 6&quot; Dobsonian reflector, and the next telescope I buy, I want it to have the equivlant of about a 11/12&quot; dobsonian reflector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking at this &lt;a href=&quot;http://www.cloudynights.com/item.php?item_id=1528&quot; rel=&quot;noreferrer&quot;&gt;Nexstar VX 9.25&lt;/a&gt; , it says 235mm aperature, but is thats more than what a 235mm reflector would acheive right? How close is to a 11/12&quot; reflector?&lt;/p&gt;&#xA;" OwnerUserId="2063" LastActivityDate="2014-07-17T20:02:50.343" Title="How to compare a Catadioptrics light asorbtion with a reflector?" Tags="&lt;telescope&gt;&lt;photography&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5898" PostTypeId="2" ParentId="5894" CreationDate="2014-07-15T14:13:57.927" Score="13" Body="&lt;p&gt;Depends on the interpretation of your question...&#xA;The best places not to observe the moon are the north and south pole. On the north pole you will only be able to see objects above the celestial equator. As the moon orbits the Earth in one month its orbit is inclined from the celestial equator. This inclination is almost the same as the inclination of the ecliptic (path of the Sun) with the celestial equator. The ecliptic crosses the equator at two opposite points on the celestial sphere. This means that for about half its orbit, the Sun, and as the moon's orbit is near the ecliptic, also the Moon, will be above the ecliptic and therefore visible from the north pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, the Moon does not follow the ecliptic precisely as the moon's orbit is inclined from the ecliptic by about 5°. The inclination of the ecliptic is 23°, so during very special circumstances the maximum altitude of the Moon above the horizon on the North pole will be 18° during one month. The duration that the Moon will be above the horizon on the north pole will be &lt;del&gt;about 10 days (a guess)&lt;/del&gt; about 13.6 days (edit:see comments below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we interpret your question as: Is there a place where the Moon will be below the horizon for a long period of time (&gt; 1 month), then the answer is NO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if this happens near June, then the Sun will also be above the horizon at the north pole (for six months), and as the Moon will be close to the Sun (as it follows more or less the ecliptic), it will be very hard to see the Moon during that time. So if you specifically ask whether the Moon will &lt;strong&gt;not&lt;/strong&gt; be &lt;strong&gt;visible&lt;/strong&gt; for an extended period of time, then the answer is YES.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there are of course also places with perpetual cloud cover ;-)&lt;/p&gt;&#xA;" OwnerUserId="864" LastEditorUserId="864" LastEditDate="2014-07-16T11:30:46.093" LastActivityDate="2014-07-16T11:30:46.093" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5899" PostTypeId="2" ParentId="3704" CreationDate="2014-07-15T22:20:42.133" Score="1" Body="&lt;p&gt;Vizier is a good resource for finding catalogs. Searching for &quot;SMC+cepheid&quot; I found a related paper (Udalski et al. 1999) with additional data &lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR?-source=J/AcA/49/437&quot; rel=&quot;nofollow&quot;&gt;http://vizier.u-strasbg.fr/viz-bin/VizieR?-source=J/AcA/49/437&lt;/a&gt;. You can also look for &quot;Similar Catalogs&quot; though it may or may not be useful in this case. &lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-07-15T22:20:42.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5900" PostTypeId="1" AcceptedAnswerId="24415" CreationDate="2014-07-16T03:18:24.010" Score="4" ViewCount="828" Body="&lt;p&gt;The moon faces the earth with one fixed side. However there is a small vibration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am not sure the south and north poles of the moon is fixed. Even if they are fixed points on the moon, the pole axis should point at different locations in the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the general positions of the two points in equatorial coordinate system of our earth?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="332" LastEditDate="2014-07-21T02:40:30.850" LastActivityDate="2018-01-08T02:07:38.520" Title="The RA and Dec of lunar poles" Tags="&lt;the-moon&gt;&lt;pole&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5901" PostTypeId="2" ParentId="5900" CreationDate="2014-07-16T05:45:37.380" Score="2" Body="&lt;p&gt;The &amp;quot;small vibration&amp;quot; is Libration, and it does not imply that lunar axis changes orientation. It is caused by Moon's movement not being a perfectly circular orbit parallel to Earth's Equator with aligned axis.&lt;/p&gt;&#xA;&lt;p&gt;Here's &lt;a href=&quot;http://www.rmg.co.uk/gcse-astronomy/sun-and-moon/moon-movement/the-moon%27s-movement&quot; rel=&quot;nofollow noreferrer&quot;&gt;a description of the causes of libration&lt;/a&gt; from the Royal Observatory (check the linked page for graphics of each source):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Synchronous rotation and libration&lt;/p&gt;&#xA;&lt;p&gt;The 27.3 days it takes the Moon to complete one orbit around the Earth is about the same as the time taken for it to complete one rotation. This synchronous rotation means that it always shows the same face to the Earth.&lt;/p&gt;&#xA;&lt;p&gt;However, libration means that it is possible to see around 59% of the surface. It has three components:&lt;/p&gt;&#xA;&lt;p&gt;i) Diurnal libration&lt;/p&gt;&#xA;&lt;p&gt;As the Earth rotates the view of the Moon changes. An observer on the surface of the Earth sees slightly around the eastern limb of the Moon at moonrise. At moonset the same observer sees slightly more around the western limb of the Moon.&lt;/p&gt;&#xA;&lt;p&gt;This arises from the changed position of the observer because of the rotation of the Earth. A parallax effect means that the Moon looks slightly different at moonrise and moonset.&lt;/p&gt;&#xA;&lt;p&gt;ii) Longitudinal libration&lt;/p&gt;&#xA;&lt;p&gt;The Moon moves in an elliptical orbit around the Earth. As a result it does not move at a constant speed (this is Kepler's second law)&lt;/p&gt;&#xA;&lt;p&gt;However the Moon rotates at a constant speed. When it is moving faster in its orbit, this allows observers on Earth to see around the trailing edge of the Moon. When it is moving slowly it is possible to see around the Moon's leading edge.&lt;/p&gt;&#xA;&lt;p&gt;iii) Libration in latitude&lt;/p&gt;&#xA;&lt;p&gt;The Moon's axis is tilted by 6.7° to the plane of its orbit around the Earth. This means that sometimes the Moon's North Pole is tilted towards the Earth and sometimes tilted away.&lt;/p&gt;&#xA;&lt;p&gt;As a result it is sometimes possible to see beyond the lunar North Pole and the South Pole is hidden. Conversely it is sometimes possible to see beyond the South Pole and the North Pole is hidden.&lt;/p&gt;&#xA;&lt;p&gt;The combination of the three kinds of libration allows observers on Earth to see 59% of the lunar surface. However 41% remained hidden from view until the space age.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So in order to get Moon's poles position you just need to know where the Moon's center is, which is quite standard, and the position of the poles referenced to that center, which is always the same (not in RA/Dec but in true space).&lt;/p&gt;&#xA;&lt;p&gt;There is a secondary effect of &amp;quot;wobbling&amp;quot; called Nutation, which is caused by the Moon on the Earth, not a movement on Moon itself.&lt;/p&gt;&#xA;" OwnerUserId="473" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2014-07-16T22:17:42.600" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="5902" PostTypeId="2" ParentId="5900" CreationDate="2014-07-16T07:52:56.557" Score="3" Body="&lt;p&gt;Here's a NASA movie showing Libration: &lt;a href=&quot;http://svs.gsfc.nasa.gov/cgi-bin/details.cgi?aid=4000&quot; rel=&quot;nofollow&quot;&gt;http://svs.gsfc.nasa.gov/cgi-bin/details.cgi?aid=4000&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As given in the details here, the &lt;strong&gt;position angle&lt;/strong&gt; is the angle between the Moon's north and celestial north. This changes because of the inclination of ecliptic to equator + inclination of moon's orbit to ecliptic and plus, there might be some contribution from the inclination of moon's rotation axis to its orbital plane. Note that all these changes are &lt;strong&gt;apparent&lt;/strong&gt;. The pole need NOT physically shift at all for these effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, whether the pole does shift, requires one to correct for all those effects and the remaining effects would be cause by some precession or nutation of the lunar poles, which is likely to be negligible given that it is tidally locked to the earth.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2014-07-16T07:52:56.557" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5903" PostTypeId="1" AcceptedAnswerId="5904" CreationDate="2014-07-16T14:18:02.717" Score="10" ViewCount="1194" Body="&lt;p&gt;If two comets come together in space, will they bounce apart, merge into a single body, or could they travel together through space, either touching or orbiting a shared center of gravity? &lt;/p&gt;&#xA;" OwnerUserId="513" OwnerDisplayName="James Jenkins" LastEditorUserId="513" LastEditDate="2014-07-16T14:34:14.213" LastActivityDate="2014-07-16T15:43:33.453" Title="Can two comets travel together as one?" Tags="&lt;comets&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5904" PostTypeId="2" ParentId="5903" CreationDate="2014-07-16T14:40:23.183" Score="14" Body="&lt;p&gt;Yes, binary comets do exist; 8P/Tuttle is an example of one. This is an area that is still being heavily researched, but it is suspected that binary comets can &quot;form from collision, mutual capture, or fission.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In recent news, the comet 67P/Churyumov-Gerasimenko has also been revealed to be a contact-binary by the European Space Agency's &lt;em&gt;Rosetta&lt;/em&gt; spacecraft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sources:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://echo.jpl.nasa.gov/asteroids/harmon.etal.comet.tuttle.pdf&quot;&gt;Radar observations of 8P/Tuttle: A contact-binary comet - Harmon, Nolan, Giorgini, Howell&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.bbc.com/news/science-environment-27110882&quot;&gt;Rosetta's target is 'double' comet - BBC&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6" LastEditorUserId="6" LastEditDate="2014-07-16T14:50:14.477" LastActivityDate="2014-07-16T14:50:14.477" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5905" PostTypeId="2" ParentId="5903" CreationDate="2014-07-16T14:42:08.470" Score="5" Body="&lt;p&gt;If the closing velocity is very low, the energy can be dissipated by the comets squishing.  You might consider them to merge into one comet in that case.  You might be left with a bunch of pieces all travelling together.  Whether you call that one comet, the original two, or many is a matter of terminology-all the stuff will be in the same orbit.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2014-07-16T14:42:08.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5907" PostTypeId="1" AcceptedAnswerId="5911" CreationDate="2014-07-16T16:35:44.007" Score="9" ViewCount="286" Body="&lt;p&gt;How was it decided which hemisphere was North and which hemisphere was South on Mars? Clearly, the designations &quot;North&quot; and &quot;South&quot; apply only to the hemispheres which contain the axis of rotation at their center, but what determined whether a hemisphere received &quot;North&quot; or &quot;South&quot; as its designation?&lt;/p&gt;&#xA;" OwnerUserId="6" LastActivityDate="2014-07-17T13:34:34.707" Title="How were the designations of &quot;North&quot; and &quot;South&quot; applied to the hemispheres of Mars?" Tags="&lt;mars&gt;&lt;naming&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5908" PostTypeId="2" ParentId="5907" CreationDate="2014-07-16T18:00:50.863" Score="3" Body="&lt;p&gt;North and south are defined according to the planet's rotation about its axis. A &quot;right hand&quot; rule is used: If you make a fist with your right hand, and orient your curled fingers in the direction the planet rotates, your thumb points north. Alternatively: Standing on the equator, facing the direction the planet rotates, North is to your left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Axial_tilt&quot; rel=&quot;nofollow&quot;&gt;Axial Tilt&lt;/a&gt; on wikipedia has this information (and more.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hey! Don't up vote this answer; It's wrong... see the comments where the OP has actually researched and discovered the real answer. (I don't want to delete this answer because we'll lose the comments too!)&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="7" LastEditDate="2014-07-17T12:51:17.130" LastActivityDate="2014-07-17T12:51:17.130" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="5910" PostTypeId="1" AcceptedAnswerId="5913" CreationDate="2014-07-17T08:26:02.310" Score="1" ViewCount="199" Body="&lt;p&gt;I am wondering, whether is it possible to use vega-altair-deneb asterism for navigational purpose? Or if there is any geometric relationship of it with Polaris?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea. &#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="1842" LastActivityDate="2014-07-18T07:09:21.090" Title="Can Summer Triangle asterism be used for navigation?" Tags="&lt;navigation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5911" PostTypeId="2" ParentId="5907" CreationDate="2014-07-17T13:34:34.707" Score="7" Body="&lt;p&gt;The research in other answers helped me come across the actual International Astronomical Union (IAU) standard:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The north pole is that pole of rotation that lies on the north side of the invariable plane of the solar system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The north of the invariable plane is the side that Earth's North pole points to. In fact, one of our other users has observed:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The North Pole of any system being the pole on the same side as earth's north is indeed the correct definition. This applies to the Milky Way galaxy as well, in which case the North Galactic Pole is the pole towards the hemisphere of the galactic sphere which also contains the celestial north pole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Dwarf planets, minor planets, their satellites, and comets do follow the right-hand rule mentioned in other answers (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Axial_tilt&quot;&gt;Axial tilt&lt;/a&gt; on Wikipedia).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sources:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://download.springer.com/static/pdf/49/art%253A10.1007%252Fs10569-007-9072-y.pdf?auth66=1405707429_bc7c5277c4389320d62d623be73d0bf0&amp;amp;ext=.pdf&quot;&gt;Report of the IAU/IAG Working Group on cartographic&#xA;coordinates and rotational elements: 2006 - P. Kenneth Seidelmann, B. A. Archinal,&#xA;M. F. A’hearn, et al.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Poles_of_astronomical_bodies&quot;&gt;Poles of astronomical bodies - Wikipedia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6" LastActivityDate="2014-07-17T13:34:34.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5912" PostTypeId="2" ParentId="5897" CreationDate="2014-07-17T20:02:50.343" Score="3" Body="&lt;p&gt;Comparing optical instruments is very difficult. There are numerous variables involved, the performance is not a single number, and the theory is complex. Perfect scenario to baffle amateurs. Also, it sounds like you may have been listening to the wrong rumors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The architecture of a telescope (refractor, reflector, catadioptric) is one of the least important factors in the performance analysis, regardless of how you define &quot;performance&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't forget that how you maintain and where you use the scope are very important factors. Collimation will strongly impact resolving power - so make sure it's always perfect. Light pollution will impact you much more than light absorption within the instrument. Etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of amount of light gathered, aperture is the dominant factor. 2x the aperture is 4x the area, so it's obviously very important. The central obstruction may factor in a little, but the effect is small. Going by diameter, a typical newtonian will have a 20% central obstruction; a typical catadioptric scope will have a 40% c.o. But if you translate that into areas, the newtonian is at 4%, whereas the catadioptric is at 16% - both are very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of resolving power, again aperture is king. Bigger aperture = more resolving power, no matter what the architecture. But poor collimation will quickly destroy your gains, so keep that scope collimated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of internal losses, the newtonian will do a bit better - it's two reflections and that's it. The catadioptric will have two or more reflections (depending on design), plus absorption / reflection on the diopters. Again, with modern instruments, the difference is tiny.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, at the same aperture, in theory, the newtonian pure catoptric scope delivers a bit more light to the sensor, compared to a catadioptric scope. In practice, the difference is far outweighed by other factors and you should completely ignore it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should make the purchase based on more important things. What's the size of the usable field? What kind of adapters can you use? Does it provide enough back-focus? What's the aberration in different points within the field? Is the telescope too heavy for the driven mount? Is it long focal ratio, or short? (each kind has different applications) Finally but not least important, price?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's more common to see catadioptrics used as astrographs. That's not to say that a newtonian reflector cannot do the job - but if you're still new to the field, just do what most people do and get a catadioptric, with a good mount.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Final note:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When doing prime focus photography, the brightness of the image is given by both aperture and focal ratio - but not all objects behave the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For stars and other point-like objects (smaller than the resolving power), brightness is driven by aperture. Bigger aperture = brighter stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For extended objects (bigger than the resolving power), such as nebulae, many galaxies, the Moon, the planets, etc., the brightness of the object is given by the focal ratio. An f/4 scope will make brighter nebulae in prime focus, compared to an f/8 scope. Aperture doesn't matter. It's just that, at the same focal ratio, the bigger scope will also make a bigger image in prime focus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is true only for prime focus photography. If you're doing eyepiece projection, or something else, different rules apply.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2014-07-17T20:02:50.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5913" PostTypeId="2" ParentId="5910" CreationDate="2014-07-18T07:09:21.090" Score="2" Body="&lt;p&gt;Yes, it can be used, but it is not easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the latitude, you usually need to know the Alt of Polaris. You can derive that if you know the Alt of two of the three, but the (spherical geometry) formulas for that are not simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the longitude, you need Greenwich time and the RA of any star. Any star includes Vega, Altair and Deneb ;)&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-07-18T07:09:21.090" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5914" PostTypeId="1" AcceptedAnswerId="5915" CreationDate="2014-07-18T08:09:28.267" Score="8" ViewCount="8234" Body="&lt;p&gt;As observed by many researches our sun will be exhausted in the distant future. So does anyone think that driving our planet (or any planet) closer to the sun for more light and heat is a solution for life to survive on earth? Or would moving it out of the sun's gravitational field and taking it to some other solar system with a healthy star be better, possibly with some kind of thrusting mechanism that we may invent in future?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware of the sun's gravity and the mass of the earth. &lt;/p&gt;&#xA;" OwnerUserId="1369" LastEditorUserId="10437" LastEditDate="2016-06-22T13:16:53.760" LastActivityDate="2016-06-22T13:16:53.760" Title="Is it possible to move a planet out of its orbit? At least a lighter planet?" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;solar-system&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5915" PostTypeId="2" ParentId="5914" CreationDate="2014-07-18T08:56:28.183" Score="15" Body="&lt;p&gt;Yes. Even with todays technology we can get started.&#xA;In &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0102126&quot; rel=&quot;noreferrer&quot;&gt;this article&lt;/a&gt; by Korycansky et al 2001 it is suggested to use rockets to move an asteroid (like a larger version of the NASA ARM Asteroid Redirect Mission) to a cycling orbit between Earth and Jupiter. The asteroid would pass near infront of the Earth in order to give the Earth a slight gravity assist which moves it outwards. The asteroid then passes near behind Jupiter to itself get a gravity assist. This transfers energy from Jupiter's to Earth's orbit. I've also heard of a proposal to use Venus instead.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2014-07-18T08:56:28.183" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5916" PostTypeId="2" ParentId="5914" CreationDate="2014-07-18T14:07:22.903" Score="3" Body="&lt;p&gt;yes it is possible by very few different ways. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.uwgb.edu/dutchs/pseudosc/flipaxis.htm&quot; rel=&quot;nofollow&quot;&gt;https://www.uwgb.edu/dutchs/pseudosc/flipaxis.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing acting solely from on or within the Earth could change its orbit or seriously alter its rotation.&#xA;One way to move an object is to throw mass in the opposite direction, the way jets or rockets do.&#xA;If we think really big and imagine blasting a chunk out of the Earth as big as North America and 100 miles thick so that its final speed, after escape, with respect to the Earth is 25,000 miles an hour, we will have expelled only 1/500 of the total mass of the Earth. The Earth would move in the opposite direction 1/500 as fast or 50 miles an hour. The speed of the Earth in its orbit is about 67,000 miles an hour. We will not change the orbit of the Earth very much--if we apply the impulse to speed up the earth in its orbit we would put the Earth into a new orbit with its most distant point about 70,000 miles further from the Sun than now--and the Earth's distance from the Sun varies now by three million miles over the course of a year! Exactly the same arguments apply to changing the orbit of the Earth through the impact of a large asteroid. The largest asteroid, Ceres, about 600 miles in diameter, is only about as massive as our hypothetical chunk of Earth above. Changing the orbit of a planet is a tall order. An impact big enough to have even a tiny effect on the Earth's orbit or rotation would almost certainly destroy all life on Earth as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://usatoday30.usatoday.com/news/science/astro/2001-02-15-orbit.htm&quot; rel=&quot;nofollow&quot;&gt;http://usatoday30.usatoday.com/news/science/astro/2001-02-15-orbit.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hope this two links provides enough satisfactory explanation to you. if not do say so i will make further search to clear you the answer.&lt;/p&gt;&#xA;" OwnerUserId="2034" LastActivityDate="2014-07-18T14:07:22.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5917" PostTypeId="1" AcceptedAnswerId="5920" CreationDate="2014-07-18T17:04:47.580" Score="3" ViewCount="82" Body="&lt;p&gt;If Neptune would align Uranus, would the radio frequency perceived on Earth increase?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the radio antennas on NM capable of getting the 2 different radio frequencies feom both planets or would they get a bigger range of frequencies?&lt;/p&gt;&#xA;" OwnerUserId="2082" LastActivityDate="2014-07-18T19:06:59.677" Title="Do planetary radio frequency change upon alignments?" Tags="&lt;planet&gt;&lt;radio-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5918" PostTypeId="1" AcceptedAnswerId="5919" CreationDate="2014-07-18T18:16:54.107" Score="3" ViewCount="235" Body="&lt;p&gt;In the answer &lt;a href=&quot;https://astronomy.stackexchange.com/questions/5903/can-two-comets-travel-together-as-one/5904#5904&quot;&gt;here&lt;/a&gt; multiple comets are identified that do not form a single round body.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A drop of water will form a round shape in the time it takes to free fall a couple of feet.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the moons and planets in our solar system are essentially round.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on observable data, I assume that there is a relationship between the size, rigidity and the passage of time; that will result in all objects subject to only their own gravitational influences (Given: &lt;em&gt;no body is ever truly uninfluenced by others&lt;/em&gt;) becoming spherically shaped (&lt;em&gt;round&lt;/em&gt;).  Any object that is in microgravity and not round, therefore must be either very rigid and/or very newly formed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the ratio or formula for calculating any one factor when the other two are known?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt; There would be other factors as well, speed of rotation, elasticity and energy of areas moving in opposition.  But I think these (&lt;em&gt;and others?&lt;/em&gt;) would be minor compared to the other three.  &lt;/p&gt;&#xA;" OwnerUserId="513" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-07-18T19:24:20.263" Title="What is the time/size/rigidity ratio for a space object to become mostly round?" Tags="&lt;gravity&gt;&lt;hydrostatic-equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5919" PostTypeId="2" ParentId="5918" CreationDate="2014-07-18T18:55:30.223" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Based on observable data, I assume that there is a relationship&#xA;  between the size, rigidity and the passage of time; that will result&#xA;  in all objects subject to only their own gravitational influences&#xA;  (Given: no body is ever truly uninfluenced by others) becoming&#xA;  spherically shaped (round).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's called hydrostatic equilibrium. That's one of the main factors that distinguish a dwarf planet from a smaller random piece of rock out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Dwarf_planet&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Dwarf_planet&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;A dwarf planet is an object the size of a planet (a planetary-mass object) but that is neither a planet nor a moon or other natural satellite. More explicitly, the International Astronomical Union (IAU) defines a dwarf planet as a celestial body in direct orbit of the Sun that is massive enough for its shape to be controlled by gravity, but that unlike a planet has not cleared its orbit of other objects.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no fixed size limit, because it depends on the composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrostatic_equilibrium#Planetary_geology&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Hydrostatic_equilibrium#Planetary_geology&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;It had been thought that icy objects with a diameter larger than roughly 400 km are usually in hydrostatic equilibrium, whereas those smaller than that are not. Icy objects can achieve hydrostatic equilibrium at a smaller size than rocky objects. The smallest object that appears to have an equilibrium shape is the icy moon Mimas at 397 km, whereas the largest object known to have an obviously non-equilibrium shape is the rocky asteroid Pallas at 532 km (582×556×500±18 km). However, Mimas is not actually in hydrostatic equilibrium for its current rotation. The smallest body confirmed to be in hydrostatic equilibrium is the icy moon Rhea, at 1,528 km, whereas the largest body known to not be in hydrostatic equilibrium is the icy moon Iapetus, at 1,470 km.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the range of transition between hydrostatic equilibrium and non-equilibrium is between approx 400 and 1500 km diameter and depends on a number of factors such as composition. There is no simple formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dwarf planets are a good related topic, and I used them in the discussion, because many of them are above the limit of HE but close to it. But any cosmic body is subject to the same laws. Planets proper, moons, stars, etc - these all may be placed above or below the HE limit, depending mostly on size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. the Earth is clearly above the HE limit. But probably all comets are below it. A rock sitting in the dirt below your feet is clearly below it.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2014-07-18T19:24:20.263" LastActivityDate="2014-07-18T19:24:20.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5920" PostTypeId="2" ParentId="5917" CreationDate="2014-07-18T19:06:59.677" Score="1" Body="&lt;p&gt;If you're talking about the electromagnetic radiation (in the radio spectrum) produced by the planets themselves, then the answer is no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, not all planets generate a lot of radio EM. Jupiter does generate some amount, but the other planets are quite a bit more quiet. Still, a very small amount is probably generated by most planets out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, this is a strictly local phenomenon. Local conditions within and on that planet predominate in driving the production of radio EM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By comparison, distances between planets are HUGE. Whether they are &quot;aligned&quot; or not is of no importance to what happens on the surface. Anyway, they appeared &quot;aligned&quot; only to us, on Earth, and even then the &quot;alignment&quot; is always imperfect (they are never exactly on a line).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shortly: no.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2014-07-18T19:06:59.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5921" PostTypeId="1" CreationDate="2014-07-18T20:47:56.683" Score="3" ViewCount="1746" Body="&lt;p&gt;Google's definition of a comet is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a celestial object consisting of a nucleus of ice and dust and, when&#xA;  near the sun, a ‘tail’ of gas and dust particles pointing away from&#xA;  the sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This doesn't cut it for me. Neither does wikipedia's similar entry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It pertains to the sun, even though comets are present in other parts of the galaxy other than the solar system and probably other galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whats more dwarf planet can be classified as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; * orbits a host star&#xA; * rounded by its own gravity&#xA; * not massive enough to induce thermonuclear fusion&#xA; * not massive enough to clear its region of debris&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which pretty much fits the bill of a comet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you think the following is a fair / recognised classification of a comet:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Comet:&#xA;* not rounded by its own gravity&#xA;* not massive enough to induce thermonuclear fusion&#xA;* not massive enough to clear its region of debris&#xA;* composition of ice and dust&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But then that's pretty similar to an asteroid:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Asteroid:&#xA;* not rounded by its own gravity&#xA;* not massive enough to induce thermonuclear fusion&#xA;* not massive enough to clear its region of debris&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However, this means if an asteroid is comprised of ice and dust, it becomes a comet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or are asteroid &lt;strong&gt;never&lt;/strong&gt; comprised of ice and dust? Does my asteroid classification need beefing up, regarding an asteroid's composition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I'm feeling the comet and the asteroid both need some classification pertaining to their orbit as well? Though thats tricky because only some comets and asteroid, so they're similar in that regard as well! Maybe more details on each objsts composition is needed, like I say...&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2084" LastActivityDate="2014-11-05T18:40:23.133" Title="Classification of a Comet" Tags="&lt;asteroids&gt;&lt;comets&gt;&lt;classification&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5922" PostTypeId="1" CreationDate="2014-07-19T10:04:05.383" Score="1" ViewCount="265" Body="&lt;p&gt;complete beginner here, only got into star gazing a few days ago using my friend's telescope and I don't really understand at all what any of the measurements mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, talk about jumping in the deep end, but I'm poking around Nasa's Intergalactic Database, but sadly it's all greek to me: &lt;a href=&quot;http://ned.ipac.caltech.edu/&quot; rel=&quot;nofollow&quot;&gt;http://ned.ipac.caltech.edu/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a galaxy: &lt;a href=&quot;http://ned.ipac.caltech.edu/cgi-bin/objsearch?objname=NGC+7590&amp;amp;extend=no&amp;amp;hconst=73&amp;amp;omegam=0.27&amp;amp;omegav=0.73&amp;amp;corr_z=1&amp;amp;out_csys=Equatorial&amp;amp;out_equinox=J2000.0&amp;amp;obj_sort=RA+or+Longitude&amp;amp;of=pre_text&amp;amp;zv_breaker=30000.0&amp;amp;list_limit=5&amp;amp;img_stamp=YES&quot; rel=&quot;nofollow&quot;&gt;NGC 7590&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's its basic data:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EquJ2000.0 RA&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EquJ2000.0 DEC&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Velocity km/s&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Redshift / z&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Magnitude / Qual Filter&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Separ arc min&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm afraid I don't really understand what any of these measurements and their values actually mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember from school what Red Shift is, its the wavelength increasing becoming &#xA;'stretched' making the light seem to appear more reddish the faster an object is accelerating away from us. But what is that z unit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm afraid I really don't have a clue regarding anything else... spear arc min, magnitude / qual filter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a book for a brief understanding of these units? Not out to solve problems regarding celestial mechanics, I just want the NED to make a bit more sense to me so I can visualize these objects.&lt;/p&gt;&#xA;" OwnerUserId="2084" LastActivityDate="2014-07-24T23:01:10.200" Title="Basic celestial data" Tags="&lt;fundamental-astronomy&gt;&lt;redshift&gt;&lt;units&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5923" PostTypeId="1" CreationDate="2014-07-19T13:23:38.400" Score="2" ViewCount="81" Body="&lt;p&gt;Using the &lt;a href=&quot;http://ned.ipac.caltech.edu/forms/byparams.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;NED&lt;/a&gt;, you can search via frequency range corresponding to the selected pass band filter. I'm a beginner here and don't really understand. Here's the quote:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;NOTES: Objects with observations in the frequency range corresponding to the selected passband filter will be returned. For&#xA;example, selection of Spitzer MIPS 24 µm will also return objects with&#xA;photometry from IRAS 25 µm because these filters overlap. This feature&#xA;permits the location of objects with photometry in the selected region&#xA;of the spectrum, while cutting across multiple surveys, detectors and&#xA;filters that have been cross-correlated by NED. A future upgrade will&#xA;include an option to restrict the search to only the selected passband&#xA;filter.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;There are then a list of 'passbands', all the way up to gamma. What is this actually mean? The amount of radiation given off by a celestial body?&lt;/p&gt;&#xA;&lt;p&gt;I'm looking at the categories I'm more familiar with:&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Helio. Radial Velocity&lt;br /&gt;&#xA;Redshift&lt;br /&gt;&#xA;Major Diameter (arcmin)&lt;br /&gt;&#xA;Minor Diameter (arcmin)&lt;br /&gt;&#xA;Magnitude and Filter&lt;br /&gt;&#xA;Classifications&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;And I'm not sure where this radiation fits in. Is it&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;Spectral Energy Distribution (SED) &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;? Could someone explain the following terms to me if this is the case:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Observed Passband&lt;/li&gt;&#xA;&lt;li&gt;Uncertainty Units&lt;/li&gt;&#xA;&lt;li&gt;Freq (Hz)&lt;/li&gt;&#xA;&lt;li&gt;Mode Aperture/Qualifiers&lt;/li&gt;&#xA;&lt;li&gt;Reference Code   No.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Wnad what about published and homogenised variables? Are homogenised variables the avarage of all observations?&lt;/p&gt;&#xA;" OwnerUserId="2084" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-01-25T12:50:22.637" Title="What is frequency range?" Tags="&lt;radio-astronomy&gt;&lt;units&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5924" PostTypeId="1" AcceptedAnswerId="5989" CreationDate="2014-07-19T14:04:23.610" Score="3" ViewCount="288" Body="&lt;p&gt;If a space ship travels at near the speed of light towards a star, wouldn't the light from the star doppler shift into gamma rays (or worse?) and become lethal to life and destructive to space ship structure?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2014-12-09T11:01:04.387" Title="Does near light-speed travel doppler-shift the light from the target into gamma rays?" Tags="&lt;light&gt;&lt;speed&gt;&lt;interstellar-travel&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5925" PostTypeId="1" CreationDate="2014-07-20T17:03:50.723" Score="3" ViewCount="549" Body="&lt;p&gt;I did a bit of googling but couldn't find the exact answer, so are the following 3 points collinear or not ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1_center of earth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2_ascending(or descending) node of sun&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3_point of intersection of prime meridian and equator(point of zero longitude on equator).&lt;/p&gt;&#xA;" OwnerUserId="2089" LastActivityDate="2016-09-29T08:59:56.200" Title="Are ascending node of sun, the point of intersection of prime meridian and equator and center of earth all collinear?" Tags="&lt;the-sun&gt;&lt;orbital-elements&gt;&lt;satellite&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5926" PostTypeId="2" ParentId="253" CreationDate="2014-07-20T18:28:53.190" Score="2" Body="&lt;p&gt;In general: you make models of the sun, and then you see which one agrees with all observations, and check which temperature this model predicts for the core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very simple model that gives a good approximation: fusion happens within a small volume in the core, and a part of the released energy is transported to the surface afterwards until it can escape as light. We know how much light the sun emits, and you can calculate the necessary temperature and density gradients inside that is required to transport this power and to keep the sun stable. Work from the surface inwards and you get an estimate for the core temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another nice approach is the fusion rate - this is known from the total power as well, and it can be compared to the fusion rate the sun would have at different temperatures.&lt;/p&gt;&#xA;" OwnerUserId="2090" LastActivityDate="2014-07-20T18:28:53.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5927" PostTypeId="1" CreationDate="2014-07-20T21:12:16.903" Score="4" ViewCount="87" Body="&lt;p&gt;Can a star's luminosity and its activity, solar flares and sunspots, be predicted?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, with what accuracy &amp;amp; precision?&lt;/p&gt;&#xA;" OwnerDisplayName="user1098" LastActivityDate="2014-08-21T19:52:31.137" Title="Can solar luminosity &amp; activity be predicted?" Tags="&lt;star&gt;&lt;solar-flare&gt;&lt;luminosity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5928" PostTypeId="2" ParentId="2461" CreationDate="2014-07-20T21:20:40.230" Score="1" Body="&lt;p&gt;Very good question. The number of quasars must be less than the number of SMBHs, since many galaxies, such as our own, contain SMBHs at their core (Sagittarius A*) and they are not classified as quasars (i.e., some galaxies are quiescent for whatever reason). Quasars represent an ultra-luminous active phase of gas accretion onto the SMBH. Such larger luminosities is believed to be caused by intense gas accretion triggered by major massive scale mergers between galaxies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, Quasars are short lived events, and the SMBHs outlive the quasar (the lifetime of a quasar is of the order of 10$^6$-10$^9$ yr whereas the lifetime of a SMBH is much greater than the Hubble time). Hence, once the gas is all but consumed by such intense accretion the Quasar will slowly become quiescent. Hence why many galaxies contain SMBHs at their cores, but are no longer active. &lt;/p&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2014-07-20T21:20:40.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5929" PostTypeId="2" ParentId="5927" CreationDate="2014-07-20T21:28:43.647" Score="1" Body="&lt;p&gt;I will be specific to answering the part regarding the prediction of solar flares. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solar flares are very volatile and dynamic phenomena associated to the breaking and reconnecting of magnetic field lines in the Sun's photosphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a phenomena associated with solar flares, called quasi-periodic pulsations, which sometimes precede a solar flaring event. As such, monitoring of such pulsations may give rise to a plausible prediction method. However, the accuracy and reliability is still to be debated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few good papers on QPPs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1107.2399&quot; rel=&quot;nofollow&quot;&gt;Quasi-Periodic Pulsations in Solar Flares: new clues from the Fermi Gamma-Ray Burst Monitor&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1203.6223&quot; rel=&quot;nofollow&quot;&gt;Time delays in quasi-periodic pulsations observed during the X2.2 solar flare on 2011 February 15&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2014-07-20T21:28:43.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5930" PostTypeId="1" CreationDate="2014-07-21T00:08:34.150" Score="3" ViewCount="64" Body="&lt;p&gt;I'm currently working on an independent project that involves discerning starburst galaxies that are themselves within active galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would assume radio observations would be best to discern star formation rates, and X-ray to understand the AGN in greater detail. However, assuming we have only optical data could I use the relative strength of the OII flux to the OIII flux to better understand the star formation rate of an active galaxy? Spectroscopic surveys of distant galaxies, particularly for redshifts between $z$ ≈ $0.4$ to $1$, routinely use [O II] $λ3727$, a prominent nebular emission line in H II regions, to estimate SFRs. If (high-ionisation) AGNs experience substantial levels of ongoing star formation, the integrated contribution from H II regions will boost the strength of the [O II] line (compared to, say, [O III] $λ5007$, which can be largely ascribed to the AGN itself).   &lt;/p&gt;&#xA;" OwnerUserId="2091" LastEditorUserId="37191" LastEditDate="2021-04-16T08:35:38.250" LastActivityDate="2021-04-16T08:35:38.250" Title="What are possible methods of discerning a starburst AGN composite galaxy?" Tags="&lt;star-formation&gt;&lt;active-galaxy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5931" PostTypeId="1" CreationDate="2014-07-21T06:30:12.983" Score="3" ViewCount="227" Body="&lt;p&gt;I'm looking for a very rough, order-of-magnitude approach to the main sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I mean is, I have a spherically-symmetric hydrogen distribution. I'm looking to get a rough approximation for density $\rho(r)$, temperature $T(r)$, and pressure $P(r)$ given the total mass $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this, I need an equation of state. Then I need an approximation for the rate of fusion, which I can balance against the heat lost due to radiation (for which I can use Stefan-Boltzmann).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how can I throw this into a back-of-the-envelope calculation that tells me that, e.g. solar mass stars should live ~10^10 years, without crunching on some differential equations?&lt;/p&gt;&#xA;" OwnerUserId="1945" LastActivityDate="2014-08-21T07:49:37.163" Title="What's an order-of-magnitude main sequence star look like?" Tags="&lt;star&gt;&lt;stellar-astrophysics&gt;&lt;hydrostatic-equilibrium&gt;&lt;main-sequence&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5932" PostTypeId="1" AcceptedAnswerId="5933" CreationDate="2014-07-21T14:19:07.957" Score="9" ViewCount="891" Body="&lt;p&gt;Suppose that someone lives in the neighborhood of Jupiter. He may live in a space station around Jupiter, or in any ot it's moons (I believe that it doesn't matter for this question). If this person look to the Sun he can see the asteroid belt, since it is between them? If generally not, then he would be able to see larger bodies such as Ceres?&lt;/p&gt;&#xA;" OwnerUserId="2093" LastEditorUserId="58" LastEditDate="2014-07-21T16:11:45.403" LastActivityDate="2014-07-21T16:11:45.403" Title="Can the asteroid belt be seen from Jupiter?" Tags="&lt;jupiter&gt;&lt;asteroid-belt&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5933" PostTypeId="2" ParentId="5932" CreationDate="2014-07-21T15:23:48.153" Score="11" Body="&lt;p&gt;No. The asteroid Belt is on average about 2.6 AU from the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth is 1 AU from Sun, 1.6 AU from the asteroid belt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter is 5.2 from Sun, 2.6 AU from the asteroid belt. I.e. much further away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it should be easier to see asteroids from inside their orbit since they reflect more sunlight towards us. From Jupiter you would see their shadowed sides. Ceres is to Jupiter what Venus is to Earth. Venus is inside of us just like Ceres is inside of Jupiter. When it is at its nearest, we can only see a thin rim of sunlight on one half of its edge. But when an outer object like Mars is at its closest to us, we see its entire hemisphere lit up by the Sun. Add to that that when Venus is at its closest to us, as when Ceres is at its closets to Jupiter, it is very near to the Sun on the sky and therefor lost in the bright Solar light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ceres wasn't discovered until year 1801. That's after the discovery of Uranus 20 AU away. You'd need an even better telescope to find Ceres from Jupiter than from Earth.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2014-07-21T15:29:39.557" LastActivityDate="2014-07-21T15:29:39.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5934" PostTypeId="1" AcceptedAnswerId="5935" CreationDate="2014-07-21T15:37:53.773" Score="15" ViewCount="1949" Body="&lt;p&gt;From what I know of elliptical orbits, an object speeds up near the periapsis and slows down at the apoapsis, much like we learned in high school physics how a sphere would roll down and back up a valley in a frictionless vacuum: The height is inversely proportional to the speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;gravity slingshot&quot; maneuver we've seen in sci-fi and even used by our own spacecraft relies on the physics of hyperbolic orbits, where an object both enters and exits the orbit before making a single lap around the planet/moon/etc. Since gravity pushes the craft toward that body both while they're heading towards and away from it, &lt;strong&gt;shouldn't the craft's velocity be the same at&lt;/strong&gt; (for example) &lt;strong&gt;1 megameter before the periapsis as 1 megameter after?&lt;/strong&gt; If so, then the gravity slingshot maneuver should only have the end purpose of redirecting the craft's trajectory, not of increasing its speed, as the name implies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding in a simple diagram:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/8fFCA.png&quot; alt=&quot;Three gray dots signify a single object as it travels along a path that dips below a large blue circle signifying a large body. The first and third gray dots has a label declaring a speed of x, and the second dot has a label declaring a speed much greater than x&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2006" LastEditorUserId="2006" LastEditDate="2015-10-14T15:52:20.543" LastActivityDate="2020-11-08T18:11:49.873" Title="How does a gravity slingshot actually work?" Tags="&lt;orbit&gt;&lt;gravity&gt;&lt;astrophysics&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5935" PostTypeId="2" ParentId="5934" CreationDate="2014-07-21T15:48:58.323" Score="5" Body="&lt;p&gt;The diagram is in the rest frame of the planet. Now suppose a spacecraft is slowing down in the frame of the solar system. A planet is nearby, so it now starts accelerating due to its gravity and gains speed. Now, this speed increase is added to some component of the speed of the planet's motion when it comes out on the other side (this added component can be changed by changing the angle from which it approaches the planet, in order to maximize the slingshot effect). Once out of the planet's influence, the spacecraft has the same velocity as before, plus a component of the planet's motion, which allows it to travel farther out. This is the slingshot effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to look at this in another way, consider the angular momentum of the spacecraft. As long as it is only under sun's gravitational influence, its angular momentum cannot change. However, once it's under the influence of another planet, the two angular momenta - one w.r.t. the sun and one w.r.t. the planet (due to their relative motion) - add, and once out of the gravitational influence of the planet, their relative components can be adjusted (based on the angle of approach towards the planet and the angle at which it flies away after the slingshot) in order to increase the angular momentum w.r.t. the sun, which in turn puts it in a larger orbit, allowing it to travel farther away than before.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2014-07-21T15:48:58.323" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5936" PostTypeId="2" ParentId="5934" CreationDate="2014-07-21T17:53:53.197" Score="7" Body="&lt;p&gt;Here's an intuitive understanding without math or physics explanations (others will provide that stuff here):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are right that approaching and leaving the vicinity of a planet in itself adds up to zero effect. Gravity assist is the effect of being &quot;dragged along&quot; with the movement of the planet. If a spaceship approaches the planet from behind in its orbit, it will be dragged along and accelerated. If a spaceship approaches from infront of the planet in its orbit, the spaceship will slow down as the moving gravity field of the meeting planet pulls it backwards.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2014-07-21T17:53:53.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5937" PostTypeId="1" AcceptedAnswerId="6277" CreationDate="2014-07-21T19:31:42.700" Score="6" ViewCount="168" Body="&lt;p&gt;Many asteroids are today known to be binary, to have &quot;moons&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what &lt;strong&gt;comets&lt;/strong&gt; are known to be binary? Or what fraction of them are theoretically thought to be binary? And if possible, then separate Oort cloud comets from comets with more local asteroid like orbits in this respect.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2014-10-08T19:53:53.620" Title="What known comets are binary?" Tags="&lt;comets&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5938" PostTypeId="1" CreationDate="2014-07-21T19:36:38.103" Score="2" ViewCount="901" Body="&lt;p&gt;Mars, Venus and Titan have an atmosphere and we can &lt;strong&gt;theoretically&lt;/strong&gt; step on them (and die quickly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as an example, let's say we can survive on these objects without a space suit. I say: “Hi” on any of these objects. Would I hear what I said or not?&lt;/p&gt;&#xA;" OwnerUserId="1890" LastEditorUserId="7982" LastEditDate="2021-02-28T11:55:34.650" LastActivityDate="2021-02-28T11:55:34.650" Title="Can we hear something on Venus, Mars and Titan?" Tags="&lt;mars&gt;&lt;planetary-atmosphere&gt;&lt;venus&gt;&lt;atmospheric-effects&gt;&lt;titan&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5939" PostTypeId="1" CreationDate="2014-07-21T21:09:06.443" Score="5" ViewCount="96" Body="&lt;p&gt;Is it ever possible for starlight to be distorted in a noticeable way from something as small as a planet? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing that the chances of this working are better on a dense terrestrial planet than a gas giant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you calculate this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I was unclear, sorry. I'm specifically thinking about lensing that would distort the shape of a star, not just move its position.&lt;/p&gt;&#xA;" OwnerUserId="158" LastActivityDate="2014-07-21T22:23:36.943" Title="How massive does a planet need to be to create gravitational lensing?" Tags="&lt;gravity&gt;&lt;light&gt;&lt;gravitational-lensing&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5940" PostTypeId="2" ParentId="5939" CreationDate="2014-07-21T22:23:36.943" Score="3" Body="&lt;p&gt;You could google &quot;Maccone focal&quot; for the most ardent proponent of this project (and I love it too, the most furthest meaningful mission possible within a life time, and what a view!) But others may have variations of his idea. One should be a bit wary about enthusiasts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a link to some article where Maccone is cited to explain that planets need 6-17 greater distance to become gravitaional lenses than the Sun needs. Neptune surprisingly happens to be next best after Jupiter: &lt;a href=&quot;http://www.centauri-dreams.org/?p=15290&quot; rel=&quot;nofollow&quot;&gt;http://www.centauri-dreams.org/?p=15290&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paper (might not be available for free):&#xA;Maccone, “A New Belt Beyond Kuiper’s: A Belt of Focal Spheres Between 550 and 17,000 AU for SETI and Science.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are slides from one of his presentations where he explains his calculations more generally, but I think only for the Sun:&#xA;&lt;a href=&quot;http://www.spaceroutes.com/astrocon/AstroconVTalks/Maccone-AstroconV.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.spaceroutes.com/astrocon/AstroconVTalks/Maccone-AstroconV.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;And then there's micro gravity lenses! A very different thing in terms of our observation.&lt;/strong&gt; I now think my answer above misunderstood what you were looking for.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2014-07-21T22:23:36.943" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5941" PostTypeId="1" AcceptedAnswerId="5942" CreationDate="2014-07-22T01:00:34.807" Score="4" ViewCount="405" Body="&lt;p&gt;There is a sci-fi television episode that shows a man looking out the window into the night sky. There up in the sky is the moon, and it's extraordinarily bright.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The man spends the rest of the night trying to do all the things he wish he had done in his life. A race to experience life before the morning sunrise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;His conclusion&lt;/em&gt;; A giant solar flare has erupted, and the side of the Earth facing the sun has been burned away. The world is ending....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How plausible is this story?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could a solar flare burn the Earth? and if you happen to be on the dark side when it happens. Would the moon be a reliable indicator?&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2014-07-22T02:07:25.360" Title="Would an ultra bright moon indicate that the world is ending?" Tags="&lt;the-moon&gt;&lt;solar-flare&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5942" PostTypeId="2" ParentId="5941" CreationDate="2014-07-22T02:07:25.360" Score="5" Body="&lt;p&gt;I remember that well. I presume you're referring to an episode of &quot;The Outer Limits,&quot; which was based on a short story by Larry Niven. Having looked into the idea back when I read the story in the 1990s, I can say it's plausible in the sense that it's not strictly prohibited by any law of physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, a solar flare would have to be of precisely the right size, coming in at precisely the right trajectory for precisely the right duration, and with the moon in precisely the right spot for the scenario in the story to play out. More likely, the entire planet would be completely roasted, and all higher life forms extinguished. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, most solar flares are not nearly that powerful. Typically, we experience transient disturbances in global telecommunications, spectacular aurorae, and little else when a flare hits. In the late 1800s a more powerful flare hit, and some people speculate that a flare that size could affect not only telecommunications but also power grids. However, the planet would not even be singed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are in fact several ways to burn the Earth, as you put it. Astronomer Phil Platt describes these in wonderful detail in his book &lt;em&gt;Death from the Skies&lt;/em&gt;. In addition to a solar flare, a gamma ray burst from a nearby supernova (&amp;lt; 20 ly) would do it. I won't mention the others since I don't want to spoil the fun of the book.&lt;/p&gt;&#xA;" OwnerUserId="1991" LastActivityDate="2014-07-22T02:07:25.360" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5943" PostTypeId="1" AcceptedAnswerId="5948" CreationDate="2014-07-22T03:28:40.760" Score="2" ViewCount="431" Body="&lt;p&gt;I don't have much knowledge of Astronomy but today I was googling randomly and found an article, according to which:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;The moon is of the wrong size for its current orbit.&lt;/li&gt;&#xA;  &lt;li&gt;The moon rings like a bell when hit by an object, which could signal that it's hollow.&lt;/li&gt;&#xA;  &lt;li&gt;Metal shavings were found in samples taken by the Apollo crew. &lt;/li&gt;&#xA;  &lt;li&gt;UFO sightings were common on NASA moon missions.&lt;br&gt;&#xA;  &lt;strong&gt;All these phenomena and more point toward the suggestion that the moon is actually an artificially created spaceship. It's a bit far fetched, but it's odd that astronomers have been discouraged from testing this theory.&lt;/strong&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can this theory be true?&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2014-07-22T14:09:23.833" Title="Are these facts about moon true?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5944" PostTypeId="2" ParentId="5934" CreationDate="2014-07-22T06:14:11.163" Score="3" Body="&lt;p&gt;You are correct that an hyperbola's outgoing speed is the same as the incoming speed with regard to the body lieing at hyperbola's focus. The direction is changed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But with regard to another body, the change of direction can mean a change of speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a diagram of how the moon might be employed in the capture of an asteroid to reduce it's hyperbolic orbit with regard to the earth to a capture orbit about the earth:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QD1G4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1232" LastActivityDate="2014-07-22T06:14:11.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5945" PostTypeId="2" ParentId="5931" CreationDate="2014-07-22T07:27:06.743" Score="1" Body="&lt;p&gt;To get an order of magnitude estimate you can just use the total mass $M$ and luminosity $L$ of the star and an assumption of your fusion process. Main sequence stars fuse Hydrogen in to Helium through the &lt;a href=&quot;http://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction&quot; rel=&quot;nofollow&quot;&gt;proton-proton chain&lt;/a&gt;, which converts 0.7% of mass into energy. So the estimated lifetime of the star would just be: $0.007\frac{Mc^2}{L}$ ($c$ is the speed of light). For example, the sun would have a lifetime ~$10^{11}$ years, but that's assuming that all Hydrogen is being converted to Helium via the proton-proton chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the scaling from observations of the Sun, you can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mass%E2%80%93luminosity_relation&quot; rel=&quot;nofollow&quot;&gt;mass-luminosity relation&lt;/a&gt; to estimate lifetimes for other main sequence stars.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-07-22T07:27:06.743" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5947" PostTypeId="2" ParentId="5930" CreationDate="2014-07-22T07:53:12.430" Score="2" Body="&lt;p&gt;For starbursts, their star formation rates are best estimated by a combination of their ultraviolet and infrared emission. [O II] and [O III] as well as Hydrogen recombination lines can be used to estimate star formation, but like the UV, these &lt;a href=&quot;http://ned.ipac.caltech.edu/level5/March05/Dopita/Dopita2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;suffer from extinction&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/E1fl6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-07-22T07:53:12.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5948" PostTypeId="2" ParentId="5943" CreationDate="2014-07-22T10:00:39.617" Score="7" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Orbital radius depends on speed of the body. Size is largely irrelevant. Imagine a  space craft and much larger space station, able to orbit meters apart at the same speed, with engines switched off. An astronaut going on a spacewalk would be launched away from the station, being much smaller and so on wrong orbit. No. It's as relevant as &quot;Boeing 747 shouldn't be able to fly because it's painted the wrong color.&quot; &lt;strong&gt;False&lt;/strong&gt; and based on wrong premise.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For anything to ring you need atmosphere, a medium for sound to travel through. Moon, floating in vacuum, doesn't ring, nor create any other sound for that matter. Even the most powerful meteor impacts are entirely silent. &lt;strong&gt;False&lt;/strong&gt; and based on wrong premise.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Countless iron meteorites hit the surface of the Moon, often shattering on impact. Specks of iron are very common in lunar soil, and I have no doubt it's entirely possible shavings-shaped pieces of lunar meteor might have found their way into the samples. &lt;strong&gt;Likely True&lt;/strong&gt; and nothing odd about it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;...how does that even relate?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="83" LastActivityDate="2014-07-22T10:00:39.617" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5949" PostTypeId="1" CreationDate="2014-07-22T11:25:36.827" Score="2" ViewCount="533" Body="&lt;p&gt;I was enjoying a casual naked eye view of the summer sky last night around midnight in a new jersey beach town. The sky was pretty clear for around here.The Milkyway was just visible and there was no could cover or buildings higher than me so I saw from Cassiopeia to Sagittarius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was hoping for a meteor and I got one I have never seen before. I will try to describe it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Appearance: .It appeared out of no where like any other meteor but it had no streak AT ALL. As it traveled it shimmered. It was blue colored and not particularly bright perhaps fuzzy. The shimmering was the best way I can describe it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trajectory: It began high in the sky slightly to the east around the &quot;southern&quot; part of the constellation Aquila. It traveled west. It was as fast as any other meteor but as I said had no tail. It lasted very long... say 4 or 5 seconds. It didn't have an arching trajectory like something lower. It traveled straight and as it disappeared, it seemed to travel off into vs down to the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was so astonished that I practiced the speed it went with my finger after it finished. It moved from say Vega to Altair in less around a second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It couldn't have been any kind of craft since it appeared and disappeared and traveled so quickly. I have seen 100's of meteors of many kinds but not any like this. What did i see?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to be as detailed as possible. If I missed anything, I'll add it in the comments. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the record, I am not a believer in visiting aliens or supernatural events. BUT you will laugh at this. Years ago,I had read about an Ozark superstition that if you say &quot;Money&quot; three times when a meteor appears, you will get rich. I liked that one and since I stargaze a lot, I figured I'd get the opportunity. Well I had my chance and then some but I was so astonished, I forgot to say it. Looks like I  will have to keep working for a living :)   &lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2014-08-29T08:55:49.500" Title="What did I see streaking across the sky last night?" Tags="&lt;meteor&gt;&lt;star-gazing&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5952" PostTypeId="2" ParentId="5927" CreationDate="2014-07-22T19:03:58.380" Score="4" Body="&lt;p&gt;All of these effects are related to the 11 year &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_cycle&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar cycle&lt;/a&gt;. And while we know there are times when the sun is &quot;active&quot; versus &quot;inactive&quot; we aren't necessarily predicting exactly where sunspots or solar flares will occur, but how many we see in total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/naJH2.gif&quot; alt=&quot;Sunspot activity from solar cycle.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8l4Cs.png&quot; alt=&quot;Fluctuation in solar irradiance&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-07-22T19:03:58.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5953" PostTypeId="2" ParentId="2185" CreationDate="2014-07-22T19:30:18.993" Score="0" Body="&lt;p&gt;To estimate galaxy stellar mass with broadband photometry, you want to compare your observations to the &lt;a href=&quot;http://ned.ipac.caltech.edu/level5/March10/Walcher/Walcher6.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;spectral energy distribution&lt;/a&gt; (SED) of the galaxy. There are synthetic SED libraries which will generate SEDs for different types of galaxies with different masses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9rnNT.png&quot; alt=&quot;An example galaxy SED&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each template you can figure out the expected luminosity in the I-band and find the best match. For this method, it's really best to have more than one point for comparison. Since the near IR will likely be a blackbody from your evolved stellar population, you might want to see if you can supplement your I-band data with &lt;a href=&quot;http://irsa.ipac.caltech.edu/applications/2MASS/PubGalPS/&quot; rel=&quot;nofollow noreferrer&quot;&gt;2MASS&lt;/a&gt; observations.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-07-22T19:30:18.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5954" PostTypeId="2" ParentId="5922" CreationDate="2014-07-22T20:28:03.710" Score="1" Body="&lt;p&gt;DEC :Declination&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RA :Right ascension&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Redshift / z  :use to calculate   velocity  over earth z axis (azimuth) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnitude/ Qual Filter : zoom into image filter quality (??)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Separ arc min   : if is a group as cloud magallans&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Equatorial_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;here more information&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2014-07-22T20:28:03.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5955" PostTypeId="1" CreationDate="2014-07-22T22:42:04.700" Score="1" ViewCount="124" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PhynR.png&quot; alt=&quot;graph&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is from the &lt;a href=&quot;http://ned.ipac.caltech.edu/cgi-bin/objsearch?objname=M83&amp;amp;extend=no&amp;amp;hconst=73&amp;amp;omegam=0.27&amp;amp;omegav=0.73&amp;amp;corr_z=1&amp;amp;out_csys=Equatorial&amp;amp;out_equinox=J2000.0&amp;amp;obj_sort=RA%20or%20Longitude&amp;amp;of=pre_text&amp;amp;zv_breaker=30000.0&amp;amp;list_limit=5&amp;amp;img_stamp=YES&quot; rel=&quot;nofollow noreferrer&quot;&gt;NED&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do these three titles mean?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;log fb (Jy)&lt;/li&gt;&#xA;&lt;li&gt;MRK 1014&lt;/li&gt;&#xA;&lt;li&gt;Log v (HZ)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Why is there &lt;code&gt;LINE&lt;/code&gt;,&lt;code&gt;LINE&lt;/code&gt;,&lt;code&gt;LINE&lt;/code&gt;,&lt;code&gt;LINE&lt;/code&gt;,&lt;code&gt;LINE&lt;/code&gt; written along the bottom? What does that signify?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the dots and lines showing? A range of information? In that chase, what's the relevance of the dot? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it like this: The dot is the &lt;code&gt;log v&lt;/code&gt; and then the horizontal line shows the range of &lt;code&gt;MRK 1014&lt;/code&gt; recorded for that &lt;code&gt;log v&lt;/code&gt; and the vertical line showing the &lt;code&gt;log fv&lt;/code&gt; recorded for it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the &lt;code&gt;log v&lt;/code&gt; between 12 - 16.3 have small amounts of &lt;code&gt;log fv&lt;/code&gt; and &lt;code&gt;MRK 1014&lt;/code&gt;, but the &lt;code&gt;log v&lt;/code&gt; between 15 - 19 have no &lt;code&gt;log fv&lt;/code&gt; at all, but a large amount of &lt;code&gt;MRK 1014&lt;/code&gt;. The same goes for the 2 &lt;code&gt;log v&lt;/code&gt; points between 10.4 - 12, except these two have a large amount of &lt;code&gt;log fv&lt;/code&gt; but no &lt;code&gt;MRK 1014&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe if you could explain the units to me and give me links to further reading it would all become a lot clearer...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt; Okay &lt;code&gt;MRK 1014&lt;/code&gt; is the same of the object. (Doh!) However, what is that titless axis showing?&lt;/p&gt;&#xA;" OwnerUserId="2084" LastEditorUserId="1862" LastEditDate="2014-07-23T02:16:16.240" LastActivityDate="2014-07-23T02:16:16.240" Title="What is this graph showing?" Tags="&lt;observational-astronomy&gt;&lt;data-analysis&gt;&lt;spectra&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5956" PostTypeId="2" ParentId="5955" CreationDate="2014-07-22T23:14:52.043" Score="5" Body="&lt;p&gt;This is a &lt;a href=&quot;http://coolwiki.ipac.caltech.edu/index.php/SED_plots&quot;&gt;spectral energy distribution&lt;/a&gt; (SED). Since astronomical objects don't emit light a single frequency an SED tells you how much emission you're getting across a range of frequencies.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;log&quot; indicates it's on a logarithmic scale&lt;/li&gt;&#xA;&lt;li&gt;$\nu$ represents frequency in Hz (e.g. $log~\nu=12$ is $10^{12}$ Hz)&lt;/li&gt;&#xA;&lt;li&gt;Jy is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Jansky&quot;&gt;unit&lt;/a&gt; of specific flux density $f_{\nu}$ is the specific flux &lt;/li&gt;&#xA;&lt;li&gt;MRK 1014 is the name of the object&lt;/li&gt;&#xA;&lt;li&gt;Each of the points are observations there is a link on the page for &quot;photometric data points&quot; to see where everything comes from. All the positions have a point, which is measurement of the specific flux density at that given frequency. The vertical lines are the error bar for the flux measurement and the horizontal is the range of frequencies a measurement covers (e.g. a filter might have a central frequency but it filters light over a range of frequencies).&lt;/li&gt;&#xA;&lt;li&gt;There are a few upper limits which are indicated by vertical lines with arrows pointing down.&lt;/li&gt;&#xA;&lt;li&gt;LINE indicates that this is an estimate from a spectrum and it is the measurement of the flux of that line, not from a imaging filter.   &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-07-22T23:14:52.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5957" PostTypeId="1" CreationDate="2014-07-23T01:01:49.450" Score="11" ViewCount="2744" Body="&lt;p&gt;In a hypothetical solar system there exist:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a sun of radius $r_s$ and absolute visual magnitude $V$&lt;/li&gt;&#xA;&lt;li&gt;the Earth, with radius $r_e$ and distance from the sun of $1\,AU$&lt;/li&gt;&#xA;&lt;li&gt;another planet, wit radius $r_p$, albedo $a$, distance from Earth $d_e$ (in units of $AU$), and distance from the sun $d_s$ ($AU$)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Assume that the second planet is magically lit up entirely by the sun (so it does not have phases when viewed from Earth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how can I compute the apparent visual magnitude of the second planet with respect to Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are other parameters required, please tell me.&lt;/p&gt;&#xA;" OwnerUserId="2112" OwnerDisplayName="feralin" LastEditorUserId="2112" LastEditDate="2014-07-23T14:08:20.357" LastActivityDate="2020-08-10T11:02:26.150" Title="Compute Planet's Apparent Visual Magnitude" Tags="&lt;solar-system&gt;&lt;planet&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5958" PostTypeId="2" ParentId="5922" CreationDate="2014-07-23T09:05:51.310" Score="2" Body="&lt;p&gt;If you browse further down the page, there are details for each of the top table data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So RA and Dec give you coordinates (Right Ascension and Declination, like longitude and latitude). There is a designation J2000.0 which tells you the zero point to which the coordinates relate (remember, the Earth wobbles, so any coordinate system tied to Earth changes over time, hence a zero point), in this case the year 2000. The Equ. in front of that means equatorial, which ties is directly to the earth axis (poles &amp;amp; equator; the latter explains the name).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Redshift (z is the symbol for redshift) for a distance estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Qual probably stands for qualification&lt;strike&gt;, for example the type of galaxy&lt;/strike&gt;. It's likely not related to the filter. The type of galaxy would be the classification, which is further down the page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnitude is the brightness, and Filter the filter it is measured in (most astronomy is done in a filter, to obtain a brightness in a specific wavelength range. Eg, red, blue, green, uv, infrared, H-alpha, Oxygen filters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Separ arc min: the separation (or more likely, the diameter) in minutes of arc (so in 1/60 of a degree). You can't really measure the actual size (unless you trust the distance estimate), but you can measure the size as a angle projected on the sky. That's what it says here (e.g., the sun and the full moon are about 0.5 degree, or 30 arcminutes, &quot;wide&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned at the start, browse down to get more in depth information, which may shed a bit more light on what the top table abbreviations actually mean and indicate.&lt;/p&gt;&#xA;" OwnerDisplayName="user51" LastActivityDate="2014-07-23T09:05:51.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5959" PostTypeId="2" ParentId="2183" CreationDate="2014-07-23T16:54:53.150" Score="0" Body="&lt;p&gt;You can look at the code of the Free application Kstars. All formulas you may need are already programmed there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the good about Free Software.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-07-23T16:54:53.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5960" PostTypeId="1" CreationDate="2014-07-23T17:00:59.420" Score="6" ViewCount="2147" Body="&lt;p&gt;I've heard the universe is just an empty cavity. Does the universe have any edge or stop or is it just an infinite empty cavity?&lt;/p&gt;&#xA;" OwnerDisplayName="user73898" LastEditorUserId="6" LastEditDate="2014-07-23T18:23:43.113" LastActivityDate="2020-02-10T22:01:55.930" Title="Does the universe have an edge?" Tags="&lt;universe&gt;" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5963" PostTypeId="2" ParentId="5960" CreationDate="2014-07-24T07:09:49.923" Score="10" Body="&lt;p&gt;As far as theories I've heard of, we can't see the edge of the universe (if truly exists) because the farther you look out into space through a telescope, the farther back in time you are actually looking. For example, if you are looking at a star 20 light-years away... you're actually currently observing what was happening on that star 20 years ago... the theory is that there is only so far you can look &lt;em&gt;out&lt;/em&gt; into the universe because there is only so much history post Big Bang. &lt;img src=&quot;https://i.stack.imgur.com/Bppor.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image above: What is the furthest we can see? In 2003, NASA's WMAP satellite took images of the most distant part of the universe observable from Earth. The image shows the furthest we can see using any form of light. The patterns show clumps of matter that eventually formed into galaxies of stars. Credit: NASA/WMAP Science Team&lt;/p&gt;&#xA;" OwnerUserId="2067" LastEditorUserId="2067" LastEditDate="2014-07-24T10:13:24.200" LastActivityDate="2014-07-24T10:13:24.200" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5964" PostTypeId="1" AcceptedAnswerId="5966" CreationDate="2014-07-24T07:29:52.387" Score="4" ViewCount="92" Body="&lt;p&gt;If a star is 20 light-years away and I look at it through a telescope, what I'm observing is what the star was doing 20 years ago yes? So, if I fix my sight on the star and move toward it (for example's sake, let's say 0.5C or half the speed of light in a vacuum) in a linear path (which is not practical for real space travel) it would take me 40 years to reach it. By the time I reach it I would be observing in (almost) real-time what is happening. So to me that says that I've watched 60 years of the star's history in 40 years worth of time. So then, would it look to me as if I was watching a movie of the star's history in 1.5 times 'normal' speed? (60/40 = 3/2 or 1 1/2) &lt;/p&gt;&#xA;" OwnerUserId="2067" LastEditorUserId="2067" LastEditDate="2014-07-24T09:26:09.537" LastActivityDate="2017-07-19T10:45:04.317" Title="Perception of an objects history passing as an observer moves toward it" Tags="&lt;observational-astronomy&gt;&lt;light&gt;&lt;space-time&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5965" PostTypeId="2" ParentId="2150" CreationDate="2014-07-24T10:07:37.447" Score="0" Body="&lt;p&gt;Couldn't comment as I do not have the proper rep yet, but regarding @user3355020 and the answer provided by @Gerald, (and me being far less educated than I ever could have imagined before reading this), I interpret it like this: (and I wouldn't be surprised if this is wrong)&#xA;the speed of light is constant, but the expansion rate of the universe is not. The over-simplified observed effect is similar to the feeling of having a dream where you're in a hallway and you're running away from something as fast as you possibly can,focused on the end of the hallway, but the hallway seems to stretch more and more as time goes by. Much in the same way, light is the person running and 'space' is the hallway and earth is the opening at the end. Some light was able to get far enough in the beginning that the stretch did not impact it so much as to make it impossible to reach earth (it's just going to take it 13+ billion yrs), and some light (and from my understanding of the answer given by @Gerald all light after a certain 'time') will never reach earth.&lt;/p&gt;&#xA;" OwnerUserId="2067" LastActivityDate="2014-07-24T10:07:37.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5966" PostTypeId="2" ParentId="5964" CreationDate="2014-07-24T15:13:21.983" Score="3" Body="&lt;p&gt;Yes, it will be 20 Julian years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, it will take 40 years to reach there with speed half the speed of light through vacuum without any obstacles and star must be stationary (though you can't travel with speed of light since physically it's not possible to achieve that speed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as you are travelling with speed half to that of the speed of light you will experience a phenomenon of time dilation. Check all the details about time dilation here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Time_dilation&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Time_dilation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will see time dilated by $\gamma = 1/\sqrt{1-v^2/c^2}$, factor so if you substitute this values it will take you 60 years to reach that star. And it squares all the things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this explanation helps you. If it is not understandable do let me know; I will explain it in a more mathematical way.&lt;/p&gt;&#xA;" OwnerUserId="2034" LastEditorUserId="7472" LastEditDate="2017-07-19T10:45:04.317" LastActivityDate="2017-07-19T10:45:04.317" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5967" PostTypeId="2" ParentId="5960" CreationDate="2014-07-24T15:53:44.827" Score="3" Body="&lt;p&gt;Galaxies extend as far as we can detect... with no sign of diminishing.There is no evidence that the universe has an edge. The part of the universe we can observe from Earth is filled more or less uniformly with galaxies extending in every direction as far as we can see - more than 10 billion light-years, or about 6 billion trillion miles. We know that the galaxies must extend much further than we can see, but we do not know whether the universe is infinite or not. When astronomers sometimes refer (carelessly!) to galaxies &quot;near the edge of the universe,&quot; they are referring only to the edge of the OBSERVABLE universe - i.e., the part we can see&lt;/p&gt;&#xA;" OwnerUserId="2034" LastActivityDate="2014-07-24T15:53:44.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5968" PostTypeId="1" AcceptedAnswerId="5971" CreationDate="2014-07-24T21:17:32.307" Score="4" ViewCount="1689" Body="&lt;p&gt;I've been reading about the reionization epoch of the universe, and it isn't entirely clear to me.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, very shortly after the big bang the universe was ionized, with the electrons stripped from the protons of hydrogen. The way that it's described is that the universe was &quot;opaque&quot; (since it was filled with ionized gas). Then the universe cooled enough for neutral hydrogen to form again, and the universe became &quot;transparent&quot;. When the first stars and galaxies began to form they reionized the universe.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand is that in illustrations of this reionization, during the reionization epoch, it seems that the universe isn't transparent, but filled with unionized IGM (neutral hydrogen). Then the ionizing radiation from the newly formed stars ionizes the medium and ... these bubbles of ionized regions grow and merge until the entire universe is ionized again (and is therefore transparent).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have I misunderstood when the universe is transparent? Or the definition of transparent? Would it be accurate to say that the present day universe is completely ionized? (Does this just mean that most matter is in the form of stars which can be considered ionized? Since it doesn't really seem immediately evident in everyday life that &quot;everything is ionized&quot;)&lt;/p&gt;&#xA;" OwnerUserId="2119" LastActivityDate="2014-08-24T14:34:33.647" Title="How did the reionization epoch make the universe &quot;transparent&quot;?" Tags="&lt;cosmology&gt;&lt;early-universe&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5969" PostTypeId="2" ParentId="5922" CreationDate="2014-07-24T23:01:10.200" Score="2" Body="&lt;p&gt;&lt;strong&gt;RA and DEC&lt;/strong&gt; are the coordinates on sky.   &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;RA (Right Ascension) is the angular position measured eastward in a full circle along the celestial equator.&lt;/li&gt;&#xA;&lt;li&gt;Dec (Declination) is the angular position measured from the celestial equator in either the north (positive) or south (negative) direction. $+90\,^{\circ}$ is the north celestial pole, $-90\,^{\circ}$ is the south.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Velocity (km/s)&lt;/strong&gt; is the radial velocity that the galaxy has away from or towards the earth.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Redshift (z)&lt;/strong&gt; is basically a distance estimate based on observing the amount that the light of the object is redshifted due to it's movement away from us. (Since galaxies farther away will appear to move away faster according to Hubble's Law) If you understand the Doppler effect, the equation for its definition is pretty simple: (from Wikipedia's &quot;Redshift&quot; page)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fr2D4.png&quot; alt=&quot;From wikipedia on &amp;#39;redshift&amp;#39;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The higher z is, the further away in distance (and time) the galaxy is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Magnitude / Filter :&lt;/strong&gt; Magnitude is the measure of brightness of the object. This depends on which filter (at which wavelength range) the observation is made in. For this measurement: smaller values = brighter. The sun is something like -27, the brightest star Sirius, is around -1.5, the faintest stars visible to the naked eye are around +6 to +8.&lt;br&gt;&#xA;The galaxy page you linked only lists the magnitude - I don't know why the filter doesn't seem to be there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Separ. arcmin :&lt;/strong&gt; The separation in arc minutes. (Probably meaning size, or diameter of the galaxy) An arc minute is an angular unit of size on the sphere of the sky (the same units as declination). So one arc minute is 1/60th of a degree, and there are 360 degrees spanning the entire celestial sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, I don't know the &quot;Qual&quot; measurement.&lt;/p&gt;&#xA;" OwnerUserId="2119" LastActivityDate="2014-07-24T23:01:10.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5970" PostTypeId="1" CreationDate="2014-07-25T00:41:44.983" Score="1" ViewCount="41" Body="&lt;p&gt;I always read/hear &quot;This galaxy is X thousand light years away&quot; or some similar statement (for example see &lt;a href=&quot;http://en.wikipedia.org/wiki/Light-year#Distances_in_light-years&quot; rel=&quot;nofollow&quot;&gt;this table&lt;/a&gt;). Please excuse the naivety of question. How do we determine that some astronomical body is &quot;x&quot; light years away. Surely we don't send light beams and wait for them to come back. :-P&lt;/p&gt;&#xA;" OwnerUserId="2121" LastActivityDate="2014-07-25T00:41:44.983" Title="How do we determine something is &quot;x&quot; light years away?" Tags="&lt;distances&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2014-07-25T12:41:36.780" ContentLicense="CC BY-SA 3.0" />
  <row Id="5971" PostTypeId="2" ParentId="5968" CreationDate="2014-07-25T14:34:15.673" Score="5" Body="&lt;p&gt;Nowadays, the universe is mostly ionized, and transparent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transparent means that photons are able to travel long distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there comes the confusion: &lt;em&gt;How could it be that recombination made the Universe transparent, and reionization keep it transparent?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the Big Bang we had a very small Universe, filled with ionized matter. That amount of ionized matter caused the photons to be short-lived, and thus the Universe were opaque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then it came the recombination, and most baryonic matter became neutral, the photons were able to travel long distances, and the Universe thus became transparent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;And it was expanding. And at the same time clusters of matter were collapsing.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The combination of these two factors made the intergalactic medium density to fall a lot. Not only were there much more space for the same matter, but also the matter was concentrated in small bunchs making almost all the space almost empty.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The second phase change occurred once objects started to condense in the early universe that were energetic enough to ionize neutral hydrogen. As these objects formed and radiated energy, the universe reverted from being neutral, to once again being an ionized plasma. This occurred between 150 million and one billion years after the Big Bang (at a redshift 6 &amp;lt; z &amp;lt; 20). At that time, however, matter had been diffused by the expansion of the universe, and the scattering interactions of photons and electrons were much less frequent than before electron-proton recombination. Thus a universe full of low density ionized hydrogen will remain transparent, as is the case today. [1]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So when the clusters of matter collapsed enough to become galaxies with stars and these stars reionized the intergallactic medium, there were almost no matter there, so photons continued being free to travel without obstacles, and so the Universe remained transparent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it has been transparent for two different reasons: neutrality of matter, after the recombination, and scarcity of matter, after the reionization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: &lt;a href=&quot;http://en.wikipedia.org/wiki/Reionization#Background&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Reionization#Background&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-07-25T14:34:15.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5972" PostTypeId="1" CreationDate="2014-07-26T17:28:43.583" Score="3" ViewCount="183" Body="&lt;p&gt;I have heard of years with longer summers and winters and shorter springs and falls and vice versa because of a change in the earths axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this from the procession that causes the season switch every 25,800 years or is it from some other change in the earth's axis like angle change and not revolution at the same angle(which I think the procession is)?&lt;/p&gt;&#xA;" OwnerUserId="1138" LastActivityDate="2014-08-08T06:54:46.863" Title="seasons elongating and shortening" Tags="&lt;earth&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5973" PostTypeId="2" ParentId="5938" CreationDate="2014-07-26T20:50:46.457" Score="1" Body="&lt;p&gt;Yes, you and everyone around you would hear you. The anatomy of speech might complicate speaking in exotic environments, but we can be heard breathing helium, and from underwater. Sound is vibration of a substance, or medium. The medium a sound is transmitted across will make no difference in whether or not it is transmitted, but will impact how far it audibly travels (i.e. the rate of diffusion of the sound wave).&lt;/p&gt;&#xA;" OwnerUserId="2127" LastActivityDate="2014-07-26T20:50:46.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5974" PostTypeId="2" ParentId="5938" CreationDate="2014-07-26T21:56:35.873" Score="1" Body="&lt;p&gt;Just as an addition to the previous answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some spacecraft that landed on other cellestial objects carried microphones to record what can be heard. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example this video &lt;a href=&quot;http://youtu.be/36ffV-CI3Mo&quot; rel=&quot;nofollow&quot;&gt;http://youtu.be/36ffV-CI3Mo&lt;/a&gt; contains a sound clip recorded by the Huygens probe during its descent on the moon Titan. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I remember there are similar recordings from Venus, as short a time as some probe can survive on this beautiful hell hole :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, there is definitely sound on other planets if they have sufficiently dense athmospheres. &lt;/p&gt;&#xA;" OwnerUserId="1483" LastActivityDate="2014-07-26T21:56:35.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5975" PostTypeId="2" ParentId="2249" CreationDate="2014-07-26T23:57:18.860" Score="1" Body="&lt;p&gt;First of all, trajectories of interplanetary missions are not designed to minimize the TRAVEL TIME but to minimize the COST, which is directly related to the fuel required to transfer the probe between the Earth and another planet. One can easily find a solution on how fast and in what direction should the probe travel, and how much time and energy will be spent after departure from the near Earth orbit by solving the gravitational force equations. But that's far not enough to launch a real probe. Scientists have built accurate kinetic models and databases of the motions of solar system objects, such as DE405, by NASA Jet Propulsion Laboratory. Some of them even take account of the perturbations of all major planets in the solar system. Engineers should also build models for the probe itself. For example, how much acceleration will the spacecraft get when turn on the booster, and how does the solar radiation pressure change the attitude angle during the trip. What's more, every probe carries at least one gyro system. Optical guiding telescopes are also widely used to calculate the attitude angle. The speed of the probe can be calculated by the Doppler shift of the radio frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details, see the book &lt;a href=&quot;http://www.ltas-vis.ulg.ac.be/cmsms/uploads/File/InterplanetaryMissionDesignHandbook.pdf&quot; rel=&quot;nofollow&quot;&gt;Interplanetary Mission Design Handbook&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2105" LastActivityDate="2014-07-26T23:57:18.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5976" PostTypeId="1" AcceptedAnswerId="5980" CreationDate="2014-07-27T04:02:24.047" Score="3" ViewCount="879" Body="&lt;p&gt;I do not understand his timing method. &lt;a href=&quot;http://en.wikipedia.org/wiki/James_Bradley&quot; rel=&quot;nofollow noreferrer&quot;&gt;James&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The magnitude of the aberration could be &lt;code&gt;v/c&lt;/code&gt;? &lt;br/&gt;&lt;/p&gt;&#xA;&lt;p&gt;The speed of earth is about &lt;code&gt;30km/s&lt;/code&gt;.  That means the deviation from the real position of a star is about &lt;code&gt;30/300000 rad =180*3600/3.14159/10000 ~20arcsecond&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;p&gt;That is a horrible value.&lt;br/&gt;&lt;/p&gt;&#xA;&lt;p&gt;Could the effect of parallex, proper motion and light aberration of a star be separated clearly?&#xA;－－－－－－－－－－－－－－－－－－－－－&lt;br/&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Sure. Aberration is always the same. Parallax depends on the distance, and the phase is different.&lt;/h1&gt;&#xA;&lt;p&gt;Why is their phase different? They all have a one year period.&lt;/p&gt;&#xA;&lt;p&gt;Generally we should not worry about the light aberration, right? Even space telescopes have reference stars. The position of the reference stars are fixed. So although there are detectable wobble(20&amp;quot;) in one year , because we use reference stars, this kind of effect is removed every day immediately. Every night, when the night assistant lets a telescope point at some direction, he has to use reference stars to calibrate that night's pointing.&lt;/p&gt;&#xA;&lt;p&gt;Am I right?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2014-07-29T23:39:50.660" Title="how did James Bradley discover light aberration?" Tags="&lt;light&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5977" PostTypeId="2" ParentId="5972" CreationDate="2014-07-27T13:40:47.850" Score="1" Body="&lt;p&gt;I think there is a cycle where the inclination of the axis of earth also changes periodically.  This is different than precession where the orientation of the axis changes and not the inclination. That will cause the season length change.  Also there are Interstellar clouds through which the solar system passes.  Which reduces the solar radiation reaching earth.  They are responsible for Ice ages. &lt;/p&gt;&#xA;" OwnerUserId="2128" LastActivityDate="2014-07-27T13:40:47.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5978" PostTypeId="1" AcceptedAnswerId="5979" CreationDate="2014-07-28T15:50:34.543" Score="2" ViewCount="1154" Body="&lt;p&gt;I find this baffling. If we can observe objects moving away from us and each other, than it stands to reason that we can track their paths (relative to each other and ourselves) backward to find a point of origin. &lt;a href=&quot;https://astronomy.stackexchange.com/questions/669/what-is-in-the-center-of-the-universe&quot;&gt;In this question&lt;/a&gt; answers given suggest that there is no center of the universe. I'm looking for clarification as to why we cannot reverse engineer (to some degree) where this point of origin exists, even if a simple general direction relative to us. I understand that paths may be distorted and skewed by forces as time goes on, but I still don't grasp the concept that we have no way to determine a point of origin or general sense thereof. Help wrapping my inferior brain around this is much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2067" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-07-28T16:06:45.267" Title="Why can't we determine the center of the universe" Tags="&lt;observational-astronomy&gt;&lt;space&gt;&lt;big-bang-theory&gt;&lt;movement&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2014-07-28T18:33:58.397" ContentLicense="CC BY-SA 3.0" />
  <row Id="5979" PostTypeId="2" ParentId="5978" CreationDate="2014-07-28T16:06:45.267" Score="3" Body="&lt;p&gt;There is no point of origin! It is a misconception that the universe started as a single point &lt;em&gt;within space&lt;/em&gt;. Indeed, measurements since the 1990s seem to show that the universe is infinite in size, which means that the very early universe must have also been infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has not really been established that the universe originated from a &quot;singularity&quot; at all. We know that in the early universe everything was much closer together, and hence the universe was much denser. The &quot;Big Bang Theory&quot; goes back to the universe being extremely dense, but it is not required that the universe started from a single point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really, the Big Bang happened &lt;em&gt;everywhere&lt;/em&gt;. Even at these early times it looks like it was infinite in size with basically constant density, and so there's no sensible way for it to have a centre.&lt;/p&gt;&#xA;" OwnerUserId="2081" LastActivityDate="2014-07-28T16:06:45.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5980" PostTypeId="2" ParentId="5976" CreationDate="2014-07-28T18:53:52.690" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;That is a horrible value.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That value for aberration of starlight is &quot;horrible&quot; in two ways: One, in that's is very small. And two, in that it's large enough to interfere with parallax measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the late 1600s, scientists had already adopted the heliocentric model, and so were looking to measure the distance to the stars via parallax. Robert Hooke was among the first to attempt it, and probably the first to claim a measurement of the parallax. His observations were not very convincing at that time, but he probably did observe some effects from aberration of starlight (though they didn't know such a thing existed back then).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the early 1700s, Bradley attempted to make a more precise measurement of parallax, and did obtain significant data, but the variations in the star positions were out of phase with the model. Using the speed of light estimate by Ole Roemer from Hooke's time, he correctly proposed that this was not parallax, but a speed-of-light phenomenon, now known as the aberration of starlight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More info here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/pdf/1208.2061.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://arxiv.org/pdf/1208.2061.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It took another century, until early 1800, for Bessel to measure the first parallax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, aberration of starlight produces the same variation in the positions of all stars, a periodic elliptic &quot;wobble&quot; that is the same for all stellar objects. Parallax produces a similar kind of wobble, but the amplitude depends on the distance to the star, and the phase is different from the aberration's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also other effects, such as nutation, which introduce further perturbations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cseligman.com/text/history/bradley.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://cseligman.com/text/history/bradley.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Aberration per se is easy to explain via the old analogy to running through the rain. When you're standing, rain falls vertically. When you're running, rain appears to come from a different direction. Same with starlight. Like in this diagram (by Professor Courtney Seligman, linked above):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Yynv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, small as that value is, it's still 20 arcsec. The resolving power of early telescopes (and modern amateur telescopes) is 1 arcsec or better - which is more than enough. What you need is a very solid mount with good setting circles. Lacking that, you could build a telescope with a fixed position, very rigid, forever pointing at the same place in the sky - that's what Hooke did - and then you're only limited by resolving power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, the precision required was quite doable with 1700s technology. An amateur could do it today, at very low cost - what you need is persistence to follow through with a long term project.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Could the effect of parallex, proper motion and light aberration of a&#xA;  star be separated clearly?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sure. Aberration is always the same. Parallax depends on the distance, and the phase is different.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is their phase different? They all have a one year period.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Go to the site by Prof. Seligman, link above, and look at the two diagrams showing the aberration and the parallax. In point B (in April), the star is in the middle of the ellipse due to parallax, but it's at the right-hand tip of the ellipse due to aberration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Broadly speaking, parallax is a position-based effect, whereas aberration is a speed-based effect. It makes sense that they are out of phase by 90 degrees (remember trigonometry?)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Proper motion appears as a long-term drift of stellar positions that remains even after you subtract aberration, parallax, nutation, etc. There's a back-and-forth-back-and-forth from aberration and parallax, whereas proper motion keeps going in one direction only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could measure proper motion with relatively short term observations, but it only works for stars that are nearby and are moving fast. If you're willing to wait decades, proper motion is much more easy to measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Generally we should not worry about the light aberration, right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yeah, sort of. You know, at any given moment, the size and the direction of it. So you could simply subtract it from the measured position of the star you're watching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OTOH, it was an additional source of error, until speed of light and the Earth's orbital parameters were very precisely determined. Perhaps it still contributes to the noise added to the data, to some extent - I'm not entirely sure here.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2014-07-29T23:39:50.660" LastActivityDate="2014-07-29T23:39:50.660" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5981" PostTypeId="2" ParentId="4803" CreationDate="2014-07-29T07:17:03.947" Score="2" Body="&lt;p&gt;The local environment is very important to star formation because it requires material as well as something to perturb the nebula to start collapsing and forming star(s). Therefore, both the type of galaxy as well as the location in the galaxy are very important to creating stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, once a star moves from protostar onto the main sequence, external influences don't much affect its development. This is how we can identify &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_kinematics#Stellar_streams&quot; rel=&quot;nofollow&quot;&gt;stellar streams&lt;/a&gt; because the stars come from a similar population when they are created, but some tidal forces cause them to spread out around the galaxy.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-07-29T07:17:03.947" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5982" PostTypeId="2" ParentId="328" CreationDate="2014-07-29T07:37:38.557" Score="4" Body="&lt;p&gt;The first order theoretical limit on stellar size is from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Eddington_luminosity&quot; rel=&quot;nofollow&quot;&gt;Eddington Limit&lt;/a&gt;. As the star collapses it is balances by radiation pressure from fusion. However, the fusion rate scales strongly with density (which is why the most massive stars have extremely short lifetime) so if the star was massive enough, the radiation pressure would probably blow it apart. In fact, this could lead to a &lt;a href=&quot;http://en.wikipedia.org/wiki/Pair-instability_supernova&quot; rel=&quot;nofollow&quot;&gt;pair-instability supernova&lt;/a&gt; and there wouldn't even be a black hole remnant even though the star is so massive.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-07-29T07:37:38.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5983" PostTypeId="2" ParentId="5957" CreationDate="2014-07-29T08:16:14.570" Score="6" Body="&lt;p&gt;Lets assume we're dealing with a superior planet, which is to say, the planet is orbiting at a greater distance than the Earth. This effectively ensures that there is no planetary phase to deal with.&lt;/p&gt;&#xA;&lt;p&gt;Now, the Sun has a luminosity of &lt;span class=&quot;math-container&quot;&gt;$L_{sun}$&lt;/span&gt; such that the Solar flux as seen by the planet is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA; f_{planet} = \frac{ L_{sun} }{ 4 \pi d_s^2 }.&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The cross section of the planet is about &lt;span class=&quot;math-container&quot;&gt;$A = \pi r_p^2$&lt;/span&gt;, so for a given albedo &lt;span class=&quot;math-container&quot;&gt;$a_p$&lt;/span&gt; the reflective luminosity of the planet will be&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{aligned}&#xA; L_{planet} &amp;amp;= a_p f_{planet} A \\ &#xA;            &amp;amp;= a_p \pi r_p^2 \frac{ L_{sun} }{ 4 \pi d_s^2 }.&#xA;\end{aligned}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Given that we know the absolute magnitude of the Sun, the absolute magnitude of the planet follows as&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{aligned}&#xA; V_{planet} &amp;amp;= -2.5 \log_{10}\left[ \frac{ L_{planet} }{ L_{sun} } \right] - V_{sun} \\&#xA;            &amp;amp;= -2.5 \log_{10}\left[ a_p \frac{ r_p^2 }{ 4 d_s^2 } \right] - V_{sun} &#xA;\end{aligned}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The apparent magnitude of the planet as seen from Earth can then be calculated as&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA; m_{planet} = V_{planet} + 5 \log_{10}\left[ d_{e-p} \right] - 5&#xA;$$&lt;/span&gt;&#xA;where the distance between Earth and the planet, &lt;span class=&quot;math-container&quot;&gt;$d_{e-p}$&lt;/span&gt;, should be in parsecs and of course depends on the orbital phase of the two respective planets.&lt;/p&gt;&#xA;&lt;p&gt;So the one additional parameter required was the Solar luminosity. There are probably a bunch of subtle effects with the albedo and the geometry of the system that are not taken into account here, but it should be a fair approximation.&lt;/p&gt;&#xA;" OwnerUserId="494" LastEditorUserId="10483" LastEditDate="2020-07-31T14:58:37.703" LastActivityDate="2020-07-31T14:58:37.703" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="5984" PostTypeId="2" ParentId="4761" CreationDate="2014-07-29T21:05:55.533" Score="3" Body="&lt;p&gt;This is a tough question to answer because the dimensions of your proverbial cube would be affected by the mass inside (as dimensions can only exist in space, so anything that affects space will also bend your cube), and mass does not bend space, it bends spacetime. You have to keep in mind that when talking about special and general relativity, you are talking about four-dimensional spacetime. This is what is curved by mass. For example, due to the curvature of spacetime due to Earth's gravity, you feel the same effects on its surface that you would if you were accelerating upwards at 9.8 m/s^2 through deep space (where we assume there is no gravity). This doesn't mean the physical space is curved - if you draw a straight line on a piece of paper in space and then travel down to Earth's surface, it will remain straight. Instead, spacetime is curved, which can affect objects' paths of motion, but not the objects themselves. Now to your questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The volume inside of the cubes would be exactly the same, but two identical objects flying through each cube would follow two different paths, and if either looked at the other, it would see that the other's clock is moving at a different speed than its own due to time dilation (due to the curvature of spacetime towards mass, time moves more slowly closer to massive objects). We can see evidence of the effect of gravity on paths with gravitational lensing. Light travels in a straight line through spacetime, but when looking at the space around a star, you will actually see objects that should be directly behind the star and blocked from sight. This is because light rays that, in the absence of gravity, would pass next to the star and continue off at an angle are pulled towards the star by it's gravitational field (in reality they're just following the curvature of spacetime around the massive star), so they curve around the star, becoming visible to us, even though, in space alone, they should be blocked by the star.&lt;/li&gt;&#xA;&lt;li&gt;The distribution of the mass has no effect on the gravitational field around it (same with electromagnetic fields). If you look at any equation that has to do with gravity (gravitational potential energy, gravitational force, etc.), you will find a term for the total mass of the object, but unless you are within the bounds of the control volume containing the mass, the distribution does not matter. This is one of the reasons why black holes are so hard to study - we can tell how much mass is inside of them, but without the ability to see inside, we have no other way of detecting their properties.&lt;/li&gt;&#xA;&lt;li&gt;Like I said above, the volume is not affected because space is not warped by gravity, only spacetime. If you made a physical box (or we'll say a cube frame) around empty space and then moved it to a space that contained a star, it would remain the same size and shape (assuming that it was strong enough to withstand the force of gravity pulling it towards the massive star).&lt;/li&gt;&#xA;&lt;li&gt;Neither, it simply curves spacetime.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2024" LastActivityDate="2014-07-29T21:05:55.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5985" PostTypeId="2" ParentId="1965" CreationDate="2014-07-30T05:40:09.317" Score="6" Body="&lt;p&gt;I see two real questions here. First, whether it's possible to have a black dwarf with a companion object. For a given black dwarf, this is unlikely, since the orbits would likely be unstable at the time scale required to produce a black dwarf. Given the size of the universe, however, it's not out of the question. A black dwarf could even capture a companion star after cooling down, although this would be extremely rare.&lt;/p&gt;&#xA;&lt;p&gt;Assuming it's possible, the next question would be whether the supernova produced would be the same as a normal type Ia supernova. The supernova is generally triggered based on the critical (Chandrasekhar) mass which triggers carbon fusion. Since it's dependent on mass, it would likely be a &amp;quot;normal&amp;quot; type 1A supernova. Keep in mind not all 1A supernovae are identical to begin with, although they can still be &amp;quot;normalized&amp;quot; to function as a fairly reliable standard candle.&lt;/p&gt;&#xA;&lt;p&gt;One thing to note is, the accretion process would cause the black dwarf to heat up, meaning it would not remain black.&lt;/p&gt;&#xA;" OwnerUserId="2138" LastEditorUserId="31410" LastEditDate="2020-12-11T00:25:43.360" LastActivityDate="2020-12-11T00:25:43.360" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="5986" PostTypeId="2" ParentId="2230" CreationDate="2014-07-30T07:35:30.790" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/3595/is-there-a-ceiling-for-stable-l4-or-l5-masses/3617#3617&quot;&gt;Is there a ceiling for stable L4 or L5 masses?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your link says orbits about Lagrangian points L4 and L5 are dynamically stable  so long as the ratio of the masses of the two main objects is greater than about 25. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth &amp;lt; 4% Sun = Stable L4,L5 points&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Giant_impact_hypothesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Giant_impact_hypothesis&lt;/a&gt; says that the Lagrangian object would destabilize once above 10% of the Earth's mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;L4 or L5 object &amp;lt; 10% Earth = Stable object&lt;/p&gt;&#xA;" OwnerUserId="1810" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-07-30T07:35:30.790" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5987" PostTypeId="2" ParentId="5949" CreationDate="2014-07-30T07:57:47.953" Score="2" Body="&lt;p&gt;It could have been an Iridium Satellite.  Their solar panels are at an angle such that the sun reflects off them causing them to look like a bright flash that crosses the sky for a few seconds.  I use a site called Heavens Above that allows you to enter your exact location so you can see what objects appear above, at what time and in what direction.  Check that site out and see if any Iridium pass over there.&lt;/p&gt;&#xA;" OwnerUserId="2139" LastActivityDate="2014-07-30T07:57:47.953" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5988" PostTypeId="1" CreationDate="2014-07-30T08:50:00.733" Score="2" ViewCount="117" Body="&lt;p&gt;This may not belong here in astronomy and my apologies if that's the case.&lt;br&gt;&#xA;The hypothetical:&lt;br&gt;&#xA;We find, and are able to travel to (in a reasonably short amount of time), a planet that is for all intents and purposes a second earth: it has close to the same atmosphere, gravitational force, a stable orbit, land and liquid water, preferable temperatures, vegetation and other life , etc...  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a 'War of the Worlds' approach to this, wouldn't it stand to reason that we &lt;em&gt;still&lt;/em&gt; would not be able to land on the planet and exit our spacecraft without suits to protect us because of the fundamental aspect of our immune systems? That is, our bodies have adapted and continue to adapt to Earth's lifeforms but have absolutely no tolerance to another planet of &lt;em&gt;almost&lt;/em&gt; identical properties and therefore would perish shortly after exposure, yes? So I wonder, does anyone know if there is a plan for the hypothetical scenario to gradually expose individuals to the new environment and if so, what is the plan? For example, we could try to collect at least &lt;em&gt;some&lt;/em&gt; of the potentially hazardous bacteria and viruses for vaccination? (maybe, not sure how that works even on Earth)&lt;/p&gt;&#xA;" OwnerUserId="2067" LastActivityDate="2014-07-30T08:50:00.733" Title="Landing on an &quot;earth-like&quot; planet (practical application)" Tags="&lt;earth-like-planet&gt;&lt;life&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2016-12-31T15:30:14.327" ContentLicense="CC BY-SA 3.0" />
  <row Id="5989" PostTypeId="2" ParentId="5924" CreationDate="2014-07-30T09:18:33.487" Score="7" Body="&lt;p&gt;The equation for Doppler shifting towards the blue spectrum (i.e. when you are traveling towards it) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda_b = \lambda_c/(c + v_b)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\lambda_b$ is the shifted wavelength, $\lambda_c$ is the original wavelength, $c$ is the speed of light and $v_b$ is your velocity towards the source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BUT! When you get closer to the speed of light you need to go relativistic!  When moving closer to the speed of light you also need to take other effects into account like time dilation.  So we need to move to an equation that looks more like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_{ob} = v_{source}\times\sqrt{(1+B)/(1-B)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $v_{ob}$ is the observed wavelength, $v_{source}$ is the emitted source radiation and $B$ is your velocity/speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So lets say that someone is shining some UV light on us at about $4\times10^{-7}~\rm{m}$ and we want to see at what velocity this light is shifted into gamma i.e. about $1\times10^{-11}~\rm{m}$.  We can re-arange that formula to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B = [(\frac{v_o}{v_s})^2-1]/[(\frac{v_o}{v_s})^2+1]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $B$ will be a decimal of the speed of light.&#xA;The term $(v_o/v_s)^2$ is ridiculously small, we are talking magnitudes of $10^{-10}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then when you add or subtract 1, you really dont move that far from 1. Just under 1/ just over 1 you get the answer that You have to be going ridiculously close to the speed of light for this to happen.  Like so ridiculously close MATLAB wont calculate the difference and just gives me the value of 1, i.e. the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IN SUMMARY:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes but you have to be going so close to the speed of light it would be physically impossible.&lt;/p&gt;&#xA;" OwnerUserId="2139" LastEditorUserId="2674" LastEditDate="2014-12-09T11:01:04.387" LastActivityDate="2014-12-09T11:01:04.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5990" PostTypeId="1" AcceptedAnswerId="5997" CreationDate="2014-07-30T10:17:17.537" Score="4" ViewCount="435" Body="&lt;p&gt;I have set a science fiction story on a moon, orbiting a gas giant (which orbits its star at approximately the same orbit as the Earth around the Sun), and given this moon its own satellites.  The moon is Earth-sized, and the moon’s satellites are on the order of a large asteroid each (perhaps up to half the size of the Moon … Luna if there was any confusion with the number of times I said ‘moon’ in this post.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a computer nerd, not an astronomer, so I’m wondering if this is something that is at least possible?  Could the orbits of the gas giant around the star, the moon around the gas giant, and the subsatellites around the moon remain stable for the requisite billion or so years for life to develop on the moon?  Could the moon be habitable?&lt;/p&gt;&#xA;" OwnerUserId="2140" LastEditorUserId="464" LastEditDate="2014-07-30T19:05:51.900" LastActivityDate="2014-07-30T19:05:51.900" Title="Could a habitable satellite of a gas giant have a stable subsatellite?" Tags="&lt;orbit&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5992" PostTypeId="2" ParentId="5960" CreationDate="2014-07-30T13:58:25.400" Score="-2" Body="&lt;p&gt;Dark energy is driving the accelerated expansion of our universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it accelerating into itself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it accelerating into nothing, An outer void?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we could see the furthest away accelerating object that would be the edge of the Universe.  Thus there is an edge to the Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, just because we can't look and see today doesn't mean it doesn't exist.&lt;/p&gt;&#xA;" OwnerUserId="2141" LastActivityDate="2014-07-30T13:58:25.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5994" PostTypeId="1" AcceptedAnswerId="5996" CreationDate="2014-07-30T08:25:47.310" Score="17" ViewCount="2546" Body="&lt;p&gt;I was reading this &lt;a href=&quot;http://zeenews.india.com/news/space/milky-way-is-way-smaller-than-thought_951214.html&quot;&gt;article&lt;/a&gt;&#xA; and it says the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Researchers measured the mass of the Milky Way and found that our galaxy is approximately half the weight of a &lt;strong&gt;neighbouring galaxy&lt;/strong&gt; known as Andromeda which has a similar structure to our own. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I was thinking, if we consider a galaxy neighbor to us, then our galaxy should be having an endpoint and the neighboring galaxy should be having a start point. So how do we know which is the starting and ending point of a galaxy, and how do we calculate it?&lt;/p&gt;&#xA;" OwnerUserId="379" OwnerDisplayName="SpringLearner" LastEditorUserId="464" LastEditDate="2014-07-30T19:05:31.400" LastActivityDate="2015-02-18T10:06:00.320" Title="Where does the Milky Way end?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;intergalactic-space&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5995" PostTypeId="2" ParentId="5994" CreationDate="2014-07-30T09:21:15.207" Score="5" Body="&lt;p&gt;Update for clarity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the visible part of a standalone galaxy, the stars can all be measured to orbit that galaxy's core. So if you wanted to measure the furthest extent at this simplified level, it would be very easy. The problem is that there is a lot of mass which is not stars, and which is dark matter. Some of it is so far out, it may be impossible to calculate what it is orbiting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you read the paper, you will see that they chose a limit beyond which the effect of the negligible mass further out was not relevant to this calculation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A galaxy can be thought of as a flattened sphere, as regards mass. Stars, at least for spiral galaxies, tend to lie in a plane, but there is mass orbiting the common centre in all planes. So there is no start point, there is just a need to decide how far out we want to say the galaxy extends. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Milky Way and Andromeda, a decision has been made that is the same for both (the exact decision doesn't appear to have been published for this paper, but as long as it is consistent, then the relative masses will be correct)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For galaxies that are colliding or near to each other it is much more difficult - do you calculate which galaxy a star belongs to by the direction of its motion? Stars may swap from one to the other. From &lt;a href=&quot;https://en.wikipedia.org/wiki/Milky_Way&quot;&gt;Wikipedia's Milky Way page:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Surrounding the Galactic disk is a spherical Galactic Halo of stars and globular clusters that extends further outward, but is limited in size by the orbits of two Milky Way satellites, the Large and the Small Magellanic Clouds, whose closest approach to the Galactic center is about 180,000 ly (55 kpc).[51] At this distance or beyond, the orbits of most halo objects would be disrupted by the Magellanic Clouds. Hence, such objects would probably be ejected from the vicinity of the Milky Way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;tl;dr - it's a pretty arbitrary decision :-)&lt;/p&gt;&#xA;" OwnerUserId="43" OwnerDisplayName="Rory Alsop" LastActivityDate="2014-07-30T10:56:29.997" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5996" PostTypeId="2" ParentId="5994" CreationDate="2014-07-30T15:11:33.850" Score="24" Body="&lt;p&gt;Here's a rough sketch of the &lt;a href=&quot;//en.wikipedia.org/wiki/Milky_Way&quot; rel=&quot;noreferrer&quot;&gt;Milky Way&lt;/a&gt; and the &lt;a href=&quot;//en.wikipedia.org/wiki/Andromeda_Galaxy&quot; rel=&quot;noreferrer&quot;&gt;Andromeda galaxy&lt;/a&gt;,&lt;sup&gt;*&lt;/sup&gt; showing their approximate sizes and distance from each other to scale:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp; &lt;img src=&quot;https://i.stack.imgur.com/mZkT8.png&quot; alt=&quot;Sketch of the Milky Way and Andromeda galaxies, to scale&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the picture (hopefully) illustrates is the incredibly vast gap of empty space &amp;mdash; around 2.5 million light years, to be exact &amp;mdash; between the galaxies, each of which has a diameter of only(!) around 100 thousand light years or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While both galaxies are shaped pretty much like &quot;fuzzy whirlpools&quot; of dust and gas (and, apparently, dark matter), with no precisely defined sharp outer edge, they are still pretty unambiguously separated.  While we may not be able to point to a specific line in space and say that &quot;the Milky Way ends right here&quot;, it's still clear that the Milky Way is over &lt;em&gt;here&lt;/em&gt;, while the Andromeda galaxy is over &lt;em&gt;there&lt;/em&gt;, and there's a huge gap of pretty much absolutely nothing at all in between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, there is &lt;em&gt;some&lt;/em&gt; stuff even in &lt;a href=&quot;//en.wikipedia.org/wiki/Intergalactic_space&quot; rel=&quot;noreferrer&quot;&gt;intergalactic space&lt;/a&gt;, including some &lt;em&gt;very&lt;/em&gt; diffuse gas, a little bit of dust and even the occasional &lt;a href=&quot;//en.wikipedia.org/wiki/Intergalactic_star&quot; rel=&quot;noreferrer&quot;&gt;stray star&lt;/a&gt;.  Still, compared to the galaxies themselves &amp;mdash; which, from a human viewpoint, are &lt;em&gt;already&lt;/em&gt; pretty full of empty space &amp;mdash; the intergalactic medium can be pretty well described as empty.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;*) The little specks in the picture around each major galaxy are meant to represent their smaller satellite galaxies, such as the lesser and greater &lt;a href=&quot;//en.wikipedia.org/wiki/Magellanic_Clouds&quot; rel=&quot;noreferrer&quot;&gt;Magellanic clouds&lt;/a&gt;. Their relative positions and distances, alas, are probably not very accurate.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="464" OwnerDisplayName="Ilmari Karonen" LastActivityDate="2014-07-30T15:11:33.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5997" PostTypeId="2" ParentId="5990" CreationDate="2014-07-30T17:38:30.670" Score="9" Body="&lt;p&gt;This is a fun question, so I spend some time thinking about it. This is what I came up with.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Is this possible?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Yes, this is certainly possible. Just consider the Solar System itself. We have a massive central body, the Sun, and several tiny subsystems (planets with their moons) orbiting it.  &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Is it stable?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Again, from the simple observation that the Solar System exists in its current form, we know that such a system can remain more or less stable over a significant portion of the stellar lifetime. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more difficult problem, I think, is if can it be formed in the first place.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Could it be habitable?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This is a tricky question as it critically depends distances between the several orbiting bodies. Lets consider the orbital mechanism and derive from what we know of the Solar System.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the following I've assumed the gas planet is Jupiter and the fictional planetary system consists of a Sun', Jupiter', Earth' and Moon' such that the apostrophe denotes the fake object. These fake objects are identical to the real equivalents, except for their orbital configuration and relative distances. If you want a different system you can follow the same line of reasoning, but plug in different numbers for the fake objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we know that the Earth-Moon system exists in a stable orbit around the Sun, which means that at 1 AU the gravitational force of the Sun is not strong enough to destabilize the Moon orbit. From this we can find a lower limit on the gravitational force that Jupiter' may exert on the Moon'&#xA;\begin{aligned}&#xA;  F_{J'M'} &amp;amp;= F_{SM} \\&#xA;  \frac{ M_{J'} M_{M'}}{r_{J'M'}^2} &amp;amp;= \frac{ M_{S} M_{M}}{r_{SM}^2}&#xA;\end{aligned}&#xA;Solving this equation we get&#xA;\begin{aligned}&#xA; r_{J'M'} &amp;amp;= r_{SM} \left( \frac{ M_{J'} M_{M'} }{ M_{S} M_{M} } \right)^{1/2} \\&#xA;	  &amp;amp;= r_{SM} \left( \frac{ M_{J'} }{ M_{S} } \right)^{1/2}&#xA;\end{aligned}&#xA;which evaluates to about 0.03 AU. So the Earth'-Moon' system should have an orbital radius around Jupiter' of at least 0.03 AU. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the question is if the Earth-Moon system can remain bound in its orbit around Jupiter. To answer this we can safely ignore the Moon' and assume it remains tightly bound to the Earth'. The potentially destabilizing force on the Jupiter'-Earth' system is again the gravitational pull of the Sun', but now on the Earth'. We can do a similar trick as before, but now we need to account for a different mass ratio. So lets require that the fraction of forces be equal in both cases:&#xA;$$&#xA;  \frac{F_{SM}}{F_{EM}} = \frac{F_{S'E'}}{F_{J'E'}}&#xA;$$&#xA;with some algebra this gives&#xA;$$&#xA; r_{S'J'} = r_{J'E'} \frac{r_{SM}}{r_{EM}} \left(\frac{M_E}{M_J}\right)^{1/2}&#xA;$$&#xA;which for $r_{J'E'} = $0.03 AU evaluates to a minimum orbital radius of about 0.7 AU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This greatly surprised me as it means you can comfortably fit the entire Jupiter'-Earth'-Moon' system inside the habitable zone, meaning that in principle life should be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the seasons on such a planet would be rather extreme.&lt;/p&gt;&#xA;" OwnerUserId="494" LastActivityDate="2014-07-30T17:38:30.670" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5998" PostTypeId="1" AcceptedAnswerId="5999" CreationDate="2014-07-31T07:23:21.643" Score="7" ViewCount="499" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble_Ultra-Deep_Field&quot;&gt;Hubble ultra-deep field&lt;/a&gt; image is a photograph of an area of sky equivalent to a 1mm x 1mm piece of paper held a metre away from the eye (one thirteen millionth of the entire sky). It contains an estimated 10,000 galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If galaxies were uniformly distributed, and the HUDF were representative of the entire sky, that would mean there were approximately 130 billion galaxies, which falls between estimates I have read of between 100 and 200 billion as the total number of galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, my understanding is that galaxies occur in clusters, which would seem to imply that some areas of the sky should be more densely populated than others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is the HUDF representative of the entire sky, or are some areas much more and much less densely populated?&lt;/p&gt;&#xA;" OwnerUserId="2144" LastEditorUserId="1862" LastEditDate="2014-09-02T19:39:28.687" LastActivityDate="2014-09-05T01:59:18.047" Title="Is the Hubble Ultra-Deep Field image representative of the entire sky?" Tags="&lt;galaxy&gt;&lt;universe&gt;&lt;galaxy-cluster&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5999" PostTypeId="2" ParentId="5998" CreationDate="2014-07-31T08:17:10.400" Score="8" Body="&lt;p&gt;Short answer: both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HUDF is representative on the type of galaxies we can find at that distance (and remember that distance also means time ago).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, we do not really &lt;em&gt;know&lt;/em&gt; how the galaxies are grouped beyond certain level, but we &lt;em&gt;think&lt;/em&gt; they are grouped on so-called filaments, leaving enormous empty voids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BRhGf.gif&quot; alt=&quot;Structure of the Universe at large scale, z=0 (simulation)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this sense, HUDF's density does not represent the entire sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cosmicweb.uchicago.edu/filaments.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://cosmicweb.uchicago.edu/filaments.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-07-31T08:17:10.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6000" PostTypeId="2" ParentId="5938" CreationDate="2014-07-31T16:56:11.553" Score="4" Body="&lt;p&gt;It depends on the composition of the atmosphere. In a helium (or other light gas) atmosphere, your voice would have a higher pitch. If the gasses are heavier, your voice drops.&lt;br&gt;&#xA;The density is also a factor: on Mars, sounds can't be as loud as on Earth. Sound is transmitted as pressure differences, and the largest pressure difference you can get is that between the atmospheric pressure and a vacuum.&lt;br&gt;&#xA;The speed of sound also changes with the composition and pressure of the atmosphere. This would change the way you perceive the distance and direction of an object that makes a sound: we normally use the tiny differences in time between a sound reaching our left ear and our right ear. A different speed of sound screws up those time differences. &lt;/p&gt;&#xA;" OwnerUserId="361" LastActivityDate="2014-07-31T16:56:11.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6001" PostTypeId="1" AcceptedAnswerId="6009" CreationDate="2014-07-31T16:56:15.217" Score="6" ViewCount="1424" Body="&lt;p&gt;Why do the planets have roughly the same inclination with respect to Earth?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the inclinations, which I got from &lt;a href=&quot;http://www.astronomynotes.com/tables/tablesb.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.astronomynotes.com/tables/tablesb.htm&lt;/a&gt;, we see that Mercury has the highest inclination at about 7 degrees with respect to Earth.  Why are there no inclinations at approximately 90 degrees, or even slightly less, like 45 or 30?  My guess is after some long time, gravitational influence causes all the planets to line up.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question two is then: Do we know of any planetary systems where the planets have high degrees of inclination with respect to each other?&lt;/p&gt;&#xA;" OwnerUserId="2151" LastEditorUserId="25988" LastEditDate="2019-05-22T00:40:44.477" LastActivityDate="2019-05-22T00:40:44.477" Title="Inclination of planets" Tags="&lt;earth&gt;&lt;planet&gt;&lt;inclination&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="6002" PostTypeId="2" ParentId="4732" CreationDate="2014-07-31T19:48:47.743" Score="6" Body="&lt;p&gt;I can attempt to address the second part of your initial question (*&quot;Is it a particle, a wave,...?&quot;) Einstein's theory of general relativity states that mass and energy bend space-time. Space-time, in turn, tells matter how to move (John Wheeler put this more elegantly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This concept is completely different from the theories of the other three fundamental forces (electromagnetism, and the strong and weak nuclear forces). In these quantum theories, forces are mediated by particles called gauge bosons. Electromagnetism is carried by photons, the strong nuclear force by gluons, and the weak nuclear force by W+, W-, and Z bosons. There have been many attempts to find a quantum theory of gravity - that is, to use quantum principles to construct a field theory of gravity. In these theories, gravity would indeed be mediated by a particle, dubbed the &lt;em&gt;graviton&lt;/em&gt;. String theory is one example of these theories; the physics community is divided about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You probably have heard other terms tossed around that embody interesting concepts. A &lt;em&gt;gravitational wave&lt;/em&gt; is essentially a ripple in space-time emitted by an object or system of objects. There are strict restrictions over what sort of objects can emit these waves; binary neutron stars are one consistently cited example. These waves should &lt;em&gt;not&lt;/em&gt; be confused with the aforesaid hypothesized gravitons; while gravitational waves carry energy, they do not &quot;mediate&quot; gravity. They should also not be confused with the unrelated &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravity_wave&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravity waves&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, basically, the predominant view in the physics community is that general relativity is the best description of gravity; at the moment, gravity is considered to be the bending of space-time, so it is indeed &quot;something else entirely.&quot; Many theories of quantum gravity, including string theory, however, attempt to create particles called gravitons as force-carrying bosons. If evidence is found relating to these theories, then we may well learn whether or not gravitons do exist. One more thing about waves: Because of the quantum concept of wave-particle duality, any particle can be described as a wave, which has a wave function. So if gravity is carried by a particle, then it is also due carried by a wave!&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-11-03T19:45:35.620" LastActivityDate="2016-11-03T19:45:35.620" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6003" PostTypeId="2" ParentId="2240" CreationDate="2014-07-31T20:23:12.020" Score="10" Body="&lt;p&gt;As others have said, mathematically, a singularity is when there is an attempt to divide by zero. Take, for example a Schwarzschild black hole. This is a black hole that has no electric charge or angular momentum; tt is the simplest type of black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to general relativity, gravity is the bending of spacetime. The curvature of space can be expressed using something called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Metric_tensor_(general_relativity)&quot; rel=&quot;noreferrer&quot;&gt;metric&lt;/a&gt;, which is similar to the distance formula we use in geometry. &lt;a href=&quot;https://en.wikipedia.org/wiki/Minkowski_space#Minkowski_metric&quot; rel=&quot;noreferrer&quot;&gt;A basic metric describing &quot;flat&quot; Minkowski four-dimensional spacetime&lt;/a&gt; looks like this&#xA;$$\mathrm{d}s^2 = -c^2\mathrm{d}t^2 + \mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2$$&#xA;This is the metric used in special relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, metrics can also be expressed in a variety of coordinate systems. Take, for instance, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_metric&quot; rel=&quot;noreferrer&quot;&gt;Schwarzschild metric&lt;/a&gt;, describing spacetime outside a massive body. In spherical coordiantes, it is&#xA;$$\mathrm{d}s^2 = -\left(1-\frac{r_s}{r}\right)c^2\,\mathrm{d}t^2 + \left(1-\frac{r_s}{r}\right)^{-1}\mathrm{d}r^2 + r^2\left(\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\phi^2\right)$$&#xA;There are other ways of writing this metric that make my next point better, but this form still works. There are two singularities in this metric: $r = 0$ and $r = r_s$, where $r_s$ is called the Schwarzschild radius. A singularity is basically a point in the equation where you are forced to divide by zero (Think of the graph of $y = 1/x$. You reach a singularity at $x = 0$). However, only one is a true physical singularity, at $r = 0$, while $r = r_s$ is called a &lt;em&gt;coordinate singularity&lt;/em&gt;, which arises from our choice of coordinates. There are different ways of expressing the Schwarzschild metric that make this clearer; one example is in Eddington-Finkelstein coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r = 0$ is the only true physical singularity, as I stated above; we can find this from analysis of the Riemann curvature tensor, which I won't get into because I have learned that I'm not exactly excellent at explaining tensors, and because it isn't central to my answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, basically, the solutions to Einstein's equations for black holes only tell us that a singularity is a point where there is infinite curvature in space and time. We only know of their existence through mathematical equations, as opposed to direct experimental discovery. However, singularities can also simply be results of the equations that don't relate to the &quot;infinite curvature&quot; singularities that are generally referred to. Singularities in equations are not always singularities in the physical world.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-11-03T19:49:36.130" LastActivityDate="2016-11-03T19:49:36.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6004" PostTypeId="2" ParentId="6001" CreationDate="2014-08-01T04:04:51.397" Score="2" Body="&lt;p&gt;It starts with the formation of the solar system.  The solar system started off in a large cloud of gas, under gravity, this cloud started to collapse.  Most of the mass was concentrated at the centre which eventually formed into the sun.  Due to the conservation of angular momentum, as the cloud collapse it started to spin and form into a disk with a bulge at the centre where the sun was forming.  The resulting planets were formed in this disk and that's why most of the planets have roughly the same inclination and why they all orbit in the same direction.  We do have some bodies that we have captured that came from outside of the solar system that has an extreme inclination with respect to the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your second question, As far as I know we have not found any other systems that have extreme inclinations as our methods of detection aren't that advance to tell much detail about other planetary system other than probably number of planets, size of planets and so forth.  But it could be possible if a rouge planet that has no star is captured by a star.&lt;/p&gt;&#xA;" OwnerUserId="2155" LastActivityDate="2014-08-01T04:04:51.397" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6005" PostTypeId="1" AcceptedAnswerId="10260" CreationDate="2014-08-01T07:05:43.360" Score="13" ViewCount="1191" Body="&lt;p&gt;Most stars which are visible to the naked eye are within 1,000 light years. The Sun is inside the Orion arm which has a diameter of about 3,500 light years. Thus, all stars (with very few exceptions) we can see unaided are inside the Orion arm and should be equally distributed in all directions. The structure of the galaxy is too large to affect the distribution of visible stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, we do see the Milky Way disk. What is it we see? The dim light of stars we can't make out individually? Or star light reflected by gas clouds? And is the density of stars to the unaided eye really not higher in the plane of the Milky Way? (Since I'm living in a light polluted city I can't easily check it out myself)&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-03-25T23:10:34.160" Title="Aren't there more naked-eye-visible stars in the Milky Way plane?" Tags="&lt;star-gazing&gt;&lt;naked-eye&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6006" PostTypeId="1" CreationDate="2014-08-01T07:49:56.147" Score="6" ViewCount="1847" Body="&lt;p&gt;Images from the Hubble Space Telescope have widely different angles of view. Is this only the result of composing multiple images, or can the telescope itself change its &quot;zooming&quot; optically, mechanically? How does the ability to change the angle of view differ between the major scientific telescopes of today and the near future, space and ground based?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what about cameras on interplanetary probes? Can for example New Horizons optically change its zooming as it passes Pluto?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2014-08-01T08:44:38.350" LastActivityDate="2014-08-01T20:58:22.923" Title="How do telescopes &quot;zoom&quot; and change angle of view?" Tags="&lt;telescope&gt;&lt;optics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6007" PostTypeId="1" AcceptedAnswerId="6008" CreationDate="2014-08-01T11:47:04.033" Score="4" ViewCount="55" Body="&lt;p&gt;How can the &lt;a href=&quot;http://www.srl.caltech.edu/ACE/&quot; rel=&quot;nofollow&quot;&gt;Advanced Composition Explorer&lt;/a&gt; (ACE) warn of incoming solar storms when is is invested with radiation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Discovery.com article &lt;a href=&quot;http://news.discovery.com/space/history-of-space/solar-storm-warning-satellite-120309.htm&quot; rel=&quot;nofollow&quot;&gt;Solar Storm Warning Satellite on Last Legs&lt;/a&gt; states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Until the sun's free-flying and highly energetic outbursts, known as&#xA;  coronal mass ejections, hit the ACE spacecraft, forecasters don't know&#xA;  the orientation of their embedded magnetic fields.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I read that it can give 15 minutes in advance warning to Earth of the presence of an incoming solar storm. However to do that, it sends a radio signal which goes at the same speed of the solar storm so it will presumably take 15 minutes as well?&lt;/p&gt;&#xA;" OwnerUserId="2158" LastEditorUserId="13" LastEditDate="2014-08-01T19:21:57.387" LastActivityDate="2014-08-01T19:21:57.387" Title="How can the Advanced Composition Explorer (ACE) warn of incoming solar storms?" Tags="&lt;the-sun&gt;&lt;artificial-satellite&gt;&lt;solar-storm&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6008" PostTypeId="2" ParentId="6007" CreationDate="2014-08-01T13:38:25.557" Score="6" Body="&lt;p&gt;Solar flare X-rays travel at light speed, but the solar wind from CME's travels at merely 600 to 2000 kilometers per second. It's the wind that causes aurora and &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_wind#Coronal_mass_ejection&quot;&gt;solar storms&lt;/a&gt;, depending on the orientation of its magnetic filed.&#xA;Here's a &lt;a href=&quot;http://www.spaceweatherlive.com/en/help/how-do-we-know-if-a-cme-is-earth-directed-and-when-its-going-to-arrive&quot;&gt;SOHO view of a CME&lt;/a&gt;. It's clear the gas is moving slower than light. You can tell if the earth is in the path of an ejection by how near to symmetrically the gas cloud expands around the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to answer your question, the satellite measures magnetic field direction, and can give 15 minutes warning because the CME gas is moving much slower than light.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2014-08-01T13:38:25.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6009" PostTypeId="2" ParentId="6001" CreationDate="2014-08-01T20:08:44.893" Score="8" Body="&lt;p&gt;While the other answer here and the link provided by Jeremy give excellent explanations, I believe a bit more nuanced reasoning is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the theory of planet formation is currently still incomplete, it is generally accepted that planets form in a so-called proto-planetary disk as a part of stellar formation process. This is backed up by several observation of such disk and even by&#xA;directly observed planets in systems where remnants of the disk are still present (most notably &lt;a href=&quot;http://en.wikipedia.org/wiki/Fomalhaut_b&quot;&gt;Fomalhaut-b&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, exactly how these planets form is still unclear. A popular theory is that these planets form in place and therefore keep the orientation and inclination of the disk. This beautifully explains the Solar System and specifically why all planets have roughly the same orbital inclination. But that's not surprising as the model was specifically designed to do just that! Yet the Solar System does not have to be the only possible type of planetary  system, or even the most common for that matter. Indeed the study of exoplanets has confirmed that very different planetary systems are possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative theory is that planets are created far away from the star and migrate inwards. During this migration multiple planets can interact with one-another and be pushed out of the disk plane to higher inclinations. They could even flip over completely and become retrograde.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to come back to the actual question; the Solar planets probably have about the same inclination simply because they were created that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are also right in saying that gravity will tend to align the planets. How this works physically is that the star can exerts a torque on the planet, which produces a tidal effect (like how the Moon creates &lt;a href=&quot;http://en.wikipedia.org/wiki/Tide&quot;&gt;tides&lt;/a&gt; on Earth). Over time this aligns the angular momentum of the planet to that of the star by pushing the planet towards the equatorial plane of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your second question, yes, there are probably planetary systems which are misaligned with respect to the star. At least there are certainly planets that are not aligned with the equatorial plane of the star. The well named &lt;a href=&quot;http://en.wikipedia.org/wiki/HAT-P-11b&quot;&gt;HAT-P-11b&lt;/a&gt; is a known example, but there are a lot more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that for exoplanets the important parameter is not the inclination (which for exoplanets is the angle between the normal of the orbit and the line of sight), but rather the &lt;a href=&quot;http://en.wikipedia.org/wiki/Axial_tilt&quot;&gt;obliquity&lt;/a&gt; or tilt of the planet rotation with respect to the rotation of the star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for a third twist to this story, very recently there was a &lt;a href=&quot;http://www.nature.com/nature/journal/v511/n7511/full/nature13521.html#ref3&quot;&gt;discovery&lt;/a&gt; of an entire proto-planetary disk that is misaligned with respect to the star, which opens the door to even stranger configurations.&lt;/p&gt;&#xA;" OwnerUserId="494" LastActivityDate="2014-08-01T20:08:44.893" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6010" PostTypeId="2" ParentId="6006" CreationDate="2014-08-01T20:58:22.923" Score="10" Body="&lt;p&gt;Telescopes tend to have a fixed focal length. What changes is the size of the sensor in the instrument used. If a small sensor is used, then a smaller section of the field of view is exposed, resulting in a narrower field of view being imaged than the equipment is capable of. If a larger sensor is used, more of the field of view of the telescope is utilised. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional optics in the light path that are associated with the specific instrument being used on the telescope will affect the field of view also. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Telescopes do not offer the capability of 'zooming' as you understand it from your consumer digital camera.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-08-01T20:58:22.923" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6011" PostTypeId="1" AcceptedAnswerId="6012" CreationDate="2014-08-02T01:59:16.453" Score="10" ViewCount="416" Body="&lt;p&gt;There's been another report of &lt;a href=&quot;http://hubblesite.org/newscenter/archive/releases/2014/33/full/&quot;&gt;gravitational lensing&lt;/a&gt; in the news.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_lens#mediaviewer/File:A_Horseshoe_Einstein_Ring_from_Hubble.JPG&quot;&gt;Wikipedia&lt;/a&gt; has a good image of an example of gravitational lensing taken by the Hubble Space Telescope.  This brings me to wonder that, since they understand the math behind the lensing, shouldn't they be able to reconstitute the original image of the galaxy &lt;em&gt;behind&lt;/em&gt; the galaxy that is acting as a lens?  That is, take the image and run it through a computer program that reconstructs the image using known parameters of the mass and the position of the lensing galaxy?&lt;/p&gt;&#xA;" OwnerUserId="1962" LastActivityDate="2014-09-02T01:47:51.857" Title="With gravitational lensing, is it possible to mathematically compute the correct image of the galaxy that is being distorted?" Tags="&lt;gravitational-lensing&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6012" PostTypeId="2" ParentId="6011" CreationDate="2014-08-02T07:18:02.100" Score="6" Body="&lt;p&gt;That would be possible theoretically. However, in practice, the lensed images are highly non-linearly distorted and due to the limits in resolution, they are often limited to a thickness of a few pixels at best. Due to the non-linear transformations, this would cause the reconstructed original image to be even blurrier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, I don't see any real need of reconstructing the original image. You can get most parameters (like a rough estimate of the size and type of the background galaxy) from the distorted image anyway.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2014-08-02T07:18:02.100" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6013" PostTypeId="2" ParentId="6011" CreationDate="2014-08-02T10:33:56.380" Score="4" Body="&lt;p&gt;The method of doing this is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Deconvolution&quot; rel=&quot;nofollow&quot;&gt;Deconvolution&lt;/a&gt;. &lt;strong&gt;In theory&lt;/strong&gt;, yes, it's possible if you know the exact distortion function, and an exact result image - but, &lt;strong&gt;in reality&lt;/strong&gt;, the image you have is not exact, so the result of deconvolution may not be very useful.&lt;/p&gt;&#xA;" OwnerUserId="1437" LastEditorUserId="1437" LastEditDate="2014-09-02T01:47:51.857" LastActivityDate="2014-09-02T01:47:51.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6014" PostTypeId="1" CreationDate="2014-08-02T12:40:12.290" Score="5" ViewCount="3575" Body="&lt;p&gt;I have a mind puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can there be an infinity of stars in the Universe ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 2 opposed reasonings, yet I don’t find any flaw in them. Can you help me here ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer A : No, the number of stars cannot be infinite.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the whole Universe, at an instant &lt;em&gt;t&lt;/em&gt;, there are W molecules of water. This number may be very big, but it has a definite value. The same goes for stars. Each of all the stars at the instant &lt;em&gt;t&lt;/em&gt; has her own size, and we can give her a name. We can list all of them, and we can sort them from nearest to farthest — if we are omniscient, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer B : Yes, the number of stars can be infinite.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space has no finite volume. Space has no ending wall. If I advance into space, I can always go further. Wherever I go, space is &lt;em&gt;by default&lt;/em&gt; filled with… vacuum. But it could be otherwise. [I am not sure of this point.] Space could be &lt;em&gt;by default&lt;/em&gt; filled with… air. So there would be +∞ molecules of air in the Universe. We can apply this reasoning to stars. Let’s say there is, in average, 1 star / 10&lt;sup&gt;60&lt;/sup&gt; m&lt;sup&gt;3&lt;/sup&gt;. So, in the total infinite volume of space, there would be +∞ stars. When I advance into space, I would keep meeting new stars, without end. Like in these video games in which mountains keep appearing as far as you walk.&lt;/p&gt;&#xA;" OwnerUserId="1580" LastActivityDate="2017-04-16T18:59:16.663" Title="Can there be an infinity of stars in the Universe?" Tags="&lt;star&gt;&lt;universe&gt;&lt;space&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" ClosedDate="2019-03-12T21:31:10.597" ContentLicense="CC BY-SA 3.0" />
  <row Id="6015" PostTypeId="2" ParentId="6014" CreationDate="2014-08-02T12:57:44.127" Score="5" Body="&lt;p&gt;Theoretically yes, there can be infinitely many stars. Since this space you're talking about is better described by general relativity and Einstein's field equations. One of its solutions describes the universe as being spatially infinite. &lt;/p&gt;&#xA;" OwnerUserId="2167" LastActivityDate="2014-08-02T12:57:44.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6017" PostTypeId="2" ParentId="6001" CreationDate="2014-08-02T14:21:05.507" Score="4" Body="&lt;p&gt;To answer your second question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upsilon Andromedae c and d have mutual inclination of 30 degrees&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Upsilon_Andromedae&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Upsilon_Andromedae&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you cannot calculate mutual inclination by subtracting one inclination from another from exoplanet data catalogs because the &quot;inclination&quot; used in exoplanet studies is a 2D line-of-sight-to Earth inclination. Mutual inclination must be got from full 3D measurements of a system using astrometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The GAIA telescope will probably find more such non-coplanar systems with large mutual inclinations.&lt;/p&gt;&#xA;" OwnerUserId="2169" LastEditorUserId="2169" LastEditDate="2014-08-03T00:21:37.627" LastActivityDate="2014-08-03T00:21:37.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6018" PostTypeId="1" AcceptedAnswerId="6019" CreationDate="2014-08-02T17:26:09.307" Score="4" ViewCount="116" Body="&lt;p&gt;In an array telescope configuration, like &lt;a href=&quot;http://en.wikipedia.org/wiki/Very_Large_Array&quot; rel=&quot;nofollow&quot;&gt;VLA&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Allen_Telescope_Array&quot; rel=&quot;nofollow&quot;&gt;ATA&lt;/a&gt;, what dictates the distance between each dish?&lt;/p&gt;&#xA;" OwnerUserId="2048" LastActivityDate="2014-08-03T09:58:34.563" Title="Distance between each dish in an array telescope configuration" Tags="&lt;telescope&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6019" PostTypeId="2" ParentId="6018" CreationDate="2014-08-02T19:25:26.977" Score="3" Body="&lt;p&gt;Physically there is nothing dictating the distances between dishes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These telescope arrays are &lt;a href=&quot;http://en.wikipedia.org/wiki/Interferometry&quot; rel=&quot;nofollow&quot;&gt;interferometers&lt;/a&gt;. By combining the signal of the two dishes you can achieve a spacial resolution of a giant virtual dish that is about as large as the distance (called the baseline) between the two real telescopes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nice example of the power of this technique is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Very_long_baseline_interferometry&quot; rel=&quot;nofollow&quot;&gt;VLBI&lt;/a&gt;, which is essentially such an array only the individual dishes are actually just observatories all over the world. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more detail on how this all works exactly take a look at the wiki pages on &lt;a href=&quot;http://en.wikipedia.org/wiki/Aperture_synthesis&quot; rel=&quot;nofollow&quot;&gt;aperture synthesis&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Astronomical_interferometer&quot; rel=&quot;nofollow&quot;&gt;astronomical interferometry&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="494" LastEditorUserId="494" LastEditDate="2014-08-03T09:58:34.563" LastActivityDate="2014-08-03T09:58:34.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6020" PostTypeId="2" ParentId="6018" CreationDate="2014-08-02T20:41:03.053" Score="2" Body="&lt;p&gt;Increasing the separation of the dishes makes the effective aperture of the array larger, so it increases the spatial resolution. However, the overall collection area of the aperture has not increased, so the signal-to-noise ratio will not increase.&lt;/p&gt;&#xA;" OwnerUserId="2170" LastActivityDate="2014-08-02T20:41:03.053" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6021" PostTypeId="2" ParentId="1997" CreationDate="2014-08-02T21:41:59.513" Score="-5" Body="&lt;p&gt;Ok, it makes a lot of sense that the gravity of the disc of the Milky Way is pulling stars up and down as they go about their circum-galactic orbit.  But this wouldn't explain why recent 3-D observations of the nearest stars using the FLAMES-GIRAFFE spectrograph on ESO’s Very Large Telescope and the IMACS spectrograph at the Las Campanas Observatory showed a definite wave-structure in the motions of stars orbiting the Galactic plane.  In other words, most of the stars in the disc are following each other in a chain or train, as if they are bobbing up and down on a CURRENT.  Which means they most certainly are oscillating on a current.  And this was predicted in 1978!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our Sun produces a magnetic field stretching out along the equatorial plane in the heliosphere.  This field stretches throughout the Solar System where it is called the “Interplanetary Magnetic Field”.  In 1965 John M., Wilcox and Norman F. Ness published their finding of the “Heliospheric Current Sheet” which showed that the Sun’s rotating magnetic field is constantly producing waves in the plasma of the interplanetary medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These waves form a “Parker Spiral” and are described in terms of an electromagnetic current but they are also mechanical waves which cause the planets themselves to oscillate up and down as they orbit the Sun.  In 1978 Hannes Alfven and Per Carlqvist suggested that there is a similar “Galactic Current Sheet” carrying an electric current of 10^17 to 10^19 amperes through the plane of symmetry of the Galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok?  That pretty much solves the mystery of the oscillation of the Milky Way's stars.  But the problem here is that (ahem, cough cough) OUR SOLAR SYSTEM IS NOT PART OF THE MILKY WAY.  In 1994 it was discovered that we are actually part of the Sagittarius Dwarf Elliptical Galaxy, or Sag-DEG for short, which is in a 500 million year POLAR ORBIT around the Milky Way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you ever wondered why they say our Solar Apex is near Vega, but Vega itself is moving towards us almost twice as fast as we are moving towards it???  Well, in the late 80's it was discovered that nearly all of the stars orbiting the Milky Way seem to be &quot;raining&quot; down on our position.  Which could only mean that our Solar System is moving upwards, out of the Milky Way.  Sorry to inform you guys, but even though the stars of the Milky Way oscillate up and down during their 250 million year orbit, we aren't part of that dance.  Our own path will take us high above the Galaxy, with a spectacular view at apogalacticon, and then back down again.  &lt;/p&gt;&#xA;" OwnerUserId="2171" LastActivityDate="2014-08-02T21:41:59.513" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6022" PostTypeId="2" ParentId="1830" CreationDate="2014-08-03T00:19:54.417" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://planetplanet.net/2014/05/23/building-the-ultimate-solar-system-part-5-putting-the-pieces-together/&quot; rel=&quot;nofollow&quot;&gt;Here's a blog post&lt;/a&gt; with an exploration of how many Earth-like planets you could pack into the habitable zone. &lt;/p&gt;&#xA;" OwnerUserId="2169" LastEditorUserId="1469" LastEditDate="2014-08-03T20:18:52.503" LastActivityDate="2014-08-03T20:18:52.503" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6023" PostTypeId="2" ParentId="987" CreationDate="2014-08-03T02:22:47.377" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If a star passes through a molecular cloud then a new disc could form providing another chance for planet formation. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Old pre-main-sequence Stars: Disc reformation by Bondi-Hoyle accretion.&#xA;&lt;a href=&quot;http://arxiv.org/abs/1405.6051&quot; rel=&quot;nofollow&quot;&gt;Reference 1&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If two stars merge then they throw off an excretion disc which could form new planets.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A binary merger origin for inflated hot Jupiter planets.&#xA;&lt;a href=&quot;http://arxiv.org/abs/1102.3336&quot; rel=&quot;nofollow&quot;&gt;Reference 2&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do you count brown-dwarfs and sub-brown-dwarfs as stars, and their satellites as planets? &lt;a href=&quot;http://en.wikipedia.org/wiki/Planets_orbiting_brown_dwarfs&quot; rel=&quot;nofollow&quot;&gt;Reference 3&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If a star has a high mass or high metal content it would burn through the main-sequence very quickly to become a white dwarf while still being within the star cluster it formed in. It then could pass through a cloud, as per item 1 on this list, and form a new disc.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Some Stars are Totally Metal.&#xA;&lt;a href=&quot;http://arxiv.org/abs/1406.5509&quot; rel=&quot;nofollow&quot;&gt;Reference 4&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In a supernova some of the ejecta may fail to escape and so create a fallback disc. This could form planets around neutron stars. (and black holes)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A debris disk around an isolated young neutron star.&#xA;&lt;a href=&quot;http://www.nature.com/nature/journal/v440/n7085/full/nature04669.html&quot; rel=&quot;nofollow&quot;&gt;Reference 5&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As you mention in an answer to your own question, the common envelope phase in a binary star can form a new disc. This can be extended to a multiple-star system: whenever one of the stars in the system turns into a red-giant, AGB-star, or planetary-nebula this creates a common envelope if the stars are close enough together, allowing new discs to form creating new generations of planets.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Planets in evolved binary systems.&#xA;&lt;a href=&quot;http://arxiv.org/abs/1012.0572&quot; rel=&quot;nofollow&quot;&gt;Reference 6&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2169" LastEditorUserId="1469" LastEditDate="2014-08-03T20:16:31.123" LastActivityDate="2014-08-03T20:16:31.123" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6026" PostTypeId="1" AcceptedAnswerId="6028" CreationDate="2014-08-03T13:30:53.390" Score="8" ViewCount="739" Body="&lt;p&gt;I wasn't around when man landed on the moon in 1969. When I see the moon, I always wonder, were people able to see the rocket?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yesterday, I looked at the moon in daylight and wondered again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: how far were people able to see the rocket after it launched?&lt;/p&gt;&#xA;" OwnerUserId="2177" LastEditorUserId="67" LastEditDate="2014-08-04T19:59:23.323" LastActivityDate="2015-07-12T01:32:38.183" Title="Visibility of Apollo 11 module" Tags="&lt;the-moon&gt;&lt;moonlanding&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6028" PostTypeId="2" ParentId="6026" CreationDate="2014-08-03T15:26:15.777" Score="8" Body="&lt;p&gt;No, the orbiting rocket was not visible to the naked eye.&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Apollo_Command/Service_Module&quot;&gt;Orbiter&lt;/a&gt; plus &lt;a href=&quot;http://en.wikipedia.org/wiki/Apollo_Lunar_Module&quot;&gt;lem&lt;/a&gt; were about 65 square meters in size, about 25 X 25 feet.&#xA;At lunar distance they subtended about 0.000001 square degrees, and be at most 5 to 10 times as bright as the lunar surface; albedo 0.5 to 1 vs 0.1.&#xA;A speck 1/1000th of a degree on a side, and a little brighter than the moon is not visible in daylight. Likely it's well below the noise level in a daylight adapted eye.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2014-08-03T15:26:15.777" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6030" PostTypeId="2" ParentId="2468" CreationDate="2014-08-04T15:46:56.010" Score="4" Body="&lt;p&gt;One way to figure out if one (or both) of the objects is a black hole, neutron star, white dwarf, or other compact object would be to try to measure its mass. For example, a neutron star and a white dwarf are both compact stellar remnants. However, there is one decisive factor that determines which type of stellar remnant a progenitor star will become: the remnant's mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Chandrasekhar_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chandrasekhar limit&lt;/a&gt; ($\sim1.39M_{\odot}$) is the maximum mass of a white dwarf; the &lt;a href=&quot;https://en.wikipedia.org/wiki/Tolman%E2%80%93Oppenheimer%E2%80%93Volkoff_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tolman-Oppenheimer-Volkoff limit&lt;/a&gt; ($\sim1.5-3.0M_{\odot}$) is the maximum mass of a neutron star. The large uncertainty in the latter means that it's not particularly useful for identifying compact objects that could be neutron stars or low-mass black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you know the mass of one of the objects - and let's assume that it's well within the mass range of one of the classes of stellar remnants - you might be able to figure out which type it is. How would you measure the mass? Well, you could study the orbit of its companion star to try and determine the stellar remnant's effect on its orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way would be to study gravitational waves emitted by the system. These waves can only be emitted under certain circumstances - for example, in a system of binary neutron stars - see, for example, the Hulse-Taylor binary system, also known as PSR B1913+16. The power radiated as these waves, as well as the orbital decay, depend on the masses of the objects. While detecting gravitational waves is an incredibly difficult task, there are several detectors planned or already in operation, such as eLISA and LIGO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also attempt to look for other effects that characterize certain stellar remnants. For instance, a certain pattern of radio waves would characterize a pulsar (white dwarfs and black holes are not normally strong radio sources). However, if the objects are not luminous, this may not be effective.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-11-03T20:02:59.070" LastActivityDate="2016-11-03T20:02:59.070" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6031" PostTypeId="1" CreationDate="2014-08-04T16:21:24.077" Score="2" ViewCount="423" Body="&lt;p&gt;I was reading this &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6014/can-there-be-an-infinity-of-stars-in-the-universe&quot;&gt;question&lt;/a&gt; about if the number of stars can or cannot be infinite.&#xA;The answer to this question is: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&quot;if the universe is considered infinite, the number of stars can be infinite, otherwise not.&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So, I came across a mental problem: If the universe is considered infinite, and therefore we can consider that the number of stars is too, then, we can consider that the number of humans (example) is also infinite, or the number of anything is also infinite?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the answer is &lt;strong&gt;No&lt;/strong&gt;, please describe why u think this is not a real affirmation. &lt;/p&gt;&#xA;" OwnerUserId="2180" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-08-05T01:30:05.527" Title="Can there be an infinity of humans in the Universe?" Tags="&lt;universe&gt;&lt;space&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" ClosedDate="2014-08-05T15:46:24.210" ContentLicense="CC BY-SA 3.0" />
  <row Id="6032" PostTypeId="1" AcceptedAnswerId="6059" CreationDate="2014-08-04T16:32:01.070" Score="6" ViewCount="252" Body="&lt;p&gt;The image below is a small region of a twilight flat field. A grid of darker pixels with 728x36 cell dimensions can be seen.&#xA;&lt;img src=&quot;https://i.imgur.com/6JHEh40.png?1&quot; alt=&quot;Flat Field with dark grid artifact&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The camera used was the SBIG STL-6303 according to the FIT headers, which uses a Kodak KAF-6303E CCD according to the online data sheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CCD dimensions, these cell dimensions, and all the binning options are all perfectly divisible, so I'm assuming this artifact is due to the structure of the CCD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I plotted the mean pixel value of each row to make sure this wasn't an artifact from the image viewing software. The plot clearly shows a drop every 36 pixels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'm hoping for some confirmation: is this decreased sensitivity due to the structure of the CCD?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, what about the CCD structure actually causes this?&lt;/p&gt;&#xA;" OwnerUserId="2181" LastActivityDate="2014-08-07T01:22:34.763" Title="Why does this twilight sky flat field have a grid of dark pixels?" Tags="&lt;photography&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6033" PostTypeId="2" ParentId="6031" CreationDate="2014-08-04T18:35:01.483" Score="3" Body="&lt;p&gt;This is relatively an assumption so theoretically you can say Yes as you dont know if there are other planets on which life exists and there are humans on these planets. If humans on those planets exist then what is the count of those humans. So assuming that there are other planets on which humans exist you can say that there are infinite number of humans on this planet but since we only have a proof that only earth has humans so practically the number of humans on earth is countable&lt;/p&gt;&#xA;" OwnerUserId="2183" LastActivityDate="2014-08-04T18:35:01.483" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6034" PostTypeId="2" ParentId="138" CreationDate="2014-08-04T18:46:35.287" Score="15" Body="&lt;p&gt;Although its too late to reply to this interesting question but trying to add few more points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes the sun rotates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now the question arises as to how we can check that?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can observe this by observing &lt;strong&gt;sunspots&lt;/strong&gt;. All sunspots move across the face of the Sun. This motion is part of the general rotation of the Sun on its axis. Observations also indicate that the Sun does not rotate as a solid body, but it spins differentially. That means that it rotates faster at the equator of the Sun and slower at its poles. (The gas giants Jupiter and Saturn also have differential rotation.) The movements of the sunspots indicate that the Sun rotates once every 27 days at the equator, but only once in 31 days at the poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may refer this interesting article:- &lt;a href=&quot;http://www.universetoday.com/60192/does-the-sun-rotate/&quot;&gt;Does The Sun Rotate?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2183" LastActivityDate="2014-08-04T18:46:35.287" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6035" PostTypeId="2" ParentId="4776" CreationDate="2014-08-04T18:55:32.213" Score="10" Body="&lt;p&gt;This is a tricky question to answer, because it has some false and imprecise assumptions baked into it. The short version is: I don't know at what rate stuff is currently crossing the cosmic Event Horizon. But future telescopes will not see these galaxies redshift themselves into oblivion, because this will happen in the infinitely far future, literally. Cosmological redshift is equivalent to relativistic time dilation; events that are redshifted from our point of view also play out slower, and infinite redshift means they come to a complete halt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also not correct that the radius of the observable Universe is expanding at the speed of light. There are two different radii that could be candidates for the title &quot;limit of the observable Universe&quot;; they are very different but both expand &lt;em&gt;faster&lt;/em&gt; than light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your intuition is right, however, that the distance at which galaxies recede from us at exactly light speed - the one we call the &lt;strong&gt;Hubble Sphere&lt;/strong&gt; - has been expanding and since approaching us again. It reached its maximum when the cosmic expansion went from decelerating to accelerating, when the Universe was around half its current age. The Hubble Sphere, however, is not the limit of anything. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the two horizons, the one that seems to be the one you have in mind in the beeginning of your question, is the &lt;strong&gt;Event Horizon&lt;/strong&gt;. This is the current distance, beyond which light emitted at this cosmological time will never have the chance to reach us. This horizon receding from us in proper distance, and since it is located outside the Hubble Sphere, these regions of Space are actually receding from us faster than light.&#xA;In &lt;strong&gt;co-moving coordinates&lt;/strong&gt;, the event horizon is &lt;em&gt;approaching&lt;/em&gt; us. Co-moving coordinates is a coordinate system that stretches along with the expansion of the Universe, such that a galaxy that is not influenced by anything else than cosmic expansion will have constant coordinates in time. Because the cosmological expansion is accelerating, it gets harder and harder for light at a fixed co-moving distance to reach us until it gets impossible and the Galaxy redshifts to infinity. This  is equivalent to saying that the Event Horizon has moved inwards past this galaxy in co-moving coordinates. In proper coordinates, it amounts to saying that the galaxy overtakes the Event Horizon.&#xA;Ours, by the way, is currently at a redshift of around 2, around 18 billion light years away. The light we receive from this distance is only slightly redshifted now, but the light that is emitted now will be redshifted to  infinity when it reaches us at infinite time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also another horizon, however, which does expand forever, also in co-moving coordinates. That is the current co-moving distance at which light emitted at the time of Big Bang reaches us just now. In co-moving coordinates, this horizon expands slower and slower, which is why the CMB gets more and more redshifted. Objects currently on the Particle Horizon are a long time out of our Event Horizon - we can only ever see the very beginning of their history, before they redshift into oblivion. And the objects shifting into our particle horizon will show us still less of their history before having left the Event Horizon for good (yes, language gets hairy when talking relativistic time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is hard say what the limit of the observable Universe is, it is simply not well defined. The Event Horizon has always been growing in proper distance but shrinking in co-moving distance. The Particle Horizon has always been growing both in co+moving and proper distance. The Event Horizon has always been expanding faster than light, and always will. The particle horizon started out slower,  but very soon overtook lightspeed in expansion velocity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, all objects with a redshift higher than $z \sim 1.5$ are not just receding faster than light, but they always have been! Yet we observe them with no trouble.&lt;/p&gt;&#xA;" OwnerUserId="2182" LastActivityDate="2014-08-04T18:55:32.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6036" PostTypeId="2" ParentId="6031" CreationDate="2014-08-04T18:59:25.643" Score="-1" Body="&lt;p&gt;There are different infinities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could say that there are an infinite number of planets, so probability says that there could be humans on a small percentage of them. Say for arguments sake humans exist on 0.001% of those planets.0.001% of infinity is infinity, right? Well, sort of, but it's obviously a smaller infinity than the number of planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely there are 99.999% of planets without humans, which is a much bigger infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So to all intents and purposes the answer is no, for this mental exercise.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in reality it's a no, as any humans on other planets would pretty much have to not be humans any more, as they are so far away that unless we have instantaneous or FTL travel, evolution would have changed them/us.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2014-08-04T18:59:25.643" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6037" PostTypeId="1" CreationDate="2014-08-04T19:33:14.333" Score="2" ViewCount="207" Body="&lt;p&gt;Supposedly, if humans were to establish a colony on one of the moons of a planet (such as gas giants in our own solar system), how could we possibly design date and time system? On Earth, we define date based on rotation about Sun and time on Earth's revolution. But, with a moon of another planet, this will add another factor to date/time system. So, how can we cope with this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I am not sure about what tags this question should carry. Feel free to correct them.&lt;/p&gt;&#xA;" OwnerUserId="2184" LastEditorUserId="1678" LastEditDate="2014-08-05T18:14:59.027" LastActivityDate="2014-08-11T19:31:20.060" Title="Date and time system on moons" Tags="&lt;the-moon&gt;&lt;time&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6038" PostTypeId="2" ParentId="6031" CreationDate="2014-08-04T19:54:50.237" Score="5" Body="&lt;p&gt;IMO, this sounds more like a philosophical question rather than an astronomical question, but since I like both subjects, I'll give it a shot.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; This is &lt;em&gt;hypothetical&lt;/em&gt; question, so this answer is purely subjective (and &lt;em&gt;hypothetical&lt;/em&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Infinity&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the number of stars is infinite, so would the numbers of humans.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I &lt;em&gt;believe&lt;/em&gt; that in every system where an entity of infinity is introduced, all other entities will (eventually) be infinite. &lt;em&gt;But&lt;/em&gt;, (and this is the fun &quot;part&quot;) you cannot measure infinity. It would give you infinite results over a time span of infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take your example for instance. &lt;code&gt;If there were infinite numbers of stars.&lt;/code&gt; Picture youself flying around in a spaceship with the goal to find out if &lt;em&gt;this&lt;/em&gt; question is true or not. What do you see? &quot;Worse case&quot; scenario; you see only stars. Wherever you go. Forever. The opposite;  only humans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One might think that it's the &lt;em&gt;&quot;in between&quot;&lt;/em&gt; that is the most likely scenario. But given the &lt;a href=&quot;http://en.wikipedia.org/wiki/Probability_theory#Law_of_large_numbers&quot;&gt;probability theory&lt;/a&gt;, we should flatten out at approx. 50%.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Finite&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;If there &lt;em&gt;is&lt;/em&gt; a finite numbers of stars then I think it's &quot;safe&quot; to conclude that there ain't an infinite number of humans as &lt;a href=&quot;http://en.wikipedia.org/wiki/Nucleosynthesis#History_of_nucleosynthesis_theory&quot;&gt;we're all made of starstuff&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Conclusion&lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Yes&lt;/strong&gt;: If there's an infinite numbers of stars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;No&lt;/strong&gt;: If there's a finite numbers of stars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2048" LastEditorUserId="2048" LastEditDate="2014-08-04T20:03:11.203" LastActivityDate="2014-08-04T20:03:11.203" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6039" PostTypeId="2" ParentId="6032" CreationDate="2014-08-04T20:43:41.287" Score="4" Body="&lt;p&gt;This looks like readout noise to me. Such noise is a consequence of the electric machinery used to convert the individual pixel detections into a digital picture, and really is an unavoidable part of CCD technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at &lt;a href=&quot;http://spiff.rit.edu/classes/phys373/lectures/readout/readout.html&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt; if you want to know more about the technical details.&lt;/p&gt;&#xA;" OwnerUserId="494" LastActivityDate="2014-08-04T20:43:41.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6040" PostTypeId="2" ParentId="6037" CreationDate="2014-08-04T22:02:49.547" Score="1" Body="&lt;p&gt;I take it you're looking for something absolute to measure time with, such as revolutions around a planet or the planet around a star? If so, you could use a &quot;year&quot; to be the time it takes the planet to orbit the star, or the moon to orbit the planet. Those could also be used as separate, unrelated measures of time. A &quot;day&quot; could be construed as the time it takes for the moon to revolve, or the time it takes for the planet to revolve. It would be interesting if the planet and moon were tidally locked, because that would mean that only one side of the moon would ever face the planet. This is also a social issue, which would have to be decided on by the occupants of such a moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smaller measures of time, such as the minute and hour could still be used, because they are defined by the second, which in turn is defined by the behavior (I don't know the specifics) of a cesium atom. So the second, minute, and hour could be used anywhere - assuming you have an accurate clock or a cesium atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I think you misunderstood the tag &quot;space-time&quot;. Here, it refers to the combined 4-dimensional (in our universe) entity consisting of space and time, as used in special and general relativity.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-08-04T22:02:49.547" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6041" PostTypeId="2" ParentId="429" CreationDate="2014-08-04T22:17:39.590" Score="5" Body="&lt;p&gt;Reading &lt;a href=&quot;http://adsabs.harvard.edu/abs/1960Sci...131.1667D&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dyson's original argument&lt;/a&gt; gives some useful information. He says that such a sphere would have a surface temperature of 200-300 Kelvin, as it would reradiate some of the energy it absorbs at infrared wavelengths. That's a temperature comparable to cool &lt;a href=&quot;https://en.wikipedia.org/wiki/Brown_dwarf#Spectral_class_Y&quot; rel=&quot;nofollow noreferrer&quot;&gt;Y-type brown dwarfs&lt;/a&gt;. If we can assume a &lt;a href=&quot;https://en.wikipedia.org/wiki/Black_body&quot; rel=&quot;nofollow noreferrer&quot;&gt;black body approximation&lt;/a&gt;, and a radius of about 1 AU, we get a &lt;a href=&quot;https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law#Temperature_of_stars&quot; rel=&quot;nofollow noreferrer&quot;&gt;luminosity&lt;/a&gt; for a 200 K sphere of&#xA;$$L_{\text{DS}}\simeq4\pi\sigma R^2T^4\sim10^{25}\text{ Watts}\simeq0.1L_{\odot}$$&#xA;That's substantially more luminous than a cool brown dwarf of the same temperature. Additionally, &lt;a href=&quot;https://en.wikipedia.org/wiki/Planck%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck's law&lt;/a&gt; gives a peak black body wavelength of about 14,500 nm, placing the emissions in the mid-infrared range, as expected. Relatively hot brown dwarfs should be expected to have high infrared emissions. In other words, a Dyson sphere should look something like, from a distance, a large, luminous brown dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dyson spheres, in the classic picture of a solid shell enveloping a star, are actually unstable, so it's unlikely that they would ever be built. We should expect other variants - Dyson swarms, rings, bubbles, etc. - that are composed of large arrays of smaller objects are much more likely to be created.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fermilab used data from the &lt;a href=&quot;https://en.wikipedia.org/wiki/IRAS&quot; rel=&quot;nofollow noreferrer&quot;&gt;Infrared Astronomical Satellite (IRAS)&lt;/a&gt; to look for Dyson spheres as &lt;a href=&quot;http://home.fnal.gov/~carrigan/infrared_astronomy/Fermilab_search.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;a secondary mission&lt;/a&gt;, searching a band in infrared wavelengths centered around 300 K (see &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0004-637X/698/2/2075/meta;jsessionid=8CC048BBAB18C31EE60139CC33C9AC98.c4.iopscience.cld.iop.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carrigan (2009)&lt;/a&gt;), covering 100 K to 600 K. The study started with 250,000 sources over 96% of the sky, then whittled them down to about 6,000 based on temperature range and flux. 17 &quot;ambiguous&quot; candidates were finally left. This is, to my knowledge, the most complete survey yet undertaken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have, of course, been &lt;a href=&quot;http://home.fnal.gov/~carrigan/infrared_astronomy/Other_searches.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;others&lt;/a&gt;, most using IRAS data. &lt;a href=&quot;https://en.wikipedia.org/wiki/Wide-field_Infrared_Survey_Explorer&quot; rel=&quot;nofollow noreferrer&quot;&gt;WISE&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/2MASS&quot; rel=&quot;nofollow noreferrer&quot;&gt;2MASS&lt;/a&gt; also contributed data. Very few candidates have been found, and it has been estimated that any within about 1 kpc in reasonable temperature ranges (less than 400 K) should have been among these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are indeed natural sources that could prove to be false positives. Carrigan lists several:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Dust shells around stars possibly undergoing mass loss&lt;/li&gt;&#xA;&lt;li&gt;Stars embedded in dusty areas, including dense nebulae&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Mira_variable&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mira variables&lt;/a&gt;, which undergo mass loss towards the end of their lives&lt;/li&gt;&#xA;&lt;li&gt;Planetary nebulae&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Asymptotic_giant_branch&quot; rel=&quot;nofollow noreferrer&quot;&gt;AGB stars&lt;/a&gt;, which may create circumstellar envelopes, as well as post-AGB stars&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-11-04T15:06:04.760" LastActivityDate="2016-11-04T15:06:04.760" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6042" PostTypeId="2" ParentId="6031" CreationDate="2014-08-05T01:30:05.527" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/0102010&quot; rel=&quot;nofollow&quot;&gt;See this article&lt;/a&gt;. There are only a finite number of states the observable part of the universe can be in, therefore you will end up with an infinite number of copies of the entire visible part of the universe. And that leads to the conclusion that all histories which are not forbidden by conservation laws will occur. This means that not only are there an infinite number of copies of you, there are also an infinite number of the fictional characters that appear in soap series in sectors of the universe where the story they appear in is reality. &lt;/p&gt;&#xA;" OwnerUserId="2185" LastActivityDate="2014-08-05T01:30:05.527" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6043" PostTypeId="1" CreationDate="2014-08-05T09:59:59.270" Score="-5" ViewCount="249" Body="&lt;p&gt;I read a question posted by a community member in my community that shocked me, I think it contains some pretty cool astronomy questions that I cant wait for you scientists to answer. I am asking this from stack exchange, because this is where all the real scientists are , and I want to see the truth , a real explanation , brief but simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wont copy the exact question details because it will make it duplicate, but the question is posted here, please take time to read the details and explain it to me it is posted here- &#xA;&lt;a href=&quot;http://srilanka.answercup.com/question/215/how-can-i-convince-my-atheist-friend-that-god-exists/&quot; rel=&quot;nofollow&quot;&gt;http://srilanka.answercup.com/question/215/how-can-i-convince-my-atheist-friend-that-god-exists/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am fascinated by his explanation, I cant wait for the astronomy community to tell me the answers this guy is asking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok this is the question I am pasting it here after being requested by a member on the comment section-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His tite is - &#xA;How can i convince my atheist friend that God exists???&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently he seems to be religious person, who has some points to make.&#xA;This is the description of his question-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is this friend that I have who believes in science and doesnt believe in religion. We have had many arguements, and it raises many interesting questions. I am posting this to see Answercup users opinion on it. If scientists say that Big bang is proven I have some simple questions . What happened before the big bang according to these scientists? And this big bang theory came from scientists thinking that the universe expands so if we reverse the time, it will do the opposite and come to one single point....but then if it did come from a point, I found that the expansion is INCREASING which is weird. How on earth can it be a blast , because obv like any blast it should decreases and come to a hault , here instead of the universe expansion decelerating its actually accelerating. How can these so called scientists tell something is right when common sense which can be understood by primary school children shows clearly that they are wrong ??? Plus how can you seriously get a blast from empty space ? If scientist says thats the beginning of time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With his question, I remembered watching the vedio &quot;Does god exist&quot; on Discovery channel by stephen hawking. His explanations wasnt complete and he bragged about how there is no need for god, but his explanation never explained many parts. And then there are scientists who gives the excuse laws of physics break at the point of the big bang so we can't explain. So this brought my mind a new question, if scientists still cant understand what has happens in the big bang , how can they tell that God doesnt exist, and stuff like that and make them appear as facts, if they knew the physics at the big bang and talked then its fine, but here they are talking without even knowing those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://srilanka.answercup.com/question/215/how-can-i-convince-my-atheist-friend-that-god-exists/&quot; rel=&quot;nofollow&quot;&gt;http://srilanka.answercup.com/question/215/how-can-i-convince-my-atheist-friend-that-god-exists/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2187" LastEditorUserId="2187" LastEditDate="2014-08-05T10:19:21.187" LastActivityDate="2014-08-05T10:42:15.413" Title="Astronomy questions I wondered for hours" Tags="&lt;big-bang-theory&gt;&lt;early-universe&gt;" AnswerCount="1" CommentCount="14" ClosedDate="2014-08-05T21:03:30.753" ContentLicense="CC BY-SA 3.0" />
  <row Id="6044" PostTypeId="2" ParentId="6043" CreationDate="2014-08-05T10:42:15.413" Score="6" Body="&lt;p&gt;If you think on Big Bang as an explosion &lt;strong&gt;IN&lt;/strong&gt; space, you are plain wrong. It was not. It was an explosion &lt;strong&gt;OF&lt;/strong&gt; space itself. And time, for that matter.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What happened before the big bang according to these scientists?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Common misconception. There is no &quot;before&quot;. Time starts at Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How on earth can it be a blast , because obv like any blast it should decreases and come to a hault , here instead of the universe expansion decelerating its actually accelerating.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Common misconception. It is not a blast in space. It is the space itself expanding, and there is no outside from where to see it decelerate and stop.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;how can they tell that God doesnt exist, and stuff like that and make them appear as facts, if they knew the physics at the big bang and talked then its fine, but here they are talking without even knowing those.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;God is not part of Science, this question is out of topic here.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-08-05T10:42:15.413" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6045" PostTypeId="2" ParentId="6005" CreationDate="2014-08-05T15:24:22.787" Score="0" Body="&lt;p&gt;This is known as &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Olbers%27_paradox&quot; rel=&quot;nofollow&quot;&gt;Olbers' paradox&lt;/a&gt;. An (in my opinion) better explanation can be found &lt;a href=&quot;http://curious.astro.cornell.edu/question.php?number=375&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that these links answer the question why the universe (not the milky way alone) isn't blazing our night sky. Now if the universe (including the milky way) can't do it, the milky way can't do it on its own... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are just not enough stars giving enough light. Brightness is reduced quadratically by distance. That counts up very quickly with astronomical distances. Even for stars that are our neighbors. &lt;/p&gt;&#xA;" OwnerUserId="2189" LastActivityDate="2014-08-05T15:24:22.787" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6046" PostTypeId="1" AcceptedAnswerId="6047" CreationDate="2014-08-05T16:44:04.147" Score="3" ViewCount="265" Body="&lt;p&gt;I am currently designing star and planet systems in the RPG framework &lt;a href=&quot;http://www.sjgames.com/gurps/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;GURPS&lt;/strong&gt;&lt;/a&gt;. This game system has a very elaborate &lt;a href=&quot;http://www.sjgames.com/gurps/books/space/&quot; rel=&quot;nofollow noreferrer&quot;&gt;set of rules&lt;/a&gt; to set up a star system, complete with planets and the like. However, one (for me at least) critical issue is not addressed in the book: The argument of perihelion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since I am writing a program to generate these systems for me, I would like to know what steps I can take to make a planet system as realistic as possible. Looking at data from our own planet system, the argument of perihelion looks to be pretty evenly distributed over all angles $[0, 2\pi)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there scientific evidence that this is actually the case, or are there any good suggestions on how I could proceed with my program?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The related question on RPG.SE can be found &lt;a href=&quot;https://rpg.stackexchange.com/questions/45558/how-do-i-determine-the-argument-of-periapsis-in-gurps-space&quot;&gt;here&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="2191" LastEditorUserId="-1" LastEditDate="2017-04-13T12:45:51.587" LastActivityDate="2016-05-23T17:57:00.530" Title="Is the argument of perihelion random?" Tags="&lt;solar-system&gt;&lt;planetary-systems&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6047" PostTypeId="2" ParentId="6046" CreationDate="2014-08-05T16:59:54.583" Score="5" Body="&lt;p&gt;Argument of perihelion or most planets/bodies changes very slowly over time due to higher order perturbations from other planets' motions (mostly Jupiter and Saturn for the solar system). General relativistic effects also cause the perihelion to advance over time, though this effect is smaller than the others for most purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So given enough time, the arguments of the planets' perihelia would likely end up being fairly randomly distributed at any given time. Also, the planets' orbits are often dynamic on such timescales due to the effects of higher order perturbations showing up, which can cause eccentricity changes, which also affect the positions of perihelia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my guess would be that it is okay to have the arguments of perihelia randomly distributed, since it is inconceivable (at least for me) that any sort of resonances would be visible at this level in order for the distributions to be non-random.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2014-08-05T16:59:54.583" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6048" PostTypeId="2" ParentId="6005" CreationDate="2014-08-05T22:52:07.240" Score="3" Body="&lt;p&gt;Firstly, the galaxy is only about &lt;a href=&quot;http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/980317b.html&quot; rel=&quot;nofollow&quot;&gt;1000ly thick&lt;/a&gt;. We are fairly close to the galactic plane, &lt;a href=&quot;http://www.nature.com/nature/journal/v316/n6030/abs/316706a0.html&quot; rel=&quot;nofollow&quot;&gt;maybe around 65 ly 'above' it&lt;/a&gt; if we call the direction we are moving away from 'down'. So on your assumption that visible objects are all within about 1000ly, we can suppose that we should see more stars in the plane than above and below the plane, as there is only about 500ly of stars in the galaxy above and below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are perhaps &lt;a href=&quot;http://iopscience.iop.org/1538-4357/597/2/L121/&quot; rel=&quot;nofollow&quot;&gt;25kly&lt;/a&gt; from the galactic centre. Therefore there are about 25kly of stars outwards, and 75kly of stars inwards. If we could see all of them with the naked eye, we would see more stars to one side than the other. While many of the stars that we see are within 1000ly, that doesn't mean that the stars within this region of visibility are evenly distributed. &lt;a href=&quot;http://www.astronomy.ohio-state.edu/~ryden/ast162_7/notes31.html&quot; rel=&quot;nofollow&quot;&gt;Stars are packed closer together nearer the centre&lt;/a&gt;, which implies that we will see more towards one side than the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know where you get the idea you seek confirmation for that the density of stars is not higher in the galactic plane, because it isn't true. Density is greater toward the centre of the galaxy; and orthogonal to the galactic plane, density decreases as you move away from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plus, we see far more than merely dots of light from other stars. We see other objects that are aggregations of stars - clusters, and galaxies like Andromeda, the LMC and the SMC.  We see nebulae, clouds of gas and dust. In fact, dust even obscures what we might otherwise see toward the galactic centre. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when you ask 'what is it we see?' the answer is: lots of things. Because there is more of that 'stuff' in the plane of the galaxy, we see the majority of that stuff as a ring around us. Because there is more of that stuff towards the centre of the galaxy, we see that the ring is brighter, fuller, more complex, and dominant on one side.&lt;/p&gt;&#xA;" OwnerUserId="67" LastEditorUserId="67" LastEditDate="2014-08-06T11:05:20.680" LastActivityDate="2014-08-06T11:05:20.680" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6049" PostTypeId="1" CreationDate="2014-08-06T06:55:54.053" Score="2" ViewCount="124" Body="&lt;p&gt;It is said that due to CMB radiation our universe has been expanding? Firstly what lead the universe to expand? What do scientists mean by dark energy? Will energy convert into matter?&lt;/p&gt;&#xA;" OwnerUserId="2195" LastEditorUserId="50" LastEditDate="2014-08-10T07:22:18.433" LastActivityDate="2014-08-10T07:22:18.433" Title="What is CMB radiation doing to the universe?" Tags="&lt;universe&gt;&lt;expansion&gt;&lt;cmb&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6050" PostTypeId="2" ParentId="6049" CreationDate="2014-08-06T10:24:30.897" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;It is said that due to the CMB radiation our universe has been expanding?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. It's the the other way round. The CMB is the result of the expansion of space. In the initial stages after the birth of the universe, there was hot plasma everywhere and light could not travel much farther. The universe was opaque at that time. But slowly as the Universe cooled and atoms began to form, light could travel farther. This light got redshifted due to the expansion of space and can be now detected as the CMB radiation. It's in the microwave region of the spectrum and just above absolute zero(about 2.7 K). &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What do we mean by Dark Energy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After Hubble's discovery, we came to the conclusion that the Universe is expanding. We expected the expansion to slow down as time progresses but what we observed as that the expansion was accelerating and is still doing so. Physicists and cosmologists did not have any explanations for this and so it was named &quot;Dark Energy&quot;. Dark in the sense that it is unknown.&lt;/p&gt;&#xA;" OwnerUserId="1156" LastActivityDate="2014-08-06T10:24:30.897" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6051" PostTypeId="1" CreationDate="2014-08-06T14:35:29.870" Score="5" ViewCount="198" Body="&lt;p&gt;What measurements and observations of the Hulse-Taylor Pulsar agreed precisely with the predictions of general relativity, and how did these measurements support the theory?&lt;/p&gt;&#xA;" OwnerUserId="2198" LastActivityDate="2016-11-02T22:24:20.273" Title="How did the Hulse-Taylor Pulsar Provide Confirming Evidence of General Relativity?" Tags="&lt;general-relativity&gt;&lt;pulsar&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6052" PostTypeId="1" AcceptedAnswerId="6097" CreationDate="2014-08-06T14:42:22.303" Score="20" ViewCount="25344" Body="&lt;p&gt;In Clarke's book 2010, the monolith and its brethren turned Jupiter into the small star nicknamed &lt;code&gt;Lucifer&lt;/code&gt;.  Ignoring the reality that we won't have any magical monoliths appearing in our future, what would the effects be on Earth if Jupiter was turned into a star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At it's closest and furthest:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How bright would the &quot;back-side&quot; of the earth be with light from Lucifer?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much heat would the small star generate on earth?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many days or months would we actually have night when we circled away behind the sun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much brighter would the sun-side of earth be when Lucifer and the sun both shine on the same side of the planet?&lt;/p&gt;&#xA;" OwnerUserId="2199" LastEditorUserId="2153" LastEditDate="2014-08-08T18:54:27.970" LastActivityDate="2017-10-13T22:12:17.363" Title="What would the effects be on Earth if Jupiter was turned into a star?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;light&gt;&lt;jupiter&gt;&lt;heat&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6053" PostTypeId="2" ParentId="6052" CreationDate="2014-08-06T15:10:00.223" Score="3" Body="&lt;ol&gt;&#xA;&lt;li&gt;Sun-Earth distance: 1AU    &lt;/li&gt;&#xA;&lt;li&gt;Earth-Jupiter distance (at the conjunction): 4AU&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So Lucifer will be four times farther than Sun when it is nearer (six times when it is farthest), and at the same time it is &lt;a href=&quot;https://en.wikipedia.org/wiki/Jupiter_mass&quot; rel=&quot;nofollow&quot;&gt;a thousand times smaller&lt;/a&gt;. This is approx 40 times more light than full moon concentrated in a tiny point on sky.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-08-06T15:10:00.223" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="6054" PostTypeId="2" ParentId="6037" CreationDate="2014-08-06T16:16:34.327" Score="2" Body="&lt;p&gt;I really love this question. Were we to establish a colony on our moon, we would probably want a calendar/time system which actually corresponds to physical patterns and cycles on the moon, but also something which is easy to track and keep in sync with Earth, since the two civilizations would certainly be in close contact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a way, we already do something this here on Earth with time zones, though time zones are just offsets within the same second/minute/hour/day system, so people throughout the world are in different times, but they stay in sync. But beyond that, our entire calendar system (we currently use the Gregorian Calendar), is mostly based on our own planet's positions relative to the sun. It just wouldn't make sense on the moon. Earth days on the moon would just be arbitrary 24-hour blocks of time; they wouldn't correlate at all to whether it's light or dark outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we really wanted to keep some logical synchronization between earth and moon calendars, we would probably want to adjust here on earth to a lunar-based calendar. Most older civilizations did this, using a full lunar cycle to indicate a &quot;month.&quot; But here's the (cool) thing: the moon is a little lopsided and way gravity works, the exact same side is always facing us. We never see the back of the moon. So, it's rotating at exactly the same speed as it's revolving around us, meaning a &quot;day&quot; on the moon lasts one full lunar cycle: just over 29 earth days. So even the idea of a day, how long a day is, and what you could accomplish in a day is so far from Earth's that it's hard to even compare the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe one idea is to have weeks or months here on Earth which would line up with days on the moon. Earth would still have its 24-hour days and the Moon would have its 709-hour days, but at least we could predict Moon time using our own calendars. One proposal is the &lt;a href=&quot;http://www.hermetic.ch/cal_stud/hlwc/hlwc.htm&quot; rel=&quot;nofollow&quot;&gt;The Hermetic Lunar Week Calendar&lt;/a&gt; which continually adjusts the length of a week to keep the months in sync with the moon, so that a full month is always one lunar cycle (or lunar day). It would be weird but in a future where humans are legitimately split between Earth and Moon, it would probably make sense for both to keep a calendar like that, which takes into account the length of a day on both bodies.&lt;/p&gt;&#xA;" OwnerUserId="2197" LastActivityDate="2014-08-06T16:16:34.327" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6055" PostTypeId="2" ParentId="6051" CreationDate="2014-08-06T16:57:26.230" Score="3" Body="&lt;p&gt;One of the predictions of general relativity is that certain objects can give off energy in the form of gravitational radiation. This means that over time, the orbits of the two neutron stars should decay, and they should come closer to each other. Using relativistic formulae, it is possible to predict the changes in the orbits and the energy of the emitted gravitational waves; analysis of the system showed that general relativity's predictions were correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Studying gravitational waves requires linearizing the Einstein field equations. However, these can be reduced to certain simpler formulae when it comes to detecting waves from a binary system, as was the case with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hulse%E2%80%93Taylor_binary&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hulse-Taylor binary&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-11-02T22:24:20.273" LastActivityDate="2016-11-02T22:24:20.273" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6056" PostTypeId="2" ParentId="6052" CreationDate="2014-08-06T21:49:48.183" Score="6" Body="&lt;p&gt;Ignoring the impossibility of Jupiter going solar:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that Jupiter turns into  duplicate of the Sun in terms of energy output.  Energy transmitted to the earth follows an inverse-square law.  Since Jupiter is, at best, 4 times farther from the Earth than the Sun, Jupiter will supply the Earth with, at most, 1/16 the energy that the Sun supplies, for an increase of a bit more than 6%, at the most.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By comparison, between aphelion and perihelion, the Sun-Earth distance increases from around 147 million kilometers to around 152 million kilometers.  This implies a seasonal energy input change of about 7%, that we experience now every year... &lt;/p&gt;&#xA;" OwnerUserId="420" LastActivityDate="2014-08-06T21:49:48.183" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6058" PostTypeId="2" ParentId="5972" CreationDate="2014-08-07T01:08:11.870" Score="4" Body="&lt;p&gt;You seem to have two types of seasons confused.  Astronomical seasons are relatively stable in their durations, while seasons as defined by the weather, can vary by several weeks.  For example, snow 3 weeks after spring begins or freezing weather before winter begins.  This is what is meant when we say an early summer or a late winter.  It simply means that the weather usually associated with that season is later or earlier than usual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, the lengths of the seasons ranges from about 88 to 93 days every year.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;For 2014 (Northern hemisphere season lengths)&#xA;Length of spring = 92d 17h 54m&#xA;Length of summer = 93d 15h 37m&#xA;Length of autumn = 89d 20h 34m&#xA;Length of winter = 88d 23h 42m&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The axis of the earth has nothing to do with it unless we are considering a time span extending to many thousands of years.  The Earth's axis hasn't changed enough in recorded history to significantly effect the seasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The astronomical seasons are defined by the geocentric position of the sun and have little to do with the weather.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;March Equinox&lt;/b&gt; &lt;br&gt;Beginning of spring in the northern hemisphere = Beginning of autumn in the southern hemisphere, when the sun reaches declination 0 degrees while crossing the celestial equator northward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;June Solstice&lt;/b&gt;&lt;br&gt; Beginning of summer in the northern hemisphere = Beginning of winter in the southern hemisphere, when the sun reaches right ascension 90 degrees (6h).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;September Equinox&lt;/b&gt;&lt;br&gt; Beginning of autumn in the northern hemisphere = Beginning of spring in the southern hemisphere, when the sun reaches declination 0 degrees while crossing the celestial equator southward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;December Solstice&lt;/b&gt;&lt;br&gt; Beginning of winter in the northern hemisphere = Beginning of summer in the southern hemisphere, when the sun reaches right ascension 270 degrees (18h).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These astronomical season lengths vary by only a matter of minutes either way each year, while the seasonal weather can vary considerably and winter-type weather can persist even after astronomical spring has begun (a late or short spring), but it is not yearly changes in the axis of the Earth that causes these annual weather variations.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;SEASONAL STATISTICS FOR THE YEAR 2014 @ TZone = UT-05:00 (New York City)&#xA;BEGINNING DATES, TIMES TT AND LENGTHS OF THE SEASONS AND SEASONAL YEARS&#xA;&#xA;Equinox or Solstice      Season Length     Since Previous     Until Following&#xA;E 2014 Mar 20 11:57:58   92d 17h 54m 23s   365d 05h 54m 51s   365d 05h 48m 15s&#xA;S 2014 Jun 21 05:52:21   93d 15h 37m 26s   365d 05h 47m 17s   365d 05h 46m 41s&#xA;E 2014 Sep 22 21:29:47   89d 20h 34m 21s   365d 05h 44m 56s   365d 05h 52m 40s&#xA;S 2014 Dec 21 18:04:08   88d 23h 42m 05s   365d 05h 52m 01s   365d 05h 44m 56s&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Perhaps these links will help:&lt;br&gt;&#xA;&lt;a href=&quot;http://neoprogrammics.com/equinoxes_and_solstices/&quot; rel=&quot;nofollow&quot;&gt;http://neoprogrammics.com/equinoxes_and_solstices/&lt;/a&gt; &lt;br&gt;&#xA;&lt;a href=&quot;http://neoprogrammics.com/equinoxes_and_solstices/seasonal_statistics/&quot; rel=&quot;nofollow&quot;&gt;http://neoprogrammics.com/equinoxes_and_solstices/seasonal_statistics/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2192" LastEditorUserId="2192" LastEditDate="2014-08-08T06:54:46.863" LastActivityDate="2014-08-08T06:54:46.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6059" PostTypeId="2" ParentId="6032" CreationDate="2014-08-07T01:22:34.763" Score="6" Body="&lt;p&gt;Yes, it's some sort of CCD structure originated from the manufacturing process. CCD chips with large area were produced by chunks of several hundred pixels. see &lt;a href=&quot;http://heasarc.gsfc.nasa.gov/Images/astroe/about/xis_config.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://heasarc.gsfc.nasa.gov/Images/astroe/about/xis_config.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And sometimes position offsets between these chunks can be detected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fjbSl.png&quot; alt=&quot;CCD pixel blocks&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from &lt;a href=&quot;http://adsabs.harvard.edu/abs/2010MNRAS.405L..16W&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wilken et al., 2010, MNRAS, 405, L16&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2105" LastActivityDate="2014-08-07T01:22:34.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6060" PostTypeId="2" ParentId="6052" CreationDate="2014-08-07T03:01:04.577" Score="5" Body="&lt;p&gt;In reality, Jupiter doesn't have nearly enough mass to initiate stellar ignition or sustain it if we could somehow start it going.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even the smallest star would require on the order of some 80 to 90 times the mass of Jupiter just to put out a faint red glow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even to become a brown dwarf proto-star, Jupiter would require a mass increase on the order of at least 10-fold or so. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lucifer is simply not possible unless Jupiter collides with something to provide the extra mass it needs to go stellar and even then, it would be a red dwarf at best and quite faint, like a red-hot nail glowing in the dark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one can dream.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;:)&lt;/p&gt;&#xA;" OwnerUserId="2192" LastActivityDate="2014-08-07T03:01:04.577" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6062" PostTypeId="1" CreationDate="2014-08-07T04:37:21.847" Score="5" ViewCount="12316" Body="&lt;p&gt;Where can a gallery of actual unaltered photographic images taken in (or of) space be found? Specifically ones that are untouched, not colorized (not necessarily black and white, but they usually are), and taken by natural light photography? Pictures and videos claiming &quot;actual image&quot; are few and far between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g., this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LlCt7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LlCt7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not false color:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vFuPa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vFuPa.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of these pictures are from NASA's (Voyager) &lt;a href=&quot;http://voyager.jpl.nasa.gov/image/saturn.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Saturn Images&lt;/a&gt; gallery. Some of the other ones there are listed as false color, some aren't (but obviously they are). Or maybe not so obviously, hence the question: &lt;em&gt;what does it really look like out there?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've a pretty good idea of what Saturn looks like IRL, because I've seen it in a telescope (exactly like the first picture, except it's more colorful - absolutely nothing like the second). For most other celestial objects, I have no such baseline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The title of the website I'm looking for would be along the lines of: &lt;em&gt;View of our solar system through the eyes of a human.&lt;/em&gt; Decidedly, &lt;em&gt;not&lt;/em&gt; containing any pictures from the HST, as all of them are photoshopped.&lt;/p&gt;&#xA;" OwnerUserId="1810" LastEditorUserId="1810" LastEditDate="2015-10-10T05:45:15.117" LastActivityDate="2020-10-21T19:06:00.123" Title="Gallery of 'actual images' from space?" Tags="&lt;photography&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6063" PostTypeId="2" ParentId="6062" CreationDate="2014-08-07T05:48:02.260" Score="4" Body="&lt;p&gt;I guess, this is among the best you can find: &lt;a href=&quot;http://heritage.stsci.edu/&quot; rel=&quot;nofollow&quot;&gt;Hubble Heritage&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are visible picture, real, and not modified, taken from space, and super-amazing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you meant pictures in wavelengths other than visible, please just ask.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastActivityDate="2014-08-07T05:48:02.260" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6064" PostTypeId="1" CreationDate="2014-08-07T12:07:27.760" Score="3" ViewCount="278" Body="&lt;p&gt;There are two kinds of cadence time for Kepler, 5 minutes and 30 minutes.&lt;br/&gt;&#xA;I just know there is no shutter and the CCDs of Kepler can be locked after exposure. &lt;a href=&quot;https://archive.stsci.edu/kepler/manuals/KSCI-19033-001.pdf&quot; rel=&quot;nofollow&quot;&gt;handbook&lt;/a&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how to realize that? Remember we have to not affect the photometry.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other question is so far how many unique sources has Kepler observed? 160,000? &lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2014-08-07T15:07:10.683" Title="how does Kepler stop exposure after a 5 minuate or 30 minute cadence time?" Tags="&lt;kepler&gt;&lt;exposure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6065" PostTypeId="2" ParentId="6064" CreationDate="2014-08-07T15:07:10.683" Score="4" Body="&lt;p&gt;They don't stop. The Kepler CCDs read out the signals collected during the  accumulated time of 6.02 seconds. The fixed read out time is 0.52 seconds. So each CCD gets one frame every (6.02 + 0.52) = 6.54 seconds. Then Kepler sums up every 9 frames (short cadence) and 270 frames (long cadence). The time between two short cadences is (6.54 x 9) = 58.9 seconds, and (6.54 x 270) = 1766 seconds between two long cadences, but the exposure time is (6.02 x 9) = 54.2 seconds for a short cadence, and (6.02 x 270) = 1625 seconds for a long cadence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are totally 95 million pixels available on Kepler CCDs, but the spacecraft can store only 5.4 million. The maximum number of targets is about 170,000. see &lt;a href=&quot;http://archive.stsci.edu/mast_faq.php?mission=KEPLER&quot; rel=&quot;nofollow&quot;&gt;http://archive.stsci.edu/mast_faq.php?mission=KEPLER&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2105" LastActivityDate="2014-08-07T15:07:10.683" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6066" PostTypeId="1" CreationDate="2014-08-07T18:32:20.817" Score="6" ViewCount="1158" Body="&lt;p&gt;What is the direct observational evidence that black holes exist? I have read that black holes have been detected through X-ray emission. What is some other observational evidence for black holes? &lt;/p&gt;&#xA;" OwnerUserId="2198" LastEditorUserId="1891" LastEditDate="2020-04-07T13:13:50.863" LastActivityDate="2020-04-07T13:13:50.863" Title="Observational evidence for the existence of black holes" Tags="&lt;black-hole&gt;&lt;general-relativity&gt;&lt;astrophysics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ClosedDate="2020-04-08T07:29:07.157" ContentLicense="CC BY-SA 4.0" />
  <row Id="6067" PostTypeId="2" ParentId="6066" CreationDate="2014-08-07T19:54:02.430" Score="5" Body="&lt;p&gt;There are a few different features of black holes that researchers look for:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;An accretion disk&lt;/li&gt;&#xA;&lt;li&gt;Mass transfer from companion star&lt;/li&gt;&#xA;&lt;li&gt;X-ray emission&lt;/li&gt;&#xA;&lt;li&gt;Absorption of gas and other matter around it&lt;/li&gt;&#xA;&lt;li&gt;Gravitational lensing&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You mentioned #3 in your question; quite a few black holes, including Cygnus X-1, emit X-rays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 is a common feature around black holes. The material can be pulled from surrounding gases, or, in many cases, a companion star. Accretion disks generally emit electromagnetic radiation; the exact type depends on the type of material and the object it surrounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 is related to 1 and 3, but it deserves its own explanation because of the effects on the donor star. Lets take mass transfer on a large scale. Take the Whirlpool galaxy. It is interacting with its neighbor, NGC 5195, via a bridge of dust and matter. What are the effects on NGC 5195? Distortion, as well as loss of structure in the area near the bridge. The same thing happens on the stellar level. A star losing mass can become distorted in the direction of its companion, potentially causing changes in brightness or other features.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 is, I suppose related to the others, but I should note that if there were merely gas clouds surrounding the black hole, it could also swallow those up. If a black hole is not in a binary system, it may not come in contact with another star, and will instead begin swallowing other bits of matter around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 is the best characteristic; a better way of detecting black holes. General Relativity predicts that massive objects bend light; using Einstein's equations it is possible to determine just how much light will be bent by such an object. This has been used in many cases, such as during the solar eclipse of 1919, to confirm Einstein's theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I should note two things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;These are not the only methods of detecting black holes (other effects, such as frame dragging around a Kerr black hole, offer distinctive descriptions of the black hole), but are rather straightforward&lt;/li&gt;&#xA;&lt;li&gt;1-4 can also be signs of other compact objects such as white dwarfs and neutron stars, and are often only found in binary systems. Only by using a multitude of different observations, and careful analysis, can an object finally be designated a black hole.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-01-10T22:47:51.647" LastActivityDate="2015-01-10T22:47:51.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6068" PostTypeId="1" CreationDate="2014-08-07T22:49:42.530" Score="2" ViewCount="166" Body="&lt;p&gt;How are millisecond pulsars formed, and what is the evolutionary track of a millisecond pulsar? &lt;/p&gt;&#xA;" OwnerUserId="2198" LastActivityDate="2014-08-09T12:30:20.510" Title="Birth and Evolution of Millisecond Pulsars" Tags="&lt;neutron-star&gt;&lt;pulsar&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6069" PostTypeId="2" ParentId="6066" CreationDate="2014-08-07T23:04:05.927" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://www.eso.org/public/usa/news/eso0846/&quot; rel=&quot;nofollow&quot;&gt;16-Year Long Study Tracks Stars Orbiting Milky Way Black Hole&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Undoubtedly the most spectacular aspect of our long term study is that it has delivered what is now considered to be the best empirical evidence that supermassive black holes do really exist. The stellar orbits in the Galactic Centre show that the central mass concentration of four million solar masses must be a black hole, beyond any reasonable doubt,&quot; says Genzel. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Turns out &lt;a href=&quot;http://en.wikipedia.org/wiki/S2_%28star%29&quot; rel=&quot;nofollow&quot;&gt;one of the stars (S2) orbiting&lt;/a&gt; this 4 million solar mass object orbits at &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/S2_%28star%29&quot; rel=&quot;nofollow&quot;&gt;a pericenter distance of 17 light hours (18 Tm or 120 AU) — about four times the distance of Neptune from the Sun.&lt;/a&gt;&quot;. That doesn't leave room for much &lt;em&gt;besides&lt;/em&gt; a black hole to account for the mass within its orbit.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2014-08-07T23:04:05.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6070" PostTypeId="2" ParentId="1184" CreationDate="2014-08-08T08:08:58.577" Score="-2" Body="&lt;p&gt;We DO have 2 summers and two winters - and 2 springs and autumns too.  The varying distance between the sun and earth does not cause the seasons.  They are caused by the tilt of the Earth towards the sun in summer and away from the sun in winter.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The northern and southern hemispheres both have the same seasons, but at opposite times of the year.&lt;br&gt;&#xA;For example, when it is winter in the United States, it is summer in Argentina and vice versa.&#xA;&lt;br&gt;&lt;br&gt;&#xA;Since the seasons in the northern and southern hemispheres occur at opposite times of the year, it means that there are 2 of each season or 8 distinct seasons each year, including 2 separate summers and 2 separate winters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps this link might help:&lt;br&gt;&#xA;&lt;a href=&quot;http://neoprogrammics.com/equinoxes_and_solstices/&quot; rel=&quot;nofollow&quot;&gt;http://neoprogrammics.com/equinoxes_and_solstices/&lt;/a&gt; &lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="2192" LastEditorUserId="2192" LastEditDate="2014-08-08T08:16:41.180" LastActivityDate="2014-08-08T08:16:41.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6071" PostTypeId="2" ParentId="6062" CreationDate="2014-08-08T09:23:26.370" Score="4" Body="&lt;p&gt;You may find unaltered images difficult to obtain. Firstly, most images are made from a combination of many short exposures, leading to the colours being 'built up'. Secondly, all but the newest astrophotographers tend to use imaging devices which are more sensitive to the IR part of the spectrum. Further to this, many will use filters to accentuate H - alpha regions, further obscuring 'true' colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that these techniques are used is to aid you, the viewer. Unedited images are often grey, faint and show little detail.&lt;/p&gt;&#xA;" OwnerUserId="2209" LastActivityDate="2014-08-08T09:23:26.370" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6072" PostTypeId="1" CreationDate="2014-08-08T11:36:59.597" Score="8" ViewCount="2436" Body="&lt;p&gt;&lt;a href=&quot;http://science.blurtit.com/154646/are-the-moon-phases-the-same-all-over-the-world-at-the-same-time&quot; rel=&quot;noreferrer&quot;&gt;It is known that theoretically&lt;/a&gt; we see the same moon phase everywhere on the earth, but practically I see that people are unable to observe the 0.8% waxing crescent at different places on earth. This has some grave consequences as in some cultures like Jewish,Muslim and hindu moon sighting can mark beginnings of festive months. &lt;a href=&quot;http://www.ibtimes.co.in/when-eid-al-fitr-2014-end-ramadan-how-muslims-celebrate-it-605439&quot; rel=&quot;noreferrer&quot;&gt;This link claims that different countries have different eid beginnings&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In India, Muslims will celebrate the festival on Monday or Tuesday, depending on the moon's sighting. In Pakistan, Eid will be celebrated on Tuesday as there are no chances of moon to be sighted on 27 July, The News Tribe reported.&#xA;  In the United Kingdom, Eid al-Fitr is likely to be celebrated on 29 July, according to timeanddate.com.&#xA;  In the United States, Muslims will mark the beginning of Shawwal on 28 July, the Fiqh Council of North America, an association of Muslims which is affiliated with the Islamic Society of North America (ISNA), stated in a press release.&#xA;  Muslims in Philippines will celebrate Eid on Tuesday as the official date of the end of Ramadan has been set as 29 July. The Philippine Government in their proclamation had declared 29 July as holiday throughout the country in observance of Eid al-Fitr.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the question is , is it really possible for people living in different part of the world to see or not see the new moon (0.8% waxing crescent) to mark their festive seasons at different dates?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words do Automated calculations \ Lunar calendars &lt;strong&gt;give an exact picture of the&lt;/strong&gt; moon sighting as &lt;strong&gt;viewed by a person on earth&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note : These cultures put personal observation of the new moon a pre requisite to mark the months hence the motivation of this question.  &lt;/p&gt;&#xA;" OwnerUserId="2215" OwnerDisplayName="earth101" LastEditorUserId="2215" LastEditDate="2014-08-08T20:15:59.287" LastActivityDate="2014-08-17T09:21:05.403" Title="Is it possible for a person to not see the new moon at different places on earth?" Tags="&lt;the-moon&gt;&lt;earth&gt;&lt;time&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6073" PostTypeId="1" AcceptedAnswerId="6085" CreationDate="2014-08-08T17:46:04.657" Score="4" ViewCount="860" Body="&lt;p&gt;In our Solar System, all the gas giants have ring systems. However, why do these rings form rings and not a cloud of debris around the planet? &lt;/p&gt;&#xA;" OwnerUserId="2212" LastActivityDate="2016-11-04T15:12:35.497" Title="Why do the rings around the Gas Giants get their 'ring' shape?" Tags="&lt;solar-system&gt;&lt;planetary-ring&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6074" PostTypeId="1" CreationDate="2014-08-08T19:26:29.503" Score="3" ViewCount="69" Body="&lt;p&gt;I have been looking on line for Collimator for my Meade Dobsonian. The only Collimators I can find are made for Newtonian Lenses. It's a 8&quot; mirror with a 1.25&quot; lens. Will this work for my telescope?&lt;/p&gt;&#xA;" OwnerUserId="2213" LastEditorUserId="1378" LastEditDate="2014-09-17T00:10:54.347" LastActivityDate="2014-09-17T00:10:54.347" Title="Collimator for Dobsonian, same as Newtonian?" Tags="&lt;telescope&gt;&lt;collimation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6075" PostTypeId="2" ParentId="6072" CreationDate="2014-08-08T21:03:44.047" Score="1" Body="&lt;p&gt;You can imagine that if you hopped into a rocket and blasted off in a lateral direction, you would be able to view the moon from a different perspective and see more and more of the side illuminated by the sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the earth is only about 12,700 km across whereas the earth-moon distance averages about 385,000 km. That means that the angle you are viewing the moon from the (nearly) opposite sides of the earth is a less than two degrees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;em&gt;(drawings not to scale)&lt;/em&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lOboN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, those 2 degree would vary over the &lt;em&gt;edge&lt;/em&gt; of the moon (where you are viewing the crescent), so you would see significantly, &lt;em&gt;significantly&lt;/em&gt; less difference in the crescent size over those two degrees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UxPeh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while it is &lt;em&gt;mathematically&lt;/em&gt; possible that two observers of a perfectly spherical illuminated body might see just the tiniest bit more and less than than a 0.8% crescent due to parallax&amp;hellip; in reality, the precision by which we make these observations is far, far less than what can be reasonably determined by the human eye or the margin of error due to the irregular surface of the moon, deciding what's &quot;illuminated&quot; in the twilight regions, etc, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discrepancy in dates is typically due to differing decrees of when the criteria have actually been met rather than any observational differences due to their position on the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;[I'm certain I saw a religious authority dispelling such differences in a Q&amp;amp;A interview;  but I cannot find the link, so perhaps someone can find a reference.]&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="58" LastEditorUserId="58" LastEditDate="2014-08-08T21:55:50.597" LastActivityDate="2014-08-08T21:55:50.597" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6078" PostTypeId="2" ParentId="5894" CreationDate="2014-08-08T23:13:14.617" Score="2" Body="&lt;p&gt;If you go far enough north or south, you will reach a latitude where the moon can be below the horizon for several days, just like the 'midnight sun'.  It can also be above the horizon for several days for the same reason.&#xA;&lt;br&gt;&lt;br&gt;&#xA;At the north or south poles, the moon may be either continuously above or below the horizon for nearly 2 weeks at a time.&lt;/p&gt;&#xA;" OwnerUserId="2192" LastActivityDate="2014-08-08T23:13:14.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6079" PostTypeId="1" AcceptedAnswerId="6117" CreationDate="2014-08-09T00:51:16.877" Score="4" ViewCount="1195" Body="&lt;p&gt;Can you tell me about Flamsteed numbers? I just know that it is a specific number used for identification of stars. But what is the theory behind it? In what way is the number given to stars?&lt;/p&gt;&#xA;" OwnerUserId="2195" LastEditorUserId="50" LastEditDate="2014-08-09T06:35:23.337" LastActivityDate="2015-04-16T10:01:50.210" Title="What are Flamsteed numbers?" Tags="&lt;milky-way&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6080" PostTypeId="2" ParentId="1531" CreationDate="2014-08-09T05:08:47.953" Score="2" Body="&lt;p&gt;As we view it, the moon moves westward across the sky due to the rotation of the earth, while at the same time it's moving eastward in its orbit at about 1 lunar diameter in about 57 minutes, or about 1 lunar diameter per hour or 12-14 degrees per day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why the moon can be seen blocking out (or occulting) a bright star or planet as it slowly moves eastward in front of it for several minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual motion of a star is too small to notice and it takes years to notice it without powerful telescopes.  For general purpose, day-to-day astronomical observations, the actual motions of the stars in 3D space can be ignored.&lt;/p&gt;&#xA;" OwnerUserId="2192" LastActivityDate="2014-08-09T05:08:47.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6081" PostTypeId="2" ParentId="6074" CreationDate="2014-08-09T06:12:31.877" Score="2" Body="&lt;p&gt;Yes, of course it will work. A Dobsonian is merely a Newtonian telescope on a special mount. &lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2014-08-09T06:12:31.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6082" PostTypeId="2" ParentId="6062" CreationDate="2014-08-09T06:22:08.803" Score="6" Body="&lt;p&gt;You can use one of the digital sky surveys. Examples include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sloan digital sky survey &lt;a href=&quot;http://www.sdss.org/&quot; rel=&quot;noreferrer&quot;&gt;SDSS&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://archive.eso.org/dss/dss&quot; rel=&quot;noreferrer&quot;&gt;ESO online digitized sky survey&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Their images contain visible (red or blue) wavelengths as well as infrared. All images are monochromatic, as are almost all professional astronomy photographs. You could build your own composite color image from the different channels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example image from the ESO DSS2 archive in red wavelength:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/o0Hv6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2014-08-09T06:22:08.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6083" PostTypeId="2" ParentId="6068" CreationDate="2014-08-09T12:30:20.510" Score="3" Body="&lt;p&gt;The standard model for the formation of millisecond pulsars is the &lt;strong&gt;recycling scenario&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The starting point of this model is a stellar binary system consisting of a neutron star and a secondary 'regular' star, where the neutron star is a normal pulsar with a rotational period of roughly $\sim7$ seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because a neutron star is such a compact object, its gravitational influence is very large ($F_g \propto M/R^2$). If the two stars are close enough, the outer layers of gas may be pulled of the companion star and flow towards the neutron star. All this matter will settle into an disk structure around the neutron star and start orbiting faster and faster as it draws closer to the surface ($v_{orbit} = \sqrt{GM/R}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the time this matter touches down on the neutron star surface its orbital period can be as fast as a millisecond, certainly significantly faster than the neutron star. This bit of motion (or more precisely, the angular momentum) is then transferred to the neutron star (this process is called accretion), causing it to rotate slightly faster (spin-up).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some $0.1\sim0.2 M_{\odot}$ (where $M_{\odot}= 2 \times 10^{30}$ kg gives the mass of the Sun) of material is transferred to the neutron star, its rotational period will have decreased from a few second down to milliseconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transfer of this much matter takes a very long time, roughly $10^8$ to $10^9$ years. After this time, and with all the redistribution of mass, the binary system is expected to widen which causes the mass transfer to stop. As the region around the neutron star clear up, the radio emission mechanism turns back on (this mechanism does not work during the accretion stage) and we are left with a millisecond pulsar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this point on the neutron star will behave like a regular pulsar: over the course of a very long time it will convert its rotational energy in radio emission and slow down until it becomes to slow to act as a pulsar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the evidence for this model; such binary systems containing an accreting neutron star emit massive amounts of X-ray emission (hence called &lt;a href=&quot;http://en.wikipedia.org/wiki/X-ray_binary&quot; rel=&quot;nofollow&quot;&gt;X-ray binaries&lt;/a&gt;) and have been known to exist since the 60's, long before &lt;a href=&quot;http://en.wikipedia.org/wiki/PSR_B1937%2B21&quot; rel=&quot;nofollow&quot;&gt;the first millisecond radio pulsar&lt;/a&gt; was discovered (in 1982). This entire scenario was proposed quickly after, but could not be confirmed for a very long time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was only in the late 90's that a neutron star was found whose X-ray emission was pulsed at millisecond periods (see &lt;a href=&quot;http://en.wikipedia.org/wiki/SAX_J1808.4-3658&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, and &lt;a href=&quot;http://www.nature.com/nature/journal/v394/n6691/full/394344a0.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;), confirming that these neutron stars are indeed spinning up. The &lt;a href=&quot;http://www.nature.com/nature/journal/v501/n7468/full/nature12470.html&quot; rel=&quot;nofollow&quot;&gt;final piece of evidence&lt;/a&gt;; a millisecond pulsar which switches back and forth between X-ray and radio emission was only found very recently.&lt;/p&gt;&#xA;" OwnerUserId="494" LastActivityDate="2014-08-09T12:30:20.510" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6084" PostTypeId="1" CreationDate="2014-08-09T17:02:42.923" Score="7" ViewCount="161" Body="&lt;p&gt;I am studying ellipticity distributions of galaxies and am having trouble moving from the second central moment matrix (covariance matrix) to properties for an object such as its half-light radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already know how to obtain $e1$ and $e2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are given by the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e = \frac{Q_{xx} - Q_{yy} + 2iQ_{12}}{N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $N = N(Q)$ and is just the normalization. The real component is $e1$ and the imaginary component is $e2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering how to obtain the half-light radius, as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2r^2 = Q_{xx} + Q_{yy}$ for a circular profile, then one can obtain the HLR from $r^2$ but what about for non-circular profiles? Is the above equation even valid?&lt;/p&gt;&#xA;" OwnerUserId="2076" LastActivityDate="2015-09-19T23:08:14.070" Title="Going from the second moments of an object to its ellipticities and half-light-radius" Tags="&lt;cosmology&gt;&lt;gravitational-lensing&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6085" PostTypeId="2" ParentId="6073" CreationDate="2014-08-09T19:23:30.603" Score="4" Body="&lt;p&gt;There are different theories as to how planetary rings form, and so there are different answers to your question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One theory of planetary ring formation is that a small (or large, depending on the ring size) moon wandered into the gas giant's &lt;a href=&quot;https://en.wikipedia.org/wiki/Roche_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Roche limit&lt;/a&gt; (the radius from the celestial body inside which no second celestial body can stay intact due to tidal disruptions by the larger body). The moon would have been orbiting the planet, and would thus have been moving in a flat . The resulting debris continued to move in a plane; if it had been angled, &lt;a href=&quot;https://physics.stackexchange.com/a/6552/56299&quot;&gt;collisions would have flattened it&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another theory is that the debris is left over from the early Solar System. A &lt;a href=&quot;https://en.wikipedia.org/wiki/Protoplanetary_disk&quot; rel=&quot;nofollow noreferrer&quot;&gt;protoplanetary disk&lt;/a&gt; formed around the Sun; it contained all the material that makes up the planets. Over time (and through a lot of collisions!), bits of the material slammed together and formed planets. The trouble is, not all of that material became planets. Some became asteroids, some became comets, and some became moons (although the theory of the formation of the Earth's moon is a bit different). Some of this material could also have coalesced around planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the protoplanetary disk would already have been in a disk shape. Why? Think of a rotating object - like a planet. You might have read that the Earth is not a sphere. Aside from the obvious deviations of mountains and valleys, some of the deformation comes from the Earth's rotation. Any rotating body will bulge a little around its equator; a spherical body will become an &lt;a href=&quot;https://en.wikipedia.org/wiki/Spheroid#Oblate_spheroids&quot; rel=&quot;nofollow noreferrer&quot;&gt;oblate spheroid&lt;/a&gt;. The same thing happened for the protoplanetary disk - it became a disk because it was rotating! Any material that coalesced around planets would already be in a plane - that's why the planets' orbits aren't at odd angles to each other. So this material would be in a flat shape if it was captured by the planets, and would become a circular ring because of rotation. If, at that point, there was material that &lt;em&gt;wasn't&lt;/em&gt; in a flat plane, it would become flat by the same process that made the protoplanetary disk flat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are stellar systems where the protoplanetary disk isn't in the same plane as the &lt;em&gt;star's rotation&lt;/em&gt;. This could affect planetary orbits; however, such systems have not been thoroughly studied because of their rarity.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-11-04T15:12:35.497" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6086" PostTypeId="2" ParentId="3626" CreationDate="2014-08-09T20:24:38.900" Score="4" Body="&lt;p&gt;Let's start answering your question in reverse. Ceres could not be &quot;the missing Theia&quot;, because of its shape. If an object hit the Earth at an angle (as is currently thought), it would be pretty deformed, if it managed to stay together. If it hit the Earth head on . . . Well, it would almost certainly not survive. So if it did survive the impact, it would be in pretty bad shape. Where would such an object go?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's investigate the idea that it went into the asteroid belt. That begs a question: Where is it today? Ceres is the largest object in the belt (and we've already ruled it out). Theia had a mass about that of Mars (see &lt;a href=&quot;http://science.sciencemag.org/content/early/2012/10/16/science.1226073&quot; rel=&quot;nofollow noreferrer&quot;&gt;Canup (2012)&lt;/a&gt;). It's safe to say that if Theia survived with much of its mass intact and then disappeared into the asteroid belt, we should have observed it by now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for your belt-formation theory. It is, admittedly a very cool idea. The mass of the asteroid belt is about 4% that of the Moon - so it could be the result of the planetary crash. And some theories say that the belt is, in fact, the remains of a planet. There is, of course, one problem - the different composition of the asteroids. There are three types, differentiated by their composition - C, S, and M. So it is very unlikely that they came from the same planet. &lt;a href=&quot;https://arxiv.org/abs/1410.3819&quot; rel=&quot;nofollow noreferrer&quot;&gt;Meier et al. (2014)&lt;/a&gt; show evidence that indicates Theia's composition would have been more like Earth, containing enstatite chondrites instead of carbonaceous chondrites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, let's look at your idea about the Earth having an orbit farther out. This one seems to have a good chance of being right. The early protoplanets in the Solar System were very - very - unstable in their orbits because of the frequency of collisions. So the Earth could have been in a different orbit early on, and then migrated inwards or outwards, though planet-planet scattering or perhaps gas disk migration.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-11-04T15:28:15.573" LastActivityDate="2016-11-04T15:28:15.573" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6089" PostTypeId="1" AcceptedAnswerId="6094" CreationDate="2014-08-10T06:53:52.280" Score="7" ViewCount="377" Body="&lt;p&gt;Do geostationary satellites need to have the equator as the plane of rotation, and the earth's centre to be the centre of rotation? Can it rotate over, say, the Tropic of Cancer, focusing on a single city?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, in which case, we need a chain of geostationary satellites to do a function, why employ geostat satellites at all? Aren't there satellites at lower altitudes which could do a better job?&lt;/p&gt;&#xA;" OwnerUserId="2223" LastEditorUserId="16462" LastEditDate="2017-04-01T06:58:00.660" LastActivityDate="2017-04-01T06:58:00.660" Title="Are multiple geostationary satellites required to handle arbitrary locations?" Tags="&lt;orbit&gt;&lt;artificial-satellite&gt;&lt;satellite&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6090" PostTypeId="1" CreationDate="2014-08-10T08:17:56.143" Score="3" ViewCount="13936" Body="&lt;p&gt;From what I've read, white stars are hotter than red ones. But a white dwarf would have just heavy elements to fuse, so shouldn't it be less bright?&lt;/p&gt;&#xA;" OwnerUserId="1156" LastActivityDate="2016-05-11T18:39:18.210" Title="Is a white dwarf hotter than a Red Giant?" Tags="&lt;star&gt;&lt;astrophysics&gt;&lt;stellar-evolution&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6091" PostTypeId="2" ParentId="6090" CreationDate="2014-08-10T09:46:39.833" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://en.m.wikipedia.org/wiki/White_dwarf&quot; rel=&quot;nofollow&quot;&gt;White dwarfs&lt;/a&gt; start hotter when they are created (up to billions degrees Kelvin), but in the end, they end as black dwarfs, which only happens after a few billion years. As per the age of the universe, it is currently the assumption that there are no black dwarfs yet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.m.wikipedia.org/wiki/Red_giant&quot; rel=&quot;nofollow&quot;&gt;Red giants&lt;/a&gt; are (on the surface) typically below 5000 K. Their core is up to a billion degrees Kelvin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://curiosity.discovery.com/question/red-giant-difference-white-dwarf&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; might be an interesting link too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a final note: white dwarfs are the end stage of a star's life. Depending on the mass of the star, red giants can end up as white dwarfs. About 97% of the milky way stars will end as white dwarfs (including our sun).&lt;/p&gt;&#xA;" OwnerUserId="2189" LastEditorUserId="2189" LastEditDate="2014-08-10T11:31:21.980" LastActivityDate="2014-08-10T11:31:21.980" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6092" PostTypeId="2" ParentId="6089" CreationDate="2014-08-10T12:32:17.307" Score="1" Body="&lt;p&gt;You cannot point an antenna with a fixed orientation at a satellite that is not in a geostationary orbit. So to get satellite tv you'd not only need a bunch of satellites in orbit instead of one, but every receiver would need probably at least a couple of dishes on complex and highly accurate alt-az mounts.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-08-10T12:32:17.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6093" PostTypeId="2" ParentId="6089" CreationDate="2014-08-10T14:46:12.300" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Do geostationary satellites need to have the equator as the plane of rotation, and the earth's centre to be the centre of rotation? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To be stationary above a point, yes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can it rotate over, say, the Tropic of Cancer, focusing on a single city?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the satellite's orbit touched the Tropic of Cancer, it would not be geostationary since the orbit about the center of the Earth would move the satellite north and south to reach the Tropic of Capricorn as well. Any geostationary satellite can be placed so that a designated city is always in line of sight, it doesn't need to be directly overhead.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If not, in which case, we need a chain of geostationary satellites to do a function, why employ geostat satellites at all? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only one geostationary satellite is needed for some functions, such as TV broadcasts to the Eastern Seaboard.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Aren't there satellites at lower altitudes which could do a better job?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It depends on what the job is. For continuous broadcast and reception of signals, a geostationary satellite is best. For obtaining crop information, a low Earth orbit satellite might be better.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-08-10T14:46:12.300" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6094" PostTypeId="2" ParentId="6089" CreationDate="2014-08-10T15:33:42.593" Score="6" Body="&lt;p&gt;Earth's center of mass must be at one of the two focus points of a satellite's elliptical orbit, or at the center of a circular orbit such as a geostationary orbit. One cannot orbit a certain latitude, except for the equator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are clever alternatives for different purposes. Geosynchronous (as opposed to geostationary) orbits mean that the satellite will always be within sight from the same location on Earth, but it would move around and probably need antennas which continuously aim towards them. &lt;a href=&quot;http://en.wikipedia.org/wiki/Quasi-Zenith_Satellite_System&quot; rel=&quot;noreferrer&quot;&gt;Japan&lt;/a&gt; is building its own regional GPS-system in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_orbits&quot; rel=&quot;noreferrer&quot;&gt;semi-synchronous orbits&lt;/a&gt; which put the satellite at the same place on the sky at the same time every day. This is used for some Earth observation purposes because the shadows on the ground remain the same for the same location on Earth everytime an image is taken.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2014-08-10T15:33:42.593" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6096" PostTypeId="1" CreationDate="2014-08-11T13:54:35.407" Score="10" ViewCount="142" Body="&lt;p&gt;I would like to know how I can estimate the &lt;strong&gt;tangential&lt;/strong&gt; and &lt;strong&gt;cross&lt;/strong&gt; component of the galaxy's &lt;strong&gt;shear&lt;/strong&gt; using &lt;a href=&quot;https://en.wikipedia.org/wiki/Gnomonic_projection&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gnomonic projection&lt;/a&gt;&#xA;of the right ascension and declination the galaxies relate to the center of galaxy cluster and estimate?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Edit.&lt;/strong&gt; &lt;em&gt;Shear&lt;/em&gt; is the second order moment of a galaxy's surface brightness. You can find more on the topic in the following &lt;a href=&quot;http://gravitationallensing.pbworks.com/w/page/15553259/Weak%20Lensing&quot; rel=&quot;nofollow noreferrer&quot;&gt;wiki on weak lensing&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2231" LastEditorUserId="37191" LastEditDate="2021-03-11T12:40:48.880" LastActivityDate="2021-03-11T12:40:48.880" Title="Estimating the tangential and cross component of the galaxy's shear using Gnomonic projection" Tags="&lt;cosmology&gt;&lt;mathematics&gt;&lt;galaxy-cluster&gt;&lt;gravitational-lensing&gt;&lt;weak-lensing&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="6097" PostTypeId="2" ParentId="6052" CreationDate="2014-08-11T17:23:36.460" Score="13" Body="&lt;p&gt;Before I start, I'll admit that I've criticized the question based on its improbability; however, I've been persuaded otherwise. I'm going to try to do the calculations based on completely different formulas than I think have been used; I hope you'll stay with me as I work it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's imagine that Lucifer becomes a main-sequence star - in fact, let's call it a low-mass red dwarf. Main-sequence stars follow the mass-luminosity relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{L}{L_\odot} = \left(\frac{M}{M_\odot}\right)^a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $L$ and $M$ are the star's luminosity and mass, and $L_\odot$ and $M_\odot$ and the luminosity and mass of the Sun. For stars with $M &amp;lt; 0.43M_\odot$, $a$ takes the value of 2.3. Now we can plug in Jupiter's mass ($1.8986 \times 10 ^{27}$ kg) into the formula, as well as the Sun's mass ($1.98855 \times 10 ^ {30}$ kg) and luminosity ($3.846 \times 10 ^ {26}$ watts), and we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{L}{3.846 \times 10 ^ {26}} = \left(\frac{1.8986 \times 10 ^ {27}}{1.98855 \times 10 ^ {30}}\right)^{2.3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This becomes $$L = \left(\frac{1.8986 \times 10 ^ {27}}{1.98855 \times 10 ^ {30}}\right)^{2.3} \times 3.846 \times 10 ^ {26}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which then becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L = 4.35 \times 10 ^ {19}$$ watts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can work out the apparent brightness of Lucifer, as seen from Earth. For that, we need the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = m_\odot - 2.5 \log \left(\frac {L}{L_\odot}\left(\frac {d_\odot}{d}\right) ^ 2\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ is the apparent magnitude of the star, $m_\odot$ is the apparent magnitude of the Sun, $d_\odot$ is the distance to the Sun, and $d$ is the distance to the star. Now, $m = -26.73$ and $d(s)$ is 1 (in astronomical units). $d$ varies. Jupiter is about 5.2 AU from the Sun, so at its closest distance to Earth, it would be ~4.2 AU away. We plug these numbers into the formula, and find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = -6.25$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is a lot less brighter than the Sun. Now, when Jupiter is farthest away from the Sun, it is ~6.2 AU away. We plug &lt;em&gt;that&lt;/em&gt; into the formula, and find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = -5.40$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is dimmer still - although, of course, Jupiter would be completely blocked by the Sun. Still, for finding the apparent magnitude of Jupiter at some distance from Earth, we can change the above formula to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = -26.73 - 2.5 \log \left(\frac {4.35 \times 10 ^ {19}}{3.846 \times 10 6 {26}}\left(\frac {1}{d}\right) ^ 2\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By comparison, the Moon can have an average apparent magnitude of -12.74 at full moon - much brighter than Lucifer. The apparent magnitude of both bodies can, of course, change - Jupiter by transits of its moon, for example - but these are the optimal values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the above calculations really don't answer most parts of your question, I hope it helps a bit. And please, correct me if I made a mistake somewhere. LaTeX is by no means my native language, and I could have gotten something wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The combined brightness of Lucifer and the Sun would depend on the angle of the Sun's rays and Lucifer's rays. Remember how we have different seasons because of the tilt of the Earth's axis? Well, the added heat would have to do with the tilt of Earth's and Lucifer's axes relative to one another. I can't give you a numerical result, but I can add that I hope it wouldn't be too much hotter than it is now, as I'm writing this!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like I said in a comment somewhere on this page, the mass-luminosity relation really only works for main-sequence stars. If Lucifer was not on the main sequence. . . Well, then none of my calculations would be right.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="525" LastEditDate="2017-05-11T13:57:13.627" LastActivityDate="2017-05-11T13:57:13.627" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6098" PostTypeId="2" ParentId="6090" CreationDate="2014-08-11T17:27:15.370" Score="6" Body="&lt;p&gt;&quot;White&quot; stars are typically much brighter than Red stars, as both the &quot;color &amp;amp; brightness&quot; of a star are directly proportional to the temperature. The only reason there are &quot;bright&quot; red stars is that their radius is incredibly large. Note that the &quot;&lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/f/ff/BlackbodySpectrum_loglog_150dpi_en.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;color&lt;/a&gt;&quot; of a star is directly linked to the temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation that best demonstrates this is the luminosity equation of a black body. Stars aren't perfect black bodies, but they are close enough that they are treated as such.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;L = 4πR²σT⁴&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;this equation tells us that the Luminosity (L) is proportionate to the Radius Squared (R²) and the Temperature to the Fourth power (T⁴). The bigger the brighter, or, the hotter the brighter. Meaning that for a given radius the hotter the star, the more luminous, and the same goes for stars of the same temperature, the larger the radius the more luminous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;White dwarfs on the other hand are &lt;strong&gt;not stars&lt;/strong&gt; in the sense that they do not fuse anything, they simply glow due to the lingering heat that was generated during their time as stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As shown in the &lt;a href=&quot;http://chandra.harvard.edu/graphics/edu/formal/variable_stars/HR_diagram.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;HR-Diagram&lt;/a&gt;, White Dwarfs are some of the hottest objects in the universe, and as stated by &lt;a href=&quot;https://astronomy.stackexchange.com/users/2189/agtoever&quot;&gt;agtoever&lt;/a&gt; there has not been enough time for even the oldest white dwarf to have cooled passed something like 4800K. &lt;/p&gt;&#xA;" OwnerUserId="2232" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-08-12T06:00:59.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6099" PostTypeId="1" AcceptedAnswerId="6108" CreationDate="2014-08-11T18:59:12.607" Score="6" ViewCount="7601" Body="&lt;p&gt;Are there any differences between quasars and Active Galactic Nuclei (AGN)?  As I understand it, they are both caused by extremely luminous electromagnetic emission from material accreting onto a super-massive black hole.  If they have the same source, is there still some fundamental difference between them (and if so what is it) or are they two names for the same thing?&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2014-08-12T07:02:20.033" Title="Difference between quasar and Active Galactic Nuclei?" Tags="&lt;quasars&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6100" PostTypeId="1" AcceptedAnswerId="6102" CreationDate="2014-08-11T19:10:13.213" Score="9" ViewCount="691" Body="&lt;p&gt;If I understand correctly, the Earth lies in an area of the galaxy - the &quot;local bubble&quot; - that is relatively free of dust and gases and where no new stars are being born.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the current theory on the relation between life on Earth and this location? Or in other words: Would we expect life to exist anywhere in the galaxy, e.g. also inside some nebula, or only in specific areas, and how are they characterised?&lt;/p&gt;&#xA;" OwnerDisplayName="user2233" LastEditorUserId="2153" LastEditDate="2014-08-11T19:58:50.180" LastActivityDate="2014-08-12T05:44:17.250" Title="Do we expect life to develop only in certain regions of the galaxy?" Tags="&lt;galaxy&gt;&lt;life&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6101" PostTypeId="2" ParentId="6037" CreationDate="2014-08-11T19:31:20.060" Score="2" Body="&lt;p&gt;Since human beings appear to have a biological sleep cycle close to an Earth day, I would strongly suspect that extraterrestrial human colonies would keep this rhythm. People living close to the poles of the Earth, where it is light all summer and dark all winter, don't stay awake for half a year and then sleep the other half. Similarly, it doesn't make sense to have a 29 Earth-day long Moon-day.&lt;/p&gt;&#xA;" OwnerDisplayName="user2233" LastActivityDate="2014-08-11T19:31:20.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6102" PostTypeId="2" ParentId="6100" CreationDate="2014-08-11T19:38:07.793" Score="3" Body="&lt;p&gt;There's a short answer here and a long answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Short Answer:&lt;/strong&gt; It isn't special.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Long Answer:&lt;/strong&gt; Yes and No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Local Bubble really isn't so special with regards to where it's placed in the galaxy. Location-wise, it's perched on the inner edge of the Orion Arm, one of the Milky Way's spiral arms. The Orion Arm, too, isn't incredibly special. It's relatively small - it's often referred to as the &quot;Orion Spur&quot; - and near 1/4 of the way out from the galactic center. Not too far away, but not too near. The donut-shaped area 4,000-10,000 parsecs out from the galactic center is, however, sometimes called the &lt;em&gt;Galactic Habitable Zone&lt;/em&gt;. In theory. life as we know it could exist quite nicely any where in this area (I should note that the scientific community is divided on whether the Zone is important, with some saying that it is arbitrary).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let's dive in deeper. Actually, we're in a decent spot in the &quot;Arm.&quot; We're on the edge of it - not too near large groups of stars that could effect our system via gravity, or near any supernova candidates. And the &quot;Arm&quot; itself isn't really a spiral arm. It is, as I said earlier, more of a spur, or bulge. There aren't as many stars to watch out for. Also, we're near the &lt;em&gt;corotation circle&lt;/em&gt; - i.e., we orbit the galaxy at such a distance that we move at a similar speed to the arms, reducing further the likelihood of crossing them. Is this necessary for life? Nope. But does it help? Oh, yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are, of course, other factors that make this area a nice spot: No black holes nearby, gamma-ray bursts, etc. But could life survive elsewhere? Yep. These same characteristics can be found in many other places in the galaxy - and life can probably survive in even more dangerous spots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, I'd avoid using the term &quot;nebula&quot;. Technically, we're not in one, or anywhere near one. The term can refer to a multitude of objects - planet-forming areas, &quot;clouds&quot; surrounding white dwarfs, and other rather photogenic places - none of which we inhabit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-08-11T19:54:11.077" LastActivityDate="2014-08-11T19:54:11.077" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6103" PostTypeId="2" ParentId="6100" CreationDate="2014-08-11T19:38:23.930" Score="3" Body="&lt;p&gt;As my username indicates, I'm eager to point out that we are currently inside the &quot;local fluff&quot;, which is &quot;stuff&quot; which has been ejected long ago by some disturbingly active young stars in that evil &lt;a href=&quot;http://en.wikipedia.org/wiki/Scorpius%E2%80%93Centaurus_Association&quot; rel=&quot;nofollow&quot;&gt;Scorpius-Centaurus association&lt;/a&gt; through the &lt;a href=&quot;http://en.wikipedia.org/wiki/Loop_I_Bubble&quot; rel=&quot;nofollow&quot;&gt;local bubble&lt;/a&gt;. The waves of their stellar winds just happen to pass us by in our current million year era.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, our Sun does circle the entire Milky Way every 250 000 000 years. Dinosaurs appeared about one galactic orbit ago, but died out a quarter of an orbit ago. Earth has had quite advanced life during at least one galactic orbit. That's why I argue that there is little dependence of a solar systems habitability on its orbital position in a galactic scale. (Or that life on Earth has been incredibly lucky, and finally someone will, right? So what do we know from observing no one but ourself?)&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2014-08-11T19:55:49.503" LastActivityDate="2014-08-11T19:55:49.503" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6104" PostTypeId="1" CreationDate="2014-08-11T20:47:36.010" Score="5" ViewCount="234" Body="&lt;p&gt;In documentaries the Big Bang is often represented as blackness followed by a big explosion with loud noises. This always feels cheap and falling short of explaining the origin of the Universe. Is this really the best way to get green-thumbs thinking about the origin of the Universe? Are there other, better, representations or analogies of the Big Bang? If so, how do they better describe the current cosmological model than the &quot;explosion&quot; analogy?&lt;/p&gt;&#xA;" OwnerUserId="966" LastActivityDate="2015-02-23T08:05:04.113" Title="Why are explosions always used to represent the Big Bang?" Tags="&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6105" PostTypeId="1" AcceptedAnswerId="6110" CreationDate="2014-08-11T20:57:59.387" Score="7" ViewCount="3036" Body="&lt;p&gt;When I've been to Astronomy-nights for the public at the local observatory, they always used green lasers to point things out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is there such a preference for green lasers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's wrong with red lasers? (Which seem cheaper and more readily available.)&lt;/p&gt;&#xA;" OwnerUserId="2234" LastEditorUserId="1891" LastEditDate="2014-08-27T11:15:41.060" LastActivityDate="2017-07-15T21:29:51.433" Title="Why do astronomers like green laser pointers?" Tags="&lt;observatory&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="1" ClosedDate="2014-08-25T16:23:40.483" ContentLicense="CC BY-SA 3.0" />
  <row Id="6106" PostTypeId="2" ParentId="6100" CreationDate="2014-08-12T05:44:17.250" Score="1" Body="&lt;p&gt;This may not be a very scientific answer. I think it gives some incredible insight, so I thought I'd share it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What &lt;a href=&quot;https://www.youtube.com/watch?v=sNGUkdovn_8&quot; rel=&quot;nofollow&quot;&gt;&quot;Vlad the Astrophysicist&quot;&lt;/a&gt; said:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If we condense the entire time that has passed since the Big Bang into 5 min.&lt;/li&gt;&#xA;&lt;li&gt;And we consider the vastness of space and the time each civilization lasts.&lt;/li&gt;&#xA;&lt;li&gt;Then life and civilizations that develop from it rise and fall in less than a split second. Likely never coming into contact with any others.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We can only assume life is quite possible anywhere where the conditions happen to right. But it's very hard to actually &lt;em&gt;quantitatively&lt;/em&gt; put a number to this since the universe is so HUGE. All we can say is basically, life (as we define it) is more likely develop in the &quot;Goldilock&quot; zones found in the galaxy, which are bound to exist. Notice this may vary depending our definition or understanding of &quot;life&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2238" LastActivityDate="2014-08-12T05:44:17.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6108" PostTypeId="2" ParentId="6099" CreationDate="2014-08-12T06:32:36.813" Score="5" Body="&lt;p&gt;A quasar is a type of AGN. Its main property is that it's galaxy axis points towards the earth and this way we receive its light and radio signals. That's why they are one of the most energetic AGN. Here you can see a relation of all the AGN types: &lt;a href=&quot;http://en.wikipedia.org/wiki/Active_galactic_nucleus#Summary&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Active_galactic_nucleus#Summary&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1678" LastActivityDate="2014-08-12T06:32:36.813" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6109" PostTypeId="2" ParentId="6099" CreationDate="2014-08-12T07:02:20.033" Score="7" Body="&lt;p&gt;All quasars are AGN, but not all AGN are quasars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AGN is a terminology that comes later than quasars. Quasars is the term applied at the beginning, when the first objects of this type have been discovered. They were radio-loud and point-like (the so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Quasar&quot; rel=&quot;nofollow noreferrer&quot;&gt;quasi-stellar radio sources&lt;/a&gt;). This characterization still holds nowadays. Another property is that quasars are cosmological, that is they are objects in the distant universe (redshift $z &amp;gt; 1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AGN is the most general term we can use to refer to active galaxies. AGN include all of them (Seyfert galaxies, Quasars, Blazars, etc.). In the end, it seems that the difference among various types of AGN is due to a &lt;a href=&quot;https://astronomy.stackexchange.com/a/2492/1038&quot;&gt;mix of angle-of-sight/epoch-of-observation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-08-12T07:02:20.033" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6110" PostTypeId="2" ParentId="6105" CreationDate="2014-08-12T09:38:32.447" Score="9" Body="&lt;p&gt;Mostly general purpose laser pointers are used for pointing things at smaller distances eg. Diagram or Equations in Powerpoint presentations, so the power of such laser pointers is quite limited/restricted to 5mw ( Class 3A or IIIa) or 10mW in some regions.&#xA;Because of the low power and small aperture of laser pointers if you point them through empty space, you can only see them impact spot where it hits a surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to point stars/planets as in astronomy club during observation sessions these low power laser pointers are not useful as there is no surface where the beam can hit &amp;amp; you can see the spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some higher-powered laser pointers project a visible beam via scattering from dust particles or water droplets along the beam path.&#xA;Higher-power ( Class 3B or IIIb lasers: greater than 5mW) and higher-frequency (Green or Blue) lasers may produce a beam visible even in clean air because of &quot;Rayleigh Scattering&quot; from air molecules, especially when viewed in moderately-to-dimly lit conditions. The strong wavelength dependence of the scattering (~λ^ −4 ) means that shorter (green &amp;amp; blue) wavelengths are scattered more strongly than longer (red) wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[ Rayleigh scattering is the mainly the elastic scattering (in lay term collision) of light or other electromagnetic radiation by particles much smaller than the wavelength of the light.&#xA;Rayleigh scattering results from the electric polarizability of the particles (locally separating positive &amp;amp; negative charges by small distance apart inside the atom. i.e. temporarily separating centre of positive charge and centre of negative charge). The oscillating electric field of a light wave acts on the charges within a particle, causing them to move at the same frequency (hence the green colour). The particle therefore becomes a small radiating dipole (or small sources of light) whose radiation we see as scattered light (the beam passing from pointer to star pointed). ]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intensity of such scattering increases when these beams are viewed from angles near the beam axis. (That's the reason, if you are standing close to the person pointing stars in astronomy club, you can see the beam bright and clearly whereas the people standing away can barely see it. So always stand near the presenter But avoid contact with your eye. :-) )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The apparent brightness of a spot from a laser beam depends on the optical power of the laser, the reflectivity of the surface, and the chromatic response of the human eye. For the same optical power, Green Laser light will seem brighter than other colours because the human eye is most sensitive at low light levels in the green region of the spectrum (wavelength 520–570 nm). The most sensitive pigment, rhodopsin, has a peak response at 500 nm. Sensitivity decreases for redder or bluer wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, Green laser pointers are of moderate power (sufficient for Rayleigh  Scattering in clean air), they are compact and relatively cheaper than blue ones (even though human eye is less sensitive to blue colour, blue laser are a little bit more expensive than green one.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Avoid their contact with your eyes and don't point them at aircraft.&#xA;People have been given up to five years in jail for aiming a green laser at an aircraft.&lt;/p&gt;&#xA;" OwnerUserId="2088" LastEditorUserId="13681" LastEditDate="2017-07-15T21:29:51.433" LastActivityDate="2017-07-15T21:29:51.433" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6111" PostTypeId="2" ParentId="6105" CreationDate="2014-08-12T09:54:18.080" Score="2" Body="&lt;p&gt;Have you ever tried shining a red laser up into the sky? If you have, you see nothing. Now, if you try shining a green laser up into the sky, you see the beam streaking off into infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because green lasers are much stronger than red lasers. Red lasers are used so often in everyday life because they aren't very dangerous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The colour red is near the infrared spectrum, which is a lower energy than the visible spectrum. This means that, when a photon of red laser light hits a particle of air or dust, it is just absorbed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The colour green, however, is much further along the spectrum, meaning that it is a higher energy than the colour red. When a photon of green laser light hits a particle of air, it is absorbed. Then, because it is so high energy, it is re-emitted. Or, in lay mans terms, the beam reflects off the air particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This scattering is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Rayleigh_Scattering&quot; rel=&quot;nofollow&quot;&gt;Rayleigh Scattering&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reflection is used so astronomers can trace the beam up to where is being pointed. This is useful for finding stars, galaxies and a whole manner of other celestial objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;P.S.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted the most powerful visible laser, you would have to go for a near ultraviolet laser. Of course, to generate such a beam, you would need power that is illegal in most countries.&lt;/p&gt;&#xA;" OwnerUserId="1494" LastEditorUserId="1494" LastEditDate="2014-08-24T08:04:29.833" LastActivityDate="2014-08-24T08:04:29.833" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6112" PostTypeId="1" CreationDate="2014-08-12T10:15:54.060" Score="12" ViewCount="1259" Body="&lt;p&gt;The following image is a map of the area surrounding the &lt;a href=&quot;http://en.wikipedia.org/wiki/Local_Bubble&quot; rel=&quot;noreferrer&quot;&gt;Local Bubble&lt;/a&gt;. One side of the image is 1700 lightyears. From the Sun to the Hyades (immediately below the sun) it is 150 lightyears.&lt;/p&gt;&#xA;&lt;p&gt;The local bubble, with the sun in its center, has three immediate neighboring bubbles, numbered in roman numerals, &lt;a href=&quot;http://en.wikipedia.org/wiki/Loop_I_Bubble&quot; rel=&quot;noreferrer&quot;&gt;Loop I&lt;/a&gt;, II, and III. At the lower right edge we can see a section of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Orion-Eridanus_Superbubble&quot; rel=&quot;noreferrer&quot;&gt;Orion-Eridanus-Superbubble&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;h3&gt;What is the name of the area marked in green?&lt;/h3&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vNYaI.gif&quot; alt=&quot;Image of Local the Bubble at a scale of 50 Lightyears&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Image source: Henbest/Couper, The Guide to the Galaxy. Found online at &lt;a href=&quot;http://www.astronomy.ohio-state.edu/%7Epogge/TeachRes/Ast162/ISM/Local500pcAnn.gif&quot; rel=&quot;noreferrer&quot;&gt;http://www.astronomy.ohio-state.edu/~pogge/TeachRes/Ast162/ISM/Local500pcAnn.gif&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user2233" LastEditorUserId="-1" LastEditorDisplayName="user2233" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2014-08-13T14:42:18.703" Title="Name of area close to Local Bubble?" Tags="&lt;terminology&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6113" PostTypeId="1" AcceptedAnswerId="6114" CreationDate="2014-08-12T12:29:48.250" Score="1" ViewCount="173" Body="&lt;p&gt;The Schwarzschild Radius is a theoretical radius, that if the mass of the object was compressed into that certain radius, the escape speed from the surface of the compressed object would equal the speed of light, therefore creating a black hole. But I wanted to know, under Earth's gravitational pull, how dense would an object have to be to reach its Schwarzschild Radius?&lt;/p&gt;&#xA;" OwnerUserId="2241" LastEditorUserId="2153" LastEditDate="2014-08-13T16:18:32.870" LastActivityDate="2014-08-13T16:18:32.870" Title="Under Earth's gravitational pull, how dense would an object have to be to reach its Schwarzschild Radius?" Tags="&lt;gravity&gt;&lt;black-hole&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6114" PostTypeId="2" ParentId="6113" CreationDate="2014-08-12T15:18:21.157" Score="2" Body="&lt;p&gt;Earth's gravity is quite weak. It manages to compress air (luckily for us), but that's about it. Since the Earth is not a black hole it can never compress something else to be a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, lets consider &lt;em&gt;some object&lt;/em&gt; and assume &lt;em&gt;something&lt;/em&gt; compresses it to a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Schwarzschild radius is defined as&#xA;$$&#xA;R_s = \frac{2GM}{c^2}.&#xA;$$&#xA;Now lets say the object is a sphere of radius $R_s$, then its uniform density is&#xA;$$&#xA; \rho = \frac{M}{ \frac{4\pi}{3} R_s^3 }.&#xA;$$&#xA;Using this to get rid of the mass term leaves us with&#xA;$$&#xA;R_s = \frac{2G \rho \frac{4\pi}{3} R_s^3}{c^2}&#xA;$$&#xA;Inverting this expression gives the &lt;em&gt;black hole density&lt;/em&gt; for a given radius&#xA;$$&#xA;\rho = \frac{3 c^2}{8 \pi G R_s^2}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that if you want a really small black hole, say $R=1$ m, the density has to be about $1.6\times10^{26}$ kg/m$^3$. That's about 20 times the mass of Earth stuffed in a cubic meter, so quite a lot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if the black hole can be as large as we please, then the density can get extremely low. A great example of this are &lt;a href=&quot;http://en.wikipedia.org/wiki/Supermassive_black_hole&quot; rel=&quot;nofollow&quot;&gt;supermassive black holes&lt;/a&gt;, which have about the same density as water (that's $10^3$ kg/m$^3$).&lt;/p&gt;&#xA;" OwnerUserId="494" LastActivityDate="2014-08-12T15:18:21.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6115" PostTypeId="1" AcceptedAnswerId="6116" CreationDate="2014-08-12T17:38:02.210" Score="2" ViewCount="1136" Body="&lt;p&gt;I have read that what keeps white dwarfs from gravitational collapse is electron degeneracy pressure. How does this pressure prevent further collapse, and how is it related to the Pauli Exclusion Principle?&lt;/p&gt;&#xA;" OwnerUserId="2198" LastEditorUserId="2365" LastEditDate="2018-07-27T22:34:49.583" LastActivityDate="2018-07-27T22:34:49.583" Title="Electron Degeneracy Pressure and the Pauli Exclusion Principle" Tags="&lt;star&gt;&lt;gravity&gt;&lt;degenerate-matter&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6116" PostTypeId="2" ParentId="6115" CreationDate="2014-08-12T18:12:08.773" Score="3" Body="&lt;p&gt;Basically, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Pauli_exclusion_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pauli exclusion principle&lt;/a&gt; says that two fermions (in this case, electrons) can't be in the same quantum state. To expand: No two electrons in an atom can share the same numbers for their &lt;a href=&quot;https://en.wikipedia.org/wiki/Quantum_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;four quantum numbers&lt;/a&gt;, properties that help describe the state of a particle. What are quantum numbers? The important consequence here is that no two electrons can have the same spin and energy level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a white dwarf or a neutron star, the fermions are packed very close together, and there's quite a lot of force due to gravity. However, the exclusion principle triumphs. Fermions near each other must have different energy levels; this leads to energy differences and &lt;a href=&quot;https://en.wikipedia.org/wiki/Degenerate_energy_levels&quot; rel=&quot;nofollow noreferrer&quot;&gt;degeneracy&lt;/a&gt; pressure pressure, which counteracts the force of gravity. Above a certain mass limit (the &lt;a href=&quot;https://en.wikipedia.org/wiki/Chandrasekhar_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chandrasekhar limit&lt;/a&gt;, roughly $\sim1.40M_{\odot}$), electron degeneracy pressure is no longer sufficient; the white dwarf collapses to a neutron star. There appears to be a similar limit for neutron stars, where neutron degeneracy pressure cannot support the remnant against gravity, and it collapses into a black hole.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-11-08T18:36:21.827" LastActivityDate="2016-11-08T18:36:21.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6117" PostTypeId="2" ParentId="6079" CreationDate="2014-08-12T19:34:45.300" Score="2" Body="&lt;p&gt;Flamsteed designations are assigned by John Flamsteed. He was an English Astronomer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flamsteed Designations (Flamsteed Numbers) are numbers assigned to the stars for the purpose of identification and cataloguing.&#xA;Flamsteed designation contained 2554 stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flamsteed designations for stars are similar to Bayer designations, except that they use numbers instead of Greek letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[ Bayer designation is a stellar designation in which a specific star is identified by a Greek letter, followed by the genitive form of its parent constellation's Latin name. The original list contained 1,564 stars.&#xA;For the most part, Bayer assigned Greek and Latin letters to stars in rough order of apparent brightness, from brightest to dimmest, within a particular constellation. Since in a majority of constellations the brightest star is designated Alpha (α), followed by beta for the 2nd brightest and so on.&#xA;But this was not the rule always followed by Bayer.&#xA;Somtimes by position of a particular star in that constellation, sometimes the order of rising up in the sky (the star rises up early will get earlier number), sometimes he follows mythological stories or data or sometimes he even uses his own way of assigning the number.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Flamsteed designations, Each star is assigned a number and the Latin genitive of the constellation it lies in.&#xA;e.g. 52 Cancri is star in the Constellation Cancer and assigned a number 52 according to star's right ascension. i.e. there are 51 stars in the Cancer constellation to the West direction of 52 Cancri star or there are 51 stars that rise before/earlier of 52 Cancri star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numbers were originally assigned in order of increasing right ascension within each constellation. &#xA;[Right ascension is the angular distance measured eastward along the celestial equator from the vernal equinox (it is one of the point of intersection in the space of two imaginary circles: Celestial Equator and Earth's orbit around the Sun. Our Earth is at this point on 21/22 March each year while orbiting around sun.) to the the position of the star. It is measured in degrees  but for simplicity and ease of use, mentioned in the form of hours. That the 360 circle is divided into 24 hours (1 hour = 15 degrees). ]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But due to the effects of precession (Top like rotation of Earth's axis of rotatation in the space mataining a constant angle of 23.5 degrees with verticle) they are now slightly out of order in some places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flamsteed catalogue don't cover all the stars visible from Earth with naked eye. Flamsteed's catalogue covered only the stars visible from Great Britain, and therefore stars of the far southern constellations have no Flamsteed numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference/Information Source/Quoted From:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) List of constellations for a list of constellations and the genitive forms of their names: &lt;a href=&quot;https://en.wikipedia.org/wiki/88_modern_constellations&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/88_modern_constellations&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)Bayer Designation: &lt;a href=&quot;https://en.wikipedia.org/wiki/Bayer_designation&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Bayer_designation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)Flamsteed Designation: &lt;a href=&quot;https://en.wikipedia.org/wiki/Flamsteed_designation&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Flamsteed_designation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2088" LastEditorUserId="2088" LastEditDate="2015-04-16T10:01:50.210" LastActivityDate="2015-04-16T10:01:50.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6118" PostTypeId="1" AcceptedAnswerId="6121" CreationDate="2014-08-12T20:10:31.787" Score="1" ViewCount="354" Body="&lt;p&gt;I recently moved to southern Mexico (16 degrees N) and I wonder what would be the best way to align my EQ telescope when the polar star is not visible?&lt;/p&gt;&#xA;" OwnerUserId="2244" LastActivityDate="2014-08-13T01:21:05.023" Title="How to polar align an equatorial mounted telescope when you are close to the equator?" Tags="&lt;amateur-observing&gt;&lt;telescope&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6119" PostTypeId="2" ParentId="5887" CreationDate="2014-08-12T22:06:44.160" Score="1" Body="&lt;p&gt;I think you're not too far from understanding the formula. However, there are a couple details you're missing: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;From what I understand from the document you linked to, $t_0 = t_{oe}$ is the time at the beginning of the week, or reference epoch.&lt;/li&gt;&#xA;&lt;li&gt;When computing an angle from a angular velocity, you usually need to compute it with respect to an initial angle and an initial time: when the Earth and the satellite rotate between $t_0$ and $t$, the angle they rotated of is $\dot{\Omega}(t - t_0)$ or $\dot{\Omega}_E(t - t_0)$.&lt;br&gt;&#xA;The only exception to this rule seems to be the initial Greenwich longitude with respect to the vernal equinox, which seems to be computed from $t = 0$.&lt;/li&gt;&#xA;&lt;li&gt;$w_0$, the initial Greenwich longitude wrt the vernal equinox, is $\dot{\Omega}_E t_{oe}$. From this formula, it seems like the document considers an origin of time $t = 0$ when the longitude of Greenwich wrt to the vernal equinox at this time is 0.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To summarize: the formula would have been more explicit and comprehensible written as:&#xA;$$&#xA;\Omega(t) = \Omega_0 - \dot{\Omega}_E t_{oe} + (\dot{\Omega} - \dot{\Omega}_E)(t - t_{oe})&#xA;$$&#xA;with:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\Omega_0$ = LAN wrt the vernal equinox at $t_{oe}$&lt;/li&gt;&#xA;&lt;li&gt;$\dot{\Omega_E}$ = angular velocity of the Earth wrt the vernal&#xA;equinox&lt;/li&gt;&#xA;&lt;li&gt;$-\dot{\Omega}_Et_{oe}$ = Longitude of Greenwich at $t_{oe}$&lt;/li&gt;&#xA;&lt;li&gt;$\dot{\Omega}$ = angular velocity of the satellite wrt the vernal&#xA;equinox&lt;/li&gt;&#xA;&lt;li&gt;$(\dot{\Omega} - \dot{\Omega}_E)(t - t_{oe})$ = relative angular&#xA;displacement between the satellite and Greenwich, between time $t$ and&#xA;$t_{oe}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That said, it seems like really bad practice to name $\Omega(t)$ the longitude with respect to Greenwich and $\dot{\Omega}$ the angular velocity with respect to the vernal equinox. This causes quite a bit of confusion, since mathematically speaking, for any linear function $f$, $f(t) = f(t_0) + \dot{f}\times(t - t_0)$&lt;/p&gt;&#xA;" OwnerUserId="2245" LastActivityDate="2014-08-12T22:06:44.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6120" PostTypeId="1" AcceptedAnswerId="6126" CreationDate="2014-08-12T22:13:29.030" Score="3" ViewCount="140" Body="&lt;p&gt;I am looking for basic data regarding red-shifting that  comes with reliable measure of distance of the emitting star.&lt;/p&gt;&#xA;" OwnerUserId="1247" LastEditorUserId="37191" LastEditDate="2021-05-11T06:46:39.483" LastActivityDate="2021-05-11T06:46:39.483" Title="Data for red-shifting" Tags="&lt;observational-astronomy&gt;&lt;redshift&gt;&lt;raw-data&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6121" PostTypeId="2" ParentId="6118" CreationDate="2014-08-13T01:21:05.023" Score="2" Body="&lt;p&gt;Mount a camera on the tripod and take a long exposure. You should be able to determine the concentric circles of the stars paths. The center of the circle is the North pole.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-08-13T01:21:05.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6122" PostTypeId="1" CreationDate="2014-08-13T04:16:52.760" Score="6" ViewCount="529" Body="&lt;p&gt;How do astronomers find out the correct weight of a planet even though there isn't any direct means to weigh them? What technique do they use? Just curious!&lt;/p&gt;&#xA;" OwnerUserId="662" LastEditorUserId="6959" LastEditDate="2015-04-12T13:52:53.893" LastActivityDate="2015-04-12T13:52:53.893" Title="Weight of a celestial body" Tags="&lt;star&gt;&lt;planet&gt;&lt;astrophysics&gt;&lt;mass&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6123" PostTypeId="2" ParentId="6105" CreationDate="2014-08-13T04:20:04.150" Score="3" Body="&lt;p&gt;Green laser pointers are great for astronomers because at the most common power output (5-15mw for consumer grade 500-532nm laser) the beam is visible in most &quot;dark sky&quot; settings. Cost for a simple green laser with a visible beam ~$15 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Red lasers on the other hand have an average power output of about 3mw, most commonly denoted with the consumer in mind as, &amp;lt; 5mw. For a red (632-660nm) to have a visible (to the unaided eye) beam, the laser would need to have a significantly higher output, on the order of 50-150mw. Cost for a &quot;simple&quot; red laser with a visible beam ~$200&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with that is that any laser with &gt; 15mw output is considered class IV and there are heavy restrictions on these in most countries. There are a few places that get around the class IV restrictions and sell them as the lesser class IIIb, however to my knowledge there are no red lasers floating around that fit this category. Cost for &quot;high powered class IIIb&quot; red laser ~$70&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that being said, there is a preference for green lasers for both cost and safety reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally prefer my red laser, however making everyone put on dark protective glasses at night is both over complicating a simple task of pointing out stellar objects, and at the same time hilarious, watching people try to figure out what the laser is pointing at, as the beam is all that they will see with the eye-ware on. Without safety goggles looking indirectly at the beam from the 12Mw laser can cause serious damage... again, that is why astronomers use a simple green laser. &lt;/p&gt;&#xA;" OwnerUserId="2232" LastEditorUserId="2232" LastEditDate="2014-08-23T15:26:45.480" LastActivityDate="2014-08-23T15:26:45.480" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6124" PostTypeId="2" ParentId="6112" CreationDate="2014-08-13T07:18:04.307" Score="10" Body="&lt;p&gt;&lt;strong&gt;I wonder if the area I am looking for is GSH 238+00+09.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;In the color image quoted in my question, the galactic core lies beyond the top center. This would be 0° &lt;a href=&quot;http://en.wikipedia.org/wiki/Galactic_coordinate_system&quot; rel=&quot;noreferrer&quot;&gt;galactic longitude&lt;/a&gt;. Going &lt;em&gt;counterclockwise&lt;/em&gt;, 238° is somewhere around Beteigeuze.&lt;/p&gt;&#xA;&lt;p&gt;Here is a &amp;quot;[p]rojection of the Local Cavity onto the Galactic plane&amp;quot;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/buXa7.gif&quot; alt=&quot;Figure 9 from Lallement et al. (2003)&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Figure 9 from Lallement et al. (2003)&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;But if you try to superimpose this image on the other one (I matched the positions of the Sun, the Ophiuchus clouds and the Pleiades bubble, the Lupus cloud is a bit off), it seems as if the area marked in green was actually &lt;strong&gt;part of our Local Bubble!&lt;/strong&gt; And GSH 238+00+09 then appears to be identical to the Orion-Eridanus-Superbubble:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/waCQv.gif&quot; alt=&quot;Superimposed images&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;I'm sure I didn't match this correctly. Especially Loop III would be in totally the wrong place. But I found an article that used the same black and white image to illustrate possible locations of Loop III, and the second alternative to the one shown here was in the upper right hand corner, where it would then overlap with &amp;quot;Loop II&amp;quot; of the color image. But probably we should not worry too much over Loops II and III. The creator of the &lt;a href=&quot;http://www.3dgalaxymap.com/Scene2.html&quot; rel=&quot;noreferrer&quot;&gt;3D Galaxy Map&lt;/a&gt; writes: &amp;quot;I haven't added Loop II, III or IV because I couldn't find any accurate information about their size and location.&amp;quot; I'm sure he has been looking more thoroughly than I have, so I feel the mismatch of Loop III is not a problem.&lt;/p&gt;&#xA;&lt;p&gt;But then the color image is from 1994, and understanding and cartography of our Galactic neighborhood has advanced quite a bit in the ten years between the two publications, so maybe the placement of the Local Bubble in the color map is only a rough equivalent and the mismatch between the two images is not my fault but a result of scientific progress.&lt;/p&gt;&#xA;&lt;p&gt;Here is another image (rotated 180° in relation to the color image from my question) of local loops, bubbles, and superbubbles, where it becomes clear that GSH 238+00+09 (termed &amp;quot;New Superbubble&amp;quot; (NSB) in this image) is &lt;em&gt;not&lt;/em&gt; identical to the Orion-Eridanus-Superbubble:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fa2py.png&quot; alt=&quot;Figure 8 from Heiles et al., (1998)&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Figure 8 from Heiles et al., (1998)&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;If you compare the scales of this last and my first color image, you will note that the color image falls completely inside the circle denoting the Local Bubble, supporting the thought that the green area lies within it (and not within GSH 238+00+09 or the Orion-Eridanus-Bubble). But this second image also supports the insight that the color image might not be that accurate, and that &lt;strong&gt;maybe the area whose name I am looking for does not exist in that shape at all&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;In the following image (which shows Loop I as the Scorpius-Centraurus Shells and the Orion-Eridanus-Loop as the Orion Shell) the red blot immediately rimward (lower right) of the sun is the Taurus Dark Cloud. This is not from a scientific but a popular publication (just as my color image), so we should not expect it to be totally accurate, but it still suports the notion that the green area might be part of the Local Bubble.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/P5A4o.gif&quot; alt=&quot;Unnumbered 3rd figure from Frisch (2000)&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Unnumbered 3rd figure from Frisch (2000)&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here is a final image, apparently a different rendering of the image from Lallement et al. (2003), the first image given in this answer. In this image the location of some important stars is marked. Most notably &lt;strong&gt;ε CM&lt;/strong&gt;, which is the scientific name for the Arab name Adhara, with which the same star is marked in the color image from my question and there lies far out in the green area, **lies still within the Local Bubble in this image, close to the edge.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oJsYK.png&quot; alt=&quot;Figure 3 from Sfreir et al. (1999)&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Figure 3 from Sfreir et al. (1999)&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here I overlayed the previous image and the image from my question. I matched the sun an Adhara. Other objects are still a bit off, but not much, and we get a good idea of where the Local Bubble ends.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JSz7n.gif&quot; alt=&quot;Second overlay image&quot; /&gt;&lt;/p&gt;&#xA;&lt;h1&gt;So it seems the local bubble extends to a bit beyond the star Adhara, halfway from the Sun to the right edge of the image in my question and does indeed include the closer part of the green area.&lt;/h1&gt;&#xA;&lt;p&gt;This thought is supported in the abstract of Heiles (1998), where it says: &amp;quot;[GSH 238+00+09] is at least partially responsible for the &lt;strong&gt;apparent extreme elongation of the Local Bubble toward &lt;em&gt;l&lt;/em&gt; ~ 230°&lt;/strong&gt; ...&amp;quot; (my emphasis).&lt;/p&gt;&#xA;&lt;p&gt;I'll leave my first thoughts and the badly matched image in this answer, to document how I came to the final conclusion.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;I'd welcome any thoughts and corrections of this attempt at an answer.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Sources:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Frisch, P. (2000). The Galactic Environment of the Sun The heliosphere appears to protect the inner solar system from the vagaries of the interstellar medium. American Scientist, 52-59. Available online at &lt;a href=&quot;https://www.americanscientist.org/issues/issue.aspx?id=862&amp;amp;y=0&amp;amp;no=&amp;amp;content=true&amp;amp;page=5&amp;amp;css=print&quot; rel=&quot;noreferrer&quot;&gt;https://www.americanscientist.org/issues/issue.aspx?id=862&amp;amp;y=0&amp;amp;no=&amp;amp;content=true&amp;amp;page=5&amp;amp;css=print&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Heiles, C. (1998). Whence the Local Bubble, Gum, Orion? GSH 238+ 00+ 09, a nearby major superbubble toward Galactic longitude 238. The Astrophysical Journal, 498(2), 689. Available online at &lt;a href=&quot;http://iopscience.iop.org/0004-637X/498/2/689/pdf/0004-637X_498_2_689.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://iopscience.iop.org/0004-637X/498/2/689/pdf/0004-637X_498_2_689.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Lallement, R., Welsh, B. Y., Vergely, J. L., Crifo, F., &amp;amp; Sfeir, D. (2003). 3D mapping of the dense interstellar gas around the Local Bubble. Astronomy and Astrophysics, 411(3), 447-464. Available online at &lt;a href=&quot;http://www.aanda.org/articles/aa/full/2003/46/aa3635/aa3635.right.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.aanda.org/articles/aa/full/2003/46/aa3635/aa3635.right.html&lt;/a&gt; (under &amp;quot;Article&amp;quot; in the sidebar)&lt;/li&gt;&#xA;&lt;li&gt;Sfeir, D. M., Lallement, R., Crifo, F., &amp;amp; Welsh, B. Y. (1999). Mapping the contours of the Local bubble: preliminary results. Astronomy and Astrophysics, 346, 785-797. Available online at &lt;a href=&quot;http://adsabs.harvard.edu/full/1999A%26A...346..785S&quot; rel=&quot;noreferrer&quot;&gt;http://adsabs.harvard.edu/full/1999A%26A...346..785S&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user2233" LastEditorUserId="-1" LastEditorDisplayName="user2233" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2014-08-13T09:16:51.920" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6125" PostTypeId="1" AcceptedAnswerId="6131" CreationDate="2014-08-13T14:32:37.770" Score="2" ViewCount="233" Body="&lt;p&gt;I'm looking for the parameters $r_c$, $\rho_0$ and $\beta$ in a $\beta$-profile fit of the Virgo cluster's ICM density. I just can't find a reference for it. Unlike for Coma (Abell 1656) where it's in Mohr et. al. 1999.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beta profile:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho(r) = \rho_0 \left[ 1+ \left(\frac{r}{r_c}\right)^2 \right]^{-\frac{3\beta}{2}}$$&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="1862" LastEditDate="2014-09-03T04:04:40.190" LastActivityDate="2014-09-03T04:04:40.190" Title="Beta profile fit of Virgo cluster gas?" Tags="&lt;cosmology&gt;&lt;gas&gt;&lt;galaxy-cluster&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6126" PostTypeId="2" ParentId="6120" CreationDate="2014-08-13T15:59:43.757" Score="3" Body="&lt;p&gt;Red shift is usually measured for galaxies rather than individual stars. Unless a star has just gone supernova, it's usually not bright enough to be seen even w the world's most powerful telescopes at the distances where cosmological redshift comes into play. &lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble%27s_law&quot; rel=&quot;nofollow&quot;&gt;Hubble's law&lt;/a&gt; operates over &lt;strong&gt;large&lt;/strong&gt; distances; the expansion constant being 67.8 km/sec per megaparsec (3.3 million light years) &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Andromeda_Galaxy&quot; rel=&quot;nofollow&quot;&gt;Andomeda galaxy (M31)&lt;/a&gt; is 2.54 million light years away, and although some individual stars are visible at that distance, the galaxy has a net &lt;strong&gt;blueshift&lt;/strong&gt; of 0.001001 (303 km/sec) due to its &lt;a href=&quot;http://en.wikipedia.org/wiki/Peculiar_velocity#Cosmology&quot; rel=&quot;nofollow&quot;&gt;peculiar (i.e. non-Hubble) velocity.&lt;/a&gt;&#xA;Further out is where you start to see redshifts in excess of peculiar motion; and there you need to start using standard candles of some sort.&#xA;Cepheid variables when visible, Supernova, globular cluster brightness functions, are all used, as well as some spectral methods: &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_distance_ladder#Standard_candles&quot; rel=&quot;nofollow&quot;&gt;Cosmic distance ladder&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supernova measurements go the furthest out, highest redshift, but are &lt;a href=&quot;http://www.astro.ucla.edu/~wright/sne_cosmology.html&quot; rel=&quot;nofollow&quot;&gt;still a bit of a can of worms as far as interpretation goes.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="2031" LastEditDate="2014-08-14T16:29:39.787" LastActivityDate="2014-08-14T16:29:39.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6127" PostTypeId="1" AcceptedAnswerId="6276" CreationDate="2014-08-13T19:16:18.123" Score="4" ViewCount="345" Body="&lt;p&gt;Rogue planets can be designated as rogue planets due to one of two possible scenarios: 1) the &quot;rogue planet&quot; formed as a sub-brown dwarf, or 2) the planet was ejected from its home star system. I want to make it clear that here, I'm interested in objects of the second category - objects that are similar to the planets we find in our solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could a rogue planet, having been ejected from its home star system, be gravitationally re-captured by another star system? If so, would it be likely to disrupt other bodies in the stellar system (I know this depends on size; let's say that the rogue planet is a &quot;Hot Jupiter&quot;)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that we don't know a whole lot about these objects (let's face it, we only know of a handful), so I'm not looking for examples, but any logical answers would be pretty helpful.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-06T20:04:56.193" LastActivityDate="2015-05-31T17:00:47.690" Title="Potential Re-Capture of Rogue Planets" Tags="&lt;star-systems&gt;&lt;rogue-planet&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6129" PostTypeId="1" CreationDate="2014-08-13T22:27:45.617" Score="1" ViewCount="286" Body="&lt;p&gt;So it seems the overall thought is that the universe is infinite. I don't get how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most depictions I read mention something to the effect of &quot;at 1^-20 seconds the universe was the size of &lt;strong&gt;_&lt;/strong&gt;&quot;. (Essentially mentioning a finite size at a very small length of time). So it came from a singularity, and had a finite size at some point. No matter how fast inflation went, any multiple of that finite size is NOT infinity though. If it was finite once, it is still finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's infinite then it was always infinite which means it was infinite density, infinitely large, etc. That's too crazy to swallow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't it more likely that the universe is really really big but not infinite? And we are at the center of the observable universe (to us), however we are off center in some grander universe that's maybe 1 trillion light years across? And nothing is past that because space and time are the same thing. Therefore if there is no light/matter/events, there is no time and hence no space.&lt;/p&gt;&#xA;" OwnerUserId="2249" LastActivityDate="2014-08-14T18:49:38.513" Title="Infinite universe? How?" Tags="&lt;universe&gt;&lt;observable-universe&gt;" AnswerCount="2" CommentCount="12" ClosedDate="2014-08-15T13:52:40.390" ContentLicense="CC BY-SA 3.0" />
  <row Id="6130" PostTypeId="1" CreationDate="2014-08-14T00:38:01.600" Score="2" ViewCount="169" Body="&lt;p&gt;this one bothering me,to buy a telescope.&#xA;I live in a city,unfortunately with dust and polluted air, is there any affect in viewing a nice capture of our galaxy? Or i need to buy a special telescope?&lt;/p&gt;&#xA;" OwnerUserId="2237" LastActivityDate="2014-08-16T13:07:54.493" Title="Telescope on polluted environment" Tags="&lt;telescope&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6131" PostTypeId="2" ParentId="6125" CreationDate="2014-08-14T00:47:46.513" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://arxiv.org/pdf/astro-ph/9811464v1.pdf&quot; rel=&quot;nofollow&quot;&gt;Schindler et al. 1999&lt;/a&gt;: Morphology of the Virgo Cluster: Gas versus Galaxies has details for $\beta$ model fits for Virgo and its subclusters.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-08-14T00:47:46.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6133" PostTypeId="1" AcceptedAnswerId="6136" CreationDate="2014-08-14T09:04:10.720" Score="2" ViewCount="294" Body="&lt;p&gt;I just read &lt;a href=&quot;https://physics.stackexchange.com/q/130918/21740&quot;&gt;When a star becomes a black hole, does its gravitational power become stronger?&lt;/a&gt; and it confused me a bit;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I always thought black holes are supermassive and that is why they bend space-time. But it is said therein that the black hole is less massive than the star that created it (and it seems logical if we think about that - indeed, the star exploded and released a lot of matter). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus my question: How can a black hole bend space-time more than the star which created it?&lt;/p&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2014-08-14T11:39:20.810" Title="How is space-time bent by black holes?" Tags="&lt;black-hole&gt;&lt;space-time&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6134" PostTypeId="1" AcceptedAnswerId="6139" CreationDate="2014-08-14T09:42:14.437" Score="1" ViewCount="121" Body="&lt;p&gt;I want to start observing the night sky, what advice would you give to get started?&lt;/p&gt;&#xA;" OwnerUserId="2226" LastEditorUserId="2226" LastEditDate="2014-08-16T03:50:20.033" LastActivityDate="2014-08-16T03:50:20.033" Title="Starting out in practical astronomy" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2014-08-15T13:49:20.283" ContentLicense="CC BY-SA 3.0" />
  <row Id="6135" PostTypeId="1" AcceptedAnswerId="6142" CreationDate="2014-08-14T10:05:18.983" Score="7" ViewCount="355" Body="&lt;p&gt;As per the &lt;a href=&quot;http://io9.com/5986723/using-the-kardashev-scale-to-measure-the-power-of-extraterrestrial-civilizations&quot;&gt;Kardashev Scale&lt;/a&gt;, type-3 civilizations are defined as &quot;civilization(s) in possession of energy on the scale of its own galaxy, with energy consumption at ~ 4 x 10^44 erg/sec&quot;. Considering that we have detected &lt;em&gt;none&lt;/em&gt;, despite projects like SETI - can we conclude with reasonable certainty that there is &quot;&lt;strong&gt;currently&lt;/strong&gt;&quot; no type-3 civilization in the milky way? Or Is it perhaps possible that there could be such civilization(s) primarily harnessing dark matter / energy and hence gone undetected? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am assuming there could yet be type-2 civilizations that extract fusion energy (&quot;normal/known&quot; sources of energy), information, and raw-materials from multiple solar systems - within our own galaxy - yet undetected.&lt;/p&gt;&#xA;" OwnerUserId="474" LastActivityDate="2016-01-25T19:56:57.607" Title="Can we conclude with reasonable certainty that there is/are no Type 3 Civilizations in our galaxy?" Tags="&lt;extra-terrestrial&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6136" PostTypeId="2" ParentId="6133" CreationDate="2014-08-14T11:39:20.810" Score="2" Body="&lt;p&gt;It doesn't bend it more. It is just that the matter of the star is now concentrated in a very compact region of space. So, previously you could get only to the surface of the star but now as the star is very compact, you can get much nearer and thus you will experience a greater gravitational force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had a similar discussion &lt;a href=&quot;http://www.physicsforums.com/showthread.php?t=730313&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Check it out and maybe your question will be answered.&lt;/p&gt;&#xA;" OwnerUserId="1156" LastActivityDate="2014-08-14T11:39:20.810" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6137" PostTypeId="2" ParentId="6135" CreationDate="2014-08-14T16:10:12.297" Score="2" Body="&lt;p&gt;As the definition says, a typeIII civilization &lt;em&gt;does&lt;/em&gt; change the energy of the galaxy. This doesn't happen. Thus, the obvious answer is: yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true for typeII civilizations in our solar system, and for typeI in our planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Extension:&lt;/em&gt; Intelligence means always structures, although structures don't always mean intelligence. We see the energy processes of the galaxy, and they doesn't show any meaningful structure.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="1891" LastEditDate="2015-01-31T00:34:28.923" LastActivityDate="2015-01-31T00:34:28.923" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6138" PostTypeId="2" ParentId="6129" CreationDate="2014-08-14T16:21:43.190" Score="0" Body="&lt;p&gt;Space and time are total different things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These citations are about the observable universe, and not from the whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see only a 13.7billion light year sphere from our spacetime. It is not very big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/88786/lower-limit-of-the-size-of-the-universe-wmap&quot;&gt;The current measurement of the WMAP satellties&lt;/a&gt; says, the universe is probably infinite big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your questions can be explained with the Friedman equations. These equations are the solutions of the Einstein Field Equations for the infinite vacuum of our spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And its spacelike sections of our Friedman-spacetime is planar&lt;/em&gt;, at least in the limits of the precision of the measurement. What currently can be said, is that &quot;if the whole universe is finite, and has a (hyper-)spherical geometry, then its radius must be at least 300billion light years&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastActivityDate="2014-08-14T16:21:43.190" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6139" PostTypeId="2" ParentId="6134" CreationDate="2014-08-14T17:44:22.083" Score="1" Body="&lt;p&gt;If you're dead broke (like most people) I would suggest you take look at a small refractor telescope, you could pick up a cheap-o Celestron Powerseeker 60/70 for something like 40 bucks. The useability of such a small telescope is very limited, but you'll be able to get a nice clear image of the rings of Saturn, four of Jupiter's moons, or nice views of the moon, maybe even a few Messier objects if you're patient enough. &#xA;-but- &#xA;If you're willing to drop a bit more coin, then I would suggest nothing smaller than an 8&quot; reflector. For less than 200 bucks you can find something that will really impress you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used to take my crappy little telescope and drive up into the hills where light pollution was at a minimum, where I would just sit for hours trying to find Messier objects. Loads of fun!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as joining an amateur astronomy society, I would suggest you check with your local Community College's astronomy professors. At least here in California, one of the requirements of the astronomy classes is that there be opportunities to participate in actual observing. This means that there will more than likely be an astronomy group having &quot;star parties&quot; and other observing events, which will give you some good connections to competent groups. &lt;/p&gt;&#xA;" OwnerUserId="2232" LastActivityDate="2014-08-14T17:44:22.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6140" PostTypeId="1" CreationDate="2014-08-14T17:50:09.047" Score="3" ViewCount="139" Body="&lt;p&gt;Most descriptions of star ignition go something like... The star reaches a critical mass and ignites, &lt;b&gt;blowing away the surrounding material&lt;/b&gt;. The most obvious question would be &quot;Why are stars not all the same size?&quot;, given that the critical mass for ignition would appear to be the same, but I understand that stars can accumulate mass from collisions from other stars, planetary bodies etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is also based on the fact that larger stars also burn more energetically, I assume blowing away more of the the lighter elements from the surrounding cloud after formation and gathering more heavy elements,... do larger stars have a significantly different composition from smaller stars? --- or is the simple description of star formation misleading?&lt;/p&gt;&#xA;" OwnerUserId="2251" LastActivityDate="2014-08-20T05:55:04.540" Title="What is star composition after formation" Tags="&lt;star&gt;&lt;star-formation&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6141" PostTypeId="2" ParentId="6129" CreationDate="2014-08-14T18:49:38.513" Score="1" Body="&lt;p&gt;Depending on the mass of the universe, the universe has three different possible shapes.&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Shape_of_the_universe&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Shape_of_the_universe&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be finite or infinite depending on the shape it has, and there are some ways we can discard possible shapes (if it started to shrink instead of expand).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps&lt;/p&gt;&#xA;" OwnerUserId="2252" LastActivityDate="2014-08-14T18:49:38.513" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6142" PostTypeId="2" ParentId="6135" CreationDate="2014-08-14T20:31:34.370" Score="7" Body="&lt;p&gt;&lt;strong&gt;Short Answer:&lt;/strong&gt; We won't have to worry about the Galactic Empire any time soon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Long Answer:&lt;/strong&gt; First we have to consider some of the properties of a Type III civilization - or, to maybe put it better, &lt;em&gt;what kinds of things it would do&lt;/em&gt;. Some are shared properties of Type II civilizations, but on a larger scale. You can probably list some off the top of your head:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Advanced interstellar travel&lt;/li&gt;&#xA;&lt;li&gt;Large-scale engineering projects&lt;/li&gt;&#xA;&lt;li&gt;The ability to capture energy from stars, black holes, and pretty much any other source you can think of.&lt;/li&gt;&#xA;&lt;li&gt;Lots and lots and lots of signals&lt;/li&gt;&#xA;&lt;li&gt;Colonies on most star systems&lt;/li&gt;&#xA;&lt;li&gt;The intelligence to not contact us&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can extrapolate #1 from just about every science fiction movie ever made. But I don't mean anything like the starship &lt;em&gt;Enterprise&lt;/em&gt;, or the Death Star. To become a Type III civilization, a civilization first has to reach Type II status. What does that entail? Well, among a lot of other things, shunning war. I know that might be hard to believe, but think about this: Could a single nation get off this planet on a large scale all by itself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good example is the International Space Station. To get the project going, it needed the combined help of the U.S., Russia, Japan, China, and the E.S.A. Admittedly, we're entering a new age of space travel. SpaceX, Blue Origin, Virgin Galactic, Reaction Engines, and other companies are showing that they can get people and things to orbit or suborbital trajectories. Maybe an entire country isn't needed to do something grand; maybe a single company can do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I can conclude that any spacecraft we find from such a civilization would most likely be peaceful - a passenger ship, perhaps, or a freighter. Such a craft would be hard to detect - admittedly, we have trouble finding anything the size of a planet or smaller, once we get outside our Solar System! But that brings us to #2: Large-scale engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture the Death Star. &quot;That's no moon!&quot; Yep, and it looks - from afar - a lot like one. Type III civilizations would have the capacity to build such structures - for peace, likely (see #1) - and these would be a lot easier to detect. Wait. Didn't I just say that anything the size of a planet or smaller would be hard to detect? Yep. But a Type III civilization wouldn't be limited to making small faux moons. Planet-sized objects, or bigger (we'll get to that in #3). How could we detect such objects? Well, if they're big enough, we could find them the same way we find (if we're lucky) exoplanets - transits of the primary star, wobbles in the star's orbit, etc. There is another way, though - for specialized creations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ah, 3#. Energy-collecting devices. A quick Internet search of &quot;Dyson sphere&quot; should give you a good idea of what I mean (and if that's not enough, you can see Dyson's original paper or see &lt;a href=&quot;https://astronomy.stackexchange.com/questions/429/what-are-the-current-observational-limits-on-the-existence-of-dyson-spheres-swar&quot;&gt;this&lt;/a&gt; question). Dyson spheres/rings/any-other-configuration-you-can-think-of would affect the star whose light they were collecting, and we could figure out if the object in question was a Dyson sphere (see the &quot;Dyson sphere&quot; question). But a Type III civilization could transcend that (Dyson spheres are often cited as the trademark of a Type II civilization). Imagine collecting the energy of jets emitted from the accretion disk of a black hole - not an easy job, but still possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up next: signals. Humans have been emitting radio waves since Marconi (okay, he wasn't the only one) turned on his transmitter. Over 100 years later, we're still going at a pretty good rate. A civilization would need millennia to reach Type III status, and should have emitted a bunch of signals along the way. Also, such a civilization would have spread out, and should be emitting them today. Now, the galaxy's a pretty big place. It is tens and tens of thousands of light years in diameter (although there is a debate about where the galaxy ends), so our radio waves won't have reached much of the galaxy. But a Type III civilization should, theoretically, be almost everywhere, and we should have gotten messages (even inadvertent ones) long ago. Unless you think the &quot;Wow&quot; signal was actually a message, there's not a lot of hope. You hit the nail on the head with your SETI remark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the next (#5) property, we turn to . . . well, just about any sci-fi show ever made (again). What would &lt;em&gt;Star Wars&lt;/em&gt; or &lt;em&gt;Star Trek&lt;/em&gt; be without colonies all over the galaxy? Technically, the definition of a Type III civilization is one that merely &lt;em&gt;uses&lt;/em&gt; all the galaxy's energy; it doesn't have to be all over. But it's not much of a conceptual leap to realize that such a civilization would actually have to be pretty much everywhere in the galaxy in order to collect all that energy. So we should, in theory, have a decent chance of stumbling upon a colony if we headed to any star. Is that going to happen any time soon? Nope. So there goes that method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last property is something of a joke, but also something of a parting message. As Calvin (from the comic strip &lt;em&gt;Calvin and Hobbes&lt;/em&gt;) &lt;a href=&quot;http://www.wanderings.net/notebook/Main/CalvinAndHobbesOnSurestSignOfIntelligentLife&quot; rel=&quot;nofollow noreferrer&quot;&gt;once said&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sometimes I think the surest sign that intelligent life exists elsewhere in the universe is that none of it has tried to contact us.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But think about it. We're a pretty violent race. Just turn on the evening news. Remember what I said before: advanced civilizations may not want to contact us. So even if we don't receive any messages, or find any Dyson spheres, of fail to find evidence of any kind, doesn't mean that we're alone on the Milky Way. Unlikely? Yep. But plausible? Yep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to summarize: There are a bunch of different properties that would scream &quot;Advanced civilization&quot; at you. We could detect them by looking at their effects on stars, by searching the sky for radio waves, or by simply hopping over to a few nearby star systems. But being as we really haven't found anything yet: We can conclude, with reasonable certainty, that there is/are no Type III civilizations in our galaxy.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-01-25T19:56:57.607" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6143" PostTypeId="1" AcceptedAnswerId="6151" CreationDate="2014-08-15T03:14:32.510" Score="2" ViewCount="627" Body="&lt;p&gt;Newton's Law of Universal Gravity states that an object will accelerate constantly as it falls. Let's assume an object falling into a black hole. According to Newton's Law, when it reaches the event horizon it must be traveling at &lt;code&gt;c&lt;/code&gt;, because c is the escape velocity at the event horizon. Therefore, after it passes the event horizon, it must continue to accelerate past &lt;code&gt;c&lt;/code&gt;, yet theory of relativity states that it's never gonna happen. &lt;strong&gt;So, in this condition, what will the object do?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2014-08-16T03:48:45.680" Title="Contradiction between Theory of Relativity and Newton's Law of Universal Gravitation?" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6144" PostTypeId="2" ParentId="6130" CreationDate="2014-08-15T06:48:55.400" Score="4" Body="&lt;p&gt;If you are going to buy a telescope in the city let me suggest a large aperture and a reflective telescope.  I live in a city as well and I use a reflector telescope.  I did have trouble with some dimmer galaxies to start with but if you buy a light pollution filter you can see them a lot better.  Also take a series of images of the same object and then stack them together!  There are many free astronomy image stacking programs on the web.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have fun!&lt;/p&gt;&#xA;" OwnerUserId="2155" LastActivityDate="2014-08-15T06:48:55.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6147" PostTypeId="5" CreationDate="2014-08-15T17:43:03.713" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-08-15T17:43:03.713" LastActivityDate="2014-08-15T17:43:03.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6148" PostTypeId="4" CreationDate="2014-08-15T17:43:03.713" Score="0" Body="Questions regarding the currently prevalent cosmological model for the origin of the universe." OwnerUserId="2258" LastEditorUserId="2258" LastEditDate="2014-08-16T00:27:16.990" LastActivityDate="2014-08-16T00:27:16.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6149" PostTypeId="2" ParentId="1901" CreationDate="2014-08-15T20:14:08.190" Score="4" Body="&lt;p&gt;Stars actually do this on their own, to an extent.  The material inside a star (in some layers) does convect, and at certain points in the star's life, the material in the core can actually rise up and mix with the rest of the star.  This might replenish some H and He in the core, but the main effect that can be observed is that the fusion products become visible at the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the size and location of the convective layers depends very much on stellar size - in a red dwarf the convective zone is most of the star.  That's one reason red dwarf stars live so long - they continually &quot;recycle&quot; their available Hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Star#Structure&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Star#Structure&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dredge-up&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Dredge-up&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2259" LastActivityDate="2014-08-15T20:14:08.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6150" PostTypeId="2" ParentId="1901" CreationDate="2014-08-16T00:11:59.510" Score="3" Body="&lt;p&gt;When stars transition from the main sequence core Hydrogen burning to core Helium burning they can experience &lt;a href=&quot;http://www.atnf.csiro.au/outreach/education/senior/astrophysics/stellarevolution_postmain.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hydrogen shell burning&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZxefW.jpg&quot; alt=&quot;http://www.atnf.csiro.au/outreach/education/senior/astrophysics/stellarevolution_postmain.html&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when you burn the Hydrogen in the shell to Helium, it will fall on the core creating a hotter core where the radiation pressure from the Helium burning core will push the Hydrogen further out into the outer envelope. From here you can get a &lt;a href=&quot;http://en.wikipedia.org/wiki/Helium_flash&quot; rel=&quot;nofollow noreferrer&quot;&gt;Helium flash&lt;/a&gt; which is the cause of &lt;a href=&quot;http://en.wikipedia.org/wiki/Horizontal_branch&quot; rel=&quot;nofollow noreferrer&quot;&gt;Horizontal Branch&lt;/a&gt; stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were able to somehow force more Hydrogen to shell burn then you could possibly extend the life for a short while, though the lifetimes post main sequence are much shorter &amp;lt; 1 Gyr compared to the lifetimes on the main sequence.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-08-16T00:11:59.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6151" PostTypeId="2" ParentId="6143" CreationDate="2014-08-16T02:37:46.227" Score="4" Body="&lt;p&gt;&lt;strong&gt;A) Don't take Newton's law too seriously in the relativistic regime.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;According to Newton's Law, when it reaches the event horizon it must be traveling at $c$, because $c$ is the escape velocity at the event horizon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is true, assuming the particle freefalls from rest at infinity, but the details are quite different from Newtonian theory. For orbits in Schwarzschild spacetime, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Effective_potential&quot; rel=&quot;nofollow&quot;&gt;effective potential&lt;/a&gt; is&#xA;$$V = -\frac{GM}{r} + \frac{l^2}{2r^2} - \frac{GMl^2}{c^2r^3}\text{,}$$&#xA;where $l\equiv L/m$ is the specific angular momentum of the orbit. The last term offers relativistic corrections that have no analogue in Newtonian gravity, so in general Newton's law cannot be counted on when near the horizon of the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for radial freefall, $l = 0$, and we a total orbital specific energy:&#xA;$$\mathcal{E} = \frac{1}{2}\left(\frac{\mathrm{d}r}{\mathrm{d}\tau}\right)^2 - \frac{GM}{r}\text{,}$$&#xA;which has exactly Newtonian form. In particular, a particle freefalling from rest at infinity will have&#xA;$$\left|\frac{\mathrm{d}r}{\mathrm{d}\tau}\right| = \sqrt{\frac{2GM}{r}}\text{,}$$&#xA;so at as it nears the horizon ($r\to 2GM/c^2$), we have $|\mathrm{d}r/\mathrm{d}\tau|\to c$, just as Newtonian theory would predict.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the motion is only roughly analogous to what's prescribed by Newton's law, because there are very important differences in the interpretation of those quantities. Specifically, (1) the $\tau$ refers to the proper time of the orbiting particle, not coordinate time $t$, and certainly not any absolute time as Newtonian theory proper, and (2) the $r$ refers to the Schwarzschild radial coordinate, which is defined such that a sphere at this coordinate have surface area $4\pi r^2$, and this is &lt;strong&gt;not&lt;/strong&gt; the radial distance to the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;B) Coordinates have no intrinsic meaning.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Therefore, after it passes the event horizon, it must continue to accelerate past $c$, yet theory of relativity states that it's never gonna happen. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suppose I fling a particle along a straight ruler, and that ruler is marked in feet. So if, for example, my particle goes five marked units along the ruler in a nanosecond, that's clearly a problem, because the particle must be superluminal ($c\approx 1\,\mathrm{ft}/\mathrm{ns}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose the ruler is not marked in feet, and I don't tell you how it's marked. Is it a problem that the particle goes five units in a nanosecond? Without knowing what the marks are, the statement of this sort of &quot;coordinate speed&quot; is not even meaningful. Finally, suppose I tell you how the ruler is marked, but its marks do not actually measure the distance along the ruler. Then is it a problem that it goes through five of those marks in a nanosecond?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moral is simply this: by themselves, coordinates are simply labels to identify events. One can pick coordinates such that the coordinate speed is arbitrarily large or small, and it doesn't matter, because coordinates are not physical things. The universe does not come with its own coordinates; they're just labels we put on things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what can give coordinates meaning is the &lt;em&gt;metric tensor&lt;/em&gt;, which relates coordinates to actual lengths or durations, so that, for example, the Schwarzschild radial coordinate has a set meaning in terms of surface areas. Finally, the Schwarzschild coordinates are pathological near the horizon. They are not defined across the horizon (technically, the Schwarzschild coordinates are actually two entirely different coordinate charts disconnected by the horizon). You can see this in the Schwarzschild metric (units of $G = c = 1$):&#xA;$$\mathrm{d}s^2 = -\left(1-\frac{2M}{r}\right)\mathrm{d}t^2 + \left(1-\frac{2M}{r}\right)^{-1}\,\mathrm{d}r^2 + r^2(\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\phi^2)\text{,}$$&#xA;where the metric coefficients in the Schwarzschild chart are undefined at the horizon $r = 2M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;C) &quot;So, in this condition, what will the object do?&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will cross the event horizon and meet its end at the singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the same Schwarzschild geometry in &lt;a href=&quot;http://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates&quot; rel=&quot;nofollow&quot;&gt;Gullstrand-Painlevé coordinates&lt;/a&gt;:&#xA;$$\mathrm{d}s^2 = -\mathrm{d}t^2 + \left(\mathrm{d}r + \sqrt{\frac{2M}{r}}\,\mathrm{d}t\right)^2 + r^2(\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\phi^2)\text{.}$$&#xA;In this coordinate chart, space is moving towards the singularity at the estate velocity $\sqrt{2GM/r}$, and the particles are carried along with it. In a related coordinate chart, &lt;a href=&quot;http://en.wikipedia.org/wiki/Lema%C3%AEtre_coordinates&quot; rel=&quot;nofollow&quot;&gt;Lemaître coordinates&lt;/a&gt;, particles freefalling from rest at infinity have zero coordinate speed, with $\mathrm{d}r/\mathrm{d}t = 0$, where here $r$ is the Lemaître radial coordinate. (Edit: corrected example mistakenly attributing this property to the GP chart.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can pick any coordinate chart that's well-behaved at the horizon, and any such chart will show it moving across the horizon with no trouble. But the Schwarzschild coordinates do not give such a chart, because they're undefined at the horizon. &lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditDate="2014-08-16T03:48:45.680" LastActivityDate="2014-08-16T03:48:45.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6152" PostTypeId="2" ParentId="6130" CreationDate="2014-08-16T13:07:54.493" Score="2" Body="&lt;p&gt;user3814836's answer seems to be right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the choice of the telescope depends on what you want.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;With a refractor telescope (I have one, a SkyWatcher BK 707AZ2). It's maybe a good start for a beginner, if you want to discover the brightest objects of the deep sky before going further. However, there is a defect: chromatic aberration. Also, you need to get used to the inversion of the image.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;With a reflector telescope, you can see the dimmer objects. Maybe it will help if you have a lot of pollution.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As for the light pollution, the objects will look dimmer without an appropriate filter and you won't probably see the colours (personal experience with my telescope). The Moon doesn't help sometimes. If you have the chance to escape the pollution, you should be more amazed.&lt;/p&gt;&#xA;" OwnerUserId="1890" LastActivityDate="2014-08-16T13:07:54.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6153" PostTypeId="1" AcceptedAnswerId="6219" CreationDate="2014-08-17T02:05:54.873" Score="3" ViewCount="166" Body="&lt;p&gt;Figure 1 from &lt;a href=&quot;http://apod.nasa.gov/diamond_jubilee/d_1996/hub_1929.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble 1929&lt;/a&gt; (original paper with &lt;a href=&quot;http://www.pnas.org/content/15/3/168&quot; rel=&quot;nofollow noreferrer&quot;&gt;open access pdf&lt;/a&gt;) has velocity with units of km. Is this just a typo or was there a reason the plot is printed this way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/clIHd.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1862" LastEditorUserId="7982" LastEditDate="2017-12-31T08:46:56.020" LastActivityDate="2017-12-31T08:46:56.020" Title="Original Hubble diagram units incorrect?" Tags="&lt;cosmology&gt;&lt;history&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6154" PostTypeId="1" CreationDate="2014-08-17T03:34:38.040" Score="8" ViewCount="144" Body="&lt;p&gt;Rosetta is now arriving at Comet 67P. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other comets have been analyzed to some degree (Giotto, Deep Space 1, Deep Impact, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is... where I can study the up-to-date theory of comets? Much of what was previously thought (eg. dirty snowball) has been discarded now, but I haven't found a good source describing the current thinking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any firm predictions that will be tested as Rosetta studies the comet up close? &lt;/p&gt;&#xA;" OwnerUserId="296" LastEditorUserId="2153" LastEditDate="2014-08-17T14:54:40.480" LastActivityDate="2014-08-17T14:54:40.480" Title="Is Rosetta changing the theory of comets?" Tags="&lt;comets&gt;&lt;space-probe&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ClosedDate="2016-12-31T08:19:00.333" ContentLicense="CC BY-SA 3.0" />
  <row Id="6155" PostTypeId="2" ParentId="6072" CreationDate="2014-08-17T09:21:05.403" Score="3" Body="&lt;p&gt;+taupunkt basically answered this: the thin lunar crescent after New Moon is only visible for a short while after sunset.  While the sun is up, the moon is invisible and it becomes visible when, after sunset, the sky turns dark enough.  However, the moon will set typically 1-2 hours after sunset, and will become invisible some time before that due to atmospheric extinction.  Whether or not, and if so for how long, you can &lt;em&gt;actually see&lt;/em&gt; the moon's crescent depends on the local cirumstances: the distance between the sun and moon (which correlates with the lunar phase and depends on your geographical longitude), the angle between the ecliptic and the horizon (depends on your latitude), whether you're in a mountainous area (and if so, whether you're on a mountain top or in a valley), atmospheric humidity, the weather (a few clouds at the western horizon is all it takes), whether you use your naked eye, binoculars or a telescope, your eyesight and experience, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of this, it is difficult to predict when you can see the first/last moon crescent.  I use &lt;a href=&quot;http://www.phys.uu.nl/~vgent/islam/downloads/naotn_69.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; to do that for &lt;a href=&quot;https://translate.googleusercontent.com/translate_c?depth=1&amp;amp;hl=en&amp;amp;ie=UTF8&amp;amp;prev=_t&amp;amp;rurl=translate.google.com&amp;amp;sl=auto&amp;amp;tl=en&amp;amp;u=http://hemel.waarnemen.com/maan/maansikkels_2014.html&amp;amp;usg=ALkJrhh-976zT2Uth1iD31jg7dp0_BbNkg&quot; rel=&quot;nofollow&quot;&gt;the Netherlands&lt;/a&gt; (Google translation, should be ok since it's a table) and for the naked eye.  One thing that strikes me is that of the 25 cases of a first/last crescent sighting this year, the lunar phase ranges from 1.2-9.2%, with a mean of 4.0%, so 0.8% seems to be rather optimistic.  You may allow the use a telescope though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The table doesn't list &lt;em&gt;how long&lt;/em&gt; the moon is visible for (the prediction of &lt;em&gt;whether&lt;/em&gt; the moon is visible or not is sufficiently uncertain by itself), but it will definitely not be more than the time difference between the rise/set times of the sun and moon (~1-2h).  If we use half of the average of those numbers, we get 45 minutes, which may well be optimistic but is not too unrealistic either (for a ~4% crescent).  The other ingredient is how long the lunar phase is at 0.8%.  Assuming that this means between 0.75% (below which it would be rounded down to 0.7%) and 0.85% (above which it would be 0.9%) this was the case for ~97 minutes for the last time this happened (July 27).  Together with the 45-minute observation window, this gives a ~142-minute window of opportunity, or about 10% of a day, indicating that the crescent would be visible from ~10% of the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that we assumed a 4% crescent here, not 0.8%, so this 10% is an overestimation.  My guess is that it is a &lt;em&gt;huge&lt;/em&gt; overestimation, since in most places you will not be able to see the moon with the unaided eye at all.  Even if the moon were somehow visible for an instant &lt;em&gt;everywhere&lt;/em&gt;, the fraction of the earth the 0.8% crescent is visible from would drop to ~6.7%.  If you want a more precise definition of 0.8%, &lt;em&gt;e.g.&lt;/em&gt; 0.80%, or between 0.795% and 0.805%, the moon spends about 10 minutes at the desired phase and this would be visible from 0.7% of the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then again, you may allow the use of a telecope, so that you can see the moon even during day time.  If you also don't worry about the weather, the window of observation would be 12 hours (since the moon is up for observers at about 50% of the earth's surface) and with the 97-minute time window for which the crescent is between 0.75% and 0.85%, the 0.8% crescent would be visible from about 57% of our planet's surface.  For 0.795%-0.805%, this would be 51% of the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conclusion is that a 0.8% crescent of the moon is visible from beteen 0% and 60% of the surface of the earth, depending on your exact definitions and your choice of accuracy and optical aid.&lt;/p&gt;&#xA;" OwnerUserId="286" LastActivityDate="2014-08-17T09:21:05.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6156" PostTypeId="1" AcceptedAnswerId="6160" CreationDate="2014-08-17T14:30:42.637" Score="2" ViewCount="840" Body="&lt;p&gt;It appears that the face of the moon &quot;rotates&quot; clockwise from the time the moon rises to the time it sets. Tycho's crater seems more to the left and the &quot;Rabbits head&quot; looks like it's facing down more.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that is not actually the case but I am having trouble visualizing why this appears so. Can anyone explain?&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2019-01-04T19:14:12.343" Title="The face of the moon at moonset" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6157" PostTypeId="1" CreationDate="2014-08-18T05:32:55.380" Score="1" ViewCount="105" Body="&lt;p&gt;we measure time based on earth and earth mass\gravity plays a role in the measurement ? so how and what would be time if not earth time on other exo planets since they will be having different mass/gravity and probably more variables affecting time measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can say Gliesa has a day of 30 hrs(assumed) , &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So will the length of second, minute and hour, likeewise day or month will fall under its own time system?&lt;/p&gt;&#xA;" OwnerUserId="538" LastActivityDate="2014-08-18T09:57:08.610" Title="the length of second, minute or hour, what defines the time of exo planetary bodies" Tags="&lt;earth&gt;&lt;exoplanet&gt;&lt;space-time&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6158" PostTypeId="2" ParentId="6157" CreationDate="2014-08-18T09:57:08.610" Score="3" Body="&lt;p&gt;You have 2 separate concepts here. The length of a day and a year are unconnected with how you measure seconds, minutes and hours.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The length of a day is defined as how long it takes for the planet to rotate on its axis, and a year is how long it takes to orbit its primary.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Seconds, minutes and hours are a measurement chosen by us back in history, and chances are we will still use the same wherever we go in the universe. The way we measure it is designed to let us always measure seconds consistently. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Even on Earth, we say that a day is 24 hours, but it isn't really, as the two sets of concepts don't match up. We haven't defined a second as a fraction of the time it takes the planet to rotate - it is defined based on physical constants. So if you were to say that Gliesa has a day that is 30 units long - it would be astonishing if those units were exactly hours!&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2014-08-18T09:57:08.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6160" PostTypeId="2" ParentId="6156" CreationDate="2014-08-18T10:15:10.567" Score="10" Body="&lt;p&gt;Actually, it is rotating relative to your perspective. (The other answer relating to liberation occurs over a longer period of time). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a simulation from Sky Safari for later tonight (moonrise) and tomorrow (moonset) for my location. Note how the angle of the lit hemisphere rotates between the pictures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moonrise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bxBQX.jpg&quot; alt=&quot;moonrise&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moonset:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/frukl.jpg&quot; alt=&quot;moonset&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like this for me because I am at 43 south. If I was at the equator, the moon would appear to have rotated 180 as it travelled from horizon to horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easier to think about on the equator. Say the moon rises, and you consider the bit at the top to be &quot;up&quot;. It continues on up, right over your head, and down to the horizon... But on the way down, the &quot;up&quot; side of the moon is now looking down. It was &quot;up&quot; until the zenith, but then it is &quot;down&quot; all the way to the horizon. It doesn't rotate at all, but the result is a 180 degree flip in appearance from horizon to horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or consider the perspective at a pole, while the moon is up. The moon travels right around the sky, always facing the same direction, with no apparent rotation (and while it is up, no moonrise or moonset, it is up all day).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Latitudes in between see an in-between effect; some apparent rotation of the moon between moonrise and moonset.&lt;/p&gt;&#xA;" OwnerUserId="67" LastEditorUserId="67" LastEditDate="2014-08-18T10:48:57.587" LastActivityDate="2014-08-18T10:48:57.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6161" PostTypeId="2" ParentId="2373" CreationDate="2014-08-18T12:57:26.197" Score="3" Body="&lt;p&gt;A month after I asked this question, &lt;a href=&quot;https://www.youtube.com/user/KISSCaltech&quot; rel=&quot;nofollow&quot;&gt;Keck Institute for Space Studies&lt;/a&gt; kindly arranged three lectures uploaded June 12, 2014, to provide some good answers :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=ORUkiyoYy0E#t=153&quot; rel=&quot;nofollow&quot;&gt;Major science questions for Venus&lt;/a&gt; Jeffery Hall, JPL&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=OwywK8zihV4&quot; rel=&quot;nofollow&quot;&gt;Options for a seismic investigation of Venus&lt;/a&gt; David Mimoun, ISAE Toulouse&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=RFeLQR5MEuI&quot; rel=&quot;nofollow&quot;&gt;Exploring Venus with Landers, orbiters and ballons&lt;/a&gt; Dave Stevenson, Caltech&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'd better watch their presentations of 38-50 minutes each if you're interested. My conclusions from watching them are that many basic things are unknown about Venus and that a ground mission would need more time than it could survive. But on the other hand that surprisingly good answers (even seismologically!) could be achieved in very clever ways from atmospheric or even orbiting missions.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2014-08-18T13:05:25.543" LastActivityDate="2014-08-18T13:05:25.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6162" PostTypeId="1" AcceptedAnswerId="6168" CreationDate="2014-08-19T00:33:17.053" Score="1" ViewCount="134" Body="&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/d/d5/WISE_full_sky.jpg/1280px-WISE_full_sky.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/d/d5/WISE_full_sky.jpg/1280px-WISE_full_sky.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;(click to enlarge)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How was this picture of the supposed entire universe taken in infrared? Also, why does it seem as though almost all of the matter is in a line formation over the center? It makes sense why most matter would be concentrated in the center, but why is it in a line?&lt;/p&gt;&#xA;" OwnerUserId="2274" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:47.220" LastActivityDate="2014-09-01T19:10:57.233" Title="How was an infrared picture of the entire universe taken?" Tags="&lt;universe&gt;&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6163" PostTypeId="1" AcceptedAnswerId="6166" CreationDate="2014-08-19T01:56:37.943" Score="6" ViewCount="836" Body="&lt;p&gt;Question is rather self-explanatory. An answer touching upon how/why atmospheres are formed in the first place would be ideal as well. An example of such astral bodies would be The Earth and it's moon. Earth has an atmosphere, while the moon does not. &lt;strong&gt;Why is that so?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="294" LastActivityDate="2014-08-27T07:56:31.300" Title="Why do some celestial bodies have atmospheres, and not others?" Tags="&lt;planetary-atmosphere&gt;&lt;formation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6164" PostTypeId="1" AcceptedAnswerId="6167" CreationDate="2014-08-19T02:03:35.910" Score="5" ViewCount="1004" Body="&lt;p&gt;Since we know that stars burn out, collapse, or blow up when too many of their atoms fuse into iron, then how come, with all the ancient galaxies we've been observing, there isn't one galaxy whose stars have become giant balls of iron, which would eventually collide into one galaxy-sized iron globule in space?&lt;/p&gt;&#xA;" OwnerUserId="2006" LastEditorUserId="2365" LastEditDate="2016-10-01T19:32:56.863" LastActivityDate="2016-10-01T19:32:56.863" Title="Why aren't there galaxy-sized balls of iron out there?" Tags="&lt;star&gt;&lt;galaxy&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6165" PostTypeId="2" ParentId="6164" CreationDate="2014-08-19T02:47:00.307" Score="-2" Body="&lt;p&gt;Our sun will eventually become a white dwarf. A star 10 times it mass will become a neutron star. A star 100 times the mass of the sun will become a black hole. So if 100 stars came together into one mass, it would collapse into a black hole. It is currently believed that a super massive black hole (at least 1000 times the mass of our sun) is at the center of our galaxy and also in the Andromeda galaxy. This is where all the iron might be found from this region of space.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-08-19T02:47:00.307" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6166" PostTypeId="2" ParentId="6163" CreationDate="2014-08-19T06:43:19.443" Score="7" Body="&lt;p&gt;The amount and kind of gases a body can trap depends on the object's &lt;strong&gt;surface temperature&lt;/strong&gt;, and its &lt;strong&gt;density &amp;amp; radius&lt;/strong&gt; (which refers to it's &lt;strong&gt;gravity&lt;/strong&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An object with high gravity and low surface temperature will be able to hold more gases in it's atmosphere. In the case of the &lt;strong&gt;Moon&lt;/strong&gt;, due to its low gravity it could barely trap an atmosphere of &lt;strong&gt;Xenon&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fUln9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can check this thanks to this amazing plot: &lt;a href=&quot;http://astro.unl.edu/naap/atmosphere/animations/gasRetentionPlot.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://astro.unl.edu/naap/atmosphere/animations/gasRetentionPlot.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is not all you need to hold an stable atmosphere like the Earth does. The object will also need some &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_wind#Atmospheres&quot; rel=&quot;nofollow noreferrer&quot;&gt;protection against the solar wind&lt;/a&gt;&lt;/strong&gt;. In the Earth we have a &lt;strong&gt;magnetic field&lt;/strong&gt; that prevents the solar wind to reach our atmosphere, but in the case of &lt;strong&gt;Mars it has a weak and poor magnetic field&lt;/strong&gt; to defend itself against the solar wind that rips it's atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="1678" LastEditorUserId="1678" LastEditDate="2014-08-27T06:36:26.900" LastActivityDate="2014-08-27T06:36:26.900" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6167" PostTypeId="2" ParentId="6164" CreationDate="2014-08-19T06:49:58.697" Score="20" Body="&lt;p&gt;In short, you can't make a galaxy-mass ball of iron because there's no way to support the ball against the inward crush of its own gravity. First, it would collapse into a neutron star, but even neutron stars can't hold themselves up beyond about 2-3 solar masses.  Your galaxy-sized object would collapse into a black hole.  But we do see these!  All massive galaxies are believed to have supermassive black holes at their centres. These are up to 10 billion solar masses in size, which is comparable to smaller galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any star, be it a main-sequence star like the Sun, a giant, or even a white dwarf or neutron star, the inward force of gravity must be balanced by some outward force or pressure.  (In fact, we basically assume this when we construct stellar models, although it's a very good assumption.) In main-sequence stars, this pressure is provided by the nuclear reactions happening inside them but as a star evolves, it gets a bit more complicated. In some situations, the nuclei get so close together that they their electrons begin to share quantum states. And because no two electrons can be in the same state, they have to occupy higher energy states. This in effect exerts a kind of pressure (called &lt;em&gt;degeneracy pressure&lt;/em&gt;), because the electrons are forced to move around faster than they otherwise would. This is what supports white dwarfs and the cores of some evolved stars. e.g. low-mass red giants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Degeneracy pressure is fine up to a limit known as the &lt;em&gt;Chandrasekhar limit&lt;/em&gt;, which is about 1.44 solar masses (depending on the composition of the object in question). Anything above this would collapse. First, this collapse would yield a neutron star, where things are a bit different. But a similar principle holds, and there's a degeneracy pressure because the neutrons themselves also occupy one quantum state each.  The details are &lt;em&gt;much&lt;/em&gt; more difficult here, but the overall consensus is that neutron degeneracy pressure can support objects up to about 2-3 solar masses. After that, there's nothing left to balance gravity, and the star collapses into a black hole.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastEditorUserId="1577" LastEditDate="2014-08-19T08:55:22.900" LastActivityDate="2014-08-19T08:55:22.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6168" PostTypeId="2" ParentId="6162" CreationDate="2014-08-19T06:54:33.490" Score="2" Body="&lt;p&gt;This picture was taken by the &lt;strong&gt;WISE (Wide-field Infrared Survey Explorer)&lt;/strong&gt; space telescope. It is in deed an infrared picture of the entire universe but what you see in this line is our home galaxy, the Milky Way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This telescope has taken infrared pictures of the entire Universe (as we see from the Earth) and joined them in this &lt;strong&gt;full-sky view&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More info: &lt;a href=&quot;http://en.wikipedia.org/wiki/Wide-field_Infrared_Survey_Explorer&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Wide-field_Infrared_Survey_Explorer&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1678" LastActivityDate="2014-08-19T06:54:33.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6169" PostTypeId="2" ParentId="6164" CreationDate="2014-08-19T08:25:03.340" Score="3" Body="&lt;p&gt;If you had a galaxy sized ball of iron the gravitational force emitted would be immense, and without any force pushing out from it the ball would collapse into itself to form a black hole. In stars such as the sun (main-sequence stars) the outward force of nuclear reactions taking place in its core keep it from collapsing into itself. However, with a ball of iron the only force acting on it would be the pull of gravity, eventually crushing the ball into itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ball of iron wouldn't even be able to reach the size of a galaxy. There is a limit known as the Chandrasekhar limit, which says anything that is about 1.44 solar masses would collapse. This would most likely form a neutron star, but this also has a limit to the amount of mass it can hold. Eventually, even this will collapse upon itself to form a black hole. &lt;/p&gt;&#xA;" OwnerUserId="2212" LastActivityDate="2014-08-19T08:25:03.340" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6170" PostTypeId="2" ParentId="1901" CreationDate="2014-08-19T09:28:55.173" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If something crashed into it rearranging its content a bit, could its life be extended?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Yes!&lt;/strong&gt; This is precisely what we think causes the creation of &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Blue_straggler&quot; rel=&quot;noreferrer&quot;&gt;blue stragglers&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we look at a cluster of stars, we expect them to all be of roughly the same age.  This is usually borne out by observations.  Because larger stars evolve faster, the age of a cluster can then be roughly determined by working out the mass of the most massive star still on the main sequence (i.e. still burning hydrogen in the core).  We call this the &lt;em&gt;main-sequence turnoff&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in some clusters, we see and handful of stars that are more massive, but still on the main sequence, even though all the evidence says they should have already evolved into giants and other more-evolved objects.  The consensus view is that these stars are at the very least interacting with binary companions, and many are thought to be &lt;em&gt;merger products&lt;/em&gt;. i.e. something left over from two stars merging into one.  Either two companions in a binary have orbited so close that they've become one star, or two stars have simply collided (which is quite possible in the dense cores of clusters).  Depending on what types of star collide, you can basically get a new star where everything is mixed up again, allowing the star to be &quot;born again&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2014-08-19T09:28:55.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6171" PostTypeId="1" CreationDate="2014-08-19T13:05:04.573" Score="1" ViewCount="2774" Body="&lt;p&gt;Why don't the planets glow like stars?&lt;/p&gt;&#xA;" OwnerUserId="2195" LastEditorUserId="633" LastEditDate="2014-08-22T15:40:24.503" LastActivityDate="2014-08-22T15:40:24.503" Title="Why don't planets give off their own light?" Tags="&lt;star&gt;&lt;planet&gt;&lt;fundamental-astronomy&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2014-08-20T13:26:12.067" ContentLicense="CC BY-SA 3.0" />
  <row Id="6173" PostTypeId="2" ParentId="6171" CreationDate="2014-08-19T15:20:00.003" Score="0" Body="&lt;p&gt;Stars' light is produced by thermonuclear fusion of hydrogen, helium and other elements at their nucleus. This processes are impossible in planets since you need high pressures and temperatures &gt;10,000 k&lt;/p&gt;&#xA;" OwnerUserId="1678" LastActivityDate="2014-08-19T15:20:00.003" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6174" PostTypeId="1" AcceptedAnswerId="6176" CreationDate="2014-08-19T18:47:57.643" Score="4" ViewCount="456" Body="&lt;p&gt;Following &lt;a href=&quot;http://www.space.com/26874-supermassive-black-hole-star-death.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt;, it is stated that: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As a star drifts too close to a supermassive black hole, intense tidal&#xA;  stresses rip the star to shreds. As this happens, the shredded&#xA;  material will be dragged into the black hole’s accretion disk — a hot&#xA;  disk of gas that is gradually pulled into the black hole’s event&#xA;  horizon, bulking up the black hole’s mass or blasted as energetic&#xA;  jets from its poles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Questions are, &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the energy type that escapes from the poles? &lt;/li&gt;&#xA;&lt;li&gt;What is the magnitude of the power? &lt;/li&gt;&#xA;&lt;li&gt;Can this energy be potentially harnessed? &lt;/li&gt;&#xA;&lt;li&gt;How far into space do these energy jets blow?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Also, if you feel like discussing this: if mass &lt;strong&gt;m&lt;/strong&gt; goes into the black hole and comes out as energy &lt;strong&gt;E&lt;/strong&gt; from the poles, assuming the speed of light &lt;strong&gt;c&lt;/strong&gt; has no play in this transaction of going in or out, what happened to &lt;span class=&quot;math-container&quot;&gt;$c^2$&lt;/span&gt; from &lt;img src=&quot;https://i.stack.imgur.com/lnMm7.png&quot; alt=&quot;E=m*c^2&quot;&gt;? Could a black hole be hypothetically faster than light? Or maybe the reverse direction of light? (I even got confused from this thought...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if those jets are not pure energy, but are energized matter, &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What type/kind of matter it would be? (gas? solid? liquid?)&lt;/li&gt;&#xA;&lt;li&gt;Is all the matter that comes in, also goes out? I am thinking if the black hole works like a car's engine - does it have conversion efficiency? &lt;/li&gt;&#xA;&lt;li&gt;Can this matter be cultivated, harnessed, or converted back to what it was or something new all together?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This question is presented here after it was &lt;a href=&quot;https://space.stackexchange.com/questions/5201/what-type-of-energy-is-escaping-from-black-holes-poles&quot;&gt;asked on the space stackexchange&lt;/a&gt;. I want to leave it there because it has some relevant discussion to this question overall.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1305" LastEditorUserId="25994" LastEditDate="2019-03-03T17:09:57.870" LastActivityDate="2019-03-03T17:09:57.870" Title="What type of energy is escaping from black-hole's poles?" Tags="&lt;black-hole&gt;&lt;mass&gt;&lt;high-energy-astrophysics&gt;&lt;pole&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="6175" PostTypeId="2" ParentId="6140" CreationDate="2014-08-20T05:55:04.540" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The star reaches a critical mass and ignites, &lt;strong&gt;blowing away the surrounding material&lt;/strong&gt;. (OP emphasis)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think the misunderstanding is that it's not that explosive when the star &quot;lights up&quot;. The process is in some sense continuous, so the nuclear reactions start slowly and gradually increase in strength as the star continues to contract, grow, and heat up inside.  Once the amount of radiation from the surface becomes sufficiently hot, we can speak of &quot;blowing away&quot; material, but even then it's not violent. The absorption of light in the gas creates a pressure that pushes the gas away. It picks up pace as the star settles down (and depending on how massive the star is), but it's broadly a consistent stream. (Astronomers are probably unwise to use the word &quot;ignite&quot; because all we mean is that the nuclear reaction rates become non-negligible. It's not like a rocket taking off or something.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, star formation is a poorly-understood process and a hot topic for research.  For stars larger than about 7 solar masses, we think that nuclear reactions begin before all the material has gathered in the star. i.e. while some is still flowing on. If this happens through a disk, then the light can stream out everywhere but the equator without much affecting the growth of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, at &lt;em&gt;very&lt;/em&gt; high masses, you do run into a problem of blowing mass away. For this reason, we don't really expect to see many stars over a few hundred solar masses, and those that are can be violently unstable (like Eta Carina, which threw off about 10 solar masses of material in the mid-19th century).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as for stars genuinely &quot;blowing up&quot;, this does happen in other circumstances. For example, in the core of a solar mass red giant, matter becomes nearly isothermal.  It eventually gets hot enough to ignite helium, and it nearly all lights up at nearly the same time in a &lt;em&gt;helium flash&lt;/em&gt;.  This briefly produces as much light as a small galaxy! But only very briefly, and it doesn't make it to the surface for us to see. Incidentally, this is also the same mechanism as Type Ia supernovae: an isothermal carbon-oxygen white dwarf accretes matter until the carbon lights up, and the whole white dwarf basically ignites at the same time.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2014-08-20T05:55:04.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6176" PostTypeId="2" ParentId="6174" CreationDate="2014-08-20T07:48:39.683" Score="6" Body="&lt;p&gt;The article you've read is not quite accurate/correct. A more correct pictue is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A star may approach a super-massive black hole (SMBH) so closely that the tidal forces of the SMBH tear it appart. The distance to the SMBH at which this happens is often referred to as the &lt;em&gt;tidal radius&lt;/em&gt;. For a (non-rotating) SMBH with a mass in excess of about $10^8$M$_\odot$, the tidal radius is within the event horizon and hence this process is invisible.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;About half (depending on the initial orbit of the star) of the stellar material is caught by the SMBH onto a elliptic orbits (the other half escapes on hyperbolic orbits), but due to gas interactions and dissipation (shocks transfer orbital energy into heat which is then radiated away) it soon circularises to form a (or contribute to an existing) accretion disc. (Note that feeding from stars is not thought to be the main mechanism for the formation of accretion discs; a more likely process is the direct infall of gas clouds).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Material in the accretion disc orbits the BH on nearly circular orbits. It cannot simply fall into the SMBH because its angular momentum is conserved. However, gas at adjacent radii orbits at different velocities (&lt;em&gt;differential rotation&lt;/em&gt;), i.e. the disc is &lt;em&gt;shearing&lt;/em&gt;. Viscosity in the gas then causes angular momentum to be transported outwards, mass being transported inwards, and the material to heat up. The material in the disc is a hot (presumably) magnetized plasma, and the viscosity is thought to be provided by the so-called &lt;em&gt;magneto rotational instability&lt;/em&gt;. The time scale for this process is, however, very long and only becomes efficient ($\sim10^6$yr) rather close ($\sim100$AU) to the SMBH.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The temperature of the disc material is extremely high, such that it radiates in the X-ray and UV, visible as quasars. This radiation is the main energy output from the accreting BH and constitutes about 10% of the gravitational energy gain $mc^2$ ($m$ being the mass accreted). This radiation interacts with any surrounding material (including the outer accretion disc itself, in particular if it is warped), generating an outflowing wind&#xA;(by its radiation pressure) and possibly preventing further accretion (this corresponds to the &lt;em&gt;Eddington limit&lt;/em&gt;).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The rotating magnetized disc may also generate collimated bi-polar outflows, commonly referred to as jets. However, IMHO, only a small fraction of the energy can be emitted in these jets, already because the jets have low entropy (they are dominated by ordered motion), while the hot disc has high entropy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2014-08-20T08:14:09.073" LastActivityDate="2014-08-20T08:14:09.073" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6177" PostTypeId="2" ParentId="6171" CreationDate="2014-08-20T08:10:24.997" Score="5" Body="&lt;p&gt;All matter radiates (except if it's at absolute zero temperature), regardless of its composition (you got that of Mercury badly wrong). The most important form of radiation is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_body_radiation&quot; rel=&quot;nofollow&quot;&gt;black-body radiation&lt;/a&gt; which only depends on the temperature of the material, but line emission and absorption may also be important (but depends on the composition and ionisation state of the material) and other emission and absorption processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars are hot enough (the Sun has surface temperator $\sim5700$K) for the black-body radiation to peak in the visible part of the electro-magnetic spectrum. As a star shines, it looses energy, i.e. it cools, reducing the gas pressure that stabelizes it against gravitational collapse. In stars, this energy loss is balanced by the energy production from thermonuclear fusion in the core (requiring temperatures $\sim10^9$K). The transport of this energy to the surface makes stars non-trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In planets and brown dwarves, there is (by definition) no thermonuclear energy source. Therefore, these objects must shrink, which generates energy from gravity. However, they cannot shrink indefinitely, as ultimately quantum mechanics becomes important: Pauli's exclusion principle demands that the electrons cannot be arbitrarily closely packed. Thus, for brown dwarves and giant gas planets (but also white dwarves) further shrinking is halted at a radius comparable to that of Jupiter (Jupiter itself is still shrinking at a very small rate). These as well as all smaller objects then merely cool down very much like a piece of glowing coal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation is a often more complicated by sources of energy. Planets, for example are&#xA;irradiated by their host star, which may dominate the energy gains at their surface (in addition, the Earth gains energy from nuclear fission in its core). The balance between this energy gain and the loss by black-body radiation determines the temperature of a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth, for example, radiates in the infrared. However, the radiation losses are also regulated by line absorption of that infrared in the higher atomsphere by so-called green-house gases, in particular CO$_2$.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2014-08-21T07:55:58.467" LastActivityDate="2014-08-21T07:55:58.467" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6178" PostTypeId="2" ParentId="6164" CreationDate="2014-08-20T08:20:56.587" Score="5" Body="&lt;p&gt;Apart from the good reasons given in the other answers as to why a ball of iron in excess of about $1.4$M$_\odot$ cannot be stable, there is another reason. Namely, there is no way to form that much pure iron. Iron is produced in supernovae, but only a fraction of the matter expelled by a supernova is actually iron and there is no natural way to select is. The remaining matter is locked in the stellar remnant, either a white dwarf, neutron star, or (stellar-mass) black hole. The latter two are already to massive for the material to be in the form of iron, while the a white dwarf may only contain some iron, but never is a ball of iron.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-08-20T08:20:56.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6180" PostTypeId="1" CreationDate="2014-08-20T17:02:39.140" Score="0" ViewCount="210" Body="&lt;p&gt;I saw something very strange last night that I'm hoping someone can help me explain.&#xA;I was sitting on my roof with a night-cap at about 3am last night. The moon, about the last quarter, was low in the sky in the east. Something truly strange happened. The moon started to disappear. There were clouds in the sky, but it wasn't this, I'm quite sure. The moon was being eclipsed. The dark part (at the top of the moon, as the moon was lying on its back - a sailor moon? That's what it looked like, like you could sit in it.) grew, the crescent closed, until it was just a spot of cream colored light and then winked out entirely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was a bit amazed - I'd stumbled onto my roof in the dead of the night and by complete chance witnessed a total lunar eclipse? I sat there in awe, staring. That's when the moon started to reappear, in the reverse procedure as its disappearance. Becoming a gleaming point of light, then a thin crescent, then the quarter moon which I suppose is what it is &quot;supposed to look like.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AND THEN IT HAPPENED AGAIN. The moon was eclipsed and revealed again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both times the whole process took about 6-10 minutes. I'm not sure if this cycling continued, it was late and I had to sleep. I checked the NASA site for eclipses for my time and location, but no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone ever seen anything like this, or have any plausible explanation as to what is going on here?&lt;/p&gt;&#xA;" OwnerUserId="2280" LastActivityDate="2016-08-17T14:16:04.607" Title="Multiple apparent lunar eclipses last night" Tags="&lt;the-moon&gt;&lt;lunar-eclipse&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6182" PostTypeId="1" CreationDate="2014-08-21T17:12:35.687" Score="7" ViewCount="172" Body="&lt;p&gt;NASA has a plan called &quot;&lt;a href=&quot;http://www.planetary.org/blogs/jason-davis/2014/the-latest-on-nasas-asteroid.html&quot;&gt;Asteroid Redirect Mission option A&lt;/a&gt;&quot;. It would robotically put an asteroid of a few meters diameter in a bag and push it into reach of astronauts for thorough examination. &lt;a href=&quot;http://solarsystem.nasa.gov/planets/profile.cfm?Object=Meteors&amp;amp;Display=OverviewLong&quot;&gt;44 tonnes of meteorites&lt;/a&gt; falls on Earth every day and almost &lt;a href=&quot;http://www.lpi.usra.edu/meteor/metbull.php&quot;&gt;50,000 meteorites&lt;/a&gt; have been found on Earth. So the unique advantage of capturing a pristine asteroid, however tiny, is that it will not have been deformed by entering Earth's atmosphere. It might be a rubble pile and/or its surface might be covered by regolith or maybe even volatiles since it might have formed and always existed in microgravity. Even the Moon is covered by hoovering electrically charged dust near above its surface, ionized by cosmic or Solar radiation. And surface properties are important because it is only the surface which can be observed by telescopes and Earth based radar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But wouldn't bagging and pushing an asteroid destroy such delicate surface and rubble pile properties?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2014-08-21T17:19:24.467" LastActivityDate="2015-01-10T22:52:39.330" Title="Would bagging an asteroid destroy valuable science about it?" Tags="&lt;asteroids&gt;&lt;surface&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6183" PostTypeId="1" CreationDate="2014-08-21T17:46:55.930" Score="33" ViewCount="13736" Body="&lt;p&gt;Referring to the mechanisms explaining the solar system formation and to the initial rotation of the gaseous cloud that collapsed, I understand easily why the planets orbit the Sun the same way this one rotate (say counterclockwise) but I can't figure out why this apply to planets rotation too. Thinking about that from Kepler's laws and angular momentum conservation point of view, I might conclude that the planets should rotate clockwise because the velocity of the particles that aggregated during the planets formation was higher closer to the Sun...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from a short explanation, I would like to have a good reference from the literature if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit, to make my reasoning more explicit:&lt;/em&gt; following Kepler's laws, the particles that aggregate on the &quot;day side&quot; of the proto-planets in the east-west direction relative to the ground are faster than the ones hitting on the &quot;night side&quot; in the west-east direction. If we add all of these contributions, the planets should rotate in the opposite direction relative to the initial cloud (i.e. relative to the actual Sun rotation). I guess something is wrong or missing there (to counterbalance the phenomenon I just described) but I can't see what it is...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;New edit:&lt;/em&gt; &lt;strong&gt;References&lt;/strong&gt; I found some published articles dealing with this kind of question but I don't have the time right now to read them carefully. If someone is motivated to do so, do not hesitate ;-) If I find the answer to my question amongst these papers, I will post it there later. Of course, you may need to use the network of an institution with a subscription to these editors to access them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R.T. Giuli (1968a) in Icarus: &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0019103568900821&quot;&gt;http://www.sciencedirect.com/science/article/pii/0019103568900821&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R.T. Giuli (1968b) in Icarus: &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0019103568900122&quot;&gt;http://www.sciencedirect.com/science/article/pii/0019103568900122&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A.W. Harris (1977) in Icarus: &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0019103577900793&quot;&gt;http://www.sciencedirect.com/science/article/pii/0019103577900793&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J.J. Lissauer, D.M. Kary (1991) in Icarus: &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/001910359190145J&quot;&gt;http://www.sciencedirect.com/science/article/pii/001910359190145J&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="993" LastEditorUserId="993" LastEditDate="2014-08-28T17:22:29.910" LastActivityDate="2021-02-16T04:15:05.713" Title="Why do (most of) the planets rotate counterclockwise, i.e. the same way the Sun does?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;rotation&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6184" PostTypeId="1" AcceptedAnswerId="6214" CreationDate="2014-08-22T08:09:41.133" Score="7" ViewCount="180" Body="&lt;p&gt;The Gaia space telescope is in a Lissajous orbit around the Sun-Earth Lagrange point 2. &lt;a href=&quot;http://www.spaceflight101.com/gaia-mission-and-orbit-design.html&quot;&gt;The orbit&lt;/a&gt; period is about 180 days and the size of the orbit is 263,000 x 707,000 x 370,000 km. It has been theoretically concluded that there should be &lt;a href=&quot;http://arxiv.org/abs/1112.3781&quot;&gt;a population of temporarily captured orbiters&lt;/a&gt; (TCO's) around that Lagrange point. All but a handful of them should be less than a meter in diameter and too small to be observed from Earth. Only one TCO, &lt;a href=&quot;http://en.wikipedia.org/wiki/2006_RH120&quot;&gt;2006 RH120&lt;/a&gt;, has ever been observed, a 5 meter diameter one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Gaia will be in a somewhat similar orbit as the TCO's and will scan the entire sky several times over the coming years, and an IR telescope is sensitive to asteroids, I wonder if it will be able to detect such TCO's. And if so, what could we find out about them? Gaia will not track any asteroids nearby, if it catches the same TCO more than once it would be by chance, I suppose. Could Gaia learn about their sizes, orbits, surface compositions?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2014-08-22T11:08:26.623" LastActivityDate="2014-08-28T17:07:51.933" Title="Can the Gaia telescope detect small temporarily captured asteroids near its Lagrange orbit?" Tags="&lt;asteroids&gt;&lt;space-telescope&gt;&lt;lagrange-point&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6185" PostTypeId="2" ParentId="71" CreationDate="2014-08-22T08:23:47.587" Score="1" Body="&lt;p&gt;The confusion about finite or infinite time dilation at the Schwarzschild radius began since Karl Schwarzschild in 1915 used a space-time coordinate system where coordinate time went to +∞ when a falling object approached the Schwarzschild radius. But &lt;em&gt;t&lt;/em&gt; → +∞ for a world line that approaches the horizon is a thing different from &lt;em&gt;t&lt;/em&gt; → +∞ far from the black hole. An object that is about to cross the horizon is in no position to synchronize clocks with an outside world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;No,&lt;/strong&gt; completely opposite situation to the OP’s expectations. If a falling observer will &lt;em&gt;transmit&lt;/em&gt; a powerful beam of light out of the hole, then photons will reach increasingly far future of the exterior universe. But the future &lt;em&gt;seen&lt;/em&gt; by the falling observer will be very limited; see &lt;a href=&quot;http://jila.colorado.edu/~ajsh/insidebh/penrose_schw.html&quot; rel=&quot;nofollow&quot;&gt;http://jila.colorado.edu/~ajsh/insidebh/penrose_schw.html&lt;/a&gt; for graphic depiction.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastActivityDate="2014-08-22T08:23:47.587" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6186" PostTypeId="2" ParentId="6183" CreationDate="2014-08-22T08:26:18.520" Score="6" Body="&lt;p&gt;The reason most of the planets in our solar system (with the exception of Venus and Uranus) rotate in a anti-clockwise direction is due to the proto-planetary disk that formed the Sun and all the planets around it. Since the proto-planetary disk was spinning in a anti-clockwise direction any objects that formed out of the same disk would have shown the same rotation. The fact that it is rotating in a anti-clockwise direction is nothing special. It all depends on the initial spin of the disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason Venus and Uranus have a spin opposite to most other planets is most likely because they were hit by another large object some time during their formation, causing their spin to change directions. This also explains why Uranus' axis is tilted to such a severe extent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you would like to read more here are some links:&#xA;&lt;a href=&quot;http://scienceblogs.com/startswithabang/2010/10/07/counterclockwise-but-there-are/&quot; rel=&quot;noreferrer&quot;&gt;http://scienceblogs.com/startswithabang/2010/10/07/counterclockwise-but-there-are/&lt;/a&gt;&#xA;&lt;a href=&quot;http://hubblesite.org/hubble_discoveries/discovering_planets_beyond/how-do-planets-form&quot; rel=&quot;noreferrer&quot;&gt;http://hubblesite.org/hubble_discoveries/discovering_planets_beyond/how-do-planets-form&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.astronomycafe.net/qadir/q50.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.astronomycafe.net/qadir/q50.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2212" LastEditorUserId="2212" LastEditDate="2014-08-22T08:32:12.760" LastActivityDate="2014-08-22T08:32:12.760" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6187" PostTypeId="2" ParentId="6183" CreationDate="2014-08-22T11:25:13.420" Score="1" Body="&lt;p&gt;I can't give you an explicit ant mathematical explanation but maybe this simulations give you a hint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the bottom one with &lt;strong&gt;beta=0.01&lt;/strong&gt; (faster initial rotation speed) you will see how possibly a binary or even ternary system is formed. I guess this could be picked as an example for planetary formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astro.ex.ac.uk/people/mbate/Animations/prestellar_discs.html&quot; rel=&quot;nofollow&quot;&gt;http://www.astro.ex.ac.uk/people/mbate/Animations/prestellar_discs.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1678" LastActivityDate="2014-08-22T11:25:13.420" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6188" PostTypeId="1" AcceptedAnswerId="6190" CreationDate="2014-08-22T17:07:36.983" Score="23" ViewCount="4574" Body="&lt;p&gt;While browsing through Physics SE, I noticed a question about satellites in geostationary orbit (unrelated to the one I'm asking here), and for a moment I interpreted it as referring to &lt;em&gt;natural&lt;/em&gt; satellites (e.g. a moon). So I wondered: Could a natural satellite exist in geostationary orbit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I stopped and thought. For large gas giants, such as Jupiter, having moons too close to the planet can be fatal (for the moon). If it ventures inside the planet's Roche limit, it's toast. But there is good news: the Roche limit depends on both the masses and densities of the primary body and the satellite. So perhaps this reason is non-applicable, as a high-mass natural satellite might be able to survive. So the question changes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Could a sufficiently high-mass, high-density natural satellite occupy geostationary orbit over its primary body?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2017-02-20T14:35:51.667" LastActivityDate="2018-02-15T13:03:03.383" Title="Can a natural satellite exist in a geostationary orbit?" Tags="&lt;orbit&gt;&lt;natural-satellites&gt;" AnswerCount="6" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6189" PostTypeId="2" ParentId="6188" CreationDate="2014-08-22T19:32:01.590" Score="7" Body="&lt;p&gt;It would require a very precise trajectory for an asteroid to end up in &lt;a href=&quot;http://en.wikipedia.org/wiki/Geostationary_orbit&quot; rel=&quot;noreferrer&quot;&gt;geostationary orbit&lt;/a&gt;. It doesn't happen by chance. Space flight providing companies have to make a real effort to put their customers' communication satellites there. And geostationary isn't a very stable kind of orbit. The varying gravity of the Moon pulls satellites out of their geostationary orbits as the satellites move closer to and further away from it daily as the Earth rotates. GEO is about a tenth of the distance to the Moon. Satellites need their small rocket engines to do recurring &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_station-keeping&quot; rel=&quot;noreferrer&quot;&gt;station keeping&lt;/a&gt; maneuvers in order to stay there. The Earth has no lasting &lt;a href=&quot;http://en.wikipedia.org/wiki/Other_moons_of_Earth&quot; rel=&quot;noreferrer&quot;&gt;natural satellite&lt;/a&gt; in any orbit, except for the Moon.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2014-08-22T19:45:07.183" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6190" PostTypeId="2" ParentId="6188" CreationDate="2014-08-22T20:07:27.717" Score="20" Body="&lt;p&gt;Of course, a natural satellite (moon) could have an orbital period equal to the spin period of its host (provided such an orbit would be accessible). However, the tidal friction that may generate such a locking is quite weak, so this would have to be a rare chance. Moreover, perturbations to the orbit from other moons or their host star may put the moon out of such an orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, what is rather common is that a moon's orbital period equals their own (rather than their host's) spin period. This is exactly the case for the Earth Moon (you may say that Earth is on a &quot;selenostationary&quot; orbit) and naturally occurs form the tidal interaction of the planet with its moon.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-08-22T20:07:27.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6191" PostTypeId="2" ParentId="6183" CreationDate="2014-08-22T20:19:57.913" Score="16" Body="&lt;p&gt;Even though I'm a professional astronomer (but not an expert on this field), I don't have a good answer. AFAIK, this question is still open. Actually, the fact that most planets in the Solar system rotate in the same sense as they orbit the Sun is an important constraint on possible planet formation scenarios (of which still several are presently discussed by scientists).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the idea that planets form out of dust collected at the centre of vortices in the proto-planetary disc can be almost ruled out, because retrograde vortices are much more stable and long-lived than prograde ones (Jupiter's red spot is a retrograde vortex), and hence planets should be retrograde, but aren't.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-08-22T20:19:57.913" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6192" PostTypeId="2" ParentId="4749" CreationDate="2014-08-24T02:21:13.490" Score="7" Body="&lt;p&gt;1977, Arecibo: &lt;a href=&quot;http://www.sciencemag.org/content/196/4290/650.abstract&quot; rel=&quot;noreferrer&quot;&gt;Galilean Satellites of Jupiter: 12.6-Centimeter Radar Observations&lt;/a&gt; Radar echoes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1973, 12.6 cm: &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/001910357390002X&quot; rel=&quot;noreferrer&quot;&gt;Radar Observations of the rings of Saturn&lt;/a&gt; Content is hidden behind Elsevier's paywall, so this might not be radar echoes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nasa.gov/vision/universe/solarsystem/asteroid-20061219.html&quot; rel=&quot;noreferrer&quot;&gt;NASA, 2006 implies&lt;/a&gt; the limit for radar echoes is at the moons of Saturn: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Radar astronomy employs the world's most massive dish-shaped antennas, which beam directed microwave signals at their targets, which can be as close as our moon and as far away as the moons of Saturn. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Looked for studies w radar echoes from moons of Uranus and Neptune, as well as Pluto/Charon, none found. The data &lt;em&gt;may&lt;/em&gt; be out there.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2014-08-24T02:21:13.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6193" PostTypeId="2" ParentId="6188" CreationDate="2014-08-24T04:46:55.943" Score="23" Body="&lt;p&gt;Yes. &lt;a href=&quot;http://en.wikipedia.org/wiki/Charon_%28moon%29#Orbital_characteristics&quot;&gt;Charon&lt;/a&gt; is in Pluto synchronous orbit. Pluto and Charon are mutually tide locked.&lt;/p&gt;&#xA;" OwnerUserId="1232" LastActivityDate="2014-08-24T04:46:55.943" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6194" PostTypeId="1" AcceptedAnswerId="6195" CreationDate="2014-08-24T08:07:40.497" Score="16" ViewCount="1324" Body="&lt;p&gt;The observable universe is constantly expanding as more light from the Big Bang reaches us. This light has been travelling for billions of years, so we are looking at the universe as it was a few billion years ago. This being said, would it be possible to look at the Big Bang in action by looking deep into the universe?&lt;/p&gt;&#xA;" OwnerUserId="2212" LastActivityDate="2014-08-26T17:08:33.657" Title="Can we see the Big Bang happen if we look far enough?" Tags="&lt;big-bang-theory&gt;&lt;observable-universe&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6195" PostTypeId="2" ParentId="6194" CreationDate="2014-08-24T10:33:08.480" Score="12" Body="&lt;p&gt;&lt;strong&gt;No.&lt;/strong&gt; The furthest we can see is the cosmic microwave background radiation (CMB). Early on (after the big bang), matter was fully ionised and the electrons frequently interacted with the photons. That has two consequences. First, the radition was that of a blackbody at the same temperature as the matter. Second, the universe was opaque, i.e. photons couldn't travel very far. &#xA;Due to the explosive expansion of the universe, the temperature decreased all the time until eventually atoms formed. This is called the epoch of re-combination, though the &quot;re&quot; makes little sense. At that point the universe became suddenly transparent and we can see most radiation emitted ever since. In particular the radiation field from the epoch of recombination has been redshifted and appears to us as the CMB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt;&#xA;To answer the speculation in the comments what if the universe was not opaque. Well, all we would be able to so is some (nearly) isotropic radiation field emitted very early on (and redshifted by the cosmic expansion since the big bang). The CMB is exactly that, but emitted at redshift $z\sim$1000 (the epoch of re-combination) rather than larger $z$.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2014-08-26T17:08:33.657" LastActivityDate="2014-08-26T17:08:33.657" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6196" PostTypeId="1" CreationDate="2014-08-25T01:29:33.117" Score="2" ViewCount="264" Body="&lt;p&gt;Will black holes evaporate, if they evaporate? When and why do they evaporate? What are the conditions for evaporation? Under what principle do they evaporate?&lt;/p&gt;&#xA;" OwnerUserId="2195" LastEditorUserId="67" LastEditDate="2014-08-25T20:00:25.023" LastActivityDate="2014-09-01T12:45:06.080" Title="How do black holes evaporate?" Tags="&lt;black-hole&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6197" PostTypeId="1" AcceptedAnswerId="6199" CreationDate="2014-08-25T12:40:18.507" Score="4" ViewCount="144" Body="&lt;p&gt;I'm making a game in which I'd like to randomly generate star systems. Which is cool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trouble I'm having is working out exactly what limits I need to place on the generation. So far, the wikipedia hasn't been all that helpful. So some questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Rocky planets form within the frostline of a protostar. What's the radius of this line? What I'm sort of asking is that given a particular mass and density, what radius from that star is the line.&lt;/li&gt;&#xA;&lt;li&gt;If the jovian planets form outside of this line; why have so many been found within this line in other star systems from our own? Is this only selection bias or is it more likely that our models are incorrect at this point?&lt;/li&gt;&#xA;&lt;li&gt;How does the creation of the gas giants differ from the creation of the ice giants? Is it simply like a second &quot;frost line&quot; where there wasn't enough gaseous material to form a large jovian planet but plenty of ices to form ice giants? If so, how is the creation of the ice giants different to Dwarf planets?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2300" LastActivityDate="2014-08-26T21:11:52.793" Title="Star System Formation questions" Tags="&lt;formation&gt;&lt;planetary-formation&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6198" PostTypeId="2" ParentId="6196" CreationDate="2014-08-25T16:06:03.987" Score="2" Body="&lt;p&gt;The evaporation occurs through Hawking radiation. This is a very slow and low energy process. So low that the cosmic microwave background radiation, which is just a few degrees above absolute zero, pours far more energy into the black hole than Hawking radiation takes away. So in principle a black hole cannot evaporate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the exception of conjectured atomic sized black holes, that is, as Hawking radiation would be more pronounced then. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: See this answer on the physics SE: &lt;a href=&quot;https://physics.stackexchange.com/a/26607/55483&quot;&gt;https://physics.stackexchange.com/a/26607/55483&lt;/a&gt;&#xA;The black hole would need to be less massive than the moon to radiate more energy than it absorbs. Bigger than atomic size, as I suggested. &lt;/p&gt;&#xA;" OwnerUserId="2188" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2014-08-25T18:07:39.340" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6199" PostTypeId="2" ParentId="6197" CreationDate="2014-08-25T16:41:05.770" Score="3" Body="&lt;p&gt;I will edit this with the appropriate equations and timeframes in a few hours when I get back to my office. For now I will address the foundations of your questions with some basics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: The radius of the frost line is determined by the star itself (edit: More precisely the nebula that the star forms from.), more in-depth answer will come with edit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: The fact that most of the data for these systems is from methods that favour large planets close to the parent star, means that our selection isn't biased, but the method instrument combination is. Another possibility is that migration and loss of angular momentum could be the culprits in this scenario. It is also worth noting, that in the next decade there should be significant data to state whether our solar system is actually common, an anomaly or just younger than many of the other solar systems observed thus far. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3: Larger gas giants &quot;Jovian&quot; planets like all orbiting objects migrate inwards as they loose angular momentum. During the oligarchic growth phase of a soon to be large planet the forming core accrets a lot of material while clearing &quot;all&quot; of the material within it's orbit and associated jeans radius, as the core migrates inwards the supply of material is once again increased. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short that is why there are &quot;Gas Giants&quot; within the frost line, things migrate inward as they loose angular momentum, and &quot;Ice Giants&quot; form beyond the line because they took too long to migrate inward and gather enough material to move to the next level. Also, if the protoplanet takes too long to accret a ~10-15 earth-mass core, all of the accretable material will be ejected from the system by the star before any significant atmosphere can be obtained. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dwarf planet part, has to do with the amount of material and how long the protoplanet took to accret it. In that respect, the ice giants and the dwarf planets share a common shortcoming. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find myself overwhelmed at work and unable to fully explain the entireties of the process at this time, please forgive me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I do offer a &lt;a href=&quot;http://ay201b.wordpress.com/the-snow-line-in-protoplanetary-disks/&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; to an upper division astronomy class' notes that gives a very good explanation of what I believe will help you with your question. &lt;/p&gt;&#xA;" OwnerUserId="2232" LastEditorUserId="2232" LastEditDate="2014-08-26T21:11:52.793" LastActivityDate="2014-08-26T21:11:52.793" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6200" PostTypeId="1" AcceptedAnswerId="6201" CreationDate="2014-08-25T16:53:37.293" Score="4" ViewCount="454" Body="&lt;p&gt;I am trying to find out which spiral arm of the Milky Way the star Kepler-62 and its planetary system are in. All I can find by Google searching is that it is part of the Lyra constellation, but I don't know if that entire constellation is within 1 spiral arm or not, nor which arm it is in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is Kepler-62 in the Carina–Sagittarius Arm, or in the Orion Arm, or...?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only map picture I can find that is somewhat related is this (I assume that the yellow light area is the area in which Kepler-62 was sighted by the Kepler spacecraft): &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/b/be/LombergA1024.jpg&quot; rel=&quot;nofollow&quot;&gt;http://upload.wikimedia.org/wikipedia/commons/b/be/LombergA1024.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="2301" LastActivityDate="2014-08-25T17:47:52.803" Title="Which spiral arm of the Milky Way is Kepler-62 in?" Tags="&lt;galaxy&gt;&lt;exoplanet&gt;&lt;milky-way&gt;&lt;habitable-zone&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6201" PostTypeId="2" ParentId="6200" CreationDate="2014-08-25T17:13:16.447" Score="5" Body="&lt;p&gt;The Kepler space telescope had a field of view along the so called Orion arm, or spur, of the galaxy. The same structure which we ourselves are inside. It basically looks along a line where the star density is the greatest within the distance that it could detect transiting planets. Less than 3% of the galactic diameter, I think. Note that the illustration your question linked to seems to overestimate that distance. &lt;a href=&quot;http://certificate.ulo.ucl.ac.uk/modules/year_one/NASA_Kepler/fov.html&quot; rel=&quot;nofollow&quot;&gt;One link&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Kepler_(spacecraft)#Spacecraft_orbit_and_orientation&quot; rel=&quot;nofollow&quot;&gt;another link&lt;/a&gt;. Kepler-62 and all another Kepler stars are certainly in that FOV. It would not be able to see transiting planets beyond the local arm. However, if the Kepler K2 mission gets going, it will use several different field of views.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2014-08-25T17:47:52.803" LastActivityDate="2014-08-25T17:47:52.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6202" PostTypeId="1" AcceptedAnswerId="6204" CreationDate="2014-08-26T03:25:25.223" Score="2" ViewCount="1096" Body="&lt;p&gt;How many light years is Earth away from the closest outer edge of the supermassive black hole in Sagittarius A* at the center of the Milky Way?  I don't mean the distance to the center of the galaxy, but to the point in the black hole's circumference where it is closest to us.&lt;/p&gt;&#xA;" OwnerUserId="2301" LastActivityDate="2014-08-26T11:58:51.803" Title="How many light years away is Earth from the closest outer edge of the black hole at the center of the Milky Way?" Tags="&lt;black-hole&gt;&lt;distances&gt;&lt;supermassive-black-hole&gt;&lt;galaxy-center&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6203" PostTypeId="1" CreationDate="2014-08-26T09:51:34.103" Score="7" ViewCount="313" Body="&lt;p&gt;I'm looking for something to cite about galaxy mergers in clusters of galaxies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need numbers on the frequency of mergers in the outer parts of galaxy clusters vs. the cluster center. In the center number densities are higher but so is the velocity dispersion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are mergers of galaxies more likely to occur in the outskirts or the central region of galaxy cluster?&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="1862" LastEditDate="2014-09-03T04:07:29.627" LastActivityDate="2014-09-26T19:54:19.563" Title="Is there a paper on galaxy mergers in clusters of galaxies?" Tags="&lt;galaxy&gt;&lt;cosmology&gt;&lt;galaxy-cluster&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6204" PostTypeId="2" ParentId="6202" CreationDate="2014-08-26T11:58:51.803" Score="3" Body="&lt;p&gt;As @Py-ser already said, there is the very clear information about that, although his link correctly is &lt;a href=&quot;https://en.wikipedia.org/wiki/Sagittarius_A*&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we can read there, the distance of this is around 25000 light years, with a precision of 1400 light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally, the size of most galaxies is around some ten thousands of light years. Compared to the black holes, their size (the diameter of their event horizon) is some kilometers (stellar black holes) or some astronomical units (some millions - hundred millions of kilometers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even the size of the event horizon of the &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_most_massive_black_holes&quot; rel=&quot;nofollow&quot;&gt;greatest known black hole&lt;/a&gt; is around the size of the solar system, which is around some thousandth of light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't need to ask, &quot;how far we are from the nearest edge of Sagittarius A*&quot;, you can simply ask &quot;how far we are from Sagittarius A&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastActivityDate="2014-08-26T11:58:51.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6205" PostTypeId="1" CreationDate="2014-08-27T03:19:20.890" Score="10" ViewCount="2604" Body="&lt;p&gt;Given a star's apparent location and apparent magnitude, how how many degrees below the horizon must the sun be for the star to be visible to the naked eye for an observer on Earth at a specified location and elevation? You can ignore all light pollution, though I understand there will be some error due to air quality/weather/atmospheric aberrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there such a formula? If not, why not and what methods of approximation are available?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I figure the brightness of the afterglow of the sun at a certain angle above the western horizon can be calculated for a certain time and place, and if it is less than the apparent brightness of the star in question, the star will be able to be seen.)&lt;/p&gt;&#xA;" OwnerUserId="2310" LastActivityDate="2019-12-29T12:47:31.353" Title="Time after sunset until star can be seen" Tags="&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6206" PostTypeId="2" ParentId="6163" CreationDate="2014-08-27T07:56:31.300" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Atmosphere_of_the_Moon&quot; rel=&quot;nofollow&quot;&gt;There is a certain sense in which the moon does have an atmosphere.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even Mercury has an atmosphere in this sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though the sense is mostly in a &quot;technically there are particles floating around the solid surface&quot;, but at densities which would make for very good vacuums on earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more practical/intuitive  definition of atmosphere excludes such meager densities.  What it takes to acquire one of those is addressed in the other answer.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2014-08-27T07:56:31.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6207" PostTypeId="2" ParentId="6205" CreationDate="2014-08-27T10:49:49.723" Score="7" Body="&lt;p&gt;It depends on the star's brightness. Astronomical darkness (full night, essentially) begins when the sun is &lt;a href=&quot;http://aa.usno.navy.mil/faq/docs/RST_defs.php&quot;&gt;18 degrees below the horizon&lt;/a&gt;. Then, you can see all of the ones you're going to be able to see. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will see stars before that point; which ones and how many will depend on how bright they are relative to the brightness of the twilight sky.&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-08-27T10:49:49.723" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6208" PostTypeId="1" CreationDate="2014-08-27T16:35:44.880" Score="3" ViewCount="71" Body="&lt;p&gt;If you trace two particles' world line backwards in time, according to current theory, both objects should converge at big bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would both objects arrive there simultaneously?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of asking the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any evidence that supports the idea that new matter is &quot;introduced&quot; into the universe as a continuous stream, as opposed to everything coming from a singularity both in time and space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analogy would be from the inside of a black hole. For each particle in that singularity, they would all have a world line that starts at the same spacial coordinate, but the particles would not be &quot;introduced&quot; into the black hole simultaneously. You can argue that time stops having meaning in a black hole, but that's only true for an &quot;outer observer&quot; in my reasoning.&lt;/p&gt;&#xA;" OwnerUserId="749" LastEditorUserId="2153" LastEditDate="2014-08-27T17:12:33.693" LastActivityDate="2014-08-28T18:15:32.213" Title="World line coordinate finiteness" Tags="&lt;space-time&gt;&lt;big-bang-theory&gt;&lt;general-relativity&gt;" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="6209" PostTypeId="2" ParentId="6182" CreationDate="2014-08-27T20:43:03.157" Score="3" Body="&lt;p&gt;The simple answer, unfortunately, is yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, it is a resounding &quot;yes&quot;. Option &quot;A&quot; uses a large &quot;bag&quot; to enclose the asteroid, and then to tow it to lunar orbit (or another feasible location). There's a problem, though, which is that simply surrounding an asteroid with a cylinder won't capture it. You have to &quot;tighten&quot; the &quot;bag&quot;. This is implied in a graphic &lt;a href=&quot;http://www.space.com/20610-nasa-asteroid-capture-mission-infographic.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and is explained in an article in Popular Mechanics (&quot;How to Mine an Asteroid&quot;, Aug. 2012), and shown in a video on NASA's website. Basically, a series of &quot;arms&quot; constrict the &quot;bag&quot; to form-fit the bag to the asteroid's surface. The procedure could create many indentations on the asteroid's surface, and, as you said, destroy some of the most valuable information. The footprints of the astronauts who walked on the moon will stay there for many years; any indentations on an asteroid will be impossible to fix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this entire idea is far from finished. NASA will not even decide between options &quot;A&quot; and &quot;B&quot; until &lt;a href=&quot;http://www.nasa.gov/content/what-is-nasa-s-asteroid-redirect-mission/&quot; rel=&quot;nofollow&quot;&gt;&quot;late 2014&quot;&lt;/a&gt;. So we won't know the precise details for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why go ahead with the whole thing if we're going to lose some of the best information we could get? Well, the Initiative isn't motivated purely by surface features. Part of it is spurred on by plans to pioneer new technologies needed to get humans to Mars. Other tangential benefits include diverting an asteroid that could be headed for Earth (although the asteroid candidates for this mission are much smaller than any of their more dangerous cousins; also, the Asteroid Initiative is under &quot;Asteroid and Comet Watch&quot; on NASA's website - right near plans to blow up a dangerous asteroid!), or even mining an asteroid for metals or resources for astronauts (such as water).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, this mission isn't the only way to study an asteroid (although it might be the best). We could always send a lander to an asteroid - after all, we can land on a comet (erm, well, we're trying - I'm rooting for Rosetta and Philae!), which would make it simple to collect samples before heading home . . . although the distance to travel would be a lot longer than just to lunar orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this comes together to mean that NASA won't worry itself over losing data on the surface of the asteroid. Would that information be nice? Oh, yes. But given the other good things coming out of the mission, the pros here outweigh the cons. It's simply one of the trade-offs that might have to be made for this mission.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Sources:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.planetary.org/blogs/jason-davis/2014/the-latest-on-nasas-asteroid.html&quot; rel=&quot;nofollow&quot;&gt;Planetary.org&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.space.com/20610-nasa-asteroid-capture-mission-infographic.html&quot; rel=&quot;nofollow&quot;&gt;Space.com&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nasa.gov/content/what-is-nasa-s-asteroid-redirect-mission/&quot; rel=&quot;nofollow&quot;&gt;NASA&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;How to Mine an Asteroid&quot;. Popular Mechanics. August, 2012.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-01-10T22:52:39.330" LastActivityDate="2015-01-10T22:52:39.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6210" PostTypeId="1" CreationDate="2014-08-28T15:48:17.800" Score="12" ViewCount="3300" Body="&lt;p&gt;Are there stars that do not emit any light in the visible part of the EM spectrum?&lt;/p&gt;&#xA;" OwnerDisplayName="user2317" LastEditorUserId="2153" LastEditDate="2014-08-29T14:14:54.270" LastActivityDate="2020-11-23T23:49:28.240" Title="Are there stars that don't emit visible light?" Tags="&lt;star&gt;&lt;light&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6211" PostTypeId="2" ParentId="6184" CreationDate="2014-08-28T15:56:05.467" Score="3" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/pdf/1212.0268.pdf&quot; rel=&quot;nofollow&quot;&gt;Detection of inner Solar System Trojan Asteroids by Gaia&lt;/a&gt; (pdf):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Gaia satellite, planned for launch by the European Space Agency (ESA) in 2013, is the next generation astrometry mission following Hipparcos. While mapping the whole sky, the Gaia space mission is expected to discover thousands of Solar System Objects. The se will include Near-Earth Asteroids and objects at Solar elongations as low as 45 degrees, which are difficult to observe with ground-based telescopes. We present the results of simulations for the &lt;strong&gt;detection of Trojan asteroids in the orbits of Earth and Mars by Gaia&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2014-08-28T15:56:05.467" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6214" PostTypeId="2" ParentId="6184" CreationDate="2014-08-28T17:07:51.933" Score="3" Body="&lt;p&gt;&lt;strong&gt;No&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Near-Earth objects (NEOs) may be temporarily captured to oribt Earth as discussed in the reference quoted in the question. Such a capture must invariably involve the NEO to pass by the L1 or L2 points, which are the saddle points in the co-rotating binary potential of the Earth-Sun system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, these objects have just high enough co-rotating energy (Jacobi constant) to enter the region around the Earth via those saddle points. Since they can only escape that region by exiting through the same saddle points, it may take a while before they &quot;find&quot; the exit, i.e. they are temporarily captured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as I read that paper, there is no expectation of NEOs captured on orbits around the L1/L2 points. In fact, there are no stable orbits around those Lagrange points and hence no such captured satellites are expected. Indeed, this is one of the major benefits of putting an artificial satellite there. It also means that, in order to keep Gaia at L2, it will need to correct its orbit from time to time. Otherwise, it will eventually drift off the L2 region. This has the advantage that no space debris from missions like this will accumulate there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, to &lt;strong&gt;answer your question&lt;/strong&gt;: there are no satellites orbiting near L2 to be discovered. The chance of any asteroid passing by (on its way to or from temporary capture) is rather small or, equivalently, the typical NEO passing there on a regular basis is tiny (1m or so), i.e. hardly detectable. Moreover, any NEO moving near Gaia will have a huge proper motion and be highly unlikely to be seen on more than one scan (and different sightings cannot be related).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, NEOs elsewhere can and will be detected by Gaia, as discussed in another answer,&#xA;in particular Trojan asteroids (orbiting L4/L5 on stable orbits) of Earth and Mars.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-08-28T17:07:51.933" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6215" PostTypeId="2" ParentId="6210" CreationDate="2014-08-28T18:08:53.720" Score="11" Body="&lt;p&gt;There are two possible reasons why a star would not be detectable in the visible part of the spectrum (even with the most powerful telescopes not yet invented), apart from the trivial (too far away, hidden behind screens of dust).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It has too low a temperature. To not emit much in the visible part of the spectrum, a star has to be very cold indeed, at most a few 100K, when radiation is mostly in the infrared (which is still detectable). By definition, a star is an object that undergoes (or has undergone) hydrogen burning (H$\to$He fusion). This sets an lower mass limit of 0.08M$_\odot$. All objects of that mass (or more) ever formed in the universe are still much hotter than 100K. However, in the very distant future, some 'dead' stars (without remaining fusion energy source) will cool down to such temperatures, this includes white dwarves.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It has too high a gravitational redshift for any light to appear in the visible. In fact, the so-called stellar-mass black holes (remnants of supernova explosions of massive stars) may in fact be such objects: strange stars denser than a nucleus (consisting of a quark-gluon plasma as dense as a neutron) but with such a small size and high mass that any radiation emitted from their surface is redshifted by a factor 1000 or larger.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So if you consider those (still hypothetical) strange stars, then these are possible candidates.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2014-09-15T07:59:48.953" LastActivityDate="2014-09-15T07:59:48.953" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6216" PostTypeId="2" ParentId="6208" CreationDate="2014-08-28T18:15:32.213" Score="2" Body="&lt;p&gt;Particles are generated and destroyed all the time. This is obvious for photons, but also holds for massive particles. Today most particles are stable and long-living, but shortly after the big bang the universe was so hot that particle generation and destruction were in some equlibrium and &lt;em&gt;all&lt;/em&gt; particles participated in this process (for photons this continued until the epoch of re-ionisation at redshift $z\sim1000$. There is definitely no particle around today that existed all the time since $t=0$.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-08-28T18:15:32.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6217" PostTypeId="1" CreationDate="2014-08-29T03:37:51.337" Score="5" ViewCount="3804" Body="&lt;p&gt;Just curious to know what will be the time difference between someone living on the outer layer of our galaxy and us, considering the &lt;em&gt;known&lt;/em&gt; facts that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Time is relative to gravity&lt;/li&gt;&#xA;&lt;li&gt;Our sun goes around at 483,000 miles/hour(792,000 km/hr) to complete its galactic year &lt;/li&gt;&#xA;&lt;li&gt;Star systems at outer layer travel faster than ours due to Dark Matter and Dark Energy - couldn't find speed of those star systems though!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We can assume that both (Solar system and outer layer of star system) are on the same side of the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;In order to re-phrase this question more clearly, as per Joan.bdm, it is related to time dilation - what is the aging rate difference if we left one of two twins on the Earth and the other one on a planet at outskirts of the Milky Way.&#xA;Conversely, what is aging rate difference between closer to center of our galaxy and us?&lt;/p&gt;&#xA;" OwnerUserId="2323" LastEditorUserId="2323" LastEditDate="2014-09-01T03:08:06.117" LastActivityDate="2014-09-01T03:08:06.117" Title="What is the relative time difference between us and a star system in outer layer of our galaxy?" Tags="&lt;galaxy&gt;&lt;space-time&gt;&lt;general-relativity&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6218" PostTypeId="2" ParentId="6217" CreationDate="2014-08-29T08:41:08.690" Score="2" Body="&lt;p&gt;Your third question has a very well known answer &lt;a href=&quot;https://en.wikipedia.org/wiki/Galaxy_rotation_curve&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nJKWw.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What means, that although a big time difference had to exist, there is a much smaller one, and in the opposite direction. This is considered the effect of the dark matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our Sun is around 30000 ly from the galactic center, the edge of the galaxy is around 50000 ly, so we can read on the diagram, that the actual speed difference is around 15 km/s, which is 1/20000 of the speed of light. This causes a (not really big) time dilation due to the special relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_time_dilation&quot; rel=&quot;nofollow noreferrer&quot;&gt;another source of the time dilation&lt;/a&gt;, which is caused by the gravity of the galaxy, which differs in the case of us and in the case of somebody living on the outer edge. On such weak gravitation (i.e. for from any black hole, neutron star, etc) it practically depends on the newtonian gravitational acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this acceleration is very small - it takes a half &lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_year&quot; rel=&quot;nofollow noreferrer&quot;&gt;galactic year&lt;/a&gt; (some hundred millions of year) to revert the orbital speed of the Sun around the galactic core! So, we can consider that actually negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus only the special relativistic effect remains. 15 km/s is around 1/20000 of the speed of the light. In cases of speed which are &lt;em&gt;much&lt;/em&gt; smaller as the light speed, we can use the speed dilation formula 1/(2*(v/c)^2). Substituting 1/20000 into this formula, we get 1:800 000 000 .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;For laymans, we can say, it takes around 25 years for the clocks of this people living on the outer edge of our galaxy to dilate 1 second.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="1891" LastEditDate="2014-08-29T08:51:28.003" LastActivityDate="2014-08-29T08:51:28.003" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6219" PostTypeId="2" ParentId="6153" CreationDate="2014-08-29T15:12:36.870" Score="4" Body="&lt;p&gt;It is in fact supposed to be in $\mathrm{\frac{km}{s}}$ (KILOMETRES per second).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that whoever authored the paper, probably took it as a random shortcut and assumed anyone reading it would either have read the paper and understand it is supposed to be $\mathrm{\frac{km}{s}}$, or just assume it from the fact it says velocity.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="9301" LastEditDate="2017-12-31T03:16:21.310" LastActivityDate="2017-12-31T03:16:21.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6220" PostTypeId="2" ParentId="6217" CreationDate="2014-08-29T16:08:03.957" Score="2" Body="&lt;p&gt;It's not really clear what you're aksing. There is no &quot;relative time difference&quot; between anywhere and somewhere else in the universe. There are, of course, time differences between different events (space-time points).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motion of stars in the Milky Way (or any other) galaxy are sub-relativistic, implying that GR effects (such as time dilatation) are negligible. Of course, the Galaxy is rather big and it takes much longer than a human life time for light, let alone space craft, to travel accross it.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-08-29T16:08:03.957" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6221" PostTypeId="1" AcceptedAnswerId="6222" CreationDate="2014-08-30T04:45:38.543" Score="3" ViewCount="684" Body="&lt;p&gt;I am not able to visualize the position of the LMC and have not found a satisfactory explanation...&#xA;Can someone please answer it this way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relative to a ray joining the Earth to the Galactic-Center(Milky Way's), please describe the position in the plane of the ray as- to the South/East/West or their combinations. And will the galaxy be below/above the plane of the ray?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Please consider the direction of the ray(Earth-&gt;Center) as the axis of a compass that points towards the north.&#xA;E.g://Answer format that i was hoping for:&#xA;   To the South-West of the Segment, and below its plane.&lt;/p&gt;&#xA;" OwnerUserId="2330" LastEditorUserId="9527" LastEditDate="2020-09-25T08:10:12.663" LastActivityDate="2020-09-25T08:10:12.663" Title="What is the exact position of the Large Magellanic Cloud?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;magellanic-cloud&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6222" PostTypeId="2" ParentId="6221" CreationDate="2014-08-30T07:49:41.660" Score="2" Body="&lt;p&gt;First of all: your coordinate system (as I understand it) is not &quot;well defined&quot;. You only provide a line (Earth --&gt; Milky way centre) and define that as the North-South axis. You don't give a reference for the east-west axis or the up-down axis. So, I'll assume that the East-West axis lies along the plane of the Milky way and up is where the Boötes Dwarf system is. This way, our coordinate system is equal to this image from the &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Large_Magellanic_Cloud%23&quot; rel=&quot;nofollow noreferrer&quot;&gt;LMC Wikipedia site&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KmQZW.jpg&quot; alt=&quot;LMC position&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with this picture is that the position of the Earth is only implicitly given. It is &quot;behind&quot; the center of the Milky Way. So from that position, the LMC is almost to our right (east) and a bit below the plane of the Milky way; about 30 to 40 degrees down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also nicely validated by this image, acquired &lt;a href=&quot;http://asterisk.apod.com/viewtopic.php?f=31&amp;amp;t=21327&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; (note that the grid is in 30 degree angles). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ghef1.jpg&quot; alt=&quot;position of the LMC&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, finally we can answer your question in the way you wanted: in the north-east to east segment, and below its plane. &lt;/p&gt;&#xA;" OwnerUserId="2189" LastEditorDisplayName="user1569" LastEditDate="2018-03-11T18:08:28.020" LastActivityDate="2018-03-11T18:08:28.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6223" PostTypeId="2" ParentId="6217" CreationDate="2014-08-31T01:53:49.233" Score="2" Body="&lt;p&gt;In the slowly-changing, weak-field regime, the effects of general relativity can be approximated by the following metric (in units of $c=1$):&#xA;$$\mathrm{d}s^2 = -\left(1+2\Phi\right)\mathrm{d}t^2 + \left(1-2\Phi\right)\mathrm{d}S^2\text{,}$$&#xA;where $\mathrm{d}S^2$ is the metric is Euclidean $3$-space and $\Phi$ is the Newtonian gravitational potential. Thus, the rate of proper time vs coordinate time is&#xA;$$\frac{\mathrm{d}\tau}{\mathrm{d}t} = \sqrt{1+2\Phi-\left(1-2\Phi\right)\frac{\mathrm{d}S^2}{\mathrm{d}t^2}} = 1+\Phi - \frac{1}{2}v^2 + \mathcal{O}(v^4,\Phi^2,v^2\Phi)\text{.}$$&#xA;At low speeds, we can drop the higher-order terms to infer a relative time dilation between two stars to be&#xA;$$\frac{\mathrm{d}\tau_1}{\mathrm{d}\tau_0} \approx \frac{1+\Phi_1-v_1^2/2}{1+\Phi_0-v_0^2/2} \approx 1+(\Phi_1 - \Phi_0) - \frac{1}{2}(v_1^2-v_0^2)\text{.}$$&#xA;The gravitational potential of the Milky Way can be modeled as the sum of an spherically symmetric bulge, an axisymmetric disk, and a spherically symmetric dark matter halo. Many galactic models exist, but for simplicity I will be using ones from Irrgang &lt;em&gt;et al.&lt;/em&gt; [arXiv:&lt;a href=&quot;http://arxiv.org/abs/1211.4353&quot; rel=&quot;nofollow&quot;&gt;1211.4353&lt;/a&gt;] for the Sun:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Model I&lt;/strong&gt; (after Allen &amp;amp; Santillan (1991)): $v_0 = 8.072\times 10^{-4}$, $\Phi_0 = -2.113\times 10^{-6}$.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Model II&lt;/strong&gt; (after Wilkinson &amp;amp; Evans (1999) and Sakamoto &lt;em&gt;et al.&lt;/em&gt; (2003)): $v_0 = 8.019\times 10^{-4}$, $\Phi_0 = -1.845\times 10^{-6}$.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Model III&lt;/strong&gt; (after Navarro &lt;em&gt;et al.&lt;/em&gt; (1997)): $v_0 = 7.996\times 10^{-4}$, $\Phi_0 = -3.664\times 10^{-6}$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Because the velocity-dependent contribution is squared while the potential contribution is not, the gravitational time dilation could be more significant. Note that this does not include the contribution from the Sun itself, the vicinity to which is going to be more significant. However, if we're comparing time dilation for observers near similar stars at similar orbits from their stars, that difference will cancel out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the gravitational potential is rather model-dependent due to the many uncertainties involved in the observations, but you can play with some of the values given in that paper if you wish.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-08-31T01:53:49.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6224" PostTypeId="1" AcceptedAnswerId="6251" CreationDate="2014-08-31T14:13:42.363" Score="10" ViewCount="555" Body="&lt;p&gt;In a few billion years, the Milky Way and Andromeda will collide. Chances are, there won't be many spectacular stellar collisions, and there's a pretty good chance that the Solar system (and most other stellar systems) won't be flung out into deep space, or be torn apart by incoming stars. But what about on a larger scale?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The product of the collision - dubbed &quot;Milkomeda&quot; - will be an elliptical galaxy, unlike the Milky Way and Andromeda, which are both spiral galaxies. My question is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How will the structure of the spiral arms be effectively destroyed by the collision? What mechanisms will cause them to coalesce, and what structures will take their place?&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2365" LastEditDate="2015-11-04T20:27:12.860" LastActivityDate="2015-11-04T20:27:12.860" Title="Fate of the Spiral Arms of the Milky Way and Andromeda" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;m31&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6225" PostTypeId="2" ParentId="6224" CreationDate="2014-08-31T14:42:59.980" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In a few billion years, the Milky Way and Andromeda will collide. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;About 5 billion years, but it will take more than a thousand years for the galaxies to overlap.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Chances are, there won't be any spectacular stellar collisions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the cores of the 2 galaxies collide, I would think it to be very likely that star systems will be effected and the central black holes to join.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The product of the collision will be an elliptical galaxy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It may take a few million years for the stars to find stability in their new orbits. After a few billion years more, there might be spiral arms again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How will the structure of the spiral arms be effectively destroyed by the collision? What mechanisms will cause them to coalesce, and what structures will take their place?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When the galaxies approach together, gravity will effect the orbits of the stars. Since momentum needs to be conserved, the orbit of most of the stars will be changed radically. The different orbits will hide the spiral arms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that I have thought more about it, it is very likely that a few stars will collide, maybe it will take a few billion years. &lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-08-31T14:42:59.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6226" PostTypeId="1" CreationDate="2014-08-31T16:41:43.303" Score="0" ViewCount="228" Body="&lt;p&gt;Suppose I aim at some object that is 13 billion light years away from Earth. Let's assume I have a twin that pilots a ship that travels in infinite speed, meaning that he can reach that object  and get back here in a split second. On the other hand I'm the poor brother, and I own a ship with top speed of only at 99.9% of c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We leave earth at the same time, with the same destination. My ship accelerates at constant rate for a full day until we reach max speed, then we enjoy the travel without accelerating for a second time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) how long will it take me to meet my brother, and how older will he be? &#xA;2) what will be the age of the universe, according to me? &#xA;3) what will be my current speed relative to surrounding objects before decelerating? Would I be able to accelerate and reach max speed again? If yes, what's the point of accelerating later rather than sooner, and having a top speed after all?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) And a plot twist: suppose that while in travel I build a second ship exactly the same as mine. When we're about to arrive we choose a distant point 13 billion light years away in the same direction, and I send a friend on it, who accelerates until he reaches 99.9% of c relative to me. If we repeat that for N times, will we ever reach &quot;free space&quot; where matter didn't reach yet, or some other cosmologically interesting place? How does it look like there? If not, a) can we tinker with any parameters in order to do it &quot;right&quot;? b) any other plans on getting there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand relativity from what I learned in college (long time ago), is there any other theory needed to understand such traveling questions? &lt;/p&gt;&#xA;" OwnerUserId="2305" LastEditorUserId="633" LastEditDate="2014-09-02T20:40:17.960" LastActivityDate="2014-09-02T20:40:17.960" Title="Thought experiment about the outskirts of the universe" Tags="&lt;big-bang-theory&gt;&lt;expansion&gt;&lt;special-relativity&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="1" ClosedDate="2014-09-04T21:20:12.357" ContentLicense="CC BY-SA 3.0" />
  <row Id="6227" PostTypeId="1" AcceptedAnswerId="6236" CreationDate="2014-08-31T20:44:47.967" Score="3" ViewCount="92" Body="&lt;p&gt;If we assume that the Single Impact Hypothesis regarding the martian dichotomy (&lt;a href=&quot;http://en.wikipedia.org/wiki/Martian_dichotomy#Single_Impact_Hypothesis&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;) and the Giant Impact Hypothesis regarding the Moon's formation (&lt;a href=&quot;http://en.wikipedia.org/wiki/Giant_impact_hypothesis&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;) are true, then is it possible that these two events were related?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, by related I mean that they were not independent from one another.&lt;/p&gt;&#xA;" OwnerUserId="2333" LastActivityDate="2014-09-01T21:46:55.220" Title="Can the martian dichotomy (SIH) and the moon's formation (GIH) be related?" Tags="&lt;the-moon&gt;&lt;mars&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6228" PostTypeId="2" ParentId="6226" CreationDate="2014-09-01T00:25:43.620" Score="3" Body="&lt;p&gt;I will answer all of your points with zero tact and/or precision in order to fit them all in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) It will take you about a &quot;split second&quot; given you stated that it only takes your brother a split second to travel from Earth to the point 13Bly away and back again. He will be exactly said split second older and no more, because as he is traveling faster than the speed of light, the rules governing his and your aging no longer apply, meaning you have created a scenario where both his and your time's are connected and simultaneous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) At the point when you meet up with your brother a split second later, the universe will be a split second older... At the time you've reached your destination the universe will be [ &gt; 13By + (the age of the universe when you left) ]. The fact that your ship takes one year to accelerate to and from the top speed is irrelevant given the duration of the travel time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) .999c ... Why not, assuming that you have enough &quot;fuel&quot; your imaginary ship should be more than capable than making a single one way trip. There given the distance and time it would take you to get there, there is close to zero reason to accelerate either sooner or later, although the whole near speed of light travel is sort of the point of your question, it would kind of be necessary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) The N times scenario is redundant and a little weirder than your other questions, as the speed of light is the same in all reference frames, you would never breach that boundary, however to each previous ship the rate of travel would appear exponential even though there would only be infinitesimal gains. Would you ever reach free space... no, without faster than light travel it is not possible. I would venture a guess that space would look somewhat similar to the here and now regardless where you ended up within your lifetime. a) by tinker, you could assume that you were able to travel as fast as your twin in part (1). b) Think Star Trek, impulse power was sub light speed travel, all warp speeds are some large multiple of c. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look into relative simultaneity and space-time diagrams for reference on the whole twin paradox issue. Also you could read something like ... the physics of star trek for faster than light travel scenarios with actual physics explanations or problems.&lt;/p&gt;&#xA;" OwnerUserId="2232" LastActivityDate="2014-09-01T00:25:43.620" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6229" PostTypeId="1" AcceptedAnswerId="6244" CreationDate="2014-09-01T09:21:15.570" Score="3" ViewCount="144" Body="&lt;p&gt;The last years we find some objects like this : &lt;a href=&quot;http://en.wikipedia.org/wiki/MACS0647-JD&quot; rel=&quot;nofollow&quot;&gt;MACS0647-JD&lt;/a&gt; which are more than 13 billion light-years away from us. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the picture we get was created only some million years after the Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why can we see objects like this in all directions? Shouldn't those objects be close to each other?&lt;/p&gt;&#xA;" OwnerUserId="2335" LastActivityDate="2014-09-02T16:41:40.100" Title="Why aren't the farthest objects close to each other?" Tags="&lt;big-bang-theory&gt;&lt;observable-universe&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6230" PostTypeId="2" ParentId="6196" CreationDate="2014-09-01T12:45:06.080" Score="1" Body="&lt;p&gt;hawking radiation. electrons and positrons created from the vacuum must 'repay' their energy, usually through annihilation of each other. however, if one crosses through the event horizon of a black hole and the other escapes, the energy from the black hole as well as that of the particle will 'repay' and the escaped particle becomes real. this is a very slow process and in most circumstances I would assume that a black hole would accumulate more energy than it loses.&lt;/p&gt;&#xA;" OwnerUserId="2337" LastActivityDate="2014-09-01T12:45:06.080" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6231" PostTypeId="2" ParentId="6188" CreationDate="2014-09-01T12:51:33.007" Score="0" Body="&lt;p&gt;charon and pluto are tidally locked, as is the moon to the earth. so you could say that earth is in a geostationary orbit with the moon. in fact, the earth rotation has been and will continue to be slowed by the moon until the moon is in a geostationary orbit with earth.&lt;/p&gt;&#xA;" OwnerUserId="2337" LastActivityDate="2014-09-01T12:51:33.007" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6232" PostTypeId="1" CreationDate="2014-09-01T13:06:59.670" Score="19" ViewCount="1292" Body="&lt;p&gt;I have 2 questions related to spiral galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, how did the arms form? Why would the stars accumulate into those specific areas? And secondly, why are they still intact?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inner parts of the arms must be revolving more rapidly than the outer parts, so surely the would become stretched and then no longer resemble arms, but simply a disk around the galactic center.&lt;/p&gt;&#xA;" OwnerUserId="2337" LastEditorUserId="363" LastEditDate="2014-09-02T05:15:49.373" LastActivityDate="2016-11-30T16:41:03.203" Title="Questions about spiral galaxy arms" Tags="&lt;galaxy&gt;&lt;galactic-dynamics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6233" PostTypeId="2" ParentId="6229" CreationDate="2014-09-01T13:13:21.437" Score="3" Body="&lt;p&gt;the big bang did not happen in one place, it happened everywhere, and since then space has been expanding. so the stars that where formed shortly after the big bang where everywhere in space, but space was smaller. therefore you can see them everywhere if you look 13 billion light-years away, because they where everywhere.&lt;/p&gt;&#xA;" OwnerUserId="2337" LastActivityDate="2014-09-01T13:13:21.437" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6234" PostTypeId="2" ParentId="6232" CreationDate="2014-09-01T15:00:39.483" Score="20" Body="&lt;p&gt;Actually, the stars and nebulae that make up the spiral arm are only temporarily part of that spiral arm. Spiral arms are more like sound waves where individual particles move around a more or less stationary position. (Look for instance at the animation of longitudinal waves from &lt;a href=&quot;http://www.acs.psu.edu/drussell/demos/waves/wavemotion.html&quot;&gt;Dan Russel&lt;/a&gt;, the red dots move a bit to the left and to the right around a stationary position). Dust, gas and stars move towards or away from another just as longitudinal waves. Where the dust, gas and stars come close together (and where, therefore, the density increases), spiral arms can be seen as more stars are close together increasing the brightness at that position in the galaxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This effect is, furthermore, much increased because the increased density of dust and gas in the spiral arm causes protostars to form. The brightest stars burn up their energy so fast that they will cease to exist even before the longitudinal wave (the spiral arm) has passed. These very bright stars only exist for a small portion of their orbital period around the galaxy's centre, and only while they are in the spiral arm. The large majority of stars exist much longer, but are also much dimmer and contribute only little to the overal brightness of the galaxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This causes the spiral arms to be so much brighter than the rest of the disk, where also a lot of stars exist. But these can hardly be seen as they are much dimmer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the stars do not revolve around a stable position in the galaxy (as the red dots in the wave animation) but follow their own orbits around the centre of the galaxy. Sometimes a bit faster, and sometimes a bit slower depending on the position relative to the spiral arms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the spiral arms are waves, it does not matter that stars near the centre move faster than the stars at the edge. It just means that they will be part of the spiral arm for a shorter period of time.&lt;/p&gt;&#xA;" OwnerUserId="864" LastEditorUserId="864" LastEditDate="2014-09-01T15:08:25.203" LastActivityDate="2014-09-01T15:08:25.203" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6236" PostTypeId="2" ParentId="6227" CreationDate="2014-09-01T18:59:08.823" Score="3" Body="&lt;p&gt;Early in the solar system's history, the Sun was surrounded by a protoplanetary disk, full of gas, dust and rock. Eventually, &lt;em&gt;protoplanets&lt;/em&gt; began to form. These were small, rocky bodies - smaller than the rocky planets today, but bigger than asteroids. Think of them as dwarf-planet-sized objects. All the terrestrial planets formed from them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, as you might know, the object that slammed into the Earth (according to the GIH) was one such protoplanet, lovingly named Theia. The resulting debris coalesced into the Moon. The SIH tells of a similar story: Once again, a protoplanet headed on a collision course, this time towards Mars. It slammed into Mars, creating a large basin. Simple but effective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's one thing that can relate these two disaster tales together, and that is the protoplanetary disk. Remember, the protoplanets formed from the same disk. The differences in them stemmed from the materials they formed from. So we could relate the two suicidal protoplanets to each other via that connection. Also, most collisions in the inner solar system were due to the erratic orbits of these protoplanets, so it is likely that the fault for the collisions lies solely on the protoplanets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you want something a bit different from random collisions, though; the previous reason is a bit of a cop-out. Well, there is something more interesting: the rocky planets weren't alone in the early solar system. There were still their big siblings: the gas giants. The gas giants hadn't yet grown to their full sizes, but their cores and early atmospheres were still pretty massive, and commanded a strong presence in the outer solar system. It is possible that the two objects that hit the Earth and Mars formed together farther away from the inner planets, and were gravitationally perturbed by the gas giants, swinging in from an outer orbit towards the inner solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might have heard of the Nice Model. Basically, this says that the gas giants formed closed to the inner solar system than they are now. They encountered some of these early protoplanets and flung them into the inner solar system, moving outwards in the process. thus, collisions increased in the inner solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a &quot;nice&quot; explanation. Trouble is, the majority of this &quot;flinging&quot; happened later on in the solar system's history; it has been suggested that the gas giants did this and thus caused the Late Heavy Bombardment. Still this idea shows that there could have been interaction between the protoplanets and the gas giants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how does this relate the two suicidal protoplanets? Well, it means that they could have formed together, further out in the solar system, and have been brought in by a gas giant. They could even have been part of a larger object, which broke apart after various collisions. But if the two objects had formed together (even somewhere in the inner solar system), or been part of a larger object, they should be made of the same materials, right? So if we were to travel to Mars we should be able to find similar materials as we did on the Moon. So far there hasn't been much luck along that avenue, but the search will continue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize: The early solar system was chaotic. It is likely that the impacts were unrelated, and that the two suicidal protoplanets formed apart in the inner solar system, but they could have formed together further our and been perturbed by one of the gas giants. They may or may not have been formed neary each other, or even been fragments of the same object. Detailed analysis in the future could give us some clues as to their formation.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-01T21:46:55.220" LastActivityDate="2014-09-01T21:46:55.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6237" PostTypeId="2" ParentId="6162" CreationDate="2014-09-01T19:10:57.233" Score="1" Body="&lt;p&gt;As other people said, what you see in that picture is a mosaic of images, centered in our Galaxy, produced with WISE observations in &lt;a href=&quot;http://wise.ssl.berkeley.edu/gallery_infraastro.html&quot; rel=&quot;nofollow&quot;&gt;infrared light&lt;/a&gt;. Astronomers have used a projection method, called Aitoff, to project the spherical sky into a 2-dimensional map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more detailed explanation of the image, please follow this link:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://wise.ssl.berkeley.edu/gallery_thesky.html&quot; rel=&quot;nofollow&quot;&gt;http://wise.ssl.berkeley.edu/gallery_thesky.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HTH,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Germán.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Sorry that my reputation doesn't allow me to put more than two links.&lt;/p&gt;&#xA;" OwnerUserId="2339" LastActivityDate="2014-09-01T19:10:57.233" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6239" PostTypeId="2" ParentId="632" CreationDate="2014-09-01T19:27:42.863" Score="5" Body="&lt;p&gt;You may want to use &lt;strong&gt;perturbation theory&lt;/strong&gt;. This only gives you an &lt;em&gt;approximate&lt;/em&gt; answer, but allows for analytic treatment. Your force is considered a small perturbation to the Keplerian elliptic orbit and the resulting equations of motion are expanded in powers of $K$. For linear perturbation theory, only terms linear in $K$ are retained. This simply leads to integrating the perturbation along the unperturbed original orbit. Writing your force as a vector, the perturbing acceleration is&#xA;$$&#xA;\boldsymbol{a} = K \frac{GM}{r^2c^2}v_r\boldsymbol{v}_t&#xA;$$&#xA;with $v_r=\boldsymbol{v}{\cdot}\hat{\boldsymbol{r}}$ the radial velocity &#xA;($\boldsymbol{v}\equiv\dot{\boldsymbol{r}}$) and&#xA;$\boldsymbol{v}_t=(\boldsymbol{v}-\hat{\boldsymbol{r}}(\boldsymbol{v}{\cdot}\hat{\boldsymbol{r}}))$ the rotational component of velocity (the full velocity minus the radial velocity). Here, the dot above denotes a time derivative and a hat the unit vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it depends what you mean with '&lt;em&gt;effect&lt;/em&gt;'. Let's work out the changes of the orbital semimajor axis $a$, eccentricity $e$, and direction of periapse. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To &lt;strong&gt;summarise the results below&lt;/strong&gt;: semi-major axis and eccentricity are unchanged, but the direction of periapse rotates in the plane of the orbit at rate&#xA;$$&#xA;\omega=\Omega \frac{v_c^2}{c^2}&#xA;\frac{K}{1-e^2},&#xA;$$&#xA;where $\Omega$ is the orbital frequency and $v_c=\Omega a$ with $a$ the semi-major axis. Note that (for $K=3$) this &lt;strong&gt;agrees with the general relativity (GR) precession rate&lt;/strong&gt; at order $v_c^2/c^2$ (given by Einstein 1915 but not mentioned in the original question).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;change of semimajor axis&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the relation $a=-GM/2E$ (with $E=\frac{1}{2}\boldsymbol{v}^2-GMr^{-1}$ the orbital energy) we have for the change of $a$ due to an external (non-Keplerian) acceleration&#xA;$$&#xA;\dot{a}=\frac{2a^2}{GM}\boldsymbol{v}{\cdot}\boldsymbol{a}.&#xA;$$&#xA;Inserting $\boldsymbol{a}$ (note that $\boldsymbol{v}{\cdot}\boldsymbol{v}_t=h^2/r^2$ with angular momentum vector $\boldsymbol{h}\equiv\boldsymbol{r}\wedge\boldsymbol{v}$), we get&#xA;$$&#xA;\dot{a}=\frac{2a^2Kh^2}{c^2}\frac{v_r}{r^4}.&#xA;$$&#xA;Since the orbit average $\langle v_r f(r)\rangle=0$ for any function $f$ (see below), $\langle\dot{a}\rangle=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;change of eccentricity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From $\boldsymbol{h}^2=(1-e^2)GMa$, we find&#xA;$$&#xA;e\dot{e}=-\frac{\boldsymbol{h}{\cdot}\dot{\boldsymbol{h}}}{GMa}+\frac{h^2\dot{a}}{2GMa^2}.&#xA;$$&#xA;We already know that $\langle\dot{a}\rangle=0$, so only need to consider the first term. Thus,&#xA;$$&#xA;e\dot{e}=-\frac{(\boldsymbol{r}\wedge\boldsymbol{v}){\cdot}(\boldsymbol{r}\wedge\boldsymbol{a})}{GMa}&#xA;         =-\frac{r^2\;\boldsymbol{v}{\cdot}\boldsymbol{a}}{GMa}&#xA;         =-\frac{Kh^2}{ac^2}\frac{v_r}{r^2},&#xA;$$&#xA;where I have used the identity&#xA;$(\boldsymbol{a}\wedge\boldsymbol{b}){\cdot}(\boldsymbol{c}\wedge\boldsymbol{d})&#xA;=\boldsymbol{a}{\cdot}\boldsymbol{c}\;\boldsymbol{b}{\cdot}\boldsymbol{d}-&#xA; \boldsymbol{a}{\cdot}\boldsymbol{d}\;\boldsymbol{b}{\cdot}\boldsymbol{c}$ and the fact $\boldsymbol{r}{\cdot}\boldsymbol{a}_p=0$.&#xA;Again $\langle v_r/r^2\rangle=0$ and hence $\langle\dot{e}\rangle=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;change of the direction of periapse&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;eccentricity vector&lt;/em&gt;&#xA;$&#xA;\boldsymbol{e}\equiv\boldsymbol{v}\wedge\boldsymbol{h}/GM - \hat{\boldsymbol{r}}&#xA;$&#xA;points (from the centre of gravity) in the direction of periapse, has magnitude $e$, and is conserved under the Keplerian motion (validate all that as an exercise!). From this definition we find its instantaneous change due to external acceleration&#xA;$$&#xA;\dot{\boldsymbol{e}}=&#xA;\frac{\boldsymbol{a}\wedge(\boldsymbol{r}\wedge\boldsymbol{v})&#xA;     +\boldsymbol{v}\wedge(\boldsymbol{r}\wedge\boldsymbol{a})}{GM}&#xA;     =\frac{2(\boldsymbol{v}{\cdot}\boldsymbol{a})\boldsymbol{r}&#xA;       -(\boldsymbol{r}{\cdot}\boldsymbol{v})\boldsymbol{a}}{GM}&#xA;     =\frac{2K}{c^2}\frac{h^2v_r\boldsymbol{r}}{r^4}&#xA;     -\frac{K}{c^2}\frac{v_r^2\boldsymbol{v}_t}{r}&#xA;$$&#xA;where I have used the identity &#xA;$\boldsymbol{a}\wedge(\boldsymbol{b}\wedge\boldsymbol{c})=(\boldsymbol{a}{\cdot}\boldsymbol{c})\boldsymbol{b}-(\boldsymbol{a}{\cdot}\boldsymbol{b})\boldsymbol{c}$&#xA;and the fact $\boldsymbol{r}{\cdot}\boldsymbol{a}=0$. The orbit averages of these expression are considered in the appendix below. If we finally put everything together, we get&#xA;$&#xA;\dot{\boldsymbol{e}}=\boldsymbol{\omega}\wedge\boldsymbol{e}&#xA;$&#xA;with [&lt;em&gt;corrected again&lt;/em&gt;]&#xA;$$&#xA;\boldsymbol{\omega}=\Omega K \frac{v_c^2}{c^2}&#xA;(1-e^2)^{-1}\, \hat{\boldsymbol{h}}.&#xA;$$&#xA;This is a rotation of periapse in the plane of the orbit with angular frequency $\omega=|\boldsymbol{\omega}|$. In particular $\langle e\dot{e}\rangle=\langle\boldsymbol{e}{\cdot}\dot{\boldsymbol{e}}\rangle=0$ in agreement with our previous finding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't forget that due to our usage of first-order perturbation theory these results are only strictly true in the limit $K(v_c/c)^2\to0$. At second-order perturbation theory, however, both $a$ and/or $e$ may change. In your numerical experiments, you should find that the orbit-averaged changes of $a$ and $e$ are either zero or scale stronger than linear with perturbation amplitude $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;disclaimer&lt;/strong&gt; No guarantee that the algebra is correct. Check it!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Appendix: orbit averages&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbit averages of $v_rf(r)$ with an abitrary (but integrable) function $f(r)$ can be&#xA;directly calculated for any type of periodic orbit. Let $F(r)$ be the antiderivative of $f(r)$, i.e. $F'\!=f$, then the orbit average is:&#xA;$$&#xA;\langle v_r f(r)\rangle = \frac{1}{T}\int_0^T v_r(t)\,f\!\left(r(t)\right) \mathrm{d}t&#xA;= \frac{1}{T} \left[F\left(r(t)\right)\right]_0^T = 0&#xA;$$&#xA;with $T$ the orbital period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the orbit averages required in $\langle\dot{\boldsymbol{e}}\rangle$, we must dig a bit deeper. For a Keplerian elliptic orbit&#xA;$$&#xA;\boldsymbol{r}=a\left((\cos\eta-e)\hat{\boldsymbol{e}}+\sqrt{1-e^2}\sin\eta\,\hat{\boldsymbol{k}}\right)\qquad\text{and}\qquad&#xA;r=a(1-e\cos\eta)&#xA;$$&#xA;with eccentricity vector $\boldsymbol{e}$ and $\hat{\boldsymbol{k}}\equiv\hat{\boldsymbol{h}}\wedge\hat{\boldsymbol{e}}$ a vector perpendicular to $\boldsymbol{e}$ and $\boldsymbol{h}$. Here, $\eta$ is the eccentric anomaly, which is related to the mean anomaly $\ell$ via&#xA;$&#xA;\ell=\eta-e\sin\eta,&#xA;$&#xA;such that $\mathrm{d}\ell=(1-e\cos\eta)\mathrm{d}\eta$ and an orbit average becomes&#xA;$$&#xA;\langle\cdot\rangle = (2\pi)^{-1}\int_0^{2\pi}\cdot\;\mathrm{d}\ell = (2\pi)^{-1}\int_0^{2\pi}\cdot\;(1-e\cos\eta)\mathrm{d}\eta.&#xA;$$&#xA;Taking the time derivative (note that $\dot{\ell}=\Omega=\sqrt{GM/a^3}$ the orbital frequency) of $\boldsymbol{r}$, we find for the instantaneous (unperturbed) orbital velocity&#xA;$$&#xA;\boldsymbol{v}=v_c\frac{\sqrt{1-e^2}\cos\eta\,\hat{\boldsymbol{k}}-\sin\eta\,\hat{\boldsymbol{e}}}{1-e\cos\eta}&#xA;$$&#xA;where I have introduced $v_c\equiv\Omega a=\sqrt{GM/a}$, the speed of the circular orbit with semimajor axis $a$. From this, we find the radial velocity $v_r=\hat{\boldsymbol{r}}{\cdot}\boldsymbol{v}=v_c e\sin\eta(1-e\cos\eta)^{-1}$&#xA;and the rotational velocity&#xA;$$&#xA;\boldsymbol{v}_t = v_c\frac{\sqrt{1-e^2}(\cos\eta-e)\,\hat{\boldsymbol{k}}-(1-e^2)\sin\eta\,\hat{\boldsymbol{e}}}{(1-e\cos\eta)^2}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these, we have [&lt;em&gt;corrected again&lt;/em&gt;]&#xA;$$&#xA;\left\langle \frac{h^2v_r\boldsymbol{r}}{r^4}\right\rangle = &#xA;\Omega v_c^2\,\hat{\boldsymbol{k}}\,&#xA;\frac{e(1-e^2)^{3/2}}{2\pi}\int_0^{2\pi}\frac{\sin^2\!\eta}{(1-e\cos\eta)^4}\mathrm{d}\eta&#xA;=\frac{\Omega v_c^2e}{2(1-e^2)}\hat{\boldsymbol{k}}&#xA;\\&#xA;\left\langle \frac{v_r^2\boldsymbol{v}_t}{r}\right\rangle = \Omega v_c^2\,&#xA;\hat{\boldsymbol{k}}\,&#xA;\frac{e^2(1-e^2)^{1/2}}{2\pi}\int_0^{2\pi}\frac{\sin^2\!\eta(\cos\eta-e)}{(1-e\cos\eta)^4}\mathrm{d}\eta=0,&#xA;$$&#xA;in particular, the components in direction $\hat{\boldsymbol{e}}$ average to zero. Thus [&lt;em&gt;corrected again&lt;/em&gt;]&#xA;$$\left\langle 2\frac{h^2v_r\boldsymbol{r}}{r^4}-\frac{v_r^2\boldsymbol{v}_t}{r}\right\rangle&#xA;=\frac{\Omega v_c^2e\,\hat{\boldsymbol{k}}}{(1-e^2)}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2014-09-09T08:35:49.607" LastActivityDate="2014-09-09T08:35:49.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6240" PostTypeId="1" AcceptedAnswerId="6241" CreationDate="2014-09-01T21:52:12.663" Score="14" ViewCount="8545" Body="&lt;p&gt;I can look at a star chart and identify things like constellations in the sky.  But if I take a picture with my DSLR camera (35mm with a decent zoom lens, no astronomy specific optics), I run into difficulty. I find that the camera can see a lot more stars than my eye, and I find that I start to have great difficulty even identifying simply constellations.  For instance, when I look for the constellation Cassiopeia in the night sky it is very easy to spot: it looks like a sideways &quot;W&quot;.  But when I take a picture I don't see a W at all in the picture, I see about 16 stars around where the &quot;W&quot; should be a weird angles to each other, and it's hard to figure out which stars to connect to form the actual constellation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it just goes downhill from there, as other constellations don't even have a shape as easily recognizable as Cassiopeia, and as a result I have a great deal of difficult even recognizing what I am looking at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A photograph with a regular camera is not quite the same as looking through a telescope, because the camera has a much wider field of view.  And at the same time, there does not seem to be a linear relationship between the actual magnitude of the star and how bright it looks.  As I mentioned above, I see 16 stars which look somewhat comparable in brightness, but I know that they differ by several magnitudes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some kind of trick or technique that I can use to quickly identify stars in pictures I take?  In looking for clues, I have not seen anybody mentioning this problem.&lt;/p&gt;&#xA;" OwnerUserId="2341" LastActivityDate="2020-11-05T19:52:36.070" Title="How to identify stars in photographs?" Tags="&lt;photography&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6241" PostTypeId="2" ParentId="6240" CreationDate="2014-09-02T02:10:25.410" Score="12" Body="&lt;p&gt;Take another photograph of the same field that is less exposed. (Doesn't matter if you do this by shortening the exposure time, decreasing the ISO, etc.) This will give you an image with far fewer stars so you can easily pattern match your images. &lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-09-02T02:10:25.410" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6242" PostTypeId="1" AcceptedAnswerId="6265" CreationDate="2014-09-02T13:32:11.653" Score="5" ViewCount="86" Body="&lt;p&gt;The LOFAR project open a new window in the universe's study because it allows to receive low frequences of the universe that we could not get because of the ionosphere. &#xA;But I'm wondering how can LOFAR pass through the ionosphere? Is it because of the large fields of antennas? Or it is something else?&#xA;Thanks for any answers which can clear my mind.&lt;/p&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="1792" LastEditDate="2014-09-02T13:46:58.163" LastActivityDate="2014-09-04T14:22:33.267" Title="How LOFAR pass through the ionosphere?" Tags="&lt;radio-astronomy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6243" PostTypeId="1" AcceptedAnswerId="6249" CreationDate="2014-09-02T13:51:32.160" Score="7" ViewCount="846" Body="&lt;p&gt;I've read (at &lt;a href=&quot;http://deepblue.lib.umich.edu/bitstream/handle/2027.42/55288/index.html?sequence=1&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;, among other places) that during the Degenerate Era, star formation will end and the last stars will go out. But it was noted that there is still the possibility of star birth, coming from the collision of two stellar remnants, such as brown dwarfs (although white dwarfs also have the potential for such a collision). When two brown dwarfs collide, if they have &lt;em&gt;enough combined mass&lt;/em&gt;, fusion can begin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is much different from how normal stars form today, and how they fuse hydrogen and other elements. So my question is this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will there be any major differences between the fusion process in the new &quot;star&quot; and a star with the same mass that formed via traditional star formation (i.e. collapse of a gas cloud)? Also, how will this affect the new &quot;star's&quot; life cycle?&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-01-10T22:44:37.553" LastActivityDate="2017-05-28T17:10:07.327" Title="Mechanism for Brown Dwarf Fusion" Tags="&lt;star&gt;&lt;star-formation&gt;&lt;brown-dwarf&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6244" PostTypeId="2" ParentId="6229" CreationDate="2014-09-02T16:36:36.867" Score="2" Body="&lt;p&gt;I'm going to try to expand a little on my comment regarding inflation. Let me know if it is in any way useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just after the Big Bang, the universe was quite small. Tiny. Smaller than the head of a pin. Then the universe reached the grand old age of $1 \times 10^{-36}$ seconds old, and it had a little bit of a growth spurt. Between $1 \times 10^{-36}$ seconds and $1 \times 10 ^{-33}$ seconds, it expanded at an incredible rate, reaching a size closer to its present-day size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before inflation, the universe was - well, we don't really know. Prior to $1 \times 10^{-43}$ seconds, all four fundamental forces were unified. But after inflation, they slowly (relative to the time scales we're talking about) became the distinct forces we know today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another curious thing happened after the inflationary epoch: baryogenesis. Prior to this, the universe had been made out of an exotic quark-gluon plasma. Now, some of the matter (now spread out throughout space) formed quarks which grouped together to form baryons - some of which are the protons and neutrons we know today. Later on (but still within the first second), electrons formed, and, gradually, the particles we know today came into being.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objects you're talking about formed 420 million years after the Big Bang - way after inflation! They were never close together; however, their constituent particles were. The reason they are so far away now is mostly because of inflation, but also partly because of the current expansion of space, which comes courtesy of dark energy. The reason we can see these objects in all directions is because inflation made space expand in every direction, so matter is, therefore, in every direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-02T16:41:40.100" LastActivityDate="2014-09-02T16:41:40.100" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6245" PostTypeId="5" CreationDate="2014-09-02T18:50:29.530" Score="0" Body="&lt;p&gt;Often called a &amp;quot;failed star&amp;quot;, brown dwarfs are sub-stellar objects that never achieve permanent nuclear fusion. They can be up to 80 Jupiter masses, but are still much smaller than the Sun.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="34459" LastEditDate="2020-07-18T20:12:25.133" LastActivityDate="2020-07-18T20:12:25.133" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6246" PostTypeId="4" CreationDate="2014-09-02T18:50:29.530" Score="0" Body="Questions about sub-stellar objects that are like stars but never achieve permanent nuclear fusion because they are too low-mass." OwnerUserId="34459" LastEditorUserId="34459" LastEditDate="2020-07-18T20:12:52.240" LastActivityDate="2020-07-18T20:12:52.240" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6247" PostTypeId="1" AcceptedAnswerId="6248" CreationDate="2014-09-02T18:58:05.517" Score="1" ViewCount="151" Body="&lt;p&gt;Given the initial conditions at the Big Bang, could our universe have evolved any differently? In other words, did the initial conditions predetermine every stage in the evolution of the universe to this day, or - if we &quot;roll the tape backward&quot; and play it again - could the universe be entirely different on the macro scale (eg. no stars and galaxies)?&lt;/p&gt;&#xA;" OwnerUserId="2348" LastActivityDate="2014-09-04T21:14:18.460" Title="Could the universe have evolved differently?" Tags="&lt;big-bang-theory&gt;&lt;early-universe&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6248" PostTypeId="2" ParentId="6247" CreationDate="2014-09-02T20:14:27.047" Score="1" Body="&lt;p&gt;I don't think your attempt to distinguish the micro and macro scales is helpful. As it stands, the physical theories (quantum mechanics etc) are &lt;em&gt;deterministic&lt;/em&gt;, that is the initial conditions pre-determine all future: there is no random element (&quot;Gott würfelt nicht&quot; -- god does not throw dice -- to quote Albert Einstein).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is debatable, but certainly beyond astronomy, better (meta)phyiscs and philosophy ...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edited&lt;/strong&gt; to respond to the comment, effectively rephrasing the question to consider a &quot;&lt;em&gt;slighly different&lt;/em&gt;&quot; early state of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, it pretty much depends what you mean by &quot;&lt;em&gt;slighly different&lt;/em&gt;&quot;. If the laws of physics are &lt;em&gt;completely&lt;/em&gt; untouched, I would think that the universe would &lt;strong&gt;evolve statistically in exactly the same way&lt;/strong&gt;. This is because all the statistical characteristics of the early universe (the power spectra and correlation functions of the various energy densities, for example) come about from the laws of physics (via natural instabilities, for example) rather than the initial state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However 1&lt;/strong&gt;, our understanding of the laws of physics and their deeper origin is still very limited. This means that it is not clear whether a &lt;em&gt;slightly different&lt;/em&gt; early state with &lt;em&gt;exactly&lt;/em&gt; the same laws of physics is actually at all possible. This is particularly true for the nature of gravity and the vacuum, neither of which is currently properly understood at the quantum mechanical level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However 2&lt;/strong&gt;, one has to bear in mind that we don't know whether the universe is finite or infinite. If it is infinite (and the observational evidence from our finite patch of the universe indicates that it has a flat or open, but not closed, spacetime), we cannot really make any sensible conclusions from our finite observable patch. This includes this answer!&#xA;Except, of course, if we invoke an extended version of the cosmological principle, something like &quot;an infinite universe must be the same everywhere&quot;, which is (meta)physical non-sense. This is the main reason why I don't like cosmology too much.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2014-09-03T07:59:46.953" LastActivityDate="2014-09-03T07:59:46.953" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6249" PostTypeId="2" ParentId="6243" CreationDate="2014-09-02T20:24:02.293" Score="6" Body="&lt;p&gt;Your scenario is, of course, rather speculative and considers the very distant future. I don't like that article you're referencing, some stuff seems inaccurate (for example the fraction of baryons locked into stars is tiny and will always remain tiny).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two brown dwarves collide (extremely unlikely, but if you wait long enough, anything with some tiny non-zero probability eventually will happen), they form a new star with their combined mass. Since the brown dwarves have not undergone any hydrogen burning, the new star has the same composition as a new-born star of the same mass would have had (except for lithium) and, if massive enough, will evolve just like an ordinary M dwarf (or so). In other words, it will fuse hydrogen into helium as all stars on the main sequence.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2014-09-11T18:36:16.450" LastActivityDate="2014-09-11T18:36:16.450" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6250" PostTypeId="2" ParentId="6232" CreationDate="2014-09-02T20:38:35.397" Score="12" Body="&lt;p&gt;To add to Dieudonné's excellent answer, I'd like to say that spiral arms are only really prominent in the blue part of the spectrum (massive stars tend to be blue and short-lived), while in infrared wavebands, for example, spiral arms only appear as mild over-densities of 10-20%. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some galaxies have clear arms winding for almost 360$^\circ$ or even more (&lt;em&gt;grand-design spirals&lt;/em&gt;), but most spirals arms much are shorter and less clear. Many galaxies in fact have only so-called &lt;em&gt;flocculent&lt;/em&gt; spiral arms, which look like short pieces of spiral arms all over the galaxy, but which cannot be linked together into one large coherent spiral pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation for the Milky is not so clear, as we don't have an external view and measuring accurate distances to objects within the Milky Way is notoriously difficult. &#xA;The hope is that ESA's Gaia satellite will give us some more accurate picture.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-09-02T20:38:35.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6251" PostTypeId="2" ParentId="6224" CreationDate="2014-09-02T21:04:15.547" Score="6" Body="&lt;p&gt;It indeed appears the Andromeda galaxy (M31) and The Milky Way (MW) are en route to a collision. This will lead to a merger of the two galaxies forming an elliptical galaxy. The flattened disc structure of M31 and MW comes about because their gas dissipated energy but conserved momentum: a disc is the minimum energy state at given angular momentum. The gas in the disc then forms stars, which are more or less on the same orbit as the gas, i.e. circling the respective galaxies. A disc galaxy is dynamically cold, i.e. the speed of the mean motion of the star around the galaxy is much larger than their random motions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The merger destroys all these structures and replaces them with a much smoother and dynamically hot structure, when the density is roughly constant on (concentric) triaxial ellipsoids. There are several types of stellar orbits in such galaxies, but most important are the so-called box orbits, when the stars oscillate in a large box-like volume and have no preferred mean direction of motion. In the early phases, the elliptical galaxy will have some temporary structures (so-called shells), which are remnants of the merger process itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gas in M31 and MW will most likely be swept into the inner parts of the elliptical where it may form many new stars and contribute to the feeding of the AGN which emerges from the coalescence of the two supermassive black holes of M31 and MW. But energy fed back from the new formed stars via supernovae and stellar winds as well as from the AGN will drive the remaining gas out of the elliptical galaxy, leaving it &quot;red and dead&quot;, i.e. without star formation and young blue stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The merger of M31 and MW will hardly increase the already neglible rate of stellar collisions. Given the vastness of galaxies compared to the size of stars, such collisions are extremely unlikely. The only places where such collisions occur are the very dense cores of globular clusters.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-09-02T21:04:15.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6252" PostTypeId="5" CreationDate="2014-09-03T03:25:26.773" Score="0" Body="&lt;p&gt;The cosmic microwave background, or CMB, is radiation formed in the early stages of the universe. Currently at a temperature of 2.7 Kelvin, it is hailed as strong evidence for the Big Bang theory. The cosmic microwave background was first observed in detail by Arno Penzias and Robert Wilson in 1964. It is highly redshifted, which is why it is currently in the microwave region of the electromagnetic spectrum.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2020-12-18T15:05:18.560" LastActivityDate="2020-12-18T15:05:18.560" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6253" PostTypeId="4" CreationDate="2014-09-03T03:25:26.773" Score="0" Body="Questions regarding the cosmic microwave background, or CMB, the leftover radiation that is taken by many as proof of the Big Bang." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-16T15:34:47.167" LastActivityDate="2014-09-16T15:34:47.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6254" PostTypeId="5" CreationDate="2014-09-03T03:36:02.083" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-09-03T03:36:02.083" LastActivityDate="2014-09-03T03:36:02.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6255" PostTypeId="4" CreationDate="2014-09-03T03:36:02.083" Score="0" Body="Waves in spacetime formed by the acceleration of massive bodies. A passing wave causes spacetime to be stretched and squashed by a very small amount. Typically we can only detect the most energic gravitational events such as black hole mergers." OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-02-17T23:02:19.597" LastActivityDate="2016-02-17T23:02:19.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6256" PostTypeId="2" ParentId="6247" CreationDate="2014-09-03T08:48:45.290" Score="3" Body="&lt;p&gt;Given chaos theory, things could have been different. Given the laws of physics, there are limits to how much difference there could have been.&lt;br/&gt;&#xA;&lt;br/&gt;&#xA;To put it bluntly, a subatomic particle could have zigged instead of zagging, causing a fusion event that led to the formation of a star not to take place in one spot but another.&lt;br/&gt;&#xA;But it would not prevent stars from forming.&lt;/p&gt;&#xA;" OwnerUserId="534" LastActivityDate="2014-09-03T08:48:45.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6257" PostTypeId="1" CreationDate="2014-09-03T10:50:41.640" Score="5" ViewCount="91" Body="&lt;p&gt;What exactly does p-wave mean when referring to particles, like charmonia states. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, see the following reference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-ph/9208254&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-ph/9208254&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My doubt is - p wave in scattering means l=1. But how is that used to label a particle state?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. &lt;/p&gt;&#xA;" OwnerUserId="2351" LastEditorUserId="6" LastEditDate="2016-05-25T15:38:09.380" LastActivityDate="2016-05-25T15:38:09.380" Title="meaning of p-wave charmonia" Tags="&lt;high-energy-astrophysics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6258" PostTypeId="1" CreationDate="2014-09-03T15:07:39.240" Score="2" ViewCount="266" Body="&lt;p&gt;I have got a question as I am trying to figure out what the real difference between the birth date of my friend and me is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am born on 1983 6th February 7 am. My Friend is born on 1984 7th February 1 am. &#xA;1982 was a leap year as the real duration of earths orbit around sun is ca. 365 1/4 days. Okay... But I am struggling to figure out if need to add/subtract 8 hours to my or my friends birthday to figure out what the real difference in our birth dates is, in relation to how long earth orbits around the sun. That leap year is at end of February makes it even harder to me as I don't know how that fits in into the calculation.&lt;/p&gt;&#xA;" OwnerUserId="2353" LastEditorUserId="13681" LastEditDate="2017-07-15T21:34:04.693" LastActivityDate="2017-07-15T21:34:04.693" Title="Birthdate differences calculation relative to earths orbit duration" Tags="&lt;orbit&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6259" PostTypeId="5" CreationDate="2014-09-03T18:23:31.833" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-09-03T18:23:31.833" LastActivityDate="2014-09-03T18:23:31.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6260" PostTypeId="4" CreationDate="2014-09-03T18:23:31.833" Score="0" Body="Questions regarding gravitationally bound systems of hundreds to thousands of galaxies as well as interactions between such systems." OwnerUserId="1862" LastEditorUserId="1862" LastEditDate="2014-09-03T20:44:54.143" LastActivityDate="2014-09-03T20:44:54.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6261" PostTypeId="2" ParentId="6258" CreationDate="2014-09-03T19:57:30.217" Score="3" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Sidereal_year&quot; rel=&quot;nofollow&quot;&gt;sidereal year&lt;/a&gt; is 6 hours and 9 minutes longer than a 365 day calendar year. To simplify your question, assuming you an your friend were born on the same date and time in 1983 and 1984 you would be 365 days, 6 hours, and 9 minutes older than your friend. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there is no leap day between the two times you are considering, it doesn't affect the calculation.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-09-03T19:57:30.217" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6262" PostTypeId="2" ParentId="6258" CreationDate="2014-09-03T22:38:32.027" Score="2" Body="&lt;p&gt;I'll use YYYY-MM-DD notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You were born 1983-02-06 07:00.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friend was born 1984-02-07 01:00.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was no leap day between your birthdays, which makes the calculation a little simpler. The most recent leap day before your birthday was 1980-02-29; the earliest leap day after your friend's birthday was 1984-02-29.  (No, 1982 was not a leap year, and even if it had been that leap day still wouldn't have been between your birthdays).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moment exactly one calendar year after your birthday was 1984-02-06 07:00; that calendar year is exactly 365 days.  Your friend was born 1 day minus 6 hours after that moment. So the difference between your births is 365 days plus 1 day minus 6 hours, or 365 days, 18 hours (365.75 days).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here's a bash script that computes the results:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#!/bin/bash&#xA;&#xA;date0='1983-02-06 07:00'&#xA;date1='1984-02-07 01:00'&#xA;&#xA;# Compute date as seconds since 1970-01-01 00:00:00 UTC&#xA;seconds0=$(date +%s -d &quot;$date0&quot;)&#xA;seconds1=$(date +%s -d &quot;$date1&quot;)&#xA;&#xA;# Do some math.  Note that this does only integer arithmetic.&#xA;# It happens to work in this case since we're dealing with whole&#xA;# hours, but this is not a general solution.&#xA;(( seconds = seconds1 - seconds0 ))&#xA;(( days      = seconds / 86400 ))&#xA;(( hours = seconds % 86400 / 3600 ))&#xA;&#xA;# Show the results.&#xA;echo &quot;$date0 = $seconds0&quot;&#xA;echo &quot;$date1 = $seconds1&quot;&#xA;echo &quot;The dates differ by $seconds seconds or $days days and $hours hours&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The output is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1983-02-06 07:00 = 413391600&#xA;1984-02-07 01:00 = 444992400&#xA;The dates differ by 31600800 seconds or 365 days and 18 hours&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="53" LastActivityDate="2014-09-03T22:38:32.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6263" PostTypeId="1" AcceptedAnswerId="6264" CreationDate="2014-09-04T03:49:20.217" Score="2" ViewCount="469" Body="&lt;p&gt;During exactly what dates of the year can the Centaurus constellation be seen from Tokyo, Japan? And what is the ideal date for viewing it from Tokyo?  I found on a website that it can been seen May-June in New Zealand, best visible at 21:00 (9pm) during May and that the midline is May 20th.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(According to the Japanese Wikipedia page for Centaurus, only from Okinawa can the entire constellation be seen; in Tokyo, apparently only part of it can be seen.)&lt;/p&gt;&#xA;" OwnerUserId="2301" LastEditorUserId="1862" LastEditDate="2014-09-05T19:36:59.540" LastActivityDate="2014-09-05T19:36:59.540" Title="During what time of the year can Centaurus be seen from Tokyo?" Tags="&lt;star-gazing&gt;&lt;constellations&gt;&lt;time&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6264" PostTypeId="2" ParentId="6263" CreationDate="2014-09-04T05:57:16.687" Score="3" Body="&lt;p&gt;You can use an application that shows the stars at any time and location. &lt;a href=&quot;http://www.stellarium.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellarium&lt;/a&gt; is free, has nice features and isn't too complicated. You can select a star/constellation and flip through the time of day and through the days to see how its elevation changes over time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, everyone with the same &lt;em&gt;latitude&lt;/em&gt; sees the same sky at a certain date (assuming an equal elevation, which counts if you're standing on Mt. Fuji). So those dates for New Sealand are irrelevant for you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the &lt;a href=&quot;http://www.star-map.fr/&quot; rel=&quot;nofollow noreferrer&quot;&gt;StarmapPro&lt;/a&gt; application, I made the following chart for your location (Tokyo) showing the elevation of Centaur at midnight for each month. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5Phfe.jpg&quot; alt=&quot;centaur elevation for Tokyo&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, it's visible from halfway march until the first half of may, with its maximum around the first week of april. It comes barely above the horizon, so you'll need a clear view to the horizon and try to avoid light pollution (good luck with that in Tokyo ;-)&lt;/p&gt;&#xA;" OwnerUserId="2189" LastActivityDate="2014-09-04T05:57:16.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6265" PostTypeId="2" ParentId="6242" CreationDate="2014-09-04T14:22:33.267" Score="3" Body="&lt;p&gt;LOFAR does not go 'through the ionosphere' as it's ground based. Rather, it is able to receive signals from outside the ionosphere due to the very low frequencies involved. These frequencies (naturally) have very long wavelengths, which means that LOFAR must be very large to obtain a decent level of resolution. The number of antennae will affect sensitivity and reliability.&lt;/p&gt;&#xA;" OwnerUserId="2209" LastActivityDate="2014-09-04T14:22:33.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6266" PostTypeId="1" AcceptedAnswerId="6269" CreationDate="2014-09-04T15:21:33.113" Score="4" ViewCount="54" Body="&lt;p&gt;I was doing research about ion tails of planetary bodies and noticed that ion tails composed of sodium were common. For example, Mercury and the Moon both have ion tails made of sodium. Why is sodium, as opposed to other ions, commonly seen in ion tails?&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2014-09-05T07:28:35.583" Title="Why is sodium such a common ion for in ion tails?" Tags="&lt;the-moon&gt;&lt;planet&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6267" PostTypeId="5" CreationDate="2014-09-04T20:51:10.037" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-09-04T20:51:10.037" LastActivityDate="2014-09-04T20:51:10.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6268" PostTypeId="4" CreationDate="2014-09-04T20:51:10.037" Score="0" Body="Questions about the source or effects of the matter that makes up 85% of the matter in the universe but appears to only interact gravitationally." OwnerUserId="1862" LastEditorUserId="1862" LastEditDate="2014-09-05T19:36:35.513" LastActivityDate="2014-09-05T19:36:35.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6269" PostTypeId="2" ParentId="6266" CreationDate="2014-09-04T21:01:27.547" Score="3" Body="&lt;p&gt;From &lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2000/ast26oct_1/&quot; rel=&quot;nofollow&quot;&gt;http://science.nasa.gov/science-news/science-at-nasa/2000/ast26oct_1/&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;When a Leonid meteoroid hits the Moon it vaporizes some dust and&#xA;  rock,&quot; explains Jody Wilson of the Boston University Imaging Science&#xA;  Team. &quot;Some of those vapors will contain sodium (a constituent of Moon&#xA;  rocks) which does a good job scattering sunlight. If any of the impact&#xA;  vapors drift over the lunar limb, we may be able to see them by means&#xA;  of resonant scattering. They will glow like a faint low-pressure&#xA;  sodium street lamp.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, relative to other elements, sodium has a &lt;a href=&quot;http://www.lenntech.com/periodic-chart-elements/ionization-energy.htm&quot; rel=&quot;nofollow&quot;&gt;low ionization energy&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-09-04T21:01:27.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6270" PostTypeId="2" ParentId="6247" CreationDate="2014-09-04T21:14:18.460" Score="1" Body="&lt;p&gt;The current universe appears to have evolved from some extremely well-tuned parameters for which the long-term state of the universe is extremely sensitive in changes to. For example, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Flatness_problem&quot; rel=&quot;nofollow&quot;&gt;observed curvature&lt;/a&gt; is so close to flat that the initial conditions must have been absurdly flat: to within an accuracy of 1 in $10^{64}$. Any further from that and there would be drastic differences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This absurd level of fine tuning is one of the motivations for inflationary models. In these the initial conditions can take a wide range of values, and then the inflation smoothes everything out to the perfect tuning we've observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, yes, the universe is very sensitive to a number of things. Which causes &lt;a href=&quot;http://en.wikipedia.org/wiki/Fine-tuning&quot; rel=&quot;nofollow&quot;&gt;a lot of problems&lt;/a&gt;.&#xA; Finding reasons why we hit the perfect combination is a matter of some debate (and some would say is philosophical, not scientific, in nature).&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2014-09-04T21:14:18.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6271" PostTypeId="2" ParentId="5998" CreationDate="2014-09-05T01:59:18.047" Score="1" Body="&lt;p&gt;HUDF suffers from &lt;a href=&quot;http://astrobites.org/2012/11/05/how-to-avoid-cosmic-variance-in-your-cosmological-survey/&quot; rel=&quot;nofollow&quot;&gt;cosmic variance&lt;/a&gt;. So while you could choose what you believe is a random and unbiased region of sky, your uncertainty will be larger than just the standard statistical sampling error. There would be additional uncertainty because of large scale structure since galaxies are not randomly distributed on the sky.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-09-05T01:59:18.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6272" PostTypeId="1" AcceptedAnswerId="6279" CreationDate="2014-09-05T05:09:36.003" Score="4" ViewCount="320" Body="&lt;p&gt;After fooling around with a FITS file(s) (e.g. setting scale, changing color) is it possible to save the session such that there is a file associated with all those tweaks the next time I open it up? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Save and Save Image just seem to keep whatever I set it to, but when I load those new saved files back into ds9 the settings are all back to whatever default load settings I have enabled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also like to be able to do this if I have FITS files loaded into multiple frames so I don't have to go through the process of loading each individual image.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastEditorUserId="950" LastEditDate="2014-09-05T15:04:44.060" LastActivityDate="2014-09-05T15:04:44.060" Title="Save current state in ds9?" Tags="&lt;observational-astronomy&gt;&lt;software&gt;&lt;ds9&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6273" PostTypeId="2" ParentId="6266" CreationDate="2014-09-05T07:28:35.583" Score="2" Body="&lt;p&gt;Extending @Aaron: Which also does matter, that it is relatively common in the universe ( &lt;a href=&quot;https://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements&lt;/a&gt; ). For example, rubidium has a much lower ionization energy, but it can only be found in traces, while sodium is one of the most frequent elements in the stony planets.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastActivityDate="2014-09-05T07:28:35.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6274" PostTypeId="1" AcceptedAnswerId="6375" CreationDate="2014-09-05T07:37:17.477" Score="13" ViewCount="24048" Body="&lt;p&gt;AFAIK it was possible long before the first interplanetary probes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who did it?&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="1862" LastEditDate="2014-09-05T19:37:12.490" LastActivityDate="2020-06-25T08:47:47.077" Title="How was the Earth-Sun distance originally calculated?" Tags="&lt;distances&gt;&lt;history&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6276" PostTypeId="2" ParentId="6127" CreationDate="2014-09-05T09:56:15.957" Score="2" Body="&lt;p&gt;For a start, the hot jupiters (HJ) are &lt;em&gt;not similar&lt;/em&gt; to the planets of the Solar system (they are much hotter and very close to their host star). It's essentially impossible for HJs to be ejected away from their star's sphere of influence. The reason is that they are gravitationally rather tightly bound to their star, so that any gravitational interaction with a third body that would gain them enough energy for ejection needs to be closer than is possible without it leading to a collision with that third body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So rogue planets that didn't form independent of a star must have been much less bound to their parent star, similar to the outer planets of the Solar system. Of course, once ejected (free-floating), such an object can in principle be re-captured into another planetary system or a binary system. However, this is quite unlikely, since the planet must move with a velocity very close to that of its capturer. Also, it must involve a third body (either another planet or star bound to its capturer) to absorb (into its orbit) the orbital energy (and angular momentum) released by the capture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only possibility is that the planet was ejected very early on, when the stellar association / group / open cluster within which it was born had not yet been dissolved. Then it may find another home with one of the other members of that association. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think that anybody has estimated the actual probabilities for this to happen, but why are you interested?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As a side remark, it seems clear that orbiting planets form from / in gas discs surrounding protostars, but it's not clear whether brown dwarves cannot also form this way. Whether or not an object formed this way becomes a gas planet, a brown dwarf, or even a star depends entirely on its mass and not the formation process.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="13" LastEditDate="2014-09-07T02:27:26.237" LastActivityDate="2014-09-07T02:27:26.237" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6277" PostTypeId="2" ParentId="5937" CreationDate="2014-09-05T10:05:11.717" Score="4" Body="&lt;p&gt;&lt;strong&gt;None&lt;/strong&gt;. and that doesn't surprise me much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comets are thought to be Oort-cloud objects which by interaction either with other Oort-cloud objects or with objects passing the Solar system have been perturbed to venture into the gravitational reach of the outer planets, whence they are&#xA;flung into the inner Solar system via close encounters (gravitational slingshots) with one of the outer planets, in particular Jupiter and Saturn. For such a slingshot to be efficient, the encounter must be close when any binary (proto-)comet will most likly be disassociated by the gravitational tides of the sling-shooting planet.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2014-10-08T19:53:53.620" LastActivityDate="2014-10-08T19:53:53.620" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6278" PostTypeId="2" ParentId="5925" CreationDate="2014-09-05T11:03:42.767" Score="3" Body="&lt;p&gt;Not exactly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;During either of the equinoxes there is a moment when the line between center of earth and the sun aligns with the equator. This doesn't coincide with prime meridian in any way though; it may happen at any meridian whatsoever that happens to coincide with the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course if it happens the sun is in zenith above prime meridian at that moment, your conditions will be satisfied. But depending on precision you allow, that may be a very short time window. It's all about how precisely you want that to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Near Equinox Earth tilts by about 0.25 degree of latitude per day, meaning that is the maximum angle by which the line will be off from equator while crossing the prime meridian on the same day as when Equinox happens. 0.25 degree of latitude is about 27 kilometers, so this is the maximum error. Of course on some years, that will be much less, it's just that Earth tilt is correlated with length of year, zenith line crossing of prime meridian is correlated with time of day, and time of day is not really correlated with length of year.&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="83" LastEditDate="2016-09-29T08:59:56.200" LastActivityDate="2016-09-29T08:59:56.200" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6279" PostTypeId="2" ParentId="6272" CreationDate="2014-09-05T12:12:12.883" Score="3" Body="&lt;p&gt;Perhaps you are looking for the backup function&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;File -&gt; Backup...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and to restore the settings&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;File-&gt;Restore...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="950" LastActivityDate="2014-09-05T12:12:12.883" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6280" PostTypeId="1" AcceptedAnswerId="6281" CreationDate="2014-09-05T17:12:21.857" Score="8" ViewCount="2961" Body="&lt;p&gt;I heard in a documentary that, in Svalbard (Spitsberg), Norwegian archipelago in the Arctic Ocean, the Moon never sets. Why? A drawing would certainly help.&lt;/p&gt;&#xA;" OwnerUserId="1580" OwnerDisplayName="Nicolas Barbulesco" LastEditorUserId="1580" LastEditDate="2014-09-11T13:56:46.060" LastActivityDate="2020-09-13T02:23:52.570" Title="Why does the Moon never set in Svalbard, Norway?" Tags="&lt;the-moon&gt;&lt;pole&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6281" PostTypeId="2" ParentId="6280" CreationDate="2014-09-05T18:42:24.973" Score="14" Body="&lt;p&gt;You must have misheard it, or the documentary you watched wasn't presenting very precise information. It does set but it also stays on the night sky for several days during the polar winter (polar night) when the Moon if full. This is relatively simple to imagine, so I'll describe it;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what's happening is that the &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_tilt&quot; rel=&quot;noreferrer&quot;&gt;Earth's axial tilt&lt;/a&gt; during the polar winters leans the whole Northern hemisphere towards the &lt;em&gt;night side&lt;/em&gt;, away from the Sun. This tilt is big enough (~ 23.4°) that the night sky objects aligned with the Earth's equatorial plane stay visible relatively low on the horizon. With those regions being either relatively flat and/or with a view towards the sea, there's not many obstructions limiting the viewing angle, so the Moon (and analogous also the Sun during polar summers) stays &quot;locked&quot; low above the horizon. To help a bit with imagining this, here's an animation of the Earth's axial tilt, &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Globespin-tilt-23.4.gif&quot; rel=&quot;noreferrer&quot;&gt;courtesy of Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/pJA82.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we imagine this animation of the Earth with the Sun in the distant left of the image, so during Northern hemisphere's winter (winter solstice to be precise), and the Moon to the distant right of the image (roughly 25 widths of the image away), so when it's either full or close to this lunar phase, it's not too difficult to appreciate that the northernmost polar regions have a direct line of sight of the Moon during Earth's full rotation on its axis, or a day. If you keep in mind that other celestials, including the Moon, are oblivious to the Earth's axial tilt (well, not exactly, but let's not nitpick about tidal effects that might take millions of years to make a difference), as the Moon moves farther in its orbit, in our case towards the viewer, this observation angle decreases further still and those northernmost latitudes hide to us for some part of the day. At lunar last quarter, it would be directly towards us relative to the image, so this direct line of sight relationship between the Earth and the Moon becomes reciprocal to how we're seeing places on the Earth on the animation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why when the Moon is full? Simply because that's when the Moon is also &lt;em&gt;behind the Earth&lt;/em&gt; (but not in its shadow), so the relative angle between the observation point and the Moon would stay high enough to observe it. As it moves in lunar phase and in orbit around the Earth farther, this angle becomes lower and the Moon indeed does set also in the arctic region. For what is worth, this goes exactly the same for South pole, only with a half a year difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One other effect that plays a role here is the Earth's &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_refraction&quot; rel=&quot;noreferrer&quot;&gt;atmospheric refraction&lt;/a&gt; which also adds to the duration during which the Moon appears not to set. Meaning, that even when the Moon wouldn't be in direct line of sight, but only marginally so, it would still appear low on the skies due to optical effect (displacement) of the atmosphere. This effect would somewhat offset observing the Moon from lowlands with possibly shallower observation angle when compared to higher altitude observation points with less direct line of sight obstructions, due to denser atmosphere and thus higher refraction index.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditDate="2014-09-06T23:00:43.077" LastActivityDate="2014-09-06T23:00:43.077" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6282" PostTypeId="2" ParentId="1530" CreationDate="2014-09-06T10:32:53.980" Score="1" Body="&lt;p&gt;Such a group is called a discordant redshift group.&#xA;It could be the case that the exceptional galaxy is not a true member of the cluster. The famous example of this is Seyfert's sextet. Five of the members show very similar redshifts, from 4000 to 4500 km/s, while the sixth is measured at nearly 20000 km/s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is usually interpreted as being a compact cluster of 5 galaxies, and one background galaxy. It is not taken as evidence that contradicts the general expansion of the universe.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2014-09-06T10:32:53.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6283" PostTypeId="2" ParentId="5937" CreationDate="2014-09-06T11:10:45.567" Score="1" Body="&lt;p&gt;Contact binaries, in which two objects are touching may be fairly common. This is one interpretation of the &quot;rubber duck&quot; shape of the &quot;Rosetta comet&quot; 67P/Churyumov–Gerasimenko, and radar observation of 8P/Tuttle suggest it is also formed of two objects in contact with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The low mass and so weak gravity of comets means that have a very small sphere of influence. It would have a radius of only a few km. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2014-09-06T11:10:45.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6284" PostTypeId="1" AcceptedAnswerId="6287" CreationDate="2014-09-06T12:05:45.143" Score="5" ViewCount="3157" Body="&lt;p&gt;The star Polaris is commonly known as the Northern Star because it always points North. This being said is there a South star that always points south?&lt;/p&gt;&#xA;" OwnerUserId="2212" LastActivityDate="2014-09-07T02:07:35.200" Title="Is there a 'South Star'?" Tags="&lt;star-gazing&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6285" PostTypeId="1" CreationDate="2014-09-06T12:58:40.550" Score="4" ViewCount="125" Body="&lt;p&gt;I've always wondered what would happen if a hypothetical scenario, two black holes meet, with similar proportions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe one swallow the other one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or would make a bigger one between them.&lt;/p&gt;&#xA;" OwnerUserId="2367" LastActivityDate="2014-09-06T19:00:34.617" Title="Hypothetical scenario, two nearby black holes" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-12-15T23:34:14.417" ContentLicense="CC BY-SA 3.0" />
  <row Id="6286" PostTypeId="1" CreationDate="2014-09-06T14:43:10.530" Score="2" ViewCount="644" Body="&lt;p&gt;I have found two answers for the radius of the observable universe.  Wikipedia (and other places, including this site) say approximately 46 billion LY.  I am researching this to study for the National Science Bowl, and their question database says 200 yottameters, which is about 23 billion LY (half the size that wikipedia says). There are other places which cite 200 yottameters as the radius.  Which is the more accurate answer?  &lt;/p&gt;&#xA;" OwnerUserId="2368" LastActivityDate="2014-09-06T16:26:50.887" Title="What is the radius of observable universe- 46 billion LY or 200 yottameters?" Tags="&lt;distances&gt;&lt;observable-universe&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6287" PostTypeId="2" ParentId="6284" CreationDate="2014-09-06T15:10:58.527" Score="6" Body="&lt;p&gt;There is a &quot;South Star&quot; called Sigma Octanis located in the constellation Octans, but it is so dim that virtually nobody calls it the south star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just another side note: Polaris doesn't always point North. The reason is that our Earth has a precession effect which make the North pole circulate around the sky. This picture illustrates my point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VRemk.png&quot; alt=&quot;Picture showing the precession effect and the changing of the pole star.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Just to clarify, the picture is showing the circulation of the south celestial pole around the south ecliptic pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT2 : References: &lt;a href=&quot;http://en.wikipedia.org/wiki/Pole_star#mediaviewer/File:Precession_S.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Pole_star#mediaviewer/File:Precession_S.gif&lt;/a&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pole_star&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Pole_star&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1931" LastEditorUserId="1931" LastEditDate="2014-09-07T02:07:35.200" LastActivityDate="2014-09-07T02:07:35.200" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6288" PostTypeId="1" CreationDate="2014-09-06T15:41:22.243" Score="10" ViewCount="4608" Body="&lt;p&gt;The Great Attractor is described as a location to towards which the Milky Way, along with all other galaxies in the &lt;a href=&quot;http://www.space.com/27016-galaxy-supercluster-laniakea-milky-way-home.html?cmpid=514630_20140903_30948526&quot;&gt;Laniakea Supercluster&lt;/a&gt; are moving. How long will it take for the Milky Way to &quot;arrive&quot; there, and what will happen when it does — what, in fact does &quot;arrive&quot; mean in this context?&lt;/p&gt;&#xA;" OwnerUserId="2369" LastEditorUserId="2153" LastEditDate="2014-09-06T20:52:41.383" LastActivityDate="2020-02-06T06:10:59.330" Title="When will the Milky Way &quot;arrive&quot; at the Great Attractor, and what all happen then?" Tags="&lt;cosmology&gt;&lt;milky-way&gt;&lt;galaxy-cluster&gt;&lt;local-group&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6289" PostTypeId="2" ParentId="6286" CreationDate="2014-09-06T16:26:50.887" Score="2" Body="&lt;p&gt;This radius of the observable universe is measured by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Comoving_distance&quot; rel=&quot;nofollow&quot;&gt;comoving distance&lt;/a&gt; which is the same in every direction. Using &lt;a href=&quot;http://www.astro.ucla.edu/~wright/CosmoCalc.html&quot; rel=&quot;nofollow&quot;&gt;CosmoCalc&lt;/a&gt; (a common tool for astronomers measuring cosmological parameters) you get 46 Gly for the comoving radial distance. &lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-09-06T16:26:50.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6290" PostTypeId="2" ParentId="6285" CreationDate="2014-09-06T19:00:34.617" Score="4" Body="&lt;p&gt;This is actually a very nice example of why black holes are such extreme objects, they even break the basic rules arithmatic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black hole math: $$1 \text{ black hole} + 1 \text{ black hole} = 1 \text{ black hole}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(this is a little insider science joke)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A black hole is a region of space from which nothing, not even light, can escape. To achieve that it must have very strong gravity, so a lot of mass in a &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6113/under-earths-gravitational-pull-how-dense-would-an-object-have-to-be-to-reach/6114#6114&quot;&gt;small region of space&lt;/a&gt;. So we may call it a &quot;hole in spacetime&quot;, but it still has a mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In physical sense black holes can be described by only three quantities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Its mass M (how much stuff was used to make it)&lt;/li&gt;&#xA;&lt;li&gt;Its total angular momentum, J (how fast is it spinning)&lt;/li&gt;&#xA;&lt;li&gt;Its electric charge, Q (does it have an electric/magnetic field)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All these three properties can change if the black hole is fed with 'stuff'. If you add matter the black hole mass increases. If that matter has (aligned) angular momentum, the black hole will spin faster and if the matter is charged, then the black hole charge changes accordingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When dealing with merging black holes, it is the mass property that is important. If I repeat the black-hole-math for its mass you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 1 \text{ black hole mass} + 1 \text{ black hole mass} = 2 \text{ black hole mass} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a merger of two identical black holes just gives a single black hole with twice the mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such black hole mergers are probably less hypothetical than you might think. They are considered to be likely method of creating &lt;a href=&quot;http://en.wikipedia.org/wiki/Supermassive_black_hole&quot; rel=&quot;nofollow noreferrer&quot;&gt;supermassive black holes&lt;/a&gt; and are also considered as a potential source for &lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_black_hole&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravitational wave&lt;/a&gt; emission. &lt;/p&gt;&#xA;" OwnerUserId="494" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-09-06T19:00:34.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6291" PostTypeId="2" ParentId="6288" CreationDate="2014-09-06T20:43:25.223" Score="6" Body="&lt;p&gt;I don't want to make any assumptions here regarding the Milky Way's presence in the Laniakea Supercluster simply because of how recent the discovery is. The findings could very well be accurate, but I don't want to base this answer off of them. Fortunately, I've found a few papers that get us around that little issue, as well as &lt;a href=&quot;http://www.ifa.hawaii.edu/info/press-releases/kocevski-1-06/&quot; rel=&quot;nofollow noreferrer&quot;&gt;the University of Hawaii's site&lt;/a&gt;, that led me to them . Coincidentally, one paper bears the name of one of the researchers who recently announced the Laniakea findings, R. Brent Tully, who was also involved with the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two papers (&lt;a href=&quot;http://arxiv.org/pdf/astro-ph/0510106v2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://arxiv.org/pdf/astro-ph/0512321v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;) suggest that, no matter whether the Virgo Supercluster is or is not part of the Laniakea Supercluster, the Local Group (and by extension the Milky Way) is &lt;em&gt;not&lt;/em&gt; being pulled toward the Great Attractor. Instead, it is headed for other galaxy superclusters &lt;em&gt;beyond&lt;/em&gt; the Great Attractor, including the Shapely Supercluster. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea that the Milky Way is not moving &lt;em&gt;directly&lt;/em&gt; towards the Great Attractor could render the question moot, except that it simply shifts the destination: What will happen when the Local Group reaches the further superclusters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, we don't [yet] know. As the papers say, there are a number of sources &quot;pulling&quot; on the Local Group, and so we can't say for sure just where the Local Group will wind up. This means that we also can't set a definitive timetable for the arrival. Finally, modeling galaxies with precision is not easy. Some theorists have used models in general relativity that treat many galaxies as grains of dust, so getting more specific is tough! We won't know just what will happen during after the &quot;arrival&quot; until we get a lot more data. Even the composition of the Great Attractor itself is rather mysterious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll end with that odd term, &quot;arrival&quot;. As the article you referenced (which, unfortunately, misses a lot of the better points of the findings, such as actually describing Laniakea) says, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The giant structures making up the universe often have unclear boundaries.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's true on a smaller scale, too (see &lt;a href=&quot;https://astronomy.stackexchange.com/questions/5994/where-does-the-milky-way-end&quot;&gt;Where does the Milky Way end?&lt;/a&gt; for an interesting discussion). We do know which galaxies are in the Local Group, but not where the Group's gravitational influence ends and other galaxy groups begin to dominate. We could define &quot;arriving&quot;, though, at the instant that the destination supercluster beings to attract one of the galaxies in the Local group that are further out, instead of the now-probably-merged Milky Way and Andromeda galaxies. At that point, the Local Group could be distorted by the gravity of the galaxies in the supercluster, or could simply merge into the supercluster without any problems. But like I said earlier, it's too early to do anything but speculate.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-01-10T22:54:06.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6292" PostTypeId="1" CreationDate="2014-09-06T23:05:21.297" Score="2" ViewCount="211" Body="&lt;p&gt;Is the Milky Way a collection of merged galaxies, or has it only been one galaxy that was created from the interstellar media?&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2015-01-10T22:55:17.220" Title="Milky Way Formation" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;formation&gt;" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6293" PostTypeId="2" ParentId="6292" CreationDate="2014-09-07T03:42:10.247" Score="2" Body="&lt;p&gt;This &lt;a href=&quot;https://www.youtube.com/watch?v=n0jRObc7_xo&quot; rel=&quot;nofollow&quot;&gt;simulation&lt;/a&gt; probably gives the best answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:&lt;br&gt;&#xA;The simulation is a 2 1/2 min video of the creation of a large galaxy from interstellar dust by collecting smaller galaxies. The time span is probably in hundreds of millions of years.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="1322" LastEditDate="2014-09-18T01:06:45.107" LastActivityDate="2014-09-18T01:06:45.107" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6294" PostTypeId="1" CreationDate="2014-09-07T11:54:53.780" Score="5" ViewCount="240" Body="&lt;p&gt;What can a lunar-based telescope do? If the band is ultraviolet and it can do high time sampling exposure, maybe it can be used to get distance for a star with stable pulses. However, the star should be bright and its distance should not be large.&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="2153" LastEditDate="2014-09-07T20:13:44.333" LastActivityDate="2014-09-07T20:32:21.060" Title="What can a lunar-based telescope do?" Tags="&lt;the-moon&gt;&lt;telescope&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6295" PostTypeId="2" ParentId="6292" CreationDate="2014-09-07T12:31:36.467" Score="3" Body="&lt;p&gt;Great link in your answer, @LCD3. In addition, I'm going to cover some things about the continuing evolution of the Milky Way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Milky Way is currently merging with one or more of its satellite galaxies (I say &quot;one or more&quot; because several objects in its vicinity, such as the Virgo Stellar Stream, may or may not be galaxies). The major merger is with the Sagittarius Dwarf Galaxy, which undergoing a 100 million year+ merger with the Milky Way. It sill eventually be completely torn apart, and become part of the Milky Way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also evidence that this has happened in the past (see &lt;a href=&quot;http://arxiv.org/pdf/1107.2923v2.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; paper) and it could happen in the future. The Milky Way has a lot of satellite galaxies, and even though they are far away, closer satellite galaxies could have been consumed in the past. The eventual merger with Andromeda will also add to the Milky Way's growth, as the two become a new galaxy.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Other sources:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_Milky_Way%27s_satellite_galaxies&quot; rel=&quot;nofollow&quot;&gt;List of Milky Way satellite galaxies&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sagittarius_Dwarf_Spheroidal_Galaxy&quot; rel=&quot;nofollow&quot;&gt;Sagittarius Dwarf Spheroidal Galaxy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I am looking for more non-Wikipedia sources, and I will update this answer as soon as I can.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-01-10T22:55:17.220" LastActivityDate="2015-01-10T22:55:17.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6296" PostTypeId="2" ParentId="6294" CreationDate="2014-09-07T20:32:21.060" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What can a lunar-based telescope do?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First, you would have no atmospheric distortions to interfere with the viewing.&lt;br&gt;&#xA;Second, there would be no atmospheric absorption of ultraviolet or infrared light.&lt;br&gt;&#xA;Third, there would be no cloud cover, so viewing can be almost continuous (the sun may interfere).  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, the star should be bright and its distance should not be large.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know what you reason for saying that, but it is unimportant since the Hubble telescope looks at all the stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disadvantages:&lt;br&gt;&#xA;The telescope will need to be self cleaning. There is a lot of dust kicked up when as meteoroid strikes.&lt;br&gt;&#xA;An extended manned mission to the moon to set up a large array.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-09-07T20:32:21.060" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6297" PostTypeId="1" AcceptedAnswerId="6300" CreationDate="2014-09-08T18:30:28.877" Score="3" ViewCount="1506" Body="&lt;p&gt;Quite simply, is there any used term for stars that don't have any planets? While contemplating this and doing numerous searches to no avail, I think I found my own answer, but I'm asking here for more input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I think there is no real term used as such is because there is yet to be confirmed of any stars that do not consist of any planets. I'm aware that we know of stars at this point in time where we haven't identified any planets YET, but that is not the same as actually confirming if planets do or do not exist around the star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct in assuming this is the reason why no term has came to fruition yet? Or is there a term used in relation to talking about the possibility of none existing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've searched a fair bit for this but haven't been able to find anything. I know that current models show that it's highly unlikely for this to occur, but I'm curious.&lt;/p&gt;&#xA;" OwnerUserId="2377" LastActivityDate="2020-11-21T11:06:42.090" Title="Is there a term for stars without any planets?" Tags="&lt;star&gt;&lt;planet&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6298" PostTypeId="1" CreationDate="2014-09-09T00:13:35.127" Score="6" ViewCount="802" Body="&lt;p&gt;Have there been any observations of objects entering or leaving the observable universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When looking at the physical limit of observing something like this I will to assume that the &quot;object&quot; would be an galaxy. I think it would be a good assumption to choose the size of such an galaxy, $G$, to be in the order of $10^{5}$ light years and the radius of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Observable_universe#Size&quot;&gt;observable universe&lt;/a&gt;, $OU$, is $46 \times 10^{9}$ light years. Using the limit of &lt;a href=&quot;http://en.wikipedia.org/wiki/Angular_resolution&quot;&gt;angular resolution&lt;/a&gt; and the fact that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Extremely_large_telescope&quot;&gt;current biggest telescope&lt;/a&gt; has a diameter of 22.8 meters, then the longest wave length which would still yield would be about $40 \mu m$. When reversing the &lt;a href=&quot;http://en.wikipedia.org/wiki/Redshift&quot;&gt;redshift&lt;/a&gt;, assuming this would be around $z=9$, then the longest emitted wave length which would be detected would be $4 \mu m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So observing such an event would technically be possible, however relative to the astronomical timescale we have only been able to observe with such a high resolution for a very short time, so the probability will not be very high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if such an event would be observed, would such an object suddenly appear/disappear. It will take along time for an entire galaxy due it size, so this would translate to more of a fade. Or will something else happen?&lt;/p&gt;&#xA;" OwnerUserId="1689" LastEditorUserId="1689" LastEditDate="2014-09-09T01:14:35.773" LastActivityDate="2015-01-10T22:59:00.300" Title="Objects entering or leaving the observable universe" Tags="&lt;observable-universe&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6300" PostTypeId="2" ParentId="6297" CreationDate="2014-09-09T13:04:43.700" Score="5" Body="&lt;p&gt;A star is described as such whether or not it has planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There doesn't appear to have ever been a need, and to be honest, I can't think of any need for one. When it was assumed (in historical times) that we were on the only planet, stars were not thought to have planets. The title worked then. Now, as we know most stars are likely to have planets as part of normal formation, we can still call them stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically words are invented as there is a need for them.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2014-09-09T13:04:43.700" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6301" PostTypeId="1" CreationDate="2014-09-09T13:44:34.943" Score="1" ViewCount="247" Body="&lt;p&gt;For one particular kind of element, what is the distribution in the whole universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are most of the metals located in ISM, IGM, or stars? Maybe different metals have different distributions?&lt;br/&gt;&#xA;For example, could most of the Mg be found in white dwarfs?&lt;br/&gt;&#xA;Have &lt;em&gt;all&lt;/em&gt; the natural elements have been found in stars?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorDisplayName="user1569" LastEditDate="2016-09-29T11:50:47.470" LastActivityDate="2018-03-21T17:35:32.053" Title="Where/how are metals distributed in our universe?" Tags="&lt;metal&gt;&lt;element&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6302" PostTypeId="2" ParentId="6301" CreationDate="2014-09-09T19:02:35.957" Score="7" Body="&lt;p&gt;From the &lt;a href=&quot;http://en.wikipedia.org/wiki/Big_Bang_nucleosynthesis&quot; rel=&quot;noreferrer&quot;&gt;big bang nucleosynthesis&lt;/a&gt; (BBN) page is a handy chart of the sources of elements:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AiCYN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since basically all metals come from some stellar process, the question of elemental distribution can vary greatly between different environments depending on when you're observing that particular galaxy, star/galaxy cluster, etc. Galaxy mergers, supernova, and stellar winds can be particularly good at moving metals around a galaxy or galaxy cluster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly Wolfram has element data with &lt;a href=&quot;http://reference.wolfram.com/language/ref/ElementData.html&quot; rel=&quot;noreferrer&quot;&gt;Universe abundances&lt;/a&gt; but I'm not sure where they derive those.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-09-09T19:02:35.957" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6303" PostTypeId="2" ParentId="2171" CreationDate="2014-09-09T19:15:28.977" Score="0" Body="&lt;p&gt;I believe your confusion is from combining two popular simplifications of our Universe. As we look further away we see further back in time because of the finite speed of light. So these distant objects are also evolving with us, but that light hasn't hit Earth yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might help to know that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Observable_universe&quot; rel=&quot;nofollow&quot;&gt;observable universe&lt;/a&gt; may only be 14 Gyr old, but its radius is 46 Gly, not 14 Gly. If the speed of light was infinite, then we wouldn't be observing back in time as we look at more distant objects.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastEditorUserId="1862" LastEditDate="2014-09-11T00:53:13.457" LastActivityDate="2014-09-11T00:53:13.457" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6305" PostTypeId="1" AcceptedAnswerId="6307" CreationDate="2014-09-09T20:17:55.700" Score="13" ViewCount="1402" Body="&lt;p&gt;Trying to wrap my head around some concepts involving the very large gravitational fields of black holes, and what the gravitational gradients look like on a cosmic scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm familiar with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Great_Attractor&quot; rel=&quot;noreferrer&quot; title=&quot;Great Attractor&quot;&gt;Great Attractor&lt;/a&gt;, and wondered if it were possible if the Milky Way and it's neighbors are already &quot;doomed&quot; to fall into whatever the anomaly is.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Per recommendation, I've moved the second half of my question to a new Question here: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6318/assuming-a-sufficient-amount-of-mass-above-the-density-threshold-does-the-actua&quot;&gt;Assuming a sufficient amount of mass above the density threshold, does the actual concentration of the mass matter in creating a black hole?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Original Text:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Extrapolating further, I've heard that the actual density of a black hole (within the volume of the event horizon) is pretty low, on the order of thin atmosphere - does that mean that any sufficiently large volume of mass over that density is also a black hole? Or does the actual concentration within the event horizon matter?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2018-01-22T23:16:50.163" Title="Is it possible that a ultra-large portion of the space we live in is already inside a black hole? How could we refute this?" Tags="&lt;black-hole&gt;&lt;astrophysics&gt;&lt;supermassive-black-hole&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6306" PostTypeId="1" AcceptedAnswerId="6311" CreationDate="2014-09-09T22:22:44.703" Score="2" ViewCount="71" Body="&lt;p&gt;I get the general idea of adaptive optics. The light from an object distorted by differences in the earth's atmosphere, and a telescope with AO tries to compensate for this distortion by various mechanisms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a good resource for more in depth overview on the systems and the physics behind them?&lt;/p&gt;&#xA;" OwnerUserId="2384" LastActivityDate="2014-09-10T09:01:41.553" Title="Adaptive Optics?" Tags="&lt;observational-astronomy&gt;&lt;optics&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6307" PostTypeId="2" ParentId="6305" CreationDate="2014-09-10T00:00:48.863" Score="9" Body="&lt;p&gt;This isn't anything near a proof that we &lt;em&gt;don't&lt;/em&gt; live inside a black hole, but it's a bundle of evidence that certainly goes against it, and that the Great Attractor is not, in fact, the singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off: Expansion of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you no doubt know, the universe is expanding. In fact, the expansion is accelerating. Do black holes expand? Yes. As they suck in more matter, they can get bigger. But if that was the case, we should notice more matter coming into the universe (well, I suppose it could be from outside the visible universe, but we should still see lots of matter coming towards us). Also, the universe's expansion is driven by dark energy, and it pushes (on a large scale) everything away from each other. In an expanding black hole, there would be no reason for the matter inside the event horizon to move away from each other; only the event horizon expands. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also made a good point in a comment below about Hawking radiation. Eventually, in the far future, when there is nothing left in the universe but black holes, black holes will evaporate via Hawking radiation (okay, they do that now, but they can still take in more matter). If our universe is a black hole, it should then contract. But we see no reason why it should. In fact, the theory that predicts the universe's eventual collapse into a singularity (i.e. the opposite of the Big Bang), the Big Crunch theory, predicts that the universe's contraction will match its expansion. Contraction due to Hawking radiation wouldn't necessarily mirror the black hole's growth. Also, the Big Crunch theory is supported only by a minority of scientists due to the evidence against it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second off: The motion of galaxies by the Great Attractor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, see &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6288/when-will-the-milky-way-arrive-at-the-great-attractor-and-what-all-happen-the&quot;&gt;When will the Milky Way &amp;quot;arrive&amp;quot; at the Great Attractor, and what all happen then?&lt;/a&gt; (and not just my answer! I dearly wish @LCD3 would expand his/her comment into an answer!). The general gist of things there, as pertains to your remark about the Great Attractor, is that &lt;em&gt;the galaxies aren't all moving towards it&lt;/em&gt;. There are doubts (see the papers I mentioned) that the galaxies previously thought to be moving towards it are, in fact, moving towards more distant objects - other superclusters. If the Great Attractor was indeed the singularity, a) all the galaxies in the universe should be accelerating towards it, which is not the case, and b) we should be moving directly towards it, and due to its gravity, not that of the superclusters beyond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like I said in my comments, I don't quite think your last part relates to your first part, but I'll try to address it. First, I'm not sure where you got your sources, but I can say that we don't really know what goes on inside a black hole, and I don't now how someone got that density figure (I am, of course, not the authority on black holes - see @JohnRennie for that, on Physics SE, and I could be wrong about this). The density inside the event horizon would not be very thin, however, in regions where there is lots of matter. For example, in a black hole with an accretion disk, the material that is being absorbed might not have that low a density. Also, a large volume of that density would not necessarily collapse to become a black hole, because it would not be compact enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope it helps.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="11264" LastEditDate="2018-01-22T23:16:50.163" LastActivityDate="2018-01-22T23:16:50.163" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6308" PostTypeId="1" CreationDate="2014-09-10T06:47:09.267" Score="3" ViewCount="153" Body="&lt;p&gt;The mean surface temperature of Venus is about 460 C and the pressure is about 92 atmospheres. Assuming this situation (or similar) has prevailed for millions of years what changes (if any) would there have been in the structure and chemistry of the rocks on the surface of Venus? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia(Venus) &lt;a href=&quot;http://en.wikipedia.org/wiki/Venus&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Venus&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2386" LastEditorUserId="2386" LastEditDate="2014-09-11T01:49:00.137" LastActivityDate="2014-09-30T12:19:55.097" Title="Long term effect on rocks of high pressure and temperature?" Tags="&lt;venus&gt;&lt;geology&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6309" PostTypeId="1" CreationDate="2014-09-10T07:50:52.130" Score="1" ViewCount="154" Body="&lt;p&gt;Imagine a line passing through the cores of two planets (in the same planetary system) and their star. Have we ever witnessed this - either with both planets on the same side of the star or on opposite sides? If so, was there a unique phenomenon associated with the event?&lt;/p&gt;&#xA;" OwnerUserId="2387" LastEditorUserId="6" LastEditDate="2016-05-23T17:56:02.527" LastActivityDate="2016-05-23T17:56:02.527" Title="any unique phenomenon when the cores of two planets are aligned with their star's core?" Tags="&lt;star&gt;&lt;planet&gt;&lt;gravity&gt;&lt;planetary-systems&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6310" PostTypeId="2" ParentId="6309" CreationDate="2014-09-10T08:14:28.963" Score="4" Body="&lt;p&gt;Referring to your #gravity tag this would only create a peak in the planets tidal forces. But this would be a visible effect in case the planets would be big and close enough... Last night happened that with Sun-Earth-Moon although they were not aligned as you say. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum up: &lt;strong&gt;a proper alignment would increase the effect of tidal forces.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1678" LastActivityDate="2014-09-10T08:14:28.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6311" PostTypeId="2" ParentId="6306" CreationDate="2014-09-10T09:01:41.553" Score="3" Body="&lt;p&gt;There are of course books about adaptive optics. For instance:&#xA;&lt;a href=&quot;http://books.google.nl/books?id=x1PUYBvHHqcC&amp;amp;printsec=frontcover&amp;amp;dq=Principles%20of%20Adaptive%20Optics&amp;amp;hl=&amp;amp;cd=1&amp;amp;source=gbs_api&quot; rel=&quot;nofollow&quot;&gt;Tyson, R. Principles of Adaptive Optics, (2010).&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2014-09-10T09:01:41.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6312" PostTypeId="1" CreationDate="2014-09-10T12:50:49.230" Score="6" ViewCount="180" Body="&lt;p&gt;Anomalous (i.e. not predicted from Newtonian theory) advances of the perihelion direction have been observed for many solar system planet orbits and have been accounted for by Einstein's General Relativity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basic Orbital elements have been obtained and published for many Asteroids and Comets.  But are the existing observational data on any Asteroids or Comets sufficient (in terms of accuracy and extent of observations) to determine/estimate rates of non-Newtonian perihelion advance and are any such determinations publically available?&lt;/p&gt;&#xA;" OwnerUserId="233" LastActivityDate="2018-03-22T03:05:38.003" Title="Are there sufficient observational data to measure non-Newtonian perihelion advances of any Asteroid and Comet orbits?" Tags="&lt;gravity&gt;&lt;observational-astronomy&gt;&lt;general-relativity&gt;&lt;orbital-migration&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6313" PostTypeId="1" CreationDate="2014-09-10T13:14:42.337" Score="1" ViewCount="135" Body="&lt;p&gt;If the speed at which the universe expands is constantly increasing (just like a derivative), then the opposite must also be true: the speed at which the universe contracts in reverse time is constantly decreasing, but, never reaches zero (like an asymptote). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is no exact point at which the speed of the contraction of the universe (in reverse time) reaches zero (because an asymptote goes on infinitely), then how did scientists arrive at a &quot;date&quot; for the &quot;big bang&quot;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the speed of the expansion of the universe will go on and on ad infinitum, how can one arrive at a specific beginning point of an infinite function? &lt;/p&gt;&#xA;" OwnerUserId="2389" LastEditorUserId="2153" LastEditDate="2014-09-11T01:34:21.900" LastActivityDate="2014-09-11T09:59:24.330" Title="If the speed at which the universe contracts in reverse time decreases over time, then how did scientists arrive at a date for &quot;The Big Bang&quot;?" Tags="&lt;big-bang-theory&gt;&lt;expansion&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6314" PostTypeId="1" AcceptedAnswerId="6315" CreationDate="2014-09-10T14:17:47.217" Score="2" ViewCount="12852" Body="&lt;p&gt;My friend and I both noticed that the Moon looked yellow Yesterday, but it wasn't like that the same time the day before, as far as I remember. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the Moon sometimes look Yellow? If it does, why does it?&lt;/p&gt;&#xA;" OwnerUserId="2390" LastActivityDate="2014-09-12T06:56:31.947" Title="Why does the Moon sometimes look Yellow?" Tags="&lt;the-moon&gt;&lt;naked-eye&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6315" PostTypeId="2" ParentId="6314" CreationDate="2014-09-10T14:24:06.723" Score="4" Body="&lt;p&gt;Because light with higher frequency, that is bluer colors, scatters more in Earth's atmosphere and celestial objects thus look redder (yellow is towards red) while the atmosphere looks blue. See &lt;a href=&quot;https://en.wikipedia.org/wiki/Rayleigh_scattering&quot; rel=&quot;nofollow&quot;&gt;Rayleigh scattering&lt;/a&gt;. &#xA;Here is video with a lengthy and somewhat entertaining explanation:&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=SRh75B5iotI&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=SRh75B5iotI&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2014-09-10T14:24:06.723" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6316" PostTypeId="1" CreationDate="2014-09-10T14:42:37.537" Score="8" ViewCount="233" Body="&lt;p&gt;Crosslisted question from &lt;a href=&quot;https://physics.stackexchange.com/questions/131876/sedna-vp113-and-the-likelihood-of-the-px-tyche-thelistos-hypotheses&quot;&gt;https://physics.stackexchange.com/questions/131876/sedna-vp113-and-the-likelihood-of-the-px-tyche-thelistos-hypotheses&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The recent discoveries in exoplanetary science (specially those findings of planets orbiting far far away from its parent star/stars) raise questions about how much we know about the (true, AIU definition) number of planets in our solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: does the discovery of VP113 and Sedna increase the odds of some kind of planetary object in the edge of the known solar system (Oort's cloud)? How could we find it if it is very cold and dark? I mean, how many Sedna or VP113 objects we would need in order to guess its position and location?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reading about this topic, I don't know how to quantify the odds of finding&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;PX(Planet X): An Earth sized body perturbing the orbits of Sedna, VP113 and other similar objects.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Thelistos: A super-Earth (about 5-10 Earth masses times a O(1) factor)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Tyche: (Now disfavoured by WISE data, I think it can not be excluded if it is extremely cold and &quot;dark&quot;) Gas giant (Neptune-like, Saturn-like, etc)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Nemesis: A &quot;dark star&quot;/brown dwarf object.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2391" LastEditorUserId="11264" LastEditDate="2018-08-23T07:57:30.787" LastActivityDate="2018-08-23T07:57:30.787" Title="Sedna, VP113 and the likelihood of the PX/Tyche/Thelistos hypotheses" Tags="&lt;exoplanet&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="6317" PostTypeId="1" AcceptedAnswerId="6322" CreationDate="2014-09-10T15:03:43.443" Score="16" ViewCount="904" Body="&lt;p&gt;I think that the answer to this question will be useful as a scale to appreciate the size of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &lt;strong&gt;If we count Avogadro's Number of stars that are closest to Earth, how big that space would be?&lt;/strong&gt; And how many galaxies/clusters will be engulfed in that space? And for a start, let's just consider only three dimensions.&lt;/p&gt;&#xA;" OwnerUserId="2324" LastEditorUserId="2324" LastEditDate="2016-09-30T08:03:14.073" LastActivityDate="2016-09-30T08:03:14.073" Title="If we count Avogadro's Number of stars that are closest to Earth, how big that space would be?" Tags="&lt;star&gt;&lt;distances&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6318" PostTypeId="1" AcceptedAnswerId="7926" CreationDate="2014-09-10T16:09:46.343" Score="7" ViewCount="279" Body="&lt;p&gt;I've heard that the actual density of a black hole (within the volume of the event horizon) is pretty low - does that mean that any sufficiently large volume of mass over that density is also a black hole? Or does the actual concentration of mass within the event horizon matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forked from &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6305/is-it-possible-that-a-ultra-large-portion-of-the-space-we-live-in-is-already-ins&quot;&gt;&quot;Is it possible that a ultra-large portion of the space we live in is already inside a black hole? How could we refute this?&quot;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-11-09T23:57:08.580" Title="Assuming a sufficient amount of mass above the density threshold, does the actual concentration of the mass matter in creating a black hole?" Tags="&lt;black-hole&gt;&lt;astrophysics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6319" PostTypeId="2" ParentId="6318" CreationDate="2014-09-10T17:04:56.070" Score="1" Body="&lt;p&gt;&quot;Concentration of matter&quot; and &quot;density&quot; are equivalent terms, i.e. density is measured in mass per unit volume, so yes, it is exactly the concentration of matter within a given volume that determines whether or not an object will form a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any body has a so-called Schwarzchild radius - the radius within which, should all its matter become concentrated, the body will form a black hole. AFAIK Black Holes do not have a low density of matter within the black hole, in fact the density there is for all intents and purposes infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further to this, any reliable description of the environment within a black hole is, at best, educated conjecture. Bar Hawking Radiation and some other possible forms of minor 'bleeding', black holes give up no information about their contents. The only indication we have is of their mass (due to their effect on spacetime) which indicates (as far as observations are concerned) that they are very dense indeed.&lt;/p&gt;&#xA;" OwnerUserId="2209" LastActivityDate="2014-09-10T17:04:56.070" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6320" PostTypeId="2" ParentId="6292" CreationDate="2014-09-10T17:14:44.627" Score="2" Body="&lt;p&gt;If we look at M31 (the Andromeda galaxy) and consider that it has most likely had a similar history to our own (similar size, age and make-up) then I would say that it is a galaxy which initially formed by itself and later collected other galaxies. M31 has a distinct distortion on its inner region, like a kink. This is quite possibly the remnant of an ealier interaction with another galaxy. Some evidence has been proposed that this interaction was with the Milky Way, although this remains unverified.&lt;/p&gt;&#xA;" OwnerUserId="2209" LastActivityDate="2014-09-10T17:14:44.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6322" PostTypeId="2" ParentId="6317" CreationDate="2014-09-10T18:33:55.700" Score="22" Body="&lt;p&gt;It's going to be all (or perhaps &lt;em&gt;nearly&lt;/em&gt; all) of the observable Universe. Roughly speaking, there are &lt;a href=&quot;http://science.howstuffworks.com/milky-way5.htm&quot;&gt;several hundred billion stars&lt;/a&gt; in the Milky Way. And extrapolating the number of galaxies in deep Hubble images suggests &lt;a href=&quot;http://www.space.com/25303-how-many-galaxies-are-in-the-universe.html&quot;&gt;something like&lt;/a&gt; a hundred billion galaxies in the observable Universe.  Put these together and &lt;a href=&quot;http://www.esa.int/Our_Activities/Space_Science/Herschel/How_many_stars_are_there_in_the_Universe&quot;&gt;you get about&lt;/a&gt; $10^{22}$ to $10^{24}$ stars in the observable Universe.  Avogadro's number is $6.022\times 10^{23}$, which is in the upper end of that range.  So, given current estimates, you'd need at least most, if not all, of the observable Universe to get $N_A$ stars.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastEditorUserId="1322" LastEditDate="2014-09-12T01:02:35.963" LastActivityDate="2014-09-12T01:02:35.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6324" PostTypeId="1" CreationDate="2014-09-11T03:51:38.653" Score="5" ViewCount="447" Body="&lt;p&gt;I want to make some observations with a telescope in Hawaii on Mauna Kea but I am living somewhere else (not Hawaii) and I want to plan a night's viewing on the Hawaiian telescope. How do I find the RA range of the night? In other words; how do I find the RA range of the first object observable to the latest object observable for a particular time (of year) for a specific location (in this case Hawaii)?&lt;/p&gt;&#xA;" OwnerUserId="2396" LastActivityDate="2018-05-12T03:53:25.033" Title="How do I find the RA of sunset and sunrise in a specific location?" Tags="&lt;telescope&gt;&lt;observational-astronomy&gt;&lt;coordinate&gt;&lt;ascension&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6325" PostTypeId="2" ParentId="6324" CreationDate="2014-09-11T06:28:10.713" Score="1" Body="&lt;p&gt;Use &lt;strong&gt;Stellarium&lt;/strong&gt;. You can set your location and date as needed and simulate all the conditions (ignoring weather...) :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.stellarium.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.stellarium.org/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1678" LastActivityDate="2014-09-11T06:28:10.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6326" PostTypeId="2" ParentId="6313" CreationDate="2014-09-11T09:59:24.330" Score="3" Body="&lt;p&gt;Under the assumptions of large-scale homogeneity and isotropy, cosmic evolution can be described by the &lt;em&gt;scale factor&lt;/em&gt; $a(t)$ that describes by what ratio the large-scale structure of the universe has expanded or contracted. The scale factor is unitless, and by convention taken to be $1$ in the present epoch, and the rate of change of $a$ can be denoted $\dot{a}$, so the the universe is expanding whenever $\dot{a}&amp;gt;0$ and contracting whenever $\dot{a}&amp;lt;0$. The existence of a past singularity is then the claim that $a\to 0$ for some finite amount of time in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scale factor is related to the &lt;em&gt;Hubble parameter&lt;/em&gt; by $H \equiv \dot{a}/a$. The Hubble parameter is very frequently and erroneously also called the &lt;em&gt;Hubble constant&lt;/em&gt;, which is probably the source of your confusion. It is true that if $H$ is taken to be constant, then trivially $a = a_0e^{Ht}$ is the only possible solution, which only asymptotically tends to $0$ in the infinite past. If spatially flat, as our universe is, this is the &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/De_Sitter_universe&quot; rel=&quot;nofollow&quot;&gt;de Sitter solution&lt;/a&gt;&lt;/em&gt; describing a universe dominated by a cosmological constant and negligible matter or radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that's simply not our universe. In the past, matter was dominant (and prior to that, radiation was dominant), and the de Sitter model simply isn't applicable. Simply put, the Hubble constant is not constant, and so should properly be called Hubble parameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the evolution of the scale factor or the Hubble parameter is described by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;nofollow&quot;&gt;Friedmann equations&lt;/a&gt;:&#xA;$$\begin{eqnarray*}&#xA;H^2\equiv\left(\frac{\dot{a}}{a}\right)^2 &amp;amp;=&amp;amp; \frac{8\pi G}{3}\rho - \frac{kc^2}{a^2}\text{,}\\&#xA;\dot{H}+H^2\equiv\frac{\ddot{a}}{a} &amp;amp;=&amp;amp; -\frac{4\pi G}{3}\left(\rho+\frac{3p}{c^2}\right)\text{,}&#xA;\end{eqnarray*}$$&#xA;where $\ddot{a}$ denotes the second time-derivative of $a$, the cosmological constant (dark energy) is included in the density $\rho$ and pressure $p$ of the universe, and for our spatially flat universe, $k=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also wish to look up the &lt;a href=&quot;http://en.wikipedia.org/wiki/Deceleration_parameter&quot; rel=&quot;nofollow&quot;&gt;deceleration parameter&lt;/a&gt; $q\equiv -\ddot{a}a/\dot{a}^2$ (so defined because the universe was once thought to be decelerating). But the main point is that $H$ is in fact not a constant, so it is consistent for the scale factor to become $0$ a finite amount of time in the past. How much time depends on the changing density $\rho$ and pressure $p$, which is studied in detail in the standard &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambda-CDM_model&quot; rel=&quot;nofollow&quot;&gt;ΛCDM&lt;/a&gt; model.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-09-11T09:59:24.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6327" PostTypeId="2" ParentId="6280" CreationDate="2014-09-11T14:57:09.240" Score="5" Body="&lt;p&gt;For the practical side of TidalWave's excellent answer, here's a moon almanac for the place:&#xA;&lt;a href=&quot;http://www.timeanddate.com/moon/norway/longyearbyen&quot; rel=&quot;noreferrer&quot;&gt;http://www.timeanddate.com/moon/norway/longyearbyen&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 2014 the moon is above the horizon for a maximum of about 9 days at a time.&lt;/p&gt;&#xA;" OwnerUserId="2362" LastActivityDate="2014-09-11T14:57:09.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6328" PostTypeId="1" CreationDate="2014-09-11T15:12:00.033" Score="7" ViewCount="8443" Body="&lt;p&gt;How many now living individuals have a PhD in astrophysics, or by some other definition can be reasonably classified as astronomers? I think Richard Feynman once said that only one in a million is an astronomer, I would think there are many more today.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2018-12-04T17:12:58.337" Title="How many astronomers are there in the world today?" Tags="&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6329" PostTypeId="2" ParentId="6328" CreationDate="2014-09-11T15:52:36.413" Score="6" Body="&lt;p&gt;&quot;The worldwide community of professional astronomers is only about 10,000; most are located in the us (with about 1,000 in the UK and 250 in Australia).&quot; From So You Want to Be an Astronomer by Duncan Forbes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another source stated that the number of professional astronomers is about the size of a small town.&lt;/p&gt;&#xA;" OwnerUserId="1356" LastActivityDate="2014-09-11T15:52:36.413" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6330" PostTypeId="2" ParentId="6292" CreationDate="2014-09-11T18:48:51.040" Score="3" Body="&lt;p&gt;The subject of galaxy (including the Milky Way) formation is still hotly debated by the professionals. But is is quite clear that the process is a hierchical mess. The standard picture involves cold dark matter (CDM), which first forms structures (sheets, filaments, haloes) into which the baryons fall. Since the baryons can dissipate and cool by radiation), some of them accumulate at the centre of the haloes, where they form a rotating disc (since angular momentum cannot be radiated away). Stars form (roughly) whenever the gas density is high enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This all is happening on various scales. First, smaller galaxies form, but mergers quickly lead to the formation of larger ones. Most stars in galactic discs actually formed in situ (rather than in a progenitor galaxies). Accreted stars (from merged progenitors) are typically not in a disc-structure, but form a spheroidal structure (stellar halo, bulge).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A major merger (mass ratios 1:3 to 1:1) destroys the stellar discs and leads to the formation of an elliptial galaxy. The Milky Way galaxy cannot have had a merger with mass ratio 1:1 to 1:4 in the last few Gyr, and even a 1:10 merger should have left some detectable traces.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-09-11T18:48:51.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6331" PostTypeId="2" ParentId="6309" CreationDate="2014-09-11T19:16:47.150" Score="4" Body="&lt;p&gt;There is &lt;strong&gt;no gravitational phenomenon associated with &lt;em&gt;exact&lt;/em&gt; alignment&lt;/strong&gt;. (moreover, exact alignment never happens, there is always a small deviation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The directions of tidal forces are aligned and hence sum-up for alignment, but this maximum is quite broad and nothing spectacular happens near exact alignment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need close to exact alignment (depending on the sizes of the objects) to get an &lt;strong&gt;eclipse&lt;/strong&gt; of one sort or another (a Solar complete eclipse is quite spectacular). This presumably requires the closest to exact alignment of all possible effects, but is not gravitational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;added in edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the planetary orbits were &lt;em&gt;exactly&lt;/em&gt; co-planar, then &lt;em&gt;exact&lt;/em&gt; alignment of the three bodies would occur regularly. However, the &lt;em&gt;chance&lt;/em&gt; for the planetary orbits to be &lt;em&gt;exactly&lt;/em&gt; (to infinite precision) co-planar is zero. This is just simple probability of continuous (as opposed to discrete) random variables: The &lt;em&gt;probability density&lt;/em&gt; $p(x)$ for a continuous variable $x$ is finite, hence finding a value between $x$ and $x+\delta x$ has probability $p(x)\delta x$, which goes to zero as $\delta x\to0$. The chance of hitting &lt;em&gt;close to exact&lt;/em&gt; alignment is the smaller the closer you want it to be.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2014-09-12T09:01:50.020" LastActivityDate="2014-09-12T09:01:50.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6332" PostTypeId="2" ParentId="6314" CreationDate="2014-09-12T06:56:31.947" Score="3" Body="&lt;p&gt;Good previous answer, just to add.  The Moon can also look shades of yellow and red when there is a high amount of pollution in the air.  I live in Australia and the moon can look Yellow, orange and red when the bush fire season is about us and there is a lot of smoke in the air.  I have also noticed in cities like New York on windless days that the moon can look Yellow there too&lt;/p&gt;&#xA;" OwnerUserId="2155" LastActivityDate="2014-09-12T06:56:31.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6333" PostTypeId="1" CreationDate="2014-09-12T07:51:16.743" Score="6" ViewCount="3251" Body="&lt;p&gt;The sun dominates in our solar system.&lt;br/&gt;&#xA;I wonder whether &lt;strong&gt;in every aspect&lt;/strong&gt; the sun plays the most important role in our system.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, all kinds of metals are mainly located in the sun instead of the rest combined together in our system? &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any complementary relation between the sun and its children (all the planets, asteroids etc..)? &lt;br/&gt; &#xA;I mean: for one kind of metal, if the sun has more, her children should have less.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.periodictable.com/Properties/A/SolarAbundance.html&quot; rel=&quot;nofollow&quot;&gt;Abundance in the Sun of the elements&lt;/a&gt;, at least Li and Tb abundance of the sun are very low ($\approx1e-8$). Furthermore, these abundances are from the solar atmosphere which is near the surface of the sun.&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorDisplayName="user1569" LastEditDate="2016-09-29T11:51:12.707" LastActivityDate="2017-12-30T22:07:27.037" Title="Metal distribution in our solar system" Tags="&lt;the-sun&gt;&lt;metal&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6334" PostTypeId="2" ParentId="6333" CreationDate="2014-09-12T08:49:17.513" Score="-3" Body="&lt;p&gt;The metals that are forming in the sun come from nuclear fusion. Everything that now exists in the sun's body came from hydrogen, which due to high temperatures is being fused into helium and helium to lithium and so on. Once hydrogen is over, that means the start of the sun's death, where the sun will shrink (meaning temperature increase) and will start expanding again and the helium will continue to be used as a fuel and there is gonna be a point were all hydrogen and helium from the core will go, due to fusion, and the sun is gonna be cool so no further fusion will occur. Eventually what will be left is a planetary nebula with a white dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am answering to your second question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the beginning of creation the planets (small dust grains) where revolving around the protostar sun. Gravitational force made dust particles to come together. The inner system which includes planets up to Mars, was very hot and molecules like water could not condense so the metals and the silicates dominated and formed the inner planets. The outer planet area (after Mara) was cool enough so the water and other elements could be in a solid state. Now, about the asteroids, these are rock leftovers that did not form into planets. Asteroids mostly found in the so called &quot;asteroid belts&quot;, which is a region between Mars and Jupiter.&lt;/p&gt;&#xA;" OwnerUserId="2397" LastActivityDate="2014-09-12T08:49:17.513" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6335" PostTypeId="2" ParentId="6333" CreationDate="2014-09-12T09:05:59.137" Score="14" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;The metal content of the Solar system is completely dominated by the Sun&lt;/strong&gt;. The Sun contains $\sim1\%$ of 'metals' (in astronomical language anything but hydrogen and helium is a 'metal'), but all the other bodies of the Solar system combined have less mass than that. So even if they were only made of metals (but the outer planets are mostly made of H and He) the Sun would still dominate the metal budget.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;The Sun does not dominate the angular momentum of the Solar system&lt;/strong&gt;, which is dominated by the orbital angular momentum of Jupiter. The angular momentum due to the spin of the Sun is rather modest. This is easy to estimate: the orbital angular momentum of a planet is $\sim m\sqrt{GM_\odot a}$ which increases with semi-major axis $a$, while the spin of the Sun contributes $\beta \omega M_\odot R_\odot^2$ with $\omega$ the Solar spin frequency, $R_\odot$ the Solar radius, and $\beta\sim0.1$ a factor depending on the inner structure of the Sun. The ratio of the latter to the former is&#xA;$$&#xA;\beta \frac{M_\odot}{m} \frac{R_\odot^2}{a^2} \frac{\omega}{\Omega}&#xA;\sim0.01&#xA;$$&#xA;For Jupiter, the first factor is $\sim10^3$, the second $\sim10^{-6}$, and the third $\sim180$ is the ratio of the Solar spin frequency (once in 25 days) to the orbital frequency $\Omega=\sqrt{GM_\odot/a^3}$ of Jupiter (once in 12 years).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2014-09-12T09:40:59.967" LastActivityDate="2014-09-12T09:40:59.967" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6336" PostTypeId="1" AcceptedAnswerId="6338" CreationDate="2014-09-12T14:33:36.400" Score="5" ViewCount="245" Body="&lt;p&gt;I am currently living in Bangkok and I love astronomy. However, I haven't been able to find an official astronomy exam that I can do to gain college credit. Could someone help me out please?&lt;/p&gt;&#xA;" OwnerUserId="2404" LastEditorUserId="616" LastEditDate="2020-11-27T15:32:08.580" LastActivityDate="2020-11-28T01:03:34.983" Title="Is there an astronomy exam I can take?" Tags="&lt;education&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6337" PostTypeId="1" AcceptedAnswerId="6382" CreationDate="2014-09-12T14:41:28.850" Score="2" ViewCount="837" Body="&lt;p&gt;I am looking for these... functions. I want to be able to say, &quot;at a 1873. second there were around 54230000000 K temperature and 0.00435 kg/m^3 density&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the net I could find only tables or diagrams, as this: &lt;img src=&quot;https://i.stack.imgur.com/JJF6m.jpg&quot; alt=&quot;here&quot;&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Here is a link to the page it accompanies: &lt;a href=&quot;http://www.astro.ucla.edu/~wright/BBhistory&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.astro.ucla.edu/~wright/BBhistory&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am practically interested in the interval between the first seconds and the first some ten millions of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible? Some simple formula were also enough (although I think, there are different formulas for the different eras after the Big Bang).&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="7472" LastEditDate="2018-12-24T15:49:57.203" LastActivityDate="2018-12-24T15:49:57.203" Title="Time-density and time-temperature functions of the Universe since the Big Bang" Tags="&lt;early-universe&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="6338" PostTypeId="2" ParentId="6336" CreationDate="2014-09-12T17:08:44.367" Score="3" Body="&lt;p&gt;There is a &lt;a href=&quot;http://getcollegecredit.com/&quot; rel=&quot;nofollow&quot;&gt;Dantes Standardized Subjet Test&lt;/a&gt; (DSST) in &lt;a href=&quot;http://getcollegecredit.com/assets/pdf/dsst_fact_sheets/DSST_Astronomy.pdf&quot; rel=&quot;nofollow&quot;&gt;astronomy&lt;/a&gt;. I have no idea if and how widely colleges give credit for these exams. Typically, you will start with physics and choose to specialize in astronomy. Therefore, most standardized exams are for physics which may have a handful of astronomy questions.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-09-12T17:08:44.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6340" PostTypeId="2" ParentId="6257" CreationDate="2014-09-12T19:46:06.873" Score="3" Body="&lt;p&gt;You're correct that &quot;$p$-wave&quot; in this context means that the charmonium has orbital angular momentum $L=1$. The principal observable effect is that two-particle states with even $L$ have even &lt;a href=&quot;https://en.wikipedia.org/wiki/Parity_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;parity&lt;/a&gt;, while those with odd $L$ change sign under parity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to skim the &lt;a href=&quot;http://pdg.lbl.gov/2014/tables/rpp2014-tab-mesons-c-cbar.pdf&quot; rel=&quot;nofollow&quot;&gt;Particle Data Group's list of charmonia&lt;/a&gt; (or the $c\bar c$ section of &lt;a href=&quot;http://pdg.lbl.gov/2014/tables/rpp2014-qtab-mesons.pdf&quot; rel=&quot;nofollow&quot;&gt;the short list&lt;/a&gt;) you would find that each particle has a value listed for spin, $J$, and parity, $P$. The lightest few states are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\eta_c(1S)$, with $J^P = 0^-$. This is a charm-anticharm bound state where the two quarks have opposite spins and the angular momentum $L=0$. (The total parity is negative because the intrinsic parities of the $c$ and $\bar c$ have opposite sign.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$J/\psi(1S)$, with $J^P=1^-$. This has the same parity as the spinless $\eta_c$, but a unit of spin. So it must be that here the two quarks have the same spin and $L=0$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\chi_{c0}(1P)$, with $J^P=0^+$. The change in parity suggests that this state must have a different $L$ than the $\eta_c$ and $J/\psi$, and it's the first with a $P$ in its name. There's also a $\chi_{c1}$ and $\chi_{c2}$ with $J^P=1^+, 2^+$, respectively. That suggests these are the three spin combinations of a charm-anticharm pair with spin $S=1$ and orbital angular momentum $L=1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are also some charmonia with $(2S)$ and $(2P)$ in their names. There don't appear to be any definitively labeled $d$-wave charmonia.&lt;/p&gt;&#xA;" OwnerUserId="1498" LastActivityDate="2014-09-12T19:46:06.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6341" PostTypeId="1" CreationDate="2014-09-13T04:38:59.243" Score="3" ViewCount="6243" Body="&lt;p&gt;I'm interested in observing the moon and planets, and maybe some nebulae and star clusters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a pair of binoculars:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Nikon OceanPro 7x50 binoculars (around \$300)&lt;/li&gt;&#xA;&lt;li&gt;Celestron 20x80 binoculars (around $100)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm looking at a step up, to 100mm. Specifically, I'm trying to choose between  25x100 binoculars (either a &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B00008Y0VU&quot; rel=&quot;nofollow&quot;&gt;Celestron&lt;/a&gt; or an &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B00066ELB2&quot; rel=&quot;nofollow&quot;&gt;Orion&lt;/a&gt;) and a 100mm &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B0002862UW&quot; rel=&quot;nofollow&quot;&gt;spotting scope&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only reason I'm considering a spotting scope is that it has a magnification of 22-66x, compared to the binoculars' 25x. But I want to check if such high magnifications really work for astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I'm observing the full moon. What's the highest usable magnification, without resulting in too dim an image?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An Amazon &lt;a href=&quot;http://www.amazon.com/review/R906C4U1258CK/ref=cm_cr_pr_perm?ie=UTF8&amp;amp;ASIN=B0002862UW&quot; rel=&quot;nofollow&quot;&gt;review&lt;/a&gt; says that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have found that while the readability improves from 22x to about 38x&#xA;  the resolution actually decreases after that and I never use it from&#xA;  40-66x, there is no point. Maybe it would have been better to have&#xA;  optimised it for 22-40x.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the scope is limited to 40x in daytime use, I expect it to be worse at night, even observing the full moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm considering buying the scope only for the increased magnification over the binoculars, but if I end up using the scope only at 22x, it defeats the point of buying the scope in the first place. Should I stick with the binoculars?&lt;/p&gt;&#xA;" OwnerUserId="962" LastActivityDate="2014-09-13T12:10:49.143" Title="Is a spotting scope or binoculars a better choice for astronomy?" Tags="&lt;the-moon&gt;&lt;planet&gt;&lt;binoculars&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6342" PostTypeId="2" ParentId="6341" CreationDate="2014-09-13T12:10:49.143" Score="1" Body="&lt;p&gt;The main point is that it all depends on the optics quality. Nikon is a good brand, but $300  for a 50mm binocular is way too much. Remember the golden rule while buying any kind of telescope or binocular: Aperture matters, not the Magnification. In a binocular, say &quot;7x50&quot; models, you get 7x magnification with 50mm aperture, which restricts how far you can see. But the Celestron 80mm scope with a 20x magnification is perfectly fine. It would be advisable if you can get hold of a good tripod, get it. The beauty of the heavens is unimaginable; they shall unfurl as you explore. A 80mm scope will let you see even Dark Sky Objects like Nebulas and Distant Galaxies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a nutshell, it would be best for you at this stage to grab the Celestron 20x80 scope, which is both better and less costly. &lt;/p&gt;&#xA;" OwnerUserId="2409" LastActivityDate="2014-09-13T12:10:49.143" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6343" PostTypeId="1" AcceptedAnswerId="6349" CreationDate="2014-09-13T20:57:16.573" Score="8" ViewCount="2245" Body="&lt;p&gt;I found this image on calculations of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;noreferrer&quot;&gt;Hill sphere&lt;/a&gt; for planets/dwarf planets of the Solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XE4mL.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA; From &lt;a href=&quot;http://en.wikipedia.org/wiki/File:Hill_sphere_of_the_planets.png&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/File:Hill_sphere_of_the_planets.png&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found it interesting that variation of Hill sphere is intuitive for the first five planets, as the variation is similar to variation of mass/radius of those planets. Mercury has the smallest Hill sphere, Venus/Earth/Mars quite similar, and a giant leap from Mars to Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, Saturn has its Hill sphere bigger than Jupiter, even it is smaller than Jupiter. and this anomaly continues to Uranus and Neptune: They have progressively larger Hill spheres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And Hill spheres of Pluto and Eris are quite larger than Mercury, Venus, Earth, and Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was quite surprising for me. Could someone explain why this—for the lack of a better word—&lt;em&gt;anomalies&lt;/em&gt; are there?&lt;/p&gt;&#xA;" OwnerUserId="2324" LastEditorUserId="2324" LastEditDate="2014-09-15T01:20:46.507" LastActivityDate="2014-09-19T03:00:00.467" Title="Can you explain the pattern of Hill sphere sizes of the objects of the Solar system?" Tags="&lt;orbit&gt;&lt;gravity&gt;&lt;natural-satellites&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6344" PostTypeId="1" CreationDate="2014-09-13T21:17:15.753" Score="3" ViewCount="433" Body="&lt;p&gt;How can we know how does the Milky Way look like if we are in it?&#xA;Sorry if the answer is evident, I am not an expert.&lt;/p&gt;&#xA;" OwnerUserId="2411" LastEditorUserId="2153" LastEditDate="2014-09-13T21:35:23.500" LastActivityDate="2014-09-29T17:04:57.913" Title="Milky way: How do we know its appearance?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;" AnswerCount="3" CommentCount="2" ClosedDate="2014-09-29T18:46:12.240" ContentLicense="CC BY-SA 3.0" />
  <row Id="6345" PostTypeId="2" ParentId="6343" CreationDate="2014-09-13T21:19:40.443" Score="2" Body="&lt;p&gt;I take it there isn't exactly a coincidence between you finding my question regarding geostationary orbits and you asking this question about Hill Spheres? :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chart you found does seem counter-intuitive at first. But consider this chart:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/en/timeline/5fb1322f537f8a55d85170976c150191.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://upload.wikimedia.org/wikipedia/en/timeline/5fb1322f537f8a55d85170976c150191.png&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/iOWnv.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I wish I could add it here, but I can't manage to add it as an image, just as a link).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's another pattern, and it has to do with distances from the Sun, and from nearby bodies. As you go out into the outer solar system, planets start to get farther apart. For example, Uranus is twice as far away as Saturn, separated by 10 AU at their closest, and Neptune is, at the closest position to Neptune, 10 AU away. That means that each planet is separated by a large margin from other planets, and Uranus and Neptune have virtually nothing else to contend with in the outer solar system, because they are so far away from any other massive bodies that could take over (i.e. Jupiter and/or Saturn).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pluto, Ceres, and Eris are interesting cases. As far as I can tell, they have large Hill spheres because they are the largest of a collection of similar bodies. Ceres dominates the asteroid belt, and Pluto is so big it was once (in what now seems like ancient times) considered to be a planet. Eris, too, is quite large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only anomaly here is, actually, Pluto - and that's only for a [relatively] short portion of a time. It comes closer towards the Sun than Neptune for a portion of its orbit, which would seem to indicate that Neptune shortens Pluto's Hill sphere, but in reality, the two are rarely near where their orbits intersect at the same time.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="1030" LastEditDate="2014-09-14T01:52:39.740" LastActivityDate="2014-09-14T01:52:39.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6346" PostTypeId="1" AcceptedAnswerId="6369" CreationDate="2014-09-13T22:47:41.407" Score="8" ViewCount="1651" Body="&lt;p&gt;I hope to observe planets like Jupiter and Saturn, and moons. I am a novice when it comes to telescopes and I know images in magazines are not taken with at least 12-inch telescope in a middle of nowhere. Is it possible to view Saturn in little yellowish and Mars in little reddish using following telescopes?  I am going to buy one of them. &lt;strong&gt;Which one is worth more for the money with the price difference?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am living in Sri Lanka (a country close to the equator) in a suburban city.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Option 1&lt;/strong&gt;&#xA;EQ-Reflecting telescope&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Aperture: 135mm&lt;/li&gt;&#xA;&lt;li&gt;Focal length: 800mm Focal ratio: f/6 &lt;/li&gt;&#xA;&lt;li&gt;Max power: 265x when the air is steady&lt;/li&gt;&#xA;&lt;li&gt;Price: 300USD approx &lt;/li&gt;&#xA;&lt;li&gt;Link: &lt;a href=&quot;http://www.ksonoptics.com:8081/Product/PrdDtls/121/&quot;&gt;http://www.ksonoptics.com:8081/Product/PrdDtls/121/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Option 2&lt;/strong&gt;&#xA;EQ-Reflecting telescope&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Aperture: 114mm&lt;/li&gt;&#xA;&lt;li&gt;Focal length: 900mm Focal ratio: f/8 &lt;/li&gt;&#xA;&lt;li&gt;Max Power: 228x when the air is steady.&lt;/li&gt;&#xA;&lt;li&gt;Price: 220USD approx&lt;/li&gt;&#xA;&lt;li&gt;Link: &lt;a href=&quot;http://www.ksonoptics.com:8081/Product/PrdDtls/119/&quot;&gt;http://www.ksonoptics.com:8081/Product/PrdDtls/119/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2410" LastEditorUserId="2153" LastEditDate="2014-09-13T22:57:11.420" LastActivityDate="2017-07-15T21:49:17.190" Title="Selecting a Telescope for Viewing Planets" Tags="&lt;planet&gt;&lt;telescope&gt;&lt;mars&gt;&lt;jupiter&gt;&lt;saturn&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6347" PostTypeId="2" ParentId="6343" CreationDate="2014-09-13T23:27:55.030" Score="6" Body="&lt;p&gt;The definition of Hill sphere is the region where the given object's gravity is dominant. In this area the object's gravity pulls more strongly than anything else; and everything else combined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary competition for a planet is the sun. The further you get from the sun then the weaker its gravity is. This means it's easier for Neptune's gravity to exceed the sun's than it is for Jupiter's gravity. And it so happens that the masses are just right that Neptune has the larger Hill sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you stuck Jupiter further out in the solar system, then its Hill sphere would increase as a result.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2014-09-13T23:27:55.030" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6348" PostTypeId="1" AcceptedAnswerId="7693" CreationDate="2014-09-14T09:59:47.670" Score="14" ViewCount="3148" Body="&lt;p&gt;Wikipedia's definition of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hill_sphere&quot;&gt;Hill sphere&lt;/a&gt; is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An astronomical body's Hill sphere is the region in which it dominates the attraction of satellites. To be retained by a planet, a moon must have an orbit that lies within the planet's Hill sphere. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And then there is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hill_sphere&quot;&gt;sphere of influence&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A sphere of influence (SOI) in astrodynamics and astronomy is the oblate-spheroid-shaped region around a celestial body where the primary gravitational influence on an orbiting object is that body. This is usually used to describe the areas in the Solar System where planets dominate the orbits of surrounding objects (such as moons), despite the presence of the much more massive (but distant) Sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To me, a planets Hill sphere and its sphere of influence appear to be the same. But the formulas to calculate the two sphere are different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what &lt;em&gt;is&lt;/em&gt; exactly the difference between Hill sphere and the sphere of influence?&lt;/p&gt;&#xA;" OwnerUserId="2324" LastActivityDate="2014-10-15T08:50:14.533" Title="What is the difference between Sphere of Influence and Hill sphere?" Tags="&lt;orbit&gt;&lt;gravity&gt;&lt;satellite&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6349" PostTypeId="2" ParentId="6343" CreationDate="2014-09-14T10:42:09.273" Score="12" Body="&lt;p&gt;Hill sphere is the region of space around a satellite where the satellite wins the gravitational tug-of-war with its primary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the mass of the primary object is $M$, mass of the satellite is $m$, semi-major axis of satellite is $a$, and eccentricity of the orbit of the satellite is $e$, then the radius $r$ of the Hill sphere for satellite is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;r \approx a (1-e) \sqrt[3]{\frac{m}{3 M}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this formula does &lt;em&gt;not&lt;/em&gt; take into account the other objects in the  vicinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The anomaly pointed out in the question is not &lt;em&gt;really&lt;/em&gt; an anomaly. The contributing factor for the surprise values is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Semi-major_axis&quot; rel=&quot;noreferrer&quot;&gt;semi-major axis&lt;/a&gt; of the planets ($a$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take Jupiter and Saturn for example: Saturn has only around $30\%$ of the mass of Jupiter, and if the two gas giants had the same semi-major axis, this mass reduction will make Saturn's Hill sphere around $68\%$ than that of Jupiter. But Saturn is around $84\%$ farther from Sun than Jupiter. This is just enough to make Saturn's Hill sphere slightly larger than that of Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking along the same lines, we can also explain why Uranus, Neptune, Pluto, and Eris have surprisingly large Hill spheres.&lt;/p&gt;&#xA;" OwnerUserId="2324" LastEditorUserId="2324" LastEditDate="2014-09-19T03:00:00.467" LastActivityDate="2014-09-19T03:00:00.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6350" PostTypeId="2" ParentId="6298" CreationDate="2014-09-14T14:34:05.427" Score="2" Body="&lt;p&gt;I don't know if there have been any observations of objects leaving the observable universe, but I'll admit that I have a hard time keeping up with the latest discoveries in observational astronomy. I'll try to approach your question from a theoretical viewpoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you pointed out, the radius of the observable universe is $4.6 \times 10^{49}$ light-years. To figure out how fast an object is moving away from Earth (or any observer), we can use Hubble's law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = H_0D$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $v$ is the recessional velocity, $H_0$ is Hubble's constant, and $D$ is the proper distance from the observer to the object (see Wikipedia for the difference between proper distance and commoving distance). The trouble here is that the exact value of Hubble's constant isn't exactly well-known. There have been many different observations since Hubble proposed his law that attempted to find values for the constant, but they range widely. Here I'll use $82$ $km$ $s^{-1} /Mpc$ (where $Mpc$ is a megaparsec - 1 million parsecs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that the object in question is $13$ gigaparsecs away (so $13,000$ megaparsecs away). The light we see from the galaxy would be from when it was very young. One of the most distant observed galaxies is this far away, and only $2,000$ light-years across (note: do not confuse the &quot;light-travel distance&quot; with the &quot;proper distance&quot;), so let's say that this galaxy is smaller in size - negligibly small, in fact. We can then use Hubble's law to calculate a recessional velocity $v$ of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(82)(13,000) = 1.066 \times 10^6$$ kilometers per second, or $3.36 \times 10^{13}$ kilometers per year. That's pretty fast! An object $13$ gigaparsecs away would be $42.38$ billion light-years away, not far from the edge of the observable universe. Yet if the edge of the observable universe is $46$ billion light-years away, the galaxy would still be $3.62 \times 10^{21}$ kilometers away, and it would take a long time to get there. A galaxy this young would be maybe $1,000$ light-years across, so, moving at its current speed, it would take it many years to cross a distance its own length - that is, if the furthest edge of it were at the edge of the observable universe, it would still take many years to fully disappear. And that's even factoring in that it would be moving away faster!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I don't have the object being further away is that any objects at the edge of the observable universe would be so young when they released the light we see today. That means that galaxies would be small, and wouldn't emit a lot of light. One of the furthest objects we have detected is only $30$ billion light-years away, and $2,000$ light-years across. So it would be terribly hard to observe an object at the edge of the observable universe (ironic, isn't it?). But here I have shown that it would take years for the object to completely disappear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My sources for the data and conversion factors (i.e. light-years to kilometers):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble%27s_law&quot; rel=&quot;nofollow&quot;&gt;Hubble's Law&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://scienceworld.wolfram.com/physics/HubbleConstant.html&quot; rel=&quot;nofollow&quot;&gt;Wolfram Science World&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Observable_universe#Most_distant_objects&quot; rel=&quot;nofollow&quot;&gt;Most distant objects&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.spacetelescope.org/news/heic0404/&quot; rel=&quot;nofollow&quot;&gt;Spacetelescope.org&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Light-year&quot; rel=&quot;nofollow&quot;&gt;Light-year&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-01-10T22:59:00.300" LastActivityDate="2015-01-10T22:59:00.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6351" PostTypeId="1" AcceptedAnswerId="6366" CreationDate="2014-09-14T15:04:01.797" Score="5" ViewCount="129" Body="&lt;p&gt;Hot Jupiters are close to their parent star - after all, that's why they're so hot. Their atmospheres are thought to be slowly blown away by the parent star, and they will eventually become chthonian planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other effects on Hot Jupiters because of their proximity to their parent star? Examples I'm looking for include issues with solar flares or coronal mas ejections, or even difficulties relating to the solar wind. I would also be tangentially interested in any effects the parent star would have on moon formation (or lack thereof).&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-14T19:35:00.870" LastActivityDate="2014-09-14T21:16:50.977" Title="Conditions on Hot Jupiters" Tags="&lt;exoplanet&gt;&lt;solar-flare&gt;&lt;solar-wind&gt;&lt;coronal-mass-ejection&gt;&lt;hot-jupiter&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6352" PostTypeId="5" CreationDate="2014-09-14T19:09:09.277" Score="0" Body="&lt;p&gt;Space-time is a concept popularized in Albert Einstein's theory of special relativity and intimately woven into his later theory of general relativity. In our universe, it consists of three dimensions of space and one of time, and so scientists sometimes to the space-time we observe as 3+1 dimensional.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-19T15:07:23.920" LastActivityDate="2014-09-19T15:07:23.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6353" PostTypeId="4" CreationDate="2014-09-14T19:09:09.277" Score="0" Body="Questions regarding the 4-dimensional background structure of our universe." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-16T15:34:33.913" LastActivityDate="2014-09-16T15:34:33.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6354" PostTypeId="5" CreationDate="2014-09-14T19:12:22.977" Score="0" Body="&lt;p&gt;Dark energy is the mysterious cause of the acceleration expansion of the universe. It makes up around three quarters of the total mass and energy of the universe, yet its properties and fundamental nature are unknown. Not to be confused with dark matter.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-24T15:13:01.037" LastActivityDate="2014-09-24T15:13:01.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6355" PostTypeId="4" CreationDate="2014-09-14T19:12:22.977" Score="0" Body="Questions regarding the as-yet-unknown energy permeating the universe that is responsible for its expansion." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-23T13:36:08.967" LastActivityDate="2014-09-23T13:36:08.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6356" PostTypeId="5" CreationDate="2014-09-14T19:27:27.137" Score="0" Body="&lt;p&gt;Neutrinos are described in the Standard Model of particle physics. They are leptons,but carry no electric charge. The existence of neutrinos was first predicted to account for missing energy in neutron decay. Neutrinos are extremely low-mass, and there is an ongoing debate in the scientific community as to whether they have any mass at all. They were also notoriously publicized by the supposed &quot;Faster-than-light&quot; experiment by the OPERA team that was later shown to have incorrect results.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-23T13:36:20.107" LastActivityDate="2014-09-23T13:36:20.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6357" PostTypeId="4" CreationDate="2014-09-14T19:27:27.137" Score="0" Body="Questions regarding a family of extremely low-mass, electrically neutral particles described in the Standard Model of particle physics." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-23T13:36:26.670" LastActivityDate="2014-09-23T13:36:26.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6358" PostTypeId="5" CreationDate="2014-09-14T19:30:35.337" Score="0" Body="&lt;p&gt;The Oort Cloud is a region of space outside the Kuiper Belt. Named after Jan Oort, it is filled with long-period comets and other icy bodies.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-19T15:05:55.053" LastActivityDate="2014-09-19T15:05:55.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6359" PostTypeId="4" CreationDate="2014-09-14T19:30:35.337" Score="0" Body="Questions regarding the large &quot;cloud&quot; of small bodies that forms a sphere around the solar system." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-23T13:36:34.533" LastActivityDate="2014-09-23T13:36:34.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6360" PostTypeId="5" CreationDate="2014-09-14T19:33:42.497" Score="0" Body="&lt;p&gt;Hydrogen is the most common element in the universe, and is largely contained in stars and gas clouds. It is used a fuel for nuclear fusion inside stars, turning into helium and releasing energy. In its simplest form, it consists of one proton and one electron. However, isotopes with more neutrons, such as deuterium and tritium, do exist.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-19T15:05:49.787" LastActivityDate="2014-09-19T15:05:49.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6361" PostTypeId="4" CreationDate="2014-09-14T19:33:42.497" Score="0" Body="Questions about the simplest and most abundant element in the Universe." OwnerDisplayName="user2449" LastEditorDisplayName="user2449" LastEditDate="2014-11-03T12:45:10.963" LastActivityDate="2014-11-03T12:45:10.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6362" PostTypeId="5" CreationDate="2014-09-14T19:38:31.733" Score="0" Body="&lt;p&gt;The electron is a lepton, with an electric charge of -1. It is an elementary particle described by the Standard Model of particle physics. Electrons are constituents of atoms, existing in various orbitals; the removal of one or more electrons turns a neutral atom into an ion.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-23T13:35:25.660" LastActivityDate="2014-09-23T13:35:25.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6363" PostTypeId="4" CreationDate="2014-09-14T19:38:31.733" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-09-14T19:38:31.733" LastActivityDate="2014-09-14T19:38:31.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6364" PostTypeId="5" CreationDate="2014-09-14T19:45:25.613" Score="0" Body="&lt;p&gt;The National Aeronautics and Space Administration, frequently abbreviated NASA, is the American civilian space agency. Formed from the National Advisory Committee for Aeronautics (NACA) in 1958, it has undergone projects such as the Apollo program, the space shuttles, and the American contributions to the International Space Station.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-23T13:35:36.843" LastActivityDate="2014-09-23T13:35:36.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6365" PostTypeId="4" CreationDate="2014-09-14T19:45:25.613" Score="0" Body="Questions regarding the National Aeronautics and Space Administration, the American civilian space agency." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-23T13:35:51.240" LastActivityDate="2014-09-23T13:35:51.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6366" PostTypeId="2" ParentId="6351" CreationDate="2014-09-14T21:16:50.977" Score="2" Body="&lt;p&gt;I think we can broadly distinguish two classes of effects here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Flares directly affecting the planet&lt;/li&gt;&#xA;&lt;li&gt;The by-products of the flare affecting the planet&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I can loosely tell you what I know (although my knowledge here is far from being complete)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Direct effects include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Magnetic field geometries: Far-away from their host star (&gt;0.05AU) planets usually have isolated dipoles. The solar wind then must penetrate the magnetosphere of a planet to erode its atmosphere. This is however much easier for the wind if due to small distance both star and planet-magnetospheres connect. What this will lead to, is quasi-free flow of a significant portion of the stars total ejected mass towards the planet and huge energy input into the planet's upper atmosphere.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Indirect physics can be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An active young star can have frequent outbreaks of CMEs, those are accompanied by strong UV / X-ray emission and this will affect the planet:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Increased photochemistry: This can generate a very active photochemistry in the upper atmospheres. Doubly ionized molecules like $O_2^{++}$ or $N_2^{++}$ can then be produced more efficiently and eject their constituents into space upon dissociation due to high dissociation energies&lt;/li&gt;&#xA;&lt;li&gt;UV-X-heating: Additional heat through thermalized high-energy photons can add to the already happening 'natural' atmospheric erosion.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We don't know however, to which extent those effects play a role. I'd add tomorrow some references if you're interested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There could be signs however of such activity in young stellar systems, as verious combinations or extensions of the above have been proposed to explain the &lt;a href=&quot;http://adsabs.harvard.edu/abs/2011ApJ...729L...7L&quot; rel=&quot;nofollow&quot;&gt;radius anomaly&lt;/a&gt; in super-close extrasolar giant planets.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2014-09-14T21:16:50.977" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6367" PostTypeId="1" CreationDate="2014-09-15T02:58:11.623" Score="5" ViewCount="58" Body="&lt;p&gt;Not sure that this is the correct place to ask this question, but here goes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to find tools to accumulate image data (preferably in FITS format) on a given (RA, Dec)-coordinate of sky. Right now there are several image services that offer cutouts, for example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;irsa.ipac.caltech.edu holds data from Spitzer, Planck, WMAP, IRAS, WISE&lt;/li&gt;&#xA;&lt;li&gt;hla.stsci.edu holds Hubble Space Telescope data&lt;/li&gt;&#xA;&lt;li&gt;skyserver.sdss.org/dr7/en/tools/chart/list.asp holds SDSS data&lt;/li&gt;&#xA;&lt;li&gt;etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I wonder if anyone has written any code for accumulating these images given an argument of RA and Dec? Or if they can point me in the right direction of tools that they know exist? Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2417" LastActivityDate="2014-09-15T05:22:09.837" Title="Tools for creating a multiwavelength view of the sky" Tags="&lt;telescope&gt;&lt;space&gt;&lt;software&gt;&lt;observatory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6368" PostTypeId="2" ParentId="6367" CreationDate="2014-09-15T05:22:09.837" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://aladin.u-strasbg.fr/aladin.gml&quot; rel=&quot;nofollow&quot;&gt;Aladin&lt;/a&gt; is a good way to do visualization from multiple catalogs. MAST also offers a cross-mission &lt;a href=&quot;https://archive.stsci.edu/&quot; rel=&quot;nofollow&quot;&gt;search&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-09-15T05:22:09.837" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6369" PostTypeId="2" ParentId="6346" CreationDate="2014-09-15T07:16:24.883" Score="4" Body="&lt;p&gt;Go with Option 1 (135mm) as it has bigger light collecting area, i.e. diameter of primary mirror. It will enable you to see faint objects and it will also help you to see the dim objects under a greater zoom eyepiece, e.g. 4mm or 10mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, don't fool yourself with the bigger numbers advertized by telescope vendors like 238X zoom or 300X zoom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, these are zooming power using Barlow lenses also the more you zoom in on any object, the less light you gather on the mirror and hence objects will look much fainter in appearance. I will suggest the 10mm eyepiece with a 135mm telescope (Option 1) will be sufficient for you to observe planets like Jupiter, its Galilean satellites, Saturn's rings and also many Messier objects like Andromeda Galaxy M31, the Honey Bee Cluster, and open star clusters in the Auriga constellation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also bear in mind that you live in a country area closer to sea like here in India and the weather will be hot and humid. &#xA;So ask the vendor about special coating which protects the mirror from rusting (spots on mirror due to moisture) during the humid monsoon season as at least for 4 months it will be humid.&lt;/p&gt;&#xA;" OwnerUserId="2088" LastEditorUserId="7472" LastEditDate="2017-07-15T21:49:17.190" LastActivityDate="2017-07-15T21:49:17.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6370" PostTypeId="1" CreationDate="2014-09-15T15:44:35.087" Score="3" ViewCount="407" Body="&lt;p&gt;We have discovered some incredibly massive stars. &lt;a href=&quot;http://en.wikipedia.org/wiki/R136a1&quot; rel=&quot;nofollow&quot;&gt;R136a1&lt;/a&gt;, the most massive known star is estimated to have 265 times mass as our Sun. Yet it has been burning for at least a million years, and must have shed a considerable mass as stellar wind. In its infancy, R136a1 must have been much more massive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which got me thinking about the question: What is the theoretical limit of mass that a newborn star can have?&lt;/p&gt;&#xA;" OwnerUserId="2324" LastActivityDate="2014-09-16T01:52:47.023" Title="How massive can a star be at birth?" Tags="&lt;mass&gt;&lt;star-formation&gt;&lt;hypergiants&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2014-09-19T11:40:58.463" ContentLicense="CC BY-SA 3.0" />
  <row Id="6371" PostTypeId="5" CreationDate="2014-09-15T15:46:13.523" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-09-15T15:46:13.523" LastActivityDate="2014-09-15T15:46:13.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6372" PostTypeId="4" CreationDate="2014-09-15T15:46:13.523" Score="0" Body="A hypergiant (luminosity class 0 or Ia+) is a star with an enormous mass and luminosity, showing signs of a very high rate of mass loss." OwnerUserId="2324" LastEditorUserId="2324" LastEditDate="2014-09-23T13:36:01.827" LastActivityDate="2014-09-23T13:36:01.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6373" PostTypeId="2" ParentId="6370" CreationDate="2014-09-15T19:48:20.970" Score="1" Body="&lt;p&gt;I found a &lt;a href=&quot;http://ned.ipac.caltech.edu/level5/ESSAYS/Carr/carr.html&quot; rel=&quot;nofollow&quot;&gt;text&lt;/a&gt; that suggests a Population III star can have a mass between 200 and 10^5$M_{Sun}$. There you can read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;...This would apply for VMOs larger than $M_c$ $\approx$ 200 $M_{Sun}$. Stars larger than 150$M_{Sun}$ are termed supermassive objects (SMOs)...&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems supermassive stars like R136a1 (with masses in excess 150$M_{Sun}$, the accepted upper limit for a non-pop III star) can be formed by &lt;a href=&quot;http://www.astro.uni-bonn.de/de/start/neues/detail/n/2012/august/article/astronomers-crack-mystery-of-the-monster-stars-ras-pr-1254/&quot; rel=&quot;nofollow&quot;&gt;the collision of massive stars&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, It is not difficult to imagine, say, two 150$M_{Sun}$ monsters colliding and forming a 300$M_{Sun}$ object.&lt;/p&gt;&#xA;" OwnerUserId="910" LastEditorUserId="2324" LastEditDate="2014-09-16T01:52:47.023" LastActivityDate="2014-09-16T01:52:47.023" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6374" PostTypeId="2" ParentId="6210" CreationDate="2014-09-15T19:56:57.230" Score="3" Body="&lt;p&gt;Maybe a very old Neutron Star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A black dwarf would not emit any visible light, but the universe is not old enough for that.&#xA;Even the oldest and coolest white dwarfs still have a temperature between 2500-4000K (sorry for not remembering the reference for this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brown dwarfs (or planemos/sub-brown dwarfs) like WISE 0855–0714 could be as cool as ice.&#xA;but they don't count as stars anyway.&lt;/p&gt;&#xA;" OwnerUserId="910" LastActivityDate="2014-09-15T19:56:57.230" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6375" PostTypeId="2" ParentId="6274" CreationDate="2014-09-15T21:12:53.100" Score="25" Body="&lt;p&gt;The book &lt;em&gt;The Transits of Venus&lt;/em&gt;, by Sheehan and Westfall, describes how Aristarchus used Hipparchus' calculation of the Earth-Moon distance, who in turn used Eratosthenes' calculation of the Earth's circumference, to calculate the Earth-Sun distance.&lt;/p&gt;&#xA;&lt;p&gt;Aristarchus of Samos was the first to seriously calculate the distance to the Sun, using geometry.  When the Moon is exactly half illuminated when seen from the Earth (first or last quarter phase), then there is a right triangle between the Earth, Moon, and Sun, with the Moon at the right angle.  Then he could measure the angular distance in the sky between the Sun and the Moon, plus the Earth-Moon distance and geometry, to get the Earth-Sun distance.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The most famous ancient estimate of the earth's circumference as made by Eratosthenes of Cyrene (c. 276-196 BCE), the librarian at the great library at Alexandria.  By using a simple gnomon, he found that at Syene, ... the sun at the summer solstice cast no shadow at all: it was exactly overhead.  ... At the same moment, at Alexandria, the shadow cast by the sun shows that it stood 7.2 degrees from vertical.  This difference is equal to 1/50 of a circle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Using the distance between the cities, Earth's circumference could be calculated.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Once the earth's radius is known, the earth itself can be used as a baseline for determining still greater distances -- the distance to the moon.&lt;/p&gt;&#xA;&lt;p&gt;[I]t becomes possible to work out the earth-moon distance indirectly from the geometry of [lunar] eclipses.  Using this method, Hipparchus of Rhodes (fl. 140 BCE) worked out that the distance of the moon was 59 earth radii.  It's a good approximation - with 1 1/2 or 2 earth radii of the modern value.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Using the Earth-Moon distance and the separation of the Moon from the Sun in the sky when the Moon was at exactly half-phase, Aristarchus calculated the Earth-Sun distance.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Aristarchus put forward a geometrical argument, based on determining the sun-earth-moon angle at the time the moon's phase is exactly half.  For this angle, which is actually 89.86 degrees, Aristarchus used 87 degrees; the disagreement is more significant that it might appear because the critical quantity is the &lt;em&gt;difference&lt;/em&gt; between the angle and 90 degrees.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Because of this Aristarchus only got a value of the equivalent of &amp;quot;5 million miles&amp;quot;, much too small.&lt;/p&gt;&#xA;&lt;p&gt;Phil Plait has, on his old Bad Astronomy site, an article answering a question about &lt;a href=&quot;http://www.badastronomy.com/mad/1996/au.html&quot; rel=&quot;noreferrer&quot;&gt;how astronomers originally calculated the distance from the Earth to the Sun&lt;/a&gt; (the AU, or astronomical unit).&lt;/p&gt;&#xA;&lt;p&gt;Huygens was the first to calculate this distance with any kind of accuracy at all.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;So how did Huygens do it? He knew that Venus showed phases when viewed through a telescope, just like our own Moon does. He also knew that the actual phase of Venus depended on the angle it made with the Sun as seen from the Earth. When Venus is between the Earth and Sun, the far side is lit, and so we see Venus as being dark. When Venus is on the far side of the Sun from the Earth, we can see the entire half facing us as lit, and Venus looks like a full Moon. When Venus, the Sun and Earth form a right angle, Venus looks half lit, like a half Moon.&lt;/p&gt;&#xA;&lt;p&gt;Now, if you can measure any two internal angles in a triangle, and know the length of one of its sides, you can determine the length of another side. Since Huygens knew the Sun-Venus-Earth angle (from the phases), and he could directly measure the Sun-Earth-Venus angle (simply by measuring Venus' apparent distance from the Sun on the sky) all he needed was to know the distance from Earth to Venus. Then he could use some simple trigonometry to get the Earth-Sun distance.&lt;/p&gt;&#xA;&lt;p&gt;This is where Huygens tripped up. He knew that if you measured the apparent size of an object, and knew its true size, you could find the distance to that object. Huygens thought he knew the actual size of Venus using such unscientific techniques as numerology and mysticism. Using these methods he thought that Venus was the same size as the Earth. As it turn out, that is correct! Venus is indeed very close to being the same size as the Earth, but in this case he got it right by pure chance. But since he had the right number, he wound up getting the about the correct number for the AU.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Basically, Huygens used good methods, except for using &amp;quot;numerology and mysticism&amp;quot; to determine the size of Venus.  He was lucky that Venus was almost the size of Earth; that made his estimate for the AU pretty close.&lt;/p&gt;&#xA;&lt;p&gt;Not long after, Cassini used the parallax of Mars to determine the AU.  (Same article as linked above.)&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In 1672, Cassini used a method involving parallax on Mars to get the AU, and his method &lt;em&gt;was&lt;/em&gt; correct.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Parallax is the apparent difference in angle observed due to the differing observing positions.  The smaller the parallax, the larger the distance.&lt;/p&gt;&#xA;&lt;p&gt;However, the precision of the resultant calculation depends on the precision of the observations, and measurements of the parallax aren't that precise.&lt;/p&gt;&#xA;&lt;p&gt;In 1716, Edmond Halley published a way to use a transit of Venus to accurately measure the solar parallax, i.e. the difference in the Sun's position in the sky due to observers at different latitudes.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Because of the latitude difference of the observers, Venus would appear to move along chords of different length over the disk of the sun.  The motion of Venus being nearly uniform, the length of each chord would be proportional with the duration of the transit.  Thus, observers would not actually have to &lt;em&gt;measure&lt;/em&gt; anything; they would only have to &lt;em&gt;time&lt;/em&gt; the transit.  Fortunately, existing pendulum clocks were more than sufficiently accurate for this purpose.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;They could time the transit, which would last hours, with great precision.  But they had to wait until the next transit of Venus in 1761.  Then, observers observed the &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_drop_effect&quot; rel=&quot;noreferrer&quot;&gt;black drop effect&lt;/a&gt;, which made it very hard to time the event from start to finish precisely.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The black drop effect cannot be eliminated altogether, but it is much more ever in observations made with telescopes of imperfect optical quality (as many of those used at the 1761 transit were) and in boiling or unsteady air.  Confusion about the times of the internal contacts ... yielded contact times that differed among observers, because of the black drop, by as much as 52 seconds.&lt;/p&gt;&#xA;&lt;p&gt;In the end, there as a wide range of published values, from 8.28 arc-seconds to 10.60 arc-seconds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;But then there was the transit of 1769.  Observations in Norway and in the Hudson Bay were made for the northern observations, and Captain James Cook was sent to what is now Tahiti to make a southerly observation.  Jérôme Lalande compiled the figures and calculated a solar parallax of 8.6 arc-seconds, close to the modern figure of about 8.794 arc-seconds.  That calculation yielded the first fairly accurate calculation of the Earth-Sun distance, of 24,000 Earth radii, which given the Earth's radius of 6,371 km, about 153,000,000 km, the accepted value being about 149,600,000 km.&lt;/p&gt;&#xA;" OwnerUserId="1082" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-05-14T16:29:40.333" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6376" PostTypeId="1" CreationDate="2014-09-16T02:44:34.467" Score="3" ViewCount="355" Body="&lt;p&gt;It is very common that we meet cosmic ray particles in optical images recorded by CCDs.&#xA;You can see the &quot;snowflakes&quot; in the hubble images below:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/vJLQK.jpg&quot; alt=&quot;snowflake&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally we should remove them in order to get photometry or spectroscopic info.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there may be useful info we can derive, for example, can we get their directions or energies statistically before they arrive at a CCD?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is some correlation between the ADU value of a cosmic ray point and its real energy?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditDate="2015-07-13T14:32:27.563" LastActivityDate="2019-05-15T19:14:20.603" Title="Which kind of properties can we get for cosmic ray particles hitting on an optical ccd?" Tags="&lt;cosmic-ray&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6379" PostTypeId="1" CreationDate="2014-09-16T12:05:05.163" Score="5" ViewCount="93" Body="&lt;p&gt;As I know, for example the modelling of the collapse of a neutron star (to a black hole) wasn't done correctly until now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? Yes, I know, the Einstein Field Equations aren't really easy to solve. But, for example, the situation is very similar in the case of the QCD, but there are already fruitful results existing ( &lt;a href=&quot;http://physicsworld.com/cws/article/news/2008/nov/21/proton-and-neutron-masses-calculated-from-first-principles&quot;&gt;http://physicsworld.com/cws/article/news/2008/nov/21/proton-and-neutron-masses-calculated-from-first-principles&lt;/a&gt; ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any theoretical obstacle in the way? Or we simply didn't have enough fast computers/physicist/programmer manyear until now?&lt;/p&gt;&#xA;" OwnerUserId="2421" LastActivityDate="2016-04-28T14:33:34.827" Title="What are the biggest problems about the numerical, finite-element GR models?" Tags="&lt;black-hole&gt;&lt;general-relativity&gt;&lt;n-body-simulations&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6380" PostTypeId="1" CreationDate="2014-09-16T23:05:13.700" Score="3" ViewCount="88" Body="&lt;p&gt;Please forgive my ignorance. I dare say that these questions arise from me trying to understand the nature of the Universe using Newtonian physics. There are a few things that really bother me when topics such as red shift and expansion are discussed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) We can view a Galaxy that is 30bn light years away because the light we are seeing it as it was 13bn years ago, and at that time it was only 13bn light years away. So, it has travelled 17bn years further in the last 13bn years? Faster than the speed of light?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) 13bn years ago, it was 13bn light years away. That is around the time of the big bang. How did it manage to get that far away so soon after the Universe began. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) How is the known Universe even as big as 40bn light years, after only 13bn years in existence. &lt;/p&gt;&#xA;" OwnerUserId="2428" LastActivityDate="2014-09-18T12:20:26.270" Title="Expansion of the Universe" Tags="&lt;expansion&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2014-09-19T11:41:40.750" ContentLicense="CC BY-SA 3.0" />
  <row Id="6381" PostTypeId="2" ParentId="6346" CreationDate="2014-09-17T00:43:18.190" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it possible to view Saturn in little yellowish and Mars in little&#xA;  reddish using following telescopes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is definitely possible to observe the rings of Saturn with telescopes this size. Even the Cassini division should sometimes appear visible, if the instruments are well collimated and seeing is not too bad. In terms of color, Saturn is just a boring buttery-yellow even in bigger scopes, so I wouldn't worry about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But Saturn is getting lower in the sky these days. If you hurry up and get the scope quickly, you may catch it for a few weeks at sunset, low in the western sky. After this, you'll have to wait until next year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars is a different animal. Most of the time, all you'll see is a bright brick-red round dot, even in a bigger scope than these ones. But every couple years Mars is at opposition, when it's closest to Earth. We just had one a few months ago. Then you can see some of the big features, such as Syrtis Major, or the polar ice caps, or Hellas Basin full of frost or fog, like a big, bright white area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, that's only doable briefly around oppositions. The scope must be in perfect collimation, and seeing must cooperate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Astronomical_seeing&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Astronomical_seeing&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If everything is at maximum parameters, I'm sure you could see Syrtis Major in a scope this size. During the last opposition, I've seen all of the above features, plus more (Utopia Planitia, Sinus Sabaeus, etc), in as low as 150 mm of aperture, in a scope with great optics, perfectly collimated, during nights with excellent seeing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, for Mars you'll have to wait until the next opposition, in May 2016.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later this year, in December, Jupiter will start rising in the East, and you could use your scope to watch it - an aperture like this is enough to see the 4 big moons and at least 2 equatorial belts. It will be high in the sky at a comfortable time in the evening early next year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until then, you can always observe the Moon, two weeks out of every four.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the planets and the Moon are not the only things accessible with this aperture. Most of the Messier objects are visible in a 100 ... 150 mm scope, even in suburban areas. M13 is spectacular at any aperture above 100 mm. The Great Orion Nebula is awesome even with binoculars. The Pleiades are great too. Most of these deep space objects require low magnification for the best view, but every case is a bit different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plenty of double stars out there, too: Mizar, Albireo, even Polaris. All visible in small apertures.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am going to buy one of them. Which one is worth more for the money&#xA;  with the price difference?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The instruments are about the same. In theory, the bigger one has a bit more resolving power and a bit larger collecting area, so theoretically it should be slightly better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, with mass-produced instruments like these, it usually depends on the build quality, which can vary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smaller instrument is an f/8. The longer focal ratio means less aberrations; it also makes it easier for cheap eyepieces to function well, whereas at f/6 ... f/5 a cheap eyepiece may start to exhibit aberrations of its own (independent of telescope aberrations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, an f/8 is easier to collimate than an f/6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, I would look at it as a matter of price. If you can easily afford the bigger one, get it. Otherwise, the smaller instrument might be a bit easier to maintain, is less demanding in terms of optics, and it should perform pretty close to the other one - all else being equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since you're focused on planetary observations, remember this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is far more important to learn to correctly collimate your telescope, and develop it into a routine whereby you do a quick collimation check every time before you observe - it only takes a couple minutes. For planetary observations, the smaller telescope, in perfect collimation, will perform far better than the larger one, uncollimated. Heck, the little scope, perfectly collimated, will perform better on planets than a MUCH larger telescope, uncollimated - that's how important collimation is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Improper collimation, or lack thereof, is one of the major factors for lackluster performance for a majority of amateur telescopes (along with poor quality optics - but there's nothing you can do about that, whereas collimation can be improved).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Search this forum, or just google, the term collimation, and read the numerous documents you'll find. Or start here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cloudynights.com/documents/primer.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.cloudynights.com/documents/primer.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/150/how-can-i-collimate-a-dobsonian-telescope-with-a-laser-collimator&quot;&gt;How can I collimate a dobsonian telescope with a laser collimator?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The owner's manual should also provide some recommendations regarding collimation (I hope).&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2014-09-17T00:49:31.620" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6382" PostTypeId="2" ParentId="6337" CreationDate="2014-09-17T01:16:50.457" Score="2" Body="&lt;p&gt;Both the temperature and the density of a toy universe can, in a way, be linked to the age of the universe ($t$). Certain properties of the universe are dependent upon a scale factor $a(t)$ that is a function of time. The expansion of such a universe can be written in the Friedmann-Lemaitre-Robertson-Walker (FLRW) metric, generally written as&#xA;$$ds^2 = dt^2 - a(t)^2(d\Sigma^2)$$&#xA;where, in spherical coordinates, $d\Sigma$ is given by&#xA;$$d\Sigma^2 = \frac{dr^2}{1-kr^2} + r^2d\theta^2 + r^2\sin^2\theta d\phi^2$$&#xA;where $k$ is a constant. This shows that, in a toy universe, $r$ and $dr$ is proportional to $a(t)$, and so in a perfectly spherical toy universe, you could relate the volume to $t$. We know that, for a sphere,&#xA;$$V=\frac{4}{3}\pi r^3$$&#xA;and because $d\Sigma^2$ (and by extension $r^2$) is related to $a(t)^2$, we can relate volume to time. If you knew the mass-energy of the universe, you could then relate the change in density to $t$. Unfortunately, our universe does not appear to be perfectly spherical, and so this may not apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can relate something else to the age of the universe: the temperature of the cosmic microwave background. It has cooled over time, and is presently at about 2.7 Kelvin. While the temperature of the CMB may not be governed by the scale factor $a(t)$, it is certainly governed by time, and thus one you could in fact write the temperature of the universe as function of time. Unfortunately, I haven't been able to find data on the evolution of the CMB, but I can try, and perhaps this will lead you to an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize: In a perfectly spherical universe, assuming you know the matter-energy content and the scale factor $a(t)$, you could write an equation for the density of the universe. In any universe, you can also write an equation for the temperature of the CMB - the &quot;temperature of the universe&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source for FLRW metric: &lt;a href=&quot;http://www.universe-galaxies-stars.com/Robertson-Walker_coordinates.html&quot; rel=&quot;nofollow&quot;&gt;http://www.universe-galaxies-stars.com/Robertson-Walker_coordinates.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-09-17T01:16:50.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6383" PostTypeId="1" CreationDate="2014-09-17T14:32:30.007" Score="4" ViewCount="237" Body="&lt;p&gt;If cosmological inflation occurred at speeds less than $c$, wouldn't we see the CMB right in front of our noses (or not at all) instead of at 13.7 billion light years?&lt;br&gt;&#xA;And &lt;em&gt;if&lt;/em&gt; it happened faster than $c$, does that mean inflation occurred before the current laws of physics (read the four basic forces) precipitated, was $c$ greater then than it is now?&lt;/p&gt;&#xA;" OwnerUserId="1050" LastActivityDate="2015-01-10T22:59:40.287" Title="Did cosmological inflation occur at speeds greater than $c$?" Tags="&lt;cosmological-inflation&gt;&lt;cmb&gt;" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="6384" PostTypeId="1" AcceptedAnswerId="6388" CreationDate="2014-09-17T15:50:17.093" Score="5" ViewCount="17675" Body="&lt;p&gt;Is it about the same as Earth's orbital speed?&lt;/p&gt;&#xA;" OwnerUserId="2431" LastEditorUserId="1891" LastEditDate="2014-09-17T20:53:02.847" LastActivityDate="2020-05-27T01:04:15.217" Title="How fast is a comet moving when it crosses Earth's orbit?" Tags="&lt;orbit&gt;&lt;speed&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6385" PostTypeId="2" ParentId="6384" CreationDate="2014-09-17T16:11:14.310" Score="0" Body="&lt;p&gt;There are relatively big varieties, but most of them is between 10 and 70 km/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a comet is a periodic comet, that means it needs to have an elliptic orbit around the Sun. That gives an upper limit to its speed of the escape speed from the solar system on the orbit of the Earth. That is around 40 km/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this 40 km/s is in the reference frame of the Sun. The Earth is moving in this reference frame with around 30 km/s, on a nearly circular orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Between the escape speed and the mean speed of a circular orbit there is always a $\sqrt{2}$ relation. It is a physical law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically it were possible to find extrasolar comets (if the speed of it were bigger as around 70 km/s, it were a clear signature of its remote origin), but they aren't coming.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastActivityDate="2014-09-17T16:11:14.310" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6386" PostTypeId="2" ParentId="6383" CreationDate="2014-09-17T22:28:03.067" Score="1" Body="&lt;p&gt;Yes, although we usually do not say the &quot;expansion occurs faster than light.&quot;  In General Relativity space can be created between all points and that can result in points that were in causal contact (ie, they were able to see each other) before inflation, being, after inflation, farther away than light could travel in the age of the universe.  Basically, the rules in General Relativity are a bit more complicated than that of special relativity (which forbids accelerating an object to speeds greater than the speed of light).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now, since the local spacetime is not expanding as quickly as is the visible universe (that which we can see because it is within the distance that light travels in the age of the universe, c*t0), we are coming back into contact with regions of space that were out of contact since the inflationary period.   As time passes, we will be able to see a larger cosmic microwave background (CMB) surface. Spots on the CMB surface were not in contact with us (or each other) if they are more than 7 degrees apart since before the inflationary period.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2015-01-09T20:36:05.060" LastActivityDate="2015-01-09T20:36:05.060" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6387" PostTypeId="2" ParentId="6383" CreationDate="2014-09-17T23:01:22.177" Score="8" Body="&lt;p&gt;The simple answer to your question is &quot;yes&quot; - the universe expanded at much greater speeds than $c$ during the inflationary epoch. This period of time was very quick but very dramatic, lasting from about $10^{-36}$ to $10^{-32}$ seconds. The universe expanded, in this very short period, by a factor of $10^{26}$. That's pretty incredible, when you think about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inflation was originally proposed to solve, among other things, the horizon problem - that is, why the universe is isotropic and homogenous (on a large scale). This would mean that all the parts of the universe were in &quot;causal contact&quot; at one point in time. Inflation is the explanation for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what does this all have to do with the CMB? Well, the temperature of the CMB is the same throughout the universe - a toasty 2.7 Kelvin. For the temperature to be uniform, all the regions of the universe would have had to be in causal contact in some point in time; hence, inflation explains the uniform temperature of the CMB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the CMB was not around during the inflationary epoch. Far from it. It formed a lot later, when the universe was at the ripe old age of $379,000$ years. But the reason that it was formed (photon decoupling) equally throughout the universe is because the conditions were roughly equal, because of inflation. The CMB was, and still is, everywhere. It was never around during inflation, and as such was not otherwise affected by it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My sources for the times:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Inflationary_epoch&quot; rel=&quot;nofollow&quot;&gt;Inflationary epoch&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://map.gsfc.nasa.gov/universe/bb_cosmo_infl.html&quot; rel=&quot;nofollow&quot;&gt;NASA&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-01-10T22:59:40.287" LastActivityDate="2015-01-10T22:59:40.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6388" PostTypeId="2" ParentId="6384" CreationDate="2014-09-18T09:31:36.423" Score="8" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Comets don't cross Earth's orbit really. Orbits are one-dimensional objects and their chance of crossing in 3D space is 0. Henceforth, I consider a comet at distance 1AU from the Sun.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;What's the maximum speed of a returning comet at 1AU from the Sun?&lt;/strong&gt; This can be easily worked out from the &lt;em&gt;orbial energy&lt;/em&gt;&#xA;$$&#xA;E = \frac{1}{2}v^2 - \frac{GM_\odot}{r},\qquad\qquad(*)&#xA;$$&#xA;which is conserved along the orbit ($v$ and $r$ the Heliocentric speed and distance). For a returning comet, $E&amp;lt;0$ and the speed cannot exceed the escape speed (which occurs for $E=0$)&#xA;$$&#xA;v_{\rm escape}^2 = 2\frac{GM_\odot}{r}.&#xA;$$&#xA;The speed of the Earth can be worked out from the &lt;strong&gt;Virial theorem&lt;/strong&gt;, according to which the orbital averages of the kinetic and potential energies, $T=\frac{1}{2}v^2$ and $W=-GM_\odot/r$, satisfy&#xA;$&#xA;2\langle T\rangle + \langle W\rangle=0.&#xA;$&#xA;For a (near-)circular orbit (such as Earth's), $r$ is constant and we have&#xA;$&#xA;v^2_{\rm Earth} = GM_\odot/r.&#xA;$&#xA;Thus, at $r$=1AU&#xA;$$&#xA;v_{\rm escape} = \sqrt{2} v_{\rm Earth}&#xA;$$&#xA;as already pointed out by Peter Horvath. Non-returning comets have local speed exceeding the escape speed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Can a comet near Earth have a speed similar to Earth's orbital speed?&lt;/strong&gt;. Let's assume a comet with the same speed as Earth at $r$=1AU and work out the consequences. Such a comet must have the same orbital energy as the Earth and, since&#xA;$$&#xA;E = -\frac{GM_\odot}{2a}\qquad\qquad(**)&#xA;$$&#xA;with $a$ the orbital semimajor axis, must also have $a=1AU$ and the same orbital period as Earth, i.e. one year. Moreover, the comet's apohelion satisfies&#xA;$$&#xA;r_{\rm apo}\le 2a = 2{\rm AU}.&#xA;$$&#xA;&lt;strong&gt;Such comets don't exist&lt;/strong&gt; AFAIK. Most returning comets have much longer periods than 1 year.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;What's the typical speed of a returning comet when at distance 1AU from the Sun?&lt;/strong&gt; To work out this question, let's parameterise the comet's orbit by its period $P=2\pi\sqrt{a^3/GM_\odot}$. From this relation we immediately get&#xA;$$&#xA;\frac{a_{\rm comet}}{\rm AU} = \left(\frac{P_{\rm comet}}{\rm yr}\right)^{2/3}.&#xA;$$&#xA;From equations ($*$) and ($**$), we can then find&#xA;$$&#xA;v_{\rm comet}(r=1{\rm AU}) = \sqrt{2-\left(\frac{P_{\rm comet}}{\rm yr}\right)^{-2/3}}\,v_{\rm Earth}.&#xA;$$&#xA;In the limit of $P\to\infty$, this recovers our previous result $v_{\rm comet}\to v_{\rm escape}$. For typical period of $\sim$70yr, the speed of the comet is close to this number.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, I shall coment that all this only relates to the magnitude of the orbital velocity (speed), but not to its direction. Comets are typically on highly eccentric orbits and, when at $r$=1AU, move in quite a &lt;strong&gt;different direction&lt;/strong&gt; than Earth, even if their speed is only slightly larger. So the &lt;strong&gt;relative speed&lt;/strong&gt; $|\boldsymbol{v}_{\rm comet}-\boldsymbol{v}_{\rm Earth}|$ of a comet with respect to Earth can be anyting between about 10 and 70 km/s.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2014-09-18T20:12:08.390" LastActivityDate="2014-09-18T20:12:08.390" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6389" PostTypeId="2" ParentId="6383" CreationDate="2014-09-18T09:40:59.400" Score="2" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;does that mean inflation occurred before the current laws of physics (read the four basic forces) precipitated, was c greater then than it is now?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NO&lt;/strong&gt; Already the very concept of an &lt;em&gt;expansion velocity&lt;/em&gt; is flawed. Expansion means that elements of the medium move away from each other with a relative velocity that is proportional to their distance (for small distances). The constant of proportionality, the &lt;strong&gt;expansion rate&lt;/strong&gt;, is the sensible concept here and has dimension of a frequency. Currently the universe expands at about 73km/s/Mpc, which is the smallest natural frequency known. At the epoch of infation, the expansion rate was just much larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of, course, over large sufficiently distances, any expansion results in superluminal velocities, but the laws of phyiscs only constrain the relative velocities of objects in close vicinity.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-09-18T09:40:59.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6390" PostTypeId="1" CreationDate="2014-09-18T10:29:52.950" Score="11" ViewCount="1413" Body="&lt;p&gt;A star consumes quite a lot of hydrogen in its life, and is pretty much &quot;vacuuming&quot; everything in its vicinity. After it dies (eventually by supernova which will spread all its composition over light years), is there enough hydrogen left in that area to light up a new star? And will that star be more short-lived compare to its predecessor?&lt;/p&gt;&#xA;" OwnerUserId="2434" LastEditorUserId="13" LastEditDate="2014-09-18T18:06:36.597" LastActivityDate="2014-11-18T20:39:37.143" Title="Is there enough hydrogen left after a star dies so another star will have enough to light up?" Tags="&lt;star&gt;&lt;supernova&gt;&lt;stellar-evolution&gt;&lt;hydrogen&gt;&lt;protostar&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6391" PostTypeId="2" ParentId="6380" CreationDate="2014-09-18T10:36:20.863" Score="-1" Body="&lt;p&gt;The current explanation is that &quot;space&quot; is continuously created in the intergalactic void, &quot;pushing&quot; the galaxies apart. From what i understood ( i am also ignorant ), the current rate at which &quot;space&quot; is created is expressed by the Hubble's constant 67.15 ± 1.2 km/s/Mpc ): for every million parsecs of distance from the observer, the rate of expansion increases by about 67 kilometers per second, which means that the distance between 2 points that are 1 megaparsec apart grows with 67km every second; between 2 points that are 2 megaparsecs apart with 134km/s, and so on so the relative speed between 2 points that are 5000 Mpc apart would be 335000 km/s which is more than the speed of light. In fact, two objects at this distance would not move, but 335000km worth of &quot;extra space&quot; would be &quot;created&quot; between them every second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in the last week there was an article depicting the shape of the universe, and it is hard for me to understand how it got to look like this : &lt;a href=&quot;http://ifreepress.com/wp-content/uploads/2014/09/laniakea.png&quot; rel=&quot;nofollow&quot;&gt;http://ifreepress.com/wp-content/uploads/2014/09/laniakea.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Laniakea_Supercluster&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Laniakea_Supercluster&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2434" LastEditorUserId="2434" LastEditDate="2014-09-18T12:20:26.270" LastActivityDate="2014-09-18T12:20:26.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6393" PostTypeId="1" AcceptedAnswerId="6430" CreationDate="2014-09-18T13:42:56.453" Score="6" ViewCount="514" Body="&lt;p&gt;The vast majority of energy from a supernova is emitted as neutrinos and high-energy radiation, both of which being invisible to the naked eye. Yet a supernova can outshine its galaxy in the visible spectrum for weeks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is a supernova generating so much visible light that nuclear fusion pales in comparison?&#xA;What is the ratio of visible and invisible energy emitted by a supernova, and how does it compare to stars' normal operation?&lt;/p&gt;&#xA;" OwnerUserId="2435" LastEditorUserId="2435" LastEditDate="2014-09-18T20:53:01.207" LastActivityDate="2016-11-14T10:42:32.803" Title="Why are (Type II) supernovae so bright?" Tags="&lt;light&gt;&lt;supernova&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6394" PostTypeId="1" CreationDate="2014-09-18T13:52:14.600" Score="3" ViewCount="117" Body="&lt;p&gt;Both the R-process and the S-process of stellar nucleosynthesis rely on absorbing 1 (S) or more (R) neutrons. Where do they come from? I thought there are no free neutrons in stars. (Also the half-life of a free neutron is only 15 minutes, so the source has to be inside the star)&lt;/p&gt;&#xA;" OwnerUserId="1050" LastActivityDate="2014-12-03T21:12:18.260" Title="Free neutrons and stellar nucleosynthesis" Tags="&lt;nucleosynthesis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6396" PostTypeId="2" ParentId="6203" CreationDate="2014-09-18T18:42:18.977" Score="2" Body="&lt;p&gt;Galaxy mergers in clusters explain the large central galaxies in clusters.  In fact, the centers of clusters often host a special type of galaxy called a cD which is usually extremely massive because of galactic cannibalism.  Dynamical friction makes galaxies spiral down toward the center where they get eaten by the central galaxy and that in turn makes the central galaxy more massive and more efficient at this process.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the rest of the cluster, the velocity dispersion is significantly higher than the escape velocities from individual galaxies.  So a) galaxies cannot gravitationally attract other galaxies because they go by too fast and b) even if there is a head on collision by sheer coincidence, they would pass through and not capture one another.  There may well be huge consequences to both galaxies, such as stellar rings and shells formed and tidal tails extracted, but the two remnants would separate and perhaps never meet again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an oversimplification though because clusters often have subgroups of galaxies which have lower velocity dispersion wrt each other and these could merge much as galaxies in groups merge.  But the tidal field of the cluster actually works against this and so the rate would be lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a lot of literature on cD or central galaxies since it is one of the standard types of galaxies. Here is a recent one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Growth of brightest cluster galaxies via mergers since z=1&quot;, Claire Burke and Chris A. Collins, 2014, MNRAS, v. 434, p. 2856.&#xA;&lt;a href=&quot;http://mnras.oxfordjournals.org/content/434/4/2856&quot; rel=&quot;nofollow&quot;&gt;http://mnras.oxfordjournals.org/content/434/4/2856&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A study of merger rates in the outer parts of clusters may take a major effort of going through the results of N-body simulations on your own and your answer may depend on which simulation you use.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2014-09-26T19:54:19.563" LastActivityDate="2014-09-26T19:54:19.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6397" PostTypeId="1" AcceptedAnswerId="6401" CreationDate="2014-09-18T22:48:59.523" Score="5" ViewCount="674" Body="&lt;p&gt;One thing that I can't quite wrap my head around is how light is traveling to Earth from 14 billion light years away, aka the beginning of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I see it, the universe itself was very small 14 billion years ago and has been steadily accelerating in size since.  So the photons that left a star that existed 14 billion years ago couldn't travel for 14 billion years because the universe itself was probably only a few million light years total in diameter.  The universe is not expanding faster than the speed of light, so that photon should have hit the end of the universe before now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What piece of this puzzle am I missing?&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2016-02-24T13:51:27.877" Title="How can light reach us from 14 billion light years away?" Tags="&lt;light&gt;&lt;dark-energy&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6398" PostTypeId="1" AcceptedAnswerId="6400" CreationDate="2014-09-18T23:00:23.410" Score="5" ViewCount="960" Body="&lt;p&gt;If the mass of a black hole is creating so much gravity that light cannot escape, why isn't the mass of the star that created the black hole (before it went supernova) trapping light as well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By all accounts, that pre-supernova star should have boatloads more mass than the black hole post supernova, right?  Doesn't the star lose most of it's mass when going supernova?&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2015-01-15T21:43:08.383" Title="Why isn't the star that created the black hole a black hole?" Tags="&lt;gravity&gt;&lt;black-hole&gt;" AnswerCount="6" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6399" PostTypeId="2" ParentId="6390" CreationDate="2014-09-18T23:11:08.443" Score="2" Body="&lt;p&gt;Our sun is a 3rd or 4th generation star, so yes, there is enough hydrogen left over to create more stars.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know this because our solar system is fairly rich in heavy elements, which means that there must have been at least 1, and probably 2 or 3 supernovas that created these heavier elements that created all of the rocky planets, asteroids, comets, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is doubtful that our sun will shed enough hydrogen to create yet another star.  It's too small now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you look at the pillars of creation, which is a nebula created by a supernova, you can see the early stages of star formation happening right now.&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2014-09-18T23:11:08.443" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6400" PostTypeId="2" ParentId="6398" CreationDate="2014-09-18T23:13:00.780" Score="7" Body="&lt;p&gt;You are correct in saying that a star loses a lot of its mass in a supernova. However, there &lt;em&gt;is&lt;/em&gt; a reason why the star still becomes a black hole. Actually, I suppose the question here is &quot;Why doesn't a star become a black hole before it even undergoes a supernova?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a reason for a supernova (I'll assume you're talking about type II supernovae, which result from incredibly massive stars). Stars undergo nuclear fusion, and this leads to &quot;thermal pressure&quot;, which counteracts the force of gravity. Without this pressure, gravity would indeed make a sufficiently large star collapse upon itself. Gravitational collapse occurs when there is not enough pressure to counteract gravity; the result is a spectacular supernova. So stars only become black holes (or other compact objects, such as neutron stars) when they cannot produce enough energy to counteract the force of gravity due to their own mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the first part of your question (sorry for answering in reverse), light in the area of a black hole cannot escape if it is inside its event horizon or on a trajectory towards it. The radius of the event horizon for a non-rotating black hole is its Schwarzschild radius, which is proportional to the mass of the black hole. The reason this is non-applicable in stars is because the Schwarzschild radius in stars is deep inside its interior, and there is not a strong enough gravitational field to produce an event horizon to trap any light near it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermal pressure reference: &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_collapse&quot;&gt;https://en.wikipedia.org/wiki/Gravitational_collapse&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-09-18T23:13:00.780" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="6401" PostTypeId="2" ParentId="6397" CreationDate="2014-09-18T23:22:59.713" Score="6" Body="&lt;p&gt;If a star 300,000 km away from you explodes, it'll normally take the radiation from that explosion one second to start vaporizing you.&#xA;If the space between you and that exploding star inflates to 2.6X10^23km (28 billion ly) one half second after the explosion, the radiation will take 14 billion years, plus a half second, to reach you, and won't vaporize you.&#xA;Of course the expansion need not be instantaneous. You can plug in about any function you want there. The important thing is not the distance between you and the star at the time of explosion, but the total distance the photon must travel on its path to you. That distance must take into account the expansion of space between the emitted radiation and you, the observer.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="2031" LastEditDate="2014-09-21T01:49:43.810" LastActivityDate="2014-09-21T01:49:43.810" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6402" PostTypeId="2" ParentId="6390" CreationDate="2014-09-18T23:41:37.610" Score="2" Body="&lt;p&gt;First, thanks to @LCD3 for leading me on the right path here. My original answer was inaccurate, and so I got rid of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A supernova occurs when a very massive star can no longer sustain enough nuclear fusion to combat the force of its own gravity pushing inwards on it. This happens after the star has gone through different stages of fusion. Typically, it begins with fusing hydrogen into helium. This is the type of fusion you have probably heard about the most because stars are largely hydrogen and helium. However, there are other fusion processes that are equally important when it comes to prolonging a star's life, which fuse together heavier elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A star begins by fusing hydrogen nuclei into helium nuclei deep in its core. This is how the star produces energy, and is indirectly responsible for the star shining. However, there is only so much of this fusion a star can undergo in its core. When the core hydrogen is depleted, the star beings fusing helium there. It continues hydrogen fusion in its outer layers, where there is still hydrogen. Eventually, the star runs out of helium in its core, and begins to fuse even heavier elements. Hydrogen fusion continues in the outermost layers, with helium fusion occurring in lower layers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the process can only go on for so long, and eventually the star can't fight gravity any longer. In very massive stars, this leads to a supernova, which flings off much of a star's mass into space. In all the cast-away matter, is there enough hydrogen left to form a new star? Well, there isn't nearly as much hydrogen as there was at the star's birth. In relatively low-mass supernova progenitors, there might not be enough hydrogen to form a new star. In very high-mass stars, however, there will still be a substantial amount left. Could &lt;em&gt;this&lt;/em&gt; form a new star? Probably not for a long time, because the hydrogen will have been thrown off into space by the supernova, and the wouldn't be very dense. It wouldn't be easy for it to collapse into a gas cloud to form a protostar. I wouldn't rule this out for very high-mass stars, but in the remnants of many stars, there probably would not be enough hydrogen to form a new star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source for the layer explanation: &#xA;&lt;a href=&quot;http://www.astronomynotes.com/evolutn/s5.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.astronomynotes.com/evolutn/s5.htm&lt;/a&gt;. Also, many thanks to @LCD3.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-19T01:01:44.813" LastActivityDate="2014-09-19T01:01:44.813" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6403" PostTypeId="1" CreationDate="2014-09-19T07:40:23.133" Score="2" ViewCount="75" Body="&lt;p&gt;Following the answers and the line of thinking from here &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6390/is-there-enough-hydrogen-left-after-a-star-dies-so-another-star-will-have-enough&quot;&gt;Is there enough hydrogen left after a star dies so another star will have enough to light up?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if is there any possibility to know if, where new space is created between the galaxies, is there also created any kind of new matter ( i.e. hydrogen, or subatomic particles that could eventually form atoms, which will lead to forming new stars and galaxies ) ?&lt;/p&gt;&#xA;" OwnerUserId="2434" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-09-19T14:04:25.547" Title="Is there matter forming in the inflationary space?" Tags="&lt;cosmological-inflation&gt;&lt;dark-matter&gt;&lt;star-formation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6405" PostTypeId="1" CreationDate="2014-09-19T13:29:12.270" Score="3" ViewCount="93" Body="&lt;p&gt;The universe is thought to be flat: $\Omega = 1 \pm 1\%$. As I understand it we can determine this by measuring triangles against the CMB.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet during inflation dark energy made the universe grew much larger than the radiation universe, of which the CMB is the limit. Isn't it possible then that at the &quot;smaller&quot; scale of our visible universe it appears flat, but at the universe's full size it's curved? A bit like a square of 1 meter sides on the surface of the earth also looks flat, while at larger scale the earth is curved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the chances that the 1% uncertainty could lead us to a universe with $\Omega \ne 1$?&lt;/p&gt;&#xA;" OwnerUserId="1050" LastActivityDate="2014-09-20T08:03:39.253" Title="How certain are we about the universe's flatness?" Tags="&lt;space-geometry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6406" PostTypeId="2" ParentId="6403" CreationDate="2014-09-19T13:36:11.993" Score="2" Body="&lt;p&gt;&lt;em&gt;A note: don't talk about inflation for the current expansion. Inflation is the brief period immediately after the BB when the universe expanded at an outlandish rate.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matter was formed when the universe was so compact that its temperature was billions and even trillions of Kelvin hot. The energy of the vacuum in any given volume was so high that matter-antimatter pairs were created, which usually quickly annihilated. Sometimes, however those short-lived particles collided, releasing large amounts of energy. For instance, to create an electron you need an energy of 0.5 MeV. The graph below shows that at a temperature of $10^{16}$ K collision energy could be as high as 1 TeV.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of its expansion the universe has cooled so much that its temperature is less than 3 K, which isn't nearly enough to create any new particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/db1If.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graph shows how temperature and energy drop as the universe expands in the first seconds after the BB. Note that at 1000 seconds (about 20 minutes) the energy is too low to create electrons (0.5 MeV), let alone heavier elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What &lt;em&gt;is&lt;/em&gt; created in the expansion, though, is dark energy. Its density is constant, so when the universe expands to double its size the amount of dark energy will also be doubled. (That's the reason why the expansion goes ever faster and faster, with nothing to stop it.)&lt;/p&gt;&#xA;" OwnerUserId="1050" LastEditorUserId="1050" LastEditDate="2014-09-19T14:04:25.547" LastActivityDate="2014-09-19T14:04:25.547" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6407" PostTypeId="1" CreationDate="2014-09-19T21:54:33.510" Score="3" ViewCount="1614" Body="&lt;p&gt;I have a &lt;a href=&quot;https://web.archive.org/web/20061216215456/http://plettstone.com/telescopes/products.htm&quot; rel=&quot;nofollow&quot;&gt;Plettstone 18&quot;&lt;/a&gt; Dobsonian telescope with GOTO and clock drive that runs off a 12 volt cigarette lighter adapter.  What is an appropriate battery to use to run this overnight while viewing all night long and then charge during the day for the following night's use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the past I've used deep cycle lead acid batteries, but these are heavy and can fail if not carefully maintained with a float charge.  Also, I don't need cranking power of a car, just sustained energy to drive the scope.  A typical &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B0013ZGZ9Q&quot; rel=&quot;nofollow&quot;&gt;Optima YellowTop&lt;/a&gt; deep cycle is rated 75 ampHours.  I don't know what the draw in Amps of my telescope is, but imagine it is less than an amp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also seen these &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B00F9LPJ8S&quot; rel=&quot;nofollow&quot;&gt;Battery Tender Lithium Iron Phosphate&lt;/a&gt; types but wonder if their 10-12 ampHour rating is enough and whether they are meant to be drained over a whole night?  Some say if you get below 8 volts the battery becomes broken, but I doubt the telescope would still run at such a low voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would a reasonable ampHour rating be?  What kind of battery is a good fit for astronomy equipment overnight?&lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2015-07-11T06:54:37.957" Title="Appropriate battery for overnight viewing to drive telescope and GOTO?" Tags="&lt;dobsonian-telescope&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6408" PostTypeId="1" AcceptedAnswerId="6510" CreationDate="2014-09-20T03:44:27.167" Score="11" ViewCount="522" Body="&lt;p&gt;I am leading a project for my High School NASA Club, and am looking for some advice. The goal of our project is to use a Raspberry Pi to control a camera that takes photos of the sky, and uses the photo to calculate sky glow, and create a map of where it in the sky is there the most sky glow. After calculating sky glow the device would add it to a plot that shows how sky glow varies throughout the year, and creates a nightly twitter post rating the night for Astronomy. I currently have working code that when given a picture, it will convert down to greyscale, and then find the brightness map, and find an average sky glow. The part that I am stuck on is in figuring out what kind of camera to use. I don't know if the RPi camera module would work for taking photos of the stars, or if I would be better suited using a USB webcam with it. I am not sure what factors are important in this decision at all. Also, once the camera is set up, what would be the best process for calibrating it to get values that agree most closely with accepted ones? I couldn't decide if this should go on the Astronomy, Raspberry Pi, or Photography Stack Exchange, but I chose this one since sky glow may be something astronomers calculate a lot. Thank you for your help.&lt;/p&gt;&#xA;" OwnerUserId="2444" LastActivityDate="2014-10-03T20:52:16.270" Title="Sky Glow calculations" Tags="&lt;photography&gt;&lt;light-pollution&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6409" PostTypeId="2" ParentId="6407" CreationDate="2014-09-20T04:29:38.480" Score="1" Body="&lt;p&gt;Many electronic systems today use standard electric power formats, as 12 Volts, 9 Volts, currents batteries has different balance in power,some ones hold  a level in volts but not electric current (12 volts 0.005 Amp), or hold a level in current Ampers (8 volts but a continue current in 0.01 Amp), but not a level in volts, the key is power in Watts/hour , check your data sheet , if you system only required power, use battery type Lithium , but if you system you can see a data like a 12 Volts + - 1 Volt, use a battery non Lithium. &lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2014-09-20T04:29:38.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6410" PostTypeId="2" ParentId="6407" CreationDate="2014-09-20T05:05:42.303" Score="1" Body="&lt;p&gt;You could buy a sealed lead acid battery (12V 7.2Ah) for a reasonable price and try it out. If it fails to run all night, you can connect 2 together for a longer run time (connected in parallel to give 12 volts at double the number of Ahs). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually sealed lead acid batteries are designed so they can run down completely before being recharged (they are used in emergency lighting, which have a minimum on time).&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="4085" LastEditDate="2015-07-11T06:54:37.957" LastActivityDate="2015-07-11T06:54:37.957" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6411" PostTypeId="2" ParentId="6405" CreationDate="2014-09-20T08:03:39.253" Score="2" Body="&lt;p&gt;It's entirely possible that our seemingly flat observable universe sits inside a much larger and very curved universe. Since there is, by definition, no way to ever measure anything beyond the observable universe, such possibilities are essentially non-scientific musings.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2014-09-20T08:03:39.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6412" PostTypeId="1" CreationDate="2014-09-20T09:46:08.350" Score="6" ViewCount="199" Body="&lt;p&gt;I've just finished my masters degree (to be exact previous week ^_^). And I'm completing my collection of applications of my studies. During previous year I started reading about Robotics and Coding theory (but in free times as a hobby). I knew that my field of study has application in Astrophysics but I didn't have enough time to go after this topic as well. But my main question; these days I felt into mood of having a general knowledge in Astronomy etc. everything related to stars and space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I'm not seeking the application of my field of study at astrophysics (if I were it was better to put this question at math stack instead of hear) and certainly almost all members here are not familiar with pure mathematics, I will be pleased if someone here introduce one or several nice texts or books which satisfy one of the two following set of conditions:&#xA;1- Some elementary sources which people who doesn't have any background can read them. The notes be exciting and the reader wants to continue it until its end.&#xA;2- Sources which also use mathematics (specially if the level of the math won't be only some calculus) but the dominant text be about stars and space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume you are suggesting a text to a person as his first thing he will read in this topic. And if you think a series or a number of texts are suitable, please give an order to which should be read first and then which for the next and so on.&#xA;(Also please note that I don't like to read guesses! So texts contains we guessed x should be because of y or we think the world is ... are not interesting to me)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your attention.&lt;/p&gt;&#xA;" OwnerUserId="2445" LastEditorUserId="1386" LastEditDate="2014-09-21T15:16:04.173" LastActivityDate="2014-09-21T15:16:04.173" Title="Requesting book references for a non-expert person with math background or just a non-expert person" Tags="&lt;star&gt;&lt;space&gt;&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6413" PostTypeId="2" ParentId="6394" CreationDate="2014-09-20T21:12:55.097" Score="2" Body="&lt;p&gt;I have yet to find good information regarding the s-process, so I'll just talk about the r-process here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key difference between the two is the conditions under which they take place. The r-process takes place during supernovae, specifically those due to gravitational collapse. In a collapsing star, many of the protons and electrons are squeezed tightly together - so tightly, in fact, that a process called &lt;em&gt;neutronization&lt;/em&gt; begins. The protons and electrons actually combine to form neutrons (and neutrinos). This creates a large &lt;em&gt;neutron flux&lt;/em&gt;, and so there are now lots of neutrons available for the r-process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of this lasts incredibly long, though, because the &quot;explosion&quot; doesn't last long. It soon propels much of the outer material of the star out into space, and the remains become either a neutron star or a black hole. That said, an interesting follow-up might address the relationship between the r-process, neutronization and the formation of neutrons stars in more detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/R-process&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/R-process&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ucolick.org/~bolte/AY4_00/week8/SNeII.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ucolick.org/~bolte/AY4_00/week8/SNeII.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-09-20T21:12:55.097" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6414" PostTypeId="1" CreationDate="2014-09-15T15:01:38.143" Score="11" ViewCount="747" Body="&lt;p&gt;During a lunar or solar eclipse, from our perspective when we look towards the moon and the sun is behind the moon, does it fit exactly?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone calculated the sizes of each in a relative context?  Obviously the sun is thousands of times larger, but I am talking about the size of the sun/moon when looking from our perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope my question makes sense.&lt;/p&gt;&#xA;" OwnerDisplayName="public static" LastActivityDate="2014-09-21T07:45:55.720" Title="During an eclipse, does the size of the moon and sun match perfectly?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6415" PostTypeId="2" ParentId="6414" CreationDate="2014-09-19T20:53:04.283" Score="8" Body="&lt;p&gt;To expand on what taupunkt commented, on average, according to the article &lt;a href=&quot;http://www.space.com/15584-solar-eclipses.html&quot; rel=&quot;noreferrer&quot;&gt;Solar Eclipse: What is a Total Solar Eclipse &amp;amp; When is the Next One?&lt;/a&gt; (Rao, 2014), it is a unique quirk of nature that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The sun's 864,000-mile diameter is fully 400 times greater than that of our puny moon, which measures just 2,160 miles.  But the moon also happens to be about 400 times closer to the sun than the Earth &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is often stated in middle and high school science textbooks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the orbits of the Earth around the sun and the moon around the Earth are elliptical, as illustrated in the diagram below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nx5VR.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://oceanservice.noaa.gov/education/kits/tides/media/supp_tide06b.html&quot; rel=&quot;noreferrer&quot;&gt;Image source: NOAA&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This results in the exact ratio mentioned above not occurring resulting in an &lt;strong&gt;annular eclipse&lt;/strong&gt;, where the moon is too 'small' to fully cover the solar disc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A comparison between a total and annular eclipse is shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3KzrY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://sciencejedi.com/science/eclipse2012/&quot; rel=&quot;noreferrer&quot;&gt;Image source: ScienceJedi&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://eclipse.gsfc.nasa.gov/solar.html&quot; rel=&quot;noreferrer&quot;&gt;NASA's Eclipse data page&lt;/a&gt;, in particular, their &lt;a href=&quot;http://eclipse.gsfc.nasa.gov/SEhelp/moonorbit.html&quot; rel=&quot;noreferrer&quot;&gt;Eclipses and moon orbit&lt;/a&gt; information page discuss in detail (with considerable data), the &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;interaction and harmonics of the synodic, anomalistic, and draconic months not only determine how frequently eclipses occur, but they also control the geometric characteristics and classification of each eclipse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Within the NASA link, data about how much of the sun's disk that would be obscured by the moon has been calculated (from observations and predicted). For example, using the data for &lt;a href=&quot;http://eclipse.gsfc.nasa.gov/SEdecade/SEdecade2011.html&quot; rel=&quot;noreferrer&quot;&gt;solar eclipses from 2011 to 2020&lt;/a&gt;, The average fraction of the sun's disk that is and is expected to be obscured (this is referred to as '&lt;a href=&quot;http://eclipse.gsfc.nasa.gov/SEdecade/SEdecade2011.html#3&quot; rel=&quot;noreferrer&quot;&gt;Eclipse magnitude&lt;/a&gt;') in an annular eclipse varies from about 0.95-0.99 (95% to 99%) for that time period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A catalog of eclipses up to the year 3000 is provide on &lt;a href=&quot;http://eclipse.gsfc.nasa.gov/SEcat5/SEcatalog.html&quot; rel=&quot;noreferrer&quot;&gt;this NASA page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The greatest difference would occur if the moon is at its furthest from the Earth (apogee) and the Earth closest to the sun (perihelion).&lt;/p&gt;&#xA;" OwnerDisplayName="user2449" LastEditorDisplayName="user2449" LastEditDate="2014-09-21T07:45:55.720" LastActivityDate="2014-09-21T07:45:55.720" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6416" PostTypeId="1" AcceptedAnswerId="6461" CreationDate="2014-09-21T01:09:53.337" Score="7" ViewCount="124" Body="&lt;p&gt;If an Earth-like exoplanet had a sufficiently extensive plant biosphere (i.e large forests, aquatic algal mats etc), is it at all conceivable that scientists could spectroscopically detect the presence of biochemicals, such as chlorophyll, anthocyanins etc?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If possible, how would the spectra of these biochemicals be distinguished from other chemicals?&lt;/p&gt;&#xA;" OwnerDisplayName="user2449" LastEditorUserId="1322" LastEditDate="2014-09-26T00:48:43.287" LastActivityDate="2014-10-26T01:42:11.397" Title="Is it conceivable that the presence of chlorophyll (or similar) could be distinguishable from other chemicals on an exoplanet?" Tags="&lt;exoplanet&gt;&lt;earth-like-planet&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6417" PostTypeId="2" ParentId="6412" CreationDate="2014-09-21T01:21:44.223" Score="7" Body="&lt;p&gt;I personally found Stephen Hawking's &quot;A Breif History of Time&quot; to be very interesting and informative.  It is a little bit advanced, but is put forward in plain English.  Here is a link to it on Amazon, and you can &quot;look inside&quot; to see part of the book and see what you think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0553380168&quot; rel=&quot;noreferrer&quot;&gt;http://www.amazon.com/gp/product/0553380168/ref=pd_lpo_sbs_dp_ss_1?pf_rd_p=1535523722&amp;amp;pf_rd_s=lpo-top-stripe-1&amp;amp;pf_rd_t=201&amp;amp;pf_rd_i=055371449X&amp;amp;pf_rd_m=ATVPDKIKX0DER&amp;amp;pf_rd_r=1Z0JYCKHWXEFPK3TS6TM&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading your comments / clarifications, you might like &quot;A Field Guide to Stars and Planets.&quot;  It appears to be a good introduction into astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0395934311&quot; rel=&quot;noreferrer&quot;&gt;http://www.amazon.com/Field-Guide-Planets-Peterson-Guides/dp/0395934311/ref=sr_1_1?s=books&amp;amp;ie=UTF8&amp;amp;qid=1411293771&amp;amp;sr=1-1&amp;amp;keywords=stars+and+planets&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as covering astronomy from a more mathematical perspective, you might like &quot;A Student's Guide to the Mathematics of Astronomy.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1107610214&quot; rel=&quot;noreferrer&quot;&gt;http://www.amazon.com/Students-Guide-Mathematics-Astronomy/dp/1107610214/ref=sr_1_1?s=books&amp;amp;ie=UTF8&amp;amp;qid=1411294172&amp;amp;sr=1-1&amp;amp;keywords=math+of+astronomy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can take a look inside each of these books to see what you think.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="1386" LastEditDate="2014-09-21T10:13:43.583" LastActivityDate="2014-09-21T10:13:43.583" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6419" PostTypeId="2" ParentId="4816" CreationDate="2014-09-21T17:24:37.873" Score="3" Body="&lt;p&gt;The answer to your question is relatively straightforward: Not much. I'll delve into it in two parts: Star formation and the activity of the supermassive black hole at the center.&lt;/p&gt;&#xA;&lt;h3&gt;Star formation&lt;/h3&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Messier_87&quot; rel=&quot;nofollow noreferrer&quot;&gt;M87&lt;/a&gt; home to many Population II stars. They have little hydrogen and helium in them, and the clouds of gas and dust in the galaxy don't have much, either. In fact, many dust clouds can be destroyed by the intense radiation emitted by the accretion disk of the central black hole. This combines to make it seem unlikely that there will be much star formation in the galaxy.&lt;/p&gt;&#xA;&lt;p&gt;That said, just because a galaxy doesn't have a lot of materials for star formation doesn't mean it can't form lots of new stars - at least, no under certain special circumstances. Galaxy collisions can induce high rates of star formation - starburst galaxies are good examples of this. While M87 might not be at a high risk for a collision, there is still the possibility of one. &lt;a href=&quot;https://en.wikipedia.org/wiki/Messier_84&quot; rel=&quot;nofollow noreferrer&quot;&gt;M84&lt;/a&gt; is nearby, and may have had an encounter with M87 in the past. While it is a long shot (and M84 has a low star-forming rate), an interaction with the two could spawn new stars.&lt;/p&gt;&#xA;&lt;h3&gt;The supermassive black hole&lt;/h3&gt;&#xA;&lt;p&gt;You seemed particularly interested in the black hole at M87's center. It is a supermassive black hole, with a mass billions of times that of the Sun. Quite a monster, by many standards. Yet it is not doomed to swallow the galaxy whole anytime soon. It is sucking in gas and dust at a rate of &lt;span class=&quot;math-container&quot;&gt;$\sim10^{-1}M_{\odot}$&lt;/span&gt; each year (see &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0202238v3&quot; rel=&quot;nofollow noreferrer&quot;&gt;Di Matteo et al. (2002)&lt;/a&gt;) - and there are trillions of solar masses of gas, dust, and stars in the galaxy. For comparison, &lt;a href=&quot;https://en.wikipedia.org/wiki/Sagittarius_A*&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sagittarius A*&lt;/a&gt; accretes matter at a rate of between &lt;span class=&quot;math-container&quot;&gt;$\sim10^{-8}M_{\odot}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\sim10^{-4}M_{\odot}$&lt;/span&gt; per year, depending on the exact model (see e.g. &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/312035&quot; rel=&quot;nofollow noreferrer&quot;&gt;Quataert et al. (1999)&lt;/a&gt; and &lt;a href=&quot;http://adsabs.harvard.edu/abs/2012ApJ...755..133S&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shcherbakov et al. (2012)&lt;/a&gt;. Still, the supermassive black hole in M87 couldn't accrete all of the matter in the galaxy in the age of the universe, if it continued at this rate.&lt;/p&gt;&#xA;&lt;p&gt;In summary: Star formation in M87 is at a low rate, and while a galactic interaction could create a new wave of star formation, it is unlikely. The central black hole is sucking in lots of gas and dust, but it will take billions upon billions of years before it eats up a substantial amount of matter.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-11-08T20:22:52.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6420" PostTypeId="1" CreationDate="2014-09-21T17:39:09.267" Score="5" ViewCount="460" Body="&lt;p&gt;Why are the natural satelites (moons) of all planets - including the moons of the gas giants - solid or rocky, and not gaseous?&lt;/p&gt;&#xA;" OwnerUserId="2088" LastEditorUserId="2153" LastEditDate="2016-11-27T19:02:59.487" LastActivityDate="2016-11-27T19:02:59.487" Title="Why are the natural satellites (moons) of all planets solid?" Tags="&lt;solar-system&gt;&lt;natural-satellites&gt;&lt;gas-giants&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6421" PostTypeId="2" ParentId="6420" CreationDate="2014-09-21T17:52:02.030" Score="6" Body="&lt;p&gt;To answer this, we have to consider the definition of an atmosphere. A popular way of looking at it is to think of an atmosphere as a layer of gases surrounding a body. by that definition, we can say that gas giants are really just planets that are massive enough to accrete substantially large atmospheres (because deep down, they have a rocky core - although not a core similar to the &quot;rocky&quot; inner planets in our solar system). So now we can reduce the question to &quot;Why do most natural satellites not have extremely large atmospheres?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This comes down to mass. The more massive an object is, the more gravitational pull it has on objects around it. This means that a very massive object traveling through a region of gas would attract the gas more than a less massive object would. In the early Solar System, the more massive bodies far enough out in the protoplanetary disk managed to gain a substantial amount of matter via &lt;em&gt;core accretion&lt;/em&gt;, as gas and dust were collected by a rocky core. Moons are unable to do this because they are very low-mass; therefore, they cannot become gas giants (in fact, many moons around giant planets may be the remains of the material the planets never accreted).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Gas_dwarf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gas dwarfs&lt;/a&gt;, low-mass gas planets with small envelopes, are certainly possible. However, they would have to have radii at least 1.5-2 times the radius of the Earth, and masses scaled appropriately. Moons are nowhere near that size.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-11-27T19:01:43.960" LastActivityDate="2016-11-27T19:01:43.960" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6423" PostTypeId="1" CreationDate="2014-09-21T19:53:30.147" Score="3" ViewCount="139" Body="&lt;p&gt;If the rate of expansion of the universe is increasing , won't it eventually become greater than c? Also, after inflation, did the rate of expansion slow down and then start to accelerate again?&lt;/p&gt;&#xA;" OwnerUserId="2452" LastActivityDate="2014-09-22T14:58:51.527" Title="Is the expansion of the universe greater than the speed of light?" Tags="&lt;cosmological-inflation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6424" PostTypeId="2" ParentId="6423" CreationDate="2014-09-21T20:21:32.183" Score="4" Body="&lt;p&gt;After inflation, the expansion of the universe did indeed slow down. During the inflationary epoch (lasting roughly $1 \times 10^{-33}$ seconds), the universe expanded by a factor of $10^{26}$. That's incredible! However, the inflationary epoch didn't last long, and that incredible expansion ended pretty soon after it started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe is, as you said, still expanding. In fact (like you also said), that expansion is increasing. However, we have to be careful when we talk about the rate of this expansion. Currently, the rate of expansion between two objects depends on the distance between them, which is encoded in Hubble's law:&#xA;$$v=H_0D$$&#xA;where $v$ is the recessional velocity, $H_0$ is Hubble's constant, and $D$ is the proper distance between the objects. This relation proves that objects that are farther away are moving away at a greater speed. We can extrapolate from that that, at a sufficiently far distance, two objects would be moving away from each other at the speed of light or greater! However, this is only true for objects very far from each other. But yes, eventually, any two objects sufficiently far from each other will move apart at the speed of light - and then greater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source (for length of inflationary epoch): &lt;a href=&quot;http://www.universe-galaxies-stars.com/Cosmic_inflation.html&quot; rel=&quot;nofollow&quot;&gt;http://www.universe-galaxies-stars.com/Cosmic_inflation.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source (for Hubble's law): &lt;a href=&quot;http://map.gsfc.nasa.gov/universe/uni_expansion.html&quot; rel=&quot;nofollow&quot;&gt;http://map.gsfc.nasa.gov/universe/uni_expansion.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-22T14:58:51.527" LastActivityDate="2014-09-22T14:58:51.527" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6426" PostTypeId="1" CreationDate="2014-09-21T23:58:31.833" Score="4" ViewCount="83" Body="&lt;p&gt;All the calendars in my house say the fall equinox is Tuesday the 23rd. All the astronomical programs and astronomical sources say it is it is Monday the 22nd in Ohio. Since the time given is 10:29 p.m. EDT all of the US starts fall on Monday. I’m guessing the calendars use Greenwich Time. This same discussion also applies to when calendars show the full moon dates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this isn't exactly an astonomy question, but I thought this site was the closest I could get.&lt;/p&gt;&#xA;" OwnerUserId="2453" LastActivityDate="2014-11-27T22:53:04.970" Title="Is the location that published calendars use to calculate the seasons one specific location on the earth?" Tags="&lt;star-gazing&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6428" PostTypeId="2" ParentId="1845" CreationDate="2014-09-22T05:08:11.307" Score="0" Body="&lt;p&gt;i use this site and it works perfectly. You just enter your coordinates and you can get times ( UTC ) and cardinal directions ( i.e. from SW to NW ) for all passes of ISS and a number of satellites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://heavens-above.com/&quot; rel=&quot;nofollow&quot;&gt;http://heavens-above.com/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2014-09-22T05:08:11.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6429" PostTypeId="2" ParentId="1888" CreationDate="2014-09-22T05:17:46.353" Score="0" Body="&lt;p&gt;Centrifugal force is too small to cancel gravity. The ISS, as all the satellites, are actually &quot;always falling&quot;, but the horizontal component of their speed will always take them over the horizon line, keeping the distance to Earth constant. You have a very good explanation here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.lasalle.edu/~smithsc/Astronomy/Orbits/orbits.html&quot; rel=&quot;nofollow&quot;&gt;http://www.lasalle.edu/~smithsc/Astronomy/Orbits/orbits.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2014-09-22T05:17:46.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6430" PostTypeId="2" ParentId="6393" CreationDate="2014-09-22T05:37:39.280" Score="3" Body="&lt;p&gt;The difference is based on the different &lt;strong&gt;efficiency&lt;/strong&gt; of the processes.&#xA;We can describe the luminosity by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L = \eta m c^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\eta$ is the conversion efficiency, and describes how much matter can be converted into luminosity (photons).&#xA;Main sequence stars (if you mean this by &quot;normal operation&quot;) extract energy from matter by nuclear fusion.&#xA;The efficiency of nuclear fusion in stars is of the order of $\eta = 1\%$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other competitive mechanism, much more efficient than nuclear fusion, is &lt;a href=&quot;http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/000822a.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;conversion from gravitational potential&lt;/a&gt;.&#xA;In core-collapse SNe, this is the mechanism that brings to the explosion.&#xA;Gravitational potential energy is defined as $U=-\frac{GMm}{r}$ where &lt;em&gt;m&lt;/em&gt; is the mass of a particle, at a distance &lt;em&gt;r&lt;/em&gt;, from a large object of mass &lt;em&gt;M&lt;/em&gt;; &lt;em&gt;G&lt;/em&gt; is the gravitation constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider the SN scenario, $r$ is the distance of a particle that is falling from the external layer towards the center (center at $r=0$). So the particle goes from $r_1$ to $r_2$ with $r_2 &amp;lt; r_1$. Then the system lost part of the potential energy, converted to kinetic energy.&#xA;All the external matter collapse in a very brief time (order of seconds), from the external shell to the compressed nucleus, bringing more compression and then generating energetic shock in the explosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Virial_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;virial theorem&lt;/a&gt; says that half the change in gravitational energy stays with the star, while the other half is emitted as luminosity.&#xA;The efficiency coefficient for gravitational conversion is $\eta\sim10\%$, i.e., 10 times larger than the nuclear fusion efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://jila.colorado.edu/~ajsh/astr2030_12/sn/baade.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this reference&lt;/a&gt;, you can check that the ratio between the invisible against visible emission is of the order $\sim10^7$ (I don't know what kind of reference is that, but the authors are totally trusty).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something more &lt;a href=&quot;http://www.google.com/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=30&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0CF0QFjAJOBQ&amp;amp;url=http%3A%2F%2Fwww.astro.uvic.ca%2F~robb%2Fa102%2Fstarform.pdf&amp;amp;ei=2LAfVL2aL9Gi8AX65ICgBw&amp;amp;usg=AFQjCNH2fBPW6bpbTd6_yw3MrcfJDIaPYA&amp;amp;sig2=hfhzZYSeTt5UaXcK01HlXA&amp;amp;bvm=bv.75775273,d.dGc&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.google.com/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=15&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0CEQQFjAEOAo&amp;amp;url=ftp%3A%2F%2Fftp.astro.keele.ac.uk%2Fpub%2Fjnr%2Fgalaxies%2FAGN2_notes.pdf&amp;amp;ei=rrQfVJz8B4zg8AWT24KIAQ&amp;amp;usg=AFQjCNGbB_Fr7rwvyp3fLSfEICRmEGcjqA&amp;amp;sig2=DB5G_17my_YJc1j_v4iSBA&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastEditorUserId="1038" LastEditDate="2016-11-14T10:42:32.803" LastActivityDate="2016-11-14T10:42:32.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6432" PostTypeId="1" AcceptedAnswerId="6496" CreationDate="2014-09-22T07:39:15.047" Score="-8" ViewCount="457" Body="&lt;p&gt;What more can happen for the end of the humans?&lt;/p&gt;&#xA;" OwnerUserId="2456" LastEditorUserId="2456" LastEditDate="2021-04-28T14:27:21.070" LastActivityDate="2021-04-28T14:27:21.070" Title="The end of the Earth and human history" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;solar-system&gt;&lt;meteor&gt;" AnswerCount="3" CommentCount="11" ClosedDate="2015-11-02T16:15:51.243" ContentLicense="CC BY-SA 4.0" />
  <row Id="6433" PostTypeId="2" ParentId="1845" CreationDate="2014-09-22T07:45:07.370" Score="0" Body="&lt;p&gt;I use this app for it. Thanks to it I could be able to take a picture of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://play.google.com/store/apps/details?id=com.runar.issdetector&amp;amp;hl=es&quot; rel=&quot;nofollow&quot;&gt;https://play.google.com/store/apps/details?id=com.runar.issdetector&amp;amp;hl=es&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1678" LastActivityDate="2014-09-22T07:45:07.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6435" PostTypeId="1" CreationDate="2014-09-22T14:15:10.880" Score="8" ViewCount="151" Body="&lt;p&gt;Dream for a while and imagine that there will be a flagship+ mission within a decade to put a large radio telescope in space:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What kind of different tasks could the same radio equipment feasibly be used for?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How do the requirements for frequency sensitivity, electric power supply, dish size, and other radio technology (such as applicability of phased array?) and what else vary most importantly between such purposes?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For example, could one and the same big space radio telescope &lt;em&gt;in the Asteroid Belt&lt;/em&gt; be used for any, some or all of the below, what would the synergies and contradictions be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;a)&lt;/strong&gt; Radio astronomy on its own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;b)&lt;/strong&gt; Interferometry together with a telescope on Earth to create an interplanetary baseline for huge resolution, like the &lt;a href=&quot;https://en.wikipedia.org/wiki/Spektr-R&quot;&gt;spektr-R&lt;/a&gt; does cis-lunar today, AFAIK.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;c)&lt;/strong&gt; Radar characterization of the orbit, shape, size et cetera of any large enough asteroid in its current section of the asteroid belt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;d)&lt;/strong&gt; Ground penetrating radar to characterize the interior composition of near flyby asteroids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;e)&lt;/strong&gt; Communication with Earth so powerful that it need not engage the Deep Space Network (DSN), but can work with smaller Earth based radio antenna.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;f)&lt;/strong&gt; Extending the DSN by working as a relay between Earth and outer space probes, or between Earth and the backside of Mars or when Mars is in apposition behind the Sun from Earth.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2014-09-22T14:30:01.370" LastActivityDate="2014-12-08T17:38:38.450" Title="Could a space radio telescope fill multiple purposes?" Tags="&lt;radio-astronomy&gt;&lt;space-telescope&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6437" PostTypeId="2" ParentId="6432" CreationDate="2014-09-22T18:51:53.570" Score="5" Body="&lt;p&gt;The end of the Earth and the end of a habitat that can support life are 2 different things.  The Earth will not be able to support life as we know it for 4 billion more years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun is slowly heating up, and estimates are that in 500 million years, the Earth will be too hot to support life.  That being said, life is resilient, and who knows what evolutionary changes might occur that would allow life to survive 500 million years from now.  After all, the Sun heating up will be such a gradual process that evolution could most definitely keep up with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is the fact that in 500 million years, even if we aren't destroyed by cataclysm, humans as we are today won't exist.  We will have evolved into something else that wouldn't even remotely resemble humans.  Probably &lt;a href=&quot;http://en.wikipedia.org/wiki/Morlock&quot; rel=&quot;nofollow&quot;&gt;Morlocks&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technology also might allow us to escape our solar system and colonize across the universe.&lt;/p&gt;&#xA;" OwnerUserId="2439" LastEditorUserId="38" LastEditDate="2014-10-01T18:10:07.190" LastActivityDate="2014-10-01T18:10:07.190" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6439" PostTypeId="1" AcceptedAnswerId="6441" CreationDate="2014-09-23T08:42:32.223" Score="3" ViewCount="1609" Body="&lt;p&gt;An alternate explanation rather than from wikipedia is preferable.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="38" LastEditDate="2014-09-24T15:18:50.530" LastActivityDate="2014-09-24T15:18:50.530" Title="What is the Fundamental plane for Elliptical Galaxies?" Tags="&lt;galaxy&gt;&lt;astrophysics&gt;&lt;luminosity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6440" PostTypeId="2" ParentId="6344" CreationDate="2014-09-23T08:46:48.257" Score="-1" Body="&lt;p&gt;Milky Way is a young galaxy, filled with neutral hydrogen along its arms.&#xA;This hydrogen can undergo state transition, a rare event, which is compensated by the huge amount of hydrogen itself, i.e., even if this transition is rare it is still strong because a large amount of atoms can exhibit it.&#xA;The basics of this transition is explained by quantum physics through state-change in the energy levels of the hydrogen atoms, and can be summarized in the following picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fumwe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electron in the atom transits from a level with parallel spin (with respect to the nuclear proton), to the lower state, where the electron spin is antiparallel.&#xA;When this happens, the atom emits a photon, known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_line&quot; rel=&quot;nofollow noreferrer&quot;&gt;21 cm radiation&lt;/a&gt;, or &lt;em&gt;hydrogen line&lt;/em&gt;.&#xA;This line falls into the radio domain, which can be observed by radio telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this radiation comes from rotating arms, &lt;a href=&quot;http://csep10.phys.utk.edu/astr162/lect/light/doppler.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;doppler shift&lt;/a&gt; of the line can trace the distance of the emitter region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And voilà, you have your map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: ops, I missed the previous links where the answer was already given.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastActivityDate="2014-09-23T08:46:48.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6441" PostTypeId="2" ParentId="6439" CreationDate="2014-09-23T11:21:23.490" Score="8" Body="&lt;p&gt;I am not an expert, but it seems that the fundamental plane is a relation among characteristic quantities of the galaxy, showing a correlation that is the analogous of the Tully-Fisher relation for spiral galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can start from the virial theorem: &#xA;$\frac{M}{R}\sim v^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($M$ mass contained in the radius $R$, $v$ velocity dispersion of the stars, also indicated with $\sigma$ in these cases)&#xA;meaning that the stars behave like an isothermal sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that all galaxies have the same mass-to-light ratio&#xA;$L/M$, and that all galaxies have the same surface brightness&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Sigma=L/R^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(with $L$ luminosity), you obtain the Tully-Fisher relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L\sim v^4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, all galaxies do not have the same surface brightness.&#xA;So if you take $\Sigma=L/R^2$ and substitute that into the virial theorem (keeping the mass-to-light ratio assumption), then a relation between luminosity, surface brightness, and velocity dispersion is obtained:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L \sim \sigma^4 \Sigma^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can plot the stars distributions in elliptical galaxy according to this relation, and will note that the stars do not distribute randomly, but are more concentrated along the fundamental plane (more or less the same happens for the HR diagram):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/y7Qrg.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Here $R_e$ is the so called &lt;a href=&quot;http://en.wikipedia.org/wiki/Effective_radius&quot;&gt;effective radius&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For completeness, another illuminating picture from &lt;a href=&quot;http://www.astro.virginia.edu/class/whittle/astr553/Topic07/t7_FP_projections.html&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vuvdn.gif&quot; alt=&quot;enter image description here&quot;&gt;&#xA;References &lt;a href=&quot;http://www2.astro.psu.edu/users/rbc/a480/lec10n.pdf&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://star-www.st-and.ac.uk/~spd3/Teaching/AS3011/AS3011_6.pdf&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastEditorUserId="1038" LastEditDate="2014-09-23T11:26:31.677" LastActivityDate="2014-09-23T11:26:31.677" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6442" PostTypeId="2" ParentId="649" CreationDate="2014-09-23T12:00:17.640" Score="4" Body="&lt;p&gt;There have been some studies based on observations on the best preserved impact craters of that time in the inner solar system, namely those on Mercury and the Moon.  In the paper, &lt;a href=&quot;http://www.planetary.brown.edu/pdfs/4943.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Impact-induced compositional variations on Mercury&lt;/a&gt; (Edgard et al. 2014), state that based on the nature of Mercury's craters, they have developed a&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;New dynamical simulations of the early evolution of the asteroid belt hint at the presence of additional&#xA;  asteroids in a region interior to the present-day belt, known as the “E-belt”.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Through their modelling and simulations, they have determined that the 'E-belt'&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The recent reevaluation of the early history of the inner Solar System suggests the primordial asteroid belt could have originally extended well into the Mars-crossing zone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The 'Hungarias' are thought to be a remnant of this belt - as shown in the diagram below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lHZV4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image attribution: &#xA;By Agmartin (Own work) [&lt;a href=&quot;http://creativecommons.org/licenses/by/3.0&quot; rel=&quot;nofollow noreferrer&quot;&gt;CC-BY-3.0&lt;/a&gt;], &lt;a href=&quot;http://commons.wikimedia.org/wiki/File%3AE-belt_Asteroids.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;via Wikimedia Commons&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user2449" LastEditorDisplayName="user2449" LastEditDate="2014-09-23T12:39:22.163" LastActivityDate="2014-09-23T12:39:22.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6444" PostTypeId="2" ParentId="533" CreationDate="2014-09-23T12:08:26.293" Score="2" Body="&lt;p&gt;Recent studies and simulations reported in the paper &lt;a href=&quot;http://www.planetary.brown.edu/pdfs/4943.pdf&quot; rel=&quot;nofollow&quot;&gt;Impact-induced compositional variations on Mercury&lt;/a&gt; (Edgard et al. 2014), suggest that the Late Heavy Bombardment (LHB) was&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;triggered by giant planet migration during which the main belt and E-belt are excited to higher impact velocities when resonances with the giant planets sweep across the primordial asteroid belt. These simulations indicate the E-belt is the principal source of LHB impactors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The 'E-belt' is a simulated and theorised primordial extension of the asteroid belt that extended&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the primordial asteroid belt could have originally&#xA;  extended well into the Mars-crossing zone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the paper &lt;a href=&quot;http://www.gps.caltech.edu/classes/ge133/reading/asteroids.pdf&quot; rel=&quot;nofollow&quot;&gt;The Primordial Excitation and Clearing of the Asteroid Belt&lt;/a&gt; (Petit et al. 2001), state an important point to be considered when answering this question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;most of the asteroids&#xA;  we see today are not primordial, but fragments of larger asteroids&#xA;  destroyed in a collision. Only the largest asteroids retain characteristics&#xA;  that relate to the formation of the asteroid belt and were&#xA;  not drastically changed by the later evolution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In their simulations, they suggest that the asteroid belt could have been a major feature of the entire inner solar system where the 'planetary embryos are forming (from 0.5 to 4AU), stating:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the presence&#xA;  of large embryos in the inner Solar System for about 100 to&#xA;  200 My after Jupiter has reached its present-day mass provides&#xA;  an efficient mechanism for depleting the asteroid belt of most of&#xA;  its primordial mass and for dynamically exciting the remaining&#xA;  small bodies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Related to the Late Heavy Bombardment:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;few percent of the particles end up on these long-lived orbits,&#xA;  equivalent to several times the mass of the present asteroid belt.&#xA;  These orbits are unstable on a long timescale and represent a&#xA;  potential source of impactors for the Late Heavy Bombardment&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user2449" LastActivityDate="2014-09-23T12:08:26.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6445" PostTypeId="1" CreationDate="2014-09-23T04:30:51.347" Score="3" ViewCount="142" Body="&lt;p&gt;According to Ptolemy's (c. 150 CE) account of the motions of planets, planets moved in circular paths (&quot;epicycles&quot;) around center points that in turn moved around the center of the earth along a path called a &quot;deferent&quot;, or around point near the center of the earth, in a path called an &quot;eccentric&quot;.  (I think I have this right.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a general term for these three kinds of paths?  &quot;Orbit&quot; doesn't seem correct, since none of the three are orbits in the modern sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd also be interested in knowing of a general term of the centers of these circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I originally posted this on History.SE, because I think of it as a question for historians of science, and the History of Science SE site is in Area 51.  Someone at History.SE suggested it would be better to ask this question on English.SE.  I did, but I left the original question up at History.SE.  The question got downvoted at English.SE, and was migrated here.  There is now an &lt;a href=&quot;https://history.stackexchange.com/questions/15329/is-there-a-general-term-for-epicycles-deferents-and-eccentrics-in-ptolemaic-as/15333#15333&quot;&gt;answer&lt;/a&gt; at History.SE that I've accepted. Perhaps this question should be deleted from Astronomy.SE?  Or I should not have posted it in two places?  I'm reluctant to delete a question for which someone has posted an answer.&lt;/p&gt;&#xA;" OwnerUserId="2465" OwnerDisplayName="Mars" LastEditorUserId="-1" LastEditDate="2017-04-13T12:47:53.117" LastActivityDate="2015-09-18T20:47:36.957" Title="Is there a general term for epicycles, deferents, and eccentrics in Ptolemaic astronomy?" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;history&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6446" PostTypeId="2" ParentId="6445" CreationDate="2014-09-23T06:22:28.353" Score="1" Body="&lt;p&gt;Orbit can also mean &quot;the usual course of one's life or range of one's activities&quot; so I would find &lt;em&gt;Ptolemic orbit&lt;/em&gt; quite acceptable if you hesitate to use just &lt;em&gt;orbit&lt;/em&gt; when the path isn't elliptical.&lt;/p&gt;&#xA;" OwnerDisplayName="Deke Thomas" LastActivityDate="2014-09-23T06:22:28.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6447" PostTypeId="5" CreationDate="2014-09-23T15:51:02.807" Score="0" Body="&lt;p&gt;The theory of inflation was first proposed by Alan Guth and Andrei Linde to explain the large-scale isotropy of the universe. It posits that in less than in a tiny fraction of a second, the universe expanded by a factor of roughly 10^26. Not to be confused with the current expansion of the universe thought to be caused by dark energy.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="38" LastEditDate="2014-09-24T15:15:18.597" LastActivityDate="2014-09-24T15:15:18.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6448" PostTypeId="4" CreationDate="2014-09-23T15:51:02.807" Score="0" Body="Questions regarding the phenomenon in the early universe characterized by an extreme expansion of space." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-24T15:12:29.380" LastActivityDate="2014-09-24T15:12:29.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6449" PostTypeId="5" CreationDate="2014-09-24T01:14:36.673" Score="0" Body="&lt;p&gt;A satellite is an object that orbits another body (typically much more massive than the satellite), which is known as the primary. Satellites can be natural (e.g. moons) or artificial (e.g. a man-made object orbiting another body). Natural satellites may have formed with the primary, or may have been captured by it in the past.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-24T15:13:34.217" LastActivityDate="2014-09-24T15:13:34.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6450" PostTypeId="4" CreationDate="2014-09-24T01:14:36.673" Score="0" Body="Questions regarding an object orbiting around another object due to the force of gravity." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-24T15:12:40.360" LastActivityDate="2014-09-24T15:12:40.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6451" PostTypeId="1" AcceptedAnswerId="6501" CreationDate="2014-09-24T04:05:32.370" Score="18" ViewCount="430" Body="&lt;p&gt;In the article, &lt;a href=&quot;http://www.space.com/25218-dwarf-planet-discovery-solar-system-edge.html&quot;&gt;New Dwarf Planet Found at Solar System's Edge, Hints at Possible Faraway 'Planet X'&lt;/a&gt; (Wall, 2014) where they report on research and observations performed by astronomers that discovered the dwarf planet &lt;a href=&quot;https://solarsystem.nasa.gov/planets/profile.cfm?Object=Dwa_2012VP113&quot;&gt;2012 VP113&lt;/a&gt; which is smaller than, but has an orbit severely elongated like that observed for Sedna.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article reports the claim from the astronomers who discovered 2012 VP113 that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Based on the amount of sky the scientists searched, Trujillo and Sheppard &lt;strong&gt;estimate that about 900 bodies larger than Sedna may exist in this faraway realm&lt;/strong&gt;, which the astronomers dub the inner Oort Cloud.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How was the estimate, in bold above, calculated?&lt;/p&gt;&#xA;" OwnerDisplayName="user2449" LastActivityDate="2015-07-26T01:57:28.697" Title="How did scientists determine an estimate of the number of planets greater than Sedna's size to exist in the Inner Oort Cloud?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;oort-cloud&gt;&lt;dwarf-planets&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6452" PostTypeId="1" AcceptedAnswerId="6453" CreationDate="2014-09-24T04:51:19.777" Score="2" ViewCount="511" Body="&lt;p&gt;Can anyone explain by looking at the solution in the pic, how did the cos (i) came about?&#xA;I guess they are getting this from the major-minor axis info in the question, but I am not sure about the derivation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also v=300/sin(i), what is this formula?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ABPqs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="1038" LastEditDate="2014-09-24T05:31:43.703" LastActivityDate="2014-09-24T06:44:20.377" Title="Using Tully Fisher to measure Distance Problem" Tags="&lt;galaxy&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6453" PostTypeId="2" ParentId="6452" CreationDate="2014-09-24T06:44:20.377" Score="2" Body="&lt;p&gt;With the Tully Fisher relation, the &lt;a href=&quot;http://www2.astro.psu.edu/~caryl/a480/lecture8_10.pdf&quot; rel=&quot;nofollow&quot;&gt;inclination angle&lt;/a&gt; is determined from the major-minor axis ratio where $sin(i) = \sqrt{1-q^2} $ and q is the inverse of your ratio. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to apply the sine of the angle to the rotational velocity because the measure is based on our line of sight observation which must be converted to the true 3D velocity.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-09-24T06:44:20.377" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6455" PostTypeId="2" ParentId="1195" CreationDate="2014-09-24T20:33:16.333" Score="4" Body="&lt;p&gt;There are several types of supernova and ways that the core can collapse.  Lets take an extreme case in which gamma-ray photodisintegration destroys all of the heavy elements (Si, Fe and Ni, etc) and breaks them all up into protons, neutrons and electrons.  Each nucleus releases all of its binding energy, about 9 MeV per nucleon mass or 0.9% of the rest mass.  Most of the energy, I believe, comes out in the form of relativistic neutrinos (the rest in kinetic energy of the protons, neutrons, and electrons).  So, an upper limit is that 0.9% of the mass of the core ends up in neutrinos.  The rest mass of the neutrinos is much less, but the relativistic mass is probably the more relevant number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only a small fraction of closure density ($\Omega$) is in stars, $\Omega_{stars}$ = 0.0027 (Fukugita &amp;amp; Peebles, 2004), about 7% of mass in stars goes into supernova, ~10% is in core-collapse, 0.9% comes out in relativistic neutrinos.  So altogether $\Omega$ in hot dark matter from SN is less than 0.0027*0.07*0.1*0.009, roughly speaking.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2014-09-24T21:55:01.553" LastActivityDate="2014-09-24T21:55:01.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6456" PostTypeId="2" ParentId="6398" CreationDate="2014-09-24T21:18:06.617" Score="1" Body="&lt;p&gt;Black holes are created because the core of the star becomes very &lt;strong&gt;dense&lt;/strong&gt;, not just because the star is massive. Prior to the creation of the black hole, the core is able to create enough outward pressure to prevent the core from gravitationally collapsing to the density needing to create a black hole.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-09-24T21:18:06.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6458" PostTypeId="1" CreationDate="2014-09-25T03:48:09.943" Score="0" ViewCount="134" Body="&lt;p&gt;What is the exact size of the universe.is it infinite?&lt;/p&gt;&#xA;" OwnerUserId="2476" LastActivityDate="2014-09-29T12:13:33.707" Title="What is the size of the universe.is it infinite?" Tags="&lt;universe&gt;&lt;astrophysics&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2014-09-29T13:53:02.520" ContentLicense="CC BY-SA 3.0" />
  <row Id="6459" PostTypeId="2" ParentId="6458" CreationDate="2014-09-25T05:42:18.470" Score="0" Body="&lt;p&gt;&quot;No one knows if the universe is infinitely large, or even if ours is the only universe that exists.&quot; - NASA&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, the &quot;observable universe&quot; is a sphere roughly 92 billion light-years in diameter. &lt;/p&gt;&#xA;" OwnerUserId="1356" LastActivityDate="2014-09-25T05:42:18.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6460" PostTypeId="2" ParentId="6398" CreationDate="2014-09-25T06:56:19.273" Score="0" Body="&lt;p&gt;The star before it turns into a black hole has what is known as radiation pressure, i.e. the fusing of elements that creates nuclear explosions. This outward radiation pressure counter balances the inward force of gravity but when the star finally runs out of fuel that radiation pressure stops.  Thus the only remaining force is gravity.  Gravity then attracts what is left of the star after the super nova inwards into a deep dense core which then forms the black hole.&lt;/p&gt;&#xA;" OwnerUserId="2155" LastActivityDate="2014-09-25T06:56:19.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6461" PostTypeId="2" ParentId="6416" CreationDate="2014-09-25T14:17:28.783" Score="2" Body="&lt;p&gt;Take a look at this article: &lt;a href=&quot;http://www.nature.com/nature/journal/v513/n7519/full/nature13785.html&quot; rel=&quot;nofollow&quot;&gt;http://www.nature.com/nature/journal/v513/n7519/full/nature13785.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every more or less stable compound has its identifying absorption lines, based, well, on what the compound is made of. The question is, what is the minimum amount for a compound in an given atmosphere at a given distance with a certain sun to be still detectable. In deed, this also depends on how &quot;empty&quot; the light's way to us is. But guessing, you want to know, if we propably could detect an exo-planet with vegetation on it - yes, this is possible under circumstances. Most of all, the area covered by the possible organisms plays an important role.&lt;/p&gt;&#xA;" OwnerUserId="2479" LastActivityDate="2014-09-25T14:17:28.783" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6462" PostTypeId="2" ParentId="6408" CreationDate="2014-09-25T14:27:33.617" Score="4" Body="&lt;p&gt;I would recommend the &lt;a href=&quot;http://www.raspberrypi.org/tag/pi-noir/&quot; rel=&quot;nofollow&quot;&gt;PiNoir Camera&lt;/a&gt;. Since it has no IR-filter attached, it's perfectly made for nightly observations. If you need the Raspberry just to calculate Skyglow in general, you could try to collect all light and bundle it to a beam, then measure its intensity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What also came to mind is: When calculating the rating of the night, or a point in the sky, consider moon phases in your calculations. A bright and full moon will significantly raise average sky-glow and vice versa.&lt;/p&gt;&#xA;" OwnerUserId="2479" LastEditorUserId="38" LastEditDate="2014-09-25T15:46:38.820" LastActivityDate="2014-09-25T15:46:38.820" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6463" PostTypeId="1" AcceptedAnswerId="6464" CreationDate="2014-09-25T16:32:14.430" Score="4" ViewCount="235" Body="&lt;p&gt;Often I see news articles claiming astronomers found exoplanets with water. If we are capable of detecting water on such distant objects, why don't we employ similar methods to find water on Mars or Titan moon?&lt;/p&gt;&#xA;" OwnerUserId="1087" LastActivityDate="2014-09-25T18:08:08.757" Title="Why can't we detect water on Mars in the same way as in exoplanets?" Tags="&lt;exoplanet&gt;&lt;mars&gt;&lt;water&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6464" PostTypeId="2" ParentId="6463" CreationDate="2014-09-25T17:53:35.560" Score="8" Body="&lt;p&gt;We can look up for water in the atmosphere of the planet, not at the planet itself. We can do that when a planet passes in front of a star. That tehnique is called transmission spectroscopy.&lt;/p&gt;&#xA;" OwnerUserId="2481" LastEditorUserId="2481" LastEditDate="2014-09-25T18:08:08.757" LastActivityDate="2014-09-25T18:08:08.757" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6465" PostTypeId="5" CreationDate="2014-09-25T20:38:39.890" Score="0" Body="&lt;p&gt;General relativity predicts that matter and energy bend spacetime. As light travels through spacetime, it is influenced by other objects and follows a &quot;bent&quot; path. This phenomenon is known as gravitational lensing. It was first observed during the 1919 solar eclipse, by Sir Arthur Eddington, and was accepted as evidence of general relativity.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-29T13:49:54.350" LastActivityDate="2014-09-29T13:49:54.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6466" PostTypeId="4" CreationDate="2014-09-25T20:38:39.890" Score="0" Body="Questions regarding the bending of light by mass or energy." OwnerUserId="-1" LastEditorUserId="2153" LastEditDate="2015-05-25T14:34:03.943" LastActivityDate="2015-05-25T14:34:03.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6467" PostTypeId="5" CreationDate="2014-09-25T23:00:16.763" Score="0" Body="&lt;p&gt;The observable universe contains all the objects that we can detect using any method of observation. It currently has a diameter of 46 billion light-years, because cosmological expansion is moving objects apart at enormous speed. The size of the observable universe is constantly changing.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-29T13:51:59.780" LastActivityDate="2014-09-29T13:51:59.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6468" PostTypeId="4" CreationDate="2014-09-25T23:00:16.763" Score="0" Body="Questions regarding the region of space containing all the objects that we can detect using any method of observation." OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-08-15T13:15:57.887" LastActivityDate="2016-08-15T13:15:57.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6469" PostTypeId="5" CreationDate="2014-09-25T23:03:46.883" Score="0" Body="&lt;p&gt;In 1905, Albert Einstein published a series of papers. In one of them, he proposed his celebrated theory of special relativity. It posits that mass and energy are the same thing. It also predicts effects at high-speeds, such as mass increase, time dilation, and length contraction.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-09-29T13:52:02.557" LastActivityDate="2014-09-29T13:52:02.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6470" PostTypeId="4" CreationDate="2014-09-25T23:03:46.883" Score="0" Body="Questions regarding Albert Einstein's theory that posits, among other things, the equivalence of mass and energy." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-10-01T14:34:48.313" LastActivityDate="2014-10-01T14:34:48.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6471" PostTypeId="1" CreationDate="2014-09-26T17:43:13.277" Score="5" ViewCount="2209" Body="&lt;p&gt;As I understand it, a gamma ray burst is created when a massive star collapses into a black hole.  The poles of the black hole eject a massive burst of energy which we know as gamma ray bursts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question... Are gamma ray burst more or less cylindrical in shape?  Somewhat like a quasar, but in a spectrum that is invisible to our visible spectrum?  And if so, what would the diameter of it be?&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2017-02-10T18:19:39.740" Title="What is the shape and size of a gamma ray burst?" Tags="&lt;gamma-ray-bursts&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6475" PostTypeId="2" ParentId="6471" CreationDate="2014-09-26T22:07:51.707" Score="3" Body="&lt;p&gt;When &quot;shape&quot; is used in the context of gamma-ray bursts, it is typically used to refer to the jets emitted on opposite ends of an axis through the source. These jets contain huge amounts of energy, most in the form of photons so high in energy that they are classified as gamma-rays, some of the highest-frequency photons known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The jets are typically emitted along an axis through the progenitor object (I'm not going to talk about the object itself, as the causes of GRBs are not yet fully understood). A quick search engine search (I prefer duckduckgo, but you might like Google better) can give you hundreds of pictures (mostly artists' impressions) of GRBs. Notice how thin the jets are. They look like long cylinders, but closer inspection shows that they aren't. They are actually cones (truncated ones, but cones nonetheless).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that these jets spread out is &lt;a href=&quot;http://en.wikipedia.org/wiki/Relativistic_beaming&quot; rel=&quot;nofollow&quot;&gt;relativistic beaming&lt;/a&gt;, which also affect the beam's luminosity. The angle at which the beams diverge is so slim - in fact, &lt;a href=&quot;http://arxiv.org/pdf/astro-ph/9903339v1.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; pre-print says that one observed GRB had a jet emitted at an angle of one tenth of a radian! That's only 18 degrees. (Note: On page 21, &lt;a href=&quot;http://www2.astro.psu.edu/users/nnp/araa_pm.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; calls an angle of 3 degrees &quot;typical&quot; for some jets ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to your question is that gamma-ray bursts are typically conical, although the jets are often nearly cylindrical. The first paper I mentioned also did an analysis of spherical models, but concluded that they did not fit observations.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-01-11T17:42:57.677" LastActivityDate="2015-01-11T17:42:57.677" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="6476" PostTypeId="2" ParentId="6344" CreationDate="2014-09-27T12:01:49.640" Score="1" Body="&lt;p&gt;There are lots of reasons. First of all because we can see that a typical galaxy is a spiral galaxy and we seem to live somewhere pretty typical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly because we can see the arms of the spiral through close examination of the Milky Way (ie the band in the sky) - through stellar distances and measurements of hydrogen cloud red shifts and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you your question is a perfectly good one because, ultimately, we are guessing a bit as we cannot see the galaxy from anywhere other than one vantage point.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2014-09-27T12:01:49.640" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6477" PostTypeId="1" AcceptedAnswerId="6479" CreationDate="2014-09-27T21:46:31.477" Score="4" ViewCount="269" Body="&lt;p&gt;I've read that a gas giant, like Jupiter, doesn't have a rocky surface. But planets must start as a rocky conglomerate of flotsam. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is the interior so hot that all the material is molten and there is not a crust?  &lt;/li&gt;&#xA;&lt;li&gt;Does the atmosphere creates enough pressure to create a supercritical fluid?   &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is the situation on Jupiter?&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-09-27T22:56:02.523" Title="Planetary gas giants" Tags="&lt;planet&gt;&lt;gas-giants&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6478" PostTypeId="1" AcceptedAnswerId="6480" CreationDate="2014-09-27T22:11:49.640" Score="4" ViewCount="122" Body="&lt;p&gt;It is said that the cosmic microwave radiation (CMB) was formed when the universe was 379,000 years old. How is this calculated?&lt;/p&gt;&#xA;" OwnerUserId="2493" LastEditorUserId="2493" LastEditDate="2014-09-29T00:54:10.017" LastActivityDate="2014-09-29T00:54:10.017" Title="Formation of Cosmic Microwave Background" Tags="&lt;cosmology&gt;&lt;big-bang-theory&gt;&lt;cmb&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6479" PostTypeId="2" ParentId="6477" CreationDate="2014-09-27T22:56:02.523" Score="6" Body="&lt;p&gt;Short answer: 1) Yes and no; 2) Yes, there is a supercritical fluid, of hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's fairly hot deep down in Jupiter; estimates range from 10,000 K to 24,000 K. You would think that anything in the core would be liquefied, and you could be right. Many models predict that Jupiter's core is rocky, but others predict that it is liquid. Still others say that it is &quot;liquefying&quot; - i.e. parts are liquid and parts are solid. We don't know for sure which are right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the early solar system, Jupiter would indeed have to have had a large core of rock (and potentially ice). It would have accreted gases, and eventually the planet we know today would have formed, with thick cloud layers complete with hydrogen, ammonia, and lots of other elements. But after Jupiter accumulated its mantle, there would have been a lot of pressure on the original core. It seems like much of the matter inside it would have been liquefied and carried away (&quot;redistributed&quot;) into other parts of the planet. Still, some might have remained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surrounding the core, there is a layer of liquid hydrogen. Yep, &lt;em&gt;liquid&lt;/em&gt; hydrogen, just as you said in your question. Temperatures and pressures are so high that hydrogen has passed its critical temperature (33 K) and become a supercritical fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in summary, we don't know for sure the state of Jupiter's core is, but parts or all of it may be liquefied, and we also think that there is indeed a layer of hydrogen as a supercritical fluid surrounding the core. The same may also be said for other gas giants, although, still, we don't know for sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.dailygalaxy.com/my_weblog/2013/02/jupiters-dissolving-core-sheds-light-on-alien-planets.html&quot;&gt;http://www.dailygalaxy.com/my_weblog/2013/02/jupiters-dissolving-core-sheds-light-on-alien-planets.html&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.universetoday.com/47966/jupiters-core/&quot;&gt;http://www.universetoday.com/47966/jupiters-core/&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.nasa.gov/audience/forstudents/5-8/features/what-is-jupiter-58_prt.htm&quot;&gt;http://www.nasa.gov/audience/forstudents/5-8/features/what-is-jupiter-58_prt.htm&lt;/a&gt;&#xA;&lt;a href=&quot;http://phys.org/news/2011-12-jupiter-core-liquefying.html&quot;&gt;http://phys.org/news/2011-12-jupiter-core-liquefying.html&lt;/a&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Jupiter#Internal_structure&quot;&gt;https://en.wikipedia.org/wiki/Jupiter#Internal_structure&lt;/a&gt;&#xA;&lt;a href=&quot;http://arxiv.org/pdf/1111.6309v1.pdf&quot;&gt;http://arxiv.org/pdf/1111.6309v1.pdf&lt;/a&gt; (You might find this particularly interesting, as it considers gas-giant exoplanets)&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-09-27T22:56:02.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6480" PostTypeId="2" ParentId="6478" CreationDate="2014-09-28T17:59:28.723" Score="5" Body="&lt;p&gt;We observe the temperature of the CMB as a ~2.7 K blackbody, but that's the redshifted version we observe. The CMB is also know as the &quot;surface of last scattering&quot; at the point of  &lt;a href=&quot;http://en.wikipedia.org/wiki/Recombination_(cosmology)&quot;&gt;recombination&lt;/a&gt; when nuclei and electrons combined to form neutral atoms the universe went from opaque to transparent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This happens at a &lt;a href=&quot;http://ned.ipac.caltech.edu/level5/Sept02/Kinney/Kinney3.html&quot;&gt;temperature&lt;/a&gt; of ~3000 K. From this we can estimate the redshift (z~1100) of the CMB which corresponds to an age given our cosmology.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2014-09-28T17:59:28.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6481" PostTypeId="2" ParentId="6458" CreationDate="2014-09-29T12:13:33.707" Score="0" Body="&lt;p&gt;This isn't quite a question from metaphysics (i.e., beyond scientific enquiry) but it is close to it and so therefore very difficult to answer - especially in &quot;exact&quot; terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, we can begin with a discussion of what you mean by &quot;the Universe&quot;. Of course, by dictionary definition &quot;the Universe&quot; is everything, but it may be that we exist as a &quot;bubble&quot; inside a bigger Universe - the bigger Universe shares some common physical laws with ours but many physical constants (and hence behaviours) could be different and so that which we think of as &quot;the Universe&quot; might just be one little fragment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, in the &quot;eternal inflation&quot; cosmology - which is highjly regarded scientifically, our &quot;universe&quot; is in a different quantum phase from an infinite surrounding universe - and there are an infinitre number of similar &quot;universes&quot; to ours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, we could be in a sole Universe, which began with the Big Bang some 14 or so billion years ago and has been expanding ever since. But it might be that the Big Bang was the result of some interaction between some other physical aspects of a &quot;higher&quot; universe and so on...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Designing experiments to test these theories is tough, though not quite impossible. But as of today the answer to your question is either &quot;opinions vary&quot; or &quot;we do not know&quot;.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2014-09-29T12:13:33.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6482" PostTypeId="1" CreationDate="2014-09-29T16:46:20.193" Score="8" ViewCount="774" Body="&lt;p&gt;Newly born stars often have bilateral jets of gas ejected from them at high speeds (hundreds of km per seconds), often ending in Herbig-Haro objects. As I understand it those jets would have something to do with the star's magnetic field.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this work? How can a magnetic field cause the high-energy ejection of matter?&lt;/p&gt;&#xA;" OwnerUserId="1050" LastActivityDate="2017-10-06T23:43:12.233" Title="What causes jets from newly born stars?" Tags="&lt;star-formation&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6483" PostTypeId="2" ParentId="6344" CreationDate="2014-09-29T17:04:57.913" Score="0" Body="&lt;p&gt;Measure the distance to a star and its direction, and plot it based on these numbers in a 3D plot. Repeat for a gazillion other stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From its appearance in the sky we can already infer that the Milky Way is rather flat; most stars (and gas clouds) can be found in a rather narrow band in the sky. That the Milky Way is a spiral galaxy is something you can't see with the naked eye (nor with a telescope), but it's something you can derive from the 3D model.&lt;br&gt;&#xA;The 3D model also tells us where the sun is in the Milky Way: there are far more stars in one direction than the other, and the former ones are also farther way. This way we know we're about 2/3 from the center of the galaxy.&lt;/p&gt;&#xA;" OwnerUserId="1050" LastActivityDate="2014-09-29T17:04:57.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6484" PostTypeId="2" ParentId="6432" CreationDate="2014-09-29T22:03:08.013" Score="3" Body="&lt;p&gt;Read &quot;Death From the Skies!&quot; by Phil Plait.  His list is quite comprehensive, going into what can kill humanity, what could destroy all life on Earth, and what could destroy the Earth itself.  I don't have my copy on me at the moment, but IIRC he goes into what damage a meteor of various sizes will do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To avoid further confusion, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0143116045&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; is a link to the book on Amazon.  And if you don't know who Dr. Phil Plait is (shame on you), &lt;a href=&quot;http://en.wikipedia.org/wiki/Phil_Plait&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; is a link to his Wikipedia entry.&lt;/p&gt;&#xA;" OwnerUserId="2503" LastEditorUserId="2503" LastEditDate="2014-10-01T17:05:48.643" LastActivityDate="2014-10-01T17:05:48.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6485" PostTypeId="2" ParentId="4771" CreationDate="2014-09-29T22:29:49.090" Score="10" Body="&lt;p&gt;As evidence of caverns detected on Mars, consider the following recent image taken from HiRISE instrument on the Martian Reconnaissance Orbiter :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q7vds.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image source: &lt;a href=&quot;http://apod.nasa.gov/apod/ap120718.html&quot; rel=&quot;noreferrer&quot;&gt;NASA&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scientists believe that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The hole appears to be an opening to an underground cavern, partly illuminated on the image right. Analysis of this and follow-up images revealed the opening to be about 35 meters across, while the interior shadow angle indicates that the underlying cavern is roughly 20 meters deep. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which accompany larger openings, reported in the National Geographic article &lt;a href=&quot;http://news.nationalgeographic.com.au/news/2007/03/070321-mars-caves.html&quot; rel=&quot;noreferrer&quot;&gt;Mars Has Cave Networks, New Photos Suggest&lt;/a&gt; (Norris, 2007) of several larger pits detected by Odyssey, and they offer a possible origin:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The surface of Mars is strewn with craters from meteor impacts and depressions formed by the collapse of underground chambers formed by flowing lava, the experts said.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are many examples of lave tubes (and indeed, in &lt;a href=&quot;http://www.vulcanospeleology.org/sym06/ISV6x34.pdf&quot; rel=&quot;noreferrer&quot;&gt;other places in the Solar System&lt;/a&gt;), caverns etc reported in the article &lt;a href=&quot;http://www.space.com/18546-mars-caves-sample-return-mission.html&quot; rel=&quot;noreferrer&quot;&gt;Mars Cave-Exploration Mission Entices Scientists&lt;/a&gt; (Wall, 2012), who also state, in regards to detection and exploration of these caverns that robotic means are being explored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, according to &lt;a href=&quot;http://quest.nasa.gov/projects/spacewardbound/docs/ThermalBehaviorCaves.pdf&quot; rel=&quot;noreferrer&quot;&gt;this NASA article&lt;/a&gt;, a method to find and characterise caves on Mars (or indeed Titan, and maybe other places, including the &lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2010/12jul_rabbithole/&quot; rel=&quot;noreferrer&quot;&gt;Moon&lt;/a&gt;) is to detect them via thermal differences caves cause in comparison to the surrounding surface.  This method is used on Earth, and &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Techniques developed&#xA;  through this research will ultimately be applied to&#xA;  locating subterranean cavities on the Martian surface.&#xA;  We anticipate one of the best techniques for systematically&#xA;  finding caves on Mars will be via remotely&#xA;  sensed thermal imagery.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user2449" LastActivityDate="2014-09-29T22:29:49.090" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6486" PostTypeId="2" ParentId="6482" CreationDate="2014-09-29T23:49:27.917" Score="2" Body="&lt;p&gt;Did you get this from Wikipedia? If you did, I sympathize; all the entries related to this give no explanation for the phenomenon. If you didn't go there . . . Well, it won't answer your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think I've found the answer somewhere else. I'll go over some background stuff for anyone out there who doesn't know much about this, to make it a bit clearer (although you should be able to skip all this). Protostars are small pre-pre-main-sequence objects. At first, they aren't more than small clumps of gas and dust in a stellar nursery (a special nebula, perhaps). Soon, though, more matter begins to come towards a clump, and gradually the clump builds up enough matter that it begins a slight gravitational collapse into a definite object. Temperatures build, and eventually the object becomes a pre-main-sequence star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These pre-main-sequence stars begin to accrete matter (here's where you should stop dozing off) into a circumstellar disk. At the moment, it doesn't matter if the disk will become a protoplanetary disk or something completely different; for now, we can treat it as a circumstellar accretion disk. Matter &quot;falls&quot; into the disk and towards the star; the star absorbs most of it. However, something else comes into play, and that's the star's magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of objects in the universe have magnetic fields, and a young pre-main-sequence star is no exception. It's field may be stronger than that of the Earth - so strong, in fact, that some of the matter in the circumstellar accretion disk follows slightly different paths than it normally would - paths along the &lt;em&gt;field lines&lt;/em&gt; of the star's magnetic field. While there isn't enough evidence (scarcely any at all, really) to support or, more importantly, disprove this idea, it is though that the magnetic field lines lead the matter towards a point on the star, perhaps at its poles, and then compress it together into an astrophysical jet (or bipolar outflow) that shoots out into space. Such jets are often associated with other cool phenomena, such as Herbig-Haro objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for everyone who considers this post a &quot;Too long; didn't read&quot; answer, here's  a summary:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The truth is that &lt;em&gt;nobody really knows&lt;/em&gt;. The current explanations for this phenomenon are that they are caused by matter from accretion disks being concentrated by magnetic field lines onto a single (well, double, one on either end of an axis) point, which then becomes an astrophysical jet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note: &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1996swhs.conf..491B&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt;, specifically the second page, was extremely helpful. Other pages that might interest you (or provide more information for those who are curious about this and related topics) are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Protostar&quot; rel=&quot;nofollow&quot;&gt;Protostar&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Herbig_Ae/Be_star&quot; rel=&quot;nofollow&quot;&gt;Herbig Ae/Be star&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Herbig%E2%80%93Haro_object&quot; rel=&quot;nofollow&quot;&gt;Herbig-Haro object&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1996swhs.conf..491B&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow&quot;&gt;Paper #1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1987IAUS..115....1L&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow&quot;&gt;Paper #2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Pre-main-sequence_star&quot; rel=&quot;nofollow&quot;&gt;Pre-main-sequence star&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Circumstellar_disk&quot; rel=&quot;nofollow&quot;&gt;Circumstellar disk&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Astrophysical_jet&quot; rel=&quot;nofollow&quot;&gt;Astrophysical jet&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Bipolar_outflow&quot; rel=&quot;nofollow&quot;&gt;Bipolar outflow&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://news.softpedia.com/news/Protostar-Found-Spewing-Cosmic-Jets-168830.shtml&quot; rel=&quot;nofollow&quot;&gt;Softpedia.com&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sci-news.com/astronomy/science-binary-protostars-01661.html&quot; rel=&quot;nofollow&quot;&gt;Sci-news.com&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://iopscience.iop.org/0004-637X/779/2/93/pdf/0004-637X_779_2_93.pdf&quot; rel=&quot;nofollow&quot;&gt;IOP&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-01-10T23:05:26.493" LastActivityDate="2015-01-10T23:05:26.493" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6488" PostTypeId="2" ParentId="6308" CreationDate="2014-09-30T12:19:55.097" Score="1" Body="&lt;p&gt;Consider not only the physical attributes of the Venusian atmosphere, but its composition, from the &lt;a href=&quot;http://nssdc.gsfc.nasa.gov/planetary/factsheet/venusfact.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA Venus fact sheet&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Major:       96.5% Carbon Dioxide (CO2), 3.5% Nitrogen (N2) &#xA;Minor (ppm): Sulfur Dioxide (SO2) - 150; Argon (Ar) - 70; Water (H2O) - 20;&#xA;             Carbon Monoxide (CO) - 17; Helium (He) - 12; Neon (Ne) - 7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The 1st and 3rd most abundant components - carbon dioxide and and sulphur dioxide, at the pressures and temperatures of the surface would result in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The reactions probably include the deterioration of silicates by carbon dioxide to produce carbonates and quartz, as well as the deterioration of silicates by sulfur dioxide to produce anhydrate calcium sulfate and carbon dioxide.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the &lt;a href=&quot;http://en.wikipedia.org/wiki/Geology_of_Venus&quot; rel=&quot;nofollow noreferrer&quot;&gt;Venus Wikipedia page&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A study reported in the paper &lt;a href=&quot;http://www.lpi.usra.edu/meetings/venus2009/pdf/2011.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Venus Geochemistry: Progress, Prospects, and New Missions (2009)&lt;/a&gt;, focusing on the basalt-atmosphere interaction, determined that the conditions on Venus is roughly the equivalent of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a metamorphic environment of ‘amphibolite&#xA;  grade,’ ‘hornfels facies;’ the paucity of H2O&#xA;  in Venus’ atmosphere (30±15 ppm) prevents formation&#xA;  of amphibole&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The metamorphic facies for Earth that are used in the above quote are shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0goAF.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://homepage.usask.ca/~mjr347/prog/geoe118/geoe118.029.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The implications of the surface-atmosphere chemistry, pressure and temperature would, according to the Venus geochemistry article (linked above), result in the formation of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;anhydrite from essentially&#xA;  all the Ca in basaltic minerals and rocks. Iron in the&#xA;  rocks and minerals will likely react to either oxides or&#xA;  sulfides, depending on the oxidation state of the atmosphere.&#xA;  The fate of alkalis is not clear – our calculations&#xA;  yield only alkali feldspar (albite), but alkalis&#xA;  could be partitioned into ionic (carbonate-sulfate)&#xA;  melts. Some Venus basalts are so rich in potassium&#xA;  that phases besides feldspar may be present (e.g.,&#xA;  leucite, nepheline-group, or sodalite-group).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user2449" LastActivityDate="2014-09-30T12:19:55.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6489" PostTypeId="1" CreationDate="2014-09-30T16:34:22.137" Score="2" ViewCount="236" Body="&lt;p&gt;I was watching an old Ancient Aliens last night and there was a whole episode dedicated to how the &quot;Gods&quot; (aliens) were spreading disease/plagues.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good portion of the show was dedicated to how aliens may be sending meteors to Earth with bacteria to spread disease.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of this was several scientists talking about an experiment with a rock set in orbit (forgive me that I don't have a link to this).  Said rock had many forms of bacteria on it and after a certain amount of time they grabbed the rock and confirmed that some bacteria can survive space conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there were other scientists saying that some forms of bacteria/virus may even get worse in zero gravity or space conditions - used salmonella as an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my question is does this even matter?  Wouldn't any bacteria or virus colonies on a meteorite just burn up in our atmosphere? &lt;/p&gt;&#xA;" OwnerUserId="2506" LastActivityDate="2014-10-01T22:56:32.377" Title="Would bacteria on incoming meteors burn before impacting?" Tags="&lt;meteor&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6490" PostTypeId="2" ParentId="6489" CreationDate="2014-09-30T18:22:22.397" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Svante_Arrhenius&quot; rel=&quot;nofollow noreferrer&quot;&gt;Svante Arrhenius&lt;/a&gt; was probably the first scientist to propose the idea of panspermia, that life may have come from, well, that black thing we all have over our heads at night, and all whatever hides behind that dotted blackness.&lt;/p&gt;&#xA;&lt;p&gt;Since it is genuinely unknown how life appeared, I think your question cannot have any good answer. Yeah, maybe microgravity doesn't seem to be any problem to microbes. Au contraire, those little bugs seem to love microgravity and multiply, and mutate, even faster there than here.&lt;/p&gt;&#xA;&lt;p&gt;That burning up of a rock in the atmosphere thing would've happened in many different ways, in different angles and speeds throughout the billions of years. The argument is that maybe once or twice... And maybe that's enough. You get a cold once and you got it.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2014-09-30T18:31:27.840" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6492" PostTypeId="1" CreationDate="2014-09-30T22:33:48.997" Score="7" ViewCount="276" Body="&lt;p&gt;It is often possible to find a list of the materials that compose the cores and the mantle of plantes (often they are iron and/or nickel).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can these materials be determined in our Solar system? In other words, for example, how can we know that Jupiter's core certanly contains iron?&lt;/p&gt;&#xA;" OwnerUserId="1423" LastActivityDate="2014-11-03T11:17:16.917" Title="Composition of planets' core" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;jupiter&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6496" PostTypeId="2" ParentId="6432" CreationDate="2014-10-01T06:43:50.817" Score="5" Body="&lt;p&gt;This has been the subject of many documentaries, many of which are based on scientific observations and associated assumptions of the past in the rocks and out in space. Some are outright bizarre imaginings of people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://discovermagazine.com/2000/oct/featworld&quot; rel=&quot;noreferrer&quot;&gt;Discover Magazine article&lt;/a&gt; about this topic lists several possibilities, including the following examples that scientists have evidence of causing mass extinctions in Earth's geological history (note: this is not an exhaustive list):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;An asteroid/comet impact - which is the widely accepted theory of the cause of the &lt;a href=&quot;http://www2.jpl.nasa.gov/sl9/back3.html&quot; rel=&quot;noreferrer&quot;&gt;KT extinction event&lt;/a&gt; that wiped out 70% of life 65 million years ago, including most of the dinosaurs.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Gamma ray burst - which some scientists hypothesise caused a &lt;a href=&quot;http://news.nationalgeographic.com.au/news/2009/04/090403-gamma-ray-extinction.html&quot; rel=&quot;noreferrer&quot;&gt;mass extinction at the end of the Ordovician period&lt;/a&gt;, around 440 million years ago.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mass volcanism - such as the what is thought by many scientists to have caused the phenomenally bad &lt;a href=&quot;http://en.wikipedia.org/wiki/Permian%E2%80%93Triassic_extinction_event&quot; rel=&quot;noreferrer&quot;&gt;extinction event (over 90% life forms extinct) at the end of the Permian period&lt;/a&gt; (around 250 million years ago) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It should be noted that some extinction level events are not one single event and some are not sudden - for example, the extinction event at the end of the Permian ('&lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2002/28jan_extinction/&quot; rel=&quot;noreferrer&quot;&gt;The Great Dying&quot;)&lt;/a&gt;is theorised to have had multiple causes - volcanism, impact, methane clathrate release etc and the extinction occurred in pulses separated by thousands of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other causes could include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Collisions with rogue planets&lt;/li&gt;&#xA;&lt;li&gt;Epidemic&lt;/li&gt;&#xA;&lt;li&gt;War&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and so on.&lt;/p&gt;&#xA;" OwnerDisplayName="user2449" LastActivityDate="2014-10-01T06:43:50.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6497" PostTypeId="2" ParentId="6489" CreationDate="2014-10-01T08:39:02.743" Score="0" Body="&lt;p&gt;While the surface of a meteor will indeed be pretty sterilised by re-entry, remember that a meteor also has an inside. If it doesn't burn up completely, then parts of the inside may remain cool enough for bacteria/viruses to survive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The range of environmental conditions life can survive in is incredibly wide, so this is not impossible.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2014-10-01T08:39:02.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6498" PostTypeId="2" ParentId="6492" CreationDate="2014-10-01T11:54:00.957" Score="2" Body="&lt;p&gt;With the volume and mass of a planet concludes are possible of its material composition. But there is more, for example auroras been observed at Jupiter. This is only possible if Jupiter has a magnetic field just like our earth. A magnetic field is a strong evidence for an iron core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next point is, heavy materials accumulate in the center, so one could ask why there isn’t gold or even heavier stuff in the core of a planet. The most common material in the universe is hydrogen. Hydrogenis burned through atomic fusion into helium. All heavier materials are produced if a star dies in an explosion. So really heavy materials like gold are less likely and iron or nickel are more common so cores more likely made of these materials. &lt;/p&gt;&#xA;" OwnerUserId="2457" LastEditorUserId="2457" LastEditDate="2014-10-04T10:38:37.293" LastActivityDate="2014-10-04T10:38:37.293" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6499" PostTypeId="1" CreationDate="2014-10-01T14:33:02.337" Score="2" ViewCount="47" Body="&lt;p&gt;Hypothetically, if Dyson spheres were already present, how could we observe them? My layman understanding of the reason we believe dark matter exists is that our current estimates of the total mass of the universe is very different from the observed stars. But what if a significant proportion of those stars were covered? My obvious hypothesis is that this would explain the missing mass of the universe, and solve both the question of dark matter and the existence of aliens.&lt;/p&gt;&#xA;" OwnerUserId="2515" LastEditorUserId="2153" LastEditDate="2014-10-01T15:22:43.010" LastActivityDate="2014-10-01T15:22:43.010" Title="How could we observe Dyson spheres?" Tags="&lt;extra-terrestrial&gt;" AnswerCount="0" CommentCount="6" ClosedDate="2014-10-01T19:10:18.117" ContentLicense="CC BY-SA 3.0" />
  <row Id="6500" PostTypeId="2" ParentId="6489" CreationDate="2014-10-01T22:56:32.377" Score="1" Body="&lt;p&gt;Any bacteria present in a meteor would not necessarily burn.  According to the Bad Astronomer's &amp;quot;Bad Addendum&amp;quot; to another unrelated point, &lt;a href=&quot;http://www.badastronomy.com/bad/misc/meteoric.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&amp;quot;A Meteoric Rise&amp;quot;&lt;/a&gt;, meteors sure do get extremely hot, but that outside part usually ablates away, leaving the inner part that was extremely cold for who knows how long, which may not have warmed up much during the fall.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Bad Addendum: many people think that a meteorite, after it hits the ground, is very hot and glows red. Actually, meteorites found shortly after impact tend to be warm, but not hot at all! It turns out that it certainly is hot enough to glow while it is in the part of the atmosphere that decelerates it the strongest, but any part that actually melts will be blown off (&amp;quot;ablated&amp;quot;) by the wind of its passage. That leaves only the warm part. Even more, the meteor is slowed down so strongly as it moves through the atmosphere that the impact speed is typically only a few hundred kilometers an hour at most. Only the very large (and we're talking meters across) meteors are still moving at thousands of kilometers an hour or more when they impact. Small ones aren't moving that fast at all. Not to say you'd want to be under one: a car in New York was struck by a small meteorite and had a hole punched through it, and the whole back end crushed in. Ouch!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.amsmeteors.org/fireballs/faqf/#9&quot; rel=&quot;nofollow noreferrer&quot;&gt;This American Meteor Society resource&lt;/a&gt; seems also to indicate that meteorites can fall to the ground without being extremely hot.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol start=&quot;9&quot;&gt;&#xA;&lt;li&gt;Are meteorites “glowing” hot when they reach the ground?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Probably not. The ablation process, which occurs over the majority of the meteorite’s path, is a very efficient heat removal method, and was effectively copied for use during the early manned space flights for re-entry into the atmosphere. During the final free-fall portion of their flight, meteorites undergo very little frictional heating, and probably reach the ground at only slightly above ambient temperature.&lt;/p&gt;&#xA;&lt;p&gt;For the obvious reason, however, exact data on meteorite impact temperatures is rather scarce and prone to hearsay. Therefore, we are only able to give you an educated guess based upon our current knowledge of these events.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This study, &lt;a href=&quot;http://link.springer.com/chapter/10.1007%2F3-540-25736-5_3#page-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;&amp;quot;Bacterial Spores Survive Simulated Meteorite Impact&amp;quot;&lt;/a&gt;, seems to indicate that it's possible, though unlikely, for a bacterium to survive the pressure shock from such an impact.&lt;/p&gt;&#xA;&lt;p&gt;So it's plausible that an extremophile bacterium could somehow survive this impact by being on the inside of the meteoroid, surviving the ablation process as a meteor, and not even get warmed up to the surrounding atmospheric temperature when it crashes as a meteorite.&lt;/p&gt;&#xA;&lt;p&gt;There is nothing to indicate that this has actually happened.  But what a discovery it would be if something like this were to happen and be confirmed.&lt;/p&gt;&#xA;" OwnerUserId="1082" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2014-10-01T22:56:32.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6501" PostTypeId="2" ParentId="6451" CreationDate="2014-10-01T23:45:54.233" Score="13" Body="&lt;p&gt;Unfortunately, the paper is not available on ArXiv (oh, what hardships we must overcome!), but I &lt;em&gt;have&lt;/em&gt; found it &lt;a href=&quot;http://www.slideshare.net/APODman2k/nature13156&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. In it, where the &quot;900&quot; figure is mentioned (2nd page, I believe), the authors (Trujillo and Sheppard) say that they ran simulations with the data already found and their additional findings, and found that 900$^{+800}_{-500}$ bodies larger than 1,000 km in diameter could exist. They also found that 430$^{+400}_{-240}$ relatively &quot;bright&quot; bodies could exist, as well. Notice that there is quite a lot of uncertainty there; they are rather cautious (and rightfully so) when announcing the results of their simulations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly enough, they also mention &lt;a href=&quot;http://www.astro.yale.edu/mschwamb/Site/Publications_files/2010Schwamb.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;a paper&lt;/a&gt; in that section, written partly by none other than &lt;a href=&quot;http://en.wikipedia.org/wiki/Michael_E._Brown&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mike Brown&lt;/a&gt; (!) that also used simulation methods. The authors, Schwamb et al, concluded (in 5.4, as well as the conclusion) that 393$^{+1286}_{-264}$ and 74$^{+249}_{-47}$ objects greater than or equal to Sedna &lt;em&gt;in brightness&lt;/em&gt;, not size, could exist (393 is for the &quot;hot&quot; population, while 74 is for the &quot;cold&quot; population). The reason Trujillo and Sheppard cited this paper is that their simulation results fit the results of the simulations by Schwamb et al., which is encouraging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summary: Trujillo and Sheppard arrived at 900 via simulations based on prior observations. There is a large amount of uncertainty in this number, but it does agree with prior simulations by Schwamb et al. While &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6451/how-did-scientists-determine-an-estimate-of-the-number-of-planets-greater-than-s/6501#comment7011_6451&quot;&gt;zibadawatimmy 's comment&lt;/a&gt; is perhaps the funniest I have seen, it is, fortunately, inaccurate. Those statisticians must be relieved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-07-26T01:57:28.697" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6502" PostTypeId="2" ParentId="2528" CreationDate="2014-10-03T00:24:52.417" Score="2" Body="&lt;p&gt;I'm going to build off of chris's original comment. I haven't had a response from him during the 24 hours since I asked the question, and there doesn't appear to have been any activity from him recently (Note: don't confuse him with &lt;em&gt;another&lt;/em&gt; user who goes by &lt;em&gt;C&lt;/em&gt;hris), so I might as well expand upon what he said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational waves do appear to be what you're looking for. They are emitted by systems with varying quadrupole moments (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_wave&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Gravitational_wave&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Quadrupole&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Quadrupole&lt;/a&gt; for more information); commonly cited examples are binary neutron stars. In fact, one such binary, the Hulse-Taylor binary, was the first discovered system to emit gravitational waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational waves carry energy away from the system, at a rate of&#xA;$$\frac{dE}{dt}=-\frac{32}{5}\frac{G^4}{c^5}\frac{(m_1m_2)^2(m_1+m_2)}{r^5}$$&#xA;where $E$ is eneryg, $t$ is time, $m_1$ and $m_2$ are the masses of the objects in the system, $r$ is the distance between them, and $G$ and $c$ are the constants, the universal gravitational constant and the speed of light. I invite you to do the calculations for a given system, if you please. I can assure you that it's one of the easier calculations in general relativity! This release of energy causes the orbits of the two neutron stars to gradually decay, and it is thought that eventually the two will merge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer boils down to this: Yes, gravitational waves can carry [angular] momentum, just like many other types of waves. They also have frequency, amplitude, wavelength, and speed, just like the &quot;normal&quot; waves we are familiar with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-10-03T00:34:30.387" LastActivityDate="2014-10-03T00:34:30.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6505" PostTypeId="1" CreationDate="2014-10-03T09:05:07.457" Score="5" ViewCount="294" Body="&lt;p&gt;If I'm correctly informed, only three stars: Sol, Betelgeuse and Altair have surfaces which have been resolved by telescopes. All other stars are only point sources of light, even in the greatest of telescopes. Is it just the huge distance to the stars which reduce them to dots? Or is there some other optical explanation? I mean, the Sun is about as large as the Moon on the sky, but I cannot see any details on its surface because it blinds me. However, the next nearest star is invisible to the naked eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some exoplanets have been directly imaged by removing the light from their star. It seems funny that we cannot detect any features on the photosphere of such a star, such as spots or flares or its shape. Plenty of photons are available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, there are several stars which have been directly imaged, maybe my sopposition is quite wrong:&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_stars_with_resolved_images&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/List_of_stars_with_resolved_images&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2014-10-03T09:23:14.797" LastActivityDate="2014-12-26T19:37:21.977" Title="Why can't the surfaces of stars be observed?" Tags="&lt;star&gt;&lt;optics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6506" PostTypeId="1" AcceptedAnswerId="6509" CreationDate="2014-10-03T13:34:56.977" Score="2" ViewCount="291" Body="&lt;p&gt;If the universe is expanding, then if you were to be close to the edge of that expansion, what would it look like? What would be happening? Would it be chaotic? Would objects be appearing or disappearing? I don't even know if it's possible to have any knowledge of this, but I'd like to know if there are any theories.&lt;/p&gt;&#xA;" OwnerUserId="2523" LastActivityDate="2014-10-03T15:07:48.743" Title="What is it like at the edge of the universe?" Tags="&lt;universe&gt;" AnswerCount="3" CommentCount="1" ClosedDate="2014-10-06T11:48:30.797" ContentLicense="CC BY-SA 3.0" />
  <row Id="6507" PostTypeId="2" ParentId="6506" CreationDate="2014-10-03T14:33:52.047" Score="0" Body="&lt;p&gt;The major cosmological models of the universe say that the universe as no edge. The universe is either infinite (if you travel in one direction you would keep going forever) or it backs up on itself (if you travel in one direction you end up where you started, kind of like in pacman). With this picture in mind there is no edge where there stuff being added to the universe. The expansion is happening everywhere in the universe. If you want to know how it looks like, just look around!&#xA;Objects would not be appearing and disappearing as the expansion of the universe involve only space and not matter.&lt;/p&gt;&#xA;" OwnerUserId="2524" LastActivityDate="2014-10-03T14:33:52.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6508" PostTypeId="2" ParentId="6506" CreationDate="2014-10-03T15:00:55.873" Score="0" Body="&lt;p&gt;It's simple - there is no edge to the universe. It is thought that the universe is infinite in size. Data from WMAP seems to support this, because it shows that the universe is very close to &quot;flat&quot; - i.e. the overall curvature is zero. This does support the &quot;infinite universe&quot; models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, there &lt;em&gt;is&lt;/em&gt; an edge to our (or any other observer's) &lt;em&gt;observable&lt;/em&gt; universe. This consists of all the objects we can see - or, rather, all the objects whose emitted light, gravity, or other effects we can detect. There isn't anything special going on at the edge of our observable universe, either, for those observers there. Nothing special happens for them, although any observer's observable universe is expanding.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-10-03T15:00:55.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6509" PostTypeId="2" ParentId="6506" CreationDate="2014-10-03T15:07:48.743" Score="2" Body="&lt;p&gt;The edge of the Universe in space-time can only be the moment of its creation (or end if the Universe is closed). What it would be like at that time depends on what you would consider close to the edge. Your question is about the edge of the expansion so I will presume that you mean during the Inflationary era (until about $10^{-32}\textrm{s}$ after the creation of the Universe).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There would be no objects as matter as we know it today did not exist (according to the big bang model). The major interacting forces (gravity, strong and weak nuclear force, and the electromagnetic force) were unified at the time of the Inflationary era. There would be no protons or neutrons as the strong interaction has not decoupled yet from the other forces. The temperature would be extremely high $&amp;gt;10^{27}\textrm{K}$.&lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2014-10-03T15:07:48.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6510" PostTypeId="2" ParentId="6408" CreationDate="2014-10-03T20:52:16.270" Score="3" Body="&lt;p&gt;Well, light, and the resulting sky glow comes in a lot of different wavelengths, so the first question you need to ask yourself is what range are you looking for?  If this project is intended to give someone a good idea as to whether it is worth their while to drag out the telescope on a given night, chances are you only care about human visible wavelengths (but don't underestimate the amateur radio astronomers in your area!).  My phone's camera will pick up light in the near-infrared, so you want to be careful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, you need a camera with very low noise.  If you go into a windowless room with the lights off, and take a picture, you want to have as close to 0x000000 rgb measurement from every resulting pixel as possible.  A magnitude 0 star will only produce 2.08 microlux, so electrical noise can alter your results significantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, and this may be obvious, make yourself a cardboard &quot;horizon shield&quot; (a name I made up) to block any direct sources of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if it isn't of vital importance to use an actual camera, since you are using a raspberry pi, you might just want to go with a lux meter instead. &lt;a href=&quot;https://www.modmypi.com/adafruit-digital-luminosity-sensor&quot; rel=&quot;nofollow&quot;&gt;This one&lt;/a&gt; has a separate meter for the human visible wavelength, and the noise should be very low, depending upon how you wire it and what voltage you supply.&lt;/p&gt;&#xA;" OwnerUserId="2503" LastActivityDate="2014-10-03T20:52:16.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6511" PostTypeId="1" AcceptedAnswerId="6512" CreationDate="2014-10-03T21:12:01.313" Score="6" ViewCount="1530" Body="&lt;p&gt;Just wondering if we can examine individual stars from other galaxies, or if we are pretty much stuck with the billions that are in the Milky Way?&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2014-10-12T19:21:50.340" Title="Can we see individual stars in other galaxies?" Tags="&lt;star&gt;&lt;galaxy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6512" PostTypeId="2" ParentId="6511" CreationDate="2014-10-03T21:49:56.697" Score="8" Body="&lt;p&gt;Yes, Edwin Hubble did that for the first time in 1919. Before that time, it was thought that the galaxies we can observe were just nearby gas nebulae located inside our Milky Way. But Hubble was able to resolve the nearby galaxies like the Andromeda nebula into individual stars. By measuring the brightness of so-called Cepheid variable stars, he was able to calculate the distance to the Andromeda galaxy. Here one uses the fact that the total power radiated by the star is related to the period of the brightness oscillations, so by observing such stars you can deduce the distance to these stars and hence the distance to the galaxy. But later it was found that there were two different types of Cepheid stars and the wrong relation had been used; the distances were actually about twice as large.&lt;/p&gt;&#xA;" OwnerUserId="2185" LastEditorUserId="2185" LastEditDate="2014-10-04T18:00:01.010" LastActivityDate="2014-10-04T18:00:01.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6514" PostTypeId="1" CreationDate="2014-10-04T05:10:11.493" Score="2" ViewCount="220" Body="&lt;p&gt;Since dark energy is causing everything to speed up infinitely, wouldn't everything near the speed of light? And since it's nearing the speed of light, wouldn't all kinds of strange things start happening such as time dilation?&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2015-01-10T23:08:26.450" Title="If the expansion of the universe is speeding up, how long until everything nears the speed of light, and what would happen?" Tags="&lt;universe&gt;&lt;light&gt;&lt;expansion&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6515" PostTypeId="2" ParentId="6505" CreationDate="2014-10-04T13:08:19.827" Score="9" Body="&lt;p&gt;There are many different ways to get spatial information about the surface of a star besides direct imaging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Direct imaging is difficult because the angular resolution available goes as $\lambda/D$.&#xA;For a 8-m telescope and light at 500 nm, one can resolve $6\times10^{-8}$ radians (assuming the blurring of the atmosphere can be overcome by adaptive optics or similar).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nearest stars are a couple of parsecs away, so the smallest spatial scales that could be resolved are $\sim 2\times 3.1\times10^{16} \times 6\times10^{-8} =  3.7\times10^{9}$ m, or about 500 solar radii. Hence no surface features or even a disk could be resolved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course you could use interferometric techniques to effectively increase the size of $D$ and measurements of the angular radii are now possible for many nearby stars or giant stars at larger distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surface imaging is harder. Indirect techniques are much more common. These include &lt;a href=&quot;http://en.wikipedia.org/wiki/Doppler_imaging&quot; rel=&quot;noreferrer&quot;&gt;doppler imaging&lt;/a&gt; and eclipse mapping. The former uses the fact that there is a relationship between the position of a bright/dark feature on a &lt;em&gt;rotating&lt;/em&gt; star and the doppler shift of the light from that feature. By observing a time-series of spectra, the lumps and bumps in spectral lines can be inverted to produce a &quot;doppler map&quot; of the surface. The technique is usually limited to stars that are considerably more rapidly rotating than the Sun. There is a lot of ambiguity in the image reconstruction process - many surfaces could lead to the same observable sigature and clever statistical techniques (and even philosphies) have to be deployed to choose between them. Many stars have published &quot;doppler maps&quot; of their surfaces. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2005MNRAS.356.1501B&quot; rel=&quot;noreferrer&quot;&gt;Here&lt;/a&gt; is a typical example of such a study and below I show an example of a &quot;doppler image&quot; for the star II Peg (a K-type subgiant), from &lt;a href=&quot;http://www.aanda.org/articles/aa/full/2003/26/aa3086/aa3086.right.html&quot; rel=&quot;noreferrer&quot;&gt;Gu et al. (2003)&lt;/a&gt;, showing dark spotted regions. A typical resolution for such an image is about 10 degrees on the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/78USH.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eclipse mapping, for which I can't easily locate a good link, uses the fact that a star/accretion disk is orbited by another star or planet that periodically eclipses it. What happens to the light from the system in and out of eclipse can be used to probe the surface of the eclipsed object. There are of course limitations to the spatial resolution that can be obtained, depending on the size of the eclipsing object, how long the eclipse takes and how wide the orbit is. But useful constraints can be made on the structure of accretion disks, sizes of starspots etc, although &quot;maps&quot; are not usually produced. A recent example using the transits of a planet to probe starspot structure is &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013ApJ...767...60R&quot; rel=&quot;noreferrer&quot;&gt;Roettenbacher et al. 2013&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is rotational modulation. Features on the surface that rotate around are self-eclipsed by the star and produce modulation of the observed light. This can be used to try and estimate e.g. the size and location of starspots. Again there are many degeneracies and ambiguities, but this has become a growth industry since the delivery of thousands of extremely high quality light curves from the Kepler satellite.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2014-12-26T19:37:21.977" LastActivityDate="2014-12-26T19:37:21.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6516" PostTypeId="2" ParentId="6514" CreationDate="2014-10-04T13:41:01.757" Score="1" Body="&lt;p&gt;If space expands, objects embedded in that space move relative to each other at velocities that are only linearly increasing with their distance. So, nearby such objects never move at the speed of light relative to each other.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-10-04T13:41:01.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6517" PostTypeId="2" ParentId="6514" CreationDate="2014-10-04T13:55:55.617" Score="2" Body="&lt;p&gt;This is  a tricky question, because dark energy does not always affect the distances between any two given objects in the universe. Consider this question: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1665/galaxies-are-moving-away-from-each-other-then-how-milkyway-and-andromeda-galaxy&quot;&gt;Galaxies are moving away from each other then how Milkyway and Andromeda galaxy coming towards each other?&lt;/a&gt;. The Milky Way and Andromeda are incredibly far apart, yet they are moving &lt;em&gt;towards&lt;/em&gt; each other because of gravity. Likewise, the other galaxies in the Local Group are gravitationally bound to the Group, and won't be leaving it. The gist of this is that it's difficult to determine just where dark energy becomes the dominant player in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can calculate the speed (recessional velocity) at which objects move away from each other by using &lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Hubble's law&lt;/em&gt;&lt;/a&gt;, $v=H_od$, where $v$ is recessional velocity, $H_o$ is the Hubble constant, and $d$ is the distance between the objects. We can use the Hubble constant, along with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Friedmann equations&lt;/a&gt;, to derive the &lt;em&gt;Hubble parameter&lt;/em&gt;, which can then be used to calculate the expansion of the universe over time. The Hubble parameter is a function of time, while the Hubble constant is constant throughout space-time; it is the current value of the Hubble parameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while we can use the Hubble constant to calculate how fast objects are moving away from each other &lt;em&gt;now&lt;/em&gt;, we can use the Hubble parameter to calculate at what time two objects will be moving away from each other at a given speed. I invite you to use the following pages to get you started on some calculations, but I will warn you to not use them for any objects inside the Local Group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the aforementioned pages:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Deceleration_parameter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Deceleration parameter&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Scale_factor_(cosmology)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Scale factor&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-01-10T23:08:26.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6518" PostTypeId="2" ParentId="6324" CreationDate="2014-10-04T14:27:41.773" Score="0" Body="&lt;p&gt;There is a very nice web-based utility called &lt;a href=&quot;http://catserver.ing.iac.es/staralt/index.php&quot; rel=&quot;nofollow&quot;&gt;Staralt&lt;/a&gt; that I frequently use before observing runs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has lots of observatories pre-programmed or you can put in your own latitude and longitude AND altitude (can you do that in Stellarium?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To quote the website - &quot;you can plot altitude against time for a particular night (Staralt), or plot the path of your objects across the sky for a particular night (Startrack), or plot how altitude changes over a year (Starobs), or get a table with the best observing date for each object (Starmult).&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also do handy thing like produce plots, charts and tables or show where the moon is with respect to your targets.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2014-10-04T14:27:41.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6519" PostTypeId="1" AcceptedAnswerId="6521" CreationDate="2014-10-04T15:32:43.693" Score="44" ViewCount="7206" Body="&lt;p&gt;This is a question I've heard many times in the past, and a quick search of the site says it hasn't been asked here, so I figured I might as well ask (and answer) it. I know that it is rare for someone to ask and answer their own question, but I think it could work here, and I welcome input (including other answers) from anyone and everyone here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun is roughly 4 light-years away from the closest star system, the Alpha Centauri system. The planets in our Solar System, however, aren't even close to that far away from the Sun. Where does our Solar System end? Is the edge considered to be the orbit of Neptune, the Kuiper Belt, the Oort Cloud, or something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: &lt;a href=&quot;https://physics.stackexchange.com/questions/25784/how-big-is-the-solar-system&quot;&gt;this&lt;/a&gt; question on Physics SE is similar, but the answers posted here go in different directions.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2021-05-24T04:33:25.183" Title="Where does the Solar System end?" Tags="&lt;the-sun&gt;&lt;solar-system&gt;&lt;oort-cloud&gt;&lt;kuiper-belt&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6520" PostTypeId="2" ParentId="6519" CreationDate="2014-10-04T16:00:03.917" Score="20" Body="&lt;p&gt;Here's my answer. I'll try to make it as comprehensive as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's pretty hard to define the edge of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_System&quot; rel=&quot;noreferrer&quot;&gt;Solar System&lt;/a&gt;. Most people would probably define it as where objects are no longer gravitationally bound to the Sun. That just shifts the question a little, though: Where is that dividing line? To try to answer this, I'll go over the regions of the Solar System.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first region is the domain of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Terrestrial_planet&quot; rel=&quot;noreferrer&quot;&gt;inner planets&lt;/a&gt; - basically everything from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Asteroid_belt&quot; rel=&quot;noreferrer&quot;&gt;asteroid belt&lt;/a&gt; inwards. It is comprised of Mars, Earth, Venus, Mercury, their moons, and all the smaller objects that surround them. The inner Solar System is very rocky, as one can imagine. The terrestrial planets are primarily made of rock, as are the asteroids and the inner planets' moons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second region is the domain of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gas_giant&quot; rel=&quot;noreferrer&quot;&gt;gas giants&lt;/a&gt;. It consists of Jupiter, Saturn, Uranus, Neptune, their moons, ring systems, and assorted smaller bodies, such as Trojan asteroids. The gas giants had a big influence on the Solar System when it was first formed, pulling in chunks of rocks, grabbing moons, and possibly stabilizing or de-stabilizing orbits. Some may have migrated outwards (as per the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nice_model&quot; rel=&quot;noreferrer&quot;&gt;Nice model&lt;/a&gt;), but their orbits are currently stable. The gas giants are made largely of gases, but it is thought they have solid or molten cores. The composition of their moons is familiar - more like objects in the inner Solar System.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next up is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kuiper_belt&quot; rel=&quot;noreferrer&quot;&gt;Kuiper Belt&lt;/a&gt;. It's sometimes introduced as a cousin of the asteroid belt, but that's not accurate. The bodies that make up the Kuiper Belt are chunks of rock and ice. Notable examples of Kuiper Belt bodies and/or trans-Neptunian objects are the dwarf planets Pluto, Sedna, Makemake and Haumea. There are also lots of smaller objects, including some short-period comets (although these are more properly part of the lesser-known &quot;scattered disk&quot;). While there have been theories for years about another planet out there, it is not considered likely. The Belt extends from 30 to 50 AU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further out still is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Oort_cloud&quot; rel=&quot;noreferrer&quot;&gt;Oort Cloud&lt;/a&gt;, named after Jan Oort. Observations of objects in the Oort Cloud are extremely difficult, if not impossible, so its existence has not yet been verified. It is populated by long-period comets and smaller objects. These are also composed of rock and ice. The Oort Cloud is thought to extend up to an incredible 50,000 AU. While the other regions so far mentioned are roughly in planes, the Oort Cloud is spherical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some consider the far edge Oort Cloud to be the edge of the Solar System, because the majority of the mass of the Solar System is within it, but the boundary between the Solar System and interstellar space is actually thought to be within its inner reaches: the heliopause. &lt;em&gt;This&lt;/em&gt; is generally accepted as the Solar System's boundary because it is where the solar wind meets the interstellar medium. This is often placed at 121 AU - which is where Voyager 1 passed through in 2013. The heliopause is the far boundary of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heliosphere&quot; rel=&quot;noreferrer&quot;&gt;heliosphere&lt;/a&gt;, beyond which the interstellar medium takes control. Inside &quot;layers&quot; are bounded by the termination shock and the heliosheath.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, while the Solar System is made of many regions, the heliopause is considered to be its outer boundary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once again, I welcome any and all input regarding this question and answer.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-08-26T22:48:49.530" LastActivityDate="2015-08-26T22:48:49.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6521" PostTypeId="2" ParentId="6519" CreationDate="2014-10-04T16:10:01.410" Score="32" Body="&lt;p&gt;According to the Case Western Reserve University webpage &lt;a href=&quot;http://burro.astr.cwru.edu/stu/advanced/solarsystem_edge.html&quot; rel=&quot;noreferrer&quot;&gt;The Edge of the Solar System&lt;/a&gt; (2006) an important consideration is that&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The whole concept of an &amp;quot;edge&amp;quot; is somewhat inaccurate as far as the solar system is concerned, for there is no physical boundary to it - there is no wall past which there's a sign that says, &amp;quot;Solar System Ends Here.&amp;quot; There are, however, specific regions of space that include outlying members of our solar system, and a region beyond-which the Sun can no longer hold any influence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The last part of that definition appears to be a viable definition of the edge of the solar system.  Specifically,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;valid boundary region for the &amp;quot;edge&amp;quot; of the solar system is the heliopause. This is the region of space where the sun's solar wind meets that of other stars. It is a fluctuating boundary that is estimated to be approximately 17.6 billion miles (120 A.U.) away. Note that this is within the Oort Cloud.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Though the article above is a bit dated, the notion of the heliopause has been still of interest to scientists, particularly how far away it is - hence, the interest in the continuing &lt;a href=&quot;http://voyager.jpl.nasa.gov/mission/interstellar.html&quot; rel=&quot;noreferrer&quot;&gt;Voyager missions&lt;/a&gt;, which states on the website, that it has 3 phases:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Termination Shock&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Passage through the termination shock ended the termination shock phase and began the heliosheath exploration phase. Voyager 1 crossed the termination shock at 94 AU in December 2004 and Voyager 2 crossed at 84 AU in August 2007.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;(AU = Astronomical Unit = mean Earth-sun distance = 150,000,000km)&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Heliosheath&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;the spacecraft has been operating in the heliosheath environment which is still dominated by the Sun's magnetic field and particles contained in the solar wind.&lt;/p&gt;&#xA;&lt;p&gt;As of September 2013, Voyager 1 was at a distance of 18.7 Billion Kilometers (125.3 AU) from the sun and Voyager 2 at a distance of 15.3 Billion kilometers (102.6 AU).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A very important thing to note from the Voyager page is that&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;The thickness of the heliosheath is uncertain&lt;/strong&gt; and could be tens of AU thick taking several years to traverse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Interstellar space, which NASA's Voyager page has defined as&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Passage through the heliopause begins the interstellar exploration phase with the spacecraft &lt;strong&gt;operating in an interstellar wind dominated environment.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The Voyager mission page provide the follow diagram of the parameters listed above&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/B3cQr.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;It is a bit complicated as we do not know the full extent of what the dynamics are like out there, a recent observation reported in the article &lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2011/09jun_bigsurprise/&quot; rel=&quot;noreferrer&quot;&gt;A big surprise from the edge of the Solar System&lt;/a&gt;, reveal that the edge may be blurred by&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;a strange realm of frothy magnetic bubbles,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Which is suggested in the article could be a mixing of solar and interstellar winds and magnetic fields, stating:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;On one hand, the bubbles would seem to be a very porous shield, allowing many cosmic rays through the gaps. On the other hand, cosmic rays could get trapped inside the bubbles, which would make the froth a very good shield indeed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user2449" LastEditorUserId="-1" LastEditorDisplayName="user2449" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2014-10-04T16:15:51.500" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6522" PostTypeId="1" CreationDate="2014-10-05T02:06:15.087" Score="1" ViewCount="98" Body="&lt;p&gt;I'm reading material that is seemingly contradictory.  Some sources indicate that the evolution of a protostar to a main sequence star is characterised by a stellar wind that precludes the accretion of further in-falling material.  That is, the (young) star now has a constant mass.  However, other sources suggest that material may continue to accrete for a (brief) period after the protostar has become a main sequence star.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please confirm the actual process?&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2014-10-05T17:22:52.667" Title="Accretion of in-falling material for a young main sequence star" Tags="&lt;protostar&gt;&lt;main-sequence&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6523" PostTypeId="1" CreationDate="2014-10-05T03:20:36.630" Score="2" ViewCount="616" Body="&lt;p&gt;Note:  I have figured out one case:  The meteor is splash from an impact on a close-in moon.  I'm looking for any path from deep space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that anything from deep space will inherently pick up the escape velocity in it's fall and thus there is no possibility in the simple case.  I'm not so sure about what could happen after an encounter with one or more moons, though.&lt;/p&gt;&#xA;" OwnerUserId="2036" LastActivityDate="2016-04-25T01:20:07.747" Title="Is there any way a meteor can hit at less than escape velocity?" Tags="&lt;gravity&gt;&lt;speed&gt;&lt;meteor&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6524" PostTypeId="1" CreationDate="2014-10-04T21:47:54.843" Score="4" ViewCount="4716" Body="&lt;p&gt;My research has told me astronomers have labeled the galaxy arm that our system is located the Orion Arm but I have yet to find out if that also applies to Zeta Reticuli or not (not asking about aliens or anything, just trying to get my locations straight).&lt;/p&gt;&#xA;" OwnerDisplayName="captain-scarlett" LastEditorUserId="13" LastEditDate="2014-10-05T09:10:42.670" LastActivityDate="2020-01-08T10:51:36.893" Title="Is Zeta Reticuli within the Orion Arm?" Tags="&lt;milky-way&gt;&lt;local-group&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6525" PostTypeId="1" AcceptedAnswerId="6532" CreationDate="2014-10-05T14:21:37.343" Score="27" ViewCount="4771" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Given our knowledge and the standard cosmological model, we estimate that the age of the universe is about 13.7 billion years old.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How much sense does it make to talk about the age of the universe as a whole?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can observe &lt;a href=&quot;http://en.wikipedia.org/wiki/Time_dilation&quot;&gt;time dilation&lt;/a&gt; so we know time passes differently for observers moving at different speeds (special relativistic time dilation) or located at different distances from centres of strong gravitational fields (as explained by general relativity). Is the estimated age of the universe a property of the universe or of us as its observers? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would a different observer perceive the age of the universe to be different? Do different parts of the universe have different ages? Do the principles of special relativity even apply when thinking on a universal scale?&lt;/p&gt;&#xA;" OwnerUserId="211" LastActivityDate="2015-03-06T17:47:59.467" Title="Age of the universe and time dilation" Tags="&lt;universe&gt;&lt;space-time&gt;&lt;general-relativity&gt;&lt;special-relativity&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6526" PostTypeId="1" CreationDate="2014-10-05T14:59:26.310" Score="15" ViewCount="2250" Body="&lt;p&gt;Recent news seem to suggest that astronomers are arguing about whether Pluto should become a planet again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I cannot find an official source for this. Is this actually true?&lt;/p&gt;&#xA;" OwnerUserId="2539" LastEditorUserId="2153" LastEditDate="2015-05-25T14:40:07.403" LastActivityDate="2015-06-05T04:54:31.093" Title="Is Pluto still a dwarf planet?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;pluto&gt;&lt;dwarf-planets&gt;&lt;classification&gt;" AnswerCount="6" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="6527" PostTypeId="2" ParentId="6526" CreationDate="2014-10-05T15:13:27.003" Score="12" Body="&lt;p&gt;Yes, Pluto is still a dwarf planet. According to the &lt;a href=&quot;http://www.iau.org/public/themes/pluto/&quot; rel=&quot;nofollow&quot;&gt;IAU website&lt;/a&gt;, it still fits the criteria for a dwarf planet, fails to meet the criteria for a planet, and still carries the &quot;dwarf planet&quot; label, whatever its future status may be. I'm sorry I can't provide a longer or more detailed answer, but this is really a yes-or-no question.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-06-03T12:27:12.603" LastActivityDate="2015-06-03T12:27:12.603" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6528" PostTypeId="2" ParentId="6526" CreationDate="2014-10-05T15:16:10.280" Score="5" Body="&lt;p&gt;A lot of the push to have Pluto reinstated as the 9th planet is coming from Harvard, from their press release &lt;a href=&quot;http://www.cfa.harvard.edu/news/2014-25&quot;&gt;Is Pluto a Planet? The Votes Are In&lt;/a&gt; (Released September, 2014), they state the following outcomes from a debate:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Science historian Dr. Owen Gingerich, who chaired the IAU planet definition committee, presented the historical viewpoint. Dr. Gareth Williams, associate director of the Minor Planet Center, presented the IAU's viewpoint. And Dr. Dimitar Sasselov, director of the Harvard Origins of Life Initiative, presented the exoplanet scientist's viewpoint.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Gingerich argued that &quot;a planet is a culturally defined word that&#xA;  changes over time,&quot; and that Pluto is a planet. Williams defended the&#xA;  IAU definition, which declares that Pluto is not a planet. And&#xA;  Sasselov defined a planet as &quot;the smallest spherical lump of matter&#xA;  that formed around stars or stellar remnants,&quot; which means Pluto is a&#xA;  planet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We will have a better understanding of Pluto, hence its classification when NASA's &lt;a href=&quot;http://www.nasa.gov/mission_pages/newhorizons/main/#.VDFev_mSwVY&quot;&gt;Horizons&lt;/a&gt; mission reaches it.  But, at this stage, Pluto is still classified as a dwarf planet.&lt;/p&gt;&#xA;" OwnerDisplayName="user2449" LastActivityDate="2014-10-05T15:16:10.280" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6529" PostTypeId="1" CreationDate="2014-10-05T15:40:36.947" Score="-3" ViewCount="141" Body="&lt;p&gt;The number of neurons in the human brain is equated with the number of stars in the Milky Way. Could the Milky Way act as a consciousness of its own, with the gravitational strength between the stars acting as weights of the network?&lt;/p&gt;&#xA;" OwnerUserId="2491" LastActivityDate="2014-10-05T18:44:07.680" Title="Brain vs galaxy" Tags="&lt;galaxy&gt;&lt;gravity&gt;&lt;universe&gt;" AnswerCount="2" CommentCount="5" ClosedDate="2014-10-06T06:45:13.683" ContentLicense="CC BY-SA 3.0" />
  <row Id="6530" PostTypeId="2" ParentId="6523" CreationDate="2014-10-05T16:00:55.960" Score="5" Body="&lt;p&gt;Edited.&#xA;&lt;strong&gt;No&lt;/strong&gt; if you talk about the escape velocity form Earth. This follows simply from the fact that energy $E$ is conserved. An object that is not gravitationally bound to Earth has $E&amp;gt;0$ and hence $v&amp;gt;v_{\mathrm{esc}}$ when hitting ground.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Yes&lt;/strong&gt;, if you meant the escape velocity from the Solar system, because the Earth moves with $v_{\rm Earth}=v_{\rm escape}/\sqrt{2}$ relative to (but not towards) the Sun. Here $v_{\rm escape}=\sqrt{2GM_{\odot}/1{\rm AU}}$ is the &lt;em&gt;local escape speed&lt;/em&gt; from the Sun, while $v_{\rm Earth}=\sqrt{GM_{\odot}/1{\rm AU}}$ is the speed of the local circular orbit. An object at 1AU form the Sun and bound to the Sun cannot have speed greater than $v_{\rm escape}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the impact speed of an object that moves at $v_{\rm escape}$ can be as low $v_{\rm escape}-v_{\rm Earth}=v_{\rm escape}(1-1/\sqrt{2})$ if it hits Earth &quot;from behind&quot;, i.e. moving in the same direction as Earth at the time of impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that meterors typically move not faster than $v_{\rm escape}$, for they don't come from outer space, but from the Solar system.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2016-04-22T19:22:12.610" LastActivityDate="2016-04-22T19:22:12.610" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6531" PostTypeId="2" ParentId="6514" CreationDate="2014-10-05T16:11:02.287" Score="3" Body="&lt;p&gt;The answer to your question is that if we assume the Universe is infinite then there will always be objects receding from us at the speed of light, simply because recession velocity depends linearly on distance. Indeed there are objects that we can see (i.e. they are within our observable Universe) that recede from us at the speed of light (this is a very misunderstood issue, so I'd advise anyone taking issue with this statement to make sure they've read &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0310808&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; first).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume dark energy takes the form of a cosmological constant, then the Universe asymptotically becomes like De Sitter space over time. This means that the sphere of objects that are receding from us at c (the Hubble sphere) settles down to a constant radius which is inversely proportional to the square root of the cosmological constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited to add: recession velocities though are purely a function of cosmological coordinates so don't confuse them with local velocities which are always constrained by c.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastEditorUserId="2092" LastEditDate="2014-10-05T16:57:53.027" LastActivityDate="2014-10-05T16:57:53.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6532" PostTypeId="2" ParentId="6525" CreationDate="2014-10-05T16:20:26.860" Score="18" Body="&lt;p&gt;The answer is yes time dilation does affect how much time an observer experiences since the big bang until the present (cosmological) time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there is a certain set of special observers called comoving observers, these are the observers to which the Universe appears isotropic to. For example we can tell the Earth is moving at about 350 km/s relative to a nearby comoving observer by measuring the anistropy in the cosmic microwave background (in fact it causes a relatively large anistropy, the pictures of the CMB you see from WMAP, etc are with this anistropy factored out).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One special property of a comoving observer in the Universe is that they maximize the age of the Universe, i.e. no other observer can experience more time since the big bang than a comoving observer (in case you ask there is no observer that minimizes the time since the big bang). When we talk about the age of the Universe we are talking about the age of the Universe from the pov of a comoving observer.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastEditorUserId="2092" LastEditDate="2014-10-05T16:54:44.100" LastActivityDate="2014-10-05T16:54:44.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6533" PostTypeId="2" ParentId="6524" CreationDate="2014-10-05T16:30:03.443" Score="9" Body="&lt;p&gt;Let's take a look!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I can't quite track down the source of this image, I've found it throughout the web, so &lt;em&gt;hopefully&lt;/em&gt; it's roughly accurate. I'm pulling it from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Orion_Arm&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article on the Orion Arm&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i1RTH.jpg&quot; alt=&quot;Sun&amp;#39;s location in Orion Arm&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this image you can also see the Orion Nebula sitting (relatively) cozily to the Sun.  The distance between the two is believed to be around 1350 light years.  Zeta Reticuli, on the other hand, is believed to be a mere 39 light years from Earth. Given the orientation of these two locations and the relative distances, it would be hard to place Zeta Reticuli anywhere but firmly within the Orion Arm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not exactly rock solid evidence, but sometimes some back of the envelope deduction will do.&lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="1288" LastEditDate="2014-10-05T21:19:16.333" LastActivityDate="2014-10-05T21:19:16.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6534" PostTypeId="2" ParentId="6525" CreationDate="2014-10-05T16:52:11.267" Score="8" Body="&lt;p&gt;In the standard model, the universe looks the same for all locations moving in the local rest frame.  This includes its apparent age.  You can tell if you are in the local rest frame if the expansion of galaxies around you is symmetric in all directions and the microwave background also is the same in all directions.  Simply put, any civilization on any galaxy in the visible universe would measure the same age of the universe that we do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, however, some object managed to move at nearly the speed of light relative to its local rest frames, which would have been changing continuously, for a significant fraction of the age of the universe, there would be relativistic time dilation and that object would not have aged as much as the rest of the universe.  That is, if there were a clock onboard, it would record less time elapsed.  However, since doing this requires continuous accelerations and vast energy, no natural phenomenon that we know of could have experienced this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an object accelerates to the speed of light and then cuts off its engines, then because of Hubble expansion, it will travel to locations where relative to the new local galaxies it is no longer going near the speed of light and hence the time dilation as seen by local observers drops off.  Cosmologists usually use expanding coordinates, which expand in time with the global scale, to do calculations.  In expanding coordinates this effect appears as a fictitious force (arising from the transformation of coordinate system) that acts to dampen velocities.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2015-03-06T17:47:59.467" LastActivityDate="2015-03-06T17:47:59.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6535" PostTypeId="1" CreationDate="2014-10-05T17:05:31.830" Score="6" ViewCount="147" Body="&lt;p&gt;On what day does Mercury reach its greatest elevation (in degrees from &#xA;the horizon) at sunset a given location? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious answer is the day of Mercury's greatest elongation from &#xA;the Sun, but, since the ecliptic is slanted with respect to the &#xA;horizon, I'm not convinced this is correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, on the day after greatest elongation, Mercury's total &#xA;angular distance from the Sun will be smaller, but it's vertical &#xA;distance in elevation (for a given location) at sunset might be &#xA;higher. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same question for Venus, and for when the sun is 6 degrees below the &#xA;horizon (ie, civil twilight), and for sunrise/dawn. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing the date might vary based on position (mostly latitude) since the ecliptic's slant varies at different locations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I googled and found nothing. My (preliminary and possibly wrong) &#xA;expierments with stellarium show that Mercury's elevation at sunset IS &#xA;higher 1-3 days after its greatest elongation, but by less than 1/2 &#xA;degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it's possible that the date of greatest elongation is a close &#xA;enough approximation. &lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2016-01-27T15:40:57.993" Title="When do Mercury/Venus reach greatest elevation at sunset/twilight for a given location?" Tags="&lt;observational-astronomy&gt;&lt;venus&gt;&lt;mercury&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6536" PostTypeId="2" ParentId="6529" CreationDate="2014-10-05T17:14:36.827" Score="2" Body="&lt;p&gt;If we assume the computationalist point of view &lt;a href=&quot;http://www.nyu.edu/gsas/dept/philo/faculty/block/papers/msb.html&quot; rel=&quot;nofollow&quot;&gt;see e.g. here&lt;/a&gt;, then because it's hard to see any nontrivial computations being performed on the local level, you would not think that the galaxy can be a conscious entity in a local sense (i.e. that it generates a consciousness that is experiencing the local environment). But this leaves open the possibility that a consciousness is generated in a non-local sense. E.g. consider ourselves sitting in front of our computers discussing this topic right now here on Stackexchange. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the computationalist theory of mind, my subjective conscious experience would be unaffected if my brain were replaced by any other machine that runs the same software. Now, whatever is going on in my brain today is only affected by the things that were going on yesterday but in a sphere with a radius of one light day. So, if you consider that sphere and apply the time evolution operator one second forward, then this is the same as running my brain as it is today one second forward, up to applying the fixed time evolution operator that maps the state of the universe one day forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that my consciousness right now could be argued to have existed already yesterday, albeit in a non-local form which simply experiences the World as it is now and not the local state of yesterday. By this same logic, you could argue that if the galaxy later would give rise to conscious organisms, then these conscious experiences already exist, albeit that they are subjectively experiencing the future. However, this is then not simply due to the gravity between the stars but also all the relevant physics that needs to be invoked to calculate how these organisms would arise from the initial state.&lt;/p&gt;&#xA;" OwnerUserId="2185" LastEditorUserId="2185" LastEditDate="2014-10-05T18:44:07.680" LastActivityDate="2014-10-05T18:44:07.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6537" PostTypeId="2" ParentId="6522" CreationDate="2014-10-05T17:22:52.667" Score="2" Body="&lt;p&gt;At the end of the protostar phase a vigorous outflow from the star develops called the T-Tauri Wind and this could cut off accretion.  Eventually, it develops into a normal star and the strong wind dies down.  Material that was not totally blown away could then continue to  fall down and be accreted&lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2014-10-05T17:22:52.667" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6538" PostTypeId="2" ParentId="6529" CreationDate="2014-10-05T17:34:26.187" Score="1" Body="&lt;p&gt;This looks like it's going to be a &quot;No, it's not possible, because . . .&quot; answer, but hopefully it answers your question. Your proposal is that the Milky Way is a brain, and that the stars are like neurons, correct? I'm not a biologist or neuroscientist, but I &lt;em&gt;do&lt;/em&gt; know one characteristic of neurons that helps them do their job: They rely on chemical and electrical impulses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Neuron&quot; rel=&quot;nofollow&quot;&gt;Neurons&lt;/a&gt; work by transmitting electrical signals along pathways called &lt;a href=&quot;https://en.wikipedia.org/wiki/Synapse&quot; rel=&quot;nofollow&quot;&gt;synapses&lt;/a&gt;. These signals can be either (nearly) continuous or in regular or irregular &quot;pulses&quot; - i.e. the strength of the signal goes up, down, or disappears. The point here, though, is that signals between neurons are generally not continuous. The gravitational force between two stars, on the other hand, is generally the same, barring a catastrophic event. In other words, the gravitational &quot;signals&quot; from stars don't function like those from neurons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other differences between stars and neurons, too. One is that neurons cannot regrow in significant quantities, whereas &lt;a href=&quot;https://en.wikipedia.org/wiki/Star_formation&quot; rel=&quot;nofollow&quot;&gt;stars are forming all the time&lt;/a&gt;. Generally in an organism, the neurons in a brain are of similar ages; this is clearly not the case for stars. Another difference is that neural networks are relatively fixed, whereas stars are constantly moving around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll conclude by repeating what I mentioned in a comment, which is that the idea that the universe is a living organism has been around for some time. Another idea is that the universe is actually a giant computer (or just &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Hitchhiker%27s_Guide_to_the_Galaxy&quot; rel=&quot;nofollow&quot;&gt;Earth&lt;/a&gt;, if you're a sci-fi fan). The problem is that while most people would disagree with these large-scale hypotheses, we may never know for sure if they are right or wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-10-05T17:34:26.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6539" PostTypeId="2" ParentId="2103" CreationDate="2014-10-05T17:48:27.117" Score="2" Body="&lt;p&gt;I realise that the following is not a great answer to the question, but was a bit too much to fit into a comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The official word on exoplanet naming is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.iau.org/public/themes/naming_exoplanets/&quot; rel=&quot;nofollow&quot;&gt;http://www.iau.org/public/themes/naming_exoplanets/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rules for the scientific designation of an exoplanet are clear and I think will remain so in the published literature for the forseeable future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, here is an earnest contribution to the literature which sees the need to change things, especially to cope with planets in multiple star systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2010arXiv1012.0707H&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/2010arXiv1012.0707H&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But perhaps the big news is that the IAU is setting up to allow &quot;popular&quot; names to be given to exoplanets and are running a competition to name a batch. These names will not replace the scientific designations, but will be recognised by the IAU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.iau.org/news/pressreleases/detail/iau1404/&quot; rel=&quot;nofollow&quot;&gt;http://www.iau.org/news/pressreleases/detail/iau1404/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2014-10-05T17:48:27.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6540" PostTypeId="2" ParentId="1975" CreationDate="2014-10-05T18:06:02.517" Score="2" Body="&lt;p&gt;The short answer is no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't even think the census of stars within 30pc is complete, let along have all the other information you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all bright stars (say V&amp;lt;10) and close, there are decent parallaxes and proper motions in the Hipparcos and Tycho catalogues, so you can compile tangential velocities for these. A lot of these will also have radial velocities, but there is no recent uniform catalogue of these. Things like the General catalogues of radial velocities are getting a bit old. The Hipparcos input catalogue might be a reasonable source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnetic fields have been measured on relatively few stars and usually those which are much younger and more active than the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no way of directly measuring stellar ages. This is one of the big problems in astronomy. Estimates can be made for stars in the field but they are rarely more precise than 20-30% and their accuracy, especially for older stars is open to question. I know of no uniform compilation that would help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to restrict your sample to even closer to the Earth, then you might find that the &lt;a href=&quot;http://www.recons.org/&quot; rel=&quot;nofollow&quot;&gt;RECONS&lt;/a&gt; database for stars closer than 10pc might be of use. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2014-10-05T18:06:02.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6542" PostTypeId="1" CreationDate="2014-10-05T19:39:14.740" Score="2" ViewCount="167" Body="&lt;p&gt;Where can I find a list of observed (not computed) sun/moon rise/set &#xA;times to the nearest second, along with the location (latitude, &#xA;longitude, altitude [or enough information to find these three &#xA;values]) and, ideally, weather conditions (temperature/pressure)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it's hard to tell where the &quot;middle&quot; of the sun/moon is when &#xA;it's setting, these would be the setting times for the lower or upper &#xA;limb of the sun/moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reason: I'm trying to compute sun/moon rise/set to the nearest second &#xA;using high-precision data, and I'm curious to see if I can get &#xA;anywhere near the actual observed time. My technique: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Use NASA's Chebyshev approximations to find the positions of the &#xA;Sun and center of the Earth. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use an elliptical (not spherical) model of the Earth to find the &#xA;ICRF xyz coordinates of a given location at a given time. This &#xA;includes compensating for altitude and precession. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Model refraction based on weather conditions, including adjusting pressure for altitude (weather stations report air pressure normalized to sea level, which isn't the same as actual air pressure, especially at higher altitudes).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I realize down-to-the-second computations of sun/moon rise/set are &#xA;probably impossible, in part because refraction depends on more than &#xA;just local conditions, but I'm curious to try. &lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2014-10-05T19:39:14.740" Title="Observed sun/moon rise/set times to nearest second?" Tags="&lt;the-moon&gt;&lt;the-sun&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6543" PostTypeId="1" CreationDate="2014-10-05T19:49:37.277" Score="3" ViewCount="242" Body="&lt;p&gt;Cosmic horizon in the de Sitter space is a sphere, centered at the observer with finite radius where the red shift due to cosmic expansion becomes infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that no information can be transmitted from behind the cosmic horizon, I wonder whether any matter can ever pass behind the event horizon, because this will give rise to an information paradox, similar to the black hole information paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matter approaching the cosmic horizon seems to experience time dilation, so will that time dilation prevent it from crossing the horizon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To work around the paradox, one have to postulate that the information about outgoing matter should be encoded in the de Sitter radiation.&lt;/p&gt;&#xA;" OwnerUserId="2541" LastActivityDate="2014-10-06T00:54:04.763" Title="Can matter leave the cosmic horizon?" Tags="&lt;cosmology&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6544" PostTypeId="2" ParentId="6543" CreationDate="2014-10-05T21:08:40.473" Score="2" Body="&lt;p&gt;A little bit of caution is needed when approaching this subject as there are &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmological_horizon&quot; rel=&quot;nofollow&quot;&gt;several surfaces in cosmological solutions that can be interpreted (or misinterpreted in some cases) as horizons&lt;/a&gt;. Not all of them appear in all comsological solutions and in some cosmological solutions some of these surfaces are only present for part of the evolution of the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to mean the &lt;a href=&quot;http://en.wikipedia.org/wiki/Particle_horizon&quot; rel=&quot;nofollow&quot;&gt;particle horizon&lt;/a&gt;, which is the limit of the observable Universe. The cosmological red shift at the particle horizon is always infinite, when there is a particle horizon. No matter can ever leave the particle horizon (regardless of the cosmological model, as long as there is a particle horizon) as it expands radially outwards at a local speed of c relative to any matter on the horizon (of course to us it expands even faster than c as it is also receding due to expansion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An ever-expanding Universe with a cosmological constant approaches a state similiar to de Sitter space in late times, however de Sitter space itself is not a realistic cosmological solution as it is devoid of matter. De Sitter space does not have a particle horizon and for that matter it doesn't have any singularities either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In de Sitter space the &lt;a href=&quot;http://en.wikipedia.org/wiki/Event_horizon#Cosmic_event_horizon&quot; rel=&quot;nofollow&quot;&gt;cosmic event horizon&lt;/a&gt;, which is the surface beyond which events happening now cannot affect us in the future &lt;em&gt;providing we stay at the same spot&lt;/em&gt; (this should not be confused with the particle horizon), sits at a constant distant to the observer that is inversely proportional to the square root of the cosmological constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In answer to your question then, de Sitter space is symmetric in time and space which actually gives rise to seemingly different physical interpretations of it. However the black hole information paradox is specifically the loss of unitariness in the time evolution of a black hole and you have not justified why you think an equivalent thing should happen in de Sitter space. The loss of unitariness in actually arises because  the event horizon disappears, but (naively at least) the information behind it also disappears. In de Sitter space the event horizon remains and even if it were to disappear by the cosmological constant miraculously going to zero we would still find the information behind it, so there is no analog to the information paradox in de Sitter space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To take another example, consider Rindler coordinates in Minkowski spacetime, there is a horizon (the Rindler horizon) and there is radiation (Unruh radiation), however there is no information paradox as we know that unitariness is preserved in Minkowski spacetime and when the Rindler horizon is made to disappear by selecting a different coordinate system, all the information that was behind the horizon is still there.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastEditorUserId="2092" LastEditDate="2014-10-06T00:54:04.763" LastActivityDate="2014-10-06T00:54:04.763" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6545" PostTypeId="2" ParentId="6526" CreationDate="2014-10-06T00:59:54.643" Score="1" Body="&lt;p&gt;Pluto will continue to be exactly Pluto no matter how we choose to categorize it. Fretting about the &quot;proper&quot; category is the &lt;a href=&quot;https://richarddawkins.net/2013/01/the-tyranny-of-the-discontinuous-mind-christmas-2011/&quot; rel=&quot;nofollow&quot;&gt;tyranny of the discontinuous mind&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="2542" LastActivityDate="2014-10-06T00:59:54.643" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6546" PostTypeId="1" CreationDate="2014-10-06T02:12:27.007" Score="6" ViewCount="3036" Body="&lt;p&gt;When one lives away from the equator, say at around 37° latitude, does the sky get darker faster after the sun sets in the winter as compared to the summer?  Why is this the case?  I believe that if you were to draw the path the sun traverses at different times of the year, the paths are parallel to each other but shifted, such that in summertime, the path is longer with a shorter section below the horizon and in winter it is shorter with a longer section below the horizon. I imagine that the angle of the path relative to the horizon is approximately the complement of your latitude.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2020-11-25T18:21:24.613" LastActivityDate="2020-11-25T18:21:24.613" Title="Does the sky get darker faster during the winter?" Tags="&lt;the-sun&gt;&lt;rotation&gt;&lt;sunrise&gt;&lt;space-geometry&gt;&lt;sunset&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6547" PostTypeId="2" ParentId="6546" CreationDate="2014-10-06T06:19:34.623" Score="3" Body="&lt;p&gt;The further north you go, the time between sunset and darkness becomes longer, no matter the season. The reason is due to the velocity at that latitude. If there is 10 min of twilight on the equator, then there is $10\sqrt{2}$ min at 45° latitude, 20 min at 60° latitude, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:&lt;br&gt;&#xA;I used the &lt;a href=&quot;http://aa.usno.navy.mil/data/docs/RS_OneYear.php&quot; rel=&quot;nofollow&quot;&gt;website&lt;/a&gt; that @barrycarter listed above and discovered that there are 2 maxima and minima during the year. The solstices have the longest twilight times and the equinoxes have the shortest twilight times.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="1322" LastEditDate="2014-10-10T01:45:39.443" LastActivityDate="2014-10-10T01:45:39.443" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6548" PostTypeId="1" CreationDate="2014-10-06T08:25:49.640" Score="1" ViewCount="124" Body="&lt;p&gt;What would it feel like on Earth if the Earth accelerated due to sudden changes in the solar system like ejection of another planet? Would there be giant tidal waves and other cataclysmic events?&lt;/p&gt;&#xA;" OwnerUserId="2491" LastActivityDate="2014-10-06T16:23:22.717" Title="Apocalypse on earth" Tags="&lt;the-sun&gt;&lt;gravity&gt;&lt;planet&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6549" PostTypeId="1" AcceptedAnswerId="6553" CreationDate="2014-10-06T09:14:08.750" Score="3" ViewCount="220" Body="&lt;p&gt;Well it is said that during the Big Bang, things were created in pairs: one matter and one anti-matter. I think its a part of the Big-Bang Theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also said that matter + anti-matter - Energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then shouldn't it be so that there would have been no matter or antimatter in existence now since all of them should have reacted to form pure Energy. I mean how were we created then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also where are all the antimatter gone now? Are they still reacting with us?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2519" LastActivityDate="2020-11-19T22:37:38.217" Title="Matter and Antimatter Interaction in the early Universe" Tags="&lt;big-bang-theory&gt;&lt;early-universe&gt;&lt;antimatter&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6550" PostTypeId="1" AcceptedAnswerId="6551" CreationDate="2014-10-06T03:45:06.270" Score="5" ViewCount="2216" Body="&lt;p&gt;Professor Oliver Manuel,  a professor of nuclear chemistry at the University of Missouri-Rolla, claims in the article &lt;a href=&quot;http://www.spacedaily.com/news/iron-02b.html&quot;&gt;Sun Is Made Of Iron, Not Hydrogen, Professor Says&lt;/a&gt; that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;iron, not hydrogen, is the sun's most abundant element.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Specifically, he and his colleagues postulate that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;We think that the solar system came from a single star, and the sun formed on a collapsed supernova core,&quot; Manuel says. &quot;The inner planets are made mostly of matter produced in the inner part of that star, and the outer planets of material form the outer layers of that star.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is there evidence to support or refute Professor Manuel's theories?&lt;/p&gt;&#xA;" OwnerDisplayName="user2449" LastEditorDisplayName="user2449" LastEditDate="2014-10-06T09:24:12.840" LastActivityDate="2015-01-12T19:11:54.700" Title="What scientific evidence is there to support or refute the Iron Sun Hypothesis?" Tags="&lt;the-sun&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6551" PostTypeId="2" ParentId="6550" CreationDate="2014-10-06T07:20:24.677" Score="11" Body="&lt;p&gt;If the sun was principally composed of Iron, then this would be apparent in the spectrum of light from the sun (which is how we know what stars are made of) (but see comment by @Keith Thompson below, the spectra only tells you about the surface, rather than the core).  The fact that all solar physicists bar one seem to think it is made up of hydrogen and helium strongly suggests this is not the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sun were made of iron, then I suspect that would require a revision of our understanding of nuclear fusion, as iron is the product of the final fusion reaction that stars can &lt;a href=&quot;http://en.wikipedia.org/wiki/Silicon-burning_process&quot;&gt;support&lt;/a&gt;, unless of course the theory is that the sun is no longer undergoing fusion reactions and all we have been seeing for the last &lt;em&gt;4 billion years+&lt;/em&gt; is the residual heat of the core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0370157307004565&quot;&gt;heliosiesmology&lt;/a&gt; suggests there isn't that much iron in the suns core (although the physics is beyond me, so caveat lector).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a further point, if the sun is primarily made of iron, it becomes difficult to explain why the planets (especially the gas giants) have so little of it.  It is reasonably straightforward to explain why the inner planets have lost much of their original hydrogen and helium, but rather more difficult to explain how the outer planets could have become hydrogen enriched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update.  Dr Manuel's website claims that &lt;a href=&quot;http://www.thesunisiron.com/&quot;&gt;&quot;About 99% of the sun has the same composition as ordinary meteorites.&quot;&lt;/a&gt;.  The density of carbonaceous chondrite meteorites is about &lt;a href=&quot;http://meteorites.wustl.edu/id/density.htm&quot;&gt;2100-34700 kg/m3&lt;/a&gt; (with porosity being an issue).  The mean density of the sun is &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/solar/sun.html&quot;&gt;1410 kg/m3&lt;/a&gt;.  Given the temperature and pressure within the suns' core, if it had a similar composition to a meteorite its' porosity would be zero, and hence even more dense than an ordinary meteorite.  There is a problem with accounting here, if 99% of the sun had a composition similar to that of a meteorite, it would have to be full of voids to have the suns observed density.  The figures just don't add up as far as I can see (although how materials may behave at ultra high temperatures and pressures means intuition may be invalid here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Manuel to be right, pretty much all we know about stars must be wrong.&lt;/p&gt;&#xA;" OwnerUserId="2548" OwnerDisplayName="Dikran Marsupial" LastEditorUserId="2548" LastEditDate="2014-10-07T17:12:07.493" LastActivityDate="2014-10-07T17:12:07.493" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="6552" PostTypeId="2" ParentId="6548" CreationDate="2014-10-06T16:23:22.717" Score="2" Body="&lt;p&gt;We wouldn't feel the ejection of another planet off the solar system, since the attractive force of distant planets to Earth is very low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only close encounters of Earth with other planets would cause noticeable up to severe changes on Earth.&#xA;This would be caused mainly by tidal forces due to different acceleration for different parts of Earth, or by changing the orbit of the Moon. A homogeneous acceleration wouldn't cause immediate damage, but with some delay it may cause temperature changes due to the change of Earth's orbit and the changing distance to the Sun.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-10-06T16:23:22.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6553" PostTypeId="2" ParentId="6549" CreationDate="2014-10-06T16:37:23.977" Score="3" Body="&lt;p&gt;Matter and antimatter aren't perfectly symmetric, when following the &quot;forward&quot; direction of time. This is commonly called &lt;a href=&quot;http://en.wikipedia.org/wiki/CP_violation&quot; rel=&quot;nofollow&quot;&gt;charge-parity violation&lt;/a&gt;.&#xA;It's assumed, that this asymmetry caused an excess of matter in the early stages of the Universe after the Big Bang, annihilating almost all antimatter, and leaving behind matter.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2014-10-06T16:37:23.977" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6554" PostTypeId="2" ParentId="6550" CreationDate="2014-10-06T18:17:39.187" Score="2" Body="&lt;p&gt;I think Mr. Marsupial's answer should get a star, but to put some of this in layman's terms, here's my answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process of fusing iron into other elements, whether within a star or within a laboratory pulls more energy from the reaction than it creates.  It is the first element in star evolution that does this, and is therefore associated closely with the end-stage of a star's life.  The energy has to come from somewhere, and the models I have seen have the star eating up its outer layers very rapidly.  Perhaps within a day, which we would have noticed.  [Edit: I previously stated this as fusing elements into iron whereas it has been correctly pointed out that it is iron fusing into other elements that is the problem.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further evidence we can look at other stars whose makeup we know to be primarily iron through spectroscopy and compare them and their systems to ours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, any useful theory needs to make definite predictions (and bonus points if they differ from the standard model) that can be either confirmed or refuted by observation and experimentation.  If I knew more about this 'Iron Sun Hypothesis', I could refute it better, but as it stands, it disagrees with enough of what we believe we know that it requires overwhelming evidence before most will pay it any attention.&lt;/p&gt;&#xA;" OwnerUserId="2503" LastEditorUserId="2503" LastEditDate="2014-10-07T16:54:05.860" LastActivityDate="2014-10-07T16:54:05.860" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6555" PostTypeId="1" CreationDate="2014-10-06T16:36:02.960" Score="4" ViewCount="310" Body="&lt;p&gt;I was looking for accurate and up-to-date data regarding the advancement of the perihelion of the planets in the Solar System, the only ones I've found so far are at least 90 years old bout I really doubt that no progress has been made so far in this field... does anyone know where to find newer ones?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to this, can anybody make an estimation about the difference between the Moon-Earth system precession around the Sun and a Moonless Earth system? It should be very small, too small to be detected maybe...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance!&lt;/p&gt;&#xA;" OwnerUserId="2553" OwnerDisplayName="Klest Dedja" LastEditorUserId="2153" LastEditDate="2014-10-07T00:23:45.373" LastActivityDate="2019-08-12T16:21:22.063" Title="Advancement of perihelion, data" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;planet&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6556" PostTypeId="1" CreationDate="2014-10-06T21:05:29.330" Score="2" ViewCount="86" Body="&lt;p&gt;When the set of official dwarf planets is expected to increase or new bodies considered for inclusion?&lt;/p&gt;&#xA;" OwnerUserId="2541" LastActivityDate="2014-10-07T14:55:50.510" Title="When the set of official dwarf planets is expected to increase?" Tags="&lt;dwarf-planets&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6558" PostTypeId="2" ParentId="2549" CreationDate="2014-10-06T22:09:06.080" Score="5" Body="&lt;p&gt;The dipole in the microwave background indicates motion of the Milky Way and thus of the whole Local Group, at least, at about 600 km/s in a certain direction.  The straightforward explanation is that the density irregularities nearby from superclusters and voids result in a net gravitational acceleration that, over the age of the universe, resulted in this velocity.  A number of superclusters (Hydra-Centaurus, Shapley, and Norma) happen to line up along this direction and reasonably fit the picture.  Also there are some big voids in the opposite direction.&lt;br&gt;&#xA;There are two basic ways to quantitatively understand what is the cause.  One can make measurement of local flows around these superclusters and voids to determine their masses and thus their net effect on the MW velocity.  Or one can try to measure the &quot;dipole motion&quot; on larger scales, that is at what size scale does this motion drop to 300 km/s and what scale does it drop to near 0?  A number of groups have tried to do this and the results are in disagreement.  In particular, one group found no drop off in motion out to fairly large distances.  Therefore, an alternate explanation was proposed that the dipole motion is primordial and due to objects beyond the horizon that interacted with us before inflation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will we ever settle this?  I am sure that we will eventually.  There is much work to do measuring the distances of galaxies out to 100 Mpc or so with greater and greater accuracy using new methods (SN Type 1a, surface brightness fluctuations, Tip of the Red Giant Branch) and larger telescopes with better instrumentation and then modeling the peculiar flow on all scales.  If the local distribution of matter cannot explain fully the dipole moment in the MWB, then perhaps Dark Flow from before inflation will be necessary to explain it.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2014-10-06T22:09:06.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6559" PostTypeId="2" ParentId="4800" CreationDate="2014-10-06T22:38:24.190" Score="6" Body="&lt;p&gt;Most of the light energy in the universe is still in the cosmic microwave background (CMB). &lt;a href=&quot;http://www.int.washington.edu/PHYS554/2011/chapter2_11.pdf&quot; rel=&quot;nofollow&quot;&gt;Spring 2011 UC Berkeley Physics 250 class materials&lt;/a&gt; calculate from the fact that $T=2.73$ for the CMB:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It follows that photons contribute only $0.0000485$ of the closure density.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Closure density is the density needed to close the universe if there were no cosmological constant. The observed density of the universe has been determined to be about 0.3 x closure density.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2016-03-31T16:22:11.720" LastActivityDate="2016-03-31T16:22:11.720" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6560" PostTypeId="2" ParentId="6556" CreationDate="2014-10-06T22:45:06.123" Score="3" Body="&lt;p&gt;As I noted in a previous answer, the official body for naming and classification of astronomical objects is the &lt;a href=&quot;http://www.iau.org/&quot; rel=&quot;nofollow&quot;&gt;International Astronomical Union&lt;/a&gt;, or IAU. They are the ones responsible for designating bodies as &quot;planets,&quot; &quot;dwarf planets,&quot; etc. The scientific community recognizes them and their decisions as &quot;official.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The IAU meets periodically for General Assemblies to discuss general issue; &lt;a href=&quot;http://astronomy2015.org/&quot; rel=&quot;nofollow&quot;&gt;the next such meeting&lt;/a&gt; is scheduled for 2015. The topics discussed may range widely, covering all sub-fields of astronomy. Unfortunately, I don't know the exact topics for discussion, and while I imagine there will be a wide range, issues relating to dwarf planets may not be covered. However, the IAU &lt;em&gt;does&lt;/em&gt; meet in between these meetings to discuss a variet of topics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That might answer your question, but if you're talking about the possible &lt;em&gt;discovery&lt;/em&gt; of dwarf-planet-like bodies, I can't help you. There are probably many on-going efforts to search for small bodies within our solar system, but I don't know of any recent discoveries of possible dwarf planet candidates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-10-07T14:55:50.510" LastActivityDate="2014-10-07T14:55:50.510" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6561" PostTypeId="2" ParentId="6435" CreationDate="2014-10-07T03:02:21.350" Score="3" Body="&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/HALCA&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/HALCA&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Spektr-R&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Spektr-R&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HALCA was a Japanese 8 meter radio telescope satellite used for Very Long Baseline Interferometry (VLBI) from 1997 to 2003.  Spektr-R is a Russian 10 m dish launched into space in 2011 with apogee nearly as far away as the moon, and still going.&#xA;So, yes a radio telescope in space is quite useful for long baseline interferometry.  You probably do not want to also use it for deep space communications because that would interfere with the scientific use.  It is also not very useful for asteroid work because the odds of one going by is not greater than the odds of one going by the Earth and there are more powerful radio dishes on the ground.  Typically, one would not bother having a strong source for bouncing off asteroids on a radio telescope meant for extragalactic work.  And you sure would not want to risk sending it into the asteroid belt!&lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2014-10-07T03:02:21.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6562" PostTypeId="1" CreationDate="2014-10-07T11:26:31.613" Score="7" ViewCount="161" Body="&lt;p&gt;I heard a claim that due to holographic principle, the surface area of the cosmic horizon corresponds to the universe's total entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such the initial state had zero surface area and later expanded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this, I wonder whether any increase in entropy (such as producing heat by means of electric power) causes the universe to expand?&lt;/p&gt;&#xA;" OwnerUserId="2541" LastActivityDate="2016-04-18T19:34:31.860" Title="Does the radius of the Universe correspond to its total entropy?" Tags="&lt;cosmology&gt;&lt;cosmological-horizon&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6563" PostTypeId="2" ParentId="2533" CreationDate="2014-10-07T11:36:24.107" Score="-1" Body="&lt;p&gt;Let's try the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equate the forces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xA;F_c &amp;amp;=&amp;amp; \frac{mv^2}R\\&#xA;F_g &amp;amp;=&amp;amp; \frac{GMm}{R^2}\\&#xA;\newline&#xA;mv^2/R &amp;amp;=&amp;amp; \frac{GMm}{R^2}\\&#xA;v^2 &amp;amp;=&amp;amp; \frac{GM}R&#xA;\end{eqnarray*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the radius is $R_s = 2GM/c^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xA;v^2 &amp;amp;=&amp;amp; \frac{GM}{2GM/c^2}\\&#xA;v &amp;amp;=&amp;amp; \sqrt{c^2/2}\\&#xA;v &amp;amp;=&amp;amp; \frac{c}{\sqrt2}\\&#xA;v &amp;amp;=&amp;amp; 0,707c&#xA;\end{eqnarray*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if I made all the considerations, but as far as my knowledge goes, if the blackhole is spinning faster than $0,707c$, the event horizon won't be able to maintain itself on the current radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when the radius expands, the rotation will slow down by conservation of angular momentum... so I don't think it will rip apart... perhaps become a &quot;grey hole&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please forgive me if any fundamental mistake was made, I'm new to all this... :P&lt;/p&gt;&#xA;" OwnerUserId="2556" LastEditorUserId="188" LastEditDate="2014-10-07T15:02:07.043" LastActivityDate="2014-10-07T15:02:07.043" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6564" PostTypeId="2" ParentId="5890" CreationDate="2014-10-07T21:46:39.713" Score="2" Body="&lt;p&gt;I see two problems.  1) Any particular source in the sky is above the horizon for a limited region of the Earth at a given time, so one would need to relay the signal around the globe.  2) It is hard to think of an astronomical source that is truly random all of the time.  Astronomical sources are physical systems reacting to internal and environmental changes that are causal, so they all tend to be partially quasiperiodic to some degree or become temporarily nearly periodic.  So it depends on just how random you need them to be, but I doubt any astronomical source is dependable for this purpose.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2014-10-07T21:46:39.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6565" PostTypeId="2" ParentId="6122" CreationDate="2014-10-07T23:31:28.923" Score="0" Body="&lt;p&gt;they use Newton's law of universal gravitation, example gravity interaction between moon and earth , jupiter Jupiter and its moons, or sun and planets, &#xA;you can find &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_constant&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; , play with m1 and m2, G and r , to calculate , the key is r and G , get m1 for earth and m2 for any body in free fall , after get distance between earth and sun, and get m for the sun , if they required most exact I guess that they use relativity and Lorenz transformations.  &lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2014-10-07T23:31:28.923" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6566" PostTypeId="2" ParentId="6164" CreationDate="2014-10-08T00:47:26.190" Score="2" Body="&lt;h2&gt;Iron spheres&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There is a hypothesis that after about $10^{1500}$ years (many star generations) iron stars (iron sphere star remnants) will form. This would require all the elements lighter than Fe to fuse into Fe and all the elements heavier than Fe to decay into Fe. Also proton is required not to decay for this hypothesis to work.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Mass limit&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Only stars under certain mass limit can possibly form a sphere of iron. As others pointed out after the nuclear fusion ceases stars heavier than 1.44 solar masses will collapse into a &lt;a href=&quot;https://en.wikipedia.org/wiki/Neutron_star&quot; rel=&quot;nofollow&quot;&gt;neutron star&lt;/a&gt; and above several solar masses they will further collapse into a &lt;a href=&quot;https://en.wikipedia.org/wiki/Black_hole&quot; rel=&quot;nofollow&quot;&gt;black hole&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that iron spheres considerably heavier than the solar mass cannot form because the iron nuclei of heavier objects will be crushed to mostly neutrons or some other state of matter not consisting of iron.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;References for the iron star hypothesis&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Iron_star&quot; rel=&quot;nofollow&quot;&gt;iron star&lt;/a&gt; on Wikipedia&lt;/li&gt;&#xA;&lt;li&gt;Dyson, Freeman J. (1979). &quot;Time without end: Physics and biology in an open universe&quot;. Reviews of Modern Physics 51 (3): 447–460. Bibcode:1979RvMP...51..447D. doi:10.1103/RevModPhys.51.447.&lt;/li&gt;&#xA;&lt;li&gt;Dyson, Freeman J. (1996). &quot;&lt;a href=&quot;http://books.google.cz/books?id=nnyNUidX1OMC&amp;amp;lpg=PA534&amp;amp;ots=T-eVa_TpyS&amp;amp;dq=%22iron%20star%22%20fusion&amp;amp;pg=PA534#v=onepage&amp;amp;q=%22iron%20star%22%20fusion&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Selected Papers of Freeman Dyson with Commentary&lt;/a&gt;&quot;, American Mathematical Society&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="188" LastEditorUserId="188" LastEditDate="2014-10-08T17:11:05.043" LastActivityDate="2014-10-08T17:11:05.043" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6568" PostTypeId="2" ParentId="6122" CreationDate="2014-10-08T01:04:06.963" Score="2" Body="&lt;p&gt;I don't know if this is what most astronomers use, but it is certainly a method that &lt;em&gt;could&lt;/em&gt; be used. It also makes for some interesting photos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_lens&quot; rel=&quot;nofollow&quot;&gt;Gravitational lensing&lt;/a&gt; was one of the great predictions of Einstein's theory of general relativity, and was actually one of the first pieces of evidence for it - see the solar eclipse of 1919. Gravitational lensing is the bending of light by a massive object. Because mass and energy warp space-time, we know that if you can calculate how much the light is bent, you can calculate the mass of the body bending it. Wikipedia gives the following formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta=\frac{4GM}{rc^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $M$ is mass, $r$ is the distance the object is from the light, and $G$ and $c$ are, of course, constants. So if you can measure how much light is bent by an object, you can calculate its mass. It's hard, and you would need good conditions for it to work, but it could work, nonetheless.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-10-08T01:04:06.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6570" PostTypeId="1" AcceptedAnswerId="6573" CreationDate="2014-10-08T09:56:56.893" Score="-1" ViewCount="184" Body="&lt;p&gt;I am not an astrophysicist and not its student.All my knowledge about astronomy comes from you tube videos and discovery channel.From them i knew that to solve the mysteries of black hole both relativity and quantum theories should be combined as quantum gravity.But it does not exist.What is the problem in quantum gravity concept?&lt;/p&gt;&#xA;" OwnerUserId="2557" LastActivityDate="2014-10-31T18:02:44.737" Title="why quantum gravity doesnot exist?" Tags="&lt;black-hole&gt;&lt;general-relativity&gt;&lt;quantum-field-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6571" PostTypeId="1" AcceptedAnswerId="6582" CreationDate="2014-10-08T14:06:37.347" Score="21" ViewCount="7182" Body="&lt;p&gt;As I understand it, the asteroid belt exists because the gravitational force of Jupiter prevents the asteroids from accreting (is that a word?) into a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, however, Jupiter didn't exist and they did create a planet, how big would that planet be?&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2021-01-03T18:28:48.830" Title="How big would the asteroid belt planet be?" Tags="&lt;jupiter&gt;&lt;asteroid-belt&gt;" AnswerCount="6" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6572" PostTypeId="2" ParentId="6571" CreationDate="2014-10-08T15:09:10.690" Score="18" Body="&lt;p&gt;The mass of the asteroid main belt is estimated at &lt;a href=&quot;https://en.wikipedia.org/wiki/Asteroid_belt&quot;&gt;4% the mass of our moon&lt;/a&gt; according to Wikipedia so any object formed from the aggregation of that mass would not be a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be the size of a very small moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if all the asteroids in the solar system were combined, the total mass would be below a third of the moon's mass. &lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2014-10-08T15:09:10.690" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6573" PostTypeId="2" ParentId="6570" CreationDate="2014-10-08T15:33:07.273" Score="8" Body="&lt;p&gt;I'm not sure what you mean by saying that quantum gravity &quot;doesn't exist&quot;. But because this is the Astronomy SE, I will interpret your question as primarily asking why astronomy hasn't found evidence of quantum gravity. This is a reasonable question; after all, nineteenth-century astronomers have found evidence of funny business in the perihelion precession of Mercury, which in the twentieth century was understood as the relativistic corrections to the Newtonian orbits. Then astronomy one-upped that and found more evidence for general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why not now? Because gravity is &lt;em&gt;weak&lt;/em&gt;. Very weak. Weakness &lt;em&gt;per se&lt;/em&gt; isn't a big problem theoretically; it makes it possible to calculate quantum gravity corrections when they aren't too large, which can be interpreted as a running Newton's constant:&#xA;$$G(r) = G\left[1 - \frac{167}{30\pi}\frac{G\hbar}{r^2c^2} + \ldots\right]\text{.}$$&#xA;For example, for the Sun's gravitational field near the Sun's surface (mass $M_\odot\sim 2\times 10^{30}\,\mathrm{kg}$, radius $R_\odot\sim 7\times 10^8\,\mathrm{m}$), the relativistic corrections are of order $GM_\odot/rc^2\sim 10^{-6}$, a few parts per million, while the quantum gravity corrections are on the order of $G\hbar/r^2c^3\sim 10^{-88}$, which is so much smaller as to be completely hopeless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making the Sun a black hole would not improve things by much, since in that case $r = 2GM_\odot/c^2$ means while the relativistic corrections become appreciable, the quantum gravity corrections are on the order of $10^{-76}$. Note that larger black holes are actually &lt;em&gt;worse&lt;/em&gt;, which is sensible because large-scale quantum gravity must agree with general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting overview by Cliff P. Burgess can be found &lt;em&gt;Living Rev. Relativity&lt;/em&gt; &lt;strong&gt;7&lt;/strong&gt; (2004) 5, &lt;a href=&quot;http://relativity.livingreviews.org/Articles/lrr-2004-5/&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-10-08T15:33:07.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6574" PostTypeId="1" CreationDate="2014-10-08T16:34:17.300" Score="0" ViewCount="218" Body="&lt;p&gt;I have heard that when our solar system formed that all planets except Venus were revolving around the sun in the same direction because of the normal rotation of their axes. Venus was rotating and revolving backwards or retrograde. There was a collision with Venus that caused it to rotate and revolve normally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why would a planet revolve backwards around a star relative to all the other planets in the stellar system?&lt;/p&gt;&#xA;" OwnerUserId="1138" LastEditorUserId="6" LastEditDate="2016-05-23T17:55:48.883" LastActivityDate="2016-05-23T17:55:48.883" Title="Venus started with retrograde rotation?" Tags="&lt;solar-system&gt;&lt;planetary-systems&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-10-13T01:39:43.140" ContentLicense="CC BY-SA 3.0" />
  <row Id="6575" PostTypeId="2" ParentId="6574" CreationDate="2014-10-08T17:30:08.387" Score="2" Body="&lt;p&gt;You've confused its rotation and revolution. &lt;a href=&quot;http://en.wikipedia.org/wiki/Venus#Orbit_and_rotation&quot; rel=&quot;nofollow&quot;&gt;Venus&lt;/a&gt; revolves around the Sun in the same way as all the other planets, but its &lt;em&gt;rotation&lt;/em&gt; is in the opposite direction of theirs. It has retrograde rotation, but not revolution. I think this solves your paradox.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-10-08T17:30:08.387" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6576" PostTypeId="1" CreationDate="2014-10-08T18:14:52.807" Score="6" ViewCount="257" Body="&lt;p&gt;I was enamored by the &lt;a href=&quot;http://en.wikipedia.org/wiki/SN_1006&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia article&lt;/a&gt; of SN1006.  Esp. the fact that it was so bright, it could cast shadows on the ground.  (Even this idea that you could have possibly read parchment with just the light from the supernova.  &lt;a href=&quot;http://news.discovery.com/space/astronomy/astronomers-pinpoint-cause-of-sn-1006-121008.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Question is basically what is the width (radius?) of the remnant as seen in this picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8O1c1.jpg&quot; alt=&quot;Picture from WikiSky of SN1006&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2561" LastActivityDate="2014-10-09T00:36:12.627" Title="What is the current size of the SN1006 supernova remnant?" Tags="&lt;supernova&gt;&lt;size&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6577" PostTypeId="1" AcceptedAnswerId="6581" CreationDate="2014-10-08T18:45:51.597" Score="1" ViewCount="128" Body="&lt;p&gt;What would happen if two galaxies collided very quickly (e.g. at near the speed of light)?  Would they mostly just pass through eachother unscathed, or would there be very energetic collisions that would rip both galaxies apart, or would something else happen?&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2014-10-08T20:08:32.670" Title="What happens if two galaxies collide very quickly?" Tags="&lt;galaxy&gt;&lt;special-relativity&gt;&lt;n-body-simulations&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6578" PostTypeId="2" ParentId="6577" CreationDate="2014-10-08T19:23:14.933" Score="3" Body="&lt;p&gt;Have 2 people stand in opposite corners of a large empty room each with a jar of marbles.  Have them roll these at each other all at once as fast as they can, and you will get a pretty good answer to your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;97% (or thereabouts) of a given galaxy is empty space.  Most of one is going to pass harmlessly through the other, though you are likely to get a few collisions.  At or near the speed of light, gravity will have very little effect as well, so a few orbits will be perturbed, but any given star will hardly notice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect (though I haven't the expertise to say definitively) that the radiation increase during the ~100,000 years or more that this collision would take (assuming the galaxies were of approximately the size of the Milky Way and that they hit exactly edge on) would be significant to any creatures living in your galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If by the off chance (I'm talking about lightning-striking-a-shark-currently-eating-a-recent-lottery-winner kind of chance) the supermassive black holes at the center of these two galaxies collided, the speed would be no obstacle, and they would combine.  The angular momentum would certainly rip the new black hole from one of, or likely both, of the colliding galaxies.  This would be bad for them.  I propose sending this as a &quot;What If?&quot; question to XKCD.  He seems to have the time to do the math on questions of this nature and follow as far as it leads.&lt;/p&gt;&#xA;" OwnerUserId="2503" LastActivityDate="2014-10-08T19:23:14.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6580" PostTypeId="1" CreationDate="2014-10-08T19:53:11.350" Score="0" ViewCount="3330" Body="&lt;p&gt;If the Universe is flat and expanding at the speed of light, that must mean that we could never leave it, because one would have to travel faster than the speed of light which is not possible. However, what about going straight up or down ? The distance is less and if the Universe is not expanding in depth but only sideways then it would be possible to exit. Could we then find another Universe once we have gone out of this one ?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastActivityDate="2018-01-11T20:54:22.360" Title="How to leave the Universe" Tags="&lt;universe&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6581" PostTypeId="2" ParentId="6577" CreationDate="2014-10-08T20:08:32.670" Score="3" Body="&lt;p&gt;As MBurke said, the star in each galaxy will hardly be perturbed. Moreover, the chance for any two stars to collide is miniscule (there is about 1 star per pc$^3$ in the denser part of a galaxy, so the star fill about $10^{-30}$ of the volume). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, most galaxies are also filled with gas in various phases (the so-called ISM -- inter-stellar medium and a warm/hot gaseous halo). The gas contents of each galaxy will crash into each other with dramatic effects: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;the gas will shock highly super-sonicly and get heated up to enormous temperatures ($&amp;gt;10^{8-10}$K), such that is glows in the X-rays or even $\gamma$ rays;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the gas will get stuck in the place of the collision while the stars (and dark matter) travel on after the encounter.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Since gas makes only a small fraction of the total mass of a galaxy, the removal of their gas will have little effect on the dynamics of the galaxies, though they will stop forming new stars or feeding an AGN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, such collisions are observed for not for galaxies, but for galaxy clusters (when the individual galaxies are mostly unscathed, but the cluster gas gets stuck in the middle), such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bullet_Cluster&quot; rel=&quot;nofollow&quot;&gt;bullet cluster&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-10-08T20:08:32.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6582" PostTypeId="2" ParentId="6571" CreationDate="2014-10-08T20:42:33.943" Score="17" Body="&lt;p&gt;The largest main belt asteroid is &lt;a href=&quot;https://en.wikipedia.org/wiki/Ceres_%28dwarf_planet%29&quot;&gt;1 Ceres&lt;/a&gt;, which alone contains almost a third of the total mass of the whole main asteroid belt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ceres is large enough to be in hydrostatic equilibrium, i.e. its own gravity is strong enough to pull it into a roughly spherical shape.  Since the mass of a spherical planet scales as the cube of the diameter (assuming constant density), piling all the other main belt asteroids together onto Ceres would only increase its diameter by a bit under 50%.  It would still be a roughly similar type of body &amp;mdash; a small sphere of partially differentiated rock and ice, with no atmosphere to speak of (as it'd be way too small to hold onto one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, to answer your question, a hypothetical planet containing all the matter currently making up the main asteroid belt would look pretty much like Ceres already does, just a bit bigger.&lt;/p&gt;&#xA;" OwnerUserId="464" LastActivityDate="2014-10-08T20:42:33.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6583" PostTypeId="1" AcceptedAnswerId="6594" CreationDate="2014-10-08T21:12:33.943" Score="8" ViewCount="5237" Body="&lt;p&gt;I've been wondering this for quite some time.  Due to the variations in tilt from the earth we know that &lt;a href=&quot;https://astronomy.stackexchange.com/questions/240/how-does-moonrise-moonset-azimuth-vary-with-time&quot;&gt;moonrise/moonset times vary&lt;/a&gt;.  Is it possible that in &lt;em&gt;any&lt;/em&gt; location on the earth one might see more than one moonrise or moonset in a given 24 hour period?&lt;/p&gt;&#xA;" OwnerUserId="2350" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2021-05-28T00:37:35.187" Title="Is it possible to see a moonrise or moonset twice in a day?" Tags="&lt;the-moon&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6584" PostTypeId="1" AcceptedAnswerId="6602" CreationDate="2014-10-08T21:15:05.243" Score="5" ViewCount="309" Body="&lt;p&gt;As I understand it, one object can orbit another at a variety of altitudes, and the stability of the orbit is determined by (among other things) the speed of the orbiting object.  Go too slowly and you'll fall into the larger object, go too quickly and you'll fly off into space, but go at just the right speed and you'll orbit indefinitely.  And while an object could be in a stable orbit at Altitude X by going Speed A, it could also be in an equally stable orbit at Altitude Y by going Speed B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's true (and given my limited understanding, feel free to say it simply isn't), is there a lower limit?  An atmosphere would obviously cause drag, but given a lack of atmosphere, could you theoretically have an asteroid stably orbiting an Earth-sized planet at an altitude of 50 feet, if it were going fast enough?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, why not?  What forces limit the closeness of an orbiting object?  Also, simple explanations would be most appreciated: I've heard of things like Lagrange points and such, but I don't have a good understanding of them.&lt;/p&gt;&#xA;" OwnerUserId="529" LastActivityDate="2014-10-09T13:13:41.003" Title="Is there a lower limit for the altitude of orbiting objects?" Tags="&lt;orbit&gt;&lt;astrophysics&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6585" PostTypeId="2" ParentId="6580" CreationDate="2014-10-08T21:26:17.757" Score="8" Body="&lt;p&gt;Welcome to Stack Exchange: Astronomy!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When scientists say the universe is &quot;flat,&quot; they don't mean that it's shaped like a dinner plate.  It's no easier to travel &quot;up&quot; than &quot;forward.&quot;  In fact, off-planet, those terms don't even have any meaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you're referring to has to do with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Shape_of_the_universe&quot; rel=&quot;noreferrer&quot;&gt;shape of the universe&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When physicists describe the universe as being flat or nearly flat, they're talking geometry: how space and time are warped according to general relativity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Think of it like this: if you draw a triangle on a balloon, is that triangle flat?  In one sense yes, because it never rises above the surface of the balloon.  Relative to the balloon, it is one-dimensional; you wouldn't be wrong to say that the image of the triangle is &quot;flat.&quot;  But relative to the rest of the room, it is curved, because it follows the curvature of the balloon's shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When most people talk about the universe, they're talking about the triangle.  The discussion you're referring to, of the &quot;flatness&quot; of the universe, is discussing the shape of the balloon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for flying &quot;out&quot; of the universe and potentially into another, we're like an ant walking on the surface of that balloon: no matter what direction the ant walks, it'll just keep walking around the balloon.  The surface of his universe may be curved, but there's no direction he can go that takes him off of that curve.  For all he cares, it's essentially a flat surface, because he can only move in two dimensions along it.  Similarly, the geometry of our universe may be one thing or another, flat or curved, but either way that doesn't mean we can point a ship in any direction and &quot;leave.&quot;&lt;/p&gt;&#xA;" OwnerUserId="529" LastActivityDate="2014-10-08T21:26:17.757" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6587" PostTypeId="2" ParentId="6580" CreationDate="2014-10-08T23:04:06.303" Score="0" Body="&lt;p&gt;Good answer by Nerrolken. I'll also add that our universe is thought to be infinite in size - which is thought to be a consequence of a flat universe. The data from &lt;a href=&quot;http://map.gsfc.nasa.gov/&quot; rel=&quot;nofollow&quot;&gt;WMAP&lt;/a&gt; seems to indicate very strongly that we live in a flat universe, and that, therefore, in an infinite one. It is impossible to leave an infinite universe, no matter how fast you're going.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-10-08T23:04:06.303" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="6588" PostTypeId="2" ParentId="6584" CreationDate="2014-10-08T23:13:06.763" Score="0" Body="&lt;p&gt;This isn't necessarily a full answer to your question, but this could cover at least part of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you ever heard of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Roche_limit&quot; rel=&quot;nofollow&quot;&gt;Roche limit&lt;/a&gt;? It's the distance inside which an object cannot orbit another object because tidal forces will tear it apart. That's thought to be the reason behind the &lt;a href=&quot;https://en.wikipedia.org/wiki/Planetary_ring&quot; rel=&quot;nofollow&quot;&gt;rings&lt;/a&gt; of the gas giants: In each case, a moon could have wandered too close to the planet, and it might have been torn apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Roche limit varies between the parent object and the orbiting object, depending on factors such as mass and density. A simple way to calculate it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d=1.26R_m \left(\frac{M_M}{M_m}\right)^{1/3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R_m$ is the radius of the orbiting object, $d$ is the Roche limit, and $M_M$ is the mass of the parent and $M_m$ is the mass or the orbiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While there are other factors that influence just where an object can and cannot orbit, the Roche limit is a good place to start investigating.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-10-08T23:13:06.763" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6589" PostTypeId="1" AcceptedAnswerId="6592" CreationDate="2014-10-08T23:56:22.863" Score="12" ViewCount="4480" Body="&lt;p&gt;So black holes are created by certain dying stars and when the star runs out of nuclear energy gravity wins out and the star implodes. An entire star’s mass collapses down into a smaller and smaller volume of space. Which then creates a black hole, so my question is, do black holes actually have energy to create that type of vacuum to suck light in?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a black hole was created by a huge star running out of fuel and then collapsing in on it self, would the black hole have energy and if so where would it come from, or is it just like the vacuum in space with no energy at all?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Please correct me if I said anything wrong about the black holes as I am very young and don't quiet grasp the concept of it yet.)&lt;/p&gt;&#xA;" OwnerUserId="2567" LastActivityDate="2021-01-25T04:04:46.093" Title="Do black holes have energy?" Tags="&lt;black-hole&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6590" PostTypeId="2" ParentId="6589" CreationDate="2014-10-09T00:21:01.250" Score="5" Body="&lt;p&gt;In outer space, a vacuum isn't what sucks in matter to a black hole. That may be the case for your household vacuum cleaner, but everywhere in outer space is essentially a vacuum. Instead, it's the black hole's powerful gravitational pull.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, black hole can have energy. One type that comes to mind is &lt;em&gt;angular kinetic energy&lt;/em&gt;. This is the energy that a spinning object has - in this case, a &lt;a href=&quot;https://en.wikipedia.org/wiki/Rotating_black_hole&quot; rel=&quot;noreferrer&quot;&gt;rotating black hole&lt;/a&gt; (see also &lt;a href=&quot;https://en.wikipedia.org/wiki/Kerr_metric&quot; rel=&quot;noreferrer&quot;&gt;Kerr black hole&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Kerr%E2%80%93Newman_metric&quot; rel=&quot;noreferrer&quot;&gt;Kerr-Newman black hole&lt;/a&gt;). Black holes can also have &lt;em&gt;translational kinetic energy&lt;/em&gt; - the energy of an object in motion. Also black holes have gravitational potential energy - potential energy due to gravity - and &lt;a href=&quot;https://en.wikipedia.org/wiki/Reissner%E2%80%93Nordstr%C3%B6m_metric&quot; rel=&quot;noreferrer&quot;&gt;black holes with electric charge&lt;/a&gt; can have electric potential energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So black holes do have energy, although perhaps not in the form you thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep asking good questions!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-10-09T00:21:01.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6591" PostTypeId="2" ParentId="6576" CreationDate="2014-10-09T00:36:12.627" Score="3" Body="&lt;p&gt;You were so close! The answer was actually given lower down on the Wikipedia page: roughly 20 &lt;a href=&quot;https://en.wikipedia.org/wiki/Parsec&quot; rel=&quot;nofollow&quot;&gt;parsecs&lt;/a&gt;. Since one parsec is about 3.26 &lt;a href=&quot;https://en.wikipedia.org/wiki/Light-year&quot; rel=&quot;nofollow&quot;&gt;light-years&lt;/a&gt;, we can calculate that that comes out to about 64 light-years, as &lt;a href=&quot;http://amazing-space.stsci.edu/resources/fastfacts/sn1006.php.p=Capture+the+cosmos%40%2Ccapture%2C%3EStellar+evolution%40%2Ccapture%2Cstars%2C&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; seems to corroborate. The outer layers of the remnant are expanding outward at an outrageous rate - 11 million miles per hour, according to &lt;a href=&quot;http://www.nasa.gov/mission_pages/chandra/multimedia/tapestry.html&quot; rel=&quot;nofollow&quot;&gt;NASA&lt;/a&gt;. I invite you to do the calculations to figure out how large it will be at a certain point in time.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-10-09T00:36:12.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6592" PostTypeId="2" ParentId="6589" CreationDate="2014-10-09T01:08:21.637" Score="4" Body="&lt;p&gt;An isolated black hole is a vacuum solution of general relativity, so in a very direct sense it does not contain any energy anywhere in spacetime. But perhaps somewhat counter-intuitively, that does not imply that such a black hole has no energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Defining the total amount of energy is usually very problematic in general relativity, but in some special cases it is possible. In particular, the usual black hole solutions are all asymptotically flat, i.e., spacetime is just the usual flat Minkowski when far away from the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here (or in general when we have a prescribed asymptotic form of the spacetime), we can calculate the total energy-momentum, by essentially measuring the gravitational field of the black hole at infinity. The energy just be one component of energy-momentum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are actually two relevant different kinds of 'infinity' here: spatial infinity and null (lightlike) infinity, depending on whether we are 'far away' from the black hole in a spacelike or lightlike direction. There's also timelike infinity, but that just corresponds to waiting an arbitrarily long time, so it's not relevant here. The two different infinities beget different definitions of energy-momentum, giving the the &lt;strong&gt;ADM energy&lt;/strong&gt; and the &lt;strong&gt;Bondi energy&lt;/strong&gt;, respectively. In a vacuum, the intuitive difference between the two is that Bondi energy excludes gravitational waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the short answer is 'yes', with the caveat that in a more complicated situation, where we can't attribute everything to the black hole itself, the answer to &lt;em&gt;how much&lt;/em&gt; energy is due to the black hole may be ambiguous or ill-defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the ADM and Bondi energy-momenta also define their corresponding measures of &lt;strong&gt;mass&lt;/strong&gt;, as the norm of those energy-momenta ($m^2 = E^2-p^2$), but for a black hole we can also define mass more operationally in terms of orbits around the black hole. There are also other alternatives for addressing mass specifically.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-10-09T01:08:21.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6593" PostTypeId="2" ParentId="6550" CreationDate="2014-10-09T05:33:29.423" Score="4" Body="&lt;p&gt;Even if the answer has already been accepted, more evidences can contribute to this thought-provoking topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think, as it has been mentioned, eliosismology is a good way to map the interior of the star, but I am not able to argue about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another disproof, comes from neutrinos.&#xA;Interior of Sun can be investigated by neutrinos detection, which do not interact passing by the outer shells, and can directly say where they come from, and how.&#xA;Standard solar theory predict neutrinos production by &lt;a href=&quot;http://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction&quot; rel=&quot;nofollow&quot;&gt;p-p chain&lt;/a&gt;.&#xA;This has been observed, even if the neutrino flux is about $1/3$ than the expected one, because of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutrino_oscillation&quot; rel=&quot;nofollow&quot;&gt;neutrino oscillation&lt;/a&gt;.&#xA;Now, this theory become solid after Bruno Pontecorvo, but I do not think we have any other evidence of observed neutrino oscillation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, Earth's age is helping us. We know Earth is ${\sim}4$ billion years old, that is roughly the age of the Solar System and the Sun itself.&#xA;From &lt;a href=&quot;http://www.google.com/imgres?imgurl=http%3A%2F%2Fwww2.astro.psu.edu%2Fusers%2Fniel%2Fastro485%2Flectures%2Flecture16-overhead02.jpg&amp;amp;imgrefurl=http%3A%2F%2Fwww2.astro.psu.edu%2Fusers%2Fniel%2Fastro485%2Flectures%2Flectures485.html&amp;amp;h=1910&amp;amp;w=1422&amp;amp;tbnid=jSKMq6LUcvTQqM%3A&amp;amp;zoom=1&amp;amp;docid=47-VrcbdG02qgM&amp;amp;ei=Axo2VPK0BtP08QXz7oLIBA&amp;amp;tbm=isch&amp;amp;client=firefox-a&amp;amp;ved=0CDsQMygWMBY&amp;amp;iact=rc&amp;amp;uact=3&amp;amp;dur=626&amp;amp;page=1&amp;amp;start=0&amp;amp;ndsp=33&quot; rel=&quot;nofollow&quot;&gt;Niels Brendt website&lt;/a&gt;, we know that after such a long time, the white dwarf luminosity has become much fainter then Solar (where Solar is referred not only to the Sun, but to general G2 main sequence stars). Neutron stars cool &lt;a href=&quot;http://www.google.com/imgres?imgurl=http%3A%2F%2Fwww2.astro.psu.edu%2Fusers%2Fniel%2Fastro485%2Flectures%2Flecture17-overhead02.jpg&amp;amp;imgrefurl=http%3A%2F%2Fwww2.astro.psu.edu%2Fusers%2Fniel%2Fastro485%2Flectures%2Flectures485.html&amp;amp;h=1867&amp;amp;w=1373&amp;amp;tbnid=6HW9j1eYfqh9yM%3A&amp;amp;zoom=1&amp;amp;docid=47-VrcbdG02qgM&amp;amp;ei=Xh02VMDCNdjv8gWXk4LYCA&amp;amp;tbm=isch&amp;amp;client=firefox-a&amp;amp;ved=0CCAQMygCMAI&amp;amp;iact=rc&amp;amp;uact=3&amp;amp;dur=153&amp;amp;page=1&amp;amp;start=0&amp;amp;ndsp=33&quot; rel=&quot;nofollow&quot;&gt;even faster&lt;/a&gt;. And this is obviously not observed: if Manuel's conjecture was real, Sun should be much fainter now. To this, add that they say that their supposed SN happened something like 5 billion years ago...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a final note, I would highlight that, not only this author has never been mentioned in literature, that is (as we know) synonym of poor quality, but he actually never published (at least since 10years or so) in any refereed paper, which is automatically translated, to me and I think to the community, as not science.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastEditorUserId="1038" LastEditDate="2014-10-09T06:13:09.640" LastActivityDate="2014-10-09T06:13:09.640" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6594" PostTypeId="2" ParentId="6583" CreationDate="2014-10-09T07:20:30.997" Score="8" Body="&lt;p&gt;This can only happen if the moonrise at a certain date is earlier than the moonrise at the previous day. There are two reasons why this could happen:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The body (Moon, Mars,...) moves in the opposite direction of the Sun viewed along the right ascension (or ecliptical longitude) axis. The sidereal day is 4 minutes shorter than 24 hours which means that a static celestial body (a star) will rise 4 minutes earlier than the day before. However planetary bodies move (more or less) along the ecliptic, which is the Sun's path during the year along the celestial sphere. The Sun moves on average 4 minutes in ecliptical longitude (and hence in right ascension with a bit more variation) every day. So for a body to rise earlier, the body would have to move slower along the celestial sphere than the Sun (or move in a retrograde direction). This is true for the outer planets (including mars) but not for the Moon, which moves in the same direction as the Sun and much faster (one orbit per month).&lt;/li&gt;&#xA;&lt;li&gt;If the observer is at a sufficient high latitude on Earth (either north or south of the equator), the inclination of the celestial equator will be small enough so that a body that increases its declination rapidly enough will be able to rise earlier on the second day.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;See for instance the figure below where each drawing is at the same time of day. The moon will increase its right ascension during the day (along the celestial equator downwards, the red arrow in the figure). If the observer is on the Earth's equator, moonrise will be later the next day. It does not matter whether the declination (the position above or below the celestial equator (to the left or right in the figure)) will increase or decrease. All objects with the same right ascension will rise at the same time on the equator. If the observer is not on the equator but near one of the poles (for instance at 65° latitude north), the increase in declination may be sufficient to counteract the increase in right ascension (see bottom panel). Note that even though  the right ascension increases (the moon moves left along the celestial equator axis), the Moon will rise earlier during the day.&lt;/p&gt;&#xA;&lt;p&gt;This only works (in the Northern hemisphere) when the Moon increases its declination (the height above the celestial equator) very rapidly. This happen only when the Moon is near the Vernal Equinox (i.e. in Aries or Pisces). And the observer needs to be at a high latitude (e.g. near the North pole such as Thule as mentioned by @barrycarter)&#xA;&lt;img src=&quot;https://i.stack.imgur.com/x9kYT.png&quot; alt=&quot;Moonrise at two days on the equator and at 60° latitude&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="864" LastEditorUserId="864" LastEditDate="2021-05-27T12:17:18.473" LastActivityDate="2021-05-27T12:17:18.473" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="6595" PostTypeId="1" CreationDate="2014-10-09T08:47:04.467" Score="6" ViewCount="786" Body="&lt;p&gt;Let's assume hypothetically that Earth has a twin planet on the opposite side of Earth's orbit. Its orbital period would be exactly the same as Earth's and it would always be behind the Sun so directly observing it from Earth would not be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would we detect the planet's existence short of sending out a spacecraft that could look &quot;behind&quot; the sun and observe it visually? Have we confirmed that there is no such planet?&lt;/p&gt;&#xA;" OwnerUserId="2569" LastActivityDate="2015-08-05T13:02:30.660" Title="How would we detect a planet behind the Sun?" Tags="&lt;planet&gt;&lt;observational-astronomy&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6596" PostTypeId="2" ParentId="6595" CreationDate="2014-10-09T08:56:06.250" Score="1" Body="&lt;p&gt;It would be detectable due to its gravity effect on other planets, comets and the Sun. Check these sites for further info:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Counter-Earth#Scientific_analysis&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Counter-Earth#Scientific_analysis&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrangian_point#L3&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lagrangian_point#L3&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1678" LastActivityDate="2014-10-09T08:56:06.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6597" PostTypeId="2" ParentId="6595" CreationDate="2014-10-09T09:00:33.797" Score="11" Body="&lt;p&gt;According to the Wikipedia article &lt;a href=&quot;http://en.wikipedia.org/wiki/Counter-Earth#Scientific_analysis&quot; rel=&quot;noreferrer&quot;&gt;Counter-Earth&lt;/a&gt;, this idea has been floated around for millennia.  However, modern day earth-bound techniques would easily detect a 'Counter-Earth' via the following (but not limited to) these methods:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Gravitational perturbations would have been observed to have occurred on the other planets (terrestrial)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As Earth's orbit is not circular, it is elliptical, it would not be consistently 180 degrees around the orbit, as below&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/phnRJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.soes.soton.ac.uk/staff/ejr/DarkMed/ch5.html&quot; rel=&quot;noreferrer&quot;&gt;Image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Most of all, it would have been detected/observed by the satellites we have around the Earth, Sun and other planets (many examples &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_active_Solar_System_probes&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user2449" LastEditorUserId="7882" LastEditorDisplayName="user2449" LastEditDate="2015-08-05T13:02:30.660" LastActivityDate="2015-08-05T13:02:30.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6598" PostTypeId="1" AcceptedAnswerId="6600" CreationDate="2014-10-09T11:41:13.187" Score="14" ViewCount="1672" Body="&lt;p&gt;What causes the dimensions of a star increase when its hydrogen fuel is exhausted? For example, the Sun is expected to increase its radius 250 times. What causes this if its temperature is expected to fall? How gas can expand if the temperature falls?&lt;/p&gt;&#xA;" OwnerUserId="2541" LastActivityDate="2017-09-19T01:38:04.117" Title="What causes the dimensions of a star increase when its hydrogen fuel is exhausted?" Tags="&lt;star&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6599" PostTypeId="1" AcceptedAnswerId="6613" CreationDate="2014-10-09T11:59:23.190" Score="2" ViewCount="175" Body="&lt;p&gt;The age of the universe is 13.798±0.037 billion years, yet the age of HD 140283 is 14.46±0.8 billion years, how this can be the case?&lt;/p&gt;&#xA;" OwnerUserId="2541" LastActivityDate="2014-10-10T00:04:33.757" Title="Age of the universe and age of stars" Tags="&lt;star&gt;&lt;cosmology&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6600" PostTypeId="2" ParentId="6598" CreationDate="2014-10-09T12:05:14.573" Score="12" Body="&lt;p&gt;The Sun will never run out of Hydrogen. This is a common misconception.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this moment the Sun is fusing Hydrogen into Helium. This fused Helium remains at the core until it will reach a &lt;strong&gt;critical mass&lt;/strong&gt;. At this point the core will begin to collapse. This collapse &lt;strong&gt;increases the temperature and pressure around the core&lt;/strong&gt; where Hydrogen is being fused causing the &lt;strong&gt;increase the Hydrogen fusion&lt;/strong&gt; which also causes an &lt;strong&gt;increase of the radiation pressure&lt;/strong&gt; making the outer layers to expand and cool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the core reaches a temperature of about 100 million Kelvin Helium fusion starts dramatically &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Helium_flash&quot; rel=&quot;nofollow&quot;&gt;(Helium Flash)&lt;/a&gt;&lt;/strong&gt; with high radiation pressure in the core. Here is where the Sun will reach 250 times it's current radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the Helium fusion stabilizes the radiation pressure at the cores decreases making the Sun's radius to shrink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;temperature will only drop at the surface&lt;/strong&gt; because as it's expands it's further from the core.&lt;/p&gt;&#xA;" OwnerUserId="1678" LastEditorUserId="1678" LastEditDate="2014-10-10T06:10:47.300" LastActivityDate="2014-10-10T06:10:47.300" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6601" PostTypeId="1" AcceptedAnswerId="6610" CreationDate="2014-10-09T12:36:17.833" Score="4" ViewCount="213" Body="&lt;p&gt;A person at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Photon_sphere&quot; rel=&quot;nofollow&quot;&gt;photon sphere&lt;/a&gt; of a black hole will observe the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The black hole surface will cover exactly half of the visible sky (say, the left half), and cosmic horizon will cover the other half.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Looking in any direction between these hemispheres the person will see oneself.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The person will observe no centrifugal force however fast he moves. Any object the person thowing to the left will experience the centrifugal force pulling it further left, while any object thrown to the right also will experience the centrifugal force pulling it to the right. Approaching the black hole in such circumstances may look quite like actually escaping it to the crew.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The space to the right as well the space to the left seemingly includes objects of any length and dimensions, the further objects are, the more they are time-dilated and length-contracted.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is there any way for a spacecraft commander to decide (by an experiment conducted in a small$^1$ amount of time) which hemisphere is the sky and which hemisphere is black hole in such circumstances?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^1$ Here &lt;em&gt;small&lt;/em&gt; means small compared to characteristic time scales of the problem.&lt;/p&gt;&#xA;" OwnerUserId="2541" LastActivityDate="2014-10-10T00:48:37.007" Title="Can a person at a photon sphere of a black hole decide where the black hole is?" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6602" PostTypeId="2" ParentId="6584" CreationDate="2014-10-09T13:13:41.003" Score="6" Body="&lt;p&gt;For really low orbits over an atmosphereless body, the body needs to have a uniform density. Otherwise the gravity field is not symmetric, the orbit changes shape over time, and you end up with a crater.&#xA;&lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2006/06nov_loworbit/&quot;&gt;NASA had trouble with this&lt;/a&gt;, lunar Mascons, during the Apollo era. In one case: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The Moon has no atmosphere to cause drag or heating on a spacecraft, so you can go really low: Lunar Prospector spent six months orbiting only 20 miles (30 km) above the surface.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In another: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The orbit of PFS-2 rapidly changed shape and distance from the Moon. In 2-1/2 weeks the satellite was swooping to within a hair-raising 6 miles (10 km) of the lunar surface at closest approach. As the orbit kept changing, PFS-2 backed off again, until it seemed to be a safe 30 miles away. But not for long: inexorably, the subsatellite's orbit carried it back toward the Moon. And on May 29, 1972—only 35 days and 425 orbits after its release—PFS-2 crashed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now that the gravity field of the moon is mapped, it turns out that there are &lt;em&gt;some&lt;/em&gt; low orbits that are reasonably stable: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;What counts is an orbit's inclination,&quot; that is, the tilt of its plane to the Moon's equatorial plane. &quot;There are actually a number of 'frozen orbits' where a spacecraft can stay in a low lunar orbit indefinitely. They occur at four inclinations: 27º, 50º, 76º, and 86º&quot;—the last one being nearly over the lunar poles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2014-10-09T13:13:41.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6603" PostTypeId="1" CreationDate="2014-10-09T13:23:19.897" Score="-9" ViewCount="595" Body="&lt;p&gt;I read an article recently that said NASA has discovered a brown dwarf that has frozen. &#xA;If we had the technology, resources, and supplies, could we actually freeze a star and stop it from burning its fuels? If so, how would we do it?&lt;/p&gt;&#xA;" OwnerUserId="2567" LastEditorUserId="2153" LastEditDate="2014-12-29T14:16:50.663" LastActivityDate="2014-12-29T14:16:50.663" Title="Could we actually freeze a star?" Tags="&lt;star&gt;&lt;brown-dwarf&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6604" PostTypeId="1" CreationDate="2014-10-09T13:49:13.303" Score="3" ViewCount="145" Body="&lt;p&gt;In the context of possible colonization of Ceres, would not such colony be totally impossible because of high probability of collision of Ceres with other asteroids in the Asteroid Belt?&lt;/p&gt;&#xA;" OwnerUserId="2541" LastActivityDate="2014-10-11T20:20:31.423" Title="What is the danger for Ceres of collision with other asteroids?" Tags="&lt;asteroids&gt;&lt;dwarf-planets&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6605" PostTypeId="2" ParentId="6571" CreationDate="2014-10-09T14:06:43.563" Score="2" Body="&lt;p&gt;Ceres constitutes about 1/3 of the mass of the asteroid belt. Assuming the same density, the whole asteroid belt combined into one planet would have 3 times greater volume than Ceres and $\sqrt[3]{3}=1.44...$ times greater radius.&lt;/p&gt;&#xA;" OwnerUserId="2541" LastActivityDate="2014-10-09T14:06:43.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6606" PostTypeId="1" AcceptedAnswerId="6607" CreationDate="2014-10-09T15:02:04.797" Score="5" ViewCount="212" Body="&lt;p&gt;Where can I observe the furthest horizon in the Solar system ? And what the length of it ?&#xA;The furthest horizon on Earth can be observed from mount Everest ~ 336 km, On Mars - from mount Olympus ~ 386 km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: If the walls of the Rheasilvia crater can be seen from the opposite sides (it appears so) you can see the surface as far as 505 km.&lt;/p&gt;&#xA;" OwnerUserId="2571" LastEditorUserId="2571" LastEditDate="2014-10-09T15:44:43.010" LastActivityDate="2014-10-09T17:34:33.520" Title="The furthest horizon in the Solar system" Tags="&lt;solar-system&gt;&lt;horizon&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6607" PostTypeId="2" ParentId="6606" CreationDate="2014-10-09T16:13:25.387" Score="3" Body="&lt;p&gt;The distance depends on the diameter of the planet and of your height above the surface (such as on a mountain). The greater the diameter, the farther away the horizon will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see this in the figure below. $d_3 &amp;gt; d_1 &amp;gt; d_2$. On a large planet your horizon (at distance $d_1$) will be farther away than on a smaller planet with horizon distance $d_2$. But if you stand on a mountain on the same planet then the distance will be even larger ($d_3$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can calculate the height from Pythagoras' rule because we have a triangle with one unknown edge. The distance from the planet's centre to the point where you see the horizon is the radius of the planet and the distance of the eye of the observer is the radius of the planet plus the height of the observer above the surface. &#xA;$$(r+h)^2 = r^2+d^2$$&#xA;where $r$ is the radius of the planet, $h$ the height above the surface, and $d$ is the horizon distance. If we rewrite this equation we get:&#xA;$$d^2 = (r+h)^2 - r^2 $$&#xA;$$d^2 = r^2+h^2+2rh-r^2$$&#xA;$$d^2 = h^2 + 2rh$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Venus we have $r=6052\textrm{km}$ and $h=11\textrm{km}$ (&lt;a href=&quot;http://en.wikipedia.org/wiki/Maxwell_Montes&quot; rel=&quot;nofollow noreferrer&quot;&gt;Maxwell Montes&lt;/a&gt;), we therefore get a horizon distance of 365km. This is still smaller than the horizon distance from Olympus Mons. Olympus Mons is the highest mountain and all other rocky planets are smaller, &lt;strong&gt;the horizon distance from Olympus Mons is, therefore, the greatest horizon distance.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether the horizon distance in the Rheasilvia crater is greater depends on whether you would be able to see the crater edge from the middle of the crater. Vesta's radius is &lt;a href=&quot;http://en.wikipedia.org/wiki/4_Vesta&quot; rel=&quot;nofollow noreferrer&quot;&gt;262km&lt;/a&gt;. If you stand on the crater edge, which is &lt;a href=&quot;http://en.wikipedia.org/wiki/Rheasilvia&quot; rel=&quot;nofollow noreferrer&quot;&gt;13km&lt;/a&gt; above the crater surface, you would only have a horizon distance of &lt;strong&gt;83km&lt;/strong&gt;. This means that from the centre of the crater you would not be able to see the crater's edge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; Above I assume that the 13km is the height of the crater edge above the centre of the crater as measured from the geodesic. If the height is the height above a flat (not spherical flat) surface than the horizon is indeed 505km (see comments below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/14f7G.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="864" LastEditorUserId="864" LastEditDate="2014-10-09T17:34:33.520" LastActivityDate="2014-10-09T17:34:33.520" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6608" PostTypeId="2" ParentId="6603" CreationDate="2014-10-09T18:06:26.470" Score="1" Body="&lt;p&gt;If you'd somehow be able to freeze a star and stop nucleosynthesis (fusion) then it would not be a star anymore. By definition, a star needs a temperature high enough to support fusion. This is why brown dwarfs are not true stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are of course stellar remnants (white dwarfs, neutron stars) that have actually crystallised (and are therefore in a solid state). See for instance &lt;a href=&quot;http://www.universetoday.com/112769/an-earth-sized-diamond-in-the-sky-the-coolest-known-white-dwarf-detected/&quot; rel=&quot;nofollow&quot;&gt;PSR J2222-0137&lt;/a&gt;. One might say they have 'frozen'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How you might actually do that artificially, I have no idea...&lt;/p&gt;&#xA;" OwnerUserId="864" LastEditorUserId="864" LastEditDate="2014-10-09T18:13:32.623" LastActivityDate="2014-10-09T18:13:32.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6609" PostTypeId="1" CreationDate="2014-10-09T20:04:29.440" Score="12" ViewCount="912" Body="&lt;p&gt;This is relevant for the definition of a dwarf planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I presume the answer will be, well, if we can tell the mass of the body and guess the material.   I don't find this very satisfactory because (1) may be impossible and (2) will have large error.&lt;/p&gt;&#xA;" OwnerUserId="1106" LastEditorUserId="1106" LastEditDate="2015-10-30T14:27:12.547" LastActivityDate="2016-12-29T05:08:01.847" Title="How can astronomers determine the difference between &quot;hydrostatic equilibrium&quot; and &quot;just happens to be spherical&quot;?" Tags="&lt;gravity&gt;&lt;asteroids&gt;&lt;dwarf-planets&gt;&lt;hydrostatic-equilibrium&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6610" PostTypeId="2" ParentId="6601" CreationDate="2014-10-09T21:56:30.037" Score="4" Body="&lt;p&gt;For an observer hovering [I say hover because by definition there are no orbits for an observer at the photon sphere] at the photon sphere for a non-maximal de Sitter-Schwarzschild black hole, a simple accelerometer will tell an observer which horizon is which. The reason for this is that in order to hover at the photon sphere the observer must take up a non-inertial local frame with a directional bias decided by the direction of the BH event horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the maximal de Sitter-Schwarzschild spacetime the BH singularity disappears and the observer will not be able to tell the difference between the two horizons as there is an automorphism of that spacetime which will map one horizon to the other.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastEditorUserId="2092" LastEditDate="2014-10-10T00:48:37.007" LastActivityDate="2014-10-10T00:48:37.007" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6611" PostTypeId="1" CreationDate="2014-10-10T00:02:02.310" Score="9" ViewCount="164" Body="&lt;p&gt;Suppose there is a tidally locked planet orbiting a star. The planet's surface consists of a global ocean, that is, liquid water. At the inner hemisphere the temperature is so high that the water is constantly boiling, creating the atmosphere of water vapor. But the vapor does not reach the dark side of the planet, and precipitates around the terminator line. The further side has no atmosphere and an icy surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this setup possible? Can it be such thing that one side of a planet has liquid (and boiling) water surface while the other side has no atmosphere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible at all that a part of an ocean was boiling while the other one was icy?&lt;/p&gt;&#xA;" OwnerUserId="2541" LastActivityDate="2014-10-10T08:46:15.330" Title="Is it possible that only one hemisphere of a planet has an atmosphere?" Tags="&lt;planet&gt;&lt;exoplanet&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6612" PostTypeId="1" CreationDate="2014-10-10T00:03:41.447" Score="4" ViewCount="189" Body="&lt;p&gt;Alpha Centauri Bb is an exoplanet orbiting Alpha Centauri B. It is asserted that given the close distance to the star the planet should be tidally locked. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbiting period of the planet is about 3.2 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the planet has no atmosphere (which is very possible due to proximity to the star), its dark side should experience low temperatures like permanently-shadowed areas of Mercury do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same time the planet is about 11 AU from the other star, Alpha Centauri A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This possibly means that the planet should experience quasi-day/night cycle each 3.2 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;To what temperatures the dark surface of such planet could be heated? Is there possibility of liquid water?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Will the radiation of the second star be enough to provide normal day-like illumination and heating?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Will the calendar on such a planet differ sufficiently from a calendar of a planet that experiences day/night cycles from the nearest star?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Will such planet experience seasons and how they would be arranged?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2541" LastActivityDate="2014-10-10T00:56:18.823" Title="What day/night cycles, climate and seasons would experience Alpha Centauri Bb inhabitants?" Tags="&lt;planet&gt;&lt;exoplanet&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6613" PostTypeId="2" ParentId="6599" CreationDate="2014-10-10T00:04:33.757" Score="5" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/HD_140283&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; reports the age of HD 140283 as the figure you cited, 14.46±0.8 billion years. Now look at &lt;a href=&quot;https://en.wikipedia.org/wiki/HD_140283#cite_note-arxiv-2&quot; rel=&quot;noreferrer&quot;&gt;the reference it refers to&lt;/a&gt;. This is the paper by Bond et al that presents measurements of some characteristics of the star and the subsequent calculations, as well as a lengthy section on where the error came from (800 million years is, after all, quite a lot - even in Astronomy!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper describes an approach to determining ages of stars without all that tedious mucking around with globular clusters. It suggests determining &quot;ages of extreme Population II subgiants in the solar neighborhood based on direct trigonometric parallaxes, combined with state-of-the-art theoretical isochrones appropriate to the  detailed composition of each star.&quot; In short, the idea is to test stars to see if there ages are consistent with the calculated age of the universe (13.77±0.06 billion years - note the much smaller uncertainty).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper cites some previous estimates of the star's age - ~14 billion years and 13.5±1.5 billion years. It's worth noting this because the authors appear to pride themselves on how they tried to reduce the error of their measurements. The team next measured the star's &lt;a href=&quot;https://en.wikipedia.org/wiki/Parallax&quot; rel=&quot;noreferrer&quot;&gt;parallax&lt;/a&gt;, also attempting to reduce error. With that in hand, as well as measurements of temperature and visual magnitude, they used stellar evolution models to determine its age. They also compensated for &lt;a href=&quot;https://en.wikipedia.org/wiki/Extinction_(astronomy)&quot; rel=&quot;noreferrer&quot;&gt;extinction&lt;/a&gt; from their measurements. The authors list the error from parallax, photometry, and extinction as 310 million years. The measurements then (page 11) list the age &quot;implied&quot; by the models and measurements to be 14.46 billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is the extra 490-million year-uncertainty? Take a look at Figure 1, near the end. It is due to a variety of factors, the largest of which is the uncertainty of the &quot;initial oxygen abundance&quot; - although more properly the uncertainty in the ratio of oxygen to hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does come down to what David H said - the error measurements due comply with a universal age of 13.77 billion years. I would note, though, that the uncertainty in the measurement of the age of HD 140283 is much greater (800 million years) than the uncertainty of the age of the universe (37 million years). If the uncertainty due to the oxygen/hydrogen ratio could be eliminated, a more accurate age could be calculated, but until this error is addressed (preferably by another team), I think it's safe to say that it's more likely that the measurements for HD 140283 are off than that the measurements for the universe, and so what could have been a paradox is somewhat resolved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR&lt;/strong&gt; - There's a heck of a lot of error in the age of HD 140283, which fits reasonably with current models of the age of the universe, and I would trust the measurements of the age of the universe much more than the measurements of the age of this star, especially because the method used here is not widely used.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-10-10T00:04:33.757" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6614" PostTypeId="2" ParentId="6612" CreationDate="2014-10-10T00:56:18.823" Score="1" Body="&lt;p&gt;This is just a partial answer (so far), but it will address at least part of your question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1.To what temperatures the dark surface of such planet could be heated? Is there possibility of liquid water?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Alpha_Centauri_Bb&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; gives a good estimate for this: 1,200 degrees Celsius. For a planet, that's pretty hot! Any water there would evaporate very quickly. In fact, it is expected that a lot of the planet's surface is molten. Now, does this apply to the back side of the planet? I would think so. While only the front side of the planet would bask in the warmth of Alpha Centauri B, the heat should dissipate throughout the entire planet.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2.Will the radiation of the second star be enough to provide normal day-like illumination and heating?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The short answer here is no. In fact, when compared to our solar system, &lt;em&gt;nothing&lt;/em&gt; about the radiation &lt;a href=&quot;https://en.wikipedia.org/wiki/Alpha_Centauri_Bb&quot; rel=&quot;nofollow&quot;&gt;Alpha Centauri Bb&lt;/a&gt; gets is normal. It is situated extremely close to Alpha Centauri B - 0.04 AU. Given that both &lt;a href=&quot;https://en.wikipedia.org/wiki/Alpha_Centauri&quot; rel=&quot;nofollow&quot;&gt;Alpha Centauri A and Alpha Centauri B&lt;/a&gt; are similar to the Sun, I think it's clear that this exoplanet will &lt;em&gt;not&lt;/em&gt; receive a &quot;normal&quot; (normal here meaning similar to Earth) amount of radiation from Alpha Centauri B. Conversely, Alpha Centauri Bb is 11 AU away from Alpha Centauri A - much greater than the distance between it and Alpha Centauri B (a bit more than the distance between &lt;a href=&quot;https://en.wikipedia.org/wiki/Saturn&quot; rel=&quot;nofollow&quot;&gt;Saturn&lt;/a&gt; and the Sun). This means that the planet will receive only a tiny fraction of light from Alpha Centauri A relative to Alpha Centauri B.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;4.Will such planet experience seasons and how they would be arranged?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seasons depend on the tilt of a planet's axis. Unfortunately, it is hard (if not impossible) to measure the tilt of the axis of Alpha Centauri Bb (I haven't been able to find any measurements). I would assume that it would experience &lt;em&gt;some&lt;/em&gt; seasons, but I don't know how much they would extend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for question 3 - I'm not quite sure what you mean by &quot;calendar.&quot; If you mean an artificial calendar, I would think that any beings on the planet (life there is &lt;em&gt;extremely&lt;/em&gt; unlikely) would only use Alpha Centauri B to determine their years (just over three days and five hours).&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-10-10T00:56:18.823" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6615" PostTypeId="2" ParentId="6603" CreationDate="2014-10-10T01:22:20.403" Score="1" Body="&lt;p&gt;The reason that this &lt;a href=&quot;https://en.wikipedia.org/wiki/Brown_dwarf&quot; rel=&quot;nofollow&quot;&gt;brown dwarf&lt;/a&gt; is so cold compared to other stars is that it &lt;em&gt;was never on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Main_sequence&quot; rel=&quot;nofollow&quot;&gt;main sequence&lt;/a&gt;&lt;/em&gt;. More specifically, it never reached the hydrogen-1-burning phase of its life. Most stars (including our Sun) fuse elements together (beginning with hydrogen-1), which generates &lt;a href=&quot;https://en.wikipedia.org/wiki/Star#Temperature&quot; rel=&quot;nofollow&quot;&gt;heat&lt;/a&gt;. Brown dwarfs do not do this, and thus are much colder than other stars - although, as caters pointed out, some of the more massive ones &lt;em&gt;can&lt;/em&gt; fuse deuterium, and possibly lithium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that brown dwarf is cold because it was never &quot;hot&quot; in the first place. I highly doubt that we could ever &quot;freeze&quot; a main sequence star, however, because we would have to somehow halt nuclear fusion.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-10-16T01:15:14.483" LastActivityDate="2014-10-16T01:15:14.483" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6616" PostTypeId="1" CreationDate="2014-10-10T01:41:00.463" Score="2" ViewCount="48" Body="&lt;p&gt;What is the furthest object from which non-electromagnetic cosmic rays were detected?&lt;/p&gt;&#xA;" OwnerUserId="2541" LastActivityDate="2014-10-10T01:41:00.463" Title="What is the furthest object from which fermion rays were detected?" Tags="&lt;cosmic-ray&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2018-02-26T13:49:44.960" ContentLicense="CC BY-SA 3.0" />
  <row Id="6617" PostTypeId="1" AcceptedAnswerId="6622" CreationDate="2014-10-10T04:58:00.820" Score="15" ViewCount="2454" Body="&lt;p&gt;Would an Earth analog exoplanet which rotates around a solar analog, but without axial tilt, have no seasons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be similar to how Los Angeles does not have very noticeable seasons compared to other parts of the Earth?&lt;/p&gt;&#xA;" OwnerUserId="2301" LastActivityDate="2014-10-10T20:54:16.977" Title="Would an exoplanet without axial tilt have no seasons?" Tags="&lt;exoplanet&gt;&lt;rotation&gt;&lt;earth-like-planet&gt;&lt;weather&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6618" PostTypeId="2" ParentId="6617" CreationDate="2014-10-10T06:30:19.233" Score="6" Body="&lt;p&gt;Exactly. Seasons and its difference between northern and southern hemisphere are caused due to the tilt of the Earth's axis. But this is not the only thing involved. An other important fact is the &lt;strong&gt;orbit's eccentricity&lt;/strong&gt;: a very eccentric orbit would cause a whole planet &quot;summer&quot; when around the perihelion and a whole planet &quot;winter&quot; near the aphelion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There would be also &lt;strong&gt;extreme permanent hot zones near the equator&lt;/strong&gt; and &lt;strong&gt;extreme permanent cold zones at poles&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1678" LastActivityDate="2014-10-10T06:30:19.233" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6619" PostTypeId="2" ParentId="6617" CreationDate="2014-10-10T06:32:17.317" Score="6" Body="&lt;p&gt;If it would not have any axial tilt, and it rotates in an orbit thats not really eccentric, it would have no seasons. &#xA;If it has (really) high eccentricity, it would have seasons, but they would be hot all over the planet a part of the year, and than cold all over the planet the other part of the year. On earth, due to the axial tilt, we have winter in the northern hemisphere, at the moment that there is summer in the southern hemisphere. &lt;/p&gt;&#xA;" OwnerUserId="1062" LastActivityDate="2014-10-10T06:32:17.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6620" PostTypeId="2" ParentId="6611" CreationDate="2014-10-10T06:38:17.587" Score="5" Body="&lt;p&gt;That situation would lead to heat transfer between the cold and the hot zones just like it happens here at the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a paper on this that may help you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] heat transport from the day side to the night side of the&#xA;  planets efficiently reduces temperature contrasts [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://arxiv.org/abs/1001.5117&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1001.5117&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1678" LastEditorUserId="1678" LastEditDate="2014-10-10T08:46:15.330" LastActivityDate="2014-10-10T08:46:15.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6621" PostTypeId="1" CreationDate="2014-10-10T08:08:49.353" Score="1" ViewCount="93" Body="&lt;p&gt;Black holes decrease as they evaporate and their radius decreases as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is with a cosmological horizon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If cosmological horizon is just a black hole centered at the opposite side of the universe, we should see the radius of the external space groving as the radius of the BH decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if the radiation decreases the horizon entropy we should see the area of the horizon decreasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the correct conclusion?&lt;/p&gt;&#xA;" OwnerUserId="2541" LastActivityDate="2014-10-10T15:19:41.333" Title="Does cosmological horizon grow or decrease as it radiates?" Tags="&lt;cosmology&gt;&lt;cosmological-horizon&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6622" PostTypeId="2" ParentId="6617" CreationDate="2014-10-10T09:34:07.027" Score="12" Body="&lt;p&gt;There are several factors that cause seasons.  In approximate order from least to most speculative:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Axial Tilt&lt;/strong&gt;:  This is by far the significant factor in determining what the season is on Earth.  As you may know, axial tilt affects the seasons because the light is &quot;spread out&quot; when it falls on an angled surface.  Note that tilt causes different seasons in different hemispheres.  An untilted planet could still have seasons for any of these other reasons.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Eccentricity&lt;/strong&gt;:  The Earth's orbit is very close to circular, so eccentricity has next to no effect on causing its seasons (and in fact, IIRC, currently Earth's eccentricity's tiny effect is actually &lt;em&gt;opposing&lt;/em&gt; its seasons).  An exoplanet with a high eccentricity could easily have &lt;em&gt;planet-wide&lt;/em&gt; seasons due to varying distance from its sun.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Star Systems&lt;/strong&gt;:  As a commenter mentioned, another factor might be how many stars a system has.  If such a solar system is stable (which, surprisingly, it might be, although I can't find the article at the moment), you could get seasons depending on where the stars are in their mutual orbits.  A binary star system, for example, would have periods of winter when the stars align with the planets (so one blocks the other) and periods of summer when the stars are side-by-side.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Star Variation&lt;/strong&gt;:  (From a comment): believe it or not, the sun actually has &quot;seasons&quot; (&lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/1999/ast22jul99_1/&quot;&gt;here&lt;/a&gt;) that are on the order of a decade or so.  These don't have much effect on Earth's climate since on the sun they mainly cause magnetic disturbances, but on an exoplanet the star's magnetic disturbances could be relatively larger, which would lead to significant temperature/brightness variations that would translate into seasons.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Internal Energy&lt;/strong&gt;:  Planets could have varying/cyclic sources of energy within them, just like solar masses.  In our own solar system, Jupiter's seasons are partly due to this.  On a terrestrial planet, variations might come from, for example, volcanic activity.  Negative feedback loops of this sort will exhibit oscillations.  These sorts of changes might not count as &quot;seasons&quot; for you, though, as they occur over larger timescales.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Stored Energy&lt;/strong&gt;: Earth's seasons are complicated by the oceans, which greatly affect the intensity of the seasons, especially in the Americas.  In particular, there are various cycles that work like subseasons underlying the main ones.  The Pacific Decadal Oscillation, for example, affects seasons years apart.  I don't see why an exoplanet couldn't have similar cycles that, unlike in the case of Earth, are the main factor in determining seasons.  This could be stored gravitational energy too; for example in a tidally-locked exoplanet/moon system.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you mean a true Earth-like analogue (in the sense of being exactly-like-Earth-in-every-way-except-the-axial-tilt), then no--it would not have noticeable seasons, since none of the above factors really apply.&lt;/p&gt;&#xA;" OwnerUserId="2576" LastEditorUserId="2576" LastEditDate="2014-10-10T18:44:00.700" LastActivityDate="2014-10-10T18:44:00.700" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6624" PostTypeId="1" AcceptedAnswerId="6626" CreationDate="2014-10-10T11:54:19.663" Score="1" ViewCount="3601" Body="&lt;p&gt;Is there a theoretical limit on what we can see with a telescope of fixed size based on Earth's orbit or the Moon, that is in Earth's region of Solar system outside the Earth's atmosphere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why not instead of sending probes to Pluto and Ceres, just invest efforts into building better telescopes, so to photograph the surfaces of these bodies in greater detail? I suspect that all necessary information in fact reacheds Earth's region, it is just matter of sensibility and exposure to fix it, am I right?&lt;/p&gt;&#xA;" OwnerUserId="2541" LastActivityDate="2014-10-10T13:25:10.897" Title="Is there a theoretical limit on telescope's resolution?" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6625" PostTypeId="1" AcceptedAnswerId="6628" CreationDate="2014-10-10T12:02:08.520" Score="1" ViewCount="509" Body="&lt;p&gt;Why Earth's perihelion occurs 3rd January rather than 1st January? Is there any effort to correct this discrepancy?&lt;/p&gt;&#xA;" OwnerUserId="2541" LastEditorUserId="2541" LastEditDate="2014-10-10T14:34:50.080" LastActivityDate="2018-08-02T12:12:32.790" Title="Why Earth's perihelion occurs 3rd January rather than 1st?" Tags="&lt;orbit&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6626" PostTypeId="2" ParentId="6624" CreationDate="2014-10-10T13:25:10.897" Score="4" Body="&lt;p&gt;The absolute limit of a telescope resolution is given by diffraction. &#xA;No matter how perfectly built and aligned is a telescope, you cannot resolve angles smaller than&#xA;$$\theta \propto \frac{\lambda}{D}$$&#xA;where $\lambda$ is the wavelength of interest and $D$ is the diameter of the telescope. This is why a number of millimeters and radio telescope (and also some -antennas) are huge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first figure &lt;a href=&quot;http://en.wikipedia.org/wiki/Angular_resolution&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; shows the basic principle. Imagine that you can divide what you observe in tiny squares and consider each one as a point like source. Each one will generate a diffraction pattern when passing into the telescope and if two points are too near, you cannot distinguish between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wisely you put your telescope in space: turbulence in the atmosphere degrade the signal, and the best/biggest telescopes have hard times to go below $0.5 arc seconds$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Astronomical_interferometer&quot; rel=&quot;nofollow&quot;&gt;Interferometry&lt;/a&gt; comes to the rescue increasing $D$ from the telescope size to the distance of two (or more) telescopes (called baseline). Interferometry has been used in radio astronomy since decades. &#xA;From what I hear optical interferometry is much much more complicated and as far as I know the only large scale attempt is the &lt;a href=&quot;http://www.eso.org/paranal/telescopes/vlti/&quot; rel=&quot;nofollow&quot;&gt;VLTI project&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you could imagine to have a constellation of relatively small telescopes spread over hundred thousands or millions of kilometers. But this have the huge problem that you have to know the position and timing of every one of them with an impressive precision (my guess is that the precision in position is of the order of $\lambda$).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The other problem is light collection. If you want to see something very faint you have two options: 1)you observe for a lot of time or 2) you build a bigger telescope (&lt;a href=&quot;http://jwst.nasa.gov/facts.html&quot; rel=&quot;nofollow&quot;&gt;6&lt;/a&gt; to &lt;a href=&quot;http://www.eso.org/sci/facilities/eelt/&quot; rel=&quot;nofollow&quot;&gt;40&lt;/a&gt; meters in diameter). &#xA;And here also the largest baseline interferometers cannot do much, as the amount of light that they collect is just the sum of the light collected by the single telescopes. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To conclude: to observe Pluto or Ceres with high enough accuracy you would need a large number of large space telescopes very far away one from the other with perfect telemetry. It's far easier and cheaper to got there to take pictures.  &lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2014-10-10T13:25:10.897" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6627" PostTypeId="2" ParentId="6625" CreationDate="2014-10-10T13:40:48.390" Score="4" Body="&lt;p&gt;The beginning of the year (1st of January) has been chosen for historical reasons (near the winter solstice) and has absolutely nothing to do with Earth's perihelion passage. Earth's perihelion could as well have occurred on the 10th of October. That it is near the 1st of January is pure chance (see for instance a post by &lt;a href=&quot;http://earthsky.org/space/are-the-december-solstice-and-the-january-perihelion-related&quot; rel=&quot;nofollow&quot;&gt;EarthSky&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is, therefore, not a discrepancy that needs to be corrected.&lt;/p&gt;&#xA;" OwnerUserId="864" LastEditorUserId="864" LastEditDate="2014-10-10T13:50:45.090" LastActivityDate="2014-10-10T13:50:45.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6628" PostTypeId="2" ParentId="6625" CreationDate="2014-10-10T13:50:35.440" Score="6" Body="&lt;p&gt;Back in 1246 CE, the perihelion coincided with the December solstice. The Earth was at perihelion at 1 Jan in 2005 and will be on 6 Jan in 2096, GMT time. The drift is not monotonic, but it does progress overall throughout the centuries. Eventually, it will drift towards the March equinox and beyond, to come back to the winter solstice in around twenty thousand more years. This is primarily due to Milanković cycles of Earth's eccentricity and precession index.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the problem isn't just effectively unfixable; it's just not a problem in the first place. The apsides were seldom a concern for calendars. The practice of 1 Jan as the new year may have begun with the Romans, but it was not continuously practiced throughout Europe, e.g., the early medieval new year began at several different days from early to late March at different times in history, and sometimes at Christmas.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-10-10T13:50:35.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6629" PostTypeId="2" ParentId="6595" CreationDate="2014-10-10T14:39:50.160" Score="2" Body="&lt;p&gt;One of the &lt;a href=&quot;https://en.wikipedia.org/wiki/STEREO&quot; rel=&quot;nofollow&quot;&gt;STEREO&lt;/a&gt; satellites has been at the earth-sun L4 point, so we have &quot;looked behind the sun&quot;, and verified that there is no counter-earth there.&#xA;That was not the purpose of their mission though.&lt;/p&gt;&#xA;" OwnerUserId="1062" LastEditorUserId="2153" LastEditDate="2014-10-11T18:26:37.803" LastActivityDate="2014-10-11T18:26:37.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6630" PostTypeId="1" CreationDate="2014-10-10T15:00:36.440" Score="-3" ViewCount="756" Body="&lt;p&gt;At the time the Big Bang happened the matter had enormous density. According the GR (I may be wrong here) such density dilates time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, could it be that the time periods just after Big Bang which are usually considered happening in small part of a second (such as the Planck epoch), in reaity took billons of year (or may be, infinity) but due to time dilation appear to us as spanning only microscopic parts of a second? Could it be that the age of the universe is dramatically underestimated?&lt;/p&gt;&#xA;" OwnerUserId="2541" LastActivityDate="2014-10-10T16:59:39.787" Title="Time dilation at the Big Bang" Tags="&lt;cosmology&gt;&lt;big-bang-theory&gt;" AnswerCount="2" CommentCount="8" ClosedDate="2014-10-13T01:57:34.800" ContentLicense="CC BY-SA 3.0" />
  <row Id="6631" PostTypeId="2" ParentId="6621" CreationDate="2014-10-10T15:19:41.333" Score="3" Body="&lt;p&gt;I think there's a misinterpretation here that's causing your confusion. There's a difference between the event horizon of a black hole and the cosmological horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Event_horizon&quot; rel=&quot;nofollow&quot;&gt;event horizon&lt;/a&gt; of a black hole is the place inside which nothing can escape from the black hole. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmological_horizon&quot; rel=&quot;nofollow&quot;&gt;cosmological horizon&lt;/a&gt;, on the other hand, is the place beyond which an observer cannot &quot;retrieve information.&quot; The difference between the two is basically that the event horizon of a black hole is in the same place for observers in different locations, while the cosmological horizon differs for observers in different locations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if the cosmological horizon &lt;em&gt;was&lt;/em&gt; the edge of a black hole, the black hole would have to be centered right where we are - which is &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6305/is-it-possible-that-a-ultra-large-portion-of-the-space-we-live-in-is-already-ins/6307&quot;&gt;highly unlikely&lt;/a&gt;!&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-10-10T15:19:41.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6632" PostTypeId="2" ParentId="6630" CreationDate="2014-10-10T16:12:32.237" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;there are certain laws that established by experiment (e.g. time diliation in presence of high density).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no such law.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So what the basis for claiming this did not happen in early universe as well?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fact there is no such law, for one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, let's step back a bit and take the simpler examples of time dilation. In special relativity, the ur-example of time dilation is between inertial frames, which correspond to a family of comoving intertial observers. So in making a statement involving the time dilation factor $\gamma$ such as&#xA;$$\frac{\mathrm{d}t}{\mathrm{d}\tau} = \frac{1}{\sqrt{1-v^2}} = \gamma\text{,}$$&#xA;we're comparing one inertial observer (with time $t$) with some other, not necessarily inertial, observer with time $\tau$. This is also generalizable to comparing arbitrarily-moving observers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For gravitational time dilation, the ur-example is a comparison of the passage of time for two different observers stationary at two different elevations in a gravitational field. The moral is just the same: any talk of time dilation is at least indirectly referring to a comparison between different clocks, and thus different observers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It basically boils down to: time dilation of &lt;em&gt;what&lt;/em&gt; relative to &lt;em&gt;what&lt;/em&gt;? Once you understand this, you will see how your question is actually reducible to the previously linked one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It seems to me that after performing this the Big Bang will be at infinite past (am I wrong?) or in other words, there would be no big bang at all. I understand that this amounts to &quot;interpretation&quot; by I still do not see why this time scale is worse than the stand one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can always formally define a time $t'$ such that the big bang occurs at $t'\to-\infty$, or some such thing, e.g., by $t' = t_0\log(t/t_0)$, where $t$ is the usual cosmological time and $t_0$ is some arbitrary time scale. But doing this is merely a relabeling of coordinates, and has no physical significance. In particular, it doesn't change the length of the worldline of any observer, which is the duration measured by a clock along that worldline, aka &lt;em&gt;proper time&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in the usual Big Bang cosmology (though not all FLRW solutions), the proper time of any observer, however idealized, is finite in the past. Relabeling to infinite time is no different then taking the usual Euclidean space in Cartesian coordinates, $(x,y,z)$, defining a new coordinate $\xi = \log x$, and then saying that the origin is now infinitely far away, being at $\xi\to-\infty$. You can define the coordinate, but it doesn't change actual distances.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-10-10T16:12:32.237" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6633" PostTypeId="2" ParentId="6630" CreationDate="2014-10-10T16:59:39.787" Score="2" Body="&lt;p&gt;If I remember Stephen Hawking alludes to some sort of &quot;quickening of time&quot; near the big bang in his book &lt;em&gt;A Brief History of Time&lt;/em&gt; (though I read it 20 years ago, so don't hold me to the exact details). He makes an argument that if time is measured by the number of collisions in a given volume then the number of collisions should go to infinity as you approach the big bang, thus meaning the big bang was infinitely far in the past. Note this isn't really time dilation (though maybe it could be reframed as such) which the comparison of time like coordinates between different spacetime coordinate systems that are set up to represent different observers' points of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact I was thinking about this the other day and I think this argument doesn't hold water. It is more sensible to talk about the number collisions in a given comoving volume and use the conformal time as the coordinates, because the number of collisions in a unit comoving volume per unit conformal time should be constant throughout the life of the Universe. As @Stan Liou alludes to though, in the models of the Universe best thought to represent our reality the big bang happens at some finite past value of conformal time.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastActivityDate="2014-10-10T16:59:39.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6634" PostTypeId="2" ParentId="6305" CreationDate="2014-10-10T17:13:07.827" Score="1" Body="&lt;p&gt;If Great Attractor were the center of a black hole to which we belong, we would approach it at speed greater than speed of light and there would be no way to move away from it, even temporarily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also we would observe it in any direction.&lt;/p&gt;&#xA;" OwnerUserId="2541" LastEditorUserId="2541" LastEditDate="2014-10-10T18:51:16.543" LastActivityDate="2014-10-10T18:51:16.543" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6635" PostTypeId="1" AcceptedAnswerId="6639" CreationDate="2014-10-10T18:30:40.207" Score="20" ViewCount="1056" Body="&lt;p&gt;I was reading the answers to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6617/would-an-exoplanet-without-axial-tilt-have-no-seasons&quot;&gt;this question&lt;/a&gt; about an exoplanet having seasons without axial tilt, and several responders mention that orbital eccentricity could cause a similar effect, but that the orbit would need to be &quot;very eccentric,&quot; as opposed to Earth's, which is &quot;very close to circular.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Earth had no axial tilt at all, would its eccentricity be high enough to have any discernible effect for the inhabitants?  Even if it didn't cause major shifts in temperature, would it be enough for pre-scientific people to track the passing of years?&lt;/p&gt;&#xA;" OwnerUserId="529" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2017-03-26T02:27:33.347" Title="Is Earth's orbital eccentricity enough to cause even minor seasons, without axial tilt?" Tags="&lt;orbit&gt;&lt;rotation&gt;&lt;weather&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6636" PostTypeId="2" ParentId="6617" CreationDate="2014-10-10T20:54:16.977" Score="1" Body="&lt;p&gt;I considered this as an explanation for George R. R. Martin's books, where several years could go by during each season.  If the planet had no axial tilt and very little eccentricity in its orbit, there would be no annual seasons.  But if solar energy output would vary sufficiently, you could have seasons, with inconsistent cycle times that measured in decades.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I thought, &quot;Ha, it is possible.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then some character in the book noticed &quot;the days were getting shorter&quot; and I had to throw it out.  In such a world, day length would always be the same, and it would always match night length.&lt;/p&gt;&#xA;" OwnerUserId="2583" LastActivityDate="2014-10-10T20:54:16.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6637" PostTypeId="1" AcceptedAnswerId="6640" CreationDate="2014-10-10T21:42:15.040" Score="5" ViewCount="189" Body="&lt;p&gt;Can we interpret the de Sitter universe as a spherical cosmic horizon null surface of finite radius, centered at Earth, and containing the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble_volume&quot; rel=&quot;noreferrer&quot;&gt;Hubble volume&lt;/a&gt; of space where time is dilated and radial dimensions contract closer to the edge in such a way that objects closer to the edge do not recognize that they are radially contracted?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything is attracted to the edge, but the total radius remains more or less constant and emits de Sitter radiation at finite temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xgtGt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I know real universe is not de Sitter but it is usually asserted that it approaches de Sitter state asymptotically.&lt;/p&gt;&#xA;" OwnerUserId="2541" LastActivityDate="2014-10-11T00:28:43.957" Title="Can Hubble red shift be interpreted as time dilation?" Tags="&lt;cosmology&gt;&lt;redshift&gt;&lt;cosmological-horizon&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6639" PostTypeId="2" ParentId="6635" CreationDate="2014-10-10T22:50:13.760" Score="14" Body="&lt;p&gt;&lt;em&gt;Very&lt;/em&gt; cool question. I want to get into a little bit of detail here because otherwise there would be a one-paragraph answer, and I don't think that would cut it. So here goes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The planets in the solar systems have orbits with pretty low &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_eccentricity#Examples&quot; rel=&quot;noreferrer&quot;&gt;eccentricities&lt;/a&gt; (see &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_gravitationally_rounded_objects_of_the_Solar_System&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; for more eccentricity values). At the upper end is Mercury, with an eccentricity of 0.2056. At the lower end is Venus, at 0.00677. Earth is in between but moderately low, at 0.0167. The distance between &lt;a href=&quot;https://en.wikipedia.org/wiki/Perihelion_and_aphelion&quot; rel=&quot;noreferrer&quot;&gt;perihelion and aphelion&lt;/a&gt; is 5 million kilometers - in an orbit with an average radius of about 150 million kilometers. Note, though, that eccentricities are always changing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would certainly have &quot;seasons&quot; if there was no axial tilt, but they most likely would not be dramatic. As &lt;a href=&quot;https://en.wikipedia.org/wiki/Perihelion_and_aphelion&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because of the increased distance at aphelion, only 93.55% of the solar radiation from the Sun falls on a given area of land as does at perihelion. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But on &lt;a href=&quot;https://en.wikipedia.org/wiki/Season#Elliptical_Earth_orbit&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; page, it says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Orbital eccentricity can influence temperatures, but on Earth, this effect is small and is more than counteracted by other factors; research shows that the Earth as a whole is actually slightly warmer when farther from the sun. This is because the northern hemisphere has more land than the southern, and land warms more readily than sea.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So more land means that the planet can absorb heat better, which counteracts the change in distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think that's pretty boring. Don't you? So let's calculate how eccentric Earth's orbit would have to be for it to have seasons without axial tilt. From the chart &lt;a href=&quot;https://en.wikipedia.org/wiki/Season#Modern_mid-latitude_meteorological&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; we can see that the highest average hemispherical temperature is 22 degrees Celsius, while the lowest is 8 degrees Celsius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2nptu.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2nptu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about other planets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_gravitationally_rounded_objects_of_the_Solar_System#Planets&quot; rel=&quot;noreferrer&quot;&gt;steady downward trend&lt;/a&gt; as we get further away from the Sun (Venus is an exception because of its runaway greenhouse effect). For Earth, the average surface temperature is 287 K (15 degrees C). So our maximum would be 295 K and our minimum would be about 281 K (Kelvin = Celsius + 273). Compare that to the other planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made a graph, which unfortunately I cannot paste here, that shows that to have a surface temperature of 295 K, the planet would have to be - well, not much closer than we are. Relatively, that is. Roughly 0.975 AU from the star. To have a surface temperature of 281 K, it would have to be at 1.05 AU from the star. This neglects, by the way, the different levels of land/sea absorption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, actually, if the Earth's orbit had an eccentricity of about 0.35, it could have some moderate seasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I used &lt;a href=&quot;http://www.1728.org/ellipse.htm&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; eccentricity calculator for the sake of accuracy and time. I would still be obliged if anyone could check my overall calculations for this answer.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="6841" LastEditDate="2017-03-25T00:35:49.477" LastActivityDate="2017-03-25T00:35:49.477" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6640" PostTypeId="2" ParentId="6637" CreationDate="2014-10-11T00:28:43.957" Score="4" Body="&lt;p&gt;The de Sitter spacetime is maximally symmetric, and so having a full set of Killing vector fields, it must have a static form produced by a timelike Killing vector field. &#xA;One way to get without worrying about coordinate transformations is to take the $M\to 0$ limit of the general spherically symmetric lambdavacuum solution, which is the the Schwarzschild–de Sitter spacetime:&#xA;$$\mathrm{d}s^2 = -\left(1-\frac{1}{3}\Lambda r^2\right)\mathrm{d}t^2 + \left(1-\frac{1}{3}\Lambda r^2\right)^{-1}\mathrm{d}r^2 + r^2\,\mathrm{d}\Omega^2\text{,}$$&#xA;where $\mathrm{d}\Omega^2 = \mathrm{d}\theta^2 + \sin^2\theta\,\mathrm{d}\phi^2$ is the usual unit $2$-sphere. Except for minor notational differences, this is also identical to the static de Sitter slicing presented &lt;a href=&quot;https://en.wikipedia.org/wiki/De_Sitter_space&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; in terms of different slices of a hyperboloid in Minwkoski $\mathrm{E}^{1,4}$, with $\alpha\equiv\sqrt{3/\Lambda}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In these coordinates, the timelike Killing field is simply $\partial_t$, which corresponds to a family of static observers with time dilation relative to the static observer at the origin given by&#xA;$$\dot{t}\equiv\frac{\mathrm{d}t}{\mathrm{d}\tau} = \frac{1}{1-r^2/\alpha^2}\text{,}$$&#xA;which diverges as $r\to\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;em&gt;geodesics&lt;/em&gt;, the Killing field also generates a conserved specific energy $\epsilon = \left(1-r^2/\alpha^2\right)\dot{t}$, which makes solving for radial geodesics not difficult. What's interesting about them is they take an infinite amount of coordinate time $t$ to reach $r=\alpha$, but only a finite amount of proper time $\tau$, and they have no trouble at all continuing beyond it. Additionally, time dilation for static observers becomes infinite there. This behavior is analogous to the event horizon of a black hole, except that the location of the de Sitter horizon is dependent on the choice of origin.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can Hubble red shift be interpreted as time dilation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. If we wish, we can think of the redshift due to a particle (e.g., galaxy) at some radial coordinate as a combination of two things: the special-relativistic time dilation of a galaxy with respect to a local static observer and the gravitational time dilation between static observers given above. In fact, the $\epsilon$ parameter above is exactly the (per-mass) energy of the particle as measured by a local static observer, i.e., the relative Lorentz gamma.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-10-11T00:28:43.957" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6641" PostTypeId="1" CreationDate="2014-10-11T03:19:20.897" Score="2" ViewCount="589" Body="&lt;p&gt;My name is Chris, I am a junior and I'm taking an Astronomy class in my high school.&#xA;For a homework assignment, I have to explain how the Sun would move across the sky from September 1st through September 30th (assuming that the weather allows viewing)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&#xA;In this picture the &lt;img src=&quot;https://i.stack.imgur.com/v6Hga.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sun is facing west.&#xA;If I were to take a picture of this same area at the same time every day for the whole month, &#xA;How would the sun change? Would it go east? &#xA;Would it go lower in the sky? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I, being a new student to Astronomy, honestly have no clue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you and if you need any more info please leave a comment. &lt;/p&gt;&#xA;" OwnerUserId="2585" LastEditorUserId="2585" LastEditDate="2014-10-13T22:30:20.167" LastActivityDate="2017-06-28T09:36:42.220" Title="Question about the Sun's Motion" Tags="&lt;the-sun&gt;" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6643" PostTypeId="1" AcceptedAnswerId="6651" CreationDate="2014-10-11T10:02:16.530" Score="6" ViewCount="370" Body="&lt;p&gt;Most of the celestial bodies we know are rotating or spinning on its axis.Is there any chance for the universe to rotate on its axis?&lt;/p&gt;&#xA;" OwnerUserId="2557" LastActivityDate="2019-04-06T12:02:36.033" Title="Is the whole universe is rotating on an axis?" Tags="&lt;universe&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6644" PostTypeId="1" CreationDate="2014-10-11T10:54:18.700" Score="7" ViewCount="275" Body="&lt;p&gt;In Astronomia Nova (1609) Johannes Kepler used observations of Mars in order to refute circular orbits. Throughout Astronomia Nova he hardly even mentions other planets than Mars, Earth and the Sun (and the Moon in a different context). Were there any astronomical reasons for him &lt;em&gt;not&lt;/em&gt; to continue, or even beginning, his investigation by using data about other planets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tycho Brahe had collected corresponding data about the other planets too, right? 20 years of data is enough for Saturn to make 2/3 of its period. Jupiter and Saturn have almost twice as frequent oppositions as does Mars. And Mercury has twice the eccentricity and its much shorter period should help getting data about its apsides when it is best seen.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Would it had helped him to use several planets during the process as he describes it in Astronomia Nova?&lt;/li&gt;&#xA;&lt;li&gt;Was it simply the calculation capacity which limited him from dragging several planets into it?&lt;/li&gt;&#xA;&lt;li&gt;Or &lt;em&gt;could he somehow conclude that it is enough to solve for Mars, to know that his physical conclusions of interests are valid for all planets?&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2014-10-12T08:22:56.773" LastActivityDate="2015-01-20T15:47:05.670" Title="Why didn't Johannes Kepler use data about more planets?" Tags="&lt;fundamental-astronomy&gt;&lt;history&gt;&lt;eccentric-orbit&gt;&lt;johannes-kepler&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6645" PostTypeId="2" ParentId="122" CreationDate="2014-10-11T11:15:02.380" Score="3" Body="&lt;p&gt;Parallel universes are a theory, we just don't know yet.  No one has slid into one.  (slid coming from the series Sliders )  According to the Big Bang theory, multiverses are possible.  &lt;a href=&quot;https://www.universeguide.com/facts/multiverse&quot; rel=&quot;nofollow&quot;&gt;https://www.universeguide.com/facts/multiverse&lt;/a&gt; Britains' most well known physics scientist/presenter Prof Brian Cox @ProfBrianCox recently said that multiverses makes sense ( &lt;a href=&quot;http://www.bbc.co.uk/news/science-environment-29321771&quot; rel=&quot;nofollow&quot;&gt;http://www.bbc.co.uk/news/science-environment-29321771&lt;/a&gt; ).  &lt;/p&gt;&#xA;" OwnerUserId="2587" LastActivityDate="2014-10-11T11:15:02.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6647" PostTypeId="2" ParentId="5921" CreationDate="2014-10-11T11:33:52.450" Score="2" Body="&lt;p&gt;You've answered your own question there, the difference between the two is that asteroids are rock whereas comets are made up of ice and dust.  An asteroid doesn't heat up when it gets close to the Sun and doesn't produce a tail.  Comets can orbit the Sun or any star in a long elliptical (Comet West) or short elliptical (Comet Encke 3.3 Years) period whereas asteroids are found in the Asteroid Belt between Mars and Jupiter and in the Kuiper Belt.  Sedna is a long period asteroid further out any other object in the solar system but it is still an asteroid according to what we know. &lt;a href=&quot;https://www.universeguide.com/fact/comets&quot; rel=&quot;nofollow&quot;&gt;https://www.universeguide.com/fact/comets&lt;/a&gt;  &lt;a href=&quot;https://www.universeguide.com/fact/asteroids&quot; rel=&quot;nofollow&quot;&gt;https://www.universeguide.com/fact/asteroids&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2587" LastEditorUserId="2587" LastEditDate="2014-11-05T18:40:23.133" LastActivityDate="2014-11-05T18:40:23.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6648" PostTypeId="2" ParentId="6643" CreationDate="2014-10-11T14:07:48.613" Score="2" Body="&lt;p&gt;As long as we are in the universe, we'll be unable to determine if the universe is rotating or not. &#xA;To decide whether an object is rotating, we need some other object for reference. Since universe is a term which includes everything we know in space, we don't have a reference to watch the universe relative to, to answer this question.&lt;/p&gt;&#xA;" OwnerUserId="2589" LastActivityDate="2014-10-11T14:07:48.613" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6649" PostTypeId="1" CreationDate="2014-10-11T14:30:00.660" Score="4" ViewCount="776" Body="&lt;p&gt;Which scenario will be worse for the ultimate fate of the Universe, forever expanding or the &quot;Big Crunch&quot;? Will the Universe be destroyed in both scenarios? At the present time does it appear that it will expand forever or will the expansion stop sometime in the future?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastEditorUserId="1891" LastEditDate="2019-12-19T15:28:27.603" LastActivityDate="2019-12-19T15:28:27.603" Title="Fate of the Universe" Tags="&lt;universe&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6650" PostTypeId="2" ParentId="6649" CreationDate="2014-10-11T15:22:27.883" Score="1" Body="&lt;p&gt;I'll be brief in my answer but I suggest you look at the wikipedia page on dark energy for a good starting point for further research.&#xA;(&lt;a href=&quot;http://en.wikipedia.org/wiki/Dark_energy&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Dark_energy&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the present time it is believed that the universe will continue to expand at an ever increasing rate due to the driving force of &quot;dark energy&quot; (which is as mysterious as it sounds!). Since space itself is expanding there is no limit to the rate of change in the distance between them, and it is quite possible this speed exceed that of light - meaning observers at each point will not be able to see each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind, eventually (in many billions of years) the universe will be expanding so fast that we will not be able to see any other galaxies, and eventually any other stars after much more time has passed. If the acceleration continues at the increasing rate Earth and the Sun will be &quot;pulled&quot; apart, and at this point the situation on Earth is pretty bleak - no Sun, no stars and no external light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the extremely far future it is possible the acceleration of space will cause Earth itself to be torn apart, and the eventual fate of the universe is for every molecule, atom and particle to be broken apart until all that remains are stray photons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that's a pretty bad scenario as far as the ultimate fate of the universe goes, but at the moment it seems like that is what may happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very important to note that extending the increasing rate of acceleration into the future that far is not good practice and probably won't provide very accurate predictions.&lt;/p&gt;&#xA;" OwnerUserId="1263" LastActivityDate="2014-10-11T15:22:27.883" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6651" PostTypeId="2" ParentId="6643" CreationDate="2014-10-11T15:32:17.807" Score="5" Body="&lt;p&gt;From what we observe it is extremely unlikely the universe is rotating, but it is a good question nonetheless. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you thought that since we see galaxies moving away from us they could be being pulled outwards as a result of the rotation. All the galaxies we observe (outside the local group) are accelerating away from us, and this includes galaxies in all parts of the universe. If the universe were like a giant spinning cylinder (or whatever object you like) then the galaxies &quot;above&quot; and &quot;below&quot; us in the cylinder would appear stationary since they are rotating at the same radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This observation does not support the claim that the universe is spinning on some invisible axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As always we should remember that our local universe may not represent the entire cosmos, and we would in fact be spinning around some axis. As more observational evidence is gathered we understand more and more about the universe, and one would be foolish to rule out a theory based on current observational evidence.&lt;/p&gt;&#xA;" OwnerUserId="1263" LastActivityDate="2014-10-11T15:32:17.807" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6652" PostTypeId="2" ParentId="6649" CreationDate="2014-10-11T15:37:44.623" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which scenario will be worse for the ultimate fate of the universe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, that's a wee bit opinionated. It depends on your point of view - that is, how you take a liking to the following two scenarios. Everyone has their own opinion; from the details I give below, I'll let you answer that yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To infinity - and . . . well, more infinity: The perpetually expanding universe&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are actually three types of universes that could fit this model: The accelerating expansion one, the constant expansion one, and the decelerating expansion one. At this point in time, we appear to be in a universe undergoing accelerated expansion, thanks to &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_energy&quot; rel=&quot;nofollow&quot;&gt;dark energy&lt;/a&gt;. However, all three scenarios give us a similar end result: A dead universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the far future, as long as the universe does not curl back into a Big Crunch, slowly all processes as we know them will begin to end. Star formation will eventually cease, and black holes will slowly start to gobble up a lot of matter. In the even farther future, black holes are evaporate to decay via &lt;a href=&quot;https://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow&quot;&gt;Hawking radiation&lt;/a&gt;, leaving the universe a collection of various subatomic particles. (If &lt;a href=&quot;https://en.wikipedia.org/wiki/Proton_decay&quot; rel=&quot;nofollow&quot;&gt;protons decay&lt;/a&gt;, then perhaps there will no longer be &lt;em&gt;any&lt;/em&gt; atoms). The temperature of the universe will drop until everything is about thermodynamically equivalent. Not a great place to be (well, there won't be any life, so nobody will be around to see it, but still. . .).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond this. . . well, there are a few different ideas regarding what could happen. The Big Freeze idea states that the universe will eventually just cool off to this low-temperature state. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Rip&quot; rel=&quot;nofollow&quot;&gt;Big Rip&lt;/a&gt; idea says that matter itself could be torn apart bye extreme expansion of space. Cosmological &lt;a href=&quot;https://en.wikipedia.org/wiki/False_vacuum&quot; rel=&quot;nofollow&quot;&gt;false vacuum&lt;/a&gt; theories suggest that quantum tunneling effects could make some interesting expansion possible. However, I emphasize that we do not know whether any of these long-term theories are correct. They are too far in the future, and there is too much uncertainty among the relevant factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the short term, however, it appears our universe will &lt;a href=&quot;https://en.wikipedia.org/wiki/Future_of_an_expanding_universe&quot; rel=&quot;nofollow&quot;&gt;expand forever&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Back to the beginning: The Big Crunch universe and all its friends&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Crunch&quot; rel=&quot;nofollow&quot;&gt;Big Crunch&lt;/a&gt; scenario says that if the density of the universe is greater than the &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann_equations#Density_parameter&quot; rel=&quot;nofollow&quot;&gt;critical density&lt;/a&gt;, the universe will fall back onto itself into a singularity - just like the &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Bang&quot; rel=&quot;nofollow&quot;&gt;Big Bang&lt;/a&gt;. This leads to other theories, such as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cyclic_model&quot; rel=&quot;nofollow&quot;&gt;epykrotic universe&lt;/a&gt;, or a &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Bounce&quot; rel=&quot;nofollow&quot;&gt;Big Bounce&lt;/a&gt;, which postulate that our current universe is just undergoing one of infinitely many stages of expansion and collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't exactly a great outcome, either, but again, I'll leave you to judge.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-10-11T19:49:09.387" LastActivityDate="2014-10-11T19:49:09.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6653" PostTypeId="1" AcceptedAnswerId="8057" CreationDate="2014-10-11T16:17:21.157" Score="10" ViewCount="270" Body="&lt;p&gt;There have been several accounts in media articles of asteroid near misses, many describe that the asteroid was not seen before the near miss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading the question and answer to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/702/how-can-we-tell-if-the-asteroid-2013-tv135-will-hit-earth-in-2032&quot;&gt;How can we tell if the asteroid 2013 TV135 will hit Earth in 2032?&lt;/a&gt;, where it was discussed about determining the orbits of known asteroids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question here is what is the furthest practical distance that a previously undetected asteroid, capable of an extinction level event, can be observed to have a trajectory that will result in a collision course with Earth?&lt;/p&gt;&#xA;" OwnerDisplayName="user2449" LastEditorUserId="-1" LastEditorDisplayName="user2449" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2014-11-25T21:33:03.330" Title="Detecting Deep Impact" Tags="&lt;asteroids&gt;&lt;impact&gt;&lt;impact-probability&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6654" PostTypeId="2" ParentId="6641" CreationDate="2014-10-11T18:03:21.170" Score="1" Body="&lt;p&gt;The cause of this phenomenon is the tilt of the Earth's axis. As the Earth revolves around the Sun, the tilt changes, the seasons progress, and more or less sunlight hits a given region of Earth, depending on where it is and which way the axis is tilted. This tilt means that the Sun appears to be at a different angle in the sky, which in turn means that in winter, the Sun appears lower in the sky. In summer, the Sun appears higher in the sky. So as September progresses, the Sun should start getting lower and lower. This graphic sums it up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/1/16/Solar_altitude.svg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check &lt;a href=&quot;https://en.wikipedia.org/wiki/Sun_path&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for more information (which I wish I had found sooner!).&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-10-11T23:07:45.997" LastActivityDate="2014-10-11T23:07:45.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6655" PostTypeId="1" CreationDate="2014-10-11T19:45:38.837" Score="4" ViewCount="1176" Body="&lt;p&gt;I've just started studiyng Astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understood how I would measure Right Ascention and Declination: for the first one I'd trace an Hour cirlce to the Celestial Equator, and then, compute the angle from this point and the Vernal Equinox; while the second one is the angle subtended by the object and the intersection of the Hour Circle with the Celestial Equator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, during the night, stars will change their Right Ascension value? Does this means this coordinate system is not fixed with respect to the stars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, why people store the RA/Dec to find objects?&#xA;For instance, I know  the RA/Dec of the NGC 5585 Galaxy ( &lt;a href=&quot;http://en.wikipedia.org/wiki/NGC_5585&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/NGC_5585&lt;/a&gt; ), but how can I use this information, if the RA changes during the night? &lt;/p&gt;&#xA;" OwnerUserId="2595" LastActivityDate="2014-10-13T15:55:19.487" Title="Why objects are uniquely defined by their right ascention and declination?" Tags="&lt;coordinate&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7655" PostTypeId="2" ParentId="6604" CreationDate="2014-10-11T20:20:31.423" Score="3" Body="&lt;p&gt;Sadly, a well referenced answer to this question likely lies behind academic publisher paywalls.&#xA;However, &lt;a href=&quot;http://en.wikipedia.org/wiki/Asteroid_belt#Collisions&quot; rel=&quot;nofollow&quot;&gt;Wikipedia claims&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Collisions between main-belt bodies with a mean radius of 10 km are expected to occur about once every 10 million years.[63]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;-Source for citation 63 has disappeared from the internet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.boulder.swri.edu/recent/iras_dust.pdf&quot; rel=&quot;nofollow&quot;&gt;However, the Veritas asteroid cluster, and the Karin cluster&lt;/a&gt; reportedly formed by the collisional disruption of 140 km diameter and 25 km diameter diameter objects only 8.3 and 5.8 million years ago. These numbers obviously cast doubt on the Wikipedia claim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This piece &lt;a href=&quot;http://articles.adsabs.harvard.edu//full/2002MmSAI..73..743G/0000746.000.html&quot; rel=&quot;nofollow&quot;&gt;Small asteroid collisional characteristics&lt;/a&gt; from 2002 states: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Orbits of nearly 88,000 astreroids from the main belt are needed to find statistically one encounter within 1000 km distance during 50 years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Naturally the more abundant smaller asteroids collide more frequently. In fact Hubble seems to have &lt;a href=&quot;http://www.nasa.gov/mission_pages/hubble/news/asteroid-collision.html&quot; rel=&quot;nofollow&quot;&gt;caught the aftermath of such an event back in 2010&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the question. It seems Ceres will remain intact for some millions of years to come. But it &lt;em&gt;does&lt;/em&gt; get hit fairly regularly, &lt;a href=&quot;http://www.space.com/24789-moon-meteorite-impact-brightest-lunar-explosion.html&quot; rel=&quot;nofollow&quot;&gt;like the moon&lt;/a&gt;. If Ceres regularly took impacts large enough to preclude habitation, we'd expect to see co-orbiting debris clouds like the Veritas cluster, or in the 2010 Hubble images. We do not see such debris clouds near Ceres.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2014-10-11T20:20:31.423" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7656" PostTypeId="2" ParentId="6655" CreationDate="2014-10-11T23:12:09.750" Score="4" Body="&lt;p&gt;Stars do not change their right ascension or declination values as the &#xA;Earth turns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position of the stars in the night sky (called azimuth and &#xA;elevation) &lt;em&gt;will&lt;/em&gt; change, but that's because the lines of right &#xA;ascension also move. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, right ascension and declination do not uniquely define an &#xA;object. Over time, stars move with respect to the sun (called &quot;proper &#xA;motion&quot;), and the direction of the Earth's north pole (currently &#xA;pointing near Polaris, the North Star) also changes (called &#xA;&quot;precession&quot;). &lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2014-10-11T23:12:09.750" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7657" PostTypeId="1" AcceptedAnswerId="7660" CreationDate="2014-10-12T09:30:13.860" Score="4" ViewCount="156" Body="&lt;p&gt;This question will likely be answered &quot;in retrospect&quot; in about 8 days.  So it's a great opportunity to &quot;lay it out there&quot; and then let reality prove you right/wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the 19th of October, Comet Siding is expected to pass Mars very closely (1/3 the distance of the moon from Earth). My assumption is that this means its nucleus.  But comet tails can be very long (100s of millions of miles).  For the sake of this questions let's keep it to &quot;obviously observable&quot; tail and coma.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally &quot;going through&quot; may need a definition.  So let's say that if a line drawn from the outer (obviously observable) coma closest to Mars and the end of the (obviously observable) tail were to intersect with-in 20 miles of the surface of Mars, then we would say that Mars has &quot;gone through the tail&quot; of the comet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is Mars expected (or even more fun, do YOU expect Mars) to go through the tail of Comet Siding on the 19th of October?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, what will be observable from the surface of Mars? Large number of shooting stars?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;NASA is preparing to hide its satellites (repositioning them to the &quot;far side&quot; of Mars)  from comet dust traveling at 35 miles/sec.    Is there any idea of how long the encounter will last? &lt;/p&gt;&#xA;" OwnerUserId="1826" LastEditorUserId="1826" LastEditDate="2014-10-13T15:21:47.223" LastActivityDate="2014-10-22T17:03:40.697" Title="Is Mars expected to go through the tail of Comet Siding?" Tags="&lt;mars&gt;&lt;comets&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7658" PostTypeId="2" ParentId="6641" CreationDate="2014-10-12T09:54:32.800" Score="13" Body="&lt;p&gt;Over the year, the sun follows a path known as an &lt;em&gt;annalemma&lt;/em&gt;. This is a figure eight shape and is caused by the combination of the Earth's rotation around the Sun and its tilt relative to the plane of the equator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analemma can be observed by pointing a camera at the sun, and at the same time each month, take a picture and over lay the images:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/8/8c/Analemma_fishburn.tif/lossy-page1-476px-Analemma_fishburn.tif.jpg&quot; alt=&quot;Analemma&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking a picture every day for a month (in September) would mean that you'd see it lower in the sky, curving slightly as it followed the figure of eight shape of the analemma.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Note:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;An analemma can be different on different planets. For example, a Martian analemma is different because its tilt and shape of orbit is different to that of Earth's:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://apod.nasa.gov/apod/image/0612/MarsAnalemma_Mammana.jpg&quot; alt=&quot;Martian Analemma&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mercury: Because orbital resonance makes the day exactly two years long, the method of plotting the Sun's position at the same time each day would yield only a single point. However, the equation of time can still be calculated for any time of the year, so an analemma can be graphed with this information. The resulting curve is a nearly straight east-west line.&lt;/li&gt;&#xA;&lt;li&gt;Venus: There are slightly less than two days per year, so it would take several years to accumulate a complete analemma by the usual method. The resulting curve is an ellipse.&lt;/li&gt;&#xA;&lt;li&gt;Mars: teardrop&lt;/li&gt;&#xA;&lt;li&gt;Jupiter: ellipse&lt;/li&gt;&#xA;&lt;li&gt;Saturn: technically a figure 8, but the northern loop is so small that it more closely resembles a teardrop&lt;/li&gt;&#xA;&lt;li&gt;Uranus: figure 8 (Uranus is tilted past &quot;sideways&quot; to an angle of 98 degrees. Its orbit is about as elliptical as Jupiter's and more elliptical than Earth's. )&lt;/li&gt;&#xA;&lt;li&gt;Neptune: figure 8&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For more, read &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Analemma&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1494" LastEditorUserId="1494" LastEditDate="2017-06-28T09:36:42.220" LastActivityDate="2017-06-28T09:36:42.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7659" PostTypeId="1" CreationDate="2014-10-12T11:51:04.113" Score="1" ViewCount="213" Body="&lt;p&gt;I came across this sentence in a magazine:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We sit outside the Lorne Community Hall, Victoria, watching the MIA satellite overhead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wonder if the author is talking about an artificial satellite. Can one recognize an artificial satellite with naked eyes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for response.&lt;/p&gt;&#xA;" OwnerUserId="2598" LastEditorUserId="2153" LastEditDate="2014-10-12T12:34:35.377" LastActivityDate="2014-10-15T06:04:17.907" Title="what is a MIA satellite" Tags="&lt;satellite&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="7660" PostTypeId="2" ParentId="7657" CreationDate="2014-10-12T15:00:48.277" Score="4" Body="&lt;p&gt;&lt;strong&gt;Short answer: Sort of&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Long answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One week before the encounter, &lt;a href=&quot;https://duckduckgo.com/l/?kh=-1&amp;amp;uddg=https%3A%2F%2Fen.wikipedia.org%2Fwiki%2FComet_Siding_Spring&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; says no. The comet's nucleus will pass by Mars at a distance somewhere of about 139,000 km from the center of Mars - way too far for any predicted collision. The main tail of the comet, too, will most likely miss Mars by about &quot;10 Mars diameters&quot; - roughly 64,000 km. However, small quantities of dust particles could (and most likely will) stray away from the main tail and hit Mars, Deimos and Phobos. This graphic gives a good picture of the encounter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/e/e5/Comet_Siding_Spring_Close_Approach.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, Mars &lt;em&gt;may&lt;/em&gt; pass through the comet's coma and be hit by any dust particles there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will certainly be observable from the Martian surface: The comet will have an apparent magnitude of -6 at its peak. It will be quite a show, but nothing extraordinarily special.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The damage to orbiting spacecraft around Mars will likely be minimal. 90 minutes after the comet passes Mars, the worst of the dust particles will hit the spacecraft. The tiny barrage will only last about 20 minutes, though, and there is a low probability of damage. In fact, &lt;a href=&quot;http://mars.nasa.gov/news/whatsnew/index.cfm?FuseAction=ShowNews&amp;amp;NewsID=1727&quot; rel=&quot;nofollow&quot;&gt;the various orbiters&lt;/a&gt; will take pictures of the comet and try to analyze it to learn more about its properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other sources:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mars.nasa.gov/comets/sidingspring/&quot; rel=&quot;nofollow&quot;&gt;http://mars.nasa.gov/comets/sidingspring/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Mars&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Mars&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As of about an hour after the encounter, Mars still exists. Read &lt;a href=&quot;http://news.yahoo.com/robotic-eyes-comet-mars-flyby-182243457.html&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; for some not-so-juicy details.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-10-19T19:49:17.447" LastActivityDate="2014-10-19T19:49:17.447" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7662" PostTypeId="2" ParentId="6511" CreationDate="2014-10-12T19:21:50.340" Score="2" Body="&lt;p&gt;In addition to Count Iblis wonderful answer, consider that supernovae in other galaxies &lt;strong&gt;can sometimes be seen with low-powered binoculars&lt;/strong&gt;. I seem to remember mention of a naked-eye extra-galactic supernova, but I cannot find the reference.&lt;/p&gt;&#xA;" OwnerUserId="209" LastActivityDate="2014-10-12T19:21:50.340" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7663" PostTypeId="1" CreationDate="2014-10-13T06:25:26.470" Score="10" ViewCount="2100" Body="&lt;p&gt;I understand that once a star starts fusing iron, it's doomed to collapse because iron fusion requires more energy than it releases in the process, allowing the opposing gravity of the star to cause it to collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why?  What makes iron special in this way?  It seems to occupy a pretty inconsequential spot in the middle of the period table among the transition metals.  So, why is it that iron breaks the rule in play for all the elements before it when it comes to fusion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I noticed that on another question someone said that it wasn't iron, but nickel that was the first element that required more energy to fuse than it released- but every documentary and book I've read claims that it's iron.  So, if you're answer is &quot;iron isn't the first element that requires more energy...&quot; please explain why every other source I've ever heard is wrong!&lt;/p&gt;&#xA;" OwnerUserId="2568" LastEditorUserId="1678" LastEditDate="2014-10-13T07:22:15.163" LastActivityDate="2020-09-12T08:57:11.223" Title="Why does iron consume more energy in the fusion process than it produces?" Tags="&lt;star&gt;&lt;supernova&gt;&lt;nucleosynthesis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7664" PostTypeId="2" ParentId="7663" CreationDate="2014-10-13T11:31:36.210" Score="6" Body="&lt;p&gt;It would be good if you referenced your sources, because you may be misunderstanding them. We'd be able to see what they actually say, and help you understand them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nucleosynthesis of iron does not use more energy than it produces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is, however often referred to as the heaviest element created in fusion that results in more energy produced than consumed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, that isn't quite true. Heavier elements can be produced by fusion that produce more energy than is used, except these fusion reactions don't occur in stars. (Eg 40Ca + 40Ca)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it is possible for heavier nuclei to be fused in stars that result in more energy being produced than is used, but these are unstable isotopes and they decay quickly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, more accurately, iron is the heaviest element produced in stellar nucleosynthesis in any significant quantity that produces more energy in fusion than the fusion consumes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is called the alpha process ladder. Keep adding alpha particles to the newly generated nuclei, until you stop getting more energy out than you put in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/astro/nucsyn.html#c1&quot; rel=&quot;noreferrer&quot;&gt;last step in the alpha process that does produce energy is 52Fe + 4He =&gt; 56Ni&lt;/a&gt;(excuse the rubbish notation; if this answer is considered helpful at all I will try to tidy up the notation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;56Ni + 4He =&gt; 60Zn uses more energy than it produces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;56Ni has a very short half-life of just 6 days, decaying to 56Co which has a half life of 77 days, which decays to 56Fe, which is stable. So when a star is on the limit of collapsing, it will be producing a lot of iron in it's final stages - some of it from decaying heavier radioactive isotopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the alpha process stop producing energy at this point? It is because that is where the &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin2.html#c1&quot; rel=&quot;noreferrer&quot;&gt;peak&lt;/a&gt; &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin.html#c1&quot; rel=&quot;noreferrer&quot;&gt;binding energy&lt;/a&gt; is. &lt;a href=&quot;http://www.physics.smu.edu/scalise/quarknet2008/FewellAJP000653.pdf&quot; rel=&quot;noreferrer&quot;&gt;More.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2014-10-13T11:31:36.210" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7665" PostTypeId="1" CreationDate="2014-10-13T13:51:18.127" Score="3" ViewCount="147" Body="&lt;p&gt;Is it possible that a binary star doesn't exceed its Roche lobe and so when the second star becomes a red giant, instead of accreting some of its plasma onto the white dwarf that it becomes a white dwarf itself which causes there to be no supernova?&lt;/p&gt;&#xA;" OwnerUserId="1138" LastActivityDate="2014-10-13T15:04:53.473" Title="Binary star not exceeding Roche Lobe?" Tags="&lt;binary-star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7666" PostTypeId="1" AcceptedAnswerId="7676" CreationDate="2014-10-13T13:51:58.620" Score="3" ViewCount="2831" Body="&lt;p&gt;What would happen if gravity completely stopped for 1 second?&lt;/p&gt;&#xA;" OwnerUserId="2557" LastEditorUserId="-1" LastEditDate="2018-01-29T19:48:39.227" LastActivityDate="2018-01-29T19:48:39.227" Title="What happens if gravity shuts down for only one second?" Tags="&lt;gravity&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" ClosedDate="2018-01-31T15:50:08.810" ContentLicense="CC BY-SA 3.0" />
  <row Id="7667" PostTypeId="1" CreationDate="2014-10-13T14:05:43.717" Score="1" ViewCount="296" Body="&lt;p&gt;A mercury day if you consider its solar day is 2 mercury years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you consider just 1 half of mercury, in 1 of those years it is facing towards the sun and in another one of those years it is facing away from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury is tidally locked. This tidal locking happens when the rotation rate slows down until it reaches 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is it possible for a planet to not rotate and yet have 1 half of it facing towards the sun each year with that half changing every year? It would seem like it would have to rotate 180 degrees in a fraction of a second which I know is not possible. &lt;/p&gt;&#xA;" OwnerUserId="1138" LastActivityDate="2014-10-15T07:49:48.957" Title="How can a planet not rotate at all and yet have 1 year that is day and 1 year that is night?" Tags="&lt;planet&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7668" PostTypeId="2" ParentId="2427" CreationDate="2014-10-13T14:16:54.153" Score="0" Body="&lt;p&gt;There is a much easier way to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Look up the length of the solar year in earth days&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) multiply the length of the years like this: Mercury year * Venus year * Earth year * Martian year * Jovian year * Saturn year * Uranus year * Neptune year&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Divide by 365 to get earth years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you have a time when they will align again longitudinally(meaning the angles will be different but from a top view they would form a line). It won't align at any higher of a frequency because some of these planets have a decimal number of earth days in their year.&lt;/p&gt;&#xA;" OwnerUserId="1138" LastActivityDate="2014-10-13T14:16:54.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7669" PostTypeId="1" CreationDate="2014-10-13T14:34:52.347" Score="2" ViewCount="290" Body="&lt;p&gt;In a star, the fact that not all the hydrogen is used up, not all the helium is used up etc. is a little counterintuitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shouldn't it be that a star looses the hydrogen layer passively, in other words all the hydrogen eventually fuses into helium forming a helium star which then fuses all its helium into carbon forming a carbon star etc. all the way to iron? In this way you get up to the temperature required to form iron and nickel in stars with a much lower mass and in massive stars you might even have Iron being fused to form heavier metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why isn't this so?&lt;/p&gt;&#xA;" OwnerUserId="1138" LastActivityDate="2014-10-14T02:24:12.377" Title="length of Star lifetime" Tags="&lt;star&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7670" PostTypeId="2" ParentId="7665" CreationDate="2014-10-13T14:59:27.970" Score="2" Body="&lt;p&gt;Of course this is possible. Whether it will fill its Roche lobe depends on the masses and radii of the stars and the distance between the stars. If the distance is big enough then the second (less massive) star will not fill its Roche lobe and continue evolving as a single star would. The (large) majority of binaries will fall in this category, only close binaries will evolve into cataclysmic variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance the Alpha Centauri/Proxima Centauri system will not evolve into a system where mass transfer takes place via the Roche lobes as the distance between Alpha Cen A/B and Proxima Centauri is much too big. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even when the stars are very close, a supernova might not occur if the masses of the stars are low. The supernova occurs when the mass of the existing white dwarf increases (because of the accretion) to a mass surpassing the Chandrasekhar limit (1.4 solar masses). If the transferred mass is too low then no supernova will happen.&lt;/p&gt;&#xA;" OwnerUserId="864" LastEditorUserId="864" LastEditDate="2014-10-13T15:04:53.473" LastActivityDate="2014-10-13T15:04:53.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7671" PostTypeId="1" CreationDate="2014-10-13T15:22:26.703" Score="5" ViewCount="203" Body="&lt;p&gt;Copernicus put the Sun in the center of the Solar system. This turns the Moon's orbit into an epicycle around Earth orbit. And the Jovian moons (later discovered) have orbits which are epicycles around Jupiter. Since Copernicus created a physical epicycle in the Moon, it maybe wasn't so strange to think about &quot;epicycles of epicycles&quot; as he did to improve the model's fit to calendar data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, Ptolemy and Brahe did not put Earth in the center of their Solar system models. They used a fixed point in empty space near the Earth which they called &quot;mean Earth&quot;. And Copernicus used a corresponding &quot;mean Sun&quot;. And as a matter of fact, planets do turn around a barycenter which is not in the center of the Sun, but nearby.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Do epicycles and &quot;mean Earth/Sun&quot; actually have physical correspondents in the orbits of moons and the barycenter respectively?&lt;/strong&gt; That'd be a bit funny since they are purely geometric constructions with the purpose to make data fit better, created even without a thought about gravity or anything physical at all.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2014-10-15T01:04:37.353" Title="Are epicycles and mean Earth/Sun actually useful astrophysical concepts?" Tags="&lt;johannes-kepler&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7672" PostTypeId="2" ParentId="7666" CreationDate="2014-10-13T15:30:17.973" Score="2" Body="&lt;p&gt;Nothing much. If it is only for one second, then depending on where you are on Earth you would start floating for one second going up somewhat (but not very much).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are on the equator you will have a velocity (parallel to the Earth's rotation) of about &lt;a href=&quot;http://en.wikipedia.org/wiki/Earth&quot; rel=&quot;nofollow&quot;&gt;465m/s&lt;/a&gt;. In that second the Earth will have rotated 0.004°. This means that you will have increased your elevation above the surface by $$465 \times \sin 0.004 = 0.034\textrm{m}$$&#xA;i.e. about 3cm. Some objects will probably break when the gravity is back on and they fall back to the surface. Some people will probably get sick because of the sudden weightlessness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbits of the planets will change a tiny bit and our theories in celestial mechanics may need some minor changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For larger structures like stars, and galaxies, the perturbations will not be noticeable as 1 second is much too short a time scale.&lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2014-10-13T15:30:17.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7673" PostTypeId="1" CreationDate="2014-10-13T15:41:47.750" Score="5" ViewCount="742" Body="&lt;p&gt;Well surely the eye has different sensitivity to various wavelengths... will the pogson's equation  not be affected if say I compare Betelgeuse and Rigel( both having a marked difference in spectral lines)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also, how can we ever draw a standard in stellarnmagnitude observed if both location and individual visual acuity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;does atmosphere also have any effect??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please refer some good resources for further study in this topic.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2014-10-13T15:41:47.750" Title="A question about stellar magnitude and the pogson's equation" Tags="&lt;magnitude&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7674" PostTypeId="2" ParentId="7669" CreationDate="2014-10-13T15:42:45.870" Score="1" Body="&lt;p&gt;The temperature in the core of the star may be sufficient to fuse hydrogen into helium but the temperature near the surface will not be high enough. During the main sequence the stellar core will be transformed into a helium core but the hydrogen in the outer layers will not have undergone fusion and will remain hydrogen. &lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2014-10-13T15:42:45.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7675" PostTypeId="2" ParentId="6655" CreationDate="2014-10-13T15:55:19.487" Score="0" Body="&lt;p&gt;The coordinate system you are using is not actually the equatorial coordinate system that is expressed in right ascension and declination but the coordinate system that uses Hour-angle and declination. This system is not fixed with respect to the stars (or the vernal equinox).&#xA;The hour-angle is the angle between the intersection of the celestial equator and your local meridian (the line from south to the zenith). It can be calculated by:&#xA;$$H = \Theta - \alpha$$&#xA;where $\Theta$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sidereal_time&quot; rel=&quot;nofollow&quot;&gt;local sidereal time&lt;/a&gt; and $\alpha$ is the right ascension. The declination is the same in both systems.&lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2014-10-13T15:55:19.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7676" PostTypeId="2" ParentId="7666" CreationDate="2014-10-13T17:51:42.123" Score="7" Body="&lt;p&gt;I expect stars would belch mightily, neutron stars and black holes would explode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars are only held at their steady state size by the opposed forces of gravity and radiation, so eliminating gravity for 1 second &lt;em&gt;will&lt;/em&gt; cause expansion, followed by collapse as gravity kicks in again.&#xA;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Surface_gravity&quot;&gt;Neutron star's compactness gives it a surface gravity of up to 7×10^12 m/s²&lt;/a&gt;, and expanding from 20km to ~70 billion km (or 300000km when c limit is considered) over the course of 1 second constitutes an explosion.&#xA;Black holes tend to have higher gravity values at the event horizon (same ref) so the result would be even more cataclysmic.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="2031" LastEditDate="2014-10-13T18:58:14.000" LastActivityDate="2014-10-13T18:58:14.000" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="7677" PostTypeId="2" ParentId="7659" CreationDate="2014-10-13T19:03:34.283" Score="2" Body="&lt;p&gt;As @usethedeathstar said, MIA satellite most likely means &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Missing_in_action&quot; rel=&quot;nofollow&quot;&gt;'Missing In Action'&lt;/a&gt;. The following lyrics from the song 'MIA satellite' seems to solidify this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;MIA Satellite-&lt;br&gt;&#xA;  You've got no path in sight.&lt;br&gt;&#xA;  You leave me hanging in the dead of night-&lt;br&gt;&#xA;  In the dead of night.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Regarding the ability to identify artificial satellites, it is possible to identify some satellites. Some identification guides:&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Iridium&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Iridium satellites are usually faster than most satellites, but their tell-tale sign is that they go from dim to bright to dim as they rotate. This bright flash is known as an Iridium Flare.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;ISS&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This moves East to West and is brighter than the average satellite.&lt;/p&gt;&#xA;" OwnerUserId="1494" LastEditorUserId="1494" LastEditDate="2014-10-15T06:04:17.907" LastActivityDate="2014-10-15T06:04:17.907" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7678" PostTypeId="2" ParentId="7671" CreationDate="2014-10-13T21:53:15.893" Score="1" Body="&lt;p&gt;None of these things are epicycles. Epicycles are perfect circles. These orbits are ellipses: it's a fundamental difference and so the answer is a very firm 'no'.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2014-10-13T21:53:15.893" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7679" PostTypeId="1" CreationDate="2014-10-13T22:44:08.657" Score="6" ViewCount="240" Body="&lt;p&gt;Since life evolved here on Earth does that mean that life had to have existed before or in the future elsewhere in the Universe or could it be a one time only freak accident ?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastActivityDate="2017-05-02T16:37:36.787" Title="Evolution of Life" Tags="&lt;life&gt;" AnswerCount="6" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7680" PostTypeId="2" ParentId="7679" CreationDate="2014-10-13T23:00:29.843" Score="3" Body="&lt;p&gt;It almost surely did, is, and will exist elsewere in the Universe. Despite being an improbable event for a single planet, there are something like $10^{18}$ planets in the observable universe.&#xA;We already have found traces of proteins typical of life on Mars and comets coming from outer space. Probably Venus could host life a few billion years ago, when it was much warmer than today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see more here, about the probability of the existence of other (intelligent) creatures in our Galaxy (notice that we're not even talking about the whole Universe!):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Drake_equation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Drake_equation&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2553" LastActivityDate="2014-10-13T23:00:29.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7681" PostTypeId="2" ParentId="7669" CreationDate="2014-10-14T02:24:12.377" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why isn't this so?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the mass of the star is about 1/4 the mass of our sun, then the core's (where fusion occurs) temperature and pressure will never be enough to fuse anything other than hydrogen into helium. The death of this star may be something like a fire dying. The flames disappear for a while, then you may get one or 2 popping back into existence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our sun will start to fuse helium into carbon before all the hydrogen in the core is used. In the description of a heavy star, it is described as a onion with its layers with the higher fusion products occurring in the lower layers. They use this model since the iron fusion requires temperature and pressures above a certain level. Neon has lower requirements, but still greater than carbon fusion, and so on. What is not mentioned, is that there is still hydrogen fusion occurring in the center, as well as all the other fusion, helium, carbon, neon, etc.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-10-14T02:24:12.377" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7682" PostTypeId="1" CreationDate="2014-10-14T07:21:27.460" Score="4" ViewCount="188" Body="&lt;p&gt;I have created a 3D rendering of the moon, which is represented as a sphere with a radius of N pixels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also have selenographic coordinates.  For example, the coordinates of the Apollo 11 landing site are 0.67408, 23.47297.  I'd like to show this coordinate on my sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's some possibly relevant code, which is used for conversion with Earth coordinates:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function latLongToVector3(lat, lon, radius, height) {&#xA;    var phi = (lat)*Math.PI/180;&#xA;    var theta = (lon-180)*Math.PI/180;&#xA;&#xA;    var x = -(radius+height) * Math.cos(phi) * Math.cos(theta);&#xA;    var y = (radius+height) * Math.sin(phi);&#xA;    var z = (radius+height) * Math.cos(phi) * Math.sin(theta);&#xA;&#xA;    return new THREE.Vector3(x,y,z);&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is this the correct way to map selenographic coordinates onto a sphere?&lt;/p&gt;&#xA;" OwnerUserId="2605" LastEditorUserId="2605" LastEditDate="2019-02-07T06:41:56.837" LastActivityDate="2019-02-07T06:41:56.837" Title="Mapping selenographic coordinates onto a sphere" Tags="&lt;the-moon&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2016-10-04T17:42:42.197" ContentLicense="CC BY-SA 4.0" />
  <row Id="7683" PostTypeId="1" AcceptedAnswerId="7688" CreationDate="2014-10-14T08:18:01.393" Score="7" ViewCount="132" Body="&lt;p&gt;I notice that Wikipedia give two contradictory terms for the time period in which a planet rotates in relation to an infinitely-distant object. In one place this is called a &quot;sidereal day&quot; and in another &quot;stellar day&quot;. &lt;strong&gt;Which of these are correct?&lt;/strong&gt; For that point, is this an &lt;a href=&quot;http://www.iau.org/&quot; rel=&quot;nofollow&quot;&gt;IAU&lt;/a&gt; definition? I cannot find the definitions on the IAU website.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the &lt;a href=&quot;http://simple.wikipedia.org/wiki/Sidereal_day&quot; rel=&quot;nofollow&quot;&gt;Sidereal Day&lt;/a&gt; page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A sidereal day is the length of time it takes a planet to rotate from the perspective of a distant star.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the &lt;a href=&quot;http://en.wikipedia.org/wiki/Day&quot; rel=&quot;nofollow&quot;&gt;Day&lt;/a&gt; page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;stellar day - an entire rotation of a planet with respect to the distant stars&lt;/li&gt;&#xA;  &lt;li&gt;sidereal day - a single rotation of a planet with respect to the vernal equinox&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="209" LastEditorUserId="209" LastEditDate="2014-10-14T21:33:36.773" LastActivityDate="2014-10-14T21:33:36.773" Title="Time period in which a planet rotates" Tags="&lt;terminology&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7684" PostTypeId="2" ParentId="7666" CreationDate="2014-10-14T08:29:49.807" Score="2" Body="&lt;p&gt;Immediate expansion. I would like to break it down into a steps starting with smaller magnitude and growing larger as we go.&#xA;However, a few assumptions must be made for this to make sense (since your question does break the laws of physics):  Since gravity is mass-dependent in the equation F=GMm/r^(2) we are not assuming that everything suddenly has no mass (and therefore no energy) as the origin of the lack of gravity. We are only putting at rest the force between the two masses that would have originally interacted. You can think of it as neutralizing two opposing charged objects so that they neither attract nor repel, however they are still present as objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As humans on the surface of the Earth we stand approximately 6371 kilometers or 3958 miles above the center of the Earth. Even as you're standing still reading these words from your computer or electronic device you are rotating around the Earth at about 1000 kph or 620 mph since we travel a circumference of 2*Pi*radius units of distance per 24 hours. In addition to flying off of Earth faster than sound (with respect to a stationary point - our Sun) the Earth itself would be flung off of the orbit of the sun just as we were tossed from the Earth. The sun itself would additionally be tossed from the orbit around the milky way. The calculated speed of the Earth around the sun (using the same method as before) is 108 000 kph or 67 000 mph.  The speed of the sun around the Milky Way is 486 000 mph or 780 000 kph. If the vector components of a person or object on Earth were to be aligned in the right way with the Earth, the Earth the right way with the Sun and the Sun the right way with the Milky Way a person or object on Earth could be ejected into intergalactic space at a speed of 890 000 kph or 550 000 mph which is equivalent to 245 kilometers or 150 miles per SECOND with respect to the center of the Milky Way. If gravity didn't come back after 1 second that would be the fate of objects. However anticlimactically, since gravity returns after 1 second, the height elevation change would range from very minimal to exactly zero depending on your position on the Earth. That is just the release of kinetic energy from the universes gravitational grasp and keep in mind that this is happening to every atom of every object in the universe both larger and smaller than our Sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that gravity allows kinetic energy to be maintained while gaining potential energy is also a troublesome violation of thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars in our universe maintain a balance between the extreme outwards force of thermonuclear explosions and the immense gravity to arrive at it's given radius (our Sun's is 695 800 km and has a mass of 1.989x10^30 kg (1989 with 27 zeros after)). When gravity disappears for 1 second the outwards force balanced by the gravity would be released causing a massive explosion. In other star systems with more immense stars and natural phenomena such as pulsars and and especially black holes the explosions and expansions would be greater. The subject of a black hole is a very interesting one indeed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without going into the physics behind a black hole it is obvious that the super-massive black holes would possess the most barbaric explosions as they are so heavy &lt;em&gt;light cannot escape&lt;/em&gt;. This release of energy would be unparalleled. Even though the release would be for only one second the decrease in density of the black whole from the rapid expansion might not be able to wield in the ejected mass from the explosion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, to which object would the one second be with respect to (it sounds funny, doesn't it?)? As we draw closer to a black hole (or any heavy object) time slows down and essentially becomes infinitely slow (zero) as we approach the event horizon of a black hole. If you were to look out of a black hole from the center you would be able to see the entire life of the universe as time for you inside the black hole is zero while the time of the entire universe outside the black hole ticks away infinitely fast. Thus if the second were to be respect to inside the black hole everything for the rest of time would happen in that one second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, with respect to outside the black hole it would cause the carefully balanced orbits of planets to be slightly stretched from and to ellipses which could possibly cause Earth to get hit by another planet from our or other solar systems (keep in mind the countless asteroids present in the belt between Mars and Jupiter though for one second this is a small chance). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no definitive answer on what would truly happen. Without gravity every theory would essentially be destroyed. Every answer I have given to your question has to of had an assumption in order to give an idea of what is going on because if there truly was no gravity, if space-time was truly uniform for every cubic angstrom of our universe for one second the only possible explanation would be that there is no mass. If there is no mass then there is no energy as described by Einstein's e=mc^(2). &#xA;As we know it, nothing would exist at all. &lt;/p&gt;&#xA;" OwnerUserId="2606" LastActivityDate="2014-10-14T08:29:49.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7685" PostTypeId="2" ParentId="7666" CreationDate="2014-10-14T09:11:20.467" Score="1" Body="&lt;p&gt;Gravity is force that comes from one object pulling on another object. This causes a rotation around a center of masses. In a case of the sun pulling the earth, the center of mass is on the sun's surface and thus the earth rotates around the sun. The force pulling the earth towards the center of mass between the sun and earth is called a centripetal force or the gravitation pull of the sun. The centripetal force cause the change in direction towards the center of the mass. The rate at which the sun changes, angular velocity is constant. The tangential velocity is also constant because the tangential acceleration is zero. The only thing changing is the direction. This principle applies to all objects pulling on each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case the gravitation going off and on for a second, I would expect that the centripetal force would go to zero and then back to normal on all the objects in the universe. The consequence of this on and off would mean that all the objects in the universe would not have an angular rate and so will move in the direction and rate of the tangential velocity at the initial moment of the gravity switching off. Once the gravity return, the objects will start pulling on each other and thus will have to establish new center of masses and might restore or create new couples of rotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/N4Top.jpg&quot; alt=&quot;drawing example of gravity versus no gravity&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2607" LastActivityDate="2014-10-14T09:11:20.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7686" PostTypeId="2" ParentId="7667" CreationDate="2014-10-14T10:09:21.520" Score="1" Body="&lt;p&gt;Like any other planet day/night will not flip immediately in a seconds. It will have its sunrise , noon and then sunset . The 1st year day refers to the time between morning and evening, and the 2nd half year of night (i.e. day in the other half) refers to the time between sunset and sunrise. Something similar happens in earth's noth pole and south.&lt;/p&gt;&#xA;" OwnerUserId="2609" LastActivityDate="2014-10-14T10:09:21.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7688" PostTypeId="2" ParentId="7683" CreationDate="2014-10-14T15:40:17.010" Score="3" Body="&lt;p&gt;Since etymologically &lt;em&gt;sidereal&lt;/em&gt; references the stars (Latin &lt;em&gt;sidus&lt;/em&gt;), one might expect a sidereal day to correspond to the rotation of the Earth with respect to distant &quot;fixed&quot; stars. But this does not seem to be the case, and because the the difference is not significant, the two definitions are sometimes used interchangeably.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know of an &lt;em&gt;explicit&lt;/em&gt; IAU definition of &lt;em&gt;sidereal day&lt;/em&gt;. However, since &lt;strong&gt;sidereal time&lt;/strong&gt; is defined as the hour angle vernal equinox (which is a local definition, although Greenwich is conventional), defining the (mean) sidereal day in terms in reference to the equinox is practically the only sensible choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Effective 1985, UT1 is now computed using very long baseline interferometry of distant quasars, and so can be taken to be authoritative regarding the &quot;fixed stars&quot;. The coordinated universal time (UTC) time approximates UT1 with atomic clocks. Anyway, UT1 derived the rotational period of the Earth as $p = 86164.09890369732\,\mathrm{s}$ of UT1 time and the mean sidereal day (in 2000) would be $86164.090530833\,\mathrm{s}$ of UT1 time, following the explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The length of one sidereal day is defined by two successive transits of the mean equinox; while the Earth is rotating eastward, the mean equinox is moving westward due to precession. Therefore, one sidereal day is shorter than the Earth's rotational period by about $0.008\,\mathrm{s}$, the amount of precession in the right ascension in one day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See: Aoki, S., et al., &lt;em&gt;The new definition of universal time&lt;/em&gt;, &lt;a href=&quot;http://adsabs.harvard.edu/full/1982A%26A...105..359A&quot; rel=&quot;nofollow&quot;&gt;Astron. Astrophys. 105, 359-361 (1982)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-10-14T15:40:17.010" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7689" PostTypeId="2" ParentId="7679" CreationDate="2014-10-14T15:49:26.610" Score="8" Body="&lt;p&gt;Lets say you put a billion balloons on a dart board.  Now lets say you throw a million darts at that dart board.  If you were to pick a specific balloon and calculate the odds that THIS particular balloon would be hit by a dart, the odds would be astronomically low.  However, if you were to say that at least SOME of the balloons would be popped, the odds are astronomically high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why I get frustrated at those who claim that the number of coincidences that had to happen (and there are many) are too many.  No, we were just one of the lucky planets out of the billions that got hit by a dart.&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2014-10-14T15:49:26.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7690" PostTypeId="1" AcceptedAnswerId="7691" CreationDate="2014-10-14T15:55:05.073" Score="3" ViewCount="366" Body="&lt;p&gt;Long ago, it was theorized that light had to travel through a medium, until the particle/wave duality of light was discovered.  Then it changed to the idea that light can travel through the vacuum of space as a particle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light travels at different speeds depending on the medium that the light is traveling through.  It travels slower in water than it does through air.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, is there any possibility that dark matter is a medium that light travels through?  Could it be conceivable that what we recognize as light speed is really just the speed at which light travels through this medium?  If we were to remove dark matter, would light go faster?&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2020-07-06T15:30:47.977" Title="Could dark matter be considered a medium?" Tags="&lt;light&gt;&lt;dark-matter&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7691" PostTypeId="2" ParentId="7690" CreationDate="2014-10-14T16:27:58.333" Score="7" Body="&lt;p&gt;I don't see any reason why dark matter couldn't be considered a medium just as air or water are, but this does not mean that your conclusion is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark matter does not interact with photons, however the reason we know dark matter exists is because of the measurable effect of its gravity.  Galaxies have a tendency to clump around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light travels unimpeded at its standard rate between galaxies where we don't detect any dark matter, so interesting question though it is, I'm afraid the answer is no.&lt;/p&gt;&#xA;" OwnerUserId="2503" LastActivityDate="2014-10-14T16:27:58.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7692" PostTypeId="2" ParentId="7667" CreationDate="2014-10-14T19:01:17.840" Score="5" Body="&lt;p&gt;Your premise is incorrect. We used to think Mercury was tidally locked, but since 1965 we now know it is in a 3:2 spin-orbit resonance, which gives it (long) days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything you read that says it is tidally locked is old, or itself using old reference material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some quick sketches to show the orbit, relative orientation of the planet, and the sidereal and solar days. If you like it as an answer I'll redraw with a graphics tool instead of my iPad Pen app...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uy0R2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/C5MMH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/82p26.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="67" LastEditorUserId="67" LastEditDate="2014-10-15T07:49:48.957" LastActivityDate="2014-10-15T07:49:48.957" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7693" PostTypeId="2" ParentId="6348" CreationDate="2014-10-14T19:53:42.703" Score="10" Body="&lt;p&gt;The two terms are used in answering different questions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hill Sphere: given a large mass (eg Sun) and a small mass (eg Earth), can a tiny mass (eg Moon) find a stable orbit around the small mass? (If the tiny mass goes outside the Hill Sphere of the small mass, no.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SOI: given two large mass objects and a small object between them, (eg sending a probe from Earth to Mars), which massive object should we use as the origin of the frame of reference? (The small object is within which massive object's SOI?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the questions are different, different approximations are used to answer the questions.&lt;/p&gt;&#xA;" OwnerUserId="67" LastEditorUserId="67" LastEditDate="2014-10-15T08:50:14.533" LastActivityDate="2014-10-15T08:50:14.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7694" PostTypeId="2" ParentId="7671" CreationDate="2014-10-15T01:04:37.353" Score="2" Body="&lt;p&gt;I'll say yes here, but only if the epicycles are used as approximations. Here's how they would be useful:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;The eccentricity of the celestial body's orbit is low, i.e. $e&amp;lt;&amp;lt;0$.&lt;/strong&gt; This means that it would be nearly circular, and epicycles could therefore be reasonably accurate. For example, it would not be a good idea to use epicycles to model Mercury's motion, because it has a high &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_eccentricity&quot; rel=&quot;nofollow&quot;&gt;eccentricity&lt;/a&gt;. Comets would also be poor choices. Triton, on the other hand, has low eccentricity.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Only a vague approximation is needed.&lt;/strong&gt; If you aren't looking for calculations with a high degree of accuracy, you could use epicycles. It would be a bad idea to use epicycles to put a spacecraft into a precision orbit, but a better idea to use them in a toy simulator. NASA, however, has higher standards.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;You don't need to use general relativity.&lt;/strong&gt; One of general relativity's most important predictions was the &lt;a href=&quot;https://en.wikipedia.org/wiki/Tests_of_general_relativity#Perihelion_precession_of_Mercury&quot; rel=&quot;nofollow&quot;&gt;precession of Mercury's orbit&lt;/a&gt;. Over long periods of time, general relativity triumphs over classical theories when it comes to this accuracy. It would be hard to use epicycles to model this - although potentially possible.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-10-15T01:04:37.353" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7695" PostTypeId="2" ParentId="7679" CreationDate="2014-10-15T03:53:54.207" Score="5" Body="&lt;p&gt;A &quot;one time only freak accident&quot; is only going to be considered as a likely possibility when scientists gain a deep understanding how life actually arises and on the basis of that can rule out that life will typically arise on Earth like planets in habitable zones. We haven't arrived at that point yet. Scientists will not invoke &quot;freak accidents&quot; to explain something unless there is a lot of evidence that points to that; clearly such assumptions amount to making the assumption that there is no natural explanation, but that has to be justified by other evidence first.&lt;/p&gt;&#xA;" OwnerUserId="2185" LastActivityDate="2014-10-15T03:53:54.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7696" PostTypeId="1" AcceptedAnswerId="7697" CreationDate="2014-10-15T07:18:56.060" Score="6" ViewCount="110" Body="&lt;p&gt;We call dark matter dark matter, because of the way it acts on galaxies, but why do we look at dark energy as an energy? With $E = m c^2$, can't dark energy be mass as well? Why dont we look at dark energy as a mass, in the same way as we look at dark matter as being a missing mass (as halo around galaxies)? &lt;/p&gt;&#xA;" OwnerUserId="1062" LastActivityDate="2014-10-15T08:49:59.327" Title="why do we look at dark energy as an energy?" Tags="&lt;dark-energy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7697" PostTypeId="2" ParentId="7696" CreationDate="2014-10-15T08:49:59.327" Score="2" Body="&lt;p&gt;Since relativistic physics almost always takes units of $c = 1$, there is no difference between units of energy density and mass density. Conceptually, however, the gravitational charge is energy, not mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general relativity, a perfect fluid has no viscuous shear stresses and no heat flux, and can be described by the stress-energy tensor $T^{\mu\nu} = (\rho + p)u^\mu u^\nu + pg^{\mu\nu}$, where $\rho$ is density, $p$ is pressure, and $u^\mu$ is the fluid four-velocity. In the case of vanishing pressure, the result is called a &lt;em&gt;dust&lt;/em&gt;, and is appropriate for modeling cold dark matter, galaxies on the cosmological scale, etc. On the other hand, a &lt;em&gt;radiation fluid&lt;/em&gt; has $\rho = 3p$, and is appropriate for modeling incoherent electromagnetic radiation, or a fluid made of strongly relativistic particles (e.g., hot dark matter), and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, in the specific context of cosmology, the &lt;em&gt;matter-dominated&lt;/em&gt; regime has $p\sim 0$ and &lt;em&gt;radiation-dominated&lt;/em&gt; regime has $p\sim\rho/3$. Note that in the latter case, the fluid particles have zero (or at least negligible) mass. In-between, we can talk about a fluid of particles that are not strongly relativistic, and hence have some mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the standard model of dark energy is a cosmological constant, which is equivalent to a perfect fluid $p = -\rho$. It does not fit into the above scale; it is not matter in the above sense, nor is radiation, nor anything in-between. Furthermore, it has the peculiar property of being exactly the same in every local inertial frame. This frame-independence enables us to think of the cosmological constant as an &lt;em&gt;intrinsic&lt;/em&gt; energy density of spacetime, but it also makes it inappropriate to think of it as a collection of fluid particles that have some rest frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your purposes, this important because $E = mc^2$ is only valid in a rest frame.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-10-15T08:49:59.327" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7698" PostTypeId="2" ParentId="7679" CreationDate="2014-10-15T10:20:46.740" Score="2" Body="&lt;p&gt;Lacking both observation of ET life and a good understanding of how life originated, this question remains a great mystery. One can never completely disprove the existence of extraterrestrial life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two basic approaches to potentially answer this: one theoretical and one empirical.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If a good theoretical understanding of the origin of life, abiogenesis, is achieved in laboratories, then the frequency of it happening given different observed exoplanetary conditions could be estimated.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Empirically, of course we might be lucky enough to discover some statistics of actually observed living exoplanets.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But there's a third way too, the SETI way. The evolutionary steps to intelligence, to technology, to space travel all seem to benefit the proliferation of life and thus would tend to become more common over time, if they at all occur. Considering the small size of the Milky Way compared to its age and reasonable space traveling and settling speeds, one could put upper limits on how common life is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone anytime anywhere in the Milky Way developed space travel to its nearest stars, they (their offspring) should be everywhere today! They won't need to travel further than a few light years to &quot;seed&quot; the entire galaxy, because their Sun does the traveling through the galaxy for them. Our Sun makes a revolution every 0.25 billion years, 18 full revolutions since it was formed. We know that dinosaurs lived on the other side of the galactic center. Traveling to the nearest stars over time spreads a space traveling civilization to all over the place. If we don't find any signs of artificiality during the next few decades, given the rapid advancements in telescopy, we can say that space traveling civilizations never happened anywhere (or got too exotic for us to discover even if under our nose). And that might be extrapolated to how frequent biological evolution is overall. And most importantly, what unique event happened here to make us walk on the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Earth, life has settled every habitable spot on Earth (everywhere where liquid water is found, life is also found). Maybe life tends to settle on interstellar scales too?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2014-10-15T10:26:33.963" LastActivityDate="2014-10-15T10:26:33.963" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7700" PostTypeId="1" AcceptedAnswerId="7701" CreationDate="2014-10-15T14:31:43.103" Score="7" ViewCount="303" Body="&lt;p&gt;For a hypothetical orbital system (Sun + single planet) the Newtonian model and the General Relativity (GR) model produce different expressions for the gravitational effect of the Sun on the planet. This is well known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio between the Newtonian and GR effects is expressed in different ways by different writers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having trouble reconciling two such expressions of the Newtonian:GR ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly &lt;a href=&quot;http://books.google.co.uk/books?id=V4_s91-Vun4C&amp;amp;pg=PA482&amp;amp;lpg=PA482&amp;amp;dq=we+recall+from+section+7.6.1&amp;amp;source=bl&amp;amp;ots=RyU1Adp8h7&amp;amp;sig=qQ5_KyjoDhaSaOadgP8zyy8ubIc&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=Hlc-VJDJK9TzaoHlgKgL&amp;amp;ved=0CCEQ6AEwAA#v=onepage&amp;amp;q=we%20recall%20from%20section%207.6.1&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Walter (2008)&lt;/a&gt; (equation 12.7.6, page 482) presents  the following expression for the equation of motion produced from the GR model&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d^2\,u}{d\theta^2} + u = \frac{GM}{h^2} + \frac{3GM}{c^2}u^2$$&#xA;where $u= (1/r)$, $h=vr$, $G$ is the universal constant of gravitation, $M$ is mass of the Sun, $c$ is speed of light.  Here the term $GM/h^2$ is the ordinary Newtonian term and the term $3GMu^2/c^2$ is the additional term introduced by GR.&#xA;&lt;p&gt;From this Walter derives the approximate ratio between Newtonian and GR effects as&#xA;$(1)$ to $(1 + 3v^2/c^2)$ where $v = \sqrt{GM/r}$ is the orbital speed of the planet in a circular orbit (with distance $r$ = $a$, the semi-major axis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, an alternative presentation (referring to the so-called Schwartzchild solution) is given by Goldstein in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201657023&quot; rel=&quot;nofollow&quot;&gt;Classical Mechanics (3rd Edition)&lt;/a&gt; pages 536-538.  The GR potential $V_{GR}$ is given by&#xA;$$V = -\frac{GMm}{r} -\frac{b}{r^3}$$&#xA;where $m$ is the target body mass and $b$ is a constant (Goldstein uses $h$ instead of $b$, see below, but I have already used $h$ to mean something different in Walter above,)&#xA;&lt;p&gt; differentiating the potential with respect to distance $r$ to give force we derive&#xA;$$F_{GR} = \frac{GMm}{r^2} + \frac{3b}{r^4}$$&#xA;&lt;p&gt;Now Goldstein defines the constant $b$ thus :-&#xA;$$&#xA;b = \frac{k\,l^2}{m^2c^2} \qquad\text{ Goldstein eqtn [12.48]} &#xA;$$&#xA;where&#xA;$$&#xA;k=GMm &#xA;$$&#xA;and &#xA;$$&#xA;l^2 = mka(1-e^2) \qquad\text{ Goldstein eqtn[12.50]}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b = \frac{GMm\, m\, GMm\, a(1-e^2)}{m^2c^2}&#xA;= \frac{GMm \, L^2}{c^2}&#xA;= \frac{GMm \,m^2 v_c^2 a^2}{c^2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the GR force equation becomes&#xA;$$&#xA;F_{GR} = \frac{GMm}{r^2} + \frac{3GMm m^2 v_c^2 a^2}{r^4c^2}&#xA;$$&#xA;substituting $a$ by $r$ we get&#xA;$$&#xA;F_{GR} = \frac{GMm}{r^2} + \frac{3GMm m^2 v_c^2}{r^2c^2}&#xA;= \frac{GMm}{r^2} \left( 1 + \frac{3 v_c^2 \, m^2}{c^2} \right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the Newtonian:GR ratio derived from Goldstein is the same as the ratio derived by Walter except that the former has the additional term of $m^2$ in the numerator.&#xA;Even if we tried to fudge this numerically by invoking a unit mass target, it would still be dimensionally incorrect.&#xA;&lt;p&gt; So what is the correct ratio?&#xA;$$$$&#xA;&lt;p&gt;UPDATE ---------------------------------------------------------------------&#xA;&lt;p&gt;In the refactoring of $b$ I used angular momentum $L$ when I should have used specific angular momentum $\mathfrak{l}$. After correction the extra $m^2$ disappears. Goldstein agrees with Walter. My thanks to Stan Liou for illumination.&#xA;&lt;p&gt;Corrected analysis:-&#xA;$$b = \frac{GMm\, m\, GMm\, a(1-e^2)}{m^2c^2}&#xA;= \frac{GMm \, \mathfrak{l}^2}{c^2}&#xA;= \frac{GMm \,v_c^2 a^2}{c^2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the GR force equation becomes&#xA;$$&#xA;F_{GR} = \frac{GMm}{r^2} + \frac{3GMm v_c^2 a^2}{r^4c^2}&#xA;$$&#xA;substituting $a$ by $r$ we get&#xA;$$&#xA;F_{GR} = \frac{GMm}{r^2} + \frac{3GMm v_c^2}{r^2c^2}&#xA;= \frac{GMm}{r^2} \left( 1 + \frac{3 v_c^2}{c^2} \right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the correct ratio of Newtonian to GR gravitational force is:-&#xA;$$&#xA;F_{Newtonian}:F_{GR}&#xA;\approx  1: \left( 1 + \frac{3 v_c^2}{c^2} \right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTES&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This ratio is approximate and only applies in the &quot;low velocity, weak field&quot; sub-domain of the GR model.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Goldstein also emphasises that the GR effect is not a velocity effect (presumably as in the velocity of the target body through any kind of aether or flux).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coincidentally (in the same sub-domain e.g. Mercury orbitting the Sun) a modified Newtonian radial force of magnitude $f=GMm/r^2 * [1 + 3v_t^2/c^2]$, where $v_t$ is the instantaneous transverse velocity of a small target planet, produces non-Newtonian apsidal rotation (&quot;perihelion precession&quot;) of the same magnitude (within 1%) as GR.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Goldstein needs to be read with care. Here he uses $l$ to denote angular momentum elsewhere (e.g. eqtn [1.7]) he uses $L$. He often refers to $V$ as &quot;potential&quot; when he is clearly referring to &quot;potential energy&quot; (e.g. eqtn [3.49]).&lt;/p&gt;&#xA;" OwnerUserId="233" LastEditorUserId="233" LastEditDate="2014-10-19T21:21:03.683" LastActivityDate="2019-05-24T18:30:10.690" Title="What is the correct ratio of Newtonian to General Relativistic gravitational effects for Sun + single planet orbital system" Tags="&lt;orbit&gt;&lt;general-relativity&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7701" PostTypeId="2" ParentId="7700" CreationDate="2014-10-15T17:50:55.947" Score="6" Body="&lt;p&gt;Orbits in Schwarzschild spacetime can be described by the effective potential&#xA;$$V_\text{eff} = -\frac{GM}{r} + \frac{\mathfrak{l}^2}{2r^2} - \frac{GM\mathfrak{l}^2}{c^2r^3}\text{,}$$&#xA;where $\mathfrak{l} = r^2\dot{\phi}$ is the specific angular momentum of the orbit, which is a conserved quantity. The first two terms match the form of the Newtonian effective potential, except that here we're referring to the Schwarzschild radial coordinate $r$ and proper time of the orbiting particle, instead of radial distance and coordinate time. The first term being the usual gravitational potential and the second being the centrifugal potential, so Goldstein's $V$ does make some sense as a gravitational potential energy term instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $l^2 = mka(1-e^2)$ with $k = GMm$ means that $l$ is the angular momentum, $l = m\mathfrak{l}$, and&#xA;$$\frac{GM\mathfrak{l}^2}{c^2r^3}m = GMm\frac{l^2}{m^2}\frac{1}{c^2r^3} = \underbrace{\frac{kl^2}{m^2c^3}}_b\frac{1}{r^3}\text{,}$$&#xA;just as Goldstein says. If we differentiate $\frac{1}{2}m\dot{r}^2 + mV_\text{eff} = \mathcal{E}$ with respect to proper time, then&#xA;$$\begin{eqnarray*}m\ddot{r} - \frac{l^2}{mr^3} &amp;amp;=&amp;amp; -\frac{k}{r^2} - \frac{3b}{r^4}\\&#xA;&amp;amp;=&amp;amp; -\frac{k}{r^2}\left(1 + 3\frac{l^2}{m^2c^2}\frac{1}{r^2}\right)\\&#xA;&amp;amp;=&amp;amp; -\frac{k}{r^2}\left(1 + 3\frac{m(GMm)a(1-e^2)}{m^2c^2}\frac{1}{r^2}\right)\\&#xA;&amp;amp;=&amp;amp; -\frac{k}{r^2}\left(1 + 3\frac{v_c^2}{c^2}\frac{a(1-e^2)}{r}\right)\text{.}&#xA;\end{eqnarray*}$$&#xA;This is dimensionally correct, as both $v_c/c$ and $a/r$ are dimensionless, while&#xA;$$\frac{l^2}{mr^3} = \frac{k}{r^2}\frac{a(1-e^2)}{r}\text{.}$$&#xA;The left-hand side has the Newtonian form.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditDate="2014-10-15T21:23:12.217" LastActivityDate="2014-10-15T21:23:12.217" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="7702" PostTypeId="1" CreationDate="2014-10-15T18:06:59.810" Score="16" ViewCount="596" Body="&lt;p&gt;If one of the rules to be a planet is that it needs to clear ALL objects from their orbit, does this also make Neptune a non-planet? Since it has thus far failed to clear Pluto (and based on an answer from the physics board, the kuiper belt) from it's orbit. Or does this rule not really apply to planets and we should welcome Pluto and a few other dwarf-planets into our family of planets? &lt;/p&gt;&#xA;" OwnerUserId="2613" LastActivityDate="2014-11-26T17:57:11.647" Title="Planets and Pluto? Neptune?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;dwarf-planets&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7703" PostTypeId="2" ParentId="6641" CreationDate="2014-10-15T18:25:40.697" Score="4" Body="&lt;p&gt;A good way to visualize the apparent movement of the sun across time is using the free stellarium program.  First input your location and time you would like to view because location will determine what the apparent movement of the sun.  For example, if you live in Seattle WA and facing the west observing the position of the sun for the periods of September 1 - 30 you will notice the sun appears to move downward and to the left in the sky (see screenshots).&#xA;&lt;img src=&quot;https://i.stack.imgur.com/fgPFo.jpg&quot; alt=&quot;September 1 2014 Seattle WA&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KKfAW.jpg&quot; alt=&quot;September 30 2014 Seattle WA&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2612" LastEditorUserId="2612" LastEditDate="2014-10-15T18:30:41.173" LastActivityDate="2014-10-15T18:30:41.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7704" PostTypeId="2" ParentId="7702" CreationDate="2014-10-15T20:14:34.687" Score="3" Body="&lt;p&gt;There is actually disagreement on this matter (within the IAU?). Dr. Alan Stern (lead of the New Horizons mission) for instance points out that &quot;this rule is inconsistent&quot; (e.g. see &lt;a href=&quot;http://news.bbc.co.uk/2/hi/science/nature/5283956.stm&quot; rel=&quot;nofollow&quot;&gt;Pluto vote 'hijacked' in revolt&lt;/a&gt;). Not only has Neptune not cleared its path, but the same holds for Earth, Mars and even Jupiter. Jupiter has a set of asteroids (the Trojans) which follow the same orbit (in the Lagrangian point L4 and L5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe the definition should be adjusted to include only those bodies as planets in the solar systems that have cleared all objects from their path except those objects that are in resonant orbits with the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_resonance&quot; rel=&quot;nofollow&quot;&gt;Resonant objects&lt;/a&gt; are those objects whose Periods are related to the Planet's period by a ration of &quot;small integer numbers&quot;. Pluto has a resonance with Neptune of 2:3. This means that for every two orbits completed by Pluto, Neptune will complete three orbits with the result that Neptune and Pluto never come close enough for Neptune to clear Pluto from its orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jupiter_Trojan&quot; rel=&quot;nofollow&quot;&gt;Trojan asteroids&lt;/a&gt;, the resonance is 1:1. the Trojans are alway either 60° in front of or behind of Jupiter and therefore never come close to Jupiter.&lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2014-10-15T20:14:34.687" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7705" PostTypeId="1" AcceptedAnswerId="7728" CreationDate="2014-10-15T22:13:49.227" Score="7" ViewCount="167" Body="&lt;p&gt;I was watching the new Cosmos and Neil DeGrasse Tyson said something that made me go hmmm...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He mentioned that the only reason that the Earth did not experience the runaway greenhouse effect like Venus is that life had evolved to the point at which it could feed on carbon and take it out of the atmosphere and deposit it in solid form as limestone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, if we were to find a planet in the Goldilocks zone, how likely is it that it would be habitable for life without already harboring life?  Would we expect that a similar scenario would have had to taken place on that planet to curb the greenhouse gasses in the atmosphere, thereby turning it into another Venus?  Can a carbon cycle exist without life?&lt;/p&gt;&#xA;" OwnerUserId="2439" LastEditorUserId="37191" LastEditDate="2021-02-25T08:47:57.950" LastActivityDate="2021-02-25T11:59:14.093" Title="Is early life required for life?" Tags="&lt;astrobiology&gt;&lt;abiogenesis&gt;&lt;carbon-cycle&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7706" PostTypeId="2" ParentId="7702" CreationDate="2014-10-16T10:57:29.030" Score="6" Body="&lt;p&gt;I answered this same question at &lt;a href=&quot;https://physics.stackexchange.com/a/141532/52112&quot;&gt;physics.SE&lt;/a&gt;. I specifically joined this part of the SE network to address this duplicate question at this site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The astronomy community faced two crises with regard to what constitutes a &quot;planet&quot;, first in the mid 19th century, and more recently at the start of the 21st century. The first crisis involved the asteroids. The second involved trans-Neptunian objects. Both crises challenged astronomers to question what a &quot;planet&quot; was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 Ceres, 2 Pallas, 3 Juno, and 4 Vesta were discovered in quick succession during the first decade of the 19th century. There was no international astronomical organization at the time of these discoveries; the International Astronomical Union wouldn't be formed for another century. Instead, the designation of what constituted a &quot;planet&quot; fell on the major astronomical almanacs such as the Berliner Astronomisches Jahrbuch (BAJ). Those discoveries at the start of the 19th century were treated as newly discovered &quot;planets&quot;. This situation remained static for about 40 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That changed in 1845 with the discovery of 5 Astraea. During the 1850s, the list of objects orbiting the Sun grew to 50, and during the 1860s, the list grew to over 100. The response of the BAJ and others was to demote Ceres, Pallas, Juno, and Vesta from planethood status to some lesser status, either minor planet or asteroid. Astronomers didn't have a clear-cut concept of what constituted a planet other than that they should somehow be large. Ceres, the largest of the bunch, is not very large. The end result of all of these discoveries starting in 1845 was that the first four discovered asteroids were demoted from planethood status.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second crisis started in 1992 with the discovery of (15760) 1992 QB&lt;sub&gt;1&lt;/sub&gt;. By 2006, the number of trans-Neptunian objects had grown significantly. Were these things &quot;planets&quot;, or something else? Some astronomers, notably Alan Stern, wanted the term &quot;planet&quot; to be extremely inclusive. Most astronomers balked at this idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paradoxically, it was Alan Stern himself, along with Harold Levison, who provided the key criterion of &quot;clearing the neighborhood&quot; that lies at the heart of what the IAU deems to constitute a &quot;planet.&quot; Their paper, &lt;a href=&quot;http://adsabs.harvard.edu/full/2002HiA....12..205S&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stern and Levison, &quot;Regarding the criteria for planethood and proposed planetary classification schemes,&quot; &lt;em&gt;Highlights of Astronomy&lt;/em&gt; 12 (2002): 205-213&lt;/a&gt; suggested splitting &quot;planet&quot; into two categories, &quot;überplanet&quot; (Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune) and &quot;unterplanet&quot; (Pluto+Charon, Eris, Ceres, Sedna, and a host of others).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stern is being quite hypocritical when he rants that there is no clear-cut boundary between &quot;planets&quot; and &quot;dwarf planets.&quot; The boundary is huge, and Stern knows this. The ratio of the square of an object's mass to its orbital radius about the Sun is key in determining whether an object can clear most of the junk from the vicinity of the object's orbit. There is a five order of magnitude difference between the smallest of the planets and the largest of the dwarf planets in terms of this ratio. This five order of magnitude difference figures predominantly in that paper by Stern and Levison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only difference between the proposal by Stern and Levison versus the voted-upon IAU resolution is that while Stern and Levison wanted to designate hundreds (and perhaps thousands) of objects into subcategories of &quot;planet&quot; (&quot;überplanet&quot; and &quot;interplant&quot;). On the other hand, the IAU chose to designate those objects as the mutually exclusive terms &quot;planets&quot; and &quot;dwarf planets&quot;. This is consistent with how astronomers dealt with that first crisis. Planets should be &quot;large.&quot; Stern and Levison provided the necessary ammunition to distinguish large from not so large.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2014-10-16T10:57:29.030" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7707" PostTypeId="2" ParentId="7679" CreationDate="2014-10-16T13:04:02.667" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Could it be a one time only freak accident?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's rather dubious, but there's no way to rule this out until we have seen signs of life elsewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the numbers one plugs into the Drake equation, or variants thereof, the number of intelligent species ranges from hundreds of millions just within the Milky Way galaxy to but one in the entire observable universe. Needless to say, that's a rather sizable range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If our nearest intelligent neighbor is in the next galaxy cluster, that still leaves room for billions of intelligent species in the observable universe, but it also leaves us essentially all alone. There's essentially no difference between a one-time freak accident where we truly are alone in the universe versus a billion intelligent species in the universe. There would need to be billions upon billions of intelligent species in the universe to have neighbors with whom we can communicate. Are there? The only possible answer science can currently offer is &quot;that's a good question!&quot;&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2014-10-16T13:04:02.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7708" PostTypeId="1" AcceptedAnswerId="7709" CreationDate="2014-10-16T16:47:38.977" Score="2" ViewCount="1509" Body="&lt;p&gt;I have a curious question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why astronauts are weightless in space is because they are orbiting Earth. If they suddenly stopped orbiting they would fall directly into Earth. Earth is doing the same around the Sun, so it would fall directly into the Sun if the Earth stopped orbiting the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Earth stopped rotating and orbiting, obviously ignoring the adverse effects of doing this and having Earth hold it's current position in it's orbit, how much would a 150 pound person weigh if they stood on suitable position on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is kinda out there, but it's a thought-provoking question for me and hopefully others.&lt;/p&gt;&#xA;" OwnerUserId="1177" LastActivityDate="2014-10-16T19:31:59.367" Title="How much would you weigh if Earth stopped orbiting the Sun?" Tags="&lt;gravity&gt;" AnswerCount="3" CommentCount="4" ClosedDate="2014-10-29T15:43:36.253" ContentLicense="CC BY-SA 3.0" />
  <row Id="7709" PostTypeId="2" ParentId="7708" CreationDate="2014-10-16T17:04:19.697" Score="4" Body="&lt;p&gt;In a sense, this is one of those &quot;If I break the laws of physics, what do the laws of physics say will happen&quot; kinds of questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring that, a 150 pound person would still weigh about 150 pounds if the Earth miraculously stopped orbiting the Sun. Almost all of a person's weight results from resistance to the Earth's gravity field. The tidal gravitational forces from the Moon and Sun are on the order of 10&lt;sup&gt;-4&lt;/sup&gt; ounces of force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I'm answering the question from the perspective of &quot;what would a bathroom scale say&quot;. A doctor's scale would say a 150 pound person weighs 150 pounds regardless of whether the person is standing on the Moon, the Earth, or a very massive planet. A doctor's scale measures mass. A bathroom scale measures apparent weight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what &lt;em&gt;would&lt;/em&gt; a bathroom scale say? If the Earth is at one astronomical unit from the Sun, whether the Earth is orbiting at 30 km/s or falling straight toward the Sun doesn't matter one iota. All that matters is that the Earth is one AU from the Sun. In both cases, that the Sun is 1 AU distant makes a 150 pound person weigh about $1.8 \times 10^-4$ ounces less at noon or at midnight than at sunrise or sunset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between the orbiting Earth and the in-falling Earth is that these tidal forces are more or less constant in the case of the orbiting Earth. They grow ever larger in the case of the in-falling Earth as it gets closer and closer to the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These tidal forces become quite large by the time the surface of the Earth just touches the surface of the Sun. Now our 150 pound person would weigh only 73.5 pounds per a bathroom scale if he was standing at that point where the surface of the Earth just touches the Sun. He would also weigh about the same at the point on the Earth furthest from the Sun. He would however weigh considerably more, about 188 pounds, if he happens to be at a point where the Sun is on the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, on the other hand, some mysterious, magical force makes the Earth stop orbiting the Sun and makes this change permanent (i.e., the Earth stops moving with respect to the Sun), a 150 pound person closest to the Sun would weigh about 1.4 ounces less, while a 150 pound person furthest from the Sun would weigh about 1.4 ounces more. A 150 pound person who sees the Sun at the horizon would still weigh about 150 pounds in this scenario.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2014-10-16T18:54:15.837" LastActivityDate="2014-10-16T18:54:15.837" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7710" PostTypeId="2" ParentId="7708" CreationDate="2014-10-16T17:07:46.940" Score="2" Body="&lt;p&gt;Weighlessness is a property of gravitational freefall. When you're standing on the surface of the Earth, the weight you feel is the reaction force of the Earth's surface stopping you from falling, i.e., weight is caused by the &lt;em&gt;failure&lt;/em&gt; of gravitational freefall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no relevant difference between a round orbit and a radial plunge orbit. They're both completely valid orbits as far as the laws of physics are concerned, and the fact that one will eventually smack you into the Earth does not matter until it actually happens.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If they suddenly stopped orbiting they would fall directly into Earth. Earth is doing the same around the Sun, so it would fall directly into the Sun if the Earth stopped orbiting the Sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So if the Earth were suddenly plunging into the Sun as you describe, there would be no discernible difference in weight, barring fairly minor tidal effects when one gets closer to the Sun. Conceptually, it would be a &lt;em&gt;different orbit&lt;/em&gt;, rather than a lack of orbiting.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the Earth stopped rotating and orbiting, obviously ignoring the adverse effects of doing this and having Earth hold it's current position in it's orbit, how much would a 150 pound person weigh if they stood on suitable position on Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Sun has $GM = 1.3271244\times 10^{20}\,\mathrm{m}^3/\mathrm{s}^2$, so at about $r \approx 1\,\mathrm{AU}$, the Earth holding still would give about $GM/r^2 = 5.9\,\mathrm{mm}/\mathrm{s}^2$, which is about $0.06\%$ of Earth's gravity. So they weigh more by about that proportion if on the far side from the Sun, and less if on the near side. This works out to be about $1.4\,\mathrm{oz}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The centrifugal force provided by the Earth's rotation is a few times more significant, up to about $34\,\mathrm{mm}/\mathrm{s}^2$, depending on the location on the Earth.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditDate="2014-10-16T17:46:03.037" LastActivityDate="2014-10-16T17:46:03.037" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7711" PostTypeId="2" ParentId="7708" CreationDate="2014-10-16T19:31:59.367" Score="2" Body="&lt;p&gt;Let me see if I can ask a different question that will help you understand why these other answers are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you weigh yourself at night (when the Earth's gravity and the Sun's gravity are pulling in the same direction), do you weigh significantly more than you do during the day (when the Sun would be 'pulling you up'?  The answer is no, because the Sun doesn't have a noticeable effect on your weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Earth suddenly stopped rotating around its axis, everything not nailed to the Earth would get flung West at about 1000 mph including you and your scale.  But a 150 lb human would still weigh 150 lbs as they splattered against the nearest building/mountain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Earth suddenly stopped revolving around the Sun, anything on the leading hemisphere would launch off the Earth at roughly 67,000 mph (well above escape velocity), and this includes things like oceans and atmosphere.  Anything on the trailing side would simply smash into the ground at 67,000 mph, effectively burying the goo that used to be you miles underground.  Worse things would happen as well, such as the Earth's molten core erupting through the crust, etc., but no one would be alive long enough to lament it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But just as your mass doesn't change when you go on a trampoline, it wouldn't change based on the Earth's rotation or revolution.  Notice I said mass this time and not weight.  If you had presence of mind during your last fleeting moments to hold your trusty scale under your feet, it would probably register '0', but that is because of your momentum, not because of the gravitational interaction between you and the Earth, which is what we call weight.  You would still be 150 pounds, you would just be a 150 pound pile of goo somewhere unexpected.&lt;/p&gt;&#xA;" OwnerUserId="2503" LastActivityDate="2014-10-16T19:31:59.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7712" PostTypeId="1" AcceptedAnswerId="7713" CreationDate="2014-10-17T04:47:37.433" Score="1" ViewCount="628" Body="&lt;p&gt;Studying the stars classes for my homework (school) i learned that stars are classified by the spectral types (Harvard) and luminosity class (MKK) .. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the sun is a star class &lt;code&gt;G2V&lt;/code&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But can not understand that clasificacion (far in the future) changes to &lt;code&gt;MIII&lt;/code&gt; when his main sequence ends and begins to burn helium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the relationship there? &#xA;Would not it be more logical to classify stars according to their initial mass and any change that is made in the classification is related to the lifetime of the star?&#xA;sub classify them according to: &#xA;Fusion element: Deuterium, Hydrogen, Helium, etc. &#xA;And Remaining time fusing current element?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are glued to the story?&lt;/p&gt;&#xA;" OwnerUserId="2475" LastActivityDate="2014-10-17T10:04:51.420" Title="Star class according to initial mass" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;classification&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7713" PostTypeId="2" ParentId="7712" CreationDate="2014-10-17T09:50:24.897" Score="5" Body="&lt;p&gt;The classification of stars using spectral class is a very useful classification when considering the properties of (the atmosphere of) a star &lt;strong&gt;at that moment&lt;/strong&gt;. If you consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Human_development_(biology)&quot; rel=&quot;nofollow noreferrer&quot;&gt;different stages in human development&lt;/a&gt; (embryo, fetus, infant, toddler, etc...), for instance, here one person also continuously changes its class. &#xA;So it is not strange/uncommon/forbidden to change a classification of a thing whether it is a human or a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also does not mean that other classifications are not useful. It depends on how you want to use the classification. In literature, concepts such as low-mass star, or Sun-like star are also used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classification according to the evolutionary stage (i.e. the type of fusion going on) of a star is also &lt;strong&gt;very common&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the image below from &lt;a href=&quot;http://www.atnf.csiro.au/outreach/education/senior/astrophysics/stellarevolution_postmain.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ATNF&lt;/a&gt; in which you can see how stars of different zero-age main-sequence mass move along a colour-magnitude diagram. If we follow the track of a star of one solar mass then we get the following subsequent evolutionary stages:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;MS - Main sequence stars (spectral type V) fuse hydrogen into helium in the core. When the hydrogen in the core is depleted, the star becomes an RGB star.&lt;/li&gt;&#xA;&lt;li&gt;RGB - Red Giant Branch stars (V-&gt;II/III) fuse hydrogen into helium in a shell around the core while the core is inert helium. At some point (Helium flash) the core temperature becomes high enough to start fusing helium into carbon and oxygen, the star then becomes an HB star.&lt;/li&gt;&#xA;&lt;li&gt;HB - Horizontal Branch stars (II/III) fuse helium into carbon/oxygen in the core. A shell with inert helium surrounds the core, which in turn is surrounded by a hydrogen burning shell.&lt;/li&gt;&#xA;&lt;li&gt;AGB - Asymptotic Gian Branch stars (I) have a carbon/oxygen core with several inert or fusion shells around it. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These types (MS/RGB/HB/AGB) are classes in a very useful and common classification scheme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end it just depends on what you want to do with your classification scheme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB. The classification of the Sun changes from G2 to M because the outer envelope and atmosphere expand enormously, causing the atmosphere to cool down. A cooler atmosphere means that the Sun changes from a yellow (6000K) to a red (4000K) star. The wavelength, and therefore colour, at which most energy is transmitted depends on the temperature of the star's atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5JRQS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="864" LastEditorUserId="864" LastEditDate="2014-10-17T10:04:51.420" LastActivityDate="2014-10-17T10:04:51.420" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7714" PostTypeId="2" ParentId="5869" CreationDate="2014-10-18T07:35:32.733" Score="4" Body="&lt;p&gt;None of us would know that the Sun had actually disappeared until the approximate eight minute point had passed. At that point, we would all be very quickly screaming and panicking at the absolute and total darkness which would suddenly replace all normal daylight. The gravitational concerns would be the least of our worries, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it happened at night, those of us in artificial light would be spared the surprise briefly. Those outside might be wondering why the Moon suddenly winked out... as eyes became used to the total darkness, stars in clear areas of the sky would be all that is visible, anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth would freeze, PDQ; we would not last long.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2014-10-18T07:35:32.733" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7715" PostTypeId="2" ParentId="7657" CreationDate="2014-10-18T16:49:22.487" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/2014/10171350-watching-siding-springs.html&quot; rel=&quot;nofollow&quot;&gt;Emily Lakdawalla at the Planetary Society as of October 17&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The nucleus of comet Siding Spring passes close by Mars on Sunday, October 19, at 18:32 UTC. Mars passes through the densest part of the comet's tail about 100 minutes later. All five orbiters at Mars will be in constant contact with the Madrid and then Goldstone Deep Space Network facilities, just to make sure they're okay.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Article also includes multiple links to live webcasts of tomorrow's events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/2014/10201007-status-update-all-mars.html&quot; rel=&quot;nofollow&quot;&gt;Opportunity rover comet image here&lt;/a&gt; as well as orbiter status update.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mars.jpl.nasa.gov/mro/news/whatsnew/index.cfm?FuseAction=ShowNews&amp;amp;NewsID=1739&quot; rel=&quot;nofollow&quot;&gt;Mars Reconnaissance Orbiter comet images here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="2031" LastEditDate="2014-10-22T17:03:40.697" LastActivityDate="2014-10-22T17:03:40.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7716" PostTypeId="1" AcceptedAnswerId="7717" CreationDate="2014-10-18T18:10:45.733" Score="39" ViewCount="10137" Body="&lt;p&gt;Do all stars have an Oort cloud like ours that will be filled with comets and other objects? If not, why are they not around every star?&lt;/p&gt;&#xA;" OwnerUserId="2629" LastEditorUserId="2531" LastEditDate="2017-06-13T15:02:46.993" LastActivityDate="2017-06-13T15:02:46.993" Title="Do all stars have an Oort cloud or is it a rare occurence?" Tags="&lt;solar-system&gt;&lt;oort-cloud&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7717" PostTypeId="2" ParentId="7716" CreationDate="2014-10-18T18:34:27.890" Score="32" Body="&lt;p&gt;Awesome question, especially since we know so little of the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nobody knows for sure how the Oort Cloud formed - I'll put that out there right now - but the current hypothesis is that it was originally part of the Sun's &lt;a href=&quot;https://en.wikipedia.org/wiki/Formation_and_evolution_of_the_Solar_System&quot; rel=&quot;noreferrer&quot;&gt;protoplanetary disk&lt;/a&gt;. All of the ice and rock coalesced into small bodies - proto-comets, if you will. While these bodies were much closer in to the Sun than they are today, they were tossed far out by gravitational interactions with the gas giants. Other &lt;a href=&quot;https://en.wikipedia.org/wiki/Interstellar_comet&quot; rel=&quot;noreferrer&quot;&gt;interstellar comets&lt;/a&gt; could also have been captured by the Sun, adding to the population.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why is the Oort Cloud spherical? After all, the protoplanetary disk was just a flat disk. Why were the orbits of the objects perturbed? Well, the Oort Cloud objects are only loosely bound to the Sun - relatively, that is. They can be influenced by passing stars or other objects. It appears that galactic-scale tidal forces, combined with the influence of passing stars, molded the Cloud into its current spherical shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what does this all tell us? Well, we know other stars have &lt;a href=&quot;https://en.wikipedia.org/wiki/Protoplanetary_disk&quot; rel=&quot;noreferrer&quot;&gt;protoplanetary disks&lt;/a&gt;, right? Some also have &lt;a href=&quot;https://en.wikipedia.org/wiki/Exoplanet&quot; rel=&quot;noreferrer&quot;&gt;exoplanets&lt;/a&gt; - gas giants like Jupiter. They are also subject to tidal forces and the passing of nearby stars. So, theoretically, there's no reason why other stars shouldn't have Oort Clouds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can we find them? The answer is, most likely, no. Here's why. According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Oort_cloud#Structure_and_composition&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The outer Oort cloud may have trillions of objects larger than 1 km (0.62 mi), and billions with solar system absolute magnitudes brighter than 11&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An &lt;a href=&quot;https://en.wikipedia.org/wiki/Absolute_magnitude&quot; rel=&quot;noreferrer&quot;&gt;absolute magnitude of a solar system object&lt;/a&gt; of 11 is very dim. Now, the object's &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_magnitude&quot; rel=&quot;noreferrer&quot;&gt;apparent magnitude&lt;/a&gt; is how it would look from a given distance; the absolute magnitude is how it looks from a distance of 1 AU (in the case of Solar System objects, this quantity is denoted &lt;a href=&quot;https://en.wikipedia.org/wiki/Absolute_magnitude#Solar_System_bodies_.28H.29&quot; rel=&quot;noreferrer&quot;&gt;$H$&lt;/a&gt;). Oort-cloud objects are 2,000 - 50,000 (or more) AU away-- so these objects, to us in the same solar system, have an apparent magnitude much fainter than 11.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point of that poorly-explained interlude is that these objects are faint. Very faint. And objects in Oort Clouds around other stars would appear even fainter. Using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Distance_modulus&quot; rel=&quot;noreferrer&quot;&gt;distance modulus&lt;/a&gt;, we can calculate the apparent magnitude of an object if the distance to that object and its absolute magnitude are known:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m-M=5(\log_{10}d-1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(from &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnitude_(astronomy)#Apparent_and_absolute_magnitude&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ is apparent magnitude, $M$ is a scaling of $H$ normally used for stars, and $d$ is the distance in AU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an Oort Cloud object $x$ light-years away, you can figure out how bright (or dim) it would appear, given that 1 light-year is 63241 AU. Try this with the distances of nearby stars, and you'll realize how dim objects in these stars' Oort Clouds would be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a final note: We don't know for sure if other Oort Clouds exist. From what I've been able to find, we don't have sufficiently powerful telescopes to observe these hypothetical Clouds, and so we don't (and may never) know if they exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/pdf/astro-ph/0512256v1.pdf&quot; rel=&quot;noreferrer&quot;&gt;This paper&lt;/a&gt; was instrumental in this answer. Start at page 38 for the relevant information. &lt;a href=&quot;http://www.solstation.com/stars/oort.htm&quot; rel=&quot;noreferrer&quot;&gt;This page&lt;/a&gt;, too, has some good information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I found from &lt;a href=&quot;http://www.universetoday.com/11283/kuiper-belt-like-disks-around-two-nearby-stars/&quot; rel=&quot;noreferrer&quot;&gt;a link&lt;/a&gt; from an answer to &lt;a href=&quot;https://physics.stackexchange.com/questions/26132/what-are-the-facts-that-allow-accepting-the-oort-cloud-theory&quot;&gt;this question&lt;/a&gt; on Physics, we've found Kuiper-Belt-like disks around other stars. This means it is certainly plausible for these stars to have Oort Clouds, too. And &lt;a href=&quot;https://en.wikipedia.org/wiki/Exocomet&quot; rel=&quot;noreferrer&quot;&gt;exocomets&lt;/a&gt; have been detected, which is another good sign.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2017-06-12T23:09:29.357" LastActivityDate="2017-06-12T23:09:29.357" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="7718" PostTypeId="1" AcceptedAnswerId="7741" CreationDate="2014-10-18T22:30:51.133" Score="4" ViewCount="2424" Body="&lt;p&gt;Researching a sci-fi story involving a &quot;twin-planet&quot; system. Is such an arrangement (however unlikely) physically possible? And if so, must they be tidally locked with each other, similar to the moon, month-long days, etc?  &lt;/p&gt;&#xA;" OwnerUserId="2630" LastActivityDate="2019-02-21T08:14:34.320" Title="What would be the dynamics of a double-planet system, similar to Earth / Moon, but with both bodies nearly Earth-sized?" Tags="&lt;orbit&gt;&lt;planet&gt;&lt;earth-like-planet&gt;" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="7719" PostTypeId="2" ParentId="7716" CreationDate="2014-10-18T23:10:29.423" Score="21" Body="&lt;p&gt;All stars could have their own Oort cloud, but all stars don't. As HDE says the Oort cloud was formed by material in the sun's protoplanetary disk and interstellar comets that were caught by the sun. Some theories say that almost all comets were formed around the Sun, and this wouldn't allow us to say much about comets around other stars. However there are others such as Levison et al. in &lt;em&gt;Capture of the Sun's Oort Cloud from Stars in Its Birth Cluster&lt;/em&gt; that contend that a majority of the comets have to have their origin from other stars' protoplanetary disks. This must be so because the current models fail to explain the amount of comets in the Oort cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun was born in a star cluster in close proximity to other stars. These stars would have to be the source of any significant contribution to the Oort cloud by stars other than the Sun. That is why Levison et al. tested their hypothesis by constructing a young star cluster and simulating its dynamics with an n-body simulator. They found that their hypothesis was confirmed by the simulation. If every star in the star cluster starts with some comets in their respective protoplanetary disks, then some stars collect many more comets from their neighbors before they leave the star cluster, as the Sun did long ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently wrote an N-body simulator to replicate their paper. The result is not available in English, but the experiment is described briefly &lt;a href=&quot;http://cekdahl.github.io/cometsimulator/starcluster.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. There is an animation there that will help you understand what I am getting at. My simulator ended up being a lot slower than theirs so I couldn't have nearly the same number of objects in my simulations, but I observed the same tendencies as they did. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the question, I did see that some stars lost all their comets and I also saw some stars that gained comets. It all depended on the dynamics in the star cluster. Some left it very early with just the comets that formed in their own protoplanetary disks, some left it with more comets than that and some left it with no comets at all. Based on these simulations I would say that, as I started by saying, all stars could have their own Oort cloud, but all stars don't.&lt;/p&gt;&#xA;" OwnerDisplayName="user2631" LastEditorDisplayName="user2631" LastEditDate="2015-08-03T20:45:17.797" LastActivityDate="2015-08-03T20:45:17.797" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7720" PostTypeId="1" CreationDate="2014-10-19T12:09:53.547" Score="3" ViewCount="6744" Body="&lt;p&gt;I've been trying to figure out how to find the diameter in kpc of any galaxy given that one knows the distance to it in Mpc (d) and also the angular diameter/size (A) in arcsec. I had a formula before to find the diameter of a planet: $D=\frac{2\pi dA}{360degrees}$ but this is only if the celestial object is round. Then I asked my friend who gave me another formula $D=d*10^6 *sin (\frac{A}{3600})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, is the second formula the correct one if I want to find the diameter of any galaxy no matter what shape it is or is there another one which I do not know of?&lt;/p&gt;&#xA;" OwnerUserId="2626" LastActivityDate="2014-10-19T15:19:14.720" Title="Diameter of any galaxy" Tags="&lt;galaxy&gt;&lt;distances&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7721" PostTypeId="1" CreationDate="2014-10-19T13:56:47.907" Score="1" ViewCount="920" Body="&lt;p&gt;Siding Spring now passes about 10 Earth diameters away from Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the estimates of how often comets fly by Earth at different distances? Are the distances to some famous historical comets known?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2014-10-19T15:42:36.800" Title="How near to Earth do comets pass?" Tags="&lt;distances&gt;&lt;comets&gt;&lt;impact-probability&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7722" PostTypeId="2" ParentId="7720" CreationDate="2014-10-19T13:57:21.393" Score="1" Body="&lt;p&gt;I'm not sure I understand the second formula (could you explain the variables?), but I &lt;em&gt;do&lt;/em&gt; know that you can use &lt;a href=&quot;https://en.wikipedia.org/wiki/Parallax&quot; rel=&quot;nofollow&quot;&gt;parallax&lt;/a&gt; in certain situations. Parallax works by using the idea that objects appear to be in different places when you (the observer) are at different places - i.e. at different points in the Earth's orbit. This graphic gives a good description of the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/e/e3/Stellarparallax2.svg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;Credit: &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/e/e3/Stellarparallax2.svg&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I think your friend is trying to get at is that if a galaxy is tilted at a certain angle, it will appear &lt;a href=&quot;https://en.wikipedia.org/wiki/Perspective_(graphical)#Foreshortening&quot; rel=&quot;nofollow&quot;&gt;foreshortened&lt;/a&gt;, and it will seem smaller than it should at that distance. This could lead you to make an incorrect measurement of its &lt;em&gt;angular&lt;/em&gt; diameter and therefore an incorrect measurement of its actual diameter. You could correct this using trigonometry if you knew just at what angle the galaxy was tilted, but it would be hard to measure that angle. You &lt;em&gt;could&lt;/em&gt; avoid this whole dilemma if the galaxy was perfectly circular, because you would always see the longer side even if it was foreshortened, but that could be an invalid assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, the reason this isn't a problem with ideal, perfectly spherical objects is that they are perfectly spherical and cannot appear foreshortened.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-10-19T13:57:21.393" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7723" PostTypeId="2" ParentId="7721" CreationDate="2014-10-19T14:12:18.867" Score="2" Body="&lt;p&gt;The answer is that they don't come very close. As &lt;a href=&quot;https://en.wikipedia.org/wiki/Near-Earth_object#Near-Earth_comets&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; notes,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is rare for a comet to pass within 0.1 AU (15,000,000 km; 9,300,000 mi) of Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even better, though, is an actual &lt;a href=&quot;http://www.minorplanetcenter.org/iau/lists/ClosestComets.html&quot; rel=&quot;nofollow&quot;&gt;list&lt;/a&gt; of some of the closest approaches of comets to Earth. The closest one listed, Comet Lexell in 1770, came 0.0151 AU away from Earth. The list only shows 20 comets have come with 0.1 AU of Earth. So Earth has not had a close encounter with a comet (like Mars is having) in a long time, if ever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cometography.com/nearcomet.html&quot; rel=&quot;nofollow&quot;&gt;This site&lt;/a&gt; says that C/1491 came within an incredible 0.0094 AU in 1491, but I'll take the IAU site over it. And it does admit that that comet has an uncertain orbit, so that figure could be wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when comets come even a &lt;em&gt;tiny&lt;/em&gt; bit near to Earth, there are some articles written about them. See &lt;a href=&quot;http://www.nydailynews.com/news/national/comet-pass-close-earth-article-1.1280062&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.huffingtonpost.com/2012/10/07/comet-c2012-s1-earth_n_1945462.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the distances to famous comets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Closest Ever&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Halley's - 0.0334 AU&lt;/li&gt;&#xA;&lt;li&gt;Great Comet of 1760 - 0.0682 AU&lt;/li&gt;&#xA;&lt;li&gt;Tempel-Tuttle - 0.0229 AU&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Also, some &lt;a href=&quot;https://en.wikipedia.org/wiki/Near-Earth_object#Near-Earth_asteroids&quot; rel=&quot;nofollow&quot;&gt;near-Earth asteroids&lt;/a&gt; used to be comets. &lt;a href=&quot;https://en.wikipedia.org/wiki/Extinct_comet&quot; rel=&quot;nofollow&quot;&gt;Extinct comets&lt;/a&gt; have lost a lot (if not all) of their ice and so may be classified as asteroids. This blurs the lines between the two groups. Some extinct comets have &lt;em&gt;become&lt;/em&gt; near-Earth asteroids. &lt;a href=&quot;http://www.boulder.swri.edu/~bottke/Reprints/Morbidelli-etal_2002_AstIII_NEOs.pdf&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; explores the issue pretty well. So if you include extinct comets as &quot;comets,&quot; you may have comets that have come even closer to Earth.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-10-19T15:42:36.800" LastActivityDate="2014-10-19T15:42:36.800" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7724" PostTypeId="2" ParentId="7721" CreationDate="2014-10-19T14:17:07.867" Score="2" Body="&lt;p&gt;Wired: &lt;a href=&quot;http://www.wired.com/2009/05/dayintech_0519/&quot; rel=&quot;nofollow&quot;&gt;May 19, 1910: Halley’s Comet Brushes Earth With Its Tail&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The 1910 pass of Earth was especially close and, thanks to expansive newspaper coverage, eagerly anticipated by the general public. In fact, Earth’s orbit carried it through the end of the comet’s 24-million-mile-long tail for six hours on May 19, earning the story the day’s banner headline in The New York Times.&#xA;  While most reporters of the day turned to astronomers to get the facts straight, the yellow press chose to pursue the story in more fanciful ways, helping to fuel the fears of the impressionable that the end of the world was nigh.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No zombies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Halley's &lt;a href=&quot;http://en.wikipedia.org/wiki/Halley%27s_Comet&quot; rel=&quot;nofollow&quot;&gt;got within 5.1 million kilometers of earth&lt;/a&gt; in 837 CE&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2014-10-19T14:17:07.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7725" PostTypeId="2" ParentId="7720" CreationDate="2014-10-19T14:40:41.590" Score="5" Body="&lt;p&gt;Given the angular diameter $a$ in &lt;strong&gt;radians&lt;/strong&gt; and the distance $d$ in Mpc, you can get the actual diameter $D$ from:&#xA;$$D = d\tan{a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the small &lt;a href=&quot;http://en.wikipedia.org/wiki/Small-angle_approximation&quot; rel=&quot;nofollow&quot;&gt;angle approximation&lt;/a&gt;, you get:&#xA;$$D = da$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a$ is in radians, so to get the distance in Mpc from the angular diameter in arc seconds you'd need to convert the angle in arc seconds to the angle in radians: $a = \frac{2\pi A}{360\times3600}$, where the factor $3600$ is used to convert arc seconds to degrees, and $\frac{2\pi}{360}$ to convert from degrees to radians:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the diameter in kpc:&#xA;$$D = 1000\times d \frac{2\pi A}{360\times3600}$$&lt;br&gt;&#xA;$$D = d \frac{\pi A}{648}$$&#xA;$$D \approx \frac{dA}{206} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $d$ is in $\textrm{Mpc}$, $D$ is in $\textrm{kpc}$, and $A$ is in $\textrm{arcsec}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $D$ is the diameter for round objects (even for disk galaxies seen at an angle). If the object is not round then this would normally be the maximum diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT (see answer and comments from HDE 226868)] For irregular galaxies you may also need to have more information (viewing angle) to find the &lt;strong&gt;real&lt;/strong&gt; maximum diameter of the galaxy. But that information is (I think) only available for galaxies in the local group.&lt;/p&gt;&#xA;" OwnerUserId="864" LastEditorUserId="864" LastEditDate="2014-10-19T15:19:14.720" LastActivityDate="2014-10-19T15:19:14.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7726" PostTypeId="2" ParentId="922" CreationDate="2014-10-19T15:28:18.197" Score="2" Body="&lt;p&gt;&lt;strong&gt;TL;DR&lt;/strong&gt; In response to your comment that &quot;note that [link to answer] supports the assertion that passing stars can influence Oort cloud object&quot; I will talk about whether this could happen to comets in the Oort cloud that surrounds the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can happen but the stars that exist today don't pass by close enough to yank away a comet at once. However many star passages could eventually do it. In this answer I attempt to present a way to think about this problem. Skip to the last paragraph to get the directly to my answer to your question without the extra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my answer &lt;a href=&quot;https://astronomy.stackexchange.com/a/7719/2631&quot;&gt;here&lt;/a&gt; I clearly state that many stars have their own Oort cloud and that if they pass by each other close enough the stars will exchange comets. This is a direct answer to your question. It is believed to happen a lot in young star clusters, but you have to realize that older stars are often separated from other stars by a great distance which prohibits this type of exchange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr /&gt;&#xA;Now I will discuss influence by stars on the comets in the Oort cloud (the usual one, that surrounds the solar system). This is the topic of chapter of 5.2, &lt;em&gt;Stellar Perturbations&lt;/em&gt;, in Julio Angel Fernández's book &lt;em&gt;Comets&lt;/em&gt;. It is possible to approximate the influence of a passing star with some reasonable simplifications. I will try to retell Fernández argument here below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that a comet is located at a heliocentric distance $r$. Since Oort cloud comets travel very slowly compared to stars, $0.1\rm{km\cdot  s^{-1}}$ versus  $30 \rm{km\cdot  s^{-1}}$, we can assume that the comet is at rest in the heliocentric frame. If we neglect any influence of the star when it is further away than $10^5 \rm{AU}$ from the closest approach to the Sun we only have to be concerned about the time it takes for a star to  travel $2\times 10^5\rm{AU}$ (imagine the star moving through the Sun), and during this time the comet has only travelled approximately $10^3\rm{AU}$. The star can be taken to travel in a straight line since it is only slightly perturbed by the Sun. This leads, and as with everything else this comes from Fernández text, to the integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta v=\int_{-\infty}^\infty F\times \rm{dt}=-\frac{2GM}{VD}&#xA;$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Delta v$ is the change of velocity of the comet, $G$ is the universal gravitational constant, $M$ is the mass of the star, $V$ is the velocity of the star and $D$ is the distance of closest approach between the star to the comet. However, we can't forget either that the Sun is also influencing the comet. If the comet is much closer to the Sun than the star the influence of the star can be neglected and vice versa. Since in this question we are dealing with the case &quot;is it possible&quot; I will assume that the comet is far out in the Oort cloud. Under these conditions we get another expression (after taking the Sun into account), i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;| \Delta v | \approx \frac{2GMr\rm{cos}(\beta)}{VD_\odot^2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\beta$ is the angle between the vector between the sun and the closest point of approach to the sun of the star and the vector from the sun to the comet. $D_\odot$ is the distance between the sun and the star at the closest point of approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this math is somewhat superfluous in the current context. I wanted to show you that it is possible to reason analytically about these things. Your question is whether a comet can be yanked away from its orbit in the Oort cloud and be captured by a passing star. The last formula presented here shows that for stars that actually exist now (not to say that stars or other small bodies have never passed by close to the solar system or even gone through it) the change of velocity imparted on the comet by the star is far to small for this to happen. However the change of velocity will accumulate over many star passages, and over a long time it will change the orbit of the comet in a meaningful way. Long periodic comets (LP comets) are comets that travel into the solar system in an orbit that is a very narrow ellipsis so that its perihelion (closest approach to the sun) is small but the aphelion (furthest point away from the sun) can be a great distance outside Oort cloud. Long periodic comets meet their end in different ways. Some passes to close to the sun and melt, others collide with planets, especially the big gas planets, and some get catapulted out of the solar system by a close approach to for example Jupiter. It is possible however, because long periodic comets can have orbits that extend beyond the Oort cloud where they are less influenced by the sun and more influenced by passing stars, that they might be yanked away and join another star eventually, although I still don't think it is likely. It would be possible I think to use the same kind of math to approximate the change of velocity that stars incur on LP comets to see if it is feasible, but I haven't done it.&lt;/p&gt;&#xA;" OwnerDisplayName="user2631" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2014-10-19T15:28:18.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7727" PostTypeId="2" ParentId="7721" CreationDate="2014-10-19T15:30:42.867" Score="2" Body="&lt;p&gt;On the &lt;a href=&quot;http://neo.jpl.nasa.gov/ca/&quot; rel=&quot;nofollow&quot;&gt;Near Earth Object Program&lt;/a&gt; web page from NASA, it lists about 80 objects that passes near the Earth over the next 4 months. Many of them pass within 0.01 AU, the closest one will pass on Oct 24 and be only 0.0038 AU (or 1.5 times the Earth Moon distance) away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_asteroid_close_approaches_to_Earth&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Scientists estimate that several dozen asteroids in the 6 to 12 meter size range fly by Earth at a distance closer than the moon every year ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is also predicted that in April 2029, &lt;a href=&quot;http://en.wikipedia.org/wiki/99942_Apophis&quot; rel=&quot;nofollow&quot;&gt;99942 Apophis&lt;/a&gt; will pass by the Earth at a distance of 38,300 km (inside the orbit of geostationary satellites).&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="1322" LastEditDate="2014-10-19T15:41:11.277" LastActivityDate="2014-10-19T15:41:11.277" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7728" PostTypeId="2" ParentId="7705" CreationDate="2014-10-20T11:45:18.343" Score="4" Body="&lt;p&gt;To answer your question directly, it is quite unlikely that planets would be habitable without already having life. It would have to be in a near-perfect condition, which would be statistically unlikely. In case of planets already having life, the existence of life provides robustness to small changes and makes habitability possible in a larger range of conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am assuming 'habitable' here means habitable for man, trees and an ecosystem etc. Remember that there are extremophiles (or certain other hardy species) which completely change the perspective on habitability, and are often the key in making those places habitable for other forms of life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also true for parts of earth which undergo a sudden change and are not 'habitable' for most forms of life till some hardy species takes over and changes the conditions, and consequently allows other species to be able to survive. See, for example, ecological succession (&lt;a href=&quot;https://en.wikipedia.org/wiki/Ecological_succession&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Ecological_succession&lt;/a&gt;), esp. primary succession and pioneer species (follow the links within the article).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, this is a long process. In case of the earth, there was also evolution that needed to happen, so it probably won't take as long as a few billion years, but even on earth, it takes a few years to make small disturbed microhabitats habitable. The atmosphere and other factors (soil, moisture etc., for the plants) need to be made suitable for living. Check this interesting article on terraforming: &lt;a href=&quot;https://microbewiki.kenyon.edu/index.php/Terraforming&quot; rel=&quot;nofollow&quot;&gt;https://microbewiki.kenyon.edu/index.php/Terraforming&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this was helpful. I don't really know if a carbon cycle is possible outside of living organisms in some way.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2014-10-20T11:45:18.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7729" PostTypeId="2" ParentId="3668" CreationDate="2014-10-20T13:59:25.467" Score="0" Body="&lt;p&gt;Galaxy Map has detailed information on 5000+ stars.&lt;br&gt;&#xA;You can download the exel spreadsheet &lt;a href=&quot;http://galaxymap.org/drupal/node/5&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; &#xA;&lt;br&gt;Don't know if it helps. Good luck ;)&lt;/p&gt;&#xA;" OwnerUserId="2640" LastActivityDate="2014-10-20T13:59:25.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7731" PostTypeId="1" AcceptedAnswerId="7732" CreationDate="2014-10-20T17:30:08.300" Score="8" ViewCount="316" Body="&lt;p&gt;After comet Sliding Spring's Oct. 19, 2014 close flyby of Mars, &lt;a href=&quot;http://www.nasa.gov/jpl/comet-siding-spring/#.VEVDVkQRhtg&quot; rel=&quot;noreferrer&quot;&gt;NASA&lt;/a&gt; reports that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;All three NASA orbiters around Mars confirmed their healthy status Sunday after each took shelter behind Mars during a period of risk from dust [from the comet's tail].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Took shelter behind Mars&quot;?  These spacecraft are in orbit around Mars, constantly in motion, so they can't very well just pack up and move to the other side of Mars to avoid getting hit with the dust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my questions are: What exactly qualifies as taking shelter behind Mars?  Does it simply mean being on the other side of the planet from the tail of the comet as dust from the tail passed Mars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did NASA get the (constantly in motion) spacecraft to hide behind Mars?  I assume that the orbits had to be modified in advance of the comet's arrival. I picture some orbital modifications that changed the phase of the orbits of the satellites so that they would be on the other side of Mars as the dust flew by, perhaps also making the orbits more elliptical with the apsis behind Mars, allowing the satellites to spend additional time in Mars' protection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will these orbital modifications affect the ability of the spacecraft to fulfil their primary missions as they relate to Mars?  I am not asking whether dust from the comet will affect the quality of the instruments on the spacecraft (indeed, this whole question focuses on steps NASA is taking to prevent that from happening) but whether the altered orbits the spacecraft have after their orbital modifications will affect their ability to pursue their primary missions (such as flying over different parts of Mars than was originally intended, which may or may not be beneficial to answering the questions the spacecraft were originally launched to answer.  Although scientists are nothing if not resourceful and I'm sure any data that come from the spacecraft, whether intended or not, will lead to important discoveries).&lt;/p&gt;&#xA;" OwnerUserId="128" LastEditorUserId="128" LastEditDate="2014-10-21T14:14:52.990" LastActivityDate="2014-10-21T22:37:34.823" Title="How did the Mars satellites &quot;take cover&quot; from the recent comet flyby?" Tags="&lt;mars&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7732" PostTypeId="2" ParentId="7731" CreationDate="2014-10-21T00:24:33.800" Score="5" Body="&lt;p&gt;I've written an answer or two dealing with this within the last week, and I can't believe I never thought about this. Great question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;These spacecraft are in orbit around Mars, constantly in motion, so they can't very well just pack up and move to the other side of Mars to avoid getting hit with the dust.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Believe it or not, they can just pack up and move. All they have to do is alter their orbital speed - i.e. either slow down, speed up, or change their orbit so they will automatically go a different speed, and so be in a different place. They execute a series of orbital maneuvers to ensure that &lt;em&gt;they are in the part of their orbit that temporarily carries them behind Mars&lt;/em&gt;. Orbital maneuvers aren't hard; practically any spacecraft can modify its orbit slightly, if need be. According to &lt;a href=&quot;http://www.viralglobalnews.com/science/comet-siding-spring-endanger-nasa-rovers-satellites-mars/13888/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The orbiters will be repositioned to a “safe zone,” where Mars will actually help shield the satellites from the cloud of dust particles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, the spacecraft will be moved so they are on the &lt;em&gt;opposite&lt;/em&gt; side of Mars as the comet. They can't be there for long - after all, they orbit Mars extremely quickly. But, fortunately, they don't have to be there for long. As &lt;a href=&quot;http://www.viralglobalnews.com/science/comet-siding-spring-endanger-nasa-rovers-satellites-mars/13888/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; says,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The satellites will remain in this region for a period of up to 40 minutes, until the threat has subsided.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why do short a period? Well, the comet will only be in the area for a short time: &lt;a href=&quot;http://mars.nasa.gov/news/whatsnew/index.cfm?FuseAction=ShowNews&amp;amp;NewsID=1727&quot; rel=&quot;nofollow&quot;&gt;NASA&lt;/a&gt; estimates the worst part will last roughly 20 minutes. After that, the danger will have subsided. Any particles that could damage the spacecraft will be long gone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Did these orbital modifications affect the quality of research these satellites will be able to perform in relation to Mars?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't find a source for the answer to this, but I would expect there wouldn't be any problems. You should be able to undo any maneuvers that would change the orbit of the craft.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-10-21T22:37:34.823" LastActivityDate="2014-10-21T22:37:34.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7733" PostTypeId="2" ParentId="6407" CreationDate="2014-10-21T00:54:43.067" Score="2" Body="&lt;p&gt;Before buying anything, get out your Amp-meter and measure what the motors pull in clock drive and GOTO mode, starting from a variety of scope positions.&#xA;An 18 incher will have a lot of mass, and Dobsonians are not known for their excellent bearings, so the amount of juice you need for a night of observing could be considerable.&#xA;Do you track most of the time, or GOTO...GOTO...GOTO?&#xA;How many hours is your typical observing session.&#xA;Multiply your average tracking amperage by the approximate number of hours you spend tracking, then do the same for your GOTO work.&#xA;Add the two numbers (amp hours) together, and double the result.&#xA;That should tell you about how big a battery you need.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The doubling is done to provide leeway, and because taking a standard lead acid below 50% is a sure way to kill it after only a few recharges. If you go Li, you can probably cut the multiplier.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="189" LastEditDate="2014-10-22T17:00:38.840" LastActivityDate="2014-10-22T17:00:38.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7734" PostTypeId="2" ParentId="7731" CreationDate="2014-10-21T06:22:35.127" Score="0" Body="&lt;p&gt;Since HDE 226868 already answered the other parts, i might add some ideas around the question related to the quality of the research of the satellites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best option to test this would be to do a calibration measurement before the comet passes, and one right after. This way values can be compared before the event, and right after the event. This way it can be investigated which (parts of) instruments might or might not be compromised. Normally if the orbital maneuvers were done correctly this should result in the same results before and after the comet passing, and if not, the differences can be tracked. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example for a satellite with a spectrograph on it, is looking at a standard-star, which 'does not vary' over time, each day (or on a regular basis), since the satellite was launched, and taking a spectrum. This way you can characterize the reliability of the instrument, and detect if there is any significant degradation of the instrument quality. Of course, these type of calibration measurements can (and will) be very specific from one mission to the other, and (obviously) from one instrument to the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to conclude, we expect that the comet passing by will not affect the quality of the research, but if it does, this can be characterized, and taken into account, so that the mission can still continue after such an event taking place.&lt;/p&gt;&#xA;" OwnerUserId="1062" LastActivityDate="2014-10-21T06:22:35.127" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7735" PostTypeId="1" AcceptedAnswerId="7775" CreationDate="2014-10-21T14:12:30.187" Score="17" ViewCount="994" Body="&lt;p&gt;If our solar system were to somehow be created outside of a galaxy as a single star in the vast nothingness between galaxies, would life on Earth change any?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does being a member of a galaxy actually provide some kind of benefit to our solar system or the creation/sustaining of life?&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2014-10-25T16:37:46.330" Title="Do we get any benefit from being in a galaxy?" Tags="&lt;galaxy&gt;&lt;solar-system&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7736" PostTypeId="1" AcceptedAnswerId="7737" CreationDate="2014-10-21T14:38:27.517" Score="9" ViewCount="22649" Body="&lt;p&gt;...today I must have left my brain home. I know it's a simple calculation but I keep failing simplest calculations today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the angular diameter of Earth as seen from the surface of the Moon?&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2017-11-29T09:44:06.610" Title="What is the angular diameter of Earth as seen from the Moon?" Tags="&lt;the-moon&gt;&lt;angular-diameter&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7737" PostTypeId="2" ParentId="7736" CreationDate="2014-10-21T16:32:34.283" Score="10" Body="&lt;p&gt;You can calculate the angular diameter of the Earth using the equation:&#xA;$$a = \arctan \frac{D}{d}$$&#xA;where $a$ is the angular diameter, $D$ is the physical diameter of the Earth, and $d$ is the distance from the Moon to the Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equatorial radius of the Earth is $r_E = 6378.1 \textrm{km}$, the diameter is therefore $D= 2 \times r_E = 12756.2$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Moon&quot; rel=&quot;noreferrer&quot;&gt;mean distance to the Moon is $d=384399\textrm{km}$&lt;/a&gt;. This gives a (mean) angular diameter of $a=1.90065$°.&lt;/li&gt;&#xA;&lt;li&gt;At &lt;a href=&quot;http://en.wikipedia.org/wiki/Moon&quot; rel=&quot;noreferrer&quot;&gt;Perigee the distance is $d=362600\textrm{km}$&lt;/a&gt;. This gives an angular diameter of $a=2.01482$°&lt;/li&gt;&#xA;&lt;li&gt;At &lt;a href=&quot;http://en.wikipedia.org/wiki/Moon&quot; rel=&quot;noreferrer&quot;&gt;Apogee the distance is $d=405400\textrm{km}$&lt;/a&gt;. This gives an angular diameter of $a=1.80226$°&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These Moon-Earth distances are as seen from the &lt;strong&gt;centre&lt;/strong&gt; of the Moon. To calculate the diameter from the surface of the Moon, you'll have to subtract the position of the observer along the Moon-Earth axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the observer is on the Moon's equator and the Earth is at zero hour angle (i.e. on the local meridian), the distance to the Earth needs to be subtracted by $r_M=1738.14\textrm{km}$. This gives the following values: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The mean distance to the Moon is $d=384399-1738 = 382661\textrm{km}$. This gives a (mean) angular diameter of $a=1.90928$°.&lt;/li&gt;&#xA;&lt;li&gt;At &lt;a href=&quot;http://en.wikipedia.org/wiki/Moon&quot; rel=&quot;noreferrer&quot;&gt;Perigee the distance is $d=362600-1738 = 360862\textrm{km}$&lt;/a&gt;. This gives an angular diameter of $a=2.02452$°&lt;/li&gt;&#xA;&lt;li&gt;At &lt;a href=&quot;http://en.wikipedia.org/wiki/Moon&quot; rel=&quot;noreferrer&quot;&gt;Apogee the distance is $d=405400-1738 = 403662\textrm{km}$&lt;/a&gt;. This gives an angular diameter of $a=1.81001$°&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The angular diameter of the Earth from the &lt;strong&gt;surface&lt;/strong&gt; of the Moon is, therefore, between $a=1.80226$° (at apogee and the Earth is near the horizon) and $a=2.02452$° (at perigee and for an observer at the equator and when the Earth is at maximum altitude on the meridian).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or about 2 degrees.&lt;/p&gt;&#xA;" OwnerUserId="864" LastEditorUserId="864" LastEditDate="2014-10-23T14:44:57.223" LastActivityDate="2014-10-23T14:44:57.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7738" PostTypeId="2" ParentId="7736" CreationDate="2014-10-21T16:33:15.073" Score="5" Body="&lt;p&gt;The average angular diameter of the Moon, as seen from the Earth, is &lt;a href=&quot;https://lcogt.net/spacebook/using-angles-describe-positions-and-apparent-sizes-objects&quot; rel=&quot;noreferrer&quot;&gt;about 31 arcminutes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angular diameter depends on the distance between the two objects and the diameter of the object being viewed.  Specifically, &lt;a href=&quot;http://en.wikipedia.org/wiki/Angular_diameter&quot; rel=&quot;noreferrer&quot;&gt;for small angles&lt;/a&gt;, it is the diameter divided by the distance.  When the distance is the same, the angular size is proportional to the diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance remains the same when viewing the Earth from the Moon, but the Earth is larger.  &lt;a href=&quot;http://www.nasa.gov/pdf/180561main_ETM.Distance.Moon.pdf&quot; rel=&quot;noreferrer&quot;&gt;According to NASA&lt;/a&gt;, the diameter of the Moon is 3,476 km, and the diameter of the Earth is 12,756 km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, because it's proportional, the angular diameter can be calculated as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{Earth} = a_{Moon} \times {d_{Earth}\over d_{Moon}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ = 31 arcminutes \times {12,756 km \over 3,476 km}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \approx 114 arcminutes$, or just under 2 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is approximate, because not only is this valid only for small degrees, where the tangent of an angle can be approximated by the angle itself (in radians), the Earth-Moon distance varies because the Moon's orbit around the Earth is an ellipse.&lt;/p&gt;&#xA;" OwnerUserId="1082" LastActivityDate="2014-10-21T16:33:15.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7739" PostTypeId="1" AcceptedAnswerId="7740" CreationDate="2014-10-21T17:46:57.333" Score="3" ViewCount="183" Body="&lt;p&gt;Its probably small, but is there a theoretical gravitational attraction between all objects in the universe? Light can move pretty far, so does that mean gravity can as well, and is the gravity blocked by the same way light is blocked by an object?&lt;/p&gt;&#xA;" OwnerUserId="36" LastActivityDate="2014-10-22T23:30:09.283" Title="Is it accurate to say that we have a gravitational attraction towards all objects in the known universe?" Tags="&lt;gravity&gt;&lt;light&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7740" PostTypeId="2" ParentId="7739" CreationDate="2014-10-21T18:23:36.337" Score="2" Body="&lt;p&gt;&lt;strong&gt;Yes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational field from an object &lt;em&gt;does&lt;/em&gt; extend infinitely far out into space, if there is an infinite space for it to fill. There isn't any sort of limit to how far an object's gravitational influence can travel. However, there is a catch: Gravity travels at a finite speed (the speed of light). If gravity propagated infinitely fast, any two objects would be &lt;em&gt;guaranteed&lt;/em&gt; to be affected by each other. However, because it is &lt;em&gt;not&lt;/em&gt; infinitely fast, two  objects that have never been less than 13.8 billion light-years apart (the universe is only 13.8 billion years old) could only now be feeling each other's influence. If the Sun were to suddenly explode, it might not feel the influence of an object 13.8 billion that has never been less than 13.8 billion light-years away.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the gravity blocked by the same way light is blocked by an object?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An gravitational field extending infinitely (eventually) in all directions means that one object should always have an influence on another. But the gravitational influence of one object can sort of be &quot;blocked&quot; by another object if its field in one spot is strong enough compared to the  other. For example, we still feel the force of Mars' gravity, but, to us, it is largely overwhelmed by that of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can show this easily using Newtonian gravity. &lt;a href=&quot;https://en.wikipedia.org/wiki/Newton&amp;#39;s_law_of_universal_gravitation&quot; rel=&quot;nofollow&quot;&gt;Newton's law of universal gravitation&lt;/a&gt; is formulated&#xA;$$F=G\frac{m_1m_2}{r_{12}^2}$$&#xA;where $m_1$ and $m_2$ are the masses of the two objects - we'll say $m_1$ is the mass of a human, $m_h$, and $m_2$ is the mass of Earth, $m_E$. A human on Earth's surface is roughly &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth&quot; rel=&quot;nofollow&quot;&gt;$6,371 \text { km}$&lt;/a&gt; from it's center, or $6,371,000 \text { meters}$. &lt;a href=&quot;https://en.wikipedia.org/wiki/Mars&quot; rel=&quot;nofollow&quot;&gt;Mars&lt;/a&gt;, on the other hand, is, at its closest, $0.3814 \text { AU}$ from Earth - or roughly $57,210,000 \text { km}$, or $57,210,000,000 \text { meters}$. Care to do the math? Well, what with this distance being squared ($r^2$), you can tell at a glance that the force from Mars is many orders of magnitude less than the force from Earth.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-10-22T23:30:09.283" LastActivityDate="2014-10-22T23:30:09.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7741" PostTypeId="2" ParentId="7718" CreationDate="2014-10-21T20:35:43.317" Score="3" Body="&lt;p&gt;This is a duplicated question but I think it fits better into astronomy than into physics. I'll try to answer all aspects of this question including the consequences.&lt;/p&gt;&#xA;&lt;p&gt;First, is it possible? As already pointed out it is, like Pluto and Charon. But is it possible for bigger planets, with the size of Earth? Well we didn't find this planet system yet, but the universe is big. There is no physical law that would forbid this system, it might be unlikely and very rare but it would be possible.&lt;/p&gt;&#xA;&lt;p&gt;Let's assume somewhere is a planet system with two Earth-sized twins. What would be the dynamics of this systems on the consequences for possible living beings on one of this planets?&lt;/p&gt;&#xA;&lt;p&gt;One consequence is that the gravity center of this system would be between both planets, if they exactly are the same size (same mass) the gravity center would be in the exact middle of both. That means both planets are rotating around this point (while they are rotating around a sun). This would lead to some effects like in a centrifuge and lead to some weird appearances. The gravity point of our Earth – Moon system is already shifted from the geometrical center point of our earth. And this leads into tides. The tides on our double Earth system would be vast (assuming at least one of this planets is just like our earth and has water on it). Complete areas would be periodically covered with water. Most living beings would probably be amphibians. This of course means the planet would rotate. In the beginning, the planets would rotate but they would slow each other down until they show the same side to each other. Just like our Earth did to the Moon. The Moon continually slows down the rotation of our Earth, but with his minor mass this takes a while. This processes would be much quicker on the double earth system. But even this would take some time. So it would be a matter of time when you look at this system. If it's a young system they would still rotate. If it's an older system they would show the same side each other.&lt;/p&gt;&#xA;&lt;p&gt;The last point would fix the tides problem, because the water level would be stationary. But this would lead into new problems. The day length would be as long as they need to rotated around their common gravity center. I'll be honest. I am not sure how long this would be. For a stable double Earth system the distance between both planets would be further then the Earth-Moon distance. Let's just assume a day in that system would take 2 Earth months for simplification. This would mean on daylight it would be very hot and at night it would be very cold. One side of the planet would be heat up for hundreds of hours (our earth hours) and the other half would cool down for the same time. Temperatures over a period would jump up and down in big steps. There is no living being known on earth that can handle both extreme hot and extreme cold in such a short period of time. Also this would lead into very heavy and constant storms in the area where the sun sets or rises, making living there even harder.&lt;/p&gt;&#xA;&lt;p&gt;Also your weight (stepping on the bath scale) would vary depend on where on the planet you are. If you on the side facing the other planet you would be heavier than on the other side of the planet.&#xA;Another point that would be possible, but I will point out I am not completely sure about this. The planets would struggle keeping their atmosphere for a long period of time because of the rotation. The side facing away from each other would have thicker atmosphere than on the sides (see my sketch).&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DDCJ5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DDCJ5.jpg&quot; alt=&quot;Sketch&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://s14.directupload.net/images/141021/5edwnqih.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;directupload.net&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;At this point the gravitational pull of the planet is also lower then everywhere else. This could help the molecules to escape the Earth, in a long period of time.&lt;/p&gt;&#xA;" OwnerUserId="2457" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-02-21T08:14:34.320" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="7742" PostTypeId="1" CreationDate="2014-10-22T05:43:18.530" Score="2" ViewCount="914" Body="&lt;p&gt;It is estimated that the sun's core is about 1/5 of the radius of the sun (from &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_core&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;). I know that the density of the plasma increases substantially the near the center, and that the volume of the core is 0.008 times the total volume. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac {V_{core}}{V_{sun}}=\frac {4/3\pi (0.2)^3}{4/3\pi (1)^3}$$&#xA;$$\frac {V_{core}}{V_{sun}}=\frac {(0.008)}{(1)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the mass of the core compared to the total mass?&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="1322" LastEditDate="2014-10-22T06:02:54.573" LastActivityDate="2014-10-23T23:46:37.793" Title="Mass of sun's core" Tags="&lt;the-sun&gt;&lt;mass&gt;&lt;core&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7743" PostTypeId="2" ParentId="7742" CreationDate="2014-10-22T06:08:16.770" Score="0" Body="&lt;p&gt;I found out it is 34% of the Sun's mass on &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_core&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-10-22T06:08:16.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7744" PostTypeId="1" AcceptedAnswerId="7745" CreationDate="2014-10-22T07:31:05.270" Score="5" ViewCount="1290" Body="&lt;p&gt;A simple question: is there light on the surface of Venus?&lt;/p&gt;&#xA;" OwnerUserId="1494" LastActivityDate="2014-10-22T09:09:07.713" Title="Is there light on the surface of Venus?" Tags="&lt;light&gt;&lt;venus&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7745" PostTypeId="2" ParentId="7744" CreationDate="2014-10-22T09:02:52.747" Score="6" Body="&lt;p&gt;The best way to answer this question is from images taken from probes that landed and took pictures from the surface of the planet.  According to the &lt;a href=&quot;http://www.lpi.usra.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lunar and Planetary Institute&lt;/a&gt;, images from the Russian Venera 13 lander revealed&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The effect of sunlight filtering through the dense atmosphere appears to give the surface an orange tint. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Despite some uncertainty of the actual colour due to possible chromatic distortions because of the atmospheric composition etc. It is clear in the image from Venera 13 (below) that there is indeed light on the surface f Venus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YBYGf.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image attribution (from the link above): Images courtesy of James Head (Brown University), in collaboration with USSR (now Russian) Academy of Sciences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observations, reported in the NASA article &lt;a href=&quot;http://www.nasa.gov/vision/universe/solarsystem/venus-20071128.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA Scientist Confirms Light Show on Venus&lt;/a&gt; reveal another potential source of light - lightning, and theorised to be &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;maybe even more activity than there is here on Earth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user2449" LastEditorDisplayName="user2449" LastEditDate="2014-10-22T09:09:07.713" LastActivityDate="2014-10-22T09:09:07.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7746" PostTypeId="1" AcceptedAnswerId="7766" CreationDate="2014-10-22T14:46:31.607" Score="3" ViewCount="571" Body="&lt;p&gt;Close to the bottom of page 4 of &lt;a href=&quot;http://link.springer.com/article/10.1023/A:1005161325181#page-1&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; (marked as page 164 in the upper left corner) states &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Values are given in the logarithmic scale usually adopted by astronomers, A$_{e\ell}$ = log N$_{e\ell}$/N$_H$ + 12.0, where N$_{e\ell}$ is the abundance by number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have never used stellar elemental abundances in this way, so I want to check a few things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, &lt;strong&gt;is the logarithm used by this paper log$_{10}$ or log$_e$ (often written ln)?&lt;/strong&gt;  My guess is log$_e$ because of what follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I want to calculate the elemental abundance of He relative to H.  I believe the correct math would be to do the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\textrm{N}_{He}}{\textrm{N}_H} = \textrm{exp}(\textrm{A}_{He} - 12.0)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is this the correct way to calculate the relative abundances?&lt;/strong&gt;  This equation gives a value of 0.343 for the helium abundance relative to hydrogen, which seems to be the correct value for the sun (the sun is usually quotes as being about 70% H and about 25% He, which gives a relative abundance of 0.357).  This correct value is what leads me to believe that the log is log$_e$.&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2015-12-25T00:05:08.940" Title="How are stellar elemental abundances quoted?" Tags="&lt;the-sun&gt;&lt;hydrogen&gt;&lt;elemental-abundances&gt;&lt;helium&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7747" PostTypeId="1" AcceptedAnswerId="23454" CreationDate="2014-10-22T18:58:23.330" Score="6" ViewCount="5910" Body="&lt;p&gt;Given an Earth station (antenna) coordinates (in lat,long) ant its Elevation (El) and Azimuth (Az), how to compute the satellite coordinates (lat,long), known its height?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplification purposes, the Earth can be considered as an sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So its given:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Earth Radius = 6371000 meters&#xA;Satellite Height, in meters (perpendicular to the Earth's)&#xA;&#xA;Station lat, long (as parameters)&#xA;Station height (0 = Zero, for simplification purposes)&#xA;Station Elevation (angle, radius or degrees)&#xA;Station Azimuth (angle, radius or degrees)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In summary: I'm trying to calculate satellite's coordinates projected on celestial sphere. I.e. it's current ground track position from the horizontal coordinate system values and the position of the observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem can be shown in this two figures: &#xA;&lt;img src=&quot;https://i.stack.imgur.com/VdpRP.jpg&quot; alt=&quot;Earth&amp;#39;s Sphere&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/4mYYd.jpg&quot; alt=&quot;Antenna&amp;#39;s Horizon Schema&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2658" LastActivityDate="2017-11-09T06:02:47.337" Title="How to compute satellite coordinates (lat,long) given antenna's coordinates, angles and satellite height" Tags="&lt;orbit&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7748" PostTypeId="2" ParentId="7742" CreationDate="2014-10-22T22:48:45.023" Score="3" Body="&lt;p&gt;I'm getting a slightly different figure from &lt;a href=&quot;http://www.space.com/58-the-sun-formation-facts-and-characteristics.html/&quot; rel=&quot;nofollow&quot;&gt;space.com&lt;/a&gt; (not my favorite source, but a source nonetheless), which says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The core extends from the sun's center to about a quarter of the way to its surface. Although it only makes up roughly 2 percent of the sun's volume, it is almost 15 times the density of lead and holds nearly half of the sun's mass. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would expect it to be fairly dense because of the extreme conditions at the center, primarily the force of gravity pushing in on it. The &quot;15 times the density of lead&quot; is not constant throughout the Sun, and only applies in the &lt;em&gt;very center&lt;/em&gt; of the core.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-10-23T17:29:51.457" LastActivityDate="2014-10-23T17:29:51.457" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="7749" PostTypeId="1" CreationDate="2014-10-22T23:45:42.173" Score="2" ViewCount="107" Body="&lt;p&gt;If I am talking about the surface of Mars, is this still nature, comparable to uninhabited places on earth, or does nature exclusively describe not man-made things on earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asking this question in Astronomy, because I would like to know what astronomers or other people dealing with extraterrestrial science on a regular basis think about it.&lt;/p&gt;&#xA;" OwnerUserId="2659" LastActivityDate="2014-10-23T00:12:49.267" Title="Do extraterrestrial landscapes or planets count as nature?" Tags="&lt;extra-terrestrial&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7750" PostTypeId="2" ParentId="7749" CreationDate="2014-10-22T23:51:00.120" Score="1" Body="&lt;p&gt;ARTFL &lt;a href=&quot;http://machaut.uchicago.edu/?action=search&amp;amp;resource=Webster%27s&amp;amp;word=nature&amp;amp;quicksearch=on&amp;amp;sourceid=Mozilla-search&quot; rel=&quot;nofollow&quot;&gt;Nature&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;The existing system of things; the world of matter, or of matter and mind; the creation; the universe.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That definition includes the surface of Mars, and the core of &lt;a href=&quot;http://en.wikipedia.org/wiki/Betelgeuse&quot; rel=&quot;nofollow&quot;&gt;Betelgeuse&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2014-10-22T23:51:00.120" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7751" PostTypeId="2" ParentId="7749" CreationDate="2014-10-23T00:12:49.267" Score="1" Body="&lt;p&gt;I'm not a professional astronomer, but I still think I may be able to help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In response to your comment to WayfaringStranger's answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do you naturally think of nature if you think about Mars? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, not really. I think that this is also partly due to society's perception of nature as a whole. &lt;a href=&quot;https://en.wikipedia.org/wiki/Nature&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; gives a decent explanation of this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Natura&lt;/em&gt; was a Latin translation of the Greek word &lt;em&gt;physis&lt;/em&gt; (φύσις), which originally related to the intrinsic characteristics that plants, animals, and other features of the world develop of their own accord.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But it goes on to say&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The concept of nature as a whole, the physical universe, is one of several expansions of the original notion; it began with certain core applications of the word φύσις by pre-Socratic philosophers, and has steadily gained currency ever since.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the original concept of nature was life, but we've expanded since then into everything that isn't made by humans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I suppose it depends partly on your personal viewpoint (or time period!), but it does seem like &lt;em&gt;nature&lt;/em&gt; today refers to the entire universe.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-10-23T00:12:49.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7752" PostTypeId="1" AcceptedAnswerId="7776" CreationDate="2014-10-23T04:24:35.920" Score="6" ViewCount="238" Body="&lt;p&gt;The giant impact hypothesis is a theory to explain how the moon was formed. What evidence is there that supports the theory and led us to find the current theory?&lt;/p&gt;&#xA;" OwnerUserId="2212" LastActivityDate="2014-10-25T20:09:13.983" Title="Evidence behind Giant Impact hypthesis" Tags="&lt;the-moon&gt;&lt;formation&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7753" PostTypeId="1" AcceptedAnswerId="7756" CreationDate="2014-10-23T11:39:37.527" Score="6" ViewCount="391" Body="&lt;p&gt;In a recent holiday, a friend made the following picture. It contains a curious green &quot;thing&quot; in the nightsky. The picture was made in Ladakh, India, 6weeks ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;thing&quot; only appeared in one of the pictures (others were made a few minutes earlier and later). He didn't see it with the naked eye (but didnt look at the sky during the exposure).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Camera: Canon EOS 350D DIGITAL&lt;/li&gt;&#xA;&lt;li&gt;7 sec exposure&lt;/li&gt;&#xA;&lt;li&gt;ISO: 1600&lt;/li&gt;&#xA;&lt;li&gt;f-number: 4&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What could that be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sGsxa.jpg&quot; alt=&quot;Full pic&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/T3JDH.jpg&quot; alt=&quot;Zoom&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2662" LastActivityDate="2017-07-20T20:04:12.060" Title="Pictures of a curious astronomical phenomenon" Tags="&lt;observational-astronomy&gt;&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7755" PostTypeId="1" CreationDate="2014-10-23T15:37:39.197" Score="2" ViewCount="100" Body="&lt;p&gt;When I look at the night sky I seem to resolve points which are local stars and milkiness at further regions of the galaxy. But what is between? I think there must be a lot of stars which are too faint to see individually, and yet not perceivable as part of a dense region of stars. Is this at all accurate? Does anybody know how much we fail to see when we think we are looking?&lt;/p&gt;&#xA;" OwnerUserId="2663" LastActivityDate="2014-10-23T22:17:22.983" Title="Does naked eye miss a lot between local stars and distant milkiness?" Tags="&lt;star-gazing&gt;&lt;naked-eye&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7756" PostTypeId="2" ParentId="7753" CreationDate="2014-10-23T15:43:30.417" Score="15" Body="&lt;p&gt;In your friends picture are more artifacts than the one you showed in the 2nd picture a little bigger. I marked more of them in the picture below (click to enlarge).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ggkIY.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ggkIYm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are all in a perfect line to the bright light. So these artifacts are caused by the bright light and the lens of the camera. A lens is not flat it's, well lenticular (it's where the name came from). The main light source is shining from an angle on the lens. When light hits a surface two things happen (at least in most cases ignoring Brewster's angle). The two things are reflection and transmission (or absorption, but we are talking about a lens so I let this one drop here). Every time the light hits the surface of the lens, a part is reflected and another part is transmitted. But the lens has a thickness so when the light that was first transmitted now hits the back of the lens, a part is transmitted to the CCD sensor of your camera and a (smaller) part is reflected back into the lens. This can happen several times. Normally this effect is small enough it will not be noticed. But this picture is not a normal case. Most of the picture (dark sky) emits no light, the picture is very dark. In contrast to this, there is a bright light source dominating everything (related to luminescence). And we have a 7 sec exposure the picture is taken with. 7 sec are enough time for the CCD sensor to capture some light from the effect described above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you now want to know why is it green and not white. White light includes all wave lengths (colors) and every wave length is reflected in another angle on the surface of air-lens. The green wave length just had the best angle to be reflected severals time and still reach the CCD sensor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the most accurate answer to your question &quot;what is this green thing“ is: It's the flashlight (or what ever this light source is) that is seen in the bottom of the picture.&lt;/p&gt;&#xA;" OwnerUserId="2457" LastEditorUserId="2365" LastEditDate="2017-07-20T20:04:12.060" LastActivityDate="2017-07-20T20:04:12.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7757" PostTypeId="2" ParentId="7755" CreationDate="2014-10-23T16:00:50.237" Score="0" Body="&lt;p&gt;I am not quite sure if i understand your question correctly. But I think the answer to your question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It depends on where you stand and look into the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light pollution is the biggest issue observing anything in the sky. If you are out in the night in a big city you just see a few stars (at a clear night). The further you are away from any light source the more stars you are able to see with eyes only. &lt;/p&gt;&#xA;" OwnerUserId="2457" LastActivityDate="2014-10-23T16:00:50.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7758" PostTypeId="2" ParentId="7755" CreationDate="2014-10-23T16:26:38.250" Score="1" Body="&lt;p&gt;Even the Hubble telescope cannot &quot;see&quot; many of the stars and galaxies in the dark parts of space at first glance.  If they fix the Hubble on a dark patch for several days, the photons from extremely distant stars and galaxies will fill the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble_Deep_Field&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hubble_Deep_Field&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the Earth's atmosphere will obscure and distort many stars that are visible from low Earth orbit and beyond.  Astronauts can see many more stars than we can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your question, &quot;how many are we missing&quot;, that's a tough one to answer. It depends on what level of detail you are talking about. &lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2014-10-23T16:26:38.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7759" PostTypeId="1" CreationDate="2014-10-23T19:28:03.430" Score="0" ViewCount="241" Body="&lt;p&gt;If Andromeda is overhead at midnight, what is the right ascension of the sun? (andromeda is ar RA =1hr, Dec = +40).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought I understood what Dec and RA were, but I can't think of how to start this question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2665" LastActivityDate="2015-02-21T12:27:36.470" Title="Declination and Ascension - the Sun and Andromeda" Tags="&lt;the-sun&gt;&lt;ascension&gt;&lt;declination&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7760" PostTypeId="1" AcceptedAnswerId="7761" CreationDate="2014-10-23T19:36:06.137" Score="2" ViewCount="257" Body="&lt;p&gt;Is there a standard mapping of symbols to terms for celestial and orbital mechanics (and physics in general)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example here are some ambiguities&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Angular Momentum: L, H, J, $\Gamma$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbital period: T or P,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M: mean anomaly or mass of the Sun,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E: energy or eccentric anomaly,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p: linear momentum or length of semi-latus rectum,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a: acceleration or length of semi-major axis,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e: eccentricity or base of natural logarithms,&lt;/p&gt;&#xA;" OwnerUserId="233" LastEditorUserId="233" LastEditDate="2014-10-23T22:57:30.720" LastActivityDate="2014-10-24T10:38:08.690" Title="Is there a standard mapping of symbols to terms for celestial and orbital mechanics" Tags="&lt;orbit&gt;&lt;orbital-elements&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7761" PostTypeId="2" ParentId="7760" CreationDate="2014-10-23T21:55:28.887" Score="1" Body="&lt;p&gt;There are a &lt;em&gt;lot&lt;/em&gt; of instances (such as those you pointed out) in all of science and mathematics where a symbol can mean two different things. &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_mathematical_symbols&quot; rel=&quot;nofollow&quot;&gt;In mathematics&lt;/a&gt;, there's actually a good deal of this (and the article says that the list is incomplete!). Many are &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_mathematical_symbols#Letter-based_symbols&quot; rel=&quot;nofollow&quot;&gt;based on letters&lt;/a&gt;, although they may have deviated from the original shape in order to make a distinction regarding just what the heck the thing actually means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, you run out of letters. You can opt to use symbols from &lt;a href=&quot;https://en.wikipedia.org/wiki/Greek_letters_used_in_mathematics,_science,_and_engineering&quot; rel=&quot;nofollow&quot;&gt;another language&lt;/a&gt;, the first choice for many being Greek. That roughly doubles the amount of letter-based symbols available to you. But there are still a lot of duplicates, as you can see by browsing through the lists &lt;a href=&quot;https://en.wikipedia.org/wiki/Physical_constant&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_common_physics_notations&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my knowledge, this is the most comprehensive list easily available. And yes, there are loads of duplicates. But the great thing is that you almost certainly won't have to use the same variable to mean multiple things in one equation, or set of equations. For example, I doubt you'll need to use $V$ for voltage, volume and shear force all in the same equation!&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-10-23T21:55:28.887" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="7762" PostTypeId="2" ParentId="7755" CreationDate="2014-10-23T22:17:22.983" Score="0" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Naked_eye&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Theoretically, in a typical dark sky, the dark adapted human eye would see the about 5,600 stars brighter than +6m while in perfect dark sky conditions about 45,000 stars brighter than +8m might be visible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(By the way, &quot;+6m&quot; means stars of apparent magnitude 6 or brighter.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the best conditions, a person could see 45,000 stars. That's quite a lot. However, &lt;a href=&quot;https://en.wikipedia.org/wiki/Milky_Way&quot; rel=&quot;nofollow&quot;&gt;the Milky Way&lt;/a&gt; alone has over &lt;em&gt;200 billion&lt;/em&gt; stars (and that's a low estimate!). So we can only see a small fraction of the total stars in the Milky Way. Earth-based optical telescopes can do a lot better, though.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-10-23T22:17:22.983" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7763" PostTypeId="1" AcceptedAnswerId="7764" CreationDate="2014-10-23T22:27:44.333" Score="4" ViewCount="1593" Body="&lt;p&gt;Viewing my Facebook feed today, my local news station posted regarding a solar eclipse taking place today:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/At6y5.png&quot; alt=&quot;Facebook post - Solar Eclipse&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the line about using a leafy tree as a filter:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Scientists say NOT to directly look at the sun (if skies are clearing where you are), but looking at the eclipse through a leafy tree creates a natural filter!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've heard a lot of people recommend viewing a solar eclipse by looking at the projection of the eclipse caused by light passing between leaves in a tree, in this case looking at the ground or a wall, where ever the light hits after passing through the tree.  However, the Facebook posting here seems to recommend looking up at the solar eclipse &lt;em&gt;through&lt;/em&gt; the tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that they have misunderstood the projection method (or inadequately explained), and their advice is not only wrong, but potentially dangerous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this method recommendable? Should it be considered dangerous?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Follow-up&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Almost immediately upon having seen the posting I had commented on it noting my belief that such advice could be dangerous, including a more clear explanation of how viewing using trees should be done. The staff member who originally posted the status later took note of this and edited the status to less ambiguously explain viewing the projection. Unfortunately, this was done &lt;em&gt;after&lt;/em&gt; the solar eclipse had concluded. Hopefully no damage was done.&lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="1288" LastEditDate="2014-10-24T23:00:31.223" LastActivityDate="2014-10-24T23:00:31.223" Title="Viewing a solar eclipse through a leafy tree" Tags="&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7764" PostTypeId="2" ParentId="7763" CreationDate="2014-10-23T22:51:34.187" Score="8" Body="&lt;p&gt;It seems that the station has, in fact, given out extremely dangerous advice - simply because it was poorly phrased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Viewing a solar eclipse is incredibly dangerous - mostly under circumstances where you're looking towards the Sun. As &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_eclipse#Viewing&quot;&gt;Wikipedia&lt;/a&gt; warns,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Looking directly at the photosphere of the Sun (the bright disk of the Sun itself), even for just a few seconds, can cause permanent damage to the retina of the eye, because of the intense visible and invisible radiation that the photosphere emits. This damage can result in impairment of vision, up to and including blindness.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You couldn't pay me enough to so much as face toward the Sun and look at my feet and attempt to view a solar eclipse out of the corner of my eye!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are safe ways to view an eclipse, the most common being using a &lt;a href=&quot;https://en.wikipedia.org/wiki/Pinhole_camera&quot;&gt;pinhole camera&lt;/a&gt; to indirectly view the eclipse. Not only are you not looking directly at the Sun, but you can actually turn away and look at the projection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here's where I think the station messed up. According to &lt;a href=&quot;http://www.cnn.com/2014/10/22/tech/innovation/partial-solar-eclipse/index.html&quot;&gt;CNN&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Or let a tree do the work for you. &quot;Overlapping leaves create a myriad of natural little pinhole cameras, each one casting an image of the crescent-sun onto the ground beneath the canopy,&quot; NASA says.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So NASA actually advocates using the leaves as a pinhole and using indirect viewing methods - i.e. looking anywhere &lt;em&gt;but&lt;/em&gt; the Sun - as opposed to watching the eclipse &lt;em&gt;through&lt;/em&gt; the leaves.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-10-23T22:51:34.187" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7765" PostTypeId="2" ParentId="7742" CreationDate="2014-10-23T23:46:37.793" Score="5" Body="&lt;p&gt;Here are the results of some arbitrary cutoffs for the &quot;core&quot; based on $2010$ solar models calculated by &lt;a href=&quot;http://www.ap.smu.ca/~guenther/evolution/solar_modeling.html&quot; rel=&quot;noreferrer&quot;&gt;Guenther et al&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Contributing $99\%$ of the total luminosity:&lt;br&gt;$R = 25.5\%$, $M = 9.88\times10^{29}\,\mathrm{kg}$ ($49.7\%$).&lt;/li&gt;&#xA;&lt;li&gt;Nuclear reaction rate falls below $1\%$ of central rate:&lt;br&gt;$R= 27.2\%$, $M = 1.07\times10^{30}\,\mathrm{kg}$ ($54.0\%$).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Overall, $\sim 1/4$ the radius ($\sim 1/64$ the volume) and $\sim 1/2$ the mass decently with some other sources. Unfortunately, there does not seem to be a clear dividing line demarcating the &quot;core&quot; like there is between the radiative zone and the convection zone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps an alternative condition would be a local maximum of $\mathrm{d}(\ln P)/\mathrm{d}(\ln\rho)$, which occurs under $20\%$ of solar radius and is &lt;em&gt;roughly&lt;/em&gt; constant until the convection zone.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-10-23T23:46:37.793" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7766" PostTypeId="2" ParentId="7746" CreationDate="2014-10-24T03:32:25.677" Score="2" Body="&lt;p&gt;First of all, your first question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.iisc.ernet.in/currsci/apr102004/921.pdf&quot; rel=&quot;nofollow&quot;&gt;This source&lt;/a&gt; clearly state that &lt;code&gt;Values are given in the usual logarithmic (dex) scale&lt;/code&gt;, for the same formula that you quoted (similar job).&#xA;It is a bit tricky as the article &quot;explains&quot; the values, but you have to pay attention to the exact definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is better to work out with an example. Let's take the &lt;code&gt;He&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good enough, you can better read the paper from &lt;a href=&quot;http://astro.uni-tuebingen.de/~rauch/TMAP/grevessesauval2001.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the table, we have $A_{el}=10.93$. This is &lt;em&gt;the &lt;strong&gt;abundance&lt;/strong&gt; of He relative to H&lt;/em&gt; (in logarithmic scale).&#xA;From this you find out that $\frac{N_{el}}{N_H}=0.08 = 8\%$.&#xA;Indeed the work confirms this value (see the last page).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you quote as &lt;code&gt;about 25% He&lt;/code&gt;, is what they call &lt;code&gt;abundances by mass of [...] Helium (Y)&lt;/code&gt;, which means $Y = mass\ of\ Helium / mass\ of\ Hydrogen$, and this is indeed about $25\%$.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastActivityDate="2014-10-24T03:32:25.677" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7767" PostTypeId="2" ParentId="7760" CreationDate="2014-10-24T09:34:56.907" Score="1" Body="&lt;p&gt;There is no formal set of symbols for physical quantities that is recognised by all scientists. Units on the other hand are more formally recognised, e.g. &lt;a href=&quot;http://www.bipm.org/en/measurement-units/&quot; rel=&quot;nofollow noreferrer&quot;&gt;International System of Units&lt;/a&gt;. Authors may select their own symbols to denote specific quantities (they should, however, always specify the meaning of a symbol). On the other hand, I imagine that it will be more difficult to get your paper published if you use different symbols then generally used ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem is a problem of disambiguation and is not limited to the physical sciences. In computer science, the W3C (Tim Berners Lee) has initiated the &lt;a href=&quot;http://www.w3.org/standards/semanticweb/&quot; rel=&quot;nofollow noreferrer&quot;&gt;semantic web&lt;/a&gt;, where concepts on the web are tagged with unique identifiers (URIs). Ontologies (vocabularies of concepts) are used to link concepts to URIs, which can be reused for disambiguation. For instance a paper may use symbol &lt;a href=&quot;http://www.wurvoc.org/vocabularies/om-1.8/Angular_momentum&quot; rel=&quot;nofollow noreferrer&quot;&gt;$L$&lt;/a&gt; for angular momentum and is linked to a URI for the concept &lt;code&gt;angular momentum&lt;/code&gt;. A website may use the symbol &lt;a href=&quot;http://www.wurvoc.org/vocabularies/om-1.8/Angular_momentum&quot; rel=&quot;nofollow noreferrer&quot;&gt;$H$&lt;/a&gt; for angular momentum and link to the same concept &lt;code&gt;angular momentum&lt;/code&gt; using the same URI. As both symbols are linked to the same concept, they are both disambiguated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been working together with Hajo Rijgersberg (&lt;a href=&quot;https://www.cs.vu.nl/en/Images/H%20Rijgersberg%2021mei2013_tcm75-342356.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;see his PhD thesis&lt;/a&gt;) on the &lt;a href=&quot;http://www.wurvoc.org/vocabularies/om-1.8/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ontology of Units of Measure&lt;/a&gt;, which is an ontology with a large number of quantities and units.&#xA;&lt;strong&gt;This ontology can be used as a list of quantities and their symbols.&lt;/strong&gt; But it is by no means complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way we could use such an ontology is to insert the identifier (URI) into the text. At the &lt;a href=&quot;http://swaie2014.wordpress.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;SWAIE workshop&lt;/a&gt;, I outlined a method to do just that in LaTeX (&lt;a href=&quot;http://ceur-ws.org/Vol-925/paper_5.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper&lt;/a&gt;). The physical symbols in the PDF output are then linked to the concept. Clicking on the symbol will cause a browser to open at the URL identifying the physical concept. At the moment I'm trying to implement this in MathJax, but I am finding some &lt;a href=&quot;https://stackoverflow.com/questions/24514895/how-to-keep-rdfa-attributes-in-mathml-using-mathjax?noredirect=1#comment41599340_24514895&quot;&gt;technical difficulties&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="864" LastEditorUserId="-1" LastEditDate="2017-05-23T12:37:21.700" LastActivityDate="2014-10-24T10:38:08.690" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7768" PostTypeId="5" CreationDate="2014-10-24T09:42:06.463" Score="0" Body="&lt;p&gt;Questions regarding an object's movement, or its distance traveled relative to the time taken. Speed is different than velocity.&lt;/p&gt;&#xA;" OwnerUserId="31410" LastEditorUserId="31410" LastEditDate="2020-12-16T04:37:51.147" LastActivityDate="2020-12-16T04:37:51.147" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="7769" PostTypeId="4" CreationDate="2014-10-24T09:42:06.463" Score="0" Body="Questions regarding an object's movement, or its distance traveled relative to the time taken." OwnerDisplayName="user2449" LastEditorUserId="2153" LastEditorDisplayName="user2449" LastEditDate="2015-04-12T22:01:56.553" LastActivityDate="2015-04-12T22:01:56.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7770" PostTypeId="2" ParentId="3672" CreationDate="2014-10-24T11:15:05.823" Score="1" Body="&lt;p&gt;Florin Andrei what evidence can you supply to charge the oldest scientific body on this planet (The Royal Society at Greenwich) to be a profit-driven media entity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bishop, if you are asking when the 2 specific points in time match, it is true that they have a 'astronomical' infrequency to look at. But if you look at what the common terms for the definition of a solstice and full moon occurring at the same time, then it is not too infrequent. Matching a 24 hour period (solstice day) against a 72 hour period (a full moon) then they occur with a much more reasonable frequency. There are lots of lunar cycle calculators around. The current cycle we are in has the same middle day of the full moon  period In 2004, 2007, 2015, 2026, 2034, 2042 and 2045. further dates can be found using this &lt;a href=&quot;http://aa.usno.navy.mil&quot; rel=&quot;nofollow&quot;&gt;http://aa.usno.navy.mil&lt;/a&gt;.&#xA;Just chose any date, calculate the drift on your 400 year cycle then calculate how often that drift occurs at the same time as the 3 day phase of the full moon drift. There will be large periods when the match occurs rarely (The center point of each phase occurring furthest apart). As the centre points close on each other the Frequency will increase. The recurring pattern of all astronomic movements. &lt;/p&gt;&#xA;" OwnerUserId="2667" LastEditorUserId="2667" LastEditDate="2014-10-24T12:52:55.013" LastActivityDate="2014-10-24T12:52:55.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7771" PostTypeId="2" ParentId="7759" CreationDate="2014-10-24T11:57:53.037" Score="1" Body="&lt;p&gt;If you mean &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_time&quot; rel=&quot;nofollow&quot;&gt;solar time&lt;/a&gt; midnight than the answer is simple (as commented by @rgettman). The Sun should then be at the opposite side of the sky at $\alpha = 13^{\textrm{h}}$, where $\alpha$ is the right ascension. There is a problem, however, when you use local time (using a timezone). If, for instance, you mean midnight GMT in Liverpool, which is at 3° west then solar midnight will happen at 12 minutes past midnight GMT and the Sun would have $\alpha \approx 13^{\textrm{h}}12^\prime$.&lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2014-10-24T11:57:53.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7772" PostTypeId="2" ParentId="4725" CreationDate="2014-10-24T12:12:46.997" Score="4" Body="&lt;p&gt;This is only a rough estimate. But the number of supernovae in the Milky Way is approximately 2-5 / 100 years (&lt;a href=&quot;http://www.skyandtelescope.com/astronomy-news/milky-way-supernova-rate-confirmed/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/980309a.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;). There is approximately $3 \times 10^{11}$ stars in the Milky Way. Provided the supernovae rate did not change much through the history (which is a big if that probably doesn't work), the total number of supernovae would be $2$-$5\times 10^8$. I do not know how much supernovae will result in a black hole, but if we estimate 10% - 30%, it might not be that wrong. This would lead to $2\times 10^7$ - $1.5\times 10^8$ stellar black holes in the Milky Way. By other words, one star from 2000 - 15000 should be a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is 1000 stars in 50.9 ly around the Sun, with this density there would be one stellar black hole per 100 - 200 ly.&lt;/p&gt;&#xA;" OwnerUserId="2529" LastActivityDate="2014-10-24T12:12:46.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7773" PostTypeId="1" AcceptedAnswerId="7774" CreationDate="2014-10-24T17:36:28.013" Score="5" ViewCount="289" Body="&lt;p&gt;When you look at the sky without any utilities like binocular or telescope, what is the rough ratio of &quot;real&quot; stars to other objects like planets, the moon, asteroids, satellites?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;90% stars - 10 % others?&lt;/li&gt;&#xA;&lt;li&gt;99% stars - 1% others?&lt;/li&gt;&#xA;&lt;li&gt;Almost 100% stars - apart from a very few you can easily name as a hobby astronomer?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1838" LastEditorUserId="2365" LastEditDate="2016-09-17T06:16:58.147" LastActivityDate="2016-09-17T06:16:58.147" Title="What is the ratio of &quot;real&quot; stars in the sky?" Tags="&lt;star&gt;&lt;near-earth-object&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7774" PostTypeId="2" ParentId="7773" CreationDate="2014-10-24T18:34:23.977" Score="16" Body="&lt;p&gt;It's almost 100% stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In good conditions, you can see perhaps 2000 stars. (There are about 6000 naked-eye visible stars; of these, 3000 are above the horizon at any time, and about 1000 are hidden because they're too close to the horizon and blocked by the atmosphere.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of non-star objects you can see without assistance is tiny in comparison:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Sun (obviously not at night, and of course it's a star)&lt;/li&gt;&#xA;&lt;li&gt;The Moon&lt;/li&gt;&#xA;&lt;li&gt;The International Space Station, but only a tiny fraction of the time&lt;/li&gt;&#xA;&lt;li&gt;The visible planets: Mercury, Venus, Mars, Jupiter, Saturn (and &lt;em&gt;maybe&lt;/em&gt; Uranus if you have excellent conditions and better eyes than mine)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A handful of other artificial satellites might be visible, but only rarely. Comets can be &lt;em&gt;very&lt;/em&gt; visible, but again that's rare. Visible asteroids are even rarer. Vesta, a large asteroid, may be barely visible, but I've never seen it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that some of these objects are quite obviously not distant stars just based on their appearance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some galaxies (Andromeda, and the two Magellanic Clouds if you're far enough south) are visible, but they don't look like stars. A few of the brighter nebulas and globular clusters might be visible; the latter are groupings of stars, so I don't know how they'd count.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meteors and aircraft can be visible, but they're within the atmosphere, and probably not covered by your question.&lt;/p&gt;&#xA;" OwnerUserId="53" LastActivityDate="2014-10-24T18:34:23.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7775" PostTypeId="2" ParentId="7735" CreationDate="2014-10-25T16:37:46.330" Score="7" Body="&lt;p&gt;I've had my eye on this question for some time, but I haven't had the time to answer it. Sorry for not getting to it sooner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As zibadawa timmy said, you'd be hard-pressed to find the materials necessary for a star system if you're outside a galaxy. For one thing, there aren't any &lt;a href=&quot;https://en.wikipedia.org/wiki/Molecular_cloud&quot; rel=&quot;nofollow noreferrer&quot;&gt;stellar nurseries&lt;/a&gt; in the intergalactic void (well, none that we know of. But I'd guess it's highly unlikely that they exist plentifully). So you have no hydrogen or helium, or dust, or other elements - in fact, you have nothing that you could use. That's a wee bit of a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a way you could get around this, though - if your star system is in a &lt;a href=&quot;https://en.wikipedia.org/wiki/Globular_cluster&quot; rel=&quot;nofollow noreferrer&quot;&gt;globular cluster&lt;/a&gt;. They're groups of stars that orbit a galaxy. We don't know just how they got out there. They could be from dwarf galaxies, which then lost them, or they could have been partially chucked out of the larger galaxy but stayed in an orbit. The trouble is, as Wikipedia says,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, they are not thought to be favorable locations for the survival of planetary systems. Planetary orbits are dynamically unstable within the cores of dense clusters because of the perturbations of passing stars. A planet orbiting at 1 astronomical unit around a star that is within the core of a dense cluster such as 47 Tucanae would only survive on the order of $10^8$ years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yep. Not a great place to live.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I take it you didn't intend for our solar system to be in a globular cluster, so I'll go back into &lt;a href=&quot;https://en.wikipedia.org/wiki/Outer_space#Intergalactic_space&quot; rel=&quot;nofollow noreferrer&quot;&gt;intergalactic space&lt;/a&gt;. It's not actually a vacuum - not that most parts of outer space are. There's a lot of plasma and hot gas there (the intergalactic medium), sometimes getting to a toasty $10^5$ or $10^7 \text { K}$. Could it harbor a stellar system? That depends on just what material the star brings with it. If &lt;a href=&quot;http://www.cfa.harvard.edu/COMPLETE/learn/planets/planets.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;planets haven't formed&lt;/a&gt; around it yet, and the star has a &lt;a href=&quot;https://en.wikipedia.org/wiki/Protoplanetary_disk&quot; rel=&quot;nofollow noreferrer&quot;&gt;protplanetary disk&lt;/a&gt;, then perhaps that disk could dominate over the intergalactic medium. Remove the ICM and maybe planetary formation could proceed normally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens then? Well, there would be some interesting effects. For example, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Oort_cloud&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oort Cloud&lt;/a&gt; might not have become spherical, because it was partly deformed by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_tide&quot; rel=&quot;nofollow noreferrer&quot;&gt;galactic tide&lt;/a&gt;. Here, tidal forces probably wouldn't be as strong (although this depends in part on just how far away we are from the galaxy). We probably wouldn't have a lot of negative effects, because we don't have a lot of extreme benefits here in the Milky Way. Our galaxy does not appear to have an &lt;a href=&quot;https://en.wikipedia.org/wiki/Active_galactic_nucleus&quot; rel=&quot;nofollow noreferrer&quot;&gt;AGN&lt;/a&gt;, there aren't any monstrous black holes nearby, and we aren't undergoing any catastrophic &lt;a href=&quot;https://en.wikipedia.org/wiki/Interacting_galaxy&quot; rel=&quot;nofollow noreferrer&quot;&gt;galactic collisions&lt;/a&gt;. And the &lt;a href=&quot;https://en.wikipedia.org/wiki/Andromeda%E2%80%93Milky_Way_collision&quot; rel=&quot;nofollow noreferrer&quot;&gt;Milky Way-Andromeda collision&lt;/a&gt; &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1911/what-will-happen-to-life-on-earth-when-andromeda-and-milky-way-galaxies-collide&quot;&gt;won't be too bad&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We probably wouldn't be in too bad of a position if we were &lt;a href=&quot;https://en.wikipedia.org/wiki/Intergalactic_star&quot; rel=&quot;nofollow noreferrer&quot;&gt;in intergalactic space&lt;/a&gt;. There's not a &lt;em&gt;lot&lt;/em&gt; of benefits from being in a galaxy - well, no significant ones, at least. Well, besides star formation depositing all the materials we need for the solar system to form. That's a biggie.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2014-10-25T16:37:46.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7776" PostTypeId="2" ParentId="7752" CreationDate="2014-10-25T20:09:13.983" Score="3" Body="&lt;p&gt;The smoking guns:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The ratio of oxygen isotopes of lunar rocks are almost identical to those of Earth.&lt;/li&gt;&#xA;&lt;li&gt;Lunar recession due to the tides which causes the Earth's rotation to slow down, means that just after the Earth formed, the Moon was very close to the Earth and the Earth was rotating very fast. This situation can be reached due to an oblique impact that transfers a lot of angular momentum to the Earth.&lt;/li&gt;&#xA;&lt;li&gt;The formation of the planets is known to have proceeded via small objects growing larger by colliding with each other, this means that the last stages of planet formation would have involved collisions between large objects. It is to be expected that around the time the Earth was taking shape, there could also be a proto-planet at the Lagrange point 60 degrees away from Earth in the same orbit. On the long term this is an unstable situation which ends with a collision that according to simulations is of the sort that typically leads to the formation of the Moon.&lt;/li&gt;&#xA;&lt;li&gt;The asymmetry in the geology between the far side and the near side of the Moon. (there are many more maria on the near side compared to the far side). If the Moon had indeed been very close to the Earth after it formed, then the near side would have stayed very hot due to the Earth's surface still being molten and the radiant heat affecting the near side of the Moon. An impact on the near side of the Moon would far more easily be able to penetrate the solid crust and cause magma to flow to the surface compared to the far side of the Moon.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2185" LastActivityDate="2014-10-25T20:09:13.983" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7777" PostTypeId="1" CreationDate="2014-10-26T09:08:32.323" Score="14" ViewCount="528" Body="&lt;p&gt;CMB polarization spectrum can tell us about the primordial scalar and tensor perturbation. By analyze B and E mode angular spectrum power spectrum  and temperature power spectrum we can determine the scalar-to-tensor ratio r, as many articles implies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that r is &lt;strong&gt;defined&lt;/strong&gt; by the ration of &lt;strong&gt;amplitude&lt;/strong&gt; of primordial tensor and scalar perturbation. But what we in fact &lt;strong&gt;measure&lt;/strong&gt; is the power spectra. So I wonder how can we get the value of r from the spectra we measure thus we can study the physics of primordial perturbations and inflation.&lt;/p&gt;&#xA;" OwnerUserId="2671" LastActivityDate="2014-10-26T09:08:32.323" Title="How to determine scalar-to-tensor ratio r from CMB polarization spectrum?" Tags="&lt;cosmology&gt;&lt;cmb&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7778" PostTypeId="1" AcceptedAnswerId="7802" CreationDate="2014-10-26T15:46:07.557" Score="17" ViewCount="2476" Body="&lt;p&gt;I was inspired by &lt;a href=&quot;https://physics.stackexchange.com/questions/143166/what-is-the-theoretical-lower-mass-limit-for-a-gravitationally-stable-neutron-st&quot;&gt;this question&lt;/a&gt; on Physics, as well as &lt;a href=&quot;https://astronomy.stackexchange.com/questions/2533/can-we-theoretically-spin-the-black-hole-so-strong-that-it-will-be-broken-apar&quot;&gt;this question&lt;/a&gt; right here on Astronomy. Neutron stars are tightly bound together as neutron degenerate matter. They're very massive and have a strong gravitational field. Is it possible to break one apart into sizable chunks? How would you do this?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The answers given are good and answer my question; I'll just clarify some one thing (based on the comments) for posterity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would define &quot;broken&quot; as when any significant amount of mass is removed from the neutron star, as in mass shedding, as &lt;a href=&quot;https://astronomy.stackexchange.com/questions/7778/is-it-possible-to-break-apart-a-neutron-star/7802#7802&quot;&gt;Mitch Goshorn wrote&lt;/a&gt;. The resulting object, however, should contain a significant amount of neutron matter - that is, it should largely retain its prior composition.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2017-10-17T23:47:00.227" Title="Is it possible to break apart a neutron star?" Tags="&lt;neutron-star&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7779" PostTypeId="2" ParentId="2125" CreationDate="2014-10-26T17:29:18.823" Score="0" Body="&lt;p&gt;The what if is superficial it could never exist, but if one were to imagine then; the out come would be very bleak for Earth and Mankind alike. Jupiter has the strongest gravitational pull of any planet in our solar system. It is one of our solar system's gas giants,or Jovian planets; most of it is made of metally hydrogen, the gas succumbs to the ultimate power of Jupiter's gravity, mass and temperature. It then turns from a gas to a liquid metal. These immense elements make it impossible for such an event to ever occur, but still cool to wonder about.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/PcY0M.jpg&quot; alt=&quot;Jupiter and Io&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2672" LastEditorUserId="1891" LastEditDate="2019-05-28T16:12:39.147" LastActivityDate="2019-05-28T16:12:39.147" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="7780" PostTypeId="1" CreationDate="2014-10-27T01:03:44.510" Score="8" ViewCount="645" Body="&lt;p&gt;I have ~1000 snapshots of a spiral galaxy simulation (from Molecular cloud perturbations t=0 to t&gt;1Gyr. What I need to do is determine whether any position at any of the snapshot times meets criteria for &lt;a href=&quot;http://en.wikipedia.org/wiki/Jeans_instability&quot;&gt;Jeans instability&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simulation is 2D, and the snapshot script can read position, velocity and surface density ∑, but not temperature. The simulation also has no gas, instead the simulation was modeled by giving each individual star extra mass which corresponds to gas.&#xA;I looked online and several site state the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jeans_instability#Jeans_mass&quot;&gt;Jeans mass&lt;/a&gt; in 2d equal to c$_{s}$$^4$ / G$^2$*∑. But I have no clue how to determine speed of sound c$_{s}$.&#xA;Also I am not really interested in Jeans mass, more so whether any position can collapse to make stars, i.e. satisfy Jeans instability.&lt;/p&gt;&#xA;" OwnerUserId="2675" LastEditorUserId="2674" LastEditDate="2014-10-28T07:03:16.633" LastActivityDate="2018-09-13T10:30:43.883" Title="Python 2D Jean Instability in Spiral Galaxy Simulation" Tags="&lt;galactic-dynamics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7781" PostTypeId="1" AcceptedAnswerId="7819" CreationDate="2014-10-27T04:07:33.740" Score="5" ViewCount="261" Body="&lt;p&gt;Halo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in how would Special relativity comment on this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there are 2 objects, &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt;. They are 13 billion light years away from each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the light from the &lt;strong&gt;B&lt;/strong&gt; object reaches the &lt;strong&gt;A&lt;/strong&gt;, the observer at the &lt;strong&gt;A&lt;/strong&gt; object sees what the (suppose planet, people, objects) were like 13 billion years ago on that object. Let's say it is also an Earth-like planet where the living creatures like exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if the observer at the &lt;strong&gt;A&lt;/strong&gt; object would like to imagine, what is now happening there, what would be &lt;strong&gt;the result:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Surely, he would have to wait another 13 billion years to actually&#xA;see&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Would that mean that the inhabitants (if any) or just matter and&#xA;objects at object B exist in future relative to A ones. And vice&#xA;versa? If yes, then they both live in present, past and future&#xA;tenses at the same time?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Or have I completely lost my way of properly understanding the Special relativity?&lt;/p&gt;&#xA;" OwnerUserId="2677" LastEditorUserId="2153" LastEditDate="2014-10-27T15:48:18.660" LastActivityDate="2020-04-28T00:18:41.937" Title="How would Special relativity explain past-present-future if thinking about 13-billion-light-years distanced objects" Tags="&lt;space-time&gt;&lt;special-relativity&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7783" PostTypeId="2" ParentId="1089" CreationDate="2014-10-27T04:26:22.367" Score="-3" Body="&lt;p&gt;It has retained it's atmosphere because there is enough water (or ice) to keep it still. That's also how Earth can retain it's atmosphere. If it is a planet or moon like Venus, then the clouds and the greenhouse effect would block the atmosphere, and therefore retain the atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="2678" LastActivityDate="2014-10-27T04:26:22.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7784" PostTypeId="1" AcceptedAnswerId="7785" CreationDate="2014-10-27T09:46:57.237" Score="3" ViewCount="307" Body="&lt;p&gt;Regarding the photo of the earth &lt;a href=&quot;http://www.reddit.com/r/spaceporn/comments/sam2o/uk_and_ireland_by_night_with_the_aurora_borealis/&quot; rel=&quot;nofollow noreferrer&quot;&gt;described here&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RFjk0.jpg&quot; alt=&quot;photo in question&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the green tinge on the left hand side is to do with the aurora borealis. What's causing the bright white light on the right hand side?&lt;/p&gt;&#xA;" OwnerUserId="2679" LastActivityDate="2014-10-27T13:11:42.193" Title="What causes the bright white area in this photo of the earth?" Tags="&lt;earth&gt;&lt;photography&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7785" PostTypeId="2" ParentId="7784" CreationDate="2014-10-27T12:25:20.133" Score="4" Body="&lt;p&gt;Taken on &lt;a href=&quot;http://www.flickr.com/photos/astro_andre/6890185644/in/photostream&quot; rel=&quot;nofollow&quot;&gt;March 28 2012 by André Kuipers&lt;/a&gt;, the version &lt;a href=&quot;http://randelirious.blogspot.com/2012/04/andre-kuipers-photos.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; is labeled &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Amazing aurora lights and &lt;strong&gt;atmospheric glare&lt;/strong&gt; with the countries Ireland, Netherlands, Scotland, England and Wales in the foreground.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Off to the north and east like that, it's likely refracted early sunrise. That's a &lt;strong&gt;lot&lt;/strong&gt; of streetlights for that time of the day; likely a longish exposure to capture aurora. That'd increase sunrise glare as well.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="2031" LastEditDate="2014-10-27T13:11:42.193" LastActivityDate="2014-10-27T13:11:42.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7786" PostTypeId="2" ParentId="687" CreationDate="2014-10-27T13:10:52.723" Score="5" Body="&lt;p&gt;Perhaps after the launch of ESA's &lt;a href=&quot;http://en.wikipedia.org/wiki/PLATO_(spacecraft)&quot; rel=&quot;nofollow noreferrer&quot;&gt;PLATO space observatory&lt;/a&gt;, currently slated to be launched in 2024.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Planetary Transits and Oscillations of stars (PLATO) is a planned European Space Agency space observatory that will use a group of photometers to discover and characterize rocky extrasolar planets of all sizes around red dwarf stars, yellow dwarf stars like our Sun, and subgiant stars where water can exist in liquid state. . . . The goal is to find planets like Earth, not just in terms of their size but in their potential for habitability.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to this &lt;a href=&quot;http://www.bbc.com/news/science-environment-25911893&quot; rel=&quot;nofollow noreferrer&quot;&gt;BBC article&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Critically, Plato would be tuned to seek out rocky worlds orbiting in the &quot;habitable zone&quot; - the region around a star where water can keep a liquid state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See also Jeremy's answer to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1970/what-are-the-differences-between-tess-and-plato-exoplanet-telescopes&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2674" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-06-19T00:12:50.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7788" PostTypeId="1" AcceptedAnswerId="7813" CreationDate="2014-10-27T22:22:50.490" Score="2" ViewCount="258" Body="&lt;p&gt;Do black holes form by when the massive star dies and all the gravity falls in on itself, or something else. Please let me know, I'm really desperate!!&lt;/p&gt;&#xA;" OwnerUserId="2678" LastActivityDate="2014-10-29T17:04:53.497" Title="Black Holes, how do they form?" Tags="&lt;gravity&gt;&lt;black-hole&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7790" PostTypeId="2" ParentId="687" CreationDate="2014-10-27T22:46:42.747" Score="2" Body="&lt;p&gt;We can find Earth like Extra-Solar Planets, but they just might be too close to their star, and therefore, life might not be in the planet. We are right now living in a &quot;Goldilocks Planet&quot;. Goldilocks planets are in the habitable zone, where there is life, and H2O that is at the right temperature to be a flowing liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, a planet such as Gliese 876 d, is WAY too close to the star it orbits, (Gliese 876) Which it is highly unlikely that life exists, considering it is 377'C (710.6'F)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to see a full table of exoplanets, then go to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_nearest_terrestrial_exoplanet_candidates&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/List_of_nearest_terrestrial_exoplanet_candidates&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2678" LastEditorUserId="2678" LastEditDate="2014-10-27T23:16:30.463" LastActivityDate="2014-10-27T23:16:30.463" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7791" PostTypeId="1" AcceptedAnswerId="7792" CreationDate="2014-10-27T23:17:26.160" Score="-2" ViewCount="326" Body="&lt;p&gt;Is there a chance if any planets in our solar system survive when the milky way and Andromeda collide?&lt;/p&gt;&#xA;" OwnerUserId="756" LastActivityDate="2014-10-27T23:34:52.577" Title="What will happen to our solar system if the milky way collided with Andromeda galaxy, would any planets survive?" Tags="&lt;galaxy&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-10-28T07:04:01.327" ContentLicense="CC BY-SA 3.0" />
  <row Id="7792" PostTypeId="2" ParentId="7791" CreationDate="2014-10-27T23:34:52.577" Score="-3" Body="&lt;p&gt;Their is a very poor chance of survival for us, but we might survive for a little longer if it hits from the opposite direction form us, but then again, humans will probably be extinct  by then. Also, try to have more detail in your questions.&lt;/p&gt;&#xA;" OwnerUserId="2678" LastActivityDate="2014-10-27T23:34:52.577" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7794" PostTypeId="1" AcceptedAnswerId="7796" CreationDate="2014-10-28T03:53:00.233" Score="2" ViewCount="2238" Body="&lt;p&gt;I was interested in finding out if there was life on any planets or moons. I saw ReNiSh A R's question, and I saw the answer about there being life on Europa, but it didn't accurately answer what I wanted to find out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Drake Equation is on: &lt;a href=&quot;http://en.wikipedia.org/wiki/Drake_equation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Drake_equation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also heard that there could possibly be life in the upper atmosphere of Jupiter, which could have floating beasts, and also on Titan, which does have surface liquids, but I am pretty sure most of it is in solid form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Drake Equation is equal to a low number, then why is it pretty common if there is a possibility of life on 2 moons and 1 planet? Does that make the Drake Equation a non-reliable source?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also looking for an accurate answer, so can you please give it to me in detail!&lt;/p&gt;&#xA;" OwnerUserId="2678" LastEditorUserId="7982" LastEditDate="2021-02-19T01:26:42.403" LastActivityDate="2021-02-19T01:26:42.403" Title="Is the Drake Equation an accurate way of finding the probability of life on planets?" Tags="&lt;life&gt;&lt;drake-equation&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7795" PostTypeId="2" ParentId="6526" CreationDate="2014-10-28T04:25:11.727" Score="0" Body="&lt;p&gt;Yes, Pluto is a dwarf planet, along with Ceres and Eris which are in the Solar System. It was classified a dwarf planet in 2006 or 2007. Sorry for my inaccurate answer.&lt;/p&gt;&#xA;" OwnerUserId="2678" LastActivityDate="2014-10-28T04:25:11.727" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7796" PostTypeId="2" ParentId="7794" CreationDate="2014-10-28T08:19:53.140" Score="9" Body="&lt;p&gt;The Drake equation is little more than a randomly assembled guess of relevant factors for intelligent life, and any use of it requires a number of hard to justify guesses for the various probabilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is otherwise an attempt to give a foundation for making Fermi estimations on the number of civilizations, and the associated Fermi paradox. A Fermi estimation is an order of magnitude estimate. It doesn't try to get a particularly accurate estimate, it only tries to tell you how many digits you can expect the answer to have. And the Fermi paradox is simply stated as &quot;There is an absurd number of stars. So absurd that finding extraterrestrial life should be easy. But in fact it is hard. Why?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resolution to the paradox is the second sentence: it assumes we have reasonable basis to suppose life is easy enough for the sheer number of stars to give us lots of observable life. We do not. The Drake equation is in one sense a way of singling out as many factors as possible so we have a concrete goal to figure those factors out, and so the entire problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But mostly it is used by pop science and sensationalists to rile us up with lines like &quot;there should be thousands of civilizations more advanced than ours in our own galaxy!&quot; and &quot;there could be life in Jupiter's atmosphere.&quot; These things have no solid science behind them, and are just attention grabbers dressed in pseudoscience.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2014-10-28T08:19:53.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7797" PostTypeId="5" CreationDate="2014-10-28T10:19:03.187" Score="0" Body="&lt;p&gt;Questions about planets that do not orbit around a star. These planets travel through space. Rogue planets probably were ejected from their former parent star.&lt;/p&gt;&#xA;" OwnerUserId="31410" LastEditorUserId="31410" LastEditDate="2021-02-11T07:41:23.153" LastActivityDate="2021-02-11T07:41:23.153" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="7798" PostTypeId="4" CreationDate="2014-10-28T10:19:03.187" Score="0" Body="Questions about planets that do not orbit around a star. These planets travel through space.&#xD;&#xA;" OwnerDisplayName="user2449" LastEditorDisplayName="user2449" LastEditDate="2014-10-30T19:05:47.733" LastActivityDate="2014-10-30T19:05:47.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7799" PostTypeId="5" CreationDate="2014-10-28T10:21:03.510" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-10-28T10:21:03.510" LastActivityDate="2014-10-28T10:21:03.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7800" PostTypeId="4" CreationDate="2014-10-28T10:21:03.510" Score="0" Body="Questions about the complete or partial blocking of the sun by the moon." OwnerDisplayName="user2449" LastEditorDisplayName="user2449" LastEditDate="2014-11-02T14:25:32.353" LastActivityDate="2014-11-02T14:25:32.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7801" PostTypeId="1" AcceptedAnswerId="7803" CreationDate="2014-10-28T14:25:56.870" Score="4" ViewCount="747" Body="&lt;p&gt;Say a photon left one of the first stars ever created in our universe.  This photon has been traveling for 13+billion years through the vastness of space without ever hitting anything.  Sure, it's probably been tugged at by gravitational interactions, but it's still happily traveling at the speed of light without direct interaction with matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we know, most of the universe is composed of mostly empty space.  The odds of any single photon hitting something (like the Earth as starlight, etc) would seem to be very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What will eventually happen to this photon?  Will it eventually reach the other end of the universe, and if so, what would happen to it there?&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2017-06-23T02:46:54.967" Title="What happens to photons that don't interact with something?" Tags="&lt;universe&gt;&lt;photons&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7802" PostTypeId="2" ParentId="7778" CreationDate="2014-10-28T14:28:37.430" Score="14" Body="&lt;p&gt;It would appear theoretically possible (to some degree) through extreme applications of recycling to trigger mass shedding in pulsars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pulsars are rapidly spinning neutron stars, the fastest class of which are millisecond pulsars. The current belief is that they build up rotational speed through accretion, a process known as &lt;strong&gt;recycling&lt;/strong&gt;. One study, &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1994ApJ...423L.117C&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Recycling Pulsars to Millisecond Periods in General Relativity&lt;/a&gt; (Cook, et al), explores the limitations of this process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following chart shows their results:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EUr4Q.png&quot; alt=&quot;Plot of gravitational mass to central energy density&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At the point where the dotted lines meet the two plots, you can see a reduction in mass at those energy levels. This is due to the angular velocity of the body creating instability which results in &lt;strong&gt;mass shedding&lt;/strong&gt; - essentially mass at the equator of our neutron star being flung off the star due to the body's angular velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, this is not exactly an easy process.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The timescale to accrete the required rest mass, ~0.1 &lt;em&gt;M&lt;/em&gt;&lt;sub&gt;☉&lt;/sub&gt;, at the Eddington limit, ~10&lt;sup&gt;-8&lt;/sup&gt; &lt;em&gt;M&lt;/em&gt;&lt;sub&gt;☉&lt;/sub&gt;yr&lt;sup&gt;-1&lt;/sup&gt;, is ~10&lt;sup&gt;7&lt;/sup&gt; yr. This timescale is largely insensitive to the adopted nuclear equation of state. If other astrophysical considerations require a considerably shorter time scale, then the simple recycling scenario described here will have to be modified beyond the variations explored in this paper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Note however that the research here is actually attempting to avoid such instabilities, and they accomplish this by adding even more mass, such that the body can support even greater rotational velocity without encounter instability. Additionally, they're trying to create millisecond pulsars, but we don't need to do this as they exist naturally, so we could save ourselves a lot of time by (very carefully) approaching an existing millisecond pulsar&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think this would exactly be &lt;em&gt;breaking apart&lt;/em&gt; (despite &lt;a href=&quot;http://en.wikipedia.org/wiki/Millisecond_pulsar&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia's&lt;/a&gt; use of that exact verbiage to describe it), but it allows for the return of mass that was at one point in a neutron star.  Of course, chances are our theoretical neutron star miners are very likely to be the ones who put that mass on the neutron star to begin with. On the other hand, this (hopefully) accomplishes the task without reducing the object to a quark star or black hole.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cook, G. B.; Shapiro, S. L.; Teukolsky, S. A. (1994). &quot;Recycling Pulsars to Millisecond Periods in General Relativity&quot;. Astrophysical Journal Letters 423: 117–120.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="1288" LastEditDate="2017-10-17T20:46:34.857" LastActivityDate="2017-10-17T20:46:34.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7803" PostTypeId="2" ParentId="7801" CreationDate="2014-10-28T15:45:41.153" Score="2" Body="&lt;p&gt;Your statement &quot;most of the universe is composed of mostly empty space&quot; is very vague. In reality, the Universe is embedded in diffuse background photon fields, from low energy (like the so-called cosmic microwave background which is relic from the big bang) to very high energy (from the extreme compact objects, like AGNs). Of course, there is also unknown dark matter. It seems that dark matter probably only takes part in weak interaction. Thus, it probably won't interact with photons (the same argument with cosmic background neutrinos). The chance of photon-photon interaction and photon-electromagnetic-fields interactions depends on the energy of the photons. For high energies, it is not low as you expected. Even if a photon gets lucky and doesn't interact with the other particles, still it won't reach the edge of the Universe. The reason is that the Universe is expanding.  &lt;/p&gt;&#xA;" OwnerUserId="2684" LastEditorUserId="58" LastEditDate="2014-10-29T13:45:26.970" LastActivityDate="2014-10-29T13:45:26.970" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7804" PostTypeId="2" ParentId="7801" CreationDate="2014-10-28T16:01:40.797" Score="1" Body="&lt;p&gt;A photon like that will simply travel forever, since what it find around (other photons and non-baryonic matter) is hard to interact with. But since Universe is expanding, the photon frequency is decreasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at microwave cosmic background we observe today: those radio photons have been travelling for the entire Age of the Universe until they hit Earth, but they were gamma rays once and are microwaves today.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-10-28T16:01:40.797" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7805" PostTypeId="2" ParentId="7801" CreationDate="2014-10-28T17:18:13.803" Score="2" Body="&lt;p&gt;Even vacuum is seething with virtual particle pairs, so photons &lt;em&gt;always&lt;/em&gt; interact with something. Check Wikipedia on &lt;a href=&quot;http://en.wikipedia.org/wiki/Scharnhorst_effect&quot; rel=&quot;nofollow&quot;&gt;the Scharnhorst effect&lt;/a&gt; for a bit of an explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Owing to Heisenberg's uncertainty principle, an empty space which appears to be a true vacuum is actually filled with virtual subatomic particles. These are called vacuum fluctuations. As a photon travels through a vacuum it interacts with these virtual particles, and is absorbed by them to give rise to a virtual electron-positron pair. This pair is unstable, and quickly annihilates to produce a photon like the one which was previously absorbed. The time the photon's energy spends as subluminal electron-positron pairs lowers the observed speed of light in a vacuum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2014-10-28T17:18:13.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7806" PostTypeId="1" CreationDate="2014-10-28T20:24:11.027" Score="13" ViewCount="18613" Body="&lt;p&gt;Just a little disclaimer beforehand: I have never studied astronomy or any exact sciences for that matter (not even IT), so I am trying to fill this gap by self-education. Astronomy is one of the areas that has captured my attention and my idea of self-education is head on applied approach. So, straight to the point - this is orbital simulation model that I am casually working on when I have time/mood. My major goal is to create complete solar system in motion and ability to plan spacecraft launches to other planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are all free to pick this project up at any point and have fun experimenting!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;update!!! (Nov10)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;velocity is now proper deltaV and giving additional motion now calculates&#xA;sum vector of velocity &lt;/li&gt;&#xA;&lt;li&gt;you can place as many static objects as you&#xA;like, on every time unit object in motion checks for gravity vectors&#xA;from all sources (and checks for collision)&lt;/li&gt;&#xA;&lt;li&gt;greatly improved the performance of calculations&lt;/li&gt;&#xA;&lt;li&gt;a fix to account for interactive mod in matplotlib. Looks like that this is default option only for ipython. Regular python3 requires that statement explicitly. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Basically it is now possible to &quot;launch&quot; a spacecraft from the surface of the Earth and plot a mission to the Moon by making deltaV vector corrections via giveMotion(). Next in line is trying to implement global time variable to enable simultaneous motion e.g. Moon orbits Earth while spacecraft tries out a gravity assist maneuver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comments and suggestions for improvements are always welcome!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Done in Python3 with matplotlib library&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import matplotlib.pyplot as plt&#xA;import math&#xA;plt.ion()&#xA;&#xA;G = 6.673e-11  # gravity constant&#xA;gridArea = [0, 200, 0, 200]  # margins of the coordinate grid&#xA;gridScale = 1000000  # 1 unit of grid equals 1000000m or 1000km&#xA;&#xA;plt.clf()  # clear plot area&#xA;plt.axis(gridArea)  # create new coordinate grid&#xA;plt.grid(b=&quot;on&quot;)  # place grid&#xA;&#xA;class Object:&#xA;    _instances = []&#xA;    def __init__(self, name, position, radius, mass):&#xA;        self.name = name&#xA;        self.position = position&#xA;        self.radius = radius  # in grid values&#xA;        self.mass = mass&#xA;        self.placeObject()&#xA;        self.velocity = 0&#xA;        Object._instances.append(self)&#xA;&#xA;    def placeObject(self):&#xA;        drawObject = plt.Circle(self.position, radius=self.radius, fill=False, color=&quot;black&quot;)&#xA;        plt.gca().add_patch(drawObject)&#xA;        plt.show()&#xA;&#xA;    def giveMotion(self, deltaV, motionDirection, time):&#xA;        if self.velocity != 0:&#xA;            x_comp = math.sin(math.radians(self.motionDirection))*self.velocity&#xA;            y_comp = math.cos(math.radians(self.motionDirection))*self.velocity&#xA;            x_comp += math.sin(math.radians(motionDirection))*deltaV&#xA;            y_comp += math.cos(math.radians(motionDirection))*deltaV&#xA;            self.velocity = math.sqrt((x_comp**2)+(y_comp**2))&#xA;&#xA;            if x_comp &amp;gt; 0 and y_comp &amp;gt; 0:  # calculate degrees depending on the coordinate quadrant&#xA;                self.motionDirection = math.degrees(math.asin(abs(x_comp)/self.velocity))  # update motion direction&#xA;            elif x_comp &amp;gt; 0 and y_comp &amp;lt; 0:&#xA;                self.motionDirection = math.degrees(math.asin(abs(y_comp)/self.velocity)) + 90&#xA;            elif x_comp &amp;lt; 0 and y_comp &amp;lt; 0:&#xA;                self.motionDirection = math.degrees(math.asin(abs(x_comp)/self.velocity)) + 180&#xA;            else:&#xA;                self.motionDirection = math.degrees(math.asin(abs(y_comp)/self.velocity)) + 270&#xA;&#xA;        else:&#xA;            self.velocity = self.velocity + deltaV  # in m/s&#xA;            self.motionDirection = motionDirection  # degrees&#xA;        self.time = time  # in seconds&#xA;        self.vectorUpdate()&#xA;&#xA;    def vectorUpdate(self):&#xA;        self.placeObject()&#xA;        data = []&#xA;&#xA;        for t in range(self.time):&#xA;            motionForce = self.mass * self.velocity  # F = m * v&#xA;            x_net = 0&#xA;            y_net = 0&#xA;            for x in [y for y in Object._instances if y is not self]:&#xA;                distance = math.sqrt(((self.position[0]-x.position[0])**2) +&#xA;                             (self.position[1]-x.position[1])**2)&#xA;                gravityForce = G*(self.mass * x.mass)/((distance*gridScale)**2)&#xA;&#xA;                x_pos = self.position[0] - x.position[0]&#xA;                y_pos = self.position[1] - x.position[1]&#xA;&#xA;                if x_pos &amp;lt;= 0 and y_pos &amp;gt; 0:  # calculate degrees depending on the coordinate quadrant&#xA;                    gravityDirection = math.degrees(math.asin(abs(y_pos)/distance))+90&#xA;&#xA;                elif x_pos &amp;gt; 0 and y_pos &amp;gt;= 0:&#xA;                    gravityDirection = math.degrees(math.asin(abs(x_pos)/distance))+180&#xA;&#xA;                elif x_pos &amp;gt;= 0 and y_pos &amp;lt; 0:&#xA;                    gravityDirection = math.degrees(math.asin(abs(y_pos)/distance))+270&#xA;&#xA;                else:&#xA;                    gravityDirection = math.degrees(math.asin(abs(x_pos)/distance))&#xA;&#xA;                x_gF = gravityForce * math.sin(math.radians(gravityDirection))  # x component of vector&#xA;                y_gF = gravityForce * math.cos(math.radians(gravityDirection))  # y component of vector&#xA;&#xA;                x_net += x_gF&#xA;                y_net += y_gF&#xA;&#xA;            x_mF = motionForce * math.sin(math.radians(self.motionDirection))&#xA;            y_mF = motionForce * math.cos(math.radians(self.motionDirection))&#xA;            x_net += x_mF&#xA;            y_net += y_mF&#xA;            netForce = math.sqrt((x_net**2)+(y_net**2))&#xA;&#xA;            if x_net &amp;gt; 0 and y_net &amp;gt; 0:  # calculate degrees depending on the coordinate quadrant&#xA;                self.motionDirection = math.degrees(math.asin(abs(x_net)/netForce))  # update motion direction&#xA;            elif x_net &amp;gt; 0 and y_net &amp;lt; 0:&#xA;                self.motionDirection = math.degrees(math.asin(abs(y_net)/netForce)) + 90&#xA;            elif x_net &amp;lt; 0 and y_net &amp;lt; 0:&#xA;                self.motionDirection = math.degrees(math.asin(abs(x_net)/netForce)) + 180&#xA;            else:&#xA;                self.motionDirection = math.degrees(math.asin(abs(y_net)/netForce)) + 270&#xA;&#xA;            self.velocity = netForce/self.mass  # update velocity&#xA;            traveled = self.velocity/gridScale  # grid distance traveled per 1 sec&#xA;            self.position = (self.position[0] + math.sin(math.radians(self.motionDirection))*traveled,&#xA;                             self.position[1] + math.cos(math.radians(self.motionDirection))*traveled)  # update pos&#xA;            data.append([self.position[0], self.position[1]])&#xA;&#xA;            collision = 0&#xA;            for x in [y for y in Object._instances if y is not self]:&#xA;                if (self.position[0] - x.position[0])**2 + (self.position[1] - x.position[1])**2 &amp;lt;= x.radius**2:&#xA;                    collision = 1&#xA;                    break&#xA;            if collision != 0:&#xA;                print(&quot;Collision!&quot;)&#xA;                break&#xA;&#xA;        plt.plot([x[0] for x in data], [x[1] for x in data])&#xA;&#xA;Earth = Object(name=&quot;Earth&quot;, position=(50.0, 50.0), radius=6.371, mass=5.972e24)&#xA;Moon = Object(name=&quot;Moon&quot;, position=(100.0, 100.0), radius=1.737, mass = 7.347e22)  # position not to real scale&#xA;Craft = Object(name=&quot;SpaceCraft&quot;, position=(49.0, 40.0), radius=1, mass=1.0e4)&#xA;&#xA;Craft.giveMotion(deltaV=8500.0, motionDirection=100, time=130000)&#xA;Craft.giveMotion(deltaV=2000.0, motionDirection=90, time=60000)&#xA;plt.show(block=True)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How it works&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It all boils down to two things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Creating object like &lt;code&gt;Earth = Object(name=&quot;Earth&quot;, position=(50.0, 50.0), radius=6.371, mass=5.972e24)&lt;/code&gt; with parameters of position on grid (1 unit of grid is 1000km by default but this can be changed too), radius in grid units and mass in kg.&lt;/li&gt;&#xA;&lt;li&gt;Giving object some deltaV such as &lt;code&gt;Craft.giveMotion(deltaV=8500.0, motionDirection=100, time=130000)&lt;/code&gt; obviously it requires &lt;code&gt;Craft = Object(...)&lt;/code&gt; to be created in the first place as mentioned in previous point. Parameters here are &lt;code&gt;deltaV&lt;/code&gt; in m/s (note that for now acceleration is instantaneous), &lt;code&gt;motionDirection&lt;/code&gt; is direction of deltaV in degrees (from current position imagine 360 degree circle around object, so direction is a point on that circle) and finally parameter &lt;code&gt;time&lt;/code&gt; is how many seconds after the deltaV push trajectory of the object will be monitored. Subsequent &lt;code&gt;giveMotion()&lt;/code&gt; start off from last position of previous &lt;code&gt;giveMotion()&lt;/code&gt;. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is this a valid algorithm to calculate orbits?&lt;/li&gt;&#xA;&lt;li&gt;What are the obvious improvements to be made?&lt;/li&gt;&#xA;&lt;li&gt;I have been considering &quot;timeScale&quot; variable that will optimize calculations, as it might not be necessary to recalculate vectors and positions for every second. Any thoughts on how it should be implemented or is it generally a good idea? (loss of accuracy vs improved performance)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Basically my aim is to start a discussion on the topic and see where it leads. And, if possible, learn (or even better - teach) something new and interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Feel free to experiment!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try using:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Earth = Object(name=&quot;Earth&quot;, position=(50.0, 100.0), radius=6.371, mass=5.972e24)&#xA;Moon = Object(name=&quot;Moon&quot;, position=(434.0, 100.0), radius=1.737, mass = 7.347e22)&#xA;Craft = Object(name=&quot;SpaceCraft&quot;, position=(43.0, 100.0), radius=1, mass=1.0e4)&#xA;&#xA;Craft.giveMotion(deltaV=10575.0, motionDirection=180, time=322000)&#xA;Craft.giveMotion(deltaV=400.0, motionDirection=180, time=50000)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;With two burns - one prograde at Earth orbit and one retrograde at Moon orbit I achieved stable Moon orbit. Are these close to theoreticaly expected values?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suggested exercise: Try it in 3 burns - stable Earth orbit from Earth surface, prograde burn to reach Moon, retrograde burn to stabilize orbit around Moon. Then try to minimize deltaV. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I plan to update the code with extensive comments for those not familiar with python3 syntax. &lt;/p&gt;&#xA;" OwnerUserId="2685" LastEditorUserId="2685" LastEditDate="2014-11-20T18:58:03.970" LastActivityDate="2020-09-28T08:59:24.413" Title="Exercise: 2D orbital mechanics simulation (python)" Tags="&lt;orbit&gt;&lt;gravity&gt;&lt;n-body-simulations&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="7807" PostTypeId="1" AcceptedAnswerId="7809" CreationDate="2014-10-28T23:15:25.457" Score="19" ViewCount="6083" Body="&lt;p&gt;A naive question. When we look at Jupiter, we see that its features didn't change largely over many years, for instance, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Atmosphere_of_Jupiter#Great_Red_Spot&quot;&gt;red-spot&lt;/a&gt;. If it is composed of gases and liquids, then why aren't the effects of mixing of these fluids visible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intuition is that due to very low temperatures ($-145\, ^{\circ}$C), diffusion of fluids doesn't occur and therefore the superficial appearence of Jupiter remains the same.&lt;/p&gt;&#xA;" OwnerUserId="1842" LastEditorUserId="1038" LastEditDate="2014-10-29T03:49:56.210" LastActivityDate="2015-02-16T23:11:03.063" Title="If Jupiter is a gas-giant then why don't its features change?" Tags="&lt;jupiter&gt;" AnswerCount="4" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7809" PostTypeId="2" ParentId="7807" CreationDate="2014-10-29T00:07:39.073" Score="25" Body="&lt;p&gt;Believe it or not, Jupiter isn't &lt;em&gt;too&lt;/em&gt; consistent. Take a look at these pictures, &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/4/43/Jupiter_on_2009-07-23_%28captured_by_the_Hubble_Space_Telescope%29.jpg&quot; rel=&quot;noreferrer&quot;&gt;the first&lt;/a&gt; taken in 2009 and &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/1/1a/Jupiter_on_2010-06-07_%28captured_by_the_Hubble_Space_Telescope%29.jpg&quot; rel=&quot;noreferrer&quot;&gt;the second&lt;/a&gt; taken in 2010:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/4/43/Jupiter_on_2009-07-23_%28captured_by_the_Hubble_Space_Telescope%29.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/1/1a/Jupiter_on_2010-06-07_%28captured_by_the_Hubble_Space_Telescope%29.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Quite the difference, eh? Why?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Jupiter&quot; rel=&quot;noreferrer&quot;&gt;Jupiter's atmosphere&lt;/a&gt; is made of &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Jupiter#Zones.2C_belts_and_jets&quot; rel=&quot;noreferrer&quot;&gt;zones and belts&lt;/a&gt;. Zones are colder and are composed of rising gases; they are dark-colored. Belts are warmer and are composed of falling gases; they are light-colored. The reason the two don't intermix is because of constant flows of wind, similar to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Jet_stream&quot; rel=&quot;noreferrer&quot;&gt;Jetstream&lt;/a&gt;. These winds make it hard for bands to mix.&lt;/p&gt;&#xA;&lt;p&gt;There are two types of explanations for the jets. &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Jupiter#Shallow_models&quot; rel=&quot;noreferrer&quot;&gt;Shallow models&lt;/a&gt; say that the jets are caused by local disturbances. &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Jupiter#Deep_models&quot; rel=&quot;noreferrer&quot;&gt;Deep models&lt;/a&gt; say that they are the byproduct of rotating cylinders comprising the mantle. At the moment, we don't know which explanation is correct.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2014-10-29T00:07:39.073" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7810" PostTypeId="2" ParentId="7807" CreationDate="2014-10-29T07:56:40.460" Score="14" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;we see that its features didn't change largely over many years&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Jupiter is huge. It is 11 earths across, and 1300 times our volume. The clouds/bands we can see are vastly larger than Earth's entire ecosystem and that means they have a lot more inertia. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also consider our observation timespan. We have been watching Jupiter* for under 2 centuries, watching it in detail for under 60, and watching it in the above picture's detail for under 30. Given the scale of Jupiter's weather this is a &lt;em&gt;very&lt;/em&gt; short timespan and we should not expect to see major upheavals within one human lifespan. Picture a nice summer day in California, from an insect's perspective - it won't change much in 72 hours, and that's a long time for many bugs.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Galileo saw a spot of light. Cassini (the person, not the probe) didn't see all that much detail.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2690" LastActivityDate="2014-10-29T07:56:40.460" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7811" PostTypeId="1" AcceptedAnswerId="7826" CreationDate="2014-10-29T10:12:28.570" Score="10" ViewCount="1095" Body="&lt;p&gt;I got this image while checking weather data for a city in North India using  Mathematica's Wolfram Alpha query &#xA;&lt;img src=&quot;https://i.stack.imgur.com/jyybk.jpg&quot; alt=&quot;WEATHER DATA&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I noticed one feature in the graph which i could not explain .&#xA;Why does the encircled part 'A' which denotes  sunrise has slightly higher slope as compared to encircled part 'B' which denotes sunset ! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also checked the graph for summer month and the pattern was exactly opposite &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summer Month &#xA;&lt;img src=&quot;https://i.stack.imgur.com/jOR7f.jpg&quot; alt=&quot;july&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am i interpreting the data wrong or is it so that sun achieves its highest intensity during the day faster in winter months than in summers ? What could be the reason behind this ? &lt;/p&gt;&#xA;" OwnerUserId="1475" LastActivityDate="2019-02-08T16:37:15.103" Title="Why does this graph for sunlight intensity on land has a steeper slope during sunrise as compared to sunset?" Tags="&lt;earth&gt;&lt;weather&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7812" PostTypeId="2" ParentId="7794" CreationDate="2014-10-29T15:08:44.183" Score="2" Body="&lt;p&gt;I think it's important to note that finding life and finding intelligent life are 2 different things.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We might expect there to be unintelligent life on a significant number of planets, but the number of factors to create intelligent life are exponentially less plausible than just creating life.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We must also factor in the very real possibility that any life intelligent enough to discover E=MC2 will quite probably destroy themselves.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cosmic window of any intelligent life could be very small, and the chances of any 2 planets with intelligent life at the same time could be remote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although homo-sapiens have been intelligent for a few thousand years, we've only really discovered how to send our message into space in the last 50 or so years.  There is a good chance that our species will die out in the next couple of hundred years (global warming, nuclear threat, super-bacteria, etc).  250 years is not even close to a blink of an eye in cosmic terms.&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2014-10-29T15:08:44.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7813" PostTypeId="2" ParentId="7788" CreationDate="2014-10-29T17:04:53.497" Score="3" Body="&lt;p&gt;There are several major types of black hole, and each forms differently:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Quantum black holes - these aren't currently proven to exist, but many scientists think they are likely. These are microscopic black holes, which are believed to have formed in the Big Bang, though they can be created wherever there is enough energy.&lt;/li&gt;&#xA;&lt;li&gt;Stellar-mass black holes. These are the remains of dead stars, and the main type of black hole people talk about. They form when very large stars collapse to the point where even light cannot escape at the end of their lives.&lt;/li&gt;&#xA;&lt;li&gt;Supermassive black holes. These, as their name suggests, are extremely massive - with masses of millions to billions of times that of the Sun. Scientists aren't quite sure as to how these formed, but the main theory at the moment is that they started as several stellar-mass black holes, left behind by the first generation of stars, which fell into the centre of galaxies that were forming around them. As they did so, they came together, to form one large black hole. Another theory is that they started life as 'dark stars', a theoretical type of star held together by dark matter, which could be far larger than normal stars. These would have been extremely short-lived, and collapsed to create a supermassive black hole.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2702" LastActivityDate="2014-10-29T17:04:53.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7814" PostTypeId="2" ParentId="7806" CreationDate="2014-10-29T17:55:34.067" Score="11" Body="&lt;p&gt;Let the two bodies involved have masses &lt;span class=&quot;math-container&quot;&gt;$m_1, m_2$&lt;/span&gt;.  Start with Newton's second law &lt;span class=&quot;math-container&quot;&gt;$$F = ma$$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is acceleration.  The gravitational force on body 2 from body 1 is given by&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F_{21} = \frac{G m_1 m_2}{|r_{21}|^3}r_{21}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$r_{21}$&lt;/span&gt; is the relative position vector for the two bodies in question.  The force on body 1 from body two is of course &lt;span class=&quot;math-container&quot;&gt;$F_{12}=-F_{21}$&lt;/span&gt; since &lt;span class=&quot;math-container&quot;&gt;$r_{12}=-r_{21}$&lt;/span&gt;.  If we denote the positions by &lt;span class=&quot;math-container&quot;&gt;$(x_1,y_1)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$(x_2,y_2)$&lt;/span&gt;, then&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$r_{21} = \begin{pmatrix} x_1-x_2 \\ y_1-y_2 \end{pmatrix}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$|r| = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}.$$&lt;/span&gt;&#xA;Then Newton's law &lt;span class=&quot;math-container&quot;&gt;$a=F/m$&lt;/span&gt; becomes&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;x_1''(t) &amp;amp; = \frac{G m_2 (x_2-x_1)}{|r|^3} \\&#xA;y_1''(t) &amp;amp; = \frac{G m_2 (y_2-y_1)}{|r|^3} \\&#xA;x_2''(t) &amp;amp; = \frac{G m_1 (x_1-x_2)}{|r|^3} \\&#xA;y_2''(t) &amp;amp; = \frac{G m_1 (y_1-y_2)}{|r|^3}.&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Together with the initial positions and velocities, this system of ordinary differential equations (ODEs) comprises an initial value problem.  The usual approach is to write this as a first-order system of 8 equations and apply a Runge-Kutta or multistep method to solve them.&lt;/p&gt;&#xA;&lt;p&gt;If you apply something simple like forward Euler or backward Euler, you will see the Earth spiral out to infinity or in toward the sun, respectively, but that is an effect of the numerical errors.  If you use a more accurate method, like the classical 4th-order Runge-Kutta method, you'll find that it stays close to a true orbit for a while but still eventually spirals inward (as long as the time step is small enough) or goes to infinity (if the time step is too large).  The right approach is to use a symplectic method, which will keep the Earth in the correct orbit -- although its phase will still be off due to numerical errors.&lt;/p&gt;&#xA;&lt;p&gt;For the 2-body problem it's possible to derive a simpler system by basing your coordinate system around the center of mass.  But I think the formulation above is clearer if this is new to you.&lt;/p&gt;&#xA;" OwnerUserId="2696" LastEditorUserId="2696" LastEditDate="2020-09-28T08:59:24.413" LastActivityDate="2020-09-28T08:59:24.413" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="7815" PostTypeId="1" AcceptedAnswerId="7816" CreationDate="2014-10-29T18:21:05.950" Score="7" ViewCount="1101" Body="&lt;p&gt;If Jupiter is made of gas, could we fly or drive through it or would its center be too dense?&lt;/p&gt;&#xA;" OwnerUserId="2704" LastEditorUserId="53" LastEditDate="2014-10-31T00:40:36.573" LastActivityDate="2014-10-31T00:40:36.573" Title="Could we fly/drive through Jupiter?" Tags="&lt;jupiter&gt;&lt;core&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7816" PostTypeId="2" ParentId="7815" CreationDate="2014-10-29T20:05:25.743" Score="13" Body="&lt;p&gt;&lt;a href=&quot;http://spaceplace.nasa.gov/review/dr-marc-solar-system/gas-giants.html&quot;&gt;http://spaceplace.nasa.gov/review/dr-marc-solar-system/gas-giants.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We think of a gas as something very . . . well, airy. After all, air is the gas we all know and love. We breathe it and fly planes right through it with no trouble. So it makes sense to think that a gas planet must be like a big, airy cloud floating out in space.&#xA;Saturn in true color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bigger a planet becomes, the heavier is the material weighing down on its center. Think of how it feels to dive under water. If you are wearing a face mask, you notice that as you dive deeper, the mask presses harder and harder on your face. Also, your ears start feeling the pressure even at 2 or 3 meters (5 or 10 feet) below the surface. The pressure you feel on your body is due to the weight of the water above you. The deeper you go, the heavier the water above you and so the greater the pressure on your body. Even on Earth's surface, each square inch of your body experiences 14.7 pounds of pressure due to the weight of the atmosphere above you. If you could dive down to the center of Earth, the pressure on your body would be about 3.5 million times as great! The center of Jupiter is more than 11 times deeper than Earth's center and the pressure may be 50 million to 100 million times that on Earth's surface!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tremendous pressure at the center of planets causes the temperatures there to be surprisingly high. At their cores, Jupiter and Saturn are much hotter than the surface of the Sun!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strange things happen to matter under these extraordinary temperatures and pressures. Hydrogen, along with helium, is the main ingredient of Jupiter's and Saturn's atmospheres. Deep in their atmospheres, the hydrogen turns into a liquid. Deeper still, the liquid hydrogen turns into a metal!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what's at the very center of these planets? The material becomes stranger and stranger the deeper you go. Scientists do not understand the properties of matter under the extreme environments inside Jupiter and Saturn. Many different forces and laws of nature are at work, and the conditions inside these planets are very difficult to create in a laboratory here on Earth. But you can be sure that you wouldn't be able to fly through these bizarre materials! As we now know, the gas giants are much more than just gas.&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2014-10-29T20:05:25.743" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7817" PostTypeId="2" ParentId="7807" CreationDate="2014-10-29T22:35:35.187" Score="3" Body="&lt;p&gt;Jupiter is highly dynamic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nasa.gov/centers/goddard/multimedia/largest/EduVideoGallery.html&quot; rel=&quot;nofollow&quot;&gt;Jupiter Animated Video Gallery&lt;/a&gt; (be sure to look at storm merger video)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.space.com/2686-differences-spotted-jupiter-big-red-storms.html&quot; rel=&quot;nofollow&quot;&gt;Differences Spotted in Jupiter's Big Red Storms&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.bibliotecapleyades.net/ciencia/ciencia_jupiternewredspot.htm&quot; rel=&quot;nofollow&quot;&gt;Jupiter's New Red Spot&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://news.nationalgeographic.com/news/2010/05/100514-science-space-jupiter-lost-belt-great-red-spot/&quot; rel=&quot;nofollow&quot;&gt;Jupiter Loses Big Belt; Great Spot Left Hanging&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The persistence of Jupiter's Red spot is in fact a bit of a mystery.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2014-10-29T22:35:35.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7818" PostTypeId="2" ParentId="7794" CreationDate="2014-10-29T22:57:45.607" Score="3" Body="&lt;p&gt;Because the Drake equation has so many variables that have to be supplied out of whole cloth, there is no way that one can say it is accurate.  As we learn more about the cosmos, we are getting a better range of values for some of those variables though.  We know fairly accurately (read: within an order of magnitude) how many stars there are in the Milky Way.  We are getting more refined estimates of what percentage of them have planets, and it is higher than anyone would have guessed previously (the highest estimate I saw before we started using doppler effects to find planets was 10%, but now we see ~22% of surveyed stars have planets).  We also know that stars which have planets almost always have more than one.  We also seem to find life pretty much everywhere we look on earth, but have still found none beyond our planet.  Still, I think most educated guesses nowadays put this percentage in the 90-100% range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the better way to phrase this question is, is the Drake Equation a &lt;em&gt;reasonable&lt;/em&gt; way of &lt;em&gt;estimating&lt;/em&gt; the probability of finding intelligent life?  Since 1961 when it was first proposed, have we found any reason to call into question any of the assumptions Dr. Drake made?  I think it is reasonable.  I think the greatest room for modification is in the definition of &quot;technology that releases detectable signs of their existence into space&quot; and what signals we search for.  But the variable itself is broad enough to encompass whatever we might come up with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One further variable might be added though:  Percentage of likelihood that we would recognize an alien signal for what it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:  We could potentially find signs of life without receiving technologically created signals.  During stellar transits we have the opportunity to examine the spectroscopy of their atmosphere and check for methane and other organic gasses that are unlikely to have formed without life, but I think we'll have to find life on Europa (where we know there is methane) before that test will be commonly accepted.&lt;/p&gt;&#xA;" OwnerUserId="2503" LastActivityDate="2014-10-29T22:57:45.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7819" PostTypeId="2" ParentId="7781" CreationDate="2014-10-29T23:43:07.950" Score="4" Body="&lt;p&gt;First off, if Earth were point B, and you were an observer at point A looking at it with the most magnificent telescope ever imagined, you would still not see the Earth, because it didn't exist 13 billion years ago.  I assume you picked 13 billion years because it is roughly the age of the universe, so you'd see the universe as it existed then, but that doesn't get you what I think you want.  I will instead pick 4.5 billion years ago (or objects A and B 4.5 billion light years apart), because then you will have an actual Sun and Earth to look at but long before life evolved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll also wave our arms over the tracking software of your telescope being able to locate Sun and Earth in their orbits over the course of 4.5 billion years as seen from point B.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes.  You would have to wait 4.5 billion years to see the observer at point A looking back at you through his telescope.  You would now be looking at a cooling hunk of molten rock with no moon or oceans.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No.  If you waited 4.5 billion years and saw Observer A, he would have been dead for 4.5 billion years by the time you see him.  In fact, his Sun would likely be expanding to the point of burning his planet to a cinder if not engulfing it entirely.  But of course, you would have to wait another 4.5 billion years to see that.  You would be observing the photons that bounced off Observer A all that time ago and have been traveling ever since.  It is important to note that this is not him, and does not mean he still exists.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hopefully this answers your question, though it doesn't have much to do with Special Relativity, it's still good blow-your-mind stuff.  It's just speed of light stuff which we really knew and measured (albeit inaccurately) long before Einstein.&lt;/p&gt;&#xA;" OwnerUserId="2503" LastActivityDate="2014-10-29T23:43:07.950" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7820" PostTypeId="1" CreationDate="2014-10-29T23:45:11.090" Score="1" ViewCount="107" Body="&lt;p&gt;On television, I heard a scientist who believes that our Sun has a companion (brown dwarf) Sun, that is responsible for comets destroying life here on Earth every 26 million years. He proposed the theory that when this companion Sun travels through the Ort Cloud, every 26 million years, it brings billions of Comets with it into our Solar System and inevitably one crashes into the Earth. Any thoughts on this?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastEditorUserId="2153" LastEditDate="2014-10-30T00:15:22.237" LastActivityDate="2014-10-30T00:15:22.237" Title="Companion Sun the cause of comet impacts with Earth" Tags="&lt;comets&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7821" PostTypeId="2" ParentId="7820" CreationDate="2014-10-30T00:14:25.737" Score="10" Body="&lt;p&gt;Ahh, I see you've met our friend, the hypothetical star known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Nemesis_(hypothetical_star)&quot;&gt;Nemesis&lt;/a&gt;. I don't know which scientist(s) you're referring to, but the idea goes back quite a while. The original theory was created by &lt;a href=&quot;https://en.wikipedia.org/wiki/David_M._Raup&quot;&gt;Raup&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Jack_Sepkoski&quot;&gt;Sepkoski&lt;/a&gt; way back in 1984. &lt;a href=&quot;http://www.pnas.org/content/81/3/801.full.pdf+html&quot;&gt;Their paper&lt;/a&gt; analyzed mass extinctions in the past and concluded that there was a pattern. Note, though, that the paper does not specifically say that a star orbiting the Sun is the cause; it merely favors general impacts of &quot;extraterrestrial&quot; origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;However&lt;/em&gt;, two more teams analyzed the data and posited that there was a celestial body responsible. &lt;a href=&quot;http://www.nature.com/nature/journal/v308/n5961/abs/308713a0.html&quot;&gt;Whitmire and Jackson&lt;/a&gt; and &lt;a href=&quot;http://www.nature.com/nature/journal/v308/n5961/pdf/308715a0.pdf&quot;&gt;Davis, Hut, and Muller&lt;/a&gt; (Regrettably, you must pay to see the papers, which I have not done) suggested that the Sun had a companion in the far reaches of the solar system, and it was perturbing comets and other bodies, which came towards Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Honestly, I (and quite a few others) don't think there's much truth to the idea. Why? Because we haven't found any evidence of its existence. As &lt;a href=&quot;https://en.wikipedia.org/wiki/Nemesis_(hypothetical_star)#Past.2C_current.2C_and_pending_searches_for_Nemesis&quot;&gt;Wikipedia&lt;/a&gt; explains,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Searches for Nemesis in the infrared are important because cooler stars shine in infrared light. The University of California's Leuschner Observatory failed to discover Nemesis by 1986. The Infrared Astronomical Satellite (IRAS) failed to discover Nemesis in the 1980s. The 2MASS astronomical survey, which ran from 1997 to 2001, failed to detect a star, or brown dwarf, in the Solar System.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and, later on,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In particular, if Nemesis is a red dwarf star or a brown dwarf, the WISE mission (an infrared sky survey that covered most of our solar neighborhood in movement-verifying parallax measurements) was expected to be able to find it. WISE can detect 150 kelvin brown dwarfs out to 10 light-years. But the closer a brown dwarf is the easier it is to detect. Preliminary results of the WISE survey were released on April 14, 2011. On March 14, 2012, the entire catalog of the WISE mission was released. In 2014 WISE data ruled out a Saturn or larger-sized body in the Oort cloud out to ten thousand AU.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fact,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;According to NASA, &quot;recent scientific analysis no longer supports the idea that extinctions on Earth happen at regular, repeating intervals, and thus, the Nemesis hypothesis is no longer needed.&quot; And, indeed, a recent sky survey by NASA's WISE mission found no star or brown dwarf orbiting the Sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;NASA's conclusions are laid out in &lt;a href=&quot;http://www.nasa.gov/jpl/wise/planet-x-20140307/#.UyY_qdxIhwY&quot;&gt;this page&lt;/a&gt;, which talks about how WISE found no brown dwarf in the Oort Cloud . . . but found other brown dwarfs further out.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-10-30T00:14:25.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7822" PostTypeId="1" AcceptedAnswerId="7847" CreationDate="2014-10-30T00:45:34.810" Score="1" ViewCount="778" Body="&lt;p&gt;I have recently done some research into black holes, and realized that big stars form black holes, whilst smaller ones don't. Is this because the gravity isn't strong enough for it to fall in on itself, or something else?&lt;/p&gt;&#xA;" OwnerUserId="2678" LastEditorUserId="2531" LastEditDate="2017-10-27T05:52:22.897" LastActivityDate="2017-10-27T05:52:22.897" Title="Why don't small stars end up as a black holes?" Tags="&lt;star&gt;&lt;gravity&gt;&lt;black-hole&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7823" PostTypeId="1" CreationDate="2014-10-30T11:05:28.733" Score="6" ViewCount="244" Body="&lt;p&gt;So, I've googled and wiki'd on this, but no clue, so I turn my question to the community:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given Titan's nice orange haze that is supposedly composed of &lt;a href=&quot;http://en.wikipedia.org/wiki/Tholin&quot; rel=&quot;noreferrer&quot;&gt;Tholins&lt;/a&gt; and strangely coincides with its stratosphere (taken from wikipedia)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TBjvv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got to think whether the haze takes the role of clouds on earth when it comes to energetics. From this a bunch of questions arose:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Do we know whether the haze is responsible for heat trapping and thus, the formation of a stratosphere on Titan?&lt;/li&gt;&#xA;&lt;li&gt;What about radiation transport modelling with those particles, is this successful?&lt;/li&gt;&#xA;&lt;li&gt;Also I'd be interested in what we know about the stability of Tholins. Could they survive harsh irradiation that atmospheres on Hot Jupiters do encounter?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would like to be able to make a statement such as &quot;Dust&quot; or &quot;Haze-controlled&quot; energetics for the stratosphere of Titan, if this is so. It would be nice if you helped to clear this up.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="7982" LastEditDate="2021-02-28T11:55:06.790" LastActivityDate="2021-02-28T11:55:06.790" Title="Energetics of Titans Tholin haze" Tags="&lt;solar-system&gt;&lt;planetary-atmosphere&gt;&lt;atmospheric-effects&gt;&lt;titan&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7824" PostTypeId="1" CreationDate="2014-10-30T11:48:27.877" Score="3" ViewCount="2459" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Dawn_simulation&quot; rel=&quot;nofollow&quot;&gt;&quot;Dawn simulators&quot;&lt;/a&gt; are used to simulate a natural sunrise, or at least the physiological effects of one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What temporal sequence of color values (e.g., hue, saturation, brightness, and timing) corresponds to a typical spring or summer sunrise? Are the methods for calculating changes in these values based on season?&lt;/p&gt;&#xA;" OwnerUserId="2369" LastActivityDate="2014-11-02T09:51:02.013" Title="What are the color progression of a typical sunrise?" Tags="&lt;the-sun&gt;&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7825" PostTypeId="2" ParentId="7794" CreationDate="2014-10-30T13:48:19.680" Score="3" Body="&lt;p&gt;The Drake Equation is a formula for gestimating the number of intelligent extraterrestrial civilizations we might be able to detect, which will most likely be a small subset of planets in the universe that host lifeforms.&lt;/p&gt;&#xA;&lt;p&gt;Let's see if we can break this down a bit. From the Wikipedia article:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The Drake equation is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$N = R_{\ast} \cdot f_p \cdot n_e \cdot f_{\ell} \cdot f_i \cdot f_c \cdot L$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; = the number of civilizations in our galaxy with which radio-communication might be possible (i.e. which are on our current past light cone);&lt;/p&gt;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$R_{\ast}$&lt;/span&gt; = the average rate of star formation in our galaxy&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$f_p$&lt;/span&gt; = the fraction of those stars that have planets&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$n_e$&lt;/span&gt; = the average number of planets that can potentially support life per star that has planets&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$f_{\ell}$&lt;/span&gt; = the fraction of planets that could support life that actually develop life at some point&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$f_i$&lt;/span&gt; = the fraction of planets with life that actually go on to develop intelligent life (civilizations)&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$f_c$&lt;/span&gt; = the fraction of civilizations that develop a technology that releases detectable signs of their existence into space&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; = the length of time for which such civilizations release detectable signals into space&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;When &lt;a href=&quot;https://en.wikipedia.org/wiki/Frank_Drake&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dr. Frank Drake&lt;/a&gt; first proposed this formula for estimating potential &lt;a href=&quot;https://en.wikipedia.org/wiki/Search_for_extraterrestrial_intelligence&quot; rel=&quot;nofollow noreferrer&quot;&gt;SETI&lt;/a&gt; signals in 1961, very few, if any, of these factors had enough reliable data upon which to form a good estimate. In the intervening decades, astronomers have done much research on the process of &lt;a href=&quot;https://en.wikipedia.org/wiki/Star_formation&quot; rel=&quot;nofollow noreferrer&quot;&gt;star formation&lt;/a&gt;. Also in recent decades, a great number of &lt;a href=&quot;https://en.wikipedia.org/wiki/Exoplanet&quot; rel=&quot;nofollow noreferrer&quot;&gt;extrasolar planets&lt;/a&gt; have been discovered; the massive amount of data returned from the Kepler space telescope is especially helpful when estimating &lt;span class=&quot;math-container&quot;&gt;$f_p$&lt;/span&gt;. In addition, the developing theory of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Circumstellar_habitable_zone&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goldilocks zone&lt;/a&gt;, the orbital band in which an exoplanet can have water in liquid form, gives us a perspective on the third factor. Thus we are beginning to get some idea of the magnitude of the first three factors.&lt;/p&gt;&#xA;&lt;p&gt;The last four parameters, however, are entirely speculative. As we only know of one planet which supports life, Earth, and know only one intelligent civilization in the universe, our own, we have insufficient data upon which to form any reasonable conclusions. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Drake_equation#Current_estimates&quot; rel=&quot;nofollow noreferrer&quot;&gt;current estimates&lt;/a&gt; for each factor are summarized in the Wikipedia article.&lt;/p&gt;&#xA;" OwnerUserId="2674" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2014-10-31T13:20:05.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7826" PostTypeId="2" ParentId="7811" CreationDate="2014-10-30T16:13:55.693" Score="4" Body="&lt;p&gt;Too long for a comment, alas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.mesonet.org/index.php/site/about/other_measurements/#srad&quot; rel=&quot;nofollow&quot;&gt;https://www.mesonet.org/index.php/site/about/other_measurements/#srad&lt;/a&gt; notes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Solar radiation measurements in the morning and evening are sensitive&#xA;  to obstruction (e.g., trees) on the east and west horizons. A delay of&#xA;  sunrise or early arrival of sunset at a particular station may be&#xA;  explained by examining panoramic site photos available at&#xA;  &lt;a href=&quot;http://www.mesonet.org/index.php/site/sites/mesonet_sites&quot; rel=&quot;nofollow&quot;&gt;http://www.mesonet.org/index.php/site/sites/mesonet_sites&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also dig around a bit starting at &lt;a href=&quot;http://mesowest.utah.edu/cgi-bin/droman/meso_graph_climo_ndb.cgi?stn=SEAM5&amp;amp;unit=0&amp;amp;hours=24&amp;amp;day1=0&amp;amp;month1=&amp;amp;year1=2014&amp;amp;hour1=00&amp;amp;windred=&amp;amp;time=LOCAL&amp;amp;var=SOLR&amp;amp;lastyear=1&amp;amp;vnamev=Solar%20Radiation&amp;amp;stationname=SEAGULL&amp;amp;vlabel=%C2%A0W/m&quot; rel=&quot;nofollow&quot;&gt;http://mesowest.utah.edu/cgi-bin/droman/meso_graph_climo_ndb.cgi?stn=SEAM5&amp;amp;unit=0&amp;amp;hours=24&amp;amp;day1=0&amp;amp;month1=&amp;amp;year1=2014&amp;amp;hour1=00&amp;amp;windred=&amp;amp;time=LOCAL&amp;amp;var=SOLR&amp;amp;lastyear=1&amp;amp;vnamev=Solar%20Radiation&amp;amp;stationname=SEAGULL&amp;amp;vlabel=%C2%A0W/m&lt;/a&gt;*m and you should find more data.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2014-10-30T16:13:55.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7827" PostTypeId="2" ParentId="7807" CreationDate="2014-10-30T21:00:27.097" Score="2" Body="&lt;p&gt;Jupiter's features change constantly, thanks to new leaps in the field of Astronomy, i.e Hubble, Spitzer and soon to come in 2018 the James Web telescope, we have been able to see our Universe in ways never imagined.  July 1994, when more than 20 pieces of Comet P/Shoemaker-Levy 9 (SL9) plunged into Jupiter's atmosphere it caused a flash of energy bright enough to be seen 400 million miles away.  A shock wave generated by ram pressure as the meteor speeds into the planet's atmosphere heats the impacting body to a very high temperature, and as the hot object streaks through the atmosphere, it leaves behind a glowing trail of superheated atmospheric gases and vaporized meteor material, leaving dark smudges behind  that rapidly cools and fades in just a few seconds.Therefore changing its atmosphere if only briefly. For more info on Jupiter visit &lt;a href=&quot;http://ccs.infospace.com/&quot; rel=&quot;nofollow&quot;&gt;http://ccs.infospace.com/&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="2672" LastEditorUserId="2672" LastEditDate="2015-02-16T23:11:03.063" LastActivityDate="2015-02-16T23:11:03.063" CommentCount="1" CommunityOwnedDate="2014-10-30T21:00:27.097" ContentLicense="CC BY-SA 3.0" />
  <row Id="7828" PostTypeId="1" CreationDate="2014-10-31T02:27:38.763" Score="2" ViewCount="334" Body="&lt;p&gt;I've been doing some research on white holes, and there has not been enough information to provide the answer. Some websites are saying that they might exist, but it is not yet proven. So, is it true that they exist or not.&lt;/p&gt;&#xA;" OwnerUserId="2719" LastActivityDate="2015-01-02T18:35:44.407" Title="Is the white hole a myth or not?" Tags="&lt;white-hole&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7830" PostTypeId="2" ParentId="7828" CreationDate="2014-10-31T02:53:22.983" Score="4" Body="&lt;p&gt;Well, currently, we don't have a shred of evidence that they exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/White_hole&quot; rel=&quot;nofollow&quot;&gt;White holes&lt;/a&gt; are hypothetical objects predicted by certain solutions to Einstein's theory of general relativity. To be specific, they  come from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kruskal%E2%80%93Szekeres_coordinates#The_maximally_extended_Schwarzschild_solution&quot; rel=&quot;nofollow&quot;&gt;maximally extended Schwarzschild solution&lt;/a&gt;. We can deal with them a little mathematically, but nobody will really believe in objects like these unless we have experimental evidence for them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we? I quote NASA &lt;a href=&quot;http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/970301.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;White holes are VERY hypothetical.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;a href=&quot;http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/011027a.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is no observational evidence for white holes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.about.com/od/physicsutoz/g/WhiteHole.htm&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; supports this position:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Unlike black holes, there is no real astronomical evidence to suggest that white holes actually exist. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, the idea is put nicely &lt;a href=&quot;http://www.universetoday.com/76909/white-holes/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A white hole is only a concept for higher levels of thinking. No one has every observed one and no one probably ever will.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why? &lt;a href=&quot;http://curious.astro.cornell.edu/question.php?number=108&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; gives an interesting answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Once even the tiniest speck of dust enters the part of space-time which includes the black hole, the part which includes the white hole disappears.&lt;/strong&gt; The universe has been around for a long time and so even if it did start with white holes, they would have all disappeared by now. (Emphasis mine)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Essentially, white holes are unstable. Even if a particle did try to enter one from a black hole via an &lt;a href=&quot;https://en.wikipedia.org/wiki/Wormhole#Schwarzschild_wormholes&quot; rel=&quot;nofollow&quot;&gt;Einstein-Rosen&lt;/a&gt; bridge, the bridge would collapse before the particle could cross, and the particle would hit the singularity.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-10-31T02:53:22.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7833" PostTypeId="1" AcceptedAnswerId="7834" CreationDate="2014-10-31T10:46:44.020" Score="4" ViewCount="1936" Body="&lt;p&gt;Given our understanding of both black holes and white holes, what would be the outcome if they were both to suddenly collide?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Black Holes:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A black hole is a region of spacetime from which gravity prevents&#xA;  anything, including light, from escaping. The theory of general&#xA;  relativity predicts that a sufficiently compact mass will deform&#xA;  spacetime to form a black hole. The boundary of the region from which&#xA;  no escape is possible is called the event horizon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;White Holes:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In general relativity, a white hole is a hypothetical region of&#xA;  spacetime which cannot be entered from the outside, although matter&#xA;  and light can escape from it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="2722" LastEditDate="2014-11-04T16:49:58.740" LastActivityDate="2018-04-30T14:04:10.187" Title="What would happen if a Black Hole and White Hole Collided?" Tags="&lt;black-hole&gt;&lt;white-hole&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7834" PostTypeId="2" ParentId="7833" CreationDate="2014-10-31T11:50:13.163" Score="3" Body="&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/White_hole&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In quantum mechanics, the black hole emits Hawking radiation and so can come to thermal equilibrium with a gas of radiation. Because a thermal-equilibrium state is time-reversal-invariant, Stephen Hawking argued that the time reverse of a black hole in thermal equilibrium is again a black hole in thermal equilibrium. This implies that black holes and white holes are the same object&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The excerpt above was an argument made by Stephen Hawking, explaining that a white hole and a black hole are essentially the same object. With this in mind, we can assume that the collision will be between two black holes. The result will then be the radius of the black hole added to the radius of the white hole.&lt;/p&gt;&#xA;" OwnerUserId="2212" LastActivityDate="2014-10-31T11:50:13.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7836" PostTypeId="1" AcceptedAnswerId="7838" CreationDate="2014-10-31T14:40:56.217" Score="5" ViewCount="179" Body="&lt;p&gt;I understand that there are magnificent tidal forces that come with getting too close to Jupiter, but what is causing comets like Shoemaker-Levy to get torn apart?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't expect that the gravitational forces are so strong that spaghetification occurs, is it?&lt;/p&gt;&#xA;" OwnerUserId="2439" LastEditorUserId="2424" LastEditDate="2014-10-31T15:33:00.217" LastActivityDate="2014-10-31T15:41:32.287" Title="Why does Jupiter break comets apart?" Tags="&lt;asteroids&gt;&lt;comets&gt;&lt;jupiter&gt;&lt;tidal-forces&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7837" PostTypeId="2" ParentId="7778" CreationDate="2014-10-31T15:14:40.517" Score="8" Body="&lt;p&gt;One problem is if you split a NS into too small pieces, the gravity of each piece becomes weaker, and with the pressure reduced, it may no longer be highly degenerate.  The minimum neutron star mass is roughly 0.1 M$_\odot$ but this is for a totally cold NS, as described in &lt;a href=&quot;https://physics.stackexchange.com/questions/143166/what-is-the-theoretical-lower-mass-limit-for-a-gravitationally-stable-neutron-st&quot;&gt;this physics stackexchange answer&lt;/a&gt;. The neutrons will then decay into protons and electrons in 886 seconds and you get new entirely hydrogen stars.  Perhaps, at the high end of NS mass range it can be split into two NS, but one needs to do the calculation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem is that a NS is not a solid in the interior, so the concept of cleaving simply does not apply. The very center is gas-like and the outer core is liquid-like.  So, you can't very well cut it with a knife no matter how sharp; just as you cannot cleave a star.  So, while a high energy relativistic beam could cut through the solid crust, the rest of NS would heal instantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem is that NS is the densest material we know of, therefore to do any damage to it one would need a more dense NS (ie a more massive one).  But, if one tries to smash it or press it with a NS, the two will merge into a more massive NS which might then collapse into a black hole if the mass threshold is reached.  There might be a few chunks of debris that get away, but again these would instantly become hydrogen gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I conclude that the answer to this question is that it cannot be done with anything known today. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a simple way to totally undo a NS though.  The process of creating a NS is a reversible process. That is, if you simply heat a NS sufficiently, it becomes non-degenerate.  Eventually the neutrons decay and it becomes a hydrogen star. &lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2014-11-07T19:18:17.693" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7838" PostTypeId="2" ParentId="7836" CreationDate="2014-10-31T15:28:57.360" Score="3" Body="&lt;p&gt;It is.  Tidal disruption occurs when $\rho_{Jup}(R_{Jup}/R)^3$ &gt; $\rho_{comet}$ where R is the separation and $\rho$ is the density.  The density of Jupiter is 1.33 g/$cm^3$ and the comet was estimated to have a density of only 0.5 g/$cm^3$ based on the radius at which it disrupted.  That is, disruption took place at separation R ~ $(1.33/0.5)^{1/3} R_{Jup}$.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2014-10-31T15:41:32.287" LastActivityDate="2014-10-31T15:41:32.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7839" PostTypeId="1" CreationDate="2014-10-31T16:32:38.930" Score="3" ViewCount="4236" Body="&lt;p&gt;What is the frequency of meteor strikes on the moon per day?&lt;/p&gt;&#xA;" OwnerUserId="2727" LastActivityDate="2015-11-06T15:50:50.430" Title="How many meteors hit the moon every day or in how many days does a new meteor hit the moon?" Tags="&lt;the-moon&gt;&lt;meteor&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7840" PostTypeId="2" ParentId="7839" CreationDate="2014-10-31T17:16:46.200" Score="4" Body="&lt;p&gt;We generally like you to check google before posting questions here.  I posed your question to google and amongst other links, I got these:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2006/13jun_lunarsporadic/&quot; rel=&quot;nofollow&quot;&gt;http://science.nasa.gov/science-news/science-at-nasa/2006/13jun_lunarsporadic/&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2014/02/24/lunar_impact_video_of_an_asteroid_hitting_the_moon.html&quot; rel=&quot;nofollow&quot;&gt;http://www.slate.com/blogs/bad_astronomy/2014/02/24/lunar_impact_video_of_an_asteroid_hitting_the_moon.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per Bill cook in the top link:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;No one knows exactly how many meteoroids hit the Moon every day. By monitoring the flashes, we can learn how often and how hard the Moon gets hit.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And in the second one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Moon is smaller and has less gravity than Earth, so it gets hit less often than we do.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Both articles mention MIDAS (the Moon Impacts Detection and Analysis System).  Since the first link is from 2006, perhaps soon we will have a good estimate for you.&lt;/p&gt;&#xA;" OwnerUserId="2503" LastActivityDate="2014-10-31T17:16:46.200" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7841" PostTypeId="1" CreationDate="2014-10-31T17:51:31.457" Score="4" ViewCount="780" Body="&lt;p&gt;I recently started trying to understand the theory of relativity and I have the following question.  Because earth is orbiting the sun at approx 67,000 mph (and its my understanding time slows down the faster you are moving) what are the effect of time on earth compared to a completely stationary observer watching earth's orbit, from outer space?  How would the twin paradox work with one twin on earth and the other twin completely stationary in outer space?&lt;/p&gt;&#xA;" OwnerUserId="2728" LastActivityDate="2014-12-02T08:22:37.773" Title="time on earth compared to completely stationary in space" Tags="&lt;space-time&gt;&lt;special-relativity&gt;" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="7842" PostTypeId="2" ParentId="6570" CreationDate="2014-10-31T18:02:44.737" Score="1" Body="&lt;p&gt;Quantum gravity could be the reason protons overcome the coulomb force and bond at the nucleus of an atom (strong force).  If you extrapolate the mass energy of the plank units that fill up the volume of the proton nucleus it could satisfy the Schwarzschild condition, which is indeed, relatively quite a bit of gravity.&lt;/p&gt;&#xA;" OwnerUserId="2729" LastActivityDate="2014-10-31T18:02:44.737" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="7843" PostTypeId="2" ParentId="7828" CreationDate="2014-10-31T23:17:56.930" Score="4" Body="&lt;p&gt;A white hole is simply the time-reversed version of a black hole. For any physical situation, its time-reversal is, formally speaking, possible. It does not mean that it is possible in any physically reasonable sense of the term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you release a macroscopic amount of gas in the corner of a large vacuum container, it will spread throughout. There is a physical state that's the exact time-reverse of this, with a gas spread through spontaneously arranging itself in a corner. Yet it does not happen. It's so overwhelmingly unlikely that one can say it is impossible. The reason it doesn't happen is the second law of thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the connection between black hole dynamics and thermodynamics demonstrated by Hawking radiation and similar insights, white holes are are as impossible as cracked eggs spontaneously re-assembling themselves. Or rather, &lt;em&gt;more&lt;/em&gt; impossible, because black holes are the most entropic objects in the universe, and so require a much larger violation of the second law.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-10-31T23:17:56.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7844" PostTypeId="1" AcceptedAnswerId="8121" CreationDate="2014-11-01T00:49:30.557" Score="12" ViewCount="312" Body="&lt;p&gt;In reading the European Space Organisation paper &lt;a href=&quot;http://www.eso.org/public/archives/announcements/pdf/ann1046.pdf&quot;&gt;Lithium depletion in solar-like stars: no planet connection&lt;/a&gt; (Baumann et al. 2010), several conclusions based on their observations include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Lithium depletion in Sun-like stars is independent of having planets (an earlier theory), but rather is a function of the star's age.  From the article:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For solar-like stars, the lithium vs age trends for planet-hosts&#xA;  and stars where no planets have been found are statistically&#xA;  identical. Thus, the presence of a planet does not influence&#xA;  the observed surface lithium abundance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, they observed that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A number of solar-like stars with unusually high lithium&#xA;  abundance for their age are present in the field.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and that these stars also have an anomalously low surface gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the current theory as to why some Sun-like stars have a relatively high lithium abundance and low surface gravity?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user2449" LastActivityDate="2014-12-03T18:53:18.193" Title="What is the reason for high lithium concentrations in Sun-like stars?" Tags="&lt;stellar-evolution&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7846" PostTypeId="2" ParentId="7841" CreationDate="2014-11-01T18:00:21.833" Score="1" Body="&lt;p&gt;Relativity (special and general) is founded on the theory - which has been well tested - that there is no fundamental reference frame in space and hence there can be no &quot;completely stationary&quot; observers at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All we can say is that observers move relative to one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time dilation effect of the Earth's motion at 67,000 mph &lt;em&gt;relative&lt;/em&gt; to another observer is actually very small, being inversely proportional to the square root of 1 minus the relative velocity squared divided by the speed of light squared - essentially this is a ratio of about 1.000000005 - a second in every seven years or so.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2014-11-01T18:00:21.833" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7847" PostTypeId="2" ParentId="7822" CreationDate="2014-11-01T23:09:34.200" Score="-1" Body="&lt;p&gt;Gravitational collapse occurs when an object's internal pressure is insufficient to resist the object's own gravity. In the cases of stars, it normally usually happens because of one of these 2 reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The star has too little &quot;fuel&quot; left to maintain its temperature &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The star that would have been stable receives extra matter in a way that does not raise its core temperature.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In either of these cases, the star's temperature is no longer high enough to prevent it from collapsing under its own weight (gravity). The collapse may be stopped by various factors condensing the matter into a denser state. The result is one of the various types of compact star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As said above, it needs a lot of mass to actually collapse into either a black hole, or implode as a supernova, and smaller stars simply don't have this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Black_hole#Formation_and_evolution&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="2732" LastActivityDate="2014-11-01T23:09:34.200" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7849" PostTypeId="2" ParentId="7841" CreationDate="2014-11-02T08:15:45.013" Score="2" Body="&lt;p&gt;Time dilation is a &lt;em&gt;relative&lt;/em&gt; phenomenon, not a local phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As adrianmcmenamin mentioned, Earth's motion around the sun is cause for no more time dilation than any other point in the universe. Consider that the sun is travelling around the galaxy at 200 km/s, and the Milky Way is approaching Andromeda at 130 km/s. We do not know at what speed we are approaching the great attractor, and I hope that at this point it becomes obvious that there is no difference in &quot;us approaching the great attractor&quot; to &quot;the great attractor approaching us&quot;. &lt;strong&gt;Nothing is considered &quot;stationary to the background&quot; in an infinite, expanding universe as there is no &quot;background&quot; to measure against.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the gedankenexperiment of comparing a clock on Earth to a clock orbiting the galaxy. Now compare those clocks to clocks outside the galaxy. You might consider the clock that ticks the most time away to be the &quot;most at rest clock&quot;. &lt;strong&gt;Now I ask you, how will you compare the clocks?&lt;/strong&gt; You must move at a significant portion of the speed of light for a very long time to get information from one clock to the next. Thus, the messenger would suffer the effects of Special Relativity, rendering the comparison meaningless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To further the point, consider the development of &lt;a href=&quot;http://en.wikipedia.org/wiki/International_Atomic_Time&quot; rel=&quot;nofollow&quot;&gt;International Atomic Time&lt;/a&gt;. I'll quote directly from wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the 1970s, it became clear that the clocks participating in TAI were ticking at different rates due to gravitational time dilation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Consider, those were clocks &lt;em&gt;on Earth&lt;/em&gt; suffering from measurable time dilation due only to their latitude and altitude differences and the effect of moving information across the distances between them!&lt;/p&gt;&#xA;" OwnerUserId="209" LastActivityDate="2014-11-02T08:15:45.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7850" PostTypeId="2" ParentId="7824" CreationDate="2014-11-02T09:51:02.013" Score="2" Body="&lt;p&gt;&lt;em&gt;These steps are in there for credit&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used Stellarium to simulate a sunrise, and took screenshots of the progression. Each screen shot was 2 seconds apart, and stellarium was accelerated to 10 minutes a second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, using &lt;a href=&quot;http://www.imagemagick.org/script/index.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;imagemagik&lt;/a&gt; to crop batch those 13 images with this command:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;mogrify -crop 1280 1024 *.png&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(those numbers are the size of my screen. I cropped to get rid of my second screen)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next I put the images into &lt;a href=&quot;http://lokeshdhakar.com/projects/color-thief/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Color Thief&lt;/a&gt; and got the pallet of each image. This is the result of those 13 images:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ynjc6.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0ETHd.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xYZ46.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hoYbs.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/U81BG.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QcgiU.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kpTBE.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mvqfi.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vycg3.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SAIRI.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MdWa8.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nesvm.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gqbAg.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those are the colour pallets. You can pick whatever colours you want from them with a colour picker, such as KColorChooser.&lt;/p&gt;&#xA;" OwnerUserId="2732" LastActivityDate="2014-11-02T09:51:02.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7851" PostTypeId="1" AcceptedAnswerId="7852" CreationDate="2014-11-02T12:39:47.957" Score="3" ViewCount="115" Body="&lt;p&gt;As I understand it, gravity cannot be attenuated by any medium (in the way that EM radiation can be, for instance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this, then, not make it a source (theoretically, I am not talking of practicalities) of infinite energy - if we assume the universe itself is infinite and looks the same everywhere (ie there are objects with mass everywhere)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it more correct to say that gravity has a net contribution of nothing to the universe's energy density because the attractional energy is balanced by a negative potential energy?&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2014-11-02T17:54:29.837" Title="Is gravity a source of infinite energy at a cosmological scale" Tags="&lt;gravity&gt;&lt;cosmology&gt;&lt;general-relativity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7852" PostTypeId="2" ParentId="7851" CreationDate="2014-11-02T13:11:47.360" Score="1" Body="&lt;p&gt;Gravity travels at light speed (or less, possibly), so even in an infinite non-expanding universe of finite age you'd only be gravitationally interacting with a finite mass in a finite volume. Our universe is observed to be expanding, further inhibiting us from coming into contact with new objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, a common assumption of cosmological models is that the universe is homogeneous and isotropic on large scales. Meaning it is pretty much identical in all directions. Observations support this assumption. One net effect of this is that the gravitational potential energy you experience from objects outside (roughly) your supercluster is zero (and really close to it from things outside your own galaxy, actually). They just balance each other out.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2014-11-02T13:11:47.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7853" PostTypeId="1" AcceptedAnswerId="7855" CreationDate="2014-11-02T14:52:56.413" Score="7" ViewCount="584" Body="&lt;p&gt;Is it possible for a gas giant (A very big one) to have an other smaller gas planet as satellite?&lt;/p&gt;&#xA;" OwnerUserId="2734" LastActivityDate="2014-11-02T17:53:11.110" Title="Can a gas giant have an other gas planet as satellite?" Tags="&lt;planet&gt;&lt;natural-satellites&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7854" PostTypeId="2" ParentId="7853" CreationDate="2014-11-02T17:00:06.850" Score="2" Body="&lt;p&gt;Video games think so: &lt;a href=&quot;http://en.spaceengine.org/forum/10-1762-1&quot; rel=&quot;nofollow&quot;&gt;http://en.spaceengine.org/forum/10-1762-1&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, we know this can work for stars in a stable way: there are ternary stars where a close binary orbits a large, distant third sun. So replace the binary stars by gas giants, and there you go. And the Pluto-Charon system tells us that similarly sized bodies can orbit each other around a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the physics does not preclude the configuration in and of itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are the usual problems then with likelihood: gravitational capture is hard, and planet formation is generally expected to be noticeably different from star formation. The latter we can maybe get around: the gas giants are nearly to the brown dwarf stage, so that star formation mechanisms are more applicable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the Selected Examples &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Roche_limit&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, Jupiter has density comparable to the Sun, so I would expect this to mean we can scale the binary part of the stellar system to gas giant levels without making the system any more unstable than its stellar counterpart. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A precise knowledge of the relevant physics to assert that confidently, and to know if the stability could last for millions or billions of years, is currently beyond me.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2014-11-02T17:00:06.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7855" PostTypeId="2" ParentId="7853" CreationDate="2014-11-02T17:53:11.110" Score="3" Body="&lt;p&gt;It would seem so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about &lt;a href=&quot;https://en.wikipedia.org/wiki/Brown_dwarf&quot; rel=&quot;nofollow&quot;&gt;brown dwarfs&lt;/a&gt;. At the lower end of the mass spectrum, they're only a couple dozen times that mass of Jupiter, and only a couple times the mass of &lt;a href=&quot;https://en.wikipedia.org/wiki/Hot_Jupiter&quot; rel=&quot;nofollow&quot;&gt;Hot Jupiters&lt;/a&gt;. Some brown dwarfs have been found to have &lt;a href=&quot;https://en.wikipedia.org/wiki/Brown_dwarf#Planets_around_brown_dwarfs&quot; rel=&quot;nofollow&quot;&gt;planets&lt;/a&gt;. Taking a look at some examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/2M1207&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;2M1207:&lt;/strong&gt;&lt;/a&gt; Roughly 25 Jupiter masses, this brown dwarf has a planetary-mass object orbiting it, &lt;a href=&quot;https://en.wikipedia.org/wiki/2M1207b&quot; rel=&quot;nofollow&quot;&gt;2M1207b&lt;/a&gt;. This object has between 3 and 10 times that mass of Jupiter, and is fairly hot, although it orbits far from the star.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/2MASS_J04414489%2B2301513&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;2MASS J04414489+2301513:&lt;/strong&gt;&lt;/a&gt; Roughly 20 Jupiter masses, with a companion on the scale of 5 to 10 Jupiter masses that could be a planet.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that these brown dwarfs are very low-mass, only a bit more massive than hot Jupiters. They can have orbiting gas giants - why couldn't high-mass hot Jupiters have the same?&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-11-02T17:53:11.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7856" PostTypeId="2" ParentId="7851" CreationDate="2014-11-02T17:54:29.837" Score="1" Body="&lt;p&gt;Gravity plus dark energy reportedly can &lt;a href=&quot;http://www.sciencedaily.com/releases/2010/06/100613212708.htm&quot; rel=&quot;nofollow&quot;&gt;combine to do awful things to the amount of energy&lt;/a&gt; in the universe:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If dark energy does exist, then it ultimately causes the expansion of the Universe to accelerate. On their journey from the CMB to the telescopes like WMAP, photons (the basic particles of electromagnetic radiation including light and radio waves) travel through giant superclusters of galaxies. Normally a CMB photon is first blueshifted (its peak shifts towards the blue end of the spectrum) when it enters the supercluster and then redshifted as it leaves, so that the two effects cancel. However, if the supercluster galaxies are accelerating away from each other because of dark energy, the cancellation is not exact, so photons stay slightly blueshifted after their passage. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A blueshifted photon is a more energetic photon, and nothing seems to be losing energy to make up the difference. Article &lt;em&gt;is&lt;/em&gt; from 2010 though. This may all be sorted out by now.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2014-11-02T17:54:29.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7857" PostTypeId="1" CreationDate="2014-11-02T19:49:22.183" Score="3" ViewCount="98" Body="&lt;p&gt;I have asked this on Physics Stack Exchange but Astronomy is probably more appropriate. I have been given the task of synthesising an elliptical galaxy, a starburst galaxy and a spiral galaxy (considering the arms and the bulge separately and adding them) using stellar spectra which I have classified. I need to add different weightings of five types of stars to produce a normalised spectrum of each of these galaxies. I have spectra for giants, main sequence stars and sub-giants with varying stellar types (O, B, A, F, G, K, M). I know that Spiral arms and the starburst are going to be primarily young stars, while the spiral bulge and the elliptical will be of older stars, but I have no idea which types or luminosity classes of stars I would use and their weightings in each. Can anyone offer any information or resources where I would find good information on adding different types of star to 'create' a galaxy?&lt;/p&gt;&#xA;" OwnerUserId="2736" LastActivityDate="2014-11-02T19:49:22.183" Title="Synthesising types of galaxies using various stellar spectra" Tags="&lt;galaxy&gt;&lt;star-systems&gt;&lt;classification&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7858" PostTypeId="1" CreationDate="2014-11-02T20:31:04.687" Score="4" ViewCount="76" Body="&lt;p&gt;What is the most accurate method for SMBH mass estimation? In the literature I could find 5 methods : (1) &lt;a href=&quot;http://adsabs.harvard.edu/abs/2003ApJ...589L..21M&quot; rel=&quot;noreferrer&quot;&gt;Marconi &amp;amp; Hunt 2003&lt;/a&gt;, (2) Hu 2008, (3) &lt;a href=&quot;http://adsabs.harvard.edu/abs/2009ApJ...698..198G&quot; rel=&quot;noreferrer&quot;&gt;Gultekin et al. 2009&lt;/a&gt;, (4) Graham 2007, (5) &lt;a href=&quot;http://adsabs.harvard.edu/abs/2002MNRAS.331..795M&quot; rel=&quot;noreferrer&quot;&gt;McLure &amp;amp; Dunlop 2002&lt;/a&gt;. Please tell me which is more accurate? Also, can you suggest other methods and concern references? Thank you.    &lt;/p&gt;&#xA;" OwnerUserId="2737" LastEditorUserId="2674" LastEditDate="2014-11-08T21:18:49.563" LastActivityDate="2014-11-08T21:18:49.563" Title="Super massive black hole mass estimation in literature" Tags="&lt;supermassive-black-hole&gt;&lt;mass&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7859" PostTypeId="1" AcceptedAnswerId="7862" CreationDate="2014-11-02T23:16:52.090" Score="4" ViewCount="2812" Body="&lt;p&gt;I was listening to Skeptics Guide to the Universe the other day, and Steve Novella mentioned that the escape velocity for the galaxy is 544 km/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an object fails to reach escape velocity on Earth, it will return to Earth.  What will be the fate of an object, say the Voyager spacecraft which is currently traveling at 17 km/s, that fail to meet the 544 km/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having a hard time wrapping my head around what galactic escape velocity means.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that Voyager is incrementally slowing down by the pull of the galaxy?  Given enough time, will it end up in the supermassive black hole in the center?&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2014-12-23T11:35:43.057" Title="What happens if you fail to reach escape velocity for the galaxy?" Tags="&lt;galaxy&gt;&lt;escape-velocity&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7860" PostTypeId="2" ParentId="7859" CreationDate="2014-11-03T00:18:36.140" Score="2" Body="&lt;p&gt;Voyager is traveling at 17 km/s with respect to the sun. Since the sun is orbiting the galaxy, it is likely that Voyager will also orbit the galaxy for 5 billion years or so, until the collision with Andromeda happens.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-11-03T00:18:36.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7861" PostTypeId="1" AcceptedAnswerId="7875" CreationDate="2014-11-03T17:45:02.097" Score="7" ViewCount="819" Body="&lt;p&gt;One of the ways they measure the (supposed?) supermassive black hole at the galactic center of the milky way is to measure those tens of stars right at the galactic center that are orbiting what appears to be nothing, but what is said to be a supermassive black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why there is a distinction between all other stars in the galaxy orbiting (for instance, our own sun) and those few right near the galactic center orbiting.  My point is, if you ask what our solar system orbits, an astronomer wouldn't say &quot;the supermassive black hole at the center,&quot; he would say &quot;the galactic center of mass.&quot;  My assumption is that you couldn't even calculate the mass of the supermassive black hole by knowing our sun's mass and orbit - people would probably say we are too far out, or we are orbiting the galactic center of mass rather than the black hole, yadda yadda.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Why the distinction?  How do we know that the stars right at the galactic center aren't also orbiting the galactic center of mass vs. the supermassive black hole?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What would happen if you magically removed the supermassive black hole in an instant?  Would the stars right at the center continue to orbit the center?  If so, how is that different than what they were originally doing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I know that center of mass of the galaxy and the supermassive black hole are two different things.  I also know that the center of mass of the galaxy is either superimposed on the supermassive black hole, or quite close to it.  I just want to know how or why some objects feel the center of mass of the galaxy and respond to it, while others feel the pull of the supermassive black hole and respond to it.  How far away do you have to be to be considered &quot;orbiting the center of mass&quot; rather than &quot;orbiting the supermassive black hole?&quot;&lt;/p&gt;&#xA;" OwnerUserId="2740" LastActivityDate="2015-02-26T09:23:20.057" Title="Orbiting supermassive black hole or galactic center of mass?" Tags="&lt;orbit&gt;&lt;supermassive-black-hole&gt;&lt;galaxy-center&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7862" PostTypeId="2" ParentId="7859" CreationDate="2014-11-03T17:57:50.527" Score="9" Body="&lt;p&gt;If you fail to reach escape velocity from the earth, you won't break free of its gravitational well, in other words, you will keep &lt;em&gt;orbiting&lt;/em&gt; the center of mass of the earth. If the path of your orbit intersects the surface of the earth, then you will, you know, crash.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you fail to reach escape velocity for the galaxy, you'll keep orbiting the center of mass of the galaxy, just like the other couple hundred billions of stars in the galaxy. The galaxy has no hard surface to crash into, so you'll keep orbiting until you come close enough to another orbiting body to perturb your orbit (which could, in the right circumstances, leave you with escape velocity and kick you out of the galaxy.&lt;/p&gt;&#xA;" OwnerUserId="366" LastActivityDate="2014-11-03T17:57:50.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7863" PostTypeId="1" CreationDate="2014-11-03T18:23:23.073" Score="33" ViewCount="9403" Body="&lt;p&gt;Questions I've often wondered about:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If all the matter and energy were concentrated at a single point at the big bang, why wasn't that a black hole, or why didn't it form one?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the reason #1 above didn't form a black hole is one of several explanations such as inflation or whatever else, then why didn't all the mass and energy form a big black hole at some finite time &lt;em&gt;after&lt;/em&gt; this big bang happened?  For instance, I've (possibly incorrectly) heard that inflation made the universe the size of an orange.  Well why didn't it form a black hole then?  Or once the universe expanded to, say, the size of the moon. Why not then? Just insert whatever reasonable size you want in place of &amp;quot;orange&amp;quot; or &amp;quot;moon.&amp;quot;  The question is why didn't a black hole form out of all the matter and energy after the Big Bang?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2740" LastEditorUserId="1891" LastEditDate="2020-10-23T09:35:50.210" LastActivityDate="2020-12-10T20:29:29.793" Title="Why did the big bang not just produce a big black hole?" Tags="&lt;black-hole&gt;&lt;big-bang-theory&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="7865" PostTypeId="2" ParentId="7863" CreationDate="2014-11-03T19:28:36.253" Score="27" Body="&lt;p&gt;Your problem essentially arises from trying to apply Schwarzschild black hole logic, the assumptions of which are pretty much maximally violated at the big bang.&lt;/p&gt;&#xA;&lt;p&gt;The following were true at the big bang, and violate the usual black hole formation logic.&lt;/p&gt;&#xA;&lt;p&gt;The event occurred everywhere in space, not actually a point. In particular, the energy was uniformly distributed everywhere. The net gravitational potential was therefore near zero, and there was no one point to which everything could collapse. Furthermore, since stuff was everywhere, there was no expanse of vacuum (in a flat spacetime, no less) outside the collapsing region. And furthermore, things were moving rapidly, were in a highly excited state, and were not in thermal equilibrium (until inflation hit, and then things were too diluted and causally disconnected to collapse en masse).&lt;/p&gt;&#xA;&lt;p&gt;As for what we could describe the universe as at time &lt;span class=&quot;math-container&quot;&gt;$t=0$&lt;/span&gt;, we have no idea. General relativity has a spacelike singularity there, and it subsequently cannot say anything beyond that. It is expected that general relativity is not a correct description of spacetime in the pre-inflation era, in large part because quantum mechanics introduces significant effects in those conditions, and it is well known that the two theories are incompatible.&lt;/p&gt;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&lt;p&gt;This question has been asked on the physics SE many times.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/q/20394/55483&quot;&gt;https://physics.stackexchange.com/q/20394/55483&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/q/3294/55483&quot;&gt;https://physics.stackexchange.com/q/3294/55483&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/q/26435/55483&quot;&gt;https://physics.stackexchange.com/q/26435/55483&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Perhaps the answers there will be illuminating.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastEditorUserId="31410" LastEditDate="2020-12-10T20:29:29.793" LastActivityDate="2020-12-10T20:29:29.793" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="7866" PostTypeId="1" CreationDate="2014-11-03T22:59:07.227" Score="2" ViewCount="276" Body="&lt;p&gt;Whenever we hear an explanation about the big bang, it is always phrased in such a way that it was an explosion which used some kind of pressure to expanded the universe out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder, however, would it be possible that our universe bubble started as a point that wrapped around the singularity and contained it.  Then, for some reason, this bubble started to expand, which caused the energy in the singularity to fill the newly created empty space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it be that this bubble, which is now our universe, is still expanding, and the universe is still trying to fill the space, which is why we see dark energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could we even know, or would either way look exactly the same to us?&lt;/p&gt;&#xA;" OwnerUserId="2439" LastEditorUserId="25088" LastEditDate="2020-01-28T00:59:27.850" LastActivityDate="2020-01-28T00:59:27.850" Title="How do we know the big bang expanded space and there wasn't a pre-existing volume for the universe to expand into?" Tags="&lt;big-bang-theory&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="7867" PostTypeId="2" ParentId="7863" CreationDate="2014-11-03T23:50:15.933" Score="10" Body="&lt;p&gt;A black hole is a region of spacetime separated by an event horizon, which means no signals from the interior can propagate outward, no matter how long one waits. Locally, there is nothing special about the event horizon; if you fall in a black hole, there is nothing marking that you've crossed and no local experiment (short in space and duration) that will tell you that you are already doomed. The most important conceptual observation here is that a 'black hole' means is defined not by the local conditions, but by the structure of spacetime on a larger scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that thinking of black holes as essentially determined by some particular density is a mistake. This bears out if you look at the density of a simple case of a Schwarzschild black hole: the &lt;em&gt;larger&lt;/em&gt; the black hole is, the &lt;em&gt;less dense&lt;/em&gt; it is (though for volume, &lt;a href=&quot;https://physics.stackexchange.com/questions/45963/do-black-holes-have-infinite-areas-and-volumes&quot;&gt;some caveats&lt;/a&gt; apply). There is not magical 'density point' for black holes; whether something does or doesn't form a black hole is determined by global conditions of spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: @zibadawa timmy's point regarding uniformity is very relevant. Since all points in space are equivalent, there is no special point around which an absolute event horizon could form to enclose it in an observer-independent manner, and thus no black hole. This is the most important difference in which the large-scale structure spacetime in Big Bang solutions is very different from stellar collapse scenarios.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) If all the matter and energy were concentrated at a single point at the big bang, why wasn't that a black hole, or why didn't it form one?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Matter and energy wasn't necessary concentrated at a single point. There only Big Bang cosmologies for which that is even a workable analogy is those involving a closed universe, which is definitely not all of them. But that's a separate misconception.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as far as we know, the local density at every point &lt;em&gt;did&lt;/em&gt; diverge to infinity in the finite past. So it still makes sense to ask why didn't &lt;em&gt;that&lt;/em&gt; cause the formation of a black hole. But the answer to that is simple: it didn't because there is no reason for it to do so, as the magnitude of the local density is not relevant.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The question is why didn't a black hole form out of all the matter and energy after the big bang?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We don't need a special mechanism for preventing it because there's no general reason for it do become a black hole in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I qualify with 'general' here because there is a sense in which a closed universe cosmology is already &lt;em&gt;like&lt;/em&gt; the interior of a black hole, and the universe as a whole could even recollapse as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Crunch&quot; rel=&quot;nofollow noreferrer&quot;&gt;Big Crunch&lt;/a&gt;, mimicking the more ordinary kind of stellar collapse into a black hole. The Big Crunch is empirically ruled out by the discovery that the cosmological expansion is accelerating, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, again, whether or not it forms (something like) a black hole depends on the large-scale structure of spacetime, not however large or small the local density becomes.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2014-11-04T17:14:42.207" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="7868" PostTypeId="1" AcceptedAnswerId="7874" CreationDate="2014-11-04T01:41:07.713" Score="5" ViewCount="156" Body="&lt;p&gt;Supermassive black holes often exist in large galaxies (e.g. the Milky Way, Andromeda). They make up a fair portion of the galaxy's mass. Is it possible for a supermassive black hole to form in a smaller galaxy - a dwarf galaxy (i.e. the Sagittarius Dwarf)?&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2015-01-03T21:53:06.690" Title="Could supermassive black holes form in dwarf galaxies?" Tags="&lt;supermassive-black-hole&gt;&lt;dwarf-galaxy&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7869" PostTypeId="1" AcceptedAnswerId="7872" CreationDate="2014-11-04T02:00:02.853" Score="19" ViewCount="1560" Body="&lt;p&gt;The Sun formed 4.5 billion years ago, in a molecular cloud. I assume that there were other stars in the vicinity (as is common in molecular clouds). Which stars are they? Where are they now? Are they similar to the Sun?&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2019-11-22T00:38:47.160" Title="Which stars did the Sun form with?" Tags="&lt;the-sun&gt;&lt;star-formation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7870" PostTypeId="2" ParentId="722" CreationDate="2014-11-04T03:31:34.620" Score="5" Body="&lt;p&gt;&lt;strong&gt;The denser rings are actually very comparable &lt;em&gt;in mass density&lt;/em&gt; to that of air&lt;/strong&gt;.  Because of their larger particle/particulate sizes, the number density is much much smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g.&lt;br&gt;&#xA;&lt;strong&gt;Ring A&lt;/strong&gt;: Radius ~ 137,000km, thickness ~ 10m, surface density ~ 20 g/cm^2&lt;br&gt;&#xA;so a density of about 0.02 g/cm^3, which is about 10 times more than air, and 100 times less than water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much, much more info here:&lt;br&gt;&#xA;&lt;a href=&quot;http://nssdc.gsfc.nasa.gov/planetary/factsheet/satringfact.html&quot;&gt;http://nssdc.gsfc.nasa.gov/planetary/factsheet/satringfact.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="918" LastActivityDate="2014-11-04T03:31:34.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7872" PostTypeId="2" ParentId="7869" CreationDate="2014-11-04T11:54:40.420" Score="10" Body="&lt;p&gt;Most stars form in clusters, so it is very likely that the Sun was part of a star cluster when it formed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in &lt;a href=&quot;http://vambartsumian.org/lib/pdf/dynamics_of_open_clusters.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;On the Dynamics of Open Clusters&lt;/a&gt;, the relaxation time of a cluster is calculated to be in the order of &lt;span class=&quot;math-container&quot;&gt;$\tau=4\times10^7 \textrm{yr}$&lt;/span&gt;. During that time, about one hundredth of the stars will escape from the cluster (i.e. reach escape velocity). The dissipation time of a star cluster is therefore in the order of a few billion years. These figures were calculated for an idealised star cluster containing stars of equal mass. For more realistic clusters the dissipation time may be much shorter. Especially dwarf stars seem to dissipate faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is, therefore, likely that after 4.5 billion years, the Sun has escaped from its star cluster. It would be very difficult to identify the stars with which it was formed, unless you can trace back the position of the Sun and other stars before dissipation, which is impracticable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I am aware, no sibling stars have been positively identified. The sibling stars should be similar in composition (element abundances), although they do not have to be solar twins as the masses of these stars may vary widely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: @adrianmcmenamin mentions an article in Sky &amp;amp; Telescope, where a sibling has been proposed. This refers to an &lt;a href=&quot;http://arxiv-web3.library.cornell.edu/abs/1405.1723&quot; rel=&quot;nofollow noreferrer&quot;&gt;article on arXiv by Ramirex et al.&lt;/a&gt; in which the element abundances have been analysed of stars that were earlier identified as possible siblings using galactic dynamics (i.e. tracing back the movement of these stars). Of the 30 stars identified as possible solar siblings using dynamical methods, only 2 stars had similar element abundances, which in itself says something about the accuracy of these dynamical methods...&lt;/p&gt;&#xA;" OwnerUserId="864" LastEditorUserId="9527" LastEditDate="2019-11-22T00:38:47.160" LastActivityDate="2019-11-22T00:38:47.160" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="7873" PostTypeId="1" CreationDate="2014-11-04T18:53:19.280" Score="10" ViewCount="1535" Body="&lt;p&gt;I've always been interested by the pattern within Venus' orbit around the sun. I found this image and formula of its retrograde motion in the upper right hand corner which confused me. What does it mean? can someone break this down in layman's terms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e96UQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="2746" LastEditorDisplayName="user2449" LastEditDate="2014-11-05T01:18:32.437" LastActivityDate="2014-11-05T10:56:27.090" Title="Explanation for the mathematics behind Venus' retrograde motion" Tags="&lt;orbit&gt;&lt;venus&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7874" PostTypeId="2" ParentId="7868" CreationDate="2014-11-04T20:07:39.603" Score="4" Body="&lt;p&gt;The prediction was that dwarf galaxies have SMBHs, but with smaller masses than regular galaxies.  This was based on the idea that regular galaxies were built from dwarf galaxies and the BHs need to have started growing early on.  Also the correlation between bulge mass and BH mass requires the BHs to come from the building blocks.  A recent &lt;a href=&quot;http://arxiv.org/pdf/1408.4451.pdf&quot; rel=&quot;nofollow&quot;&gt;paper at arxiv&lt;/a&gt; detects 28 AGN in dwarf galaxies with BHs in the $10^3 - 10^4 M_{\odot}$ mass range.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2014-11-04T20:07:39.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7875" PostTypeId="2" ParentId="7861" CreationDate="2014-11-04T20:16:09.413" Score="1" Body="&lt;p&gt;1) It is the mass internal to the orbit that determines the orbital speeds. These stars are so close to the center that the amount of bulge star mass contained within the orbit would be insignificant compared to the BH.&#xA;2) If the BH disappeared these stars would go into highly radial orbits out to much greater distances.  That distance would be about where the mass of the bulge/nucleus stars and gas equals the mass of the BH. &lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2014-11-04T20:16:09.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7877" PostTypeId="1" AcceptedAnswerId="7882" CreationDate="2014-11-05T10:42:02.840" Score="6" ViewCount="509" Body="&lt;p&gt;Seven probes have passed by or orbited outer planets. I wonder how their photos differ from the naked eye view one would have if one were actually there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn is 10 AU from the Sun, which means that it has just 1% of the sunlight we have on Earth. Would an astronaut orbiting Saturn, looking out of the window, even notice that Saturn is there? A clouded winter night on Earth can be really very dark, so you can't see where you put your foot while walking. But the outer planets at least have star light and a wage glimpse of the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the light conditions out there be compared to a clear night on Earth with a full moon, half moon or new moon?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2021-04-22T16:08:14.277" Title="How bright are the outer planets, when at close range?" Tags="&lt;solar-system&gt;&lt;light&gt;&lt;photography&gt;&lt;naked-eye&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7878" PostTypeId="2" ParentId="7873" CreationDate="2014-11-05T10:56:27.090" Score="3" Body="&lt;p&gt;The really simple explanation is that Venus' orbital period is 225 days. Earth's is 365 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That ratio is 13:8 (well, more accurately 12.9777777777..:8) so you can see that your complete cycle is 13 orbits of Venus to 8 of Earth, but within that cycle our view of Venus travelling round the sun gives the pretty pattern as shown in your image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://www.youtube.com/watch?v=4cgQNUhtmHM&quot; rel=&quot;nofollow&quot;&gt;YouTube video&lt;/a&gt; is quite a nice visualisation - should make it easy to understand.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2014-11-05T10:56:27.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7879" PostTypeId="1" AcceptedAnswerId="7880" CreationDate="2014-11-05T16:40:22.417" Score="7" ViewCount="3047" Body="&lt;p&gt;A common representation of a black hole is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_hole&quot;&gt;black circle&lt;/a&gt; which if I'm correct, indicates its breadth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nasa.gov/audience/forstudents/k-4/stories/what-is-a-black-hole-k4.html&quot;&gt;This representation&lt;/a&gt; depicts it as a 'plate', so a really small depth compared to its breadth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On this &lt;a href=&quot;http://phys.org/news154627589.html&quot;&gt;other representation&lt;/a&gt;, its depth has a funnel shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How deep and what shape has the depth of a black hole ?&lt;/p&gt;&#xA;" OwnerUserId="1630" LastActivityDate="2021-05-17T08:30:17.573" Title="How deep and shaped is the depth of a black hole?" Tags="&lt;black-hole&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7880" PostTypeId="2" ParentId="7879" CreationDate="2014-11-05T16:52:03.693" Score="11" Body="&lt;p&gt;The first representation isn't depicting the black hole; it's depicting its &lt;a href=&quot;http://en.wikipedia.org/wiki/Accretion_disc&quot; rel=&quot;nofollow&quot;&gt;accretion disk&lt;/a&gt;, which is indeed plate-shaped. The accretion disk is around the black hole, and - if it's a rotating (Kerr) black hole - it's perpendicular to the black hole's axis of rotation. Accretion disks are disk-shaped, and so the representation is disk-shaped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other is a common representation of a black hole, but it's misleading. It's an extension of the &quot;rubber sheet&quot; analogy, which describes the bending of spacetime. A heavy object, placed on a semi-taut sheet, will curve it a little bit. This represents ordinary matter curving spacetime. A black hole is represented in that image as bending spacetime by an infinite amount. The issue is that this representation embeds the two-dimensional representation of spacetime in three dimensional space, making it appear as if the black hole has this extra depth. Black holes are not embedded in an extra spatial dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that the small object moving near the edge of the black hole is also represented by a tiny funnel. The creator seems to have tried to take the analogy to the extreme, portraying the object's funnel as finite, and the black hole's as either extremely large or infinite, in an attempt to explain that light cannot escape from a black hole. It's not too accurate, but it works to get the point across.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, black holes do have size (the central point singularities of non-rotating black holes are infinitely tiny, while the ring singularities of rotating black holes are in the shapes of infinitely thin rings). You might want to read up on &lt;a href=&quot;http://en.wikipedia.org/wiki/Event_horizon&quot; rel=&quot;nofollow&quot;&gt;event horizons&lt;/a&gt;. The radius of the event horizon of a non-rotating black hole is a finite number, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;nofollow&quot;&gt;Schwarzschild radius&lt;/a&gt;. It is calculated as&#xA;$$r_s=\frac{2GM}{c^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the event horizon is spherical. In rotating black holes, it may be in the shape of an &lt;a href=&quot;http://en.wikipedia.org/wiki/Oblate_spheroid&quot; rel=&quot;nofollow&quot;&gt;oblate spheroid&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-04-22T00:46:01.393" LastActivityDate="2015-04-22T00:46:01.393" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7882" PostTypeId="2" ParentId="7877" CreationDate="2014-11-05T17:54:37.533" Score="4" Body="&lt;p&gt;There is a rule about average surface brightness: it is conserved as you change distance or magnification.  For the gory details &lt;a href=&quot;https://www.cv.nrao.edu/%7Esransom/web/Ch2.html#S1&quot; rel=&quot;nofollow noreferrer&quot;&gt;see&lt;/a&gt;.  Telescopes make unresolved objects brighter, but for resolved objects, a telescope shows you the actual surface brightness (assuming that the exit pupil diameter is well matched to your eye's pupil (~7mm) and is not throwing away light, dimming the view).   The surface brightness of any resolved object through a telescope is the same as it would be if you were there.&lt;/p&gt;&#xA;&lt;p&gt;Also, the 0.01 of earth sunlight received at Saturn is way more than the lighting provided by a full moon &lt;a href=&quot;http://en.wikipedia.org/wiki/Apparent_magnitude&quot; rel=&quot;nofollow noreferrer&quot;&gt;which is 400,000 times fainter than the Sun&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;What you also need to take into account, is that when you look at Saturn through a telescope, it is generally (not always) done at night when your eyes are dark adapted.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2021-04-22T16:08:14.277" LastActivityDate="2021-04-22T16:08:14.277" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="7883" PostTypeId="1" AcceptedAnswerId="8000" CreationDate="2014-11-05T18:46:03.500" Score="5" ViewCount="1687" Body="&lt;p&gt;I am wondering how many free electrons per baryon and how many free electrons per atom there are in the solar photosphere.  This number depends on the abundances of the various atoms found in the photosphere as well as each atom's ionization potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book &quot;Introduction to Stellar Astrophysics&quot; by Boehm-Vitense, vol. 2, 1997 reprint, on page 76 claims that one out of every $10^4$ hydrogen atoms are ionized, so hydrogen is mostly neutral in the photosphere; however, there are other heavier elements with lower ionization potentials.  Boehm-Vitense goes through and calculates the fraction of ionization of iron but calculations for only two species present in the photosphere can hardly be considered comprehensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether you calculate these numbers yourself or provide a summary of a reference (with a link to that reference if possible) that has more details about ionization in the photosphere it will be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that, since hydrogen has such a low degree of ionization, the number of electrons per baryon and per atom will be small, since hydrogen is the majority constituent of the photosphere.&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2014-11-18T22:13:07.697" Title="What is the degree of ionization is the solar photosphere?" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7885" PostTypeId="1" AcceptedAnswerId="33089" CreationDate="2014-11-05T22:11:06.213" Score="4" ViewCount="144" Body="&lt;p&gt;VSOP87 (and the subsequent VSOP2000 and VSOP2013) provide reasonably &#xA;accurate planetary positions while still maintaining a small file size &#xA;compared to the Chebyshev polynomial files. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;VSOP includes our moon, but has anyone created something similar for &#xA;other planets' moons (such as Jupiter's Io, Saturn's Titan, and &#xA;Uranus' Ariel)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking about trying to create something like this and wanted to &#xA;check that it hadn't already been done. &lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2019-08-19T04:57:41.993" Title="VSOP for exosatellites? (Io, Titan, Ariel, etc)" Tags="&lt;positional-astronomy&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7886" PostTypeId="2" ParentId="7879" CreationDate="2014-11-05T22:28:08.393" Score="-1" Body="&lt;p&gt;Look at this picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;!black-hole shape][1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;www.astro.umd.edu&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the star gets bigger, the bigger the black hole will go, it's event horizon goes lower, and its inside grows thinner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if the star is small, you will fell the gravitational pull from much further away. But with a much larger one like a super-massive black-hole, you will get closer without noticing anything much.&lt;/p&gt;&#xA;" OwnerUserId="2678" LastActivityDate="2014-11-05T22:28:08.393" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7887" PostTypeId="2" ParentId="460" CreationDate="2014-11-06T03:07:05.557" Score="1" Body="&lt;p&gt;No special techniques are required that I am aware of, but several observations must be taken to make a reasonable estimate of the orbit. Once a sufficient number of observations are made so that the &lt;a href=&quot;http://www.braeunig.us/space/orbmech.htm&quot; rel=&quot;nofollow&quot;&gt;orbital elements&lt;/a&gt; can be determined, then anything that has an elliptical orbit (eccentricity &gt;0 but &gt;1) can be assumed to be orbiting as part of the solar system and thus not a transiting rogue object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An object with an eccentricity to its orbit greater than 1 is following a hyperbolic escape trajectory, and, unless substantially perturbed by the gravity of another planet, is on its way out of the solar system. I am not aware of any such objects that are thought to have originated from outside the system, but comets have been known to have been flung from the system along hyperbolic paths from the influence of Jupiter. &lt;/p&gt;&#xA;" OwnerUserId="2674" LastEditorUserId="2674" LastEditDate="2014-12-10T11:48:15.667" LastActivityDate="2014-12-10T11:48:15.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7888" PostTypeId="1" CreationDate="2014-11-06T03:37:19.137" Score="4" ViewCount="157" Body="&lt;p&gt;Several observations of a distant solar object must be taken before determining its orbit, but in the case of objects discovered beyond Neptune's orbit, just how many are required over what period of time (or what portion of arc) before a reasonably accurate orbit can be determined, sufficient, say, to be added to the Minor Planet Database or a JPL ephemeris?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be interested to know to what degree of certainty orbital elements for objects beyond 30 AU can be determined when we have only observed them for (so far) short portions of their orbits. What about 60 AU, or 90 AU?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, the Wiki for &lt;a href=&quot;https://en.wikipedia.org/wiki/VNH0004&quot; rel=&quot;nofollow&quot;&gt;VNH004&lt;/a&gt;, a TNO that is a photo target for the New Horizons probe, states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The asteroid only was observed 12 times by the Mauna Kea and Las Campanas Observatory over a period of about 33.8 days between May 29th and July 2nd, 2011, so its current orbit is extremely uncertain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.boulder.swri.edu/~buie/kbo/astrom/VNH0004.html&quot; rel=&quot;nofollow&quot;&gt;How uncertain&lt;/a&gt;? Here are orbital elements along with +/- values:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Heliocentric elements and errors&lt;br&gt;&#xA;  Epoch . . . . . .  2455710.5000  =  2011/05/29&lt;br&gt;&#xA;  Mean Anomaly: . . .       309.75557 +/-    31.234&lt;br&gt;&#xA;  Argument of Peri: .     112.06593 +/-    54.010&lt;br&gt;&#xA;  Long of Asc Node:  .    247.74234 +/-     3.336&lt;br&gt;&#xA;  Inclination: .  . . . . . . . .   3.81106 +/-     0.396&lt;br&gt;&#xA;  Eccentricity: . . . . .     0.29886369 +/-    0.8050&lt;br&gt;&#xA;  Semi-Major Axis: .    37.47217640 +/-   14.2156&lt;br&gt;&#xA;  Time of Perihelion: 2467404.0720 +/-    2926.4&lt;br&gt;&#xA;  Perihelion: . . . . . . .   26.27310354 +/-   31.7695&lt;br&gt;&#xA;  Aphelion: . . . . . . . .    48.67124926 +/-   35.3678&lt;br&gt;&#xA;  Period (y) . . . . . . . . . . 229.3885 +/-    130.53  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An orbital period of over 229 years, plus or minus over 130 years. How many more observations would be required to half the margin of error?&lt;/p&gt;&#xA;" OwnerUserId="2674" LastEditorUserId="2674" LastEditDate="2014-11-09T02:54:58.217" LastActivityDate="2015-08-03T16:40:21.490" Title="How many observations does it take to determine the orbit of a TNO" Tags="&lt;observational-astronomy&gt;&lt;orbital-elements&gt;&lt;oort-cloud&gt;&lt;kuiper-belt&gt;&lt;tno&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7889" PostTypeId="5" CreationDate="2014-11-06T03:43:51.607" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-11-06T03:43:51.607" LastActivityDate="2014-11-06T03:43:51.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7890" PostTypeId="4" CreationDate="2014-11-06T03:43:51.607" Score="0" Body="A Trans-Neptunian object is any minor planet in the Solar System that orbits the Sun at a greater average distance (semi-major axis) than Neptune." OwnerUserId="2674" LastEditorUserId="2674" LastEditDate="2014-11-07T15:23:03.970" LastActivityDate="2014-11-07T15:23:03.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7891" PostTypeId="1" AcceptedAnswerId="7892" CreationDate="2014-11-06T14:40:05.057" Score="5" ViewCount="705" Body="&lt;p&gt;I have seen several references that say that helium burning begins in a star once the core temperature reaches $10^8$K (&lt;a href=&quot;http://en.wikipedia.org/wiki/Triple-alpha_process&quot; rel=&quot;nofollow&quot;&gt;such as here&lt;/a&gt;) but nowhere that says what density that corresponds to.  Does anybody know a reference that has this value?  And if it makes a difference, I would like to know the density for quiescent helium burning, not helium flash helium burning.&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2016-11-28T07:44:16.860" Title="At what density does helium burning start in a star?" Tags="&lt;star&gt;&lt;helium&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7892" PostTypeId="2" ParentId="7891" CreationDate="2014-11-06T15:35:50.073" Score="6" Body="&lt;p&gt;The short answer: $\sim10^4$ grams per cubic centimeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.astro.cornell.edu/academics/courses/astro201/highmass.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this webpage&lt;/a&gt;, I have a few statistics regarding required mean density and during each phase of fusion (the fusion of lighter elements may happen in the &quot;shells&quot; outside the core).&#xA;$$\begin{array}{|c|c|}&#xA;\hline \text{Fusion phase}&amp;amp;\text{Mean density (g/cm}^3)\\&#xA;\hline \text{Hydrogen} &amp;amp; 5\\&#xA;\hline \text{Helium} &amp;amp; 700\\&#xA;\hline \text{Carbon} &amp;amp; 200,000\\&#xA;\hline \text{Neon} &amp;amp; 4,000,000\\&#xA;\hline \text{Oxygen} &amp;amp; 10,000,000\\&#xA;\hline \text{Silicon} &amp;amp; 30,000,000\\&#xA;\hline&#xA;\end{array}$$&#xA;This is data for a star of $25M_{\odot}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, however, that this is &lt;em&gt;mean&lt;/em&gt; density in the star, &lt;em&gt;not&lt;/em&gt; the density in the core, where fusion is taking place. That is on the order of $\sim10^4$ grams per cubic centimeter (see e.g. &lt;a href=&quot;https://astro.uni-bonn.de/~nlanger/siu_web/ssescript/new/chapter9.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;). There may be a sharp density dropoff starting from the center of the star, showing that only the hot, dense inner regions can fuse the main element in each stage.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-11-27T19:13:49.047" LastActivityDate="2016-11-27T19:13:49.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7893" PostTypeId="2" ParentId="4724" CreationDate="2014-11-06T20:57:10.733" Score="3" Body="&lt;p&gt;What we see at the outer reaches of the observable universe is a surface of the cosmic microwave background (CMB), which is at the time when the universe, still quite homogeneous (perturbations were a few parts in $10^5$), cooled enough for the electrons to combine with the protons and make hydrogen atoms.  Each day, we can see a light-day farther and so we slowly see a different (larger) surface of the CMB (but there were no galaxies at this epoch).  Therefore, as we observe the CMB over many years, we will always see the same time epoch, but at different spherical shells.  Will we see new galaxies?  Yes. Galaxies are forming continuously in view at high redshifts (2-20?).  In a few million years, the small perturbations that we see today in the CMB will develop into young galaxies within our view.  One cannot provide numbers for how many we might be seeing because small galaxies merge into bigger ones and those merge again.  At high redshifts there is a lot of merging going, so the rate is particularly ill defined there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this changed by the acceleration from the cosmological constant (aka dark energy)? From the Friedmann equation one sees that the Hubble constant reaches a final constant value of $c\sqrt{\Lambda/3}$ or about 40% higher than its present value.  Eventually, in billions of years, the CMB will shift from $3^oK$ to much lower and then the CMB will fade (redshift) away.  After that, those new distant galaxies will also fade away.  And then more and more of the universe will go dim.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2014-11-07T16:57:33.693" LastActivityDate="2014-11-07T16:57:33.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7894" PostTypeId="5" CreationDate="2014-11-07T16:31:41.620" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://www.iau.org/science/scientific_bodies/divisions/A/&quot; rel=&quot;nofollow noreferrer&quot;&gt;IAU Division A&lt;/a&gt; is responsible for fundamental astronomy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The scientific community of Division A:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;provides definitions and models that describe reference systems and frames used in astronomy to determine positions and motions of celestial objects in space and time;&lt;/li&gt;&#xA;  &lt;li&gt;investigates dynamical behavior of celestial bodies including both high-accuracy descriptions for shorter periods of time and evolution over extended periods of time;&lt;/li&gt;&#xA;  &lt;li&gt;obtains physical information on celestial objects and investigates physical laws using the methods of astrometry and celestial mechanics;&lt;/li&gt;&#xA;  &lt;li&gt;offers services that provide data and ephemerides of solar system bodies, Earth orientation data, time scales, astronomical constants, models including relevant software procedures, etc. for users within the astronomical community as well as within society.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="616" LastEditDate="2019-06-16T05:26:54.063" LastActivityDate="2019-06-16T05:26:54.063" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="7895" PostTypeId="4" CreationDate="2014-11-07T16:31:41.620" Score="0" Body="Questions involving reference frames, time scales, and how they are determined. Not for general Astronomy 101 questions." OwnerUserId="2674" LastEditorUserId="616" LastEditDate="2021-04-28T15:56:19.633" LastActivityDate="2021-04-28T15:56:19.633" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="7896" PostTypeId="5" CreationDate="2014-11-07T16:37:08.247" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-11-07T16:37:08.247" LastActivityDate="2014-11-07T16:37:08.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7897" PostTypeId="4" CreationDate="2014-11-07T16:37:08.247" Score="0" Body="Questions involving the physics of the universe, especially the nature of astronomical objects, energy fields, and/or regions, rather than their positions or motions in space. " OwnerUserId="2674" LastEditorUserId="2674" LastEditDate="2014-11-07T20:20:08.163" LastActivityDate="2014-11-07T20:20:08.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7898" PostTypeId="5" CreationDate="2014-11-07T16:41:29.327" Score="0" Body="&lt;p&gt;There are two types of mass: gravitational mass and inertial mass. By a curious coincidence, the two are exactly the same in our universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational mass determines how an object interacts with other objects via the force of gravity. Inertial mass determines how an object accelerates when a force is applied to it.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="2153" LastEditDate="2015-04-12T22:01:33.597" LastActivityDate="2015-04-12T22:01:33.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7899" PostTypeId="4" CreationDate="2014-11-07T16:41:29.327" Score="0" Body="Questions regarding the property of astronomical objects that determines their inertial resistance and gravitational interaction with other bodies." OwnerUserId="2674" LastEditorUserId="2153" LastEditDate="2015-04-12T22:01:33.597" LastActivityDate="2015-04-12T22:01:33.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7900" PostTypeId="1" AcceptedAnswerId="7904" CreationDate="2014-11-07T17:24:54.693" Score="5" ViewCount="346" Body="&lt;p&gt;In a previous question about &lt;a href=&quot;https://astronomy.stackexchange.com/questions/632/determining-effect-of-small-variable-force-on-planetary-perihelion-precession&quot;&gt;differences in Newtonian and GTR gravitional force&lt;/a&gt; for the case of star-planet gravitational interactions an &lt;em&gt;approximate&lt;/em&gt; relationship was noted between the expressions for the gravitational force exerted by a star on an orbiting planet (the solution is valid situations with low gravity, slow orbital speed, and spherical source).  This was based on texts in &lt;a href=&quot;http://books.google.co.uk/books?id=V4_s91-Vun4C&amp;amp;pg=PA482&amp;amp;lpg=PA482&amp;amp;dq=we+recall+from+section+7.6.1&amp;amp;source=bl&amp;amp;ots=RyU1Adp8h7&amp;amp;sig=qQ5_KyjoDhaSaOadgP8zyy8ubIc&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=Hlc-VJDJK9TzaoHlgKgL&amp;amp;ved=0CCEQ6AEwAA#v=onepage&amp;amp;q=we%20recall%20from%20section%207.6.1&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;Walter 2008&lt;/a&gt; and &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201657023&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goldstein et al 2001&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Walter derived an approximate relationship assuming a circular orbit.  Goldstein focussed on deriving an orbit-average expression for perihelion precession.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On re-examining these texts it seems to me that GTR (General Theory of Relativity) provides more than just an orbit-averaged approximation.  Rather it provides a phase-specific formula for total acceleration (in Scwharzschild space-time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation of motion for a Newtonian orbit is&#xA;$$u''_\theta + u_\theta =\frac{\mu}{h^2}$$&#xA;where $u_\theta = 1/r_\theta$ and $u'_\theta =$ d$(u_\theta)/$d$\theta$ and $h$ is specific angular momentum which is constant, $\mu = GM$ the gravitational parameter, $G$ is Newton's Universal gravitational constant, $M$ is mass of the star, $r$ is distance from star to planet, $\theta$ is the True anomaly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $h^2 = Vt_\theta^2 \,r_\theta^2$ where $Vt$ is the instantaneous transverse component of velocity (in vector terms = full velocity minus radial velocity) we can  obtain&#xA;$$u''_\theta + u_\theta = \frac{1}{Vt_\theta^2} \left( \frac{\mu}{r_\theta^2} \right)$$ where the term in brackets is the Newtonian acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Walter presents the following equation for a GTR orbit (Schwarzschild model)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u''_\theta + u_\theta =\frac{\mu}{h^2} + \frac{3\mu}{c^2}\,u_\theta^2$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, using $h^2 = Vt_\theta^2 \,r_\theta^2$ and $u_\theta^2 = 1/r_\theta^2$ we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u''_\theta + u_\theta &#xA;=\frac{\mu}{Vt_\theta^2 \,r_\theta^2} + \frac{3\mu}{c^2 \,r_\theta^2}&#xA;=\frac{1}{Vt_\theta^2 } &#xA;\left( \frac{\mu}{r_\theta^2} \,+ \frac{\mu}{r_\theta^2} \,\frac{3\,Vt_\theta^2}{c^2 } \right)$$ &#xA;where the terms in brackets are the Newtonian acceleration and the extra acceleration according to GTR.  And so the ratio of Newtonian acceleration to GTR-specific acceleration, at any angle $\theta$ is &#xA;$$1 \, \mathrm{to} \, \frac {3 Vt_\theta^2}{c^2}$$.&#xA;Goldstein emphasises that the GR acceleration does not indicate a velocity-dependence, so an alternative, more palatable, distance-dependent form of the ratio of accelerations, at any angle $\theta$, would be:-&#xA;$$1 \, \mathrm{to} \, \frac {3 h^2}{c^2 \, r_\theta^2} \equiv &#xA;1 \, \mathrm{to} \, \frac {3 GM.P}{c^2 \, r_\theta^2}&#xA;$$&#xA;where $h$ (specific angular momentum) and $P$ (semi-latus rectum) are the values for the orbit of the particular subject planet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so (ignoring other massive perturbing bodies) the magnitude of the total instantaneous radial force on the planet (of mass $m$) towards the Sun is given by:-&#xA;$$ \mathrm{F/m} \,=  \, \frac {GM}{r_\theta^2} + \frac {3 GM.GM.P}{c^2 \, r_\theta^4}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N.B. The GTR/Schwarzschild equations relate to proper time and Schwarzschild radial distance, not their Newtonian equivalents so strictly the ratio of accelerations is still an approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this analysis valid, or have I missed something?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Update&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I have accepted Stan Liou's answer as being very helpful in (a) providing a derivation from GTR/Schwarzschild of the formulae presented by Walter and Goldstein et al. and (b) indicating the imperfect correspondence between GTR terms/concepts and Newtonian terms/concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is as follows. In a Newtonian central-force elliptical-orbit model the addition of an extra centre-directed acceleration ($dV/dt$), which varies through the orbit as $(+3 V_t^2/c^2\,\equiv \, +3 h^2/r^2c^2)$ times the co-temporal standard Newtonian centre-directed gravitational acceleration, can be shown (by first order perturbation analysis or numerical modelling) to produce apsidal precession at a rate (radians per orbit) defined by a Newtonian formula $$\epsilon = 24 \pi^3 \frac{a^2}{T^2 c^2(1-e^2)}$$ .  This formula is well-known (see for example &lt;a href=&quot;http://en.wikipedia.org/wiki/Apsidal_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia Apsidal precession&lt;/a&gt;). According to, but not clearly referenced by, &lt;a href=&quot;http://www.mathpages.com/home/kmath527/kmath527.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt;, the formula (or an algebraic equivalent using other terms) was well known c. 1895 (i.e. &lt;em&gt;before&lt;/em&gt; the publications of Gerber 1898 and Einstein 1915).  The formula predicts very well the long-term values of apsidal precession determined using Newtonian models from observations of solar planets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various writers (&lt;a href=&quot;http://www.gsjournal.net/old/eeuro/vankov.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein&lt;/a&gt;, Goldstein, Walter, presumably many others) present mathematical arguments indicating how an identical formula can be derived from Einstein's GTR.  The presented arguments may involve approximations (e.g. Walter's use of near-circular orbits, Goldstein's use of orbit-averaged precession) and non-mathematical &quot;correspondences&quot; between the concepts/terms of the GTR model and the concepts/terms of the Newtonian model.&lt;/p&gt;&#xA;" OwnerUserId="233" LastEditorUserId="233" LastEditDate="2018-01-31T03:30:45.177" LastActivityDate="2018-03-08T01:14:40.910" Title="Can General Relativity indicate phase-dependent variations in planetary orbital acceleration?" Tags="&lt;orbit&gt;&lt;general-relativity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7902" PostTypeId="2" ParentId="1259" CreationDate="2014-11-07T22:40:27.437" Score="1" Body="&lt;p&gt;In addition to the requirements that it be night, and the ISS be overhead, the space station must also not be eclipsed by the earth's shadow. With earth taking up roughly 88 degrees of sky from the ISS's &lt;a href=&quot;http://en.wikipedia.org/wiki/International_Space_Station&quot; rel=&quot;nofollow&quot;&gt;typical altitude of 420 km&lt;/a&gt;, eclipses correspond to night-time on the space station. They're each &lt;a href=&quot;http://www.theguardian.com/science/2010/oct/24/international-space-station-nasa-astronauts&quot; rel=&quot;nofollow&quot;&gt;45 minutes long, and happen 16 times a day&lt;/a&gt;.&#xA;Harmonics between earth day/night cycle and ISS day/night cycle account for those mysterious gaps in visibility noted by aman207. The start of such eclipses are oten observed from the ground: You'll see the ISS come over the horizon like a ball of fire, get halfway across the sky, then fade to nothing in a minute or two. &lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2014-11-07T22:40:27.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7904" PostTypeId="2" ParentId="7900" CreationDate="2014-11-08T04:42:48.733" Score="7" Body="&lt;p&gt;Since I don't have Walter's book, I'm uncertain as the context of the derivation of the equation you quote. Therefore, I've simply re-derived it here; apologies if there's some repetition of things you already know, but perhaps it'll be useful for anyone else reading this regardless.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Constants of Motion&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The Schwarzschild solution is the unique nontrivial spherically symmetric vacuum solution of general relativity. In the Schwarzschild coordinate chart and units of $G = c = 1$, the metric takes the form&#xA;$$\mathrm{d}s^2 = -\left(1-\frac{2M}{r}\right)\mathrm{d}t^2 + \left(1-\frac{2M}{r}\right)\mathrm{d}r^2 + r^2\left(\mathrm{d}\theta^2 + \sin^2\theta\,\mathrm{d}\phi^2\right)\text{,}$$&#xA;and one can immediately note that that the metric coefficients are completely independent of $t$ and $\phi$, which implies that $\partial_t$ and $\partial_\phi$ are &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Killing_vector_field&quot; rel=&quot;noreferrer&quot;&gt;Killing vector fields&lt;/a&gt;&lt;/em&gt;. They are important here because along with generating symmetries of the geometry, they also produce conserved orbital quantities in the following way: given an orbit with four-velocity $u^\mu = (\dot{t},\dot{r},\dot{\theta},\dot{\phi})$, the inner product with a Killing vector field is conserved:&#xA;$$\epsilon = -\langle\partial_t,u\rangle = \left(1-\frac{2M}{r}\right)\frac{\mathrm{d}t}{\mathrm{d}\tau}\text{,}$$&#xA;$$h = \langle\partial_\phi,u\rangle = r^2\sin^2\theta\,\frac{\mathrm{d}\phi}{\mathrm{d}\tau}\text{.}$$&#xA;The overdot indicates differentiation with respect to any affine parameter of the orbit, which for timelike geodesics appropriate for massive particles we can take without loss of generality to be the proper time $\tau$. An alternative way to find these constants of motion is to integrate the $t$ and $\phi$ components of the geodesic equation, but in this way they can read off immediately from the metric. These are the specific energy and specific angular momentum of the orbit, respectively. Also note that the coordinates are analogues of the spherical coordinates for Euclidean space, where $\theta$ is the zenith angle while $\phi$ is the azimuth; if we take the orbital plane to be the equatorial plane ($\theta = \pi/2$), then $\phi$ would represent the true anomaly.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Effective Potential&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Substituting the above constants of motion into the timelike worldline condition $\langle u,u\rangle \equiv g_{\mu\nu} u^\mu u^\nu = -1$, i.e.,&#xA;$$-\left(1-\frac{2M}{r}\right)\dot{t}^2 + \left(1-\frac{2M}{r}\right)^{-1}\dot{r}^2 + r^2\dot{\phi}^2 = -1\text{,}$$&#xA;one can immediately derive the effective gravitational potential:&#xA;$$\frac{1}{2}(\epsilon^2-1) = \frac{1}{2}\dot{r}^2 + \underbrace{\left[-\frac{M}{r}+\frac{h^2}{2r^2} - \frac{Mh^2}{r^3}\right]}_{V_\text{eff}}\text{,}$$&#xA;or if one insists on a &lt;em&gt;formal&lt;/em&gt; comparison with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Effective_potential&quot; rel=&quot;noreferrer&quot;&gt;Newtonian effective potential&lt;/a&gt; ($L\equiv mh$),&#xA;$$E = \underbrace{\frac{1}{2}m\dot{r}^2 + \frac{L^2}{2mr^2} - \frac{GMm}{r}}_{\text{Newtonian form}} - \frac{GML^2}{mr^3c^2}\text{.}$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Orbit Equation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Differentiation of the above effective potential gives&#xA;$$\ddot{r} + \frac{M}{r^2} - \frac{h^2}{r^3} + 3\frac{Mh^2}{r^4} = 0\text{.}$$&#xA;In terms of $u \equiv 1/r$ with prime denoting differentiation with respect to $\phi$,&#xA;$$u''= \frac{\mathrm{d}\tau}{\mathrm{d}\phi}\frac{\mathrm{d}}{\mathrm{d}\tau}\left(\frac{\mathrm{d}\tau}{\mathrm{d}\phi}\dot{u}\right) = \frac{r^2}{h}\frac{\mathrm{d}}{\mathrm{d}\tau}\left(\frac{r^2}{h}\left(-r^{-2}\dot{r}\right)\right) = -\frac{\ddot{r}r^2}{h^2}\text{,}$$&#xA;this gives, after multiplication through by $-r^2/h^2$,&#xA;$$u'' + u = \frac{M}{h^2} + 3Mu^2\text{.}$$&#xA;However, there is really no need to consider a second-order at any point; there's a simpler one in terms of $V \equiv V_\text{eff} - h^2/2r^2$, the effective potential sans the centrifugal potential term:&#xA;$$\begin{eqnarray*}&#xA;\frac{2}{h^2}\left[\frac{E}{m}-V\right] &amp;amp;=&amp;amp; \frac{\dot{r}^2}{h^2} + \frac{1}{r^2}&#xA;\\&amp;amp;=&amp;amp; \frac{1}{r^4}\left[\frac{\mathrm{d}r}{\mathrm{d}\phi}\right]^2 + u^2&#xA;\\&amp;amp;=&amp;amp; (u')^2 + u^2\text{.}&#xA;\end{eqnarray*}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Walter derived an approximate relationship assuming a circular orbit. Goldstein focused on deriving an orbit-average expression for perihelion precession. On re-examining these texts it seems to me that GR provides more than just an orbit-averaged approximation.&#xA;  ...&#xA;  Walter presents the following equation for a GR orbit (Schwarzschild model)&#xA;  $$u''_\theta + u_\theta =\frac{\mu}{h^2} + \frac{3\mu}{c^2}\,u_\theta^2$$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One can immediately see that Walter's equation is the above second-order equation, just in normal units rather than $G = c = 1$. I don't know what Walter's argument is (I'm willing to bet the approximation is because Walter substituted a circular-orbit case for $L^2$ or $h^2$ somewhere, though), but that particular relationship holds &lt;em&gt;exactly&lt;/em&gt; for massive test particles in Schwarzschild spacetime. It does not even have to be a bound orbit, although of course if one is interested in precession specifically, it would have to be at least bound for precession to make sense. Lightlike geodesics are described by nearly the same equation, just without the $M/h^2$ term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, we can also restate it as&#xA;$$u'' + u = \frac{M}{h^2}\left[1 + 3\frac{h^2}{r^2}\right] \leadsto \frac{\mu}{h^2}\left[1+3\frac{h^2}{r^2c^2}\right]\text{,}$$&#xA;which after substitution of $V_t\equiv r\dot{\phi} = h/r$ is what you have.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Conclusion&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... so an alternative, more palatable, distance-dependent form of the ratio of accelerations ... would be:-&#xA;  $$1\;\text{to}\;\frac {3 h^2}{c^2 \, r_\theta^2}\text{.}$$&#xA;  The GR/Schwarzschild equations relate to proper time and Schwarzschild radial distance, not their Newtonian equivalents so strictly the ratio of accelerations is still an approximation.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Is this analysis valid, or have I missed something?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is mostly valid, but I would like you to caution you on several points regarding the way you frame the problem and interpret the result, although you are likely already aware of some of them:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The Schwarzschild time coordinate $t$ is quite different from the proper time $\tau$. The former is a special coordinate in which the Schwarzschild geometry is time-independent. It defines the worldlines of a family of observers that are stationary with respect to the geometry, and its scaling matches a stationary observer at infinity. On the other hand, proper time is simply the time measured along some particular worldline; in this context, by the orbiting test particle.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Schwarzschild radial coordinate $r$ is not a radial distance. It could be called an areal radius in the sense that it is chosen to make a sphere of constant $r$ have area of exactly $4\pi r^2$, but usually it is simply called the &lt;em&gt;Schwarzschild radial coordinate&lt;/em&gt;. In the Schwarzschild coordinate chart, the radial distance between Schwarzschild radial coordinates $r = r_0$ and $r = r_1$ would be given by&#xA;$$D = \int_{r_0}^{r_1}\frac{\mathrm{d}r}{\sqrt{1-\frac{2GM}{rc^2}}}\text{,}$$&#xA;and would be the distance one would measure if one slowly crawled along the radial direction from $r = r_0$ to $r = r_1$ with some ideal meter-stick, in the limit of zero speed. Of course, $r$ could serve as an approximation to the radial distance in appropriate contexts, but the point is that not only does $r$ fail to be the Newtonian radial distance, it's not actually the 'Schwarzschild radial &lt;em&gt;distance&lt;/em&gt;' either.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Acceleration&lt;/em&gt; is a bit of a loaded word here. If we mean the second derivative of our radial coordinate with respect to proper time, then no, $\ddot{r}_\text{GTR}/\ddot{r}_\text{Newtonian}$ does not simplify quite that nicely, but you calculate it from the above anyway. On the other hand, if we mean the second derivative of the inverse radial coordinate with respect to the azimuthal angle, then yes, the above correct.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But then, it doesn't really make sense to actually call it 'acceleration', does it? This explains (if your previous question was accurate in this phrasing) why Walter uses a more vague term of 'effects' when talking about the above ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead (once again using the intentional conflation between $r$,$\tau$ and their Newtonian counterparts as an approximation or analogy), it would probably be better to simply think of the Schwarzschild geometry as introducing a new term in the potential that is analogous to a quadrupole moment, which would also put a $\propto 1/r^3$ term into the potential, with the corresponding Newtonian equation being&#xA;$$(u')^2 + u^2 = \frac{2}{h^2}\left(\frac{E}{m} - \Phi(u)\right)\text{.}$$&#xA;Both the effective potential and the first-order equation in $u$ provide a much more straightforward analogy between the Newtonian and Schwarzschild cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is actually pretty interesting: if one assumes that the Sun does indeed have a quadrupole moment, e.g., caused by solar oblateness, then one can easily account for the perihelion advance of Mercury. However, because this is simply an analogy, blaming Mercury's behavior on this would simultaneously mess up the behavior of other planets (since the new term depends on orbital angular momentum) and be even more inconsistent for orbits outside the equatorial plane (since actual oblateness should have the quadrupole term dependent on zenith angle, whereas GTR's is not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also possible to think of the Schwarzschild geometry itself as a scalar field, which we can similarly decompose into spherical harmonic components. Naturally, like most of the above, this peculiarity is specific to the niceness of the spherically symmetric vacuum.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditDate="2014-11-08T18:36:31.063" LastActivityDate="2014-11-08T18:36:31.063" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="7905" PostTypeId="1" CreationDate="2014-11-08T10:08:54.470" Score="5" ViewCount="885" Body="&lt;p&gt;The number density of stars (e.g. the number of stars per cubic parsec) is given by &#xA;$$N(z) = N(0) e^{-|z|/h}$$&#xA;where $z$ is the height above the Galactic plane, $h$ is the scale height for a specific component of the Galaxy, and $N(z)$ is the number density at different heights above the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scale heights for the thin disk and thick disks are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Halo stars: $h=3\textrm{kpc}$&lt;/li&gt;&#xA;&lt;li&gt;Thick disk: $h=1.5\textrm{kpc}$&lt;/li&gt;&#xA;&lt;li&gt;Thin disk stars: $h=300\textrm{pc}$&lt;/li&gt;&#xA;&lt;li&gt;Thin disk gas: $h=100\textrm{pc}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I cannot find the values for the number density at the galactic plane: $N(0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have any references in which I can find these numbers?&lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2014-11-10T08:45:42.563" Title="Number density of stars on the Galactic plane" Tags="&lt;milky-way&gt;&lt;disk&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7906" PostTypeId="5" CreationDate="2014-11-08T11:19:59.120" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-11-08T11:19:59.120" LastActivityDate="2014-11-08T11:19:59.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7907" PostTypeId="4" CreationDate="2014-11-08T11:19:59.120" Score="0" Body="Questions on astronomical objects orbiting a star massive enough to be rounded, not massive enough to cause fusion, and which have cleared its orbit of planetesimals." OwnerUserId="2674" LastEditorUserId="2674" LastEditDate="2014-11-09T14:22:01.300" LastActivityDate="2014-11-09T14:22:01.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7908" PostTypeId="5" CreationDate="2014-11-08T11:24:24.853" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-11-08T11:24:24.853" LastActivityDate="2014-11-08T11:24:24.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7909" PostTypeId="4" CreationDate="2014-11-08T11:24:24.853" Score="0" Body="Questions on systems to uniquely determine position of a point or other geometric element on a manifold such as Euclidean space." OwnerUserId="2674" LastEditorUserId="2674" LastEditDate="2014-11-09T14:21:15.163" LastActivityDate="2014-11-09T14:21:15.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7910" PostTypeId="5" CreationDate="2014-11-08T11:28:05.787" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-11-08T11:28:05.787" LastActivityDate="2014-11-08T11:28:05.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7911" PostTypeId="4" CreationDate="2014-11-08T11:28:05.787" Score="0" Body="Questions on celestial bodies that orbit planets." OwnerUserId="2674" LastEditorUserId="2674" LastEditDate="2014-11-09T14:20:48.973" LastActivityDate="2014-11-09T14:20:48.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7912" PostTypeId="5" CreationDate="2014-11-08T11:34:43.750" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-11-08T11:34:43.750" LastActivityDate="2014-11-08T11:34:43.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7913" PostTypeId="4" CreationDate="2014-11-08T11:34:43.750" Score="0" Body="Questions involving a star along with all celestial material in its sphere of influence." OwnerUserId="2674" LastEditorUserId="2674" LastEditDate="2014-11-09T14:21:32.410" LastActivityDate="2014-11-09T14:21:32.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7914" PostTypeId="5" CreationDate="2014-11-08T11:35:45.790" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-11-08T11:35:45.790" LastActivityDate="2014-11-08T11:35:45.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7915" PostTypeId="4" CreationDate="2014-11-08T11:35:45.790" Score="0" Body="Questions about the outermost solid or liquid layer of a celestial body." OwnerUserId="2674" LastEditorUserId="36852" LastEditDate="2020-11-22T14:16:09.340" LastActivityDate="2020-11-22T14:16:09.340" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="7917" PostTypeId="1" AcceptedAnswerId="7985" CreationDate="2014-11-09T04:58:34.997" Score="6" ViewCount="1268" Body="&lt;p&gt;I have a &lt;a href=&quot;http://www.celestron.com/browse-shop/astronomy/astronomy-binoculars/skymaster-20x80-binocular&quot; rel=&quot;noreferrer&quot;&gt;Celstron SkyMaster 20x80&lt;/a&gt; binoculars, and a tripod — &lt;a href=&quot;http://www.vanguardworld.com/index.php/en/pv/products/photo-video/detail-1-4-15-88.html&quot; rel=&quot;noreferrer&quot;&gt;Vanguard 263AT&lt;/a&gt; with a &lt;a href=&quot;http://www.vanguardworld.com/index.php/en/pv/products/photo-video/detail-1-4-23-144.html&quot; rel=&quot;noreferrer&quot;&gt;SBH-100 head&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The binoculars weighs 2.2kg, while the head is rated to a loading capacity of 10kg, and the legs to a tilted loading capacity of 5kg. This means that the tripod should work well, but it doesn't — it vibrates a lot even when my face gently touches the eyepiece, even without the center column extended, so much so that it becomes a pain to use the binoculars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm considering upgrading my tripod legs. I'd like this tripod to work well not just with the 20x80 binoculars I have, but also with a 25x100 pair of binoculars I'm considering buying (weight: 4.6kg), and maybe a small telescope (say with a 6-inch aperture).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct in understanding that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I need to upgrade my tripod legs.&lt;/li&gt;&#xA;&lt;li&gt;I don't need to upgrade the head, since it's rated to twice the capacity of the legs, and to twice the weight of the biggest binoculars I'll use.&lt;/li&gt;&#xA;&lt;li&gt;I should look for legs that are rated to twice the weight of the biggest binoculars I'll use.&lt;/li&gt;&#xA;&lt;li&gt;I should look for a set of wooden legs, since these have greater stability than aluminium legs while being affordable, unlike carbon fiber.&lt;/li&gt;&#xA;&lt;li&gt;What else should I keep in mind when I look for an astronomy tripod?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I &lt;a href=&quot;http://thewirecutter.com/reviews/best-tripod/&quot; rel=&quot;noreferrer&quot;&gt;read&lt;/a&gt; that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] the most popular materials are aluminum, carbon fiber, and,&#xA;  somewhat surprisingly, wood. It’s one of those common compromises—pick&#xA;  two of the following three: weight, price, and/or vibration dampening.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Wood is extremely good at absorbing vibrations, and is pretty&#xA;  affordable, but weighs a ton. Carbon fiber is light and stable, but&#xA;  you’ll pay for it. Aluminum is affordable, and fairly light, but is&#xA;  prone to channeling vibrations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I'll prioritise stability and affordability over weight.&lt;/p&gt;&#xA;" OwnerUserId="962" LastEditorUserId="962" LastEditDate="2014-11-09T05:44:55.630" LastActivityDate="2016-12-06T06:52:42.690" Title="What should I look for in an astronomy tripod?" Tags="&lt;telescope&gt;&lt;binoculars&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7919" PostTypeId="2" ParentId="7917" CreationDate="2014-11-09T14:58:35.823" Score="3" Body="&lt;p&gt;I have the exact same binocular, and in fact I also have a very similar tripod and &lt;a href=&quot;http://www.bhphotovideo.com/c/product/743889-REG/Benro_BH1_BH1_Ballhead_with_Quick.html&quot; rel=&quot;nofollow&quot;&gt;head&lt;/a&gt;. I'll take a photograph this evening to demonstrate my setup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am very happy with the setup, and I find it very stable. I usually sit the tripod on grass, but I do not recall ever having a problem with it on concrete or stone. I do try to keep the neck as short as possible, and use the legs for height. Also, the binocular can have it's tripod mount moved fore or aft, I think that mine is in the full aft position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be able to hang a large mass (bag, perhaps) from the underside of the tripod; you should find a hook already there made specifically for this purpose.&lt;/p&gt;&#xA;" OwnerUserId="209" LastEditorUserId="209" LastEditDate="2014-12-24T11:40:05.977" LastActivityDate="2014-12-24T11:40:05.977" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7925" PostTypeId="1" AcceptedAnswerId="8668" CreationDate="2014-11-09T21:29:25.320" Score="3" ViewCount="879" Body="&lt;p&gt;I have a lot of data taken in R, B, V, Luminance and Halpha-bands which I want to analyse photometrically (one target in the frame).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The frames are already reduced (with flats, darks etc.) and have a correct wcs-header. I also already have the instrumental magnitude of my target and other stars in the frame (using APT).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question now is, how do I calculate the real magnitude of my targets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most simple way would be comparing the other stars' instrumental magnitude and real magnitude and then applying the (logarithmic/linear?) transformation to the instrumental magnitude of my target. But I don't know how exact that would be and how I could calculate the error of the photometric solution (by applying the same transformation to the error of the instrumental magnitude APT gave me?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more exact photometry I don't have enough data, aka air mass or calibration fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have any ideas or solutions?&lt;/p&gt;&#xA;" OwnerUserId="2777" LastActivityDate="2015-01-24T22:08:49.737" Title="Instrumental magnitude to &quot;real&quot; magnitude - Photometry with not enough data?" Tags="&lt;star&gt;&lt;photography&gt;&lt;luminosity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7926" PostTypeId="2" ParentId="6318" CreationDate="2014-11-09T23:57:08.580" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;... does that mean that any sufficiently large volume of mass over that density is also a black hole? Or does the actual concentration of mass within the event horizon matter?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not completion sure what distinction you're drawing between &lt;em&gt;concentration&lt;/em&gt; and &lt;em&gt;density&lt;/em&gt;, but I will assume that what you mean by the former is the details of the matter distribution, e.g., whether it's concentrated at the center, spread throughout, or whatnot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a spherically symmetric isolated body of mass $M$, it is completely irrelevant. The reason is &lt;a href=&quot;https://en.wikipedia.org/wiki/Birkhoff%27s_theorem_%28relativity%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Birkhoff's theorem&lt;/a&gt;: outside the gravitating body, spacetime geometry is necessarily Schwarzschild. This is the general-relativistic analogue of Newton's &lt;a href=&quot;https://en.wikipedia.org/wiki/Shell_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;shell theorem&lt;/a&gt;. Therefore, it doesn't matter whether the (radial) distribution is uniform, concentrated at the center, or some kind of shell, or anything else: once it is compact enough that its outer surface gets to the Schwarzschild radius $2GM/c^2$ or below, it is fully enclosed by an event horizon, and is therefore a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So under those assumptions, the answers to your two questions are 'yes' and 'no', respectively, although you might want to be &lt;a href=&quot;https://physics.stackexchange.com/questions/45963/do-black-holes-have-infinite-areas-and-volumes&quot;&gt;careful&lt;/a&gt; about how you define 'volume' when comparing overall densities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens if we get rid of the assumption of spherical symmetry is a bit more complicated. If we're in an asymptotically flat universe, then we can think of a black hole as all the event from which an ideal light ray fails to escape to infinity, and the boundary would be the event horizon; more generally, we might have to be more careful about how we define 'inside' and 'outside'. Note that this makes the event horizon depend on the future, i.e. it depends on what light rays escape or don't escape even if you wait an arbitrarily long time for them. Hence, in a dynamic situation (such as a collapse to a black hole), where the location of the event horizon depends on not only on the past and present, but also on what &lt;em&gt;will&lt;/em&gt; fall into the black hole in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes general statements about density quite difficult in situations that don't have some simplifying assumptions. Density is too simplistic; the general notions of a black hole and event horizon are highly non-local.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, there is a general result that is morally similar to the above that is very relevant to your second question: the &lt;a href=&quot;https://en.wikipedia.org/wiki/No-hair_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;no-hair theorem&lt;/a&gt;. In general relativity, any isolated black hole is fully characterized by conserved quantities at infinity (mass, angular momentum, electric charge...). That means that the details of the matter distribution inside the event horizon do not matter at all. Of course, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Penrose%E2%80%93Hawking_singularity_theorems&quot; rel=&quot;nofollow noreferrer&quot;&gt;singularity theorems&lt;/a&gt; guarantee that at least under some general assumptions about the behavior of the matter, it will collapse to a singularity, but that's a separate issue.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2014-11-09T23:57:08.580" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7927" PostTypeId="2" ParentId="7905" CreationDate="2014-11-10T08:45:42.563" Score="2" Body="&lt;p&gt;The number density of stars is not well constrained, because it hinges critically on the behaviour of the stellar mass function at low stellar masses. Low-mass stars tend to be dim and hard to detect. What is reasonably well constrained, instead is the mass density of stars, where low-mass objects contribute little, even if their number is large.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-11-10T08:45:42.563" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7928" PostTypeId="1" CreationDate="2014-11-10T15:23:27.673" Score="3" ViewCount="545" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/2245/2734&quot;&gt;This answer&lt;/a&gt; says that some models describes the universe as finite. How do those models describe the universe's border?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the border (theoretically) exist?&#xA;Is it a solid border? Do they predict what happens if matter crosses this border (maybe it simply can't)?&lt;/p&gt;&#xA;" OwnerUserId="2734" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2014-11-11T03:18:40.617" Title="How is the universe bordered?" Tags="&lt;universe&gt;&lt;big-bang-theory&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7929" PostTypeId="2" ParentId="7928" CreationDate="2014-11-10T19:53:54.263" Score="2" Body="&lt;p&gt;Your intuition is correct.  It does not really make any sense for the universe to have a border.  Would it be an unbreakable wall? What would be on the other side?  Models with a finite universe would not have a border.  The universe would be curved into a closed loop like the surface of a sphere.  In other words, if you go far enough in any one direction you come back to the starting point.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2014-11-10T19:53:54.263" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7930" PostTypeId="1" CreationDate="2014-11-10T21:00:07.587" Score="1" ViewCount="293" Body="&lt;p&gt;In films such as Interstellar, 2001: A Space Odyssey, and Contact (and also Star Trek Deep Space Nine) wormholes are depicted as tunnels along the edges of which fantastical colours and shapes lie. For example, this is a picture from 2001:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://1.bp.blogspot.com/--KJvwpd7xXM/UUXkAKsi0QI/AAAAAAAAJI0/lCbtiNc4Pn0/s1600/2001+a+space+odyssey6.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would passing through a wormhole be really like though?&lt;/p&gt;&#xA;" OwnerUserId="1494" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:47.247" LastActivityDate="2016-11-30T22:38:40.887" Title="Is a Wormhole a tunnel?" Tags="&lt;wormhole&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2014-11-18T19:56:10.690" ContentLicense="CC BY-SA 3.0" />
  <row Id="7931" PostTypeId="5" CreationDate="2014-11-10T21:01:52.100" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-11-10T21:01:52.100" LastActivityDate="2014-11-10T21:01:52.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7932" PostTypeId="4" CreationDate="2014-11-10T21:01:52.100" Score="0" Body="Tag for questions asking if fictional ideas are possible or correct" OwnerUserId="1494" LastEditorUserId="1494" LastEditDate="2014-11-17T10:32:09.250" LastActivityDate="2014-11-17T10:32:09.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7933" PostTypeId="2" ParentId="7928" CreationDate="2014-11-11T02:29:02.477" Score="2" Body="&lt;p&gt;Take a sheet of paper. Glue/tape/whatever pairs of opposing edges together without twisting. The first such gluing will give a cylinder. The second then gives you a torus (a doughnut). This is a closed 2-d surface. It has no border. Things can just loop back to where they started by heading in a straight line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Practical note: you're likely to get a bit of a sharp bend in your torus, but that's the fault of the paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, try drawing curves on the paper to see what happens. If it's unfolded, connect points on borders exactly opposite each other (straight up/down or left/right on the paper). Fold it into a torus, and what do you see? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When folded draw some lines on the torus that pass through the edges of the paper. As curvy as you want. Labelling each edge entry and exit will help in the next part. Now unfold it and what do you see?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, if you've ever played the old arcade game Asteroids then you are already familiar with life on a torus. The lines you're drawing will behave exactly as your ship in the game will when you think of the unfolded paper as the screen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is basically what is meant by a closed universe. A more precise description requires the use of topology and the concept of a (smooth) manifold. Manifolds can have boundaries (the ends of the cylinder, for example), though as far as I know all serious cosmological models are boundaryless. There would be interesting and distinctive physics on a boundary, to the point that it should have been pretty obvious if one existed in the visible universe.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2014-11-11T02:29:02.477" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7934" PostTypeId="2" ParentId="7928" CreationDate="2014-11-11T03:18:40.617" Score="2" Body="&lt;p&gt;The models we're talking about are in the Friedmann–Robsetson–Walker family of solutions of general relativity, which are used because they have the properties of being spatially &lt;em&gt;homogeneous&lt;/em&gt;, i.e. they are the same at every point in space, and also &lt;em&gt;isotropic&lt;/em&gt;, i.e. every direction is equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many possible finite yet boundaryless spatial geometries, and I recommend &lt;strong&gt;zibadawa timmy&lt;/strong&gt;'s answer for a simple illustration of a toroidal two-dimensional universe and the accompanying analogy of the classic arcade game &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Asteroids_%28video_game%29&quot; rel=&quot;nofollow&quot;&gt;Asteroids&lt;/a&gt;&lt;/em&gt;. It is probably the most straightforward illustration that being finite does not imply having a border while at the same time making it intuitive that we don't need to actually consider a higher-dimensional space to embed it it in, because the rules of &lt;em&gt;Asteroids&lt;/em&gt; don't actually need an extra dimensions. Such embeddings are just convenient visualizations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, where the toroidal universe differs from the FRW models is that it's not isotropic. You can see this by the fact that on a torus, going in some directions will get you back where you started, whereas going in others will wind you endlessly along the torus, never quite making it exactly where you've started at. Thus, not all directions behave the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are only four kinds of spatial geometries for the three-dimensional space of our universe that are homogeneous and isotropic: the Euclidean space $\mathbf{E}^3$, the hyperbolic space $\mathbf{H}^3$, the sphere $\mathbf{S}^3$, and the projective real space $\mathbf{RP}^3$, the last of which is like a sphere but with a different global topology.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So it's some kind of 4 dimension sphere (sphere in any direction you go in 3D) if the universe is finite?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A three-dimensional sphere, actually. Probably the most important intuitive leap here is that any particular embedding, or even whether there exists any embedding, is completely irrelevant. The surface of an ordinary beach-ball is actually a two-dimensional sphere, and as a manifold it makes sense whether or not we think of it a surface of some three-dimensional object. When we're talking about the universe, an embedding is a tool for visualization, not reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe could be a $3$-sphere, which we could think of as the surface of a ball in flat four-dimensional space, but there is no need to. That four-dimensional space wouldn't be part of our universe (not in general relativity, anyway), and such embeddings aren't unique. Actually, for completely general spacetime manifolds (i.e. if we try to embed spacetime in flat higher-dimensional spacetime), they're not even guaranteed to exist, which is a good thing because we don't need them--all our measurements are within the universe.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-11-11T03:18:40.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7936" PostTypeId="1" CreationDate="2014-11-11T08:55:22.440" Score="10" ViewCount="1952" Body="&lt;p&gt;Searching is easy to find terminator line (frontier between day and night) or the position of the sun in the sky given a position on the earth and a time; but I can't find how to obtain where is the zenith of the sun given a date (and time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to obtain the center of the illuminated zone of the earth (latitude and longitude) at a given time. (Well, actually I need the opposite, the Nadir, but with one you can calculate easily the other).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any knows the function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2784" LastActivityDate="2015-06-23T23:29:15.973" Title="Given a date obtain latitude and longitude where is the sun zenith" Tags="&lt;the-sun&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7938" PostTypeId="1" CreationDate="2014-11-11T15:52:58.277" Score="9" ViewCount="664" Body="&lt;p&gt;The following equation (which I shall term the Planetary Precession Formula, PPF for short) famously appeared in a 1915  &lt;a href=&quot;http://www.gsjournal.net/old/eeuro/vankov.pdf&quot; rel=&quot;noreferrer&quot;&gt;publication&lt;/a&gt; by Einstein where he indicated how it could be derived from his General Theory of Relativity (GTR).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\epsilon = \frac{24 \, \pi^3a^2}{c^2 T^2(1-e^2)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\epsilon$ is the (anomalous, non-Newtonian) angular precession per orbit, $a$ is the orbit semi-major axis, $c$ is the speed of light, $T$ is the orbital period, $e$ is the orbit ellipticity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The PPF formula accurately predicts the (anomalous, non-Newtonian) precession of Mercury and other Solar planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula was known in scientific circles well before 1915.  For example Gerber (1898) derived it from his own (widely-derided) model of gravity.  In the internet article &lt;a href=&quot;http://www.mathpages.com/home/kmath527/kmath527.htm&quot; rel=&quot;noreferrer&quot;&gt;Gerber's Gravity&lt;/a&gt; it is written that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It became a fairly popular activity in the 1890s for physicists to propose various gravitational potentials based on finite propagation speed in order to account for some or all of Mercury's orbital precession.  Oppenheim published a review of these proposals in 1895.  The typical result of such proposals is a predicted non-Newtonian advance of orbital perihelia per revolution of... &gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$k\,\frac {\pi\,m}{L \,c^2} =  k \frac{4 \, \pi^3a^2}{c^2 T^2(1-e^2)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $L = a(1 - e^2)$  is the semi-latus rectum of an ellipse, $m$ is a function of the angular speed $\omega$ of an orbiting planet: $m = a^3 \omega^2$ with $\omega = 2\pi/T$ and $k$ is a constant which might be derived from theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly with $k = 6$ we get the PPF formula given above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish to know where the $k\pi m/Lc^2$ expression comes from. From the article it would appear to come from the 28-page review paper by Oppenheim, 1895 which is scanned &lt;a href=&quot;http://daten.digitale-sammlungen.de/~db/0007/bsb00075099/images/index.html?id=00075099&amp;amp;fip=193.174.98.30&amp;amp;no=&amp;amp;seite=1&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. I have been through the scans of this paper but without finding that equation explicitly (the paper is in German which I know very poorly, Google Translate helps a bit but leaves a lot of ambiguity). It might be that the anonymous author of the article extracted the expression from a review of Oppenheim's paper or even the original (French &amp;amp; German) papers themselves, but he is not contactable.  Maybe someone here is familiar with this era of astrophysical history and can point me in the right direction?&lt;/p&gt;&#xA;" OwnerUserId="233" LastEditorUserId="233" LastEditDate="2014-11-12T09:40:36.597" LastActivityDate="2014-11-12T09:40:36.597" Title="Where did this famous Planetary Precession Formula come from?" Tags="&lt;orbit&gt;&lt;general-relativity&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7939" PostTypeId="1" AcceptedAnswerId="7944" CreationDate="2014-11-11T17:33:53.670" Score="5" ViewCount="308" Body="&lt;p&gt;If the expansion of the universe (the metric) continues (and perhaps is accelerating), in a very large but finite time the expansion of the metric will clash with the effects of the strong nuclear force. What happens then? Would the energy required halt the expansion and if so what is the slowing effect of the &quot;stretch&quot; on the strong force today - too small to be measured?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Update&lt;/em&gt;: Thanks for the answers so far - I hadn't even considered the big rip scenario. I was merely thinking that as the metric expanded the strong force inside baryons would require unlimited resources of energy to be overcome - but I now understand that is wrong however it still looks to me like a potential perpetual motion machine - continually creating matter/anti-matter pairs. Perhaps that should be a separate question though...&lt;/p&gt;&#xA;" OwnerUserId="823" LastEditorUserId="823" LastEditDate="2014-11-12T17:38:54.940" LastActivityDate="2014-11-13T17:19:26.560" Title="Strong force and metric expansion" Tags="&lt;cosmology&gt;&lt;general-relativity&gt;" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7941" PostTypeId="2" ParentId="7939" CreationDate="2014-11-11T21:54:25.080" Score="3" Body="&lt;p&gt;Definitely too small to measure. The expansion isn't even enough to measurably impact galaxy superclusters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are otherwise thinking of is known as &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Big_Rip&quot; rel=&quot;nofollow&quot;&gt;the Big Rip&lt;/a&gt;. This is not guaranteed to happen, though. It depends on a constant describing the density of dark energy over time, usually denoted $w$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $w&amp;lt;-1$ the Big Rip occurs in finite time, separating all parts of a structure in the sense that lightspeed signals can no longer be transmitted between the parts. This includes hadrons like protons, so in the very last moments the quarks will no longer be able to communicate and will thus cease to be bound together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disclaimer: That ignores quantum gravity and QCD in highly curved spacetime. We can barely manage QCD in the flat case, and quantum gravity has been a holy grail for decades. Since a definitive answer to your question requires both of these things, the only definitive answer is &quot;nobody knows for sure&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $w=-1$ we have constant density and structures smaller than galaxies survive. Our solar system, assuming it survives a very long time, will remain bound together. Star clusters will remain together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, however, that we would still, technically, be able to see everything we could see before. Telescopes can already see galaxies which are currently receding at superluminal speeds. We just see those last moments before such speeds were hit get smeared over ever longer time frames, and heavily redshifted. But we will never receive new signals from the moment it hit superluminal speeds. Only ever the faint remnants of those last moments. The redshifts will eventually be so extreme that everything at galactic distances and beyond becomes effectively impossible to detect these remnants from. At which point it's just you and your closest neighbors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current observations put $w$ close to -1, but we cannot currently distinguish between $w&amp;lt;-1$, $w=-1$, and $w&amp;gt;-1$&lt;/p&gt;&#xA;" OwnerUserId="2188" LastEditorUserId="2188" LastEditDate="2014-11-11T23:34:35.020" LastActivityDate="2014-11-11T23:34:35.020" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7943" PostTypeId="2" ParentId="7930" CreationDate="2014-11-12T00:51:50.710" Score="1" Body="&lt;p&gt;A wormhole, or Einstein-Rosen bridge, is inherently unstable. It will collapse before anything can traverse it. Keeping it stable enough for even a photon to get through requires negative energy densities. And that requires exotic and suspicious new physics in most cases. Technically there are well-known effects which can produce negative energy densities, such as the Casimir effect. But being able to harness such an effect to stabilize a wormhole is beyond current technology (to say nothing of creating the wormhole in the first place). We have no observational evidence to suggest wormholes exist, especially not on the scales you see in Sci-Fi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the rapidly collapsing case it looks like nothing except possibly a spray of subatomic particles (which you might not even know aren't the usual background ones), and in the stable case it's impossible to say because you'd have to know what the exotic matter or other negative densities look like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for what it's like to traverse a stable wormhole, as long as the exotic matter doesn't mess with you then it would be pretty much the same as any other spaceship travel. You could just have interesting time travel effects and beat light rays that went around the wormhole. A number of these are mentioned on the wiki page.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2014-11-12T00:51:50.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7944" PostTypeId="2" ParentId="7939" CreationDate="2014-11-12T02:52:54.843" Score="4" Body="&lt;p&gt;Excepting a &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Rip&quot; rel=&quot;nofollow&quot;&gt;Big Rip&lt;/a&gt; scenario, there is no eventual 'clash'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann%E2%80%93Lema%C3%AEtre%E2%80%93Robertson%E2%80%93Walker_metric&quot; rel=&quot;nofollow&quot;&gt;Friedmann–Lemaître–Robertson–Walker universe&lt;/a&gt;:&#xA;$$\mathrm{d}s^2 = -\mathrm{d}t^2 + a^2(t)\left[\frac{\mathrm{d}r^2}{1-kr^2} + r^2\left(\mathrm{d}\theta^2 + \sin^2\theta\,\mathrm{d}\phi^2\right)\right]\text{,}$$&#xA;where $a(t)$ is the scale factor and $k\in\{-1,0,+1\}$ corresponds to a spatially open, flat, or closed cases, respectively. In a local orthonormal frame, the nonzero Riemann curvature components are, up to symmetries:&#xA;$$\begin{eqnarray*}&#xA;R_{\hat{t}\hat{r}\hat{t}\hat{r}} = R_{\hat{t}\hat{\theta}\hat{t}\hat{\theta}} = R_{\hat{t}\hat{\phi}\hat{t}\hat{\phi}} =&amp;amp; -\frac{\ddot{a}}{a} &amp;amp;= \frac{4\pi G}{3}\left(\rho + 3p\right)\text{,} \\&#xA;R_{\hat{r}\hat{\theta}\hat{r}\hat{\theta}} = R_{\hat{r}\hat{\phi}\hat{r}\hat{\phi}} = R_{\hat{\theta}\hat{\phi}\hat{\theta}\hat{\phi}} =&amp;amp; \frac{k+\dot{a}^2}{a^2} &amp;amp;= \frac{8\pi G}{3}\rho\text{,}&#xA;\end{eqnarray*}$$&#xA;where overdot denotes differentiation with respect to coordinate time $t$ and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;nofollow&quot;&gt;Friedmann equations&lt;/a&gt; were used to rewrite them in terms of density $\rho$ and pressure $p$. From the link, note in particular that $\dot{\rho} = -3\frac{\dot{a}}{a}\left(\rho+p\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If dark energy is described by a cosmological constant $\Lambda$, as it is in the standard &lt;a href=&quot;https://en.wikipedia.org/wiki/Lambda-CDM_model&quot; rel=&quot;nofollow&quot;&gt;ΛCDM&lt;/a&gt; model, then it contributes a constant density and pressure $\rho = -p = \Lambda/8\pi G$, and so no amount of cosmic expansion would change things. Locally, things look just the same as they ever did: for a universe dominated by dark energy, the curvature stays the same, so the gravitational tidal forces do too. Through those tiny tidal forces, the dark energy provides some immeasurably tiny perturbation on the behavior of objects, including atomic nuclei, forcing a slightly different equilibrium size than would be otherwise. But it is so small that that it has no relevance to anything on that scale, nor do those equilibrium sizes change in time. The cosmological constant holds true to its name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if dark energy has an equation of state $p = w\rho$, then a flat expanding universe dominated by dark energy has&#xA;$$\dot{\rho} = -3\frac{\dot{a}}{a}\left(\rho+p\right) = -3\sqrt{\frac{8\pi G}{3}}\left(1+w\right)\rho^{3/2}\text{,}$$&#xA;and immediately one can see that there is something special about $w&amp;lt;-1$, leading to an accumulation of dark energy, while the cosmological constant's $w = -1$ leads to no change. This leads to a Big Rip more generally, as &lt;strong&gt;zibadawa timmy&lt;/strong&gt;'s answer explains.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;? If the metric expands surely objects get further away from one another and that would include the stars inside galaxaies as well as the galaxies themselves?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not at all. It wouldn't even make any sense: if you have an object like an atom or a star in gravitational freefall, by the equivalence principle only tidal forces across it are relevant. The tidal forces stretch the object until the internal forces balance them. But for a Λ-driven accelerated expansion, dark energy contribution to tidal forces is &lt;em&gt;constant&lt;/em&gt;. Hence, an object already in equilibrium has no reason to further change its size, no matter how long the cosmic acceleration occurs. This also applies to galaxies, only that the internal forces are also gravitational and balance the dark energy contribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at this in more detail, imagine a test particle with four-velocity $u$, and a nearby one with the same four-velocity, separated by some vector $n$ connecting their worldlines. If they're both in gravitational freefall, then their relative acceleration is given by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Geodesic_deviation&quot; rel=&quot;nofollow&quot;&gt;geodesic deviation equation&lt;/a&gt;, $\frac{D^2n^a}{d\tau^2} = -R^\alpha{}_{\mu\beta\nu}u^\mu u^\nu n^\beta$. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Bel_decomposition&quot; rel=&quot;nofollow&quot;&gt;gravitoelectric&lt;/a&gt; part of the Riemann tensor, $\mathbb{E}^\alpha{}_\beta = R^\alpha{}_{\mu\beta\nu}u^\mu u^\nu$, represents the static tidal forces in a local inertial frame comoving with $u$, which will drive those particles apart (or together, depending). Hence, keeping those particles at the same distance would require a force between them, but it's not necessary for this force to change unless the tidal forces &lt;em&gt;also&lt;/em&gt; change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxies don't change size through cosmological expansion. Stars don't either, nor do atoms. Not for Λ-driven expansion, at least. It would take an increase in tidal forces, such as those provided by a Big Rip, for them to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A related way of looking at the issue is this: according to the Einstein field equation, the initial acceleration of a small ball of initially comoving test particles with some four-velocity $u$ (in the comoving inertial frame), which is given by the Ricci tensor, turns out to be:&#xA;$$\lim_{V\to 0}\left.\frac{\ddot{V}}{V}\right|_{t=0}\!\!\!\!= -\underbrace{R_{\mu\nu}u^\mu u^\nu}_{\mathbb{E}^\alpha{}_\alpha} = -4\pi G(\rho + 3p)\text{,}$$&#xA;where $\rho$ is density and $p$ is the average of the principal stresses in the local inertial frame. For a positive cosmological constant, $\rho = -p&amp;gt;0$, and correspondingly a ball of test particles will expand. That's cosmic expansion on a local scale; because of uniformity of the FLRW universe, the same proportionality works on the large scale as well, as we can think of distant galaxies as themselves test particles if their interactions are too minute to make a difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, we are led to the same conclusion: if the ball has internal forces that prevent expansion, then those forces don't need to change in time unless the dark energy density also changes, which doesn't happen for cosmological constant. If they're not 'clashing' &lt;em&gt;now&lt;/em&gt;, then they won't need to in the future either.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditDate="2014-11-13T17:19:26.560" LastActivityDate="2014-11-13T17:19:26.560" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7945" PostTypeId="2" ParentId="7938" CreationDate="2014-11-12T07:26:49.663" Score="3" Body="&lt;p&gt;I do not know where that formula was first published in full, but Oppenheim at least does something &lt;em&gt;very close&lt;/em&gt; to it. First, keep in mind some of the relevant symbols in Oppenheim, though they're rather standard:&#xA;$$\def\anode{☊}\begin{eqnarray*}&#xA;k &amp;amp;=&amp;amp; \small\sqrt{G} = \small\text{Gaussian gravitational constant}\\&#xA;\anode &amp;amp;=&amp;amp; \small\text{longitude of the ascending node}\\&#xA;\omega &amp;amp;=&amp;amp; \small\text{argument of perihelion}\\&#xA;\varpi &amp;amp;=&amp;amp; \omega+\anode = \small\text{longitude of perihelion}\\&#xA;n &amp;amp;=&amp;amp; \small{k\sqrt{(m_0+m_1)/a^3}} = \small\text{mean motion}&#xA;\end{eqnarray*}$$&#xA;We can see that if we massage the notation, the proportionality we're looking for is equivalently stated as:&#xA;$$\frac{\pi m}{Lc^2} \leadsto \frac{\pi GM}{a(1-e^2)c^2} \leadsto \frac{\pi n^2a^2}{(1-e^2)c^2}\text{.}$$&#xA;Since the mean motion is $n\equiv 2\pi/T$, where $T$ is the orbital period,&#xA;$$\frac{n^3a^2}{c^2} = 2\pi\frac{n^2a^2}{c^2}\frac{1}{T}\text{,}$$&#xA;which means that if we want to talk about a perihelion advance of $\delta\varpi \propto \frac{\pi n^2a^2}{(1-e^2)c^2}$ per orbit, it is equivalent to talk about a term in the form&#xA;$$\frac{\mathrm{d}\varpi}{\mathrm{d}t} \propto \frac{n^3a^2}{(1-e^2)c^2} = \frac{n^3a^2}{c^2}\left(1+e^2+\mathcal{O}(e^4)\right)\text{.}$$&#xA;I cannot find where, if indeed &lt;em&gt;anywhere&lt;/em&gt;, Oppenheim considers the missing factor of $(1-e^2)$ as belonging to the anomalous precession, but otherwise the formula is definitely there. I suspect that he only uses a circular orbit approximation, $e^2\approx 0$, because it's nowhere to be found in Sec. IV (Weber's 1846 theory) and performing his calculation (sans $1-e^2$) gives me $\delta\varpi = 13.72''$ per century for Mercury, in good agreement with his stated result of $\delta\varpi = 13.65''$, whereas putting in a factor of $(1-e^2)$ by hand gives $\delta\varpi = 14.32''$ instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps Oppenheim did not consider it explicitly and the author of the &lt;code&gt;MathPages&lt;/code&gt; took it as obvious that the eccentricity factor should be there. Or perhaps there's a side comment in the text that I don't see; unfortunately, I'm nowhere fluent enough in German to understand much of what's going on.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-11-12T07:26:49.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7946" PostTypeId="1" AcceptedAnswerId="7954" CreationDate="2014-11-12T08:31:57.463" Score="2" ViewCount="726" Body="&lt;p&gt;I'm trying to build a 3D visualization of the Supergalactic coordinate system and couldn't find any reference to where the center point should be (the Sun, galactic center, the Earth, etc). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I already tried to look into some articles and books, but they only refer to the zero point related to the galactic coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where should it be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could I use this model as the reference for the Supergalactic Coordinate system, starting to count the SGL (L) and SGB (B) from the yellow line?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is still unclear to me if (l=137.37 b=0) indicates just the direction of the center of the Supergalactic Coordinates System.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The Sun is being used as the center point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6QsVC.png&quot; alt=&quot;Galactic Coordinates&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plano Galactico translates into Galactic Plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Centro Galactico translates into Galactic Center.&lt;/p&gt;&#xA;" OwnerUserId="2790" LastEditorDisplayName="user5434" LastEditDate="2017-04-22T01:20:38.733" LastActivityDate="2017-04-22T01:20:38.733" Title="Where is the center point for the Supergalactic coordinate system?" Tags="&lt;observational-astronomy&gt;&lt;visualization&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7947" PostTypeId="1" AcceptedAnswerId="7948" CreationDate="2014-11-12T12:57:03.843" Score="5" ViewCount="473" Body="&lt;p&gt;Each object in the universe has its own gravitational influence on all other objects in the universe. What distance do they have to be from each other to create only one gravitational influence? Example - Every human does in my opinion contribute to the earth gravitation thanks to the mass of our bodies. Now would there be a difference if I distanced myself from the surface by any amount?&lt;/p&gt;&#xA;" OwnerUserId="2724" LastEditorUserId="66" LastEditDate="2014-11-14T23:06:41.660" LastActivityDate="2014-11-14T23:06:41.660" Title="From what distance can one object influence gravity of another object?" Tags="&lt;gravity&gt;&lt;earth&gt;&lt;mass&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7948" PostTypeId="2" ParentId="7947" CreationDate="2014-11-12T17:04:43.737" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What id like to know, in what distance do they have to be from each other to create only one gravitational influence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At whichever point you decide to call them two objects rather than one object. It's a completely arbitrary choice that depends on you rather than gravitational physics. What's going on is that gravity can be described by a mass density distribution, and which part of that distribution corresponds to &quot;one object&quot; and which to &quot;another object&quot; isn't important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may be confused by Newton's law of gravity that says the gravitational force force is proportional to the product of the masses and inversely proportional to the distance between them squared. But this law only applies to spherically symmetric objects. It only applies to you exactly if you're a &lt;a href=&quot;https://en.wikipedia.org/wiki/Spherical_cow&quot; rel=&quot;nofollow&quot;&gt;spherical cow&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This kind of arbitrariness actually applies to the Earth as well. Even if the Earth were perfectly spherically symmetric, one could say that Earth's gravity is due to the influence of the northern hemisphere &lt;em&gt;and&lt;/em&gt; the influence of the southern hemisphere, etc. Whether you consider the Earth to be one object, or two, or a trillion, depends on you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth isn't spherically symmetric. It's closer to an oblate spheroid, since it bulges at the equator. But it isn't exactly that either, having mountains or other topographical feature, to say nothing of mass density variations &lt;em&gt;inside&lt;/em&gt; the Earth. In principle, one could describe its gravity &lt;em&gt;including&lt;/em&gt; people and trees or whatnot. Gravitationally, there is no fundamental difference between you and some rock. That you consider yourself to not be &quot;part of Earth&quot; is a choice you make for other reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you distance yourself from the Earth, there is a difference because the mass distribution would change. But again, whether you consider this to be &quot;only one gravitational influence&quot; or &quot;the Earth plus you&quot; is up to you, and this holds true &lt;em&gt;regardless&lt;/em&gt; of whether you distanced yourself or not.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditDate="2014-11-12T21:31:34.943" LastActivityDate="2014-11-12T21:31:34.943" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7950" PostTypeId="1" AcceptedAnswerId="7952" CreationDate="2014-11-12T21:59:46.937" Score="4" ViewCount="235" Body="&lt;p&gt;Following on from &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6051/how-did-the-hulse-taylor-pulsar-provide-confirming-evidence-of-general-relativit&quot;&gt;an earlier question&lt;/a&gt; about the very interesting &lt;a href=&quot;https://en.wikipedia.org/wiki/Hulse%E2%80%93Taylor_binary&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hulse-Taylor binary pulsar&lt;/a&gt;.  The high-frequency (radio) beam from the spinning pulsar sweeps across Earth about 17 times per second.  The total power of the gravitational radiation (waves) emitted by the binary system is calculated from GTR to be $7.35 × 10^{24}$ watts at present (declining as the orbital period and radius diminish).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/pdf/astro-ph/0407149v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Weisberg &amp;amp; Taylor, 2004&lt;/a&gt; report that the beam has &quot;a flux density of about 1 mJy at 1400 MHz.&quot;  (mJy = milliJansky, thanks Stan Liou). Assuming that this flux is uniform across a conical beam with cross-sectional radius 5 arc degrees  and assuming that the source is 21,000 light years from Earth and has mass = 1.44 Solar Masses:-  How much energy is emitted (per second) from the source in the beam? Also (if it is possible to estimate a reasonable range of values) what might be the rate of steady mass loss from such a source?  &lt;/p&gt;&#xA;" OwnerUserId="233" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-11-13T16:54:32.770" Title="Hulse-Taylor binary pulsar - what is the rate of mass/energy loss from the source?" Tags="&lt;general-relativity&gt;&lt;pulsar&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7951" PostTypeId="2" ParentId="7939" CreationDate="2014-11-13T00:33:35.597" Score="2" Body="&lt;p&gt;I think there's several points that need addressing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly how expansion operates on scales below the intergalactic is very much an open question. The approximation of spatial homogeneity and isotropy fails before you reach these scales. Whilst it may be tempting to think that cosmic expansion could be modeled as something akin to a small repulsive force between objects on these scales that isn't necessarily the case. For example in the, admittedly not very realistic, Swiss-cheese model the regions around stars experience no expansion. This is before we even get to scales when we would expect to need more than general relativity to describe what is going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In de Sitter space w = -1 and it is essentially models a Universe that purely contains dark energy in the form of a cosmological constant and so it acts as a sort of limiting case for Universe with dark energy in the form of a cosmological constant. However de Sitter space, whilst it has accelerated expansion, is also static and there is no essential difference between how test particles behave in it at any given time. Consequently if a structure can exist in de Sitter space at a given time it can exist at all times, and as it is the limiting case, if a structure (like an atom for example) can exist in a Universe with a dark energy described by a cosmological constant at a given time, it can also exist at all later times too. Therefore you need to go into the realms of phantom energy for expansion that can rip apart atoms at later times in the Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An additional point I would make though is that even in a Universe where dark energy is described by a cosmological constant structures such as galaxies will come apart. However this is due to effects such galactic evaporation rather than as a direct result of expansion. &lt;/p&gt;&#xA;" OwnerUserId="2092" LastEditorUserId="2092" LastEditDate="2014-11-13T01:00:48.887" LastActivityDate="2014-11-13T01:00:48.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7952" PostTypeId="2" ParentId="7950" CreationDate="2014-11-13T02:31:11.860" Score="2" Body="&lt;p&gt;For reference, the relevant bit from the paper is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The observable pulsar is a weak radio source with a flux density of about $1\,\mathrm{mJy}$ at $1400\,\mathrm{MHz}$. ... Our most recent data have been gathered with the Wideband Arecibo Pulsar Processors (“WAPPs”), which for&#xA;  PSR B1913+16 achieve $13\,\mathrm{\mu s}$ time-of-arrival measurements in each of four $100\,\mathrm{MHz}$ bands, using $5$-minute integrations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those are millijanskys, aka milli flux units, so that the flux density is about&#xA;$$1\,\mathrm{mJy} = 10^{-29}\,\frac{\mathrm{W}}{\mathrm{m}^2\cdot\mathrm{Hz}}\text{,}$$&#xA;and hence the detected irradiance is on the order of $10^{-27}\,\mathrm{W}/\mathrm{m}^2$. Since we're about to make rather uncertain assumptions anyway, I won't bother worrying about doing more than an order-of-magnitude calculation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Assuming that this flux is uniform across a conical beam with cross-sectional radius 5 arc degrees and assuming that the source is 21,000 light years from Earth ... - How much energy is emitted (per second) from the source in the beam?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A spherical cap has surface area of $A = 2\pi Rh$, and here $R = 21\,\mathrm{kly}$. Now, I'm unclear what &lt;em&gt;cross-sectional radius&lt;/em&gt; means if measured as an angle, but I take it to mean that the opening half-angle of the cone is $\frac{\vartheta}{2} = 5^\circ$, in which case&#xA;$$A = 2\pi R^2\left(1-\cos\frac{\vartheta}{2}\right) \sim 10^{39}\,\mathrm{m}^2\text{.}$$&#xA;Thus, the power would be $P\sim 10^{12}\,\mathrm{W}$, but note that in addition to the assumptions you've just listed, we're only talking about a particular radio band.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-11-13T02:31:11.860" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7954" PostTypeId="2" ParentId="7946" CreationDate="2014-11-13T09:39:07.830" Score="0" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Supergalactic_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, the origin of the system is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;The zero point (SGB=0°, SGL=0°) lies at (l=137.37°, b=0°). In J2000 equatorial coordinates, this is approximately (2.82 h, +59.5°).&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same article there is &lt;a href=&quot;http://cdsads.u-strasbg.fr/cgi-bin/nph-data_query?bibcode=2000MNRAS.312..166L&amp;amp;db_key=AST&amp;amp;link_type=ARTICLE&quot; rel=&quot;nofollow&quot;&gt;a recent referred paper&lt;/a&gt; on this topic, which explains &lt;code&gt;[The zero point] is one of the two regions where the SGP is crossed by the Galactic plane.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you already know &lt;a href=&quot;http://fisherka.csolutionshosting.net/astronote/Structure/Supergal/Supergalacticcoords.html&quot; rel=&quot;nofollow&quot;&gt;this website&lt;/a&gt;, which shows many good pictures. I was trying to find a better figure, but I guess you have to push your imagination according to the definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: According to &lt;a href=&quot;http://books.google.co.jp/books?id=f-l9bB-7AQMC&amp;amp;pg=PA378&amp;amp;lpg=PA378&amp;amp;dq=supergalactic%20center&amp;amp;source=bl&amp;amp;ots=Z1c-n36an9&amp;amp;sig=VSjHder9IA4-DVvr2samzWw7lGA&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=rVBlVO4WotqaBZCQgMgG&amp;amp;ved=0CBwQ6AEwADgK#v=onepage&amp;amp;q=supergalactic%20center&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, SG coordinates are similar to the galactic ones, so that the North Supergalactic Pole is defined in Galactic coordinates, which means that they have the same origin (i.e., the Sun).&lt;/p&gt;&#xA;" OwnerUserId="1038" LastEditorUserId="1038" LastEditDate="2014-11-14T00:52:06.553" LastActivityDate="2014-11-14T00:52:06.553" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7955" PostTypeId="1" AcceptedAnswerId="7956" CreationDate="2014-11-13T16:01:04.217" Score="6" ViewCount="311" Body="&lt;p&gt;In many science fiction stories, we hear the term &lt;strong&gt;star system&lt;/strong&gt; that mostly refers to a star and its planets. This is often used interchangeably with the term &lt;strong&gt;solar system&lt;/strong&gt;. But after some research I understand this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Planetary system&lt;/strong&gt; refers specifically to anything orbiting a star that is not a star itself. According to Wikipedia, this generally refers to the planets, but may include other objects like asteroids and dwarf planets. It does not refer to the star itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Star system&lt;/strong&gt; refers to any system with two (binary) or more (multiple) stars, but only the stars themselves, not planets or anything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solar system&lt;/strong&gt; generally refers to our own Solar system but there are situations where &quot;solar system&quot; (lowercase) refers to a star and the planets or other objects orbiting it (as long as something orbits the star).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is it incorrect to use the term solar system/star system to refer to a star and its planets?&lt;/p&gt;&#xA;" OwnerUserId="2798" LastEditorUserId="2153" LastEditDate="2014-11-13T16:19:13.833" LastActivityDate="2014-11-13T23:06:17.447" Title="The defintion of star/planetary/solar system" Tags="&lt;solar-system&gt;&lt;space&gt;&lt;terminology&gt;&lt;star-systems&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7956" PostTypeId="2" ParentId="7955" CreationDate="2014-11-13T16:18:56.360" Score="4" Body="&lt;p&gt;I think it's largely a matter of personal preference. Some people will want you to be technically correct and only use &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_System&quot; rel=&quot;nofollow&quot;&gt;&quot;solar system&quot;&lt;/a&gt; to refer to the Sun and the objects that are under its gravitational influence. Wikipedia defines it as such:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Solar System comprises the Sun and the objects that orbit it, whether they orbit it directly or by orbiting other objects that orbit it directly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://solarsystem.nasa.gov/planets/&quot; rel=&quot;nofollow&quot;&gt;NASA&lt;/a&gt; says that the term is more general:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A solar system refers to a star and all the objects that travel in orbit around it. Our solar system consists of the sun - our star - eight planets and their natural satellites (such as our moon); dwarf planets; asteroids and comets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As an astronomy enthusiast, I do differentiate between &lt;a href=&quot;http://en.wikipedia.org/wiki/Star_system&quot; rel=&quot;nofollow&quot;&gt;&quot;stellar system&quot;/&quot;star system&quot;&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Planetary_system&quot; rel=&quot;nofollow&quot;&gt;&quot;planetary system&quot;&lt;/a&gt;. I'd use &quot;planetary system&quot; to refer solely to the planets in a system, as well as their moons, while I'd use &quot;stellar system&quot; to refer to the whole thing. That's just my opinion, though, and I'm unsure what professionals use. In casual conversation (especially with a non-expert), you can probably use whatever term you want. If you're using it formally, you should probably do some more research. And while you're at it, check whether or not you should capitalize &quot;Solar&quot; and &quot;System&quot; in &quot;Solar/solar System/system&quot;. If people nag you about terminology, they'll nag you about capitalization.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-11-13T16:18:56.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7957" PostTypeId="1" AcceptedAnswerId="8027" CreationDate="2014-11-13T19:23:06.520" Score="8" ViewCount="347" Body="&lt;p&gt;I am familiar with the theory on generating the positions of the members of the solar system, but have been unable to find out the standard practices and methods of professional astronomers in acquiring this information. The basic methods of &lt;a href=&quot;http://en.wikipedia.org/wiki/Numerical_model_of_the_Solar_System&quot;&gt;generating an ephemeris of the solar system&lt;/a&gt; can be found on the Wikipedia. Also, I have the book Astronomical Algorithms by Jean Meeus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do practicing astronomers do it though? Are there standard programs or software that everyone uses, or do they access servers at some international body that returns positions at a given time, or does each astronomer have his own custom program that he made? What is the normal way an astronomer gets planetary positions?&lt;/p&gt;&#xA;" OwnerUserId="2800" LastActivityDate="2015-09-03T19:44:00.307" Title="Standard practices and models for generating the ephemerides of the solar system" Tags="&lt;solar-system&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7959" PostTypeId="1" AcceptedAnswerId="7962" CreationDate="2014-11-13T20:45:10.153" Score="11" ViewCount="730" Body="&lt;p&gt;I was wondering why every picture we saw on internet about Rosetta's landing on 67p just black and white ? Is there something i should know ?&lt;/p&gt;&#xA;" OwnerUserId="2801" LastActivityDate="2014-11-14T02:03:07.757" Title="Why all the photos from 67p are black and white?" Tags="&lt;comets&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7960" PostTypeId="2" ParentId="7955" CreationDate="2014-11-13T23:06:17.447" Score="2" Body="&lt;p&gt;&quot;we have detailed knowledge of only one planetary system, the Solar System. Data from other planetary systems around both main sequence stars and pulsars ...&quot;[1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above quote, one can see that the authors say the Solar System is a planetary system and in the very next sentence they say &quot;planetary systems around...stars&quot; so the question of whether the term planetary system includes the central star or just the planets seems to be a linguistic issue and it changes, even by the same authors, from one sentence to the next depending on context. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Linguistics aside, from an astronomy perspective it is clear that one cannot sensibly discuss a planetary system without reference to its central star. Everything about a system depends on the star. The concept of habitable zone doesn't even make sense without the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a red giant star that has engulfed its inner planets. If the planets are massive enough they will survive. e.g. Kepler-70b. How can one exclude the star from the planetary system if there are planets inside the star, except by linguistic hair-splitting which as the opening quote shows would be contrary to actual usage which is context-dependent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Planetary Sciences by Imke de Pater and Jack J. Lissauer&lt;/p&gt;&#xA;" OwnerUserId="2169" LastActivityDate="2014-11-13T23:06:17.447" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7961" PostTypeId="2" ParentId="7959" CreationDate="2014-11-13T23:27:33.043" Score="6" Body="&lt;p&gt;True fact is that most of the images seen so far are taken from the navigation camera (NAVCAM) of the orbiter, Rosetta, and that takes black&amp;amp;white pictures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rosetta.jpl.nasa.gov/news/navcam%E2%80%99s-shades-grey&quot;&gt;Here&lt;/a&gt; is a very interesting article about how these pictures are taken (ironically it is on a NASA blog). They state that they show pretty much what you would see with a human eye from there, even thought the comet is actually all very dark. The white parts are due to the contrast with the very dark surroundings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure about the images received by the lander. They are captured by the CIVA cameras, where CIVA stands for &lt;a href=&quot;http://www.open.ac.uk/science/research/rosetta/mission/philae-lander/civa&quot;&gt;Comet Infrared and Visible Analyser&lt;/a&gt;. So I guess they might show some infrared light not visible to the human eye, then processed to be in b&amp;amp;w. Unfortunately, I could not find a precise explanation on ESA's website so far. (&lt;a href=&quot;http://sci.esa.int/rosetta/31445-instruments/?fbodylongid=892&quot;&gt;here&lt;/a&gt; their page for Philae instrumentation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, &lt;a href=&quot;http://www.universetoday.com/11863/true-or-false-color-the-art-of-extraterrestrial-photography/&quot;&gt;as a general rule you should not expect that all images from spacecraft are in true colors&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="623" LastActivityDate="2014-11-13T23:27:33.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7962" PostTypeId="2" ParentId="7959" CreationDate="2014-11-13T23:37:10.320" Score="5" Body="&lt;p&gt;Regular digital cameras &lt;a href=&quot;http://en.wikipedia.org/wiki/Color_filter_array&quot; rel=&quot;noreferrer&quot;&gt;color filter arrays&lt;/a&gt; that put the pixel sensors so that it only detects a particular color. Therefore, there is a trade-off, because without the filter array, you would have a significantly higher-resolution black-and-white image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, most scientific digital cameras work differently. Rather, the usual technique is to put filters over the entire image, and switch out the filters if a different color is required. Subsequent differently-filtered images can then be combined into one image. But there's still a trade-off: putting in filters decreases the dynamic range of the camera significantly. Additionally, this approach does not require one to use the usual &lt;em&gt;red&lt;/em&gt;, &lt;em&gt;green&lt;/em&gt;, &lt;em&gt;blue&lt;/em&gt; scheme; e.g., and IR filter would allow one to emphasize different features.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very likely that they simply didn't bother to put color filters in the cameras. Why would they? The colors are probably rather drab on the comet anyway, and it unnecessarily increases the complexity of the cameras (which is already a quite complicated panoramic array).&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-11-13T23:37:10.320" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7965" PostTypeId="1" AcceptedAnswerId="7968" CreationDate="2014-11-13T17:04:21.750" Score="23" ViewCount="562" Body="&lt;p&gt;Most descriptions of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Oort_cloud&quot;&gt;Oort cloud&lt;/a&gt; depict it as a mostly spherical distribution of planetesimals, with occasional allowance for an inner component that is more donut-shaped. This is slightly at odds with the fact that most protoplanetary clouds and their derivative objects - planets, asteroids, comets and dust - will collapse to a fairly well-defined plane relatively early in a stellar system's evolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What evidence is used to postulate this? Does it come from numerical simulations of the solar system? Or does it help account for observed orbital inclinations of real comets?&lt;/p&gt;&#xA;" OwnerUserId="207" OwnerDisplayName="Emilio Pisanty" LastActivityDate="2014-11-14T17:29:06.950" Title="Why is the Oort cloud presumed to be spherical?" Tags="&lt;oort-cloud&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7966" PostTypeId="2" ParentId="7965" CreationDate="2014-11-13T17:41:43.193" Score="8" Body="&lt;p&gt;The latter.  Long-period comets appear to come from random directions.&lt;/p&gt;&#xA;" OwnerUserId="512" OwnerDisplayName="Mark Adler" LastActivityDate="2014-11-13T17:41:43.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7967" PostTypeId="1" CreationDate="2014-11-14T10:56:55.260" Score="5" ViewCount="146" Body="&lt;p&gt;What are the experiments, or measurements, that can detect, or account for, dark energy involved in making calculations concerning Earth and its only natural satellite? &lt;/p&gt;&#xA;" OwnerUserId="2809" LastActivityDate="2014-11-14T23:12:23.943" Title="Is the dark energy between the moon and Earth measurable in any capacity?" Tags="&lt;dark-energy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7968" PostTypeId="2" ParentId="7965" CreationDate="2014-11-14T12:03:47.083" Score="5" Body="&lt;p&gt;Nobody has &quot;seen&quot; the Oort-cloud (yet). The Oort cloud is simply a concept that can explain why long-period comets appear to come from random directions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the current instruments, we are not able to detect any of these comets &quot;at the source&quot;. It is also not even possible to show with measurements that at that distance there might be a companion-object for the sun (long-period binary, with a brown dwarf companion going through the oort-cloud, which could explain some things about periodical mass-extinctions). We can however put some limits on the mass and distance of such an object, but we can not yet prove with measurements that it is impossible. This just to show how small the amount of information is that we have on these distances. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing we know about objects that are there, is what we see from objects that come our way, and when we calculate the orbit, we notice that it comes from the same region of the solar system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &lt;a href=&quot;http://iopscience.iop.org/0004-637X/781/1/4/&quot; rel=&quot;noreferrer&quot;&gt;this publication &lt;/a&gt; shows that the WISE mission has been able to narrow it down, showing that if a Jupiter-mass brown dwarf exists in our solar system, it has to be at least at a distance of 26,000 AU, to stay under the detection limits of WISE. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is not to say wether or not such an object exists or not, but to point out that at those distances, we can only detect things that are massive, compared to the average comet. This shows that the only information we have about the Oort-cloud, is indirect information from objects that are on orbits going through the Oort-cloud, and near enough to earth for us to detect them.&lt;/p&gt;&#xA;" OwnerUserId="1062" LastEditorUserId="1062" LastEditDate="2014-11-14T14:58:19.687" LastActivityDate="2014-11-14T14:58:19.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7969" PostTypeId="2" ParentId="1041" CreationDate="2014-11-14T16:59:41.303" Score="-1" Body="&lt;p&gt;For some very large (and hence relatively cool) red giants you might be able to ascertain something from their spectra, as emission lines are sometimes seen - these are typically brighter central patches seen the middle of the more typical absorption (dark) spectral lines - caused by the large size of (realtively!) hot gas clouds that surround the giants. But that would not be a reliable method of red giant detection.&lt;/p&gt;&#xA;" OwnerUserId="823" LastEditorUserId="823" LastEditDate="2014-11-16T14:26:40.037" LastActivityDate="2014-11-16T14:26:40.037" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7970" PostTypeId="1" AcceptedAnswerId="8128" CreationDate="2014-11-14T17:16:23.890" Score="9" ViewCount="840" Body="&lt;p&gt;If I'm not mistaken, it is believed that the reason for such turbulent weather on the 4 outer gas giant planets is that the internal pressure is so great that it is generating heat, which is causing convection, which causes extreme weather.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will these planets forever generate heat, or at some point will they freeze up?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What will Jupiter look like a trillion years from now?&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2014-12-04T13:06:56.293" Title="What is the long term fate of the gas giants?" Tags="&lt;gas-giants&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7971" PostTypeId="2" ParentId="7965" CreationDate="2014-11-14T17:23:51.497" Score="5" Body="&lt;p&gt;In addition to Mark's answer, we also have reasons to expect a spherical distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following makes some assumptions on how our solar system formed. They are standard, but we are not completely certain on its correctness. What I use is usually considered noncontroversial--it's how the planets themselves arose that is most problematic, but is not needed here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Early in the formation of the solar system the gas and dust would have had a fairly uniform and spherical distribution. It is unlikely that the cloud would have exactly 0 net angular momentum, meaning it would have net angular momentum in some direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the gas that is sufficiently close to the sun will be dense enough that the particles will be interacting and colliding regularly. This causes the angular momentum of the particles to align in the direction of the original net angular momentum. This is due to the conservation of angular momentum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process creates the dominant protoplanetary disc you are familiar with, leaving a thin layer of low density gas and dust in the same sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Low density particle distributions will be essentially collisionless. They will therefore not align themselves into a disc, whether they have a net angular momentum or not. Each particle orbits on whatever plane it just happens to be aligned on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to the Oort cloud...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Get far enough from the center of the formation of our sun and the gas becomes less dense. As such the gas becomes mostly collisionless, and preferential alignment on a disc becomes less likely. Stay just close enough and sufficient interactions and random inhomogeneities arise for planetesimals to build up, each one aligned essentially independently of the others. They remain sparsely distributed and collisionless as a whole (basically the particles just got bigger), and so do not align.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The models you see with a doughnut-like region are ones that are expecting a region where the dust and gas was still interacting enough with itself, and the rest of the solar system as we know it, to still (partially) fall into the preferred alignment.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastEditorUserId="2188" LastEditDate="2014-11-14T17:29:06.950" LastActivityDate="2014-11-14T17:29:06.950" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7972" PostTypeId="2" ParentId="7970" CreationDate="2014-11-14T17:29:59.607" Score="1" Body="&lt;p&gt;From what I know, the heat was mostly generated when the gas giants are created. Some of this was from the friction caused by internal pressure. However, this heat is not being generated anymore, as it was only generated as matter fell into the planet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They presumably do generate heat from radioactive elements in the core (though nobody has ever been down and checked if there are any :P), and will also receive a 'boost' from solar heating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over time, however, each heat source will diminish. The latent heat from birth will be dissipated into space in the form of radiation, the radioactive elements will decay, and the star it's orbiting will die off. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I guess that Jupiter will no longer have its dramatic weather in a trillion years. &lt;/p&gt;&#xA;" OwnerUserId="2702" LastActivityDate="2014-11-14T17:29:59.607" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="7973" PostTypeId="2" ParentId="7957" CreationDate="2014-11-14T17:59:44.127" Score="1" Body="&lt;p&gt;I have found the answer to this question. Most people use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jet_Propulsion_Laboratory_Development_Ephemeris&quot; rel=&quot;nofollow&quot;&gt;Jet Propulsion Laboratory Development Ephemeris&lt;/a&gt; or JPL-DE as it is known. The book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1891389858&quot; rel=&quot;nofollow&quot;&gt;Explanatory Supplement to the Astronomical Almanac&lt;/a&gt; explains the methods of the JPL-DE and how to use it.&lt;/p&gt;&#xA;" OwnerUserId="2800" LastEditorUserId="2800" LastEditDate="2015-09-03T19:44:00.307" LastActivityDate="2015-09-03T19:44:00.307" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7974" PostTypeId="1" AcceptedAnswerId="7982" CreationDate="2014-11-14T18:42:47.460" Score="5" ViewCount="2255" Body="&lt;p&gt;Following on from a &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6051/how-did-the-hulse-taylor-pulsar-provide-confirming-evidence-of-general-relativit&quot;&gt;previous question&lt;/a&gt; about the interesting &lt;a href=&quot;https://en.wikipedia.org/wiki/Hulse%E2%80%93Taylor_binary&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hulse Taylor Binary Pulsar&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BACKGROUD&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/pdf/astro-ph/0407149v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Weisberg &amp;amp; Taylor, 2004&lt;/a&gt; present a graph showing the change in &quot;Cumulative Periastron Time Shift&quot; over a 30 year period.  Here is a later version of that graph&#xA;&lt;img src=&quot;https://i.stack.imgur.com/992mC.gif&quot; alt=&quot;orbit period decay&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graph indicates excellent agreement between the observed data points and the theoretical curve predicted by the General Theory of Relativity (GTR).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors determined the theoretical change in orbit period &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; over time (due to orbital energy loss from emission of gravitational waves) from a formula (Peters&#xA;&amp;amp; Matthews (1963)) of the form:-&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ dP/dt = K/P^{(5/3)}$$&lt;/span&gt; &#xA;where K is an (effectively constant) function of the masses of the two stars and the orbit eccentricity. For the Hulse-Taylor Binary, the period &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; is approximately 7.75 hours and the value of K is approximately &lt;span class=&quot;math-container&quot;&gt;$(-6 * 10^{-5})$&lt;/span&gt;, giving an approximate value of &lt;span class=&quot;math-container&quot;&gt;$$ dP/dt = 2.34 × 10^{−12} s/s. $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$dP/dt$&lt;/span&gt; is constant then obviously we can determine the period &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; at a future time by applying the formula&#xA;&lt;span class=&quot;math-container&quot;&gt;$$P_t = P_0 + t\,(dP/dt).$$&lt;/span&gt;&#xA;As a rough initial estimate over one year the approximate change in period is: 365.25 x 24 x 3600 x -2.34 x 10^{-12} = 0.074 ms. (Wikipedia article indicates 0.076 ms.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;span class=&quot;math-container&quot;&gt;$dP/dT$&lt;/span&gt; is not constant because it depends on the value of &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; which obviously changes.  So we would expect the period to decay over time and the rate of decay to increase with time. However, the change in P is currently so very small that over the period of 30 years the change in dP/dt (which is a function of &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;) is extremely small also. (It is important to appreciate that the graph shown above does &lt;strong&gt;not&lt;/strong&gt; plot &lt;span class=&quot;math-container&quot;&gt;$dP/dt$&lt;/span&gt; against time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;zibadawa timmy&lt;/strong&gt; (thanks) provided the formula &lt;span class=&quot;math-container&quot;&gt;$P=(8/3 \int K dt)^{3/8}$&lt;/span&gt; giving:- &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;P_t = \left( \frac{8}{3} (Kt+C)\right)^{3/8} \qquad \mathrm{with}\qquad C=\frac{3}{8} P_o^{8/3}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;P_t = &#xA; \left( \frac{8}{3} Kt+ P_o^{8/3})\right)^{3/8} &#xA;=&#xA; \left(  \frac{-1.6\,t}{10^4}+ P_o^{8/3}\right)^{3/8} &#xA;=&#xA; \left(  \frac{-1.6\,t}{10^4}+ 27900^{8/3}\right)^{3/8} &#xA;$$&lt;/span&gt;&#xA;for 1 year , t = 31,557,600 s., so -1.6 t/10^4 = -5,049.12 = c. 5e3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for 10 years t = 310,557,600 s., so -1.6 t/10^4 = -50,491.2 = c. 5e4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while 27900^{8/3} = 716,051,966,959.181 = c. 7e11.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;P_{1 year} = (-5049.12 + 716,051,966,959.181)^{3/8} = 27,899.999 926 s,\qquad deltaP_{1 year} = -0.0000738 s.&#xA;$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;P_{10 years} = (-50491.2 + 716,051,966,959.181)^{3/8} =  27,899.999 262 s,\qquad deltaP_{10 years} = -0.000738 s.&#xA;$$&lt;/span&gt;&#xA;These &lt;span class=&quot;math-container&quot;&gt;$deltaP$&lt;/span&gt; values indicate that the value of &lt;span class=&quot;math-container&quot;&gt;$dP/dt$&lt;/span&gt; is effectively constant ($=-0.000 073 8 s/s) over 10years.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;QUESTION 1 - What is plotted on the graph? (&lt;span class=&quot;math-container&quot;&gt;$\equiv$&lt;/span&gt; What is Cumulative Perihelion Time Shift?) (ANSWERED).&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially I was confused about what the Weisberg &amp;amp; Taylor graph was actually plotting.  The 2004 paper has &quot;The observed and theoretical decays are compared graphically in Figure 1&quot;.  The wikipedia article labels the y-axis &quot;Cumulative Period Shift&quot; but the caption reads &quot;observed change in the epoch of periastron with date&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An earlier &lt;a href=&quot;http://www.nature.com/scientificamerican/journal/v245/n4/pdf/scientificamerican1081-74.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;1981 Scientific American article&lt;/a&gt; indicates that&#xA;the vertical axis of the graph indicates the difference (round data points) between (A) the calculated epoch of a periastron event in a hypothetical &quot;steady&quot; system whose orbital period remains constant and (B) the observed epoch of the corresponding actual periastron event in the &quot;decaying&quot;-period system. Also shown is the continuous curve showing the difference between (A) steady system periastron epochs and (C) theoretical predictions of decayed periastron epochs.   The periastron's of the actual and hypothetical orbits are synchronized at &lt;span class=&quot;math-container&quot;&gt;$t=0$&lt;/span&gt; (in late 1974).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;QUESTION 2 - How are the theoretical values of CPTS Calculated? (ANSWERED).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several (similar) methods have now been identified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method 1 The method(s) used by Weisberg &amp;amp; Taylor to calculate theoretical values of CPTS have not been identified. These following methods seem reasonable candidates, based on the agreement with their results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method 2 (see Stan Liou's answer)&#xA;&lt;strong&gt;Stan Liou&lt;/strong&gt; has provided a formula &lt;span class=&quot;math-container&quot;&gt;$\Delta t = 0.5(\dot P/P_0)t^2$&lt;/span&gt; that produces values which closely match the Weisberg &amp;amp; Taylor graph.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method 3 (see my answer Method 3)&#xA;I have derived a formula identical to Stan Liou's but derived using phase instead of mean anomaly.  Another formula is derived from the same trunk expressing &lt;span class=&quot;math-container&quot;&gt;$\Delta t$&lt;/span&gt; in terms of &lt;span class=&quot;math-container&quot;&gt;$\dot P, P and N$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\Delta t = -0.5 P_o \dot P N^2$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method 4 (see my answer Method 4)&#xA;I have derived a formula derived using a coarsely approximated  binomial series which gives similar results &#xA;&lt;span class=&quot;math-container&quot;&gt;$ \Delta t_{N} = P_0{\dot P} \frac{N^2+N}{2}  $&lt;/span&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method 5 (not shown)&#xA;A finely approximated series expansion approach has been assessed. For the given binary system it produces a formula that is effectively indistinguishable from that of the other methods for the given time range.&lt;/p&gt;&#xA;" OwnerUserId="233" LastEditorUserId="7472" LastEditDate="2018-12-24T15:50:04.243" LastActivityDate="2018-12-24T15:50:04.243" Title="Hulse Taylor Binary Pulsar - How is &quot;Cumulative Periastron Time shift&quot; calculated?" Tags="&lt;general-relativity&gt;&lt;binary-star&gt;&lt;pulsar&gt;" AnswerCount="4" CommentCount="14" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="7975" PostTypeId="2" ParentId="7967" CreationDate="2014-11-14T19:42:05.833" Score="4" Body="&lt;p&gt;Yes and no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the &lt;a href=&quot;https://astronomy.stackexchange.com/q/7939/2188&quot;&gt;answers here&lt;/a&gt; should give you some understanding of dark energy and its effects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net effect of dark energy is tiny over small distances. If it is given by a cosmological constant, then its effects on gravitationally bound systems is independent of time (hence 'constant'). Observational evidence puts us at least close to the constant case. Which means that even if we are not actually in the constant case, then it will be many billions of years until there would be a measurable difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can, however, estimate the mass equivalent of dark energy in a given area by measuring the apparent amount in the universe itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As estimated &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/astro/dareng.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If we take the WMAP value for critical density at $$\rho_{c,0} = 9.47 \times 10^{-27}\, kg/m^3$$&#xA;  and presume that dark energy makes up about 73% of that, then the effective density of the dark energy would amount to just over 4 hydrogen atoms ($m = 1.67 \times 10^{-27}$ kg) in a cubic meter of space. If we take $5.9 \times 10^9$ m as a mean radius of Pluto and calculate the volume of a sphere of that radius, then the dark energy in that sphere would be equivalent to just under 6000 kg distributed throughout a space representing the solar system. The density of the asteroid Ida has been measured to be about 2.7 g/cm$^3$. So all the dark energy in the solar system would amount to only about 2.2 m$^3$ of the material of Ida, or a sphere of about 0.8 m radius. This is a hundred times smaller than Ida's tiny moon Dactyl! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course, that assumes dark energy is essentially the same on small scales as it is large scales, which is unknown. Nevertheless I think it is pretty clear that it would be extremely difficult to observe dark energy directly within the Earth-Moon system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also Stan's comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our understanding of dark energy is minimal at best, to boot, giving us no solid ideas on how to detect it in the first place. We infer its existence from the large scale expansion of the universe, but we're otherwise short on solid ideas on its nature.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-11-14T23:12:23.943" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7977" PostTypeId="2" ParentId="7974" CreationDate="2014-11-14T21:27:47.047" Score="2" Body="&lt;p&gt;As for solving the equation, $dP/dt=K/P^{5/3}$ is a separable differential equation. See the &lt;a href=&quot;http://en.wikipedia.org/wiki/Examples_of_differential_equations&quot; rel=&quot;nofollow&quot;&gt;first section here&lt;/a&gt;. Google also brings up several pages describing how to solve such equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, we rewrite the equation in differential form as:&#xA;$$P^{5/3}\, dP = K\, dt.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Loosely speaking, I've cleared the denominators, with a goal to get the $P$ stuff all together on one side, and the $t$ stuff on the other. Technically it's not that at all, but pretending like it is works out. I am implicitly assuming that $K$ does not depend on $P$, which your post supports.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we integrate both sides:&#xA;$${3\over8}P^{8/3}=\int P^{5/3}\, dP = \int K\, dt,$$&#xA;where we let the right hand integral cover the constant of integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So our solution is $$P=\left({8\over3}\int K\, dt \right)^{3/8},$$ with initial condition setting the constant of integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note I took an 8th root in finding $P$, so we need the inside function to be nonnegative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your post suggests $K$ is independent of time, in which case $\int K\, dt= K t+C$ for some constant $C$ which is determined by the initial conditions. If it does depend on time, you have to do the integration accordingly, or use data to estimate it.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastEditorUserId="2188" LastEditDate="2014-11-14T21:32:56.253" LastActivityDate="2014-11-14T21:32:56.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7978" PostTypeId="1" CreationDate="2014-11-14T23:30:45.137" Score="1" ViewCount="93" Body="&lt;p&gt;In some fields of astrophysics, where distances are measured in parsecs (pc), velocities in kilometers per second (km/s) and masses in solar units (), it is useful to express G as approximately 4.302 x 10^-3 pc M^-1 (km/s)^2 ...&#xA;Is G not approximately 6.67 x 10^-11 N(m/kg)^2? Any reasonable explanation would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2809" LastEditorUserId="2153" LastEditDate="2014-11-14T23:58:08.700" LastActivityDate="2014-11-15T23:56:24.067" Title="What are some applicable problems with the correct usage of G?" Tags="&lt;units&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7979" PostTypeId="2" ParentId="7978" CreationDate="2014-11-14T23:57:08.193" Score="1" Body="&lt;p&gt;I think one of the reasons for using different units is how you take the measurements for the other variables in the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let's say you have two marbles, and you want to calculate the gravitational force between them (not a lot!). You place them a certain distance apart, and measure it with your meter stick. You find the distance between them to be $0.146 \text{ meters}$. Now you weigh them, and find their masses to be $0.00156 \text{ kilograms}$ apiece. You decide to use the value of $G$ with units including meters and kilograms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's say you have two stars far, far away. You take measurements based on parallax and find that they're roughly $35,400 \text{ light-years}$ away from each other. Your instruments can't measure this to any greater accuracy. You measure their masses by calculating how much they deflect light, and come up with a figure of $10 \text { solar masses}$ apiece. You therefore decide to use some form of $G$ that, for some reason, has solar masses and light-years as some of its units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, you choose different forms of a constant depending on the units you're able to measure with your instruments. You won't be able to figure out exactly how many meters your stars are apart from each other, but you can figure out just about how many light-years apart they are, with less than a light-year of uncertainty.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-11-15T23:56:24.067" LastActivityDate="2014-11-15T23:56:24.067" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7980" PostTypeId="1" AcceptedAnswerId="7991" CreationDate="2014-11-15T01:04:29.810" Score="17" ViewCount="503" Body="&lt;p&gt;The Great Red Spot is a persistent anticyclonic storm, 22° south of Jupiter's equator, but why is it reddish?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Atmosphere_of_Jupiter#Great_Red_Spot&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is not known exactly what causes the Great Red Spot's reddish color.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Are there updated data?&lt;/p&gt;&#xA;" OwnerUserId="2475" LastEditorUserId="9527" LastEditDate="2020-10-09T09:57:03.800" LastActivityDate="2020-10-09T09:57:03.800" Title="Why Jupiter's Great Red Spot is Reddish?" Tags="&lt;jupiter&gt;&lt;great-red-spot&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7981" PostTypeId="2" ParentId="7978" CreationDate="2014-11-15T03:23:59.793" Score="1" Body="&lt;p&gt;Usually, using different units is done for convienience, as &lt;strong&gt;HDE 226868&lt;/strong&gt;'s answer explains. If we're talking about stars, then measuring masses relative to the Sun is sensible. The Sun is the closest star to us, and astronomers understand it much better than any other, making it the de facto standard of comparison. Similarly, on the interstellar scales the parsec is more convenient than the meter. The conversion is trivial and doesn't change anything physically. Recall $F = ma$ to see how the newton corresponds to base SI units:&#xA;$$\mathrm{N}\cdot\left(\frac{\mathrm{m}}{\mathrm{kg}}\right)^2 = \left(\frac{\mathrm{kg}\cdot\mathrm{m}}{\mathrm{s}^2}\right)\frac{\mathrm{m}^2}{\mathrm{kg}^2} = \frac{\mathrm{m}^3}{\mathrm{kg}\cdot\mathrm{s}^2}\text{,}$$&#xA;so we can just multiply the SI value of $G$ by&#xA;$$\left(\frac{1.989\times 10^{30}\,\mathrm{kg}}{1\,\mathrm{M}_\odot}\right)&#xA;\left(\frac{1\,\mathrm{pc}}{3.086\times 10^{16}\,\mathrm{m}}\right)&#xA;\left(\frac{1\,\mathrm{km}}{1000\,\mathrm{m}}\right)^2\text{.}$$&#xA;But there could be another reason to use solar masses in particular. The value of Newton's gravitational constant is:&#xA;$$\begin{eqnarray*}&#xA;G &amp;amp;=&amp;amp; 6.674\times 10^{-11}\,\frac{\mathrm{m}^3}{\mathrm{kg}\cdot\mathrm{s}^2}\\&#xA;  &amp;amp;=&amp;amp; 2.959122083\times 10^{-4}\,\frac{\mathrm{AU}^3}{\mathrm{M}_\odot\cdot\mathrm{day}^2}\\&#xA;%  &amp;amp;=&amp;amp; 3.964015993\times 10^{-14}\,\frac{\mathrm{AU}^3}{\mathrm{M}_\odot\cdot\mathrm{s}^2}\\&#xA;  &amp;amp;=&amp;amp; 4.300917271\times 10^{-3}\,\frac{\mathrm{pc}}{\mathrm{M}_\odot}\frac{\mathrm{km}^2}{\mathrm{s}^2}\text{.}&#xA;\end{eqnarray*}$$&#xA;The difference in precision is real. The 2010 CODATA value of $G$ is $&#xA;(6.67384\pm0.00080)\times 10^{-11}$ in the SI units, which has a relative uncertainty of $10^{-4}$, so we can't expect more than about $4$ significant figures. On the other hand, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Standard_gravitational_parameter&quot; rel=&quot;nofollow noreferrer&quot;&gt;standard gravitational parameter&lt;/a&gt; $\mu = GM$ of the Sun is known to a much greater precision than either $G$ or their masses individually, so expressing $G$ in solar masses gives a more precise value. The basic reason for that is simple: it's much easier to compare the orbits of planets to each other than, say, to compare the Sun to the International Prototype of the Kilogram kept in a French vault.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, masses of other stars and interstellar distances tend to have uncertainties larger than that, so this is irrelevant on that scale. Still, &lt;em&gt;within&lt;/em&gt; the solar system, using units of solar masses (or just $GM_\odot$ directly) allows one to make much more precise measurements, and similarly Earth masses for satellite orbits around the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measuring $G$ in terms of parsecs (or light-years) and $(\mathrm{km}/\mathrm{s})^2$ would be more convenient any time one is dealing with galactic-scale distances, such as in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Galaxy_rotation_curve&quot; rel=&quot;nofollow noreferrer&quot;&gt;galactic rotation curve&lt;/a&gt;:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/dnJhU.gif&quot; alt=&quot;Galactic rotation curve&quot;&gt;&#xA;(Image from Wikipedia).&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-11-15T03:23:59.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7982" PostTypeId="2" ParentId="7974" CreationDate="2014-11-15T10:17:00.620" Score="2" Body="&lt;p&gt;Write the mean anomaly as a Taylor series ($n\equiv 2\pi/P$):&#xA;$$\begin{eqnarray*}&#xA;M(t) \equiv \int_0^tn\,\mathrm{d}t &amp;amp;=&amp;amp; M_0 + \dot{M}_0t + \frac{1}{2}\ddot{M}_0t^2 + \mathcal{O}(t^3)\\&#xA;&amp;amp;=&amp;amp;n_0t + \frac{1}{2}\dot{n}_0t^2 + \mathcal{O}(t^3)\\&#xA;&amp;amp;=&amp;amp;\frac{2\pi}{P_0}t - \pi\frac{\dot{P}_0}{P_0^2}t^2+\mathcal{O}(t^3)\text{.}&#xA;\end{eqnarray*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The periastron time (epoch) happens when $M(t) = 2\pi N$, where $N$ counts number of orbits, while $NP_0$ &lt;em&gt;would&lt;/em&gt; be the periastron time if the orbital period stayed constant. Therefore, treating $N$ as continuous, the cumulative difference is just&#xA;$$t - \frac{M(t)}{2\pi}P_0 \approx \frac{1}{2}\frac{\dot{P}_0}{P_0}t^2\text{.}$$&#xA;If we want more accuracy, we can calculate the next term in the Taylor series,&#xA;$$\frac{1}{6}\ddot{n}_0t^3 = \frac{\pi}{3P_0^2}\left[\frac{2\dot{P}_0^2}{P_0}-\ddot{P}_0\right]t^3\text{,}$$&#xA;and if $\dot{P} = KP^{-5/3}$ for $\dot{K}\approx 0$, we can eliminate the $\ddot{P}_0$ term to give a cumulative shift of&#xA;$$t - \frac{M(t)}{2\pi}P_0 \approx \frac{1}{2}\frac{\dot{P}_0}{P_0}t^2 - \frac{11}{18}\left(\frac{\dot{P}_0}{P_0}\right)^2t^3\text{.}$$&#xA;According to the Weisberg &amp;amp; Taylor paper, $|\dot{P}_0/P_0| \sim 10^{-16}\,\mathrm{s}^{-1}$, so the cubic term is negligible over the thirty-year time scale, and we can safely ignore anything past the quadratic term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for kicks, we can solve $\dot{P} = KP^{-5/3}$ for constant $K$, which yields a cumulative periastron time shift of&#xA;$$\frac{3}{5x}\left[1+\frac{5}{3}xt-\left(1-\frac{8}{3}xt\right)^{5/8}\right]\text{.}$$&#xA;$K$ is not actually constant, being a function of the eccentricity $e$, although the rate of change of $e$ is tiny, on the order of $10^{-11}\,\mathrm{yr}^{-1}$, and we can ignore it on the scales we're talking about. Additionally, the proportionality formula is based on a orbital average of radiation reaction at 2.5-th post-Newtonian order.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The values presented in the paper are for the epoch $T_0 = 52144.90097844$ (MJD):&#xA;$$\small{\begin{eqnarray*}&#xA;P_0 &amp;amp;=&amp;amp; 0.322997448930\,(4)\,\mathrm{d}\\&#xA;%\dot{P}_0 &amp;amp;=&amp;amp; −(2.4184\pm0.0009)\times 10^{-12}\\&#xA;%\dot{P}_0^{\scriptsize\text{corr}} &amp;amp;=&amp;amp; -(2.4056\pm 0.0051)\times 10^{-12}\\&#xA;\dot{P}_0^{\scriptsize\text{GTR}} &amp;amp;=&amp;amp; −2.40242\,(2)\times 10^{-12}\\&#xA;\end{eqnarray*}}$$&#xA;Thus, we can calculate the ratio $x_0 = \dot{P}_0/P_0 = -8.60867\,(7)\times 10^{-17}\,\mathrm{s}^{-1}$. The difference between $T_0$ and the beginning of 1975 is $\delta t = 9731\,\mathrm{d}$, which is not long enough for this ratio to change appreciably, since relative first-order correction would be around $|\dot{x}_0(\delta t)/x_0| \sim |x_0(\delta t)| \sim 10^{-7}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, following &lt;strong&gt;zibadawa timmy&lt;/strong&gt;, we can also solve $K = \dot{P}P^{5/3}$ explicitly for constant $K$, which is not exactly true but approximates well:&#xA;$$\newcommand{\constKeq}{\overset{\scriptsize\dot{K}\approx 0}{=}}\begin{eqnarray*}&#xA;P &amp;amp;\constKeq&amp;amp; \left[P_0^{8/3}+\frac{8}{3}\dot{P}_0P_0^{5/3}t\right]^{3/8}\!\!\!\!\!\text{,}\\&#xA;x &amp;amp;\constKeq&amp;amp; \frac{\dot{P}}{P} = \frac{x_0}{1+\frac{8}{3}x_0t}\text{,}&#xA;\end{eqnarray*}$$&#xA;and therefore $\left.x\right|_{1975}$ is indeed indistinguishable from $x_0$ at this level of accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've graphed (in light blue) the result from the quadratic shift formula with this $x$  and superimposed the graph from the Weisberg &amp;amp; Taylor paper (cropped on outer borders and then rescaled to be $500\times 500\,\mathrm{px}$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jteR1.png&quot; alt=&quot;Hulse–Taylor graph reconstruction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's an exact match except for a constant time shift, which is just a matter of when we set the $t = 0$ time. PSR B1913+16 was actually discovered in 1974, and the first data point is at the end of 1974. That is why my graph is shifted slightly by a fraction of a year.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am unclear whether your formula excludes the effect of periastron advance (4.226595 degrees per year):- (A) If not then the periastron advance effect is so large that the change in period due to gravitational wave emission is not visible on the graph and the agreement with theory cannot be assessed ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The GTR prediction of the cumulative periastron time shift is made through through equation (1) in the paper, which prescribes $\dot{P}_0 = \left.\dot{P}_b\right|_{t=0}$. The periastron time is well-defined because it depends on the closest approach, so the magnitude of the positional periastron advance $\langle\dot{\omega}\rangle = 4.226595^\circ/\mathrm{yr}$ is not directly relevant. GTR &lt;em&gt;also&lt;/em&gt; predicts $\langle\dot{\omega}\rangle$, but that's not what we're considering here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mean anomaly is not actually an angle, despite counting off each periastron as $2\pi$. It isn't an angle even in exact Keplerian elliptic orbits, even though they have the exact same periastron at each period. Thus, $\langle\dot{\omega}\rangle$ is a different issue.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(B) If so then must we conclude that gravitational wave emission causes change in period of approx 4s in 10 years? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The period changes by a bit less than $1\,\mathrm{ms}$ over a decade, as you have already calculated above.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditDate="2014-11-16T15:08:46.497" LastActivityDate="2014-11-16T15:08:46.497" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="7983" PostTypeId="1" CreationDate="2014-11-15T11:22:27.897" Score="2" ViewCount="1519" Body="&lt;p&gt;If the Moon has a magnetic field and can possibly receive solar wind, then does it have an aurora? If Earth has an aurora, and Saturn has an aurora, then could the Moon possibly have an aurora?&lt;/p&gt;&#xA;" OwnerUserId="2809" LastEditorUserId="211" LastEditDate="2014-11-16T01:11:35.137" LastActivityDate="2017-12-04T16:13:15.487" Title="Does the Moon have an aurora?" Tags="&lt;solar-wind&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7984" PostTypeId="2" ParentId="7983" CreationDate="2014-11-15T14:15:50.563" Score="1" Body="&lt;p&gt;No, &lt;a href=&quot;http://en.wikipedia.org/wiki/Aurora&quot; rel=&quot;nofollow&quot;&gt;Auroras&lt;/a&gt; result from emissions of photons in the Body's upper atmosphere gases. The Moon &lt;a href=&quot;http://en.wikipedia.org/wiki/Lunar_Atmosphere_and_Dust_Environment_Explorer&quot; rel=&quot;nofollow&quot;&gt;may&lt;/a&gt; have a tenuous atmosphere of moving particles, called &quot;dust atmosphere&quot;. But is composed of dust particles in constant motion, not of gases. &lt;/p&gt;&#xA;" OwnerUserId="2475" LastActivityDate="2014-11-15T14:15:50.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7985" PostTypeId="2" ParentId="7917" CreationDate="2014-11-15T22:52:11.093" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I need to upgrade my tripod legs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, this seems to be the case&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I don't need to upgrade the head, since it's rated to twice the capacity of the legs, and to twice the weight of the biggest binoculars I'll use.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seems right to me. However, if you have some money to spare, I think you should consider investing in a Parallelogram Mount (P-mount). These usually look like long arms with counterweights and a place to attach your bino. They make it easier to adjust the height, observe objects close to zenith and are usable from a chair. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I should look for legs that are rated to twice the weight of the biggest binoculars I'll use.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't have this problem with binoculars as the ones I own are pretty light but with photographic equipment, I try not to exceed half the declared capacity. The better the brand, the closer you can get, I suppose. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I should look for a set of wooden legs, since these have greater stability than aluminium legs while being affordable, unlike carbon fiber.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've used wooden legs for a small telescope and I wouldn't like to do it again. Unless you're only planning to use a tripod around the house. Wooden tripods are usually heavy, clunky and not very portable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon fiber tripods, on the other hand, tend to be very light so you'll have to be a little more careful with a big and heavy bino than in case of an aluminium or wooden one. If you get one, make sure it has a hook under the column, on which you can hang something to shift the center of mass closer to the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I once had a carbon tripod almost tip over after a strong gust of wind when I had a large-ish telephoto lens mounted on it. Managed to catch it just in the nick of time.&lt;/p&gt;&#xA;" OwnerUserId="211" LastActivityDate="2014-11-15T22:52:11.093" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7986" PostTypeId="2" ParentId="7983" CreationDate="2014-11-16T04:16:42.507" Score="0" Body="&lt;p&gt;Auroras require an atmosphere, which the moon does not have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Auroras occur as solar wind and other particles that interact or are associated with Earth's magnetic field are funneled by the magnetic field and accelerated until they collide with atoms/molecules in Earth's atmosphere.  These collisions cause the atoms/molecules to ionize or excite the the subsequent recombination/de-excitation of the electrons is what causes the light seen in the auroras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without at atmosphere, though, there are no particles to ionize and excite to create an aurora.&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2014-11-16T04:16:42.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7987" PostTypeId="2" ParentId="7983" CreationDate="2014-11-16T04:24:24.043" Score="6" Body="&lt;p&gt;Technically, yes, the moon does have an atmosphere that gets ionized by the sun. &#xA;NASA: &lt;a href=&quot;http://www.nasa.gov/mission_pages/LADEE/news/lunar-atmosphere.html#.VGgX79Yi10Q&quot; rel=&quot;noreferrer&quot;&gt;Is There an Atmosphere on the Moon?&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;our moon does indeed have an atmosphere consisting of some unusual gases, including sodium and potassium, which are not found in the atmospheres of Earth, Mars or Venus. It's an infinitesimal amount of air when compared to Earth's atmosphere. At sea level on Earth, we breathe in an atmosphere where each cubic centimeter contains 10,000,000,000,000,000,000 molecules; by comparison the lunar atmosphere has less than 1,000,000 molecules in the same volume.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pretty thin soup. Sodium and potassium get kicked up by photons hitting the surface. There's also &lt;a href=&quot;http://en.wikipedia.org/wiki/Atmosphere_of_the_Moon&quot; rel=&quot;noreferrer&quot;&gt;argon, helium and other minor components&lt;/a&gt;.&#xA;As shown in this figure from the first link: &lt;img src=&quot;https://i.stack.imgur.com/oXJfk.jpg&quot; alt=&quot;enter image description here&quot;&gt; The sodium in that atmosphere does glow; (&lt;a href=&quot;http://en.wikipedia.org/wiki/Rayleigh_%28unit%29&quot; rel=&quot;noreferrer&quot;&gt;Rayleigh&lt;/a&gt; unit definition). At least some of the &lt;a href=&quot;http://meetingorganizer.copernicus.org/EGU2009/EGU2009-6734.pdf&quot; rel=&quot;noreferrer&quot;&gt;glow is due to full fledged &lt;strong&gt;ions&lt;/strong&gt;&lt;/a&gt; rather than charge-neutral &lt;a href=&quot;http://en.wikipedia.org/wiki/Fluorescence&quot; rel=&quot;noreferrer&quot;&gt;fluorescence&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is confirmed that the ions include oxygen, sodium and potassium. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, whether by ionic or fluorescent emission, the moon's atmosphere does glow. Unless one wants to argue about the glow not being &lt;a href=&quot;http://en.wikipedia.org/wiki/Aurora&quot; rel=&quot;noreferrer&quot;&gt;restricted to the poles by a magnetic field&lt;/a&gt;, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Airglow&quot; rel=&quot;noreferrer&quot;&gt;airglow&lt;/a&gt;, that glow counts as an aurora.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="2031" LastEditDate="2014-11-16T06:27:32.277" LastActivityDate="2014-11-16T06:27:32.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7991" PostTypeId="2" ParentId="7980" CreationDate="2014-11-16T17:44:34.187" Score="21" Body="&lt;p&gt;It's red, because it's a 'sunburn'. The clouds in the red spot reach to higher altitudes than the surrounding and are more exposed to Solar UV radiation, which in turn changes the structure of some of the organic molecules etc. This is at least the explanation suggested by recent data from NASA's Cassini mission, see &lt;a href=&quot;http://www.nasa.gov/jpl/cassini/jupiters-red-spot-is-likely-a-sunburn-not-a-blush/#.VGjhQ4dwRFU&quot;&gt;this 5 day old press release&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="1378" LastEditDate="2014-12-18T00:38:37.547" LastActivityDate="2014-12-18T00:38:37.547" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7992" PostTypeId="1" CreationDate="2014-11-16T19:04:31.870" Score="5" ViewCount="114" Body="&lt;p&gt;How many times have we broadcast the location of Earth to the outer space? Are there any regulation in doing so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know so far we have send the Arecibo Message, and any receiver can trace the message sender's direction by the direction with which the message is received. But have we ever pin point our exact location of the galaxy and send it as a message?&lt;/p&gt;&#xA;" OwnerUserId="2825" LastActivityDate="2014-11-16T20:51:27.957" Title="Is broadcasting the location of Earth to potential extraterrestial civilization regulated?" Tags="&lt;radio-astronomy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7993" PostTypeId="2" ParentId="7992" CreationDate="2014-11-16T20:51:27.957" Score="2" Body="&lt;p&gt;No, except in the sense that national broadcasting and so on is regulated. Our location can be fixed, as you suggest, from simply looking at where the signals come from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps, though, you mean is it sent out &quot;regularly&quot; (i.e., at specified intervals). In truth we have been sending very powerful signals continuously into space since the dawn of the television age. Those signals may now start to wane in strength as we move to digital terrestrial television but that won't bother any aliens for some time yet.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2014-11-16T20:51:27.957" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7994" PostTypeId="1" AcceptedAnswerId="8043" CreationDate="2014-11-16T21:14:30.743" Score="0" ViewCount="301" Body="&lt;p&gt;Two things we take, for this purpose, to be axiomatic:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The CMB is the oldest thing we can observe directly&lt;/li&gt;&#xA;&lt;li&gt;The cosmological red shift tells us how &quot;old&quot; something we are observing is&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Yet we also have a third axiom:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;There are some parts of the universe we can never observe because they are receding away from us at a superluminary speed.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But this surely also implies that we can see beyond the CMB if we see anything which has a red shift indicating an expansion speed very close to $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This suggests axiom 2 is incorrect - so what should it be instead?&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2014-11-24T06:19:14.537" Title="How can there be anything &quot;beyond&quot; the CMB?" Tags="&lt;cmb&gt;&lt;cosmological-horizon&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7995" PostTypeId="2" ParentId="1041" CreationDate="2014-11-17T06:28:08.900" Score="2" Body="&lt;p&gt;I'm no expert on stellar atmospheres, so I have a limited idea of how things like $\log g$ affect the lines. But I work with stellar models, so I can take a stab at that part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall principle is that computing stellar model ages is a kind of optimization problem. We model the structure of stellar interiors by constructing a system of differential equations based on a few simple assumptions. (When I teach stellar structure and evolution, I usually recommend the outstanding and free lecture notes by &lt;a href=&quot;http://www.astro.ru.nl/~onnop/education/stev_utrecht_notes/&quot; rel=&quot;nofollow&quot;&gt;Onno Pols&lt;/a&gt; and &lt;a href=&quot;http://astro.phys.au.dk/~jcd/evolnotes/LN_stellar_structure.pdf&quot; rel=&quot;nofollow&quot;&gt;Jørgen Christensen-Dalsgaard&lt;/a&gt;.) These models depend on many parameters. Some are familiar: the mass, composition, and age. Some less-so: there's usually at least one parameter for how convection is parametrized. e.g. the &lt;em&gt;mixing length&lt;/em&gt;. Some are discrete: which opacity data is used, what solar abundances are chosen. And some are relatively inconsequential: there a dozens (or even hundreds!) of numerical parameters used in solving the equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's just say we have a magical black box that takes five parameters—mass, initial metallicity, initial helium abundance, age and mixing length—and produces $T_\text{eff}$ and $\log g$. What we have to do is select values of the parameters to match the observations, which is a standard problem in optimization, inference, parameter estimation, or whatever you want to call it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that age is a special parameter. There are ways of measuring things like mass, radius or luminosity relatively directly. But choosing the sequence of models that produces the appropriate star is always depends on which stellar models you use in the first place.  Ages are uncertain both because of the uncertainties in the observations, but also because of the intrinsic uncertainty in the models. Although something like interferometry can potentially give an independent radius, we can only get indirect measures of age, and converting these indirect measures &lt;em&gt;to&lt;/em&gt; ages also introduces uncertainty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick now is how much data you have...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given a single, high-resolution (R≳50000) spectrum as your only data point, how easy is it to accurately infer the age and evolutionary status of a star?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd say it's &lt;em&gt;very&lt;/em&gt; hard to get an accurate (or even precise) age just given a single spectrum.  Currently, the spectrum would probably first be used to determine $T_\text{eff}$ and $\log g$, and thus values would then be used as inputs in the stellar model.  Remember: I'm talking about interior models, so they don't typically produce a model atmosphere to compare. You've then already got the issue that there are more parameters than observables. This is resolved by supposing that the mixing length parameter is the same as the best-fitting values for the Sun (for which we have &lt;em&gt;much&lt;/em&gt; more data) and that the abundances of helium and metals are correlated. (We call this an &lt;em&gt;enrichment law&lt;/em&gt;.) This makes the problem tractable, because the high-resolution spectrum should also tell us the metal content.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing the evolutionary state is easier, I think, because the surface gravity should help you to distinguish, especially given a high resolution spectrum.  As said, I'm no expert here, and I'm aware that misclassification can happen with multi-colour photometry, but I don't expect it to happen with high-res spectra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you'd like to read further, here are some quick resources that might be of interest. First, some lecture notes on determining stellar ages recently appeared on arXiv:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/1410.5336&quot; rel=&quot;nofollow&quot;&gt;How accurate are stellar ages based on stellar models? I. The impact of stellar models uncertainties&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/1410.5337&quot; rel=&quot;nofollow&quot;&gt;How accurate are stellar ages based on stellar models? II. The impact of asteroseismology&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Second, you can play around with synthetic line profiles and other atmospheric data with &lt;a href=&quot;http://www.ap.smu.ca/~ishort/GrayStar/&quot; rel=&quot;nofollow&quot;&gt;GrayStar&lt;/a&gt;, a web app that computes basic atmosphere data.  (I'm not experienced with it, so I'm not exactly sure how it works, but you can play around to get the information you want about e.g. the difference between line profiles in giants and dwargs, I think.)&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2014-11-17T06:28:08.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7996" PostTypeId="1" CreationDate="2014-11-17T23:05:08.120" Score="5" ViewCount="377" Body="&lt;p&gt;I am creating a virtual solar system model and I want it to be as realistic as possible (e.g. orbits are ellipses, not circles, and orbits are oriented correctly, not all coplanar).  In order for me to do this, I need to know the eccentricity of the orbit, the perihelion, the semi-major-axis, inclination and azimuth.  I have all of this information, except for the azimuth of the orbit, which is the angle from the sun to the aphelion.  Are there any reliable sources with this information?&lt;/p&gt;&#xA;" OwnerUserId="2830" LastActivityDate="2016-02-19T06:03:16.343" Title="What are the azimuths of the planets' orbits?" Tags="&lt;orbit&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7997" PostTypeId="1" AcceptedAnswerId="8662" CreationDate="2014-11-18T03:59:06.953" Score="3" ViewCount="158" Body="&lt;p&gt;I've been reading about the Leonid meteor showers and am struggling to understand one piece of information. On Wikipedia (&lt;a href=&quot;http://en.wikipedia.org/wiki/55P/Tempel%E2%80%93Tuttle&quot; rel=&quot;nofollow&quot;&gt;Tempel-Tuttle&lt;/a&gt;), it states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The orbit of 55P/Tempel–Tuttle intersects that of Earth near exactly,&#xA;  hence streams of material ejected from the comet during perihelion&#xA;  passes do not have to spread out over time to encounter Earth. This&#xA;  coincidence means that streams from the comet at perihelion are still&#xA;  dense when they encounter Earth, resulting in the 33 year cycle of&#xA;  Leonid meteor storms. For example, in November 2009, the earth passed&#xA;  through meteors left behind mainly from the 1466 and 1533 orbit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why are we going through the meteors left behind by the orbits from 1466 and 1533? What happened to the orbit of 1500 and every subsequent orbit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that the perihelion moves slightly from orbit to orbit so the earth moves through a different one each time but I'd love to get an expert explanation!&lt;/p&gt;&#xA;" OwnerUserId="1167" LastEditorUserId="1167" LastEditDate="2014-11-18T05:25:15.490" LastActivityDate="2016-03-10T21:32:28.303" Title="Leonid meteor showers and the Tempel-Tuttle comet" Tags="&lt;orbit&gt;&lt;comets&gt;&lt;meteor&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7998" PostTypeId="2" ParentId="6390" CreationDate="2014-11-18T18:06:31.877" Score="5" Body="&lt;p&gt;There are several misconceptions in your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, a star does not &lt;em&gt;vacuum everything in its vicinity&lt;/em&gt;. Rather it forms from a condensation in a gas cloud, which in turn collapses to a proto-star surrounded by a gas disc, which can contribute further material. Once formed in this way, a star typically does not acquire more gas (exceptions are symbiotic binary stars etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, a star with mass in excess of $\sim8$M$_\odot$ will (typically after a long time) suffer from a supernova, when most of its envelope is flung back into space. That gas is still mostly hydrogen, though enriched by 'metals' (non-primordial elements). However, the gas is hot and fast moving and hence not in a condition to form another star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, the gas from the supernova will eventually mix with other gas and dissolve into the general pool of interstellar medium (ISM). Some of that may cool to form a molecular cloud (as gas cloud where molecular $H_2$ dominates), which in turn may become the site of new star formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the Sun has been formed from enriched material, which is a mixture of primordial gas with the ejections of several supernovae.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2014-11-18T20:39:37.143" LastActivityDate="2014-11-18T20:39:37.143" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7999" PostTypeId="1" AcceptedAnswerId="8576" CreationDate="2014-11-18T21:19:12.700" Score="4" ViewCount="99" Body="&lt;p&gt;I remember one time a while back looking at the night sky through night vision goggles.  One interesting thing I noticed is that there appeared to be many very faint meteors in the sky practically all the time.  I assume this is dust burning up in the atmosphere, or possibly an unusual effect produced by the night vision goggles themselves.  I more suspect that it is dust burning faintly in the atmosphere because it was seen only in the sky.  My main question is: Is space dust constantly entering the atmosphere as small meteors at a fairly high rate?  If I remember right, it was a few every second, which is much more frequent than normal &quot;brighter&quot; meteors, even during a meteor shower.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="1386" LastEditDate="2014-11-19T13:56:36.900" LastActivityDate="2015-01-15T22:34:26.703" Title="Are there many faint meteors that are too faint to see with the naked eye?" Tags="&lt;planetary-atmosphere&gt;&lt;meteor&gt;&lt;dust&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8000" PostTypeId="2" ParentId="7883" CreationDate="2014-11-18T22:13:07.697" Score="2" Body="&lt;p&gt;The degree of ionization in the photosphere varies with depth of course, but overall it is small.   Table 1 of the &lt;em&gt;Bilderberg Continuum Atmosphere&lt;/em&gt;  &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1968SoPh....3....5G&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow&quot;&gt;(Solar Physics, 3, 5, 1968)&lt;/a&gt; gives the pressure and the electron pressure at various optical depths in a comprehensive model.  The ratio of the pressure gives the Ne/Ntotal.  At optical depth = 1 at 5000 Angstroms, Pgas = 1.412E5 (cgs) and Pe = 6.239E1.  The ratio is 4.4E-4. The hydrogen ionization, also given in the table is 4.07E-4.  The model assumes Helium abundance of 0.1 which is off a bit from the &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/tables/suncomp.html&quot; rel=&quot;nofollow&quot;&gt;modern number&lt;/a&gt; of 0.087.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2014-11-18T22:13:07.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8001" PostTypeId="1" AcceptedAnswerId="8003" CreationDate="2014-11-18T22:41:21.803" Score="17" ViewCount="20666" Body="&lt;p&gt;Ive read on some sites and saw on youtube videos that the moon is getting away from earth by 1-3 cm a year. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this enough to make the Earth lose the Moon before the Sun goes into Supernova?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Im asking because I would like to do the calculations for Earths magnetic pull on this subject. It appears to me that it should be a non-linear function (there are also other factors, like if the Moon could be captured by another planet or having asteroids messing with its orbit on Earth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2790" LastActivityDate="2017-12-22T06:25:52.660" Title="Will Earth lose the Moon before the Sun goes into supernova?" Tags="&lt;the-moon&gt;&lt;the-sun&gt;&lt;gravity&gt;&lt;earth&gt;&lt;supernova&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8002" PostTypeId="2" ParentId="8001" CreationDate="2014-11-18T23:21:29.423" Score="13" Body="&lt;p&gt;The Sun does not have nearly enough mass to become a supernova. Instead, it will swell to become a &lt;a href=&quot;https://en.wikipedia.org/wiki/Red_giant&quot;&gt;red giant&lt;/a&gt;, enveloping Mercury, Venus, and possibly Earth. After that, it will shed its outer layers as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Planetary_nebula&quot;&gt;planetary nebula&lt;/a&gt;, and settle down to become a &lt;a href=&quot;https://en.wikipedia.org/wiki/White_dwarf&quot;&gt;white dwarf&lt;/a&gt;. &lt;a href=&quot;https://en.wikipedia.org/wiki/Sun#After_core_hydrogen_exhaustion&quot;&gt;Wikipedia&lt;/a&gt;, apparently, says the exact same things I had though of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Sun does not have enough mass to explode as a supernova. Instead it will exit the main sequence in approximately 5.4 billion years and start to turn into a red giant. It is calculated that the Sun will become sufficiently large to engulf the current orbits of the Solar System's inner planets, possibly including Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Still, that does put a deadline for the Earth and the Moon to evolve to the point you're looking for. &lt;a href=&quot;https://en.wikipedia.org/wiki/Formation_and_evolution_of_the_Solar_System#Moon.E2.80.93ring_systems&quot;&gt;Wikipedia&lt;/a&gt; addresses the possibility that the Earth and Moon survive:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Today, the Moon is tidally locked to the Earth; one of its revolutions around the Earth (currently about 29 days) is equal to one of its rotations about its axis, so it always shows one face to the Earth. The Moon will continue to recede from Earth, and Earth's spin will continue to slow gradually. In about 50 billion years, if they survive the Sun's expansion, the Earth and Moon will become tidally locked to each other; each will be caught up in what is called a &quot;spin–orbit resonance&quot; in which the Moon will circle the Earth in about 47 days and both Moon and Earth will rotate around their axes in the same time, each only visible from one hemisphere of the other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the Moon will certainly still be around by the time the Sun becomes a red giant, and for many billions of years after that.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's the math:&#xA;$$\frac{1 \text{ cm}}{1 \text{ year}} \times \frac{1 \text{ m}}{100 \text{ cm}} \times \frac{1 \text{ km}}{1000 \text{ m}} \times 5,400,000,000 \text{ years}=54,000 \text{ km}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's about one-seventh the current distance to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Moon&quot;&gt;Moon&lt;/a&gt; - at its closest pass.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-11-18T23:36:40.247" LastActivityDate="2014-11-18T23:36:40.247" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8003" PostTypeId="2" ParentId="8001" CreationDate="2014-11-20T08:50:42.690" Score="28" Body="&lt;p&gt;As HDE 226868 noted in his answer, the Sun is not going to go supernova. That's something only large stars experience at the end of their main sequence life. Our Sun is a dwarf star. It's not big enough to do that. It will instead expand to be a red giant when it burns out the hydrogen at the very core of the Sun. It will continue burning hydrogen as a red giant, but in a shell around a sphere of waste helium. The Sun will start burning helium when it reaches the tip of the red giant phase. At that point it will shrink a bit; a slight reprieve. It will expand to a red giant once again on the asymptotic red giant branch when it burns all the helium at the very core. It will then burn helium in a shell surrounding a sphere of waste carbon and oxygen. Larger stars proceed beyond helium burning. Our Sun is too small. Helium burning is where things stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun has two chances as a red giant to consume the Earth. Some scientists say the Sun will consume the Earth, others that it won't. It's all a bit academic because the Earth will be dead long, long before the Sun turns into a red giant. I'll have more to say on this in the third part of my answer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The current lunar recession rate is 3.82 cm/year, which is outside your one to three centimeters per year window. This rate is anomalously high. In fact, it is extremely high considering that dynamics says that&#xA;$$\frac {da}{dt} = (\text{some boring constant})\frac k Q \frac 1 {a^{11/2}}$$&#xA;Here, $a$ is the semi major axis length of the Moon's orbit, $k$ is the Earth-Moon tidal Love number, and $Q$ is the tidal dissipation quality factor. Qualitatively, a higher Love number means higher tides, and a higher quality factor means less tidal friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That inverse $a^{5.5}$ factor indicates something seriously funky must be happening to make the tidal recession rate so very high right now, and this is exactly the case. There are two huge north-south barriers to the flow of the tides right now, the Americas in the western hemisphere and Afro-Eurasia in the eastern hemisphere. This alone increases $k/Q$ by a considerable amount. The oceans are also nicely shaped so as to cause some nice resonances that increase $k/Q$ even further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If something even funkier happens and the Moon recedes at any average rate of four centimeters per year over the next billion years, the Moon will be at a distance of 425,000 km from the Earth (center to center). That's less than 1/3 of the Earth's Hill sphere. Nearly circular prograde orbits at 1/3 or less of the Hill sphere radius should be stable. Even with that over-the-top recession rate, the Moon will not escape in the next billion years.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What about after a billion years? I chose a billion years because that's about when the Moon's recession should more or less come to a standstill. If the Earth hasn't already died before this billion year mark, this is when the Earth dies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dwarf stars such as our Sun get progressively more luminous throughout their life on the main sequence. The Sun will be about 10% more luminous than it is now a billion years into the future. That should be enough to trigger a moist greenhouse, which in turn will trigger a runaway greenhouse. The Earth will become Venus II. All of the Earth's oceans will evaporate. Water vapor will reach well up into what is now the stratosphere. Ultraviolet radiation will photodissociate that water vapor into hydrogen and oxygen. The hydrogen will escape. Eventually the Earth will not only be bare of liquid water on the surface, it will be bare of water vapor in the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost all of the Moon's recession is a consequence of ocean tides. Without oceans, that tunar recession will more or less come to a standstill.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="-1" LastEditDate="2017-05-29T19:38:30.647" LastActivityDate="2017-05-29T19:38:30.647" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8004" PostTypeId="1" AcceptedAnswerId="8011" CreationDate="2014-11-20T10:53:16.287" Score="6" ViewCount="174" Body="&lt;p&gt;What can be said about 67P/Churyumov-Gerasimenko axial tilt?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it fixed (if so, at what value), only changing with precession (at what rate?) or does 67P &quot;tumble&quot; with axial tilt changing over time? (and if so, at what rate?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in estimating the 'polar night' area of the comet, and how it changes over time - but without knowing the rotation properties this is quite hard to do.&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2014-11-25T16:51:00.047" Title="How does 67P rotate?" Tags="&lt;comets&gt;&lt;67p&gt;&lt;obliquity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8005" PostTypeId="1" AcceptedAnswerId="8006" CreationDate="2014-11-20T15:38:40.343" Score="0" ViewCount="158" Body="&lt;p&gt;To my (admittedly limited) understanding a human being in outer space wouldn't last very long, mainly due to three factors:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the extreme cold&lt;/li&gt;&#xA;&lt;li&gt;the intense radiation&lt;/li&gt;&#xA;&lt;li&gt;and the vacuum of space&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So if my understanding is correct, humans would die within seconds (if anyone could put an approximate number on this that would be great!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to know is that if a human being was exposed to space for half that time, would they suffer lasting and life threatening after effects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, is there any amount of exposure to space that wouldn't result in your imminent or eventual death?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would exposure of just a hand kill you, or would you just lose that hand?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would being exposed to space from inside a room with a single open door help much (I am assuming here the room would shield you from much of the radiation in space), or would the other two factors still kill you off?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please feel free to correct my assumptions wherever I am wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks for any help offered to my slightly morbid musings on space.&lt;/p&gt;&#xA;" OwnerUserId="2844" LastActivityDate="2014-11-21T01:38:00.370" Title="Humans surviving in space" Tags="&lt;space&gt;&lt;exposure&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ClosedDate="2014-11-23T11:47:02.943" ContentLicense="CC BY-SA 3.0" />
  <row Id="8006" PostTypeId="2" ParentId="8005" CreationDate="2014-11-20T16:45:08.037" Score="2" Body="&lt;p&gt;One would not, under normal conditions, die within seconds. This is a common trope in science fiction, but there has been quite a bit of real research into the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's speak on each point.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The extreme cold&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This isn't really a huge problem. Space isn't necessarily &lt;em&gt;cold&lt;/em&gt;. The extraordinarily low density of matter in space means that the movement of energy through induction is extremely limited. If one were to step into 'space' one wouldn't &lt;em&gt;feel&lt;/em&gt; cold, or even cool.  For engineers, often the greater problem is not how to deal with cold, but how to deal with heat, as even though one can't easily release heat through induction, one can certainly gain it through radiation or internal chemical processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A person, however, is likely to experience cooling due to the process of liquids in the body evaporating and escaping, taking some of their heat with them. It's unlikely that this would prove fatal though, or at least it's unlikely to be the &lt;em&gt;first thing&lt;/em&gt; to prove fatal.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Intense Radiation&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Radiation is a problem in space, but it's not going to kill you so quickly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;During a 360-day round trip, an astronaut would receive a dose of about 662 millisieverts (mSv), according to RAD measurements. National space agencies limit exposure to about 1000 mSv or less during an astronaut's entire career; NASA's limit corresponds to a 3% risk of exposure-induced death from cancer. (&lt;a href=&quot;http://www.sciencemag.org/content/340/6136/1031.full.pdf&quot; rel=&quot;nofollow&quot;&gt;Kerr, 1031&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Currently, orbital missions get some protection from radiation as they are within the Earth's magnetic field, but for those travelling beyond this there isn't a lot of protection. Part of the reason is that things that provide good protection are also very heavy, which can be quite difficult in the economies of reaching escape velocity. As such, one wouldn't necessarily be exposed to that much more radiation outside of a shuttle or their suit in space than they would otherwise.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Vacuum&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is where things can get messy, but there has been some research on the subject. Obviously one can't breath in a vacuum, in fact the lungs actually begin to dump oxygen, essentially working in reverse due to the lack of air pressure. Our victim is likely to lose consciousness within 10-15 seconds. If they happen to be holding their breath when they become exposed, the pressure difference could cause some real damage to their lungs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People often theorize that blood would boil in space. Geoffrey Landis disagrees with this on his &lt;a href=&quot;http://www.geoffreylandis.com/vacuum.html&quot; rel=&quot;nofollow&quot;&gt;blog&lt;/a&gt;, in which he speaks on this subject. I highly recommend reading through it as I would consider him both thorough and authoritative on the subject. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't find the official report, but there are frequent references to a 1965 (Earth based) pressurization incident, in which a technician's suit depressurized within a chamber. He remarked afterwords that the last thing he recalled before blacking out was feeling his saliva begin to boil on his tongue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While your blood arguably won't boil, you would experience the boiling of other liquids in your body, water particularly.  This can result in rapid swelling of the body. While not pretty, this isn't likely to kill you. However, nitrogen can also become gaseous due to this, potentially resulting the bends, a condition more closely associated with pressure differences in diving. This condition could prove fatal, but AFAIK there hasn't been a lot of research into this in reference to vacuum exposure (This condition tends to become dangerous during the return to &lt;em&gt;normal pressures&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this can be survived, as long as pressure is returned relatively quickly. Within a minute is probably enough to save a person's life with little lasting damage. After 2-3 minutes, it's likely too late. The difficulty is that the unfortunate victim may be able to survive 1-2 minutes of exposure, they really only get 10-15 seconds to do something about the problem. Of those who have survived vacuum pressurization incidents, their survival is mostly credited to the quick thinking of those around them.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To respond to your more specific questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Exposure for &lt;em&gt;half&lt;/em&gt; of what would generally be fatal&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This could vary a lot, like depending on whether they were holding their breath. However, generally it's believed that a person could recover from this with little to no permanent damage. Probably the biggest risks are lung damage due to pressure differences, the bends and other related complications, and brain damage due to oxygen deprivation.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Exposure to space that wouldn't result in imminent or eventual death&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A little exposure without other complications will generally be survivable. We don't have a ton of data on the subject, so it's possible that unpredictable complications could arise, and frankly it's not exactly advisable, especially considering the person will lose consciousness quite quickly.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Partial exposure&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I can't really rely on any specific source for this, but presumably exposure of an extremity would be perfectly survivable. I'm not quite sure how this would work as it would be very difficult to partially expose someone to a vacuum. The human body isn't exactly leak proof. In theory, you would begin to exhibit symptoms in the exposed body part, but you would remain conscious (assuming your respiratory system isn't exposed) and generally be in control of what's going on. Eventually other complications would likely set in and need to be managed carefully.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Exposed to space from inside a room with a single open door&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As noted the radiation and temperature is not terribly dangerous, but with no way to repressurize the room this can quickly turn fatal. The Russian Soyuz II mission ended catastrophically when the capsule carrying the cosmonauts depressurized on way for reentry, killing all three crew members on board.  A valve malfunctioned during the reentry, causing depressurization. The craft was recovered after reentry and had not taken any significant damage. It was believed that Patsayev had made attempts to fix the valve in order to save the crew, but unfortunately there was simply too little time to accomplish this. Initial reports suggested that the crew had died of asphyxiation, however autopsies showed that the cause was hemorrhaging in the brain - reportedly caused by oxygen and nitrogen in the blood boiling. Not sure how this works out with Landis's claim that blood does not boil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Radiation Will Make Astronauts' Trip to Mars Even Riskier&lt;/strong&gt;&#xA;Richard A. Kerr&#xA;Science 31 May 2013: 340 (6136), 1031. [DOI:10.1126/science.340.6136.1031]&lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="1288" LastEditDate="2014-11-21T01:38:00.370" LastActivityDate="2014-11-21T01:38:00.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8007" PostTypeId="2" ParentId="7974" CreationDate="2014-11-20T17:17:37.053" Score="0" Body="&lt;p&gt;&lt;strong&gt;METHOD 4:Predicting CPTS using a Coarse Binomial Series approximation.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method does not make any assumptions about Mean Anomaly. Orbit is defined as the cycle from one periapsis to the next periapsis. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SUMMARY&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time from the $0$th to the $N$th periastron can be obtained from the following sum, from which a binomial expression can be derived&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{T_N}{P_0} &#xA;\qquad = \sum_{i=1}^{i=N} (1+\dot P)^i&#xA;\qquad = N + \sum_{j=2}^{N+1}\binom{N+1}{j}\dot P^{j-1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For our purposes terms in $\dot P$ greater than first order can be dropped leading to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T_{N} = N{P_0} + P_0 \frac{N^2+N}{2} {\dot P} .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time difference between the $N$th periapses of a steady period system and a decaying period system is then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta t_{N} =  0.5 P_0 {\dot P}  (N^2+N) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives results which agree closely with the Wesiberg &amp;amp; Taylor graph and the formula in Stan Liou's answer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definitions and Assumptions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us consider the variation over time in the &lt;strong&gt;period&lt;/strong&gt; of orbit. Weisberg &amp;amp; Taylor provide a formula for the variation in &quot;period&quot; with time.  However the meaning of &quot;period at a point in time&quot; is ambiguous when the period is changing with time. We therefore have to define how &quot;period at a point in time&quot; translates into a time interval between the start and end of a particular orbital cycle (e.g. from periapsis to periapsis).  In this method we will &quot;fudge&quot; the translation by assuming (A1) that the time interval between successive periapses is equal to the &quot;hypothetical period at the point in time of the end of the cycle&quot;. Also to &quot;boot&quot; the model we will assume (A2) that at time zero ($t=0$) the hypothetical period is $H_o$ and thus the hypothetical period at the end of the first cycle is given by $H_1 = H_o + H_o \dot H$.  Therefore, by (A1) we have the following definition of the duration of the first cycle $D_1 = H_1 = H_o + H_o \dot H$.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We further assume (A3) that, over the period of consideration (~ 40 years) the rate of change of period with time is effectively constant ($\ddot H = 0$ and  $\dot H(t) = constant = \dot H$ ). (See Stan Liou's answer for justification). Thus we have&#xA;$$&#xA;H(t) = H(0) + t \dot H .&#xA;$$&#xA;It is emphasized that these assumptions are not strictly mappable to a consistent physical model.  However given that we are dealing with very small changes in period over a very large number of orbits the ruleset may be useful.  We shall proceed and see what comes out.  Henceforth I will not distinguish between hypothetical period $H$ and cycle duration $D$.  I will assume that they are both represented by the same term period $P$ and that the duration $P$ of a particular periapsis to periapsis cycle is equal to the hypothetical period $P(t)$ where $t$ is the epoch of the end of the cycle.  Also $\dot P$ will represent both the rate of change of hypothetical period and also the rate of change of cycle duration between cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_N$ is the period of the $N$th orbit (orbit $O_N$). Orbit number $0$ terminates at $t=T_0=0$ at perihelion number $0$ and has period $Po = P(0)$.   The first considered orbit ($O_1$) starts at time $t=0$ at perihelion number $0$ and finishes at perihelion number $1$ at time $t=T_1$ and has period duration $P_1 = P_0+ P_0\dot P =P(T_1)$. Therefore we can write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_1 = P_0(1+\dot P)^1$$&#xA;$$P_2 = P_1(1+\dot P) = P_0(1+\dot P)(1+\dot P)= P_0(1+\dot P)^2$$&#xA;$$P_3 = P_2(1+\dot P) = P_0(1+\dot P)(1+\dot P)(1+\dot P)= P_0(1+\dot P)^3$$&#xA;In general then, the period of any orbit is given by the formula:&#xA;$$P_N = P_0(1+\dot P)^N.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let us consider the &lt;strong&gt;time&lt;/strong&gt; $T$ at which each orbit ends. We have &#xA;$$T_1 = T_0+P_1 = P_1            = P_0(1+\dot P)$$ &#xA;$$T_2=T_1+P_2 = P_1+P_2  = P_0(1+\dot P)+P_0(1+\dot P)^2$$&#xA;$$T_3=T_2+P_3 = P1+P2+P3 = P_0(1+\dot P)+P_0(1+\dot P)^2+P_0(1+\dot P)^3$$&#xA;$$T_N=P_1+P_2+...+P_N = P_0(1+\dot P)+P_0(1+\dot P)^2+...+P_0(1+\dot P)^N.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&#xA;$$T_N = {P_0}\sum_{i=1}^{i=N} (1+\dot P)^i\, .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The Note at the end of this answer explores a way of proceeding which uses an exact equivalent of this expression in a non-serial form.  However, that route is not productive due to difficulties with exponentials.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We shall proceed as follows,  (using $k$ in place of $\dot P$)&#xA;$$ \frac{T_N}{P_0}= S_N = \sum_1^N (1+k)^i $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can multiply out the first few terms and examine the patterns in the following &lt;strong&gt;pyramid&lt;/strong&gt;...&#xA;$$(1+k)^1 =               k +1  $$&#xA;$$(1+k)^2 =        k^2  +2k +1  $$&#xA;$$(1+k)^3 = k^3  +3k^2  +3k +1  $$&#xA;$$(1+k)^4 =k^4+4 k^3+6 k^2+4 k+1$$&#xA;$$(1+k)^5 =k^5+5 k^4+10 k^3+10 k^2+5 k+1$$&#xA;$$(1+k)^6 =k^6+6 k^5+15 k^4+20 k^3+15 k^2+6 k+1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for example, for $N=3$ we would obtain the sum&#xA;$$&#xA;S_3= k^3  +4k^2 +6k +1&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The results suggest a solution with a pattern of the form&#xA;$$&#xA;S_N = a + bk^1 +ck^2+dk^3...&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see straightaway that $a=N$.  The other coefficients increase with $N$. It might be possible to determine a formula for the coefficients from the pattern. Although the general pattern is not convergent, it is possible that for certain restricted ranges of $N$ and $k$ a convergent formula could be obtained.  If so then it might be possible to obtain a useful approximation of $S_N$  by evaluating the first few terms in the series.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The following binomial equation was derived in an answer to &lt;a href=&quot;https://math.stackexchange.com/questions/1029172/how-do-i-express-the-sum-1k1k21k3-1kn-for-k1-as-a-se&quot;&gt;this question&lt;/a&gt; at Maths Stack Exchange by User73985 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_1^N (1+k)^i \qquad = \qquad N + \sum_{j=2}^{N+1}\binom{N+1}{j}k^{j-1}$$&#xA;So&#xA;$$ S_N = N + \sum_{j=2}^{N+1}\frac{(N+1)!}{(N+1-j)!j!}k^{j-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then&#xA;$$ S_N = N &#xA;+  \frac{(N+1)!}{(N-1)!2!} k^{1} &#xA;+  \frac{(N+1)!}{(N-2)!3!} k^{2}&#xA;+  \frac{(N+1)!}{(N-3)!4!} k^{3} +...&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;giving&#xA;$$ S_N = N &#xA;+  \frac{(N+1)(N)}{2!} k^{1} &#xA;+  \frac{(N+1)(N)(N-1)}{3!} k^{2}&#xA;+  \frac{(N+1)(N)(N-1)(N-2)}{4!} k^{3} +...&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus&#xA;$$ S_N= N &#xA;+  \frac{N^2+N}{2} k^{1} &#xA;+  \frac{N^3-N}{6} k^{2}&#xA;+  \frac{N^4-2 N^3-N^2+2 N}{24} k^{3} +...&#xA;$$&#xA;(Switching back from using $k$ to using $\dot P$...)&#xA;For $N= 10,000$ and ${k=\dot P}= 2.40242 * 10^{-12}$ the ratio of each term to the next term is greater than $1 : N{\dot P} \approx 1:2.4 * 10^{-8}$ so for our purposes this formula can be approximated by truncating to the first order power of ${\dot P}$ to give&#xA;$$ S_{N} = \frac{T_{N}}{P_0} = N + \frac{N^2+N}{2} {\dot P}^{1} $$&#xA;So&#xA;$$ T_{N} = N{P_0} + P_0 \frac{N^2+N}{2} {\dot P} $$&#xA;and so the time difference between the Nth periapses (fixed $P$ - decaying $P$) is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta t_{N} = P_0{\dot P} \frac{N^2+N}{2}  $$&#xA;and for $N=10,000$ we obtain (from Excel) $\Delta t = 3.35255 s$ which is very close to the value $(3.35221 s)$ obtained from Stan Liou's formula ($\Delta t= 0.5 {\dot P}/P_0 t^2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over 30 years (33,924.42 cycles) $\Delta t$ = 38.5806 $s$ which is very close to the amount indicated by the graph and by Stan Liou's formula $(38.5794 s)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WolframAlpha indicates the equality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_N = {P_0}\sum_{i=1}^{i=N} (1+\dot P)^i&#xA;\qquad =&#xA;\frac{P_0}{\dot P}&#xA;\,&#xA;(1+\dot P)[(1+\dot P)^N -1]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For small $\dot P$ this can be approximated by:-&#xA;$$T_N =\frac{P_0}{\dot P}\,(1+\dot P)[(1+N\dot P) -1]&#xA;=\frac{P_0}{\dot P}\,(1+\dot P)(N\dot P) &#xA;=N{P_0}\,(1+\dot P). &#xA;$$&#xA;The end time of $N$ cycles of period $P_{constant}=P_0$ will be $T_{NC}=N P_0$. And so the time difference between the end times of  (i) N cycles of constant $P$ and (ii) $N$ cycles of decaying $P$ is approximated by &#xA;$$\Delta t = NP_0 - NP_0(1-\dot P) = NP_0 \dot P.&#xA;$$&#xA;For  $P_0$ = 0.322997448930 d = 27906.979587552 s and $\dot P$ = -2.40242 x 10$^{-12}$ we obtain $(P_0\,dot P) =$ -6.70442859007267$^{-08}$. For  $N=10,000$ cycles (approximately 10 years) this indicates a time difference of 6.70442859007267 $^{-03}$ s, about $7 ms$.  This time difference is very small compared to the value roughly indicated by the original graph and the value calculated by Stan Liou's formula, from which we would expect a time difference of about 3$s$.  The explanantion is that the first-order approximation is not suitable in this case because the true value of $\Delta t$ is so small compared to $t$.&lt;/p&gt;&#xA;" OwnerUserId="233" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:35.350" LastActivityDate="2014-11-25T14:13:31.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8008" PostTypeId="2" ParentId="6519" CreationDate="2014-11-20T17:23:25.337" Score="11" Body="&lt;p&gt;Whenever I see this question discussed, it seems that the heliopause, or some variation thereof, is given as an answer -- and then it's mentioned that the Oort Cloud extends beyond it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more correct answer, therefore, should be that it ends at that distance at which objects are, for all practical purposes, no longer bound to the solar system barycenter. This is usually defined by the &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;noreferrer&quot;&gt;Hill Sphere&lt;/a&gt;, which approximates the gravitational sphere of influence.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One simple view of the extent of the Solar System is the Hill sphere of the Sun with respect to local stars and the galactic nucleus.(1)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This extends out to two hundred thirty thousand AU, about 3.6 light years. Again, not a wall. According to (1) &lt;a href=&quot;http://adsabs.harvard.edu/full/1965SvA.....8..787C&quot; rel=&quot;noreferrer&quot;&gt;Cherbatov (1965)&lt;/a&gt;, the radii of the gravitational spheres of the sun may be subdivided into:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Sphere of attraction out to 4500 AU (sun's attraction &gt; galactic center's attraction),  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Sphere of action 60,000 AU (more convenient to use sun as central body &amp;amp; galactic center as pertubating body in orbital calculations), and finally  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Hill sphere 230,000 AU (object must orbit within this limit to be retained by the Sun).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2674" LastEditorUserId="2674" LastEditDate="2015-08-26T23:37:47.887" LastActivityDate="2015-08-26T23:37:47.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8009" PostTypeId="5" CreationDate="2014-11-20T19:36:57.797" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-11-20T19:36:57.797" LastActivityDate="2014-11-20T19:36:57.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8010" PostTypeId="4" CreationDate="2014-11-20T19:36:57.797" Score="0" Body="Questions related to techniques and equipment related to amateur to professional astronomical photography. " OwnerDisplayName="user2449" LastEditorDisplayName="user2449" LastEditDate="2014-11-21T05:16:33.413" LastActivityDate="2014-11-21T05:16:33.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8011" PostTypeId="2" ParentId="8004" CreationDate="2014-11-20T21:00:45.187" Score="1" Body="&lt;p&gt;Looking at &lt;a href=&quot;http://sci.esa.int/rosetta/14615-comet-67p/&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt;, there is a table, which has some information.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Rotation period: 12.4043 ± 0.0007 * hours    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It also rotates around a fixed axis&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;recent observations suggest that the axis is tilted about 40 degrees&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are actually only two objects known to rotate chaotically in the solar system - one is Saturn's moon Hyperion, and the other is a small asteroid called 4179 Toutatis (source: &lt;a href=&quot;http://www.guinnessworldrecords.com/records-12000/largest-object-chaotically-rotating/&quot; rel=&quot;nofollow&quot;&gt;this link from Guinness World Records&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="2702" LastActivityDate="2014-11-20T21:00:45.187" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8012" PostTypeId="1" AcceptedAnswerId="8015" CreationDate="2014-11-21T01:01:25.760" Score="7" ViewCount="812" Body="&lt;p&gt;In a model of the solar system, given the Sun is at the point $(0, 0, 0)$, an given the six orbital elements for each object in orbit ($a$, $\epsilon$, $i$, $\Omega$, $\omega$, $M_0$), how can I calculate the relative position to the sun at a given time $t$?&lt;/p&gt;&#xA;" OwnerUserId="2830" LastActivityDate="2015-11-30T22:22:09.850" Title="How do I calculate the positions of objects in orbit?" Tags="&lt;orbit&gt;&lt;orbital-elements&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8013" PostTypeId="2" ParentId="8012" CreationDate="2014-11-21T01:42:54.890" Score="1" Body="&lt;p&gt;I'm not sure how much this will help, but I can give you &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_elements#Orbit_prediction&quot; rel=&quot;nofollow&quot;&gt;this passage&lt;/a&gt; from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Under ideal conditions of a perfectly spherical central body, and zero perturbations, all orbital elements, with the exception of the Mean anomaly are constants, and Mean anomaly changes linearly with time[dubious – discuss], scaled by the Mean motion, $n=\sqrt{\frac{\mu } {a^3}}$. Hence if at any instant $t_0$ the orbital parameters are $[e_0,a_0,i_0,\Omega_0,\omega_0,M_0]$, then the elements at time $t_0+\delta t$ is given by $[e_0,a_0,i_0,\Omega_0,\omega_0,M_0+n\delta t]$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, most of the orbital elements (save $M_0$) are constant under very short intervals.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-11-21T01:42:54.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8014" PostTypeId="1" CreationDate="2014-11-21T04:02:24.403" Score="-2" ViewCount="100" Body="&lt;p&gt;When a radio telescope listen to the Universe; a constant sound is recorded. Is it the noise &quot;of the mic&quot; : interferences? Scientists says that when they will notice variations on this basic constant same noise I call &quot;the interference coming from the mic&quot; they will conclude on an extra terrestrial source of sound; but in fact if is waves that are &quot;listen&quot; to. So, any news?&lt;/p&gt;&#xA;" OwnerUserId="2850" LastActivityDate="2014-11-21T15:07:51.803" Title="As of now did a radio telescope ever recorded something else than the silence of space?" Tags="&lt;radio-astronomy&gt;&lt;extra-terrestrial&gt;&lt;space-telescope&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ClosedDate="2014-11-21T15:57:08.447" ContentLicense="CC BY-SA 3.0" />
  <row Id="8015" PostTypeId="2" ParentId="8012" CreationDate="2014-11-21T05:03:00.717" Score="2" Body="&lt;p&gt;You must solve the &lt;a href=&quot;http://mathworld.wolfram.com/KeplersEquation.html&quot; rel=&quot;nofollow&quot;&gt;Kepler's equation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M = E - e \sin E$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $M$ is the &lt;em&gt;mean anomaly&lt;/em&gt;, $e$ the &lt;em&gt;eccentricity&lt;/em&gt; of the orbit, $E$ the &lt;em&gt;eccentric anomaly&lt;/em&gt;).&#xA;As you can see, this is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Transcendental_equation&quot; rel=&quot;nofollow&quot;&gt;transcendental function&lt;/a&gt;, so you will probably need to iterate through a calculator, in order to solve it.&#xA;Many of useful formulas to understand the parameters dependence are reported on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Formulary&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt; as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.bogan.ca/orbits/kepler/keplerex.html&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is another page with some examples. Just use the formulas and try to play with numbers, at some point you will realize at least orders of magnitude of the binary system elements.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastEditorUserId="2365" LastEditDate="2015-11-30T22:22:09.850" LastActivityDate="2015-11-30T22:22:09.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8016" PostTypeId="1" AcceptedAnswerId="8019" CreationDate="2014-11-21T13:36:04.917" Score="4" ViewCount="275" Body="&lt;p&gt;Earth comes closest to the Sun every year around &lt;a href=&quot;http://en.wikipedia.org/wiki/Perihelion_and_aphelion&quot; rel=&quot;nofollow&quot;&gt;January 3&lt;/a&gt;. Winter solstice is at &lt;a href=&quot;http://en.wikipedia.org/wiki/Solstice&quot; rel=&quot;nofollow&quot;&gt;21-22 December&lt;/a&gt;, merely some 12-13 days away. That means projection of Earth rotation axis onto the plane of its orbital movement almost overlaps with its orbit's semi-major axis; the axial tilt direction coincides with the direction Earth orbit is elongated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't find corresponding data regarding other bodies in the Solar System.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this merely a coincidence, or is there a regularity, with direction of the axial tilt coinciding with eccentricity of the orbit? How does that look like in case of other bodies than Earth?&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2014-11-21T19:00:31.960" Title="Is axial tilt direction correlated with semi-major axis direction?" Tags="&lt;orbit&gt;&lt;obliquity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8017" PostTypeId="2" ParentId="8014" CreationDate="2014-11-21T15:07:51.803" Score="1" Body="&lt;p&gt;Space is not quiet at all and is actually very noisy when listening via a radio telescope.  There are many sources of radio emissions in the universe, with pulsars being a very common one.  They can be quite interesting to listen when the signals are converted into sound frequencies we can hear. They range from slow clicks every few seconds to high pitched buzzing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a link with some examples:&#xA;&lt;a href=&quot;http://www.parkes.atnf.csiro.au/people/sar049/eternal_life/supernova/pulsars.html&quot; rel=&quot;nofollow&quot;&gt;http://www.parkes.atnf.csiro.au/people/sar049/eternal_life/supernova/pulsars.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When using a wide angle antenna or one that uses a wide band of frequencies the mixture of all these emissions at once would just sound like background static.&lt;/p&gt;&#xA;" OwnerUserId="2856" LastActivityDate="2014-11-21T15:07:51.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8019" PostTypeId="2" ParentId="8016" CreationDate="2014-11-21T19:00:31.960" Score="4" Body="&lt;p&gt;It's mere coincidence. In about 10,000 years, perihelion will coincide with the northern hemisphere summer solstice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two precessions of interest here. One is the 26,000 year period axial precession of the Earth's rotation axis, caused mostly by the Moon and the Sun; the other is the 112,000 year period apsidal precession of the Earth's orbit, caused mostly by Jupiter and Venus (but also with a pinch of help from general relativity). This result in three different concepts of a year:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The sidereal year is the mean time it takes the Earth to make a complete orbit with respect to inertial space (&quot;the fixed stars&quot;).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The tropical year (aka solar year) is the mean time it takes for the Sun's ecliptic longitude to increase by 360 degrees, and also the time between one vernal equinox to the next. This is the year used as the basis of our calendar. This year is shorter than the sidereal year by about 20 minutes and 25 seconds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The anomalistic year is the mean time between one perihelion passage of the Earth to the next. This year is longer than the sidereal year by about 4 minutes and 43 seconds, and hence longer than the tropical year by about 25 minutes and 8 seconds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That 25+ minute difference between the anomalistic and tropical years means that time of perihelion passage advances by about a day every 57 years.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2014-11-21T19:00:31.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8023" PostTypeId="1" AcceptedAnswerId="8024" CreationDate="2014-11-22T01:07:47.713" Score="9" ViewCount="1947" Body="&lt;p&gt;Kepler's first law states that planets (and all celestial bodies orbiting another body) travel in elliptical orbits, which have well-known formulas that make it relatively easy to calculate the orbital elements and associated behavior. However, ongoing precession means that the orbit is constantly changing - and so the planet isn't actually traveling in the ellipse it originally set out on! You can calculate precession and its related effects (&lt;a href=&quot;https://astronomy.stackexchange.com/questions/7938/where-did-this-famous-planetary-precession-formula-come-from&quot;&gt;this question and answer&lt;/a&gt; are helpful), but is there any way to calculate how the elliptical orbit will be &quot;deformed&quot; by precession?&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2018-12-18T10:27:23.803" Title="How do you calculate the effects of precession on elliptical orbits?" Tags="&lt;orbit&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8024" PostTypeId="2" ParentId="8023" CreationDate="2014-11-22T02:51:08.367" Score="9" Body="&lt;p&gt;A good starting point would be &amp;lt;insert name of some scientist from long ago&amp;gt; planetary equations of motion. For example, there are Lagrange's planetary equations (sometimes called the Lagrange-Laplace planetary equations), Gauss' planetary equations, Delaunay's planetary equations, Hill's planetary equations, and several more. The common theme amongst these various planetary equations is that they yield the time derivatives of various orbital elements as a function of the partial derivatives of the perturbing force / perturbing potential with respect to some generalized position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the only words that can describe the result of this process at first is &quot;hot mess.&quot; A hot mess did not deter those brilliant minds of old. Via various simplifying assumptions and long term time averaging, they came up with fairly simple descriptions of, for example, $\left \langle \frac{d\omega}{dt} \right\rangle$ (apsidal precession) and $\left \langle \frac{d\Omega}{dt} \right\rangle$ (planar precession). You can see some of this in the cited 1900 work by Hill below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While these techniques are old, these planetary equations are still used today. That sometimes you do get a &quot;hot mess&quot; is okay now that we have computers. People are using planetary equations coupled with geometric integration techniques to yield integrators that are fast, accurate, stable, and conserve angular momentum and energy over long spans of time. (Normally, you can't have all of these. You're lucky if you get just two or three.) Another nice feature of these planetary equations is that they let you see features such as resonances that are otherwise obscured by the truly &quot;hot mess&quot; of the cartesian equations of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Selected reference material, sorted by date:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ams.org/journals/tran/1900-001-02/S0002-9947-1900-1500534-3/S0002-9947-1900-1500534-3.pdf&quot; rel=&quot;noreferrer&quot;&gt;Hill (1900), &quot;On the Extension of Delaunay's Method in the Lunar Theory to the General Problem of Planetary Motion,&quot; &lt;em&gt;Transactions of the American Mathematical Society&lt;/em&gt;, 1.2:205-242.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vallado (1997 and later), &quot;Fundamentals of Astrodynamics and Applications&quot;, various publishers.&#xA;Other than the hole it punches through your wallet, you can't go wrong with this book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://silver.ima.umn.edu/preprints/feb02/1844.pdf&quot; rel=&quot;noreferrer&quot;&gt;Efroimsky (2002), &quot;Equations for the keplerian elements: hidden symmetry,&quot; &lt;em&gt;Institute for Mathematics and its Applications&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/astro-ph/0305344&quot; rel=&quot;noreferrer&quot;&gt;Efroimsky and Goldreich (2003), &quot;Gauge symmetry of the N-body problem in the Hamilton–Jacobi approach.&quot; &lt;em&gt;Journal of Mathematical Physics&lt;/em&gt;, 44.12:5958-5977.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.ast.cam.ac.uk/~wyatt/lecture2_planetarysystemdynamics.pdf&quot; rel=&quot;noreferrer&quot;&gt;Wyatt (2006-2009), Graduate lecture course on planetary systems, Institute of Astronomy, Cambridge.&lt;/a&gt;&lt;br&gt;&#xA;The results of the Lagrange planetary equations are presented on slide 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1211.3078&quot; rel=&quot;noreferrer&quot;&gt;Ketchum et al. (2013), &quot;Mean Motion Resonances in Exoplanet Systems: An Investigation into Nodding Behavior.&quot; &lt;em&gt;The Astrophysical Journal&lt;/em&gt; 762.2.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2014-11-22T02:51:08.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8025" PostTypeId="1" CreationDate="2014-11-22T09:20:50.160" Score="-4" ViewCount="177" Body="&lt;p&gt;On a related note, why didn't Venus remain habitable for longer than it did? Is the current  hypothesis as to the time when Venus's atmosphere became hostile and what event triggered it, proven true?&lt;/p&gt;&#xA;" OwnerUserId="2846" LastEditorUserId="2846" LastEditDate="2014-11-23T15:21:20.077" LastActivityDate="2014-11-23T15:21:20.077" Title="Why does Earth enjoy a significantly longer time as a habitable planet as compared to Mars or Venus?" Tags="&lt;earth&gt;&lt;venus&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" ClosedDate="2014-12-02T01:34:11.603" ContentLicense="CC BY-SA 3.0" />
  <row Id="8026" PostTypeId="2" ParentId="8025" CreationDate="2014-11-22T16:28:39.867" Score="2" Body="&lt;p&gt;By survived, do you mean remained habitable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, then there are several reasons: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Venus, the problem was that it was just barely on the edge of the habitable zone when the Sun was younger and less bright. As it aged and brightened towards the values we see today, any oceans began to evaporate. The resultant water vapour is a very powerful greenhouse gas, and so would have trapped yet more heat, causing a runaway greenhouse effect, until it is the baked, hellish world we see today. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars, on the other hand, was believed to be too small. It (probably, assuming modern theories are correct) started off as another habitable world, but since it was smaller, it would've had less residual heat left over from its birth, and would've cooled quicker. This would've caused the core to solidify, removing the magnetic field. This meant that the atmosphere was vulnerable to the solar wind, which would've gradually blown it - and any warming greenhouse gases - away, and with the lower pressure, the water would evaporate away as well, rendering Mars uninhabitable.&lt;/p&gt;&#xA;" OwnerUserId="2702" LastActivityDate="2014-11-22T16:28:39.867" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8027" PostTypeId="2" ParentId="7957" CreationDate="2014-11-22T20:24:02.817" Score="4" Body="&lt;p&gt;There are a number of different organizations that produce ephemeris models of the solar system. Even within one organization, there a number of different ephemeris models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three leading organizations that produce solar system ephemerides are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The Jet Propulsion Laboratory (JPL) in the USA, which produces the Development Ephemeris (DE) series of solar system ephemerides (and also ephemerides for many of the smaller bodies in the solar system);&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Institute of Applied Astronomy at the Russian Academy of Sciences, which produces the Ephemerides of the Planets and Moon (EPM) series of solar system ephemerides; and&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Institut de Mecanique Celeste et de Calcul de Ephemerides (IMCCE) in France, which produces the INPOP series of solar system ephemerides.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There appears to exist a friendly spirit of competition and cooperation amongst these three groups. They apparently share measurements and ideas amongst one another. They do not exactly agree on results (which is a good thing). Each group regularly updates their respective ephemeris models as new measurements are gathered, and as numerical integration techniques and computing capabilities are improved.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2014-11-22T20:24:02.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8028" PostTypeId="1" CreationDate="2014-11-23T00:37:12.150" Score="2" ViewCount="75" Body="&lt;p&gt;Can someone perhaps assist me in identifying the yellow object between δ Ori and η Ori? I took an image of this object, but I can't identify it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be seen on the image below, hosted on the Eagle Creek Obs. site. (&lt;a href=&quot;http://www.eaglecreekobservatory.org/eco/doubles/images/ori.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.eaglecreekobservatory.org/eco/doubles/images/ori.gif&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the orange object on the constellation line between 28 and 34.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vaQQZ.gif&quot; alt=&quot;Orion Constellation&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="155" LastActivityDate="2014-11-23T01:27:28.647" Title="Yellow object between δ Ori and η Ori" Tags="&lt;star&gt;&lt;constellations&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8029" PostTypeId="2" ParentId="8028" CreationDate="2014-11-23T01:27:28.647" Score="2" Body="&lt;p&gt;Someone suggested this object to be 31 Ori. After further research, equipped with the name, I confirmed this object to be 31 Ori. Source of confirmation: &lt;a href=&quot;http://www.astrostudio.org/xhip.php?hip=25737&quot; rel=&quot;nofollow&quot;&gt;http://www.astrostudio.org/xhip.php?hip=25737&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="155" LastActivityDate="2014-11-23T01:27:28.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8031" PostTypeId="1" CreationDate="2014-11-23T10:35:50.067" Score="2" ViewCount="141" Body="&lt;p&gt;I am working through Liddle's &quot;An Introduction to Modern Cosmology&quot;, and in a newtonian derivation of the Friedman equation he states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;in a spherically symmetric distribution of matter, a particle feels no force at all from the material at greater radii, and the material at smaller radii gives exactly the force which one would get if all the material was concentrated at the central point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He attributes this result to Newton, Can anyone give a mathematical or physical explanation as to why this is true? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2869" LastEditorUserId="2153" LastEditDate="2014-11-23T14:52:49.397" LastActivityDate="2014-11-23T14:52:49.397" Title="Why does a particle feel no force at radii greater than itself?" Tags="&lt;gravity&gt;&lt;newtonian-gravity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8032" PostTypeId="2" ParentId="8031" CreationDate="2014-11-23T11:37:26.167" Score="2" Body="&lt;p&gt;The original result is Newton's &lt;a href=&quot;https://en.wikipedia.org/wiki/Shell_theorem&quot; rel=&quot;nofollow&quot;&gt;shell theorem&lt;/a&gt;. Since we can break up a spherically symmetric distribution into spherically symmetric concentric shells, it is sufficient to consider the corresponding statement for one such shell: for each shell taken &lt;em&gt;individually&lt;/em&gt;, there is no force on a particle inside, and a force on a particle outside as if all of the mass was concentrated at the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be derived directly from Newton's law of gravitation through careful integration (one such derivation in the wiki link above), but probably the cleanest method is rather through Gauss's law, which states that the divergence of the gravitational field is proportional to the mass density:&#xA;$$\nabla\cdot\mathbf{g} = -4\pi G\rho\text{.}$$ It's just the same as the law for the electric field because Coulomb's law and Newton's law of gravitation have the same form (with $\epsilon_0\mapsto-(4\pi G)^{-1}$). By the &lt;a href=&quot;https://en.wikipedia.org/wiki/Divergence_theorem&quot; rel=&quot;nofollow&quot;&gt;divergence theorem&lt;/a&gt;, the flux through a surface enclosing volume $V$ is proportional to the enclosed mass:&#xA;$$\int_{\partial V}\mathbf{g}\cdot\mathrm{d}\mathbf{S} = \int_V\nabla\cdot\mathbf{g}\,\mathrm{d}V = -4\pi GM_\text{enc}\text{,}$$&#xA;since then we are just integrating the density over the entire volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But each shell should have a spherically symmetric gravitational field, so $\mathbf{g} = g\hat{\mathbf{r}}$, where $g = g(r)$ is a function of radius only. Since the Gaussian surface has constant $r$, all we get is $g(r)$ times its area:&#xA;$$-4\pi GM_\text{enc} = g(r)\int_{\partial{V}}\mathrm{d}\mathbf{S} = 4\pi r^2 g(r)\text{,}$$&#xA;and it follows that&#xA;$$\mathbf{g} = -\frac{GM_\text{enc}}{r^2}\text{.}$$&#xA;Obviously, if the particle is inside the shell, the Gaussian surface encloses no mass, since again we are considering each shell individually. The total effect of a spherically symmetric distribution would just be the sum of the effects of each individual shell.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-11-23T11:37:26.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8033" PostTypeId="1" CreationDate="2014-11-23T11:57:58.540" Score="4" ViewCount="811" Body="&lt;p&gt;I'd like to one day take the Astronomy Olympiad, however I'm at a loss to know where to begin. As it currently stands I have no experience whatsoever with Astronomy. If there are people here with some expertise on the matter, I'd like to know what books you would recommend to study? The syllabus is at the following link: &lt;a href=&quot;http://www.ioaa.info/page2/page7/page7.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ioaa.info/page2/page7/page7.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2871" LastEditorUserId="2871" LastEditDate="2014-11-24T12:41:46.083" LastActivityDate="2014-12-02T16:54:57.787" Title="Astronomy Olympiad Books" Tags="&lt;resource-request&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8034" PostTypeId="2" ParentId="8031" CreationDate="2014-11-23T13:41:35.667" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;in a spherically symmetric distribution of matter, a particle feels no force at all from the material at greater radii, and the material at smaller radii gives exactly the force which one would get if all the material was concentrated at the central point&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This may be poorly worded. The idea is not that the distance from the center of the body must be smaller than the radius of the particle. Rather, those parts of the (uniform, spherical) body which are further away from the center than is the particle have all their respective gravitational attractions cancelled out.&lt;/p&gt;&#xA;" OwnerUserId="209" LastActivityDate="2014-11-23T13:41:35.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8035" PostTypeId="2" ParentId="8023" CreationDate="2014-11-23T18:47:25.800" Score="6" Body="&lt;p&gt;The only truly confocal elliptical orbit is that of a bound test particle in the central potential $-k/r$ or, equivalently, that of two point-like (with spherically symmetric internal mass distributions) masses attracting each other with Newtonian gravity (and having negative total energy, i.e. being bound to each other).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything else is non-elliptic (unbound orbits are parabolic or hyperbolic), but most deviations are small. Small deviations can arise from a number of sources, including quadrupole terms in the mass distribution of the bodies (in particualr the Sun), non-gravitational forces (radiation pressure and gas drag on dust grains), non-Newtonian (GR) effects, perturbations from other objects (all the other planets). Newton himself was well aware of this last effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the deviations are small, then the traditional way to estimate them is &lt;strong&gt;perturbation theory&lt;/strong&gt;, where one integrates the perturbing force along the unperturbed (elliptical) orbit. For example, in order to obtain the precession of the periapse, one could integrate the changes to the eccentricity vector. A rotation of that vector corresponds to periapse precession. See my answer to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/632/determining-effect-of-small-variable-force-on-planetary-perihelion-precession&quot;&gt;this question&lt;/a&gt;, for an example of exactly that.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-11-23T18:47:25.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8037" PostTypeId="1" CreationDate="2014-11-23T19:08:14.837" Score="-2" ViewCount="123" Body="&lt;p&gt;I suspect the scope of this question may be to broad, but: what are some methods for determining distances between two objects in Space? Please present the methods in historical order, so as to make the evolution from primitive methods to modern ones clear.&lt;/p&gt;&#xA;" OwnerUserId="2873" LastActivityDate="2014-11-23T19:22:59.703" Title="Determining Distances in Space" Tags="&lt;distances&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2014-12-02T07:10:22.340" ContentLicense="CC BY-SA 3.0" />
  <row Id="8038" PostTypeId="2" ParentId="8037" CreationDate="2014-11-23T19:22:59.703" Score="2" Body="&lt;p&gt;There are a few methods that could be used to measure distances to objects in space. I'm not sure of the ages of these methods, but in many cases, the smaller-scaled methods tend to be older (as we simply didn't know about the larger scale).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scale of the solar system was initially calculated using the parallax effect - measuring the position of Venus relative to the Sun at different times from several places across the Earth, during a transit. We can use orbital velocities nowadays to calculate average distances from the Sun. In some cases, we can use the light travel time to orbiting satellites to calculate distances very accurately, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For nearby stars, the main method is to measure the change in position of the star as the Earth moves across its orbit. This also uses the parallax effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more distant stars, scientists often need to use specific variable stars (such as Cepheids, and RR Lyrae type stars), whose variations are related to their mass. Scientists can then use this data to calculate distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cepheid variables are also used for measuring distances to nearby galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more distant galaxies, scientists can use redshift to calculate distance, as more distant galaxies are receding faster than nearer ones, causing a larger doppler effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method for measuring extremely distant objects is to look for Type 1a supernovae, which, since they always have the same luminosity (their mass is always the Chandrasekhar limit, or about 1.4 solar masses). This was used to calculate that the expansion of the universe was accelerating. However, this method isn't used very often, as these supernovae don't happen very often.&lt;/p&gt;&#xA;" OwnerUserId="2702" LastActivityDate="2014-11-23T19:22:59.703" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8039" PostTypeId="2" ParentId="2585" CreationDate="2014-11-23T22:25:31.880" Score="4" Body="&lt;p&gt;I'll address WR104 first. The National Geographic article calls it a &amp;quot;potential threat.&amp;quot; Yet that potential may be low. There are a slew of articles quoting astronomer Grant Hill on the subject. Hill studied the star and found that it looks like it isn't pointing straight at us. Its axis might be up to 45 degrees in another direction, meaning that we'd be fine if it underwent a burst.&lt;/p&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.universetoday.com/23342/wr-104-wont-kill-us-after-all/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;It would appear the original Keck imagry may not have been as straight-forward as it seemed. Spectroscopic emission lines from the binary pair strongly suggest the system is in fact inclined 30°-40° (possibly as much as 45°) away from us.&lt;/p&gt;&#xA;&lt;p&gt;So, Earth doesn’t appear to be in the firing line of WR 104 after all…&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.foxnews.com/science/2013/05/28/gamma-ray-burst-from-star-wr-104/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; and &lt;a href=&quot;http://www.news.com.au/technology/science/lethal-gamma-ray-burst-from-star-wr-104-could-reach-earth/story-fn5fsgyc-1226651950006&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; Hill is quoted as saying that the star &lt;em&gt;looks&lt;/em&gt; like it could be pointed at us - which goes against the evidence he found! Finally, &lt;a href=&quot;http://www.forbes.com/sites/brucedorminey/2013/05/27/earth-may-still-lie-in-path-of-potential-gamma-ray-burst-grb-say-astronomers/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; is another example of Hill's cautious attitude: The evidence says we're good, but he's not making any assumptions.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Sabre Tooth, I know you said that you didn't want Wikipedia, but my primary goal here isn't to collect the bounty, so I'm going to reference it in this section. I hope you're okay with that. I may withdraw it, though, if Jonathan indicates that he doesn't want it.&lt;/p&gt;&#xA;&lt;p&gt;Okay, back to the topic at hand. Wikipedia, of course, has a short tidbit on the frequency of gamma-ray bursts effecting Earth:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Estimating the exact rate at which GRBs occur is difficult, but for a galaxy of approximately the same size as the Milky Way, the expected rate (for long-duration GRBs) is about one burst every 100,000 to 1,000,000 years. Only a small percentage of these would be beamed towards Earth. Estimates of rate of occurrence of short-duration GRBs are even more uncertain because of the unknown degree of collimation, but are probably comparable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;'A small percentage' isn't too specific. &lt;a href=&quot;http://www.bbc.com/news/science-environment-21082617&quot; rel=&quot;nofollow noreferrer&quot;&gt;The BBC article&lt;/a&gt; that is cited in this passage says the following:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Observations of deep space suggest that gamma ray-bursts are rare. They are thought to happen at the most every 10,000 years per galaxy, and at the least every million years per galaxy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;However, the 'small percentage' is never elaborated on.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;That's all I have for this edit; more to come later. By the way, &lt;a href=&quot;https://physics.stackexchange.com/questions/26580/what-will-happen-if-a-gamma-ray-burst-hits-the-earth&quot;&gt;this&lt;/a&gt; Physics.SE question may interest anyone reading this. . .&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2014-11-26T20:31:57.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8040" PostTypeId="1" CreationDate="2014-11-23T23:50:46.747" Score="4" ViewCount="202" Body="&lt;p&gt;I am trying to integrate the four stellar model equations from the core to a fitting point and from the surface to the fitting point ($x_f$). When I start the outward integration, however, I have to use a huge mass shell step $dx$ where $x_1=dx$ too and the callling routine has the form $rk4(dx,x,y,dy)$(Runge-Kutta 4th order with mass step $dx,x$ core). As a result I am getting a bus error. I do not know what is the appropriate step since if we consider that I want to integrate from $x_1$ to $x_f=10^{33}$, it means that I need a huge step $dx$ to make it work. I have tried it in many different ways and I cannot figure out how to properly do that and get a result when dealing with such huge numbers. If you have any ideas I would appreciate it, since I am stuck! Thank you very much!&lt;/p&gt;&#xA;" OwnerUserId="2875" LastEditorUserId="2153" LastEditDate="2014-12-19T21:11:44.433" LastActivityDate="2014-12-19T21:11:44.433" Title="stellar evolution fortran code-runge kutta mass step from center to fitting point" Tags="&lt;stellar-evolution&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2014-12-20T23:10:37.010" ContentLicense="CC BY-SA 3.0" />
  <row Id="8041" PostTypeId="2" ParentId="7974" CreationDate="2014-11-24T00:15:39.560" Score="0" Body="&lt;p&gt;&lt;strong&gt;METHOD 3 Predicting Periastron Times and Cycle Durations from Orbital Phase ($\Phi$)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the answer by &lt;strong&gt;Stan Liou&lt;/strong&gt; he uses a Taylor Series approximation of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mean_anomaly&quot; rel=&quot;nofollow&quot;&gt;Mean Anomaly&lt;/a&gt; to derive a nice formula which determines the CPTS (Cumulative Perihelion Time Shift) value as a function of $t^2$.  This formula produces results very close to those graphed by Weisberg &amp;amp; Taylor.  As it took me a while to understand how the Mean Anomaly can be applied for a decaying-period orbit I thought it useful to note here what I learnt and present a slightly different way of obtaining a formula to predict periastron times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mean Anomaly basically indicates the phase of the orbit at a particular epoch, e.g. what time fraction of the orbit period has been completed.  For Mean Anomaly the fraction is scaled by $2\pi$. Let $\Phi(t)$ be the phase of the orbit at time $t$ such that at the start of the orbit $t=0$ and $\Phi(0)$ = 0 and at the end of the orbit when $t=P$ (the period of orbit) $\Phi(P)$ = 1.  I will assume that an orbit starts at one periapsis and finishes at the next periapsis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a decaying-period orbital system the value of $P(t)$ changes with $t$. Here $P$ is the period of orbit.  In a decaying-period orbit I will assume that orbital phase $\Phi(t)$ is completely specified by time $t$ and either $P(t)$ or $F(t)$, i.e. aspidial precession, progressive changes in semi-major axis length or eccentricity do not lead to significant additional changes in phase.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a decaying-period system the concept of period at a given epoch is somewhat abstract, it might be construed as a hypothetical period that would occur thereafter if the force causing period change ceased at that particular moment. Instead of period we can think in terms of $F(t)=1/P(t)$ where $F$ is the orbital frequency.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like to think of $F(t)$ as &quot;the rate of change of phase with time&quot;, i.e. $F(t) = \dot{\Phi}(t)$.  Picture an imaginary phase clock comprising a circular dial whose circumference has markings running from $\Phi=0$ around in a full circle to $\Phi=1$. The phase $\Phi(t)$ of our subject system at any given epoch $t$ will be indicated by a marker at a particular point on the circumference of the dial. Then $F(t)$ can be thought of as the speed at which the marker is moving around the circumference of the dial at a given epoch $t$.  In a particular short interval of time $\delta t$ the (system state) marker will move a certain distance around the dial and this distance travelled will indicate the change in phase. The &quot;travel&quot; (change of phase) will depend on the value of $F$ during that time as per $\delta \Phi \approx \delta t * F(t)$. This is only approximate because F(t) changes during the interval $\delta t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let us assume that the time interval $T$ between succesive periaspes is known and that the time of the first periapsis is at $t=0$. If $P(t)$ (and therefore $F(t)$ too) changed in a step-wise manner between time intervals, then we could write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sum_{i=1}^{i=N=T/\delta t} \frac{\delta t}{P(t)}&#xA;=&#xA;\sum_{i=1}^{i=N=T/\delta t}  \delta t F(t)&#xA;=&#xA;\sum_{i=1}^{i=N=T/\delta t}  \delta \Phi(t)&#xA;= 1.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To represent continuously-changing $P$ and $F$ we reduce $\Delta t$ towards zero and obtain these integral functions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\int_{t=0}^{t=T} \frac{1}{P(t)}\, \mathrm{d}t&#xA;=&#xA;\int_{t=0}^{t=T} F(t)\, \mathrm{d}t&#xA;=&#xA;\int_{t=0}^{t=T} \dot{\Phi}(t)\, \mathrm{d}t&#xA;= 1.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any time $\tau$ during the orbit ($0\leq \tau \leq 1$) the current instantaneous phase is given by&#xA;$$&#xA;\Phi(\tau) = &#xA;\int_{t=0}^{t=\tau} \dot{\Phi}(t)\, \mathrm{d}t&#xA;=&#xA;\int_{t=0}^{t=\tau} F(t)\, \mathrm{d}t&#xA;=&#xA;\int_{t=0}^{t=\tau} \frac{1}{P(t)}\, \mathrm{d}t.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and given a constant value of $\dot P$ (the rate of change of period), and $P_o=P(0)$ we obtain&#xA;$$&#xA;\Phi(\tau) &#xA;=&#xA;\int_{t=0}^{t=\tau} \frac{1}{P_o + \dot P t}\, \mathrm{d}t&#xA;\qquad \mathrm{and} \qquad &#xA;\dot{\Phi}(\tau) = F(\tau) = \frac{1}{P(\tau)} = \frac{1}{P_o + \dot P \tau}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{\mathrm{d}}{\mathrm{d}t} \left( &#xA;\frac{1}{f(x)} &#xA;\right)&#xA;=&#xA;\frac{-\dot f(x)}{f(x)^2}&#xA;\qquad \mathrm{and} \qquad&#xA;f(x) = P_o + \tau \dot P&#xA;\qquad \mathrm{and} \qquad&#xA;\dot f(x) =  \dot P&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we get&#xA;$$&#xA;\ddot \Phi(\tau)&#xA;= &#xA;\frac{\mathrm{d}}{\mathrm{d}t} \left( &#xA;\frac{1}{P_o + \tau \dot P}&#xA;\right)&#xA;=&#xA;\frac{-\dot P}{P_o^2 + (\tau \dot P)^2 + 2 P_o \tau \dot P}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and for $\tau = 0$ we get&#xA;$$&#xA;\Phi(0) = 0&#xA;\qquad \mathrm{and} \qquad&#xA;\dot{\Phi}(0)  = \frac{1}{P_o }&#xA;\qquad \mathrm{and} \qquad&#xA;\ddot \Phi(0)&#xA;= \frac{-\dot P}{P_o^2}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PREDICTING PERIASTRON TIMES&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can obtain an expression for CPTS by copying Stan Liou's approach , but using phase ($\Phi$) instead of Mean Anomaly ($M$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write the phase at time $\tau$ as a Maclaurin series (a Taylor series with $a=0$):&#xA;$$\begin{eqnarray*}&#xA;\Phi(\tau) \equiv \int_{t=0}^{t=\tau} \dot{\Phi}(t)\,\mathrm{d}t &#xA;&amp;amp;=&amp;amp; \Phi(0) + \dot{\Phi}(0) \tau + \frac{1}{2}\ddot{\Phi}(0) \tau^2 + \mathcal{O}(\tau^3)\\&#xA;&amp;amp;=&amp;amp;  0 +\frac{1}{P_0}\tau - \frac{\dot{P}}{2P_o^2}\tau^2+\mathcal{O}(\tau^3)\text{.}&#xA;\end{eqnarray*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dropping the third order and residual error terms (Stan Liou's answer explains justification for this) we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Phi(\tau) \approx \frac{1}{P_0}\tau- \frac{\dot{P}}{2P_o^2}\tau^2.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $N$ count the number of completed orbits at some time $\tau_N$. Then the phase at the end of orbit number $N$ is given by $\Phi(\tau_N) = N$. In a steady system (A) the orbital period $P_o$ stays constant $\dot P=0$ and so the time $\tau_{NA}$ at which the $N$th orbit completed is given exactly by $\tau_{NA} = NP_o$ and thus $N=\tau_{NA}/P_o$.  In a decaying-period system (B) the Nth orbit completes at time $\tau_{NB}$ when $\Phi(\tau_{NB}) =  N$ but with $\Phi(\tau_{NB})$ depending on a non-zero value of $\dot P$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;&amp;amp;\Phi(\tau_{NA}) = N = \frac{1}{P_0}\tau_{NA}\\&#xA;&amp;amp;\Phi(\tau_{NB}) = N \approx \frac{1}{P_0}\tau_{NB}- \frac{\dot{P}}{2P_o^2}\tau_{NB}^2.&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can obtain a formula for $\Delta t_{N}$ which is the difference in time between $\tau_{NB}$ (the epoch of the decaying system $N$th periastron) and $\tau_{NA}$ (the epoch of the steady system $N$th periastron) as follows.  Note that following Weisberg &amp;amp; Taylor we define $\Delta t_{N}\,(=$ CPTS $)=\tau_{NB} - \tau_{NA}$. The $N$th periastron in the decaying period system occurs earlier than the $N$th periastron in the steady system. Therefore   $\Delta t_{N}$ (=$ CPTS $)  will become increasingly negative as time passes (as $t$ increases).&#xA;$$&#xA;\frac{\tau_{NA}}{P_o} = N \approx  \frac{\tau_{NB}}{P_0}- \frac{\dot P}{2P_o^2} \,{\tau_{NB}}^2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;{\tau_{NA}}  \approx  {\tau_{NB}} - \frac{\dot P}{2P_o} \,{\tau_{NB}}^2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta t_N &#xA;= \tau_{NB} - \tau_{NA} \approx + \frac{\dot P}{2P_o} \,{\tau_{NB}}^2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is Stan Liou's approximation for $\Delta t_N$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what value is this equation?&#xA;In general assume that we have determined the time $t_0$ of the $0$th periastron and measured accurately an initial orbital period $P_o$ and we keep track of the observed periastra. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1 - we observe that the $N$th periastron occurs at a particular time ($\tau_{NB}$). Using $\tau_{NA}=N P_o$, we can easily calculate $\Delta t_N = \tau_{NB} - \tau_{NA}$.  Then, using Stan Liou's approximation of $\Delta t_N$ we can obtain an empirical estimate of $\dot P$ from&#xA;$$&#xA;{\dot P} = \frac{2 P_o \,\Delta t_N} {\tau_{NB}}^2.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2 - we have a theoretical formula which, given our measurement of $P_o$ predicts the value of  $\dot P$.  Given various values of $N$ and using $\tau_{NA}=N P_o$ we can easily calculate the hypothetical epochs ($\tau_{NA}$ for various $N$) of the hypothetical steady-system periastra.  Now we wish to plot a curve showing the theoretical values of $\Delta t_N$ at various hypothetical periastron epochs ($\tau_{NA}$).  But Stan Liou's equation for $\Delta t_N$ uses $\tau_{NB}$ not $\tau_{NA}$.  We need to find $\Delta t_N$ as a function of $\tau_{NA}$.  Therefore we would need to solve the following quadratic equation for $\Delta t_N$ in terms of $\tau_{NA}$ and $K$, where $K=\frac{\dot P}{2 P_o}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta t_{N} = K {\tau_{NB}}^2 = K ({\tau_{NA}}^2 + 2 \tau_{NA}\Delta t_{N} +{\Delta t_{N}}^2)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA; 0 =  (K{\tau_{NA}}^2 + 2K \tau_{NA}\Delta t_{N} +K{\Delta t_{N}}^2) -\Delta t_{N}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA; 0 =  (K){\Delta t_{N}}^2    +(2K \tau_{NA} -1)\Delta t_{N}      +(K{\tau_{NA}}^2) &#xA;.$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The following way of getting an equation for $\Delta t$ is more protracted but it will provide a formula for $\Delta t$ as a function of $N$ (which can easily be converted into a function of $\tau_{NA}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with &#xA;$$&#xA;\frac{\tau_{NA}}{P_o} =\frac{1}{P_0}\tau_{NB}- \frac{\dot P}{2P_o^2} \,{\tau_{NB}}^2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\tau_{NA} =\tau_{NB}- \frac{\dot P}{2P_o}{\tau_{NB}}^2&#xA;\qquad\rightarrow \qquad&#xA; 0 = -\tau_{NA} + \tau_{NB} - \frac{\dot P}{2P_o} \,{\tau_{NB}}^2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ with $a=- \frac{\dot P}{2P_o} , b=1, c=-\tau_{NA}$ we obtain&#xA;$$&#xA;\tau_{NB} = \frac{-(1) \pm \sqrt{(1)^2-4(- \frac{\dot P}{2P_o}).(-\tau_{NA})}}{2(- \frac{\dot P}{2P_o})}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\tau_{NB} = \frac&#xA;{1 \pm \sqrt {1 - 2 \frac{ \dot P  }{P_o}.\tau_{NA}     }}&#xA;{\dot P  /{P_o}}&#xA;\qquad \longrightarrow \qquad&#xA;\tau_{NB} = &#xA; \frac{P_o}{ \dot P } \left( 1 \pm \sqrt {1 - 2 \frac{\dot P  }{P_o}.\tau_{NA}     }\,\, \right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\tau_{NA} = P_o \,N$ so we can write&#xA;$$&#xA;\tau_{NB} = &#xA;\frac{P_o}{ \dot P } \left(1 \pm \sqrt {1 - 2  \dot P  N } \right)&#xA;\qquad \longrightarrow \qquad&#xA;-\Delta t = \tau_{NA}-\tau_{NB} &#xA;= P_o \left( N -\frac{1}{ \dot P}  \left(1 \pm \sqrt {1 - 2  \dot P  N } \right) \right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= = = = = = =&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can try a simple approximation of this using $(1+\alpha)^{0.5} = (1+ 0.5\alpha)$ so that $\sqrt {1 - 2  \dot P   N } \approx 1- \dot P  N$ and by inspection the &quot;$\pm$&quot; becomes a &quot;$+$&quot; &#xA;$$&#xA;-\Delta t &#xA;= P_o \left( N -\frac{1}{  \dot P  } \left(1 \pm 1 - \dot P   N  \right) \right)&#xA;= P_o ( N - N ) &#xA;= 0&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a valid approximation but it goes a bit too far for our purposes! We know that $\tau_{NA}-\tau_{NB} \neq 0$. So let us try expanding the term $[1 -2 \dot P  N]^{0.5}$ in a series expansion,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;[1 + (-2 \dot P  N)]^{0.5} &#xA;= 1 +\frac{1}{2}(-2\dot P  N)^{1} -\frac{1}{8}(-2 \dot P  N)^{2} &#xA;+\frac{1}{16}(-2\dot P  N)^{3}+...&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;[1+ (-2\dot P  N)]^{0.5} &#xA;= 1 - \dot P N                    -\frac{1}{2}(\dot P  N)^{2} &#xA;-\frac{1}{2}( \dot P  N)^{3}+...&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us limit ourselves to terms with $ \dot P  $ to the power of $2$ or less thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;[1-2 \dot P  N]^{0.5} \approx &#xA;1 -  \dot P  N -\frac{1}{2}\dot P ^2 N^2 .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can insert this into the time difference equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;-\Delta t&#xA;\approx P_o \left( N -\frac{1}{ \dot P  } &#xA;\left(1 \pm (1 - \dot P N -\frac{1}{2} \dot P ^2N^2 \right)  \right) &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by inspection the &quot;$\pm$&quot; becomes a &quot;$-$&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\Delta t\approx P_o \left( N -\frac{1}{ \dot P  } &#xA;\left(  + \dot P  N +\frac{1}{2}\dot P  ^2 N^2 \right) \right) &#xA;\qquad \longrightarrow \qquad&#xA;-\Delta t\approx P_o \left( N - \left( N +\frac{1}{2} \dot P N^2) \right) \right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;-\Delta t \approx -\frac{1}{2}P_o \dot P  \, N^2&#xA;\qquad \longrightarrow \qquad&#xA;\Delta t \approx \frac{1}{2}P_o \dot P  \, N^2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From the Weisberg &amp;amp; Taylor paper we are given  $P_o=27906.979587552 s$ and $\dot P  =  -2.40242 *10^{-12} s/s$ &#xA;so $\frac{1}{2}P_o  \dot P  = -0.5 * 27906.979587552 * 2.40242 *10^{-12}  = -3.352216747 * 10^{-08} s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $N= 10,000$ cycles we get $\Delta t = -3.352217 s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we can make the substitution ${\tau_{NA}}^2 = N^2 Po^2$ into the equation for $\Delta t$ to give&#xA;$$&#xA;\Delta t \approx \frac{1}{2}P_o \dot P  \, N^2  = \frac{1}{2} \frac {\dot P}{P_o}  \,{\tau_{NA}}^2&#xA;.$$&#xA;The difference between this $\Delta t$ and the value of $\Delta t$ from Stan Liou's equation is&#xA;$$&#xA;\frac{1}{2} \frac {\dot P}{P_o}  \left( {\tau_{NA}}^2 -{\tau_{NB}}^2   \right)&#xA;$$&#xA;$$&#xA;= &#xA;\frac{1}{2} \frac {\dot P}{P_o}  \left( &#xA;{{\tau_{NA}}^2 &#xA;-\tau_{NA}^2 +2 \tau_{NA} \Delta t + \Delta t^2}   \right)&#xA;$$&#xA;$$&#xA;= &#xA;\frac{1}{2} \frac {\dot P}{P_o}  \left( &#xA;{2 \tau_{NA} \Delta t + \Delta t^2}   \right)&#xA;$$&#xA;and the difference as a fraction of $\Delta t$&#xA;$$&#xA;= &#xA;\frac{1}{2} \frac {\dot P}{P_o}  \left( &#xA;{2 \tau_{NA}  + \Delta t}   \right)&#xA;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the present case $(\dot P/P_o \approx 8.6 * 10^{-17})$ and so, after 34,000 periastron cycles (948,838,000 s, about 30 years) the difference between the two approximations of $\Delta t$ would be only $0.8 * 10^{-8} s$ which is much smaller than the time resolution of the observations.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PREDICTING CYCLE DURATIONS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also derive an expression for the cycle duration $D$, as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The start periaston is at $t=0=T0$ and the first following periaston is at $t=T1$ then&#xA;$$&#xA;\Phi(T1)=\int_{t=0}^{t=T1}\frac{1}{P(t)}\,\mathrm{d}t = 1&#xA;$$&#xA;this can be expressed as &#xA;$$&#xA;1 =\int_{t=0}^{t=T1}\frac{1}{ \dot{P} t + P_0 }\,\mathrm{d}t&#xA;= \left[ \frac{1}{\dot{P}} ln  ( C\dot{P}t + CP_0 )\right]_0^{T1}&#xA;$$&#xA;hence&#xA;$$&#xA;\dot{P} = &#xA;ln ( C\dot{P}T1 + CP_0 ) - ln  ( CP_0 )&#xA;=&#xA;ln \left(   \frac{ C ( \dot{P}T1 + P_0 )} { C ( P_0 )}   \right)&#xA;=&#xA;ln \left(   \frac{ \dot{P}T1 + P_0 } {P_0 }       \right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\rightarrow exp(\dot{P}) &#xA;= \frac{ \dot{P}T1 + P_0 } {P_0 }&#xA;\rightarrow&#xA;P_0 exp(\dot{P}) -P_0&#xA;= \dot{P}T1 &#xA;$$&#xA;giving us&#xA;$$&#xA;\rightarrow&#xA;T1 =&#xA;P_0 \frac{exp(\dot{P}) -1} {\dot{P}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the duration of the first orbit (from start periastron to first following periastron) is $D_1 = T1-T0 = T1$ thus&#xA;$$&#xA;D_1 =&#xA;D_0 \frac{exp(\dot{P}) -1} {\dot{P}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we can generalize this to&#xA;$$&#xA;D_{N} =&#xA;D_{N-1} \left( \frac{exp(\dot{P}) -1} {\dot{P}} \right)&#xA;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example using the made-up value $\dot P = -2.34\,10^{-8}$ we obtain&#xA;$$&#xA;D_{N+1} =&#xA;D_{N} \frac{exp(-2.34*10^{-5}) -1} {-2.34*10^{-8}}&#xA;= 0.999 999 988 D_{N}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, using standard arithmetical software (such as Excel) when we try to calculate using $\dot P = -2.34\,10^{-10}$ we get nonsense results because of truncation errors. The approximate value of $\dot P$ reported for the Hulse-Taylor system is about $-2.34\,10^{-12}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can analyze the formula for $D_{N+1}$ using series expansions.  A Taylor Series expansion of $exp(x)/x -1/x$  is given by WolframAlpha as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;exp(x)/x -1/x= 1 + &#xA;  \frac{x}{2}&#xA;+ \frac{x^2}{6}&#xA;+ \frac{x^3}{24}&#xA;+ \frac{x^4}{120}&#xA;+ \frac{x^5}{720}&#xA;+ \frac{x^6}{5040}&#xA;+ \frac{x^7}{40320}&#xA;+ \frac{x^8}{362880}&#xA;+ \frac{x^9}{3628800}&#xA;+ \frac{x^{10}}{39916800}&#xA;+O(x^{11})&#xA;$$&#xA;or &#xA;$$1 + \frac{x^{2-1}}{2} +  \frac{x^{3-1}}{3*2} + ... \frac{x^{i-1}}{i!} + ...$$&#xA;No easilly computable expression or approximation is obvious at present.&#xA;Proceeding anyway, it is clear that the duration of any subsequent orbit $N$ can be computed from &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;D_N &#xA;= &#xA;D_0 \left(\frac{exp^{\dot P} -1} {\dot P} \right) ^N &#xA;=&#xA;D_0 \left(\frac{1}{\dot P} \right)^N \left(exp^{\dot P} -1 \right) ^N &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is interesting to compare this expression for cycle duration with that which was used as the basis of the coarse binomial solution (see other answer: Method 4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D_1=D_0(1+\dot P)^1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio of durations Phase-based to Coarse binomial based is&#xA;$$&#xA;D_0 \left( \frac{exp(\dot P)-1}{\dot P} \right)^{1}&#xA;:D_0(1+\dot P)^1&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;becoming&#xA;$$&#xA;\frac{exp(\dot P)-1}{\dot P} &#xA;:1+\dot P&#xA;\qquad \rightarrow&#xA;exp(\dot P)-1&#xA;:\dot P +\dot P^2&#xA;\qquad \rightarrow&#xA;exp(\dot P)&#xA;: 1+\dot P +\dot P^2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;applying the series expansion of $exp(\dot P)$ we obtain&#xA;$$&#xA;\qquad \rightarrow&#xA;1+\dot P + \frac{\dot P^2}{2} + \frac{\dot P^3}{6} + \frac{\dot P^4}{24} +\,...\,&#xA;: 1+\dot P +\dot P^2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly the two expressions give different values for cycle duration $D$ and the difference appears at the term in $\dot P^2$ and seems to be no bigger than $\frac{\dot P^2}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="233" LastEditorUserId="233" LastEditDate="2014-11-27T13:46:16.533" LastActivityDate="2014-11-27T13:46:16.533" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8042" PostTypeId="1" AcceptedAnswerId="8044" CreationDate="2014-11-24T04:18:42.303" Score="6" ViewCount="63517" Body="&lt;p&gt;How to define moonless night? There is no moon at all during some night? &lt;br/&gt; &#xA;How to calculate and know whether the moon would appear during one night?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the lunar phase is full moon, is it possible that the moon will not appear ?&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2018-01-04T01:51:13.253" Title="moonless night and lunar phase" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8043" PostTypeId="2" ParentId="7994" CreationDate="2014-11-24T06:19:14.537" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Yet we also have a third axiom:&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;There are some parts of the universe we can never observe because they are receding away from us at a superluminary speed.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is simply false, so of course it gets you in trouble if you insist on taking it as axiomatic. The average recession velocity goes luminal at redshift $z\approx 1.4$, while there are observed galaxies and objects at $z\sim 8$, e.g., &lt;a href=&quot;https://en.wikipedia.org/wiki/Z8_GND_5296&quot; rel=&quot;nofollow&quot;&gt;Z8 GND 5296&lt;/a&gt; dwarf galaxy and the &lt;a href=&quot;https://en.wikipedia.org/wiki/GRB_090423&quot; rel=&quot;nofollow&quot;&gt;GRB 090423&lt;/a&gt; gamma ray burst. Additionally, there are at least &lt;a href=&quot;https://en.wikipedia.org/wiki/UDFj-39546284&quot; rel=&quot;nofollow&quot;&gt;some candidates&lt;/a&gt; for objects much farther than even that, possibly $z\sim 12$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that other than the CMB itself, the most distant and most ancient object we &lt;em&gt;might&lt;/em&gt; have observed hails from when the universe as around $370\,\mathrm{Myr}$ old, which is around a thousand times older than the &lt;a href=&quot;https://en.wikipedia.org/wiki/Recombination_%28cosmology%29&quot; rel=&quot;nofollow&quot;&gt;recombination&lt;/a&gt; epoch when the universe first became transparent and hence when the cosmic background radiation was emitted. In terms of redshift, the CMB has $z\gtrsim 1100$ or so.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But this surely also implies that we can see beyond the CMB if we see anything which has a red shift indicating an expansion speed very close to $c$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Something at redshift $z\gtrsim 1100$ has a recession velocity of $v\gtrsim 3.2c$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This suggests axiom 2 is incorrect - so what should it be instead?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since (3) says incorrect things, it's best to throw it out instead.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-11-24T06:19:14.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8044" PostTypeId="2" ParentId="8042" CreationDate="2014-11-24T06:27:53.193" Score="9" Body="&lt;p&gt;A moonless night is, as you suspect, &lt;strong&gt;a night in which the Moon does not appear visible in the sky&lt;/strong&gt;. This happens once per month, when the Moon is near the Sun. Due to the proximity of the Moon and the Sun in the sky, at that time the Moon is the smallest sliver possible, and therefore &lt;strong&gt;not&lt;/strong&gt; a full moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because it is actually the Sun that illuminates the Moon, and when the Sun and the Moon are in the same direction in the sky we are seeing the non-illuminated side of the Moon. &lt;strong&gt;Note the direction of the sunlight in this image:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3S4RZ.jpg&quot; alt=&quot;Moon phases&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, the direction of the sunlight is the direction of &quot;up&quot; during the day. If you look at the horizon slightly after sunset or slightly before sunrise, you might actually catch a glimpse of the sliver of Moon before it set or rises slightly after or before the Sun.&lt;/p&gt;&#xA;" OwnerUserId="209" LastEditorUserId="209" LastEditDate="2014-11-24T08:13:19.820" LastActivityDate="2014-11-24T08:13:19.820" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="8045" PostTypeId="1" CreationDate="2014-11-24T11:33:18.170" Score="1" ViewCount="153" Body="&lt;p&gt;In a discussion elsewhere about binary planets, it came the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How near can a planet like Earth (maybe Earth itself) be to another planet like Earth before both of them tearing apart due to tidal forces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words: Which is the Roche limit for Earth, if it were orbiting (another) Earth?&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-11-24T17:38:17.610" Title="Roche Limit for Earth-like body around Earth-like body" Tags="&lt;roche-limit&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8046" PostTypeId="2" ParentId="8045" CreationDate="2014-11-24T17:38:17.610" Score="2" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Roche_limit&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;, the roche limit for an object with the same density as the Moon would be around 10,000 kilometres. Since the density of the Earth is about 166% of that of the Moon, if you don't take the atmosphere, or oceans into account, then the binary planets could be much closer to each other. However, when you take into account the oceans, and atmosphere, it would probably be a lot further away.&lt;/p&gt;&#xA;" OwnerUserId="2702" LastActivityDate="2014-11-24T17:38:17.610" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8047" PostTypeId="2" ParentId="703" CreationDate="2014-11-24T19:39:45.417" Score="1" Body="&lt;p&gt;A &quot;(very) bright star is still up even though it's midnight (6 hours after sunset).&quot; in the east at this time of year is likely &lt;a href=&quot;http://en.wikipedia.org/wiki/Sirius&quot; rel=&quot;nofollow&quot;&gt;Sirius, the brightest star in the sky&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2014-11-24T19:39:45.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8048" PostTypeId="2" ParentId="703" CreationDate="2014-11-24T20:00:50.540" Score="4" Body="&lt;p&gt;As other people have pointed out, it is hard to work out which star it is, without knowing your general location. However, after checking on Stellarium, there seem to be a couple of likely suspects:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sirius - the brightest star in the sky. I've seen it myself - and on a good, dark night, it can really stand out. &lt;/li&gt;&#xA;&lt;li&gt;Jupiter - the king of the planets is also rising at about the same time. It is brighter than any star in the sky, by a wide margin (though fainter than Venus), and it can &lt;i&gt;really&lt;/i&gt; stand out.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Other than that, there aren't really that many objects rising in the East at the time you specify that could really stand out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a couple of useful ways to tell the two apart:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sirius is a bright white object - perhaps with a subtle bluish tinge to it, whereas Jupiter has a slight yellow tint to it.&lt;/li&gt;&#xA;&lt;li&gt;Jupiter is currently rising in the North-East, and can get very high in the sky at the moment from the northern hemisphere, whereas Sirius rises in the South-East, and doesn't get that high (though that does depend on location).&lt;/li&gt;&#xA;&lt;li&gt;Sirius tends to twinkle, and 'flicker', as its light is disturbed by air currents, whereas Jupiter remains very steady - perhaps not twinkling at all.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As mentioned earlier, the best method is usually to use software like &lt;a href=&quot;http://www.stellarium.org/en_GB/&quot; rel=&quot;nofollow&quot;&gt;Stellarium&lt;/a&gt;, which will tell you &lt;i&gt;exactly&lt;/i&gt; where everything is, and hopefully give you a definitive answer to which object it is.&lt;/p&gt;&#xA;" OwnerUserId="2702" LastActivityDate="2014-11-24T20:00:50.540" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8049" PostTypeId="1" AcceptedAnswerId="8051" CreationDate="2014-11-24T20:11:41.830" Score="2" ViewCount="885" Body="&lt;p&gt;As it is obvious from the question title &#xA;is there any restriction or cases that we can see the binary stars as not in pairs &lt;/p&gt;&#xA;&#xA;&lt;p&gt;also is there any binary stars doesn't linked together by gravity ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the definition of binary stars that it is  a pair of stars that revolve around their common center of mass but my question is about special cases if it is exist &lt;/p&gt;&#xA;" OwnerUserId="1970" LastActivityDate="2014-11-24T22:36:35.620" Title="can we see all binary stars as pairs?" Tags="&lt;binary-star&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8050" PostTypeId="2" ParentId="8049" CreationDate="2014-11-24T20:36:48.873" Score="1" Body="&lt;p&gt;As you've said, the definition of binary star systems would mean that the cases you are describing would not be considered binary star systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, there are other arrangements which exceed binary, but are often thought of as including binary relationships.  This is due to star systems which reach beyond binary often resulting in &lt;strong&gt;hierarchical&lt;/strong&gt; arrangements. For instance, many triple star systems involve what is largely a close binary star system with a third star that orbits the pair at a much greater distance. For an extreme example of this, consider &lt;a href=&quot;http://en.wikipedia.org/wiki/Castor_(star)&quot; rel=&quot;nofollow&quot;&gt;Castor&lt;/a&gt;, which is a sextuple star system containing three binary star pairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These systems wouldn't technically be said to be binary star systems, but rather to contain binary pairs, but given the question I can't think of a better way of having a binary star system that isn't a binary star system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the hierarchical nature of super-binary systems is a matter of system evolution. Systems which would not be considered hierarchical are believed to be unstable, which would eventually result in the ejection of a star from the system.&lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="1288" LastEditDate="2014-11-24T20:42:27.330" LastActivityDate="2014-11-24T20:42:27.330" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8051" PostTypeId="2" ParentId="8049" CreationDate="2014-11-24T22:06:35.923" Score="2" Body="&lt;p&gt;To answer the main part of your question: Yes, there do exist such systems. They're called &lt;a href=&quot;https://en.wikipedia.org/wiki/Visual_binary&quot; rel=&quot;nofollow&quot;&gt;visual binaries&lt;/a&gt;. We generally need a telescope to tell them apart. Most binary systems look like a single star when viewed with only the naked eye; many cannot be resolved without the aid of a telescope. But visual binaries can.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;also is there any binary stars doesn't linked together by gravity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, there aren't. You were right when you said that the definition of a binary system is basically two stars orbiting a common center of mass. If the stars don't orbit each other, they aren't a binary system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, check out &lt;a href=&quot;https://en.wikipedia.org/wiki/Proxima_Centauri&quot; rel=&quot;nofollow&quot;&gt;Proxima Centauri&lt;/a&gt;, the nearest star to the Sun. It's one of the three stars in the triple-star Alpha Centauri system. The reason I mention it is that even though it's seemed for years like it was gravitationally bound to the system, &lt;a href=&quot;https://en.wikipedia.org/wiki/Proxima_Centauri#Distance_and_motion&quot; rel=&quot;nofollow&quot;&gt;that idea&lt;/a&gt; is now under debate.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let me address something that Mitch pointed out. The classification of star systems as &quot;visual binaries&quot; is based solely on our ability to observe them. If we had better telescope, this class could change. An analogy might be our naming certain wavelengths of light &quot;visible&quot;. They only depend on our perceptions, not some objective characteristic that all observers in the universe could agree on.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-11-24T22:36:35.620" LastActivityDate="2014-11-24T22:36:35.620" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8052" PostTypeId="2" ParentId="8033" CreationDate="2014-11-25T14:13:48.870" Score="4" Body="&lt;p&gt;I happen to have represented my nation at IOAA&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with &quot;Roy and Clarke's&quot; book : &quot;Astronomy: Principles and Practice&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Move on to the references mentioned in the book.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2014-11-25T14:13:48.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8053" PostTypeId="2" ParentId="8004" CreationDate="2014-11-25T14:44:29.797" Score="1" Body="&lt;p&gt;Some useful sources:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instructions:&#xA;&lt;a href=&quot;http://naif.jpl.nasa.gov/pub/naif/self_training/individual_docs/25_shape_model_preview.pdf&quot; rel=&quot;nofollow&quot;&gt;http://naif.jpl.nasa.gov/pub/naif/self_training/individual_docs/25_shape_model_preview.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3d model: &lt;a href=&quot;http://naif.jpl.nasa.gov/pub/naif/ROSETTA/kernels/dsk/former_versions/&quot; rel=&quot;nofollow&quot;&gt;http://naif.jpl.nasa.gov/pub/naif/ROSETTA/kernels/dsk/former_versions/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;Ancient&quot;, preliminary data from ros_67pcg_v00.tpc:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;BODY1000012_POLE_RA = ( 270.0 0.0 0.0 )  &lt;/li&gt;&#xA;&lt;li&gt;BODY1000012_POLE_DEC = (  66.560709 0.0 0.0 )  &lt;/li&gt;&#xA;&lt;li&gt;BODY1000012_PM = ( 0.0 720.0 0.0 )&lt;/li&gt;&#xA;&lt;li&gt;BODY1000012_RADII = ( 2.5 1.5 1.5 )  (ellipsoid, kilometers)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2012 data from ROS_CGS_RSOC_V03.TPC:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;BODY1000012_POLE_RA       = (  220.0            0.0            0.0 )&lt;/li&gt;&#xA;&lt;li&gt;BODY1000012_POLE_DEC      = (  -70.0            0.0            0.0 )&lt;/li&gt;&#xA;&lt;li&gt;BODY1000012_PM            = (  -86.369364827  679.245283019    0.0 )&lt;/li&gt;&#xA;&lt;li&gt;BODY1000012_RADII         = (    2.398          1.887          1.532   )&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&quot;679.245283019&quot; should define rotation period in terms of degrees: 24*360/679.245283019  gives 12.72 (hours?)&#xA;Don't know about the &quot;-86.369364827&quot;, anyway &quot;PM&quot; should stay for &quot;Prime Meridian&quot;.&#xA;Anyway the xxx_PM row corresponds to this definition (from pck00010.tpc ):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;W  =  -86.369364827  + 679.245283019 d&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;No nutation/precession data available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For xxx_RADII:&#xA;&quot;The first number is the largest equatorial radius (the length of the semi-axis containing the prime meridian), the second number is the smaller equatorial radius, and the third is the polar radius.&quot;&lt;/p&gt;&#xA;" OwnerUserId="2889" LastEditorUserId="2889" LastEditDate="2014-11-25T16:51:00.047" LastActivityDate="2014-11-25T16:51:00.047" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8054" PostTypeId="1" AcceptedAnswerId="8067" CreationDate="2014-11-25T16:18:51.813" Score="2" ViewCount="1124" Body="&lt;p&gt;What is the effect of absence of atmospheric refraction on sin rise and sunset time?&#xA;Is it will be delayed or earlier for sunrise and sunset, and why?&#xA;Thank you:)&lt;/p&gt;&#xA;" OwnerUserId="2787" LastEditorUserId="7982" LastEditDate="2021-02-28T11:54:42.507" LastActivityDate="2021-02-28T11:54:42.507" Title="Effect of the absence of atmospheric refraction to sun rise and sunset times" Tags="&lt;the-sun&gt;&lt;atmospheric-effects&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8055" PostTypeId="1" AcceptedAnswerId="8059" CreationDate="2014-11-25T17:19:12.957" Score="3" ViewCount="600" Body="&lt;p&gt;Is a Low Mass Black Hole also called a Stellar Mass Black Hole?&#xA;Can Low Mass BH be also called Primordial BH?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If low mass BH are Stellar Mass BH-What is their size in terms of the mass of the sun?)&lt;/p&gt;&#xA;" OwnerUserId="2891" LastEditorUserId="2153" LastEditDate="2014-11-26T00:59:17.197" LastActivityDate="2014-11-26T01:41:27.563" Title="What is a low mass Black Hole?" Tags="&lt;black-hole&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8056" PostTypeId="2" ParentId="8055" CreationDate="2014-11-25T17:32:40.727" Score="3" Body="&lt;p&gt;Stellar Mass black holes are (generally) black holes which form from collapsed stars. They get their name because the mass of the black hole is in the order of the mass of stars. I don't believe they would fit your definition of &quot;Low Mass,&quot; as they are believed to be fairly standard in size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Primordial black holes are theoretical black holes which were believed to have been created during the early moments of the universe. My understanding is that these would be a subset of &lt;em&gt;quantum black holes&lt;/em&gt;, which also seem to be called &lt;em&gt;micro black holes&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See this &lt;a href=&quot;https://astronomy.stackexchange.com/questions/7788/black-holes-how-do-they-form/7813#7813&quot;&gt;related question&lt;/a&gt;, in which the answer covers types of blackholes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not quite sure what the lower-end masses of observed black holes are, but in theory we could say that the lowest mass of a stellar black hole is just beyond the upper limit of mass for neutron stars, as above that limit they would be collapse into a stellar black hole. This is known as the TOV limit, and is approximated to be about 1.5-3 solar masses&lt;sup&gt;1&lt;/sup&gt;, so we could say that the smallest stellar black holes could have a mass of about 1.5-3 times that of the Sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note however, that this doesn't mean that the Sun is nearly massive enough to become a black hole. The mass of the resultant stellar black hole is only a small portion of the original mass of the star which created it.&#xA;&lt;hr&gt;&#xA;&lt;sup&gt;1&lt;/sup&gt; I. Bombaci (1996). &quot;&lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1996A%26A...305..871B&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Maximum Mass of a Neutron Star&lt;/a&gt;&quot;. &lt;em&gt;Astronomy and Astrophysics&lt;/em&gt; 305: 871–877.&lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-11-26T01:41:27.563" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8057" PostTypeId="2" ParentId="6653" CreationDate="2014-11-25T21:16:48.557" Score="4" Body="&lt;p&gt;Something to consider might be a large very dark (low albedo) asteroid.  I would say it is quite possible a large dark asteroid (e.g. P-type) &quot;could&quot; be headed for us, and we would not know it until it hit us (e.g. distance 0).  I think you are asking for the closest distance.  It appears the main method of detecting asteriods is with telescopes, but we are starting to scan the skies in infrared too.  Hopefully infrared will reveal these darker asteriods better.  Sources for further research include:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/P-type_asteroid&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/P-type_asteroid&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.space.com/3126-asteroids-data-sheet.html&quot; rel=&quot;nofollow&quot;&gt;http://www.space.com/3126-asteroids-data-sheet.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nbcnews.com/science/space/how-we-scan-solar-system-find-dangerous-asteroids-f8C11480423&quot; rel=&quot;nofollow&quot;&gt;http://www.nbcnews.com/science/space/how-we-scan-solar-system-find-dangerous-asteroids-f8C11480423&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the furthest distance, that question is trickier.  It would largely depend on what is considered an asteroid, for example, if we consider Pluto to be an asteroid, we can at least detect very large asteroids out to the distance of Pluto, and we could track its path and see if it were on a collision course with earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On average, it is said that we have already detected most of the globally threatening asteroids and it does not appear we are in immediate danger.  Still, I'd say we need to stay on the lookout, and especially look out in infrared to detect any dark asteroids we may have missed :-).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summary: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Closest distance before large threatening asteriod detection: 0 (e.g. upon impact)&lt;/li&gt;&#xA;&lt;li&gt;Furthest distance before large threatening asteroid detection: Beyond the orbit of Pluto (and most asteroids tracked are in the asteroid belt between Mars and Jupiter)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="1386" LastEditDate="2014-11-25T21:33:03.330" LastActivityDate="2014-11-25T21:33:03.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8058" PostTypeId="2" ParentId="4846" CreationDate="2014-11-25T21:32:29.747" Score="2" Body="&lt;p&gt;With normal matter, the strength of gravity depends on the stress-energy tensor, which, in an isotropic homogeneous universe, has trace of &lt;span class=&quot;math-container&quot;&gt;$\rho + 3p$&lt;/span&gt;.  Note that the pressure from relativistic particles (photons and neutrinos) enhances the mass density in slowing down expansion.   In inflation theory, it is assumed that there was a scalar field (perhaps) in a small section of the universe that began in a false vacuum state. That is, the field's energy density remained roughly constant as the universe began to inflate, and therefore the total energy of this field grew as the volume of the universe grew.  The field's pressure was negative and greatly overwhelmed the inward force of the density of normal matter and pressure. It is this negative pressure that rapidly expanded this section of the universe by many factors.&lt;/p&gt;&#xA;&lt;p&gt;Energy was conserved because work was done by the gravitational field as it pressed inward against the field's outward pressure; the gravitational potential became very negative, while the energy density in the field became very large.  Eventually, the field decayed away, heating the particles, which resulted in the traditional Big Bang conditions of a hot expanding universe.&lt;/p&gt;&#xA;&lt;p&gt;Throughout this theory, standard physics is expected to hold with just the addition of this new scalar field.  It is introduced because it helps to explain why the entire visible universe today is so homogeneous even though there are subregions that were previously thought to have not been in causal contact until quite recently.  It also explains why the universe is very nearly flat.  It converted a small section of the universe to extremely large radii and thus small curvature.&lt;/p&gt;&#xA;&lt;p&gt;You can &lt;a href=&quot;http://ned.ipac.caltech.edu/level5/Guth/Guth3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;read about this from one of its inventors here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2014-11-25T21:32:29.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8059" PostTypeId="2" ParentId="8055" CreationDate="2014-11-25T21:47:22.963" Score="3" Body="&lt;p&gt;Don't use the term 'low-mass black hole'. It is ambiguous and not used by astrophysicists. Rather, one distinguishes between 'stellar-mass', 'supermassive' and 'intermediate-mass' black holes. The first are remnants of stellar collapse in a supernova and typically several Solar masses. The second are the beast in the cores of most massive galaxies and have masses in the range $10^{6-10}$M$_\odot$. Finally, 'intermediate black holes' are speculative black holes in the range $10^{2-4}$M$_\odot$, which are postulated to exist as an evolutionary stage for supermassive black holes, but we have no firm evidence for their existence (though many an observation is interpreted as such).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, there are even more speculative primordial black holes.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-11-25T21:47:22.963" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8060" PostTypeId="1" AcceptedAnswerId="8068" CreationDate="2014-11-26T01:19:33.317" Score="3" ViewCount="56" Body="&lt;p&gt;I'm working with some equations to model the evolution of a circumstellar disk. One of the equations is&#xA;$$\rho(r)=Ce^{-\frac{(r-r_{peak})^2}{2 \sigma ^2}}$$&#xA;where $\rho$ is density, $r$ is the distance from the center, $C$ is a constant, $\sigma$ is one standard deviation, and $r_{peak}$ is the radius at which the density is at a maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the function was of the form&#xA;$$\rho(r)=Ce^{f(r)}$$&#xA;where $f(r)$ is a function of $r$, I could find the maximum easily by finding&#xA;$$\rho'(r)=Cf'(r)e^{f(r)}=0$$&#xA;and solving for $r$. However, this appears to be impossible in the current case because $\rho(r)_{peak}$ is already in the equation, at $r_{peak}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is $r_{peak}$ determined in a given scenario? Is it determined experimentally?&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-11-26T16:58:25.907" Title="How is the maximum density in a circumstellar disk determined?" Tags="&lt;disk&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8061" PostTypeId="1" CreationDate="2014-11-26T03:24:00.927" Score="6" ViewCount="501" Body="&lt;p&gt;I'm observing the full moon, from a major city, with heavy light pollution and dust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objective size is fixed, for this comparison, something between 4 and 6 inches. The question is about the tradeoff between magnification and dimness of the image. I'd like to know what range of exit pupils works best. Again, the goal is to observe the full moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it differently:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What's the smallest exit pupil at which the moon gets so dim that you can't take advantage of the increased magnification? (&quot;Empty magnification&quot;).&lt;/li&gt;&#xA;&lt;li&gt;What's the largest exit pupil at which the moon is bright enough, so a larger exit pupil doesn't help?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The only reference I could find is &lt;a href=&quot;http://www.telescope.com/Choosing-Eyepieces/p/99803.uts&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt;, which claims that 1-2mm works best for the moon. Is that correct?&lt;/p&gt;&#xA;" OwnerUserId="962" LastActivityDate="2015-01-20T19:26:01.227" Title="What range of exit pupils work for observing the full moon?" Tags="&lt;the-moon&gt;&lt;telescope&gt;&lt;binoculars&gt;&lt;light-pollution&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8062" PostTypeId="1" AcceptedAnswerId="8065" CreationDate="2014-11-26T05:29:53.960" Score="3" ViewCount="1627" Body="&lt;p&gt;We never see the dark side of the moon. It rotates so that the same side always faces us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I heard two (dinner party) theories on this:&#xA;1. The moon was made by smashing out a chunk of the earth, and so somehow preserved some orbital rotation from the earth &#xA;2. The gravitational drag from the earth to the moon is lowest when the moon is in 'sync' to only show one face to the earth &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't trust either of these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &quot;Why does the face of the moon 'sync' with the earth?&quot;&lt;/p&gt;&#xA;" OwnerUserId="2894" LastActivityDate="2014-11-27T14:19:45.037" Title="Why does the face of the moon 'sync' with the earth?" Tags="&lt;the-moon&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2014-11-27T20:01:28.783" ContentLicense="CC BY-SA 3.0" />
  <row Id="8064" PostTypeId="1" CreationDate="2014-11-26T09:33:59.453" Score="3" ViewCount="470" Body="&lt;p&gt;In Randall Munroe's What If? blog he tackles &lt;a href=&quot;https://what-if.xkcd.com/42/&quot; rel=&quot;nofollow&quot;&gt;this question&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the longest possible sunset you can experience while driving, assuming we are obeying the speed limit and driving on paved roads?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;His final answer is the following, which allows you to experience about a 95 minute sunset. But the diagram isn't the most self-evident. What does the diagram mean, and how does the strategy work?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you're in northern Norway on a day when the Sun just barely sets and then rises again, the terminator (day-night line) moves across the land in this pattern:&lt;br&gt;&#xA;  &lt;img src=&quot;https://what-if.xkcd.com/imgs/a/42/sunset_terminator.png&quot; alt=&quot;Randall&amp;#39;s diagram of the terminator line&quot;&gt;&lt;br&gt;&#xA;  To get a long sunset, the strategy is simple: Wait for the date when the terminator will just barely reach your position. Sit in your car until the terminator reaches you, drive north to stay a little ahead of it for as long as you can (depending on the local road layout), then u-turn and drive back south fast enough that you can get past it to the safety of darkness.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2896" LastActivityDate="2018-11-12T02:03:27.133" Title="How does this strategy for chasing the sunset work?" Tags="&lt;the-sun&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8065" PostTypeId="2" ParentId="8062" CreationDate="2014-11-26T11:31:14.957" Score="3" Body="&lt;p&gt;It is called &quot;Tidal locking&quot;, or &quot;gravitational locking&quot; or &quot;captured rotation&quot;. According the the Wikipedia page on &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_locking#Mechanism&quot; rel=&quot;nofollow&quot;&gt;tidal locking&lt;/a&gt;&quot; (check the references for sources), it is due to Earth's gravity causing a small tidal bulge on the Moon, which affects its rotation. Over time, the Moon's rotation, affect by Earth's gravity, makes it's orbital rotation synchronise with its lunar rotation.&lt;/p&gt;&#xA;" OwnerUserId="2880" LastActivityDate="2014-11-26T11:31:14.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8066" PostTypeId="2" ParentId="8064" CreationDate="2014-11-26T11:33:42.710" Score="1" Body="&lt;p&gt;I would go to the Arctic circle, where you can experience an almost perpetual sunset, without having to drive around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited to answer the question: My interpretation of the diagram, is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The terminator at a certain time near sunset, is the one shown towards the east (darkness approaches), running NE/SW.&lt;/li&gt;&#xA;&lt;li&gt;As time progresses, it moves into the top horizontal location, running E/W; this consistent with the more northerly latitudes having continual daytime, coinciding with a certain parallel polar circle.&lt;/li&gt;&#xA;&lt;li&gt;As morning approaches, the terminator moves towards the west (bringing daylight with it), and runs NW/SE.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I think the arrows would be better if the tip of the one on the right, joined the base of the one on the left, showing your car journey as a continuous one.&lt;/p&gt;&#xA;" OwnerUserId="2880" LastEditorUserId="2880" LastEditDate="2014-11-26T11:49:37.097" LastActivityDate="2014-11-26T11:49:37.097" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8067" PostTypeId="2" ParentId="8054" CreationDate="2014-11-26T15:58:44.547" Score="0" Body="&lt;p&gt;The simple answer is that atmospheric refraction makes the disk of the sun visible at sunrise or sunset, when it's actually &lt;a href=&quot;http://piecubed.co.uk/atmospheric-refraction/&quot; rel=&quot;nofollow noreferrer&quot;&gt;half a degree or so below the horizon&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mhGJu.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, both the extent of refraction of the sun's disk and the refraction of the apparent horizon come into play. Both depend on the atmosphere's density and temperature profile; which changes irregularly. That makes it &lt;a href=&quot;http://mintaka.sdsu.edu/GF/explain/sunset_time.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;impossible to predict exact sunrise or sunset times&lt;/a&gt; when there is an atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2014-11-26T15:58:44.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8068" PostTypeId="2" ParentId="8060" CreationDate="2014-11-26T16:58:25.907" Score="1" Body="&lt;p&gt;If your $r_{peak}$ is not known a priori then you have two unknowns and only one equation: this cannot be solved. At least not without a second equation or a measured $(\rho,r)$-pair (say the central density $\rho_0 = \rho(r=0)$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, this definition of radial density looks like an ad-hoc assumption to me, and not something that came out of a proper hydrodynamic model. So without knowing why you chose this form or the context of your disk model I can't comment on what equation is best suited to determine $r_{peak}$. A well-informed guess seems like the way to go.&lt;/p&gt;&#xA;" OwnerUserId="494" LastActivityDate="2014-11-26T16:58:25.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8069" PostTypeId="2" ParentId="7702" CreationDate="2014-11-26T17:42:40.673" Score="1" Body="&lt;p&gt;IAU failed to understand that the concept &quot;planet&quot; is a cultural concept, not a scientific concept, not any more anyway. Of course there's no distinct objective scientific definition between what is a planet and what is not. Mercury and Jupiter in the same category? Yeah, right, as if that clarifies what we are talking about when we use the word &quot;planet&quot;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pluto was discovered in 1930. Back then a planet was a dot in the sky. Since then our knowledge has advanced. IAU should've decided that there were 9 planets discovered in our historical culture. And now we discover many objects which are much the same as some planets. Another plutoid here, a jupiteoid there. But there are only 9 planets in the entire universe, because &quot;planet&quot; is a cultural concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what shall we call Ceres? It is not a planet, it is not an asteroid, it is not a plutoid. Are we back in 1801?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so called &quot;exoplanets&quot;, are they really planets? How do we know that they have cleared their orbits? IAU should consequently decide to classify them as &quot;exodwarfplanets&quot;. No exoplanets have been confirmed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How long is a string according to the confused IAU?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2014-11-26T17:57:11.647" LastActivityDate="2014-11-26T17:57:11.647" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8070" PostTypeId="1" CreationDate="2014-11-27T09:44:11.610" Score="4" ViewCount="93" Body="&lt;p&gt;While searching for different methods of speed of light measurement, i came across one of the method of fizeau discussed below which i cannot fully understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short,in Fizeau’s apparatus, a&#xA;beam of light was shone between the teeth of&#xA;a rapidly rotating toothed wheel, so the&#xA;“lantern” was constantly being covered&#xA;and uncovered. Fizeau had a&#xA;mirror, reflecting the beam back, where&#xA;it passed a second time between the&#xA;teeth of the wheel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do understand the idea i.e. to measure the time during the course of light &quot;from the wheel to mirror and then back to the mirror&quot;.But where are the mirror,lantern and wheel  located? What kind of wheel is that? Does light pass through the holes (or teeth) in the wheel or light gets refracted through the wheel? Finally if a group of light particles goes through one of the hole (or teeth),then how do we know that the same group of light particles came back through which hole (or teeth) after its journey,since there are soo many light particles leaving and entering different holes (or teeth)?&lt;/p&gt;&#xA;" OwnerUserId="2901" LastActivityDate="2014-11-27T15:03:13.013" Title="speed of light measurement" Tags="&lt;light&gt;&lt;speed&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8071" PostTypeId="1" CreationDate="2014-11-27T12:13:46.550" Score="6" ViewCount="67" Body="&lt;p&gt;What are the altitude differences in the high resolution part of &lt;a href=&quot;https://i.stack.imgur.com/1GjpQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this image&lt;/a&gt; of Europa? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1GjpQ.jpg&quot; alt=&quot;Ridges on Europa&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How high do those ridges rise over their neighboring valleys? And how large is the area shown in this image?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And are the craters believed to have been formed by impacts or some kind of &quot;eruptions&quot; from below? It looks to me as if the areas which have the most (tiny) craters is on top of and thus newer than the less cratered areas. Well, here and there anyway.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="7472" LastEditDate="2018-12-24T15:50:36.990" LastActivityDate="2018-12-24T15:50:36.990" Title="How tall are the &quot;fractured&quot; linear structures on Europa's surface?" Tags="&lt;photography&gt;&lt;europa&gt;&lt;galileo-probe&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8072" PostTypeId="2" ParentId="8071" CreationDate="2014-11-27T13:49:04.900" Score="4" Body="&lt;p&gt;Apparently they have been inferred to be &quot;220 meters&quot;. See the references in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S. D. Kadel, S. A. Fagents, &quot;TROUGH-BOUNDING RIDGE PAIRS ON EUROPA&quot;, , Lunar and Planetary Science XXIX. (&lt;a href=&quot;http://www.lpi.usra.edu/meetings/LPSC98/pdf/1078.pdf&quot; rel=&quot;nofollow&quot;&gt;PDF&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="2880" LastEditorUserId="2153" LastEditDate="2014-11-28T16:42:39.423" LastActivityDate="2014-11-28T16:42:39.423" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8073" PostTypeId="2" ParentId="8062" CreationDate="2014-11-27T14:19:45.037" Score="1" Body="&lt;p&gt;The Earth and Moon have been &quot;together&quot; billions of years and so even realitively weak forces such as gravitation have had a chance to cause profound effects. One of those is the Moon's tidal lock. The Earth too is being driven towards tidal lock - ie at a point in the future only one face of the Earth will face the Moon, but as the Moon's tidal drag on the Earth is relatively lower (compared to the energy contained in the Earth's spin) than Earth's on the Moon, this is not going to happen for very many millions of years.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2014-11-27T14:19:45.037" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8074" PostTypeId="2" ParentId="8070" CreationDate="2014-11-27T15:03:13.013" Score="2" Body="&lt;p&gt;Here is a diagram of his setup:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/rjspc.jpg&quot; alt=&quot;http://en.wikipedia.org/wiki/Fizeau–Foucault_apparatus&quot;&gt;&lt;br&gt;&#xA;See the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fizeau%E2%80%93Foucault_apparatus&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; page for a discription.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-11-27T15:03:13.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8076" PostTypeId="2" ParentId="6426" CreationDate="2014-11-27T22:53:04.970" Score="1" Body="&lt;p&gt;Astronomical events such as the solstices and equinoxes occur at a specific instance in time on a specific day, often given in &lt;a href=&quot;https://en.wikipedia.org/wiki/Universal_Time&quot; rel=&quot;nofollow&quot;&gt;Universal Time&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, this winter's solstice occurs on 21 Dec 2014 at 23:03 Universal Time. Since most of Europe is an hour ahead, this may occur on 22 Dec 2014 at 00:03 (and subsequent hours) given in local time based on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Time_zone&quot; rel=&quot;nofollow&quot;&gt;time zone&lt;/a&gt;. So British and American calendars will show the solstice occurring on 21 Dec, and in countries east of &lt;a href=&quot;https://en.wikipedia.org/wiki/Royal_Observatory,_Greenwich&quot; rel=&quot;nofollow&quot;&gt;Greenwich&lt;/a&gt; on 22 Dec.&lt;/p&gt;&#xA;" OwnerUserId="2880" LastActivityDate="2014-11-27T22:53:04.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8079" PostTypeId="2" ParentId="8023" CreationDate="2014-11-28T16:23:38.923" Score="5" Body="&lt;p&gt;David Hammen wrote&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;People are using planetary equations coupled with geometric integration techniques...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You could also try (what I call) a simple finite-step simulation using Newton's laws to operate on object masses, positions, velocities and accelerations. I'm not sure if this falls within what David calls &quot;geometric integration techniques&quot;. My point is that you can do it without incorporating the planetary equations. Disadvantage = the simulator &quot;cuts corners&quot; by using approximations and this leads to behaviours in the model which are artefacts.  These disadvantages can be overcome by using other techniques. Advantage = it makes the code design easier, it avoids the suspicion that the planetary equations (and their assumptions) are driving the show. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You do not need to be an expert in numerical methods to use the simple &lt;a href=&quot;http://en.wikipedia.org/wiki/Leapfrog_integration&quot; rel=&quot;nofollow noreferrer&quot;&gt;Leapfrog Integration technique&lt;/a&gt; (described in detail in &lt;a href=&quot;http://www.feynmanlectures.caltech.edu/I_09.html#Ch9-S7&quot; rel=&quot;nofollow noreferrer&quot;&gt;Feynman Lectures vol I&lt;/a&gt;) to model Newtonian Precession in Solar system orbits over periods of up to a few centuries.  By running simulations at various time steps (e.g. &lt;span class=&quot;math-container&quot;&gt;$dt=1200s, 600s, 300s, 100s$&lt;/span&gt;) plotting the results in Excel, fitting a curve and extrapolating to &lt;span class=&quot;math-container&quot;&gt;$dt=0$&lt;/span&gt; you can obtain results for long-term average Newtonian precession which are within 1% of the accepted figures.  Another advantage over analytical methods which produce long-term average results is that you can examine behaviours at shorter time-scales.  For example if you graph perihelion direction vs. time for a certain planet (e.g. Mercury) you can see the &lt;span class=&quot;math-container&quot;&gt;$\approx 11.9 $&lt;/span&gt; year periodic fluctuations in precession rate resulting from the movement of Jupiter around the Sun.  It is also a lot of fun (and very easy once you have written the basic code) to play &quot;what if?&quot; simulations by varying the number and properties of the bodies in the system and even by adding additional Non-Newtonian forces.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To quote Feymnan:-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It may be that in one cycle of calculation, depending on the problem, we may have 30 multiplications, or something like that, so one cycle will take 300 microseconds. That means that we can do 3000 cycles of computation per second. In order to get an accuracy, of, say, one part in a billion, we would need 4×10^5 cycles to correspond to one revolution of a planet around the sun. That corresponds to a computation time of 130 seconds or about two minutes. Thus it takes only two minutes to follow Jupiter around the sun, with all the perturbations of all the planets correct to one part in a billion, by this method! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you do have to think carefully about what you can reliably infer from the simulations - for example if your time-step is longer than a few hundred seconds the simulation will indicate precession in the opposite direction to that which really occurs (i.e. retrograde when it should be prograde).&lt;/p&gt;&#xA;" OwnerUserId="233" LastEditorUserId="233" LastEditDate="2018-12-18T10:27:23.803" LastActivityDate="2018-12-18T10:27:23.803" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="8080" PostTypeId="1" AcceptedAnswerId="8089" CreationDate="2014-11-28T16:42:35.447" Score="8" ViewCount="1433" Body="&lt;p&gt;NASA's Galileo orbiter at Jupiter ended its mission in 2003 by intentionally dive into the gas planet. Did it take and transmit any close up images before it ceased to function? If not, why? Are there any images from Galileo, or its separate impactor probe, which show the Jovian clouds from close enough range that their topology is discernible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On all images I've seen, Jupiter looks like a perfect sphere because of its huge size and far distances from which it has been imaged. But if two spacecrafts have dived into it, there should be close up images of its clouds, right?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2014-11-30T13:54:13.980" Title="What are the last images from the Galileo orbiter before impacting Jupiter?" Tags="&lt;jupiter&gt;&lt;photography&gt;&lt;galileo-probe&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8081" PostTypeId="1" CreationDate="2014-11-28T17:08:07.257" Score="5" ViewCount="785" Body="&lt;p&gt;Does NASA's New Horizons flyby probe to Pluto have instruments to measure Pluto's (or Charon's) magnetic field, if any?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a bit confused by the &lt;a href=&quot;https://en.wikipedia.org/wiki/New_Horizons&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; which says that New Horizons, during its flyby of Jupiter, on one hand:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Travelling through the planet's magnetosphere New Horizons collected&#xA;  valuable particle readings.[45] &quot;Bubbles&quot; of plasma which are believed&#xA;  to be formed from material ejected by the moon Io were noticed in the&#xA;  magnetotail.[47]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But then on the other hand:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An objective to measure any magnetic field of Pluto was dropped. A&#xA;  magnetometer instrument could not be implemented within a reasonable&#xA;  mass budget and schedule, and SWAP and PEPSSI could do an indirect job&#xA;  detecting some magnetic field around Pluto.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And how true is this story which I read or heard somewhere? That when New Horizons was being built and launched, no one really believed that Pluto could have a magnetic field. But since then planetary science has advanced and it is now plausible that it might have a magnetic field, because of tidal forces between Pluto and Charon which might make both objects somewhat fluid and possibly generating magnetic fields.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-01-24T10:24:07.180" Title="What will New Horizons be able to say about a magnetic field at Pluto?" Tags="&lt;magnetic-field&gt;&lt;pluto&gt;&lt;charon&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8082" PostTypeId="1" CreationDate="2014-11-28T21:27:32.120" Score="3" ViewCount="98" Body="&lt;p&gt;Whenever I read any Astro papers in &lt;a href=&quot;http://arxiv.org/&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/&lt;/a&gt; they usually talk about Data fitting or modelling or sentences like 'we fitted the spectral region by analytic functions  i.e. by multi-component Gaussian model plus continuum'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any websites, online books or maybe YouTube videos which teaches these techniques in detail?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any info will be much appreciated!&lt;/p&gt;&#xA;" OwnerUserId="2891" LastEditorUserId="2891" LastEditDate="2014-11-29T01:07:32.630" LastActivityDate="2014-11-29T04:23:16.263" Title="Where can I learn these Astrophysical techniques? [Read below]" Tags="&lt;astrophysics&gt;&lt;data-analysis&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8083" PostTypeId="1" CreationDate="2014-11-29T02:27:36.333" Score="9" ViewCount="181" Body="&lt;p&gt;It is reasonably well known that Mars has a greater obliquity range than Earth, due to Mars lacking a stabilising influence of a large moon.  However, the Martian obliquity seems to have gone through periods of high and low &lt;em&gt;mean&lt;/em&gt; obliquity, as shown in the top chart below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JvVWu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image source: &lt;a href=&quot;http://spacescience.arc.nasa.gov/mars-climate-modeling-group/past.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mars Climate Modeling Group&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the current accepted theory as to why there are periodic high and low mean obliquities on Mars?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Please note, I am not asking about orbital eccentricity&lt;/em&gt;)&lt;/p&gt;&#xA;" OwnerDisplayName="user2449" LastActivityDate="2016-09-28T17:54:51.130" Title="What is the cause of the variation from high and low mean obliquity periods of Mars?" Tags="&lt;mars&gt;&lt;obliquity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8084" PostTypeId="2" ParentId="8082" CreationDate="2014-11-29T04:23:16.263" Score="1" Body="&lt;p&gt;If you are going to do data analysis, you need to understand how the fitting procedure works. This means a lot of statistics, and new terminology and so on, which is hard and takes much time.&lt;/p&gt;&#xA;&lt;p&gt;If you want to start to fit, e.g., a spectrum, you should definitely read the &lt;a href=&quot;http://heasarc.gsfc.nasa.gov/xanadu/xspec/manual/manual.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;XSPEC&lt;/a&gt; guide. You can find a pdf online as well.&lt;/p&gt;&#xA;&lt;p&gt;In short, fitting is to take a model and measure how well this model fits to your data. To quantify this &amp;quot;how well&amp;quot;, &lt;strong&gt;usually&lt;/strong&gt; the chi-squared distribution is considered:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\chi^2 =  \sum\limits_{i=1}^n(\frac{X_i - \mu_i}{\sigma_i})^2$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; is your data (observed value), &lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt; is your expected value (which, in this case, corresponds with the prediction of the model), and &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; is the variance on your data point (the error).&lt;/p&gt;&#xA;&lt;p&gt;This is the most general information that you need.&#xA;You can read more &lt;a href=&quot;http://physics.ucsc.edu/%7Edrip/133/ch4.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, and especially on the &lt;a href=&quot;http://www2.units.it/ipl/students_area/imm2/files/Numerical_Recipes.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Numerical Recipes&lt;/a&gt; (don't be scared by the huge format of the last document, the pages you need are only few around Chapter 15).&lt;/p&gt;&#xA;&lt;p&gt;For fun, you can just try to play with XSPEC, to take confidence, and see at least how things change when you change your model or your data (especially if you know which model is the best-fit for your data). To understand everything will take years, but if you never start, you'll never arrive ;)&lt;/p&gt;&#xA;" OwnerUserId="1038" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2014-11-29T04:23:16.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8087" PostTypeId="1" AcceptedAnswerId="8088" CreationDate="2014-11-29T22:31:50.293" Score="2" ViewCount="456" Body="&lt;p&gt;If there are many universes, they must have a gravity effect on each other. Since dark matter is measured through gravity, but we don't know the mass behind it, isn't it possible that it comes from another universe?&lt;/p&gt;&#xA;" OwnerUserId="2914" LastEditorUserId="7472" LastEditDate="2017-07-15T21:47:53.227" LastActivityDate="2017-07-15T21:47:53.227" Title="Can the theory of multiverse explain dark matter?" Tags="&lt;gravity&gt;&lt;universe&gt;&lt;dark-matter&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8088" PostTypeId="2" ParentId="8087" CreationDate="2014-11-29T22:56:44.200" Score="7" Body="&lt;p&gt;This is a bit of a speculative question, but I can answer it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark matter has been observed in galaxies, and the distributions of dark matter in galaxies have also been measured. It seems clear that the matter is firmly in our universe - we just can't detect it with electromagnetic radiation.&#xA;*However, there are ideas (extremely speculative) that dark matter &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_matter#Mass_in_extra_dimensions&quot;&gt;is from a different spatial dimension&lt;/a&gt;. It sort of explains why gravity is so weak (it &quot;leaks off&quot; through the other dimension(s)). Note, though, that this highly, &lt;em&gt;highly&lt;/em&gt; speculative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are ideas like this in string theory, too. Some say that we are trapped on a &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Brane_cosmology&quot;&gt;brane&lt;/a&gt;&lt;/em&gt;, a section of space-time with three spatial dimensions. However, there are one or more additional dimensions. All of these dimensions are known as the &lt;em&gt;bulk&lt;/em&gt;. Gravity can &quot;leak off&quot; through these extra dimensions. Some theories take this to the extreme, and say that there exist many universes (of a certain number of dimensions each, but with less dimensions that those of the bulk) that are branes, just like ours. These have been considered the source of phenomena such as dark energy - in which case your scenario is within the realm of plausibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these ideas, however, are speculative and have no evidence in their favor.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-11-29T22:56:44.200" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="8089" PostTypeId="2" ParentId="8080" CreationDate="2014-11-30T13:54:13.980" Score="4" Body="&lt;p&gt;The &lt;a href=&quot;http://spaceref.com/news/viewpr.html?pid=12593&quot; rel=&quot;nofollow&quot;&gt;wonderful article&lt;/a&gt; linked by Wayfaring Stranger in the comments alludes to the reason for not sending photographs, but does not answer the question directly. As with most spacecraft outside Earth or Lunar orbit, the data rate is extraordinarily low. &lt;a href=&quot;http://blogs.esa.int/rosetta/2014/08/05/tracking-the-spacecraft-following-a-comet/&quot; rel=&quot;nofollow&quot;&gt;This page&lt;/a&gt; gives data rates for the current Rosetta mission, which are on the order of a dozen or two KiB/s. So a 10 MiB uncompressed image file would take 15 minutes to send. Though it would be an undoubtedly interesting photograph, it would come at the &lt;strong&gt;opportunity cost&lt;/strong&gt; of magnetic readings, temperature readings, radiation readings, and other important scientific data which would help plan our next probe to survive the environment.&lt;/p&gt;&#xA;" OwnerUserId="209" LastActivityDate="2014-11-30T13:54:13.980" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8090" PostTypeId="1" CreationDate="2014-11-30T17:19:51.830" Score="13" ViewCount="3722" Body="&lt;p&gt;Carbon is the fourth &lt;a href=&quot;https://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements#Earth.27s_crustal_elemental_abundance&quot;&gt;most common element&lt;/a&gt; in the universe and in the Solar system. It is about the ninth most common element in Earth's crust. It doesn't seem to be part of any of the ten &lt;a href=&quot;http://education.jlab.org/glossary/abund_com.html&quot;&gt;most common compounds&lt;/a&gt; in the Earth's crust. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So where is all the carbon hiding?&#xA;Why isn't it more common in planetary and lunar crusts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it maybe mostly occurring in volatiles together with hydrogen and oxygen, and has left bodies which lack atmosphere or freezing temperatures? Is it primarily biological processes which binds carbon to solid molecules? Having a look around on Earth's natural surface today, there's lots of solid carbon compounds everywhere, so the contrast is interesting.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2019-02-06T00:23:43.437" Title="Why is carbon so rare on the Moon and on Mars?" Tags="&lt;earth-like-planet&gt;&lt;carbon-cycle&gt;&lt;elemental-abundances&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8091" PostTypeId="1" CreationDate="2014-11-30T18:56:32.980" Score="2" ViewCount="334" Body="&lt;p&gt;Could someone explain what is &quot;median stellar mass&quot; and how is it calculated? &#xA;I have seen it written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log(M)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or as: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log\left(\frac{M_*}{M_{\text{Sun}}}\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I do not understand what it means.&lt;/p&gt;&#xA;" OwnerUserId="2913" LastEditorUserId="2153" LastEditDate="2014-11-30T19:31:04.293" LastActivityDate="2014-12-01T15:13:05.210" Title="What is &quot;Median Stellar Mass&quot;?" Tags="&lt;star&gt;&lt;mass&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8092" PostTypeId="2" ParentId="8091" CreationDate="2014-11-30T20:02:46.433" Score="2" Body="&lt;p&gt;just check what 'median' means. the median stellar mass is such that half of all stars have lower and the other half hihger mass. it has nothing to do with the logarithm. I havn't seen the term (median stellar mass) in the scientific literature.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Looking at the paper you're referring to, they never use the expression &quot;median stellar mass&quot;. Moreover, this study is not concerned about the masses of individual stars, but with the total stellar mass of a galaxy, which they donote by $M$. They then consider the typical or characteristic value for $M$ (as obtained by fitting a Schechter function to the observed distribution of $M$), and denote it by $M^\ast$. Since this is a rather large number, of the order of $\sim10^{10}$M$_\odot$, they prefer to use its logarithm to base ten: $$\log\left(\frac{M^\ast}{\mathrm{M}_\odot}\right).$$&#xA;I think the paper explains all that quite clearly and am puzzled by your difficulty.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2014-12-01T15:13:05.210" LastActivityDate="2014-12-01T15:13:05.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8093" PostTypeId="1" AcceptedAnswerId="8095" CreationDate="2014-12-01T12:32:30.577" Score="3" ViewCount="224" Body="&lt;p&gt;What are the nearest galaxies I can observe using my Telescope? Does it require to be out of the city lights?&lt;/p&gt;&#xA;" OwnerUserId="2923" LastActivityDate="2014-12-04T08:40:28.397" Title="What are the nearest galaxies I can observe?" Tags="&lt;galaxy&gt;&lt;telescope&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8094" PostTypeId="1" CreationDate="2014-12-01T13:24:29.640" Score="8" ViewCount="215" Body="&lt;p&gt;I keep thinking of various planetary system configurations, and would like to know:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What are the fundamental references based on &lt;strong&gt;Monte Carlo long-term simulation of planetary systems' evolution&lt;/strong&gt; that put testable restrictions on the distribution of stable configurations around Sol-like stars (&lt;strong&gt;books, papers, software packages&lt;/strong&gt;)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How rare is the configuration wherein a dwarf planet (&lt;strong&gt;Ceres&lt;/strong&gt;, I'm looking at you) is located in the inner system?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;The main question is the first one, in the vein of &quot;teach a man to fish...&quot;&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Related Q&amp;amp;A's here that do not answer my question:&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;sub&gt; &lt;a href=&quot;https://astronomy.stackexchange.com/questions/872/is-there-a-upper-limit-to-the-number-of-planets-orbiting-a-star&quot;&gt;Is there a upper limit to the number of planets orbiting a star?&lt;/a&gt;&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;sub&gt; &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6343/can-you-explain-the-pattern-of-hill-sphere-sizes-of-the-objects-of-the-solar-sys&quot;&gt;Can you explain the pattern of Hill sphere sizes of the objects of the Solar system?&lt;/a&gt;&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;sub&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/5990/could-a-habitable-satellite-of-a-gas-giant-have-a-stable-subsatellite&quot;&gt;Could a habitable satellite of a gas giant have a stable subsatellite?&lt;/a&gt;&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="793" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-08-27T17:08:43.847" Title="Likelihood of a stable system with a dwarf planet's orbit inside that of a gas giant" Tags="&lt;exoplanet&gt;&lt;dwarf-planets&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8095" PostTypeId="2" ParentId="8093" CreationDate="2014-12-01T13:43:13.433" Score="3" Body="&lt;p&gt;The nearest one is the one you are on: Milky Way. It can be seen from any place, as it is all around us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to actually look at a whole galaxy in a simple view, then the nearest galaxies easy to be observed are the Small and Large Magellan Clouds, and the Galaxy of Andromeda.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are galaxies nearer that those, but harder to be observed. A quite complete list is at &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_nearest_galaxies&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/List_of_nearest_galaxies&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-12-01T13:43:13.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8096" PostTypeId="1" AcceptedAnswerId="8098" CreationDate="2014-12-01T13:52:57.453" Score="2" ViewCount="328" Body="&lt;p&gt;In the following partial solar eclipse, the moon is &quot;invisible&quot; rather than being dark.  By contrast, if someone stands between my view and the sun, they will appear as a dark figure instead of invisible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why doesn't the moon appear as a dark circle instead of being invisible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yx3qp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2925" LastActivityDate="2014-12-01T17:28:04.023" Title="Why isn't the moon dark in a solar eclipse?" Tags="&lt;solar-eclipse&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8097" PostTypeId="2" ParentId="8096" CreationDate="2014-12-01T14:19:01.150" Score="2" Body="&lt;p&gt;Sunlight reflects through the clouds, making them visible in front of the dark Moon.&lt;/p&gt;&#xA;" OwnerUserId="2880" LastActivityDate="2014-12-01T14:19:01.150" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8098" PostTypeId="2" ParentId="8096" CreationDate="2014-12-01T15:29:48.747" Score="1" Body="&lt;p&gt;What you see is clouds which are &lt;em&gt;in front&lt;/em&gt; of the Moon. If there were neither clouds nor atmosphere between you and the Moon and you had amazing eye sight, you couldn't see any stars where the Moon is, but the Moon itself illuminated from the Earth (which is in full Sun ligth). Thus &lt;strong&gt;the Moon is neither invisible nor dark&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-12-01T15:29:48.747" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8099" PostTypeId="2" ParentId="8096" CreationDate="2014-12-01T15:31:50.007" Score="3" Body="&lt;p&gt;Light refracts in the atmosphere, causing the light to scatter in many directions. This light scattering is what causes our day to be blue, our sunsets to be their red-orange-violetish colors and allows us to see the light of the Sun shortly before sunrise and after sunset. As this only begins happening in any significant amount &lt;em&gt;after&lt;/em&gt; the light enters Earth's atmosphere, there should not be any notable difference in the color or intensity of light between you and the moon and any other part of the sky at a relative angle from the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the foreground does not have direct light on the photographer's side (because it is nearly between the sun and the photographer) and there is a very limited amount of atmosphere to scatter light. As a result, less light will be coming to the camera from that direction.&lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="1288" LastEditDate="2014-12-01T17:28:04.023" LastActivityDate="2014-12-01T17:28:04.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8100" PostTypeId="1" AcceptedAnswerId="8114" CreationDate="2014-12-01T19:15:42.050" Score="68" ViewCount="8820" Body="&lt;p&gt;I just want to be sure I am visualizing this correctly, because it seems odd. The Moon is tidally locked to the Earth but there are wobbles to its motion due to libration. So from a point on the surface of the near side of the Moon, the Earth would always be near the same place in the sky? It would describe a small circle or a side-to-side wobble over the course of a month, but never move far from that point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That would seem very strange, like it was a gigantic stage prop or something. We are so conditioned that everything rises and sets (except for a few stars near the poles). &lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="9186" LastEditDate="2015-10-16T13:40:15.067" LastActivityDate="2017-02-14T07:24:37.030" Title="How does the Earth move in the sky as seen from the Moon?" Tags="&lt;the-moon&gt;&lt;earth&gt;&lt;observational-astronomy&gt;&lt;orbital-elements&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="8101" PostTypeId="2" ParentId="8100" CreationDate="2014-12-01T21:27:33.030" Score="42" Body="&lt;p&gt;You are correct. The Earth would always appear in approximately the same location in the sky, when viewed from a point on the lunar surface. And it would be seen to spin, the continents coming in and out of view over the course of an Earth day (24 hours). The sun would make it's way across the sky, from one horizon to the other over a period of about two weeks, to rise again in another two weeks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might sound odd, because the rock we live on spins pretty quickly relative to the most obvious points of reference. In general (assuming that you are viewing from a place not particularly close to one of the poles) the bodies we see in the sky rise in the East, and set in the West about 12 hours later. But if you take a little bit closer look, you start to see that this is not the whole story. Each night, the moon rises and sets almost an hour later than the night before. The stars that you can see at night shift over a period of one year. And the planets move in such strange patterns that before Copernicus (at least in Europe), they seemed to just be &lt;em&gt;&lt;a href=&quot;http://www.etymonline.com/index.php?term=planet&quot;&gt;wandering&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And not everything appears to move accross the sky. As you mentioned, Polaris, the north star, seems to hang in one spot all of the time, though you can't see it during the day. But even taking all of this into account, the idea of a large planetary body just hanging in one spot all of the time in the sky, may still seem... um, alien. And to that, I might just say, we've been here on this one planet an awfully long time. It's bound to feel a bit like home.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditDate="2014-12-02T17:28:48.683" LastActivityDate="2014-12-02T17:28:48.683" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8102" PostTypeId="1" CreationDate="2014-12-02T00:16:47.260" Score="2" ViewCount="74" Body="&lt;p&gt;&lt;a href=&quot;http://www.chromoscope.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chromoscope&lt;/a&gt; is a nice toy. Pressing 'L' shows the constellations as guide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a bright X-ray source between Ophiuchus, Libra and Scorpius. In Hydrogen-Alpha there are two circular formations to the right and left of it. Are the latter formed by polar jets from the former, and what is that object?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="2153" LastEditDate="2016-11-27T21:24:51.607" LastActivityDate="2016-11-28T07:55:39.183" Title="In Chromoscope, what is the X-ray source between Ophiuchus, Libra and Scorpius?" Tags="&lt;light&gt;&lt;milky-way&gt;&lt;spectra&gt;&lt;x-ray&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8105" PostTypeId="1" CreationDate="2014-12-02T04:49:39.017" Score="25" ViewCount="4276" Body="&lt;p&gt;This question is prompted from a comment on &lt;a href=&quot;https://astronomy.stackexchange.com/a/8101/2934&quot;&gt;this answer&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could a person on the surface of the moon see man-made lights on the dark side of the Earth with the naked eye? If not, how much magnification would be required to do see them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest evidence I could find from a quick search was this picture from &lt;a href=&quot;http://www.slcschools.org/departments/curriculum/science/Grade-3-to-6/Grade-3/earth-and-moon-appearance.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Salt Lake City Schools' Website&lt;/a&gt;:&lt;img src=&quot;https://i.stack.imgur.com/IEF4Q.jpg&quot; alt=&quot;Picture of Earth from the moon&quot;&gt;&lt;br&gt;&#xA;The dark part of Earth is a bit hard to see here, though it can be seen more easily in &lt;a href=&quot;https://i.stack.imgur.com/IEF4Q.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;the full resolution image&lt;/a&gt;. You can certainly see &lt;em&gt;something&lt;/em&gt; toward the bottom edge in the dark area, but I can't tell for sure if it is lights from Southeast Asia/Oceania or something else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: As pointed out in the comments, the image of Earth in this photo wasn't actually taken from the moon, but from NASA's Terra and Aqua satellites and combined with the lunar imagery in a &lt;a href=&quot;http://visibleearth.nasa.gov/view.php?id=63447&quot; rel=&quot;nofollow noreferrer&quot;&gt;2002 visualization created by NASA&lt;/a&gt; to commemorate the 30th anniversary of Apollo 17.&lt;/p&gt;&#xA;" OwnerUserId="2934" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-07-12T21:33:54.807" Title="Can you see man-made lights on the dark side of the Earth from the surface of the moon with the naked eye?" Tags="&lt;the-moon&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="10" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8106" PostTypeId="2" ParentId="8100" CreationDate="2014-12-02T10:49:30.510" Score="20" Body="&lt;p&gt;To add to Mark Bailey's answer; the Earth would indeed hang in the sky and rotate, but it would also &lt;em&gt;wax and wane&lt;/em&gt; over the course of a lunar day (27.3 Earth-days).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting at lunar dawn, the Earth would be half-full. The Earth would then wane (more shadow) towards lunar noon. At lunar noon, the Earth would be all in shadow (New Earth) and quite close to the sun. With a telescope (to eliminate the sun), a lunian would see city lights and atmospheric glow around the edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the lunar afternoon, the Earth would wax (less shadow) and be half-full again (on the other side) at lunar sunset. Full Earth would occur at lunar midnight. This would be about 16 times as bright as a full Moon on Earth, so it never gets really dark on the lunar Near Side. The Earth would then wane again until it was half-full at lunar dawn. &lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="2938" LastEditDate="2014-12-02T16:43:24.273" LastActivityDate="2014-12-02T16:43:24.273" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8107" PostTypeId="1" CreationDate="2014-12-02T14:33:20.093" Score="7" ViewCount="68" Body="&lt;p&gt;I have a friend that watches a lot of TV news and they have the impression that impact events are becoming more common than in recent memory. Does the evidence support or refute this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking it would answer the question, I've tried to find data on the annual rate of impact events. But maybe there is another way to look at it.&lt;/p&gt;&#xA;" OwnerUserId="2943" LastEditorUserId="2153" LastEditDate="2014-12-02T15:55:28.827" LastActivityDate="2020-05-13T14:23:03.030" Title="Where can I find historic rates of meteor impact events?" Tags="&lt;asteroids&gt;&lt;meteor&gt;&lt;impact-probability&gt;&lt;resource-request&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8108" PostTypeId="1" AcceptedAnswerId="8110" CreationDate="2014-12-02T17:21:43.113" Score="5" ViewCount="1587" Body="&lt;p&gt;Well, we all know Sun gravity , makes the  Space time curvature and space pushes the earth towards the Sun. But my question is &quot;why the distance between the Sun and Earth is Same? Earth should have moved closer to Sun and obviously Earth should fall into the Sun.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;Selva&lt;/p&gt;&#xA;" OwnerUserId="2948" LastEditorUserId="633" LastEditDate="2014-12-03T08:56:59.100" LastActivityDate="2014-12-04T13:49:28.067" Title="Why does the distance between Sun and Earth stay the same?" Tags="&lt;gravity&gt;&lt;general-relativity&gt;&lt;space-time&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8109" PostTypeId="2" ParentId="8100" CreationDate="2014-12-02T17:34:36.943" Score="12" Body="&lt;p&gt;To expand a little more, yes the Earth would hang in the same spot in the sky, moving around in a small circle as the moon rotated around it over the course of each of its 28 day orbits. It would have phases, full Earth when the moon is between it and the sun, new Earth when the Earth is between the moon and the sun, and wax and wane between these two points. Each time the Earth was full, a different part of it would be visible from the moon, depending on the season and how it happened to line up when full Earth occurred.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth would roll in the sky according to the seasons on Earth and the precession of the moon's orbit. The moon tilts only 1.5 degrees from the ecliptic and so in essence has no seasons. The orbit of the moon is tilted 5 degrees from the ecliptic, so when it is farthest to the south of the plane of the solar system, more of the south of our planet is visible, and vice-versa when it is farthest north. But the Earth is tilted 23 degrees from the ecliptic, so this is what would mostly determine what part of it was visible from the moon. Here is a good summary of the relationship of the Earth and Moon. The distance between the two bodies is not to scale, but the relative size of the one to the other is to scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6UQHz.jpg&quot; alt=&quot;the orbit of the moon around the Earth&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In December, Antarctica would be visible - as it was during the famous Blue Marble photo from Apollo 17, taken on Dec. 7, 1972. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gcqly.jpg&quot; alt=&quot;Apollo 17 photograph of Earth, showing Africa, Arabia, and Antarctica.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This version of the photo is supposed to be as it was in the original photograph. They took it on the way to the moon, so the orientation of the camera was the only thing that determined which way was 'up'. If you were on the moon at far southern latitudes, this is in fact how it would look from your point of view.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In July, the Arctic would be visible. Apollo 11 took this photo on July 16, 1969, when they were half-way to the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sygmF.jpg&quot; alt=&quot;Apollo 11 photo of Earth, showing Africa, Europe, and the Middle East&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here actually you don't see much of the Arctic. The moon's orbit had carried it a bit south of the Earth at that point. Heading to it meant going south a bit, so when they looked back at the Earth, they saw less of the far north.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to be clear, something you would NEVER see is this, which has been widely misunderstood as a photo of the whole planet from space. Actually it is a super-wide-angle photo taken from low earth orbit - the horizon line is for the part of the world the satellite could see, it is not one hemisphere of the Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PqlfX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think about it for a second you realize this can't possibly be true. The United States is gigantic, if it was really this size Canada would occupy the whole arctic and roll over onto the other side of the world, displacing most of Russia. Argentina and Chile would take the place of Antarctica.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2014-12-02T17:34:36.943" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8110" PostTypeId="2" ParentId="8108" CreationDate="2014-12-02T17:34:43.867" Score="5" Body="&lt;p&gt;While the Sun and Earth attract each other, they cannot fall into each other because of &lt;strong&gt;angular momentum conservation&lt;/strong&gt;. In a central field (where the force is acts in the direction of the distance vector and depends on distance only), the specific angular momentum vector $\boldsymbol{L}=\boldsymbol{r}\times\boldsymbol{v}$ is conserved ($\boldsymbol{r}$ is position and $\boldsymbol{v}$ the velocity). In particular $L=|\boldsymbol{L}|=rv_t$ (with $v_t$ the tangential velocity: the component of velocity perpendicular to the direction Earth-Sun). The specific orbital energy $E=\tfrac{1}{2}|\boldsymbol{v}|^2-GM/r$ is also conserved and must be negative for a bound orbit (such as Earth's). Combining these two we have&#xA;$$&#xA;E = \tfrac{1}{2}v_r^2 + \frac{L^2}{2r^2} - \frac{GM}{r}&#xA;$$&#xA;with $v_r$ the radial component of velocity. Since $v_r^2\ge0$, but $E&amp;lt;0$, not all radii are reachable. In particular there are two radii at which $v_r=0$: the apo- and peri-apse of the orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, for the Earth, the peri- and apo- apse are quite similar, and the orbit is nearly circular. The gravitational pull of the Sun is very nearly balanced by the centrifugal force due to the rotating orbit.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In GR, the picture doesn't really change much. Non-Newtonian (GR) effects are tiny in the Solar system and completely unimportant for Earth's orbit.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note that the picture of pushing by ripples in space time (vs. pull from Sun's gravity) doesn't make a difference either. The point is that just because you're pushed some way doesn't imply that you're falling that way. All bodies also have something called inertia which makes them want to follow their path rather being pushed around. In case of a circular orbit (similar to Earth's around the Sun), gravity and inertia balance in such a way as to create a circular orbit: inertia want's to go straight, gravity want's to pull/push you into the Sun, but the result is something between those. &lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2014-12-03T09:02:51.550" LastActivityDate="2014-12-03T09:02:51.550" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="8111" PostTypeId="2" ParentId="8105" CreationDate="2014-12-02T22:00:22.420" Score="13" Body="&lt;p&gt;It's highly doubtful you could see any normal light source on the surface of the earth.&#xA;Using &lt;a href=&quot;http://zebu.uoregon.edu/~soper/Light/luminosity.html&quot;&gt;$$\text{brightness} = \frac{\text{luminosity}}{4 \pi \times \text{distance}^2}$$&lt;/a&gt;&#xA;(with brightness in watts, and luminosity in watts per square meter.&#xA;and &lt;a href=&quot;http://en.wikipedia.org/wiki/Moon&quot;&gt;distance to moon&lt;/a&gt; of $3.84 \times 10^8$ meters.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Try&lt;/strong&gt; a hypothetical light source 100 megawatts output, all visible light, no heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{brightness} = \frac{100  \times 10^6}{4 \pi \times 1.474 \times 10^{17}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{brightness} = 5.4 \times 10^{-11} \text{ watts per square meter}$$ at the lunar surface.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;That's pretty dim. By, contrast sunlight at earth's surface runs about 1300 watts per square meter.&#xA;In reality it'd take about a gigawatt to produce 100 megawatts of light in the visible range.&#xA;That's about what it takes to power a city of a million homes.&#xA;Cities also bounce most of the light they do produce off the ground, which'll have an albedo of somewhere around 0.3. So with ordinary city lights it'll take over 3 gigawatts to reach $5.4 \times 10^{-11}$ watts per square meter on the lunar surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might fare better with a &lt;strong&gt;big&lt;/strong&gt; laser. &lt;a href=&quot;http://en.wikipedia.org/wiki/Apache_Point_Observatory_Lunar_Laser-ranging_Operation&quot;&gt;The Apache Point Observatory Lunar Laser-ranging Operation&lt;/a&gt; picks up multi-photon signals from the Apollo retroreflectors using only a &lt;a href=&quot;http://www.spacetoday.org/SolSys/Moons/TheMoon/Retroreflectors.html&quot;&gt;1 gigawatt laser&lt;/a&gt; and a 3.5 meter telescope. As the article states, the laser beam only expands to 9.3 miles in diameter on the way to the moon, so you &lt;em&gt;might&lt;/em&gt; see it wink at you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 10 parsecs, the sun has a &lt;a href=&quot;http://en.wikipedia.org/wiki/Absolute_magnitude&quot;&gt;magnitude of 4.83&lt;/a&gt;. It'd be visible on an average night. That magnitude corresponds to &lt;a href=&quot;http://books.google.com/books?id=wkIOzagyQmAC&amp;amp;pg=PA72&amp;amp;lpg=PA72&amp;amp;dq=starlight%20watts%20per%20square%20meter&amp;amp;source=bl&amp;amp;ots=aaCRrQ-Fbm&amp;amp;sig=UqNIcbgyJdYAQZw0eGswpS57IQo&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=5Dx-VNDNF5L2yQSLsYKwBQ&amp;amp;ved=0CC4Q6AEwAg#v=onepage&amp;amp;q=starlight%20watts%20per%20square%20meter&amp;amp;f=false&quot;&gt;a brightness of 3X10e-10 watts per square meter&lt;/a&gt;, about 5.6 fold brighter than our hypothetical earth based light source. &lt;a href=&quot;http://en.wikipedia.org/wiki/Apparent_magnitude&quot;&gt;That puts our light at magnitude 6.5 to 7&lt;/a&gt;. Naked eye visibility runs to about 6.0&lt;/p&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="2031" LastEditDate="2014-12-03T08:11:35.807" LastActivityDate="2014-12-03T08:11:35.807" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8112" PostTypeId="1" AcceptedAnswerId="8369" CreationDate="2014-12-02T23:25:25.823" Score="21" ViewCount="2540" Body="&lt;p&gt;I'm surprised that this question hasn't been asked before (here or on Physics), to the best of my knowledge. It's one that I might have asked when I was a bit younger, and one that I think other people will ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, it's clear that &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1509/will-saturn-rings-become-a-moon&quot;&gt;Saturn's rings won't form a moon&lt;/a&gt;, and the same is likely to be true for other ring systems. However, I'm guessing that they won't last forever (it's just a guess).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How long do planetary rings in general last? What mechanisms could cause them to dissipate/fall apart/end? I'm guessing the &lt;a href=&quot;https://en.wikipedia.org/wiki/Poynting%E2%80%93Robertson_effect&quot; rel=&quot;noreferrer&quot;&gt;Poynting-Robertson effect&lt;/a&gt; could come into play, but I'm not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for anyone curious, yes, I checked just for the fun of it, and Yahoo Answers had a bunch of really, really bad, unsourced and most likely inaccurate answers (given that there was no consensus), ranging from '3 million years' to '13-18 billion years' to 'forever'.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2019-10-24T15:53:09.313" Title="How long do planetary rings last?" Tags="&lt;saturn&gt;&lt;planetary-ring&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8113" PostTypeId="2" ParentId="8090" CreationDate="2014-12-03T04:22:12.423" Score="5" Body="&lt;p&gt;Although carbon is highly abundant in the universe, it is not homogeneously distributed.  Some regions of the interstellar medium could be rich in carbon and others rich in silicon or oxygen, depending on the source of the heavier elements (&quot;metals&quot;).  The region could be enriched from mass loss of evolved stars, or from a SN of a massive star, or from a SN from a white dwarf.  Each of these bring different elemental abundances.&lt;br&gt;&#xA;Also, if the region of the protoplanetary disk where the Earth formed was warm, then the carbon would be in the gaseous phase or in gaseous phase molecules like CO, CO$_2$ and CH$_4$.  Perhaps, where the terrestial planets formed much of the carbon was in gaseous form rather than in dust particles, and therefore only a small fraction of the carbon was incorporated into the rocky planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, much of the Earth's carbon is not in the crust. &lt;a href=&quot;http://science.howstuffworks.com/environmental/green-science/earth-carbon-stored.htm&quot; rel=&quot;nofollow&quot;&gt;It is in trees, the atmosphere and mostly in the oceans, dissolved as carbonates and in shells etc.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2014-12-09T18:50:35.953" LastActivityDate="2014-12-09T18:50:35.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8114" PostTypeId="2" ParentId="8100" CreationDate="2014-12-03T06:13:38.973" Score="48" Body="&lt;p&gt;Still new at stellarium but here are some quick capture gif  &lt;strong&gt;lasting one month&lt;/strong&gt;. Sorry about the quality- limited to 256 colors for smaller gifs. Date on lower left corner. By the way the sun is of course the brightest and i use it as reference for recording (start record when sun is in frame then stop when it appears again in the same position which is roughly one month)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Location on Moon : Sea of Tranquility&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are looking straight up&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yellow lines are azimuth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KAsZz.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;second picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zoomed view of earth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;One whole day 24 hours&lt;/strong&gt; (give or take a few minutes)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Location on Moon: Sea of Tranquility&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are looking straight up&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yellow lines are azimuth (gif itself rotated to approximately match the orientation of the 260 degree azimuth line in the first picture)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vKVxv.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer for Emilio Pisanty comment on whether the oscillation is detectable by eye&lt;/p&gt;&#xA;&#xA;&lt;p&gt;stellarium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Field of view 60 degrees (default view when stellarium first opened)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shows one end of the oscillation near the 200 azimuth line (see yellow arrows)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;other end of ellipse near 185 azimuth line&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can use Gemini (lower right corner) for reference (see yellow arrows) from head to crotch of one Gemini twin would be a good reference for &lt;strong&gt;how wide the oscillation&lt;/strong&gt; is (not sure if your sundial can detect the difference)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yeVIA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;picture 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;shows other end of the oscilation near the 185 azimuth line (see yellow arrows)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wG4wM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;picture 5 shows my stellarium location settings on the moon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nvBUb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2957" LastEditorUserId="2957" LastEditDate="2014-12-04T14:29:28.700" LastActivityDate="2014-12-04T14:29:28.700" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8116" PostTypeId="1" AcceptedAnswerId="8118" CreationDate="2014-12-03T03:25:31.543" Score="5" ViewCount="121" Body="&lt;p&gt;I am an amateur astronomer please correct if any of my understandings are wrong. We all know that from the Hubble's ultra deep field image the most distant galaxies which were created shortly after big bang looked to be red shifted. This is because of the process of expansion of the universe these lights are reached its highest visible wavelength. So in near future (in terms of astronomy) the galaxies which appears to be normal today become red shifted?  and those are already appeared as red shifted become invisible? If it become invisible still can we see it through infrared telescopes right? &lt;/p&gt;&#xA;" OwnerUserId="2958" OwnerDisplayName="Anand Asir" LastEditorUserId="13" LastEditDate="2014-12-03T11:47:49.650" LastActivityDate="2014-12-03T15:39:04.507" Title="Red shifting galaxies" Tags="&lt;fundamental-astronomy&gt;&lt;redshift&gt;&lt;cosmological-inflation&gt;&lt;observable-universe&gt;&lt;deep-sky-observing&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8117" PostTypeId="1" AcceptedAnswerId="8119" CreationDate="2014-12-03T15:35:24.523" Score="9" ViewCount="7139" Body="&lt;p&gt;Here's an image that describes the phenomenon I'm asking about.  The very thin sliver to the left is of course the surface of the moon as lighted by the sun.  But the rest of the moon is also faintly lighted -- by what?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing that some amount of &quot;earthshine&quot; -- sunlight reflected from the earth -- is the culprit here, but I've found no authoritative confirmation online. It would be great to have a link to a source that explains what's really going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7IGKg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2962" LastActivityDate="2020-12-15T14:55:04.053" Title="Why can I see the whole moon during various non-full-moon phases?" Tags="&lt;the-moon&gt;&lt;moon-phases&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8118" PostTypeId="2" ParentId="8116" CreationDate="2014-12-03T15:39:04.507" Score="1" Body="&lt;p&gt;Assuming an &lt;a href=&quot;http://en.wikipedia.org/wiki/Ultimate_fate_of_the_universe#Open_universe&quot; rel=&quot;nofollow&quot;&gt;open model of the universe&lt;/a&gt; and a long enough timeline galaxies which appear &quot;normal&quot; today would become more red shifted over the course of observation.  I put normal in quotes because there are no galaxies which are relatively static to ours.  Any galaxy which has movement will have some &quot;shift&quot; to them, and all galaxies &lt;a href=&quot;http://abyss.uoregon.edu/~js/cosmo/lectures/lec14.html&quot; rel=&quot;nofollow&quot;&gt;are moving away from us&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that eventually the light from them could move into the infrared and even the microwave in extreme cases.&lt;/p&gt;&#xA;" OwnerUserId="2963" LastActivityDate="2014-12-03T15:39:04.507" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8119" PostTypeId="2" ParentId="8117" CreationDate="2014-12-03T15:40:18.450" Score="13" Body="&lt;p&gt;The dark parts of the Moon are partially illuminated by &amp;quot;&lt;a href=&quot;http://earthsky.org/space/what-is-earthshine&quot; rel=&quot;noreferrer&quot;&gt;Earthshine&lt;/a&gt;&amp;quot;. That is sunlight which is &lt;a href=&quot;https://en.wikipedia.org/wiki/Earthlight_%28astronomy%29&quot; rel=&quot;noreferrer&quot;&gt;reflected from Earth&lt;/a&gt; to the Moon. Just like the ground on Earth is lit up a bit by the Moon at night, so is the ground on the Moon lit up a bit by the Earth in the lunar night. And even more so because Earth is much larger in the lunar sky, than is the Moon in our sky. And the surface of Earth is actually more reflective per area unit than is the surface of the Moon.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-02-08T13:28:27.780" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8120" PostTypeId="2" ParentId="8117" CreationDate="2014-12-03T16:20:54.670" Score="5" Body="&lt;p&gt;It's poetically called &lt;a href=&quot;http://apod.nasa.gov/apod/ap020419.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Old Moon in the New Moon's Arms&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.bbso.njit.edu/Research/EarthShine/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Project Earthshine&lt;/a&gt; used measurements of the brightness of the non-sunlit portion of the moon to accurately determine earth's albedo:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A global and absolutely calibrated albedo can be determined by measuring the amount of sunlight reflected from the Earth and, in turn, back to the Earth from the dark portion of the face of the Moon (the &amp;quot;earthshine&amp;quot; or &amp;quot;ashen light&amp;quot;).  For more than a decade we have been measuring the Earth’s large-scale reflectance from BBSO.  The observations are now done remotely utilizing our earthshine coronagraph under the small dome appearing to the right of the NST dome in the Figure 1. To get full coverage of the Earth, we have installed a carefully calibrated copy of the BBSO earthshine telescope in Tenerife.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="23881" LastEditDate="2020-12-15T14:55:04.053" LastActivityDate="2020-12-15T14:55:04.053" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8121" PostTypeId="2" ParentId="7844" CreationDate="2014-12-03T18:06:57.373" Score="3" Body="&lt;p&gt;I could offer a couple of speculations - but there is no &quot;theory&quot;. Incidentally, I was looking at the refereed journal paper by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2010A%26A...519A..87B&quot; rel=&quot;nofollow&quot;&gt;Baumann et al. (2010)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The atmospheric parameters of these stars are not correct. Given that the ages are derived from the gravities and temperatures of the stars and that the lithium abundances are also dependent on the gravities and particularly the temperatures, there will be correlated errors. I'm actually struggling to understand how you get stars which are &quot;solar analogues&quot;, all of which have similar gravities, yet seem to have a spread of ages (estimated from the Teff vs log g plot) from 1-10Gyr? It is notable that there appears to be just 1/117 in Baumann's sample, compared with 10/82 in Israelian's work...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Maybe they are actually PMS stars, so the estimated ages are completely spurious since they assume that the stars are on the main sequence. I would however have expected the abundances to be 3 or higher.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Neither of these explanations seems completely satisfactory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mention in passing that the question of planets and lithium is far from settled. The latest salvo in the war was fired by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014A%26A...570A..21F&quot; rel=&quot;nofollow&quot;&gt;Figueira et al. (2014)&lt;/a&gt; who claim there &lt;em&gt;is&lt;/em&gt; statistical evidence for enhanced Li depletion in solar analogues with exoplanets.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2014-12-03T18:53:18.193" LastActivityDate="2014-12-03T18:53:18.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8122" PostTypeId="2" ParentId="1041" CreationDate="2014-12-03T18:50:33.900" Score="5" Body="&lt;p&gt;The spectra of a red giant and a red dwarf are &lt;em&gt;completely different&lt;/em&gt;, so there isn't really too much to say about this and distinguishing giants and dwarfs is simple. For example, alkali lines are almost non-existent in red giants, but strong in red dwarfs. The theory as to why this happens is to do with the surface gravity and pressure broadening; it is the stuff of a standard graduate/undergraduate course on stellar atmospheres, not an SE answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact is that a R=50,000 spectrum with decent signal to noise ratio will quite easily&#xA;give you the temperature (to 100K), surface gravity (to 0.1 dex) and metallicity (to 0.05 dex), plus a host of other elemental abundances (including Li) to precisions of about 0.1 dex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can you do with this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can plot the star in the log g vs Teff plane and compare it with theoretical isochrones appropriate for the star's metallicity. This &lt;em&gt;is&lt;/em&gt; the best way to estimate the age of a solar-type (or more massive) star, even if you don't have a distance and is the most-used method. How well this works and how unambiguously depends on the star's evolutionary stage. For stars like the Sun, you get an age precision of maybe 2 Gyr. For lower mass stars, well they hardly move whilst on the main sequence in 10Gyr, so you can't estimate the age like this unless you know the object is a pre-main sequence star (see below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can look at the Li abundance. Li abundance falls with age for solar-mass stars and below. This would work quite well for sun-like stars from ages of 0.3-2Gyr and for K-type stars from 0.1-0.5 Gyr and for M-dwarfs between 0.02-0.1 Gyr - i.e. in the range from where Li  starts to be depleted in the photosphere until the age where it is all gone. Typical precision might be a factor of two. A high Li abundance in K and M dwarfs usually indicates a pre main sequence status.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gyrochronology is not much help - that requires a rotation period. However you can use the relationship between rotation rate (measured in your spectrum as projected rotation velocity) and age. Again, the applicability varies with mass, but in the opposite way to Li. M-dwarfs maintain fast rotation for longer than G-dwarfs. Of course you have the problem of uncertain inclination angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That brings us to activity-age relations. You can measure the levels of chromospheric magnetic activity in the spectrum. Then combine this with empirical relationships between activity and age (e.g. Mamajek &amp;amp; Hillenbrand 2008). This can give you the age to a factor of two for stars older than a few hundred Myr. Its poorly calibrated for stars less massive than the Sun though. But in general a more active M-dwarf is likely to be younger than a less active M dwarf. It should certainly distinguish between a 2Gyr and 8Gyr M dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you measure the line of sight velocity from your spectrum, this can give you at least a probabilistic idea of what stellar population the star belongs to. Higher velocities would tend to indicate an older star. This would work better if you had the proper motion (and preferably the distance too, roll on the Gaia results).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, in a probabilistic sense, low metallicity stars are older than high metallicity stars. If you were talking about stars as old as 8Gyr, these would be quite likely to have low metallicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary. If you are talking about G-dwarfs you can ages to precisions of about 20% using log g and Teff from the spectrum. For M dwarfs, unless you are fortunate enough to be looking at a young PMS object with Li, then your precision is going to be a few Gyr at best for an individual object, though combining probabilistic estimates from activity, metallicity and kinematics simultaneously might narrow this a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an add-on I'll also mention radio-isotope dating. If you can measure the abundances of isotopes of U and Th with long half lives and then make some guess at their initial abundances using other r-process elements as a guide then you get an age estimate - &quot;nucleocosmochronology&quot;. Currently, these are very inaccurate - factors of 2 differences for the same star depending on what methods you adopt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read &lt;a href=&quot;http://adsabs.harvard.edu/abs/2010ARA%26A..48..581S&quot; rel=&quot;nofollow noreferrer&quot;&gt;Soderblom (2013)&lt;/a&gt;; &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014EAS....65..289J&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jeffries (2014)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Since I wrote this answer, there is at least one more promising method that has emerged. It turns out that the abundance of certain s-process elements (e.g. barium, yttrium) are enriched quite slowly during the lifetime of the Galaxy (by the winds of dying asymptotic giant branch stars), and more slowly than enrichment with iron and much more slowly than alpha elements like Mg and Si. Thus a measurement of the relative fractions of these elements, like [Y/Mg] can give the age to precisions of a billion years or so (e.g. &lt;a href=&quot;https://arxiv.org/abs/1604.05733&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tucci Maia et al. 2016&lt;/a&gt;; &lt;a href=&quot;https://arxiv.org/abs/1912.02800&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jofre et al. 2020&lt;/a&gt;). This method is probably the best for solar-type stars older than a Gyr, but remains unexplored/uncalibrated for lower mass stars.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2020-02-20T11:21:39.963" LastActivityDate="2020-02-20T11:21:39.963" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8123" PostTypeId="1" AcceptedAnswerId="8124" CreationDate="2014-12-03T19:09:11.090" Score="8" ViewCount="459" Body="&lt;p&gt;According to &lt;a href=&quot;https://astronomy.stackexchange.com/a/8118/1142&quot;&gt;this&lt;/a&gt; answer it is possible for galaxies' light to move beyond the visible frequencies due to redshift:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;It is possible that eventually the light from them could move into the infrared and even &#xA;the microwave in extreme cases.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Could this have already happened?  Have we already looked for it?  Is there a telescope with the instrumentation to detect galaxies' 'light' at these wavelengths?&lt;/p&gt;&#xA;" OwnerUserId="1142" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2014-12-03T20:00:30.553" Title="Have we detected galaxies which have red-shifted beyond the visible light range?" Tags="&lt;redshift&gt;&lt;deep-sky-observing&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8124" PostTypeId="2" ParentId="8123" CreationDate="2014-12-03T20:00:30.553" Score="7" Body="&lt;p&gt;Yes, of course. Many, many examples. Telescopes work in the infrared, far-infrared and there are even samples of galaxies that are selected on the basis of their mm emission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most distant galaxies detected now have redshifts of 7 or more. This means the wavelength of their light has been stretched by a factor $1+z$ - i.e. by a factor of 8. Thus light in the visible range, say 500nm, now appears at wavelength of 4 microns, in the infrared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Telescopes that work in this range include the Spitzer space telescope; and many ground-based telescopes. Observations of highly redshifted galaxies are routinely made at infrared wavelengths on telescopses all around the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxies are also detected in the far infrared by the Herschel satellite or at mm (getting on for microwave) wavelengths by JCMT or the ALMA telecope.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2014-12-03T20:00:30.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8125" PostTypeId="2" ParentId="8090" CreationDate="2014-12-03T21:01:21.193" Score="11" Body="&lt;p&gt;There is not much doubt that the abundance of carbon in the protosolar nebula was not abnormal. We can tell that by looking at the carbon abundance in the atmosphere of the Sun - it has an abundance of 8.43 on a logarithmic (base 10) scale where hydrogen has an abundance of 12 (&lt;a href=&quot;http://arxiv.org/pdf/0909.0948v1.pdf&quot; rel=&quot;noreferrer&quot;&gt;Asplund et al. 2009&lt;/a&gt;). This is typical for stars in the solar neighbourhood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However carbon in the solids of the solar system - i.e. in the Earth, moon, meteorites etc. is underabundant with respect to this by something like a factor of 10. Nitrogen is similarly underabundant in the Earth and oxygen to a lesser extent (a factor of two). (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2009LanB...4B...44L&quot; rel=&quot;noreferrer&quot;&gt;Lodders et al. 2009)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this is that C,N,O are all capable of remaining in the gaseous phase at the kinds of condensation temperatures that are typical for forming most of the &quot;rocks&quot; that make up the Earth and other solids in the solar system. In particular, carbon monoxide is readily formed at high temperatures and cannot easily be captured into terrestrial planets and other rocky materials.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2014-12-09T21:26:27.843" LastActivityDate="2014-12-09T21:26:27.843" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8126" PostTypeId="2" ParentId="6394" CreationDate="2014-12-03T21:12:18.260" Score="2" Body="&lt;p&gt;The main sources of neutrons for the s-process (can be found in under a minute on &lt;a href=&quot;http://en.wikipedia.org/wiki/S-process&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; and in this case is quite reliable) are carbon 13 and neon 22. These undergo reactions with alpha particles yielding oxygen and magnesium respectively plus a free neutron. The first reaction takes place in low-mass (1-8$M_{\odot}$) stars whilst they are in the AGB phase (hydrogen and helium shell burning); the second takes place in higher mass stars at the end of carbon burning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the r-process, HDE226868's explanation is the consensus explanation.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2014-12-03T21:12:18.260" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8128" PostTypeId="2" ParentId="7970" CreationDate="2014-12-03T23:56:32.943" Score="3" Body="&lt;p&gt;The timescale on which Jupiter cools is reasonably well understood and predicted by the current generation of evolutionary models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter's luminosity is provided mostly by gravitational contraction. For a planet that only contains gas governed by the perfect gas law, the appropriate timescale for this contraction (or indeed for the luminosity to fall significantly) is given by the Kelvin Helmholtz timescale.&#xA;$$ \tau = \eta \frac{GM^2}{RL},$$&#xA;where $M$ and $R$ are Jupiter's mass and radius and $L$ is its current power output (or luminosity), and the parameter $\eta \sim 1$. This timescale is a few $10^{11}$ years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, giant planets like Jupiter are not governed by perfect gas laws. The gas in the centre of Jupiter is dense enough that electrons become degenerate. Degenerate electrons fill the available energy levels up to the Fermi energy. Their consequent non-zero momenta of the electrons exerts a degeneracy pressure that is &lt;em&gt;independent of temperature&lt;/em&gt;. As a result, the rate of contraction slows and the release of gravitational potential energy slows; the planet is able to cool and remain in hydrostatic equilibrium without the same degree of contraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can express this change using the $\eta$ parameter. For Jupiter $\eta \simeq 0.03$ (&lt;a href=&quot;http://arxiv.org/abs/1405.3752&quot; rel=&quot;nofollow&quot;&gt;Guillot &amp;amp; Gautier 2014&lt;/a&gt;) - i.e. the timescale for the luminosity to fade is 30 times quicker than the naive Kelvin-Helmholtz time and Jupiter's luminosity will scale as the reciprocal of its age and will fall by a factor of a few in $10^{10}$ years. In a trillion years, the luminosity of Jupiter will be lower than it is now by roughly a factor of 250.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2014-12-03T23:56:32.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8129" PostTypeId="1" CreationDate="2014-12-04T00:19:19.410" Score="3" ViewCount="126" Body="&lt;p&gt;Wouldn't a droplet of liquid (say water) floating around in space become spherical and &quot;in hydrostatic equilibrium&quot; due to surface tension but not gravity?&lt;/p&gt;&#xA;" OwnerUserId="1106" LastEditorUserId="1678" LastEditDate="2014-12-04T07:16:59.100" LastActivityDate="2015-01-05T02:59:18.560" Title="Can objects enter hydrostatic equillibrium through processes other than the influence of gravity?" Tags="&lt;hydrostatic-equilibrium&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8131" PostTypeId="2" ParentId="8129" CreationDate="2014-12-04T07:53:19.007" Score="1" Body="&lt;p&gt;Strictly speaking (as far as I know), hydrostatic equilibrium applies whenever a fluid balances external body forces with the pressure gradient.  From &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrostatic_equilibrium&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In continuum mechanics, a fluid is said to be in hydrostatic equilibrium or hydrostatic balance when it is at rest, or when the flow velocity at each point is constant over time. This occurs when external forces such as gravity are balanced by a pressure gradient force.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think the concept happens to be most frequently used in areas where gravity is the external force, but it could in principle be anything else. So, though I stand to be corrected, I think a droplet isolated in space long enough could be said to be in hydrostatic equilibrium, even though the most relevant force is the surface tension, rather than gravity.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2014-12-04T07:53:19.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8132" PostTypeId="2" ParentId="8093" CreationDate="2014-12-04T08:40:28.397" Score="0" Body="&lt;p&gt;The obvious answer is our own, the Mikyway. The only  problem is, Where you are in the world? We all will see it differently, many factors play into this, your elevation,city lights,e.t.c One of the biggest problems is light, it has plagued astronomers for years. Depending on how good your telescope is, the next closest galaxy to us is the Andromada galaxy with a good telescope you will be able to see it. You may also be able to view the sm. And lg. Magellan clouds with a larger telescope, remember the better equipment you have,the more you will be able to see. More farther,clearer and be able enjoy more than just galaxies.&lt;/p&gt;&#xA;" OwnerUserId="2672" LastActivityDate="2014-12-04T08:40:28.397" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8133" PostTypeId="1" CreationDate="2014-12-04T11:56:37.180" Score="12" ViewCount="4159" Body="&lt;p&gt;At lunar midnight (i.e. the new moon as seen from Earth), the Earth is in its full phase with its entire disk in sunlight, and it is the brightest object in the lunar sky. How bright is it, and how variable is its brightness?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, I'm interested in specific figures for the magnitude and in comparisons to similar objects. Would I be able to read in that light? To drive? Would I notice the variations in brightness? What are the main variations due to?&lt;/p&gt;&#xA;" OwnerUserId="207" LastEditorUserId="15415" LastEditDate="2018-09-30T12:38:33.443" LastActivityDate="2018-09-30T12:38:33.443" Title="How bright is the full Earth during the lunar midnight?" Tags="&lt;the-moon&gt;&lt;earth&gt;&lt;light&gt;&lt;apparent-magnitude&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8134" PostTypeId="2" ParentId="7970" CreationDate="2014-12-04T13:06:56.293" Score="1" Body="&lt;p&gt;In &quot;one trillion&quot; years Jupiter's fate will be affected by how violent our Sun becomes when it transforms in Red Giant, 5000 million years (5Gy) from now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With our Sun being so big and brilliant, it will heat Jupiter quite a lot more than now. But also the mass lose will make Jupiter spiral out towards a bigger orbit, while capturing some extra mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in &quot;one trillion&quot; years Jupiter will be a bigger, colder and denser (as well pointed by Rob Jeffries in his answer), more external planet around a white dwarf.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2014-12-04T13:06:56.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8135" PostTypeId="2" ParentId="6609" CreationDate="2014-12-04T13:11:42.807" Score="-10" Body="&lt;p&gt;For a celestial body to be a sphere in hydrostatic equilibrium, it needs to be a fluid. Hydrostatic equilibrium does not make sense for solid bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Earth and Mars &lt;strong&gt;are not&lt;/strong&gt; in hydrostatic equilibrium. It is spherical for the good reason of being a massive body where its own gravity is enough to avoid big irregularities, but it is not supported by (fluid) pressure, but by (solid) incompressibility and material resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, Jupiter and the Sun &lt;strong&gt;are&lt;/strong&gt; in hydrostatic equilibrium since the force that avoids them collapsing is actually (fluid) pressure.&lt;/p&gt;&#xA;" OwnerUserId="473" LastEditorUserId="9301" LastEditDate="2016-12-29T05:08:01.847" LastActivityDate="2016-12-29T05:08:01.847" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="8136" PostTypeId="1" AcceptedAnswerId="8142" CreationDate="2014-12-04T13:40:57.640" Score="9" ViewCount="241" Body="&lt;p&gt;As I learned, that the bigger gravity source you are influenced by the more slow time ticks for you, the farther away you are from a gravity source the faster times ticks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Imagine two different planets where on each people live, one planet is massively big (time ticks slow), the other one very small compared (time ticks faster). How would it look like for bigger planet people to observe smaller planet and for smaller planet people to observe them. Hope it makes some sense.&lt;/p&gt;&#xA;" OwnerUserId="2980" LastActivityDate="2014-12-04T23:02:08.023" Title="What it would look like to observe people with a different time flows?" Tags="&lt;gravity&gt;&lt;universe&gt;&lt;observational-astronomy&gt;&lt;space-time&gt;&lt;general-relativity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8137" PostTypeId="2" ParentId="8108" CreationDate="2014-12-04T13:49:28.067" Score="3" Body="&lt;p&gt;Technically, the distance between Sun and Earth changes throughout the year. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth%27s_orbit&quot; rel=&quot;nofollow&quot;&gt;Earth moves around the Sun&lt;/a&gt; on an elliptical orbit. It's distance from the Sun &lt;a href=&quot;https://en.wikipedia.org/wiki/Apsis&quot; rel=&quot;nofollow&quot;&gt;changes&lt;/a&gt; between 152m km (aphelion) and 147m km (perihelion), a difference of 5 million kilometres, twice per year.&lt;/p&gt;&#xA;" OwnerUserId="2880" LastActivityDate="2014-12-04T13:49:28.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8138" PostTypeId="2" ParentId="8129" CreationDate="2014-12-04T13:55:29.350" Score="1" Body="&lt;p&gt;Video: &lt;a href=&quot;https://www.youtube.com/watch?feature=player_embedded&amp;amp;v=9ZEdApyi9Vw&quot; rel=&quot;nofollow&quot;&gt;Space Station Astronauts Grow a Water Bubble in Space&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surface tension tends to draw the water into a nice sphere.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2014-12-04T13:55:29.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8139" PostTypeId="1" AcceptedAnswerId="8157" CreationDate="2014-12-04T16:56:11.380" Score="5" ViewCount="668" Body="&lt;p&gt;When I was 20, I went to Naxos (Greece) with friends. I was driving my car during the night on the mount Zeus (1003 meters - 3,291 ft) listening Pink Floyd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up there we stopped the car to take a walk, looked above and ... wooooooowwwwww!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sky was dense! Full of stars. We didn't know it could have been like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How dense is the deep sky from outside the atmosphere?&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For example, on the ISS during the night (the Earth that cover the Sun) looking at the deep sky, how many stars can you see? I saw a lot of pictures of the stars taken by the Hubble telescope), but I don't think it can be compared, right?&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="490" LastEditorUserId="7982" LastEditDate="2021-02-28T11:54:16.883" LastActivityDate="2021-02-28T11:54:16.883" Title="How many stars can be seen naked eye from outside the atmosphere? How dense is the &quot;sky&quot; from there?" Tags="&lt;atmospheric-effects&gt;&lt;naked-eye&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8140" PostTypeId="1" AcceptedAnswerId="8141" CreationDate="2014-12-04T17:25:14.577" Score="2" ViewCount="68" Body="&lt;p&gt;I understand that astronomers once thought that the material in the disc of a galaxy was moving around the galactic centre (where most of the mass was thought to be) in roughly circular orbits. The circular motions would be explained by the combination of inertia and gravitational acceleration towards the centre. This follows the Kepler/Newton model which describes/explains the orbits of planets around the Sun.  In a Kepler/Newton system the tangential or transverse velocity of a low-mass orbitting object depends on the radial distance of the high-mass object from the centre.  Objects further out must have lower velocities so that the weaker centrally-directed acceleration at that distance can keep the object moving in a roughly circular track. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the centre of the galaxy the velocities are low and increase rapidly with distance away from the centre.  This is understandable as the mass of the central bulge of the galaxy  is not concentrated at the centre, it is spread out over a relatively large volume.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it is now well-known that the measured velocities of visible material in the disks of discoidal galaxies do not vary as expected for a Kepler/Newton system.  At  a certain distance where the velocities would be expected to start decreasing (as per curve A) they either level out or continue increasing at a slow rate (as per curve B). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/b/b9/GalacticRotation2.svg/200px-GalacticRotation2.svg.png&quot; alt=&quot;galaxy velocity rotation curve&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current explanantion is that a large mass of Dark Matter extends throughout the galaxy in such a way as to produce non-Keplerian behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;QUESTION&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But isn't there another possible explanantion of the non-Keplerian velocity profile namely that the ordinary detectable material (e.g. gas, dust, stars) in the disk is not completely gravitationally bound to the galactic centre and is actually gradually moving outwards along a spiral path?&lt;/p&gt;&#xA;" OwnerUserId="233" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:47.257" LastActivityDate="2014-12-06T20:39:43.303" Title="Could discoid galaxies be expanding?" Tags="&lt;galactic-dynamics&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8141" PostTypeId="2" ParentId="8140" CreationDate="2014-12-04T18:18:36.603" Score="5" Body="&lt;p&gt;No, for several reasons.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;there are no radial velocities in the required range observed, for example in the Milky Way galaxy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the reservoir of stars moving out from the centre would quickly be drained, so you would need to magically generate them at the centre -- this is much worse than postulating dark matter.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;you'd expect huge amounts of stars and gas that have moved out to large radii.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The only serious alternative to dark matter is some modification to Einstein's/Newton's theory of gravity. There are several candidates, but none is really convincing (though arguably, dark matter is not too convincing either). &lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-12-04T18:18:36.603" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8142" PostTypeId="2" ParentId="8136" CreationDate="2014-12-04T19:41:09.427" Score="5" Body="&lt;p&gt;&lt;strong&gt;Normal case:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect would not be very noticeable unless the large planet was extremely large / high gravity (e.g. it would look &quot;almost&quot; normal) looking at one another.  If the planet was that large, it would probably not be able to support life, but to see what would happen, lets assume it can support life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Relativistic case, from the large planet, observing the small planet&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the large planet were extremely &quot;heavy / massive&quot; (think more like a neutron star), the people on the large planet would observe the people on the small planet moving much faster.  It would be like watching a movie with &quot;fast forward&quot; on.  However, light would also be blue shifted, which would produce other interesting visual effects (e.g. what was red might now be blue, and what was in invisible &quot;infrared&quot; might now be red depending on the amount of the shift.)  The amount of blue shift (light wave compression) and &quot;fast forwarding&quot; would increase if the planet were made bigger.  The rest of the universe would have similar effects when observed from the large planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Relativistic case, from the smaller planet, observing the large planet&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The people on the smaller planet would observe the people on the extremely large planet moving in &quot;slow motion&quot;.  It would be like watching a movie in slow motion.  The people on the large planet would also be &quot;red shifted&quot;.  For example, something that was blue might appear red, and something in the ultraviolet range might now appear blue.  The &quot;slow motion&quot; effect and red shift (light wave expansion) would increase if the extremely large planet were made bigger.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="1386" LastEditDate="2014-12-04T23:02:08.023" LastActivityDate="2014-12-04T23:02:08.023" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8143" PostTypeId="1" AcceptedAnswerId="8147" CreationDate="2014-12-04T21:05:02.660" Score="7" ViewCount="146" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/8142/1142&quot;&gt;This&lt;/a&gt; answer made me wonder if matter/energy/events near a black hole are viewed (by us) in a very different time frame, how might that affect our measurements?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The people on the smaller planet would observe the people on the extremely large planet &#xA;moving in &quot;slow motion&quot;. (note: due to gravitational time dilation)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;On a related note, does light being 'bent' around a black hole experience time dilation or red/blue shift from our viewing perspective? (as possibly suggested by &lt;a href=&quot;https://astronomy.stackexchange.com/a/1331/1142&quot;&gt;this&lt;/a&gt; answer).&lt;/p&gt;&#xA;" OwnerUserId="1142" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-12-05T19:33:26.713" Title="Are our 'observations' of or near a black hole affected by gravity induced time dilation?" Tags="&lt;gravity&gt;&lt;general-relativity&gt;&lt;gravitational-lensing&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8144" PostTypeId="1" AcceptedAnswerId="8152" CreationDate="2014-12-04T21:32:48.507" Score="8" ViewCount="175" Body="&lt;p&gt;Could the Philae comet lander be recharged with a laser (or possibly many lasers) from earth?  My thought would be to aim one or more powerful lasers from earth at the Philae lander to recharge it through its solar panels.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thought would be to heat up a small part of the comet with one or more lasers, possibly causing the comet to slowly rotate into a position where the sun light falls onto the lander.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would either of these ideas be feasible?&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="1386" LastEditDate="2014-12-05T17:45:09.593" LastActivityDate="2014-12-06T00:47:30.240" Title="Could the Philae comet lander be recharged by laser?" Tags="&lt;earth&gt;&lt;comets&gt;&lt;space-probe&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8145" PostTypeId="2" ParentId="8144" CreationDate="2014-12-04T23:09:20.030" Score="3" Body="&lt;p&gt;I only have time for a partial answer, so I'll discuss the bit about rotating the comet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to do some calculations with &lt;a href=&quot;https://en.wikipedia.org/wiki/Radiation_pressure&quot; rel=&quot;nofollow&quot;&gt;radiation pressure&lt;/a&gt;, the force on an object from light.$^1$ &lt;a href=&quot;http://www.webassign.net/question_assets/buelemphys1/chapter22/section22dash3.pdf&quot; rel=&quot;nofollow&quot;&gt;This pdf&lt;/a&gt; has some good starting calculations, although they assume that the source is spherical - the Sun. Our laser clearly is not. However, the beam &lt;em&gt;will&lt;/em&gt; diverge over such a long distance, so I'll have to use a substitution (given by CountIblis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force due to radiation pressure is&#xA;$$F_{\text{rad}}=\frac{2I}{c}A$$&#xA;where $I$ is the intensity, $c$ is the ubiquitous &lt;a href=&quot;https://en.wikipedia.org/wiki/Speed_of_light&quot; rel=&quot;nofollow&quot;&gt;speed of light&lt;/a&gt;, and $A$ is the area onto which the force is applied. However,&#xA;$$I=\frac{\text{power}}{4 \pi r^2}$$&#xA;so we get&#xA;$$F_{\text{rad}}=\frac{\text{power}\times A}{2 \pi r^2c}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $4 \pi$ has to be replaced (by CountIblis' substitution) with $$\Omega=\pi \left(\frac{\alpha}{2}\right)^2$$&#xA;For an ideal laser, $$\alpha=2.44\frac{\lambda}{d}$$&#xA;where $\lambda$ is the wavelength and $d$ is the initial beam diameter. Let's assume that our laser has a power output of one megawatt - the &lt;a href=&quot;https://en.wikipedia.org/wiki/Chemical_oxygen_iodine_laser&quot; rel=&quot;nofollow&quot;&gt;COIL&lt;/a&gt; laser, from a &lt;a href=&quot;https://en.wikipedia.org/wiki/Boeing_YAL-1&quot; rel=&quot;nofollow&quot;&gt;Boeing YAL-1&lt;/a&gt;, as per &lt;a href=&quot;https://what-if.xkcd.com/13/&quot; rel=&quot;nofollow&quot;&gt;Randall Munroe's idea&lt;/a&gt;. $1,000,000 \text{ watts}$ is a &lt;em&gt;lot&lt;/em&gt; from a laser. Let's also assume that no power is lost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For COIL, $\lambda=1.315 \times 10^{-6}$ and &lt;a href=&quot;http://www.globalsecurity.org/military/systems/aircraft/systems/atl.htm&quot; rel=&quot;nofollow&quot;&gt;$d=0.1016 \text{ meters}$&lt;/a&gt;. This gives an $\alpha$ of $3.158070866 \times 10^{-5}$, and thus an $\Omega$ of $7.83309915 \times 10^{-10}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Philae is currently on comet &lt;a href=&quot;https://en.wikipedia.org/wiki/67P/Churyumov%E2%80%93Gerasimenko&quot; rel=&quot;nofollow&quot;&gt;67P/Churyumov-Gerasimenko&lt;/a&gt;, which has, on its longest side, a surface area (combined lobes) of about $19,370,000 \text{ meters}$. However, it's about &lt;a href=&quot;http://edition.cnn.com/2014/11/12/world/comet-landing-countdown/index.html&quot; rel=&quot;nofollow&quot;&gt;$310,000,000 \text{ miles}$&lt;/a&gt; from Earth, which is $4.9879 \times 10^{11} \text{ meters}$. Knowing that, to rotate the comet, we have to apply force on only half the area of one side, we find that this laser would apply a force of&#xA;$$F_{\text{rad}}=\frac{1,000,000 \text{ watts}\times 9,685,000\text{ meters}^2}{7.83309915 \times 10^{-10} \times 2.487914641 \times 10^{23} \text{ meters} \times 299,792,458 \text{ meters/second}}$$&#xA;$$=1.65771483 \times 10^{-10} \text{ Newtons}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yep, we're not rotating a comet any time soon. I'll try to add in the bit about recharging Philae by laser sometime later. I do think that will be a lot more plausible.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^1$ Unfortunately, I can't find any calculations on &lt;a href=&quot;https://en.wikipedia.org/wiki/Laser_propulsion&quot; rel=&quot;nofollow&quot;&gt;laser propulsion&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-12-05T00:31:32.580" LastActivityDate="2014-12-05T00:31:32.580" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8146" PostTypeId="1" AcceptedAnswerId="8155" CreationDate="2014-12-04T23:17:23.183" Score="23" ViewCount="3713" Body="&lt;p&gt;My guess is that life bearing planets are too far apart to be detected. I think we can only find the ones within a sphere around our planet that is 100s of light years in diameter but I suspect life bearing planets may be much farther away than that.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to estimate the diameter of the sphere within which we could detect life on another planet and then estimate the probability that there is life within that sphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, give our current technology what is the furthest distance would be able to detect life on Earth? &#xA;How many stars like our sun are in that sphere? &#xA;How long would it take for &lt;a href=&quot;http://www.seti.org/&quot; rel=&quot;noreferrer&quot;&gt;SETI&lt;/a&gt; to rule out each of those stars? &lt;/p&gt;&#xA;" OwnerUserId="2984" LastEditorUserId="14021" LastEditDate="2016-09-15T17:03:20.883" LastActivityDate="2017-01-31T12:21:54.590" Title="how far away could we detect that Earth has life?" Tags="&lt;exoplanet&gt;&lt;life&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8147" PostTypeId="2" ParentId="8143" CreationDate="2014-12-05T04:09:59.020" Score="3" Body="&lt;p&gt;The Schwarzschild black hole is the simplest, being just an isolated, nonrotated, uncharged black hole. If $\tau$ is the proper time measured by a clock moving along an arbitrary path in this spacetime, then in Schwarzschild coordinates&#xA;$$-\mathrm{d}\tau^2 = -\left(1-\frac{R}{r}\right)\mathrm{d}t^2 + \left(1-\frac{R}{r}\right)^{-1}\mathrm{d}r^2 + r^2\left(\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\varphi^2\right)\text{,}$$&#xA;where $R = 2GM/c^2$ is the Schwarzschild radius. If the observer is stationary, then neither the radial nor angular coordinates change ($\mathrm{d}r = \mathrm{d}\theta = \mathrm{d}\varphi = 0$). So, for example, relative the rate of time of stationary observers at some radial coordinates $r_1$ and $r_2$ is just&#xA;$$\frac{\mathrm{d}\tau_1}{\mathrm{d}\tau_2} = \frac{\mathrm{d}\tau_1/\mathrm{d}t}{\mathrm{d}\tau_2/\mathrm{d}t} = \sqrt{\frac{1-R/r_1}{1-R/r_2}}\text{.}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;how might that affect our measurements?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the sense of the measurement process, it doesn't, since a measurement apparatus is not time-dilated relative to itself. In the sense of what results we get, then of course it does.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So it would logically follow that events happening near a black hole (a very large gravitational field) would seem to us to happen much more slowly than if they happened outside that field.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. If the first stationary observer is shifted closer to the horizon ($r_1\to R$), then $(1-R/r_1)\to 0$, so the above time dilation goes to $0$ as well. So, for example, if that observer shines a laser beam toward the other, it will be redshifted when it arrives to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is really the same thing as time dilation. The laser beam has a certain frequency, so it takes a certain amount of time for one oscillation. Thus a frequency shift is just another way of talking about time dilation, except that time dilation means that this happens to every process, not just light: e.g, if the first observer taps his foot at a certain frequency (say, a tap per second), the second observer will see him tapping at a lower frequency (say, a tap per two seconds). Slowed. We can can still call this &quot;redshift&quot; in the sense that the frequency was shifted in the same way, although the term is usually used specifically for electromagnetic radiation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On a related note, does light being 'bent' around a black hole experience time dilation or red/blue shift from our viewing perspective?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Light itself &lt;em&gt;experiences&lt;/em&gt; only trivial time dilation in the sense that the proper time along any lightlike trajectory is exactly $0$. Above, we were talking about &lt;em&gt;stationary&lt;/em&gt; observers, and light is not stationary. Rather, $\mathrm{d}\tau = 0$ for light beams, always. So strictly speaking, light doesn't experience anything at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you mean what the stationary observers would get when measuring light bent around the black hole, then it is blueshifted when coming closer to the black hole and redshifted when moving away--in the sense that this would be what stationary observers closer or farther away would measure. The details of the path it took between the observers do not matter; the ratio of measured frequencies will just be the inverse of the time dilation factor between the observers.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-12-05T04:09:59.020" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8148" PostTypeId="1" AcceptedAnswerId="8154" CreationDate="2014-12-05T13:56:16.213" Score="8" ViewCount="380" Body="&lt;p&gt;What kind of reasoning is appropriate to understand the as of today unanswered question of whether there are (other) interstellar space travelling civilizations in the Milky Way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have already sent probes towards the border of the Solar system. And even landed human beings on another celestial body and brought them home alive and well. If we extrapolate the 50 years of space travel, the 100 years of electronics (radio), the 400 years of physical science, to just a fraction of the biological age of humankind into the future (like a few thousands of years), interstellar travel is not out of the question for us or at least our artefacts. So I imagine two possible alternatives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The Milky Way is cluttered by lots of space travelling civilizations like us and our future. Once one of them/us gets going, they'll soon be everywhere. The Sun orbits the Milky Way every 250 million years, about 2% of the age of the galaxy. Going to the nearest stars is enough to soon be everywhere. But if they are everywhere since almost always, they should be here, we should be their seed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) We are the only space travelling civilization in the entire galaxy, ever. But then what makes us unique? We consist of the most common elements and volatiles of the universe and our planet and star and galactic location all seem to be very typical. There's no known trace of any uniqueness here. Whatever could it be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there more alternatives?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While we cannot say today which alternative is true, we should be able to at least specify the possible alternatives. But to me they all seem to be absurd! What would be a rational logical scientific approach to this apparent paradox?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1378" LastEditDate="2014-12-05T20:11:55.767" LastActivityDate="2015-04-11T16:21:11.130" Title="How should one rationally deal with the issue of space travelling alien civilizations?" Tags="&lt;extra-terrestrial&gt;&lt;interstellar-travel&gt;&lt;origin-of-life&gt;&lt;space-travel&gt;&lt;fermi-paradox&gt;" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8149" PostTypeId="2" ParentId="8087" CreationDate="2014-12-05T16:55:53.130" Score="0" Body="&lt;p&gt;I have a theory about this and she is new, she explains quantum phenomena, but is in Portuguese. this here =&gt; &lt;a href=&quot;http://forum.intonses.com.br/viewtopic.php?f=77&amp;amp;t=147809&quot; rel=&quot;nofollow&quot;&gt;http://forum.intonses.com.br/viewtopic.php?f=77&amp;amp;t=147809&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here =&gt; &lt;a href=&quot;http://forum.intonses.com.br/viewtopic.php?f=77&amp;amp;t=287255&quot; rel=&quot;nofollow&quot;&gt;http://forum.intonses.com.br/viewtopic.php?f=77&amp;amp;t=287255&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2989" LastActivityDate="2014-12-05T16:55:53.130" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8150" PostTypeId="1" AcceptedAnswerId="8151" CreationDate="2014-12-05T17:27:40.540" Score="1" ViewCount="6371" Body="&lt;ol&gt;&#xA;&lt;li&gt;The universe is around 13.8 billion years old.&lt;/li&gt;&#xA;&lt;li&gt;We know how fast the light is.&lt;/li&gt;&#xA;&lt;li&gt;We also know how fast our galaxy moving around inside observable universe.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is it possible to calculating distance between Earth and the place where The Big Bang happened ? If so how far away our planet ? and is there a something special about this place ? (A black hole or something else really crazy?)&lt;/p&gt;&#xA;" OwnerUserId="2801" LastEditorUserId="1386" LastEditDate="2014-12-06T01:12:32.487" LastActivityDate="2014-12-06T01:12:32.487" Title="Distance between Earth and where The Big Bang's happened?" Tags="&lt;earth&gt;&lt;distances&gt;&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ClosedDate="2014-12-07T01:53:27.213" ContentLicense="CC BY-SA 3.0" />
  <row Id="8151" PostTypeId="2" ParentId="8150" CreationDate="2014-12-05T17:37:39.637" Score="5" Body="&lt;p&gt;The distance to where the big bang theoretically happened would be &quot;0&quot; (and this should be true for any point within the universe).  The entire universe was theoretically produced in the big bang, with every point in space being &quot;inside&quot; the big bang.  The whole universe is very special, it is amazing what is out there (including black holes, earth, stars, planets, nebulae, etc...).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether or not there is even a &quot;center&quot; to the big bang / the universe, however is another related interesting question.  Current models seem to indicate that there is no center.  It would be interesting if there is a true center of the universe, and would be interesting to know if there is anything special about the center if it exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question of whether or not there is a center of the universe is addressed in the following links:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/669/what-is-in-the-center-of-the-universe&quot;&gt;What is in the center of the universe?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.spaceanswers.com/deep-space/is-there-a-centre-of-the-universe/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.spaceanswers.com/deep-space/is-there-a-centre-of-the-universe/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2014-12-06T00:05:57.367" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8152" PostTypeId="2" ParentId="8144" CreationDate="2014-12-05T19:26:47.103" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Could the Philae comet lander be recharged with a laser (or possibly many lasers) from earth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could make a laser beam with a pretty low divergence - meaning a beam that remains tight and parallel over a very long distance. However, &quot;a very long distance&quot; is nothing on the cosmic scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lunar Laser Ranging experiment is using high-power lasers focused by telescopes, shooting a beam into the Moon. The light is reflected back to Earth by mirrors placed there by the Apollo program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, even with all the tight focusing at the origin, of the initial 10^17 photons in a pulse, only 1 makes it back to the detector. It's very hard to tell that something makes it back at all, let alone charging any solar panels with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The comet and Philae are much, much further away than the Moon. You couldn't possibly project any useful energy density at that distance. Not with the current technology.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2014-12-06T00:47:30.240" LastActivityDate="2014-12-06T00:47:30.240" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8153" PostTypeId="2" ParentId="8143" CreationDate="2014-12-05T19:33:26.713" Score="1" Body="&lt;p&gt;The first experiment that proved Einstein's theory of general relativity was the bending of light around our sun. So if the sun can effect our measurement, in this case the light from a star behind the sun, a black hole sure can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, gravitational lensing is observed by galaxy clusters where multiple images of a background galaxy are formed around the foreground lens. Each of these images are red-shifted or blue-shifted by different amounts, the specific value which can be measured by studying the variability of the background source observed in the different images formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in order to really say whether the light will be red-shifted or blue-shifted, we will have to understand the gravitational potential of the black hole. The position of the background source with respect to the foreground lensing object (black hole in this case) also matters.&lt;/p&gt;&#xA;" OwnerUserId="1278" LastActivityDate="2014-12-05T19:33:26.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8154" PostTypeId="2" ParentId="8148" CreationDate="2014-12-05T19:52:50.860" Score="14" Body="&lt;p&gt;That's basically the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermi_paradox&quot;&gt;Fermi paradox&lt;/a&gt;. It seems likely that there are numerous civilizations in the galaxy, and yet we see no trace of them anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Drake_equation&quot;&gt;Drake equation&lt;/a&gt; is often invoked to calculate the probability of existence of other civilizations, by compounding several other, more simple probabilities: the probability that a star has planets, that the planets are habitable, that life could arise on such a planet, etc. The proposed values so far indicate that there should be many civilizations in our galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomy, just plain observations, might be one way to solve the paradox by detecting alien societies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Broadcasting messages via radio has also been attempted, in the hope that someone might reply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the only constant in this search has been the negative answer. There are some hypotheses trying to explain that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the rare Earth hypothesis: maybe there's something very special about Earth that allows the evolution of intelligent life&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the destruction hypotheses: maybe life tends to self-destruct, or maybe there are space marauders out there killing everyone&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;disasters: maybe things like gamma ray bursts tend to wipe out life periodically&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the young Universe hypothesis: if there are very, very many universes in the Multiverse, then the vast majority of them are young. If so, in the vast majority of them there will be 0, or at most 1 civilizations. It's simply a matter of probability that we are in such a universe.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;religious arguments such as the creation of Man was a special, unique act by a higher entity&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Others argue that many civilizations might exist, but we don't see them. Possible reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;we're just too far from each other, in space, in time, or both&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;it's just too hard to travel between stars&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;we haven't been searching long enough, or we're looking or listening to the wrong things&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;they're not interested in us&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;they all go into a Technological Singularity pretty quickly and become god-like, undetectable and incomprehensible to us&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;they are too busy browsing their internets to be interested in the outer space (yes, that has been &lt;a href=&quot;http://books.google.co.uk/books?id=-vZ0BVSHix4C&quot;&gt;proposed seriously&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;they are too alien for us to detect or comprehend&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;they are like the Na'vi in Avatar - a non-technological society&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;they deliberately avoid us (the Earth-as-a-zoo hypothesis)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;they do try to talk to us, but evidence is being supressed by the government (yes, that is also &lt;a href=&quot;http://adsabs.harvard.edu/abs/1990AcAau..21...97T&quot;&gt;a serious proposal&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;they are here already, hiding amongst us (proposed by Iosif Shklovskii and Carl Sagan, so I guess it qualifies as serious - albeit highly hypothetical at best)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Truth is, we really don't know. All we can do now is speculate.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="2153" LastEditDate="2014-12-05T23:52:49.957" LastActivityDate="2014-12-05T23:52:49.957" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="8155" PostTypeId="2" ParentId="8146" CreationDate="2014-12-05T19:53:28.853" Score="6" Body="&lt;p&gt;Depends on what you mean by detecting life. As is explained in &lt;a href=&quot;https://what-if.xkcd.com/47/&quot; rel=&quot;noreferrer&quot;&gt;this what-if post by Randall Munroe&lt;/a&gt;, the algae on earth will tell the aliens about us before we can tell them about us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider the presence of liquid water or the presence of $O_2$ as detecting life, then such detection can be made by studying the spectrum of extra-solar planets, measurements that we can currently make. The furthest extra-solar planet discovered so far is &lt;a href=&quot;http://exoplanets.co/extrasolar-planets/farthest-exoplanet-from-earth.html&quot; rel=&quot;noreferrer&quot;&gt;at a distance of 27,700 light years&lt;/a&gt;. So, a partial answer to your questions would be to study the spectrum of every extra-solar planet found within the circumstellar habitable zone to look for signatures of tell-tale signs of life. We currently do have the technology to measure the optical reflection spectrum of an extra-solar planet, for example &lt;a href=&quot;http://www.eso.org/public/news/eso1002/&quot; rel=&quot;noreferrer&quot;&gt;ESO's VLT&lt;/a&gt;, the &lt;a href=&quot;http://spie.org/x109304.xml&quot; rel=&quot;noreferrer&quot;&gt;Gemini Observatory&lt;/a&gt; and the &lt;a href=&quot;http://astrobites.org/2012/11/11/detecting-exoplanet-atmospheres-from-the-ground/&quot; rel=&quot;noreferrer&quot;&gt;OSIRIS instrument on GTC&lt;/a&gt; but I don't know if SETI has that capability. You could further look up the work of &lt;a href=&quot;http://seagerexoplanets.mit.edu/research.htm&quot; rel=&quot;noreferrer&quot;&gt;Dr. Sara Seager&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1278" LastActivityDate="2014-12-05T19:53:28.853" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8156" PostTypeId="2" ParentId="8129" CreationDate="2014-12-05T19:55:18.680" Score="1" Body="&lt;p&gt;Yes, a droplet of water would do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, at the planetary scale, the contribution of superficial tension is negligible. At that scale, gravity dominates all other forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, you could model a planet as a loose mass or rubble, kept together by gravity alone, and the model would be very close to reality.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2014-12-05T19:55:18.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8157" PostTypeId="2" ParentId="8139" CreationDate="2014-12-06T11:17:25.623" Score="4" Body="&lt;p&gt;Depending on your eyesight, the faintest stars you can see in a perfect environment have an apparent magnitude (brightness) of 6-6.5 (though some people have managed to see magnitude-7). According to &lt;a href=&quot;http://www.stargazing.net/david/constel/howmanystars.html&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt;, that translates into about 10,000 stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for how it would look, I have seen the sky from the stratosphere (about 3 miles up, in a plane, where there is a lot less atmosphere to worry about, and no light pollution), and it is a surreal sight - with the sky absolutely littered with stars. However, it doesn't  quite look like a Hubble image, though - there's very little colour in these fainter stars, and Hubble tends to focus on narrower fields of view, at higher zoom.&lt;/p&gt;&#xA;" OwnerUserId="2702" LastActivityDate="2014-12-06T11:17:25.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8159" PostTypeId="1" AcceptedAnswerId="8163" CreationDate="2014-12-06T14:29:47.633" Score="4" ViewCount="87" Body="&lt;p&gt;I'm solving the axisymmetric Jeans equations to determine the initial conditions for an exponential disk inside an NFW halo. The density profiles of the two components are &#xA;\begin{equation}&#xA;\rho_d(R,z)=\rho_{0,d}\exp\left(-\frac{R}{R_0}\right)\exp\left(-\frac{|z|}{z_0}\right)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\rho_h(r)=\frac{\rho_{0,h}}{r/r_s(1+r/r_s)^2}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the axial symmetry, for the meridional velocity dispersion we have&#xA;\begin{equation}&#xA;\sigma^2(R,z)=\frac{1}{\rho}\int_z^\infty \rho \frac{\partial \Phi}{\partial z'}dz'&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the galaxy is symmetric about the disk plane ($z=0$), we would expect the distribution function at a distance $h$ above the disk plane to be identical to that at $z=-h$; in other words, if we flip the galaxy, the only difference will be a change in direction of particle motion, but not speed or dispersion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, can't we replace the $z$ in the limit with $|z|$?&lt;/p&gt;&#xA;" OwnerUserId="2992" LastEditorUserId="2153" LastEditDate="2014-12-06T21:14:13.953" LastActivityDate="2014-12-06T23:56:42.553" Title="Exploiting symmetry in the axisymmetric Jeans equations" Tags="&lt;astrophysics&gt;&lt;galactic-dynamics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8160" PostTypeId="1" CreationDate="2014-12-06T16:52:29.963" Score="5" ViewCount="60" Body="&lt;p&gt;During solar flare, electromagnetic energy is released at a wide range of wavelengths but the most dominating are x-rays and extreme ultraviolet rays. Why is that intensity of these wavelengths is greatest?&lt;/p&gt;&#xA;" OwnerUserId="2823" LastActivityDate="2016-08-05T16:42:12.210" Title="Spectra of electromagnetic radiation emitted during solar flare" Tags="&lt;solar-flare&gt;&lt;the-sun&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8161" PostTypeId="1" CreationDate="2014-12-06T21:41:50.707" Score="2" ViewCount="104" Body="&lt;p&gt;Will the Voyager space crafts actually reach another Star System in 63,000 years? If the answer is yes, what will be  powering the crafts to keep them going for so long?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastEditorUserId="2153" LastEditDate="2014-12-06T21:54:44.427" LastActivityDate="2014-12-06T22:07:32.327" Title="Voyager spacecrafts" Tags="&lt;interstellar-travel&gt;&lt;space-probe&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8162" PostTypeId="2" ParentId="8161" CreationDate="2014-12-06T22:07:32.327" Score="5" Body="&lt;p&gt;&lt;strong&gt;Sort of. But not the same system.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/5/5c/Outersolarsystem-probes-4407b.svg&quot;&gt;Here's a photo&lt;/a&gt; of the directions the two &lt;em&gt;Voyager&lt;/em&gt; probes (and a couple &lt;em&gt;Pioneers&lt;/em&gt;) are traveling:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/5/5c/Outersolarsystem-probes-4407b.svg&quot; alt=&quot;*Voyager* directions&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They're also not traveling in a flat plane, as &lt;a href=&quot;http://voyager.jpl.nasa.gov/science/planetary.html&quot;&gt;this page says&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Voyager 1 has crossed into the heliosheath and is leaving the solar system, rising above the ecliptic plane at an angle of about 35 degrees at a rate of about 520 million kilometers (about 320 million miles) a year. (Voyager 1 entered interstellar space on August 25, 2012.) Voyager 2 is also headed out of the solar system, diving below the ecliptic plane at an angle of about 48 degrees and a rate of about 470 million kilometers (about 290 million miles) a year.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was able to find some good information &lt;a href=&quot;http://spectrum.ieee.org/aerospace/robotic-exploration/plotting-the-destinations-of-4-interstellar-probes&quot;&gt;here&lt;/a&gt;. &lt;em&gt;Voyager 1&lt;/em&gt; is headed towards &lt;a href=&quot;https://en.wikipedia.org/wiki/Gliese_445&quot;&gt;Gliese 445&lt;/a&gt;. However, it will be 1.6 light-years away at its closest approach - not exactly close by! It will reach that star, 17.6 light-years away, in 40,000 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Voyager 2&lt;/em&gt; will go by &lt;a href=&quot;https://en.wikipedia.org/wiki/Ross_248&quot;&gt;Ross 248&lt;/a&gt;, a red dwarf currently 10 light-years away. The craft will take 40,000 years to get there. 256,000 years later, &lt;em&gt;Voyager 2&lt;/em&gt; will come within 4.3 light-years of &lt;a href=&quot;https://en.wikipedia.org/wiki/Sirius#Visibility&quot;&gt;Sirius&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probes will have absolutely no power by that time.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-12-06T22:07:32.327" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="8163" PostTypeId="2" ParentId="8159" CreationDate="2014-12-06T23:56:42.553" Score="0" Body="&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;: If $f(\mathbf{x},\mathbf{v},t)$ is the distribution function of the stars in phase space and $n = \int f\,\mathrm{d}^3\mathbf{v}$ is the star density, then in the cylindrical coordinates $(R,\varphi,z)$, the $z$-component of the second of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Jeans_equations&quot; rel=&quot;nofollow&quot;&gt;Jeans equations&lt;/a&gt; applied to an axisymmetric system is&#xA;$$\renewcommand{\expv}[1]{\langle{#1}\rangle}\partial_t(n\expv{v_z}) + \partial_R(n\expv{v_Rv_z}) + \partial_z(n\expv{v_z^2}) + \frac{n\expv{v_Rv_z}}{R} + n\partial_z\Phi = 0\text{.}$$&#xA;In a steady state, the first term vanishes, so if the positive and negative terms of the $z$-component of velocity terms balance, then so do the second and fourth, in which case:&#xA;$$\partial_z(n\expv{v_z^2}) = -n\partial_z\Phi\text{,}$$&#xA;which we can then integrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the the star and mass distributions are axisymmetric and symmetric about the equatorial ($z = 0$) plane, then the potential has even symmetry in $z$, $\Phi(R,z) = \Phi(R,-z)$, and so does $n$. Thus their derivatives with respect to $z$ are odd functions, and $\Phi_{,z}(R,z) = -\Phi_{,z}(R,-z)$ implies&#xA;$$\frac{1}{n}\int_{-z}^z n\Phi_{,z'}\,\mathrm{d}z' = 0\text{,}$$&#xA;since the integrand has odd symmetry. Hence,&#xA;$$\frac{1}{n}\int_z^\infty n\Phi_{,z'}\,\mathrm{d}z' = \frac{1}{n}\int_{|z|}^\infty n\Phi_{,z'}\,\mathrm{d}z'\text{.}$$&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-12-06T23:56:42.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8164" PostTypeId="5" CreationDate="2014-12-07T00:04:03.020" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-12-07T00:04:03.020" LastActivityDate="2014-12-07T00:04:03.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8165" PostTypeId="4" CreationDate="2014-12-07T00:04:03.020" Score="0" Body="Questions about spacecraft journeying to other stellar systems." OwnerUserId="2674" LastEditorUserId="2674" LastEditDate="2014-12-07T14:43:12.123" LastActivityDate="2014-12-07T14:43:12.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8166" PostTypeId="1" AcceptedAnswerId="8170" CreationDate="2014-12-07T03:51:41.717" Score="2" ViewCount="73" Body="&lt;p&gt;I am doing some calculations, and I wanted to know that after using the gravity equation, $$F = G \frac{m_1m_2}{r^2}$$ if you can use $F$ and convert it into a speed, such as km/year, and also a direction or angle from the original direction. I am just looking into astronomy, and any help is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2996" LastEditorUserId="2153" LastEditDate="2014-12-07T14:42:57.487" LastActivityDate="2014-12-07T14:42:57.487" Title="How to convert a gravitational force to speed and direction" Tags="&lt;orbit&gt;&lt;gravity&gt;&lt;solar-system&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8167" PostTypeId="1" AcceptedAnswerId="8168" CreationDate="2014-12-07T04:32:32.677" Score="2" ViewCount="1015" Body="&lt;p&gt;Inspired by this &lt;a href=&quot;https://astronomy.stackexchange.com/questions/8100/how-does-the-earth-move-in-the-sky-as-seen-from-the-moon&quot;&gt;question&lt;/a&gt;. I am curious whether earth, besides being nearly fixed on one place on the moon's sky, is it visible during the day-time on moon too?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that earth should be visible as moon has no atmosphere. Also, if the NASA didn't edit the following photograph, it suggests that the day-time sky of moon is all black and earth should be visible in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TJNET.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think, Stellarium don't take into account the atmosphere once you are on other planet.&lt;/p&gt;&#xA;" OwnerUserId="1842" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-05-24T15:50:30.787" Title="Visibility of earth from moon during day-time of moon" Tags="&lt;the-moon&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8168" PostTypeId="2" ParentId="8167" CreationDate="2014-12-07T04:58:15.170" Score="2" Body="&lt;p&gt;Yes, the Earth would be visible even during day on the Moon. Without a (significant) atmosphere diffusing light, the day sky would be much like the night sky on Earth. If you were to look straight up such that you couldn't see the ground, you may not even notice that it was day at all (save for the Sun shining down at you).  This could actually be rather hazardous as the Sun would be more intense, but your eyes may not be adjusted to high light like they are on bright Earth days.&lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="1288" LastEditDate="2014-12-07T05:08:56.363" LastActivityDate="2014-12-07T05:08:56.363" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8169" PostTypeId="1" CreationDate="2014-12-07T07:39:18.473" Score="10" ViewCount="534" Body="&lt;p&gt;I read that solar flares are customarily viewed in H-alpha light, as a temporary brightening of a small portion of chromosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What all can be interpreted from this? Is it because, energy of the radiation contained by the flare lies around this wavelength? And why chromosphere ?&lt;/p&gt;&#xA;" OwnerUserId="2823" LastEditorUserId="661" LastEditDate="2014-12-07T10:15:53.607" LastActivityDate="2014-12-08T21:21:08.867" Title="Typical wavelength of solar flare" Tags="&lt;the-sun&gt;&lt;solar-flare&gt;&lt;spectra&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8170" PostTypeId="2" ParentId="8166" CreationDate="2014-12-07T10:07:34.000" Score="1" Body="&lt;p&gt;You must apply Newton's law&#xA;$$ m\frac{\mathrm{d}\boldsymbol{v}}{\mathrm{d}t} = \boldsymbol{F}$$&#xA;which related the force $\boldsymbol{F}$ to the acceleration (=change of velocity). Note that positions, velocities, and forces are all &lt;strong&gt;vector&lt;/strong&gt; quantities. Also note that as the objects move (change position), their mutual forces change (both direction and strength).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an enormous amound of literature on the resulting &lt;strong&gt;orbits&lt;/strong&gt; for celestial bodies, going all the way back to Kepler and Newton. I suggest you look into elementary books or start with some Wikipedia articles.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-12-07T10:07:34.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8171" PostTypeId="2" ParentId="8169" CreationDate="2014-12-07T10:10:41.673" Score="1" Body="&lt;p&gt;In the transition from a higher electron energy level to a lower one, say $m\mapsto n$, a hydrogen atom emits a photon of wavelength $\lambda$ satisfying&#xA;$$\frac{1}{\lambda} = R_\infty\left[\frac{1}{n^2}-\frac{1}{m^2}\right]\text{,}$$&#xA;where $R_\infty = 1.09737315685\,\mathrm{m}^{-1}$ is the Rydberg constant. For $n=1$, i.e. the destination energy level is the ground state, varying $m$ forms the &lt;em&gt;Lyman series&lt;/em&gt;: $\text{Ly}_\alpha$ ($2\mapsto 1$), $\text{Ly}_\beta$ ($3\mapsto 1$), $\text{Ly}_\gamma$ ($4\mapsto 1$), etc. The $n=2$ destination energy level forms the &lt;em&gt;Balmer series&lt;/em&gt;: $\text{Ba}_\alpha$ ($3\mapsto 2$), $\text{Ba}_\beta$ ($5\mapsto 2$), etc., which was actually the first series to be discovered, and is frequently labeled simply with hydrogen instead.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What all can be interpreted from this? Is it because, energy of the radiation contained by the flare lies around this wavelength? And why chromosphere ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A solar flare is a very hot, violent event that radiates energy across the electromagnetic spectrum. The importance of the H-α line is due to the conveniences of observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spectral lines of hydrogen are outside the visible band &lt;em&gt;except&lt;/em&gt; for the first four of the Balmer series, from the red H-α line to the violet H-δ line. When an hydrogen ion and an electron recombine into an atom, the result is generally a hydrogen atom in an excited state. Eventually, it decays to the ground state, but it doesn't have to transition directly there, and typically does so in a random sequence of transitions. A very sizable fraction of those transitions, however, include the $3\mapsto 2$ jump that produces the H-α line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the presence of the H-α line is an easy way to identify ionized hydrogen, and in particular, a sudden brightening of the H-α line in an emission-line spectrum is an indicator that something energetic is happening to ionize the hydrogen (moreso than usual, that is). And that's where the chromosphere, the low-density &quot;atmosphere&quot; surrounding the Sun, comes in: it has an emission-line spectrum, i.e., its spectrum is bright in narrow bands that correspond to its atomic or molecular composition. This is unlike the photosphere, which has an absorption-line spectrum instead.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-12-07T10:10:41.673" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8172" PostTypeId="1" AcceptedAnswerId="8366" CreationDate="2014-12-07T14:43:03.237" Score="5" ViewCount="597" Body="&lt;p&gt;The well-known &lt;strong&gt;galaxy rotation problem&lt;/strong&gt; is described &lt;a href=&quot;http://en.wikipedia.org/wiki/Galaxy_rotation_curve&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. Basically (as one moves outward from the galactic centre, $R$ increasing) the &lt;strong&gt;observed&lt;/strong&gt; pattern of variation in orbital speed $V$ is very different from the pattern &lt;strong&gt;expected&lt;/strong&gt; on the basis of the distribution of mass inferred from detectable radiation. This is illustrated by the two curves in the following figure.&#xA;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/e/ec/M33_rotation_curve_HI.gif/420px-M33_rotation_curve_HI.gif&quot; alt=&quot;galactic rotation curve&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Feng and Gallo&lt;/strong&gt; in several papers such as &lt;a href=&quot;http://arxiv.org/abs/1007.3778&quot; rel=&quot;nofollow noreferrer&quot;&gt;this 2010 article&lt;/a&gt; have proposed a method for modelling mass distributions in discoidal galaxies.  The method can be applied to an individual galaxy to produce a &quot;tailor-made&quot; mass distribution which (by applying basic Newtonian physics) predicts a galactic rotation curve matching the curve observed for the galaxy. They explain the failure of other attempts, using &quot;ordinary&quot; physics, to predict observed galactic rotations on the use of crude luminosity:mass assumptions and on the inappropriate application of Keplerian models to galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Mysterious&quot; explanations of the galaxy rotation problem such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Dark_matter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dark Matter&lt;/a&gt;  and &lt;a href=&quot;http://en.wikipedia.org/wiki/Modified_Newtonian_dynamics&quot; rel=&quot;nofollow noreferrer&quot;&gt;MOND&lt;/a&gt; are still being considered seriously. Presumably therefore Feng and Gallo's ideas are not widely accepted.  What are the reasons for the rejection of their explanations?&lt;/p&gt;&#xA;" OwnerUserId="233" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:47.267" LastActivityDate="2018-10-19T16:09:34.537" Title="What are the arguments against the Feng and Gallo thin disk explanation of galactic rotation curves?" Tags="&lt;gravity&gt;&lt;galactic-dynamics&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8173" PostTypeId="2" ParentId="874" CreationDate="2014-12-07T23:00:29.820" Score="-4" Body="&lt;p&gt;Even we could try to calculate the distance of a center, it would change every per seconds of the lights speed away that would always make the calculation inaccurate at all time.. It is really hard to predict why, what, how the unlimited (as we know) space of the universe evolved and is being formed and will happen. &#xA;However its spectacular feeling to make hypothesis such things that makes our brains immortal to create our modern earth even more interesting place.. &#xA;But first we should understand how to get along with our own planet and become more sensitive with it.. &#xA;As it is told everywhere we look we see the &lt;strong&gt;past&lt;/strong&gt; which includes the newly discoveries (as we say) like galaxies, orphan or solar planets from the hubble. &#xA;I am just wondering what if there are more explosions supernovas or someothers that creates meteors that  are routed on us from these explosions and maybe we can have a chance to see it 20-30 years later(the most positive elapsed time). &#xA;What I am saying we might be discovering stars that could be exploded by an other star already a billion years ago, on the other hand we observe these discoveries as our current UTC or GMT time coordinate, which makes no sense on this time theorem of the space ?!?&lt;/p&gt;&#xA;" OwnerUserId="2997" LastEditorUserId="2997" LastEditDate="2014-12-07T23:08:21.053" LastActivityDate="2014-12-07T23:08:21.053" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8174" PostTypeId="1" AcceptedAnswerId="8353" CreationDate="2014-12-07T23:15:38.853" Score="6" ViewCount="446" Body="&lt;p&gt;I use MESA right now to play around with stellar dynamics, but it can't do mass ejections and therefore, I can't think of a way to let the simulated stars explode as supernovae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What simulation software would I use for these?&lt;/p&gt;&#xA;" OwnerUserId="422" LastActivityDate="2015-01-22T23:23:47.233" Title="Is there a star simulation software that can handle mass ejections and supernovae?" Tags="&lt;supernova&gt;&lt;stellar-evolution&gt;&lt;stellar-astrophysics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8175" PostTypeId="1" AcceptedAnswerId="8197" CreationDate="2014-12-08T01:45:17.110" Score="1" ViewCount="92" Body="&lt;p&gt;There is a region called &lt;a href=&quot;http://en.wikipedia.org/wiki/Baade%27s_Window&quot; rel=&quot;nofollow&quot;&gt;Baade's window&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember there is a similar region called &quot;somebody's window&quot;, which also has a low amount of dust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know the name?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="2153" LastEditDate="2014-12-09T21:51:32.120" LastActivityDate="2014-12-09T22:07:09.860" Title="The name of a particular low extinction region" Tags="&lt;dust&gt;&lt;extinction&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8177" PostTypeId="1" AcceptedAnswerId="8189" CreationDate="2014-12-08T11:23:54.520" Score="13" ViewCount="2068" Body="&lt;p&gt;While reading reports about the &lt;a href=&quot;http://www.nasa.gov/newhorizons/on-plutos-doorstep-new-horizons-spacecraft-awakens-for-encounter/#.VIVccGOy7qs&quot; rel=&quot;noreferrer&quot;&gt;New Horizons misson&lt;/a&gt;, I noticed an odd vertical, black stripe in the images of Pluto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/83fw7.jpg&quot; alt=&quot;Hubble Discovers a Fifth Moon Orbiting Pluto&quot;&gt;&#xA;&lt;em&gt;Source: &lt;a href=&quot;http://www.nasa.gov/mission_pages/hubble/science/new-pluto-moon.html&quot; rel=&quot;noreferrer&quot;&gt;Hubble Discovers a Fifth Moon Orbiting Pluto&lt;/a&gt; (07.11.12). Credit: NASA; ESA; M. Showalter, SETI Institute&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is that black stripe there?&lt;/p&gt;&#xA;" OwnerUserId="3001" LastEditorUserId="3007" LastEditDate="2014-12-09T01:11:55.857" LastActivityDate="2014-12-09T01:11:55.857" Title="Why is there a black stripe in the Hubble images of Pluto?" Tags="&lt;photography&gt;&lt;hubble-telescope&gt;&lt;pluto&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8178" PostTypeId="2" ParentId="8177" CreationDate="2014-12-08T13:17:26.370" Score="5" Body="&lt;p&gt;I'm pretty sure that the stripe is there because the two sections of the image have different contrasts.  Pluto and Charon are so bright relative to Nix, Hydra, P4 and P5 that to have one contrast, you'd either (a) not be able to see the smaller four, or (b) totally wash out the image with the light from the brighter two.  So we simply superimpose two images at different brightnesses (or gains, I guess).&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2014-12-08T13:17:26.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8179" PostTypeId="2" ParentId="8177" CreationDate="2014-12-08T13:17:58.857" Score="7" Body="&lt;p&gt;Pluto itself so bright that Charon would not be visible in the image if it were exposed in a way to show the remaining moons. Likewise, the remaining moons as so faint as to not be visible in an image that resolves Charon. Thus, the photo that you see is a composite of two image processing techniques: one designed to showcase Charon and one designed to showcase the remaining moons.&lt;/p&gt;&#xA;" OwnerUserId="209" LastActivityDate="2014-12-08T13:17:58.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8180" PostTypeId="1" AcceptedAnswerId="8188" CreationDate="2014-12-08T15:23:20.093" Score="3" ViewCount="1510" Body="&lt;p&gt;There are some very vast spinning objects out there.   How high can the eccentricity get, or, which I think is easier to understand, how small can the ratio between the polar and equatorial diameter be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For clarification, I'm asking about objects composed composed of particles (not some boundary or field etc) and that are devoid of non-microscopic empty spaces. I.e. a planet or star etc; not a galaxy or rings or a heliosphere etc. &#xA;For the purpose of this question, I just don't find galaxies or nebulae interesting because they can have almost any shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm talking about structures that are oblate spheroids essentially.&lt;/p&gt;&#xA;" OwnerUserId="1106" LastEditorUserId="1106" LastEditDate="2014-12-11T16:17:25.550" LastActivityDate="2014-12-11T16:17:25.550" Title="What is the most oblate astronomical object known?" Tags="&lt;star&gt;&lt;gravity&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8181" PostTypeId="2" ParentId="8180" CreationDate="2014-12-08T17:24:45.633" Score="1" Body="&lt;p&gt;I think it is going to depend on the type of object being examined. Planetary objects tend to be more spherical than stars, which in turn tend to be more spherical than galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My vote goes toward the &lt;a href=&quot;https://en.wikipedia.org/wiki/Heliospheric_current_sheet&quot; rel=&quot;nofollow&quot;&gt;Heliospheric current sheet&lt;/a&gt;, which is thought to extend 10-20 astronomical units (about 1.5x10^9km to 3x10^10km) from the Sun, and is thought to be about 10,000 km near the orbit of the Earth. But it tends to be influenced more by electromagnetic forces, than due to spin forces.&lt;/p&gt;&#xA;" OwnerUserId="2880" LastActivityDate="2014-12-08T17:24:45.633" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8182" PostTypeId="2" ParentId="6435" CreationDate="2014-12-08T17:38:38.450" Score="2" Body="&lt;p&gt;Jodrell Bank's 76m Lovell Telescope is sometimes &lt;a href=&quot;http://www.jb.man.ac.uk/news/2014/WatchtheSkies/&quot; rel=&quot;nofollow&quot;&gt;used as a movie screen&lt;/a&gt;, having shown 2001: A Space Odyssey.&lt;/p&gt;&#xA;" OwnerUserId="2880" LastActivityDate="2014-12-08T17:38:38.450" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8183" PostTypeId="2" ParentId="8177" CreationDate="2014-12-08T18:45:14.540" Score="0" Body="&lt;p&gt;Without knowledge of this particular image, I'm pretty sure it is because of the use of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Coronagraph&quot; rel=&quot;nofollow&quot;&gt;Coronagraph&lt;/a&gt;. One of a set of different methods  to block or divert or phase shift or in some other way get rid of unwanted light reflected from Pluto and Charon which would blind the light detector from the much weaker light reflected by its tiny moons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.stsci.edu/hst/stis/strategies/pushing/coronagraphic_bars&quot; rel=&quot;nofollow&quot;&gt;This link&lt;/a&gt; about the coronagraph on Hubble is way beyond me, but maybe it is helpful to someone:&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2014-12-08T18:45:14.540" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8184" PostTypeId="2" ParentId="8175" CreationDate="2014-12-08T19:56:48.327" Score="1" Body="&lt;p&gt;I think you are talking about &lt;strong&gt;Plaut's window&lt;/strong&gt;, which is located at &lt;code&gt;l=0°&lt;/code&gt; and &lt;code&gt;b=-8°&lt;/code&gt;. It has a reddening of &lt;code&gt;E(B-V)=0.25&lt;/code&gt; magnitudes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;HTH,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Germán.&lt;/p&gt;&#xA;" OwnerUserId="2339" LastActivityDate="2014-12-08T19:56:48.327" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8185" PostTypeId="1" CreationDate="2014-12-08T20:36:15.737" Score="-2" ViewCount="237" Body="&lt;p&gt;Lack of contact could be due to one of three things ?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Interstellar space travel is just not possible .&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;There is no other Alien Life in the Galaxy .&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There exists Alien life and interstellar space travel is possible but with tens of billions of Planets in our Galaxy, the odds of Aliens detecting Earth is like picking the winning lottery numbers .&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Which of these three is the correct answer ?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastEditorUserId="2153" LastEditDate="2014-12-08T23:28:05.730" LastActivityDate="2015-01-05T22:47:57.507" Title="Lack of contact with Aliens" Tags="&lt;life&gt;&lt;extra-terrestrial&gt;&lt;fermi-paradox&gt;" AnswerCount="2" CommentCount="5" ClosedDate="2014-12-09T20:56:07.797" ContentLicense="CC BY-SA 3.0" />
  <row Id="8186" PostTypeId="2" ParentId="8148" CreationDate="2014-12-08T21:01:32.287" Score="0" Body="&lt;p&gt;I propose the &quot;rare Earth&quot; theory. Life could have existed in the past or could exist again in the future because if it happened here why could it not happen again ? Maybe at this point in time we are all there is. It would be very strange to imagine that once we are gone that life will never exist again.&lt;/p&gt;&#xA;" OwnerUserId="2562" LastActivityDate="2014-12-08T21:01:32.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8187" PostTypeId="2" ParentId="8169" CreationDate="2014-12-08T21:21:08.867" Score="3" Body="&lt;p&gt;Solar flares are observed at wavelengths right across the electromagnetic spectrum, not just H alpha.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic model for a solar flare starts with the magnetic field in the corona. You can think of the topology of the magnetic field to consist of loops that poke up out of the photosphere and extend into the corona. However, the photosphere of the Sun is turbulent and constantly in motion due to convection and differential rotation. Whilst a loop may be formed in a minimum energy state, it can get twisted and stressed by these motions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some point an instability is reached and the magnetic field can undergo a &quot;reconnection&quot; event, to flip back to a lower energy configuration. During this event, charged particles are accelerated and travel down the magnetic field lines towards the photosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before they get there, they encounter the chromosphere, which is where the bulk of the particle kinetic energy is deposited. i.e. the density increases as you go down towards the photosphere and once a certain column density is reached, the accelerated electrons are stopped and deposit their kinetic energy. This results in heating and excess H alpha emission from material at about 10 thousand kelvin in the flare footpoints. Any hotter than this and all the hydrogen is ionised. The H alpha is in emission because the only material above it is optically thin to the H alpha radiation. There is hotter, ionised material produced too, and much of this is evaporated such that it fills the magnetic loops with X-ray emitting plasma at temperatures of more than a million kelvin. Some of the flare energy may also be used to accelerate material away from the Sun in &quot;coronal mass ejections&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2014-12-08T21:21:08.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8188" PostTypeId="2" ParentId="8180" CreationDate="2014-12-08T22:00:34.420" Score="4" Body="&lt;p&gt;Saturn is the most oblate planet in the solar system. If the equatorial diameter is &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; and the polar diameter is &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; then its oblateness, &lt;span class=&quot;math-container&quot;&gt;$(a-b)/a = 0.1$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;We do not know the oblateness values for more than one or two exoplanets and even these are somewhat uncertain, but are thought to be lower than Saturn's value. For example &lt;a href=&quot;http://adsabs.harvard.edu/abs/2010ApJ...709.1219C&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2010ApJ...709.1219C&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The oblateness of a star depends on the ratio of it gravitational acceleration to centrifugal acceleration at the equator. So the most oblate stars ought to be the ones with big radii and fast rotation rates (i.e. types of stars that would have low surface gravities if non-rotating).&lt;/p&gt;&#xA;&lt;p&gt;Giant stars come in two basic types - red giants, which are evolved intermediate and low-mass stars. These tend to be slowly rotating, though there is a class called FK Com variables which are fast-rotating red giants.&lt;/p&gt;&#xA;&lt;p&gt;The other class is the blue giants, which are high mass main sequence, or slightly evolved, stars.&lt;/p&gt;&#xA;&lt;p&gt;For maximum oblateness you want the star with the highest value of:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ f = \frac{R \omega^2}{GM/R^2} = \frac{R^3 \omega^2}{GM}$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is the stellar mass and &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; its angular velocity. Often we don;t know &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt;, but can estimate the rotation velocity at the equator &lt;span class=&quot;math-container&quot;&gt;$v=R\omega$&lt;/span&gt;. Hence&#xA;&lt;span class=&quot;math-container&quot;&gt;$$f = \frac{R^3 v^2}{GMR^2} = \frac{Rv^2}{GM}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;A possible record holder is the O-type star &lt;a href=&quot;http://www.eso.org/public/news/eso1147/&quot; rel=&quot;nofollow noreferrer&quot;&gt;VTFS 102&lt;/a&gt; in the Large Magellanic Cloud.&#xA;This probably has a mass of &lt;span class=&quot;math-container&quot;&gt;$25M_{\odot}$&lt;/span&gt; a radius of about &lt;span class=&quot;math-container&quot;&gt;$10R_{\odot}$&lt;/span&gt; and has a measured rotation speed of &lt;span class=&quot;math-container&quot;&gt;$\geq 600$&lt;/span&gt; km/s. Thus its value of &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt; is 0.75, compared with a value of &lt;span class=&quot;math-container&quot;&gt;$f=0.15$&lt;/span&gt; for Saturn.&lt;/p&gt;&#xA;&lt;p&gt;EDIT: Due to Michael B - what about &amp;quot;millisecond&amp;quot; pulsars? The shortest periods are about 1.4 ms, the radii are about 10 km and the masses are likely to be about 1.4&lt;span class=&quot;math-container&quot;&gt;$M_{\odot}$&lt;/span&gt;. Do the sums and you find &lt;span class=&quot;math-container&quot;&gt;$f=0.1$&lt;/span&gt;. So, (at least in Newtonian mechanics) they should be no more oblate than Saturn.&lt;/p&gt;&#xA;&lt;p&gt;A further contender might by the dwarf planet Haumea that orbits at 40-50 au from the Sun. It is thought large enough to have achieved its shape via hydrostatic equilibrium between gravity and internal pressures (and rotation). Its mass is &lt;span class=&quot;math-container&quot;&gt;$4\times10^{21}$&lt;/span&gt; kg, (mean) radius is 700 km and it has a rotation period of 3.91 hours. These numbers give &lt;span class=&quot;math-container&quot;&gt;$f=0.25$&lt;/span&gt;. It has been modelled using this, and matching its light curve, to be a triaxial ellipsoid (rather than oblate) with axis ratios 2 : 1.5 : 1 (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2006ApJ...639.1238R&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2006ApJ...639.1238R&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2014-12-10T22:11:25.867" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8189" PostTypeId="2" ParentId="8177" CreationDate="2014-12-08T23:59:22.677" Score="8" Body="&lt;p&gt;That photograph is a composite of two images taken with different &lt;strong&gt;exposure times&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;To be correct we'd have to say that the exposure of the two photographs is different, i.e. the outer photo was created by absorbing more light. In this case we can assume that the focal ratio (derived from Hubble's lens aperture) and the luminance of the scene (how much light is travelling in the lens' direction) are identical for both photographs, which leaves only the exposure time as a free variable when it comes to determining &lt;a href=&quot;http://en.wikipedia.org/wiki/Exposure_%28photography%29&quot; rel=&quot;noreferrer&quot;&gt;exposure&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;This is necessary because we're photographing objects with very different brightness. For Pluto to show up a relatively short exposure time is required, but its moons reflect much less light and would need a longer exposure time to visible. As long as the sensor is exposed, Pluto would continue to increase in brightness to the point that it becomes washed out. Objects that are significantly brighter become over-exposed resulting in a loss of detail and fidelity, known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Clipping_%28photography%29&quot; rel=&quot;noreferrer&quot;&gt;blown-out highlights&lt;/a&gt; in photography. In our case Pluto would turn into a solid white dot compared to the more detailed picture that is now possible. You can draw a parallel with false color images rendered from infrared: this composite is not what the human eye would see if it was capable of picking up this level of light and detail.&lt;/p&gt;&#xA;&lt;p&gt;In another Hubble image, NASA explained the reason why composite imaging is used:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;This is a composite image because a single exposure of the stellar background, comet Siding Spring, and Mars would be problematic. Mars actually is 10,000 times brighter than the comet, so it could not be properly exposed to show detail in the Red Planet. The comet and Mars also were moving with respect to each other and could not be imaged simultaneously in one exposure without one of the objects being motion blurred. Hubble had to be programmed to track on the comet and Mars separately in two different observations.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Source: &lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2014/23oct_marscomet/&quot; rel=&quot;noreferrer&quot;&gt;Hubble Sees Comet Next to Mars&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Very long exposure times are often necessary since relatively little light is coming our way from distant planets and stars. As the Hubble website explains for its &lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble_Deep_Field&quot; rel=&quot;noreferrer&quot;&gt;Deep Fields&lt;/a&gt; images:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Hubble has made a series of very deep observations taken in very dark parts of the sky. Like using a long exposure on a digital camera, these long exposure shots (up to several weeks) reveal very faint details that are not normally visible in shorter exposures.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Source: &amp;quot;What are the Hubble Deep Fields?&amp;quot;, &lt;a href=&quot;http://www.spacetelescope.org/about/faq/&quot; rel=&quot;noreferrer&quot;&gt;Spacetelescope.org FAQ&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Wikipedia summarises a paper by &lt;a href=&quot;http://arxiv.org/abs/astro-ph/9607174&quot; rel=&quot;noreferrer&quot;&gt;Robert E. Williams and the HDF team&lt;/a&gt;, &amp;quot;The Hubble Deep Field: Observations, Data Reduction, and Galaxy Photometry&amp;quot; as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Between December 18 and December 28, 1995—during which time Hubble orbited the Earth about 150 times—342 images of the target area in the chosen filters were taken. The total exposure times at each wavelength were 42.7 hours (300 nm), 33.5 hours (450 nm), 30.3 hours (606 nm) and 34.3 hours (814 nm), divided into 342 individual exposures to prevent significant damage to individual images by cosmic rays, which cause bright streaks to appear when they strike CCD detectors. A further 10 Hubble orbits were used to make short exposures of flanking fields to aid follow-up observations by other instruments.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble_Deep_Field#Observations&quot; rel=&quot;noreferrer&quot;&gt;Hubble Deep Field&lt;/a&gt;, Wikipedia, retrieved 2014-12-09&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3007" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2014-12-08T23:59:22.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8190" PostTypeId="5" CreationDate="2014-12-09T03:57:28.703" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-12-09T03:57:28.703" LastActivityDate="2014-12-09T03:57:28.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8191" PostTypeId="4" CreationDate="2014-12-09T03:57:28.703" Score="0" Body="Questions on any of the 5 locations where a small object can maintain a stable orbital configuration with respect to two co-orbiting larger bodies." OwnerUserId="2674" LastEditorUserId="2674" LastEditDate="2014-12-10T08:53:34.227" LastActivityDate="2014-12-10T08:53:34.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8192" PostTypeId="1" CreationDate="2014-12-09T04:11:45.237" Score="14" ViewCount="1093" Body="&lt;p&gt;The usual examples of Lagrange points one most commonly encounters, Sun-Earth  and Earth-Moon Lagrange points, are examples of 3-body problems where $M_1\gg M_2\gg M_3$. The Pluto-Charon system, however, are much closer in their relative masses, so much so that their barycenter is outside Pluto's surface. From &lt;a href=&quot;https://en.wikipedia.org/wiki/Pluto&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pluto and Charon are sometimes considered a binary system because the barycenter of their orbits does not lie within either body. The IAU has not formalized a definition for binary dwarf planets, and Charon is officially classified as a moon of Pluto.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How does this affect the orbital stability of the five Pluto-Charon Lagrange points?&lt;/p&gt;&#xA;" OwnerUserId="2674" LastEditorUserId="2153" LastEditDate="2016-11-19T15:36:57.183" LastActivityDate="2018-05-21T20:54:11.060" Title="Do Pluto and Charon have unusual Lagrange points?" Tags="&lt;pluto&gt;&lt;lagrange-point&gt;&lt;charon&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8193" PostTypeId="2" ParentId="8185" CreationDate="2014-12-09T14:35:24.733" Score="3" Body="&lt;p&gt;I think you're question is incorrect, and what you should be asking is &quot;Which of these is plausible&quot;, in which case, all of them are. I think you are also missing other options:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;There is alien life, who have already visited, but have not made formal contact with us, for any one of a number of reasons.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We are the alien life, who has forgotten how we arrived here, and doesn't know how we get back.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Some of us are alien life, and we just keep a low profile.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2880" LastActivityDate="2014-12-09T14:35:24.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8194" PostTypeId="2" ParentId="3621" CreationDate="2014-12-09T15:28:29.103" Score="1" Body="&lt;p&gt;We can't see how those galaxies look right now. If a galaxy appears to be 10 billion light years away, that also means that the light took 10 billion years to reach us. It's a bit confusing that &quot;distance&quot; and &quot;time&quot; are sometimes the same thing in astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the light which the galaxies emit today (if they still exist) will reach us in another 10 billion years. What we see right now is light that was emitted 10 billion years in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also means that by looking at the farthest possible distance, we can see light which was created shortly after the big bang. One such source of light is the ubiquitous &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow&quot;&gt;background radiation&lt;/a&gt; which was probably created by the big bang itself. It's easy to see because it fills all the gaps between the celestial objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, this also means that we always get an &quot;outdated&quot; view of the universe. If aliens started to blowing stars at the far side of our galaxy, it would take us 100'000 years to see the light.&lt;/p&gt;&#xA;" OwnerUserId="3001" LastEditorUserId="3001" LastEditDate="2014-12-09T15:33:34.520" LastActivityDate="2014-12-09T15:33:34.520" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8195" PostTypeId="2" ParentId="8185" CreationDate="2014-12-09T17:26:07.473" Score="-1" Body="&lt;p&gt;Perhaps alien contact is so mundane and commonplace that you dismiss it for lack of congruity with Hollywood presentations. Documentation of visitations and encounters is as old as writing and history. I'll give you just one example, among many. I like to read regularly from a collection of accounts which for some cultures has been around for so long, and has been so central to these cultures, that they just call it The Book, or The Holy Book. They began calling it this in Latin and Greek, and over time, that word has morphed into the word bible. This &quot;Holy Bible&quot; as these human cultures call it describes interactions of various kinds with a super-being who claims to have created this entire world, and us. It also describes interactions with other beings possessed of various great powers (technologies perhaps) who certify that they represent this same super-being. All of these accounts are given in the most sober and earnest language by men who were known by their peers to be very honest and forthright.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astonishingly, these texts even describe detailed instructions for communicating with this super-being using nothing more than the biology that this very super-being engineered into our bodies. Over the centuries, many of your own ancestors may have even written of their own encounters with this &quot;God&quot;. I know mine have. In fact, if you read these texts, you can make direct contact yourself by following the instructions, and certify for yourself that His claims, as wild as they may seem initially, are in fact true. The copyright of His creation is, in a sense, stamped right into the nature of the created in that this communication mechanism which He built into our nature does in fact function as described. Amazing, but true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But anyway, are you sure that Astronomy is the right stack exchange for this inquiry?&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditDate="2015-01-05T22:47:57.507" LastActivityDate="2015-01-05T22:47:57.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8197" PostTypeId="2" ParentId="8175" CreationDate="2014-12-09T21:48:19.310" Score="2" Body="&lt;p&gt;The Lockman hole?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lockman_Hole&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lockman_Hole&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has a column density of neutral hydrogen of $6\times10^{19}$ cm$^{-2}$, which will correspond to very little dust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two others I've heard of towards the Galactic bulge are known as the Sgr I and Sgr II windows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I offer &quot;Stanek's window&quot; towards the bulge, which has been referred to in a number of the Chandra Galactic plane X-ray surveys but appears to have no refereed publication associated with its discovery. I found this on arXiv.  &lt;a href=&quot;http://arxiv.org/abs/astro-ph/9802307&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/astro-ph/9802307&lt;/a&gt;  It is at $l=0.25$, $b=-2.15$ and has $E(B-V) \sim 0.6$.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2014-12-09T21:54:08.680" LastActivityDate="2014-12-09T21:54:08.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8198" PostTypeId="2" ParentId="8175" CreationDate="2014-12-09T22:07:09.860" Score="2" Body="&lt;p&gt;Puppis Window?&#xA;(&lt;a href=&quot;http://adsabs.harvard.edu/full/1982A%26AS...50..377S&quot; rel=&quot;nofollow&quot;&gt;Neutral hydrogen observations towards the Puppis Window of the Milky Way&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="1142" LastActivityDate="2014-12-09T22:07:09.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8199" PostTypeId="2" ParentId="7823" CreationDate="2014-12-10T00:47:37.723" Score="3" Body="&lt;p&gt;This is a part answer to your question, based on some recent research of the photochemical behaviour modelled and observed for Titan's tholin haze and modelling of Titan's stratosphere.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process appears to begin, according to the paper &lt;a href=&quot;http://adsabs.harvard.edu/abs/2012DPS....4431210B&quot; rel=&quot;nofollow&quot;&gt;Ice condensation layers in Titan’s Stratosphere&lt;/a&gt; (Barth, 2012) (Abstract only - paywalled), with&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Photochemical destruction of methane along with the destruction of nitrogen molecules from energetic electrons in Titan’s upper atmosphere result in the production of a number of hydrocarbon and nitrile compounds which may be capable of condensing at the colder temperatures of Titan’s lower stratosphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then, according to the paper &lt;a href=&quot;http://yly-mac.gps.caltech.edu/Titan/Imanaka%202004.pdf&quot; rel=&quot;nofollow&quot;&gt;Laboratory experiments of Titan tholin formed in cold plasma at various&#xA;pressures: implications for nitrogen-containing polycyclic aromatic&#xA;compounds in Titan haze&lt;/a&gt; (Imanaka et al. 2004), in particular in reference to Titan's stratosphere, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;. In the stratosphere (100–&#xA;  300 km), further chemical reactions are induced by the catalytic&#xA;  $CH_4$ dissociation by such molecules as $C_2H_2$ and&#xA;  $C_4H_2$ absorbing the long UV (&gt; 155 nm) irradiation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The significance of these UV absorbing molecules is explained in the article &lt;a href=&quot;http://www.nature.com/ncomms/journal/v4/n4/full/ncomms2649.html&quot; rel=&quot;nofollow&quot;&gt;Photochemical activity of Titan’s low-altitude condensed haze&lt;/a&gt; (Gudipati et al. 2013) (Abstract only - paywalled), they state that tholin haze could form on condensed aerosols in Titan's atmosphere, demonstrating that, at least part of Titan's atmosphere is photochemically active. Through modelling, they found that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Detected in Titan’s atmosphere, dicyanoacetylene ($C_4N_2$) is used in our laboratory simulations as a model system for other larger unsaturated condensing compounds. We show that $C_4N_2$ ices undergo condensed-phase photopolymerization (tholin formation) at wavelengths as long as 355 nm pertinent to solar radiation reaching a large portion of Titan’s atmosphere, almost close to the surface.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and evidence of these ices is suggested in the article &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0019103511000352&quot; rel=&quot;nofollow&quot;&gt;Titan’s aerosol and stratospheric ice opacities between 18 and 500 μm: Vertical and spectral characteristics from Cassini CIRS&lt;/a&gt; (Anderson and Samuelson, 2011) stating that the ices and aerosols&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;appear to be located over a narrow altitude range in the stratosphere centered at ∼90 km. Although most abundant at high northern latitudes, these nitrile ice clouds extend down through low latitudes and into mid southern latitudes, at least as far as 58°S.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user2449" LastActivityDate="2014-12-10T00:47:37.723" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8200" PostTypeId="2" ParentId="8192" CreationDate="2014-12-10T01:19:30.747" Score="12" Body="&lt;p&gt;The L1, L2, and L3 points are unstable in any orbital system. (&lt;a href=&quot;http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node126.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The L4 and L5 points of a pair of bodies are only stable if the larger of the bodies is at least 25 times as massive than the smaller (&lt;a href=&quot;http://wmap.gsfc.nasa.gov/media/ContentMedia/lagrange.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;).  The ratio of the Pluto/Charon system is only 8.7.  Because of this, none of the Lagrange points are stable, and an object orbiting at any of them will require active station-keeping to compensate for perturbations in the orbit.&lt;/p&gt;&#xA;" OwnerUserId="1559" LastEditorUserId="4091" LastEditDate="2018-05-21T20:54:11.060" LastActivityDate="2018-05-21T20:54:11.060" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="8201" PostTypeId="1" CreationDate="2014-12-10T01:56:27.520" Score="7" ViewCount="231" Body="&lt;p&gt;Wikipedia has &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:First_photo_from_space.jpg&quot; rel=&quot;noreferrer&quot;&gt;this photo&lt;/a&gt; (see below) labelled &quot;First photo from space&quot; which was taken in 1946 from a U.S. launched German V-2 rocket. Approximately two thirds of the image is a section of the Earth showing some clouds and their shadows. The photo is very grainy so it's hard to see much more detail than this (I can't make out any land masses as all).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the large black section occupying the top-left corner there are some very faint dots. Are these dirt on the lens, marks on the film, JPG artifacts or are they actually stars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hDu0g.jpg&quot; alt=&quot;The first photo from space&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1435" LastActivityDate="2014-12-10T18:38:13.440" Title="Are there stars in the first photo from space?" Tags="&lt;photography&gt;&lt;history&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8202" PostTypeId="2" ParentId="8201" CreationDate="2014-12-10T03:46:50.927" Score="7" Body="&lt;p&gt;I really doubt it. (&lt;em&gt;Note: I'm answering this more as a photographer than an astronomer&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this vantage point, Earth is far brighter than any stars you would see beyond it (save the Sun). When you set a camera to expose Earth, if you set it such that you can expose Earth without blowing out the details you need to set the camera accordingly. This, however, would mean that it's unlikely the camera settings would be sensitive enough to also get stars. As we can see rough details of the atmosphere in your photo (Despite the grain), it would suggest that the exposure settings would not be sensitive enough to depict stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, here is a photo of Jupiter:&lt;img src=&quot;https://i.stack.imgur.com/nkSuC.jpg&quot; alt=&quot;Jupiter&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few things to note here:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Jupiter is entirely blown out. It's over exposed and you can't see any details of the planet.&lt;/li&gt;&#xA;&lt;li&gt;You can see several moons. If you properly expose Jupiter, you generally won't also get the moons, because your camera settings won't be sensitive enough to pick them up. (at one time I had a more properly exposed photo that lacks the moons. Apologies, I'm not sure where it's gone off to)&lt;/li&gt;&#xA;&lt;li&gt;You aren't seeing any stars in the background. Again, the exposure settings to pick up the moon are too sensitive to properly expose Jupiter, but not sensitive enough to pick up the stars behind it. (A brighter star, had it been in the frame would be visible in the photo, but there are certainly &lt;em&gt;some&lt;/em&gt; stars that were in a shot this wide. Regardless, there are stars far brighter than the Galilean moons, so this isn't a particularly solid point)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There is certainly something in the photo. Judging by the amount of grain in the photo and by the fact that JPG images will always contain some artifacts it would seem more likely that what you're seeing is the affect of noise or artifacts. Dirt on the lens is unlikely - this usually doesn't show up as specks in the final photo, as this is before the light has not been distributed yet. You see dust if it's on the lens exit or between the lens and the sensor, but in that situation you're going to see it generally reducing light from the final image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer may change if, by chance, the photo was taken when it was &lt;em&gt;night&lt;/em&gt; on Earth. However, this seems unlikely for an early rocket launch and certainly for a first attempt at a photo of Earth from space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this is really more about analyzing a photograph than astronomy, if you want to probe further you may want to inquire in &lt;a href=&quot;https://photo.stackexchange.com/&quot;&gt;Photography.SE&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="-1" LastEditDate="2017-04-13T12:44:06.413" LastActivityDate="2014-12-10T04:05:44.853" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8203" PostTypeId="1" CreationDate="2014-12-10T05:32:29.033" Score="9" ViewCount="147" Body="&lt;p&gt;In the paper &lt;a href=&quot;http://www.nature.com/ngeo/journal/v7/n11/full/ngeo2252.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Evidence for basaltic volcanism on the Moon within the past 100 million years&lt;/a&gt; (Braden et al. 2014), suggest that features such as Ina (image below) represent a far more recent age of volcanism (relatively speaking, around 100 million years).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CFlrz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image source: &lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2006/09nov_moonalive/&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA Science News&lt;/a&gt;, where they have a different non-volcanic theory.  However, this question is &lt;em&gt;not&lt;/em&gt; about whether the features are volcanic or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the Braden et al. article abstract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The morphology of the features is also consistent with small basaltic eruptions that occurred significantly after the established cessation of lunar mare basaltic volcanism. Such late-stage eruptions suggest a long decline of lunar volcanism and constrain models of the Moon’s thermal evolution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming the latest models indicate volcanism as recent as within the 100 million years ago, &lt;strong&gt;what are the implications for lunar thermal history of younger volcanism?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user2449" LastEditorDisplayName="user2449" LastEditDate="2014-12-10T13:32:27.800" LastActivityDate="2016-10-24T16:52:33.030" Title="What implications does younger volcanism have on the Moon's thermal evolution?" Tags="&lt;the-moon&gt;&lt;volcanism&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8205" PostTypeId="2" ParentId="8201" CreationDate="2014-12-10T18:32:54.997" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://www.harrisonruess.com/first-view-of-earth-from-space-in-october-1946/&quot; rel=&quot;nofollow&quot;&gt;Newsreel video&lt;/a&gt;, put together from the camera footage at the time shows several persistent white spots towards the top left of the camera frame. You can see them moving with the frame as the camera pans through the blackness of space, and later even against the limb of the earth. If there are any stars visible in the original linked image,they are no brighter than any of several obvious pointlike lens artifacts. I don't know what the smudgy objects are, but doubt that they're comets or the clouds of Magellan, as seen from New Mexico. There were &lt;a href=&quot;http://www.fallofathousandsuns.com/brightest-comets-in-the-20th-century.html&quot; rel=&quot;nofollow&quot;&gt;no bright comets in 1946&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="2031" LastEditDate="2014-12-10T18:38:13.440" LastActivityDate="2014-12-10T18:38:13.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8206" PostTypeId="1" CreationDate="2014-12-10T20:58:00.703" Score="3" ViewCount="156" Body="&lt;p&gt;It seems from the recent Science paper by &lt;a href=&quot;http://www.sciencemag.org/content/early/2014/12/09/science.1261952&quot; rel=&quot;nofollow&quot;&gt;Altwegg et al (2014)&lt;/a&gt; that the commonly accepted source of Earth's water being the comets might not be (completely) true. Their study suggest that the water in the comet they studied is different from Earth's (higher Deuterium content). The alternative source of water that Altwegg suggests is from asteroids. My perception is that there is not that much water in asteroids or at least not sufficient to create Earth's oceans. How much water is in asteroids now? How much water was there in asteroids reaching Earth when the oceans were formed? &lt;/p&gt;&#xA;" OwnerUserId="1578" LastActivityDate="2014-12-11T19:29:08.023" Title="Do/did the asteroids contain enough water to create Earth's oceans?" Tags="&lt;comets&gt;&lt;asteroids&gt;&lt;water&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8207" PostTypeId="1" CreationDate="2014-12-10T21:01:57.303" Score="4" ViewCount="308" Body="&lt;p&gt;Is there a clear cut between them?&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How long and how bright will the moon be for a dark night?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I googled but could not find a definition for them. People just talk about them empirically? &lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2014-12-13T17:31:05.973" Title="where is the boundary between dark night and grey night" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8208" PostTypeId="1" AcceptedAnswerId="8209" CreationDate="2014-12-10T21:05:51.757" Score="7" ViewCount="373" Body="&lt;p&gt;We can observe Pillars of Creation with 7000 years delay, but it is destroyed only 6000 years ago. How do we know that? How the information about explosion reached the Earth before light do?&lt;/p&gt;&#xA;" OwnerUserId="3015" LastActivityDate="2018-07-16T15:47:58.140" Title="How is it known that Pillars of Creation are destroyed?" Tags="&lt;light&gt;&lt;supernova&gt;&lt;speed&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8209" PostTypeId="2" ParentId="8208" CreationDate="2014-12-10T21:46:28.030" Score="5" Body="&lt;p&gt;Much of the evidence for the destruction of the Pillars of creation lies in observations of dust clouds in the region (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2009AAS...21332401F&quot; rel=&quot;nofollow noreferrer&quot;&gt;Flagey et al. (2009)&lt;/a&gt;). The morphology and structure of a cloud in the dust are suggestive of a shock wave, which seems to be indicative of a recent supernova. The shock waves from supernovae do often interact with both the interstellar medium and local gas clouds. The original observations supporting this hypothesis were made by the Spitzer Space Telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.newscientist.com/article/dn10925-pillars-of-creation-destroyed-by-supernova.html#.VIi-cWdMvIU&quot; rel=&quot;nofollow noreferrer&quot;&gt;It should be noted that the SST uses infrared light&lt;/a&gt;, and as the dust wasn't seen in the visible portion of the spectrum, we wouldn't necessarily have realized from the Hubble photos that there were any effects from a hypothetical supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some dissenters believed that something completely different is responsible for the behavior of the dust - possibly the winds of massive stars. Now, more recent x-ray observations (see e.g. &lt;a href=&quot;https://arxiv.org/abs/1205.2111&quot; rel=&quot;nofollow noreferrer&quot;&gt;Guarcello et al. (2012)&lt;/a&gt;) have failed to see either the x-ray spectra expected of O-type stars or the x-ray emission that would be indicative of a supernova, leading some to conclude that &lt;a href=&quot;https://www.forbes.com/sites/startswithabang/2018/07/16/the-pillars-of-creation-havent-been-destroyed-say-new-nasa-images/#4a9abe2f4d55&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Pillars of Creation have &lt;em&gt;not&lt;/em&gt; yet been destroyed&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2018-07-16T15:47:58.140" LastActivityDate="2018-07-16T15:47:58.140" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="8210" PostTypeId="2" ParentId="8207" CreationDate="2014-12-10T22:29:07.257" Score="5" Body="&lt;p&gt;What a nightmare - there is no accepted definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common method is to use the fraction of the night (between the times of astronomical twilight) that the moon spends above the horizon. The problem is that this does not correlate perfectly with fractional lunar illumination or sky brightness. It does provide a good method of deciding when a night is &quot;dark&quot;, but you get a  bit of a mixed bag of &quot;grey&quot; nights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is found that the sky brightness is pretty much determined by fractional lunar illumination (FLI - the fraction of the moon's visible hemisphere illuminated by the Sun) once it exceeds about 0.15; so this is the definition that is currently used at the Isaac Newton Group of Telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark $0\leq$ FLI $&amp;lt;0.25$; Grey $0.25\leq$ FLI $&amp;lt;0.65$; Bright $0.65 \leq$ FLI $\leq 1$, where the FLI is calculated at 0h UT (close to midnight at the ING).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See  &lt;a href=&quot;http://www.ing.iac.es/PR/newsletter/news6/tel1.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ing.iac.es/PR/newsletter/news6/tel1.html&lt;/a&gt; for further discussion (in which I only just noticed I got a mention in the acknowledgements ;) ).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2014-12-13T17:31:05.973" LastActivityDate="2014-12-13T17:31:05.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8211" PostTypeId="2" ParentId="8160" CreationDate="2014-12-10T22:41:55.950" Score="3" Body="&lt;p&gt;Flares involve releasing huge amounts of potential energy that are stored in stressed magnetic structures in the solar corona. So-called reconnection events can release this stress but at the same time accelerate charged particles to high energies. These are trapped along the magnetic field lines, spiralling around them, but travelling until they hit the footpoints of the loop that are anchored in the solar chromosphere and photosphere. The much denser plasma here is sufficient to stop the charged particles and this is where they dump all their kinetic energy, some of which is promptly released in the form of non-thermal hard X-rays, which are insufficient to cool the plasma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plasma heats up, increases its pressure and it expands to fill the overlying magnetic structures - a.k.a. chromospheric evaporation. This hot plasma radiates and because it is at temperatures between $10^{5}$ and $10^7$ K, most of that radiation is emitted at EUV and X-ray wavelengths in free-free and transitions in highly ionised species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the root cause as to why the plasma can get up to these temperatures is that the radiative and conductive cooling processes are much slower than the heating processes caused by the accelerated charged particles.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-08-05T16:42:12.210" LastActivityDate="2016-08-05T16:42:12.210" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8212" PostTypeId="2" ParentId="1799" CreationDate="2014-12-10T23:24:06.343" Score="1" Body="&lt;p&gt;For me, the best and most reliable source of absolute magnitudes and spectral types is the book &lt;a href=&quot;http://adsabs.harvard.edu//abs/2000asqu.book.....C&quot; rel=&quot;nofollow&quot;&gt;Allen's astrophysical quantities&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chapter 15, called Normal Stars, contains spectral types, absolute magnitudes, colors and effective temperatures for main sequence, giant and supergiant stars. The references are at the end of the chapter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HTH,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Germán.&lt;/p&gt;&#xA;" OwnerUserId="2339" LastActivityDate="2014-12-10T23:24:06.343" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8213" PostTypeId="2" ParentId="8180" CreationDate="2014-12-11T09:05:51.913" Score="0" Body="&lt;p&gt;If you're only interested in &lt;em&gt;objects devoid of non-microscopic empty spaces&lt;/em&gt;, then we're pretty much into the realm of stars, planets, and asteroids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objects of this type larger than $\sim100$km are all nearly spherical, simply because of the effect of their own gravity (the spherical shape minimises the gravitational energy).&#xA;However, small asteroids can have all sort of sometimes quite irregular shapes. Of course, your concepts of &lt;em&gt;polar&lt;/em&gt; and &lt;em&gt;equatorial&lt;/em&gt; diameter then become somewhat ill-defined, but one can still define the three principal axes (for example of the moment of intertia) and consider their ratios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eros, for example, has extents 34.4 x 11.2 x 11.2 km. It is thus prolate, but the semi-minor to -major axis ratio is less than 1/3: pretty small, though not flat.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-12-11T09:05:51.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8214" PostTypeId="2" ParentId="8180" CreationDate="2014-12-11T13:51:02.080" Score="1" Body="&lt;p&gt;The dwarf planet Haumea two equatorial diameters, it is triaxial. The longest equatorial diameter is about twice the length of the polar diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Google and Wikipedia, the most oblate star is Achernar which has an equatorial diameter 56% larger than the polar diameter due to it rapid spin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discussion at &lt;a href=&quot;http://cosmoquest.org/forum/showthread.php?80772-Oblate-Neutron-Stars&quot; rel=&quot;nofollow&quot;&gt;http://cosmoquest.org/forum/showthread.php?80772-Oblate-Neutron-Stars&lt;/a&gt; says that if the oblateness is extreme enough then an oblate spheroid isn't stable and it turns into a triaxial ellipsoid and that none of the known neutron stars are oblate as that. &lt;/p&gt;&#xA;" OwnerUserId="2169" LastActivityDate="2014-12-11T13:51:02.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8215" PostTypeId="2" ParentId="6398" CreationDate="2014-12-11T15:35:38.523" Score="2" Body="&lt;p&gt;A supernova &lt;em&gt;may&lt;/em&gt; actually be necessary in the creation of a stellar black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the ends of their lives the cores of massive stars are made mostly of iron-peak nuclei from which you cannot extract more fusion energy. To support their weight, these stars rely on electron degeneracy pressure - the pressure caused by the Pauli exclusion principle allowing no more than one electron to share the same quantum state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle a star might be supported by degeneracy pressure forever as it gradually cools - this is the fate of most white dwarfs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the core of a massive star is just too big for that to work. The density increases until all the electron are moving at close-to the speed of light and that's as high as the degeneracy pressure can get. If the core exceeds the Chandrasekhar mass, it will collapse and as it does so, the rest of the star collapses with it (a little more slowly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The collapse is triggered by the removal of electrons by electron capture into nuclei to form neutrons. At some point enough neutrons are produced for neutron degeneracy pressure to halt or at least slow the collapse. This and the release of a lot of gravitational potential energy are ultimately what power a supernova explosion. But if the collapse is not halted then even neutron degeneracy pressure will not support the star and collapse to a black hole becomes inevitable. A black hole status is reached once a proportion of its mass is compressed inside its Schwarzschild radius $r_s = 2GM/c^2$. i.e. once its density achieves&#xA;$$ \rho &amp;gt; \frac{3M}{4\pi r_s^{3}}$$&#xA;i.e. when a central mass $M$ has a density that exceeds&#xA;$$ \rho &amp;gt; \frac{3}{32\pi} \frac{c^6}{G^3 M^2} = 1.8\times10^{19} \left(\frac{M}{M_{\odot}}\right)^{-2}\ {\rm kg/m}^3$$&#xA;This is a ball park figure and assumes spherical symmetry and neglects any detailed GR treatment, but is more or less correct - a few times higher than typical neutron star densities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words it is the &lt;em&gt;density&lt;/em&gt; of the material that largely determines whether something becomes a black hole. The mass is only an indirect parameter.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-01-15T21:43:08.383" LastActivityDate="2015-01-15T21:43:08.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8216" PostTypeId="2" ParentId="8206" CreationDate="2014-12-11T19:29:08.023" Score="4" Body="&lt;p&gt;This is far too soon to draw sweeping conclusions. The Philae probe has only measured deuterium for one comet. But comets are diverse, of diverse origins. We need to take a lot more samples, from many different comets, before we can conclude firmly that water has, or has not, come from comets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only conclusion we could draw so far from this study is that earthly water has not come exclusively from comets similar to 67P/C-G.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you're seeing now is the usual distortion that occurs in the media after some hugely popular experiment is completed - its conclusions are blown out of proportion. But that's not how science works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's write this one down, and wait for further science to be put forth. We don't know enough yet about these objects.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2014-12-11T19:29:08.023" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8217" PostTypeId="1" AcceptedAnswerId="8246" CreationDate="2014-12-11T20:08:21.193" Score="9" ViewCount="4589" Body="&lt;p&gt;So, this is mere musing, but it seems that stars are quite extremely far apart. I tried to determine the mean distance between nearest neighbors for stars (in just our galaxy) but I'm not sure what it is, though a &lt;a href=&quot;http://www.wolframalpha.com/input/?i=list+of+closest+stars&quot; rel=&quot;noreferrer&quot;&gt;first approximations&lt;/a&gt; I would guess that the mean is around 5 ly with some 0.5 standard deviation or so!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, assuming I didn't screw up, the distance between two sun-massed objects required to generate 1,000N of force between them is an &lt;a href=&quot;http://www.wolframalpha.com/input/?i=sqrt%28%28G+*+%28mass+of+sun%29%5E2%29%2F%281000+newton%29%29&quot; rel=&quot;noreferrer&quot;&gt;incredible 54.3 million ly&lt;/a&gt;. [Edit] If you change that to 1 billion Newtons, the distance is one galactic radius -- 54000~ ly!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm probably mis-estimating how much even 1kN of force would effect solar trajectories over time, but I still can't help but feel that our solar neighbors are very far away given the spheres of interaction that I would expect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any insights?&lt;/p&gt;&#xA;" OwnerUserId="4019" LastEditorUserId="2153" LastEditDate="2014-12-11T23:33:27.783" LastActivityDate="2020-08-04T16:05:16.390" Title="Why are stars so far apart?" Tags="&lt;star&gt;&lt;gravity&gt;&lt;distances&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8219" PostTypeId="2" ParentId="6398" CreationDate="2014-12-12T02:15:49.037" Score="0" Body="&lt;p&gt;Whether an object is a black hole is not just determined by its mass.  It's determined by whether that mass is entirely within its &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;nofollow&quot;&gt;Schwarzschild radius&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, any object can be a black hole &lt;em&gt;if&lt;/em&gt; all its mass is concentrated into a sufficiently small volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the Sun's Schwarzschild radius is about 3.0 km -- but its actual radius is about 700,000 km.  It could become a black hole only if it were compressed down to a radius of 3.0 km.&lt;/p&gt;&#xA;" OwnerUserId="53" LastActivityDate="2014-12-12T02:15:49.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8221" PostTypeId="2" ParentId="925" CreationDate="2014-12-12T06:27:16.490" Score="0" Body="&lt;p&gt;Warning: citations above about IC 1101 being the largest known galaxy are probably wrong! Wikipedia's claimed 5 Mly size is certainly incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The visible light extent is about 360 kly (&lt;a href=&quot;http://ned.ipac.caltech.edu/cgi-bin/objsearch?objname=IC%201101&amp;amp;extend=no&amp;amp;hconst=73&amp;amp;omegam=0.27&amp;amp;omegav=0.73&amp;amp;corr_z=1&amp;amp;out_csys=Equatorial&amp;amp;out_equinox=J2000.0&amp;amp;obj_sort=RA%20or%20Longitude&amp;amp;of=pre_text&amp;amp;zv_breaker=30000.0&amp;amp;list_limit=5&amp;amp;img_stamp=YES#ObjNo1&quot; rel=&quot;nofollow&quot;&gt;NED's summary page&lt;/a&gt; gives a major diameter of 1.2' and a redshift of z=0.077947, for a physical size of around 115 kpc, or 370 kly). The largest measurement one could reasonably suggest for IC 1101 (around 500 kpc, or 1.5 Mly) comes from the maximal measured extent of the diffuse light from Uson, Bough &amp;amp; Kuhn (1991). That's not what one usually thinks of when talking about the &quot;size&quot; of a galaxy. Usually we refer to the radius in which most of the visible light is enclosed, not the maximum distance we could possibly measure light that might be associated with the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, NGC 262 is not 1.3 Mly across. It has an extended cold Hydrogen gas halo of around 600 kly (~10' HI measurement from Morris &amp;amp; Wannier 1979, distance ~65 Mpc), but the physical extent of its stars is around 66 kly: large for a spiral, but not huge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the question: I'd go with galaxy superclusters as being the largest currently gravitationally-bound objects in the universe.&lt;/p&gt;&#xA;" OwnerUserId="4023" LastActivityDate="2014-12-12T06:27:16.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8222" PostTypeId="1" CreationDate="2014-12-12T09:40:11.820" Score="7" ViewCount="54" Body="&lt;p&gt;I asked a question &lt;a href=&quot;https://space.stackexchange.com/questions/6319/sneakernet-would-data-return-together-with-sample-return-be-useful&quot;&gt;here&lt;/a&gt; about returning to Earth a physical memory capsule along with soil samples, as a complement to radio operations as today. This might return a much larger amount of data compared with radio today, but of course only at the end of the mission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How useful would it be for a sample return mission like Mars2020 or Hayabusa2, to be able to return much more data? Would the choice of instruments change to allow for higher sampling rates, higher resolution, more filters et cetera? Is Big Data science instruments from an interplanetary mission even a possibility, or is today's bandwidth good enough for any conceivable payload?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-03-25T20:30:10.770" Title="How does radio bandwidth restrict the choice and use of science payload on interplanetary probes?" Tags="&lt;data-analysis&gt;&lt;space-probe&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8223" PostTypeId="2" ParentId="8217" CreationDate="2014-12-12T17:42:44.360" Score="2" Body="&lt;p&gt;Most of the universe is pretty empty in terms of the density you're used to in daily life. It's perhaps not that stars are far apart, but that they are pretty compact. This is because baryonic matter (as opposed to dark matter) can lose energy via electromagnetic radiation and hence condense to smaller and denser objects. This is only opposed by angular momentum (which cannot be simply radiated away) forcing disc-like structures such as the Galaxy and proto-stellar and -planetary discs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Milky Way, the density of stars varies a lot between the stellar halo, the disc, the central bulge, and the cores of star clusters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your concept of &lt;em&gt;spheres of interaction&lt;/em&gt; is flawed, however. Each star feels the combined force of &lt;em&gt;all&lt;/em&gt; other stars, even those on the far side of the Galaxy. The resulting total gravitational potential of the Galaxy can be approximated as smooth in space and time. This is because individual star-star gravitational interactions are comparatively weak and have negligible effect. Combined they would eventually scatter a star off its orbit, but the time scale for that is much longer than the Hubble time. For the dense cores of star clusters, the situation is different and so-called two-body relaxation effects their structure over their life time.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2014-12-12T17:42:44.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8224" PostTypeId="1" AcceptedAnswerId="8241" CreationDate="2014-12-12T18:47:26.997" Score="1" ViewCount="125" Body="&lt;p&gt;I noticed this in a thread from world building, what is the continuous line of light in MACS J1206 slightly right of center in this picture? The only thing I could think of would be the side view of a galaxy, but it's much longer than any other galaxies pictured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gc7bu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4026" LastActivityDate="2015-01-13T02:35:00.090" Title="What is the line of light in galaxy cluster MACS J1206?" Tags="&lt;galaxy&gt;&lt;deep-sky-observing&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8225" PostTypeId="1" CreationDate="2014-12-12T04:23:48.493" Score="0" ViewCount="89" Body="&lt;p&gt;In this &lt;a href=&quot;http://zeenews.india.com/news/sci-tech/astronomers-spot-two-stars-beginning-to-merge-to-create-a-megastar_1513391.html&quot; rel=&quot;nofollow&quot;&gt;news article&lt;/a&gt;, it says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The image captured by the telescope at the Calar Alto Observatory in southern Spain, reveal the physical properties of each of the stars including temperature and shape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I Googled to see the actual image but unfortunately all are graphical images.So can I see the actual image as captured by the Calar Alto Observatory?&lt;/p&gt;&#xA;" OwnerUserId="379" OwnerDisplayName="SpringLearner" LastEditorUserId="2153" LastEditDate="2015-05-25T14:50:51.767" LastActivityDate="2015-05-25T14:50:51.767" Title="Where can I find the image of MY Camelopardalis?" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8226" PostTypeId="2" ParentId="8225" CreationDate="2014-12-12T07:39:13.880" Score="3" Body="&lt;p&gt;The work on &lt;a href=&quot;http://arxiv.org/pdf/1410.5575v1.pdf&quot; rel=&quot;nofollow&quot;&gt;the recent paper&lt;/a&gt; was done using spectra and light curves, not via direct imaging.  There may not be a good image to view.  I checked the &lt;a href=&quot;http://hla.stsci.edu/hlaview.html#&quot; rel=&quot;nofollow&quot;&gt;Hubble Legacy Archive&lt;/a&gt; and could not find a Hubble image of the object.&lt;/p&gt;&#xA;" OwnerUserId="5373" OwnerDisplayName="BowlOfRed" LastActivityDate="2014-12-12T07:39:13.880" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8227" PostTypeId="1" CreationDate="2014-12-12T22:23:12.783" Score="6" ViewCount="125" Body="&lt;p&gt;I'm trying to find some data that I can use to derive dark matter (in a loosey-goosey sort of way, I won't be too rigorous). I'm helping a friend out with a final project for his astronomy course and I'm tasked with doing the maths. Any idea where I can find a table of various stars distance from the center of our galaxy and their speed (how fast they are going around the center of the galaxy)? This does not need to be comprehensive but would ideally contain several data sources from various representative radii. Also some sort of formatted text would be SUPER cool so I could just tell my computer to do the calculations!&lt;/p&gt;&#xA;" OwnerUserId="4029" LastEditorUserId="2153" LastEditDate="2018-08-14T19:35:48.600" LastActivityDate="2018-08-15T01:12:46.553" Title="Deriving Dark Matter; specifically looking for a table of stellar speed versus distance from center of galaxy to derive dark matter" Tags="&lt;dark-matter&gt;&lt;data-analysis&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8229" PostTypeId="1" AcceptedAnswerId="8231" CreationDate="2014-12-13T15:21:24.320" Score="6" ViewCount="1119" Body="&lt;p&gt;Is this a consequence of planet formation in accretion disks ?&lt;/p&gt;&#xA;" OwnerUserId="4033" LastActivityDate="2014-12-13T23:10:28.823" Title="Why, in the Solar System, all the mass seems to be concentrated at the centre?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;formation&gt;&lt;planetary-formation&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8230" PostTypeId="2" ParentId="8229" CreationDate="2014-12-13T15:48:05.123" Score="2" Body="&lt;p&gt;&lt;strong&gt;It's because of the Sun.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be good if I give a quick overview on &lt;a href=&quot;https://en.wikipedia.org/wiki/Star_formation&quot; rel=&quot;nofollow&quot;&gt;star formation&lt;/a&gt; before I get to the meat of the issue. Here's star formation in a few simple steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Molecular_cloud&quot; rel=&quot;nofollow&quot;&gt;Giant Molecular Cloud&lt;/a&gt; forms.&lt;/strong&gt; A large region of gas and dust, essentially a dense version of the interstellar medium, coalesces into an interstellar cloud. GMCs can be tens or hundreds of light-years across, enough to give birth to many stars. Within the GMC, some regions will be slightly denser than others.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;A portion of the cloud collapses.&lt;/strong&gt; A certain region of the GMC collapses, generally due to an outside disturbance. The most commonly cited cause is a supernova shockwave that compresses portions of the GMC, although &lt;a href=&quot;https://en.wikipedia.org/wiki/Interacting_galaxy&quot; rel=&quot;nofollow&quot;&gt;close passes between galaxies&lt;/a&gt; have been known to &lt;a href=&quot;https://en.wikipedia.org/wiki/Starburst_galaxy&quot; rel=&quot;nofollow&quot;&gt;incite star formation&lt;/a&gt;. My favorite example is &lt;a href=&quot;https://en.wikipedia.org/wiki/Cartwheel_Galaxy&quot; rel=&quot;nofollow&quot;&gt;the Cartwheel galaxy&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;The region heats up.&lt;/strong&gt; There is quite a lot of matter pressing in on what has now become a protostar, and so it heats up. Eventually, conditions become such that hydrogen fusion is possible. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Protostar&quot; rel=&quot;nofollow&quot;&gt;protostar&lt;/a&gt;, now a &lt;a href=&quot;https://en.wikipedia.org/wiki/Pre-main-sequence_star&quot; rel=&quot;nofollow&quot;&gt;pre-main-sequence star&lt;/a&gt;, begins to shine.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Protoplanetary_disk&quot; rel=&quot;nofollow&quot;&gt;protoplanetary disk&lt;/a&gt; forms.&lt;/strong&gt; At this point, the star dominates this region of the GMC. Matter nearby is pulled towards it by the force of gravity, and a circumstellar disk forms. It may be composed of gas and dust. Eventually, small grains of dust collide and form bigger grains. &lt;a href=&quot;https://en.wikipedia.org/wiki/Planetesimal&quot; rel=&quot;nofollow&quot;&gt;Planetesimals&lt;/a&gt; form, then &lt;a href=&quot;https://en.wikipedia.org/wiki/Protoplanet&quot; rel=&quot;nofollow&quot;&gt;protoplanets&lt;/a&gt;, and finally planets.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The reason that there isn't more matter in a given stellar system is that the star dominates the surrounding area. It pulls in nearly everything around it during its early life. Much of the region of the collapsing cloud is made of molecular $H_2$, and so it is pulled in and used for fusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the question translates to 'Why isn't the protoplanetary disk more massive'? The answer is that when the disk formed, much of the matter that was in its inner reaches spiraled into the Sun. This is partly due to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Poynting%E2%80%93Robertson_effect&quot; rel=&quot;nofollow&quot;&gt;Poynting-Robertson effect&lt;/a&gt;, where photons from the Sun pull dust grains in. Over billions of years, the star can accumulate much of the matter than was originally close to it in the disk.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-12-13T15:48:05.123" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8231" PostTypeId="2" ParentId="8229" CreationDate="2014-12-13T16:15:07.330" Score="2" Body="&lt;p&gt;I feel that HDE gives a good start of an answer, but stops short of the important part. We have seen in HDE's answer the formation of a star at the center of the collapsing molecular cloud. When the star begins to fuse lighter elements, the protoplanetary disk has several forces acting on it:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The momentum of the particles in the disk.&lt;/li&gt;&#xA;&lt;li&gt;The gravity of the star in the center and other particles of the protoplanetary disk on the other side.&lt;/li&gt;&#xA;&lt;li&gt;The gravity of the particles of the protoplanetary disk opposite the sun (outward).&lt;/li&gt;&#xA;&lt;li&gt;The radiation pressure of the new star.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Interestingly, the second and third forces pretty much cancel out when the disk is still uniform in distribution. But once large clumps of matter accumulate those large masses (protoplanets) gravitationally pull at the other matter in inconsistent and sometimes violent ways, especially when multiple protoplanets align (once per orbit of the inner body).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the area of the protoplanetary disk is being 'swept up' of all the mass: some is &lt;strong&gt;pulled into the sun&lt;/strong&gt; by gravity and loss of momentum due to collisions, some is &lt;strong&gt;pushed out of the solar system&lt;/strong&gt; by radiation, and whatever matter was not disturbed by one of those processes is then subject to being disturbed by the gravity of the protoplanets themselves. The protoplanets will, over time, either absorb that matter or gravitationally slingshot that matter &lt;strong&gt;out of the system or too it's very edges&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, the protoplanetary disk within a few tens of AU of the star in the center is &lt;strong&gt;a chaotic mess of a place&lt;/strong&gt;. Not much matter can form a stable orbit there.&lt;/p&gt;&#xA;" OwnerUserId="209" LastActivityDate="2014-12-13T16:15:07.330" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8232" PostTypeId="2" ParentId="8229" CreationDate="2014-12-13T17:06:10.217" Score="1" Body="&lt;p&gt;Are you asking why there isn't more mass in the planetary system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reasons go back to the collapse and fragmentation of the protostellar cloud and the subsequent accretion (from a pseudo-spherical envelope) of the bulk of the protostellar material. It has &lt;em&gt;little&lt;/em&gt; to do with subsequent processes occurring in the accretion disc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the disc had too much mass it becomes unstable to further fragmentation and this is one of the ways in which binary systems are formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general discs around stars rarely exceed about a tenth of the stellar mass. But of course there are lots of binary systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If instead you mean why didn't more of the protoplanetary disc end up in the planets - the main two reasons are: (i) a large fraction of the protoplanetary disc was accreted. We see class II T-Tauri stars with mass accretion rates of $10^{-9}-10^{-8}$ solar masses per year and this appears to persist for a few million years. (ii) The disc gets photoevaporated by the high energy radiation coming from the star and possibly from external sources too. This drives a disc wind that will deplete the disc of material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planet formation is in competition with these processes. The popular &quot;core accretion&quot; model for the formation of giant planets (there aren't enough rocks in the protoplanetary disc for rocky planets/asteroids et al. to make any difference) has a timescale of 5-10 million years for the accretion of gas, so most of the raw materials can easily have disappeared before much accretion of material into giant planets has occurred.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2014-12-13T23:10:28.823" LastActivityDate="2014-12-13T23:10:28.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8234" PostTypeId="2" ParentId="1737" CreationDate="2014-12-13T17:58:14.993" Score="5" Body="&lt;p&gt;If we take neutron star material at say a density of $\sim 10^{17}$ kg/m$^{3}$ the neutrons have an internal kinetic energy density of $3 \times 10^{32}$ J/m$^{3}$. So even in a teaspoonful (say 1 cc), there is $3\times10^{26}$ J of kinetic energy (similar to what the Sun emits in a second, or 10 billion or so H-bombs) and this will be released instantaneously.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy is in the form of around $10^{38}$ neutrons travelling at around 0.1-0.2$c$. So roughly speaking it is like half the neutrons (about 50 million tonnes) travelling at 0.1$c$ ploughing into the Earth. If I have done my Maths right, that is roughly equivalent to a 30 km radius near-earth asteroid hitting the Earth at 30 km/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this material would instantly vapourise and take a large chunk of the Earth with it, probably destroying most of life on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation for a white dwarf is much less extreme. The density would be more like $10^{9}$ kg/m$^3$ and the energy density more like $10^{22}$ J/m$^3$ - so 10 orders of magnitude less kinetic energy density. Nevertheless that is still $10^{16}$ J, which is like a 2.5 megatonne H-bomb.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2014-12-13T17:58:14.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8235" PostTypeId="2" ParentId="7822" CreationDate="2014-12-13T18:11:46.760" Score="4" Body="&lt;p&gt;The actual answer has nothing at all to do with the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even low-mass stars would form black holes if they ran out of nuclear fuel to burn, and simply cooled whilst being supported by &quot;standard&quot; gas pressure in their centres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is because that gas pressure would be proportional to the temperature, but the star is able to cool so it would need to shrink and use gravitational energy to stay hot, but of course eventually it would disappear inside its event horizon and become a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real reason this does not occur is &lt;strong&gt;electron degeneracy pressure&lt;/strong&gt;. This is a quantum mechanical effect, related to the Pauli exclusion principle, that does not allow two electrons to occupy the same quantum state. As the gas gets squashed in the core, the electrons are forced to fill higher and higher momentum states in order to &quot;avoid&quot; each other. Because these electrons have large momentum they also exert a large pressure - degeneracy pressure. This degeneracy pressure &lt;em&gt;does not depend on temperature&lt;/em&gt; and so the star (it is a white dwarf when a low-mass star arrives at this stage) can cool and remain in hydrostatic equilibrium without shrinking any further. Thus it avoids becoming a black hole (or even a neutron star) and simply cools at nearly constant radius.   &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2014-12-13T18:11:46.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8238" PostTypeId="2" ParentId="1700" CreationDate="2014-12-13T21:41:07.713" Score="5" Body="&lt;p&gt;The question was whether there were any &quot;Earth-like&quot; planets, not whether any were habitable. I believe the answer is currently, not very many.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the 31 (as of 6th Aug 2015) confirmed &lt;a href=&quot;http://phl.upr.edu/projects/habitable-exoplanets-catalog&quot; rel=&quot;nofollow&quot;&gt;potentially habitable planets&lt;/a&gt; most orbit K and M-dwarfs and are much closer to their Sun than we are; all are probably bigger and more massive than the Earth. There are 3 that orbit G-type stars: tau Cet e, Kepler 22b and Kepler 452b, all are thought to be at least 50% bigger than Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it depends how far you are prepared to stretch your definition of &quot;Earth-like&quot;, but I would say there are about 3. Kepler 452b is the only one with an orbital &quot;year&quot; greater than 300 days.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-08-06T20:47:54.630" LastActivityDate="2015-08-06T20:47:54.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8239" PostTypeId="2" ParentId="2032" CreationDate="2014-12-13T21:57:12.610" Score="1" Body="&lt;p&gt;There are (at least) two ways that you can do it, both of which work best for massive planets that are located far away from their parent star (which may be quite rare).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is &lt;em&gt;direct imaging&lt;/em&gt;. Several planets have been found using this technique (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013EPJWC..4713004N&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/2013EPJWC..4713004N&lt;/a&gt; ). It obviously works best if you have very good spatial resolution, which these days is accomplished using adaptive optics. It also works best in the infrared, because in that part of the spectrum, the contrast between the brightness of a star and the brightness of a companion giant planet is big, but not insurmountable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second is using the astrometric wobble (change in position) seen in a star's position as it orbits its common centre of gravity with a massive planet. The amplitude of this wobble will be bigger for planets that are more massive and further away from the star. However, if they are too far away, then the wobble period will be too long to sensibly detect it on timescales of years. At least a couple of previously known planetary systems have had this wobble measured (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014ApJ...795...41M&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/2014ApJ...795...41M&lt;/a&gt; , &lt;a href=&quot;http://adsabs.harvard.edu/abs/2010ApJ...715.1203M&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/2010ApJ...715.1203M&lt;/a&gt;), but it thought that the Gaia astrometry satellite will discover lots of planetary systems in this way (see &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013EPJWC..4715005S&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/2013EPJWC..4715005S&lt;/a&gt; for details and estimates of numbers).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2014-12-13T21:57:12.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8240" PostTypeId="2" ParentId="918" CreationDate="2014-12-13T22:29:43.960" Score="10" Body="&lt;p&gt;The currently accepted answer is not relevant for finding the distance to a star like Proxima Centauri.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how parallax works. You measure the position of a star in a field of stars that are (presumably) much further way. You do this twice, separated by 6 months. You then calculate the angle that the star has moved against its background stars. This angle forms part of a large triangle, with a base that is equal to the &lt;em&gt;diameter&lt;/em&gt; of the Earth's orbit around the Sun. Trigonometry then tells you what the distance is as a multiple of the distance from the Earth to the Sun. [In practice you perform many measurements with any separation in time and combine them all.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;parallax angle&quot; is actually &lt;em&gt;half&lt;/em&gt; this angular displacement, and a star is said to be 1 parsec away if the parallax angle is 1 second of arc. So 1pc is 1 AU/$\tan (\theta) = 3.08\times10^{16}$ m. The larger the parallax, the closer the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gaia satellite is currently mapping the entire sky and will estimate tiny parallaxes &#xA;with precisions of $10^{-5}$ to $10^{-4}$ arcseconds (depending on target brightness) for about a billion stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Parallax - as illustrated at &lt;a href=&quot;http://www.bbc.co.uk/schools/gcsebitesize/science/21c/earth_universe/earth_stars_galaxiesrev4.shtml&quot; rel=&quot;noreferrer&quot;&gt;http://www.bbc.co.uk/schools/gcsebitesize/science/21c/earth_universe/earth_stars_galaxiesrev4.shtml&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mr5ad.gif&quot; alt=&quot;Parallax, taken from the BBC &amp;quot;bitesize website&amp;quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in reality, it is a bit more difficult than this because stars also have a &quot;proper motion&quot; across the sky due to their motion in our Galaxy relative to the Sun. This means you have to do more than two measurements to separate out this component of motion on the sky. In the case of Proxima Centauri the motion against the background stars due to proper motion is larger than the parallax. But the two components can clearly be seen and separated (see below). It is (half) the amplitude of the curved motion in the picture below that corresponds to the parallax. The proper motion is just the constant linear trend with respect to the background stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;HST images of the path of Proxima Centauri against background stars. The green curve shows the measured and predicted path of the star against the background field over the next few years.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GFQkp.jpg&quot; alt=&quot;HST images of Proxima Centauri&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Parallax measurements work best for nearby stars, because the parallax angle is larger. For more distant stars or those without a parallax measurement, there are a battery of techniques. For isolated stars, the most common is to attempt to establish what type of star it is, either from its colour(s) or preferably from a spectrum that can reveal its temperature and gravity. From this one can estimate what the absolute luminosity of the object is and then from its observed brightness one can calculate the distance. This is known as a &lt;em&gt;photometric parallax&lt;/em&gt; or &lt;em&gt;spectroscopic parallax&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-04-05T10:53:34.603" LastActivityDate="2015-04-05T10:53:34.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8241" PostTypeId="2" ParentId="8224" CreationDate="2014-12-14T02:01:09.643" Score="2" Body="&lt;p&gt;I have highlighted the feature I believe you are talking about. It looks like a galaxy that has been repeated (smaller green oval). There are several foreground objects that appear to be causing additional gravitational lensing (red arrows). Due to the shapes of the objects causing the lensing effect, the apparent image can be affected in many different ways, such a image duplication (multiple identical images of a galaxy, etc.), magnification, distortion, etc. The universe is massive and three-dimensional and we still don't understand that much of it yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yunIR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1435" LastActivityDate="2014-12-14T02:01:09.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8246" PostTypeId="2" ParentId="8217" CreationDate="2014-12-14T21:03:34.283" Score="12" Body="&lt;p&gt;Most stars are of a solar-mass or below. The average number of companions that each stars has (in the sense of being part of binary or higher multiple systems) systems ranges from 0.75 for stars of a solar mass to approximately 0.35 (not a well-established number) for the more numerous M-dwarfs. Let's take a compromise value, say 0.5.&#xA;The separation distribution of these multiples peaks at around 50 AU for solar-type stars, reducing to about 5 AU for low-mass M dwarfs. Again, lets take a compromise value of 20 AU. See &lt;a href=&quot;http://arxiv.org/abs/1303.3028&quot;&gt;Duchene &amp;amp; Kraus (2013)&lt;/a&gt; for all the details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we take 1000 stars, then 333 of them (roughly speaking) are companions to another 333 stars, while 333 are isolated single stars. (NB This does not mean the frequency of multiple systems is 50%, because some of the companions will be in higher order multiple systems)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, taking your calculation of the separation between stellar systems of 5 light years ($= 3.2\times10^{5}$ AU), then the mean separation is:&#xA;$$\bar{D} = 0.667\times 20 + 0.333 \times 3.2\times10^{5} \simeq 10^{5} AU, $$&#xA;but the &lt;em&gt;median&lt;/em&gt; separation is 20 AU!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is of course sophistry, because I'm sure your question is really, why are stellar systems so far apart?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars (and stellar systems) are born in much denser environments. The number density of stars in the Orion Nebula Cluster (ONC - the nearest very large stellar nursery) is about 1 per cubic light year. The equivalent number for the solar neighbourhood is 0.004 stars per cubic light year. Thus the average interstellar separation in the ONC is 1 light year, but in the solar neighbourhood it is about 6.3 light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this separation at birth is the Jeans length - the critical radius at which a clouds self-gravity will overcome its thermal energy and cause it to collapse. It can be expressed as &#xA;$$\lambda_J = c_s \left( \frac{\pi}{G\rho } \right)^{1/2},$$&#xA;where $c_s$ is the sound speed in a molecular cloud and $\rho$ its density. For star forming giant molecular clouds $c_s = 0.2$ km/s and $\rho=10^{-23}$ kg/m$^3$. So clouds of scale hundreds of light years could collapse. As they do, the density increases and the Jeans length becomes smaller and allows the cloud to fragment. Exactly how far the fragmentation goes and the distribution of stellar masses it produces is an area of intense research, but we know observationally that it can produce things like the ONC or sometimes even more massive and dense clusters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there we know that a new born cluster of stars tends not to survive very long. For various reasons - outflows, winds and ionising radiation from newborn stars are able to heat and expel the remaining gas; star formation appears to have an average efficiency of a few to perhaps 20-30%. Expelling the gas, plus the tidal field of the galaxy breaks up the cluster and disperses it into the field, which gives us the lower field star (or system) density that we see around us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the stars are part of the field they essentially don't interact with each other; they are too far apart to feel the influence of individual objects and move subject to the overall gravitational potential of the Galaxy.&#xA; so your consideration of the force between stars is not really relevant.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2014-12-14T22:11:00.787" LastActivityDate="2014-12-14T22:11:00.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8247" PostTypeId="2" ParentId="2593" CreationDate="2014-12-14T22:33:18.973" Score="2" Body="&lt;p&gt;There have of course been extensive observations of clusters of stars. To all intents and purposes, to the limits of experimental accuracy, it looks like stars that are born in the same &lt;em&gt;open cluster&lt;/em&gt; or star-forming region are all born with the same composition. e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014A%26A...567A..55S&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/2014A%26A...567A..55S&lt;/a&gt; , &lt;a href=&quot;http://adsabs.harvard.edu/abs/2002AJ....124.2799W&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/2002AJ....124.2799W&lt;/a&gt; , &lt;a href=&quot;http://adsabs.harvard.edu/abs/2007AJ....133.1161D&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/2007AJ....133.1161D&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are of course various &lt;em&gt;photospheric&lt;/em&gt; abundance anomalies in some stars, but these are normally ascribed to mixing processes or nucleosynthesis within the stars, not their birth composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation in globular clusters is not so clear cut. Multiple populations distinguished by their chemical compositions have been found. Possibly as a result of ancient mergers. So this does not perhaps represent a violation of Tobler's law since these stars would actually have been born in different environments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once stars leave their birth environment then they very quickly become mixed around the Galaxy. A star with a peculiar velocity of only 1km/s will travel a parsec in a million years. Stars born in the same place are probably widely distributed around the galaxy in a billion years or so. Most of the mixing is in azimuth or vertically, rather than Galactocentric radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless there is a &lt;em&gt;radial&lt;/em&gt; dependence (Galactocentric radius) of chemical composition (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2010A%26A...511A..56P&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/2010A%26A...511A..56P&lt;/a&gt; ). Thus stars with similar Galactocentric radii have similar gross chemical compositions (with some scatter). But then this does violate Tobler's law because it means there is a similarity between objects that are on opposite sides of the Galaxy but with similar distance from the Galactic centre.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2014-12-14T22:33:18.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8248" PostTypeId="2" ParentId="3601" CreationDate="2014-12-14T23:01:04.843" Score="16" Body="&lt;p&gt;Here is the very study you are looking for by &lt;a href=&quot;http://arxiv.org/abs/1412.3648&quot;&gt;Bailer-Jones (2014)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the re-reduction of the Hipparcos astrometry, he has integrated orbits for 50,000 stars to look for objects that might come or might have come close to the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The K-dwarf &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=Hip%2085605&amp;amp;NbIdent=1&amp;amp;Radius=2&amp;amp;Radius.unit=arcmin&amp;amp;submit=submit%20id&quot;&gt;Hip 85605&lt;/a&gt; is the winner on that timescale, with a &quot;90% probability of coming between 0.04 and 0.20pc between 240,000 and 470,000 years from now&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next best is GL710 a K-dwarf that will come within about 0.1-0.44pc in 1.3 million years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a statistical basis, some work has been done by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2001A%26A...379..634G&quot;&gt;Garcia Sanchez et al. (2001)&lt;/a&gt;. They estimate, using the Hipparcos data, that encounters within 1pc occur every 2.3 million years. However, the Hipparcos data isn't complete for low-mass, faint objects. Making a correction for this, the authors estimate a $&amp;lt;1pc$ encounter every 100,000 years. The probability for closer approaches scales as the inverse of the square of the separation. i.e. the timescales for approaches closer than 0.1pc is 100 times longer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are plenty of big error bars in all this work - recall that an error of 1km/s in velocity leads to a 1pc error in position after 1 million years. All this should be nailed down really well by Gaia results in the next 2-3 years.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2014-12-15T00:00:11.427" LastActivityDate="2014-12-15T00:00:11.427" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8251" PostTypeId="1" CreationDate="2014-12-15T12:04:41.223" Score="20" ViewCount="10483" Body="&lt;p&gt;I heard that Jupiter is made out of gas. But in school I learned that Jupiter has gravity which is 2.5 times that of Earth (Gravity that can tear apart a comet) and gravity is proportional to mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if Jupiter is made out of gas alone, how does it have such high gravity and maintain so many moons around it?&lt;/p&gt;&#xA;" OwnerUserId="4046" LastEditorUserId="2531" LastEditDate="2014-12-17T11:08:31.673" LastActivityDate="2014-12-17T11:08:31.673" Title="Is Jupiter made entirely out of gas?" Tags="&lt;gravity&gt;&lt;jupiter&gt;&lt;mass&gt;&lt;gas-giants&gt;&lt;core&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8252" PostTypeId="2" ParentId="8251" CreationDate="2014-12-15T12:14:31.223" Score="13" Body="&lt;p&gt;It doesn't matter if the body is made of gas, rocks, liquid or plasma, the four states of matter all have mass. So, as we know, mass create a gravitational field, and the more mass the stronger the gravity - and Jupiter has 317x Earth mass.&lt;/p&gt;&#xA;" OwnerUserId="4047" LastEditorUserId="2153" LastEditDate="2014-12-16T02:09:43.357" LastActivityDate="2014-12-16T02:09:43.357" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8254" PostTypeId="1" CreationDate="2014-12-15T12:32:38.177" Score="3" ViewCount="4175" Body="&lt;p&gt;I know gravity is the result of mass. Gravity is nothing but the attractive force on a body due to its mass. I have a question that may be silly. In magnetism we have attraction and repulsion. Earth is a big magnet, so I'm assuming that like Earth, other celestial objects also are magnetic. I'm bringing in the same concept of magnetism here into gravity. Since planets can attract each other they should repel something. What is that something that planets repel? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pardon me if I'm wrong anywhere. &lt;/p&gt;&#xA;" OwnerUserId="4046" LastEditorUserId="2153" LastEditDate="2015-02-21T15:02:57.967" LastActivityDate="2019-05-20T06:11:43.650" Title="Do planets repel?" Tags="&lt;gravity&gt;&lt;magnetic-field&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8255" PostTypeId="1" AcceptedAnswerId="8264" CreationDate="2014-12-15T13:47:13.763" Score="5" ViewCount="431" Body="&lt;p&gt;I have read about &lt;a href=&quot;http://en.wikipedia.org/wiki/Liquid_mirror_telescope&quot; rel=&quot;nofollow&quot;&gt;Liquid Mirror Telescopes&lt;/a&gt; in a number of places, and according to these sources, a major disadvantage of these telescopes is that they can only &quot;see&quot; straight up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a nutshell, the usually very expensive primary mirror is replaced with a rotating mercury container. The mercury surface will become parabolic with the rotation and earth's gravity, and a camera is placed in the focal point of the parabola (straight above the container). Since the parabolic shape depends on gravity, the primary cannot be tilted and the telescope can only point straight up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: couldn't one use a mirror scheme to steer the sight of this telescope? For example a pair of big plain mirrors a bit above the focal point, working similarly to a periscope. Then one could change the angles of these two plain mirrors and point to different directions in the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am assuming a large plain mirror is much cheaper than a parabolic mirror of the same size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has this been done? If not, why?&lt;/p&gt;&#xA;" OwnerUserId="4049" LastEditorUserId="2153" LastEditDate="2014-12-15T22:12:41.693" LastActivityDate="2016-05-26T22:43:28.260" Title="Is it possible to steer the sight of a Liquid Mirror Telescope using a plain mirror scheme? If yes, why hasn't it been done?" Tags="&lt;telescope&gt;&lt;optics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8256" PostTypeId="2" ParentId="8251" CreationDate="2014-12-15T14:59:38.533" Score="12" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Comet_Shoemaker%E2%80%93Levy_9#Chemical_studies&quot;&gt;Comet Shoemaker–Levy 9&lt;/a&gt; crashed into Jupiter a few years back. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As well as these molecules, emission from heavy atoms such as iron, magnesium and silicon was detected, with abundances consistent with what would be found in a cometary nucleus. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those heavy elements are consistent with the comet being at least being partially composed of rock. So Jupiter is known to contain at least &lt;em&gt;some&lt;/em&gt; rock. In fact, gas giants are thought to form around an initial small rocky or metallic core, which has sufficient gravity to draw in hydrogen, helium, water etc. from the protoplanetary nebula. By 'small rocky core' something with the mass of two or more earths is usually implied. The gravity field of such a body makes it difficult for even hydrogen to achieve escape velocity.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2014-12-15T14:59:38.533" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8257" PostTypeId="2" ParentId="8254" CreationDate="2014-12-15T15:37:41.027" Score="4" Body="&lt;p&gt;Gravity does not have polarity, it only attracts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An analogy with magnetic or electric fields is appealing (because they are all field forces, decay with the square of the distance, etc...) but science is not made of analogies, it is made of observations. And no one has observed gravity repulsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity is in fact much much weaker than the electric force (the factor has 42 zeros), and the only reason we feel gravity is because there is no repulsion, so all those little tiny mass pulls add up to something sensible. Electrical forces, albeit much stronger, usually cancel each other out due to an equilibrium of positive and negative charges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is of course, as far as we have seen in nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is however, &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_interaction_of_antimatter&quot; rel=&quot;nofollow noreferrer&quot;&gt;a very interesting speculation&lt;/a&gt; about what should happen with anti-matter. Would it repel &quot;normal&quot; matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the physics page there is a more in deep answer using the &lt;a href=&quot;https://physics.stackexchange.com/questions/11542/why-is-gravitation-force-always-attractive&quot;&gt;results of quantum&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recommended reading: &lt;a href=&quot;http://www.feynmanlectures.caltech.edu/I_07.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Feynman Lectures: Theory of Gravitation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4049" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2014-12-15T15:37:41.027" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8260" PostTypeId="1" AcceptedAnswerId="8263" CreationDate="2014-12-15T17:20:04.360" Score="4" ViewCount="496" Body="&lt;p&gt;It is known that the Sun is 4.6 billion years old, and the complete Solar System is of a similar age. &lt;a href=&quot;https://en.wikipedia.org/wiki/G-type_main-sequence_star&quot; rel=&quot;nofollow noreferrer&quot;&gt;The class of stars to which the Sun belongs&lt;/a&gt; seems to be quite common. Stars of its class can live for 8 or 9 billion years before moving to the red giant stage. Also, the whole universe is less than 14 billion years old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, not many generations of Sun-like stars could have existed before the first stages of our Solar System. And heavy elements are created when stars collapse. While there could be a lot of star classes with much shorter lifespans, I'm wondering: where do the heavier elements came from? Not only for our Solar System, but many others. I think it is very likely that our Solar System is of average age; do the numbers match between dead stars and remote stellar systems? Are there studies about this?&lt;/p&gt;&#xA;" OwnerUserId="4055" LastEditorUserId="2531" LastEditDate="2016-11-19T10:13:49.713" LastActivityDate="2016-11-19T10:13:49.713" Title="Solar System formation, considering its and the universe's age" Tags="&lt;the-sun&gt;&lt;solar-system&gt;&lt;universe&gt;&lt;star-formation&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8261" PostTypeId="2" ParentId="8251" CreationDate="2014-12-15T17:21:40.803" Score="6" Body="&lt;p&gt;According to Newton's Law of Universal Gravitation, you simply need interacting masses in order to generate a gravitational force between them.  Gases have mass and they therefore can contribute to gravity.  So even if Jupiter is entirely gaseous, it is so incredibly massive besides (so much gas!), that it has a much stronger gravitational pull than Earth.  The Sun is gaseous too, after all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be cautious, however, when someone compares gravitational forces so simply as &quot;2.5 times&quot;.  There is always a hidden assumption/reference in this because the force depends on more than just mass (e.g. distance).  Earth's gravity is way stronger where I sit!  In your teacher's case, it is probably meant that the gravitational force felt at the surface of Jupiter is 2.5 times what you would feel on the surface of Earth.  To choose a &quot;surface&quot; for a gaseous planet, you need to make more assumptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow, for the titled question, it is highly likely that Jupiter is not entirely gaseous.  With all of that matter (not just hydrogen and helium, but everything else everything in our solar system is made of) and all of that gravitational pull, you are bound to have precipitated solids that condense into a solid planetary core.&lt;/p&gt;&#xA;" OwnerUserId="4056" LastEditorUserId="4056" LastEditDate="2014-12-15T17:36:40.363" LastActivityDate="2014-12-15T17:36:40.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8263" PostTypeId="2" ParentId="8260" CreationDate="2014-12-15T18:39:58.417" Score="5" Body="&lt;p&gt;Early in the life of the universe it is thought that star formation in &quot;metal-free&quot; gas probably favoured larger stars. These &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_population#Population_III_stars&quot; rel=&quot;nofollow noreferrer&quot;&gt;Population III&lt;/a&gt; stars had short lives and very quickly enriched the interstellar and intergalactic medium with nucleosynthesis products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact the main enrichment of the interstellar medium (ISM) in our own Galaxy is thought to have taken place some billions of years before the Sun was born. Stars that are being formed now have basically the same composition as the Sun. The plot below shows the iron abundance (on a logarithmic scale) as a function of deduced age for stars in the Gaia-ESO spectroscopic survey (from &lt;a href=&quot;https://arxiv.org/abs/1401.4437&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bergemann et al. 2014&lt;/a&gt;). As you can see, there has not been too much change in iron abundance over the last 10 billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/K3ND3.png&quot; alt=&quot;Bergemann et al. 2014&quot;&gt;&#xA;I think the key to this is that most of the heavy elements in the gas in our Galaxy come from stars more massive than the Sun, and which have much shorter lifetimes as a result. [Stellar lifetime is roughly proportional to $M^{-2.5}$]. So several &quot;generations&quot; of these have lived and died (and must have done so).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To add some more detail. Almost all the elements heavier than Helium (known as &quot;metals&quot;) are produced inside stars. To get back into the ISM you have to get the &quot;processed&quot; material out of the centres of the stars. This occurs in basically three ways. (i) Supernovae - the explosions associated with the final collapse of very massive stars ($&amp;gt;10 M_{\odot}$). The supernova progenitors have short lives $&amp;lt;50$ million years, so many generations of such stars can have lived and died before the Sun was formed, and probably of order a few hundred million to a billion had done so. (ii) Giant star winds. These occur towards the ends of the lives of stars with mass of $&amp;lt;10M_{\odot}$. Some fraction of processed material gets dredged out of the centre and expelled into the ISM. The lifetimes of these stars is 50 million to $&amp;gt;10$ billion years. Most of the enrichment is caused by stars about in the middle of this range. These stars go on to fade as white dwarfs. (iii) Type Ia supernovae. Explosions triggered by mass transfer on to a white dwarf (probably in binary systems) causing a detonation and the scattering of metals into the ISM. Progenitors have lifetimes similar to the giant stars in (ii).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look carefully at the colour scheme of the plot above, you will notice that ancient stars have a higher ratio of magnesium to iron. This supports the ideas in the paragraph above. Magnesium is an alpha-capture element, produced mainly in short-lived massive stars and gets into the interstellar medium through type II supernovae. Conversely, although iron is also produced in massive stars, it is primarily produced and dispersed in type Ia supernovae that have much longer lived progenitors. Thus Mg enrichment occurs more rapidly than Fe enrichment in the early Milky Way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numbers of white dwarfs we see in the Galaxy and the deduced rate of supernova explosions that are occurring (e.g. from counting the number of neutron star remnants) also matches up with what is required to produce the metallicity of the ISM and the Sun. But there are plenty of variables and unknowns and not everything is solved. For instance it looks quite likely that star formation needed to be higher in the past; and some parts of the Galaxy (e.g. the bulge) have a higher metallicity, thought to be due to a period of very vigorous high mass star formation 10 billion years ago. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-11-19T10:10:41.357" LastActivityDate="2016-11-19T10:10:41.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8264" PostTypeId="2" ParentId="8255" CreationDate="2014-12-15T20:53:48.630" Score="8" Body="&lt;p&gt;Designs where the curved primary mirror is fixed, and steering is achieved via a moving flat mirror (siderostat) in front of it (or a set of flats) have been done before. One example is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pfund_telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pfund telescope&lt;/a&gt;. The main design constraint that led to the invention of the Pfund telescope was the need to have a focal point which is fixed in space and easily accessible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a variation of that design; the need for the fixed focal plane is driven by accessibility concerns here (the user cannot follow a moving focal plane):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aVn1b.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also see the related concept of &lt;a href=&quot;http://en.wikipedia.org/wiki/Heliostat&quot; rel=&quot;nofollow noreferrer&quot;&gt;heliostat&lt;/a&gt; - same idea, just applied to sunlight. A moving flat reflects sunlight in a fixed direction. Some solar power plants use heliostats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in principle, fixed curved primary + moving flat is doable, and has been implemented a number of times. However, there are substantial issues with this design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, the size and the quality of the flat mirror are not less than the size and quality of the parabolic primary. Let's say you build a liquid mirror telescope because it's easy. Then you have to make a flat mirror of the same diameter, with the same precision, and all the ease of construction is now lost. Flat mirrors are not easier to make than parabolic. In fact, as a mirror maker, I prefer to work on curved catopters - in some ways they are more predictable during manufacturing than large, high precision flats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are further issues down the road for the liquid mirror telescope. You would actually need 2 flats, both very large, both very precise. Both would have to be suspended in the air, above the liquid primary, and supported by complex mirror cells just like curved primaries in any telescope. Overall, it will be &lt;em&gt;more&lt;/em&gt; difficult to make such a telescope, not less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why is the Pfund design used? Not for ease of manufacturing; as I said above, it's actually more difficult to make it than a regular scope. But in some situations you just need that fixed focal plane for some reason. In that case, you don't have much of a choice. Of course, it's not a design that scales well.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am assuming a large plain mirror is much cheaper than a parabolic&#xA;  mirror of the same size.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That was the incorrect assumption. Once that's taken care of, everything else should be clear.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2016-05-26T22:43:28.260" LastActivityDate="2016-05-26T22:43:28.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8265" PostTypeId="2" ParentId="8251" CreationDate="2014-12-15T21:18:36.260" Score="9" Body="&lt;p&gt;I concur with everyone else here (of course) that the gravity at the &quot;surface&quot; of Jupiter is entirely determined by the mass contained within that surface. The composition makes no difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I differ with some on the answer to the headline title question. We simply &lt;strong&gt;do not know&lt;/strong&gt; whether Jupiter has a rocky core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A popular theory for the formation of giant planets is that they must begin by forming a rocky/icy core with perhaps a mass of 10-20 times the mass of the Earth. This then slowly accretes for a few million years until it grows massive enough for a runaway short period of gas accretion that builds its entire mass. See  &lt;a href=&quot;http://blog.planethunters.org/tag/core-accretion/&quot;&gt;http://blog.planethunters.org/tag/core-accretion/&lt;/a&gt; for another popular account. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Jupiter formed like this then it should have a core. However, this is not the only game in town. Planets &lt;em&gt;could&lt;/em&gt; also form in the protoplanetary disk by collapsing directly, in which case there would be no rocky/icy core. The recent discoveries of possible planet formation around the $&amp;lt;1$ million-year old HL Tau has led to a resurgence of this idea - it is difficult to form gas giants by core-accretion so quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer may come reasonably soon. In July 2016, NASA's Juno spacecraft will arrive in orbit around Jupiter. One of its prime mission objective is to gather more information on how massive the core is at the centre of Jupiter using a battery of instruments including an experiment very sensitive to tiny gravitational field variations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Juno_(spacecraft)&quot;&gt;http://en.wikipedia.org/wiki/Juno_(spacecraft)&lt;/a&gt; &lt;a href=&quot;http://www.nasa.gov/mission_pages/juno/overview/index.html#.VI9PuiusVSg&quot;&gt;http://www.nasa.gov/mission_pages/juno/overview/index.html#.VI9PuiusVSg&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2014-12-15T21:18:36.260" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8267" PostTypeId="1" AcceptedAnswerId="8270" CreationDate="2014-12-16T16:13:05.323" Score="0" ViewCount="986" Body="&lt;p&gt;Is this the following statement true?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An observer on earth only may see half of sky from the northern or southern hemisphere, and even if the observer stands on the equator on top of a very high mountain, he will hardly be able to see all the stars in the sky in given period!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Do most globe-like objects in space have this property?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd appreciate any more remarks on the subject.&lt;/p&gt;&#xA;" OwnerUserId="862" LastEditorUserId="2153" LastEditDate="2014-12-16T19:35:33.703" LastActivityDate="2014-12-16T19:35:33.703" Title="Can an observer on Earth only see half of the sky?" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8268" PostTypeId="1" AcceptedAnswerId="8271" CreationDate="2014-12-16T16:32:42.010" Score="2" ViewCount="406" Body="&lt;p&gt;I am learning orbital elements and one thing isn't clear to me. When I look at the Horizon's ephemeris data for Mars, &lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi#results&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; gives me argument of perifocus equal to roughly 286.53 degrees on epoch J2000. When I look up its longitude of perihelion for the same epoch &lt;a href=&quot;http://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, I get roughly 336.04 degrees. I thought these two parameters are the same thing. What is the difference between them? Why are the two numbers not the same?&lt;/p&gt;&#xA;" OwnerUserId="4070" LastEditorUserId="2153" LastEditDate="2014-12-16T19:01:33.887" LastActivityDate="2014-12-26T19:26:17.897" Title="Orbital elements: Difference between longitude of perihelion and argument of perifocus" Tags="&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8269" PostTypeId="1" CreationDate="2014-12-16T18:03:56.940" Score="3" ViewCount="106" Body="&lt;p&gt;I understand that &lt;a href=&quot;http://aa.usno.navy.mil/faq/docs/eqtime.php#solartime&quot; rel=&quot;nofollow&quot;&gt;UT1&lt;/a&gt; (and for that matter UT0, UT2, etc.) are based on &lt;em&gt;averages&lt;/em&gt; of actual earth rotation, and serve as a form of &lt;em&gt;mean&lt;/em&gt; solar time. However it’s not clear to me when these averages are taken and over what range? Certainly the average is taken over at least a year; but beyond that I can’t tell. Daily over the year? Yearly over the year; over several years?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How often and over what period is Earth’s rotation averaged to compute UT1?&lt;/p&gt;&#xA;" OwnerUserId="2369" LastEditorUserId="2369" LastEditDate="2014-12-17T15:47:02.213" LastActivityDate="2016-01-07T04:23:06.727" Title="How often and over what period is Earth’s rotation averaged to compute UT1?" Tags="&lt;observational-astronomy&gt;&lt;units&gt;&lt;time&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8270" PostTypeId="2" ParentId="8267" CreationDate="2014-12-16T18:22:20.977" Score="4" Body="&lt;p&gt;At any given instant of time, in any place on Earth, if the sky is clear and the horizon is low and flat, you see half of the celestial sphere - at that very instant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as the Earth keeps turning, you may end up seeing more, depending on where you are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're at the North or South Pole, you see exactly half of the sky no matter how long you wait. That's all you'll ever see from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you move closer to the Equator, you'll end up seeing more than half, if you're willing to wait.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the Equator, you see pretty much all the sky, if you wait as the Earth keeps spinning around, revealing all the sky to you eventually.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2014-12-16T18:22:20.977" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8271" PostTypeId="2" ParentId="8268" CreationDate="2014-12-16T18:41:00.707" Score="1" Body="&lt;p&gt;I know the answer now. Turns out that longitude of perihelion ($\overline\omega$) is defined as the sum of longitude of ascending node ($\Omega$) on the ecliptic plane and argument of perifocus ($\omega$) on the orbital plane: ($\Omega$ + $\omega$). It's strange to add up two non-coplanar angles, even when the inclination between the two planes is very small, but that's what it is.&lt;/p&gt;&#xA;" OwnerUserId="4070" LastEditorUserId="362" LastEditDate="2014-12-26T19:26:17.897" LastActivityDate="2014-12-26T19:26:17.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8272" PostTypeId="1" CreationDate="2014-12-16T19:36:40.327" Score="11" ViewCount="280" Body="&lt;p&gt;What is the method to determine the amount of reflected starlight necessary for an exoplanet to be visible from a given distance/angle? (Not from occlusion but actually visible on its own.) Further, how can you quantify this and calculate/predict/measure natural and artificial variations in the amount of reflected light?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to see equations that take into account/assume the following things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the properties of the detection equipment, such as telescope mirror diameter, optical resolving power, digital resolution and sensitivity (I'm a newbie to astronomy so I don't know the proper terms or units for these things; I come from the world of photography with ISO, pixel count, noise levels, etc.). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the properties of the exoplanet such as diameter, distance from host star, surface reflectance coefficient on a hypothetical maximally &quot;clear day&quot;, atmospheric thickness and reflected light absorption due to gasses, angles of a triangle drawn between the host star, exoplanet, and Earth.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;assume a space-based telescope on Earth and assume no interstellar dust of significance.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;assume no significant warping of spacetime by intense gravitational fields (i.e. we are looking at something in our stellar neighborhood, no further than, say, 50-100 LY)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Based on my nooblet research so far this equation should result in a figure expressed in Janksy units, I think? But I should like to know the average (RMS) photons per second that come from the exoplanet into the telescope, along with the peak value, during the periods of maximum and minimum intensity (i.e. when the planet is brightest and dimmest).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please give an example using figures for the best equipment we currently have available for this purpose and provide links if you can to the website for that project. Please in the example, just for the sake of argument, use a planet and star exactly like Earth and the Sun. If a planet like Earth would not be detectable at these parameters, assume more powerful detection equipment (just scale up the telescope's properties until it would work at a 10 LY distance and say how much you had to scale it up, and feel free to opine on how unrealistic such a device is).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, and here's the kicker: now assume an artificial object with a perfectly flat reflecting surface exists upon one side of this exoplanet. How would you incorporate the precise amount of extra light added by such a surface to the above equation as a separate part of the equation? Assume that you know the following properties of the reflecting surface in advance:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;area of the flat reflecting surface&lt;/li&gt;&#xA;&lt;li&gt;reflectance coefficient of the material used for constructing the surface &lt;/li&gt;&#xA;&lt;li&gt;geometric shape (assume a contiguous primitive geometric shape like a disc or square or triangle, or disregard if this property is irrelevant. If it's not irrelevant, then why is it not? What difference would the shape make on what gets detected?)&lt;/li&gt;&#xA;&lt;li&gt;angle between the reflecting surface plane and a plane tangent to the average sphere of the planet.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For an example lets assume a disc of polished white stone (0.8 reflectance) 20,000 sq. m. in size, positioned on at the part of the planet with the brightest insolation, at a 45-degree angle from the tangent plane. It's polished to a degree, but not a mirror. When the planet rotates, at some point, this surface reflects the host star's light directly towards Earth and would cause a glint of light for a small amount of time. How do we calculate the glint's duration? How would we mathematically distinguish such a glint? How could you tell if such a glint was reflected starlight as opposed to a laser pulse or other artificial light source?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What size would such an artificial reflecting surface (0.8 reflectance) need to be in order to make a planet visible with today's best telescopes from various distances up to 10 LY, where under current technology, the planet is not visible? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realized that in the case of flat reflecting surfaces then the inverse square law would not apply from the distance of the exoplanet to Earth, due to &lt;a href=&quot;https://en.wikipedia.org/wiki/Specular_reflection&quot; rel=&quot;nofollow&quot;&gt;Hero's rule&lt;/a&gt; (aka the &lt;strong&gt;law of reflection&lt;/strong&gt; aka &lt;strong&gt;specular reflection&lt;/strong&gt;). That means a totally still, large, flat body of water reflecting starlight, or a highly polished surface (like that of the Great Pyramid after its construction) would reflect a signal into space from the host star that would be detectable from a much greater distance away than the diffuse, incident light being reflected off of the planetary surface (which light would be subject to the inverse square law from the point of incidence).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Variables that would come into play would be the reflectance of the surface, the absorption of the atmosphere, flatness of the surface, and the level of insolation provided by the star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By my preliminary calculations the reflections off of the Great Pyramid would be visible from 20 LY. &lt;/p&gt;&#xA;" OwnerUserId="4071" LastEditorUserId="4071" LastEditDate="2016-03-18T15:14:52.270" LastActivityDate="2016-03-18T15:14:52.270" Title="Method to determine the amount of reflected starlight necessary for an exoplanet to be visible from a given distance/angle?" Tags="&lt;telescope&gt;&lt;exoplanet&gt;&lt;life&gt;&lt;extra-terrestrial&gt;&lt;space-telescope&gt;" AnswerCount="0" CommentCount="5" ClosedDate="2016-12-30T11:50:12.257" ContentLicense="CC BY-SA 3.0" />
  <row Id="8273" PostTypeId="2" ParentId="8269" CreationDate="2014-12-16T22:37:06.843" Score="5" Body="&lt;p&gt;Conceptually, the average is over an average year. Practically, that's not what's done anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sidereal time is much more readily measured than is solar time. Radio astronomy can measure the position of remote quasars much, much more precisely than they can measure the position of the Sun. The Sun is a big fiery blob with a not quite-well defined surface. Quasars are pinpoints. Very long baseline interferometry makes the observations of those quasars extremely precise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation of the equinoxes converts between apparent and mean sidereal time. A very simple relation converts between mean sidereal and mean solar time. This makes variations in UT1 observable on a rapid basis. Sub-arc second astronomy depends on these rapid updates. IERS Bulletin A is updated daily, with the data for the most recent week or so subject to subsequent small updates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Universal Time is not based on apparent solar time. Time is now effectively divorced from the the apparent revolution of the Earth with respect to the Sun. It took several thousands of years to make this divorce final. (And the divorce is still not quite final. There's the matter of leap seconds. There's a proposal to eliminate leap seconds that has been shelved/postponed multiple times since 2003. A vote is scheduled for 2015.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ancient Egyptians divided daytime into 12 equal parts and nighttime into 12 equal parts. Daytime and nighttime hours were of different lengths, and these lengths varied over the course of a year. Hipparchus (190 – 120 BCE) suggested dividing one day into 24 equal parts, the basis for our modern hour. Ptolemy, using information already garnered by ancient Babylonians, asked &quot;what day&quot;? The length of a solar day (local noon to local noon) varies over the course of a year thanks to what we now call the equation of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was of mostly academic interest until Christopher Huygens' development of the pendulum clock in 1656. Pendulum clocks tick at a rate proportional to the mean solar day rather than the apparent solar day. Those early pendulum clocks weren't particularly accurate. They had to be reset on a regular basis. The equation of time was central in determining the time to which those clocks needed to be set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the mid 1800s it was becoming apparent that even a mean solar day was not that good a basis for defining time. The clock kept by the solar system didn't agree with a clock kept by the Earth's mean rotation. A clock based on the solar system was more accurate and more consistent than was a clock based on the Earth's rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By 1900, the divorce between Earth rotation and time was well on its way to being finalized. The basis for time was switched from the Earth's rotation to the clock kept by the solar system, with the second based on the length of a year rather than the length of a day. About 150 years of observations of the length of the mean solar day went into that circa 1900 definition of the second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1967, time was further divorced from the clock kept by the solar system. By that time, atomic clocks had become extremely accurate devices, better than observations of the solar system. The International Atomic Time second was defined to be consistent with that circa 1900 definition of a second, which in turn was defined to be consistent with that ~150 years of observations of the length of an apparent solar day, centered on about 1821.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That 1821-based definition of a second as 1/86,400 of a day is no longer correct. A day is now about 86400.002 seconds long, and it will grow longer with passing millennia. This leads to the need to insert leap seconds to keep UTC in synchronization with the Earth's mean rotation. There is considerable debate with regard to whether even that should be done. There is a serious proposal to abandon leap seconds. This is now scheduled for a vote in the upcoming year (2015).&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2014-12-17T18:25:13.197" LastActivityDate="2014-12-17T18:25:13.197" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="8275" PostTypeId="1" AcceptedAnswerId="8278" CreationDate="2014-12-17T12:52:10.190" Score="7" ViewCount="2963" Body="&lt;p&gt;So, the theory (briefly) about accretion disks is that stuff falling into the black hole will most likely not fall straight into it but will miss it and go into orbit, perhaps for a long time, perhaps for a short time, before finally falling in.  That makes sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, why disks?  Wouldn't the matter falling into the BH be coming from all directions, and thus fall into orbits in all planes around the BH, not just one?  Or, is it that once there's a hint of &quot;plane preference&quot; in the debris, this will tend to be self-reinforcing, and anything subsequently falling into orbit around the BH will hit something in the disk, and so end up (perhaps after many collisions) being swept around in the disk as well?&lt;/p&gt;&#xA;" OwnerUserId="4076" LastActivityDate="2014-12-17T16:47:47.197" Title="Accretion disks - why are they disk-shaped, rather than spherical?" Tags="&lt;black-hole&gt;&lt;accretion-discs&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8276" PostTypeId="2" ParentId="133" CreationDate="2014-12-17T14:26:45.117" Score="1" Body="&lt;p&gt;Micro black holes are believed to exist, coming into existence for a very brief period of time following high energy particle collisions.  &lt;a href=&quot;http://arxiv.org/abs/1303.4723&quot; rel=&quot;nofollow&quot;&gt;Some theorists&lt;/a&gt; believe that this may occur naturally in Earth's atmosphere due to cosmic rays. I read a statistic somewhere which said that at any given time there would be three black holes in our atmosphere due to this process (if it exists).&lt;/p&gt;&#xA;" OwnerUserId="4076" LastEditorUserId="2153" LastEditDate="2014-12-17T16:15:50.700" LastActivityDate="2014-12-17T16:15:50.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8278" PostTypeId="2" ParentId="8275" CreationDate="2014-12-17T16:47:47.197" Score="5" Body="&lt;p&gt;There's two different effects here, and they're both related to viscuous forces in the accreting matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, if the infalling matter has some nonzero angular momentum vector $\mathbf{L}$, then consider a plane perpendicular to this vector. Due to its conservation, we can't get rid of rotation along this plane, but the component of angular momentum orthogonal to the plane is zero. As the accreting matter heats up and radiates away energy, that energy is taken from the gravitational potential, which is lessened by bringing the matter particles closer together. Thus, such interactions tend to flatten the matter into the plane orthogonal to its angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there's something more that happens for rotating black holes, as a combination of its gravitomagnetic field and viscous forces in the matter, that drives the matter to the plane of the black hole's rotation, regardless of its initial angular momentum. This is called the &lt;em&gt;Bardeen–Petterson effect&lt;/em&gt; &lt;sup&gt;[1]&lt;/sup&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In an accretion disk, viscosity causes each ring of gas to spiral inward toward the hole with a radial speed $|u^r|\ll\text{(orbital velocity)} = (M/r)^{1/2}$. If this were the only effect of viscosity, the warping of the disk would consist of a differential precession. Successive rings, each of smaller radius than the preceding, would have interacting longer with the [gravitomagnetic] field and thus would have precessed through successively larger angles... . As a result the disk would be radially twisted, but it would not be driven into the equatorial plane. However, the precession changes the disk's shear $\sigma_{jk}$ and thus changes the viscous forces; and these altered forces drive the disk toward the equatorial plane of the black hole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is only significant closer to the black hole, at $r\lesssim 100M$. At larger radii, the disk shape is due to the direction of the angular momentum of the matter instead.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;[1] &lt;em&gt;Black Holes: The Membrane Paradigm&lt;/em&gt;, ed. by Kip S. Thorne, Richard H. Price, and Douglas A. Macdonald.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-12-17T16:47:47.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8279" PostTypeId="1" CreationDate="2014-12-17T18:07:32.117" Score="2" ViewCount="207" Body="&lt;p&gt;The planets &lt;em&gt;approximately&lt;/em&gt; orbit the sun, but a more precise statement is that the planets &lt;em&gt;and&lt;/em&gt; the sun all orbit around a common barycenter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the sun is so massive, this barycenter is very close to the sun.  If we take the sun to be the origin, the barycenter travels in a complex, nonlinear trajectory (from Wikipedia):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/9XqRL.png&quot; alt=&quot;barycenter&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modeling this accurately is critically important to predicting celestial events.  I wrote a simulator to predict the &lt;a href=&quot;http://en.wikipedia.org/wiki/Transit_of_Venus,_2012&quot; rel=&quot;nofollow noreferrer&quot;&gt;2012 transit of Venus&lt;/a&gt;, based on estimates of orbits and the assumption that the sun is at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The error introduced by these assumptions was enough to miss the transit by a solar diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This data must exist, since the transit actually did happen.  Additionally, errors in the orbital data of the planets could have affected it.  Where can I find data on where the planets and the sun are and what trajectories they're moving in (relative to the solar system's barycenter)?&lt;/p&gt;&#xA;" OwnerUserId="2576" LastEditorUserId="286" LastEditDate="2016-07-17T13:29:55.393" LastActivityDate="2016-07-17T13:29:55.393" Title="Data for Sun's Orbit" Tags="&lt;orbit&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8280" PostTypeId="2" ParentId="748" CreationDate="2014-12-17T20:22:17.710" Score="20" Body="&lt;p&gt;Just to focus on one part of your question. Whilst it might be possible for a neutron star to accrete material, or for two neutrons stars to collide, in order to form black holes, this kind of event must be quite rare (although see below)&lt;/p&gt;&#xA;&lt;p&gt;The distribution of measured neutrons star and black holes masses can be fitted with an estimated true distribution. Here it is, from &lt;a href=&quot;http://adsabs.harvard.edu/abs/2012ApJ...757...55O&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ozel et al. (2012)&lt;/a&gt;. You can see there is a distinct gap between the highest mass neutron stars (currently the record holder has a mass of about &lt;span class=&quot;math-container&quot;&gt;$2M_{\odot}$&lt;/span&gt; and the smallest black holes (about &lt;span class=&quot;math-container&quot;&gt;$5M_{\odot}$&lt;/span&gt;). This confirmed slightly earlier work by &lt;a href=&quot;http://iopscience.iop.org/0004-637X/741/2/103/article#apj403391s5&quot; rel=&quot;nofollow noreferrer&quot;&gt;Farr et al. (2011)&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wkZQn.png&quot; alt=&quot;Neutron star and black hole mass distributions from Ozel et al. (2012).&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The merging of neutron stars must happen though. The obvious example is the Hulse-Taylor binary neutron star system, where the two object are spiralling together, presumably by the emission of gravitational waves, and will merge in about 300 million years. The combined mass of the 2 neutron stars is &lt;span class=&quot;math-container&quot;&gt;$2.83M_{\odot}$&lt;/span&gt;, but the mass of any black hole they create would be lower, with the difference radiated away as neutrinos and gravitational waves.&lt;/p&gt;&#xA;&lt;p&gt;Merging neutron stars (or merging neutron star + black hole binaries) are thought to be the progenitors of &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma-ray_burst#Short_gamma-ray_bursts&quot; rel=&quot;nofollow noreferrer&quot;&gt;short duration gamma ray bursts&lt;/a&gt; or so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Kilonova&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kilonova events&lt;/a&gt;that are generally seen in high redshift galaxies. These typically last a second or less, but involve an energy release of about &lt;span class=&quot;math-container&quot;&gt;$\sim 10^{44}$&lt;/span&gt; J. They may produce a black hole, or perhaps a more massive neutron star. There will also be a gravitational wave signature (a &amp;quot;chirp&amp;quot;) that could be detected by the next generation of gravitational wave experiments (now a reality). These black holes may be isolated and hence not represented in the mass distribution above. A further observational signature of these events may be in the form of the current levels of a number of heavy r-process elements, like Iridium and Gold, that &lt;em&gt;may&lt;/em&gt; mostly be produced in these events.&lt;/p&gt;&#xA;&lt;p&gt;As for accretion onto an existing neutron star - well it looks quite rare because there may be a large gap between the highest masses at which neutron stars are produced in supernovae (maybe &lt;span class=&quot;math-container&quot;&gt;$1.5M_{\odot}$&lt;/span&gt;) and the maximum mass of a neutron star. We know that the latter is at least &lt;span class=&quot;math-container&quot;&gt;$2M_{\odot}$&lt;/span&gt;, but it could be higher, perhaps &lt;span class=&quot;math-container&quot;&gt;$3M_{\odot}$&lt;/span&gt;, the maximum allowed by General Relativity. As to the outcome of this hypothetical event, well sticking to non-speculative physics, the most likely thing to happen would be the production of massive hyperons in the neutron star core at sufficiently high densities (&lt;span class=&quot;math-container&quot;&gt;$&amp;gt;10^{18}$&lt;/span&gt; kg/m&lt;span class=&quot;math-container&quot;&gt;$^3$&lt;/span&gt;), which would lead to an instability (due to the removal of degenerate neutrons that are providing the majority of support); the neutron star may then slip inside its event horizon (about 6km for a &lt;span class=&quot;math-container&quot;&gt;$2M_{\odot}$&lt;/span&gt; neutron star) and become a black hole. Some sort of explosion seems unlikely, though a gravitational wave signature might be possible.&lt;/p&gt;&#xA;&lt;p&gt;EDIT: An update on the NS/BH mass distribution above. I saw a talk recently at a conference - the explanation of the distribution has two broad thrusts; either the black holes are not produced in this mass range because of the physics of the progenitors, or there is a strong observational bias against seeing them. An example of the former explanation can be found in &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014ApJ...785...28K&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kochanek (2014)&lt;/a&gt;, who proposes that there is a class of &amp;quot;failed supernovae&amp;quot; between 16 and 25&lt;span class=&quot;math-container&quot;&gt;$M_{\odot}$&lt;/span&gt; that do manage to eject their envelopes in &lt;em&gt;weak&lt;/em&gt; transient events, but leave behind their helium cores to form the lowest mass 5-8&lt;span class=&quot;math-container&quot;&gt;$M_{\odot}$&lt;/span&gt; black holes. Lower mass progenitors are then responsible for the neutron stars.&lt;/p&gt;&#xA;&lt;p&gt;The observational bias is that the companions to the lowest mass black holes in binary systems may be &lt;em&gt;always&lt;/em&gt; overflowing their Roche lobes. The resultant accretion signature swamps the companion spectrum and prevents a dynamical mass estimate (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/1999ApJ...522..413F&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fryer 1999&lt;/a&gt;). The &lt;a href=&quot;https://personal.sron.nl/%7Epeterj/gbs/survey.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chandra Galactic Bulge Survey&lt;/a&gt;&#xA;is attempting to find examples of quiescent, relatively low X-ray luminosity, eclipsing compact binaries, with which to measure a more unbiased black hole mass distribution.&lt;/p&gt;&#xA;&lt;p&gt;Further Edit: There continue to be challenges and claims that there are &amp;quot;low-mass&amp;quot; black holes that could be formed via accretion-induced collapse of a neutron star (pointed out by Alexandra Veledina). For instance Cygnus-X3 has a claimed mass of &lt;span class=&quot;math-container&quot;&gt;$2.4^{+2.1}_{-1.1}\ M_{\odot}$&lt;/span&gt; according to &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013MNRAS.429L.104Z&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zdziarski et al. 2013&lt;/a&gt;, but these observations lack the precision to be really sure yet.&lt;/p&gt;&#xA;&lt;p&gt;Addition 30/6/20:&lt;/p&gt;&#xA;&lt;p&gt;A compact object has now been found with a precise mass measurement between &lt;span class=&quot;math-container&quot;&gt;$2.5-2.67 M_{\odot}$&lt;/span&gt;. It was detected through a gravitational wave signature during its merger with a much more massive black hole. This object was either the most massive neutron star found yet, or the least massive black hole (&lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/ab960f&quot; rel=&quot;nofollow noreferrer&quot;&gt;Abbott et al. 2020&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2020-06-30T08:22:10.643" LastActivityDate="2020-06-30T08:22:10.643" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8281" PostTypeId="1" AcceptedAnswerId="8305" CreationDate="2014-12-17T21:08:51.733" Score="3" ViewCount="243" Body="&lt;p&gt;Is there a stable polar solar orbit such that the Earth is always in view and not eclipsed by the Sun?&lt;/p&gt;&#xA;" OwnerUserId="4041" LastEditorUserId="13" LastEditDate="2014-12-19T16:40:05.937" LastActivityDate="2014-12-19T18:30:28.870" Title="Stable polar solar orbit with the Earth continuously observable" Tags="&lt;orbit&gt;&lt;solar-system&gt;" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8282" PostTypeId="1" CreationDate="2014-12-17T23:37:26.443" Score="1" ViewCount="168" Body="&lt;p&gt;I've heard from a number of different places that the density in the interstellar medium can have an average of 1 atom per centimeter cubed. Perhaps I have a wrong understanding of what 'nothing' is (I am very knew to astronomy/astrophysics), but how would a vehicle, like the Voyager for example, be able to travel through it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My (perhaps very flawed) thinking behind this is that if there are so few atoms not in close proximity, there couldn't be any (meaningful) forces interacting between them to provide any sort of way to travel through it. &lt;/p&gt;&#xA;" OwnerUserId="3012" LastActivityDate="2014-12-17T23:50:05.787" Title="How would vehicles travel through the interstellar medium with its such low density?" Tags="&lt;interstellar-travel&gt;&lt;interstellar-medium&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8283" PostTypeId="2" ParentId="8281" CreationDate="2014-12-17T23:42:30.447" Score="0" Body="&lt;p&gt;I could imagine a polar orbit that is in a plane which rotates to stay orthogonal to the direction of the Earth.  The rotation would be very slow and match the revolution rate of the Earth.  The satellite path would be similar to how a ball of yarn is wound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's nothing too fancy about this (you could do the same switching Earth and the Sun), so I wouldn't expect any abnormal forces that would decay the orbit quicker than any other.  Also, with a little ion engine, it'd be that much easier to maintain.&lt;/p&gt;&#xA;" OwnerUserId="4056" LastActivityDate="2014-12-17T23:42:30.447" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8284" PostTypeId="2" ParentId="8282" CreationDate="2014-12-17T23:49:25.593" Score="2" Body="&lt;p&gt;For certain methods of propulsion, you need a medium.  In space, you can use a medium that includes cosmic wind in order to drive something like a solar sail.  In that case, the energy of particles is important even if they are sparse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, often you use a rocket engine in space and it doesn't matter what you are moving through.  Not to mention that inertia is a valuable way to get around in low friction environments (once you get moving, keep going).&lt;/p&gt;&#xA;" OwnerUserId="4056" LastActivityDate="2014-12-17T23:49:25.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8285" PostTypeId="2" ParentId="8282" CreationDate="2014-12-17T23:50:05.787" Score="6" Body="&lt;p&gt;The answer is in your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You said&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;there couldn't be any (meaningful) forces interacting between them to provide any sort of way to travel through it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's the great thing about outer space. There normally aren't many particles around that could slow down a spacecraft via drag or friction. If you give something a push in outer space - somewhere that unrealistically is free from any gravitational interaction with other objects - it will travel forever. Objects travel through the ISM via Newton's first law - an object in motion tends to stay in motion.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-12-17T23:50:05.787" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8286" PostTypeId="1" AcceptedAnswerId="8297" CreationDate="2014-12-18T01:07:21.650" Score="8" ViewCount="178" Body="&lt;p&gt;There are essentially two configurations of orbits &lt;a href=&quot;https://en.wikipedia.org/wiki/Binary_star#Calculating_the_center_of_mass_in_binary_stars&quot; rel=&quot;noreferrer&quot;&gt;a system with two bodies&lt;/a&gt; can have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A configuration where the paths do not cross:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/f/f2/Orbit2.gif&quot; alt=&quot;Paths don&amp;#39;t cross&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A configuation where the paths cross:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/0/0e/Orbit5.gif&quot; alt=&quot;Paths cross&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: Size of the bodies relative to one another is unimportant here.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What determines which of the two configurations a binary system will develop into?&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-12-18T16:39:00.860" LastActivityDate="2014-12-19T17:30:19.953" Title="What determines the configuration of orbits in a binary system?" Tags="&lt;orbit&gt;&lt;binary-star&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8287" PostTypeId="2" ParentId="8281" CreationDate="2014-12-18T02:49:25.547" Score="0" Body="&lt;p&gt;Sure. Imagine an object orbiting the Sun in a plane normal to the plane of Earth's orbit. Let AB the the intersection of these two planes. (Imagine that A and B are points on the celestial sphere. We're not interested for now in distance from the Sun.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the only time the Sun hides the Earth from the object is when the object is at A and the Earth is at B, or vice versa. The Earth will be at A or B every 6 months, so we only need to arrange that the object is not at the other point at either of these times. Put it in a circular orbit whose period is a multiple of 6 months, and time it so that whenever the Earth is at A or B the object is due north (or south) of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every six months, the Earth will be at A or B, and the object will be north of the Sun and can see Earth. At all other times, the Earth is not at A or B, and the object can see Earth no matter where the object is.&lt;/p&gt;&#xA;" OwnerUserId="4077" LastActivityDate="2014-12-18T02:49:25.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8288" PostTypeId="1" AcceptedAnswerId="8290" CreationDate="2014-12-18T04:53:02.597" Score="5" ViewCount="2282" Body="&lt;p&gt;According to wikipedia and other sources, a planet and a star always move in a circular orbit around the common center of mass of the both bodies, in case  of Earth and sun, this center of mass lies inside the sun. Therefore, Earth should orbit the sun in a circular orbit. So, my question is, why does the Earth orbits the sun in an Elliptical orbit?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, if it does, and one focus is the sun, then what is the other focus?&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2014-12-18T15:41:43.643" LastActivityDate="2018-11-25T21:23:12.143" Title="What is the orbit of Earth around the Sun?" Tags="&lt;orbit&gt;&lt;the-sun&gt;&lt;gravity&gt;&lt;solar-system&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8289" PostTypeId="2" ParentId="8288" CreationDate="2014-12-18T12:04:53.013" Score="-5" Body="&lt;p&gt;The trajectory of Earth's orbit is shaped not only by Sun's gravity. Earth's orbit is being changed by different sources of gravity as well, for example center of our galaxy. This is why it's not perfectly round.&lt;/p&gt;&#xA;" OwnerUserId="4078" LastEditorUserId="4078" LastEditDate="2014-12-18T13:03:19.327" LastActivityDate="2014-12-18T13:03:19.327" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8290" PostTypeId="2" ParentId="8288" CreationDate="2014-12-18T15:36:55.920" Score="10" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;According to wikipedia and other sources, a planet and a star always move in a circular orbit around the common center of mass of the both bodies ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is not true. In the absence of other gravitational sources, a planet and a star move in elliptical orbits about the common center of mass. Ancient scientists assumed circular orbits, but only because circles were somehow most pure. There were problems with this: It didn't match what those ancient scientists saw. So they modified these circular orbits by tacking on other circles. It was a mess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kepler cleaned this mess up by noting that the orbits of the planets are very close to elliptical rather than circular. This empirical result was later strengthened by Newton, who derived that elliptical orbits (not circular orbits) are a natural consequence of a force that follows an inverse square law. Gravitation is such a force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbits of the planets about the solar system are not quite elliptical because planets are attracted to one another as well as to the Sun. Nonetheless, the orbits are still very close to elliptical. The Sun is more than 1000 times more massive than Jupiter. Jupiter, Saturn, and the other planets represent tiny perturbations on the nearly elliptical orbits of the planets about the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about other perturbing effects, such as nearby stars and the Milky Way as a whole? These are immeasurably small. A passing star will perturb planetary orbits over the course of the tens of millions of years or more. The Milky Way as a whole has essentially no effect whatsoever. Theoretically, the effect most certainly does exist, but it is so small that it can be ignored, even over the lifespan of the solar system.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2014-12-18T15:36:55.920" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8291" PostTypeId="1" AcceptedAnswerId="8293" CreationDate="2014-12-18T20:14:46.833" Score="9" ViewCount="2777" Body="&lt;p&gt;I have been looking for some kind of collection of all known stars and their individual respective mass/radius/temperature/and luminosity, can you link me to one if it exists?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal is to write an algorithm that creates polynomial functions that calculate a stars temp/lum/radius based on mass. I know there are existing equations, however the ones I have found are more troublesome and imprecise compared to if I could use polynomial functions I create myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hertzsprung-Russel diagram ( &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/1/17/Hertzsprung-Russel_StarData.png&quot; rel=&quot;noreferrer&quot;&gt;http://upload.wikimedia.org/wikipedia/commons/1/17/Hertzsprung-Russel_StarData.png&lt;/a&gt; ) is a great example of representation of what data I need, however it's axes are not constant, so I can not visually gather accurate sample data to deduce any functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Essentially even a large quantity of estimates will suffice. I intend to use this data to generate a semi-realistic galaxy for an MMO I am developing, so it doesn't by any means need to be perfect.&lt;/p&gt;&#xA;" OwnerUserId="4080" LastEditorUserId="4080" LastEditDate="2014-12-18T20:38:47.440" LastActivityDate="2014-12-18T20:47:43.570" Title="Is there anywhere I can find some kind of database on all known stars and their properties (mass, surface temperature, radius, and luminosity)?" Tags="&lt;star&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8292" PostTypeId="2" ParentId="418" CreationDate="2014-12-18T20:23:21.110" Score="2" Body="&lt;p&gt;If you stick to observing &quot;hot Jupiters&quot;, then this is very much in reach of &quot;amateur&quot; technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would agree that a 10-inch+ telescope is probably needed, along with a cooled CCD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hot Jupiters (giant planets orbiting close to their parent stars) produce transit signals of about 0.01-0.02 mag amplitude. The transits last of order a couple of hours, occur every 1-10 days and the times of transit are well predicted. In principle you could collect all the data you need in about 6 hours of observing. But, the amplitude of the transit dip is small, so you need to get very precise &lt;em&gt;differential&lt;/em&gt; photometry. Your best bet is to observe targets which have plenty of other stars in the same CCD field that can act as comparisons - this may mean you need a CCD with a wide field of view. On the other hand, you &lt;em&gt;must&lt;/em&gt; make sure that the seeing disk of the star is well-sampled by the CCD pixels (e.g. in your typical seeing is 2 arcseconds, each CCD pixel must not image more than 1 arcsec on the sky, and preferably less).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other strategies for success involve observing at low airmasses, which should improve the quality of the differential photometry, and don't bother with anything observed even through the thinnest cirrus. Observing multiple transits will allow you to improve your data by &quot;phase-folding&quot; on the known planet orbital period.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2014-12-18T20:23:21.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8293" PostTypeId="2" ParentId="8291" CreationDate="2014-12-18T20:44:33.247" Score="2" Body="&lt;p&gt;I did something similar with the HRD about a week ago. I found that &lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR&quot; rel=&quot;nofollow&quot;&gt;VizieR&lt;/a&gt; has a very large database with observations. You can download the tables in different formats (like csv or plain text), but you should first check if the Temperature (mostly log.Tegg) and luminosity (logL) are available. A very detailed one is table &lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR?-source=V/19&quot; rel=&quot;nofollow&quot;&gt;V/19&lt;/a&gt;. It has those values for 68 clusters, enough to keep you busy for a while!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also masses in that list, but you can also create an isochrone yourself, and determine the age of each cluster. Well that was my assignment, and we got pretty close to the expected values for some clusters!&lt;/p&gt;&#xA;" OwnerUserId="1309" LastActivityDate="2014-12-18T20:44:33.247" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8294" PostTypeId="2" ParentId="8291" CreationDate="2014-12-18T20:47:43.570" Score="4" Body="&lt;p&gt;There really isn't anything as comprehensive as you want. And, we don't have accurate information for all of these variables for a large number of stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two main problems are masses and radii.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really the only set of stars that have both measured masses and radii are those in eclipsing binary systems. If these binary stars also have distances, measured by parallax, then the components could also have luminosity estimates and estimates of effective temperature.&#xA;The best catalogue I can recommend is that compiled by John Southworth, containing perhaps a 100 systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astro.keele.ac.uk/jkt/debcat/&quot; rel=&quot;nofollow&quot;&gt;http://www.astro.keele.ac.uk/jkt/debcat/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just want a mass, but can cope without a radius, then you can also include components of astrometric binary systems. There is a well-known paper by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2000A%26A...364..217D&quot; rel=&quot;nofollow&quot;&gt;Delfosse et al. (2000)&lt;/a&gt; containing stellar masses and luminosities.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2014-12-18T20:47:43.570" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8295" PostTypeId="1" CreationDate="2014-12-18T21:42:50.323" Score="4" ViewCount="1682" Body="&lt;p&gt;I am working on an orbital dynamics code and, for fun, I would like to model the Sun-Earth-Moon system with my code.  I can look up the masses of each object just fine, as well as average distances and relative velocities (which I can also calculate).  However, I would like to model the Sun-Earth-Moon system in great detail.  I would like to know the positions and velocities of each object on any given day (today works just fine) so that I can represent the orbits of these objects with great accuracy, hopefully even reproducing their actual ellipticity.  Where can I find values for:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;current Sun-Earth distance&lt;/li&gt;&#xA;&lt;li&gt;current Earth-Moon distance&lt;/li&gt;&#xA;&lt;li&gt;current Sun-Moon angular separation&lt;/li&gt;&#xA;&lt;li&gt;current velocity of Earth relative to Sun (or relative to Sun-Earth(-Moon?) center of mass)&lt;/li&gt;&#xA;&lt;li&gt;current velocity of Moon relative to Earth&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and, maybe even&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;current velocity of Sun relative to Sun-Earth(-Moon?) center of mass.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is there any resource that provides these current values?&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2014-12-18T22:01:32.593" Title="Look up current positions and velocities of solar system objects" Tags="&lt;the-moon&gt;&lt;the-sun&gt;&lt;solar-system&gt;&lt;earth&gt;&lt;n-body-simulations&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8296" PostTypeId="2" ParentId="8295" CreationDate="2014-12-18T22:01:32.593" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there any resource that provides these current values?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. The &lt;a href=&quot;http://ssd.jpl.nasa.gov/?horizons&quot; rel=&quot;nofollow&quot;&gt;JPL HORIZONS on-line solar system data and ephemeris computation service&lt;/a&gt; provides these values, and much more.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2014-12-18T22:01:32.593" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8297" PostTypeId="2" ParentId="8286" CreationDate="2014-12-18T23:58:22.923" Score="6" Body="&lt;p&gt;Following the convention $\mathbf{r} = \mathbf{r}_2-\mathbf{r}_1$, $M = m_1+m_2$, in the center-of-mass frame we have, by definition,&#xA;$$\begin{eqnarray*}\mathbf{r}_1 = -\frac{m_2}{M}\mathbf{r}\text{,}\quad&amp;amp;\mathbf{r}_2 = \frac{m_1}{M}\mathbf{r}\text{.}\end{eqnarray*}$$&#xA;Hence, $\ddot{\mathbf{r}} = -GM\hat{\mathbf{r}}/r^2$ implies that the individual orbits are similar conic sections in this frame, and moreover in the bound case they are ellipses that share a common focus at the center of mass, with all three of the distinct foci being collinear and with the center between them. Although there is a degenerate case of a circular orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, there's a geometrically simple condition for the intersecting configuration: there is intersection if, and only if, the distance to apoapsis of the larger mass (smaller orbit) is greater than or equal to the distance to periapsis of the smaller mass (larger orbit). Since a general ellipse in polar coordinates about a focus can be described by&#xA;$$r = \frac{p}{1+e\cos(\phi-\phi_0)}\text{,}$$&#xA;where $e$ is the eccentricity, and the individual orbits are proportional, we have intersection if, and only if,&#xA;$$\frac{1-e}{1+e}\leq \frac{m_1}{m_2}\leq\frac{1+e}{1-e}\text{,}$$&#xA;as the apsides occur when the cosine term is $\pm 1$.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditDate="2014-12-19T17:30:19.953" LastActivityDate="2014-12-19T17:30:19.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8298" PostTypeId="2" ParentId="418" CreationDate="2014-12-19T03:18:47.910" Score="3" Body="&lt;p&gt;IEEE Spectrum recently ran a piece on detecting exoplanets:&#xA;&lt;a href=&quot;http://spectrum.ieee.org/geek-life/hands-on/diy-exoplanet-detector&quot; rel=&quot;nofollow&quot;&gt;DIY Exoplanet Detector -&#xA;You don’t need a high-powered telescope to spot the signature of an alien world&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Canon EOS Rebel XS DSLR. With old manual-focus lenses now useless to most photographers, I was able to acquire a 300-millimeter Nikon telephoto lens...&#xA;   An Arduino-controlled star tracker...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More info here: &lt;a href=&quot;http://www.cloudynights.com/topic/462161-detect-known-exoplanet-with-dslrtelephote-lens/#entry6010854&quot; rel=&quot;nofollow&quot;&gt;Detect known exoplanet with DSLR/telephote lens&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The star in question, &lt;a href=&quot;http://en.wikipedia.org/wiki/HD_189733&quot; rel=&quot;nofollow&quot;&gt;HD 189733A&lt;/a&gt;, has a visual magnitude of 7.6. Dimmer targets would naturally require more light gathering power than an old telephoto lens can supply.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2014-12-19T03:18:47.910" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8299" PostTypeId="2" ParentId="1065" CreationDate="2014-12-19T09:55:46.420" Score="4" Body="&lt;p&gt;I thought I'd contribute an answer because there's a &lt;em&gt;very&lt;/em&gt; recent paper on the subject:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://iopscience.iop.org/0004-637X/798/1/48&quot; rel=&quot;nofollow&quot;&gt;Measuring the solar radius from space during the 2012 Venus Transit&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appeared in my RSS feeds this morning!  A related writeup is online &lt;a href=&quot;http://hmi.stanford.edu/hminuggets/?p=522&quot; rel=&quot;nofollow&quot;&gt;at the HMI website&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the question, this measurement uses the transit of Venus to fit the limb-darkening law of the Sun. That is, the Sun is a bit fainter the further from the centre that you look. As you reach the optically thinner layers near the &quot;surface&quot;, the brightness falls off rapidly, towards zero in the vacuum of space. The inflection point of the curve (as a function of distance from the centre of the disk) is a reasonable estimate of the &quot;radius&quot;.  As pointed out elsewhere, the value changes depending on which wavelength you use, but only by a few hundred km, compared to the Sun's overall radius of about 700 000km (actually more like 695 946 km), so the uncertainty is at or below the 0.1% level. Phil Plait &lt;a href=&quot;http://blogs.discovermagazine.com/badastronomy/2012/03/21/the-sun-is-1392684-65-km-across/#.VJP1EuDkA&quot; rel=&quot;nofollow&quot;&gt;wrote&lt;/a&gt; about a similar measurement (by the same team, I believe) that used the transits of Mercury in 2003 and 2006.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the team also used the limb-darkening (I think) to &lt;a href=&quot;http://www.sciencemag.org/content/337/6102/1638.abstract&quot; rel=&quot;nofollow&quot;&gt;measure how round the Sun is&lt;/a&gt;. i.e. the diameter from top-to-bottom versus left-to-right. Answer: the Sun is &lt;em&gt;very very&lt;/em&gt; round, with the radii differing by a few parts per million.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2014-12-19T09:55:46.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8300" PostTypeId="1" AcceptedAnswerId="8303" CreationDate="2014-12-19T10:42:55.713" Score="23" ViewCount="3680" Body="&lt;p&gt;I was just wondering what are the chances that there is a small object (say less than 1&amp;nbsp;km but more than few meters) that orbits the Earth but has remained undetected by us? Are we actually scanning the space around the Earth continuously for orbiting bodies?&lt;/p&gt;&#xA;" OwnerUserId="4084" LastEditorUserId="1501" LastEditDate="2014-12-19T16:39:01.393" LastActivityDate="2015-06-29T07:35:07.490" Title="Does the Earth have another moon?" Tags="&lt;the-moon&gt;&lt;earth&gt;&lt;near-earth-object&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8301" PostTypeId="1" CreationDate="2014-12-19T10:49:31.107" Score="4" ViewCount="144" Body="&lt;p&gt;What is the minimum size of an object that can be observed from Earth with our most powerful telescope? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the answer depends on how far the object is from Earth. For the sake of  discussions let us consider distances less the orbit of Neptune.&lt;/p&gt;&#xA;" OwnerUserId="4084" LastActivityDate="2014-12-20T10:00:06.337" Title="Smallest body that can be observed from Earth?" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8302" PostTypeId="2" ParentId="8300" CreationDate="2014-12-19T11:33:23.800" Score="14" Body="&lt;p&gt;One kilometer, no way! That would've been known since long ago. Most asteroids of that size have already been found, all the way out to the asteroid belt beyond Mars. Earth has &lt;a href=&quot;https://en.wikipedia.org/wiki/Other_moons_of_Earth&quot;&gt;no second Moon&lt;/a&gt;. But there are always some tiny asteroids around, which are temporarily captured by Earth's gravity. Here's a &lt;a href=&quot;http://www.space.com/25321-minimoons-asteroids-human-spaceflight.html&quot;&gt;funny illustration&lt;/a&gt; of such an orbit, it is not what we would like to call a moon. I think that only one of them have been found, the 5 meter diameter &lt;a href=&quot;https://en.wikipedia.org/wiki/2006_RH120&quot;&gt;2006RH120&lt;/a&gt;. Smaller objects would not be detectable today, but there are at any moment likely many meter and submeter sized asteroids visiting the Sun-Earth Lagrange points. &lt;a href=&quot;http://arxiv.org/pdf/1112.3781v1.pdf&quot;&gt;A paper about it&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space is not yet being scanned, but telescopes are now being built to do that. It will be a new kind of astronomy, looking for the unexpected, and who knows what will be found out there in the blackness. Still today amateur astronomers can discover asteroids with the telescope in their backyard. I'm afraid that space telescopes like &lt;a href=&quot;http://sci.esa.int/gaia/&quot;&gt;Gaia&lt;/a&gt; will kill their hobby and that this classic astronomical mapping of the sky will finally be finished.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2014-12-19T11:33:23.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8303" PostTypeId="2" ParentId="8300" CreationDate="2014-12-19T12:52:39.343" Score="25" Body="&lt;p&gt;Not strictly satellites/moons, but certainly companions are &lt;a href=&quot;https://en.wikipedia.org/wiki/2010_TK7&quot;&gt;2010 TK&lt;sub&gt;7&lt;/sub&gt;&lt;/a&gt; with a diameter of ~300&amp;nbsp;m, an &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth_trojan&quot;&gt;Earth trojan&lt;/a&gt; at the &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_objects_at_Lagrangian_points#L4&quot;&gt;L4&lt;/a&gt; point, and the ~5&amp;nbsp;km &lt;a href=&quot;https://en.wikipedia.org/wiki/3753_Cruithne&quot;&gt;3753 Cruithne&lt;/a&gt; in a peculiar orbit locked to the Earth's.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="332" LastEditDate="2014-12-20T02:44:47.110" LastActivityDate="2014-12-20T02:44:47.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8304" PostTypeId="2" ParentId="2305" CreationDate="2014-12-19T18:11:15.367" Score="0" Body="&lt;p&gt;Maybe move your planar restricted problem to a 3D orbital restricted one.&#xA;(i.e. first two bodies in fixed orbit then solve for third body off the plane)&lt;/p&gt;&#xA;" OwnerUserId="4041" LastEditorUserId="4041" LastEditDate="2014-12-19T18:16:57.110" LastActivityDate="2014-12-19T18:16:57.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8305" PostTypeId="2" ParentId="8281" CreationDate="2014-12-19T18:30:28.870" Score="2" Body="&lt;p&gt;As ganbustein says, this is not too difficult to imagine.  The simplest case (approximating with circular orbits and only the Sun, Earth and Satellite) would have the satellite orbit orthogonally to the Earth with a 1 year orbit.  The Satellite will pass the Earth orbit plane in two places, call these &quot;down crossing&quot; and &quot;up crossing&quot; points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To minimize Earth - Satellite iterations, keep them 90 degrees apart.  Have the Satellite over the north solar pole when the Earth is at the &quot;down crossing&quot; point.  Then when the satellite gets to the &quot;down crossing point&quot; the Earth will be furthest from the Satellite plane.  When the earth is at the &quot;up crossing point&quot; the satellite will under the Suns south pole. And so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would not be completely stable when we include Jupiter and the mutual interactions, but I think they should be small, allowing this to work generally.  If someone &quot;does the math&quot; and proves me wrong, I will accept that.&lt;/p&gt;&#xA;" OwnerUserId="4090" LastActivityDate="2014-12-19T18:30:28.870" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8307" PostTypeId="1" CreationDate="2014-12-20T08:06:22.833" Score="3" ViewCount="433" Body="&lt;p&gt;I have been getting mixed opinions on this. I have a 15x70 Pair of Binoculars. I have tried staring at Jupiter for several minutes with it, I was able to see the Galilean moons, in fact the last time was so long that I noticed the Moon's position changing a little bit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the glare of Jupiter is so high that I could not see the bands. Is it really possible to see any of Jupiter's bands using a 15x70 pair when the angular size is about 45 arcseconds ?&lt;/p&gt;&#xA;" OwnerUserId="4099" LastEditorUserId="4099" LastEditDate="2014-12-30T10:17:42.530" LastActivityDate="2014-12-30T10:17:42.530" Title="Can Jupiter's bands be made out using a 15x70 pair of binoculars?" Tags="&lt;jupiter&gt;&lt;star-gazing&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2015-01-07T16:03:02.463" ContentLicense="CC BY-SA 3.0" />
  <row Id="8308" PostTypeId="2" ParentId="8307" CreationDate="2014-12-20T09:53:54.657" Score="2" Body="&lt;p&gt;I use a &lt;a href=&quot;http://www.celestron.com/browse-shop/astronomy/astronomy-binoculars/skymaster-20x80-binocular&quot; rel=&quot;nofollow&quot;&gt;20x80&lt;/a&gt; binocular on a tripod and I cannot make out Jupiter's bands. I believe that a 15x70 binocular will not be adequate.&lt;/p&gt;&#xA;" OwnerUserId="209" LastActivityDate="2014-12-20T09:53:54.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8309" PostTypeId="2" ParentId="8301" CreationDate="2014-12-20T10:00:06.337" Score="3" Body="&lt;p&gt;The smallest &lt;strong&gt;natural&lt;/strong&gt; object that I know of that has been detected from an Earth-bound telescope is &lt;a href=&quot;http://en.wikipedia.org/wiki/2011_CQ1&quot; rel=&quot;nofollow&quot;&gt;2011 CQ1&lt;/a&gt;, a one-meter asteroid observed while &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalina_Sky_Survey&quot; rel=&quot;nofollow&quot;&gt;specifically looking for NEOs&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As DeerHunter mentions in a comment, smaller artificial objects are identified and tracked, mostly by &lt;a href=&quot;http://en.wikipedia.org/wiki/North_American_Aerospace_Defense_Command&quot; rel=&quot;nofollow&quot;&gt;NORAD&lt;/a&gt; and some other national defence centers.&lt;/p&gt;&#xA;" OwnerUserId="209" LastActivityDate="2014-12-20T10:00:06.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8312" PostTypeId="2" ParentId="4741" CreationDate="2014-12-20T15:03:07.653" Score="4" Body="&lt;p&gt;The Hertzsprung-Russell diagram is an &lt;em&gt;observational&lt;/em&gt; tool. The axes are things that can be observed, or at least estimated reasonably well, for most stars (well, those with known distance anyway). We cannot in general measure the masses of stars - only those in &#xA;some binary systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The schematic that you show, with an arrow indicating that mass increases with increasing luminosity and temperature is valid &lt;em&gt;for stars on the main sequence&lt;/em&gt;. It is not generally true. Whereas the arrow showing increasing size to the top right is always correct, since it merely depends on the other two axes as $L \propto R^2 T^4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one were to add a third axis to the HR diagram - and this is not a bad idea at all - then the possible axes I would choose are: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) Something representing photospheric chemical composition, probably overall metallicity [M/H]. This has the advantage of being measurable for large numbers of stars, and increasingly so given the very large spectroscopic surveys now taking place on many telescope (SDSS, Gaia-ESO, RAVE, LAMOST etc.) On the other hand, stars with similar masses but different compositions already separate in the luminosity-temperature plane reasonably well, so another possibility might be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) Surface gravity. You are correct that stars of different mass but similar radius can end up appearing in more or less the same place in the HR diagram. However, they have different surface gravities. This can be measured from detailed analysis of spectra and would provide an interesting third axis.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2014-12-20T15:03:07.653" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8313" PostTypeId="2" ParentId="8133" CreationDate="2014-12-20T18:02:00.480" Score="9" Body="&lt;p&gt;Let's take an average albedo for the Earth of 0.3 (it depends, which hemisphere is visible, how much cloud cover etc.). That means the Earth reflects 30% of the light incident upon it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flux $f$ falling on the Earth is given by&#xA;$$ f_{\odot} = \frac{L_{\odot}}{4\pi d^2} = 1.369\times10^{3}\ Wm^{-2}$$&#xA;where $L_{\odot}=3.85\times10^{26}\ W$ from the Sun and $d= 1$AU. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integrated luminosity from the illuminated hemisphere will be&#xA;$$L_{earth} = 0.3\pi R^2 f = 5.2\times10^{16}\ W$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we can compare this with the Sun. One hemisphere of the Sun radiates $1.93\times10^{26}\ W$, and produces a flux of $1.369\times10^{3}\ Wm^{-2}$ at 1 AU. Therefore the illuminated hemisphere of the Earth results in a flux of approximately $f_E=0.056\ Wm^{-2}$, assuming the average Earth-Moon distance of 384,400 km. This calculation assumes isotropic emission, but it is quite likely that the albedo is higher for light reflected through 180 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun has an apparent magnitude of -26.74, so the magnitude of the &quot;full-earth&quot; at the moon is&#xA;$$ m_{Earth} = 2.5\log_{10}\left( \frac{f_{\odot}}{f_{E}}\right) - 26.74 = \underline{-15.77}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer will of course vary with the albedo of the visible hemisphere, which in turn depends on the time of year and how much of the polar regions can be seen (e.g. &lt;a href=&quot;http://www.climatedata.info/Forcing/Forcing/albedo.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.climatedata.info/Forcing/Forcing/albedo.html&lt;/a&gt; ). Variations of a few hundredths seem possible, which will lead to apparent magnitude variations in $m_{Earth}$ of $\sim \pm 0.1-0.2$ mag. The albedo may also vary in detail with the exact angle at which the sunlight hits the Earth - an &quot;opposition surge&quot; in brightness, when the Sun-Earth and Moon are almost aligned is possible. The Earth-Moon distance varies from 363,000 to 405,000 km. This will lead to magnitude variations of $\pm 0.12$ mag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A further way to check this is that the albedo of the Moon is 0.12 and it has a radius of 0.273 times that of the Earth. Therefore the Earth seen from the Moon ought to be $(0.3/0.12)\times(1/0.273)^2 = 33.5$ times brighter. This is 3.81 magnitudes brighter. The mean magnitude of the full Moon is -12.74 (maximum is -12.92), so the brightness of the &quot;full Earth&quot; should be -16.55 on average. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure why these figures don't agree; I suspect it is that the albedo for reflection when the Sun's light is normally incident on the Moon is quite a bit larger than 0.12. The so-called &quot;opposition surge&quot;. If the Earth's albedo behaves in the same way, then the latter figure may be more accurate than my first calculation. My gut instinct is that the answer is somewhere between the two.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-05-12T20:33:34.830" LastActivityDate="2017-05-12T20:33:34.830" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8314" PostTypeId="1" AcceptedAnswerId="8318" CreationDate="2014-12-20T18:14:46.190" Score="8" ViewCount="1236" Body="&lt;p&gt;Are the stars in constellation located in a plane or they are in different distances from each other and when we observe them from the Earth we think they are all in a plane?&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="2987" LastEditDate="2014-12-20T18:51:13.367" LastActivityDate="2020-04-15T09:08:12.187" Title="Are the stars in constellation located in a plane or they are in different distances" Tags="&lt;star&gt;&lt;orbit&gt;&lt;distances&gt;&lt;constellations&gt;" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8315" PostTypeId="2" ParentId="8314" CreationDate="2014-12-20T19:05:14.733" Score="1" Body="&lt;p&gt;The stars are not in the same plane and move in different directions. See this &lt;a href=&quot;https://www.youtube.com/watch?v=D-Ko9edZYXE&quot; rel=&quot;nofollow&quot;&gt;youtube&lt;/a&gt; video.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2014-12-20T19:05:14.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8316" PostTypeId="1" AcceptedAnswerId="8319" CreationDate="2014-12-20T19:06:46.083" Score="6" ViewCount="4332" Body="&lt;p&gt;I am having trouble finding data on this. So far, I have found at least seven values:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1: 1.9885  e30 kg&#xA;2: 1.98855 e30 kg&#xA;3: 1.9891  e30 kg&#xA;4: 1.991   e30 kg&#xA;5: 1.989   e30 kg&#xA;6: 1.99    e30 kg&#xA;7: 2       e30 kg&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The first value can be found from &lt;a href=&quot;http://nssdc.gsfc.nasa.gov/planetary/factsheet/sunfact.html&quot; rel=&quot;noreferrer&quot;&gt;NASA GSFC&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The second value can be found from Wikipedia in (at least) two places.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Sun#cite_note-nssdc-1&quot; rel=&quot;noreferrer&quot;&gt;first Wikipedia occurence&lt;/a&gt; references the GSFC page--which is erroneous since that value does not occur there.&lt;/li&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_mass&quot; rel=&quot;noreferrer&quot;&gt;second Wikipedia occurence&lt;/a&gt; references &lt;a href=&quot;http://asa.usno.navy.mil/static/files/2014/Astronomical_Constants_2014.pdf&quot; rel=&quot;noreferrer&quot;&gt;USNO's 2014 Astronomical Constants&lt;/a&gt;, which appear to themselves be copy-pasted from data in 2009 or 2012.  The second Wikipedia occurrence also sources &lt;a href=&quot;http://physics.nist.gov/cgi-bin/cuu/Value?bg&quot; rel=&quot;noreferrer&quot;&gt;NIST&lt;/a&gt; (presumably &lt;a href=&quot;http://physics.nist.gov/cuu/Constants/Table/allascii.txt&quot; rel=&quot;noreferrer&quot;&gt;this table&lt;/a&gt;).  I couldn't find the cited mass from either reference.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The third (and also the fifth) value (which, it is interesting to note, are both outside the error bound given in the second Wikipedia occurrence) can be found on this &lt;a href=&quot;http://solarsystem.nasa.gov/planets/profile.cfm?Object=Sun&amp;amp;Display=Facts&quot; rel=&quot;noreferrer&quot;&gt;generic NASA&lt;/a&gt; site.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The fourth value occurs in the book &lt;em&gt;Handbook of Chemistry and Physics&lt;/em&gt; (Robert C. Weast, 1980)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The fifth value apparently also occurs in the book &lt;em&gt;Astrophysical Data&lt;/em&gt; (Kenneth R. Lang, 1990ish).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The sixth value occurs in the book &lt;em&gt;Physics--3rd Edition&lt;/em&gt; (Cutnell et al. 1995).  &lt;a href=&quot;http://hypertextbook.com/facts/2000/LeonVaysburd.shtml&quot; rel=&quot;noreferrer&quot;&gt;This site&lt;/a&gt; lists several of the values and does a calculation to obtain the sixth value.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Occurs in many places and is pretty clearly rounded.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The fifth, sixth, and seventh values (also) occur in various non-primary sources online, and I assume are rounded values of others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: what is the most recent, best estimate, to the greatest precision, of the Sun's mass?  Obviously, citable, primary sources only!&lt;/p&gt;&#xA;" OwnerUserId="2576" LastEditorUserId="13" LastEditDate="2014-12-21T04:43:56.923" LastActivityDate="2014-12-23T11:38:31.723" Title="What is the exact mass of the Sun?" Tags="&lt;the-sun&gt;&lt;mass&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8317" PostTypeId="1" AcceptedAnswerId="8320" CreationDate="2014-12-20T19:21:16.273" Score="7" ViewCount="236" Body="&lt;p&gt;Firstly, I have to admit that I don't fully comprehend the concept of &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_microwave_background#Polarization&quot; rel=&quot;noreferrer&quot;&gt;&quot;CMB Polarization&quot;&lt;/a&gt;. When I think of &quot;Polarization&quot;, I normally think of the optical definition of polarization, as defined by the electric field. What I currently think is that CMB Polarization is simply the temperature gradient of the CMB. The Wikipedia page also said that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The cosmic microwave background is polarized at the level of a few microkelvin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since microkelvin is a unit of temperature, CMB Polarization must be somehow related to temperature. Is CMB Polarization simply the temperature gradient of the CMB? &lt;/p&gt;&#xA;" OwnerUserId="1931" LastEditorUserId="2153" LastEditDate="2014-12-20T19:40:40.433" LastActivityDate="2014-12-21T00:50:34.063" Title="Is CMB Polarization simply the temperature gradient of the CMB?" Tags="&lt;cmb&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8318" PostTypeId="2" ParentId="8314" CreationDate="2014-12-20T20:17:44.820" Score="10" Body="&lt;p&gt;Constellations and asterisms are generally not proximate in space, but rather happen to be nearby only when viewed from Earth. From Wikipedia's article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Asterism_(astronomy)&quot;&gt;asterism&lt;/a&gt;s:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Like constellations, asterisms are in most cases composed of stars which, while they are visible in the same general direction, are not physically related, often being at significantly different distances from Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A recognizable example of this would be the brightest star in Taurus, &lt;a href=&quot;http://en.wikipedia.org/wiki/Aldebaran&quot;&gt;Aldebaran&lt;/a&gt;.  Aldebaran appears in the V-shaped bull's head, mostly formed by stars in the Hyades star cluster. However, while those stars are actually close to one another, Aldebaran is less than half as far away.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since the star is located (by chance) in the line of sight between the Earth and the Hyades, it has the appearance of being the brightest member of the more scattered Hyades open star cluster that makes up the bull's-head-shaped asterism; however, the star cluster is actually more than twice as far away, at about 150 light years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Similarly, the stars of &lt;a href=&quot;http://en.wikipedia.org/wiki/Orion%27s_Belt&quot;&gt;Orion's belt&lt;/a&gt; are actually hugely separated in space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, it's worth noting that the stars have different velocities across the sky, so the constellations will look increasingly different over long periods of time. &lt;a href=&quot;http://news.discovery.com/space/history-of-space/changing-constellations-50000-years.htm&quot;&gt;Here&lt;/a&gt; is one prediction for 50,000 years time, and the effect will get proportionately larger as time goes on.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2014-12-20T20:17:44.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8319" PostTypeId="2" ParentId="8316" CreationDate="2014-12-20T20:48:39.787" Score="10" Body="&lt;p&gt;The mass of the Sun is determined from Kepler's laws:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{4\pi^2\times(1\,\mathrm{AU})^3}{G\times(1\,\mathrm{year})^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each term in this component contributes to both the value of the solar mass and our uncertainty. First, we know to very good precision that &lt;a href=&quot;http://hpiers.obspm.fr/eop-pc/models/constants.html&quot; rel=&quot;noreferrer&quot;&gt;the (sidereal) year is&lt;/a&gt; 365.256363004 days. We have also &lt;a href=&quot;http://www.iau.org/static/resolutions/IAU2012_English.pdf&quot; rel=&quot;noreferrer&quot;&gt;defined the astronomical unit (AU)&lt;/a&gt; to be 149597870700 m. Strictly speaking, the semi-major axis of the Earth's orbit is slightly different, but by very little in the grand scheme of things (see below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, we can solve for the product $GM$, known as the &lt;em&gt;gravitational parameter&lt;/em&gt;, sometimes denoted $\mu$. &lt;a href=&quot;http://ssd.jpl.nasa.gov/?constants&quot; rel=&quot;noreferrer&quot;&gt;For the Sun,&lt;/a&gt;&#xA;$$\mu_\odot=132712440018\pm9\,\mathrm{km}^3\cdot\mathrm{s}^{-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So solve for $M_\odot$, we need the gravitational constant $G$, which, as it turns out, is by far the largest contributor to the uncertainty in the solar mass. The &lt;a href=&quot;http://physics.nist.gov/constants&quot; rel=&quot;noreferrer&quot;&gt;current CODATA value&lt;/a&gt; is $6.67384\pm0.00080\times10^{-11}\,\mathrm{N\cdot m}^2\cdot\mathrm{kg}^{-2}$, which all combines to give&#xA;$$M_\odot=1.98855\pm0.00024\times10^{30}\,\mathrm{kg}$$&#xA;where my uncertainty is purely from the gravitational constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value $1.9891\times10^{30}\,\mathrm{kg}$ (and nearby values) probably come from an older value of the gravitational constant of $6.672\times10^{-11}\,\mathrm{N\cdot m}^2\cdot\mathrm{kg}^{-2}$, which is still supported by &lt;a href=&quot;http://www.nature.com/nature/journal/v510/n7506/full/nature13507.html&quot; rel=&quot;noreferrer&quot;&gt;some measurements&lt;/a&gt; of $G$.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastEditorUserId="13" LastEditDate="2014-12-21T04:39:30.500" LastActivityDate="2014-12-21T04:39:30.500" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8320" PostTypeId="2" ParentId="8317" CreationDate="2014-12-21T00:50:34.063" Score="4" Body="&lt;p&gt;Polarization of an electromagnetic wave refers to the extent to which the electric field (which is always perpendicular to the direction of wave motion) oscillates in one particular direction, rather than randomly in the plane perpendicular to the wave motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complete linear polarization means the electric field oscillates to and fro along one particular axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polarization of the CMB is produced by Thomson scattering from free electrons just before they combine with protons and the Universe becomes transparent to the CMB radiation.&#xA;The polarization occurs because the radiation field seen by the electrons is not uniform in space (not isotropic), it varies according to which direction the radiation is coming from. These anisotropies are the fingerprints of what went on in the early Universe and what the values of the cosmological parameters are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an excellent primer on this topic - &lt;a href=&quot;http://background.uchicago.edu/~whu/intermediate/polar.html&quot; rel=&quot;nofollow&quot;&gt;http://background.uchicago.edu/~whu/intermediate/polar.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CMB is a blackbody spectrum at a temperature of 2.7K. Differences in this temperature are seen on a variety of scales at levels of 30-60 microKelvin ($\mu K$). These are the &quot;ripples&quot; first seen by COBE, then WMAP and Planck with increasing precision and sensitivity to smaller angular scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas expressing these spatial variations as differences in temperature seems straightforward, it is not so obvious why you would express polarization signals in the same terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In brief, what is going on here is that when a polarization signal is measured it is actually finding the &lt;em&gt;difference&lt;/em&gt; between two perpendicular polarization intensities. Because these intensities are measured as changes of temperature in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bolometer&quot; rel=&quot;nofollow&quot;&gt;bolometer arrays&lt;/a&gt; used as detectors, this is how the results of the polarization are expressed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The polarization anisotropies are expected to be an order of magnitude smaller than the CMB temperature variations. That's because repeated scattering from free electrons can dilute the signal, so it is only light that has been scattered once that contributes.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2014-12-21T00:50:34.063" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8321" PostTypeId="2" ParentId="8314" CreationDate="2014-12-21T03:27:33.367" Score="3" Body="&lt;p&gt;Only very rare instances would get close to such stars being on a plane, if any can be found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nice graphic image showing how the Big Dipper stars appear on an imaginary plane as viewed from Earth combined with a side-view showing approximate distances can be seen at &lt;a href=&quot;http://earthsky.org/favorite-star-patterns/big-and-little-dippers-highlight-northern-sky#history&quot; rel=&quot;nofollow&quot;&gt;this earthsky.org page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2543" LastActivityDate="2014-12-21T03:27:33.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8322" PostTypeId="2" ParentId="8314" CreationDate="2014-12-21T07:26:32.040" Score="2" Body="&lt;p&gt;Like the above: they can be significantly separated by space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to visualize this is to turn on constellations in Celestia then travel to a different star.   The constellation outlines can shift significantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Celestia is an open source visualization thing, it is available at: &lt;a href=&quot;http://www.shatters.net/celestia/&quot; rel=&quot;nofollow&quot;&gt;http://www.shatters.net/celestia/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4107" LastActivityDate="2014-12-21T07:26:32.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8323" PostTypeId="2" ParentId="8316" CreationDate="2014-12-22T03:28:44.640" Score="-1" Body="&lt;p&gt;The answer to your question is a very simple one but, a good one at that. The exact mass is 1.9891×10&lt;sup&gt;30&lt;/sup&gt; kg. Math and new advances in solar exploration have given us so much more information about our Sun. We have been able to see solar eruptions, (CMA's), and gained information about the Sun never known.&lt;/p&gt;&#xA;" OwnerUserId="2672" LastEditorUserId="53" LastEditDate="2014-12-22T23:25:14.103" LastActivityDate="2014-12-22T23:25:14.103" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8324" PostTypeId="1" AcceptedAnswerId="8330" CreationDate="2014-12-22T15:45:38.733" Score="3" ViewCount="826" Body="&lt;p&gt;We know that there is a limit to which an object can be cooled down that is absolute zero, and getting there is also not very easy we may have to wait infinite years to get to that point. But is there any limit to how hot can any object get, or, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the hottest temperature it can reach in the universe, and how can this temperature be maintained? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I did some research and found that hottest temperature we have ever recorded is by the collision of gold atoms in Large Hadron Collider (LHC), that was 40000000000255.375 K !!! (that is ridiculously hot), but it was for a really short time. How can this extreme temperature be maintained for a significant amount of time?&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2014-12-23T05:04:46.447" LastActivityDate="2014-12-23T05:04:46.447" Title="Limit of hotness!" Tags="&lt;temperature&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8325" PostTypeId="1" AcceptedAnswerId="8329" CreationDate="2014-12-22T16:36:17.633" Score="6" ViewCount="2008" Body="&lt;p&gt;I have been looking at large astronomy binoculars in the 25x100 and 20x80 range. Clearly, these monsters cannot be held steady by hand, or even be held at all for any reasonable length of time using one's arms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somewhat related question &lt;a href=&quot;https://astronomy.stackexchange.com/questions/7917/what-should-i-look-for-in-an-astronomy-tripod&quot;&gt;What should I look for in an astronomy tripod?&lt;/a&gt; isn't helpful as a telescope tripod needs to hold a much bigger weight, so it stands to reason that one would get away with something lighter and cheaper for big binoculars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's recommended, tripod-wise? Any related pro tips from the binocular crowd?&lt;/p&gt;&#xA;" OwnerUserId="4114" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2014-12-28T12:01:59.980" Title="What to look for in a tripod for binoculars?" Tags="&lt;binoculars&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8326" PostTypeId="1" AcceptedAnswerId="8328" CreationDate="2014-12-22T18:23:11.417" Score="4" ViewCount="204" Body="&lt;p&gt;if we have spherical mass held by its self-gravity, there will be a limit to distance - let's say it is $r$ - that the center of this body may approach a much more massive body (lets we call $M$ body). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our spherical mass here would be held - as I understand - by gravitational tidal force. But the question is what is the distance $r$ between to masses his effect will hold?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r=\left(\frac{2M}{m} \right)^{1/3} \times R$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m$ - Spherical body mass &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R$ - Spherical body radius &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M$ - Massive body &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess this is the equation for this limit, but I don't understand the conditions of this limitation.&lt;/p&gt;&#xA;" OwnerUserId="1970" LastEditorUserId="2153" LastEditDate="2014-12-29T14:54:39.940" LastActivityDate="2014-12-29T14:54:39.940" Title="Gravitational tidal force problem" Tags="&lt;tidal-forces&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8327" PostTypeId="2" ParentId="8324" CreationDate="2014-12-22T18:41:59.417" Score="1" Body="&lt;p&gt;Perhaps more fundamentally than temperature, the 'hotness' or 'coldness' of a system can be described by the &lt;em&gt;thermodynamic beta&lt;/em&gt; instead, which just like temperature is determined by the rate of change of entropy with respect to energy (at constant volume and particle number):&#xA;$$\beta\equiv\frac{1}{k_\text{B}}\left.\frac{\partial S}{\partial E}\right|_{V,N}\!\!=\frac{1}{k_\text{B}T}\text{.}$$&#xA;Intuitively, the thermodynamic beta measures the coldness of a system from $-\infty$ to $+\infty$. Since temperature per se is just the reciprocal of $\beta$, up to a factor of Boltzmann's constant $k_\text{B}$, that means the coldest is at $T\to 0^+$, approaching zero from positive temperature side, while the hottest is $T\to 0^-$, from the negative. Treating $+0\,\mathrm{K}$ as different from $-0\,\mathrm{K}$ may seem bizarre, but that's what we get for taking a reciprocal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Negative temperatures ($\beta&amp;lt;0$) are &lt;em&gt;hotter&lt;/em&gt; than any positive temperature ($\beta &amp;gt; 0$). They can happen when a system has more high-energy states occupied than low-energy states, such as during &lt;em&gt;population inversion&lt;/em&gt; of any laser, or during similar &lt;a href=&quot;https://en.wikipedia.org/wiki/Astrophysical_maser&quot; rel=&quot;nofollow&quot;&gt;astrophysical phenomena&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2014-12-22T18:41:59.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8328" PostTypeId="2" ParentId="8326" CreationDate="2014-12-22T20:17:56.540" Score="3" Body="&lt;p&gt;That equation is the Roche limit. It pertains to objects that are held together by self-gravitation only. No chemical bonds allowed! For example, an artificial satellite in low Earth orbit is almost inevitably going to be well within the Roche limit for that satellite. For example, the Roche limit for the ISS is almost halfway to the Moon. The ISS isn't being torn apart by tidal forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinds of bodies that this limit does apply to include rubble pile asteroids and comets. For example, comet 67P/Churyumov-Gerasimenko, currently being investigated by the Rosetta spacecraft, has some rubble pile characteristics to it. An even better example is comet Shoemaker–Levy 9, which was torn apart into more than 20 pieces before it collided into Jupiter.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2014-12-22T20:17:56.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8329" PostTypeId="2" ParentId="8325" CreationDate="2014-12-22T20:31:08.210" Score="6" Body="&lt;p&gt;Get the sturdiest tripod that you can!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My &lt;a href=&quot;http://www.celestron.com/browse-shop/astronomy/astronomy-binoculars/skymaster-20x80-binocular&quot; rel=&quot;nofollow noreferrer&quot;&gt;20x80 binoculars&lt;/a&gt; are sitting on a ball and socket mount on a very heavy tripod. I use the &lt;a href=&quot;http://www.bhphotovideo.com/c/product/743889-REG/Benro_BH1_BH1_Ballhead_with_Quick.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Benro BH1&lt;/a&gt; ball mount, which is an absolute joy to aim. This mount is rated for 6 KG, almost three times the actual weight of the binoculars. With high-powered binoculars the aim must be adjusted every minute or so, therefore you will want an easy to adjust mount that is far from the limits of its capacity. You really can't beat the ball and socket design for this purpose in my opinion. I need to photograph the setup for &lt;a href=&quot;https://astronomy.stackexchange.com/questions/7917/what-should-i-look-for-in-an-astronomy-tripod/7919#7919&quot;&gt;another question&lt;/a&gt; and when I do I'll check what model the tripod is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look also at the tripod of the user who asked the above-linked question. These large binoculars need the absolute heaviest duty tripod that you can reasonably afford. You will need &lt;strong&gt;all-metal construction&lt;/strong&gt; and I would suggest a hook as well to hang a weight from, this will make the tripod steadier. You will have a hard time going overboard, so if you find two tripods with a similar price but one is heavier-duty, get that one.&lt;/p&gt;&#xA;" OwnerUserId="209" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2014-12-28T12:01:59.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8330" PostTypeId="2" ParentId="8324" CreationDate="2014-12-22T21:05:34.137" Score="4" Body="&lt;p&gt;I'm going to assume two things. (I) That we are tarking about a material/object at least in local thermodynamic equilibrium, otherwise it is difficult to give much physical meaning to the temperature. (II) That when you say &quot;we can produce&quot; I take that (on Astronomy SE) to mean &quot;we can observe or deduce&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious winner, in the sense that we can still &quot;observe&quot; what happened (the outcome) is from the first fraction of a second of the universe. The electroweak transition occurred at temperatures of around $10^{15}$K and left us with the weak force, electromagnetic force, baryon masses and possibly the matter/anti-matter asymmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The production of helium and lithium in the universe took place at temperatures of $&amp;gt;10^{10}$K and we can still see the evidence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutrinos are produced copiously in the cores of supernovae at temperatures of $10^{11}$K. These neutrinos have been observed from Supernova 1987A.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2014-12-23T00:11:53.940" LastActivityDate="2014-12-23T00:11:53.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8331" PostTypeId="1" CreationDate="2014-12-22T22:21:26.193" Score="1" ViewCount="120" Body="&lt;p&gt;Wouldn't it be nice to be able to render any viewpoint from all the data transformed into a point cloud, an augmented virtual reality of moving objects in our solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would just like to be able to chose any point in the solar system and view exactly what is there at a specific time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's possible to extrapolate 3D data from multiple viewpoints of an object, don't we have enough satellite data to do it?&lt;/p&gt;&#xA;" OwnerUserId="4118" LastEditorUserId="7472" LastEditDate="2017-10-14T06:45:37.903" LastActivityDate="2017-10-14T06:45:37.903" Title="Why don't we have a browsable spacetime augmented virtual solar system?" Tags="&lt;solar-system&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8332" PostTypeId="2" ParentId="755" CreationDate="2014-12-22T22:28:52.810" Score="5" Body="&lt;p&gt;The Galactic rotation period at the Sun's Galactocentric radius is about 230 million years (with a five percent uncertainty) and the Sun, as stated in user8's answer is 4.57 billion years old - giving $\sim 20\pm 1$ orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the idea of &lt;strong&gt;a&lt;/strong&gt; Galactic year is misleading.  For instance, the Milky Way rotation curve is quite flat between 1 and 10 kpc from the centre (see picture), so the &quot;year&quot; also varies by a factor of 10 over this range, with the year being much shorter in the inner part of the Galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore to know how many rotation periods the Sun has executed requires us to know at what Galactocentric radius it has spent its life. It is an &lt;em&gt;ongoing debate&lt;/em&gt; as to whether the Sun has migrated to its position from outside or inside its current radius, or whether it has been where it is all along. As a result, how many &quot;Galactic years&quot; our Sun has experienced is still quite uncertain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;eg. see &#xA;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2015MNRAS.446..823M&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2015MNRAS.446..823M&lt;/a&gt;&#xA;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2002MNRAS.336..785S&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2002MNRAS.336..785S&lt;/a&gt;&#xA;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2013A%26A...558A...9M&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2013A%26A...558A...9M&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mgeUD.gif&quot; alt=&quot;Milky Way rotation curve&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2014-12-23T00:13:36.873" LastActivityDate="2014-12-23T00:13:36.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8333" PostTypeId="2" ParentId="8331" CreationDate="2014-12-22T23:54:28.573" Score="2" Body="&lt;p&gt;There is NASA/JPL's &lt;a href=&quot;http://space.jpl.nasa.gov/&quot; rel=&quot;nofollow&quot;&gt;Solar System Simulator&lt;/a&gt;. It doesn't cover all points, but does include many of the interesting ones.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2014-12-22T23:54:28.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8334" PostTypeId="2" ParentId="8316" CreationDate="2014-12-23T00:01:07.910" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://ilrs.gsfc.nasa.gov/docs/2014/196C.pdf&quot; rel=&quot;nofollow&quot;&gt;http://ilrs.gsfc.nasa.gov/docs/2014/196C.pdf&lt;/a&gt; gives G*m for the Sun as &lt;code&gt;132712440041.939400 km^3/s^2&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.nist.gov/cgi-bin/cuu/Value?bg&quot; rel=&quot;nofollow&quot;&gt;http://physics.nist.gov/cgi-bin/cuu/Value?bg&lt;/a&gt; gives G as &lt;code&gt;6.67384*10^-11 m^3*kg^-1*s^-2&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If these numbers were exact (they're not), the sun's mass (after converting m^3 to km^3 above) would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;165890550052424250000000000000000000/83423 kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or about&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1988546924138717739712069812881.3396785059276218789 kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.9885469241387177397120698128813396785059276218789*10^30 kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, that level of precision is ridiculously unwarranted, but I would argue that's the most precise value you can get, if you accept NASA's definitions of the sun's Gm and the universe's G.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2014-12-23T00:01:07.910" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8335" PostTypeId="2" ParentId="8331" CreationDate="2014-12-23T00:09:57.740" Score="1" Body="&lt;p&gt;I don't know if I understand your question correctly, but can't &lt;a href=&quot;http://www.shatters.net/celestia/&quot; rel=&quot;nofollow&quot;&gt;Celestia&lt;/a&gt; do what you want?&lt;/p&gt;&#xA;" OwnerUserId="2182" LastActivityDate="2014-12-23T00:09:57.740" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8336" PostTypeId="2" ParentId="822" CreationDate="2014-12-23T00:35:09.747" Score="23" Body="&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/full/1995AJ....110.2183H&quot; rel=&quot;noreferrer&quot;&gt;Humphreys &amp;amp; Larsen (1995)&lt;/a&gt; suggest, using star count information, a distance of $20.5 \pm 3.5$ pc &lt;strong&gt;above the Galactic plane&lt;/strong&gt;; consistent with, but more precise than the Bahcall paper referred to by Schleis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2007MNRAS.378..768J&quot; rel=&quot;noreferrer&quot;&gt;Joshi (2007)&lt;/a&gt; is more guarded, investigating some systematic uncertainties in the estimation techniques and ends up with distances between &lt;strong&gt;13 and 28 pc above the plane&lt;/strong&gt;.&#xA;This paper gives an excellent review of the topic in its first couple of pages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun moves at about 15-20 km/s with respect to a local standard of rest defined by the general motion of stars in our vicinity around the Galaxy. In three-dimensions, this &quot;peculiar velocity&quot; is $U=10.00 \pm 0.36$ km/s (radially inwards), $V=5.25 \pm 0.62$ km/s (in the direction of Galactic rotation) and $W=7.17 \pm 0.38$ km/s (&lt;strong&gt;up and out of the plane&lt;/strong&gt;).  &lt;a href=&quot;http://adsabs.harvard.edu/abs/1998MNRAS.298..387D&quot; rel=&quot;noreferrer&quot;&gt;(Dehnen &amp;amp; Binney 1998)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun executes oscillations around its mean orbit in the Galaxy, periodically crossing the Galactic plane. I borrowed this illustration (not to scale!) from &lt;a href=&quot;http://www.visioninconsciousness.org/Science_B08.htm&quot; rel=&quot;noreferrer&quot;&gt;http://www.visioninconsciousness.org/Science_B08.htm&lt;/a&gt; to show this oscillatory motion.&#xA;As the Sun is currently above the plane and moving upwards, and each cycle takes about 70 million years with an amplitude of 100pc (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1995Icar..116..255M&quot; rel=&quot;noreferrer&quot;&gt;Matese et al. 1995&lt;/a&gt;), it will be roughly 30 million years before we cross the plane again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dv1Ii.jpg&quot; alt=&quot;Sun&amp;#39;s motion around the Galaxy&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: An update. A new analysis by &lt;a href=&quot;https://arxiv.org/abs/1610.08125&quot; rel=&quot;noreferrer&quot;&gt;Karim &amp;amp; Mamajek (2016)&lt;/a&gt; yields a distance above the plane of $17.1 \pm 5$ pc and they provide a meta-analysis of previous analyses that gives a median distance of $17.4 \pm 1.9$ pc. They also refer to a paper by  &lt;a href=&quot;https://arxiv.org/abs/0912.3693&quot; rel=&quot;noreferrer&quot;&gt;Schonrich et al. (2010)&lt;/a&gt; that gives the velocity &lt;em&gt;away&lt;/em&gt; from the plane as $7.25 \pm 0.37$ km/s (though with an additional systematic uncertainty of 0.5 km/s).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-10-27T12:14:35.063" LastActivityDate="2016-10-27T12:14:35.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8338" PostTypeId="1" CreationDate="2014-12-23T06:53:38.180" Score="4" ViewCount="105" Body="&lt;p&gt;I've a question on Habitable zone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does Habitable zone represent the region where we humans or the living beings of earth can survive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it a region where life can exist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is second I want to know what are the parameters that are considered for existing of life. &lt;/p&gt;&#xA;" OwnerUserId="4046" LastActivityDate="2014-12-23T12:46:16.567" Title="What is the Habitable zone all about?" Tags="&lt;habitable-zone&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8339" PostTypeId="2" ParentId="8338" CreationDate="2014-12-23T07:20:44.530" Score="1" Body="&lt;p&gt;Nothing that complex. Habitable zone merely defined a circumstellar volume where a planet (if any) can receive an amount of radiative energy to keep H&lt;sub&gt;2&lt;/sub&gt;O in the liquid phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, it has been based on the requirements for life from Earth, lacking any wider samples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I encourage you to play around with &lt;a href=&quot;http://astro.unl.edu/naap/habitablezones/animations/stellarHabitableZone.html&quot; rel=&quot;nofollow&quot;&gt;interactive illustration&lt;/a&gt; of concept (Flash applet).&lt;/p&gt;&#xA;" OwnerUserId="4108" LastEditorUserId="4108" LastEditDate="2014-12-23T12:46:16.567" LastActivityDate="2014-12-23T12:46:16.567" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="8340" PostTypeId="1" CreationDate="2014-12-23T08:21:53.930" Score="8" ViewCount="1651" Body="&lt;p&gt;In 1939, Robert Oppenheimer and other concluded that a certain neutron stars could collapse to black holes and no known law of physics would intervene.  As far as I can tell, there is no observational data that shows that an event horizon of a black hole exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that a black hole with an event horizon exists, then there are all kinds of mathematics that explain what is observed externally, approaches the event horizon.  It can be explained what an infalling observer sees as they approach the event horizon.  However, what happens when an infalling observer passes the event horizon, the result is a mathematical absurdity, in that the observer would observe the universe outside the event horizon as beyond &quot;the end of time.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that passing through an event horizon is as mathematically absurd as an object with non-zero mass reaching and surpassing the speed of light, as it would require more than infinite amount of energy in the latter, or more than an infinite amount of time in former.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, can an event horizon even &quot;exist&quot; in the first place, if &quot;existing&quot; means within the bounds of any moment after the big bang and before an infinite amount of time from any point within the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that the existence of black holes seems to be well accepted, however faster than than light travel does not under the current understanding of the physics of relativity.  &lt;/p&gt;&#xA;" OwnerUserId="4122" LastActivityDate="2020-01-31T15:31:20.763" Title="Can A Black Hole Exist?" Tags="&lt;black-hole&gt;&lt;general-relativity&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8341" PostTypeId="2" ParentId="8340" CreationDate="2014-12-23T10:55:16.513" Score="13" Body="&lt;p&gt;There &lt;em&gt;is&lt;/em&gt; evidence that both black holes and their event horizons exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary evidence for stellar-mass black holes arises from observations of the dynamics of binary systems. What has been found, for at least 20 binary systems, is that the optically visible star has a dark companion, that is usually more massive, and more massive than can possibly be supported as a neutron star or any other degenerate configuration ($&amp;gt;3M_{\odot}$ and mostly $&amp;gt;5M_{\odot}$ - see &lt;a href=&quot;http://arxiv.org/abs/1312.6698&quot; rel=&quot;noreferrer&quot;&gt;Narayan &amp;amp; McClintock (2014)&lt;/a&gt; for an excellent, accessible review).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supermassive black holes at the centres of galaxies can also be &quot;weighed&quot; using the dynamics of gas, or in the case of our Galaxy, actually watching the stars orbiting something that must have a mass of $4.4\times10^{6}M_{\odot}$, yet is remarkably compact and emits little or no radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The evidence for event horizons is more circumstantial, but not absent. In any case, absence of evidence would not be evidence of absence - event horizons are difficult to prove; they are small and far away, with elusive observational signatures. The lack of an event horizon could only be accommodated by inventing something even more bizarre than a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most compelling argument for an event horizon is found in sources that accrete from their surroundings, either in a binary system or at the centres of galaxies (see &lt;a href=&quot;http://arxiv.org/abs/0803.0322&quot; rel=&quot;noreferrer&quot;&gt;Narayan &amp;amp; McClintock 2008; read particularly section 4)&lt;/a&gt;. In some circumstances, the accretion flow can become radiatively inefficient (dark), in which case one expects that if the accretion flow reaches a &quot;surface&quot;, it will radiate its kinetic energy strongly as it thermalises on impact. On the other hand, a black hole event horizon can swallow such flows without trace. The prediction is, that in their low-states, transient black hole accretors will be much less luminous than their neutron star counterparts, and this is what is found (by a factor of 100).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other evidence is that quiescent neutron star binaries show hot, thermal emission, presumed to be from the neutron star surface. None has been seen from the black hole binaries. &lt;a href=&quot;http://en.wikipedia.org/wiki/X-ray_burster&quot; rel=&quot;noreferrer&quot;&gt;Type I X-ray bursts&lt;/a&gt; are seen in neutron star binaries, caused by an accumulation of matter and subsequent nuclear ignition on their surface. The black hole candidates do not show these bursts. The accretion rate onto the supermassive black hole in the centre of the Galaxy should produce thermal radiation at any &quot;surface&quot; that would be quite visible in the infrared part of the spectrum. No such detection has been made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these observations are best explained if the black hole has no surface and can simply make the accreted energy &quot;disappear&quot; inside its event horizon.&#xA;Narayan &amp;amp; McClintock conclude that these lines of evidence are &quot;impervious to counterarguments that invoke strong gravity or exotic stars&quot;.&lt;/p&gt;&#xA;&#xA;&#xA;" OwnerUserId="2531" LastEditorUserId="362" LastEditDate="2014-12-26T19:22:00.033" LastActivityDate="2014-12-26T19:22:00.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8342" PostTypeId="2" ParentId="7859" CreationDate="2014-12-23T11:35:43.057" Score="0" Body="&lt;p&gt;Escape velocity is only relevant for unpowered objects, ie. a lump of space rock. Any powered craft can move at any speed, and escape the gravity of any object (black holes excepted).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the escape velocity of the Sun is around 617.5 km/s, yet the &quot;slow&quot; solar wind has a speed of only around 400km/s, but manages to leave the Sun with ease. Since the solar wind accelerates from the Sun, there must be a constant force acting on the solar wind particles.&lt;/p&gt;&#xA;" OwnerUserId="2880" LastActivityDate="2014-12-23T11:35:43.057" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8343" PostTypeId="2" ParentId="8316" CreationDate="2014-12-23T11:38:31.723" Score="0" Body="&lt;p&gt;The Sun is losing the equivalent of around 4.26 million metric tons per second of hydrogen, as it is converted into energy. The Sun does not have an exact mass.&lt;/p&gt;&#xA;" OwnerUserId="2880" LastActivityDate="2014-12-23T11:38:31.723" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8344" PostTypeId="2" ParentId="496" CreationDate="2014-12-23T12:44:40.760" Score="4" Body="&lt;p&gt;The galaxy that is receding away from us fastest is whichever galaxy currently holds the record for the highest measured redshift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This galaxy obviously changes as more and more discoveries are made. This web page &lt;a href=&quot;http://en.wikipedia.org/wiki/Redshift&quot; rel=&quot;nofollow noreferrer&quot;&gt;that describes redshift&lt;/a&gt;, may keep an up-to-date record of this. It lists &lt;a href=&quot;http://en.wikipedia.org/wiki/UDFy-38135539&quot; rel=&quot;nofollow noreferrer&quot;&gt;UDFy-38135539&lt;/a&gt;, discovered in the Hubble Ultra Deep Field, as a previous record holder (see below), with a spectroscopically confirmed redshift of $z=8.5549 \pm 0.0002$.  That is, lines in the spectrum appear at wavelengths that are a factor of $(1+z)=9.55$ times longer (redder). But there is apparent controversy over the claimed detection of spectral features in this galaxy. (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013MNRAS.430.3314B&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bunker et al. 2013&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2013ApJ...765L...2B&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bramnmer et al. (2013)&lt;/a&gt; report the &quot;tentative detection&quot; of a possible Lyman alpha emission line at $z=12.12$ for UDFj-39546284, another ultra deep field galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way it seems only a matter of time before $z&amp;gt;9$ galaxies are confirmed, since there are many candidates that have so far been selected only by virtue of their peculiar colours (Lyman break galaxies at near infrared wavelengths, including MACS0647-JD mentioned in comments to other posts). e.g. &lt;a href=&quot;http://arxiv.org/abs/1211.2230&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bouwens et al. (2014)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt; &lt;a href=&quot;http://adsabs.harvard.edu/abs/2016ApJ...819..129O&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oesch et al. (2016)&lt;/a&gt; report HST grism spectroscopy of a high redshift galaxy candidate, GN-z11, identified in the CANDELS/GOODS survey. They confirm the presence of a Lyman break indicating a spectroscopic redshift of 11.1. This appears to be the &lt;em&gt;current&lt;/em&gt; record holder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reasons that these galaxies are receding from us is simply the expansion of the universe. Galaxies that are further away appear to recede faster. The galaxies discussed above must be $&amp;gt;13$ billion light years away according to the current &quot;concordance&quot; cosmological model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Translating a redshift to a recession velocity requires knowledge of the cosmological parameters. Fig.2 of &lt;a href=&quot;https://www.mso.anu.edu.au/~charley/papers/DavisLineweaver04.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Davis &amp;amp; Lineweaver (2004)&lt;/a&gt; shows the relationship. For our current best estimates of cosmological parameters, a redshift of 11 corresponds to a recession velocity of &lt;strong&gt;2.3c&lt;/strong&gt;. That is, this galaxy is &lt;em&gt;currently&lt;/em&gt; receding from us at 2.3 times the speed of light.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-06-19T07:01:11.187" LastActivityDate="2017-06-19T07:01:11.187" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8345" PostTypeId="1" CreationDate="2014-12-23T12:52:32.200" Score="19" ViewCount="8123" Body="&lt;p&gt;Titan, which is smaller than Mars, has an atmosphere but Mars is not able to maintain its atmosphere. Even Luna (the Moon) doesn't have an atmosphere. Why is Titan able to hold an atmosphere, despite being smaller than Mars?&lt;/p&gt;&#xA;" OwnerUserId="4046" LastEditorUserId="31410" LastEditDate="2020-11-26T18:51:22.503" LastActivityDate="2020-11-26T18:51:22.503" Title="How does Titan maintain its atmosphere?" Tags="&lt;planetary-atmosphere&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="8346" PostTypeId="1" AcceptedAnswerId="8352" CreationDate="2014-12-23T14:39:48.730" Score="5" ViewCount="676" Body="&lt;p&gt;As far as I understand, during the formation of a planetary disc, grains of dust stick together due to collisions and chemical bonding. Considering the size of the grain - say, 1 mm - and the speed - 1cm per sec - it would seem that gravity plays a minor role during accretion. My question is, to what extent is gravity a factor during accretion? To what extent is it related to fragmentation? In other words - how does gravity affect whether the fragments return and stick to the main body? Is there a size barrier, or is it dependent only on the speed?&lt;/p&gt;&#xA;" OwnerUserId="4126" LastEditorUserId="2531" LastEditDate="2014-12-23T16:01:44.930" LastActivityDate="2014-12-23T23:00:19.937" Title="The role of gravity during planetesimal accretion" Tags="&lt;gravity&gt;&lt;planet&gt;&lt;accretion-discs&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8348" PostTypeId="2" ParentId="8345" CreationDate="2014-12-23T15:16:52.173" Score="12" Body="&lt;p&gt;You are right that it's surprising that Titan, being just a moon, has a thick atmosphere. Usually, the answer includes magnetism: Earth has an atmosphere because the liquid magma inside the planet produces a magnetic field. This magnetic field changes the paths of the particles in the solar wind, thus preserving the volatile gases intact. Mars did use to have an atmosphere, just like Earth, but it being farther away from the Sun, the magma froze and lost its magnetic properties.&lt;/p&gt;&#xA;&lt;p&gt;Titan itself doesn't have a magnetic field, but Saturn does. Saturn's magnetosphere is produced by the motion of the super-compressed hydrogen gas inside the planet (metallic fluid). The field is so strong it encompasses the satellites, including Titan.&lt;/p&gt;&#xA;&lt;p&gt;Saturn doesn't have the same magma as Earth. Earth is a rocky planet. This means that it was formed so close to the sun that light gases (such as hydrogen) could not coalesce due to the high temperature and the solar wind. Therefore, the inner planets (Mercury to Mars) are composed mostly of rock and metal. It's the liquid iron in the outer core that makes Earth magnetic.&lt;/p&gt;&#xA;&lt;p&gt;Gaseous planets instead have a small solid metal/rocky core (therefore, no magnetic field from there), and a huge layer of light gases (hydrogen and helium). The gases are not usually magnetic, but under such an enormous pressure they take a &amp;quot;metallic&amp;quot; structure, which means they can conduct electricity like a metal. This same property allows them to generate a magnetic field.&lt;/p&gt;&#xA;" OwnerUserId="4126" LastEditorUserId="31410" LastEditDate="2020-11-26T18:50:56.610" LastActivityDate="2020-11-26T18:50:56.610" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="8349" PostTypeId="2" ParentId="8345" CreationDate="2014-12-23T18:42:49.940" Score="6" Body="&lt;p&gt;It's cold out by Saturn, which reduces the tendency of gases to evaporate off into space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://claesjohnson.blogspot.com/2010/04/black-body-temperatures-of-planets.html&quot; rel=&quot;nofollow&quot;&gt;Black body temperature of solar system objects&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Earth: distance D = 1 AU (288 K) 16°C&lt;/li&gt;&#xA;&lt;li&gt;Mars: D =1.5 (232 K) -40°C&lt;/li&gt;&#xA;&lt;li&gt;Jupiter: D = 5.2 ( 134 K) -138°C&lt;/li&gt;&#xA;&lt;li&gt;Saturn: D = 9.5 (103 K) -169°C&lt;/li&gt;&#xA;&lt;li&gt;...........Titan  (94 K) -178°C -measured vs calc for others&lt;/li&gt;&#xA;&lt;li&gt;Uranus: D = 19.2 ( 73 K) -199°C&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Neptune: D = 30.1 ( 63 K) -209°C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ganymede, massing 1.5X10^23 kg, vs Titan at 1.3X10^23 kg, has no substantial atmosphere, but it's also a lot warmer due to its proximity to the sun. Nitrogen boils at −196 °C, so it's not impossible that cold days on Titan involve nitrogen rainstorms. Methane and ethane have even higher boiling points&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="2031" LastEditDate="2015-01-16T14:34:39.953" LastActivityDate="2015-01-16T14:34:39.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8350" PostTypeId="2" ParentId="8331" CreationDate="2014-12-23T20:46:39.133" Score="0" Body="&lt;p&gt;Hayden Planetarium / AMNH's &lt;a href=&quot;http://www.amnh.org/our-research/hayden-planetarium/digital-universe&quot; rel=&quot;nofollow&quot;&gt;Digital Universe&lt;/a&gt; might be what you are looking for. However, the free version does not include the solar system, only Milky Way stars and extragalactic objects. If if somehow you could get or create the Solar data, it can definitely show them. It is not super user friendly though, but it can do some pretty impressive things. So... It exists. But it is not free, and I am not even sure it is available if you are not a Planetarium.&lt;/p&gt;&#xA;" OwnerUserId="2182" LastActivityDate="2014-12-23T20:46:39.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8351" PostTypeId="2" ParentId="8345" CreationDate="2014-12-23T21:23:21.507" Score="6" Body="&lt;p&gt;Solar wind (a flux of charged particles emitted by the star) is the main cause for &lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2008/21nov_plasmoids/&quot; rel=&quot;nofollow noreferrer&quot;&gt;celestial body losing its atmosphere&lt;/a&gt;. So, to keep an atmosphere celestial body would need a magnetosphere, that it, a magnetic field which deflects protons and electrons of solar wind and prevents them from giving molecules an energy to escape from upper layers of atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to current studies, the source of planetary magnetic field are eddy currents in liquid metal of rotating outer core, caused by convection and Coriolis force, so called &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dynamo_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;geomagnetic dynamo&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, Luna is pretty much geologically dead now. For the case of Mars, it has been hypothesized what its &quot;dynamo&quot; stopped by some reason, in any case it is a fact what Mars has weak and irregular magnetic field. But in case of Titan (which doesn't even have much metal in core), a &lt;a href=&quot;http://www.newscientist.com/article/dn14717-saturn-magnetises-its-moon-titan.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;gas giant with powerful magnetic field comes to rescue&lt;/a&gt; and protects its atmospheres from adverse effect of solar wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/5/54/Plasma_magnet_saturn.jpg&quot; alt=&quot;https://upload.wikimedia.org/wikipedia/commons/5/54/Plasma_magnet_saturn.jpg&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an interesting trivia note, it has been hypothesized what this neighboring with radiation belt is the cause of higher hydrocarbons abundance on Titan.&lt;/p&gt;&#xA;" OwnerUserId="4108" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:47.280" LastActivityDate="2014-12-23T21:23:21.507" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8352" PostTypeId="2" ParentId="8346" CreationDate="2014-12-23T23:00:19.937" Score="3" Body="&lt;p&gt;The very first stage of planet formation involves purely inelastic (i.e., sticky) collisions between pairs of small particles of dust to make slightly larger particles of dust, pairwise collisions between these slightly larger particles to make even larger particles, and so on. So far, the only gravitation that comes into play is that of the central protostar and the disk as a whole. The gravitation of these little objects is very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then some magic happens that makes the little rocks that result from this very first stage grow into planetesimals that do gravitate. How do little rocks (a few centimeters in diameter) grow into little planetesimals (a few hundred meters in diameter)?  This is the key open problem in the theory of planet formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The core of the problem is that objects the size of a spec of dust to an object the size of a small pebble move with the gas that forms the bulk of the disk. The outward pressure from the gas in the disk makes the gas and small objects move at a slightly less than orbital speed. Once an object reaches the centimeter scale or so, the pressure from the gas no longer provides the support needed to keep those objects moving at that slightly suborbital speed. The objects start to become separated from the gas. By the time an object reaches a meter or so in diameter, the gas is equivalent to a hurricane force headwind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One consequence of this is that collisions become ever more violent as objects grow beyond a centimeter in size, and hence are more likely to break colliding objects apart than to make them coalesce. An even bigger problem is that that headwind makes centimeter- to meter-sized objects spiral into the protostar, and very quickly (a few hundred years from 1 AU). A number of proposals have been put forth to overcome this meter barrier problem. So far, none appears to have stuck. (Sorry for the pun.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once objects reach some critical size, the drag from the gas reduces in importance thanks to the square-cube law and the gravitational attraction increases. That critical size varies from hundreds of meters to tens of kilometers, depending on which paper one reads (experts vary). These objects are &quot;planetesimals,&quot; and there are lots of them. These once again combine readily to make planetary embryos. From then on, its just a mopping up process as gravitation either makes these planetesimals / planetary embryos combine or ejects them from the star system.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2014-12-23T23:00:19.937" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8353" PostTypeId="2" ParentId="8174" CreationDate="2014-12-23T23:14:44.870" Score="3" Body="&lt;p&gt;&lt;strong&gt;Regarding Terminology&lt;/strong&gt;:&lt;br&gt;&#xA;'&lt;strong&gt;&lt;em&gt;Mass ejections&lt;/em&gt;&lt;/strong&gt;' (at least semantically) are very different from supernovae.  A &quot;mass ejection&quot; would generally refer to something like &lt;a href=&quot;http://en.wikipedia.org/wiki/Coronal_mass_ejection&quot; rel=&quot;nofollow&quot;&gt;what the sun does as part of its normal activity --- ejecting very small amounts of plasma&lt;/a&gt;, or on the other end of the spectrum, &lt;a href=&quot;http://en.wikipedia.org/wiki/Supernova_impostor&quot; rel=&quot;nofollow&quot;&gt;the ejection of massive shells of material by very massive (e.g. $M \gtrsim 40 M_\odot$) stars&lt;/a&gt;$^{[1]}$.  &lt;strong&gt;&lt;em&gt;Supernovae&lt;/em&gt;&lt;/strong&gt;, on the other hand, are &lt;a href=&quot;http://en.wikipedia.org/wiki/Supernova&quot; rel=&quot;nofollow&quot;&gt;the explosion and destruction of an entire star (or at least everything outside of the core)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Simulations&lt;/strong&gt;&lt;br&gt;&#xA;The sad fact of the matter is that there are &lt;em&gt;no simulations&lt;/em&gt; which can fully self-consistently explode a star.  In other words, there is no code that can start with a star and have it naturally explode$^{[2]}$.  There are many specialized codes, however, which deal with particular aspects of the explosions.  The initial collapse and explosion is usually done with very specialized, complex hydrodynamic codes by people like Christian Ott, Adam Burrows, Chris Fryer, etc.  These codes are not public.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More general hydrodynamic codes are usually used to study the effects of supernovae, i.e. how the blast waves evolve under different circumstances.  In these simulations, the user artificially deposits a huge amount of energy in the center of a stellar model (mimicking the energy produced by an explosion), and then sees how the system evolves.  One of the most popular and advanced codes for this is called &lt;a href=&quot;http://www.flash.uchicago.edu/site/&quot; rel=&quot;nofollow&quot;&gt;FLASH, produced and managed at the U of Chicago&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;[1] For a technical article see, e.g.: &lt;a href=&quot;http://arxiv.org/abs/1010.3718&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1010.3718&lt;/a&gt;&lt;br&gt;&#xA;[2] There are numerous reasons for this, first and foremost that it is an incredibly difficult computational task.  Stars are generally about $10^6$ km in size, while their cores are only a few km --- to understand the explosion, the code needs to model this entire dynamic range which is incredibly challening.  Additionally, the densities and temperatures involved in the cores of supernovae are &lt;strong&gt;&lt;em&gt;way&lt;/em&gt;&lt;/strong&gt; outside of the range of whats ever been explored in a laboratory --- so information about that materials behavior is still largely uncertain.  There are lots of other big challenges (e.g. incorporating general relativity, advanced nuclear reactions, etc) but these are some of the key issues people are exploring.&lt;/p&gt;&#xA;" OwnerUserId="918" LastActivityDate="2014-12-23T23:14:44.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8355" PostTypeId="1" CreationDate="2014-12-24T16:21:12.167" Score="1" ViewCount="64" Body="&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=FG0fTKAqZ5g&amp;amp;feature=share&quot; rel=&quot;nofollow&quot;&gt;This video&lt;/a&gt;, which shows time-lapse views of Earth from the ISS, consistently shows some sort of hard boundary in the atmosphere, below which there seems to be significant reflection from sunlight and/or Earth's cities and above which there seems to be nothing impeding the view of black space.  What is this boundary?  I don't think its the edge of the atmosphere, partly because the atmosphere has no hard edge and partly because the aurorae (an atmospheric phenomenon) seen in the video exist, at least in part, above this boundary.&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2014-12-24T18:49:32.363" Title="What is the boundary in the atmosphere seen in this video?" Tags="&lt;earth&gt;&lt;planetary-atmosphere&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8356" PostTypeId="1" CreationDate="2014-12-24T17:20:23.093" Score="0" ViewCount="115" Body="&lt;p&gt;So, I hear people say blah blah blah is 3 light years away, and here's a photo. So does that mean the photo that we see are actually at least 3 years ago since that's how long it takes that light to travel to Earth for us to take the picture?&lt;/p&gt;&#xA;" OwnerUserId="4134" LastEditorUserId="2153" LastEditDate="2014-12-24T18:00:55.440" LastActivityDate="2015-01-08T13:24:00.390" Title="Photos from galaxy" Tags="&lt;star&gt;&lt;distances&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8357" PostTypeId="2" ParentId="8355" CreationDate="2014-12-24T18:49:32.363" Score="2" Body="&lt;p&gt;Yes this is neither aurora, nor reflection of any light.&#xA;This is airglow, that comes from upper layers of atmosphere.&#xA;The brightest emission is green 558nm light from excited atomic oxygen in a layer 90-100 km high. At that height most of the oxygen is in atomic state, not molecular, and is constantly being exposed to solar UV radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you can find more details.&#xA; &lt;a href=&quot;http://www.atoptics.co.uk/highsky/airglow2.htm&quot; rel=&quot;nofollow&quot;&gt;airglow formation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4135" LastActivityDate="2014-12-24T18:49:32.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8358" PostTypeId="2" ParentId="8345" CreationDate="2014-12-24T19:03:05.843" Score="1" Body="&lt;p&gt;Simply said, the ability of planet or other body to maintain atmosphere depends on three factors:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Its gravity&lt;/li&gt;&#xA;&lt;li&gt;Atmosphere temperature&lt;/li&gt;&#xA;&lt;li&gt;Chemical composition of atmosphere&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can estimate that using the simple formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$kT \ll \frac{GMm}{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $k$ is Boltzmann's constant, $T$ is absolute temperature of the gas, $G$ is the gravitational constant, $M$ is planet mass, and $m$ is the mass of particular molecule.&#xA;If gas molecules satisfy the formula above, their kinetic energy is much less than the depth of potential well of planet's gravity, therefore they cannot escape. So the planet will keep its atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of Earth, for example, its gravity is not enough to keep Hydrogen and Helium, which have smallest molecular masses ($m$=2 proton mass).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of cold atmosphere, the mass of celestial body able to keep it may be quite small. This is the reason, that Titan or Pluto have an atmosphere but Mercury or Moon have not.&lt;/p&gt;&#xA;" OwnerUserId="4135" LastEditorUserId="2153" LastEditDate="2014-12-24T19:07:53.130" LastActivityDate="2014-12-24T19:07:53.130" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8359" PostTypeId="1" CreationDate="2014-12-24T18:42:43.893" Score="1" ViewCount="415" Body="&lt;p&gt;I haven't studied space that much, but I know that Pluto has very little gravity, but just enough to keep a human on its surface. Would Pluto keep a regular orbit around the sun without some extra support gravity, such as its moon?&lt;/p&gt;&#xA;" OwnerDisplayName="Mr Squer" LastActivityDate="2014-12-25T00:05:52.917" Title="Would Pluto keep an orbit without its moon?" Tags="&lt;orbit&gt;&lt;gravity&gt;&lt;pluto&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8360" PostTypeId="2" ParentId="8359" CreationDate="2014-12-24T19:05:19.630" Score="0" Body="&lt;p&gt;It would is a short answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The long answer is it depends on how you eject the mass of the moon from Pluto's sphere of influence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good way to think of this is if we launch a satellite into solar orbit, it can orbit the sun in the same orbit as the Earth. Now, if we launch far to much mass from the earth then we could either move closer to, or further from the sun. This is the same way rockets propel themself.&lt;/p&gt;&#xA;" OwnerUserId="4136" OwnerDisplayName="FraserOfSmeg" LastActivityDate="2014-12-24T19:05:19.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8361" PostTypeId="2" ParentId="8359" CreationDate="2014-12-24T19:11:47.773" Score="5" Body="&lt;p&gt;The orbit of a planet does not in any way depend in its moons, mass or own gravity. The orbit is the same for a grain of dust as for a giant planet. It only depends on the mass on the Sun and the distance to the Sun. The eccentricity of the orbit is then given by the initial angular momentum of the orbiting object. &lt;a href=&quot;http://www.qrg.northwestern.edu/projects/vss/docs/space-environment/1-what-causes-an-orbit.html&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; might explain it a bit better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was a really BIG breakthrough in physics, actually the foundation of physics, about 400 years ago. Big because it is irrefutable although it is counter-intuitive. It is really difficult to believe unless one tries to accept the actual observations and logic.&lt;/p&gt;&#xA;" OwnerUserId="1030" OwnerDisplayName="LocalFluff" LastEditorUserId="1030" LastEditDate="2014-12-25T00:05:52.917" LastActivityDate="2014-12-25T00:05:52.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8362" PostTypeId="2" ParentId="2054" CreationDate="2014-12-24T20:26:46.170" Score="1" Body="&lt;p&gt;The velocity of the Earth going around the Sun changes such that the sign changes every six months.  Therefore, the part of the aberration due solely to the Earth's motion about the Sun also changes sign.  In fact, the aberration from the Earth's orbit makes the positions of stars go into ellipses that take 1 year to complete.  The motion of the Sun about the Galaxy also makes galaxy positions move around in an ellipse, but one that takes 250 million years to complete. &lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2017-06-28T20:12:32.333" LastActivityDate="2017-06-28T20:12:32.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8363" PostTypeId="2" ParentId="8172" CreationDate="2014-12-24T21:34:09.360" Score="2" Body="&lt;p&gt;Feng and Gallo and others before them ask the question, what is the expected mass distribution derived from the rotation profile, assuming that the matter is distributed entirely in the disk (ie. in a plane).  There is a solution with the density falling roughly as an exponential.  The problem is that it does not follow the density law that we get from the light distribution.   From the light distribution, including spectral information from UV to radio we know the stellar distribution and the gas distribution.  The Feng and Gallo finds a density falloff that is shallower.  In other words, it also needs a &quot;missing mass&quot; component (albeit less than dark matter), but now it has to be distributed in a flat disk.  Is this preferred?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is the issue that it would not help to explain the high velocity dispersion in groups and clusters.   And, on top of that, it would not help to explain how galaxies and clusters formed so quickly given that the density of the universe started out so highly uniform.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2014-12-24T22:02:28.523" LastActivityDate="2014-12-24T22:02:28.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8365" PostTypeId="2" ParentId="8359" CreationDate="2014-12-24T23:36:42.687" Score="3" Body="&lt;p&gt;Yes, a planetoid such as Pluto will be able to orbit no matter how small its mass, so long as its angular momentum is within a range determined by its distance from the sun. In the case of Pluto and its primary moon, Charon, however, things are even more interesting.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Charon were to magically disappear (ejection would be another story), the point on the line of Pluto's orbit at any one time would be very near the center of Pluto. There are other moons, but their gravitational tug is minimal. The orbital trace of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pluto#Charon&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pluto/Charon system&lt;/a&gt;, however, is not even inside Pluto.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most planets have satellites that are very much less massive than they are, and thus the center of their systemic rotation is fairly close to the primary's center. The masses of Pluto and Charon, however, are significantly closer in size.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Pluto–Charon system is noteworthy for being one of the Solar System's few binary systems, defined as those whose &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth-Moon_barycenter&quot; rel=&quot;nofollow noreferrer&quot;&gt;barycenter&lt;/a&gt; lies above the primary's surface . . . &lt;a href=&quot;http://adsabs.harvard.edu/abs/2006AREPS..34...47R&quot; rel=&quot;nofollow noreferrer&quot;&gt;[108]&lt;/a&gt; This and the large size of Charon relative to Pluto has led some astronomers to call it a &lt;a href=&quot;https://en.wikipedia.org/wiki/Double_planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;dwarf double planet&lt;/a&gt;.&lt;a href=&quot;http://adsabs.harvard.edu/abs/2006Natur.439...52S&quot; rel=&quot;nofollow noreferrer&quot;&gt;[109]&lt;/a&gt; . . . This also means that the rotation period of each is equal to the time it takes the entire system to rotate around its common center of gravity.&lt;a href=&quot;http://dx.doi.org/10.1007%2F978-1-4020-5544-7&quot; rel=&quot;nofollow noreferrer&quot;&gt;[66]&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, for practical purposes, with respect to the orbit of Pluto, it's far simpler to determine the trajectory of the Pluto/Charon barycenter -- which is completely above the surface of Pluto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/b08aD.png&quot; alt=&quot;Pluto/Charon system&quot;&gt;&lt;br&gt;&#xA;&lt;em&gt;An oblique view of the Pluto–Charon system showing that Pluto orbits a point outside itself. Pluto's orbit is shown in red and Charon's orbit is shown in green.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2674" LastActivityDate="2014-12-24T23:36:42.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8366" PostTypeId="2" ParentId="8172" CreationDate="2014-12-25T10:45:05.623" Score="6" Body="&lt;p&gt;Feng &amp;amp; Gallo have published a series of extremely similar papers, all of which essentially claim that they have &quot;discovered&quot; a major flaw in the way (some) astrophysicists think about rotation curves. Instead of assuming spherical symmetry, they try to solve for the mass distribution, using a rotation curve, without assuming spherical symmetry, instead adopting a planar geometry with cylindrical symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course they &lt;em&gt;do&lt;/em&gt; have a point; statements that the flat rotation curve can be compared with a Keplerian prediction (that assume spherical symmetry, or that all the mass is concentrated at the centre) are overly simplistic. So far so good, but they then go on to claim that their analysis is compatible with the total &lt;em&gt;stellar&lt;/em&gt; mass of galaxies and that dark matter is not required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in their planar model (and obviously this is open to criticism too) they invert rotation curves to obtain a radially dependent surface density distribution that drops pseudo-exponentially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem 1: They concede (e.g. in &lt;a href=&quot;http://arxiv.org/abs/1104.3236&quot; rel=&quot;nofollow noreferrer&quot;&gt;Feng &amp;amp; Gallo 2011&lt;/a&gt;) that &quot;the surface mass density decreases toward the galactic periphery at a slower rate than that of the luminosity density. In other words, the mass-to-light ratio in a disk galaxy is not a constant&quot;. This is an understatement! They find exponential scale lengths for the mass that are &lt;em&gt;around twice&lt;/em&gt; (or more for some galaxies) the luminosity scalelengths, so this implies a &lt;em&gt;huge, unexplained&lt;/em&gt; increase in the average mass to light ratio of the stellar population with radius. For the Milky way they give a luminosity scalelength of 2.5 kpc and a mass scale length of 4.5 kpc, so the &lt;span class=&quot;math-container&quot;&gt;$M/L$&lt;/span&gt; ratio goes as &lt;span class=&quot;math-container&quot;&gt;$\exp[0.18r]$&lt;/span&gt;, with radius in kpc (e.g. increases by a factor 4 between 2 kpc and 10 kpc). They argue this may be due to the neglect in their model of the galactic bulge, but completely fail to explain how this could affect the mass-to-light ratio in such an extreme way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem 2: In their model they derive a surface mass density of the disk in the solar vicinity as between 150-200 &lt;span class=&quot;math-container&quot;&gt;$M_{\odot}/pc^2$&lt;/span&gt;. Most (&lt;span class=&quot;math-container&quot;&gt;$\sim 90$&lt;/span&gt;%) of the stars in the solar neighbourhood are &quot;thin disk&quot; stars, with an exponential scale height of between &lt;span class=&quot;math-container&quot;&gt;$z_0= 100-200$&lt;/span&gt;pc. If we assume the density distribution is exponential with height above the plane and that the Sun is near the plane (it is actually about 20pc above the plane, but this makes little difference), a total surface mass density of &lt;span class=&quot;math-container&quot;&gt;$\sigma = 200M_{\odot}/pc^2$&lt;/span&gt; implies a local volume mass density of &lt;span class=&quot;math-container&quot;&gt;$\rho \simeq \sigma/2z_0$&lt;/span&gt;, which is of order &lt;span class=&quot;math-container&quot;&gt;$0.5-1 M_{\odot}/pc^3$&lt;/span&gt; for the considered range of possible scale heights.  The &lt;em&gt;total&lt;/em&gt; mass density in the Galactic disk near the Sun, derived from the dynamics of stars observed by Hipparcos, is actually only &lt;span class=&quot;math-container&quot;&gt;$0.076 \pm 0.015 M_{\odot}/pc^3$&lt;/span&gt;  (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1998A%26A...329..920C&quot; rel=&quot;nofollow noreferrer&quot;&gt;Creze et al. 1998&lt;/a&gt;), &lt;em&gt;which falls short of Feng &amp;amp; Gallo's requirements by an order of magnitude&lt;/em&gt;. (This does not bother the cold, non-baryonic dark matter model because the additional (dark) mass is not concentrated in the plane of the Galaxy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem 3: For the most truncated discs that they consider, with an edge at &lt;span class=&quot;math-container&quot;&gt;$r=15$&lt;/span&gt; kpc, the total Galaxy mass is &lt;span class=&quot;math-container&quot;&gt;$1.1\times10^{11}\ M_{\odot}$&lt;/span&gt; (again from &lt;a href=&quot;http://arxiv.org/abs/1104.3236&quot; rel=&quot;nofollow noreferrer&quot;&gt;Feng &amp;amp; Gallo 2011&lt;/a&gt; ). The claim is then that this &quot;is in very good agreement with the Milky Way &lt;em&gt;star counts&lt;/em&gt; of 100 billion (Sparke &amp;amp; Gallagher 2007)&quot;. I would not agree. Assuming &quot;stars&quot; covers the full stellar mass range, then I wouldn't dissent from the 100 billion number; but the average stellar mass is about &lt;span class=&quot;math-container&quot;&gt;$0.2\ M_{\odot}$&lt;/span&gt; (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2003PASP..115..763C&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chabrier 2003&lt;/a&gt;), so this implies &lt;span class=&quot;math-container&quot;&gt;$\sim 5$&lt;/span&gt; times as much mass as there is in stars (i.e. essentially the same objection as problem 2, but now integrated over the Galaxy). Gas might close this gap a little, white dwarfs/brown dwarfs make minor/negligible contributions, but we still end up requiring some &quot;dark&quot; component that dominates the mass, even if not as extreme as the pseudo-spherical dark matter halo models. Even if a factor of 5 additional baryonic dark matter (gas, molecular material, lost golf balls) were found this still leaves the problem of points 1 and 2 - why does this dark matter not follow the luminous matter and why does it not betray its existence in the kinematics of objects perpendicular to the disc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem 4: Feng &amp;amp; Gallo do not include any discussion or consideration of the more extended populations of the Milky Way. In particular they do not consider the motions of distant globular clusters, halo stars or satellite galaxies of the Milky Way, which can be at 100-200 kpc from the Galactic centre (e.g. &lt;a href=&quot;http://iopscience.iop.org/0004-637X/785/1/63?rel=sem&amp;amp;relno=2&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bhattachargee et al. 2014&lt;/a&gt;). At these distances, any mass associated with the luminous matter in the disk at &lt;span class=&quot;math-container&quot;&gt;$r \leq 15$&lt;/span&gt; kpc &lt;em&gt;can&lt;/em&gt; be well approximated using the Keplerian assumption. Proper consideration of these seems to suggest a much larger &lt;em&gt;minimum&lt;/em&gt; mass for the Milky way independently of any assumptions about its distribution, though perhaps not in the inner (luminous) regions where dark matter appears not to be dominant and which is where F&amp;amp;G's analysis takes place. i.e the factor of 5-10 &quot;missing&quot; mass referred to above may be quite consistent with what others say about the total disk mass and the required dark matter &lt;em&gt;within 15kpc&lt;/em&gt; of the Galactic centre (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014ApJ...794...59K&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kafle et al. 2014&lt;/a&gt;). To put it another way, the dynamics of these very distant objects require a large amount of mass in a spherical Milky Way halo, way more than the luminous matter and way more even than derived by Feng &amp;amp; Gallo. For instance, Kafle et al. model the mass (properly, using the Jeans equation) as a spheroidal bulge, a disk and a spherical (dark) halo using the velocity dispersions of halo stars out to 150 kpc. They find the total Galaxy mass is &lt;span class=&quot;math-container&quot;&gt;$\sim 10^{12} M_{\odot}$&lt;/span&gt; and about 80-90% is in the spherical dark halo. Yet this dark halo makes almost no contribution to the mass density in the disk near the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem 5: (And to be fair I do think this is beyond the scope of what Feng &amp;amp; Gallo are doing) Feng &amp;amp; Gallo treat this problem in isolation without considering how their rival ideas might impact on all the other observations that non-baryonic dark matter was brought in to solve. Namely, the dynamics of galaxies in clusters, lensing by clusters, the CMB ripples, structure formation and primordial nucleosynthesis abundances to state the obvious ones. A new paradigm needs to do at least as well as the old one in order to be considered competitive.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-10-19T16:09:34.537" LastActivityDate="2018-10-19T16:09:34.537" CommentCount="15" ContentLicense="CC BY-SA 4.0" />
  <row Id="8367" PostTypeId="1" AcceptedAnswerId="9915" CreationDate="2014-12-25T21:24:50.007" Score="4" ViewCount="515" Body="&lt;p&gt;I would like to define both initial velocities and masses of components so that orbits may be either elliptical of hyperbolic. Is there any software to make such simulation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, Binary Star Combined Solution Package is out of date and NASA online plotter supports only circular orbits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4139" LastActivityDate="2018-07-15T18:14:56.660" Title="What is a possible software for simulating binary star systems?" Tags="&lt;binary-star&gt;&lt;n-body-simulations&gt;&lt;software&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8368" PostTypeId="2" ParentId="4847" CreationDate="2014-12-26T04:16:03.363" Score="2" Body="&lt;p&gt;The answer here is deceptively simple, and has very little to do with gravity directly. You only really need to know a bit of quantum mechanics, and the answer comes almost for free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any quantum mechanical theory, we have a Hilbert space $\mathcal H$. (Actually, we really want a rigged Hilbert space, but this distinction isn't particularly relevant here.) This can be as simple as the space of a single q-bit, or it can be as complicated as you need it to be (e.g. the Fock spaces which arise in quantum field theory). The Hilbert space describes all the possible configurations of any system described by this theory; an individual vector is a specific configuration. We also have a linear operator $H$ on $\mathcal H$, called the Hamiltonian, which describes the dynamics of such a system via the Schrödinger equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;information&quot; is just knowing what state $| \psi \rangle \in \mathcal H$ your system is in. This state vector tells you the result of every possible measurement, and thus contains all the information of your system. If you really wanted to, it's possible to express any such state uniquely as a sequence of zeros and ones, but that's a very classical way of thinking, and we're dealing with quantum information, which means that the fundamental objects aren't zeros and ones, but state vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when we talk about holography, what we're really saying is that we can determine the state $| \psi \rangle$ that our system (e.g. the universe) is in simply from knowing the results of experiments we perform on the boundary of the spacetime (e.g. infinitely far away). The holographic principle alone doesn't say how this reconstruction works, only that it is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is a bit broad, it may be helpful to see how it works in the only really well-understood example, the AdS/CFT correspondence. In this case, we have a theory of quantum gravity  in $d+1$ dimensions with Hilbert space $\mathcal H$ and Hamiltonian $H$, constrained to have spatially-asymptotically AdS$_{d+1}$ metric (AdS is just a special, maximally symmetric solution to the vacuum Einstein field equations which has nice properties that make this work). It turns out that, at least morally speaking (there is ongoing work to understand the full extent of this), when we collect all the observables in this theory and shuffle them around in well-defined ways, we can construct out of them vectors in a different Hilbert space $\mathcal H'$. That is, we have a linear map $T: \mathcal H \rightarrow \mathcal H'$. This map is 1-to-1, meaning that we don't lose information. Strictly speaking, it doesn't need to be onto, but this distinction isn't crucial at first pass, so you can think of $T$ as a 1-to-1 correspondence (i.e. a linear isomorphism) if you like. In addition, the Hamiltonian $H$ can be mapped to $H'$, which describes compatible dynamics to $H$ on $\mathcal H'$ (in the sense that one can first evolve in the original theory and then map to the new one, or first map and then evolve, and the results will be the same). When we look at the pair $\mathcal H', H'$, we recognize this not as a $d+1$-dimensional theory of quantum gravity, but as a $d$-dimensional theory without gravity, but with extra symmetries that turn it into a so-called &quot;conformal field theory&quot; (which come from the extra symmetries of the AdS spacetime). In some sense, this new theory can be thought of as living on the boundary of AdS. So when we say that all the information is contained on the boundary, we mean that given a state $| \psi' \rangle \in \mathcal H'$ which describes the boundary theory observables, we can reconstruct the full state in the quantum gravity theory simply by $T^{-1}(|\psi'\rangle)$. What I've given here is just the 2-minute summary; AdS/CFT is still an active area of research and everything I said above is only approximately and/or morally true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might wonder why we need to rely on quantum mechanics. It turns out that (at least in the AdS/CFT correspondence), we can't do it classically. Highly quantum mechanical behavior in one theory is recovered by the classical limit of the other, and vice-versa. This is both a blessing and a curse, but at any rate there's no real classical equivalent. This is suspected to be true in any nontrivial example of the holographic principle, so we're pretty stuck describing things in terms of quantum information.&lt;/p&gt;&#xA;" OwnerUserId="2028" LastEditorUserId="2028" LastEditDate="2014-12-26T04:22:12.837" LastActivityDate="2014-12-26T04:22:12.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8369" PostTypeId="2" ParentId="8112" CreationDate="2014-12-26T12:33:46.983" Score="10" Body="&lt;p&gt;It appears (and I am no expert) that Saturn's ring evolution is governed mainly by &quot;viscous spreading&quot; - collisions between ring particles; and also by interactions with Saturn's moons (resonances); and bombardment with meteoritic material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There appears to be no consensus on how old the ring system is. Most theories of their formation have this taking place early in the life of the solar system, but there are observations that suggest little contamination by meteoritic material and hence that the rings are young.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://adsabs.harvard.edu/abs/2010Icar..209..771S&quot;&gt;Salmon et al. (2010)&lt;/a&gt;  the viscosity in the ring is critical to its evolution and therefore estimates of ages and lifetimes. As the disc spreads due to viscosity, the viscosity itself will reduce, slowing the evolution and thus the ring system may not empty for many billions of years. On the other hand, comparing viscous spreading models with the current density profile of Saturn's rings suggests they are much less than a billion years old. However, it should be noted that other authors have used similar models to argue that Saturn's rings did indeed form 4.5 billion years ago and were much more massive (by a factor $&amp;gt;100$) then they are today! &lt;a href=&quot;http://adsabs.harvard.edu/abs/2011Icar..216..535C&quot;&gt;(Charnoz et al. 2011)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2014-12-26T19:06:50.787" LastActivityDate="2014-12-26T19:06:50.787" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8370" PostTypeId="2" ParentId="1509" CreationDate="2014-12-26T19:21:15.730" Score="11" Body="&lt;p&gt;The currently leading answer is correct to say that moon formation inside the Roche limit is unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the disk is evolving due to viscosity between the particles, and as a consequence it &quot;spreads&quot;, so that material is able to move to outside the Roche limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact this is a leading possible explanation for the formation of the inner moons of Saturn - that an initially much more massive ring system underwent viscous evolution and spreading, and that material spread outside the Roche limit was able to condense into the inner moons. See for example &lt;a href=&quot;http://arxiv.org/abs/1109.3360&quot;&gt;http://arxiv.org/abs/1109.3360&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether such a process can continue is doubtful. Models for viscous disk evolution show that the initial evolution is very rapid and that subsequent evolution is very much slower, so that the rate of mass transferral to outside the Roche limit is now quite small. It maybe that it is too slow to form anything new and that any mass would be just accreted onto existing Saturnian satellites.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2014-12-26T19:21:15.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8371" PostTypeId="1" CreationDate="2014-12-26T22:21:39.050" Score="0" ViewCount="315" Body="&lt;p&gt;From TV shows I have known that we still get some radio signals of the big bang and they are every where. My question is is the universe expanding so fast that even electromagnetic wave has failed to cross it after big bang?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it possible that there is another dimension that we don't know of? Like the earth is nearly flat and in small portion it's surface can be called 2d. But it possible to go &quot;straight&quot; and come back at the same spot as it is in &quot;3d space&quot;. So is it possible that the space is 4d and somehow light has crossed the cosmos several times?&lt;/p&gt;&#xA;" OwnerUserId="4117" LastActivityDate="2014-12-29T16:22:03.100" Title="How do we get radio signals of the big bang?" Tags="&lt;big-bang-theory&gt;&lt;space&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8372" PostTypeId="2" ParentId="8371" CreationDate="2014-12-27T00:11:05.193" Score="3" Body="&lt;p&gt;The best estimates at the moment are that the Universe is &quot;flat&quot;. By that we mean that a pair of parallel laser beams will travel as far as you like and they will always remain the same distance apart and parallel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ironically, the best evidence for this comes from detailed measurements of the cosmic microwave background by the WMAP and now Planck satellites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem you are having is you (I think) are imagining that the big bang happened at a point in space 13.7 billion years ago, and that radiation has been travelling outwards from that point. This is &lt;em&gt;incorrect&lt;/em&gt;. Every point in space that we see now was, about 13.7 billion years ago, part of the big bang. As the space in the Universe expanded, its contents cooled and due to the trapping of free electrons by positively charged protons to form hydrogen atoms, the light that was emitted by the hot gas &lt;em&gt;everywhere&lt;/em&gt; in the Universe at a time about 400,000 years after the big-bang was able to travel unabsorbed and unscattered. This radiation has travelled (at the speed of light) in all directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when we observed the microwave background, we are seeing light that has been travelling in a straight line for just short of 13.7 billion years. That light was in the near infrared and visible part of the spectrum when it was emitted, but because of the expansion of space, the wavelength has also been stretched by a factor of 1100 to the microwave/short-wave radio part of the spectrum.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2014-12-29T16:22:03.100" LastActivityDate="2014-12-29T16:22:03.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8373" PostTypeId="1" CreationDate="2014-12-27T07:02:32.900" Score="2" ViewCount="99" Body="&lt;p&gt;How do we know an event is a gamma-ray burst (grb)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two kinds of grbs, long burst and shot burst, which are caused by supernova explosions explosion and the merger of two neutron stars, respectively. However, at least one grb, J1644, is a tidal disruption event. Is there any other grb which was not caused by a supernova or the merger of two neutron stars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When can we say that a supernova explosion is a grb after observations? How do we know an event is a grb especially when we missed its explosion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we tell that the afterglow is caused by a grb?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please give me some reference papers since I am a layman of this field.&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="2153" LastEditDate="2014-12-27T22:20:26.593" LastActivityDate="2015-01-24T04:36:36.903" Title="How do we know an event is a gamma-ray burst?" Tags="&lt;gamma-ray-bursts&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8374" PostTypeId="2" ParentId="2445" CreationDate="2014-12-27T11:33:27.270" Score="1" Body="&lt;p&gt;The answer to this, and other questions of its kind about exoplanets, can easily be found at the website &lt;a href=&quot;http://exoplanets.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://exoplanets.org/&lt;/a&gt;  This site contains a very authoritative catalogue of exoplanet discoveries and has tools to enable tables and plots of many variables to be constructed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, to handle your question, I produced the following plot for exoplanets &lt;em&gt;discovered using the doppler radial velocity technique&lt;/em&gt;. This shows effective temperature of the parent star on the x-axis, versus projected exoplanet mass $M\sin i$. The the hottest star with an RV-identified exoplanet (on the website, you can click on the points) is HD113337, an F6V-type star discovered by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013arXiv1310.1994B&quot; rel=&quot;nofollow noreferrer&quot;&gt;Borginiet et al. (2013)&lt;/a&gt;. The next hottest is HD103774, which is given a slightly earlier spectral type of F5V. These two are joint hottest within their uncertainties. Just a touch cooler is Tau Boo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is difficult to find exoplanets around high-mass &lt;em&gt;main sequence&lt;/em&gt; stars using the doppler method because of the paucity of strong, narrow spectral lines. Most of what we know is from observations of subgiant stars - i.e. stars that &lt;em&gt;would have been&lt;/em&gt; early-type stars on the main sequence, but which have cooler photospheres after they have left the main sequence. HD102956, mentioned in another answer, is an example if this - i.e. it &lt;em&gt;was&lt;/em&gt; an early-type A star on the main sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZTYyv.png&quot; alt=&quot;Host temperature vs planet mass&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2014-12-27T11:33:27.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8375" PostTypeId="1" CreationDate="2014-12-27T11:45:41.800" Score="1" ViewCount="386" Body="&lt;p&gt;If there are indeed multiple universes, where do they exist? We &quot;know&quot; that outside our universe, consisted of spacetime, there is nothing - i.e. 0 dimensions, no space and no time. So &quot;where&quot; are other universes, how would the distance between our space and other be measured inside nothing?&lt;/p&gt;&#xA;" OwnerUserId="692" LastActivityDate="2016-06-01T17:46:17.723" Title="The multiverse spacetime paradox?" Tags="&lt;space-time&gt;&lt;multiverse&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2016-06-02T21:06:47.743" ContentLicense="CC BY-SA 3.0" />
  <row Id="8376" PostTypeId="1" CreationDate="2014-12-27T13:52:31.303" Score="3" ViewCount="80" Body="&lt;p&gt;I'm wondering if the data in star catalogues are heliocentric equatorial coordinates or geocentric equatorial coordinates. Also given the distance between stars compared to the distance between the sun and the earth, would this be a big deal to mix up the two?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Celestial_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Celestial_coordinate_system&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1190" LastActivityDate="2014-12-27T13:52:31.303" Title="Is Nomad data heliocentric or geocentric?" Tags="&lt;star&gt;&lt;coordinate&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8377" PostTypeId="1" AcceptedAnswerId="8383" CreationDate="2014-12-27T23:41:51.283" Score="1" ViewCount="335" Body="&lt;p&gt;I'm currently taking a course called Introduction to Astronomy at Coursera. It's not very formal so they didn't present a proof of the statement in the title. I understand why logically one could argue that when a star lies on the projection of the observer's local meridean onto the celestial sphere (I understand that this projection defines the observer's local sidereal time) then this star is at its highest point on the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't found any mathematical proof of this statement and also do not know how to approach the problem geometrically (either euclidean geometry or analytic geometry).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I appreciate any hint or direct answer on how to approach the problem. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2877" LastActivityDate="2014-12-28T14:05:15.827" Title="Why is a star at its highest point when local sidereal time = RA (right ascension)?" Tags="&lt;positional-astronomy&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8378" PostTypeId="1" CreationDate="2014-12-28T00:56:56.170" Score="2" ViewCount="473" Body="&lt;p&gt;Comets form where stars form. And they are then ejected by the orbital changes of giant planets. But why are they ejected to form specifically a local Oort cloud instead of going interstellar and becoming vagabonds? The fact that comets come from all directions doesn't discriminate between them coming from an Oort cloud or from interstellar sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main argument against the idea of one star capturing a comet from another star, seems to be that the relative velocity is too high for that. But consider the huge age of the Sun which is now on its 18th or so orbit through this crowded galactic star field disc. And which by chance or co-formation might have now, or at some earlier time, roughly the same velocity as some other millions of other stars. And what if there are many (many) more comets formed and/or ejected from young stars than currently supposed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is it really unlikely that the Oort &quot;cloud of comets&quot; instead largely represents an interstellar source?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2014-12-28T01:58:21.367" LastActivityDate="2014-12-31T13:19:18.827" Title="Why do comets come from our local Oort cloud instead of from interstellar sources?" Tags="&lt;comets&gt;&lt;formation&gt;&lt;oort-cloud&gt;&lt;interstellar&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8379" PostTypeId="2" ParentId="8378" CreationDate="2014-12-28T01:14:32.600" Score="3" Body="&lt;p&gt;Since most comets are on a predictable orbit that has them circling the sun (several times), then they cannot be from outside the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any comet that originates outside the solar system, will pick up enough velocity approaching the sun to be able to leave the solar system. Therefore, we would only see the comet once. It is unlikely that the comet approaches Jupiter (Saturn, Neptune or Uranus) where the trajectory shifts enough so that it will remain in the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the solar system is believe to be formed from the gravitational attraction of interstellar gas and debris, it is very unlikely that comets will be ejected unless their orbits are perturbed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there is a low probability of a comet being ejected from one system, another low probability that the comet will encounter another system (after millions and billions of years) and yet another low probability that the comet gets captured in the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:&lt;br&gt;&#xA;If the object was pushed out to 1 LY and it then returned as a comet, it was still under the influence the sun's gravity and would not be considered as extra-solar. The object would need to come from another system to be called extra-solar. Since the ejected objects are in a nearby orbit to the solar system, they will either return, or travel outward. Think about a satellite with a period of 12 hours. A slight push will put it in orbit 2 meters higher, and will increase the period significantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a lot of space at Neptune's orbital distance, and the intersection with Neptune's gravity is quite small (ratio of area of influence (0.1g) / surface area = ${8.6\times 10^{-11}}$). The reason the Jupiter captures asteroids is because they are in the same plane and are moving in the same direction. This increases the probability of changing the orbit significantly. It could also take several hundred passes before the orbit changes significantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of influence is the cross section area where the gravity is greater than 0.1g. The surface area is the spherical surface area of the mean distance to the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter - ${5.325\times 10^{-8}}$ ~= 0.0000053%&lt;br&gt;&#xA;Saturn  - ${4.703\times 10^{-9}}$ ~= 0.00000047%&lt;br&gt;&#xA;Uranus  - ${1.791\times 10^{-10}}$ ~= 0.000000018%&lt;br&gt;&#xA;Neptune - ${8.605\times 10^{-11}}$ ~= 0.0000000086%  &lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="1322" LastEditDate="2014-12-28T16:28:24.697" LastActivityDate="2014-12-28T16:28:24.697" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8380" PostTypeId="1" AcceptedAnswerId="8382" CreationDate="2014-12-28T07:24:40.650" Score="2" ViewCount="5460" Body="&lt;p&gt;I am trying to figure out the position of sun with respect to moon on a full moon day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an image from certain &lt;a href=&quot;http://dockastronomy.wikispaces.com/Monn%27s+Phases&quot; rel=&quot;nofollow noreferrer&quot;&gt;wiki&lt;/a&gt; which I do not find convincing.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Pz9xR.jpg&quot; alt=&quot;enter image description here&quot;&gt;1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sun rays point as shown in the image and the earth stands between the moon and sun on Full moon's day, I dont see any reason why the moon should be shining in its full glory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly when the moon is directly in between earth and sun, it should have maximum brightness but the image shows it as the new moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, I cannot deduce how at all can moon receive light across the entire sphere at any given angle from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting it specifically, what would the position of sun w.r.t to earth and moon on a full moon day ?&lt;/p&gt;&#xA;" OwnerUserId="1636" LastActivityDate="2019-08-14T13:25:56.010" Title="Full Moon Position of Sun" Tags="&lt;solar-system&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8381" PostTypeId="2" ParentId="8375" CreationDate="2014-12-28T07:30:42.940" Score="1" Body="&lt;p&gt;Read Edwin Abbot's book &lt;a href=&quot;http://www.amazon.co.uk/gp/product/048627263X/ref=pd_lpo_sbs_dp_ss_1?pf_rd_p=569136327&amp;amp;pf_rd_s=lpo-top-stripe&amp;amp;pf_rd_t=201&amp;amp;pf_rd_i=1623750318&amp;amp;pf_rd_m=A3P5ROKL5A1OLE&amp;amp;pf_rd_r=1XNZ9DBB7DGFZ6P8N6YP&quot; rel=&quot;nofollow&quot;&gt;Flatland&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this story we visit a vision of a two dimensional &quot;Universe&quot; embedded in a higher dimensional space. Now by analogy we may suppose that our Universe is embedded in a higher dimension space with plenty of room for other (3+1)-Universes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not claim this is a viable model of &quot;reality&quot; just a thought experiment to show &quot;where&quot; other universes might be if there are indeed multiple universes. Also Flatland is a fun read.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2014-12-28T07:30:42.940" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8382" PostTypeId="2" ParentId="8380" CreationDate="2014-12-28T10:15:36.463" Score="6" Body="&lt;p&gt;The orbits are not all in exactly the same plane, that is why we do not have eclipses every month. With the orbits at a slight skew it is possible to have the geometry very near to that in your diagram without the Earth blocking the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lunar_eclipse&quot; rel=&quot;noreferrer&quot;&gt;Lunar eclipses&lt;/a&gt;, the Moon passing through the Earth's shadow which is what your post is about, do occur relatively frequently compared to &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_eclipse&quot; rel=&quot;noreferrer&quot;&gt;Solar eclipses&lt;/a&gt; because the Earth is bigger than the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At New Moon the lit side of the Moon is facing away from the Earth and so we do not see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will find a fairly good explanation, and a better diagram, of all this &lt;a href=&quot;http://www.moonconnection.com/moon_phases.phtml&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2014-12-28T10:21:17.957" LastActivityDate="2014-12-28T10:21:17.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8383" PostTypeId="2" ParentId="8377" CreationDate="2014-12-28T14:05:15.827" Score="2" Body="&lt;p&gt;At some geocentric latitude $\phi$ and longitude $\lambda$, the elevation or altitude $a$ of a star with right ascension $\alpha$ and declination $\delta$ is given by&#xA;$$\begin{aligned}&#xA;\sin a = \sin\phi \sin \delta + \cos \phi \cos\delta \cos h&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;(1)&#xA;\end{aligned}$$&#xA;where&#xA;$$\begin{aligned}&#xA;h=\theta_\phi - \alpha\qquad&amp;amp;&amp;amp;\qquad&amp;amp;&amp;amp;\qquad&amp;amp;&amp;amp;\quad\,&amp;amp;&amp;amp;(2)&#xA;\end{aligned}$$&#xA;is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hour_angle&quot; rel=&quot;nofollow&quot;&gt;hour angle&lt;/a&gt; to the star, expressed in terms of the local sidereal time $\theta_{\phi}$ and the right ascension $\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiating (1) with respect to time yields&#xA;$$\begin{aligned}&#xA;\cos a \frac{da}{dt} = &amp;amp;&#xA;\phantom{+}\,(\cos\phi\sin\delta - \sin\phi \cos\delta \cos h)\,\frac{d\phi}{dt} \\&#xA;&amp;amp; + \,(\sin\phi \cos\delta - \cos\phi \sin\delta \cos h)\,\frac{d\delta}{dt} \\&#xA;&amp;amp; - \cos\phi \cos\delta \sin h\,\frac{dh}{dt} &amp;amp;\!\!\!(3)&#xA;\end{aligned}&#xA;$$&#xA;While not quite zero, the time derivatives of latitude $\phi$ and declination $\delta$ are negligibly small compared to the $2\pi$ radians per sidereal day time derivative of hour angle $h$. For all practical purposes, the above thus reduces to&#xA;$$\begin{aligned}&#xA;\cos a \frac{da}{dt} = - \cos\phi \cos\delta \sin h\,\frac{dh}{dt} \quad\!&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;(4)&#xA;\end{aligned}&#xA;$$&#xA;The left hand side of (4) is zero at extrema of elevation angle. Thus we're looking for conditions that make the right hand side of (4) equal zero. Since $dh/dt$ is nonzero, the right hand  side is zero only if one or more of $\cos \phi$, $\cos \delta$, or $\sin h$ is zero. The first two cases ($\cos \phi = 0$ and $\cos \delta = 0$) represent conditions where elevation is constant. The only condition of interest is $\sin h = 0$, which means an hour angle of 0° or 180° (hour angle is constrained to lie between 0° (inclusive) and 360° (exclusive)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since latitude $\phi$ and declination $\delta$ both lie between -90° and +90°, the condition $h=180^{\circ}$ represents the minimum possible elevation, while $h=0$ represents the maximum possible elevation. From (2), $h=0$ means $\theta=\alpha$, or local sidereal time being equal to right ascension.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2014-12-28T14:05:15.827" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8384" PostTypeId="2" ParentId="8380" CreationDate="2014-12-28T17:41:27.430" Score="1" Body="&lt;p&gt;I found this little video from youtube as pretty good explanation for my question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=wz01pTvuMa0&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=wz01pTvuMa0&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1636" LastActivityDate="2014-12-28T17:41:27.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8385" PostTypeId="1" CreationDate="2014-12-28T19:37:51.590" Score="5" ViewCount="733" Body="&lt;p&gt;Can someone recommend some Stellarium settings to mimic the amount of light pollution, and magnitude of stars that should be visible on a clear night in London, UK?&lt;/p&gt;&#xA;" OwnerUserId="4146" LastActivityDate="2014-12-30T00:09:28.423" Title="How can I make Stellarium match the magnitude of stars visible in London, UK?" Tags="&lt;star-gazing&gt;&lt;software&gt;&lt;light-pollution&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8386" PostTypeId="1" CreationDate="2014-12-28T22:09:35.737" Score="6" ViewCount="369" Body="&lt;p&gt;I am looking for a design of a cheap and easy telescope to build with my 9yo daughter with the only basic requirement being that I'd love to be able to see detail on the moon.  Looking around, there seems to be a wide variety of either extremely simple or very advanced (to me) options.  I've seen &lt;a href=&quot;https://astronomy.stackexchange.com/questions/2212/which-type-of-telescope-should-i-make&quot;&gt;this post&lt;/a&gt; and &lt;a href=&quot;https://astronomy.stackexchange.com/questions/3714/any-cheap-less-than-100-euro-telescope-for-observing-jupiter&quot;&gt;this post&lt;/a&gt;, but they look a bit advanced.  I am hoping for something as basic as possible (i.e. that we can construct in an afternoon or two) that does not come from a kit.  &lt;a href=&quot;http://www.nymetroparents.com/article/diy-telescope-from-national-geographic-kids-get-outside-guide#.VKB92v8SAA&quot; rel=&quot;nofollow noreferrer&quot;&gt;This site&lt;/a&gt; has an option, but I imagine that's simply unpractical.  Is there anything else that anyone would recommend to get started?&lt;/p&gt;&#xA;" OwnerUserId="1525" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-01-14T19:10:41.200" Title="Basic DIY Telescope" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8388" PostTypeId="1" CreationDate="2014-12-29T09:40:28.037" Score="5" ViewCount="266" Body="&lt;p&gt;If Universe's analogy is 2D surface of the sphere, could we come around in circle? Say on Earth you decided to travel straight in one direction, after completing one full circle you'd find yourself at exact spot where you started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if we up that up by 1 dimension and apply it to our Universe, would the same thing happen if we pointed the rocket in a straight line? I.e. would the rocket come back to earth?&lt;/p&gt;&#xA;" OwnerUserId="692" LastActivityDate="2014-12-29T13:51:35.867" Title="Universe as 2d sphere surface - a round trip?" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8389" PostTypeId="2" ParentId="1692" CreationDate="2014-12-29T10:12:09.907" Score="7" Body="&lt;p&gt;I'm afraid the only correct answer to this question is that nobody knows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cosmology is based on a number of simplifying assumptions (sometimes called principles) - the most basic of which is that the universe is &quot;isotropic and homogeneous on large scales&quot;. Isotropic means that is (or looks) the same in any direction; homogeneous means that what we see does not depend on our position in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &quot;cosmological principle&quot; is reasonably well tested and would certainly make sense if the universe were very much bigger than we are capable of seeing. The most distant thing we can &quot;see&quot; in our universe is the cosmic microwave background (CMB), the light of which has been travelling for about 13.6 billion years. However, if we were to think about a parcel of gas that emitted the CMB and where it is &lt;em&gt;now&lt;/em&gt;, we could calcuate that it is about 46 billion light years away due to the expansion of space in the universe. Thus the &quot;edge of the observable universe&quot; is about 46 billion light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we can say, is that on these scales the Universe is very isotropic, and homogeneous, because the CMB is very smooth and looks the same in all directions to a very good approximation (but see below). This suggests that if you were to travel 46 billion light years in any direction, that you would probably have a very similar view to what we have now (that is, if you were able to miraculously travel instantaneously; if you travelled more slowly then of course the universe would continue to get older).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are talking about travelling more than a factor 10000 further. On these scales we do not know anything outside of speculation. The inflationary model for the big bang might suggest that our observable universe is just a tiny part of a much greater entity and that within this very much larger entity everything will be homogeneous and isotropic at a given cosmic time. However, the recent very precise &lt;a href=&quot;http://sci.esa.int/planck/51551-simple-but-challenging-the-universe-according-to-planck/&quot; rel=&quot;noreferrer&quot;&gt;Planck measurements&lt;/a&gt; of the CMB, suggest there may be small, but real, anisotropies on scales as big as the observable universe, which would then suggest that any extrapolation by a factor of 10000 might be wishful thinking.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-01-01T22:06:58.677" LastActivityDate="2015-01-01T22:06:58.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8390" PostTypeId="1" AcceptedAnswerId="8415" CreationDate="2014-12-29T11:42:36.253" Score="11" ViewCount="1604" Body="&lt;p&gt;I have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Dobsonian_telescope&quot; rel=&quot;noreferrer&quot;&gt;Dobsonian telescope&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;It is using &lt;a href=&quot;http://en.wikipedia.org/wiki/Altazimuth_mount&quot; rel=&quot;noreferrer&quot;&gt;Altazimuth mount&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Basic idea of using it is to target the object by moving telescope vertical axis perpendicular to the ground, and an elevation axis that is parallel to the ground.&lt;/p&gt;&#xA;&lt;p&gt;I have installed two step motors to automate the movement along both vertical and elevation axis.&lt;/p&gt;&#xA;&lt;p&gt;I would like to find out how can I cancel out earth rotation speed by simultaneously moving both vertical axis and elevation axis motors.&lt;/p&gt;&#xA;&lt;p&gt;Idea behind it is to point telescope at the object and press the button.&#xA;Then the step motors driver software would follow the object as the earth rotate.&lt;/p&gt;&#xA;&lt;p&gt;I will quote few lines from &lt;a href=&quot;http://www.astronomyhints.com/mounts.html&quot; rel=&quot;noreferrer&quot;&gt;The Basic Astronomical Telescope Mounting Designs&lt;/a&gt; to help me explain what I am trying to achieve:&lt;/p&gt;&#xA;&lt;p&gt;[Using altazimuth telescope...:]&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If you happen to be observing from the North or South Pole, the vertical axis would be aligned with the Earth's spin axis. The nice thing about that would be that when you found an object to observe, rotation in only the vertical axis would be needed to keep the object in the field of view. Rotating at the Earth's spin rate in the opposite direction as the Earth's rotation would keep and object motionless in the eyepiece.&lt;/p&gt;&#xA;&lt;p&gt;However, for any other latitude on the planet, the vertical axis is not aligned with the Earth's spin axis. This means that to keep an object in the field of view requires motion in both axes. The motion rates will change over time as the elevation angle changes. Tracking objects near the horizon requires mostly changes in elevation, and tracking objects more straight up requires mostly changes in azimuth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I need to find mathematical algorithm that will help me solve the problem described in the second paragraph.&lt;/p&gt;&#xA;&lt;p&gt;Hope this is clear.&lt;/p&gt;&#xA;" OwnerUserId="4150" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-09-18T21:02:43.883" Title="Cancelling out earth rotation speed, Altazimuth mount" Tags="&lt;rotation&gt;&lt;dobsonian-telescope&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8392" PostTypeId="2" ParentId="8388" CreationDate="2014-12-29T13:46:28.973" Score="4" Body="&lt;p&gt;This is another case of taking an analogy too far (the other one I can think of is assuming that the &lt;a href=&quot;http://theory.uwinnipeg.ca/users/gabor/black_holes/slide5.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;rubber sheet&lt;/a&gt; analogy indicates that there is some extra spatial dimension$^1$). In this case, I'm assuming you're talking about the &quot;balloon&quot; analogy - the universe is expanding, like a balloon, but since there's no center, all bits of matter are on the surface of the balloon. This naturally leads to the idea that if you travel in one direction, you'll eventually come back to where you started. This is, unfortunately, not true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scientists often talk about the &lt;a href=&quot;https://en.wikipedia.org/wiki/Shape_of_the_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;shape of the universe&lt;/a&gt; - how it is curved on a local or global scale. There are three possible categories, depending on one parameter, $\Omega$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Positive curvature $\to$ $\Omega&amp;gt;1$&lt;/li&gt;&#xA;&lt;li&gt;Zero curvature $\to$ $\Omega=1$&lt;/li&gt;&#xA;&lt;li&gt;Negative curvature $\to$ $\Omega&amp;lt;1$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/9/98/End_of_universe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; diagram gives a good visual of each curvature:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/9/98/End_of_universe.jpg&quot; alt=&quot;Curvature&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should add that these are merely examples of universes (more accurately, &lt;em&gt;manifolds&lt;/em&gt;) with each $\Omega$. There are other cases with these curvatures, and so other universe where you can start at one point and end at the same point, as in a &lt;a href=&quot;https://en.wikipedia.org/wiki/Three-torus_model_of_the_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;3-torus&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these shapes &lt;em&gt;could&lt;/em&gt; exist. However, in our universe, the global curvature is one where $\Omega \approx 1$ - a flat universe. The measurements that essentially clinch this are from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Wilkinson_Microwave_Anisotropy_Probe&quot; rel=&quot;nofollow noreferrer&quot;&gt;WMAP&lt;/a&gt; probe, which has analyzed the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow noreferrer&quot;&gt;CMB&lt;/a&gt; and determined the age of the universe and its curvature. Measurements show, to a decent degree of accuracy, that our universe is flat - and so if we started in one direction, we would never return to where we started. Check out &lt;a href=&quot;http://wmap.gsfc.nasa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;the NASA page&lt;/a&gt; for a full rundown of WMAP's results.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^1$ Good questions dealing with this can be found &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1451/spacetime-curvature-illustration-accuracy&quot;&gt;here&lt;/a&gt;, &lt;a href=&quot;https://physics.stackexchange.com/questions/82383/universe-being-flat-and-why-we-cant-see-or-access-the-space-behind-our-univer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://astronomy.stackexchange.com/questions/7879/how-deep-and-shaped-is-the-depth-of-a-black-hole/7880#7880&quot;&gt;here&lt;/a&gt;; John Rennie's answer on Physics is especially good.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2014-12-29T13:51:35.867" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8394" PostTypeId="2" ParentId="1715" CreationDate="2014-12-29T16:39:26.070" Score="7" Body="&lt;p&gt;Arguably, &quot;rogue planets&quot; have already been discovered by direct imaging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Giant planets when first formed are big and hot. They radiate their own light, mostly in the infrared. So &lt;em&gt;young&lt;/em&gt; isolated planets can be seen directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been various claims in the literature that objects as small as a few Jupiter masses have been identified in young star forming regions. See various papers by the IAC brown dwarf research group&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2000Sci...290..103Z&quot; rel=&quot;noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2000Sci...290..103Z&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2002ApJ...578..536Z&quot; rel=&quot;noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2002ApJ...578..536Z&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2014A%26A...568A..77Z&quot; rel=&quot;noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2014A%26A...568A..77Z&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2013MmSAI..84..926Z&quot; rel=&quot;noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2013MmSAI..84..926Z&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These claims are open to criticism - sometimes it is hard to tell whether a faint object really belongs to the star forming region observed, rather than being an unassociated background object. The claimed masses also depend heavily on models for the luminosity-mass relation as a function of age, and the ages of these objects are not easily constrained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless it would not be surprising if, in the maelstrom of the formation of a cluster of stars, some planetary systems were stripped from their parent stars by close encounters with other objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chances of seeing older, isolated, planetary mass objects are slim, but microlensing appears to be the only technique presently available. The microlensing signature of a free-floating planet is of course unrepeatable so a discovered planet could not be followed up in any way. However, surveys of microlensing events could be a way of saying something statistically about how common such objects are. See for example &lt;a href=&quot;http://astrobites.org/2011/05/24/free-floating-planets-might-outnumber-stars/&quot; rel=&quot;noreferrer&quot;&gt;http://astrobites.org/2011/05/24/free-floating-planets-might-outnumber-stars/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: It is also worth noting that the whether these things really are &quot;planets&quot; at all is disputed. They could either be genuine planets, formed in the same way that is hypothesised for most giant planets - that is by accretion onto a rocky core that formed around a star. They could then have been displaced from their parent star by dynamical interactions with other bodies in their system or with a third body. As I said above, N-body simulations do predict that this will happen (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013ApJ...772..142L&quot; rel=&quot;noreferrer&quot;&gt;Liu et al. 2013&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand they could represent the very lowest mass gas fragments that are able to form during the collapse and fragmentation of a molecular cloud and that for some reason were unable to accrete further gas (i.e. they are really more like low-mass brown dwarfs). This so-called &quot;fragmentation limit&quot; is of order 10 Jupiter masses, but if it were a little lower it might explain the free-floating planets that have been seen so far.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-05-19T08:32:52.787" LastActivityDate="2015-05-19T08:32:52.787" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8396" PostTypeId="1" AcceptedAnswerId="8399" CreationDate="2014-12-29T23:43:51.497" Score="2" ViewCount="445" Body="&lt;p&gt;Is it possible for a red dwarf to orbit a gas giant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OR&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has this happened and it is just assumed that the gas giant is orbiting the star?&lt;/p&gt;&#xA;" OwnerUserId="4160" LastEditorUserId="2531" LastEditDate="2014-12-30T09:05:43.720" LastActivityDate="2014-12-31T05:50:25.680" Title="Is it possible for a star to orbit a planet?" Tags="&lt;star&gt;&lt;orbit&gt;&lt;gas-giants&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8397" PostTypeId="1" AcceptedAnswerId="8412" CreationDate="2014-12-29T23:57:55.757" Score="-2" ViewCount="179" Body="&lt;p&gt;I know that it can vary A LOT if you are using the Drake Equation but using realistic figures, how likely is it to discover a civilisation that is at least semi-intelligent (stone age - medieval era, comparing to human history).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I heard that there are &lt;a href=&quot;https://en.wikipedia.org/wiki/Kardashev_scale&quot; rel=&quot;nofollow&quot;&gt;various classes of civilization&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Us (Humans), doesn't reach the energy milestones in the classes. &lt;strong&gt;(Level 0)&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Controls 100% of the energy from their home planet (1,000 - 10,000 yrs away). &lt;strong&gt;(Level 1)&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Controls 100% of the energy from their home star (5,000 - 20,000 yrs away). &lt;strong&gt;(Level 2)&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Controls 100% of the energy from their local group of stars (never or 10,000 - 30,000 yrs away). &lt;strong&gt;(Level 3)&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Controls 100% of the energy from their home galaxy (NEVER! just deal with it). &lt;strong&gt;(Level 4)&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The last paragraph is &lt;strong&gt;NOT PART&lt;/strong&gt; of the question; it is just for your knowledge!&lt;/p&gt;&#xA;" OwnerUserId="4160" LastEditorUserId="2153" LastEditDate="2014-12-31T00:48:49.947" LastActivityDate="2014-12-31T00:48:49.947" Title="What are the odds of (Semi) Intelligent Life in the Milky Way?" Tags="&lt;life&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ClosedDate="2014-12-31T00:56:04.397" ContentLicense="CC BY-SA 3.0" />
  <row Id="8398" PostTypeId="2" ParentId="8385" CreationDate="2014-12-30T00:09:28.423" Score="6" Body="&lt;p&gt;Ok go on Stellarium then hover over the left side of the screen afterwards click on the location and select London, England (City Of London is the original walled city), then go and hover over the left side of the screen and select sky and viewing options and in the sky section and select the air pollution level from location database then it will automatically set the air pollution depending on where you are on Earth, cool right?&lt;/p&gt;&#xA;" OwnerUserId="4160" LastActivityDate="2014-12-30T00:09:28.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8399" PostTypeId="2" ParentId="8396" CreationDate="2014-12-30T01:40:37.580" Score="6" Body="&lt;p&gt;Pretty much by definition, &lt;strong&gt;no&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for &quot;object A&quot; to orbit &quot;object B&quot;, &quot;object B&quot; needs to be substantially more massive than &quot;object A&quot;.  In order for your red dwarf to orbit something, that &quot;something&quot; needs to be more massive than a red dwarf.  With the current composition of the universe, that &quot;something&quot; will be mostly hydrogen, and once you reach a certain level of mass (about 90 times that of Jupiter), you can't prevent nuclear fusion from starting.  The definition of &quot;planet&quot; is a bit fuzzy, but everyone agrees something lit by nuclear fusion is a star, not a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could theoretically change in the distant future, when there's enough non-hydrogen mass around for stellar-scale rocky bodies.&lt;/p&gt;&#xA;" OwnerUserId="1559" LastActivityDate="2014-12-30T01:40:37.580" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8400" PostTypeId="2" ParentId="8396" CreationDate="2014-12-30T06:38:57.967" Score="4" Body="&lt;p&gt;Well, technically, a planet and a star orbit each other. In that sense, yes. If a gas giant is orbiting a red dwarf, then the red dwarf is also orbiting the gas giant. Always.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you quibble that it doesn't count if the star doesn't move much, I throw the quibble back by pointing to Jupiter and Sol. The barycenter of the system is outside the sun, and that means the sun's orbit is not exactly tiny. (Smaller than Jupiter's orbit, but now we're no longer talking absolutes.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see how it might be possible for a gas giant to out-mass a red dwarf. The gas giant could be really a protostar, on its way to becoming a red dwarf, but not there yet. That would mean the gas giant would have to be a lot younger than the star, meaning that they formed separately and one of them captured the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, the red dwarf could be a gas giant that got heated by some external source, say by a nearby normal star. -- No, scratch that. External heating would keep it from contracting, and its core would never become hot and dense enough to initiate fusion. -- Unless the red dwarf was a gas giant that swallowed a neutron star or black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All pretty unlikely, but not downright impossible. And in astronomy, whatever is not prohibited is mandatory. It it can happen, it has.&lt;/p&gt;&#xA;" OwnerUserId="4077" LastActivityDate="2014-12-30T06:38:57.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8402" PostTypeId="2" ParentId="8396" CreationDate="2014-12-30T09:04:30.727" Score="6" Body="&lt;p&gt;Neither statement is true, just like the statement that the &quot;Earth orbits the Sun&quot; is not strictly speaking true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In truth, if we are talking about a system with two masses, what happens is that they both orbit their common centre of gravity. Whether we perceive that one object orbits another is really just a question of their mass ratio. In the example shown below, the smaller, less massive object might be said to orbit the larger object and follows the red path. However, the larger object also executes a smaller circular orbit shown in green, with the same orbital period. Both objects actually orbit the centre of the picture, which is the centre of gravity of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many cases - e.g. a planet like Earth and a star like the Sun or a satellite around the Earth, the mass ratio is so small, that it does make sense to make the approximation that one object orbits another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qagHK.gif&quot; alt=&quot;Orbits&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2014-12-30T09:04:30.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8404" PostTypeId="1" AcceptedAnswerId="8417" CreationDate="2014-12-30T10:23:03.917" Score="5" ViewCount="779" Body="&lt;p&gt;I have a pair of 15x70 binoculars. I would not say what manufacturer to avoid being the subject of 'hold'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current angular diameter is 15.5 arcseconds. I am wondering if that condition is enough to be able to see rings. My eyesight is fair.&lt;/p&gt;&#xA;" OwnerUserId="4099" LastActivityDate="2014-12-31T06:13:47.823" Title="How to see Saturn's rings through a pair of binoculars?" Tags="&lt;amateur-observing&gt;&lt;saturn&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8405" PostTypeId="2" ParentId="601" CreationDate="2014-12-30T10:33:03.537" Score="2" Body="&lt;p&gt;I think it is incorrect to call this object a &quot;free floating planet&quot;, or at least it is not clear at all that it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the object is based on (a) evolutionary models of the luminosity-mass relationship and (b an assumed age for the beta Pic moving group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first point gives at least factors of two uncertainty in the claimed mass. The second point contains (in my view) a systematic error in that the authors have used an outdated age of 12Myr for the beta Pic group. It has been revised (by myself and others) to about 21-24 Myr. This would make the estimated mass about 30-50% higher and so it would be $\geq 10$ times the mass of Jupiter and could easily be at a deuterium burning mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as an &quot;orbit&quot; goes, as this object is unassociated with another star directly, but is probably part of a co-moving kinematic group (the Beta Pic moving group), it will share the orbit of that group &lt;em&gt;around our Galaxy&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2014-12-30T10:33:03.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8406" PostTypeId="2" ParentId="8094" CreationDate="2014-12-30T11:22:20.220" Score="1" Body="&lt;p&gt;This is just a repository of useful links to software and papers along the way.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://bima.astro.umd.edu/nemo/&quot; rel=&quot;nofollow&quot;&gt;NEMO&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On Toolboxes and Telescopes , by Hut and Sussman, (1986) in: The Use of Supercomputers in Stellar Dynamics, Springer Verlag, p 193-198.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="793" LastActivityDate="2014-12-30T11:22:20.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8407" PostTypeId="1" CreationDate="2014-12-30T17:51:08.877" Score="-1" ViewCount="136" Body="&lt;p&gt;I watched a program where four astrophysicists were debating this topic. I couldn't fully understand what they were saying but wanted to ask others' opinions on whether there is only one Universe or multiple universes. I believe that there are other universes as well but I can't go along with the theory that we exist in other universes as well because this is just too far-fetched. What do others have to say about this? Also they said that there was no &quot;Big Bang&quot; but instead it was a ball of light that expanded out to form the Universe.&lt;/p&gt;&#xA;" OwnerUserId="2562" LastEditorUserId="2153" LastEditDate="2014-12-30T17:58:02.853" LastActivityDate="2014-12-30T19:08:08.850" Title="One Universe or Multiverse?" Tags="&lt;multiverse&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ClosedDate="2014-12-31T01:02:15.607" ContentLicense="CC BY-SA 3.0" />
  <row Id="8408" PostTypeId="1" AcceptedAnswerId="8422" CreationDate="2014-12-30T18:26:52.370" Score="5" ViewCount="114" Body="&lt;p&gt;This might seem like a backwards question, but I'm interested in what wavelength to select in a (hypothetical) ground-based radio telescope observation to expect detecting as &lt;em&gt;little&lt;/em&gt; as possible! :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should still be reachable from space, i.e. a wavelength with a high atmospheric absorption is probably a bad candidate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess this could be answered by looking at a broad averaged spectrum collected by a radio telescope, and comparing against an atmospheric absorption spectrum chart.&lt;/p&gt;&#xA;" OwnerUserId="4171" LastActivityDate="2014-12-31T22:43:51.103" Title="Which wavelength is the most quiet, for a ground-based radio telescope?" Tags="&lt;radio-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8409" PostTypeId="2" ParentId="8407" CreationDate="2014-12-30T19:08:08.850" Score="1" Body="&lt;p&gt;One instance of &quot;many universes&quot; come from interpretation of quantum mechanics. Quantum mechanics' formalism works surprisingly well, however what it actually means in the real world is not known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several &lt;a href=&quot;https://en.wikipedia.org/wiki/Interpretations_of_quantum_mechanics&quot; rel=&quot;nofollow&quot;&gt;interpretations of the formalism&lt;/a&gt;. Interpretation means that different ways of looking at the mathematical phenomenon always gives the same results, thus they can not be distinguished by any known experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the interpretation is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Copenhagen_interpretation&quot; rel=&quot;nofollow&quot;&gt;Copenhagen interpretation&lt;/a&gt;, the &quot;standard&quot; interpretation, which's main idea is that the properties only become part of reality when they are measured.&#xA;Another famouse one is &lt;a href=&quot;https://en.wikipedia.org/wiki/De_Broglie%E2%80%93Bohm_theory&quot; rel=&quot;nofollow&quot;&gt;Bohm Theory&lt;/a&gt;, where is explicitly nonlocal (Events at point A can influence a very distant point B faster than speed of light).&#xA;And now, back to your question, there is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Many-worlds_interpretation&quot; rel=&quot;nofollow&quot;&gt;many-worlds-interpretation&lt;/a&gt;, which says that everytime a particle in a superposition is measured, the universe splits and in the one unverse it has the first property, and in the second universe it has the second property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said, there is no known way yet to distinguish between those different interpretations via experiments, thus the question cannot be part of physics, but of philosophy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiverse also appear in other speculative contexts in physics. One instance is string theory, where it has been found a decade ago that the theory has $10^{500}$ different solutions. Instead of giving up, string theoretists predicted that all these different implementations exist. That leaded to alot of controversy, people argued that it shows that string theory is &lt;em&gt;not even wrong&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other instances where multiverse appear, I would suggest to have a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Multiverse&quot; rel=&quot;nofollow&quot;&gt;the corresponding wikipedia page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2662" LastActivityDate="2014-12-30T19:08:08.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8410" PostTypeId="1" AcceptedAnswerId="8411" CreationDate="2014-12-30T19:18:41.300" Score="8" ViewCount="1636" Body="&lt;p&gt;As we know atmospheres of celestial bodies don't just stop at a given distance. They gradually become less dense as you move away from the center. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the diameter of stars is typically given to be that of their photosphere (i.e. the boundary where a laser shined from behind the star toward an observer would just barely be visable). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assumed that a similar method using optical depth also applied to gas giants. However wikipedia currently has the somewhat confusing explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As Jupiter has no surface, the base of its atmosphere is usually&#xA;  considered to be the point at which atmospheric pressure is equal to 1&#xA;  MPa (10 bar), or ten times surface pressure on Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which isn't quite the same thing as diameter, but it got me thinking: When wikipedia or a textbook or whatever lists a diameter for a gas giant how is that figure arrived at?&lt;/p&gt;&#xA;" OwnerUserId="4172" LastActivityDate="2015-01-03T17:12:04.537" Title="How is the diameter of a gas giant calculated?" Tags="&lt;jupiter&gt;&lt;gas-giants&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8411" PostTypeId="2" ParentId="8410" CreationDate="2014-12-30T23:15:24.163" Score="6" Body="&lt;p&gt;I won't argue with the wikipedia definition (although the &lt;a href=&quot;http://nssdc.gsfc.nasa.gov/planetary/factsheet/jupiterfact.html&quot; rel=&quot;nofollow&quot;&gt;NASA Jupiter fact sheet&lt;/a&gt; lists it as the radius at 1 bar), but just to point out that the scale height of the atmosphere of Jupiter is given by, $h \sim kT/m g$, where $T$ is the temperature, $m$ is the mean molecular mass and $g$ is local gravity. Putting in some numbers: $ g \simeq 24.8 m/s^2$, $m \simeq 2.2 m_u$, $T \simeq 165K$, gives $h \simeq 25\ km$. Thus the pressure changes extremely rapidly compared with the actual radius of a gas giant,and is very small compared to the difference for instance between the polar and equatorial radius for a gas giant (like Jupiter) with significant rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, whether you give a planet's radius at 1 bar or 10 bar isn't going to make a lot of difference $&amp;lt;100\ km$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However your comment about whether the definition is analogous to the photosphere of a star does raise an interesting question. The effective opacity radius of a gas giant can vary with the wavelength of light. It is this property that can be used in conjunction with transiting exoplanetary systems to learn something about their atmospheres. For instance, it requires less physical depth of a gas to block wavelengths corresponding to the strong 589nm sodium doublet and thus the exoplanet is opaque at larger radii at this wavelength and thus has a (slightly) deeper transit. This provides the opportunity for transit transmission spectroscopy. e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2012MNRAS.426.1663S&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/2012MNRAS.426.1663S&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is that at UV and X-ray wavelengths there have been observations suggesting that hot Jupiters have very extended (and opaque to high energy photons) atmospheres and may be being photoevaporated. e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013ApJ...773...62P&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/2013ApJ...773...62P&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: So this raises an important distinction. A definition based on the radius at a particular pressure is mostly used by theorists and solar system scientists. However, we still don't know what the atmospheric structures of exoplanets are and can't measure pressures. The radii quoted for exolanets are more akin to how the radius of a star is measured - based on its opacity at the wavelengths being observed. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-01-03T17:12:04.537" LastActivityDate="2015-01-03T17:12:04.537" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8412" PostTypeId="2" ParentId="8397" CreationDate="2014-12-31T00:46:17.217" Score="2" Body="&lt;p&gt;The odds of detecting a civilization at the technology level you describe are exceedingly remote. The reason is suggested by your mention of the Kardashev scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Search_for_extraterrestrial_intelligence&quot; rel=&quot;nofollow&quot;&gt;search for extraterrestrial intelligence (SETI)&lt;/a&gt; generally involves looking for various electromagnetic signatures, such as radio waves, etc. At the level of stone age through medieval era, about the best one might hope for is to see fires on the night side of a planet. This seems hardly likely at interstellar distances (though such signatures could certainly be able to be picked up within a particular stellar system).&lt;/p&gt;&#xA;" OwnerUserId="2674" LastActivityDate="2014-12-31T00:46:17.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8413" PostTypeId="1" AcceptedAnswerId="8414" CreationDate="2014-12-31T01:08:35.707" Score="7" ViewCount="4133" Body="&lt;p&gt;What is the temperature of an accretion disc surrounding a supermassive black hole? Is there plasma in the disc?&lt;/p&gt;&#xA;" OwnerUserId="4175" LastEditorUserId="2153" LastEditDate="2014-12-31T01:40:18.683" LastActivityDate="2017-04-04T16:25:30.780" Title="What is the temperature of an accretion disc surrounding a supermassive black hole?" Tags="&lt;supermassive-black-hole&gt;&lt;temperature&gt;&lt;accretion-discs&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8414" PostTypeId="2" ParentId="8413" CreationDate="2014-12-31T01:59:45.963" Score="5" Body="&lt;p&gt;It depends on the distance from the central body. &lt;a href=&quot;http://jila.colorado.edu/~pja/astr3830/lecture26.pdf&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; gives the temperature $T$ at a given point as a function of the distance from that point to the center ($R$):&#xA;$$T(R)=\left[\frac{3GM \dot{M}}{8 \pi \sigma R^3} \left(1-\sqrt{\frac{R_{\text{inner}}}{R}} \right) \right]^{\frac{1}{4}}$$&#xA;where &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_constant&quot; rel=&quot;nofollow&quot;&gt;$G$&lt;/a&gt;, $\pi$, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_constant&quot; rel=&quot;nofollow&quot;&gt;$\sigma$&lt;/a&gt; are the familiar constants, $M$ is the mass of the central body (and $\dot{M}$ is the rate of accretion onto the body), and $R_{\text{inner}}$ is the inner radius of the disk - possibly (if the object is a black hole) the Schwarzschild radius $R_s$, in which case we can simplify this a little more. So the temperature in the accretion disk is far from constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether or not there is plasma depends on the exact nature of the disk, the central object and the region around it. For example, a supermassive black hole may have different matter in its disk than that of a stellar-mass black hole. I should think, though, that black holes in binary systems accreting mass from a companion should have plasma in their accretion disks, and supermassive black holes might also have plasma from nearby stars.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2014-12-31T14:59:44.707" LastActivityDate="2014-12-31T14:59:44.707" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8415" PostTypeId="2" ParentId="8390" CreationDate="2014-12-31T03:48:06.763" Score="7" Body="&lt;p&gt;Given ra, dec, lat, lon in radians, and d in number of fractional days &#xA;since '1970-01-01 00:00:00 UTC', the azimuth of a star is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-\cot ^{-1}(\tan (\text{dec}) \cos (\text{lat}) \sec (\text{d1}+\text{lon}-\text{ra})+\sin (\text{lat}) \tan (\text{d1}+\text{lon}-\text{ra}))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the elevation is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tan ^{-1}\left(\frac{\sin (\text{dec}) \sin (\text{lat})-\cos (\text{dec}) \cos (\text{lat}) \sin (\text{d1}+\text{lon}-\text{ra})}{\sqrt{(\cos (\text{dec}) \sin (\text{lat}) \sin (\text{d1}+\text{lon}-\text{ra})+\sin (\text{dec}) \cos (\text{lat}))^2+\cos ^2(\text{dec}) \cos ^2(\text{d1}+\text{lon}-\text{ra})}}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where d1 is: $\frac{\pi  (401095163740318 d+11366224765515)}{200000000000000}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(you will need to resolve the ambiguity in the arc(co)tangent, but &#xA;this isn't difficult). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These formulas aren't as daunting as they seem, since, for you, lat, &#xA;lon, ra, and dec will be fixed, and the only thing that changes is d. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps, but I'm worried that it just demonstrates how complicated these formulas are.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2014-12-31T03:57:39.067" LastActivityDate="2014-12-31T03:57:39.067" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8416" PostTypeId="2" ParentId="8396" CreationDate="2014-12-31T05:50:25.680" Score="2" Body="&lt;p&gt;Simply said if a star would orbit a planet, the planet would become the star and viceversa. Hm... never seen that. But who knows. The universe is so vast and the physics we know only covers an insignificant percent of all phenomena in the universe. You just imagined that, so in another universe this is true. It is your universe!&lt;/p&gt;&#xA;" OwnerUserId="4177" LastActivityDate="2014-12-31T05:50:25.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8417" PostTypeId="2" ParentId="8404" CreationDate="2014-12-31T06:13:47.823" Score="6" Body="&lt;p&gt;You will not be able to see Saturn's rings directly with a 15x70 telescope, but you will be able to notice that Saturn appears &lt;strong&gt;elliptical, not round&lt;/strong&gt;. This is due to the shape of the rings around the planet, but you will not be able to resolve the hole between the rings and the planet (much less the Cassini gap).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as an &lt;strong&gt;impractical suggestion&lt;/strong&gt; in the interest of giving to you a more complete answer, you &lt;em&gt;might&lt;/em&gt; be able to mount a camera and use &lt;a href=&quot;http://www.instructables.com/id/Astrophotography-Star-Photo-Stacking/&quot;&gt;image stacking&lt;/a&gt; to see a smidge of ring and differentiate it from the sphere of the planet. If you do go this route, &lt;strong&gt;please&lt;/strong&gt; post online about your setup and results, whether it succeeds or not.&lt;/p&gt;&#xA;" OwnerUserId="209" LastActivityDate="2014-12-31T06:13:47.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8418" PostTypeId="1" CreationDate="2014-12-31T08:40:29.780" Score="3" ViewCount="208" Body="&lt;p&gt;This question is about finding the elevation and bearing of a given star at a particular time at Latitude 52N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I look up the RA and Declination of my star. How do I calculate its bearing and elevation at hourly intervals?&lt;/p&gt;&#xA;" OwnerUserId="4180" LastEditorUserId="2365" LastEditDate="2015-11-16T19:37:17.110" LastActivityDate="2016-09-12T02:22:05.520" Title="How do I get from Star Atlas to real life?" Tags="&lt;positional-astronomy&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8419" PostTypeId="2" ParentId="8378" CreationDate="2014-12-31T13:19:18.827" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;And they are then ejected by the orbital changes of giant planets. But why are they ejected to form specifically a local Oort cloud instead of going interstellar and becoming vagabonds?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Per multiple models, most comets &lt;strong&gt;did&lt;/strong&gt; become vagabonds. Estimates vary, but the lowest I've seen is 65% being ejected, and some estimate that well over 90% were ejected. Moreover, comets in the Oort cloud are ejected as well over time. The long period comets and Halley-style comets that we see are but a small fraction of the perturbed comets in the Oort cloud. Most perturbations result in escape rather than diving deep into the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If other stars are like our Sun, interstellar space should be awash with escaped comets. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The fact that comets come from all directions doesn't discriminate between them coming from an Oort cloud or from interstellar sources.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are ignoring velocity and eccentricity, which do distinguish between comets that originated from within the gravitational bounds of the solar system versus comets from interstellar sources. This is the key reason Jan Oort hypothesized the Oort cloud. All comets observed to date are consistent with having an intra-solar system origin and are inconsistent with having an extra-solar system origin. An interstellar comet should have a large excess velocity. That is not what has been observed. The few comets that apparently are on a hyperbolic trajectory are deemed to have originated within the gravitational bounds of the solar system, with the hyperbolic orbit a result of non-gravitational forces and gravitational interactions with planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If interstellar space is awash with ejected comets, estimates on how often we should see an interstellar comet range from once per 25 years (which has not been observed) to once every 450 years (which is consistent with not seeing any). Moreover, there are reasons to think that interstellar space is not awash with comets. If interstellar space was awash with comets, we should see a lot more gamma ray bursts than are observed, and we shouldn't see any white dwarfs depleted of hydrogen (which are observed; 16% of the white dwarfs are DB white dwarfs). This suggests to some that the solar system is an oddity, and that most stars don't produce Oort clouds.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2014-12-31T13:19:18.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8421" PostTypeId="1" AcceptedAnswerId="10006" CreationDate="2014-12-31T20:11:24.733" Score="2" ViewCount="119" Body="&lt;p&gt;When I open the 5 degree data of Green Coronal Emission line from &lt;a href=&quot;http://www.ngdc.noaa.gov/stp/space-weather/solar-data/solar-indices/solar_corona/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, I get a weird table which is not simple like the one for Coronal Index. I am only providing the table for 1939.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ngdc.noaa.gov/stp/space-weather/solar-data/solar-indices/solar_corona/5-degree-data/data/1939/1939&quot; rel=&quot;nofollow&quot;&gt;http://www.ngdc.noaa.gov/stp/space-weather/solar-data/solar-indices/solar_corona/5-degree-data/data/1939/1939&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't provide the link for the formats used due to the restriction (max of 2 links until I have a rep above 10) but it's in the folder for 5 degree data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I searched extensively connecting the formats (like I4, I2) to the datasheet but couldn't even scratch the surface of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want the data for green coronal intensities and positional degrees in that respective year but can't find from this table. How do I find them?&lt;/p&gt;&#xA;" OwnerUserId="4185" LastEditorUserId="2153" LastEditDate="2014-12-31T20:47:15.260" LastActivityDate="2015-04-28T10:58:54.243" Title="How to read 5 degree data of green line intensity" Tags="&lt;the-sun&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8422" PostTypeId="2" ParentId="8408" CreationDate="2014-12-31T22:43:51.103" Score="3" Body="&lt;p&gt;You may be looking for the &quot;water hole&quot;. See &lt;a href=&quot;http://www.setileague.org/general/waterhol.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.setileague.org/general/waterhol.htm&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4077" LastActivityDate="2014-12-31T22:43:51.103" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8423" PostTypeId="1" AcceptedAnswerId="8424" CreationDate="2015-01-01T08:52:25.597" Score="6" ViewCount="150" Body="&lt;p&gt;Given that SETI was created before the discovery of exoplanets, has this changed how they search? Would it make more sense to focus efforts on discovered earth like exoplanets?&lt;/p&gt;&#xA;" OwnerUserId="4146" LastActivityDate="2015-01-01T14:38:06.127" Title="How should exoplanet discoveries affect SETI?" Tags="&lt;exoplanet&gt;&lt;earth-like-planet&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8424" PostTypeId="2" ParentId="8423" CreationDate="2015-01-01T14:38:06.127" Score="6" Body="&lt;p&gt;Well you are correct and of course that kind of research has been going on. The SETI programme targeted a set of Kepler stars with evidence for a planet (or planet candidate) in the &quot;habitable zone&quot;. The initial searches for narrow-band radio emission (at 1.1-1.9 Hz) appears to have concentrated on those systems containing multiple planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The (null) results for $\sim 100$ system were published by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013ApJ...767...94S&quot;&gt;Siemion et al. (2013)&lt;/a&gt;. The observations were not sensitive enough to detect the sort of &quot;radio chatter&quot; emitted from our planet, but might have detected some sort of deliberate signalling beacon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A slightly more imaginative approach, which I'm struggling to find the details of, is that one can observe Kepler multiple planet systems when one calculates that two planets are in conjunction as seen from the Earth. There is then the chance of seeing communications from one planet to the other. (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014ebi..conf..5.2K&quot;&gt;http://adsabs.harvard.edu/abs/2014ebi..conf..5.2K&lt;/a&gt; ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a more general sense, the exoplanetery discoveries have provided great impetus to the SETI programme. It is now clear that planetary systems are common and present around most stars. It is also now apparent that small planets are common. New possibilities are also being considered. Most exoplanets in habitable zones are orbiting stars smaller and less luminous than the Sun. No exact Earth analogues have yet been found. This has led to an explosion of activity investigating planetary atmospheres and habitability around K- and M-dwarf stars.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-01-01T14:38:06.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8425" PostTypeId="1" CreationDate="2015-01-01T16:42:05.887" Score="9" ViewCount="8014" Body="&lt;p&gt;Why is that Neutron Stars are never depicted in an H R diagram? They can be placed in the bottom right corner but you will never find any diagram in literature showing neutron stars.&lt;/p&gt;&#xA;" OwnerUserId="2823" LastActivityDate="2020-01-23T20:20:12.690" Title="Position of Neutron stars in H R diagram" Tags="&lt;neutron-star&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8426" PostTypeId="2" ParentId="8425" CreationDate="2015-01-01T17:36:07.340" Score="10" Body="&lt;p&gt;The HR diagram is an &lt;em&gt;observational diagram&lt;/em&gt;. Whilst neutron stars could be placed in the HR diagram in the same way as white dwarf stars are, it turns out to be impractical to do so because the &lt;em&gt;photospheric&lt;/em&gt; luminosity and &lt;em&gt;photospheric&lt;/em&gt; temperature of neutron stars is next to impossible to determine. The reason for this is two-fold: (i) Neutron stars start off very hot (interior temperatures of $\sim 10^{10}$K and photospheric temperatures of $\sim 10^{7}$K, but they cool very rapidly. Within 10,000 years after the originating supernova they will have cooled below a million degrees, then photon cooling takes over from neutrino losses and they cool to a few thousand degrees within 10 million years (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2004ARA%26A..42..169Y&quot; rel=&quot;noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2004ARA%26A..42..169Y&lt;/a&gt; ). There are many uncertainties and unknowns in these processes. (ii) the photospheric emission is usually dwarfed by emission from the magnetosphere or luminosity due to accretion from a companion or the interstellar medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can theoretically work out where neutron stars should be by assuming that the emission is like that from a blackbody and that the radius $R \sim 10$ km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case neutron stars lie on a locus defined by&#xA;$$ \frac{L}{L_{\odot}} = 1.9\times 10^{-9} \left(\frac{T}{10^{4}K}\right)^4 $$&#xA;So, &lt;em&gt;contrary&lt;/em&gt; to what you you say in your question, most neutron stars could be cool and very, very faint and spend the majority of their (cooling) lives at the bottom-left or even bottom-right of the HR diagram. There is actually a huge uncertainty over where neutron stars would appear on this locus. Their very low heat capacities means that any &quot;reheating processes&quot; could very effectively raise their temperatures. Such processes include Ohmic dissipation of the magnetic field, some kind of thermalisation of their rotational energy or accretion from the interstellar medium. For the latter, luminosities of $10^{20}-10^{21}$ W may be possible, implying effective temperatures of tens of thousands of Kelvin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A neutron star at the same temperature as Sirius would have an absolute visual magnitude that was about 22 magnitudes fainter, $M_V \sim 23$. Another way of visualising this is that the neutron star cooling sequence is roughly parallel with the white dwarf cooling sequence but about 13 magnitudes fainter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You never see this locus shown on an HR diagram because it is usually way off the bottom of the plot. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-01-01T20:36:12.723" LastActivityDate="2015-01-01T20:36:12.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8427" PostTypeId="1" AcceptedAnswerId="8429" CreationDate="2015-01-01T23:34:06.763" Score="2" ViewCount="916" Body="&lt;p&gt;Is there any known pair of objects tidally locked with each other?  I know the moon is tidally locked with the earth (e.g. one side always facing earth), but I don't know of any pair of objects that are both tidally locked with each other.  For example, the planet would be tidally locked with the moon, and the moon would be tidally locked with the planet.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2015-01-02T07:00:18.627" Title="Are there any known objects in a &quot;dual&quot; tidally locked orbit?" Tags="&lt;orbit&gt;&lt;tidal-forces&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8428" PostTypeId="2" ParentId="8427" CreationDate="2015-01-02T00:11:46.667" Score="3" Body="&lt;p&gt;Consider the case of similar-mass, close binary systems. If the orbital periods are less than a few days they are expected to become tidally locked on a timescale much shorter than the stellar lifetimes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many cases the synchronisation can be established by comparing the rotation periods of the stars using modulation by starspots with the orbital period from eclipses e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/1983A%26A...123..316E&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/1983A%26A...123..316E&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other cases you can find the projected equatorial velocities of eclipsing binary star components and show that these are consistent with the orbital period, the measured radii and the orbital inclination e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2009ApJ...691.1400M&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/2009ApJ...691.1400M&lt;/a&gt; ; &lt;a href=&quot;http://adsabs.harvard.edu/abs/2002ApJ...567.1140T&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/2002ApJ...567.1140T&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-01-02T00:11:46.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8429" PostTypeId="2" ParentId="8427" CreationDate="2015-01-02T01:37:23.413" Score="8" Body="&lt;p&gt;Pluto and its largest moon &lt;a href=&quot;http://en.wikipedia.org/wiki/Charon_(moon)#Orbital_characteristics&quot; rel=&quot;noreferrer&quot;&gt;Charon&lt;/a&gt; are tidally locked to each other.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Charon and Pluto revolve about each other every 6.387 days. The two objects are both gravitationally locked to the other, so each keeps the same face towards the other. This is a mutual case of tidal locking . . .  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because of Charon's large size compared to Pluto, and because its barycenter lies above the primary's surface, some astronomers have referred to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pluto#Charon&quot; rel=&quot;noreferrer&quot;&gt;Pluto/Charon system&lt;/a&gt; as a dwarf double planet.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The system is also unusual among planetary systems in that each is tidally locked to the other: Charon always presents the same face to Pluto, and Pluto always presents the same face to Charon: from any position on either body, the other is always at the same position in the sky, or always obscured.&lt;a href=&quot;http://www.boulder.swri.edu/~layoung/projects/talks03/IfA-jan03v1.ppt&quot; rel=&quot;noreferrer&quot;&gt;(110)&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2674" LastEditorUserId="2674" LastEditDate="2015-01-02T07:00:18.627" LastActivityDate="2015-01-02T07:00:18.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8430" PostTypeId="2" ParentId="8375" CreationDate="2015-01-02T02:08:30.583" Score="0" Body="&lt;p&gt;I'm not about to suggest that multiverse(s) do not exist, but merely that it is a fascinating, but highly speculative, topic. The Wikipedia article is a good start. From the &lt;a href=&quot;http://en.wikipedia.org/wiki/Multiverse&quot; rel=&quot;nofollow&quot;&gt;wiki&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The structure of the multiverse, the nature of each universe within it and the relationships among the various constituent universes, depend on the specific multiverse hypothesis considered. Multiple universes have been hypothesized in cosmology, physics, astronomy, religion, philosophy, transpersonal psychology, and fiction, particularly in science fiction and fantasy. In these contexts, parallel universes are also called &quot;alternate universes&quot;, &quot;quantum universes&quot;, &quot;interpenetrating dimensions&quot;, &quot;parallel dimensions&quot;, &quot;parallel worlds&quot;, &quot;alternate realities&quot;, &quot;alternate timelines&quot;, and &quot;dimensional planes,&quot; among others.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the way you phrase your question, you may want particularly to look at the sections on &lt;a href=&quot;https://en.wikipedia.org/wiki/Multiverse#M-theory&quot; rel=&quot;nofollow&quot;&gt;M-theory&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2674" LastEditorUserId="2674" LastEditDate="2016-05-30T18:17:37.270" LastActivityDate="2016-05-30T18:17:37.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8431" PostTypeId="1" CreationDate="2015-01-02T03:23:31.970" Score="5" ViewCount="241" Body="&lt;p&gt;Philae landed on comet 67P this year. And it means a big breakthrough for the science and technology of human being.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However Hayabusa mission of Japan not only landed on an asteroid but also returned an asteroid sample to Earth in 2010. &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technologically ,which is more difficult? &lt;br/&gt;&#xA;And why is Philae more important? &lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2018-03-07T07:47:32.967" Title="landing on a comet is more important than landing on an asteroid?" Tags="&lt;comets&gt;&lt;asteroids&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8432" PostTypeId="2" ParentId="8431" CreationDate="2015-01-02T06:14:00.970" Score="7" Body="&lt;p&gt;This is a very brief answer but I hope it helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Distance&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hayabusa&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hayabusa&lt;/a&gt; collected samples from a near-earth asteroid while &lt;a href=&quot;http://en.wikipedia.org/wiki/Philae_%28spacecraft%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Philae&lt;/a&gt; landed on a comet after traveling a distance of over 6.4 billion kilometers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Type Of Mission&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hayabusa's did not literally land on the asteroid. It just touched down and collected samples. While, Philae had to land, steady itself, drill and study the composition. Plus, the asteroid on which Hayabusa landed was a near-earth asteroid and so it's nature could have been known easily. On the other hand, the Rosetta team came to know about the exact nature of &lt;a href=&quot;http://en.wikipedia.org/wiki/67P/Churyumov%E2%80%93Gerasimenko&quot; rel=&quot;nofollow noreferrer&quot;&gt;67P&lt;/a&gt; just weeks before the landing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Complexity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Rosetta mission used about 6 gravity assists and performed 2 flybys. It was not an easy task to plan a journey of 10 years. Plus, when the probe reaches the comet, radio signals would take a lot of time to reach it and thus it would be very difficult to control the probe if something went wrong. Plus, the surface of an asteroid is vastly different from the surface of a comet. The Hayabusa Mission was not the first to touch down on an asteroid. It had been done once by &lt;a href=&quot;http://en.wikipedia.org/wiki/NEAR_Shoemaker&quot; rel=&quot;nofollow noreferrer&quot;&gt;NEAR Shoemaker&lt;/a&gt; and so we knew what to expect. See &lt;a href=&quot;http://science.nasa.gov/missions/near/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; for a detailed report.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this sums up how the Rosetta mission was a major breakthrough.&lt;/p&gt;&#xA;" OwnerUserId="1156" LastEditorUserId="1156" LastEditDate="2018-03-07T07:47:32.967" LastActivityDate="2018-03-07T07:47:32.967" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8435" PostTypeId="1" AcceptedAnswerId="8441" CreationDate="2015-01-02T20:11:09.540" Score="18" ViewCount="9816" Body="&lt;p&gt;My understanding is that the Earth's axis points in the same direction in space during its entire orbit around the sun. And this is what causes our seasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why doesn't the axis follow the orbital path (kind of like a car driving around a circle)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine the Moon plays a part in this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a secondary question, do the other planets in our solar system orbit the same as Earth (its axis points in the same direction in space for its entire orbit)?&lt;/p&gt;&#xA;" OwnerUserId="4197" LastActivityDate="2016-03-29T00:25:56.247" Title="Why doesn't Earth's axis change during the year?" Tags="&lt;orbit&gt;&lt;the-sun&gt;&lt;earth&gt;" AnswerCount="7" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8436" PostTypeId="2" ParentId="8435" CreationDate="2015-01-02T20:59:59.100" Score="8" Body="&lt;p&gt;Picture the Earth as a small ball suspended in midair, not moving, although it's rotating on its axis. Unless forces are applied to it, absolutely nothing will happen. That's conservation of energy (or momentum; you can work with it either way). Earth will not spontaneously start moving in one direction because that would violate conservation of translational energy and conservation of linear momentum. If that did happen, energy would be gained out of nowhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the same reason, Earth will not start tilting to one side. This is conservation of rotational kinetic energy and conservation of angular momentum at work. Savvy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, yep, Keith is spot-on. You'd need some sort of torque to turn the Earth's axis like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why &lt;em&gt;does&lt;/em&gt; the Earth's axis &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_precession&quot;&gt;move&lt;/a&gt; over 26,000 years (it does, by the way, just not one cycle per year)? Just like with the tides, as well as how Earth's rotation is slowing down, it's primarily the tidal forces of the Moon and the Sun at work, although the other planets do make some contributions to the overall changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can add in some math if you want, but be warned, I'm learning it myself, so it won't be explained as well as you might like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other planets obey the same laws as the Earth, so their axes should point in the same direction throughout the year, ignoring the effects of axial precession.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-01-02T21:34:04.390" LastActivityDate="2015-01-02T21:34:04.390" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8437" PostTypeId="2" ParentId="8435" CreationDate="2015-01-02T21:35:00.820" Score="6" Body="&lt;p&gt;The changing angle between sun and moon does cause some slight &lt;a href=&quot;http://en.wikipedia.org/wiki/Nutation#Earth&quot; rel=&quot;noreferrer&quot;&gt;nutation in the earth's axis of rotation over an 18.6 year period&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the case of the Earth, the principal sources of tidal force are the Sun and Moon, which continuously change location relative to each other and thus cause nutation in Earth's axis. The largest component of Earth's nutation has a period of 18.6 years, the same as that of the precession of the Moon's orbital nodes.&lt;a href=&quot;http://en.wikipedia.org/wiki/Nutation#Earth&quot; rel=&quot;noreferrer&quot;&gt;1&lt;/a&gt; However, there are other significant periodical terms that must be calculated depending on the desired accuracy of the result. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The principal term of nutation is due to the regression of the moon's nodal line and has the same period of 6798 days (18.61 years). It reaches plus or minus 17″ in longitude and 9″ in obliquity. All other terms are much smaller; the next-largest, with a period of 183 days (0.5 year), has amplitudes 1.3″ and 0.6″ respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not a large effect (from link):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UuDlj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-01-02T21:35:00.820" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8438" PostTypeId="2" ParentId="8435" CreationDate="2015-01-03T03:48:48.313" Score="4" Body="&lt;p&gt;The Earth acts like a massive gyroscope. A tremendous force would be required to change the orientation of the axis. While forces are certainly exerted, they are not large enough to cause appreciable change in the orientation of the axis as the Earth falls/orbits about the Sun.  As you indicate, the orientation of the axis is responsible for the change in season.  The areas that receive the most direct exposure to sunlight are in summer; areas receiving glancing sunlight are in winter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earths rotation about the Sun is not circular but elliptical.  During its orbit the Earths distance from the Sun changes.  Many sources report that while this is true, it has little effect on the season.  These sources indicate that the orientation of the Earths axis combined with the &quot;tilt&quot; of the Earths orbit above or below the Suns equator determines the seasons.   &lt;/p&gt;&#xA;" OwnerUserId="4199" LastActivityDate="2015-01-03T03:48:48.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8439" PostTypeId="1" CreationDate="2015-01-03T04:11:36.083" Score="1" ViewCount="84" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;A satellite which is moving around a planet of mass &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; in a circular orbit of radius &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; is given a sudden thrust into the center of the planet, so that it deviates an angle &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Find the latus-rectum &lt;span class=&quot;math-container&quot;&gt;$l$&lt;/span&gt; and the eccentricity &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt; of the new orbit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4200" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-01-04T01:09:34.260" Title="Question: calculating orbit using another orbit" Tags="&lt;orbit&gt;&lt;gravity&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8440" PostTypeId="1" AcceptedAnswerId="8444" CreationDate="2015-01-03T05:33:54.560" Score="7" ViewCount="1725" Body="&lt;p&gt;I'm working on a star system generator for a game; I'd like its results to be plausible but they needn't be super-realistic. I've got the orbital distances and masses of each body in the system, and now want to render them, so I need to determine the radius of each body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For gas giant and ice giant planets, what's the relationship between mass and density? What are the minimum and maximum densities expected? &lt;/p&gt;&#xA;" OwnerUserId="2573" LastActivityDate="2015-01-03T17:08:42.050" Title="Maximum and minimum gas giant &amp; ice giant densities" Tags="&lt;planet&gt;&lt;gas-giants&gt;&lt;planetary-formation&gt;&lt;density&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8441" PostTypeId="2" ParentId="8435" CreationDate="2015-01-03T09:42:30.570" Score="15" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My understanding is that the Earth's axis points in the same direction in space during its entire orbit around the sun. And this is what causes our seasons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second statement is correct. The axial tilt is the primary driver of the seasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first statement is not exactly correct. There is a small but persistent change in the orientation of the Earth's axis. The change over the course of a year is small. The change over the course of 13,000 years is large. In 13,000 years the Earth's axis will be tilted by about 47 degrees compared to the current orientation. In yet another 13,000 years the Earth will be oriented very similarly to how it is oriented now. I'll write more about this later.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My question is why doesn't the axis follow the orbital path (kind of like a car driving around a circle)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is not a good mental model. The only contact between your car and the ground are the bottoms of the four wheels on your car. The forces at these four small contact areas are what change your car's orientation and change your car's velocity vector. This means that the torque on your car as you turn the steering wheel and the centripetal force on your car that makes your care negotiate a turn are intimately connected. Turn the steering wheel gently and the heavy front end of your car follows suit. Assuming a gentle turn, the light rear end of you car soon follows suit. If you turn too hard, your car starts to skid instead of turn. Turn even harder yet and you risk having your car flip and roll. The motion of your car depends on that intimate coupling between force and torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitation doesn't work like that. It instead acts on each and every bit of the Earth, all together, all at once, all the time. There is a slight variation in the gravitationally-induced acceleration across the Earth due to variations in distance to the Moon (or Sun), but these variations are small. The net gravitational force on the Earth is very much decoupled from the small net gravity gradient torque on the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principal result of the gravity gradient torque exerted by the Moon and the Sun on the Earth is a very slow but large precession of the Earth's rotation axis (for more, read this &lt;a href=&quot;http://en.wikipedia.org/wiki/Axial_precession&quot;&gt;wikipedia article on axial precession&lt;/a&gt;). This precession makes the Earth's rotation axis rotate about the normal to the ecliptic at a rate of one revolution per 25772 years. There are lesser effects collectively called nutations and polar motion that result from gravity gradient torque. These lesser effects are much, much smaller in magnitude but have much, much shorter periods. The largest of these lesser effects results from the 18.61 year cyclical variations in the lunar node. Even this largest component of nutation is small, two orders of magnitude smaller than the large but slow precession.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-01-03T09:42:30.570" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8442" PostTypeId="1" AcceptedAnswerId="8558" CreationDate="2015-01-03T10:17:33.230" Score="4" ViewCount="940" Body="&lt;p&gt;I've been reading some of Stephen Hawking's work on the theory of the expansion (and ultimate contraction) of the universe and was curious if anyone has every hypothesized that the universe has possibly gone through this cycle before (or many many times before).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know it would be impossible to determine because the idea of the universe contracting back to the size it was at the point of the Big Bang (or at the end of a Big Crunch) suggests that all of the laws of physics would break down but has anyone hypothesized and written work on this. I'd be curious to read more about the idea if so.&lt;/p&gt;&#xA;" OwnerUserId="4201" LastActivityDate="2015-01-14T21:54:34.290" Title="Big Bang / Big Crunch cycle?" Tags="&lt;universe&gt;&lt;big-bang-theory&gt;&lt;expansion&gt;&lt;early-universe&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8443" PostTypeId="2" ParentId="8440" CreationDate="2015-01-03T12:30:25.817" Score="3" Body="&lt;p&gt;With regard to gas giants, there is no mass limit. Conceptually, what's the difference between a star plus a Jupiter-class gas giant versus a star plus a brown dwarf companion versus a star plus a red dwarf versus a binary pair of stars of more or less the same mass? This is a spectrum with no clear boundary, no clear limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regard to density, there's a funny thing about Jupiter: It's about as big, radius-wise, as gas giants and brown dwarfs can get. Add mass to a Jupiter-sized gas giant and you get a more massive gas giant that is close to the same size as Jupiter. Add even more mass and you get a brown dwarf that is close to the same size as Jupiter. Add even more mass and you get a small red dwarf that is close to the same size as Jupiter. The mass-radius relationship is very, very flat from Jupiter-mass gas giant to smallest red dwarf.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-01-03T12:30:25.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8444" PostTypeId="2" ParentId="8440" CreationDate="2015-01-03T13:50:34.170" Score="7" Body="&lt;p&gt;Here is a plot I generated in 5 minutes at the site &lt;a href=&quot;http://exoplanets.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;exoplanets.org&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To construct this I took planets discovered by the &lt;em&gt;transit&lt;/em&gt; method &lt;em&gt;and&lt;/em&gt; which had a $M \sin i$ measured using radial velocities. I divided the $M \sin i$ by the sine of the measured inclination angle (this is required to avoid using masses that have been estimated using an assumed mass-radius relation). The y-axis is density, which comes directly (and quite precisely) from transit measurements. Of course transiting planets are the only ones with radii and density measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to do this to avoid some very uncertain values that are given for planetary masses that have been simply assumed from a theoretical mass-radius relationship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, there is a wide spread (factor of three) in density at a given mass for hot Jupiter's (most of the &lt;em&gt;transiting&lt;/em&gt; giant planets are hot Jupiters), but there is a strong correlation. The density is at a minimum for a few tenths of Jupiter mass, but then smaller planets (presumably rocky and icy, rather than gas giants) appear to show higher densities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, these are all transiting exoplanets, and therefore predominantly orbiting close to their parent stars. There could be biases and selection effects at work! For example although the cores of gas giants are governed by degeneracy pressure and this makes the &lt;em&gt;theoretical&lt;/em&gt; mass-radius relationship quite flat, there is the perturbing effect of radiation from the parent star (&quot;insolation&quot;) that can make some objects larger. Even beyond this there appears to be scatter that is difficult to understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oWHbA.png&quot; alt=&quot;Density vs planet mass for transiting planets&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: For an easy-to-use empirical formulation you could try the relationships proposed by &lt;a href=&quot;http://phl.upr.edu/library/notes/standardmass-radiusrelationforexoplanets&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Planetary Habitability Laboratory&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-01-03T17:08:42.050" LastActivityDate="2015-01-03T17:08:42.050" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8445" PostTypeId="2" ParentId="8435" CreationDate="2015-01-03T16:52:05.500" Score="3" Body="&lt;p&gt;Because the Sun's gravitational pull on Earth is (nearly) uniform, it doesn't tilt the Earth, it only pulls it as a whole, without affecting the Earth's spinning around its own axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth's radius is ~ 6,400 km, the distance to the Sun is ~ 150,000,000 km, and the gravitational force diminishes as a square of a distance, so the ratio in the pull on some two small volumes of Earth, the closest and the farthest from the Sun at a given moment, is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{(150000+6.4)^2}{(150000-6.4)^2} = 1.00017068$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the tilting force is &lt;em&gt;very&lt;/em&gt; small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Were the variation exactly 0 (i.e. the ratio exactly 1) the tilting force would be exactly 0 and the Earth's spinning wouldn't have been affected, at all. To get that, remove the Sun from the picture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And by the general theory of relativity, if Earth were moving in a uniform force, it would be as if there was no force at all, from Earth's point of view.&lt;/p&gt;&#xA;" OwnerUserId="4204" LastEditorUserId="2153" LastEditDate="2015-01-04T01:33:28.567" LastActivityDate="2015-01-04T01:33:28.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8446" PostTypeId="1" AcceptedAnswerId="8454" CreationDate="2015-01-03T16:54:03.137" Score="2" ViewCount="229" Body="&lt;p&gt;The question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you stand on the equator of the Moon, at midday your position occurs when the Moon is located at the line of the node. What is the altitude of the sun at that moment?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Information: This is when the Moon is located at the line of node (line of node is a line that occurs from crossing between the ecliptic plane and the Moon's orbital plane, which makes an angle of approximate 5 degrees). The equatorial plane of the Sun makes an angle of 1 degrees, 32 arc minutes and 40 arc seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is 88 degrees, 27 arc minutes and 20 arc seconds through 90 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't understand why the answer isn't 1 degree, 32 arc minutes and 20 arc seconds, because the ecliptic plane where the sun is located is above the horizon or celestial equator. equal to its altitude.&lt;/p&gt;&#xA;" OwnerUserId="2787" LastEditorUserId="2153" LastEditDate="2015-01-05T00:23:52.253" LastActivityDate="2015-01-05T00:23:52.253" Title="How do you find the altitude of the Sun if you are on the Moon?" Tags="&lt;the-moon&gt;&lt;the-sun&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8447" PostTypeId="2" ParentId="8435" CreationDate="2015-01-03T23:32:30.650" Score="4" Body="&lt;p&gt;Lets Keep things simple.&#xA;Below is a picture of the orbit of the Earth.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/j7Dxi.jpg&quot; alt=&quot;Orbit of the Earth&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I take the question to mean why the axis doesn't point toward the Sun all the time.&#xA;The answer comparing the Earth to a Gyroscope is the easiest to use. The spinning effect on the axis does cause a force that helps keep the Earth's axis at a 23.5° angle (Approx) there is a slight wobble of a few meters in the earths axis. &#xA;Try looking from above might make it clearer.&#xA;There is a red circle of rotation at the pole if you look closely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MwFwD.jpg&quot; alt=&quot;Rotation from above&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth is not driving around the Sun it is in effect being pulled to the Sun but is going fast enough around it to counteract gravity and not get closer, the same way satellites stay orbiting in space. The car would be more like a spaceship as it would have a force/thrust pushing it around the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the second question the other planets do the same but at different angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jcOyq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4209" LastActivityDate="2015-01-03T23:32:30.650" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8448" PostTypeId="2" ParentId="8439" CreationDate="2015-01-03T23:49:39.097" Score="1" Body="&lt;p&gt;As a hint: a force directed inward does not change the angular momentum, and therefore does not change the semi-major axis nor the period. It's not clear which angle you're talking about, but knowing that the semi-major axis did not change should let you set the problem up as a simple geometry problem.&lt;/p&gt;&#xA;" OwnerUserId="4077" LastActivityDate="2015-01-03T23:49:39.097" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8450" PostTypeId="1" CreationDate="2015-01-04T07:21:56.397" Score="3" ViewCount="3068" Body="&lt;p&gt;How do scientists and astronomers recognize the perfect path or orbit (if it's orbiting an object) of an asteroid or a comet?&lt;/p&gt;&#xA;" OwnerUserId="662" LastEditorUserId="16038" LastEditDate="2017-03-06T01:24:13.137" LastActivityDate="2017-03-06T01:24:13.137" Title="How is the path of an asteroid calculated?" Tags="&lt;orbit&gt;&lt;comets&gt;&lt;asteroids&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8451" PostTypeId="1" CreationDate="2015-01-04T07:24:41.880" Score="3" ViewCount="1025" Body="&lt;p&gt;Can we calculate the area of sky visible to us from the point we are standing?I mean is there any idea or experiment to calculate it?&lt;/p&gt;&#xA;" OwnerUserId="662" LastActivityDate="2015-06-08T13:04:34.790" Title="Calculating area of visible sky" Tags="&lt;observational-astronomy&gt;&lt;deep-sky-observing&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8452" PostTypeId="2" ParentId="8451" CreationDate="2015-01-04T14:02:42.207" Score="0" Body="&lt;p&gt;For the area of sky as defined by &quot;areas visible above your head inside Earth's atmosphere&quot; then the answer depends on the height of the objects that you would like to observe. I like to observe airplanes, so without loss of generality I'll use ~10,000 meters as my &quot;sky&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see airplanes up to about 40 KM away.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;π * (40)^2 ~= 5000 KM^2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The &lt;code&gt;altitude:area&lt;/code&gt; relationship is a squared relationship, therefore if you decide that you are interested in only 1000 meters as your &quot;sky&quot; (maybe you like to photograph birds) then you have about 3 KM^2 of sky available to you.&lt;/p&gt;&#xA;" OwnerUserId="209" LastEditorUserId="209" LastEditDate="2015-06-08T13:04:34.790" LastActivityDate="2015-06-08T13:04:34.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8453" PostTypeId="2" ParentId="8451" CreationDate="2015-01-04T15:10:17.953" Score="1" Body="&lt;p&gt;Approximating the Earth as a sphere with radius $R$, then when viewing from a height $h$ above the surface, the Earth blocks out a cone of some opening angle $2\vartheta$, where $\csc\vartheta = 1+\frac{h}{R}$. Thus, the visible portion has a solid angle of&#xA;$$\Omega = 2\pi\left(1+\cos\vartheta\right) = 2\pi\left(1+\frac{\sqrt{h^2+2Rh}}{R+h}\right)$$&#xA;steradians. Divide this by $4\pi$ to obtain the &lt;em&gt;fractional area&lt;/em&gt; of the the visible sky, compared to what you could have if the Earth wasn't blocking your view, since a full sphere subtends a solid angle of $4\pi$ steradians. That is probably a more natural measure of the visible sky than a literal area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an actual area, you need some sort of reference distance $r$ to measure from, with the visible sky a distance $r$ away having area $A = \Omega r^2$.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2015-01-04T15:10:17.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8454" PostTypeId="2" ParentId="8446" CreationDate="2015-01-04T17:49:03.740" Score="2" Body="&lt;p&gt;@DavidHammen is essentially correct. According to stellarium, the Sun gets as high as 89 58' 48&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jbgQs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2015-01-04T17:49:03.740" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8455" PostTypeId="2" ParentId="8450" CreationDate="2015-01-04T19:02:57.300" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How do scientists and astronomers recognize the perfect path or orbit (if it's orbiting an object) of an asteroid or a comet?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They don't, and they can't. The best they can do is estimate the object's trajectory based on observations and based on techniques used to propagate the object (and other objects in the solar system) over time. These estimates are always imperfect:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The measurements upon which the estimates are based are limited in number. This is particular true for comets and asteroids. The historical record on observations of the Sun and the planets go back thousands of years. The number of observations for a newly observed comet or asteroid might go back a few months.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The measurements upon which the estimates are based inherently have some error associated with them. These measurement errors limit the ability to determine the object's true trajectory.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The physical models used to propagate trajectories are simplified and incomplete versions of reality.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;General relativity, our best model of how gravitation works, is exceeding complex and highly non-linear. Even the very best precision orbit determination programs use a linear &lt;a href=&quot;http://en.wikipedia.org/wiki/Parameterized_post-Newtonian_formalism&quot; rel=&quot;noreferrer&quot;&gt;parameterized post-Newtonian expansion&lt;/a&gt; of general relativity. The non-linear effects? They're ignored.&lt;/li&gt;&#xA;&lt;li&gt;Nearby stars and the galaxy as a whole perturb the trajectories of the Sun, the planets, etc. These effects are ignored.&lt;/li&gt;&#xA;&lt;li&gt;Comets vent gas when they approach the Sun. When you blow up a balloon and let it go, the balloon veers this way and that. There is no predicting where the balloon will go. Comets do the same thing. After time, there is no predicting where a comet will be or where it is going.&lt;/li&gt;&#xA;&lt;li&gt;Asteroids (and comets) are subject to &lt;a href=&quot;http://en.wikipedia.org/wiki/Radiation_pressure&quot; rel=&quot;noreferrer&quot;&gt;radiation pressure&lt;/a&gt;, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Yarkovsky_effect&quot; rel=&quot;noreferrer&quot;&gt;Yarkovsky effect&lt;/a&gt;, and the &lt;a href=&quot;http://en.wikipedia.org/wiki/YORP_effect&quot; rel=&quot;noreferrer&quot;&gt;YORP effect&lt;/a&gt;. These are very hard to estimate. One of the key purposes of the upcoming &lt;a href=&quot;http://www.asteroidmission.org&quot; rel=&quot;noreferrer&quot;&gt;Osiris-REx mission&lt;/a&gt; is to get a better handle on the Yarkovsky effect.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The numerical techniques used to propagate objects' trajectories are inevitably flawed to some extent. They introduce errors that grow with time into the propagated trajectories. There's always a tradeoff between precision, accuracy, fidelity to the physics, and computational expense. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The solar system is ultimately chaotic. This is particularly the case for asteroids and comets on highly elliptical trajectories. Consider a comet whose orbit takes it from beyond Neptune's orbit to inside Venus's orbit. Over the centuries, that comet has ample opportunities to make a close pass with a planet. There are but slight differences between the planet making a minor change in the comet's orbit, the comet colliding into the planet, the planet sending the comet toward the Sun or another planet, or the planet ejecting the comet from the solar system. This is chaos at its worst.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2015-01-04T19:17:36.790" LastActivityDate="2015-01-04T19:17:36.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8456" PostTypeId="1" AcceptedAnswerId="8463" CreationDate="2015-01-04T21:19:50.217" Score="3" ViewCount="252" Body="&lt;p&gt;I was recently watching a video on the physics and chemistry in planetary accretion disks, and the speaker at a certain point mentioned 'freeze-out and UV dissociation'. Is freeze-out a purely chemical term, or is it specific for this process? Is it very important in planetary formation? &lt;/p&gt;&#xA;" OwnerUserId="4126" LastEditorUserId="4126" LastEditDate="2015-01-04T21:51:19.520" LastActivityDate="2015-01-05T15:59:24.203" Title="Astrochemistry - what does freeze-out mean?" Tags="&lt;accretion-discs&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8458" PostTypeId="1" CreationDate="2015-01-04T22:13:11.860" Score="7" ViewCount="217" Body="&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Questions are at the bottom. The rest had simply been written in a format to better organize my thoughts. The formulas used to construct this fictional solar system, had been borrowed from &lt;a href=&quot;https://www.youtube.com/channel/UCeh-pJYRZTBJDXMNZeWSUVA&quot;&gt;Artefxian&lt;/a&gt; and are presented in several of his videos. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theoretical:&lt;/strong&gt;&lt;br&gt;&#xA;A P-Type binary star system featuring a gas giant on or near the habitable zone, with a habitable moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Considerations:&lt;/strong&gt;   &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gas giant would need a stable magnetosphere. Jupiter and Saturn may be useful examples. &lt;/li&gt;&#xA;&lt;li&gt;The moon's mass must be great enough to sustain an atmosphere. In this instance, a nitrogen/oxygen atmosphere. It is estimated that a moon with a mars-like density, would need at least 7% of Earth's mass in order to support such an atmosphere for several billion years.&lt;/li&gt;&#xA;&lt;li&gt;Both gas giant and the habitable moon must maintain stable orbit. Simulations would suggest that, to maintain a stable orbit to a gas giant, or a brown dwarf that orbits 1 AU from a sun-like star, would require a moon orbital period of less than 45 - 60 days.   &lt;/li&gt;&#xA;&lt;li&gt;The moon itself must be capable of generating it's own magnetosphere in order to deflect stellar wind and the gas giants' naturally generated radiation belts.  &lt;/li&gt;&#xA;&lt;li&gt;There is a high likelihood that the moon would be tidally locked with it's parent world. Monoj Joshi, Robert Haberle, and their colleagues suggest that the effect of tidal heating could support conditions amenable to habitability. Additionally, tidal effects may allow for plate tectonics, causing volcanic activity and a regulation of the moon's surface temperature. The potential, resulting geodynamo effect would allow for a strong magnetic field.  &lt;/li&gt;&#xA;&lt;li&gt;Balance: The moon should be large enough to support tectonic activity, dense enough to support a strong protective magnetosphere, close enough to the gas giant to maintain stable orbit, and be far enough away that it's own magnetosphere may better protect from sputtering caused by it's parent worlds' radiation belts.  &lt;/li&gt;&#xA;&lt;li&gt;It is suggested that the larger and more dense a terrestrial, water-rich world, the further out it's habitable zone extends.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fictional Data:&lt;/strong&gt;&lt;br&gt;&#xA;Star A: Mass = 1.1; Luminosity = 1.39; Diameter = 1.07; Surface Temperature = 1.04&lt;br&gt;&#xA;Star B: Mass = 0.9; Luminosity = 0.69; Diameter = 0.92; Surface Temperature = 0.94&lt;br&gt;&#xA;Separation: 0.3 AU&lt;br&gt;&#xA;Barycenter: 0.45/0.225 AU&lt;br&gt;&#xA;Eccentricity: 0.4/0.5&lt;br&gt;&#xA;Separation Minimum: 0.27/0.1125&lt;br&gt;&#xA;Separation Maximum: 0.63/0.3375&lt;br&gt;&#xA;Separation Overall: 0.9/0.45&lt;br&gt;&#xA;Inner Boundary: 0.2 AU&lt;br&gt;&#xA;Outer Boundary: 80 AU&lt;br&gt;&#xA;Frostline: 6.984 AU&lt;br&gt;&#xA;Inner Habitable Zone: 1.976 AU&lt;br&gt;&#xA;Outer Habitable Zone: 2.808 AU&lt;br&gt;&#xA;Ideal Position: 1.8 AU&lt;br&gt;&#xA;Inner Forbidden &quot;Hell&quot; Zone: 0.3 AU&lt;br&gt;&#xA;Outer Forbidden &quot;Hell&quot; Zone: 1.35 AU  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Goal:&lt;/strong&gt; Create a fictional, theoretical system which can be demonstrated as stable, in which an earth analog may safely orbit a gas giant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Further Considerations:&lt;/strong&gt;   &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The moon does not necessarily need to be an earth analog, and may simply be demonstrated as habitable to human life.&lt;/li&gt;&#xA;&lt;li&gt;The gas giant does not necessarily need to be within the habitable zone, and may cradle the outer limits of the CMZ, or be further out provided it can be demonstrated that the orbiting moon could feasibly support human life unassisted by technology. i.e. Robin Crusoe could become stranded on the moon, and survive. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How far would the moon have to orbit from it's parent body, to prevent tidal locking?  &lt;/li&gt;&#xA;&lt;li&gt;How can I find a balance, between the binary stars' gravitational pull, the gas giant's gravitational pull, and the chaos it may play on the moon?  &lt;/li&gt;&#xA;&lt;li&gt;If no balance can be met, I will accept tidal locking with the moon. In this instance, how can I go about determining a safe and stable orbiting distance for the moon?  &lt;/li&gt;&#xA;&lt;li&gt;How can I determine the effect of the gas giants' magnetosphere on the moon? What would the math behind it look like?  &lt;/li&gt;&#xA;&lt;li&gt;Are there any considerations I may be missing, and how would they affect this scenario?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4207" LastEditorUserId="2153" LastEditDate="2015-01-04T22:19:20.307" LastActivityDate="2015-01-04T22:19:20.307" Title="Questions about a fictional binary system, and habitability" Tags="&lt;the-moon&gt;&lt;binary-star&gt;&lt;gas-giants&gt;&lt;habitable-zone&gt;" AnswerCount="0" CommentCount="6" FavoriteCount="1" ClosedDate="2016-04-25T16:58:07.133" ContentLicense="CC BY-SA 3.0" />
  <row Id="8459" PostTypeId="2" ParentId="8129" CreationDate="2015-01-05T02:59:18.560" Score="0" Body="&lt;p&gt;We have found instances of asteroids that are mostly rubble piles, but spinning fast enough that they would fly apart if bound only by gravity. The most likely binding force is static charge, although some sort of glue is not out of bounds. I don't know if either of these would be called &quot;hydrostatic&quot;, but if the question is really &quot;Is gravity the only force that can bind a non-solid object together?&quot; the answer is no.&lt;/p&gt;&#xA;" OwnerUserId="4077" LastActivityDate="2015-01-05T02:59:18.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8460" PostTypeId="1" AcceptedAnswerId="8466" CreationDate="2015-01-05T09:04:41.973" Score="32" ViewCount="11809" Body="&lt;p&gt;Does every star become a black hole? Is there any probability that our sun can become a black hole? If yes then is it on its way to become a black hole? what is the current state of sun as per the black hole life cycle? What will be the effect on all the planetary objects in the solar system if the sun turns into a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for so many questions but i cannot miss them as these are some questions on my mind.&lt;/p&gt;&#xA;" OwnerUserId="4046" LastActivityDate="2020-09-14T05:58:35.900" Title="Can our Sun become a black hole" Tags="&lt;the-sun&gt;&lt;solar-system&gt;&lt;black-hole&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8462" PostTypeId="1" AcceptedAnswerId="8465" CreationDate="2015-01-05T15:28:45.140" Score="6" ViewCount="2347" Body="&lt;p&gt;I've read the &lt;a href=&quot;https://www.iau.org/public/themes/naming/#stars&quot; rel=&quot;noreferrer&quot;&gt;a IAU page&lt;/a&gt; on the subject but i am no closer to understanding how this works. Also did search the already asked question, but didn't see anything on the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does bolded item in the names below stand for? How do they choose them?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;VY&lt;/strong&gt; Canis Majoris&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;NML&lt;/strong&gt; Cygni&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;UY&lt;/strong&gt; Scuti&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2020-12-28T21:14:47.637" Title="What does designation VY, NML, UU in star names stand for?" Tags="&lt;star&gt;&lt;naming&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8463" PostTypeId="2" ParentId="8456" CreationDate="2015-01-05T15:59:24.203" Score="2" Body="&lt;p&gt;Freeze-out is indeed a chemical term. It happens in the midplane of accretion disks where density is high enough to block UV ionization/chemical dissociation and limit heating from the central star. No one seems to have gotten around to writing up a layman friendly article on the subject yet, but these refs give the gist:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://books.google.com/books?id=EF1LOOgc8IcC&amp;amp;pg=PA753&amp;amp;lpg=PA753&amp;amp;dq=freeze-out%20and%20UV%20dissociation%20accretion&amp;amp;source=bl&amp;amp;ots=YYFql7SwLC&amp;amp;sig=xBE6cEgjLWiccc1qtIDKqSxx98U&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=LLCqVMPrC5SoyASBs4KoAw&amp;amp;ved=0CDcQ6AEwBA#v=onepage&amp;amp;q=freeze-out%20and%20UV%20dissociation%20accretion&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;Protostars and planets V&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.stsci.edu/institute/conference/jwst2007/presentations-posters/dominik_talk.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Accretion Disks 1: During Star Formation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://books.google.com/books?id=EWURgHzF3DQC&amp;amp;pg=PA11&amp;amp;lpg=PA11&amp;amp;dq=freeze-out%20and%20UV%20dissociation&amp;amp;source=bl&amp;amp;ots=B5EMwewH6m&amp;amp;sig=IqQjpyuoQYFTYQ_Ih8cFmugShUo&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=Ya-qVIHLOYKZyAST7YKoAw&amp;amp;ved=0CDQQ6AEwBA#v=onepage&amp;amp;q=freeze-out%20and%20UV%20dissociation&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;Organic Matter in Space (IAU S251) Evolution of Organic Matter&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From ref 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/41zQX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-01-05T15:59:24.203" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8464" PostTypeId="1" AcceptedAnswerId="8467" CreationDate="2015-01-05T16:11:06.727" Score="-1" ViewCount="149" Body="&lt;p&gt;I've recently read about dark matter and dark energy. But unfortunately, there were not enough answers for me. The article I read was more of a hypothesis than a real answer. So I'm asking for all the information that you have and all the things you know about dark matter and dark energy. For example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is it?&lt;/li&gt;&#xA;&lt;li&gt;Why does it exist?&lt;/li&gt;&#xA;&lt;li&gt;Where is it?&lt;/li&gt;&#xA;&lt;li&gt;From what does it form?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4214" LastEditorUserId="2153" LastEditDate="2015-01-06T01:12:14.730" LastActivityDate="2015-01-06T01:12:14.730" Title="What are dark matter and dark energy?" Tags="&lt;dark-matter&gt;&lt;dark-energy&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-01-06T16:53:11.177" ContentLicense="CC BY-SA 3.0" />
  <row Id="8465" PostTypeId="2" ParentId="8462" CreationDate="2015-01-05T16:21:22.087" Score="5" Body="&lt;p&gt;VY Canis Majoris and UY Scuti are variable star designations. The first discovered variable star in a constellation is called R, the second, S, and then unto Z. After Z comes RR and so on. For a full description see &lt;a href=&quot;http://en.wikipedia.org/wiki/Variable_star_designation&quot; rel=&quot;noreferrer&quot;&gt;this wikipedia page&lt;/a&gt;. NML Cygni is also a variable star, but its variable star designation is V1489 Cygni. &lt;a href=&quot;http://en.wikipedia.org/wiki/NML_Cygni#Observational_history&quot; rel=&quot;noreferrer&quot;&gt;The letters NML come from its discoverers: Neugebauer, Martz, and Leighton.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2015-01-05T16:21:22.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8466" PostTypeId="2" ParentId="8460" CreationDate="2015-01-05T16:29:42.233" Score="41" Body="&lt;h2&gt;No, the sun won't ever become a black hole.&lt;/h2&gt;&#xA;&lt;p&gt;The choice between the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_evolution#Stellar_remnants&quot; rel=&quot;noreferrer&quot;&gt;three fates of stars&lt;/a&gt; (white dwarf, neutron star, black hole) is entirely determined by the star's mass.&lt;/p&gt;&#xA;&lt;p&gt;A star on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Main_sequence&quot; rel=&quot;noreferrer&quot;&gt;main sequence&lt;/a&gt; (like most stars, including our sun) is constantly in a balance between the inward pressure of gravity and the outward pressure of the energy generated by the hydrogen fusion that makes it &amp;quot;burn&amp;quot;.&lt;sup&gt;&lt;strong&gt;1&lt;/strong&gt;&lt;/sup&gt;  This balance stays relatively stable until the star runs out of whatever its current fuel is - at that point, it stops burning, which means there's no longer outward pressure, which means it starts collapsing.  Depending on how much mass there is, it might get hot enough as it collapses to start fusing helium together. (If it's &lt;em&gt;really&lt;/em&gt; massive, it might continue on to burn carbon, neon, oxygen, silicon, and finally iron, which &lt;em&gt;can't&lt;/em&gt; be usefully fused.)&lt;/p&gt;&#xA;&lt;p&gt;Regardless of what its final fuel is, eventually the star will reach a point where the collapse from gravity is insufficient to start burning the next fuel in line.  This is when the star &amp;quot;dies&amp;quot;.&lt;/p&gt;&#xA;&lt;h3&gt;White dwarfs&lt;/h3&gt;&#xA;&lt;p&gt;If the star's remains&lt;sup&gt;&lt;strong&gt;2&lt;/strong&gt;&lt;/sup&gt; mass less than 1.44 solar masses (the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chandrasekhar_limit&quot; rel=&quot;noreferrer&quot;&gt;Chandrasekhar limit&lt;/a&gt;&lt;sup&gt;&lt;strong&gt;3&lt;/strong&gt;&lt;/sup&gt;), eventually gravity will collapse the star to the point where each atom is pushed right up against the next.  They can't collapse further, because the electrons can't overlap.  While white dwarfs &lt;em&gt;do&lt;/em&gt; shed light, they do so because they are extremely hot and slowly cooling off, not because they're generating new energy.  Theoretically, a white dwarf will eventually dim until it becomes a black dwarf, although the universe isn't old enough for this to have happened yet.&lt;/p&gt;&#xA;&lt;h3&gt;Neutron stars&lt;/h3&gt;&#xA;&lt;p&gt;If the collapsing star is above the Chandraskhar limit, gravity is so strong that it can overcome the &amp;quot;electrons can't overlap&amp;quot; restriction.  At that point, all the electrons in the star will be pushed into combining with protons to form neutrons.  Eventually, the entire star will composed primarily of neutrons pushed right up next to each other.  The neutrons can't be pushed into occupying the same space, so the star eventually settles into being a single ball of pure neutrons.&lt;/p&gt;&#xA;&lt;h3&gt;Black holes&lt;/h3&gt;&#xA;&lt;p&gt;Black holes are the step beyond neutron stars, although they're worth discussing in a bit more detail.  Everything, in theory, has a &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;noreferrer&quot;&gt;Schwarzschild radius&lt;/a&gt;.  That's the radius where a ball of that mass would be so dense that light can't escape.  For example, the Schwarzschild radius for Earth is about 9mm.  However, for all masses smaller than somewhere between 2-3 times the mass of the sun, it's impossible to squeeze the matter small enough to get it inside that radius.  Even a neutron star isn't massive enough.&lt;/p&gt;&#xA;&lt;p&gt;But a star that becomes a black hole is.  We don't actually know what happens to a star once it's become a black hole - the edges of the &amp;quot;hole&amp;quot; itself is simply the Schwarzschild radius - the point light can't escape.  From outside, it doesn't matter whether the matter collapsed to the point that the neutrons started overlapping, whether it stopped just inside the radius, or whether it continued collapsing until it broke all known physical laws.  The edges are still the same, because they're just a cutoff based on the escape velocity.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;sup&gt;&lt;strong&gt;1&lt;/strong&gt;&lt;/sup&gt;  I'm ignoring the red giant phase here, since it's just a delay in the &amp;quot;run out of fuel&amp;quot; step.  Basically, the core is helium &amp;quot;ash&amp;quot;, while the hydrogen fusion process takes place further and further out.  Once that runs out, you get a nova and the collapse continues.&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;&lt;strong&gt;2&lt;/strong&gt;&lt;/sup&gt; Likewise, I'm ignoring the mass that stars shed in their various nova phases.  All given masses are based on the remnants left behind.&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;&lt;strong&gt;3&lt;/strong&gt;&lt;/sup&gt; &lt;a href=&quot;http://www.britannica.com/EBchecked/topic/105468/Chandrasekhar-limit&quot; rel=&quot;noreferrer&quot;&gt;Every&lt;/a&gt; &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/astro/whdwar.html#c5&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt; &lt;a href=&quot;http://www.fisica.edu.uy/%7Esbruzzone/FlexPaper_1.4.2_flash/prueba.pdf&quot; rel=&quot;noreferrer&quot;&gt;I've&lt;/a&gt; &lt;a href=&quot;http://farside.ph.utexas.edu/teaching/sm1/lectures/node88.html&quot; rel=&quot;noreferrer&quot;&gt;found&lt;/a&gt; &lt;a href=&quot;http://scienceworld.wolfram.com/physics/ChandrasekharLimit.html&quot; rel=&quot;noreferrer&quot;&gt;for&lt;/a&gt; Chandrasekhar mass, except Wikipedia, gives 1.44 or 1.4 solar masses (which are compatible).  Wikipedia gives 1.39, and gives at least &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0702351v1&quot; rel=&quot;noreferrer&quot;&gt;one source&lt;/a&gt; to back that number.&lt;/p&gt;&#xA;" OwnerUserId="2945" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-03-31T04:47:02.907" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="8467" PostTypeId="2" ParentId="8464" CreationDate="2015-01-05T17:19:43.610" Score="2" Body="&lt;p&gt;If you read an article that presents a hypothesis on dark matter and dark energy, that is about as much as you will find right now. No one on this planet &lt;em&gt;knows&lt;/em&gt; what they are. We only know from the behavior we see in the cosmos that there must be something we cannot detect out there influencing the movement of the heavens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you read and understand everything that has currently been written by an authority on the subject right now (January 2015), you will still not know the answers to the questions you have posed. If you truly must know, I suggest you begin by applying to enter into a PhD program in cosmology or astronomy.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2015-01-05T17:19:43.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8468" PostTypeId="1" CreationDate="2015-01-05T20:03:29.213" Score="7" ViewCount="223" Body="&lt;p&gt;Since stars are an ideal example of macroscopic objects, is their evolution  determined by their initial mass and metallicity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any non-trivial random &lt;strong&gt;internal&lt;/strong&gt; processes going on that may significantly affect a star's development and eventual fate? (It is quite evident that capture of another star or a brown dwarf is an external, exogenous event.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT/clarification: I'm looking for things that may shorten/extend the lifetime of a star beyond the &quot;norm&quot; prescribed by nuclear physics/gravity, change its composition, surface temperature. Solar flares/quakes and other instabilities come to mind first.&lt;/p&gt;&#xA;" OwnerUserId="793" LastEditorUserId="793" LastEditDate="2015-01-05T22:36:44.877" LastActivityDate="2015-01-06T18:43:45.520" Title="Is there an element of chance/chaos in stellar evolution?" Tags="&lt;stellar-evolution&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8469" PostTypeId="2" ParentId="8468" CreationDate="2015-01-05T20:35:46.153" Score="1" Body="&lt;p&gt;If you want to go into truly random stuff, you should look into quantum. The internal processes inside a star are not random, however they are bviously undetectable. Some factors other than mass and metallicity are temperature of the original dust cloud, or angular velocity. The temperature obviously affects the brightness, while the angular velocity will affect the gravity. I am not an expert, but I'd assume that any minor perturbation inside the star could cause a butterfly effect and result in, say, solar flares, so on average there would be stars who would be losing more mass than others with no  particular reason.&#xA;I assume there would be other factors linked with more complicated physics, too.&lt;/p&gt;&#xA;" OwnerUserId="4126" LastActivityDate="2015-01-05T20:35:46.153" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8470" PostTypeId="1" CreationDate="2015-01-05T22:15:32.713" Score="1" ViewCount="104" Body="&lt;p&gt;My telescope (Newtonian reflector) has a focal length of 900 mm. So, using a 25 mm eye-piece, I get 900 mm / 25 mm = 36X magnification. Using a 10 mm eye-piece, the magnification would be 90X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is that I cannot see a thing with the 10 mm eye-piece. If I would insert the 25 mm eye-piece, find a bright star, and adjust the eye-piece distance from the secondary mirror, I would get the things ready after some trial and error. With the higher magnification eye-piece, it's just all black. What should I do to get around this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aperture is 130 mm if it matters.&lt;/p&gt;&#xA;" OwnerUserId="4229" LastActivityDate="2015-01-06T00:12:04.023" Title="Working with high-magnification eye-pieces" Tags="&lt;amateur-observing&gt;&lt;newtonian-telescope&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8471" PostTypeId="2" ParentId="8468" CreationDate="2015-01-05T22:28:58.680" Score="1" Body="&lt;p&gt;This is not an answer that stems from vast cosmological knowledge and experience, but from an intuitive mathematical understanding of stochasticity. I would point out that stars do not have the visual appearance of stability or a predictable nature. They are churning, belching balls of nuclear plasma. On a macroscopic scale, I would expect stability to look like smoother surfaces or uniform activity. Any object that you might describe as &quot;constantly exploding&quot; or even &quot;on fire&quot;, or particularly, &quot;undergoing nuclear reaction&quot;, I would ordinarily say involves an element of chance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flare and CME activity of our Sun is governed by a roughly 11 year cycle. But it's not exactly old faithful. The activity of the Sun follows patterns, but they are noisy patterns. However, even though stars are the very image of chaos to our view, it seems that they are actually quite predictable. One reason is that they live in a vacuum. Literally. Or very near a vacuum. Particularly since your question eliminates consideration of external interactions, we are talking about stars like ours, surrended by mostly a lot of nothing. So when plasma erupts out of the star at less than escape velocity, we know that ultimately that plasma is coming right back home to the star. It's not going to smear onto some other object. And the goop from other objects isn't going to smear onto the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the time-scale of a star's evolution is enormous. Imagine watching a timelapse movie of our Sun where each frame of the movie shows the average (mean) appearance of the sun over 100 eleven-year solar cycles. Now you would see something that looks quite smooth, stable. And you could watch that movie of a stable glowing ball all day long without seeing any dramatic change. (Assuming 24 frames per second, watching the movie all day long would take you through 2.3 billion years of the Sun's life.) Because the timescale in question is so long, the bubbling, churning, and exploding we see on our timescale amounts to unfathomably insignificant blips in a stable, predictable, burn process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, any stochastic (random) event reproduced enough times, in the same way (with the same kind of randomness), will lead to stable, predictable outcomes. A man has solar panels on his roof. Some days are sunny, and some are overcast. Some days the panels get covered up completely in snow. But over the course of a year, we can still predict approximately how much power his panels will produce with about 98% accuracy, assuming there is no external influence, like the house burning down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, excluding external interactions, if you know the total mass and composition of a star at conception, it makes sense that the chaos and noise of the atomic fireball would have a net zero effect on the outcome.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditDate="2015-01-06T18:43:45.520" LastActivityDate="2015-01-06T18:43:45.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8472" PostTypeId="1" AcceptedAnswerId="8475" CreationDate="2015-01-06T00:02:40.820" Score="2" ViewCount="394" Body="&lt;p&gt;This comes from &lt;a href=&quot;https://astronomy.stackexchange.com/questions/8468/is-there-an-element-of-chance-chaos-in-stellar-evolution#comment10132_8468&quot;&gt;a comment&lt;/a&gt; on &lt;a href=&quot;https://astronomy.stackexchange.com/q/8468/1909&quot;&gt;this question&lt;/a&gt;, to quote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The death of (large mass) stars is also based on quantum events with probabilities technically not 1 (and is very fast), so it is technically possible for a star not to die in a way it really ought to by sheer chance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The question is what quantum events are involved in the death of a star (likely and unlikely)? If many of these very unlikely quantum events happened at the same time all over the star, how could the death of a star differ form the typical nova or simply cooling down?&lt;/p&gt;&#xA;" OwnerUserId="1909" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-01-06T03:51:07.710" Title="Quantum death of stars" Tags="&lt;star&gt;&lt;supernova&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8473" PostTypeId="2" ParentId="8470" CreationDate="2015-01-06T00:06:57.800" Score="1" Body="&lt;p&gt;Try the moon first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you see nothing but black, assuming that you don't have a lens cap on or something, then most likely, you are zoomed in on, well, relative blackness. The star you were viewing is probably off to the side now. Or, you may just be looking through the eye-piece at the wrong angle or something. The moon is too big a target to miss, and you have nice rough features to use to tune in your focus. And if you see black when you are looking at the moon, it's much easier to troubleshoot. You should definitely see &lt;em&gt;something&lt;/em&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, secondarily the reason you expect to see something, but instead see nothing may be a viewfinder problem. If you have a small, low magnification viewfinder scope attached to your telescope, you usually can't just trust that what you see in the center of that view is exactly what your telescope will be aimed at. You may have to align it. &lt;a href=&quot;https://www.google.com/webhp?sourceid=chrome-instant&amp;amp;ion=1&amp;amp;espv=2&amp;amp;ie=UTF-8#q=align%20telescope%20viewfinder&quot; rel=&quot;nofollow&quot;&gt;Search &quot;align telescope viewfinder&quot;&lt;/a&gt; for help with this. From your description, it sounds like you are not using a viewfinder, but this seems to be a common problem, so I thought I should mention it.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditDate="2015-01-06T00:12:04.023" LastActivityDate="2015-01-06T00:12:04.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8474" PostTypeId="2" ParentId="8472" CreationDate="2015-01-06T03:33:16.177" Score="3" Body="&lt;p&gt;Suppose you collect 11.0114 grams of carbon-11, come back 27.11 hours (80 half lives) later, and see what your sample has become. The most likely outcome: You'll find that you have 11.0093 grams of boron-11 and absolutely no carbon 11. You might find an atom or two of carbon-11 amongst those 11.0093 grams of boron-11.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about other results, for example, what is the probability that not a single one of those Avogadro's number of carbon-11 atoms will decay into boron-11 over a 27.11 hour interval?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is &quot;technically possible&quot; that none of those Avogadro's number of carbon-11 atoms will decay in 27.11 hours. The probability that a single atom of carbon-11 won't have decayed in that time interval is $2^{-80}$. That's a tiny number. The probability that not a single one of those Avogadro's number of carbon-11 atom won't have decayed is $\left(2^{-80}\right)^{6\times 10^{23}}$, or less than one out of one in $10^{10^{25}}$. That's not just a tiny number. The difference between &quot;technically possible&quot; and &quot;can't happen&quot; is essentially zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this applicable? This result is, for example, applicable to a white dwarf stealing mass from a binary pair and turning into a neutron star rather than becoming a type 1A supernova as the white dwarf's mass approaches the Chandrasekhar limit. All it takes is a tiny number (maybe just one) carbon-carbon fusion reactions. The odds this won't happen as the mass gets ever closer to the Chandrasekhar limit is not just small. It is smaller than smaller than small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it is &quot;technically possible.&quot;&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-01-06T03:33:16.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8475" PostTypeId="2" ParentId="8472" CreationDate="2015-01-06T03:51:07.710" Score="4" Body="&lt;p&gt;Stellar fusion and supernovae are governed by quantum particle interactions (on massive scales).  In general there are many possible ways for particles to interact, decay, etc. at various probabilities.  To understand the physics of a system correctly you must take account of all possibilities (up to an error tolerance in actual practice).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One often sees phrases of the form &quot;situation X favors reaction Y&quot;.  This means the conditions (X) make the the probability of Y greater than any other possible reaction, but it doesn't equate to Y having probability 1.  Some may prefer to use this phrasing for when Y's probability is, in some sense, much larger than any other event.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction&quot; rel=&quot;nofollow&quot;&gt;proton-proton chain&lt;/a&gt; that governs the main sequence stage of a star is a prime example of why you cannot discount the reactions which aren't favored.  If you did, you would conclude that our own sun is incapable of fusion.  The 'favored' reaction for two interacting protons in a star like our sun is to simply bounce off of each other thanks to &lt;a href=&quot;http://en.wikipedia.org/wiki/Coulomb%27s_law&quot; rel=&quot;nofollow&quot;&gt;Coulomb repulsion&lt;/a&gt;.  The protons simply don't have enough energy to overcome this repulsion in order to undergo fusion.  Technically this statement is also not probability 1, but rather very close to it.  But an improbable &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_tunnelling&quot; rel=&quot;nofollow&quot;&gt;quantum tunneling&lt;/a&gt; event is possible to bypass this potential barrier, allowing the interacting protons to fuse when they collide just right and the dice favor them.  But a diproton is unstable.  The heavily favored reaction in this state is for the diproton to immediately &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_decay&quot; rel=&quot;nofollow&quot;&gt;decay&lt;/a&gt; into two protons.  But there is a small chance that one of the protons will undergo a &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_decay&quot; rel=&quot;nofollow&quot;&gt;beta decay&lt;/a&gt; into a neutron, instead, at which point you have a stable helium nucleus.  That these things are so unlikely is a major factor in why our sun needs billions of years to exhaust enough of its hydrogen (aka protons) in order to initiate helium fusion and exit the main stage.  Even very large stars still need millions of years because of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while the proton-proton reaction is rather unlikely, stars are simply so immense that it is not only a virtual certainty that some proton-proton reaction will yield helium, but in fact that it is a virtual certainty that the reactions will occur in large numbers and continuously until the available reactants dwindle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the other extreme, we have &lt;a href=&quot;http://en.wikipedia.org/wiki/Supernova&quot; rel=&quot;nofollow&quot;&gt;supernovae&lt;/a&gt;.  There are quite a lot of varieties of supernovae, but the commonly understood types--the outer layers collapse and then violently rebound off the core--feature a sudden loss of radiation pressure due to certain endothermic reactions becoming favorable.  Iron production from fusion is the culprit here: it takes more energy to fuse it than is released.  The final split second of this sort of star's life is when conditions favor the production of iron via fusion.  Ridiculously favors it, as I really do mean it is the final split second.  There is an inconceivably tiny chance that this iron production would fail to occur for some chosen period of time, and if it happens to stall long enough then the star would likely undergo some other event.  See also David's answer for another wildly improbable, but technically possible, example of random chance altering a star's death.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2015-01-06T03:51:07.710" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8477" PostTypeId="1" CreationDate="2015-01-06T08:46:09.287" Score="1" ViewCount="1458" Body="&lt;p&gt;Assume wormholes exist. One hole allows matter to pop out of another hole somewhere else in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one end of the wormhole is swallowed by a black hole, then no information can get from that end to the other hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the wormhole acts as some kind of gateway to another part of the universe, wouldn't that mean that the &lt;em&gt;entire&lt;/em&gt; universe would be swallowed by the black hole? ie. the wormhole end is an entry portal to the entire universe, so if that is inside the black hole, wouldn't that mean the entire universe is then in the black hole?&lt;/p&gt;&#xA;" OwnerUserId="4238" LastEditorUserId="2153" LastEditDate="2016-01-29T03:11:28.513" LastActivityDate="2016-08-08T21:33:55.673" Title="What would happen if a wormhole gets swallowed by a black hole?" Tags="&lt;black-hole&gt;&lt;wormhole&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="8478" PostTypeId="1" AcceptedAnswerId="8479" CreationDate="2015-01-06T10:08:41.153" Score="8" ViewCount="1678" Body="&lt;p&gt;I have a basic understanding of the sinusoidal relationship of the rate of change of daylight as the year goes on. But it seems intuitive that the changes should be equally split between sunrise and sunset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For my location, Bend Or., todays sunrise/sunset is: &lt;a href=&quot;http://www.accuweather.com/en/us/bend-or/97701/daily-weather-forecast/335268?day=1&quot;&gt;07:40 and 16:43&lt;/a&gt; And for January 31st it is: &lt;a href=&quot;http://www.accuweather.com/en/us/bend-or/97701/daily-weather-forecast/335268?day=26&quot;&gt;07:24 and 17:14&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So sunrise will be 16 minutes earlier and sunset will be 31 minutes later. It appears that there is twice the added daylight to sunset than to sunrise. Why aren't the differences equivalent?&lt;/p&gt;&#xA;" OwnerUserId="4240" LastActivityDate="2015-01-08T10:51:32.383" Title="As days get longer, why isn't the added daylight split evenly between sunrise and sunset?" Tags="&lt;earth&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8479" PostTypeId="2" ParentId="8478" CreationDate="2015-01-06T12:02:20.627" Score="5" Body="&lt;p&gt;It's a consequence of the &lt;a href=&quot;http://aa.usno.navy.mil/faq/docs/eqtime.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;equation of time&lt;/a&gt;, which represents the time difference between apparent solar time (time as measured by a sundial) and mean solar time (a fictitious device; essentially mean solar time is time as measured by a clock). Below are two plots from the referenced website.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://web.archive.org/web/20130217020355/http://aa.usno.navy.mil/graphics/EqT_graph_2014.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure 1: Equation of time. Currently (early January), a sundial will be slightly behind a clock and is getting slower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://web.archive.org/web/20130217025859/http://aa.usno.navy.mil/graphics/EqTdot_graph_2014.jpg&quot; alt=&quot;&quot;&gt;&lt;br&gt;&#xA;Figure 2: Time derivative of the equation of time. Apparent solar time changes most rapidly from early December to early January, with the maximum change occurring on December 25.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;Days are currently getting longer in the northern hemisphere, but noon is advancing by 20 to 25 seconds per day. Sunset advances by half of the increase in the length of day plus that 20 to 25 second advance in noon. Sunrise advances by half the increase in length of day less that 20 to 25 second advance in noon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The increase in the length of day (time from sunrise to sunset) depends on latitude. In Bogota Columbia (4°35′53″N latitude), length of day increases by only 3 to 10 seconds per day during January, growing throughout the month. That tiny change in length of day means sunrise continues to advance throughout the month of January in Bogota. On the other hand, in Reykjavík, Iceland (64°08′N latitude), length of day increases drastically from day to day, even at this time of year. Sunrise started becoming earlier in Reykjavík on Christmas day.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:47.290" LastActivityDate="2015-01-08T10:51:32.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8480" PostTypeId="1" AcceptedAnswerId="8481" CreationDate="2015-01-06T14:46:31.310" Score="3" ViewCount="186" Body="&lt;p&gt;It would be nice to get a relation between the blueshifted regions and the distance to earth. I'm searching especially for objects in other &lt;a href=&quot;https://en.wikipedia.org/wiki/Milky_Way#mediaviewer/File:Milky_Way_Arms.svg&quot; rel=&quot;nofollow&quot;&gt;galactic arms&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4242" LastActivityDate="2015-01-06T16:23:15.437" Title="Wher can I find a list of blueshifted regions in the Milky Way?" Tags="&lt;milky-way&gt;&lt;redshift&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8481" PostTypeId="2" ParentId="8480" CreationDate="2015-01-06T16:23:15.437" Score="2" Body="&lt;p&gt;I don't think there is a catalogue containing information about blue/red shift per region of the sky. You probably need a catalogue containing blue/red shifts per object and then calculate the average blue/red shift per region.&#xA;Most catalogues with blue/red shift are however targeted towards galaxies. They may also contain stars but you will have to filter the stars from the background galaxies. If I remember correctly, the &lt;a href=&quot;http://www.sdss.org&quot; rel=&quot;nofollow&quot;&gt;Sloan Digital Sky Survey (SDSS)&lt;/a&gt; contains a flag specifying whether an object is a galaxy or a star. So you might use the SDSS data to get a large set of stars with blue/red shift. This flag has been set automatically using pattern recognition. I do not know how good the classification was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can access the data at the &lt;a href=&quot;http://skyserver.sdss.org/dr12/en/home.aspx&quot; rel=&quot;nofollow&quot;&gt;SDSS SkyServer&lt;/a&gt;. There are a lot of objects in SDSS so you will need time to acquire the data and knowledge about SQL to query the database.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even then, you still don't have information about the spiral arm in which the object is located although you might use a method such as was used by &lt;a href=&quot;http://stacks.iop.org/1538-4357/642/i=2/a=L137&quot; rel=&quot;nofollow&quot;&gt;Belokurov et al. (2006)&lt;/a&gt; to detect the field of streams to separate the different spiral arms.&lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2015-01-06T16:23:15.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8482" PostTypeId="1" AcceptedAnswerId="8493" CreationDate="2015-01-06T17:52:04.820" Score="4" ViewCount="400" Body="&lt;p&gt;Recently, I got around to seeing the movie &lt;em&gt;Interstellar&lt;/em&gt;. In it, the characters of the movie visit a stellar system that appears to be built around a black hole instead of a star. On top of this, their mission is to find a habitable planet in this system. Question: &lt;strong&gt;I was wondering how stable the system, as a whole, would be in terms of operation and would any of the planets be habitable at all due to the obvious differences between a star and a black hole.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;In the movie the black hole is supposed to be a 'supermassive rotating black hole'. As for being a binary star system, it's never stated as such nor depicted as such. It's depicted visually as the black hole alone being the center. However one of the characters does mention a &amp;quot;Neutron Star&amp;quot; as part of the system so it could possibly be a binary star system. The ambient lighting for the planets is generated by the accretion disk of the black hole. The size and rotation speed of the event horizon are not defined in the movie. At least not that I can remember.&lt;/p&gt;&#xA;&lt;p&gt;As for the planets, their proximity varies, with the first planet depicted as being so close to the event horizon that it is effected by time dilation. The distances of the other two are not specified directly, but traveling to the second planet out seemingly takes days, while traveling to the third planet out is stated to take months, if I remember correctly. As for their surface gravity, the first planet is depicted as having higher surface gravity than Earth but not so much higher that movement is impossible, just strained. The second planet is depicted as being '80% Earth's gravity', if memory serves me correctly.&lt;/p&gt;&#xA;&lt;p&gt;To better help define certain variables relevant to the question I found an info-graphic related to the movie that illustrates the size of the black hole and it's rotational speed: &lt;a href=&quot;http://tinyurl.com/pqph8wl&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://tinyurl.com/pqph8wl&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4246" LastEditorUserId="11045" LastEditDate="2020-10-13T12:44:38.897" LastActivityDate="2020-10-14T16:02:54.007" Title="Ramifications of black hole stellar system" Tags="&lt;orbit&gt;&lt;black-hole&gt;&lt;exoplanet&gt;&lt;habitable-zone&gt;" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="8483" PostTypeId="1" AcceptedAnswerId="8484" CreationDate="2015-01-06T18:48:51.117" Score="24" ViewCount="2672" Body="&lt;p&gt;This, of course, is a newbie question, as I am nothing more than a hobbyist. But I was quite surprised to see the recently-released &quot;before and after&quot; pictures of the so-called &quot;Pillars of Creation&quot; in the Eagle Nebula (&lt;a href=&quot;http://phenomena.nationalgeographic.com/2015/01/05/hubble-revisits-an-icon-the-pillars-of-creation/&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.space.com/28175-pillars-of-creation-hubble-photos-2015.html&quot;&gt;here&lt;/a&gt;). For something that looks like such a &quot;nebulous&quot; cloud of gas (pardon the pun), I would have expected at least some visible sign of change in the 25 years between the two photos. But at the pixel resolution which I had available, I could not detect even the slightest difference between them. Of course, I am accustomed to terrestrial clouds which are in constant motion, so I (mistakenly?) expected something analogous at the astronomical level. Can someone provide, in layman's terms, how my understanding of this cosmological feature is deficient? In other words, how might I adjust my instinctive intuition that this object should be more dynamic? Is it just the sheer scale that I am not comprehending? (By the way: what is the distanse between the three spires, as compared to, say, our solar system?)&lt;/p&gt;&#xA;" OwnerUserId="4247" LastActivityDate="2015-01-07T18:22:32.343" Title="Why is the Eagle Nebula so &quot;static&quot;?" Tags="&lt;distances&gt;&lt;nebula&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8484" PostTypeId="2" ParentId="8483" CreationDate="2015-01-06T19:04:12.230" Score="30" Body="&lt;p&gt;It appears to be static because it's huge beyond your imagination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance to the nebula is 7,000 light years. Its apparent size is 7 arc minutes. Therefore its linear size is about 14 light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about that. The whole nebula is so big, it takes light 14 years to cross it. Any motion therein must necessarily be much, much slower. No wonder you're not seeing much change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Eagle_Nebula&quot;&gt;Data source: wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=7000%20light-years%20*%20tangent%287%20arcmin%29&quot;&gt;Calculation using Wolfram Alpha&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2015-01-06T19:04:12.230" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8485" PostTypeId="1" CreationDate="2015-01-06T19:07:04.600" Score="11" ViewCount="216" Body="&lt;p&gt;My question is about the implications of the observations recently made by the Cosmic Infrared Background Experiment, or CIBER, from Caltech. I've read at &lt;a href=&quot;http://www.caltech.edu/news/caltech-rocket-experiment-finds-surprising-cosmic-light-44399&quot;&gt;Caltech web site&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The total light produced by these stray stars is about equal to the background light we get from counting up individual galaxies.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does that suggest that about a half of the stars in the observable universe could not belong to any galaxy?&lt;/p&gt;&#xA;" OwnerUserId="4248" LastActivityDate="2016-04-01T18:57:16.047" Title="Does CIBER Experiment from Caltech suggest that there can be lots of stars which are not in any galaxy?" Tags="&lt;galaxy&gt;&lt;observational-astronomy&gt;&lt;observable-universe&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8487" PostTypeId="1" CreationDate="2015-01-06T19:58:24.347" Score="3" ViewCount="1902" Body="&lt;p&gt;So let's say that an asteroid the twice the size of Earth passed by going relatively slow say in between the Moon and Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among so many other things (and perhaps you can touch on some of the biggest 'other' things), would anything on the asteroid's side of Earth be essentially 'sucked' towards the asteroid. That is if I am standing outside and it passes would I lift off the ground and float towards it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, would things on the opposite side experience two times the standard weight? That is if I am standing on the other side, would I feel heavier as the asteroid passes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realise there would be many other devastating effects but I'm most interested in this one...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possible answer I suspect is that objects on Earth will not 'feel' anything as the Earth as a whole will move towards the asteroid and not every little thing inside Earth...&lt;/p&gt;&#xA;" OwnerUserId="4250" LastEditorUserId="2153" LastEditDate="2015-01-06T23:07:09.747" LastActivityDate="2019-02-22T14:18:30.353" Title="If an asteroid twice the size of Earth passed super close would half of the Earth be pulled towards it?" Tags="&lt;gravity&gt;&lt;asteroids&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8488" PostTypeId="2" ParentId="2330" CreationDate="2015-01-06T21:17:29.183" Score="2" Body="&lt;p&gt;Contracting objects can and do exhibit a measurable red shift, and this has actually been measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cepheid_variable&quot; rel=&quot;nofollow&quot;&gt;Cepheid variable&lt;/a&gt; are stars that expand on contract in a cycle lasting days to months.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/full/1998ASPC..135..201B&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; by R. Paul Butler discusses the measurement of Doppler shifts in Cepheid variables. According to the paper, radial velocities as small as 3 meters/second have been measured. Cepheids have been measured to have Doppler velocities of 30 to 1000 meters/second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper was published in 1997. It says that the precision of such measurements had improved by more than two orders of magnitude (a factor of 100) in the preceding decade. I presume the technology has continued to improve in the nearly two decades since then.&lt;/p&gt;&#xA;" OwnerUserId="53" LastActivityDate="2015-01-06T21:17:29.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8489" PostTypeId="2" ParentId="8483" CreationDate="2015-01-06T21:52:18.853" Score="19" Body="&lt;p&gt;To add to Florin Andrei's answer, with an image height of 7,000 pixels for 14 light years, that's 17.5 light hours per pixel. That's 20 billion kilometres &lt;strong&gt;per pixel&lt;/strong&gt;. To make a change in a single pixel over that time, something of that size must have either changed composition dramatically (to give a different colour or opacity) or it must have moved by a comparable distance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the timeframe, that's a billion kilometres per year, or 123,000 kilometres per hour. (77,000 miles per hour, if you prefer) Few things that large are moving that fast relative to their neighbours. &lt;/p&gt;&#xA;" OwnerUserId="4251" LastActivityDate="2015-01-06T21:52:18.853" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8490" PostTypeId="2" ParentId="8487" CreationDate="2015-01-06T22:17:41.663" Score="1" Body="&lt;p&gt;The Earth wouldn't simply split in half, it would be pulled towards the asteroid as a whole. To simplify matters, let's assume that the mass is equal to twice that of Earth (size alone is not enough, you must also consider density). You can then use Newton's law of gravitational attraction for the calculations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F=G \frac{m_1m_2}{r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G$ is a constant, $r$ is half the distance between the moon and Earth, $m_1$ is the mass of Earth, $m_2$ is twice that. The gravitational force would be about $1.3 \times 10^{29} \text{ N}$. In comparison, The gravitational attraction between Earth and the moon is about $2.0 \times 10^{26} \text{ N}$. Which means that the attraction Earth-meteorite will be 2000 times larger than Earth-Moon. The Earth will speed rapidly toward the asteroid and crash into it. During that flight Earth indeed is to be expected to have gravity fluctuations and probably lose major parts of its atmosphere. However it is unlikely that Earth, being a telluric planet (made of rock and metal) would subdue major deformation. After all, Jupiter's moons are round, even though they are in much more extreme conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would however need to consider the velocity of the meteorite for any more precise estimations. The speed and the trajectory count. An object that passes faster would tear the Earth out of its orbit but it would continue moving, whilst if you just place that meteorite between Earth and Moon, Earth (and Moon) would simply fall on a perpendicular trajectory.&lt;/p&gt;&#xA;" OwnerUserId="4126" LastEditorUserId="2153" LastEditDate="2015-01-06T23:11:18.787" LastActivityDate="2015-01-06T23:11:18.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8491" PostTypeId="2" ParentId="8477" CreationDate="2015-01-06T22:19:47.530" Score="3" Body="&lt;p&gt;The concept of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Wormhole&quot; rel=&quot;nofollow&quot;&gt;wormhole&lt;/a&gt; is, of course, highly speculative.  None the less, such a phenomenon could take many forms - most simple something which to outside observers looks like a blackhole on each end of the wormhole.  Alternatively, a &lt;strong&gt;traversable wormhole&lt;/strong&gt; (see for example &lt;a href=&quot;http://adsabs.harvard.edu/abs/1988AmJPh..56..395M&quot; rel=&quot;nofollow&quot;&gt;Morris &amp;amp; Thorne 1988&lt;/a&gt;; which has the same basic topology but no horizon), could still behave for the most part* like a normal, massive object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In either case, if one end were 'swallowed' by a blackhole, it is unclear to me whether it would remain stable (i.e. remain a wormhole).  In the case of a non-traversable wormhole, &lt;em&gt;if it remained stable&lt;/em&gt;, it would continue to look like a blackhole on each end --- one end of which now merged with another blackhole, and subsequently more massive.  A &lt;strong&gt;traversable wormhole&lt;/strong&gt; on the other hand could (extremely hypothetically) still remain traversable if there were enough exotic matter, or it could retain the horizon from the blackhole it merged with and be &lt;em&gt;no-longer&lt;/em&gt; traversable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All wormhole metrics &lt;strong&gt;&lt;em&gt;that I've seen&lt;/em&gt;&lt;/strong&gt; are more or less symmetric, and thus either have no horizon, or horizons on both ends... &lt;strong&gt;I don't &lt;em&gt;think&lt;/em&gt; it would be possible to have a horizon on only one side&lt;/strong&gt; --- allowing something to fall into an apparent blackhole on one side, and come out of a wormhole on the other.  Even so, however, remember that a blackhole isn't a magic vacuum --- and wouldn't 'suck in' the 'entire universe'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*A traversable wormhole is believed to require some sort of &lt;em&gt;'exotic matter'&lt;/em&gt; in the neck, which may behave somewhat differently... &lt;/p&gt;&#xA;" OwnerUserId="918" LastActivityDate="2015-01-06T22:19:47.530" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8492" PostTypeId="2" ParentId="8487" CreationDate="2015-01-06T23:18:43.290" Score="4" Body="&lt;p&gt;In a close approach, Earth would feel &lt;a href=&quot;http://en.wikipedia.org/wiki/Tidal_force&quot; rel=&quot;nofollow&quot;&gt;tidal forces&lt;/a&gt;: the near side to the approach would be affected by the vistor's gravity more than the far side, because the force of gravity decreases with distance from a mass. If you were standing on the far side, though, your effective weight wouldn't increase -- it would decrease! This is because the visitor would be pulling the center of the Earth more strongly than it would be pulling on you, again because of the gravitational falloff with distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a very close approach, closer than the &lt;a href=&quot;http://en.wikipedia.org/wiki/Roche_limit&quot; rel=&quot;nofollow&quot;&gt;Roche limit&lt;/a&gt;, the tidal forces would overcome the Earth's gravity completely, and Earth would begin to break apart. So it matters very much how close an approach we're talking about. Assuming the visitor has the same density as the Earth, the critical distance is about 10 Earth radii, or 64000km -- about a sixth of the distance from the Earth to the moon. &lt;/p&gt;&#xA;" OwnerUserId="2573" LastEditorUserId="2573" LastEditDate="2015-01-06T23:25:54.833" LastActivityDate="2015-01-06T23:25:54.833" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8493" PostTypeId="2" ParentId="8482" CreationDate="2015-01-06T23:59:04.163" Score="5" Body="&lt;p&gt;For those who haven't seen it:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;Some human explorers land on a planet orbiting a black hole. The black hole is surrounded by a large accretion disk. The planet orbits at a distance such that going any closer to the black hole will mean that your odds of getting out are slim; it's also composed of water. Finally, time dilation from the black hole means that even though the characters spend about two hours on the planet, a decade or so passes for their colleague on board.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The basic answer is that a planet can orbit a black hole. There are stable circum-black-hole orbits, just as there are stable orbits around just about any celestial body. There's a problem: A black hole typically forms as a result of a supernova. This will eject most nearby planets out of the stellar system. Alternatively, it's unlikely that a planet could be captured by a black hole and be in a stable orbit, so the whole premise - while possible - is highly unlikely. Then again, it's improbable that a planet will be made out of water, a wormhole will open up near Jupiter, or Matthew McConaughey will star in a decent sci-fi movie, so why should anything else in the story be normal?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;However, you can't just put a planet anywhere near a black hole, give it a strong enough push, and hope it orbits. The innermost orbit is at the boundary of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Photon_sphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;photon sphere&lt;/a&gt;. On this sphere, only photons can orbit. Inside it, nothing can orbit. However, the only stable orbit is twice as far away, at &lt;span class=&quot;math-container&quot;&gt;$2r_p$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The radius of the sphere is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$r_p=\frac{3GM}{c^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;We'll assume that the object is not rotating (I don't remember exactly if it is or isn't rotating, but it's simpler in this demo to say it isn't.). The formula for &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_time_dilation&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravitational time dilation&lt;/a&gt; is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$t_0=t_f \sqrt{1-\frac{3r_0}{2r_f}}$$&lt;/span&gt;&#xA;where&#xA;&lt;span class=&quot;math-container&quot;&gt;$$r_0=\frac{2GM}{c^2}$$&lt;/span&gt;&#xA;Assuming that &lt;span class=&quot;math-container&quot;&gt;$t_0$&lt;/span&gt; (the time for the observer inside the field) is two hours (7200 seconds) and &lt;span class=&quot;math-container&quot;&gt;$t_f$&lt;/span&gt; is ten years (315360000 seconds),&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{t_0}{t_f}=\frac{1}{43800}=1-\frac{(3)2GM}{(2)r_fc^2}$$&lt;/span&gt;&#xA;Simplifying, and saying that &lt;span class=&quot;math-container&quot;&gt;$\frac{2GM}{c^2}=\frac{2}{3}r_p$&lt;/span&gt;,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{r_p}{r_f}=\frac{43799}{43800}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$r_f \approx 1.0000228315715 r_p$$&lt;/span&gt;&#xA;which is outside the photon sphere, but just barely. However, it's well inside &lt;span class=&quot;math-container&quot;&gt;$2r_p$&lt;/span&gt;, and so most likely instable. The planet is, in short, not going to survive for long. And so giant waves - mini-spoiler - are the least of Matthew McConaughey's problems.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Post-question-edit modifications:&lt;/p&gt;&#xA;&lt;p&gt;It couldn't have been a &lt;a href=&quot;https://en.wikipedia.org/wiki/Supermassive_black_hole&quot; rel=&quot;nofollow noreferrer&quot;&gt;supermassive black hole&lt;/a&gt;; these form at the center of galaxies. It could have been a &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_black_hole&quot; rel=&quot;nofollow noreferrer&quot;&gt;stellar-mass black hole&lt;/a&gt;, though an &lt;a href=&quot;https://en.wikipedia.org/wiki/Intermediate-mass_black_hole&quot; rel=&quot;nofollow noreferrer&quot;&gt;intermediate-mass black hole&lt;/a&gt; is also likely - if not likelier, if the massiveness is emphasized.&lt;/p&gt;&#xA;&lt;p&gt;The existence of the neutron star is interesting. If the black hole were intermediate-mass, I would expect that it would have gobbled up the neutron star by now - and the planets, too. So I'd bet the black hole is a slightly-more-massive-than-average stellar-mass black hole. I highly doubt that multiple planets could orbit a black hole - for the reason I gave above; the supernova would have destroyed them or flung them out of the system.&lt;/p&gt;&#xA;&lt;p&gt;Would any of the planets be habitable? I doubt it. The accretion disk could heat up enough to provide some light, but there probably wouldn't be a lot. I'll write up the calculations either later today or possibly tomorrow, as I'm a bit LaTeXed-out after writing a math-heavy answer on Worldbuilding to find the luminosity, but I suspect it'll be negligible - as will Hawking radiation, in case any smart-aleck was planning on bringing that up.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="35106" LastEditDate="2020-10-14T16:02:54.007" LastActivityDate="2020-10-14T16:02:54.007" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="8494" PostTypeId="2" ParentId="8483" CreationDate="2015-01-07T18:22:32.343" Score="10" Body="&lt;p&gt;If you go to this site, &lt;a href=&quot;http://heritage.stsci.edu/2015/01/supplemental.html&quot;&gt;http://heritage.stsci.edu/2015/01/supplemental.html&lt;/a&gt; , there is a set of comparison photos. The movement that is detectable is very slight but it is there. &lt;/p&gt;&#xA;" OwnerUserId="5261" LastActivityDate="2015-01-07T18:22:32.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8495" PostTypeId="1" AcceptedAnswerId="8496" CreationDate="2015-01-07T22:28:16.640" Score="0" ViewCount="72" Body="&lt;p&gt;Did the Neptune and Pluto trajectories ever cross or was it a mere reclassification? Do studies say that I can predict an intersection between 2 trajectories  or movements based on data now? Do I have to consider at least 3 solid or larger space bodies moving in a way like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Three-body_problem&quot; rel=&quot;nofollow&quot;&gt;3-body problem&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="5265" LastActivityDate="2015-01-08T03:06:46.580" Title="How can I predict space directions?" Tags="&lt;orbit&gt;&lt;pluto&gt;&lt;n-body-simulations&gt;&lt;neptune&gt;&lt;newtonian-gravity&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8496" PostTypeId="2" ParentId="8495" CreationDate="2015-01-08T03:06:46.580" Score="3" Body="&lt;p&gt;If you project the orbits onto a plane, for example the plane of the ecliptic, the projections will cross. But that's only because you're looking at a 3D problem in 2D. If you look at the orbits in 3D, you'll see that Pluto's orbit is highly inclined (17º) from the ecliptic, so it never actually passes through Neptune's orbit. Each time it seems to cross (in the 2D view), it's actually well above or well below Neptune's orbit. Pluto and Neptune never come closer than about 17AU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least for now. Minor perturbations over tens of millions of years may change that. Or might not. Pluto and Neptune currently have a 3:2 resonance, and such resonances tend to be stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Pluto&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Pluto&lt;/a&gt;, especially the section &quot;Relationship with Neptune&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4077" LastActivityDate="2015-01-08T03:06:46.580" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8497" PostTypeId="1" AcceptedAnswerId="8502" CreationDate="2015-01-08T03:58:04.427" Score="0" ViewCount="261" Body="&lt;p&gt;Assume that two Earth-like planets orbited the same star, with elliptic orbits. Eventually, something occurs that caused the two planets to collide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the outcome of such a catastrophe? I figure that both planets (and any life if it exists) would be destroyed, but exactly how? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would both planets explode, combine together into a larger planet, or be able to rebound and either reestablish an orbit or be pulled into the sun?&lt;/p&gt;&#xA;" OwnerUserId="5267" LastActivityDate="2015-01-08T15:56:54.733" Title="Collision of two Earths" Tags="&lt;orbit&gt;&lt;astrophysics&gt;&lt;earth-like-planet&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8498" PostTypeId="1" CreationDate="2015-01-08T04:52:48.993" Score="0" ViewCount="221" Body="&lt;p&gt;What kinds of astronomical events can pose a serious threat to life on Earth in the near future (1000-10,000 years). If there are none in the near future, when is the soonest event that is predicted to threaten life on Earth? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am ignorant about astrononomy and so far I have only thought of an asteroid collision, solar flares, and effects from the sun dying in several billion years (and a visit by some hostile aliens).&lt;/p&gt;&#xA;" OwnerUserId="2873" LastActivityDate="2015-01-09T16:02:55.207" Title="Near-future astronomic events that could threaten Earth" Tags="&lt;earth&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8500" PostTypeId="2" ParentId="8356" CreationDate="2015-01-08T13:24:00.390" Score="1" Body="&lt;p&gt;As the comments to your question say, the short answer is &quot;yes&quot;. A lightyear is, by definition, the distance traveled by a photon in one year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, however, that at large distances, the answer becomes more complicated. Due to the expansion of the Universe, the light received from a galaxy far, far away was emitted when that galaxy was closer to us. That means that in the beginning, the photons traverse a larger &quot;fraction&quot; of the Universe per time than at a later time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, to calculate the total pathlength you need to integrate its distance in an expanding coordinate system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if a galaxy was a 4 billion lightyears away from the Milky Way when the Universe was only 1 billion years old, it is now 28 billion lightyears away. But as the Universe is 13.8 billion years old, the light received today from that galaxy has been traveling for &quot;only&quot; 12.8 billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objects in our local neighborhood, however, such as stars in our own galaxy and the nearest galaxies (e.g. Andromeda) are gravitationally bound and thus &quot;resist&quot; the expansion of space, so here your assumption is correct.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2015-01-08T13:24:00.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8501" PostTypeId="2" ParentId="2085" CreationDate="2015-01-08T13:38:17.350" Score="6" Body="&lt;p&gt;In general, the answer given by @Jeremy is completely correct. However, transient phenomena such as supernovae (exploding stars) and the even more energetic gamma-ray bursts can for a brief period be visible to the naked eye, despite distances of billions of lightyears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 2008, the gamma-ray burst &lt;a href=&quot;http://arxiv.org/abs/0803.3215&quot; rel=&quot;noreferrer&quot;&gt;GRB 080319B&lt;/a&gt; reached a magnitude of 5.8, and was visible to human eyes for half a minute. The distance to GRB 080319B is 10 billion lightyears!&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2015-01-08T13:38:17.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8502" PostTypeId="2" ParentId="8497" CreationDate="2015-01-08T15:56:54.733" Score="1" Body="&lt;p&gt;To start with, such two orbits are extremely unlikely. First of all, any object orbiting a star and deviating strongly from a circular orbit has a high risk of crashing into something. As you notice, there are very few planets in our solar system, and all of them except Pluto have a very regular orbit. Planets which have intercrossing orbits would have crashed into each other and fallen apart on the planetesimal stage, much earlier than becoming a planet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, if you suppose that such an event does occur, the consequences will depend on multiple factors. For one, the composition of the planets - if the two planets are terrestrial (rocky), one would expect them to fragmentate and lose a large amount of their mass. Their cores would maybe fuse into one planet, and the debris would form one or several satellites. The exact outcome would be dependent on their speed, their density, and the angle of their collisions. A similar (but not the same) scenario happened during the late stage of Earth formation, when an object close to Earth's size crashed into it to create the Moon. If the sizes are the same and the formation is complete, chances are that most of the matter would fly away and the planet that would be left would be tiny.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be a different story if the two planets are gaseous. They have almost no solid matter in them, so chances are that the planets will fuse whilst losing (proportionally) significantly less matter than terrestrial planets. Still, there would be losses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, if a terrestrial and a gaseous planet meet, the outcome is quite easy to predict. Gas planets are much larger than rocky planets, so the rocky planet would simply be engulfed. It will maybe cause a disturbance on the surface of the gas giant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To wrap it up let me just note that such a collision would almost definitely change the orbit of whatever is left, so that it would either get pushed into the sun or away from it, causing it to float eternally in outer space. Have a nice day.&lt;/p&gt;&#xA;" OwnerUserId="4126" LastActivityDate="2015-01-08T15:56:54.733" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8503" PostTypeId="1" AcceptedAnswerId="8507" CreationDate="2015-01-08T16:57:41.837" Score="12" ViewCount="2984" Body="&lt;p&gt;Consider black hole A, a super massive black hole at the center of the galaxy. Orbiting it is black hole B, a much less massive black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If some passing body were to modify black hole B's orbit such that it fell within the Roche limit of black hole A, what would happen to black hole B?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were to turn into a ring, would the black hole matter re-inflate since it wouldn't be under such high gravity? Do black holes even respond to Roche limits like regular matter does?&lt;/p&gt;&#xA;" OwnerUserId="4026" LastEditorUserId="1435" LastEditDate="2018-07-15T11:56:59.470" LastActivityDate="2020-11-28T08:51:04.890" Title="Do Roche limits apply to black holes?" Tags="&lt;black-hole&gt;&lt;roche-limit&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8504" PostTypeId="2" ParentId="8498" CreationDate="2015-01-08T18:04:37.327" Score="2" Body="&lt;p&gt;Essentially, anything could happen! We can't really predict events such as you've described.&#xA;The only event we can really be certain of is that the Sun will inevitably die in about 5 billion years. When it's almost run out of hydrogen to fuse it will expand into a red giant, becoming so large it will take out the Earth. There's a good video demonstrating this here: &lt;a href=&quot;http://science.howstuffworks.com/life/evolution/life-on-earth-end1.htm&quot; rel=&quot;nofollow&quot;&gt;http://science.howstuffworks.com/life/evolution/life-on-earth-end1.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could add geomagnetic reversal and gamma ray bursts to your list. Also see: &lt;a href=&quot;http://science.howstuffworks.com/life/evolution/life-on-earth-end1.htm&quot; rel=&quot;nofollow&quot;&gt;http://science.howstuffworks.com/life/evolution/life-on-earth-end1.htm&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="1801" LastActivityDate="2015-01-08T18:04:37.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8505" PostTypeId="2" ParentId="8503" CreationDate="2015-01-08T19:50:58.190" Score="3" Body="&lt;p&gt;Because the density of the matter at the center of a black hole is infinite (or nearly so) the Roche Radius is 0 (or nearly so).  Two black holes in orbit spiral in towards each other because they radiate gravity waves and form a mutual event horizon without any disturbance to the matter at their centers.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2015-01-09T19:00:03.850" LastActivityDate="2015-01-09T19:00:03.850" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8506" PostTypeId="2" ParentId="8503" CreationDate="2015-01-08T19:59:40.983" Score="7" Body="&lt;p&gt;It's different. Black holes are not objects, like planets or stars. Rather, they are powerful distortions of spacetime, maintained by the concentration of mass/energy therein (which itself is kept there prisoner by the spacetime distortion - a vicious cycle broken only slowly by the Hawking radiation). As such, they cannot be &quot;ripped apart&quot;, since there is nothing to be ripped apart there - just a giant warped tangle of spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, two black holes could merge, if they get close enough and lose enough mutual orbital motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=4m-ZVsLf070&quot; rel=&quot;nofollow&quot;&gt;merging black holes&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is a single, larger black hole.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-01-09T19:27:34.590" LastActivityDate="2015-01-09T19:27:34.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8507" PostTypeId="2" ParentId="8503" CreationDate="2015-01-09T00:44:09.410" Score="7" Body="&lt;p&gt;The Roche limit applies when a smaller body that would be held together by its own self-gravity is in the gravitational field of another, such that the tidal forces of the latter are stronger than the self-gravity of the latter, thus destroying the smaller body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the gravitational tidal forces of a black hole are always finite, except at the internal singularity. This is a problem because the self-gravity of a black hole, in the sense of the acceleration a mass would need to remain stationary on its surface, is infinite&lt;sup&gt;1&lt;/sup&gt;. Thus, we shouldn't expect for a large black hole to destroy another through gravitational tidal forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put another way, the Roche limit occurs when particles from the smaller body can escape them... but they can't escape the event horizon of a black hole. Thus, the black holes will either orbit or merge, which is what happens in numerical simulations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;sup&gt;1&lt;/sup&gt;There is a separate concept of &lt;em&gt;surface gravity&lt;/em&gt; of a black hole that's essentially this re-scaled by the gravitational time dilation factor, and thus kept finite.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2015-01-09T00:44:09.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8508" PostTypeId="2" ParentId="8503" CreationDate="2015-01-09T02:41:23.040" Score="3" Body="&lt;p&gt;(Yes, this should be a comment. It's too big, though.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To address Sidney's comment to Ed Shaya's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Roche limit equation can be expressed as&#xA;$$1.26 \times R_{\text{secondary}} \times \sqrt[3]{\frac{\text{Primary}}{\text{Secondary}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the radius of the secondary is zero and zero times anything is zero, the Roche limit is likewise zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you think about what the Roche limit really means this is self-evident. The Roche limit is the point where the primary is raising a tide on the secondary that's stronger than it's own gravity. Tides are based on the fact that while the object is at a certain distance it really occupies space, parts are closer (which experience a greater pull) and parts are farther (which experience a lesser pull.) There is no point on the black hole that's closer or farther, thus there is no tide, thus it won't be torn apart even if it had no self-gravity.&lt;/p&gt;&#xA;" OwnerUserId="2036" LastEditorUserId="2153" LastEditDate="2015-01-09T03:27:43.867" LastActivityDate="2015-01-09T03:27:43.867" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="8510" PostTypeId="1" CreationDate="2015-01-09T10:20:17.877" Score="6" ViewCount="264" Body="&lt;p&gt;If the Earth's Moon had a water ocean of depth 2-4 Km, how high would the tides rise due to the Earth's gravity? (Just a hypothetical question.)&lt;/p&gt;&#xA;" OwnerUserId="662" LastEditorUserId="2424" LastEditDate="2015-01-22T03:14:30.133" LastActivityDate="2015-01-24T05:05:41.953" Title="Tide on the Moon" Tags="&lt;the-moon&gt;&lt;gravity&gt;&lt;earth&gt;&lt;tidal-forces&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8511" PostTypeId="2" ParentId="8498" CreationDate="2015-01-09T16:02:55.207" Score="2" Body="&lt;p&gt;Adding to Oceanescence's mention of Gamma Ray Bursts, someone's recently done some calculations: &lt;a href=&quot;http://www.universetoday.com/117259/gamma-ray-bursts-limit-the-habitability-of-certain-galaxies-says-study/&quot; rel=&quot;nofollow&quot;&gt;Gamma Ray Bursts Limit The Habitability of Certain Galaxies, Says Study&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In fact, according to the authors of the new paper, there is a 90% chance that a GRB powerful enough to destroy Earth’s ozone layer occurred in our stellar neighborhood some time in the last 5 billion years, and a 50% chance that such an event occurred within the last half billion years. These odds indicate a possible trigger for the second worst mass extinction in Earth’s history: the Ordovician Extinction. This great decimation occurred 440-450 million years ago and led to the death of more than 80% of all species.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gamma ray bursts are thought to emerge from the &lt;em&gt;poles&lt;/em&gt; of supernova, so a star's spin axis would have to be in the right direction for it to hit us. Betelgeuse, a nearby supernova candidate has it's spin axis at &lt;a href=&quot;http://en.wikipedia.org/wiki/Betelgeuse&quot; rel=&quot;nofollow&quot;&gt;an inclination of about 20° to the direction of Earth&lt;/a&gt;, so that star is not likely to get us when she blows. However, we don't have accurate information on the direction of the spin axis of &lt;em&gt;most&lt;/em&gt; stars.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-01-09T16:02:55.207" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8512" PostTypeId="1" CreationDate="2015-01-10T11:15:50.767" Score="5" ViewCount="816" Body="&lt;p&gt;Current and upcoming telescopes which find exoplanets and take spectra of their atmospheres seem to be optical or infrared. What about radio telescopes applied to exoplanets? They are larger and more easy to use with interferometry. Why aren't radio frequencies as useful as optical/IR for exoplanet studies?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-01-10T11:44:30.967" LastActivityDate="2015-01-11T12:52:02.873" Title="How do or could radio telescopes contribute to exoplanet studies?" Tags="&lt;telescope&gt;&lt;exoplanet&gt;&lt;planetary-atmosphere&gt;&lt;radio-astronomy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8513" PostTypeId="2" ParentId="8512" CreationDate="2015-01-10T17:32:55.230" Score="2" Body="&lt;p&gt;Some radio telescopes have been used to observe young stars and protoplanetary disks, if that counts.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Atacama_Large_Millimeter_Array#Planetary_formation&quot; rel=&quot;nofollow&quot;&gt;Atacama Large Millimeter Array (ALMA):&lt;/a&gt; Though it's used for a variety of things, ALMA was used in late 2014 &lt;a href=&quot;https://public.nrao.edu/static/pr/planet-formation-alma.html&quot; rel=&quot;nofollow&quot;&gt;to observe a young star&lt;/a&gt;, HL Tau. Among the data gathered was information about HL Tau's circumstellar disk. It found a series of gaps in the disk. According to ALMA Deputy Director Stuartt Corder,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;These features are almost certainly the result of young planet-like bodies that are being formed in the disk. This is surprising since HL Tau is no more than a million years old and such young stars are not expected to have large planetary bodies capable of producing the structures we see in this image.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;ALMA was used because it could see through gas and dust surrounding the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The telescope was also used in December &lt;a href=&quot;http://alma.mtk.nao.ac.jp/e/news/info/2014/1215alma_detects_pluto-size_objects_kicking-up_dust_around_sun-like_star.html&quot; rel=&quot;nofollow&quot;&gt;to find evidence of &quot;Pluto-size objects&quot;&lt;/a&gt; around a different star, HD 107146. Only dust grains in the circumstellar disk were detected, but their behavior was considered consistent with the existence of Pluto-sized objects nearby. A paper was published on the findings, entitled &quot;ALMA observations of the debris disk around the young Solar Analog HD 107146.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://public.nrao.edu/images/non-gallery/2014/c-blue/11-05-HL-Tau/HLTau_nrao.jpg&quot; rel=&quot;nofollow&quot;&gt;Here's&lt;/a&gt; one of ALMA's pictures, of the first star, and &lt;a href=&quot;http://alma.mtk.nao.ac.jp/e/news/files/editor/141211_ALMA_swarms_02_uxga.jpg&quot; rel=&quot;nofollow&quot;&gt;a picture&lt;/a&gt; of the second star.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Very_Large_Array&quot; rel=&quot;nofollow&quot;&gt;VLA:&lt;/a&gt; The &lt;a href=&quot;https://science.nrao.edu/facilities/vla&quot; rel=&quot;nofollow&quot;&gt;VLA&lt;/a&gt;, too, has been used to look at young stars and protoplanetary disks. &lt;a href=&quot;http://images.nrao.edu/Miscellaneous/Protoplanetary_Disks/61&quot; rel=&quot;nofollow&quot;&gt;Astronomers looked at star-forming region LDN 1551&lt;/a&gt; and made observations of what appear to be protoplanetary disks around two young stars; the abstract from the resulting paper can be found &lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1998Natur.395..355R&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. In another instance, the VLA was used &lt;a href=&quot;http://images.nrao.edu/Miscellaneous/Protoplanetary_Disks/59&quot; rel=&quot;nofollow&quot;&gt;to observe protoplanetary disk NGC 2071&lt;/a&gt;; the abstract from the resulting paper can be found &lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1998ApJ...505..756T&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Green_Bank_Telescope&quot; rel=&quot;nofollow&quot;&gt;Green Bank Observatory:&lt;/a&gt; Within the last decade or so, there have been a handful of &lt;a href=&quot;http://www.gb.nrao.edu/science/GBT_Pubs.html&quot; rel=&quot;nofollow&quot;&gt;observations&lt;/a&gt; of young stars and protostars, though exoplanets do not appear to have been studied.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would include the images for both telescopes, but it appears they are copyrighted. To my knowledge, though, these two telescopes have not been used for observing full-fledged exoplanets. I'm also investigating other major radio telescopes to see if they've been involved.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-01-10T17:48:13.277" LastActivityDate="2015-01-10T17:48:13.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8514" PostTypeId="1" AcceptedAnswerId="8515" CreationDate="2015-01-10T22:14:55.867" Score="0" ViewCount="34" Body="&lt;p&gt;When I listen to the Planetary Radio podcast, there is a section at the end where they suggest objects to look out for in the night sky. If I am in the UK, but they are talking about the USA, will I still be able to find the objects they are referring to?&lt;/p&gt;&#xA;" OwnerUserId="4146" LastActivityDate="2015-01-10T23:08:52.767" Title="US observations relevant in UK?" Tags="&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8515" PostTypeId="2" ParentId="8514" CreationDate="2015-01-10T23:08:52.767" Score="3" Body="&lt;p&gt;If they mention times of night then you have to figure out to what extent the time zone (BST/GMT vs the US time zones) makes any difference, but it shouldn't make more than an hour difference either way for an object that is moving sidereally with the rest of the night sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main difference would be that the USA is at lower latitudes than the UK. Therefore there could be instances where an object with a small or southerly declination is better seen from the USA than the UK. Conversely, more of the sky is circumpolar from the UK.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if you are trying to see a specific event at a &lt;em&gt;given time (UT)&lt;/em&gt;, then of course it may well make a difference. The Sun could be up in the UK and down in the USA, or vice-versa.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-01-10T23:08:52.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8516" PostTypeId="2" ParentId="6482" CreationDate="2015-01-10T23:31:49.677" Score="2" Body="&lt;p&gt;A protostar star in the class I phase is surrounded by an optically thick envelope, but accretes matter through a disk. The protostar will have some magnetic field - it may be mostly primordial at this stage, or could be generated by an internal dynamo. Material accreting from the disk is thought to be locked onto the field lines -  a well known phenomenon caused by charged particles spiralling around the field lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens net to drive the jets/outflows is still very actively debated. A very popular baseline model is the so-called Uchida-Shibata theory. Imagine an initially radial magnetic field. As the material locks onto the field is must pull it into a spiral pattern because the accreting material has angular momentum and cannot fall directly into the stellar surface. This action will twist the field lines into a helical shape, with an axis similar to that of the accretion disk of the protostar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The helix will be tighter closer to the stellar rotation pole and will open out further away from the protostar. This gradient in magnetic field strength provides a pressure gradient (because a force acts on charged particles perpendicularly to the field lines) which accelerates material along the polar axis. At the same time, the accelerating material is collimated by punching its way out through the envelope - once some material has opened up a path, it becomes easier for subsequent material to follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is an extract from a paper (unfortunately behind a paywall) by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2009pjc..book..185M&quot; rel=&quot;nofollow noreferrer&quot;&gt;Matsumoto (2009)&lt;/a&gt; which has a picture that gives you the general idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ELM9h.png&quot; alt=&quot;Uchida-Shibata mechanism&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-01-10T23:31:49.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8517" PostTypeId="2" ParentId="8512" CreationDate="2015-01-10T23:49:40.687" Score="4" Body="&lt;p&gt;Radio telescopes are frequently used to observe the births of stars and their planetary systems. The longer wavelengths are able to penetrate from beneath the envelope of gas and dust that shrouds any attempts to view these events at optical or infrared wavelengths. Unfortunately, the smallest angular resolution of a telescope goes as $\lambda/D$, where $D$ is the telescope diameter and $\lambda$ the observation wavelength. As radio waves are $10^5$ to $10^{6}$ times longer than optical waves, then one needs enormous telescopes to get similar resolving power. This is attempted/accomplished by constructing arrays that can act as a single aperture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whilst interferometric techniques to gain spatial resolution are certainly possible, if one wants to directly image exoplanets then optical/IR is still the way to go because this is where most of the intrinsic light from an exoplanet is found.&#xA;It is possible that planets orbiting stars may emit radio waves in their own right, through charged particles accelerated in their magnetospheres. Jupiter, for instance, is a strong source of radio waves in our solar system, but would be very difficult to spot at the distances of exoplanets. Some searches have been attempted, but nothing yet found (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2007MNRAS.382..455G&quot; rel=&quot;nofollow&quot;&gt;George &amp;amp; Stevens 2007&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course &lt;em&gt;transit&lt;/em&gt; observations do not require superb angular resolution, but they do require lots of signal and precision. Levels of precision that are routinely achieved by optical telescopes (especially in space - e.g. Kepler) are not possible with radio telescopes, and the parent stars themselves are &lt;em&gt;weak, variable&lt;/em&gt; radio sources, which makes studying any transit signature very difficult. In fact the main use of transits has been the other way around - to search for intrinsic planetary emission by looking for a drop in flux when the planet is eclipsed by the star (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2011A%26A...533A..50L&quot; rel=&quot;nofollow&quot;&gt;Lecavelier Des Etangs et al. 2011&lt;/a&gt;).  Radio emission might be enhanced in hot Jupiters if the exoplanetary magnetic field interacts with the stellar magnetic field.&#xA;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2013A%26A...552A..65L&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; does claim to have detected a weak exoplanetary signal from a hot Jupiter, identified by its disappearance when eclipsed by the parent star. However, the result was not repeatable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The SETI programme targeted a set of Kepler stars with evidence for a planet (or planet candidate) in the &quot;habitable zone&quot;. The initial searches for narrow-band radio emission (at 1.1-1.9 Hz) appears to have concentrated on those systems containing multiple planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The (null) results for $\sim 100$ system were published by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013ApJ...767...94S&quot; rel=&quot;nofollow&quot;&gt;Siemion et al. (2013)&lt;/a&gt;. The observations were not sensitive enough to detect the sort of &quot;radio chatter&quot; emitted from our planet, but might have detected some sort of deliberate signalling beacon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A slightly more imaginative approach is to observe Kepler multiple planet systems when two planets are in conjunction as seen from the Earth. There is then the chance of seeing communications from one planet to the other. (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014ebi..conf..5.2K&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/2014ebi..conf..5.2K&lt;/a&gt; ).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-01-11T12:52:02.873" LastActivityDate="2015-01-11T12:52:02.873" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8518" PostTypeId="1" CreationDate="2015-01-11T13:19:04.807" Score="7" ViewCount="182" Body="&lt;p&gt;The latest information suggests that Earth's water came from Meteorites . The Moon was also bombarded by Meteorites and yet it has no water. Is this because it lacks an atmosphere and its water was lost in space ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On further thought on this matter, it occurred to me that Mars has an atmosphere and yet it has no oceans either, just like the Moon . So Mars I assume was bombarded by meteorites, just like the Earth was and it has a stronger gravity than the Moon, yet it doesn't have oceans either ? So what happened ?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastEditorUserId="2562" LastEditDate="2015-01-12T16:53:00.247" LastActivityDate="2017-04-15T14:12:11.280" Title="Meteorites bring water" Tags="&lt;the-moon&gt;&lt;solar-system&gt;&lt;earth&gt;&lt;water&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8519" PostTypeId="1" AcceptedAnswerId="8520" CreationDate="2015-01-11T15:02:08.320" Score="3" ViewCount="323" Body="&lt;p&gt;There are exoplanets called Gliese 581c, Gliese 581d etc. I assume their star is &lt;a href=&quot;https://en.wikipedia.org/wiki/Gliese_581&quot; rel=&quot;nofollow&quot;&gt;Gliese 581&lt;/a&gt;. So where does this Gliese come from?&lt;/p&gt;&#xA;" OwnerUserId="5287" LastEditorUserId="2153" LastEditDate="2015-01-11T15:51:30.287" LastActivityDate="2015-01-11T15:51:30.287" Title="Why stars are named Gliese?" Tags="&lt;star&gt;&lt;exoplanet&gt;&lt;naming&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8520" PostTypeId="2" ParentId="8519" CreationDate="2015-01-11T15:08:48.010" Score="6" Body="&lt;p&gt;They come from Gliese's catalog of nearby stars (various editions and co-author), the number is the star's catalog number.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-01-11T15:14:15.797" LastActivityDate="2015-01-11T15:14:15.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8521" PostTypeId="1" CreationDate="2015-01-11T16:40:23.240" Score="-3" ViewCount="225" Body="&lt;p&gt;Astronomers don't face much of an ordinary jobs market in their &quot;trade&quot;. What does the process look like from choosing a topic for the PhD to doing research in a specific topic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that you're very interested in exoplanets, would you as a new PhD on the topic be offered and accept a position in a project which does galaxies instead? And could you choose weather to do the physical theory or the observations or coding simulations? Can you pick and choose?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="6" LastEditDate="2019-01-29T14:44:52.127" LastActivityDate="2019-01-29T14:44:52.127" Title="How do astronomers choose their research topic?" Tags="&lt;planetary-science&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ClosedDate="2015-01-14T03:28:36.510" ContentLicense="CC BY-SA 3.0" />
  <row Id="8523" PostTypeId="1" AcceptedAnswerId="8524" CreationDate="2015-01-12T00:53:36.797" Score="3" ViewCount="1057" Body="&lt;p&gt;I've been wanting to create a simulator to run tests based on properties of a star to find patterns in how planets form. Are there any already made equations for any of these countless relationships? Here are a few examples to further explain what sort of information I am looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, how does the mass of a star relate to the distance of the frost line, and is there an already existing equation to estimate this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More general example; If I wanted to find out how to estimate, say, a planets surface temperature based on a stars temperature and distance, what is the best way to do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example; To find the most likely average mass of a potential planet, what information would I need? Distance to star? Stars mass? Distance to nearest planets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another; How would I go about finding likely measurements of compositions of any hypothetical planets (surface/atmosphere/etc)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Absolutely any information is very welcome. I do not require any form of proven or precise measurements. If there is no equations, then all I need is what factors are involved in specific attribute development.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to emphasize my understanding that this is an extremely complex subject to go into detail. All I am asking for is any information or recommendations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6197/star-system-formation-questions&quot;&gt;related question&lt;/a&gt; to my dilemma, in addition to this information, I am looking for more. Absolutely anything that may help, probably will. I am attempting to make as highly detailed of a model as possible.&lt;/p&gt;&#xA;" OwnerUserId="4080" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-05-23T17:55:10.410" Title="Are there equations to simulate creation of a solar system?" Tags="&lt;planetary-systems&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8524" PostTypeId="2" ParentId="8523" CreationDate="2015-01-12T01:22:22.800" Score="5" Body="&lt;p&gt;I may have to answer this in parts, a bit at a time, so I apologize if I'm not able to answer everything at the moment.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If I wanted to find out how to estimate, say, a planets surface temperature based on a stars temperature and distance, what is the best way to do this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ah, I just studied this about a week ago! What you're looking for is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Effective_temperature&quot; rel=&quot;nofollow&quot;&gt;effective temperature&lt;/a&gt; of the planet, which can be calculated as&#xA;$$T=\left(\frac{L(1-A)}{16 \pi \sigma D^2} \right)^{\frac{1}{4}}$$&#xA;where $L$ is the star's luminosity, $A$ is the planet's albedo, $\sigma$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_constant&quot; rel=&quot;nofollow&quot;&gt;Stefan-Boltzmann constant&lt;/a&gt;, and $D$ is the distance from the star.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;how does the mass of a star relate to the distance of the frost line, and is there an already existing equation to estimate this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wasn't able to find an explicit equation for this, but it's calculable. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Frost_line_(astrophysics)&quot; rel=&quot;nofollow&quot;&gt;frost line&lt;/a&gt; is a certain distance from a star in a stellar nebula such that the temperature is about 150 Kelvin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know the mass of the star, you can use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Mass%E2%80%93luminosity_relation&quot; rel=&quot;nofollow&quot;&gt;mass-luminosity relation&lt;/a&gt;:&#xA;$$\left(\frac{L}{L_{\odot}} \right)=\left(\frac{M}{M_{\odot}} \right)^a$$&#xA;Re-arrange that to solve for the luminosity. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Inverse-square_law#Light_and_other_electromagnetic_radiation&quot; rel=&quot;nofollow&quot;&gt;inverse-square law&lt;/a&gt; says that the intensity of power drops off as the distance from the source increases:&#xA;$$I=\frac{P}{4 \pi r^2}$$&#xA;You can solve for $r$ if you can figure out the required intensity:&#xA;$$r=\frac{1}{2}\sqrt{\frac{P}{\pi I}}$$&#xA;Here, $P \approx L$. Figure out the necessary intensity to heat the nebula to 150 K and you've found the frost line.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To find the most likely average mass of a potential planet, what information would I need?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't help you much there, but I can give you a starter. The density $\rho$ of a circumstellar disk at a distance $r$ (as given by Michael Woolfson in &lt;em&gt;On the Origin of Planets&lt;/em&gt;) is&#xA;$$\rho (r)=C \exp \left[-\frac{(r-r_{\text{peak}})^2}{2 \sigma ^2} \right]$$&#xA;where $\sigma$ is one standard deviation and $r_{\text{peak}}$ is the peak density, and $C$ is a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As promised, here are some more, set forth by Woolfson.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The disk decays over time; its density at time $t$ is approximated by&#xA;$$\rho_t=\rho e^{- \gamma t}$$&#xA;where $\gamma$ is a decay constant. If you've taken certain math course, you've probably seen this equation many times before, just in with $\rho=y$ and $\gamma=k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another equation - really, a couple equations - deal with accretion by a small dust particle in a circumstellar disk. The variables involved are: $m$ (mass of the particle), $t$ (time), $s$ (the radius of the particle), $\rho_s$ (the density of the particle), $\rho$ (the density of the disk), $f$ (the fraction of the medium that is dust), $T$ (the temperature), and $k$ (Boltzmann's constant, not be confused with the Stefan-Boltzmann constant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relevant pair of equations are:&#xA;$$\frac{dm}{dt}=4 \pi s^2 \rho_s \frac{ds}{dt}$$&#xA;$$\frac{ds}{dt}=\frac{3f \rho}{4} \left( \frac{kT}{4 \pi \rho_s^3} \right)^{1/2}s^{-3/2}$$&#xA;Using these, you can find the size and mass of a given dust particle at any given time.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-01-12T23:01:13.080" LastActivityDate="2015-01-12T23:01:13.080" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8525" PostTypeId="1" AcceptedAnswerId="8534" CreationDate="2015-01-12T14:20:57.403" Score="9" ViewCount="4186" Body="&lt;p&gt;I know light, and practically nothing but gravity can escape a black hole.  My question is: can magnetism escape a black hole?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple things that convince me it can are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Jupiter's magnetic field shape compared with the jets that can come from near black holes (I would think this might push material falling toward a black hole into or out of the poles):&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2JUEx.jpg&quot; alt=&quot;IO&amp;#39;s plasma and Jupiter&amp;#39;s magnetic field&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Apparently black holes have very strong magnetic fields:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.iflscience.com/space/magnetic-fields-can-be-strong-black-holes-gravity&quot; rel=&quot;noreferrer&quot;&gt;http://www.iflscience.com/space/magnetic-fields-can-be-strong-black-holes-gravity&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="1386" LastEditDate="2015-01-13T15:07:48.637" LastActivityDate="2019-04-07T02:52:40.860" Title="Can magnetism escape a black hole?" Tags="&lt;black-hole&gt;&lt;jupiter&gt;&lt;supermassive-black-hole&gt;&lt;magnetic-field&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8526" PostTypeId="1" AcceptedAnswerId="8533" CreationDate="2015-01-12T14:35:11.907" Score="8" ViewCount="476" Body="&lt;p&gt;My main question is: Is there a strong galactic magnetic field, perhaps driven by the supermassive black hole at the center of our galaxy?  I am also wondering if this field would be strong enough to make it so that the galaxy rotates in the way it does (with the outer stars moving faster than would be expected), and if this would be an alternate explanation for dark matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing that led me to ask this question is reading about Jupiter's magnetic field interactions with the plasma emitted by IO.  Jupiter's magnetic field forces the plasma to orbit Jupiter about as fast as Jupiter spins, and I am wondering if likewise, the supermassive black hole at the center of our galaxy &quot;herds&quot; the rest of the galaxy in a similar manner as per the article and image below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetosphere_of_Jupiter#Role_of_Io&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Magnetosphere_of_Jupiter#Role_of_Io&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2JUEx.jpg&quot; alt=&quot;IO&amp;#39;s plasma and Jupiter&amp;#39;s magnetic field&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2016-01-01T18:51:48.923" Title="Is there a strong galactic magnetic field?" Tags="&lt;galaxy&gt;&lt;jupiter&gt;&lt;supermassive-black-hole&gt;&lt;dark-matter&gt;&lt;magnetic-field&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8527" PostTypeId="1" AcceptedAnswerId="8536" CreationDate="2015-01-12T14:44:38.223" Score="3" ViewCount="260" Body="&lt;p&gt;Do the jets sometimes emitted by a black hole result from its magnetic field?  It would seem Jupiter's magnetic field tends to be concentrated at the poles, and I would imagine this might extend to black holes as well.  I could see the concentrated fields near the poles accelerating particles either into or away from the black hole at high speeds.  I have included an image of Jupiter's magnetic field below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetosphere_of_Jupiter#Role_of_Io&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Magnetosphere_of_Jupiter#Role_of_Io&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2JUEx.jpg&quot; alt=&quot;IO&amp;#39;s plasma and Jupiter&amp;#39;s magnetic field&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also appears that a black hole's magnetic field is indeed very strong:&#xA;&lt;a href=&quot;http://www.iflscience.com/space/magnetic-fields-can-be-strong-black-holes-gravity&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.iflscience.com/space/magnetic-fields-can-be-strong-black-holes-gravity&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="1386" LastEditDate="2015-01-13T15:16:04.853" LastActivityDate="2015-02-06T14:59:09.737" Title="Are a black hole's jets caused by the black hole's magnetic field?" Tags="&lt;black-hole&gt;&lt;jupiter&gt;&lt;magnetic-field&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8528" PostTypeId="2" ParentId="8525" CreationDate="2015-01-12T15:55:29.633" Score="3" Body="&lt;p&gt;Nothing can escape a black hole, not even gravity. What black holes being inescapable means is just this: if you have any sort of system inside a black hole, there's nothing it can do to send a signal outside. This is true regardless of whether the attempt is made through the gravitational, electric, or magnetic fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes obviously do have a gravitational field, and may indeed have a nonzero electric or magnetic fields, but that doesn't mean anything is &quot;escaping&quot; them. On the contrary, one can view their having gravitational or electromagnetic fields as a &lt;em&gt;consequence&lt;/em&gt; of them being inescapable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, as matter collapses to a black hole, the external gravitational field takes on some value. The collapsing matter crosses the horizon and is later crushed out of existence. Does that mean the gravitational field should disappear? No, because if that happened, that would constitute a signal from inside to outside! So in fact, it's the inescapable nature of black holes that keeps the gravitational field from changing in response to whatever happens to the matter inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, no, magnetic fields can't &quot;escape&quot; black holes, but that doesn't imply that black holes can't have magnetic fields.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2015-01-12T15:55:29.633" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8529" PostTypeId="1" CreationDate="2015-01-12T16:13:06.977" Score="-2" ViewCount="113" Body="&lt;p&gt;Do any theories exist as to where this bright light and matter came from that expanded to form our Universe ?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastActivityDate="2015-01-14T12:47:12.730" Title="Origin of the original bright light and matter" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2015-01-14T14:37:10.503" ContentLicense="CC BY-SA 3.0" />
  <row Id="8530" PostTypeId="2" ParentId="8521" CreationDate="2015-01-12T16:16:29.537" Score="1" Body="&lt;p&gt;Let me answer what I think you're asking (parts of your question aren't that clear to me).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Background: I am currently an astrophysics graduate student.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, from you question it seems to me that you are proceeding under an assumption that you are accepted into a research group when you are accepted into a PhD program.  Everywhere I looked as I applied for graduate school, though, this certainly was not the case.  Not only was I not accepted into a specific research group at the schools I was accepted to but I was also encouraged to try out different research projects before settling on a specific topic to pursue for my dissertation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, I do think there are some programs which may choose to accept or not accept you based on what you are thinking you would like to do for research (if you know).  This may be because of resources (for instances, if all the professors who are experts in what you want to do research in already have as many graduate students as they want/have money for, that may impact your acceptance and will impact your chances of doing the research you want should you be accepted).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can you pick and choose?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A specific answer to your question is going to vary significantly from school to school.  For example, some programs are very accommodating in letting students &quot;pick and choose&quot; what there want to do research in partly because there are currently resources available to allow that.  As explained above, because of resources or other considerations some programs may not have the luxury of opening all doors for its graduate students. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And could you choose whether to do the physical theory or the observations or coding simulations?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, this is very school-specific.  Some schools have immense computing resources and getting computer time to run simulations/heavy calculations is easy.  Other schools may have no computing resources available (other than your desktop workstation) and it will be more difficult to get the computing time you'll need for simulations since you will have to look outside your university for resources.  Some schools have access to a lot of time on some very big telescopes; some have access to telescopes that are specialized for specific topics in astrophysics; and other schools may have no access to telescopes at all, meaning you will have to find telescopes with allocation time you can apply for without having to belong to an institution affiliated with that telescope.  It varies from school to school.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, your ability to pursue specific research topics in astrophysics can be enhanced or hampered by the computing and telescope resources available at your department.  Take that into consideration as you apply to schools.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SUMMARY&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers to your questions vary greatly from department to department and are largely dependent on the resources available to a specific department.  I have found generally that there is great flexibility in astrophysics programs to choose a PhD dissertation topic but that certain research topics have more support and resources at some institutions than at others.  Also, at many institutions professors are limited in how many students they can advise by the amount of funding they have for PhD students, which can significantly affect the research options available to you.  There are, however, some institutions that have the resources to cover for an under-funded professor if a student does want to work for them, but again it just boils down to &lt;strong&gt;it varies from school to school&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2015-01-12T16:16:29.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8531" PostTypeId="1" AcceptedAnswerId="8535" CreationDate="2015-01-12T16:25:07.277" Score="10" ViewCount="259" Body="&lt;p&gt;In the &lt;a href=&quot;http://news.nationalgeographic.com/news/2015/01/150106-kepler-goldilocks-exoplanets-universe-space-science/&quot;&gt;National Geographic article&lt;/a&gt; about NASA's announcement of new planets discovered by the Kepler spacecraft, it states that there are 4,175 candidate planets. What is the process for vetting these candidate planets? How do we know they exist without knowing whether they are planets? Or are they candidates because the data hints that a planet might be there but we need more detailed data or more analysis before we can confirm it?&lt;/p&gt;&#xA;" OwnerDisplayName="user5291" LastActivityDate="2015-01-12T22:28:43.480" Title="What is the process for vetting the 4,175 candidate planets Kepler has discovered?" Tags="&lt;planet&gt;&lt;exoplanet&gt;&lt;kepler&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8532" PostTypeId="2" ParentId="6550" CreationDate="2015-01-12T19:11:54.700" Score="2" Body="&lt;p&gt;Standard solar models reproduce the sun's mass, radius, luminosity, spectral energy distribution, line spectrum, neutrino luminosity, and vibration spectrum (helioseismological) with great accuracy.  The age inferred for the sun lines up nicely with the age of the solar system inferred from isotope ratios in meteors.  The solar models are based on the same physics that produces stellar evolution models, which fit the color-magnitude diagrams of clusters of various ages and compositions in impressive detail. In other words, the standard picture of solar and stellar structure is accepted because it explains, in detail, a stupendous amount of richly textured observational detail.  Manuel's theory, by contrast, explains nothing in particular.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a reference, I'd point the interested reader at any of the many excellent astronomy textbooks aimed at the distribution-requirement college courses.  &lt;/p&gt;&#xA;" OwnerUserId="5293" LastActivityDate="2015-01-12T19:11:54.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8533" PostTypeId="2" ParentId="8526" CreationDate="2015-01-12T20:01:04.740" Score="9" Body="&lt;p&gt;No, the galactic magnetic field is very weak, about 0.1nT.&#xA;It is able to bend the trajectory of highly-energetic charged particles and also to align dust grains across the magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, is too weak to affect the rotation of a galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the origin of galactic magnetic field is not clear yet, the supermassive black holes do not significantly influence the galactic magnetic field far from them.&lt;/p&gt;&#xA;" OwnerUserId="4135" LastEditorUserId="1386" LastEditDate="2015-01-16T15:57:52.547" LastActivityDate="2015-01-16T15:57:52.547" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8534" PostTypeId="2" ParentId="8525" CreationDate="2015-01-12T20:09:26.620" Score="16" Body="&lt;p&gt;Nothing &quot;escapes&quot; a BH - in the sense that a signal originating inside the event horizon remains forever inside. If something is observed moving away from the BH, then it was generated outside the event horizon. If it was generated inside, it would never be observed at all, forever and ever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity itself does not &quot;escape&quot; a BH - and neither does &quot;not escape&quot;. Gravity is simply a characteristic of the metric of spacetime. If spacetime is warped in a certain way, gravity can be measured to exist. A BH is simply a very powerful distortion of spacetime, nothing more, nothing less. It is generated by a concentration of mass/energy, which warps spacetime, and then that concentration becomes trapped by this distortion that it has produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that sense, gravity is simply part of the BH, because gravity is spacetime being warped, and because a BH is essentially just that - warped spacetime. A BH's gravitational field is part of the BH itself, extending to infinity (but getting weaker with distance). It doesn't &quot;escape&quot; because there's nothing there in the process of escaping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's like having a plastic bag tied into a knot to keep water inside, and someone asks &quot;so how does the plastic escape the knot?&quot; The plastic does not &quot;escape&quot; the knot, the knot is part of the plastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This all becomes easier to understand when you realize that gravity is not a thing, it's just an effect of spacetime being distorted.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: I think what you were really asking was - can a BH have its own magnetic field? The answer is yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A BH can have &lt;a href=&quot;http://en.wikipedia.org/wiki/No-hair_theorem&quot; rel=&quot;noreferrer&quot;&gt;3 characteristics: mass, spin (rotation), and electric charge (a.k.a. the no-hair theorem)&lt;/a&gt;. All other characteristics of the matter falling in it are lost, except these three. If you drop a proton into a neutral BH, then the BH acquires a charge equal to one proton, and that's a measurable electric field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider a spinning BH with an electric charge, &lt;a href=&quot;http://en.wikipedia.org/wiki/Kerr%E2%80%93Newman_metric&quot; rel=&quot;noreferrer&quot;&gt;the Kerr-Newman metric&lt;/a&gt;. You have a charge, and you have spin. That means you have magnetism. So, yes, a BH can have a magnetic dipole. However, the rotation axis and the magnetic dipole axis must be aligned - a BH cannot be seen as &quot;pulsing&quot;. Again, no signal from inside the event horizon can be observed outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you should not imagine the electric (or magnetic, same thing) field as &quot;escaping&quot; the BH. It does not escape. What happens is, when the charges were swallowed by the BH, the lines of electric field remain &quot;glued&quot; to the BH, which then acquires a charge. Those lines of electric field have existed forever, they don't &quot;escape&quot; anything, and continue to exist after the charge is trapped by the BH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: electric fields and magnetic fields are one and the same. One could appear to be the other, depending on the motion of the observer.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2017-04-26T22:58:27.433" LastActivityDate="2017-04-26T22:58:27.433" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8535" PostTypeId="2" ParentId="8531" CreationDate="2015-01-12T22:28:43.480" Score="6" Body="&lt;p&gt;Planet &quot;candidates&quot; are Kepler Objects of Interest (KOIs)that have a transit-like light curve and have passed a number of observational tests. They are candidates, because although they do show a transit in the light curve of the star in question, there is no independent confirmation of a planetary mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One problem to overcome is that of &quot;false positives&quot;. There are other astrophysical phenomena besides planets that can mimic an exoplanetary transit in a Kepler light curve. For instance a grazing incidence stellar binary can produce what looks like a transit in a light curve; so too can a chance alignment of a star with a background eclipsing binary star. There are also a number of instrumental anomalies, such that a signal in one part of a CCD or the Kepler field can produce a &quot;ghost&quot; at another position that might look like a transit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can obtain detailed spectroscopy including accurate radial velocity measurements you can usually rule out most of these false positives, largely by getting a mass constraint for the companion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even where you have a very clean transit signal and can estimate a planetary radius, a further problem is that a wide range of masses produce objects with very similar radii. ie. more massive brown dwarfs have very similar radii to exoplanets. Again, only a mass estimate, either through radial velocity measurements or sometimes through &quot;transit timing variations&quot; if the object is in a multiple exoplanet systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the problem with the KOIs is many of these stars are way too faint ($V&amp;gt;14$) to do the kind of detailed spectroscopy that is easily possible on the much brighter exoplanets around stars found in say the HATNET or WASP ground-based surveys of bright stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what you can do is tackle the problem statistically, by identifying the kinds of false-positives you might have, quantifying their influence, and throwing away suspect objects (see for instance &lt;a href=&quot;http://arxiv.org/pdf/1202.5852v1.pdf&quot;&gt;Batalha 2012&lt;/a&gt;). Section 4 of this paper describes in detail some of the tests that are done: e.g. comparing the depths of odd and even numbered transits to look for asymmetries that would indicate grazing incidence binaries; looking between transits for a secondary eclipse that would also indicate a stellar companion; the shape of the transit is diagnostic but cannot easily rule out a planetary candidate; searching for motion of the &quot;photocenter&quot; of the source - if the photocenter moves during a transit it could indicate a background, diluted, eclipsing binary lightcurve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An early paper by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2011ApJ...738..170M&quot;&gt;Morton &amp;amp; Johnson (2011)&lt;/a&gt; claimed on the basis of a population synthesis approach, that that astrophysical false positives were limited to less than 10%. However, a recent paper by &lt;a href=&quot;http://arxiv.org/abs/1401.1240&quot;&gt;Coughlin et al. (2014)&lt;/a&gt; discusses how instrumental effects can be tested for by comparing the transit periods with the periods of other known objects in the Kepler field of view. They claim that around 30% of the KOIs may in fact be false positives. Either way it looks like the big majority of the KOIs are indeed exoplanets, but identifying which ones aren't will require detailed follow-up.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-01-12T22:28:43.480" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8536" PostTypeId="2" ParentId="8527" CreationDate="2015-01-12T22:47:26.520" Score="2" Body="&lt;p&gt;You appear to have a thing about linking the magnetic field generated by Jupiter to that around a black hole (note that a field does not &quot;belong&quot; to an object) - the two things are entirely different, the only connection being that magnetic fields are generated in both cases by currents caused by the motion of charged particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Jupiter's case, the magnetic field is (likely) generated via a dynamo process, involving currents that flow in its metallic hydrogen interior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a black hole, as material travels inwards, it forms an accretion disk around the black hole (because angular momentum must be conserved), in which magnetic field can be generated and amplified. The twisting of the magnetic field lines around the rotation axis of the black hole &lt;em&gt;could&lt;/em&gt; create a kind of magnetic funnel lined up with the rotation axis, along which charged particles can be accelerated by a strong magnetic field gradient. The field lines would not look like you have shown for Jupiter, they would be twisted into a helical topology along the rotation axis. This is part of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Blandford%E2%80%93Znajek_process&quot; rel=&quot;nofollow&quot;&gt;Blandford-Znajek effect&lt;/a&gt;, which I do not claim to properly understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to your question is maybe, but its very much a topic of contemporary research.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-02-06T14:59:09.737" LastActivityDate="2015-02-06T14:59:09.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8537" PostTypeId="2" ParentId="8518" CreationDate="2015-01-13T11:45:55.023" Score="1" Body="&lt;p&gt;It would have to be a big meteorite to supply the quantaties we have here and therefore support the nuclear break down theory.ie- hydrogen to helium etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In respect of delivery by cosmic objects, the atmospheric pressure on the moon would result in a low boiling point and the evaporation velocity(due to low gravitational attraction) would exceed the escape velocity of the moon and meet with no atmospheric resistance.&#xA;This is the way I see it.&#xA;The body of water, although spread over the surface, would consider itself as an independent body and have its own &quot;existence parameters&quot; with regards to its environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ability of Mars to retain water vapour is much reduced by its lower gravitational strength and atmospheric density. Can water exist at these parameters continuously, the evidence shows not.&lt;/p&gt;&#xA;" OwnerUserId="5295" LastEditorUserId="2365" LastEditDate="2017-04-15T11:28:40.050" LastActivityDate="2017-04-15T11:28:40.050" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8538" PostTypeId="2" ParentId="1611" CreationDate="2015-01-13T15:15:43.510" Score="1" Body="&lt;p&gt;Water molecules are bipolar, a bit like a bar magnet. For this reason they form non-covalent bonds with other molecules. This makes both the breaking up and making of strong covalent chemical bonds easier. Water is not only a fluid in the meaning that things float in it. Water also increases the chemical turnover and this is important for biological processes where different molecules need to find each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bipolarity of water helps make the right amino acids find their place along an RNA molecule, simply by facilitating more tests per millisecond until the right one comes along and sticks. Methane, for example, is not a bipolar molecule. It does not facilitate chemical recombinations in the way water does. Life would have a hard time to originate and survive in liquid methane. There are of course other bipolar volatile molecules than water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really know what I'm talking about here, but I heard someone who seems to do so... And &lt;a href=&quot;https://www.youtube.com/watch?v=41_Ne5mS2ls&amp;amp;list=PLC049FDC3EE7B9B78&amp;amp;index=5&quot; rel=&quot;nofollow&quot;&gt;animations like this&lt;/a&gt; help my impression that bipolarity is important in order to make the right stuff come together sooner rather than later.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-01-13T15:21:12.493" LastActivityDate="2015-01-13T15:21:12.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8539" PostTypeId="2" ParentId="8525" CreationDate="2015-01-13T16:36:58.253" Score="6" Body="&lt;p&gt;In &lt;em&gt;theory&lt;/em&gt;, a charged and rotating black hole can generate its own magnetic field. The magnetic (and electric) field can exist and can be measured outside the event horizon of the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I completely agree with both existing answers that magnetic field does not &quot;escape&quot; from black holes, however I would argue that it is &lt;em&gt;extremely unlikely&lt;/em&gt; that any real astrophysical black hole generates a significant magnetic field. The simple reason for this is that is is extremely difficult to see how any realistic physical process would deposit material with a net charge inside the black hole. i.e. &lt;strong&gt;Most astrophysical black holes are expected to be uncharged and have no magnetic field.&lt;/strong&gt; (Though there are at least a couple of astronomers who think otherwise - see  &lt;a href=&quot;http://adsabs.harvard.edu/abs/2003ApJ...596L.203R&quot; rel=&quot;noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2003ApJ...596L.203R&lt;/a&gt; ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnetic fields you are thinking of, and which are referred to in the link that you provide, are fields that are generated within the accretion disk of material that is spiralling in towards the event horizon. i.e. They are generated &lt;em&gt;outside&lt;/em&gt; the black hole, and are completely unrelated to the magnetic field that you show for a planet like Jupiter, where the field is generated by processes inside the planet.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-01-13T21:50:29.253" LastActivityDate="2015-01-13T21:50:29.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8540" PostTypeId="1" CreationDate="2015-01-13T16:58:49.090" Score="7" ViewCount="191" Body="&lt;p&gt;NASA measured 43,13 arc seconds per century. General relativity predicts 42,98 arc seconds per century.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I try to find out what the parameters' values such as $G$, $M_{sun}$, $\omega_{min}$ at aphelion radius, etc. used to get this exact 42,98.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a new model developed by R. Plamondon, we get exactly 43,13 and also it matched perfectly with Mercury's period of revolution around Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is the following. When I put my parameters into Einstein metric, I got something unexpected.&lt;/p&gt;&#xA;" OwnerUserId="5298" LastEditorUserId="2153" LastEditDate="2015-01-13T19:06:33.483" LastActivityDate="2015-02-03T01:36:54.607" Title="What are the exact physical parameters used to calculate Mercury precession with Einstein theory?" Tags="&lt;general-relativity&gt;&lt;mercury&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8541" PostTypeId="1" AcceptedAnswerId="8546" CreationDate="2015-01-13T20:33:58.510" Score="5" ViewCount="4547" Body="&lt;p&gt;So last night I was about to sleep with the window in front of me. there was a 3 inch gap between the curtains that let me see a small portion of the sky. I could see something magnificently bright in the sky that couldn't be anything but Venus. So while I was looking at it for about 1-2 minutes it gradually moved out of my sight. That seemed pretty fast. I mean I kinda estimated that with such a speed it could move from east to west in less than 2 hours.&#xA;So is that normal?&lt;/p&gt;&#xA;" OwnerUserId="5300" LastActivityDate="2020-01-29T15:07:01.790" Title="How fast does Venus move as seen from the earth?" Tags="&lt;venus&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8542" PostTypeId="2" ParentId="8510" CreationDate="2015-01-13T20:56:41.023" Score="0" Body="&lt;p&gt;There would be a tide of approximately 28 days due to centrifuge effect,the water being free flowing,about the mid point centre of gravity of the 2 masses.&#xA;Our tides are monthly but appear daily due to earth 24hr rotation.Variation in sub-marine land masses and ocean baison cross flow have sub tidal effect that can lag or lead the moon phased tide.On a water depth of 2-4km it may,as an independent body, go towards the outside of the rotational system leaving one side dry so to speak.&lt;/p&gt;&#xA;" OwnerUserId="5295" LastActivityDate="2015-01-13T20:56:41.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8546" PostTypeId="2" ParentId="8541" CreationDate="2015-01-14T06:31:22.370" Score="5" Body="&lt;p&gt;The Earth rotates on its axis once every 24 hours. (Well, actually a little less, by about 17 minutes, but close enough.) That's $360^\circ$ in 24 hours, or about $15^\circ$ per hour. $15^\circ$ is 30 times the width of the Sun or Moon. The Sun moves exactly that fast (on average), the stars a smidgen faster, and the Moon a tad slower. (The Moon lags behind the stars by approximately its own width every hour.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Venus is never more than about $45^\circ$ from the Sun, it always sets at most 3 hours after the Sun does, and rises at most 3 hours before.&lt;/p&gt;&#xA;" OwnerUserId="4077" LastActivityDate="2015-01-14T06:31:22.370" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8550" PostTypeId="2" ParentId="8386" CreationDate="2015-01-14T10:56:30.827" Score="1" Body="&lt;p&gt;There are many refractors/reflectors that can be obtained ready made and be more useful as the interest develops&#xA;You would be wise to get the largest objective you can afford,although an interesting project kit it is minimal in aperture(ref: Galscope)&lt;/p&gt;&#xA;" OwnerUserId="5295" LastActivityDate="2015-01-14T10:56:30.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8551" PostTypeId="2" ParentId="270" CreationDate="2015-01-14T11:25:03.893" Score="0" Body="&lt;p&gt;We did an exercise on this particular subject at Secondary Modern School in 1958.&#xA;With reference to&quot;Patrick Moor&quot; publications a 3-dimentional model wasmade of the nearest 50 stars.&#xA;Particular reference to the &quot;well known&quot; ie:the plough.As the model evolved,those we are familiar with loose their familiarity because of the distance.The plough is familar if assumed all the same distance,they are not and the model highlights the lack of pattern if viewed from Orion's Belt.&#xA;Our Sun did not appear to fit into any&quot;Constellation type pattern&quot;from any angle and all familiar ones were lost when viewed anywhere else.&#xA;An interesting &quot;Experiment&quot; though.Try it.Cheers Colin&lt;/p&gt;&#xA;" OwnerUserId="5295" LastActivityDate="2015-01-14T11:25:03.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8552" PostTypeId="2" ParentId="8529" CreationDate="2015-01-14T12:41:56.157" Score="1" Body="&lt;p&gt;I assume that what you want to know is &quot;where did everything come from?&quot; As for the light and the matter itself, we have some pretty good theories: &lt;strong&gt;Matter&lt;/strong&gt; consists of many different particles, each of which were created from energy at different epoch during the first few moments of the Big Bang (BB). For instance, leptons were created from 1 to 10 seconds after BB, hadrons from 10&lt;sup&gt;-6&lt;/sup&gt; to 1 seconds after BB, and quarks even earlier. &lt;strong&gt;Light&lt;/strong&gt; was created when the different particles annihilated with their corresponding antiparticles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to how the original energy &amp;mdash; from which light and matter came &amp;mdash; was created in the first place, there definitely is no well-grounded theory here, but either we must accept that&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;  &lt;li&gt;the Universe came into existence out of nothing, or else&lt;/li&gt;&#xA;  &lt;li&gt;there was something before that created the Universe, in which case we either&#xA;    &lt;ol&gt;&#xA;      &lt;li&gt;just push the question further back, or&lt;/li&gt;&#xA;      &lt;li&gt;must accept that whatever was there before has been there for eternity.&lt;/li&gt;&#xA;    &lt;/ol&gt;&#xA;  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Option 2.2 is boring because I don't think there's any way of testing this theory, so option 1 or 2.1 seem most appealing, especially option 1, since time itself is thought to have been created along with space (though I think there are models where this is not necessarily the case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Creating something out of nothing seems magical, but in fact &lt;strong&gt;it happens all the time&lt;/strong&gt;. A vacuum is the lowest state of energy, but as a consequence of Heisenberg's uncertainty principle, in a finite time, the lowest state of energy cannot be zero, since &amp;Delta;E &amp;Delta;t &amp;#8819; &amp;hbar;. This results in so-called quantum fluctuations. Usually, the energy popping up out of nowhere annihilates immediately, but there are ways of making it survive (e.g. on the border of a black hole). In a similar way, universes can be thought to pop up out of nowhere. Such a universe could annihilate immediately, but if inflated fast enough to a large enough size, it could grow to macroscopic size before it has time to annihilate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to read further, Wikipedia has some nice articles on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chronology_of_the_universe&quot; rel=&quot;nofollow&quot;&gt;timeline of Big Bang&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmogony&quot; rel=&quot;nofollow&quot;&gt;cosmogony&lt;/a&gt; (the theory of stuff coming into being)&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-01-14T12:47:12.730" LastActivityDate="2015-01-14T12:47:12.730" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8553" PostTypeId="1" AcceptedAnswerId="8556" CreationDate="2015-01-14T17:42:54.910" Score="4" ViewCount="796" Body="&lt;p&gt;According to Draine in his book &quot;Physics of the interstellar and intergalactic medium&quot; (page 210/211) and &lt;a href=&quot;http://ned.ipac.caltech.edu/level5/March01/Dinerstein2/Diner2.html#2.2&quot; rel=&quot;nofollow&quot;&gt;Caltech&lt;/a&gt; [O III] line ratio's (amongst others) are a good density probe, but I cannot really understand why that is the case. Can someone explain me how this works? It doesn't have to be numerical or with detailed formulas, just the idea what is going on. &lt;/p&gt;&#xA;" OwnerUserId="1309" LastActivityDate="2015-01-14T20:52:21.617" Title="Why is [O III] a good density probe in interstellar medium?" Tags="&lt;interstellar-medium&gt;&lt;density&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8556" PostTypeId="2" ParentId="8553" CreationDate="2015-01-14T20:52:21.617" Score="3" Body="&lt;p&gt;The key to understanding this is the concept of &lt;a href=&quot;http://en.wikipedia.org/wiki/Forbidden_mechanism&quot; rel=&quot;nofollow&quot;&gt;forbidden lines and forbidden transitions&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A forbidden transition is one that cannot occur radiatively via an electric dipole interaction. Instead, it must proceed either through magnetic dipole or electric quadrupole emission, with &lt;em&gt;much&lt;/em&gt; lower probability, or the transition may be accomplished through collisional (de)excitation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A forbidden line is the radiation at a wavelength corresponding to a forbidden transition. In laboratory plasmas these are usually not seen - hence the term forbidden lines - because the transitions are normally accomplished through collisional de-excitation on a timescale much shorter than the radiative lifetime through magnetic dipole/electric quadrupole emission. However, in astrophysical plasmas the densities can be many orders of magnitude lower than even the best laboratory vacuums. In this case, forbidden lines &lt;em&gt;are&lt;/em&gt; seen and indeed are often an important means of radiative cooling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The property of forbidden lines which makes them an excellent density-sensitive tool is that they can be &quot;quenched&quot; if the density becomes large enough to make collisional de-excitation more probable than radiative de-excitation. The strength of the forbidden line is thus sensitive to the electron density (which is what dominates the collisions) between densities where &quot;quenching&quot; starts to become effective and higher densities where the line essentially becomes unobservable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the specific example of OIII (doubly ionised oxygen) optical line emission. There are three forbidden transitions of interest between the $^1D_2 \rightarrow\ ^3P_2$ (501 nm), $^1D_2 \rightarrow\ ^3P_1$ (496 nm) and $^1S_0 \rightarrow\ ^1D_2$ (436 nm) states.&#xA;These transitions produce optical forbidden lines which are quenched at different characteristic densities. Measuring a line strength &lt;em&gt;ratio&lt;/em&gt; is important because the ratio will be &lt;em&gt;independent&lt;/em&gt; of the abundance of OIII. The ratio will depend on electron density, providing the electron density is $&amp;gt;10^{5}\ cm^{-3}$, &lt;em&gt;and&lt;/em&gt; temperature (at lower densities, the ratio is just temperature dependent).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the density in this case requires more information - usually supplied by a similar line ratio for something like NII, which has a similar set of forbidden lines, but with differing density and temperature dependence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more straightforward route is to use the line ratio for forbidden lines of OII or SII where there are a closely spaced pair of energy levels, both undergoing forbidden transitions to the same lower level. In this case the ratios are density sensitive, but not temperature sensitive, again over a range of densities where one or other of the transitions is in the process of being quenched (e.g. for OII 372.6/372.8nm, $10^2 &amp;lt;n_e &amp;lt; 10^{6}\ cm^{-3}$.)&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-01-14T20:52:21.617" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8558" PostTypeId="2" ParentId="8442" CreationDate="2015-01-14T21:54:34.290" Score="2" Body="&lt;p&gt;I figured I might as well expand my earlier comment into an answer, as I've cobbled together some more information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're talking about the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cyclic_model&quot; rel=&quot;nofollow&quot;&gt;cyclic model&lt;/a&gt;, which states that the universe goes through cycles of Big Bangs and Big Crunches. We live in the middle of one of those cycles. Albert Einstein was one of the first to investigate it, but his efforts came to nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory has been reincarnated, most recently in 2002 by &lt;a href=&quot;http://www.space.com/2372-recycled-universe-theory-solve-cosmic-mystery.html&quot; rel=&quot;nofollow&quot;&gt;Paul Steinhardt and Neil Turok&lt;/a&gt;. They propose that the cosmological constant has changed over time, which could make the accelerating expansion of the universe fit into the theory. Another good article on their idea can be found &lt;a href=&quot;http://www.nature.com/news/2006/060501/full/news060501-8.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Roger Penrose has also come up with a theory - &lt;a href=&quot;http://www.ideasroadshow.com/issues/sir-roger-penrose-2013-07-12&quot; rel=&quot;nofollow&quot;&gt;Conformal Cyclic Cosmology&lt;/a&gt; - along similar lines as Einstein's. You can use &lt;a href=&quot;https://en.wikipedia.org/wiki/Conformal_cyclic_cosmology&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; as a starting point for more reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A related topic is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Bounce&quot; rel=&quot;nofollow&quot;&gt;Big Bounce&lt;/a&gt;, which attempts to address the Big Bangs and Big Crunches.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2015-01-14T21:54:34.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8561" PostTypeId="1" AcceptedAnswerId="8563" CreationDate="2015-01-15T01:35:01.537" Score="5" ViewCount="234" Body="&lt;p&gt;I've always been stumped with the question of what lies beyond the universe, and I've come to the firm conclusion - void (i.e. completely empty).  Does it make sense then to think of the universe as a finite 'blob' of atoms (the completely full) flying around each other in nothing?  Furthermore, and more importantly, does it have a shape like a sphere?&lt;/p&gt;&#xA;" OwnerUserId="5306" LastEditorUserId="5306" LastEditDate="2015-01-15T02:02:40.213" LastActivityDate="2015-01-15T07:59:49.853" Title="Is the universe simply a mass of atoms in 'void'; what does it look like?" Tags="&lt;universe&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8562" PostTypeId="2" ParentId="8561" CreationDate="2015-01-15T07:50:02.237" Score="2" Body="&lt;p&gt;It is not known. The universe has a finite age so we cannot see beyond the speed of light times that age (and taking the Hubble expansion into account as well). This means there is a 'visible universe' which defines a boundary of what we can observe. Information does not travel faster than light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole universe is bigger than the visible universe, but we don't know how much bigger (it could even be infinite). However, as the universe contains all space ánd time, the stuff outside the universe does not contain both (try to wrap your head around that!) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I mentioned it could be infinite, we could live on the three dimensional surface of a four dimensional ball (like an ant on a balloon) or our universe could be embedded in a higher dimensional multiverse (like string theory tells us). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We probably will never know because we cannot see past the visible universe.&lt;/p&gt;&#xA;" OwnerUserId="5308" LastActivityDate="2015-01-15T07:50:02.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8563" PostTypeId="2" ParentId="8561" CreationDate="2015-01-15T07:53:15.517" Score="1" Body="&lt;p&gt;No, beyond the universe does not lie anything. Not even space itself. You cannot be outside the universe because there is nothing to be in. The three (or four) dimensions you think of as space, do not exist outside of the Universe. According to the Big Bang theory not only matter was created during the big bang but most importantly &lt;strong&gt;space&lt;/strong&gt; was created during the Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.exploratorium.edu/origins/cern/ideas/bang.html&quot; rel=&quot;nofollow&quot;&gt;CERN&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There's another important quality of the Big Bang that makes it&#xA;  unique. While an explosion of a man-made bomb expands through air, the&#xA;  Big Bang did not expand through anything. That's because there was no&#xA;  space to expand through at the beginning of time. Rather, physicists&#xA;  believe the Big Bang created and stretched space itself, expanding the&#xA;  universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is also why an inflationary phase where space expanded faster than the speed of light was possible. Matter cannot travel faster than the speed of light, but space can expand faster than the speed of light.&#xA;So it is possible for two galaxies to 'appear' to have a relative velocity larger than the speed of light because in actual fact the space between those galaxies is expanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the shape of the Universe: You can only assign a shape to something relative to the space it is in. As there is no space outside the Universe, the answer to your question would be that the Universe has no shape.&lt;/p&gt;&#xA;" OwnerUserId="864" LastEditorUserId="864" LastEditDate="2015-01-15T07:59:49.853" LastActivityDate="2015-01-15T07:59:49.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8564" PostTypeId="1" CreationDate="2015-01-15T08:46:59.310" Score="0" ViewCount="464" Body="&lt;p&gt;Generally a telescope can memorize how to point to a target with a particular RA&amp;amp;Dec. There is no need to calibrate every day.&lt;br/&gt;&#xA;How can it happen?&lt;br/&gt;&#xA;How to let the telescope(non-amateur) know the position of a star after it was built? For an amateur telescope, how to realize it? There should be a zero point? &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How often should the calibration be?&lt;br/&gt;&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2015-01-26T13:55:39.313" Title="the initial astrometry calibration for a telescope" Tags="&lt;telescope&gt;&lt;astrometry&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8565" PostTypeId="1" AcceptedAnswerId="8566" CreationDate="2015-01-15T08:56:22.210" Score="2" ViewCount="7837" Body="&lt;p&gt;I've recently been seeing a bunch of pictures of nebulae and I'm just fascinated by their beauty and complexity. Is there any kind of telescope that would make it possible to view it from here on Earth?&lt;/p&gt;&#xA;" OwnerUserId="5310" LastActivityDate="2020-10-27T20:09:42.877" Title="How can I see a nebula?" Tags="&lt;universe&gt;&lt;observational-astronomy&gt;&lt;observable-universe&gt;&lt;nebula&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8566" PostTypeId="2" ParentId="8565" CreationDate="2015-01-15T10:17:42.183" Score="12" Body="&lt;p&gt;Yes, indeed! Many nebulae are visible from Earth in a small and cheap telescope, and even to the naked eye (if you are standing in a sufficiently dark place).&lt;/p&gt;&#xA;&lt;p&gt;In fact, yesterday I was watching the Orion Nebula with my 4.5&amp;quot; telescope (which is worth $200 or so) from my apartment in the middle of Copenhagen.&lt;/p&gt;&#xA;&lt;p&gt;The term &amp;quot;nebula&amp;quot; is a bit of a… well, nebulous term, as it covers several distint phenonomena, for instance:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Galaxies&lt;/strong&gt;,           e.g. the Andromeda Galaxy (though this is an old expression)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;(Open) stellar clusters&lt;/strong&gt;,   e.g. the Eagle Nebula&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Globular clusters&lt;/strong&gt;,  e.g. 47 Tucanae&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Planetary nebulae&lt;/strong&gt;,  e.g. the Eskimo Nebula&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Supernova remnants&lt;/strong&gt;, e.g. the Crab Nebula&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;However, you should be aware that even in a large and expensive telescope, the nebulae do not look anything like the beautiful images you find on the internet. With your eye, you will merely see diffuse, whitish clouds. The beautiful colors arise only in images taken through telescopes with long exposure times. If you want to make such pictures, you will need a telescope with a motor (since the telescope must follow the rotation of the sky) and the possibility of attaching a camera to it.&lt;/p&gt;&#xA;&lt;p&gt;People are often disappointed when shown a nebula through a telescope. I think the beauty of it lies in knowing what it is that you are seeing, knowing for instance that it lies 600 lightyears away, that you are looking 600 years back in time, and that stars are right now being formed there.&lt;/p&gt;&#xA;&lt;p&gt;Googling something like &amp;quot;nebulae visible in a small telescope&amp;quot; should get you started. Good luck!&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2020-10-27T08:21:46.190" LastActivityDate="2020-10-27T08:21:46.190" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="8567" PostTypeId="1" CreationDate="2015-01-15T11:21:33.647" Score="24" ViewCount="12139" Body="&lt;p&gt;If light keeps travelling in a straight line, why can't we see distant galaxies with the naked eye? Surely if you stared long enough, the light from them would eventually hit your eye? I apologize if this is a dumb question:)&lt;/p&gt;&#xA;" OwnerUserId="5313" LastActivityDate="2018-09-23T22:11:23.333" Title="Why can't we see distant galaxies with the naked eye?" Tags="&lt;galaxy&gt;&lt;observational-astronomy&gt;&lt;light&gt;&lt;observable-universe&gt;" AnswerCount="8" CommentCount="3" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8568" PostTypeId="2" ParentId="8567" CreationDate="2015-01-15T12:22:27.323" Score="4" Body="&lt;p&gt;Your reasoning would be valid not only for galaxies, but also for stars and anything it shines in the Universe, but there is an important effect which invalidates it: absorption of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intergalactic and interstellar medium is filled with dust and gas, which contributes to absorb and scatter the light from distant objects.&#xA;Especially on the plane of our Galaxy, we still have plenty of gas and dust (Milky Way is a relatively young galaxy): indeed, to look at distant object we try to orient our telescopes towards the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lockman_Hole&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lockman Hole&lt;/a&gt;, whenever it is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is especially valid for low frequencies light: at higher energies, scattering and absorption of X-ray and gamma-rays from standard amount of absorbing material is negligible (even if, the more distant you look at, the younger the objects, the more is the dust and gas available which is still not locked in stars).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Olbers%27_paradox&quot; rel=&quot;nofollow noreferrer&quot;&gt;Olbers' paradox&lt;/a&gt;, which indicates for an expanding Universe to account for the &quot;dark sky&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastActivityDate="2015-01-15T12:22:27.323" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8569" PostTypeId="2" ParentId="8567" CreationDate="2015-01-15T12:41:37.753" Score="14" Body="&lt;p&gt;Not at all a dumb question, but actually you &lt;em&gt;can&lt;/em&gt; see distant galaxies with the naked eye. From the northern hemisphere, the Andromeda Galaxy, our biggest neighboring galaxy, is visible if you know where to look, and is at a reasonably dark place. From the southern hemisphere, the two smaller, but nearer, irregular galaxies called the Small and Large Magellanic Clouds are visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that more distant galaxies are not visible, is due to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Inverse-square_law&quot;&gt;inverse-square law&lt;/a&gt;: As the light particles (photons) recede from the galaxy (or any other light source), they are distributed over an ever-increasing surface. That means that a detector (e.g. your eye) of a given area will catch less photons, the farther it is placed from the galaxy. The law says that if in a time interval &amp;Delta;t on average it detects, say, 8 photons at a distance D, then in the same time interval, at a distance 2D it will detect 8/2&lt;sup&gt;2&lt;/sup&gt; = 2 photons. At a distance of 4D, it will detect 8/4&lt;sup&gt;2&lt;/sup&gt; = 0.5 photons. Or, equivalently, it will need twice the time to detect a single photon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line is that in principle you can see the very distant galaxies, but the photons are so few and arrive so rarely, that your eye is not a good enough detector. The benefit of a telescope is that 1) it has a larger area than your eye, and 2) you can put a camera at its focal point instead of your eye and take a picture with a large exposure time, i.e. increasing the &amp;Delta;t.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2015-01-15T12:41:37.753" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8570" PostTypeId="2" ParentId="8567" CreationDate="2015-01-15T13:48:25.577" Score="21" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Surely if you stared long enough, the light from them would eventually hit your eye?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Collecting light over a long span of time is how telescopes can see very dim objects. The human visual system doesn't work that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one thing, even when you think you are staring at something, your eyes still dance around a bit. It's a built-in response called ocular microtremors. These microtremors appear to be an essential part of making the vision system functional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For another, your eye does not and cannot collect light for arbitrarily long periods of time (the way a photographic telescope can). There's an immense amount of signal processing that happens in the eye and along the way to the brain. This signal processing depends on light being collected for short spans of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our vision system evolved to see food, friends, and dangers under well-lit conditions. We are very good at seeing motion in broad daylight. We're not so good at seeing stationary objects, and we're not good at all at seeing barely visible sources under very dark skies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naked eye astronomy is limited by the nature of the human vision system. The most remote object we can see is the Triangulum Galaxy, and that's only under conditions of extremely dark and extremely clear skies.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-01-15T13:48:25.577" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8571" PostTypeId="1" AcceptedAnswerId="8572" CreationDate="2015-01-15T14:54:05.483" Score="3" ViewCount="705" Body="&lt;p&gt;As far as I know, scientists seeking extraterrestial life are simply looking for plantets containing water in liquid state. To be honest, I find this weird.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that life - in the form we know on Earth - cannot exists without water. But how can one proof that such life isn't but a special case out of a bunch of other different types of life?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, I think we can come out with at least one example of a &quot;organism-like thingy&quot; that would not require liquid water. Well, we could build a simple robot, whose &quot;natural ecosystem&quot; would be a junkyard (for mechanical, electric and electronic garbage). That robot would would fuel itself with solar energy, and reproduce by seeking out appropiate parts in the junkyard and building a copy of itself from them. Of course, that's just a theoretical example, I know it has many faoults, but I still hope the idea is clear - it should be indeed possible for life to exists without water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also heard (sorry, I can't remember the source) that theoretically, there could be an organism whose metabolism would rely on sulphur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, unless scientists can proove that no life can exists without water (any life; not just Earth-like life), why do they keep looking just for water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless, of course, we define &quot;life&quot; as &quot;any organism functioning like Earth organisms; therefore, metabolising organic structures with enzymes, built of protein, requiring water, etc&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW. This question is a little bit different than &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1611/why-is-liquid-water-considered-a-requirement-for-life&quot;&gt;Why is liquid water considered a requirement for life?&lt;/a&gt; . The answers there were pointing out that water is a good solvent. And I'm (also) asking why is any similar solvent considered needed. See my example above.&lt;/p&gt;&#xA;" OwnerUserId="5314" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-01-16T20:24:03.927" Title="Why do scientists think life requires water?" Tags="&lt;life&gt;&lt;extra-terrestrial&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2015-01-18T14:56:50.457" ContentLicense="CC BY-SA 3.0" />
  <row Id="8572" PostTypeId="2" ParentId="8571" CreationDate="2015-01-15T15:20:10.917" Score="6" Body="&lt;p&gt;When you lose your keys on a dark night you look where the probability of finding them is highest (which is a combination of being likely to see them with their probability of being there), which is not necessarily where they are most likely to be. In the case of the search for extra-terrestrial life we have a fairly good idea what the signature of carbon based life using water as its primary solvent (conventional life) looks like. We also know that such life is possible. Other unconventional forms of life could exist but we have a very poor idea of what signature to look for, so we devote most of our effort to looking for the signature we will recognise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do not need to prove that unconventional life is impossible in order to spend most effort on searching for the conventional. It is sufficient to suspect that the probability of finding such life is higher than unconventional just because the probability of detection is so much higher. If we find nothing looking for conventional life, then a larger proportion of the effort will be devoted to searching for the unconventional.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-01-15T21:24:23.933" LastActivityDate="2015-01-15T21:24:23.933" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8573" PostTypeId="2" ParentId="6398" CreationDate="2015-01-15T20:24:33.123" Score="0" Body="&lt;p&gt;A much easier way to think about this is to consider clouds in the sky. They contain hundreds to thousands of gallons of water molecules but they are very spread out. Same as a hydrogen cloud before it creates a star. When you compact molecules in a tighter space you get a stronger magnetic field in the near vicinity of the object. The amount of molecules in a specific space dictate the &quot;strength&quot; of the magnetic field. When a star explodes it does lose mass but what remains is compacted into an infinitely small space to create a stronger magnetic field. A Swarzchild radius is calculated by using existing mass of a star and seeing how much space do we need to cram that amount of mass into to overcome the speed of light but it does not consider how much mass remains after a supernova explosion. And as for &quot;forever and ever&quot;... Our ideas are constantly changing. Remember, we used to think the planet was flat...I hope this helps, aloha.&lt;/p&gt;&#xA;" OwnerDisplayName="user5320" LastActivityDate="2015-01-15T20:24:33.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8574" PostTypeId="1" CreationDate="2015-01-15T20:28:35.347" Score="4" ViewCount="223" Body="&lt;p&gt;Ignoring any current beliefs, is it possible that the planets could have been captured by the Sun's gravitational force after drifting through space while retaining their axis of spin and speed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notably, Venus uniquely has retrograde rotation; how would this have come about based on current theories?&lt;/p&gt;&#xA;" OwnerUserId="5295" LastEditorUserId="2153" LastEditDate="2015-01-16T00:48:10.093" LastActivityDate="2015-01-17T03:45:38.053" Title="Could the planets in the Solar System have been captured?" Tags="&lt;solar-system&gt;&lt;rotation&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8575" PostTypeId="2" ParentId="2241" CreationDate="2015-01-15T20:30:47.367" Score="9" Body="&lt;p&gt;I think the source of confusion between the two concepts - the Big Bang singularity and an infinite universe - is the misconception that the universe &lt;em&gt;began&lt;/em&gt; as a finite expanse originally. This misconception easily arises from analogies using present-day logic and numbers that were not applicable in the early universe. For example, I've heard it said that shortly after the Big Bang, the entire observable universe was the size of a grapefruit, but that explanation neglects to mention that &lt;em&gt;grapefruits would have been much larger then.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that space is where we can measure how large something is, but space expands, so something that is a certain distance away currently was a lot closer a long time ago, even if neither object has moved in the normal sense. As an analogy to help illustrate the effect:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You and I are standing on a preposterously large deflated balloon. You set down a meter stick, make a mark on the balloon at each end and we each stand on one mark and are now a meter apart. Then I turn on a pump and start inflating the balloon. As the balloon inflates, the surface stretches out and you and I appear to get farther from each other, when though we're not 'moving' (e.g. walking away from each other): now we have conflicting sets of information to consider; according to the marks on the balloon surface we're still one meter apart, but according to the meter stick in your hand (which is not expanding) the distance is greater than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that while I called the balloon &quot;preposterously large,&quot; it could have been &lt;em&gt;infinitely&lt;/em&gt; large and still behave the same way. I point this out because I've seen in comments on other answers that you don't see how space could be both infinite and expanding - that if it's expanding, then it must have been previously finite. That is incorrect: in fact, because infinity is the quality of unboundedness, something that is infinitely large can always get bigger, because &lt;em&gt;by definition&lt;/em&gt; there is no upper bound on its size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that if you recorded the earlier analogy in reverse, it would appear that space was shrinking such that a several-meters distance between us reduced over time to one meter. If you continue shrinking the universe in such a manner, it eventually becomes the case that there is &lt;em&gt;zero&lt;/em&gt; distance between us. And if you apply that to a scenario where there are people infinitely distributed across the balloon, all of them would come closer together as the balloon deflated, until there was zero distance between any two people... in theory, at least, since real human beings have size. Energy and space don't have size, however, so at the point of the Big Bang, space was still infinite (since an infinite/unbounded space cannot shrink to become finite/bounded) but the distance between any two points in space was zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you could go back in time to the Big Bang you'd see an infinite ocean of energy, since all the energy was &quot;shoulder-to-shoulder&quot; (infinitely dense) but it rapidly expands (and therefore cools) to the point that basic particles can form, then later matter and molecules. Of course since your size would depend on the metric of space, it wouldn't necessarily look like space was expanding, but simply like the energy and matter were cooling down. In fact we still see this as an effect of spatial expansion in the redshift of light from distant sources: the light &quot;cools down&quot; or loses energy along the way because it is stretched out on its journey through space.&lt;/p&gt;&#xA;" OwnerUserId="4042" LastActivityDate="2015-01-15T20:30:47.367" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8576" PostTypeId="2" ParentId="7999" CreationDate="2015-01-15T22:15:04.583" Score="2" Body="&lt;p&gt;The answer is yes. The mass (and hence brightness) of meteorites follows some sort of power-law relationship such that there are many more small particles than big lumps of rock. This means that there are indeed many more faint meteorites than bright ones, though I am struggling to find any detailed study that goes below the magnitude range you can see with the naked eye. This paper by &lt;a href=&quot;http://adsabs.harvard.edu/abs/1966CoSka...3...75K&quot; rel=&quot;nofollow&quot;&gt;Kresakova (1966)&lt;/a&gt; seems thorough and representative. It suggests that the number of meteorites goes up by a factor of 3 for each unit increase in astronomical magnitude (i.e. decreasing brightness).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper referred to by Barry Carter in his comments is &lt;a href=&quot;http://adsabs.harvard.edu/abs/1980MNRAS.193..645C&quot; rel=&quot;nofollow&quot;&gt;Cook et al. (1980)&lt;/a&gt;, which finds that log of the &lt;em&gt;cumulative&lt;/em&gt; number of meteors is proportional to about half the astronomical magnitude. i.e. &#xA;$$\frac{d \log \phi}{dm} \simeq 0.5 $$&#xA;This means that the number of meteorites increases by a factor of $10^{0.5}$ for each unit increase in magnitude - i.e. also about 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a separate issue as to whether you see more with night vision goggles on. I guess you would because these would have light amplification properties that would enable you to see fainter objects - that's the whole point of them.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-01-15T22:34:26.703" LastActivityDate="2015-01-15T22:34:26.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8577" PostTypeId="2" ParentId="16" CreationDate="2015-01-16T04:24:31.630" Score="8" Body="&lt;p&gt;RE: Always seeing the same side of the moon.&#xA;It may also be attributed to a difference in density of the moon.  It was established recently that there are gravitational variations around the moon that would support the possibility that there is a higher density of material on one side of the moon.  The higher density area of moon would then always face the earth.  This would be similar to taping a heavy weight to one side of a ball on a hard surface.  The heavier side of the ball will always face down towards the floor and towards the gravitational force.  This is the same result of the heavier side of the moon facing towards the gravitational force emanating from the earth.  It's too much of a coincidence that the moon rotates perfectly to always keep the same side facing towards the earth.  Especially since the moon is moving away from the earth and is changing it's distance and time to revolve around the earth constantly.  I came up with this theory about 10 years ago and I have not seen any other theories that would convince me that this is not a real possibility. This theory would also explain why Mercury always faces the same side towards the sun.  There are no tides on Mercury to affect how it is orientated and it would appear that both bodies are affected by the same forces in keeping the same sides facing the same direction. &lt;/p&gt;&#xA;" OwnerUserId="5325" LastActivityDate="2015-01-16T04:24:31.630" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8579" PostTypeId="1" CreationDate="2015-01-16T06:04:02.373" Score="7" ViewCount="12642" Body="&lt;p&gt;I have a little 76mm Newtonian Reflector. I want to see the moon at daytime. Is it safe to look at sky at daytime when sun is shinning or there is any risk of sun.&lt;/p&gt;&#xA;" OwnerUserId="1441" LastEditorUserId="1309" LastEditDate="2015-01-16T14:11:18.433" LastActivityDate="2015-01-16T20:16:13.923" Title="Is it safe to view moon at daytime through telescope" Tags="&lt;the-moon&gt;&lt;telescope&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8580" PostTypeId="2" ParentId="8579" CreationDate="2015-01-16T06:30:21.250" Score="4" Body="&lt;p&gt;As long as you don't directly look at the sun through your telescope, then there is no problem. But don't try to see the moon during solar eclipse without special filters (easily available for your telescope's eyepieces)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In  my telescope, I have seen the moon day time when there was very light cloud cover. It was awesome scene. You can feel that the moon is spherical by observing its terminator and the lighted edge. You can easily make out the dark side edge.&lt;/p&gt;&#xA;" OwnerUserId="4084" LastActivityDate="2015-01-16T06:30:21.250" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8581" PostTypeId="2" ParentId="8579" CreationDate="2015-01-16T09:49:24.853" Score="1" Body="&lt;p&gt;I have found a good time to view the moon is before the sky is fully light,say 8am.The pupil of the eye will no be fully dilated and better eye/lens will give improved resolution(f ratio).In addition to this an improved contrast will be achieved as the atmosphere has less reflected light relative to the optical system of the telescope.&#xA;A friend and I agreed more detail  of the moon could be seen to the naked eye under these conditios&lt;/p&gt;&#xA;" OwnerUserId="5295" LastActivityDate="2015-01-16T09:49:24.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8583" PostTypeId="2" ParentId="8567" CreationDate="2015-01-16T11:04:41.620" Score="4" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Few photons -- You have tiny pupils. Only photons that manage to travel that far over that much distance along a path that manages to intersect with your tiny pupils will have a chance of being seen. And only some photons that reach your retina actually interact with molecules that register their arrival.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Interference -- The molecules of the atmosphere, dust in the atmosphere, reflection/refraction of and in your eye, dust in the solar system, the Oort cloud, interstellar dust in our galaxy, dust in intergalactic space, any molecule at all along the path, all may absorb any of the few photons and re-emit them in a different direction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Stability -- Telescopes, especially such as Hubble, can be really, really still compared to your eyes. Not only do your eyes constantly make tiny shifts, but you breathe and your heart beats and other things keep very dim images from being able to form.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Exposure -- The very first Hubble Deep Field image was collected over &lt;a href=&quot;http://www.spacetelescope.org/science/deep_fields/&quot; rel=&quot;nofollow&quot;&gt;around 100 hours of exposure&lt;/a&gt;. You might find that difficult with your eyes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Retention -- Exposure time affects how much 'data' is retained about where photons have hit the recording surface. Your eyes will not remember that a photon registered at a receptor even a minute earlier. Your eyes are not good at all for 'still photography'.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Light pollution/universal expansion -- The universe has been expanding for billions of years. As it expands, light propagating through space 'stretches' more to the red end of the visible spectrum. For distant galaxies this effectively means that the visible light from them has shifted far enough to be infrared and invisible by the time it gets here. Now, ultraviolet light would also shift, some of it becoming 'visible'. But it then starts getting mixed with any scattered 'light pollution' effects once it gets to our atmosphere. Your eyes aren't good at all in keeping track of which photons come from what sources.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are probably other factors, but maybe those are more than enough to indicate how big the problem is. Note that the early 100-hour Hubble image was a big surprise to astronomers. Even with the large light-gathering telescopes previously available, they were unable to get enough light for useful data. That earlier equipment had much larger pupils than yours, more sensitive imaging surfaces and could 'sit still' much longer than you; and it still had difficulty with distant galaxies.&lt;/p&gt;&#xA;" OwnerUserId="2543" LastActivityDate="2015-01-16T11:04:41.620" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8584" PostTypeId="2" ParentId="8567" CreationDate="2015-01-16T16:26:10.987" Score="4" Body="&lt;p&gt;Just because you can keep your eyelids open for $x$ seconds doesn't mean you are collecting light for $x$ seconds and using it to form a single image in your brain. How would you &quot;save&quot; the photo? How would you decide when to end light collection? You know as well as I do that you can't simply lift your finger off your brain's shutter release!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that's on top of all the other factors as explained in other answers (but I wanted to belabour this particular point a little further than do the other answers.).&lt;/p&gt;&#xA;" OwnerUserId="4155" LastActivityDate="2015-01-16T16:26:10.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8585" PostTypeId="2" ParentId="8574" CreationDate="2015-01-16T18:27:48.807" Score="2" Body="&lt;p&gt;I'll address the second question first, &lt;em&gt;Venus uniquely has retrograde rotation; how would this have come about based on current theories?&lt;/em&gt; The answer is that there are a number of hypotheses related to Venus's rotation. See the related question &lt;a href=&quot;https://astronomy.stackexchange.com/questions/26&quot;&gt;What is the current accepted theory as to why Venus has a slow retrograde rotation?&lt;/a&gt;. None of them involve capture of a rogue planet. Note that this also applies to Uranus; it too has an odd orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why not? Note that I am now addressing your first question, &lt;em&gt;is it possible that the planets could have been captured by the Sun's gravitational force after drifting through space while retaining their axis of spin and speed?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is simple. Capture is extremely unlikely (and that's putting it mildly), and it is not needed to explain the odd orientations of Venus and Uranus.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-01-17T03:45:38.053" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8586" PostTypeId="1" CreationDate="2015-01-16T19:50:48.143" Score="0" ViewCount="283" Body="&lt;p&gt;With reference to the Ebb and Flow lunar satellites, was there any data about its geology? Is there a central core, or is it homogenous?&lt;/p&gt;&#xA;" OwnerUserId="5295" LastEditorDisplayName="user2449" LastEditDate="2015-01-17T06:54:29.993" LastActivityDate="2015-01-19T02:40:15.273" Title="Do we have knowledge of the geology of the moon?" Tags="&lt;the-moon&gt;&lt;geology&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8587" PostTypeId="2" ParentId="8579" CreationDate="2015-01-16T19:53:32.083" Score="1" Body="&lt;p&gt;It is safe to observe anything during the day, as long as you don't point the telescope close to the Sun. Just stay away from the Sun and you'll be fine. Be careful to not accidentally swing the tube in the direction of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. Mercury is pretty hard to observe at other times than during the day (or dusk), and yet astronomers have been observing it for a very long time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One trick that works is to stay in the shade of a tall building, therefore screening off the Sun, but obviously in such a manner that the object you want to observe is still visible. You can almost always do this, unless the object is very close to the Sun, in which case you may want to cancel that observation altogether.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-01-16T20:16:13.923" LastActivityDate="2015-01-16T20:16:13.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8588" PostTypeId="2" ParentId="8571" CreationDate="2015-01-16T20:24:03.927" Score="2" Body="&lt;p&gt;All biological life requires an electrolyte.Water is the only none reactive solution that can carry the solute,(salts,acids,alkali,) that would cause reactive change.&#xA;Water has limits in its ability to remain the solution(evaporation:thereby restricting chemical mobility,)or (crystalisation/freezing:thus causing precipitation of chemicals in the &quot;solid state&quot;.)&#xA;Biological life ceases to evolve.&lt;/p&gt;&#xA;" OwnerUserId="5295" LastActivityDate="2015-01-16T20:24:03.927" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8589" PostTypeId="1" AcceptedAnswerId="8597" CreationDate="2015-01-17T02:09:52.843" Score="3" ViewCount="4035" Body="&lt;p&gt;I'd like to calculate the distance (in arcsec) from the object nearest to ra='08h55m10s' dec='-7d14m42s' to that point on the celestial sphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am actually having trouble identifying the object. I have python code to run a cone search: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from astropy.coordinates import SkyCoord&#xA;from astroquery.irsa import Irsa&#xA;import astropy.units as u&#xA;&#xA;loc = SkyCoord('08h55m10s',' -7d14m42s','icrs')&#xA;table = Irsa.query_region(coordinates=loc,catalog=&quot;wise_allsky_4band_p3as_psd&quot;, radius= 1 * u.arcminute)&#xA;print(table)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This prints:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;designation        ra     dec   ...       angle         id&#xA;                  deg     deg   ...        deg            &#xA;------------------- ------- ------- ... ------------------ ---&#xA;J085507.72-071428.4 133.782  -7.241 ... 291.84348499999999   0&#xA;J085511.22-071522.2 133.797  -7.256 ... 155.64652000000001   1&#xA;J085508.65-071354.0 133.786  -7.232 ... 337.39116000000001   2&#xA;J085509.94-071534.8 133.791  -7.260 ... 180.87171799999999   3&#xA;J085511.28-071528.1 133.797  -7.258 ... 157.48249000000001   4&#xA;J085509.47-071502.9 133.789  -7.251 ...         200.548395   5&#xA;J085513.60-071440.0 133.807  -7.244 ... 87.949470000000005   6&#xA;J085510.83-071442.5 133.795  -7.245 ... 92.439341999999996   7&#xA;J085511.39-071427.5 133.797  -7.241 ...          55.110833   8&#xA;J085513.30-071420.4 133.805  -7.239 ... 66.332006000000007   9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And this is where I am stumped. Which of these objects is nearest to the point? (I assume it must be the object in the zero index). And why? And how can I calculate the distance from the object to that point on the celestial sphere?&lt;/p&gt;&#xA;" OwnerDisplayName="user5341" LastActivityDate="2015-01-18T05:57:17.483" Title="How to calculate distance from object to point on celestial sphere" Tags="&lt;distances&gt;&lt;coordinate&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8590" PostTypeId="2" ParentId="8586" CreationDate="2015-01-17T05:19:19.353" Score="4" Body="&lt;p&gt;We know a considerable amount about the moon, but the picture is not complete.  There are considerable resources available and so much data available that it would take too long to list every aspect of it, so instead the following are examples of resources (with annotations):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our knowledge of lunar geology comes from the results of direct sampling (from rocks brought back) and orbiting satellite observations.  Also, with inferences and comparisons with the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://lunar.ksc.nasa.gov/science/geologys.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA's Lunar Geology page&lt;/a&gt;, this page approaches lunar geology by focussing on the 2 main sub-branches, the geochemistry and geophysics of the moon.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The geochemistry focuses on the crustal composition, volcanic petrology and the existence of polar ice, hydrogen and helium.  The geophysics focuses on the crustal magnetism, &lt;a href=&quot;https://earthscience.stackexchange.com/q/989/889&quot;&gt;moonquakes&lt;/a&gt;, outgassing, crustal features.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of which help scientists develop a picture of the geological history and the geological structure of the moon.  Exploration of the moon's geological features in comparisons with Earth, are described as described in the online chapter &lt;a href=&quot;http://explanet.info/Chapter04.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Geology of the Moon&lt;/a&gt; as a geological 'Rosetta Stone'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of the internal structure of the moon, observations from direct sampling, sensors on the lunar surface and from satellite data have built up a reasonable picture of what the internal geology of the moon looks like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the NASA page &lt;a href=&quot;http://www.nasa.gov/topics/moonmars/features/lunar_core.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA Research Team Reveals Moon Has Earth-Like Core&lt;/a&gt;, re-analysis of Apollo mission seismic data has revealed that there are several similarities between the lunar and Earth internal structure (as seen below):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/pw1Pz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Caption from NASA: An artist's rendering of the lunar core as identified in new findings by a NASA-led research team. (NASA/MSFC/Renee Weber)&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As can be seen in the diagram (and explained in all the links), the moon's internal structure is thermochemically layered, &lt;em&gt;broadly similar&lt;/em&gt; to Earth (there are considerable differences).  But the main point, in answer to your main question, the interior of the moon is not homogenous, but layered with a central core.&lt;/p&gt;&#xA;" OwnerDisplayName="user2449" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:29.267" LastActivityDate="2015-01-17T05:19:19.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8591" PostTypeId="2" ParentId="8589" CreationDate="2015-01-17T09:17:25.763" Score="2" Body="&lt;p&gt;The shortest angular distance between two points on the celestial sphere is measured along a great circle that passes through both of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation your program is using is&#xA;$$ \cos \theta = \sin \delta_1 \sin \delta_2 + \cos \delta_1 \cos \delta_2 \cos (\alpha_1 - \alpha_2), $$&#xA;where $(\alpha_1, \delta_1)$ and $(\alpha_2, \delta_2)$ are the right ascension and declination of the two points on the sky (expressed in degrees, then converted to radians).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of output from your program its harder to help, especially as you have not given the full output. The first three columns should be self explanatory - columns 2 and 3 are just the coordinates expressed in degrees. I am guessing that the third column from last, that you haven't shown, gives the separation between your object and the catalogue objects and that the penultimate column is the angle that the great circle between the points makes on the sky. That looks right because the fourth object in the list has almost the same RA as your input object but  is 67 arcseconds due south (i.e. at 180 degrees).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-01-17T09:17:25.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8592" PostTypeId="1" CreationDate="2015-01-17T12:21:21.817" Score="2" ViewCount="181" Body="&lt;p&gt;I am reading Roobert Hooke's paper &lt;em&gt;An attempt to prove the motion of the earth from observations&lt;/em&gt; (1674, &lt;a href=&quot;http://echo.mpiwg-berlin.mpg.de/ECHOdocuView?url=/mpiwg/online/permanent/library/XXTBUC3U/pageimg&amp;amp;start=11&amp;amp;mode=imagepath&amp;amp;pn=18&amp;amp;ww=0.7142&amp;amp;wh=0.7142&amp;amp;wx=0.0277&amp;amp;wy=0.1065&quot; rel=&quot;nofollow&quot;&gt;faksimile&lt;/a&gt;). Hooke writes&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The principal dayes of doing which will be about the 4 of April, when our Zenith passeth by the said Star at midnight&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The said star is Gamma Draconis, and the observations are made in London. As far as I see it, the star is closest to the Zenit at 6 in the morning in London at April 4th.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why does Hooke say &lt;em&gt;at midnight&lt;/em&gt;? Is this an expression of the 17th century? &lt;/p&gt;&#xA;" OwnerUserId="1293" LastActivityDate="2015-11-22T02:36:34.853" Title="When is Gamma Draconis closest to the zenith in London on April 4th" Tags="&lt;telescope&gt;&lt;positional-astronomy&gt;&lt;parallax&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8593" PostTypeId="1" AcceptedAnswerId="8594" CreationDate="2015-01-17T12:41:39.367" Score="5" ViewCount="822" Body="&lt;p&gt;It's big. It's bright. It's white. It's so intense. But what is at the center of the Milky Way?&lt;/p&gt;&#xA;" OwnerUserId="5344" LastEditorUserId="2153" LastEditDate="2015-01-17T15:20:46.780" LastActivityDate="2015-01-18T20:25:25.027" Title="What is supposedly in the center of the Milky Way?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8594" PostTypeId="2" ParentId="8593" CreationDate="2015-01-17T15:17:12.993" Score="15" Body="&lt;p&gt;Photos of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_Center&quot; rel=&quot;nofollow noreferrer&quot;&gt;galactic center&lt;/a&gt; aren't too bright because of all the gas and dust between us and it. &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/1/11/Galactic_Cntr_full_cropped.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;For example&lt;/a&gt; (in infrared):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/1/11/Galactic_Cntr_full_cropped.jpg&quot; alt=&quot;Galactic center&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing, though, that you're talking about other galaxies, because there are no views of the galactic center of the Milky Way face-on. Although the galactic center is pretty luminous, just not in the wavelengths we're used to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the galactic center has lots of stars - many massive, but a handful like our Sun. They're young, though, which is odd, as there isn't much star formation happening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main attraction, though, is &lt;a href=&quot;https://en.wikipedia.org/wiki/Sagittarius_A*&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sagittarius A*&lt;/a&gt;, a radio source that is apparently a &lt;a href=&quot;https://en.wikipedia.org/wiki/Supermassive_black_hole&quot; rel=&quot;nofollow noreferrer&quot;&gt;supermassive black hole&lt;/a&gt;. It's actually a subset of the more complex radio source &lt;a href=&quot;https://en.wikipedia.org/wiki/Sagittarius_A&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sagittarius A&lt;/a&gt;. So far, astronomers think that most galaxies have supermassive black holes in their centers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that it has to be really massive because of how it perturbs the &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/c/c1/Galactic_centre_orbits.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;orbits of stars nearby&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/c/c1/Galactic_centre_orbits.svg/1000px-Galactic_centre_orbits.svg.png&quot; alt=&quot;Star orbits near Sagittarius A*&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know it's massive and we know it's a very strong radio source. A supermassive black hole is just about the only thing it could be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Courtesy of WayfaringStranger: &lt;a href=&quot;https://apod.nasa.gov/apod/ap130906.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;An awesome picture from NASA&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://apod.nasa.gov/apod/image/1309/sgra_gasChandra.jpg&quot; alt=&quot;Awesome picture from NASA&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://arxiv.org/pdf/0808.2870v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt; by Ghez et al., the orbits of stars were monitored and a mass of the central object was determined to be $4.1 \pm 0.6 \times 10^6 M_{\odot}$. &lt;a href=&quot;http://www.aanda.org/articles/aa/pdf/2009/28/aa10922-08.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; paper discusses the properties of the star cluster itself.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:47.307" LastActivityDate="2015-01-18T20:25:25.027" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="8596" PostTypeId="2" ParentId="8564" CreationDate="2015-01-17T23:04:42.283" Score="2" Body="&lt;p&gt;After installation set up,the coordinates Ra and Dec are locked in to the ra of GMT(0 astronomically) and installation latitude.These are input requirements of installation.&#xA;From that point,an onboard clock remembers all points from that reference.&#xA;It is not a sat-nav base unit and installation would have to be repeated if repositioned in another location.&lt;/p&gt;&#xA;" OwnerUserId="5295" LastActivityDate="2015-01-17T23:04:42.283" CommentCount="1" CommunityOwnedDate="2015-01-17T23:04:42.283" ContentLicense="CC BY-SA 3.0" />
  <row Id="8597" PostTypeId="2" ParentId="8589" CreationDate="2015-01-18T05:10:22.783" Score="1" Body="&lt;p&gt;You can use &lt;code&gt;SkyCoord.separation&lt;/code&gt; to compute the separation between coordinates:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;c1 = SkyCoord('08h55m10s',' -7d14m42s', frame='icrs')  # your coords&#xA;c2 = SkyCoord(133.782, -7.241, unit='deg', frame='icrs')  # first object in table&#xA;sep = c1.separation(c2)&#xA;&#xA;print c1&#xA;&amp;lt;SkyCoord (ICRS): ra=133.791666667 deg, dec=-7.245 deg&amp;gt;&#xA;print c2&#xA;&amp;lt;SkyCoord (ICRS): ra=133.782208333 deg, dec=-7.24123611111 deg&amp;gt;&#xA;print sep&#xA;0d00m36.3947s&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The table you printed is truncated in between. If you enlarge your terminal emulator, you will be able to see a column called &lt;code&gt;dist&lt;/code&gt;, which is the distance between your input coordinates and the corresponding object in arcsecs:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;print table['dist']&#xA;       dist       &#xA;------------------&#xA;36.420772999999997&#xA;44.138542000000001&#xA;51.987606999999997&#xA;         52.838003&#xA;49.937266999999999&#xA;22.372699000000001&#xA;53.621690000000001&#xA;         12.375482&#xA;          25.27176&#xA;53.707365000000003&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then the nearest object is the one whose separation is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;print table['dist'].min()&#xA;12.375482&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Or equivalently, the one whose index is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;print table['dist'].argmin()&#xA;7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The separation is calculated using the equation that @Rob Jeffries pointed out.&lt;/p&gt;&#xA;" OwnerUserId="2339" LastEditorUserId="2339" LastEditDate="2015-01-18T05:57:17.483" LastActivityDate="2015-01-18T05:57:17.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8598" PostTypeId="1" CreationDate="2015-01-18T10:33:47.643" Score="9" ViewCount="264" Body="&lt;p&gt;Given the studies of the Sun and similar stars and the most impressive understanding of its physics, is there any kind of probability distribution for its dangerous behavior? There are also some historic data about events in &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_storm_of_1859&quot;&gt;1859&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/774%E2%80%93775_carbon-14_spike#cite_note-usoskin12-7&quot;&gt;775&lt;/a&gt;. What is the estimated probability of solar activity destroying our electric grid within this century?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-01-19T23:11:28.837" Title="How likely is it that the Sun will destroy our electric society?" Tags="&lt;the-sun&gt;&lt;solar-flare&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8599" PostTypeId="2" ParentId="8565" CreationDate="2015-01-18T11:23:27.520" Score="0" Body="&lt;p&gt;Some filters also help improve the view as seen through the telescope.&#xA;Even a simple and cheap red filter would help,as it will remove most part of the light pollution spectrum and will let Ha reach your eye :)&lt;/p&gt;&#xA;" OwnerUserId="5355" LastActivityDate="2015-01-18T11:23:27.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8601" PostTypeId="1" AcceptedAnswerId="8603" CreationDate="2015-01-18T15:48:40.360" Score="1" ViewCount="1216" Body="&lt;p&gt;I am to compute the temperature of planet--in fact I've already found out it's Earth--knowing only: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the surface temperature of Mars (210 K) and its distance from Sun (1.524 AU)&lt;/li&gt;&#xA;&lt;li&gt;and of course distance of Earth from Sun (1 AU). &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I have no data concerning the temperature of Sun, surfaces of the planet, etc. What formula should I use? Thank you.&lt;/p&gt;&#xA;" OwnerUserId="5358" LastActivityDate="2015-01-20T12:59:47.380" Title="Finding the temperature of Earth from temperature of Mars and its distance from Sun" Tags="&lt;solar-system&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8602" PostTypeId="1" CreationDate="2015-01-18T17:57:48.607" Score="3" ViewCount="86" Body="&lt;p&gt;I think it would be an interesting exercise to measure the latiude and longitude of my house using N occultation timings. I would like to see how precisely I can measure it using that technique, and compare it with the GPS readings from my phone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a reference which will tell me how to do the calculation of inferring my position from occultation timings, as accurately as possible.&lt;/p&gt;&#xA;" OwnerUserId="5359" LastActivityDate="2015-09-07T06:52:15.110" Title="How do I deduce my latitude and longitude from N obervations of occultations from the same place?" Tags="&lt;the-moon&gt;&lt;amateur-observing&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8603" PostTypeId="2" ParentId="8601" CreationDate="2015-01-18T18:55:26.943" Score="2" Body="&lt;p&gt;I might as well write up my comment into an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of my new favorite tools in my equations toolkit is the formula for &lt;a href=&quot;https://en.wikipedia.org/wiki/Effective_temperature&quot; rel=&quot;nofollow&quot;&gt;effective temperature&lt;/a&gt;:&#xA;$$T=\left( \frac{L(1-a)}{16 \pi \sigma D^2} \right)^{\frac{1}{4}}$$&#xA;where $T$ is the temperature of the planet, $L$ is the star's luminosity, $a$ is the planet's albedo, $\sigma$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_constant&quot; rel=&quot;nofollow&quot;&gt;Stefan-Boltzmann constant&lt;/a&gt;, and $D$ is the distance between the star and the planet. To find the star's luminosity, simply solve for $L$:&#xA;$$L=\frac{T^4 16 \pi \sigma D^2}{1-a}$$&#xA;where all the terms are for the first planet (Mars). Now, knowing $D$ for Earth, solve for $T$ on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All you have to know for this that &lt;em&gt;isn't&lt;/em&gt; in the question is albedo. That could be tricky, because it's not given. You &lt;em&gt;could&lt;/em&gt; assume that it's the same for both planets - and so solve not for $L$ but for $L(1-a)$ - but that might not be better than an approximation.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2015-01-18T18:55:26.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8604" PostTypeId="2" ParentId="8564" CreationDate="2015-01-18T20:11:11.693" Score="5" Body="&lt;p&gt;First you have enter your coordinates into the mount's software and then align it(e.g. 3 star alignment).After having done that, when you finish the observation you have to park the telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually &lt;a href=&quot;https://astronomy.stackexchange.com/a/8596/43&quot;&gt;Colin's answer&lt;/a&gt; is really helpful.They keyphrase is &quot;After installation set up&quot;.&#xA;The whole procedure should be explained in your telescope's manual.&#xA;If you tell us which model you have we might be able to help more.&lt;/p&gt;&#xA;" OwnerUserId="5355" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-01-19T12:12:05.947" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8605" PostTypeId="1" CreationDate="2015-01-19T00:19:16.353" Score="30" ViewCount="1602" Body="&lt;p&gt;Suppose there was an exact replica of our solar system 4.4 ly away (people included). What would we be able to detect and with what telescope(s)? Which planets? Could we detect radio transmissions and/or any atmospheres?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume detection would be optimal if we were co-planar with the other star's ecliptic, so what would we see in the best- and worst-case (90°-view?) scenarios?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Post Script: a few months later I asked something like this during a &lt;a href=&quot;http://www.ustream.tv/recorded/72967023#to01:02:14&quot; rel=&quot;noreferrer&quot;&gt;von Karman lecture by Neil Turner&lt;/a&gt;.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="2098" LastEditorUserId="2098" LastEditDate="2018-06-09T20:27:14.793" LastActivityDate="2020-12-16T18:18:44.917" Title="If Alpha Centauri A's solar system exactly mirrored our own, what would we be able to detect?" Tags="&lt;exoplanet&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="8606" PostTypeId="2" ParentId="8586" CreationDate="2015-01-19T02:27:45.800" Score="2" Body="&lt;p&gt;We actually know a number of elements present on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Moon&quot; rel=&quot;nofollow noreferrer&quot;&gt;lunar&lt;/a&gt; surface including oxygen, silicon, iron, magnesium, calcium, aluminium, manganese and titanium.  We have also observed that carbon and nitrogen have deposited over time from &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_wind&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar wind&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/e/e7/Composition_of_lunar_soil.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CmYLt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5362" LastEditorUserId="2153" LastEditDate="2015-01-19T02:40:15.273" LastActivityDate="2015-01-19T02:40:15.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8607" PostTypeId="1" AcceptedAnswerId="8608" CreationDate="2015-01-19T09:04:40.263" Score="3" ViewCount="64" Body="&lt;p&gt;As we (laymen) are told, comets lose gas and dust when they descent into the denser regions of the solar system. Seemingly so much and so large parts that they can even cause meteor showers like in the case of the Leonids from 55P/Temple-Tuttle. &#xA;Nevertheless we were also told, that the surface of 67P is covered with several centimeres of dust over a much harder surface. Where does the dust come from? The solar wind should have blown away any lose dust long ago, no? Is there a process which regenerates dust on the surface every cycle?&lt;/p&gt;&#xA;" OwnerUserId="844" LastEditorUserId="2153" LastEditDate="2015-01-19T21:23:51.243" LastActivityDate="2015-01-19T21:23:51.243" Title="Where does the dust on 67P/Churyumov-Gerasimenko come from?" Tags="&lt;comets&gt;&lt;dust&gt;&lt;67p&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8608" PostTypeId="2" ParentId="8607" CreationDate="2015-01-19T13:26:31.923" Score="1" Body="&lt;p&gt;Your assumption that solar wind should have blown the dust away is only valid for comets that have a close perihelion. But you are correct in that there is a process which regenerates dust.&lt;/p&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Comet_dust&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A comet will experience a range of diverse conditions as it traverses its orbit. For long period comets, most of the time it will be so far from the Sun that it will be too cold for evaporation of ices to occur. When it passes through the terrestrial planet region, evaporation will be rapid enough to blow away small grains, but the largest grains may resist entrainment and stay behind on the comet nucleus, beginning the formation of a dust layer. Near the Sun, the heating and evaporation rate will be so great, that no dust can be retained. Therefore, the thickness of dust layers covering the nuclei of a comet can indicate how closely and how often a comet's perihelion travels are to the Sun. If a comet has an accumulation of thick dust layers, it may have frequent perihelion passages that don't approach the Sun too closely.&lt;/p&gt;&#xA;&lt;p&gt;A thick accumulation of dust layers might be a good description of all of the short period comets, as dust layers with thicknesses of order meters are thought to have accumulated on the surfaces of short period comet nuclei. The accumulation of dust layers over time would change the physical character of the short-period comet. A dust layer both inhibits the heating of the cometary ices by the Sun (the dust is impenetrable by sunlight and a poor conductor of heat), and slows the loss of gases from the nucleus below. A comet nucleus in an orbit typical of short period comets would quickly decrease its evaporation rate to the point that neither a coma or a tail would be detectable and might appear to astronomers as a low-albedo near-Earth asteroid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="43" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-01-19T13:42:07.503" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8609" PostTypeId="2" ParentId="8564" CreationDate="2015-01-19T14:52:34.747" Score="4" Body="&lt;p&gt;Telescopes will generally come with a library of objects, which includes information on where they are located on the celestial sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two important pieces of information needed for the telescope to be able to locate objects from your Earthly location - Your position so that it knows its orientation relative to the celestial sphere, and the time so it can predict how that orientation will change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you initially set up your telescope, it will likely ask you first for the time, and then you will locate some easy to find objects. By finding three objects, the telescope will generally be able to understand its orientation relative to the celestial sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, it becomes trivial to find any object in its library, because it knows where the object is located, it now knows where it is, and it knows the time to adjust to changing in the Earth's orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if power down your telescope, it takes very relatively little energy to track time and saving your location information is quite cheap. If the telescope can assume you're viewing from a relatively similar location, and it has kept track of the time, it has all the information it will need to find objects as soon as it starts up. As Colin said, if you change locations, you would then need to recalibrate as one of the key pieces of information the telescope is using would have changed.&lt;/p&gt;&#xA;" OwnerUserId="1288" LastActivityDate="2015-01-19T14:52:34.747" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8611" PostTypeId="2" ParentId="8061" CreationDate="2015-01-19T20:50:29.983" Score="-1" Body="&lt;p&gt;Since the magnification is determined by focal length of the objective/focal length of eyepiece it is difficult to have an eye-piece with the same f ratio as the objective.&#xA;Assuming pupil aprature 0.25&quot; then the ideal eye-piece aperature will be 0.25&quot; for most effective light transmission of the eye.A generally acceptable f8 would have a focal length of (0.25x8)=2&quot;.&#xA;A 2&quot; eye piece x 100 magnification would require an objective focal length of 200&quot;,for f8 this means 200/8 which is 25&quot; diameter.This would give comparable brilliance at that magnification.&#xA;Any deviation from the ideal means loss of resolution due to lens errors of manufacture,the combination of these errors result higher arc&quot; resolution.&#xA;You will find a lower magnification more useful as the lower mag system may effect finer arc&quot;,res and additional light to eye/optical system efficiency.&#xA;There are no easy answers and sadley cost raises the barriers&lt;/p&gt;&#xA;" OwnerUserId="5295" LastActivityDate="2015-01-19T20:50:29.983" CommentCount="1" CommunityOwnedDate="2015-01-19T20:50:29.983" ContentLicense="CC BY-SA 3.0" />
  <row Id="8612" PostTypeId="2" ParentId="8598" CreationDate="2015-01-19T23:03:44.027" Score="6" Body="&lt;p&gt;I should give credit here to @honeste_vivere, who pointed out to me today that there have been recent studies excited by some extremely large coronal mass ejections that were classed as &quot;near misses&quot; in terms of causing major disruption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of particular interest to you would be the event of July 2012 discussed by &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/swe.20097/abstract&quot;&gt;Baker et al. (2013)&lt;/a&gt;. I quote from the paper&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot; Had the season and time of day for this CME passage been right on striking the Earth, the world would have witnessed a storm larger (possibly much larger) than the 1859 Carrington event. This most likely would have had devastating consequences for many technological systems &quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then in the discussion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&#xA;It is the opinion of the authors that our advanced technological society was very fortunate, indeed, that the 23 July solar storm did not occur just a week or so earlier.&#xA;Had the storm occurred in mid-July 2012, the Earth would have been directly targeted by the CME and an unprecedentedly large space weather event would have resulted. In fact, there is very legitimate question of whether our society would still be “picking up the pieces” from such as severe event [see NRC, 2008].&#xA;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks likely that such  events are perhaps things that occur every few decades. &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1029/2011SW000734/abstract&quot;&gt;Riley et al. (2012)&lt;/a&gt; suggest something as powerful as the Carrington event has a 12% chance of occurring in the next decade, with a 1% chance of something several times bigger. Even more powerful events are seen on other stars, where both flare and CME energies follow a power law relationship of the form $dN/dE \propto E^{-\alpha}$, where $\alpha \sim 2.5$ (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2013ApJ...764..170D&quot;&gt;Drake et al. 2013&lt;/a&gt;). Thus much larger, but rarer flares are possible and also seen on (usually) much younger and faster rotating stars. There may however be a very significant sporadic tail to the solar flare energy distribution too. Recent observations of candidate &quot;superflares&quot; on otherwise unremarkable solar-type stars have been reported, though the reality of these and the mechanisms are still being explored. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013ApJS..209....5S&quot;&gt;Shibayama et al. (2013)&lt;/a&gt; There appear to be a group, consisting of 1% of stars that show (repeated) superflares &lt;em&gt;more than 100 times the size of the Carrington event&lt;/em&gt;. These appear a little less likely around stars rotating as slow as the sun, but the bottom line is that, if there is nothing &quot;special&quot; about these stars, then these events occur every 800-5000 years (for events of energy 1e34 - 1e35 ergs) on a slow-rotating G-star. (Though you need to read carefully - it is probably a function of temperature and rotation). The authors note that superflares may be associated with the formation of very large starspots (or starspot groups). So that may be our clue to duck and cover.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-01-19T23:11:28.837" LastActivityDate="2015-01-19T23:11:28.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8613" PostTypeId="2" ParentId="8061" CreationDate="2015-01-20T03:02:36.710" Score="2" Body="&lt;p&gt;Light pollution does not matter for the Moon. Even transparency doesn't matter that much. What does matter is &lt;a href=&quot;http://en.wikipedia.org/wiki/Astronomical_seeing&quot; rel=&quot;nofollow&quot;&gt;seeing, a.k.a. air turbulence&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very rare that an exit pupil smaller than 0.5 mm is useful for anything - perhaps for some tight double stars, but that's about it. So take that as a hard lower limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of a &quot;soft&quot; limit, it depends. If seeing is excellent, if the optics are very good, if the instrument is in &lt;a href=&quot;http://www.cloudynights.com/documents/primer.pdf&quot; rel=&quot;nofollow&quot;&gt;perfect collimation&lt;/a&gt; and at &lt;a href=&quot;http://www.garyseronik.com/?q=node/55&quot; rel=&quot;nofollow&quot;&gt;perfect thermal equilibrium&lt;/a&gt;, then you can push e.p. below 1 mm, especially for a small aperture like the range you mention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If seeing is less than perfect, if optics are so-so, if the instrument is miscollimated or too hot, then the minimum usable e.p. goes up. Each one of these factors influences the observation, so it's hard to pin a number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line is, use what works for you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;claims that 1-2mm works best for the moon. Is that correct?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2 mm seems too big for a small 4...6&quot; aperture when observing the Moon. It might be okay if you want to capture the whole disk, but it seems large if you're interested in small details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, it's all relative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a 6&quot; reflector with a self-made primary mirror of excellent quality. I always take care to collimate the scope, and to keep it at thermal equilibrium. I live in California, so seeing is often good. It's not uncommon for me to use less than 1 mm exit pupil when observing the Moon. I use 0.84 mm as a starting point, sometimes pushing it down as low as 0.6 mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I use an e.p. of 1.1 mm for the Moon that means seeing is pretty bad - but in those cases I'd rather do something else, instead of being frustrated by bad seeing. But at that e.p. in a wide eyepiece (82 degrees apparent field) you see the whole Moon at once, which is always nice when you do a public demo (sidewalk astronomy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For much larger apertures, the ideal exit pupils tend to be somewhat larger even when observing the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To predict seeing, go on the &lt;a href=&quot;http://cleardarksky.com/csk/&quot; rel=&quot;nofollow&quot;&gt;Clear Sky Chart&lt;/a&gt; site, find a location near your home, and look at the third row in the chart (labeled Seeing); when it's dark blue, seeing is good.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the largest practical e.p., don't worry about it. The Moon is so bright, losing light is not something you should pay any attention to at all.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-01-20T19:26:01.227" LastActivityDate="2015-01-20T19:26:01.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8614" PostTypeId="2" ParentId="8601" CreationDate="2015-01-20T12:22:56.580" Score="1" Body="&lt;p&gt;If you use no additional data you assume that in all respects other than distance from the star the planets are the same. Then assuming thermal equilibrium the energy radiated bt the planet is proportional to the stellar intensity at the planets distance. Also assume that the Stefan–Boltzmann law applies to the energy radiated from the planets (so it's proportional to the fourth power of the temperature). The intensity of the stellar radiation at the planets distance is inversely proportional to the square of the distance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us the equation:&#xA;$$&#xA;\frac{T_1^4}{T_2^4}=\frac{I_1}{I_2}=\frac{R_2^2}{R_1^2}&#xA;$$&#xA;which simplifies to:&#xA;$$&#xA;T_1=\sqrt{\frac{R_2}{R_1}}T_2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now using the data provided we get:&#xA;$$&#xA;T_1=\sqrt{1.524}\times 210 \approx 259\ \text{K}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is of course much less than the average temperature of the Earth since not all else is the same in reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the significant differences is the Earth has a more massive atmoshpere than Mars, and it has a significant grenhouse effect on the temprature of the Earth. Also the Earth has a higher albedo than Mars which would reduce the equilibrium temperature from that calculated. I expect there are other factors that also contribute but I will leave those to others to mention.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-01-20T12:59:47.380" LastActivityDate="2015-01-20T12:59:47.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8615" PostTypeId="2" ParentId="6644" CreationDate="2015-01-20T15:40:47.873" Score="6" Body="&lt;p&gt;I know that quoting Wikipedia is frowned upon here, but as there has been no other answer posted, this is what the Wikipedia article on Kepler has to say about the matter:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;He then set about calculating the entire orbit of Mars, using the geometrical rate law and assuming an egg-shaped ovoid orbit. After approximately 40 failed attempts, in early 1605 he at last hit upon the idea of an ellipse, which he had previously assumed to be too simple a solution for earlier astronomers to have overlooked. Finding that an elliptical orbit fit the Mars data, he immediately concluded that all planets move in ellipses, with the sun at one focus—Kepler's first law of planetary motion. &lt;strong&gt;Because he employed no calculating assistants, however, he did not extend the mathematical analysis beyond Mars.&lt;/strong&gt; By the end of the year, he completed the manuscript for Astronomia nova, though it would not be published until 1609 due to legal disputes over the use of Tycho's observations, the property of his heirs.[Caspar][Koyré]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[1] Caspar M., &lt;em&gt;Kepler&lt;/em&gt;, pp. 131–140;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3] Koyré A., &lt;em&gt;The Astronomical Revolution&lt;/em&gt;, pp. 277–279&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-01-20T15:47:05.670" LastActivityDate="2015-01-20T15:47:05.670" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8616" PostTypeId="1" CreationDate="2015-01-20T16:07:16.997" Score="4" ViewCount="100" Body="&lt;p&gt;Saturn was formed around billion years ago along with Jupiter. After which it had enough time to form moon which can break into rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then why is it said that Saturn's ring are of recent origin - about few million years ago?&lt;/p&gt;&#xA;" OwnerUserId="1981" LastActivityDate="2015-01-20T16:46:49.720" Title="Recent origin of Saturn's rings" Tags="&lt;planet&gt;&lt;saturn&gt;&lt;planetary-ring&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8617" PostTypeId="2" ParentId="8616" CreationDate="2015-01-20T16:46:49.720" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Rings_of_Saturn#Formation_of_main_rings&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; claims that Saturn's rings may be as old as Saturn itself; &lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2002/12feb_rings/&quot; rel=&quot;nofollow&quot;&gt;NASA&lt;/a&gt; says that they could be only a few hundred million years old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 4.something-billion-years-old fits in with one of the two theories of the rings' formation - that Saturn swept up material from the protoplanetary disk. The date makes sense, because there's no way the disk would last for billions of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The newer date &lt;em&gt;does&lt;/em&gt; fit in with the moon-torn-apart hypothesis, which is perhaps the dominant hypothesis. The hypothetical progenitor moon could have been torn apart at any point between Saturn's birth and the present day. &quot;A few hundred million years ago&quot; fits in reasonably well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should add that all the theories surrounding the rings' formation have reasonable evidence in their favor, though there's quite a lot for the 4.something-billion-years theory (see &lt;a href=&quot;http://www.space.com/23949-saturn-rings-age-cassini-spacecraft.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, &lt;a href=&quot;http://www.space.com/20424-saturn-rings-moons-age.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.nasa.gov/mission_pages/cassini/media/cassini20071212.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;). Thank you, Cassini!&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2015-01-20T16:46:49.720" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8618" PostTypeId="1" CreationDate="2015-01-20T17:30:29.927" Score="9" ViewCount="12970" Body="&lt;p&gt;Compared to &lt;a href=&quot;https://en.wikipedia.org/wiki/VY_Canis_Majoris&quot;&gt;VY Canis Majoris&lt;/a&gt;, our Sun is more like a speck of space dust. Is there any planet or star which is &lt;strong&gt;known&lt;/strong&gt; to us that is bigger than the VY Canis Majoris?&lt;/p&gt;&#xA;" OwnerUserId="5344" LastEditorUserId="2153" LastEditDate="2015-01-26T22:45:38.347" LastActivityDate="2016-01-08T10:57:04.403" Title="Is there any planet/star bigger than VY Canis Majoris?" Tags="&lt;star&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8620" PostTypeId="2" ParentId="8618" CreationDate="2015-01-20T20:13:49.170" Score="22" Body="&lt;h1&gt;Part 1: Assuming &quot;larger&quot; means greater in diameter.&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stars&lt;/strong&gt;&lt;br&gt;&#xA;Estimates on the size of stars are just that, estimates, and estimates based on rather fuzzy observations. VY Canis Majoris has been bumped down to size. The current thinking is that there are seven known stars larger than VY Canis Majoris, the largest of which is UY Scuti.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current models indicate that the first generation of stars were much, much larger than anything we see now. It will be quite some time before we can resolve a first generation star. So far, they are just theoretical objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Planets&lt;/strong&gt;&lt;br&gt;&#xA;Jupiter-mass planets are about as large as a planet can get. There are some exoplanets that are larger than Jupiter, but that's because they orbit much closer to their parent star than does Jupiter. This makes them puff up a bit. The reason Jupiter-mass planets are deemed to be the largest possible is that planets of this mass are presently assumed to have a core of degenerate hydrogen. A funny thing happens to degenerate masses when mass is added to them: They shrink in diameter. (The shrinkage becomes catastrophic as the mass approaches the Chandrasekhar limit.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that assuming all other things are equal (temperature, composition), a planet more massive than Jupiter will be small in diameter than Jupiter is. Even if all other things aren't equal, a Jupiter-diameter planet is (give or take) about as large as planet can get.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Part 2: Assuming &quot;larger&quot; means greater in mass.&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stars&lt;/strong&gt;&lt;br&gt;&#xA;In terms of mass, VY Canis Majoris doesn't even make the top ten, not even close! The most massive known star is R136a1. Again, these are estimates, but mass is a bit easier to pin down than is radius (or diameter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As is the case with physical extent, the first generation stars are presently modeled as being much, much more massive than anything we see now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Planets&lt;/strong&gt;&lt;br&gt;&#xA;There's not much difference between the largest planet and the smallest brown dwarf. I would argue there's very little difference. It's a spectrum with no distinguishing characteristic that lets one say &quot;this is a planet&quot; and &quot;that is a brown dwarf&quot;. Ignoring the distinction between super-Jupiters and brown dwarfs the largest is about 80 Jupiter masses. V1581 Cygni C is 79 Jupiter masses. (More massive than that and they start burning hydrogen, thus making them a small red dwarf.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current factor that is used to distinguish between brown dwarfs and super Jupiters is mass. Anything larger than 13 Jupiter masses is a brown dwarf, anything smaller, a planet. That boundary is very arbitrary.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-01-20T20:13:49.170" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8621" PostTypeId="2" ParentId="420" CreationDate="2015-01-20T21:08:25.463" Score="15" Body="&lt;p&gt;It's pretty simple, actually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lFYWs.jpg&quot; alt=&quot;tidal bulging&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon creates tides. Due to tides, the water bulges out towards the Moon (and also on the opposite side).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the Earth also rotates pretty fast (once a day), faster than the Moon orbits the Earth (once a month). There's friction between the rotating Earth, and the watery bulge created by tides. The rotation of the Earth &quot;wants&quot; to rotate the bulge faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In effect, the rotation of the Earth drags the tidal bulge forward - the bulge is always a bit ahead of the Moon. When the Moon is at meridian, the tide is already decreasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there's a bit of extra watery mass on Earth, a little bit ahead of the Moon. This watery bulge interacts gravitationally with the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has two effects:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;it slows down the rotation of the Earth, gradually sucking energy from it (the Moon pulls the bulge, and therefore the Earth, &quot;back&quot;)&lt;/li&gt;&#xA;&lt;li&gt;that energy is dumped into the Moon's orbital motion, effectively &quot;pulling&quot; it forward&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When you dump energy of motion into an orbiting body, it settles into a higher orbit - higher orbit means more energy. Therefore, the transfer of energy from Earth' spin to Moon's orbit gradually makes Moon's orbit larger and larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This only happens because the Earth is spinning faster than the Moon orbits it. If the Earth was tidally locked to the Moon (spinning exactly as fast as the Moon orbits it), then no transfer would happen. If the Earth was spinning slower than the Moon's orbit, then the transfer would be opposite (from Moon's orbital motion to Earth's spin).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Counterintuitively, a satellite with more energy actually moves slower, but in a higher orbit. The extra energy goes into raising the orbit, not into making its speed faster. Why this happens exactly is a whole 'nother discussion.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2015-01-20T21:08:25.463" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8622" PostTypeId="2" ParentId="8510" CreationDate="2015-01-20T23:13:04.107" Score="1" Body="&lt;p&gt;The tide would be locked in place, roughly, because the Moon always shows the same side to the Earth (tidally locked). But the height of the water at the location facing the Earth and the opposite side should be greater than the mean height, and at the limb, as we see it, the height will be lower than the mean but by half the magnitude of the high tide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine installing a pipe with cross-section 1 cm$^2$ from the top of the ocean at its high point down to the center of the moon and then back up to the top of the ocean at a point where the net tidal field from the Earth is 0.  Fill it with water until one side is equal with the water level on one side and its level on the other side will automatically be equal to the water level there.  The weights of the water on both sides need to be equal if it is static.   For now, we will ignore that g changes when one gets deep into the moon. Therefore, roughly speaking $\rho$gh should be the same in both sections of the pipe, and:&#xA;$$&#xA;\rho(g_M - g_t)(h_t + r_M) = \rho g_M r_M&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $g_t$ is tidal acceleration &#xA;$$&#xA;  g_t(Moon) = 2\frac{GM_Er_M}{d^3}.&#xA;$$&#xA;Subscript M means Moon and E means Earth and d is the separation between the Earth and Moon.  Solving for h$_t$ gives:&#xA;$$&#xA;h_t = \frac{g_tr_M}{g_M - g_t} \sim \frac{g_t}{g_M}r_M&#xA;$$&#xA;We could substitute values in here, but if we do this in comparison with the height of the tide on the Earth, 54 cm, we cancel out some of the error in the approximation and get an even simpler formula, namely,&#xA;$$&#xA;\frac{h_t(Moon)}{h_t(Earth)} = \frac{M_E}{M_M} \frac{g_E}{g_M} \Big(\frac{r_M}{r_E}\Big)^2&#xA;$$&#xA;The mass of the Earth is 81.3 time greater, the acceleration at the surface of the Moon is 1/6.25 of Earth's g force, and the diameter of the Moon is 0.2725 of the Earth.  So, that is 81.3*6.25*.2725$^2$ = 37.7 time higher than the Earth's tide or 37.7*54 cm = 20.4 meters.  Note that the height of the ocean does not matter as long as it is considerably more than 20 meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the water would begin boiling since it is at 0 pressure until a water atmosphere is created, but the moon is too small to hold an atmosphere for long and the whole ocean would be gone in a while.  &lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2015-01-24T05:05:41.953" LastActivityDate="2015-01-24T05:05:41.953" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8623" PostTypeId="1" AcceptedAnswerId="8632" CreationDate="2015-01-21T04:22:58.037" Score="8" ViewCount="353" Body="&lt;p&gt;My current interest is black holes and quasars. I was wondering how often they erupt--specifically ours? Would it be blindingly bright on Earth? Finally, would there be any effects that would alter our life on Earth that we know of?&lt;/p&gt;&#xA;" OwnerUserId="5381" OwnerDisplayName="Scott" LastEditorUserId="-1" LastEditDate="2015-01-22T01:00:59.250" LastActivityDate="2015-01-22T02:10:18.340" Title="How often does the supermassive black hole in the Milky Way become a quasar?" Tags="&lt;supermassive-black-hole&gt;&lt;quasars&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8625" PostTypeId="2" ParentId="5994" CreationDate="2015-01-21T10:55:40.933" Score="3" Body="&lt;p&gt;As already answered, the definition of the size of a galaxy must always to some extend be arbitrary. In astronomy, several definitions are used, according to the context in which it's used, e.g.:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$R_{\mathrm{vir}}$ (the virial radius): Used when considering the galaxy's dynamics; defined by $GM_{\mathrm{vir}}/R_{\mathrm{vir}} \sim V^2$, where $G$ is the gravitational constant, $M_{\mathrm{vir}}$ is the mass inside the virial radius, and $V$ is the circular velocity (for disk galaxies) or velocity dispersion (for elliptical or irregular galaxies). For instance, if an initial density perturbation is smaller than its virial radius, it will collapse to form a galaxy. This radius is also the radius within particles will be gravitationally bound if their velocity does not exceed $V$.&lt;/li&gt;&#xA;&lt;li&gt;$R_{1/2}$ (the half-light radius): The radius within which half of the total observed light is emitted. This definition is useful when comparing the luminosity of galaxies.&lt;/li&gt;&#xA;&lt;li&gt;$R_{200}$: The radius within which the average density is equal to 200 times the average density of the Universe. Similarly, sometimes $R_{500}$ or $R_{1000}$ are used. These definitions make sense when discussing the mass of a galaxy, since beyond they sort of blend in to the intergalactic medium.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Although these definition give different sizes, they are all of the same order of magnitude.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-02-18T10:06:00.320" LastActivityDate="2015-02-18T10:06:00.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8627" PostTypeId="1" AcceptedAnswerId="38975" CreationDate="2015-01-21T22:14:09.980" Score="5" ViewCount="171" Body="&lt;p&gt;Last night (January 21, 2015, possibly 5:30 UTC), from my house in northern California, I saw a moving point of light in the sky, going west to east, crossing about 120 degrees of sky in 5 seconds or so.&lt;/p&gt;&#xA;&lt;p&gt;It didn't look like my experience of meteors - thicker and slower, and quite bright. I think it survived until it was out of sight entirely, rather than burning out.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.aerospace.org/cords/reentry-predictions/upcoming-reentries-2-2/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aerospace.org&lt;/a&gt;'s reentry predictions doesn't show anything for that time.&lt;/p&gt;&#xA;&lt;p&gt;I've seen ISS and other satellites; this was a lot larger and brighter, and my favorite &lt;a href=&quot;https://in-the-sky.org/satpasses.php?country=240&amp;amp;reg1=3627&amp;amp;reg2=34639&amp;amp;town=17913&quot; rel=&quot;nofollow noreferrer&quot;&gt;satellite pass schedule&lt;/a&gt; also shows nothing.&lt;/p&gt;&#xA;&lt;p&gt;Was this more likely a rocket body re-entry, or a large meteor?&lt;/p&gt;&#xA;" OwnerUserId="2573" LastEditorUserId="2573" LastEditDate="2020-09-19T04:58:12.837" LastActivityDate="2020-09-19T08:53:43.833" Title="Did I see a meteor or a re-entry?" Tags="&lt;meteor&gt;&lt;artificial-satellite&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8628" PostTypeId="2" ParentId="8605" CreationDate="2015-01-21T22:51:31.267" Score="26" Body="&lt;p&gt;This is a broad question and too broad for me to answer comprehensively. It should be broken down into doppler methods, transits and direct imaging; and that's before we get to questions of detecting Kuiper belts, radio emission etc.&lt;/p&gt;&#xA;&lt;p&gt;I'll stick for the moment with what I know about detection of planets using the doppler wobble technique.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Doppler Technique&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The reflex radial velocity semi-amplitude of a star for the case of a planet of mass &lt;span class=&quot;math-container&quot;&gt;$m_2$&lt;/span&gt; orbiting a star of mass &lt;span class=&quot;math-container&quot;&gt;$m_1$&lt;/span&gt;, in an elliptical orbit with eccentricity &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt;, and orbital period &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; and with an orbital axis inclined at &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; to the line of sight from Earth is:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$  \left( \frac{2\pi G}{P}\right)^{1/3}\frac{m_2 \sin i}{m_1^{2/3}} (1-e^2)^{-1/2}.      $$&lt;/span&gt;&#xA;A (very) detailed derivation is given by &lt;a href=&quot;http://astro.berkeley.edu/%7Ekclubb/pdf/RV_Derivation.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Clubb (2008)&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;So I built myself a little spreadsheet and assumed that all the planets were seen optimally at &lt;span class=&quot;math-container&quot;&gt;$i=90^{\circ}$&lt;/span&gt; (they could not all be seen optimally, but the smallest inclination would be about &lt;span class=&quot;math-container&quot;&gt;$i=83^{\circ}$&lt;/span&gt; for Mercury, so it doesn't make too much difference) I'll also assume the mass of Alpha Cen A is about &lt;span class=&quot;math-container&quot;&gt;$M \simeq 1.1M_{\odot}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The results are&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;Planet&lt;/th&gt;&#xA;&lt;th&gt;RV semi-amplitude (m/s)&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Mercury&lt;/td&gt;&#xA;&lt;td&gt;&lt;span class=&quot;math-container&quot;&gt;$8.3\times 10^{-3}$&lt;/span&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Venus&lt;/td&gt;&#xA;&lt;td&gt;&lt;span class=&quot;math-container&quot;&gt;$8.1\times 10^{-2}$&lt;/span&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Earth&lt;/td&gt;&#xA;&lt;td&gt;&lt;span class=&quot;math-container&quot;&gt;$8.4\times 10^{-2}$&lt;/span&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Mars&lt;/td&gt;&#xA;&lt;td&gt;&lt;span class=&quot;math-container&quot;&gt;$7.5\times 10^{-3}$&lt;/span&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Jupiter&lt;/td&gt;&#xA;&lt;td&gt;&lt;span class=&quot;math-container&quot;&gt;$11.7$&lt;/span&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Saturn&lt;/td&gt;&#xA;&lt;td&gt;&lt;span class=&quot;math-container&quot;&gt;$2.6$&lt;/span&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Uranus&lt;/td&gt;&#xA;&lt;td&gt;&lt;span class=&quot;math-container&quot;&gt;$0.28$&lt;/span&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Neptune&lt;/td&gt;&#xA;&lt;td&gt;&lt;span class=&quot;math-container&quot;&gt;$0.26$&lt;/span&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;The limits of what are possible are well illustrated by a planet around Alpha Cen B, claimed to be in a 3 day orbit and with a mass similar to the Earth (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2012Natur.491..207D&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dumusque et al. 2012&lt;/a&gt;, and see &lt;a href=&quot;http://exoplanets.org/detail/alpha_Cen_B_b&quot; rel=&quot;nofollow noreferrer&quot;&gt;exoplanets.org&lt;/a&gt;). The radial velocity semi-amplitude detected here was &lt;span class=&quot;math-container&quot;&gt;$0.51\pm 0.04$&lt;/span&gt; m/s, and some spectrographs, notably the HARPS instruments, are routinely delivering sub 1 m/s precision. Thus Jupiter and Saturn would be detectable, Uranus and Neptune are right on the edge of detectability (remember you can average over many RV observations), but the terrestrial planets would not be found (Earth detections would require precisions below 10 cm/s. Remember also that the weaker signals would have to be dug out from the larger signals due to the Jupiter- and Saturn-like planets.&lt;/p&gt;&#xA;&lt;p&gt;However, there is a second limitation: to find a planet using the doppler method you need to observe for at least a significant fraction of the orbital period. Given that current m/s precisions have been available for only &lt;span class=&quot;math-container&quot;&gt;$\sim 5$&lt;/span&gt; years, it is unlikely that Saturn would &lt;em&gt;yet&lt;/em&gt; have been detected.&lt;/p&gt;&#xA;&lt;p&gt;A picture that illustrates the situation can be obtained from the exoplanets.org website, to which I have added lines that approximate where RV semi-amplitudes would be for 10 m/s and 1 m/s precision (assuming the Alpha Cen A mass and circular orbits). I've marked on the Earth, Jupiter and Saturn. Note that few objects have been discovered below the 1 m/s line. Also note the lack of planets between the 1 and 10m/s lines with periods longer than a couple of years - the recent increase in sensitivity has yet to feed through to lower mass, longer period exoplanet discoveries.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pu5Zg.png&quot; alt=&quot;Planet mass vs period&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;In conclusion: only Jupiter would have been so far found by the doppler technique.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Transit techniques&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;I'll also add a few comments about the transit technique. Transit detection will only work if the exoplanets orbit such that they cross in front of the star. So high inclinations are mandatory. Someone who is better at spherical trigonometry should use the published data for the solar system to work out how many (and which) planets transit in some highly optimal orientation. Given that the planets have orbital inclinations with a scatter of a few degrees, then some straightforward trigonometry and a comparison with the solar radius, tells you that these orbits will generally not all transit for any particular viewing angle. Indeed a number of the Kepler-discovered multiple transit systems are much &amp;quot;flatter&amp;quot; than the solar system.&lt;/p&gt;&#xA;&lt;p&gt;The Kepler satellite is/was capable of detecting very small transiting planets thanks to its very high photometric precision (the dip in flux is proportional to the square root of the exoplanet radius). The picture below, presented by the NASA Kepler team (slightly out of date now), shows that planetary &lt;em&gt;candidates&lt;/em&gt; have been discovered that are down to the size of Mars. However these tend to be in short period orbits because a transit signal needs to be seen a number of times, and Kepler studies this patch of sky for about 2.5 years (when this plot was produced).&lt;/p&gt;&#xA;&lt;p&gt;So from this point of view, possibly Venus would have been seen, but none of the other planets could be confirmed.&lt;/p&gt;&#xA;&lt;p&gt;However, there is a wrinkle. Alpha Cen A is way too bright for these kinds of studies and way brighter than the Kepler stars. You would have to build a special instrument or telescope to look for transits around very bright stars. Some of this work has been done by ground-based surveys (mainly finding hot Jupiters). A new satellite called &lt;a href=&quot;http://tess.gsfc.nasa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;TESS&lt;/a&gt; (Transiting Exoplanet Survey Satellite, launched in April 2018) is a two year mission, focused on finding small planets (Earth-sized and bigger) around bright stars. However, most of its targets (including Alpha Cen) are only observed for a 1-2 months, so only the inner parts of their planetary systems will be probed.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uOWoB.png&quot; alt=&quot;Kepler discovered planetary candidates&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2365" LastEditDate="2020-12-16T18:18:44.917" LastActivityDate="2020-12-16T18:18:44.917" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8629" PostTypeId="1" CreationDate="2015-01-21T23:01:37.843" Score="0" ViewCount="235" Body="&lt;p&gt;Can someone please explain this phenomenon? A solid explanation seems to be missing from the entire internet. &lt;/p&gt;&#xA;" OwnerUserId="5384" LastEditorUserId="5384" LastEditDate="2015-01-22T05:46:29.363" LastActivityDate="2015-01-22T07:38:28.047" Title="Why does the sun rise north of east between the vernal and autumnal equinox?" Tags="&lt;orbit&gt;&lt;the-sun&gt;&lt;azimuth&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8630" PostTypeId="1" CreationDate="2015-01-22T01:18:03.547" Score="2" ViewCount="81" Body="&lt;p&gt;I just purchased a powerseeker 114q by Celestron. This is my very first telescope. I believe I assembled it correctly however I can't see anything out of it. Well I can see a small corner of an image. When I look at the reflector in the back of the telescope, I can see a small part of my face. It is the same when I look through the lens. I only can see a small piece of the sky. I think its the sky anyway. The rest is completely black. All of the lens caps are off. I followed the setup directions down to the letter. Can someone please help.&lt;/p&gt;&#xA;" OwnerUserId="5385" LastActivityDate="2015-04-24T03:47:55.667" Title="PowerSeeker 114Q" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8631" PostTypeId="1" CreationDate="2015-01-22T02:07:37.067" Score="2" ViewCount="839" Body="&lt;p&gt;I'd like to find the distance at which a 2.4 cm coin is subtended by an angle of 10&quot;. I've drawn my diagram and found that $D=\dfrac{d}{\alpha}$ using the small angle approximation. Since $10&quot; = 4.85\times10^{-5}$ rad then $ D = \dfrac{2.4 \times 10^{-2} m}{4.85 \times 10^{-5}rad} = 495$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the units of distance in this calculation meters? In other words why would the units of radians be 'dropped'?&lt;/p&gt;&#xA;" OwnerUserId="5387" LastEditorUserId="5387" LastEditDate="2015-01-22T02:19:34.037" LastActivityDate="2015-01-22T04:21:46.357" Title="What are the units of distance in this subtended angle calculation?" Tags="&lt;distances&gt;&lt;units&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8632" PostTypeId="2" ParentId="8623" CreationDate="2015-01-22T02:10:18.340" Score="2" Body="&lt;p&gt;To answer your question, the black hole in the milky way was never a quasar, and as it currently stands will never become one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Info about how earth and life on earth could be affected if this ever happened, can be found &lt;a href=&quot;https://www.youtube.com/watch?v=-pm1WzeTkAg&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5362" LastActivityDate="2015-01-22T02:10:18.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8633" PostTypeId="2" ParentId="8631" CreationDate="2015-01-22T03:29:17.013" Score="0" Body="&lt;p&gt;The equation you are using is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\tan(\theta) = \frac{d}{D}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which in the small angle approximation $\tan(\theta) \sim \theta$. But you have to take into account that $\theta$ has units of radians, where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;1 \: radian = 3600 \times \frac{180}{\pi} \sim 206265 \: arcsecs.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, the small angle approximation formula is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{\theta}{206265} = \frac{d}{D} \: \Rightarrow \: \theta_{arcsecs} = 206265 \times \frac{d}{D},&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;provided you use the same units for $d$ and $D$.&lt;/p&gt;&#xA;" OwnerUserId="2339" LastEditorUserId="2339" LastEditDate="2015-01-22T03:35:54.870" LastActivityDate="2015-01-22T03:35:54.870" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8634" PostTypeId="2" ParentId="8631" CreationDate="2015-01-22T04:21:46.357" Score="2" Body="&lt;p&gt;The units of radians are 'dropped' because unlike most units, they are dimensionless. Recall that the definition of radian angle measure is the ratio of the length of a circular arc to its radius. Thus, radian angle measures have units of $[\mathrm{Length}]/[\mathrm{Length}] \equiv 1$, i.e. dimensionless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radians are units in the sense that they give information about what standard the angle quantity is measured by. Thus, the convention of writing $\mathrm{rad}$ is useful in that it distinguishes it from other ways of measuring angles (e.g., in your question, arc-seconds), but in terms of dimensional analysis, $\mathrm{rad} \equiv 1$, so $\mathrm{m}/\mathrm{rad} = \mathrm{m}$.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2015-01-22T04:21:46.357" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8635" PostTypeId="2" ParentId="8629" CreationDate="2015-01-22T07:38:28.047" Score="2" Body="&lt;p&gt;You know, the first time someone told me this, I was absolutely certain they were confused, ignorant, or otherwise mistaken, and I told him he had his facts wrong. To be precise, we were talking about an observer on the equator. I maintained that the Sun would rise due east and set due west, all year round. He said no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What gave me pause, though, is that this person was far from ignorant, so maybe the other two appellations didn't fit either. I went home and tried to construct a convincing proof of my side, and in the middle of the night a very simple and convincing proof did come to mind. Except that it was a proof that I was wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that you're on equator at sunrise on the summer solstice. Due east is along the equator. But the plane of the equator is tilted relative to the Earth's orbit, and the Sun is well above (north) of it. You cannot see it due east at sunrise. It has to be $23.5^\circ$ north of east. (Remember, this is for an observer on the equator.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the same token, the Sun must rise $23.5^\circ$ south of east on the winter solstice, and gradually transition between these extremes during the year.&lt;/p&gt;&#xA;" OwnerUserId="4077" LastActivityDate="2015-01-22T07:38:28.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8636" PostTypeId="1" CreationDate="2015-01-22T10:50:09.343" Score="2" ViewCount="366" Body="&lt;p&gt;Would twin planets orbiting around each other present a solution to tidal locking of habitable planets in red dwarf systems? It's an awfully easy equation to set up. I'm fibbing; I can't do it. I was hoping someone might show me when they get a moment.&lt;/p&gt;&#xA;" OwnerUserId="5390" LastEditorUserId="2153" LastEditDate="2015-05-14T00:07:54.310" LastActivityDate="2015-06-13T00:54:52.027" Title="The tidal locking problem concerning Earth sized planets in habitable zone around Red Dwarfs" Tags="&lt;orbit&gt;&lt;tidal-forces&gt;&lt;red-dwarf&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8637" PostTypeId="1" CreationDate="2015-01-22T16:22:52.723" Score="9" ViewCount="290" Body="&lt;p&gt;This may overlap with other questions, but I'm asking what is unique about Saturn in location, properties, etc that is has far more pronounced rings?   It's between Jupiter and Uranus and Neptune in size and position.  So what's special?  Is there some Goldilocks zone of rings formation?   Or is it just an accident of history?&lt;/p&gt;&#xA;" OwnerUserId="1106" LastEditorUserId="1106" LastEditDate="2015-01-22T19:34:15.903" LastActivityDate="2015-01-24T01:48:26.593" Title="What's special about Saturn that it has distinctive rings?" Tags="&lt;formation&gt;&lt;saturn&gt;&lt;planetary-ring&gt;&lt;planetary-formation&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8638" PostTypeId="1" CreationDate="2015-01-23T18:04:29.803" Score="2" ViewCount="272" Body="&lt;p&gt;We linked a physical constant (the time of one second) to a physical non constant (the rotation of the earth) so there will have to be adjustments - or not? Is it important that the UTC in one line on earth is the same as the astronomical time? As the earth slows down, it means that this line slowly changes position. When the line travels 1 hour away we can bring it back by an hour change if necessary. If currently we add 1 second say every 3 years, it would take thousands of years to change by 1 hour. The last time a second was added, in 2012, problems were caused in some systems. Why do we need to do it?&lt;/p&gt;&#xA;" OwnerUserId="5397" LastActivityDate="2015-01-23T19:22:04.370" Title="Why do we need to add a second to 30 June and risk upsetting Internet?" Tags="&lt;time&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8639" PostTypeId="2" ParentId="8638" CreationDate="2015-01-23T19:22:04.370" Score="2" Body="&lt;p&gt;There's quite a few things here that need comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, the &quot;linking&quot; you describe was actually the first definition of the second (day, hour, and all the way down to second). Remember, since the dawn of civilization we've used celestial bodies to keep time for us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phenomena of the Sun and the Moon had good enough precision for all practical purposes until very recently. It was then that the definition of the time unit was rebased to a different phenomenon. BTW, the second of time is not a physical constant. It is a unit of measure currently defined as a multiple of the frequency coming out of the transitions of a certain atom between two different states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the motion of the Earth follows its own clock, different from atomic phenomena, so it ought to drift off after a while. Do we &quot;need&quot; to make adjustments then? That's debatable. We still prefer to keep the Earth's clock and our clock in sync, and therefore we make adjustments when necessary. But is it needed in an absolute sense? No, not really. It's just convenient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the risk of &quot;upsetting the Internet&quot;, this is an alarmist rumor the likes of the &quot;Y2K bug&quot;. I actually happen to work in a related field - I've done infrastructure at a variety of tech companies in the Silicon Valley; the NTP infrastructure (Network Time Protocol - what keeps Internet services in time sync) is something I'm very familiar with. A jump of 1 second should be and will be tolerated by essentially everything that's out there now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will there be stragglers here and there that will hiccup when the time takes a tiny leap? Yes. And that's a bug, the engineers need to look at the software and fix it. Remember, a lot of existing services (like Google Search, or Instagram) have servers and instances right now that are off by more than 1 second (just the way there are cars coming out of the assembly line with missing bolts once in a while), and usually nothing bad happens as long as the drift is not too big); fix NTP, the time base jumps on that instance, and again nothing bad happens in the vast majority of cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relax, nothing important will be &quot;upset&quot;. In fact, I doubt that the vast majority of people will be affected in any perceivable way.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2015-01-23T19:22:04.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8640" PostTypeId="1" AcceptedAnswerId="8643" CreationDate="2015-01-23T23:48:51.950" Score="3" ViewCount="269" Body="&lt;p&gt;From another question I posted, I learned that not only was matter created from the Big Bang, but more importantly, space itself.  What does it mean for space to be created?  Does it exist in the same way substances exist?  In other words, what do we mean by 'space exists?'&lt;/p&gt;&#xA;" OwnerUserId="5306" LastEditorUserId="5306" LastEditDate="2015-01-23T23:57:32.630" LastActivityDate="2015-01-28T15:58:04.013" Title="What does it mean to say 'space exists?'" Tags="&lt;space&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8642" PostTypeId="1" CreationDate="2015-01-24T01:23:18.220" Score="7" ViewCount="1509" Body="&lt;p&gt;I am standing in an open area and can see the sky for a full 360 degrees. I have the coordinates of an object (&lt;a href=&quot;https://en.wikipedia.org/wiki/Right_ascension&quot; rel=&quot;nofollow noreferrer&quot;&gt;RA&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Declination&quot; rel=&quot;nofollow noreferrer&quot;&gt;DEC&lt;/a&gt;). Where do I look to find RA 0?&lt;/p&gt;&#xA;" OwnerUserId="5399" LastEditorUserId="37191" LastEditDate="2021-03-14T14:16:59.607" LastActivityDate="2021-03-14T14:16:59.607" Title="How to find RA 0" Tags="&lt;amateur-observing&gt;" AnswerCount="6" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="8643" PostTypeId="2" ParentId="8640" CreationDate="2015-01-24T01:27:56.743" Score="2" Body="&lt;p&gt;This is a very complex question, and perhaps more appropriate for the Physics or Philosophy stacks. I will attempt to give you an initial direction, but really I think this is too broad for this stack. I'm merely writing here because comments are too short.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a very practical way, space is simply that which allows objects and events to have relative position, direction, and extent. There is still disagreement over whether space is a thing in itself, or merely a relationship between things, or part of a conceptual framework. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Space&quot; rel=&quot;nofollow&quot;&gt;wiki&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In relativity, the dominant concept is not space, but spacetime - closely related to a fundamental constant of the universe (speed of light) and to gravity. Space itself, or time itself, are less important, but the 4-dimensional metric of spacetime is far more fundamental, in a sense.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does it exist in the same way substances exist?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We don't know. See the wiki link for the philosophical debate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;what do we mean by 'space exists?'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It means this universe is built in such a way as to allow relationships of position or direction. This is a bit of a circular definition, but it's the best we have.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What does it mean for space to be created?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Actually, that happens all the time, even now. The so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Dark_energy&quot; rel=&quot;nofollow&quot;&gt;dark energy&lt;/a&gt; actually makes more space, continuously, even now. That simply means that, whereas a while ago there was X amount of distance between galaxies A and B, now there's X + Y amount of distance. The low-level mechanism that allows this creation is not well understood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It could be imagined that something similar happened during the Big Bang, except it started from zero. Again, the low-level mechanisms around the origin are not well understood. To understand that we first must be capable of understanding something far more simple - the physics of the center of a black hole. To do that, we would need something like &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_gravity&quot; rel=&quot;nofollow&quot;&gt;quantum gravity&lt;/a&gt;, which is a part of Physics that has not been invented / built yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this seems overly speculative, it's because you're asking a very profound and difficult question - so difficult that it exceeds the limits of current science and even philosophy.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2015-01-24T01:27:56.743" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8644" PostTypeId="2" ParentId="8637" CreationDate="2015-01-24T01:48:26.593" Score="2" Body="&lt;p&gt;Saturn's rings are unique in that a lot of the rings compositions contain new and recycled material. Much of the uniqueness is complementary from Saturn's moons, which play a critical role in shaping and recycling the rings materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A great example is the moon Enceladus, which contributes largely to the E Ring. Icy material is ejected from Enceladus, a process called cryovolcanism (spewing plumes of icy materials instead of silicate rock); due to the lack of atmospheric pressure and low surface gravity, Enceladus doesn't hold the material that it ejects, but rather feeds material into the E Ring. Discovered by Cassini, the material is comprised of simple hydrocarbons such as methane, propane, acetylene and formaldehyde.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A contributing factor for the unique shaping of the rings and much of their orbital stability stems from the moons also. Another well known example is the Cassini Division and a process called orbital resonance. Essentially the moon Mimas orbits once for every two orbits for material inside the Cassini Division. As a result of that orbital resonance the material receives accumulation of radial momentum, the effect of such accumulation forces a tug of war with the particles. Mimas attracts the material pulling it into the A Ring as the material increases in velocity, resulting in the division.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other moons have also been documented to have this effect, such as the moons Pan and Daphnis in the A Ring, however it is thought that the effect from these smaller moons only create a division around 10km in width. Stability can be exaggerated from the F Ring, with the moons Pandora and Prometheus. These moons twist the material into a spiraling vortex, like a horizontal tornado encircling the planet. Processes of stability arise from moon ringlets, such as the Titan Ringlet. Titan has orbital resonance with material in the C Ring and provides radial momentum with the Colombo Gap, the gap is slightly elliptical rather than circular (just as Titan is) and the material is attracted towards Titan in the same pattern, in a process called apsidal precession. This creates a compression and expansion of the rings giving them stability while accumulating velocity due to orbital resonance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distortions from the moons on an independent inclination to the rings as they come close to the rings, also pull material from the rings and spread it around in chaos, additionally recycling material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each ringlet has an individual story as to the forces which make that ringlet unique, comprising of the orbital parameters of the moons and the unique positioning of the forces around them. Along with material that is recycled in a unique fashion not evident with other planetary rings system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The uniqueness stems from these factors.&lt;/p&gt;&#xA;" OwnerUserId="5398" LastActivityDate="2015-01-24T01:48:26.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8645" PostTypeId="2" ParentId="8642" CreationDate="2015-01-24T02:24:15.690" Score="3" Body="&lt;p&gt;A brief overview of RA is the angular distance measured eastward from the vernal equinox along the celestial equator. Another way to say that refers to where the celestial equator intersects the horizon at a right angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are able to follow the celestial equator to the point known as the vernal equinox, 0Deg RA is at the intersection point, with RA increasing directly east. The vernal equinox can be found at the first point of Aries, on a celestial sphere it is the point where the Sun crosses the celestial equator from south to north during the March equinox. The point during the March equinox is located in Pisces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are methods to determine the vernal equinox at other times, which can be dependent on location and the time. A great way that I personally use, is the same method as 'star hopping'. Using reference points, you can generally determine a line of sight between two or more points, for example if you are aware of a point and it's approximate RA, you can follow the path through another known RA point and evaluate the distance between the points. Following the line of sight you can them determine how far along the line the point of 0 would be, using this method takes much practice and you could even use visual guides such as a thumb as a measuring tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using my thumb, I can say held at .5 meters is equivalent to 2.5 degrees, using stars as reference points I could approximate the line of sight and have a reliant measuring tool to judge distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method you may require, could be entirely dependent on the reason you need to determine the point. If using electronic devices you can follow the celestial equator west along the celestial equator to find the point etc. I can explain other methods in more detail if required, others may have developed their own methods also.&lt;/p&gt;&#xA;" OwnerUserId="5398" LastActivityDate="2015-01-24T02:24:15.690" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8646" PostTypeId="2" ParentId="8630" CreationDate="2015-01-24T02:46:11.377" Score="1" Body="&lt;p&gt;The PowerSeeker will come with a manual on collimation of the mirrors. As the light entering the telescope hits the primary lens, the light reflects and compresses onto the secondary mirror, in turn reflects and compresses the light onto the eyepiece. When the mirrors are misaligned, the light rays aren't correctly focused onto a point where they are projected through the eyepiece, as such like turning a mirror around your point of view, you will not see directly behind yourself but rather an angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what is happening even as you look directly down the tube at the primary mirror; if it is not directly facing you, you will not see your face (perhaps part of it) and you will see the sides inside the tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various method for collimation, personally I use a laser collimater and a barlow to increase the sensitivity. The methods are dependent on what you have available however, you don't need any additional tools to collimate apart from the telescope itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The PowerSeeker does not have a central marking on it's primary mirror to indicate the center of the mirror, if you are confident in disassembling the rear of the tube you can remove the mirror and measure the central point and mark it. If not, you will not get a collimation as precise, however generally it will suffice until you gain experience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When looking through an empty eyepiece barrel, if the primary mirror doesn't appear central in the view, as in you can see the whole mirror and the spacing around the mirror is equal, then the secondary mirror needs adjustment. In the same respect, if the central point of the primary mirror does not line up in the center of view, the primary mirror needs adjusting. You do not need to get the line of sight with the primary mirror 100%, typically users of telescopes may use a star itself for collimation; too lengthy to explain however when viewing a star outside of focus (best to use largest magnification possible/smallest diameter eyepiece) the airy disk of a star will appear distorted to one side, centralizing that airy disk will provide an accurate collimation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further information will be available in the manual, or there are a many websites to help finetune collimation. If you need further advice just edit your above post or post a comment.&lt;/p&gt;&#xA;" OwnerUserId="5398" LastEditorUserId="5398" LastEditDate="2015-01-24T02:51:50.750" LastActivityDate="2015-01-24T02:51:50.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8647" PostTypeId="2" ParentId="1896" CreationDate="2015-01-24T03:43:56.180" Score="1" Body="&lt;p&gt;Much like Jeremy has answered, Mars will appear just larger than a point, you will be able to discern a disk, however with minimal detail. For an overview of the disk you will see, I two versions of a virtual telescope in Wolfram Mathemtica; one with an orange disk and another using models of the planets. Anyway, using the orange disk variant, I input the CPC1100 details and the resulting view will match my output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The field of view using a 40mm eyepiece without a barlow results in a 0.74degree apparent field of view, this equates to 2664 arcseconds. For reference Mars right now has an apparent diameter of 4.5 arcseconds, in your field of view that is .0017% of the apparent field of view (as represented by the orange disk).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;eg. 40mm eyepiece @70x magnification w/no barlow&lt;img src=&quot;https://i.stack.imgur.com/F7Miw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You telescope has a maximum useful magnification of 560x (2x telescope aperture) however it is airy disk limited to 481.2x. This means that a star will only appear 'point like' up to 481.2x magnification, above that the disk becomes increasingly blurrier to a point where you can no longer discern a point it appears as a cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a point of reference to the size of Mars using different equipment;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;40mm eyepiece @210x magnification w/3x barlow&lt;img src=&quot;https://i.stack.imgur.com/cqDti.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10mm eyepiece @280x magnification w/no barlow&lt;img src=&quot;https://i.stack.imgur.com/3AbcL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;17mm eyepiece @494x magnification w/3x barlow&lt;img src=&quot;https://i.stack.imgur.com/af1Nd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On another note apart from what you 'should see', is what you can see. Aperture plays the ONLY role in telescope design which defines your clarity. The primary mirror diameter is a light bucket, a larger mirror is a higher image resolution (more discernible features/clearer image). Magnification also plays a role in clarity, each time you increase the magnification by 2x, you reduce the brightness by 4x; increase by 3x, you reduce brightness by 9x. So brightness is directly proportional to magnification. The highest possible brightness you can achieve, is your lowest possible magnification. How aperture works by collecting more light; allows a brighter image for a larger telescope than a smaller one at the same magnification, which is why so many people get 'aperture fever' and think big is best (for clarity yes). Your focal length is the determining factor for image size, meaning a 20inch telescope won't see an object larger than a 5inch at the same focal length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are correct in saying that both Mars and Sirius appear brighter with the naked eye than your lowest magnification; referring above, if your pupil diameter is 7mm, then the object appears at a factor of hundreds of times dimmer through your telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can provide material to calculate your own virtual telescope, it's parameters and viewing expectations if you are an avid user of Mathematica. Hope this has cleared up some information for you.&lt;/p&gt;&#xA;" OwnerUserId="5398" LastActivityDate="2015-01-24T03:43:56.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8649" PostTypeId="2" ParentId="8373" CreationDate="2015-01-24T04:36:36.903" Score="1" Body="&lt;p&gt;Yes, on the right track where the energy produced in a supernova can indicate if it was a LGRB or SGRB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An indicative tools stems from the light curves of the remnants, both the afterglow brightness and it's region of explosion can indicatively be traced to how long the event lasted. Large gamma-ray remnants are typically associated with rapid star formation regions and are a more luminous remnant, where short gamma-rays are typically associated with little or no star formation regions and are a less luminous remnant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not an exact science, as reproducing the effects and duplicating results have proven vastly difficult, however using distribution plots can definitively separate the differences and provide a theoretical model to work on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some minor information can be found at &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma-ray_burst#Classification&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, however further reading and a more in depth view can be found at &lt;a href=&quot;http://arxiv.org/abs/1309.3869&quot; rel=&quot;nofollow&quot;&gt;Cornel University Library&lt;/a&gt;. It is a bit of a read and attempts to explain the physics behind emission through mathematics, I'm not entirely sure whether is will provide the answer you are seeking.&lt;/p&gt;&#xA;" OwnerUserId="5398" LastActivityDate="2015-01-24T04:36:36.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8650" PostTypeId="2" ParentId="8367" CreationDate="2015-01-24T04:52:33.967" Score="1" Body="&lt;p&gt;I use Wolfram Mathematica to produce model of systems, bodies and parameters. There are probably a plethora of math programs which may also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good resource are 'programmable notebooks' produced from Mathematica, demonstrations of these notebooks can be found at the &lt;a href=&quot;http://demonstrations.wolfram.com/index.html&quot; rel=&quot;nofollow&quot;&gt;Wolfram Demonstrations Project&lt;/a&gt;. The system Wolfram has integrated is a very advanced programmable environment, however from first glance there was nothing available in the demonstrations project, at least nothing simple.&lt;/p&gt;&#xA;" OwnerUserId="5398" LastActivityDate="2015-01-24T04:52:33.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8651" PostTypeId="2" ParentId="8146" CreationDate="2015-01-24T05:22:35.980" Score="2" Body="&lt;p&gt;I'm finding this very tough to answer, the method of detection is critical in how far we can detect. There are two probable methods I can think of, one superior to the other. The first method involves the speed of light and our production of waves. The second involves how we have tailored our atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our production of waves (radio) started late 19th century, if we use a point of reference, say 1900; we have been broadcasting for 115 years, at light speed a species no further than 115 light years away could detect us. Hence the idea of the SETI program as Rahul has suggested, with the intention of broadcasting ourselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best method, and the one I can see working for humans in their search for others, is atmospheric poisoning. There are specific hydrocarbons in our atmosphere which are thought to only be produced by man, if we think like this, then it's plausible we may also detect atmospheric poisoning around an exoplanet. Detecting oxygen simply is not enough, as it's not indicative that life exists, oxygen can be produced naturally in limited quantities as found elsewhere in the solar system however to sustain carbon based lifeforms such as ourselves there would have to be a major abundance. Detecting pollutants is the more logical way to conceive detection. If we are able to produce elements not naturally found, it is a clear indication that a species put it there. This also relies on the speed of light, however man made pollutants have existed pre-wave era, and have had longer to transmit light than our production of waves. The downside is detection method of the pollutants, currently as humans we rely on either using a star with a transiting planet to determine composition, or less precisely spectrum data (which doesn't indicate atmospheric material).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another viewpoint is looking at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kardashev_scale&quot; rel=&quot;nofollow&quot;&gt;Kardashev scale&lt;/a&gt;, one could put forth that we have the technology to determine that answer based on energy consumption. If we could detect a massive gravitational field and no apparent source of energy, the energy may well be harvested by another species; such as a Dyson sphere. Such a detection I believe would be too easy to overlook as it's not something our species is actively searching for. While this holds truer for more of a theoretical detection, another species may be able to detect energy consumption on our planet, through lighting of our planet and atmosphere along with increasing surface temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe at best, as for human interference, we may be looking in the range of 100-150 light years. As for detection of life in general, I can't imagine pre-modern era if there was a simple way to determine that life existed if viewed from elsewhere apart from the fact we had a stable system containing liquid water and atmospheric oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may be too reliant on providing the argument from our point of view being carbon based lifeforms, if another species advanced or more than us wasn't carbon based, it could very well be they are looking for other indications more localized to their own species, in the same way we look for indications which we imagine detecting ourselves with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; As requested by Rob Jeffries; NO, using transit photometry using today's current technology is not yet possible. At &lt;code&gt;1ly&lt;/code&gt; Earth would appear as &lt;code&gt;2.776*10^-4″&lt;/code&gt;-&gt;&lt;code&gt;3600*(180/π)*(12734/9.460*10^12)&lt;/code&gt; or &lt;code&gt;2.776mas&lt;/code&gt;, which is possible by &lt;a href=&quot;http://www.eso.org/public/teles-instr/vlt/&quot; rel=&quot;nofollow&quot;&gt;ESO's Very Large Telescope&lt;/a&gt; which has an angular resolution able to image in milliarcseconds. At &lt;code&gt;10ly&lt;/code&gt; Earth would appear as &lt;code&gt;2.776*10^-5″&lt;/code&gt;-&gt;&lt;code&gt;3600*(180/π)*(12734/9.460*10^13)&lt;/code&gt; or &lt;code&gt;277.6μas&lt;/code&gt;, possible after completion of the &lt;a href=&quot;http://arxiv.org/abs/1205.5743&quot; rel=&quot;nofollow&quot;&gt;Cherenkov Telescope Array&lt;/a&gt; which has an angular resolution able to image in microarcseconds. Whilst the Cherenkov Telescope Array, is limited to &lt;code&gt;100μas&lt;/code&gt; at &lt;code&gt;400nm&lt;/code&gt; and not able to image &lt;code&gt;1μas&lt;/code&gt;, at this next level it's we are imaging at &lt;code&gt;100ly&lt;/code&gt;. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaia_%28spacecraft%29&quot; rel=&quot;nofollow&quot;&gt;Gaia spacecraft&lt;/a&gt; can resolve up to &lt;code&gt;20μas&lt;/code&gt; however is not able to image at this level. The NASA Ames Research Center is demonstrating resolution abilities down to &lt;code&gt;5μas&lt;/code&gt; at the attempt to resolve down to &lt;code&gt;1μas&lt;/code&gt;, however again that is not imaging resolution. For radio waves, truly enough I had not mentioned the inverse square law and wave degradation. For us as humans, yes a few light-years may be possible with a realm of possibility opening up with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Square_Kilometre_Array&quot; rel=&quot;nofollow&quot;&gt;Square Kilometer Array&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want me to retract my guesstimate from the first time, pollution and transit photometry are in fact possible using today's existing technology within &lt;code&gt;1ly&lt;/code&gt;, on par with existing radio receivers within &lt;code&gt;1yr&lt;/code&gt;. If you deter from the fact that the new instruments are however not built yet you can vastly increase this up to &lt;code&gt;100ly&lt;/code&gt;, just because something isn't built does not make the technology non-existent (Is the SKA technology feasible? Yes, we have the technology to make it happen right now, we just haven't done so. That does &lt;strong&gt;not&lt;/strong&gt; make it technology that does not exist).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.seti.org/seti-institute/kepler-186f-first-earth-sized-planet-orbiting-in-habitable-zone-of-another-star&quot; rel=&quot;nofollow&quot;&gt;Seti Home&lt;/a&gt; has published finding of the the first earth sized planet detected from transiting. Further publication by &lt;a href=&quot;http://arxiv.org/abs/1404.5667&quot; rel=&quot;nofollow&quot;&gt;Cornell University Library&lt;/a&gt; claims that the planet is within the habitable zone and implies it's within possibility of having an atmosphere and liquid H20 on it's surface. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Kepler_%28spacecraft%29&quot; rel=&quot;nofollow&quot;&gt;Kepler&lt;/a&gt; spacecraft detected this finding, in case you are unaware, Kepler maps light curves as a body transits across the face of another body, this is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Transit_%28astronomy%29&quot; rel=&quot;nofollow&quot;&gt;Transit&lt;/a&gt;. To even suggest that this technology doesn't already exist is preposterous, if you want a true analogue to Earth as is, with technology already existing; &lt;code&gt;1ly&lt;/code&gt;, if you want to use technology possible however not built; &lt;code&gt;100ly&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5398" LastEditorUserId="5398" LastEditDate="2015-01-25T01:00:09.773" LastActivityDate="2015-01-25T01:00:09.773" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="8652" PostTypeId="2" ParentId="8081" CreationDate="2015-01-24T05:51:07.440" Score="3" Body="&lt;p&gt;I believe, I don't know if there is evidence of it however, that Pluto does not have an active core. An active core as I'm sure your aware produces the magnetic field, without a magnetic field neither Pluto nor Charon have a magnetosphere. If there was sufficient or any evidence at all to suggest otherwise, I believe it would have become a priority.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretical models of Pluto are based typically on it's orbital speed, which suggest there is no field. However as you have stated we have witnessed tidal stresses producing activity, such as Io, which should be frozen like rock however is the most geological active body we know of it the solar system. One major difference, is we don't have images of neither Plutos nor Charons surfaces, if we had images of a smooth surface we could suffice the surface at least is active through tidal forces. Being all cratered and looking ~5 billion years old indicates the opposite. There just isn't any data as yet to say otherwise, and certainly wasn't when New Horizons was being designed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for what the craft could detect with it's instruments, the most indicative feature would be optical imagery and the dwarfs' surfaces. In essence, all of it's instruments could be used; ALICE for detecting atmospheric properties could certainly indicate a magnetic field, whether it's able to hold onto any material spewing from it's surface for example and any ionization of any atmosphere, which in turn is what PEPSSI does. The other great tool would be Ralph, mapping thermal imagery of the dwarfs. If there is indication of internal heating unexpected from gravitational compression of the core, or veins of thermal radiation could also help determine any activity in turn producing the magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this was what you were asking.&lt;/p&gt;&#xA;" OwnerUserId="5398" LastEditorUserId="2531" LastEditDate="2015-01-24T10:24:07.180" LastActivityDate="2015-01-24T10:24:07.180" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8653" PostTypeId="2" ParentId="6104" CreationDate="2015-01-24T07:08:48.220" Score="1" Body="&lt;p&gt;No, it's not the best reproduction of explaining the big bang. Generally when you think logically, it was an explosion of matter, however it was not a fireworks show.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To best explain it simply, it is conceived as a singularity that rapidly expanded. An important thing to understand is that within the beginning of the expansion, the matter was quark–gluon plasma and opaque, meaning there was no 'BANG'. Intensity of the heat was too great to allow quarks to form hadrons. Hadrons comprise of baryons such as protons and neutrons, at the earliest stage of inflation protons had not formed, being a representation of the 'big bang' they aren't implicitly going to introduce these factors. A general representation if the bang factor itself, if the bang is presented as an inflating balloon filled with gas, it's not going to attract. To be a true analogue, the balloon would simply appear into existence (between 10^-6 sec and 1 sec), already expanded and have pockets of matter now cool enough to begin producing light, within the next 10 seconds electrons begin to form attracting photons and neutrons, the first atomic bonds (hydrogen) would continue for the next 380,000 years. Much of the earlier matter has been annihilated by antimatter and we begin to see matter in the form it is today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a nutshell, it's hard to present this theory when something just 'becomes', so I believe they use the explosion as a way of demonstrating the nature of forces rather than an analogue of mere inflation.&lt;/p&gt;&#xA;" OwnerUserId="5398" LastActivityDate="2015-01-24T07:08:48.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8654" PostTypeId="2" ParentId="5879" CreationDate="2015-01-24T07:35:56.953" Score="1" Body="&lt;p&gt;Some can, some can't. Viewing really is determined by ability, experience and technique when you leave viewing conditions aside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of DSOs have a lower surface brightness, one thing I always notice is when an observer continuously increases their magnification to see the object larger. Surface brightness is proportional to magnification, the higher the magnification the less bright an object appears, to see an object as bright as possible you need to use the lowest possible magnification. Using a smaller diameter eyepiece (larger magnification), you will need to develop techniques such as averted vision to truly see the beauty of some objects that otherwise appear dull.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding them can be exceptionally hard if your not familiar with the viewing region, easily overlooked or thought to be a tuft of cloud. Eye adaption is key when viewing so you need to spend significant time on an object before the details appear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Messier objects contain some well known DSOs, and can show some amazing details with patience. Understanding that the Messier objects were cataloged with poor optics and small apertures makes them prime targets for beginners.&lt;/p&gt;&#xA;" OwnerUserId="5398" LastActivityDate="2015-01-24T07:35:56.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8656" PostTypeId="1" AcceptedAnswerId="8657" CreationDate="2015-01-24T12:54:56.027" Score="4" ViewCount="2828" Body="&lt;p&gt;At the center of the milky way, there must be some giant mass that keeps the galaxy from just floating away from itself. So, obviously, we feel the immense gravitational pull of whatever this is. But, on a smaller scale we can see this with the earth and the sun. The sun keeps us from spinning into oblivion with its gravitational pull. The question I am asking is, why do we orbit the sun, but not the center of the galaxy, which has a much stronger gravitational pull?&lt;/p&gt;&#xA;" OwnerUserId="5400" LastActivityDate="2018-05-30T22:46:13.657" Title="Why don't we orbit the center of our galaxy?" Tags="&lt;gravity&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8657" PostTypeId="2" ParentId="8656" CreationDate="2015-01-24T13:08:56.453" Score="6" Body="&lt;p&gt;The so called &lt;a href=&quot;https://en.wikipedia.org/wiki/Inverse-square_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;inverse square law&lt;/a&gt; makes gravity dominate locally. The Moon orbits Earth, Earth orbits the Sun, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_year&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Sun orbits the Galactic center&lt;/a&gt;! Because the distance from Earth to the Moon is 1/500 the distance from Earth to the Sun, it is Earth's gravitational field which dominates the orbit of the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it is not a single thing which attracts mass to orbit the galaxy. It is the sum of mass of everything in it. The center is of course the galactic center of mass.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorDisplayName="user1569" LastEditDate="2018-05-23T07:24:00.030" LastActivityDate="2018-05-23T07:24:00.030" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="8658" PostTypeId="2" ParentId="8146" CreationDate="2015-01-24T15:33:19.923" Score="11" Body="&lt;p&gt;I had put off answering this question because it seems too broad without specifying what sort of detection methods are proposed. But if you answer it directly from the perspective of - if we were to take the solar system and put it at some distance from us, would we able to detect signs of life on planet Earth - then the answer is probably not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &lt;em&gt;current technology&lt;/em&gt; (and by that I mean experiments and telescopes that are available now) we would probably be unable to detect life on Earth even if observed from a distance of a few light years. Therefore there are no stars within this sphere (other than the Sun).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;No planets quite like the Earth have yet been detected around another star. That is to say, none that have a similar mass, radius and orbit at 1 au (or close to it) from a solar-type star [EDIT: There is of course now a close contender in Kepler-452b, though it is 60% bigger than Earth; &lt;a href=&quot;http://arxiv.org/abs/1507.06723&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jenkins et al. 2015&lt;/a&gt;.]. With current technology, it is just about in reach. Therefore any directed search for life on Earth has a limited number of places in which to start. If you can't detect the planet &lt;em&gt;at all&lt;/em&gt; then there is absolutely no chance of looking at its atmospheric composition to look for biomarkers (e.g. oxygen along with a reducing gas like methane, or chlorofluorocarbons from an industrial civilisation - &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014ApJ...792L...7L&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lin et al. 2014&lt;/a&gt;). The only exoplanets for which atmospheric compositions have been (crudely and tentatively) measured are &quot;hot Jupiters&quot;. - giant exoplanets orbiting very close to their parent stars. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A &quot;blind&quot; search could look for radio signatures and of course this is what SETI has been doing. If we are talking about detecting &quot;Earth&quot;, then we must assume that we are not talking about deliberate beamed attempts at communication, and so must rely on detecting random radio &quot;chatter&quot; and accidental signals generated by our civilisation.&#xA;The SETI Phoenix project was the most advanced search for radio signals from other intelligent life. Quoting from &lt;a href=&quot;http://adsabs.harvard.edu/full/2000ASPC..213..451C&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cullers et al. (2000)&lt;/a&gt;: &quot;&lt;em&gt;Typical signals, as opposed to our strongest signals, fall below the detection threshold of most surveys, even if the signal were to originate from the nearest star&lt;/em&gt;&quot;. Quoting from &lt;a href=&quot;http://adsabs.harvard.edu/abs/2001ARA%26A..39..511T&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tarter (2001)&lt;/a&gt;: &quot;&lt;em&gt;At current levels of sensitivity, targeted microwave searches could detect the equivalent power of strong TV transmitters at a distance of 1 light year (within which there are no other stars)...&lt;/em&gt;&quot;. The equivocation in these statements is due to the fact that we &lt;em&gt;do&lt;/em&gt; emit stronger beamed signals in certain well-defined directions, for example to conduct metrology in the solar system using radar. Such signals have been calculated to be observable over a thousand light years or more. But these signals are brief, beamed into an extremely narrow angle and unlikely to be repeated. You would have to be very lucky to be observing in the right direction at the right time if you were performing targeted searches. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hence my assertion that with current methods and telescopes there is not much chance of success. But of course technology advances and in the next 10-20 years there may be better opportunities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step in a directed search would be to find planets like Earth. The first major opportunity will be with the &lt;a href=&quot;http://tess.gsfc.nasa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;TESS spacecraft&lt;/a&gt;, launching in 2017, capable of detecting earth-sized planets around the brightest 500,000 stars. However, it's 2-year mission would limit the ability to detect an Earth-analogue. The best bet for finding other Earths will come later (2024 perhaps) with the launch of &lt;a href=&quot;http://www.esa.int/Our_Activities/Space_Science/ESA_selects_planet-hunting_PLATO_mission&quot; rel=&quot;nofollow noreferrer&quot;&gt;Plato&lt;/a&gt;, a six-year mission that again, studies the brightest stars. However, there is then a big leap forward required to perform studies of the atmospheres of these planets. Direct imaging and spectroscopy would probably require space-borne nulling interferometers; indirect observations of phase-effects and transmission spectroscopy through an exoplanet atmosphere does not require great angular resolution, just massive precision and collecting area. Spectroscopy of something the size of Earth around a normal star will probably require a bigger &lt;em&gt;successor&lt;/em&gt; to the James Webb Space Telescope (&lt;a href=&quot;http://www.jwst.nasa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;JWST&lt;/a&gt; - launch  2018), or even more collecting area than will be provided by the E-ELT in the next decade. For example &lt;a href=&quot;http://www.epj-conferences.org/articles/epjconf/abs/2013/08/epjconf_hpcs2012_11001/epjconf_hpcs2012_11001.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Snellen (2013)&lt;/a&gt; argues it would take 80-400 transits-worth of exposure time (i.e. 80-400 years!) to detect the biomarker signal of an Earth-analogue with the E-ELT!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has been suggested that new radio telescope projects and technology like the Square Kilometre Array may be capable of serendipitously detecting radio &quot;chatter&quot; out to distances of 50 pc ($\sim 150$ light years) - see &lt;a href=&quot;http://adsabs.harvard.edu/abs/2007JCAP...01..020L&quot; rel=&quot;nofollow noreferrer&quot;&gt;Loeb &amp;amp; Zaldarriaga (2007)&lt;/a&gt;. This array, due to begin full operation some time after 2025 could also monitor a multitude of directions at once for beamed signals. A good overview of what might be possible in the near future is given by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2009astro2010S.294T&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tarter et al. (2009)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-01-31T12:21:54.590" LastActivityDate="2017-01-31T12:21:54.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8659" PostTypeId="2" ParentId="8656" CreationDate="2015-01-24T16:22:57.357" Score="8" Body="&lt;p&gt;There are a number of misconceptions with this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, the &quot;gravitational pull&quot; (which I'm interpreting as gravitational force) by the Milky Way on the Earth is seven orders of magnitude smaller than that exerted by the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, gravitational force is the wrong metric. The Newtonian gravitational force exerted by the Sun on the Moon is more than twice that exerted by the Earth. Yet all but a tiny, tiny minority of professionals who deal with the solar system will say that the Moon does indeed orbit the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thirdly, it's not correct to look at &quot;orbit&quot; as being a mutually exclusive concept. That the Moon does indeed orbit the Earth doesn't mean that it doesn't orbit the Sun. It does. The Moon not only orbits the Earth and Sun, but it also orbits the Milky Way, the Local Group, and the Local Supercluster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the right way of looking at orbits amongst a hierarchy of masses? One issue that needs to be addressed is what the word &quot;orbit&quot; means. Note that people do write and talk about parabolic and hyperbolic &quot;orbits&quot;. I'm assuming that &quot;orbit&quot; in the context of this question means a bounded trajectory, which would rule out calling a hyperbolic or parabolic trajectory an &quot;orbit&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That leads to the first of three characteristics of whether object A can be said to be orbiting object B: Object A can be said to be orbiting object B only if object A is gravitationally bound to object B. The appropriate metric here is energy, at least for cosmologically small distances. (Note: The distance between the Milky Way and the Andromeda Galaxy is very small in a cosmological sense.) From the perspective of a frame centered at object B, the total mechanical energy of object A needs to be negative he mechanical energy to be able to say that object A is (at least temporarily) orbiting object B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bounded orbits remain bounded forever in the two body point mass problem. That may not be the case when other larger objects are involved. This leads to a second criterion: Object A can be said to be orbiting object B only if the gravitational influences of larger objects on object A are but small perturbative effects that don't result in instabilities. The appropriate metric here is an energy-based sphere of influence. The two most widely used spheres are the Hill sphere and the Laplace sphere. For example, because the Moon is gravitationally bound to the Earth and because the Moon is well inside the Earth's Hill and Laplace spheres with respect to the Sun, it is appropriate to say that the Moon does indeed orbit the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the Sun and the Milky Way (and even larger objects such as the Local Group)? Can the Moon also be said to orbit those? The answer is yes. One way to look at it is to ask what would happen if the Earth suddenly disappeared. Would the Moon's path around the Sun change much? The answer is no. The Moon would still be orbiting the Sun at about one astronomical unit. The same applies to the Earth with respect to the Milky Way. The Earth's trajectory about the galaxy would not change by much at all if the rest of the solar system magically disappeared. This point of view however leads to some troubling issues. Planets (and stars and galaxies) don't magically disappear. Moreover, if Saturn somehow did magically disappear, its rings and all but its very outermost moons would be on hyperbolic trajectories with respect to the Sun. A way around this magic is to say that if object A orbits object B, and object B orbits object C, then object A also orbits object C. &quot;Orbit&quot; is not a mutually exclusive concept.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-01-24T16:22:57.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8660" PostTypeId="2" ParentId="874" CreationDate="2015-01-24T16:25:50.037" Score="3" Body="&lt;p&gt;Yes, we do. The exact spot is: All of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Big Bang, space-time itself expanded. To say that it took place only at a certain location, you would need something which is external to the entire universe. There is no such thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relative to us, we are inside the region that the Big Big started at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, hypothetically there were some sort of reference frame external to this universe, like some sort of 8-dimensional (6 spatial, 2 time) Super Universe (SU), the people in the SU could say, &quot;That 4 dimensional sub universe began expanding outward from location X.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;we ourselves are still inside the Big Bang.&lt;/p&gt;&#xA;" OwnerUserId="5402" LastActivityDate="2015-01-24T16:25:50.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8661" PostTypeId="2" ParentId="8222" CreationDate="2015-01-24T19:28:20.813" Score="1" Body="&lt;p&gt;Sample returns are expensive and risky.  Radio is simple and reliable.  Radio bandwidth on a particular mission is determined by several engineering tradeoffs.  Higher bandwidth requires either larger antennas, more output power, or more complicated processing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spacecraft are typically limited in mass, power and compute capabilities. What most missions have plenty of, is time.  Lower bitrate takes longer, but allows more mass and power to be used by other experiments.  This is often a highly desirable tradeoff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Missions like New Horizons that have a fairly limited time to record lots of observations do so to a data storage device, and then spend months or years relaying the data back to Earth during the long coast to the next target.&lt;/p&gt;&#xA;" OwnerUserId="5404" LastActivityDate="2015-01-24T19:28:20.813" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8662" PostTypeId="2" ParentId="7997" CreationDate="2015-01-24T20:01:42.857" Score="3" Body="&lt;p&gt;When material is shed from a comet, it mostly continues to follow the same orbit as the comet, but drifts out ahead or behind the comet in a complex way due to interactions with solar wind and the gravity of other objects.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you imagine the elliptical orbit of 55P/Tempel-Tuttle, it will have patches of denser dust in clumps spaced out around the ellipse corresponding to different perihelion approaches.  The 1466 clump will be in a different place than the 1500 clump, and they will both be travelling more or less around the comet's orbit.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the comet's orbit only crosses Earth's at one place, and we only cross that spot once a year, we will only encounter whatever clumps of dust happen to be at that point at the same time the Earth is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating exactly where these clumps of comet dust are is non-trivial.  And that is a big reason why predicting how good a meteor shower will be is a rather tricky business.&lt;/p&gt;&#xA;" OwnerUserId="5404" LastActivityDate="2015-01-24T20:01:42.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8663" PostTypeId="2" ParentId="7936" CreationDate="2015-01-24T20:47:02.963" Score="2" Body="&lt;p&gt;What you're looking for is the latitude and longitude of the subsolar point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating it with accuracy is somewhat complicated.  Here is a python library that has routines to do it for you:&#xA;&lt;a href=&quot;http://rhodesmill.org/pyephem/&quot; rel=&quot;nofollow&quot;&gt;http://rhodesmill.org/pyephem/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also possibly query this information from other sites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really want to work it out on your own, I suggest &quot;Spherical Astronomy&quot; by Robin M. Green published by Cambridge University Press as a starting point.&lt;/p&gt;&#xA;" OwnerUserId="5404" LastActivityDate="2015-01-24T20:47:02.963" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8664" PostTypeId="2" ParentId="7888" CreationDate="2015-01-24T21:05:04.320" Score="3" Body="&lt;p&gt;The main problem with determining an object's orbit is we only know the position with certainty in two dimensions.  The distance to the object is largely unknown.  This accounts for the large uncertainty in the period of newly discovered TNOs.  Many possible orbits could fit the early observations, and therefore the uncertainty is large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As time goes on, and more observations are made, more and more of these possible orbits are excluded and the orbit fit becomes more exact.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer your question, it really depends on how exact you want the answer, and how good your observations are.  The more observations you take, and the better they are, the closer your orbital fit becomes.  And more importantly, the bigger the arc of the orbit your observations cover, the more accurate the orbit becomes. It is impossible to give a specific number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give a practical example, Pluto takes 248 years to orbit the Sun.  It was discovered in 1930, so we've only observed it for 85 years, or about 1/3rd of one orbit.  One of the problems the New Horizons probe sent to Pluto had to overcome was that we didn't know, at launch, exactly how far away Pluto was, so therefore we didn't know exactly when New Horizons would pass the dwarf planet. That made it difficult to tell the probe where to point to take pictures at closest approach.&lt;/p&gt;&#xA;" OwnerUserId="5404" LastEditorUserId="1288" LastEditDate="2015-08-03T16:40:21.490" LastActivityDate="2015-08-03T16:40:21.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8665" PostTypeId="2" ParentId="7996" CreationDate="2015-01-24T21:24:04.477" Score="0" Body="&lt;p&gt;The Keplerian elements for the planets provides all of the information needed to define the orbit and where the planet is currently along it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Approximate elements are available from JPL and can be found here: &lt;a href=&quot;http://ssd.jpl.nasa.gov/txt/p_elem_t1.txt&quot; rel=&quot;nofollow&quot;&gt;http://ssd.jpl.nasa.gov/txt/p_elem_t1.txt&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't want to do the math, you can just ask JPL Horizons to do it for you and calculate an ephemeris telling you which direction each planet is in right now: &#xA;&lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow&quot;&gt;http://ssd.jpl.nasa.gov/horizons.cgi&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5404" LastActivityDate="2015-01-24T21:24:04.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8666" PostTypeId="2" ParentId="5891" CreationDate="2015-01-24T21:31:45.150" Score="1" Body="&lt;p&gt;With the wedge in place, you will only do polar alignments.  If you want to go back to az/el, you must remove the wedge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For me polar alignments are much faster and easier than az/el alignments.&#xA;For polar:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;level the mount&lt;/li&gt;&#xA;&lt;li&gt;Set your latitude on the wedge&lt;/li&gt;&#xA;&lt;li&gt;Let the auto alignment point the scope at dec 90 and manually spin the mount until you align on Polaris.  You use the manual controls on the wedge for this, not the hand controller.&lt;/li&gt;&#xA;&lt;li&gt;Do a one star alignment on an overhead star.  I always use Vega in the summertime.  You use the hand controller for this.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you haven't moved the mount since last time (or if it is permanently mounted) all you have to do is power up and do a one star alignment.  Easy!&lt;/p&gt;&#xA;" OwnerUserId="5404" LastActivityDate="2015-01-24T21:31:45.150" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8667" PostTypeId="1" CreationDate="2015-01-24T21:47:12.843" Score="1" ViewCount="209" Body="&lt;p&gt;So I was watching a documentary on television about this and a number of theories that are proposed as to what the fate of the Universe will be:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&quot;Big Crunch&quot;&lt;/li&gt;&#xA;&lt;li&gt;&quot;Big Freeze&quot;&lt;/li&gt;&#xA;&lt;li&gt;&quot;Big Rip&quot; &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I understand these, but there was also another scenario which I don't understand:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;A bubble forms in the Universe and expands at the speed of light. It annihilates everything in the Universe that it comes into contact with, until eventually at a very, very distant time the Universe ceases to exist. Does anyone have any clue as to what they were talking about (because I don't)?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2562" LastEditorUserId="2153" LastEditDate="2015-01-24T22:21:20.323" LastActivityDate="2019-07-13T11:25:07.370" Title="The destruction of the Universe by a bubble" Tags="&lt;universe&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8668" PostTypeId="2" ParentId="7925" CreationDate="2015-01-24T22:08:49.737" Score="2" Body="&lt;p&gt;The easiest method of photometry is by comparing the measured instrumental intensity of the unknown target with one or more other known targets in the same field and then converting the difference into magnitudes.  This is called differential photometry, and is easier because you don't have to worry about things like atmospheric extinction.  The more comparison stars you use, the better your result.  There are tools that will automatically do this analysis and calculate your error bars.  I use MaximDL.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Visit &lt;a href=&quot;http://www.aavso.org&quot; rel=&quot;nofollow&quot;&gt;http://www.aavso.org&lt;/a&gt; and use the Variable Star Plotter (VSP) tool to generate a chart for your star of interest.  This will give known magnitudes of comparison stars in the same field.  You can use those to determine the magnitude of your target.  While you're there, check out some of the FAQs and documentation on variable star observing, and submit your results to the online database.&lt;/p&gt;&#xA;" OwnerUserId="5404" LastActivityDate="2015-01-24T22:08:49.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8669" PostTypeId="2" ParentId="8667" CreationDate="2015-01-24T22:19:46.937" Score="4" Body="&lt;p&gt;I think I know what they mean. They're talking about a &lt;a href=&quot;https://en.wikipedia.org/wiki/False_vacuum&quot; rel=&quot;nofollow noreferrer&quot;&gt;false vacuum scenario&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;A vacuum state is a state of lowest energy. It's thought that the vacuum of our universe is in a lowest-energy state and is stable, and so nothing special will happen to spacetime. &lt;em&gt;However&lt;/em&gt;, if our universe is actually a false vacuum, then it could merely be &lt;a href=&quot;https://en.wikipedia.org/wiki/Metastability&quot; rel=&quot;nofollow noreferrer&quot;&gt;metastable&lt;/a&gt;, and some tiny perturbation could cause it to fall into a lower energy state - a true vacuum, or else a false vacuum of lower energy. The &amp;quot;bubble&amp;quot; that would appear would be a region of this lower-energy vacuum that would expand across the universe without stopping.&lt;/p&gt;&#xA;&lt;p&gt;Scientists aren't sure if our universe is a false vacuum. &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/en/7/70/Higgs-Mass-MetaStability.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's&lt;/a&gt; a chart of the masses of the Higgs boson and the top quark:&#xA;&lt;a href=&quot;https://web.archive.org/web/20150426132139/http://upload.wikimedia.org:80/wikipedia/en/7/70/Higgs-Mass-MetaStability.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://web.archive.org/web/20150426132139/http://upload.wikimedia.org:80/wikipedia/en/7/70/Higgs-Mass-MetaStability.svg&quot; alt=&quot;Higgs boson vs. top quark&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The latest measurements suggest that their masses lie in a metastable region of the graph (which could be bad), although it's towards the stable end of the region (which is good). This vacuum catastrophe could still happen, but the odds aren't in its favor. I also suggest reading some of the excellent papers in the references section of the Wikipedia article. They're quite comprehensive.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Better explanation:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The universe can be thought of as space containing a variety of quantum fields. A quantum field might be best thought of as something that has a value at every point in space. Put together a bunch of these fields and you can describe particles, particle interaction, and all the matter and energy in a given region of space - in fact, in the whole universe!&lt;/p&gt;&#xA;&lt;p&gt;Picture a region of space with absolutely nothing in it. Nothing. (I'm ignoring &lt;a href=&quot;https://en.wikipedia.org/wiki/Vacuum_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;vacuum energy&lt;/a&gt;, even though I really should discuss it) That's a true vacuum. Remove all the fields, particles, and other interesting stuff from our universe and that's what you'll get. Nothing.&lt;/p&gt;&#xA;&lt;p&gt;The false vacuum scenario happens if there isn't actually &amp;quot;nothing&amp;quot; but &amp;quot;something&amp;quot;. If our &amp;quot;vacuum&amp;quot; has some extra energy that it shouldn't (again, I'm not talking about normal vacuum energy). In other words, there's something where there should be nothing.&lt;/p&gt;&#xA;&lt;p&gt;In this end-of-the-universe scenario, the universe goes from having this something to having nothing - from a false vacuum to a true vacuum. The region of space actually having nothing starts at a certain region and expands outward at the speed of light. That's what the scientists were talking about.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;For anyone who wants a further explanation and/or could tell how many mistakes I made there: Yes, I'm aware that there were mistakes in that incomplete explanation. I didn't properly explain vacuum energy or an energy state, nor did I discuss stability or even properly touch on quantum fields. But quite frankly, I don't think this explanation needs that baggage. Does it add to the richness of the concept? Yes. But is it confusing? Also yes. I don't think working that in would be productive.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-07-13T11:25:07.370" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="8671" PostTypeId="2" ParentId="8667" CreationDate="2015-01-25T00:42:52.550" Score="3" Body="&lt;p&gt;Perhaps this analogy is apt, or perhaps not...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes when you go out to the car on a cold winter morning, the windscreen looks clear, but there is actually a supercooled film of water on it. As soon as you start up the engine, the vibration disturbs the unstable equilibrium, the water freezes on the windscreen, the phase transition occurring at points and spreading. At the same time latent heat of crystallisation is released.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the idea is meant to be that our universe could be in that supercooled state and that there is actually a more energetically favourable phase that the (false) &quot;vacuum&quot; could be in, but the transition to that new phase would release sufficient energy to drive a new bubble of cosmic inflation.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-01-25T00:42:52.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8672" PostTypeId="2" ParentId="5923" CreationDate="2015-01-25T03:32:58.040" Score="2" Body="&lt;p&gt;Take for instance radio, when listening to radio you gather static on the frequency. A passband filters the static which in turn provides a clearer signal. Objects radiate in multiple frequencies/wavelengths, for instance heat/thermal (infrared) and visible. Not all object radiate in a range of frequencies, humans for example can be seen well in infrared and visible, but our skin and circulatory system are opaque in x-ray and radio. &lt;strong&gt;Spitzer MIPS 24µm&lt;/strong&gt; is the wavelength corresponding to mid-infrared (between 3µm-50µm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the statement is saying, is multiple imaging sources have been used for the same frequency. When choosing the Spitzer MIPS 24µm result, it additionally includes results from other imaging sources such as the IRAS 25µm result. It does this because the filters of the passband overlap and return the same result, as a result the process is changing where it only displays the result of the filter you choose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All radiation is apart of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_spectrum&quot; rel=&quot;nofollow&quot;&gt;electromagnetic spectrum&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectral_energy_distribution&quot; rel=&quot;nofollow&quot;&gt;spectral energy distribution&lt;/a&gt; is a way to present the data results as a plot, in turn that can be used to identify the source of the radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The observed passband, is just stating which frequencies are being filtered, such as perhaps within a small range. For example, a Quasar is a highly active radio source typically found in the center of young galaxies. In correlation, you can determine what kind of stellar objects you would expect given a certain range of frequencies or the observed passband.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frequency is the simple measure of how frequent something happens, for example if you flash a torch once a second, the time between each flash is 1 second long, the frequency 1Hz. If you flash a second torch twice a second, the time between each flash is a 1/2 second long, the frequency is 2Hz. More relevant in physics, is the correlation between wavelength, frequency and photon energy. Waves oscillate and light acts as waves, and the frequency between each wave correlates to how fast the wave is moving and what wavelength the radiation is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Homogenized results are just a collection of results which are made to fit a pattern, sometimes to better represent the data (this may be filtering results in a database to exclude results outside a range of study). Published results on the other hand are raw results which include a full spectrum of the data. I am unsure about the mode aperture, this may refer to the sensitivity and correlate to how much interference/static is captured with the wave. Best if someone else can explain this.&lt;/p&gt;&#xA;" OwnerUserId="5398" LastActivityDate="2015-01-25T03:32:58.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8673" PostTypeId="1" AcceptedAnswerId="8679" CreationDate="2015-01-25T09:39:31.553" Score="2" ViewCount="769" Body="&lt;p&gt;I've downloaded hipparcos catalog from the Strasbourg Universe but whilst there is a lot of information, I can't find the names of the stars that the HipID relates to.  There is the wikipedia page which lists some but not all the stars, for instance Zosma (54872) is missing from the list and don't fancy searching for each and every star page to find out the missing details such as what the HIP's name is, how far away it is, etc.  Is there a downloadable list which lists the HipID, the name, how far away it is, the constellation etc?&lt;/p&gt;&#xA;" OwnerUserId="5401" LastActivityDate="2015-01-29T10:37:33.920" Title="How to complete the Hipparcos Catalog?" Tags="&lt;star&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8674" PostTypeId="1" AcceptedAnswerId="8680" CreationDate="2015-01-25T12:21:53.863" Score="4" ViewCount="1967" Body="&lt;p&gt;I stumbled upon &lt;a href=&quot;http://ned.ipac.caltech.edu/level5/March02/Sahni/Sahni4_5.html&quot; rel=&quot;nofollow&quot;&gt;this text&lt;/a&gt; which illustrates and says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We find that as the object is moved to higher redshifts its angular&#xA;  size first decreases (as naively expected) but soon begins to increase&#xA;  after passing through a minimum value.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is this correct for the mainstream standard model? What would an English language explanation be?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-01-28T22:39:57.130" Title="Do objects look larger the further away they are, beyond z=1?" Tags="&lt;distances&gt;&lt;expansion&gt;&lt;redshift&gt;&lt;angular-diameter&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8675" PostTypeId="1" CreationDate="2015-01-25T12:36:06.330" Score="8" ViewCount="330" Body="&lt;p&gt;Space radio interferometers could have a baseline of millions of kilometers, but is there a point where a larger baseline doesn't improve the resolution anymore because the photons observed are distorted before they arrive? &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6624/is-there-a-theoretical-limit-on-telescopes-resolution&quot;&gt;This question&lt;/a&gt; deals with technological limits of resolution. I'm instead asking about cosmic limitations due to for example interstellar and extragalactic gas which scatters light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1501.04449&quot; rel=&quot;noreferrer&quot;&gt;This paper&lt;/a&gt; about results from the RadioAstron space/Earth interferometer is well above my pay grade, but it seems to be about this problem. The executive summary says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At longer baselines of up to 235,000 km, where no interferometric&#xA;  detection of the scattering disk would be expected, significant&#xA;  visibilities were observed with amplitudes scattered around a constant&#xA;  value. These detections result in a discovery of a substructure in the&#xA;  completely resolved scatter-broadened image of the pointlike source,&#xA;  PSR B0329+54. They fully attribute to properties of the interstellar&#xA;  medium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2018-06-01T04:23:39.833" Title="Is there a cosmic, rather than technological, upper limit to what a telescope can resolve?" Tags="&lt;telescope&gt;&lt;radio-astronomy&gt;&lt;space-telescope&gt;&lt;angular-resolution&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8676" PostTypeId="2" ParentId="5923" CreationDate="2015-01-25T12:50:22.637" Score="1" Body="&lt;p&gt;Ashley's comments are all fine. I'll add that homogenisation just means that where you have results from different instruments at the same frequency, they have been averaged in some way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reference code refers to something like this  1995MNRAS.273..559J&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That can be used to look up the paper referred to at somewhere like the &lt;a href=&quot;http://www.adsabs.harvard.edu/&quot; rel=&quot;nofollow&quot;&gt;NASA ADS&lt;/a&gt; catalogue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you add a sample of the output you are looking at: uncertainty would usually mean it indicates an &quot;error bar&quot; on the flux or magnitude in question - i.e. what is the estimated uncertainty in the measurement. The units are probably just referring to whether this is quoted in magnitudes or Janskys right?&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-01-25T12:50:22.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8677" PostTypeId="2" ParentId="8673" CreationDate="2015-01-25T13:01:08.280" Score="2" Body="&lt;p&gt;Stars have lots of names; it would make the catalogue unwieldy to include them all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in particular objects it is reasonably simple to work this out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do a &quot;search by identifier&quot; at &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-fid&quot; rel=&quot;nofollow&quot;&gt;CDS SIMBAD&lt;/a&gt;. If I put in your identifier - Zosma  - it comes up with nothing. Zosma is in fact the arabic name for delta Leo. Anyhow, if I input delta Leo I find that one of its alternate names (in a long list) is HIP 54872  i.e. it is star 54872 in the Hipparcos catalogue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can of course do the the other way around by inputting HIP 54872, but I'm afraid the arabic name is not part of the results, not least I guess because not everyone agrees on what they are or what the spelling is for many.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are ways of automating this process for lists of queries if you wish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely the Hipparcos catalogue gives you the distance (or parallax) to the star!&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-01-25T13:08:09.277" LastActivityDate="2015-01-25T13:08:09.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8678" PostTypeId="2" ParentId="942" CreationDate="2015-01-25T13:18:15.153" Score="6" Body="&lt;p&gt;Measuring the ages of stars is an incredibly difficult process. It is not something that can be summarised adequately in an SE answer. Please take a look at these references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2013arXiv1311.7024S&quot; rel=&quot;nofollow noreferrer&quot;&gt;Soderblom et al. (2013)&lt;/a&gt;; &lt;a href=&quot;http://arxiv.org/abs/1404.7156&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jeffries (2014)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methods split into -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fundamental - e.g. radioisotope dating - only possible for the Sun&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Semi-fundamental - kinematic traceback of stars to their birthplace; observing the depletion of lithium in a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Model-dependent - comparing the luminosity and temperature (and composition) of a star with that predicted by stellar evolutionary models. Or equivalently, its surface gravity and temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Empirical - looking at how things like rotation and magnetic activity decline with age and using this to estimate the age of a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statistical - using the metallicity or galactic space motions of a star to provide a (crude) estimate of its age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a lot more detail given in this SE answer.&#xA;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/1041/without-using-absolute-magnitudes-or-isochrones-how-might-we-tell-a-stars-age/8122#8122&quot;&gt;Without using absolute magnitudes or isochrones, how might we tell a star&amp;#39;s age and evolutionary status?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-05-12T17:06:03.297" LastActivityDate="2017-05-12T17:06:03.297" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8679" PostTypeId="2" ParentId="8673" CreationDate="2015-01-25T14:46:05.437" Score="3" Body="&lt;p&gt;Most of the stars in the Hipparcos catalogue do not have a common name. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the main catalogue file you will also find the Henry Draper (HD) number of the star (if it has one) at columns 391-396. You can use this ID to find the name in the Bright Star Catalogue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Bright Star Catalogue contains all stars brighter than magnitude 6.5 (naked eye stars) together with the HD number (columns 26-31) and a Name (columns 5-14), which in fact is the Bayer (greek letter - constellation) or Flamsteed (number - constellation) name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using both catalogues you can get the Bayer or Flamsteed name from the HIP identifier. I do not know of any catalogue to get the common name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance of the star can of course be found in the Hipparcos catalogue, which is the reason the Hipparcos catalogue exists. Columns 80-86 give the trigonometric parallax $\varpi$ in milliarcseconds. You can get the distance from: $$d=\frac{1}{\varpi}$$ where the distance $d$ is in parsec and the parallax $\varpi$ is in arc seconds. For the Hipparcos catalogue you should divide the parallax by 1000 as the parallax is given in milliarcseconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get the constellation for any position in the sky from the algorithm and data in &lt;a href=&quot;http://cdsarc.u-strasbg.fr/viz-bin/Cat?VI/42&quot; rel=&quot;nofollow&quot;&gt;http://cdsarc.u-strasbg.fr/viz-bin/Cat?VI/42&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2015-01-25T14:46:05.437" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8680" PostTypeId="2" ParentId="8674" CreationDate="2015-01-25T15:39:39.273" Score="5" Body="&lt;p&gt;Yes this is correct for the standard model. I believe the English language explanation would be that the angular size of a galaxy does not depend on its current distance to us, but its distance when the light we currently see from it was emitted (here using &quot;distance&quot; synonymous with &quot;proper distance&quot;). Though I also believe there are other complicating factors such as the spatial geometry of the Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the first diagram in Figure 1 in &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0310808&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;, notice how the size of the past light cone in terms of proper distance firstly increases as we go back in time, before reaching a maximum and decreasing to zero at t = 0.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastEditorUserId="2092" LastEditDate="2015-01-25T16:15:31.923" LastActivityDate="2015-01-25T16:15:31.923" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8681" PostTypeId="1" AcceptedAnswerId="8683" CreationDate="2015-01-25T18:50:04.757" Score="3" ViewCount="269" Body="&lt;p&gt;It is my understanding that when a star, a planetary disk, or a galaxy forms, the rotational momentum of the whole system is conserved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the smaller size of the resulting object, it will spin with a significantly higher speed than the original nebula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I do not understand is where the original rotation comes from. Why should a random dust cloud have an overall spin? Wouldn't the impulses of all particles in the cloud tend to average each other out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some alternative source of spin or a reason why nebula have an inherent spin?&lt;/p&gt;&#xA;" OwnerUserId="5407" LastEditorUserId="2531" LastEditDate="2015-01-25T19:49:39.757" LastActivityDate="2015-01-26T08:42:22.980" Title="How do stars or galaxies get their spin?" Tags="&lt;star&gt;&lt;galaxy&gt;&lt;rotation&gt;&lt;protostar&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8682" PostTypeId="2" ParentId="8681" CreationDate="2015-01-25T19:11:43.000" Score="1" Body="&lt;p&gt;Any gaseous object has some spin, usually acquired by interactions with other objects. For example, (proto-)galaxies torque each other to acquire a low rate of angular momentum. Initially, this spin is rather low in the sense that it does not dominate the dynamics: the energy in rotational motion is small compared to other energies, typically by a factor $\sim100$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, energy can be lost via dissipation (and ultimately radiated away), while angular momentum (spin) is much harder to get rid off. That's why rotating gaseous objects eventually form a disc-like configuration (galactic and proto-stellar discs). In these discs, the kinetic energy is dominated by the rotation. Such systems can only significantly evolve if angular momentum can be exchanged and/or transported. For example, the formation of a star from a proto-stellar disc is promoted by (outwards) angular-momentum transport within the disc. The newly born star retains some residual spin, but that is no longer dominating its energy (otherwise the star wouldn't be near-spherical by disc-like). The same essentially holds for planets.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2015-01-26T08:42:22.980" LastActivityDate="2015-01-26T08:42:22.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8683" PostTypeId="2" ParentId="8681" CreationDate="2015-01-25T19:44:39.627" Score="4" Body="&lt;p&gt;You could start from the premise that there was no net angular momentum in the universe at all; but it would still be the case that everything of interest was spinning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the scales of stars and planets there are (at least) two important mechanisms that result in individual systems having angular momentum. The first is turbulence. If you take a parcel of turbulent gas from a giant molecular cloud it will always possess &lt;em&gt;some&lt;/em&gt; angular momentum, even if the total cloud does not. As the parcel collapses to form a star/planets conservation of angular momentum $J$ and dissipative interactions result in an ncrease in spin rate and collapse towards a planar geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, stars form in clusters. There is interaction between stellar systems early in their lives. Again, the cluster may have little net J, but groups of stars can, relative to their own centre of mass frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On bigger scales (galaxies) the second of these explanations becomes more important. The interaction and accretion of galaxies is what gives individual galaxies a spin, even if the clusters they are born in have much less or even no net angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example of how turbulent velocity fields lead to gravitational condensations containing angular momentum you could do worse than study the star formation simulation performed by Matthew Bate and collaborators. These simulations start off in clouds with zero net angular momentum, yet produce a host of stars with swirling accretion disks, binary systems of all shape and sizes etc. An example journal paper can be found here: &lt;a href=&quot;http://adsabs.harvard.edu/abs/2009MNRAS.392..590B&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2009MNRAS.392..590B&lt;/a&gt;  Here is a web page where you can download the animations and study them at length  &lt;a href=&quot;http://www.astro.ex.ac.uk/people/mbate/Cluster/cluster500RT.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.astro.ex.ac.uk/people/mbate/Cluster/cluster500RT.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turbulent clouds are by their nature random and stochastic in terms of their motions. Often the velocity field is defined in terms of a power law dependence on spatial scale. The formation of &lt;a href=&quot;http://en.wikipedia.org/wiki/Vortex&quot; rel=&quot;nofollow noreferrer&quot;&gt;vortices&lt;/a&gt; is a characteristic of turbulent media. They can be produced in the absence of external forces. The vortices contain angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also worth noting that not all galaxies have an appreciable spin. Spiral galaxies do, but many elliptical galaxies have little net rotation. See &lt;a href=&quot;https://physics.stackexchange.com/questions/93830/why-the-galaxies-forms-2d-plane-or-spiral-like-instead-of-3d-ball-or-spherica&quot;&gt;https://physics.stackexchange.com/questions/93830/why-the-galaxies-forms-2d-plane-or-spiral-like-instead-of-3d-ball-or-spherica&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-01-25T19:44:39.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8685" PostTypeId="2" ParentId="254" CreationDate="2015-01-26T12:53:07.383" Score="19" Body="&lt;p&gt;The angular resolution of the telescope &lt;em&gt;really has no direct bearing on our ability to detect Oort cloud objects&lt;/em&gt; beyond how that angular resolution affects the depth to which one can detect the light from faint objects. Any telescope can detect stars, even though their actual discs are way beyond the angular resolution of the telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The detection of Oort cloud objects is simply a question of detecting the (unresolved) reflected light in exactly the same way that one detects a faint (unresolved) star. Confirmation of the Oort cloud nature of the object would then come by observing at intervals over a year or so and obtaining a very large ($&amp;gt;2$ arcseconds) parallax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question amounts to how deep do you need to go? We can do this in two ways (i) a back of the envelope calculation assuming the object reflects light from the Sun with some albedo. (ii) Scale the brightness of comets when they are distant from the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) The luminosity of the Sun is $L=3.83\times10^{26}\ W$. Let the distance to the Oort cloud be $D$ and the radius of the (assumed spherical) Oort object be $R$.&#xA;The light from the Sun incident on the object is $\pi R^2 L/4\pi D^2$.&#xA;If we now assume that a fraction $f$ of this is reflected uniformly into a $2\pi$ solid angle. This latter point is an approximation, the light will not be reflected isotropically, but it will represent some average over any viewing angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To a good approximation, as $D \gg 1$ au, we can assume that the distance from the Oort object to the Earth is also $D$. Hence the flux of light received at the Earth is&#xA;$$F_{E} = f \frac{\pi R^2 L}{4\pi D^2}\frac{1}{2\pi D^2} = f \frac{R^2 L}{8\pi D^4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting some numbers in, let $R=10$ km and let $D= 10,000$ au. Cometary material has a very low albedo, but let's be generous and assume $f=0.1$.&#xA;$$ F_E = 3\times10^{-29}\left(\frac{f}{0.1}\right) \left(\frac{R}{10\ km}\right)^2 \left(\frac{D}{10^4 au}\right)^{-4}\ Wm^{-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To convert this to a magnitude, assume the reflected light has the same spectrum as sunlight. The Sun has an apparent visual magnitude of -26.74, corresponding to a flux at the Earth of $1.4\times10^{3}\ Wm^{-2}$. Converting the flux ratio to a magnitude difference, we find that the apparent magnitude of our fiducial Oort object is &lt;strong&gt;52.4&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) Halley's comet is similar (10 km radius, low albedo) to the fiducial Oort object considered above. Halley's comet was observed by the &lt;a href=&quot;http://www.eso.org/~ohainaut/nice/phot-27-03.html&quot; rel=&quot;noreferrer&quot;&gt;VLT in 2003&lt;/a&gt; with a magnitude of 28.2 and at a distance of 28 au from the Sun. We can now just scale this magnitude, but it scales as distance to the power of &lt;em&gt;four&lt;/em&gt;, because the light must be received and then we see it reflected.&#xA;Thus at 10,000 au, Halley would have a magnitude of $28.2 - 2.5 \log (28/10^{4})= 53.7$, in reasonable agreement with my other estimate. (Incidentally my crude formula in (i) above suggests a $f=0.1$, $R=10\ km$ comet at 28 au would have a magnitude of 26.9. Given that Halley probably has a smaller $f$ this is excellent consistency.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The observation of Halley by the VLT represents the pinnacle of what is possible with today's telescopes. Even the Hubble deep ultra deep field only reached visual magnitudes of about 29. Thus a big Oort cloud object remains &lt;strong&gt;more than 20 magnitudes below&lt;/strong&gt; this detection threshold!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most feasible way of detecting Oort objects is when they occult background stars. The possibilities for this are discussed by &lt;a href=&quot;http://arxiv.org/abs/0912.0948&quot; rel=&quot;noreferrer&quot;&gt;Ofek &amp;amp; Naker 2010&lt;/a&gt; in the context of the photometric precision provided by Kepler. The rate of occultations (which are of course single events and unrepeatable) was calculated to be between zero and 100 in the whole Kepler mission, dependent on the size and distance distribution of the Oort objects. As far as I am aware, nothing has come of this (yet).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-02-21T18:39:56.010" LastActivityDate="2015-02-21T18:39:56.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8687" PostTypeId="2" ParentId="5879" CreationDate="2015-01-26T19:23:45.643" Score="2" Body="&lt;p&gt;It's a pretty broad topic. It depends a bit on the type of DSO. Speaking in general, most DSOs lack brightness, and so to observe them you need two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;lack of light pollution&lt;/li&gt;&#xA;&lt;li&gt;lots of aperture&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Light pollution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observing from the city, DSOs are challenging. The further away you are from city lights, the better your views. Deserts, national parks, sparsely populated areas are all good. Driving at least 1 hour away from the city usually provides very noticeable improvements. Light pollution maps help in finding places with a dark sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jshine.net/astronomy/dark_sky/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.jshine.net/astronomy/dark_sky/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.lightpollutionmap.info/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.lightpollutionmap.info/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Aperture&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bigger the aperture, the better you can see DSOs. If you're a DSO hunter, you must be focused on increasing aperture before you do anything else. What helps here is having an instrument with a good aperture / cost ratio, such as a dobsonian reflector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's essentially no limit here - more is always better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted that a larger aperture will always perform better than a smaller one, even when light pollution is very heavy - but of course it's best if both aperture and dark sky cooperate.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Other factors that may help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Filters&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the most over-utilized and over-hyped accessories, especially for beginners (most people are too much focused on filters and not enough focused on things that really matter). That being said, some nebulae do look better in some filters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dave Knisely is one of the foremost experts in the amateur astronomy community in terms of DSOs and filters. Read anything he writes on the subject, e.g. this long article:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.prairieastronomyclub.org/resources/by-dave-knisely/filter-performance-comparisons-for-some-common-nebulae/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.prairieastronomyclub.org/resources/by-dave-knisely/filter-performance-comparisons-for-some-common-nebulae/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a nutshell: If you can only get one DSO filter, get a UHC. If you can get two, also get an OIII filter. Of course, dark sky and large aperture should always come first. Don't bother with filters if you're observing from the city using a small aperture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Filters are useful for nebulae. They don't help with galaxies, star clusters, individual stars, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Books&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find a book called Turn Left At Orion. It's a fantastic introduction for beginners to DSOs. It will help you find the easier targets, and essentially open up the sky for you. Later you'll be able to figure things out on your own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521153972&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.amazon.com/Turn-Left-Orion-Hundreds-Telescope/dp/0521153972/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It should be noted that the entire Messier catalog can be observed in a relatively small telescope even in the city - and all Messier objects are DSOs. I've seen the Dumbbell Nebula (M27) and the Ring Nebula (M57) in 6&quot; of aperture in a very light-polluted place (Silicon Valley). So, no, it's not that hard to observe DSOs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if the sky is dark and/or the aperture large, all DSOs look better. This is especially true of globular clusters, such as M13. They already look pretty spectacular in any aperture, even in a small scope - but the view is mind boggling in a very large dob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KwNkN.png&quot; alt=&quot;M13 aperture comparison&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-01-26T19:36:50.223" LastActivityDate="2015-01-26T19:36:50.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8688" PostTypeId="2" ParentId="8674" CreationDate="2015-01-26T19:26:13.240" Score="3" Body="&lt;p&gt;Here's a drawing that may help understand (edited to take into account the fact that the Universe expands not only along the line of sight, but in &lt;i&gt;all&lt;/i&gt; directions &amp;mdash; thanks to @Ed Shaya):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&#xA;In the distant past, at $t = t_1$, you (or your ancestors, or the Milky Way seed) was close to the other galaxy, and photons were emitted from the edges in the shown direction (as well as all other directions). If the Universe didn't expand, the paths of the two photons would cross at the point $p_1$, at an angle $\theta_1$.&lt;/li&gt;&#xA;&#xA;&lt;li&gt;At a later time, $t = t_2$, while the Universe is expanding, those photons are traveling in your direction, but still haven't caught up with you. At the same time, the expansion have pulled the photons farther apart, so that their linearly extended point of crossing $p_2$ lies farther away, and their angle of cross is $\theta_2 &amp;gt; \theta_1$.&lt;/li&gt;&#xA;&#xA;&lt;li&gt;At present time, $t = t_3$, you receive the photons at the point $p_3$. The galaxy is now far, far away (and has evolved significantly), but the angle you measure between the photons has $\sim$nothing to do with the present size of the galaxy. Rather, it is given by its original angular size, plus a further expanded angle $\theta_3$, so that it now spans a large angular diameter (and looks as it did at $t = t_1$).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VFy0l.jpg&quot; alt=&quot;Angular size time evolution slices&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-01-28T22:39:57.130" LastActivityDate="2015-01-28T22:39:57.130" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="8689" PostTypeId="1" CreationDate="2015-01-26T22:02:39.177" Score="3" ViewCount="101" Body="&lt;p&gt;Is it possible to view asteroid 2004 BL86 using Stellarium or the iOS Star Walk app?&lt;/p&gt;&#xA;" OwnerUserId="4146" LastActivityDate="2015-01-27T02:45:29.233" Title="Calculating 2004 BL86 position using software" Tags="&lt;asteroids&gt;&lt;software&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8690" PostTypeId="2" ParentId="8689" CreationDate="2015-01-27T02:26:41.553" Score="3" Body="&lt;p&gt;I'm going to describe the steps that I followed to show 2004 BL86 in Stellarium in my notebook:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Open the &lt;em&gt;Configuration Window&lt;/em&gt; by pressing F2&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Select the &lt;em&gt;Plugins&lt;/em&gt; tab and from the left list select &lt;em&gt;Solar System Editor&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Click in the &lt;em&gt;configure&lt;/em&gt; button at the bottom&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Go to &lt;em&gt;Solar System&lt;/em&gt; tab in the opened window&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Click the button &lt;em&gt;Import orbital elements in MPC format...&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Click in the drop-down menu called &lt;em&gt;Select bookmark...&lt;/em&gt; and select &lt;em&gt;MPCORB: near-Earth asteroids (NEAs)&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Click the button &lt;em&gt;Get orbital elements&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the search field, type 2004 BL86 and select it once it appears below&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Click the last button called &lt;em&gt;Add object&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Close the windows and then press F3 to open the &lt;em&gt;Search window&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Start typing 2004 BL86 and once it is showed below, click in the magnifying glass to the right, that will get you to the position of 2004 BL86&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Please let me know if that answers your question. I'm using Stellarium version 0.13.1... and sorry if my English is not very precise :)&lt;/p&gt;&#xA;" OwnerUserId="2339" LastEditorUserId="2339" LastEditDate="2015-01-27T02:45:29.233" LastActivityDate="2015-01-27T02:45:29.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8691" PostTypeId="2" ParentId="8674" CreationDate="2015-01-27T04:34:30.267" Score="-2" Body="&lt;p&gt;This is an interesting thought,but red shift was a theory that could be answered by assuming near light velocities.The belief in,this unproven idea,invokes argument of&quot;If it is coming toward us then we will already have seen it&quot;&lt;/p&gt;&#xA;" OwnerUserId="5295" LastActivityDate="2015-01-27T04:34:30.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8692" PostTypeId="1" CreationDate="2015-01-27T16:35:50.993" Score="-1" ViewCount="144" Body="&lt;p&gt;Forgive me for the naive question. Besides taking pictures of tons of objects in space, what kinds of revolutionary discoveries were made as a result of data from Hubble? Did Hubble shed light on any open problems or did it lead astronomers to conclude the existence of something that they had not known to have existed before? &lt;/p&gt;&#xA;" OwnerUserId="2873" LastActivityDate="2015-01-27T18:50:58.437" Title="What &quot;revolutionary&quot; discoveries were made by Hubble Telescope" Tags="&lt;hubble-telescope&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-01-27T22:22:33.947" ContentLicense="CC BY-SA 3.0" />
  <row Id="8693" PostTypeId="2" ParentId="8692" CreationDate="2015-01-27T17:41:53.977" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble_Space_Telescope#Important_discoveries&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; has some handy-dandy discoveries. I don't know if they're &amp;quot;revolutionary&amp;quot; &lt;em&gt;per se&lt;/em&gt;, but they're pretty interesting.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Hubble measured the distance to some &lt;a href=&quot;https://en.wikipedia.org/wiki/Cepheid_variable&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cepheid variables&lt;/a&gt; with very good accuracy. Cepheids are variable stars that periodically oscillate between larger and more luminous states and smaller and more dimmer states. They are used as &amp;quot;standard candles&amp;quot; in astronomy because their luminosity and periods can be used to easily calculate how far away they are. In fact, Edwin Hubble used Cepheids to establish &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble's law&lt;/a&gt;, &lt;span class=&quot;math-container&quot;&gt;$v=H_0D$&lt;/span&gt;. It states how fast objects are moving away from us (&lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt;) as a function of distance (&lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt;). &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt; is Hubble's constant. It was originally written as &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;, but it was determined that it isn't actually a constant at all! &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt; is used as the value in the present era.&lt;/p&gt;&#xA;&lt;p&gt;Anyway, the measurements from Hubble were used to establish the distance to the Cepheid variables and their recessional velocities. With that, &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt; could be determined more accurately than ever. &lt;a href=&quot;http://arxiv.org/pdf/astro-ph/0012376v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Freedman et al. published a paper&lt;/a&gt; detailing the results. The team found that &lt;span class=&quot;math-container&quot;&gt;$H_0 = 72 \pm 8 \text{ kms}^{-1}\text{ Mpc}^{-1}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Hubble had a tiny role in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Supernova_Cosmology_Project&quot; rel=&quot;nofollow noreferrer&quot;&gt;Supernova Cosmology Project&lt;/a&gt;, as described by Gerson Goldhaber &lt;a href=&quot;http://arxiv.org/pdf/0907.3526v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;in this paper&lt;/a&gt;. Under Saul Perlmutter's leadership, the team gathered more evidence that the universe is expanding at an accelerating rate, which earned Perlmutter a Nobel Prize in 2011. Two members of the &lt;a href=&quot;https://en.wikipedia.org/wiki/High-Z_Supernova_Search_Team&quot; rel=&quot;nofollow noreferrer&quot;&gt;High-Z Supernova Search Team&lt;/a&gt; shared the prize with Perlmutter, and they, too, briefly used the telescope. Hubble played a minor role, as Goldhaber explains, but it was still important: looking at supernovae to determine their light curves, data which was then used in the final results.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;I'm not sure I'd describe any of Hubble's other achievements as &amp;quot;revolutionary&amp;quot;, though it has an amazing track record. &lt;a href=&quot;http://www.nasa.gov/mission_pages/hubble/main/accomplishments_index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA discusses some of Hubble's successes here&lt;/a&gt;, including the discovery of black holes and supermassive black holes in other galaxies; see &lt;a href=&quot;http://arxiv.org/pdf/astro-ph/0006289v2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt; and &lt;a href=&quot;http://arxiv.org/pdf/astro-ph/0006053v2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;. There's also a decent page &lt;a href=&quot;http://hubblesite.org/newscenter/archive/releases/1997/01/text/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; that discusses those discoveries.&lt;/p&gt;&#xA;&lt;p&gt;One last field is &lt;a href=&quot;http://www.nasa.gov/vision/universe/newworlds/0112_missing_link.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;planet formation&lt;/a&gt;, one area of study that's been hot recently. &lt;a href=&quot;http://hubblesite.org/the_telescope/nuts_.and._bolts/instruments/stis/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; is one of the better links from that NASA page.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-01-27T17:41:53.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8694" PostTypeId="1" CreationDate="2015-01-27T18:01:44.610" Score="6" ViewCount="342" Body="&lt;p&gt;Is there an inherent instability or limit that prevents natural satellites like the moons of Jupiter from having their own natural satellites?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked at other questions such as &lt;a href=&quot;https://astronomy.stackexchange.com/questions/856/do-moons-have-moons&quot;&gt;do moons have moons&lt;/a&gt; but they focus on discovery rather than theoretical limits analogous to the Roche limit for satellite formation.&lt;/p&gt;&#xA;" OwnerUserId="5403" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-01-27T23:41:19.883" Title="Theoretical limits for natural satellites having natural satellites" Tags="&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8695" PostTypeId="2" ParentId="8694" CreationDate="2015-01-27T18:14:44.117" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Natural_satellite#Satellites_of_satellites&quot; rel=&quot;nofollow&quot;&gt;Wikipedia says there may be a reason (emphasis mine):&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;No &quot;moons of moons&quot; (natural satellites that orbit a natural satellite of another body) are currently known as of 2014. &lt;strong&gt;In most cases, the tidal effects of the primary would make such a system unstable.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This seems to be because the secondary would have to be very close to the primary - close enough that it doesn't become simply another satellite of the &lt;em&gt;actual&lt;/em&gt; primary (i.e. the planet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The page then goes on to talk about &lt;a href=&quot;https://en.wikipedia.org/wiki/Rhea_(moon)&quot; rel=&quot;nofollow&quot;&gt;Rhea&lt;/a&gt;, a moon of Saturn that may or may not have a ring system. &lt;a href=&quot;http://www.sciencemag.org/content/319/5868/1380.abstract?maxtoshow=&amp;amp;HITS=10&amp;amp;hits=10&amp;amp;RESULTFORMAT=&amp;amp;fulltext=rhea&amp;amp;searchid=1&amp;amp;FIRSTINDEX=0&amp;amp;resourcetype=HWCIT&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; gives some more information.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2015-01-27T18:14:44.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8698" PostTypeId="1" CreationDate="2015-01-28T15:16:13.457" Score="3" ViewCount="488" Body="&lt;p&gt;As seen from Earth, planets such as Mars and Jupiter exhibit retrograde motion when they are near opposition (from Earth).  I am wondering how this effect extends to very distant objects, such as those in the Kuiper Belt.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since objects in the Kuiper Belt orbit the Sun at a velocity much slower than Earth (or Mars or Jupiter for that matter) because of Kepler's third law I'd imagine that retrograde motion for these objects is more extreme in the sense that these objects are seen to be in retrograde motion for a longer time and over a larger portion of the sky (in an angular sense) than would be seen for any of the planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an extreme case, is it possible that a very distant Kuper Belt Object (KBO) would never be observed in prograde motion?  I imagine that if you were to observe a KBO that was in the direction of Earth's motion then there would be no contribution from the Earth's velocity on the observation and you would be able to see the object in prograde motion, no matter how slow it was travelling. However, such an object would likely be obscured by sunlight, since objects that lie in the direction of Earth's motion lie above the day/night interface on the Earth.  My question here is that, given real-life observational constraints (like sunlight) how far from the Sun would an object have to be to never be seen in prograde motion?  You can be mathematical.&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2015-04-28T18:07:02.967" Title="Retrograde motion and Kuiper Belt Objects" Tags="&lt;solar-system&gt;&lt;observational-astronomy&gt;&lt;kuiper-belt&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8699" PostTypeId="2" ParentId="8640" CreationDate="2015-01-28T15:58:04.013" Score="2" Body="&lt;p&gt;Space is considered to be the volume in which stuff (eg. stars) exists. It is also considered that it can expand everywhere, resulting in more space between the stars, without the stars having moved away from each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual analogy is the expanding balloon, with dots representing stars or galaxies, and the surface of the balloon representing empty space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you blow up the balloon, its surface area expands (ie. like space expands), and the space between the dots (stars) increases. If you were living on a planet around one of the stars (dots on the balloon), you would observer all the other dots/stars appear to move away from you, and if you were on any of the other dots, you would observer the same thing.&lt;/p&gt;&#xA;" OwnerUserId="2880" LastActivityDate="2015-01-28T15:58:04.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8700" PostTypeId="2" ParentId="8698" CreationDate="2015-01-28T16:10:54.380" Score="1" Body="&lt;p&gt;No, this is not possible.&#xA;Kuiper belt objects are in orbit around the Sun and their average distance (semi-major axis) will be beyond the orbit of Neptune. This means that their orbital period will be much longer than Earth's (following Kepler's third law). In the most extreme case (further away) the motion of the object will be virtually non observable and the object will appear not to move with respect to the background stars apart from the apparent motion due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_parallax&quot; rel=&quot;nofollow noreferrer&quot;&gt;parallax&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These objects will still show apparent retrograde and prograde motion because of the movement of the Earth around the Sun. This prograde and retrograde motion can also be observed in nearby stars where it is used to measure the parallax and thereby the distance to the star. See the figure below for the apparent motion of a star as measured by &lt;a href=&quot;http://en.wikipedia.org/wiki/Hipparcos&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hipparcos&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I imagine that if you were to observe a KBO that was in the direction of Earth's motion then there would be no contribution from the Earth's velocity on the observation and you would be able to see the object in prograde motion, no matter how slow it was travelling.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The direction of the Earth's motion is constantly changing (changes 360° in a year) so the time at which a KBO will be in the direction of the motion of the Earth will be very short.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2J176.gif&quot; alt=&quot;This figure shows prograde and retrograde motion of a star as measured by Hipparcos.&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2015-01-28T16:10:54.380" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8701" PostTypeId="1" CreationDate="2015-01-29T08:18:00.227" Score="14" ViewCount="630" Body="&lt;p&gt;I read a &lt;a href=&quot;http://phys.org/news202456660.html&quot;&gt;science report&lt;/a&gt; recently that mentioned an accidental discovery where radioactive decay rates shifted slightly slower in advance ( about a day and a half) of solar storms or in sync with sun's 33-day pattern/rotation of the sun's core. Seems that neutrinos emitted from within the sun or at surface of the sun arrive in advance of other wave/particle radiations AND effect earthly atomic natures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from predicting solar flares/storms, does this imply decaying atoms/elements absorb or block neutrino emissions to any substantial degree that might of use in other areas of science? &lt;/p&gt;&#xA;" OwnerUserId="825" LastActivityDate="2020-06-11T10:19:59.477" Title="How Are Radioactive Decay Rates Influenced by Neutrinos - On Earth and Other Dense Planets" Tags="&lt;neutrinos&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8703" PostTypeId="2" ParentId="8094" CreationDate="2015-01-29T13:22:49.287" Score="1" Body="&lt;p&gt;I have an answer for the first part of your question, because I looked it up in answering &lt;a href=&quot;https://physics.stackexchange.com/questions/8827/question-on-the-stability-of-the-solar-system/161973#161973&quot;&gt;https://physics.stackexchange.com/questions/8827/question-on-the-stability-of-the-solar-system/161973#161973&lt;/a&gt; on Physics SE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to see what the current state of the art on solar system simulations is you could do worse than watch &lt;a href=&quot;https://www.youtube.com/watch?v=7dRLvSzDHo8&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sean Raymond's presentation&lt;/a&gt; at &quot;Protostars and Planets VI&quot; from 2013. You can find the actual write up &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014prpl.conf..595R&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. Or from the same conference there is &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014prpl.conf..787D&quot; rel=&quot;nofollow noreferrer&quot;&gt;Melvyn Davies'&lt;/a&gt; review of the long-term dynamics of planetary systems. The talk can &lt;a href=&quot;https://www.youtube.com/watch?v=3wKY0LuDuIk&quot; rel=&quot;nofollow noreferrer&quot;&gt;be seen here.&lt;/a&gt; This review does contain the sort of information you are looking for. It discusses the past and future evolution of our solar system, as well as planetary systems in general. It presents and reviews simulations and discusses the relevant issues. Both of these guys are excellent speakers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A brief summary would be that the solar system is &lt;em&gt;probably&lt;/em&gt; stable for the remaining lifetime of the Sun. However, there is the intriguing &lt;em&gt;possibility&lt;/em&gt; that Mercury could fall into the Sun or collide with Venus in the next billion years or that Mars &lt;em&gt;could&lt;/em&gt; be ejected from the solar system on a similar timescale (e.g. from the N-body simulations of &lt;a href=&quot;http://adsabs.harvard.edu/abs/2008ApJ...683.1207B&quot; rel=&quot;nofollow noreferrer&quot;&gt;Battygin &amp;amp; Laughlin 2008&lt;/a&gt;). &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-01-29T13:22:49.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8704" PostTypeId="2" ParentId="107" CreationDate="2015-01-29T13:46:37.170" Score="2" Body="&lt;p&gt;I completely agree with the answer from MBR. The number is actually $1.20\pm 0.38$ per cent, is published by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2012ApJ...753..160W&quot; rel=&quot;nofollow&quot;&gt;Wright et al. (2012)&lt;/a&gt; and is the fraction of F, G, K stars that have a hot Jupiter defined as being larger than 0.1 Jupiter masses and having an orbital period less than 10 days. Table 2 of that paper summarises results from other workers, who obtain between 0.5 and 1.5 per cent. The paper also discusses observational biases, including &lt;em&gt;metallicity&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has long been known that close-in planet incidence is higher around more metal-rich stars. There is also a bias whereby it is easier to find planets around metal-rich stars, whereas the average star in the solar neighborhood is slightly metal-poor compared with the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A study by &lt;a href=&quot;http://arxiv.org/abs/1406.0861&quot; rel=&quot;nofollow&quot;&gt;Gonzalez (2014)&lt;/a&gt; accumulates our current knowledge of exoplanetary systems and their metallicities, deriving a planetary incidence rate&#xA;$$ P_{planet} = \alpha 10^{\beta[Fe/H]},$$&#xA;with $\alpha= 0.022 \pm 0.007$ (i.e. 2.2 per cent), $\beta=3.0\pm 0.5$ and where [Fe/H] is the usual logarithmic ratio of the metallicity of the star to the metallicity of the Sun. (i.e. the Sun has [Fe/H]=0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This calculation is done for giant planets with orbital periods less than 4 years, so not all of them would be classed as hot Jupiters. Bottom line, the number given by Wright et al. is about right on average, but it is higher for higher metallicity stars.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-03-17T18:21:34.820" LastActivityDate="2016-03-17T18:21:34.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8705" PostTypeId="1" CreationDate="2015-01-29T20:51:45.833" Score="0" ViewCount="1911" Body="&lt;p&gt;Not sure if this question belongs here, or on &lt;a href=&quot;https://physics.stackexchange.com/&quot;&gt;Physics&lt;/a&gt;. Would someone please describe the &quot;anatomy&quot; of a light cone and provide an intuitive example?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read the Wikipedia article (&lt;a href=&quot;http://en.wikipedia.org/wiki/Light_cone&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Light_cone&lt;/a&gt;) and I think I have a basic grasp of the concept. However, I still have questions: is a cone emitted every time there is a disturbance in the electromagnetic spectrum? Does every electron changing energy states emit a light cone? I understand the &quot;future&quot; light cone (I think). However, what of the &quot;past&quot; light cone? Is that what is observed when detecting redshifts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your answers/comments. They have helped my understanding of the light cone concept.&lt;/p&gt;&#xA;" OwnerUserId="2757" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-02-01T03:57:22.150" Title="Light Cone Explanation" Tags="&lt;space-time&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8706" PostTypeId="2" ParentId="8705" CreationDate="2015-01-29T21:44:16.320" Score="1" Body="&lt;p&gt;Think of dropping a pebble into a pond, and the circular pattern the resulting wave makes. Now take a photo every second, and stack these one on top of another, and you have a &quot;cone&quot; of sounds waves. The oldest image of the wave is also the biggest, and furthest back in time.&lt;/p&gt;&#xA;" OwnerUserId="2880" LastActivityDate="2015-01-29T21:44:16.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8707" PostTypeId="1" AcceptedAnswerId="8720" CreationDate="2015-01-29T23:47:56.760" Score="5" ViewCount="2421" Body="&lt;p&gt;Could an object from outer space with the right velocity and orbit come into contact with the surface of our planet in a &quot;slowish&quot; manner? &lt;/p&gt;&#xA;" OwnerUserId="5432" LastEditorUserId="2153" LastEditDate="2015-01-30T16:39:38.917" LastActivityDate="2020-09-03T17:57:02.300" Title="How gently could a comet/asteroid/meteorite &quot;hit&quot; Earth?" Tags="&lt;impact&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8708" PostTypeId="1" CreationDate="2015-01-30T02:39:39.967" Score="2" ViewCount="346" Body="&lt;p&gt;HESS, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cherenkov_Telescope_Array&quot; rel=&quot;nofollow noreferrer&quot;&gt;CTA&lt;/a&gt;, and VERITAS are used to study cosmic rays. &lt;br/&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/shFLa.jpg&quot; alt=&quot;veritas&quot;&gt;&#xA;What they receive are optical (maybe very blue) photons.&lt;br/&gt;&#xA;So my question is: Why don't we use optical telescopes to study cosmic rays directly?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="2153" LastEditDate="2015-01-30T21:42:26.483" LastActivityDate="2019-08-20T16:34:08.987" Title="Why do not we use optical telescopes to study cosmic rays directly?" Tags="&lt;cosmic-ray&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8709" PostTypeId="1" CreationDate="2015-01-30T02:52:25.823" Score="1" ViewCount="99" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutrino&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; most neutrinos passing through the Earth emanate from the Sun. About 65 billion (6.5^10) solar neutrinos per second pass through every square centimeter perpendicular to the direction of the Sun in the region of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the &lt;a href=&quot;http://en.wikipedia.org/wiki/IceCube_Neutrino_Observatory&quot; rel=&quot;nofollow&quot;&gt;IceCube Neutrino Observatory&lt;/a&gt; or other neutrino detectors only detect a few neutrinos every year. Why?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorDisplayName="user1569" LastEditDate="2015-01-30T12:25:30.513" LastActivityDate="2015-01-30T12:25:30.513" Title="Why do we have so few neutrinos to study?" Tags="&lt;neutrinos&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8710" PostTypeId="2" ParentId="8709" CreationDate="2015-01-30T04:00:06.357" Score="6" Body="&lt;p&gt;It's not that the neutrino detectors don't have a chance of detecting a neutrino, but that the neutrinos do not interact with matter (well hardly ever). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are $6.5\times 10^{10}$ neutrinos per second per $cm^2$ and if the detector has an area of $10^{6} cm^2$, and it detects 1 neutrino per year, then 1 out of $2.05\times 10^{24}$ neutrinos interacts with the detector. $2.05\times 10^{24}$ pass through the detector unnoticed.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2015-01-30T04:00:06.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8712" PostTypeId="1" CreationDate="2015-01-30T10:00:54.503" Score="-5" ViewCount="948" Body="&lt;p&gt;I know earth is spherical in shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this means earths atmosphere is spherical in shape or the ground/(land+water combined) is in spherical shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some doubts if the second is true.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;since the earth is rotating around itself and  revolving around sun in the same time. When we see sun in the afternoon on top of your head. it can be actually you are upside down looking onto sun which is under you.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'll add my other doubts once some one answers my question.&lt;/p&gt;&#xA;" OwnerUserId="4046" LastActivityDate="2015-01-30T18:15:24.367" Title="earth is spherical , does it mean the ground on earth is like a ball?" Tags="&lt;orbit&gt;&lt;the-sun&gt;&lt;gravity&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2015-02-02T13:33:53.047" ContentLicense="CC BY-SA 3.0" />
  <row Id="8713" PostTypeId="2" ParentId="8709" CreationDate="2015-01-30T11:56:20.050" Score="4" Body="&lt;p&gt;It is because the neutrinos &lt;a href=&quot;http://en.wikipedia.org/wiki/Cross_section_%28physics%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;cross section&lt;/a&gt; is very small.&lt;/p&gt;&#xA;&lt;p&gt;Roughly, the cross section is a measure of the interaction probability (for a given force) of the given particle with your target.&lt;/p&gt;&#xA;&lt;p&gt;Since the probability for interaction &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/crosec.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;depends upon the strength and range of the interaction&lt;/a&gt;, the cross-sections for neutrino interactions are &lt;a href=&quot;http://cupp.oulu.fi/neutrino/nd-cross.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;very small since they interact only by the weak interaction&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Hence, it is not that we receive few neutrinos, rather it is our (in)ability to detect them, due to our current knowledge of their intrinsic nature, which prevent us from measuring high fluxes of neutrinos.&lt;/p&gt;&#xA;&lt;p&gt;For this very reason, neutrinos detectors are made of huge volume targets, in order to maximize the detection probability.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-01-30T11:56:20.050" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8716" PostTypeId="2" ParentId="8707" CreationDate="2015-01-30T12:52:55.363" Score="6" Body="&lt;p&gt;It is difficult to see how. Most comets and asteroids would encounter the Earth on a crossing orbit and the encounter velocity would be roughly the vector sum of the Earth's velocity around the Sun (of order 30 km/s) and the individual velocity of the rogue object. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you were to arrange it so that the asteroid/comet was diverted by something else so that it &quot;caught up&quot; with the Earth from behind travelling at an initially low relative velocity, there is the additional influence of the Earth's own gravitational potential. This would accelerate the approaching object to something like the order of the escape velocity from the Earth's surface - about 11 km/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This fairly authoritative site on calculating the &lt;a href=&quot;http://impact.ese.ic.ac.uk/ImpactEffects/&quot; rel=&quot;nofollow noreferrer&quot;&gt;effect of asteroid/comet impacts&lt;/a&gt; suggests a minimum impact velocity of 11 km/s, which is indeed the Earth's escape velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: (And credit to Keith Thompson for pointing it out). These are the speeds at the top of the atmosphere. If the objects are smaller than 20-30m, then the atmosphere will basically take out most of that kinetic energy, so for &lt;em&gt;small&lt;/em&gt; objects, a relatively slow impact is de riguer. But for anything larger than &lt;span class=&quot;math-container&quot;&gt;$\sim$&lt;/span&gt;50 m, it is basically the full 11 km/s or more. &lt;a href=&quot;http://www.lsst.org/lsst/science/scientist_near_earth_objects_neoquant&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.lsst.org/lsst/science/scientist_near_earth_objects_neoquant&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-10-05T07:26:32.090" LastActivityDate="2018-10-05T07:26:32.090" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="8717" PostTypeId="2" ParentId="8708" CreationDate="2015-01-30T13:12:03.270" Score="9" Body="&lt;p&gt;The telescopes in your picture &lt;em&gt;are&lt;/em&gt; optical telescopes. They are used to detect cosmic rays using the   &lt;a href=&quot;http://en.wikipedia.org/wiki/Cherenkov_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cherenkov radiation&lt;/a&gt; caused by charged particles moving faster than the speed of light (in air) in the atmosphere. These are normally secondary particles that have been produced in collisions between cosmic rays and nuclei in the upper atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radiation produced by particles moving at close to light speed is &lt;a href=&quot;http://en.wikipedia.org/wiki/Relativistic_beaming&quot; rel=&quot;nofollow noreferrer&quot;&gt;beamed&lt;/a&gt; in the direction of motion. An array of optical telescopes picks up the beam, and the orientation of the image is a projection of the original track of the particle. By combining images taken with telescopes covering a wide area, and taken at almost the same time, one can reconstruct the paths of the secondary particles to work out which direction the original cosmic ray came from. The intensity of the images tells you something about the original energy of the cosmic ray.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possibly this is the answer to your question - a &lt;em&gt;single&lt;/em&gt; optical telescope might detect the Cherenkov radiation from a single secondary particle, but it would be difficult to reconstruct the path or energy of the original cosmic ray on its own - at least a pair of telescopes, or better still, a network is preferable. The properties of these &quot;cosmic ray&quot; telescopes are also quite different from &quot;ordinary&quot; astronomical telescopes. Cosmic ray telescopes need as much collecting area as possible, because the Cherenkov radiation is faint - but they can do this at the expense of image quality. They use very wide angle cameras (5 degrees or more) and use huge faceted/sgemented mirrors that give massive collecting area, but image quality (maybe 3 arminutes full width half maximum for a point source) that would be completely unacceptable for conventional optical light astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or (thanks Conrad Turner) you are asking why we cannot use the mirrors of an optical telescope to focus the cosmic rays onto a detector and image their source? The reason for that is simply that cosmic rays are not reflected from glass/silvered surfaces in the same way that light is. They are extremely energetic particles that either pass straight through the mirrors or are absorbed within them. i.e. They do not reflect or refract in the same way as light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other circumstances, cosmic rays are a nuisance when conducting observations with optical telescopes. They are a source of background noise in CCD detetor images. Cosmic rays (or the secondary particles) are capable of liberating electrons in the silicon and therefore simulating small intense light sources in the sky. Often these are seen as small bright pixels, groups of pixels or trails on the CCD image, confusing what you were originally trying to look at. However, these cosmic rays are not focused by the telescope, they are essentially being picked up by the detector itself and would have been so if the detector was not even attached to a telescope at all. They are nearly impossible to shield against because they have very high energies and your detector needs to have a hole to let the optical light in!&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-08-20T16:34:08.987" LastActivityDate="2019-08-20T16:34:08.987" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="8718" PostTypeId="2" ParentId="8712" CreationDate="2015-01-30T17:26:30.993" Score="3" Body="&lt;p&gt;The form of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Earth&quot; rel=&quot;nofollow&quot;&gt;Earth&lt;/a&gt; is a approximately (to a very good approximation) an &lt;a href=&quot;http://en.wikipedia.org/wiki/Oblate_spheroid&quot; rel=&quot;nofollow&quot;&gt;oblate spheroid&lt;/a&gt; (difference between the polar diameter and equatorial diameter is ~20km which is ~1/300 of the mean diameter). The shape is distorted by tides (both in the solid body as well as the sea) and by topography. Topographical variation is about +10 to -10km (height of &lt;a href=&quot;http://en.wikipedia.org/wiki/Mount_Everest&quot; rel=&quot;nofollow&quot;&gt;Everest&lt;/a&gt; to depth of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Challenger_Deep&quot; rel=&quot;nofollow&quot;&gt;Challenger Deep&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up and down are defined by the local gravity vector, which to a good approximation is pointing into the Earth towards its centre. Hence wherever the sun is, down is into the Earth's surface (in the direction colloquially know as down)&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-01-30T17:26:30.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8719" PostTypeId="2" ParentId="8712" CreationDate="2015-01-30T18:15:24.367" Score="1" Body="&lt;p&gt;As noted by Conrad Turner, the approximate shape of the Earth is an oblate spheroid, though it is so close to spherical that you would be pretty hard-pressed to see the difference without precise scientific equipment. &lt;a href=&quot;http://blogs.discovermagazine.com/badastronomy/2008/09/08/ten-things-you-dont-know-about-the-earth/#.VMu8tysa6VM&quot; rel=&quot;nofollow&quot;&gt;This web page&lt;/a&gt; has some pretty good information on this topic that you might be interested in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we talk about this shape, we are generally speaking of the land and sea. Gas in general has the property of not only having fluid shape, like a liquid, but also having indefinite volume. Gas expands to fill the container it is in. The atmosphere is not contained per se. What keeps it from simply expanding all the way out to the moon is Earth's gravity. But this means that faster moving air particles get further away from the surface than slower ones before being pulled back by gravity. So where exactly is the edge of the atmosphere? The outer-most layer of the atmosphere, &lt;a href=&quot;http://en.wikipedia.org/wiki/Ionosphere&quot; rel=&quot;nofollow&quot;&gt;the ionosphere&lt;/a&gt;, is very thin, and the further out toward space you go, the thinner it is. Where exactly its edge is, and what shape it has is... &lt;a href=&quot;http://iopscience.iop.org/0370-1301/68/8/302/pdf/0370-1301_68_8_302.pdf&quot; rel=&quot;nofollow&quot;&gt;complicated&lt;/a&gt;. Anyone who wants to make any authoritative statement on the shape or height of the edge of our atmosphere will have a few other authoritative statements to argue against.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the exact shape of the atmosphere is difficult to define, it's pretty safe to say that it is not &lt;em&gt;exactly&lt;/em&gt; spherical due mostly to the planet's rotation, and solar winds. The Earth's surface is also not &lt;em&gt;exactly&lt;/em&gt; spherical due mostly to the planet's rotation, and tidal forces. Even if the entire surface was covered in water, rotation would flatten the planet's shape ever so slightly. And even if our rotation became tidally locked with the sun, and the moon... um... went away somehow, the tidal forces of the sun, would distort the spherical perfection ever so slightly. But laying all of that aside, the Earth is still probably closer to a sphere than most ball-shaped objects you have ever handled, whether you measure the &quot;edge&quot; of the atmosphere, the land and sea, or whatever.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2015-01-30T18:15:24.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8720" PostTypeId="2" ParentId="8707" CreationDate="2015-01-30T20:59:28.443" Score="8" Body="&lt;p&gt;Yes, spacecraft do it all the time by using air resistance (and sometimes rockets) to slow down.&lt;/p&gt;&#xA;&lt;p&gt;Meteorites enter the atmosphere at high speed, typically 10-70 km/sec, but the smaller ones are slowed by air resistance, so they typically hit the surface at just a few hundred kilometers per hour. Reference: &lt;a href=&quot;http://csep10.phys.utk.edu/astr161/lect/meteors/impacts.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://csep10.phys.utk.edu/astr161/lect/meteors/impacts.html&lt;/a&gt; (that link is now dead).&lt;/p&gt;&#xA;&lt;p&gt;As gerrit points out in comments, a sufficiently small micrometeroid, like a dust particle, probably &lt;em&gt;could&lt;/em&gt; hit the surface gently.&lt;/p&gt;&#xA;&lt;p&gt;Larger bodies are not affected nearly as much by the atmosphere. For any such body, the velocity at impact is going to be at least several kilometers per second. A body moving obliquely in the same direction as the Earth's rotation would have a slightly lower speed at impact, but the impact still could not be described as &amp;quot;gentle&amp;quot;; the Earth's rotational speed is still a small fraction of orbital or escape velocity.&lt;/p&gt;&#xA;&lt;p&gt;Here's one way to think of it. Freefall trajectories are reversible. If you watch a movie of an incoming body in reverse, it still makes physical sense (ignoring air resistance). Any meteorite hitting the surface has to have been in deep space at some point before the impact. If there were a trajectory that allowed such a body to have a &amp;quot;gentle&amp;quot; impact speed, then it would be possible to &lt;em&gt;start&lt;/em&gt; with the same body near the surface at the same &amp;quot;gentle&amp;quot; speed, but in the opposite direction, and have it reach deep space. Unless the meteorite has its own propulsion system, that's just not going to happen.&lt;/p&gt;&#xA;&lt;p&gt;You &lt;em&gt;can&lt;/em&gt; get to space with a lower starting speed by moving along with the Earth's rotation - which is why most rockets are launched to the east, to take advantage of that. Reversing such a trajectory can result in a slightly slower impact, but only slightly.&lt;/p&gt;&#xA;" OwnerUserId="53" LastEditorUserId="53" LastEditDate="2020-09-03T17:57:02.300" LastActivityDate="2020-09-03T17:57:02.300" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="8721" PostTypeId="2" ParentId="1443" CreationDate="2015-01-30T23:12:47.063" Score="4" Body="&lt;p&gt;Here are some thoughts adapted to an answer I placed on Phyiscs SE to a similar question some time ago.  In order to observe the past we need to detect light from the Earth, reflected back to us from somewhere distant in space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average albedo of the Earth is about 0.3 (i.e. it reflects 30 percent of the light incident upon it). The amount of incident radiation from the Sun at any moment is the solar constant ($F \sim 1.3 \times 10^3$ Wm$^{-2}$) integrated over a hemisphere. Thus the total reflected light from the Earth is about $L=5\times 10^{16}$ W.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this light from the Earth has the same spectrum as sunlight and it gets reflected from something which is positioned optimally - i.e. it sees the full illuminated hemisphere. then, roughly speaking, the incident flux on a reflecting body will be $L/2\pi d^2$ (because it is scattered roughly into a hemisphere of the sky).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have to explore some divergent scenarios.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There just happens to be a large object at a distance that is highly reflective. I'll use 1000 light years away as an example, which would allow us to see 2000 years into the Earth's past.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Let's be generous and say it is a perfect reflector, but we can't assume specular reflection. Instead let's assume the reflected light is also scattered isotropically into a $2\pi$ solid angle. Thus the radiation we get back will be&#xA;$$ f = \frac{L}{2\pi d^2} \frac{\pi r^2}{2\pi d^2} = \frac{L r^2}{4\pi d^4},$$&#xA;where $r$ is the radius of the thing doing the reflecting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To turn a flux into an astronomical magnitude we note that the Sun has a visual magnitude of $-26.74$. The apparent magnitude of the reflected light will be given by&#xA;$$ m  = 2.5\log_{10} \left(\frac{F}{f}\right) -26.74 = 2.5 \log_{10} \left(\frac{4F \pi d^4}{L r^2}\right) -26.74 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's put in some numbers. Assume $r=R_{\odot}$ (i.e. a reflector as big as the Sun) and let $d$ be 1000 light years. From this I calculate $m=85$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put this in context, the Hubble space telescope ultra deep field has a magnitude limit of around $m=30$ (&lt;a href=&quot;http://arxiv.org/abs/1305.1931&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1305.1931&lt;/a&gt; ) and each 5 magnitudes on top of that corresponds to a factor of 100 decrease in brightness. So $m=85$ is about 22 orders of magnitude fainter than detectable by HST. What's worse, the reflector also scatters all the light from the rest of the universe, so picking out the signal from the earth will be utterly futile.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;A big, flat mirror 1000 light years away.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;How did it get there? Let's leave that aside. In this case we would just be looking at an image of the Earth as if it were 2000 light years away (assuming everything gets reflected). The flux received back at Earth in this case:&#xA;$$ f = \frac{L}{2\pi [2d]^2} $$&#xA;with $d=1000$ light years, which will result in an apparent magnitude at the earth of&#xA;$m=37$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, this is more promising, but still 7 magnitudes below detection with the HST and perhaps 5 magnitudes fainter than might be detected with the James Webb Space Telescope if and when it does an ultra-deep field. It is unclear whether the sky will be actually full of optical sources at this level of faintness and so even higher spatial resolution than HST/JWST  might be required to pick it out even if we had the sensitivity.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Just send a telescope to 1000 light years, observe the Earth, analyse the data and send the signal back to Earth.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Of course this doesn't help you see into the past because we would have to send the telescope there. But it could help those in the future see into &lt;em&gt;their&lt;/em&gt; past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming this is technically feasible, the Earth will have a maximum brightness corresponding to $m \sim 35$ so something a lot better than JWST would be required and that ignores the problem of the brightness contrast with the Sun, which would be separated by only 0.03 arcseconds from the Earth at that distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that these calculations are merely to detect the light from the &lt;em&gt;whole Earth&lt;/em&gt;. To extract anything meaningful would mean collecting a spectrum at the very least! And all this is for only 2000 years into the past.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-01-30T23:12:47.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8722" PostTypeId="2" ParentId="8705" CreationDate="2015-01-31T01:59:04.343" Score="1" Body="&lt;p&gt;Regarding the recent edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A light cone is not a physical object, nor is it made of light. &lt;a href=&quot;http://en.wikipedia.org/wiki/Light_cone&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; calls it&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the path that a flash of light, emanating from a single event (localized to a single point in space and a single moment in time) and traveling in all directions, would take through spacetime.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More accurately, it is the path that a massless particle traveling at the &lt;em&gt;speed&lt;/em&gt; of light would take.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to clear up your confusion: It's just the imaginary boundary causally separating two regions of space. Nothing more.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2015-01-31T01:59:04.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8723" PostTypeId="1" AcceptedAnswerId="8728" CreationDate="2015-01-31T03:21:45.957" Score="1" ViewCount="177" Body="&lt;p&gt;My professor used the term 'local star'. What range of distance is generally considered 'local' for a star? And what would be considered distant?&lt;/p&gt;&#xA;" OwnerDisplayName="user5341" LastActivityDate="2015-01-31T17:20:53.130" Title="What qualifies as a local star?" Tags="&lt;distances&gt;&lt;star-systems&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8724" PostTypeId="2" ParentId="8705" CreationDate="2015-01-31T04:18:51.123" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;However, I still have questions: is a cone emitted overtime there is a disturbance i the electromagnetic spectrum?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In relativity, a light cone is not actually conceptually dependent on light or electromagnetism, though those things are related. It's a bit more abstract that that, and it exists regardless of whether or not anything is actually going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you get a bit used to thinking about spacetime as one thing, and the sequence-of-snapshots is a reasonable analogy, you can ask: given an event (a location and a time), what events could it be affected by, and what events could it affect? Thus, we can define an $O$'s &lt;em&gt;causal future&lt;/em&gt; as all events that a signal originating from $O$ could reach, and its &lt;em&gt;causal past&lt;/em&gt; as all events that could from which a signal that reaches $O$ could originate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because in there is a maximum speed any signal could propagate, the boundary of the causal future of any event looks like a cone in spacetime. That speed is indeed the speed of light (as well as that of any massless particle), which is why it's called a (future) light cone, and similarly for the past light cone. But that's incidental--relativity would work even if light was massive. In that case, we'd just have to call the invariant speed $c$ something else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this means that light cone is defined in terms of &lt;em&gt;all possible&lt;/em&gt; signals that &lt;em&gt;could&lt;/em&gt; be sent, not just those that are &lt;em&gt;actually&lt;/em&gt; sent. Therefore, no, it's not something that's 'emitted' by disturbances in the electromagnetic field or anything else. Some disturbances might travel &lt;em&gt;along&lt;/em&gt; the light cone, but that's not what the light cone is about, and it exists regardless of whether there are such disturbances.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2015-01-31T04:18:51.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8725" PostTypeId="2" ParentId="8723" CreationDate="2015-01-31T06:35:40.130" Score="1" Body="&lt;p&gt;When your professor refers to &lt;strong&gt;your&lt;/strong&gt; &quot;local star&quot; he is referring to the Sun.By &lt;strong&gt;a&lt;/strong&gt; &quot;local star&quot; he may mean any star within the &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Local_Bubble&quot; rel=&quot;nofollow&quot;&gt;local bubble&lt;/a&gt;&quot; which has a size ~100 parsecs (300 or so ly). Though he may also just be using the term loosely with no defined meaning other than nearby.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-01-31T10:41:51.877" LastActivityDate="2015-01-31T10:41:51.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8726" PostTypeId="5" CreationDate="2015-01-31T08:28:28.890" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-31T08:28:28.890" LastActivityDate="2015-01-31T08:28:28.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8727" PostTypeId="4" CreationDate="2015-01-31T08:28:28.890" Score="0" Body="Questions related to earth processes or related processes and principles." OwnerDisplayName="user2449" LastEditorDisplayName="user2449" LastEditDate="2015-02-01T00:10:56.953" LastActivityDate="2015-02-01T00:10:56.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8728" PostTypeId="2" ParentId="8723" CreationDate="2015-01-31T09:15:52.410" Score="2" Body="&lt;p&gt;This is too vague a term to be useful, because the Sun does not have any kind of sphere of influence (beyond the solar system) and nor do the nearby stars affect us with their light or gravity (Oort cloud perturbations aside).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead it would normally refer to the stellar population in the solar vicinity. This is a mixed bag of objects, with few shared characteristics. For example, objects within 10 light years of the Sun now, will not be so in 10 million years time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A working definition might be those stars that are close enough to the Sun that we think the census is complete to that distance. Unfortunately, incompleteness remains in all such samples at low masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.recons.org/&quot; rel=&quot;nofollow&quot;&gt;RECONS&lt;/a&gt; programme has been working hard for the last decade to complete the census of objects that are within 10pc (33 light years).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of a distant star is even more vague - all stars are distant. The Sun is 5 orders of magnitude closer than the next stars, so if anything, that is the only natural division between local and distant!&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-01-31T17:20:53.130" LastActivityDate="2015-01-31T17:20:53.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8729" PostTypeId="2" ParentId="8705" CreationDate="2015-01-31T09:56:36.120" Score="1" Body="&lt;p&gt;You are on the surface of the Earth now (I assume), as are almost all other humans. The surface of the Earth can be thought of as a two dimensional plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, let's say you do something amazing and you want to tell everyone on Earth about it, so you set up a means of broadcasting your news, at the speed of light, across the surface of the Earth. The speed of light is quick, but it still takes nearly two tenths of a second for your signal to make it round the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before that signal gets to them, nobody knows your news. You can imagine a circle, defined by the speed of light times the travel time, spreading outward from your location, containing those regions that have heard the news.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what you do is imagine those circles stacked on top of each other, each drawn after a small increment in time, with a larger radius, and separated vertically by an amount that represents that increment in time. The circles begin at your position, when you start to broadcast. This stack of circles of increasing radius forms a light cone. Inside the cone are regions of space and time that it is &lt;em&gt;possible&lt;/em&gt; for you to communicate your news to. Outside the cone are regions (of space and time) that can never hear the news because no signal can get there faster than the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A concrete example. It is about 5000 km between London and New York. If the stock market crashes at 17:00 (UT) in New York, it is impossible for traders in London to hear about it for at least 0.016 seconds. For that period, they lie outside the light cone of a signal produced at 17:00 in New York. More distant cities take even longer to intercept the light cone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The light cone idea also works in reverse. You can invert the cone to mark regions of space and time from which you could have heard some news. In the example above, the inverted light cone of the London traders does not contain New York until 0.016 seconds after 17:00.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A point of confusion is the idea of a cone, which is really only appropriate if space is defined as two dimensional - e.g. points on the surface of the Earth. Conceptually it is harder to work with in 3D (though mathematically equivalent). In 1D (points along a line), the light cone approximates to a light triangle.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-01-31T09:56:36.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8731" PostTypeId="1" CreationDate="2015-01-31T14:17:19.317" Score="6" ViewCount="855" Body="&lt;p&gt;The development of telescopes enabled the discovery of moons orbiting around Jupiter, and the existence of a full set of phases of Venus, as opposed to a limited set which would be predicted by geocentrism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is it possible to disprove, or at least make implausible, geocentrism without a telescope? Preferably without calculus or more advanced maths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious for two reasons. One is knowing whether it was lack of technology, or humanity's attitude towards the universe, that was responsible for geocentrism. The other is that I've heard that some astronomical facts, such as the size of the earth, were calculated before the advent of telescopes, and I'd find it interesting if more facts could have been deduced even with a limited set of technology.&lt;/p&gt;&#xA;" OwnerUserId="2679" LastActivityDate="2015-02-01T21:51:16.417" Title="Is it possible to disprove geocentrism without telescopes?" Tags="&lt;solar-system&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8732" PostTypeId="1" AcceptedAnswerId="8738" CreationDate="2015-01-31T14:34:06.107" Score="7" ViewCount="2339" Body="&lt;p&gt;In geocentric models, did the Earth rotate around its axis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that if the Earth rotated around its axis, then the Sun would not have to orbit around the Earth at a very fast rate, whereas if the Earth didn't rotate, then the Sun would have to orbit the Earth every day.&lt;/p&gt;&#xA;" OwnerUserId="2679" LastEditorUserId="7411" LastEditDate="2016-09-16T16:54:20.913" LastActivityDate="2016-09-17T09:11:48.460" Title="In geocentric models, did the Earth rotate?" Tags="&lt;solar-system&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8733" PostTypeId="2" ParentId="8731" CreationDate="2015-01-31T15:35:14.433" Score="4" Body="&lt;p&gt;I would argue that the &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_retrograde_motion&quot; rel=&quot;nofollow&quot;&gt;Apparent retrograde motion&lt;/a&gt; of the planets over time, is evidence against geocentrism, and probably why &lt;a href=&quot;https://en.wikipedia.org/wiki/Aristarchus_of_Samos&quot; rel=&quot;nofollow&quot;&gt;Aristarchus of Samos&lt;/a&gt; proposed a Heliocentric model of the Solar System.&lt;/p&gt;&#xA;" OwnerUserId="2880" LastActivityDate="2015-01-31T15:35:14.433" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8734" PostTypeId="1" CreationDate="2015-01-31T15:50:22.740" Score="0" ViewCount="245" Body="&lt;p&gt;Could there be another planet between Mercury and the Sun?&#xA;Kepler found a lot of planets orbiting its star even 20 times closer than Mercury.&#xA;If there were such planets in our system - would it be possible to discover them?&lt;/p&gt;&#xA;" OwnerUserId="5441" LastActivityDate="2015-01-31T17:32:43.840" Title="Could there be another planet between Mercury and the Sun?" Tags="&lt;solar-system&gt;&lt;planet&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8735" PostTypeId="2" ParentId="8734" CreationDate="2015-01-31T16:10:11.797" Score="6" Body="&lt;p&gt;It is very unlikely. While the hypothetical planet &lt;a href=&quot;http://en.wikipedia.org/wiki/Vulcan_%28hypothetical_planet%29&quot; rel=&quot;noreferrer&quot;&gt;Vulcan&lt;/a&gt; between Mercury and the Sun was ruled out many decades ago searches for &lt;a href=&quot;http://en.wikipedia.org/wiki/Vulcanoid_asteroid&quot; rel=&quot;noreferrer&quot;&gt;Vulcanoid asteroids&lt;/a&gt; would very likely have detected anything that might qualify as a intra-Mercurial planet.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-01-31T16:10:11.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8738" PostTypeId="2" ParentId="8732" CreationDate="2015-01-31T16:45:25.083" Score="8" Body="&lt;p&gt;In the historic geocentric models, and in particular in the Ptolemaic model, the earth is immobile in the centre of the cosmos, and the sphere of the fixed stars rotates around it once a day, carrying the sun, moon and planets with it. There were astronomers who realised that this apparent motion could also be explained, from a purely mathematical point of view, by assuming that the earth rotates on its axis, but they rejected this option. By the way, neither Ptolemy nor any other serious astronomer believed that the earth was flat. &lt;/p&gt;&#xA;" OwnerUserId="5442" LastActivityDate="2015-01-31T16:45:25.083" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8739" PostTypeId="2" ParentId="8732" CreationDate="2015-01-31T17:10:39.297" Score="2" Body="&lt;p&gt;The idea of a rotating Earth was the biggest hurdle for the non-geocentric ideas to overcome. If the Earth were rotating, wouldn't we notice it? Riding a horse is a motion which one cannot help but notice. And a rotating Earth would be much more violent, at 465 meters per second, as was estimated already in ancient times. Ten times a hurricane. That is obviously not the case. I can stand up on the ground without falling over, but I can't stand up on a galloping horse's back, people argued quite reasonably.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It requires careful cleverness to apply our gardening-evolved intuition to the reality beyond the fence. I personally find the &lt;a href=&quot;https://en.wikipedia.org/wiki/Oberth_effect&quot; rel=&quot;nofollow&quot;&gt;Oberth effect&lt;/a&gt; to be unintuitive, that in empty space a push gives more acceleration to a fast object than to a slow one. With friction from roads, rails and air the opposite is true in our everyday life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resolution &lt;em&gt;(certainly with ancient predecessors if documents are preserved)&lt;/em&gt; is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Galileo%27s_ship&quot; rel=&quot;nofollow&quot;&gt;Gallileo's ship&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2016-09-17T09:11:48.460" LastActivityDate="2016-09-17T09:11:48.460" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8740" PostTypeId="2" ParentId="8734" CreationDate="2015-01-31T17:32:43.840" Score="3" Body="&lt;p&gt;No, that is not possible. There are quite a few capable &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_solar_telescopes&quot; rel=&quot;nofollow&quot;&gt;telescopes studying the Sun&lt;/a&gt; since some decades. Already Galileo stared at the Sun until he got blind. Don't you think a planet would've been detected if it passed by in &lt;a href=&quot;https://www.youtube.com/watch?v=8Sm8PBr3ysc#t=34&quot; rel=&quot;nofollow&quot;&gt;images like these&lt;/a&gt;? But &lt;a href=&quot;https://www.youtube.com/watch?v=M2JCKiSO7AA&quot; rel=&quot;nofollow&quot;&gt;Sun grazing comets&lt;/a&gt; are pretty common visitors in the corona.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a bit weird that a couple of percent of exoplanetary systems have hot Jupiters. They probably didn't form there, but migrated. And still stay there somehow.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-01-31T17:32:43.840" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8741" PostTypeId="1" AcceptedAnswerId="8743" CreationDate="2015-01-31T21:05:25.990" Score="0" ViewCount="1487" Body="&lt;p&gt;Using data from the Hipparcos Catalog download file, is there a way to determine the Visual or Apparent Magnitude from the data within?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I obtain the Mass, Radius, colour, temperature, type (dwarf, sun-like, hypergiant, etc) of the star from the catalog?  Which column relates to which item i'm looking for?  Any examples?&lt;/p&gt;&#xA;" OwnerUserId="5401" LastActivityDate="2015-02-01T00:35:10.553" Title="Mass, Radius, Colour, Size, Type of a Star from the Hipparcos Catalog" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8743" PostTypeId="2" ParentId="8741" CreationDate="2015-02-01T00:35:10.553" Score="6" Body="&lt;p&gt;The apparent magnitude (in V ~ visual) is in column 42-46. Other magnitudes are BT magnitude (columns 218-223) and VT magnitude (columns 231-236). These are magnitudes recorded with different (colour) filters. &#xA;The colour ($B-V$) can be found in columns 246-251, expressed in magnitudes. This is the difference between V (visual) magnitude and B (blue) magnitude also known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Color_index&quot;&gt;colour index&lt;/a&gt;. Blue stars have a low colour index ($B-V&amp;lt;0$) as the star will be brighter in the blue (B) passband (filter), and will therefore have a low B magnitude, while red stars will have a higher $B-V$ colour index.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass, Radius, and Temperature are not in the catalogue. Mass can only be directly determined for multiple star systems and requires a lot of observations per multiple star. Radius is determined from stellar models. And Temperature can be determined from the spectrum of the star. A crude indication can be gotten from the colour index of the star. (see for instance this &lt;a href=&quot;https://docs.kde.org/stable/en/kdeedu/kstars/ai-colorandtemp.html&quot;&gt;webpage&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally the type can be found in the star's spectral type. The spectral type is in columns 436-447. These spectral types are not always easy to read but they consist of a spectral class (e.g. O, or F type stars) and the luminosity class (and some other information). The luminosity class gives the type of class in roman numerals. Below is the list from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_classification&quot;&gt;wikipedia page on stellar classification&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;0 or Ia+ (hypergiants or extremely luminous supergiants). Example:&#xA;Cygnus OB2#12 (B3-4Ia+)[12] &lt;/li&gt;&#xA;&lt;li&gt;Ia (luminous supergiants). Example: Eta&#xA;Canis Majoris (B5Ia)[13] &lt;/li&gt;&#xA;&lt;li&gt;Iab (intermediate luminous supergiants).&#xA;Example: Gamma Cygni (F8Iab)[14] &lt;/li&gt;&#xA;&lt;li&gt;Ib (less luminous supergiants).&#xA;Example: Zeta Persei (B1Ib)[15] &lt;/li&gt;&#xA;&lt;li&gt;II bright giants. Example: Beta&#xA;Leporis (G0II)[16] &lt;/li&gt;&#xA;&lt;li&gt;III normal giants. Example: Arcturus (K0III)[17]&lt;/li&gt;&#xA;&lt;li&gt;IV subgiants. Example: Gamma Cassiopeiae (B0.5IVpe)[18] &lt;/li&gt;&#xA;&lt;li&gt;V main-sequence stars (dwarfs), Example: Achernar (B6Vep)[15] &lt;/li&gt;&#xA;&lt;li&gt;sd (prefix) subdwarfs. Example: HD 149382 (sdB5)[19] &lt;/li&gt;&#xA;&lt;li&gt;D (prefix) white&#xA;dwarfs.[nb 2] Example: van Maanen 2 (DZ8)[20]&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The Sun's spectral type is G2V, which means that the spectral class is G2 and the luminosity class is V, which means that the Sun is a main-sequence star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following record is for the star Arcturus (α Boötis):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;H|       69673|H|14 15 40.35|+19 11 14.2|-0.05|1|G|213.91811403|+19.18726997|*|  88.85|-1093.45|-1999.40|  0.92|  0.42|  0.74|  0.60|  0.52|-0.52| 0.16|-0.13|-0.07| 0.05|-0.01| 0.03|-0.12|-0.03|-0.31|  7| 2.01| 69673| 1.629|0.010| 0.286|0.009|*| 1.239|0.006|G|1.22|0.02|A|*| 0.1114|0.0020|0.015| 70|*| 0.09| 0.14|       |U|2| |14157+1911|H| 1| 2|C| |A|AB|198|  0.255|0.039| 3.33|0.31|S| | |124897|B+19 2777 |          |          |1.22|K2IIIp      |X &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The apparent visual is(V) magnitude $-0.05$. The colour index is $B-V = 1.239$, which means that it is probably a red K type star with temperature $T\approx 4000 \textrm{K}$ (&lt;a href=&quot;https://docs.kde.org/stable/en/kdeedu/kstars/ai-colorandtemp.html&quot;&gt;from figure 2 on this webpage&lt;/a&gt;). The spectral type is K2IIIp, i.e. the spectral class is K2 and the luminosity class is III, a normal giant star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An indication for the Radius can be found from the Hertzsprung-Russel Diagram in which radii are added (see &lt;a href=&quot;http://astro.wsu.edu/worthey/astro/html/im-Galaxy/hr-radii.gif&quot;&gt;Figure here&lt;/a&gt;). The luminosity of Arcturus can be calculated as follows:&#xA;$$ d = \frac{1}{\varpi}$$&#xA;where $d$ is in parsec and $\varpi$ is the parallax in arsceconds. The parallax is in columns 80-86 and is $\varpi = 88.85mas = 0.08885^{\prime\prime}$ for Arcturus. The distance is, therefore, $d=11.25\textrm{pc}$. You can then calculate the &lt;a href=&quot;http://en.wikipedia.org/wiki/Absolute_magnitude&quot;&gt;absolute magnitude ($M_V$):&lt;/a&gt;&#xA;$$ M_V = m_V -5(\log{d}-1)$$&#xA;where $m_V$ is the apparent visual (V) magnitude. The absolute magnitude of Arcturus is: $M_V=-0.05 - 5(\log{11.25}-1)=-0.31$.&#xA;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnitude_(astronomy)&quot;&gt;luminosity&lt;/a&gt; is then given by:&#xA;$$M-M_{\textrm{sun}} = -2.5\log{\left( \frac{L}{L_{\textrm{sun}}}\right)}$$&#xA;where the absolute magnitude of the Sun is $M_{\textrm{sun}} = 4.83$ and $L$ is the luminosity of Arcturus. This gives:&#xA;$$\log{\left( \frac{L}{L_{\textrm{sun}}}\right)} = 2.056 $$&#xA;$$L = 114  \textrm{L}_{\textrm{sun}}$$&#xA;From the &lt;a href=&quot;http://astro.wsu.edu/worthey/astro/html/im-Galaxy/hr-radii.gif&quot;&gt;Hertzsprung-Russel diagram&lt;/a&gt; we get a radius for Arcturus of about $R \approx 10  \textrm{R}_{\textrm{sun}}$.&lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2015-02-01T00:35:10.553" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8745" PostTypeId="2" ParentId="8731" CreationDate="2015-02-01T21:01:07.353" Score="3" Body="&lt;p&gt;You can't disprove geocentricism. It is a perfectly valid way to look at the universe. After all, all frames of reference are equally valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can disprove is that geocentricism is the one and only way to look at the universe. Once again, all frames of reference are equally valid. This means that a Highlander point of view (&quot;in the end, there is only one&quot;) is nonsense, which in turn means that claims that geocentricism is the one and only valid point of view is nonsense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That all frames of reference are equally valid does not mean that frame X is just as good as frame Y when it comes to explaining motion. Frame Y might well need a slew of fictitious forces while frame X needs none. The absence of fictitious forces in frame X makes the equations of motion take their simplest form when viewed from the perspective of frame X. Other frames of reference such as Frame Y (e.g., geocentricism) will result in fictitious forces. These fictitious forces vanish with a smarter choice of reference frames.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2015-02-01T21:06:10.747" LastActivityDate="2015-02-01T21:06:10.747" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8746" PostTypeId="2" ParentId="8731" CreationDate="2015-02-01T21:51:16.417" Score="1" Body="&lt;p&gt;Johannes Kepler actually disproved geocentrism only a few years before the telescope was invented. He used Tycho Brahe's surprisingly precise naked eye observations of Mars' and the Sun's relative positions on the sky. He first proved that Ptolemy's, Copernicus' and Brahe's Solar system models were all mathematically identical, they just chose different frames of reference. Then he painstakingly founded astrophysics and proved genuine heliocentrism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's &lt;a href=&quot;http://science.larouchepac.com/kepler/newastronomy/&quot; rel=&quot;nofollow&quot;&gt;a free online study guide&lt;/a&gt; for Johannes Kepler's work &quot;New Astronomy&quot;, with some interactive java animations. He was a contemporary with William Shakespeare, so it should not come as a surprise that he was pretty easy to read (in English translation, first available only in the 1980's!) for being a 400 years old astronomer, making personal comments and even some jokes between the lines.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-02-01T21:51:16.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8748" PostTypeId="2" ParentId="1273" CreationDate="2015-02-01T23:10:33.607" Score="0" Body="&lt;p&gt;Celestron 127slt is a great way to start and reasonable price for a beginner. &#xA;It has star align system that it can guide you through the sky objects during the time zone you are in after you calibrate it.&#xA;It has also pretty good vision with its spectacular lens ! &lt;/p&gt;&#xA;" OwnerUserId="2997" LastActivityDate="2015-02-01T23:10:33.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8750" PostTypeId="1" AcceptedAnswerId="8757" CreationDate="2015-02-02T12:54:39.583" Score="0" ViewCount="1394" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If the observer observes Arcturus passed through the meridian about 23:00:00 of 28 April. Find the time that Arcturus passed the meridian in 29 April. And what day does the Arcturus will pass the meridian about 01:00:00?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think that when Arcturus passed through the meridian, the hour angle will be 0 degree. Sidereal time equal hour angle plus right ascension. right ascention of any stars always be the same. So Sidereal time will be as same as every times while the star located at the meridian. And it doesn't matter because solar time equal sidereal time minus 12hr. Do I understand something wrong? And how do I do this question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding question;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your answer was very clear. But if I consider about the moon instead ex; the observer is at the equator and saw the moon which has a declination of 0 degree set at 11:00 PM yesterday, what time the observer will see the moon set today? I try to use the thinking process like this again. I know that the sidereal month of the moon is 27.32166 days. I calculate the solar month of the moon from 365.24219 of solar year divided by 12, and I get 30.43684 days per solar month. From subtracting both of these numbers, I get 3 days plus about 2 hours 45 minutes and 52 seconds. So to answer this question I will get 2 hours 45 minutes 52 seconds behind. My answer is that I will see the moon set at 13:45:52 PM. Do I right? I don't understand why the time I get is behind instead of earlier as the question about the Arcturus I asked before. I'm also not sure that does the declination of the moon and the position of the observer relate or not too. Thank you very much:)&lt;/p&gt;&#xA;" OwnerUserId="2787" LastEditorUserId="2787" LastEditDate="2015-02-07T05:22:27.210" LastActivityDate="2015-02-07T05:22:27.210" Title="Calculate the time when Arcturus passed through the meridian" Tags="&lt;time&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8751" PostTypeId="1" AcceptedAnswerId="8762" CreationDate="2015-02-02T13:11:23.000" Score="1" ViewCount="479" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;One night, you and your friend were talking to each other. Your friend far about 20 km. from you. You told your friend that you see the shooting star pass through the sky at altitude of 75 degrees. Your friend also see this shooting star, but has difference apparent location by 6 degrees. Find the location of this shooting star, what is it hight from earth ground.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think 6 degrees is the different of azimuth angle. But as I live in different lattitude, it may be effect the distance between friend and me. Or the answer is just 75degrees turn to the height from the earth ground. But how? Which and how does the correct way to solve this question?&lt;/p&gt;&#xA;" OwnerUserId="2787" LastEditorUserId="2153" LastEditDate="2015-02-02T14:00:10.353" LastActivityDate="2015-02-04T01:25:37.513" Title="Find the height of the shooting star from the Earth's ground" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8752" PostTypeId="1" AcceptedAnswerId="8753" CreationDate="2015-02-02T13:29:31.933" Score="8" ViewCount="2251" Body="&lt;p&gt;Assume there were a roughly spherical object in space like a meteorite or a comet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I weighed 80kg on Earth, how much mass would be required for an object in space so that I could stay on its surface without &quot;flying away?&quot; It's not necessary have the same gravity as Earth, but I am wondering what minimal mass the object would require for it to have &lt;em&gt;meaningful&lt;/em&gt; gravity for someone to remain on the surface.&lt;/p&gt;&#xA;" OwnerUserId="5451" LastEditorUserId="58" LastEditDate="2015-02-02T17:46:21.033" LastActivityDate="2015-04-05T08:48:16.907" Title="How much mass does an object in space need to keep a human on its surface?" Tags="&lt;gravity&gt;&lt;space&gt;&lt;planet&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8753" PostTypeId="2" ParentId="8752" CreationDate="2015-02-02T15:14:40.327" Score="9" Body="&lt;p&gt;Any object with mass (even you) has gravity. The mutual attractive force between two objects is given by the formula&#xA;$$ F = G \frac{M_1 M_2}{R_{12}^2}, $$&#xA;where the two mass are $M_1$ and $M_2$ and $R_{12}$ is the separation of their centres of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer your question we need to define some sort of parameter that specifies what you mean by &quot;meaningful gravity&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way of doing this would be to demand that the forces due to tides from the Sun are greater than the gravitational force holding you to the body in question. Even here though we need to make an assumption about how far away the object is from the Sun - let it be $r$. Now let your mass be $m$, the mass of the body be $M$, the radius of the body be $R$, and the mass of the Sun be $M_{\odot}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To stay attached to the object when you are at the &quot;subsolar point&quot; - i.e. the body rotates so that you are closest to the Sun, then we require&#xA;$$G \frac{M m}{R^2} &amp;gt; G \frac{m M_{\odot}}{(r-R)^2} - G\frac{m M_{\odot}}{r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can assume that $r \gg R$, so cancelling out $Gm$ and performing a binomial expansion on the middle term, keeping only the first two terms of the expansion:&#xA;$$\frac{M}{R^{2}} &amp;gt; \frac{M_{\odot}}{r^{2}}(1 + \frac{2R}{r}) - \frac{M_{\odot}}{r^{2}}$$ &#xA;$$ \frac{M}{R^{2}} &amp;gt;  2M_{\odot}\frac{R}{r^3}$$&#xA;Thus this sets a limit on the &lt;em&gt;density&lt;/em&gt; of the object in question&#xA;$$ \frac{3M}{4\pi R^3} = \rho &amp;gt; \frac{3}{2\pi} \frac{M_{\odot}}{r^3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At $r=1\ au$, this means the density only needs to exceed $3 \times 10^{-4}$ kg/m$^3$, which will be satisfied by any solid body. NB: This limit would be a limit where you would actually be pulled from the surface by the Sun's tides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more stringent requirement might be to ensure that you can't &lt;em&gt;jump&lt;/em&gt; off the surface. On Earth, an average person might be able to jump vertically about 50cm. Using the equations of uniform acceleration (SUVAT), we know $u^2 = 2g s$, where $g$ is the gravitational acceleration, $s$ is the height jumped and $u$ is the initial upward velocity. This tells us that you can jump upwards at about 3m/s. Assuming this is the same on any other body (its hard to say how well you could jump in much lower gravity) you could equate this to the &lt;em&gt;escape velocity&lt;/em&gt; of the object, given by $v_{esc} = (2GM/R)^{1/2}$. Thus this gives a constraint of $M/R &amp;gt; u^2/2G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we set a realistic density of $\rho = 5000$ kg/m$^{3}$ for an asteroid, we can replace $M$ with $4\pi R^{3} \rho/3$, to say that you would be able to jump off an asteroid if it were smaller than:&#xA;$$ R &amp;lt; u \left(\frac{3}{8\pi G \rho}\right)^{1/2}$$&#xA;&lt;strong&gt;For the numbers discussed this means $R&amp;lt; 1.8$ km&lt;/strong&gt; and &lt;strong&gt;$M&amp;lt;1.2\times 10^{14}$ kg&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many more details at &#xA;&lt;a href=&quot;https://physics.stackexchange.com/questions/46318/is-there-a-small-enough-planet-or-asteroid-you-can-orbit-by-jumping&quot;&gt;https://physics.stackexchange.com/questions/46318/is-there-a-small-enough-planet-or-asteroid-you-can-orbit-by-jumping&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, as you will have noticed, the result doesn't depend on &lt;em&gt;your&lt;/em&gt; mass.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-04-05T08:48:16.907" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8755" PostTypeId="2" ParentId="1011" CreationDate="2015-02-02T16:47:28.010" Score="3" Body="&lt;p&gt;A comet that travels to the Sun from the Oort cloud will be travelling roughly at its escape velocity at perihelion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fastest comets will be those that travel from the Oort cloud and plunge directly into the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the comet to emerge from around the Sun, then its perihelion distance must be no lower than $\sim 7 \times 10^{8}$ m. At that point it will have approximately the escape velocity at that radius $v_{esc} = (2GM_{\odot}/r)^{1/2} = 617$ km/s or 1.39 million miles per hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore I expect that lots of comets have attained something like or exceeding this speed before. &lt;a href=&quot;http://en.wikipedia.org/wiki/C/2011_W3_%28Lovejoy%29&quot; rel=&quot;nofollow&quot;&gt;Comet Lovejoy&lt;/a&gt; for example achieved a perihelion distance of $8.36\times10^{8}$ m and was travelling at 536 km/s (1.2 million miles per hour, and in fairly good agreement with the formula I quoted above - the aphelion was not infinity, but $\sim 157$ au, which largely accounts for the difference).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-02-02T16:47:28.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8757" PostTypeId="2" ParentId="8750" CreationDate="2015-02-03T01:05:40.293" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the time that Arcturus passed the meridian in 29 April.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With respect to the Sun, the Earth rotates once every 24 hours.  But because the Sun's apparent position changes in the sky due to Earth's revolution around the Sun, the Earth rotates once every &lt;a href=&quot;http://www.universetoday.com/14700/how-long-is-a-day-on-earth/&quot; rel=&quot;nofollow&quot;&gt;23 hours, 56 minutes, 4 seconds with respect to the stars&lt;/a&gt;.  Because of this 3 minutes, 56 seconds difference, all stars rise and set 3 minutes, 56 seconds earlier each succeeding day.  This applies to the time of a star passing through the meridian as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, on the next day, April 29, Arcturus will pass the meridian at 22:56:04.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And what day does the Arcturus will pass the meridian about 01:00:00?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Over the course of an entire year, about 365 days and 6 hours, these 3 minutes and 56 seconds adds up to an entire day, so that exactly one year later, the same star will pass through the meridian at the same time of day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Arcturus to pass the meridian exactly 22 hours earlier, it should be 22/24 of a year before then, or about 335 days later, or about 30 days earlier, or March 29.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;solar time equal sidereal time minus 12hr&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Generally not correct.  Solar time is based on the average time it takes for the sun to cross the meridian (24 hours), and &lt;a href=&quot;http://earthsky.org/astronomy-essentials/what-is-sidereal-time&quot; rel=&quot;nofollow&quot;&gt;sidereal time&lt;/a&gt; is based on the time it takes for a star to cross the meridian.  Sidereal time counts 24 &quot;hours&quot; in 23 hours, 56 minutes, and 4 seconds, meaning that it runs slightly faster than our normal solar time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one day in which solar and sidereal time are 12 hours off, on the northward equinox in March.  The Sun is on the northward equinox, but solar time is measured since midnight where sidereal time is measured since passing the meridian (local noon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one day in which solar and sidereal time coincide, on the southward equinox in September.  The Sun is on the southward equinox, 180 degrees in the sky from the northward equinox, so the 12 hours off from midnight and the Sun being 12 hours off from the northward equinox cancel.&lt;/p&gt;&#xA;" OwnerUserId="1082" LastActivityDate="2015-02-03T01:05:40.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8758" PostTypeId="2" ParentId="8540" CreationDate="2015-02-03T01:36:54.607" Score="4" Body="&lt;p&gt;In very a general post-Newtonian metric for a two-body system with the first body oblate, where $M\equiv m_1+m_2$ is the total mass, $\mu\equiv m_1m_2/M$ is the reduced mass, and $p\equiv a(1-e^2)$ is the semi-latus rectum of the orbit, the perihelion advance per orbit is&#xA;$$\small\delta\varpi = 6\pi\frac{GM}{pc^2}\left[\underbrace{\frac{2-\beta+2\gamma}{3}}_{\text{GTR}=1} + \underbrace{\frac{2\alpha_1-\alpha_2+\alpha_3+2\zeta_2}{6}}_{\text{GTR}=0}\frac{\mu}{M} + \frac{J_2R^2c^2}{2GMp}\right]\text{.}$$&#xA;According to GTR, $\beta=\gamma=1$ exactly. The second term contains various parameters relating to preferred-frame effects and energy-momentum non-conservation. In GTR, they are all identically zero, while experimentally, $\alpha_1\lesssim10^{-4}$ and the rest are many orders of magnitude less than that. Finally, in the third term, $J_2$ is the quadrupole moment of the first body; $J_2R^2 = (C-A)/m_1$ where $C$ and $A$ are moments of inertia about rotational and equatorial axes, respectively. A derivation can be found Will §7.3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I try to find out what the parameters' values such as $G$, $M_{sun}$, $\omega_{min}$ at aphelion radius, etc. used to get this exact 42,98.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is just the front coefficient in the above: if $T$ is the orbital period of Mercury, then&#xA;$$6\pi\frac{GM/c^2}{a(1-e^2)}\frac{1}{T} = \frac{42.98''}{\mathrm{century}}\text{,}$$&#xA;and we can restate the GTR prediction of Mercury as&#xA;$$\dot\varpi_{\text{☿}}^{\scriptsize\text{GTR}} = \left.\frac{42.98''}{\mathrm{century}}\right.\left[1+2958J_2\right]\text{,}$$&#xA;where $J_2$ is the gravitational quadrupole moment of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;NASA measured 43,13 arc seconds per century. General relativity predicts 42,98 arc seconds per century.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you're referring to one of Anderson &lt;em&gt;et al.&lt;/em&gt;'s analyses, the value is $43.13\pm0.14$. Under GTR, this value is compatible with any $J_2$ up to about&#xA;$$\begin{eqnarray*}\frac{J_2R^2c^2}{2GMp}\lesssim 0.0067 &amp;amp;\Longleftrightarrow &amp;amp; J_2\lesssim 2.3\times10^{-6}\text{.}\end{eqnarray*}$$&#xA;But what is the Sun's quadrupole moment? This is a somewhat sticky question, but the results of lunar laser ranging experiments are incompatible with $J_2$ of more than about $3\times10^{-6}$, while using helioseismology, Pijpers (1998) estimates $J_2 = (2.18\pm0.06)\times 10^{-7}$. Using Pijper's value yields a GTR prediction of $43.01''$ per century, which is compatible with Anderson &lt;em&gt;et al.&lt;/em&gt;'s analysis. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;With a new model developed by R. Plamondon, we get exactly 43,13 and also it matched perfectly with Mercury's period of revolution around Sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no idea what Plamondon's model is, but I would be rather suspicious of any model that gives &quot;exactly&quot; $43.13$ when that's just the middle of an interval of statistical uncertainty according to one of the many prior analyses of Mercury's orbit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When I put my parameters into Einstein metric, I got something unexpected.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we ignore the post-Newtonian parameters other than $\beta$ and $\gamma$, the static post-Newtonian metric is&#xA;$$\mathrm{d}s^2 = -\left(1+2\Phi+2\beta\Phi^2\right)\mathrm{d}t^2 + \left(1-2\gamma\Phi\right)\mathrm{d}S_\text{Euclid}^2\text{,}$$&#xA;where the Newtonian potential of the Sun is modeled up to the quadrupole term&#xA;$$\Phi = -\frac{M_\odot}{r}\left[1-J_2\frac{R_\odot^2}{r^2}\frac{3\cos^2-1}{2}\right]\text{.}$$&#xA;Deriving the perihelion advance of Mercury with this metric is Exercise 40.5 in MTW, and it matches the general result referenced above with $M\equiv M_\odot+M_{☿} \approx M_\odot$ and sans the middle term. For the Sun-Mercury system, $\mu/M\sim 10^{-7}$, so the middle term is even more irrelevant than bounds on the PPN parameters would suggest.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Will, C. M., &lt;em&gt;Theory and experiment in gravitational physics&lt;/em&gt;, (Cambridge University Press, Cambridge, 1993), 2nd edition.&lt;/li&gt;&#xA;&lt;li&gt;Anderson, J. D., Campbell, J. K., Jurgens, R. F., Lau, E. L., Newhall, X. X., Slade, M. A., Standish Jr, E. M., in &lt;em&gt;Proceedings of the Sixth Marcel Grossmann Meeting on General Relativity&lt;/em&gt;, ed. by Sato, H. and Nakamura, T., (World Scientific, Singapore, 1992).&lt;/li&gt;&#xA;&lt;li&gt;Pijpers, F. P., &lt;em&gt;Mon. Not. R. Astron. Soc.&lt;/em&gt; &lt;strong&gt;297&lt;/strong&gt;, L76 (1998). [arXiv:&lt;a href=&quot;http://arxiv.org/abs/astro-ph/9804258&quot; rel=&quot;nofollow&quot;&gt;astro-ph/9804258&lt;/a&gt;]&lt;/li&gt;&#xA;&lt;li&gt;Misner, C. W., Thorne, K. S., and Wheeler, J. A., &lt;em&gt;Gravitation&lt;/em&gt;, (Freeman,&#xA;San Francisco, 1973).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="661" LastActivityDate="2015-02-03T01:36:54.607" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8760" PostTypeId="1" AcceptedAnswerId="8761" CreationDate="2015-02-03T12:55:11.963" Score="5" ViewCount="497" Body="&lt;p&gt;I am reading &quot;The Essential Cosmic Perspective&quot; by Jeffrey O. Bennett, Megan O. Donahue, Nicholas Schneider, Mark Voit.&#xA;In Chapter 14, it is stated that an evidence of the presence of dark matter in our galaxy is that the rotational curve does not match that obtained from calculation.&#xA;In the calculation, the following formula is used&#xA;$$M_r=\frac{rv^2}{G}$$&#xA;where $M_r$ is the mass enclosed inside a radius $r$ from the galactic center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that the above equation is a result of the shell theorem. But we know that shell theorem applies to spherically symmetric mass distribution only, and most galaxies are disks. So why can we still do this?&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2015-02-06T01:53:24.057" LastActivityDate="2015-02-06T12:50:06.133" Title="Galaxy rotation curve and dark matter" Tags="&lt;galaxy&gt;&lt;dark-matter&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8761" PostTypeId="2" ParentId="8760" CreationDate="2015-02-03T16:11:59.160" Score="5" Body="&lt;p&gt;We can't. That is an over simplification only used in elementary treatments simply to provide the the flavour of the argument. If you see it done somewhere in the refereed literature, it is probably incorrect. Of course it may be true that the mass is almost spherically symmetric, especially if it is dominated by a spherically symmetric dark matter component, but that the visible light is not. Or, it may be true that at large distances from even an asymmetric mass distribution, a Keplerian potential is a good approximation for the orbits of distant objects (e.g. satellite galaxies to the Milky Way). A measured galaxy rotation curve makes no such assumption, it is merely a measurement of rotation speed as a function of radius. It is only the interpretation and modelling that needs to deal with the mass distribution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real situation is much more complex. See for example &lt;a href=&quot;http://ned.ipac.caltech.edu/level5/March01/Battaner/revision.html&quot; rel=&quot;nofollow&quot;&gt;http://ned.ipac.caltech.edu/level5/March01/Battaner/revision.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, even if one assumed all the mass was concentrated into a disk-like shape, the only way you can get flat rotation curves is to assume that the mass in the disk does not &quot;follow the light&quot; - that the mass-to-luminosity ratio increases vastly with radius - which is essentially still saying that you have &quot;dark matter&quot;, just in a disk.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-02-06T12:50:06.133" LastActivityDate="2015-02-06T12:50:06.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8762" PostTypeId="2" ParentId="8751" CreationDate="2015-02-04T01:25:37.513" Score="4" Body="&lt;p&gt;This is basically a trigonometry question.  I wasn't sure from the question if the apparent location from your friend's position was six degrees higher or lower.  I chose higher for the example, but you could change it.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ysmtn.jpg&quot; alt=&quot;meteor altitude&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have two triangles with the same altitude, but different bases and angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{altitude} = x \tan(81^{\circ}) = (x + 20km) \tan (75^{\circ})$$  Do you see how to calculate it from there?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-02-04T01:25:37.513" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8763" PostTypeId="1" AcceptedAnswerId="8770" CreationDate="2015-02-04T01:50:41.003" Score="9" ViewCount="3159" Body="&lt;p&gt;Does the cosmic microwave background pattern change over time?  I would assume it is getting &quot;cooler&quot; as in more redshifted as time passes, but I am more interested if the pattern on the background, as depicted below changes.  I would expect we would see something like the cosmic microwave background retreating one light year (or possibly further due to inflation) every year, and it would be neat if we could make a &quot;3D map&quot; of the background as time passes (e.g. take an image of it every year).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7nGzm.jpg&quot; alt=&quot;Cosmic Microwave Background&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://lambda.gsfc.nasa.gov/product/map/dr5/map_images/wmap_planck/wmap_planck_ilc_300uK_med.png&quot; rel=&quot;noreferrer&quot;&gt;http://lambda.gsfc.nasa.gov/product/map/dr5/map_images/wmap_planck/wmap_planck_ilc_300uK_med.png&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="1386" LastEditDate="2015-02-04T13:13:16.453" LastActivityDate="2020-03-07T11:17:51.367" Title="Does the cosmic microwave background change over time?" Tags="&lt;cmb&gt;&lt;data-analysis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8764" PostTypeId="1" AcceptedAnswerId="8765" CreationDate="2015-02-04T02:00:23.517" Score="1" ViewCount="403" Body="&lt;p&gt;Does the cosmic microwave background recede at the speed of light?  Is it possible it recedes even faster because of the stretching of space?&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2015-02-04T03:16:29.383" Title="Does the cosmic microwave background recede at the speed of light?" Tags="&lt;light&gt;&lt;speed&gt;&lt;cmb&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8765" PostTypeId="2" ParentId="8764" CreationDate="2015-02-04T02:49:19.590" Score="3" Body="&lt;p&gt;Well obviously the CMBR arrives (towards us, not receding) at us at the speed of light as it is a form of electromagnetic radiation and local velocities are not affected by expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I guess you may mean though is the surface of last scattering, which is the surface from which the CMBR that we receive at a given instant was emitted in the early Universe. This surface of course is no longer emitting CMBR in the present cosmological time and it is not a fixed (comoving) surface either as which surface we receive the CMBR from changes with time, however we can give a recession velocity to the comoving set of points which make up the surface of last scattering at any given instant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the standard model of cosmology, the Lambda-CDM model, the instantaneous surface of last scattering is receding from us at approximately 3.2c (i.e. a recession velocity 3.2 times the speed of light).&lt;/p&gt;&#xA;" OwnerUserId="2092" LastEditorUserId="2092" LastEditDate="2015-02-04T03:16:29.383" LastActivityDate="2015-02-04T03:16:29.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8766" PostTypeId="2" ParentId="8763" CreationDate="2015-02-04T03:01:37.003" Score="6" Body="&lt;p&gt;When we observe the CMBR we are observing the surface of last scattering, however the comoving points that make up the surface of last scattering (which infact will actually have a comparitvely very small, but non-zero thickness rather than being a 2D surface) change with time, which in theory should correspond to a change in the observed pattern (of anistropy) over time. On human time scales and given the limitations on our ability to observe the minute details of the CMBR, the observable effect on the pattern will be negligible.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastEditorUserId="2092" LastEditDate="2015-02-04T04:22:56.670" LastActivityDate="2015-02-04T04:22:56.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8767" PostTypeId="1" AcceptedAnswerId="8769" CreationDate="2015-02-04T05:16:57.990" Score="8" ViewCount="4670" Body="&lt;p&gt;Stars which exceed the Tolman-Oppenheimer-Volkoff limit can become a black hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens to star after it becomes a black hole?&#xA;Does it regain its status of star?&lt;/p&gt;&#xA;" OwnerUserId="4046" LastEditorUserId="9301" LastEditDate="2016-10-23T02:09:39.957" LastActivityDate="2019-08-15T08:30:19.880" Title="Does a black hole become a normal star again?" Tags="&lt;black-hole&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8769" PostTypeId="2" ParentId="8767" CreationDate="2015-02-04T06:25:35.137" Score="17" Body="&lt;p&gt;There's several ways this question could be answered, but they all come down to an emphatic &quot;no&quot; - a black hole will not return to being a main sequence star. The simplest way to see this is probably that a black hole has a much higher entropy than a star or even another type of stellar remnant of even vaguely similar mass and so there simply could not exist a spontaneous process by which a black hole develops back into a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A black hole once formed will stay a black hole, however it is believed that the Hawking process will lead to the black hole eventually evaporating. The time scale for evaporation though of a stellar remnant black hole is mindbogglingly long and they will not evaporate until long after stellar formation has ceased in the Universe.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastActivityDate="2015-02-04T06:25:35.137" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8770" PostTypeId="2" ParentId="8763" CreationDate="2015-02-04T10:41:48.560" Score="14" Body="&lt;p&gt;The CMB patterns do indeed change over time, although statistically they remain the same, and although it will not be noticeable on human timescales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CMB we observe now comes from a thin shell with us in the center, and with a radius equal to the distance that the light has traveled from the Universe was 379,000 years old and until now. As time passes, we will receive CMB from a shell with an increasingly larger radius. As that light has traveled farther through space, it will, as you say, be more redshifted, or &quot;cooler&quot;. But it will also have been emitted from more distant regions in the early Universe that, although statistically equivalent, simply will be other regions and hence look different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The patterns that change the fastest are the smallest patterns we can observe. The angular resolution of the Planck satellite is &lt;a href=&quot;http://nssdc.gsfc.nasa.gov/nmc/spacecraftDisplay.do?id=2009-026B&quot; rel=&quot;nofollow noreferrer&quot;&gt;5-10 arcmin&lt;/a&gt;. Since the CMB comes from a redshift of ~1100, the &lt;em&gt;angular diameter distance&lt;/em&gt; defining the physical distance spanned by a given angle — is ~13 Mpc, so 5 arcmin corresponds to a physical scale of roughly 19 kpc in physical coordinates, or 21 Mpc in comoving coordinates (that is, a structure spanning 5 arcmin today were ~19 kpc across at the time of emission, but have now expanded to a size of ~21 Mpc, with 1 Mpc = 1000 kpc = 3261 · 10³ lightyears).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming an isotropic Universe, if the smallest observable parcels of gas were 19 kpc across &lt;em&gt;perpendicular&lt;/em&gt; to our line of sight, they are also on average 19 kpc across &lt;em&gt;along&lt;/em&gt; our line of sight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question of how fast the CMB changes comes down to how much time did it take light to travel 19 kpc when the Universe was 379,000 years old. This is not simply 19 kpc divided by the speed of light, since the Universe expands as the light travels, but it's pretty close. Hence, it took light roughly 62,000 years to traverse such a patch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we see events at redshift &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; time dilated by a factor &lt;span class=&quot;math-container&quot;&gt;$1+z$&lt;/span&gt;, we will have to wait &lt;span class=&quot;math-container&quot;&gt;$62\,\mathrm{kyr}\times1100$&lt;/span&gt;, or roughly &lt;strong&gt;70 million years&lt;/strong&gt; (assuming that Planck will not get replaced by better instruments within that time which is, um, dubious).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So you're right, you could make a 3D image of the CMB, but since the patterns are much larger than a light-year, you don't have to take a new picture every year.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="1808" LastEditDate="2020-03-07T11:17:51.367" LastActivityDate="2020-03-07T11:17:51.367" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="8771" PostTypeId="2" ParentId="8767" CreationDate="2015-02-04T12:21:33.573" Score="4" Body="&lt;p&gt;Definitely not. When a star collapses, it may stop at the stage of a neutron star. This thing is still a physical &quot;real&quot; object, living within our space time. But when the mass of the original star is too big, the thing will go right through the neutron star stage and collaps further, because gravity is even stronger than the repulsive forces between the neutrons. In fact, no one knows where this thing goes. From a classical view, the collaps never ends, the thing vanishes out of the classical space time and cannot longer be reached. From a classical point of view there is no way back into space time. &quot;Black hole&quot; is a synonyme for a thing we do not understand.&lt;/p&gt;&#xA;" OwnerUserId="5465" LastActivityDate="2015-02-04T12:21:33.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8772" PostTypeId="1" AcceptedAnswerId="8773" CreationDate="2015-02-04T12:36:26.903" Score="9" ViewCount="495" Body="&lt;p&gt;A line from one of the &lt;a href=&quot;https://astronomy.stackexchange.com/a/8769&quot;&gt;answers&lt;/a&gt; on a different question got me thinking:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The simplest way to see this is probably that a black hole has a much&#xA;  higher entropy than a star or even another type of stellar remnant of&#xA;  even vaguely similar mass and so there simply could not exist a&#xA;  spontaneous process by which a black hole develops back into a star.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, I agree a black hole turning into a star seems far-fetched since it's a one-way trip (like you can't recover a block of sugar from a glass of water to that exact form). But as far as I know entropy is the amount of disorder. A black hole is denser than a star. For a density that high, I assume a certain amount of order (inverse entropy?) is required. It's an enormous amount of mass in a small amount of space, keeping itself together. Sounds to me like a system, not like a random collection of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can the amount of order necessary for such dense objects as black holes be lower than that of the star they originate from?&lt;/p&gt;&#xA;" OwnerUserId="5464" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-02-04T18:13:15.470" Title="Entropy of black hole" Tags="&lt;star&gt;&lt;black-hole&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8773" PostTypeId="2" ParentId="8772" CreationDate="2015-02-04T18:13:15.470" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But as far as I know entropy is the amount of disorder. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Entropy is a measure of the number of possible microscopic states consistent with an observed macroscopic state&lt;sup&gt;1&lt;/sup&gt;, $S = k_\text{B}\ln N$. Fundamentally it has nothing to do with disorder, although as an analogy it sometimes works. For example, in simple situations like an $n$ point-particle gas in a box: there many more ways to put point-particles in a box in disorderly manner than an orderly one. However, the exact the opposite may be true if they have a positive size and the box is crowded enough. Overall, disorder is just a bad analogy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;sup&gt;1&lt;/sup&gt; Even that's not quite true, but it's better than disorder. Specifically, it's a simplification under the assumption that all microstates are equally likely.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A black hole is denser than a star. For a density that high, I assume a certain amount of order (inverse entropy?) is required.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If an object is crushed inside ideal box that isolates it and prevents any leaks to the outside, the crushed object still has information about what it was before. And an event horizon is about as an ideal box as there can be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classically, &lt;a href=&quot;https://en.wikipedia.org/wiki/No-hair_theorem&quot; rel=&quot;nofollow&quot;&gt;black holes have no hair&lt;/a&gt;, meaning that the spacetime of an isolated black hole is characterized by mass, angular momentum, and electric charge. So there are two possible responses to this: either the black hole really has no structure other than those few parameters, in which case the information is destroyed, or it does have structure that's just not externally observable classically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if information is not destroyed, we should expect the number of microstates of a black hole to be huge simply because there's a huge number of ways to produce a black hole. Roughly, &lt;em&gt;at least&lt;/em&gt; the number of microstates of possible collapsing star remnants of the same mass, angular momentum, and charge (though this is idealized because a realistic collapsing process sheds a lot).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For a density that high, I assume a certain amount of order (inverse entropy?) is required.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Quite the opposite; black hole are the most entropic objects for their size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the early 1970s, physicists have noticed an interesting analogies between how black holes behave and the laws of thermodynamics. Most relevantly here, the surface gravity $\kappa$ of a black hole is constant (paralleling zeroth law of thermodynamics) and the area $A$ of a black hole is classically nondecreasing (paralleling second law). This is extended further with analogies of the first and third laws of thermodynamics with $\kappa$ acting like temperature and $A$ as the entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that for this to be more than an analogy, black holes should radiate with temperature that's (some multiple of) their surface gravity. But they do; this is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow&quot;&gt;Hawking radiation&lt;/a&gt;. So the area can shrink as long as there is a compensating entropy emitted to the outside:&#xA;$$\delta\left(S_\text{outside} + A\frac{k_\text{B}c^3}{4\hbar G}\right)\geq 0\text{.}$$&#xA;Thus, semi-classically, the entropy of a black hole is proportional to its surface area. In natural units, it is simply $S_\text{BH} = A/4$, which is huge because Planck areas are very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we know that in a semi-classical approximation, a black hole must radiate with temperature proportional to its surface gravity and entropy proportional to its area. It's natural to wonder the next step: if a black hole has all this entropy, where is the structure? How can it have so many possible microstates if it's classically just a vacuum? But going there takes us to the land of quantum gravity, which is not yet firmly established, and is outside the scope for astronomy.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2015-02-04T18:13:15.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8774" PostTypeId="1" AcceptedAnswerId="8775" CreationDate="2015-02-04T20:56:29.630" Score="12" ViewCount="20558" Body="&lt;p&gt;Jupiter is a gas giant, so landing on it will not be like landing on Earth, our Moon or Mars etc., as it &lt;strong&gt;does not&lt;/strong&gt; have a &lt;em&gt;solid&lt;/em&gt; surface like these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have a hypothetical spaceship or probe landing on Jupiter, and it can withstand the enormous pressure, what or how will be the sequence of events?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already read &lt;a href=&quot;https://en.wikipedia.org/wiki/Exploration_of_Jupiter&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A major problem in sending space probes to Jupiter is that the planet&#xA;  has no solid surface on which to land, as there is a smooth transition&#xA;  between the planet's atmosphere and its fluid interior. Any probes&#xA;  descending into the atmosphere are eventually crushed by the immense&#xA;  pressures within Jupiter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I do not understand:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is this enormous or immense pressure of atmosphere or Jupiter's &lt;em&gt;surface&lt;/em&gt;?&lt;/li&gt;&#xA;&lt;li&gt;While descending, will we know that we have transitioned from the atmosphere to the &lt;em&gt;surface&lt;/em&gt;?&lt;/li&gt;&#xA;&lt;li&gt;Why does animated footage (real or conceptual) of Jupiter appear like a high viscous liquid (e.g. lava) rotating in an opposite direction, rather that gas(es)?&lt;/li&gt;&#xA;&lt;li&gt;Does Jupiter have an inner solid core?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I've already gone through these:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/q/8251/698&quot;&gt;Is Jupiter made entirely out of gas?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/q/7815/698&quot;&gt;Could we fly/drive through Jupiter?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/q/7807/698&quot;&gt;If Jupiter is a gas-giant then why don't its features change?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="698" LastEditorUserId="14094" LastEditDate="2018-09-19T08:11:00.387" LastActivityDate="2018-09-19T08:11:00.387" Title="What will happen when landing on Jupiter?" Tags="&lt;jupiter&gt;&lt;gas-giants&gt;&lt;surface&gt;&lt;core&gt;&lt;gas&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="8775" PostTypeId="2" ParentId="8774" CreationDate="2015-02-04T21:13:37.820" Score="17" Body="&lt;p&gt;Jupiter does not have a &quot;surface&quot; and nor is there anything but an arbitrary division between interplanetary space and where its atmosphere begins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crushing pressure is its atmospheric pressure. The deeper into the atmosphere you go, the greater the column of gas that lies above you. It is the weight of this column of gas that is responsible for the rapid increase in pressure with depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to your last question is most definitely addressed in the duplicate question about whether Jupiter is entirely made of gas. There is quite likely to be a liquid phase nearer the centre and there may be a solid core of order ten times the mass of the Earth. It is not a settled question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gas motions you talk about are essentially belts of weather systems in the upper layers of Jupiter's atmosphere. It is all most definitely gas that you can see.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="6" LastEditDate="2016-07-07T16:11:03.090" LastActivityDate="2016-07-07T16:11:03.090" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8776" PostTypeId="1" AcceptedAnswerId="8800" CreationDate="2015-02-04T23:11:18.893" Score="1" ViewCount="783" Body="&lt;p&gt;The earth's axis is titled 23.4 degrees with respect to the ecliptic. Since this tilt is responsible for the seasons, it's clear that the tilt is in the direction away from the sun on the shortest day of year in the winter (on the northern hemisphere). That uniquely defines the tilt direction for me. How do we define the tilt direction for the other planets, though? What is the proper way of determining in what direction the tilt of a planet occurs?&lt;/p&gt;&#xA;" OwnerUserId="4070" LastEditorUserId="4070" LastEditDate="2015-02-08T17:36:53.067" LastActivityDate="2015-02-08T17:36:53.067" Title="How precisely is planetary tilt defined (the tilt direction, not just the angle)?" Tags="&lt;planet&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8777" PostTypeId="2" ParentId="8776" CreationDate="2015-02-04T23:59:07.353" Score="4" Body="&lt;p&gt;A bit too simply (but not too much), it's the angle between the planet's rotational angular velocity vector and the planet's orbital angular momentum vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where it gets tricky is dealing with little wobbles and such. Planets don't quite orbit in a plane because of gravitational interactions amongst the planets. This means the instantaneous (osculating) orbital angular momentum vector isn't quite constant, both in magnitude and direction. There are a number of different mean orbital elements that smooth out most of those tiny wobbles. One of those mean orbital elements sets (I'm not sure which one exactly) is used to determine the  mean orbital angular momentum vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets don't quite rotate nice and smoothly because planets aren't perfect spheres and aren't rigid bodies. This gives the Moon, the Sun, and other planets a handle by which they can exert tiny little torques in the planet in question. The response of the planet to these torques is somewhat arbitrarily divided into two categories based on frequency. Very slow responses are called precession; faster responses, nutation. The non spherical nature of a planet means the planet undergoes a small torque-free nutation as well as the torque-induced precession, nutation, and polar motion. In addition to these, there are a number of terms (all small) that don't yet have a very good model behind them. This oddball terms, along with the torque-free nutation, are collectively called polar motion. The motion is fairly smooth if one ignores nutation and polar motion, and the effects are all small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precession, while very slow, can be quite large in magnitude. The rotational angular velocity vector that is used in determining the axial tilt incorporates precession but smooths out nutation and polar motion.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-02-04T23:59:07.353" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8778" PostTypeId="1" AcceptedAnswerId="8779" CreationDate="2015-02-05T03:16:15.220" Score="1" ViewCount="3370" Body="&lt;p&gt;As I understand it, Jupiter, Saturn, Uranus, and Neptune are all made primarily from varying proportions of hydrogen and helium.  Despite this, Jupiter is very red, Saturn is yellow, and Uranus and Neptune are blue.  Why the difference in colors for planets with similar chemical makeups?  Or do they not have similar chemical makeups?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If such an explanation is available, I would like specific descriptions of why each planet is colored the way it is.&lt;/p&gt;&#xA;" OwnerUserId="128" LastEditorUserId="128" LastEditDate="2015-02-06T16:12:40.697" LastActivityDate="2015-02-06T16:12:40.697" Title="Why are gas giants colored the way they are?" Tags="&lt;solar-system&gt;&lt;jupiter&gt;&lt;gas-giants&gt;&lt;saturn&gt;&lt;uranus&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8779" PostTypeId="2" ParentId="8778" CreationDate="2015-02-05T06:04:09.820" Score="1" Body="&lt;p&gt;You could just have &lt;a href=&quot;https://www.google.co.uk/search?hl=en&amp;amp;q=colour%20of%20gas%20giants&amp;amp;meta=&amp;amp;rlz=1I7ADBF_en-GB&amp;amp;gws_rd=ssl&quot; rel=&quot;nofollow&quot;&gt;Googled this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Post #5 from the &lt;a href=&quot;http://cosmoquest.org/forum/showthread.php?97951-Gas-Giant-Colors&quot; rel=&quot;nofollow&quot;&gt;first hit&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;First off, by &quot;wondering what colors different gas giants can be&quot;, you&#xA;  are presumably asking about their light spectra through the visible&#xA;  range of wavelengths (380-720 nm), right?** &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Light interacts primarily with electrons. It is scattered or absorbed&#xA;  in the presence of electrons, which come in a variety of &quot;phases&quot;.&#xA;  Here are the most relevant:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1) free within an ionized gas (can absorb in the presence of the&#xA;  electric field of an ion)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2) attached to atoms and ions&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;3) attached to molecules&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;4) attached to molecules that have condensed into solid state aerosols and grains, or liquid droplets.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;The most important thing to take away from this is that every type of&#xA;  material, in terms of composition and &quot;phase&quot;, absorbs and scatters&#xA;  light uniquely.&lt;/strong&gt; &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The prevalence and importance of each of the above four &quot;phases&quot;&#xA;  depend on (a) the elemental composition of the giant planet's&#xA;  atmosphere (defined as that layer responsible for light&#xA;  reflected/emitted by planet), and (b) its equation of state (how&#xA;  pressure changes as a function of density and temperature). The first&#xA;  of these provides the raw materials, and the second arranges them in&#xA;  &quot;phase&quot;. Very roughly speaking, one may assign decreasing temperatures&#xA;  (T) to the above 4 &quot;phases&quot; moving down the list 1--&gt;4. Pressure (P)&#xA;  also plays a role, and in general one may place the above phases on a&#xA;  P-T diagram. Physics and Chemistry are at work to determine what kind&#xA;  of &quot;stuff&quot; is present as a function of depth through the giant planet&#xA;  atmosphere. One rule of thumb is that chemistry is much more effective&#xA;  at higher temperatures (to a point) and/or in the presence of&#xA;  moderately energetic light. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Next, before proceeding, go back and read the &lt;strong&gt;bold statement&lt;/strong&gt;, above.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;To finish off this overly long post: Two giant planets of equal bulk&#xA;  compositions will almost certainly appear differently if their P vs. T&#xA;  profiles differ, or if their atmospheric compositions differ (e.g.,&#xA;  due to convective mixing from the interior, mixing due to wind&#xA;  currents, heterogeneous settling of heavier matter towards the center&#xA;  over time). Two giant planets of equal bulk compositions, but&#xA;  differing ages will appear differently, since a planet's interior&#xA;  cools over time, affecting P-T relation within the planet as well as&#xA;  its thermally emitted spectrum. The intensity and spectral shape of&#xA;  the light incident from the parent star will affect the P-T diagram,&#xA;  the chemistry and phase of the matter, the thermally emitted spectrum,&#xA;  as well as the distribution of photons available for scattering.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the &lt;a href=&quot;http://coolcosmos.ipac.caltech.edu/ask/203-Why-do-the-planets-have-different-colors-&quot; rel=&quot;nofollow&quot;&gt;second hit&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Jupiter is a giant gas planet with an outer atmosphere that is mostly&#xA;  hydrogen and helium with small amounts of water droplets, ice&#xA;  crystals, ammonia crystals, and other elements. Clouds of these&#xA;  elements create shades of white, orange, brown and red. Saturn is also&#xA;  a giant gas planet with an outer atmosphere that is mostly hydrogen&#xA;  and helium. Its atmosphere has traces of ammonia, phosphine, water&#xA;  vapor, and hydrocarbons giving it a yellowish-brown color. Uranus is a&#xA;  gas planet which has a lot of methane gas mixed in with its mainly&#xA;  hydrogen and helium atmosphere. This methane gas gives Uranus a&#xA;  greenish blue color Neptune also has some methane gas in its mainly&#xA;  hydrogen and helium atmosphere, giving it a bluish color&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-02-05T06:04:09.820" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8780" PostTypeId="2" ParentId="8776" CreationDate="2015-02-05T08:33:21.987" Score="2" Body="&lt;p&gt;The tilt doesn't depend at all on where one is on the planet, and whether or not it is to the left or right is irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle (excluding the wobbles and longer term effects as David mentioned) doesn't vary depending on where you measure it. See this pic from &lt;a href=&quot;http://www.srh.noaa.gov/abq/?n=clifeatures_wintersolstice&quot; rel=&quot;nofollow noreferrer&quot;&gt;NOAA&lt;/a&gt; for a simple perspective:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qr8iw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2015-02-05T08:33:21.987" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8781" PostTypeId="1" AcceptedAnswerId="8839" CreationDate="2015-02-05T20:17:27.907" Score="3" ViewCount="246" Body="&lt;p&gt;How long was the hyper inflation phase? Haw can we know that it lasted for a period of time and not more or les time?&lt;/p&gt;&#xA;" OwnerUserId="5471" LastActivityDate="2015-02-11T15:02:59.837" Title="How long was the hyper inflation phase?" Tags="&lt;cosmological-inflation&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8782" PostTypeId="1" AcceptedAnswerId="9903" CreationDate="2015-02-05T20:22:38.307" Score="16" ViewCount="5603" Body="&lt;p&gt;Why are all quasars so far away?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the universe is homogeneous, we could expect to have a homogeneous distribution of quasars, but all of then seem to be far away from Earth. Why is that?&lt;/p&gt;&#xA;" OwnerUserId="5471" LastEditorUserId="2531" LastEditDate="2015-02-06T00:14:34.827" LastActivityDate="2020-08-12T00:25:37.257" Title="Why are all quasars so far away?" Tags="&lt;quasars&gt;" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8784" PostTypeId="2" ParentId="8782" CreationDate="2015-02-05T22:27:50.187" Score="25" Body="&lt;p&gt;There are essentially two reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, quasars are rare objects, so even though they are homogenously distributed on large scales, the average distance is large. Moreover, the brightest quasars are even rarer, but visible to large distances, so their average distance is even larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, most quasars were most active at redshift $z\sim2$. This is thought to be because quasar activity is triggered by galaxy mergers, which bring gas into galactic centres so that the supermassive black holes can be fed. Galaxy mergers and star formation peaked at $z\sim2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, there is some bias in the sense that astronomers are looking for the most distant (and hence oldest) of objects, because they are more interesting than nearby objects when it comes to learn about the formation history of the universe.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2015-02-06T09:31:39.587" LastActivityDate="2015-02-06T09:31:39.587" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8785" PostTypeId="2" ParentId="8778" CreationDate="2015-02-05T22:31:19.787" Score="2" Body="&lt;p&gt;The simple answer is that the composition in terms of elements doesn't matter for the colour, since that is determined by the molecular composition as well as state (pressure and temperature) of the visual layers (which contain a small but significant fraction of other elements than H and He). Given the differences in the giant planets temperature and pressure, this molecular composition is most likely different.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2015-02-05T22:31:19.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8786" PostTypeId="2" ParentId="8760" CreationDate="2015-02-05T22:35:37.987" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;Your claim that sphericity is assumed is incorrect &lt;em&gt;when referred to hard scientific evidence&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Even if that was assumed, the resulting error in the rotation speed is small for spheroidal galaxies and smallish for flat galaxies, provided they are centrally concentrated (which their stellar distribution is).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Scientists are, of course, well aware of the errors emanating from this (and other) assumptions (in particular if the situation is as simple as here) and care is taken (if not by every individual scientist, then certainly by the concensus reached by the scientific community) that these errors have no bearing on the scientific conclusions. Your quote from a undergraduate-level textbook proves nothing in this respect.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2015-02-06T09:23:55.180" LastActivityDate="2015-02-06T09:23:55.180" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8787" PostTypeId="2" ParentId="8782" CreationDate="2015-02-06T00:29:29.980" Score="16" Body="&lt;p&gt;You have stumbled across a profound observation and almost grasped one of its most important consequences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two forms of the so-called cosmological principle. There is the more limited &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmological_principle&quot; rel=&quot;noreferrer&quot;&gt;cosmological principle&lt;/a&gt;, which to paraphrase, says that the universe will look the same in all directions to any observer anywhere in the universe &lt;em&gt;at the same time&lt;/em&gt; (i.e. at the same cosmological epoch). There is also a &lt;a href=&quot;http://en.wikipedia.org/wiki/Perfect_Cosmological_Principle&quot; rel=&quot;noreferrer&quot;&gt;Perfect Cosmological Principle&lt;/a&gt;, that says the universe is homogeneous and isotropic in &lt;em&gt;both&lt;/em&gt; space and time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Perfect Cosmological Principle was the underpinning for the steady state theory of the universe. However, one of the most obvious objections to this was that we &lt;em&gt;can&lt;/em&gt; see that the universe has evolved in time. One of the first realisations of this was indeed the observation that quasars were more common at large distances and hence more common in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus this observation tells us that the characteristics of the universe are changing with time and therefore that the &lt;em&gt;Perfect&lt;/em&gt; cosmological principle is incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more limited Cosmological Principle meanwhile remains. It ony asserts that everything should look the same to all observers at a given cosmological epoch; it does not require the universe to look the same at all times and therefore does not require the density of particular types of astronomical objects to be constant with distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quasar activity peaked at moderate redshifts due to the required feeding processes of active galactic nuclei and the competition between the merger activity of gas-rich galaxies and the quenching caused by massive star formation and negative feedback from the AGN themselves. It appears that the &quot;sweet spot&quot; for comparatively short-lived phases of &quot;quasar activity&quot; is at redshifts of 2-3 where there was significant merger activity and the transport of gas into the central regions of galaxies, but that there had been insufficient time to fully exhaust the gas in galaxies with central black holes.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-05-30T14:12:23.447" LastActivityDate="2015-05-30T14:12:23.447" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="8788" PostTypeId="1" AcceptedAnswerId="8797" CreationDate="2015-02-06T07:09:00.500" Score="25" ViewCount="141418" Body="&lt;p&gt;I have a 3inch Newtonian reflector telescope with 300 mm focal length. I can use highest magnification of 75x using a 4mm eyepiece. But in 75x I can't see the details of Jupiter what was expected. Instead I see a little blurry image. Now I would like to know how much magnification is necessary to see a good details of Jupiter and other planets. And one more question: Is there any way to improve the vision of my 3 inch telescope?&lt;/p&gt;&#xA;" OwnerUserId="1441" LastEditorUserId="1441" LastEditDate="2020-10-14T15:44:43.073" LastActivityDate="2020-10-14T15:44:43.073" Title="How much magnification is needed to see the planets of solar system?" Tags="&lt;solar-system&gt;&lt;telescope&gt;&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="8789" PostTypeId="1" AcceptedAnswerId="8796" CreationDate="2015-02-06T09:03:29.577" Score="4" ViewCount="588" Body="&lt;p&gt;Most small-scale N-body simulations (e.g., planetary systems, solar system, stellar clusters, ...) use  classical Newtonian gravity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most large-scale N-body simulations (e.g., galaxy clusters, ...) use general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: if I were to make a &lt;strong&gt;large-scale&lt;/strong&gt; N-body simulation using &lt;strong&gt;Newtonian&lt;/strong&gt; gravity, would it be enough to &quot;correct&quot; that to first-order by modifying the equations of motion to take into account the fact that gravitational fields propagate at the speed of light?&lt;/p&gt;&#xA;" OwnerUserId="599" LastEditorUserId="599" LastEditDate="2015-02-06T10:52:41.803" LastActivityDate="2019-07-13T00:47:48.440" Title="Does one need to take into account finite gravity speed in N-body simulations?" Tags="&lt;general-relativity&gt;&lt;n-body-simulations&gt;&lt;newtonian-gravity&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8790" PostTypeId="1" AcceptedAnswerId="8822" CreationDate="2015-02-06T09:36:18.253" Score="-5" ViewCount="6467" Body="&lt;p&gt;Earth receives sunlight from Sun; in a similar way, do we receive any kind of energy from other planets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: We are dependent on sunlight, which is energy from Sun; similarly, are we dependent on other planets as well? What would happen if they weren't there?&lt;/p&gt;&#xA;" OwnerUserId="1598" LastEditorUserId="8001" LastEditDate="2015-10-01T09:46:35.070" LastActivityDate="2015-10-01T09:46:35.070" Title="Why there are other planets in our solar system?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;life&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" ClosedDate="2015-03-09T13:12:15.893" ContentLicense="CC BY-SA 3.0" />
  <row Id="8791" PostTypeId="2" ParentId="8790" CreationDate="2015-02-06T09:43:17.077" Score="1" Body="&lt;p&gt;Without just disregarding your question as one of a philosophical nature, the significance is irrelevant in the context of what you are asking. Other planets are a consequence of the evolution of our solar system - accretion of matter, gravitational influence, energy from the sun etc. - they aren't there for any 'reason'. Energy from other planets (and indeed other objects in our solar system) comes in the form of gravitational energy and light which is reflected off them emanating from the sun.  &lt;/p&gt;&#xA;" OwnerUserId="5472" LastActivityDate="2015-02-06T09:43:17.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8792" PostTypeId="2" ParentId="8790" CreationDate="2015-02-06T09:58:31.520" Score="6" Body="&lt;p&gt;Did you mean to ask&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Would life on Earth be possible if Earth was the only planet in the Solar system?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is &lt;strong&gt;perhaps&lt;/strong&gt; &lt;em&gt;, we can only speculate and know too little about the evolution of the planetary system to give any firm answer.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon, for example, (not a planet I know) stabilises the Earth spin axis, preventing it from flipping, which would otherwise happen (on long time scales) with drastic consequences for the Earth climate and hence any life forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter (and to a much lesser degree the other outer planets) plays a significant role in the dynamics of smaller bodies (asteroids and meteroids), including absorbing them. If, for example, the meteroid and asteroid impacts on Earth were more frequent, that too would have drastic consequences for the climate and life.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2015-02-06T09:58:31.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8793" PostTypeId="2" ParentId="8788" CreationDate="2015-02-06T10:41:21.153" Score="2" Body="&lt;p&gt;In typical seeing condition you should be able to use a magnification (&lt;a href=&quot;https://www.astronomics.com/eyepiece-highest-useful-magnification_t.aspx&quot; rel=&quot;nofollow&quot;&gt;see here&lt;/a&gt;) of about 25-30x per inch of apperture, so for your telescope that is about 100x, in exceptional condition you could push that up to maybe double that. Also the more magnification you use the less contrast you will have in the image, so really you want the lowest magnification that gives an image size compatible with being able to see the bands since you will be contrast limited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will find simulated images of Jupiter through a small telescope &lt;a href=&quot;http://astronomyhints.com/simviews.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and Saturn &lt;a href=&quot;http://www.rocketroberts.com/astro/scopeview.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Though personal experience suggests that the simulated image of Jupiter through a 3&quot; aperture is optimistic. IIRC a suggestion of banding is just about at the limit of what I can see on Jupiter in the small scope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am guessing that your telescope is &lt;a href=&quot;http://stargazerslounge.com/topic/166500-celestron-firstscope-76-iya/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-02-06T11:30:50.037" LastActivityDate="2015-02-06T11:30:50.037" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8795" PostTypeId="1" AcceptedAnswerId="9970" CreationDate="2015-02-06T15:45:53.333" Score="11" ViewCount="1941" Body="&lt;p&gt;A moon-sized object is running loose in the Solar System, perhaps after a planetary collision.  As it approaches a planet, it's presumably following an approximately hyperbolic path.  If it goes on past, it's still on the same hyperbola, on a curve mirroring its approach (presumably).  How can the planet ever capture it, whatever the body's velocity?  Why doesn't it either collide or go on past?&lt;/p&gt;&#xA;" OwnerUserId="5457" LastEditorUserId="2701" LastEditDate="2017-06-13T17:13:41.427" LastActivityDate="2017-06-13T17:13:41.427" Title="How do moons get captured?" Tags="&lt;jupiter&gt;&lt;natural-satellites&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8796" PostTypeId="2" ParentId="8789" CreationDate="2015-02-06T19:02:19.023" Score="7" Body="&lt;p&gt;If you're asking whether it's sufficient to use a retarded (time-delayed) positions to calculate gravitational forces, then no, that would be much worse than Newtonian gravity. For example, that would predict that the Earth should spiral into the Sun on the order of about 400 years. See also answers to &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://space.stackexchange.com/q/21467/12102&quot;&gt;Besides retarded gravitation, anything else to worry about when calculating MU69's orbit from scratch?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/q/5869/7982&quot;&gt;Is the influence of gravity instantaneous?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Most small-scale N-body simulations (e.g., planetary systems, solar system, stellar clusters, ...) use classical Newtonian gravity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is true, but some simulations use relativistic corrections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The post-Newtonian expansion works in order &lt;span class=&quot;math-container&quot;&gt;$\epsilon\sim v^2\sim U$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt; is (the negative of) Newtonian potential, and including the first-order post-Newtonian correction to the N-body equations of motion forms the &lt;a href=&quot;https://en.wikipedia.org/wiki/Einstein%E2%80%93Infeld%E2%80%93Hoffmann_equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein–Infeld–Hoffmann equations&lt;/a&gt;. Because they contain &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt;-body terms, the simulation would now be &lt;span class=&quot;math-container&quot;&gt;$\mathcal{O}(N^3)$&lt;/span&gt; instead of the Newtonian simulation &lt;span class=&quot;math-container&quot;&gt;$\mathcal{O}(N^2)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The post-Newtonian three-body equations are known to 2PN, while the two-body problem to 3.5PN. Effects beyond the first order can be important in e.g. binary simulations because gravitational radiation losses are seen at 2.5PN order. However, they are not translatable to an N-body problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A technique of making the EIH equations more tractable by making a second expansion that depends on the problem at hand can be found in Will (2014), which is applicable when most the relativistic dynamics are dominated by a few contribution (e.g., a supermassive central black hole in galaxy simulations).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Most large-scale N-body simulations (e.g., galaxy clusters, ...) use general relativity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As &lt;strong&gt;Walter&lt;/strong&gt; says, most of them use Newtonian gravity on a expanding background. Not necessarily directly computer particle-particle Newtonian force, but perhaps modeling Poisson's equation through mesh methods, or perhaps some combination thereof (e.g., direct force calculations for close particles, mesh methods for larger scales).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My question: if I were to make a &lt;strong&gt;large-scale&lt;/strong&gt; N-body simulation using &lt;strong&gt;Newtonian&lt;/strong&gt; gravity, would it be enough to &quot;correct&quot; that to first-order by modifying the equations of motion to take into account the fact that gravitational fields propagate at the speed of light?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you discount cosmological expansion, it will simply be wrong. However, first-order corrections to the Newtonian gravity part of simulations do exist in the &lt;em&gt;post-Friedmann formalism&lt;/em&gt;. I'm not really familiar with PFF, but according to Bruni &lt;em&gt;et al.&lt;/em&gt; (2014), the leading-order correction to Newtonian gravity introduces a vector potential in the &lt;span class=&quot;math-container&quot;&gt;$g_{0i}$&lt;/span&gt; term in the metric that can produce weak lensing effects but &lt;em&gt;doesn't&lt;/em&gt; affect matter dynamics. Which is entirely sensible because that kind of gravitomagnetic effects should be suppressed by a further &lt;span class=&quot;math-container&quot;&gt;$v/c$&lt;/span&gt; term for matter, just like magnetism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For references on how low order GR corrections to Newtonian gravity simulations are often made, see answers to &lt;a href=&quot;https://space.stackexchange.com/q/23408/12102&quot;&gt;How to calculate the planets and moons beyond Newtons's gravitational force?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Will, C. M. &quot;Incorporating post-Newtonian effects in N-body dynamics&quot;. Phys. Rev. D &lt;strong&gt;89&lt;/strong&gt;, 044043 (2014) [arXiv:&lt;a href=&quot;http://arxiv.org/pdf/1312.1289v3&quot; rel=&quot;nofollow noreferrer&quot;&gt;1312.1289&lt;/a&gt;]&lt;/li&gt;&#xA;&lt;li&gt;Bruni, M., Thomas, D. B., Wands, D. &quot;Computing general-relativistic effects from Newtonian N-body simulations: Frame dragging in the post-Friedmann approach.&quot; Phys. Rev. D &lt;strong&gt;89&lt;/strong&gt;, 044010 (2014) [arXiv:&lt;a href=&quot;http://arxiv.org/abs/1306.1562&quot; rel=&quot;nofollow noreferrer&quot;&gt;1306.1562&lt;/a&gt;]&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="661" LastEditorUserId="7982" LastEditDate="2019-07-13T00:47:48.440" LastActivityDate="2019-07-13T00:47:48.440" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8797" PostTypeId="2" ParentId="8788" CreationDate="2015-02-06T20:51:57.240" Score="40" Body="&lt;p&gt;You're probably asking the wrong question - which I am going to answer anyway, and after that I am going to answer the question you should have asked instead.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As a general rule, there isn't much point in pushing the magnification above 2x the diameter of the instrument, measured in mm. 3 inch, that's 75mm, that's 150x max. Beyond that limit, even under ideal skies the image is large but blurry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, &lt;a href=&quot;http://en.wikipedia.org/wiki/Astronomical_seeing&quot; rel=&quot;noreferrer&quot;&gt;seeing (or air turbulence)&lt;/a&gt; pushes that limit further down. Your aperture is small enough that it almost never suffers from seeing, but larger instruments are often affected. It varies greatly with time, place and season. There are times when a 12&quot; dobsonian, that in theory could do 600x, is clamped down by seeing to 150 ... 180x. There are times when you could take a 20&quot; dobsonian all the way up to 1000x - but that's very, VERY rare, it's the stuff of legends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming average seeing conditions and instruments of usual size (refractors of 3...4&quot; aperture, reflectors 6&quot; or larger), here are some rules of thumb:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter is seen best under mid-high magnification. It's rare that more than 200x is beneficial. This is because it's a very low contrast object, and additional magnification comes at the cost of less contrast, which makes things worse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn works best at high-ish magnification, bit more than Jupiter but maybe not much more. Around 200 ... 250x usually works. It depends on what you do - if you're trying to see the ring divisions, push it a bit higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars can use the highest magnification that you could generate, given the instrument and the conditions. It's a very small object, contrast is not bad, so crank it all the way up. Most instruments are limited by seeing when observing Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moon is the same as Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, magnification is never an issue for you. More magnification will not make it better. In fact, more magnification always means the image is more blurry, not more crisp - it's always a compromise between size and blurriness that decides the optimal magnification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't worry, everyone begins thinking that more is always better. Soon enough, experience shows them what's really going on.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;That being said, I believe it's not magnification that's giving you trouble, but the general condition of the optical stack that you're using. These are things that are extremely important, and yet are ignored by many, many amateurs - and the results are not optimal. Here are a few things that you should investigate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Collimation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is your scope collimated? In other words, are all optical elements aligned on the same axis? The likely answer is no. It makes a huge difference in the scope's performance, especially for planets. Here's a collimated scope, compared to the same scope out of collimation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dz4N8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yQaDk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further information on &lt;a href=&quot;http://www.astrophoto.fr/collim.html&quot; rel=&quot;noreferrer&quot;&gt;Thierry Legault's site&lt;/a&gt;, which is extremely informative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A series of articles and documents regarding collimation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cloudynights.com/documents/primer.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://www.cloudynights.com/documents/primer.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://garyseronik.com/a-beginners-guide-to-collimation/&quot; rel=&quot;noreferrer&quot; title=&quot;A Beginner’s Guide to Collimation&quot;&gt;Gary Seronik: A Beginner’s Guide to Collimation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://garyseronik.com/collimation-tools-what-you-need-what-you-dont/&quot; rel=&quot;noreferrer&quot; title=&quot;Gary Seronik: Collimation Tools: What You Need and What You Don’t&quot;&gt;Gary Seronik: Collimation Tools: What You Need and What You Don’t&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://garyseronik.com/no-tools-telescope-collimation/&quot; rel=&quot;noreferrer&quot; title=&quot;No-Tools Telescope Collimation&quot;&gt;Gary Seronik: No-Tools Telescope Collimation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Some telescopes (e.g. pretty much all refractors) do not require collimation; they are collimated from factory and hold collimation pretty well. But most reflectors (SCTs, all newtonians including dobsonians, etc) do require this periodic maintenance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thermal equilibrium&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 3&quot; aperture, this is probably not a big issue, but there's no reason why you should add another problem to the existing ones. Your scope should be at the same temperature as the air around it, otherwise its performance decreases. Take it outside 1 hour before you start observing, and that should be enough for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FkCIj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Larger telescopes (around 10&quot; ... 12&quot; and larger) should use active ventilation for better cooling (a fan on the back of the mirror). More details here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-1/&quot; rel=&quot;noreferrer&quot; title=&quot;Beat the Heat: Conquering Newtonian Reflector Thermals — Part 1&quot;&gt;Gary Seronik: Beat the Heat: Conquering Newtonian Reflector Thermals — Part 1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-2/&quot; rel=&quot;noreferrer&quot; title=&quot;Beat the Heat: Conquering Newtonian Reflector Thermals — Part 2&quot;&gt;Gary Seronik: Beat the Heat: Conquering Newtonian Reflector Thermals — Part 2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, simple passive cooling for 1 hour should be enough, but it's worth reading those articles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Focal ratio&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 3&quot; scope, at 300mm focal length, that's an f/4 instrument. That's a pretty steep f/ ratio. Most eyepieces will not do well with such a blunt cone of light, and will start to exhibit aberrations that blur the image. Only very expensive eyepieces work well at such low focal ratios - things like TeleVue Ethos, or Explore Scientific 82 degree eyepieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try and keep the planet in the center - most aberrations are lower there. Even very simple eyepieces do better in the middle of the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the stars. Are they tiny and round in the center, and large and fuzzy at the edge? Those are aberrations from various sources (eyepiece, primary mirror, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Coma&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, at f/4 even the best eyepieces out there cannot do anything about coma - an aberration coming out of any parabolic mirror, which becomes pretty obvious around f/5, very obvious at f/4, and a major problem at f/3. Again, coma is zero in the center of the image, and increases towards the edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QiILH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A coma corrector is used in some cases, such as the TeleVue Paracorr, but I strongly recommend that you DO NOT use one - I suspect your instrument is aberrating in ways that overwhelm coma anyway. Jupiter would not be too blurry even at full f/4 coma at the edge. This paragraph is for informational purposes only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coma should become a concern with large telescopes, using high quality optics, with a focal ratio of around f/5 and less. E.g., you have a 20&quot; dob with an f/4 mirror, then you should worry about coma - provided that collimation and so on are taken care of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Optics quality&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An f/4 parabola is not super easy to make at any size. I've made my own optics, and the lower the f/ ratio, the more difficult the process is. Many small, cheap telescopes are made in a hurry, and the difficult focal ratio poses additional problems - as a result, many manufacturers do a poor job. There are even cases where the primary mirror is left spherical, with disastrous results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is something you can do nothing about. If the primary mirror is bad, then that's just the way things are. An optician might try to correct it, but it's a difficult process, and quite expensive. I only added this here so you are informed.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is what I would do in your case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would take the scope out 1 hour before observing, every time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would try and learn how to collimate the scope. I would try to figure out a few simple collimation techniques, and a few simple tests. I would spend a few days / weeks practicing that. I would keep reading about collimation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When collimation is at least partially under control, I would learn how to properly focus the scope. Seems simple, but it can be tricky. Use a bright star, and try and make it as small as possible. Use the Moon when it's visible, and try and make it crisp and clear. Do not try this with a miscollimated scope, since it's pointless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a few months, when I gain confidence that the scope is in better shape, very well collimated, very well focused, I might try to borrow a better eyepiece from a friend. I said borrow, not buy. Something like a 3 ... 4mm eyepiece, good quality, that would give me a comparison for the existing eyepieces. This ONLY makes sense with a scope that is in perfect collimation, perfect temperature, perfect focus. If an improvement is seen, then get a better eyepiece - but do not spend hundreds of dollars for an expensive eyepiece that will then be used in a tiny cheap scope. Second-hand eyepieces often work exactly as well as new ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know someone in your area who makes mirrors, see if they agree to put your primary mirror on the Foucault tester, and assess its condition. But beware: the results might be very disappointing. Or not. You kind of never know with these little scopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: After the scope is collimated and so on, you could try to increase magnification by using a 2x barlow with your eyepieces, but do not expect miracles - the image will be bigger, but probably rather &quot;mushy&quot;. More magnification is not always better, there's always a trade-off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck, and clear skies to you!&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2019-04-02T06:27:35.360" LastActivityDate="2019-04-02T06:27:35.360" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="8798" PostTypeId="2" ParentId="8795" CreationDate="2015-02-06T22:27:38.987" Score="1" Body="&lt;p&gt;There are two effects that alter the simple hyperbolic (or elliptic) relative orbit of any minor body (&quot;moon&quot;) and a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the gravity of the Sun (and to a much lesser degree of Jupiter). To good approximation the planet-Sun system is a circular binary and the moon a test particle (its mass negligible). The orbits of test particles in such a system (known as the restricted three-body problem) are complicated, but the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobi_integral&quot; rel=&quot;nofollow&quot;&gt;Jacobi energy&lt;/a&gt;, preventing capture (similar to the angular-momentum conservation for the hyperbolic orbit). Hence, capture requires deviation from this&#xA;approximation, in particular the moon's mass must not be too small and/or another  interacting body participates (the Wikipedia page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Asteroid_capture&quot; rel=&quot;nofollow&quot;&gt;asteroid capture&lt;/a&gt; is quite disappointing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, tidal forces can transfer orbital energy into internal energy (of the planet and/or moon), which is then dissipated (converted into heat). Under lucky circumstances this process can be sufficient to convert an unbound to a bound orbit. Once bound, tides will continue to bind the moon more and more.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2015-02-06T22:27:38.987" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8799" PostTypeId="1" AcceptedAnswerId="8801" CreationDate="2015-02-07T06:30:15.633" Score="0" ViewCount="204" Body="&lt;p&gt;According to recent news, Kepler-444, a 11.2 billion year old star, has a system of five planets. It has a mass of 3/4 of solar mass. So I believe both Kepler 444 and the Sun will share the fate of 'orange main sequence stars'. If so, while scientists fear for life on Earth in the coming billion years, which is only one sixth for our solar system lifetime, how is it that the Kepler-444 planets have existed for a longer period of time?&lt;/p&gt;&#xA;" OwnerUserId="5478" LastEditorUserId="2531" LastEditDate="2015-02-07T11:42:19.007" LastActivityDate="2015-02-09T06:15:47.870" Title="If Kepler-444 planets existed for 11.2 billion years, why fear for life on Earth after six billion years?" Tags="&lt;exoplanet&gt;&lt;stellar-evolution&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8800" PostTypeId="2" ParentId="8776" CreationDate="2015-02-07T07:19:14.327" Score="0" Body="&lt;p&gt;I think I found the answer to my question. The rotational axis of each planet is identical with the vector pointing from the planet's south pole to its north pole. Therefore, if I find a table that defines positions of north poles of the planets, I will also find the unique vector pointing in the direction of the planets' tilts. This page &lt;a href=&quot;http://en.wikipedia.org/wiki/Poles_of_astronomical_bodies&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Poles_of_astronomical_bodies&lt;/a&gt; shows the positions of the north poles of major bodies in the solar system (as well as their tilt angles) projected onto the celestial sphere. The coordinates are given relative to Earth's celestial equator and the vernal equinox as they existed at the J2000 epoch. These coordinates uniquely define the tilt of each of the planets and other relatively stable bodies within the solar system.&lt;/p&gt;&#xA;" OwnerUserId="4070" LastActivityDate="2015-02-07T07:19:14.327" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8801" PostTypeId="2" ParentId="8799" CreationDate="2015-02-07T11:24:46.860" Score="4" Body="&lt;p&gt;The rate of evolution of main sequence stars is highly dependent on their mass. Roughly speaking, the time on the main sequence is proportional to $M^{-5/2}$, where $M$ is the mass of the star. Thus if Kepler 444 has $M=0.75\ M_{\odot}$, it can live for $0.75^{-5/2} = 2.05$ times as long as the Sun on the main sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of saying the same thing is that the Sun evolves 2.05 times as fast as Kepler 444. This is important because even whilst on the &quot;stable&quot; main sequence, burning hydrogen in their cores, stars are still changing because the average mass of a particle in their cores is increasing due to the build up of a Helium ash. This slow change means that the temperature of the core must gradually increase, the nuclear burning rate increases and the star becomes &lt;em&gt;more luminous&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of the Sun this means, depending on the atmospheric composition at the time, that the Earth could become &lt;em&gt;much&lt;/em&gt; hotter in a billion years or so, and probably too hot for life (as it currently exists on Earth). Thus it is not the &lt;em&gt;existence&lt;/em&gt; of Earth that is in jeopardy, it is the surface temperature that may destroy life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of the discovered planets in the Kepler 444 system are (or ever were) far enough from their parent star that they are in the &quot;habitable zone&quot; where liquid water could exist. The significance of Kepler 444 is that it shows that rocky planetary systems can have formed as long ago as 11 billion years, giving such systems more time, and therefore perhaps more chance, to have formed life.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-02-09T06:15:47.870" LastActivityDate="2015-02-09T06:15:47.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8802" PostTypeId="1" CreationDate="2015-02-07T19:49:36.297" Score="4" ViewCount="110" Body="&lt;p&gt;I just watched an iMax planetarium video that claimed that no matter where you are in the universe, it will appear that you are at the center of the universe and all things are expanding away from you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the heck?? Can someone please explain this? &lt;/p&gt;&#xA;" OwnerUserId="5484" LastActivityDate="2015-02-07T20:59:53.870" Title="Why would all places appear to be at the center of the universe's expansion?" Tags="&lt;expansion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8803" PostTypeId="1" AcceptedAnswerId="8823" CreationDate="2015-02-07T20:00:09.140" Score="2" ViewCount="1807" Body="&lt;p&gt;I guess theoretically if we could go faster than light, which we clearly cannot, at some point we'd be able to see the big bang itself. Just curious - how far away is that light? &lt;/p&gt;&#xA;" OwnerUserId="5484" LastEditorUserId="661" LastEditDate="2015-02-08T04:19:25.877" LastActivityDate="2015-09-25T01:52:24.240" Title="How far away is the light that would reveal the Big Bang?" Tags="&lt;light&gt;&lt;big-bang-theory&gt;" AnswerCount="5" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8804" PostTypeId="2" ParentId="8803" CreationDate="2015-02-07T20:27:55.873" Score="5" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Particle_horizon&quot;&gt;particle horizon&lt;/a&gt; is the point where light arriving at us now would've had to have been emitted at the big bang (though note this is hypothetical as the Universe did not become transparent to light until it was several 100,000 years old).. How far away it is depend on the model you use (not to mention there are several different distance measures in cosmology), but it's proper distance is reckoned to be about 46.9 billion light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A note of caution though, when calculating the particle horizon cosmic inflation is usually ignored. What is light arriving (again hypothetically) from usually called the particle horizon therefore would be actually light that was emitted from just the end of the inflationary epoch. If you were to take a model with a big bang singularity and inflation the &quot;true particle horizon&quot; would in fact be much further away.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastActivityDate="2015-02-07T20:27:55.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8805" PostTypeId="1" AcceptedAnswerId="9904" CreationDate="2015-02-07T20:45:57.420" Score="3" ViewCount="325" Body="&lt;p&gt;NASA developed SPIKE: Intelligent Scheduling of Hubble Space Telescope Observations &lt;a href=&quot;http://comanche.stsci.edu/institute/software_hardware/spike/documents/spike-chapter3.pdf&quot; rel=&quot;nofollow&quot;&gt;(HST)&lt;/a&gt; and &lt;a href=&quot;http://www.stsci.edu/~miller/papers-and-meetings/93-Intelligent-Scheduling/spike/spike-chapter3.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is simple, is there a programming library, preferably in python or C#, to implement the algorithm? AFAIK  NASA never released their code.&lt;/p&gt;&#xA;" OwnerUserId="5485" LastActivityDate="2015-02-17T17:54:21.300" Title="Hubble HST scheduling algorithm" Tags="&lt;telescope&gt;&lt;hubble-telescope&gt;&lt;nasa&gt;&lt;algorithm&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8806" PostTypeId="2" ParentId="8802" CreationDate="2015-02-07T20:59:53.870" Score="2" Body="&lt;p&gt;This is a famous analogy, which was prbably given to you in the video. Imagne that you are on a ball of dough full of raisins. You obviously perceive yourself as the center. Then, the dough starts to grow. The raisins are still the same size, but there is constantly more and more dough between them. From your position it would seem like there is always more and more space between you and everything else. You don't see yourself moving, but you see everything else moving away from you. This effect would happen anywhere in the universe, not only at the center. In fact, you would think - correctly, relatively speaking - that the center (if you can see it) is being pushed away from you. So basically, it all comes down to your point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This expansion is thoroughly studied presently. The most common theory is that the universe contains a giant amount of energy (called dark energy) that is so far undetectable and that is causing the the space to expand. Think of it as yeast. There are other theories that are much more complicated, which a non-professional such as I wouldn't be able to explain. (For instance, there is one that suggests that our universe adheres to other universes, like soap bubbles, which adds space). But dark energy is the most likely candidate, so far.&lt;/p&gt;&#xA;" OwnerUserId="4126" LastActivityDate="2015-02-07T20:59:53.870" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8808" PostTypeId="2" ParentId="8803" CreationDate="2015-02-08T04:05:25.033" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I guess theoretically if we could go faster than light, which we clearly cannot, at some point we'd be able to see the big bang itself. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we travel instantaneously across the universe (as measured by cosmological time), then no, theoretically that's not the case. Our theoretical assumptions involve large-scale homogeneity and isotropy, meaning that the view different observers is basically the same regardless of where they are, at least at the same cosmological epoch. This is called the &lt;em&gt;cosmological principle&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mind, because of the way relativity works, an FTL drive might also be capable of taking you back in time, so perhaps you could theoretically see the Big Bang after all. But that's obviously more to do with time travel than going to distant places.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Seems like the moment something happened, light was dispatched and unless particles traveled faster than the speed of light to obstruct that light, if I were to appear in front of that light, for a single tiny moment, I'd see that little white dot that would've appeared the moment the big bang began (and then see everything that happened afterward). Why not? Light point A -&gt; same light at point B [observer]. Where's the complication?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You seem to have an implicit assumption about &lt;em&gt;where&lt;/em&gt; the Big Bang occurred. It actually occurred &lt;em&gt;everywhere&lt;/em&gt;. Including &lt;em&gt;here&lt;/em&gt;. See &lt;a href=&quot;https://astronomy.stackexchange.com/questions/874/do-we-know-the-exact-spot-where-big-bang-took-place&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine looking out to the most distant galaxy that you can barely see. Because of the finite speed of light, you are looking at it not as it is, but as it was in the distant past. You are just now catching light from that distant place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were instantly transported to that place, and look back to where the Milky Way should be, you won't see it, but rather as this area of space was in the distant past.  You'll be just catching light from somewhere the Milky Way area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, you will see light from the distant past no matter where you go. Because the Big Bang happened everywhere, light from it (well, actually from recombination epoch later, because the universe wasn't transparent before then) is present everywhere. And hence it'll look about the same no matter where you go, again &lt;em&gt;assuming&lt;/em&gt; isotropy and homogeneity.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2015-02-08T04:05:25.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8809" PostTypeId="5" CreationDate="2015-02-08T04:22:03.160" Score="0" Body="&lt;p&gt;Questions about the processes, theories and evidence for the formation of planetary bodies.&lt;/p&gt;&#xA;" OwnerUserId="31410" LastEditorUserId="31410" LastEditDate="2021-01-08T00:27:20.357" LastActivityDate="2021-01-08T00:27:20.357" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8810" PostTypeId="4" CreationDate="2015-02-08T04:22:03.160" Score="0" Body="Questions about the processes, theories and evidence for the formation of planetary bodies." OwnerDisplayName="user2449" LastEditorDisplayName="user2449" LastEditDate="2015-02-08T14:16:46.787" LastActivityDate="2015-02-08T14:16:46.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8811" PostTypeId="5" CreationDate="2015-02-08T04:26:49.880" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-02-08T04:26:49.880" LastActivityDate="2015-02-08T04:26:49.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8812" PostTypeId="4" CreationDate="2015-02-08T04:26:49.880" Score="0" Body="Questions about rings of debris that encircle the equatorial regions of some planets." OwnerDisplayName="user2449" LastEditorDisplayName="user2449" LastEditDate="2015-02-08T15:10:49.877" LastActivityDate="2015-02-08T15:10:49.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8813" PostTypeId="5" CreationDate="2015-02-08T04:28:28.910" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-02-08T04:28:28.910" LastActivityDate="2015-02-08T04:28:28.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8814" PostTypeId="4" CreationDate="2015-02-08T04:28:28.910" Score="0" Body="Questions about observations performed without the aid of any optical instrument." OwnerDisplayName="user2449" LastEditorDisplayName="user2449" LastEditDate="2015-02-08T15:10:53.323" LastActivityDate="2015-02-08T15:10:53.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8815" PostTypeId="1" CreationDate="2015-02-08T05:27:23.413" Score="-2" ViewCount="181" Body="&lt;p&gt;Recently the sugar glycoaldehyde was detected in a star system 400 light years from Earth. How exactly are molecules detected in space? I am aware that spectroscopy is used to detect them, but I don't know the details. Please explain the methods involved.&lt;/p&gt;&#xA;" OwnerUserId="2873" LastEditorUserId="2531" LastEditDate="2015-02-14T20:22:17.313" LastActivityDate="2015-02-14T20:22:17.313" Title="How are molecules detected in space?" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8817" PostTypeId="2" ParentId="8815" CreationDate="2015-02-08T11:48:25.097" Score="1" Body="&lt;p&gt;This is a broad question. The energy levels occupied by molecules consist of a mixture of rotational and vibrational energy states. The values of the energies of these states are characteristic of particular molecules - they can be measured in the laboratory or occasionally need to be calculated if the molecules can only be formed in conditions found in space. A molecule may make a transition from a higher to a lower energy level. When it does so, the energy difference is emitted as a photon with a frequency corresponding to the energy level difference: $\nu = (E_2 - E_1)/h$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally speaking, the differences in energy between these quantum states is smaller than of electronic transitions in atoms. This means that the photons emitted or absorbed, corresponding to these transitions are in the infrared, microwave or even radio part of the spectrum. What one measures is intensity as a function of frequency. If one is looking at a warm gas, then a pattern of emission lines will be seen at characteristic frequencies depending on what molecules are present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case you mention, the observations were conducted with the Atacama Large Millimetre Array - ALMA, a microwave telescope. The molecules exist in a warm gas (a few hundred Kelvin) which excites certain transitions in the molecules and the emission of mm-wave (100s of GHz) radiation. This is then received by the ALMA instruments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking for a treatise on mm-wave astronomical techniques, this is not the place. ALMA is able to detect microwaves with a high angular resolution, but also able to separate the microwaves out into narrow frequency windows of tens of kHz (i.e. it has good spectral resolution too).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-02-08T16:35:34.433" LastActivityDate="2015-02-08T16:35:34.433" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8818" PostTypeId="2" ParentId="8803" CreationDate="2015-02-08T17:19:43.683" Score="0" Body="&lt;p&gt;There is no &quot;far enough away&quot; to see the big bang.&#xA;The bang was the creation of space and time, so every view within the universe is a view from &lt;em&gt;inside&lt;/em&gt; the bang. To see it from outside, you'd have to view the bang from whatever space the universe is embedded in; assuming it is embedded in a higher space, and that space has something like a time-axis.&#xA;The 3.7°K microwave background photons hitting us right here, right now is the best view of the consequences of the bang that we'll ever get; barring exotic SF scenarios of course.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-02-08T17:19:43.683" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8819" PostTypeId="1" AcceptedAnswerId="8821" CreationDate="2015-02-09T02:46:30.447" Score="2" ViewCount="606" Body="&lt;p&gt;It seems to me that nowadays the more proper term is &quot;Trans-Neptunian Object&quot;.  Is there any difference between Kuiper Belt Objects and Trans-Neptunian Objects?  If not, is one term preferred over the other?&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2015-02-10T15:33:48.590" Title="Is it proper to refer to objects beyond Neptune as &quot;Kuiper Belt Objects?&quot;" Tags="&lt;solar-system&gt;&lt;kuiper-belt&gt;&lt;tno&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8820" PostTypeId="2" ParentId="8803" CreationDate="2015-02-09T06:32:11.960" Score="1" Body="&lt;p&gt;Seeing (or not) the &quot;light from the big bang&quot; has little to do with how fast we can travel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cosmic microwave background is here and now; you can pick it up as short wave radio interference. This light was emitted when the universe had cooled to about 3000 degrees, some 400,000 years after the big bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that before that time the Universe was opaque to radiation. That is, light that was emitted was quickly reabsorbed or scattered before it could travel very far. If that were not the case then we &lt;em&gt;would&lt;/em&gt; be able to see light from an even earlier epoch. It would be red shifted because of the expansion of the universe by factors of well over a thousand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact there &lt;em&gt;is&lt;/em&gt; (or is predicted to be) &quot;radiation&quot; all around us from the first second after the big bang, in the form of low energy neutrinos. Neutrinos interact extremely weakly, so the universe became transparent to them at much higher densities than it did for light. These particles have been travelling the universe since, in all directions, gradually losing energy to the expanding space. Efforts are underway to detect these neutrinos, which would be a superb vindication of the big bang model. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-02-09T16:41:59.250" LastActivityDate="2015-02-09T16:41:59.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8821" PostTypeId="2" ParentId="8819" CreationDate="2015-02-09T09:07:19.473" Score="5" Body="&lt;p&gt;Yes there is a difference, Kuiper Belt Objects (KBOs) are a subset of Trans Neptunian Objects (TNOs). Other subsets are Oort-cloud objects (OCOs) and scattered disk objects (SDOs). These are not KBOs but they are TNOs. See for instance this &lt;a href=&quot;http://en.wikipedia.org/wiki/Trans-Neptunian_object#Distribution_and_classification&quot; rel=&quot;noreferrer&quot;&gt;wikipedia page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2015-02-09T09:07:19.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8822" PostTypeId="2" ParentId="8790" CreationDate="2015-02-09T11:36:56.873" Score="2" Body="&lt;p&gt;Well without your big brother jupiter looking out for poor little earth then life on earth wouldn't be possible (most likely, but definitely not intelligent life as it has taken us millions of years to evolve to this point, and meteor collisions would increase at an alarming rate). You receive a force on your body from all the planets in the solar system and everything in the universe for that matter, but they are neglible compared to the force you feel from earth. &quot;But Jupiter is so much bigger than earth how could its gravity be negligible?&quot;  Well the intensity of gravity is proportional to 1/distance^2 (one over distance squared) so the farther you are away from something the less gravitational effect it has on you, and earth is big enough that it is most likely the only gravitational effect your body will ever truly be aware of. However you can see the tidal effects caused by the moon. High tide low tide are created by the moon's gravity. I once heard a man say astrology (the arch nemesis of the astronomer) is based in the tidal affects on the heavy metals in one's body caused by the positioning of the planets at one's birth, which I had to admit made for a pretty convincing argument though I still don't read my horoscope. &lt;/p&gt;&#xA;" OwnerUserId="5494" LastActivityDate="2015-02-09T11:36:56.873" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8823" PostTypeId="2" ParentId="8803" CreationDate="2015-02-09T12:07:35.093" Score="-2" Body="&lt;p&gt;the light from the big bang is in your living room... because that is where the big bang took place... did you know you were so special? Just kidding, but maybe I misunderstood your question. I think what you are asking is, well if you look into deep space you are looking back in time since it takes light X amount of years to travel to the observer. So look far enough back and you would see the big bang right? We already thought of it, and the problem is you can't see the moment of the big bang, because there were no such thing as photon's at that moment. It took about 400,000 years (an afternoon nap on the scale of the universe) for the intensity of the moment of creation to simmer down enough atoms could form removing the opacity of the universe freeing photons to begin their cosmic journeys.. If you were to look out that far you would just see a big opaque blob like trying to look into the center of the sun.  The light from that moment still exists and you do see it... you just don't SEE it. The spectrum of the sun's light that has the highest intensity is in the visible light wavelength.  One would imagine that all species of the universe see in the same wavelength and thus we would all agree on what visible light is. This is not true. We see in the visible wavelengths because of evolution.  Our eyes adapted to the highest intensity light available. Think about natural selection. Ancient primates who could see better would be better hunters. So those who were utilizing the highest intensity light the best would be ideal mating candidates giving birth to more off spring with eyes like there own (a little theory of mine). If you grew up on a planet with a sun that gave off a lot of infrared light you would see in infrared.  Anyway, what I am getting at is if we somehow managed to evolve to this point on a rock with no companion star (an impossibility for all intents and purposes) in some dark corner of the universe where there was no other really intense light around, theoretically eyes would have evolved to see in the cosmic microwave background because it would be the only light available.  CMB radiation (the afterglow of the big bang) is ubiquitous. It is everywhere at once. It gets cooler as time goes on as space time stretches, stretching the wave length of the light making it harder and harder to detect. I hope that answered your question.&lt;/p&gt;&#xA;" OwnerUserId="5494" LastActivityDate="2015-02-09T12:07:35.093" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8824" PostTypeId="1" AcceptedAnswerId="8828" CreationDate="2015-02-09T13:26:50.407" Score="1" ViewCount="195" Body="&lt;p&gt;Mars landings, Moon landings. . . Wherever I see a documentary or still pictures in Wikipedia, there are parachutes in the landing of rovers. Whenever I see them, I ask myself, &quot;How they can it decelerate the speed of rovers, when there is no / thin air?&quot; Something needs to be there to create friction!  &lt;/p&gt;&#xA;" OwnerUserId="277" LastEditorUserId="2153" LastEditDate="2015-02-10T00:06:49.550" LastActivityDate="2015-02-10T00:06:49.550" Title="How are parachutes usable in other places than Earth?" Tags="&lt;mars&gt;&lt;moonlanding&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8826" PostTypeId="1" CreationDate="2015-02-09T14:22:56.677" Score="1" ViewCount="93" Body="&lt;p&gt;Ok, bear with me, while I try to explain the question I have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read that time slows as we approach a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, suppose,theoretically, as astronaut is approaching a Black Hole. And he gets near enough now that time starts slowing down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you remember this equation: v=d/t  So, at the moment, t is going infinity, so it means &#xA;V is going zero.&#xA;So the astronaut can never the reach the Black Hole because time is slowing down and its taking longer and longer for him to approach the Black Hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then again, we know, a Black Hole sucks everything and that light cannot even escape it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what's happening to the equation above then? [ this time it means v is going infinity and so, t is almost zero], so the astronaut will be sucked almost immediately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which one is the correct theoretical answer?&lt;/p&gt;&#xA;" OwnerUserId="5496" LastActivityDate="2015-02-09T14:22:56.677" Title="Can an astronaut ever reach a Black Hole theoretically?" Tags="&lt;black-hole&gt;&lt;astrophysics&gt;&lt;fundamental-astronomy&gt;&lt;supermassive-black-hole&gt;" AnswerCount="0" CommentCount="7" ClosedDate="2015-02-10T22:12:53.527" ContentLicense="CC BY-SA 3.0" />
  <row Id="8827" PostTypeId="1" CreationDate="2015-02-09T16:57:57.833" Score="5" ViewCount="58" Body="&lt;p&gt;The &lt;a href=&quot;http://www.minorplanetcenter.net/iau/lists/TNOs.html&quot; rel=&quot;noreferrer&quot;&gt;Minor Planet Center&lt;/a&gt; shows that we know of about 1350 Trans-Neptunian Objects (TNO's).  I think it is safe to assume that we have not found all the TNO's there are to find, even the ones that we are capable of detecting with current telescope technology (simply because we haven't looked at them yet).  Are there any estimates for how many TNO's exist (not just how many we've discovered)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know such estimates probably are for numbers of objects with a radius greater than a certain amount and for objects belonging to a certain group (Plutinos, Kuiper Belt Objects, scattered disk objects, etc.).  I am fine with estimates that make such size cuts and group specifications.&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2015-02-09T16:57:57.833" Title="Estimates for how many Trans-Neptunian Objects there are" Tags="&lt;tno&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ClosedDate="2015-02-12T07:30:51.283" ContentLicense="CC BY-SA 3.0" />
  <row Id="8828" PostTypeId="2" ParentId="8824" CreationDate="2015-02-09T17:01:29.070" Score="1" Body="&lt;p&gt;Parachutes have never been used on the moon, but &lt;a href=&quot;http://mars.nasa.gov/mer/mission/spacecraft_edl_parachute.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;they are viable for Mars&lt;/a&gt; because &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Mars&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mars &lt;em&gt;does have an atmosphere&lt;/em&gt;&lt;/a&gt; - albiet one much lighter than that of Earth.  For that reason, parachutes cannot be the only means of slowing down on Mars --- for example, in the case of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Mars_Science_Laboratory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Curiosity Rover&lt;/a&gt;, that's why an elaborate booster and crane combination was used (see figure).  Another approach, taken by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Mars_Pathfinder&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pathfinder Rover&lt;/a&gt;, was to use large airbags to bounce on landing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One benefit to using parachutes on Mars, is that they can be used at higher velocities --- unlike on Earth where the heavier atmosphere (and thus stronger drag force) would shred a parachute at comparable speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ESk80.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="918" LastEditorUserId="918" LastEditDate="2015-02-09T17:06:48.127" LastActivityDate="2015-02-09T17:06:48.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8829" PostTypeId="1" CreationDate="2015-02-09T18:52:55.543" Score="5" ViewCount="95" Body="&lt;p&gt;How are orbits calculated for Trans-Neptunian Objects?  I understand there are six standard parameters for defining an orbit (&lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_elements&quot; rel=&quot;noreferrer&quot;&gt;link&lt;/a&gt;) but I am not sure how telescope observations translate into these parameters.  Telescope observations tell us where the object is on the Celestial Sphere but can't even tell us (without a good spectroscopic line to do a redshift analysis with) how fast the object is moving towards or away from us.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, the motion of these objects on the sky is dominated by Earth's motion relative to the objects rather than the motion of the objects themselves.  This seems to me to offer additional complications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: how do 2-D observations get fit with a reasonable degree of accuracy to a 6-D parameter space with the additional complication of Earth's motion relative to the object?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't be afraid to be mathematical in your response.&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2015-02-10T13:14:58.883" Title="Determination of orbital elements for Trans-Neptunian Objects, how?" Tags="&lt;orbit&gt;&lt;tno&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8830" PostTypeId="1" CreationDate="2015-02-09T23:22:28.713" Score="1" ViewCount="184" Body="&lt;p&gt;One billion years into the future and the Sun has swollen in size and it is now not possible to live on Earth due to the heat. Mankind has relocated to Mars where the temperature is more favorable now than on Earth. My question is how much time has this gotten us before we can't live on Mars any longer because the Sun is dying ? The next step I guess will be to relocate to another Solar System ?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastActivityDate="2015-02-12T06:16:44.887" Title="Relocation to Mars" Tags="&lt;the-sun&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8831" PostTypeId="2" ParentId="8830" CreationDate="2015-02-10T05:48:04.287" Score="1" Body="&lt;p&gt;The Sun is not dying when the Earth supposedly becomes uninhabitable, it is just getting hotter as it evolves on the main sequence (core hydrogen burning, the Sun has something like 5+ billion years before reaching the red giant stage of its evolution)). It will continue to get more luminous and at a guess if Mars could be made habitable it would remain so for millions, possibly another billion years. But then do you really think humanity (or rather our descendant civilization/s) if it survives for another billion years will need a planetary surface to survive?&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-02-10T05:48:04.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8833" PostTypeId="2" ParentId="8830" CreationDate="2015-02-10T09:30:04.847" Score="2" Body="&lt;p&gt;As it gets older, the core of the Sun starts to fill with Helium ash. This increases the average mass per particle and hence the core temperature must increase to maintain the pressure. This increases the nuclear reaction rate and the Sun becomes more luminous, at almost a constant surface temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The habitable zone is controlled not only by the luminosity of the star, but also by the atmosphere of the planet. It's doubtful Mars would ever become &quot;habitable&quot; in that sense (without our intervention), but what I will assume is that you want the equilibrium temperature to be warmer than 263K, but say cooler than 303K (i.e. between -10 and 30 Celsius).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The details of finding the blackbody equilibrium temperature can be found &lt;a href=&quot;http://en.wikipedia.org/wiki/Planetary_equilibrium_temperature&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. The formula we need is&#xA;$$ T = \left(\frac {L_{\odot} (1-a)}{16\pi \sigma D^2}\right)^{1/4},$$&#xA;where $L_{\odot}$ is the luminosity of the Sun at any time, $D$ is the distance to the planet, $a$ is the albedo and $\sigma$ is the Stefan-Boltzmann constant. $T$ is in Kelvin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will assume that the average &lt;a href=&quot;http://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html&quot; rel=&quot;nofollow&quot;&gt;albedo&lt;/a&gt; of Mars is 0.25 (though it varies considerably with wavelength, depends on icecap coverage etc.) and that $D=2.27\times10^{11} m$.&#xA;We can then rearrange the formula above to give the luminosity of the Sun for a given equilibrium temperature. &#xA;$$L_{\odot} =21.3\ \pi \sigma D^2 T^4$$&#xA;This means that for $T&amp;gt;263 K$, then $L_{\odot}&amp;gt;9.35\times 10^{26}\ W$, but for $T&amp;lt;303\ K$, we require $L_{\odot}&amp;lt;1.65\times10^{27}$\ W$. That is the Sun's luminosity should be somewhere between 2.44 and 4.30 times its current luminosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step is to look at a stellar evolutionary model for a star like the Sun.You can &lt;a href=&quot;http://astro.df.unipi.it/stellar-models/HZ/index.php?c=1&quot; rel=&quot;nofollow&quot;&gt;generate one here&lt;/a&gt;. I find that the Sun will have a luminosity in this range from ages 8.9 billion (at the cool end) to 10.0 billion years (at the hot end).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously you can play with the numbers (upper and lower temperature bound, albedo) to get different answers (the big assumption was to just use the equilibrium temperature, but an atmosphere could warm things up a bit), but you should be able to follow this prescription using whatever numbers you wish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, at the &quot;cool end&quot; of the Mars calculation - the Earth's equilibrium temperature would be 315K (42 Celsius).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-02-10T09:30:04.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8834" PostTypeId="2" ParentId="8829" CreationDate="2015-02-10T13:03:18.783" Score="2" Body="&lt;p&gt;You've asked a big question, too big perhaps for a Q&amp;amp;A forum such as this. Your question is the sole subject of graduate level aerospace engineering classes, e.g., &lt;a href=&quot;http://ccar.colorado.edu/ASEN5070&quot; rel=&quot;nofollow&quot;&gt;University of Colorado ASEN 5070, Introduction to Statistical Orbit Determination&lt;/a&gt;, and is the subject of multiple graduate level texts, e.g., &lt;a href=&quot;https://books.google.com/books?id=Ct3qN1VCHewC&quot; rel=&quot;nofollow&quot;&gt;Statistical Orbit Determination by Bob Schutz, Byron Tapley and George H. Born&lt;/a&gt;. To have a chance in that class, you'll need to already be well-versed in multivariate calculus, linear algebra, probability and statistics, numerical methods, and computer programming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, you might want to look at &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0008348&quot; rel=&quot;nofollow&quot;&gt;Bernstein and Khushalani (2000), &quot;Orbit fitting and uncertainties for Kuiper belt objects,&quot; &lt;em&gt;The Astronomical Journal&lt;/em&gt;, 120.6:3323&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The preferred approach in statistical orbit determination is to collect data over multiple orbits. That's a luxury that is not possible with Trans-Neptunian Objects that have only been seen a handful of times, and only over a small arc of the multi-hundred year long orbits of those objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing Bernstein and Khushalani did to overcome this was to realize that TNOs are nearly inertial (Newton's first law) objects. Gravitation is but a small perturbation of inertial behavior at such distances. Another thing they did was to take advantage of the fact that for observations separated by a short span of time (e.g., a day or two), almost all of the apparent motion is due to the Earth rather than proper motion of the target object. This gives a good estimate of the distance to the target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their approach involves doing some of the regression in Cartesian space, with gravitation being a small perturbation, and then switching to orbital element space to complete the regression. Along the way, they worry about whether they have enough information to do the full orbital element space regression.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2015-02-10T13:14:58.883" LastActivityDate="2015-02-10T13:14:58.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8835" PostTypeId="1" AcceptedAnswerId="8851" CreationDate="2015-02-10T14:00:28.890" Score="0" ViewCount="69" Body="&lt;p&gt;I know that, in L4&amp;amp;L5, the distances to the two main bodies should be equal. Still, how can I calculate that distance with accuracy? How can I know that that distance is inside the equipotential surface?&#xA;Let's take, as an example, the Sun-Earth system and the Earth-Moon system.&#xA;Thank you in advance!&lt;/p&gt;&#xA;" OwnerUserId="987" LastEditorDisplayName="user2764" LastEditDate="2015-02-10T22:12:22.780" LastActivityDate="2015-02-11T06:25:55.693" Title="L4&amp;L5 positions?" Tags="&lt;distances&gt;&lt;lagrange-point&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8836" PostTypeId="2" ParentId="8830" CreationDate="2015-02-10T14:06:20.137" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;One billion years into the future and the Sun has swollen in size and it is now not possible to live on Earth due to the heat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The problem isn't so much the Sun swelling in size. The key problems are that the Sun's luminosity increases over time and that the Earth is covered with oceans. The Earth will become uninhabitable long before the Sun turns into a red giant. Ever increasing luminosity will result in an increased surface temperature, which in turn will result in increased atmospheric water vapor. Right now this isn't a problem because increased atmospheric water vapor means more clouds, which increases albedo, which counters the effect of increased luminosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another effect to consider, however. Water vapor is an extremely powerful greenhouse gas. At some point, increased surface temperature will increase atmospheric water vapor to the point where the increased greenhouse effect increases the surface temperature, which increases vaporization even more, and so on. A loop! And a loop of the very worst kind, a positive feedback loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When this will happen is subject to debate, perhaps as short as 500 million years (Goldblatt 2013), perhaps a billion years (Kasting 1988), or perhaps as long as 1.5 to 2 billion years (Leconte 2013) in the future.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Mankind has relocated to Mars where the temperature is more favorable now than on Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Speculating what humanity will do or become 500 million years from now is a bit much. It's not science; nobody will live to see whether there speculations turn out to be true. But it is a bit fun. Why would we relocate to Mars? Why not relocate Earth instead (Korycansky 2001)? We might eventually have to worry about Mars getting in the way. In that case, simply move Mars, too!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nature.com/ngeo/journal/v6/n8/abs/ngeo1892.html&quot; rel=&quot;nofollow&quot;&gt;Goldblatt, Colin, et al (2013), &quot;Low simulated radiation limit for runaway greenhouse climates,&quot; &lt;em&gt;Nature Geoscience&lt;/em&gt;, 6.8:661-667.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chriscunnings.com/uploads/2/0/7/7/20773630/runaway_greenhouse_venus.pdf&quot; rel=&quot;nofollow&quot;&gt;Kasting, James F. (1988), &quot;Runaway and moist greenhouse atmospheres and the evolution of Earth and Venus,&quot; &lt;em&gt;Icarus&lt;/em&gt;, 74.3:472-494.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/astro-ph/0102126&quot; rel=&quot;nofollow&quot;&gt;Korycansky, D. G., Gregory Laughlin, and Fred C. Adams (2001), &quot;Astronomical engineering: a strategy for modifying planetary orbits,&quot; &lt;em&gt;Astrophysics and Space Science&lt;/em&gt;, 275.4:349-366.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1312.3337&quot; rel=&quot;nofollow&quot;&gt;Leconte, Jérémy, et al. (2013), &quot;Increased insolation threshold for runaway greenhouse processes on Earth-like planets,&quot; &lt;em&gt;Nature&lt;/em&gt;, 504.7479:268-271.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-02-10T14:06:20.137" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8837" PostTypeId="2" ParentId="8819" CreationDate="2015-02-10T15:33:48.590" Score="0" Body="&lt;p&gt;So to refer to objects beyond Neptune you use 'trans-Neptunian objects'. It really sounds simple, there are several sub-sets of trans-Neptunian objects: referring to minor-planets (neither Planets nor comets) that exist in this region of the Solar System.  &lt;/p&gt;&#xA;" OwnerUserId="5505" LastActivityDate="2015-02-10T15:33:48.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8838" PostTypeId="1" AcceptedAnswerId="8843" CreationDate="2015-02-10T16:06:56.553" Score="4" ViewCount="5050" Body="&lt;p&gt;First of all, I'm studying orbits for a hobby: world building. Unfortunately, my mathematical abilities approach a ridiculous low threshold, which means I am stuck with reading the simplest explanations, which in turn leave me asking tons of fairly basic questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Allow me to start with a simple point. I know that Kepler's Laws state that planetary orbits must always be elliptical. I also know that Earth's orbit varies from more elliptical to less elliptical, and that its less elliptical stage is nearly circular. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... what would happen if Earth did have a circular orbit? Why is it impossible for any planet (or moon, by the way) to orbit another body in a perfectly circular path?&lt;/p&gt;&#xA;" OwnerUserId="5504" LastActivityDate="2020-10-24T18:02:23.800" Title="Circular orbits" Tags="&lt;orbit&gt;" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8839" PostTypeId="2" ParentId="8781" CreationDate="2015-02-10T16:07:26.843" Score="4" Body="&lt;p&gt;Googling &quot;hyperinflation&quot; mostly returns articles about Zimbabwe and the Weimar Republic, so I'm going to assume that you are referring to what is usually just called (cosmological) inflation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are still pretty uncertain about when and for how long inflation occured, but we do however have some sort of constraints. In general, inflation is thought to be caused by some sort of scalar field, usually written as $\phi(\mathbf{r},t)$, and its associated potential energy $V(\phi)$. Various forms of how this field changes as a function of time exist, but a popular one is, or has been at least, the &lt;em&gt;slow-roll&lt;/em&gt; field, which makes it behave like a cosmological constant, enabling it to drive an exponential inflation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inflation can be initiated by a symmetry break when the unified forces of electromagnetism and the weak and strong forces are separated in the strong and the electroweak force (the &quot;GUT&quot; era, for Grand Unified Theory), at temperatures of $T\sim10^{28}$ K. It will continue until the field has &quot;rolled&quot; from its metastable false vacuum state to a true vacuum. During this time, the Universe expands something like $e^{100}$ times, and its temperature drops by the same factor, i.e. to a freezing temperature of $T\sim10^{-15}$ K. This doesn't last long, though, because the energy of the field is converted to photons and other relativistic particles. This is called &lt;em&gt;reheating&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The factor by which the Universe expands during inflation is essentially set by the minimum factor needed to explain in particular three cosmological &quot;problems&quot;, viz. &lt;a href=&quot;http://en.wikipedia.org/wiki/Flatness_problem&quot; rel=&quot;nofollow&quot;&gt;the flatness of the Universe&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Horizon_problem&quot; rel=&quot;nofollow&quot;&gt;the causality of the horizon&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetic_monopole&quot; rel=&quot;nofollow&quot;&gt;the missing monopoles&lt;/a&gt;, and the maximum factor allowed in order not to blow stuff too far apart that no structure would have had the time to form before gravitational attraction was too small, so that we wouldn't be here to ask the question. This requires at least 60 e-foldings (i.e. expanding by a factor of $e^{60}$), but could be 100 or maybe more. The factor is discussed e.g. in &lt;a href=&quot;http://www.google.dk/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=6&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0CEAQFjAF&amp;amp;url=http%3A%2F%2Fstaff.ustc.edu.cn%2F~wzhao7%2Fc_index_files%2Fmain.files%2FHJMo.pdf&amp;amp;ei=PFHbVOyvI4znasvdgdgP&amp;amp;usg=AFQjCNG-TI9yOchvK8hjCdgpGXQjKZS-nw&amp;amp;sig2=IbZWtf4193BGdywMIrG8Qw&quot; rel=&quot;nofollow&quot;&gt;Galaxy Formation And Evolution by Mo, Bosch &amp;amp; White, Sec. 3.6.2&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the size evolution of the Universe during inflation is exponential, the scale factor can be written as&#xA;$$a(t) \propto e^{Ht},$$&#xA;where $H$ is the Hubble constant during inflation, so for an expansion of, say, $e^{100}$ we have&#xA;$$e^{100} = \frac{a(t_{\mathrm{end}})}{a(t_{\mathrm{begin}})}&#xA;          = \frac{e^{H t_{\mathrm{end}}}}{e^{H t_{\mathrm{begin}}}}&#xA;          = e^{H(t_{\mathrm{end}} - t_{\mathrm{begin}})},$$&#xA;or&#xA;$$\Delta t \equiv t_{\mathrm{end}} - t_{\mathrm{begin}} = 100/H,$$&#xA;where $t_{\mathrm{begin}}$ and $t_{\mathrm{end}}$ marks the time of beginning and end of inflation, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During inflation, the Hubble constant should not evolve, but have a constant value of $H=(\Lambda/3)^{1/2}$, where $\Lambda$ is a cosmological constant. If inflation started when the GUT era ended, which was around $t_{\mathrm{GUT}} \sim 10^{-36}$ s, the Hubble constant was of the order $H \sim t_{\mathrm{GUT}}^{-1} \sim 10^{36}$ s. Then a hundred e-foldings will take you to $t_{\mathrm{end}} \sim 10^{-34}$ s. &lt;a href=&quot;http://en.wikipedia.org/wiki/Inflationary_epoch&quot; rel=&quot;nofollow&quot;&gt;Wikipedia states&lt;/a&gt; that inflation should last somewhat longer, something like $10^{-33}$ to $10^{-32}$ s. This might be if you do a more careful analysis, or maybe a different model. For instance, H is not exactly constant during inflation, but does evolve a little.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various observations can constrain the physical parameters that enter the equations, e.g. cosmic microwave background and baryon acoustic oscillations. The latest results from the Planck satellite can be found &lt;a href=&quot;http://www.cosmos.esa.int/documents/387566/522789/Planck_2015_Results_XX_Constraints_Inflation.pdf/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Wikipedia also has a &lt;a href=&quot;http://en.wikipedia.org/wiki/Inflation_%28cosmology%29#Observational_status&quot; rel=&quot;nofollow&quot;&gt;summary&lt;/a&gt; of the observational status of inflation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I guess the answer to your question is, &quot;&lt;strong&gt;We don't, really, but we have several theories that together with several observational constraints tell us that the duration of inflation should be of this order of magnitude. If it were much shorter, it wouldn't solve the cosmological problems that gave rise to the idea of inflation, and if it were much longer, the Universe would be blown apart&lt;/strong&gt;&quot;.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-02-11T15:02:59.837" LastActivityDate="2015-02-11T15:02:59.837" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8840" PostTypeId="2" ParentId="8838" CreationDate="2015-02-10T16:23:37.763" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is it impossible for any planet (or moon, by the way) to orbit another body in a perfectly circular path?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One way to look at it is from the perspective of probability and statistics. Think of position and velocity as random variables drawn from some continuous probability distributions. Given some position vector, the velocity vector has to have a very specific value to yield a circular orbit. The probability of drawing a specific value from a well-behaved continuous probability distribution is identically zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An even better way to look at it: Even perfectly elliptical orbits aren't possible. Kepler's laws are an approximation that result from assuming a universe that obeys Newtonian mechanics and comprises but two point masses. Newtonian mechanics is only approximately valid in the real universe, objects are lumpy and can only approximately be treated as point masses, and there are a lot more than two objects in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that by some fluke chance, an object appears to have a perfectly circular orbit at some point in time (to within measurement error). The non-Newtonian nature of the universe, the non-spherical mass distributions of the objects, and the multiplicity of objects means that a moment later the object will no longer appear to have a perfectly circular orbit.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-02-10T16:23:37.763" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="8841" PostTypeId="1" CreationDate="2015-02-10T17:41:08.633" Score="5" ViewCount="849" Body="&lt;p&gt;When an astronomer talks about her/his topic and shows an X/Y-plot with error bars. What should one assume that those error bars represent?&#xA;1 standard deviation? Or 2? Or some specific significance level like 95% or 99%?&#xA;Is there a generally understood convention for this, or does it vary depending on topic? I've noticed that it is very rarely explicitly stated what the meaning is of the range which the error bars cover.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And how are the cases treated, where exact observations reveal a variation in the actual population? Not &quot;errors&quot;, but natural true certain variation. Are there other bars to illustrate that range of true values, as opposed to the variation due to observational uncertainty?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-05-31T18:12:57.190" Title="What uncertainty does an error bar signify in astronomy?" Tags="&lt;observational-astronomy&gt;&lt;astrophysics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8842" PostTypeId="2" ParentId="8841" CreationDate="2015-02-10T18:30:51.957" Score="0" Body="&lt;p&gt;The most common way to represent uncertainty is with symmetric error bars around a central point. This is in turn commonly interpreted as the 95 % confidence interval. Ie, the actual data point is the centre of a Gaussian which has a width at 95 % of it's height as the size of the error bars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is &lt;strong&gt;only&lt;/strong&gt; statistical uncertainty and often not explicitly stated. One also refers to measurement and discovery with different confidence intervals... &lt;strong&gt;discovery&lt;/strong&gt; is commonly only claimed when there is a a 5-sigma confidence interval. Ie, if the measurement lies more than 5 widths of the peak away from the theory or prediction, you've made a discovery. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, we are leaving out here systematic uncertainty and instument bias, which can only increase the total uncertainty. Usually it is assumed that there is no correlation, so they are combined using the sum of their squares. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long story short - always ask what the error bars represent, especially if they look too &quot;clean&quot;.&lt;/p&gt;&#xA;" OwnerUserId="5506" LastActivityDate="2015-02-10T18:30:51.957" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8843" PostTypeId="2" ParentId="8838" CreationDate="2015-02-10T20:15:13.977" Score="7" Body="&lt;p&gt;You've been given an answer, and it's perfectly valid, but here's something from a different perspective (less strict).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A circle is really just a particular case of an ellipse. Take an ellipse, and change it, by moving its focal points closer together. When those two points coincide, what you get is a circle. It's still an ellipse, technically - one that happens to have both focal points in the same place, is all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0T43P.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, you can actually have planetary orbits, or any orbits, circular. There's nothing forbidding that. It's just pretty unlikely that this will occur via a natural process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As indicated elsewhere, in the real world, all orbits and trajectories are a bit imperfect due to perturbations - whether they be elliptical, circular, parabolic or hyperbolic, they are always a bit perturbed by external factors. In many cases, perturbations are so tiny that you can ignore them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a planet is orbiting the Sun, and the orbit is elliptical, the Sun will be in one of those two focal points; the other point has no particular signification. If you could circularize that orbit, then the Sun would be in the center of the circle, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kepler's laws remain valid for a circular orbit:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The orbit of every planet is an ellipse with the Sun at one of the two foci.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Still true. A circle is an ellipse where the foci coincide.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;A line joining a planet and the Sun sweeps out equal areas during equal intervals of time.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Still true. On a circular orbit, the planet moves at constant speed, so the swept area remains constant per time.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Still true. The semi-major axis becomes the radius of the circle.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You must understand that Kepler's laws now have more of a historic interest. They are not exactly at the bleeding edge of science anymore. During Kepler's time, it seemed reasonable to state that all orbits must be elliptical (in the strict sense of the term), but now we know that trajectories (including orbits, or closed trajectories) can be circular, elliptical, parabolic or hyperbolic, depending on a few factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also know that perturbations actually deflect all these trajectories a little bit from ideal shapes (but it's usually a very tiny effect).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also know that relativity makes all &quot;elliptical&quot; orbits more complex - they remain close to elliptic, but the whole ellipse keeps turning around the central star very slowly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this stuff was not known during Kepler's time, so take his laws for what they are - a snapshot of the development of our understanding in time.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-02-11T00:21:04.460" LastActivityDate="2015-02-11T00:21:04.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8844" PostTypeId="2" ParentId="8838" CreationDate="2015-02-10T21:30:16.117" Score="3" Body="&lt;p&gt;A circle is an ellipse with eccentricity zero. And in fact tidal evolution can drive orbital eccentricity to values negligibly close to zero. See &lt;a href=&quot;http://iopscience.iop.org/1538-3881/125/2/902/fulltext/&quot; rel=&quot;nofollow&quot;&gt;Regarding the Putative Eccentricity of Charon's Orbit&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From observations we already know Pluto and Charon move about each other in very nearly circular orbits. I expect when &lt;a href=&quot;http://en.wikipedia.org/wiki/New_Horizons&quot; rel=&quot;nofollow&quot;&gt;New Horizons&lt;/a&gt; flies by the system in July, 2015, we will know their orbits more precisely.&lt;/p&gt;&#xA;" OwnerUserId="1232" LastActivityDate="2015-02-10T21:30:16.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8845" PostTypeId="1" CreationDate="2015-02-11T01:38:06.467" Score="-7" ViewCount="392" Body="&lt;p&gt;First of all &lt;strong&gt;&lt;em&gt;not to consider me a conspiracy theorist&lt;/em&gt;&lt;/strong&gt;, but isn't landing on the moon a questionable issue? I am really not an expert in astronomy but let's assume that a moon landing has happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- &lt;strong&gt;How would the Gamma rays and cosmic rays affect the equipments on the lunar surface and would these equipments function normally like on earth?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2-&lt;strong&gt;Could the enormous amounts of solar energy bursts be avoided somehow for protecting human flesh?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3-&lt;strong&gt;How is it possible for an FM or any kind of data transmission type be achieved in an environment that has nothing but an empty space? How would the electromagnetic waves travel and enter the earth's atmosphere to be captured by the receivers?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4-&lt;strong&gt;What is the bright lunar surface real temperature and is there a way to equip an astronaut to protect him from the over-heat?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for sharing any ideas that could explain these points not just for me, but for interested people who might view this question.&lt;/p&gt;&#xA;" OwnerUserId="5512" LastActivityDate="2015-02-11T03:41:56.633" Title="What are the Gamma rays and Cosmic rays effects on humans and equipments?" Tags="&lt;gamma-ray-bursts&gt;&lt;solar-wind&gt;&lt;solar-storm&gt;&lt;the-moon&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ClosedDate="2015-02-13T13:09:29.647" ContentLicense="CC BY-SA 3.0" />
  <row Id="8846" PostTypeId="2" ParentId="8845" CreationDate="2015-02-11T03:41:56.633" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;First of all *not to consider me a conspiracy theorist(, but isn't landing on the moon a questionable issue?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only to conspiracy theorists. To everyone else, no, it's not a questionable issue. My father in law helped send men to the Moon. I have worked with a number of people who sent men to the Moon. I was once called on the carpet in Gene Kranz's office. I find it extremely insulting to think that it is a questionable issue. So excuse me if my response might seem be a bit insulting.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How would the Gamma rays and cosmic rays affect the equipments on the lunar surface and would these equipments function normally like on earth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Equipment and humans did not suffer immediate damage given the short period of time that humans did spent on the Moon. One of the effects of mild radiation is increased risk of cancer (but that's a long-term effect). The men who went to the Moon (and they did go to the Moon) did indeed suffer increased cancer rates compared to the Earth-bound population.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Could the enormous amounts of solar energy bursts be avoided somehow for protecting human flesh?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By luck, there were no large solar energy bursts when men were in space on the way to the Moon, on the Moon, or coming back from the Moon. A large coronal mass ejection event did happen on August 7, 1972, but that was (by luck) sandwiched between the Apollo 16 and Apollo 17 lunar missions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How is it possible for an FM or any kind of data transmission type be achieved in an environment that has nothing but an empty space? How would the electromagnetic waves travel and enter the earth's atmosphere to be captured by the receivers?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This question makes no sense. Look up in the sky during the day. What do you see? You see the Sun. Look up in the sky at night. What do you see? You see the Moon, the planets, stars, and if you live in an area with low humidity and limited light pollution, you even see other galaxies. With your naked eye. The only difference between light and FM is frequency. Both are forms of electromagnetic radiation. Electromagnetic radiation travels unfettered through empty space.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the bright lunar surface real temperature and is there a way to equip an astronaut to protect him from the over-heat?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first defense against the temperature extremes of the space environment is very simple: It's coloration. A spacesuit that was jet black in the visible range but white in the thermal infrared would have quickly killed the NASA astronauts on the Moon due to overheating. On the flip side, a spacesuit that was white in the visible range but jet black in the thermal infrared would have resulted in overcooling. The space suits worn by the NASA astronauts on the Moon were white in the visible range but grayish in the thermal infrared. NASA spent a lot (a whole lot) of money investigating different fabrics, different dyes, and different paints. Passive thermal control is the first step in any space operation against the extremes of space. Active thermal control addresses what little passive thermal control can't address.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-02-11T03:41:56.633" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8847" PostTypeId="1" CreationDate="2015-02-11T04:16:28.160" Score="4" ViewCount="139" Body="&lt;p&gt;I have read that immensley powerful telescopes such as the European Extremley Large Telescope will be able to directly image the atmospheres of exoplanets, and even determine their rotation rates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering, what are the obstacles that prevent current telescopes from achieving this; how will these obstacles be overcome exactly by the new telescopes such as the EELT; and tentativley when can we expect these kinds of telescopes to come online?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is it theoretically possible to build a telescope (earth or space based) that can discern fine structures such as clouds, or surface features of exoplanets (or is this in fact the level of detail that will be achieved by telescopes such as the EELT)?&lt;/p&gt;&#xA;" OwnerUserId="2873" LastActivityDate="2015-03-18T00:53:21.643" Title="What kinds of technologies are required to directly image exoplanets" Tags="&lt;telescope&gt;&lt;exoplanet&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8848" PostTypeId="1" AcceptedAnswerId="8857" CreationDate="2015-02-11T05:01:04.767" Score="8" ViewCount="534" Body="&lt;p&gt;There are several ways we are coming closer to answering the question &quot;is there life elsewhere in the universe?&quot;. One is by first understanding very well the origin life on our own planet. Another is by trying to look for radio signals from extraterrestrial civilizations. Another is to try to search for habitable exoplanets, and eventually examine the habitable ones in detail via direct imaging. Another is too look for evidence of life or life that existed in the past right in our own solar system (for exmaple on Mars, Titan, Europa, etc). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Have I missed any other astronomy-related approaches to answering this question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Short of radio contact from an ET civilization, or a visit by aliens, or the discovery of life/past-life in our solar system, what would give conclusive evidence of alien life in the galaxy? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 20 or so years when we develop the technology to directly image exoplanets, we will be able to determine their chemical composition and characterize their structure. If astronmers who study exoplanets found what they are looking for -evidence of oxygen and other biomarker gases - this might be highly suggestive of life, but not conclusive evidence. What kinds of observations of an exoplanets could enable us to say with ceartainty &quot;there is life on this planet&quot;? Short of telescopic observations of truly unbelievable detail, I cannot think of a way that we can look at a planet and say &quot;yes there is life on it, or no there is not&quot; other than perhaps if seeing lights on the dark side of the planet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Actually, I forgot the obvious scenario where there are trees/grass on the exoplanet.  But I don't think the plant life is required to be a different color from the rest of the terrain, and a planet with aliens does not have to have plant life. So we might not be able to detect life visually.&lt;/p&gt;&#xA;" OwnerUserId="2873" LastActivityDate="2020-09-28T04:03:23.967" Title="What astronomical observations would give conclusive proof of alien life?" Tags="&lt;planet&gt;&lt;exoplanet&gt;&lt;extra-terrestrial&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8849" PostTypeId="1" AcceptedAnswerId="8850" CreationDate="2015-02-11T05:19:47.110" Score="24" ViewCount="6884" Body="&lt;p&gt;It is amazing that Kepler determined his three laws by looking at data, without a calculator and using only pen and paper. It is conceivable how he proved his laws described the data after he had already conjectured them, but what I do not understand is how he guessed them in the first place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will focus in particular on Kepler's third law, which states that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of the orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that Kepler was working with data about the planets only, plus our own moon, and the sun. I make this assumption because I don't think Kepler had data about other moons, comets, or asteroids, which had not been observed by telescope yet. If this is true, knowing that Neptune, Uranus, and Pluto were not yet discovered when Kepler was alive, this means Kepler had less than 9 data points to work with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My friend claims that it is totally concievably how Kepler guessed this relationship (although he provides no method of how Kepler might have done it), and also that Kepler's observations are &quot;not that hard&quot;.  As a challenge, I gave my friend a data table with one column labeled $x$, the other $y$, and 9 coordinates $(x,y)$ which fit the relationship $x^4=y^3$. I said &quot;please find the relationship between $x$ and $y$&quot;, and as you might expect he failed to do so. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please explain to me how in the world did Kepler guess this relationship working with so few data points. And if my assumption that the number of data points Kepler had at his disposal is small, is wrong, then I still think its quite difficult to guess this relationship without a calculator. &lt;/p&gt;&#xA;" OwnerUserId="2873" LastEditorUserId="2153" LastEditDate="2015-07-08T21:08:44.233" LastActivityDate="2019-06-19T09:34:51.397" Title="How did Kepler &quot;guess&quot; his third law from data?" Tags="&lt;orbit&gt;&lt;history&gt;&lt;johannes-kepler&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="8850" PostTypeId="2" ParentId="8849" CreationDate="2015-02-11T06:01:35.887" Score="26" Body="&lt;p&gt;Kepler's third law is trivial (in my opinion) compared to his first law. I am quite impressed that he was able to deduce that the orbits were ellipses. To get that, he had to go back and forth plotting Mars' direction from Earth and Earth's direction from Mars. He knew the length of both planets' years, so observations taken one Mars year apart would differ only because Earth had moved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But maybe not so trivial. He published his first two laws in 1609. The third law didn't come along until ten years later, in 1619. With ten years to work on it, even the most obscure relationship will eventually be found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To discover a ratio-of-powers relationship, plot the logarithms of the numbers. In your example with $x^4 = y^3$, the logs would plot on a straight line with a slope of $3/4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The timing is right. Napier published his book on logarithms in 1614. Kepler may have on a whim applied this shiny new mathematical tool to his crusty old data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The major hurdle was that at the time there were only six known planets, so he didn't have an abundance of data points, and the ones he had were by no means precise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kepler's other problem is that none of his laws made any sense to him. They fit the data, but he had no idea why. He didn't have Newton's laws of motion to work from, he had no understanding of force, momentum, angular momentum, and certainly not gravity. So far as he knew, the planets moved the way they did because God decreed it, and angels were tasked with pushing the planets along their orbits. The outer planets moved slower because they were being pushed by lesser angels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Feynman makes the comment that we understand so much more now. We now know that the angels are on the outside pushing in toward the Sun.)&lt;/p&gt;&#xA;" OwnerUserId="4077" LastActivityDate="2015-02-11T06:01:35.887" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="8851" PostTypeId="2" ParentId="8835" CreationDate="2015-02-11T06:25:55.693" Score="1" Body="&lt;p&gt;The distances to the two main bodies are not only equal, they are equal to the distance between the two main bodies. In other words, the two main bodies and L4 (or L5) form an equilateral triangle.&lt;/p&gt;&#xA;" OwnerUserId="4077" LastActivityDate="2015-02-11T06:25:55.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8852" PostTypeId="2" ParentId="8849" CreationDate="2015-02-11T06:29:29.370" Score="31" Body="&lt;p&gt;Kepler's account of how the third law came to be is as follows (Caspar p.286; emphasis mine):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On the 8th of March of this year 1618, if exact information about the time is desired, it appeared in my head. But I was unlucky when I inserted it into the calculation, and rejected it as false. Finally, on May 15, it came again and with a new onset conquered the darkness of my mind, whereat there followed such an excellent agreement between my seventeen years of work at the Tychonic observations and my present deliberation that I at first believed that I had dreamed and assume the sought for in the supporting proofs. &lt;em&gt;But it is entirely certain and exact that the proportion between the periodic times of any two planets is precisely&lt;/em&gt; &lt;strong&gt;one and a half&lt;/strong&gt; &lt;em&gt;times the proportion of the mean distances&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although Kepler does not actually describe inspiration that led him to believe this, the curious phrasing provides a very strong clue when combined with some background biographical information:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;John Napier published &lt;em&gt;Mirifici Logarithmorum Canonis Descripto&lt;/em&gt; in 1614, which contained the then-new invention of logarithms. Kepler was aware of Napier's work by 1617 (Caspar p. 308), perhaps earlier.&lt;/li&gt;&#xA;&lt;li&gt;Joost Bürgi published work on logarithms almost at the same time as Napier, and Kepler was similarly aware of Bürgi, even praising his mathematical abilities as surpassing most professors of mathematics.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thus, Kepler's statement is is equivalent to saying that the data makes a slope of 1.5 on a log-log graph, a which is a very simple linear relationship on this scale.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Caspar, Max, &lt;em&gt;Kepler&lt;/em&gt;, (Dover, New York, 1993).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="661" LastActivityDate="2015-02-11T06:29:29.370" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8853" PostTypeId="2" ParentId="8848" CreationDate="2015-02-11T12:49:01.993" Score="0" Body="&lt;p&gt;I don't see why this question would be down voted. I think that the most we could determine is that the conditions for life are present on a Planet. The only way to know for certain would be to actually go there . However we could perhaps detect certain gases that could only be emitted from living organisms and this would be be some proof.&lt;/p&gt;&#xA;" OwnerUserId="2562" LastActivityDate="2015-02-11T12:49:01.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8854" PostTypeId="1" CreationDate="2015-02-11T12:53:48.227" Score="3" ViewCount="2888" Body="&lt;p&gt;Does the Earth receive any heat at all from the millions of other Stars in our Galaxy ? Is it light that is bringing the heat and perhaps it cools down on the long journey in Space getting to the Earth and that's why no significant amount of heat is getting here ?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastActivityDate="2015-02-11T17:50:25.167" Title="Heat from other Stars" Tags="&lt;heat&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8855" PostTypeId="2" ParentId="8854" CreationDate="2015-02-11T13:33:09.393" Score="5" Body="&lt;p&gt;Yes, the Earth does receive energy from the stars, but not much. The effective temperature of the night sky is about 3 Kelvin which is not much more that that of the cosmic microwave background at 2.7 Kelvin. The difference being due to the total energy of starts etc.(room temperature is ~295 kelvin)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coldness of the night sky is interesting in itself see &lt;a href=&quot;http://en.wikipedia.org/wiki/Olbers%27_paradox&quot;&gt;Olber's Paradox&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-02-11T13:33:09.393" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8856" PostTypeId="2" ParentId="8854" CreationDate="2015-02-11T13:37:30.533" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the Earth receive any heat at all from the millions of other Stars in our Galaxy ? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Effectively, no. Stars are too few and far between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Qualifying that &quot;effectively, no&quot;: From &lt;a href=&quot;http://stjarnhimlen.se/comp/radfaq.html#10&quot;&gt;http://stjarnhimlen.se/comp/radfaq.html#10&lt;/a&gt;, the stellar magnitude from total starlight is -5. Compare that to the -26.7 magnitude of the Sun as viewed from the Earth. That difference of 21.7 means that starlight is responsible for one part in 10&lt;sup&gt;22&lt;/sup&gt; of the heating of the Earth. Another way to express one part in 10&lt;sup&gt;22&lt;/sup&gt; is &quot;effectively none&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to look at it: The Earth would eventually cool to 2.7 kelvins if the Sun and stars magically turned off. If it was only the Sun magically turned off, the Earth would cool to 3 kelvins. Compare that to the nice balmy 287 kelvins we experience thanks to the Sun.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it light that is bringing the heat and perhaps it cools down on the long journey in Space getting to the Earth and that's why no significant amount of heat is getting here ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The stars in our galaxy are extremely close to us in a cosmological sense. Even the Andromeda galaxy is extremely close. The light we see from stars in our galaxy is more or less the same as emitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over very, very long distances (much, much longer than the distance to Andromeda), the cosmological expansion of space means that light is redshifted. How much light is redshifted offers a clue as to the distance to some remote object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do receive a minuscule amount of energy from the cosmic microwave background. That radiation was not emitted by stars. It marked the transition from the very early hot and opaque universe to a cooler and transparent universe. The universe transitioned from opaque to transparent when the temperature dropped below 3000 K or so. Now that the light has an effective temperature of only 2.725 K.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2015-02-11T17:50:25.167" LastActivityDate="2015-02-11T17:50:25.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8857" PostTypeId="2" ParentId="8848" CreationDate="2015-02-11T14:50:30.097" Score="6" Body="&lt;p&gt;Lots of frontier science lectures to watch on nearby topics here:&#xA;&lt;a href=&quot;https://astrobiology.nasa.gov/seminars/&quot; rel=&quot;noreferrer&quot;&gt;https://astrobiology.nasa.gov/seminars/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your very question seems to be the hottest topic in astronomy today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My impression is that if oxygen gas and methane were on the same exoplanet, then it certainly must be living biological processing creating that there. Nothing else could explain the coexistence of the two. And even if one of those two gasses could occur abiologically, and a range of exoplanets with one of them would be found, statistically one could deduce a frequency of planets which must have life, because the abiotic processes could only explain a certain maximum part of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a pitty that most exoplanets are in size between Venus and Neptune, which are at the same time the least explored planets in our Solar system. When a sub-Neptune or a super-Venus has its atmosphere examined for the first time, we won't have as good references as we could've had.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-02-11T14:50:30.097" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8858" PostTypeId="2" ParentId="8830" CreationDate="2015-02-12T06:16:44.887" Score="1" Body="&lt;p&gt;If you  postulate some things that began as humans are still around in 1,000,000 years, I would also guess that way off topic of astronomy (but not for exploration) that &quot;people&quot; will be safely basically immortal in virtual worlds and have sought out the safest place to do their thinking and playing. Perhaps headed outside the galaxy or the plane of the galaxy to wherever the least junk is flying about.&lt;/p&gt;&#xA;" OwnerUserId="5526" LastActivityDate="2015-02-12T06:16:44.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8859" PostTypeId="1" CreationDate="2015-02-12T06:24:02.800" Score="6" ViewCount="2153" Body="&lt;p&gt;If the Earth and its host star were located exactly in-between the Milky Way and Andromeda would the night sky be completely void of light? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would stars be visible to the unaided eye?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there could be a civilization that exists in which the night is completely void of light? …that would suck &lt;/p&gt;&#xA;" OwnerUserId="4175" LastActivityDate="2015-07-11T10:24:56.053" Title="What would the night sky look like if Earth orbited an intergalactic star?" Tags="&lt;star&gt;&lt;distances&gt;&lt;intergalactic-space&gt;&lt;light-pollution&gt;&lt;rogue-planet&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8860" PostTypeId="2" ParentId="8859" CreationDate="2015-02-12T09:41:14.357" Score="6" Body="&lt;p&gt;You can see the Andromeda galaxy with a naked eye, even with some level of light pollution. So if the star you are asking about would be even closer to Andromeda, you would see at least that galaxy. There are some other galaxies that can be seen from Earth with a naked eye or with binoculars, so sky of such a lonely planet would still have some night lights on it (and of course its own &quot;sun&quot; that it would orbit). &#xA;You would not see any stars, as all the stars we see on our sky are from our galaxy. You might be able to see some Active Galactic Nuclea, which are far away galaxies that look like stars (point sources), and are extremely luminous, so we are able to see them even though they are really far away. &#xA;But if you consider the probability of a stellar system that would be far away from any other galaxies, such probability is defnitely nonzero, as the universe is very inhomogenous. The issue of any planets having civilizations on them is a separate topic, so I will leave that to your personal opinion. &lt;/p&gt;&#xA;" OwnerUserId="5530" LastActivityDate="2015-02-12T09:41:14.357" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8861" PostTypeId="2" ParentId="8859" CreationDate="2015-02-12T10:07:46.740" Score="11" Body="&lt;p&gt;Since Andromeda is already visible to the naked eye, to a civilization located at half the distance from the Milky Way, Andromeda would be still be visible. Its total brightness would be four times higher, but since its area would grow by the same factor, its surface brightness would stay constant. The Milky is less bright by a factor of ~2.5, but also smaller, so the surface brightness and hence visibility would be similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Individual stars would not be visible except if 1) one exploded as a supernova or 2) the civilization has evolved into beings with extremely large pupils, so that more light is being recorded by their retinas. In fact, we can calculate how large:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_most_luminous_known_stars&quot;&gt;this list&lt;/a&gt;, the brightest star known in the Milky Way is called WR 25, and has an &lt;a href=&quot;http://en.wikipedia.org/wiki/Absolute_magnitude&quot;&gt;absolute magnitude&lt;/a&gt; of $M = -12.25$. Half of the distance from the MW to Andromeda is $d \sim 390,000$ pc, so the &lt;a href=&quot;http://en.wikipedia.org/wiki/Distance_modulus&quot;&gt;distance modulus&lt;/a&gt; is&#xA;$$\mu = 5\log d - 5 = 22.95,$$&#xA;and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Apparent_magnitude&quot;&gt;apparent magnitude&lt;/a&gt; of WR 25 would be&#xA;$$m = M + \mu = 10.7.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the darkest conditions, humans are able to see objects down to an apparent magnitude of ~6, i.e. a factor of&#xA;$$f = 2.5^{10.7 - 6} = 75$$&#xA;smaller. Thus, to be able to see WR 25, the diameter of the aliens' pupils would have to be a factor of $\sqrt{75} \sim 9$ larger. Since human pupils are roughly 6 mm wide, their pupils would be 5 cm wide, and the eyeballs somewhat larger, &lt;em&gt;incidentally&lt;/em&gt; roughly the size of a teacup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, it seems that my compatriot H. C. Andersen predicted the existence of these beings in his fairytale &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Tinderbox&quot;&gt;The Tinderbox&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-02-15T19:25:06.110" LastActivityDate="2015-02-15T19:25:06.110" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="8862" PostTypeId="1" AcceptedAnswerId="8863" CreationDate="2015-02-12T12:41:19.517" Score="7" ViewCount="4496" Body="&lt;p&gt;How would you estimate the angle covered by the star trails and deduce how long the exposure lasted in the following image?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fh0iZ.jpg&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the photo's caption:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Star trails beyond the Gemini Observatory on Mauna Kea, from which a laser beam forms a guide star on the Earth's ionosphere that enables the use of adaptive optics. In this long exposure, the laser beam tracks across the sky, making the visible arc. (The object in the sky at which the laser was pointed changed several times during the night.) Note that stars rise and set at an angle (not perpendicular) relative to the horizon, because Hawaii is about 20 degrees north of Earth's equator. Also, the farther a star is from the pole, the more its path looks straight, over a short distance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I read &lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20081024132334AAP642v&quot; rel=&quot;noreferrer&quot;&gt;the answer to this question about how to measure star trail angles&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The only star trail angles I'm familiar with deal with astrophotography.&#xA;  You set up a camera, put the film in, open the shutter for a time exposure of the stars. As the stars appear to revolve around the earth, they make &quot;arcs&quot; on the film.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;To calculate the angle of the arc, use the formula: 1 minute of time = 15 minutes of arc. If the answer is over 60, divide by 60 to get degrees of arc. So: 12 minute exposure = 180 minutes of arc or 3˚ of arc length for your star trail. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to approach this problem since the answer uses time to calculate the angle of the arc, but no time is given here.&lt;/p&gt;&#xA;" OwnerUserId="5532" LastActivityDate="2015-02-17T08:45:08.713" Title="Estimating the angle covered by the star trails and deducing how long the exposure lasted" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8863" PostTypeId="2" ParentId="8862" CreationDate="2015-02-12T14:14:18.020" Score="9" Body="&lt;p&gt;The answer &lt;em&gt;is&lt;/em&gt; given here. One minute of time corresponds to 15 arcminutes (written as 15'). This is because in 24 h the Earth revolves 360º, so&#xA;$$\textrm{angle per time} = \frac{360º}{24 \textrm{ h}} =\frac{21,600'}{1440 \textrm{ min}} = 15'/\textrm{min}.$$&#xA;If you turn this fraction upside down, you see that 1' corresponds to 1/15 min, or 4 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, you measure the angle (let's call it $\theta$) of any of the star traces, as seen from the center (notice that the Northern Star is not exactly at the center, so that it itself traces a tiny arc instead of a dot). From the picture below, I get roughly $\theta = 135º$. The exposure time is thus:&#xA;$$t_\mathrm{exp} = \frac{\theta}{\textrm{angle per time}} = \frac{135º}{360º/24 \textrm{ h}} \sim 9\textrm{ h}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, if you mark the position of the ends of the trails, you can recover the stellar sky. I found Ursa Major, marked by the yellow dots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kASpl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-02-17T08:45:08.713" LastActivityDate="2015-02-17T08:45:08.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8864" PostTypeId="1" AcceptedAnswerId="12762" CreationDate="2015-02-12T15:41:45.327" Score="5" ViewCount="366" Body="&lt;p&gt;Note 1: I've studied downvoted questions and decided that I either asked too many questions or I sound like I'm setting a fantasy world. So I'm editing my question accordingly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 2: I'm a language teacher. I learnt equations with variables but that was over 20 years ago and I remember little of it; I never learnt trigonometry and other more advanced maths, that's a fact, but if there's no way around it I'll do my best to tackle it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 3: My approach to understanding eclipses, due to the above stated handicap, is removing all the complicating factors (namely, the axial tilt and the moon's 'ondulating' orbit). I assume that, if I can fully understand such a simplistic scenario, I can then follow the mechanics of the real Earth scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, I have written below the points I am confident are true. The ones I don't know are asked as questions (Q) and formatted in bold. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Every new moon would be a total solar eclipse.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The path of totality would always cover the same area: a 250km  wide corridor with its centre at the equator (so 125km over the northern hemisphere and 125km over the southern hemisphere). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I think the path of totality wouldn't always cover the same area, e.g. always over Africa, so &lt;B&gt;Q1: Is there a simple formula to predict which areas would be in the path of totality?&lt;/B&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In real life, the moment of totality can vary from seconds to seven minutes. But in my study example, &lt;B&gt;Q2: the period of totality should always be the same, right?&lt;/B&gt; Because the crossing path of moon and sun is always the same (while in real life the Moon can overlap the Sun at different angles and going in different directions)&#xA;&lt;B&gt;Q3: How can I calculate this period of time?&lt;/B&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What about the latitudes for whom the solar eclipse would only be seen as partial? I understand that the nearest to the path of totality, the more covered the sun will be, and vice-versa. &lt;B&gt;Q4: Is there a set value that says e.g. for each km, another degree is visible?&lt;/B&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If this question is still not ideally adjusted, please leave a comment so I can further improve it.&lt;/p&gt;&#xA;" OwnerUserId="5504" LastEditorUserId="5504" LastEditDate="2015-02-14T13:02:41.443" LastActivityDate="2015-12-04T19:10:53.780" Title="Solar Eclipses for dummies: Step 1 - moon equatorial orbit around a planet without axial tilt" Tags="&lt;solar-eclipse&gt;&lt;lunar-eclipse&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8865" PostTypeId="2" ParentId="8592" CreationDate="2015-02-12T16:29:11.277" Score="0" Body="&lt;p&gt;One of us has misread the paper because I think he says he plans to use &quot;Benanaim&quot; which seems to be an alternative name for Alkaid, $\eta$ Ursae Majoris and very much &quot;ultima cauda ursae majoris&quot; and which would, indeed, transit the meridian at around 9pm on that date.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2015-02-12T16:29:11.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8866" PostTypeId="2" ParentId="8859" CreationDate="2015-02-12T19:44:26.537" Score="9" Body="&lt;p&gt;&lt;strong&gt;Planets&lt;/strong&gt;, if any in that star system, would be visible to the naked eye, the way you can see Mercury, Venus, Mars, Jupiter, Saturn and Uranus in our system. On a virtually empty sky, they would draw that much more attention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you had a &lt;strong&gt;Moon&lt;/strong&gt; (or several moons), that would be visible too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very rarely and briefly, &lt;strong&gt;asteroids&lt;/strong&gt; passing by very close to the planet might also be visible. In an empty sky, that would be a big event - and difficult to explain with an under-developed astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Comets&lt;/strong&gt; would be visible as usual, if the system has an Oort cloud and enough perturbations within it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than planets, moons, asteroids and comets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a desert, or a farm way out in the boondocks, you will see the nearest &lt;strong&gt;galaxies&lt;/strong&gt; as faint &quot;clouds&quot; of luminescence in the sky, the way you can see Andromeda now from Earth. Any light pollution from cities nearby would kill it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There would be no individual &lt;strong&gt;stars&lt;/strong&gt; visible to the naked eye, since all the stars you can see that way must be very, very close (within your own galaxy, if you're located in one). It goes without saying that naked-eye &lt;strong&gt;star clusters&lt;/strong&gt; that you can see from Earth (like the Pleiades) would not be visible there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;strong&gt;nebulae&lt;/strong&gt; visible to the naked eye from a place far away from cities on Earth, like the Great Orion Nebula, but there would be no such thing visible from your star system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So most people (city-dwellers) in a modern civilization in a place like that would not see anything in the night sky except planets (if any).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that the Milky Way and Andromeda are fairly close, as galaxies go. If your system was outside a cluster, and had no planets, then the sky would be completely empty to the naked eye no matter where you're looking from. Only telescopes would be able to see anything.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So there could be a civilization that exists in which the night is completely void of light?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;unlikely but yes (unless they make telescopes)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;that would suck&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may also retard the development of astronomy, cosmology and fundamental physics - especially if they had no other planets and moons.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-02-12T20:02:46.093" LastActivityDate="2015-02-12T20:02:46.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8867" PostTypeId="2" ParentId="888" CreationDate="2015-02-13T11:41:12.580" Score="2" Body="&lt;p&gt;There are two faintly visible areas of nebulosity on the above image taken from the Palomar Sky Survey, that correlate with the remnants of Supernova 1572. Sidney van den Bergh imaged this object in 1970 with the Palomar 200 inch telescope, comparing this with similar images taken by Walter Baade in 1949, 1955 and 1957. His findings were published in the Astrophysical Journal (168: 37-39, 1971 August 15). Using my 32&quot; telescope, two areas of nebulosity were recovered visually in August, 2006. These were at: 00h 25m 06.9s, +64d 11m 37s; and 00h 25m 51.4s, +64d 09m 19s. &lt;/p&gt;&#xA;" OwnerUserId="5539" LastActivityDate="2015-02-13T11:41:12.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8868" PostTypeId="1" AcceptedAnswerId="8870" CreationDate="2015-02-13T14:10:20.640" Score="3" ViewCount="3245" Body="&lt;p&gt;Given that Jupiter is 5 AU from our sun and a remote observer viewing our solar system from some other part of our galaxy looks at it. We assume that Jupiter's radius is $11 \times 6700$ km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the relative brightness (luminosity?) of Jupiter relative to our sun's? &lt;/p&gt;&#xA;" OwnerUserId="5542" LastEditorDisplayName="user2449" LastEditDate="2015-02-14T14:09:03.107" LastActivityDate="2015-02-14T14:09:03.107" Title="How would Jupiter's brightness relative to our sun seem to a remote observer (observing from a remote star)" Tags="&lt;astrophysics&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8869" PostTypeId="2" ParentId="8868" CreationDate="2015-02-13T19:41:56.077" Score="2" Body="&lt;p&gt;The absolute magnitude of Jupiter is at best +26 (give or take, depending on how you look at it). The absolute magnitude of the Sun is +4.8. There is a 21 magnitudes difference between them, or a ratio of 2.5 * 10^8 (250 million times brighter), which is huge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a 10 parsec distance (basically in our galactic neighborhood, where most visible stars are), the Sun and Jupiter would be approx 0.5 arcsec from each other at best, which is a very tiny angular distance, and there would also be the mentioned 250 mil brightness ratio (or more). It would be very difficult to tell Jupiter from the glare of the Sun with current technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At that distance, apparent magnitude and absolute magnitude are equal. So the Sun would be a +5 star (difficult to see with the naked eye, but doable in a dark sky), whereas Jupiter would be a +26 object, impossible to see except in a very large telescope (assuming it would be far from any star, which is NOT the case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TLDR: Even from not far away, Jupiter would be quickly overwhelmed by Sun's glare, and would be pretty weak anyway.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2015-02-13T19:41:56.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8870" PostTypeId="2" ParentId="8868" CreationDate="2015-02-13T19:47:45.517" Score="6" Body="&lt;p&gt;The absolute visual magnitude of the Sun is about 4.8. (This is the Sun's brightness viewed from 10 parsecs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The brightest visual magnitude for Jupiter is -2.7, when it is about 4 au from the Earth. Using the usual formula we can therefore calculate the absolute magnitude for Jupiter as 25.9. (Note that this is way brighter than the faintest objects seen for instance with the Hubble Space Telescope; but as Jupiter would be separated from the Sun by only 0.5 arcseconds, a telescope with even better angular resolution than the HST would be needed to pick it out of the glare from the Sun). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus to first order Jupiter is 21 visual magnitudes fainter than the Sun (a factor of 250 million) . The situation is slightly worse than this because as seen from outside the solar system, Jupiter could not be &lt;em&gt;fully&lt;/em&gt; illuminated, because it must have an angular separation from the Sun to be visible at all. A half-illuminated Jupiter would be at least a magnitude fainter (actually a factor $\pi$ for a Lambertian reflector).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But all is not lost. If we we to switch to the infrared, the situation is far more favourable. The approximate ratio of Jupiter/Sun brightness improves from about $1.4 \times10^{-9}$ in the visible to about $2.8\times10^{-8}$ at a wavelength of 10 microns (&lt;a href=&quot;http://www.amnh.org/content/download/53052/796511/file/DirectImagingChapter.pdf&quot; rel=&quot;nofollow&quot;&gt;Traub &amp;amp; Oppenheimer (2010)&lt;/a&gt;. But this is still a 19 magnitude difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The technology to achieve these sorts of contrasts is not quite there yet, even if we were viewing from nearby stars.  An example of what is currently possible can be seen with the discovery of a few Jupiter mass planet around the star GJ 504 by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013ApJ...774...11K&quot; rel=&quot;nofollow&quot;&gt;Kuzuhara et al. (2013)&lt;/a&gt;. The star-planet contrast was about 15-16 magnitudes at near-infrared wavelengths of 1-4 microns and with a star-planet separation of 2.5 arcseconds (equivalent to Jupiter as seen from only 2 parcsecs).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-02-14T09:44:35.163" LastActivityDate="2015-02-14T09:44:35.163" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8871" PostTypeId="2" ParentId="3595" CreationDate="2015-02-13T20:47:12.663" Score="2" Body="&lt;p&gt;Narrow answer: if all masses were equal, it would be a 3-body Klemperer rosette. It is known that such trivial KR configurations are not stable in the long term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Klemperer_rosette&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Klemperer_rosette&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2015-02-13T20:47:12.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8872" PostTypeId="2" ParentId="8847" CreationDate="2015-02-14T00:20:58.223" Score="2" Body="&lt;p&gt;When the light from two telescopes is combined to make an interferometer, the resolution of the combined instrument is equal to the resolution of a single telescope with a diameter equal the the distance between the two (or more) smaller telescopes. This is the technology you need to view exo-planets to any resolution you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has been incredibly successful in radio astronomy. Long baseline radio astronomy gives the resolution of a radio telescope the size of the diameter of the Earth. Optical telescopes in space can be even bigger. Arrays of relatively small telescopes with specially calculated spacings can do even better. Big mirrors are still better in order to get enough light on a sensor to be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The twin Keck telescopes in Hawaii can form an interferometer but funding has dried up at the moment. It has an effective size of 280 feet. I don't know why there is no funding, but at this point it may be likely that a space-based interferometer would work so well that finishing the ground based instrument is not cost effective. Or the time required to set up and get data is a poor trade-off compared to other demands for telescope time. Also, longer IR is better for the planet imaging and space is a lot better place for that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some reason, this is a hot topic among PhD candidates, but a delay in getting pictures of these objects is not going to mean much for the rest of us.&lt;/p&gt;&#xA;" OwnerUserId="5526" LastActivityDate="2015-02-14T00:20:58.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8874" PostTypeId="2" ParentId="3595" CreationDate="2015-02-14T16:25:13.367" Score="3" Body="&lt;p&gt;Note:  Answering from a comment posted on  &lt;a href=&quot;https://space.stackexchange.com/questions/233/how-deep-is-the-force-well-of-l4-and-l5-lagrangian-points-of-earth-sun-set/4007?noredirect=1#comment21322_4007&quot;&gt;Space Exploration&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classical stability analysis of these libration points assumes that we are examining the motion of a particle whose dynamics are perturbed by the gravitational impacts of a primary and secondary mass, so as a bottom-line-up-front type of answer, the mass of T is negligible - so any large increases in mass will negate these assumptions.  Further, the stability analysis is &lt;em&gt;linear stability analysis&lt;/em&gt;, implying that the stability is only valid within a neighborhood of the equilibrium point, and very little information can be said about the non-linear behavior (However, an unstable equilibrium point will be unstable in the non-linear dynamics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that said, the critical mass value in the circular restricted three body problem (CR3BP) can be found from the following development, summarized from most major astrodynamics texts to include Vallado (1), Roy (2), Schaub (3), or the essential 1967 CR3BP text by Szebehely (4).  The linear variational equations of motion for small in-plane perturbations about the triangular libration points can be found as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ddot{\xi}=2\dot{\eta}+U_{xx}^*\xi + U_{xy}^{*}\eta\\&#xA;\ddot{\eta}=-2\dot{\xi}+U_{yx}^{*}\xi+U^{*}_{yy}\eta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\xi,\eta$ are the perturbations in the $x$ and $y$ directions in the CR3BP synodic frame, and $U^{*}_{..}$ is are partials of an artificial pseudo-potential function.  Essentially, the characteristic equation for this linear system is found as $\Lambda^2 + \Lambda + \frac{27}{4}\mu(1-\mu)=0$, where $\Lambda = \lambda^2$, $\lambda$ being an eigenvalue of the actual characteristic equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we let $g = 1-27\mu(1-\mu)$, the four roots of the system can be expressed as slightly complicated functions of $g$, but the eigenvalue behavior can be classified according to the value of $g$ as below:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$0 &amp;lt; g \leq 1$:  Pure imaginary eigenvalues, marginal stability&lt;/li&gt;&#xA;&lt;li&gt;$g = 0$:  Repeated eigenvalues; secular terms present; unstable&lt;/li&gt;&#xA;&lt;li&gt;$g &amp;gt; 0$:  Eigenvalues with positive reals; unstable&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The critical $\mu$ value ($\mu_c$) comes from setting $g=0$.  Solving this, we find that $\mu_c=\frac{1}{2}\left(1\pm\frac{\sqrt{69}}{9}\right)\approx0.0385$.  Again, &lt;em&gt;a key assumption in this development is that the mass of the third body is negligible&lt;/em&gt;.  A lot of systems of interest are below this critical mass value to include the Earth-Moon, Sun-Earth, Sun-Jupiter, etc.; however, some systems are definitely above this value -- consider the Pluto-Charon system with a $\mu$ value of approximately 0.1101. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1:  Vallado, D.A.  Fundamentals of Astrodynamics and Applications.  30 Jun, 2001.  Springer Science &amp;amp; Business Media.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2:  Roy, A.E.  Orbital Motion, 4th Ed. 31 Dec 2004.  CRC Press.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3:  Schaub, H.P.  Analytical Mechanics of Space Systems.  2003.  AIAA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4:  Szebehely, V.G.  Theory of Orbits in the Restricted Problem of Three Bodies.  June 1967.  Academic Pr.&lt;/p&gt;&#xA;" OwnerUserId="1349" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2015-02-14T16:25:13.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8875" PostTypeId="1" CreationDate="2015-02-14T19:26:01.033" Score="3" ViewCount="130" Body="&lt;p&gt;I am new to astronomy. Jupiter is now visible to the naked eye but I have difficulties in locating it with my (new Newtonian reflector) telescope. The lowest magnification I have is 36X, which I guess is part of the problem(?). Any tips for the beginner..? I feel that if I knew any star constellations, it would help in locating objects.&lt;/p&gt;&#xA;" OwnerUserId="5550" LastEditorUserId="5550" LastEditDate="2015-02-14T22:20:58.503" LastActivityDate="2015-02-15T23:34:01.317" Title="Difficulties in finding Jupiter" Tags="&lt;planet&gt;&lt;telescope&gt;&lt;newtonian-telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8877" PostTypeId="2" ParentId="8875" CreationDate="2015-02-14T23:58:08.030" Score="4" Body="&lt;p&gt;Go out in the daytime and practice lining up on the leaves of trees on a distant hill or some such. It's easier to find targets when they aren't against a nearly featureless black backdrop. &#xA;If you have a finder scope, likewise align it to the scope's view using a distant daytime target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At night, look first for the moon to get a feel for how aligning works.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="2031" LastEditDate="2015-02-15T00:57:59.060" LastActivityDate="2015-02-15T00:57:59.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8879" PostTypeId="2" ParentId="2545" CreationDate="2015-02-15T09:39:17.160" Score="2" Body="&lt;p&gt;The &lt;a href=&quot;https://astronomy.stackexchange.com/a/2574/5524&quot;&gt;answer by Florin Andrei&lt;/a&gt; comprehensively explains scientific and Earth based observation support for the heliocentric model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we include interplanetary probes in the discussion, then &lt;strong&gt;yes, we have observed the Earth [orbiting] the Sun&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to receive commands from controllers on Earth the probes track (read calculate and observe) the location of Earth in order to point their radio dishes. Without accurate pointing, they are not able to receive the radio signal from Earth. Accurate pointing is also required to transmit results to Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And some of those results have been photos of the Earth, in space right where the heliocentric model says it should be.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;While not showing the Earth orbiting the Sun, there is a movie showing the Moon orbiting Earth &lt;a href=&quot;http://sploid.gizmodo.com/first-video-ever-to-capture-the-moon-rotating-earth-is-1480576442&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://www.youtube.com/watch?v=_CzBlSXgzqI&amp;amp;feature=youtu.be&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5524" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-02-18T03:45:26.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8881" PostTypeId="1" CreationDate="2015-02-15T22:13:20.810" Score="2" ViewCount="247" Body="&lt;p&gt;It's been known that the Earth's axial tilt varies from 22.5 to 24.5. Does this result in a wobbling while the value of the tilt varies about these two limits? If so, how does it affect the mechanics of the Earth's movement (its rotation and revolution about the Sun)?&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2015-11-14T07:44:30.370" Title="Precessing of the Earth" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;earth&gt;&lt;rotation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8882" PostTypeId="2" ParentId="8881" CreationDate="2015-02-15T22:57:40.253" Score="1" Body="&lt;p&gt;The Wikipedia &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Precession&quot; rel=&quot;nofollow&quot;&gt;Precession&lt;/a&gt; article explains this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precession of the Earth's axis is caused mainly by the Sun's gravity acting on the oblate spheriod of the Earth. The Earth bulges at the equator and the force of the Sun's gravity on the bulge makes the axis wobble like a child's spinning top - the axis traces out a cone shape in space. It has no effect on the Earth's orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perihelion precession is caused by the gravity of Jupiter and the other outer planets.&lt;/p&gt;&#xA;" OwnerUserId="5524" LastActivityDate="2015-02-15T22:57:40.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8887" PostTypeId="2" ParentId="8567" CreationDate="2015-02-16T15:19:18.220" Score="1" Body="&lt;p&gt;I think your question is a reframing of what is known as &quot;Obler's paradox&quot; - namely if the universe is infinite why is the night sky not white, as sooner or later our line of sight hits a star, and even if very far away there would be infinite stars out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to this is either (a) the Universe is not infinite or (b) the Universe has not been here for ever, so even if it is infinite, light from very far away is yet to reach us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case (b) is generally accepted - ie the Universe began a finite time ago in the &quot;big bang&quot; - though (a) is disputed - ie it may be that the universe is not infinite in any case.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2015-02-16T15:19:18.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8888" PostTypeId="1" AcceptedAnswerId="8889" CreationDate="2015-02-16T17:01:52.110" Score="1" ViewCount="3986" Body="&lt;p&gt;Telescope: Orion 09843 SpaceProbe 3,3 inch aperture(76mm), 700mm focal length, eyepieces - 25mm and 10mm focal lengths&#xA;Hi, I am very happy and excited to   see moon craters, Jupiter and Saturn. And planets like Mars, Venus, and Uranus are just looks &#xA;like a brighter star but not able to see its surface color. Similarly Jupiter looks like white small ball &#xA;instead of an orange surface. I am very much excited and  want to go bit closer. So please guide me list of eyepieces that I can use with this scope at least to see Jupiter  surface and big moon craters.&lt;/p&gt;&#xA;" OwnerUserId="5568" LastEditorUserId="7472" LastEditDate="2019-01-31T22:00:32.223" LastActivityDate="2019-01-31T22:00:32.223" Title="Which eyepieces I can use it for best viewing experience using my existing telescope?" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8889" PostTypeId="2" ParentId="8888" CreationDate="2015-02-16T18:48:45.393" Score="1" Body="&lt;p&gt;Every scope has a minimum and maximum useful magnification (minimum on your scope is 18X and maximum is 152X - from specs on Telescope.com). Anything outside of those numbers will compromise your ability to see well through the scope. To figure out how best to make use of your scope within the restrictions of useful magnification, you need to start with how to calculate magnification from focal length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnification is equal to the focal length of the telescope divided by the focal length of the eyepiece. For example, your 25mm eyepiece would allow you to get 28X magnification (700mm / 25mm = 28X). Using the same formula, your 10mm eyepiece would allow you to get 70X magnification. For an easy way to get close to 152X, you could purchase a simple 2X Barlow. (700mm / 10mm) * 2X = 140X. For now, start with your 25mm eyepiece and get used to the night sky through it. You can always step it up a bit with other eyepieces later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you want to see Jupiter? I think you should just barely be able to make out the dark bands in the atmosphere around 100X (so maybe a 6mm or 7mm eyepiece for your scope). You will definitely be able to see the Galilean Moons though. And speaking of moons, you should get a moon filter to cut down on the brightness when looking at our Moon. I've found it's best to observe the edge of the Moon where light turns to dark. The shadows created along the craters and mountains are magnificent to see. Enjoy!&lt;/p&gt;&#xA;" OwnerUserId="5569" LastActivityDate="2015-02-16T18:48:45.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8890" PostTypeId="2" ParentId="8642" CreationDate="2015-02-16T19:43:49.387" Score="2" Body="&lt;p&gt;I'd suggest a good star atlas and look for stars that are close to that meridian. Depending on how accurate you want to be you can use a telescope to get closer (ie to observe stars that are not visible to the naked eye).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what is it that you are trying to do? Set setting circles on a telescope? You don't need to look for RA=0h0'00&quot; then - simply look for an object with a known RA and Dec.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2015-02-16T19:43:49.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8891" PostTypeId="2" ParentId="8418" CreationDate="2015-02-16T19:52:15.430" Score="1" Body="&lt;p&gt;The Right Ascension of a star is fixed (ignoring precession of the equinoxes). What you need or want to know is the &quot;sidereal time&quot; - that is to say the &quot;time&quot; based on which RA is at your local meridian at a given time. The sidereal day is slightly shorter than the mean solar day (ie the 24 hour day) and so sidereal and solar times drift apart - only being equivalent at the vernal equinox.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2015-02-16T19:52:15.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8894" PostTypeId="1" AcceptedAnswerId="8895" CreationDate="2015-02-16T23:00:22.900" Score="8" ViewCount="6985" Body="&lt;p&gt;Partly inspired by &lt;a href=&quot;https://astronomy.stackexchange.com/q/7663/1909&quot;&gt;this question&lt;/a&gt;, the end result of stellar fusion is Iron-56 (eventually, after some decay). Then it explodes. How/where then to the heavier elements come from. The answer &lt;a href=&quot;https://astronomy.stackexchange.com/a/1411/1909&quot;&gt;here&lt;/a&gt; mentioned two colliding stars form gold. What process is taking place to form this gold (and I'd assume the rest of the heavier elements)?&lt;/p&gt;&#xA;" OwnerUserId="1909" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2019-01-29T08:49:51.320" Title="Where do the heavier elements come from?" Tags="&lt;star&gt;&lt;element&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8895" PostTypeId="2" ParentId="8894" CreationDate="2015-02-16T23:38:40.380" Score="14" Body="&lt;p&gt;The processes at work are the &lt;a href=&quot;https://en.wikipedia.org/wiki/R-process&quot;&gt;r-process&lt;/a&gt; and the &lt;a href=&quot;https://en.wikipedia.org/wiki/S-process&quot;&gt;s-process&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The r-process occurs in supernovae. Under the extreme conditions leading up to it, heavy elements are initially blocked from undergoing beta decay to other elements. If a heavy atom captures a nearby neutron, it can go to a heavier isotope or element. This happens multiple times to an atom, producing heavy elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/a/a7/Rapid_neutron_capture.svg&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/a/a7/Rapid_neutron_capture.svg&quot; alt=&quot;r-process&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The s-process is slower, and occurs under less dramatic conditions in asymptotic giant branch (AGB) stars - a certain type of red giant. It also uses neutron capture and the subsequent decay of that neutron into a proton. Over thousands of years, atoms climb to become heavier elements. The s-process, though, requires that there already be some heavy elements in existence, unlike the r-process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/4/4e/S-process-elem-Ag-to-Sb.svg&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/4/4e/S-process-elem-Ag-to-Sb.svg&quot; alt=&quot;s-process&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information, &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/Astro/nucsyn.html&quot;&gt;Hyperphysics&lt;/a&gt;, &lt;a href=&quot;http://www.mso.anu.edu.au/~akarakas/Kodai_Lecture4.pdf&quot;&gt;this&lt;/a&gt; lecture and &lt;a href=&quot;https://www-alt.gsi.de/forschung/kp/kp2/nuc-astro/HeavyElements_e.html&quot;&gt;this&lt;/a&gt; German site (in English) are very useful.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2015-02-16T23:38:40.380" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8899" PostTypeId="2" ParentId="8894" CreationDate="2015-02-17T08:47:46.353" Score="10" Body="&lt;p&gt;HDE226868's answer is perfectly correct. Just to add the following important information though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The s-process is limited to elements lighter than lead and produces little peaks of abundance at certain elements along the s-process path - eg. Ba, Sr, Eu, Y. These are often called s-process elements. All the elements heavier than lead are produced by explosive r-process nucleosynthesis in supernova explosions, colliding neutron stars etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The split between r-process and s-process production of heavier than iron (peak) elements is about 50:50. ie They weren't mainly made in supernovae, which is a frequent, incorrect claim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An update is also in order regarding the site of the r-process. In the past it has been thought that most of the r-processing occurs in the early stages of a supernova explosion. However, theoretical models have really struggled to get the environment as neutron-rich as required to produce the heavier elements - certainly those up around gold, platinum, osmium etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the last few years it has become more popular to think that merging neutron star are the main site of the r-process. This hypothesis received a boost with the observation of a &lt;a href=&quot;https://en.wikipedia.org/wiki/GW170817&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravitational wave source&lt;/a&gt; source that appeared to be that of a merging neutron star system coinciding with a &lt;a href=&quot;https://en.wikipedia.org/wiki/Kilonova&quot; rel=&quot;nofollow noreferrer&quot;&gt;kilonova&lt;/a&gt; explosion seen in gamma rays, visible and infrared light. The infrared spectrum of the decaying light from this explosion betrays the presence of &quot;lanthanide&quot; elements that are produced in the r-process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note however that the rate and yield of such events is highly uncertain. There is plenty of good work which suggests that rare types of core-collapse supernovae (aka &quot;collapsars&quot;) are still in fact the dominant source of r-process elements in the Milky Way (&lt;a href=&quot;http://ttps://arxiv.org/abs/1901.09044&quot; rel=&quot;nofollow noreferrer&quot;&gt;Siegel 2019&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relative contributions of various sites to the r-process remains an unsettled matter. You could also read my answers on this topic in Physics Stack Exchange.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/231981/heavy-element-production-from-supernova&quot;&gt;https://physics.stackexchange.com/questions/231981/heavy-element-production-from-supernova&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/7131/origin-of-elements-heavier-than-iron-fe&quot;&gt;https://physics.stackexchange.com/questions/7131/origin-of-elements-heavier-than-iron-fe&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-01-29T08:49:51.320" LastActivityDate="2019-01-29T08:49:51.320" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8900" PostTypeId="2" ParentId="4741" CreationDate="2015-02-17T09:01:34.627" Score="0" Body="&lt;p&gt;I am not from astro background. but from what I understood about the diagram, why not try the third axis to be time?. hope u he some data for the time-age for stars in the main sequences and the age for few stars in the giants /WD region. try plotting the available star-age as third axes and try to see what data u get for the other star. I think u will get something like a tilted plane with a wavy pattern near the unstable region.&lt;/p&gt;&#xA;" OwnerUserId="5579" LastActivityDate="2015-02-17T09:01:34.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8901" PostTypeId="1" AcceptedAnswerId="8902" CreationDate="2015-02-17T10:51:17.520" Score="0" ViewCount="207" Body="&lt;p&gt;I have read quite different opinions about that, so can someone tell me what is the sign of the acceleration of the stars in the Universe if we accept that the center of the Universe is the beginning of the coordinate system? At least what is considered as correct finally :)&lt;/p&gt;&#xA;" OwnerUserId="5580" LastActivityDate="2015-02-17T12:12:18.960" Title="What is the acceleration of the stars' speed in the Universe? Positive or negative?" Tags="&lt;star&gt;&lt;universe&gt;&lt;speed&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ClosedDate="2015-02-19T15:19:03.620" ContentLicense="CC BY-SA 3.0" />
  <row Id="8902" PostTypeId="2" ParentId="8901" CreationDate="2015-02-17T12:12:18.960" Score="2" Body="&lt;p&gt;This is the classic misconception about the expansion of the Universe: &lt;strong&gt;there is no center of the Universe&lt;/strong&gt;. The whole universe expands as one, &lt;strong&gt;not from a single point nowhere&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing this it makes no sense speaking of the direction of an object due to the Universe. You could compare the directions of two objects in it though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore star's direction depend on where and how they where created and also the interaction with other stars and their home galaxy.&lt;/p&gt;&#xA;" OwnerUserId="1678" LastActivityDate="2015-02-17T12:12:18.960" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8903" PostTypeId="1" AcceptedAnswerId="9907" CreationDate="2015-02-17T12:56:19.247" Score="2" ViewCount="2462" Body="&lt;p&gt;The Hercules–Corona Borealis Great Wall is &quot;to date (February 16, 2015), it is the largest and the most massive structure known in the observable universe&quot; (&lt;a href=&quot;https://en.wikipedia.org/wiki/Hercules%E2%80%93Corona_Borealis_Great_Wall&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;). How many galaxies is the Hercules–Corona Borealis Great Wall believed to contain?&lt;/p&gt;&#xA;" OwnerDisplayName="user5581" LastActivityDate="2021-03-23T21:53:33.200" Title="How many galaxies are there in the Hercules–Corona Borealis Great Wall?" Tags="&lt;galaxy-cluster&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9902" PostTypeId="1" CreationDate="2015-02-17T14:15:00.077" Score="1" ViewCount="598" Body="&lt;p&gt;How difficult is it to make a telescope oneself? How much time would it take for a first-timer? What are the typical pitfalls?&lt;/p&gt;&#xA;" OwnerUserId="6581" LastActivityDate="2015-02-17T23:02:44.687" Title="How long does it typically take to make a telescope yourself for the first time?" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-02-19T15:18:40.577" ContentLicense="CC BY-SA 3.0" />
  <row Id="9903" PostTypeId="2" ParentId="8782" CreationDate="2015-02-17T17:12:05.803" Score="11" Body="&lt;p&gt;The discussion of the Cosmological Principle above is very relevant, but it is possible that so is a (weak) application of the anthropic principle - in other words if we were in a region of extremely energetic physical phenomena, such as quasars, we would be unlikely to exist - as the evidence suggests that the development of intelligent life takes a considerable time and highly energetic events are likely to disrupt that.&lt;/p&gt;&#xA;" OwnerUserId="823" LastEditorUserId="823" LastEditDate="2015-02-18T09:15:42.557" LastActivityDate="2015-02-18T09:15:42.557" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="9904" PostTypeId="2" ParentId="8805" CreationDate="2015-02-17T17:54:21.300" Score="1" Body="&lt;p&gt;There is apparently an option on the STSI (Space Telescope Science Institure) website to &lt;a href=&quot;http://www.stsci.edu/institute/software_hardware/spike/try&quot; rel=&quot;nofollow&quot;&gt;try their Spike implementation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't clearly answer your question, as it's not open, and I don't know the language that has been used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is a lot of information in the documents you provided and on the linked website, so maybe you can start your own implementation of the algorithm? I'd be happy to help!&lt;/p&gt;&#xA;" OwnerUserId="4188" LastActivityDate="2015-02-17T17:54:21.300" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9905" PostTypeId="2" ParentId="8642" CreationDate="2015-02-17T18:14:48.293" Score="1" Body="&lt;p&gt;It depends on the your current date, time of day and your location since, from your point of view, Right Ascension (in hours) is always progressing in the same direction as the stars are moving. Remember that RA &quot;0&quot; is the same as RA &quot;24&quot; and is a line that runs from the north celestial pole to the south celestial pole. If we are trying to find the actual &quot;line&quot; in the sky that corresponds to RA &quot;0&quot; we have a couple of options. One is to obtain a star chart, find any star that is on that line, locate it in the sky and you are there. If you can't find a star chart, figure out the &quot;Local Apparent Sidereal Time&quot; (LAST) which will be the same as whatever RA is directly overhead (at your location) at that moment and you can estimate from that or wait until your &quot;LAST&quot; time is &quot;0&quot;, at which time it will be directly overhead. Star charts as well as your current &quot;LAST&quot; time can be found directly on many astronomy smartphone apps or can be calculated (another question).&lt;/p&gt;&#xA;" OwnerUserId="5497" LastActivityDate="2015-02-17T18:14:48.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9907" PostTypeId="2" ParentId="8903" CreationDate="2015-02-17T21:04:13.453" Score="7" Body="&lt;p&gt;I'm sorry for the long answer. The result is in the bottom. :)&lt;/p&gt;&#xA;&lt;p&gt;The number of galaxies in the &amp;quot;wall&amp;quot;, as in any volume of space, is the number density &lt;span class=&quot;math-container&quot;&gt;$n_{\mathrm{gal}}$&lt;/span&gt; times its volume &lt;span class=&quot;math-container&quot;&gt;$V_{\mathrm{wall}}$&lt;/span&gt;. However, the number density of galaxies depends on their masses/sizes: Small galaxies are much more common than large galaxies. Usually we describe this by the &lt;em&gt;halo mass function&lt;/em&gt; (HMF), which gives the (differential) number density dN/dM of dark matter halos hosting galaxies as a function of halo mass M. Depending on your favorite cosmology, it looks something like this (created with &lt;a href=&quot;http://hmf.icrar.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; tool):&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/R7qSk.png&quot; alt=&quot;Halo mass function&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The solid line shows the HMF &lt;em&gt;today&lt;/em&gt; in the local Universe, i.e. at redshift zero, while the dashed and dotted line shows how it was at redshift 1.6 and 2.1, respectively, roughly 10 billion years ago, which mark the beginning and end of the wall (according to &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014A%26A...561L..12H&quot; rel=&quot;nofollow noreferrer&quot;&gt;Horváth et al. 2014, A&amp;amp;A, 561, L12&lt;/a&gt;). Notice how the number of large halos grow with time, at the expense of small halos. This is due to the smaller ones accreting onto larger ones.&lt;/p&gt;&#xA;&lt;p&gt;The HMFs are characterized by a power law at low masses with an exponential cut-off at high masses. To get the &lt;em&gt;total&lt;/em&gt; number density we need to integrate over the entire range of masses, and here comes the problem: Whereas the upper limit is set by the maximum size a galaxy can have (roughly &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_galaxies&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$3\times10^{13}M_\odot$&lt;/span&gt;&lt;/a&gt;, but the exact number isn't very important, as these galaxies are so rare), the lower limit depends on your definition of a galaxy. Large galaxies have smaller galaxies orbiting as satellites, and there isn't a well-defined threshold for when a clump of gas, stars, and dark matter is called a galaxy. &lt;a href=&quot;http://www.mnn.com/earth-matters/space/stories/just-how-big-is-the-smallest-galaxy-in-the-universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;These guys&lt;/a&gt; report on the detection of a &lt;span class=&quot;math-container&quot;&gt;$\sim10^3M_\odot$&lt;/span&gt; clump which they call a galaxy. That's probably not what you think of as a galaxy, so for the sake of this exercise, let's use Small Magellanic Cloud-sized galaxies as our lower minimum, i.e &lt;span class=&quot;math-container&quot;&gt;$\sim 6.5\times10^9 M_\odot$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The total number density of galaxies is thus found by integrating the HMF. Luckily, the online HMF tool can do that as well. The result, i.e. &lt;em&gt;the number density of halos above a given mass&lt;/em&gt;, looks like this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DqQlq.png&quot; alt=&quot;Cumulative HMF&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Note that the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; axis is given in terms of &lt;span class=&quot;math-container&quot;&gt;$h^{-1}M_\odot$&lt;/span&gt;. Since &lt;span class=&quot;math-container&quot;&gt;$h \simeq 0.7$&lt;/span&gt;, the mass of the SMC is &lt;span class=&quot;math-container&quot;&gt;$M_\mathrm{SMC} \simeq 5\times10^9 h^{-1}M_\odot$&lt;/span&gt;. From the graph, the number density of these objects is &lt;span class=&quot;math-container&quot;&gt;$\sim0.5\,h^3\mathrm{Mpc}^{-3} = 0.17\,\mathrm{Mpc}^{-3}$&lt;/span&gt; (1 Mpc, or megaparsec, equals 3.26 million lightyears). In principle we should subtract the number density of halos above &lt;span class=&quot;math-container&quot;&gt;$3\times10^{13}M_\odot$&lt;/span&gt;, but since these are four orders of magnitude smaller, they can be neglected.&lt;/p&gt;&#xA;&lt;p&gt;The wall starts at redshift ~1.6 and ends at redshift ~2.1. This translates into starting at a distance of 4.65 Gpc (billion parsec) and ending at 5.47 Gpc, i.e. a depth of ~800 Mpc. These numbers are in &lt;em&gt;comoving&lt;/em&gt; coordinates, i.e. what it has expanded to today, as opposed to &lt;em&gt;physical&lt;/em&gt; coordinates, i.e. what it was at the time it emitted the light that we see today. But the HMF is given in terms of comoving volume, so that's cool.&lt;/p&gt;&#xA;&lt;p&gt;To get the volume, we also need the area. The finders report that it's roughly 2 Gpc times 3 Gpc, but they don't seem to state whether this is in physical or comoving coordinates (which differ by a factor (1+z) ~ 3). I suspect it's comoving, though.&lt;/p&gt;&#xA;&lt;p&gt;Thus, the total volume is &lt;span class=&quot;math-container&quot;&gt;$V_{\mathrm{wall}} = 0.8 \times 2 \times 3\,\mathrm{Gpc}^3 = 4.8\,\mathrm{Gpc}^3 = 4.8\times10^9\,\mathrm{Mpc}^3$&lt;/span&gt;, or 166 cubic gigalightyears. Here, I have ignored the fact that the physical area spanned by the angle of observation is a bit different at the front and the back of the wall, but since these numbers are already quite uncertain, it's of lesser importance.&lt;/p&gt;&#xA;&lt;p&gt;The considerations above are all for the average Universe. But the wall represents an &lt;em&gt;overdensity&lt;/em&gt; in space, so the number will be larger. I can't find any estimates of the overdensity of this particular supercluster, but in general galaxies in superclusters are more numerous by a factor of &lt;span class=&quot;math-container&quot;&gt;$\delta \sim 5$&lt;/span&gt;–&lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt;. Let's use 5.&lt;/p&gt;&#xA;&lt;p&gt;That is, &lt;strong&gt;the total number of galaxies the size of the Small Magellanic Cloud or above is&lt;/strong&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$N_\mathrm{gal,tot} = n_\mathrm{gal} \times V_\mathrm{wall} \times \delta$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$= 0.17\,\mathrm{Mpc}^{-3} \times 4.8\times10^9\,\mathrm{Mpc}^3 \times 5$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\sim 4 \mathrm{\,billion\,galaxies}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;But if you go through the same exercise for halos of masses above, say, &lt;span class=&quot;math-container&quot;&gt;$10^3\,M_\odot$&lt;/span&gt;, you'll find a number which is a million times larger.&lt;/p&gt;&#xA;&lt;p&gt;In reality, the number density of &lt;em&gt;luminous&lt;/em&gt; galaxies is smaller. At the high-mass end, the gas has trouble cooling sufficiently to condense and create stars, and active galactic nuclei in massive galaxies heat the gas. At the low-mass end, gas is more easily blown out of the potential well by stellar feedback from supernovae.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2021-03-23T21:53:33.200" LastActivityDate="2021-03-23T21:53:33.200" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="9909" PostTypeId="2" ParentId="9902" CreationDate="2015-02-17T23:02:44.687" Score="1" Body="&lt;p&gt;Making your own telescope is very do-able given basic working skills. Usually a DOB or Newtonian reflector is a good choice. You can buy a mirror and components all ready to assemble or you can choose to grind your own (which is more difficult). If you have an astronomy club in your area check to see if they can help putting your project together. If not, check online ( Youtube, etc) for guidance. Websites like Edmonds Optics may still be a good choice for components. Once you get all the components together, they can be assembled in a pretty short time. Good Luck!  &lt;/p&gt;&#xA;" OwnerUserId="5497" LastActivityDate="2015-02-17T23:02:44.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9910" PostTypeId="2" ParentId="8888" CreationDate="2015-02-17T23:16:18.667" Score="1" Body="&lt;p&gt;Often it helps to use colored filters that screw into the base of your eyepieces. They can increase the contrast of the details you are looking for. They are not very expensive and work quite well for this. You can also find used ones on many of the amateur astronomy sites that have a &quot;classified&quot; section. Different colors work better depending on the colors of the object you are observing. &lt;/p&gt;&#xA;" OwnerUserId="5497" LastActivityDate="2015-02-17T23:16:18.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9911" PostTypeId="1" AcceptedAnswerId="9960" CreationDate="2015-02-17T23:41:05.977" Score="2" ViewCount="1299" Body="&lt;p&gt;I'd like to estimate the typical resolving power of the largest optical telescopes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've calculated the theoretical resolving power of the VLA for 21 cm light, $$R=\frac{\lambda}{B}=\frac{2.1 \times 10^{-1} m}{2.7 \times 10^4 m}=1.6 ''$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I know that the largest optical telescopes on Earth are GTC, Keck 1 &amp;amp; 2, SALT, VLT, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't know how to calculate a typical resolving power in practice. How can I make this estimate taking into account astronomical seeing and other effects?&lt;/p&gt;&#xA;" OwnerDisplayName="user5341" LastActivityDate="2015-03-24T11:22:06.180" Title="What is the typical resolving power of the largest optical telescopes?" Tags="&lt;telescope&gt;&lt;angular-resolution&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9912" PostTypeId="1" AcceptedAnswerId="9913" CreationDate="2015-02-17T23:56:20.627" Score="6" ViewCount="733" Body="&lt;p&gt;According to &lt;a href=&quot;https://web.njit.edu/~gary/202/Lecture6.html&quot;&gt;this lecture&lt;/a&gt;,&quot;Radio astronomy has several advantages over optical astronomy...We can make the highest resolution images, and see things happening on the smallest scales.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this? If resolving power, $r \propto \frac{\lambda}{B}$ then shouldn't shorter wavelength light result in smaller angular resolution i.e. better resolution?&lt;/p&gt;&#xA;" OwnerDisplayName="user5341" LastActivityDate="2015-08-28T18:14:56.687" Title="Why does radio astronomy offer higher resolution images than optical?" Tags="&lt;radio-astronomy&gt;&lt;angular-resolution&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9913" PostTypeId="2" ParentId="9912" CreationDate="2015-02-18T03:09:07.853" Score="5" Body="&lt;p&gt;The relation you cited holds for a single telescope. But, as also noted in the lecture you linked &#xA;&quot;One thing that is possible in radio astronomy is to use interferometry, which combines the signals from an array of antennas as if they were all part of the same aperture. That means that the resolving power of a radio telescope is not just what it would be for each antenna alone, but is as for a single telescope the size of the separation between antennas.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This basically means you use multiple telescopes - which in case of radio astronomy are called antennas for historical reasons - and combine their measurements into a single one. To get an intuition of how big of arrays we are talking about, google images of ALMA, LOFAR (which has stations in several countries) or SKA. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle other wavelengths, such as optical, can also be observed using interferometry, but it is technically much more complicated to build such instruments. On top of that, optical light is the one most affected by the atmosphere and it is hard enough to take it into account for one telescope, combining such effects to create an image from multiple ones is a very daunting task. &lt;/p&gt;&#xA;" OwnerUserId="5530" LastActivityDate="2015-02-18T03:09:07.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9914" PostTypeId="2" ParentId="9912" CreationDate="2015-02-18T04:42:29.103" Score="2" Body="&lt;p&gt;If you look at raw resolving power you get for a 10m optical telescope: $\theta_{optical}\approx \frac{5\times 10^{-7}}{10}\approx 5 \times 10^{-8}$ radian. For a synthetic aperture ground based radio telescope operating at 21cm the limiting aperture is of the order of the diameter of the Earth so: $\theta_{21cm}\approx \frac{21\times 10^{-2}}{12\times 10^{6}}\approx 1.75 \times 10^{-8}$ radian. So in this case the resolution of the radio telescope is somewhat better than that of the optical. In the absence of adaptive optics the situation is even more favorable to the radio telescope since we can achieve its theoretical resolution, but ground based optical telescopes are constrained to lower resolutions by atmospheric seeing. Also adaptive optics still do not improve the resolution of optical systems fully to the diffraction limit of the aperture.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-08-28T18:14:56.687" LastActivityDate="2015-08-28T18:14:56.687" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9915" PostTypeId="2" ParentId="8367" CreationDate="2015-02-18T15:12:15.807" Score="2" Body="&lt;p&gt;Mathematica. For details take a look at my simulation &lt;a href=&quot;http://demonstrations.wolfram.com/preview.html?draft/21067/000006/FourBody3DSimulation&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for gravity and &lt;a href=&quot;http://yukterez.ist.org/4kp2&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://yukterez.ist.org/4kp2&lt;/a&gt; for gravity plus electric charge.&lt;/p&gt;&#xA;" OwnerUserId="5431" LastEditorUserId="7472" LastEditDate="2018-07-14T18:09:46.880" LastActivityDate="2018-07-14T18:09:46.880" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9917" PostTypeId="1" AcceptedAnswerId="9918" CreationDate="2015-02-18T20:24:54.580" Score="10" ViewCount="3157" Body="&lt;p&gt;A while ago I saw Neil deGrasse Tyson comparing the number of stars in the universe with the number of words spoken by all of humankind, ever since.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize both of these numbers are not strictly defined, but still, we can use our best observations together with our best guesses to find out which number is larger, and by how many magnitudes. It's a good example of making you realize the size of &lt;em&gt;astronomical&lt;/em&gt; numbers in the true sense of the word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is your answer or guesstimate?&lt;/p&gt;&#xA;" OwnerUserId="6593" LastEditorUserId="6" LastEditDate="2016-05-25T15:38:00.647" LastActivityDate="2017-05-25T22:42:06.910" Title="Are there more stars than all the words ever spoken by humans?" Tags="&lt;fundamental-astronomy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9918" PostTypeId="2" ParentId="9917" CreationDate="2015-02-18T22:04:36.827" Score="16" Body="&lt;h3&gt;Number of stars in the observable universe&lt;/h3&gt;&#xA;&lt;p&gt;There are about 100 to 200 billion galaxies in the observable universe. Galaxies range in size from a few billion to hundreds of trillions stars. Using 100 billion galaxies and 1 trillion stars in a galaxy yields 10&lt;sup&gt;23&lt;/sup&gt; stars in the observable universe as a rough order of magnitude estimate. (After getting this result, I found multiple sources that give answers ranging from 10&lt;sup&gt;21&lt;/sup&gt; to 10&lt;sup&gt;24&lt;/sup&gt; stars in the universe.)&lt;/p&gt;&#xA;&lt;h3&gt;Number of words ever spoken&lt;/h3&gt;&#xA;&lt;p&gt;We're talkative, but we also sleep and eat and listen. Assuming we talk for the equivalent of 3 hours per day and we speak at about 150 words per minute, or about 10 million words per year. I recall 100 billion as being a rough estimate of the number of modern humans who have ever lived. Many of those people died during or shortly after  childbirth. Assuming an average life expectancy at birth of 25 years (generous), that means humans have spoken 2.5×10&lt;sup&gt;19&lt;/sup&gt; words since we first gained the gift of gab.&lt;/p&gt;&#xA;&lt;p&gt;That might be off by a factor of ten, but it's not off by a factor of ten thousand. There are a lot more stars in the observable universe than the number of words ever spoken by humans.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-02-18T22:04:36.827" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9919" PostTypeId="1" AcceptedAnswerId="9922" CreationDate="2015-02-19T01:51:59.030" Score="3" ViewCount="731" Body="&lt;p&gt;We know that a black hole behaves like a whirlpool or a tornado or any of the other rotating phenomena we experience on Earth.But the thing is, all these phenomena, except the black hole, are 2-Dimensional (plus 1 dimension of time) rotating vortices, they move objects from 2D planes to other 2D planes instantly, but through a 3D medium. My question is this, since a black hole is a rotating sphere(3D for that matter, plus 1D of time), does it mean that its endpoint in the vortex is 5D?&lt;/p&gt;&#xA;" OwnerUserId="6596" LastEditorUserId="661" LastEditDate="2015-02-19T06:07:57.533" LastActivityDate="2015-02-19T06:07:57.533" Title="Is a black hole a 5 dimensional vortex?" Tags="&lt;black-hole&gt;&lt;general-relativity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9920" PostTypeId="1" CreationDate="2015-02-19T01:59:20.110" Score="6" ViewCount="699" Body="&lt;p&gt;So, I've been starting to investigate radio astronomy, and am wondering about if certain things are possible from an amateur standpoint. I was looking at this powerpoint(that discusses building a tiny radio telescope), &lt;a href=&quot;http://nightsky.jpl.nasa.gov/download-redirect.cfm?Doc_ID=347&amp;amp;Doc_Filename=IBTcom1%2Eppt&quot; rel=&quot;noreferrer&quot;&gt;http://nightsky.jpl.nasa.gov/download-redirect.cfm?Doc_ID=347&amp;amp;Doc_Filename=IBTcom1%2Eppt&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Slides 11, 12, and 13, it shows various radio emissions overlayed on the orion constellation. As well as Slide 17 and 18.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering, is the data acquired in these images, doable from an amateur standpoint? From previous research I think I can build some detection devices for the Hydrogen Line emission(1420Mhz/21cm line), and that could lead to data(per my initial/basic understanding)  that's shown in image 1. Am I right in that assumption, and are the other emissions or data doable from an amateur standpoint?&lt;/p&gt;&#xA;" OwnerUserId="4161" LastActivityDate="2015-07-09T20:14:51.240" Title="Radio Astronomy and Imaging" Tags="&lt;radio-astronomy&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9921" PostTypeId="1" CreationDate="2015-02-19T02:06:26.023" Score="4" ViewCount="302" Body="&lt;p&gt;From inflation theory and other theories explaining the birth of the universe, the universe is expanding and recently it's been discovered that the expansion has been accelerating. My question is simple, what is happening on the boundary of our inflating universe and what is that 'space' that is experiencing that boundary?&lt;/p&gt;&#xA;" OwnerUserId="6596" LastActivityDate="2018-09-03T14:26:01.867" Title="What is our universe expanding into?" Tags="&lt;general-relativity&gt;&lt;cosmological-inflation&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9922" PostTypeId="2" ParentId="9919" CreationDate="2015-02-19T04:39:39.240" Score="4" Body="&lt;p&gt;The short answer is &lt;strong&gt;no&lt;/strong&gt;, with some caveats to the effect of &lt;em&gt;sort of&lt;/em&gt;, depending on how loose an analogy you want to make.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sound propagation in a fluid is limited by the speed of sound, which can be used to define a &quot;sound cone&quot; structure analogous to the causal light cone structure in spacetime. This is a described by an &lt;a href=&quot;https://en.wikipedia.org/wiki/Acoustic_metric&quot; rel=&quot;nofollow noreferrer&quot;&gt;acoustric metric&lt;/a&gt;, which could have an acoustic horizon when the speed of the fluid exceeds the speed of sound, and even an analogue of Hawking radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A general acoustic metric for a perfect fluid has the form&#xA;$$g_{\mu\nu}=\alpha^2\begin{bmatrix}-(c^2-v^2)&amp;amp;-\vec{v}\\-\vec{v}&amp;amp;\mathbf{1}\end{bmatrix}\text{,}$$&#xA;where $\alpha$ is a conformal factor. The Schwarzschild black hole can be put in this form, as in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gullstrand–Painlevé chart&lt;/a&gt; is not just spatially conformally flat, but exactly Euclidean at every instant of time. One can imagine that a Schwarzschild black hole is like a drain sucking space down to the singularity at the local escape velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rotating Kerr black hole &lt;em&gt;cannot&lt;/em&gt; be put in this form, although its equatorial slice can be. See Visser and Winfurtner (2005) for details, as well for a discussion as to why interpreting the metric as corresponding to a &lt;em&gt;physical fluid&lt;/em&gt; is problematic even in the Schwarzschild case due to the the conformal factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, Hamilton and Lisle (2008) found that the Doran chart of Kerr spacetime can be interpreted as a six-dimensional &quot;Lorentz river&quot; characterized not only by a velocity but also a twist. This really quite different from the acoustic fluid analogy, as the 'river' does not spiral inwards, but rather has an intrinsic twist that rotates infalling objects. Still, it is interesting in its own right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e8USN.gif&quot; alt=&quot;River model of Kerr black hole, by A. Hamilton&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Image by &lt;a href=&quot;http://jila.colorado.edu/~ajsh/insidebh/waterfall.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Andrew Hamilton&lt;/a&gt;. (Twist not shown.)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Visser, M., Weinfurtner, S. &quot;Vortex analogue for the equatorial geometry of the Kerr black hole&quot;, &lt;em&gt;Class. Quant. Grav.&lt;/em&gt; &lt;strong&gt;22&lt;/strong&gt;:2493-2510 (2005) [arXiv:&lt;a href=&quot;http://arxiv.org/abs/gr-qc/0409014&quot; rel=&quot;nofollow noreferrer&quot;&gt;gr-qc/0409014&lt;/a&gt;] &lt;/li&gt;&#xA;&lt;li&gt;Hamilton, A. J. S., Lisle, J. P., &quot;The river model of black holes&quot;, &lt;em&gt;Am. J. Phys.&lt;/em&gt; &lt;strong&gt;76&lt;/strong&gt;:519-532 (2008) [arXiv:&lt;a href=&quot;http://arxiv.org/abs/gr-qc/0411060&quot; rel=&quot;nofollow noreferrer&quot;&gt;gr-qc/0411060&lt;/a&gt;]&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="661" LastActivityDate="2015-02-19T04:39:39.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9923" PostTypeId="2" ParentId="9921" CreationDate="2015-02-19T07:45:27.740" Score="-4" Body="&lt;p&gt;Most of the Astronomers and the community of Astrophysics, will agree that the fate of planet earth is &quot;The Big rip, most comparable to The Big Bang;but polar opposites.Where the universe is expanding to it's breaking point, total expansion of the know universe. &lt;/p&gt;&#xA;" OwnerUserId="2672" LastActivityDate="2015-02-19T07:45:27.740" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9924" PostTypeId="1" AcceptedAnswerId="9931" CreationDate="2015-02-19T09:07:59.993" Score="2" ViewCount="3322" Body="&lt;p&gt;Retrograde motion is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The apparent motion of the planets when they appear to move backward (from east to west) from the direction that they move ordinarily with respect to the stars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can planets inside the Earth's orbit around the Sun appear to undergo retrograde motion?&lt;/p&gt;&#xA;" OwnerUserId="5532" LastEditorUserId="2153" LastEditDate="2015-02-19T16:17:21.713" LastActivityDate="2019-05-29T10:03:49.177" Title="Can planets inside the Earth's orbit around the Sun appear to undergo retrograde motion?" Tags="&lt;orbit&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9925" PostTypeId="2" ParentId="7863" CreationDate="2015-02-19T09:49:00.853" Score="1" Body="&lt;p&gt;A black hole being created by, say, the collapse of a star has a void on one side and matter  moving in one direction (towards the center) on the other getting denser as time progressed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Big Bang represents almost the exact opposite - all matter was surrounded by equal amount of  equally dense matter and all matter was moving away from each other. In such a uniform universe there is nothing to cause a singular collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much later, when the density and expansion rate were lower, there is a possibility of random motion gathering enough mass to create a black hole ... however, in this case you are probably talking about billions of black holes - which by now would be innumerable very large black holes.&lt;/p&gt;&#xA;" OwnerUserId="6603" LastActivityDate="2015-02-19T09:49:00.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9926" PostTypeId="1" AcceptedAnswerId="9966" CreationDate="2015-02-19T10:28:10.620" Score="3" ViewCount="281" Body="&lt;p&gt;If we blow up a global killer (in the 3 - 20km range) days before impact into enough pieces that none of the them exceeds, let's say, a 30m diameter and the impact pattern is dispersed over most of the athmosphere, would that be a possibility that the impact energy is soaked up with less catastrophic consequences than the hit of the asteroid in one piece? Or would the distributed energy heat up the athmosphere so much that it would kill higher life instantly or eventually over chemical/biological chain reactions?&#xA;Or would the early impacts heat up or disturb the athmosphere so much that later (in the range of seconds or minutes) impacts would experience not enough friction so that there would be a surface bombardement with millions of little pieces?&lt;/p&gt;&#xA;" OwnerUserId="844" LastActivityDate="2015-02-23T00:01:12.957" Title="Would a killer asteroid shattered into thousands of pieces produce the same devastation?" Tags="&lt;asteroids&gt;&lt;impact&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9927" PostTypeId="2" ParentId="9921" CreationDate="2015-02-19T11:39:54.277" Score="2" Body="&lt;p&gt;There is no boundary there is no outside to be expanding into. Only slightly misleading is the usual analogy of the expansion of the universe is that of the &lt;strong&gt;surface&lt;/strong&gt; of a balloon as it is blown up, which just gets larger but has no boundary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One problem with the balloon analogy is that with a balloon we can see that it is embedded in a 3-space, but there is no need for such a 4-space embedding for the Universe. Another problem is that the surface of the balloon is very definitely finite, we don't really know if that is the case or not for the Universe.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-02-19T16:51:21.693" LastActivityDate="2015-02-19T16:51:21.693" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9928" PostTypeId="2" ParentId="9926" CreationDate="2015-02-19T13:34:54.383" Score="12" Body="&lt;p&gt;TL;DR version: Too big and way, way too late.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dispersal can't be done, even at the lower end of that 3-20 km scale. Holsapple claims 5 kilojoule/kg are needed to disrupt and disperse a solid 1 km asteroid asteroid, with energy scaling with radius&lt;sup&gt;1.65&lt;/sup&gt;. Disrupting and dispersing a solid 3 km diameter asteroid with a density of 3 g/cc would require a deeply buried two Tsar Bomba weapon. The energy needed to disrupt and disperse a 20 km diameter solid asteroid is beyond anything humanity can achieve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This disruption and dispersal will not be uniform. It will instead create a few big chunks, perhaps 1/3 the diameter of the original body, a larger number of intermediate sized chunks, and thousands upon thousands of little chunks. Those big chunks are still civilization killers. The intermediate-sized chunks will make Tunguska look small, and the thousands upon thousands of little chunks -- we recently saw what a little chunk can do, over Chelyabinsk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if it's a rubble pile or a comet? While they're not quite as dense, they need much, much more energy to disrupt and disperse. Rubble piles and comets are very good at absorbing impact energy. Disrupting and dispersing a one km diameter comet is beyond the scope of human technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time scale, two days, is extremely short. A decade of advanced notice is considered to be very short notice when it comes to diverting a one km impactor. Even larger ones require even more advance notice. A century's advance notice might be enough time to deal with a 20 km diameter object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://books.google.com/books?hl=en&amp;amp;lr=&amp;amp;id=Dw0A7T0fy6AC&amp;amp;oi=fnd&amp;amp;pg=PA113#v=onepage&amp;amp;q&amp;amp;f=false&quot;&gt;Holsapple, &quot;About deflecting asteroids and comets,&quot; &lt;em&gt;Mitigation of Hazardous Comets and Asteroids: Vol. 1&lt;/em&gt;, Cambridge Univ, 2004.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-02-19T13:34:54.383" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9929" PostTypeId="2" ParentId="9924" CreationDate="2015-02-19T14:12:00.960" Score="-1" Body="&lt;p&gt;You would need to have a planet orbit very slowly around the star. A planet moves retrograde because the planet is passing the other in its orbit so that the slower planet appears to be going backwards.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in order for this to happen you would need a planet having an orbit such that the outer planet passes the inner planet.  But I think that this would result in an unstable orbit for one of the two planets.  Either the inner one would fall into the orbiting star or the outer one would escape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inner planets have a smaller orbit so they move &quot;faster&quot; in order to not fall into the sun.  So they are always &quot;passing&quot; the planets in orbits that are outside of their orbit.&lt;/p&gt;&#xA;" OwnerUserId="281" LastEditorUserId="281" LastEditDate="2015-02-19T14:30:12.853" LastActivityDate="2015-02-19T14:30:12.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9930" PostTypeId="2" ParentId="9924" CreationDate="2015-02-19T16:29:55.520" Score="-1" Body="&lt;p&gt;&lt;strong&gt;TL;DR Not for planets, but for other stuff yes.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As explained in an earlier answer, planets are on circular orbits (quasi-circular actually but that's a detail). The closer to the star, the faster the planet is on its orbit. So, the retrograde motion observed for Mars cannot happen for Venus for instance. At most you could see Venus going retrograde when it actually turns around the sun, but you would clearly understand that the planet is just on the other side of the solar system -- which I think is not what you were wondering about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there's another type of situation where you see a retrograde orbit for an inner body: if the orbit is not circular but a nice ellipse! The orbit would have to be very eccentric though (very narrow). Planets do not have eccentric orbits (in the solar system at least) because, in short, the formation of planets favors circular orbits in the long term. But some asteroids are in this situation (for example many in the Aten family).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the perihelion (closest point) is very close to the sun and the aphelion (farthest point) is much further, then even if the asteroid remains in the interior of the Earth's orbit you would see it going backwards near aphelion. What happens is that speed on a eccentric (elliptic) orbit is slower at the apoapse. For instance, if your perihelion was grazing the surface of the sun, you would have a speed of almost 0 at your aphelion. And you would see it drifting backward because of Earth's motion.&lt;/p&gt;&#xA;" OwnerUserId="6604" LastActivityDate="2015-02-19T16:29:55.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9931" PostTypeId="2" ParentId="9924" CreationDate="2015-02-19T16:47:38.463" Score="5" Body="&lt;p&gt;Both Mercury and Venus undergo retrograde motion, but are visually close to the Sun when this happens, so we don't notice it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is Mercury's right ascension from May 15 to July 15 of this year. Note that it starts off increasing, then decreases (retrograde motion) and then increases again.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;*******************************************************************************&#xA;Ephemeris / WWW_USER Thu Feb 19 08:02:41 2015 Pasadena, USA      / Horizons    &#xA;*******************************************************************************&#xA;Target body name: Mercury (199)                   {source: DE-0431LE-0431}&#xA;Center body name: Earth (399)                     {source: DE-0431LE-0431}&#xA;Center-site name: GEOCENTRIC&#xA;*******************************************************************************&#xA;Start time      : A.D. 2015-May-15 00:00:00.0000 UT      &#xA;Stop  time      : A.D. 2015-Jul-15 00:00:00.0000 UT      &#xA;Step-size       : 1440 minutes&#xA;*******************************************************************************&#xA;Target pole/equ : IAU_MERCURY                     {East-longitude -}&#xA;Target radii    : 2439.7 x 2439.7 x 2439.7 km     {Equator, meridian, pole}    &#xA;Center geodetic : 0.00000000,0.00000000,0.0000000 {E-lon(deg),Lat(deg),Alt(km)}&#xA;Center cylindric: 0.00000000,0.00000000,0.0000000 {E-lon(deg),Dxy(km),Dz(km)}&#xA;Center pole/equ : High-precision EOP model        {East-longitude +}&#xA;Center radii    : 6378.1 x 6378.1 x 6356.8 km     {Equator, meridian, pole}    &#xA;Target primary  : Sun&#xA;Vis. interferer : MOON (R_eq= 1737.400) km        {source: DE-0431LE-0431}&#xA;Rel. light bend : Sun, EARTH                      {source: DE-0431LE-0431}&#xA;Rel. lght bnd GM: 1.3271E+11, 3.9860E+05 km^3/s^2                              &#xA;Atmos refraction: NO (AIRLESS)&#xA;RA format       : HMS&#xA;Time format     : CAL &#xA;EOP file        : eop.150218.p150512                                           &#xA;EOP coverage    : DATA-BASED 1962-JAN-20 TO 2015-FEB-18. PREDICTS-&amp;gt; 2015-MAY-11&#xA;Units conversion: 1 au= 149597870.700 km, c= 299792.458 km/s, 1 day= 86400.0 s &#xA;Table cut-offs 1: Elevation (-90.0deg=NO ),Airmass (&amp;gt;38.000=NO), Daylight (NO )&#xA;Table cut-offs 2: Solar Elongation (  0.0,180.0=NO ),Local Hour Angle( 0.0=NO )&#xA;**************************************************************************************************************&#xA; Date__(UT)__HR:MN     R.A._(ICRF/J2000.0)_DEC  APmag  S-brt            delta      deldot    S-O-T /r    S-T-O&#xA;**************************************************************************************************************&#xA;$$SOE&#xA;     2015-May-15 00:00     04 42 01.78 +24 08 32.2   1.53   4.36 0.69142867651373 -28.9445515  18.6988 /T 129.3762&#xA;     2015-May-16 00:00     04 43 23.87 +24 00 54.9   1.74   4.49 0.67512627148213 -27.4926615  18.0144 /T 132.4768&#xA;     2015-May-17 00:00     04 44 24.75 +23 51 35.3   1.96   4.63 0.65967892310678 -25.9844651  17.2466 /T 135.5875&#xA;     2015-May-18 00:00     04 45 04.61 +23 40 37.1   2.21   4.77 0.64511881456032 -24.4211059  16.3963 /T 138.7127&#xA;     2015-May-19 00:00     04 45 23.80 +23 28 04.3   2.47   4.91 0.63147747327791 -22.8037380  15.4651 /T 141.8554&#xA;     2015-May-20 00:00     04 45 22.86 +23 14 01.5   2.75   5.06 0.61878570842679 -21.1337209  14.4555 /T 145.0181&#xA;     2015-May-21 00:00     04 45 02.49 +22 58 33.7   3.05   5.20 0.60707343838108 -19.4128031  13.3707 /T 148.2014&#xA;     2015-May-22 00:00     04 44 23.61 +22 41 46.8   3.37   5.33 0.59636941752453 -17.6432855  12.2149 /T 151.4048&#xA;     2015-May-23 00:00     04 43 27.36 +22 23 47.7   3.71   5.45 0.58670087720487 -15.8281545  10.9939 /T 154.6255&#xA;     2015-May-24 00:00     04 42 15.08 +22 04 44.1   4.07   5.56 0.57809309959244 -13.9711803   9.7148 /T 157.8583&#xA;     2015-May-25 00:00     04 40 48.34 +21 44 44.8   4.46   5.63 0.57056894619759 -12.0769753   8.3871 /T 161.0937&#xA;     2015-May-26 00:00     04 39 08.89 +21 23 59.8   4.87   5.66 0.56414836550867 -10.1510071   7.0239 /T 164.3154&#xA;     2015-May-27 00:00     04 37 18.70 +21 02 40.1   5.29   5.62 0.55884790663755  -8.1995645   5.6456 /T 167.4924&#xA;     2015-May-28 00:00     04 35 19.87 +20 40 57.9   n.a.   n.a. 0.55468026744915  -6.2296724   4.2893 /T 170.5572&#xA;     2015-May-29 00:00     04 33 14.63 +20 19 06.0   n.a.   n.a. 0.55165390565664  -4.2489596   3.0439 /T 173.3313&#xA;     2015-May-30 00:00     04 31 05.30 +19 57 18.0   n.a.   n.a. 0.54977273920552  -2.2654833   2.1722 /T 175.2574&#xA;     2015-May-31 00:00     04 28 54.24 +19 35 48.1   n.a.   n.a. 0.54903595776444  -0.2875219   2.2105 /L 175.1837&#xA;     2015-Jun-01 00:00     04 26 43.78 +19 14 50.1   n.a.   n.a. 0.54943796056976   1.6766529   3.1266 /L 173.1900&#xA;     2015-Jun-02 00:00     04 24 36.24 +18 54 38.3   n.a.   n.a. 0.55096842790710   3.6190086   4.3897 /L 170.4268&#xA;     2015-Jun-03 00:00     04 22 33.81 +18 35 25.8   5.29   5.65 0.55361252502737   5.5319758   5.7564 /L 167.4122&#xA;     2015-Jun-04 00:00     04 20 38.57 +18 17 25.2   4.88   5.70 0.55735122910170   7.4086481   7.1443 /L 164.3125&#xA;     2015-Jun-05 00:00     04 18 52.42 +18 00 48.1   4.49   5.68 0.56216176249744   9.2429487   8.5180 /L 161.1913&#xA;     2015-Jun-06 00:00     04 17 17.09 +17 45 44.4   4.12   5.62 0.56801810956371  11.0297524   9.8584 /L 158.0785&#xA;     2015-Jun-07 00:00     04 15 54.09 +17 32 22.7   3.77   5.53 0.57489158958468  12.7649573  11.1530 /L 154.9902&#xA;     2015-Jun-08 00:00     04 14 44.76 +17 20 50.1   3.45   5.42 0.58275145605987  14.4455044  12.3929 /L 151.9356&#xA;     2015-Jun-09 00:00     04 13 50.19 +17 11 11.8   3.15   5.31 0.59156549247635  16.0693469  13.5713 /L 148.9204&#xA;     2015-Jun-10 00:00     04 13 11.32 +17 03 31.6   2.87   5.18 0.60130057734526  17.6353760  14.6833 /L 145.9474&#xA;     2015-Jun-11 00:00     04 12 48.85 +16 57 51.6   2.61   5.06 0.61192319592983  19.1433125  15.7250 /L 143.0178&#xA;     2015-Jun-12 00:00     04 12 43.37 +16 54 12.5   2.37   4.94 0.62339988172426  20.5935736  16.6938 /L 140.1313&#xA;     2015-Jun-13 00:00     04 12 55.28 +16 52 33.7   2.15   4.81 0.63569757625371  21.9871244  17.5876 /L 137.2863&#xA;     2015-Jun-14 00:00     04 13 24.86 +16 52 53.2   1.94   4.69 0.64878390050511  23.3253217  18.4055 /L 134.4805&#xA;     2015-Jun-15 00:00     04 14 12.31 +16 55 08.3   1.75   4.58 0.66262733523404  24.6097559  19.1469 /L 131.7103&#xA;     2015-Jun-16 00:00     04 15 17.69 +16 59 15.2   1.57   4.46 0.67719731085661  25.8420965  19.8117 /L 128.9719&#xA;     2015-Jun-17 00:00     04 16 41.05 +17 05 09.2   1.40   4.35 0.69246421084326  27.0239479  20.4002 /L 126.2604&#xA;     2015-Jun-18 00:00     04 18 22.35 +17 12 45.4   1.25   4.24 0.70839929524826  28.1567176  20.9133 /L 123.5707&#xA;     2015-Jun-19 00:00     04 20 21.53 +17 21 57.8   1.10   4.14 0.72497455264057  29.2415000  21.3516 /L 120.8972&#xA;     2015-Jun-20 00:00     04 22 38.49 +17 32 40.5   0.96   4.04 0.74216248879516  30.2789728  21.7164 /L 118.2337&#xA;     2015-Jun-21 00:00     04 25 13.15 +17 44 46.7   0.83   3.94 0.75993585914434  31.2693054  22.0086 /L 115.5738&#xA;     2015-Jun-22 00:00     04 28 05.39 +17 58 09.6   0.71   3.84 0.77826734979511  32.2120717  22.2296 /L 112.9109&#xA;     2015-Jun-23 00:00     04 31 15.13 +18 12 42.2   0.59   3.75 0.79712920966069  33.1061670  22.3805 /L 110.2380&#xA;     2015-Jun-24 00:00     04 34 42.27 +18 28 17.0   0.48   3.65 0.81649283449812  33.9497246  22.4625 /L 107.5478&#xA;     2015-Jun-25 00:00     04 38 26.75 +18 44 46.2   0.37   3.56 0.83632830264123  34.7400316  22.4769 /L 104.8329&#xA;     2015-Jun-26 00:00     04 42 28.49 +19 02 02.2   0.27   3.47 0.85660386201288  35.4734443  22.4247 /L 102.0858&#xA;     2015-Jun-27 00:00     04 46 47.47 +19 19 56.7   0.17   3.38 0.87728536859778  36.1453054  22.3070 /L  99.2988&#xA;     2015-Jun-28 00:00     04 51 23.65 +19 38 21.2   0.07   3.30 0.89833567803594  36.7498651  22.1247 /L  96.4642&#xA;     2015-Jun-29 00:00     04 56 17.02 +19 57 07.2  -0.02   3.21 0.91971399456822  37.2802130  21.8789 /L  93.5742&#xA;     2015-Jun-30 00:00     05 01 27.57 +20 16 05.6  -0.12   3.13 0.94137518555319  37.7282290  21.5704 /L  90.6210&#xA;     2015-Jul-01 00:00     05 06 55.30 +20 35 07.1  -0.20   3.05 0.96326907552925  38.0845640  21.2001 /L  87.5974&#xA;     2015-Jul-02 00:00     05 12 40.17 +20 54 01.8  -0.29   2.96 0.98533974152847  38.3386660  20.7688 /L  84.4961&#xA;     2015-Jul-03 00:00     05 18 42.16 +21 12 39.8  -0.38   2.88 1.00752484088954  38.4788698  20.2775 /L  81.3104&#xA;     2015-Jul-04 00:00     05 25 01.19 +21 30 50.5  -0.46   2.80 1.02975501346900  38.4925685  19.7269 /L  78.0344&#xA;     2015-Jul-05 00:00     05 31 37.13 +21 48 23.0  -0.54   2.73 1.05195341081967  38.3664848  19.1182 /L  74.6631&#xA;     2015-Jul-06 00:00     05 38 29.80 +22 05 05.9  -0.63   2.65 1.07403541451894  38.0870574  18.4526 /L  71.1926&#xA;     2015-Jul-07 00:00     05 45 38.91 +22 20 47.8  -0.71   2.57 1.09590861358459  37.6409536  17.7315 /L  67.6206&#xA;     2015-Jul-08 00:00     05 53 04.08 +22 35 16.7  -0.79   2.49 1.11747311584451  37.0157141  16.9564 /L  63.9467&#xA;     2015-Jul-09 00:00     06 00 44.78 +22 48 20.6  -0.87   2.41 1.13862226817234  36.2005225  16.1294 /L  60.1725&#xA;     2015-Jul-10 00:00     06 08 40.36 +22 59 47.7  -0.95   2.34 1.15924385212199  35.1870766  15.2529 /L  56.3023&#xA;     2015-Jul-11 00:00     06 16 49.99 +23 09 26.1  -1.03   2.26 1.17922180051076  33.9705111  14.3295 /L  52.3431&#xA;     2015-Jul-12 00:00     06 25 12.69 +23 17 04.6  -1.11   2.17 1.19843844390623  32.5502953  13.3626 /L  48.3050&#xA;     2015-Jul-13 00:00     06 33 47.27 +23 22 32.8  -1.19   2.09 1.21677724378686  30.9310023  12.3560 /L  44.2011&#xA;     2015-Jul-14 00:00     06 42 32.40 +23 25 41.0  -1.28   2.00 1.23412590569533  29.1228358  11.3140 /L  40.0476&#xA;     2015-Jul-15 00:00     06 51 26.58 +23 26 21.1  -1.37   1.91 1.25037969990154  27.1418048  10.2415 /L  35.8640&#xA;    $$EOE&#xA;**************************************************************************************************************&#xA;Column meaning:&#xA;&#xA;TIME&#xA;&#xA;  Prior to 1962, times are UT1. Dates thereafter are UTC. Any 'b' symbol in&#xA;the 1st-column denotes a B.C. date. First-column blank (&quot; &quot;) denotes an A.D.&#xA;date. Calendar dates prior to 1582-Oct-15 are in the Julian calendar system.&#xA;Later calendar dates are in the Gregorian system.&#xA;&#xA;  Time tags refer to the same instant throughout the universe, regardless of&#xA;where the observer is located.&#xA;&#xA;  The dynamical Coordinate Time scale is used internally. It is equivalent to&#xA;the current IAU definition of &quot;TDB&quot;. Conversion between CT and the selected&#xA;non-uniform UT output scale has not been determined for UTC times after the&#xA;next July or January 1st.  The last known leap-second is used over any future&#xA;interval.&#xA;&#xA;  NOTE: &quot;n.a.&quot; in output means quantity &quot;not available&quot; at the print-time.&#xA;&#xA; R.A._(ICRF/J2000.0)_DEC =&#xA;   J2000.0 astrometric right ascension and declination of target center.&#xA;Adjusted for light-time. Units: HMS (HH MM SS.ff) and DMS (DD MM SS.f)&#xA;&#xA; APmag S-brt =&#xA;   Target's approximate apparent visual magnitude &amp;amp; surface brightness. For&#xA;planets and satellites, values are available only for solar phase angles in the&#xA;range generally visible from Earth. This is to avoid extrapolation of models&#xA;beyond their valid (data-based) limits.&#xA;   Units: MAGNITUDE &amp;amp; VISUAL MAGNITUDES PER SQUARE ARCSECOND&#xA;&#xA; delta  deldot =&#xA;   Range (&quot;delta&quot;) and range-rate (&quot;delta-dot&quot;) of target center with respect&#xA;to the observer at the instant light seen by the observer at print-time would&#xA;have left the target center (print-time minus down-leg light-time); the&#xA;distance traveled by a light ray emanating from the center of the target and&#xA;recorded by the observer at print-time. &quot;deldot&quot; is a projection of the&#xA;velocity vector along this ray, the light-time-corrected line-of-sight from the&#xA;coordinate center, and indicates relative motion. A positive &quot;deldot&quot; means the&#xA;target center is moving away from the observer (coordinate center). A negative&#xA;&quot;deldot&quot; means the target center is moving toward the observer.&#xA;Units: AU and KM/S&#xA;&#xA; S-O-T /r =&#xA;    Sun-Observer-Target angle; target's apparent solar elongation seen from&#xA;observer location at print-time. If negative, the target center is behind&#xA;the Sun. Angular units: DEGREES.&#xA;&#xA;    The '/r' column is a Sun-relative code, output for observing sites&#xA;with defined rotation models only.&#xA;&#xA;         /T indicates target trails Sun (evening sky)&#xA;         /L indicates target leads Sun  (morning sky)&#xA;&#xA;    NOTE: The S-O-T solar elongation angle is the total separation in any&#xA;direction. It does not indicate the angle of Sun leading or trailing.&#xA;&#xA; S-T-O =&#xA;   &quot;S-T-O&quot; is the Sun-&amp;gt;Target-&amp;gt;Observer angle; the interior vertex angle at&#xA;target center formed by a vector to the apparent center of the Sun at&#xA;reflection time on the target and the apparent vector to the observer at&#xA;print-time. Slightly different from true PHASE ANGLE (requestable separately)&#xA;at the few arcsecond level in that it includes stellar aberration on the&#xA;down-leg from target to observer.  Units: DEGREES&#xA;&#xA;&#xA; Computations by ...&#xA;     Solar System Dynamics Group, Horizons On-Line Ephemeris System&#xA;     4800 Oak Grove Drive, Jet Propulsion Laboratory&#xA;     Pasadena, CA  91109   USA&#xA;     Information: http://ssd.jpl.nasa.gov/&#xA;     Connect    : telnet://ssd.jpl.nasa.gov:6775  (via browser)&#xA;                  telnet ssd.jpl.nasa.gov 6775    (via command-line)&#xA;     Author     : Jon.Giorgini@jpl.nasa.gov&#xA;&#xA;**************************************************************************************************************&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2015-02-19T16:47:38.463" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9932" PostTypeId="2" ParentId="9921" CreationDate="2015-02-19T16:56:02.980" Score="5" Body="&lt;p&gt;It's a rather difficult concept to grab but, in short, the image of expansion you have in your mind is wrong. It's not your fault, it's everywhere out there, the inflatable balloon etc. And it's terrible :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to look at it is that the universe properties are changing. Whether it is infinite or finite, its size (as measured within itself) changes. Things are pulled appart. You &lt;em&gt;could&lt;/em&gt; imagine that free vacuum keeps being generated between all objects but that would be a misrepresentation of the theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most simple way I see it is actually what the equations say. It's lame but sometimes the math are the simplest way to understand a phenomenon. The equation say &lt;strong&gt;the metric is expanding&lt;/strong&gt;. It means two point in the fabric of the universe are measured at increasing distances over time. As a result, unless objects are pulled to each other with sufficient force, they tend to drift appart. It's a property of the universe that has not been explained yet (at least not completely), and we still are on the fence as to what influences the expansion rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, this concept is very easy to accept when you cease to represent the universe from outside of itself. The universe is everything that is so if any image in your mind represents the universe from outside, it won't give you a good picture. Balloons are especially bad :). If you need an image, draw a grid or take a notebook, draw some points here and there on it and imagine the grid keeps getting bigger and bigger. But don't look at the whole page, forget the page limits, focus on the grid. And keep in mind that the points do not expand, only the grid does...&lt;/p&gt;&#xA;" OwnerUserId="6604" LastActivityDate="2015-02-19T16:56:02.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9933" PostTypeId="1" AcceptedAnswerId="9936" CreationDate="2015-02-19T17:17:00.113" Score="5" ViewCount="404" Body="&lt;p&gt;Dimitris (see below) argues that the syzygies of the Earth and Venus and those of Mercury, Earth and Jupiter distort the Sun's corona, which in some way affects climate on the scale of hundreds of years.  Is there any evidence to support the idea that the planets do have such an effect on the sun, &lt;strong&gt;&lt;em&gt;other than&lt;/em&gt;&lt;/strong&gt; the correlation noted by Dimitris?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planetary orbits’ effect to the Northern Hemisphere climate, from solar corona&#xA;formation to the Earth climate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Poulos Dimitris&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Abstract&#xA;The four planets that influence the most the solar surface through tidal forcing seem to affect the Earth climate. A simple two cosine model with periods 251 years, of the seasonality of the Earth – Venus syzygies, and 265.4 years, of the combined syzygies of Jupiter and Mercury with Earth when Earth is in synod with Venus, fits well the Northern Hemisphere temperatures of the last 1000 years as reconstructed by Jones et al (1998). Later reconstructions that give too much emphasis on multy-centenial variation are due to increased error. The physical mechanism proposed is that planetary gravitational forces derange the Solar Corona that in turn deranges the planetary geomagnetic field causing&#xA;temperature variations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.itia.ntua.gr/getfile/1486/1/documents/PoulosPaper.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://www.itia.ntua.gr/getfile/1486/1/documents/PoulosPaper.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2018-05-02T05:01:50.697" Title="Does the orbital variation in planetary gravity affect the Sun's corona" Tags="&lt;orbit&gt;&lt;the-sun&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9934" PostTypeId="1" AcceptedAnswerId="9935" CreationDate="2015-02-19T19:25:53.177" Score="3" ViewCount="352" Body="&lt;p&gt;Earlier I read an article &lt;a href=&quot;http://www.rochester.edu/newscenter/scholz-star/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; stating that a binary pair of stars passed within 0.8 light years from the sun. That made we wonder why the stars weren't captured by the Sun's gravity since the sun's gravitational influence spans over 1 light year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could a passing star be captured by the Sun's gravity? if so how close would it have to be before it was captured?&lt;/p&gt;&#xA;" OwnerUserId="2629" LastEditorUserId="2153" LastEditDate="2015-02-19T19:49:04.960" LastActivityDate="2015-02-19T19:56:33.683" Title="Could a close passing star be captured by the Sun's gravity?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;gravity&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9935" PostTypeId="2" ParentId="9934" CreationDate="2015-02-19T19:56:33.683" Score="12" Body="&lt;p&gt;Unless the stars comes so close that they actually collide, two stars will not be able to catch each other gravitationally. The reason is energy conservation: As they approach each other, their potential energy is converted into kinetic energy, increasing their velocities. When they are closest, their velocities are at their highest, but since there's nothing to take out energy from the system, their kinetic energy will be converted back to potential energy, propelling them far apart again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a third star is present, however, this may be able to extract energy from the system, so that one star is slung out while the other two start orbiting each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note however that the probability of two stars encountering each other extremely small, and that of three stars is close to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above, I have ignored the presence of planets. In principle, a large planet could help, but since their masses are much smaller than the stars', their effect is quite small.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2015-02-19T19:56:33.683" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9936" PostTypeId="2" ParentId="9933" CreationDate="2015-02-19T22:24:05.723" Score="4" Body="&lt;p&gt;The gravitational effect of the planets on the Sun can be calculated very accurately. The acceleration of a planet of mass $M$ at a distance from the Sun of $r$ is&#xA;$$a = \frac{G M}{r^2}.$$&#xA;If you plug in the numbers, you'll find that the largest effect comes from Jupiter due to its enormous mass. The gravitational acceleration of Jupiter on the Sun is $2\times10^{-7}$ m/s$^2$. Even if all planets were aligned so to add up all their gravitational accelerations, the total would only be $2.8\times10^{-7}$ m/s$^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the &lt;em&gt;acceleration&lt;/em&gt; is not what is important; rather it is the &lt;em&gt;tidal forces&lt;/em&gt;, which are proportional to $1/r^3$ instead on $1/r^2$. The tidal acceleration felt over a distance $\Delta r$ at the surface of the Sun is&#xA;$$a_{\mathrm{tidal}} \simeq \frac{2GM}{r^3} \Delta r.$$&#xA;Now I didn't really bother to read the paper in detail, but the guy doesn't seem to write what exactly in the Sun's atmosphere should be affected by the planets. But even if we say that it's all of Sun's corona, which has a thickness of $\Delta r \sim 10^6$ km, &lt;strong&gt;the combined effect of all eight planets standing on a line would be roughly $\mathbf{10^{-9}}$ m/s$\mathbf{^2}$&lt;/strong&gt;, a minor effect compared to the difference in the Sun's own gravitational field over the same distance, which is 11 orders of magnitude higher.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-03-03T13:31:07.840" LastActivityDate="2015-03-03T13:31:07.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9937" PostTypeId="1" AcceptedAnswerId="9938" CreationDate="2015-02-20T08:25:53.913" Score="2" ViewCount="2074" Body="&lt;p&gt;A hypothetical simulation starts with a planet called Earth in a perfectly circular orbit around a star called Sol. There is also a moon called Luna in a perfectly circular orbit around Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:&lt;br&gt;&#xA;Can Luna's orbit remain perfectly circular or does it become elliptical over time? At what point does it become stable? Never?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking:&lt;br&gt;&#xA;The distance between Sol and Luna is not constant, it fluctuates as Luna orbits Earth. Therefore the pull sunward also fluctuates. Shouldn't this cause all moon orbits to eventually deteriorate into longer and longer elliptical orbits until they collide with the planet they orbit?&lt;/p&gt;&#xA;" OwnerUserId="6607" LastActivityDate="2016-11-17T04:06:33.057" Title="Shouldn't all moon orbits be inherently unstable?" Tags="&lt;the-moon&gt;&lt;orbit&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9938" PostTypeId="2" ParentId="9937" CreationDate="2015-02-20T08:40:28.170" Score="4" Body="&lt;p&gt;The answer to your simple question is that while a circular orbit could remain in the absence of any other gravitational fields, in reality you'll have an ellipse and it can be stable from the beginning, remember stable has a wide range of values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you allow drag through tidal forces etc then the orbit will change, eg our Moon is slowly getting further away from us, but we still consider this stable over realistic timeframes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, the actual orbit will track out a &lt;a href=&quot;http://blogs.discovermagazine.com/badastronomy/2008/09/29/the-moon-that-went-up-a-hill-but-came-down-a-planet/#.VOb_1vmf7OE&quot; rel=&quot;nofollow&quot;&gt;sort of helix&lt;/a&gt; as the planet travels round the sun.&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="43" LastEditDate="2015-02-20T09:37:21.837" LastActivityDate="2015-02-20T09:37:21.837" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9939" PostTypeId="1" AcceptedAnswerId="9955" CreationDate="2015-02-20T11:50:22.327" Score="1" ViewCount="616" Body="&lt;p&gt;I have a &lt;a href=&quot;http://www.fotosense.co.uk/sky-watcher-skyhawk-1145p-parabolic-newtonian-reflector-telescope.html?gclid=CKu8zsu28MMCFQIGwwodRhUAfw#fo_c=77&amp;amp;fo_k=8fa4b0af4ced5cb933af545cf914fba6&amp;amp;fo_s=gplauk&quot; rel=&quot;nofollow&quot;&gt;4.5&quot; Newtonian Reflector&lt;/a&gt; and I took a look at Jupiter the other night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was able to get a reasonably sharp view of it and see a number of it's moons, but I was unable to see any surface details, it was pretty much just white glare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if it's possible to distinguish any surface details with the telescope I have, so any advice on best setup would be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a few accessories:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;x2 barlow lense&lt;/li&gt;&#xA;&lt;li&gt;10mm eye piece&lt;/li&gt;&#xA;&lt;li&gt;25mm eye piece&lt;/li&gt;&#xA;&lt;li&gt;6mm eye piece&lt;/li&gt;&#xA;&lt;li&gt;moon filter&lt;/li&gt;&#xA;&lt;li&gt;color filters&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Would anything else, other than a more powerful telescope, help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was unsure if I might have an alignment issue that was causing this, but as I could make out the moons as sharp dots on the planetary plane, I assumed not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any advice would be welcomed.&lt;/p&gt;&#xA;" OwnerUserId="2485" LastEditorUserId="2485" LastEditDate="2015-02-20T12:01:42.533" LastActivityDate="2015-02-21T21:30:02.080" Title="How to view Jupiter with 4.5&quot; Newtonian Reflector" Tags="&lt;jupiter&gt;&lt;newtonian-telescope&gt;&lt;telescope-lens&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9940" PostTypeId="1" AcceptedAnswerId="9942" CreationDate="2015-02-20T20:57:55.090" Score="1" ViewCount="116" Body="&lt;p&gt;What is the minimum size telescope required to enable clear visual detection of Ceres?&lt;/p&gt;&#xA;" OwnerUserId="4041" LastActivityDate="2015-02-20T22:21:21.530" Title="Telescope size to detect Ceres (Newtonian)" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9941" PostTypeId="1" CreationDate="2015-02-20T22:16:40.327" Score="2" ViewCount="988" Body="&lt;p&gt;This is a multiple choice question I had trouble with on an exam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will a white dwarf that has had too much mass piled on top of it&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) undergo a thermonuclear flash and explode in a type IA supernova&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) collapse to form a black hole&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) collapse to form a neutron star in a core collapse (type II) supernova&lt;/p&gt;&#xA;" OwnerUserId="6615" LastEditorUserId="6615" LastEditDate="2015-02-21T02:20:38.100" LastActivityDate="2015-02-21T02:20:38.100" Title="What happens to a white dwarf past the Chandrasekhar limit?" Tags="&lt;star&gt;&lt;black-hole&gt;&lt;astrophysics&gt;&lt;mass&gt;&lt;neutron-star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9942" PostTypeId="2" ParentId="9940" CreationDate="2015-02-20T22:21:21.530" Score="2" Body="&lt;p&gt;Ceres varies between mag 9.3 and mag 6.7 which are below the nominal naked eye limiting magnitude of ~5. The &lt;a href=&quot;http://w1.411.telia.com/~u41105032/visual/limiting.htm&quot; rel=&quot;nofollow&quot;&gt;gain in limiting magnitude&lt;/a&gt; for something like 10x50 binoculars (~4.5 magnitudes) would bring Ceres even at its faintest above the limiting magnitude of about 9.5.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-02-20T22:21:21.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9943" PostTypeId="2" ParentId="9941" CreationDate="2015-02-20T23:51:02.980" Score="6" Body="&lt;p&gt;It's not completely clear what you are asking, but if this is a multi-choice quiz, then the only option that &lt;em&gt;could&lt;/em&gt; be correct is (a).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Is not correct, because a white dwarf that just passes the Chandrasekhar mass is comfortably below the maximum mass that is supportable by a neutron star. So neutronisation followed by neutron degeneracy pressure and the strongly repulsive nuclear force between neutrons at small separations ought to be capable of preventing black hole formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) Is not correct because by definition, type II supernovae result from the collapse of massive stars. Observationally, they are distinguished by hydrogen absorption in their spectra, but since a white dwarf will contain little if any hydrogen, then this is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) &lt;em&gt;Might&lt;/em&gt; happen. As mass is added to the white dwarf it will become &lt;em&gt;smaller&lt;/em&gt; and denser. It is possible that nuclear reactions (carbon fusion) might begin. Because the white dwarf is supported by temperature-independent electron degeneracy pressure, then the nuclear reactions take off at constant density, but the increasing temperature leads to runaway nuclear reactions that detonate the star as a type Ia supernova.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-02-21T00:01:53.207" LastActivityDate="2015-02-21T00:01:53.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9944" PostTypeId="1" CreationDate="2015-02-21T11:10:50.630" Score="2" ViewCount="639" Body="&lt;p&gt;Magnetars are the most magnetic things in our universe. Because of how powerful their magnetic fields are, they could split every single atom apart from each other in our body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of this could they destroy planets (rocky or not)? If so, how close would the planet have to be?&lt;/p&gt;&#xA;" OwnerUserId="6612" LastEditorUserId="2153" LastEditDate="2015-02-21T15:05:33.567" LastActivityDate="2015-02-21T20:49:17.100" Title="Can Magnetars destroy planets?" Tags="&lt;planet&gt;&lt;magnetic-field&gt;&lt;neutron-star&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9945" PostTypeId="1" AcceptedAnswerId="9947" CreationDate="2015-02-21T11:25:11.943" Score="0" ViewCount="605" Body="&lt;p&gt;I was reading this &lt;a href=&quot;https://astronomy.stackexchange.com/questions/2315/how-small-can-a-planet-be-and-still-have-earth-like-gravity?rq=1&quot;&gt;question&lt;/a&gt; about &lt;code&gt;how small could a planet be while having earth-like gravitational pull&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This got me thinking, how dense would planet Earth have to be to have the same gravitational pull as Jupiter while all the other factors staying the same (even if it is impossible in the real world)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are any formulas could you please explain them so I understand them please?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry about some of the confusion but I meant keeping everything &lt;strong&gt;but&lt;/strong&gt; the mass the same.&lt;/p&gt;&#xA;" OwnerUserId="6612" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-02-22T18:09:34.593" Title="How dense would planet earth have to be to have the same gravity as Jupiter?" Tags="&lt;gravity&gt;&lt;earth&gt;&lt;jupiter&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9946" PostTypeId="2" ParentId="9945" CreationDate="2015-02-21T12:29:00.540" Score="4" Body="&lt;p&gt;You only need two formulae. Gravitational field of a spherically symmetric mass distribution is given by &#xA;$$g = \frac{GM}{R^2},$$&#xA;where $M$ is the mass inside a radius $R$. The second formula is the average density of a sphere is its mass divided by its volume, hence&#xA;$$\rho = \frac{M}{(4/3)\pi R^3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two formulae can obviously be put together to give the gravitational field as a function of mass and density. &#xA;$$ g = \frac{GM}{(3M/4\pi \rho)^{2/3}}$$&#xA;$$ \rho = \frac{3}{4\pi} g^{3/2} M^{-1/2} G^{-3/2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $g=24.8$ m/s$^{2}$ for the surface gravity of Jupiter and $M=6\times10^{24}$ kg for the (unchanged) mass of the Earth. We obtain $\rho = 22096$ kg/m$^{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that my answer assumes that the &lt;em&gt;mass&lt;/em&gt; of the Earth is fixed. If you instead change the mass and leave the radius fixed:&#xA;$$ \rho = \frac{3g}{4\pi G R}$$&#xA;which gives $\rho = 13930$ kg/m$^{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't leave mass and radius fixed!&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-02-21T16:22:18.097" LastActivityDate="2015-02-21T16:22:18.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9947" PostTypeId="2" ParentId="9945" CreationDate="2015-02-21T15:56:59.970" Score="6" Body="&lt;p&gt;There are at least two interpretations to this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per &lt;a href=&quot;https://en.wikipedia.org/wiki/Jupiter&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, Jupiter's surface gravity is $2.528$ times Earth's. Thus, &#xA;if the Earth were $2.528$ times denser, it would have the same surface &#xA;gravity as Jupiter. The Earth's current density is $5.514$ grams per &#xA;cubic centimeter, so the new density would be $2.528 \times 5.514$, or about &#xA;$13.9394$ grams per cubic centimeter. This assumes we change Earth's mass, but not its radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Rob_Jeffries answer assumes the Earth's mass remains constant and the radius changes. If the radius shrinks by a factor of $2$, the volume decreases by a factor of $8$, and the planet becomes 8 times more dense. The surface gravity increases by $4$, since it depends on the radius squared. In general, shrinking the planet's radius by $x$ will increase the density by $x^3$ and the gravity by $x^2$. If we want the gravity $2.528$ times higher, we choose $x = \sqrt{2.528}$ or right around $1.590$. This makes $x^3$ equal to about $4.019$. Multiplying that by Earth's current density of $5.514$, we get $22.16$ grams per cubic centimeter, pretty close to what Rob got.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you can't really change the density without changing anything else: either the mass or the volume must change.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorUserId="2153" LastEditDate="2015-02-22T18:09:34.593" LastActivityDate="2015-02-22T18:09:34.593" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="9948" PostTypeId="2" ParentId="9944" CreationDate="2015-02-21T16:20:07.647" Score="3" Body="&lt;p&gt;Completely disregarding magnetic fields for now, an extremely dense object like a magnetar (which is a kind of pulsar, which itself is a kind of neutron star) could destroy planets just with gravity (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Roche_limit&quot; rel=&quot;nofollow&quot;&gt;Roche Limit&lt;/a&gt;). Now, if the planet is far away from the star, things might not be necessarily that ominous. The problem with electromagnetic forces is that their range is not as far-reaching as gravity. In fact, the first exoplanets ever detected were found around a pulsar (&lt;a href=&quot;http://en.wikipedia.org/wiki/PSR_B1257%2B12&quot; rel=&quot;nofollow&quot;&gt;PSR B1257+12&lt;/a&gt;), and these objects already have pretty damn strong magnetic fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the thing: atoms nuclei are held up by outrageously strong &lt;a href=&quot;http://en.wikipedia.org/wiki/Strong_interaction&quot; rel=&quot;nofollow&quot;&gt;nuclear forces&lt;/a&gt;, but they are even more short-ranged than electromagnetic forces. I haven't run any numbers on this, but I think that these forces are so strong, that in order to break apart the atoms of the rocky planet just by using the magnetic fields, the planet would have to be so close to the magnetar, that its gravity would already have destroyed the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the way gravity and electromagnetism precisely work on such extraordinary environments like the regions around a magnetar are still mysteries to be solved. Magnetism is counter-intuitive and the calculations are hard (been there). Even so, I think gravity would still win.&lt;/p&gt;&#xA;" OwnerUserId="1151" LastEditorUserId="1151" LastEditDate="2015-02-21T20:49:17.100" LastActivityDate="2015-02-21T20:49:17.100" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9949" PostTypeId="2" ParentId="8254" CreationDate="2015-02-21T16:30:57.867" Score="1" Body="&lt;p&gt;As tomassch said, the analogy of gravity and electromagnetism is not valid by current observations. However, some systems, depending on how they are arranged, can have gravitational interactions that may launch other objects at extremely high speeds. This is the case with &lt;a href=&quot;http://en.wikipedia.org/wiki/Three-body_problem&quot; rel=&quot;nofollow&quot;&gt;three-body systems&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravity_assist&quot; rel=&quot;nofollow&quot;&gt;gravitational assist&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1151" LastActivityDate="2015-02-21T16:30:57.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9950" PostTypeId="1" CreationDate="2015-02-21T17:35:40.160" Score="2" ViewCount="268" Body="&lt;p&gt;Just wondering: if a magnetar was close enough to Mars would it have saved its atmosphere by protecting it from solar radiation? I was wondering this because Earth protects us via a magnetic field; the magnetar would be providing Mars one. This is if it did not destroy the planet first as suggested &lt;a href=&quot;https://astronomy.stackexchange.com/questions/9944/can-magnetars-destroy-planets&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&#xA;I meant if the magnetar was safe to live by. &lt;/p&gt;&#xA;" OwnerUserId="6612" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2020-09-28T21:00:36.943" Title="Could a magnetars magnetic field have saved Mars atmosphere?" Tags="&lt;mars&gt;&lt;planetary-atmosphere&gt;&lt;magnetic-field&gt;&lt;neutron-star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9951" PostTypeId="1" AcceptedAnswerId="9953" CreationDate="2015-02-21T20:06:47.427" Score="-8" ViewCount="3528" Body="&lt;p&gt;Like the title says, why are a few things in our universe named after chocolate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, there are the &lt;em&gt;Milky Way&lt;/em&gt;, &lt;em&gt;galaxy&lt;/em&gt; and &lt;em&gt;Mars&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a reason for this or not? Or is it that the chocolates were named after them?&lt;/p&gt;&#xA;" OwnerUserId="6612" LastEditorUserId="2153" LastEditDate="2015-02-21T20:58:47.440" LastActivityDate="2015-02-21T20:58:47.440" Title="Why are a few things in our universe named after chocolate?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;mars&gt;&lt;naming&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ClosedDate="2015-02-22T01:39:11.283" ContentLicense="CC BY-SA 3.0" />
  <row Id="9952" PostTypeId="2" ParentId="9944" CreationDate="2015-02-21T20:46:10.780" Score="6" Body="&lt;p&gt;Perhaps one way of answering this is to note that the (approximately) dipole magnetic fields of a magnetar fall off radially as something like $r^{-3}$. So at the magnetar surface the fields could be of order $10^{11}$ Teslas (at a maximum), but these are reduced by a factor of $\sim 8$ at 2 magnetar radii, a factor of $\sim 27$ at 3 magnetar radii etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now given that the radius of a magnetar is going to be $\sim 10$ km, then the fields die away pretty quickly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's say the Earth orbited a magnetar. The field strength at $1.5 \times 10^{11}$ m is going to be $\sim 10^{11} \times (10^4/1.5\times10^{11})^3 =  3 \times 10^{-11}$ Teslas - much smaller than the Earth's magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put this in context. The field at the surface of a small neodymium magnet are of order 1 Tesla and they don't disrupt your atoms when you pick them them up. To get a field this strong from a magnetar you would have to get to within about 50,000 km of it. The synchrotron radiation field would have fried you before then and any planet would have been ripped apart by tidal forces about an order of magnitude further away. The Roche limit given by&#xA;$$ R_{Roche} \sim 1.26 R_1 \left(\frac{\rho_1}{\rho_2}\right)^{1/3},$$&#xA;where $\rho_1$ is the density of the neutron star ($\sim 3 \times 10^{17}$ kg/m$^{3}$), $\rho_2$ is the density of the Earth ($\sim 5500$ kg/m$^3$) and $R_1$ is the radius of the neutron star (10 km). This yields $R_{roche} = 480,000$ km.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-02-21T20:46:10.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9953" PostTypeId="2" ParentId="9951" CreationDate="2015-02-21T20:52:15.847" Score="6" Body="&lt;p&gt;None of these things were named after chocolate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Milky_Way&quot;&gt;Milky Way&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The name comes from a Greek myth, at the end of which Hera - Zeus' wife - spills her breast milk. In places where the sky has not been affected by light pollution, the Milky Way looks like a milky-white streak. Hence the name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Milky_Way_(chocolate_bar)&quot;&gt;The chocolate of the same name&lt;/a&gt; was introduced in 1923. Apparently, &lt;a href=&quot;http://www.foodchannel.com/articles/article/sweet-milky-way-bar-celebrates-85th-anniversary/&quot;&gt;it was not named after our home galaxy&lt;/a&gt;, but after a milk shake of a similar name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Galaxy&quot;&gt;Galaxy&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Greek word &lt;em&gt;γαλαξίας&lt;/em&gt; (&lt;em&gt;galaxias&lt;/em&gt;) is closely tied to the Milky Way (because, for a long, long time, other galaxies were considered nebulae - &quot;island universes&quot;). It's English translation is &quot;milky one&quot; - a nod to the same Greek myth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Galaxy_(chocolate)&quot;&gt;The chocolate of the same name&lt;/a&gt; was introduced in 1960.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Mars&quot;&gt;Mars&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars was a Roman god, not a Greek one (his Greek equivalent is Ares, I think). The name dates back for millennia - indeed, &lt;a href=&quot;https://en.wikipedia.org/wiki/History_of_Mars_observation&quot;&gt;the Greeks associated Ares with the planet&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Mars_(chocolate_bar)&quot;&gt;The chocolate of the same name&lt;/a&gt; was introduced in 1932.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2015-02-21T20:52:15.847" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9954" PostTypeId="2" ParentId="9950" CreationDate="2015-02-21T20:56:44.637" Score="6" Body="&lt;p&gt;How would you get a magnetar close to Mars, within the solar system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, let's leave that to one side, but the Earth's magnetic field is of  order $10^{-4}$ Tesla at strongest. As the dipole field of a magnetar diminishes as $r^{-3}$ and could be as high as $10^{11}$ Teslas at the surface of the magnetar, at a radius of 10 km, then the magnetar woud need to be at a distance of $10^{6}$ km to surround Mars with a field of $10^{-4}$ Tesla.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that the radiation coming from the magnetar in the form of charged particles accelerated from the magnetic poles would &lt;em&gt;completely dwarf&lt;/em&gt; the charged particle radiation from the Sun at Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So no, a nearby Magnetar would not save Mars' atmosphere - quite the reverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: A BOTE calculations: If we look at the Crab pulsar - not quite a magnetar, but close - it is calculated that 10% of the rotational KE being lost ends up in synchrotron raiation from charged particles. That's about $10^{30}$ Watts. So there is at least $\sim 10^4$ solar luminosities, just in the kinetic energy of relativistic particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnetars are also very strong and variable sources of hard X-ray and Gamma ray radiation. Typical magnetar bursts/flares can release $10^{7}-10^{9}$ solar luminosities of hard X-rays and Gamma rays over periods of 1-40 s. That should fry any atmosphere (or anything else) that was only $10^{6}$ km away!&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-02-21T21:27:08.713" LastActivityDate="2015-02-21T21:27:08.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9955" PostTypeId="2" ParentId="9939" CreationDate="2015-02-21T21:30:02.080" Score="2" Body="&lt;p&gt;Jupiter is a very bright object, so maybe the best procedure is to use your filters, they might help with the glare. However, I think you have either a case of low resolution or distortion. Generally, the larger the diameter of the telescope, the better is the resolution (more details &lt;a href=&quot;http://www.telescope-optics.net/telescope_resolution.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;). Resolution defines how much detail you are able to resolve with the telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing the eyepieces or using Barlow lenses can amplify the images, but it doesn't improve resolution (you'll just see a bigger blob - but still, it's cooler than just a small speck).  There are also other things to check:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Was it windy? Too much wind can, usually, mean a turbulent atmosphere, and it hampers observations. Another way to check for turbulence is to look at stars with the naked eye: if they are twinkling too much, the atmosphere is very turbulent&lt;/li&gt;&#xA;&lt;li&gt;Was Jupiter near the horizon? If so, maybe you should try observing it when it's higher in the sky. When looking near the horizon, light has to pass through a large column of air (a parameter known as airmass by astronomers), and it distorts images&lt;/li&gt;&#xA;&lt;li&gt;Is the telescope well collimated? Poor collimation (mirror alignment) usually results in tail-like structures in sharp objects, almost like a comet. This is probably not the case if the scope is new&lt;/li&gt;&#xA;&lt;li&gt;What is the quality of the optics? Some telescope manufacturers ship their products with very low-cost, low-quality optical components (eyepieces, filters etc.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Also, if your objective is to observe mainly solar system objects, I suggest using a refractor telescope, they are wonderful for that. Additionally, &lt;a href=&quot;https://www.youtube.com/watch?v=mYhy7eaazIk&quot; rel=&quot;nofollow&quot;&gt;this video&lt;/a&gt; by Phil Plait on Crash Course might help you and give invaluable information.&lt;/p&gt;&#xA;" OwnerUserId="1151" LastActivityDate="2015-02-21T21:30:02.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9956" PostTypeId="1" AcceptedAnswerId="9957" CreationDate="2015-02-21T22:52:02.563" Score="-2" ViewCount="375" Body="&lt;p&gt;I'm curious about stars and I am wondering why a star would change its position in the sky due to another star right next to it? &lt;/p&gt;&#xA;" OwnerUserId="6620" LastActivityDate="2015-02-24T17:41:27.487" Title="Why would a star’s position in the sky change relative to another star right next to it?" Tags="&lt;star-systems&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2015-02-23T13:15:08.267" ContentLicense="CC BY-SA 3.0" />
  <row Id="9957" PostTypeId="2" ParentId="9956" CreationDate="2015-02-21T23:03:17.263" Score="0" Body="&lt;p&gt;The answer is gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars have a lot of mass so there for there gravity is strong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 1 star gets close to another star that has slightly more mass this would mean that it gets pulled in. As a result the orbit would change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does not affect all stars due to the distance they are from each other. Usually if it affects start it is in binary systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to &lt;code&gt;Joan.bdm&lt;/code&gt; for this.&lt;/p&gt;&#xA;" OwnerUserId="6612" LastEditorUserId="6612" LastEditDate="2015-02-24T17:41:27.487" LastActivityDate="2015-02-24T17:41:27.487" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9958" PostTypeId="1" AcceptedAnswerId="9959" CreationDate="2015-02-22T02:29:39.730" Score="5" ViewCount="374" Body="&lt;p&gt;I just read that a star called Scholz passed only 0.8 years-light away from us 70,000 years ago.  It is a red dwarf. I don´t know how bright it was in the night sky. Does someone know how bright it was? &lt;/p&gt;&#xA;" OwnerUserId="5471" LastActivityDate="2015-02-23T00:18:55.660" Title="How bright was Scholz when is passed near the Sun 70,000 years ago?" Tags="&lt;star&gt;&lt;solar-system&gt;&lt;binary-star&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9959" PostTypeId="2" ParentId="9958" CreationDate="2015-02-22T02:39:46.117" Score="7" Body="&lt;p&gt;I just found the answer to my question at this place:&#xA;&lt;a href=&quot;http://astronomynow.com/2015/02/18/suns-close-encounter-with-scholzsstar/&quot; rel=&quot;nofollow&quot;&gt;http://astronomynow.com/2015/02/18/suns-close-encounter-with-scholzsstar/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Currently, Scholz’s Star is a small, dim red dwarf in the&#xA;  constellation of Monoceros, about 20 light-years away. However, at the&#xA;  closest point in its flyby of the solar system, Scholz’s Star would&#xA;  have been a 10th magnitude star — about 50 times fainter than can&#xA;  normally be seen with the naked eye at night. It is magnetically&#xA;  active, however, which can cause stars to “flare” and briefly become&#xA;  thousands of times brighter. So it is possible that Scholz’s Star may&#xA;  have been visible to the naked eye by our ancestors 70,000 years ago&#xA;  for minutes or hours at a time during rare flaring events. The star is&#xA;  part of a binary star system: a low-mass red dwarf star (with mass&#xA;  about 8% that of the Sun) and a “brown dwarf” companion (with mass&#xA;  about 6% that of the Sun). Brown dwarfs are considered “failed stars;”&#xA;  their masses are too low to fuse hydrogen in their cores like a&#xA;  “star,” but they are still much more massive than gas giant planets&#xA;  like Jupiter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5471" LastEditorDisplayName="user2449" LastEditDate="2015-02-23T00:18:55.660" LastActivityDate="2015-02-23T00:18:55.660" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9960" PostTypeId="2" ParentId="9911" CreationDate="2015-02-22T10:16:30.243" Score="2" Body="&lt;p&gt;The angular resolving power of a telescope is approximately $1.22 \lambda/D \times 648000/\pi$, where this is the Airy disc diameter in arcseconds, $\lambda$ is the working wavelength in metres and $D$ the diameter of the telescope primary mirror in metres. For an 8-m telescope working at 500 nm, this gives an angular resolution of 0.05 arcseconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is then a separate question about whether this theoretical limit is ever achieved because of detector limitations or degradation by blurring due to atmospheric turbulence (seeing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, at visible wavelengths, seeing always trumps the intrinsic angular resolution and has a typical value of 0.5-1 arcsecond at the world's best astronomical sites. There is no set number for this. It varies with atmospheric conditions from night-to-night or even hour-to-hour. Most observatories publish statistics on this so that you can see what the mean (or median) seeing is. For instance here are some statistics for the &lt;a href=&quot;https://www.eso.org/gen-fac/pubs/astclim/paranal/seeing/adaptive-optics/statfwhm.html&quot; rel=&quot;nofollow&quot;&gt;Paranal (VLT)&lt;/a&gt; site, where the median is about 0.8 arcsec.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Special techniques can mitigate the seeing degradation. You may have heard of adaptive optics - you can analyse the incoming wavefront and send a correcting signal to a tip-tilt or fully deformable mirror that removes the distortion, based either on an adjacent star or a laser guide star. However, these techniques are most effective in the near infrared and are not usually attempted at visible wavelengths. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Adaptive_optics&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Adaptive_optics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method is Lucky Imaging - which can be used by amateur astronomers too. Take many pictures with short exposures. If there is a bright point source in the image somewhere, you can use this to select the best images and to shift and stack the images to produce a cleaner final product. The shorter the exposures the better this works. Very close-to the theoretical maximum resolution can be obtained in this way - but it needs a special high speed, low noise CCD and you end up throwing away more than 90% of the images.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-02-22T10:47:58.933" LastActivityDate="2015-02-22T10:47:58.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9963" PostTypeId="1" CreationDate="2015-02-22T22:13:41.937" Score="1" ViewCount="653" Body="&lt;p&gt;I was wondering, in the process of star formation, does the temperature of the nebula that produces a star play a role in the size of that star? I mean, it's only logical that the size would depend on the amount of matter around.But then, if the temperature is lower, that means that the net density is lower, so the star would 'absorb' less matter. Or is that not a factor, and the growing star would gobble up anything on its vicinity? I've encountered some contradicting sources and I'm trying to double check.&lt;/p&gt;&#xA;" OwnerUserId="4126" LastActivityDate="2015-02-22T23:19:38.297" Title="Relationship between temperature of nebula and size of star" Tags="&lt;temperature&gt;&lt;star-formation&gt;&lt;size&gt;&lt;nebula&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9964" PostTypeId="2" ParentId="9963" CreationDate="2015-02-22T23:19:38.297" Score="2" Body="&lt;p&gt;This is such a complex issue that I'm not going to try to be comprehensive. One major thing you are missing is that stars tend not to form in isolation, especially massive stars, but in groups and clusters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic unit in star formation is the Jeans mass.&#xA;$$ M_J \propto T^{3/2} \rho^{-1/2}$$&#xA;where $T$ is the temperature, $\rho$ is the density and $M_J$ is the minimum mass for a cloud to gravitationally collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So from this point of view, a higher temperature cloud must be more massive in order to collapse in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as the cloud collapses, the density goes up and the Jeans mass decreases - and the cloud fragments into smaller sub-units and ultimately into a cluster of stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exactly how this pans out in practice is basically the whole research field of star formation. It depends not only on relatively simple things like the temperature of the gas, its density, the equation of state; but also on much more complicated ideas like magnetic field support, competitive accretion between stars, feedback from newly-formed stars and protostellar outflows and the rate at which turbulence is generated and then dissipated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There almost certainly is no simple relationship as you propose. In fact the outcome of star formation seems to produce a mass function (number of star per unit mass) that is only weakly dependent on the initial conditions, but may differ a bit in low density and high density environments.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-02-22T23:19:38.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9965" PostTypeId="1" CreationDate="2015-02-22T23:39:36.410" Score="1" ViewCount="452" Body="&lt;p&gt;I've heard a lot about red shift and it sounds pretty legit. But it made it sound as if blue shift was a lot less likely. Can anyone help me out? &lt;/p&gt;&#xA;" OwnerUserId="6623" LastActivityDate="2015-02-23T12:45:36.500" Title="Is Blue Shift just as provable as Red Shift?" Tags="&lt;redshift&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9966" PostTypeId="2" ParentId="9926" CreationDate="2015-02-23T00:01:12.957" Score="1" Body="&lt;p&gt;I don't think it's doable to pulverize a dino-killer on such short notice. But years of movie and TV sci fi have stretched my WSOD bubble enough that I can consider this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an 3 km diameter asteroid moving 13 km/s and having a density of 3 tonnes per  cubic meter, I get kinetic energy of about 3.6e21 joules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my arithmetic's right, earth receives about 1.75e17 joules of sunlight each second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my BOTE seems to indicate about the same energy as 6 hours of sunlight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hemisphere receiving the energy equivalent of 6 hours of sunlight over a brief time -- it wouldn't be pleasant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe we'd be screwed whether the impactor was one chunk or a cloud of gravel.&lt;/p&gt;&#xA;" OwnerUserId="1232" LastActivityDate="2015-02-23T00:01:12.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9967" PostTypeId="2" ParentId="9965" CreationDate="2015-02-23T05:16:30.890" Score="1" Body="&lt;p&gt;All &lt;a href=&quot;http://en.wikipedia.org/wiki/Galaxy&quot; rel=&quot;nofollow&quot;&gt;galaxies&lt;/a&gt; other than those in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Local_Group&quot; rel=&quot;nofollow&quot;&gt;Local Group&lt;/a&gt; show red-shift (Doppler shift towards longer wavelengths which for optical frequencies is towards the red end of the spectrum, hence red-shift). This is due to the expansion of space from which we initially inferred the Big-Bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxies within the Local Group and stars in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Milky_Way&quot; rel=&quot;nofollow&quot;&gt;Milky Way&lt;/a&gt; (the Galaxy) may show either red or blue Doppler shifts depending on their motion with respect to the Earth.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-02-23T12:45:36.500" LastActivityDate="2015-02-23T12:45:36.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9968" PostTypeId="1" CreationDate="2015-02-23T05:47:46.073" Score="3" ViewCount="186" Body="&lt;p&gt;I'm working on detecting optically overlapping galaxies for future astronomical surveys. I'm looking into feature extraction for HST images and have been computing the fourier transform of images in the hope to construct a metric signifying overlapping vs non-overlapping light profiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I have the following image after having filtered out noise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gCTF5.png&quot; alt=&quot;Real Image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been having trouble trying to interpret the results after taking the fourier transform, taking the magnitude of each element, and then plotting the logarithm of the magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i37KR.png&quot; alt=&quot;Logarithm of the magnitude of the Fourier Transform&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2076" LastActivityDate="2015-02-23T05:47:46.073" Title="Fourier Transform of Galaxy Images" Tags="&lt;galaxy&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ClosedDate="2015-03-09T13:11:18.557" ContentLicense="CC BY-SA 3.0" />
  <row Id="9969" PostTypeId="1" AcceptedAnswerId="9972" CreationDate="2015-02-23T07:18:24.557" Score="1" ViewCount="1269" Body="&lt;p&gt;I've been using an IOS app to predict visibility of the Milky Way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm planning a trip to capture the Milky Way on 9th March 2015, but the app shows the visibility will be very weak on that date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon doesn't even rise until 9:12am.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the Moon phase still affect the visibility of the Milky Way even if it is not above the horizon?&lt;/p&gt;&#xA;" OwnerUserId="6626" LastEditorUserId="2485" LastEditDate="2015-02-24T10:58:37.650" LastActivityDate="2015-02-25T08:55:08.623" Title="What's the relationship between Moon phase and the Milky Way visibility?" Tags="&lt;the-moon&gt;&lt;milky-way&gt;&lt;moon-phases&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9970" PostTypeId="2" ParentId="8795" CreationDate="2015-02-23T07:41:38.337" Score="13" Body="&lt;p&gt;&lt;strong&gt;How can a planet capture a moon?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 178 moons in the Solar System, according to the &lt;a href=&quot;http://nssdc.gsfc.nasa.gov/planetary/factsheet/&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA Planetary Fact Sheet&lt;/a&gt;, so it &lt;em&gt;seems&lt;/em&gt; to be a common event. The following sections will show that moon capture is actually unlikely, but when a planet has one or more moons capture becomes easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Initial Conditions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting from the initial conditions, the planet is in orbit about the sun, and an asteroid is in a different orbit about the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for capture to become possible, the asteroid and the planet must come into proximity. When the asteroid comes inside the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sphere_of_influence_%28astrodynamics%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sphere of influence&lt;/a&gt; of the planet, the gravity of the planet is the main factor in determining the path of the asteroid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Possible Outcomes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relative to the planet, the asteroid will be following a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperbolic_trajectory&quot; rel=&quot;nofollow noreferrer&quot;&gt;hyperbolic trajectory&lt;/a&gt;, and hence has sufficient kinetic energy to avoid capture. A large variety of outcomes may occur, but the ones that lead to capture are those where the asteroid somehow loses enough kinetic energy for its velocity to fall below the &lt;a href=&quot;http://en.wikipedia.org/wiki/Escape_velocity&quot; rel=&quot;nofollow noreferrer&quot;&gt;escape velocity&lt;/a&gt; of the planet while retaining enough energy to achieve a closed (elliptical) orbit. The main (not the only) possible outcomes are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the orbit of the asteroid is perturbed, by a greater or lesser extent, and it continues on its way out of the sphere of influence of the planet.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the orbit of the asteroid is perturbed, and the asteroid impacts the planet surface. That would usually be the end of the process, but current theories on how Earth captured the Moon are that a body named &lt;a href=&quot;http://en.wikipedia.org/wiki/Theia_%28planet%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thea impacted the Earth&lt;/a&gt;, and the Moon formed from some of the collision debris.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the orbit of the asteroid is perturbed, and the path of the asteroid intersects the atmosphere of the planet, losing kinetic energy as heat in the atmosphere (similar to &lt;a href=&quot;http://en.wikipedia.org/wiki/Aerobraking&quot; rel=&quot;nofollow noreferrer&quot;&gt;aerobraking&lt;/a&gt;).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the orbit of the asteroid nears an existing moon of the planet and is accelerated (in the sense that deceleration is just acceleration with the opposite sign) by the existing moon, as used by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravity_assist#MESSENGER&quot; rel=&quot;nofollow noreferrer&quot;&gt;MESSENGER spacecraft to slow its speed&lt;/a&gt; before orbiting Mercury. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The last two cases admit the &lt;em&gt;possibility&lt;/em&gt; of capture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Possible Capture&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After losing energy in the planetary atmosphere, if the asteroid has lost enough energy it may enter a closed orbit around the planet. The problem is that the orbit will intersect the atmosphere again, losing energy each time it does so, until it impacts on the planetary surface. Capture can occur when an existing moon is present and is in just the right place for its gravity to reduce the &lt;a href=&quot;http://en.wikipedia.org/wiki/Elliptic_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;eccentricity&lt;/a&gt; of the orbit of the asteroid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the most likely case where a planet can capture a free asteroid is when there is already one or more moons present. The incoming asteroid must avoid entering the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hill sphere&lt;/a&gt; of the existing moon - the region where the moon would dominate the path of the asteroid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gravity_assist&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gravity assist&lt;/a&gt; can accelerate an asteroid when the asteroid is passing outside the orbit of the moon, but can decelerate the asteroid is passing &lt;em&gt;inside&lt;/em&gt; the orbit of the moon. In this case some of the kinetic energy of the asteroid is transferred to the moon. As is the case with aerobraking capture, gravity assisted capture requires the existing moon to be in just the right place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Another mechanism&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rather elegant paper published in Nature (mentioned below) shows how &lt;a href=&quot;http://www.nature.com/nature/journal/v441/n7090/fig_tab/nature04792_F1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;two bodies orbiting each other&lt;/a&gt; as they approach the planet could have led to one being captured by Neptune. This mechanism could apply in other cases also. This &lt;a href=&quot;http://www.astro.umd.edu/people/Theses/2010Philpott.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dissertation&lt;/a&gt; (pdf) discusses a similar process for Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Irregular bodies&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that irregular shaped bodies can be captured more easily than spherical bodies. Orbiting within the Hill sphere of the planet is not enough for capture to be permanent. Only orbits in the lower half of the Hill sphere are stable. Bodies in higher orbits can be perturbed by nearby planets, and the body can eventually be ejected. But irregular shaped bodies exert minute fluctuations in gravitational attraction on the planet, and actually orbit in a chaotic manor. When other moons or rings are present these chaotic orbits gradually transfer energy to the bodies in the lower orbits, causing the new body to orbit lower, and hence become immune to external perturbation. [citation needed]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Prograde vs retrograde orbits&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same analysis of chaotic orbits, and &lt;a href=&quot;https://www.astro.umd.edu/~dphamil/research/reprints/HamKriv97.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;earlier work&lt;/a&gt; also concluded that retrograde orbits are &lt;a href=&quot;http://en.wikipedia.org/wiki/Retrograde_and_prograde_motion#Moons_and_rings&quot; rel=&quot;nofollow noreferrer&quot;&gt;more stable than prograde orbits&lt;/a&gt;. Whereas prograde orbits are only stable in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hill_sphere#True_region_of_stability&quot; rel=&quot;nofollow noreferrer&quot;&gt;inner half&lt;/a&gt; of the Hill sphere, retrograde orbits can be stable out to &lt;a href=&quot;https://physics.stackexchange.com/a/136586&quot;&gt;100% of the Hill radius&lt;/a&gt;. Hence retrograde capture is more commonly observed (this is not the whole story, it is still a matter if research).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Multiple existing moons, rings, and the early Solar System&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the probability of a single moon being in the right place at the right time is low, when there are multiple moons the probability of an initial helpful interaction rises linearly. But the probability of additional interactions rises geometrically, so the more moons a planet has the more likely it is to capture more. The existence of rings also aids capture by exerting a drag on the new moon, taking it's energy and lowering it's orbit, in much the same way that uncaptured gas would do in the early Solar System.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The biggest planets have the most moons&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be obvious, but the biggest planets have the most moons. This is because they have deeper gravity wells, and sweep in more objects. Even though the probability of capture is low (most objects are just pulled into the planet), a steady trickle have have captured over millions of orbits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each capture mechanism requires a fortuitous set of conditions, and so is actually a quite rare event. One mechanism is a that a pair of co-orbiting asteroids become separated when one enters the planetary Hill sphere. The odds for an individual asteroid are improved when the asteroid arrives with low kinetic energy that must be given up to other bodies orbiting the planet, and when there are already many moons or a ring system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://earthsky.org/space/asteroids-accretion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Could Earth's gravity capture an asteroid? - earthsky.org&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.astro.umd.edu/~dphamil/research/reprints/HamKriv97.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dynamics of Distant Moons of Asteroids - Icarus&lt;/a&gt; (pdf)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://thewatchers.adorraeli.com/2012/10/22/hypothesis-new-evidence-on-origin-of-the-moon-support-cataclysmic-collision-theory/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hypothesis: New evidence on origin of the Moon support cataclysmic collision theory&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.nature.com/nature/journal/v441/n7090/full/nature04792.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Neptune's capture of its moon Triton in a binary-planet gravitational encounter - Nature&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://nssdc.gsfc.nasa.gov/planetary/factsheet/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planetary Fact Sheet - NASA&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://journals.cambridge.org/download.php?file=%2FIAU%2FIAU2004_IAUC197%2FS1743921304008968a.pdf&amp;amp;code=8b07a25780a2ca1828e851f47f408cbf&quot; rel=&quot;nofollow noreferrer&quot;&gt;The use of the two-body energy to study problems of escape/capture&lt;/a&gt; (pdf)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Wikipedia&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Aerobraking&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aerobraking&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Elliptic_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Elliptic orbit&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Escape_velocity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Escape velocity&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hill sphere&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperbolic_trajectory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hyperbolic trajectory&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gravity_assist&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gravity assist&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Retrograde_and_prograde_motion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Retrograde and prograde motion&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sphere_of_influence_%28astrodynamics%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sphere of influence&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Theia_%28planet%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thea&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5524" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-02-23T07:41:38.337" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9971" PostTypeId="2" ParentId="9965" CreationDate="2015-02-23T12:06:26.853" Score="2" Body="&lt;p&gt;The expansion of the Universe can be expressed using Hubble's law. This is that the &lt;em&gt;recession&lt;/em&gt; velocity that a galaxy has with respect to our own is proportional to its distance from us. i.e.&#xA;$$ v = H_0 D, $$&#xA;where $v$ is the recession velocity, $D$ is the distance and H_0 is the Hubble &quot;constant&quot;, which has a value of about 70 km/s per Megaparcsec.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hubble's law is attributed tot he overall expansion of the universe and means that once we get further away than about 10 Mpc (or about 30 million light years), that other galaxies recede from us with velocities of 700 km/s or more. This recession produces a redshift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, superimposed on Hubble's law at a more local scale, individual galaxies are gravitationally interacting with each other and hence have their own peculiar motions with respect to the general &quot;Hubble flow&quot;. In addition, don't forget our Sun rotates around the centre of our Galaxy with a velocity of some 230 km/s, so is heading in the direction of some galaxies and away from others.&#xA;These &lt;em&gt;local&lt;/em&gt; peculiar motions means that some nearby galaxies are moving towards us (at relatively modest speeds) and do exhibit a blue shift. An example would be the Andromeda galaxy at 2 million light years, which is approaching our Galaxy at about 110 km/s (or at a relative closing speed of 300 km/s with respect to the Sun).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are only about 50 galaxies in our &lt;a href=&quot;http://en.wikipedia.org/wiki/Local_Group&quot; rel=&quot;nofollow&quot;&gt;local group&lt;/a&gt; and a further few tens of galaxies in some other structures out to about 30 million light years, and it is only from this small sample that any can exhibit blueshifts. The vast majority of known galaxies are further way and are swept along in the general universal expansion and hence away from us. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-02-23T12:06:26.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9972" PostTypeId="2" ParentId="9969" CreationDate="2015-02-23T14:22:50.467" Score="1" Body="&lt;p&gt;&lt;strong&gt;Moon Phase&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;With the moon phase, it's going to be in &lt;a href=&quot;https://lunaf.com/english/moon-phases/lunar-calendar-2015/03/09/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Waning Gibbous on March 9th&lt;/a&gt;, so it will be reflecting a lot of light back to earth. Any light that interferes with the sky will make dimmer objects harder to see.&lt;/p&gt;&#xA;&lt;p&gt;&lt;kbd&gt;&lt;img src=&quot;https://lunaf.com/images/moon-phases/lunar_phase_18.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/kbd&gt;&lt;/p&gt;&#xA;&lt;p&gt;The Moon's reflectivity figures taken from &lt;a href=&quot;https://lunaf.com/english/moon-phases/lunar-calendar-2015/03/09/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Fraction of the moon: 0.61371425766053&lt;/p&gt;&#xA;&lt;p&gt;Illumination of the moon disk: 88 %&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Location&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;It will also depend a lot on your home location (hemisphere) as to the position of the Milky Way at different times. Your proximity to any light pollution will of course have an impact, as living on the outskirts of a city myself, it is impossible to view the Milky Way. Only when in very rural areas on clear nights have I had a good view of it.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;New Moon&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;On the &lt;a href=&quot;https://lunaf.com/english/moon-phases/lunar-calendar-2015/03/20/&quot; rel=&quot;nofollow noreferrer&quot;&gt;20th March there is a New Moon&lt;/a&gt;, so this may be a more opportune time to try to capture it.&lt;/p&gt;&#xA;" OwnerUserId="2485" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-02-25T08:55:08.623" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9973" PostTypeId="1" CreationDate="2015-02-23T16:44:45.120" Score="3" ViewCount="3902" Body="&lt;p&gt;I watched a video yesterday on how telescopes not only magnify images but also increase the amount of light entering into your eye, allowing you to see things not normally visible with naked eye. Would it be possible to create a &quot;telescope&quot; with no magnification, a wide field of view... say ~190 degrees left/right, ~120 degrees up/down (the average human field of view), and small enough to wear? Essentially creating glasses that allow for better star gazing? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A side question is do night vision goggles let you see more stars? or does the increase in ambient light from earthly objects negate any increase. &lt;/p&gt;&#xA;" OwnerUserId="6629" LastActivityDate="2015-04-26T12:21:44.237" Title="Enhanced Star-Gazing with Special Glasses" Tags="&lt;telescope&gt;&lt;star-gazing&gt;&lt;optics&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9974" PostTypeId="1" AcceptedAnswerId="9981" CreationDate="2015-02-23T20:23:03.677" Score="4" ViewCount="117" Body="&lt;p&gt;The Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Sphere_of_influence_(astrodynamics)&quot; rel=&quot;nofollow&quot;&gt;Sphere of influence&lt;/a&gt; states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;A &lt;strong&gt;sphere&lt;/strong&gt; of influence (SOI) in astrodynamics and astronomy is the&#xA;  &lt;strong&gt;oblate-spheroid-shaped&lt;/strong&gt; region around a celestial body where the&#xA;  primary gravitational influence on an orbiting object is that body.&quot;&#xA;  (Emphasis added.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It then gives an expression for the radius of the sphere of influence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the SOI a spherical region or a oblate-spheroid-shaped region? &#xA;If it is an oblate-spheroid-shaped region, then why?&lt;/p&gt;&#xA;" OwnerUserId="5537" LastActivityDate="2015-02-24T12:59:27.517" Title="Is the SOI a spherical region or a oblate-spheroid-shaped region?" Tags="&lt;gravity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9979" PostTypeId="2" ParentId="9973" CreationDate="2015-02-24T02:32:04.550" Score="2" Body="&lt;p&gt;Stereo Vision?&#xA;Optically, you'll need something like a couple big &lt;a href=&quot;http://en.wikipedia.org/wiki/Fisheye_lens&quot; rel=&quot;nofollow&quot;&gt;Fisheye lenses&lt;/a&gt; mounted on a helmet.&#xA;Looking straight through such optics yields horrible distortion, so you'd have to feed the lens' output into CCD/CMOS sensors, and massage that into flattened images your eyes can handle, on a video screen. Perhaps &lt;a href=&quot;https://www.google.com/get/cardboard/&quot; rel=&quot;nofollow&quot;&gt;Google Cardboard&lt;/a&gt;?&#xA;Probably doable and light enough to wear.&#xA;A straight optical approach to correcting the fisheyes would probably end up too heavy to lift.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-02-24T02:32:04.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9981" PostTypeId="2" ParentId="9974" CreationDate="2015-02-24T12:59:27.517" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the SOI a spherical region or a oblate-spheroid-shaped region?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sphere of influence is neither a sphere nor an oblate spheroid. It is a surface with no name. An approximation of this surface is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac r R\right)^{10}(1+3\cos^2\theta) = \left( \frac m M \right)^4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is neither a sphere nor an oblate spheroid, and this is but an approximation. Thefull expression is an absolute mess. Dropping the factor of $(1+3\cos^2\theta)^{1/10}$ (which is close to one) yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r = \left( \frac m M \right)^{2/5} R$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tada! The equation of a sphere!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The true surface is defined in terms of two ratios. Consider two gravitating bodies, call them body A and body B. From the perspective of an inertial frame, the acceleration of a tiny test mass toward these two bodies is given by Newton's law of gravitation. These two bodies accelerate toward one another as well, so a frame based at the origin of either body is non-inertial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the perspective of a frame at the center of body A, the acceleration of the test mass is the inertial frame acceleration of the test mass toward body A plus the inertial frame acceleration of the test mass toward body B less the acceleration of body A toward body B. Denote the acceleration of the test mass toward body A as the primary acceleration and the difference between the inertial frame accelerations of the test mass and body A toward body B as the disturbing acceleration. Finally, define $Q_A$ as the ratio of these two. Now do the same for a frame with origin at the center of body B. The sphere of influence is that surface where $Q_A = Q_B$.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-02-24T12:59:27.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9982" PostTypeId="2" ParentId="9969" CreationDate="2015-02-24T14:42:44.460" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the Moon phase still affect the visibility of the Milky Way even if it is not above the horizon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. The main culprit is the Sun. Lunar twilight is extremely weak. It essentially ends when the Moon goes below the horizon. See &lt;a href=&quot;https://physics.stackexchange.com/a/2016/52112&quot;&gt;this answer&lt;/a&gt; at the physics stackexchange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One problem is that this is the wrong time of the year to be looking at the Milky Way. The brightest part of the bands we call the Milky Way is the core, and that's best seen from May to October. Right now the core of the Milky Way is visible from the Northern Hemisphere for only the last few hours before sunrise, and it's not very high on the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another problem, however ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Moon doesn't even rise until 9:12am.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's backwards. On March 8-10 2015, the Moon rises between 10:00 PM to midnight and sets between 9:00 AM to 10:30 AM. The Moon will be big and fat in the wee hours of the morning sky, which is when the core of the Milky Way would otherwise be visible.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-02-24T14:50:15.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9983" PostTypeId="1" CreationDate="2015-02-24T21:38:14.943" Score="1" ViewCount="3945" Body="&lt;p&gt;What is the temperature of the light that leaves the Sun towards the Earth and does it cool down during it's journey here ? If it doesn't cool down then how come it isn't frying the Earth ? Is it our atmosphere that is preventing it from doing so or does it cool down during it's trip here in space ?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastEditorUserId="2531" LastEditDate="2015-02-24T23:02:09.737" LastActivityDate="2015-02-26T20:00:47.437" Title="Temperature of the Sun's rays" Tags="&lt;the-sun&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9986" PostTypeId="2" ParentId="9983" CreationDate="2015-02-24T23:08:05.057" Score="1" Body="&lt;p&gt;The Sun's radiation spectrum is close to thermal at $T_\odot = 5778\,\mathrm{K}$. That means that that power output, given by the Stefan–Boltzmann law, is around $L = \sigma A T^4$, where $\sigma = 5.6704\times 10^{-8} \,{\mathrm{W}}/{\mathrm{m}^2\mathrm{K}^4}$ is the Stefan–Boltzmann constant and $A$ is its surface area:&#xA;$$L_\odot = \sigma(4\pi R_\odot^2)T_\odot^4 = 3.85\times10^{26}\,\mathrm{W}\text{.}$$&#xA;However, at the Earth's orbital radius $r = 1\,\mathrm{AU}$, this radiation is spread across a surface area of $4\pi r^2$, so the power incident on an Earth-sized sphere would be just&#xA;$$P = L_\odot\frac{\pi R_\oplus^2}{4\pi r^2}\text{,}$$&#xA;so approximating the Earth as a spherical blackbody, the equilibrium temperature is determined by its own thermal radiation balanced by this incident radiation:&#xA;$$\sigma(4\pi R_\oplus^2)T^4 = P\text{,}$$&#xA;which gives a surface temperature of&#xA;$$T =  T_\odot\sqrt{\frac{R_\odot}{2 r}} = 279\,\mathrm{K}\text{.}$$&#xA;Therefore, no special mechanisms are necessary to prevent the Earth from frying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, the Earth is not a blackbody because of its albedo and atmosphere, but its actual global temperature is not too far off from this estimate. The average surface temperature is close to $288\,\mathrm{K}$. Earth's atmosphere actually raises the average temperature, not lowers it.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2015-02-24T23:08:05.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9987" PostTypeId="2" ParentId="9983" CreationDate="2015-02-24T23:18:55.613" Score="1" Body="&lt;p&gt;Yes, you *can *assign a temperature to the radiation from the Sun. It is approximately a blackbody radiation field with a temperature of 5800 Kelvin. But no, it doesn't &quot;cool&quot; on the way to the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radiation field is still a blackbody radiation field with a temperature of 5800K when it reaches the Earth, but its &lt;em&gt;intensity&lt;/em&gt; (or power per unit area) has decreased from $6.4 \times 10^{7}\ Wm^{-2}$ at the surface (photosphere) of the Sun, to about $1.4 \times 10^3\ Wm^{-2}$, when it reaches the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is this latter figure that is important in terms of how much the Earth is heated, not the temperature of the radiation field, because it determines how much energy per second is fed into the Earth's biosphere. An analogy would be an electric bar heater. The element will be emitting radiation at a temperature of 1000 C, but you don't burst into flames if you sit more than a metre away from it, because the power is distributed over a larger area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the absence of an atmosphere, the Earth would achieve an equilibrium temperature, such that it radiated away as much energy as it absorbed. This equilibrium temperature is &lt;em&gt;much&lt;/em&gt; lower than the temperature of the Sun's surface because the power per unit area incident on the Earth is much less than the power per unit area emitted at the Sun's surface. The calculation, which I won't bore you with can be found &lt;a href=&quot;http://en.wikipedia.org/wiki/Planetary_equilibrium_temperature&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&#xA;$$ T = T_{\odot}(1-a)^{1/4}(R_{\odot}/2D)^{1/2},$$&#xA;where $T_{\odot}$ is the temperature of the Sun, $R_{\odot}$ is the radius of the Sun and $D$ is the distance between the Earth and the Sun. $a$ is the &quot;albedo&quot; - the fraction of the Sun's light that is reflected back into space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting numbers - $a=0.3$, $L_{\odot}=3.8\times10^{26}\ W$, $R_{\odot}=7\times10^{8}\ m$, $D=1.5\times 10^{11}\ m$; we find $T= 256 K$ or $-16\ C$. Of course the presence of an atmosphere warms us over this simple calculation by some tens of degrees.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-02-26T20:00:47.437" LastActivityDate="2015-02-26T20:00:47.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9988" PostTypeId="1" CreationDate="2015-02-25T02:21:17.753" Score="3" ViewCount="664" Body="&lt;p&gt;Kelvin stated that gravitational energy released due to contraction was responsible for the energy radiated in the stars. Although the theory was wrong, was the assumption practically right? If it's right, how does contraction take place?&lt;/p&gt;&#xA;" OwnerUserId="2195" LastEditorUserId="2153" LastEditDate="2015-02-25T02:59:05.080" LastActivityDate="2015-02-27T22:24:32.797" Title="Is gravitational energy released when a body contracts?" Tags="&lt;star&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9989" PostTypeId="2" ParentId="9988" CreationDate="2015-02-25T02:34:20.433" Score="7" Body="&lt;p&gt;It is correct. Contracting stars heat up due to their own contraction. This justifies the initial temperature raise in a star, before fusion begins in earnest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kelvin was not 100% wrong, it's just that fusion, over the life time of the star, makes much, much more energy than contraction. The collapse heats up the star at the very beginning, then fusion takes over and makes essentially all the heat you observe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also true of gas giant planets. Even a tiny amount of contraction may raise their temperature measurably. E.g., Jupiter radiates more energy than it receives from the Sun - this is due to contraction, at a rate of 2 cm / year (diameter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TLDR: Gravitational energy becomes heat.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-02-25T23:42:32.857" LastActivityDate="2015-02-25T23:42:32.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9990" PostTypeId="1" AcceptedAnswerId="9991" CreationDate="2015-02-25T03:53:11.943" Score="6" ViewCount="23482" Body="&lt;p&gt;The Sun's rays hit our eyes around 8 minutes after they are emitted from the Sun.&#xA;Does this mean that the Sun that we see is always the Sun as it was some 8 minutes before? I strongly think this must be happening; is it really a fact? Do we always see the Sun's past?&lt;/p&gt;&#xA;" OwnerUserId="6640" LastEditorUserId="15770" LastEditDate="2020-05-03T18:40:30.857" LastActivityDate="2020-05-03T18:40:30.857" Title="When we see the Sun, do we actually see its past?" Tags="&lt;the-sun&gt;&lt;space-time&gt;&lt;time&gt;&lt;special-relativity&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9991" PostTypeId="2" ParentId="9990" CreationDate="2015-02-25T07:19:22.547" Score="21" Body="&lt;p&gt;&lt;strong&gt;Yes&lt;/strong&gt;, you are right. We don't only see the Sun 8 minutes in the past, we actually see the past of everything in space. We even see our closest companion, the Moon, 1 second in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The further an object is from us the longer its light takes to reach us since the speed of light is finite and distance in space are really big.&lt;/p&gt;&#xA;" OwnerUserId="1678" LastEditorUserId="2153" LastEditDate="2015-03-23T22:48:16.033" LastActivityDate="2015-03-23T22:48:16.033" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9993" PostTypeId="1" CreationDate="2015-02-25T18:33:48.263" Score="5" ViewCount="232" Body="&lt;p&gt;I am just wondering whether any man-made satellites are currently orbiting our Moon. If there is one, would I be able to see it using my telescope?&lt;/p&gt;&#xA;" OwnerUserId="5568" LastEditorUserId="2153" LastEditDate="2015-02-25T23:23:37.357" LastActivityDate="2015-02-26T11:15:13.477" Title="Is there any man-made satellite orbiting our moon?" Tags="&lt;the-moon&gt;&lt;artificial-satellite&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9994" PostTypeId="2" ParentId="9993" CreationDate="2015-02-25T18:50:09.003" Score="5" Body="&lt;p&gt;&lt;a href=&quot;http://www.astronomy.com/news/2015/02/the-lunar-reconnaissance-orbiter-discovers-lunar-hydrogen-more-abundant-on-moons-pole-facing-slopes&quot;&gt;The Lunar Reconnaissance Orbiter&lt;/a&gt; is still up there. At &lt;a href=&quot;http://en.wikipedia.org/wiki/Lunar_Reconnaissance_Orbiter&quot;&gt;a compact 2000kg&lt;/a&gt;, it's likely too small to see via earth based telescope. &lt;a href=&quot;http://www.nasa.gov/mission_pages/LRO/main/index.html#.VO4ZAkak85k&quot;&gt;Mission Page&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the &lt;a href=&quot;http://www.universetoday.com/115887/chinese-unmanned-lunar-orbiter-returns-home-safely-paves-path-for-ambitious-lunar-sample-return/&quot;&gt;Chinese orbiter is back on earth now&lt;/a&gt;, but there may be others.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-02-25T18:50:09.003" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9996" PostTypeId="2" ParentId="9988" CreationDate="2015-02-26T08:01:57.210" Score="3" Body="&lt;p&gt;Gravitational contraction will always release gravitational potential energy. &lt;strong&gt;In most systems, where this happens slowly, you can apply the virial theorem to say that half the released energy is radiated and half is used to heat the contracting gas&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real question is whether the release of gravitational PE is &lt;em&gt;significant&lt;/em&gt; compared with other sources. In most stars it is not. Nuclear fusion is responsible for extending their lives way beyond the Kelvin-Helmholtz timescale - the amount of GPE divided by their luminosity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some stars GPE is hardly released at all though. The capability to contract depends on the behaviour of gas's pressure as a function of temperature. If pressure is independent of temperature then no contraction of an object is possible even as it radiates away its energy and even if it has no other energy sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the case for white dwarfs supported by electron degeneracy pressure. They cannot contract further and will simply cool down at constant radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter is an interesting intermediate case. Yes it is contracting, but not at the rate you would expect for a perfect gas. Eventually, its contraction will slow right down and it will just cool.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-02-26T09:02:06.340" LastActivityDate="2015-02-26T09:02:06.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9997" PostTypeId="2" ParentId="7861" CreationDate="2015-02-26T09:23:20.057" Score="2" Body="&lt;p&gt;To a first approximation when we calculate how fast an object is orbiting around some mass distribution we can assume that the gravitational attraction it experiences is only that due to the mass &lt;em&gt;interior&lt;/em&gt; to its orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approximation, known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Shell_theorem&quot; rel=&quot;nofollow&quot;&gt;shell theorem&lt;/a&gt; should only be applied when the mass distribution is spherically symmetric, or most of the mass is concentrated well inside the orbit of the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bulge of our Galaxy can be considered roughly spherically symmetric, and it appears that the mass distribution is strongly concentrated towards the centre.&#xA;Hence we can say that the centripetal force on an orbiting object is provided by the gravitational force due to the mass interior to its (again, I am assuming a circular orbit for the sake of argument, but this doesn't greatly matter) orbit is&#xA;$$ m\frac{v^{2}}{r} = Gm \frac{M(r)}{r^2},$$&#xA;where $m$ is the mass of the object (which cancels out) and $M(r)$ is the mass inside radius $r$. Thus the orbital speed $v$ is given by $\sqrt{GM(r)/r}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus by observing orbital speeds we can calculate roughly what mass exists &lt;em&gt;interior&lt;/em&gt; to the orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of stars near the centre of our Galaxy, that turns out to be 4 million solar masses, packed into a space that is extremely small. The reason it is thought to be a black hole is not because of the mass, but because it is mass that is (i) densely packed and (ii) cannot be seen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if it were a bunch of massive stars shrouded in dust, so that you could not see it in visible light, you can easily calculate that the heated dust should be kicking off huge amounts of infrared radiation; and that is not seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the Sun, well, the mass interior to the Earth's orbit is &lt;em&gt;much&lt;/em&gt; larger than 4 million solar masses, it is many &lt;em&gt;billions&lt;/em&gt; of solar masses. The contribution of the black hole at the centre is completely swamped by all the other mass in our Galaxy, so we are not &lt;em&gt;directly&lt;/em&gt; influenced by the central black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you removed the central black hole, the stars that were near it would be orbiting way to fast for the new $M(r)$ that they experience. They would fly off and assume highly elliptical orbits with much larger semi-major axes, though I think they would still be part of the Galactic bulge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars further out, including our Sun would have their orbits changed imperceptibly, because $M(r)$ will hardly have changed for them.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-02-26T09:23:20.057" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9999" PostTypeId="2" ParentId="9993" CreationDate="2015-02-26T11:15:13.477" Score="1" Body="&lt;p&gt;Except for LRO, an upper stage of a payload destined to lunar orbit, might stay in lunar orbit for a while, and be much larger than the payload itself. But according to &lt;a href=&quot;https://space.stackexchange.com/questions/221/which-artificial-satellites-in-lunar-orbit-are-currently-active&quot;&gt;this answer&lt;/a&gt; it would be short lived without station keeping.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-02-26T11:15:13.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10000" PostTypeId="2" ParentId="9983" CreationDate="2015-02-26T13:44:42.343" Score="0" Body="&lt;p&gt;An attempt to clarify/answer both this question and your &#xA;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/8854/heat-from-other-stars&quot;&gt;Heat from other Stars&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Temperature is a property of matter, not of light. As others have &#xA;noted, you can assign a temperature to light, but I think that just &#xA;confuses the issue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's true the (mostly hydrogen and helium) atoms at the Sun's surface &#xA;are about 10,000F. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, these atoms don't fly through space and then strike &#xA;Earth. Even with their high temperature, most of them can't escape the &#xA;Sun's gravity and remain on the Sun. The few that do escape form the &#xA;&quot;solar wind&quot;, which has virtually no effect on the Earth's &#xA;temperature.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What travels though space is electromagnetic radiation (light), which &#xA;is measured by intensity and frequency. Light is not a physical &#xA;object, and doesn't have a temperature in the traditional sense of the &#xA;word. &lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-02-26T13:44:42.343" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10001" PostTypeId="1" CreationDate="2015-02-26T13:53:17.507" Score="1" ViewCount="1606" Body="&lt;p&gt;Does this happen when there is a full moon?&lt;/p&gt;&#xA;" OwnerUserId="5532" LastActivityDate="2015-02-27T16:37:28.407" Title="What are the positions of the Earth, Moon, and Sun at a time when there is the least difference between high and low tides?" Tags="&lt;tidal-forces&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10003" PostTypeId="2" ParentId="10001" CreationDate="2015-02-26T21:34:11.100" Score="2" Body="&lt;p&gt;'Neap' tides when there is the least difference between high and low tide, generally occur around the time of the first and third quarter moons.&#xA;'Spring' tides when there is the largest difference between high and low tide, generally occur around the time of the New and Full moons.&#xA;The exact timing is complicated by your position on earth, the shape / size and depth of the ocean, coastlines also the moons distance from the earth and earths distance from the sun.&lt;/p&gt;&#xA;" OwnerUserId="4041" LastEditorUserId="4041" LastEditDate="2015-02-26T21:40:48.853" LastActivityDate="2015-02-26T21:40:48.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10004" PostTypeId="1" AcceptedAnswerId="10005" CreationDate="2015-02-27T03:53:53.803" Score="3" ViewCount="593" Body="&lt;p&gt;I read on &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomy_on_Mars#Long-term_variations&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As on Earth, the period of rotation of Mars (the length of its day) is slowing down. However, this effect is three orders of magnitude smaller than on Earth because the gravitational effect of Phobos is negligible and the effect is mainly due to the Sun.&lt;a href=&quot;http://adsbit.harvard.edu/cgi-bin/nph-iarticle_query?1988BAICz..39..168B&quot; rel=&quot;nofollow&quot;&gt;[19]&lt;/a&gt; On Earth, the gravitational influence of the Moon has a much greater effect. Eventually, in the far future, the length of a day on Earth will equal and then exceed the length of a day on Mars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, how far into the future exactly will it be when the length of a day on Earth and Mars will be equal? Also, for how long will their day lengths be within say, one second of each other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;day&quot;, I mean the Mean Solar Day.&lt;/p&gt;&#xA;" OwnerUserId="6656" LastActivityDate="2015-02-27T05:38:03.897" Title="When will a day on Earth &amp; Mars be the same length?" Tags="&lt;mars&gt;&lt;time&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10005" PostTypeId="2" ParentId="10004" CreationDate="2015-02-27T05:38:03.897" Score="7" Body="&lt;p&gt;Since the rate of lengthening of the day on Mars is three orders of magnitude less than that of the Earth for a first go at calculating how long it will be before the day lengths are equal we can ignore the change in day length on Mars. The difference in day lengths is $\approx 40$ minutes. The Earths day lengthens by $\approx 1.7$ ms/100 years. Therefore the day lengths will be equal in: $\frac{40\times 60}{1.7\times 10^{-3}}\times 100 \approx 141 \times 10^6$ years&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-02-27T05:38:03.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10006" PostTypeId="2" ParentId="8421" CreationDate="2015-02-27T08:48:56.117" Score="0" Body="&lt;p&gt;The format of the files seems to be described in &lt;a href=&quot;http://www.ngdc.noaa.gov/stp/space-weather/solar-data/solar-indices/solar_corona/5-degree-data/data/old.data/corona.fmt&quot; rel=&quot;nofollow noreferrer&quot;&gt;this file&lt;/a&gt;, the content of which I've included below.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;The annual files (1939-1993) contain the green coronal emission line&#xA;530.3 nm observations.  The coronal intensities are given in millionths&#xA;of intensity of the solar disk (coronal units) and converted to the&#xA;photometrical scale of Lomnicky Stit Station at a height of 40&amp;quot; above&#xA;the solar limb.&#xA;&#xA;Several stations were used in this database with Lomnicky Peak being the&#xA;primary station since 1965.&#xA;&#xA;Reference:  Rybansky, et al, 1994, Solar Physics 152, p 153-159.&#xA;&#xA;Data format:&#xA; &#xA; COLUMN  FORMAT   DESCRITPION&#xA;  1-  2    I2     Year, last two digits&#xA;  3-  4    I2     Month&#xA;  5-  6    I2     Day&#xA;  7-  8    2X     Blank&#xA;      9    I1     Coronal station code:&#xA;                  0 - Interpolated data&#xA;                  1 - Lomnicky Stit&#xA;                  2 - Sacramento Peak&#xA;                  3 - Norikura&#xA;                  4 - Kislovodsk&#xA;                  5 - Pic Du Midi&#xA;                  6 - Wenselstein&#xA;                  7 - Arosa&#xA;                  8 - Kanzelhohe&#xA; 10- 13    I4     Time of the observation in hours (or zero)&#xA; 14- 17    I4     Time of the observation in minutes (or zero)&#xA; 18-305  72(I4)   Coronal intensities (72 values) from 0 degrees to 355&#xA;                  degrees of position angle in increments of 5 degrees.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Using the first row data as an example...&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;DATE        SC  H   M   1   2   3   4   5   6   7   8   9   10  11  12  13  14  15  16  17  18  19  20  21  22  23  24  25  26  27  28  29  30  31  32  33  34  35  36  37  38  39  40  41  42  43  44  45  46  47  48  49  50  51  52  53  54  55  56  57  58  59  60  61  62  63  64  65  66  67  68  69  70  71  72&#xA;-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------&#xA;19390101    0   0   0   0   0   0   0   0   0   0   0   3   3   3   8   15  22  34  41  52  46  40  52  68  59  49  35  17  8   3   3   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   3   3   6   8   12  19  31  40  50  47  43  38  53  73  71  65  56  47  38  24  17  12  6   0   0   0   0   0   0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;19390101 is the date (1939-01-01).&#xA;The first zero (Marked as SC) is the station code, the second the recorded hour, and the third the recorded minute (H/M). Most of this is 0 due to being interpolated data.&lt;/p&gt;&#xA;&lt;p&gt;After this information, there are 72 remaining columns (numbered 1-72 above), which represents collected data consolidated in groups of 5 degrees, presumably starting at 0.&lt;/p&gt;&#xA;&lt;p&gt;Given the amount of interpolation of old data the value may be questionable, but newer data appear to be more complete.&lt;/p&gt;&#xA;&lt;p&gt;The data format might be a little overwhelming to work with, but you can organize it in excel quite easily - paste it in and each line break will form a row, then use the text-to-column function to break each data into its own column.&lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-03-29T10:04:32.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10008" PostTypeId="2" ParentId="10001" CreationDate="2015-02-27T16:31:32.597" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What are the positions of the Earth, Moon, and Sun at a time when there is the least difference between high and low tides?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;TL;DR response: The simple answer to this question is twice per month, when the Moon is half full. A not so simple answer is roughly every 8.85 years, when the Moon is half full and the Moon is close to apogee. A full answer: Who knows? There are thousands of different frequencies involved, and their contributions change over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The greatest tidal variations (difference between high tide and low tide) occur when the tides induced by the Moon and the Sun are in phase. These are the spring tides, and this happens twice per month, roughly near full moon and new moon (but with some variation place to place). The smallest variations (which is what the question asks) occur when the tides induced by the Moon and Sun are out of phase. These are the neap tides, and this also happens twice per month, roughly when the Moon is half full, or first and third quarter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon's contributions to the tides is about twice that of the Sun (but this also varies from place to place). The Moon's orbit is eccentric, so tidal variation will be even greater when the Moon is full or new &lt;strong&gt;and&lt;/strong&gt; the Moon is closest to the Earth. These are the perigean spring tides. The flip side of this effect, which is what the question asks, would be the neap tides that occur when the Moon is furthest from the Earth. These are the apogean neap tides. See Donald Olson (2012), &quot;Perigean Spring Tides and Apogean Neap Tides in History,&quot; &lt;em&gt;American Astronomical Society Meeting Abstracts# 219&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've described three different frequency effects above. There are a huge number of other frequency considerations that affect the magnitudes of the tides. Modern tidal theory is the result of the combined efforts of William Thomson (Lord Kelvin), George Darwin, E.W. Brown, and A.T. Doodson. By Doodson's time, there were 388 different frequency components of the tides in use. My description above captures just three of them.  Nowadays, people use over a thousand different frequencies to fully characterize tidal behaviors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these effects are small compared to those that result in the spring tides and neap tides. The tidal variations (difference between high tide and low tide) that distinguish the spring tides from the neap tides can be huge, several feet in some places. The tidal variations that distinguish the perigean spring tides from the apogean spring tides are much, much smaller, on the order of inches. The other thousands of effects are smaller yet.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2015-02-27T16:37:28.407" LastActivityDate="2015-02-27T16:37:28.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10010" PostTypeId="2" ParentId="9988" CreationDate="2015-02-27T22:24:32.797" Score="0" Body="&lt;p&gt;In addition to the answers by Florin and Rob, I'd like to point out that the gravitational energy in the end dominates over all other sources, like fusion, at least for massive stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a massive star eventually collapses to a neutron star, the gravitational pressure is so strong that protons and electrons are pressed into neutrons and all heavy nuclei are destroyed. In other words, the energy gained by fusion was merely a credit payed back with huge interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation is even more dramatic when the star collapses to a black hole. Such objects release an energy of $\epsilon M c^2$ with efficiency $\epsilon\sim0.1$ depending on the details (e.g. the spin of the black hole). This is far more than the $\epsilon&amp;lt;0.01$ from Hydrogen fusion and by far the most efficient natural energy source. The energy emitted from the formation of supermassive black holes equals (approximately) the binding energy of their host galaxies.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2015-02-27T22:24:32.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10012" PostTypeId="1" CreationDate="2015-02-28T04:01:52.597" Score="6" ViewCount="424" Body="&lt;p&gt;Observationally, how can this be established? &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is known theoretically about the process?&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any reference papers to read?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="2531" LastEditDate="2015-02-28T15:08:15.407" LastActivityDate="2017-03-13T13:42:26.243" Title="How can we tell that a short-period binary is tidally locked?" Tags="&lt;binary-star&gt;&lt;tidal-forces&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10013" PostTypeId="1" CreationDate="2015-02-28T04:35:41.637" Score="5" ViewCount="1703" Body="&lt;p&gt;I was given an exoplanet similar in size and distance to host star to our own earth. It's orbiting a star with luminosity six times our sun, the greenhouse coefficient 0.3, bond albedo 0.3.&lt;/p&gt;&#xA;&lt;p&gt;I would like to calculate the approximate surface temperature (i.e. close enough to know whether it is in habitable range).&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Disclaimer:&lt;/strong&gt; This question was on a past exam and we never got our test back.&lt;/p&gt;&#xA;" OwnerUserId="6665" LastEditorUserId="37191" LastEditDate="2021-02-01T22:49:32.263" LastActivityDate="2021-02-01T22:49:32.263" Title="How to calculate the temperature of an exoplanet?" Tags="&lt;exoplanet&gt;" AnswerCount="0" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10015" PostTypeId="2" ParentId="10012" CreationDate="2015-02-28T10:29:18.980" Score="4" Body="&lt;p&gt;The phenomenon of tidal locking occurs primarily in short period binary systems with convective envelopes. The tides raise bulges in the stars and these interact with convective motions to dissipate energy and to synchronise the orbital and rotational periods of stars and circularise their orbits. The process is much less efficient in stars with radiative envelopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we tell observationally? Well, we have to show that the rotation period of a star is the same as its orbital period. The orbital period is easily found, either by measuring eclipses, in the case of binary systems with a high inclination, or by measuring the motions of the star(s) using the doppler shift in their spectra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measuring rotation periods is also not so difficult. Fast-rotating stars with convective envelopes have magnetic activity and starspots (analogous to sunspots, but covering a larger area). As the star rotates then the light from the star goes up and down depending on what starspots are on the visible surface. This modulation gives the rotation period. In eclipsing binaries it can be difficult to separate out spots from eclipses, but here one can use the known orbital inclination and known radii of the stars to see whether their projected rotation velocities (measured from the widths of spectral absorption lines) matches the predictions from a tidally locked orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many known tidally locked bnaries. Any low-mass binary with an orbital period of less than a few days is almost certainly tidally locked, but the situation at orbital periods greater than about about 6-8 days becomes more complicated (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2006ApJ...653..621M&quot; rel=&quot;nofollow noreferrer&quot;&gt;Meibom et al. 2006&lt;/a&gt;). The plot below is taken from a review by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2008EAS....29....1M&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mazeh (2008)&lt;/a&gt;, showing the rotation vs orbital periods from some small, but well-defined samples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ymQBO.png&quot; alt=&quot;Orbital vs rotation period from Mazeh (2008).&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absolutely classic, must-read, theoretical work on this is by &lt;a href=&quot;http://adsabs.harvard.edu/abs/1989A%26A...220..112Z&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zahn (1989)&lt;/a&gt;. The timescale for synchronisation depends on structural properties of the stars, their moments of inertia, the binary mass ratio and the ratio of their separation to their radii &lt;em&gt;to the power of 6&lt;/em&gt;. It is this latter property that means tidal locking is nearly bimodal; binaries with periods shorter than some threshold (about 6 days) are usually tidally locked, but those with slightly longer periods almost never are.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-03-13T13:42:26.243" LastActivityDate="2017-03-13T13:42:26.243" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10016" PostTypeId="1" CreationDate="2015-02-28T13:22:09.990" Score="3" ViewCount="503" Body="&lt;p&gt;I was reading an article on planetary differentiation, and apparently internal heat production plays a major role. There are several sources of such heat described, such as tidal heat, radiogenic heat, etc. What I did not understand however is how, on chemical level, does warming up a system &quot;heterogenize&quot; it. Logically, wouldn't an addition of kinetic energy mean that molecules bump around more violently, therefore becoming more mixed up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or does this 'internal heat' story mean that the heat is actually escaping the planet, therefore kinetic energy is lost, therefore the planet is actually cooling down?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;source: &lt;a href=&quot;http://geology.isu.edu/wapi/Geo_Pgt/Mod03_PlanetaryEvo/mod3_pt1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://geology.isu.edu/wapi/Geo_Pgt/Mod03_PlanetaryEvo/mod3_pt1.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4126" LastEditorUserId="7982" LastEditDate="2019-05-02T00:14:54.143" LastActivityDate="2020-12-09T07:56:19.410" Title="Internal heat and planetary differentiation" Tags="&lt;planet&gt;&lt;heat&gt;&lt;astrochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10017" PostTypeId="1" AcceptedAnswerId="10018" CreationDate="2015-02-28T14:00:55.660" Score="9" ViewCount="2450" Body="&lt;p&gt;The Moon orbits the Earth.&lt;br&gt;&#xA;The Earth orbits the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the Sun orbit another bigger star?&lt;br&gt;&#xA;If so, does this star orbit, in turn, a very big star?&lt;br&gt;&#xA;... etc ...   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are all the intermediate subsystems up to motion around the center of the Milky Way?  &lt;/p&gt;&#xA;" OwnerUserId="6668" LastEditorUserId="2365" LastEditDate="2017-05-18T16:17:57.483" LastActivityDate="2018-11-27T13:19:21.893" Title="Does the Sun turn around a big star?" Tags="&lt;star&gt;&lt;orbit&gt;&lt;the-sun&gt;" AnswerCount="5" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10018" PostTypeId="2" ParentId="10017" CreationDate="2015-02-28T15:39:54.787" Score="16" Body="&lt;p&gt;The Sun is not within the gravitational sphere of influence of any other star. The centre of mass of the solar system (which is very close to the Sun) instead orbits in the general Galactic gravitational potential. Because this has a roughly cylindrical symmetry (the Galaxy is basically a disk with a bulge in the middle), this means that it executes a (roughly) circular orbit around the centre of the Galaxy, taking about 230 million years to do so. At the same time it is wobbling in the vertical plane of the Galactic disc, up and down with a cycle of about 70 million years (see &lt;a href=&quot;https://astronomy.stackexchange.com/questions/822/how-far-is-the-earth-sun-above-below-the-galactic-plane-and-is-it-heading-towar&quot;&gt;How far is the Earth/Sun above/below the galactic plane, and is it heading toward/away from it?&lt;/a&gt; )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is really nothing intermediate because the stars in the milky way form a &quot;collisionless system&quot; they don't really interact gravitationally on an individual basis. Stars can be influenced by perturbations in the smooth Galactic potential caused by massive star clusters, giant molecular clouds and spiral arms. This is thought to be why the velocity dispersions of stars around regular circular orbits increases with age.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-02-28T15:39:54.787" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10019" PostTypeId="1" CreationDate="2015-02-28T17:01:25.753" Score="2" ViewCount="3288" Body="&lt;p&gt;Telescope: Orion 09843 SpaceProbe 3,3 inch aperture(76mm), 700mm focal length, eyepieces - 25mm and 10mm focal lengths. Please tell me whether it is possible  to see nebula and  galaxies.&lt;/p&gt;&#xA;" OwnerUserId="5568" LastActivityDate="2015-04-30T18:45:46.223" Title="Nebula and galaxies using 70mm scope" Tags="&lt;telescope&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10020" PostTypeId="2" ParentId="10019" CreationDate="2015-02-28T20:13:11.377" Score="2" Body="&lt;p&gt;You can see some of them with a naked eye. Many of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Messier_object&quot; rel=&quot;nofollow&quot;&gt;Messier objects&lt;/a&gt;, including galaxies and nebulae are observable even without any telescopes. As for others, if C&lt;a href=&quot;https://en.wikipedia.org/wiki/Charles_Messier&quot; rel=&quot;nofollow&quot;&gt;harles Messier&lt;/a&gt; could catalog them in 1771 with a 4 inch refractor, you should see most using your 3.3 inch reflector too given good observation conditions, knowing where to point and using a properly collimated and otherwise maintained telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Messier was from the Northern hemisphere and you seem to be closer to the equator judging by information you provide in your profile, so you shouldn't have problems seeing these objects either, and there are equally interesting objects also in the Southern skies that you could observe. Some of the easiest recognizable on the Northern skies are the &lt;a href=&quot;https://en.wikipedia.org/wiki/Andromeda_Galaxy&quot; rel=&quot;nofollow&quot;&gt;Andromeda galaxy&lt;/a&gt;, or Messier 31, and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Orion_Nebula&quot; rel=&quot;nofollow&quot;&gt;Orion Nebula&lt;/a&gt;, or Messier 42. You should also be able to discern quite a lot of detail of other astronomical bodies, including e.g. &lt;a href=&quot;https://en.wikipedia.org/wiki/Pleiades&quot; rel=&quot;nofollow&quot;&gt;Pleiades&lt;/a&gt;, or Messier 45 which is also visible to the naked eye and many astronomers use to determine observational conditions and their visual acuity by.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditDate="2015-02-28T20:20:32.047" LastActivityDate="2015-02-28T20:20:32.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10021" PostTypeId="2" ParentId="10017" CreationDate="2015-02-28T22:12:01.023" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the Sun turn around a big star?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. Such a star, if it existed, would easily be the brightest star in the sky. You would have been taught about it early on in school if it existed. But it doesn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a while it was conjectured that the Sun had a small companion star to explain a perceived periodicity in mass extinction events. This too has been ruled out by the Wide-field Infrared Survey Explorer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What are all the intermediate subsystems up to motion around the center of the Milky Way?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our Sun, being a single star, is a bit of an oddity. Most stars are members of multiple star systems, typically pairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some stars occur in clusters. The Pleiades is a relativity nearby (440 light years) cluster of stars. Someone with extremely keen eyesight and exceptionally good viewing conditions, might be able to see 14 stars of the over 3,000 stars that form this cluster. Open clusters such as the Pleiades don't last long. The stars in an open cluster are only weakly bound to the cluster and are eventually dispersed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A key feature of the Milky Way is its spiral arms. Our Sun is currently in a lesser arm of the Milky Way, the Orion Arm. Stars however are not gravitationally bound to spiral arms. One widely used explanation of the spiral arms is that they are gravitational traffic jams in space.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We could also ask the same beyond...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our galaxy is a member of the Local Group, which in turn is a member of the Virgo Supercluster, which in turn is a part of the Laniakea Supercluster. Even larger scale objects include galaxy filaments. And that's where the hierarchy ends. The expansion of space overtakes gravity at such immense distances.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-02-28T22:12:01.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10022" PostTypeId="1" CreationDate="2015-02-28T23:57:40.137" Score="5" ViewCount="137" Body="&lt;p&gt;In a very recent edition of New Scientist, an article &lt;a href=&quot;http://www.newscientist.com/article/dn26983-mystery-cloudlike-blobs-over-mars-baffle-astronomers.html#.VPJUo_mUcVY&quot; rel=&quot;noreferrer&quot;&gt;Mystery cloud-like blobs over Mars baffle astronomers&lt;/a&gt; (16 Feb, 2015), detailed that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;on 12 March 2012, amateur astronomers around the world noticed a strange blob rising out of the planet's southern hemisphere, soaring to 250 kilometres above the surface.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to the article, this anomaly grew to be about 1000km across and even had 'fingers' stretching to space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article, details several theories (some being a bit wild) including atmospheric processes, aurora and stretching reality quite a bit - aliens (from the article, not me).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An image of the plume is shown below (a still image from &lt;a href=&quot;https://www.youtube.com/watch?v=urpOIB6znKA&quot; rel=&quot;noreferrer&quot;&gt;this YouTube clip&lt;/a&gt;, which states that a 2nd plume was observed some months later:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/U0vvg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plumes appear to have formed in a region called &lt;a href=&quot;http://en.wikipedia.org/wiki/Terra_Cimmeria&quot; rel=&quot;noreferrer&quot;&gt;Terra Cimmeria&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What are scientific theories, models etc for the formation of the cloud-like blobs in the Martian southern hemisphere?&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerDisplayName="user2449" LastEditorDisplayName="user2449" LastEditDate="2015-03-01T06:38:38.330" LastActivityDate="2015-03-03T06:16:31.073" Title="What are the cloud-like blobs in the Martian southern hemisphere?" Tags="&lt;mars&gt;&lt;planetary-atmosphere&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10023" PostTypeId="2" ParentId="10017" CreationDate="2015-03-01T00:03:17.253" Score="1" Body="&lt;p&gt;The Sun does &lt;strong&gt;&lt;em&gt;NOT&lt;/em&gt;&lt;/strong&gt; revolve around an even bigger star. If it did, you'd see two stars, assuming the larger one didn't outshine the Sun. In that case, you'd only see one. Also, the earth would be a lot hotter, getting heat from two stars. We'd also be in danger of solar wind, more solar wind than we'd usually get from just the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, because we don't see two stars and it isn't getting any hotter than it should and we don't get double the solar wind that we do, and add the presumed fact that you weren't taught that in school, it's safe to say that the Sun orbits around nothing. &lt;/p&gt;&#xA;" OwnerUserId="6623" LastActivityDate="2015-03-01T00:03:17.253" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10024" PostTypeId="2" ParentId="10016" CreationDate="2015-03-01T09:42:22.980" Score="7" Body="&lt;p&gt;Your argument (large temperature leads to greater mixing) is correct so long as there are no other large scale forces acting on the system.  This isn't true in planet formation, because gravity plays a very important role.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not an expert on planet formation, but I think the argument goes something like this:  As a planet forms from material from the protoplanetary disc it will begin very homogenous, something like an asteroid, just rock and metal all the way through.  If the planet is heated enough in the core, some material will melt.  Buoyancy (due to gravity) will drive lighter material &quot;higher&quot; in the planet, away from the core.  The more heat you generate the more melting can happen and the more material will start to separate by density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there's a lot more to the story.  Fluid material obeys the laws of hydrodynamics, so large scale convective flows can form in some regions, mixing those areas quite well.  Planetary rotation adds centrifugal and Coriolis forces to the mix, which pull more material to the equator.  The surface of a planet is exposed to space and can radiate excess heat, cooling to a solid (like the Earth's crust).  If the core is magnetic and rotating, metals in the planet may be pushed by the magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this is just to say that planetary formation is a balancing act of a lot of competing forces.  These forces can differentiate (or not!) different types of material, but first (for rocky planets at least) they need to be freed by internal heating.&lt;/p&gt;&#xA;" OwnerUserId="6672" LastActivityDate="2015-03-01T09:42:22.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10025" PostTypeId="2" ParentId="10022" CreationDate="2015-03-01T17:00:33.997" Score="4" Body="&lt;p&gt;From the &lt;a href=&quot;http://www.nature.com/nature/journal/v518/n7540/full/nature14162.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nature article&lt;/a&gt;, the authors propose two alternative explanations:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;We used photometric measurements to explore two possible scenarios and investigate their nature.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;For particles reflecting solar radiation, clouds of CO&lt;sub&gt;2&lt;/sub&gt;-ice or H&lt;sub&gt;2&lt;/sub&gt;O-ice particles with an effective radius of 0.1 micrometres are favoured over dust.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Alternatively, the plume could arise from auroral emission, of a brightness more than 1,000 times that of the Earth’s aurora, over a region with a strong magnetic anomaly where aurorae have previously been detected&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Importantly, both explanations defy our current understanding of Mars’ upper atmosphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I think, they are the most trustworthy interpretations that one can get.&#xA;Please note that, even if the data are from &lt;strong&gt;2012&lt;/strong&gt;, the paper is published in &lt;strong&gt;2015&lt;/strong&gt;, therefore the interpretation is not outdated.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-03-03T06:16:31.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10026" PostTypeId="2" ParentId="10019" CreationDate="2015-03-01T17:55:23.967" Score="1" Body="&lt;p&gt;Whether you'll be able to see them depends on the levels of light pollution in your area. As TildalWave mentioned, a number nebulae and galaxies are perfectly observable with the naked eye so unless you live somewhere very bright, you should be fine. Under really dark skies, objects like M31 are very easy to find with the naked eye. Where I live, I hardly see it with a bright binocular due to very poor seeing, induced by a large number of shopping centres fond of pointing powerful searchlights at the sky for whatever reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to how much you'll be able to see, you can get a general idea of what you should be able to see in ideal conditions by calculating your telescope's &lt;a href=&quot;http://en.wikipedia.org/wiki/Limiting_magnitude&quot; rel=&quot;nofollow&quot;&gt;limiting magnitude&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, you'll also have to factor in the quality of the optics (mainly transmission), both the telescope and the eyepieces. Of the two you mentioned in your question, you should use the 25 mm one. It will give you a lower magnification, a bigger field of view (better for most bright nebulae and galaxies due to their often significant angular dimensions)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a highly configurable &lt;a href=&quot;http://www.cruxis.com/scope/limitingmagnitude.htm&quot; rel=&quot;nofollow&quot;&gt;calculator&lt;/a&gt; that you can use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you live in the northern hemisphere, objects from the Messier catalogue are great candidates to begin your observations with.&lt;/p&gt;&#xA;" OwnerUserId="211" LastEditorUserId="211" LastEditDate="2015-03-01T18:00:35.750" LastActivityDate="2015-03-01T18:00:35.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10027" PostTypeId="1" CreationDate="2015-03-02T01:47:49.123" Score="2" ViewCount="473" Body="&lt;p&gt;I've been looking for sources of printable electronic star charts - there seem to be many different ones available, either as part of planetarium software or just plain charting packages. But there's such a richness of offerings that I'm having trouble figuring out which of those, if any, can produce printable charts in the format that I'm looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like the general format of the Millennium Star Atlas - black stars on a white sky background, and with horizontal lines representing 1 degree of declination and vertical lines representing 4 minutes of RA (see sample page at &lt;a href=&quot;http://www.skyandtelescope.com/wp-content/uploads/GD-Sept2011-Chart.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.skyandtelescope.com/wp-content/uploads/GD-Sept2011-Chart.pdf&lt;/a&gt; ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to be able to specify a range of RA and declination and get that chart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone suggest any free star chart printing software that can print similar output for any desired sector of the sky?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking at Cartes du Ciel (&lt;a href=&quot;http://sourceforge.net/projects/skychart/&quot; rel=&quot;nofollow&quot;&gt;http://sourceforge.net/projects/skychart/&lt;/a&gt;) which might have that capability, but if so, I haven't yet figured out how to enable it.&lt;/p&gt;&#xA;" OwnerUserId="6675" LastActivityDate="2015-05-01T15:49:31.887" Title="Free source of printable star charts in the format of the Millennium Star Atlas?" Tags="&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10028" PostTypeId="1" AcceptedAnswerId="10031" CreationDate="2015-03-02T04:40:52.737" Score="5" ViewCount="1018" Body="&lt;p&gt;What's approximatively the mass of the interstellar neighborhood of the Sun ?&lt;br&gt;&#xA;(for example the neighborhood of radius 50 light-years). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Remark:&lt;/em&gt; The mass of the observable matter should be completed by the mass of the &quot;dark matter&quot;.&lt;/p&gt;&#xA;" OwnerUserId="6668" LastActivityDate="2015-03-14T11:34:34.017" Title="What's the mass of the interstellar neighborhood of the Sun?" Tags="&lt;the-sun&gt;&lt;mass&gt;&lt;dark-matter&gt;&lt;interstellar&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10029" PostTypeId="2" ParentId="10028" CreationDate="2015-03-02T06:53:47.410" Score="2" Body="&lt;p&gt;Very roughly: $3.5 \times 10^{33}kg$, or 1800 solar masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how I came by that number, it is a &lt;em&gt;very&lt;/em&gt; rough approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The major mass components of the galaxy are stars, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Interstellar_medium&quot; rel=&quot;nofollow&quot;&gt;interstellar medium&lt;/a&gt;, and dark matter.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the &lt;a href=&quot;https://github.com/astronexus/HYG-Database&quot; rel=&quot;nofollow&quot;&gt;HYG Database&lt;/a&gt; there are approximately 1000 stars within 50 light years of the Earth.  The average mass of a star is 0.2 solar masses (thanks to Rob Jeffries in the comments), where a solar mass is about $M_\odot \approx 2 \times 10^{30} kg$. This gives us $4\times 10^{32} kg$ of stars nearby.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interstellar medium (ISM) is primarily atomic hydrogen, and has an average density of 1 proton per cubic centimetre, although it can vary &lt;em&gt;widely&lt;/em&gt; in different parts of the galaxy. A proton weighs $1.6 \times 10^{-27} kg$, so a 50 light year sphere of &quot;average&quot; ISM will weigh about $7 \times 10^{32}kg$.  We can do a little bit better than that though.  Our solar system lives in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Local_Interstellar_Cloud&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;local fluff&lt;/em&gt;&lt;/a&gt; which is a cloud in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Local_Bubble&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;local bubble&lt;/em&gt;&lt;/a&gt;.  The local fluff has a radius of about 15 ly, and a density of 0.3 atoms per cubic centimetre. The local bubble is about 150 ly across, and has a density of only 0.05 atoms per cubic centimetre.  Using these figures instead we get an approximate ISM mass of $4 \times 10^{31} kg$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an order of magnitude smaller than the contribution of stars, and our estimate for the stars could easily be off by more than 10%, so let's err on the high side and say the total mass of stars + ISM is $5 \times 10^{32}kg$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't know how much dark matter exactly is in the galaxy, but if its similar to the cosmological average then there is roughly six times as much dark matter as baryonic.  If this holds true locally, then there is about $3\times 10^{33}kg$ of dark matter nearby.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, a rough estimate says there is about $3.5 \times 10^{33}kg$ of mass within 50 light years of us.  This is equivalent to 1800 solar masses, or $2 \times 10^{60}$ protons!&lt;/p&gt;&#xA;" OwnerUserId="6672" LastEditorUserId="6672" LastEditDate="2015-03-02T08:10:07.087" LastActivityDate="2015-03-02T08:10:07.087" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10031" PostTypeId="2" ParentId="10028" CreationDate="2015-03-02T09:37:50.553" Score="7" Body="&lt;p&gt;You don't have to guesstimate to come up with the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you do is look at the dynamics of stars with respect to the Galactic plane - in particular, the velocity dispersions of stars with known distances from the plane, combined with a reasonable assessment of where the Sun is with respect to the plane (close), yields an almost model-independent assessment of the mass density in the disk of our Galaxy in the vicinity of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This work was done using the Hipparcos cataogue by &lt;a href=&quot;http://adsabs.harvard.edu/abs/1998A%26A...329..920C&quot; rel=&quot;nofollow&quot;&gt;Creze et al. (1998)&lt;/a&gt;. They found that the local mass-density (all forms of matter) in the Galactic disk was $0.076 \pm 0.015 M_{\odot}/pc^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus within 50 light years (15.3 pc), because the Galactic disk has an exponential scale height much bigger than this (about 100-200 pc), we can assume a constant density and derive a total mass in the solar neighborhood of $1100\pm 200\ M_{\odot}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This mass is almost compatible with the total mass estimated to be in stars, white dwarfs, neutron stars, brown dwarfs and the interstellar medium gas. For instance, according to &lt;a href=&quot;http://articles.adsabs.harvard.edu//full/2001ASPC..228..249C/0000252.000.html&quot; rel=&quot;nofollow&quot;&gt;Chabrier (2001)&lt;/a&gt;, the total mass density is almost accounted for in the form of 60% main sequence stars plus white dwarfs and neutrons stars, about 5% brown dwarfs and 30% gas, and so &lt;strong&gt;the contribution of dark matter to the local disk density is probably very small&lt;/strong&gt;. Dark matter is not distributed in the same way as normal matter; it did not collapse to a disk, because it is dissipationless and probably exists in a spherical halo. &lt;strong&gt;So this consistency between the local baryonic matter density and the total mass density is not an argument against the dark matter hypothesis&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I found a few estimates of the local dark matter density that use the kinematics of local stars as a constraint. The range of densities quoted is from 0.008 to 0.02 $M_{\odot}$/pc$^3$ (&lt;a href=&quot;http://arxiv.org/abs/1206.0015&quot; rel=&quot;nofollow&quot;&gt;Bovey &amp;amp; Tremaine 2012&lt;/a&gt;; &lt;a href=&quot;http://arxiv.org/abs/1206.0015&quot; rel=&quot;nofollow&quot;&gt;Garbari et al. 2012&lt;/a&gt;) i.e. small compared with the contribution from stars and gas.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-03-14T11:34:34.017" LastActivityDate="2015-03-14T11:34:34.017" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10032" PostTypeId="2" ParentId="10027" CreationDate="2015-03-02T15:37:29.027" Score="1" Body="&lt;p&gt;OK, I think that I've found the answer - it looks like Cartes du Ciel will do what I'm looking for (although I haven't yet figured out the simplest workflow).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click on the icon for Equatorial Coordinates&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then click on the icon for 'configure the program'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the Colour tab, click on 'black on white'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the Lines tab, make sure that 'show grid label' and 'show coordinates grid - line' are checked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zoom in to the desired magnification&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click on printer icon&lt;/p&gt;&#xA;" OwnerUserId="6675" LastActivityDate="2015-03-02T15:37:29.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10033" PostTypeId="2" ParentId="9933" CreationDate="2015-03-02T16:36:19.150" Score="0" Body="&lt;p&gt;The evidence is not only in Earth's climate but Earth's magnetic field too. For more evidence you can search for yourself too. Of cource there is the complete physical mechanism described, not just correlations. And the mathematical documentation is overwhelming.&#xA;Pela: there is no such thing as self-tidal force, the only tidal forces to the Sun are those from the planets orbiting. That's why they play a key role.Self forces sum to zero, eliminate each other to zero.&lt;/p&gt;&#xA;" OwnerUserId="6682" LastActivityDate="2015-03-02T16:36:19.150" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10036" PostTypeId="1" AcceptedAnswerId="10040" CreationDate="2015-03-03T09:53:58.493" Score="4" ViewCount="121" Body="&lt;p&gt;In &lt;a href=&quot;http://motherboard.vice.com/read/the-puzzling-galaxy-at-the-edge-of-time-and-space-3252351&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;, the author writes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This galaxy is so distant that it would have been invisible without the gravitational lensing effects of a large galaxy cluster called Abell 1689, located nearly 2.2 billion light years away in the constellation Virgo. The cluster’s enormous mass acts as a natural telescope by bending and amplifying light from objects behind it, like A1689-zD1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To me this doesn't quite sound the same as what a telescope is doing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;strong&gt;How can a galaxy in-between our view and the galaxy behind it have a 'lensing effect'?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2894" LastActivityDate="2019-02-18T07:51:42.743" Title="How can a galaxy in-between our view and the galaxy behind it have a 'lensing effect'?" Tags="&lt;gravity&gt;&lt;galaxy&gt;&lt;telescope&gt;&lt;gravitational-lensing&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10037" PostTypeId="1" AcceptedAnswerId="10038" CreationDate="2015-03-03T10:06:15.790" Score="3" ViewCount="333" Body="&lt;p&gt;The latest big quasar find at ~12.8 bn LJ with an estimated mass of 12 bn M☉ (see e.g. &lt;a href=&quot;http://www.newscientist.com/article/mg22530104.000-ancient-black-hole-had-an-inexplicable-growth-spurt.html&quot; rel=&quot;nofollow&quot;&gt;http://www.newscientist.com/article/mg22530104.000-ancient-black-hole-had-an-inexplicable-growth-spurt.html&lt;/a&gt;) puts current black-hole-forming hypotheses into question. As the linked article explains, scientist doubt that most of the matter could pass into the hole in such a short time due to the outward pushing force of the radiation created by the material falling into it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is rather, was there enough matter in such a close vicinity of a BH that it could grow to this size? At 900 million years from the big bang I have the feeling that there is hardly &lt;em&gt;enough&lt;/em&gt; time for the required matter to even pass through the volume around the BH where accretion can happen. &lt;/p&gt;&#xA;" OwnerUserId="844" LastActivityDate="2015-03-03T12:45:41.167" Title="Mass distribution in the early universe" Tags="&lt;supermassive-black-hole&gt;&lt;early-universe&gt;&lt;matter&gt;&lt;quasars&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10038" PostTypeId="2" ParentId="10037" CreationDate="2015-03-03T10:25:38.693" Score="4" Body="&lt;p&gt;It is unfortunate that the usual poor journalism labels the growth of the black hole as &quot;inexplicable&quot; and then further down in the article refers to some possible explanations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic problem is a growth timescale one. Radiation pressure introduces a negative feedback, such that there is a &quot;theoretical&quot; maximum for spherical accretion called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Eddington_luminosity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eddington limit&lt;/a&gt;, which occurs when the quasar is radiating at its Eddington luminosity. The shortest growth timescale is achieved if the efficiency with which mass is converted to luminosity is low; but if it's too low we wouldn't see the quasar at all. This is the crux of the problem. You can look at &lt;a href=&quot;https://physics.stackexchange.com/questions/167250/is-there-a-limit-as-to-how-fast-a-black-hole-can-grow/167279#167279&quot;&gt;this Physics SE answer&lt;/a&gt; for some more of the details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is there are ways and means by which this limit can be exceeded - non spherical accretion for one - so there are lots of ideas about how this can be achieved. Another possibility is that you start off with a seed black hole that is pretty big to begin with, perhaps as a result of a merger. Or the quasar could have been less efficient in the past and is more efficient now, which is why we can see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there enough matter? Well, yes, galaxies have masses that can be much bigger than the mass of this black hole. They are rare, but of course so are &gt; 1 billion solar-mass black holes, and these tend to be the only ones that we can see at distances of &gt;10 billion light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way of assessing the feasibility would be just to ask what a freefall timescale would be. If you have say $10^{11}\ M_{\odot}$ in a sphere of radius 10 kpc (I am just using typical sorts of numbers for a big galaxy), then the average density is $5\times 10^{-22} kg/m^{3}$ and has a freefall time $\sim (G\rho)^{-1/2}$ of 200 million years. Of course there are other problems, like shedding angular momentum, but it looks like this timescale is short enough for gravity to do its thing (in the absence of radiation pressure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the short answer to your question is that yes, there must be enough time, because this is just the latest in a population of such objects. We know that quasars with supermassive black holes have formed within a billion years after the big bang.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-03-03T12:45:41.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10040" PostTypeId="2" ParentId="10036" CreationDate="2015-03-03T13:26:33.583" Score="2" Body="&lt;p&gt;It's called gravitational lensing. Here's a link to the wikipedia article on the subject: &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_lens&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Gravitational_lens&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Gravity affects everything, including light. A massive object such as a star, a galaxy, or in this case, a cluster of galaxies, bends the path of photons that pass very close to the massive object. Bending light is the basis of the lenses in an optical telescope.&lt;/p&gt;&#xA;&lt;p&gt;The behavior of a gravitational lens is different from that of an optical lens., but the end result is the same: Gravitational lenses enable astronomers to see more distant objects than they would without them. Gravitational lenses create multiple images of the remote objects behind the lens. Oftentimes these multiple images are in the form of streaks. See the below image of gravitational cluster Abell 1689, which is the cluster that let those astronomers see A1689-zD1.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZdgzS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZdgzS.jpg&quot; alt=&quot;&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://imgsrc.hubblesite.org/hu/db/images/hs-2003-01-b-print.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;hubblesite.org&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;The cluster magnified the remote galaxy A1689-zD1 by a factor of 9.3. Without that magnification, that remote galaxy would be invisible to current technology.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-02-18T07:51:42.743" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="10041" PostTypeId="1" AcceptedAnswerId="10059" CreationDate="2015-03-03T17:37:57.563" Score="4" ViewCount="222" Body="&lt;p&gt;Earth undoubtly has very good conditions for supporting life. Although it is expected that many other planets on the outer space have conditions at least as good as Earth, the vast majority doesn't, making them unhospitable to life or probably being able to support only very simple lifeforms. Earth itself for some billions of years until the Ediacaran or Cambrian could only support very simple lifeforms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many parameters that may influence the habitability of a planet and its ability to support complex life: Star type; star temperature; star luminosity; stellar activity; stellar stability; star age; planet age; planet composition; planet size; orbital excentricity; orbital length; rotation axis inclination; planet tectonics; planet magnetosphere; presence and influence of satellites; abundance of water; planet atmosphere; interactions with other planets; presence or absence of asteroids, comets and minor planets planets belts and their position, distribution and composition; galactic orbit; galactic neighborhood; mass-extinction events rate, probability and intensity; and hundred of other possible variables including some based on pure luck and random chance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of the parameters are modeled after Earth itself, since Earth is the only place so far that we know that life exists, and even if we found some alien life somewhere, it will probably be limited only to very simple forms of life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, &lt;strong&gt;what combinations of those parameters could lead to a planet with better life support than Earth itself?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, you may argue that the question is too broad, so by &quot;good life support&quot; we could say something that allows the planet to evolve plenty biodiverse multicellular life ranging from simple microscopic creatures to complex dozens-meters long creatures with many body-differentiated parts and organs in a short timespan. So, a planet that has an environment which allows the development of richly-diverse and complex plant-like and animal-like creatures in a billion years after formation and stay like this for another 10 billion years is expected to be more life-friendly than Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, lets restrict the biochemistry to what we know: water-based and carbon-based life, but not necessarily oxygen-breathing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, I am not asking anything about intelligent life or humans, just complex multicellular and biodiverse life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way: two important links about the subject: &lt;a href=&quot;http://www.nature.com/scientificamerican/journal/v312/n1/box/scientificamerican0115-32_BX1.html&quot; rel=&quot;nofollow&quot;&gt;http://www.nature.com/scientificamerican/journal/v312/n1/box/scientificamerican0115-32_BX1.html&lt;/a&gt; and &lt;a href=&quot;http://www.nature.com/scientificamerican/journal/v312/n1/full/scientificamerican0115-32.html&quot; rel=&quot;nofollow&quot;&gt;http://www.nature.com/scientificamerican/journal/v312/n1/full/scientificamerican0115-32.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="609" LastEditorUserId="609" LastEditDate="2015-03-05T17:22:21.673" LastActivityDate="2015-03-05T21:17:56.950" Title="Better than Earth habitability" Tags="&lt;habitable-zone&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" ClosedDate="2015-03-16T23:06:02.530" ContentLicense="CC BY-SA 3.0" />
  <row Id="10042" PostTypeId="1" AcceptedAnswerId="10044" CreationDate="2015-03-03T17:54:08.163" Score="1" ViewCount="897" Body="&lt;p&gt;As I understand it astronomers estimate the mass of stars by studying the orbits of their companions (other stars or planets).  This helps them to derive relationships between the luminosities and masses of stars.  This allows them to infer the mass of stars of known luminosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do astronomers estimate the mass of &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_dust&quot; rel=&quot;nofollow&quot;&gt;cosmic dust&lt;/a&gt; in particular clouds or in entire galaxies? &lt;/p&gt;&#xA;" OwnerUserId="233" LastActivityDate="2015-03-05T13:59:21.687" Title="How do astronomers estimate the total mass of dust in clouds and galaxies?" Tags="&lt;galaxy&gt;&lt;dust&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10043" PostTypeId="2" ParentId="10041" CreationDate="2015-03-03T18:13:00.937" Score="1" Body="&lt;p&gt;Your question is interesting but impossible to answer. Life adapts to its environment so suddenly the problem becomes: what type of environment creates life most suitable to itself? And that is pretty difficult to answer given that we don't really know anything about life outside Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there exists many metrics to measure habitability, and Earth does not necessarily rank as the best planet in these metrics. Wikipedia has a good table on it here:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_potentially_habitable_exoplanets&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/List_of_potentially_habitable_exoplanets&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But note that all these index put hypotheses on the expected form of life. It's also worth noting that life influences a planet's conditions: oxygen, in gaseous form, is due to the presence of plants. Imagine if Earth proto-life had never produced gaseous oxygen, Earth would now be very inhospitable to the life we know. So you analyze a closed-looped system, with a lot of randomness in it. It's a real nightmare!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could answer completely your question, you would probably know so much about life that you could define an &quot;optimal&quot; path to life. That is, conditions that lead to the emergence of the most efficient living system ever. As far as I know, we are very far from understanding anything about life. So, unfortunately this question is probably impossible to answer.&lt;/p&gt;&#xA;" OwnerUserId="6604" LastActivityDate="2015-03-03T18:13:00.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10044" PostTypeId="2" ParentId="10042" CreationDate="2015-03-03T19:48:51.327" Score="2" Body="&lt;p&gt;Dust absorbs stellar light (primarily in the ultraviolet), and is heated up. Subsequently it cools by emitting infrared, &quot;thermal&quot; radiation. Assuming a dust composition and grain size distribution, the amount of emitted IR light per unit dust mass can be calculated as a function of temperature. Observing the object at several different IR wavelengths, a Planck curve can be fitted to the data points, yielding the dust temperature. The more UV light incident on the dust, the higher the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is somewhat sensitive to the assumptions, and thus the uncertainties are sometimes quite large. The more IR data points obtained, the better. If only one IR point is available, the temperature cannot be calculated. Then there's a degeneracy between incident UV light and the amount of dust, and the mass can only be estimated to within some orders of magnitude (I think).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If lines from various atomic or molecular transitions are seen as well, the composition can be better constrained. The size distribution can be determined from fitting the theoretical spectrum of a given distribution to observed dust spectra. This information is often not available in a given high-redshift galaxy, so here we can be forced to assume that the dust is similar in nature to &quot;local&quot; dust, i.e. in the Milky Way and our nearest neighbors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're interested in the relevant equations, they can be found many places, e.g. &lt;a href=&quot;http://www.ifa.hawaii.edu/~jpw/classes/ast622.fall2009/handouts/dust.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to estimate the dust mass is to measure the metallicity of the gas with which the dust is is mixed, either from emission lines or absorption lines if a background source is available. The dust mass is then found from an assumed dust-to-metals ratio, which is pretty well-established in the local Universe, and to some extend also at higher redshifts.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-03-05T07:28:23.137" LastActivityDate="2015-03-05T07:28:23.137" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10045" PostTypeId="1" CreationDate="2015-03-03T19:56:03.723" Score="0" ViewCount="263" Body="&lt;p&gt;I had a question about the following chart. What mathematical observation or equation could be made about the data given in respect to the doppler effect? I'm confused if this means that planets are moving at an exponential rate away from us? Would these galaxies appear blue shifted? Thank you in advance for any sort of help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dQ2wC.png&quot; alt=&quot;Table&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6692" LastActivityDate="2015-03-03T21:17:37.197" Title="Help w/ Hubble's Law + Doppler Effect." Tags="&lt;galaxy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10046" PostTypeId="2" ParentId="10045" CreationDate="2015-03-03T21:17:37.197" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Look up the laboratory wavelengths of the Calcium H and K lines&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Calculate the redshift implied by the observed wavelengths of the H and K lines.&#xA;$$ v/c = (\lambda - \lambda_0)/\lambda_0,$$&#xA;where $v$ is the implied recession velocity, $c$ is the speed of light, $\lambda$ is the observed wavelength and $\lambda_0$ is the laboratory wavelength.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Plot recession velocity versus distance and see what you get.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Read about the &lt;a href=&quot;http://astro.wku.edu/astr106/Hubble_intro.html&quot; rel=&quot;nofollow&quot;&gt;Hubble redshift-distance relation&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-03-03T21:17:37.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10047" PostTypeId="1" CreationDate="2015-03-03T21:29:06.013" Score="6" ViewCount="1233" Body="&lt;p&gt;I suppose the answer would be a range as the Jupiter-facing side of Ganymede passed from Jupiter's light side to its shadow side. I'd also like to know how much light the sun projects onto the surface of Ganymede as well. I'm also interested in day night cycle characteristics. &lt;/p&gt;&#xA;" OwnerUserId="6694" LastActivityDate="2017-04-11T14:37:16.250" Title="How much light does Jupiter project onto the surface of Ganymede?" Tags="&lt;orbit&gt;&lt;jupiter&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10048" PostTypeId="1" AcceptedAnswerId="10049" CreationDate="2015-03-04T06:07:32.843" Score="5" ViewCount="389" Body="&lt;p&gt;Regardless of how big or small, how hot or cold, do all stars have a goldilocks zone, and if so, could any given planet in this zone (with the right configuration) support life? &lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2015-03-04T09:21:01.917" Title="Do all stars have the potential to have life supporting planets?" Tags="&lt;life&gt;&lt;star-systems&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10049" PostTypeId="2" ParentId="10048" CreationDate="2015-03-04T09:21:01.917" Score="5" Body="&lt;p&gt;Great question. The goldilocks zone is usually defined in terms of a region where the equilibrium temperature of the planet lies between some temperature limits (these temperature limits are somewhat debatable, but irrelevant for the purposes of this question - the boundary becomes fuzzy). This region can be calculated by working out how much flux is received from the star at a given radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that sense - all stars have a goldilocks zone - you can always work out some distance from a star where the flux is roughly equivalent to the flux we receive from the Sun for instance. It is much closer to a dim star and much further away from a luminous star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But&lt;/strong&gt;, there is a difference between a habitable zone and a &lt;em&gt;continuously&lt;/em&gt; habitable zone. If you seriously want life to evolve on a planet, then it will take time. How much time? No-one is sure but it seems to have taken a few hundred million years in our solar system. There are also good reasons to suppose that young planets are &lt;em&gt;not&lt;/em&gt; going to be habitable - either they are still extremely hot after formation or they are being bombard by debris.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus you can probably exclude any star with a main sequence lifetime of less than about 100 million years (or longer if you feel conservative about how long it takes life to get going). This rules out stars of more than about 5 solar masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also rule out stars that evolve quickly. If a star changes its luminosity quickly on a short timescale then the habitable zone moves drastically too. This will happen near the ends of the lives of all main sequence stars and for all subgiant and giant stars. So for these, yes there is an instantaneous habitable zone, but no region is in a habitable zone for hundreds of millions of years or more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to consider not only the luminosity of the star, but its spectrum as well. For instance, both hot white dwarfs and low-mass M-dwarfs are reasonably stable and long-lived; they are faint, so have habitable zones close to the star (much closer than 1 au). However, for different reasons, both these types of object emit copious UV radiation. In hot white dwarfs it would be because they have hot photospheres. In M-dwarfs they can be highly magnetically active into old age, having a hot chromosphere and corona that would strongly irradiate any nearby planet. You might consider that their &quot;habitable zones&quot; were in fact uninhabitable. Of course an atmosphere and a strong magnetic field might counteract this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of this depends on your terms of reference and definition. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Circumstellar_habitable_zone&quot;&gt;wikipedia page&lt;/a&gt; on habitable zones has quite a nice discussion, which emphasizes the debatable nature of this topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A final thought, which is not in the wiki page. Multiplicity could mess things up. Planetary systems will not be stable in the habitable zone of either star in a binary system if their separation is comparable with the orbital radius of the habitable zone. If you had two stars like the Sun, their separations would have to be less than a few tenths of an au or greater than of order 10 au in order to allow something to orbit at about 1 au around one of them. Even then, there could be all sorts of dynamical instabilities which prevent long-lived planets, especially if there were other planets in the system too.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-03-04T09:21:01.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10050" PostTypeId="2" ParentId="10047" CreationDate="2015-03-04T12:50:08.593" Score="7" Body="&lt;p&gt;How much? Well how accurately do you need it? How do you want it quantifying? And in what wavelength range?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter scatters a fraction of its incident sunlight. It also has its own luminosity (predominantly in the infrared).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick calculation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solar constant (flux at 1 au) is about 1370 W/m$^{2}$. Jupiter is situated about 5.2 au from the Sun (it varies by about +/- 5%) and has a radius of 70,000 km. The albedo is about 0.34.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus it receives about $7.8\times 10^{17}$ W from the Sun and radiates about $2.6\times 10^{17}$ W back into space. Assuming this is done more-or-less isotropically into a hemmisphere, then Ganymede, at a distance of 1,070,000 km from Jupiter, receives a flux of only 0.1 W/m$^2$ multiplied by the fraction of the sunlit hemisphere that can be seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This compares with the $\sim 50$ W/m$^2$ it receives from the Sun!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;strong&gt;surprising result&lt;/strong&gt; (to me) puts into perspective all the simulations you see of things in orbit around Jupiter. The planet is still pretty faint compared to the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be grateful if someone could double-check the sums!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[The intrinsic infrared luminosity of Jupiter is less than a fifth of what it receives from the Sun, so this would increase, but not double the received power.]&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-03-04T12:50:08.593" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10051" PostTypeId="1" AcceptedAnswerId="10052" CreationDate="2015-03-04T16:06:14.063" Score="4" ViewCount="975" Body="&lt;p&gt;When a proto-star becomes a main sequence star, is that something that happens in an instant when a certain threshold is met, or is it a process that takes a few thousand/million years?&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2018-03-24T02:21:40.077" Title="Is the conversion from proto-star to main sequence an event or a process?" Tags="&lt;protostar&gt;&lt;main-sequence&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10052" PostTypeId="2" ParentId="10051" CreationDate="2015-03-04T16:33:48.613" Score="6" Body="&lt;p&gt;Astronomers distinguish a prototstar from a star based on whether the object is visible. A protostar is hidden by the gas cloud that surrounds it. Protostars aren't visible. At some point in their evolution (and where this occurs depends on mass and metallicity), a protostar will start clearing the surrounding cloud of gas. This process happens very quickly from an astronomical point of view. (Aside: From a human point of view, this is anything but quickly.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of very massive stars, the newly emerged star is already on the main sequence. Very massive stars are &quot;stars&quot; (objects with a stable mass, stable size, and fusing hydrogen into helium) well before astronomers can see them. In contrast, very small stars spend hundreds of millions of years of evolution between being a &quot;star&quot; (visible to astronomers) and being on the main sequence. The star is a pre-main sequence star during this long span of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intermediate mass stars also spend some time as a pre-main sequence star. How much time depends on mass and metallicity. In the case of a star with a mass of ~1 solar mass and metallicity comparable to that of the Sun, the time spent as a pre-main sequence star is on the order of tens of millions of years. In contrast, the time spent as a protostar is very short, tens of thousands of years. The time taken to clear the star system of gas (the transition from protostar to PMS star) is shorter yet, hundreds to a few thousand of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no point at which one can say fusion starts. It's a probabilistic thing, with probability increasing sharply with density and temperature. What about objects at the low end of density and temperature? This is the boundary that separates red dwarfs from brown dwarfs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is very little difference between the smallest red dwarfs and the largest brown dwarfs. The smallest red dwarfs are quite cool. They fuse hydrogen into helium via the &lt;em&gt;p-p&lt;/em&gt; chain, but rather infrequently. The largest brown dwarfs also fuse hydrogen into helium via the &lt;em&gt;p-p&lt;/em&gt; chain, but here the fusion reactions are so very infrequent that the brown dwarf gets cooler and cooler. In a trillion years, we might be able to point to a marked boundary between old red dwarfs and old brown dwarfs. Right now, that boundary is rather arbitrary.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2015-03-04T16:41:09.220" LastActivityDate="2015-03-04T16:41:09.220" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10053" PostTypeId="1" CreationDate="2015-03-05T01:50:28.740" Score="0" ViewCount="130" Body="&lt;p&gt;Henrietta Leavitt discovered a relation between luminosity and period of variable stars, so it was possible to measure distance between two variable stars. I am curious to know how that was important back in time. How is the luminosity-period relation applied to measure distance? Was this method important because they finally had trusted apparent/absolute magnitude values to apply to the distance modulus equation?&lt;/p&gt;&#xA;" OwnerUserId="6700" LastEditorUserId="2153" LastEditDate="2015-03-05T16:10:37.297" LastActivityDate="2015-03-05T16:10:37.297" Title="How was Henrietta Leavitt's discovery important for star measurement?" Tags="&lt;star&gt;&lt;history&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2015-03-09T13:12:52.997" ContentLicense="CC BY-SA 3.0" />
  <row Id="10054" PostTypeId="2" ParentId="10042" CreationDate="2015-03-05T09:44:13.570" Score="1" Body="&lt;p&gt;Even if the accepted answer is well posed and clear, I would mention another method for dust measurements.&#xA;X-ray spectra can also help to infer the amount of dust, analyzing the absorption at low energy of the observed spectra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, the neutral hydrogen column density $N_H$ and the extinction $A(V)$, which is caused by the dust, is found in our Galaxy to be (Guver &amp;amp; Ozel, 2009):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N_H = 2.2\times 10^{21} A(V)\,$mag&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the extinction one can get the dust optical depth (see &lt;a href=&quot;http://astro.berkeley.edu/~ay216/08/NOTES/Lecture05-08.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here),&lt;/a&gt; and from that the volumetric dust density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good example of how to obtain $N_H$ is given in the following plot (&lt;a href=&quot;http://www.google.de/imgres?imgurl=http%3A%2F%2Fej.iop.org%2Fimages%2F0004-637X%2F692%2F1%2F608%2FFull%2Fapj296684f5_lr.gif&amp;amp;imgrefurl=http%3A%2F%2Fiopscience.iop.org%2F0004-637X%2F692%2F1%2F608%2Farticle&amp;amp;h=509&amp;amp;w=600&amp;amp;tbnid=0I-0HwtXVyw56M%3A&amp;amp;zoom=1&amp;amp;docid=mwsa7H5fLufQXM&amp;amp;ei=oyL4VMbbOoisPa-GgZgG&amp;amp;tbm=isch&amp;amp;iact=rc&amp;amp;uact=3&amp;amp;dur=347&amp;amp;page=2&amp;amp;start=22&amp;amp;ndsp=29&amp;amp;ved=0CLgBEK0DMDE&quot; rel=&quot;nofollow noreferrer&quot;&gt;S. Ikeda et al. 2009, ApJ 692 608&lt;/a&gt;):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/fOBUh.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it can be seen, the spectra is differently modified by different amounts of neutral hydrogen column density $N_H$. This, of course, can be related to different aboundances, to best-fit the spectra and find the metals amounts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, this method is tightly related to the iron K$\alpha$ line at $6.4\,$keV, which is another measurement of the metals amount (Fabian et al. 2000):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/cnnV9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To distinguish among different ionization states, high-resolution spectroscopy is very common.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastEditorUserId="1038" LastEditDate="2015-03-05T13:59:21.687" LastActivityDate="2015-03-05T13:59:21.687" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10055" PostTypeId="2" ParentId="1145" CreationDate="2015-03-05T10:09:10.077" Score="5" Body="&lt;p&gt;Galaxies are disc shaped because they are gas rich and dynamically young. Stars are also gas rich but they are dynamically old so they have had time to rid themselves of their discs. Young protostars (which are dynamically young) are surrounded by proto-stellar discs. The reason many young gas rich objects are disc shaped has to do the fact that circular orbits are compatible with no orbit crossing, hence no shocks. In some sense, many young astronomical objects are disc shaped because they contain(ed) gas which can radiate away non circular motion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a disc is not the most likely state of a gravitational system: given time, torquing, instabilities or viscous processes it will tend towards a more likely compact state, where mass flows inwards and angular momentum outwards. This is why proto-stellar discs become stars. Galaxies on the other hand have not had time to turn themselves into gigantic black holes, or been given the opportunity to do so via torquing with their environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When two gas poor disc galaxies collide they produce an elliptical which is not disc like.  When two gas rich  disc galaxies collide they produce a disc like galaxy with a bulge.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Update&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Another way to think about this is to consider  the ratio of angular momentum to total energy.  Through shocks, if gas is present the system can radiate away some of its energy so that this ratio becomes larger. A high ratio will typically correspond to a disc-like system.&lt;/p&gt;&#xA;" OwnerUserId="1227" LastEditorUserId="1227" LastEditDate="2016-09-10T06:34:46.727" LastActivityDate="2016-09-10T06:34:46.727" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10057" PostTypeId="1" CreationDate="2015-03-05T15:54:05.213" Score="0" ViewCount="74" Body="&lt;p&gt;In the television show &quot; Ascension &quot; they are supposed to be 50 years into a 100 year journey to another Star System but get this, they never left Earth . They are here on Earth in a huge biosphere . My question is if this is possible that people can be here on Earth and not know that they are not traveling through Space in a Spacecraft ? How could they not have instruments or any other way of not knowing . This plot seems a little far fetched to me, what do the rest of you have to say about it ?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastActivityDate="2015-03-06T08:58:26.660" Title="Detecting motion" Tags="&lt;interstellar-travel&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10059" PostTypeId="2" ParentId="10041" CreationDate="2015-03-05T21:17:56.950" Score="2" Body="&lt;p&gt;&lt;em&gt;Note: I am self-answering my own question in hope that someone post another answer that beats this one.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Earth is near the inner edge of the Sun's habitable zone. And since the Sun is expected to grow and increase it luminance, &lt;a href=&quot;http://en.wikipedia.org/wiki/Future_of_the_Earth#Loss_of_oceans&quot; rel=&quot;nofollow&quot;&gt;Earth might be unhihabitable for any life somewhere between 1 or 3 billion years in the future&lt;/a&gt;. So, a planet that have longer time to develop before its parent star moves it beyond the inner edge of the habitable zone is more favorable than Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Earth itself always was inside the habitable zone since it formed 4.6 billion years ago and will still be for lets say more 1.4 billion years, with a large uncertanity factor, this gives roughly 6 billion years of time for complex life to develop. Given that it is unlikely to form early due to the time needed for evolution to take place and due to an elevated level of large bollides collisions, we could discount the first 2 billion years from any life bearing planet, including Earth, as unlikely to develop complex life. Further, it is unlikely that complex life would finally evolve out from simpler forms when the planet is already overheated and already crossing the inner edge of the habitability zone, so lets take out the finishing 10% of that period for any planet (probably something more than 10%, but lets keep this as a conservative estimative). So, for Earth, this gives a window of a size of 3.4 billions years to complex life evolve. Similar planets with larger windows have better probabilities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars larger and more luminous than the Sun tends to be more unstable and live shorter. As a result, it is expected that planets around stars larger than the Sun has less time to develop complex life, and thus a shorter time-window. On the other hand, this means that stars smaller and less luminous than the Sun gives a larger time-window to the planets to develop life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For stars smaller than the Sun (a G-type yellow star), we could consider the K-types (aka, orange dwarf) and the M-types (aka, red dwarf) as specially favorable.&#xA;An orange dwarf star &lt;a href=&quot;http://en.wikipedia.org/wiki/K-type_main-sequence_star&quot; rel=&quot;nofollow&quot;&gt;may live for 10 to 30 billions years in the main sequence&lt;/a&gt;. A red dwarf star may &lt;a href=&quot;http://en.wikipedia.org/wiki/Red_dwarf&quot; rel=&quot;nofollow&quot;&gt;live in the main sequence for trillions of years&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, planets in the habitable zone of red dwarfs are likely to become tidally lock, and we don't know if this is really that bad or not for life biodiversity. Lets assume that this is really bad, so a planet orbiting an orange dwarf in the habitable zone is likely to have a better habitability than Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accordingly to &lt;a href=&quot;http://www.nature.com/scientificamerican/journal/v312/n1/box/scientificamerican0115-32_BX1.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, a planet with two times the mass of the Earth, will have stronger gravity, and thus it is likely to be flatter. Further, it is likely to have a ticker atmosphere that would protect the surface from UV radiation better than Earth. It would be geologically active for a longer time, resulting in more carbon cycling. With the right quantity of water (not a desert nor a global very deep ocean), it might be an archipelago world, since its flatness would not allow the ocean to be very deep nor the continents to be very large. As a result, life would flourish in a number of rich biologically favourable environments significantly larger than Earth. Further it magnetic field is likely to be stronger than Earth's one, protecting the surface from cosmic rays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;As a result, a planet with two Earth masses orbiting an orange dwarf star in the habitable zone has a good chance to be more habitable to life than Earth itself.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Needless to say, near-circular orbits are more favourable than excentric ones, since excentric orbits may make the planet enter in periods of freezing or boiling. However, a reasonably excentricity that periodically changes the environment in a significant manner, but not as too much that it would extinguish non-extremophile life, might give to the planet life a selective pressure needed for developing rapid evolution to face the always changing climate.&lt;/p&gt;&#xA;" OwnerUserId="609" LastActivityDate="2015-03-05T21:17:56.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10061" PostTypeId="1" AcceptedAnswerId="35596" CreationDate="2015-03-06T03:32:29.690" Score="4" ViewCount="167" Body="&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Olbers%27_paradox&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, here's Lord Kelvin's statement of Olbers' Paradox:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Were the succession of stars endless, then the background of the sky would present us a uniform luminosity, like that displayed by the Galaxy – since there could be absolutely no point, in all that background, at which would not exist a star.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is one form of Olbers' Paradox I've heard, which is that with infinitely many stars scattered in an infinite volume, every ray must eventually intersect a star. I call this the &quot;strong form&quot; because it seems to me stronger than the statement that infinite stars imply an infinite amount of light hitting every point, though not necessarily from every possible direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strong form of the paradox doesn't hold if each star is a point mass and there are countably many stars, since the number of rays from any given point is uncountably infinite. Or imagine instead that stars are balls with small positive radius. If there is such a star at every lattice point in $R^3$ (an assumption that is consistent with my crude understanding of the cosmological principle), I can easily find a ray that intersects no star. I start from (0, 0, 0.5) and choose the ray in the direction of (1, 0, 0.5); the z coordinate will always be 0.5 and therefore my ray will never come close to a lattice point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there another, palatable assumption we can add to make the strong form of Olbers' Paradox true, perhaps an argument that the distribution of stars must be more random than what I've described?&lt;/p&gt;&#xA;" OwnerUserId="6587" LastActivityDate="2020-03-29T00:28:09.683" Title="Assumptions necessary for the strong form of Olber's Paradox" Tags="&lt;cosmology&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10062" PostTypeId="1" AcceptedAnswerId="10063" CreationDate="2015-03-06T04:08:22.007" Score="5" ViewCount="868" Body="&lt;p&gt;If not, do any of the four groups in the title lie on the same plane as other members of the group? E.g. does the asteroid belt lie on the same plane as the planets do?&lt;/p&gt;&#xA;" OwnerUserId="6710" LastActivityDate="2015-03-06T14:06:53.717" Title="Do the planets, asteroid belt, kuiper belt, and scattered disc lie on the same plane?" Tags="&lt;solar-system&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10063" PostTypeId="2" ParentId="10062" CreationDate="2015-03-06T05:11:57.137" Score="8" Body="&lt;p&gt;They are not exactly in the same plane, but roughly are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example this graph representing the centaurs, the Kuipier belt and scattered disc objects show that most objects are around the same plane, with few objects with more than 30 degrees of inclination:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/35FJA.png&quot; alt=&quot;Centaurs, Kuipier belt and Scattered disc representation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the inclination of the asteroids:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ymvgz.png&quot; alt=&quot;Asteroids inclinations&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, few have more than 30 degrees of inclination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The planets are also all roughly in the same plane. Lets take the ecliptic as a reference (which is almost the same thing as Earth orbital's plane):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Sun's equator has an inclination of 7.25 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Mercury has an inclination of 7.00 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Venus has an inclination of 3.39 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Mars has an inclination of 1.85 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Vesta has an inclination of 7.13 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Juno has an inclination of 12.98 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Ceres has an inclination of 10.59 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Pallas has an inclination of 34.84 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Interamnia has an inclination of 17.29 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Hygiea has an inclination of 3.84 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Jupiter has an inclination of 1.30 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Saturn has an inclination of 2.48 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Uranus has an inclination of 0.77 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Neptune has an inclination of 1.76 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Orcus has an inclination of 20.57 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Pluto has an inclination of 17.15 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Ixion has an inclination of 19.58 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Salacia has an inlination of 23.94 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Varuna has an inclination of 17.2 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Haumea has an inclination of 28.19 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Quaoar has an inlination of 7.99 degrees.&lt;/li&gt;&#xA;&lt;li&gt;1992 QB&lt;sub&gt;1&lt;/sub&gt; has an inclination of 2.19 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Makemake has an inclination of 29.00 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Eris has an inclination of 44.04 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Sedna has an inclination of 11.92 degrees.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Most of those bodies have a low inclination to the ecliptic, but Pallas is an interesting exception. I don't know why Pallas has such a high inclination but I would bet on a &lt;a href=&quot;http://en.wikipedia.org/wiki/Kozai_mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kozai mechanism&lt;/a&gt; due to Jupiter's influence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plutinos and Kuipier belt objects are influenced by Neptune, and part by the Kozai mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eris is a special and curious case, it is too far from Neptune, although there is still some weak gravitational Neptune's influence there, but I don't know why its inclination is so high. This contrasts with Sedna, which have a more reasonably lower inclination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, the Oort cloud is expected to be full of bodies of essentially unrestricted inclinations and possibly very excentric orbits, as a result of gravitational interactions with passing stars on the past. By the way, a passing star in a distant past might be an explanation for Eris orbit and inclination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Note: All of the data here, as well as both of the images, I found on wikipedia.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="609" LastEditorUserId="609" LastEditDate="2015-03-06T14:06:53.717" LastActivityDate="2015-03-06T14:06:53.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10064" PostTypeId="2" ParentId="10057" CreationDate="2015-03-06T06:15:13.950" Score="2" Body="&lt;p&gt;Can they see the stars? If not then motion is undetectable in a closed environment as it would constitute an inertial frame. But even if they could see the &quot;stars&quot; presumably these would be generated by some form of display technology and behave consistently with the supposed mission profile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However acceleration is detectable, it would manifest as a gravity like force. The way gravity varies withing the environment could indicate that this was a planetary field rather than an artificial one (depending on how the artificial gravity is supposedly generated). I would expect the payload will have been told some story that would explain this away though.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-03-06T06:15:13.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10065" PostTypeId="1" AcceptedAnswerId="10066" CreationDate="2015-03-06T06:56:30.387" Score="2" ViewCount="429" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZMw1a.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the rookie question, but I've got no idea about that red &quot;belt&quot; crossing over center of the galaxy, in this photo. I've actually tried all sorts of google searches but I couldn't get anywhere with keywords like &quot;The red stuff&quot; or so. Years ago I had a color book in which I saw this photo for the first time, and I can remember my attempt to remove the red dirt and clean the image with my shirt!&lt;/p&gt;&#xA;" OwnerUserId="1473" LastActivityDate="2015-07-25T18:20:05.130" Title="What's this red stuff in some astronomical photos, e.g. Centaurus A?" Tags="&lt;galaxy&gt;&lt;photography&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10066" PostTypeId="2" ParentId="10065" CreationDate="2015-03-06T07:13:10.060" Score="6" Body="&lt;p&gt;&quot;Clean the image&quot;, how apt, because most of what you can see is caused by dust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a bit difficult to comment specifically because I think you have chosen an example which is a combination of images and not all the images are in the visible part of the spectrum and so have had a &quot;false colour&quot; applied. In fact it is a combination of optical, X-ray and sub-mm images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general terms red is present in false colour astronomical images, usually used to representing the coolest or longest wavelength light in the image. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Red is also present in true-colour images of nebulae and is caused by light emitted due to transitions of atoms and ions between excited states separated by particular energies. Red light is predominantly due to Hydrogen alpha and ionised nitrogen emission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coolest stars are also red in true colour images, simply because their pseudo-blackbody radiation spectra peak in the red part of the spectrum for temperatures lower than 5000 K. The integrated light from old stellar populations (predominantly cool main sequence stars and &lt;em&gt;red&lt;/em&gt; giants) would therefore have a reddish tinge - for example in elliptical galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said, it is hard to say in the case of this particular composite image, but the band across the centre of the galaxy is dust. It obscures (makes dark), the optical image, but warm dust glows in the sub-mm image. The structures coming out of the galaxy at right angles are jets from the active galactic nucleus traced in the X-ray.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-03-06T09:39:23.313" LastActivityDate="2015-03-06T09:39:23.313" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10067" PostTypeId="2" ParentId="10057" CreationDate="2015-03-06T07:28:21.733" Score="2" Body="&lt;p&gt;It is part of General Relativity that if you are in a (small) closed box, you cannot tell the difference between acceleration and a gravitational field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose you could argue that you would explain Earth's gravity to the occupants of the biosphere as a 1g acceleration away from the Earth for 50 years, a slow swivel of the spacecraft, and then deceleration at 1g for 50 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is hard to say whether such a deception is possible without more details, but in principle it could be done if you have some way of obscuring the initial part of the journey - launch etc, where you would think some non-uniform acceleration would be inevitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the question of how far you are going and whether there is a feasible propulsion system. An acceleration of 1g for only a year gets you up to relativistic speeds. You could easily travel to other &lt;em&gt;galaxies&lt;/em&gt; in 50 years at this acceleration. So, if the destination was supposed to be a nearby star then the deception fails miserably, unless you try to convince the passengers that you are executing some sort of weird trajectory to get there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, how big is the biosphere? What about tides due to the Sun and moon? Any sizeable body of water would show up tides (and indeed possibly the coriolis force due to the Earth's rotation).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-03-06T08:58:26.660" LastActivityDate="2015-03-06T08:58:26.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10068" PostTypeId="2" ParentId="8567" CreationDate="2015-03-06T12:31:20.313" Score="1" Body="&lt;p&gt;Can a Human See a Single Photon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The human eye is very sensitive but can we see a single photon?  The answer is that the sensors in the retinacanrespond to a single photon.  However, neural filters only allow a signal to pass to the brain to trigger a conscious response when at least about five to nine arrive within less than 100 ms.  If we could consciously see single photons we would experience too much visual &quot;noise&quot; in very low light, so this filter is a necessary adaptation, not a weakness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to this paper &lt;a href=&quot;http://math.ucr.edu/home/baez/physics/Quantum/see_a_photon.html&quot; rel=&quot;nofollow&quot;&gt;http://math.ucr.edu/home/baez/physics/Quantum/see_a_photon.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this is not always possible for distant galaxies we cannot see distant galaxies.&lt;/p&gt;&#xA;" OwnerUserId="6649" LastActivityDate="2015-03-06T12:31:20.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10069" PostTypeId="2" ParentId="8567" CreationDate="2015-03-06T16:43:19.617" Score="1" Body="&lt;p&gt;The core of the question has already been answered, but it's still interesting to illustrate just how difficult it is to make a naked eye observations of a the extremely bright nearby galaxy M81. The astronomer &lt;a href=&quot;http://en.wikipedia.org/wiki/Brian_A._Skiff&quot; rel=&quot;nofollow&quot;&gt;Brian Skiff&lt;/a&gt; gives an account of his successful naked eye observation of this galaxy &lt;a href=&quot;http://messier.seds.org/xtra/supp/m81naked.txt&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, galaxies of a given brightness are more difficult to spot than stars of the same brightness, because of their extended nature. If the sky is sufficiently dark, then you can see stars as faint as magnitude 8, but you'll still struggle to spot M81 that has a brighness of magnitude 7. The magnitude 7 is an artificial figure obtained by adding up the light that comes from slightly different directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you only need very slight amount of light polution to make the sky backgound just a tiny bit grey to make the galaxy vanish from view, while the visibility of faint stars remains essentially unaffected. This is because the brighness as a function of the position in the sky in the case of a star has a very strong and narrow peak while in case fo a galaxy, due to its extended nature, doesn't show a big peak. The integrated brightness may be the same for both cases, but the amount of background light you need to make the galaxy invisible is obviously much less than what you need for the star.&lt;/p&gt;&#xA;" OwnerUserId="2185" LastActivityDate="2015-03-06T16:43:19.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10070" PostTypeId="1" CreationDate="2015-03-07T03:12:25.727" Score="2" ViewCount="520" Body="&lt;p&gt;Asteroids and comets only seem to refer to masses within the solar system, and it seems unclear whether planetesimal does. &lt;/p&gt;&#xA;" OwnerUserId="6710" LastActivityDate="2020-03-14T17:04:10.867" Title="What is the term for astronomical objects outside the solar system that are smaller than dwarf planets?" Tags="&lt;terminology&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10071" PostTypeId="2" ParentId="10070" CreationDate="2015-03-07T04:24:54.530" Score="4" Body="&lt;p&gt;Just add the prefix &quot;exo-&quot;. At least exocomet and exomoon seem to be &lt;a href=&quot;https://en.wikipedia.org/wiki/Exocomet&quot; rel=&quot;nofollow&quot;&gt;established&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-03-07T04:24:54.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10072" PostTypeId="1" CreationDate="2015-03-07T09:43:10.310" Score="3" ViewCount="53" Body="&lt;p&gt;How do I calculate the positions of above satellite in space ?&lt;/p&gt;&#xA;" OwnerUserId="2863" LastActivityDate="2015-10-06T19:50:12.880" Title="Calculating Postition of ISS / Hubble in space" Tags="&lt;artificial-satellite&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2015-10-07T13:41:15.263" ContentLicense="CC BY-SA 3.0" />
  <row Id="10073" PostTypeId="1" CreationDate="2015-03-08T10:00:47.700" Score="3" ViewCount="275" Body="&lt;p&gt;Reading about the formation of planetary disks, one of the major problems, it seems like, is the evacuation of angular momentum. Aparently planets can't form with the amount o angular momentum the system has in its early stages. I think I understand where that excess comes from, the collapse of the nebula onto itself and provoking a spin. Then there are many hypothesis on how it's evacuated, which are mostly pretty logical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question, as a beginner in the study of physics, is this: Why does the angular momentum even need to be evacuated? If the angular momentum is too big, why can't planets still form? Does this have something to do with too much kinetic energy in the system?&#xA;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="4126" LastActivityDate="2015-03-08T10:00:47.700" Title="Angular momentum in planetary disk formation" Tags="&lt;solar-system&gt;&lt;astrophysics&gt;&lt;accretion-discs&gt;" AnswerCount="0" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10074" PostTypeId="1" AcceptedAnswerId="10075" CreationDate="2015-03-08T17:01:19.123" Score="0" ViewCount="260" Body="&lt;p&gt;A nearby star has crossed Chandrasekhar's limit. How do we come to know when it will undergo supernova explosion? &#xA;Or has it already undergone supernova explosion?&lt;/p&gt;&#xA;" OwnerUserId="6720" LastActivityDate="2015-03-08T21:14:51.547" Title="Supernova explosion nearby" Tags="&lt;star&gt;&lt;supernova&gt;&lt;explosion&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10075" PostTypeId="2" ParentId="10074" CreationDate="2015-03-08T17:29:06.660" Score="2" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Chandrasekhar_limit&quot; rel=&quot;nofollow&quot;&gt;Chandrasekhar limit&lt;/a&gt; applies only for white dwarfs. Stars on the main sequence (or even off the main sequence) can easily surpass it, but if a white dwarf's mass is greater than the Chandrasekhar limit ($1.39 M_{\odot}$), it will undergo some sort of collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, though, in response to&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Or has it already undergone supernova explosion?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;White dwarfs are &lt;em&gt;not&lt;/em&gt; formed from supernovae, but from stars that (relatively) gently shed their outer layers to form planetary nebulae. So no, the stellar remnant will not have gone supernova already.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to the main question. If a white dwarf exists, its mass is less than the Chandrasekhar limit. To surpass that, it needs to gain mass. This can happen in binary systems where the white dwarf accretes matter from the more massive star (generally a giant). The white dwarf can then undergo a &lt;a href=&quot;https://en.wikipedia.org/wiki/Type_Ia_supernova&quot; rel=&quot;nofollow&quot;&gt;Type Ia supernova&lt;/a&gt; and explode. The light curve will look a bit like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/8/88/SNIacurva.png&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/8/88/SNIacurva.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second (but far less likely) scenario is that two white dwarfs merge. The result? A Type Ia supernova and, possibly, a neutron star.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2015-03-08T17:29:06.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10076" PostTypeId="2" ParentId="10074" CreationDate="2015-03-08T21:14:51.547" Score="2" Body="&lt;p&gt;The Chandrasekhar limit in general does not pertain to the mass of the star as a whole. It addresses the mass of the degenerate core. It's only in white dwarfs where the Chandrasekhar limit applies to the mass of white dwarf as a whole, but that's because white dwarfs are almost entirely degenerate matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a 1.6 solar mass that is not a member of a multiple star system. Even though this star's total mass exceeds the Chandrasekhar limit, this star will never go supernova. This star will instead live for billions of year on the main sequence, then a bit longer as a post-main sequence star, and finally end it's life as a white dwarf with considerably less mass than that original 1.6 solar masses (and considerably less than Chandrasekhar limit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The star leaves the main sequence when it burns all of the hydrogen in its core. Two things happen at this point. One is that it starts burning hydrogen in a shell around an inert core of helium. The other is that it expands into a red giant. (Note: Some people are taught that a star becomes a red giant when it starts burning helium. This is not the case. This post-main sequence star is a red giant that is not yet burning helium.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That inert core of helium increases in mass as hydrogen shell burning proceeds. A funny thing begins to happen with this increase in mass: The inert core becomes degenerate. An even funnier thing happens next: Adding even more mass makes the degenerate core shrink in size. The star is poised on the next phase of its evolution, which is helium burning. The formerly inert, degenerate core of helium is no longer inert or degenerate at this stage. Think of this phase of a star's life as a main sequence helium burning star. It's definitely not on the main sequence, but much of the physics is the same. One thing that is different: The shell burning of hydrogen is still occurring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This phase doesn't last long. The star will soon consume all of the helium in its core. At this point, the star becomes a red giant for a second time. This is the asymptotic red giant branch. The star has an inert core of carbon and oxygen surrounded by a shell of fusing helium, which in turn is surrounded by a shell of fusing hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These end of life convulsions are not good for the stuff that surrounds the shell of fusing hydrogen. The star expels a lot of that gas into nearby space. Eventually it will expel most of the burning shells of hydrogen and helium, leaving an inert, degenerate core of mostly carbon and oxygen. This is a white dwarf. A planetary nebula of expelled gases surrounds the white dwarf. The white dwarf itself has but a fraction of the star's initial mass. Most of the mass is in that planetary nebula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Larger stars are even more proficient at expelling mass than was this 1.6 solar mass star. For stars with an initial mass of about eight solar masses or less, the white dwarfs left behind at the end of the stars' lives will be less than the Chandrasekhar limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fate of stars over ten solar masses is much more violent. At the end of their lives, they have lurking inside them a series of burning shells that surround an inert, degenerate core of iron. A star has about five days to live once that core of iron starts forming. That's about how long it takes to create (from scratch) a degenerate iron core that approaches the Chandrasekhar limit. At or near the limit, the core starts to collapse, and the star forms a type II supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A white dwarf can form a type Ia supernova with a little help from a binary neighbor from which the white dwarf is stealing mass. Conditions have to be just right for this to happen, but when it does, the white dwarf can gradually accumulate mass and go supernova. Since white dwarfs have expelled all of their hydrogen, there won't be any signatures of hydrogen in the supernova explosion. In fact, it's the lack or presence of hydrogen in the supernova signature that distinguishes type I from type II supernova.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-03-08T21:14:51.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10079" PostTypeId="2" ParentId="421" CreationDate="2015-03-09T09:23:52.320" Score="1" Body="&lt;p&gt;The sizes of dust particles varies enormously, and thus the density of dust particles depend upon which sizes we are talking about. The dust size distribution can be described by a power law with a slope of roughly $-3.5$; that is, the density of 0.1 µm particles of is $10^{-3.5}$ times that of 0.01 µm particles, or roughly 3000 times smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, the discussion is complicated by the fact that we don't really have a formal definition of what is dust. But one &quot;definition&quot; takes the minumum size as a conglomeration of a few molecules, and the maximum size as &quot;what has time to grow in a formation process&quot; (these dust particles can later &quot;stick together&quot; and grow to form pebbles, rocks, asteroids, and planets. That's why any definition will be arbitrary).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a dust-dense region of the interstellar medium, this results in a typical mean density of roughly one dust particle per cubic centimeter.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2015-03-09T09:23:52.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10082" PostTypeId="2" ParentId="10070" CreationDate="2015-03-09T16:17:31.727" Score="3" Body="&lt;p&gt;I have to disagree with the accepted answer. The terms &lt;em&gt;Planets&lt;/em&gt;, &lt;em&gt;moons&lt;/em&gt;, &lt;em&gt;asteroids&lt;/em&gt;, &lt;em&gt;planetesimals&lt;/em&gt; are generic and are not implied to mean objects in the Solar system. So, these terms are completely correct when dealing with the generic objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When specifically emphasizing the fact that a certain objects is not in the Solar system, then you may want to add 'exo' for disambiguation, but this should be reserved for this purpose only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A planet is a planet no matter where. Same with a house, which is a house no matter whether it's in your home town or elsewhere on the Earth.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note in edit: My answer is based on the practice among professional astronomers and not on what some dictionaries say. I reckon the latter are somewhat behind the times, when all planets, asteroids etc. known were those in orbiting the Sun. Wikipedia is a great resource, but &lt;em&gt;anybody&lt;/em&gt; can change its contents and, not surprisingly, some of its pages are quite biased.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2020-03-14T17:04:10.867" LastActivityDate="2020-03-14T17:04:10.867" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="10083" PostTypeId="1" AcceptedAnswerId="10089" CreationDate="2015-03-09T20:37:08.033" Score="4" ViewCount="339" Body="&lt;p&gt;I haven't been able to find the orbit time of Dawn around Ceres.&#xA;Do you know how long it takes?&lt;/p&gt;&#xA;" OwnerUserId="5471" LastEditorUserId="13" LastEditDate="2015-10-24T02:47:53.223" LastActivityDate="2015-10-26T16:03:38.413" Title="How long does it take Dawn to orbit Ceres?" Tags="&lt;orbit&gt;&lt;asteroids&gt;&lt;dwarf-planets&gt;&lt;space-probe&gt;&lt;ceres&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10085" PostTypeId="1" CreationDate="2015-03-10T19:19:29.393" Score="3" ViewCount="418" Body="&lt;p&gt;I'm looking for help in understanding how black holes can move through space. This is probably a trivial question for anyone who knows astronomy, but probably due to my lack of education on the topic, I can't come up with an answer or visualization on my own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, my confusion is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I saw Dr. Kip Thorne in a documentary on black holes state that there is no longer any matter in a black hole once it forms. He said something like, the matter was there, but it has been crushed out of existence and basically has been transformed into the energy that produces such a great curvature of spacetime.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;From that description, it seems to me that within the event horizon of a black hole it is simply this enormous &quot;dent&quot; in spacetime itself that is moving around. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I also saw Dr. Andrew Hamilton in another documentary state that within the event horizon of a black hole (again paraphrasing), space is falling toward the singularity so fast that it effectively drags everything with it (including light).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Based on 1 (no matter remains in the black hole) and 2 (space is being pulled towards the gravitational singularity from all directions), I try to picture the black hole moving, and it seems to me that if, say, the black hole moves in one direction, I can see how its event horizon is also moving, bringing that bit of space into it. But how can the other end of the black hole let space &quot;escape&quot; in order to complete the motion? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure I am just missing something here, because as I have learned, black holes have been found in galaxies and solar systems and many other places where they have to be moving. But I want to understand why it is that space itself can cross the event horizon when other things can't. And along those lines, it seems amazing to me that a black hole could be in essence a kind of scar in spacetime that can also move through it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help and sorry for the wordy question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: Thanks for the answers so far. I hope someday I'll be able to comprehend them fully. :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to add some further context that better illustrates my question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you go to the video here and watch the first minute or so (up to around 31:00)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://youtu.be/Niurc_6xIYg?t=29m48s&quot; rel=&quot;nofollow&quot;&gt;http://youtu.be/Niurc_6xIYg?t=29m48s&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can see Andrew Hamilton explaining what I have a question about along with an animation to illustrate the description. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the one hand, we have spacetime curved in such a way that the black hole structure remains intact - that is, there is a constant event horizon generated by that extreme curvature. But on the other hand, Dr. Hamilton describes a flow of spacetime between the event horizon and the inner horizon, and he says it is plausible that within that area, space itself (which has no substance, in his words) can be described as flowing in and being flung back out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand how space can have these two opposing properties - (1) a curvature that is maintained in order for the black hole to exist, that is, a curvature generated because matter can interact with the space and (2) a flow of space that sounds like it is not subject to a restriction of crossing back out of the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's that apparent difference between (1) and (2) that I want to understand better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6739" LastEditorUserId="6739" LastEditDate="2015-03-11T15:54:43.357" LastActivityDate="2015-03-11T17:25:46.413" Title="Looking for help in understanding how black holes can move" Tags="&lt;black-hole&gt;&lt;cosmology&gt;&lt;general-relativity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10086" PostTypeId="2" ParentId="10085" CreationDate="2015-03-10T20:43:45.777" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;I saw Dr. Kip Thorne in a documentary on black holes state that there is no longer any matter in a black hole once it forms. He said something like, the matter was there, but it has been crushed out of existence and basically has been transformed into the energy that produces such a great curvature of spacetime.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Isolated black holes are indeed vacuum solutions to general relativity. Thus in particular, the mass and energy density is identically zero everywhere in spacetime. Because of some technical issues, this does &lt;em&gt;not&lt;/em&gt; imply that black holes have no energy or mass; rather, it means that we can't directly think of them as an integral of mass or energy density. See also &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6589/do-black-holes-have-energy&quot;&gt;this&lt;/a&gt; question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;I also saw Dr. Andrew Hamilton in another documentary state that within the event horizon of a black hole (again paraphrasing), space is falling toward the singularity so fast that it effectively drags everything with it (including light).&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For an isolated black hole in a particular frame field, this is a valid picture. For example, the Schwarzschild spacetime in Gullstrand–Painlevé coordinates can be interpreted as Euclidean space falling into the black hole. However, the spacetime is itself is stationary (even static) in the geometrical sense—there is a timelike field representing a direction in which spacetime geometry is left unchanged (this corresponds to Schwarzschild time, in fact). Similarly, rotating black hole spacetimes are stationary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But generally speaking, you really can't think of black holes as some sort of suck-holes for space. It's just an analogy that applies to simple situations (isolated black holes, with nothing else in spacetime) and then only in a particular frame or coordinates. That's not to say it's useless--e.g., &lt;a href=&quot;https://en.wikipedia.org/wiki/Acoustic_metric&quot; rel=&quot;nofollow&quot;&gt;acoustic horizons&lt;/a&gt; in fluids are interesting analogies to event horizon--but don't take it too literally. See also &lt;a href=&quot;https://astronomy.stackexchange.com/questions/9919/is-a-black-hole-a-5-dimensional-vortex&quot;&gt;this&lt;/a&gt; question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;4&quot;&gt;&#xA;  &lt;li&gt;Based on 1 (no matter remains in the black hole) and 2 (space is being pulled towards the gravitational singularity from all directions), I try to picture the black hole moving, and it seems to me that if, say, the black hole moves in one direction, I can see how its event horizon is also moving, bringing that bit of space into it.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So what's the problem?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But how can the other end of the black hole let space &quot;escape&quot; in order to complete the motion? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For realistic black holes, there is no &quot;other end&quot;. If you really want to tie everything together, then interpret Dr. Thorne's comment about stuff falling into the singularity as crushed out of existence as also applying to space. It doesn't come out of any &quot;other end&quot;. It stops existing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But really, that's past the point where the fluid analogy make sense anyway. I think Dr. Hamilton might note that the full maximally-extended Schwarzschild spacetime is a black hole, so one may be able to think as it &quot;coming out&quot; of the corresponding white hole, but he &lt;a href=&quot;http://jila.colorado.edu/~ajsh/insidebh/waterfall.html&quot; rel=&quot;nofollow&quot;&gt;would also tell you&lt;/a&gt; that this white hole is a mathematical artifact that doesn't have anything to do with actual astrophysical black holes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But I want to understand why it is that space itself can cross the event horizon when other things can't.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course they can. In the infalling-fluid analogy, they're carried along by the space that's being sucked in. Perhaps you're thinking of the fact that it takes an infinite Schwarzschild time for an object to reach the event horizon. But that doesn't imply that things can't cross the horizon; rather, it's just a symptom of Schwarzschild coordinate chart not covering the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, think of a coordinate chart as a &quot;grid&quot; drawn on a patch of spacetime. That patch may be the entire spacetime, or it might be just a piece of it. In the case of the usual Schwarzschild coordinates, it's just a piece... one that simply doesn't cover the horizon.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Regarding the last part, I meant space crossing the event horizon the other way - out of the sphere demarcating the event horizon - whereas material objects go one way - in but not out, spacetime itself does not seem so restricted, at least not in the sense of &quot;pinning&quot; a black hole to one part of itself. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;OK, your edited question provides some more context. But the answer is completely the same: what Dr. Hamilton is talking about in the video is the maximal analytic extensions of the rotating black hole solution, which does contain a passage into a completely different region (sometimes called different 'universe'). This is very much analogous to the maximally extended Schwarzschild spacetime, which contains a wormhole that comes back out of a white hole in another spacetime region, except that the extension of the rotating Kerr black hole spacetime contains an infinite chain of such connected regions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't watched the entire video, but it's clear that's what they're talking about when they say &quot;in a ship propelled by pure mathematics&quot;, because the causal structure of such spacetimes is well-known. Once again, I direct your attention to Dr. Hamilton's own page, linked above, that explains that this dual waterfall picture is an artifact of mathematical idealization and &lt;em&gt;not&lt;/em&gt; something that actually happens in reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, even if you take the maximal analytic extension of such black hole spacetimes overly seriously, it's important to emphasize that you don't come back out into the same region, but rather a different one connected by a wormhole inside the horizon. If you're up to it, I also recommend Dr. Hamilton's &lt;a href=&quot;http://jila.colorado.edu/~ajsh/insidebh/penrose.html&quot; rel=&quot;nofollow&quot;&gt;conformal diagrams&lt;/a&gt; of said black hole spacetimes, which make it quite clear that's what going on is a whole &quot;chain&quot; of black holes and white holes.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditDate="2015-03-11T17:25:46.413" LastActivityDate="2015-03-11T17:25:46.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10087" PostTypeId="1" CreationDate="2015-03-10T20:46:26.963" Score="7" ViewCount="421" Body="&lt;p&gt;I was reading about rotating pulsars and other more-massive objects spinning at absurd speeds (hundreds of revolutions per second), and there was one point of which I was not clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does a rotation of a mass change the gravitational lines of force? In other words, does a body which is rotating change the direction of gravitational pull from this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/edkbq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nIiin.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="58" LastEditorUserId="58" LastEditDate="2015-03-11T14:15:43.497" LastActivityDate="2015-03-11T21:29:53.130" Title="Does rotation affect gravitational lines of force" Tags="&lt;gravity&gt;&lt;rotation&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10088" PostTypeId="2" ParentId="10087" CreationDate="2015-03-10T22:34:27.017" Score="2" Body="&lt;p&gt;An spherical electric charge has the same electric field lines whether spinning or not. The difference between those two cases is entirely in the magnetic field. Thus, one should expect as similar thing to happen for gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Parameterized_post-Newtonian_formalism&quot; rel=&quot;nofollow&quot;&gt;parametrized post-Newtonian formalism&lt;/a&gt;, weak-field GTR has the metric&#xA;$$\mathrm{d}s^2 = -(1+2\Phi)\,\mathrm{d}t^2 + 2\mathcal{A}_j\,\mathrm{d}t\,\mathrm{d}x^j + (1-2\Phi)\delta_{ij}\,\mathrm{d}x^i\,\mathrm{d}x^j$$&#xA;where $\Phi\equiv -U$ is essentially the Newtonian gravitational potential, while $\mathcal{A}_j\equiv-\tfrac{7}{4}V_j - \tfrac{1}{4}W_j$ in terms of the other PPN potentials. For the four-velocity $U^\alpha\equiv{\mathrm{d}x^\alpha}/{\mathrm{d}\tau} = (U^0,\vec{U})$, the geodesic equation for time-independent $\Phi$ and $\mathcal{A}$ becomes, to linear order in $\Phi$,&#xA;$$\frac{\mathrm{d}\vec{U}}{\mathrm{d}\tau} = U^0(\vec{G} + \vec{U}\times\vec{H})\text{,}$$&#xA;where the gravitoelectric field is $\vec{G} = -\nabla\Phi$ and the gravitomagnetic field is $\vec{H} = \nabla\times\vec{\mathcal{A}}$, here $\nabla$ being used in the ordinary sense of $\nabla_i = \partial_i$, i.e. with respect to the Euclidean metric $\delta_{ij}$. If $\Phi$ and $\mathcal{A}$ are not time-independent, then $\vec{G} = -\nabla\Phi - \partial_t\vec{\mathcal{A}}$, paralleling electromagnetism, but there will be an extra term in the analogue of Lorentz force that has no electromagnetic counterpart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much the same thing can be done in any stationary spacetime, including the rotating Kerr black hole. See also Costa and Natário (2014) for a much more general treatment of several gravito-electromagnetic analogies.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Costa, L. F. O, Natário, J. &quot;Gravito-electromagnetic analogies&quot;. &lt;em&gt;Gen. Rel. Grav.&lt;/em&gt; &lt;strong&gt;46&lt;/strong&gt;, 1792 (2014)[arXiv:&lt;a href=&quot;http://arxiv.org/abs/1207.0465&quot; rel=&quot;nofollow&quot;&gt;1207.0465&lt;/a&gt;]&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="661" LastActivityDate="2015-03-10T22:34:27.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10089" PostTypeId="2" ParentId="10083" CreationDate="2015-03-10T22:39:59.663" Score="6" Body="&lt;p&gt;As of March 6, 2015, &lt;a href=&quot;http://www.nbcnews.com/science/space/dawn-spacecraft-slips-quietly-orbit-around-dwarf-planet-ceres-n318371&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dawn has entered orbit around Ceres&lt;/a&gt;.  But it's really so far only &quot;captured by Ceres' gravitational pull&quot;.  Then, it was still 61,000 km from Ceres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's slowly spiraling down into an orbit that &lt;a href=&quot;http://dawnblog.jpl.nasa.gov/2014/02/28/dawn-journal-february-28-2014/&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL's Dawn Journal&lt;/a&gt; calls &quot;RC3&quot;, which will be 13,500 km above Ceres.  This orbit will last 15 days per revolution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It will take about 15 days to complete a single orbital revolution at this altitude.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first news link above indicates that Dawn will reach RC3 on April 23, 2015.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dawnblog.jpl.nasa.gov/2014/02/28/dawn-journal-february-28-2014/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This picture from the Dawn Journal&lt;/a&gt; gives us a picture of Dawn's path: it's not completing any revolutions around Ceres yet; it's still approaching the RC3 orbit as of this writing (March 10, 2015).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2AQEB.jpg&quot; alt=&quot;Dawn approaching Ceres&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as of now, March 10, 2015, there is no orbital period yet.  But when RC3 is achieved, it will be a slow orbit -- 15 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other orbital altitudes planned past RC3.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dawnblog.jpl.nasa.gov/2014/05/31/dawn-journal-may-31/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;second orbital phase, survey orbit&quot;&lt;/a&gt; - Altitude 4,400 km, Orbital period 3 days, 3 hours&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dawnblog.jpl.nasa.gov/2014/06/30/dawn-journal-june-30-2/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;high altitude mapping orbit&quot;&lt;/a&gt; - Altitude 1,470 km, Orbital period 19 hours&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dawnblog.jpl.nasa.gov/2014/08/31/dawn-journal-august-31/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;low altitude mapping orbit&quot;&lt;/a&gt; - Altitude 375 km, Orbital period 5.5 hours&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.jpl.nasa.gov/news/news.php?release=2013-347&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dawn's schedule&lt;/a&gt;, the &quot;survey orbit&quot; will only last 22 days before it spirals in to the &quot;high altitude mapping orbit&quot; in August 2015, although it's unclear exactly when the &quot;survey orbit&quot; will take place.  The &quot;low altitude mapping orbit&quot; will be in November 2015.  In between orbits, when spiraling down to the next scheduled orbit, Dawn's orbital period will slowly decrease.  At the last orbit, Dawn will be left in this orbit after its batteries and hydrazine fuel are exhausted.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Update (Oct 23, 2015)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Excerpt from &lt;a href=&quot;http://dawn.jpl.nasa.gov/news/news-detail.html?id=4725&quot; rel=&quot;nofollow noreferrer&quot;&gt;latest news update from the Dawn mission (Sept 30, 2015)&lt;/a&gt;:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Dawn is currently orbiting Ceres at an altitude of 915 miles (1,470&#xA;  kilometers), and the spacecraft will image the entire surface of the&#xA;  dwarf planet up to six times in this phase of the mission. Each&#xA;  imaging cycle takes 11 days.&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Starting in October and continuing into December, Dawn will descend to&#xA;  its lowest and final orbit, an altitude of 230 miles (375 kilometers).&#xA;  The spacecraft will continue imaging Ceres and taking other data at&#xA;  higher resolutions than ever before at this last orbit. It will remain&#xA;  operational at least through mid-2016.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1082" LastEditorUserId="9186" LastEditDate="2015-10-26T16:03:38.413" LastActivityDate="2015-10-26T16:03:38.413" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10090" PostTypeId="2" ParentId="10085" CreationDate="2015-03-10T23:36:35.910" Score="1" Body="&lt;p&gt;Mostly, but not exclusively, black hole-like solutions (I append the &quot;-like&quot; as a nod to the difficulties in mathematically defining BHs in a satisfactorily general way) are studied in the idealized settings of asymptotically flat spacetimes. An asymptotically flat spacetime has the property of being like flat Minkowski spacetime at a correctly chosen notion of infinity, and can be seen as idealized if for no other reason as all mainstream cosmological models are decided not asymptotically flat. However this setting is still a good way to make predictions about real black holes as we can approximate &quot;very faraway&quot; for &quot;at infinity&quot; and &quot;almost flat&quot; for &quot;flat&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In flat spacetime if we want to change our point of view we can perform a Lorentz boost on our coordinates which will not change the spacetime-physics but will change how we perceive certain  properties such as the velocities of objects, lengths, times between events, etc. In asymptotically flat spacetime we can perform a Lorentz boost on our coordinates at infinity to achieve a similar effect without changing the spacetime-physics. For example we can perform a Lorentz boost at infinity and extend the new boosted coordinates to create new coordinates where the black hole is moving through space at some non-zero velocity. In our new coordinates some properties will be changed (e.g. the shape of the event horizon), whereas other properties will remain invariant.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastActivityDate="2015-03-10T23:36:35.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10091" PostTypeId="1" AcceptedAnswerId="10092" CreationDate="2015-03-11T00:42:54.987" Score="5" ViewCount="955" Body="&lt;p&gt;In school I learned that methane is organic matter and that it is a possible product of crude oil refinement. However, recently I read that Uranus has huge methane storms and the Saturn moon Titan even has lakes consisting only of methane or ethane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm wondering, how does all that methane get out there? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it the result of organic processes or is it created in the cores of suns like other elements? Do we even know?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methane is also a pretty complex material. Do we know sources of more complex molecules other than methane in such quantities? Except life on earth? &lt;/p&gt;&#xA;" OwnerUserId="6740" LastEditorUserId="6740" LastEditDate="2015-03-11T00:48:35.800" LastActivityDate="2015-03-11T20:32:25.890" Title="Why is there so much methane in space?" Tags="&lt;abiogenesis&gt;&lt;astrochemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10092" PostTypeId="2" ParentId="10091" CreationDate="2015-03-11T09:44:19.903" Score="3" Body="&lt;p&gt;Often for molecules to form in interstellar space, dust is used as a catalyst. The reason is that in typical interstellar environments, densities are so immensely low that even for just two atoms to meet, the probability is so small that formation time scales are very long. For 3+ atoms, the chance decreases rapidly. Instead, an atom can stick to a dust grain and wait for ages until other atoms stick. The atoms slowly &quot;crawl&quot; around on the surface of the dust grain, eventually meet and make bonds. If the formation of a bond releases energy (is exothermic), the molecule can be ejected from the grain surface. This process is call &lt;em&gt;adsorption&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dust also helps shielding the molecules from stellar radiation which would otherwise easily destroy them. This is why molecular clouds are also very dusty. But actually UV irradiation helps with the formation of very complex molecules by ionizing less complex molecules that subsequently can make bonds with other atoms and molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the environments are extremely dense, as when a star dies and ejects its gas either as a supernova or a planetary nebula, molecules can also form. This is probably also how the dust itself is formed, although it might also be formed later on (this is currently debated; the problem is that supernovae are so powerful that their shock waves tend to destroy dust shortly after it forms, and planetary nebulae are created from stars that live so long that they can't explain the abundance of dust in the very early Universe where they wouldn't have had the time to live their lives).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Stan Liou and LocalFluff says, methane is actually an &quot;easy&quot; molecule to form, both because it's rather simple, and because its constituents, hydrogen and carbon, are the most and the fourth most abundant elements in the interstellar medium, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, far more complex molecules are regularly found in interstellar space, as can be seen on &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_interstellar_and_circumstellar_molecules#Ten_or_more_atoms_.2815.29&quot; rel=&quot;nofollow&quot;&gt;this list&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-03-11T20:32:25.890" LastActivityDate="2015-03-11T20:32:25.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10093" PostTypeId="1" AcceptedAnswerId="10095" CreationDate="2015-03-11T15:58:21.593" Score="5" ViewCount="458" Body="&lt;p&gt;Whenever you take a look at a photo that is taken by hubble, &#xA;Lets say M87 or Eagle Nebula or maybe Ultra Deep space field,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Image is always &quot;Cut&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like this :&#xA; &lt;img src=&quot;https://i.stack.imgur.com/6k04d.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GNCgL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know reason for this ,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's during photo capture or processing error... or no details about elements to color?&lt;/p&gt;&#xA;" OwnerUserId="2863" LastActivityDate="2015-03-11T16:44:21.233" Title="Why are Hubble images &quot;Cut&quot;" Tags="&lt;space&gt;&lt;deep-sky-observing&gt;&lt;hubble-telescope&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10095" PostTypeId="2" ParentId="10093" CreationDate="2015-03-11T16:44:21.233" Score="6" Body="&lt;p&gt;The &quot;stair-step&quot; pattern visible in the first image is characteristic of the the Hubble's &lt;a href=&quot;https://en.wikipedia.org/wiki/Wide_Field_and_Planetary_Camera_2&quot;&gt;Wide Field and Planetary Camera 2 (WFPC2)&lt;/a&gt;. This pattern is created by the arrangement of the four CCD chips. &lt;a href=&quot;https://en.wikipedia.org/wiki/Wide_Field_and_Planetary_Camera_2#mediaviewer/File:Jfader_hubble.jpg&quot;&gt;This image&lt;/a&gt; shows how the three larger &quot;wide field&quot; chips and the smaller, single &quot;planetary camera&quot; chip are arranged to create the pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom picture appears to be a slice of the original Hubble Deep Field, which was also captured with WFPC2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer, for these two images, is that the shape of the image is an artifact of image capture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in general, finished images such as these could take up any arbitrary shape. Many of the modern images contain data from multiple views, multiple cameras, even multiple telescopes all drizzled together using specialized photo processing software.&lt;/p&gt;&#xA;" OwnerUserId="38" LastActivityDate="2015-03-11T16:44:21.233" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10096" PostTypeId="2" ParentId="10087" CreationDate="2015-03-11T19:56:37.260" Score="2" Body="&lt;p&gt;In general relativity the gravitational field of rotating mass is described by &lt;a href=&quot;http://en.wikipedia.org/wiki/Kerr_metric&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Kerr metric&lt;/em&gt;&lt;/a&gt;. It can be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ ds^2 = ds^2_{\text{Schwarzschild}} + \frac{4GJ}{c^2r} \sin^2 \theta dt d\phi $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last term is responsible for the so called &lt;a href=&quot;http://en.wikipedia.org/wiki/Lense%E2%80%93Thirring_precession&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Lense-Thiring Effect&lt;/em&gt;&lt;/a&gt;, a graviomagnetic effect as already mentioned by Stan Liou. A nice description can be found in the mission description of &lt;a href=&quot;https://einstein.stanford.edu/MISSION/mission1.html&quot; rel=&quot;nofollow&quot;&gt;Gravity Probe B&lt;/a&gt;. So if I look at your image I think your intuition is correct.&lt;/p&gt;&#xA;" OwnerUserId="6712" LastEditorUserId="2153" LastEditDate="2015-03-11T21:29:53.130" LastActivityDate="2015-03-11T21:29:53.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10097" PostTypeId="1" AcceptedAnswerId="10416" CreationDate="2015-03-11T21:21:18.413" Score="16" ViewCount="879" Body="&lt;p&gt;Recently, I encountered the concept of carbon planets - planets, which would be, unlike the Earth, formed mostly by carbon, instead of oxygen, silicon and magnesium. (I am not counting iron, which is mostly locked in the core). Now, I am interested in the planetary chemical compositions in general. I found interesting articles investigating the topic of variance of the planetary chemical compositions (for example &lt;a href=&quot;http://www.researchgate.net/profile/D_Fischer/publication/230967637_The_Planet-Metallicity_Correlation/links/00b7d52be357729458000000.pdf&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;http://arxiv.org/pdf/1405.3253.pdf&quot;&gt;here&lt;/a&gt;), but it is quite hard for me to make an overall picture of the state in the field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to ask on these things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;1. What typical compositions of terrestrial planets are expected to occur?&lt;/em&gt; (I guess there will be plenty of correlations, some groups of elements occuring together as they are bound, for example to the CNO cycle or other nuclear cycle.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;2. How much different the chemical compositions of terrestrial planets really are?&lt;/em&gt; (I.e. are all terrestrial planets mostly similar to each other, or should one rather expect that every planetary system has unique ratios of elements?)&lt;/p&gt;&#xA;" OwnerUserId="2529" LastEditorUserId="6" LastEditDate="2019-05-02T03:22:43.837" LastActivityDate="2019-05-02T03:22:43.837" Title="Statistics of elements abundance in exoplanets" Tags="&lt;exoplanet&gt;&lt;terrestrial-planets&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10099" PostTypeId="1" CreationDate="2015-03-12T07:59:40.503" Score="2" ViewCount="50" Body="&lt;p&gt;Say one has a simple two body system (and one body is much more massive than the other, for example a star, planet system) where all the quantities about it are know, that is the eccentricity, masses, perihelion, etc. Now, at some point in the orbit defined by the polar coordinates at that point, there is a instantaneous change in velocity. Now, the question is what will be the new orbit of the object around the other body? What are all the components of that orbit such as eccentricity, perihelion. Can the system even be solved exactly or only analytically?  &lt;/p&gt;&#xA;" OwnerUserId="6744" LastActivityDate="2015-03-12T07:59:40.503" Title="How to calculate the new orbit after a instantaneous change in a well defined initial one?" Tags="&lt;orbit&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10100" PostTypeId="1" CreationDate="2015-03-09T21:32:45.943" Score="2" ViewCount="121" Body="&lt;p&gt;Visibility of a celestial body on the sky during the day depends on a number of factors - if I recall correctly: the body's magnitude, &lt;em&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/101/what-are-the-metrics-of-observation-conditions&quot;&gt;Seeing&lt;/a&gt;&lt;/em&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Sky_brightness&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sky brightness&lt;/a&gt;, and probably some others I have missed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who was the first to discover/compile the set of variables responsible for determining, whether a celestial body remains visible with naked eye in certain (daylight) conditions?&lt;/p&gt;&#xA;" OwnerUserId="6772" OwnerDisplayName="Major Stackings" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-03-12T11:58:02.797" Title="Who discovered what set of factors is responsible for visibility of celestial bodies on Earth's daytime sky?" Tags="&lt;history&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10101" PostTypeId="1" AcceptedAnswerId="10104" CreationDate="2015-03-12T16:45:06.943" Score="4" ViewCount="249" Body="&lt;p&gt;in physics we were learning about the universe and the big bang. we were being told that everything came from one point and 'exploded' outwards, and is continuing to expand. I felt that this explanation was unclear, so I said that it was more like everything being one the surface of an expanding 'balloon', but with an extra dimension, a space which is the hyper surface of the expanding universe. my teacher disagreed and said it was just like a big explosion. I think this is wrong-it implies that there is an edge of some description, or that the universe is just an expanding sphere of stuff inside infinite space. what is the best way to visualise the universe?&lt;/p&gt;&#xA;" OwnerUserId="2337" LastActivityDate="2015-03-13T16:23:43.110" Title="visualisation of the universe's expansion" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10103" PostTypeId="1" CreationDate="2015-03-12T18:25:09.090" Score="1" ViewCount="23087" Body="&lt;p&gt;I've named a star and would love to be able to find it in the sky - is there an app I can use to locate stars given a specific coordinate?&lt;/p&gt;&#xA;" OwnerUserId="6749" LastActivityDate="2016-01-22T15:39:07.913" Title="app to locate star by coordinates" Tags="&lt;star&gt;&lt;star-gazing&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10104" PostTypeId="2" ParentId="10101" CreationDate="2015-03-12T22:19:01.330" Score="6" Body="&lt;p&gt;You were correct, you teacher was incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the space that expands - much as a surface of a balloon does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An explosion is a poor analogy in contrast because - as you suggest - it implies something to expand into.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For another way, not as accurate as the balloon analogy but maybe helpful - imagine being trapped inside an expanding loaf of bread in an oven. You have no knowledge of the world outside the loaf - and you may assume the loaf is infinite in extent. But as the dough rises the gaps inside the loaf get larger without the mass of the loaf itself increasing. You could compare our position in the universe to being inside such an infinite loaf.&lt;/p&gt;&#xA;" OwnerUserId="823" LastEditorUserId="823" LastEditDate="2015-03-13T16:23:43.110" LastActivityDate="2015-03-13T16:23:43.110" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10105" PostTypeId="2" ParentId="10103" CreationDate="2015-03-13T00:43:02.837" Score="3" Body="&lt;p&gt;There are apps which will help guide you to more significant stars (Google skymap comes to mind if you're looking for a simple mobile system - Stellarium may also be a good option), but it's highly unlikely that any app will include the location and names of so-called 'personal' stars, as the bodies who sell such things don't have any the authority to do so and are not generally recognized in the actual world of Astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the IAU (International Astronomy Union) is the body generally recognized to have the responsibility of naming and classification of stars and other bodies. They also come out strongly against such practices and services: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As an international scientific organization, the IAU dissociates itself entirely from the commercial practice of &quot;selling&quot; fictitious star names or &quot;real estate&quot; on other planets or moons in the Solar System. Accordingly, the IAU maintains no list of the (several competing) enterprises in this business in individual countries of the world.  Readers wanting to contact such enterprises despite the explanations given below should search commercial directories in their country of origin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.iau.org/public/themes/buying_star_names/&quot; rel=&quot;nofollow&quot;&gt;Link to IAU page on buying of star names&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="1288" LastEditDate="2015-03-13T00:55:03.007" LastActivityDate="2015-03-13T00:55:03.007" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10106" PostTypeId="1" AcceptedAnswerId="10107" CreationDate="2015-03-13T06:54:38.190" Score="4" ViewCount="381" Body="&lt;p&gt;As far as I know, a brown dwarf is a 'star' whose core never underwent a fusion reaction, so it never became a star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I was wondering if, apart from orbiting a star, is there any difference between a massive planet and a brown dwarf? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I heard that they found an exoplanet 5 times the mass of Jupiter, and say a brown dwarf was in a binary pair where the other is a star: What stops it from being classified as a planet? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess what I am really asking for is a clearer definition of a brown dwarf.&lt;/p&gt;&#xA;" OwnerUserId="2337" LastEditorUserId="11264" LastEditDate="2018-01-05T14:53:15.503" LastActivityDate="2018-01-05T14:53:15.503" Title="Brown dwarfs and planets" Tags="&lt;star&gt;&lt;planet&gt;&lt;exoplanet&gt;&lt;brown-dwarf&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10107" PostTypeId="2" ParentId="10106" CreationDate="2015-03-13T07:52:18.903" Score="5" Body="&lt;p&gt;Stanley, there really isn't a very clear definition and this is still a keenly argued point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definitions include:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Browns dwarfs burn deuterium. In models this happens if they are more massive than about 13 times Jupiter. The weakness of this that we think isolated brown dwarfs could condense from a gas cloud that are less massive than this; and young brown dwarfs won't have got around to fusing deuterium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets must form from the disk around a star. This is ok, but: brown dwarfs &lt;em&gt;may&lt;/em&gt; also form from the disk and it is also possible for planets to be tidally stripped from their stars and be found alone in space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets must have a rocky core. This used to be thought definite, but now we think maybe sometimes planets can collapse from a gas instability in the disk in some circumstances, without the need for a rocky/icy core. It is true that brown dwarfs should not have a rocky core. However as an observational definition this is fairly hopeless since we can't even tell yet if Jupiter has a rocky core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A flavour of the controversy can be gleaned from reading between the lines of the &lt;a href=&quot;http://home.dtm.ciw.edu/users/boss/definition.html&quot; rel=&quot;nofollow&quot;&gt;IAU statement&lt;/a&gt; on the definition of planets vs brown dwarfs.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-03-13T09:00:00.837" LastActivityDate="2015-03-13T09:00:00.837" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10108" PostTypeId="1" AcceptedAnswerId="26292" CreationDate="2015-03-13T09:08:36.773" Score="6" ViewCount="1042" Body="&lt;p&gt;I apologize for the way this question is worded but I don't think I know the proper verbiage. Basically, I want to know how the stars orbit one another in the two septuple star systems. For example, are there two massive stars in the middle that have the five remaining stars orbiting similar to the planets of our system; does one of the orbiting stars have one or more smaller star-&quot;moons;&quot; maybe there are stars trapped in one of the Lagrange Points. I hope that clears up the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware that the orbits of these systems may not be known/understood. If that is the case, then I would appreciate the orbital relationships between the stars in the highest-multiple star system. If that is confusing, I mean, if we know the orbits for the stars in a sextuple system; if not then a quintuple... etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, as an added bonus, I would appreciate being told the correct verbiage. Thank you, much. &lt;/p&gt;&#xA;" OwnerUserId="6753" LastActivityDate="2018-05-14T15:40:38.570" Title="How do the orbits of Nu Scorpii and AR Cassiopeiae work?" Tags="&lt;orbit&gt;&lt;star-systems&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10109" PostTypeId="1" AcceptedAnswerId="10110" CreationDate="2015-03-13T09:14:14.867" Score="8" ViewCount="198" Body="&lt;p&gt;I just recently heard that there are currently no theoretical models that support planet formation without a star, but I can't remember where. So, I thought I would request all of your expertises. Thank you much. I wanted to confirm whether this is true or not and to know if there are any theories that could account for planet formation in interstellar or intergalactic space. &lt;/p&gt;&#xA;" OwnerUserId="6753" LastEditorUserId="6753" LastEditDate="2015-03-13T21:41:17.133" LastActivityDate="2015-03-13T22:23:56.957" Title="Is there any way a planet could form independent of a star?" Tags="&lt;exoplanet&gt;&lt;planetary-formation&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10110" PostTypeId="2" ParentId="10109" CreationDate="2015-03-13T12:26:45.667" Score="5" Body="&lt;p&gt;Well you need to see the related question &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10106/brown-dwarfs-and-planets&quot;&gt;brown dwarfs and planets&lt;/a&gt; , because the answer to your question depends on how you define a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you demand that a &quot;planet&quot; has a rocky core then it seems very unlikely that a planet could form in isolation away from a parent star. The parent star is needed in order to differentiate the rocky material from the gas and allow it to condense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you wish to define a planet as simply an object below a certain mass (say the deuterium burning threshold at 13 Jupiter masses) then it seems very likely that such an object &lt;em&gt;could&lt;/em&gt; form in isolation. They would be entirely gaseous, but there would be little to distinguish them from brown dwarfs at only slightly higher masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At present there are plenty of candidate &quot;free-floating planetary mass&quot; objects. &#xA;For example see &lt;a href=&quot;http://arxiv.org/abs/1407.7864&quot; rel=&quot;nofollow noreferrer&quot;&gt;Joergens et al. (2014)&lt;/a&gt;; &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013ApJ...777L..20L&quot; rel=&quot;nofollow noreferrer&quot;&gt;Liu et al. (2013)&lt;/a&gt;; &lt;a href=&quot;http://adsabs.harvard.edu/abs/2000Sci...290..103Z&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zapatero-Osorio et al. (2000)&lt;/a&gt;. Unless we have our understanding of the physics completely wrong, then it is likely that at least some of these are lower than 13 Jupiter masses. However, their origin remains unclear. It is &lt;em&gt;possible&lt;/em&gt; they could all have formed around stars and then subsequently been ejected, but the presence of significant numbers of these objects in young star forming regions and the &lt;em&gt;lack&lt;/em&gt; of $\sim$10 Jupiter-mass objects orbiting stars, suggests that there is an alternative formation scenario that can produce such objects in isolation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could these formation scenarios be? These low-mass objects could just be an extension to lower masses of the fragementation process that forms stars; they could be ejected embryos that started their lives in multiple systems; they could be &quot;failed&quot; stellar cores that could not accrete more gas because of photoevaporation by nearby massive stars; or they could form by gravitational instability around stars with unusually massive disks and be ejected by a close encounter with another star. These possibilities are reviewed by &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0602367&quot; rel=&quot;nofollow noreferrer&quot;&gt;Whitworth et al. (2006)&lt;/a&gt; and &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014prpl.conf..619C&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chabrier et al. (2014)&lt;/a&gt;, and are all still thought plausible to some extent.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-03-13T22:23:56.957" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10112" PostTypeId="1" AcceptedAnswerId="10118" CreationDate="2015-03-13T14:27:57.607" Score="4" ViewCount="675" Body="&lt;p&gt;Could we detect an earth-like planet (e.g. similar distance from star, and similar size) with current planet finding techniques?  If an earth-like planet were orbiting our nearest star, excluding the sun :-), would we have detected it by now?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems current techniques tend to find large planets orbiting close to their star, or for direct observation, find large planets further away from their star.  Likewise, we have found many very large planets very close to their star (e.g. hot Jupiters), and some rocky planets somewhat larger than earth close to stars, but not very many (if any) that are very similar to earth (e.g. distance from star and size).  It seems the current techniques available do not readily detect earth-like planets though, so it may be that there really are many earth-like planets, but we haven't seen them yet.  Here is an article on current planet detecting techniques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Methods_of_detecting_exoplanets&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Methods_of_detecting_exoplanets&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2015-03-15T19:05:40.273" Title="Could we detect an earth-like planet around another star?" Tags="&lt;exoplanet&gt;&lt;earth-like-planet&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10113" PostTypeId="1" AcceptedAnswerId="10116" CreationDate="2015-03-13T17:42:59.357" Score="12" ViewCount="38061" Body="&lt;p&gt;I'm writing a program to generate solar systems but I'm having trouble calculating the expected temperature of a planet. I have found a formula to calculate this, but I haven't been able to get a remotely correct answer out of it as it doesn't clearly state what units your supposed to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula I found:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4 \pi R ^ 2 ơ T ^ 4 = \frac{\pi  R ^ 2  L_{\odot}(1 - a)}{(4  \pi  d ^ 2)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R$ is the planet's radius (not sure what units), $d$ is the distance from the Sun (it mentions AU), $a$ is the albedo, $L_{\odot}$ is the luminosity of the Sun (which I assume can be interchanged with the luminosity of any star), $T$ is the temperature of the planet (kelvin, this is what I'm trying to get), and $ơ$ is the Stefan-Boltzmann constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The site I found it on is notes for an astronomy college course. Here is the link:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astronomynotes.com/solarsys/s3c.htm#&quot; rel=&quot;noreferrer&quot;&gt;http://www.astronomynotes.com/solarsys/s3c.htm#&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be very much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="6755" LastEditorUserId="9301" LastEditDate="2016-10-17T16:27:03.747" LastActivityDate="2020-09-20T20:58:59.180" Title="How to calculate the expected surface temperature of a planet" Tags="&lt;planet&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10114" PostTypeId="1" CreationDate="2015-03-13T17:48:01.560" Score="8" ViewCount="264" Body="&lt;p&gt;I am interested in quick generation of planetary systems for programs similar to &lt;a href=&quot;http://en.spaceengine.org/&quot;&gt;Space Engine&lt;/a&gt; or &lt;a href=&quot;http://pioneerspacesim.net/#&amp;amp;panel1-3&quot;&gt;Pioneer Space Simulator&lt;/a&gt;. As far as I understand, formation of planets is a complex process and the planet positions are to a great extend given by the orbital resonances, particularly with the gas giant planets. It is very clearly visible in this &lt;a href=&quot;https://www.youtube.com/watch?v=7dRLvSzDHo8#t=8m00s&quot;&gt;talk of Sean Raymond&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;em&gt;How exactly do the orbital resonances affect the stability of the planet trajectories? Which resonances are stable, and which are unstable in the long term evolution of the planetary system?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2529" LastActivityDate="2016-03-13T11:23:23.393" Title="Orbital resonances and position of planets" Tags="&lt;orbit&gt;&lt;exoplanet&gt;&lt;planetary-formation&gt;&lt;orbital-migration&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10116" PostTypeId="2" ParentId="10113" CreationDate="2015-03-13T18:31:56.047" Score="12" Body="&lt;p&gt;The formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4 \pi R ^ 2 ơ T ^ 4 = \frac{\pi  R ^ 2  L_{\odot}(1 - a)}{4  \pi  d ^ 2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is correct, if you want to calculate the &lt;a href=&quot;http://en.wikipedia.org/wiki/Planetary_equilibrium_temperature&quot; rel=&quot;noreferrer&quot;&gt;radiative equilibrium temperature&lt;/a&gt;. You only need to use the right units. We can further simplify the formula to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T ^ 4 = \frac{ L_{\odot}(1 - a)}{16  \pi  d ^ 2 ơ}\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should input the luminosity in watts, the distance to the star in meters and the Stefan-Boltzmann constant as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$σ = 5.670373 × 10^{−8} \;\mathrm{W}\; \mathrm{m}^{−2}\; \mathrm{K}^{−4}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The albedo is dimensionless. The resulting temperature will be in Kelvins. Let me make an example for Earth:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d = 149,000,000,000 \;\mathrm{m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L = 3.846×10^{26} \;\mathrm{W}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Albedo of Earth is 0.29. (The &lt;a href=&quot;http://en.wikipedia.org/wiki/Bond_albedo&quot; rel=&quot;noreferrer&quot;&gt;Bond albedo&lt;/a&gt; should be used.) You will get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;T ^ 4 = \frac{ 3.846×10^{26}(1 - 0.29)}{16  \pi \times  (149,000,000,000) ^ 2 \times (5.670373 × 10^{−8})}=4,315,325,985 \;\mathrm{K}^4\;.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After powering this number to 1/4, we obtain temperature 256 K, which is -17° C. This looks reasonable. The real average temperature on Earth is closer to 15° C, but the greenhouse effect is responsible for the difference.&lt;/p&gt;&#xA;" OwnerUserId="2529" LastEditorUserId="9301" LastEditDate="2016-10-17T16:24:24.307" LastActivityDate="2016-10-17T16:24:24.307" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10117" PostTypeId="1" AcceptedAnswerId="10120" CreationDate="2015-03-13T18:59:08.883" Score="-5" ViewCount="479" Body="&lt;p&gt;Does dark matter really exist or is just a made-up theory in order to cover up the loopholes in the Big Bang theory, and also to explain why universe is expanding at an accelerated rate?&lt;/p&gt;&#xA;" OwnerUserId="6758" LastEditorUserId="2153" LastEditDate="2015-03-13T21:43:27.540" LastActivityDate="2016-10-29T16:28:17.197" Title="Does &quot;Dark Matter&quot; really exist?" Tags="&lt;dark-matter&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10118" PostTypeId="2" ParentId="10112" CreationDate="2015-03-13T20:47:53.220" Score="1" Body="&lt;p&gt;The short answer is yes, the Kepler mission has used &lt;em&gt;transits&lt;/em&gt; to detect nearly &quot;Earthlike&quot; planets, one of the classic examples is &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler-22b&quot; rel=&quot;nofollow&quot;&gt;Kepler-22b&lt;/a&gt; is only about 2x the radius of Earth and has a year that is 290 (Earth) days. Kepler has detected many planets smaller than Earth and closer to the star. With more observing time Kepler could discover similarly sized or smaller planets at comparable orbits, so technologically transits can already detect  &quot;Earthlike&quot; planets. Practically, with Kepler's lost reaction wheel this has been difficult, but more Earth sized planet detections are likely from the &lt;a href=&quot;http://arxiv.org/abs/1402.5163&quot; rel=&quot;nofollow&quot;&gt;K2 Kepler reboot&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Microlensing&lt;/em&gt; is more sensitive to planets further from stars, a next-generation space telescope like the &lt;a href=&quot;http://wfirst.gsfc.nasa.gov/learn/exoplanets/index.html&quot; rel=&quot;nofollow&quot;&gt;WFIRST space telescope&lt;/a&gt;, which might launch in the 2020s, expects to be able to detect planets smaller than Earth at slightly larger distances from stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking a photograph or spectra, or &lt;em&gt;directly imaging&lt;/em&gt;, a small planet far from its parent star is much more difficult, light from the Sun is around 10 billion times brighter than the sunlight the Earth reflects (a $10^{10}$ contrast), the sensitivity of current ground based instruments (like &lt;a href=&quot;http://planetimager.org/&quot; rel=&quot;nofollow&quot;&gt;GPI&lt;/a&gt;) and the Hubble space telescope are closer to $10^7$.  &lt;a href=&quot;http://wfirst.gsfc.nasa.gov/learn/exoplanets/index.html&quot; rel=&quot;nofollow&quot;&gt;WFIRST project&lt;/a&gt; will be be able to detect giant planets and may be able to detect Earthlike planets but the technology is not quite there yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of instrument stability and stellar variability, &lt;em&gt;radial velocity&lt;/em&gt; instruments are not yet sensitive enough to pick out the incredibly small wobbles caused by a small planet in an Earthlike orbit. Future instruments, like the &lt;a href=&quot;http://exoplanets.astro.yale.edu/science/100_Earths.php&quot; rel=&quot;nofollow&quot;&gt;100 Earths Project&lt;/a&gt; may get to this sensitivity in the moderately near future.&lt;/p&gt;&#xA;" OwnerUserId="457" LastEditorUserId="457" LastEditDate="2015-03-15T19:05:40.273" LastActivityDate="2015-03-15T19:05:40.273" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10119" PostTypeId="2" ParentId="10112" CreationDate="2015-03-13T20:54:06.757" Score="1" Body="&lt;p&gt;The short answer is no; we cannot &lt;strong&gt;quite&lt;/strong&gt; detect earth-like planets around Sun-like stars with orbital periods of 1 year. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two main planetary detection techniques are transit photometry and the radial velocity variation technique. Direct imaging of earth-like planets at 1 au from the host star is utterly impossible with &lt;em&gt;current technology&lt;/em&gt; - the problem is not the sensitivity, it is the contrast achievable at small angular separations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first demands high precision photometry (a transit by an &quot;earth&quot; across a &quot;sun&quot; produces a lightcurve dip of about 0.01%). This kind of precision &lt;em&gt;has&lt;/em&gt; been achieved (by space-based observatories), but they have not observed stars for sufficiently long to build up the requisite number of transits (you need &lt;em&gt;at least&lt;/em&gt; a few) to confirm a detection at periods of 365 days The Kepler primary mission ceased after about 4 years, meaning it will be tricky to dredge out convincing Earth-like transit signals at periods of 1 year (but not impossible) -  and even then you need to perform some sort of follow up to prove it is a planetary mass object, rather than some false positive, and actually estimate the mass to show that it is a rocky planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which brings us to the doppler radial velocity technique. The motion of the Earth-Sun system results in the Sun executing a 1-year orbit around the common centre-of-mass, with an amplitude of about 9 cm/s. This is about a factor of 5-10 smaller than the best precision that is available at any telescope in the word right now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So - although none have been confirmed yet (there are &lt;em&gt;candidates&lt;/em&gt; in the Kepler data), that does not mean that Earth-like planets are uncommon. Indeed most sensible extrapolations of the frequency of Earth-&lt;em&gt;sized&lt;/em&gt; planets found at closer orbital distances suggests that they could be quite common (e.g. $\sim 25$% &lt;a href=&quot;http://www.pnas.org/content/110/48/19273.abstract&quot; rel=&quot;nofollow&quot;&gt;Petigura et al. 2013&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-03-14T12:40:03.043" LastActivityDate="2015-03-14T12:40:03.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10120" PostTypeId="2" ParentId="10117" CreationDate="2015-03-13T22:22:43.167" Score="7" Body="&lt;p&gt;The question seems to be based on some misconceptions, so I'll try to clear them up.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Dark matter is not the same as dark energy.&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_Matter&quot; rel=&quot;nofollow&quot;&gt;Dark matter&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_energy&quot; rel=&quot;nofollow&quot;&gt;dark energy&lt;/a&gt; are two completely separate and unrelated concepts. It's unfortunate that many articles trying to entice readers with tales of the unexplored frontiers of astronomy tend to use the phrase &quot;dark matter and dark energy&quot; quite a lot, typically with the buzzwords &quot;mysterious,&quot; &quot;unknown,&quot; and even &quot;spooky.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, the best introduction here is to read the Wikipedia articles on dark matter and dark energy. You can also see the &lt;a href=&quot;/questions/tagged/dark-matter&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged 'dark-matter'&quot; rel=&quot;tag&quot;&gt;dark-matter&lt;/a&gt; and &lt;a href=&quot;/questions/tagged/dark-energy&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged 'dark-energy'&quot; rel=&quot;tag&quot;&gt;dark-energy&lt;/a&gt; tags (and the questions that use them) for more information. &lt;a href=&quot;https://physics.stackexchange.com/&quot;&gt;Physics&lt;/a&gt; also has some good questions about these topics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll give you the important points:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Dark matter . . .&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;. . . interacts via gravity&lt;/li&gt;&#xA;&lt;li&gt;. . . has no electric charge (and so does not interact with light outside of possibly &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_lens&quot; rel=&quot;nofollow&quot;&gt;gravitational lensing&lt;/a&gt;, which does not involve electric charge)&lt;/li&gt;&#xA;&lt;li&gt;. . . may be composed of one (or more) of &lt;a href=&quot;https://astronomy.stackexchange.com/questions/322/dark-matter-particle-candidates?rq=1&quot;&gt;several hypothetical (and some not-so-hypothetical) particles&lt;/a&gt;, such as axions or WIMPS (one of the more popular choices)&lt;/li&gt;&#xA;&lt;li&gt;. . . is responsible for some stars &lt;a href=&quot;https://en.wikipedia.org/wiki/Galaxy_rotation_curve&quot; rel=&quot;nofollow&quot;&gt;moving faster than they should&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Dark energy . . .&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;. . . is responsible for the acceleration of the expansion of the universe, though not the expansion of the universe&lt;/li&gt;&#xA;&lt;li&gt;. . . can be represented by &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmological_constant&quot; rel=&quot;nofollow&quot;&gt;the cosmological constant&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The one thing that the two share? We know very little about them.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Dark energy does not solve as many problems in the Big Bang theory as you may think.&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In fact, it causes more problems than it solves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can come up with many different universes that begin with a Big Bang but don't have an accelerated expansion (not counting &lt;a href=&quot;https://en.wikipedia.org/wiki/Inflation_(cosmology)&quot; rel=&quot;nofollow&quot;&gt;inflation&lt;/a&gt;, though you don't have to have that in a universe under some circumstances). They don't somehow suffer from any problems that universes with dark energy don't have. In fact, they're better because you don't have to somehow explain dark energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Courtesy of &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10117/does-dark-matter-really-exist/10120#comment13269_10120&quot;&gt;Rob Jeffries&lt;/a&gt;, here's one tidbit about the one fairly important role of dark energy in the Big Bang:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our universe is close to flat (see the many papers by the WMAP team), and quite a lot of measurements have confirmed that we're not living on some wacky-shaped universe. The issue is that the measurements indicate that the universe isn't &lt;em&gt;quite&lt;/em&gt; flat. Again, see the WMAP data. You can attribute part of the percentage offset from flatness to experimental error, or you can attribute it to the idea that perhaps the universe isn't so flat after all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark energy provides an explanation for this inherent not-quite-flat-ness that characterizes our universe. In those other toy universes (without dark matter), you need to explain this near-flatness some other way. Not in our universe, it seems.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Dark matter and dark energy aren't just theories.&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Well, sort of. Coming from someone like me (who is firmly on the &quot;theorist&quot; side of the scientific spectrum), that's quite a whopper. But from a certain point of view, it's completely accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark matter is actually a theory. It's an explanation for some screwed-up galactic rotation curves, where some stars near the outer edges of a galaxy move much faster than they should.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/e/ec/M33_rotation_curve_HI.gif&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/e/ec/M33_rotation_curve_HI.gif&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's actually a theory to explain a phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark energy, on the other hand, is an experimental phenomenon. We don't have a good theory to explain it. But it's a phenomenon responsible for &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_energy#Evidence_of_existence&quot; rel=&quot;nofollow&quot;&gt;many different problems&lt;/a&gt;. There's quite a lot of evidence that says, &quot;Something odd is going on.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I mean by all this is that there's a heck of a lot of evidence for them. Our theories of them were created to explain evidence, and not vice versa (evidence gathered to support or refute a theory).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I advise, once again, that you look at the Wikipedia articles on both concepts to start learning more (especially the sections on evidence). Physics and Astronomy have plenty of good information. &lt;a href=&quot;http://arxiv.org/&quot; rel=&quot;nofollow&quot;&gt;arXiv&lt;/a&gt; has many pre-prints that may be helpful. And there are lots of books, magazines, and (reputable) web sites that can give you even more detailed information.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-03-14T22:09:15.560" LastActivityDate="2015-03-14T22:09:15.560" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10121" PostTypeId="1" CreationDate="2015-03-13T22:29:41.510" Score="10" ViewCount="193" Body="&lt;p&gt;If Space is expanding between the Galaxies then why isn't it also expanding between the Stars within the Galaxies as well ? In fact why isn't Space expanding within our Solar System ?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastEditorUserId="5278" LastEditDate="2015-03-14T22:09:38.813" LastActivityDate="2015-03-14T22:09:38.813" Title="Expansion of Space" Tags="&lt;space&gt;&lt;expansion&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10122" PostTypeId="2" ParentId="10121" CreationDate="2015-03-13T22:41:41.627" Score="0" Body="&lt;p&gt;Actually, space is expanding everywhere. It is more noticeable between galaxies since space between galaxies is much larger than space between stars and space between the planets in the solar system.&#xA;A 1% expansion of space only represents 1.5 million kilometers between the Sun and the Earth.  But the same 1% expansion between two galaxies that are four billion light-years away means that they are now 40 light-years farther away.&#xA;And 40 light-years is ten times the distance from the Sun to the nearest star Proxima Centaury.&lt;/p&gt;&#xA;" OwnerUserId="5471" LastActivityDate="2015-03-13T22:41:41.627" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10123" PostTypeId="1" AcceptedAnswerId="10130" CreationDate="2015-03-14T04:05:39.147" Score="4" ViewCount="277" Body="&lt;p&gt;Are all galaxies moving away from us at constant speed even those that may be moving in our direction as the space is being formed? How does nothingness appear to push matter? A black hole sucks as space is compressing (being stretched without stopping) or compressed (like how our Sun pulling all planets) and is it the opposite phenomenon that show universe is expanding?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="2153" LastEditDate="2015-03-14T22:11:00.420" LastActivityDate="2016-10-07T18:00:07.283" Title="Are galaxies moving away from us faster than before?" Tags="&lt;galaxy&gt;&lt;universe&gt;&lt;cosmology&gt;&lt;expansion&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10124" PostTypeId="1" AcceptedAnswerId="10127" CreationDate="2015-03-14T04:43:12.733" Score="9" ViewCount="2858" Body="&lt;p&gt;I just recently learned about the theory that our Sun has a small companion star with a 26 million year orbit. This theory came about when it was realized that mass extinction events happened approximately every 26 million years. My question is not whether or not this is possible as that has been asked before; my question is how could we have missed it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we look for exoplanets in the Universe, we are sometimes able to detect a slight wobble in the parent stars. Along those lines, if the Sun had a companion star, how would we not notice the effects on its rotation/orbit? If the sun and this companion star were locked in a 26 million year orbit, would we be able to notice the orbit? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, when you have a star system like ours, with so many planets, is it possible to tell which planet is causing which specific motion in the Sun? For example, let's say we traveled to a different star system a few hundred light years away, and we were studying our solar system. Would we be able to look at the way the Sun wobbles and determine how many planets were in orbit around our Sun, or would we only be able to say that there was one, or simply that there may be several? I guess, the root question is how accurately can we judge the contents of a star system based on the perturbations in the star's orbit? Thank you kindly for your time. &lt;/p&gt;&#xA;" OwnerUserId="6753" LastEditorUserId="20348" LastEditDate="2019-09-26T12:16:41.987" LastActivityDate="2019-09-26T12:16:41.987" Title="Is it possible that the Sun has a binary partner (the Nemesis Theory) that has eluded detection?" Tags="&lt;the-sun&gt;&lt;orbit&gt;&lt;exoplanet&gt;&lt;nemesis&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ClosedDate="2016-11-19T19:50:18.820" ContentLicense="CC BY-SA 3.0" />
  <row Id="10125" PostTypeId="2" ParentId="10124" CreationDate="2015-03-14T05:25:46.497" Score="0" Body="&lt;p&gt;Revised:&lt;br&gt;&#xA;There are several reasons why the sun cannot be part of a binary system (even if the orbit is 26 million years). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the star has the mass of Sol, the binary system would orbit around the point equal distant from both masses. This would definitely show a change in parallax for some of the stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Sol size star in a 26 million year orbit (about 3.5 ly) would have been noticed. It would have an apparent magnitude of -0.01. At this distance, the star would be influenced (by gravity) from our neighbouring stars and would probably separate into its own system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The greatest mass that would not become a star would be less than 1/10 solar masses, which would have an orbital radius of less than 1.43 ly (90600 AU). The center of mass would be over 9000 AU from the sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Oort Cloud extends beyond 2 ly (Wikipedia states 3 ly), a Jupiter size object would probably have cleared an area around its orbit of other objects. &lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="1322" LastEditDate="2015-03-15T19:26:14.167" LastActivityDate="2015-03-15T19:26:14.167" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10126" PostTypeId="2" ParentId="10121" CreationDate="2015-03-14T08:06:22.637" Score="4" Body="&lt;p&gt;The Universe is remarkably homogeneous, i.e. the same everywhere, and isotropic, i.e. the same in every direction (NB homogeneity does not imply isotropy and there are toy cosmological models which are homogeneous but not isotropic) and this is the underlying physical assumption of big bang cosmology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When modelling the Universe the simplification of it being 100% homogeneous and isotropic is most often used as because on the very largest scale (i.e. the scale of the observable Universe) this is very close to being true and any deviations from this are small enough that they do not effect most of the properties being studied. In 100% homogeneous and isotropic Universe, expansion takes place at all places and at all scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However we know that the Universe is not 100% homogeneous and isotropic, especially on a smaller scale; for example the centre of a star is a very different place from intergalactic space. It is a fair question then to ask in view of this how does expansion manifest itself on smaller scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that in gravitationally bound systems such as our local group (which includes Andromeda) matter has the tendency to be drawn closer by gravity rather than to move away as objects do in an expanding Universe, so it seems our local group is not expanding. Still, it is very tempting to see expansion on this scale as being like a small repulsive force which acts against gravity, but is not strong enough to overcome gravity. So in this sense even the local group would be expanding&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the view of expansion acting like a small repulsive force on a small scale is not necessarily the right view. A counterexample would be the Einstein-Strauss Swiss-cheese model where expansion only takes place in deep space and not in the space around stars (modeled as vacuoles). The Einstein-Strauss model though cannot be taken as the last word on the subject as the way it models stars in space is not that realistic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall then how expansion manifests itself on a smaller scale is still very much an open question and we just don't know what tiny corrections if any it would make on the dynamics of a very small system such as the solar system.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastActivityDate="2015-03-14T08:06:22.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10127" PostTypeId="2" ParentId="10124" CreationDate="2015-03-14T09:46:29.027" Score="15" Body="&lt;p&gt;The first part of your question has been asked before:  &lt;a href=&quot;https://astronomy.stackexchange.com/questions/2531/is-sun-a-part-of-a-binary-system&quot;&gt;Is Sun a part of a binary system?&lt;/a&gt; and the current (lack of) evidence for such a companion is discussed on the relevant &lt;a href=&quot;http://en.wikipedia.org/wiki/Nemesis_%28hypothetical_star%29&quot; rel=&quot;noreferrer&quot;&gt;wikipedia page&lt;/a&gt; about &quot;Nemesis&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarise: if it were a small companion star, or even a brown dwarf that had been cooling for 4.5 billion years &lt;em&gt;and&lt;/em&gt; it had a 26 million year orbit, then Kepler's third law tells us how far away that object would be. It turns out that this is not far enough away that the object would have eluded detection in all-sky surveys. The recent WISE survey in the near-infrared should have been capable of detecting even a very cool brown dwarf (and indeed it has detected some very cool brown dwarfs, just not &lt;em&gt;that&lt;/em&gt; close to the Sun - e.g. a 250K brown dwarf only 6 light years away &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014ApJ...786L..18L&quot; rel=&quot;noreferrer&quot;&gt;Luhman et al. 2014&lt;/a&gt;) that was close enough to the Sun to be a Nemesis candidate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the details do not appear to be on the wikipedia page, I'll fill some in. If we take a very low mass brown dwarf - say 20 Jupiter masses - in a 26 million year orbit, then Kepler's 3rd law tells us this will be about 90,000 au (1.4 light years) from the Sun (assuming a circular orbit). According to the evolutionary models of &lt;a href=&quot;http://adsabs.harvard.edu/abs/2008ApJ...689.1327S&quot; rel=&quot;noreferrer&quot;&gt;Saumon &amp;amp; Marley (2008)&lt;/a&gt;, such an object has an intrinsic luminosity of $10^{-7}$ times (1 ten millionth) that of the Sun and a temperature of 400 Kelvin and would appear to have a spectral type of late T or early Y. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the calibration of absolute magnitudes versus spectral type for cool brown dwarfs in &lt;a href=&quot;http://arxiv.org/abs/1211.6977&quot; rel=&quot;noreferrer&quot;&gt;Marsh et al. (2013)&lt;/a&gt; we know that at 90,000 au, such a brown dwarf would have magnitudes of $H=14$ and $W2=8$. The former is bright enough to have been seen in the 2MASS all-sky survey and the latter is easily detected by WISE. The combination of data would also easily have revealed the large parallax of such an object. We can conclude that an object would have to have &lt;em&gt;much&lt;/em&gt; lower mass than this to remain undetected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the middle part of your question: A star will &quot;wobble&quot; in response to its companion with exactly the same period as the orbit. So, if you are prepared to wait for an appreciable fraction of 26 million years, then yes, the presence of the binary companion might be revealed in a net small oscillation of average proper motions over the whole sky, with a period of 26 million years! Otherwise not. Measuring the &quot;wobble&quot; by doppler methods is currently capable of detecting Jupiter-mass objects in orbital periods of 10-20 years. Measuring the wobble &quot;astrometrically&quot;  - that is measuring the position displacement of the star due to an unseen companion is more sensitive to distant companions, but still, one is limited by the fact that the wobble period will be the same as the orbital period of the companion. An assessment of the performance of the Gaia astrometry satellite by &lt;a href=&quot;http://arxiv.org/abs/1411.1173&quot; rel=&quot;noreferrer&quot;&gt;Perryman et al. (2014)&lt;/a&gt; suggests detection of planets with orbital periods out to 10 years might be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the final part of the question - yes, the motion/wobble of a star can be decomposed (using Fourier techniques) into components that are due to each planet. There are numerous examples of multiple planetary systems that have been discovered by the radial velocity technique in this manner. Similar techniques can and will be employed to analyse any astrometric wobble. An interesting plot is to show the motion of the Sun in the plane of our solar system, in the rest frame of the solar system centre of mass. The Sun executes a complicated trajectory in this plot due primarily to the influence of Jupiter (on an 11 year orbit), but with superimposed &quot;epicycles&quot; due to the influence of the smaller planets on different orbital periods. See below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motion of the Sun around the solar system centre of mass (from &lt;a href=&quot;http://commons.wikimedia.org/wiki/File:Solar_system_barycenter.svg&quot; rel=&quot;noreferrer&quot;&gt;http://commons.wikimedia.org/wiki/File:Solar_system_barycenter.svg&lt;/a&gt; )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NBknM.png&quot; alt=&quot;Motion of Sun around solar system barycentre&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-11-17T22:05:47.407" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10128" PostTypeId="2" ParentId="8636" CreationDate="2015-03-14T11:35:15.470" Score="2" Body="&lt;p&gt;There is quite simple &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_locking#Timescale&quot; rel=&quot;nofollow&quot;&gt;formula&lt;/a&gt; that will give you the tidal-locking half-time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T = C\, \frac{a^6 R \mu}{  M_s M_p^2}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$a$ - semi-major axis, or simply the radius of the planet circular orbital trajectory in meters&lt;/li&gt;&#xA;&lt;li&gt;$R$ - planet radius in meters&lt;/li&gt;&#xA;&lt;li&gt;$M_s$ - planet mass in kg&lt;/li&gt;&#xA;&lt;li&gt;$M_p$ - parent planet/star mass in kg&lt;/li&gt;&#xA;&lt;li&gt;$\mu$ - rigidity, approximately $3×10^{10}$ for rocky objects and $4×10^9$ for icy ones.&lt;/li&gt;&#xA;&lt;li&gt;$C$ - a prefactor dependent on units used. On Wikipedia, they write $C = 6 × 10^{10}$ if the time is to be in years, but this number is probably wrong. (&lt;a href=&quot;https://en.wikipedia.org/wiki/Talk:Tidal_locking#Contradicting_formulae_in_.22Timescale.22&quot; rel=&quot;nofollow&quot;&gt;See the discussion&lt;/a&gt;.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Solving the problem of interaction between the parent star and between the planets will be difficult. Generally, if tidal-locking half-time between the planets is much shorter than the tidal-locking half-time between the star and the planet, the planets will tidal-lock to each other rather than to the star. So we ask, whether &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{A^6}{  M_s^3} \ll \frac{a^6}{  M_s M_p^2}\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ is the semi-major axis of the orbit of the double-planet. For typical planet in the habitable zone of the red dwarf star, $a$ = 9,000,000,000 m (0.06 AU), while the distance of the double-planet could be for example $A$ = 30,000,000 m. $M_s = 6\times 10^{24}\;\mathrm{kg}$ and $M_p = 2.9\times 10^{29}\;\mathrm{kg}$. We get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\cdot 10^{-41}\ll 6.3\cdot 10^{-36}\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is well-satisfied and the planets will tidal lock to each other rather than to the parent star. Were the distance between the planets $A$ = 300,000,000 m, the inequality would yield &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\cdot 10^{-35}\ll 6.3\cdot 10^{-36}\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is not satisfied and the tidal forces of the star would probably disrupt the double-planet. (The planets would not even be in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;nofollow&quot;&gt;Hill sphere&lt;/a&gt;, which approximately gives the stable orbits.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is: a very close double planets can tidal-lock to each other rather than to the parent red-dwarf star, but they would have to be much closer than the Earth to the Moon.&lt;/p&gt;&#xA;" OwnerUserId="2529" LastEditorUserId="2529" LastEditDate="2015-03-14T12:06:31.130" LastActivityDate="2015-03-14T12:06:31.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10130" PostTypeId="2" ParentId="10123" CreationDate="2015-03-14T20:52:57.497" Score="2" Body="&lt;p&gt;The recessional velocity $v$ of an object depends on two things: firstly it depends on how faraway an object is in terms of proper distance $D$, and secondly on the rate of the Universe's expansion as a function of cosmological time $t$, which is best expressed as the Hubble parameter $H(t)$. Specifically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = D \times H(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth noting that this equation is slightly vacuous as it is merely the definition of recessional velocity, which is not something that can be directly measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As recessional velocity depends not just on a function of $t$, but also on $D$, the question as to whether objects are receding from us faster than ever before could be answered in several ways. Before I look at the different ways we could answer your question, I will note a few things. Firstly the definition of the Hubble parameter is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H^2(t) = \bigg( \frac{\dot{a}(t)}{a(t)}\bigg)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a(t)$ is the scale factor which describes how the scale of the Universe changes with $t$, and $\dot{a}(t)$ is the first derivative of the scale factor with respect to $t$. Due to cosmological observations, the Universe is said to contain dark energy which causes the Universe's expansion to accelerate. What is meant by this is that at the current time $\ddot{a}(t) &amp;gt; 0$ where $\ddot{a}(t)$ is the second derivative of the scale factor with respect to time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first way we could look at your question is we could ask whether galaxies currently at a distance $D_0$ are receding faster than other galaxies that were previously at distance $D_0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the definition of accelerating expansion and the Hubble parameter we can see that accelerating expansion does &lt;strong&gt;not&lt;/strong&gt; imply that the answer to this question is &quot;yes&quot; and in fact if we assume dark energy takes the form of a cosmological constant (ignoring cosmic inflation), and we delve into the dynamics of the Universe, we find that galaxies currently at $D_0$ must be receding from us slower than the galaxies that were previously at $D_0$ were receding when they were at $D_0$. So in this particular sense the Universe's expansion is slowing down, even though we usually describe it as accelerated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second way we could answer your question is to ask whether the recessional velocity of any given galaxy is larger now than it has ever been in the past.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to this question is more difficult as accelerated expansion does imply that the recessional velocity of a given galaxy increase with time, but the Universe's rate of expansion in previous epochs was decelerating. However, again taking dark energy as taking the form of a cosmological constant, we see that the answer is that galaxies achieve their highest recessional velocities twice: firstly at the Big Bang, and secondly in the infinite future. So the answer to this question is that galaxies are &lt;strong&gt;not&lt;/strong&gt; currently receding from us faster than they have been at all previous times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recession velocity is different from peculiar velocity (i.e. the local velocity with respect to the CMB). We could add the two to find the 'real velocity', but as I've noted recession velocity doesn't have a direct physical meaning so what this 'real velocity' actually means is not straightforward. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expansion is homogeneous, whereas the vacuum around a black hole is not homogeneous, so in this sense the 'sucking' of a black hole is not the opposite of expansion.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastEditorUserId="9301" LastEditDate="2016-10-07T18:00:07.283" LastActivityDate="2016-10-07T18:00:07.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10131" PostTypeId="1" CreationDate="2015-03-14T23:11:02.780" Score="6" ViewCount="102" Body="&lt;p&gt;According to the NASA JPL web-report, &lt;a href=&quot;http://www.jpl.nasa.gov/spaceimages/details.php?id=pia18822&quot; rel=&quot;noreferrer&quot;&gt;Gravity Gradients Frame Oceanus Procellarum&lt;/a&gt;, a rift-like ring structure surrounding the Moon's Oceanus Procellarum has been detected as a gravity anomaly by the &lt;a href=&quot;http://solarsystem.nasa.gov/grail/home.cfm&quot; rel=&quot;noreferrer&quot;&gt;GRAIL&lt;/a&gt; scientific mission (as can be as the dark blue features on the image below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IZpIf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image credit: NASA/Colorado School of Mines/MIT/GSFC/Scientific Visualization Studio&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What mechanism could have formed the Moon's Oceanus Procellarum rift-like gravity anomalies?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user2449" LastEditorDisplayName="user2449" LastEditDate="2015-03-15T09:42:01.157" LastActivityDate="2018-08-20T22:49:34.850" Title="What mechanism could have formed the Moon's Oceanus Procellarum rift-like gravity anomalies?" Tags="&lt;the-moon&gt;&lt;volcanism&gt;&lt;nasa&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10132" PostTypeId="1" AcceptedAnswerId="10153" CreationDate="2015-03-15T00:24:49.697" Score="4" ViewCount="176" Body="&lt;p&gt;Is there a spatial or gravitational relationship between black holes and dark matter/energy?&lt;/p&gt;&#xA;" OwnerUserId="6750" LastEditorUserId="6750" LastEditDate="2015-03-15T19:51:52.087" LastActivityDate="2016-12-09T19:46:50.587" Title="Is there a connection between black holes and dark matter/energy?" Tags="&lt;supermassive-black-hole&gt;&lt;dark-matter&gt;&lt;dark-energy&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10133" PostTypeId="2" ParentId="10132" CreationDate="2015-03-15T00:27:57.363" Score="-2" Body="&lt;p&gt;No there is no known particular connection between black holes, dark matter and dark energy.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastActivityDate="2015-03-15T00:27:57.363" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10134" PostTypeId="1" CreationDate="2015-03-15T03:52:33.600" Score="5" ViewCount="121" Body="&lt;p&gt;Are there certain parts of the Universe that are expanding faster than others? Is there anyway to definitively measure this?&lt;/p&gt;&#xA;" OwnerUserId="6753" LastActivityDate="2015-03-16T08:21:55.660" Title="Is dark energy evenly spaced throughout Universe?" Tags="&lt;universe&gt;&lt;expansion&gt;&lt;dark-energy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10135" PostTypeId="1" AcceptedAnswerId="10136" CreationDate="2015-03-15T04:31:26.197" Score="1" ViewCount="201" Body="&lt;p&gt;The escape velocity of an object is given by the following formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_e=\sqrt{\frac{2GM}{r}}=\sqrt{\frac{2 \mu}{r}}= \sqrt{2gr}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I am aware, the mass of the star which formed the black hole is the same as the black hole itself, the only difference is the black hole's density. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the escape velocity of the star which formed the black hole is not as high because light &lt;em&gt;can&lt;/em&gt; escape the star. Considering the equation above, $G$ is constant, $M$ is constant, however, $r$ changes resulting in a much smaller region as compared to the star. Consequently, it's the radius which determines the high escape velocity of a black hole, not its density. Then why is it that I always hear that black holes behave the way they do because they are highly dense regions in spacetime? What am I getting wrong? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1897" LastEditorUserId="1897" LastEditDate="2015-03-15T15:23:09.507" LastActivityDate="2015-03-15T15:23:09.507" Title="Why are the high density regions in outer space black holes?" Tags="&lt;black-hole&gt;&lt;density&gt;&lt;escape-velocity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10136" PostTypeId="2" ParentId="10135" CreationDate="2015-03-15T08:57:26.613" Score="1" Body="&lt;p&gt;Making a major edit to this answer, as it would seem that there is a rather significant misunderstanding in the formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially describing &lt;em&gt;r&lt;/em&gt; as the radius may have been a mistake.  While we may normally describe &lt;em&gt;r&lt;/em&gt; as the radius of Earth when describing Earth, this is because we are describing the escape velocity of Earth from the surface, which means that the difference between our location and the center of gravity of the Earth is roughly equal to the radius of Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When determining escape velocity from an arbitrary distance, &lt;em&gt;r&lt;/em&gt; should be the distance between the escaping body and the center of gravity of the body being escaped. For objects at a normal distance, changing the volume, and thus the density of the object, will not affect nearby objects, as the mass stays the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this changes when you get close, when you compress the object this allows you to get closer to the center of gravity of the object. As you know, gravity is a weak force which deteriorates quickly with distance.  As such, when you decrease the volume of the body, a body can then become closer to its center of gravity, increasing the potential maximum escape velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The body becomes a black hole when the physical volume of the body becomes small enough that the Schwarzschild Radius extends beyond it. Because we take the same mass, and decrease the volume, the result is a much denser area of spacetime, but this isn't generally the result of only a tiny volume, but also a large amount of mass.&lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="1288" LastEditDate="2015-03-15T11:01:48.673" LastActivityDate="2015-03-15T11:01:48.673" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10137" PostTypeId="2" ParentId="10134" CreationDate="2015-03-15T13:21:41.937" Score="5" Body="&lt;p&gt;If dark energy varied by location, then plots of 1a supernova brightness vs redshift should vary depending on which direction you look in the sky. AFAIK, that's not the case. For example, although coords are not accounted for, there's not a lot of scatter &lt;a href=&quot;http://phys.org/news/2011-10-supernovae-universe-expansion-understood-dark.html&quot;&gt;in this plot&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/F8YhF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the relationship between the distance and redshift of Type 1a supernovae, the data (points) agree with the equation in which light propagates through the expanding universe on the least-time path (solid line). Image credit: Annila. ©2011 Royal Astronomical Society&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Read more at: &lt;a href=&quot;http://phys.org/news/2011-10-supernovae-universe-expansion-understood-dark.html#jCp&quot;&gt;http://phys.org/news/2011-10-supernovae-universe-expansion-understood-dark.html#jCp&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-03-15T13:21:41.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10138" PostTypeId="1" CreationDate="2015-03-15T14:49:27.970" Score="2" ViewCount="429" Body="&lt;p&gt;Wikipedia , on several asteroid articles, claims that right ascension and declination of the North Pole can give you the axial tilt of the object. What formula is used to convert these 2 values into the axial tilt? (Ceres, for example, has an R.A. Of 291 degrees and a declination of 59 degrees; its axial tilt is approximately 3 degrees. How is this figured?&lt;/p&gt;&#xA;" OwnerUserId="6769" LastActivityDate="2016-02-22T17:16:44.870" Title="North Pole Right Ascension/Declination to axial tilt conversion" Tags="&lt;planet&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10140" PostTypeId="1" AcceptedAnswerId="10142" CreationDate="2015-03-15T16:27:41.320" Score="3" ViewCount="2078" Body="&lt;p&gt;Why can't the distances between our Sun and the celestial objects be measured directly and instead had to rely on looking around for supernova? Is these events frequent, evenly distributed and lasting enough for astronomers to measure any place in our observable universe? Is this method of measurement more preferable than others currently?does this method had any limitations such as an absent of supernova nearby?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastActivityDate="2017-03-28T06:36:58.743" Title="Why do astronomers use supernova to measure distance in space?" Tags="&lt;distances&gt;&lt;supernova&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10142" PostTypeId="2" ParentId="10140" CreationDate="2015-03-15T18:08:48.883" Score="7" Body="&lt;p&gt;Type Ia supernovae are not common; they are quite rare events, happening maybe once per 100-200 years in a big galaxy. Nevertheless they have two properties that make them fantastically useful for distance measurement.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;They are (very close to) &quot;standard candles&quot;. The physics of the supernova detonation, thought to be when a white dwarf accretes matter and exceeds the Chandrasekhar limit, is very &quot;standardised&quot;. The bomb goes off in exactly the same way with the same amount of identical explosive. That means to a good approximation, measuring the &lt;em&gt;apparent&lt;/em&gt; brightness of a type Ia supernova and comparing it with nearby examples means that the distances to these events can be accurately estimated.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;They are really luminous and last 2-3 weeks. This means that they can be seen at enormous distances, they almost outshine the galaxies that they are in, and they last long enough for astronomers to discover them in automated surveys and still have time to follow them up and measure their light curves and estimate their peak brightness.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The limitations are that you can't choose which galaxies you measure the distance to. You have to wait for a supernova to go off and that might take 100-200 years or more for a particular galaxy. There are also continuing debates about just how standard a candle these objects are. It is possible that more distant galaxies that we see in the very early universe made stars with a different composition (far less elements heavier than helium) and that this alters things a bit. Other problems are associated with &quot;de-reddening&quot; the supernova light curves to account for the possibility of obscuring dust in the host galaxy. It is also possible that there is more than one way to produce a type Ia supernova (accretion of additional mass or merger of two white dwarfs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am puzzled by your question &quot;Why can't the distances between our Sun and the celestial objects be measured directly?&quot;. Galaxies are much too far away to have a trigonometric parallax distance. Methods like using Cepheids or RR Lyrae variables don't work for very distant galaxies because we can't resolve the individual stars. You could estimate a distance using a measurement of redshift and a cosmological model for the expansion of the Universe, but really the point of using these supernovae was and is to &lt;em&gt;test&lt;/em&gt; and &lt;em&gt;improve&lt;/em&gt; the cosmological models.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-03-28T06:36:58.743" LastActivityDate="2017-03-28T06:36:58.743" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10144" PostTypeId="2" ParentId="10134" CreationDate="2015-03-15T18:24:04.033" Score="1" Body="&lt;p&gt;Good question but we can only see out a certain distance because of it's expansion. AFAIK acceleration of the universe will eventually reduce the distance we can see over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No one knows if the universe is infinitely large, or even if ours is the only universe that exists. And other parts of the universe, very far away, might be quite different from the universe closer to home.  The site &lt;a href=&quot;http://skyserver.sdss.org/dr1/en/astro/universe/universe.asp&quot; rel=&quot;nofollow&quot;&gt;http://skyserver.sdss.org/dr1/en/astro/universe/universe.asp&lt;/a&gt; give a good basic picture.&lt;/p&gt;&#xA;" OwnerUserId="6766" LastActivityDate="2015-03-15T18:24:04.033" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10145" PostTypeId="1" CreationDate="2015-03-15T19:06:21.803" Score="1" ViewCount="292" Body="&lt;p&gt;Simple question: I read that the axial tilt of Uranus is 97.77 degrees. I don't understand why it would be this instead of 82.23 degrees the other way. It seems like saying that the Earth's axial tilt is 157 degrees rather than 23, or something. can someone explain why Uranus's axial tilt is bigger than 90 degrees?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;this is not a question of how it came to be so tilted, I am sure that is answered elsewhere, but simply why it is accepted as being bigger than 90 degrees rather than less than 90 degrees the other way. it is the same for Venus.&lt;/p&gt;&#xA;" OwnerUserId="2337" LastEditorUserId="2337" LastEditDate="2015-03-17T07:07:53.967" LastActivityDate="2015-03-17T07:07:53.967" Title="axial tilt of uranus" Tags="&lt;uranus&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ClosedDate="2015-03-24T17:48:59.253" ContentLicense="CC BY-SA 3.0" />
  <row Id="10146" PostTypeId="2" ParentId="10145" CreationDate="2015-03-15T19:35:41.717" Score="4" Body="&lt;p&gt;I have found the answer. I just found that venus has an axial tilt of nearly 180 degrees, so I realised that the &gt;90 degree tilts are due to the fact that they are the planets with retrograde rotation, which is equivalent to an 'upside down' (&gt;90 degrees) counter clockwise, normal, rotation.&lt;/p&gt;&#xA;" OwnerUserId="2337" LastActivityDate="2015-03-15T19:35:41.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10147" PostTypeId="2" ParentId="575" CreationDate="2015-03-15T19:43:31.253" Score="4" Body="&lt;p&gt;An affordable option is to build a &lt;a href=&quot;https://en.wikipedia.org/wiki/Barn_door_tracker&quot; rel=&quot;nofollow&quot;&gt;&quot;Barn Door Tracker&quot;&lt;/a&gt;, essentially two hinged plates connected by a threaded rod, with the hinge axis pointing at the north celestial pole, if you change the separation between the plates at the appropriate rate star trails will be cancelled out.&#xA;&lt;a href=&quot;http://(http://stellafane.org/misc/links.html#OtherComponents&quot; rel=&quot;nofollow&quot;&gt;The Springfield Telescope Makers&lt;/a&gt; have a long list of links to online instructions on their website . &lt;a href=&quot;http://www.garyseronik.com/?q=node/52&quot; rel=&quot;nofollow&quot;&gt;Gary Seronick's&lt;/a&gt; is a particularly complete motorized version while &lt;a href=&quot;http://barn-door-tracker.co.uk/&quot; rel=&quot;nofollow&quot;&gt;Noctilove&lt;/a&gt; is an example of a simple motorless design (with a moderate focal length lens, a clock and a steady hand, you can can manually drive the Barn Door Tracker).&lt;/p&gt;&#xA;" OwnerUserId="457" LastActivityDate="2015-03-15T19:43:31.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10148" PostTypeId="1" AcceptedAnswerId="10149" CreationDate="2015-03-15T19:48:41.440" Score="2" ViewCount="100" Body="&lt;p&gt;When the Milky Way galaxy collides with the Andromeda galaxy, what will the mass and volume of the new galaxy be?&lt;/p&gt;&#xA;" OwnerUserId="6750" LastEditorUserId="2365" LastEditDate="2015-11-04T20:24:03.410" LastActivityDate="2015-11-04T20:24:03.410" Title="What will the mass of the new galaxy be?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;galactic-dynamics&gt;&lt;m31&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10149" PostTypeId="2" ParentId="10148" CreationDate="2015-03-15T21:07:56.420" Score="0" Body="&lt;p&gt;The mass will be slightly less than the combine masses of the 2 galaxies since some of the stars will be hurled away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the disks of the galaxies are at an angle to each other, the volume would be (roughly) the volume of the 2 galaxies as they collide. Eventually, the volume will decrease (I'm guessing to between 60 to 70%) as the stars adapt to their new environment and the central massive black holes combine.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2015-03-15T21:07:56.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10153" PostTypeId="2" ParentId="10132" CreationDate="2015-03-16T12:03:37.887" Score="3" Body="&lt;p&gt;No, not really.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Stan Liou hints at in his comment, massive objects such as black holes (BHs) were previously a candidate to dark matter (DM). But it can be shown that in order to explain the various problems that the existence of DM needs to solve, there would have to be so many, that their gravitational effect (&quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_microlensing&quot; rel=&quot;nofollow&quot;&gt;microlensing&lt;/a&gt;&quot;) on background sources would be much larger than what is observed. Today, most people think that DM is simply a particles with an extremely small interaction cross section, such that it virtually doesn't interact with other particles, except through gravitational attraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also ask if there's any &lt;em&gt;spatial&lt;/em&gt; relationship, and in a way you could say there is. Most, if not all, galaxies sport a supermassive BH, and these tend to lie in the center of the gravitational potential wells created by the galaxies' DM. &quot;Normal&quot; BHs, however, just float around like stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BHs also tend to grow by accretion of matter onto them, and of course they can also eat DM, so to some extend BHs also consist of DM. But since the DM doesn't interact, a DM particle would need to fly directly toward the BH, lest it would simply be deflected and escape. Normal matter, on the other hand, is slowed down by friction and spirals toward the BH. Thus, BHs consist mostly of normal matter. It doesn't really matter, though, since after something has fallen into a BH, everything but its mass, charge, and angular momentum will be forgotten.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark &lt;em&gt;energy&lt;/em&gt; is something completely different, better described as a negative pressure which is a property of space itself. I think. Or we think. Nobody really knows. But it presence doesn't correlate with that of BHs or DM. Ten years ago, there was a hypothesis that a collapsing star should be converted into dark energy rather than a BH, causing a so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Dark-energy_star&quot; rel=&quot;nofollow&quot;&gt;dark-energy star&lt;/a&gt;, but I don't think anybody believes this anymore.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2015-03-16T12:03:37.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10158" PostTypeId="1" CreationDate="2015-03-16T16:31:21.650" Score="1" ViewCount="116" Body="&lt;p&gt;All I know is that the spectral index is related to the CMB, and that it has to be slightly less than 1 to favor inflation.&lt;/p&gt;&#xA;" OwnerUserId="6778" LastEditorUserId="2531" LastEditDate="2015-03-16T23:07:28.430" LastActivityDate="2019-05-13T15:04:51.960" Title="What do cosmologists mean when they talk about &quot;the running of the spectral index&quot;?" Tags="&lt;cosmology&gt;&lt;cosmological-inflation&gt;&lt;cmb&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10159" PostTypeId="1" AcceptedAnswerId="10160" CreationDate="2015-03-16T18:26:40.593" Score="3" ViewCount="1586" Body="&lt;p&gt;Earth is traveling at a certain velocity.  Earth orbits the Sun at velocity X.  The sun is orbiting the center of our galaxy at velocity Y.  The galaxy is orbiting (something?) at velocity Z. On and on... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In relation to zero, Earth has a velocity of X + Y + Z + (?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it theoretically possible that there would be a point in space where nothing is moving and velocity is essentially 0?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, what would time look like at this point in space?&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2018-11-20T12:01:40.527" Title="In theory, is there anywhere in the universe where velocity=0?" Tags="&lt;special-relativity&gt;&lt;velocity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10160" PostTypeId="2" ParentId="10159" CreationDate="2015-03-16T20:02:18.750" Score="9" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The velocities you list (X, Y, …) are all velocities with respect to some reference frame. But all reference frames are arbitrary, and you can always define a reference frame where the velocity of some object is exactly zero, as long as it's not accelerating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, Earth's velocity in the Sun's reference frame is X, but in its own reference frame, it's zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not just semantics; it is probably one of the most profound concepts in physics that there is no such thing as an &quot;absolute frame of reference&quot;. Before Einstein, people thought that time and space are absolute, but Einstein taught us that everything is relative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In cosmology, there &lt;em&gt;is&lt;/em&gt; a special reference frame in which all matter &lt;em&gt;on average&lt;/em&gt; is zero. The coordinates of this frame are called &lt;em&gt;comoving coordinates&lt;/em&gt;, and in many cases it makes sense to regard this as the most natural reference frame. But it is important to remember that it is still just a choice, and that any other reference frame is equally legit, although a bad choice of reference may make certain calculations harder or impossible to solve. In comoving coordinates, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow&quot;&gt;cosmic microwave background&lt;/a&gt; (CMB) is isotropic, i.e. it looks the same no matter which direction you look. But since Earth is traveling with a non-zero velocity in comoving coordinates (at an average of &lt;a href=&quot;http://arxiv.org/abs/1303.5087&quot; rel=&quot;nofollow&quot;&gt;369 km s$^{-1}$&lt;/a&gt;), the CMB is slightly blueshifted in the direction in which we're traveling, and slightly redshifted in the other direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would time look like? &lt;a href=&quot;http://en.wikipedia.org/wiki/Time_dilation&quot; rel=&quot;nofollow&quot;&gt;Time runs slower&lt;/a&gt;, the faster you move. In any non-accelerating reference frame (an &lt;em&gt;inertial system&lt;/em&gt;), time runs &quot;as it should&quot;, but an observer in this frame who measures how time runs in a frame that has a non-zero velocity with respect to his/her frame, will see it run slower.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-03-19T14:12:19.837" LastActivityDate="2015-03-19T14:12:19.837" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="10161" PostTypeId="2" ParentId="10158" CreationDate="2015-03-16T20:51:53.797" Score="1" Body="&lt;p&gt;The spectral index &lt;span class=&quot;math-container&quot;&gt;$n_s$&lt;/span&gt; describes how the clumpiness of stuff varied on various scales just after cosmic inflation. The primordial power spectrum &lt;span class=&quot;math-container&quot;&gt;$P(k)$&lt;/span&gt; of the fluctuations (where the wave number is &lt;span class=&quot;math-container&quot;&gt;$k=2\pi/\lambda$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; being the physical scale), is predicted by many inflationary models to be:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$P(k) \propto k^{n_s-1}.$$&lt;/span&gt;&#xA;If &lt;span class=&quot;math-container&quot;&gt;$n_s=1$&lt;/span&gt;, the fluctuations are &lt;em&gt;scale invariant&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$n_s$&lt;/span&gt; is not a constant, but instead changes with &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt;, that is if&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{dn_s}{d \ln k} \neq 0,$$&lt;/span&gt;&#xA;it is called a &quot;running spectral index&quot;. And in fact it seems that that &lt;span class=&quot;math-container&quot;&gt;$n_s$&lt;/span&gt; &lt;em&gt;does&lt;/em&gt; chance with &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; (see e.g. &lt;a href=&quot;http://arxiv.org/abs/1403.4589&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cherny et al. 2014&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term &quot;&lt;em&gt;the running of the spectral index&lt;/em&gt;&quot; refers to the quantity &lt;span class=&quot;math-container&quot;&gt;$dn_s\,/\,d\ln k$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2019-05-13T15:04:51.960" LastActivityDate="2019-05-13T15:04:51.960" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="10163" PostTypeId="1" CreationDate="2015-03-17T00:52:25.390" Score="1" ViewCount="85" Body="&lt;p&gt;So, I am getting interested in gravitational collapse of large molecular clouds that fragment and collapse into multiple stars. Are there well-known popular academic simulation packages that simulate these, for research purposes? Or are there papers or references I can use to write my own?&lt;/p&gt;&#xA;" OwnerUserId="422" LastActivityDate="2015-03-17T21:07:03.543" Title="Simulation packages or theory to work with gravitational collapse of massive molecular clouds?" Tags="&lt;star-formation&gt;&lt;n-body-simulations&gt;&lt;newtonian-gravity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10164" PostTypeId="1" AcceptedAnswerId="10173" CreationDate="2015-03-17T01:17:03.110" Score="1" ViewCount="246" Body="&lt;p&gt;As the title already says, I want to know how one measures velocities of far-off, bright objects, e.g. when the mean parallax drift isn't measurable with current apparatus (this means when there is only a &quot;non-moving&quot; picture of that object available)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that one can measure the redshift of spectral lines and correct them for gravitational redshifting if the distance is already known.&#xA;But then the relation between source-frequency $f_S$ and observed frequency $f_O$ is&#xA;$$ f_O = f_S \frac{\sqrt{1-\frac{v^2}{c^2}}}{1+\frac{v}{c}\cos(\alpha)}$$,&#xA;where $\alpha$ is the angle between line of sight and the velocity vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one only has a &quot;non-moving&quot; picture of the object, there is no way to determine this angle $\alpha$ and therefore no way to get the velocity by means of redshift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an addition: At which distance does the mean parallax drift become unmeasurble with current apparatus?&lt;/p&gt;&#xA;" OwnerUserId="6775" LastActivityDate="2015-03-18T12:08:32.280" Title="How does one measure velocities of far-off, bright objects" Tags="&lt;redshift&gt;&lt;velocity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10166" PostTypeId="1" CreationDate="2015-03-17T04:48:08.700" Score="5" ViewCount="232" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_most_massive_known_stars&quot; rel=&quot;noreferrer&quot;&gt;The most massive &lt;em&gt;known&lt;/em&gt; stars are Wolf-Rayet stars&lt;/a&gt;. However, as Wolf-Rayet stars do not appear to be the first stage of any star's lifecycle, I infer that whatever these Wolf-Rayet stars &lt;em&gt;used to be&lt;/em&gt; must have been even more massive. My guess is that the most massive stars are Wolf-Rayet because of some combination of 1) their luminosity and/or 2) the potential short life of prior stages of the star's evolution. Assuming this is correct, to what group would we expect the most massive stars to belong (even if we don't know of specific examples)?&lt;/p&gt;&#xA;" OwnerUserId="6587" LastActivityDate="2015-03-17T20:18:27.140" Title="What type of star does theory predict should be the most massive?" Tags="&lt;star&gt;&lt;mass&gt;&lt;stellar-evolution&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10167" PostTypeId="2" ParentId="10166" CreationDate="2015-03-17T07:43:46.693" Score="4" Body="&lt;p&gt;There is no particular type. Yes Wolf Rayets were once more massive and have suffered extensive mass loss. Other examples of evolved massive stars are very luminous red supergiants and luminous blue variables (LBVs). The evolutionary tracks through these processes are somewhat poorly understood, as are the exact magnitudes and durations of the various mass loss phases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, they all started out as massive main sequence stars with very short hydrogen burning lifetimes. There are a number of examples of O3V main sequence stars in the Galaxy which probably have masses of around 100 Suns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the most massive stage of a massive star's life is (briefly) as a main sequence star of type early-O.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-03-17T20:18:27.140" LastActivityDate="2015-03-17T20:18:27.140" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10168" PostTypeId="1" CreationDate="2015-03-17T08:54:24.703" Score="8" ViewCount="1449" Body="&lt;p&gt;What makes the Moon shine white in the sky, although it seems to be as dark as &lt;a href=&quot;http://web.archive.org/web/20070829153207/http://eetd.lbl.gov/HeatIsland/Pavements/Albedo/&quot; rel=&quot;noreferrer&quot;&gt;asphalt&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth's average albedo is 3 and 4 times that of the Moon and Ceres respectively (0.37 vs 0.12 and 0.09). But except for the clouds, Earth doesn't shine bright white in images of Earth taken from cis-lunar space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an image of Earth seen from the Moon. Although it has 3 times as high albedo, it does not seem to be as bright as the white Moon is as seen on Earth's blue sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SUvui.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-03-17T12:26:43.417" LastActivityDate="2015-05-26T04:48:46.600" Title="Why do dark objects look white from a distance? (Moon, Ceres, but not Earth!)" Tags="&lt;the-moon&gt;&lt;apparent-magnitude&gt;&lt;albedo&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10169" PostTypeId="2" ParentId="10166" CreationDate="2015-03-17T08:58:08.673" Score="3" Body="&lt;p&gt;Since you phrase your question &quot;What type does &lt;em&gt;theory&lt;/em&gt; predict?&quot;, I guess the answer must be the so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Metallicity#Population_III_stars&quot; rel=&quot;nofollow&quot;&gt;Population III stars&lt;/a&gt;, which are thought to be the first generation of stars, born from zero-metallicity gas. With no metals, it is difficult for the gas to cool. The mass of a star is given by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jeans_instability&quot; rel=&quot;nofollow&quot;&gt;Jeans mass&lt;/a&gt; of the collapsing cloud, which is proportional to $T^{3/2}/\rho^{1/2}$. This temperature dependence means that the cooler the cloud can be, the smaller clumps the cloud is able to fragment into. Hence, if it's difficult to cool, the cloud has a large Jeans mass, i.e. only large cloud collapse and form stars (this explanation is rather simplified and doesn't take into account stuff like shocks, turbulence, etc., but captures the basic physics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, without metals to act as absorbers, the radiation may escape the star more easily, i.e. without interacting with the stellar atmosphere, so mass losses may be less significant and the star is expected to maintain its mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pop III stars are expected to have masses of several hundred to 1000 $M_\odot$. With such large masses, they burn the fuel fast ($\sim$$10^6$ yr), but their spectra are extremely hard, and they should be able to ionize helium, so one way to get a hint of the existence of these guys would be by detection of the He $\lambda$1640 Å line. &lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2015-03-17T08:58:08.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10170" PostTypeId="2" ParentId="10164" CreationDate="2015-03-17T09:42:17.643" Score="2" Body="&lt;p&gt;In general the total velocity of a faraway object is not obtainable, only the component along the line of sight. However, if a star is not too far away, and its position on the sky is measured very accurately with an interval of some years, the velocity component perpendicular to the line of sight can be determined, yielding together with the redshift the total velocity. The recently launched Gaia spacecraft is expected to be measure the proper motion of millions of stars in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In distant galaxies, this is not possible. If one is interested in, say, the 3D velocity dispersion of the gas, the 1D dispersion is measured, and assuming (fairly) an isotropic dispersion, this number is then multiplied by $\sqrt{3}$ to get the total velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance to which parallax is measurable depends on the angular resolution of your instrument. For milliarcsec resolution, parallax measurements are good out to roughly 1 kpc. If you wait long enough, the motion around the galactic center increases your baseline such that larger distance can be probed.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2015-03-17T09:42:17.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10171" PostTypeId="1" AcceptedAnswerId="10172" CreationDate="2015-03-17T15:11:57.497" Score="1" ViewCount="376" Body="&lt;p&gt;Perhaps my title is misleading. but here is my understanding and I'm looking for confirmation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot have dark without light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing faster than light is darkness (i.e. shadows)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars are in the galaxy and are inherently bright.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this make &quot;dark matter&quot; mere shadows of extremely bright objects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that also why it has to get dark before we can really see the majority of stars (subjective experience)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With all due respect to user with answer below and thank you for taking the time to answer... I think this can be expanded on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have provided a source for my claims: youtu.be/JTvcpdfGUtQ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out the video, and it more explains my understanding of things...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why I don't accept answer below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you provide sources for your claims?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I didn't say you can't have DARK MATTER without Light , I said you can't have Dark, as in Darkness. I think you're getting hung up on Dark Matter which is a big part of my question, but not the entire question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, are you claiming that all darkness in space is dark matter? I don't like to assume but for scientific purposes (not an accredited scientist) would we not agree that the measures made on dark matter were on a sample, and not the entire universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to further question this, do shadows not exist in space?&lt;/p&gt;&#xA;" OwnerUserId="6782" LastEditorUserId="6782" LastEditDate="2015-03-18T14:03:52.033" LastActivityDate="2015-03-19T02:36:45.573" Title="Shadows of Light = Space or Dark Matter?" Tags="&lt;light&gt;&lt;dark-matter&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10172" PostTypeId="2" ParentId="10171" CreationDate="2015-03-17T15:45:03.447" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You cannot have dark without light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not true in the case of dark matter (for the general case, see below). Dark matter is called &quot;dark&quot; because it appears that it doesn't absorb or emit electromagnetic radiation - light! It can interact with light via gravitational lensing, but dark matter particles have no electromagnetic charge (we think) and so are considered &quot;dark&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The only thing faster than light is dark (i.e. shadows)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is true, as the video explains.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does this make &quot;dark matter&quot; mere shadows of extremely bright objects?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the meat of the question, and the answer is a clear &quot;no&quot;.  First, any &quot;bright objects&quot;, while capable of &lt;a href=&quot;https://physics.stackexchange.com/questions/117101/can-the-sun-cast-a-shadow&quot;&gt;casting a shadow of something &lt;em&gt;else&lt;/em&gt;&lt;/a&gt;, will emit electromagnetic radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, so why can't there just be non-extremely luminous objects casting a shadow? One reason is that the dark matter simply isn't where these shadows should be. Many galaxies have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Dark_matter_halo&quot; rel=&quot;nofollow noreferrer&quot;&gt;dark matter halo&lt;/a&gt; that extends far beyond the galactic plane. Any light sources emitting light (that is then shadowed) would have to be on the opposite side of the galactic disk. This is simply not the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second, more convincing, reason is that dark matter has mass. In fact, that's the reason the idea was first conceived (to account for anomalies in &lt;a href=&quot;http://en.wikipedia.org/wiki/Galaxy_rotation_curve&quot; rel=&quot;nofollow noreferrer&quot;&gt;galactic rotation curves&lt;/a&gt;)! Shadows don't have mass; therefore, dark matter cannot be explained by shadows.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is that also why it has to get dark before we can really see the majority of stars (subjective experience)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. I don't know for sure, but I believe that that's because the Sun's light is still blocking out the light of distant stars.&lt;sup&gt;[citation needed]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum after edit to question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can you provide sources for your claims?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think that they're pretty well sourced! The first section can easily be checked via &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_matter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; (which I assumed you had seen). The rest seems to be well-sourced, too (aside from the last bit, which is, admittedly, unsourced but only tangential).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also I didn't say you can't have DARK MATTER without Light , I said you can't have Dark, as in Darkness. I think you're getting hung up on Dark Matter which is a big part of my question, but not the entire question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Okay, I may have misinterpreted this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Darkness is, by definition, the absence of light (electromagnetic radiation). So no, you don't have to have light for there to be darkness.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, are you claiming that all darkness in space is dark matter? I don't like to assume but for scientific purposes (not an accredited scientist) would we not agree that the measures made on dark matter were on a sample, and not the entire universe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I never said that, but you did. Your original question asked if dark matter could be shadows. The answer is no.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So to further question this, do shadows not exist in space?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They exist. Any object blocking a light source casts a shadow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/2/2e/Diagram_of_umbra%2C_penumbra_%26_antumbra.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/2/2e/Diagram_of_umbra%2C_penumbra_%26_antumbra.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This happens to have been used in the video, so it shouldn't be new to you.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-03-19T02:36:45.573" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10173" PostTypeId="2" ParentId="10164" CreationDate="2015-03-17T20:47:16.097" Score="1" Body="&lt;p&gt;The &quot;redshift&quot; of a distant galaxy is defined in terms of its line of sight velocity. In our model of the expanding universe, once we move away from the local group of galaxies (which have their own peculiar motions), distant galaxies follow the Hubble flow and to first order have a line of sight velocity tht is proportional to their distance way (it gets more complicated for very distant galaxies).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distant galaxies may well have a &quot;tangential&quot; velocity too, but for galaxies outside the local group these velocities will be be &lt;em&gt;negligible&lt;/em&gt; compared with the redshift. i.e. The line of sight velocity due to the expansion of the universe is dominant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess by &quot;parallax drift&quot; you actually mean &lt;a href=&quot;http://en.wikipedia.org/wiki/Proper_motion&quot; rel=&quot;nofollow&quot;&gt;proper motion&lt;/a&gt; - which is the rate at which a star's position changes with respect to the celestial coordinate system. This proper motion depends on how far away the star is &lt;em&gt;and&lt;/em&gt; how fast it is moving tangentially with respect to the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus to estimate a tangential &lt;em&gt;velocity&lt;/em&gt; you need &lt;em&gt;both&lt;/em&gt; the proper motion &lt;em&gt;and&lt;/em&gt; the distance to the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;em&gt;think&lt;/em&gt; the most distant object for which a proper motion has been determined with any accuracy is the Andromeda galaxy, which is a couple of million light years away. This was achieved by studying the position of many stars in Andromeda over a 7 year period using the Hubble Space Telescope. The details can be found in &lt;a href=&quot;http://arxiv.org/abs/1205.6863&quot; rel=&quot;nofollow&quot;&gt;Sohn et al. (2012)&lt;/a&gt;; but the headline numbers are that the proper motion is a mere $\sim 0.05$ milli-arcseconds per year(!) , implying a tangential velocity (with respect to the solar system) of about 150 km/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another candidate is measuring the velocities of material in the jet of the active galaxy M87 by &lt;a href=&quot;http://arxiv.org/abs/1308.4633&quot; rel=&quot;nofollow&quot;&gt;Meyer et al. (2013)&lt;/a&gt;. This galaxy is at 50 million light years, but the motion of the jet is only detectable here because it is moving relativistically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are quite special cases. In general, the tangential velocities of stars in our Galaxy are small and large-scale susrveys of proper motions are generally inaccurate beyond a few thousand light years. The upcoming Gaia results will improve this dramatically meaning we have good proper motions for objects out to tens of thousands of light years.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-03-18T12:08:32.280" LastActivityDate="2015-03-18T12:08:32.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10174" PostTypeId="2" ParentId="10163" CreationDate="2015-03-17T21:07:03.543" Score="1" Body="&lt;p&gt;This sounds very ambitious! I think your way in to this topic might be to have a look at the works of two authors - &lt;a href=&quot;http://www.astro.ex.ac.uk/people/mbate/&quot; rel=&quot;nofollow&quot;&gt;Matthew Bate&lt;/a&gt; and &lt;a href=&quot;https://sites.google.com/a/ucsc.edu/krumholz/&quot; rel=&quot;nofollow&quot;&gt;Mark Krumholz&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I suggest these two names is that they are representative of the two basic approaches to simulating the collapse of molecular clouds (which has little to do with n-body simulations). These approaches are known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Smoothed-particle_hydrodynamics&quot; rel=&quot;nofollow&quot;&gt;Smoothed-Particle Hydrodynamics&lt;/a&gt; (SPH) and &lt;a href=&quot;http://en.wikipedia.org/wiki/Adaptive_mesh_refinement&quot; rel=&quot;nofollow&quot;&gt;Adaptive Mesh Refinement&lt;/a&gt; (AMR).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crudely put, SPH treats the fluid as discrete particles whose properties are smoothed over the size of a kernel - the resolution is controlled by the number of particles in the simulation and the smoothing length. AMR is a more &quot;traditional&quot; grid-based approach where the size of the grid is varied adaptively with time and space to get the requisite resolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick google reveals a number of public codes -e.g. for SPH there is &lt;a href=&quot;https://wiki.manchester.ac.uk/sphysics/index.php/SPHYSICS_Home_Page&quot; rel=&quot;nofollow&quot;&gt;SPHYSICS&lt;/a&gt; or &lt;a href=&quot;http://www.ase.uc.edu/~liugr/SPH.html&quot; rel=&quot;nofollow&quot;&gt;GRLAB&lt;/a&gt;; for AMR there is &lt;a href=&quot;http://iopscience.iop.org/0067-0049/211/2/19&quot; rel=&quot;nofollow&quot;&gt;ENZO&lt;/a&gt; or &lt;a href=&quot;http://www.physics.drexel.edu/~olson/paramesh-doc/Users_manual/amr.html&quot; rel=&quot;nofollow&quot;&gt;PARAMESH&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the main activities in model simulations these days is trying to incorporate &quot;feedback&quot; into star formation. i.e. the winds and radiation from newly born high-mass stars and the outflows from protostars. Such things do not form part of standard codes.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-03-17T21:07:03.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10175" PostTypeId="1" AcceptedAnswerId="10176" CreationDate="2015-03-17T21:09:45.760" Score="16" ViewCount="8139" Body="&lt;p&gt;How many arms does the Milky Way galaxy have and how is it determined? &lt;/p&gt;&#xA;" OwnerUserId="6750" LastActivityDate="2018-12-24T15:51:01.727" Title="How many arms does the Milky Way galaxy have?" Tags="&lt;milky-way&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10176" PostTypeId="2" ParentId="10175" CreationDate="2015-03-17T22:39:13.590" Score="22" Body="&lt;p&gt;This is actually a really, really tough question.&lt;/p&gt;&#xA;&lt;p&gt;Look at this diagram:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/1/1d/Milky_Way_Arms.svg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/1/1d/Milky_Way_Arms.svg&quot; alt=&quot;&quot; /&gt;&lt;/a&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;Purple: Norma Arm and Outer Arm.&lt;/sup&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;Green: Scutum-Centaurus Arm&lt;/sup&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;Pink: Carina-Sagittarius Arm&lt;/sup&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;Cyan: 3 kpc Arm and Perseus Arm&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;So we can slightly modify this picture by saying that there are four arms, and calling them by the following names:&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Norma_Arm&quot; rel=&quot;noreferrer&quot;&gt;Norma-Outer Arm&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;This arm has one end at the center of the Milky Way; this end is called the Norma Arm. It's actually quite small. However, as you continue outward along the lanes of gas, dust and stars, the Norma Arm becomes the - wait for it - &lt;em&gt;Outer&lt;/em&gt; Arm.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Scutum%E2%80%93Centaurus_Arm&quot; rel=&quot;noreferrer&quot;&gt;Scutum-Centaurus/Crux-Scutum Arm&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;This arm also emanates from the center (well, duh!) and is distinguished by &lt;a href=&quot;http://arxiv.org/pdf/1002.1823v1.pdf&quot; rel=&quot;noreferrer&quot;&gt;large numbers of clusters of red supergiants&lt;/a&gt; near the center (where it is referred to as the Scutum Arm) of the Milky Way. The Scutum-Sagittarius Arm is one of the Milky Way's two major arms.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Carina%E2%80%93Sagittarius_Arm&quot; rel=&quot;noreferrer&quot;&gt;Carina-Sagittarius Arm&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The defining feature of this arm is that is has many &lt;a href=&quot;http://en.wikipedia.org/wiki/H_II_region&quot; rel=&quot;noreferrer&quot;&gt;H II regions&lt;/a&gt;, where ionized gas is plentiful. It is though that stars can form there. However, H II regions are also present in many other parts of the galaxy. &lt;a href=&quot;http://www.astro.wisc.edu/sirtf/Churchwell_2009.pdf&quot; rel=&quot;noreferrer&quot;&gt;Observations from the Spitzer Space Telescope&lt;/a&gt; seem to support these theories and prior observations.&lt;/p&gt;&#xA;&lt;p&gt;The Spitzer results are, I think, very important for understanding our galaxy and the things in it. It confirmed the existence of all four arms, but also confirmed that the Norma-Outer Arm and Sagittarius Arm are relatively minor compared to the other two.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Near_3_kpc_Arm&quot; rel=&quot;noreferrer&quot;&gt;Near&lt;/a&gt;/&lt;a href=&quot;http://en.wikipedia.org/wiki/Far_3_kpc_Arm&quot; rel=&quot;noreferrer&quot;&gt;Far 3 kpc Arm&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Perseus_Arm&quot; rel=&quot;noreferrer&quot;&gt;Perseus Arm&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The Near and Far 3 kpc Arms (which really are just one arm) are very close to the galactic center - about 3,000 parsecs (hence the name). The Far Arm was &lt;a href=&quot;http://arxiv.org/pdf/0807.1752v2.pdf&quot; rel=&quot;noreferrer&quot;&gt;only discovered recently&lt;/a&gt;, while the Near Arm was first observed about 50 years ago. These two mini-arms are expanding outward at an enormous rate. The combined 3 kpc Arm then becomes the Perseus Arm, which extends out to about 3.5 times the distance from the center of the 3 kpc Arm. It is the other major spiral arm of the Milky Way.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Okay, so that's simple enough. The Milky Way has four spiral arms, right? Well . . . sort of. &lt;a href=&quot;http://iopscience.iop.org/2041-8205/798/2/L27/pdf/2041-8205_798_2_L27.pdf&quot; rel=&quot;noreferrer&quot;&gt;The discovery of the &amp;quot;New Outer Arm&amp;quot;&lt;/a&gt;&lt;sup&gt;1&lt;/sup&gt; shook things up, because that would mean that the Norma-Outer Arm reaches nearly &lt;em&gt;all the way around the Milky Way.&lt;/em&gt; That's strange and borderline inexplicable. Well, not really. But it's very interesting.&lt;/p&gt;&#xA;&lt;p&gt;This image gives you a good idea of what it might look like:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NLSJF.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NLSJF.jpg&quot; alt=&quot;&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;By the way, someone at one point wondered, &amp;quot;Which are we in?&amp;quot; That, too, doesn't have a simple answer. Current observations state that we're in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Orion_Arm&quot; rel=&quot;noreferrer&quot;&gt;Orion-Cygnus Arm&lt;/a&gt;, a/the midget of the galaxy (depending on whether or not there are other &amp;quot;mini-arms&amp;quot;). It's tiny - 1.1 kpc wide and 10 kpc long - and is between the Carina-Sagittarius Arm and the Perseus Arm.&lt;/p&gt;&#xA;&lt;p&gt;Here's what it might look like:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/9/9b/OrionSpur.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/9/9b/OrionSpur.png&quot; alt=&quot;&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Putting it all together, Robert Hurt made a very famous impression of what the Milky Way might look like:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gFOwe.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gFOwe.jpg&quot; alt=&quot;&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;NASA gives a good explanation of it &lt;a href=&quot;http://apod.nasa.gov/apod/ap080711.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If you want a good overview on &lt;em&gt;everything&lt;/em&gt;, check out &lt;a href=&quot;http://www.galaxymap.org/drupal/node/171&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; blog post, which is reasonably well-sourced. &lt;a href=&quot;http://galaxymap.org/drupal/node/134&quot; rel=&quot;noreferrer&quot;&gt;Another post&lt;/a&gt; explains the four-arms-vs.-two-arms issue:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Like the proverbial blind men describing an elephant, the two groups of scientists are examining very different parts of the Milky Way. The Spitzer study detected hot objects visible in infrared. The CfA study used radio telescopes which can also detect colder objects such as supernova remnants, very young star formation regions and huge clouds of hydrogen gas. So it seems as though the older established star formation regions are mostly concentrated in two spiral arms, but that the very new star formation regions and the hydrogen clouds from which they form are also developing in two additional arms.&lt;/p&gt;&#xA;&lt;p&gt;If you were hovering above the Milky Way in a spacecraft using binoculars or a regular optical telescope, you would see two main arms. But if you also had a radio telescope with you, it would detect two more.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The other group of scientists referred to here &lt;a href=&quot;http://www.cfa.harvard.edu/news/2009-03&quot; rel=&quot;noreferrer&quot;&gt;used the Very Long Baseline Array&lt;/a&gt; to image the Milky Way with radio waves; their observations support the four-main arms model.&lt;/p&gt;&#xA;&lt;p&gt;So at the moment, I think the answer is four, though two arms are more distinct that the other two.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;&lt;sub&gt;Dammit, paywall! Thank goodness for &lt;a href=&quot;http://arxiv.org/pdf/1412.2425v1.pdf&quot; rel=&quot;noreferrer&quot;&gt;arXiv&lt;/a&gt;.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-12-24T15:51:01.727" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10177" PostTypeId="1" AcceptedAnswerId="10194" CreationDate="2015-03-18T00:13:05.523" Score="2" ViewCount="312" Body="&lt;p&gt;I know there are several theories as to how the universe might end, if it does. Is it possible for it to end by time stopping? A friend of mine told me about this theory, but her explanation that it would be caused by heat death doesn't have me convinced.  &lt;/p&gt;&#xA;" OwnerUserId="6783" LastEditorUserId="9301" LastEditDate="2016-08-22T20:54:46.973" LastActivityDate="2016-08-23T06:50:07.270" Title="Will the Universe end by time stopping?" Tags="&lt;universe&gt;&lt;cosmology&gt;&lt;fate-of-universe&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10178" PostTypeId="1" AcceptedAnswerId="10181" CreationDate="2015-03-18T01:37:25.003" Score="3" ViewCount="150" Body="&lt;p&gt;This seems to be a chicken and egg problem, the sun begins shining due to hydrogen becoming helium, but it's odd that there was no helium initially without the stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my logic flawed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: First year astronomy student)&lt;/p&gt;&#xA;" OwnerUserId="6785" LastEditorUserId="2153" LastEditDate="2015-03-18T02:09:58.560" LastActivityDate="2015-03-18T03:41:54.270" Title="What came first: the Sun shining or the existence of helium?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;helium&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10179" PostTypeId="1" CreationDate="2015-03-18T02:05:00.687" Score="1" ViewCount="92" Body="&lt;p&gt;Mathematicians much more often use radii over diameters when discussing about circles and spheres, because in mathematics the radius is more fundamental than the diameter (the sphere is defined using its radius).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about in astronomy? Do astronomers and astrophysicist more often use diameters or radii when characterizing planets, dwarf planets, exoplanets and stars?&lt;/p&gt;&#xA;" OwnerUserId="6786" LastActivityDate="2015-03-18T07:25:08.217" Title="Do astronomers and astrophysicists more often use diameters or radii when discussing about planets, dwarf planets, exoplanets and stars?" Tags="&lt;star&gt;&lt;planet&gt;&lt;exoplanet&gt;&lt;distances&gt;&lt;size&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10180" PostTypeId="1" AcceptedAnswerId="10185" CreationDate="2015-03-18T02:11:58.713" Score="6" ViewCount="141" Body="&lt;p&gt;The sun is a big ball of hot plasma which contains free electrons. However, how does the absence of metallic elements in the sun generate the magnetic field to orchestrate these free electrons to display such beautiful solar phenomenon?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="2153" LastEditDate="2015-03-18T02:28:27.193" LastActivityDate="2015-03-19T09:22:40.640" Title="What triggers solar flares?" Tags="&lt;the-sun&gt;&lt;magnetic-field&gt;&lt;solar-flare&gt;&lt;plasma-physics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10181" PostTypeId="2" ParentId="10178" CreationDate="2015-03-18T03:41:54.270" Score="4" Body="&lt;p&gt;The Big Bang theory predicts that (depending what assumptions you choose) the &lt;a href=&quot;http://en.wikipedia.org/wiki/Big_Bang_nucleosynthesis&quot; rel=&quot;nofollow&quot;&gt;initial elements&lt;/a&gt; were formed from 10 seconds to 20 minutes after the Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial elements were &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;mostly hydrogen&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;some deuterium &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;some helium-3 and helium-4&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a little lithium-7&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a couple of unstable isotopes that decayed to lithium-7 or helium-3.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As the linked Wikipedia article says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Essentially all of the elements that are heavier than lithium and beryllium were created much later, by &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_nucleosynthesis&quot; rel=&quot;nofollow&quot;&gt;stellar nucleosynthesis&lt;/a&gt; in evolving and exploding stars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the answer to your question is: the Helium came first.&lt;/p&gt;&#xA;" OwnerUserId="5524" LastActivityDate="2015-03-18T03:41:54.270" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10182" PostTypeId="1" AcceptedAnswerId="10184" CreationDate="2015-03-18T06:18:35.943" Score="4" ViewCount="2303" Body="&lt;p&gt;What techniques and tools are available to the astronomers to measure the size of any celestial objects such as, stars or perhaps black holes that doesn't emit light nor reflects starlight?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastActivityDate="2015-03-18T12:21:12.667" Title="How do astronomer measures the size of any celestial objects?" Tags="&lt;distances&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10183" PostTypeId="2" ParentId="10179" CreationDate="2015-03-18T07:25:08.217" Score="2" Body="&lt;p&gt;Definitely &lt;strong&gt;radii&lt;/strong&gt;, with one notable exception...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When observers talk about how large an object is on the sky, they usually discuss &lt;em&gt;angular size&lt;/em&gt;, which is related to the &lt;em&gt;diameter&lt;/em&gt; of an object, not the radius.  So when discussing the angular size of, say, Alpha Centauri A (a few milliarcseconds), this is related to the diameter of the star, not the radius.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2015-03-18T07:25:08.217" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10184" PostTypeId="2" ParentId="10182" CreationDate="2015-03-18T08:52:39.923" Score="2" Body="&lt;p&gt;If a celestial body is larger than the resolving power of a telescope, its size can be measured directly. This is the case for most galaxies, molecular clouds in the Milky Way and nearby galaxies, and even for a few nearby stars. EDIT: See discussion by Rob Jeffries on how these measurements are carried out for stars using interferometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more distant stars, we can rely on our understanding of stellar evolution, which tells us pretty accurately the radius, once we know its spectrum (EDIT: or just a assume blackbody radiation and use the formula given by Rob). If the star is a member of a binary system whose orbit we observe roughy edge-on, we can measure how the luminosity declines as one star occults the other, and calculate the radius. This can also be used to measure the sizes of exoplanet. And for stars, the same technique is even possible using our own Moon as occulter. See a description &lt;a href=&quot;http://spiff.rit.edu/classes/phys440/lectures/size/size.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes (BH) that don't emit light, cannot be measured (at least not until we are able to detect gravitational waves), but often BHs are surrounded by a disk of accreting gas, which is heated to million of degrees by friction as it spirals down the drain. Measuring the temperature of this gas tells us the mass of the BH, which is directly proportional to its radius ($R_{\mathrm{BH}} \simeq 3 M/M_\odot$ km).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nice technique for measuring the size of the quite small region of gas clouds around a supermassive BHs, even though they are billions of lightyears away, is called &lt;em&gt;reverberation mapping&lt;/em&gt;. Here, some of the light emitted from the BH's accretion disk travels directly in our direction, while some of if travels in other directions, illuminating the clouds around it. When we measure the light from those clouds, the signal looks like the directly observed signal, but with a delay $t$ corresponding to extra length of the path that the light has taken. Since we know the speed of light $c$, we can calculate the extra distance as $d = ct$, i.e. the size of the system.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/yitpa.png&quot; alt=&quot;Reverberation mapping&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-03-18T11:25:07.547" LastActivityDate="2015-03-18T11:25:07.547" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10185" PostTypeId="2" ParentId="10180" CreationDate="2015-03-18T10:11:32.000" Score="3" Body="&lt;p&gt;Magnetic fields are generated by currents - i.e. by the motion of charged particles. As you say, the Sun is full of freely moving charged particles, and these generate currents which in turn generate magnetic fields. No metals required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the magnetic field generation is thought to occur at the interface between the radiative interior of the Sun and an outer convective envelope. This region, called the tachocline, is subject to large shearing motions which are able to take small magnetic fields and amplify them. The stronger magnetic fields are then buoyant, because magnetic fields in a plasma exert a pressure. They therefore emerge at the solar surface in the form of loops of magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These loops are strongly coupled to the plasma in the Sun. As the plasma moves around turbulently, driven by convective motions and differential rotation on the Sun's surface, the footpoints of the loops are sheared and twisted. There comes a point where the loop will snap back into a configuration with lower magnetic potential energy through a process called reconnection. The reconnection process has a side effect of accelerating particles within the magnetic loops. These relativistic particles smash into the photosphere (actually, they are mainly stopped in the chromosphere which sits above the photosphere), where they release their energy, heating plasma to millions of degrees, which evaporates into the corona. This is a solar flare. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-03-18T10:11:32.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10186" PostTypeId="2" ParentId="10182" CreationDate="2015-03-18T10:28:52.323" Score="3" Body="&lt;p&gt;The main tool to measure the diameter of a star is interferometry combined with a parallax-based distance measurement - a brief review by &lt;a href=&quot;http://journals.cambridge.org/download.php?file=%2FIAU%2FIAU4_S252%2FS1743921308023326a.pdf&amp;amp;code=8d3977f8167140871e3a1a9fe14560df&quot; rel=&quot;nofollow&quot;&gt;Kervella (2008)&lt;/a&gt; might be useful. The principles behind interferometry are &lt;a href=&quot;http://en.wikipedia.org/wiki/Astronomical_interferometer&quot; rel=&quot;nofollow&quot;&gt;described here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interferometry involves measuring the light from a star using two (or more) telescopes that are separated by some distance. Together, the signals from these telescopes can be combined to give an angular resolution that can be (in the best circumstances) equivalent to a telescope with a diameter equal to the telescope separation. These measurements give the &lt;em&gt;angular&lt;/em&gt; size of the star, which must then be multiplied by their distances to get a physical diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the most successful experiments is the &lt;a href=&quot;http://en.wikipedia.org/wiki/CHARA_array&quot; rel=&quot;nofollow&quot;&gt;Chara array&lt;/a&gt;, which has yielded diameters for many nearby stars. Precisions can be as good as a few percent, but more usually 10% and of order 100 (predominantly nearby) stars have had their radii measured in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second main direct technique is to use eclipsing binary systems. The measured light curve can be used in an almost model-independent way to estimate the radii of the two stars involved. Of course most eclipsing binaries are close pairs with short orbital periods and with orbital inclinations that allow us to see the eclipse. They are therefore highly prized objects. Radii can be measured with precisions of 1%. A reasonably complete catalogue of the $\sim 100$ known eclipsing binaries with precise radii can be found &lt;a href=&quot;http://www.astro.keele.ac.uk/jkt/debcat/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another technique is lunar occultation. The passage of a star behind the limb of the moon results in a changing diffraction pattern that can be used to estimate the angular size of the star. Again a distance is required to convert this into an actual diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More distant stars are inaccessible - their angular diameters are simply too small. At the moment only indirect estimates of their radii are possible. For example, if we were to assume that a star radiates as a blackbody, then its luminosity ($L$), radius ($R$) and temperature ($T$) are related by Stefan's law.&#xA;$$ L = 4\pi R^2 \sigma T^4,$$&#xA;where $\sigma$ is the Stefan constant. If the star has a measured flux at the Earth and we know how far away it is, then $L$ can be estimated. If we take a spectrum and estimate its temperature, then the equation above can be rearranged to give the radius in terms of the measured luminosity and temperature. Real stars are more complicated than blackbodies, but the principle is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither of the above techniques can work for black holes, and the sizes (event horizon or Schwarzschild radius) of black holes have not yet been directly measured. The physics of a black hole is relatively simple(!) and so there is a direct relationship between their Schwarzschild radii and their masses (modified somewhat by rotation). Basically it is 3 km multiplied by the mass in solar units. Therefore a measurement of the black hole mass gives its &quot;radius&quot;. The masses of black holes are measured by looking at the motions of stars and gas around them and applying our knowledge of how gravity works.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-03-18T12:21:12.667" LastActivityDate="2015-03-18T12:21:12.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10187" PostTypeId="1" AcceptedAnswerId="10193" CreationDate="2015-03-18T13:34:39.707" Score="5" ViewCount="548" Body="&lt;p&gt;David Christian's &lt;a href=&quot;http://www.ucpress.edu/book.php?isbn=9780520271449&quot; rel=&quot;noreferrer&quot;&gt;Maps of Time&lt;/a&gt; has this to say about the period during which the first stars started to form from the diffuse clouds of hydrogen and helium that then made up the universe:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Under the pull of their own gravity, the clouds of hydrogen and helium began to collapse in on themselves...As gravity packed each cloud into ever smaller spaces, pressure built up in the center. Increasing pressure means increasing temperatures, and so, as they shrank, each gas cloud began to heat up (~Kindle location 1469).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The part I'm working on understanding is &quot;Increasing pressure means increasing temperatures.&quot; Does the pressure itself cause the molecules to heat up? Or is it the kinetic energy that the particles pick up as gravity pulls them towards the center of the cloud? Or is the phenomenon here something other than increased energy per unit mass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, once the nuclear chain reaction gets started in the star it's not hard to figure out where the energy comes from. I'm talking about before that.&lt;/p&gt;&#xA;" OwnerUserId="6587" LastEditorUserId="2153" LastEditDate="2015-03-18T17:04:14.330" LastActivityDate="2015-03-18T21:13:16.980" Title="Where does energy at the beginning of a star's lifecycle (before any nuclear reactions) come from?" Tags="&lt;stellar-evolution&gt;&lt;stellar-astrophysics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10188" PostTypeId="1" AcceptedAnswerId="10191" CreationDate="2015-03-18T13:54:18.377" Score="5" ViewCount="725" Body="&lt;p&gt;I believe that sometime after Big Bang the entire universe was filled with high energy particles. Now we fast forward to the present; most of these particles still linger and managed to avoid being absorbed, and also they lost quite a significant amount of energy and became today's &quot;noise&quot; which can be picked up by our TV antenna. Is this true, or am I mistaken?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="2153" LastEditDate="2015-03-19T00:28:55.133" LastActivityDate="2015-03-19T00:28:55.133" Title="Was the cosmic microwave background (CMB) radiation originally gamma rays?" Tags="&lt;universe&gt;&lt;cosmology&gt;&lt;cmb&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10189" PostTypeId="1" AcceptedAnswerId="10190" CreationDate="2015-03-18T16:41:06.790" Score="28" ViewCount="31280" Body="&lt;p&gt;It is often stated that the magnetosphere not only shields the planet from cosmic radiation, but also prevents atmospheric loss. Why then did Venus not lose most of its atmosphere if it doesn't have a strong magnetic field? Is there another mechanism at play, or is the statement about the importance of magnetosphere to atmospheric loss prevention wrong? &lt;/p&gt;&#xA;" OwnerUserId="2529" LastEditorUserId="31410" LastEditDate="2021-01-09T22:06:00.893" LastActivityDate="2021-01-09T22:06:00.893" Title="Why did Venus not lose its atmosphere without a magnetic field?" Tags="&lt;planetary-atmosphere&gt;&lt;magnetic-field&gt;&lt;venus&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="10190" PostTypeId="2" ParentId="10189" CreationDate="2015-03-18T17:19:04.857" Score="19" Body="&lt;p&gt;There is an interesting article on the magnetosphere of Venus on the ESA Science and Technology site. You can find the article &lt;a href=&quot;http://sci.esa.int/venus-express/50246-a-magnetic-surprise-for-venus-express/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and it will probably answer your question.&lt;/p&gt;&#xA;&lt;p&gt;The article states, like you did, that some planets, like Earth, Mercury, Jupiter and Saturn, have magnetic fields internally induced by their iron core. These magnetic fields shield the atmosphere from particles coming from solar winds. It also confirms your statement that Venus lacks this intrinsic magnetosphere to shield its atmosphere from the solar winds.&lt;/p&gt;&#xA;&lt;p&gt;The interesting thing, however, is that spacecraft observations, like the ones made by ESA's Venus Express, have shown that the Venusian ionosphere's direct interaction with the solar winds causes an externally induced magnetic field, which deflects the particles from the solar winds and protects the atmosphere from being blown away from the planet.&lt;/p&gt;&#xA;&lt;p&gt;However, the article also explains that the Venusian magnetosphere is not as protective as Earth's magnetosphere. Measurements of the Venusian magnetic field show several similarities, such as deflection of the solar winds and the reconnections in the tail of the magnetosphere, causing plasma circulations in the magnetosphere. The differences might explain the fact that some gasses and water are lost from the Venus atmosphere. The magnetic field of Venus is about 10 times smaller than the earth's magnetic field. The shape of the magnetic field is also different. Earth has a more sharp magnetotail facing away from the Sun and Venus has a more comet-shaped magnetotail. During the reconnections most of the plasma is lost in the atmosphere.&lt;/p&gt;&#xA;&lt;p&gt;The article explains therefore that although Venus does not have an intrinsic magnetic field, the interaction of the thick atmosphere with the solar winds causes an externally induced magnetic field, that deflects the particles of the solar winds. The article suggests, however, that the different magnetic field may mean that lighter gasses are not as protected and therefore are lost into space.&lt;/p&gt;&#xA;&lt;p&gt;I hope this sufficiently answers the question.&lt;/p&gt;&#xA;" OwnerUserId="6599" LastEditorUserId="10004" LastEditDate="2020-12-17T14:22:09.323" LastActivityDate="2020-12-17T14:22:09.323" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="10191" PostTypeId="2" ParentId="10188" CreationDate="2015-03-18T17:25:37.933" Score="6" Body="&lt;p&gt;The cosmic microwave background is a result of an almost perfect blackbody emitter. That means the spectrum covers a broad wavelength range with a peak that is given by Wien's law:&#xA;$$\lambda_{peak} = \frac{2.9\times 10^{-3}}{T},$$&#xA;where the wavelength is in metres and temperature in Kelvin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The microwave background is formed at the epoch of (re)combination, when protons and electrons combine to form hydrogen atoms and the Uinverse effectively becomes transparent to the radiation field within it. This occurs at approximately 3000 K. So at that time, the cosmic microwave background had a peak wavelength of 970 nm, which in the near infrared part of the spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The epoch of (re)combination occurred at a redshift of approximately $z=1100$. What this means is that since then, the wavelength of the photons has been stretched by a factor of $1+z = 1101$, so that $\lambda_{peak} \simeq 1$ mm, and this corresponds to a temperature of $T=2.7$ K. I.e It is near infrared photons emitted at the epoch of recombination that have their wavelengths stretched by the expansion of the universe, and are seen today as microwaves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, &lt;em&gt;prior&lt;/em&gt; to the epoch of recombination, the Universe would still have been full of radiation, and as we go back in time towards the big bang, that radiation field becomes hotter and hotter, and the peak wavelengths get shorter and shorter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at models for the thermal history of the Universe, then at about 1 second after the big bang, the Universe would have been at about $10^{10}$ K and the radiation field would have been in equilibrium with the matter at this temperature. The formula above then gives $\lambda_{peak}=2.9\times10^{-13}$ m, which is definitely in the gamma-ray regime. But let me emphasize that these photons are emitted then absorbed many times and we do not see them now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;particles&quot; that you are talking about are &lt;em&gt;photons&lt;/em&gt; of light.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-03-18T19:29:03.317" LastActivityDate="2015-03-18T19:29:03.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10192" PostTypeId="2" ParentId="10187" CreationDate="2015-03-18T17:47:39.353" Score="1" Body="&lt;p&gt;Interesting question that demonstrates &quot;thinking through&quot;. I will give my shot on trying to explain it by my perception. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first part holds the ideal gas laws for true, which to a certain point will hold. When the star cloud obeys the laws of ideal gasses the temperature of a gas can be described as: $$PV = nRT = NkT$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is $P$ the pressure, $V$ the volume, $T$ the temperature, $N$ Avogadro's number and $k$ the Boltzmann constant. ($n$ is the number of moles and $R$ the gas constant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This relation shows that when a star cloud collapses under its weight the volume decreases and thus pressure and/or temperature increases. This relation is based on the statistical relation that many atoms are packed together, but not physically interact. The increasing temperature and therefore increasing kynetic energy of the atoms is caused by the increasing chance that the atoms collide with each other more frequently when the volume decreases. The ideal gas law explains the first increase in temperature by increase of the kynetic energy of the atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chain reaction first appears when the ideal gas law doesn't apply anymore. The further collapse of a large star cloud causes the center of the cloud to reach those levels of pressure, that atoms start to physically interact and the gas becomes a plasma. First then the nuclear fusion will start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this answers your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kind regards,&#xA;MacUserT &lt;/p&gt;&#xA;" OwnerUserId="6599" LastEditorUserId="2153" LastEditDate="2015-03-18T21:13:16.980" LastActivityDate="2015-03-18T21:13:16.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10193" PostTypeId="2" ParentId="10187" CreationDate="2015-03-18T17:58:48.190" Score="7" Body="&lt;p&gt;This is basic thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you compress a gas, you inject energy into it. Think of the pump you use to inflate the tires on your bike. It takes some force to move the piston, right? That effort is not wasted, but goes directly into the air in the pump. Now the air has more energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what happens to a gas when you put energy into it? It's molecules jiggle around faster. Well, faster jiggling is basically the definition of higher temperature. By putting more energy into the gas, you raise its temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can actually tell that the bike pump is getting warmer if you pump quickly and forcefully - this is something you can experience yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same with stars - the whole star is the &quot;bike pump&quot;, and gravity is the one who's pushing the piston. Due to compression (shrinking) under gravity, the gas gets hotter and hotter. It turns out a star has A LOT of gravitational energy, so the gas can get VERY hot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your terms, yes, it's the acceleration that molecules experience falling into the gravity well of the star that makes them move faster. Faster moving molecules = higher temperature. Pretty straightforward phenomenon, really.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Historically, gravitational compression was thought to be the main source of energy for the stars, before the discovery of nuclear physics. Helmholtz and Lord Kelvin proposed this hypothesis in the 1800s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure-temperature relation of any gas was originally known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Gay-Lussac%27s_law&quot; rel=&quot;nofollow&quot;&gt;the Gay-Lussac law&lt;/a&gt;. Now we know it's just a particular case of more general phenomena (&lt;a href=&quot;http://en.wikipedia.org/wiki/Ideal_gas_law&quot; rel=&quot;nofollow&quot;&gt;ideal gas law&lt;/a&gt;) tying together pressure, temperature, volume, and various kinds of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A spectacular application of the p-T relation is the so-called &quot;fire piston&quot; or &quot;fire syringe&quot;, which can ignite small pieces of cotton or paper by just hitting a piston really hard (extremely strong compression = big temperature rise). Search Youtube for some videos like this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=4qe1Ueifekg&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=4qe1Ueifekg&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-03-18T19:48:25.030" LastActivityDate="2015-03-18T19:48:25.030" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10194" PostTypeId="2" ParentId="10177" CreationDate="2015-03-18T20:01:12.007" Score="4" Body="&lt;p&gt;There are several theories on how the universe might end by time stopping. The problem with cosmology and some other parts of theoretical physics is that it cannot be proven, nor reproduced or tested. This makes it often more a faith/religion than a science. So, if you notice some irony in my answer, please, it's because of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two types of theories which fantasize about the end of the universe by end of times. I freely call those type the formula type and transcendental type theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first theory I will only briefly describe, because they end up transcendental anyway. There are several models on the universe that will describe what the universe (or universes) looks like and how they will evolve in time. The problem with these models is the variable time. If you have an endless time or an endless amount of universes, you end up with an endless amount of possibilities, which will result in all things will eventually happen. So, somewhere in time or a universe we will all become filthy rich and marry a gorgeous, intelligent, funny, friendly, etc., woman/man that never says no (to you). Because of this anomaly scientists have started to theorize that time must end at some point. If you believe these models to be correct, the scientists might be right. The problem in this lies that you first must believe the model is right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transcendental theories are a bit more difficult to explain and I will briefly leap out of astronomy into philosophy. In the 18th century a great philosopher wrote a book on the critique of pure reason (Kritik der reinen Vernunft). In this book he was looking for a transcendental doctrine of method and elements. The transcendental part of his doctrine was that it had to be build up on a priori knowledge that was not based on experience. This was his pure reason, because the elements and the methods are not tainted by (personal) experience from an observer. They are true in itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this interesting? Because Kant (that great philosopher) said that there were two transcendental elements, space and time. I will skip space, but must say something on time, because someone thinks it might end. Kant states that time is an a priori element of knowledge in pure form. Kant uses five reasons, why time is universal and not empirical. They all can be brought back to the fact that we cannot see things/phenomena to coexist together or successively when there is no concept of time. Therefore, time is an a priori element of pure form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what has this to do with the ending of the universe in respect to time? The theory you mentioned as heat death is such a transcendental theory. The theory states that the universe expands until the universe is too big to be heated by the energy provided by the matter in the universe. This theory models that the universe follow the same rules as any thermo dynamic system and therefore at a certain point in time all energy, matter and temperature are evenly distributed in this huge universe. Because all is evenly distributed no stars will be created and all processes come to a grinding halt. Since the state before, now and after are the same, the scientists now determine this as the end of time. The formula theories also end up making some theory that time will end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with these theories is that time will never end. According to Kant time is an a priori element of pure reason and will be there despite of what we observe. Because the state of the universe doesn't change, doesn't mean that there will be no time. Time has just stopped. Theories that state that time will end, do not comply with Kant's critique of pure reason, which states that time is the formal condition of all phenomena whatsoever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it didn't entirely answered your question, I hope it was entertaining and educational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kind regards,&#xA;MacUserT&lt;/p&gt;&#xA;" OwnerUserId="6599" LastActivityDate="2015-03-18T20:01:12.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10195" PostTypeId="2" ParentId="718" CreationDate="2015-03-19T00:48:56.320" Score="0" Body="&lt;p&gt;Since the Andromeda Galaxy is predicted to collide (or merge very slowly) with the Milky Way in about 4 billion years, we can expect to see stars from other galaxies until then. However, it will eventually be one galaxy so then it's a problem of definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some facts: The Andromeda is moving toward the Milky Way at a rate of about 110 km/s (68 mi/s). The rest of the Local group is also destined to this fate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia: &lt;a href=&quot;https://en.wikipedia.org/wiki/Andromeda-Milky_Way_collision&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Andromeda-Milky_Way_collision&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3008" LastActivityDate="2015-03-19T00:48:56.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10196" PostTypeId="1" CreationDate="2015-03-19T01:54:36.003" Score="1" ViewCount="247" Body="&lt;p&gt;Can someone define mass resolution in terms of cosmological simulations such as the millennium simulation? Is it just the smallest mass that a dark matter particle can have and still be detected by the simulation? &lt;/p&gt;&#xA;" OwnerUserId="6798" LastEditorUserId="43" LastEditDate="2017-12-05T07:10:32.487" LastActivityDate="2017-12-05T07:10:32.487" Title="Mass resolution" Tags="&lt;galaxy&gt;&lt;cosmology&gt;&lt;dark-matter&gt;&lt;n-body-simulations&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10197" PostTypeId="1" AcceptedAnswerId="10200" CreationDate="2015-03-19T02:18:55.133" Score="5" ViewCount="7629" Body="&lt;p&gt;Does our solar system orbit around another massive celestial object? Like a massive star or black hole for instance? From my understanding our sun rotates around the centre of our galaxy, thus our solar system follows. But I'm curious to know if there is/are other(s) gravitational force(s) that affect our suns? Other than the rotation around the centre of the milky way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would make sense to me that the gravity of massive objects would affect our sun, which in turn would affect our solar system, but I cannot find any answers other than ones concerning the &quot;galactic year&quot;.&lt;/p&gt;&#xA;" OwnerUserId="6799" LastActivityDate="2015-03-19T07:45:39.550" Title="Does our sun/solar system orbit around any other celestial objects?" Tags="&lt;orbit&gt;&lt;the-sun&gt;&lt;gravity&gt;&lt;solar-system&gt;&lt;galaxy&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2015-03-24T17:49:27.767" ContentLicense="CC BY-SA 3.0" />
  <row Id="10198" PostTypeId="2" ParentId="8881" CreationDate="2015-03-19T04:03:45.840" Score="1" Body="&lt;p&gt;Based on what I can access right now, the precession of Earth's axes are in a cone shape, so there isn't much wobble. But with all of those Near-Earth Objects, and nearby planets, there is &lt;strong&gt;probably&lt;/strong&gt; some wobble (I'm no professional). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also this thing called the Milankovitch Cycle, which shows how the combined effects of precession, external forces, and etc. affect climate. It also affects the &lt;strong&gt;eccentricity&lt;/strong&gt;, or the circularity, of Earth's orbit in a cycle of 430,000 years. This affects seasonal length and season intensity. It also changes orbit orientation, so the perihelion and aphelion move to different locations. If you want to learn more, read &lt;a href=&quot;https://en.wikipedia.org/wiki/Milankovitch_cycles&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Milankovitch_cycles&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3008" LastActivityDate="2015-03-19T04:03:45.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10199" PostTypeId="2" ParentId="10189" CreationDate="2015-03-19T05:11:08.567" Score="9" Body="&lt;p&gt;There are other ways to lose atmosphere. For example &lt;a href=&quot;http://en.wikipedia.org/wiki/Atmospheric_escape#Thermal_escape_mechanisms&quot; rel=&quot;noreferrer&quot;&gt;Jean's Escape&lt;/a&gt;. If average velocity of a gas molecule exceeds escape velocity, the planet will lose atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus' atmopshere is mostly $CO_2$ which has a higher molecular weight than the $0_2$ and $N_2$ of our atmosphere. So, for a given temperature and pressure, the carbon dioxide molecules have a slower speed. Venus' gravity is about the same as earth's and about twice Mars' gravity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, Venus' steep gravity well and massive gas molecules might be helpful in letting Venus hold on to an atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="1232" LastActivityDate="2015-03-19T05:11:08.567" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10200" PostTypeId="2" ParentId="10197" CreationDate="2015-03-19T07:37:45.270" Score="1" Body="&lt;p&gt;The orbit of the Sun around the Galaxy is quite complicated, because unlike the solar system, the mass is not completely concentrated at the centre. So, in addition to the roughly circular 230 million year orbit in the plane of the Galaxy, there are superimposed motions in and out of the plane and towards and away from the Galactic centre. These roughly sinusoidal additional motions, called epicycles, do not have large amplitudes - a few hundred light years - and take about 70 million and 160 million years respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun does not systematically orbit any other Galactic structures or stars (see the flagged potential duplicates for more details - there is no evidence for any binary companion to the Sun larger than a few Jupiter masses)  and is unlikely to do so in the forseeable future. The space between the stars in our Galaxy is large enough that they are essentially non-interacting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Galaxy itself is in motion with respect to the galaxies around it. The nearest tens of galaxies form the local group and probably have complicated orbits within their summed gravitational potential. These orbits cannot be precisely determined, because although we can measure line of sight velocities using the Doppler effect, the tangential motions require extraordinarily accurate position data over many years. These are now becoming available - for instance we know our Galaxy and M31 will collide in around 4 billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further afield, the local group is part of a larger galaxy aggregate, called the Virgo supercluster, but here, determining an orbit is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even further, large galaxy clusters are arranged into a web of larger superclusters, voids and filaments that are all in motion with respect to each other.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-03-19T07:45:39.550" LastActivityDate="2015-03-19T07:45:39.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10201" PostTypeId="2" ParentId="10196" CreationDate="2015-03-19T08:16:11.090" Score="3" Body="&lt;p&gt;Usually, in an N-body or SPH simulation the term &quot;mass resolution&quot; refers to the mass of a single particle, which usually all have the same mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;em&gt;single&lt;/em&gt; particle can always be &quot;detected&quot; in the simulation, since we have control of the coordinates of all particles, but a &lt;em&gt;structure&lt;/em&gt; of several particles becomes ill-defined if the number of particles is too small&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the smallest &lt;em&gt;resolved&lt;/em&gt; structure depends on your definition of &quot;resolved&quot;. That could be, say, 10 or 100 particles, depending on what quantity you are interested in measuring, and how accurately you want it. For instance, to define the mass of the structure, you need to be able to count the number of particles in the structure. But how does one know where to stop counting in a more or less continuous field of particles? One way is to determine an approximate center of mass $x_\mathrm{CM}$ (approximate, since it can only be exact once all associated particles are defined), calculate the average density $\langle\rho\rangle$ inside a sphere centered on $x_\mathrm{CM}$ (which will be higher than the global average, since you started out on an overdensity), and increase the radius until $\langle\rho\rangle$ falls below some threshold (e.g. 200 times the global average). If the number of particles is too small, $\langle\rho\rangle$ will change a lot between each iteration, making your mass inaccurate. I think for this purpose, &lt;em&gt;at least&lt;/em&gt; ten particles are needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're interested in the structure of the interstellar medium in a galaxy in a hydrodynamic (i.e. SPH) simulation, you probably need more particles than this. But I thinks it's fair to say that people disagree on how many particles are needed to resolve a galaxy.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-03-19T08:55:35.223" LastActivityDate="2015-03-19T08:55:35.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10202" PostTypeId="1" AcceptedAnswerId="10207" CreationDate="2015-03-19T09:03:32.563" Score="11" ViewCount="672" Body="&lt;p&gt;I read from somewhere that astronomer use X-ray to observe the &quot;shaking&quot; not to be confused with wobbling of a distant neutron star, I think it is similar to what geologist using microphone to detecting sound wave (vibration) inside the earth (no pun) to find out its composition. How does a tightly packed super dense object like neutron star generate starquake? is this neutron star signature move... i mean trait?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastActivityDate="2017-02-25T00:07:00.320" Title="Is starquake unique to neutron star?" Tags="&lt;neutron-star&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10203" PostTypeId="2" ParentId="10180" CreationDate="2015-03-19T09:22:40.640" Score="0" Body="&lt;p&gt;The Sun has a very powerful magnetic field. But the rotation of the Sun on its axis is uneven. The velocity of rotation is higher at the equator and lower at the poles. Due to the large difference in the velocities, the magnetic field lines gets twisted and turned that causes some of them to come really close. The instability causes an outburst seen as solar flares.&lt;/p&gt;&#xA;" OwnerUserId="6804" LastActivityDate="2015-03-19T09:22:40.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10204" PostTypeId="1" CreationDate="2015-03-19T09:26:42.473" Score="8" ViewCount="37535" Body="&lt;p&gt;You can find in many place on the Internet that welder's glass #14 is good for looking at an eclipse. Tomorrow (March, 20th 2015 at 10:45 CET) there's a solar eclipse and yesterday I could only find glasses #11 and #9. I tried briefly this morning, and combining a #11 and a #9 was giving me a good view of the sun. Using 2 #11 gave a too dark image. Now, I read different opinions on the internet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Against (Perkins Observatory):&#xA;&lt;a href=&quot;http://perkins.owu.edu/solar_viewing_safety.htm&quot; rel=&quot;noreferrer&quot;&gt;http://perkins.owu.edu/solar_viewing_safety.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Be careful that you use the right kind of glass! Welder's glass is&#xA;  numbered from 1 to 14 with 14 being the darkest. It is only number 14&#xA;  glass that is dark enough for solar viewing! And NO STACKING! A pair&#xA;  of number 7's or a 10 and a 4 together DO NOT have the same protection&#xA;  as a single piece of number 14 (see unsafe methods for more details).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Favorable (Royal Astronomical Society of Canada):&#xA;&lt;a href=&quot;https://www.rasc.ca/tov/safety&quot; rel=&quot;noreferrer&quot;&gt;https://www.rasc.ca/tov/safety&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If SN14 filter is not available, it is possible to combine lower shade&#xA;  numbers to get roughly the same level of eye protection from solar&#xA;  radiation, e.g. combining SN 6 and SN 8 filters. However the image&#xA;  quality may be considerably poorer than that seen through the single&#xA;  SN14 filter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could not find a table or something explaining which kind of protection gives each number; according to the Canadian website, the only concern is about how much infrared light goes through, ultraviolet does not seem to be a problem in almost any case (I was surprised to read that).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: It's not my intention to open here the discussion on what could be other safe methods to watch the eclipse, this is well explained everywhere around. I read too late about the eclipse to order specific glasses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;One of the answers &lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20120603075230AAXTXwG&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; report the following formula: (more insight at &lt;a href=&quot;http://www.polytechforum.com/metalworking/how-do-welding-shade-numbers-add-up-eclipse-viewing-544529-.htm#544575&quot; rel=&quot;noreferrer&quot;&gt;this link&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;13 or darker is safe enough. Also, you CAN add up welding glass, using&#xA;  the formula S(sum) = S1 + S2 -1. S(sum) should be greater than or&#xA;  equal to 13&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2077" LastEditorUserId="2077" LastEditDate="2015-03-19T13:03:04.243" LastActivityDate="2017-08-21T01:29:47.500" Title="Is stacking welder's glasses a safe way to watch at the eclipse?" Tags="&lt;observational-astronomy&gt;&lt;amateur-observing&gt;&lt;solar-eclipse&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10207" PostTypeId="2" ParentId="10202" CreationDate="2015-03-19T13:48:31.293" Score="9" Body="&lt;p&gt;I'm not sure what the current status of neutron starquake theory is, but last time I looked there were a couple of ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outer parts of a neutron star consist of a solid crust of neutron-rich nuclei (plus electrons, and a little deeper, free neutrons). Neutron stars are born rotating extremely fast, but they spin down due to the loss of rotational kinetic energy through magnetic dipole radiation. As a result, they are born oblate, but as the spin rate decreases this is not the lowest energy configuration. It used to be thought that starquakes (pulsar glitches) were caused by cracking of the crust and rearrangement towards a more spherical configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second idea is that it could be due to changes in the very intense magnetic field of the neutron star. The magnetic field is tied very tightly to the crustal material, so if the magnetic field shifts, possibly associated with the neutron star spin down, then it will put the crust under stress, which can be relieved by a starquake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;White dwarf stars also contain solid, crystalline material, however they do not spin so fast and do not have similar ultra-strong magnetic fields. I have not heard that they experience starquakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Normal stars&quot; do not have a crust or any other solid layer, so there can be no similar starquake phenomenon.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-02-25T00:07:00.320" LastActivityDate="2017-02-25T00:07:00.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10208" PostTypeId="1" AcceptedAnswerId="10219" CreationDate="2015-03-19T15:59:38.070" Score="6" ViewCount="1157" Body="&lt;p&gt;Solar eclipses on the earth are so spectacular because the moon has roughly the same size as the sun when viewed from the earth's surface. This is an incredible coincidence, and my guess is that it is a very uncommon phenomenon in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to say anything sensible about the likelihood of a planet having a moon and a sun with same apparent size? Is it possible to make an educated guess about how many places in the universe you would be able to see an earth-like solar eclipse?&lt;/p&gt;&#xA;" OwnerUserId="6808" LastActivityDate="2015-03-20T21:53:57.350" Title="How rare are earth-like solar eclipses?" Tags="&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10209" PostTypeId="1" AcceptedAnswerId="10232" CreationDate="2015-03-19T19:00:54.303" Score="3" ViewCount="775" Body="&lt;p&gt;I'm slowly starting to interest in astronomy. Currently I'm enjoying in stargazing but unfortunately place where I live is very light polluted.&lt;/p&gt;&#xA;&lt;p&gt;I'm planing my vacation in a couple of months and one of the attractive places on my list is Tenerife because I can combine both star-gazing and going to the beaches. As per my research, base of Mountain El Teide is perfect for star-gazing so I tried to simulate El Teide location and date in the Stellarium software to see what I can expect. Based on the light pollution index (2) which Stellarium picked for me based on the location I got following picture:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/d5oml.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/d5oml.jpg&quot; alt=&quot;a stellarium render&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://s30.postimg.org/libpikia9/star_view_may_elteide.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Full resolution image&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;My question is: How this rendered picture will match my naked eye in the reality at the same location and time (we can assume that there will not be any clouds and that I'll be stargazing when moon is couple of days from New Moon phase). Can I expect something like this or even something better?&lt;/p&gt;&#xA;&lt;p&gt;I'm aware of all the tips which can maximize my experience but I'm trying to just be sure that El Teide will not disappoint me. I really want have great star gazing-experience and try to see Milky way and much stars as possible because I don't have that luxury in the country where I live.&lt;/p&gt;&#xA;&lt;p&gt;Any personal star gazing experience from Tenerife location is more than welcome. For my first time I'll plan to not use any equipment (except maybe binoculars)&lt;/p&gt;&#xA;&lt;p&gt;Btw, there are stellarium settings used for picture that I attached (I assume that this is closest what you can see with naked eye).&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7YFB8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7YFB8.png&quot; alt=&quot;stellarium settings&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;" OwnerUserId="6810" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-02-18T08:11:57.457" Title="Sky view from Stellarium software vs. Sky view with naked eye" Tags="&lt;milky-way&gt;&lt;star-gazing&gt;&lt;light-pollution&gt;&lt;naked-eye&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10210" PostTypeId="2" ParentId="10159" CreationDate="2015-03-19T20:22:13.390" Score="3" Body="&lt;p&gt;The question itself is wrong, actually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no such thing as absolute velocity, which is what you're assuming in your question. Velocity is always relative to a frame of reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your speed relative to your chair is zero, but it's not zero relative to the airplanes flying over your house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you say &quot;a point in space where nothing is moving and velocity is essentially 0&quot;, you must add &quot;relative to such-and-such frame of reference&quot;, otherwise your statement makes no sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By that token, any object in any point of space has zero speed relative to any frame of reference rigidly attached to it, and non-zero speed relative to other frames. You can't talk about speed unless you specify the frame against which it is measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speed doesn't even exist by itself. It's always relative to a frame. It's not a &lt;strong&gt;property&lt;/strong&gt; of an object or (even worse) a property of &quot;a point in space&quot;. It's a &lt;strong&gt;relation&lt;/strong&gt; between an object (the moving thing) and another object (the frame). &quot;Points in space&quot; don't have properties themselves, space is featureless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cars may seem like they have a speed as an intrinsic property, but that's just their speed relative to the ground. The car's speed relative to its driver is very different (and hopefully equal to zero). And the car's speed relative to a comet in the Solar System is yet again different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any further considerations, such as the flow of time, etc., are invalid as long as you're not asking the right question. Again, speed can only exist relative to something and, as such, it depends on the choice of the frame. And time depends on that whole causal chain.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-03-19T20:28:13.390" LastActivityDate="2015-03-19T20:28:13.390" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10212" PostTypeId="1" CreationDate="2015-03-20T09:47:21.730" Score="7" ViewCount="629" Body="&lt;p&gt;I have a simple Newtonian reflector telescope. Using it, I am able to see the Galilean moons of Jupiter. However, Jupiter has much more moons than that (Wikipedia says 67 have been discovered this far). What kind of telescope I would need to see most or all of them?&lt;/p&gt;&#xA;" OwnerUserId="6815" LastActivityDate="2015-03-20T12:25:05.387" Title="What kind of telescope do I need to see most of the Jupiter's moons?" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;jupiter&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10213" PostTypeId="1" AcceptedAnswerId="10264" CreationDate="2015-03-20T10:04:10.197" Score="8" ViewCount="1582" Body="&lt;p&gt;Planets of the solar system have been named with Roman mythology gods names. I have a few questions on my mind concerning that subject for a while:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Who decided to name them like this?&lt;/li&gt;&#xA;&lt;li&gt;When did the planets got named?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Finally, the ultimate question is: if the planets got named by the Romans, implying they had not technology to observe the solar system whatsoever, why some of the names match the color of the planets? &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mars is red - Mars is the god of war&lt;/li&gt;&#xA;&lt;li&gt;Neptune is blue - Neptune is the god of the oceans&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Or maybe I am wrong (most likely) and the planets have been named more recently.&lt;/p&gt;&#xA;" OwnerUserId="6816" LastEditorUserId="58" LastEditDate="2015-03-20T20:43:02.520" LastActivityDate="2020-01-26T05:09:01.180" Title="Naming of the planets of the solar system" Tags="&lt;solar-system&gt;&lt;history&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10215" PostTypeId="2" ParentId="10212" CreationDate="2015-03-20T12:25:05.387" Score="6" Body="&lt;p&gt;Unless you're really rich, unfortunately you won't be able to see all of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter's fifth largest Moon, Amalthea, has an apparent magnitude of $m$ = 14.1. Comparing this to the magnitude of Europa, the dimmest of the Galilean moons, which is 5.3, tells us that Amalthea is roughly 3000 times less bright. Your telescope thus needs to have an area 3000 (or radius ~55) times larger for Amalthea to have the same apparent brightness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, without a camera on your telescope, the dimmest object you can see depends on your vision, but on average, humans are able to see objects of magnitude 6. That means that in order to detect an object of magnitude $m_\mathrm{obj}$, you need a light-collecting area which is larger than you pupil by a factor of&#xA;$$f = 10^{(m_\mathrm{obj}-6)/2.5}.$$&#xA;Thus, in order to be able to just barely detect Amalthea, you need a telescope which is larger than you pupil (radius = 6 mm) by a factor of ~1738, i.e. has a radius of 25 cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It quickly becomes really difficult to see them; for instance, Elara, which is the eighth largest moon, has an apparent magnitude of 16.3 and thus requires a 70 cm telescope. And remember, this is just on the threshold of what you can see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you mount a camera to your telescope and take images with long exposure times, you can get away with smaller telescopes. In fact all moons smaller than Amalthea were discovered this way, some of them by cameras on the Voyager space probes.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2015-03-20T12:25:05.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10216" PostTypeId="1" AcceptedAnswerId="10218" CreationDate="2015-03-20T18:43:20.990" Score="4" ViewCount="76" Body="&lt;p&gt;Going through &lt;a href=&quot;http://earthsky.org/&quot; rel=&quot;nofollow&quot;&gt;http://earthsky.org/&lt;/a&gt;…/dates-of-next-lunar-and-solar-eclipses I made the (what was for me - momentous) discovery that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;March 20, 2015 - New moon - is a Solar Eclipse, and that the succeeding&#xA;April 04, 2015 - Full moon - heralds a Lunar Eclipse.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is it common that both New/Full moon in a fort-night cause an eclipse?&lt;/li&gt;&#xA;&lt;li&gt;When was the last time an Eclipse occurred back-to-back on a consecutive New Moon/Full Moon?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="259" LastEditorUserId="259" LastEditDate="2015-03-21T06:21:38.493" LastActivityDate="2015-03-21T06:21:38.493" Title=".2015: When last did both New/Full moon in a fort-night cause an Eclipse?" Tags="&lt;solar-eclipse&gt;&lt;lunar-eclipse&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10217" PostTypeId="1" AcceptedAnswerId="10223" CreationDate="2015-03-20T20:29:51.510" Score="3" ViewCount="258" Body="&lt;p&gt;Reading upon the eclipse of March 20, 2015, I stumbled upon this page: &lt;a href=&quot;http://www.timeanddate.com/eclipse/list-solar.html&quot; rel=&quot;nofollow&quot;&gt;http://www.timeanddate.com/eclipse/list-solar.html&lt;/a&gt;. What caught my eye is that for each year there are two eclipses whose paths are almost symmetrical relative to equator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just curious why that's always exactly so.&lt;/p&gt;&#xA;" OwnerUserId="6819" LastEditorUserId="6819" LastEditDate="2015-03-20T20:45:16.170" LastActivityDate="2015-03-21T03:45:42.600" Title="Why solar eclipse paths are symmetrical?" Tags="&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10218" PostTypeId="2" ParentId="10216" CreationDate="2015-03-20T20:43:01.497" Score="1" Body="&lt;p&gt;As written &lt;a href=&quot;http://www.timeanddate.com/eclipse/10-facts-solar-eclipse-march-2015.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Another oddity of nature is that solar eclipses and lunar eclipses tend to come in pairs – a solar eclipse always takes place about two weeks before or after a lunar eclipse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And &lt;a href=&quot;http://vnatsci.ltu.edu/s_schneider/astro/wbstla2k/mytalk/leclipse.shtml&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Rules of Eclipses (Solar and Lunar) ... Eclipses tend to go in pairs or threes : solar-lunar-solar. Lunar eclipse always preceeded by or followed by a solar eclipse (two weeks between them)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, seems like a totally common thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. The previous pair was first lunar on Oct 8, 2014, and then solar on Oct 23rd.&lt;/p&gt;&#xA;" OwnerUserId="6819" LastEditorUserId="2153" LastEditDate="2015-03-21T01:18:08.460" LastActivityDate="2015-03-21T01:18:08.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10219" PostTypeId="2" ParentId="10208" CreationDate="2015-03-20T20:59:54.543" Score="4" Body="&lt;p&gt;We can certainly &lt;em&gt;speculate&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are dozens of moons in our own Solar System. I've done some preliminary calculations of their apparent size as seen from the planet vs. the apparent size of the Sun at that distance. The large moons (Jupiter's 4 Galilean satellites, Saturn's Titan, Neptune's Triton) are all substantially larger in the sky than the Sun is. I &lt;em&gt;think&lt;/em&gt; there are some satellites that are fairly close to the apparent size of the Sun, but I haven't done all the calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The apparent size of our Moon has changed over time, as the Moon has gradually moved farther away. We are coincidentally in a period of history in which it happens to be very nearly the same apparent size as the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The criterion for an exoplanet to have solar eclipses like the spectacular ones we have here on Earth is a moon that happens to have an angular size large enough to cover the photosphere, but not so large that it also hides the corona.  For a larger moon, you'd have eclipses in which the corona is visible, but not all the way around the moon. For a smaller moon, you'd only have annular eclipses.  (It could also get interesting if you consider eclipses as seen from other moons.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So given the distribution of sizes and distances of moons in our Solar System, we can &lt;em&gt;guess&lt;/em&gt; that situations where the apparent size of a moon and sun closely match is fairly rare, but since there are moons whose apparent size is smaller than the Sun &lt;em&gt;and&lt;/em&gt; others whose apparent size is larger than the Sun, it probably happens sometimes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one piece of the puzzle that we're still missing, I think, is the typical sizes and distances of satellites of Earth-like planets. Moons the size of ours orbiting habitable planets might be common or very rare. If they're rare (say, if Mars-like moon systems are far more common), then Earth/Moon style eclipses might be very rare &lt;em&gt;for habitable planets&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="53" LastEditorUserId="53" LastEditDate="2015-03-20T21:53:57.350" LastActivityDate="2015-03-20T21:53:57.350" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10220" PostTypeId="1" AcceptedAnswerId="10967" CreationDate="2015-03-20T21:19:45.987" Score="7" ViewCount="876" Body="&lt;p&gt;How would the night sky look like from the Martian surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would there be more stars visible or less (compared to Earth)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would they appear brighter/dimmer/same?&lt;/p&gt;&#xA;" OwnerUserId="6820" LastEditorUserId="14097" LastEditDate="2016-10-04T10:18:02.440" LastActivityDate="2016-10-04T10:18:02.440" Title="Mars night sky - how many stars visible?" Tags="&lt;mars&gt;&lt;naked-eye&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10221" PostTypeId="2" ParentId="10213" CreationDate="2015-03-20T22:07:12.967" Score="13" Body="&lt;p&gt;Despite of the explanation of your question, it is a valid question to ask why planets all have Roman names. First of all, the Romans could, like the Greeks and Sumerians, could only see Mercury, Venus, Mars, Jupiter and Saturn. These planets can be seen with the naked eye. However, the fact that the Romans could see these planets, didn't give all the planets latin names. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Names of the planets were given by the ones that discovered them first, the planets would have Sumerian names. The answer to the question why the planets have latin names is because they don't. The first western scientist took over the Roman names, because latin was they scientific language of the middle ages and the renaissance. After the discovery of Uranus and Neptune these planets were not given these names by their discoverers. Only after longer controversy western astronomers have standardized the names of planets and moons according to Greek and Roman mytholohy, but this was only around the second half of the 19th century. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the beginning of the 20th century the IAU is setting the &lt;a href=&quot;http://www.iau.org/public/themes/naming/#inss&quot; rel=&quot;noreferrer&quot;&gt;standards&lt;/a&gt; for naming celestial bodies including planets. However, the IAU doesn't set the names, it only sets the standards. Newly discovered objects are now named according to these rules by their discoverers. Older planets are still living with their old names and for western astronomers those are still the names from Greek and Roman mythological figures. However, the IAU allows you to consistently name the planets by their Arabic or Chinese names. So, the question is valid, the text around your question is unfortunately not true, since the planets don't necessarily have Roman names. If you are An Asian person and correctly think the planets should have their Chinese names, because they were earlier in discovering these objects than the Romans (although not earlier than the Sumerians), you are allowed by the IAU (sic) to call them by their Chinese names.&lt;/p&gt;&#xA;" OwnerUserId="6599" LastEditorUserId="1106" LastEditDate="2015-04-02T22:37:06.920" LastActivityDate="2015-04-02T22:37:06.920" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10222" PostTypeId="1" AcceptedAnswerId="27407" CreationDate="2015-03-21T00:29:37.657" Score="5" ViewCount="221" Body="&lt;p&gt;&lt;a href=&quot;http://news.nationalgeographic.com/2015/03/150318-total-solar-eclipse-equinox-supermoon-astronomy-spring/&quot; rel=&quot;noreferrer&quot;&gt;Today&lt;/a&gt; (March 20, 2015) is seeing a rare combination of the spring equinox, a total solar eclipse, and a supermoon.  I am wondering if there is anything special astronomically about all three of these happening at the same time, like is the equinox a particularly more or less likely time for solar eclipses to happen (my guess is that it has no impact eclipses).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The supermoon/&lt;strong&gt;total&lt;/strong&gt; solar eclipse combination doesn't seem surprising to me, since if the opposite were to happen (moon far away and solar eclipse) we would end up with an annular eclipse instead of a total eclipse.&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2018-08-20T22:14:47.640" Title="Total solar eclipse, supermoon, and spring equinox all happening at the same time: anything special about this?" Tags="&lt;the-moon&gt;&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10223" PostTypeId="2" ParentId="10217" CreationDate="2015-03-21T03:45:42.600" Score="1" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbit_of_the_Moon&quot; rel=&quot;nofollow&quot;&gt;orbit of the moon&lt;/a&gt; is inclined by 5.14° to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ecliptic&quot; rel=&quot;nofollow&quot;&gt;ecliptic&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you may know, the ecliptic is the apparent path of the Sun across the sky, so it is also the plane of Earth's orbit around the Sun. The plane of the Moon's orbit is inclined by 5.14° to the plane of the Earth's orbit. The intersection of the two plans is a line bisecting both planes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a nice picture of this at &lt;a href=&quot;http://www.cnyo.org/tag/orbital-plane/&quot; rel=&quot;nofollow&quot;&gt;http://www.cnyo.org/tag/orbital-plane/&lt;/a&gt;. [I'll insert it here after checking the copyright]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it is only when the Earth is very near to the intersection line that an eclipse can occur, because it is only then that the Sun, Moon and Earth can be in a strait line. This occurs for slightly more than a month (called an &lt;a href=&quot;http://en.wikipedia.org/wiki/Eclipse_season&quot; rel=&quot;nofollow&quot;&gt;eclipse season&lt;/a&gt;), twice a year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another angle comes into the picture too. The axis of the Earth is tilted, so the equator is at 23.4° to the plane of the ecliptic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the pattern you have observed comes about because of that geometry, where mirror image eclipse paths occur about six months apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The linked pages explain a lot more of the complexities.&lt;/p&gt;&#xA;" OwnerUserId="5524" LastActivityDate="2015-03-21T03:45:42.600" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10227" PostTypeId="1" AcceptedAnswerId="10233" CreationDate="2015-03-21T15:08:07.083" Score="3" ViewCount="282" Body="&lt;p&gt;My wife and I were watching a TV show last night, which takes place in Salem, Massachusetts circa 1693.  One character went to a particular place in the woods at a particular time to be exactly under Saturn.  I told my wife that Saturn could never be exactly above such a Northern latitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the Northernmost Latitude that Saturn could possibly be above?  By &quot;above&quot;, I mean if you draw a line through the center of the Earth and through Saturn, the point on the Earth's surface that line passes through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how often would it reach that point?  Would it be every few years, or every few thousand years?&lt;/p&gt;&#xA;" OwnerUserId="6830" LastActivityDate="2015-03-23T16:37:08.690" Title="What is the Northernmost Latitude of Saturn?" Tags="&lt;saturn&gt;&lt;positional-astronomy&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10229" PostTypeId="1" CreationDate="2015-03-22T09:45:38.643" Score="6" ViewCount="401" Body="&lt;p&gt;With the current news about the possibility of a planet in the Oort cloud, I was thinking, why is the matter in the Oort cloud distributed so? It makes sense that inside the planet region there is not a huge concentration of comets and meteors, but why is the matter inside the cloud itself not agglomerating? It seems like there is enough matter to form not one or two, but many more planets.&lt;/p&gt;&#xA;" OwnerUserId="4126" LastEditorUserId="6" LastEditDate="2015-06-24T17:04:36.190" LastActivityDate="2015-07-22T22:04:21.237" Title="Lack of planets in the Oort Cloud" Tags="&lt;gravity&gt;&lt;planet&gt;&lt;comets&gt;&lt;oort-cloud&gt;&lt;planetary-formation&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10230" PostTypeId="1" AcceptedAnswerId="10235" CreationDate="2015-03-22T09:59:30.933" Score="3" ViewCount="484" Body="&lt;p&gt;Astrophysics can be said to have been founded by Johannes Kepler around the year 1600. He based his break-through science on data of the position of Mars in the sky and disproved the ancient ideas about circular orbits and epicycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why wasn't this done far earlier, by using observations of the Moon? Wasn't it pretty obvious to a careful astrologer a thousand years ago, that the Moon does not have a circular orbit and does not describe epi-cycles? It is the easiest celestial object to observe, visible both night and day. Moon calendars may have been designed tens of thousands of years ago, there's no lack of observational data. Kepler instead used a few oppositions of Mars which take place only once every two+ years. Since the Moon is the one object which does orbit Earth, in a geocentric world view it should've been the perfect test of circular and epi-cyclical theories about its orbit. Its nearness causes a daily parallax between moonrise and moonset, but that wouldn't be beyond a genius like Kepler or many mathematical astrologers before him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What made the orbit of the Moon difficult for the ancients to understand?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="2365" LastEditDate="2015-11-13T20:02:37.350" LastActivityDate="2015-11-13T20:02:37.350" Title="What were the challenges for the ancients to observe the orbit of the Moon (instead of Mars)?" Tags="&lt;the-moon&gt;&lt;observational-astronomy&gt;&lt;history&gt;&lt;johannes-kepler&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10232" PostTypeId="2" ParentId="10209" CreationDate="2015-03-22T17:26:55.540" Score="1" Body="&lt;p&gt;From personal experience - yes your Stellarium picture looks like a good dark sky in the Canary Islands. Certainly, the Milky Way is plainly visible, with lots of detail available to the dark-adapted eye. I more frequently view from the top of La Palma, which suffers a little less light pollution than Teide to the East.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Canarian &lt;a href=&quot;http://www.iac.es/eno.php?op1=4&amp;amp;op2=10&amp;amp;lang=en&quot; rel=&quot;nofollow noreferrer&quot;&gt;sky law&lt;/a&gt; was largely setup to protect the night sky quality at the observatory on La Palma, but covers all lighting on Tenerife in a straightline view to La Palma. So I would be guessing that the western side of Teide will be the darkest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this picture which illustrates the controlled lighting areas (in green). It is not to scale, La Palma is further away from Tenerife (on the right) than illustrated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hahNJ.jpg&quot; alt=&quot;Canary island dark sky zones&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-03-22T17:26:55.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10233" PostTypeId="2" ParentId="10227" CreationDate="2015-03-22T18:25:27.193" Score="8" Body="&lt;p&gt;Salem is at 42.5° north. At the equator, the &lt;a href=&quot;http://en.wikibooks.org/wiki/General_Astronomy/Yearly_Motions&quot; rel=&quot;noreferrer&quot;&gt;position of the sun varies by 23.5 degrees&lt;/a&gt; on either side of the vertical over the course of each year, due to earth's 23.5° axial tilt with respect to the ecliptic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn, however is not on the ecliptic. It &lt;a href=&quot;http://en.wikipedia.org/wiki/Saturn&quot; rel=&quot;noreferrer&quot;&gt;departs by about 2.5°&lt;/a&gt; over the course of its 29.45 year, 9.5 AU orbit. So &lt;strong&gt;once every 29.5 years&lt;/strong&gt; Saturn's angle to the ecliptic will add 2.5° to the max latitude the planet can be seen overhead.&#xA;That adds up to 26° as the max latitude Saturn can be seen directly overhead. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's one more factor to account for though. Earth being 1AU out from the sun, it's only 8.5 AU from Saturn at close approach. That will add another 0.2° (by trig) to the max latitude at that time of year when earth is nearest Saturn. That gets us to the possiblity of Saturn appearing directly overhead at up to &lt;strong&gt;latitude 26.2°&lt;/strong&gt;: &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_cities_by_latitude#30.C2.B0_North&quot; rel=&quot;noreferrer&quot;&gt;Miami, Jaipur, or Dubai&lt;/a&gt;, not Salem&lt;/p&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="2031" LastEditDate="2015-03-23T06:43:30.343" LastActivityDate="2015-03-23T06:43:30.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10235" PostTypeId="2" ParentId="10230" CreationDate="2015-03-22T22:41:23.297" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Wasn't it pretty obvious to a careful astrologer a thousand years ago, that the Moon does not have a circular orbit and does not describe epi-cycles?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The ancient Greek model of the motion of planetary bodies remained unchallenged for almost two millennia, so obviously not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hipparchus' model did a fairly good job dealing with the elliptical motion of the Moon; it did even better with the planets. The Moon's motion is tough to model because of perturbations by the Sun, Venus, and Jupiter. Ptolemy discovered what would eventually be called evection, the largest of the perturbations caused by the Sun. There was one problem with Ptolemy's model: It had the Moon swinging in and out by a huge amount. If Ptolemy's model was correct, we would see the Moon changing in diameter by a factor of two over the course of a bit over half an orbit. Copernicus much later came up with a scheme that fixed this problem and still relied on those old concepts of deferents, equants, epicycles, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While Newton pointed the way to describing the Moon's orbit, it wouldn't be until 200 years after Newton's death that a decent (one that matches observations) model of the Moon's orbit was developed.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-03-22T22:41:23.297" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10237" PostTypeId="1" AcceptedAnswerId="10285" CreationDate="2015-03-23T11:36:30.617" Score="0" ViewCount="249" Body="&lt;p&gt;I'm using NOVAS 3.1. I know that I can convert equatorial coordinates to horizontal coordinates using the &lt;code&gt;equ2hor&lt;/code&gt; function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to make NOVAS do the inverse transformation: from horizontal to equatorial? There is no &lt;code&gt;hor2equ&lt;/code&gt;, but maybe some other function implements this functionality?&lt;/p&gt;&#xA;" OwnerUserId="6840" LastActivityDate="2015-03-27T23:45:08.247" Title="How to convert horizontal coordinates using NOVAS?" Tags="&lt;fundamental-astronomy&gt;&lt;coordinate&gt;&lt;software&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10238" PostTypeId="2" ParentId="10227" CreationDate="2015-03-23T16:37:08.690" Score="4" Body="&lt;p&gt;To add to the other answers, if you were in a place where Saturn could be directly above, make sure you get there just over an hour and a half early. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now; do you need to be there when Saturn is directly overhead or when the light from Saturn is directly overhead? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest Saturn gets to earth is approximately 1,200,000,000 km (750,000,000 miles), light takes 1 second to travel 300,000 km therefore the light from Saturn would take a minimum of 4000 light seconds (1,200,000,000 / 300,000 = 4000) 4000 seconds = 1.11 hours. So you are seeing Saturn where it was just over an hour before. This could increase to over 1 1/2 hours depending on the distance between Earth and Saturn at the time (max approx 1,700,000,000 km).&lt;/p&gt;&#xA;" OwnerUserId="4209" LastActivityDate="2015-03-23T16:37:08.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10239" PostTypeId="2" ParentId="10204" CreationDate="2015-03-23T18:46:23.120" Score="-3" Body="&lt;p&gt;I know it's a bit late for the UK eclipse, but for future reference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Welders goggles do not cut out the same frequency of light that Mylar/Black Polymer eclipse glases do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arc welding glass #14 is much darker than brazing welders glass but there is no guarantee that they will block the Sun's ultra-violet and infrared rays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is there are no pain sensors in your retina so you don't feel your eyes cooking and the dark lenses (as do sunglasses) cause your pupils to dialate letting in more ultra-violet and infrared rays so they could make things worse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have used arc Welders glass on a camera and it worked fine but gave a picture of a green Sun. Project the image on paper to take it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do NOT look through a camera viewfinder even with the correct eclipse glasses on if you do not have the correct solar filer on the LENS end.&lt;/p&gt;&#xA;" OwnerUserId="4209" LastActivityDate="2015-03-23T18:46:23.120" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10240" PostTypeId="2" ParentId="9990" CreationDate="2015-03-24T00:33:39.723" Score="2" Body="&lt;p&gt;The speed of light in a vacuum is 299,792,458 meters per second, not infinite. Let's say, for example, particle of a beam of light, the photon, is emitted. It takes ~8 minutes to get to us; when it hits our eyes, we see it. This means that we see a photon that was emitted from the sun 8 minutes ago. We aren't, per se, looking &quot;back in time&quot;, but we're looking at a photon that is ~8 minutes old. &lt;/p&gt;&#xA;" OwnerUserId="6769" LastActivityDate="2015-03-24T00:33:39.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10242" PostTypeId="1" AcceptedAnswerId="10244" CreationDate="2015-03-24T05:10:12.270" Score="28" ViewCount="9083" Body="&lt;p&gt;Where there is fire, there is always smoke. So why there isn't any smoke near the Sun?&lt;/p&gt;&#xA;" OwnerUserId="5343" LastEditorUserId="2153" LastEditDate="2015-03-24T15:06:20.493" LastActivityDate="2017-05-06T22:18:09.163" Title="Why there is no smoke around the Sun?" Tags="&lt;the-sun&gt;" AnswerCount="4" CommentCount="10" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10243" PostTypeId="2" ParentId="10242" CreationDate="2015-03-24T05:23:34.683" Score="9" Body="&lt;p&gt;The sun is not actually on fire, like a log burning. In the core of the sun, a nuclear reaction takes place that generates heat. Because the sun is so large, it generates a lot of heat and it glows like an incandescent light bulb.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2015-03-24T05:23:34.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10244" PostTypeId="2" ParentId="10242" CreationDate="2015-03-24T05:29:06.503" Score="40" Body="&lt;p&gt;The sun isn't on fire.   Fire is actually very rare in the solar system.   It requires chemical potential energy, which happens on earth because of life.  Photosynthesis uses solar energy to build things out of carbon, oxygen, nitrogen and hydrogen (and some other elements).   It's these carbon chain structures that are ultimately flammable, and pretty much, only in an Oxygen atmosphere.   Oxygen atmosphere is produced by the same photosynthesis.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun's heat comes from nuclear fusion which doesn't make smoke.    Also, the nuclear fusion only happens deep the inside of the sun, the outside is just a thick blanket of molecules - mostly hydrogen.   The sun also isn't transparent.  The light we see from the sun is like the light we see from a red hot piece of metal or lava.   Red hot metal doesn't smoke (unless you drop water on it - which is actually steam.)   The sun, like red hot metal or lava, glows bright because of temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a certain sense, the tail of a comet is kind of like steam being made by the heat of the sun.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-03-24T05:29:06.503" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10246" PostTypeId="1" CreationDate="2015-03-24T13:29:27.000" Score="2" ViewCount="2195" Body="&lt;p&gt;Brightness of the lunar disk varies as a function of phase angle. This phenomenon is very well established, as is the increase in brightness at low phase angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a simple analytical formula that describes this function? It looks like brightness could be modeled as an exponential as a function of phase angle, modulo the opposition surge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I realize that these effects depend on the wavelength of the reflected light. I'm looking for an average curve, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CNcmj.gif&quot; alt=&quot;Lunar brightness curve, pointing out the opposition surge&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6851" LastEditorUserId="6851" LastEditDate="2015-03-24T13:46:46.157" LastActivityDate="2020-11-06T16:49:13.107" Title="Is there a simple, analytical formula for the lunar phase brightness curve?" Tags="&lt;the-moon&gt;&lt;light&gt;&lt;moon-phases&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10247" PostTypeId="1" CreationDate="2015-03-24T14:47:54.620" Score="3" ViewCount="107" Body="&lt;p&gt;How exactly, with steps and math shown, do you derive the masses of asteroids via perturbations? &lt;em&gt;please&lt;/em&gt;, no links, because I have searched and searched for years and can't find anything. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean like asteroids- how are their masses determined exactly? What equations are used?&lt;/p&gt;&#xA;" OwnerUserId="6769" LastEditorUserId="11264" LastEditDate="2017-02-03T13:36:02.153" LastActivityDate="2017-02-03T13:36:02.153" Title="Dynamic Method-Please explain" Tags="&lt;asteroids&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10248" PostTypeId="2" ParentId="10242" CreationDate="2015-03-24T18:28:02.470" Score="9" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fire&quot;&gt;Fire&lt;/a&gt; is actually the rapid oxidation of a combustible material.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Smoke&quot;&gt;Smoke&lt;/a&gt; is the airborne particulates and gases that result from the combustion, or from &lt;a href=&quot;http://en.wikipedia.org/wiki/Pyrolysis&quot;&gt;pyrolysis&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun is not undergoing an oxidation reaction, so it's not producing particulates that one might refer to as smoke.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process the sun is undergoing is &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_fusion&quot;&gt;nuclear fusion&lt;/a&gt;, where hydrogen are combined and create helium.  This reaction is very energetic and releases heat, visible radiation, and other radiation along a wide swath of the electromagnetic spectrum. The reaction is self-sustaining - as long as there is fuel, the emissions of fusion reactions cause nearby fuel to react as well.  No oxygen or oxidizing agent is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, the resulting helium is comparatively heavy, and the sun being a huge mass keeps both the hydrogen that is the fuel and the helium that is the product nearby - they don't leave like hot smoke does from a fire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is no smoke as we might consider it - just helium gas (plasma), and even if there were it would simply drop into the sun, there is no such thing as &quot;rising&quot; from it as one might consider smoke does on earth-borne fires.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might find &lt;a href=&quot;https://www.youtube.com/watch?v=sLkGSV9WDMA&quot;&gt;this music video&lt;/a&gt; provides further instruction on composition and reaction of the sun.&lt;/p&gt;&#xA;" OwnerUserId="559" LastActivityDate="2015-03-24T18:28:02.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10249" PostTypeId="1" AcceptedAnswerId="10250" CreationDate="2015-03-24T19:57:48.120" Score="5" ViewCount="347" Body="&lt;p&gt;I was just wondering what is the normal amount of planets is in a star system, because in my mind 8 planets seems like a relatively large amount. So are we above average? Can we even see all the exoplanets in a star system? Also, do we know if belts of asteroids are common? Is it even likely for a star to have planets?&lt;/p&gt;&#xA;" OwnerUserId="2337" LastEditorUserId="1891" LastEditDate="2019-07-29T17:21:18.830" LastActivityDate="2019-07-29T17:21:18.830" Title="Average number of exoplanets in a system" Tags="&lt;exoplanet&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10250" PostTypeId="2" ParentId="10249" CreationDate="2015-03-24T21:22:32.113" Score="6" Body="&lt;p&gt;Best answer is that we don't know. The techniques that are used for finding exoplanets are incomplete - that is they are capable of finding planets around stars in some circumstances but not in others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, finding stars using the transit detection technique is more sensitive to planets that are (i) big, (ii) close to their parent star and (iii) orbiting their star with an edge-on orientation from our point of view. The doppler techniques are sensitive to (i) massive exoplanets, (ii) those that are close-in to their parent star, (iii) favour detection when the line of sight lies close to the orbital plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each detection method (and each observing strategy and telescope/satellite) has their own particular &lt;em&gt;selection bias&lt;/em&gt;, but overall the picture is that the census of exoplanets is more complete for close-in massive planets, and very incomplete for planets that are either small or orbiting with periods much longer than 10 years. Then at very large separations and very long orbital periods (think 100s of years), direct imaging surveys are again capable of spotting massive exoplanets. Microlensing observations are also capable of detecting even small planets at large distances from their star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result of all this is that any figure for the &lt;em&gt;observed&lt;/em&gt; fraction of stars with exoplanets or the number of exoplanets per star is a &lt;em&gt;lower limit&lt;/em&gt;.&#xA;The exception is where researchers try to account for the various selection effects and give a figure for the fraction of stars with certain types of planet in certain types of orbit around certain types of star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment it looks like the modal number of exoplanets per star is at least 1. In other words it seems increasingly likely that most stars have exoplanets of one sort or another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some facts and figures. From Kepler transit data, &lt;a href=&quot;http://www.pnas.org/content/110/48/19273.abstract&quot; rel=&quot;nofollow&quot;&gt;Petigura et al. (2013)&lt;/a&gt; say that 22% of Sun-like stars have &quot;Earth-sized&quot; planets orbiting &lt;em&gt;in their habitable zone&lt;/em&gt; (my italics), but their plots show that the overall results for planets with orbital periods in the range 5-100 days (where Kepler is most sensitive), are that nearly 60% of stars have planets from Earth- to Jupiter- sized in this range. &lt;a href=&quot;http://arxiv.org/abs/1202.0903&quot; rel=&quot;nofollow&quot;&gt;Cassan et al. (2012)&lt;/a&gt; use microlensing survey data to claim that 17% of stars have &quot;Jupiters&quot; at 0.5-10 au from their parent star, and that Neptune-sized or &quot;super-Earth&quot; sized objects orbit a further (approximately) 50% and 60% respectively. The combination of these figures also suggests multiple planets are not rare and indeed the title of their paper is &quot;One or more bound planets per Milky Way star from microlensing observations&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The frequency of multiple planet systems is also hard to assess. All the techniques are &lt;em&gt;capable&lt;/em&gt; of detecting certain types of multiple system, but equally all are capable of missing multiple planets even when they are there.&#xA;Kepler did detect many multiple transiting systems, so they are not rare. A study by &lt;a href=&quot;http://iopscience.iop.org/1538-3881/143/4/94/&quot; rel=&quot;nofollow&quot;&gt;Tremain &amp;amp; Dong (2012)&lt;/a&gt; estimated a minimum fraction of 20-30% of stars that have one transiting planet will also have others. This is on top of the 15% of Kepler planetary systems where multiple planets can be directly observed (i.e where the orbital planes are very close together).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-03-25T20:24:50.363" LastActivityDate="2015-03-25T20:24:50.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10252" PostTypeId="1" AcceptedAnswerId="15117" CreationDate="2015-03-25T01:34:29.993" Score="2" ViewCount="6152" Body="&lt;p&gt;When we see the Milky Way on a dark night are we seeing the bulk of the galaxy, or just our local arm? How much of the milky way is visible to the naked eye on a dark night?&lt;/p&gt;&#xA;" OwnerUserId="6870" LastActivityDate="2016-07-29T23:20:22.117" Title="How much of the Milky Way is visible to the naked eye from earth?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;naked-eye&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10253" PostTypeId="2" ParentId="10252" CreationDate="2015-03-25T01:59:39.750" Score="2" Body="&lt;p&gt;An average eye can see about 4000 objects in the night sky.  The brightest are planets from the Solar system, and almost all others are stars from our galaxy, the Milky Way. Some other galaxies may be spotted, but very few.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even accounting for just half the sky, that is a very very tiny proportion of our galaxy, estimated to contain between 100 and 400 billion stars, around .000003%!  The galaxy itself being an average one among more than 100 billion galaxies in the observable universe...&lt;/p&gt;&#xA;" OwnerUserId="6871" LastEditorUserId="6871" LastEditDate="2015-03-25T02:02:04.977" LastActivityDate="2015-03-25T02:02:04.977" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10255" PostTypeId="2" ParentId="10252" CreationDate="2015-03-25T05:35:56.643" Score="1" Body="&lt;p&gt;I'm not sure exactly how to answer this, but certainly most of what you see is local.   In the southern hemisphere we see the &quot;milky Way&quot; as a kind of lighter smudge across the sky:  &lt;a href=&quot;http://s.ngm.com/2010/12/milky-way/img/milky-way-galaxy-615.jpg&quot; rel=&quot;nofollow&quot;&gt;http://s.ngm.com/2010/12/milky-way/img/milky-way-galaxy-615.jpg&lt;/a&gt;, but that's probably still only a small fraction of the Milky-way, probably mostly the closest arm.    The farthest star we can see easily from the eye (some people might be able to see further ones), but that's Deneb, about 1,550 light years away, about 2% the distance of the other side of the Milky Way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.armaghplanet.com/blog/how-far-away-is-the-farthest-star.html&quot; rel=&quot;nofollow&quot;&gt;http://www.armaghplanet.com/blog/how-far-away-is-the-farthest-star.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know where to look, you can see Andromeda, but that's cause it's a mostly unobstructed view.   &lt;a href=&quot;http://www.public.asu.edu/~rjansen/localgroup/NVWS99B_03.gif&quot; rel=&quot;nofollow&quot;&gt;http://www.public.asu.edu/~rjansen/localgroup/NVWS99B_03.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could likely see a milky way super-nova much more distant than Deneb and the Milky Way should have about 1 supernova every 50-100 years or so.   The last milky-way supernova that was seen was the Kepler supernova in 1604, about 21,000 light years away and at the time it was brighter than all the other stars in the sky briefly.  &lt;a href=&quot;http://phys.org/news/2012-09-kepler-supernovae.html&quot; rel=&quot;nofollow&quot;&gt;http://phys.org/news/2012-09-kepler-supernovae.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but however you calculate it, we only see a small percentage of the Milky Way.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-03-25T05:35:56.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10256" PostTypeId="1" AcceptedAnswerId="10258" CreationDate="2015-03-25T06:13:43.070" Score="6" ViewCount="250" Body="&lt;p&gt;How can you determine the gravitational force needed to keep a particular gas in the atmosphere of a planet (for example, carbon dioxide (CO&lt;sub&gt;2&lt;/sub&gt;))?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came across the following formula $\left(\frac{8RT}{πM}\right)^{0.5}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here: &lt;a href=&quot;http://www.tau.ac.il/~roichman/CVI/hw1/hw1.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://www.tau.ac.il/~roichman/CVI/hw1/hw1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M$ = Molecular mass &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R$ = Gas constant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T$ = Temperature&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this formula apply?&lt;/p&gt;&#xA;" OwnerUserId="6873" LastEditorUserId="2153" LastEditDate="2015-03-25T15:54:57.337" LastActivityDate="2015-03-25T15:54:57.337" Title="Gravitational pull needed to keep a gas in atmosphere" Tags="&lt;gravity&gt;&lt;astrophysics&gt;&lt;planetary-atmosphere&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10258" PostTypeId="2" ParentId="10256" CreationDate="2015-03-25T08:35:27.047" Score="5" Body="&lt;p&gt;Yes, the formula you quote applies to some extend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can also be written in terms of Boltzmann's constant $k_\mathrm{B}$ as&#xA;$$v_\mathrm{rms} = \sqrt{\frac{3k_\mathrm{B}T}{m_\mathrm{m}}},$$&#xA;with $m_\mathrm{m}$ the mass of the molecule in question,&#xA;gives the mean (&quot;root-mean-square&quot;) velocity of the gas molecules as a function of temperature $T$. Comparing this to the escape velocity&#xA;$$v_\mathrm{esc}=\sqrt{\frac{2GM_\mathrm{p}}{R_\mathrm{p}}},$$&#xA;where $M_\mathrm{p}$ and $R_\mathrm{p}$ are the mass and radius of the planet, respectively, gives an order-of-magnitude estimate of whether or not the planet will be able to maintain its atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the two velocities are equal, it would still mean that half of the atmosphere evaporated instantly, and that over time, a large fraction would disappear, although the time scale for this to happen may be very long. In order for the planet to really sustain its atmosphere, $v_\mathrm{esc}$ needs to be roughly 6 times larger than $v_\mathrm{rms}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity is not the only factor determining the stability of the atmosphere, though. Radiation pressure from the star (&quot;Solar wind&quot;) can easily rip a planet of its atmosphere. On the other hand, if the planet has an efficient magnetic field, this will shield it against the wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Atmospheric_escape&quot; rel=&quot;noreferrer&quot;&gt;Other factors&lt;/a&gt; are also involved. But you're right in that at the very least, the planet needs a certain gravitational field to keep its atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2015-03-25T08:35:27.047" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10260" PostTypeId="2" ParentId="6005" CreationDate="2015-03-25T12:31:38.183" Score="8" Body="&lt;p&gt;You can tell a lot about Galactic structure by just looking. The ~5000 stars that can be seen with the naked eye have a roughly &quot;lognormal&quot; distribution of distance. I show plots below which were generated from the most recent version of the Hipparcos parallax catalogue. Fig.1 shows results for all stars with $5.5&amp;lt;V&amp;lt;6.5$  (i.e. very faint naked eye stars). The Hipparcos catalogue is almost complete for these stars, though some stars are so far away that the distance is highly uncertain - nevertheless, the general picture should be ok.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The median distance is about 440 light years. But the premise of your question is I think, that you dispute that this is far enough for the non-spherical distribution of stars in our Galaxy to become apparent. The answer is actually that it is, but only just. The Sun is very close to the plane of the disk of our Galaxy. The scale height of stars above this disk varies depending on stellar age and mass. Very approximately, the exponential scale height is 300-500 light years for most of the stars in our Galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just small enough that if we look at the demographics of stars in two Galactic latitude regions we &lt;em&gt;do&lt;/em&gt; see a difference. Fig. 2 below shows stars with $5.5&amp;lt;V&amp;lt;6.5$ at low Galactic latitudes (within 15 degrees of the plane in green) and more than 45 degrees from the plane (in blue). There is a clear and significant difference. More stars can be seen at greater distances towards the plane, betraying the non spherically-symmetric nature of stars in our Galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words the structure of the Galaxy &lt;em&gt;is not&lt;/em&gt; too small to affect the distribution of naked eye stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we look at the Milky Way we are seeing millions of unresolved stars that are in general more distant still. The effects of the disk-like nature of the Galaxy become more apparent as one goes to greater distances. In particular, at high Galactic latitudes one simply runs out of stars and hence there is no uniformly illuminated sky - leading to Olber's paradox, as correctly stated in another answer. But at low latitudes, there are sufficient stars that in general, within the resolution of our eyes there are many stars whose light sums to provide a visual stimulus. This picture is interrupted by &lt;em&gt;dust&lt;/em&gt;. Dust in the Galaxy is even more concentrated towards the plane than the stars. It is for this reason that beyond a few thousand light years, dust plays a major role in shaping the Milky Way structures that we can see, effectively blocking light at very low Galactic latitudes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nrkJX.png&quot; alt=&quot;Distribution of stellar distances for naked eye stars&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Fig 1: Probability distribution of distance for naked eye stars in the Hipparcos catalogue&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/I5pyX.png&quot; alt=&quot;Distribution of distances for naked eye stars in two Galactic latitude zones&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Fig 2: Probability distribution of distance for naked eye stars divided into low and high Galactic latitude regions (i.e. towards and away from the Galactic plane)&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IMPORTANT EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the answer was accepted I did a bit of double checking. I split the Hipparcos sample into bright ($V&amp;lt;6$, 5000 stars) and very bright ($V&amp;lt;3$, 173 stars) and looked at the distribution of stars per unit area on the sky as a function of Galactic latitude (Fig. 3). The results are shown below and more obviously make the point. It turns out that the asymmetry between in an out of the plane is clearly seen, even in the brighter sample. The reason is that the very bright stars are not &lt;em&gt;all that much&lt;/em&gt; closer than the bright stars. Perhaps a median of 200 light years vs 350 light years. Thus the scale height of the Galactic disc must be small enough that the difference from spherical symmetry is already appearent at these kinds of distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lOpa1.png&quot; alt=&quot;Distribution of stars vs Galactic latitude&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Fig. 3: Normalised plot of number of stars per unit area versus Galactic latitude for a sample of bright and very bright stars. Note the concentration towards the Galactic plane&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FURTHER EDIT: You can even see from the plot above that there is also a slight concentration towards negative latitudes; the peak density is around -5 to -10 degrees in both cases. This is likely because the Sun is above the Galactic plane at present (though I also wonder if dust plays a role). The Sun is currently 60 light years above the plane and heading upwards (see &lt;a href=&quot;https://astronomy.stackexchange.com/questions/822/how-far-is-the-earth-sun-above-below-the-galactic-plane-and-is-it-heading-towar&quot;&gt;How far is the Earth/Sun above/below the galactic plane, and is it heading toward/away from it?&lt;/a&gt; ). The combination of the results I have shown &lt;em&gt;may&lt;/em&gt; be sufficient to conclude that the Galactic disc has a &quot;characteristic thickness* not more than a few hundred light years, but more than 60 light years!&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-03-25T23:10:34.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10262" PostTypeId="1" CreationDate="2015-03-25T22:06:10.637" Score="1" ViewCount="2413" Body="&lt;p&gt;I would like some help in finding the exact constellations, or some sort of visualization of the stars over an specific location an time. The idea is to find out the closest star or group of stars on the vertical of that position in that exact moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also aware than &quot;vertical&quot; can have some different interpretations, but a nice approximation should be good enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not an expert in physics nor astronomy, but I think providing the coordinates and the date/time should be enough for an approximation.&#xA;According to google maps, the coordinates are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;39.467062 -0.377381&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No idea what format that is, I'm familiarized with degree/minutes/seconds but I don't know exactly how to translate those. The time would be: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;March 25, 2015, 09:40 (Spain)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;, which would be &lt;strong&gt;08:40 in UTC&lt;/strong&gt;, correct me if I'm wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated, if you can't provide a direct answer, at least point me to some website form or program (Linux preferably) I can use to find this out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case someone is wandering, this involves a girl and a tattoo, so please bear in mind this is something I'm going to be carrying for the rest of my life :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6888" LastActivityDate="2015-03-27T22:32:42.687" Title="Find constellation over Earth coordinates on a specific date-time" Tags="&lt;star&gt;&lt;constellations&gt;&lt;time&gt;&lt;positional-astronomy&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10264" PostTypeId="2" ParentId="10213" CreationDate="2015-03-26T04:09:07.627" Score="-12" Body="&lt;p&gt;Cavemen were able to see the first six planets, up to Saturn. All these planets were named after the Roman and Greek mythological characters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rule they followed was something like this: The planet will be named after the Roman god and all its satellites will be named after the corresponding Greek gods. For example, Jupiter was named after the Roman King of Gods, and one of its satellites, Ganymede , was named after the man-servant of Zeus. Now Zeus is the Greek counterpart of Jupiter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, Uranus is the lone exception for this scheme. All of its satellites, Miranda, Ariel, Umbriel, Titania, and Oberon were named after Shakespearean characters. This was to please the British because, they were very powerful in that period of time and its discovery was made by William Herschel in the 1780's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Uranus was first named 'George' by Herschel because the King of England, King George, was his main source of funding and he didn't want to annoy him.     &lt;/p&gt;&#xA;" OwnerUserId="6892" LastEditorUserId="2153" LastEditDate="2015-03-26T22:16:06.540" LastActivityDate="2015-03-26T22:16:06.540" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10265" PostTypeId="1" AcceptedAnswerId="10266" CreationDate="2015-03-26T07:25:08.793" Score="1" ViewCount="420" Body="&lt;p&gt;An astronomical unit is defined as the measurement of distance between Earth and our Sun, my question is since distances between celestial objects beyond our solar system are vast and unimaginable astronomer adopted another familiar term &quot;light-year&quot; instead. My question is do people still use astronomical unit in the present days?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastActivityDate="2015-03-26T22:18:56.883" Title="Do we still use the term &quot;astronomical unit&quot; nowadays?" Tags="&lt;the-sun&gt;&lt;solar-system&gt;&lt;earth&gt;&lt;light&gt;&lt;distances&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10266" PostTypeId="2" ParentId="10265" CreationDate="2015-03-26T08:54:52.123" Score="10" Body="&lt;p&gt;Certainly. Astronomical Unit is probably one of the most used distance units used in astronomy. It is of course only used when discussing the distances within a stellar system, such as the distances between the Sun and its planets or other bodies in the solar system. It is also used to discuss distances in other stellar systems, e.g. the distances between stars and their (exo)planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easier to get a sense of the distances between bodies in the solar system if you use astronomical units. It doesn't say much when we say that the distance between Neptune and the Sun is $5\times10^{-4}$ light year. You would have to know the distance between Earth and the Sun ($1.5\times10^{-5}$ light year), and then calculate the difference (i.e. 30 times = 30AU) to get a sense of scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, in professional astronomy, the light year is hardly ever used. Professional astronomers use the parsec (=3.26 light year), kilo parsec (kpc), megaparsec (Mpc), and Gigaparsec (Gpc) to specify distances between stars, galaxies, and galaxy clusters.&lt;/p&gt;&#xA;" OwnerUserId="864" LastEditorUserId="38" LastEditDate="2015-03-26T20:38:11.837" LastActivityDate="2015-03-26T20:38:11.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10267" PostTypeId="1" AcceptedAnswerId="10269" CreationDate="2015-03-26T13:14:15.570" Score="5" ViewCount="2453" Body="&lt;p&gt;I got from &lt;a href=&quot;https://astronomy.stackexchange.com/questions/546/why-is-our-solar-system-tipped-about-63%c2%b0-with-respect-to-the-plane-of-our-gala&quot;&gt;this answer&lt;/a&gt; that the axis of the solar system is tilted of about 63 degrees in relation to that of our galaxy, so I can assume that we move through space (at least locally) roughly in a direction that is 63° tilted in relation to our solar system's plane, is that correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, then: is there also some net velocity of our solar system (probably small) in a direction that is perpendicular to the galaxy's plane, as opposed to the main direction that is &quot;parallel&quot; to the galaxy's plane?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think my question is different from &lt;a href=&quot;https://astronomy.stackexchange.com/questions/822/how-far-is-the-earth-sun-above-below-the-galactic-plane-and-is-it-heading-towar&quot;&gt;this one&lt;/a&gt; in that the latter asks about the distance from the solar system to the galactic plane, with no regards as to the speed with which this distance is changing. On the other hand, I'm interested not on the distance, but on the speed with which this distance is changing, which can promptly be approximated with respect to time with a sine function from pela's answer, which was my main intention.&lt;/p&gt;&#xA;" OwnerUserId="6661" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-03-26T23:47:42.860" Title="What is the direction of the movement of the solar system in relation to the galaxy's plane" Tags="&lt;solar-system&gt;&lt;galaxy&gt;&lt;galactic-dynamics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10268" PostTypeId="2" ParentId="9973" CreationDate="2015-03-26T13:18:40.260" Score="0" Body="&lt;p&gt;Fisheye lenses. Sure, it's doable. But does it really lead to &quot;better&quot; stargazing? It's debatable. The human field of vision is already pretty wide as it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are telescope eyepieces out there that operate on the principle of light amplification, and they do indeed raise the bar in terms of the faintest stars that can be seen in that telescope. They aren't cheap, though.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2015-03-26T13:18:40.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10269" PostTypeId="2" ParentId="10267" CreationDate="2015-03-26T13:50:27.140" Score="2" Body="&lt;p&gt;Yes indeed, the Sun (and other stars) has an oscillatory velocity perpendicular to the galactic plane. According to &lt;a href=&quot;http://www.nature.com/nature/journal/v316/n6030/abs/316706a0.html&quot; rel=&quot;nofollow&quot;&gt;this Nature article&lt;/a&gt;, the Sun crosses the galactic plane roughly every 30 million years, reaching a max height of 150-300 lightyears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on how you measure it, a more recent analysis by &lt;a href=&quot;http://arxiv.org/abs/0704.0950&quot; rel=&quot;nofollow&quot;&gt;Joshi (2007)&lt;/a&gt; finds that we are currently somewhere between 20 and 90 lightyears above the plane.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-03-26T14:05:24.150" LastActivityDate="2015-03-26T14:05:24.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10270" PostTypeId="1" AcceptedAnswerId="10282" CreationDate="2015-03-26T14:22:00.757" Score="4" ViewCount="180" Body="&lt;p&gt;It's often speculated that matter from comets played a crucial role in the development of life on Earth. For example, &lt;a href=&quot;http://smile.amazon.com/Maps-Time-Introduction-Big-History/dp/0520271440/ref=sr_1_1?ie=UTF8&amp;amp;qid=1427378942&amp;amp;sr=8-1&amp;amp;keywords=maps+of+time&quot; rel=&quot;noreferrer&quot;&gt;Maps of Time&lt;/a&gt; says,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Many of the ingredients of the early atmosphere (including much of its water), together with many of the organic chemicals that formed the first life forms, may have been brought to Earth by the comets that bombarded its surface for the first billion years of its life (~Kindle location 1922).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why the presumption that these materials must be brought by comets? Did Earth even have the right atoms to make water and those organic chemicals (hydrogen, oxygen, carbon)? And what are comets doing with heavier elements in the first place? My understanding is that comets are (for most of their lives) very far from the run, and my impression was most of the elements other than hydrogen and helium and are concentrated close to the sun, with T Tauri Wind having push hydrogen and helium further away.&lt;/p&gt;&#xA;" OwnerUserId="6587" LastActivityDate="2015-03-30T15:39:40.817" Title="Why did Earth need material from comets to assemble the raw materials for life?" Tags="&lt;solar-system&gt;&lt;comets&gt;&lt;life&gt;&lt;origin-of-life&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10271" PostTypeId="1" AcceptedAnswerId="10303" CreationDate="2015-03-26T18:54:41.997" Score="1" ViewCount="321" Body="&lt;p&gt;I know that some systems tend to tidal locking (such as earth-moon), which occurs basically because the difference in the gravitational pull on one side is significantly different from the pull on the opposite side. (There's a lot of great answers &lt;a href=&quot;https://astronomy.stackexchange.com/questions/16/why-is-only-one-side-of-the-moon-visible-from-earth&quot;&gt;here&lt;/a&gt;.) But what can be considered a significant different?&#xA;In other words, is there a threshold on distance or size in order for systems to tend to tidal lock?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6661" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-03-30T03:53:45.093" Title="Is there a threshold on distance/size for a tidal locking?" Tags="&lt;the-moon&gt;&lt;orbit&gt;&lt;gravity&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10272" PostTypeId="2" ParentId="10262" CreationDate="2015-03-26T19:10:39.927" Score="0" Body="&lt;p&gt;There's a couple of celestial coordinate systems you should learn about, which are the most commonly used:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Horizontal coordinate system, observer-centered.&lt;/li&gt;&#xA;&lt;li&gt;Equatorial coordinate system, earth-centered.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Depending on the system you use, the coordinates of your celestial object will be represented in one way or the other and should be read accordingly.&lt;/p&gt;&#xA;" OwnerUserId="6898" LastEditorUserId="2153" LastEditDate="2015-03-26T22:14:33.900" LastActivityDate="2015-03-26T22:14:33.900" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10273" PostTypeId="2" ParentId="10265" CreationDate="2015-03-26T19:14:54.137" Score="2" Body="&lt;p&gt;I guess the use of one unit over the other (AUs, parsecs,lightyears, etc.) will depend mainly on the distances of the object under study. If you work with Solar System objects, it will be easier to use AUs, whereas if you work with Galaxies, AUs are not much use and you'd probably work better with lightyears.&lt;/p&gt;&#xA;" OwnerUserId="6898" LastActivityDate="2015-03-26T19:14:54.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10274" PostTypeId="2" ParentId="10265" CreationDate="2015-03-26T22:18:56.883" Score="1" Body="&lt;p&gt;I want to expand on the use of an AU in comparing planetary systems a bit.&#xA;When we look for example at this image of the (instantly famous) protoplanetary disc imaged last year in HL Tau:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/gIRHi.png&quot; alt=&quot;HL Tau: Planetary formation in action&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In observational astronomy, when we look at an object at a distance $d$ and it has a certain angular size $a$, thanx to the use of the century old &lt;a href=&quot;http://en.wikipedia.org/wiki/Parallax&quot; rel=&quot;nofollow noreferrer&quot;&gt;parsec&lt;/a&gt; we know the objects size $s$ in AU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you make the image of an object, you always know $a$. This is simply the apparent size on the celestial sphere in arcseconds. &#xA;Nowadays we also know the distance to objects. So by use of the fundamental relation &quot;angle = size / distance&quot;, we always can have $s(AU) = a(arcseconds) * d(pc)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This way any astronomer can quickly convert the apparent angular size into physical size and immediately tell about the dimensions of his object!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check this method, we can know that the HL Tau is 450Lj away, which is 140pc and if you now look at the scale given in the image, this will be vagely familiar ;)&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2015-03-26T22:18:56.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10275" PostTypeId="2" ParentId="10267" CreationDate="2015-03-26T23:39:00.320" Score="3" Body="&lt;p&gt;I think your question is answered by the duplicate mentioned: but here are the relevant highlights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/full/1995AJ....110.2183H&quot; rel=&quot;nofollow noreferrer&quot;&gt;Humphreys &amp;amp; Larsen (1995)&lt;/a&gt; suggest, using star count information, a distance of $20.5 \pm 3.5$ pc &lt;strong&gt;above the Galactic plane&lt;/strong&gt;; consistent with, but more precise than the Bahcall paper referred to by Schleis. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2007MNRAS.378..768J&quot; rel=&quot;nofollow noreferrer&quot;&gt;Joshi (2007)&lt;/a&gt; is more guarded, investigating some systematic uncertainties in the estimation techniques and ends up with distances between &lt;strong&gt;13 and 20 pc above the plane&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun moves at about 15-20 km/s with respect to a local standard of rest defined by the general motion of stars in our vicinity around the Galaxy. In three-dimensions, this &quot;peculiar velocity&quot; is $U=10.00 \pm 0.36$ km/s (radially inwards), $V=5.25 \pm 0.62$ km/s (in the direction of Galactic rotation) and &lt;strong&gt;$W=7.17 \pm 0.38$ km/s&lt;/strong&gt; (&lt;strong&gt;up and out of the plane&lt;/strong&gt;).  &lt;a href=&quot;http://adsabs.harvard.edu/abs/1998MNRAS.298..387D&quot; rel=&quot;nofollow noreferrer&quot;&gt;(Dehnen &amp;amp; Binney 1998)&lt;/a&gt;. Different authors arrive at velocities that differ by $\sim 1-2$ km/s from these values and so this would probably be a more conservative estimate of the uncertainties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun executes oscillations around its mean orbit in the Galaxy, periodically crossing the Galactic plane. I borrowed this illustration (not to scale!) from &lt;a href=&quot;http://www.visioninconsciousness.org/Science_B08.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.visioninconsciousness.org/Science_B08.htm&lt;/a&gt; to show this oscillatory motion. The oscillations will not be exactly sinusoidal because the restoring force towards the plane does not very linearly with height above the plane.&#xA;As the Sun is currently above the plane and moving upwards, and each cycle takes about 70 million years with an amplitude of 100pc (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1995Icar..116..255M&quot; rel=&quot;nofollow noreferrer&quot;&gt;Matese et al. 1995&lt;/a&gt;), it will be roughly 30 million years before we cross the plane again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Actually I'm glad I revisited this question because I think the picture is not very good at all. As the Sun takes ~230 million years to go around the Gaaxy, it should only execute 3 complete vertical oscillations in our Galactic year, whereas the picture implies many more. Secondly, the Sun executes a radial oscillation with a period of around 160 million years, which is not even indicated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dv1Ii.jpg&quot; alt=&quot;Sun&amp;#39;s motion around the Galaxy&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-03-26T23:47:42.860" LastActivityDate="2015-03-26T23:47:42.860" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10276" PostTypeId="2" ParentId="10262" CreationDate="2015-03-26T23:59:51.633" Score="1" Body="&lt;p&gt;Ahh got it. The coordinates are given as latitude, longitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately, the longitude hardly matters (depending what precision you need), but the latitude does. A star at the zenith will have a declination equal to the latitude on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as Right Ascension goes, well around the March equinox then stars with 12 hours in Right Ascension will reach their highest point (culmination) at &quot;local midnight&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one would expect roughly speaking for stars at around 21 hours, +40 would be overhead at 9h40m later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Checking out Stellarium for that location and time it seems that this is pretty much spot on and that &lt;strong&gt;Cygnus (the Swan) is overhead&lt;/strong&gt;, with the bright star &lt;strong&gt;Deneb&lt;/strong&gt; very close to being at Zenith.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-03-27T00:13:51.223" LastActivityDate="2015-03-27T00:13:51.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10277" PostTypeId="1" CreationDate="2015-03-27T02:17:57.027" Score="11" ViewCount="1929" Body="&lt;p&gt;I recently read that water more than likely got here by comets (carrying water) hitting the Earth. However it also it says the impact of a comet hitting the Earth is much greater than an atomic bomb. So if comets brought water to Earth how was it not vaporized when the comet hit the atmosphere or in the massive explosion when it hit the Earth itself?&lt;/p&gt;&#xA;" OwnerUserId="6901" LastEditorUserId="2365" LastEditDate="2017-03-16T06:22:22.160" LastActivityDate="2017-03-16T18:39:10.893" Title="How did water get on Earth" Tags="&lt;earth&gt;&lt;comets&gt;&lt;water&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10279" PostTypeId="1" CreationDate="2015-03-27T07:50:50.633" Score="6" ViewCount="334" Body="&lt;p&gt;Since the surface pressure of a planet is determined by the mass of the column of gasses above it one would surmise that to determine the pressure you must know the volume and mass of the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing the composition and volume of the gasses in the atmosphere would allow you to calculate their mass but how do we determine what that volume is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming we are only concerned with initial conditions when the primary atmosphere (created through accretion) was formed would the volume be arbitrary?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this isn't too broad of a question. I'm trying to pin down a possible initial state that would be created when a new planet is formed and the conditions stabilize.&lt;/p&gt;&#xA;" OwnerUserId="6873" LastActivityDate="2016-01-31T04:44:56.803" Title="How can you determine the initial volume of a planet's atmosphere?" Tags="&lt;planetary-atmosphere&gt;&lt;planetary-formation&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10280" PostTypeId="2" ParentId="8460" CreationDate="2015-03-27T10:24:38.503" Score="3" Body="&lt;p&gt;I'm not an astronomer, just an enthusiast, but I believe the only way that the sun could become a black hole is if when the Andromeda galaxy and the Milky Way galaxy collide, if our star combines with another star and the mass of the two is great enough to create a black hole then it's possible; however, from what I've read, despite the enormous size of galaxies and the absurd number of stars within them, because the planets and stars (especially in the outer edges of the galaxy) are so far away that collisions are actually very unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some sources of information (none talk about the black hole scenario though):&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=2WEI8WBJkKk&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=2WEI8WBJkKk&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=7uiv6tKtoKg&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=7uiv6tKtoKg&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.space.com/15947-milky-andromeda-galaxies-collision-simulated-video.html&quot; rel=&quot;nofollow&quot;&gt;http://www.space.com/15947-milky-andromeda-galaxies-collision-simulated-video.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6909" LastEditorUserId="6909" LastEditDate="2015-03-28T19:36:49.860" LastActivityDate="2015-03-28T19:36:49.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10281" PostTypeId="2" ParentId="10277" CreationDate="2015-03-27T10:37:43.547" Score="3" Body="&lt;p&gt;I don't believe there is factual answer just yet, though many solid theories. Here are some sources I've found to be helpful. I don't necessarily have an opinion myself and while certainly everyone is entitled to one, I wouldn't put too much stock in any one source of information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was &quot;always&quot; here:&#xA;&lt;a href=&quot;http://news.nationalgeographic.com/news/2014/10/141030-starstruck-earth-water-origin-vesta-science/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://news.nationalgeographic.com/news/2014/10/141030-starstruck-earth-water-origin-vesta-science/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comets:&#xA;&lt;a href=&quot;http://space.com/27969-earth-water-from-asteroids-not-comets.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://space.com/27969-earth-water-from-asteroids-not-comets.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meteors:&#xA;&lt;a href=&quot;http://youtube&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://youtube&lt;/a&gt; .com/watch?v=_LpgBvEPozk&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ps: had to break the links due to reputation. :/&lt;/p&gt;&#xA;" OwnerUserId="6909" LastEditorUserId="11264" LastEditDate="2017-03-16T18:39:10.893" LastActivityDate="2017-03-16T18:39:10.893" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10282" PostTypeId="2" ParentId="10270" CreationDate="2015-03-27T10:44:35.253" Score="3" Body="&lt;p&gt;The earth in its early stages was far too hot for water to be in liquid form and the water vapor that would have been around at the time would have been blasted away via solar wind (&lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_wind&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Solar_wind&lt;/a&gt;). Anyone claiming to know exactly how water came to be on Earth is likely daft, but comets and asteroids seem to be the most common theory for the source. MinuteEarth, a YouTube creator has a great video about it but as a new user on this particular sub-site I cannot link multiple things at once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=_LpgBvEPozk&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=_LpgBvEPozk&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6909" LastEditorUserId="6909" LastEditDate="2015-03-30T15:39:40.817" LastActivityDate="2015-03-30T15:39:40.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10283" PostTypeId="2" ParentId="10277" CreationDate="2015-03-27T12:46:08.170" Score="10" Body="&lt;p&gt;The key finding why we think Earth's water came from Asteroids (big rocks) and not Comets (small rocks) is the Deuterium/Hydrogen ratio that we can measure in several sources.&lt;br&gt;&#xA;When a star forms, it has an initial value of D/H that came from the nucleosynthesis in its progenitor nebular / star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a protoplanetary disc, as dust grows to rocks grow to planets your nebular gas will be trapped with initial D/H ratios in gas giant atmospheres. But the way to get Water onto Asteroids and Comets (those don't have substantial mass to retain atmospheres themselves!) is sublimation and maybe adsorption.&lt;br&gt;&#xA;The latter two processes are strongly sensitive to the gas-mass and therefore different D/H ratios from the protosolar ones are expected.&#xA;And in fact we find them to be different:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cmT0G.jpg&quot; alt=&quot;D/H ratios from various sources in the solar system&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was in the news lately, as ESA managed to touch down and measure 67P's D/H ratio which gave another hint on the asteroidic origin of water on earth.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This finding, however does not resolve the question: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There maybe other isotopical tracers like D/H that give a hint on the history of water.&lt;/li&gt;&#xA;&lt;li&gt;There could have been a decrease in D/H and subsequential increase again, or the other way round.&lt;/li&gt;&#xA;&lt;li&gt;We know that the protosolar nebula must have had massive amounts of water (Oxygen is the third-most abundand element in normal stellar fusion!), so again the question is, why did Earth retain so much water, while the &lt;em&gt;lighter&lt;/em&gt; planets Venus and Mars retained the &lt;em&gt;heavier&lt;/em&gt; element $CO_2$&#xA;...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I could continue this for quite some time, but the bottom line is: We only have hints, not definitive answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To refer to the rest of your question: A hot plume in a colder atmosphere will not necesarily leave the whole gravitational potential well.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2015-03-27T12:46:08.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10284" PostTypeId="1" CreationDate="2015-03-27T16:52:24.143" Score="1" ViewCount="108" Body="&lt;p&gt;If an Asteroid, the size of the one that killed off the dinosaurs , passed by very close to the Earth but still remained outside of Earth's atmosphere, would it cause any damage ?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastActivityDate="2016-03-18T20:16:46.507" Title="damage from an Asteroid" Tags="&lt;asteroids&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10285" PostTypeId="2" ParentId="10237" CreationDate="2015-03-27T23:45:08.247" Score="0" Body="&lt;p&gt;This is the reply I got from the US Naval Observatory itself:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;No, we do not have that transformation in NOVAS.  Generally, altitude and azimuth are not determined to very high accuracy, so using them to obtain RA and Dec, which are much more precisely defined and measured, does not make a lot of sense.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6840" LastActivityDate="2015-03-27T23:45:08.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10288" PostTypeId="1" AcceptedAnswerId="10601" CreationDate="2015-03-28T05:09:07.017" Score="5" ViewCount="591" Body="&lt;p&gt;If we managed to survive in a black hole and move inside the event horizon then could we see the surroundings of the black hole inside the event horizon by source of light? Can the light not come up to the event horizon or does it have to travel through different spheres of the black hole and enter a sphere which won't allow the light to return to its previous sphere? If the we reach up to the singularity then can we ever reach up to the event horizon or would we get stuck in the point of the singularity?&lt;/p&gt;&#xA;" OwnerUserId="6916" LastEditorUserId="2153" LastEditDate="2015-03-28T21:04:45.277" LastActivityDate="2015-06-01T07:49:31.827" Title="Could someone see anything while being inside black hole?" Tags="&lt;black-hole&gt;&lt;light&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10289" PostTypeId="2" ParentId="10288" CreationDate="2015-03-28T05:30:16.780" Score="0" Body="&lt;p&gt;Sort of. There are several videos that I've watched on YouTube that explain this phenomenon but it's all theoretical obviously. The overly simplified story is that as you approach the the black hole the comes a point that the gravity is neither too strong to suck light into the event horizon (the point that light cannot escape the gravity of the black hole) nor weak enough to allow it to leave and what you end up seeing is a warped version of what entered the black hole. Vsauce does a pretty good job of explaining some of this in this video at around 3:30. Visual of a possibility of what the inside of a blackhole of at 6:00 &lt;a href=&quot;https://www.youtube.com/watch?v=3pAnRKD4raY&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=3pAnRKD4raY&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another visualization: &lt;a href=&quot;https://www.youtube.com/watch?v=GYKyt3C0oT4&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=GYKyt3C0oT4&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6909" LastEditorUserId="6909" LastEditDate="2015-03-28T08:41:14.053" LastActivityDate="2015-03-28T08:41:14.053" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10290" PostTypeId="1" CreationDate="2015-03-28T09:22:49.020" Score="1" ViewCount="112" Body="&lt;p&gt;Does the sun have a feature or prominence like the red spot on Jupiter? Is there a feature that will remain as long as the red spot?  Maybe beneath the surface?&lt;/p&gt;&#xA;" OwnerUserId="6750" LastActivityDate="2015-03-28T09:48:12.073" Title="Does the sun have a feature like the red spot on Jupiter?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;solar-system&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10291" PostTypeId="2" ParentId="10290" CreationDate="2015-03-28T09:48:12.073" Score="4" Body="&lt;p&gt;The Sun has lots of features &quot;like the red spot&quot;, but they are dissimilar too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarities: The Sun's photosphere - the bit we can see - is entirely gaseous; the photosphere rotates differentially with solar latitude; the gas is turbulent. There are features that can be seen quite easily - these are the dark magnetic sunspots, typically of size a few thousand km; and the pattern of bright points surrounded by dark lanes, known as granulation, typical sizes of order 100 km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These features are associated with the convective upwelling of hotter material from below. In sunspots that process is inhibited by magnetic fields so that the surface temperature is lower and that area looks dark in comparison to the rest of the photosphere. Granulation is the visible appearance of the tops of the upwelling convective cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, these phenomena are short-lived compared with the GRS on Jupiter. Sunspots have lifetimes of maybe a month or two and then the magnetic fields are reordered by the Sun's differential rotation and sheared apart. Granulation is ever present but the detailed pattern shifts on timescales of hours as different convective cells rise and fall.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-03-28T09:48:12.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10292" PostTypeId="5" CreationDate="2015-03-28T10:45:57.847" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-03-28T10:45:57.847" LastActivityDate="2015-03-28T10:45:57.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10293" PostTypeId="4" CreationDate="2015-03-28T10:45:57.847" Score="0" Body="Questions about computer programs that assist modeling and observing the dynamics of astronomy." LastEditorUserId="-1" LastEditDate="2015-04-06T10:19:22.560" LastActivityDate="2015-04-06T10:19:22.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10294" PostTypeId="1" AcceptedAnswerId="10305" CreationDate="2015-03-28T15:41:44.033" Score="3" ViewCount="2643" Body="&lt;p&gt;Stars can easily fuse atoms to give of heat and radiation. But at Wikipedia it said that only sub-iron atoms give of energy when fused and take energy when split, and post-iron atoms is the exact opposite. So, if enough heavy elements got together could a &quot;reverse star&quot; be made in which it gives of light and heat through means of fission? How long would such a star last, and why aren't they common, if they even exist?&lt;/p&gt;&#xA;" OwnerUserId="5533" LastEditorUserId="2153" LastEditDate="2015-03-28T20:35:01.227" LastActivityDate="2015-03-30T07:10:00.607" Title="Is a star powered by fission possible?" Tags="&lt;star&gt;&lt;nucleosynthesis&gt;&lt;elemental-abundances&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10295" PostTypeId="1" CreationDate="2015-03-28T18:53:43.713" Score="4" ViewCount="10468" Body="&lt;p&gt;By lookback time distance I mean the same as Light travel distance.&#xA;Can you calculate it by just knowing its redshift?&lt;/p&gt;&#xA;" OwnerUserId="6920" LastEditorUserId="6920" LastEditDate="2015-03-28T21:16:26.673" LastActivityDate="2019-05-18T02:21:32.957" Title="How do you calculate the lookback time distance to a given galaxy?" Tags="&lt;distances&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10296" PostTypeId="2" ParentId="10295" CreationDate="2015-03-28T22:58:41.490" Score="4" Body="&lt;p&gt;No you can't. Other information is required. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For low redshifts - let's say smaller than 0.1 - and by that I mean the wavelength increases by 10 percent, you might get away with using Hubble's law to estimate the distance and then get the look back time by dividing by the speed of light.&#xA;$$ t \simeq \frac{\lambda - \lambda_0}{H_0 \lambda_0},$$&#xA;where $H_0$ is the present day Hubble parameter of about 70 km/s per Mpc, $\lambda$ is the measured wavelength and $\lambda_0$ is the rest wavelength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far so good, you just need to know $H_0$. However for larger redshifts it gets horribly complicated because the Hubble parameter changes with time in a way that depends on the curvature of the Universe and hence on the cosmological parameters defining the matter density and dark energy density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is indeed a highly complicated, non- linear formula involving integrals that I will look up a reference for. But possibly the better way to proceed is to use a simple look-up table produced from such calculations with certain assumed values for ratios of matter and dark energy densities with respect to the critical density; a.k.a. $\Omega_M$ and $\Omega_{\Lambda}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plot below is an example taken from &lt;a href=&quot;http://www.astro.caltech.edu/~eran/MATLAB/Cosmology.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.astro.caltech.edu/~eran/MATLAB/Cosmology.html&lt;/a&gt; which shows look back time versus redshift for two different cosmologial models (but with the same value of $H_0$). The curves are very different at high redshifts, but converge at small redshifts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;http://faraday.uwyo.edu/~chip/misc/Cosmo2/cosmo.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;cosmological calculator&lt;/a&gt; that can do the job for you. Enter the redshift and your assumptions about the Hubble parameter and other cosmological parameter and it will tell you the age of the universe at that redshift as well as the lookback time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JVMNQ.jpg&quot; alt=&quot;Lookback time versus redshift&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-03-29T12:09:22.800" LastActivityDate="2015-03-29T12:09:22.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10297" PostTypeId="1" CreationDate="2015-03-28T23:07:58.440" Score="2" ViewCount="250" Body="&lt;p&gt;Today I heard on the radio that the Super Collider has discovered mini black holes. They said that this discovery is proof of the existence of Parallel Universes and that the &quot;Big Bang&quot; never happened.I don't understand this and why this discovery means that Parallel Universe exist ? Does anyone know ?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastActivityDate="2018-09-03T10:47:44.387" Title="Proof that Parallel Universes exist" Tags="&lt;black-hole&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10298" PostTypeId="1" CreationDate="2015-03-29T01:49:53.803" Score="1" ViewCount="609" Body="&lt;p&gt;I live around Houston, Texas and I'm taking advantage of Earth hour to look at the stars. I've found two objects that I think might be Mars. Both of them are in roughly the same plane as Venus and Jupiter. Both are closer to Venus than Jupiter. One of them has a reddish tint and is higher in the sky than Venus, and the other is closer to the horizon than venus. It is 8:50 as I type this. Is one of these Mars?&lt;/p&gt;&#xA;" OwnerUserId="6924" LastActivityDate="2015-03-30T09:46:19.220" Title="Where is Mars in the night sky" Tags="&lt;planet&gt;&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10299" PostTypeId="2" ParentId="10294" CreationDate="2015-03-29T15:28:58.447" Score="2" Body="&lt;p&gt;It is very unlikely with the normal fission process for most of the elements. They can be divided into 2 groups: slow reaction and fast reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elements with slow reactions do not generate enough energy in a short enough time to be able to heat sufficiently to provide light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elements with fast reactions would disappear before enough accumulated to form a large body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also a fusion chain reaction is not likely since the combination of material to maintain a self sustained reaction for thousands of years is very remote.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2015-03-29T15:28:58.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10300" PostTypeId="2" ParentId="10295" CreationDate="2015-03-29T18:31:08.703" Score="1" Body="&lt;p&gt;@RobJeffries has covered most points, but just to add a little. The lookback distance is directly proportional to the cosmological time elapsed since emission, whereas the cosmological red-shift depends on [basically] how much scale factor, which is a function of cosmological time, has changed since emission. If the Universe has been expanding for its history up to the present then the scale factor to present will be a strictly increasing function of cosmological time, so there will be a one-to-one correspondence between red-shift and lookback distance. NB this will not be true if, for example, the Universe was initially expanding but is currently contracting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However how the scale factor changes is governed by the stress-energy of the contents of the Universe so the relationship between red-shift and lookback distance will not generally be simple.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastActivityDate="2015-03-29T18:31:08.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10301" PostTypeId="1" AcceptedAnswerId="18071" CreationDate="2015-03-29T19:36:00.177" Score="4" ViewCount="575" Body="&lt;p&gt;I was thinking about the possibility of determining what phase of evolution a variable star is in when it is located in the instability strip (say, a Cepheid object). Can you differentiate whether it is evolving toward the blue (in HR diagram) or back toward the giant branch?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was looking at a figure recently in Carroll and Ostlie's Intro to Astrophysics text, which brought this to mind. It shows on the HR diagram where Cepheids lie, which is right in this strip.&lt;/p&gt;&#xA;" OwnerUserId="6928" LastEditorUserId="31410" LastEditDate="2021-05-27T13:56:41.183" LastActivityDate="2021-05-27T13:56:41.183" Title="Variables in the Instability Strip" Tags="&lt;stellar-evolution&gt;&lt;hr-diagram&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10302" PostTypeId="2" ParentId="10297" CreationDate="2015-03-29T22:09:41.407" Score="3" Body="&lt;p&gt;Your radio information source is wrong.&#xA;The Large Hadron Collider has &lt;strong&gt;not&lt;/strong&gt; discovered mini black holes.&#xA;When it comes back on line this spring, the LHC will begin looking for mini black holes: &lt;a href=&quot;http://www.sciencetimes.com/articles/4928/20150327/large-hadron-collider-prove-existence-star-treks-parallel-universe.htm&quot; rel=&quot;nofollow&quot;&gt;Large Hadron Collider Could Prove the Existence of Star Trek's Parallel Universe&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Journalists &lt;em&gt;do&lt;/em&gt; like their headlines, but the gist is this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When the Large Hadron Collider is brought back online in the spring, researchers will be looking for the existence of mini black holes. These mini black holes would lend support to string theory, which posits that different dimensions and parallel universes are possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No miniholes yet, maybe never.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-03-29T22:09:41.407" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10303" PostTypeId="2" ParentId="10271" CreationDate="2015-03-30T03:53:45.093" Score="0" Body="&lt;p&gt;You can see an example of tidal locking and atmosphere simulation for a planet closely orbiting a dim star. They show a simulation of the atmosphere and some interesting theories about the movement of gasses due to tidal locking (convection) that occurs between the bright and dark side of the planet.  The link goes directly to the discussion of tidal locking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/5HZsFMqqGJo?t=33m13s&quot; rel=&quot;nofollow noreferrer&quot;&gt;KEPLER 186F - LIFE AFTER EARTH&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar question was posted in the physics forum and it looks like that could be close to the answer you are looking for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/12541/tidal-lock-radius-in-habitable-zones&quot;&gt;Tidal Lock Radius in Habitable Zones&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also interested in tidal locking and hope someone may post a more concise answer.&lt;/p&gt;&#xA;" OwnerUserId="6873" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-03-30T03:53:45.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10305" PostTypeId="2" ParentId="10294" CreationDate="2015-03-30T07:10:00.607" Score="5" Body="&lt;p&gt;I think it's an interesting question.    The trick would be a sustained fission reaction, faster than half life, but slower than a chain reaction.   A chain reaction could hardly be considered &quot;star like&quot; - it would just explode.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say you had a planet sized object, maybe 100 parts Iron, basically inert, to 1 part Uranium  , which would generate heat through half life decay and some neutrons would spread from uranium to uranium - many wouldn't.  And as the object got hotter, more and more uranium would drift towards the center.  You'd have gradual heating, and the object, in time, would probobly begin to glow.     I'm guessing it's not possible using anything close to pure uranium, as the neutrons released in a uranium rich ball of matter would do 1 of 2 things - either blow the thing apart, or increase the rate of reaction and cause a chain reaction.   But i think, with the proper mix of uranium and inert elements, and the uranium spread out, not all in the center, you probobly could get something sort of star-like.   It would have much less available energy so it wouldn't burn nearly as long as a star.    At least, that's my thought experiment answer to this question.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with nothing but heavy, neutron rich elements is that, I don't see any way that wouldn't cascade and burn or blow up quickly.   The nuclear reaction is driven by neutrons hitting the nucleus with potential energy and as the reaction happens, more free Neutrons are produced which speeds up the reaction significantly.   A ball of of Uranium many miles across would undergo fission and explode.   A ball heavy enough to not explode - say, jupiter sized, it would heat up and burn quickly and then settle down and cool off slowly. - perhaps similarish to a star going Nova then settling into a white dwarf.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a material inside that slows the reaction, absorbs and re-admits Neutrons you could probobly generate a sustained burn - similar to what happens inside a nuclear plant while maintaining enough gravity to keep the object in one piece and over time, That would heat up and glow like a star in time and last for a while.   Planet sized, even moon sized would probobly be more than sufficient, but you'd need the material to slow the reaction and keep the Uranium spread out.   That's my take at least.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-03-30T07:10:00.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10306" PostTypeId="1" AcceptedAnswerId="10315" CreationDate="2015-03-30T04:49:01.950" Score="2" ViewCount="496" Body="&lt;p&gt;Someday before i read about a article which explained about a phenomenon called BIG CRUNCH, according to it the way Big Bang happened because of which the remains moved away from one another forming universe planets and all other celestial bodies. In the same way some day the gravitational pull caused by each of these celestial bodies will drag each other celestial bodies towards it ending the universe by making it one, is this true can this happen? &lt;/p&gt;&#xA;" OwnerUserId="6930" OwnerDisplayName="Akshay Vasu" LastActivityDate="2015-03-30T18:01:19.253" Title="BIG CRUNCH Theory" Tags="&lt;planet&gt;&lt;space&gt;&lt;universe&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10307" PostTypeId="2" ParentId="10306" CreationDate="2015-03-30T06:49:17.017" Score="-3" Body="&lt;p&gt;It is one of the three theory's they have on this moment about possible ends for the universe. So we can't be sure, they can't really test it yet and there is not enough evidence to mark it as true. It is just a logical answer to a question, but still only a possibility. &lt;/p&gt;&#xA;" OwnerDisplayName="Robin" LastActivityDate="2015-03-30T06:49:17.017" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10308" PostTypeId="2" ParentId="10298" CreationDate="2015-03-30T08:54:00.083" Score="1" Body="&lt;p&gt;You will be able to recognize Mars easily from its red color. Mars looks like a red tiny dot in the night sky.&#xA;You can always use apps such as Google sky map. You simply point your phone in the direction of the red dot and your phone will identify the object for you. &lt;/p&gt;&#xA;" OwnerUserId="1963" LastActivityDate="2015-03-30T08:54:00.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10309" PostTypeId="2" ParentId="10298" CreationDate="2015-03-30T09:46:19.220" Score="2" Body="&lt;p&gt;At that time (20:50), and that location (Houston, Tx), Mars was closer to the horizon.&#xA;You can see that in the &lt;a href=&quot;http://heavens-above.com/wholeskychart.ashx?lat=29.7604&amp;amp;lng=-95.3698&amp;amp;loc=Houston&amp;amp;alt=15&amp;amp;tz=CST&amp;amp;size=800&amp;amp;SL=1&amp;amp;SN=1&amp;amp;BW=1&amp;amp;time=57101.07639&amp;amp;ecl=0&amp;amp;cb=0&quot; rel=&quot;nofollow noreferrer&quot;&gt;sky-chart provided&lt;/a&gt; by &lt;a href=&quot;http://heavens-above.com/skychart.aspx?lat=29.7604&amp;amp;lng=-95.3698&amp;amp;loc=Houston&amp;amp;alt=15&amp;amp;tz=CST&quot; rel=&quot;nofollow noreferrer&quot;&gt;heavens-above.com&lt;/a&gt; (attached below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot tell what's the other object you describe, but if you are not sure whether something is a planet or a star, the planets doesn't twinkle whether &lt;a href=&quot;https://astronomy.stackexchange.com/questions/222/why-do-stars-appear-to-twinkle&quot;&gt;the star does&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sdV50.png&quot; alt=&quot;Sky-chart for Houston at 2015-03-29T20:50:00&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="137" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-03-30T09:46:19.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10311" PostTypeId="1" CreationDate="2015-03-30T14:00:17.333" Score="2" ViewCount="125" Body="&lt;p&gt;are all the planet in solar system the same plane in the space or not ? if yes is there any especial reason ? and if no how do they orbits then ?&#xA;as far as i know they are in the same plane , but is that mean they are exactly in same plane or its some thing near that ?&#xA;I would also like to know if they all orbit in the same direction ?&#xA;Is any of the picture below describing them true ?!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/F2VAc.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eGZ5G.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6937" LastActivityDate="2015-03-30T14:10:58.213" Title="are all the planet in the same plane?" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;planet&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2015-03-30T17:43:28.107" ContentLicense="CC BY-SA 3.0" />
  <row Id="10314" PostTypeId="2" ParentId="10306" CreationDate="2015-03-30T16:53:16.220" Score="0" Body="&lt;p&gt;To answer your question: yes it can happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three major theories exist: big crunch, big freeze (or heat death), big rip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Big crunch means everything is eventually crushed into a single point. Big freeze is that things get so far appart that &quot;nothing&quot; can happen anymore (except quantum stuff but that's another story), the universe becomes the most boring place in the ...hmm universe... Big rip means that eventually the fabric of time and space stretches so much that everything is torn appart (absolutely everything) in a rather short instant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To simplify, they really depends on a ratio being smaller than 1, equal to 1 or greater than 1. Right now, the measurement of this ratio is that it's very close to 1... + or - a small quantity. But this small quantity makes all the difference. So until we can refine this measurement, we don't know. It's like keeping a pencil on its end: whether it falls left or right can depend on a very subtle initial imbalance. Right now, we say &quot;it's sort of in the middle&quot; which is not very useful!&lt;/p&gt;&#xA;" OwnerUserId="6604" LastActivityDate="2015-03-30T16:53:16.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10315" PostTypeId="2" ParentId="10306" CreationDate="2015-03-30T18:01:19.253" Score="0" Body="&lt;p&gt;The 3 major theories are Big Crunch, Big Freeze, and Big Rip. Each of them have&#xA; reasons why they work and don't work. The future will shed some light on this, but for now it's just fun to talk about!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Big Crunch: Poetry of sort, the universe started with a big bang and will end collapsing into the same &quot;infinitely&quot; small point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Big Freeze: Everything is moving away from everything else. Eventually things will be so far away that as stars blow up, reform, and the sources of hydrogen and other things that provide stars their fuel become more rare, things will eventually become colder and colder until there's nothing but ice cubes with no scotch to put them in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Big Rip: I don't know much about this one, mostly because it doesn't seem to be likely from what I've heard others say. I'd suggest reading up on it on the wiki article (&lt;a href=&quot;http://en.wikipedia.org/wiki/Big_Rip&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Big_Rip&lt;/a&gt;) or googling it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see an episode from National Geographic that talks about these 3 theories here: &lt;a href=&quot;https://www.youtube.com/watch?v=eYc8iRyL7YU&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=eYc8iRyL7YU&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A short article on the Big Crunch: &lt;a href=&quot;http://www.universetoday.com/37018/big-crunch/&quot; rel=&quot;nofollow&quot;&gt;http://www.universetoday.com/37018/big-crunch/&lt;/a&gt;&#xA;A short article about Big Rip (includes some talk of the math that Godzy referenced): &lt;a href=&quot;http://www.universetoday.com/22382/no-big-rip-in-our-future-chandra-provides-insights-into-dark-energy/&quot; rel=&quot;nofollow&quot;&gt;http://www.universetoday.com/22382/no-big-rip-in-our-future-chandra-provides-insights-into-dark-energy/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6909" LastActivityDate="2015-03-30T18:01:19.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10316" PostTypeId="1" AcceptedAnswerId="10325" CreationDate="2015-03-30T19:16:05.960" Score="16" ViewCount="2809" Body="&lt;p&gt;For stars we see that are burning fuel at a fast rate, that are very bright, (or any star for that matter sufficiently far away from us) could it be that they are already black holes (the sufficiently large ones) and we are just seeing the light emitted during it's main phase? So, if so, is it safe to say that most of the stars are already dead that we see?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I appreciate all the answers. This has been a fascinating read. I also hate having to select a &quot;the answer&quot; when I gleaned something from all of them.&lt;/p&gt;&#xA;" OwnerUserId="1468" LastEditorUserId="1468" LastEditDate="2015-04-02T00:03:19.410" LastActivityDate="2016-09-26T13:16:42.497" Title="Is it possible that some stars are already black holes yet we see light emitted from before becoming a black hole?" Tags="&lt;black-hole&gt;" AnswerCount="6" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10317" PostTypeId="1" CreationDate="2015-03-30T19:41:51.643" Score="3" ViewCount="47" Body="&lt;p&gt;The further away we see, the older states of the world we observe. And the volume of the part of space which is observable, increases by the square of the distance to it. So there should basically be a million times more phenomena (data) observable about what happened a billion years ago, than about what happened a million years ago. Given good enough observatories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is obviously counter intuitive, and sorting it out for so called redshift effects doesn't help in that respect. Do/can we really (potentially) know more about things/events the older they are? Is there some epistemological problem involved in that?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-03-30T21:37:21.487" Title="Do astronomers, in principle, have more information about older phenomena than younger?" Tags="&lt;distances&gt;&lt;space-time&gt;&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10318" PostTypeId="2" ParentId="10316" CreationDate="2015-03-30T19:51:49.890" Score="5" Body="&lt;p&gt;No, I think that most stars huge enough to form black holes, need a few millions of years to do so from birth to death. Since we can only see stars in the Milky Way up to a few tens of thousands of light years away, only about 1% of them should've collapsed to a black hole since they emitted the light we now see here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But many of the same kind of giant stars observed in the Andromeda galaxy, which is 2+ million light years away, might have turned into black holes &quot;today&quot;. Because their light travel time is comparable with their lifetime. If one could observe such young giant stars more than 0.1 billion light years away, one could confidently say that ALL of them have turned into black holes now as the light of their past starlife reaches us.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-03-30T19:58:22.310" LastActivityDate="2015-03-30T19:58:22.310" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10319" PostTypeId="2" ParentId="10316" CreationDate="2015-03-30T20:03:55.523" Score="4" Body="&lt;p&gt;It is safe to say that a decent number of the stars we see are dead today. Space is infinite so it is a safe assumption that there are stars that are over a billion light years away that did not live a billion years. Therefore they will have died before their light reaches Earth. Additionally, the larger the mass of the star the quicker it burns its fuel, so the stars that will become black holes are less likely to still be alive. I don't have specific examples of stars that are likely to become black holes in the &quot;future&quot; but it is exceedingly likely. I put future in quotes because it happened 500 million years ago but the star is 501 million light years away so the fact that it is collapsing into a black hole won't reach us for another million years. It's hard to explain conceptually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example:&#xA;Star A is 400 million light years away. This means it takes 400 million years for the light to get here. We are seeing the star as it was 400 million years ago. If Star A officially &quot;formed&quot; (started fusion in its core) 350 million years ago and collapsed into a black hole 10 million years ago we would not yet see the light from the star in the first place, but it would already be a black hole. In 50 million years we would see the star start to form, and in 390 million years we would see the star collapse into a black hole.&lt;/p&gt;&#xA;" OwnerUserId="6662" LastActivityDate="2015-03-30T20:03:55.523" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10320" PostTypeId="2" ParentId="10317" CreationDate="2015-03-30T21:37:21.487" Score="2" Body="&lt;p&gt;There are more stars and galaxies but not necessarily more information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might choose to measure the information content in terms of photons received, in which case, for a given star/galaxy, this decreases as the inverse square of the distance, so the two effects cancel out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomy is a constant struggle with the tension between these two competing dependencies. As we look further, not only do we see more of particular types of object, but we also get to see examples of rare objects - i.e. the tails of the distributions. On the other hand as we expand our horizons, so our grasp of what is going on becomes increasingly blurred in terms of spatial, spectral and temporal resolution. It is for that reason that bigger and bigger telescopes are built!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue you raise concerning redshift is interesting. The co-moving volume versus look back time in any direction will not be a simple function (i.e. doesn't just go as the square of the lookback time) and depends critically on the adopted cosmological parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is of course a limit in any case. Once we get back to the microwave background at $z=1100$, then further probing backwards with electromagnetic radiation is stymied by the optically thick nature of the universe at early times.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-03-30T21:37:21.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10321" PostTypeId="1" AcceptedAnswerId="10621" CreationDate="2015-03-30T23:20:16.143" Score="4" ViewCount="147" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Yarkovsky_effect&quot; rel=&quot;nofollow&quot;&gt;Yarkovsky effect&lt;/a&gt; is responsible for changes in the rotation and orbit of some celestial bodies, most notably asteroids. It has been measured on asteroids, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/6489_Golevka&quot; rel=&quot;nofollow&quot;&gt;6489 Golevka&lt;/a&gt; and &lt;a href=&quot;http://earthsky.org/astronomy-essentials/the-yarkovsky-effect-pushing-asteroids-around-with-sunlight&quot; rel=&quot;nofollow&quot;&gt;1999 RQ36&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the largest object on which the Yarkovsky effect has been observed? &lt;a href=&quot;http://www.scholarpedia.org/article/Yarkovsky_and_YORP_effects#Detection_of_Yarkovsky_effect&quot; rel=&quot;nofollow&quot;&gt;Scholarpedia has an excellent article and list of asteroids&lt;/a&gt;, but the list isn't necessarily complete, and I don't know if the effect has been detected on other (non-asteroid) bodies.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2015-05-01T17:30:13.503" Title="What is the largest object on which the Yarkovsky effect has been observed?" Tags="&lt;orbit&gt;&lt;observational-astronomy&gt;&lt;rotation&gt;&lt;asteroids&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10322" PostTypeId="1" CreationDate="2015-03-31T01:08:08.100" Score="0" ViewCount="167" Body="&lt;p&gt;For a very rough guideline using healthy/corrected eyes adjusted to the dark, around how bright should an object be to expect it to be visible?&lt;/p&gt;&#xA;" OwnerUserId="6942" LastEditorUserId="7982" LastEditDate="2021-02-28T11:53:41.290" LastActivityDate="2021-02-28T11:53:41.290" Title="Around what apparent magnitude can the naked eye observe an object during full moon" Tags="&lt;observational-astronomy&gt;&lt;the-moon&gt;&lt;apparent-magnitude&gt;&lt;atmospheric-effects&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10323" PostTypeId="2" ParentId="10322" CreationDate="2015-03-31T01:39:14.133" Score="1" Body="&lt;p&gt;It may shave off 1 or 2 magnitudes, but it depends on many factors: nature of the object (star, nebula, galaxy), altitude (higher altitude has less light scatter), transparency, etc. There's no One Single Answer To Rule Them All.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light pollution from artificial sources has a greater impact in most cases.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2015-03-31T01:39:14.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10324" PostTypeId="1" AcceptedAnswerId="10351" CreationDate="2015-03-31T02:38:59.647" Score="1" ViewCount="81" Body="&lt;p&gt;I'm writing a science fiction short story which involves a group of people being in suspended animation for a very long period of time, on the order of thousands of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, would an astronomer on waking be able to determine that a significant period of time has passed purely by making observations in the night sky? How accurate would they be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the purposes of the story, I'm more interested in what an astronomer could observe unaided or with a primitive telescope of the sort Galileo might have had early in his career (around 3X magnification).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6944" LastActivityDate="2015-04-02T03:30:55.043" Title="Is it possible to use the stars to determine the passage of time?" Tags="&lt;telescope&gt;&lt;observational-astronomy&gt;&lt;star-gazing&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10325" PostTypeId="2" ParentId="10316" CreationDate="2015-03-31T03:12:01.733" Score="8" Body="&lt;p&gt;It depends what you mean by &quot;see&quot;.  I assume that you mean with the naked eye.  Let's do some rough calculations to estimate the odds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.armaghplanet.com/blog/how-far-away-is-the-farthest-star.html&quot;&gt;The farthest single star visible to the naked eye is Deneb&lt;/a&gt;, estimated distance around 1550 light years. Most stars visible to the naked eye are much closer.  Since even the sort of star that forms a black hole lives a few million years, we would expect perhaps as many as one sufficiently large star in a thousand to have formed a black hole during the time their light was on the way to us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;a href=&quot;http://www.theatlantic.com/technology/archive/2013/11/how-many-stars-are-there-in-the-sky/281641/&quot;&gt;about 5000 stars visible to the naked eye&lt;/a&gt; under good conditions, but only &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_most_massive_known_stars&quot;&gt;a hundred or so&lt;/a&gt; massive enough to form black holes.  So on that basis, the odds are that there is at most one such star, and probably none.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To estimate this another way, consider that there is &lt;a href=&quot;http://www.uni.edu/morgans/astro/course/Notes/section2/new9.html&quot;&gt;on average one supernova in a galaxy per century&lt;/a&gt;.  Comparing the distance to the farthest visible star with the diameter of the galaxy I estimate we can see at most 1/2500th of the Milky Way, so we should expect only one visible star to go supernova every 250,000 years.  Dividing that by the distance in light years to the farthest visible stars, we get only about one chance in a hundred that any of the visible stars have gone supernova during the time their light was on the way to us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One caveat is that I've gotten these figures from disparate sources, based on Google searches.  However, it seems clear that the odds are not good.  Hopefully one of the experts can provide better data.&lt;/p&gt;&#xA;" OwnerUserId="6946" LastActivityDate="2015-03-31T03:12:01.733" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10326" PostTypeId="2" ParentId="10316" CreationDate="2015-03-31T05:11:24.267" Score="1" Body="&lt;p&gt;Not all stars that die become black holes. Only the really, really massive ones, the &quot;giants&quot; they call them. They are rare. None of the stars visible to the naked eye from Earth are massive enough to be black holes. Scientists can tell the mass of a star using the Hertzsprung–Russell diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a possible scenario however, just not on Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's possible because the light emitted from a living star will travel forever until it is absorbed by matter in it's path. So, a star can be dead for eons before the light is finally absorbed by something—like an alien planet and alien eyes—far away and long after the light-bearing star has reached black hole status. It's just that specific to Earth's situation the visible stars are not black holes because their masses aren't great enough. &lt;/p&gt;&#xA;" OwnerUserId="2025" LastActivityDate="2015-03-31T05:11:24.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10329" PostTypeId="2" ParentId="6188" CreationDate="2015-03-31T08:06:12.137" Score="5" Body="&lt;p&gt;Charon and Pluto are bad examples. They have comparable mass: Pluto only 9 times heavier that Charon (Earth is 81 times more massive than Moon), so center of mass in that system lies outside main body (about 1000 km from the Pluto surface).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Main problem for satellites is Roche limit. For the Eath-Moon system Roche radius is about 15500 km from center to center (7400 km from surface to surface).&#xA;Geostationary orbit for Earth is 42 164 from the Earth center or 35 786 from geoid surface (sea level). It works only in Equatorial plain (Moon is tilted 18,3-28,6 to Earth equator). So, Earth-size planet can have Moon-sized satellite in geostationary orbit. In distant past our Moon was much closer – possibly about 50 000 km (about 60 000 from center to center).&lt;/p&gt;&#xA;" OwnerUserId="6949" LastActivityDate="2015-03-31T08:06:12.137" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10330" PostTypeId="2" ParentId="10316" CreationDate="2015-03-31T10:33:54.757" Score="4" Body="&lt;p&gt;&lt;a href=&quot;https://xkcd.com/1342/&quot; rel=&quot;nofollow&quot;&gt;https://xkcd.com/1342/&lt;/a&gt; and &lt;a href=&quot;http://www.explainxkcd.com/wiki/index.php/1342:_Ancient_Stars&quot; rel=&quot;nofollow&quot;&gt;http://www.explainxkcd.com/wiki/index.php/1342:_Ancient_Stars&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://xkcd.com/1440/&quot; rel=&quot;nofollow&quot;&gt;https://xkcd.com/1440/&lt;/a&gt; and &lt;a href=&quot;http://www.explainxkcd.com/wiki/index.php/1440:_Geese&quot; rel=&quot;nofollow&quot;&gt;http://www.explainxkcd.com/wiki/index.php/1440:_Geese&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;could it be that they are already black holes (the sufficiently large&#xA;  ones) and we are just seeing the light emitted during it's main phase?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For some stars, yes it could be. There are certain stars that we can individually observe, and that we believe could go supernova more-or-less at any moment. Betelgeuse is probably the most famous example. It's 640 light years away, it could go type II supernova any time (expected within the next million years, but could conceivably happen within the next 640 years as we observe it, such that it's already happened). It might become a black hole when it does (although a mere neutron star is also on the cards, I believe, and probably more likely).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, when we observe a distant galaxy it is certain (as far as our theories are concerned) that some of the stars contributing to the light from that galaxy have since become black holes. However, since we aren't resolving individual stars in those galaxies even with our best telescopes, you could have an argument whether we're &quot;seeing&quot; the stars or not.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is it safe to say that most of the stars are already dead that we see?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stars in our galaxy, the Milky Way, no. It's only around 100k light years across, it contains 100 billion - 1 trillion stars, of which we can individually &quot;see&quot; only a small proportion. Most of those are very close, the furthest around 1500 light years, and most of &lt;em&gt;those&lt;/em&gt; are not supernova candidates at all, let alone black hole candidates. Even allowing for our best telescopes, we still only resolve a small proportion of stars individually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The galaxy probably experiences roughly one supernova every 50-100 years. So even if all of those were visible stars (which they aren't: even of the largest stars in the galaxy only a small proportion can be seen with the naked eye), and all the supernovae resulted in black holes (which they don't), then at most 15-30 naked-eye-visible stars could already be black holes, and only up to 1000-2000 total stars in the whole galaxy regardless of visibility are in the duration between going supernova, and earth entering the future light cone of that supernova. This &lt;em&gt;extremely&lt;/em&gt; loose upper bound is by no means &quot;most of the stars that we see&quot;, and in fact the chances are that none of the stars you see when you look up at the night sky are already black holes. It might well be that none of them ever will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also unlikely that any star you observe with the naked eye is dead, since their lifetimes are of the order of 10 million years (for the largest) to billions of years, and their distances from earth are 1500 light years or less. But there are a few stars, like Betelgeuse, that we know are near the ends of their life and that therefore might be &quot;already dead&quot; in this sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using telescopes we can see some individual stars in nearby galaxies, or at least indirectly observe them since they're lighting up nebulae that we see, up to a few tens of millions of light years away (these tend to be blue supergiants, I think). Of these, those on course to turn into a black hole may well have lifetimes less than or equal to their distance from us. It's not necessarily the case that all or even most of them become black holes, I'm not sure what's typical of the types of star in question, but for those stars you &lt;em&gt;can&lt;/em&gt; say that many or most are already black holes. The most distant individually-resolved stars are all dead one way or another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you allow for us &quot;seeing&quot; each star in a very distant galaxy, even though we can't resolve any of them individually before they supernova, then the numbers change again, but my instinct is that it's not the case that &quot;most&quot; of the observable universe is already collapsed into black holes. But the stars may well be dead even though they aren't black holes. We can see galaxies 10+ billion light years away, which is about the entire lifetime of a star like the Sun. To come up with &quot;most&quot;, though, you have to firstly agree what counts as a star and secondly survey what kinds of star those galaxies contain. If you end up concluding that most stars are (by comparison with the Sun) tiny little red-brown things with lifetimes in the tens of billions of years, then most stars aren't already dead no matter how far away the galaxy they're in :-) On the other hand, the brighter objects that contribute most of what we observe via telescope in the most distant galaxies, sure, certainly die in far less than a few billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in special relativity the concept of &quot;simultaneous&quot; is a bit difficult to pin down. But I've followed the convention that if something is X light years away, then anything we observe from it between now and X years from now, has &quot;already happened&quot;. We're not necessarily entitled from where we're standing to say that it's &quot;already happened&quot;, since it's not in our past light cone, but good enough :-)&lt;/p&gt;&#xA;" OwnerUserId="6953" LastEditorUserId="6953" LastEditDate="2015-03-31T11:27:41.053" LastActivityDate="2015-03-31T11:27:41.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10331" PostTypeId="1" AcceptedAnswerId="10333" CreationDate="2015-03-31T14:34:22.773" Score="2" ViewCount="1273" Body="&lt;p&gt;I know there are two criteria to meet in order for nuclear fusion to occurs.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;High temperature (many times temperature at Sun's core)&lt;/li&gt;&#xA;&lt;li&gt;High pressure (protons are very close to each other)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[Goal]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I want to know what equations allows me to calculate the amount of hydrogen gas needed so that the internal pressure can produce the right amount of heat for nuclear fusion to occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[My own understandings]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also understand that the temperature of our Sun's core is not sufficient to overcome the coulomb barrier, it make use of quantum tunneling to achieve fusion instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[Question]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimum amount of hydrogen gas required to form a star? (also please include the name of the equation used)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[Assumptions]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;shape is perfectly spherical&lt;/li&gt;&#xA;&lt;li&gt;no angular velocity, not rotating&lt;/li&gt;&#xA;&lt;li&gt;homogenous, consist of 95% hydrogen and 5% helium&lt;/li&gt;&#xA;&lt;li&gt;sustain fusion for at least 1 million years&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="2153" LastEditDate="2015-03-31T22:26:59.903" LastActivityDate="2015-04-01T07:41:53.990" Title="What is the minimum size of a ball of gas to become a star?" Tags="&lt;star&gt;&lt;temperature&gt;&lt;star-formation&gt;&lt;hydrogen&gt;&lt;helium&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10332" PostTypeId="1" AcceptedAnswerId="10350" CreationDate="2015-03-31T14:52:19.303" Score="2" ViewCount="386" Body="&lt;p&gt;I am looking at several galactic sources in the $I$, $R$ and $V$ bands and I want to calculate their absolute magnitudes. I can get the apparent magnitude and I know the distance so I just need to account for extinction to acquire the absolute magnitude. I first used the &lt;a href=&quot;https://ned.ipac.caltech.edu/forms/calculator.html&quot; rel=&quot;nofollow&quot;&gt;NED Coordinate Transformation &amp;amp; Galactic Extinction Calculator&lt;/a&gt; but the calculated extinction values are too high. I figured that this is probably due to the fact that my sources are relatively close ($\sim 100\,\text{pc}$) and I am thus only looking through a fraction of the dust that these coefficients were calculated for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I lower the extinction coefficients to a certain fraction of NED's value (say 60% in a particular case) then I get the expected answer but I'm looking for an accurate way to choose this fraction. I've looked for a method to calculate the extinction coefficients as a function of distance but haven't been able to find anything. Does there exist a map of extinction as a function of distance (and galactic coordinates) in the galaxy from which I can extrapolate the information required? Is there a way to calculate the extinction coefficients ($A_I$, $A_R$ and $A_V$) for an object in the galaxy a known distance away? I found &lt;a href=&quot;http://www.iphas.org/extinction/#map&quot; rel=&quot;nofollow&quot;&gt;IPHAS&lt;/a&gt; which seems to be the right idea but I'm not sure if I can use it to get what I want (or how).&lt;/p&gt;&#xA;" OwnerUserId="2736" LastActivityDate="2015-04-17T13:29:21.970" Title="Galactic extinction as a function of distance" Tags="&lt;galaxy&gt;&lt;apparent-magnitude&gt;&lt;extinction&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10333" PostTypeId="2" ParentId="10331" CreationDate="2015-03-31T15:14:02.930" Score="2" Body="&lt;p&gt;A proposal for the minimum mass of a gas cloud before collapsing to become a star was made by James Jeans; because of this, it was termed the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jeans_instability#Jeans_mass&quot; rel=&quot;nofollow&quot;&gt;Jeans mass&lt;/a&gt;. It is calculated as&#xA;$$M=\frac{4 \pi}{3} \rho R^3$$&#xA;where $\rho$ is density and $R$ is the radius of the cloud - one half of the Jeans length, which is dependent on the speed of sound and the density of the cloud. It was originally thought that any mass above the Jeans mass would not be in hydrostatic equilibrium, and would collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But Jeans didn't realize that any regions &lt;em&gt;outside&lt;/em&gt; this radius would &lt;em&gt;also&lt;/em&gt; collapse. So his arguments were flawed, though they're still sound in many applications.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2015-03-31T15:14:02.930" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10334" PostTypeId="1" AcceptedAnswerId="10337" CreationDate="2015-03-31T21:22:21.090" Score="2" ViewCount="239" Body="&lt;p&gt;I was reading an article that explains why JWST is a successor to Hubble and not a replacement for Hubble. They explained that Hubble's science pushed astronomers to look at longer wavelength. And then they said:    &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In particular, more distant objects are more highly redshifted, and their light is pushed from the UV and optical into the near-infrared.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So basically to observe the first galaxies, astronomers have to observe in infrared. My question is why distant objects require observations in the infrared? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because they are at a very large distance from us, so the light has lost a lot of energy on its way so it's detectable in the infrared? &lt;/p&gt;&#xA;" OwnerUserId="1963" LastActivityDate="2015-03-31T21:47:32.280" Title="Why are distant objects observed in the near infrared?" Tags="&lt;observational-astronomy&gt;&lt;distances&gt;&lt;hubble-telescope&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10335" PostTypeId="1" AcceptedAnswerId="10338" CreationDate="2015-03-31T21:39:09.447" Score="3" ViewCount="1393" Body="&lt;p&gt;If we had a hypothetical spacecraft, would it be possible to exit our galaxy, wait for some time and enter back into another section? Will this method be more efficient that a more traditional flight path? &lt;/p&gt;&#xA;" OwnerUserId="6957" LastEditorUserId="2153" LastEditDate="2015-03-31T22:24:08.363" LastActivityDate="2015-04-01T21:51:57.637" Title="Leaving the Milky Way" Tags="&lt;milky-way&gt;&lt;galactic-dynamics&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10336" PostTypeId="2" ParentId="10334" CreationDate="2015-03-31T21:47:02.833" Score="1" Body="&lt;p&gt;This is because of something called the Doppler effect. As something moves away from us, the Electromagnetic waves it releases will have a longer wavelength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/e/e4/Redshift_blueshift.svg/2000px-Redshift_blueshift.svg.png&quot; alt=&quot;Look at the Star that is moving away&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, because of Hubble's law, galaxies that are further away have an increased velocity, making the red shift even more pronounced.&lt;/p&gt;&#xA;" OwnerUserId="6957" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:47.390" LastActivityDate="2015-03-31T21:47:02.833" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10337" PostTypeId="2" ParentId="10334" CreationDate="2015-03-31T21:47:32.280" Score="2" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When light moves through the expanding space, it loses energy and becomes redshifted. As galaxies form, they a emit a lot of UV light, and are often detected from there emission at the so-called Lyman $\alpha$ light, which has a wavelength of 1216 Å. Galaxies began forming just a few hundred million years after the Big Bang. Since then, the Universe has a expanded by more than a factor of 10 in all direction. The wavelength of the light expands by the same factor, and thus a Lyman $\alpha$ photon emitted in this epoch today has a wavelength of $\sim12,000$ Å, or 1.2 $\mu$m, which is in the infrared.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2015-03-31T21:47:32.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10338" PostTypeId="2" ParentId="10335" CreationDate="2015-03-31T21:58:19.920" Score="13" Body="&lt;p&gt;Not really, for the same reason that you cannot travel west by jumping up in the air and let Earth rotate underneath you, such that you land a little farther to the west.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is that standing on Earth's surface, you already have a velocity toward the east which matches exactly the speed of the surface. Thus, in the reference frame of Earth, you simply jump up and down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same manner, if you use your mad hypothetical spacecraft to fly &quot;up&quot;, i.e. away from the Galactic plane, you already have a velocity of $\sim$250 km s$^{-1}$ in the direction of the rotation of the plane, so that in the reference frame of the plane, you simply fly straight up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from this, in order to leave the Galactic plane, you need to fly roughly 500 lightyears. This will take a long time. You should stay home.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-04-01T08:00:57.847" LastActivityDate="2015-04-01T08:00:57.847" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10339" PostTypeId="1" AcceptedAnswerId="10370" CreationDate="2015-03-31T23:51:10.167" Score="3" ViewCount="206" Body="&lt;p&gt;Let's say I have a moon orbiting a planet at a distance $d$. Eventually, $d \leq d_R$, where $d_R$ is the Roche limit of the moon-planet system. I can figure out the mass of both bodies by experiment, as well as their radii, and thus their density (which is how I can calculate the Roche limit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious as to how the debris will spread out, though, once the moon is ripped apart. Can I calculate the inner and outer radii of the newly formed ring, based on what I know? Or is there not a direct relationship between these radii and the initial conditions?&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2015-06-03T00:46:00.123" Title="How can I calculate how the debris of an object ripped apart at the Roche limit will spread out?" Tags="&lt;gravity&gt;&lt;roche-limit&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10340" PostTypeId="2" ParentId="10331" CreationDate="2015-04-01T01:03:06.890" Score="3" Body="&lt;p&gt;There is no equation, you need a detailed model of the interior physics of very low mass stars. &lt;em&gt;Very&lt;/em&gt; roughly, you can say that&#xA;hydrogen fusion occurs when the central temperature exceeds about $10^{7}\ K$ (the density dependence is secondary) and that from the virial theorem, the central temperature is given approximately by $T \simeq 1.6\times 10^{7} M/R$, where the mass and radius are in solar units. $M/R$ decreases slowly down the main sequence towards lower mass stars, but then what happens is that electron degeneracy pressure becomes important and less massive objects do not become much smaller than Jupiter-sized and thus $M/R$ decreases and fusion stops - or rather it never starts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In detail the minimum mass for hydrogen fusion in a manner that is capable of sustaining a star in equilibrium against gravitational contraction is about $0.075M_{\odot}$. With an uncertainty of about $0.002M_{\odot}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is slightly more complicated than this, since at lower internal temperatures the deuterium in a star can fuse. This will happen during the early life of any ball of gas more massive than about 13 Jupiter masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your assumptions about the relative H and He abundances are completely wrong. Even in the early universe the gas is $\sim 25$% He by mass.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-04-01T07:41:53.990" LastActivityDate="2015-04-01T07:41:53.990" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10341" PostTypeId="1" AcceptedAnswerId="10348" CreationDate="2015-04-01T16:58:04.277" Score="19" ViewCount="2282" Body="&lt;p&gt;Although astronomy is very cool and the things we are learning are awesome, is there really any practical use to knowing the things we know about the universe?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do other fields of science draw from the current tome of astronomical knowledge?&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2018-03-28T12:52:40.033" Title="Is there any practical use for astronomy?" Tags="&lt;universe&gt;" AnswerCount="9" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10342" PostTypeId="1" AcceptedAnswerId="10349" CreationDate="2015-04-01T17:10:30.840" Score="1" ViewCount="148" Body="&lt;p&gt;I did read several of the Oort Cloud questions/answers which I found very helpful and instructive, but this question still persists for me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the Oort Cloud actually opaque, or is the &quot;cloud&quot; used euphemistically because minor planets inside of it are too dim to be seen?&lt;/p&gt;&#xA;" OwnerUserId="6959" LastEditorUserId="2153" LastEditDate="2015-04-01T17:36:08.890" LastActivityDate="2015-04-02T00:45:41.460" Title="Is the Oort Cloud actually opaque, or is the &quot;cloud&quot; used euphemistically because minor planets inside of it are too dim to be seen?" Tags="&lt;universe&gt;&lt;oort-cloud&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10343" PostTypeId="2" ParentId="10341" CreationDate="2015-04-01T17:41:33.637" Score="0" Body="&lt;p&gt;Other fields of science don't &quot;draw from Astronomy&quot; the same way they don't &quot;draw from cartography&quot;. It may not be increasing the technology we have right now, but it will help us get around much safer once we're out in the universe. Basically right now we have an extremely detailed (every single road) map of the entire United States (or other very large country) but we can only walk. We know a large quantity of information that is not currently useful. Get a car (spaceship) that can go far and suddenly all that information becomes much more useful. You know where to go to get fuel, where you want to sightsee, where you want to shop. It is not useful in and of itself, but it is extraordinarily useful at increasing the efficiency of other activities once you can use it properly. &lt;/p&gt;&#xA;" OwnerUserId="6662" LastActivityDate="2015-04-01T17:41:33.637" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10344" PostTypeId="1" AcceptedAnswerId="10346" CreationDate="2015-04-01T19:21:12.293" Score="6" ViewCount="7307" Body="&lt;p&gt;Can you explain what is a Friedmann model to a layman?&#xA;And also give some examples of Friedmann models, specially I would like to know if the lambda-CDM model is considered a Friedmann model.&lt;/p&gt;&#xA;" OwnerUserId="6920" LastActivityDate="2018-05-29T06:07:24.200" Title="what is a Friedmann model?" Tags="&lt;cosmology&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10345" PostTypeId="2" ParentId="10341" CreationDate="2015-04-01T19:40:44.957" Score="9" Body="&lt;p&gt;Astronomy has several practical applications/spin-off. For instance, in developing rocket engines, NASA accidentally invented &lt;a href=&quot;http://blogs.discovermagazine.com/badastronomy/2012/03/21/this-is-why-we-invest-in-science-this/#.VRxGL2ZIj4Y&quot; rel=&quot;noreferrer&quot;&gt;an extremely efficient fire extinguisher&lt;/a&gt;, very beneficial for firefighters. Imaging technology for telescopes is used in cameras, spaceship insulation led to teflon, figuring out how to go to the bathroom in space led to better dialysis apparatus, and &lt;a href=&quot;http://spinoff.nasa.gov/&quot; rel=&quot;noreferrer&quot;&gt;much, much more&lt;/a&gt;. Investigating the unknown always leads to new and unforeseen insight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I think that for most astronomers, this is not the reason we do astronomy. With the risk of sounding a little pretentious, I thnk that the goal of astronomy is to explore the unknown, find our place in this incredible cosmos, learn about the origin of ourselves, and simply find out as much as possible about Nature. Not unlike people did 500 years ago and before, seeking to map out the world and find out what's beyond the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, that did sound pretentious. I still think it's true, though.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2015-04-01T19:40:44.957" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10346" PostTypeId="2" ParentId="10344" CreationDate="2015-04-01T20:09:07.300" Score="18" Body="&lt;p&gt;The &quot;Friedmann model&quot; is a model of the Universe governed by the Friedmann equations, which describes how the Universe expands or contracts. These equations are a solution to Einstein's field equations, and with two very important assumptions they form the basis for our understanding of the evolution and structure of our Universe. These assumptions, together called &quot;the cosmological principle&quot;, are that the Universe is &lt;em&gt;homogeneous&lt;/em&gt;, and that it's &lt;em&gt;isotropic&lt;/em&gt;. This is a layman's explanation, so I won't write the equations here, but farther down I've added the equation and go a bit more in detail what it means.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The cosmological principle&lt;/h2&gt;&#xA;&#xA;&lt;h3&gt;Homogeneity&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;That the Universe is homogeneous means that its &quot;the same&quot; everywhere. Obviously, it's not, really. For instance, under your feet is a dense, rocky planet, whereas above your head there's thin air. We live in a galaxy full of stars and molecular clouds and what not, while 100,000 lightyears from the Milky Way, there is virtually nothing. But on very large scales, say above half a billion lightyears, the Universe actually looks the same all over.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Isotropy&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;That it's isotropic, means that it looks the same in all directions. Again, obviously it doesn't on small scales, but on large scales, it does. If it didn't, it would mean that we occupied a special place in the Universe, and we don't think we do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, neither of these assumptions &lt;em&gt;have&lt;/em&gt; to be true, but observations tell us that apparently, they are. See this image, where every dot is a galaxy (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1990MNRAS.243..692M&quot; rel=&quot;noreferrer&quot;&gt;Maddox et al. 1990&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/foijq.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/foijq.jpg&quot; alt=&quot;APM&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might think that a homogeneous Universe would also be isotropic and/or vice versa, but &lt;a href=&quot;https://astronomy.stackexchange.com/a/13193/5264&quot;&gt;that's not the case&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Three possible solutions&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It turns out that, for these assumptions, there are three possible solutions to the Friedmann equation. We call the three possible universes &lt;em&gt;flat&lt;/em&gt;, &lt;em&gt;positively curved&lt;/em&gt; (or &quot;closed&quot;), and &lt;em&gt;negatively curved&lt;/em&gt; (or &quot;open&quot;). Which of these possible universes &lt;em&gt;we&lt;/em&gt; live in, turns out to depend upon the average density in the Universe, so by measuring this, we can determine the &quot;geometry&quot; of our own Universe. And it seems that it's &quot;flat&quot;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;A flat universe&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The reason it's called flat is that the geometry is like that of a flat 2D table, only in 3D. That is, a triangle has 180º, parallel lines never meet, etc. And it's infinitely large. Intuitively, we'd think that this is the way the Universe is, and definitely on small scales (say within our own Galaxy), it's an adequate approximation. In the 2D analogy, the 2D surface of Earth seems flat &lt;em&gt;locally&lt;/em&gt;, and for all practical purposes the parking lot outside &lt;em&gt;is&lt;/em&gt; flat. But if you draw a triangle from Congo→Indonesia→the North Pole→Congo, you'd measure it's sum of angles to be roughly 270º. That's because the geometry of the surface of Earth is &lt;em&gt;not&lt;/em&gt; flat, but is &quot;closed&quot;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;A closed universe&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;If the Universe were &quot;closed&quot;, it's geometry would, in the 2D analogy, correspond to that of the surface of a ball, i.e. a triangle has &lt;em&gt;more&lt;/em&gt; than 180º, lines that start out being parallel will at some point meet, etc. But like the surface area of a ball is finite (but doesn't have a border), so is the Universe. So if you take you spaceship and fly straight away from Earth, you would end up back here (assuming the Universe doesn't collapse before you get back, or expand too fast for you).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;An open universe&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;If it were &quot;open&quot;, it's geometry would, in the 2D analogy, correspond to that of the surface of a saddle, i.e. a triangle has &lt;em&gt;less&lt;/em&gt; than 180º, lines that start out being parallel will diverge, etc. And it's infinitely large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This picture from &lt;a href=&quot;http://www.welchco.com/02/14/01/60/04/03/1228.HTM&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; visualizes the 2D analogies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OgZSA.gif&quot; alt=&quot;geo&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 3D, only a flat geometry can be pictured, and this doesn't look &quot;flat&quot; by any means; it is simply your good old 3D (&quot;Euclidian&quot;) space that you know from your everyday senses.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Expansion of the Universe&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The Friedmann equation, together with the densities of the constituents of the Universe (radiation, normal matter, dark matter, and dark energy) tell us how the Universe expands. So, again by measuring these densities, we can prediect the evolution of the Universe. And it seems that the Universe not only expands, but actually &lt;a href=&quot;http://en.wikipedia.org/wiki/Accelerating_universe&quot; rel=&quot;noreferrer&quot;&gt;expands faster and faster&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Beyond the layman's explanation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here, I'll expand a bit on how to understand the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first Friedmann equation is intuitively most understandable, I think, when written like this:&#xA;$$\frac{H^2}{H_0^2} = &#xA;\frac{\Omega_\mathrm{r,0}}{a^4} +&#xA;\frac{\Omega_\mathrm{M,0}}{a^3} +&#xA;\frac{\Omega_k}{a^2} +&#xA;      \Omega_\mathrm{\Lambda}.$$&#xA;This equation tells us the connection between the expansion rate of the Universe (the left hand side), and the density of its components and its size (the right hand side). Below, I'll go through the components of the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Hubble parameter&lt;/strong&gt;&lt;br&gt;&#xA;In the equation, $H$ is the Hubble parameter describing how fast a galaxy at a given distance recedes (or approaches, for a collapsing universe), at a given time in the history of the Universe. $H_0$ is its value &lt;em&gt;today&lt;/em&gt;, and is measured to be roughly $70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$. This means that a galaxy at a distance of, say, 10 Mpc ($\simeq33$ lightyears), moves away from us at a current speed of 700 km/s. A galaxy 20 Mpc away recedes at 1400 km/s, and so on (and galaxies farther away than roughly 4.3 Gpc recede faster than the speed of light, but &lt;a href=&quot;https://astronomy.stackexchange.com/a/19910/5264&quot;&gt;that's no problem and we can still see them&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Size&lt;/strong&gt;&lt;br&gt;&#xA;The size of the Universe is unknown, and quite possibly infinite (lest it wouldn't be homogeneous, but to be fair, we only know that the &lt;em&gt;observable&lt;/em&gt; Universe is homogeneous). Thus, we cannot talk about its absolute size. But we &lt;em&gt;can&lt;/em&gt; talk about how much a certain volume of space expands in a given time. We us the parameter $a$, called the expansion factor. Defining $a$ to be 1 today, that means that at the time when the Universe was so small that all distances between the galaxies was, say, half of today's values, $a$ was equal to 0.5 (this happens to be 8 billion years ago).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Density parameters&lt;/strong&gt;&lt;br&gt;&#xA;Whether the geometry of the Universe as described above is flat, closed, or open, depends on whether the &lt;em&gt;total&lt;/em&gt; density $\rho_\mathrm{tot}$ is exactly equal to, above, or below a certain critical threshold $\rho_\mathrm{cr} \sim 10^{-29}\,\mathrm{g}\,\mathrm{cm}^{-3}$. It is customary to parametrize the density of the $i$'th component as $\Omega\equiv \rho_i / \rho_\mathrm{cr}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Matter&lt;/strong&gt;&lt;br&gt;&#xA;The term &quot;matter&quot; includes &quot;normal&quot; matter (gas, stars, planets, bicycles, etc.), and the mysterious dark matter. As the universe expands, the volume grows as $a^3$. That means that the density falls as $\Omega_\mathrm{M}=\Omega_\mathrm{M,0}/a^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Radiation&lt;/strong&gt;&lt;br&gt;&#xA;Photons redshift as space expands, and this redshift goes as $1/a$. This is in addition to having their &lt;em&gt;number&lt;/em&gt; density decrease, so the total energy density of radiation decreases faster than matter, namely as $1/a^4$. Today, the energy density of radiation is &lt;a href=&quot;https://physics.stackexchange.com/a/196373/70207&quot;&gt;dominated by the CMB&lt;/a&gt;, and can be neglected, but in the early times, they would dominate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Curvature&lt;/strong&gt;&lt;br&gt;&#xA;If space is &lt;em&gt;not&lt;/em&gt; flat, its curvature contributes to $\Omega_\mathrm{tot}$. The reason is that the curvature affects the volume in which we measure densities (thanks to John Davis for this explanation). This scales as $1/a^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dark energy&lt;/strong&gt;&lt;br&gt;&#xA;Finally, there's the magical dark energy, of which even less is known than the dark matter. If it exists, it's thought to be a property of space itself, i.e. its energy density grows proportionally to the volume of the Universe, and thus there's no $a$-dependency.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Interpretation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;From the equation, it is readily seen that if we're able to measure all the Omegas, then we know how fast the Universe has been expanding at all times. That means that we can integrate backwards in time and calculate when $a$ was 0, i.e. we can calculate the age of the Universe. Also, from the $a$-dependencies we can see when the Universe went from being &lt;a href=&quot;http://en.wikipedia.org/wiki/Radiation-dominated_era&quot; rel=&quot;noreferrer&quot;&gt;radiation-dominated to being  matter-dominated&lt;/a&gt;. We can also see that not only is it now dominated by dark energy (since $\Omega_\mathrm{\Lambda}\simeq0.7$, but $\Omega_\mathrm{M}\simeq0.3$), but due to the $a$ factor, it will only get &quot;worse&quot;. That is, all other densities will keep decreasing, but $\rho_\Lambda$ stays the same, and since dark energy has a repulsive effect rather than attracting, the expansion of the Universe accelerates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observationally, it is found (in several &lt;em&gt;independent&lt;/em&gt; ways) that &lt;a href=&quot;https://astronomy.stackexchange.com/a/19395/5264&quot;&gt;all the $\Omega$s add up to one&lt;/a&gt;, i.e. that the total energy density of the Universe happens to be exactly equal to the critical density. This is pretty amazing. This figure, taken from &lt;a href=&quot;http://wmap.gsfc.nasa.gov/media/080998/&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;, shows the contribution from the different components now (top), and at the time of the CMB emission (380,000 yr after the Big Bang; bottom):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/94t86.jpg&quot; alt=&quot;lagkage&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2017-07-13T09:54:32.003" LastActivityDate="2017-07-13T09:54:32.003" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10347" PostTypeId="2" ParentId="10335" CreationDate="2015-04-01T21:51:57.637" Score="1" Body="&lt;p&gt;It depends greatly on the nature of your &quot;hypothetical spacecraft&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any spacecraft travelling through interstellar space will have to deal with interstellar gas and dust. At sufficiently high relativistic velocities, running into a hydrogen atom is like being hit by a high-energy cosmic ray, and running into a dust particle could seriously ruin your afternoon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intergalactic medium has substantially less matter than the interstellar medium within the galaxy. A route that exits the body of the galaxy by going &quot;galactic north&quot;, then travels parallel to the disk, then re-enters the galaxy at another point will obviously be longer than a straight-line route, but it will encounter significantly less matter along the way. If the maximum safe velocity of the USS Hypothetical depends strongly on the density of the medium through which it's travelling, then the indirect route could be faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whichever route you choose, this will take a long time. You should bring a book and plenty of snacks.&lt;/p&gt;&#xA;" OwnerUserId="53" LastActivityDate="2015-04-01T21:51:57.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10348" PostTypeId="2" ParentId="10341" CreationDate="2015-04-01T22:49:33.777" Score="16" Body="&lt;p&gt;This question begs the question, does everything need a practical use? The answer is a resounding no. What's the practical use of the Louvre, or of your local neighborhood public park where you enjoy weekend barbecues?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some things that are very worthwhile that have little or no economical gain. Your local neighborhood public park in fact has negative economic gain. Admission is free, but maintenance is not. Think of how much money your city would make if they sold it to a condominium developer, and how much money it would save by not having to pay to have the park maintained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite having no obvious economic gain, some things are nonetheless worth quite a bit. Many of the sciences fall in this category. For example, what is the practical use of archeology? (There are some, but that's not the point.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomy, like archeology, the Louvre, and your local public park, doesn't need a practical economical purpose. The purpose of the science is good enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, there are practical applications of astronomy. The key application has been and still is navigation. Knowing the location of a ship at sea or the orientation of a vehicle in space requires astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A less direct but still very important application of astronomy is in how it informs physics. Kepler was an astronomer, not a physicist. (Those two disciplines were very, very distinct in Kepler's day). Yet Kepler's work informed Newton on how to describe gravitation. More recently, astronomy has informed physics that its standard model was not quite correct. The observed neutrino flux from the Sun (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_neutrino_problem&quot;&gt;http://en.wikipedia.org/wiki/Solar_neutrino_problem&lt;/a&gt;) was a third of what physics at the time said it should be. This resulted in a change to the standard model. Neutrinos have a small but non-zero mass, and they oscillate from one form to another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomy continues to inform physics to this day. Physicists (and astronomers) remain clueless with regard to what constitutes dark matter and dark energy. But whatever they are, they certainly do exist.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2015-04-01T23:20:09.070" LastActivityDate="2015-04-01T23:20:09.070" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10349" PostTypeId="2" ParentId="10342" CreationDate="2015-04-01T23:30:30.697" Score="3" Body="&lt;p&gt;Well, given that we can see the stars and that the Oort &quot;cloud&quot; is closer than the nearest star, then the answer to your main question is obviously that the &quot;cloud&quot; is not opaque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is called a cloud because it consists of many individual, small &quot;particles&quot; that don't interact with each other. Also, the term cloud avoids giving the impression that the Oort objects are arranged in any disk-like configuration - they should be much more spherically symmetric. The word cloud also avoids giving the impression that the objects fall between two relatively tight orbital radius limits. i.e. The Oort &lt;em&gt;cloud&lt;/em&gt; is not like the Kuiper &lt;em&gt;belt&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-04-01T23:30:30.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10350" PostTypeId="2" ParentId="10332" CreationDate="2015-04-01T23:43:00.383" Score="3" Body="&lt;p&gt;If your sources are within 100pc then the best thing to do is assume the extinction is zero, unless you are looking for absolute magnitudes that are a lot more precise than say +/- 0.1 mag. I think you would be very unfortunate if the extinction was any more than about ~0.1 mag in the V-band (and less than that at R and I). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond that you are really struggling because the structure of the ISM is very non-uniform, so you would have to go with the kinds of maps that are produced (for example) using IPHAS - but these are only in the galactic plane. But even if you could find some general map covering all positions it would be unlikely to give you an accurate answer out to 100pc, because there isn't enough extinction to give an effect that is big enough to be measured with any precision (i.e. the uncertainty might be bigger than the value).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have spectral types for your objects then you could always compare the V-I colour with what you expect for a star of that spectral type. This gives you the reddening $E(V-I)$ and then (roughly speaking) $A_V \simeq 2E(V-I)$.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-04-01T23:49:21.493" LastActivityDate="2015-04-01T23:49:21.493" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10351" PostTypeId="2" ParentId="10324" CreationDate="2015-04-02T00:06:31.823" Score="1" Body="&lt;p&gt;In the night sky in 10,000 years, two things will have changed in relation to the stars. The first, the rotational axis of the Earth will have changed, shifting the celestial sphere. The second, the stars themselves will have moved a bit relative to each other due to proper motion. So, the night sky will be quite different in 10,000 years, but still recognizable, particularly the constellations which will have changed somewhat but will still be identifiable. I would posit that as long as the person or persons were at least casual star gazers or astronomy enthusiasts (not even necessarily professionals) they could estimate how much time has passed in the course of their slumber, I'd say with a margin of error of about +/-2000 years. Should they be trained astronomers that margin of error should fall. &lt;/p&gt;&#xA;" OwnerUserId="2025" LastEditorUserId="2025" LastEditDate="2015-04-02T03:30:55.043" LastActivityDate="2015-04-02T03:30:55.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10352" PostTypeId="2" ParentId="254" CreationDate="2015-04-02T00:29:39.477" Score="0" Body="&lt;p&gt;Put simply, it is because the objects that make up the Oort Cloud, remnants from the formation of our sun, are both too small and too faint for us to detect. They are faint because of their vast distance away. There is minimal absorption of the sun's light and even less is reflected back. So little light is reflected back that there is nothing for even our most advanced telescopes to see. &lt;/p&gt;&#xA;" OwnerUserId="2025" LastActivityDate="2015-04-02T00:29:39.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10353" PostTypeId="2" ParentId="10342" CreationDate="2015-04-02T00:45:41.460" Score="0" Body="&lt;p&gt;&quot;Cloud&quot; distinguishes the Oort Cloud as a 3-dimensional sphere of objects and not a disk like the orbital plane of the planets.&lt;/p&gt;&#xA;" OwnerUserId="2025" LastActivityDate="2015-04-02T00:45:41.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10354" PostTypeId="1" CreationDate="2015-04-02T03:30:07.063" Score="2" ViewCount="332" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GAhCd.jpg&quot; alt=&quot;spec&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A featureless spectrum and a spectrum low S/N are very similar. &lt;br/&gt;&#xA;Is there a way to tell the difference between them?&lt;br/&gt;&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="9533" LastEditDate="2016-08-25T01:57:45.073" LastActivityDate="2017-02-02T13:57:37.093" Title="How to tell whether an optical spectrum is featureless or just with a low S/N?" Tags="&lt;spectra&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10355" PostTypeId="1" CreationDate="2015-04-02T03:47:15.273" Score="4" ViewCount="80" Body="&lt;p&gt;Currently 22-23 magnitude could be the limit to make a spectrum for ground based 10m class telescopes, 21-22 magnitude may be easy for them.&lt;br/&gt;&#xA;4m class telescopes could possibly handle 20 magnitude, I am not sure.&lt;br/&gt;&#xA;2m could be appropriate for stars brighter than 16magnitude generally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the infrared bands: J H K W1 W2 respectively? Are there any empirical values or references?  &lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="9527" LastEditDate="2020-05-13T13:42:39.007" LastActivityDate="2020-05-13T13:42:39.007" Title="Infrared telescopes, magnitude and observations" Tags="&lt;observational-astronomy&gt;&lt;infrared&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10357" PostTypeId="1" AcceptedAnswerId="10358" CreationDate="2015-04-02T11:20:05.887" Score="5" ViewCount="2454" Body="&lt;p&gt;OK, first, I know there's a variety of sizes and types of red dwarf stars and the universe is too young for any of them to have reached the end of their main sequence phase yet, so it's all theoretical and/or modeling.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Red_dwarf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Red_dwarf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;But what is the theoretical size needed for a star to undergo the electron degeneracy process which turns a small star from at least Jupiter sized, usually bigger, into to earth sized super dense object, where, as I understand it, the electrons are squeezed off the nuclei - electron degeneracy.&lt;/p&gt;&#xA;&lt;p&gt;It would seem to me that a 7.5% solar mass star, which gradually burns hydrogen into helium but doesn't burn helium, might not have the mass to compact into a true white dwarf but might end it's life looking more like a brown dwarf / super Jupiter - well, talking appearance, not really, because super Jupiter &amp;amp; Brown dwarfs are mostly hydrogen while and an end of life red dwarf should be mostly helium - which, in and of itself, might make the difference.&lt;/p&gt;&#xA;&lt;p&gt;It's just my curiosity whether all red dwarfs turn into white dwarfs at the end of their burning phase or is there a theoretical mass that's needed for that level of shrinkage to occur?&lt;/p&gt;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="31410" LastEditDate="2020-12-11T18:09:12.287" LastActivityDate="2020-12-11T18:09:12.287" Title="death of a red dwarf star / minimum mass needed for a white dwarf?" Tags="&lt;stellar-evolution&gt;&lt;stellar-astrophysics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10358" PostTypeId="2" ParentId="10357" CreationDate="2015-04-02T11:50:09.030" Score="2" Body="&lt;p&gt;Stars that have a mass lower than about $0.5 M_{\odot}$ will not ignite helium in their cores, in an analogous fashion to the way that stars with $M&amp;lt;8M_{\odot}$ have insufficiently massive cores that never reach high enough temperatures to ignite carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cause in both cases is the onset of electron degeneracy pressure, which is independent of temperature and allows the core to cool at constant pressure and radius. [A normal gas would contract and become hotter as it loses energy!]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The end result for a $0.5M_{\odot}$ star will be a helium white dwarf with a mass (depending on uncertain details of the mass-loss process) of around $0.2M_{\odot}$. Such things do exist in nature now, but only because they have undergone some kind of mass transfer event in a binary system that has accelerated their evolution. The collapse to a degenerate state would be &lt;strong&gt;inevitable&lt;/strong&gt; even for the lowest mass stars (which would of course then be very low-mass white dwarfs). As an inert core contracts it loses heat and cools - a higher density and lower temperate eventually lead to degenerate conditions that allow the core to cool without losing pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lowest mass stars ($&amp;lt;0.3 M_{\odot}$) do get there via a slightly different route - they are fully convective, so the &quot;core&quot; doesn't exist really, it is always mixed with the envelope. They do not develop into red giants and thus I guess will suffer much less mass loss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remnant &lt;em&gt;would&lt;/em&gt; be a white dwarf in either case and is fundamentally different from a brown dwarf both in terms of size and structure, because it would be made of helium rather than (mostly) hydrogen. This should have an effect in two ways. For the same mass, the brown dwarf should end up bigger because the number of mass units per electron is smaller (1 vs 2) and also because the effects of a finite temperature are larger in material with fewer mass units per particle - i.e. its outer, non-degenerate layer would be more &quot;puffed up&quot;. NB: The brown dwarfs we see today are Jupiter-sized, but are still cooling. They will get a bit smaller and more degenerate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple size calculation could use the approximation of an ideal, cold, degenerate gas. A bit of simple physics using the virial theorem gives you&#xA;$$ \left(\frac{R}{R_{\odot}}\right) \simeq 0.013\left(\frac{\mu_e}{2}\right)^{-5/3} \left(\frac{M}{M_{\odot}}\right)^{-1/3},$$&#xA;where $\mu_e$ is the number of atomic mass units per electron.&#xA;Putting in appropriate numbers I get $0.32\ R_{Jup}$ for a $0.07M_{\odot}$ Helium white dwarf versus $1.01\ R_{Jup}$ for a $0.07M_{\odot}$ completely degenerate Hydrogen brown dwarf (in practice it would be a bit smaller because it isn't all hydrogen).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it would be interesting to see some realistic calculations of what happens to a $0.07M_{\odot}$ brown dwarf versus a $0.08M_{\odot}$ star in a trillion years or so. I will update the answer if I come across such a study.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I knew I'd seen something on this. Check out &lt;a href=&quot;http://adsabs.harvard.edu/abs/1997ApJ...482..420L&quot; rel=&quot;nofollow&quot;&gt;Laughlin et al. (1997)&lt;/a&gt;, which studies the long-term evolution of very low-mass stars. Low-mass stars do not pass through a red giant phase, remain fully convective and can thus convert almost all their hydrogen into helium over the course of $10^{13}$ years and end up cooling as degenerate He white dwarfs.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-04-02T17:22:03.463" LastActivityDate="2015-04-02T17:22:03.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10359" PostTypeId="1" CreationDate="2015-04-02T12:16:37.237" Score="5" ViewCount="156" Body="&lt;p&gt;1) Could a &lt;a href=&quot;http://www.jpl.nasa.gov/news/news.php?release=2014-089&quot; rel=&quot;noreferrer&quot;&gt;starshade&lt;/a&gt; in orbit be used by ground based telescopes? If so, what kind of orbit should it have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Could the same starshade be used by two different space telescopes, or does the positioning make it impractical?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2017-08-10T22:22:44.767" Title="Can ground based telescopes use a starshade in space?" Tags="&lt;telescope&gt;&lt;space-telescope&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10360" PostTypeId="1" CreationDate="2015-04-02T15:57:30.607" Score="4" ViewCount="210" Body="&lt;p&gt;What could be the utmost &lt;strong&gt;lowest&lt;/strong&gt; temperature in the universe/multiverse?&lt;/p&gt;&#xA;" OwnerUserId="760" LastEditorUserId="2153" LastEditDate="2015-04-02T17:36:03.747" LastActivityDate="2016-09-23T06:24:01.907" Title="What could be the utmost lowest temperature in the universe/multiverse?" Tags="&lt;universe&gt;&lt;temperature&gt;&lt;multiverse&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10362" PostTypeId="2" ParentId="10360" CreationDate="2015-04-03T05:48:45.743" Score="4" Body="&lt;h2&gt;What temperature means...&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Temperature is the measure of the energy of particles. The higher the temperature, the more energized the particles. The more energy particles have the faster they move around. This is the particles' &lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic_energy&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;kinetic energy&lt;/em&gt;&lt;/a&gt; that is rising. As it rises, the particles will begin using up more space. Moving particles need more space. In a vacuum this can be measured as pressure which is the stress of &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_expansion&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;thermal expansion&lt;/em&gt;&lt;/a&gt; on the closed system. In an open system the matter will expand freely. As the particles increase their speed they also move more erratically so the &lt;a href=&quot;http://en.wikipedia.org/wiki/Entropy&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;entropy&lt;/em&gt;&lt;/a&gt;, or measure of disorder, will also increase. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The coldest...&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now understanding all that, what would the lowest temperature, or energy state, be in the universe? The answer is a state of no energy, 0 K, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Absolute_zero&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;&lt;em&gt;absolute zero&lt;/em&gt;&lt;/strong&gt;&lt;/a&gt; in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kelvin&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Kelvin scale&lt;/em&gt;&lt;/a&gt;. It is −273.15° on the Celsius scale and −459.67° on the Fahrenheit scale. At this temperature, which is impossible to occur by only thermodynamic means, the particles are completely still and entropy drops to 0.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Temperature reference points...&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The surface temperature of the sun is 5,778 K.&lt;/li&gt;&#xA;&lt;li&gt;Water boils at 373 K.&lt;/li&gt;&#xA;&lt;li&gt;Water freezes at 273.15 K.&lt;/li&gt;&#xA;&lt;li&gt;The moon’s darkest craters that never receive sunlight are 33 K.&lt;/li&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;cosmic microwave background&lt;/em&gt;&lt;/a&gt; fluctuates around 2.8 K.&lt;/li&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Boomerang_Nebula&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Boomerang Nebula&lt;/em&gt;&lt;/a&gt;, the coolest natural place currently&#xA;known in the universe, measures 1 K.&lt;/li&gt;&#xA;&lt;li&gt;Absolute zero is 0 K.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2025" LastEditorUserId="2025" LastEditDate="2015-04-04T00:08:50.897" LastActivityDate="2015-04-04T00:08:50.897" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10363" PostTypeId="1" AcceptedAnswerId="10365" CreationDate="2015-04-03T06:34:52.100" Score="3" ViewCount="209" Body="&lt;p&gt;I read an article mentioning that when two stars orbiting each other they'll produce gravitational waves that can carry away energy into space. These phenomenon increases the speed of the two orbiting stars and draws them closer to each other until they finally collided, is it true that gravitational waves can carry away energy into space?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="6760" LastEditDate="2015-04-03T07:21:27.733" LastActivityDate="2015-04-03T11:59:15.630" Title="Binary stars system ultimate fate" Tags="&lt;binary-star&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10365" PostTypeId="2" ParentId="10363" CreationDate="2015-04-03T11:59:15.630" Score="1" Body="&lt;p&gt;Well, yes apparently that is how it works and a Nobel prize was won in 1993 by Hulse and Taylor for demonstrating this phenomena in the object &lt;a href=&quot;http://en.wikipedia.org/wiki/PSR_B1913%2B16&quot; rel=&quot;nofollow&quot;&gt;PSR B1913+16&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PSR B1916+16 is a pulsar (a rapidly rotating neutron star with a strong magnetic field that emits beamed radiation along its magnetic axis). The pulsar provides an extremely precise ticking clock with which to study the motion of stars in a binary system. The pulsar is in orbit with a companion that is also a neutron star (though not an observable pulsar). The orbit is elliptical with a period of 7.75 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The period of the orbits is decreasing. That is, the two stars are gradually getting closer together. This effect is a consequence of general relativity, that predicts that such massive objects that are so close to each other will act as a source of gravitational waves. The gravitational waves have not (yet) been directly observed, but GR makes a prediction for how much energy they carry away and hence at what rate the orbital period changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate of decrease in the distance between two bodies is given by&#xA;$$\frac{dr}{dt} = -\frac{64G^3}{5c^5}\frac{(m_1 m_2)(m_1+m_2)}{r^3}$$ &#xA;and leads to a merger between the two objects in a time of&#xA;$$t = \frac{5c^5}{256G^3}\frac{r^4}{(m_1 m_2)(m_1+m_2)}, $$&#xA;where $m_1$ and $m_2$ are the masses of the two orbiting bodies, $G$ is the gravitational constant and $c$ is the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In slightly more friendly units&#xA;$$ \frac{dr}{dt} = 7.8\times10^{-19} \frac{(M_1 M_2)(M_1+M_2)}{r^3}\ au/yr,$$&#xA;where $r$ is in astronomical units and the masses are in solar masses, and&#xA;$$ t = 3.2\times10^{17} \frac{R^4}{(M_1 M_2)(M_1+M_2)}\ yr. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus you can see that the effect gets much stronger (and the merger times shorter) if you have &lt;em&gt;two&lt;/em&gt; high mass objects (it depends on the product of the masses) with a small orbital separations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &quot;normal&quot; stars you can never get them closer than their stellar radii and generally means that the effect is too small to be important. However, compact objects (white dwarfs and neutron stars) can be brought much closer together and this is why the effect has been seen there. The Hulse-Taylor binary components &#xA;have masses of $\sim 1.4 M_{\odot}$ and are separated by an average of about 2 million km (0.013 au). Inserting this into the formula above give a merger time of 1.7 billion years. A more accurate calculation that takes account of the elliptical orbit gives 300 million years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you made the neutron stars orbit with a separation of 0.1 au (more typical for a normal stellar binary system), the merger timescale would increases to well beyond the current age of the universe.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-04-03T11:59:15.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10368" PostTypeId="1" AcceptedAnswerId="10369" CreationDate="2015-04-03T18:39:22.977" Score="2" ViewCount="190" Body="&lt;p&gt;From what I understand Earth's L3 is obscured by the Sun, so is that true for our viewpoint of the other planets? Or, are there times when other planets are on the other side of the Sun from Earth and we can see their L3 position? For the Hilda's of Jupiter have we observed them in that position or is it that we have calculated their coordinates to be in the L3 position?&lt;/p&gt;&#xA;" OwnerUserId="6959" LastActivityDate="2015-04-03T23:17:21.687" Title="Can we observe what objects exist at the L3 positiion of planets of our solar system?" Tags="&lt;lagrange-point&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10369" PostTypeId="2" ParentId="10368" CreationDate="2015-04-03T22:26:41.710" Score="2" Body="&lt;p&gt;You asked&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;L3 is obscured by the Sun, so is that true for our viewpoint of the other planets?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. Each planet orbits with it's own &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_period&quot; rel=&quot;nofollow&quot;&gt;period&lt;/a&gt; because each planet is at a different distance from the sun. So most of the time we can see the L3 point of each planet's orbit from Earth. Occasionally it goes behind the sun from our point of view.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For the Hilda's of Jupiter have we observed them in that position or is it that we have calculated their coordinates to be in the L3 position?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilda_family&quot; rel=&quot;nofollow&quot;&gt;Hilda family&lt;/a&gt; do not occupy the Jupiter L3 point. They generally orbit closer to the sun than Jupiter, and orbit three times for every two orbits of Jupiter. Their orbits are interesting; they have eccentric orbits (up to 0.3) that in some cases cross the orbit of some of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Trojan_%28astronomy%29#Trojan_asteroids&quot; rel=&quot;nofollow&quot;&gt;Trojans&lt;/a&gt;. You may have noticed that some of the Hildas also approach the Jupiter L3 point. At that point, and when other members of the family approach the Tojans at L4 and l5, the Hildas are moving more slowly. So the Sun's gravity gradually accelerates them, pulling them back to the lower, faster part of their orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have linked to the Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilda_family&quot; rel=&quot;nofollow&quot;&gt;Hilda family&lt;/a&gt; page. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrangian_point&quot; rel=&quot;nofollow&quot;&gt;Lagrangian point&lt;/a&gt; page is also helpful.&lt;/p&gt;&#xA;" OwnerUserId="5524" LastEditorUserId="5524" LastEditDate="2015-04-03T23:17:21.687" LastActivityDate="2015-04-03T23:17:21.687" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10370" PostTypeId="2" ParentId="10339" CreationDate="2015-04-04T00:07:09.360" Score="3" Body="&lt;p&gt;I think there are two key aspects to the answer.  1) Solid/rocky bodies should tend to &lt;strong&gt;collide before they reach the Roche limit&lt;/strong&gt;.  2) When gaseous bodies reach the Roche limit (and undergo 'Roche-Lobe Overflow'), &lt;strong&gt;the dynamics are basically those of test-bodies&lt;/strong&gt; and are fairly straightforward and well understood from binary stellar dynamics.  To expand on both:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1) Rocky Roche Limit&lt;/strong&gt;.  If you are thinking about a gaseous donor (the object being disrupted), then this is an irrelevant point, but It sounds like rocky is what you had in mind.  To an order of magnitude, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Roche_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Roche limit&lt;/a&gt; is the same as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hill sphere&lt;/a&gt;, or the Tidal Radius (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/1988Natur.333..523R&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rees 1988&lt;/a&gt;) --- which is simply the radius at which the density of the donor equals the average density of the primary in a sphere of that radius:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_t^3/M \approx r^3/m \rightarrow R_t \approx r \left( M/m \right)^{1/3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rocky material varies &lt;em&gt;very&lt;/em&gt; little in density (e.g. &lt;a href=&quot;http://meteorites.wustl.edu/id/density.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Iron meteorites are only about twice as dense as chrondritic ones&lt;/a&gt;), which means that the densities will only match very near the radius of the primary.  The tidal bulge in the secondary (donor) also makes it easier for the objects to collide before meeting this criteria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. &lt;img src=&quot;https://i.stack.imgur.com/HeEtv.jpg&quot; alt=&quot;http://starchild.gsfc.nasa.gov/docs/StarChild/questions/question38.html&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2) Gaseous Roche Lobe Overflow&lt;/strong&gt;.  Basically every text on stars and star systems will have a &lt;a href=&quot;http://eagle.phys.utk.edu/guidry/astro411/lectures/lecture_ch13.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;section on mass transferring binaries&lt;/a&gt; which will describe the &lt;a href=&quot;http://www-astro.physics.ox.ac.uk/~podsi/binaries.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;dynamics of mass transfer (this PDF is by Philipp Podsiadlowski who is a wizard of the field)&lt;/a&gt;.  In the type of situation you are describing, like in (semi-)stable binary systems, it is a very gradual process where material is slowly syphoned off of the donor.  This material can either form an accretion disk (high angular momentum material) and gradually accrete onto the primary, or directly impact the primary (low angular momentum material).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. &lt;img src=&quot;https://i.stack.imgur.com/lWFN9.gif&quot; alt=&quot;http://www.personal.psu.edu/mtr11/research/journey.html&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="918" LastActivityDate="2015-04-04T00:07:09.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10372" PostTypeId="1" CreationDate="2015-04-04T04:31:22.413" Score="8" ViewCount="283" Body="&lt;p&gt;An earlier answer on temperature mentioned that the temp of the Cosmic Microwave Background (CMB) is &lt;span class=&quot;math-container&quot;&gt;$2.4\,{\rm K}$&lt;/span&gt; and the temp of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Boomerang_Nebula&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boomerang nebula&lt;/a&gt; as &lt;span class=&quot;math-container&quot;&gt;${\rm 1\,K}$&lt;/span&gt;. How did the nebula cool faster than the CMB?&lt;/p&gt;&#xA;" OwnerUserId="6975" LastEditorUserId="37191" LastEditDate="2021-04-07T13:28:56.363" LastActivityDate="2021-04-07T13:28:56.363" Title="Why is the Boomerang Nebula colder than the CMB?" Tags="&lt;temperature&gt;&lt;cmb&gt;&lt;nebula&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="10373" PostTypeId="1" CreationDate="2015-04-04T04:57:01.190" Score="11" ViewCount="14991" Body="&lt;p&gt;So, I watched Interstellar and if you watched it too you know that there's a planet orbiting a black-hole, they call it Miller's Planet. According to the movie, every hour on Miller's Planet is equivalent to 7 years on Earth due to gravitational pull from the black-hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: Assuming there are other life forms in the universe, is it really possible for it to be near a black-hole? Is it possible that they came to existence thousands (or even millions) of years before us, but are not as advanced as us because time on our planet is way quicker than that of theirs? is a lot more for us? If they have an assignment due tomorrow, we have, say, 100 years more time to do it (assume what else we can do in that 100 years). Or they actually are more advanced than us, but from Earth, they somehow live in the past?&lt;/p&gt;&#xA;" OwnerUserId="6976" LastEditorUserId="25727" LastEditDate="2019-01-30T05:46:00.210" LastActivityDate="2021-02-16T15:30:07.727" Title="Life planets orbiting black-holes. Can/Do they really exist?" Tags="&lt;black-hole&gt;&lt;earth&gt;&lt;space-time&gt;&lt;gravitational-waves&gt;&lt;life&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10374" PostTypeId="1" AcceptedAnswerId="11135" CreationDate="2015-04-04T06:25:44.723" Score="8" ViewCount="2968" Body="&lt;p&gt;First do we have anyway to even estimate the mass of the entire observable universe?  And then is there any data that shows mass being gained or lost?  Would we ever know if someone was playing with the til.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I want to be clear that I am not talking about small masses on the outskirts of the &quot;universe&quot; or small discrepancies in measurement or anything of the sort.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt;  I would like to add that maybe we should define the observable universe as NOW (x-date) so that we aren't calculating a moving target.  &lt;/p&gt;&#xA;" OwnerUserId="2506" LastEditorUserId="2506" LastEditDate="2015-11-03T21:00:27.580" LastActivityDate="2016-12-28T17:39:40.010" Title="Does the mass of the observable universe ever change?" Tags="&lt;universe&gt;&lt;mass&gt;" AnswerCount="4" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="10375" PostTypeId="2" ParentId="10373" CreationDate="2015-04-04T09:11:45.357" Score="13" Body="&lt;p&gt;Well, first things first.   It's not likely to have a planet orbiting near a black hole and in significant time dilation because the tidal effects would likely tear anything that close apart.   Certainly a planet orbiting a stellar mass black hole would need to be quite far away so as to not be torn apart, so any time dilation would be pretty small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Around a super-massive black hole, the tidal effects are smaller and a nearish orbit with some measure of time dilation is possible.  (see link below for more specifics), &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/110044/time-dilation-factor-for-the-circular-orbit-at-3-2-schwarzschild-radius&quot;&gt;https://physics.stackexchange.com/questions/110044/time-dilation-factor-for-the-circular-orbit-at-3-2-schwarzschild-radius&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a stable planetary orbit, you probably max out at about 20% time dilation and only around a super-massive black hole, where there's only 1 per galaxy.    The idea of 1 day to 100 years isn't practical, it's 80 days to 100 days if you're talking about a stable planetary orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and, I'm not sure you'd want to be that close to the black hole in the center of the galaxy, not because the orbit isn't stable, but because stars are in that orbit too.  It might not be a safe place to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in reality, you'd want a stellar mass black hole and a distant orbit, where the time dilation would be quite small, and in that scenario, yes, life is possible because of tidal energy, so a planet could have liquid surface water and an atmosphere, even if the black hole gave off very little light and heat.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a planet in a tidal-energy orbit would probably be tidally locked which would protect the far side from any gamma rays the black hole spits out when it eats, so it would in theory be a good place for life.   No significant light source, unless it was a binary system, so, plants would have it harder, but there would be heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's another problem.   The creation of black holes tends to blow everything appart in a huge explosion.  It's unclear that a planet would survive a black hole's birth, so you might need a captured planet.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, intelligent life . . . we really don't know enough about how common intelligent life is on other planets.   Life might be fairly enough, but intelligent life is far less clear and there's other factors than just time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today, we simply don't know enough to predict if there's intelligent life out there or not.   There probably is life elsewhere in the universe, though even that isn't 100% certain, but regarding intelligent life, there's far too many unknowns in that equation.   I think a black hole might not be optimal for the formation intelligent life because of the lack of light so, much less photosynthesis, so, slower oxygen formation (if it follows the same pattern the earth does) and the unlikeliness that a planet would survive the black hole's creation.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="11264" LastEditDate="2019-01-30T15:57:58.197" LastActivityDate="2019-01-30T15:57:58.197" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="10376" PostTypeId="1" AcceptedAnswerId="10378" CreationDate="2015-04-04T09:28:11.610" Score="4" ViewCount="1573" Body="&lt;p&gt;Some stars such as a neutron star can spin very fast around 600 times a second, wouldn't the star be ripped apart? Although the gravitational pull is very strong but gravity is the weakest of all known forces and any good theory why neutron stars spin that fast does it have anything to do with the magnetic field and the surrounding dust clouds?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="6760" LastEditDate="2015-04-04T09:39:49.963" LastActivityDate="2020-06-04T22:56:50.693" Title="What happens when star spins too fast?" Tags="&lt;neutron-star&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10377" PostTypeId="2" ParentId="10376" CreationDate="2015-04-04T09:53:10.187" Score="2" Body="&lt;p&gt;Related: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/7778/is-it-possible-to-break-apart-a-neutron-star/7802#7802&quot; title=&quot;Is it possible to break apart a neutron star?&quot;&gt;Is it possible to break apart a neutron star?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rotation must be powerful enough to overpower the force of gravity in order for anything to happen.  I'm not sure about the 'ripping apart,' but if the rotation does overpower gravity surface material will be ejected from the body, a process known as &lt;strong&gt;mass shedding&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As neutron stars are extremely dense they can have great amounts of angular velocity, like the pulsar class neutron star mentioned in your question. If you can quickly increase the angular velocity of the body without also increasing the mass and density, perhaps you could rip it apart, but that would be very difficult to do to a neutron star.&lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-04-04T10:07:21.910" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10378" PostTypeId="2" ParentId="10376" CreationDate="2015-04-04T12:36:21.370" Score="7" Body="&lt;p&gt;The structure (mass versus radius and density profile) &lt;em&gt;is&lt;/em&gt; influenced by its rotation rate, but not by as much as you might think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in Newtonian physics you can think of a mass element &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; at the surface of a star of mass &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; and radius &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt;, rotating with angular velocity &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A condition for stability would be that the surface gravity is strong enough to provide the centripetal acceleration of the test mass.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{GMm}{R^2} &amp;gt; m R \omega^2$$&lt;/span&gt;&#xA;If this is not satisfied then the object might break up (it is more complicated than this because the object will not stay spherical and the radius at the equator will increase etc., but these are small numerical factors).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \omega &amp;lt; \left(\frac{GM}{R^3}\right)^{1/2}$$&lt;/span&gt;&#xA;or in terms of rotation period &lt;span class=&quot;math-container&quot;&gt;$P = 2\pi/\omega$&lt;/span&gt; and so&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ P &amp;gt; 2\pi \left(\frac{GM}{R^3}\right)^{-1/2},$$&lt;/span&gt;&#xA;is the condition for stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a typical &lt;span class=&quot;math-container&quot;&gt;$1.4M_{\odot}$&lt;/span&gt; neutron star with radius 10 km, then &lt;span class=&quot;math-container&quot;&gt;$P&amp;gt;0.46$&lt;/span&gt; milliseconds. A proper General Relativistic calculation of this limit would give a similar result, but depends to some extent on the equation of state of the neutron star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Happily, this is easily satisfied for all observed neutron stars - they can spin extremely fast because of their &lt;em&gt;enormous&lt;/em&gt; surface gravities and all are well below the instability limit. I believe the fastest known rotating pulsar has a period of 1.4 milliseconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also ask how pulsars can attain these speeds. There are two classes of explanation for the two classes of pulsars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most pulsars are thought (at least initially) to be the product of a core-collapse supernova. The core collapses from something a little smaller than the radius of the Earth, to about 10km radius in a fraction of a second. Conservation of angular momentum demands that the rotation rate increases as the inverse of the radius &lt;em&gt;squared&lt;/em&gt;. i.e. The spin rate increases by factors of a million or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pulsars spin down with age because they turn their rotational kinetic energy into magnetic dipole radiation. However, the fastest rotating pulsars - the &quot;millisecond pulsars&quot; are &quot;reborn&quot;, by accreting material from a binary companion. The accreted material has angular momentum and the accretion of this angular momentum is able to spin the neutron star up to very high rates because it has a relatively (for a stellar-mass object) small moment of inertia.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="16685" LastEditDate="2020-06-04T22:56:50.693" LastActivityDate="2020-06-04T22:56:50.693" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10379" PostTypeId="2" ParentId="10341" CreationDate="2015-04-04T12:59:07.700" Score="3" Body="&lt;p&gt;One practical use of astronomy (I actually don't go along with the premise that it needs to be &lt;em&gt;directly&lt;/em&gt; useful - there are many examples of blue-skies research turning into applications and I don't see why astronomy should differ; there is also the issue of inspiring and training the next generation of scientists: something that astronomy and astrophysics is particularly suited to) is in space weather forecasting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An understanding of solar activity, coronal mass ejections and the geocoronal environment has led to power generation grids, satellite operators etc. being able to take action to minimise the damage and disturbance caused by forecasted geomagnetic storms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going back a bit, astronomy used to be fairly critical for navigation! I guess we've all got used to GPS...&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-04-04T12:59:07.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10380" PostTypeId="1" CreationDate="2015-04-04T18:41:43.613" Score="4" ViewCount="257" Body="&lt;p&gt;Why aren't galaxies distributed evenly in space, but instead form those sponge-like structures, with huge voids between filaments?&lt;/p&gt;&#xA;" OwnerUserId="6920" LastActivityDate="2015-04-14T21:04:00.303" Title="what is the current explanation for the formation of cosmic voids?" Tags="&lt;cosmology&gt;&lt;galaxy-cluster&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10381" PostTypeId="2" ParentId="10380" CreationDate="2015-04-04T19:06:42.040" Score="1" Body="&lt;p&gt;Void formation is due to &lt;a href=&quot;https://en.wikipedia.org/wiki/Baryon_acoustic_oscillations&quot; rel=&quot;nofollow&quot;&gt;baryon acoustic oscillations&lt;/a&gt;. These can be thought of as fluctuations in the dense matter in the early stages of the universe; they became &lt;a href=&quot;https://en.wikipedia.org/wiki/Anisotropy&quot; rel=&quot;nofollow&quot;&gt;anisotropies&lt;/a&gt;, which can be seen in the cosmic microwave background. The fluctuations grew, until they hit the &lt;a href=&quot;http://www.sdss3.org/surveys/boss.php&quot; rel=&quot;nofollow&quot;&gt;150 Mpc scale&lt;/a&gt;, which correlates nicely to the sizes of voids in the universe today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's an interesting section (4.2) in &lt;a href=&quot;http://arxiv.org/pdf/astro-ph/9508159v2.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; by Robert Brandenberger, which gives a mathematical treatment of how fluctuations of &lt;em&gt;all&lt;/em&gt; kinds grew in the early universe. There is a better version of the explanation in &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1972MNRAS.160P...1Z&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow&quot;&gt;Zeldovich (1972)&lt;/a&gt;, which in fact puts forth the theory. The number density (if you'll pardon the pun) is lesser in that latter, but the former has a good overview, as it is meant to be a broad summary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simple explanation is that tiny fluctuations grew over time to become enormous structures.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2015-04-04T19:06:42.040" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10382" PostTypeId="1" AcceptedAnswerId="10383" CreationDate="2015-04-04T23:07:03.840" Score="2" ViewCount="326" Body="&lt;p&gt;Please forgive my ignorance, I am not an astronomer, just an interested layman.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The further away a galaxy is from us, the faster it is moving away from us.  But the galaxies we see exist in the distant past because it takes a long time for their light to reach us. The closer galaxies are, the more recent their light is, and the slower they move away from us. Doesn’t this seem to indicate that the expansion of the universe is slowing down, because the more recent light from galaxies, appears to be moving away slower? Does this make sense?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quote from &lt;a href=&quot;http://en.wikipedia.org/wiki/Big_Bang#Overview&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hubble observed that the distances to faraway galaxies were strongly correlated with their redshifts. This was interpreted to mean that all distant galaxies and clusters are receding away from our vantage point with an apparent velocity proportional to their distance: that is, the farther they are, the faster they move away from us, regardless of direction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6981" LastEditorUserId="2365" LastEditDate="2015-11-04T20:23:21.977" LastActivityDate="2015-11-04T20:23:21.977" Title="How do we know the universe's expansion is speeding up?" Tags="&lt;big-bang-theory&gt;&lt;expansion&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10383" PostTypeId="2" ParentId="10382" CreationDate="2015-04-05T01:10:35.780" Score="1" Body="&lt;p&gt;As a fellow layman, I'll give this my best shot.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The further away a galaxy is from us, the faster it is moving away&#xA;  from us. But the galaxies we see exist in the distant past because it&#xA;  takes a long time for their light to reach us. The closer galaxies&#xA;  are, the more recent their light is, and the slower they move away&#xA;  from us.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All true.   Think of a single point explosion, at any given time, the object twice as far should be moving twice as fast, cause they both started in the same place at the same time, so twice as fast moves twice s far.   There are, complications - with the explosion there's air resistance, with the universe well, I'll get to that.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Doesn’t this seem to indicate that the expansion of the universe is&#xA;  slowing down, because the more recent light from galaxies, appears to&#xA;  be moving away slower? Does this make sense?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;OK, it sounds like you have an incorrect assumption there.   Expansion means the far away is moving away faster and the closer is moving away slower, that's true with accelerating or decelerating expansion.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What they look for when studying expansion is measuring the precise speed and comparing that to distance.   In steady expansion, you expect to see the galaxy 8 billion light years away to be apparently moving away twice as fast as the galaxy 4 billion light years away.  (I say apparently cause we can only observe the relative velocity from 8 billion years in the past).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, whether the galaxy 8 billion light years is traveling slightly more or slightly less than twice as fast away from us as the galaxy 4 billion light years away - that's what tells us about acceleration vs deceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a car driving 60 MPH, and the driver puts the ever so slightest pressure on one of the pedals, but you don't know which one, and similarly the car driving 30 MPH does the same, but we can only see how fast the 60 MPH car was driving 6 minutes ago and we see how fast the car driving 30 was driving 3 minutes ago - that gives the 30 MPH car more time to apply the acceleration or deceleration.    It's that time differential that gives us the information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, like the cars above, the galaxy 4 billion light years away has been in the accelerating expansion of space longer than the one 8 billion years ago and that's what tells us that the acceleration of expansion is happening, the nearer galaxies are moving a bit faster away from us than they would in a steady expansion or a gravitationally slowed expansion.  (edited to make a bit more clear)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The newer light is moving slower. The older light moving faster. So&#xA;  from this it seems that the universe was expanding faster in the past?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;light doesn't move slower or faster.  it red-shifts when objects are moving away from each other.  The faster they move away the greater the red-shift.  That's one way relative velocity can be measured. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-04-05T04:15:50.680" LastActivityDate="2015-04-05T04:15:50.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10384" PostTypeId="1" AcceptedAnswerId="10385" CreationDate="2015-04-05T06:01:36.450" Score="10" ViewCount="2691" Body="&lt;p&gt;There seems to be not enough matter to account for the fact that the speed of some stars located far from the Galaxy center is almost the same speed as those stars found nearer to the center of the Galaxy, therefore it seems there should be even more matter: so this missing matter is termed dark matter. There is a kind of stellar object that isn't massive enough to produce its own light but is many times more massive than Jupiter, a.k.a brown dwarf. Since it isn't bright enough to be detected by our deep space telescope nor the earth bound counterparts, they should be numerous in galaxies as I believe these objects sits in the middle range between planets and stars. However it seems the scientific community had ruled out this potential dark matter candidate without scanning the unobservable Milky Way galaxy as well as billions other with different light spectrums. Why aren't brown dwarfs considered candidates for the mysterious dark matter?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="2365" LastEditDate="2015-11-04T20:23:04.217" LastActivityDate="2016-09-26T18:30:40.687" Title="Why aren't brown dwarfs the dark matter?" Tags="&lt;dark-matter&gt;&lt;brown-dwarf&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10385" PostTypeId="2" ParentId="10384" CreationDate="2015-04-05T08:21:30.087" Score="10" Body="&lt;p&gt;Two reasons.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;We know from looking at galaxy rotation curves and the motion of galaxies in clusters and from gravitational lensing, that the amount of &quot;dark matter&quot; is some 30% of the density of the universe. But on the other hand, estimates of the abundances of deuterium, helium, tritium and lithium produced in the big bang indicate that only 5% of the density of the universe can be in the form of normal matter (e.g atoms, protons, neutrons etc). As brown dwarfs are made of these things, they can only &lt;em&gt;contribute&lt;/em&gt; to this small percentage and cannot be responsible for most of the dark matter (the same is true for anything made of normal, &quot;baryonic matter&quot;).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Astronomers have &lt;em&gt;counted&lt;/em&gt; how many brown dwarfs there are, both in the local neighbourhood and in star clusters (see for example &lt;a href=&quot;http://adsabs.harvard.edu/abs/2008ApJ...683L.183A&quot;&gt;Andersen et al. 2008&lt;/a&gt;;  &lt;a href=&quot;http://adsabs.harvard.edu/abs/2011ApJS..197...19K&quot;&gt;Kirkpatrick et al. 2011&lt;/a&gt;; &lt;a href=&quot;http://adsabs.harvard.edu/abs/2011ApJS..197...19K&quot;&gt;Burningham et al. 2013&lt;/a&gt;; . In other words we &lt;em&gt;can&lt;/em&gt; see brown dwarfs - they emit most of their light in the infrared part of the spectrum, and surveys like SDSS, 2MASS and WISE have uncovered them in their thousands. It turns out that there is about 1 brown dwarf for every 4 more massive stars. So, although numerous, they contribute a very tiny fraction of the normal matter in our Galaxy. In addition, microlensing experiments towards the Galactic bulge and Magellanic clouds suggest that although there are lots of brown dwarfs elsewhere in our Galaxy (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2000ApJ...542..281A&quot;&gt;Alcock et al. 2000&lt;/a&gt;; &lt;a href=&quot;http://adsabs.harvard.edu/abs/2008A%26A...480..723C&quot;&gt;Novati et al. 2008&lt;/a&gt;), there are not enough to make much of a contribution to dark matter, and so the results from our solar neighbourhood appear representative of the Galaxy as a whole.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-04-05T15:01:38.230" LastActivityDate="2015-04-05T15:01:38.230" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10386" PostTypeId="2" ParentId="2375" CreationDate="2015-04-05T11:30:26.810" Score="4" Body="&lt;p&gt;This question is &lt;em&gt;very&lt;/em&gt; broad - there are very many techniques for estimating temperatures, so I will stick to a few principles and examples. When we talk about measuring the temperature of a star, the only stars we can actually resolve and measure are in the local universe; they do not have appreciable redshifts and so this is rarely of any concern. Stars do of course have line of sight velocities which give their spectrum a redshift (or blueshift). It is a reasonably simple procedure to correct for the line of sight velocity of a star, because the redshift (or blueshift) applies to all wavelengths equally and we can simply shift the wavelength axis to account for this. i.e. We put the star back into the rest-frame before analysing its spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gerald has talked about the blackbody spectrum - indeed the wavelength of the peak of a blackbody spectrum is inversely dependent of temperature through &lt;a href=&quot;http://en.wikipedia.org/wiki/Wien%27s_displacement_law&quot; rel=&quot;nofollow&quot;&gt;Wien's law&lt;/a&gt;. This method could be used to estimate the temperatures of objects that have spectra which closely approximate blackbodies and for which flux-calibrated spectra are available that properly sample the peak. Both of these conditions are hard to satisfy in practice: stars are in general not blackbodies, though their &lt;em&gt;effective&lt;/em&gt; temperatures - which is usually what is quoted, are defined as the temperature of a blackbody with the same radius and luminosity of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;effective temperature&lt;/em&gt; of a star is most accurately measured by (i) estimating the total flux of light from the star; (ii) getting an accurate distance from a parallax; (iii) combining these to give the luminosity; (iv) measuring the radius of the star using interferometry; (v) this gives the effective temperature from Stefan's law:&#xA;$$ L = 4\pi R^2 \sigma T_{eff}^4,$$&#xA;where $\sigma$ is the Stefan-Boltzmann constant. Unfortunately the limiting factor here is that it is difficult to measure the radii of all but the largest or nearest stars. So measurements exist for a few giants and a few dozen nearby main sequence stars; but these are the fundamental calibrators against which other techniques are compared and calibrated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second major secondary technique is a detailed analysis of the spectrum of a star. To understand how this works we need to realise that (i) atoms/ions have different energy levels; (ii) the way that these levels are populated depends on temperature (higher levels are occupied at higher temperatures); (iii) transitions between levels can result in the emission or absorption of light at a particular wavelength that depends on the energy difference between the levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use these properties we construct a model of the atmosphere of a star. In general a star is hotter on the inside and cooler on the outside. The radiation coming out from the centre of the star is absorbed by the cooler, overlying layers, but this happens preferentially at the wavelengths corresponding to energy level differences in the atoms that are absorbing the radiation. This produces absorption lines in the spectrum. A spectrum analysis consists of measuring the strengths of these absorption lines for many different chemical elements and different wavelengths. The strength of an absorption line depends primarily on (i) the temperature of the star and (ii) the amount of a particular chemical element, but also on several other parameters (gravity, turbulence, atmospheric structure). By measuring lots of lines you isolate these dependencies and emerge with a solution for the temperature of the star - often with a precision as good as +/-50 Kelvin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where you don't have a good spectrum, the next best solution is to use the colour of the star to estimate its temperature. This works because hot stars are blue and cool stars are red. The colour-temperature relationship is calibrated using the measured colours of the fundamental calibrator stars. Typical accuracies of this method are +/- 100-200 K (poorer for cooler stars).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-04-05T17:10:36.317" LastActivityDate="2015-04-05T17:10:36.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10387" PostTypeId="1" CreationDate="2015-04-05T12:20:26.120" Score="16" ViewCount="824" Body="&lt;p&gt;What are some open problems in astronomy that an amateur would have a chance of solving? Suppose the amateur has a PhD in some other field, owns a basic telescope, a set of filters, diffraction gratings, cameras, and happens to know a lot about machine learning, signal processing, spectral estimation, statistics &amp;amp; design of experiments, and basic physics and chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there better tags available (such as &quot;research&quot;)?&lt;/p&gt;&#xA;" OwnerUserId="6983" LastEditorUserId="941" LastEditDate="2017-04-10T11:59:28.617" LastActivityDate="2017-04-10T11:59:28.617" Title="Open problems in astronomy that an amateur (with a PhD in some other field) would have a chance of solving?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;astrophysics&gt;&lt;fundamental-astronomy&gt;&lt;stellar-astrophysics&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10388" PostTypeId="1" AcceptedAnswerId="10389" CreationDate="2015-04-05T13:51:10.670" Score="2" ViewCount="377" Body="&lt;p&gt;Is there a publicly available list of the Gliese catalogue of nearby stars with their constellations? or a Gliese to Hipparcos List ?&lt;/p&gt;&#xA;" OwnerUserId="5401" LastEditorUserId="2531" LastEditDate="2015-04-05T13:57:57.717" LastActivityDate="2015-10-31T20:21:07.833" Title="Is there a publicly available Gliese Star List" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10389" PostTypeId="2" ParentId="10388" CreationDate="2015-04-05T13:57:00.967" Score="1" Body="&lt;p&gt;The Gliese catalogue can be found at CDS &lt;a href=&quot;http://cdsarc.u-strasbg.fr/viz-bin/Cat?V/70A&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an RA and Dec then you can use the fortran program found &lt;a href=&quot;http://cdsarc.u-strasbg.fr/viz-bin/Cat?VI/42&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; to get the constellation.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-04-05T13:57:00.967" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10393" PostTypeId="1" AcceptedAnswerId="10407" CreationDate="2015-04-05T22:40:35.960" Score="0" ViewCount="85" Body="&lt;p&gt;do you know any website in which I could obtain orbital elements of planets at some date?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, I'm now needing the orbital elements of Mars around June, 2014.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2877" LastActivityDate="2015-04-07T14:53:14.817" Title="Orbital Elements Mars near June 2014" Tags="&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10395" PostTypeId="1" CreationDate="2015-04-06T12:12:50.500" Score="4" ViewCount="1616" Body="&lt;p&gt;What happens when asteroids (or other small bodies) &quot;collide&quot; with a gas planet? In my head it would either go through it or just stay &quot;inside&quot; it due to the strong gravity. I tend to believe in the latter one, but if this is the case, does this mean that the gas giants are full of small bodies inside them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="6661" LastEditorDisplayName="user15217" LastEditDate="2016-12-24T09:03:34.913" LastActivityDate="2016-12-24T09:03:34.913" Title="Collision of asteroid and gaseous planet - what would happen" Tags="&lt;planet&gt;&lt;gravity&gt;&lt;asteroids&gt;&lt;gas-giants&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10396" PostTypeId="2" ParentId="10395" CreationDate="2015-04-06T13:57:02.887" Score="4" Body="&lt;p&gt;We have observed impacts with Jupiter&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In 1993 comet &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Comet_Shoemaker%E2%80%93Levy_9&quot; rel=&quot;nofollow&quot;&gt;Shoemaker Levy 9&lt;/a&gt; impacted. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In 2009 an unknown object &lt;a href=&quot;http://en.m.wikipedia.org/wiki/2009_Jupiter_impact_event&quot; rel=&quot;nofollow&quot;&gt;impacted&lt;/a&gt;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In each case the object was destroyed in Jupiter's atmosphere. &lt;/p&gt;&#xA;" OwnerUserId="5524" LastActivityDate="2015-04-06T13:57:02.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10397" PostTypeId="2" ParentId="1676" CreationDate="2015-04-06T15:23:16.917" Score="5" Body="&lt;p&gt;Let's assume that what is falling onto the neutron star is &quot;normal&quot; material - i.e. a planet, an asteroid or something like that. As the material heads towards the neutron star it gains an enormous amount of kinetic energy. If we assume it starts from infinity, then the energy gained (and turned into kinetic energy) is approximately (ignoring GR)&#xA;$$ \frac{1}{2}m v^2 = \frac{GMm}{R}, $$&#xA;where $m$ is the mass of the object (which cancels) and $M$ and $R$ are the mass and radius of the neutron star (let's assume typical values of $1.4 M_{\odot}$ and 10 km respectively).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This results in a velocity as it approaches the neutron star surface of $1.9 \times 10^{8}$ m/s - i.e. big enough that you would have to do the calculation using relativistic mechanics actually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I doubt that the object would get to the surface intact, due to tidal forces. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Roche_limit&quot; rel=&quot;noreferrer&quot;&gt;Roche limit&lt;/a&gt; for the breakup of a rigid object occurs when the &#xA;object is a distance &#xA;$$d = 1.26 R \left(\frac{\rho_{NS}}{\rho_O}\right)^{1/3},$$&#xA;where $\rho_{NS}$ and $\rho_O$ are the average densities of our neutron star and object respectively. For rocky material, $\rho_O \simeq 5000$ kg/m$^{3}$. For our fiducial neutron star $\rho_{NS} \simeq 7\times10^{17}$ kg/m$^{3}$. Thus when the object gets closer than $d= 500,000$ km it will disintegrate into its constituent atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will thus arrive in the vicinity of the neutron star as an extremely hot, ionised gas. But if the material has even the slightest angular momentum it could not fall directly onto the neutron star surface without first shedding that angular momentum. It will therefore form (or join) an accretion disk. As angular momentum is transported outwards, material can move inwards until it is hooked onto the neutron star magnetic field and makes its final journey onto the neutron surface, probably passing through an accretion shock as it gets close to the magnetic pole, if the object is already accreting strongly. Roughly a few percent of the rest mass energy is converted into kinetic energy and then heat which is partly deposited in the neutron star crust along with matter (nuclei and electrons) and partly radiated away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the high densities in the outer crust the raw material (certainly if it contains many protons) will be burned in rapid nuclear reactions. If enough material is accreted in a short time this can lead to a runaway thermonuclear burst until all the light elements have been consumed. Subsequent electron captures make the material more and more neutron rich until it settles down to the equilibrium composition of the crust, which consists of neutron-rich nuclei and ultra-relativistically degenerate electrons (no free neutrons).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-04-06T15:23:16.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10399" PostTypeId="2" ParentId="10341" CreationDate="2015-04-06T21:13:38.517" Score="4" Body="&lt;p&gt;Are you willing to include all of astronomy, including the basics of mapping the positions and movements of astronomical features?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, then astronomy has been the theoretical and practical underpinning of basically all timekeeping and navigation, for all of human history, and every other technology or practice that depends on them. You can't get more practical than &quot;when do I plant the crops&quot; or &quot;which direction do I sail to get to my destination,&quot; both of which have life-and-death consequences.&lt;/p&gt;&#xA;" OwnerUserId="366" LastActivityDate="2015-04-06T21:13:38.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10400" PostTypeId="1" CreationDate="2015-04-06T21:36:00.733" Score="1" ViewCount="262" Body="&lt;p&gt;The main idea about dark matter today seems to be supersymmetry. That there exists a set of yet undiscovered fundamental particles which (in a mathematically most convenient way) mirror the particles already known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But has it been established that dark matter consists of particles at all? Couldn't it for example be a continuous property of space itself in some regions? That some parts of space are curved without the influence of any particles or waves. Is anything but particles ruled out?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-12-09T18:56:44.720" Title="Is it certain that dark matter consists of particles? (And not just curved space)" Tags="&lt;gravity&gt;&lt;dark-matter&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10401" PostTypeId="2" ParentId="10400" CreationDate="2015-04-06T23:07:44.057" Score="4" Body="&lt;p&gt;Also asked at &lt;a href=&quot;https://physics.stackexchange.com/questions/174080/how-do-we-know-dark-matter-isnt-curved-spacetime&quot;&gt;https://physics.stackexchange.com/questions/174080/how-do-we-know-dark-matter-isnt-curved-spacetime&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically no. Or at least you can't have this idea &lt;em&gt;and&lt;/em&gt; General Relativity. GR demands that you have something (matter/energy density) to cause the curvature. Curvature without cause is not part of the model. That's not to say that what you propose couldn't be true, but one would then have to ditch GR, which is a reasonably successful model (though some would argue that the dark matter problem is one of its biggest failures!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think a lot of parameter space &lt;em&gt;has&lt;/em&gt; been ruled out for dark matter. We know it can't be baryonic from our understanding of primordial element abundances and also from the way it interacts (or doesn't interact) with normal matter in forming the basic structures in our universe on the right timescales. The microlensing surveys that have been conducted rule out large (planetary sized and above) dark matter bodies (and black holes, brown dwarfs, cold white dwarfs etc.), since to account for all the mass, these would produce lots of microlensing events that simply aren't observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An excellent primer on these topics has been produced by &lt;a href=&quot;http://arxiv.org/abs/1006.2483&quot; rel=&quot;nofollow noreferrer&quot;&gt;Garrett &amp;amp; Duda (2011)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-04-06T23:07:44.057" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10403" PostTypeId="1" CreationDate="2015-04-07T06:15:30.503" Score="7" ViewCount="144" Body="&lt;p&gt;In the New Scientist article &lt;a href=&quot;http://www.newscientist.com/article/mg22630153.600-is-this-et-mystery-of-strange-radio-bursts-from-space.html?page=1#.VSNy5fmUcVY&quot; rel=&quot;noreferrer&quot;&gt;Is this ET? Mystery of strange radio bursts from space&lt;/a&gt;, it is reported that several times since 2001, astronomers have detected &lt;a href=&quot;http://www.newscientist.com/article/dn12699-extragalactic-radio-burst-puzzles-astronomers.html#.VSN1IvmUcVY&quot; rel=&quot;noreferrer&quot;&gt;fast radio bursts&lt;/a&gt; that seem to have a frequency of multiples of 187.5.  In the article, the theory that it could be a pulsar is discounted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any astronomical phenomena that could emit strong radio waves with multiples of a discrete frequency?&lt;/p&gt;&#xA;" OwnerDisplayName="user6996" LastEditorUserId="2153" LastEditorDisplayName="user6996" LastEditDate="2021-05-03T15:28:29.550" LastActivityDate="2021-05-03T15:28:29.550" Title="Are there any astronomical phenomena that could emit strong radio waves with multiples of a discrete frequency?" Tags="&lt;radio-astronomy&gt;&lt;data-analysis&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="10404" PostTypeId="2" ParentId="10341" CreationDate="2015-04-07T07:41:29.777" Score="0" Body="&lt;p&gt;Everything is in space. Everything! Where else would it be? So anything anyone imagine as &quot;practical&quot;, is in space. It is the only place to go.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-04-07T07:41:29.777" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10407" PostTypeId="2" ParentId="10393" CreationDate="2015-04-07T14:53:14.817" Score="3" Body="&lt;p&gt;As @barrycarter mentioned you can use the Horizons web interface from JPL: &lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow&quot;&gt;http://ssd.jpl.nasa.gov/horizons.cgi&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can change the setting in the web form by clicking on the [change] links for the different settings (for instance the time span or the planet). There are several different settings that can be changed to get different outputs (table columns) and such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really want the orbital elements (and not just the Right Ascension and Declination) you will have to select ELEMENTS in the Ephemeris type settings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then get a table with the following elements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;JDCT     Epoch Julian Date, Coordinate Time&lt;/li&gt;&#xA;&lt;li&gt;EC     Eccentricity, e                                                   &lt;/li&gt;&#xA;&lt;li&gt;QR     Periapsis distance, q (AU)                                        &lt;/li&gt;&#xA;&lt;li&gt;IN     Inclination w.r.t xy-plane, i (degrees)                           &lt;/li&gt;&#xA;&lt;li&gt;OM     Longitude of Ascending Node, OMEGA, (degrees)                     &lt;/li&gt;&#xA;&lt;li&gt;W      Argument of Perifocus, w (degrees)                                &lt;/li&gt;&#xA;&lt;li&gt;Tp     Time of periapsis (Julian day number)                             &lt;/li&gt;&#xA;&lt;li&gt;N      Mean motion, n (degrees/day)                                      &lt;/li&gt;&#xA;&lt;li&gt;MA     Mean anomaly, M (degrees)                                         &lt;/li&gt;&#xA;&lt;li&gt;TA     True anomaly, nu (degrees)                                        &lt;/li&gt;&#xA;&lt;li&gt;A      Semi-major axis, a (AU)                                           &lt;/li&gt;&#xA;&lt;li&gt;AD     Apoapsis distance (AU)                                            &lt;/li&gt;&#xA;&lt;li&gt;PR     Sidereal orbit period (day)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="864" LastActivityDate="2015-04-07T14:53:14.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10408" PostTypeId="2" ParentId="10400" CreationDate="2015-04-07T17:46:45.737" Score="-1" Body="&lt;p&gt;In terms of dark matter, there are two notions which are incorrect. One is that dark matter is a clump of stuff traveling with the matter. The other is that dark matter does not interact with matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark matter fills 'empty' space. Dark matter is displaced by matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Milky Way moves through and displaces the dark matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Milky Way's halo is the state of displacement of the dark matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The state of displacement of the dark matter is also known as the deformation of spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Milky Way's halo &lt;strong&gt;&lt;em&gt;is&lt;/em&gt;&lt;/strong&gt; the deformation of spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark matter is the physical manifestation of spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'Ether and the Theory of Relativity by Albert Einstein' &#xA;&lt;a href=&quot;http://www-groups.dcs.st-and.ac.uk/~history/Extras/Einstein_ether.html&quot; rel=&quot;nofollow&quot;&gt;http://www-groups.dcs.st-and.ac.uk/~history/Extras/Einstein_ether.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Think of waves on the surface of water. Here we can describe two entirely different things. Either we may observe how the undulatory surface forming the boundary between water and air alters in the course of time; or else-with the help of small floats, for instance - we can observe how the position of the separate particles of water alters in the course of time. If the existence of such floats for tracking the motion of the particles of a fluid were a fundamental impossibility in physics - if, in fact nothing else whatever were observable than the shape of the space occupied by the water as it varies in time, we should have no ground for the assumption that water consists of movable particles. But all the same we could characterise it as a medium.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if, in fact nothing else whatever were observable than the shape of the space occupied by the dark matter as it varies in time, we should have no ground for the assumption that dark matter consists of movable particles. But all the same we could characterise it as a medium having mass which is displaced by the particles of matter which exist in it and move through it.&lt;/p&gt;&#xA;" OwnerUserId="6999" LastActivityDate="2015-04-07T17:46:45.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10410" PostTypeId="2" ParentId="10380" CreationDate="2015-04-07T22:07:37.043" Score="1" Body="&lt;p&gt;You might like to watch this short movie from one of the authorities on this subject:&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=wI12X2zczqI&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=wI12X2zczqI&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The movie starts by answering the question &quot;what is the cosmic web and what does it look like?&quot;.  Simulations can reproduce such structures, but they do not explain &quot;why&quot; they look the way they do.  The clue lies in the geometric appearance of the structures: The voids resemble the elements of a Voronoi tessellation in which the walls of the Voronoi polyhedra intersect in lines that are identified with the filaments.  We need to understand the dynamics that generated this pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At around 2:30 the movie addresses the issue of why it looks the way it does using a simple ballistic model for the motion of a set of discrete particles that represent a random density field (the Zel'dovich model).  The filaments are there but too fuzzy.  So at ~3:00 it generalizes this so as to achieve greater realism by making the particles sticky.  This is the &quot;adhesion model&quot;, which is governed by the Burgers-Hopf equation, which is easily solved by geometric means.  There is no gravity in this - it's ballistics acting on density field generated by a Gaussian Random process with a known covariance function (or power spectrum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The adhesion model is a remarkable representation of the structure.  It shows how nothing &quot;special&quot; is required to generate the pattern of the cosmic web.  That's the &quot;how&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;why does the structure look like it does?&quot; is a little more complex: this is the question of emergent geometry which is understood in terms of the technical language of the Lagrangian description of the flow. But the graphics is pretty and quite didactic, so worth a look.  This happens after around 5:30.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The movie compares appearance of the flow simultaneously in both Lagrangian and Eulerian space.  In the Lagrangian picture the particles have fixed coordinates, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A short descriptive article goes with the movie:&#xA;&lt;a href=&quot;http://arxiv.org/pdf/1205.1669v1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/1205.1669v1.pdf&lt;/a&gt;&#xA;and a short explanatin of the Lagrangiuan view is at&#xA;&lt;a href=&quot;http://arxiv.org/pdf/1211.5385v1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/1211.5385v1.pdf&lt;/a&gt;&#xA;This is part of the thesis work of Johan Hidding from the Kapteyn Instutue in Groningen (supevisor Rien van de Weygaert - the father of the cosmic web according to Richard Bond).  A full presentation of the nature of the singularities in terms of Morse Theory is at &#xA;&lt;a href=&quot;http://arxiv.org/pdf/1311.7134v1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/1311.7134v1.pdf&lt;/a&gt; &#xA;and there are several other papers pending.&lt;/p&gt;&#xA;" OwnerUserId="5301" LastEditorUserId="5301" LastEditDate="2015-04-09T22:33:21.587" LastActivityDate="2015-04-09T22:33:21.587" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10412" PostTypeId="2" ParentId="10341" CreationDate="2015-04-07T22:30:47.803" Score="1" Body="&lt;p&gt;From the point of view of public interest in sciences, astronomy has immense pubic appeal, perhaps only second to dinosaurs: it has a real &quot;wow&quot; value.  It can be readily appreciated by non-scientists and provides a brilliant entry in science education.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other aspect of astronomy is that it en-trains other sciences: nuclear physics, high energy physics, plasma physics, fluid dynamics, relativity, radiation processes, and so on.  Astronomy, or astrophysics, pushes the boundaries of and contributes to all of these subjects.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the most remarkable, in terms of &quot;everyday use&quot;, is Einstein's General relativity without which our GPS navigation systems would be hopelessly wrong (and difficult to correct).&lt;/p&gt;&#xA;" OwnerUserId="5301" LastActivityDate="2015-04-07T22:30:47.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10414" PostTypeId="2" ParentId="10097" CreationDate="2015-04-09T00:29:58.850" Score="3" Body="&lt;p&gt;You may have a hard time finding definitive answers to these questions. I'm after the same answers and hope this will contribute as a starting point to something more specific and comprehensive .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally the abundance of elements appears to be closely related to their mass. In effect the heavier the element the more scarce it will be. This is due to the greatly increasing amount of energy needed for the fusion process to occur as you progress towards the heavier elements in the periodic table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elements heavier than Fe (Iron) are said to require an event such as a supernova explosion to form (or an environment with similar temperatures), which would occur once the star burns through its silicon fuel and collapses generating enough energy for the subsequent fusion reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/yQexV341t-E?t=10m22s&quot; rel=&quot;nofollow noreferrer&quot;&gt;This video describes the process&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without going into compositions of elements you could probably start with the relative abundance of individual elements to determine the odds of them existing on any given planet and from there get into the forms/chemical compositions they would form in the planet's crust and atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a graph from wikipedia showing the relative abundance of elements based on spectroscopy (based on emission and absorption spectra of the elements) measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fRpG3.png&quot; alt=&quot;Element abundance in the universe&quot;&gt;&#xA;&lt;a href=&quot;http://www.astro.princeton.edu/~gk/A403/nucsyn.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Also shown here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6873" LastEditorUserId="6873" LastEditDate="2015-04-09T16:53:40.047" LastActivityDate="2015-04-09T16:53:40.047" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10416" PostTypeId="2" ParentId="10097" CreationDate="2015-04-09T13:48:39.043" Score="5" Body="&lt;p&gt;The observational determination of the chemical abundances in exoplanets is in its infancy. In terms of &lt;em&gt;terrestrial&lt;/em&gt; type planets, i.e. those of size less than a few Earth radii, the constraints are confined to comparing the measured densities (obtained from the masses and radii of transiting planets found by Kepler and CoRoT) with models of what planets with certain assumed composition would look like. An excellent recent example of this can be found in &lt;a href=&quot;http://adsabs.harvard.edu/abs/2015ApJ...800..135D&quot; rel=&quot;noreferrer&quot;&gt;Dressing et al. (2015)&lt;/a&gt;. In this paper they make the claim that all of the low-mass&#xA;planets are consistent with a single, simple 2-component models (a mixture of 83% MgSiO$_3$ and 17% iron, but that this changes at higher masses, where more volatile elements or significant water are needed to explain their lower densities. The plot below, taken from that paper, illustrates the available data and should be quite up-to-date. Note how all the low-mass planets (and the Earth and Venus) can lie on the same family of models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0pKj2.png&quot; alt=&quot;Planet mass vs radius from Dressing et al. (2015)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think that the authors are claiming that this is exactly what all the planets are made of, but simply illustrating that at present there do not seem to be any large deviations from such a composition (for example, planets that are made solely of iron). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are relatively few planets on this diagram, because it is difficult to obtain the masses of small transiting planets (it requires detection of the doppler shift caused by the pull of the planet on its host star).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course different models yield somewhat different results. For instance, &lt;a href=&quot;http://adsabs.harvard.edu/abs/2012A%26A...541A.103W&quot; rel=&quot;noreferrer&quot;&gt;Wagner et al. (2012)&lt;/a&gt; used the same data for Kepler-10b and CoRoT-7b and their own detailed models to argue that these planets have an iron core that makes up about 60% of the planet - i.e. a lot more than makes up the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment the data for the lowest mass planets currently indicate that there &lt;em&gt;could&lt;/em&gt; only be a limited amount of diversity. But the information that we are working with, the sample size, and the fact that only masses and radii are&#xA;determined, is too sparse to be sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a theoretical point of view there are many ideas. The basic concept about the formation of the terrestrial-type planets is that they form (relatively) close in to the parent star and have compositions that are reflective of what elements and minerals can condense out of the protoplanetary disk at high temperatures. This in turn depends on the balance of elements that are present in the protoplanetary disk, where in the disk the planet forms, the detailed structure of the protoplanetary disk, how it cools and how planets migrate in the disk.  Unsurprisingly, by varying some of these conditions it is possible to create planets with a wide variety of compositions, which as I said above, appears to be mildly contradicted by the available evidence. At the most basic level we understand why the Earth and other terrestrial planets are comparatively lacking in &quot;volatile&quot; elements like hydrogen and helium (and to some extent C, N, O and noble gases), because they (or minerals containing them) could not condense at the high temperatures close to the proto-Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of these theoretical approaches can be found in &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014ApJ...787...81M&quot; rel=&quot;noreferrer&quot;&gt;Moriarty et al. (2014)&lt;/a&gt; (of which you are familiar), but also &lt;a href=&quot;http://adsabs.harvard.edu/abs/2012ApJ...747L...2C&quot; rel=&quot;noreferrer&quot;&gt;see Carter-Bond et al. (2012)&lt;/a&gt; for examples of how chemical diversity might arise. It seems that the Mg/Si and the C/O ratios have the biggest influence on the final compositions of the formed planets. A low C/O ratio favours the formation of silicates and fewer carbon-carrying compounds; but if there is more carbon than oxygen then it becomes more favourable to form carbon and silicon-carbide (I guess this is what you mean by &quot;carbon planets&quot;), but this also depends on the temperature in the region where the planet forms. For reference, the solar C/O ratio is 0.54 and the relative abundance of carbon in the Earth is much lower (than in the Sun) but the C/O ratio measured in other stars can be higher.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-04-09T13:48:39.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10417" PostTypeId="2" ParentId="10372" CreationDate="2015-04-09T14:54:30.533" Score="10" Body="&lt;p&gt;The Boomerang Nebula (or Bow Tie Nebula) is a cloud of gas being expelled from a dying low-mass star, at &lt;span class=&quot;math-container&quot;&gt;$164~\mathrm{km}~\mathrm{s}^{-1}$&lt;/span&gt; (cf. Raghvendra Sahai and Lars-Åke Nyman: &lt;em&gt;&lt;a href=&quot;http://iopscience.iop.org/1538-4357/487/2/L155/pdf/975386.web.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Boomerang Nebular: The coldes region of the universe?&lt;/a&gt;&lt;/em&gt;). In general, when a gas expands, it cools (see extended explanation below). If the gas were &lt;em&gt;optically thin&lt;/em&gt; to the CMB — that is, if it were sufficiently dilute that CMB photons could easily penetrate — it would quickly get reheated to the temperature of the CMB, i.e. 2.7 K. However, the Boomerang Nebula is optically thick (dense), so the CMB hasn't yet had the time to heat it. The temperature in its outer parts, however, is higher, and as the nebula expands further, it will eventually (on timescales of, say, 1000s of years) be heated not only by the CMB, but also by the central white dwarf, i.e. the remnant of the star that produced the nebula.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Why does an expanding gas cool?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The usual approach to explain this is to consider a gas in a piston. When the volume is increased, the gas molecules do work on the piston, and hence lose energy, so temperature decreases. However, in the case of the Boomerang Nebula, there are no walls on which the gas can do work.&lt;/p&gt;&#xA;&lt;p&gt;Cosmologically speaking, the nebula expands rather quickly (it has &amp;quot;only&amp;quot; been expanding for ~1500 yr). Assuming that it doesn't have the time to exchange energy with its surrounding, the expanding is thus &lt;em&gt;adiabatic&lt;/em&gt;. For an ideal gas undergoing a &lt;a href=&quot;http://en.wikipedia.org/wiki/Adiabatic_process#Ideal_gas_.28reversible_process.29&quot; rel=&quot;nofollow noreferrer&quot;&gt;reversible adiabatic expansion&lt;/a&gt; (or contraction), we know that&#xA;&lt;span class=&quot;math-container&quot;&gt;$$P V^\gamma = \mathrm{constant},$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; are the pressure and volume of the gas, respectively, and &lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt; is the &lt;em&gt;adiabatic index&lt;/em&gt;. For a monoatomic gas, &lt;span class=&quot;math-container&quot;&gt;$\gamma = 5/3$&lt;/span&gt;, but here there are probably also molecules, so it's probably somewhat higher. At any rate, it's higher than &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;, which is the important thing for us, as we shall see below.&lt;/p&gt;&#xA;&lt;p&gt;Now the temperature &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; of the gas can be obtained from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ideal_gas_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;ideal gas law&lt;/a&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$P V = N k_\mathrm{B} T.$$&lt;/span&gt;&#xA;Here, &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; is the total number of particles, and &lt;span class=&quot;math-container&quot;&gt;$k_\mathrm{B} = 1.38\times10^{-16}~\mathrm{erg}~\mathrm{K}^{-1}$&lt;/span&gt; is a constant (Boltzmann's, to be specific). Even for non-ideal gasses, this relation is usually a pretty good approximation. Combining these two equations, we see that&#xA;&lt;span class=&quot;math-container&quot;&gt;$$T V^{\gamma-1} = \mathrm{constant},$$&lt;/span&gt;&#xA;and since &lt;span class=&quot;math-container&quot;&gt;$\gamma \gt 1$&lt;/span&gt;, it is evident that if &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; &lt;em&gt;in&lt;/em&gt;creases, &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; must &lt;em&gt;de&lt;/em&gt;crease.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="37191" LastEditDate="2021-04-07T13:28:04.690" LastActivityDate="2021-04-07T13:28:04.690" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="10418" PostTypeId="1" CreationDate="2015-04-09T21:00:19.320" Score="2" ViewCount="201" Body="&lt;p&gt;Given that by definition of scale height an atmosphere thins by a factor of &lt;strong&gt;1/e^x&lt;/strong&gt; where &lt;strong&gt;x&lt;/strong&gt; is elevation in terms of scale height multiples (See the table here: &lt;a href=&quot;http://astro.unl.edu/naap/scaleheight/sh_bg1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Definition of Scale Height&lt;/a&gt;), can we assume that the atmosphere is effectively non-existent at the elevation of 6H?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Density at elevation 6H&#xA;1/e^6 = ~0.00248 would mean about 0.2% of density at surface level&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there is no real physical boundary but what I'm looking for is what is the assumed standard for simplifying calculations. Or is it just a bad idea to try and do it this way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*This is a followup to another question:&#xA;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/10279/how-can-you-determine-the-initial-volume-of-a-planets-atmosphere&quot;&gt;How can you determine the initial volume of a planet's atmosphere?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6873" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-05-11T06:00:42.350" Title="Can you assume atmosphere height for the purpose of surface pressure calculation?" Tags="&lt;exoplanet&gt;&lt;astrophysics&gt;&lt;planetary-atmosphere&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10419" PostTypeId="1" CreationDate="2015-04-10T00:52:39.533" Score="3" ViewCount="147" Body="&lt;p&gt;Earth receives shower of both solar and cosmic radiations every seconds. The Sun put up a heliosphere around the solar system and thus foreign high energy charged particles could not penetrate easily. Is there any other source most probably that lies within our solar system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Very high energy cosmic particle beyond 5.7 x 10^19 eV were observed&#xA;by Pierre Auger Observatory in Argentina. (rare but the GZK limit&#xA;seems to rule out distance sources)&lt;/li&gt;&#xA;&lt;li&gt;GZK limit roughly capped the upper energy level of cosmic ray as&#xA;they are slowed due to CMB over vast distance&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="6760" LastEditDate="2015-04-10T06:12:59.093" LastActivityDate="2016-11-01T23:51:00.540" Title="Source of high energy cosmic particles" Tags="&lt;the-sun&gt;&lt;solar-system&gt;&lt;cosmic-ray&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10422" PostTypeId="1" CreationDate="2015-04-10T20:14:20.283" Score="0" ViewCount="208" Body="&lt;p&gt;The gravity anomalies on the moon (generally accepted to be caused by mascons) are powerful enough to affect the orbit of satellites.  I'm trying to determine how much effect they would have on a person standing on the surface.  If we experience about 17% of Earth's gravity on the lunar surface, how much would that differ within one of these anomalies? &lt;/p&gt;&#xA;" OwnerUserId="7019" LastActivityDate="2016-04-21T15:32:14.103" Title="How much do the mascons on the moon affect surface gravity?" Tags="&lt;the-moon&gt;&lt;gravity&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10423" PostTypeId="1" AcceptedAnswerId="10424" CreationDate="2015-04-11T02:16:22.347" Score="1" ViewCount="50" Body="&lt;p&gt;I'm searching for works or individuals that consider and discuss the idea of planetary bodies being considered stand alone living creatures, rather than objects either capable or incapable of supporting life.  Anyone familiar with something along those lines?&lt;/p&gt;&#xA;" OwnerUserId="7022" LastActivityDate="2015-04-11T03:02:12.210" Title="Are there any works discussing planetary bodies being forms of life?" Tags="&lt;planet&gt;&lt;life&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10424" PostTypeId="2" ParentId="10423" CreationDate="2015-04-11T03:02:12.210" Score="2" Body="&lt;p&gt;Look up &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaia_hypothesis&quot; rel=&quot;nofollow&quot;&gt;Gaia hypothesis&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/James_Lovelock#Gaia&quot; rel=&quot;nofollow&quot;&gt;James Lovelock&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Lynn_Margulis#Gaia_hypothesis&quot; rel=&quot;nofollow&quot;&gt;Lynn Margulis&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-04-11T03:02:12.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10425" PostTypeId="2" ParentId="10418" CreationDate="2015-04-11T05:58:46.577" Score="2" Body="&lt;p&gt;It all depends on what you are trying to do. For some purposes 6H might correspond to an adequate vacuum  for others not. The density is not low enough to ignore its effects on a space craft (we are talking of $\approx 48$ km in the case of the Earth). Indeed $48$km is less than the altitude record for an unmanned balloon, though greater than the altitude records for air breathing aircraft.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-04-11T05:58:46.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10426" PostTypeId="1" CreationDate="2015-04-11T10:41:19.550" Score="3" ViewCount="566" Body="&lt;p&gt;I would like to write a program that visualizes the solar system. To do this, I would need equations for calculating the coordinates of each solar system object at a given time. Where could I obtain such equations? I hope they would be as accurate as possible. If there are such equations, it would be also nice to see the raw data they are based on and a description how they were computed. I would be grateful for all help.&lt;/p&gt;&#xA;" OwnerUserId="7025" LastActivityDate="2015-04-11T13:42:28.890" Title="Equations for coordinates of solar system objects" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;earth&gt;&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10427" PostTypeId="2" ParentId="10426" CreationDate="2015-04-11T13:27:19.423" Score="6" Body="&lt;p&gt;It depends a bit on how precise you would want to be. A very good discussion on how to calculate the orbits of solar system objects is given in the book by &lt;a href=&quot;http://www.willbell.com/math/mc1.htm&quot;&gt;Jean Meeus, Astronomical Algorithms (1999)&lt;/a&gt;, which is at an advanced amateur level. At professional level you have the &lt;a href=&quot;http://fer3.com/arc/img/127158.699962749.pdf&quot;&gt;Explanatory Supplement to the Astronomical Almanac by Urban and Siedelmann&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For precise calculations Meeus uses the VSOP theory from Bretagnon (1987) (&lt;a href=&quot;http://en.wikipedia.org/wiki/VSOP_(planets)&quot;&gt;wikipedia link&lt;/a&gt;). You can download all files needed for these calculations from &lt;a href=&quot;http://cdsarc.u-strasbg.fr/cgi-bin/Cat?VI/81&quot;&gt;VizieR&lt;/a&gt;. These files contain a lot of numerical terms that are needed in the equations. The resulting positions are however very precise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the highest accuracy, you can download the predicted positions from the Jet Propulsion Laboratory (JPL) &lt;a href=&quot;http://ssd.jpl.nasa.gov/?horizons&quot;&gt;Horizon&lt;/a&gt; system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also quite a few software library projects that implement the equations from Meeus. This might be the best option. Then you do not have to implement the equations yourself. For C/C++ you have for instance &lt;a href=&quot;http://libnova.sourceforge.net&quot;&gt;LibNova&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For highest precision you have professional software libraries such as for instance the &lt;a href=&quot;http://aa.usno.navy.mil/software/novas/novas_info.php&quot;&gt;NOVAS&lt;/a&gt; libraries from the Naval Observatory for Fortran, C, or Python. But to use that correctly you will have to have a good understanding of celestial mechanics.&lt;/p&gt;&#xA;" OwnerUserId="864" LastEditorUserId="864" LastEditDate="2015-04-11T13:42:28.890" LastActivityDate="2015-04-11T13:42:28.890" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10429" PostTypeId="2" ParentId="8148" CreationDate="2015-04-11T16:21:11.130" Score="0" Body="&lt;p&gt;I think that the problem is that the question is too poorly defined to have an answer like &lt;em&gt;&quot;1 we are alone, or 42 they are here and there&quot;&lt;/em&gt;. Two numbers cannot be true at the same time, and any single scenario suggested thus far seems pretty unreasonable, so it is frustrating. But if we cannot define a phenomena, how could we count or estimate its number of occurrences?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problems with a question of the kind: &lt;em&gt;&quot;How many living intelligent space traveling civilizations exist in the Milky Way?&quot;&lt;/em&gt; are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What is life? Biologists seem to have a hard time nailing it down. Best thing they've come up with is a loose collection of properties like a thingy which organizes energy and material from its environment into a stable structure which copies itself, but with some errors or recombinations in order to allow for evolution. Each of those components have some definition problems, and each of them exists separately in non-living things. The working definition seems to be: &quot;We know it when we see it, if it quacks like a duck...&quot;, which means: &quot;It is like we are and we can recognize ourselves&quot;. (We do have inherited some genes in common with EVERY living cell on Earth).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Intelligence is poorly defined. Those who work with developing AI do manage to make computers do some things like we do them. But if I'm not misinformed, tools like neural nets and evolutionary algorithms can be simplified to plain old 19th century math. They solve math problems, but it is not crystal clear that they create intelligence. Is a math schoolbook intelligent?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Civilization, I've never heard of a good definition of that phenomena, nor of colonization. And spaceflight can turn out to be different from how we imagine it today. &quot;Alien sociology&quot; is a common joke.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the story goes that life and intelligence are like we are, and civilization IS us! This is on Ptolemy's geocentric level, its all about us. Should we count one forest or a thousand trees? The question has to be more rigorously specified. WE are obviously alone. Others, whatever is meant by that, might be around.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-04-11T16:21:11.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10431" PostTypeId="1" AcceptedAnswerId="10432" CreationDate="2015-04-11T18:21:51.143" Score="3" ViewCount="323" Body="&lt;p&gt;In the lambda-CDM model, the Universe is spatially flat, and will expand forever. Is possible that a spatially closed Universe expands forever?&lt;/p&gt;&#xA;" OwnerUserId="6920" LastActivityDate="2015-04-11T22:00:16.817" Title="Can the Universe be spatially closed and expand forever?" Tags="&lt;cosmology&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10432" PostTypeId="2" ParentId="10431" CreationDate="2015-04-11T18:44:19.070" Score="5" Body="&lt;p&gt;Yes, if the Universe is matter-dominated and spatially closed, then a cosmological constant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Lambda &amp;gt; \frac{4\pi G \rho}{c^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\rho$ is the critical energy density (of matter) of the Universe, will cause it to expand forever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB Where $\Lambda$ is equal to this value you get the (unstable) Einstein static Universe.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastEditorUserId="2092" LastEditDate="2015-04-11T22:00:16.817" LastActivityDate="2015-04-11T22:00:16.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10433" PostTypeId="1" CreationDate="2015-04-11T20:15:06.700" Score="10" ViewCount="2125" Body="&lt;p&gt;I would like to obtain &lt;a href=&quot;http://en.wikipedia.org/wiki/Tycho_Brahe&quot; rel=&quot;noreferrer&quot;&gt;Tycho Brahe's&lt;/a&gt; data on Mars. What would be an authoritative source? One source I could find is &lt;a href=&quot;http://www.pafko.com/tycho/observe.html&quot; rel=&quot;noreferrer&quot;&gt;this page&lt;/a&gt; (the data is given as an Excel file) but I have no idea can that be trusted. Does some astronomical organization or university keep the data public? I suppose Tycho wrote his data into a book. Where is that book located at the moment?&lt;/p&gt;&#xA;" OwnerUserId="7029" LastEditorUserId="616" LastEditDate="2019-08-21T17:29:01.503" LastActivityDate="2020-12-20T20:46:52.350" Title="Where to obtain Tycho Brahe's data?" Tags="&lt;mars&gt;&lt;history&gt;&lt;resource-request&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10434" PostTypeId="1" AcceptedAnswerId="10437" CreationDate="2015-04-12T00:21:16.163" Score="1" ViewCount="603" Body="&lt;p&gt;I recently heard in a discussion that the sun is not massive enough to self-ignite via core-pressure. It simply has not enough mass to &quot;generate&quot; the gravitational force needed for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why the sun is still burning is that there is a really small chance that two Hydrogen-Atoms fuse without needing the activation-energy/the activation-energy being provided by a random effect like High-Energy-Particles. Even though the chance is really low, there are so many Hydrogen-Atoms that overall the chance is actually high (law of big numbers). Once some fusion started it provides enough energy to fusion more Hydrogen, leading to our sun shining.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The person who presented it only stated that there are sources, but sadly the ones I found from his material didn't support this (nor did they deny it). The sources where about other topic related to stars. I can't ask the person himself as he already changed the university.&lt;/p&gt;&#xA;" OwnerUserId="7031" LastEditorUserId="1322" LastEditDate="2015-04-12T00:31:00.943" LastActivityDate="2015-04-12T07:00:44.060" Title="Is the sun too small to self-ignite?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10436" PostTypeId="1" AcceptedAnswerId="10445" CreationDate="2015-04-12T05:45:10.940" Score="3" ViewCount="140" Body="&lt;p&gt;Astronomers can tell if there are any exoplanets orbiting its parent star using but not limited to the following methods:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Transit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an massive object usually a planet passing in front of a star causing the amount of light emitted from the star to dim, by measuring the dimness the data can reveal how big the planet is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Radial velocity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The star will seems to wobble as if there is a massive object usually a planet in orbit playing tug of war in a game of whose gravity is mightier. However since this wobbling effect is difficult to observe nerds prefer to measure the blueshift of the light (as star accelerates in our direction) and redshift (as star accelerates away from our direction) which seems to also tells how massive the planet is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;&#xA;Astronomers can derive the planet's density from its mass and radius(see transit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far astronomers are able to determine the size, mass as well as density thus even its composition, is there any way to find the angle of the orbital plane of the exoplanet orbiting its parent star?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="6760" LastEditDate="2015-04-12T15:16:16.887" LastActivityDate="2015-04-13T08:43:03.573" Title="Can we know the orbital planes of extraterrestrial planetary bodies?" Tags="&lt;star&gt;&lt;orbit&gt;&lt;planet&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10437" PostTypeId="2" ParentId="10434" CreationDate="2015-04-12T07:00:44.060" Score="7" Body="&lt;p&gt;Stan has essentially answered this in his comment, which I will attempt to spell out a little more laboriously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The significant majority of our Sun's energy output comes from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction&quot; rel=&quot;noreferrer&quot;&gt;proton-proton chain&lt;/a&gt;.  This was advocated by Eddington back in the 1920's, but at that time your basic concern was a very real and major problem.  Objects with like electrical charges repel each other.  In particular, protons will repel other protons since all protons have a positive charge.  What they knew of the sun then indicated that the core was far too cold for protons to overcome this repulsion&amp;mdash;at least not even remotely close to the rate that was obviously necessary to produce a brightly shining sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the development of quantum mechanics it was determined that a process known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_tunnelling&quot; rel=&quot;noreferrer&quot;&gt;quantum tunneling&lt;/a&gt; would give two protons a non-zero probability of &quot;overcoming&quot; this repulsion.  But not in the sense of they just suddenly gain enough energy (hence why &quot;overcoming&quot; is in quotations).  Instead, in the sense that the &quot;fused together&quot; state has the same energy as the &quot;just about to repel each other before fusion can occur&quot; state, and they just randomly switched from the latter to the former, despite every intermediate stage between the two of them requiring more energy than is available.  That this is possible is one of the many non-intuitive features of quantum mechanics, and I think it would be beyond the scope of this question (and site) to try to get much more precise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, even this failed to explain why our sun was very obviously fusing atoms to the extent that it was.  If you fuse two protons together, you are left with an incredibly unstable state: the diproton.  As soon as one of these forms it pretty much immediately breaks up into two distinct protons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the late 30's Hans Bethe (a man who eventually won the Nobel prize and was a part of the Los Alamos team that developed the atom bomb) proposed that yet another random quantum mechanical event would save the day: &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_decay&quot; rel=&quot;noreferrer&quot;&gt;beta decay&lt;/a&gt;, a feature of the recently discovered fundamental force known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Weak_interaction&quot; rel=&quot;noreferrer&quot;&gt;weak force&lt;/a&gt;.  In this situation one of the protons in the diproton undergoes a beta decay into a neutron before the diproton separates, at which point you have a stable deuterium nucleus: one proton, and one neutron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That so many unlikely events have to occur for two protons to produce one deuterium is the reason why the sun's lifetime is approximately 10 billion years.  That the sun shines as bright as it does is, as you suggest, a sheer numbers game: there are truly prodigious numbers of protons in the sun's core, so even though the proton-proton chain is really unlikely, we have so very many chances at it that there is a great many successful fusion events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically every other stellar fusion reaction proceeds much more rapidly than those that produce deuterium.  You may be familiar with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermonuclear_weapon&quot; rel=&quot;noreferrer&quot;&gt;hydrogen bomb&lt;/a&gt;, which fuses its fuel very rapidly.  One of the key differences (and there are many) between the hydrogen bomb and a star like our sun is that the bomb is fusing together deuterium (and tritium) into helium, and is not using the proton-proton chain.  Stars have to produce large supplies of deuterium directly, and this is a lengthy process that consumes roughly 90% of a star's life&amp;mdash;even for very massive stars, whose lifespans are orders of magnitude shorter than our Sun's.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that very massive stars are rather more complicated beasts, in fact, but this is again beyond the scope of this question)&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2015-04-12T07:00:44.060" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10439" PostTypeId="2" ParentId="10433" CreationDate="2015-04-12T13:14:28.890" Score="2" Body="&lt;p&gt;The &lt;a href=&quot;http://www.pafko.com/tycho/bib.html&quot; rel=&quot;nofollow&quot;&gt;page&lt;/a&gt; I referred to in my question lists the reference &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Brahe, Tycho. Edited by I.L.E. Dreyer Tychonis Brahe Dani Opera Omnia. (in Latin) Vol 1-15. 1913-1929. (contains the observations in Tycho's notebooks)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now &lt;a href=&quot;https://archive.org/details/operaomniaedidit02brahuoft&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt; is supposed to be an electronic version of Tycho's book. It says &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Opera omnia, edidit I.L.E. Dreyer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in the big title at the bottom of the page. However, browsing through the book, I find the title &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Tychonis Brahe Dani Scripta Astronomica.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't see words &quot;opera omnia&quot; anywhere. Is that the correct book? The first observation listed in that Excel file is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1582  DIE 12 NOUEMBRIS, MANE.     Declinatio [MS]  23  7 B&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ok. So, on page 174, it does say something about the year 1582. However, the given coordinates do not appear, and neither does the date 12th November. The Excel file also refers to the Mars sign, which I cannot find from that page at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope someone could help from here :)&lt;/p&gt;&#xA;" OwnerUserId="7036" LastActivityDate="2015-04-12T13:14:28.890" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10441" PostTypeId="2" ParentId="10433" CreationDate="2015-04-12T13:54:40.547" Score="3" Body="&lt;p&gt;This must be what you are looking for, &lt;a href=&quot;http://193.206.220.110/Teca/Viewer?an=334726#&quot; rel=&quot;nofollow&quot;&gt;&quot;Tabulae Rudolphinae&quot;&lt;/a&gt;. Tycho Brahe's star catalogue and planetary data &lt;a href=&quot;https://en.wikipedia.org/wiki/Rudolphine_Tables&quot; rel=&quot;nofollow&quot;&gt;published by Kepler&lt;/a&gt; in 1627 .&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-04-12T13:54:40.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10442" PostTypeId="1" CreationDate="2015-04-12T16:07:17.423" Score="8" ViewCount="251" Body="&lt;p&gt;How often are new astronomical objects (variable stars, supernovae, comets, etc) discovered by amateurs? Where could one report new findings?&lt;/p&gt;&#xA;" OwnerUserId="7036" LastActivityDate="2015-04-12T21:05:19.383" Title="How often are new astronomical objects (variable stars, supernovae, comets, etc) discovered by amateurs?" Tags="&lt;star&gt;&lt;observational-astronomy&gt;&lt;amateur-observing&gt;&lt;comets&gt;&lt;supernova&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10443" PostTypeId="2" ParentId="10436" CreationDate="2015-04-12T16:59:13.323" Score="0" Body="&lt;p&gt;&lt;em&gt;Is there any way to find the angle of the orbital plane of the exoplanet orbiting its parent star?&lt;/em&gt; It's unclear what precisely you're asking. The planet will be orbiting (to very good approximation) on a Keplerian orbit (closed ellipse co-focal with the parent star). Such an orbit has &lt;em&gt;two relevant angles&lt;/em&gt;. One is the angle between the orbital plane and the line of sight to Earth (often called the inclination $i$); the other is the direction of the orbital periapse (usually measured as the angle in the orbital plane between the line of nodes and the periapse). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of transit detections, $i\sim 90^\circ$ (for otherwise, there will be no ecclipse/transit), but it cannot (usually) inferred for radial-velocity detections. Since the amplitude of the observed radial velocity is proportional to $M\sin i$, this implies that the planet mass is only limited but could be larger than for $i=90^\circ$.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2015-04-12T16:59:13.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10444" PostTypeId="1" CreationDate="2015-04-12T19:52:39.663" Score="2" ViewCount="209" Body="&lt;p&gt;I was curious to how long it will be until the stars are not able to be seen due to light pollution. I started wondering this after reading Yummie's Flash comic &lt;a href=&quot;http://yuumei.deviantart.com/art/Knite-Chapter-1-151132545&quot; rel=&quot;nofollow&quot;&gt;Knite&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7040" LastEditorUserId="2153" LastEditDate="2015-04-12T21:54:47.717" LastActivityDate="2015-04-12T21:54:47.717" Title="How long until the stars are no longer visible by the human eye due to light pollution?" Tags="&lt;observational-astronomy&gt;&lt;light-pollution&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2016-10-17T14:58:00.640" ContentLicense="CC BY-SA 3.0" />
  <row Id="10445" PostTypeId="2" ParentId="10436" CreationDate="2015-04-12T20:47:47.850" Score="2" Body="&lt;p&gt;I presume what you mean is how does the plane of the orbit compare to the equatorial rotation plane of the star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is, you can sort of estimate this, by using something called the &lt;a href=&quot;https://exoplanetmusings.wordpress.com/2013/03/25/764/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rossiter-McLaughlin&lt;/a&gt; effect (see also &lt;a href=&quot;http://adsabs.harvard.edu/abs/1924ApJ....60...15R&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rossiter 1924&lt;/a&gt;; &lt;a href=&quot;http://adsabs.harvard.edu/abs/1924ApJ....60...22M&quot; rel=&quot;nofollow noreferrer&quot;&gt;McLaughlin 1924&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find plenty of information on the web - I'll add a couple of links when I have a moment - but to summarise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rotation of a star broadens its spectral absorption lines. The hemisphere coming towards us emits light that is is blue shifted, the hemisphere receding is redshifted. If we now take a &lt;em&gt;transiting&lt;/em&gt; planet, during the eclipse it crosses the disc of the star and obscures regions that are blue or red shifted by various amounts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what you do is measure the line-of-sight velocity of the star. During the transit you would not expect this to vary due to the &quot;Doppler wobble&quot; caused by the exoplanet, except that if the planet obscures a blue shifted portion of the stellar disc, the net spectral absorption line that remains shifts to the red, and vice-versa. The pattern of red, then blue shift (or vice versa) as the transit progresses is known as the Rossiter-Mclaughlin effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A schematic showing how the Rossiter-McLaughlin effect works and how a different transit geometry with respect to the rotation axis of the star leads to a different line-of-sight-velocity signature in the spectral lines of the star. (Image credit: Subaru Telescope, National Observatories of Japan.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/X5eOF.jpg&quot; alt=&quot;Schematic of Rossiter McLaughlin effect.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the planet orbits in the same plane and in the same direction as the stellar rotation (as the planets in our solar system nearly do), then the blue shifted limb of the parent star is obscured first, followed by an equal amount of redshift as the planets moves to obscure the receding stellar limb (see image above, left). Thus the stellar absorption lines show a redshift followed by a symmetric blueshift. If the planet was retrograde it would be symmetric but occur in the opposite order. A Polar orbit would show no RM effect. Inclined orbits would have an asymmetric RM effect - ie perhaps more blueshift than redshift (see image above, right).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The RM effect cannot give an exact orientation, it gives the projected angle between orbital and rotation axes on the plane of the sky. Nevertheless, that is sufficient for us to know that a lot of the transiting exoplanets have highly misaligned orbital and stellar rotation axes.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-04-13T08:43:03.573" LastActivityDate="2015-04-13T08:43:03.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10446" PostTypeId="2" ParentId="10442" CreationDate="2015-04-12T21:03:14.383" Score="5" Body="&lt;p&gt;The answer is: frequently. There are many amateur astronomers that make it their ambition to discover new supernovae or to observe and report on new variable stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, let me cite amateurs &lt;a href=&quot;http://en.wikipedia.org/wiki/Robert_Evans_%28astronomer%29&quot;&gt;Robert Evans&lt;/a&gt;, who apparently holds the record for most supernovae found by visual observation, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Tom_Boles&quot;&gt;Tom Boles&lt;/a&gt;, who holds the record for supernova discoveries by an individual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observations of variable stars can be reported to the Information Bulletin of Variable Stars (IBVS). Supernovae discovery or new comets would normally be reported to the International Astronomical Union Circulars.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2153" LastEditDate="2015-04-12T21:05:19.383" LastActivityDate="2015-04-12T21:05:19.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10447" PostTypeId="1" CreationDate="2015-04-12T21:16:39.377" Score="3" ViewCount="262" Body="&lt;p&gt;I read that a few years ago some planets were discovered orbiting Alpha Centauri. Has there been any more information since then concerning the possibility of life there? This is the closest star system to Earth, but it would still take us 63,000 years to reach it.&lt;/p&gt;&#xA;" OwnerUserId="2562" LastEditorUserId="2153" LastEditDate="2015-04-12T21:52:15.567" LastActivityDate="2019-01-06T11:54:31.383" Title="Planets orbiting Alpha Centauri" Tags="&lt;exoplanet&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10448" PostTypeId="2" ParentId="10387" CreationDate="2015-04-13T05:08:42.363" Score="5" Body="&lt;p&gt;If you have good knowledge of software development and pattern recognition, there are several problems that you could assist in solving. Much of observational astronomy requires long time series data, and removing the noise from this data. I have just left the field where some colleagues are trying to develop some software to use image subtraction techniques to isolate individual stars in the centre of clusters. The centre of the cluster is typically more dense and harder to get clear measurements of each individual star to analyse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pattern recognition would be particularly useful in pipeline analysis, where a generic pipeline is used on large amounts of data to 1: find the types of stars one is interested in; and 2: extracting some interesting information on these stars. Machine learning techniques could also be used to assist in developing the general pipelines for more specific interests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am happy to put you in touch with a few people that could provide you with some specific problems that you might be able to assist in.&lt;/p&gt;&#xA;" OwnerUserId="7042" LastActivityDate="2015-04-13T05:08:42.363" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10449" PostTypeId="1" AcceptedAnswerId="10452" CreationDate="2015-04-13T05:37:56.073" Score="0" ViewCount="348" Body="&lt;p&gt;Both Quito and Kampala lie on the Equator. The longitude of Quito is 82°30'W and that of Kampala is 37°30'E. What is the distance from Quito to Kampala?&#xA;A) along the shortest surface path&#xA;B) along a direct through the Earth path?&#xA;Please help me visualize it with a figure. How to do it?&lt;/p&gt;&#xA;" OwnerUserId="7043" LastActivityDate="2015-04-13T10:41:01.570" Title="What is the distance between 2 cities?" Tags="&lt;earth&gt;&lt;distances&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-04-13T11:53:44.713" ContentLicense="CC BY-SA 3.0" />
  <row Id="10450" PostTypeId="1" CreationDate="2015-04-13T07:01:17.573" Score="0" ViewCount="476" Body="&lt;p&gt;If the singularity had infinite mass, temperature and density, then it should curve space and time but there was no space-time before big bang. So where did the singularity come to existence without space-time ? &lt;/p&gt;&#xA;" OwnerUserId="7044" LastActivityDate="2019-07-01T23:45:10.690" Title="How was all the matter curled up inside a singularity during big bang ?" Tags="&lt;space-time&gt;&lt;big-bang-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10452" PostTypeId="2" ParentId="10449" CreationDate="2015-04-13T09:00:10.070" Score="2" Body="&lt;p&gt;This is hardly an astronomy question, but I like drawing, so here you are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1cGfH.png&quot; alt=&quot;Earth from above&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longitudes are measured from the Greenwich meridian, so the angle between Kampala and Quito is&#xA;$$\theta_\mathrm{Q} + \theta_\mathrm{K} = 82.5^\circ + 37.5^\circ = 110^\circ.$$&#xA;(remember that $0^\circ30' = 0.5^\circ$).&#xA;The shortest surface path is along Equator. Since $110^\circ$ is $\frac{110^\circ}{360^\circ} \simeq 0.3$ times the &lt;a href=&quot;https://www.google.dk/#q=circumference+of+Earth+at+Equator&quot; rel=&quot;nofollow noreferrer&quot;&gt;circumference of Earth at Equator&lt;/a&gt;, the length of path A (the dashed line) is&#xA;$$\mathrm{A}:\,\,d = 0.3 \times 40,075\,\mathrm{km} = 12,245\,\mathrm{km}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For path B (the solid line), you need a bit more  trigonometry. The radius of Earth is $R = 40,075\,\mathrm{km}\,/\,2\pi = 6,378\,\mathrm{km}$. The rest will be left as an exercise.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-04-13T10:41:01.570" LastActivityDate="2015-04-13T10:41:01.570" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10453" PostTypeId="1" CreationDate="2015-04-13T15:51:13.947" Score="4" ViewCount="124" Body="&lt;p&gt;I was wondering what parameters I need to know/measure to calculate the velocity dispersion of a galaxy, specifically an AGN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I have spectra where there are blueshifted and redshifted components to certain emission lines in units of km/s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I have [O III] 5007Å but I have a blueshifted component centre at 5002Å and a redshifted component at 5011Å. I was wondering how I calculate the doppler shift again in units of km/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks.  &lt;/p&gt;&#xA;" OwnerUserId="5505" LastEditorUserId="37191" LastEditDate="2021-04-16T08:36:08.527" LastActivityDate="2021-04-16T08:36:08.527" Title="Spectral analysis of AGN (velocity dispersion of galaxy absorption, doppler shifts)" Tags="&lt;galaxy&gt;&lt;data-analysis&gt;&lt;spectra&gt;&lt;active-galaxy&gt;&lt;spectrometry&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10454" PostTypeId="2" ParentId="10387" CreationDate="2015-04-13T17:08:17.387" Score="5" Body="&lt;p&gt;The &lt;a href=&quot;https://www.kaggle.com/c/galaxy-zoo-the-galaxy-challenge&quot; rel=&quot;noreferrer&quot;&gt;Kaggle galaxy zoo challenge&lt;/a&gt; is an example of a problem begging for ideas from outside the field. Sander Dieleman, with a background in deep learning and feature learning, bravely stepped forward, creating an image classifier utilising convolutional neural networks; his full solution is described fluently &lt;a href=&quot;http://benanne.github.io/2014/04/05/galaxy-zoo.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These kinds of techniques could be applied to any image classification problem in Astronomy, or similar techniques could be used to classify other astrophysical objects from survey or signal data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would steer clear of doing your own image capture, as there are &lt;a href=&quot;http://science.nasa.gov/astrophysics/astrophysics-data-centers/&quot; rel=&quot;noreferrer&quot;&gt;plenty of openly available data sets&lt;/a&gt; with greater depth, resolution and coverage than you could hope to carry out yourself within a reasonable time frame.&lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2015-04-13T17:08:17.387" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10455" PostTypeId="1" AcceptedAnswerId="11843" CreationDate="2015-04-13T17:43:30.327" Score="3" ViewCount="16488" Body="&lt;p&gt;I was watching Venus with the naked eye yesterday at about 7 pm and I noticed that it was flickering, almost like a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have always learned that planets don't flicker to the naked eye, only stars (indeed, every other time I've seen a planet it wasn't flickering), so I was rather confused. I even checked Stellarium to make sure that what I was seeing was, in fact, Venus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody have an explanation for that behaviour?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you all.&lt;/p&gt;&#xA;" OwnerUserId="6661" LastEditorUserId="11822" LastEditDate="2018-09-21T16:24:31.203" LastActivityDate="2018-09-21T16:24:31.203" Title="why does venus flicker?" Tags="&lt;observational-astronomy&gt;&lt;amateur-observing&gt;&lt;venus&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="10456" PostTypeId="2" ParentId="10455" CreationDate="2015-04-13T18:19:25.827" Score="3" Body="&lt;p&gt;It's true that planets usually don't flicker, or twinkle (or scincillate, as astronomers like to call it). The reason is that they are close enough that they are actually seen as a disk with a larger diameter than the atmosphere can &quot;wash out&quot; (the &lt;em&gt;seeing disk&lt;/em&gt;). Stars, on the other hand, are point sources, and this point jumps back and forth on the sky, as the light takes slightly different paths through the atmosphere. This is what we see as twinkling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can think of three reasons that may make a planet twinkle nevertheless:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If it stands very close to the horizon, so that the total column of air that the light passees (the &quot;airmass&quot;) is very high. If this is the case, it should also look a bit redder than usual.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If observed close to some building that emits heat, like an exhaust, so that the light passes through parcels of hot and cold air on small scales.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If an airplane passed by, emitting hot air.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2015-04-13T18:19:25.827" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10457" PostTypeId="1" AcceptedAnswerId="10459" CreationDate="2015-04-13T21:29:12.467" Score="1" ViewCount="237" Body="&lt;p&gt;Is the universe much like the earth, in that if I were to travel in one and only one direction, I'd eventually end up where I started from?&lt;/p&gt;&#xA;" OwnerUserId="5306" LastActivityDate="2015-04-16T23:12:34.740" Title="Is the universe like the earth?" Tags="&lt;space&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10459" PostTypeId="2" ParentId="10457" CreationDate="2015-04-13T23:06:12.770" Score="4" Body="&lt;p&gt;I'm not an astrophysicist but I'll give it a shot. I'm taking this mostly from Hawking's &quot;A brief history of time&quot; and my own personal studies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the universe is not like the earth. As far as I know this analogy only works when one considers the concept of &lt;a href=&quot;http://en.wikipedia.org/wiki/Imaginary_time&quot; rel=&quot;nofollow&quot;&gt;imaginary time&lt;/a&gt;. According to Hawking, when you consider the theory of relativity with a imaginary time (which mathematically is pretty much only considering time as a imaginary number, rather than a real one) the equations that describe space are mathematically similar to those that describe a sphere. This is a trick used to try to make a quantum mechanics concept (Feynman's &lt;a href=&quot;http://en.wikipedia.org/wiki/Path_integral_formulation&quot; rel=&quot;nofollow&quot;&gt;sum over histories&lt;/a&gt;) work with relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, however, does not mean that we can experience space (or the universe) like a sphere, because we are not capable of experiencing imaginary time. We experience only real time. If we were to experience time as it is defined by imaginary time, we would be able to to travel to the future and past as easily as we would travel from latitude -25º to +25º, and we all know that is not the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in this answer, I'm not even taking into account the expansion of the universe, which would complicate things even further for someone trying to go around the universe. Just for the sake of the argument, imagine going around a sphere that is growing more and more. Even if you take the shortest path possible, depending on the rate of expansion the task becomes either impossible of very hard to say the least.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="6661" LastActivityDate="2015-04-13T23:06:12.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10460" PostTypeId="1" AcceptedAnswerId="10468" CreationDate="2015-04-13T23:40:17.540" Score="0" ViewCount="67" Body="&lt;p&gt;I have a doubt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can stars be observed from space by x-rays, near infrared and radio wavelengths?&lt;/p&gt;&#xA;" OwnerUserId="7027" LastEditorUserId="1678" LastEditDate="2015-04-14T09:55:27.933" LastActivityDate="2015-04-15T06:19:50.793" Title="Can stars be observed from space by x-rays, near infrared and radio wavelengths?" Tags="&lt;star&gt;&lt;spectroscopy&gt;&lt;wavelength&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ClosedDate="2015-04-15T10:16:05.723" ContentLicense="CC BY-SA 3.0" />
  <row Id="10462" PostTypeId="2" ParentId="5896" CreationDate="2015-04-14T00:23:47.110" Score="3" Body="&lt;p&gt;By observation, gravity dominates the universe on large scales.  If there were a significant disparity in positive and negative charges then we would expect electromagnetism&amp;mdash;which is approximately $10^{39}$ times more powerful than gravity&amp;mdash;to dominate.  So by this observation we can conclude that the universe is approximiately electrically neutral.  Exactly why this should be so is not well-understood, and ultimately ties into the matter-antimatter problem, aka the &lt;a href=&quot;http://en.wikipedia.org/wiki/Baryon_asymmetry&quot; rel=&quot;nofollow&quot;&gt;baryon asymmetry&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Protons and electrons are produced through distinct processes.  That we have many protons is not particularly remarkable, as this would come about by conservation of baryon numbers as other particles decayed.  Why we have any electrons at all is a fair bit more mysterious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the early universe, we understand the energy was well above the electron-positron pair production threshold, and so the early universe had large amounts of both electrons and positrons.  When the universe cooled below this threshold, the production would have stopped and we would expect the electrons and positrons to then collide and annihilate each other, leaving essentially none of either.  But we obviously see large amounts of electrons: approximately as many electrons as protons by the aforementioned observational evidence.  And we also do not see large amounts of antimatter&amp;mdash;positrons in particular.  If there were large amounts of positrons out there, we would expect to see tell-tale signatures in space as they annihilate with regular matter (or in any weak force interactions, such as would be evident in supernovae), and we have no such observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So somehow the early universe must have produced significantly more electrons than positrons.  In other words, we have strong observational evidence that there must be an asymmetry in physics between matter and antimatter production.  There are ways to work this into the theory, but the Standard Model by default does not support it, and experts have not really settled on any one particular modification.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2015-04-14T00:23:47.110" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10463" PostTypeId="1" CreationDate="2015-04-14T00:33:37.417" Score="6" ViewCount="758" Body="&lt;p&gt;Imagine you're inside a black hole event horizon. The curvature of spacetime is such that the direction towards the singularity becomes timelike, as reverse movement becomes literally impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How would Hawking radiative evaporation look like from inside the black hole?&lt;/strong&gt; From the outside perspective, we know that black holes form (from our time dimension perspective) and then evaporate at a rate that is an inverse function of their mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hawking radiation is created when the vacuum state at the horizon looks as if it is transformed into a non-vacuum state for a distant observer, resulting in a negative energy 'input' to the black hole. If the energy loss this way is not compensated by inputs from outside, the black hole event horizon should gradually shrink, resulting in eventual evaporation. How if at all would that be perceptible by an observer inside the black hole horizon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/q/1723/7049&quot;&gt;Similar question&lt;/a&gt;, but no accepted answer.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="7049" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-07-10T21:50:21.150" Title="How Does Black Hole Evaporation Look From the Inside?" Tags="&lt;black-hole&gt;&lt;hawking-radiation&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10464" PostTypeId="1" CreationDate="2015-04-14T00:42:31.497" Score="6" ViewCount="382" Body="&lt;p&gt;Since Dark Matter is assumed to interact gravitationally, it follows that the gravitation of massive regular matter bodies should have an effect on it. Even if it is otherwise weakly interacting with regular matter, one might naively expect concentrations of it enter orbit around massive objects, like neutron stars, regular stars, and even (why not) our own planet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, a two-fold question:&lt;br&gt;&#xA;&lt;strong&gt;1) Could there be Dark Matter in orbit around (or within) the Earth?&lt;br&gt;&#xA;2) If that were possible, would there be any way for humans to tell if such is actually the case?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7049" LastActivityDate="2015-04-15T13:57:53.217" Title="Is there Dark Matter inside the Earth?" Tags="&lt;earth&gt;&lt;dark-matter&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10465" PostTypeId="1" AcceptedAnswerId="10466" CreationDate="2015-04-14T02:46:44.257" Score="2" ViewCount="1505" Body="&lt;p&gt;Perhaps this is a silly question but I am trying to understand Einstein's general theory of relativity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to explain my question in more detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The greater the mass of an object, the greater the curvature of spacetime. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I imagine the curvature of spacetime as a hole with a steep slope around it, I can picture that the closer I get to the center of the hole, the steeper the slope is and therefore the more I would be pushed into the center of the hole. I am not completely sure if that's a very good analogy but in general I think I get the picture of timespace curvature as depicted here: &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/2/22/Spacetime_curvature.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/2/22/Spacetime_curvature.png&quot; alt=&quot;https://upload.wikimedia.org/wikipedia/commons/2/22/Spacetime_curvature.png&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, my question is why is it that the size of the hole matters if I am not standing on the edge (side) of the hole or in the hole but rather on an object that 'covers' the hole (Earth).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say if I stand on a bridge that covers a 20-meter-wide hole and on a bridge the covers 500-meter-wide hole. Wouldn't the gravitational force experienced be the same since the surface on both bridges is flat?&lt;/p&gt;&#xA;" OwnerUserId="6862" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:47.400" LastActivityDate="2015-11-17T15:30:37.870" Title="Why does Earth have more gravitational force than the Moon if I stand on it?" Tags="&lt;gravity&gt;&lt;general-relativity&gt;&lt;space-time&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10466" PostTypeId="2" ParentId="10465" CreationDate="2015-04-14T03:59:37.487" Score="3" Body="&lt;p&gt;The Earth doesn't &lt;em&gt;cover&lt;/em&gt; the hole, the Earth &lt;em&gt;is&lt;/em&gt; the hole. Gravity is an attractive force, so were you not standing on the Earth, the Earth's gravity would cause you to accelerate towards it. As you are standing on the Earth, you feel this acceleration as your weight, just as you would feel pressure if you were to push against a wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To take a more complex view, we would say that you are being attracted to the local center of gravity, which is located within the Earth. Take your bridges example, it doesn't matter that something obstructs you from reaching the center of gravity. You will feel the attraction of gravity regardless. If you jump while standing on a bridge, you would land just the same as if you jumped while standing on the ground (&lt;em&gt;though please try not to miss the bridge&lt;/em&gt;). The graphic is meant to illustrate the strength of the force based on the mass of the attractive object and distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were to put you on a more massive object, which would have a steeper curvature, you would experience this as increased weight, which is why our lunar explorers bounced around on the moon instead of walking.&lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="1288" LastEditDate="2015-04-15T02:31:11.123" LastActivityDate="2015-04-15T02:31:11.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10467" PostTypeId="1" AcceptedAnswerId="10470" CreationDate="2015-04-14T04:49:54.787" Score="16" ViewCount="1761" Body="&lt;p&gt;This may be a really dumb question (I'm more of a Biologist than an Astronomer) so I apologize in advance for my little knowledge pertaining to Astronomy, but, if I'm not mistaken, time is effected by gravity, right? So what is Sagittarius A*'s time compared to ours since it has so much stronger gravity? Do we know specifically the difference?&lt;/p&gt;&#xA;" OwnerUserId="7051" LastEditorDisplayName="user24157" LastEditDate="2019-11-02T12:16:33.437" LastActivityDate="2019-11-02T12:16:33.437" Title="Time according to the gravity of Sagittarius A*?" Tags="&lt;space-time&gt;&lt;supermassive-black-hole&gt;&lt;time-dilation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="10468" PostTypeId="2" ParentId="10460" CreationDate="2015-04-14T07:25:42.690" Score="1" Body="&lt;p&gt;Yes, and not only from space but from the Earth surface too. Stars emit in almost all wavelengths depending on their surface temperatures. The hotter the star is the shorter (higher energy) wavelengths it'll emit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can try this simulator to check this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://astro.unl.edu/naap/blackbody/animations/blackbody.html&quot; rel=&quot;nofollow&quot;&gt;http://astro.unl.edu/naap/blackbody/animations/blackbody.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1678" LastEditorUserId="1678" LastEditDate="2015-04-15T06:19:50.793" LastActivityDate="2015-04-15T06:19:50.793" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10469" PostTypeId="2" ParentId="1930" CreationDate="2015-04-14T10:58:15.537" Score="0" Body="&lt;p&gt;Ultimately we don't know enough about exoplanets to be sure; for now all our data is skewed toward more massive planets which are easier to detect using Doppler wobble, or large diameter planets (almost certainly gas giants) which are easy to detect by their host star dimming when they eclipse it relative to us.  More data is coming in every day, and as fantastic as Keplar has been, I think we need to at least hold out for the James-Webb to be online before we draw any really hard conclusions form the data.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without the data, all we can rely on is our theories of planet formation, which we're fairly good at.&lt;br&gt;&#xA;Earth is probably more dense than an average planet of it's size, as a result of it colliding with a roughly mars-sized object (nicknamed Theia) early in it's development.  Theia's core would have been absorbed into earth's core, but the outer layers of both were stripped away, creating a ring which would coalesce into our moon.  This would leave earth with a higher mass core than a planet forming at its distance would have. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see this in the densities of the terrestrial planets; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;--Object-------Density (g cm−3)-----Semi-major axis (AU)-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-------------Mean----Uncompressedm-----------------------&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Mercury-----5.4---------5.3-------------0.39------------&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Venus-------5.2---------4.4-------------0.72------------&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Earth-------5.5----------4.4--------------1.0-------------&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Mars--------3.9----------3.8--------------1.5-------------&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Credit, Wikipedia &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets closer to their star are naturally going to have higher densities as a result of mass differentiation; denser material settling to the core of a planet or the center of a solar accretion disk.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at it from the perspective of habitability,&lt;br&gt;&#xA;We know that density is positively correlated to surface gravity, so we can expect that earth would have a slightly higher than average surface gravity for a planet in the habitable zone around a star in the category of one solar mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, most stars do not have one solar mass, most of the stars in the universe are red dwarfs, which are much dimmer and lighter than our sun, and would have a closer, narrower habitable zone.  A habitable planet around a red dwarf would probably be smaller and lighter, but denser, due to its lower mass accretion cloud and closer proximity to it's star respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think we could expect the majority of exoplanets to be planets similar to mercury orbiting red dwarfs.&lt;br&gt;&#xA;If this is the case we can expect that earth would have a high surface gravity relative to terrestrial planets (although FAR more massive terrestrial planets of similar diameter exist) and about average gravity when taking all planets into account. &lt;/p&gt;&#xA;" OwnerUserId="7054" LastActivityDate="2015-04-14T10:58:15.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10470" PostTypeId="2" ParentId="10467" CreationDate="2015-04-14T11:35:25.207" Score="25" Body="&lt;p&gt;Not at all a dumb question. As you have heard, it is true that time is affected by gravity. The stronger the gravitational field, the slower time passes. If you're far from any gravitating matter, time passes &quot;normally&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to answer your question, we must specify what is meant by &quot;the black holes's time&quot; (let's call the black hole &lt;span class=&quot;math-container&quot;&gt;$\mathrm{BH}_\mathrm{Sgr\,A^*}$&lt;/span&gt;; see note below on the nomenclature), since it depends on how far from Sgr A* we are talking. The time pace at a distance &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; from the center of a BH is given by&#xA;&lt;span class=&quot;math-container&quot;&gt;$$t = t_\infty \sqrt{1 - \frac{r_\mathrm{S}}{r}},$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$t_\infty$&lt;/span&gt; is the time &quot;at infinity&quot;, i.e. far from the BH, and&#xA;&lt;span class=&quot;math-container&quot;&gt;$$r_\mathrm{S} \equiv \frac{2GM}{c^2} \simeq 3\,\mathrm{km}\,\times \left( \frac{M}{M_\odot}\right)$$&lt;/span&gt;&#xA;is the so-called Schwarzschild radius (the &quot;surface&quot; of the BH), which is where not even light can escape. Here, &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; is the gravitational constant, &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is the mass of the BH, &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; is the speed of light, and &lt;span class=&quot;math-container&quot;&gt;$M_\odot$&lt;/span&gt; is the mass of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last equality shows that a BH with the mass of the Sun would have a radius of 3 km. The mass of &lt;span class=&quot;math-container&quot;&gt;$\mathrm{BH}_\mathrm{Sgr\,A^*}$&lt;/span&gt; is some 4.1 million Solar masses, so its radius is &lt;span class=&quot;math-container&quot;&gt;$r_\mathrm{S} = 12.1$&lt;/span&gt; million km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging in the other numbers, we can see that at a distance from &lt;span class=&quot;math-container&quot;&gt;$\mathrm{BH}_\mathrm{Sgr\,A^*}$&lt;/span&gt; of&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;1 lightyear&lt;/strong&gt;, time runs slower by a factor of 1.00000064, i.e. unnoticeably.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;1 astronomical unit&lt;/strong&gt; (the distance from Earth to the Sun), time runs 4% slower.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;1 million km from the surface&lt;/strong&gt;, time runs slower by a factor of 3.6.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;1000 km from the surface&lt;/strong&gt;, time runs slower by a factor of 110.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;1 km from the surface&lt;/strong&gt;, time runs slower by a factor of ~3500.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;1 m from the surface&lt;/strong&gt;, time runs more than a 100,000 times slower.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;At the surface&lt;/strong&gt;, time stops.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note that this time dilation is &lt;em&gt;what a distant observer&lt;/em&gt; (i.e. the guy with the &lt;span class=&quot;math-container&quot;&gt;$t_\infty$&lt;/span&gt; time) would measure for an observer at the distance &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt;. The person at &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; would just measure his/her own time as usual. For instance, according to point 5 above, if you were hovering 1 km from the surface, waving your hand every second, then I, choosing to stay at a safe distance of 1 lightyear but with a magically powerful telescope, would see you wave approximately once every hour. And when you run out of fuel and plummet into the BH, then when you cross the surface you wouldn't notice anything particular, but I would see you frozen in time. This is the concept of relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, let me use this chance to clarify something that people, including myself, often have gotten wrong: &lt;strong&gt;Sagittarius A&lt;/strong&gt; (without an asterisk) is a radio source in the center of the Milky Way. It consists of three parts: &lt;strong&gt;Sagittarius A East&lt;/strong&gt; (a supernova remnant), &lt;strong&gt;Sagittarius A West&lt;/strong&gt; (dust and gas clouds), and &lt;strong&gt;Sagittarius A*&lt;/strong&gt;, or &lt;strong&gt;Sgr A*&lt;/strong&gt;, which is a very bright and compact radio source believed to be formed by a supermassive BH. Sgr A* isn't actually the BH &lt;em&gt;itself&lt;/em&gt;. I think the BH doesn't really have a name, so I'll call it &lt;span class=&quot;math-container&quot;&gt;$\mathrm{BH}_\mathrm{Sgr\,A^*}$&lt;/span&gt;. Maybe that's a bad name…&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2019-11-01T13:18:32.240" LastActivityDate="2019-11-01T13:18:32.240" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="10471" PostTypeId="2" ParentId="10457" CreationDate="2015-04-14T19:43:34.260" Score="2" Body="&lt;p&gt;Actually, it &lt;em&gt;does&lt;/em&gt; make sense to say that a univere can be like the (surface of the) Earth. A &quot;uniform&quot; (i.e. homogeneous and isotropic) universe can have three distinct geometries. See &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10344/what-is-a-friedmann-model&quot;&gt;this post&lt;/a&gt; about the shape of a universe. The bottom line of this is that a 2D analogy of a universe can be either flat (like an infinite table), round (like the Earth), or hyperbolic (like a saddle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which of the three &lt;em&gt;our&lt;/em&gt; Universe is, depends on its average density, and observations suggest that it's flat. That means that you can fly forever in a straight line, forever increasing you're distance to Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the density were just a little higher, it would be &quot;round&quot;. This does not mean exactly like a ball, but it does mean that traveling in a straight line &lt;em&gt;away&lt;/em&gt; from Earth would eventually &lt;em&gt;in theory&lt;/em&gt; bring you back to Earth, just as traveling in a straight line on the surface of the Earth eventually brings you back home. Or to Rome, I've heard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I say in theory, because for a universe with normal and dark matter, it would actually stop expanding and collapse before you came back. Only if traveling at the speed of light and starting at the time of Big Bang is it possible to return exactly at the Big Crunch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the universe contained also dark energy, there is a range of ratios between matter and dark energy that would make such a journey possible, but if there were too much dark energy, that universe would expand faster than it would be possible to traverse the universe.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-04-14T19:43:34.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10472" PostTypeId="1" CreationDate="2015-04-14T20:21:26.167" Score="6" ViewCount="138" Body="&lt;p&gt;I am looking at how the Eddington Luminosity is derived and I do no understand why we only care about $P_{rad}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When working out $L_{Edd}$ you take the ratio between  $dP_{rad}/dr$ and $dP_{hydrostatic}/dr$ which must be less or equal to than 1. If $P_{hydro} = P_{gas} + P_{rad}$ why don't we take $P_{gas}$ into account anywhere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L_{Edd} = \frac{4\pi c G M}{\kappa}$$&lt;/p&gt;&#xA;" OwnerUserId="7058" LastActivityDate="2015-05-15T17:20:24.700" Title="What happens to the Gas Pressure when working out the Eddington Luminosity?" Tags="&lt;astrophysics&gt;&lt;luminosity&gt;&lt;hydrostatic-equilibrium&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10473" PostTypeId="2" ParentId="10380" CreationDate="2015-04-14T21:04:00.303" Score="1" Body="&lt;p&gt;The baryonic acoustic oscillations (BAO) have a present-day scale $\sim 100 h^{-1}$ Mpc.  This has little or nothing to do with the appearance of voids, which appear predominantly on scales of $\sim 20h^{-1}$ Mpc. and below.  The BAO scale is only detected in the galaxy distribution by correlation analysis and does not manifest itself by mere visual inspection, unlike the voids, filaments and clusters that dominate the appearance of the cosmic web.  If anything, the voids scales are determined by the velocity field power spectrum, which is to be expected in light of the way the Zel'dovich approximation works (as explained in the movie I referred to).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would have added this as a comment to the post except that I have insufficient reputation and so this is the only way I can make the point.  Apologies are offered if they are in order.&lt;/p&gt;&#xA;" OwnerUserId="5301" LastActivityDate="2015-04-14T21:04:00.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10474" PostTypeId="2" ParentId="10464" CreationDate="2015-04-14T23:38:20.260" Score="6" Body="&lt;p&gt;Yes, there should be dark matter within the Earth, but at very low densities - a few $10^{-22}$ kg/m$^3$ (&lt;a href=&quot;http://iopscience.iop.org/0004-637X/756/1/89&quot; rel=&quot;noreferrer&quot;&gt;Bovy &amp;amp; Tremaine 2012&lt;/a&gt;), something like one hundredth of the density of the interplanetary medium. The whole of the Earth would contain few hundred grammes!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this density averaged over a sphere a lot larger than the Galaxy adds up to a lot of mass, but unlike luminous matter, dark matter should not be particularly concentrated in the plane of the Galaxy (and observations of the dynamics of local stars confirm this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither do we expect dark matter to be especially concentrated within the Earth. Although it does interact gravitationally, the kinetic energy it receives from falling in to the Earth's gravitational potential will be exactly sufficient for it to escape again, unless it suffers some inelastic interaction inside the Earth. Some researchers &lt;em&gt;do&lt;/em&gt; think that it may be possible that weak interactions may trap dark matter in the core of the &lt;em&gt;Sun&lt;/em&gt; (e.g. &lt;a href=&quot;http://arxiv.org/abs/1411.6626&quot; rel=&quot;noreferrer&quot;&gt;Vincent et al. 2015&lt;/a&gt;) or even more likely, in neutron stars (&lt;a href=&quot;http://arxiv.org/pdf/1201.2400.pdf&quot; rel=&quot;noreferrer&quot;&gt;Guver et al. 2012&lt;/a&gt;). But the Earth is not large or dense enough for this to be a likely scenario.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-04-15T13:57:53.217" LastActivityDate="2015-04-15T13:57:53.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10476" PostTypeId="2" ParentId="10457" CreationDate="2015-04-15T05:31:56.523" Score="2" Body="&lt;p&gt;A Universe with (hyper)spherical spatial geometry is realistic, though we don't see any evidence for it but it can't be ruled out either. As a point of interest, hyperspherical geometry doesn't necessarily mean hyperspherical topology, though it does mean the Universe must be compact i.e. of finite volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to address more thoroughly though whether it is possible to circumnavigate a spherical (in both geometric and topological senses) Universe:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing can travel faster than light, so in order to find out if it is possible to circumnavigate the Universe we need to know if light travelling radially outward from some point in space will arrive at the same point in space at some later time. If $R(t_1)$ is the radius of curvature of the Universe at time $t_1$, then to circumnavigate the Universe by that time, the light must have traveled at least a proper distance of $2\pi R(t_1)$ from its starting point. Or in other words the radius of the observable Universe must be at least $2\pi R(t_1)$ by time $t_1$ in order for anything to circumnavigate the Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Particle_horizon&quot; rel=&quot;nofollow&quot;&gt;The radius of the observable Universe is given by&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r_{obs}(t_1) = R(t_1) {\LARGE{\int}}^{t_1}_{ t_ {int}} \frac{cdt}{R(t)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $t_{int}$ is when the Universe begin (e.g. the big bang)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the condition for it to be possible to circumnavigate a spherical Universe in a rocket ship is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\LARGE{\int}}^{t_{fin}}_{ t_ {int}} \frac{cdt}{R(t)} &amp;gt; 2\pi$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $t_{fin}$ is the time when the Universe ends (note in a Universe without beginning $t_{int} = -\infty$ and in a Universe without end $t_{fin} = \infty$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simple models with spherical spatial geometry the value of this integral is easy to calculate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{array} {|l|l|}&#xA;\hline \mathbf{MODEL}&amp;amp;\mathbf{VALUE}\\&#xA;\hline&#xA;\mbox{Matter-dominated closed R-W}&amp;amp;2\pi \\&#xA;\hline&#xA; \mbox{Radiation-dominated closed R-W}&amp;amp;\pi \\&#xA;\hline&#xA;\mbox{De Sitter closed-slicing}&amp;amp;\pi\\&#xA;\hline&#xA;\mbox{Einstein static}&amp;amp;\infty\\&#xA;\hline&#xA;\mbox{Eddington-Lemaitre}&amp;amp;\infty\\&#xA;\hline&#xA;\end{array}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that in the case of a Robertson-Walker closed Universe light can just circumnavigate the Universe once from the big bang to the big crunch as long as the pressure is zero. In the de Sitter closed slicing it can only make it half-way around the circumference, even though there is no big bang or big crunch. In the Einstein static Universe you can go round as many times as you like as you have an infinite amount of time to travel the circumference of fixed length. In the Eddington-Lemaitre model the expansion in early times is so slow you can circumnavigate the Universe as many times as you like.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastEditorUserId="2092" LastEditDate="2015-04-16T23:12:34.740" LastActivityDate="2015-04-16T23:12:34.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10479" PostTypeId="1" CreationDate="2015-04-15T10:23:43.703" Score="7" ViewCount="390" Body="&lt;p&gt;The &quot;Lord of the Rings&quot; has more than 60 moons, some of which are larger than Mercury. My question is, wouldn't the ring, which is mostly comprised of rocks, experience tidal effect whenever a moon is near?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="6760" LastEditDate="2015-04-16T10:37:57.580" LastActivityDate="2019-02-18T08:12:04.607" Title="Wouldn't the rings of Saturn experience tidal effect?" Tags="&lt;gravity&gt;&lt;saturn&gt;&lt;tidal-forces&gt;&lt;natural-satellites&gt;&lt;planetary-ring&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10480" PostTypeId="1" AcceptedAnswerId="10484" CreationDate="2015-04-15T11:20:01.620" Score="3" ViewCount="107" Body="&lt;p&gt;The light coming from a distant galaxy or star can be distorted en route to Earth; this distortion is caused by the curvature of space when a massive object is present. My question is, how can we tell whether these doppelgangers are from the same galaxy (especially in the case of a star)?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="2153" LastEditDate="2015-04-15T22:22:17.370" LastActivityDate="2015-04-15T22:50:12.213" Title="How to differentiate between images of a gravitationally lensed object" Tags="&lt;star&gt;&lt;galaxy&gt;&lt;light&gt;&lt;gravitational-lensing&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10481" PostTypeId="2" ParentId="10472" CreationDate="2015-04-15T14:42:37.500" Score="1" Body="&lt;p&gt;The Eddington luminosity is &lt;em&gt;defined&lt;/em&gt; in this way - but I guess that is not the answer you are looking for!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we ask what isotropic luminosity is required in order to balance the inward gravitational force on a lump of gas then yes we could consider gas pressure as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct formulation is that the pressure gradient due to gas plus radiation balances the density times local gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The justification for neglecting gas pressure would depend on the application. For example, when using the mass limit to justify an approximate upper mass limit for stars, it is reasonably easy to show that the ratio of radiation pressure to gas pressure  $\sim 0.1 (M/M_{\odot})^{2}$, where $M$ is the stellar mass in solar units. Thus for stars $&amp;gt;3M_{\odot}$, radiation pressure dominates, and because $L_{edd} \simeq 3\times10^{4} (M/M_{\odot})\ L_{\odot}$ and $L \simeq (M/M_{\odot})^{3.5}\ L_{\odot}$ for a main sequence star, then $L&amp;lt;L_{edd}$ means that&#xA;$$ (M/M_{\odot})^{3.5} &amp;lt; 3\times10^{4} (M/M_{\odot})$$&#xA;and hence $M &amp;lt; 61 M_{\odot}$. At this mass the radiation pressure will be nearly 400 times the gas pressure, justifying the use of a &quot;gas pressure-free&quot; Eddington limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect, though I can't give chapter and verse, that for the intense radiation fields normally associated with the Eddington limit this will always be the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we are talking about spherical accretion, then because the luminosity is constant at any radius, then the Eddington limit applies to material at any radius. But at greater radii from the central source, for a constant velocity inflow, the density will fall as $r^{-2}$, and the temperature will be lower at large radii. Thus the ratio of radiation to gas pressure will likely &lt;em&gt;increase&lt;/em&gt; with radius and so at some point it will always be appropriate just to consider the radiation pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this implies is that reaching the Eddington limit could stop the accretion, but that even if it does not (e.g. clumpy accretion), then the increasing gas pressure at smaller radii could limit the accretion if the compressed gas is unable to cool.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-05-15T17:20:24.700" LastActivityDate="2015-05-15T17:20:24.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10482" PostTypeId="1" CreationDate="2015-04-15T21:11:21.967" Score="3" ViewCount="185" Body="&lt;p&gt;If neutrino detectors keep improving so that a fair number of neutrinos can be observed, would they be as informative for astronomy as photons are?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are of course a very valuable complement to photons, but I'm thinking about neutrinos in and of themselves. Photons have wavelength, spectral lines, redshift, diffraction, polarization which reveal their origin and interactions on the way to us. Do neutrinos say more than just what direction they come from?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-10-07T22:13:40.867" Title="Do neutrinos have as much information as photons do?" Tags="&lt;observational-astronomy&gt;&lt;neutrinos&gt;&lt;photons&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10483" PostTypeId="2" ParentId="10482" CreationDate="2015-04-15T22:42:53.827" Score="3" Body="&lt;p&gt;This is a very broad question and though a comprehensive answer lies outside the scope of this simple Q&amp;amp;A format, I give you a couple of examples where &quot;neutrino telescopes&quot; would revolutionary.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;There is a predicted cosmic neutrino background, analogous to the cosmic microwave background. Neutrinos decoupled at 1 second after the big bang, filling the universe with neutrinos that should have now cooled to a temperature of about 1.9 K. Confirming this would be a(nother) spectacular validation of the big bang model. However, there is a wrinkle; now that we know neutrinos have mass, it turns out that these neutrinos are non-relativistic at the present epoch. That means they are capable of being diverted and concentrated by gravitational structures - potentially making them fantastic probes of such structure and perhaps more sensitive than photons. So neutrino telescopes would make a fantastic contribution to our understanding of cosmology.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Neutrinos arise from energetic processes in the cores of stars. These processes are otherwise invisible, we certainly can't see them directly with light and can only probe them indirectly using techniques such as asteroseismology or looking at the mixing of chemical elements from the core to the surface. Neutrinos potentially tell us much more; for instance giving a direct estimate of the number of nuclear reactions taking place per second. Neutrino emission is also the predominant way that cooling takes place in supernovae, and in hot neutron stars and white dwarfs. A neutrino telescope could therefore be fundamental  for a deep (literally) understanding of stellar evolution and particularly the late stages of stellar evolution.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In terms of what can be measured; you can measure the flux of neutrinos and the neutrino energy spectrum. There is also of course the timing information that you might get from transient events like supernovae. There is also the question of flavour oscillations and the mix of neutrino types that are observed. I don't understand enough about this to make any predictions about how astronomically useful this property is - but I bet it will be.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-10-07T22:13:40.867" LastActivityDate="2015-10-07T22:13:40.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10484" PostTypeId="2" ParentId="10480" CreationDate="2015-04-15T22:50:12.213" Score="2" Body="&lt;p&gt;In the case of multiple images of a background, distant object, the answer is relatively simple. You take a spectrum of the multiple images or parts of an extended lensed image and you see whether the spectrum looks the same, and in particular whether the redshift of the multiple images are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational lensing affects light of all wavelengths equally, so regardless of the path taken, the spectrum should be unaltered, &lt;em&gt;except&lt;/em&gt; that the different paths take different amounts of time to travel from the original object to us. So, if the background source, or its spectrum, are time-varying, then the lensed images could also appear different. However, distant galaxies do not change their redshift on such short timescales and so this should still be the same regardless.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-04-15T22:50:12.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10485" PostTypeId="1" CreationDate="2015-04-16T05:19:47.703" Score="1" ViewCount="106" Body="&lt;p&gt;I found this at &#xA;&lt;a href=&quot;http://www.nature.com/news/the-exoplanet-next-door-1.11605&quot; rel=&quot;nofollow&quot;&gt;http://www.nature.com/news/the-exoplanet-next-door-1.11605&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Even when launched by one of the most powerful rockets on Earth, boosted by a gravitational slingshot around Jupiter, and further accelerated by a radioisotope thruster, that probe would take about 28,000 years to reach α Centauri. Quoting British author Douglas Adams, McNutt quips: “Space is big. Really big.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is this still the fastest we can get right now?&lt;/p&gt;&#xA;" OwnerUserId="5501" LastActivityDate="2015-04-16T06:05:36.130" Title="The fastest Probe we could build now" Tags="&lt;interstellar-travel&gt;&lt;space-probe&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10486" PostTypeId="2" ParentId="10485" CreationDate="2015-04-16T06:05:36.130" Score="1" Body="&lt;p&gt;For small values of practical an interstellar &lt;a href=&quot;http://en.wikipedia.org/wiki/Project_Orion_%28nuclear_propulsion%29#Interstellar_missions&quot; rel=&quot;nofollow&quot;&gt;Orion craft&lt;/a&gt; could achieve ~3.3% c (~133 years to Alpha Centauri, one way with no deceleration at the end of the trip)). Which is in principle within reach of present technology if we chose to pursue it. But cost and aversion to nuclear technology on such an extravagant scale make it unlikely that we will see Orion space craft of any ilk built in the foreseeable future.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-04-16T06:05:36.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10487" PostTypeId="1" CreationDate="2015-04-16T19:21:17.943" Score="3" ViewCount="801" Body="&lt;p&gt;When we travel through space, we use a curved path following an orbit to reach our destinations.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Pq8fP.gif&quot; alt=&quot;Like So&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;As far as I'm aware, we could take a more straight line approach but it would require much more deltaV (both in reaching the speed and in slowing down when reaching the target to be captured.)&lt;/p&gt;&#xA;&lt;p&gt;However the path would still be slightly curved because of the curvature of space and because of orbital tracjectory when leaving Earth.&lt;/p&gt;&#xA;&lt;p&gt;Does the light escaping the Sun move fast enough to follow a straight line path until it hits Earth, or is it curved in those 8 minutes?&lt;/p&gt;&#xA;&lt;p&gt;What about to somewhere further like Pluto?&lt;/p&gt;&#xA;" OwnerUserId="2046" LastEditorUserId="37191" LastEditDate="2021-05-16T22:02:44.390" LastActivityDate="2021-05-16T22:02:44.390" Title="Does the Sun's light travel fast enough to have a straight path to Earth?" Tags="&lt;orbit&gt;&lt;light&gt;&lt;speed&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="10488" PostTypeId="1" AcceptedAnswerId="10495" CreationDate="2015-04-16T22:13:37.543" Score="3" ViewCount="3963" Body="&lt;p&gt;I want to find a star's magnitude in a triple star. I already knew the the magnitude of the triple star and the other two star, but I don't quit know how to solve it. Is there a way to find it?&lt;/p&gt;&#xA;" OwnerUserId="7076" LastActivityDate="2015-04-21T05:00:26.220" Title="How to calculate magnitude of a star in a triple star system?" Tags="&lt;star&gt;&lt;apparent-magnitude&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10489" PostTypeId="2" ParentId="10487" CreationDate="2015-04-16T22:57:15.320" Score="4" Body="&lt;p&gt;Light travels in straight lines in spacetime, but not necessarily straight lines through space, and the same is true for free-falling orbits of test particles like satellites (when thrust-less). However, in relativity what constitutes 'space' means taking some 'slice of &lt;em&gt;now&lt;/em&gt;' through spacetime, which is of course depends on the reference frame one chooses to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For weak, slowly changing gravitational fields appropriate to the Sun, a conventional choice is the static weak-field metric&#xA;$$\mathrm{d}s^2 = -\left(1+2\frac{\Phi}{c^2}\right)c^2\,\mathrm{d}t^2 + \left(1-2\frac{\Phi}{c^2}\right)\underbrace{\left(\mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2\right)}_{\mathrm{d}S^2}\text{,}$$&#xA;where $\Phi$ is the Newtonian gravitational potential. Thus, we can compare to a hypothetical flat spacetime, with a flat Euclidean space given by $\mathrm{d}S^2$. The trajectory of light follows null geodesics ($\mathrm{d}s = 0$), so the light travels &lt;em&gt;as if&lt;/em&gt; traveling through a medium of &lt;a href=&quot;https://en.wikipedia.org/wiki/Refractive_index&quot; rel=&quot;nofollow&quot;&gt;refractive index&lt;/a&gt;&#xA;$$n = c\frac{\mathrm{d}t}{\mathrm{d}S} = c\sqrt{\frac{1-2\Phi/c^2}{1+2\Phi/c^2}}\approx 1 - 2\frac{\Phi}{c^2}\text{.}$$&#xA;A changing refractive index does make light bend; this is exemplified in the simplest case by &lt;a href=&quot;https://en.wikipedia.org/wiki/Snell%27s_law&quot; rel=&quot;nofollow&quot;&gt;Snell's law&lt;/a&gt;, although the continuously varying case found here needs the more general &lt;a href=&quot;https://en.wikipedia.org/wiki/Fermat%27s_principle&quot; rel=&quot;nofollow&quot;&gt;Fermat's principle&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You shouldn't take the analogy of a refractive medium too literally. Some differences between an actual medium is that there is no dispersion as the refractive index is independent of frequency, and that there is an additional effect of &lt;em&gt;gravitational redshift&lt;/em&gt; not present in refractive media. But the latter is separate issue from the trajectories of light rays, so we can ignore it for our immediate purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light deflection is quantified by an &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Impact_parameter&quot; rel=&quot;nofollow&quot;&gt;impact parameter&lt;/a&gt;&lt;/em&gt; $b$, which can be thought of as the perpendicular distance between the Sun and a hypothetical flat-spacetime trajectory, and a &lt;em&gt;scattering angle&lt;/em&gt; $\theta$, the angle between that and its actual path in the limit of infinite distance. (Note that the illustration in wikipedia assumes a repulsive rather than attractive potential, but otherwise conceptually it's the same situation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modeling the Sun as having a spherically symmetrical gravitational potential $\Phi = -GM/r$, the leading-order light deflection is,&#xA;$$\begin{eqnarray*}&#xA;\theta = 4\frac{GM}{c^2b} &amp;amp;\quad&amp;amp;\left(\text{if}\;\frac{GM}{c^2b}\ll 1\right)\text{,}\end{eqnarray*}$$&#xA;which is twice what one would get for a particle at light speed under Newtonian mechanics. For light ray just grazing the outer edge of the Sun (or being emitted from there), this angle is about $1.7$ seconds of arc. Completely radial light rays would not be bent in this spherically-symmetric case.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2015-04-16T22:57:15.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10490" PostTypeId="1" AcceptedAnswerId="10491" CreationDate="2015-04-17T00:57:03.460" Score="10" ViewCount="14720" Body="&lt;p&gt;Why do we divide time into weeks?  Is there any celestial reason why humans do this?  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;one year: earth revolution around the sun &lt;/li&gt;&#xA;&lt;li&gt;one month: moon revolution around the earth &lt;/li&gt;&#xA;&lt;li&gt;one week: 7 days = ??? &lt;/li&gt;&#xA;&lt;li&gt;one day: earth rotation about its axis&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2547" LastEditorUserId="2547" LastEditDate="2017-11-18T21:01:40.757" LastActivityDate="2018-10-31T19:44:35.423" Title="Years, months, days, and ... weeks?" Tags="&lt;planet&gt;&lt;time&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10491" PostTypeId="2" ParentId="10490" CreationDate="2015-04-17T01:19:05.237" Score="20" Body="&lt;p&gt;The synodic period of the moon is $29.53$ days, a little shorter than a calendar month, which is on average about $30.4$ days. This is slightly longer than its orbital period, but corresponds to the periodic visual appearance of the moon as viewed from Earth. I mention this to make it clear that we should be forgiving of a little imprecision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conventionally, the moon's appearance is divided into four phases: first quarter, full, last quarter, new. That means that on average, each phase lasts about $7.4$ days. Since calendars count days in integer amounts, a $7$-day period seems to be a natural choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The social importance of the seven-day period in Western cultural probably has much more to do with its religious significance in Abrahamic religions than astronomy per se (although certainly not unique to it). But its ultimate origin probably does lie in the natural division of the moon's appearance into four phases, which correspond to an apparent geocentric celestial latitude difference between the Moon and Sun of $0^\circ$, $90^\circ$, $180^\circ$, and $270^\circ$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That, the explicit answer to your question is&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 week = 7 days = one lunar phase.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="661" LastActivityDate="2015-04-17T01:19:05.237" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10492" PostTypeId="1" CreationDate="2015-04-17T05:50:20.720" Score="5" ViewCount="135" Body="&lt;p&gt;We'll mount all of them in a same location and connect them to each other (like VLT) so they will combined all the results and will produce some staggering images however my question is, would they be able to produce an image close to the resolution of a 300m optical telescope?&lt;/p&gt;&#xA;" OwnerUserId="7078" LastEditorUserId="7078" LastEditDate="2015-04-17T05:56:20.160" LastActivityDate="2015-04-17T05:56:20.160" Title="If we built one hundred 3m telescopes and point them to a single star what would be the final resolution?" Tags="&lt;telescope&gt;" AnswerCount="0" CommentCount="6" ClosedDate="2017-01-17T00:51:41.963" ContentLicense="CC BY-SA 3.0" />
  <row Id="10494" PostTypeId="1" AcceptedAnswerId="10496" CreationDate="2015-04-17T06:09:53.737" Score="1" ViewCount="106" Body="&lt;p&gt;Is there any way I can tell from Hipparcos catalog which stars are Binary/Multiple Stars and which are Variable stars?&lt;/p&gt;&#xA;" OwnerUserId="5401" LastEditorUserId="5401" LastEditDate="2015-04-17T12:50:37.803" LastActivityDate="2015-04-17T13:05:52.290" Title="Variable and Multiple Stars in Hipparcos" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10495" PostTypeId="2" ParentId="10488" CreationDate="2015-04-17T07:47:13.010" Score="2" Body="&lt;p&gt;Basically you need to convert between luminosities (which you can add) and magnitudes using &#xA;$$M-M_\odot=-2.5\log_{10}(L/L_\odot)$$&#xA;Let's call the total luminosity $L_0$ and magnitude $M_0$ and the individual luminosities and magnitudes $L_1$, $L_2$ and $L_3$ and $M_1$, $M_2$ and $M_3$.&#xA;Then, you have the total luminosity of the system, directly&#xA;$$L_0/L_\odot=10^{-0.4(M_0-M_\odot)}$$&#xA;and as the sum of the components&#xA;$$L_0/L_\odot=(L_1+L_2+L_3)/L_\odot=10^{-0.4(M_1-M_\odot)}+10^{-0.4(M_2-M_\odot)}+10^{-0.4(M_3-M_\odot)}$$&#xA;Solving these equations for $M_3$ gives&#xA;$$M_3-M_\odot=-2.5\log_{10}\left(10^{-0.4(M_0-M_\odot)}-10^{-0.4(M_1-M_\odot)}-10^{-0.4(M_2-M_\odot)}\right)$$&#xA;I'm assuming you have absolute magnitudes, but you can rewrite the formulae in terms of apparent magnitudes using&#xA;$$M=m+5(1-\log_{10}d)$$&#xA;but I &lt;em&gt;think&lt;/em&gt; the result then also depends on the distance.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastEditorUserId="1577" LastEditDate="2015-04-21T05:00:26.220" LastActivityDate="2015-04-21T05:00:26.220" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10496" PostTypeId="2" ParentId="10494" CreationDate="2015-04-17T07:54:28.190" Score="2" Body="&lt;p&gt;Yes. If you have the Hipparcos data from ftp (&lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR?-source=I/239&amp;amp;-to=3&quot; rel=&quot;nofollow&quot;&gt;http://vizier.u-strasbg.fr/viz-bin/VizieR?-source=I/239&amp;amp;-to=3&lt;/a&gt;), you will have several data files.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;code&gt;hip_main.dat&lt;/code&gt; you have a field &lt;code&gt;MultFlag&lt;/code&gt; at position 347 that indicates whether the star is a double or multiple star:    &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Note on MultFlag: indicates that further details are given in the Double&#xA;     and Multiple Systems Annex:&#xA;     C : solutions for the components&#xA;     G : acceleration or higher order terms&#xA;     O : orbital solutions&#xA;     V : variability-induced movers (apparent motion arises from variability)&#xA;     X : stochastic solution (probably astrometric binaries with short period)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To determine whether the star is a double or multiple star system, look in one of the files for double star solutions. For instance in &lt;code&gt;h_dm_com.dat&lt;/code&gt; there is a field at bytes 24-25 that gives the number of components (2 for double and &gt;2 for multiple stars).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;24- 25  I2     ---     Ncomp    Number of components in this solution   (DCM2)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The Hipparcos number identifying the star can be found in bytes 43-48.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this only works for multiple stars with orbital solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;strong&gt;variability&lt;/strong&gt;, three fields are provided in the main Hipparcos file:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;314-320  F7.2  d       Period    ? Variability period (days)              (H51)&#xA;     322  A1    ---     HvarType *[CDMPRU]? variability type               (H52)&#xA;     324  A1    ---     moreVar  *[12] Additional data about variability   (H53)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;with two notes (NB. if &lt;code&gt;HvarType&lt;/code&gt; is &lt;code&gt;C&lt;/code&gt; then the star is constant, i.e. not variable!):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Note on HvarType: Hipparcos-defined type of variability (a blank entry&#xA;     signifies that the entry could not be classified as variable or constant):&#xA;     C : no variability detected (&quot;constant&quot;)&#xA;     D : duplicity-induced variability&#xA;     M : possibly micro-variable (amplitude &amp;lt; 0.03mag)&#xA;     P : periodic variable&#xA;     R : V-I colour index was revised due to variability analysis&#xA;     U : unsolved variable which does not fall in the other categories&#xA;Note on moreVar: more data about periodic variability are provided&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There are two additional files on variability: &lt;code&gt;hip_va_1.dat&lt;/code&gt; and &lt;code&gt;hip_va_2.dat&lt;/code&gt;, which provide information such as variable type (GCVS) and max and min magnitude and also the variable star name.&#xA;You can find a description of the variable type in the General Catalogue of Variable Stars (GCVS) in the file &lt;code&gt;vartype.txt&lt;/code&gt; if you download the GCVS at &lt;a href=&quot;http://cdsarc.u-strasbg.fr/viz-bin/Cat?B/gcvs&quot; rel=&quot;nofollow&quot;&gt;http://cdsarc.u-strasbg.fr/viz-bin/Cat?B/gcvs&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="864" LastEditorUserId="864" LastEditDate="2015-04-17T13:05:52.290" LastActivityDate="2015-04-17T13:05:52.290" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10497" PostTypeId="2" ParentId="10332" CreationDate="2015-04-17T13:29:21.970" Score="1" Body="&lt;p&gt;You might want to have a look to the &lt;a href=&quot;http://www.galextin.org/&quot; rel=&quot;nofollow&quot;&gt;GALExtin&lt;/a&gt; models. The official site is still not finished (not even sure if it's still being developed), but you can access the original article &lt;a href=&quot;http://adsabs.harvard.edu/abs/2005AJ....130..659A&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, and download the models &lt;a href=&quot;http://www.galextin.org/modextin.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a &lt;a href=&quot;http://www.jenam2010.org/clusters/Programme_files/E_Amores.pdf&quot; rel=&quot;nofollow&quot;&gt;poster&lt;/a&gt; that provides a quick introduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically this is composed of two models of the Galaxy (one with spiral arms and one without) to which you give a &lt;code&gt;(l, b)&lt;/code&gt; direction and a distance, and it gives you back the extinction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a little bit old but perhaps it can be useful to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the advice given by Rob is a good one: for such small distances perhaps the best thing to do is to assume zero extinction.&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2015-04-17T13:29:21.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10498" PostTypeId="2" ParentId="10490" CreationDate="2015-04-17T14:09:11.673" Score="1" Body="&lt;p&gt;The 7 days of the week certainly fit somewhat as a quarter of a month. But there was another argument for the ancients to use the number seven, because they knew about seven &quot;planets&quot;, seven objects which wander across the sky relative to the thousands of fixed stars. The Sun, the Moon, Mercury, Venus, Mars, Jupiter, Saturn. Earth wasn't a planet according to the IAU definition at the time. Earth was instead the center of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weekly calendar of the ancients was pretty busy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sunday&lt;/strong&gt; - &lt;a href=&quot;http://www.etymonline.com/index.php?term=Sunday&amp;amp;allowed_in_frame=0&quot; rel=&quot;nofollow&quot;&gt;Sun&lt;/a&gt; you're free, sober up and get a suntan&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Monday&lt;/strong&gt; - &lt;a href=&quot;http://www.etymonline.com/index.php?term=Monday&amp;amp;allowed_in_frame=0&quot; rel=&quot;nofollow&quot;&gt;Moon&lt;/a&gt; the day after being free&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tuesday&lt;/strong&gt; - &lt;a href=&quot;http://www.etymonline.com/index.php?term=Tuesday&amp;amp;allowed_in_frame=0&quot; rel=&quot;nofollow&quot;&gt;Jupiter&lt;/a&gt; Zeus day, be clever and figure something out&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Wednesday&lt;/strong&gt; - &lt;a href=&quot;http://www.etymonline.com/index.php?term=Tuesday&amp;amp;allowed_in_frame=0&quot; rel=&quot;nofollow&quot;&gt;Mercury&lt;/a&gt; go somewhere and talk about it&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thursday&lt;/strong&gt; - &lt;a href=&quot;http://www.etymonline.com/index.php?term=Thursday&amp;amp;allowed_in_frame=0&quot; rel=&quot;nofollow&quot;&gt;Mars&lt;/a&gt; fight&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Friday&lt;/strong&gt; - &lt;a href=&quot;http://www.etymonline.com/index.php?term=Friday&amp;amp;allowed_in_frame=0&quot; rel=&quot;nofollow&quot;&gt;Venus&lt;/a&gt; make love&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Saturday&lt;/strong&gt; - &lt;a href=&quot;http://www.etymonline.com/index.php?term=Saturday&amp;amp;allowed_in_frame=0&quot; rel=&quot;nofollow&quot;&gt;Saturn&lt;/a&gt; please take a bath!&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-04-17T14:09:11.673" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10499" PostTypeId="1" AcceptedAnswerId="10640" CreationDate="2015-04-17T18:27:22.013" Score="13" ViewCount="2960" Body="&lt;p&gt;In 4 billion years, when our Sun sheds all of its outer gas layers and turns into a white dwarf, how much mass will the white dwarf have compared to what the sun has today?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the planets still orbit in the same way, or will the reduced mass cause the planets' trajectories to change, so that they eventually leave the solar system?&lt;/p&gt;&#xA;" OwnerUserId="2439" LastEditorUserId="2531" LastEditDate="2015-05-25T14:25:18.930" LastActivityDate="2019-10-11T09:15:56.970" Title="How much mass will the Sun have when it becomes a white dwarf?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;white-dwarf&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10500" PostTypeId="1" AcceptedAnswerId="10524" CreationDate="2015-04-17T21:06:15.590" Score="0" ViewCount="250" Body="&lt;p&gt;Is there a legend for these 88 Hubble Space Telescope images?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/W91ky.jpg&quot; alt=&quot;HubbleGallery.jpg - Hubble Space Telescope&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a 6000px x 4800 px version of this image (reduced &amp;amp; rotated to 480x600 here) that I use for an &lt;a href=&quot;http://forums.aussievapers.com/club-vape-|-off-topic-discussion/31195-hubble-gallery-slide-show.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;image slide show&lt;/a&gt;.  I would like to add the description of each image into the slide show software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On this system the file name is &lt;code&gt;HubbleGallery.jpg&lt;/code&gt; but I suspect I coined that title (it was a long time ago it was downloaded).  The original image is 5,150,779 bytes in size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Searching on the smaller, rotated version of the image points to pages like &lt;a href=&quot;http://hubblesite.org/gallery/printshop/ps17/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomy Printshop&lt;/a&gt;, but there is nothing closely related to NASA in the first pages of the search, and the pages listed did not have much detail (certainly no legend).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might possible to search on the clipped, individual images, but I was hoping to avoid having to do that 88 times.  &lt;/p&gt;&#xA;" OwnerUserId="1840" LastActivityDate="2016-03-16T21:56:41.333" Title="Is there a legend for these 88 HST images?" Tags="&lt;hubble-telescope&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10502" PostTypeId="1" AcceptedAnswerId="10504" CreationDate="2015-04-18T09:48:55.647" Score="8" ViewCount="616" Body="&lt;p&gt;I was reading &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10387/open-problems-in-astronomy-that-an-amateur-with-a-phd-in-some-other-field-woul&quot;&gt;this question about open problems in astronomy&lt;/a&gt;. The bounty-receiving answer says that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have just left the field where some colleagues are trying to develop some software to use image subtraction techniques to isolate individual stars in the centre of clusters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I find this problem interesting. Where could I find databases of astronomical images for testing different kind of ML techniques to solve such problems?&lt;/p&gt;&#xA;" OwnerUserId="7089" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-04-18T14:12:56.960" Title="Astronomical databases for machine learning?" Tags="&lt;star&gt;&lt;fundamental-astronomy&gt;&lt;data-analysis&gt;&lt;photography&gt;&lt;star-cluster&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10504" PostTypeId="2" ParentId="10502" CreationDate="2015-04-18T14:12:56.960" Score="3" Body="&lt;p&gt;The European Southern Observatory has catalogues with image data available from &lt;a href=&quot;http://www.eso.org/qi/&quot; rel=&quot;nofollow&quot;&gt;http://www.eso.org/qi/&lt;/a&gt;, you will have to register before you are able to access them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd suggest you look at other observatory's websites for their data. You will have to look past the pages targeted at the general public and find links for data or science, or user portal or something like that. They are sometimes difficult to find.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally you would have a standard annotated data set of images for classification so that you will be able to compare your results with others. Unfortunately I'm not aware of any standard sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For literature on using pattern recognition on astronomical images, &lt;a href=&quot;http://astrometry.net/biblio.html&quot; rel=&quot;nofollow&quot;&gt;http://astrometry.net/biblio.html&lt;/a&gt; is a good resource. They've created an ML system that is able not only to distinguish stars from other object classes but also able to identify which stars are visible in an image! Very interesting research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB For classifying stars in clusters, the problem is likely not classification, but more segmentation.&lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2015-04-18T14:12:56.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10505" PostTypeId="1" AcceptedAnswerId="10507" CreationDate="2015-04-18T16:15:06.060" Score="1" ViewCount="745" Body="&lt;p&gt;Can anyone give me an example of a star (ideally those with a common name e.g. Rigel, Sirius, etc) of the following multiple star systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equal Mass Star System, where two binary stars orbit round a central point of space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unequal Mass Star System, where a smaller star(s) orbits round a larger binary star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Double Binary where two sets of binary stars orbit round a spot. (Regulus?)&lt;/p&gt;&#xA;" OwnerUserId="5401" LastEditorUserId="5401" LastEditDate="2015-04-18T16:36:51.040" LastActivityDate="2015-04-18T18:12:37.660" Title="Examples of Equal Mass, Unequal Mass and Double Binary Star Systems" Tags="&lt;binary-star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10506" PostTypeId="1" CreationDate="2015-04-18T17:34:46.937" Score="2" ViewCount="1090" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kSVt6.png&quot; alt=&quot;messier 44&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I know the coordinates (right ascension and declination) of the stars marked with red. How should I calculate the coordinates of an unknown star, marked with yellow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Searching over the web, I found a technique called &lt;em&gt;least squares plate constants&lt;/em&gt;. I guess &lt;em&gt;plate constants&lt;/em&gt; dates back before CCD cameras. Unfortunately there was only little explanation. However, if I understood the matter correctly, I should assume the stars are on a surface of a sphere imaged to a plane? Please help :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how should I calculate the distances between the stars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original image is from Wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="7092" LastActivityDate="2017-01-07T18:31:10.830" Title="How to calculate position of an unknown star knowing positions of some other stars from an image?" Tags="&lt;star&gt;&lt;distances&gt;&lt;fundamental-astronomy&gt;&lt;coordinate&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10507" PostTypeId="2" ParentId="10505" CreationDate="2015-04-18T18:12:37.660" Score="2" Body="&lt;h1&gt;Equal mass&lt;/h1&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Alpha_Centauri&quot; rel=&quot;nofollow noreferrer&quot;&gt;Alpha Centauri&lt;/a&gt;&lt;/strong&gt;, the nearest star system to us, is composed of a binary star and Proxima Centauri (making it a triple star system). Alpha Centauri A has a mass of 1.1 M&lt;span class=&quot;math-container&quot;&gt;$_{\odot}$&lt;/span&gt;, while Alpha Centauri B has a mass of 0.9 M&lt;span class=&quot;math-container&quot;&gt;$_{\odot}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If you count brown dwarfs, &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Luhman_16&quot; rel=&quot;nofollow noreferrer&quot;&gt;Luhman 16&lt;/a&gt; are a good target, with each being about 0.04 M&lt;span class=&quot;math-container&quot;&gt;$_{\odot}$&lt;/span&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;h1&gt;Unequal mass&lt;/h1&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Eta_Carinae&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eta Carinae&lt;/a&gt;&lt;/strong&gt; contains a Luminous Blue Variable (LBV) with a mass of 120-160 M&lt;span class=&quot;math-container&quot;&gt;$_{\odot}$&lt;/span&gt; and a smaller (but still very massive) star with a mass of 30-60 M&lt;span class=&quot;math-container&quot;&gt;$_{\odot}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sirius&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sirius&lt;/a&gt;&lt;/strong&gt; is okay, though with two stars of much smaller masses, 2 M&lt;span class=&quot;math-container&quot;&gt;$_{\odot}$&lt;/span&gt; and 1 M&lt;span class=&quot;math-container&quot;&gt;$_{\odot}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h1&gt;Double binary&lt;/h1&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mizar_and_Alcor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mizar&lt;/a&gt;&lt;/strong&gt;, in reality part of a six-star system (Mizar/Alcor), has four stars separated into two pairs. This  ensemble then interacts with the binary star Alcor.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;You can find plenty of stars &lt;a href=&quot;http://en.wikipedia.org/wiki/Star_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-04-18T18:12:37.660" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10510" PostTypeId="1" AcceptedAnswerId="10511" CreationDate="2015-04-18T22:44:13.693" Score="26" ViewCount="985" Body="&lt;p&gt;There are red stars, and orange stars, and yellow stars, and blue stars, and they are all understandable save the fact that there is a 'gap': &lt;strong&gt;There are no green stars.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this because of hydrogen's chemical properties (e.g. the emission spectrum) or some other reason? Or are there just green stars that I have no idea of? If so, I need some pictures; they must look awesome.&lt;/p&gt;&#xA;" OwnerUserId="7095" LastEditorUserId="34945" LastEditDate="2020-08-31T09:12:23.920" LastActivityDate="2020-08-31T09:12:23.920" Title="Why are there no green stars?" Tags="&lt;star&gt;&lt;stellar-astrophysics&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10511" PostTypeId="2" ParentId="10510" CreationDate="2015-04-19T00:47:49.123" Score="33" Body="&lt;p&gt;Human color vision is based on three types of &quot;cones&quot; in the eye that respond differently to different wavelengths of light. Thus, not counting overall brightness, the human color space has two degrees of freedom. In contrast, the spectra of stars are very close to a &lt;a href=&quot;https://en.wikipedia.org/wiki/Planck%27s_law&quot; rel=&quot;noreferrer&quot;&gt;black body&lt;/a&gt;, which depends only on effective temperature. As one varies the temperature, the color of a star should make a one-dimensional curve in this color space. Thus, unless some &lt;a href=&quot;https://en.wikipedia.org/wiki/Space-filling_curve&quot; rel=&quot;noreferrer&quot;&gt;perverse shenanigans&lt;/a&gt; are going on, it is intuitive that we necessarily miss most of colors, i.e. there will be no stars of those colors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our Sun actually has a &lt;a href=&quot;https://en.wikipedia.org/wiki/Wien%27s_displacement_law&quot; rel=&quot;noreferrer&quot;&gt;peak&lt;/a&gt; at about $500\,\mathrm{nm}$, which is a green. However, that's just the peak: since the Sun also radiates lots of light with shorter wavelength (bluer) and also longer wavelengths (redder), the resulting mixture doesn't look green to human eyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An image from &lt;a href=&quot;https://en.wikipedia.org/wiki/Color_temperature&quot; rel=&quot;noreferrer&quot;&gt;wikipedia&lt;/a&gt; on the color of a blackbody of a given temperature:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sps3c.png&quot; alt=&quot;Color temperature&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the colors at near edges of this color space aren't accurate, because the sRGB standard used in computers only covers a fairly small triangle portion of it. Still, that's a complication that's not very important here.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2015-04-19T00:47:49.123" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10513" PostTypeId="1" AcceptedAnswerId="10514" CreationDate="2015-04-19T07:07:09.103" Score="7" ViewCount="3957" Body="&lt;p&gt;I would like to work in the field of observational astronomy (not so much theoretical astronomy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I currently know two computer programming languages: C# and MATLAB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read that computer programming is absolutely essential to every astronomer, and that it takes a very significant part of the time of the work of most astronomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore I would like to learn a few more programming languages. Hence my question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the most common programming languages in observational astronomy?&lt;/p&gt;&#xA;" OwnerUserId="6786" LastActivityDate="2020-06-18T16:40:13.230" Title="What are the most popular computer programming languages in observational astronomy?" Tags="&lt;software&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10514" PostTypeId="2" ParentId="10513" CreationDate="2015-04-19T09:08:30.853" Score="12" Body="&lt;p&gt;Python is gaining popularity and replacing MATLAB in many fields of science as the tool for fast prototyping and writing research code.&lt;/p&gt;&#xA;&lt;p&gt;For example, have a look at:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.astropy.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.astropy.org/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.astropython.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.astropython.org/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://python4astronomers.github.io/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://python4astronomers.github.io/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.astroml.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.astroml.org/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://asteca.github.io/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://asteca.github.io/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;(I'm not working in astronomy as a researcher but in some other field.)&lt;/p&gt;&#xA;" OwnerUserId="6991" LastEditorUserId="9582" LastEditDate="2020-06-18T16:40:13.230" LastActivityDate="2020-06-18T16:40:13.230" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10516" PostTypeId="1" AcceptedAnswerId="10517" CreationDate="2015-04-19T10:16:41.827" Score="6" ViewCount="350" Body="&lt;p&gt;I read that the stars orbiting around Sagittarius A* (a.k.a the supermassive black hole located at the heart of the Milky Way) move very fast (many times faster than our Sun moves across the galaxy), and it is believed they have enough speed to wander off. My question is, is any exception to this rule, where somehow some stars may break a stellar speed limit not due to a black hole in their vicinity?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="37191" LastEditDate="2021-02-17T08:03:40.060" LastActivityDate="2021-02-17T08:03:40.060" Title="Are black holes solely responsible for hyper velocity stars?" Tags="&lt;star&gt;&lt;gravity&gt;&lt;supermassive-black-hole&gt;&lt;velocity&gt;&lt;runaway-star&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10517" PostTypeId="2" ParentId="10516" CreationDate="2015-04-19T11:40:39.427" Score="6" Body="&lt;p&gt;No, black holes are not the only cause of HVSs, although it is thought to be the most common mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hyper velocity stars are &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_kinematics#Origin_of_hypervelocity_stars&quot; rel=&quot;noreferrer&quot;&gt;believed&lt;/a&gt; to be caused &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;when binary stars come close enough to a &lt;a href=&quot;http://en.wikipedia.org/wiki/Supermassive_black_hole&quot; rel=&quot;noreferrer&quot;&gt;supermassive black hole&lt;/a&gt; for one of the pair to be captured while the other star is ejected at high velocity. This appears to the main mechanism for HVSs. See for example this &lt;a href=&quot;https://www.cfa.harvard.edu/oir/sp/hypervel.html&quot; rel=&quot;noreferrer&quot;&gt;Harvard Smithsonian Center for Astrophysics&lt;/a&gt; page, and this &lt;a href=&quot;http://unews.utah.edu/news_releases/nearest-bright-hypervelocity-star-found/&quot; rel=&quot;noreferrer&quot;&gt;University of Utah&lt;/a&gt; page.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;when one star of a binary goes super nova, the second star may be ejected at high velocity. See &lt;a href=&quot;http://www.sci-news.com/astronomy/science-us708-hypervelocity-star-supernova-galactic-speed-record-02571.html&quot; rel=&quot;noreferrer&quot;&gt;US 708: Hypervelocity Star Ejected by Supernova Breaks Galactic Speed Record&lt;/a&gt; and &lt;a href=&quot;http://www.universetoday.com/102585/how-do-hypervelocity-stars-end-up-breaking-the-speed-limit/&quot; rel=&quot;noreferrer&quot;&gt;How do Hypervelocity Stars End up Breaking The Speed Limit?&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There may be other causes, since not all HVSs seem to be explained by these mechanisms. Perhaps some of them are not from this galaxy at all - they may be just passing through.&lt;/p&gt;&#xA;" OwnerUserId="5524" LastEditorUserId="5524" LastEditDate="2015-04-19T13:33:01.380" LastActivityDate="2015-04-19T13:33:01.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10518" PostTypeId="2" ParentId="10513" CreationDate="2015-04-19T14:50:33.490" Score="3" Body="&lt;p&gt;While I'm not doing so now, for a significant part of my career I worked as a programmer supporting scientists in in some fields that the general public would consider branches of astronomy.  In that time I worked with (chronologically ordered in the sequence in which I first worked with them):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Fortran&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fortran&lt;/a&gt; - still useful because of its speed and existing code base&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/C_%28programming_language%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;C&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/C%2B%2B&quot; rel=&quot;nofollow noreferrer&quot;&gt;C++&lt;/a&gt; - useful for commanding instrumentation, as well as being a good - albeit a sometimes difficult - general purpose language &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.perl.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Perl&lt;/a&gt; - general purpose &quot;glue&quot; language&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.python.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Python&lt;/a&gt; - good general purpose language which is easy to learn and understand.  See &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10513/what-are-the-most-popular-computer-programming-languages-in-observational-astron/10514#10514&quot;&gt;mmh's answer&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.oracle.com/java/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Java&lt;/a&gt; - general purpose language&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://exelisvis.com/ProductsServices/IDL.aspx?utm_source=bing&amp;amp;utm_medium=cpc&amp;amp;utm_campaign=idl&quot; rel=&quot;nofollow noreferrer&quot;&gt;IDL&lt;/a&gt; - useful for imaging and array manipulation&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://php.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;PHP&lt;/a&gt; - for websites &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My recommendation for two or three to learn, would be: Python, Fortran, and C.  Python seems to be rising in popularity in the community and has lots of add on modules for scientific applications (once you've got a handle on Python, I recommend learning &lt;a href=&quot;http://www.numpy.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;NumPy&lt;/a&gt; and maybe &lt;a href=&quot;http://www.scipy.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;SciPy&lt;/a&gt;).  C is an always useful language to know, and I think Fortran will continue to have its uses for a while. Perl is waning in popularity in general.  IDL was once very popular, but its expensive licenses are limiting its adoption for new projects now that free alternatives exist. &lt;/p&gt;&#xA;" OwnerUserId="25" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-04-19T14:50:33.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10519" PostTypeId="1" CreationDate="2015-04-19T15:40:17.870" Score="1" ViewCount="1160" Body="&lt;p&gt;Do we know for certain that the nearest star to Earth, excluding the Sun and the theorized Nemesis, is the Alpha Centauri multiple star system, at +/-4.4 light-years away? Have we been able to identify every star that is &amp;lt;5 light-years from Earth or is there a possibilty a faint star could be nearer?  Are there stars that we haven't looked at yet that could be closer?&lt;/p&gt;&#xA;" OwnerUserId="5401" LastEditorUserId="2153" LastEditDate="2015-04-19T17:02:12.357" LastActivityDate="2015-04-20T20:14:34.190" Title="Could there be a closer star to Earth than the Alpha Centauri triple star system, excluding the Sun?" Tags="&lt;star&gt;&lt;observational-astronomy&gt;" AnswerCount="3" CommentCount="3" ClosedDate="2015-04-24T00:52:26.173" ContentLicense="CC BY-SA 3.0" />
  <row Id="10520" PostTypeId="2" ParentId="10519" CreationDate="2015-04-19T17:24:17.460" Score="3" Body="&lt;p&gt;It's begining to look like &lt;a href=&quot;http://www.nasa.gov/mission_pages/WISE/news/wise20120608.html&quot; rel=&quot;nofollow&quot;&gt;there aren't as many brown dwarfs as was once thought&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;NASA's Wide-field Infrared Survey Explorer, or WISE, has been turning up a new crowd of stars close to home: &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Previous estimates had predicted as many brown dwarfs as typical&#xA;  stars, but the new initial tally from WISE shows just one brown dwarf&#xA;  for every six stars. It's the cosmic equivalent to finally being able&#xA;  to see down a mysterious, gated block and finding only a few homes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.solstation.com/stars/pc10bd.htm&quot; rel=&quot;nofollow&quot;&gt;Reasonably current data&lt;/a&gt; puts the current closest brown dwarf at 6.5LY (2 parsec) and less than 0.08 solar masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nearer one &lt;em&gt;could&lt;/em&gt; show up, but it's not looking likely.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-04-19T17:24:17.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10521" PostTypeId="2" ParentId="10513" CreationDate="2015-04-19T17:37:50.040" Score="7" Body="&lt;p&gt;As an observational astronomer, most of your programming will be to perform data analysis, data exploration, and possibly image manipulation.  Previously much of this was done with IDL, and the analysis pipelines for several/many/all(?) telescopes still rely on IDL.  As &lt;a href=&quot;https://astronomy.stackexchange.com/a/10518/6672&quot;&gt;GreenMatt&lt;/a&gt; points out though, IDL is on its way out.  Since you have to buy a license to use it, you can only share code with other IDL users, it's not open source, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The community is in the process of switching to Python, which is absolutely what I would recommend first for a new entry into the field.  Python is free, open source, etc, etc.  Since you'll be using python for data analysis and numerics, it is &lt;em&gt;vital&lt;/em&gt; that you learn the &lt;a href=&quot;http://www.numpy.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;NumPy&lt;/a&gt; package as well.  With NumPy array operations in python are almost as fast as C or Fortran.  &lt;a href=&quot;http://www.scipy.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;SciPy&lt;/a&gt; and &lt;a href=&quot;http://matplotlib.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;MatPlotLib&lt;/a&gt; are also incredibly useful, you don't need to learn the entire library, but you should be familiar with the basics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The astronomy community is collaborating on some python packages to make more analysis techniques common.  These are the &lt;a href=&quot;http://www.astropy.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;AstroPy&lt;/a&gt; and &lt;a href=&quot;http://www.astroml.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;AstroML&lt;/a&gt; packages.  Once you get a handle on things, both would be good to be familiar with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C is great to know, but not vital for day-to-day observational astronomy.  We use it a lot in theoretical astronomy to write numerical simulations.  Mainly I see C codes used in observational either for one-off calculations from a large dataset, or used to write fast modules for Python :)  If that sounds up your alley, you should give it a shot too!&lt;/p&gt;&#xA;" OwnerUserId="6672" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-04-19T17:37:50.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10522" PostTypeId="2" ParentId="10519" CreationDate="2015-04-19T19:01:48.490" Score="3" Body="&lt;p&gt;It's pretty easy to put upper limits on how bright any nearer stars could be. As a rough guide, consider a survey like Hipparcos, which is complete to something like 9th magnitude. If there's another star closer than Alpha Centauri, the brightest possibility is if it's at the same distance.  Wolfram|Alpha tells us that &lt;a href=&quot;http://www.wolframalpha.com/input/?i=9th%20magnitude%20at%201.34%20parsec&quot; rel=&quot;nofollow&quot;&gt;a magnitude 9 star at Alpha Centauri's distance&lt;/a&gt; would have a luminosity of about 0.0004 times solar. Using the rough main-sequence mass-luminosity relation, $L\propto M^{3.5}$, that would give &lt;a href=&quot;http://www.wolframalpha.com/input/?i=3.901e-4%5E%282%2F7%29&quot; rel=&quot;nofollow&quot;&gt;a mass of about 0.11 times solar&lt;/a&gt;. So already, you can see that if there's anything out there it's barely a star even if it's as distant as Alpha Centauri, and otherwise almost certainly a brown dwarf. I'm sure more recent surveys will place even stronger limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Admittedly, the catch here is that the brightness will be concentrated at longer and longer wavelengths as the mass decreases, so things could be a bit bigger on the ground that we may have only looked in the wrong band so far. But even so, I think the sum of present observations make it nearly impossible that there are any stars closer to us than Alpha Centauri.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2015-04-19T19:01:48.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10523" PostTypeId="2" ParentId="10519" CreationDate="2015-04-19T22:47:11.030" Score="1" Body="&lt;p&gt;In this &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10124/is-it-possible-that-the-sun-has-a-binary-partner-the-nemesis-theory-that-has-e&quot;&gt;answer&lt;/a&gt; you can find a calculation for how bright &quot;Nemesis&quot; would be at near-infrared wavelengths. This calculation assumed we were looking for a 20 Jupiter mass object with a similar age to the Sun at a distance of 1.4 light years (to fit in with the Nemesis hypothesis). The calculated magnitudes were H=14 and W2=8 (in the WISE infrared satellite system).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we relax the assumptions and &lt;strong&gt;let the object be 4.4 light years away, we have to add 2.5 magnitudes to these numbers. i.e. H=16.5 and W2=10.5&lt;/strong&gt;. Whilst the former is right on the edge of detectability in the 2MASS survey, the former is comfortably detectable in the WISE survey (limit is about W2=15.6 see &lt;a href=&quot;http://wise2.ipac.caltech.edu/docs/release/allsky/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://wise2.ipac.caltech.edu/docs/release/allsky/&lt;/a&gt; ) and the object would have a large parallax/proper motion. Whether this would be detectable might depend on the presence of a good &quot;first epoch&quot; image. In the case of the brown dwarf Luhman 16 at 6.5 light years (but which is maybe 40-50 Jupiter masses), well it showed up easily in first epoch 2MASS images (H=9.6).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, your question asks whether a &lt;strong&gt;star&lt;/strong&gt; may have been missed. Such an object would be much brighter than the hypothesised brown dwarf above, or Luhman 16. The &quot;industry-standard&quot; models of &lt;a href=&quot;http://adsabs.harvard.edu/abs/1998A%26A...337..403B&quot; rel=&quot;nofollow noreferrer&quot;&gt;Baraffe et al. (1998)&lt;/a&gt; suggest that a $0.075\ M_{\odot}$ minimum mass stellar object at 4.4 light years &lt;strong&gt;would have H=7.5&lt;/strong&gt;. It is hard to imagine how such an object would have been missed, unless it has such a &lt;em&gt;small&lt;/em&gt; proper motion that it has not moved significantly between the 2MASS survey in the 90s to the WISE survey in ~2010. This is unlikely (but not impossible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once Gaia results are published in 2017, it will have complete parallax data for all stars down to about V=19. This should include even the lowest mass M-dwarf stars out to around 10 parsecs (30 light years).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-04-20T20:14:34.190" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10524" PostTypeId="2" ParentId="10500" CreationDate="2015-04-20T00:14:35.157" Score="1" Body="&lt;p&gt;To answer my own question: no, not that I can find.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an attempt to identify them by image search.  Let me know if there is an image/legend match that seems wrong.  I can upload a higher resolution (466x426 px) version of the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The images were cropped from the original image when it was oriented so the the 'HUBBLE SPACE TELESCOPE' message read from the bottom to top of image, and I will try to identify them in the same order.  As such, we do 1 column at a time from top to bottom, starting from the right most.  The column is marked in red above/below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But first, here is the &lt;a href=&quot;https://www.youtube.com/attribution_link?a=IIZIGoYF240&amp;amp;u=%2Fwatch%3Fv%3Ds33EBgtmkYA%26feature%3Dshare&quot; rel=&quot;nofollow noreferrer&quot;&gt;captioned video&lt;/a&gt; created from combining the captions with a video slideshow of the images! &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Column 1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wgAtB.png&quot; alt=&quot;marker&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/W91ky.jpg&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wgAtB.png&quot; alt=&quot;marker&quot;&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Gravitational lensing of distant galaxies.&lt;/li&gt;&#xA;&lt;li&gt;Red Rectangle Nebula.&lt;/li&gt;&#xA;&lt;li&gt;The Sombrero Galaxy: Messier 104.&lt;/li&gt;&#xA;&lt;li&gt;Doradus Nebula.&lt;/li&gt;&#xA;&lt;li&gt;Stingray Nebula.&lt;/li&gt;&#xA;&lt;li&gt;Elliptical Galaxy NGC 4881 in Coma.&lt;/li&gt;&#xA;&lt;li&gt;Galaxy NGC 4214.&lt;/li&gt;&#xA;&lt;li&gt;Glowing Eye of NGC 6751.&lt;/li&gt;&#xA;&lt;li&gt;Collection of galaxies of different types, from different ages in the universe.&lt;/li&gt;&#xA;&lt;li&gt;Uranus: Springtime in the Southern Hemisphere.&lt;/li&gt;&#xA;&lt;li&gt;The Lunar Aristarchus crater imaged using ultraviolet and visible wavelengths.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Column 2&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wgAtB.png&quot; alt=&quot;marker&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/W91ky.jpg&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wgAtB.png&quot; alt=&quot;marker&quot;&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Supernova remnant N132D, in the Large Magellanic Cloud.&lt;/li&gt;&#xA;&lt;li&gt;Ring Galaxy AM 0644-741.&lt;/li&gt;&#xA;&lt;li&gt;Bug Nebula NGC 6302.&lt;/li&gt;&#xA;&lt;li&gt;Planetary Nebula M2-9.&lt;/li&gt;&#xA;&lt;li&gt;The Cone Nebula in NGC 2264.&lt;/li&gt;&#xA;&lt;li&gt;Young planetary nebula MyCn18.&lt;/li&gt;&#xA;&lt;li&gt;A stellar jet in the Trifid nebula.&lt;/li&gt;&#xA;&lt;li&gt;Galaxy Pair NGC 3314. A face-on galaxy lying precisely in front of another spiral galaxy.&lt;/li&gt;&#xA;&lt;li&gt;Comet LINEAR dispersing after breaking up.&lt;/li&gt;&#xA;&lt;li&gt;A 5000 light year long jet from the giant elliptical galaxy M87.&lt;/li&gt;&#xA;&lt;li&gt;The Massive Stars of 30 Doradus as seen in the star cluster R136.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Column 3&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wgAtB.png&quot; alt=&quot;marker&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/W91ky.jpg&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wgAtB.png&quot; alt=&quot;marker&quot;&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;NGC 346 is an open cluster with associated nebula located in the Small Magellanic Cloud. &lt;/li&gt;&#xA;&lt;li&gt;The Cat's Eye Nebula or NGC 6543, is a relatively bright planetary nebula in the northern constellation of Draco. &lt;/li&gt;&#xA;&lt;li&gt;This infrared view of Uranus reveals that the planet is surrounded by its four major rings and by 10 of its 17 known satellites. &lt;/li&gt;&#xA;&lt;li&gt;Galaxy NGC 4013. An amazing \&quot;edge-on\&quot; view of a spiral galaxy 55 million light years from Earth. &lt;/li&gt;&#xA;&lt;li&gt;The inner ring of NGC 1512. &lt;/li&gt;&#xA;&lt;li&gt;Detail of the Omega Nebula. &lt;/li&gt;&#xA;&lt;li&gt;This close-up of the red planet Mars was taken when it was just 55,760,220 km away. &lt;/li&gt;&#xA;&lt;li&gt;Stellar Spire in the Eagle Nebula &lt;/li&gt;&#xA;&lt;li&gt;Interacting galaxies NGC 1409 &amp;amp; 1410. &lt;/li&gt;&#xA;&lt;li&gt;NGC 1850 - Star clusters in the Large Magellanic Cloud. &lt;/li&gt;&#xA;&lt;li&gt;Io, Ganymede, and Callisto casting eclipse shadows on Jupiter. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Column 4&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wgAtB.png&quot; alt=&quot;marker&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/W91ky.jpg&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wgAtB.png&quot; alt=&quot;marker&quot;&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Crab Nebula Mosaic &lt;/li&gt;&#xA;&lt;li&gt;N44F's central star is wrapped within a roughly spherical shell of cool orange gas, with fingerlike columns of cool gas and dust protruding from the interior wall. &lt;/li&gt;&#xA;&lt;li&gt;The Whirlpool Galaxy (M51) and companion galaxy. &lt;/li&gt;&#xA;&lt;li&gt;Aurora at Jupiter's north pole as seen in ultraviolet light. &lt;/li&gt;&#xA;&lt;li&gt;Bow shock near young star LL Ori. &lt;/li&gt;&#xA;&lt;li&gt;The Black Eye Galaxy (also called Evil Eye Galaxy; designated Messier 64, M64, or NGC 4826) &lt;/li&gt;&#xA;&lt;li&gt;A color-composite image of the Pleiades, or Seven Sisters (Messier 45 or M45), an open star cluster. &lt;/li&gt;&#xA;&lt;li&gt;Orion nebula mozaic. &lt;/li&gt;&#xA;&lt;li&gt;The Eskimo Nebula (NGC 2392). &lt;/li&gt;&#xA;&lt;li&gt;Galaxy collision in NGC 6745. &lt;/li&gt;&#xA;&lt;li&gt;Rich star fields and glowing hydrogen gas silhouette dense, opaque clouds of interstellar gas and dust in a bright star forming region in Centaurus &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Column 5&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wgAtB.png&quot; alt=&quot;marker&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/W91ky.jpg&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wgAtB.png&quot; alt=&quot;marker&quot;&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Aristarchus Plateau on the Moon. &lt;/li&gt;&#xA;&lt;li&gt;The Tadpole Galaxy features a trail of stars about 280,000 light-years long and massive, bright blue star clusters. &lt;/li&gt;&#xA;&lt;li&gt;Infant Stars in the Small Magellanic Cloud. &lt;/li&gt;&#xA;&lt;li&gt;The Ghost-Head Nebula: NGC 2080. &lt;/li&gt;&#xA;&lt;li&gt;The Pencil Nebula (NGC 2736) is a small part of the Vela Supernova Remnant, located near the Vela Pulsar in the constellation Vela. &lt;/li&gt;&#xA;&lt;li&gt;Flash from star V838 Monocerotis echoes through space. &lt;/li&gt;&#xA;&lt;li&gt;Rampaging Supernova Remnant N63A. &lt;/li&gt;&#xA;&lt;li&gt;The center of globular cluster Omega Centauri. &lt;/li&gt;&#xA;&lt;li&gt;The Bubble Nebula (NGC 7635), also called the Sharpless 162, or Caldwell 11, in the constellation Cassiopeia. &lt;/li&gt;&#xA;&lt;li&gt;The center of NGC 4414, an unbarred spiral galaxy about 62 million light-years away in the constellation Coma Berenices. &lt;/li&gt;&#xA;&lt;li&gt;NGC 3132, also known as the Eight-Burst Nebula, the Southern Ring Nebula, is a bright and extensively studied planetary nebula in the constellation Vela. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Column 6&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wgAtB.png&quot; alt=&quot;marker&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/W91ky.jpg&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wgAtB.png&quot; alt=&quot;marker&quot;&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Messier 22 (also known as M22 or NGC 6656) is an elliptical globular cluster in the constellation Sagittarius, near the Galactic bulge region. &lt;/li&gt;&#xA;&lt;li&gt;Star forming region N11B lies in the Large Magellanic Cloud. &lt;/li&gt;&#xA;&lt;li&gt;The Homunculus Nebula is a bipolar reflection nebula surrounding the massive star system Eta Carinae. &lt;/li&gt;&#xA;&lt;li&gt;Hoag's Object: a Ring Galaxy.  On the outside is a ring dominated by bright blue stars, while near the center lies a ball of much redder stars. &lt;/li&gt;&#xA;&lt;li&gt;Detail of the Keyhole Nebula. The small nebula to the upper left has been nicknamed \&quot;finger of God\&quot; or \&quot;God's birdie\&quot;, due to the gesture it appears to be making. &lt;/li&gt;&#xA;&lt;li&gt;Reflection nebulosity near Merope (IC 349). &lt;/li&gt;&#xA;&lt;li&gt;NGC 1569: Starburst in a small calaxy. &lt;/li&gt;&#xA;&lt;li&gt;NGC 1300 is a barred spiral galaxy about 61 million light-years away in the constellation Eridanus. &lt;/li&gt;&#xA;&lt;li&gt;The Lagoon Nebula &lt;/li&gt;&#xA;&lt;li&gt;Planetary Nebula NGC 6543, the \&quot;Cat's Eye Nebula\&quot; is thought to have formed one one star of a double system went supernova. &lt;/li&gt;&#xA;&lt;li&gt;NGC 1999 is a dust filled bright nebula with a vast hole of empty space represented by a black patch of sky. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Column 7&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wgAtB.png&quot; alt=&quot;marker&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/W91ky.jpg&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/SuyBw.png&quot; alt=&quot;spacer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wgAtB.png&quot; alt=&quot;marker&quot;&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;This glowing region reveals arcs and bubbles formed when stellar winds - streams of charged particles ejected by the Trapezium stars - collide with material. &lt;/li&gt;&#xA;&lt;li&gt;Auroras at Saturn's north and south poles. &lt;/li&gt;&#xA;&lt;li&gt;Sheets of debris from a stellar explosion (N 49, DEM L 190). &lt;/li&gt;&#xA;&lt;li&gt;The Horsehead Nebula. &lt;/li&gt;&#xA;&lt;li&gt;The Retina Nebula: dying star IC 4406 has a doughnut shaped nebula seen edge on. &lt;/li&gt;&#xA;&lt;li&gt;The center of NGC 1275 (also known as Perseus A or Caldwell 24), which is a type 1.5 Seyfert galaxy &lt;/li&gt;&#xA;&lt;li&gt;NGC 4622 has bright twisting arms containing newborn stars and lies 111 million light years away. &lt;/li&gt;&#xA;&lt;li&gt;Double Bubble' of gas and dust in the Large Magellanic Cloud. &lt;/li&gt;&#xA;&lt;li&gt;Saturn displays its familiar banded structure, with haze and clouds at various altitudes.&lt;/li&gt;&#xA;&lt;li&gt;Comet-like filaments along the inner rim of the Helix Nebula's gas ring. &lt;/li&gt;&#xA;&lt;li&gt;Galactic nebula NGC 3603 shows various stages of the life cycle of stars. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Column 8&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wgAtB.png&quot; alt=&quot;marker&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/W91ky.jpg&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/wgAtB.png&quot; alt=&quot;marker&quot;&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Thousands of stars are forming in the cloud of gas and dust known as the Orion nebula. &lt;/li&gt;&#xA;&lt;li&gt;A wheel galaxy around 500 million light-years from the Earth. &lt;/li&gt;&#xA;&lt;li&gt;Stellar spire of the Eagle Nebula &lt;/li&gt;&#xA;&lt;li&gt;The Circinus Spiral Galaxy &lt;/li&gt;&#xA;&lt;li&gt;The Egg Nebula &lt;/li&gt;&#xA;&lt;li&gt;Gaseous streamers flutter from a group of stars.&lt;/li&gt;&#xA;&lt;li&gt;The Seyfert's Sextet.&lt;/li&gt;&#xA;&lt;li&gt;Bubbly ocean of glowing hydrogen, oxygen, and sulphur gas in the extremely massive and luminous molecular nebula Messier 17. &lt;/li&gt;&#xA;&lt;li&gt;North Eastern region of NGC 6369: The Little Ghost Nebula. &lt;/li&gt;&#xA;&lt;li&gt;Double galaxies of NGC 4676 in collision. &lt;/li&gt;&#xA;&lt;li&gt;Star clusters born among the interacting galaxies of Stephan's Quintet. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1840" LastEditorUserId="1840" LastEditDate="2016-03-16T21:56:41.333" LastActivityDate="2016-03-16T21:56:41.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10525" PostTypeId="1" CreationDate="2015-04-20T15:35:19.607" Score="2" ViewCount="98" Body="&lt;p&gt;Isn't it an unlikely coincidence that we live in such special times, that we will have observatories (&lt;a href=&quot;http://www.jwst.nasa.gov/&quot; rel=&quot;nofollow&quot;&gt;JWST&lt;/a&gt;) that see all the way back to the first starlight in the universe? Are we dreaming, and is it possible that we will be surprised and keep finding evolved galaxies more than 13.8 billion years old?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-04-20T15:42:27.967" LastActivityDate="2015-05-20T16:26:09.390" Title="How come, in our lifetime, we will see the first stars which ever formed?" Tags="&lt;telescope&gt;&lt;big-bang-theory&gt;&lt;early-universe&gt;&lt;geocentrism&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10526" PostTypeId="1" CreationDate="2015-04-20T15:42:49.977" Score="3" ViewCount="176" Body="&lt;p&gt;My point of view is that first there was the Big Bang singularity, and then the period of inflation which resulted in the observable Universe becoming many times bigger. But was the Universe expanding between the Big Bang singularity and the beginning of inflation?&lt;/p&gt;&#xA;" OwnerUserId="6920" LastActivityDate="2015-07-19T21:27:01.947" Title="Was the Universe expanding before the beginning of inflation?" Tags="&lt;big-bang-theory&gt;&lt;expansion&gt;&lt;cosmological-inflation&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10527" PostTypeId="1" AcceptedAnswerId="10528" CreationDate="2015-04-20T16:56:14.703" Score="4" ViewCount="981" Body="&lt;p&gt;I have seen a large circular formation of cloud around the full moon sometimes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have anyone seen the same? What is the reason behind this formation?&#xA;Due to lunar attraction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the sky is empty and clear, and no clouds at all, only around the moon, far from the moon, not just very close, that cloud circle formed. &#xA;I saw this many times, like the below drawing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G1Euy.jpg&quot; alt=&quot;Enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7101" LastEditorUserId="332" LastEditDate="2015-04-21T13:52:28.560" LastActivityDate="2016-11-15T10:52:04.540" Title="Circular formation around the moon" Tags="&lt;the-moon&gt;&lt;astrophysics&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10528" PostTypeId="2" ParentId="10527" CreationDate="2015-04-20T18:48:01.680" Score="16" Body="&lt;p&gt;That sounds very much like a &lt;a href=&quot;http://en.wikipedia.org/wiki/22%C2%B0_halo&quot; rel=&quot;nofollow&quot;&gt;22° halo&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not an astrophysical phenomenon; it results from the refraction of light by ice crystals in Earth's atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is what it is, it should be a fuzzy but regular circle centered on the Moon, with a radius of about 22 degrees. If you hold your fist out at arm's length with the thumb extended, the angular distance between the base of your fist and the tip of your thumb is typically close to 15 degrees; you can use that for visual comparison. Remember that 22° is the radius of the halo; the diameter is 44°, close to three time the width of your fist+thumb.&lt;/p&gt;&#xA;" OwnerUserId="53" LastEditorUserId="53" LastEditDate="2015-04-21T23:47:29.050" LastActivityDate="2015-04-21T23:47:29.050" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10529" PostTypeId="2" ParentId="10527" CreationDate="2015-04-20T18:56:05.243" Score="12" Body="&lt;p&gt;I'm guessing what you see is the moonlight being scattered by the hexagonal ice crystals in cirrus or cirrostratus clouds, which lie at very high altitudes, 5-6 km and above. The light is scattered by roughly 22º, and because of the slight wavelength dependence, the halo actually has rainbow-like colors, although often they are so faint that you just perceive it as white.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read more &lt;a href=&quot;http://en.wikipedia.org/wiki/22%C2%B0_halo&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/F2XXY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F2XXY.png&quot; alt=&quot;Moonhalo&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-11-15T10:52:04.540" LastActivityDate="2016-11-15T10:52:04.540" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10531" PostTypeId="1" CreationDate="2015-04-20T21:58:44.963" Score="0" ViewCount="63" Body="&lt;p&gt;I'm looking for a way to find stars within a certain region of space, the Lagoona Nebula in this case, based on what we currently know. Are there some databases that allow you to make such queries?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that I'm not looking for stars that can be &lt;em&gt;seen&lt;/em&gt; as being near M8 from earth, but for a list of known stars that are thought to actually be near or part of M8 itself.&lt;/p&gt;&#xA;" OwnerUserId="7103" LastActivityDate="2015-07-31T15:05:01.540" Title="How can I find out which known stars lie within or close to M8" Tags="&lt;distances&gt;&lt;star-gazing&gt;&lt;nebula&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10533" PostTypeId="2" ParentId="10531" CreationDate="2015-04-20T22:32:21.293" Score="1" Body="&lt;p&gt;Have a look at the following references - found relatively easily using the following &lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/nph-abs_connect?db_key=AST&amp;amp;db_key=PRE&amp;amp;qform=AST&amp;amp;arxiv_sel=astro-ph&amp;amp;arxiv_sel=cond-mat&amp;amp;arxiv_sel=cs&amp;amp;arxiv_sel=gr-qc&amp;amp;arxiv_sel=hep-ex&amp;amp;arxiv_sel=hep-lat&amp;amp;arxiv_sel=hep-ph&amp;amp;arxiv_sel=hep-th&amp;amp;arxiv_sel=math&amp;amp;arxiv_sel=math-ph&amp;amp;arxiv_sel=nlin&amp;amp;arxiv_sel=nucl-ex&amp;amp;arxiv_sel=nucl-th&amp;amp;arxiv_sel=physics&amp;amp;arxiv_sel=quant-ph&amp;amp;arxiv_sel=q-bio&amp;amp;sim_query=YES&amp;amp;ned_query=YES&amp;amp;adsobj_query=YES&amp;amp;aut_logic=OR&amp;amp;obj_logic=OR&amp;amp;author=&amp;amp;object=&amp;amp;start_mon=&amp;amp;start_year=&amp;amp;end_mon=&amp;amp;end_year=&amp;amp;ttl_logic=OR&amp;amp;title=&amp;amp;txt_logic=AND&amp;amp;text=Lagoon%20nebula&amp;amp;nr_to_return=200&amp;amp;start_nr=1&amp;amp;jou_pick=ALL&amp;amp;ref_stems=&amp;amp;data_and=ALL&amp;amp;group_and=ALL&amp;amp;start_entry_day=&amp;amp;start_entry_mon=&amp;amp;start_entry_year=&amp;amp;end_entry_day=&amp;amp;end_entry_mon=&amp;amp;end_entry_year=&amp;amp;min_score=&amp;amp;sort=SCORE&amp;amp;data_type=SHORT&amp;amp;aut_syn=YES&amp;amp;ttl_syn=YES&amp;amp;txt_syn=YES&amp;amp;aut_wt=1.0&amp;amp;obj_wt=1.0&amp;amp;ttl_wt=0.3&amp;amp;txt_wt=3.0&amp;amp;aut_wgt=YES&amp;amp;obj_wgt=YES&amp;amp;ttl_wgt=YES&amp;amp;txt_wgt=YES&amp;amp;ttl_sco=YES&amp;amp;txt_sco=YES&amp;amp;version=1&quot; rel=&quot;nofollow&quot;&gt;query&lt;/a&gt; in the NASA ADS site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2000AJ....120..333S&quot; rel=&quot;nofollow&quot;&gt;UBVRI and Hα Photometry of the Young Open Cluster NGC 6530&lt;/a&gt;&#xA;This paper by Sung &amp;amp; Bessell (2000) contains an extensive photometric catalogue. The main table can be &lt;a href=&quot;http://vizier.cfa.harvard.edu/viz-bin/VizieR-3?-source=J/AJ/120/333/table1&quot; rel=&quot;nofollow&quot;&gt;queried interactively&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2008hsf2.book..533T&quot; rel=&quot;nofollow&quot;&gt;The Lagoon Nebula and its Vicinity&lt;/a&gt;&#xA;This is a review article by Tothill et al. (2008), from which you can get pointers to most of the significant literature (pre-2008) on the region.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-04-20T22:32:21.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10536" PostTypeId="1" CreationDate="2015-04-21T01:08:38.867" Score="5" ViewCount="2847" Body="&lt;p&gt;Considering the solar system as a closed system, how long until all the planets pass twice by the same position relative to each other? I would say that you'd have to get the length of the year in each planet and find the lowest number that can be divided by them, is that correct or are there more things to take into account?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, can one do this also considering the moons of the planets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="6661" LastActivityDate="2020-08-11T09:00:35.277" Title="How long for a complete solar system cycle" Tags="&lt;orbit&gt;&lt;solar-system&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10537" PostTypeId="2" ParentId="10536" CreationDate="2015-04-21T01:55:11.097" Score="4" Body="&lt;p&gt;It depends how precisely you mean &lt;em&gt;the same position&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Take 2&lt;/strong&gt; since my previous attempt was ... optimistic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the late 1970's, the outer planets were moving into a configuration that made a &lt;a href=&quot;http://en.wikipedia.org/wiki/Planetary_Grand_Tour&quot; rel=&quot;nofollow&quot;&gt;Planetary Grand Tour&lt;/a&gt; much cheaper and quicker. They will be in a similar configuration in about 175 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when will the planets be in the &lt;em&gt;exact&lt;/em&gt; configuration? &lt;strong&gt;Never&lt;/strong&gt;, well, not before the Sun turns into a red giant and swallows the inner planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But from time to time the planets will all be on the same side of the Sun. If the planets were all in a line, then after 495 years they would all be on the same side of the Sun, then again at 500 years, 522 years, 523 years (these last two are, of course 88 days apart, the period of Mercury), 995 years, 997 years, 1006, 1183 years, etc.&lt;/p&gt;&#xA;" OwnerUserId="5524" LastEditorUserId="5524" LastEditDate="2015-04-21T09:07:37.700" LastActivityDate="2015-04-21T09:07:37.700" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10538" PostTypeId="1" AcceptedAnswerId="10554" CreationDate="2015-04-21T03:48:54.830" Score="10" ViewCount="3571" Body="&lt;p&gt;I am wondering what exactly is meant when it is said the universe expands. Does it simply create new space for matter to fill, does it also create new matter/dark matter to fill that space, or am I way off? Thank you for any help!&lt;/p&gt;&#xA;" OwnerUserId="7109" LastActivityDate="2018-09-06T15:11:48.413" Title="When the universe expands does it create new space, matter, or something else?" Tags="&lt;universe&gt;&lt;space&gt;&lt;expansion&gt;&lt;matter&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10539" PostTypeId="1" AcceptedAnswerId="10541" CreationDate="2015-04-21T04:31:14.553" Score="0" ViewCount="137" Body="&lt;p&gt;Andromeda Galaxy is the largest galaxy in the galaxy cluster&lt;a href=&quot;http://en.wikipedia.org/wiki/Galaxy_cluster&quot; rel=&quot;nofollow&quot;&gt;[1]&lt;/a&gt; called Local Group&lt;a href=&quot;http://en.wikipedia.org/wiki/Local_Group&quot; rel=&quot;nofollow&quot;&gt;[2]&lt;/a&gt; and our Milky Way Galaxy as well as some satellite/dwarf galaxies are orbiting around each other within Local Group. My question isn't Milky Way Galaxy supposed to orbit around Andromeda Galaxy&lt;a href=&quot;http://en.wikipedia.org/wiki/Satellite_galaxy&quot; rel=&quot;nofollow&quot;&gt;[3]&lt;/a&gt; so how come the predicted merger&lt;a href=&quot;http://en.wikipedia.org/wiki/Andromeda%E2%80%93Milky_Way_collision&quot; rel=&quot;nofollow&quot;&gt;[4]&lt;/a&gt;? Whose slowing down or do the gravity work differently as things are being scaled further up?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="6760" LastEditDate="2015-04-21T06:00:16.093" LastActivityDate="2015-04-21T15:54:55.747" Title="Concerning fate of Milky Way Galaxy" Tags="&lt;gravity&gt;&lt;milky-way&gt;&lt;galaxy-cluster&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10540" PostTypeId="1" CreationDate="2015-04-21T07:27:20.563" Score="2" ViewCount="674" Body="&lt;p&gt;In spectroscopic observations, sometimes you meet grism, sometimes grating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of them could cause light dispersion, but what is the difference?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="19" LastEditDate="2015-04-21T13:40:32.640" LastActivityDate="2015-08-17T17:07:57.410" Title="What's the difference between grism and grating?" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10541" PostTypeId="2" ParentId="10539" CreationDate="2015-04-21T10:25:57.123" Score="1" Body="&lt;p&gt;The Milky-Way does not orbit the Andromeda galaxy, they both move under the influence of all the members of the local group. Even if one were orbiting the other the orbit need not be near circular but could be a very eccentric (elongated) ellipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The projected merger is because the tangential component of Andromeda's velocity with respect to the Milky-Way is small compared to its radial component That is Andromeda galaxy appears to be moving almost directly towards the Milky-Way (which is what it says in the Wikipedia page you link to, but it is not too difficult to find primary sources using Google, &lt;a href=&quot;http://arxiv.org/abs/1205.6863&quot; rel=&quot;nofollow&quot;&gt;here is an arXiv paper&lt;/a&gt; reporting a proper motion study of Andromeda and reporting such).&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-04-21T15:54:55.747" LastActivityDate="2015-04-21T15:54:55.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10542" PostTypeId="1" CreationDate="2015-04-21T11:28:53.620" Score="2" ViewCount="138" Body="&lt;p&gt;Is there the possibility for a habitable Planet around a red dwarf? It is not required that the whole surface is habitable, but it shouldn't be a tiny spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading a bit I assume the planet would be tidally locked to the star, with the side that faces the sun looking like an archery-target (from center to border: scorching, hot, warm, cold, freezing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also from space it would look like a Rugby Ball (exaggerated) pointed at the star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But would such a planet be stable enough to develop and support life more complex than the simplest forms? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I do not need a specific star, but the general answer if this is possible, however unlikely (though a value how unlikely would be nice).&lt;/p&gt;&#xA;" OwnerUserId="7031" LastActivityDate="2015-04-23T08:37:16.357" Title="Habitable Planet around Red Dwarf" Tags="&lt;habitable-zone&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10543" PostTypeId="2" ParentId="10536" CreationDate="2015-04-21T14:12:51.223" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Considering the solar system as a closed system, how long until all the planets pass twice by the same position relative to each other?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If that means when will the bodies in the solar system simultaneously repeat their positions from some time long ago, the answer &lt;a href=&quot;http://en.wikipedia.org/wiki/Almost_surely&quot; rel=&quot;nofollow&quot;&gt;almost certainly&lt;/a&gt; is never.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way to get a repeated position amongst a pair of planets is if the ratio of their orbital periods is a rational number, and if perturbations are also somehow related rationally. The only way to get a repeated position amongst all of the planets is for this to be the case amongst all pairs of planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The planets migrated from where they initially formed to their present positions. That the current orbits would somehow be representable exactly in terms of the rationals would be quite the long shot. It's a space of measure zero.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-04-21T14:12:51.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10544" PostTypeId="1" AcceptedAnswerId="10546" CreationDate="2015-04-21T14:18:56.930" Score="4" ViewCount="526" Body="&lt;p&gt;Not all galaxies are spiral in shape&lt;a href=&quot;http://en.wikipedia.org/wiki/Spiral_galaxy&quot; rel=&quot;nofollow&quot;&gt;[1]&lt;/a&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Irregular_galaxy&quot; rel=&quot;nofollow&quot;&gt;[2]&lt;/a&gt;, some nebulae are huge&lt;a href=&quot;http://en.wikipedia.org/wiki/Carina_Nebula&quot; rel=&quot;nofollow&quot;&gt;[3]&lt;/a&gt; and nebula are the nursery of stars&lt;a href=&quot;http://en.wikipedia.org/wiki/Star_formation&quot; rel=&quot;nofollow&quot;&gt;[4]&lt;/a&gt;. How to tell them apart?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have already compare the contents for galaxy and a nebula, they are very similar as both have stars, planets and dusts.&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="6760" LastEditDate="2015-04-21T14:24:25.320" LastActivityDate="2015-04-23T08:47:24.680" Title="How to tell a nebula from a galaxy?" Tags="&lt;galaxy&gt;&lt;star-formation&gt;&lt;nebula&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10546" PostTypeId="2" ParentId="10544" CreationDate="2015-04-21T15:11:34.170" Score="0" Body="&lt;p&gt;Galaxies are a large organised collection of stars (&amp;amp; nebulae) outside of our own galaxy, very distant. They shine by their own light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nebulae are clouds of gas &amp;amp; debris from, usually, a stellar explosion (Nova, supernova, etc), within our own galaxy, lit up by nearby stars (maybe internal ones) - but the gas is not glowing with it's own light, generally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a galaxy is going to be a bright clump of glowing stars, regular in shape - be that elliptical, barred or other, generally they will appear symmetrical &amp;amp; regular. A nebula will be much more amorphous &amp;amp; patchy, possibly with shapes from the clouds of gas (c.f. Horsehead, Orion), and you will be able to see individual stars in it - unlike a galaxy where they are too tightly packed &amp;amp; distant to make out individual stars.&lt;/p&gt;&#xA;" OwnerUserId="4017" LastActivityDate="2015-04-21T15:11:34.170" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10547" PostTypeId="1" AcceptedAnswerId="10548" CreationDate="2015-04-21T21:33:23.890" Score="6" ViewCount="1474" Body="&lt;p&gt;I'm using &lt;a href=&quot;ftp://ssd.jpl.nasa.gov/pub/eph/planets/&quot; rel=&quot;noreferrer&quot;&gt;JPL planetary ephemerides&lt;/a&gt; to calculate the position of planets. Using DE405 ephemeris (using a parser provided by Project Pluto with all tests passing). I am fetching the position of Mars with the observer set as the Earth (geocentric).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first assumption is that the result is a set of geocentric equatorial coordinates in rectangular form (x, y, z). This may be incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal is to calculate RA/Dec. My second assumption is that I should use a formula converting from Cartesian to Spherical:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r =\sqrt(x^2+y^2+z^2)$&#xA;$ra =\arctan(y/x)$&#xA;$dec = \pi/2 - arccos(z/r)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Julian day 2457134 (21st Apr 2015, noon UTC), and cross-checking the result with NASA's &lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;noreferrer&quot;&gt;Horizons&lt;/a&gt; service which apparently uses the same ephemeris (DE405) gets me almost there but my results are a little off:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;x&lt;/strong&gt; 1.721427968227801 &lt;strong&gt;y&lt;/strong&gt; 1.5802812301744067 &lt;strong&gt;z&lt;/strong&gt; 0.6897426775298559&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;RA&lt;/strong&gt; 02h50m12.51s &lt;strong&gt;Dec&lt;/strong&gt; +16°26'41.37&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Horizons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;RA&lt;/strong&gt; 02h50m11.52s &lt;strong&gt;Dec&lt;/strong&gt; +16°26'36.5&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what I'm missing. Do I have to factor distance of the planet due to the time it takes for light to reach us? Am I not using the correct time? Are my assumptions wrong?&lt;/p&gt;&#xA;" OwnerUserId="7112" LastActivityDate="2015-04-22T00:20:11.943" Title="Calculating RA/dec from JPL ephemeris data" Tags="&lt;planet&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10548" PostTypeId="2" ParentId="10547" CreationDate="2015-04-21T23:57:19.177" Score="7" Body="&lt;p&gt;Read the fine print under the Horizons output when you select an OBSERVER table. It says (emphasis mine)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;R.A._(ICRF/J2000.0)_DEC =&lt;br&gt;&#xA;     J2000.0 astrometric right ascension and declination of target center.&#xA;  &lt;strong&gt;Adjusted for light-time.&lt;/strong&gt; Units: HMS (HH MM SS.ff) and DMS (DD MM SS.f)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You could do the same, but it will take some extra work. The Mars you see at some time $t$ was emitted from Mars several minutes earlier than that. You need to find the $\Delta t$ such that the distance between Mars at time $t-\Delta t$ and Earth at time $t$ is equal to the speed of light times $\Delta t$. One correction will get you very close, a second will get you even closer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another issue is time. JPL uses its own time scale, JPL ephemeris time (T&lt;sub&gt;eph&lt;/sub&gt;). This is a relativistic time scale. It differs from Terrestrial Time (TT) by at most a few milliseconds, so for most purposes you can use TT in lieu of T&lt;sub&gt;eph&lt;/sub&gt; in your homebrew ephemeris calculator. TT currently differs from UTC by 67.184 seconds, and beginning in July (there's a leap second at the end of June), it will differ from UTC by 68.184 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you could just live with it. If you are using your calculations for your backyard telescope, the handful of arc seconds of error that results from not accounting for the finite speed of light is small.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2015-04-22T00:20:11.943" LastActivityDate="2015-04-22T00:20:11.943" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10549" PostTypeId="2" ParentId="10542" CreationDate="2015-04-22T04:33:25.930" Score="1" Body="&lt;p&gt;Well, it's a relatively new field of study so I don't think there's any exact answers, but I've read up on this a bit out of interest.  Red Dwarfs have 2 main disadvantages.   One, which you've mentioned.  They're relatively cold compared to our star so the planet would need to be comparatively close and therefor, probobly tidally locked.   The other problem is red dwarfs tend to have very active sunspots compared to our sun, so the planet would need a crazy strong magnetic field to protect the planet.   Neither of those are &quot;deal breakers&quot; but perhaps less likely than a sun-like star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A tidally locked planet wouldn't necessarily have a scorching hot side and a freezing cold side.  Modeling suggests that temperature would circulate.   If it had extensive oceans on the tidally locked side, for example, that could create a cloud cover and not cause intense heat.   You could also have the sun side be far enough away to be temperate and the star-side icy, or a livable section in a ring around the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also the possibility that a tidally locked planet could have winds that we would find very difficult to adapt to.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you'd ideally have the sun side be the habitable side for photosynthesis.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the flip side, if the star is too large, the star's life would probobly be too short for the planet to cool down enough to be habitable for us.  The life of a star is the inverse cube of the size, so a star twice the size of our sun would only have a main sequence of 1.2-1.3 billion years.  When you get a bit bigger than that, there's not enough time for a planet to cool off and a solar system to get past it's heavy bombardment period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'd give a red dwarf a better chance of having a habitable planet than a sun 2.5 times or larger than our sun.   That, and red-dwarfs are the most common stars in the galaxy, so I think there's a probability that some red dwarfs could have habitable planets.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-04-22T04:33:25.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10550" PostTypeId="1" CreationDate="2015-04-22T08:01:46.287" Score="4" ViewCount="114" Body="&lt;p&gt;For example: Are they soft landed meteorites? Or do they form from melted material during a violent impact? Or are they the result of some kind of erosion process?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-05-22T19:08:37.897" Title="How are boulders formed on asteroids?" Tags="&lt;asteroids&gt;&lt;formation&gt;&lt;geology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10552" PostTypeId="2" ParentId="2346" CreationDate="2015-04-22T14:19:38.150" Score="1" Body="&lt;p&gt;You need to specify on what timescales and at what wavelengths to get a proper answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite your question, the Sun does appear to be a very stable star indeed. The luminosity changes over the solar cycle are about 0.1% (or 0.001 magnitudes) and even the biggest sunspots modulate the light by a fraction of one percent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to find stars more &quot;stable&quot; than this, the key is to find stars with lower levels of magnetic activity. For &lt;em&gt;solar-type stars&lt;/em&gt; this means you have to find slower rotating objects which, because angular momentum is lost as stars get older, means you need to look for older stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;K- and M-dwarfs have longer spin-down timescales and can remain active to much older ages, therefore I would &lt;em&gt;not&lt;/em&gt; think these would generally be less active/less variable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My recommendation would therefore be to look at old, population II halo F/G-type stars. As long as they are not in binary systems (which can influence the angular momentum evolution), then these ought to be the least magnetically active stars and the least variable (in principle).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-04-22T16:15:18.713" LastActivityDate="2015-04-22T16:15:18.713" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10553" PostTypeId="2" ParentId="10550" CreationDate="2015-04-22T18:08:48.860" Score="3" Body="&lt;p&gt;It looks like you are asking about &lt;a href=&quot;http://en.wikipedia.org/wiki/Rubble_pile&quot; rel=&quot;nofollow&quot;&gt;rubble piles&lt;/a&gt;, asteroids that are made up of a large number of different sized objects that are weakly held together by gravity. A few of the component objects are large, but most are very small (down to grains of sand).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By way of analogy, think of playing pocket pool. Rack the 15 target balls but leave the rack on. The cue ball bounces off when you strike the rack with the cue ball, but that's about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's try again, but this time with the rack removed. To make the target balls a even more rubble pile-like, we'll use a piece of cardboard to add a bit of space between the target balls. Now something very different happens when the cue ball strikes the target balls. The balls are only loosely connected to one another. This makes them very good at absorbing the momentum of the cue ball and at distributing the energy and momentum throughout the rack. Given a low to moderate cue ball velocity, the collision will be close to purely inelastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that's how you get a rubble pile. It's one inelastic collision after another after another. Over the course of 4.6 billion years, you have a pile of dust and sand with some rocks and a few large boulders mixed in.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-04-22T18:08:48.860" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10554" PostTypeId="2" ParentId="10538" CreationDate="2015-04-22T20:13:40.123" Score="9" Body="&lt;p&gt;Yes, space is constantly being created. The new space does not hold any matter (like atoms) or dark matter. This means that the &lt;em&gt;density&lt;/em&gt; of normal and dark matter &lt;em&gt;de&lt;/em&gt;creases at the same rate as the volume &lt;em&gt;in&lt;/em&gt;creases. However, dark &lt;em&gt;energy&lt;/em&gt;, which is something completely different and thought to be a property of vacuum itself, is being created with the new space, so the density of dark energy stays constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This in turns mean that while the early Universe (i.e. from it was 70,000 years old and until it was almost 10 billion years) was dominated by matter, the Universe is now dominated by dark energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it will only get worse.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2015-04-22T20:13:40.123" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10555" PostTypeId="1" AcceptedAnswerId="10556" CreationDate="2015-04-22T22:05:26.223" Score="0" ViewCount="134" Body="&lt;p&gt;I am looking for a theory that shows whether or not two passing celestial bodies will go into orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume there would have to be a critical point where the gravitational attraction is stronger than the inertia of the moving bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bodies would then orbit until they collide, or be affected by a force to fall behind the critical point and escape orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a theory that states that objects past this critical point of gravitational attraction must either collide immediately, or go into orbit after falling within a certain distance.&lt;/p&gt;&#xA;" OwnerUserId="7119" LastEditorUserId="6980" LastEditDate="2015-04-24T06:12:17.900" LastActivityDate="2015-04-24T06:12:17.900" Title="Is there a theory / equation showing whether or not two passing bodies will go into orbit around each other?" Tags="&lt;orbit&gt;&lt;gravity&gt;&lt;distances&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10556" PostTypeId="2" ParentId="10555" CreationDate="2015-04-22T22:19:06.980" Score="6" Body="&lt;p&gt;If there are only two bodies, then they will never enter a mutual orbit. For two objects initially gravitationally &lt;em&gt;un&lt;/em&gt;bound, in order to become gravitationally bound you must remove energy from the system. With only two bodies (that don't collide), this does not happen. They will accelerate toward each other, change directions according to how close they get, and then leave each other again with exactly the same total energy and momentum as before, but in general shared in some other ratio (for instance, if a small body encounters a large body, the smaller will gain energy and leave with a larger velocity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you have three (or more) bodies, one may get slung out with high velocity, thus extracting energy from the two others, which can then go in orbit. But alas, there's no equation for this; the so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/N-body_problem&quot;&gt;N-body problem&lt;/a&gt; has no analytic solution, and must in general be solved numerically.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2015-04-22T22:19:06.980" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10557" PostTypeId="1" AcceptedAnswerId="10560" CreationDate="2015-04-22T23:04:27.713" Score="3" ViewCount="1736" Body="&lt;p&gt;To survive high in the atmosphere of Venus, all you would nead to wear is suit that protects you from the sulfuric acid vapors in the air and a supply of breathable air. Assuming (for simplicity) that this gear has the same density as the human body (the air supply would probably much heavier), could you float on top of denser gas layers below you?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the density of the gas would only have to be about as high as the density of water on the surface of the sea. And would that mean that when you pour out a bottle of water, the water would float in the air (given the temperature/pressure ratio is below the boiling point)?&lt;/p&gt;&#xA;" OwnerUserId="7120" LastActivityDate="2015-04-23T04:02:06.980" Title="Would a human body float in the dense atmosphere of Venus?" Tags="&lt;planetary-atmosphere&gt;&lt;venus&gt;&lt;space-travel&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10558" PostTypeId="1" AcceptedAnswerId="10561" CreationDate="2015-04-23T00:50:33.023" Score="2" ViewCount="2932" Body="&lt;p&gt;Gas is abundance in the universe, can a massive asteroid draws in these gas forming a thin atmosphere? &lt;/p&gt;&#xA;" OwnerUserId="6760" LastActivityDate="2015-04-23T08:27:39.073" Title="Can asteroids contain atmosphere?" Tags="&lt;gravity&gt;&lt;asteroids&gt;&lt;gas&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10560" PostTypeId="2" ParentId="10557" CreationDate="2015-04-23T04:02:06.980" Score="6" Body="&lt;p&gt;No. According to the Nasa &lt;a href=&quot;http://nssdc.gsfc.nasa.gov/planetary/factsheet/venusfact.html&quot; rel=&quot;noreferrer&quot;&gt;Venus fact sheet&lt;/a&gt; the density of the atmosphere at the surface is ~65 kg/m^3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparison, water is 1000 kg/m^3. We only &lt;em&gt;just&lt;/em&gt; float in water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you were there, in a suit that could somehow withstand the heat (464 C), the &quot;air&quot; would feel thick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe a person could strap on some wings and fly ...&lt;/p&gt;&#xA;" OwnerUserId="5524" LastActivityDate="2015-04-23T04:02:06.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10561" PostTypeId="2" ParentId="10558" CreationDate="2015-04-23T08:27:39.073" Score="1" Body="&lt;p&gt;At a certain size, huge asteroids get classified as dwarf planets. Pluto has an atmosphere 100,000 times thinner than Earth, and Pluto is already one of the two largest dwarf planets known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asteroids (like everything) do have gravity, so nearby gas would be drawn to them. But it would take just very tiny distrubances for that gas to drift away, so what little there is would probably be close to undetectable.&lt;/p&gt;&#xA;" OwnerUserId="7120" LastActivityDate="2015-04-23T08:27:39.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10562" PostTypeId="2" ParentId="10542" CreationDate="2015-04-23T08:37:16.357" Score="0" Body="&lt;p&gt;Gliese 581 is a red dwarf with several planets. The planet Gliese 581 c was in fact the first roughly Earth Sized planet ever discovered in the habitable zone around another star. Currently, it looks like it probably is very hot because of extreme Greenhouse effect similar to Venus. But that's already pretty good.&#xA;Gliese 581 is not a particularly big red dwarf, there are some that have almost twice the mass and their habitable zone would be farther out, making the whole system a bit more similar to ours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since red dwarves are small and live extremely long, they are actually considered pretty good candidates, since conditions on orbiting planets would be pretty stable over very long amount of time, without the sun changing its behavior.&lt;/p&gt;&#xA;" OwnerUserId="7120" LastActivityDate="2015-04-23T08:37:16.357" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10563" PostTypeId="2" ParentId="10544" CreationDate="2015-04-23T08:47:24.680" Score="0" Body="&lt;p&gt;That is a good question, and before it was known that Andromeda is a galaxy, it was indeed assumed to be a nebula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What really gives away a galaxy is its distance. Even Andromeda, with is the closest galaxy to ours, is over 30 times farther away than any object within our galaxy. If you know the distance, you can tell pretty clearly if it's something within our galaxy, or a completely different galaxy farther out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just from looking through a telescope it can be really difficult to tell them apart.&lt;/p&gt;&#xA;" OwnerUserId="7120" LastActivityDate="2015-04-23T08:47:24.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10565" PostTypeId="1" CreationDate="2015-04-23T18:10:02.473" Score="1" ViewCount="90" Body="&lt;p&gt;In the case that magnetic monopoles exist, when were they created? I thought that all particles were created after inflation, but I've read some comments recently that hint at a pre-inflationary creation of magnetic monopoles.&lt;/p&gt;&#xA;" OwnerUserId="6920" LastActivityDate="2015-04-26T09:11:16.060" Title="Were magnetic monopoles created before, during or after inflation?" Tags="&lt;cosmology&gt;&lt;cosmological-inflation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10566" PostTypeId="1" CreationDate="2015-04-23T18:27:13.757" Score="1" ViewCount="613" Body="&lt;p&gt;Considering how the our solar system, and how every other star in the galaxy is traveling through the cosmos, it seems like it's a miracle that the stars and constellations in the sky stay as constant as they are.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aOLkr.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is it that the North Star seems to be fixed in space? Is it following our star through space? are we following it? Is it actually moving? Are we getting further away from it? Closer to it? What is our relationship to Polaris in the Galaxy?&lt;/p&gt;&#xA;" OwnerUserId="7124" LastEditorUserId="7124" LastEditDate="2015-04-23T21:32:46.580" LastActivityDate="2015-04-23T21:32:46.580" Title="Is the North star going to stay in the North sky?" Tags="&lt;star&gt;&lt;orbit&gt;&lt;galaxy&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10567" PostTypeId="2" ParentId="10566" CreationDate="2015-04-23T20:34:17.980" Score="3" Body="&lt;p&gt;The relative direction to an object depends mainly on two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The location of the observer and the object&lt;/li&gt;&#xA;&lt;li&gt;The coordinate system you use to determine direction.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the first part of that concerns where we are compared with Polaris (and by extension of your question, where the other stars are as well).  Yes the stars and other components of the galaxy all have certain motions.  But this motion is not large compared to the distances between stars.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the stars we can see in the sky are quite nearby, and are orbiting the galaxy at similar speeds.  The speed of the solar system around the galaxy is around $230 km/s$.  The relative speed between our sun and nearby stars is much less than that.  If we assume that most stars are farther than $5ly$ from us, then we can calculate how long it might take to make a $1\%$ change in distance.&#xA;$$t = \frac{d}{v}$$&#xA;$$t = \frac{4.7x10^{23}m}{2.3x10^5m/s}$$&#xA;$$t = 2.06x10^{11}s$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That works out to over six thousand years.  There's a lot of caveats with that number, but it gives you a good idea that for the most part it takes too long for stars to move relative to each other for you to notice during your lifetime without very careful observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second point is that things can move if we change what we mean by &quot;north&quot; or any other direction.  North, both on earth and in astronomy, is defined by the rotational axis of the earth.  But the axis changes its direction with respect to other stars over time.  Besides some smaller scale wobbles, the largest change to the direction of the axis is due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Axial_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;precession&lt;/a&gt; and has a period of about 26,000 years.  This change will take &quot;north&quot; from a direction that passes relatively near Polaris in a large circle around the sky.  With such a long period, again this isn't something to be seen by individuals without detailed measurements.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YZPlI.gif&quot; alt=&quot;Precession sky map&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is that people alive today will be able to always think of Polaris as the pole star, but this will be less true for future generations as changes are easily visible on the scale of thousands of years.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-04-23T20:34:17.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10568" PostTypeId="2" ParentId="10566" CreationDate="2015-04-23T20:36:39.433" Score="1" Body="&lt;p&gt;The Sun is of course in motion with respect to other stars in our Galaxy, but it does not move quickly compared with the vast distances involved. For instance it takes about 220 million years for our Sun to orbit the Galaxy once, travelling at around 200 km/s. The stars that are closest to the Sun &lt;em&gt;tend&lt;/em&gt; to be orbiting in more-or-less the same direction and at a similar speed (that is why they are in the vicinity of the Sun).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking specifically about Polaris. There are three components of its motion with respect to the Sun - two tangential directions on the plane of the sky and a line of sight velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=Polaris&amp;amp;NbIdent=1&amp;amp;Radius=2&amp;amp;Radius.unit=arcmin&amp;amp;submit=submit%20id&quot; rel=&quot;nofollow&quot;&gt;SIMBAD CDS database&lt;/a&gt; we see that Polaris has a line of sight velocity of 16 km/s &lt;em&gt;towards&lt;/em&gt; the Sun and tangential motions of 28 km/s in the right ascension direction and 7 km/s in the decreasing declination direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sounds a lot, but a velocity of 1 km/s means it takes about 300,000 years for the star to move 1 light year, and Polaris is about 400 light years from Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, given its net velocity, the position of Polaris with respect to us will show significant changes (of degrees) on timescales of hundreds of thousands to millions of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It &lt;em&gt;is&lt;/em&gt; heading southward in the sky, but will take around 25 million years to cross the equator at the rate it is travelling now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say that the constellations are &quot;constant&quot;. Yes, they are &lt;em&gt;on human timescales&lt;/em&gt;, but the motions of the stars both radially and tangentially is routinely measured. If you waited some millions of years, the constellations &lt;em&gt;would&lt;/em&gt; look very different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB: I am ignoring the precession of the Earth's rotation axis, since this is just a rotation of the coordinate system and not a change of &lt;em&gt;position&lt;/em&gt; of the stars with respect to the Earth.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-04-23T20:36:39.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10569" PostTypeId="1" CreationDate="2015-04-24T11:03:16.610" Score="1" ViewCount="226" Body="&lt;p&gt;Flashes from some meteoroid impacts on the Moon can be seen with the naked eye. How much light must a future lunar base emit in order to be visible as a dot in the lunar night (like the flag of Tunisia)? I suppose it is a bit complicated and depends on the phase of the Moon, and that the new Moon is so close to the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kchuA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-04-24T11:03:16.610" Title="How strong does a light source on the Moon have to be to be visible from Earth?" Tags="&lt;the-moon&gt;&lt;light&gt;&lt;naked-eye&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2017-01-17T14:07:16.853" ContentLicense="CC BY-SA 3.0" />
  <row Id="10570" PostTypeId="1" CreationDate="2015-04-24T15:27:06.010" Score="4" ViewCount="138" Body="&lt;p&gt;I suppose that the Kepler field was well chosen. It is rich in stars and has minimal background light since it is pointing out from the disc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But other things could have been considered. For example, a field which contains the most stars which are near enough to be followed up by ground based telescopes. And why wasn't Kepler designed to change its pointing so it could cover several fields, such as all of the above? Funny enough, it does so now as K2 after two of its reaction wheels have broken, so studying only one field doesn't seem to be a technical issue, but something which astronomers prefer. Why?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-04-24T21:42:05.523" Title="What decided how the Kepler space telescope was pointed?" Tags="&lt;kepler&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10571" PostTypeId="2" ParentId="10570" CreationDate="2015-04-24T18:25:01.363" Score="6" Body="&lt;p&gt;Changing fields breaks the timeline of the observations.  If you have variation with a period similar to when you switch, you might completely miss it.  Even if the periods are different, it degrades your observations and makes it harder to draw conclusions.  The more uninterrupted viewing you get, the better.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nasa.gov/pdf/189566main_Kepler_Mission.pdf&quot;&gt;This mission document&lt;/a&gt; covers some of the reasons for the choice of the field.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Continuously viewable during the mission (you can't pick something where the earth, sun or moon constrains your viewing).&lt;/li&gt;&#xA;&lt;li&gt;Rich in stars.  Observe as many as possible.  &lt;/li&gt;&#xA;&lt;li&gt;The size of the sunshade limits the field to be more than 55 degrees from the ecliptic.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Although it can image lots of stars, the actual field isn't all that large.  Attempting to pick a single field with some fraction nearby would be difficult, and was not Kepler's mission.  In fact the resource of follow-up observations from ground telescopes was mentioned as one reason a northern field was chosen over a southern one.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-04-24T18:25:01.363" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10572" PostTypeId="2" ParentId="10570" CreationDate="2015-04-24T21:42:05.523" Score="5" Body="&lt;p&gt;The prime objective of the Kepler mission was to attempt to find &quot;Earth-like&quot; planets using the transit technique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To establish that you definitely have a transiting planet requires, at a minimum, that you see three regularly spaced transits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Kepler mission (originally) was planned for 4 years. Thus to ensure the detection of 3 transits for planets in a 1-year orbit really &lt;em&gt;requires&lt;/em&gt; that you observe a large set of stars continuously for that period (since the transits are quite brief).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to do this, you need to observe a field in which neither the Sun nor the Earth get in the way during the year. This requires you to look away from the ecliptic plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, to get a large number of stars in the fixed field of view, a direction was chosen that was &lt;em&gt;close&lt;/em&gt;, but not in the Galactic plane and viewing along a spiral arm. I believe this was done to maximise the number of stars with $V&amp;lt;16$ for which Kepler would supply good photometry. Closer to the plane would have given even more stars, but many would have been faint and one runs into more difficulties in terms of resolving which star is actually the variable when there is too much confusion with many sources. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-04-24T21:42:05.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10573" PostTypeId="1" CreationDate="2015-04-25T15:59:40.380" Score="6" ViewCount="261" Body="&lt;p&gt;I am looking for sets of astronomical images for testing different kinds of stacking algorithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is simple: if one has $N$ images of the same object, the signal-to-noise ratio of the averaged image increases with $\sqrt{N}$ under certain conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where could I obtain data for testing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. I would propose adding the tags &quot;image&quot;, &quot;ccd&quot;, &quot;research&quot;, &quot;signal processing&quot;, and &quot;machine learning&quot;.&lt;/p&gt;&#xA;" OwnerUserId="7134" LastActivityDate="2015-10-30T10:13:10.987" Title="Image sets for testing stacking algorithms?" Tags="&lt;star&gt;&lt;solar-system&gt;&lt;planet&gt;&lt;fundamental-astronomy&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10574" PostTypeId="1" AcceptedAnswerId="10616" CreationDate="2015-04-25T16:18:14.260" Score="1" ViewCount="137" Body="&lt;p&gt;I am preparing a talk on the Equation of Time, and I want to produce an animation of the movement of the earth around the sun, in an exaggerated ellipse, to illustrate the speeding up and slowing down in different parts of the orbit. Ideally I would like to have x and y as functions of time, e.g x = f(t) and y = g(t). Accuracy is not the prime consideration, a reasonable approximation will do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will use a Processing sketch to produce the animation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone please tell me what these equations are?&lt;/p&gt;&#xA;" OwnerUserId="2022" LastActivityDate="2015-05-01T18:14:59.923" Title="What are the equations for an orbit with time as parameter?" Tags="&lt;orbit&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10575" PostTypeId="2" ParentId="9973" CreationDate="2015-04-25T22:50:11.100" Score="1" Body="&lt;p&gt;The closest I've seen to what you mention in your question are wide-angle binoculars, the like of &lt;a href=&quot;http://www.vixenoptics.com/binoculars/SG_Binocular.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vixen SG 2.1x42&lt;/a&gt;. There are other similar instruments available, often of Russian or Chinese (like &lt;a href=&quot;http://www.holgermerlitz.de/opera2.3x40.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;) production. They have low magnifications and are examples of &lt;a href=&quot;http://en.wikipedia.org/wiki/Binoculars#Galilean_binoculars&quot; rel=&quot;nofollow noreferrer&quot;&gt;Galilean binoculars&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/alUOy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They don't have a fixed-size exit pupil and you can gain about 1.8 mag above what you can see with a naked eye (based on a review from a Polish astronomy forum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the field of view of about 25 degrees, this is an experience somewhat (to put it mildly) different from what you describe. However, it's enough to see quite a large piece of the sky at the same time and is probably the closest you can get to star-gazing goggles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make it easier to picture, 25 degrees is approximately the angular size of the Big Dipper, from one end of the bowl to the tip of the handle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to their size and wide FOV, these binos are also known to be used in the same manner as goggles. Pictured below is a pair of Kasai WideBino 28 in a wearable, rubber casing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1OwOl.jpg&quot; alt=&quot;Kasai WideBino&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://www.cloudynights.com/topic/463676-vixen-sg-21x42-widefield-binoculars/&quot; rel=&quot;nofollow noreferrer&quot;&gt;thread on cloudynights&lt;/a&gt; provides a nice overview of this kind of binoculars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image sources:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.vixenoptics.co.uk/Pages/sg_2.1x42.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;vixenoptics.co.uk&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.kasai-trading.jp/widebino28.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;kasai-trading.jp&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="211" LastEditorUserId="211" LastEditDate="2015-04-26T12:21:44.237" LastActivityDate="2015-04-26T12:21:44.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10576" PostTypeId="1" AcceptedAnswerId="10582" CreationDate="2015-04-26T02:38:18.693" Score="1" ViewCount="523" Body="&lt;p&gt;I have been trying hard to find the star &quot;Polaris&quot; in the following image from &lt;a href=&quot;http://www.eso.org/public/images/eso0932a/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ESO and Serge Brunier&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dO0jO.jpg&quot; alt=&quot;The Milky Way panorama from&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From peering at other maps and simulation apps, I think it should be in the top-left quarter, relatively close to the middle (vertically), however taking the brightest starts one by one, I cannot piece together the other stars belonging to Ursa Minor (small dipper).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to use this image as a background in a 3D model of Earth. The projection is perfect for that purpose, I only need to rotate it in such a way that Polaris lines up with the north pole. Once that is accomplished I can rotate it around the Earth's polar axis to set the right time of day at a certain part of the year (it does not have to be 100% accurate for my purpose, but it should at least look a bit realistic to an astronomer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that makes this task quite hard with this high resolution image is that stars seem to pop in and out of existence while zooming in and out, probably due to lossy fast re-sampling algorithms used in my image viewers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone with a trained eye could point out the Ursa Minor constellation for me it would save me a ton of time!&lt;/p&gt;&#xA;" OwnerUserId="7139" LastActivityDate="2015-04-26T20:28:45.677" Title="How to orient ESO's Milky Way Panorama in a 3D model" Tags="&lt;milky-way&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10577" PostTypeId="1" AcceptedAnswerId="10658" CreationDate="2015-04-26T05:42:24.107" Score="4" ViewCount="109" Body="&lt;p&gt;I was reading &lt;strong&gt;Appendix F&lt;/strong&gt; of Steven Weingberg's book &quot;Cosmology&quot;. In this Appendix he works out the perturbations to a cosmological fluid described by non-relativistic hydrodynamics and Newtonian gravity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that the first order perturbations satisfy,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{\partial \delta \rho }{\partial t } + 3 H \delta \rho + H \vec{X} \cdot \nabla \delta \rho + \bar{\rho} \nabla \cdot \vec{v} = 0, \qquad \tag{1}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{\partial \delta \vec{v}}{\partial t } +  H \vec{X} \cdot \nabla \delta \vec{v} + H \delta \vec{v} = - \nabla \delta \phi, \qquad \tag{2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\nabla^2 \delta \phi = 4\pi G \delta \rho. \qquad \tag{3} &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weinberg applies the following Fourier transform to these equations,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(\vec{X},t)  = \int \exp \left(  \frac{i \vec{q} \cdot \vec{X}}{a} \right)  f_{\vec{q}}(t) \ \mathrm{d}^3\vec{q} $$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f(\vec{X},t)$ is a place holder for $\delta \vec{v}, \delta \rho, $ and $\delta \phi$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting equations he gets are,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{\mathrm d \delta \rho_{\vec{q}}}{\mathrm d t } + 3 H \delta \rho_{\vec{q}} + \frac{i\bar{\rho}}{a}\ \vec{q} \cdot \delta \vec{v}_{\vec{q}} = 0 \qquad \tag{1'}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{\mathrm d \delta \vec{v}_{\vec{q}}}{\mathrm d t } +  H \delta \vec{v}_{\vec{q}} = -\frac{i}{a}\ \vec{q}  \delta \phi_{\vec{q}} \qquad \tag{2'}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\vec{q}^2 \delta \phi_{\vec{q}} = -4\pi G a^2 \delta \rho_{\vec{q}} \qquad \tag{3'}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the most part these new equations can be obtained by making the substitution $\nabla \rightarrow i \vec{q}/a$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question&lt;/strong&gt; : There doesn't seem to be any terms in the transformed equations which correspond to the terms $ H \vec{X} \cdot \nabla \delta \rho$ and $H \vec{X} \cdot \nabla \delta \vec{v}$. Weinberg makes no comment about their absence. Is anyone aware of a legitimate mathematical reason for these terms to disappear in the transformed equations?&lt;/p&gt;&#xA;" OwnerUserId="7053" LastEditorUserId="2153" LastEditDate="2015-05-03T21:02:59.427" LastActivityDate="2015-05-08T19:36:33.667" Title="Missing Terms in Weinberg's treatment of perturbations on Newtonian Cosmology" Tags="&lt;cosmology&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10578" PostTypeId="2" ParentId="9973" CreationDate="2015-04-26T07:56:32.760" Score="0" Body="&lt;p&gt;On &lt;em&gt;Murdoch's Mysteries&lt;/em&gt;, the technophile detective at the end of the Victorian era made, in one episode, night vision goggles that used mirrors surrounding the eyes, essentially a pair of reflecting telescopes with no magnification. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would not cover the same angle, though: just straight ahead. The naive design you might throw together would handle light arriving parallel to straight forward, so the actual field of view may or may not be useful. To purposfully make a field of view of a specific size and quality would require knowing something about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common SLR photography lens has an enormous opening compared to your pupil, and high quality formerly pro lenses for pure mechanical film cameras can be found cheaply. Some designs even have a lens board for mounting, rather than a fixed camera enclosure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In their original use, they project the image on a large (relative to your eye) screen, so produce an image dimmer than life.  However, if instead of making a projection 44mm in diameter (say), you added an &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Objective_(optics)#Telescope&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;objective lens&lt;/em&gt;&lt;/a&gt; made for telescopes or a handy jewler's loupe (any magnifying glass works, but for more than 5&amp;times; you need a fancy compound lens), you could project all the light into your eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; using on-hand materials it takes &lt;em&gt;3&lt;/em&gt; pieces, not 2, and is unacceptably long (unless you want it looking like an old seafarers telescope). To concentrate the entire collected light, the next lens needs to be very large around to scoop in all of it.  a hand-magnifier doesn't result in an image as large as the directly-observed target, and a powerful loupe is not large enough. &lt;/p&gt;&#xA;" OwnerUserId="7142" LastEditorUserId="7142" LastEditDate="2015-04-26T08:39:07.633" LastActivityDate="2015-04-26T08:39:07.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10579" PostTypeId="1" CreationDate="2015-04-26T08:25:42.437" Score="2" ViewCount="4748" Body="&lt;p&gt;Would a more significant curvature of spacetime relative to earth (i.e. time on jupiter) effect the aging process? Basically would you age at a slower rate?&#xA;Or age faster on the moon?&lt;/p&gt;&#xA;" OwnerUserId="7143" LastEditorUserId="31410" LastEditDate="2020-12-14T14:26:52.470" LastActivityDate="2020-12-14T23:35:26.083" Title="Space time and aging" Tags="&lt;space-time&gt;&lt;space&gt;&lt;time&gt;&lt;time-dilation&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10580" PostTypeId="2" ParentId="10565" CreationDate="2015-04-26T09:00:14.570" Score="1" Body="&lt;p&gt;Before. The point is that there were too many predicted, and no concept of inflation yet. &lt;a href=&quot;http://ned.ipac.caltech.edu/level5/Watson/Watson4_4.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is a reference that doesn't just gloss over it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?hl=en&amp;amp;lr=&amp;amp;id=Lqg5AAAAIAAJ&amp;amp;oi=fnd&amp;amp;pg=PA171&amp;amp;dq=monopole+production+phase+transition+timeline&amp;amp;ots=GHAvSVSR7v&amp;amp;sig=W8zpH5rIZUxC_dGIpc8sz4u5Ja4#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is the original paper by Guth, and &lt;a href=&quot;https://physics.stackexchange.com/questions/119342/why-didnt-electroweak-symmetry-breaking-happen-earlier-than-it-did&quot;&gt;physics.SE&lt;/a&gt; has a nice drawing of the same timeline shown on that page. (Original &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/astro/unify.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You ought to try that physics.SE thread and continue there for more details.&lt;/p&gt;&#xA;" OwnerUserId="7142" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-04-26T09:11:16.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10582" PostTypeId="2" ParentId="10576" CreationDate="2015-04-26T10:31:55.553" Score="2" Body="&lt;p&gt;Here is a star map (from &lt;a href=&quot;http://www.datasync.com/~rsf1/fun/sm-newd.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.datasync.com/~rsf1/fun/sm-newd.gif&lt;/a&gt; ) given in Galactic coordinates. It appears to be presented in a similar (but not the same, since it is rectangular) projection to the one you show above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Galactic centre is in the middle and the Large and Small Magellanic clouds are to the lower right and oriented in the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see from this map, your assumption is completely correct. Ursa Major and Polaris should be situated towards the upper left hand side of the picture. Conveniently it should be on the opposite side and roughly equi-distant from the Galactic plane as M31 (which is clearly visible in the image) and at a similar (slightly lower) Galactic longitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have added a copy of your image at the bottom and marked where I think Polaris is. I suggest you get hold of the higher resolution image available from Serge Brunier if you want to pick out individual stars. The way you transfrom this into equatorial coordinates for different times of the year is not as simple as you suggest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fOmnW.gif&quot; alt=&quot;Star map in Galactic coordinates&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oDDSk.jpg&quot; alt=&quot;Modified starmap with Polaris marked&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-04-26T20:28:45.677" LastActivityDate="2015-04-26T20:28:45.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10583" PostTypeId="1" CreationDate="2015-04-26T15:33:41.423" Score="1" ViewCount="139" Body="&lt;p&gt;For example, are there binaries with an orbital &lt;code&gt;period &amp;lt; 5 day&lt;/code&gt;  and that are very eccentric?&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of factors could slow down orbital circularisation?&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there examples of binaries with &lt;code&gt;p&amp;lt;5day&lt;/code&gt; and &lt;code&gt;e&amp;gt;0.1&lt;/code&gt;?&lt;br/&gt;&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="2531" LastEditDate="2015-05-03T10:25:19.183" LastActivityDate="2015-05-03T10:25:19.183" Title="Can close binaries have a very eccentric orbit?" Tags="&lt;binary-star&gt;&lt;tidal-forces&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10584" PostTypeId="2" ParentId="10573" CreationDate="2015-04-26T17:17:08.900" Score="0" Body="&lt;p&gt;You could use archival data, e.g., from the WISE mission: &lt;a href=&quot;http://irsa.ipac.caltech.edu/applications/wise/&quot; rel=&quot;nofollow&quot;&gt;http://irsa.ipac.caltech.edu/applications/wise/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data are reduced as well as photometrically and astrometrically calibrated, which should make the image combination a lot easier. Just type in random coordinates and download data for a field with the desired number of frames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also try other observatories, for example, Gemini, Spitzer, ... to obtain well-calibrated test data.&lt;/p&gt;&#xA;" OwnerUserId="7144" LastActivityDate="2015-04-26T17:17:08.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10585" PostTypeId="1" AcceptedAnswerId="10586" CreationDate="2015-04-26T17:37:09.117" Score="4" ViewCount="1304" Body="&lt;p&gt;In the lambda-CDM model describing an accelerating Universe, the Hubble parameter is currently decreasing with time. Will it continue to decrease forever?&lt;/p&gt;&#xA;" OwnerUserId="6920" LastActivityDate="2015-05-12T15:28:36.130" Title="Evolution of the Hubble parameter" Tags="&lt;cosmology&gt;&lt;expansion&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10586" PostTypeId="2" ParentId="10585" CreationDate="2015-04-26T18:55:29.600" Score="8" Body="&lt;p&gt;The solution to the Friedmann equation in a flat universe is&#xA;$$H^2 = \frac{8\pi G}{3}\rho + \frac{\Lambda}{3},$$&#xA;where $\rho$ is the matter density (including dark matter) and $\Lambda$ is the cosmological constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the universe expands, $\rho$ of course decreases, but $\Lambda$ remains constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the Hubble &quot;constant&quot; actually decreases from its current value $H_0$ and asymptotically tends towards $ H = \sqrt{\Lambda/3}$ as time tends towards infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $\Lambda = 3H_0^{2} \Omega_\Lambda$, and measurements suggest that $\Omega_{\Lambda} \simeq 2/3$, then $\Lambda \simeq 2H_0^2$, and the Hubble parameter will therefore decrease to approximately $\sqrt{2/3}$ of its present value &lt;em&gt;if&lt;/em&gt; the cosmological constant stays constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course if $\Lambda = \Lambda(t)$, (ie not the basic $\Lambda$-CDM model) then the behaviour will be different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Another useful form of the solution (for the case of a constant vacuum energy density) is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H^2 = H_0^2 \left(  \frac{\Omega_r}{a^4} + \frac{\Omega_M}{a^3} + \frac{\Omega_k}{a^2} + \Omega_{\Lambda}\right),$$&#xA;where $H_0$ is the Hubble parameter now, $a(t)$ is the scale factor of the universe, $\Omega_r$ is the current (i.e. $a=1$) ratio of the radiation density to the critical density and $\Omega_M$, $\Omega_k$ and $\Omega_{\Lambda}$ are the equivalent densities for the matter (baryonic and dark), curvature and (constant) vacuum energy densities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $a$ increases you can see that all three of the leading terms get smaller and the Hubble parameter decreases at all times. When $a$ is very large, $H$ approaches $\sqrt{\Omega_{\Lambda}} H_0$ as before.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-05-12T15:28:36.130" LastActivityDate="2015-05-12T15:28:36.130" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10587" PostTypeId="2" ParentId="10573" CreationDate="2015-04-26T22:06:25.543" Score="0" Body="&lt;p&gt;Try the NASA data catalog,  &lt;a href=&quot;https://data.nasa.gov/data&quot; rel=&quot;nofollow&quot;&gt;https://data.nasa.gov/data&lt;/a&gt;.&#xA;Also Try the Keck Observatory Archive, KOA, &lt;a href=&quot;http://nexsci.caltech.edu/archives/koa/index.shtml&quot; rel=&quot;nofollow&quot;&gt;http://nexsci.caltech.edu/archives/koa/index.shtml&lt;/a&gt;.&#xA;Look for raw images. of clusters or galaxies.&lt;/p&gt;&#xA;" OwnerUserId="7146" LastActivityDate="2015-04-26T22:06:25.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10591" PostTypeId="2" ParentId="1767" CreationDate="2015-04-27T06:32:26.630" Score="11" Body="&lt;p&gt;The chemical elements in our bodies are inherited from the Earth. The Earth was formed in a disc of gas and dust swirling around the protosun 4.5 billion years ago. The material that formed the Earth was a selection of the material from that protostellar nebula that was itself once part of a larger molecular cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the atoms in our body were once part of this molecular cloud, so we need to understand how they got there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the first ten minutes or so, the universe contained mainly hydrogen, helium and some traces of lithium, deuterium and tritium - and that's all. No oxygen, iron, carbon etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost all of the heavier chemical elements are made inside stars. We could stop there - the atoms of carbon, oxygen, calcium etc. in our bodies &lt;em&gt;must&lt;/em&gt; have been made in stars, and since these atoms/nuclei are stable, they must survive unchanged (you could argue about whether their electrons get swapped about in chemical reactions etc., but since electrons are indistinguishable this hardly matters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do they get into a molecular cloud and what sort of stars make these elements? A couple of answers correctly identify massive stars that explode as supernovae as important. But they are by no means the only contributor, or even the most important contributor for some elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take carbon and nitrogen, these are manufactured in nuclear reactions inside stars of even a bit less than a solar mass during the horizontal branch and asymptotic giant branch stages. These stars may be less massive and produce less C and N than massive stars, but there are many more of them. The central material is mixed to the surface during thermal pulses and the outer envelope, enriched in a variety of chemical elements, is gradually lost into space via a slow wind. This is a major source of carbon, nitrogen, fluorine, lithium and a number of heavy elements - Ba, La, Zr, Sr, Pb and many others - produced in the &lt;a href=&quot;https://en.wikipedia.org/wiki/S-process&quot; rel=&quot;noreferrer&quot;&gt;s-process&lt;/a&gt;. About 50% of the elements heavier than iron are made in the s-process, which can occur in both massive stars that explode (mainly isotopes with $A&amp;lt;90$) and the less massive AGB stars with slow, massive winds (elements up to lead and bismuth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron, nickel and many other elements such as sulphur and silicon are also produced during &lt;a href=&quot;https://en.wikipedia.org/wiki/Type_Ia_supernova&quot; rel=&quot;noreferrer&quot;&gt;type Ia supernovae&lt;/a&gt;. This is the detonation of a white dwarf, the end stage of a low-mass star, after mass transfer or merger. Milder novae explosions caused by the ignition of material accreted onto a white dwarf also enrich the interstellar medium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these different processes produce distinctive patterns of element abundances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The enriched material is swept up by neighbouring supernova explosions, by interactions with spiral arms and other molecular clouds. It cools, condenses and collapses to form a new generation of stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analysis of &quot;presolar grains&quot; found inside meteorites tells us what our solar system formed from. These analyses tell us that all of the above processes were important in making the chemical elements that made up the Earth and hence those in our bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Further details on the production of elements heavier than iron (including supernovae, low-mass AGB stars, colliding neutron stars etc.) can be found in my &lt;a href=&quot;https://physics.stackexchange.com/questions/7131/origin-of-elements-heavier-than-iron-fe&quot;&gt;Physics SE answer to this question.&lt;/a&gt; ]&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-08-14T13:19:57.437" LastActivityDate="2015-08-14T13:19:57.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10593" PostTypeId="1" AcceptedAnswerId="10596" CreationDate="2015-04-27T22:58:23.500" Score="10" ViewCount="544" Body="&lt;p&gt;I've heard people say that a tablespoon of neutron star would weigh over a billion tons. If we ever &lt;em&gt;could&lt;/em&gt; take a tablespoon of one would it still remain intact with the same density? &lt;/p&gt;&#xA;" OwnerUserId="6783" LastActivityDate="2016-12-29T17:43:53.457" Title="Would a tablespoon of a neutron star remain intact?" Tags="&lt;astrophysics&gt;&lt;neutron-star&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10594" PostTypeId="2" ParentId="10593" CreationDate="2015-04-27T23:37:24.240" Score="4" Body="&lt;p&gt;No, it's not stable without the huge pressure from gravity. There is a minimum stable size, but it's definitely much greater than a spoon.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2015-04-27T23:37:24.240" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10595" PostTypeId="1" AcceptedAnswerId="10602" CreationDate="2015-04-28T05:39:07.093" Score="4" ViewCount="295" Body="&lt;p&gt;Suppose in a binary stars system there is a dying star and a companion black hole as they orbit around each other. My question is what will possibly happen to the black hole if the partner goes supernova? I understand that the kinetic energy from the supernova cannot destroy the black hole.&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="6760" LastEditDate="2015-04-28T07:34:36.187" LastActivityDate="2016-11-26T18:08:44.203" Title="How can a supernova affect black hole in a binary system?" Tags="&lt;black-hole&gt;&lt;supernova&gt;&lt;binary-star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10596" PostTypeId="2" ParentId="10593" CreationDate="2015-04-28T08:11:43.380" Score="12" Body="&lt;p&gt;If we take neutron star material and somehow transport it somewhere for examination (say the Earth!), the results would be catastrophic. At say a density of $\sim 10^{17}$ kg/m$^{3}$ the neutrons have a number density of $\sim 6\times 10^{43}$ m$^{-3}$ an internal kinetic energy density of $3 \times 10^{32}$ J/m$^{3}$ (calculated using the relevant equations for an ideal gas of degenerate neutrons at this density). So even in a tablespoonful (say 20ml, which would have a mass of 2 billion tonnes!), there is $6\times10^{27}$ J of kinetic energy (15 times more than the Sun emits in a second, or a few billion atom bombs) and this will be released &lt;strong&gt;instantaneously&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy is in the form of around $10^{39}$ neutrons travelling at around 0.1-0.2$c$. So roughly speaking it is like half the neutrons (about a billion tonnes) travelling at 0.1$c$ ploughing into the Earth. If I have done my Maths right, that is roughly equivalent to a 50km radius near-earth asteroid hitting the Earth at 30 km/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The neutrons in a dense neutron star gas are relatively stable (beta decay is blocked by electron degeneracy). The expansion described above would allow beta decay into protons and electrons, but as this happens on timescales of 10 minutes, it is hardly relevant to the initial destruction. However, you would end up after a few tens of minutes with an expanding cloud of ionised hydrogen a few light minutes across.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimum possible size to gravitationally bind neutron star material is thought to be around $0.15 M_{\odot}$ (see &lt;a href=&quot;https://physics.stackexchange.com/questions/143166/what-is-the-theoretical-lower-mass-limit-for-a-gravitationally-stable-neutron-st/143174#143174&quot;&gt;here&lt;/a&gt;). The equilibrium &lt;em&gt;electron&lt;/em&gt; density (there are always some electrons and protons present in neutron star material) for lower masses is too low to block neutron beta-decay.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-12-29T17:43:53.457" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10597" PostTypeId="1" AcceptedAnswerId="10600" CreationDate="2015-04-28T04:27:45.883" Score="15" ViewCount="11034" Body="&lt;p&gt;I was doing my own &quot;space exploration&quot; last night with a telescope. Being a space noob I can't visually determine stars or planets (I know the moon, though), but I focused on one particularly bright one when I saw 4 smaller &quot;stars&quot; on each side. After a quick Google I learned I had discovered the moons of Jupiter!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PbMtw.jpg&quot; alt=&quot;Galilean Moons&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I now know the moons I could see were Io, Europa, Ganymede, and Callisto. But which is which?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an easy way to determine which Galilean moon is which?&lt;/p&gt;&#xA;" OwnerUserId="7156" OwnerDisplayName="Coomie" LastActivityDate="2019-06-13T09:33:29.907" Title="Determine the moons of Jupiter through a telescope" Tags="&lt;jupiter&gt;&lt;telescope&gt;&lt;europa&gt;" AnswerCount="8" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10598" PostTypeId="2" ParentId="10597" CreationDate="2015-04-28T06:57:56.780" Score="2" Body="&lt;p&gt;The easiest way is to look it up, e.g. here: &lt;a href=&quot;http://www.skyandtelescope.com/observing/a-jupiter-almanac/&quot; rel=&quot;nofollow&quot;&gt;http://www.skyandtelescope.com/observing/a-jupiter-almanac/&lt;/a&gt;&lt;br&gt;&#xA;Because the moons' positions change, and they are too small to distinguish in a small telescope, there are no general rules. &lt;/p&gt;&#xA;" OwnerUserId="361" OwnerDisplayName="Hobbes" LastActivityDate="2015-04-28T06:57:56.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10599" PostTypeId="2" ParentId="10597" CreationDate="2015-04-28T07:38:25.770" Score="6" Body="&lt;p&gt;One option is to sketch (or photograph) their approximate positions at the start of the night and again 4-5 hours later, if this is possible.  Then repeat the next night.  It should be easy to at least identify Io and Europa that way (they're the fastest moving).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(That's a pretty good picture for a beginner by the way!)&lt;/p&gt;&#xA;" OwnerUserId="7491" OwnerDisplayName="Andy" LastActivityDate="2015-04-28T07:38:25.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10600" PostTypeId="2" ParentId="10597" CreationDate="2015-04-28T08:19:06.337" Score="8" Body="&lt;p&gt;Use an orrery that will let you specify a specific epoch and vantage point. There are some quite fancy ones online, for example this &lt;a href=&quot;http://www.solarsystemscope.com/&quot; rel=&quot;noreferrer&quot;&gt;Solar System Scope&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/r2B45.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you do in this particular tool is click on the calendar bar below and enter date and time of your observation (if you forgot that, there's a good chance your photograph has a time stamp, either of the file or in its &lt;a href=&quot;https://en.wikipedia.org/wiki/Exchangeable_image_file_format&quot; rel=&quot;noreferrer&quot;&gt;EXIF&lt;/a&gt; data). You can adjust epoch later on, too. Then double-click on Jupiter, then its &lt;em&gt;Orbit&lt;/em&gt; button, then click on link &lt;em&gt;As seen from Earth&lt;/em&gt; below, then click &lt;em&gt;Go back&lt;/em&gt; button and use mouse wheel to zoom out so all four Galilean moons display. Then compare with the photograph that you took. Note that it might take a bit of rotation for you image to align perfectly. I use the old &lt;em&gt;neck twisting and turning technique&lt;/em&gt;, but photo editing tools or even rotatable screens will do just as well, if not better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this mentioned tool does offer settings to switch between orrery and realistic model and large and realistic sizes of displayed celestial bodies. But there are many such tools online or downloadable for both computers and smartphones. I think even this one is available as an app for smartphones for free, but last time I tried to install it, it didn't work for me. Maybe they fixed that. Do check because it's a lot easier if you can check orbital alignment of Galilean moons as seen from Earth directly on the field, as you're doing your observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have clear skies and I wish you happy hunting for planets and moons of our Solar system and beyond!&lt;/p&gt;&#xA;" OwnerUserId="13" OwnerDisplayName="TildalWave" LastActivityDate="2015-04-28T08:19:06.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10601" PostTypeId="2" ParentId="10288" CreationDate="2015-04-28T09:19:22.193" Score="5" Body="&lt;p&gt;The answer is most definitely yes, or at least yes, as far as our current understanding of how gravity works goes. It is observationally untestable (let's be more specific - nobody could report the results of an observational test!) since no signal can emerge from inside the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scenario is treated in some detail by Taylor &amp;amp; Wheeler (&quot;Exploring Black Holes&quot;, Addison, Wesley, Longman - highly recommended) in terms of what an observer would see on a direct radial trajectory into a non-rotating &quot;Schwarzschild&quot; black hole. I won't bore you with the maths - it is fairly complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A star situated at &lt;em&gt;exactly&lt;/em&gt; 180 degrees from the observer's radial path will always appear in that position as the observer looks back - right down to the singularity. The light will be gravitationally &lt;em&gt;blueshifted&lt;/em&gt; by an ever-increasing amount - essentially tending towards an infinite blueshift at the singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For stars at an angle to the radial path, their positions will be distorted such that they appear to move away from the point at which the observer has come from (and are also blue-shifted). In the final moments (it takes less than $1.5\times 10^{-4}$ seconds of proper time to fall from the event horizon to the singularity of a 10 solar mass black hole, but a huge $\sim 60$ seconds for the black hole at the centre of our Galaxy) the light from the external universe will flatten into an intense ring at 90 degrees to the radial direction of motion. So you would end up seeing blackness in front of you, blackness behind and the sky split in two by a dazzling ring of light (almost seems worth it!).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-06-01T07:49:31.827" LastActivityDate="2015-06-01T07:49:31.827" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10602" PostTypeId="2" ParentId="10595" CreationDate="2015-04-28T09:54:57.403" Score="2" Body="&lt;p&gt;The likely result would either be a black hole-black hole binary system; a neutron star-black hole binary system, or the black hole and the compact remnant from the second supernova explosion would go their separate ways at reasonably high speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot disrupt a black hole in this way. In fact all that will happen to the original black hole is that it will likely get a bit more massive from accreting some of the supernova ejecta.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="633" LastEditDate="2015-04-29T11:37:46.060" LastActivityDate="2015-04-29T11:37:46.060" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10604" PostTypeId="1" CreationDate="2015-04-28T12:55:28.630" Score="-4" ViewCount="287" Body="&lt;p&gt;Can we track its orbit well now?&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is any study that may indicate our sun may be a companion of a big black hole?&lt;br/&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the current constrain about its orbit now? There is a 1000 solar mass black hole within 10 pc?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="1038" LastEditDate="2015-04-28T15:24:24.780" LastActivityDate="2015-04-28T15:24:24.780" Title="Could our Sun be a companion star of a massive black hole?" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10605" PostTypeId="2" ParentId="10573" CreationDate="2015-04-28T14:16:49.887" Score="1" Body="&lt;p&gt;It's pretty easy to create a dataset for yourself using remotely controlled telescopes. I've used iTelescope in the past and had a good experience. Their &quot;starter&quot; account costs $20 and should be good to get you going.&lt;/p&gt;&#xA;" OwnerUserId="7154" LastActivityDate="2015-04-28T14:16:49.887" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10606" PostTypeId="2" ParentId="10604" CreationDate="2015-04-28T14:30:03.740" Score="2" Body="&lt;p&gt;There is no indication nor any astrophysical reason for such a scenario. The most relevant constraints are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The Solar velocity is typical for stars in our immediate Galactic neighbourhood. &lt;/li&gt;&#xA;&lt;li&gt;Soft binaries (those with orbital velocity smaller than the local velocity dispersion $\sigma$) dissolve (Heggie's law: soft binaries become softer and hard binaries harder).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These are in contradiction with a Solar binary nature: a viable binary has a large orbital velocity that would put the Sun outside of the typical velocity for local stars. This contradiction can only be temporarily avoided if the current (but not in $\sim1000$ years) orbital velocity and the velocity of the massive binary companion add up to a typical stellar velocity. This is an unlikely chance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Binaries with a stellar-mass black hole always form from an ordinary stellar binary, with one of the stars going supernova. Often, such binaries are quite compact. All this is ruled out for the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, black holes of intermediate masses (100 to $10^5$ M$_\odot$) haven't (yet?) been detected unambigously (though there are several objects that have been claimed to be such intermediate-mass black holes also IMBHs).&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2015-04-28T14:30:03.740" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10608" PostTypeId="2" ParentId="10597" CreationDate="2015-04-28T16:47:50.640" Score="1" Body="&lt;p&gt;In-the-Sky.org has a &lt;a href=&quot;https://in-the-sky.org/jupiter.php&quot; rel=&quot;nofollow&quot;&gt;dedicated calendar for the Galilean moons&lt;/a&gt;. I can't seem to reconcile your picture with the plot for the last day or two, though!&lt;/p&gt;&#xA;" OwnerUserId="2573" LastActivityDate="2015-04-28T16:47:50.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10609" PostTypeId="1" AcceptedAnswerId="10610" CreationDate="2015-04-29T10:04:54.647" Score="3" ViewCount="356" Body="&lt;p&gt;I'm making a simplified animation for college project on how a star reacts with a black hole. I had known that a black hole sucks in everything and thus should be doing the same for stars. But upon my research I found out that recently a discovery was made that a super-massive black hole &lt;em&gt;ripped&lt;/em&gt; apart a star. Now I've always been curious about cosmology and this took my attention and also worried me for my assignment. There were two conditions with the above mentioned calamity - one, the black was a super-massive one and second, the victim star was possibly already stripped off it's outer gaseous by the same black hole. My question is that the &lt;em&gt;ripping&lt;/em&gt; apart happens in all conditions and the sucking in thing was wrong or it happens only in mentioned circumstances?&lt;/p&gt;&#xA;" OwnerUserId="7164" LastActivityDate="2015-04-29T11:36:28.850" Title="Star versus Black Hole" Tags="&lt;star&gt;&lt;black-hole&gt;&lt;cosmology&gt;&lt;supermassive-black-hole&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10610" PostTypeId="2" ParentId="10609" CreationDate="2015-04-29T11:36:28.850" Score="6" Body="&lt;p&gt;A black hole (BH) &lt;strong&gt;never sucks anything in&lt;/strong&gt;. '&lt;em&gt;Sucking&lt;/em&gt;' requires gas pressure, but a black hole only acts via its gravity. A non-rotating (=Schwarzschild) black hole attracts all massive objects, very much like the Sun attracts the planets.&#xA;Yet, the planets are not falling into the Sun. This is because the planets move on near-circular orbits when the centrifugal force balances the gravitational attraction. The planets orbital &lt;em&gt;angular momemtum&lt;/em&gt; is conserved (because the Solar force field is purely radial [to good approximation]) and these orbits are stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference to a black hole is that very close to the event horizon no such stable circular orbits exist anymore. At those distances, nothing can orbit the BH without &lt;em&gt;falling&lt;/em&gt; in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, an object orbiting a BH on a stable orbit (including passing trajectories) is still subjected to the BH's &lt;em&gt;tidal forces&lt;/em&gt;. These distort the object, very much like the Moon's gravity distorts the Earth, generating tides.&#xA;In case, the trajectory passes sufficiently close to the BH and the object is rather big and fluffy, like (some) stars, the tidal forces may not merely distort the object, but rip it apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such tidal disruption of stars must happen if a star passes close to a supermassive BH. This can hardly be directly observed, but the it is thought that some of the stellar matter forms an &lt;em&gt;accretion disc&lt;/em&gt; around the BH and produces a characteristic light curve. An observed light curve in agreement with this model is often interpreted as circumstantial evidence for a stellar disruption event.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2015-04-29T11:36:28.850" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10611" PostTypeId="1" AcceptedAnswerId="10614" CreationDate="2015-04-29T13:58:20.297" Score="1" ViewCount="186" Body="&lt;p&gt;Would the expansion of space (e.g. dark energy) move the earth further away from the sun over time, possibly saving us for perhaps a few billion years as the sun heats up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scientists now tend to think that the sun is heating up very gradually (e.g. over billions of years):&#xA;&lt;a href=&quot;http://tvblogs.nationalgeographic.com/2010/07/21/hotter-and-hotter/&quot; rel=&quot;nofollow&quot;&gt;http://tvblogs.nationalgeographic.com/2010/07/21/hotter-and-hotter/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark energy is the unexplained &lt;em&gt;accelerating&lt;/em&gt; expansion of the universe, as described here:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dark_energy&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Dark_energy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put another way, my question is, is dark energy working at such a rate that it would move earth (and other planets) further from the sun and prolong life on earth as the sun heats up?&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2015-04-30T01:42:04.357" Title="Would dark energy save the earth for a while as the sun heats up?" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;life&gt;&lt;dark-energy&gt;&lt;habitable-zone&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10612" PostTypeId="2" ParentId="10611" CreationDate="2015-04-29T14:12:27.163" Score="-4" Body="&lt;p&gt;I welcome further discussion, and evidence supporting, or refuting my answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my idea is correct that antimatter is distributed fairly evenly through the voids between galaxies, and that antimatter is the cause of dark energy, then no, dark energy will not move earth out of the way of a sun that is heating up (e.g. it would have no effect, or very little effect).  This is because antimatter would not exist within the solar system.  Solar wind, and other particles would annihilate any antimatter that found its way into the solar system.  A &quot;galactic wind&quot; would also likely prevent antimatter from even entering the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be interesting to see if we can observe earth's past position in relation to the sun (e.g. perhaps by examining the rocks or fossils) and see if earth is migrating out further from the sun or not.  This analysis may show us if dark energy is indeed &quot;everywhere&quot; (true if the earth used to be closer, and is migrating away from the sun) or just in the &quot;voids&quot; (true if the earth is not migrating away from the sun).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I very much welcome further discussion and answers.  I especially welcome someone to perform the research as discussed in the prior paragraph.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="1386" LastEditDate="2015-04-29T14:21:02.263" LastActivityDate="2015-04-29T14:21:02.263" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10614" PostTypeId="2" ParentId="10611" CreationDate="2015-04-30T01:36:45.580" Score="0" Body="&lt;p&gt;I don't think this is a bad question, and I think Pela is essentially correct, the effect is very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark energy doesn't expand objects, for example, cause those are bound together, but dark energy should expand space between the sun and the earth though the effect is tiny.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take this number, 68 KM per second per 1 million light years (link below), that's 245,000 KM per hour per million light years or about 1 KM every 4 hours per lightyear.   With 63,000 Astronomical units per light year, between the sun and the earth, we can expect the earth (assuming it's in a perfectly stable orbit around the sun - which it's not, but lets say it is), a velocity of 1KM/30 years.   And that's over the age of the universe.   The earth/sun is only 1/3rd the age of the universe, so dark energy has accelerated the earth maybe 1 KM/90 years velocity since the formation of the solar system.   That's a pretty small push.  A velocity change of a bit a meter per year over the age of the solar-system.   The effect of the solar wind and the Jupiter and Venus orbital tugging are probably greater than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source:  &lt;a href=&quot;http://www.space.com/26279-universe-expansion-measurement-quasars-boss.html&quot; rel=&quot;nofollow&quot;&gt;http://www.space.com/26279-universe-expansion-measurement-quasars-boss.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, over a few billion years, a small velocity change of 1 meter per year would add up - so I think dark energy is part of the solar system's long term orbital calculation.  I might be wrong on that, but I think it has a measurable effect over billions of years even in distances as small as planetary orbits.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But would it save the earth - I doubt it.  The sun's heating up accelerates over time.   It will at some point in the future, move too fast and get too hot for the earth, so, we'll have to save ourselves, either by moving the earth or terraforming mars, or building space colonies or some combination of all 3.   The good news, we have tens of millions, maybe hundreds of millions of years to figure that stuff out.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark energy has been useful though.   It's probably kept us from crashing into Andromeda and/or the great attractor - however far away that is.   Certainly delayed those collisions (We're never going to crash into the great attractor, I gather, thanks to dark energy), and crashing into Andromeda, when it happens, might harmless anyway, but I'm just kind of thinking out-loud.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do think it's a good question as there probably is some small expansion due to dark energy between the earth and the sun.   If my numbers are wrong, I welcome correction.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-04-30T01:42:04.357" LastActivityDate="2015-04-30T01:42:04.357" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10615" PostTypeId="2" ParentId="10597" CreationDate="2015-04-30T03:57:49.153" Score="1" Body="&lt;p&gt;As mentioned previously, to be certain you need to record the day and time of the picture and compare to a program or table of moon positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short of doing that, the far right &amp;amp; left ones appear to be the smaller moons (IO &amp;amp; Europa) while the middle two appear to be Callisto &amp;amp; Ganymede (I base this upon apparent size of the moons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you forced me to &lt;strong&gt;guess&lt;/strong&gt;, I would say they were (from left to right) Io, Callisto, Ganymede, &amp;amp; Europa.&lt;/p&gt;&#xA;" OwnerUserId="7168" LastActivityDate="2015-04-30T03:57:49.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10616" PostTypeId="2" ParentId="10574" CreationDate="2015-04-30T19:02:49.023" Score="2" Body="&lt;p&gt;I have done a lot more research into this and found an answer. It comes down to solving Kepler's Equation, and my main references are: &#xA;&lt;a href=&quot;http://www.stargazing.net/kepler/kepler.html&quot; rel=&quot;nofollow&quot;&gt;Kepler's equation and the Equation of Centre&lt;/a&gt; which gives an iterative method for solving the equation, and &lt;a href=&quot;http://www.bogan.ca/orbits/kepler/e_anomly.html&quot; rel=&quot;nofollow&quot;&gt;Determination of Position in Orbits&lt;/a&gt; which has the formulae for calculating orbital coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the Processing sketch. If you are not familiar with Processing, it is a language developed for graphic designers, but much more versatile than just that, based on Java and hence C++. It expects two functions, one is &lt;code&gt;setup()&lt;/code&gt;, called once at the start up of the sketch, and a second &lt;code&gt;draw()&lt;/code&gt; that is repeated over and over until the sketch is stopped. In case you wonder: the whole is encased in a &lt;code&gt;main()&lt;/code&gt; before it is compiled. For more details see &lt;a href=&quot;https://processing.org&quot; rel=&quot;nofollow&quot;&gt;the Processing home page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;int wsize = 400;&#xA;float rp = 9.0; // radius of planet symbol&#xA;float rs = 15.0; // --&quot;-- sun&#xA;// Constants of orbital ellipse &#xA;float a = 150, b; // major and minor axes&#xA;float e = 0.5; // eccentricity&#xA;float halfa = a/2.0;&#xA;float c; // focal distance from centre&#xA;float E, E0; // eccentric anomaly, temporary values in iteration&#xA;float v; // true anomaly = polar coord angle&#xA;float M; /* mean anomaly which is effectively the ..&#xA;            .. time variable round the orbit*/&#xA;float r, x, y; // polar coord radius, rect coords&#xA;float k, cosE, cosv; // useful intermediate variables&#xA;float twopi = 2*PI;&#xA;float PIby2 = PI/2.0;&#xA;float Minc = twopi/250; // time proxy increment&#xA;&#xA;void setup()&#xA;{ size(wsize,wsize);&#xA;  //e = sqrt(((a*a)-(b*b))/(a*a)); kept for reference&#xA;  b = a*sqrt(1-e*e);&#xA;  c = e*a;&#xA;  k = 1-e*e;&#xA;  M = 0; // Mean anomaly increases linearly with time&#xA;}&#xA;&#xA;void draw()&#xA;{ background(255);&#xA;  translate(wsize/2,wsize/2); // move origin to centre of display window&#xA;  ellipse(0,0,2*a,2*b); // show orbit&#xA;  spot(c,0,rs,#FFF303); // draw sun&#xA;  translate(c,0); // move origin to focus&#xA;  // Calculate Eccentic Anomaly by iteration&#xA;  E = M; E0 = 1000; // make sure while starts&#xA;  while(abs(E-E0) &amp;gt; 0.000001)&#xA;  { E0 = M + e*sin(E);&#xA;    E = E0;&#xA;  } &#xA;  // calculate polar coordinates of planet&#xA;  cosE= cos(E);&#xA;  cosv = (cosE-e)/(1-e*cosE);&#xA;  v= acos(cosv);&#xA;  r = a*(1-e*e)/(1+e*cosv);&#xA;  // calculate x,y coordinates of planet&#xA;  x = r * cosv;&#xA;  y = -r*sin(v);// negative makes it go anti-clockwise&#xA;  if (M &amp;gt; PI){y = -y;} // show second half of orbit&#xA;  // println(v+&quot; : &quot;+x+&quot; : &quot;+y);&#xA;  // line(0,0,x,y);&#xA;  spot(x,y, rp, #03ADFF); // draw planet&#xA;  M += Minc;&#xA;  if(M &amp;gt; twopi){M = 0.0;}&#xA;}&#xA;&#xA;void spot(float x, float y, float w, color tint)&#xA;{ fill(tint);&#xA;  ellipse(x,y,w,w);&#xA;  fill(255); // restore white background&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2022" LastEditorUserId="2022" LastEditDate="2015-05-01T18:14:59.923" LastActivityDate="2015-05-01T18:14:59.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10621" PostTypeId="2" ParentId="10321" CreationDate="2015-05-01T17:30:13.503" Score="5" Body="&lt;p&gt;According to &lt;a href=&quot;http://arxiv.org/abs/1502.01249&quot; rel=&quot;noreferrer&quot;&gt;Vokrouhlicky et al. 2015&lt;/a&gt; Yarkovsky forces can be measured for small bodies with diameters up to 30-40 km. The largest object they have in their list of Yarkovsky detections is 4179 Toutatis with a diameter of (only) 2.8 km. I am not aware if Yarkovsky forces have been measured on anything larger than asteroids.&lt;/p&gt;&#xA;" OwnerUserId="7144" LastActivityDate="2015-05-01T17:30:13.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10623" PostTypeId="1" CreationDate="2015-05-02T05:38:27.083" Score="1" ViewCount="397" Body="&lt;p&gt;I posed a question or two on here a couple months or so ago about the orbital patterns of each of the seven stars in the two septenary star systems Nu Scorpii and AR Cassiopeiae. They remain unanswered so I have to assume the answer is unknown. So, now I want to what the most numerous star system is for which we actually do know the orbital pattern, and obviously I want to know the orbital pattern.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am NOT talking about a star with multiple planetary or other smaller bodies. I am talking about multiple STAR systems. Do we know the orbit for systems of multiple stars. Again, I'm not talking about every rock larger than a school bus. I mean STARS. Stars that have nuclear fusion occurring. &lt;/p&gt;&#xA;" OwnerUserId="6753" LastEditorUserId="6753" LastEditDate="2016-03-29T12:16:50.700" LastActivityDate="2016-03-30T17:17:26.917" Title="What is the most populated/numerous stellar system in which the orbits of all objects are known?" Tags="&lt;orbit&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10624" PostTypeId="1" CreationDate="2015-05-02T06:07:23.403" Score="10" ViewCount="624" Body="&lt;p&gt;What are the conditions for a planetary moon to have a satellite of its own? How far do the Solar System's bodies fall from the necessary threshold?&lt;/p&gt;&#xA;" OwnerUserId="2541" LastEditorUserId="2153" LastEditDate="2015-05-02T17:54:25.540" LastActivityDate="2015-05-02T21:40:47.457" Title="Why does the Solar System have no (natural) satellites of satellites?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;natural-satellites&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10625" PostTypeId="2" ParentId="10624" CreationDate="2015-05-02T07:30:15.867" Score="9" Body="&lt;p&gt;This is due to the mechanics of capture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one object (moon) to be captured by a another (planet), some energy has to be removed from the system. If the incoming moon has an existing satellite then it would be ejected, carrying a lot of kinetic energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a small body were to be captured by the planet/moon combination, it would usually be captured by the larger object. Again some energy has to removed from the incoming object by being transferred to one the objects already present. One way that could occur is for the existing moon to be accelerated into a higher orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, it could happen, but with a &lt;em&gt;very&lt;/em&gt; low probability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;em&gt;much&lt;/em&gt; more information, see this answer to the question &lt;a href=&quot;https://astronomy.stackexchange.com/a/9970/5524&quot;&gt;How do moons get captured?&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5524" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-05-02T07:55:33.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10626" PostTypeId="2" ParentId="10506" CreationDate="2015-05-02T07:43:32.143" Score="3" Body="&lt;p&gt;First, let assume your image geometry is homogeneous, and has no peculiar distorsion in either direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, let assume you have the resolution of your image: the number of arcseconds / pixel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, take one 'red-cross' star, call it A. It will be the origin of the triangle we will draw. Name your 'yellow-cross' star B. Now, take a new point, called 'C', that is at the same pixel-y coordinate of A, and the same pixel-x coordinates of B. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drawing lines between A, B and C gives you a right triangle. You cannot simply apply flat geometry equations, since you are on the celestial sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, one must compute the 'Bearing' angle between A, B and C. See for instance &lt;a href=&quot;http://www.mathsrevision.net/gcse-maths-revision/trigonometry/bearings&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;: (A bearing is an angle, measured clockwise from the north direction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a little piece of code you should be able to read:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;double adjacent = sqrt(pow(B.x-C.x, 2) + pow(B.y-C.y, 2)); // dBC&#xA;double opposite = sqrt(pow(A.x-C.x, 2) + pow(A.y-C.y, 2)); // dAC&#xA;&#xA;double theta = atan2(opposite, adjacent) * ONE_RAD_IN_DEGREES;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;theta&lt;/code&gt; is the bearing angle, here expressed in degrees, thanks to the conversion constant &lt;code&gt;ONE_RAD_IN_DEGREES&lt;/code&gt;. &lt;code&gt;atan2&lt;/code&gt; is the Arc-Tangent function that takes care of which quadrant you are in (it computes &lt;code&gt;arctangent(opposite / adjacent)&lt;/code&gt;, correcting for the quadrant, see the &lt;a href=&quot;http://en.wikipedia.org/wiki/Atan2&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia article&lt;/a&gt; for instance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, depending on whether you have East to the left or not (astro images have East to the left usually), you need to correct your angle. Again this little piece of code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;BOOL eastLeft = &amp;lt;true or false&amp;gt;&#xA;&#xA;if (B.x &amp;lt; A.x &amp;amp;&amp;amp; B.y &amp;gt; A.y) {&#xA;    theta = (eastLeft) ? theta : 360.0 - theta;&#xA;}&#xA;else if (B.x &amp;lt; A.x &amp;amp;&amp;amp; B.y &amp;lt; A.y) {&#xA;    theta = (eastLeft) ? 180.0 - theta : theta + 180.;&#xA;}&#xA;else if (B.x &amp;gt; A.x &amp;amp;&amp;amp; B.y &amp;lt; A.y) {&#xA;    theta = (eastLeft) ? theta + 180. : 180.0 - theta;&#xA;}&#xA;else if (B.x &amp;gt; A.x &amp;amp;&amp;amp; B.y &amp;gt; A.y) {&#xA;    theta = (eastLeft) ? 360.0 - theta : theta;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, we have the correct &lt;code&gt;theta&lt;/code&gt; value. Now, compute the distance (below, in degrees) between A and B, and call it &lt;code&gt;delta&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the R.A. and Declination of A are called &lt;code&gt;lambda1&lt;/code&gt; and &lt;code&gt;phi1&lt;/code&gt;, you can compute the R.A. and Declination of C, &lt;code&gt;lambda2&lt;/code&gt; and &lt;code&gt;phi2&lt;/code&gt;, using the formulae given in &lt;a href=&quot;http://www.movable-type.co.uk/scripts/latlong.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, under the section &quot;Destination point given distance and bearing from start point&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;double phi1 = declination_A * ONE_DEG_IN_RADIANS;&#xA;double lambda1 = rightAscension_A * ONE_HOUR_IN_RADIANS;&#xA;&#xA;double delta = degrees * ONE_DEG_IN_RADIANS;&#xA;double theta = bearing * ONE_DEG_IN_RADIANS;&#xA;&#xA;double phi2 = asin(sin(phi1)*cos(delta) + cos(phi1)*sin(delta)*cos(theta));&#xA;double lambda2 = lambda1 + atan2(sin(theta) * sin(delta) * cos(phi1), cos(delta) - sin(phi1) * sin(phi2));&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;with the usual meaning of trigonometric functions (&lt;code&gt;sin&lt;/code&gt; is sine, &lt;code&gt;asin&lt;/code&gt; is arcsine, etc).&lt;/p&gt;&#xA;" OwnerUserId="1599" LastEditorUserId="1599" LastEditDate="2017-01-07T17:31:13.073" LastActivityDate="2017-01-07T17:31:13.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10628" PostTypeId="2" ParentId="10624" CreationDate="2015-05-02T21:40:47.457" Score="7" Body="&lt;p&gt;In addition to the capture mechanisms mentioned by Andy, you also need to take into account the stability of any orbit due to perturbations from tidal effects. E.g. in case of our Moon, there are no stable orbits possible. Every satellite put in orbit around the Moon had had to implement course corrections to prevent it from prematurely crashing into the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This effect explains why you don't have a lot of dusts and rocks in orbit around the Moon due to the large number of impacts. What happens is that after an impact you do get a lot ejecta, a small fraction will then collide with each other so that they get in an orbit around the Moon. But on the long term the tidal effects of the Earth and the Sun causes perturbations until the orbit crosses the lunar surface. &lt;/p&gt;&#xA;" OwnerUserId="2185" LastActivityDate="2015-05-02T21:40:47.457" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10629" PostTypeId="1" AcceptedAnswerId="10637" CreationDate="2015-05-02T22:42:25.507" Score="2" ViewCount="740" Body="&lt;p&gt;I am playing around with data from exoplanets.org, and am interested in the plot of surface gravity to planet mass. I reproduced this plot &lt;img src=&quot;https://i.imgur.com/lUtlJkh.png&quot; alt=&quot;this plot&quot;&gt; after downloading their data and performed a non-linear regression model to fit the curve. Unfortunately,the covariance matrix of my fit has infinite values, so I am now trying out a linear fit on a log-log plot, shown below. &lt;img src=&quot;https://i.imgur.com/LPIJF1c.png&quot; alt=&quot;&quot;&gt; My fit, for $y=ax+b$, is $a=0.9511$, $b=0.8631$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am now thinking about what I am plotting. I suspect there may not be anything interesting in $\log \frac{GM}{R^2}$ vs. $\log M$, but, regardless, I am trying to understand if there is any meaning in the y-intercept. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are possible explanations for the value $b$?&lt;/p&gt;&#xA;" OwnerDisplayName="user5341" LastEditorDisplayName="user5341" LastEditDate="2015-05-03T21:56:27.160" LastActivityDate="2015-05-03T22:40:00.483" Title="On a log-log plot of surface gravity to planet mass, what is the meaning of the y-intercept?" Tags="&lt;gravity&gt;&lt;planet&gt;&lt;exoplanet&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10630" PostTypeId="1" AcceptedAnswerId="10665" CreationDate="2015-05-02T23:08:15.360" Score="0" ViewCount="392" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Inflation_%28cosmology%29&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; tells us:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In physical cosmology, cosmic inflation, cosmological inflation, or just inflation is the exponential expansion of space in the early universe. The inflationary epoch lasted from 10&lt;sup&gt;-36&lt;/sup&gt; seconds after the Big Bang to sometime between 10&lt;sup&gt;-33&lt;/sup&gt; and 10&lt;sup&gt;-32&lt;/sup&gt; seconds. Following the inflationary period, the Universe continues to expand, but at a less accelerated rate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've no idea about astronomy, I just wondered whether somebody ever published a theory about cosmological deflation, which would make the universe shrink in a fraction of a second. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a SFI novel you could travel by locally causing a deflation of space in the direction of your spaceship; the &lt;em&gt;unshrunken space&lt;/em&gt; of a lightyear would (only locally in front of the ship) measure 1000 km for the fraction of the second. Or maybe a technically very advanced civilization could keep its solar system inflated by the time of a deflation of the whole universe and thus survive, which leads to the interesting question, whether this solar system could &lt;em&gt;reenter&lt;/em&gt; the space of the next universe after its new cosmolocical inflation and if yes, does somebody look out for stellar systems at places in space where you wouldn't expect them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However: cosmological deflation? &lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="2153" LastEditDate="2015-05-02T23:41:54.350" LastActivityDate="2015-05-11T18:47:49.760" Title="Cosmological deflation?" Tags="&lt;cosmological-inflation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10631" PostTypeId="1" AcceptedAnswerId="11094" CreationDate="2015-05-03T03:40:58.170" Score="3" ViewCount="5425" Body="&lt;p&gt;I currently own a telescope, I'm an amateur astronomer and currently just trying to view the moon. I own the Celestron PowerSeeker 70AZ. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Link for information on the telescope: &lt;a href=&quot;http://www.celestron.com/browse-shop/astronomy/telescopes/powerseeker-70az-telescope&quot; rel=&quot;nofollow&quot;&gt;http://www.celestron.com/browse-shop/astronomy/telescopes/powerseeker-70az-telescope&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem: I can't view the moon! All I see is a big ball of white, and since I'm new I thought I could get some advice for eyepieces:  mine being a regular 20mm eyepiece. I don't use a barlow lens, mine broke.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! If you want any more specifications, just ask. I'm just looking for help! Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7189" LastEditorUserId="6" LastEditDate="2015-06-24T17:00:50.813" LastActivityDate="2017-12-29T08:30:07.833" Title="Why is the moon a fuzzy, white ball?" Tags="&lt;telescope&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10632" PostTypeId="2" ParentId="10629" CreationDate="2015-05-03T05:48:57.893" Score="1" Body="&lt;p&gt;For planets of constant mean density you have:&#xA;$$&#xA;M=\rho \times 4\pi r^3&#xA;$$&#xA;and the surface value of $g$ is:&#xA;$$&#xA;g(r)=\frac{GM}{r^2}=G \times \rho\times 4 \pi \times r&#xA;$$&#xA;So for bodies of constant density the surface gravity is proportional to the radius, and the slope as $r \to 0$ tells you the density. So for bodies of equal density $\log(g(r)) \to -\infty$ as $r \to 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of mass:&#xA;$$&#xA;g(M)={G(4\pi \rho)^{2/3}M^{1/3}}&#xA;$$ &#xA;So as $M\to 0$ we have $g(M)\to 0$ and again $\log(g(M))\to -\infty$ and the intercept of the $\log-\log$ plot as $M$ goes to zero alows you to calculate the mean density.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-05-03T13:19:22.230" LastActivityDate="2015-05-03T13:19:22.230" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10633" PostTypeId="2" ParentId="10623" CreationDate="2015-05-03T06:53:15.510" Score="5" Body="&lt;p&gt;&lt;em&gt;What is the most populated/numerous stellar system in which the orbits of all objects are known?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is none. Other than our own solar system, astronomers don't know if they know all of the large bodies (aka planets) orbiting any given star system. Presumably other star systems have asteroids, comets, and other stuff. The orbits of those small bodies in star systems is unknown, and will remain unknowable for a long time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't even know that for our own solar system. If we did, we wouldn't have to worry about killer asteroids and comets.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-05-03T06:53:15.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10634" PostTypeId="2" ParentId="10573" CreationDate="2015-05-03T06:58:27.747" Score="0" Body="&lt;p&gt;Data from observatory archives is a good way to go. Here is another one with tons of imaging datasets: &lt;a href=&quot;http://archive.eso.org&quot; rel=&quot;nofollow&quot;&gt;http://archive.eso.org&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1599" LastActivityDate="2015-05-03T06:58:27.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10636" PostTypeId="2" ParentId="10583" CreationDate="2015-05-03T10:23:04.233" Score="3" Body="&lt;p&gt;There are surprisingly many examples of short period, eccentric binry systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very incomplete list would include&lt;/p&gt;&#xA;&#xA;&lt;p&gt;KIC4544587 P=2.19 d  e=0.288 +/- 0.026 &lt;a href=&quot;http://arxiv.org/abs/1306.1819&quot; rel=&quot;nofollow&quot;&gt;Hambleton et al. (2013)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HD174884 P=3.66 d  e=0.2939 +/- 0.0005 &lt;a href=&quot;http://arxiv.org/abs/0910.3513&quot; rel=&quot;nofollow&quot;&gt;Maceroni et al. (2010)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CoRoT 102918586 P=4.39 d  e=0.249 +/- 0.005 &lt;a href=&quot;http://arxiv.org/abs/1302.0167&quot; rel=&quot;nofollow&quot;&gt;Maceroni et al. (2013)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HD313926 P=2.27d  e=0.209 +/-0.001 &lt;a href=&quot;http://mnrasl.oxfordjournals.org/content/380/1/L63.full&quot; rel=&quot;nofollow&quot;&gt;Rucinski et al. (2007)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each of these cases, the stars are relatively warm, with thin (if any) subphotospheric convection zones. Tidal circularisation is much more effective in stars with thick convection zones (giants, or cooler main sequence stars). Therefore it is possible that all of these objects (and, as I say, there are many other examples) are too young to have circularised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some theory is presented by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2005ASPC..333....4Z&quot; rel=&quot;nofollow&quot;&gt;Zahn et al. (2005)&lt;/a&gt;. Indeed, circularisation times are very short - basically you expect everything with a solar mass or below to be circularised if its period is less than 6 days, but the process is less efficient for higher mass stars. Circularisation will essentilly take place during pre-main-sequence evolution. Since a circular orbit represents the minimum energy configuration, the only way you are going to get non-circular orbits in solar-type and lower mass stars is if they are very young (less than some tens of millions of years), the binary system formed &lt;em&gt;after&lt;/em&gt; their pre-main-sequence evolution or if there is an injection of energy via a dynamical interaction or an eccentric third body in a much wider orbit.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-05-03T10:23:04.233" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10637" PostTypeId="2" ParentId="10629" CreationDate="2015-05-03T11:55:56.853" Score="1" Body="&lt;p&gt;I think what you have established here is just that $\rho$ tends to increase with mass. The density of planets isn't constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\rho = \rho_0 (M/M_{earth})^{\alpha}$, so that $M = (4/3)\pi R^{3} \rho_0 (M/M_{earth})^{\alpha}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&#xA;$$g = \frac{GM}{R^2} = \frac{4\pi G}{3} R \rho$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replace $R$ with $(3M/4\pi \rho)^{1/3}$ so that&#xA;$$ g = \frac{4\pi G}{3} \left(\frac{3M}{4\pi \rho}\right)^{1/3} \rho$$&#xA;$$ g = \left(\frac{4\pi}{3}\right)^{2/3} G M^{1/3} \rho_0^{2/3} (M/M_{earth})^{2\alpha/3}$$&#xA;$$ g = \left(\frac{4\pi}{3}\right)^{2/3} GM_{earth}^{1/3} \rho_0^{2/3} (M/M_{earth})^{(2\alpha+1)/3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, bar a (highly possible) algebraic slip, if you plot $\log g$ vs $\log M$, the gradient is $(2\alpha+1)/3$, which from your plot, gives $\alpha \simeq 0.92$ - i.e. the average planet density increases almost linearly with mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intercept then is &#xA;$$ b = \log \left[ \left(\frac{4\pi}{3}\right)^{2/3} GM_{earth}^{1/3} \rho_0^{2/3}\right],$$&#xA;which yields $\rho_0 \simeq 3.5$ kg/m$^3$ (NB: I subtracted 2 from your $b$ to make it SI; giving a density of around 814 kg/m$^3$ at a Jupiter mass).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that density is almost proportional to mass can be found from the same dataset. e.g. see below. Below 0.1 Jupiter masses, the relationship appears to break down, though in actual fact very few of the densities for such planets are very accurately measured (since it requires a radius from a transit), but it works well enough in the range you have plotted. The physics here is that gas giants are governed by a partially (electron) degenerate equation of state that results in them all having a similar radius from about a tenth of a Jupiter mass to about 50 Jupiter masses (albeit with considerable and largely unexplained scatter). Thus the density is proportional to mass. This relationship does not work for small rocky planets, where the radius decreases for smaller masses. Thus your nice line in log space does not continue to lower masses (see plot at bottom - NB: some of the low-mass points have huge uncertainties).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cPaWk.png&quot; alt=&quot;Planet density versus mass&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xME4A.png&quot; alt=&quot;Surface gravity versus mass&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-05-03T22:40:00.483" LastActivityDate="2015-05-03T22:40:00.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10638" PostTypeId="2" ParentId="1037" CreationDate="2015-05-03T14:25:11.047" Score="6" Body="&lt;p&gt;Using &lt;em&gt;current technology&lt;/em&gt; (and by that I mean experiments and telescopes that are available now) we have not detected an &quot;Earth-like&quot; planet and we would probably be unable to detect life on Earth even if observed from a distance of a few light years. Therefore there is currently no prospect of detecting life on an &quot;Earth doppelganger&quot;. I elaborate below:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;No planets &lt;em&gt;like the Earth&lt;/em&gt; have yet been detected around another star. That is to say, none that have a similar mass, radius and orbit at 1 au (or close to it) from a solar-type star. With current technology, it is &lt;em&gt;just&lt;/em&gt; out of reach. Therefore any directed search for life on an Earth-like planet wouldn't actually know where to start. If you can't detect the planet &lt;em&gt;at all&lt;/em&gt; then there is absolutely no chance of looking at its atmospheric composition to look for biomarkers (e.g. oxygen along with a reducing gas like methane, or chlorofluorocarbons from an industrial civilisation - &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014ApJ...792L...7L&quot; rel=&quot;noreferrer&quot;&gt;Lin et al. 2014&lt;/a&gt;). The only exoplanets for which atmospheric compositions have been (crudely and tentatively) measured are &quot;hot Jupiters&quot;. - giant exoplanets orbiting very close to their parent stars. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A &quot;blind&quot; search could look for radio signatures and of course this is what SETI has been doing. If we are talking about detecting &quot;Earth&quot;, then we must assume that we are not talking about deliberate beamed attempts at communication, and so must rely on detecting random radio &quot;chatter&quot; and accidental signals generated by our civilisation.&#xA;The SETI Phoenix project was the most advanced search for radio signals from other intelligent life. Quoting from &lt;a href=&quot;http://adsabs.harvard.edu/full/2000ASPC..213..451C&quot; rel=&quot;noreferrer&quot;&gt;Cullers et al. (2000)&lt;/a&gt;: &quot;&lt;em&gt;Typical signals, as opposed to our strongest signals fall below the detection threshold of most surveys, even if the signal were to originate from the nearest star&lt;/em&gt;&quot;. Quoting from &lt;a href=&quot;http://adsabs.harvard.edu/abs/2001ARA%26A..39..511T&quot; rel=&quot;noreferrer&quot;&gt;Tarter (2001)&lt;/a&gt;: &quot;&lt;em&gt;At current levels of sensitivity, targeted microwave searches could detect the equivalent power of strong TV transmitters at a distance of 1 light year (within which there are no other stars)...&lt;/em&gt;&quot;. The equivocation in these statements is due to the fact that we &lt;em&gt;do&lt;/em&gt; emit stronger beamed signals in certain well-defined directions, for example to conduct metrology in the solar system using radar. Such signals have been calculated to be observable over a thousand light years or more. But these signals are brief, beamed into an extremely narrow angle and unlikely to be repeated. You would have to be very lucky to be observing in the right direction at the right time if you were performing targeted searches. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hence my assertion that with current methods and telescopes there is not much chance of success. But of course technology advances and in the next 10-20 years there may be better opportunities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step in a directed search would be to find planets like Earth. The first major opportunity will be with the &lt;a href=&quot;http://tess.gsfc.nasa.gov/&quot; rel=&quot;noreferrer&quot;&gt;TESS spacecraft&lt;/a&gt;, launching in 2017, capable of detecting earth-sized planets around the brightest 500,000 stars. However, it's 2-year mission would limit the ability to detect an Earth-analogue. The best bet for finding other Earths will come later (2024 perhaps) with the launch of &lt;a href=&quot;http://www.esa.int/Our_Activities/Space_Science/ESA_selects_planet-hunting_PLATO_mission&quot; rel=&quot;noreferrer&quot;&gt;Plato&lt;/a&gt;, a six-year mission that again, studies the brightest stars. However, there is then a big leap forward required to perform studies of the atmospheres of these planets. Direct imaging and spectroscopy would probably require space-borne nulling interferometers; indirect observations of phase-effects and transmission spectroscopy through an exoplanet atmosphere does not require great angular resolution, just massive precision and collecting area. Spectroscopy of something the size of Earth around a normal star will probably require a bigger &lt;em&gt;successor&lt;/em&gt; to the James Webb Space Telescope (&lt;a href=&quot;http://www.jwst.nasa.gov/&quot; rel=&quot;noreferrer&quot;&gt;JWST&lt;/a&gt; - launch  2018), or even more collecting area than will be provided by the E-ELT in the next decade. For example &lt;a href=&quot;http://www.epj-conferences.org/articles/epjconf/abs/2013/08/epjconf_hpcs2012_11001/epjconf_hpcs2012_11001.html&quot; rel=&quot;noreferrer&quot;&gt;Snellen (2013)&lt;/a&gt; argues it would take 80-400 transits-worth of exposure time (i.e. 80-400 years!) to detect the biomarker signal of an Earth-analogue with the E-ELT!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has been suggested that new radio telescope projects and technology like the Square Kilometre Array may be capable of serendipitously detecting radio &quot;chatter&quot; out to distances of 50 pc ($\sim 150$ light years) - see &lt;a href=&quot;http://adsabs.harvard.edu/abs/2007JCAP...01..020L&quot; rel=&quot;noreferrer&quot;&gt;Loeb &amp;amp; Zaldarriaga (2007)&lt;/a&gt;. This array, due to begin full operation some time after 2025 could also monitor a multitude of directions at once for beamed signals. A good overview of what might be possible in the near future is given by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2009astro2010S.294T&quot; rel=&quot;noreferrer&quot;&gt;Tarter et al. (2009)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-05-03T15:09:19.783" LastActivityDate="2015-05-03T15:09:19.783" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10640" PostTypeId="2" ParentId="10499" CreationDate="2015-05-04T10:58:08.717" Score="11" Body="&lt;p&gt;&lt;strong&gt;Short answer:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun will lose about half of its mass on the way to becoming a white dwarf. Most of this mass loss will occur in the last few million years of its life, during the Asymptotic Giant Branch (AGB) phase. At the same time the orbital radius of the Earth around the Sun will grow by a factor of two (as will the outer planets). Unfortunately for the Earth, the radius of the Sun will also reach to about 2 au, so it will be toasted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is the possibility that the decreased binding energy and increased eccentricity of the Earth and the outer planets will lead to dynamical instabilities that could lead to planetary ejection. This is highly dependent on the exact time dependence of the late, heavy mass loss and the alignment or otherwise of the planets at the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Long answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars with mass less than about 8 solar masses will end their lives as white dwarfs on a timescale which increases as their main sequence initial mass decreases. The white dwarfs that are formed are of lower mass than their progenitor main sequence stars, because much of the initial mass of a star is lost through stellar winds (particularly during the thermally pulsating &lt;a href=&quot;http://en.wikipedia.org/wiki/Asymptotic_giant_branch&quot; rel=&quot;nofollow noreferrer&quot;&gt;asymptotic giant branch&lt;/a&gt; phase) and final ejection of a planetary nebula.  Thus, the current distribution of white dwarf masses, that peaks between &lt;span class=&quot;math-container&quot;&gt;$0.6$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$0.7 M_{\odot}$&lt;/span&gt; and with a dispersion of &lt;span class=&quot;math-container&quot;&gt;$\sim 0.2 M_{\odot}$&lt;/span&gt;, reflects the final states of all main sequence stars with &lt;span class=&quot;math-container&quot;&gt;$0.9 &amp;lt;M/M_{\odot}&amp;lt;8 M_{\odot}$&lt;/span&gt;, that have had time to evolve and die during our Galaxy's lifetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most reliable information we have about the relationship between the initial main sequence mass and final white dwarf mass (the initial-final mass relation or IFMR) comes from measuring the properties of white dwarfs in star clusters of known age. Spectroscopy leads to a mass estimate for the white dwarf. The initial mass is estimated by calculating a main sequence plus giant branch lifetime from the difference between the age of the star cluster and the cooling age of the white dwarf. Stellar models then tell us the relationship between main sequence plus giant lifetime and the initial main sequence mass, hence leading to an IFMR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A recent compilation from &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013MmSAI..84...58K&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kalirai (2013)&lt;/a&gt; is shown below. This shows that a star like the Sun, born with an initial mass of &lt;span class=&quot;math-container&quot;&gt;$1M_{\odot}$&lt;/span&gt; (or maybe a per cent or two more, since the Sun has already lost some mass), ends its life as a white dwarf with &lt;span class=&quot;math-container&quot;&gt;$M = 0.53 \pm 0.03\ &#xA;M_{\odot}$&lt;/span&gt;. i.e. The Sun should lose approximately 50% of its initial mass in stellar winds and (possibly) planetary nebula ejection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IS3SM.png&quot; alt=&quot;IFMR from Kalirai (2013)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A comprehensive treatment of what happens to solar systems when the central star loses mass in a time-dependent way is given in &lt;a href=&quot;http://arxiv.org/abs/1303.3841&quot; rel=&quot;nofollow noreferrer&quot;&gt;Adams et al. (2013)&lt;/a&gt;. The simplest cases are initially circular orbits where the mass loss takes place on much longer timescales than the orbital period. As mass loss proceeds, the gravitational potential energy increases (becomes less negative) and thus the total orbital energy increases and the orbit gets wider. Roughly speaking, &lt;span class=&quot;math-container&quot;&gt;$aM$&lt;/span&gt; is a constant, where &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is the orbital radius, which is a simple consequence of conservation of angular momentum: so the Earth would end up in a 2 au orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the presence of a non-zero eccentricity in the initial orbit, or in the case of rapid mass loss, such as that which occurs towards the end of the AGB phase, then things become altogether more unpredictable, with the eccentricity also growing as mass loss proceeds. This has a knock-on effect when considering the dynamical stability of the whole (evolved) solar system and may result in planetary ejection. The faster the mass loss, the more unpredictable things get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radius of an AGB star can be calculated using &lt;span class=&quot;math-container&quot;&gt;$ L = 4\pi R^2 \sigma T_{eff}^{4}$&lt;/span&gt;. Stars at the tip of the AGB branch have luminosities of &lt;span class=&quot;math-container&quot;&gt;$\sim 10^{4} L_{\odot}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$T_{eff} \simeq 2500\ K$&lt;/span&gt;, leading to likely radii of &lt;span class=&quot;math-container&quot;&gt;$\sim 2$&lt;/span&gt; au. So it is quite likely that unless the Earth is ejected or has its orbit significantly modified by some dynamical instability that, like the inner planets, it will end up engulfed in the outer enevelope of the AGB star and spiral inwards...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even should it narrowly escape this immediate fate, it is then quite likely that tidal dissipation will rapidly extract energy out of the orbit and the Earth will spiral in towards the envelope of the giant Sun... with the same result.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-10-11T09:15:56.970" LastActivityDate="2019-10-11T09:15:56.970" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10641" PostTypeId="2" ParentId="10597" CreationDate="2015-05-04T16:39:59.840" Score="7" Body="&lt;p&gt;Well, if &quot;last night&quot; means around 2015-04-22 at 2230 GMT, the order from left to right is: Europa, Io, Ganymede, Callisto. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a stellarium snapshot of (roughly) this configuration. The snapshot doesn't label the moons, but I clicked on each individually to confirm the order above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vOLcn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2015-05-04T17:27:07.293" LastActivityDate="2015-05-04T17:27:07.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10642" PostTypeId="1" CreationDate="2015-04-29T17:02:21.097" Score="3" ViewCount="247" Body="&lt;p&gt;Kepler-442b&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm doing a project in which I need to find a planet within our galaxy that might be habitable. I found this planet that is within its stellar system's habitable zone, and due to research I have found that this planet is one of the closest in similarity to Earth, in terms of size, and temperature. But I don't really know anything about the surface or physical features.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there is an equation to find if there is &lt;em&gt;life&lt;/em&gt; on another planet, but I cant find anything about &lt;em&gt;water&lt;/em&gt;... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can help me, thank you so much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to know the physical and chemical features on a planet's surface. I know the planet is within the habitable zone and is thought to contain water, but I need more detail, if there is any way to get the specific or at least highly probable features of the surface. Sorry I didn't make this clear enough.&lt;/p&gt;&#xA;" OwnerUserId="7212" OwnerDisplayName="Nathanael Smith" LastEditorUserId="2153" LastEditDate="2015-05-04T23:34:54.557" LastActivityDate="2015-05-13T14:20:27.060" Title="Is there a way to tell what the surface of a planet is like?" Tags="&lt;planet&gt;&lt;observational-astronomy&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10645" PostTypeId="2" ParentId="10642" CreationDate="2015-05-05T14:47:21.267" Score="0" Body="&lt;p&gt;I think the methodologies you're asking for are still very much being worked out (and you might want to think of this as looking for a &lt;em&gt;model&lt;/em&gt; rather than an &lt;em&gt;equation&lt;/em&gt; -- the latter is too simplistic). It might be best to find some relevant articles (starting with some that are geared to the layman) and work from there:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.universetoday.com/104652/how-do-you-find-the-signs-of-life-in-alien-planets/&quot; rel=&quot;nofollow&quot;&gt;http://www.universetoday.com/104652/how-do-you-find-the-signs-of-life-in-alien-planets/&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.planetary.org/explore/space-topics/exoplanets/direct-imaging.html&quot; rel=&quot;nofollow&quot;&gt;http://www.planetary.org/explore/space-topics/exoplanets/direct-imaging.html&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.nasa.gov/press-release/goddard/nasa-giss-to-help-lead-search-for-habitable-exoplanets&quot; rel=&quot;nofollow&quot;&gt;http://www.nasa.gov/press-release/goddard/nasa-giss-to-help-lead-search-for-habitable-exoplanets&lt;/a&gt;&#xA;&lt;a href=&quot;http://astrobites.org/2012/11/11/detecting-exoplanet-atmospheres-from-the-ground/&quot; rel=&quot;nofollow&quot;&gt;http://astrobites.org/2012/11/11/detecting-exoplanet-atmospheres-from-the-ground/&lt;/a&gt;&#xA;&lt;a href=&quot;http://phys.org/news/2014-07-earth-exoplanet-life-visible.html&quot; rel=&quot;nofollow&quot;&gt;http://phys.org/news/2014-07-earth-exoplanet-life-visible.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7154" LastActivityDate="2015-05-05T14:47:21.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10654" PostTypeId="2" ParentId="10642" CreationDate="2015-05-07T14:57:39.883" Score="0" Body="&lt;p&gt;It is possible that the basic colouration of the visible surface would lead you to some conclusions - the presence, or otherwise, of clouds for example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe filtering the light through a diffraction grating would give you some idea of the atmosphere or surface's constituent elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have some serious kit, you could wait and hope that the planet passes in front of either a background star, or it's own 'sun' - the change in the spectrum of that light pre and during the occultation could reveal information about the atmosphere too. That's a bit of a long shot, though! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that is useful in some way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;OSH.&lt;/p&gt;&#xA;" OwnerUserId="7216" LastActivityDate="2015-05-07T14:57:39.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10655" PostTypeId="1" CreationDate="2015-05-08T02:27:09.533" Score="3" ViewCount="252" Body="&lt;p&gt;Tonight (10:00 PM EST Middlesex County, Mass.) when looking at a mostly clear sky I observed what I at first believed was a meteor. This point of light was moving east to northeast, and that would be in the vicinity of the Lyrid meteor shower. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the app Star Walk for iOS it says that the &quot;n-Lyrids&quot; meteor shower peaks on May 9th. However, information I found online about this meteor shower says it already peaked back in April. (I did see a bright meteor a couple weeks ago.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone offer insight on possible meteors right now? I highly doubt that this was a meteor though, because it lasted way too long, probably two minutes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My next guess was the International Space Station but NASA's &lt;a href=&quot;https://spotthestation.nasa.gov/&quot; rel=&quot;nofollow&quot;&gt;Spot The Station&lt;/a&gt; did not show any viewing times for my area today. The Star Walk app also showed the ISS well below the horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads me to conclude the object was a plane, however there were no blinking lights and the only time its light changed was when it passed behind a thin cloud. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It could also have been a satellite glint such as an Iridium flare. I did not see the beginning of the light's appearance, so it is possible I missed the flare. However flares do not last for minutes, but glint from solar panels or other satellites may last this long. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What did I see?&lt;/p&gt;&#xA;" OwnerUserId="7221" LastEditorDisplayName="user1569" LastEditDate="2016-10-03T14:09:23.337" LastActivityDate="2016-10-03T14:09:23.337" Title="What would this moving point of light be?" Tags="&lt;observational-astronomy&gt;&lt;amateur-observing&gt;&lt;artificial-satellite&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10656" PostTypeId="1" CreationDate="2015-05-08T08:01:01.390" Score="3" ViewCount="68" Body="&lt;p&gt;I'm looking at the &lt;a href=&quot;http://www.iphas.org/dr2/#download&quot; rel=&quot;nofollow&quot;&gt;IPHAS DR2&lt;/a&gt; catalog. It has the Right Ascension and Declination but I'm not seeing any distance information (lightyears or parsecs). I'm wondering, given the columns in the dataset, if the distance can be calculated.&lt;/p&gt;&#xA;" OwnerUserId="7222" LastActivityDate="2016-01-29T20:18:17.670" Title="Is it possible to get the distance to a star in the IPHAS DR2 catalog?" Tags="&lt;distances&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10658" PostTypeId="2" ParentId="10577" CreationDate="2015-05-08T19:36:33.667" Score="0" Body="&lt;p&gt;The answer turns out to be embarrassingly simple, and I suspect nobody came up with it because of my poor communication in the question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $a$ is the occurring in the Fourier transform is the scale factor &lt;em&gt;which has a time dependence&lt;/em&gt;. So if we look at the term $\frac{\partial}{\partial t} \rho$ we will get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{\partial}{\partial t} \rho = \frac{\partial}{\partial t} \int  \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho_{\vec{q}}(t) \ \mathrm{d}^3 \vec{q} &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA; = \int \frac{\partial}{\partial t}\left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)} \right) \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho_{\vec{q}}(t) +  \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \frac{ \mathrm{d} \rho_{\vec{q}}}{\mathrm{d} t }  \ \mathrm{d}^3 \vec{q} &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA; = \int \left(-\frac{\dot{a}}{a} \frac{ i \vec{q} \cdot \vec{X}}{a(t)} \right) \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho_{\vec{q}}(t) +  \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \frac{ \mathrm{d} \rho_{\vec{q}}}{\mathrm{d} t }  \ \mathrm{d}^3 \vec{q} &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA; = \int \left(-H\frac{ i \vec{q} \cdot \vec{X}}{a(t)} \right) \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho_{\vec{q}}(t) +  \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \frac{ \mathrm{d} \rho_{\vec{q}}}{\mathrm{d} t }  \ \mathrm{d}^3 \vec{q} &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left hand term in the integrand matches the Fourier transform of $H (\vec{X} \cdot \nabla) \rho$. Which is why the terms cancel. &lt;/p&gt;&#xA;" OwnerUserId="7053" LastActivityDate="2015-05-08T19:36:33.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10659" PostTypeId="2" ParentId="10642" CreationDate="2015-05-09T22:26:59.803" Score="2" Body="&lt;p&gt;The answer to your question for that specific planet is No. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no way at this point in time to tell if Kepler-442b is a big barren rock or a water world or like earth for that matter. They only know it is there because it passes in front of the star it is orbiting and that dims the light. That is the only reason they know it is there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This planet is not visible.&lt;/p&gt;&#xA;" OwnerUserId="7226" LastEditorUserId="7226" LastEditDate="2015-05-09T22:48:00.333" LastActivityDate="2015-05-09T22:48:00.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10660" PostTypeId="1" AcceptedAnswerId="10662" CreationDate="2015-05-10T17:34:36.753" Score="5" ViewCount="1059" Body="&lt;p&gt;&lt;a href=&quot;http://www.space.com/14075-10-biggest-telescopes-earth-comparison.html&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt; makes the claim that the Giant Magellan Telescope (GMT, number 4 in the list) will have resolution 10 times better than that of Hubble, while the Thirty Meter Telescope (TMT, number 3 in the list) will have resolution 12 times better than that of Hubble.  These are claims I find impossible to believe.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple application of, for instance, the Raleigh Criterion ($\theta = 1.22~\lambda/D$), where $\theta$ is the angular resolution of the telescope, $\lambda$ is the wavelength of the light in question, and $D$ is the diameter of the telescope, would show that a comparison between angular resolutions is as simple as comparing the diameters of the telescope apertures.  Hubble's mirror is 2.4 m; comparing with the 24.5 m GMT and the 30 m TMT we see that the GMT will have 10.2 times the angular resolution of Hubble, while the TMT will have 12.5 times the angular resolution of Hubble.  I think a calculation similar to the one I have described is how the article linked above came up with the numbers they did about the angular resolution of these telescopes compared to Hubble. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the Raleigh criterion only applies to telescopes working at their diffraction limit.  Space telescopes (if they're designed well and built correctly) can work close to the diffraction limit (maybe even at the diffraction limit).  Ground-based telescopes, however, are limited in angular resolution by the atmosphere, which at best will limit resolution to about an arc-second at the best sites on Earth. Thus the GMT and TMT by themselves will &lt;strong&gt;not&lt;/strong&gt; have better image resolution than Hubble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question, then, is whether this article is correct (possibly because of one of the reasons I list below) or whether it seems this article naively applied the Raleigh Criterion for angular resolution with no thought about how the atmosphere will affect the resolving capabilities of these large ground-based telescopes.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possible reasons the article may still be correct:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Adaptive optics, a developing technology which can allow telescopes to correct for distortions to the image produced by the atmosphere.  Perhaps GMT and TMT will have very fancy adaptive optics systems.&lt;/li&gt;&#xA;&lt;li&gt;Another technique, such as speckle imaging or lucky imaging.&lt;/li&gt;&#xA;&lt;li&gt;Article could be referring to &lt;em&gt;spectral&lt;/em&gt; resolution, rather than &lt;em&gt;angular&lt;/em&gt; resolution (or image resolution).  However, obtaining good spectral resolution is as much the job of the spectral instrument as it is of the telescope so I don't consider &quot;spectral resolution&quot; to be an inherent property of a telescope.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="128" LastEditorUserId="37191" LastEditDate="2021-03-26T10:27:59.690" LastActivityDate="2021-03-26T10:27:59.690" Title="Claim that 30-m class telescopes will have resolution far superior to Hubble: true?" Tags="&lt;telescope&gt;&lt;hubble-telescope&gt;&lt;angular-resolution&gt;&lt;giant-magellan-telescope&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10661" PostTypeId="2" ParentId="10656" CreationDate="2015-05-10T20:08:18.413" Score="2" Body="&lt;p&gt;Only by knowing what kind of a star it is, or by inferring that information from the available colours. From there you use a photometric or spectroscopic parallax calibration. ie. There is a relationship between the absolute magnitude and colours of a main sequence star. Of course there is no guarantee you are looking at a main sequence star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get more idea by cross correlating with the 2MASS JHK catalogue, as the J-H vs H-K colours of giants are usually distinctive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specific help needs a more specific question.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-05-10T20:08:18.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10662" PostTypeId="2" ParentId="10660" CreationDate="2015-05-11T03:25:27.287" Score="5" Body="&lt;p&gt;Adaptive optics, as it says on the &lt;a href=&quot;http://www.gmto.org/overview/&quot; rel=&quot;nofollow&quot;&gt;GMT website&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One of the most sophisticated engineering aspects of the telescope is&#xA;  what is known as “adaptive optics.” The telescope’s secondary mirrors&#xA;  are actually flexible. Under each secondary mirror surface, there are&#xA;  hundreds of actuators that will constantly adjust the mirrors to&#xA;  counteract atmospheric turbulence. These actuators, controlled by&#xA;  advanced computers, will transform twinkling stars into clear steady&#xA;  points of light. It is in this way that the GMT will offer images that&#xA;  are 10 times sharper than the Hubble Space Telescope.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and on the &lt;a href=&quot;http://www.tmt.org/science-case&quot; rel=&quot;nofollow&quot;&gt;TMT web&lt;/a&gt; site:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In addition to providing nine times the collecting area of the current&#xA;  largest optical/infrared telescopes (the 10-meter Keck Telescopes),&#xA;  TMT will be used with adaptive optics systems to allow&#xA;  diffraction-limited performance, i.e., the best that the optics of the&#xA;  system can theoretically provide. This will provide unparalleled&#xA;  high-sensitivity spatial resolution more than 12 times sharper than&#xA;  what is achieved by the Hubble Space Telescope. For many applications,&#xA;  diffraction-limited observations give gains in sensitivity that scale&#xA;  like the diameter of the mirror to the fourth power, so this increase&#xA;  in size has major implications.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="128" LastEditDate="2015-05-12T06:10:11.470" LastActivityDate="2015-05-12T06:10:11.470" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10665" PostTypeId="2" ParentId="10630" CreationDate="2015-05-11T18:47:49.760" Score="1" Body="&lt;p&gt;There are a number of flaws with your idea (despite the relativity breaking transatlantic journey times that would be possible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inflation is a popular theoretical model put forward to solve three main issues with the Big Bang cosmology: the horizon problem, the flatness problem &amp;amp; the magnetic monopole problem (see the &lt;a href=&quot;http://en.wikipedia.org/wiki/Inflation_%28cosmology%29&quot; rel=&quot;nofollow&quot;&gt;wiki&lt;/a&gt; for details). It consists of a scalar field called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Inflaton&quot; rel=&quot;nofollow&quot;&gt;inflaton&lt;/a&gt;, a physical field that pervades all space. During inflation it undergoes a phase transition to a lower energy state, releasing huge amounts of energy that drive the expansion. To force the field back to a higher energy state would require inconceivable amounts of energy, and have to be applied across the entire universe. Not only is this impossible, but also annoyingly prevents its use for local galactic travel.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Or maybe a technically very advanced civilization could keep its solar system inflated by the time of a deflation of the whole universe &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I assume you mean contraction, as opposed to the current expansion that we see today due to dark energy. Again this would be unfeasible due to the reasons outlined above, and if contraction was to proceed to it's conclusion, i.e. a singularity like the Big Bang, then no information from this universe could be carried through that singularity, even if there was another universe on the other side.&lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2015-05-11T18:47:49.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10666" PostTypeId="2" ParentId="6052" CreationDate="2015-05-11T19:45:39.887" Score="7" Body="&lt;p&gt;I think it's a fun question, if impossible.    The only way to turn Jupiter into a star that's even remotely practical is to add to it's mass.   Ignoring brown dwarfs that are very limited in energy output, to get a red dwarf going, you'd need to add at least 75-80 or so Jupiter masses.   (a bit more than 24,000 earth masses).   You'd want to add a fair percentage of hydrogen, but some rocky debris wouldn't hurt the mix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, assuming the impossible is done, there's several things to consider.   The greater gravity (75-80 times) would significantly alter all the planets orbits.   Predicting exactly how is hard, but that much more mass and the planets orbits, certainly all the inner ones, would wobble a lot more and some might get pulled completely out of their orbit, likely thrown out of the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might think that the planets nearer to Jupiter would be the most affected, but it really has more to do with tidal synch than anything else.  Any of the 4 inner planets could get tugged into a new orbit.   You'd also likely see the earth's orbit elongate in resonance with Jupiter perhaps increasing the ice age/ice melt cycle.   Precise answers are hard, and none of these things would happen over 1 orbit, but over time, certainly.   Orbital changes to all the inner planets and perhaps Saturn as well would be inevitable if Jupiter becomes a red-dwarf.   Imagine if Saturn was pulled closer to the earth, into an orbit between Mars and Jupiter, or Mercury was pulled out past the earth.  Odds are it wouldn't hit us, but we might want to keep an eye on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stability_of_the_Solar_System#Mercury.E2.80.93Jupiter_1:1_resonance&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Stability_of_the_Solar_System#Mercury.E2.80.93Jupiter_1:1_resonance&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd thing to consider is magnetism and solar flares.   Young stars tend to spin very fast due to conservation of angular momentum when the stars form and this creates enormous magnetic fields and huge solar flares, much bigger than we get from the sun.   It's strange to think that a tiny red dwarf, 4 times as far from our sun as the sun would create solar flares to worry about but it's possible.  Whether it would need high angular momentum for this to happen, I'm not sure, but we could see larger solar flares from star-Jupiter than from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Flare_star&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Flare_star&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brightness, heat and visibility was covered above, but I'll touch on that.   Brightness of -6.25 would be 5-6 times brighter than Venus and you'd see it at night, Venus isn't seen in peak darkness, so it would be significantly brighter than any other star/planet in the sky, but significantly less bright than the moon, like, you couldn't make your way with just that star's light the way you can see things around you in moonlight.   But when I run the numbers, I think it would be quite a bit brighter than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass to Luminosity is to the power of 3.5 - quick estimate, so, lets say the red dwarf has a mass of 80 Jupiters.  That's 0.076 Suns.    0.076^3.5 = about 1/8,000, so 4.2 times as far away at closest point (square of that), 1/8000th as bright, we're looking at 1/140,000 times the light we get from the sun - not very much and likely less than that in it's early stages and because the smaller stars tend to fall off, so lets estimate 1/200,000 - 1/300,000 the apparent brightness of the sun as a ballpark estimate.   That's not enough to heat the earth at all, but that's still brighter, (a little bit) than the full moon which is about 1/400,000 the brightness of the sun.   it would be enough light to see your way around, but I wouldn't want to try to read by it.  It would also be distinctly reddish light.  Not the white light we're used to getting from the day or night sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally size - a red-dwarf star of 80 Jupiter masses would actually be slightly smaller than Jupiter due to the gravitation so it would appear like a planet - not quite a point in the sky, but almost a point, but a bit brighter than the full moon and red.   That's likely bright enough to see during the day too.    I don't think it would be hard to look at or hurt your eyes, but it would shine like a tiny bright red flashlight in the distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.space.com/21420-smallest-star-size-red-dwarf.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.space.com/21420-smallest-star-size-red-dwarf.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think I like star-Jupiter.    Lets not plan on doing this.   :-)&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-05-11T19:45:39.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10667" PostTypeId="2" ParentId="10660" CreationDate="2015-05-11T21:32:14.147" Score="3" Body="&lt;p&gt;The claims that you have seen refer to the possibility of attaining near diffraction-limited images using adaptive optics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both telescopes will feature wavefront sensors that can apply corrections to the telescope optics using either natural guide stars or laser guide stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the comparison is not quite what it seems. The AO systems on both telescopes will operate in the near infrared only. The HST of course operates over a broader wavelength range, from the UV to the near infrared. So, in the near infrared then indeed these new telescopes will provide images with a sharpness that should greatly exceed the NIR images from HST or JWST. However, it appears that at visible wavelengths that will not be the case.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-05-11T21:32:14.147" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10668" PostTypeId="1" CreationDate="2015-05-12T04:56:44.540" Score="1" ViewCount="60" Body="&lt;p&gt;In the two body problem, if each body is given an initial condition as a momentum vector, what is the transient part of the solution as it settles into the steady state orbit?&lt;/p&gt;&#xA;" OwnerUserId="7236" LastActivityDate="2015-11-03T17:19:35.527" Title="Two body transient solution" Tags="&lt;orbit&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10673" PostTypeId="2" ParentId="2572" CreationDate="2015-05-12T20:22:17.283" Score="-1" Body="&lt;p&gt;The OP is obviously correct.  If you knew an asteroid was headed for Earth and had the choice of either doing nothing (and letting it strike) or cutting it in half and letting both halves strike, you would split it.&#xA;Nuking it into a million pieces would spread out the impact, which would make the collision less dangerous.  Would you rather get shot in the chest with a bullet or 100 BBs?&lt;/p&gt;&#xA;" OwnerUserId="7242" LastActivityDate="2015-05-12T20:22:17.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10674" PostTypeId="1" AcceptedAnswerId="10675" CreationDate="2015-05-13T02:56:08.110" Score="4" ViewCount="638" Body="&lt;p&gt;A scientist at CERN LHC (Large Hadron Collider) was asked during an interview about the possible results and consequences of operating the LHC at full power of collision in the near future, one of the results he literally said :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One of the theories that predicts the consequences of operating the LHC at full power collision is that mini black holes might form, but do not be afraid of those black holes, they will not start expanding and destroy the earth, they will instantaneously disappear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that black holes create disturbance in space-time. And they actually appear when matter massively gets pressurized into a small size of volume in space which causes gravity to increase insanely which in turn initialize a black hole that even light its self can not escape away from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how can a mini black hole disappear anyway?&#xA;What in the world would cause a small size, a mini size, a tiny little black hole to vanish or disappear?&lt;/p&gt;&#xA;" OwnerUserId="5512" LastActivityDate="2015-11-01T13:25:19.190" Title="Mini black holes, a possible result of the LHC experiment at CERN" Tags="&lt;black-hole&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="1" ClosedDate="2020-02-11T12:24:55.517" ContentLicense="CC BY-SA 3.0" />
  <row Id="10675" PostTypeId="2" ParentId="10674" CreationDate="2015-05-13T04:58:02.813" Score="7" Body="&lt;p&gt;This result is due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;noreferrer&quot;&gt;Hawking radiation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to random quantum fluctuations, particles are radiated from a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the black hole has a mass less than the Moon, it will &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_hole#Evaporation&quot; rel=&quot;noreferrer&quot;&gt;evaporate&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should add that any black holes created can only have the mass of the particles involved in the collision - two protons. So it's not going to do any harm before it evaporates. However, we &lt;em&gt;would&lt;/em&gt; learn a lot from the event.&lt;/p&gt;&#xA;" OwnerUserId="5524" LastEditorUserId="5524" LastEditDate="2015-05-14T09:28:54.970" LastActivityDate="2015-05-14T09:28:54.970" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10677" PostTypeId="1" AcceptedAnswerId="10679" CreationDate="2015-05-13T06:54:46.440" Score="1" ViewCount="515" Body="&lt;p&gt;Is the calculated age of the universe that of the &lt;em&gt;visible&lt;/em&gt; universe or the entire universe? I dont know how the age is calculated but if it is believed that all we see visibly IS the entire universe then that means it is finite. If it is not finite then the light from the really far astronomcical objects has not reached us yet so there is a possibility it is infinite.&#xA;Can the method used to calculate the universe age since the Big Bang guarantee that it has an age or is it possible that it existed infinitely?&lt;/p&gt;&#xA;" OwnerUserId="7246" LastEditorUserId="7246" LastEditDate="2015-05-13T07:00:11.297" LastActivityDate="2015-05-13T12:34:13.730" Title="Calculating the age of the universe" Tags="&lt;universe&gt;&lt;big-bang-theory&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10678" PostTypeId="2" ParentId="10677" CreationDate="2015-05-13T09:38:44.730" Score="2" Body="&lt;p&gt;The simplest assumption about the global properties of the Universe is that it looks the same outside the part that is observable to us, as it does inside. That is, we see a finite part of a Universe that is (probably) infinite in extend. If so, then the calculated age &amp;mdash; which is finite &amp;mdash; applies to &lt;em&gt;all&lt;/em&gt; of the Universe, not just the observable part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The age is calculated on the basis of the observed expansion rate, and the observed densities of the constituents of the Universe. &lt;strong&gt;It is possible to imagine a universe with the right mixture of constituents that has existed forever$^{\dagger}$, but for our particular Universe, this just doesn't seem to be the case; it is ruled out by observations&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a first-order approximation, you can simply take the age $t_\mathrm{Uni}$ to be the reciprocal of the expansion rate $H_0 = 70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1} = 2\times10^{-18}$ s. That is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t_\mathrm{Uni} \sim \frac{1}{H_0} = 14\,\mathrm{billion\,years}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this assumes that the Universe has been expanding at the same rate throughout its entire history, which is hasn't. More generally, the age is calculated from integrating (numerically except for simplified approximations) the &lt;a href=&quot;http://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;nofollow&quot;&gt;Friedmann equation&lt;/a&gt;, yielding &lt;a href=&quot;http://en.wikipedia.org/wiki/Age_of_the_universe#Planck&quot; rel=&quot;nofollow&quot;&gt;13.819 billion years&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should say that the calculated age is the time from the Big Bang till now. I guess the safest thing to say is that we don't know what happened the first tiny fraction of a second or so after creation, and in principle it could have existed before this instant, collapsed, and then re-expanded. But no observations I know of suggest this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$An example of a temporally infinite universe is one containing energy only in the form of a cosmological constant. In this case, the Friedmann equation reduces to $da/dt=aH_0$, with $a$ the scale factor (&quot;size&quot;) of the universe, the solution of which is an exponential function with zero size only at $t = -\infty$.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-05-13T10:06:46.917" LastActivityDate="2015-05-13T10:06:46.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10679" PostTypeId="2" ParentId="10677" CreationDate="2015-05-13T12:34:13.730" Score="3" Body="&lt;p&gt;It is a common point of confusion, but an infinite universe can still have been at a unique singularity in space-time at a finite time in the past. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This singularity is not a &quot;point in space&quot;, since every point in the universe participated in the big-bang. It is merely a time at which the scale factor, which multiplies the separation between all spatial points in the universe,  was zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in cosmology governed by the Friedmann equation, the age of the universe thus calculated is the same whether you are talking about the &quot;visible&quot; universe or an infinite universe. The light travel time really doesn't come into this, because different parts of even a very large universe can be causally separated yet still have originated at the same singularity.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-05-13T12:34:13.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10680" PostTypeId="2" ParentId="10642" CreationDate="2015-05-13T14:20:27.060" Score="2" Body="&lt;p&gt;By viewing the star and the amount of cycles it went through we can determine what access material are available within its solar system. Next we look at the habitat zone of that star and the size of the planet. It the planet is in the habitat zone then there is a chance for liquid water. If it is also roughly the size of Earth or a little larger it can hold onto some lighter gases that allow us to live on Earth. If the planet is much larger it can hold onto some heavier gases that might make life on the planet impossible. Next given the age of the star and planet we can see how cool the planet has become.&#xA;A professor at Iowa State University is on the Steering Committee for the Kepler Asteroseismology Research Consortium and he adds that if you want a view of what the surface looks like that it becomes trickier. &quot;We do have some 'maps' from the phase curve of the planet  - how the combined brightness changes as the planet goes from primary transit, through quadrature, to secondary transit. If the planet's reflectivity is uneven, that can show up in the light curve if you have enough sensitivity&quot;&lt;/p&gt;&#xA;" OwnerUserId="7249" LastActivityDate="2015-05-13T14:20:27.060" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10681" PostTypeId="2" ParentId="2572" CreationDate="2015-05-13T23:45:14.257" Score="0" Body="&lt;p&gt;This is from Phil Plait's &quot;Bad Astronomy&quot; page in which he dissects some of the scenes from the movie &quot;Deep Impact.&quot;  In the interest of science and as long as I give him proper credit, I'm sure he doesn't mind my quoting him.  Here's the link if you want to read the whole thing yourself: &lt;a href=&quot;http://www.badastronomy.com/bad/movies/di2.html&quot; rel=&quot;nofollow&quot;&gt;http://www.badastronomy.com/bad/movies/di2.html&lt;/a&gt;&lt;br&gt;&lt;blockquote&gt;&#xA;Bad: Minutes before final impact, the astronauts blow up the second comet, and we are treated to a spectacular light show. &#xA;Good: Aaaaarrgg! This was the Biggest Baddest Astronomy in the movie. Blowing up a comet does no good at all, and might even make matters worse. Just because the pieces are smaller doesn't mean you have changed anything. If every piece still impacts the Earth (by that I mean actually is stopped by the Earth or its atmosphere) you are still dumping all the kinetic energy of The Comet into the Earth's atmosphere! That's a HUGE amount of energy, dumped in practically all at once. It would still create a massive explosion, dwarfing all of our nuclear bombs combined. Even if you could somehow soften the blow, all that heat would wreak havoc with our weather. Some people actually think it might be better to simply let a big one hit rather than blow it up, because the Earth itself can absorb the energy of impact better than the atmosphere can. This is still argued, though. I'd prefer not to try any experiments!&lt;/blockquote&gt;&lt;/p&gt;&#xA;" OwnerUserId="7108" LastEditorUserId="7108" LastEditDate="2015-10-01T18:16:13.177" LastActivityDate="2015-10-01T18:16:13.177" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10682" PostTypeId="2" ParentId="10373" CreationDate="2015-05-14T00:49:08.287" Score="6" Body="&lt;p&gt;Interstellar's &quot;Miller's Planet&quot; is utter rubbish...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, black holes don't start out as black holes. Black holes form at the end of the life cycle of very big star (at least 25 solar masses but more frequently over 35 or 40 SMs) when it goes supernova or hypernova. Any planets in orbit around that star are going to be obliterated before the black hole becomes a black hole. We need to remember that black holes are not stars which happen to be so dense light cannot escape their gravity. Black holes are the REMNANTS of stars. And, the process of becoming a black hole obliterates planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, that planet in the movie orbiting a black hole had liquid water. There is no such thing as a goldilocks zone around a black hole like you have around a star where water is not frozen and not boiled off. That means it must be just the right distance to a heat source. The black hole isn't going to give it that heat or light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, for that &quot;planet&quot; to have gravitational time dilation equivalent to 7 years per earth hour, it'll have to be so close to the black hole that mile high waves will be the last thing you worry about. If you are not fried by radiation in seconds if you are that close. In fact, the planet will be ripped apart, pulverized, irradiated and basically become part of the (glowing) accretion disc depicted in the movie. That disc, BTW, can be light years across in massive black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, hey, it's a movie and it's supposed to be entertainment not science. So I can forgive all that. What I could not forgive was the blatant stupidity in the plot, such as the crew KNOWING about the time dilation but not knowing that Miller would have just landed even though they had gotten his signal for many years. They landed, had their harrowing encounter with the wave and THEN realize that as far as Miller is concerned he had just landed and only just died!?!! Bugs Bunny Science can nonetheless be entertaining. But, silly and illogical plots make a movie hard to follow.&lt;/p&gt;&#xA;" OwnerUserId="7254" LastEditorUserId="11622" LastEditDate="2019-01-30T07:43:11.697" LastActivityDate="2019-01-30T07:43:11.697" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="10683" PostTypeId="2" ParentId="10674" CreationDate="2015-05-14T01:23:27.547" Score="2" Body="&lt;p&gt;Mini black holes are thought not to exist, so it's an unlikely idea anyway, and Hawking Radiation is largely accepted (but I think, not proven) . . . so maybe, &quot;what if&quot; concerns very unlikely but slightly valid, but there's also this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get particle collisions faster than Cern can produce in our upper atmosphere every day.    If there was any danger of high energy collisions, We'd have likely observed it already either in our atmosphere or by Hubble.    The fact that high energy collisions happen every day on every planet, every good sized moon and every star in the galaxy suggests that high energy collisions don't form into matter eating strangelets or tiny but hungry black holes.   If they did, we'd have seen some evidence of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have both very strong observational evidence as well as theoretical evidence.    CERN isn't playing with particle energy that doesn't happen - it happens all the time in space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for fun - the Oh-my-god particle, way more energy energy than anything that happens in CERN:   &lt;a href=&quot;http://en.wikipedia.org/wiki/Oh-My-God_particle&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Oh-My-God_particle&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-05-14T01:23:27.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10684" PostTypeId="1" AcceptedAnswerId="10687" CreationDate="2015-05-14T03:49:48.840" Score="6" ViewCount="576" Body="&lt;p&gt;A question came up today in astronomy class and it got me wondering all night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is it that if stars are so far away, we can see them so clearly but we cannot see astronauts on the moon.&lt;/p&gt;&#xA;" OwnerUserId="7255" LastActivityDate="2015-05-14T13:24:46.283" Title="Why can we see stars but not astronaut on the moon" Tags="&lt;star&gt;&lt;the-moon&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10687" PostTypeId="2" ParentId="10684" CreationDate="2015-05-14T11:27:50.960" Score="13" Body="&lt;p&gt;An astronaut on the moon could only be seen by reflecting the Sun's light towards Earth. Stars on the other hand emit their own light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To first order, the amount of flux incident upon the Moon from the Sun is the same as that at the Earth - about 1.4 kW/m$^{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us assume that an astronaut is &lt;em&gt;perfectly reflective&lt;/em&gt; and that the relevant reflective area that we can see from the Earth is 1 m$^2$. NB: If the astronaut is not lit up by the Sun, then there is obviously no way that they can be seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Treating the astronaut on the Moon as an isotropic point source emitter of reflected light, we have a light source of power 1.4 kW at a distance of 400,000 km. The flux at the Earth is therefore $7\times10^{-16}$ W m$^{-2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does that compare with starlight? Well, the total luminosity of the Sun is $3.8\times10^{26}$ W. It has an absolute magnitude of 4.8. This means that if we put the Sun at a distance of about 20 pc, it would be about as faint as the faintest naked eye star in the sky. The flux received at the Earth from such a star would be $8\times 10^{-11}$ W m$^{-2}$ and thus 100,000 times brighter than the astronaut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No need to worry about the resolution of the eye, since both the star and the astronaut (at the distance of the moon) are unresolved points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also no need to go into the problems of contrast against the moon's bright surface (which you &lt;em&gt;would&lt;/em&gt; need to consider if an astronaut's reflective area was 100,000 times bigger), the reflected light from the astronaut is just too faint to be seen at that distance.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-05-14T11:27:50.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10688" PostTypeId="2" ParentId="10668" CreationDate="2015-05-14T19:49:03.980" Score="1" Body="&lt;p&gt;Your question is very unclear.  As long as the orbit is bound (i.e. the total energy is negative), then there is a unique closed orbit solution (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Bertrand%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Bertrand's theorem&lt;/a&gt;).  There will be no 'transient' part of the solution.&lt;/p&gt;&#xA;" OwnerUserId="918" LastActivityDate="2015-05-14T19:49:03.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10689" PostTypeId="1" AcceptedAnswerId="10693" CreationDate="2015-05-15T00:17:37.583" Score="7" ViewCount="484" Body="&lt;p&gt;Are the stars we can see with the naked eye in the night sky only from our Milky Way galaxy or can we see stars from Andromeda? I am aware we can see other celestial objects like nebulas and the Andromeda Galaxy itself. &lt;/p&gt;&#xA;" OwnerUserId="7258" LastEditorUserId="1678" LastEditDate="2015-05-15T07:52:36.703" LastActivityDate="2015-05-15T08:07:38.810" Title="Stars in the sky" Tags="&lt;star&gt;&lt;amateur-observing&gt;&lt;naked-eye&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10691" PostTypeId="1" CreationDate="2015-05-15T03:41:14.057" Score="5" ViewCount="3139" Body="&lt;p&gt;Which is more rare: Lunar eclipse or Solar eclipse? Please explain why either one is more rare than the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance for your inputs.&lt;/p&gt;&#xA;" OwnerUserId="7260" LastActivityDate="2015-05-15T13:48:23.357" Title="Which is more rare: Lunar eclipse or Solar eclipse?" Tags="&lt;fundamental-astronomy&gt;&lt;solar-eclipse&gt;&lt;lunar-eclipse&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10692" PostTypeId="2" ParentId="10691" CreationDate="2015-05-15T05:23:56.697" Score="2" Body="&lt;p&gt;They are not that rare; there are &lt;a href=&quot;http://www.timeanddate.com/eclipse/list.html&quot; rel=&quot;nofollow&quot;&gt;4 to 6&lt;/a&gt; solar and lunar eclipses each year with an equal number of each (on average). &lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2015-05-15T05:23:56.697" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10693" PostTypeId="2" ParentId="10689" CreationDate="2015-05-15T07:00:58.393" Score="6" Body="&lt;p&gt;The resolved stars (those that can be seen as individuals) are all part of the Milky Way Galaxy (unless there are any interlopers that have been captured!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distances to the next nearest galaxies of any size are more than 100,000 light years. Andromeda is 2 million light years away. Unless one goes supernova, there basically aren't any types of star that are intrinsically bright enough to be seen with naked eye at these distances. Most of the stars we do see are at distances of one AU (the Sun) to a few thousand light years, with a median of a few hundred light years (See &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6005/arent-there-more-naked-eye-visible-stars-in-the-milky-way-plane?rq=1&quot;&gt;Aren&amp;#39;t there more naked-eye-visible stars in the Milky Way plane?&lt;/a&gt; for more detail).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-05-15T08:07:38.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10694" PostTypeId="2" ParentId="10691" CreationDate="2015-05-15T08:51:13.103" Score="9" Body="&lt;p&gt;Solar and lunar eclipses occur about equally often - between two and four times per year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when you do not intentionally travel around the world chasing solar eclipses, you are more likely to observe more lunar eclipses. The reason is that solar eclipses can only be observed from a comparatively small area while lunar eclipses can be observed from anywhere were the moon is visible when they occur, which is half the planet.&lt;/p&gt;&#xA;" OwnerUserId="7261" LastEditorUserId="7261" LastEditDate="2015-05-15T10:09:28.330" LastActivityDate="2015-05-15T10:09:28.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10695" PostTypeId="2" ParentId="10691" CreationDate="2015-05-15T10:15:45.967" Score="1" Body="&lt;p&gt;The &lt;em&gt;reason&lt;/em&gt; that solar and lunar eclipses are about equally common, is that they occur when the moon is &quot;close enough&quot; to the plane of the ecliptic at the point where it is full (lunar eclipse) or new (solar eclipse). Otherwise it completely misses the sun (solar eclipse) or the earth's shadow (lunar eclipse), and you don't get even a partial eclipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the moon orbited precisely in the ecliptic plane, then there would be a total lunar eclipse every full moon, and every new moon a solar eclipse that's either total or annular according to the distance of the moon from the earth at that moment. But the moon doesn't orbit in that plane, it intersects it twice a month, at a time that &lt;em&gt;most&lt;/em&gt; months isn't when it's close enough to being lined up with the earth and sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Close enough&quot; in the case of a solar eclipse is when the earth, as viewed from somewhere on the sun, is partly obscured by the moon (therefore: part of the sun is obscured by the moon when viewed from part of the earth). Whereas &quot;close enough&quot; for a lunar eclipse is when the moon is partly obscured by the earth from the somewhere on the sun (and therefore part of the earth's shadow falls on part of the moon). This is just about the same size target area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the sun &quot;sees&quot; as the moon orbits the earth is similar to what we see when we look at the moons of Jupiter: the moon moves from one side to the other and back. Sometimes it passes &quot;over&quot; the earth, sometimes &quot;under&quot;, sometimes through. What's required for a partial eclipse of either kind is that it moves through, such that those two disks intersect as viewed from the sun. The difference from Jupiter, is that Jupiter is &lt;em&gt;really&lt;/em&gt; big, so passing through is more likely for the moons of Jupiter than our moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really want to know which kind is a few percentage points more or less frequent, then you're beyond my ability to predict. Consult a table of eclipses covering a long period of time, or wait for an answer from someone with better qualifications :-)&lt;/p&gt;&#xA;" OwnerUserId="6953" LastEditorUserId="6953" LastEditDate="2015-05-15T13:42:56.670" LastActivityDate="2015-05-15T13:42:56.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10696" PostTypeId="2" ParentId="10691" CreationDate="2015-05-15T13:48:23.357" Score="2" Body="&lt;p&gt;The answer is &quot;it depends.&quot; It depends on what kinds of eclipses you count, and perhaps even more importantly, how likely you are to see the event.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are four types of solar eclipses:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Total solar eclipses, in which the Moon blocks the entirety of the Sun at some point on the surface of the Earth. About 26.7% of the solar eclipses are total eclipses.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Annular solar eclipses, in which the center of the Moon as viewed from the passes close to the center of the Sun, leaving an annular ring of sunlight around the Moon. About 33.2% of the solar eclipses are annular eclipses.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Hybrid solar eclipses, in which the eclipse changes from total to annular (or vice versa) along the path of the eclipse. About 4.8% of the solar eclipses are hybrid eclipses.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Partial solar eclipses, in which no part of the Earth sees a total or annular eclipse but some part of the Earth does see the Moon as partially blocking a portion of the Sun. About 35.3% of the solar eclipses are partial eclipses.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(Source for numbers: &lt;a href=&quot;http://eclipse.gsfc.nasa.gov/SEcat5/SEcatalog.html&quot; rel=&quot;nofollow&quot;&gt;http://eclipse.gsfc.nasa.gov/SEcat5/SEcatalog.html&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;There are three types of lunar eclipses:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Total lunar eclipses, in which the entirety of the Moon receives no sunlight because the Earth is between the Moon and the Sun. About 28.8% of the lunar eclipses are total eclipses.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Partial lunar eclipses, in which the part of the Moon receives no sunlight because the Earth is between the Moon and the Sun, but there is no time at which all of the Moon receives no sunlight. About 34.9% of the lunar eclipses are partial eclipses.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Penumbral lunar eclipses, in which the part of the Moon sees the Sun as partially eclipsed by the Earth, but no part of the Moon sees the Sun as totally eclipsed by the Earth. About 36.3% of the lunar eclipses are penumbral eclipses.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(Source for numbers: &lt;a href=&quot;http://eclipse.gsfc.nasa.gov/LEcat5/LEcatalog.html&quot; rel=&quot;nofollow&quot;&gt;http://eclipse.gsfc.nasa.gov/LEcat5/LEcatalog.html&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;From the above two sources, there have been / will be 12064 lunar eclipses and 11898 solar eclipses between 2000 BCE and 3000 CE. These are essentially the same number. In this sense, solar eclipses and lunar eclipses are equally likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regard to the likelihood of your seeing an eclipse, lunar eclipses are much more likely than solar eclipses. For any given lunar eclipse, a bit over half the world gets to see at least part of the event. For any given solar eclipse, only a tiny portion of the world gets to see part of the event, and for most of them, it's just a partial dimming.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-05-15T13:48:23.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10697" PostTypeId="1" AcceptedAnswerId="10717" CreationDate="2015-05-15T15:40:16.923" Score="2" ViewCount="141" Body="&lt;p&gt;I have a galaxy 'A', say at redshift 1. Let's suppose this galaxy has no peculiar velocity. What would be the redshift of another galaxy 'B', that has a radial velocity of 500 km/s, relative to A?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this in the standard (Planck) model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any tools to do get this easily (an python code, astropy?)&lt;/p&gt;&#xA;" OwnerUserId="1092" LastEditorUserId="9301" LastEditDate="2016-06-29T04:38:16.393" LastActivityDate="2016-06-29T04:38:16.393" Title="Difference in redshift between 2 interacting galaxies" Tags="&lt;galaxy&gt;&lt;cosmology&gt;&lt;python&gt;&lt;astropy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10698" PostTypeId="1" CreationDate="2015-05-15T19:42:25.860" Score="1" ViewCount="4980" Body="&lt;p&gt;Up to last year, I never got blinded by the sun in everyday situations (e.g. running towards it), at no time of the year I was. But since last christmas, it occured to me more often that I cannot watch in the direction where the sun is because it is so bright that it really hurts me.&#xA;My mother and my roommate have observed the same phenomenon independently of me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, of course, is only anectodal evidence, which is practically worthless. We all get older and we may have the same kind of &quot;photophobia&quot;, or we just forgot about being blinded the last few years and just now consciously felt an remembered it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore I want to ask this question: is there any objectively observable increase of brightness over the last few months and if so, why?&lt;/p&gt;&#xA;" OwnerUserId="7268" LastActivityDate="2018-09-04T16:06:02.990" Title="Has the sun become way brighter the last years?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10699" PostTypeId="2" ParentId="10698" CreationDate="2015-05-15T21:22:34.330" Score="19" Body="&lt;p&gt;The Sun's luminosity is stable to 0.1% - it varies slightly in response to, or associated with, levels of solar magnetic activity (&lt;a href=&quot;https://arxiv.org/abs/1210.5911/&quot; rel=&quot;noreferrer&quot;&gt;Solanki &amp;amp; Unruh 2012&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This variation would not be perceptible and is completely outweighed by the $\pm 3.4$% variation caused by the non-circular orbit of the Earth (the Sun is closest in January). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, if you live in the northern hemisphere, then the Sun is generally higher in the sky now (in May) than it was in December. Its direct light is less absorbed/scattered by the atmosphere. This is a much more perceptible effect. It could change the visual brightness by tens of per cent, and by even more in the short wavelength blue part of the spectrum. That is the only physics-based explanation I can offer.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-06-06T21:30:37.903" LastActivityDate="2017-06-06T21:30:37.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10700" PostTypeId="2" ParentId="3634" CreationDate="2015-05-16T00:44:24.193" Score="6" Body="&lt;p&gt;This obviously depends on which stars you choose, but in general Ursa Major is likely to win by a factor of about five(ish), since it occupies a &lt;a href=&quot;http://en.wikipedia.org/wiki/88_modern_constellations_by_area&quot; rel=&quot;nofollow noreferrer&quot;&gt;much larger chunk of the sky&lt;/a&gt; (1280 square degrees) than Ursa Minor (256 square degrees). At a fixed magnitude, and therefore at a fixed visibility radius for each intrinsic luminosity, the volume will be (very roughly) proportional to the area of each constellation on the celestial sphere. (There are subtleties, such as &lt;a href=&quot;https://physics.stackexchange.com/questions/164720/the-fainter-the-star-the-further-away-it-is&quot;&gt;these&lt;/a&gt;, and this only rigorously holds when there are many stars, but it is a good rule of thumb.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The volumes themselves can quite easily be calculated in Mathematica using the curated data. There is one easy way, which selects what Wolfram Research thinks are the 'bright stars' in each constellation. If you do that, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{Vol}(\mathrm{UMa})=9.20133\times 10^{20}\mathrm{AU}^3,&#xA;\ \mathrm{Vol}(\mathrm{UMi})=1.7105\times 10^{20}\mathrm{AU}^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the volume of the convex hull of the stars shown here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pbjwL.png&quot; alt=&quot;Mathematica graphics&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have Mathematica (v10+ only, I think) and you want to tinker with the code, here it goes:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;RegionMeasure[&#xA;   ConvexHullMesh[&#xA;    StarData[&#xA;     ConstellationData[Entity[&quot;Constellation&quot;, #], &#xA;      EntityProperty[&quot;Constellation&quot;, &quot;BrightStars&quot;]], &#xA;     &quot;HelioCoordinates&quot;]]&#xA;] &amp;amp; /@ {&quot;UrsaMajor&quot;, &quot;UrsaMinor&quot;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What one should really do is set a threshold magnitude and only count stars brighter than this, and then see the dependence of the volumes on the magnitude threshold. If you do this, you get pretty similar results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/H75wG.png&quot; alt=&quot;Mathematica graphics&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, Ursa Major has a consistently higher volume at all naked-eye magnitude thresholds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the curious, here's the code.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;nakedEyeStarProperties = &#xA; StarData[#, {&quot;Name&quot;, &quot;Constellation&quot;, &quot;ApparentMagnitude&quot;, &#xA;     &quot;HelioCoordinates&quot;, &quot;RightAscension&quot;, &quot;Declination&quot;}&#xA;] &amp;amp; /@ StarData[EntityClass[&quot;Star&quot;, &quot;NakedEyeStar&quot;]]&#xA;constellationVolume[constellation_, magnitude_] := &#xA; Block[{selectedStars},&#xA;  selectedStars = &#xA;   Select[nakedEyeStarProperties, &#xA;    And[#[[2, 2]] == constellation, #[[3]] &amp;lt; magnitude] &amp;amp;];&#xA;  RegionMeasure[ConvexHullMesh[selectedStars[[All, 4]]]]&#xA;  ]&#xA;ListLogPlot[&#xA; Table[&#xA;   {{m, constellationVolume[&quot;UrsaMajor&quot;, m]}, {m, &#xA;     constellationVolume[&quot;UrsaMinor&quot;, m]}}&#xA;   , {m, 3, 6.5, 0.1}]\[Transpose]&#xA; , AxesLabel -&amp;gt; {&quot;magnitude threshold&quot;, &#xA;   &quot;constellation volume/\!\(\*SuperscriptBox[\(AU\), \(3\)]\)&quot;}&#xA; , PlotLegends -&amp;gt; {&quot;Ursa Major&quot;, &quot;Ursa Minor&quot;}&#xA; ]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="207" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-05-16T02:50:33.040" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10701" PostTypeId="1" AcceptedAnswerId="10714" CreationDate="2015-05-16T01:17:54.207" Score="4" ViewCount="292" Body="&lt;p&gt;this is my first post on the Astronomy stack exchange site. Please let me know how this question is based on y'all's generally accepted criteria for a good question as well as possible improvements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my AP comp sci class we were given pretty much total freedom to come up with a project, I remembered hearing about it being possible to calculate a person's position or the position a picture was taken based on time of day and the position of the Sun, that is, longitudinal and latitudinal position. For the picture, I heard Sun position was determined through some implementation of shadows in the picture. I thought this was cool and decided to look into it. I found an &lt;a href=&quot;http://jsoc.stanford.edu/doc/keywords/Chris_Russel/http___www-ssc.igpp.ucla.edu_personnel_russell_papers_gct1.html_.pdf&quot; rel=&quot;nofollow&quot;&gt;article written by a prominent UCLA physicist&lt;/a&gt; and a &lt;a href=&quot;http://en.wikipedia.org/wiki/Position_of_the_Sun&quot; rel=&quot;nofollow&quot;&gt;well-researched Wikipedia page&lt;/a&gt;. I understand what the Wikipedia page requires in regards to values for calculations and what the equations calculate, but I do not understand how to implement them to find longitudinal and latitudinal location, nor do I understand why it has rectangular equatorial coordinates as that is not included in the three-step explanation in the Wikipedia page intro at the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the UCLA report, I understand none of it except for the matrix mathematics and manipulations, and I'm sketchy on those at best, but I do not understand all the coordinate systems that it mentions nor how to implement them to achieve my objective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question comes down to this: Can anyone clarify the equations, their purpose, their implementation, how to convert an answer to longitude and latitude for either the Wikipedia page or the UCLA report?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OR&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can explain how to find the latitude and longitude of a picture or person. If you choose to explain, I do need equations or an explanation that I can turn into equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please suggest question revisions and whether this should be posted on a different Stackexchange forum. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7271" LastActivityDate="2015-05-18T01:34:09.320" Title="Finding Latitude and Longitude of a person or of where a picture was taken" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10702" PostTypeId="1" AcceptedAnswerId="10716" CreationDate="2015-05-16T05:24:53.677" Score="1" ViewCount="1504" Body="&lt;p&gt;From my understanding, &quot;up&quot; and &quot;down&quot; in space is going towards (down) an object's gravitational pull or (up) which is going away from it. I get confused with this explanation, and I believe simply because of how we perceive up and down here on earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question(s) is this: Can we go in all directions in space with the &quot;Earth&quot; interpretation of up, down, left, right, north west, south east, etc if we can get past gravitational pulls of objects in space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JAxMg.png&quot; alt=&quot;Space Drawing&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at my Gimp version of the universe, I have the Earth and stars. So going up would essentially be go towards any of the stars in the picture that has the strongest gravitational pull, although each object are on different planes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In relation to my question above, I have imagined Earth in space as to like the 10 floor on an infinite elevator that can go in all directions. Say we can somehow go through star A and escape it's gravitational pull, we would be essentially &quot;going down&quot; right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my other question: Are there objects in every single direction is space? How about in the Bootes void? Since it's such a massive void and there isn't much objects around to pull you in &quot;down&quot; or go away from &quot;up&quot;, essentially can't we go in any direction? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it may seems that my question about this is kind of confusing, I apologize because I'm confused about this myself! &lt;/p&gt;&#xA;" OwnerUserId="7258" LastEditorUserId="2618" LastEditDate="2015-05-19T16:34:35.873" LastActivityDate="2015-05-20T09:53:22.853" Title="&quot;Up&quot; and &quot;down&quot; in Space" Tags="&lt;gravity&gt;&lt;space&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ClosedDate="2015-05-26T00:56:12.547" ContentLicense="CC BY-SA 3.0" />
  <row Id="10708" PostTypeId="1" CreationDate="2015-05-17T04:31:24.930" Score="0" ViewCount="172" Body="&lt;p&gt;What is the evidence that galaxies are made of billions of stars? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even faint galaxies?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can spectral analysis distinguish galaxies from individual stars?&lt;/p&gt;&#xA;" OwnerUserId="7279" LastEditorUserId="1322" LastEditDate="2015-05-17T04:40:56.700" LastActivityDate="2015-05-17T13:15:39.163" Title="Evidence that galaxies are made of billions of stars?" Tags="&lt;galaxy&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10709" PostTypeId="2" ParentId="10708" CreationDate="2015-05-17T07:41:30.317" Score="6" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The individual stars can be resolved in galaxies close to us...including our own, the Milky Way Galaxy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;More distant galaxies are morphological similar to those that can be resolved into stars, so a natural induction would be that they are similar in nature to those we can resolve into stars.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Galactic spectra are consistent with them being composed of stars in the sort of relative motion we expect.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Supernovae are observed in distant galaxies, and these are similar to those seen close up which we know are stellar in origin.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;... etc&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="1322" LastEditDate="2015-05-17T13:15:39.163" LastActivityDate="2015-05-17T13:15:39.163" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10710" PostTypeId="1" CreationDate="2015-05-17T14:11:30.250" Score="15" ViewCount="282" Body="&lt;p&gt;I am a retired engineer that has an ongoing interest in space efforts.  In my youth I did work on the Apollo program but on propulsion and vehicle thermal control: not flight dynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have reviewed/studied some astrodynamics (Vallado, Bates, etc) and can run GMAT and tudat libraries from Delft University.  I have explored Systemic Console, tap, and other software for implying the existence of exoplanets from RV curves and light transit curves.  I can download data in FITS format from various astronomical databases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where might I direct my efforts in some meaningful way by participating in a project, collaborating with an investigator, or pursuing some research oriented endeavour on my own?   Could someone point me in a direction where I might do some meaningful work?  Just picking a star and starting a transit analysis does not seem to have any probability of leading anywhere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those sites where one views a light transit curve or tries to classify a galaxy type are not very interesting to me.  I feel I could investigate things more in depth than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Directions, Ideas, offers! Anyone.&#xA;I will post this in Space Exploration forum also.&#xA;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tom Kosvic&lt;/p&gt;&#xA;" OwnerUserId="7280" LastActivityDate="2015-07-29T21:18:31.610" Title="Where might a semi proficient amateur analyst participate in meaningful astronomical efforts" Tags="&lt;planet&gt;&lt;data-analysis&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10711" PostTypeId="1" AcceptedAnswerId="10713" CreationDate="2015-05-17T17:18:45.727" Score="3" ViewCount="77" Body="&lt;p&gt;As transits of Venus come in pairs, each separated by 8 years, wouldn't that imply an accurate ratio between Earth's and Venus' revolution period? I've found the ratio to be about 0.681. I presume one could see a resonance of 2/3 in that number, but as other resonances in our solar system as for example Thethys and Enceladus (4:3) are far more accurate, could it be &quot;just&quot; coincidental? And if so, why the regularity in the transits? &lt;/p&gt;&#xA;" OwnerUserId="7281" LastEditorUserId="7472" LastEditDate="2018-02-03T18:34:57.340" LastActivityDate="2018-02-03T18:34:57.340" Title="Occurrence of Venusian transits" Tags="&lt;venus&gt;&lt;planetary-transits&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10712" PostTypeId="1" AcceptedAnswerId="10730" CreationDate="2015-05-17T18:38:48.227" Score="2" ViewCount="74" Body="&lt;p&gt;I was reading &lt;a href=&quot;http://www.sciencemag.org/content/344/6180/183.full.pdf&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Unfolding the Laws of Star Formation: The Density Distribution of Molecular Clouds&lt;/em&gt;&lt;/a&gt; by Kainulainen et al., which discusses star formation rates and efficiencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One variable used is $\varepsilon_{\text{core}}$, the core-to-star efficiency, describing how much gas above the critical value of $s$, the logarithmic mean-normalized density, forms a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using this parameter in some calculations, but I'd like to use an average value, as opposed to a value for a given molecular cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a typical value for $\varepsilon_{\text{core}}$?&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2015-05-19T14:16:29.147" Title="What is a typical value for core-to-star efficiency?" Tags="&lt;star-formation&gt;&lt;gas&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10713" PostTypeId="2" ParentId="10711" CreationDate="2015-05-17T18:39:39.863" Score="3" Body="&lt;p&gt;Quoting &lt;a href=&quot;http://eclipse.gsfc.nasa.gov/transit/catalog/VenusCatalog.html&quot; rel=&quot;nofollow&quot;&gt;http://eclipse.gsfc.nasa.gov/transit/catalog/VenusCatalog.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;When a transit of Venus occurs, a second one often follows eight years later. This is because the orbital periods of Venus (224.701 days) and Earth (365.256 days) are in an 8 year (2922 days) resonance with each other. In other words, in the time it takes Earth to orbit the Sun eight times, Venus completes almost exactly thirteen revolutions about the Sun. As a result, Venus and Earth line up in the same positions with respect to the Sun. Actually, the two orbital periods are not quite commensurate with each other since Venus arrives at the eight year rendezvous about 2.45 days earlier that Earth. After the third eight-year cycle, Venus arrives too early for a transit to occur.&quot;&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2015-05-17T18:39:39.863" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10714" PostTypeId="2" ParentId="10701" CreationDate="2015-05-17T21:31:01.240" Score="3" Body="&lt;p&gt;EDIT: It turns out I answered this question earlier (in a slightly different format):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/8390/cancelling-out-earth-rotation-speed-altazimuth-mount/8415#8415&quot;&gt;Cancelling out earth rotation speed, Altazimuth mount&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you are at 35N,106W at 5h local sidereal time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The celestial north pole is at (0,90) in the celestial frame and&#xA;(0,35) in your frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point (5h,0) is (75,0) in the celestial sphere (1h = 15 degrees),&#xA;and (180,55) in your frame&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point (11h,0) is (165,0) in the celestial sphere and (90,0)&#xA;(setting in the east) in your frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting to rectangular coordinates, we need the matrix that does this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{0,0,1} -&amp;gt; {Cos[35 Degree], 0, Sin[35 Degree]}&#xA;&#xA;{Cos[75*Degree],Sin[75*Degree],0} -&amp;gt; {-Sin[35 Degree], 0, Cos[35 Degree]}&#xA;&#xA;{Sin[-75*Degree],Cos[75*Degree],0} -&amp;gt; {0,1,0}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or, generalizing a bit (lst = local sidereal time as an angle, lat =&#xA;latitude, notice that lon is irrelevant since we're using local&#xA;sidereal time)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{0,0,1} -&amp;gt; {Cos[lat], 0, Sin[lat]}&#xA;&#xA;{Cos[lst],Sin[lst],0} -&amp;gt; {-Sin[lat], 0, Cos[lat]}&#xA;&#xA;{-Sin[lst],Cos[lst],0} -&amp;gt; {0,1,0}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Solving (Mathematica):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;m0 = Table[a[i,j],{i,1,3},{j,1,3}]&#xA;&#xA;(* Some simplifying assumptions, intentionally avoiding corner cases *)&#xA;&#xA;conds = {-Pi/2 &amp;lt; dec &amp;lt; Pi/2, -Pi/2 &amp;lt; lat &amp;lt; Pi/2, 0 &amp;lt; ra &amp;lt; 2*Pi, -Pi &amp;lt; lst &amp;lt; Pi}&#xA;&#xA;m = Simplify[m0 /. Solve[{&#xA; m0.{0,0,1} == {Cos[lat], 0, Sin[lat]},&#xA; m0.{Cos[lst],Sin[lst],0} == {-Sin[lat], 0, Cos[lat]},&#xA; m0.{-Sin[lst],Cos[lst],0} == {0,1,0}&#xA;},Flatten[m0]],conds]&#xA;&#xA;m = m[[1]]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;   \left(&#xA;   \begin{array}{ccc}&#xA;-\cos (\text{lst}) \sin (\text{lat}) &amp;amp; -\sin (\text{lat}) \sin (\text{lst})&#xA;  &amp;amp; \cos (\text{lat}) \\&#xA;-\sin (\text{lst}) &amp;amp; \cos (\text{lst}) &amp;amp; 0 \\&#xA;\cos (\text{lat}) \cos (\text{lst}) &amp;amp; \cos (\text{lat}) \sin (\text{lst}) &amp;amp;&#xA;  \sin (\text{lat}) \\&#xA;   \end{array}&#xA;   \right)&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inverse of this matrix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;   \left(&#xA;   \begin{array}{ccc}&#xA;-\cos (\text{lst}) \sin (\text{lat}) &amp;amp; -\sin (\text{lst}) &amp;amp; \cos&#xA;  (\text{lat}) \cos (\text{lst}) \\&#xA;-\sin (\text{lat}) \sin (\text{lst}) &amp;amp; \cos (\text{lst}) &amp;amp; \cos (\text{lat})&#xA;  \sin (\text{lst}) \\&#xA;\cos (\text{lat}) &amp;amp; 0 &amp;amp; \sin (\text{lat}) \\&#xA;   \end{array}&#xA;   \right)&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;will transform rectangular azimuth and elevation back to right&#xA;ascension and declination and right ascension. This would be an&#xA;interesting alternative approach to this problem, but I won't be using&#xA;it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, if we convert right ascension and declination to rectangular&#xA;coordinates, multiply by the first matrix above and convert the result&#xA;back to spherical coordinates, we will have azimuth and&#xA;elevation. Let's do that...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spherical coordinates for (ra,dec) are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{Cos[dec] Cos[ra], Cos[dec] Sin[ra], Sin[dec]}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Multiplying by the matrix, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;   \{\sin (\text{dec}) \cos (\text{lat})-\cos (\text{dec}) \sin (\text{lat})&#xA;\cos (\text{lst}-\text{ra}),-\cos (\text{dec}) \sin&#xA;(\text{lst}-\text{ra}),\cos (\text{dec}) \cos (\text{lat}) \cos&#xA;(\text{lst}-\text{ra})+\sin (\text{dec}) \sin (\text{lat})\}&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(note that lst-ra is sometimes called the &quot;hour angle&quot;, which would&#xA;simplify the above)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting back to spherical coordinates and simplifying:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{x,y,z} = Simplify[m.{Cos[dec] Cos[ra], Cos[dec] Sin[ra], Sin[dec]},conds]&#xA;&#xA;(* NOTE: the use of single argument arctangent here introduces&#xA;ambiguity, and the final answer must compensate for this *)&#xA;&#xA;az = Simplify[ArcTan[y/x],conds]&#xA;&#xA;el = Simplify[ArcSin[z],conds]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;yielding:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;   \text{azimuth}=-\tan ^{-1}\left(\frac{\cos (\text{dec}) \sin&#xA;(\text{lst}-\text{ra})}{\sin (\text{dec}) \cos (\text{lat})-\cos&#xA;(\text{dec}) \sin (\text{lat}) \cos (\text{lst}-\text{ra})}\right)&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;   \text{elevation}=\sin ^{-1}(\cos (\text{dec}) \cos (\text{lat}) \cos&#xA;(\text{lst}-\text{ra})+\sin (\text{dec}) \sin (\text{lat}))&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming I haven't made any mistakes, the next steps would be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Converting local time to local sidereal time&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Inverting these equations to get latitude and longitude from&#xA;azimuth, elevation, and time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I may or may not edit this answer to add those steps, but this&#xA;hopefully provides a start.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-05-18T01:34:09.320" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10716" PostTypeId="2" ParentId="10702" CreationDate="2015-05-18T11:46:06.033" Score="0" Body="&lt;p&gt;Your confusion is that you are treating gravity from all those objects as a bunch of discrete 'pulls' - but it doesn't manifest itself like that. Gravity from all bodies in the universe effects you as one force (I'm simplifying and excluding getting close to black holes etc where you have dramatically changing gravitational potential over the length of your body)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add up all those gravitational forces and you will be pulled in a direction. It will not feel like up or down as you will just accelerate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are in a rocket with the engine running, &quot;down&quot; will be the direction opposite to thrust. In the same way that the Earth is down, because it is pushing upwards against your feet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acceleration from your rocket engine works a lot like acceleration from the gravitational pull from a planet or star. It affects your velocity. You can go any direction - some will just be easier than others - but that is a much bigger set of questions. Read some of the posts on &lt;a href=&quot;https://space.stackexchange.com/&quot;&gt;Space Exploration Stack Exchange&lt;/a&gt; on changing orbits, travelling between planets etc.&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2015-05-18T11:46:06.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10717" PostTypeId="2" ParentId="10697" CreationDate="2015-05-18T13:17:31.977" Score="3" Body="&lt;p&gt;From eq. 10 in &lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/bib_query?arXiv:astro-ph/9905116&quot; rel=&quot;nofollow&quot;&gt;Hogg's classic paper&lt;/a&gt;, assuming that the peculiar velocity $v_\mathrm{pec} \ll c$:&#xA;$$v_\mathrm{pec} = c \frac{z_\mathrm{obs} - z_\mathrm{cos}}{1 + z_\mathrm{cos}},$$&#xA;where $z_\mathrm{obs}$ is the observed redshift, and $z_\mathrm{cos}$ is the redshift from cosmological expansion only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me invert that and wrap it up in Python for ya:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def zobs(vpec,zcos):&#xA;  c = 3e5&#xA;  return (1+zcos) * vpec/c + zcos&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-05-18T20:40:17.127" LastActivityDate="2015-05-18T20:40:17.127" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10718" PostTypeId="2" ParentId="10710" CreationDate="2015-05-18T13:30:50.910" Score="2" Body="&lt;p&gt;I recommend visiting a few university Astronomy departments, and investigating their PhD programmes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While a PhD may seem like a big thing to undertake as a retired person (and it is), you will certainly be doing real research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just discussing the possibilities with people in a few Astronomy departments will lead to other ideas, or actual research projects you can participate in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck.&lt;/p&gt;&#xA;" OwnerUserId="5524" LastActivityDate="2015-05-18T13:30:50.910" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10720" PostTypeId="1" CreationDate="2015-05-18T18:35:03.470" Score="2" ViewCount="548" Body="&lt;p&gt;The current system of constellations is historical and has kind of arbitrary boundaries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LBJjZ.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has a number of obvious downsides:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Difficult to define boundaries need complex tables to express&lt;/li&gt;&#xA;&lt;li&gt;Each constellation subtends a different solid angle (area of sky)&lt;/li&gt;&#xA;&lt;li&gt;etc&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Has anyone ever proposed, or does anyone use, a system of cutting the sky into regular chunks, that are either simple to define or similar in area?   Do any of these systems (if they exist) attempt to preserve the most important/famous constellations (as least the part people normally see) ?&lt;/p&gt;&#xA;" OwnerUserId="1106" LastEditorUserId="7472" LastEditDate="2018-12-24T15:51:15.587" LastActivityDate="2018-12-24T15:51:15.587" Title="Standardized &quot;constellation&quot; regions?" Tags="&lt;constellations&gt;&lt;standards&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="10721" PostTypeId="2" ParentId="8701" CreationDate="2015-05-18T20:56:04.550" Score="1" Body="&lt;p&gt;In the paper that this report is based on, &lt;em&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0927650510001234&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;&lt;/em&gt;,  they simply see an annual period in the $\beta$ decay rates of radioactive isotope samples in the lab.  Basically, the rate is a fraction of a percent higher in winter than in summer. They conclude that absent any simple instrumentation explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;we conclude that these results are consistent with the hypothesis that&#xA;  nuclear decay rates may be influenced by some form of solar radiation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Several things can be changing in a laboratory during a year.  Obviously, temperature and humidity changes and these were tested in the experiment.  But, also radon levels change as the amount of outside air exchanged with inside air is changed.  The solar cosmic ray flux (high energy electrons, protons, and He nuclei generated in the chromosphere of the sun) changes as the Sun angle changes, and neutrinos (produced in the core) also as Sun angle changes.  These could be affecting nuclei decay rates directly or the instrument used to measure these (subtle changes in threshold energies, false counts from ions produced in the instrument, potential shifts, etc.).&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2015-05-19T16:38:13.963" LastActivityDate="2015-05-19T16:38:13.963" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10722" PostTypeId="2" ParentId="10540" CreationDate="2015-05-18T21:27:07.550" Score="1" Body="&lt;p&gt;Transmission gratings on their own introduce chromatic aberration.  This is because they change the effective focal length and do so as a function of wavelength.  The chromatic aberration can be eliminated by introducing a prism of the correct dispersion.  Known as a grism, the grating/prism combination provides an unaberrated image at the 0th order in addition to the spectrum, although the spectrum has fairly low resolution. &lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2015-05-19T16:01:46.457" LastActivityDate="2015-05-19T16:01:46.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10723" PostTypeId="1" AcceptedAnswerId="10724" CreationDate="2015-05-19T04:04:50.700" Score="5" ViewCount="610" Body="&lt;p&gt;Stars explode when they die and blast heavy elements into space.  Do galaxies do the same thing?&lt;/p&gt;&#xA;" OwnerUserId="5306" LastActivityDate="2015-05-19T09:49:46.567" Title="What happens to galaxies when they die?" Tags="&lt;galactic-dynamics&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10724" PostTypeId="2" ParentId="10723" CreationDate="2015-05-19T04:25:42.300" Score="8" Body="&lt;p&gt;Well, it would be useful to define what a 'dead' galaxy is.  Probably the most simple method would be a galaxy that is no longer producing new stars.  We might also consider a galaxy that no longer produces significant light in the visual spectrum, or perhaps EMR across the entire spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, there's unlikely to be a firm line between living and dead, and not nearly as dramatic as larger stars. More akin to watching a camp fire burn itself out.  Star formation is largely dependent available gases, but as more and more stars fuse those gases into heavier elements, there is less gas available for star formation. For your average sized galaxy, this will eventually result in running out of gas. Eventually the galaxy will dim and go dark, a process purported to begin at the center of the galaxy, where star formation is heaviest according to research based on Hubble images of giant galaxies. (Tacchella, et al.) The matter ought to (mostly) all still be there and still orbiting the (presumed) SMBH, but with no energy coming from fusion, it's going to be a dark, cold, and barren place. Sounds dead to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some complicating factors.  It's believed that encounters with nearby galaxies can affect available gases. The gravity from a larger galaxy could potentially strip the gases from a smaller one, a fatal blow for the smaller galaxy. Fortunately, it won't suffer much as the death will come (relatively) quickly. This process has been deemed 'strangulation' by a study published in Nature several &lt;s&gt;years&lt;/s&gt; &lt;s&gt;months&lt;/s&gt; days ago. (Ping, et al.) Note that as the study indicates, the methods of death are proposed solutions - not conclusive understanding of the exact processes that result in a galaxy's death.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;S. Tacchella, C. M. Carollo, A. Renzini, N. M. Förster Schreiber, P. Lang, S. Wuyts, G. Cresci, A. Dekel, R. Genzel, S. J. Lilly, C. Mancini, S. Newman, M. Onodera, A. Shapley, L. Tacconi, J. Woo, and G. Zamorani. &lt;a href=&quot;http://www.spacetelescope.org/static/archives/releases/science_papers/heic1508a.pdf&quot;&gt;Evidence for Mature Bulges and an Inside-out Quenching Phase 3 Billion Years After the Big Bang&lt;/a&gt;&#xA;&lt;em&gt;Science&lt;/em&gt; 17 April 2015: 348 (6232), 314-317. [DOI:10.1126/science.1261094]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Y. Peng, R. Maiolino &amp;amp; R. Cochrane. &lt;a href=&quot;http://www.nature.com/nature/journal/v521/n7551/full/nature14439.html&quot;&gt;Strangulation as the primary mechanism for shutting down star formation in galaxies&lt;/a&gt; &lt;em&gt;Nature&lt;/em&gt; 521, 192–195 14 May 2015 [DOI:10.1038/nature14439]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Andrea Cattaneo. &lt;a href=&quot;http://www.nature.com/nature/journal/v521/n7551/full/521164a.html&quot;&gt;Astrophysics: The slow death of red galaxies&lt;/a&gt; &lt;em&gt;Nature&lt;/em&gt; 521, 164–165 14 May 2015 [DOI:10.1038/521164a]&lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="1288" LastEditDate="2015-05-19T05:51:54.650" LastActivityDate="2015-05-19T05:51:54.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10725" PostTypeId="2" ParentId="10723" CreationDate="2015-05-19T06:56:22.980" Score="3" Body="&lt;p&gt;On any &lt;em&gt;practical timescale&lt;/em&gt; (ie say 100 billion years) a galaxy could not be said to have died. The &lt;em&gt;majority&lt;/em&gt; of stars that ever formed in that galaxy will still be &quot;alive&quot;, but they are dim, low-mass stars. However, some of those low-mass stars will have reached the ends of their main sequence lives and, if more massive than $\sim 0.4M_{\odot}$ will go through a luminous, and highly visible, red giant phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mitch Goshorn's definition of a cessation of star formation is a reasonable one, although on that definition many of the gas-poor elliptical galaxies we see today would be &quot;dead&quot;. The reason the galaxies &quot;run out&quot; of gas is that is might be blown out by the winds and supernovae caused by higher mass stars; it might be stripped from them in tidal interactions with other galaxies, or in the long-run it is because the star formation process mostly produces stars that have the mass of the Sun or lower. Such stars, once formed, can lock away material and prevent it being recycled. The lifetimes of these stars increase rapidly with &lt;em&gt;decreasing&lt;/em&gt; stellar mass - reaching a 100 billion years for a 0.4 solar mass star, which is still a bit more massive than the average star. So the mass spectrum of stars in a galaxy should tilt towards lower masses as it gets older.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even when these low mass stars reach the ends of their lives they only return about half their gas to the galaxy, the rest is trapped in a cooling white dwarf. So new star formation will gradually expire as the amount of returned gas gets lower and lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, that the fact that the gas has been inside a star previously does not prevent new star formation. The returned gas consists mostly of unused hydrogen. It is just a question of how much of that gas there is, and whether it can be concentrated sufficiently to form new stars.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-05-19T08:02:39.050" LastActivityDate="2015-05-19T08:02:39.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10726" PostTypeId="2" ParentId="10702" CreationDate="2015-05-19T08:49:42.103" Score="1" Body="&lt;p&gt;I think that one thing confusing you, is that up/down is a 2 dimensions concept (just like your image by the way), but the Universe is in 3 physical dimensions (let's not talk about time here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when you say « going up from Earth is extracting from it's gravity », it can be in many direction. If two persons are in North Pole and South Pole of Earth, and if they are both going up. They are in fact going in opposite directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second point is that you need a &lt;a href=&quot;http://en.wikipedia.org/wiki/Frame_of_reference&quot; rel=&quot;nofollow&quot;&gt;frame of reference&lt;/a&gt;. You go up/down « to » or « from » something.&#xA;For example, you go up (going away) from the North Pole, but from Sun's or Moon's point of view, it's not going up at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The up/down/left/right concepts always need a reference. See this very simple example: If we are facing each other, what is on my left, is on your right.&#xA;Your question here is exactly like asking : « What is on my left in the universe ? ». There is no real answer to that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope I could help.&lt;/p&gt;&#xA;" OwnerUserId="204" LastActivityDate="2015-05-19T08:49:42.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10727" PostTypeId="2" ParentId="10723" CreationDate="2015-05-19T09:49:46.567" Score="1" Body="&lt;p&gt;Galaxies are not really living objects, so while they can’t die, they can in fact change. The galaxies we see in the night sky have been around for billions of years and remain unchanged for millions of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However they do evolve and they can also be disrupted by the gravity from other galaxies. In fact, our own Milky Way is on a collision with one of our neighbours, the Andromeda galaxy. In a few billion years we will pass close enough that both galaxies will be deformed and will eventually merge. At this point a huge amount of star formation will be kicked off and the resulting galaxy will be a large elliptical blob rather than the fine disc like structures we can currently see.&lt;/p&gt;&#xA;" OwnerUserId="7291" LastActivityDate="2015-05-19T09:49:46.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10728" PostTypeId="1" AcceptedAnswerId="10729" CreationDate="2015-05-19T10:08:18.420" Score="2" ViewCount="567" Body="&lt;p&gt;Iron is generated by stars in a certain part of their life cycle. Earth contains a lot of iron inside, however it is clear that this iron could have not been generated in a star in close proximity.&#xA;If we consider the current Earth position as reference, where would have lied 4.5 billions years ago the stars that generated the iron currently found in Earth core? Can we say that in the current Earth position there was a huge star 4.5 billions years ago? How did the iron formed in these old stars reach the primordial gas cloud that turned into the solar system? For how long did it travel? Can we say that it came from a set of stars that lived (again, 4.5 billions years ago) in the area that we now call the galaxy core, for example? Can we define the set of &quot;iron generating&quot; areas? Can we define something like the &quot;Earth core iron generator sphere&quot;?&lt;/p&gt;&#xA;" OwnerUserId="7292" LastEditorUserId="7292" LastEditDate="2015-05-19T11:53:43.790" LastActivityDate="2015-05-19T13:35:50.923" Title="What supernova has created the iron currently found in Earth core?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10729" PostTypeId="2" ParentId="10728" CreationDate="2015-05-19T10:49:27.427" Score="4" Body="&lt;p&gt;Iron is mainly made or is the product of decays from nuclear processed material inside supernovae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the Earth (and solar system) is around 4.5 billion years old, then the stars that manufactured the iron that is currently in the Earth's core died more than 4.5 billion years ago. Note that the solar system formed out of the gases that had been enriched by the supernova explosions of hundreds of millions of stars all mixed up together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are basically two categories of star that can have exploded as supernovae and disseminated this iron into the interstellar medium and from which the solar system could then have formed. The first is massive stars ($&amp;gt;8-10M_{\odot}$). These can produce iron and nickel in their cores as a result of silicon burning in the final stages of their evolution. There is then a brief core-collapse followed by an explosion and the resulting supernova can scatter some of this processed, iron-rich material into space.&#xA;The remnants of these ancient supernova explosions could be a neutron stars or a black holes. These are almost untraceable/unobservable now, but there should be around a billion of them in our Galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second category is the progenitors of what are called type Ia supernovae. These are thought to arise from the thermonuclear explosion of a white dwarf star. White dwarfs are the end point of the evolution of less massive stars. Iron-producing type Ia supernovae $&amp;gt;4.5$ billion years ago, would have begun as stars with masses between about 1.5 and 8$M_{\odot}$. These would have burned hydrogen, then helium to produce a degenerate carbon and oxygen core. This core then simply cools and fades away as a white dwarf star in most cases. In type Ia supernovae, some event later in their lives, either mass transfer from a companion, or merger with a companion, caused the white dwarf to exceed its Chandrasekhar mass and triggers an instability that leads to the rapid total consumption of the star in a thermonuclear explosion. The products of this explosion include a large amount of Nickel, that then radioactively decays into Iron. Nothing is left of the white dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Having established this we can begin to look at your edited question. Firstly, the gas and stars in the Galaxy are basically orbiting the Galactic centre. The orbital period at the Sun's radius is about 230 million years, so it has completed many Galactic orbits. Not only that, but it could have migrated in orbital radius as well. There are claims and counter claims in the literature and the issue is not settled. The Sun could have moved in or out by a significant fraction of its current Galactic orbital radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The high mass progenitors of core collapse supernovae will have been born (and died) very close to the Galactic plane. The same is not so true of type Ia supernovae, which had longer lived progenitors that could have moved significantly from the Galactic plane before exploding, and indeed would themselves have orbited the Galaxy many times. The gas expelled in a supernova explosion spreads out (over thousands of years) over tens of light years and becomes mixed into the interstellar medium. The interstellar medium is itself stirred and mixed by the energy input from these supernovae, but also due to the heating and winds of other stars, the tides of the galaxy and spiral arms. The interstellar medium appears to be quite homogeneous in terms of chemical composition, though radial gradients exist with scale lengths of order ten thousand light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, what you ask is almost impossible to answer. The solar system iron almost certainly came from countless supernovae with a variety of progenitors, that would have exploded any time between almost the birth of the Galaxy 11-12 billion years ago (in fact the supernova rate was likely higher then) up until the Sun's birth. The biggest contributors would come from those stars inhabiting an annulus of many thousands of light years across, centered roughly on where the Sun was born, which is itself uncertain.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-05-19T13:35:50.923" LastActivityDate="2015-05-19T13:35:50.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10730" PostTypeId="2" ParentId="10712" CreationDate="2015-05-19T14:16:29.147" Score="1" Body="&lt;p&gt;From &lt;a href=&quot;https://books.google.de/books?id=tQswBQAAQBAJ&amp;amp;pg=PA465&amp;amp;lpg=PA465&amp;amp;dq=core-to-star%20efficiency&amp;amp;source=bl&amp;amp;ots=JJOAAWFbxm&amp;amp;sig=dJQw-o9QGl8sH36DdeQe0kkRmkY&amp;amp;hl=it&amp;amp;sa=X&amp;amp;ei=gUNbVYumKISasAHk4IHwCg&amp;amp;ved=0CCIQ6AEwAA#v=onepage&amp;amp;q=core-to-star%20efficiency&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this book&lt;/a&gt; it seems you can have $\epsilon_{core}$ values raging from ~$30\%$ (a low value) up to $70\%$ (which is more &quot;usual&quot;, i.e., requested by theories).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More information can be found &lt;a href=&quot;http://iopscience.iop.org/0004-637X/545/1/364/pdf/51522.web.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://iopscience.iop.org/0004-637X/761/2/156/pdf/apj_761_2_156.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastActivityDate="2015-05-19T14:16:29.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10731" PostTypeId="2" ParentId="10720" CreationDate="2015-05-19T17:40:38.633" Score="3" Body="&lt;p&gt;Modern astronomers are not instructed on the constellations and pay little attention to them except as a naming convention for stars.  Positional information is nearly always either equatorial (ra,dec), or galactic (gl,gb), or supergalactic (SGL, SGB).  To improve the speed in which information on a region can be extracted from a database, a scheme called Hierarchical Triangular Mesh (&lt;em&gt;&lt;a href=&quot;http://skyserver.org/htm/index.html&quot; rel=&quot;nofollow&quot;&gt;HTM&lt;/a&gt;&lt;/em&gt;), is often used.  The sky is divided into three sections (0,1,2) and each of those is divided into 3 sections (00,01,02,10,11...) and this iterated until the level of finest size is reached.  One can address the finest region by using all of the digits or a larger region by dropping some of the final digits. &lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2015-05-19T20:04:15.110" LastActivityDate="2015-05-19T20:04:15.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10732" PostTypeId="2" ParentId="10479" CreationDate="2015-05-19T20:33:06.803" Score="4" Body="&lt;p&gt;Most of the 60 moons in the Saturn system are far away from the rings and very small, so their effect on the rings is negligible. But larger ones that are closer in (Enceladus) do have a rather significant effect on the rings, but as the gravitational pull of these moons is radially outward, it is hardly visible. On the other hand, small moons inside the rings that are a few kilometers in diameter cause significant changes in the ring system.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mQlbu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mQlbu.jpg&quot; alt=&quot;&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://lollitop.magicgate.eu/media/LollitopMagicgate_011/lollitop_Saturn_at_equinox_083536_s04_PIA11477.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;magicgate.eu&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;The orbits of those giant rocks are slightly inclined relative to the rings, which causes the rings to create a bulge in the direction of the small moon as seen in the picture. And these effects are fairly significant as the rings are mostly just a few hundred meters thick or less without a moon present, but with one, they are possibly hundreds of kilometers spread out.&lt;/p&gt;&#xA;" OwnerUserId="7281" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-02-18T08:12:04.607" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10733" PostTypeId="1" CreationDate="2015-05-20T01:32:35.687" Score="5" ViewCount="87" Body="&lt;p&gt;Ok, so I have read about how exoplanets are found using all sorts of genius methods.  I also understand the further an obect in a telescope is the further into the past I am looking.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So given that we are finding exoplantets with methods that go well beyond any practical visual range are we fnding these exoplanets 'in their past' as well? As far as I understand radio telescopes and other methods can't travel faster than light therfore their results are also from past events, is my reasoning in this right?&lt;/p&gt;&#xA;" OwnerUserId="7298" LastEditorUserId="2531" LastEditDate="2015-05-20T06:50:57.293" LastActivityDate="2015-05-20T14:00:52.333" Title="Exoplanets and relativity" Tags="&lt;exoplanet&gt;&lt;general-relativity&gt;&lt;special-relativity&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2015-05-21T12:27:18.543" ContentLicense="CC BY-SA 3.0" />
  <row Id="10734" PostTypeId="1" CreationDate="2015-05-20T04:11:50.870" Score="0" ViewCount="351" Body="&lt;p&gt;Ok so I was reading about the Fermi paradox, assuming life will form and thrive where possible as soon as possible and there is a very probable chance many other earths formed would they have started to form around at the same time our earth started forming.  To be clear, another question is at what point in time in the universe after the big bang was life (exactly like ours) even possible? &lt;/p&gt;&#xA;" OwnerUserId="7298" LastEditorUserId="2531" LastEditDate="2016-12-20T19:42:26.200" LastActivityDate="2016-12-20T19:42:26.200" Title="The Fermi paradox" Tags="&lt;big-bang-theory&gt;&lt;fermi-paradox&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10735" PostTypeId="2" ParentId="10734" CreationDate="2015-05-20T06:36:00.227" Score="1" Body="&lt;p&gt;Already &lt;a href=&quot;http://www.nature.com/news/life-possible-in-the-early-universe-1.14341&quot; rel=&quot;nofollow&quot;&gt;15 million years after Big Bang!&lt;/a&gt; That is 13,783 million years ago. Well, that is according to the &quot;habitable zone&quot; criteria which is the temperature where water can exist as a liquid. The entire universe had a temperature which allowed for liquid water on all rocky planets, though rocky planets themselves were quite rare back then in the old days. But maybe the origin of life depends on something more than only temperature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole thing is just a thought experiment since we have no data about alien life and no explanation of how it started on Earth. That allows for many degrees of freedom, which is the phrase physicists use when they don't know what they are talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;“Anything is habitable if you are clever enough,”&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Freeman Dyson (The immortal alien life optimist)&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-05-20T06:44:01.693" LastActivityDate="2015-05-20T06:44:01.693" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="10736" PostTypeId="2" ParentId="10734" CreationDate="2015-05-20T06:48:12.427" Score="5" Body="&lt;p&gt;I think this too broad, but I'll offer the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The star Kepler 444 is orbited by several small, assumed to be rocky, exoplanets. Kepler 444 is estimated to be a very old star, perhaps 11 billion years old, with a metal content of about one third that of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whilst the planets around Kepler 444 are small, they are too hot to be &quot;earth like&quot;, but there appears to be no reason why planets at larger orbital radii (that are much harder to detect by the transit technique) should not be there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the answer appears to be demonstrably, at least 11 billion years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the limit cannot be much longer than this, since a certain time must elapse between the formation of the first stars in the Galaxy to the enrichment of the interstellar medium with metals. These metals (all elements heavier than He are referred to as such) are required to build a &quot;rocky&quot; planet. While Kepler 444 demonstrates you don't need a solar metallicity, you still need some.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fastest place this enrichment took place in general was in the Galactic bulge. A burst of star formation probably enriched the ISM in much less than a billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus in principle I would say less than a billion years after Galaxy formation and this is probably not much more than 11 billion years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kepler 444&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.m.wikipedia.org/wiki/Kepler-444&quot; rel=&quot;noreferrer&quot;&gt;http://en.m.wikipedia.org/wiki/Kepler-444&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2015ApJ...799..170C&quot; rel=&quot;noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2015ApJ...799..170C&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second part of your question is hard. It has taken 4.5 billion years &quot;for life like ours&quot; to evolve. Since we don't fully understand the factors that lead to this, the only realistic answer is that it is probable that it takes another 4.5 billion years after the formation of planets for life &quot;exactly like ours&quot; to emerge.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-05-20T09:54:22.727" LastActivityDate="2015-05-20T09:54:22.727" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10737" PostTypeId="1" AcceptedAnswerId="10739" CreationDate="2015-05-20T06:57:38.133" Score="6" ViewCount="765" Body="&lt;p&gt;I understand a neutron star to be the densest stuff that can exist without becoming a black hole, the densest thing which directly gives off radiation we can detect. At the event horizon of a black hole, time stands still, in our frame of reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does there exist neutron stars which are very close to getting an event horizon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this &quot;neutron star horizon&quot; cause time dilation, a zone from within of which movement is very slow in our frame of reference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it for example be that the first light from a neutron star itself reaches us billions of years after the light of the supernova of the &quot;classical&quot; star which created the neutron star and its event-horizon like relativistic effects?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-05-20T07:14:13.947" LastActivityDate="2015-05-20T10:25:50.943" Title="Is radiation from neutron stars delayed by time dilation?" Tags="&lt;black-hole&gt;&lt;neutron-star&gt;&lt;time-dilation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10738" PostTypeId="2" ParentId="10702" CreationDate="2015-05-20T09:53:22.853" Score="-1" Body="&lt;p&gt;It is not only about up and down, there are deeper dimensions to this. Like infront and behind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon seems to be &quot;behind&quot; us, as in expressions like: &quot;Going BACK to the Moon&quot;. Although it does go all around and never gets anywhere, much like ones own bottom, actually. Even Ptolemy didn't argue with that fact. Might this be a clue to this geometric puzzle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your grandparents might argue against this and say that they definitely remember that the Moon was the horizon in front of them. How little they knew about what would become our behind!&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-05-20T09:53:22.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10739" PostTypeId="2" ParentId="10737" CreationDate="2015-05-20T10:25:50.943" Score="5" Body="&lt;p&gt;A typical neutron star of $1.5M_{\odot}$ is thought to have a radius of around 8-10 km. This is only a factor of 2 larger than the Schwarzschild radius for a similar mass black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that more massive neutron stars do exist. The current record holder is around $2M_{\odot}$. Most equations of state (the adopted relationship between pressure and density) for dense nuclear matter suggest that more massive neutron stars are smaller and therefore must be even closer in radius to the Schwarzschild radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the premise of you question is basically correct. It is certainly true that when you deal with neutron star spectra you do have to apply significant general relativistic corrections to measured temperatures and the same corrections would need to be applied to any temporal variations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus a time-variable signal from a neutron star surface will appear slower to an observer on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the last part, I suspect that the scenario you propose is extremely unlikely. &lt;a href=&quot;http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.32.324&quot; rel=&quot;noreferrer&quot;&gt;Rhoades &amp;amp; Ruffini (1974)&lt;/a&gt; first established that there must be a maximum mass for a neutron star under GR conditions, even if we allow the equation of state to harden to the point where the speed of sound is the speed of light. This maximum mass is around $3.2M_{\odot}$.  This sets an upper limit to the possible value of $GM/Rc^{2} \leq 0.405$ (see p.261 of Shapiro &amp;amp; Teukolsky, &lt;em&gt;Black holes, white dwarfs and neutron stars&lt;/em&gt;). This in turn sets an upper limit the possible gravitational redshift (and time dilation factor) of 2.29.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond this point the neutron star is unstable and will collapse to become a black hole. In reality the limit is probably a bit tighter than that because most proposed equations of state result in neutron stars becoming unstable at finite densities and at masses quite a bit lower than $3.2M_{\odot}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think the most time dilation you are ever going to see from a neutron star surface is a factor of $\sim 2$. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-05-20T10:25:50.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10740" PostTypeId="2" ParentId="8367" CreationDate="2015-05-20T13:05:13.000" Score="2" Body="&lt;p&gt;mikeonly,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I stumbled across PHOEBE code the other day in another response in this forum.  I started looking into it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;PHOEBE stands for PHysics Of Eclipsing BinariEs. It is a tool for modeling eclipsing binary stars based on photometric, spectroscopic, interferometric and/or polarimetric data.&quot; -- abstracted from About Phoebe webpage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps this can fulfill your needs.  I've not yet installed and looked at it.&#xA;It is developed by a group of professional astronomers from around the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tom Kosvic&lt;/p&gt;&#xA;" OwnerUserId="7280" LastActivityDate="2015-05-20T13:05:13.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10741" PostTypeId="1" CreationDate="2015-05-20T13:09:52.087" Score="2" ViewCount="937" Body="&lt;p&gt;If I fix the Earth as the center of the solar system, what would this look like? Are there any good images? How complex is this system?&lt;/p&gt;&#xA;" OwnerUserId="7300" LastEditorUserId="2531" LastEditDate="2015-05-21T09:05:52.880" LastActivityDate="2016-03-17T18:59:32.807" Title="How would the solar system look in a Geocentric model?" Tags="&lt;solar-system&gt;&lt;geocentrism&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10742" PostTypeId="2" ParentId="10733" CreationDate="2015-05-20T14:00:52.333" Score="0" Body="&lt;p&gt;Yes, you are indeed correct. We always see what was happening at a given location when the light was emitted there for us to be detected. Let's assume we observe a oscillating star 100 light years away by measuring the shift in frequency of the detected light. That means that the star was actually oscillating a hundred years earlier and we just happen to observe it now; But we (that means nobody) have no way of knowing what is happening now due to the fact that information can't travel faster than light (especially when the information is transmitted via light as in our case).&lt;/p&gt;&#xA;" OwnerUserId="7281" LastActivityDate="2015-05-20T14:00:52.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10743" PostTypeId="1" AcceptedAnswerId="10747" CreationDate="2015-05-20T14:06:48.413" Score="4" ViewCount="417" Body="&lt;p&gt;Context: I want to simulate globular clusters in a simple way, just to display the positions of stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming isotropy, what would be a reasonable model of the stellar number density as a function or r, the distance from the center of the cluster?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming that such a model would have some free parameters as well.&lt;/p&gt;&#xA;" OwnerUserId="5359" LastActivityDate="2015-05-20T20:05:26.667" Title="Globular cluster star density as a function of distance from the center" Tags="&lt;globular-clusters&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10744" PostTypeId="2" ParentId="10526" CreationDate="2015-05-20T14:12:54.010" Score="-1" Body="&lt;p&gt;Inflation has started with the bigbang. In fact for anything to take place you need an expansion right away. If you had a static universe just after the bigbang, it couldn't have formed subatomic particles or atoms, without any increase in size. &#xA;Inflation was always happening and will always happen. The process is behaving (in terms of duration) like radioactive material, that means that inflation slowy decays wth a halflife of about 10^-32 seconds. I hope this answer suffices.&lt;/p&gt;&#xA;" OwnerUserId="7281" LastActivityDate="2015-05-20T14:12:54.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10745" PostTypeId="1" AcceptedAnswerId="10751" CreationDate="2015-05-20T14:19:52.713" Score="2" ViewCount="170" Body="&lt;p&gt;If an object was a distance from a planet away that equaled the distance from the object to a moon of the said planet, would it be drawn into the direction of the barycenter as it is the common center of mass?&lt;/p&gt;&#xA;" OwnerUserId="7281" LastActivityDate="2015-05-20T18:06:39.007" Title="Attraction to barycenter" Tags="&lt;newtonian-gravity&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10746" PostTypeId="2" ParentId="10674" CreationDate="2015-05-20T14:39:35.280" Score="1" Body="&lt;p&gt;Small mass black holes are &lt;strong&gt;extremely&lt;/strong&gt; small and, if Hawking Radiation is real, &lt;strong&gt;extremely&lt;/strong&gt; short-lived. (I cannot over-emphasise how extremely!) Even a one hundred ton black hole would be smaller than an electron and evaporate in the blink of an eye (and in a multi-million megaton explosion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Large_Hadron_Collider&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Large_Hadron_Collider&lt;/a&gt;, the LHC is achieving 328 TeV per ion in its heavy ion collisions. 328 TeV is about 5x10&lt;sup&gt;-24&lt;/sup&gt;kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A black hole of that mass would be about 10&lt;sup&gt;-50&lt;/sup&gt;m in diameter and would live for about 10&lt;sup&gt;-70&lt;/sup&gt;s. &lt;em&gt;(Both of these are much less than planck length and planck time (10&lt;sup&gt;-35&lt;/sup&gt;m and 10&lt;sup&gt;-44&lt;/sup&gt;s) so such a small black hole is arguably impossible for that reason if not for others.)&lt;/em&gt;&#xA;If it were possible it would be soo many orders of magnitude too small and too short-lived to be able to acquire any extra mass before it disappeared. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process of evaporation (Hawking radiation) is virtual particle pairs forming on the event horizon and one falling in and one escaping. The one that escapes becomes a real particle with mass and that mass is taken from the black hole. (In effect the other virtual particle becomes a real negative mass falling into it). That's very simplistic and I don't understand enough myself to explain it in significantly greater detail. It is not proven to exist as UserLTK says and I doubt if it could happen at the lengths and timescales of this sub-planck-scale black hole if such a thing were possible anyway.&lt;/p&gt;&#xA;" OwnerUserId="7301" LastActivityDate="2015-05-20T14:39:35.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10747" PostTypeId="2" ParentId="10743" CreationDate="2015-05-20T15:28:34.253" Score="4" Body="&lt;p&gt;The usual thing is a &lt;a href=&quot;http://adsabs.harvard.edu/abs/1966AJ.....71...64K&quot; rel=&quot;nofollow&quot;&gt;King model&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are indeed free parameters. These are the central density, the &quot;core radius&quot; and a tidal truncation radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The background and rationale for these models is given in the link. They provide a pretty good representation of the surface density of globular clusters (or indeed open clusters). They require a numerical scheme to &quot;deproject&quot; from the plane of the sky to 3D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you find the nitty-gritty of Abel integrals too tricky in the deprojection, then you could always approximate with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Plummer_model&quot; rel=&quot;nofollow&quot;&gt;Plummer model&lt;/a&gt;. This is analytically and computationally easier to deal with, but lacks a bit of physical realism. Central density and a characteristic radius are free parameters here.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-05-20T16:58:38.550" LastActivityDate="2015-05-20T16:58:38.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10748" PostTypeId="2" ParentId="10579" CreationDate="2015-05-20T16:19:34.127" Score="3" Body="&lt;p&gt;Einstein's general theory of relativity explains time dilation. &lt;a href=&quot;https://en.wikipedia.org/wiki/Time_dilation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; provides a concise summary:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;time dilation&lt;/strong&gt; is a difference in the elapsed time measured by two&#xA;  observers, either due to a velocity difference relative to each other,&#xA;  or by being differently situated relative to a gravitational field. As&#xA;  a result of the nature of spacetime, a clock that is moving relative&#xA;  to an observer will be measured to tick slower than a clock that is at&#xA;  rest in the observer's own frame of reference. A clock that is under&#xA;  the influence of a stronger gravitational field than an observer's&#xA;  will also be measured to tick slower than the observer's own clock.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As one experiences more gravity, time seems to flow slower compared to (say) Earth. That means that &quot;standing&quot; on Jupiter, which isn't possible due to the gaseous surface, will cause you to move through time faster; but you do not age slower. Suppose your lifespan is 80 years. On Jupiter you still would live 80 years, but the time that has passed on Earth during your 80 years would be slightly more than that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational time dilation on the Moon is the opposite: the weaker gravity means that an Earth clock ticks very slightly slower than a Moon clock.&lt;/p&gt;&#xA;" OwnerUserId="7281" LastEditorUserId="31730" LastEditDate="2020-05-29T08:27:44.363" LastActivityDate="2020-05-29T08:27:44.363" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10749" PostTypeId="2" ParentId="10525" CreationDate="2015-05-20T16:26:09.390" Score="2" Body="&lt;p&gt;We will not find Galaxies that have existed before the big bang.&#xA;There's a principle in Astronomy that states: The further you look, the earlier you look. That means that the more distance the light had to travel to reach our telescopes, the more time it needed to reach us, because the speed of light is not infinite. Therefore seeing the first galaxies of the universe isn't some lucky coincidence, we are just looking the right distance away (13 billion light years). If we had lived earlier we could still see these galaxies, but they would be closer to us. And the same goes vice versa; If we had lived at an earlier time, we would still see said galaxies, but much farther away. I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="7281" LastActivityDate="2015-05-20T16:26:09.390" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10750" PostTypeId="1" CreationDate="2015-05-20T16:39:34.167" Score="2" ViewCount="3373" Body="&lt;p&gt;What is the longest and what is the shortest wavelength of light? How fast would a light-emitting object have to recede in order to shift the frequency of light to increase the wavelength from the shortest to the longest? (If there are limits)&lt;/p&gt;&#xA;" OwnerUserId="7281" LastEditorUserId="2153" LastEditDate="2015-05-21T00:11:26.637" LastActivityDate="2015-05-26T19:33:30.540" Title="Longest and shortest wavelength" Tags="&lt;light&gt;&lt;wavelength&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10751" PostTypeId="2" ParentId="10745" CreationDate="2015-05-20T18:06:39.007" Score="2" Body="&lt;p&gt;The Newtonian gravitational acceleration of a test object (a small object with negligible mass) toward a point mass is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec g =\frac {GM}{||\vec r||^3} \vec r \tag{1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\vec g$ is the acceleration vector, $G$ is the Newtonian gravitational constant, $M$ is the mass of the point mass, and $\vec r$ is the displacement vector directed from the test object toward the point mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation (1) cannot be used to compute the net gravitation acceleration of a test object toward a pair of point masses by pretending that the pair of point masses is equivalent to a single point mass located at the barycenter of the two point masses. Instead, what needs to be done is to apply equation (1) to the point masses individually and then form the vector sum of these individual accelerations. This in general will not point toward the barycenter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two classes of points where the net gravitational acceleration vector to a pair of point masses is directed toward the barycenter of the point masses:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Points along the line connecting the point masses, less for two regions where the net gravitational acceleration vector points away from the point mass, and&lt;/li&gt;&#xA;&lt;li&gt;Points away from this line and whose projection onto this line is exactly midway between the point masses.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For a test object located anywhere else, the net gravitational acceleration vector does not point toward the barycenter.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-05-20T18:06:39.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10752" PostTypeId="2" ParentId="10743" CreationDate="2015-05-20T19:24:33.593" Score="4" Body="&lt;p&gt;Check out the &lt;a href=&quot;http://bogdan.massclean.org/massclean.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;MASSCLEAN&lt;/a&gt; package, it can be used to generate artificial/synthetic clusters of arbitrary metallicities, ages, mass, radius (based on a King profile distribution) and even includes field stars contamination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example of a cluster generated with the code (right) imitating a known cluster (NGC 3603, left):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uz5Bp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the &lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/bib_query?arXiv:0811.4210&quot; rel=&quot;nofollow noreferrer&quot;&gt;original article&lt;/a&gt; for a good description of the capabilities of the code.&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="354" LastEditDate="2015-05-20T20:05:26.667" LastActivityDate="2015-05-20T20:05:26.667" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10753" PostTypeId="2" ParentId="10526" CreationDate="2015-05-20T20:14:25.183" Score="2" Body="&lt;p&gt;This is likely unanswerable.  Inflation was in part a resolution to a fine tuning problem: without it, it seemed we needed the early universe in a very specific and precisely balanced state to get to where it is today, and there was no solid scientific way to explain why things were so perfectly arranged (other than to simply assert that they were).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With inflation, the state of the universe before the inflationary epoch is fairly irrelevant.  Mostly all it needs is that the region that inflates into what we will know as the observable universe had enough time to achieve thermal equilibrium (which we need to explain why the universe looks pretty much the same in all directions, exactly as if things well outside of light speed communication had nonetheless achieved thermal equilibrium at some point).  Curvature and such get &quot;smeared out&quot; by the inflation to give us the mostly flat and homogeneous universe we see today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a pedantic note, some researchers consider inflation and the big bang to be the same thing. At least in the sense that talking about &quot;before inflation&quot; is scientifically meaningless, so if the Big Bang is the (scientific) beginning then we might as well take it to be inflation.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2015-05-20T20:14:25.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10754" PostTypeId="2" ParentId="10741" CreationDate="2015-05-21T01:16:10.073" Score="2" Body="&lt;p&gt;Only for Sun, Earth and Mars, but here's an interactive Flash that illustrates it graphically. Third link from the top &lt;em&gt;&quot;Part I, Equivalence of Hypotheses (Flash)&quot;&lt;/em&gt; Note that the planets don't move as you move the mouse over the image, only the circles (the theory) change.&#xA;&lt;a href=&quot;http://science.larouchepac.com/kepler/newastronomy/part1/MegaEquivalence.swf&quot; rel=&quot;nofollow&quot;&gt;http://science.larouchepac.com/kepler/newastronomy/part1/MegaEquivalence.swf&lt;/a&gt; for the animation. More generally: &lt;a href=&quot;http://science.larouchepac.com/kepler/newastronomy/&quot; rel=&quot;nofollow&quot;&gt;http://science.larouchepac.com/kepler/newastronomy/&lt;/a&gt;&#xA;(And I don't know or care who that LaRouge Pacman is, some rich guy for some reason financed good explanations of what Kepler did, that's what I know about it and that's all very good)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you move the mouse pointer to the upper left you get the heliocentric system. To the upper right you get the geocentric system. The point is to illustrate how Kepler proved that Ptolemy, Copernicus and Tycho all had the same mathematical model for the solar system. They just used different frames of reference to it, just choosing different fixpoints from where to view it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some dots in the illustration represent important points in that ancient model. Some of them are stand-ins for the foci of an ellipse in a model made out of perfect circles. And the Sun was actually not the center of the Copernican model, a point near the Sun was. According to Copernicus the Sun was just a decoration which randomly happened to hang around near the center. Kepler was the first to come up with the thought that maybe there is a relationship between the geometry and the physical existence of stuff. An idea later developed to the theory of gravity by Newton and Einstein, who just added to Kepler's initial formula for describing the movements of the planets.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2016-03-17T18:59:32.807" LastActivityDate="2016-03-17T18:59:32.807" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10756" PostTypeId="1" AcceptedAnswerId="10757" CreationDate="2015-05-21T07:59:02.457" Score="14" ViewCount="1275" Body="&lt;p&gt;I just was wondering and searching on the internet with little luck in the topic. On Antimatter Wiki they tell the observable universe is built up by matter. I read antimatter can be detected in cosmic rays in very small quantity. But let's go in theories and our practical knowledge, my question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could we detect if a set of objects (like an entire galaxy) is completely built up by antimatter? Since matter and antimatter look like &quot;mirror image&quot; on current, could they form a similar object which is indistinguishable from &quot;our&quot; matter objects? I mean, would spectrum analysis for example show that it is not matter, but built up by antimatter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Linked question: Are we sure about that what we see in universe is matter?&lt;/p&gt;&#xA;" OwnerUserId="1799" LastEditorUserId="1799" LastEditDate="2015-05-21T08:12:03.160" LastActivityDate="2015-05-21T20:46:48.667" Title="How can we tell the difference between matter and antimatter by observation in space?" Tags="&lt;observational-astronomy&gt;&lt;antimatter&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10757" PostTypeId="2" ParentId="10756" CreationDate="2015-05-21T12:14:29.030" Score="13" Body="&lt;h2&gt;Tl;DR&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Detection via &lt;strong&gt;polarized light&lt;/strong&gt; - Antimatter interaction with polarized light could be detected by vector rotation;&lt;/li&gt;&#xA;&lt;li&gt;We're &lt;strong&gt;&lt;em&gt;mostly&lt;/em&gt; sure&lt;/strong&gt;, because absence of gamma rays and characteristic Faraday polarization indicates absence of observable antimatter in meaningful amounts.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Long answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I do believe @userLTK to be correct on his comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my limited knowledge, the absence of gamma-ray bursts that characterizes matter-antimatter interactions hints of baryonic matter prevalence throughout the visible universe; an anti-galaxy near a baryonic galaxy would show a healthy amount of gamma rays originating from boundary particle collisions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LyAjq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This effect would be observable between structures of any scale - e.g. star/anti-star, galaxy/anti-galaxy or supercluster/anti-supercluster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://books.google.com/books?id=H3fpCAAAQBAJ&amp;amp;printsec=frontcover#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;&quot;Light from the Depths of Time&quot;&lt;/a&gt;, by Rudolph Kippenhahn, mentions that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] Even when, like a star, it is radiation, its spectrum remains&#xA;  exactly the same, quite independent of whether atoms or anti-atoms are&#xA;  responsible for the light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the light originating from anti-galaxies would look the same as ordinary galaxies. But what about gravity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quote from &lt;a href=&quot;https://books.google.com/books?id=T43ZoeIan64C&amp;amp;pg=PA317&amp;amp;lpg=PA317&amp;amp;dq=light%20originating%20from%20anti-matter&amp;amp;source=bl&amp;amp;ots=u6_o3ZWqEv&amp;amp;sig=TzYMlXCz2SzPKCmsXspvOWQ6MW4&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=lsldVfnJI4GqNvCkgagN&amp;amp;ved=0CCAQ6AEwAA#v=onepage&amp;amp;q=light%20originating%20from%20anti-matter&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;&quot;Isodual Theory of Antimatter: with applications to Antigravity, Grand Unification and Cosmology&quot;&lt;/a&gt; may help us further:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] However, the photons is invariant under charge conjugation and&#xA;  travel at the maximal causal speed in vacuum, &lt;em&gt;c&lt;/em&gt;. Therefore, the&#xA;  photon could well result to be a superposition of positive and&#xA;  negative energies, perhaps as a condition to travel at the speed &lt;em&gt;c&lt;/em&gt;,&#xA;  in which case the photon would be an isoselfdual state, thus&#xA;  experiencing attraction in both fields of matter and antimatter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Meaning that antimatter would also cause gravitational effects, such as lensing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That would make remote detection of anti-matter quite hard - it would emit and bend light in exactly the same fashion as common matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Faraday polarization rotation could give us some hope:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Polarized light, e.g., from non-thermal synchrotron sources, that&#xA;  passes through gas with a non-zero magnetic field will have its&#xA;  polarization vector rotated by the process of Faraday rotation. [...]&#xA;  Note that regions dominated by antimatter (positrons) cause a rotation&#xA;  &lt;strong&gt;opposite&lt;/strong&gt; to that caused by regions dominated by matter (electrons).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.slac.stanford.edu/econf/C040802/papers/L017.PDF&quot; rel=&quot;noreferrer&quot;&gt;SLAC Summer Institute on Particle Physics (SSI04), Aug. 2-13, 2004&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the same source mentions that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The effect has been measured many times and the amount of rotation,&#xA;  usually expressed in terms of the so-called “rotation measure” is&#xA;  given by the line-of-sight integral [...] The fact that we observe an&#xA;  effect at all means that on average, we can’t have equal amounts of&#xA;  antimatter and matter along our various lines of sight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So it seems that the visible universe is missing some antimatter after all.&lt;/p&gt;&#xA;" OwnerUserId="893" LastEditorUserId="893" LastEditDate="2015-05-21T20:46:48.667" LastActivityDate="2015-05-21T20:46:48.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10758" PostTypeId="1" CreationDate="2015-05-21T17:49:39.313" Score="3" ViewCount="187" Body="&lt;p&gt;The paper is :&#xA;&lt;a href=&quot;http://www.astro.washington.edu/users/ivezic/Publications/tomographyI.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.astro.washington.edu/users/ivezic/Publications/tomographyI.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation is equation #23 in the paper. It's a model for the density of stars in the Milky Way's disk. It has an exponential dependence on both $R$ and $Z$. $R$ is the distance from the center of the galaxy, and $Z$ is the distance above/below the plane of the disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relation is roughly this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\varrho=\text{constant}\times e^{-\frac{R}{L}-\frac{Z+Z_0}{H}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I'm seeing is the $Z$-dependence of the formula. $R$ and $Z$ here are standard cylindrical coordinates. Thus, $Z$ can be positive or negative. The problem is that the formula blows up when $Z$ is negative. This is unphysical, since the number density must generally decrease going further away from the disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I missing something? Or should the equation really have an absolute magnitude of $|Z + Z_0|$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Added Later:]&#xA;I finished making a &lt;a href=&quot;http://johnohanley.ca/sky/milky_way_stellar_density/form.html&quot; rel=&quot;nofollow&quot; title=&quot;3D demo&quot;&gt;3D demo&lt;/a&gt; of the stellar number density of the Milky Way. Note that your browser needs to support &lt;a href=&quot;https://get.webgl.org/&quot; rel=&quot;nofollow&quot;&gt;WebGL&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5359" LastEditorUserId="5359" LastEditDate="2015-06-22T20:32:48.110" LastActivityDate="2015-06-22T20:32:48.110" Title="Milky Way stellar number density : is the stated equation in this paper incorrect?" Tags="&lt;milky-way&gt;&lt;stellar-dynamics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10759" PostTypeId="1" AcceptedAnswerId="10762" CreationDate="2015-05-22T19:00:59.140" Score="6" ViewCount="1301" Body="&lt;p&gt;I have seen different pictures in a different community to showcasing the Milky Way. I can never tell whether it's edited or it's the real picture. Actually, I can never see any galaxy as such by with my naked eye. But can the Milky Way be seen with the naked eye? If yes, then I would really like to know when and how this is possible.&lt;/p&gt;&#xA;" OwnerUserId="7313" LastEditorUserId="2153" LastEditDate="2015-05-24T20:14:44.750" LastActivityDate="2015-05-26T04:41:29.943" Title="Can the Milky Way be seen with the naked eye? Does this apply to any galaxy? If yes, then how and when?" Tags="&lt;galaxy&gt;&lt;amateur-observing&gt;&lt;milky-way&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10760" PostTypeId="2" ParentId="10759" CreationDate="2015-05-22T19:58:41.383" Score="3" Body="&lt;p&gt;To see the milky way, you need merely go out at night somewhere where there is a dark, clear sky. In northern hemisphere fall and winter, you'll also be able to see the nearest large galaxy, &lt;a href=&quot;http://en.wikipedia.org/wiki/Andromeda_Galaxy&quot; rel=&quot;nofollow&quot;&gt;M31&lt;/a&gt;. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Magellanic_Clouds&quot; rel=&quot;nofollow&quot;&gt;large and small Magellanic clouds&lt;/a&gt;, visibly grace southern skies.&#xA;If you live in a city, a cloudy, hazy or very humid region, you're out of luck, unless you can get out into unspoiled countryside.&#xA;Modern digital cameras are &lt;em&gt;far&lt;/em&gt; more sensitive than the human eye.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-05-22T19:58:41.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10761" PostTypeId="2" ParentId="10759" CreationDate="2015-05-22T21:42:38.193" Score="3" Body="&lt;p&gt;In order to see the Milky Way, or the white streak across the night sky, you need to go to a VERY dark place where there is no light pollution whatsoever, and of course the sky has to be clear. So you should go to places like the desert, the woods, the mountains, and some observatories. When you look up, you will see the white streak which is apart of the Orion Arm of the Milky Way where we live ;)&lt;/p&gt;&#xA;" OwnerUserId="7258" LastActivityDate="2015-05-22T21:42:38.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10762" PostTypeId="2" ParentId="10759" CreationDate="2015-05-22T22:52:25.807" Score="3" Body="&lt;p&gt;Most photographs you will see of the Milky Way are long exposure photographs (often around 30 seconds exposure, whatever the limit is before star trails become noticeable at their focal distance; sometimes longer if they use a tracking mount) at a very wide aperture.  This is done to capture the maximum amount of light and, in fact, capture far more light than your eyes can capture. It also usually takes a little adjusting to colorize the Milky Way the way we expect it to be colored. Note, that this doesn't mean the photos aren't real - if you define photos' realism by representing exactly what people see, then no photo has ever been real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While you may not be able to see the Milky Way to this degree, you can still go out and see her. As others described, consider your location and time of year, take care to find a dark spot with clear skies and give your eyes enough time in the dark to adjust.&lt;/p&gt;&#xA;" OwnerUserId="1288" LastActivityDate="2015-05-22T22:52:25.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10763" PostTypeId="1" AcceptedAnswerId="10840" CreationDate="2015-05-23T04:40:38.063" Score="8" ViewCount="473" Body="&lt;p&gt;Some kind of ellipsoidal shape mirror could reflect to a secondary mirror which is not in the way of the infalling light. Two advantages are immediately obvious. The primary mirror would not be partly shadowed by the secondary. And the support structure of the secondary would not cause diffraction spikes in point sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the reasons for why this is not only not the standard, but quite uncommon? Especially on the greatest telescopes, one would think that the additional production difficulties, if any, are not a relevant consideration given the budget and clever engineers behind them.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-05-27T14:32:31.407" Title="Why aren't secondary mirrors offset to get rid of diffraction spikes due to the support vanes?" Tags="&lt;telescope&gt;&lt;optics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10764" PostTypeId="2" ParentId="10447" CreationDate="2015-05-23T12:19:15.427" Score="0" Body="&lt;p&gt;Alpha Centauri B is the closest star we know of that has a planet revolving around it other than the sun. We know of no planets in orbit of Proxima Centauri or Alpha Centauri A, but there are possible candidates. The planet around Alpha Centauri B (Alpha Centauri Bb) revolves in just 3.24 days which means it doesn't orbit in the habitable zone. The terrestrial planet is about 1.1 times as massive as Earth and scientist do not assume that it harbours life. &lt;a href=&quot;http://en.wikipedia.org/wiki/Alpha_Centauri#Alpha_Centauri_Bb&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Alpha_Centauri#Alpha_Centauri_Bb&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7281" LastEditorUserId="16685" LastEditDate="2019-01-06T11:54:31.383" LastActivityDate="2019-01-06T11:54:31.383" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="10765" PostTypeId="2" ParentId="10450" CreationDate="2015-05-23T12:33:28.030" Score="1" Body="&lt;p&gt;There are multipble theories that could provide an answer to your question. One of them is that the universe is a quantum fluctuation that created forces, energy, etc.. This would explain why we observe the total energy of the universe to be 0 (negative energy from repulsive gravity and positive energy from radiation, mass etc.). This theory is supported by alan guth's and Andrei Linde called inflation. &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_fluctuation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Quantum_fluctuation&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Inflation_%28cosmology%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Inflation_%28cosmology%29&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another theory called brane-theory, an idea that arose frome string theory which include multiverses etc., states that our universe was created in a collison between to branes. &lt;a href=&quot;http://en.wikipedia.org/wiki/String_cosmology&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/String_cosmology&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7281" LastActivityDate="2015-05-23T12:33:28.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10767" PostTypeId="1" AcceptedAnswerId="10777" CreationDate="2015-05-23T14:12:53.587" Score="3" ViewCount="90" Body="&lt;p&gt;We've found over a thousand exoplanets by now, but how many of them could support life? What percentage of newly found planets are potentially life-harbouring? Or haven't we found any exoplanets that meet these criteria yet?&lt;/p&gt;&#xA;" OwnerUserId="7281" LastActivityDate="2015-05-24T10:47:02.853" Title="Probability of finding life-supporting exoplanets" Tags="&lt;exoplanet&gt;&lt;life&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10768" PostTypeId="1" AcceptedAnswerId="10780" CreationDate="2015-05-23T14:47:11.383" Score="2" ViewCount="134" Body="&lt;p&gt;We discover vastly more exoplanets that are gas giants that orbit their parent star very closely, than small terrestrial planets that are more distant to their sun. Could this be due to the methods we use to find exoplanets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gas giants have stronger gravity which makes their star oscillate more. Moreover they are larger which eclipses more of the light of the star during transits?  Are big hot Jupiters far easier to detect than small rocky worlds?&lt;/p&gt;&#xA;" OwnerUserId="7281" LastEditorUserId="2365" LastEditDate="2016-07-17T09:13:43.633" LastActivityDate="2016-07-17T09:13:43.633" Title="Types of Exoplanets" Tags="&lt;exoplanet&gt;&lt;gas-giants&gt;&lt;terrestrial-planets&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10769" PostTypeId="1" CreationDate="2015-05-23T15:45:35.467" Score="2" ViewCount="326" Body="&lt;p&gt;The question is a bit vague, but let me explain:    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take for example 2 bodies of the same mass. They can orbit around their centre of mass/gravity. Is something like this possible with multiple bodies of varying mass and distance from the centre of gravity without a relatively extremely massive body there?&#xA;i.e. would it be possible to have a planetary system that orbited around the centre of mass without there being a massive star (roughly) in the middle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the Sun also has an orbit around the centre of gravity of the solar system, but that orbit is very small compared to its size, which puts it roughly in the centre.&lt;/p&gt;&#xA;" OwnerUserId="7318" LastActivityDate="2020-12-23T20:15:20.377" Title="For how many bodies can there be a stable orbit with no very heavy central body?" Tags="&lt;orbit&gt;&lt;n-body-simulations&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10771" PostTypeId="1" AcceptedAnswerId="10772" CreationDate="2015-05-23T17:54:21.423" Score="9" ViewCount="838" Body="&lt;p&gt;I understand why our sky is blue and turns red when the sun sets or rises. I know why the sky on mars is red, but why does it turn blue during sunsets and rises? Shouldn't this actually not work due to the shorter wavelenght blue has compared to red?&lt;/p&gt;&#xA;" OwnerUserId="7281" LastActivityDate="2019-02-01T10:57:28.223" Title="Sunsets: Mars/Earth" Tags="&lt;light&gt;&lt;mars&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10772" PostTypeId="2" ParentId="10771" CreationDate="2015-05-23T19:54:48.403" Score="13" Body="&lt;p&gt;&lt;strong&gt;tl;dr&lt;/strong&gt;: On Earth, atmospheric molecules scatter blue light more than red. In contrast, on Mars atmospheric, &lt;span class=&quot;math-container&quot;&gt;$\mu\mathrm{m}$&lt;/span&gt;-sized hematite dust particles happen to give the opposite effect.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Scattering theory&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;First a bit of theory. Pretty pictures later.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Scattering cross section&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Scattering of light of wavelength &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; on particles of radius &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; may be described in general using &lt;a href=&quot;https://en.wikipedia.org/wiki/Mie_scattering&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mie theory&lt;/a&gt;. Using the variable &lt;span class=&quot;math-container&quot;&gt;$x \equiv 2\pi r/\lambda$&lt;/span&gt;, the scattering cross section &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; increases steeply with &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; until &lt;span class=&quot;math-container&quot;&gt;$x\sim2\pi$&lt;/span&gt;, after which it decreases again and starts to oscillate, as seen in this figure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XHam9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XHam9.png&quot; alt=&quot;scat&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Scattering cross section (normalised to the geometrical cross section) as a function of particle size-to-wavelength ratio (from &lt;a href=&quot;https://en.wikipedia.org/wiki/Mie_scattering&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; with my own annotations).&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, when a particle is much smaller than the wavelength of the light (&lt;span class=&quot;math-container&quot;&gt;$x\ll 1$&lt;/span&gt;), the probability of scattering the light increases steeply with decreasing wavelength — in fact the cross section goes as &lt;span class=&quot;math-container&quot;&gt;$\lambda^{-4}$&lt;/span&gt;. This regime is called Rayleigh scattering. The scattering efficiency reaches a maximum for particles roughly equal to the wavelength of the light (&lt;span class=&quot;math-container&quot;&gt;$x\simeq2\pi$&lt;/span&gt;; green arrow in the figure). Hereafter, it falls again, exhibits some oscillations that have to do with interference in the particle, slowly converging to the &quot;geometric&quot; value &lt;span class=&quot;math-container&quot;&gt;$\sigma = \pi r^2$&lt;/span&gt; (i.e. what would be expected classically).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many (astrophysical) situations dust tends to be a distribution of sizes with more small than large particles (typically &lt;span class=&quot;math-container&quot;&gt;$n\propto r^{-3.5}$&lt;/span&gt;), meaning that an ensemble of dust has much more particles that scatter short wavelengths (blue) than long wavelengths (red). Thus, blue light is filtered out and red light light tends to be transmitted, in effect reddening the light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if the dust consists of particles of the same size, then the extinction may rise and fall with wavelength, as seen in the figure. For example, optical light with &lt;span class=&quot;math-container&quot;&gt;$\lambda \simeq [0.4\text{–}0.7] \,\mu\mathrm{m}$&lt;/span&gt;, scattered by particles of size &lt;span class=&quot;math-container&quot;&gt;$0.7\,\mu\mathrm{m}$&lt;/span&gt;, is scattered according to the orange region in the figure — i.e. red light is scattered more efficiently than blue light. The exact form depend on the refractive index of the scatterer.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Phase function&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;When a photon is scattered, it doesn't happen in a completely random direction, but follows a direction distribution given by the &lt;em&gt;phase function&lt;/em&gt;. In the Rayleigh regime, there is a slight preference for forward and backward scattering, but for large &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; there is a strong preference for forward scattering, as seen in this diagram (from &lt;a href=&quot;http://pages.mtu.edu/~scarn/teaching/GE4250/scattering_lecture_slides.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;these slides&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/joXoa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/joXoa.png&quot; alt=&quot;phase&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Sunsets of Earth&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;On Earth the sunlight is scattered primarily by molecules which are much smaller than the wavelength of the light, i.e. we're in the Rayleigh regime. Hence, blue light tends to be scattered a lot, creating the blue sky. As the blue light is filtered out of the line of sight (LOS) toward the Sun, its light — white to begin with — becomes more yellow. And as the Sun sets, its light has to go through a thicker layer of atmosphere, enhancing this &lt;em&gt;reddening effect&lt;/em&gt; and giving a red sunset.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Sunsets on Mars&lt;/h1&gt;&#xA;&#xA;&lt;h3&gt;Blue Sun&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;On Mars, there is almost no atmosphere, only 1% of Earth's. Instead, the Sun's light is scattered by the fine, red dust swirled up in the (thin) air. Martian dust contains &lt;a href=&quot;https://en.wikipedia.org/wiki/Hematite&quot; rel=&quot;nofollow noreferrer&quot;&gt;hematite&lt;/a&gt; (&lt;span class=&quot;math-container&quot;&gt;$\alpha\text{-}\mathrm{Fe}_2\mathrm{O}_3$&lt;/span&gt;), the mineral responsible for Mars' red color. This iron oxide has a very wavelength-dependent refractive index, and the grain sizes lie rather tightly around a few &lt;span class=&quot;math-container&quot;&gt;$\mu\mathrm{m}$&lt;/span&gt;, scattering red light more than blue as in the (ideal) example given above (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/1997JGR...102.9039O&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ockert-Bell et al. 1997&lt;/a&gt;; &lt;a href=&quot;http://adsabs.harvard.edu/abs/1999ficm.conf.6085L&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lane et al. 1999&lt;/a&gt;). This has nothing to do with the fact that the dust is red, but is merely a consequence of the size of the particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, in this case it is the &lt;em&gt;red&lt;/em&gt; light which is filtered away, making the Sun appear blue, especially as it sets.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Blue halo and red sky&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The unusual size distribution of the dust particles described above is the reason for the blue disk of the Sun, but cannot explain the blue halo around it (&lt;a href=&quot;https://www.osapublishing.org/ao/abstract.cfm?uri=ao-53-9-1808&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ehlers et al. 2014&lt;/a&gt;). This spectacular phenomenon is caused by the difference in the phase function described above: Whereas Rayleigh scattering is rather isotropic, Mie scattering tends to scatter in the forward direction, and  Martian dust in particular has a large asymmetry parameter (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2007JGRE..112.8S13V&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vincendon et al. 2007&lt;/a&gt;), meaning that light that &lt;em&gt;is&lt;/em&gt; scattered has a preference for being scattered along the LOS, i.e. toward the observer of the sunset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, the asymmetry parameter is dependent on the wavelength of the light, with the forward scattering of blue light being much more pronounced than for red light, as seen in this figure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nxnIH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nxnIH.png&quot; alt=&quot;asym&quot;&gt;&lt;/a&gt;&#xA;&lt;sup&gt;Probability of scattering at a given angle for blue light (425 nm, left) and red light (694 nm, right). The three line types correspond to different models (from &lt;a href=&quot;https://www.osapublishing.org/ao/abstract.cfm?uri=ao-53-9-1808&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ehlers et al. (2014)&lt;/a&gt;).&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the red light has a larger tendency to be scattered out of the LOS, creating the reddish sky.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Pretty picture&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I promised a pretty picture. Here's a Martian sunset:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Hy7dC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hy7dC.jpg&quot; alt=&quot;MartianSunset&quot;&gt;&lt;/a&gt;&#xA;&lt;sup&gt;Credit: &lt;a href=&quot;https://mars.nasa.gov/resources/5300/a-martian-sunset/?site=insight&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA/JPL-Caltech/Texas A&amp;amp;M/Cornell&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Blue Sun and Moon on Earth&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;On rare occasions, aerosols in Earth's atmosphere can make the Sun and Moon  appear blue. When the volcano Krakatoa erupted in 1853, the Moon looked blue in some places on Earth for almost a month (&lt;a href=&quot;http://www.mv.helsinki.fi/home/asmi/Aitken/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aitken 1892&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2019-02-01T10:57:28.223" LastActivityDate="2019-02-01T10:57:28.223" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="10774" PostTypeId="2" ParentId="10759" CreationDate="2015-05-24T03:58:43.313" Score="5" Body="&lt;p&gt;There are a few nearby galaxies that can be seen with the naked eye, but clear &amp;amp; dark skies are essential. Most notable for northern-hemisphere viewers is the Andromeda Galaxy also known as M31, best viewed in late summer and fall; from the southern hemisphere the Magellanic Clouds, irregular satellites of our own Galaxy, are famous landmarks. These all look like small fuzzy patches, rather unexciting until you consider that they‘re on the order of a million light-years away.&lt;/p&gt;&#xA;" OwnerUserId="7321" LastActivityDate="2015-05-24T03:58:43.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10775" PostTypeId="2" ParentId="10750" CreationDate="2015-05-24T04:31:34.257" Score="4" Body="&lt;p&gt;Regarding visible light, the nominal range is usually given as 400 nm (violet) to 700 nm (red). For light of the former wavelength to get Doppler-shifted to the latter means a &lt;em&gt;z&lt;/em&gt; value of (700–400)/400 = 0.75, corresponding to a recession of 152,000 km/s, just over half the speed of light.&lt;/p&gt;&#xA;" OwnerUserId="7321" LastActivityDate="2015-05-24T04:31:34.257" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10776" PostTypeId="2" ParentId="240" CreationDate="2015-05-24T05:26:12.997" Score="0" Body="&lt;p&gt;Moonrise azimuth... The Moon travels around the Ecliptic over a months time, so its azimuth angle at rising will be &quot;comparable&quot; to the Sun's rising azimuth angle pn each day - equal to the Sun's at times but other times with a sizable &quot;offset&quot; in  angle as large as +/- 30 degrees. &lt;/p&gt;&#xA;" OwnerUserId="7323" LastActivityDate="2015-05-24T05:26:12.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10777" PostTypeId="2" ParentId="10767" CreationDate="2015-05-24T10:47:02.853" Score="3" Body="&lt;p&gt;The question can be rephrased into: How many exoplanets have been found that could harbour life? As of early May 2015 the number appears to be about 30, see the &lt;a href=&quot;http://phl.upr.edu/projects/habitable-exoplanets-catalog&quot; rel=&quot;nofollow&quot;&gt;Habitable Exoplanet Catalog&lt;/a&gt; on the Planetary Habitability Laboratory's web site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should also note that exoplanet detection methods will bias this sample in favour of Super-Earths.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-05-24T10:47:02.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10778" PostTypeId="2" ParentId="10769" CreationDate="2015-05-24T14:39:49.433" Score="1" Body="&lt;p&gt;Two star systems:   &lt;a href=&quot;http://en.wikipedia.org/wiki/Circumbinary_planet&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Circumbinary_planet&lt;/a&gt; - not exactly what you asked, but I think it's as close as real systems will get.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example you give, 2 planets orbiting each other is easy, two objects will tend to orbit each other but when you have 3 objects, it gets a bit more complicated, unless the 3rd one is distant and orbits the two from a distance.    3 body orbits tend to be unstable unless you have them kind of locked in (sun/earth/moon) with the moon around the earth and the earht-moon around the sun, or in synch (Sun/Neptune/Pluto) - where Pluto and Neptune are in orbital resonance.   Neither of those would be stable if the sun wasn't so dominant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are exotic mathematical models for 3 object orbits but they are improbable in the actual universe cause they aren't very stable.   &lt;a href=&quot;http://www.ams.org/samplings/feature-column/fcarc-orbits1&quot; rel=&quot;nofollow&quot;&gt;http://www.ams.org/samplings/feature-column/fcarc-orbits1&lt;/a&gt;  and  &lt;a href=&quot;https://www.sciencenews.org/article/strange-orbits-1&quot; rel=&quot;nofollow&quot;&gt;https://www.sciencenews.org/article/strange-orbits-1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-05-24T14:39:49.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10779" PostTypeId="1" CreationDate="2015-05-24T14:41:54.237" Score="4" ViewCount="98" Body="&lt;p&gt;I was reading &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1961PASJ...13..450H&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow&quot;&gt;Stellar Evolution in Early Phases of Gravitational Contraction&lt;/a&gt;, by Chushiro Henyey, where he writes,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $L \propto R^{-\alpha}$ along the path, the age of a star from the time when $R=\infty$ is given by&#xA;  $$t=C/(1-\alpha), C=GM^2/RL=10^{7.20}\left(\frac{M}{M_{\odot}}\right)^2\frac{R_{\odot}}{R}\frac{L_{\odot}}{L}\text{ years}$$&#xA;  where $R$ and $L$ are the present values.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This seems to be just for stars on the Hayashi track (and treated here just for Population 1 stars). Is there a similar expression for stars on the Henyey track, or is the same expression valid in both scenarios?&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-24T14:50:56.810" LastActivityDate="2015-05-24T14:50:56.810" Title="How is the age of a star on the Henyey track calculated?" Tags="&lt;star&gt;&lt;stellar-evolution&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10780" PostTypeId="2" ParentId="10768" CreationDate="2015-05-24T21:00:51.067" Score="1" Body="&lt;p&gt;Any statement about the relative frequencies of exoplanets of different size, is limited to the parameter space over which they can detected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A recent analysis of the Kepler-discovered exoplanets, that takes account of observational selection biases, indicates that for planets orbiting with periods less than about 85 days, that &quot;small Neptune&quot;-sized objects (a few Earth-radii in size) are slightly more common (20%) than Earth-sized objects (17%), which are in turn much more common than Jupiter-sized objects (few %).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extending this comparison outside this range of orbital periods is not possible with present data (though the paper was in 2013 and things may have now moved on a little).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I never did attach a reference: Here it is &lt;a href=&quot;https://www.cfa.harvard.edu/news/2013-01&quot; rel=&quot;nofollow&quot;&gt;https://www.cfa.harvard.edu/news/2013-01&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-07-17T09:05:46.683" LastActivityDate="2016-07-17T09:05:46.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10781" PostTypeId="1" AcceptedAnswerId="10923" CreationDate="2015-05-25T01:18:28.157" Score="4" ViewCount="271" Body="&lt;p&gt;The telescope &lt;a href=&quot;http://en.wikipedia.org/wiki/Big_Bear_Solar_Observatory&quot; rel=&quot;nofollow noreferrer&quot;&gt;big bear&lt;/a&gt; is located in the Southern California.&#xA;According to the wikipedia page, water can cool the observatory.&lt;/p&gt;&#xA;&lt;p&gt;Is it better to build it on a cold high mountain with stable and clear atmosphere?&lt;/p&gt;&#xA;&lt;p&gt;What is the difference when building a solar and an optical telescope?  Optical telescopes are usually not build on lakes.&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="37191" LastEditDate="2021-05-19T15:22:03.017" LastActivityDate="2021-05-19T15:22:03.017" Title="Why are solar telescopes built on lakes?" Tags="&lt;the-sun&gt;&lt;telescope&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10784" PostTypeId="2" ParentId="10759" CreationDate="2015-05-25T14:02:01.697" Score="2" Body="&lt;p&gt;Yes, we certainly can see the Milky Way from Earth. I saw it through an airplane window at night. But you don't need to do that, either. Find some place with not much pollution, especially light and smoke, and gaze up at night. You should see a beautiful band of stars. Proof? In the 1990's, there was a major electricity blackout in Los Angeles. People looked up and saw it, but many of them thought that aliens were invading or something . . . so they called up 911. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All you need to know is that you can, you should, and hopefully, you will. As for other galaxies, you won't see them in quite the same way, because we're not situated in such a viewpoint. But you can see them, like the Andromeda Galaxy, if you find a dark place without pollution. Go to the countryside!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also possible at any time. Because the milky way is shaped like a bulging disc, and we're in a great spot, we can see it at any time; it spans across the whole Celestial Sphere!&lt;/p&gt;&#xA;" OwnerUserId="7332" LastEditorUserId="7332" LastEditDate="2015-05-26T04:41:29.943" LastActivityDate="2015-05-26T04:41:29.943" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10785" PostTypeId="5" CreationDate="2015-05-25T14:27:38.567" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:27:38.567" LastActivityDate="2015-05-25T14:27:38.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10786" PostTypeId="4" CreationDate="2015-05-25T14:27:38.567" Score="0" Body="Questions regarding the history of astronomy, including discoveries and scientists." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:27:38.567" LastActivityDate="2015-05-25T14:27:38.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10787" PostTypeId="5" CreationDate="2015-05-25T14:28:19.173" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:28:19.173" LastActivityDate="2015-05-25T14:28:19.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10788" PostTypeId="4" CreationDate="2015-05-25T14:28:19.173" Score="0" Body="Questions regarding interactions between galaxies or between major parts of a galaxy." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:28:19.173" LastActivityDate="2015-05-25T14:28:19.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10789" PostTypeId="5" CreationDate="2015-05-25T14:28:41.667" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:28:41.667" LastActivityDate="2015-05-25T14:28:41.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10790" PostTypeId="4" CreationDate="2015-05-25T14:28:41.667" Score="0" Body="For questions about objects or phenomena outside Earth, generally on other planets." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:28:41.667" LastActivityDate="2015-05-25T14:28:41.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10791" PostTypeId="5" CreationDate="2015-05-25T14:29:25.253" Score="0" Body="&lt;p&gt;See also the &lt;a href=&quot;/questions/tagged/amateur-observing&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;amateur-observing&amp;#39;&quot; rel=&quot;tag&quot;&gt;amateur-observing&lt;/a&gt; and &lt;a href=&quot;/questions/tagged/observational-astronomy&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;observational-astronomy&amp;#39;&quot; rel=&quot;tag&quot;&gt;observational-astronomy&lt;/a&gt; tags.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:29:25.253" LastActivityDate="2015-05-25T14:29:25.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10792" PostTypeId="4" CreationDate="2015-05-25T14:29:25.253" Score="0" Body="Questions regarding casually watching the stars." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:29:25.253" LastActivityDate="2015-05-25T14:29:25.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10793" PostTypeId="5" CreationDate="2015-05-25T14:30:31.800" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:30:31.800" LastActivityDate="2015-05-25T14:30:31.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10794" PostTypeId="4" CreationDate="2015-05-25T14:30:31.800" Score="0" Body="Questions regarding massive planets composed of layers of gases, such as hydrogen and helium, surrounding a solid/liquid core." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:30:31.800" LastActivityDate="2015-05-25T14:30:31.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10795" PostTypeId="5" CreationDate="2015-05-25T14:34:36.757" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:34:36.757" LastActivityDate="2015-05-25T14:34:36.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10796" PostTypeId="4" CreationDate="2015-05-25T14:34:36.757" Score="0" Body="Questions regarding disks of gas, dust, and plasma around a central object, such as a black hole, which takes in the matter." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:34:36.757" LastActivityDate="2015-05-25T14:34:36.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10797" PostTypeId="5" CreationDate="2015-05-25T14:34:59.653" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:34:59.653" LastActivityDate="2015-05-25T14:34:59.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10798" PostTypeId="4" CreationDate="2015-05-25T14:34:59.653" Score="0" Body="Questions regarding telescopes in orbit around Earth, such as the Hubble Space Telescope." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:34:59.653" LastActivityDate="2015-05-25T14:34:59.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10799" PostTypeId="5" CreationDate="2015-05-25T14:35:38.757" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:35:38.757" LastActivityDate="2015-05-25T14:35:38.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10800" PostTypeId="4" CreationDate="2015-05-25T14:35:38.757" Score="0" Body="Questions regarding specific terms, names, or naming conventions." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:35:38.757" LastActivityDate="2015-05-25T14:35:38.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10801" PostTypeId="5" CreationDate="2015-05-25T14:36:30.227" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:36:30.227" LastActivityDate="2015-05-25T14:36:30.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10802" PostTypeId="4" CreationDate="2015-05-25T14:36:30.227" Score="0" Body="Questions regarding two or more stars gravitationally bound to each other" OwnerUserId="2153" LastEditorUserId="9301" LastEditDate="2016-05-24T14:18:25.303" LastActivityDate="2016-05-24T14:18:25.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10803" PostTypeId="5" CreationDate="2015-05-25T14:37:21.603" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:37:21.603" LastActivityDate="2015-05-25T14:37:21.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10804" PostTypeId="4" CreationDate="2015-05-25T14:37:21.603" Score="0" Body="Questions regarding bodies smaller than planets but larger than other bodies, that do not meet at least one of the IAU's criteria for being a planet." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:37:21.603" LastActivityDate="2015-05-25T14:37:21.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10805" PostTypeId="5" CreationDate="2015-05-25T14:38:06.467" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:38:06.467" LastActivityDate="2015-05-25T14:38:06.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10806" PostTypeId="4" CreationDate="2015-05-25T14:38:06.467" Score="0" Body="Questions regarding how bright an object appears to be." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:38:06.467" LastActivityDate="2015-05-25T14:38:06.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10807" PostTypeId="5" CreationDate="2015-05-25T14:39:03.450" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:39:03.450" LastActivityDate="2015-05-25T14:39:03.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10808" PostTypeId="4" CreationDate="2015-05-25T14:39:03.450" Score="0" Body="Questions regarding the theory of gravity as proposed by Newton, in which the force of gravity depends only on the mass of bodies, a constant, and the distance between them, and depends on one equation." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:39:03.450" LastActivityDate="2015-05-25T14:39:03.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10809" PostTypeId="5" CreationDate="2015-05-25T14:39:29.623" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:39:29.623" LastActivityDate="2015-05-25T14:39:29.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10810" PostTypeId="4" CreationDate="2015-05-25T14:39:29.623" Score="0" Body="Questions regarding grouping objects together based on one or more shared property." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:39:29.623" LastActivityDate="2015-05-25T14:39:29.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10811" PostTypeId="5" CreationDate="2015-05-25T14:41:34.990" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:41:34.990" LastActivityDate="2015-05-25T14:41:34.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10812" PostTypeId="4" CreationDate="2015-05-25T14:41:34.990" Score="0" Body="Questions regarding the movement of an object at a given speed and direction, often related to the concepts of orbital velocity and escape velocity." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:41:34.990" LastActivityDate="2015-05-25T14:41:34.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10813" PostTypeId="5" CreationDate="2015-05-25T14:42:05.553" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:42:05.553" LastActivityDate="2015-05-25T14:42:05.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10814" PostTypeId="4" CreationDate="2015-05-25T14:42:05.553" Score="0" Body="Questions regarding the inner portion of a galaxy; in the Milky Way, this is the area around Sagittarius A*." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:42:05.553" LastActivityDate="2015-05-25T14:42:05.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10815" PostTypeId="2" ParentId="10499" CreationDate="2015-05-25T14:44:44.063" Score="-1" Body="&lt;p&gt;Well, put simply, the Sun will certainly lose at least a fourth of its mass. This is because most of the mass of the Sun is centered in its core. And since a white dwarf is just the remnant core of a star . . . &#xA;Oh, and before the Sun becomes a white dwarf, it goes through the &quot;red giant&quot; phase, where it grows to about the size of the orbit of mars. All the planets will burn up, or stop orbiting, and they will cease to exist when the Sun's nova occurs. Happy ending . . .&lt;/p&gt;&#xA;" OwnerUserId="7332" LastActivityDate="2015-05-25T14:44:44.063" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10816" PostTypeId="5" CreationDate="2015-05-25T14:44:57.943" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:44:57.943" LastActivityDate="2015-05-25T14:44:57.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10817" PostTypeId="4" CreationDate="2015-05-25T14:44:57.943" Score="0" Body="Questions regarding an event where the Earth goes between the Sun and the Moon and casts a shadow on the Moon." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:44:57.943" LastActivityDate="2015-05-25T14:44:57.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10818" PostTypeId="5" CreationDate="2015-05-25T14:46:36.920" Score="0" Body="&lt;p&gt;Questions regarding a manmade object sent to explore space and celestial objects. Examples include Cassini (decommissioned), Voyagers 1 and 2, etc.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="31410" LastEditDate="2020-12-16T05:55:15.273" LastActivityDate="2020-12-16T05:55:15.273" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10819" PostTypeId="4" CreationDate="2015-05-25T14:46:36.920" Score="0" Body="Questions regarding a manmade object sent to explore space and celestial objects." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:46:36.920" LastActivityDate="2015-05-25T14:46:36.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10820" PostTypeId="5" CreationDate="2015-05-25T14:48:25.643" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:48:25.643" LastActivityDate="2015-05-25T14:48:25.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10821" PostTypeId="4" CreationDate="2015-05-25T14:48:25.643" Score="0" Body="Questions regarding matter in its gaseous state, and how it behaves in astronomy (e.g. forming stars in giant molecular clouds)." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:48:25.643" LastActivityDate="2015-05-25T14:48:25.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10822" PostTypeId="5" CreationDate="2015-05-25T14:48:55.393" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:48:55.393" LastActivityDate="2015-05-25T14:48:55.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10823" PostTypeId="4" CreationDate="2015-05-25T14:48:55.393" Score="0" Body="Questions regarding the amount of energy produced by an object per unit of time." OwnerUserId="2153" LastEditorUserId="9301" LastEditDate="2016-10-04T20:21:30.100" LastActivityDate="2016-10-04T20:21:30.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10824" PostTypeId="5" CreationDate="2015-05-25T14:49:40.337" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:49:40.337" LastActivityDate="2015-05-25T14:49:40.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10825" PostTypeId="4" CreationDate="2015-05-25T14:49:40.337" Score="0" Body="Questions regarding the large-scale change of orbits over time, as in the Nice Model." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T14:49:40.337" LastActivityDate="2015-05-25T14:49:40.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10826" PostTypeId="2" ParentId="6526" CreationDate="2015-05-25T15:01:33.887" Score="0" Body="&lt;p&gt;Currently, Pluto is very much a dwarf planet. &#xA;I don't think Pluto will ever be reinstated as a planet again . . . &#xA;On the the pro-planet side, Pluto has five moons, which can possibly qualify it as a planet.&#xA;On the no-planet side, Pluto has a mass of 1.30900 × 10^22 kilograms, which is much lower than the currently smallest planet, Mercury, which weighs 328.5 x 10^21 kilograms. Also, the diameter of Mercury is 4,879.4 km across, while Pluto’s diameter is 2,360 km across. Big difference. Really.&#xA;I'm pretty sure Pluto won't be called a planet again, but I can't be certain.&#xA;Look, here's the thing about controversies like this: you never know until they're over. Yes, it's a horrible answer, but it's the candid truth. At least I've given you the data.&lt;/p&gt;&#xA;" OwnerUserId="7332" LastActivityDate="2015-05-25T15:01:33.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10827" PostTypeId="2" ParentId="10168" CreationDate="2015-05-25T15:11:45.203" Score="0" Body="&lt;p&gt;Distance is what it's all about. The distance of view. And the behavior of sunlight. Look at Jupiter in the night sky. It's a brilliant, but apparently white, dot. Look through a good telescope. It isn't white in any sense, but rather, brown and red. It is exactly what you see in those pictures. Venus? Same deal. It's a golden brown in reality, but through the naked eye on Earth, it's white. I bet Earth does appear white to, say, an observer on Pluto.&lt;/p&gt;&#xA;" OwnerUserId="7332" LastEditorUserId="7332" LastEditDate="2015-05-26T04:48:46.600" LastActivityDate="2015-05-26T04:48:46.600" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10830" PostTypeId="2" ParentId="10168" CreationDate="2015-05-25T18:07:24.317" Score="5" Body="&lt;p&gt;Here's my take on it. I'm not sure what the mystery is - there appear to be two contributing factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both the Moon and the Earth simply reflect/partially absorb the light that is incident upon them from the Sun. The overall of albedo is of little consequence, since the reflection of any light will result in an object that appears bright against the night sky.  The moon reflects quite uniformly across the whole of the visible range - it actually reflects red a bit better than blue. On the other hand the Earth's albedo is strongly wavelength dependent depending on what is doing the reflecting on the visible hemisphere. See the image below from &lt;a href=&quot;http://missionscience.nasa.gov/images/ems/emsVisible_mainContent_spectral-signatures.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;this site&lt;/a&gt;. This injects some perceived colour into reflected earthlight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Yh8j2.png&quot; alt=&quot;Reflectance from Earth features&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second point is the way the eye works at low illumination levels. Although you might think so on a dark night, the moon is &lt;em&gt;not&lt;/em&gt; very bright. It is faint enough that the eye starts to switch over to mesopic or even &lt;a href=&quot;http://en.wikipedia.org/wiki/Scotopic_vision&quot; rel=&quot;nofollow noreferrer&quot;&gt;scotopic vision&lt;/a&gt; where the light levels are too low to stimulate the colour-sensitive cone cells and just stimulate the colour-insensitive rod cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the moon, this has the effect of removing the red tinge it would have otherwise and giving the moon a &quot;silvery&quot;/white appearance. This is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Purkinje_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Purkinje shift&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How the Earth appears to the human eye from the moon I could not say. It is possible that it is bright enough to stimulate the cone cells and thus appear as coloured as the photographs suggest. Either way, you can't tell anything from a photograph other than that the light from the Earth definitely is coloured, whether we perceive it to be or not. Photographs of the moon however do not show any strong colouration. In terms of brightness you also can't tell anything from a photograph showing just the Earth. The Earth &lt;em&gt;is&lt;/em&gt; much brighter than the moon in terms of reflected light per unit area and integrated over its disc as seen from the moon.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-05-25T18:07:24.317" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10831" PostTypeId="1" CreationDate="2015-05-25T19:56:51.547" Score="5" ViewCount="63" Body="&lt;p&gt;In &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1972ApJ...177..681R&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;noreferrer&quot;&gt;Metal-Poor Stars IV: The Evolution of Red Giants&lt;/a&gt;, Rood writes&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The differences in the results of these papers are large enough to introduce appreciable uncertainties into the study of the horizontal branch. For instance, there is still disagreement even as to the sign of the variation of core mass with heavy metal abundance ($dM_c/dZ$).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the cause of the dispute?&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-05-25T22:36:59.277" LastActivityDate="2015-05-25T22:36:59.277" Title="What is the reason for the dispute over the variation of core mass with heavy metal abundance?" Tags="&lt;star&gt;&lt;metal&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10832" PostTypeId="1" AcceptedAnswerId="10836" CreationDate="2015-05-25T23:33:06.953" Score="4" ViewCount="444" Body="&lt;p&gt;The median values of the bias frames coming from our teaching telescope shift up or down with each capture. For example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import glob, numpy as np, astropy.io.fits as fio, matplotlib.pyplot as plt&#xA;&#xA;data = np.zeros((10,255,765), dtype=np.int)&#xA;for i, f in enumerate(glob.glob('bias_*.fit')):&#xA;    data[i,:,:] = fio.getdata(f)&#xA;&#xA;print(np.median(data, axis=(1,2)))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[ 106.  106.  108.  108.  110.  108.  105.  106.  107.  107.]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, during data reduction we average combine our bias frames. We also reject outliers (either minmax or sigmaclip). But if some of our frames are &quot;shifted&quot; higher than others, wouldn't this skew the averaging (and especially the outlier rejection) so that we might be introducing more noise during averaging?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give you an idea of the nature of this fluctuation, I did a quick analysis: read 10 bias frames, then iterated through them, subtracting each frame from the median combination of all other frames. This gives me something like &quot;noise frames&quot;, showing only how much each frame deviates from the median of the rest. Below is the code and the distributions of two of these noise frames.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;noiseframes = np.zeros((10,255,765), dtype=np.int)&#xA;otherframes = np.ones(10, dtype=np.bool) # just a mask&#xA;for i in xrange(10):&#xA;    otherframes[:] = True&#xA;    otherframes[i] = False # exclude current frame from mask&#xA;    median_of_others = np.median(data[otherframes,:,:], axis=0)&#xA;    noiseframes[i,:,:] = data[i,:,:] - median_of_others&#xA;&#xA;def plotdist(i):&#xA;    frame = noiseframes[i].flatten()&#xA;    x = np.arange(np.min(frame),np.max(frame)+1)&#xA;    y = np.bincount(frame-np.min(frame))&#xA;    plt.step(x, y, label=str(np.median(data[i,:,:])))&#xA;&#xA;plotdist(4)&#xA;plotdist(3)&#xA;plt.xlabel('Difference from median [ADU]')&#xA;plt.legend()&#xA;plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/Q6e1Ms2.png?1&quot; alt=&quot;Plot of 2 noise frame distributions&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the noise frames have near-perfect gaussian distributions with pretty much constant FWHM, it's just the central value that shifts around with each frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads me to a couple of conclusions: firstly, this shift truely is a shift in median value and not just a result of something like cosmic rays; secondly, the fluctuation is an additive effect and so not due to a change in gain due to a difference in voltage or something like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also could not find any correlation with temperature or time since last readout, both of which I suspected for a moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps it's slight changes in readout time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't find any mention of this kind of artefact in my own course notes. It's also not mentioned anywhere online that I can find.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can any of you help me to a) identify the possible sources of such a fluctuation and/or b) provide a proven strategy for dealing with this during data reduction?&lt;/p&gt;&#xA;" OwnerUserId="2181" LastActivityDate="2015-05-26T08:32:02.997" Title="How to deal with shifting CCD bias (zero exposure) levels?" Tags="&lt;photography&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10833" PostTypeId="1" AcceptedAnswerId="10834" CreationDate="2015-05-26T01:40:29.173" Score="0" ViewCount="141" Body="&lt;p&gt;Unfortunately, I'm not any good at figuring these things out myself so hope for help here. If the (true) north pole were over the Keweenaw Peninsula of Northern Michigan, what would the North Star be; a star, similar to Polaris that never sets? Or, what star would be directly overhead? I couldn't even pick out Venus or Mars in the night sky when I was younger and now, for personal reasons, I'm unable to go out and look. Any help would be incredibly appreciated. Thank you&lt;/p&gt;&#xA;" OwnerUserId="7335" LastEditorUserId="7335" LastEditDate="2015-05-26T01:46:33.467" LastActivityDate="2015-05-26T19:10:05.057" Title="Michigan's Pole Star" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10834" PostTypeId="2" ParentId="10833" CreationDate="2015-05-26T03:20:38.440" Score="3" Body="&lt;p&gt;It would depend on what time the Earth started to spin around Michigan. There is no way to answer this without saying what time, time of year, and which century (or at least millennium) it happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Hi Dayna. Thanks for the the time of year the Keweenaw Peninsula becomes the North Pole. One could be omniscient except for that and not know this. But we also still haven't ruled out a single star since your OP. With the time though we could tell you exactly what star. The decade is not important. What century or part of history it happens, yes but not the decade. (yes, astronomy is weird).&lt;/p&gt;&#xA;" OwnerUserId="6784" LastEditorUserId="6784" LastEditDate="2015-05-26T19:10:05.057" LastActivityDate="2015-05-26T19:10:05.057" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10835" PostTypeId="2" ParentId="10168" CreationDate="2015-05-26T03:43:48.760" Score="4" Body="&lt;p&gt;The photo in your question is -- well, not exactly fake, but a composite. The biggest clue is that Earth is too close to the horizon; it would have had to be taken from within a few degrees of the boundary between the near and far sides of the Moon, and none of the Apollo missions landed there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, take a close look at the cloud patterns. The view of Earth appears to be identical to that in famous Earthrise photo taken from lunar orbit by Apollo 8 (click to see a larger version):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Space_Race#/media/File:NASA-Apollo8-Dec24-Earthrise.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8Db5x.jpg&quot; alt=&quot;Earthrise&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see in that photo, the surface of the Moon is considerably darker than the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the photo in the question, ignoring the inserted image of the Earth, the sunlit surface of the Moon is the brightest thing in the photo. The light balance must have been adjusted to make everything in the image easy to see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for why the Moon looks white in the night sky, it has an albedo of about 0.37, whereas the night sky has an albedo of about 0.00. Human eyes are very good at adapting to varying lighting conditions. When you look at a dark object against an even darker background, it's going to look white or light gray, even if it's intrinsically dark gray.&lt;/p&gt;&#xA;" OwnerUserId="53" LastEditorUserId="53" LastEditDate="2015-05-26T03:58:03.630" LastActivityDate="2015-05-26T03:58:03.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10836" PostTypeId="2" ParentId="10832" CreationDate="2015-05-26T08:32:02.997" Score="4" Body="&lt;p&gt;Looks like you've got some flaky electronics attached to your CCD if the bias level is changing by several units in consecutive exposures. However, all is not lost if your CCD has an &lt;a href=&quot;http://www.mirametrics.com/tech_note_overscan_bias.htm&quot; rel=&quot;nofollow&quot;&gt;overscan region&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is normally several columns of &quot;pixels&quot; that appear appended to each row of your data, but which do not correspond to physical pixels on the CCD. They are actually dummy readouts to assess the actual bias level in each particular row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overscan region should be clearly visible in any real exposure of a light source, since the overscan region will be un-illuminated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the overscan in two ways. The conventional method is to overscan-correct your images &lt;em&gt;before&lt;/em&gt; doing a bias subtraction. You should also do this to your bias frames &lt;em&gt;before&lt;/em&gt; combining them into your &quot;master bias&quot; frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, take your overscan region and either (i) calculate a median (or clipped mean) within it and subtract this value from every pixel in the image before use (including the bias frames); or (ii) fit something more complicated as a function of row number and then subtract this from your actual data on a row-by-row basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method (ii) is required if you notice that there is any structure or drift/trend in the overscan values as a function of row number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://www.mirametrics.com/tech_note_overscan_bias.htm&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; gives plenty of details, some pictures and a recipe.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-05-26T08:32:02.997" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10837" PostTypeId="2" ParentId="2017" CreationDate="2015-05-26T12:09:12.107" Score="2" Body="&lt;p&gt;If you read scientific papers (at least in astronomy, as far as I am concerned), you will always read of a section which is called &lt;code&gt;Observations and data&lt;/code&gt;, or similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There you (as author) have to explain which kind of data you used and the detailed method for data reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the time, this section goes along with a table in which all used data are described in detail: the &lt;code&gt;observation log&lt;/code&gt;. The guide principle in this section is that, any reader can reproduce your results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same section, information about instrument calibration is also reported in detail. Also, error sources and error determinations. Basically, everything that is included in your published data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, mistakes can happen, and sometimes we do not consider variables which are important. I do not know the details of the two examples that you mention, but they are not the first and will not be the last. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paradoxically, what you report is exactly the reason for disproving those experiments. Both major and minor results are always tested twice or thrice or more, by different groups, different facilities/observatories, different software versions. There is no way you can guarantee for a breakgrounding result without being &quot;differently&quot; tested.&#xA;In the end, the method works well in this way! If I can't reproduce your results by following your description, either you are a bad writer, or your experiment is going to be confuted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the rest is explained well in the answer from OP @moonboy13, with the only exception that, for my experience, when data are private, they are released after 1 year.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastActivityDate="2015-05-26T12:09:12.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10838" PostTypeId="2" ParentId="10750" CreationDate="2015-05-26T14:35:47.377" Score="4" Body="&lt;p&gt;Well, I don't think this question is entirely answerable. The true answer, I think, is that there really are no limits, as Rob Jeffries commented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, using Wikipedia as my only source, the crab pulsar holds the current record for most energetic gamma ray emissions at 80 TeV (wavelength of about $1.5\times10^{-11}$nm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas the longest detected wavelength would surely only be limited by our detector sensitivity. Any imaginable wavelength could be lengthened further by an arbitrary factor via, for example, scattering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma-ray_astronomy&quot; rel=&quot;nofollow&quot;&gt;Gamma-ray Astronomy (Wikipedia)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2181" LastActivityDate="2015-05-26T14:35:47.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10839" PostTypeId="2" ParentId="2017" CreationDate="2015-05-26T15:18:18.283" Score="0" Body="&lt;p&gt;All data from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble_Space_Telescope&quot; rel=&quot;nofollow&quot;&gt;Hubble Space Telescope&lt;/a&gt; that is used in scientific research is released into the public domain (usually within one year, sometimes less) and available from the &lt;a href=&quot;http://mast.stsci.edu&quot; rel=&quot;nofollow&quot;&gt;Mikulski Archive for Space Telescopes&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="38" LastActivityDate="2015-05-26T15:18:18.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10840" PostTypeId="2" ParentId="10763" CreationDate="2015-05-26T16:05:22.433" Score="4" Body="&lt;p&gt;There is a 1.6 meter off-axis telescope for solar observation at Big Bear Observatory called the &lt;a href=&quot;http://www.bbso.njit.edu/nst_publ/NST_align.pdf&quot; rel=&quot;nofollow&quot;&gt;New Solar Telescope (NST)&lt;/a&gt;.  Off-axis is particularly helpful in solar observations because it reduces stray light and eliminates a rotating diffraction pattern which can be severe when the sun is the target.  However, for most astronomical work the additional costs are just not worth it, yet. The extra mechanical costs may be greater than the extra costs for the optics.  The total weight of optics and supporting structures of professional telescopes can be many tons.  There is additional strain required to rotate and support an asymmetric structure like this, i.e. bigger motors are needed.  Galaxies are not bright enough to produce diffraction patterns and the issue of stray light from nearby bright objects is far less of a problem than the foreground stars, zodiacal light, and night sky light actually in the line of sight.  If a nearby star is producing a bothersome diffraction spike, one can usually rotate the field to avoid this. I am not saying that stray light or diffraction effects are never a problem, just that this is not a top problem and, if we need to trade down in size at all for an off-axis design, most astronomers would not accept it. Perhaps if relative costs for these telescopes come down in the future, they will be more common.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2015-05-27T14:32:31.407" LastActivityDate="2015-05-27T14:32:31.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10841" PostTypeId="2" ParentId="10763" CreationDate="2015-05-26T18:22:25.270" Score="2" Body="&lt;p&gt;Yes that does seem like the best possible scope design. They've been built by hobbyist astronomers (even with truss tubes holding the secondary/eyepiece section and a &quot;tube&quot; made out of cloth to block stray light).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if they did over 4 times the primary mirror grinding just to use an off-axis circle-ellipse that's less than a quarter the area of the mirror. That sounds like a lot of work and waste to get rid of diffraction spikes. Imagine grinding a mirror about a yard wide just to make a 12-16&quot; telescope. Although you could cut and sell the rest of the mirror or use it to build an imaging scope for each CCD or something and still make a small regular scope from the middle. Maybe if you're really careful it'd be less work to grind the side of a parabola without grinding the rest of it. Especially if they sell off-axis mirror blanks now so you don't have to buy the costlier huge mirror blank, cut it and then be burdened with a mirror blank that has a circle cut out of it. I don't know, I know almost nothing about telescope making.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other hobbyist telescope makers have built 1-vane secondary mirror supports cause they prefer fewer and worse spikes; and curved secondary mirror vanes, which have every possible angle somewhere along it in equal amounts instead of just orthogonal or triangular one so they spread out the spikes into a less bothersome halo.&lt;/p&gt;&#xA;" OwnerUserId="6784" LastActivityDate="2015-05-26T18:22:25.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10842" PostTypeId="2" ParentId="10750" CreationDate="2015-05-26T19:33:30.540" Score="2" Body="&lt;p&gt;Someone mentioned redshift. Note that the cosmic background radiation limits redshift we can detect now but there is also a much older neutrino background radiation even though we can't do science on it (at least not yet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10^20 electron volts (approximately) is the most powerful cosmic ray ever detected. By an order of magnitude or two I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I once saw an electromagnetic spectrum poster that had DC electricity at 0 Hz.&lt;/p&gt;&#xA;" OwnerUserId="6784" LastActivityDate="2015-05-26T19:33:30.540" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10843" PostTypeId="1" CreationDate="2015-05-26T20:16:23.110" Score="2" ViewCount="110" Body="&lt;p&gt;There are Asteroids &quot;trapped&quot; in Jupiters Langrange points 4 and 5 called trojans and greeks. Are there any asteroids in the earths L4 and L5? Have we seen asteroids in Lagrange points of the earth and moon system? Why are only L4 and L5 stable and L1, L2 and L3 require corrections to remain at those positions?&lt;/p&gt;&#xA;" OwnerUserId="7281" LastActivityDate="2015-05-27T12:17:42.600" Title="Asteroids in langrangian Points 4 &amp; 5" Tags="&lt;earth&gt;&lt;jupiter&gt;&lt;orbital-elements&gt;&lt;lagrange-point&gt;&lt;trojan-asteroids&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10844" PostTypeId="1" AcceptedAnswerId="10845" CreationDate="2015-05-26T20:31:10.837" Score="1" ViewCount="165" Body="&lt;p&gt;For example, for a comet that travels on the hyperbolic orbit around the sun.&lt;/p&gt;&#xA;" OwnerUserId="7337" LastEditorUserId="5524" LastEditDate="2015-05-27T06:13:13.807" LastActivityDate="2015-05-27T06:13:13.807" Title="How to calculate full (mechanical) energy on the hyperbolic orbit?" Tags="&lt;orbit&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10845" PostTypeId="2" ParentId="10844" CreationDate="2015-05-27T01:36:03.927" Score="3" Body="&lt;p&gt;The total energy is the sum of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic_energy&quot; rel=&quot;nofollow&quot;&gt;kinetic energy&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Potential_energy&quot; rel=&quot;nofollow&quot;&gt;potential energy&lt;/a&gt;. It is constant, unless some interaction such as a collision between bodies occurs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinetic energy is easy to understand. For a mass $m$ and velocity $v$ it is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_k = ½mv^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The potential energy formula is (technically) given as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U = -μm/r$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $μ$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_gravitational_parameter&quot; rel=&quot;nofollow&quot;&gt;Standard gravitational parameter&lt;/a&gt;. (I say technically because it can be hard to grasp the concept of negative energy. See the linked articles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a hyperbolic orbit the velocity and distance from the sun must be measured; one cannot be derived from the other as is the case with a known elliptical orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Specific_orbital_energy&quot; rel=&quot;nofollow&quot;&gt;Specific orbital energy&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_energy&quot; rel=&quot;nofollow&quot;&gt;Gravitational energy&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_constant&quot; rel=&quot;nofollow&quot;&gt;Gravitational constant&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5524" LastActivityDate="2015-05-27T01:36:03.927" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10848" PostTypeId="1" AcceptedAnswerId="10850" CreationDate="2015-05-27T07:47:24.407" Score="-6" ViewCount="284" Body="&lt;p&gt;Comet on the parabola orbit flies by Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know what is comet's full energy? What is its distance from Sun at perihelion?&lt;/p&gt;&#xA;" OwnerUserId="7337" LastActivityDate="2015-05-27T10:54:28.947" Title="How much time comet is within the Earth's orbit?" Tags="&lt;orbit&gt;&lt;earth&gt;&lt;comets&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ClosedDate="2015-05-28T14:33:32.937" ContentLicense="CC BY-SA 3.0" />
  <row Id="10850" PostTypeId="2" ParentId="10848" CreationDate="2015-05-27T10:54:28.947" Score="3" Body="&lt;p&gt;If the comet is on a parabolic trajectory, then its total energy (kinetic plus gravitational potential) is zero. This enables you to solve for the perihelion distance if you know the perihelion velocity (or vice-versa). The comet always travels at the escape speed appropriate for its distance from the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This and other useful properties of parabolic trajectories are discussed on the appropriate &lt;a href=&quot;http://en.wikipedia.org/wiki/Parabolic_trajectory&quot; rel=&quot;nofollow&quot;&gt;wikipedia page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-05-27T10:54:28.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10851" PostTypeId="2" ParentId="10843" CreationDate="2015-05-27T12:17:42.600" Score="1" Body="&lt;p&gt;There is one known earth Trojan - &lt;a href=&quot;http://en.wikipedia.org/wiki/Earth_trojan&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Earth_trojan&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason L1, L2 and L3 aren't stable is covered here:   &lt;a href=&quot;https://physics.stackexchange.com/questions/36092/why-are-l4-and-l5-lagrangian-points-stable&quot;&gt;https://physics.stackexchange.com/questions/36092/why-are-l4-and-l5-lagrangian-points-stable&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason the earth has so few L4 and L5 asteroids is cause other planets like Jupiter and Venus tend to destabilize the orbits - same reason why  Jupiter has so many.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-05-27T12:17:42.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10854" PostTypeId="1" CreationDate="2015-05-28T10:47:36.953" Score="7" ViewCount="442" Body="&lt;p&gt;I wish to get a concise list of the analysis steps required to take raw light data from a Kepler data set of a star through the steps needed to get to an analytical determination of the existence of a planetary system (i.e. find a planet candidate in a light curve).  I would also like any software analysis tools that would be suggested for each data analysis step.  A good reference specifically outlining these steps would also be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would first like to replicate the complete set of steps needed to corroborate the existence of a well accepted planetary system. After that I would feel confident that I could move on to analysis of some Kepler KOI data sets that NASA has published or even some data from a non-KOI entity.  I am an amateur and I realize that astronomical knowledge needs to be applied at each data step and that this might not be an automated process but the steps of this analysis would open up new areas of study for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to assemble the basic data analysis tools from various astronomical python sites and other light transit tool sites and get a little familiar with them but I am not clear as to the actual steps as to how these programs feed to each other.  I have looked at the NASA Kepler pipeline to see the steps but it is not clear to me what specific software they apply. I have read several papers confirming exoplanetary discoveries but from them I can't get a complete picture of the data analysis steps and software employed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is disappointing that the actual code listings are not presented in the papers or in a separate reference doc.  In the past (50 years ago) , it was required to publish code listings in reports and references to the code listings in any papers.  But, this discussion is probably another subject.&lt;/p&gt;&#xA;" OwnerUserId="7280" LastEditorUserId="6" LastEditDate="2016-10-04T17:41:12.903" LastActivityDate="2016-10-04T17:41:12.903" Title="What are the analysis steps in taking raw data from Kepler to a planetary system determination" Tags="&lt;exoplanet&gt;&lt;kepler&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10857" PostTypeId="1" CreationDate="2015-05-29T06:00:20.300" Score="3" ViewCount="37" Body="&lt;p&gt;Almost all gas giants in our solar system are observed to have a planetary ring comprises of ice, dust and rocks. My question is do these rings obediently sticks around their foster parent until our starving Sun becomes red giant? Also why don't the ring forms into moon?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="6760" LastEditDate="2015-05-29T06:07:08.820" LastActivityDate="2015-05-29T06:07:08.820" Title="Do the planetary ring lasts?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;gas-giants&gt;&lt;planetary-ring&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2015-05-29T07:52:07.797" ContentLicense="CC BY-SA 3.0" />
  <row Id="10858" PostTypeId="2" ParentId="10506" CreationDate="2015-05-29T19:09:00.267" Score="0" Body="&lt;p&gt;You need at least two images from same objects in diferent angle or position for generate a depth to calculate position, or same image position in diferent dates and a possible object aditional in image , in other way , any calculatiion is wrong&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2015-05-29T19:09:00.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10859" PostTypeId="1" CreationDate="2015-05-29T19:35:53.177" Score="12" ViewCount="571" Body="&lt;p&gt;The planetary discriminant is a measure of how dominant a body is within its region of the solar system. For (true) planets, it is $&amp;gt;10000$ and for dwarf planets it is $&amp;lt;1$. (See &lt;a href=&quot;https://astronomy.stackexchange.com/a/592/&quot;&gt;this answer&lt;/a&gt; to &lt;em&gt;&quot;How many planets are there in this solar system?&quot;&lt;/em&gt; which explains the related details.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible, given what has already been observed about our solar system, for there to be an object with  &quot;planetary discriminant&quot; µ between that of Ceres and Neptune?   In other words, could the clear cut line between planets and dwarf planets get fuzzier?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously it would have to be a massive body pretty far out.  Is there a region in the solar system which could be devoid enough of many small objects, which also could be home to a large enough body that we also haven't observed yet?   How does the math work out?&lt;/p&gt;&#xA;" OwnerUserId="1106" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-04-01T21:21:40.207" Title="Can there be an object with planetary discriminant between Ceres and Neptune?" Tags="&lt;planet&gt;&lt;dwarf-planets&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10861" PostTypeId="1" CreationDate="2015-05-29T20:43:30.070" Score="4" ViewCount="192" Body="&lt;p&gt;Would it be possible to slightly change the trajectory of Ceres to make it end up orbiting the Earth without messing up the Solar System? How much energy  would be required and how long would the process take?&lt;/p&gt;&#xA;" OwnerUserId="7353" LastEditorUserId="2153" LastEditDate="2015-05-31T20:30:04.293" LastActivityDate="2015-06-01T04:54:06.503" Title="Capturing Ceres" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;asteroids&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10862" PostTypeId="1" CreationDate="2015-05-30T11:24:45.740" Score="2" ViewCount="272" Body="&lt;p&gt;In &lt;a href=&quot;https://en.wikipedia.org/wiki/Mysterium_Cosmographicum#Shapes_and_the_planets&quot; rel=&quot;nofollow&quot;&gt;Mysterium Cosmographicum (1596)&lt;/a&gt; Johannes Kepler proposed that the relative distances between the orbits of the six ancient planets (six because heliocentrism had recently added Earth as one of the planet) correspond to the geometry of the five Platonic solids. Each planetary orbit was assumed to be the grand circle of a sphere. Each Platonic solid would circumscribe an inner planet's orbitally defined sphere, and in same position inscribe the next outer planet's &quot;orbital sphere&quot;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How well does it fit?&lt;/li&gt;&#xA;&lt;li&gt;Is it somehow mathematically related to the equally spurious Bode's law?&lt;/li&gt;&#xA;&lt;li&gt;I would appreciate some hints or links to the geometric calculation itself, for me to compare with my own exercise trying to solve part of it.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Johannes Kepler proposed this order of matching planets with Platonics (I suppose because this order gives the best fit):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury &amp;lt;-- &lt;em&gt;planet&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;octahedron&lt;/strong&gt; &amp;lt;-- &lt;em&gt;Platonic solid&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;icosahedron&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;dodecahedron&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;tetrahedron&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;cube&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="2365" LastEditDate="2015-11-13T20:01:52.673" LastActivityDate="2015-11-13T20:01:52.673" Title="How well do planetary orbits fit with Johannes Kepler's in- &amp; circumscribed Platonic solids?" Tags="&lt;solar-system&gt;&lt;history&gt;&lt;johannes-kepler&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10863" PostTypeId="2" ParentId="10862" CreationDate="2015-05-30T17:38:29.683" Score="5" Body="&lt;p&gt;It is easy enough to do the calculations, formulae for the in and circum radii of the Platonic solids can be found &lt;a href=&quot;http://whistleralley.com/polyhedra/platonic.htm&quot;&gt;here&lt;/a&gt; which give ratios of circ to in radii of (note formulea for the radii have dropped common factor of the side length, which we don't need as we are interested in the ratios):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt;ri4=sqrt(6)/12,rc4=sqrt(6)/4&#xA;     0.204124 &#xA;     0.612372 &#xA;&amp;gt;&#xA;&amp;gt;ri6=1/2,rc6=sqrt(3)/2&#xA;          0.5 &#xA;     0.866025 &#xA;&amp;gt;&#xA;&amp;gt;ri8=sqrt(6)/6,rc8=sqrt(2)/2&#xA;     0.408248 &#xA;     0.707107 &#xA;&amp;gt;&#xA;&amp;gt;ri12=sqrt(250+110*sqrt(5))/20,rc12=(sqrt(15)+sqrt(3))/4&#xA;      1.11352 &#xA;      1.40126 &#xA;&amp;gt;&#xA;&amp;gt;ri20=(3*sqrt(3)+sqrt(15))/12,rc20=sqrt(10+2*sqrt(5))/4&#xA;     0.755761 &#xA;     0.951057 &#xA;&amp;gt;&#xA;&amp;gt;rho4=rc4/ri4&#xA;            3 &#xA;&amp;gt;rho6=rc6/ri6&#xA;      1.73205 &#xA;&amp;gt;rho8=rc8/ri8&#xA;      1.73205 &#xA;&amp;gt;rho12=rc12/ri12&#xA;      1.25841 &#xA;&amp;gt;rho20=rc20/ri20&#xA;      1.25841 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which may be compared with the orbital radius ratios from &lt;a href=&quot;https://solarsystem.nasa.gov/planets/charchart.cfm&quot;&gt;here&lt;/a&gt; (radii in km)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt;RMecury=57.9e6;&#xA;&amp;gt;RVenus=108.2e6;&#xA;&amp;gt;REarth=149.6e6;&#xA;&amp;gt;RMars=227.9e6;&#xA;&amp;gt;RJupiter=778.3e6;&#xA;&amp;gt;RSaturn=1426.7e6;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now we can compare the corresponding radii ratios:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt;[RVenus/RMecury,rho8]&#xA;      1.86874       1.73205 &#xA;&amp;gt;[REarth/RVenus,rho20]&#xA;      1.38262       1.25841 &#xA;&amp;gt;[RMars/REarth,rho12]&#xA;       1.5234       1.25841 &#xA;&amp;gt;[RJupiter/RMars,rho4]&#xA;      3.41509             3 &#xA;&amp;gt;[RSaturn/RJupiter,rho6]&#xA;       1.8331       1.73205 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which, as these things go, is not bad.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-05-30T18:27:57.433" LastActivityDate="2015-05-30T18:27:57.433" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10866" PostTypeId="1" CreationDate="2015-05-31T03:21:47.557" Score="4" ViewCount="281" Body="&lt;p&gt;I became interested in astronomy a few years ago and started with a Celestron AstroMaster 70AZ refractor telescope. That seemed nice, and I wanted to upgrade to a go-to telescope that was also more much more powerful, so I bought the 8 inch Celestron Schmidt-Cassegrain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's great, but honestly I'm a bit disappointed because the magnification doesn't seem all that much stronger than my 70AZ. I haven't done a side by side comparison. But for example when looking at Saturn, I was expecting to see more than just a white ball with a ring around it - I was able to see that with my 70AZ. I thought I'd see colors, and maybe be able to differentiate between some of the bigger ring bands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't tried any deep sky objects. I'm using various lenses starting from 30mm down to 6mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I doing something wrong?&lt;/p&gt;&#xA;" OwnerUserId="7361" LastEditorUserId="7361" LastEditDate="2015-05-31T12:37:54.070" LastActivityDate="2015-05-31T14:52:15.460" Title="Am I using my Celestron 8 Schmidt-Cassegrain telescope correctly?" Tags="&lt;telescope&gt;&lt;saturn&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10867" PostTypeId="2" ParentId="8523" CreationDate="2015-05-31T06:51:18.943" Score="0" Body="&lt;p&gt;This link has a lot of simulators for each individual property. I do not know if it has everything, but I hope it helps. Each simulator includes the equation it uses.&#xA;&lt;a href=&quot;http://astro.unl.edu/animationsLinks.html&quot; rel=&quot;nofollow&quot;&gt;http://astro.unl.edu/animationsLinks.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7362" LastActivityDate="2015-05-31T06:51:18.943" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10868" PostTypeId="2" ParentId="6127" CreationDate="2015-05-31T11:49:44.817" Score="4" Body="&lt;p&gt;The processes by which planets can be removed from their parent stars are discussed in some detail by &lt;a href=&quot;http://arxiv.org/abs/1311.6816&quot; rel=&quot;nofollow&quot;&gt;Davies et al. (2014)&lt;/a&gt;. These include direct ejection through interactions with other stars in dense birth environments; the ejection of planets due to planet-planet interactions, again usually taking place fairly early in a star's life; the later ejection of planets after an instability due to a passing star. There is also a possibility of exchange interactions, i.e. a planet will swap stars during an interaction. Again, this is most likely to happen in a dense birth environment rather than later in a star's life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the ejection of the planet happens while they &lt;em&gt;are&lt;/em&gt; still in the dense environs of their birth, then recapture by another star &lt;em&gt;is&lt;/em&gt; possible. A very recent paper by &lt;a href=&quot;http://arxiv.org/abs/1503.03077&quot; rel=&quot;nofollow&quot;&gt;Wang et al. (2015)&lt;/a&gt; explores this very possibility using N-body simulations, but there are other papers to look at by &lt;a href=&quot;http://arxiv.org/abs/1503.03077&quot; rel=&quot;nofollow&quot;&gt;Parker &amp;amp; Quanz (2012)&lt;/a&gt;, or &lt;a href=&quot;http://arxiv.org/abs/1202.2362&quot; rel=&quot;nofollow&quot;&gt;Perets &amp;amp; Kouwenhoven (2012)&lt;/a&gt;.&#xA;These papers seem to agree that the process of recapture is not common (but not rare either) and will happen sooner rather than later, before the cluster disperses. The latter of these papers has some particularly eye-catching results: they find that if the number of free floating planets is comparable to the number of stars then about 5% of stars recapture a planet. The capture tends to put the planet in a wide, eccentric orbit (hundreds of au or more). Capture is more likely for massive stars, but they claim that even brown dwarf-planet or planet-planet pairs could be formed in small numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the planet has left the cluster environment then, as is the case of star-star interactions in the Galactic field, the probability of encountering another star and being captured is very remote.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="633" LastEditDate="2015-05-31T17:00:47.690" LastActivityDate="2015-05-31T17:00:47.690" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10869" PostTypeId="2" ParentId="10866" CreationDate="2015-05-31T14:52:15.460" Score="2" Body="&lt;p&gt;Keep in mind that many of the photos you see with colors are composite images.  To differentiate colors try to use some various color filters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see bands on Saturn, try using a yellow.  This will bring out different details in what you are seeing.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This link provides some more details for using various filters.&#xA;&lt;a href=&quot;http://www.telescopes.com/telescopes/howdifferentfilterscanbetteryourviewarticle.cfm&quot; rel=&quot;nofollow&quot;&gt;http://www.telescopes.com/telescopes/howdifferentfilterscanbetteryourviewarticle.cfm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="281" LastActivityDate="2015-05-31T14:52:15.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10870" PostTypeId="1" CreationDate="2015-05-31T20:41:04.613" Score="5" ViewCount="295" Body="&lt;p&gt;The James Webb Space Telescope costs nearly 9 billion dollar. What parts of this project cost how much?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose that almost every component has to be developed and is not available to buy off the shelf. So most of the costs should end up as salaries to engineers. I see that the top 10% US specialist aerospace engineers &lt;a href=&quot;http://swz.salary.com/SalaryWizard/Aerospace-Engineer-V-Salary-Details.aspx?&amp;amp;hdcbxbonuse=on&quot; rel=&quot;noreferrer&quot;&gt;earn 163,000 dollar&lt;/a&gt;. The JWST budget covers 54,000 such yearly salaries. Are really thousands of the best engineers working on this project? Or are maybe some materials, like beryllium, very expensive? Like JWST's average cost of about $1½ million/kg. Launch costs are anyway negligible in this context.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-05-31T23:27:08.477" Title="How is the cost of JWST distributed on different parts of development and operations?" Tags="&lt;space-telescope&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10871" PostTypeId="1" AcceptedAnswerId="10873" CreationDate="2015-05-31T20:51:38.063" Score="15" ViewCount="1008" Body="&lt;p&gt;Measurements of Andromeda's blue shift let us conclude that the distance between the Andromeda galaxy and the Milky Way is decreasing and in a few billion years they will &quot;collide&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blue shift only yields the radial component of Andromeda's velocity vector. It is my understanding that measuring the tangential component is crucial in determining whether a &quot;collision&quot; will actually happen (in a gravitationally bound two body system, for point-like bodies to collide, the relative velocity component must point &lt;em&gt;exactly&lt;/em&gt; towards the other body, i.e. the tangential component must vanish).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, galaxies are not point-like, so some small nonzero tangential component might lead to a collision where at least some galaxy arms intersect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has the tangential velocity been measured? If so, how? How central is the collision (bulge into bulge, bulge into arms, arms into arms)?&lt;/p&gt;&#xA;" OwnerUserId="6593" LastEditorUserId="1891" LastEditDate="2018-10-14T05:15:18.523" LastActivityDate="2018-10-14T05:15:18.523" Title="How much overlap will the Andromeda galaxy and the Milky Way have when they collide?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;m31&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="10872" PostTypeId="2" ParentId="10870" CreationDate="2015-05-31T21:03:33.570" Score="4" Body="&lt;p&gt;A systems cost breakdown (as of 2011) is given &lt;a href=&quot;http://www.stsci.edu/jwst/meetings/webinar2011/program-status-smith.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/HSgw1yI.png&quot; alt=&quot;&quot;&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;The &quot;% spent&quot; refers to the status of spending at the time.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's not &lt;em&gt;completely&lt;/em&gt; helpful, because that's only broken down by instrument sections. For example, &quot;Northrop-Grumman&quot; doesn't tell you the salary breakdown, as you mentioned.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2015-05-31T21:03:33.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10873" PostTypeId="2" ParentId="10871" CreationDate="2015-05-31T22:30:36.577" Score="11" Body="&lt;p&gt;&lt;a href=&quot;http://astrobites.org/2012/06/01/the-fate-of-the-milky-way/&quot;&gt;This page&lt;/a&gt;, from June 2012, contains a fairly detailed summary of three papers that considered the tangential velocity and collision problem.  I'll quote a few choice bits from that page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;Andromeda (M31) and the Milky Way are the two largest galaxies in the small group of galaxies called the Local Group. At the moment, M31 is about 770 kpc away from our galaxy, but Doppler measurements of the line-of-sight velocites of its stars have long shown that it is inexorably falling toward us. The exact nature of the collision, however, has so far remained unkwnon. Will it be a miss, glancing blow, or head-on smashup? This depends on the proper motion (sideways motion) of Andromeda in the sky, which is extremely hard to measure.&lt;/blockquote&gt;&#xA;&#xA;&lt;blockquote&gt;At last, the authors of this study manage to collect extraordinarily accurate Hubble Space Telescope observations of the tangential motion of M31 over a five-to seven-year period, and these data remove any doubt that Andromeda is destined to strike and merge with our Milky Way in four billion years. The Triangulum galaxy M33, the third most massive galaxy of the Local group, which is bound to M31, will likely join in the cosmic boom (with a small chance that M33 will hit the Milky Way first).&lt;/blockquote&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;The final result is that M31 has a radial velocity with respect to the Milky Way of about  -109 km/s, and a tangential velocity of about 17.0 km/s (&amp;lt;34.3 km/s at 1-sigma confidence). These numbers clearly imply that the velocity vector of Andromeda Galaxy is statistically consistent with a radial (head-on collision) orbit towards the Milky Way. This means that the Milky Way-M31 system is bound, and that the two galaxies will merge, as investigated further in Paper III.&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;The simulations show that the Milky Way and the Andromeda galaxy will merge, consistently with the radial (head-on collision) orbit deduced in Paper II for M31, and  that the first “pericenter” (closest approach) will occur at around 4 Gyr from now. For our Milky Way,  the encounter has 72.2% probability of being prograde (the galaxy spins in the same direction as the flyby). In 41.0% of the Monte-Carlo orbits M31 makes a direct hit with our Milky Way, where the authors define a “direct hit” as an encounter with a first pericenter distance less than 25 kpc. The two galaxies will eventually merge after 5.9 Gyr, and the radial mass profile of the merger remnant will be significantly more extended than the original individual profiles. Roughly speaking, this profile will follow the $R^{1/4}$ law characteristic of elliptical galaxies, in agreement with the predictions from the numerical simulations of major mergers of spiral galaxies.&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll note that $25 \text{ kpc} = 25,000 \text{ pc} \cong 81,540 \text{ ly}$, and that the Milky Way is only about 100,000 ly in diameter, so the &quot;direct hit&quot; definition is quite a substantial distance here.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastEditorUserId="2188" LastEditDate="2015-05-31T22:41:14.663" LastActivityDate="2015-05-31T22:41:14.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10874" PostTypeId="2" ParentId="10870" CreationDate="2015-05-31T23:00:15.530" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So most of the costs should end up as salaries to engineers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even in a project that is solely people-based, most of the costs do not end up as salaries to engineers. Salary is typically a bit less than a half of the cost. Those engineers get benefits. Health care in the US isn't cheap. They get holidays, vacation, and sick leave. They receive matching funds on their 401K contributions. In the US, employers and employees both contribute to the &lt;strike&gt;cesspool&lt;/strike&gt; lockbox sometimes known as &quot;Social Security&quot;. The employer share is part of the cost of having an employee (and companies charge for it). The employer also pays for unemployment insurance and other items on behalf of their employees. Those engineers sit at desks in an office or a cubicle, and have a nice computer on which they do their work. They use the employer's servers, printers, copy machines, conference rooms, break rooms, and coffee machines. All of that costs money. These overhead costs are the first of several indirect costs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The accountants who write those engineers their paychecks aren't charging to any project. They're still paid by the company, as are the executives of the company. The executives also typically don't charge their time to a project, and those executives can be paid quite handsomely. This leads to another category that adds to quantity that companies charge the government, &quot;general and accounting&quot;. Finally, for-profit companies need to keep their shareholders / stakeholders happy. They need to earn a profit. This leads to a third category that companies charge the government, &quot;profit&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you add up all of those indirect costs, they come close to and oftentimes exceed direct costs (i.e., chargeable salary). Note that this is for a project where the only costs are the people working on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Building a spacecraft notches things up a lot. Cleanrooms are expensive. Control rooms are expensive. The flight computers are expensive. The bolts and fasteners are expensive. The custom, one-off sensors are expensive. The spacecraft's thermal control system is expensive. The spacecraft's structure is expensive. Launch is expensive. Every piece of hardware, every fabrication facility is expensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, cost overruns, requirements changes, and delays can represent a significant portion of the cost of a poorly managed project. JWST has experienced numerous cost overruns, requirements changes, and delays. Big contractors have learned that the only way to win big government contracts is to underbid. They would lose their shirts if the government kept to that initial contract. But it never does. The government instead changes the requirements, changes the payments, and changes the schedule. Those changes are beyond expensive.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2015-05-31T23:27:08.477" LastActivityDate="2015-05-31T23:27:08.477" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10875" PostTypeId="1" CreationDate="2015-06-01T03:25:36.567" Score="4" ViewCount="1246" Body="&lt;p&gt;I want to build a system to view my telescope without being uncomfortable. Currently I have to stand up and bend over to view the eye piece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to build a system that allows me to 'pipe' the output of the view piece in a non-digital format along a flexible fiber optic cable to my eye and possible split the view into two for a sort of VR headset. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the cable will transfer the light, but will it be just burly light or break down too much (one meter or less) or will I be able to focus it into view-able picture? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could just try it, but the cable itself is really expensive to get, and I want to know if this is worth trying before I spend. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking of using &lt;a href=&quot;http://www.wiedamark.com/14mmsolidcoreendfiberopticcable-1.aspx&quot; rel=&quot;nofollow&quot;&gt;this 14 mm 'solid core' cable&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7367" LastEditorDisplayName="user1569" LastEditDate="2016-04-30T21:54:07.777" LastActivityDate="2016-04-30T21:54:07.777" Title="Focusing light through fiber optic cable to extend telescope viewer" Tags="&lt;telescope&gt;&lt;telescope-lens&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10876" PostTypeId="2" ParentId="10861" CreationDate="2015-06-01T04:35:17.900" Score="5" Body="&lt;p&gt;It's not possible to &quot;slightly&quot; change it's orbit unless you have a very very long time to wait.    When Orbits get in synch with large bodies like Jupiter, for example, then Jupiter can give a tiny tug to Ceres each orbit and if positioned just right, it might nudge Ceres towards the inner solar-system, but only over millions, if not tens of millions of years.   That's the slow way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no fast way to move Ceres into an earth orbit without exerting a significant force on it.   Orbits are largely stable and to change an orbit that much would requires a serious amount of force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My math isn't always right, but Ceres orbits the sun at 17.9 KM per S and the earth orbits the sun at about 30 KM/S - now, you might think you need to speed Ceres up, but actually the opposite, it needs to be slowed down to fall into a lower orbit and the amount it needs to be slowed down (My math would be long and clumsy), but the actual deceleration needed is about 1/2 of that velocity difference - so imagine how much energy you'd need to move an object nearly 1,000 KM from side to side, made of ice and rock, twice as dense as water (mass of about a billion billion tons) and you need to accelerate that (decelerate it) oh about 5-6 KM per second (20,000 Kilometers per hour).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, a lot of energy could be saved by dropping Ceres to a fly-by near mars and then doing a gravity assist, and then, using earth's gravity as well to help capture it into an orbit around the earth.   Mars weighs about 700 times what Ceres weighs, so the effect on Mars would be pretty tiny (all bets are off for Mars' 2 tiny moons though).   But at minimum, you still need to decelerate Ceres by about 3 KPS (10-11,000 KPH), and at a billion billion tons that's a ungodly amount of energy, unless you're willing to do it very very slowly, perhaps or by finding a fuel source on Ceres to generate some kind of thrust or moving it so that it's in proper synch with Jupiter so that it gets a small kick towards the sun every so often, (see Mercury/Jupiter resonance below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stability_of_the_Solar_System#Mercury.E2.80.93Jupiter_1:1_resonance&quot;&gt;http://en.wikipedia.org/wiki/Stability_of_the_Solar_System#Mercury.E2.80.93Jupiter_1:1_resonance&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I love ideas like this by the way.   Ceres would be a nice source of water Though in an earth's orbit it would lose it's water over time to solar wind, so it would be a trade-off moving it closer.   Might be better to leave it where it is.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your 2nd question, Ceres weighs about 1/80th the mass of the moon.   It's not massive enough to &quot;mess up&quot; any planets in the solar system, though if it crashed into the earth for example, it would &quot;mess up&quot; life on the earth big time, but it wouldn't significantly change the earth's orbit.   It would change the earth's orbit a little, but only a tiny bit.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-06-01T04:54:06.503" LastActivityDate="2015-06-01T04:54:06.503" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10877" PostTypeId="1" CreationDate="2015-06-01T10:07:41.290" Score="1" ViewCount="181" Body="&lt;p&gt;How is Chandrashekhar Limit calculated? What happens to stars below the mass of the Chandrashekhar limit after they collapse?&lt;/p&gt;&#xA;" OwnerUserId="7369" LastActivityDate="2015-06-01T19:51:41.613" Title="Collapse of a star below the mass of Chandrashekhar Limit" Tags="&lt;star&gt;&lt;black-hole&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10878" PostTypeId="1" AcceptedAnswerId="10902" CreationDate="2015-06-01T13:29:10.173" Score="8" ViewCount="1867" Body="&lt;p&gt;There are apparently not many reasonably priced radio telescopes available for the amateur users. I only could find &lt;a href=&quot;http://www.primalucelab.com/radioastronomy/spider230-amateur-radio-telescope.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;a SPIDER 230C 2.3 meter diameter compact radio telescope&lt;/a&gt;, which costs ~10 k€. Reading the page, I don't get a good overview what kind of things I could &amp;quot;see&amp;quot; with a radio telescope that has a 2.3 meter antenna.&lt;/p&gt;&#xA;&lt;p&gt;Is the instrument already a (semi-)professional one?&lt;/p&gt;&#xA;" OwnerUserId="6991" LastEditorUserId="37191" LastEditDate="2021-05-11T10:05:53.363" LastActivityDate="2021-05-11T10:05:53.363" Title="What kind of things I could &quot;see&quot; with an amateur radio telescope?" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;radio-astronomy&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="10879" PostTypeId="1" AcceptedAnswerId="10883" CreationDate="2015-06-01T17:13:11.700" Score="4" ViewCount="227" Body="&lt;p&gt;My understanding is that there are &lt;a href=&quot;http://www.space.com/8293-universe-born-black-hole-theory.html&quot; rel=&quot;nofollow&quot;&gt;credible theories&lt;/a&gt; out there in which the formation of a singularity in a black hole also represents the beginning of a new universe via a big bang. We can't see the new universe because we can't see anything from within a black hole. Further, the apparent size of the black hole to us does not have to equal the size of the new universe from the perspective of someone inside that universe. My question is, is mass+energy conserved? That is, is the total amount of material within the new universe limited to how much stuff has fallen into the black hole, or how much stuff has reached the singularity? If so then it would seem these black hole universes are only a tiny fraction the size of our own universe.&lt;/p&gt;&#xA;" OwnerUserId="6587" LastEditorUserId="6980" LastEditDate="2015-06-01T20:07:41.463" LastActivityDate="2015-06-02T04:46:57.530" Title="Is mass+energy conserved when a new universe forms inside a black hole?" Tags="&lt;black-hole&gt;&lt;universe&gt;&lt;cosmology&gt;&lt;big-bang-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10880" PostTypeId="2" ParentId="10877" CreationDate="2015-06-01T19:51:41.613" Score="4" Body="&lt;p&gt;The Chandrasekhar mass is a theoretical maximum stellar mass that can be supported by ideal electron degeneracy pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was originally calculated by Chandrasekhar using polytropic equations of state for white dwarf stars of various compositions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is found that if the equation of state is $P \propto \rho^{4/3}$, which is appropriate for ultra relativistic electrons, then equilibrium is only obtained for $M_{CH} = 5.8 \mu_e^{2} M_{\odot}$, where the density becomes infinite and $\mu_e$ is the number of atomic mass units per electron in the gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical white dwarfs are carbon and oxygen with $\mu_e= 2$, and $M_{CH}= 1.45M_{\odot}$, but in the core of a supernova progenitor made of iron-56 it would be lower - around $M_{CH}=1.25M_{\odot}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more recent years, the Chandrasekhar limit has evolved to colloquially mean the maximum possible mass for a white dwarf. The main corrections to the ideal case considered by Chandrasekhar are Coulomb corrections, the possible onset of inverse beta decay (electron capture) and the instability at a finite density predicted by using GR rather than Newtonian gravity. The latter probably sets the &quot;Chandrasekhar mass&quot; for carbon white dwarfs to be $1.38M_{\odot}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A carbon white dwarf that had a mass less than this could be supported by electron degeneracy pressure and would not collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your question is asking for a derivation of the Chandrasekhar mass, then I suggest you look at the stability analyses that are presented in a standard text on the subject like &quot;Black holes, white dwarfs and neutron stars&quot; by Shapiro and Teukolsky. An approximate result can be obtained using the virial theorem assuming a uniform density star.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-06-01T19:51:41.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10881" PostTypeId="2" ParentId="10879" CreationDate="2015-06-01T20:13:58.420" Score="3" Body="&lt;p&gt;Under the assumption that there is some sort of &quot;Big Bang&quot; scenario, then we are assuming an expanding universe.  General relativity does not require the conservation of mass-energy on a universal scale in a changing spacetime (arguably, GR doesn't even have a concept of energy on a universal scale), so there is no &lt;em&gt;a priori&lt;/em&gt; reason why this black-hole-universe cannot be comparable to our own (in whatever senses we pick).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple way to see why energy conservation must be violated, even in a vaccuum, is to first observe that the vacuum energy density is fixed.  So if spacetime is expanding (or contracting), then we are changing the total volume without changing this vacuum energy density.  Keeping the density the same but increasing the volume means more total energy; decreasing the volume with the same density means less total energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also easily observe this violation.  You may be familiar with the concept of redshift: light coming to you from a source that is moving away from you (or emitted from a gravitating source) is shifted towards the red side of the spectrum.  This stretches out the wavelength, so the light has less energy.  The energy doesn't go anywhere, it's just gone.${}^*$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We might also note that, with Noether's theorem in hand, we know that time translation invariance implies conservation of energy.  But a non-static spacetime is inherently not time translation invariant, so conservation of energy is not required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^*$ Well, some may try to regain conservation of energy by asserting that there is an energy of the gravitational field to consider.  This isn't adopted by everyone, but it can be done if you're careful with your definitions.  It just doesn't work as generally and easily as you might naively hope for.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2015-06-01T20:13:58.420" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10882" PostTypeId="2" ParentId="10878" CreationDate="2015-06-02T02:04:06.457" Score="4" Body="&lt;p&gt;From my simplistic analysis, it's not good for much.&lt;/p&gt;&#xA;&lt;p&gt;For comparison, the first &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Radio_telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;radio telescope&lt;/a&gt; was 9 meters.&lt;/p&gt;&#xA;&lt;p&gt;One of the favorite parts of the spectrum for radio telescopes is the water hole - 21 cm.&lt;/p&gt;&#xA;&lt;p&gt;From my quick mental arithmetic, this dish would be able to &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Angular_resolution&quot; rel=&quot;nofollow noreferrer&quot;&gt;resolve&lt;/a&gt; sources of 21 cm signals of they were about 5 degrees apart.&lt;/p&gt;&#xA;" OwnerUserId="5524" LastEditorUserId="37191" LastEditDate="2021-05-11T09:18:22.460" LastActivityDate="2021-05-11T09:18:22.460" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10883" PostTypeId="2" ParentId="10879" CreationDate="2015-06-02T04:46:57.530" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;is mass+energy conserved?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes.  Even in general relativity energy and momentum are conserved, although it is a bit more complicated than in Newtonian Mechanics.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is the total amount of material within the new universe limited to how much stuff has fallen into the black hole, or how much stuff has reached the singularity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No.  Well, sort of.  The physicist who's work you linked in your question addressed this in a follow-up paper.  His original work (about which your article was based on) is: &lt;a href=&quot;http://arxiv.org/abs/1007.0587&quot; rel=&quot;nofollow&quot;&gt;Cosmology with torsion: An alternative to cosmic inflation&lt;/a&gt;.  A follow-up discussing the mass of the new universe is: &lt;a href=&quot;http://arxiv.org/abs/1103.4192&quot; rel=&quot;nofollow&quot;&gt;On the mass of the Universe born in a black hole&lt;/a&gt;.  In this paper he claims if our entire universe were inside a black hole, the black hole would only have to be 1,000 solar masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Believe it or not, this doesn't violate conservation of energy.  In fact, he explicitly uses conservation of energy in his calculations.  The resolution of this paradox is that there is A LOT of energy in the gravitational field of a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this model, the singularity inside the black hole never forms.  An event horizon forms as the matter collapses, as you would expect, but inside the horizon space-time &quot;bounces&quot; before it has a chance to make a singularity.  As the matter falls inwards its energy increases; it accelerates and increases its kinetic energy due to the immense gravitational field.  When it reaches the stationary &quot;universe&quot; inside this kinetic energy is translated into rest-mass energy: and if the matter was accelerating for long enough the increase in energy can be enormous.  This gives the &quot;universe&quot; inside potentially a lot more mass than the stuff that fell in.  This process is described in the paper as the creation of particle-antiparticle pairs by the gravitational field, which amounts to the same effect.  Either way, you take energy from the gravitational field and turn it into the energy (mass) of particles inside the horizon.  Someone sitting outside the black hole doesn't notice this, because they can only measure the total energy and that remains constant.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If so then it would seem these black hole universes are only a tiny fraction the size of our own universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not quite.  Inside the event horizon the crazy warping of space-time can make the inner &quot;universe&quot; quite large.  Like the TARDIS, something much bigger on the inside :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am obliged to say that all of this work is very theoretical, his conclusions rest upon some assumptions that we have no evidence to support, but its a neat idea!&lt;/p&gt;&#xA;" OwnerUserId="6672" LastActivityDate="2015-06-02T04:46:57.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10884" PostTypeId="1" CreationDate="2015-06-02T06:37:50.557" Score="3" ViewCount="486" Body="&lt;p&gt;As has been pointed out by some scientists, that black holes aren't stable bodies but are eternally collapsing objects. How does one come to such a conclusion? Is hawking radiation related to this in some way?&lt;/p&gt;&#xA;" OwnerUserId="7369" LastActivityDate="2015-11-30T09:27:20.657" Title="Eternally collapsing objects?" Tags="&lt;black-hole&gt;&lt;hawking-radiation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10885" PostTypeId="1" CreationDate="2015-06-02T08:00:36.300" Score="1" ViewCount="340" Body="&lt;p&gt;Australia and Russia have large land areas. However, I am only aware of &lt;a href=&quot;http://en.wikipedia.org/wiki/Siding_Spring_Observatory&quot; rel=&quot;nofollow&quot;&gt;one 3.9m telescope&lt;/a&gt; in Australia. As far as I know, there are lots of &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_mountains_in_Australia&quot; rel=&quot;nofollow&quot;&gt;mountains&lt;/a&gt; in Australia and Russia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why are there not many large optical telescopes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reasons for each country may differ because of their latitudes.&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="2153" LastEditDate="2016-05-15T13:24:25.633" LastActivityDate="2016-05-15T13:24:25.633" Title="Why don't Australia/Russia have large optical telescopes?" Tags="&lt;telescope&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10886" PostTypeId="1" CreationDate="2015-06-02T08:04:00.630" Score="3" ViewCount="855" Body="&lt;p&gt;I recently watched Interstellar with some friends and we didn't come to the same conclusion. In my opinion, the portal they use to go to the other galaxy (to visit the three planets) is &lt;strong&gt;not&lt;/strong&gt; a black hole. My friends tell me it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(warning, contains spoiler!!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, it's not because of these points:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When they are in front of the portal, they can see the other galaxy, which is not the case when Coop goes to the black hole at the end&lt;/li&gt;&#xA;&lt;li&gt;When they go to that portal, they &quot;flow&quot; in a tube like channel. Again, that's not the case when Coop goes to the black hole at the end.&lt;/li&gt;&#xA;&lt;li&gt;When Coop goes to that black hole, he goes through some sort of light dust that hits his ship, which is not the case when they go to the portal.&lt;/li&gt;&#xA;&lt;li&gt;When the robot (I forgot his name) goes in the black hole, his purpose is to collect some quantic data. If the initial portal was also a black hole, he would have made that collection at that time.&lt;/li&gt;&#xA;&lt;li&gt;When they goes to the portal, they seems &quot;confident&quot;, but when Coop goes to that black hole, it's for killing himself. If the portal was a black hole, they would behave more identical.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So I believe it's not a black-hole, but what is it exactly (maybe not in scientific terms, but in Interstellar's terms)? A Wormhole?&lt;/p&gt;&#xA;" OwnerUserId="7373" LastEditorUserId="6980" LastEditDate="2015-06-22T00:40:16.070" LastActivityDate="2015-06-22T00:40:16.070" Title="Interstellar movie: What is the &quot;portal&quot; to the other galaxy?" Tags="&lt;black-hole&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10887" PostTypeId="2" ParentId="10886" CreationDate="2015-06-02T09:05:39.993" Score="2" Body="&lt;p&gt;Yes, it is a wormhole indeed. This has been indicated quite clearly in the movie as well, when Dr. Romily explains with a pen and paper to Dr. Cooper. The explanation goes like this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a sheet of paper to be 2-D space, then a line joining two points on the sheet of paper is the shortest distance possible to reach that point, but if due to some disturbance, the space is bent ( achieved by folding the sheet of paper), you can pierce a hole in the sheet after aligning the two points together. That is a 2-D wormhole (which is a circle indeed). So what happens when you consider a 3-D wormhole? It becomes a sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you can see the other end of a 2-D wormhole (which is effectively the other hole in the sheet visible from the first hole, and one can see beyond the hole in the other direction as well). Same happens with the 3-D spherical wormhole where you can see to the other side of the wormhole as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the means to bend spacetime : Space time can be bent by having a very big mass placed inside the space time (read Einstien's &lt;a href=&quot;http://en.wikipedia.org/wiki/General_relativity&quot; rel=&quot;nofollow&quot;&gt;General Relativity&lt;/a&gt; ). So, it is safe to assume that the wormhole is a space time disturbance created due to a massive object, but it is not a blackhole.&lt;/p&gt;&#xA;" OwnerUserId="7369" LastActivityDate="2015-06-02T09:05:39.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10888" PostTypeId="2" ParentId="10885" CreationDate="2015-06-02T09:11:52.687" Score="2" Body="&lt;p&gt;Russia has &lt;a href=&quot;http://en.wikipedia.org/wiki/BTA-6&quot; rel=&quot;nofollow&quot;&gt;BTA-6&lt;/a&gt; which is a 6m optical reflector telescope (which is good enough for their research). Additionally, Australia is a member of many telescope projects throughout the world, and one of them (3.89m telescope) is in Australia itself. Have a look at the link &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_largest_optical_reflecting_telescopes&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7369" LastEditorUserId="6980" LastEditDate="2015-06-02T18:05:44.793" LastActivityDate="2015-06-02T18:05:44.793" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10889" PostTypeId="1" CreationDate="2015-06-02T09:24:54.577" Score="3" ViewCount="1263" Body="&lt;p&gt;While studying the Solar System, I found that some galaxies are either elliptical or elongated. What's the difference?&lt;/p&gt;&#xA;" OwnerDisplayName="arsh garg" LastEditorUserId="2153" LastEditDate="2015-06-02T18:05:26.253" LastActivityDate="2015-06-05T15:50:21.480" Title="Is there a difference between the terms 'elliptical' and 'elongated' for galaxies?" Tags="&lt;galaxy&gt;&lt;terminology&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10890" PostTypeId="2" ParentId="10889" CreationDate="2015-06-02T11:34:05.293" Score="3" Body="&lt;p&gt;&lt;strong&gt;tl;dr: &quot;Elliptical&quot; refers to a special kind of galaxies. &quot;Elongated&quot; is a loose term meaning that a galaxy is stretched out in one direction.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Etymologically, &quot;elliptical&quot; means having a shape that can be described by a certain mathematical function — the ellipse. For a three-dimensional object, the proper term is &quot;ellipsoidal&quot;, but still it refers to being described by a certain equation. In contrast, &quot;elongated&quot; is a more loose term, meaning something like &quot;significantly deviating from a spherical shape in one direction&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of galaxies, the term &quot;elliptical&quot; refers to a certain type of galaxies that tends to be not actively star-forming, have no or very little structure (as opposed to the beautiful spiral arms of spiral galaxies), be quite massive, as well as other characteristica. Having no ongoing star formation, the massive, short-lived stars, which have blue colors, have died long ago, leaving behind the less massive, red stars. In addition, elliptical are often quite dusty, further reddening the light. Thus, ellipticals appear red in color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape of an elliptical galaxy is described by the relative size of its three axes, $a$, $b$, and $c$. If two axes, say $a$ and $b$, are roughly the same size and are &lt;em&gt;larger&lt;/em&gt; than $c$, the galaxy is said to be &quot;oblate&quot;, whereas if they are &lt;em&gt;less&lt;/em&gt; than $c$, it is said to be &quot;prolate&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in the etymological meaning, an elongated galaxy is not a specific type, but rather refers to a galaxy departing from spherical shape in one direction. This could for instance be a prolate elliptical, but it could also by any other galaxy that has been distorted by merging with another galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, though (I think), it will refer to ellipticals. The &quot;elongation&quot; of an elliptical is defined as $10\times(1 - b/a)$, where $a$ and $b$ is now the &lt;em&gt;observed&lt;/em&gt; axes, i.e. of the two-dimensional projection on the sky. Ellipticals are classified according to this number as E0 (being spherical, such that $a=b$) to E7 (being very elongated). In principle, you could have even higher E-numbers, but that's not observed.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-06-05T15:50:21.480" LastActivityDate="2015-06-05T15:50:21.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10892" PostTypeId="2" ParentId="10889" CreationDate="2015-06-02T15:04:23.403" Score="2" Body="&lt;p&gt;&lt;em&gt;Elongated&lt;/em&gt; means made longer. An elongated shape or object is stretched primarily in one direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Elliptical&lt;/em&gt; means having an ellipse-like shape. An ellipse is an elongated circle, where the elongation is done in a constant/linear way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a limousine often has the form of an ordinary car, elongated (an elongated car). It is not at all ellipse-like, however.&lt;/p&gt;&#xA;" OwnerUserId="7378" LastActivityDate="2015-06-02T15:04:23.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10894" PostTypeId="1" AcceptedAnswerId="10895" CreationDate="2015-06-02T20:54:12.120" Score="6" ViewCount="810" Body="&lt;p&gt;A very basic question, but one to which I keep finding different answers: does Mercury have a liquid core, or is it all solid? Whatever the reason, what are the causes of it being so?&lt;/p&gt;&#xA;" OwnerUserId="4126" LastEditorUserId="2153" LastEditDate="2015-06-02T22:52:10.517" LastActivityDate="2015-06-02T23:05:21.217" Title="Is Mercury's core liquid?" Tags="&lt;planet&gt;&lt;core&gt;&lt;mercury&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10895" PostTypeId="2" ParentId="10894" CreationDate="2015-06-02T22:45:19.673" Score="6" Body="&lt;h1&gt;It's liquid.&lt;/h1&gt;&#xA;&lt;p&gt;As detailed &lt;a href=&quot;http://www.space.com/3756-surprise-slosh-mercury-core-liquid.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;To figure out whether Mercury's core was liquid or solid, a team of scientists led by Jean-Luc Margot at Cornell University measured small twists in the planet's rotation. They used a new technique that involved bouncing a radio signal sent from a ground telescope in California off the planet and then catching it again in West Virginia.&lt;/p&gt;&#xA;&lt;p&gt;After 5 years and 21 such observations, the team realized their values were twice as large as what would be expected if Mercury's core was solid.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;The variations in Mercury's spin rate that we measured are best explained by a core that is at least partially molten,&amp;quot; Margot said. &amp;quot;We have a 95 percent confidence level in this conclusion.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The NRAO has &lt;a href=&quot;http://www.nrao.edu/pr/2007/mercury/&quot; rel=&quot;nofollow noreferrer&quot;&gt;another article&lt;/a&gt; on it, which goes slightly more in-depth into the subject.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://messenger.jhuapl.edu/why_mercury/q4.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The official site of the &lt;em&gt;Messenger&lt;/em&gt; mission&lt;/a&gt; is slightly more cautious:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;However, these constraints are limited because of the low precision of current information on Mercury's gravity field from the Mariner 10 and MESSENGER flybys. Fundamental questions about Mercury's core remain to be explored, such as its composition. A core of pure iron would be completely solid today, due to the high melting point of iron. However, if other elements, such as sulfur, are also present in Mercury's core, even at a level of only a few percent, the melting point is lowered considerably, allowing Mercury's core to remain at least partially molten as the planet cooled. Constraining the composition of the core is intimately tied to understanding what fraction of the core is liquid and what fraction has solidified. Is there just a very thin layer of liquid over a mostly solid core or is the core completely molten? Addressing questions such as these can also provide insight into the current thermal state of Mercury's interior, which is very valuable information for determining the evolution of the planet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;At this point in time, though, all evidence indicates that Mercury has a molten core.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/10894/is-mercurys-core-liquid/10895#comment14549_10895&quot;&gt;As userLTK pointed out&lt;/a&gt;, lower pressure inside Mercury makes it easier for the core to stay liquid at lower temperatures.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-06-02T23:05:21.217" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10896" PostTypeId="1" CreationDate="2015-06-03T02:53:22.400" Score="9" ViewCount="10900" Body="&lt;p&gt;Ok, so it's supposed that there is a black hole in the center of any galaxy, that makes my head explode because:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There is light in the center of the galaxy, but it is supposed that black holes, also attract light.&lt;/li&gt;&#xA;&lt;li&gt;For a black hole to exist there should be a star explosion, so what size does the star needed to be to create such a galaxy.&lt;/li&gt;&#xA;&lt;li&gt;Why does the black hole doesn't eat us, I mean what objects are pulling us away from the extreme gravity of the black hole.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm not god, I didn't study Astronomy, I'm just a kid fascinated by the Universe. (This question should be dumb for anyone who studied Astronomy)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you think my question is dumb, (which isn't) or if you have any recommendation for further asking, let me know!&lt;/p&gt;&#xA;" OwnerUserId="7381" LastEditorUserId="1577" LastEditDate="2015-06-03T07:30:03.673" LastActivityDate="2018-05-04T14:32:57.220" Title="What's in the center of a Galaxy?" Tags="&lt;star&gt;&lt;galaxy&gt;&lt;light&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10897" PostTypeId="2" ParentId="10896" CreationDate="2015-06-03T04:22:29.127" Score="15" Body="&lt;p&gt;At the center of our galaxy is a powerful radio source named Sagittarius A*, which is believed to be a super massive black hole (SMBH). This blackhole would contain far more mass than your run-of-the-mill supernova remnant. Our galaxy is believed to contain a SMBH containing the mass of likely a bit above &lt;a href=&quot;http://iopscience.iop.org/0004-637X/692/2/1075/pdf/0004-637X_692_2_1075.pdf&quot;&gt;4 million times the mass&lt;/a&gt; (Gillessen) &lt;a href=&quot;http://iopscience.iop.org/0004-637X/689/2/1044/&quot;&gt;(2)&lt;/a&gt; (Ghez) of our Sun. For reference, I don't think we've ever discovered a star more massive than 600 times that of our Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also important to understand that while many people look at blackholes as mystical or all-consuming, they actually have to follow the same rules as everyone else in the stellar neighborhood. The stars that make up our galaxy don't fall into the black hole for the same reason our planet doesn't fall into the Sun. Our star orbits black hole, our star system's velocity in equilibrium with the attractive force of the galaxy's center of gravity. This should hopefully resolve point 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For point 1, we should make clear that the 'black' part of the black hole is only true once you cross the event horizon. This is the case because at this point the escape velocity to escape the gravity of the black hole because greater than the speed of light. The light that isn't within the event horizon and is moving away from it is free to escape. So we can see light around it.  But why is there so much light? Well, as it happens there are a &lt;strong&gt;lot&lt;/strong&gt; of rather young and large stars in this area. It is not completely understand why this is the case. Lots of stars, lots of light! There are other factors that can contribute to this too, such as there simply being lots of stars between us and the center, not just in the center itself. The accretion disk of a black hole can also be exceptionally bright. Hopefully that clears up part 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for part 2. As far as I know, we don't really have any way of determining exactly where our SMBH came from originally. Black holes aren't necessarily formed just from a supernova event, there are a handful of other ways they can be created in nature. What &lt;em&gt;is&lt;/em&gt; apparent, however, is that SMBHs contain far too much mass to be from a single star. It has probably consumed plenty of other black holes to grow to what it is now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One interesting and notable difference between the comparison of a star system and a galaxy is the distribution of mass. While our Sun is believed to contain 99.8% of the mass of our solar system, the SMBH at the center of the Milky Way is not nearly as massive as the total mass of the Milky Way. The ratio can vary a lot, and there are some galaxies which are believed to host no SMBH at all.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Gillessen, Stefan et al. (23 February 2009). &quot;Monitoring stellar orbits around the Massive Black Hole in the Galactic Center&quot;. The Astrophysical Journal 692 (2): 1075–1109.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ghez, A. M. et al. (December 2008). &quot;Measuring Distance and Properties of the Milky Way's Central Supermassive Black Hole with Stellar Orbits&quot;. Astrophysical Journal 689 (2): 1044–1062. &lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="1288" LastEditDate="2015-06-03T06:54:20.107" LastActivityDate="2015-06-03T06:54:20.107" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10898" PostTypeId="2" ParentId="10896" CreationDate="2015-06-03T06:41:38.413" Score="2" Body="&lt;p&gt;If you understand the facts, your head won't explode!&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You are right, light cannot escape from even a black hole, so why is there so much light in the center? Well, simply because there is &lt;strong&gt;a lot&lt;/strong&gt; of light in the center of the galaxy, because there is a high concentration of stars. These stars are very far away from us, so the light from each together create a &quot;halo&quot; effect in the center, which makes it bright. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The type of black hole you are thinking of is a stellar black hole, caused by a supernova which eventually leads to a neutron star collapsing. Not all supernovae; become black holes, look up the Chandrasekhar Limit if you want more information. So it is &lt;strong&gt;believed&lt;/strong&gt; that galaxies contain supermassive black holes at the center and the mass of the supermassive black hole is equivalent to the entire galaxy!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This black hole won't eat us. You have the typical misconception of black holes. Black holes don't have super sucking power; in fact, they do not &quot;suck&quot; at all. It is only if you get close to the event horizon of a black hole that you cannot escape it's gravitational pull. If the Sun became a black hole we wouldn't get &quot;sucked&quot; into it because it would have the same mass as itself before, and we would orbit it normally. Gravity and inertia is what keeps us in orbit and the stars in our galaxy in orbit, so as long as we stay in orbit, we will be okay. By the way, this supermassive black hole at the center of our galaxy is about 30,000 light years away, so even if we somehow get out of our normal orbit and head towards the center, we won't get &quot;sucked&quot; in for a very long time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are no dumb questions, just dumb answers.&lt;/p&gt;&#xA;" OwnerUserId="7258" LastEditorUserId="2153" LastEditDate="2015-06-03T12:32:25.537" LastActivityDate="2015-06-03T12:32:25.537" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10899" PostTypeId="2" ParentId="10884" CreationDate="2015-06-03T06:56:01.283" Score="1" Body="&lt;p&gt;With respect to &quot;eternally collapsing&quot;, they are probably referrring to the fact that in the reference frame of an external observer, gravitational time dilation prohibits matter from ever reaching the event horizon (the &quot;surface&quot;) of the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denoting the radius of the event horizon $r_\mathrm{S}$ (for &quot;Schwarzschild radius&quot;), time runs slower by a factor of $(1 - r_\mathrm{S}/r)^{-1/2}$ for an observer at a distance of $r$. As $r$ approaches $r_\mathrm{S}$, this factor goes to infinity, i.e. the observer will never reach $r_\mathrm{S}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is only for the external observer; the falling observer would cross the horizon and reach the center in a finite time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With respect to &quot;not being stable&quot;, you're probably right that they are referring to Hawking radiation which presumably makes the black hole slowly &quot;evaporate&quot;. Near $r_\mathrm{S}$, pairs of virtual particles are being created and can be turned into real particles by the gravitational field. If they avoid falling into the black hole, energy is taken away from the hole, reducing its mass.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2015-06-03T06:56:01.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10901" PostTypeId="1" CreationDate="2015-06-03T13:53:25.107" Score="1" ViewCount="2497" Body="&lt;p&gt;In &lt;em&gt;Astronomical Algorithms&lt;/em&gt; (2nd ed, ch. 27, 2009 corrected printing) Jean Meeus gives expressions to calculate the date and time (dynamical time, equivalent to Terrestrial Time) of equinoxes and solstices from the year -1000 to the year +3000. The expressions are accurate to 51 seconds or better for the years 1951-2050. First what Meeus calls the &quot;instant of the 'mean&quot; equinox or solstice&quot; is calculated using a fourth degree polynomial; there are 8 expressions. There are different expressions for each solstice or equinox, and different expressions for the year ranges -1000 to 1000 vs. 1000 to 3000. Then two corrections are applied; the corrections are calculated the same way no mater which time period or equinox or solstice is being corrected. The first step is to calculate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T = \frac{(\text{mean JD of event} - 2451545.0)}{36525}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W = 35999.373°T - 2.47°$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta \lambda = 1 + 0.334 \cos W + 0.007 \cos 2 W$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, an additional correction is computed involving 24 periodic terms with various periods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain, in general terms, how Meeus derived these expressions? I'm especially interested in understanding what the &quot;mean&quot; value represents?&lt;/p&gt;&#xA;" OwnerUserId="7385" LastEditorUserId="2153" LastEditDate="2015-06-03T20:11:39.267" LastActivityDate="2019-02-08T13:05:16.230" Title="How did Meeus calculate equinox and solstice dates?" Tags="&lt;solar-system&gt;&lt;data-analysis&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10902" PostTypeId="2" ParentId="10878" CreationDate="2015-06-03T18:25:15.837" Score="5" Body="&lt;p&gt;Rather than looking for ready-made systems, take a look at projects. Right now, plenty of amateurs are using software defined radio coupled to various antennas for astronomy. Start here:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.rtl-sdr.com/rtl-sdr-for-budget-radio-astronomy/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.rtl-sdr.com/rtl-sdr-for-budget-radio-astronomy/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;And while it has nothing to do with imaging, there's plenty of radio astronomy that amateurs can do using simple (albeit sometimes large) antennas:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://radiojove.gsfc.nasa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://radiojove.gsfc.nasa.gov/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://radio-astronomy.org/pdf/qex/radio-jove-proof.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://radio-astronomy.org/pdf/qex/radio-jove-proof.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.radiosky.com/project.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.radiosky.com/project.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7154" LastEditorUserId="37191" LastEditDate="2021-05-11T09:19:21.583" LastActivityDate="2021-05-11T09:19:21.583" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10903" PostTypeId="2" ParentId="1120" CreationDate="2015-06-03T19:56:04.217" Score="1" Body="&lt;p&gt;Any sundial that gives the same result as this is correct and any other is wrong (but sometimes close enough):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                    _      /############&#xA;                    /|    /#############&#xA;     skewer        /     /##############&#xA;(central) v       /     /###############&#xA;          |  north     /################&#xA;          | (S in S.  /#################&#xA;          | hmsphre) /##################&#xA;  ________|________ /###################&#xA;hoop      |        /##### LEVEL  #######&#xA;(from the | side) /###### GROUND #######&#xA;          |      /####### (SOIL) #######&#xA;latitude--|-    /#######################&#xA;(use      ||   /########################&#xA;protractor)|  /#########################&#xA;          |V /##########################&#xA;          | /###########################&#xA;          |/############################&#xA;          j#############################&#xA;         ,|#############################&#xA;        /#|#############################&#xA;       /##|#############################&#xA;      /###|#############################&#xA;     /####|#############################&#xA;    /#####|#############################&#xA;   /######|#############################&#xA;  /#######|#############################&#xA; /########V#############################&#xA;/#######################################&#xA;########################################&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;write noon on hoop's inside closest to ground, midnight opposite, 6 pm on the east side, 9 pm midway between the last two, and so on (hours only occurring in darkness optional).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your sundial reads 6 pm at due west all the time then you're doing it wrong. Let's say I put a vertical stick in the ground, draw a 24 hour clock face around it, put noon poleward and think it's a sundial. In New York City, it could literally be saying 6 am when a genuine sundial says 9 am. That's just middle latitudes. At the equator on the equinox, it would read 6 am all morning and 6pm all afternoon at the equator. If you go 1 mile south of where the next noon, summer solstice and Tropic of Cancer coincide, it would say about 4:30am at sunrise, go forwards at first, then backwards, finally showing the middle 6 hours of the night passing in 7 seconds. Backwards. At noon. Then it will run forwards again until it reaches 7:30pm at sunset. If you place the stick right, you can even make it stay between 4:30a and 6a all morning, stop at 6 am at the instant of noon then instantly become midnight, run infinity years per second backwards for an infinitely short amount of time, go almost 6 hours backwards in seconds, then later forwards again very slowly until it shows about 7:30pm at sunset. This is why sundials cannot be made that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(of course, this is theoretical, there are no infinitely thin, vertical, and straight sticks, shadows are fuzzy, they can be too short to see, the Earth wobbles a bit, the speed of light is not infinite, this would only be true if only the Sun and Earth existed, even a flea jumping in Russia moves the Earth etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, the sundial time can be up to 16 minutes away from mean solar time (the equation of time), easily noticeable, but if you wanted correct local clock time instead of correct sundial time then you could put as many dates as needed on another dial and rotate the hour scale until the arrows point at the current time of year. The shadow then shows mean solar time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That should be close enough to mean solar time that you wouldn't care for a number of centuries, certainly a century if you're real picky. As for clock time, what sundial disagreements are possible is limited only by the whims of man. The sundial is several hours wrong in West China. Cause they use the zone that's good for Shanghai (or Tokyo when daylight savings).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And all sundials without moving parts are latitude specific. Some are adjustible, though. Some designs are more suited for some latitudes or even become impossible in some places, like the kind with a wedge or rod on a level face. They will also not work on days with polar night. Though you could use a moondial if it's also not polar moon's below the horizon for days or weeks. Yes, moondials exist! You need to correct for moon phase and time of year or they're useless.&lt;/p&gt;&#xA;" OwnerUserId="6784" LastEditorUserId="616" LastEditDate="2016-08-26T04:05:16.770" LastActivityDate="2016-08-26T04:05:16.770" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10904" PostTypeId="1" CreationDate="2015-06-04T08:56:56.800" Score="1" ViewCount="1450" Body="&lt;p&gt;To find the best observation time for an object, I'd like to calculate the time when it is 30° or more above the horizon. Local Sideral Time would be sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To include that in my program, I need the formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&#xA;On June 4th, Jupiter has the coordinates&#xA;RA= 9h 19m 28.0s&#xA;Dekl= 16° 32' 0&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It rises at 10:32 and sets at 00:05.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After rise, when is it at altitude 30°, and after transit, when is it at altitude 30° again ?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I found this formula at &lt;a href=&quot;http://www.stjarnhimlen.se/comp/riset.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.stjarnhimlen.se/comp/riset.html&lt;/a&gt;. Although it's for the sun, it seems to be what I'm looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos (\text{LHA}) = \frac{\sin (\text{h}) - \sin (\text{lat}) \times \sin (\text{Decl})}{\cos (\text{lat}) \times \cos (\text{Decl})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applied to the sample assuming a latitude of 45° I get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KW7yI.png&quot; alt=&quot;Sample caluclation&quot;&gt;&#xA;&lt;br/&gt;Is this the correct approach?&lt;/p&gt;&#xA;" OwnerUserId="7395" LastEditorUserId="2153" LastEditDate="2015-06-04T22:32:36.823" LastActivityDate="2016-09-16T21:34:44.163" Title="Calculate time when star is above altitude 30°" Tags="&lt;observational-astronomy&gt;&lt;fundamental-astronomy&gt;&lt;positional-astronomy&gt;" AnswerCount="3" CommentCount="10" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10906" PostTypeId="2" ParentId="6026" CreationDate="2015-06-04T19:03:52.380" Score="7" Body="&lt;p&gt;You cannot see it from Earth, it is too small and too far away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lunar_Reconnaissance_Orbiter&quot; rel=&quot;noreferrer&quot;&gt;Lunar Reconnaissance Orbiter&lt;/a&gt; imaged it. &lt;img src=&quot;https://i.stack.imgur.com/iKShg.jpg&quot; alt=&quot;lunar surface with Apollo 11 lander&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7399" LastEditorUserId="2810" LastEditDate="2015-06-04T19:30:42.873" LastActivityDate="2015-06-04T19:30:42.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10907" PostTypeId="5" CreationDate="2015-06-04T19:09:39.517" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-06-04T19:09:39.517" LastActivityDate="2015-06-04T19:09:39.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10908" PostTypeId="4" CreationDate="2015-06-04T19:09:39.517" Score="0" Body="Second lightest element in the periodic table" OwnerUserId="7399" LastEditorUserId="7399" LastEditDate="2015-06-05T11:56:10.000" LastActivityDate="2015-06-05T11:56:10.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10909" PostTypeId="1" CreationDate="2015-06-04T19:46:20.987" Score="6" ViewCount="137" Body="&lt;p&gt;This is actually related to a question I recently asked on &lt;a href=&quot;https://worldbuilding.stackexchange.com/questions/18051/tomorrow-is-groundhog-day-for-everyone-how-does-society-respond&quot;&gt;Worldbuilding&lt;/a&gt;, but seemed more appropriately asked here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To keep this from being too broad in scope, let's assume that someone figured out the exact moment that Eta Carinae will go supernova (I wouldn't hold it against anyone if they suggested a more scientifically interesting event).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to this foreknowledge, we could point every telescope we have in its direction for the main event.  Is there any science we know we could do/learn from this event that we wouldn't get by reacting to it after the fact?  I know such an event would generate data for a century or more to come, but I'm interest in the moment of the explosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somewhat related, is there anything we would/could do now to prepare for it if this event were going to happen today?&lt;/p&gt;&#xA;" OwnerUserId="2503" LastEditorUserId="-1" LastEditDate="2017-04-13T12:52:24.020" LastActivityDate="2016-10-19T17:34:26.473" Title="What more could be learned from a rare astronomical event if we knew precisely when it would occur?" Tags="&lt;astrophysics&gt;&lt;supernova&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10910" PostTypeId="1" AcceptedAnswerId="10913" CreationDate="2015-06-04T20:19:21.273" Score="3" ViewCount="108" Body="&lt;p&gt;Gravitational lensing takes advantage of statistically uniform distributions of background objects in order to infer a credible map of dark matter concentrations that isn't too noisy. If there are too many background objects and is not clear there is an uniform pattern of distribution that one can compensate numerically, then the noise of the measurement makes the reading essentially worthless&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'm curious if there is any research attempt to try to pull this kind of local measurement of lensing in the &quot;backyard&quot;. If one takes the Milky Way's galactic rotation curve, and compares it on what's expected given baryonic matter, one would be able to have a radial distribution of expected dark matter in order to compare with the results of lensing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is that kind of lensing measurement something that is possible today? &lt;/p&gt;&#xA;" OwnerUserId="7400" LastActivityDate="2015-06-05T00:06:06.467" Title="How close are we to be able to detect and measure gravitational lensing inside the milky way?" Tags="&lt;dark-matter&gt;&lt;gravitational-lensing&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10911" PostTypeId="2" ParentId="2402" CreationDate="2015-06-04T20:41:36.027" Score="0" Body="&lt;p&gt;It's because it takes more energy to fuse iron into anything else than you get out of the reaction.  The whole life of a star is a delicate balance of the release of energy by the process of fusion pushing out against the force of gravity and holding the start in a somewhat stable balance.  It fuses Hydrogen into Helium which generates energy which generates pressure which pushes back against the force of gravity.  Each successive phase of fusion releases less and less energy from the fusion, so it has to work harder and harder to keep everything up against the force of gravity.  That's why each successive fusion lasts for a shorter and shorter time, because it has to fuse a lot more Helium faster to get the same amount of energy that it got from less Hydrogen fusion.  So back to the supernova.  It gets to Iron, suddenly it has no where else to go.  Nothing else it can fuse is going to produce energy, boom all that energy that was holding the core up agains the force of gravity suddnely is not there.  Gravity is left alone pulling the core together and the core collapses until it gets so dense that you can't pack anymore of into the space.  Basically everything turns into one giant ball of Neutrons pressed together as close as they can be and the laws of physics say &quot;STOP&quot;.  Well everything else is still collapsing in at this wall of neutrons at, i've seen people say 10-20% the speed of light.  It hits that wall and bounces back ripping out through the rest of the gas in the star in a shockwave.  That shockwave creates really high temperatures and high pressures in the other layers of the star which can also cause fusion.  As i understand this shockwave is the way that elements higher than Iron are actually created.&lt;/p&gt;&#xA;" OwnerUserId="7401" LastActivityDate="2015-06-04T20:41:36.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10912" PostTypeId="2" ParentId="2402" CreationDate="2015-06-04T22:11:47.843" Score="6" Body="&lt;p&gt;The answer is that in a pre-supernova star, most of its mass is still in the form of hydrogen and helium. It is only the central core where the primordial H and He has fused to heavier elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This picture of onion layers is typically what you see in elementary text books. It is &lt;em&gt;completely misleading&lt;/em&gt; in a quantitative sense. It &lt;em&gt;schematically&lt;/em&gt; represents what's at the &lt;em&gt;centre&lt;/em&gt; of the pre-supernova star, but in terms of the mass that is in each shell (it is obviously a 3d object) you get completely the wrong idea, because this diagram is only about 1 Earth diameter across, compared with the actual stellar radius of something like the distance between the Earth and the Sun!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jv0nz.gif&quot; alt=&quot;Onion layers in a pre-supernova star&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a more sophisticated plot taken from a paper by &lt;a href=&quot;http://arxiv.org/abs/1502.07779&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fuller et al. (2015)&lt;/a&gt;. It shows time &lt;em&gt;until&lt;/em&gt; the supernova along the x-axis and the y-axis represents a radial assay of the chemical composition from the centre of the star to the outside. The initial total stellar mass is $12M_{\odot}$. As you move leftwards towards the supernova explosion, notice how what is at the core changes - from being H dominated, to He dominated, to C/O dominated then Si and finally Fe (actually iron-peak elements). Note how much mass is contained within these core region for each stage of nuclear burning. The edge of the &quot;helium core&quot; encloses the central $4M_{\odot}$ of the star. The subsequent heavier element cores inside the onion ring structure enclose significantly less mass, until the iron core is around $1.3M_{\odot}$ just prior to the explosion. Blue shading indicates regions that are thoroughly mixed and homogenised by convection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the explosion, the neutron star that is produced will also have a mass of around $1.3-1.4M_{\odot}$. In other words most of the rest of the star (about $10M_{\odot}$ just prior to the explosion) gets blown out in the supernova. But of the $8.6M_{\odot}$ that makes into the interstellar medium, well over half is still in the form of hydrogen and helium; the minority will be carbon, oxygen, neon, silicon, iron etc., and only a very small fraction of that will have been transformed (by the &lt;a href=&quot;http://en.wikipedia.org/wiki/R-process&quot; rel=&quot;nofollow noreferrer&quot;&gt;r-process&lt;/a&gt;) into elements heavier than iron and nickel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus although the material injected back into the interstellar medium is enriched with heavier elements, there is still plenty of hydrogen to start a new generation of stars. It is also the case that star formation is an inefficient process, so the material from which the supernova progenitor formed will still mostly be around in the interstellar medium. The picture you should have is of a &lt;em&gt;gradual&lt;/em&gt; enrichment with heavy elements, especially as the interstellar medium gets churned up and mixed through a variety of processes (including supernova explosions!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y1ewe.jpg&quot; alt=&quot;Pre-supernova chemical evolution&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Here is an even more awesome picture. The lower plot shows the relative mass fraction of each element as a function of enclosed mass as you work your way out from a 15 solar mass star (it has shed 2 solar masses during its evolution). The really awesome thing is that it is animated, so it shows you the first few moments after the core-collapse and how things start to change. Note that the outer $5M_{\odot}$ of the envelope is about half H and half He by mass prior to the core collapse. Lots of He and then O in the layers below that. The upper plot show how the density temperature and outward velocity are behaving. The image is from the website of &lt;a href=&quot;http://2sn.org/nucleosynthesis/WH2007.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;Woosley and Heger (2007)&lt;/a&gt;, a canonical work on the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hmm. I can't upload the animated gif. &lt;a href=&quot;http://2sn.org/nucleosynthesis/movie/s25a28d-bulk.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here it is&lt;/a&gt;; well worth a look.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-06-04T23:19:07.273" LastActivityDate="2015-06-04T23:19:07.273" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10913" PostTypeId="2" ParentId="10910" CreationDate="2015-06-05T00:06:06.467" Score="3" Body="&lt;p&gt;Well, there are &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_microlensing&quot; rel=&quot;nofollow&quot;&gt;microlensing&lt;/a&gt; experiments, the results of which have largely removed &lt;a href=&quot;http://en.wikipedia.org/wiki/Massive_compact_halo_object&quot; rel=&quot;nofollow&quot;&gt;MACHOs&lt;/a&gt; as a significant source of Dark Matter.  The background field of stars is usually a nearby galaxy (a Magellanic cloud one or Andromeda, say), or the galactic bulge, and the experiment looks for lensing objects between us and there.  In particular, they're pretty much always looking for the microlensing effect within our own Milky Way or its halo.  These experiments have detected a number of microlensing events (and many, many more candidates that were due to other things), so, yes, this is not only possible, it is already done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wiki page I started with linking has this to say, in particular:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;The first two microlensing events in the direction of the Large Magellanic Cloud that might be caused by dark matter were reported in back to back Nature papers by MACHO and EROS in 1993, and in the following years, events continued to be detected. The MACHO collaboration ended in 1999. Their data refuted the hypothesis that 100% of the dark halo comprises MACHOs, but they found a significant unexplained excess of roughly 20% of the halo mass, which might be due to MACHOs or to lenses within the Large Magellanic Cloud itself. EROS subsequently published even stronger upper limits on MACHOs, and it is currently uncertain as to whether there is any halo microlensing excess that could be due to dark matter at all. The SuperMACHO project currently underway seeks to locate the lenses responsible for MACHO's results.&lt;br /&gt;&lt;br /&gt;&#xA;&#xA;Despite not solving the dark matter problem, microlensing has been shown to be a useful tool for many applications. Hundreds of microlensing events are detected per year toward the Galactic bulge, where the microlensing optical depth (due to stars in the Galactic disk) is about 20 times greater than through the Galactic halo. In 2007, the OGLE project identified 611 event candidates, and the MOA project (a Japan-New Zealand collaboration) identified 488 (although not all candidates turn out to be microlensing events, and there is a significant overlap between the two projects). In addition to these surveys, follow-up projects are underway to study in detail potentially interesting events in progress, primarily with the aim of detecting extrasolar planets. These include MiNDSTEp, RoboNet, MicroFUN and PLANET.&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2015-06-05T00:06:06.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10914" PostTypeId="1" AcceptedAnswerId="10917" CreationDate="2015-06-05T00:14:46.817" Score="5" ViewCount="251" Body="&lt;p&gt;Dark energy and the cosmological constant are often identified, but &lt;a href=&quot;http://arxiv.org/pdf/astro-ph/0207347v2.pdf&quot; rel=&quot;noreferrer&quot;&gt;Peebles and Ratra&lt;/a&gt; explain that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;em&gt;Einstein did not consider the cosmological constant to be part of the stress-energy term... a new constant of nature, &#xA;  $\Lambda$, appears in the addition to Einstein’s original field equation. One can equally well put Einstein’s new term on the right hand side of the equation, and count [...] as part of the source term in the stress-energy tensor. The distinction becomes interesting when $\rho_\Lambda$ takes part in the dynamics, and the field equation is properly written with $\rho_\Lambda$, or its generalization, as part of the stress-energy tensor.&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, Einstein treated $\Lambda$ term as a modification of GR with a &quot;new constant of nature&quot;, while dark energy may potentially manifest itself in some other ways. Later they write about dark energy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;em&gt;The idea that the universe contains close to homogeneous dark energy that approximates a time-variable cosmological “constant” arose in particle physics, ...in cosmology..., and on both sides by the thought that $\Lambda$ might be very small now because it has been rolling toward zero for a very long time. The idea that the dark energy is decaying by emission of matter or radiation is now strongly constrained by the condition that the decay energy must not significantly disturb the spectrum of the 3K cosmic microwave background radiation.&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So if I understand this correctly the idea of a dynamical effect came from suspecting that $\Lambda$ may be time dependent, but so far that was not detected. And since &quot;one can equally well put Einstein’s new term&quot; on either side of the equation mathematically both interpretations are equivalent. Still, the dynamical interpretation seems to be overwhelmingly preferred.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there empirical reasons for conjecturing dark energy, some spatial non-uniformity in expansion perhaps? If not, what are the theoretical reasons for preferring it to the cosmological constant of modified GR? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7403" LastActivityDate="2015-06-05T06:32:22.333" Title="Why is dark energy preferred to the cosmological constant?" Tags="&lt;cosmology&gt;&lt;general-relativity&gt;&lt;dark-energy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10915" PostTypeId="2" ParentId="6526" CreationDate="2015-06-05T04:54:31.093" Score="0" Body="&lt;p&gt;The 3rd requirement for a celestial object to be a planet is that it has to &quot;clear its neighborhood&quot; which means it has to be gravitationally dominant. Pluto has not enough mass to interact with other object in its orbit(consuming them or swinging them away) and it is only 0.07 times the mass of the other objects in its orbit. Earth is 1.7 million times the mass of other objects in its orbit.&lt;/p&gt;&#xA;" OwnerDisplayName="user7407" LastActivityDate="2015-06-05T04:54:31.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10916" PostTypeId="2" ParentId="10896" CreationDate="2015-06-05T05:17:16.657" Score="0" Body="&lt;p&gt;In the middle of the Milky Way is a supermassive black hole and a strong radio source in Sagittarius A*. &#xA;It is 26 thousand light years away from us.&#xA;Latest estimates estimate the black hole to be 4.31±0.38 million solar masses.&#xA;It's Schwarzschild radius is about 12Gm(0.08AU).&lt;/p&gt;&#xA;" OwnerDisplayName="user7407" LastActivityDate="2015-06-05T05:17:16.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10917" PostTypeId="2" ParentId="10914" CreationDate="2015-06-05T06:32:22.333" Score="6" Body="&lt;p&gt;A cosmological constant should be considered a special case of dark energy. The effective stress-energy tensor for a cosmological constant is proportional to the metric $g_{\mu\nu}$, so in a local inertial frame will be proportional $\mathrm{diag}(-1,+1,+1,+1)$. This is equivalent to perfect fluid with energy density and pressure directly opposite one another, but more importantly, it is the only possible form for the stress-energy that would give the exact same energy density and pressure in all local inertial frames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If by 'dark energy', we understand it to mean all the contributions to stress-energy in the above form, then there is no reason for this to be constant, and plenty of reasons why it might not be, as this situation is not exceptional in fundamental physics. For example, there could be &lt;a href=&quot;https://en.wikipedia.org/wiki/False_vacuum&quot; rel=&quot;noreferrer&quot;&gt;false vacua&lt;/a&gt; with various different energy densities, and they must be invariant across inertial frames. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the basic idea of inflation considers a flat FRW universe with expansion driven by a scalar field $\phi$ at a local extremum of its potential, $V'(\phi_0) = 0$, which yields an exponential expansion with constant energy density $T^0{}_0 = V(\phi_0)$. More refined models, such as &lt;em&gt;slow-roll inflation&lt;/em&gt;, could therefore be directly interpreted as a time-varying dark energy density, while &lt;em&gt;eternal inflation&lt;/em&gt; would also include spatial variability. There's plenty of other inflationary models besides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One the interpretational flip-side, one could always have a $\Lambda$ that corresponds to the energy density of the true vacuum, and the rest as separate contributions on top of that. It's just not as useful in a cosmological context compared to grouping all 'dark energies' together, as all stress-energy gravitates equally.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2015-06-05T06:32:22.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10918" PostTypeId="1" AcceptedAnswerId="10926" CreationDate="2015-06-05T07:43:39.457" Score="3" ViewCount="44" Body="&lt;p&gt;In all the illustrations regarding space distortion, I find an extra dimension that depicts space distortion. Is there any mathematical equation that proves the creation of an extra dimension ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7306" LastActivityDate="2015-06-05T17:19:45.070" Title="Question about extreme space distortion and creation of a new dimension" Tags="&lt;general-relativity&gt;&lt;space&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10919" PostTypeId="2" ParentId="1523" CreationDate="2015-06-05T11:45:33.177" Score="3" Body="&lt;p&gt;I suspect the main problem is that the JPEG images you're working with are &lt;em&gt;already&lt;/em&gt; log-scaled (or scaled via some other function such as a square root). Since the centers of most galaxies are orders of magnitude brighter in linear terms than the outer regions, a JPEG image (which, remember, has only 8 bits per color channel, and so can only represent 256 levels of brightness per channel) representing linear brightness would show a bright center and nothing outside. So your fit is effectively something like&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;log(log(log(I))) ~ log(R)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Programs like &lt;code&gt;GALFIT&lt;/code&gt; and &lt;code&gt;GIM2D&lt;/code&gt; (or my own &lt;code&gt;Imfit&lt;/code&gt; code) are meant to fit 2D grayscale, floating-point images with linear intensities which are in the FITS astronomical image format. (&lt;code&gt;GIM2D&lt;/code&gt; also requires the &lt;code&gt;IRAF&lt;/code&gt; software system in order to run.) They won't work with JPEG images.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2015-06-05T11:45:33.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10920" PostTypeId="2" ParentId="10889" CreationDate="2015-06-05T11:54:30.940" Score="1" Body="&lt;p&gt;More simply, &quot;elongated&quot; is a general term that can be used to describe the appearance of any galaxy, and means more or less what it means in ordinary English.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Elliptical&quot; &lt;em&gt;is&lt;/em&gt; a specific technical term for galaxies, and refers to galaxies which are in simplistic terms &quot;round and fuzzy&quot; -- made up primarily of stars on randomized three-dimensional orbits, with little or no gas and little or no current star formation, without a dominant stellar disk. They can be &quot;elongated&quot;, but they can also be (intrinsically, or when seen at the right orientation) circular. There's a decent introductory &lt;a href=&quot;https://en.wikipedia.org/wiki/Elliptical_galaxy&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2015-06-05T11:54:30.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10921" PostTypeId="1" CreationDate="2015-06-05T12:15:02.510" Score="5" ViewCount="1055" Body="&lt;p&gt;This question repetitively comes to my mind but I've never found any good scientist article to argue about it, only esoteric or speculative web articles without much valuable background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's suppose we're in the middle of a &quot;black hole&quot; ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, here is why this would make sense:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Big bang&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The word big bang itself may be inapproriate, indeed it is said to be an &quot;explosion&quot; but an &quot;explosion&quot; without any center.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the idea of being inside a black-hole, it is easier to represent this as being the matter entering the black-hole. From our position, and as we are attracted by the singularity, everything seems to expand wherever we look at. And as far as we can look at, we can only see the cosmic background radiation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Dark matter&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the contrary of matter entering the black-hole, dark matter is the matter already in front of us and we'll never be able to access it as its light (i.e. information) cannot go backward within the singularity.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Time/Gravitation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time is a dimension like any other and as we feel x,y,z dimensions as a field of view, we fell time as an irrepressible attraction toward the singularity. Gravity may be see also as a less irrepressible dimension (depending on mass around).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Other black holes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other black-hole we &quot;see&quot; may let us envision the shape of the universe not as &lt;a href=&quot;http://en.wikipedia.org/wiki/Three-torus_model_of_the_universe&quot; rel=&quot;noreferrer&quot;&gt;simple torus&lt;/a&gt;, but higher dimensional &quot;toric&quot; shape with &quot;connections&quot; everywhere (maybe shapes like high order &lt;a href=&quot;http://csi.chemie.tu-darmstadt.de/ak/immel/tutorials/orbitals/hydrogenic.jpg&quot; rel=&quot;noreferrer&quot;&gt;atomic orbitals&lt;/a&gt; ?).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is there any reasonable scientific background to argue/contradict about this ?&lt;/p&gt;&#xA;" OwnerUserId="7410" LastEditorUserId="7410" LastEditDate="2015-06-05T14:49:32.973" LastActivityDate="2015-06-15T21:22:18.277" Title="Is our universe included inside a black hole?" Tags="&lt;black-hole&gt;&lt;universe&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ClosedDate="2020-10-14T12:36:47.990" ContentLicense="CC BY-SA 3.0" />
  <row Id="10922" PostTypeId="2" ParentId="10885" CreationDate="2015-06-05T12:33:22.260" Score="4" Body="&lt;p&gt;Optical telescope sites ideally need to be somewhat remote (to avoid light pollution) yet still accessible for construction, engineering, and observing; high in altitude; dry; with exceptional &quot;seeing&quot; (stable atmospheric conditions); and clear weather. Neither Australia nor Russia really have any sites meeting all those criteria: their mountains tend to have some combination of too low altitude, poor seeing, and/or bad weather. (This is also why there aren't any really big, modern telescopes in Europe.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.noao.edu/education/gsmt/site&quot; rel=&quot;nofollow&quot;&gt;This site&lt;/a&gt; has some discussion of the different criteria that go into choosing the best sites for new, large telescopes.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2015-06-05T12:33:22.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10923" PostTypeId="2" ParentId="10781" CreationDate="2015-06-05T13:00:59.463" Score="4" Body="&lt;p&gt;To amplify andy256's comment, the problem that solar telescopes face is that heating of the surrounding ground during the day gives rise to turbulence in the air near the ground, making the observing conditions worse (think of the heat shimmer just above the surface of hot pavement or a hot road -- that's turbulence bad enough for your naked eyes to notice). Since water has a high thermal inertia, it doesn't heat up as much as land does, so you get less turbulence. (Another approach is to put the telescope on top of a tower, above the worst of the turbulence.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that Big Bear is at an altitude of 2,000 meters, so it &lt;em&gt;is&lt;/em&gt; at a moderately high altitude.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2015-06-05T13:00:59.463" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10924" PostTypeId="2" ParentId="10758" CreationDate="2015-06-05T13:45:09.997" Score="1" Body="&lt;p&gt;It may make more sense to convert the exponential into the product of the individual exponential terms:&#xA;$$&#xA;\exp(-\frac{R}{L}) \exp(-\frac{Z}{H}) \exp(\frac{Z_{\odot}}{H})&#xA;$$&#xA;This makes it clearer that the $Z_{\odot}$ term is a constant, which is just part of the overall normalization (similar to the $\exp(R_{\odot}/L)$ term earlier in the original equation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, yes, $Z$ is assumed to be always positive, so $|Z|$ would be slightly more correct.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2015-06-05T16:24:46.783" LastActivityDate="2015-06-05T16:24:46.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10925" PostTypeId="1" AcceptedAnswerId="10928" CreationDate="2015-06-05T17:01:54.767" Score="5" ViewCount="180" Body="&lt;p&gt;Part of &lt;a href=&quot;https://space.stackexchange.com/a/9367/10347&quot;&gt;this answer&lt;/a&gt; (discussing the moving center of mass of our solar system) explains that this movement is one way we know a star has planets:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bonus: We use this phenomenon to find planets outside the Solar System! If a distant star is observed to 'wobble' or oscillate about it's mean position, we can use that data to infer the presence of one or more exoplanets, and calculate their mass.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It makes sense to me that we could tell that a start has &lt;em&gt;some&lt;/em&gt; planet(s) in orbit around it, based on how they affect the star's position. But is this movement really precise and predictable enough to pin down how many exoplanets, and how large they are? Can we produce a &quot;map&quot; of a distant solar system based on this movement? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If needed, we can even restrict the discussion to planets that are &quot;big enough&quot; to detect. For example, when performing this exercise on our Sun, would we be able to accurately predict at least the four gas giants, even if we couldn't get Mercury?&lt;/p&gt;&#xA;" OwnerUserId="7413" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2015-06-06T10:18:04.690" Title="How can we tell how many exoplanets a star has?" Tags="&lt;exoplanet&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10926" PostTypeId="2" ParentId="10918" CreationDate="2015-06-05T17:19:45.070" Score="2" Body="&lt;p&gt;Just because there's an extra dimension on a diagram doesn't mean that it's real. No such thing has been created. It's just an artifact of embedding a manifold with a non-Euclidean geometry into a Euclidean space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the biggest stumbling blocks of intuition for people learning general relativity is that the physics only cares about the &lt;em&gt;intrinsic&lt;/em&gt; geometry. For example, imagine an ordinary ball in three (Euclidean) dimensions, and take its surface. In jargon, the surface is the two-dimensional sphere $\mathrm{S}^2$, and it has geometric properties that can be described by reference to it alone, e.g., lengths of curves drawn on it, angles between intersecting curves, that if one starts at some point and goes in a single direction, eventually one would be back to the starting point, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One should consider the two-sphere as a valid geometry by itself, and the fact that it can be pictured as the surface of a three-dimensional Euclidean object as purely incidental--it $\mathrm{S}^2$ can be embedded in the Euclidean space $\mathrm{E}^3$ in that simple manner, yes, but that's a fairly arbitrary choice. It can be embedded in other spaces as well, or not embedded in anything at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Riemannian manifolds (the purely &lt;em&gt;spatial&lt;/em&gt; geometries), there are some &lt;a href=&quot;https://en.wikipedia.org/wiki/Whitney_embedding_theorem&quot; rel=&quot;nofollow&quot;&gt;nice&lt;/a&gt; mathematical &lt;a href=&quot;https://en.wikipedia.org/wiki/Nash_embedding_theorem&quot; rel=&quot;nofollow&quot;&gt;results&lt;/a&gt; regarding embedding into Euclidean spaces, but nevertheless, they're usually impractical and not relevant to general relativity. Even worse, &lt;em&gt;general&lt;/em&gt; Lorentzian manifolds (&lt;em&gt;spacetimes&lt;/em&gt;) have no comparably 'nice' results regarding embedding spacetime geometries into flat pseudo-Euclidean spaces $\mathrm{E}^{n,m}$, so it's doubly useless to worry about it, except perhaps in the cases where the spacetime is extraordinarily simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, such extra dimension(s) are just artifacts of making a picture. They're not physically real, so they're not 'created' in any physical sense.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2015-06-05T17:19:45.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10927" PostTypeId="1" CreationDate="2015-06-05T20:15:42.490" Score="3" ViewCount="218" Body="&lt;p&gt;Assume two similarly sized bodies tide-locked to one another orbiting a barycenter between the two. That barycenter orbits a star. Since the two are tide-locked, their sidereal rotational period and sidereal orbital period (about one another) are the same (right?). There's no axial tilts or eccentricities to screw with things either for the sake of simplicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would I base the length of a conventional day on for a society living on these worlds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My gut's telling me to calculate the sidereal period based of the barycentral semi-major axes (or would I use the separation between the two worlds still?) for each and then compute the synodic periods and just use those, but my brain's too stupid to tell whether that's going to actually be anything close to a &quot;solar day&quot;. I don't need perfect - just close enough.&lt;/p&gt;&#xA;" OwnerUserId="7416" LastActivityDate="2015-07-06T15:58:05.833" Title="Day Length on Double Planets" Tags="&lt;orbit&gt;&lt;time&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10928" PostTypeId="2" ParentId="10925" CreationDate="2015-06-05T20:50:50.033" Score="6" Body="&lt;p&gt;What you are describing is a basic signal processing problem. The doppler shift that one observes is due to the motion of the star in the system around the system's centre of mass. The star will be influenced by the gravitational pull of each of the planets in that system, each of which exerts a gravitational pull that increases with the mass of the planet and decreases with the orbital radius of the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall motion of the star will be the &lt;em&gt;sum&lt;/em&gt; of the effects of all the planets. Importantly, the effect of each planet will have its own amplitude and will be periodic with a period equal to the orbital period of that planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's imagine that each planet is in a circular orbit (elliptical orbits are more complicated, but the principle is the same). Each planet would cause a circular motion in the star about the centre of mass of the planet-star pair, leading to an observable doppler signal which has the form of a sine wave with a period equal to the orbital period of the planet. The &lt;em&gt;amplitude&lt;/em&gt; of that signal will increase with the mass of the planet and increases with decreasing orbital radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall signal is the sum over all the planets in the system. Fortunately the decomposition of this signal back into its individual components is a well-trodden problem in physics, electronics and many other fields and is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_analysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fourier Analysis&lt;/a&gt;. Whether you can successfully recover the original signals from each of the planets depends on how long you observed the system (ideally you want to observe for longer than the longest orbital period) and the amplitude of the signals compared with the noise in your observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general it is easier to recover high-mass planets with short orbital periods and more difficult to recover low-mass planets with long orbital periods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image below might be helpful. It shows the track of the solar system centre of mass compared with centre of the Sun over a period of several decades. Notice how the Sun executes a complex trajectory (with respect to the solar system centre of mass) that is &lt;em&gt;mainly&lt;/em&gt; caused by the orbit of Jupiter, but then there are smaller, superimposed, signals caused by the smaller planets. In principle, if you observed for longer than the period of Neptune and had a detector which gave perfect measurements, you could reconstruct how many planets there were in the solar system, what their orbital periods were (and then from Kepler's 3rd law, what the planet-star sepration was) and what their masses were (multiplied by the inclination of their orbits with respect to the line of sight of observation, which is generally an unknown in doppler measurements). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XHonq.jpg&quot; alt=&quot;Motion of the Sun relative to the solar system barycentre&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of what we could currently see if we observed the Sun as a star: basically we would (assuming we observed for 20 years) detect Jupiter quite easily with a doppler amplitude of about 13 m/s. We would also see that there was a drift in Jupiter's signal due to the influence of Saturn, but we would have to observe for &gt;Saturn's orbital period in order to confirm the presence of Saturn, its orbital period and mass. The inner planets produce an amplitude that is too small to be visible using the technology currently available. e.g. The Earth would produce a doppler wobble of amplitude $&amp;lt;8$ cm/s, but the current precision of doppler measurements is limited to about 50 cm/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The doppler amplitudes in m/s due to each of the planets (assuming we view them edge on) are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury &amp;lt;0.01&#xA;Venus    0.08&#xA;Earth    0.08&#xA;Mars    &amp;lt;0.01&#xA;Jupiter  12.5&#xA;Saturn   2.6&#xA;Uranus   0.28&#xA;Neptune  0.26&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus with current technology, only Jupiter and Saturn are detectable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below I simulate what the doppler signal due to these two planets would be. I hope you can &lt;em&gt;see&lt;/em&gt; that the overall signal consists of the superposition of two sinusoidal signals with different periods and amplitudes. Myriad computational tools are available to do the fourier decomposition to establish these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uzsRg.png&quot; alt=&quot;Combined doppler signal due to Jupiter and Saturn&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-06-06T10:18:04.690" LastActivityDate="2015-06-06T10:18:04.690" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10929" PostTypeId="1" CreationDate="2015-06-05T22:25:20.767" Score="3" ViewCount="63" Body="&lt;p&gt;What kind of impact would you expect ram pressure stripping / tidal interactions / harassment / interactions with the cluster potential (etc!) to have on the h-alpha velocity fields of infalling spiral galaxies? &#xA;What I mean is, how easy is it to detect such processes at work, and how are you able to distinguish between these different mechanisms (using also h-alpha intensity maps, rotation curves, optical morphologies if available).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers!&lt;/p&gt;&#xA;" OwnerUserId="7418" LastEditorUserId="1890" LastEditDate="2016-06-29T20:01:57.177" LastActivityDate="2016-06-29T20:01:57.177" Title="H-alpha velocity fields of spirals falling into a cluster" Tags="&lt;galaxy&gt;&lt;spectroscopy&gt;&lt;galaxy-cluster&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10930" PostTypeId="2" ParentId="10927" CreationDate="2015-06-06T00:52:01.990" Score="1" Body="&lt;p&gt;Yes, if the planets are tide locked to one another, their orbital period and sidereal day are the same. That's assuming their orbital plane about each other isn't much inclined to their orbital plane about the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_period#Two_bodies_orbiting_each_other&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article on orbital period&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_d=2\pi*\sqrt{a_p^3/(G*(M_1+M_2))}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where G is gravitational constant, $M_1$ is mass of 1st body and $M_2$ mass of 2nd body. $a_p$ is semi major axis of elliptical orbit of the planets about each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the planet's year you'd use a very similar equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_y=2\pi*\sqrt{a_s^3/(G*(M_0+M_1+M_2))}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $a_s$ is the radius of their orbit about the sun and $M_0$ is the mass of the sun. If you like, you can leave off $M_1$ and $M_2$ as their masses are probably neglible in comparison to the sun's mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am guessing by &quot;conventional day&quot; you want the time an inhabitant sitting on his front porch would measure between one high noon and the next.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solar day = sidereal day * (1 + (sidereal day/year))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example our solar day is about 1/365 longer than a sidereal day.&lt;/p&gt;&#xA;" OwnerUserId="1232" LastEditorUserId="1232" LastEditDate="2015-06-06T15:54:26.760" LastActivityDate="2015-06-06T15:54:26.760" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="10931" PostTypeId="1" CreationDate="2015-06-06T06:21:29.670" Score="4" ViewCount="1311" Body="&lt;p&gt;I have two series of velocities (HJD, $v_{1}$ and $v_{2}$ ), how to fit in detail?&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is any formula on the web? $sine$ curve is easy, but I need to consider an eccentric orbit. &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we can estimate $m_{1}$, $m_{2}$, and inclination angle $\approx80^{\circ}$, if we just know velocities of both of the two companions, which parameters can me derive?&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the difference between binary velocity fitting and star+planet fitting?&lt;br/&gt;&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditDate="2015-06-06T08:34:50.670" LastActivityDate="2015-06-10T21:56:13.877" Title="radial velocity fitting of a binary" Tags="&lt;velocity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10932" PostTypeId="1" AcceptedAnswerId="10935" CreationDate="2015-06-06T09:33:04.047" Score="3" ViewCount="184" Body="&lt;p&gt;I am new to this community and not quite sure if this is an appropriate question to be asked here. Anyways its just my curiosity. As far as I could know about &lt;a href=&quot;http://www.nasa.gov/chandra/multimedia/wasp18-2014.html&quot; rel=&quot;nofollow&quot;&gt;WASP-18b&lt;/a&gt; is that, it orbits very close to its sun and takes less than 24 hours to complete one orbit. However this was not the case always, as the planet actually was formed in a cold part of the space and later got tug into of its sun's gravitational pull. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this is the case, I am curious to know how it was formed in the first place? If I am not wrong planets are formed when the debris orbiting in the gravitational pull of a star stick together overtime gets in to the form of the planetary bodies. So, if WASP-18b was not always so close to its sun and it was formed far away in cold space, then could there be a possibility that, this exoplanet was formed in an orbit of another star and overtime it got away from its original orbit and trapped in the tug of its current sun? &lt;em&gt;(Its my imagination! just bare with it :))&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also could any one point me to good documentations from where I can learn more. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7419" LastEditorUserId="7419" LastEditDate="2015-06-06T10:05:26.283" LastActivityDate="2015-06-09T06:10:59.243" Title="How did WASP-18b exoplanet form?" Tags="&lt;exoplanet&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10933" PostTypeId="2" ParentId="10921" CreationDate="2015-06-06T10:00:56.610" Score="6" Body="&lt;p&gt;There's a lot to pick apart in everything you try to propose, as it includes a lot of far fetched (or at least rather non-standard) claims.  I am frankly not up to attempting to address every one of them, if for no other reason than that it makes the question as a whole rather too broad for my taste (and perhaps more in the territory of Physics.SE, which has quite a lot of answered questions concerning black holes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is, however, the following simple and amusing observation: current estimates of the mass-energy of the observable universe tell us that it is &lt;em&gt;too&lt;/em&gt; dense to be a black hole.  That might sound a little weird if you're not familiar with black holes.  In fact, the density of a (non-rotating, Schwarzchild) blackhole is inversely proportional to the square of its mass, and the radius is directly proportional to the mass.  More explicitly:&#xA;$$r=\frac{2 G M}{c^2},$$&#xA;$$\rho(M) = \frac{3 c^6}{32 \pi G^3}\cdot \frac{1}{M^2},$$&#xA;where $r$ is the radius, $\rho$ is the density, $M$ is the mass, $c$ is the speed of light, and $G$ is the gravity constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The order of magnitude estimate for $M$ is $10^{54}$ (and $M\geq 10^{54}$ in particular), which makes the universe ~3 times too small at least.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A key fact here is that the universe is not static with respect to itself. See &lt;a href=&quot;https://physics.stackexchange.com/questions/1901/if-the-observable-universe-were-compressed-into-a-super-massive-black-hole-how/1904#1904&quot;&gt;this Physics.SE Q&amp;amp;A&lt;/a&gt; in particular.  I'll quote the end of Lubos Motl's answer, in particular:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;Our Universe, dominated by the dark energy, is already rather close to an empty de Sitter space which is, from many viewpoints, analogous to a black hole except that the interior of the visible part of the de Sitter space is analogous to the exterior of a normal black hole, and the analogy of the interior of a black hole is everything that is behind the cosmic horizon - where we don't see. It is misleading to create the analogy with the static black holes directly because our Universe is not static in the normal cosmological coordinates.&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, there are lots of important and sometimes subtle issues with the whole &quot;the universe is a black hole&quot; concept.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-06-06T10:00:56.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10934" PostTypeId="1" CreationDate="2015-06-06T11:07:24.253" Score="4" ViewCount="120" Body="&lt;p&gt;67P/Churyumov–Gerasimenko has a highly irregular dumbbell shape. But the sample of comet shapes observed is very small, so I wonder if irregular shape is the norm for comets and for small moons. Many known moons are no larger than this ~4 km diameter comet. The smallest moons imaged are modestly irregular, basically just elongated, at least they don't have a waist like 67/P. The most elongated moon is perhaps ~135/60 km Prometheus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there reason to believe that moons which are too small for hydrostatic equilibrium get more rounded than comets of similar mass and composition? Moons differ from comets in several ways. AFAIK: Moons in general have a very different gravitational environment, a much more stable distance to the Sun, experience more frequent impacts, another formation history if not captured. Composition and density depends on the formation distance from the Sun, although even ~300 km Hyperion has a similar density as 67/P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should we expect the small moons in general, and the moons of Pluto especially, to be rounded or dumbbell shaped? If 67P is a merged binary, isn't it more likely for two objects to merge if they are moons instead of comets, since neighboring moons have low relative speed? ~100-200 km Janus and Epimetheus look as if they could merge fairly calmly. Does the lack of (observed) dumbbell shaped moons tell us something about 67P, that its shape is a rare outlier for example?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-07-12T13:37:01.343" Title="Is there a mechanism that makes small moons more rounded than comets?" Tags="&lt;comets&gt;&lt;formation&gt;&lt;natural-satellites&gt;&lt;hydrostatic-equilibrium&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10935" PostTypeId="2" ParentId="10932" CreationDate="2015-06-06T11:22:23.090" Score="3" Body="&lt;p&gt;@LocalFluff's comment is more-or-less complete. The process of &quot;swapping&quot; planets between stars &lt;em&gt;could&lt;/em&gt; happen during the early lives of stars when most of them are born in relatively dense clusters. You can find some more details and references in my answer to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6127/potential-re-capture-of-rogue-planets/10868#10868&quot;&gt;this question&lt;/a&gt;, but the outcome is that a captured planet would usually be in a very &lt;em&gt;wide&lt;/em&gt; orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Approximately 1% of stars have a &quot;hot Jupiter&quot;, but even in dense cluster environments it is thought that only a few percent of stars can capture a planet, and because almost all of these end up in wide orbits there are just too many hot Jupiters to be explained in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two classes of theory to explain hot Jupiters. One is the dyanmical explanation offered by LocalFluff. This is itself divided into several different mechanisms: &quot;Secular chaos&quot;, (e.g. &lt;a href=&quot;http://arxiv.org/abs/1012.3475&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wu &amp;amp; Lithwick 2011&lt;/a&gt; - this paper also briefly reviews some of the other mechanisms) which is a gradual change in the planetary system that then suddenly drives a transition to a new configuration. Planet-planet scattering where interaction between planets causes one orbit to become highly eccentric and then tidal circularisation results in a hot Jupiter in a close orbit (e.g.&lt;a href=&quot;http://adsabs.harvard.edu/abs/2008ApJ...686..621F&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ford &amp;amp; Rasio 2008&lt;/a&gt;). The &lt;a href=&quot;http://en.wikipedia.org/wiki/Kozai_mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kozai mechanism&lt;/a&gt; is the interaction beteen two planets that results in an interchange between the inclination and eccentricity of the inner planet. This excites an increase in eccentricity that, when combined with tidal circularisation, can  also result in a close-in planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other class of mechanism involve migration through interaction with a disk. This must necessarily happen early in the star's life whilst surrounded by a disk of circumstellar material. Basically viscous drag in the disk itself causes the planet to move inwards (or sometimes outwards). There are three possible subtypes of disk migration and they are briefly defined &lt;a href=&quot;http://en.wikipedia.org/wiki/Planetary_migration&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is probable that all of these mechanisms play some role in forming hot Jupiters and one of the goals of exoplanetary study is to figure out which may be more important. Typical parameters of interest are the distribution of planetary orbits and eccentricities and whether there is evidence of previous interactions in the form of misalignments between the rotation axis of the star and the orbital axis of the planet.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-06-09T06:10:59.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10936" PostTypeId="1" CreationDate="2015-06-06T14:16:15.733" Score="2" ViewCount="528" Body="&lt;p&gt;We have Hubble's law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v=Hd$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where H is the Hubble parameter, which is decreasing in value, but it will be constant in the distant future. So, assuming the Standard Model of Cosmology holds true, will Hubble's law be accurate when the Hubble parameter becomes a constant?&lt;/p&gt;&#xA;" OwnerUserId="6920" LastEditorUserId="9301" LastEditDate="2016-10-17T15:01:41.077" LastActivityDate="2016-10-17T20:58:31.850" Title="Will Hubble's law always be accurate?" Tags="&lt;cosmology&gt;&lt;distances&gt;&lt;expansion&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10937" PostTypeId="1" AcceptedAnswerId="10942" CreationDate="2015-06-06T17:17:16.407" Score="12" ViewCount="309" Body="&lt;p&gt;The asteroid 4864 Nimoy was &lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2015/06/06/asteroid_4864_nimoy_named_for_leonard_nimoy.html&quot;&gt;recently named after Leonard Nimoy&lt;/a&gt;. It was discovered on September 2, 1988 so it went nearly 27 years without a name. Why was this asteroid chosen to be named after him? Were they trying to think of a name for it and chose to name it for Leonard Nimoy, or were they looking for an asteroid to name after him and chose that one?&lt;/p&gt;&#xA;" OwnerUserId="1980" LastActivityDate="2015-06-08T12:24:56.047" Title="Why was this asteroid (4864 Nimoy) chosen to be named after Leonard Nimoy?" Tags="&lt;asteroids&gt;&lt;naming&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10939" PostTypeId="1" CreationDate="2015-06-07T21:43:54.330" Score="5" ViewCount="78" Body="&lt;p&gt;Mars' thin atmosphere and lack of magnetosphere exposes its surface to lots of Solar UV radiation and cosmic radiation. But Mars also has some dramatic altitude gradients in for example Valles Marinaris. Are there any areas which in this way are protected from A) Solar UV radiation and flares and coronal mass ejections, and/or B) substantially protected from cosmic radiation?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-06-07T21:50:58.753" LastActivityDate="2015-06-11T07:18:05.410" Title="Are there naturally radiation shielded areas on Mars? (shadowed by terrain formations)" Tags="&lt;mars&gt;&lt;radiation&gt;&lt;topographic-map&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10942" PostTypeId="2" ParentId="10937" CreationDate="2015-06-08T12:24:56.047" Score="2" Body="&lt;p&gt;I'm not entirely sure, but I would guess that it's simply because the discoverer suggested it. After a few steps in the process of characterizing an asteroid's orbit, the discoverer gets to suggest a name (within reason) to a committee.  The full guidelines are available at the relevant &lt;a href=&quot;https://www.iau.org/public/themes/naming/#minorplanets&quot; rel=&quot;nofollow&quot;&gt;IAU webpage&lt;/a&gt;.  (It's longer than is worth reproducing here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there are plenty of &quot;celebrity&quot; asteroids, drawn from a wide range of human endeavours: Adamsmith, Balzac, Beethoven, Megryan, etc. If anything, I'm surprised that an asteroid wasn't named Nimoy sooner. There's already a Mr. Spock!&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2015-06-08T12:24:56.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10946" PostTypeId="1" CreationDate="2015-06-08T17:19:24.517" Score="11" ViewCount="19458" Body="&lt;p&gt;If the suns gravitational pull is strong enough to hold much larger masses in place (all the planets) and at much greater distances (all planets further away from the sun then earth) why does it not pull the moon away from earth?&lt;/p&gt;&#xA;" OwnerUserId="6901" LastEditorUserId="2618" LastEditDate="2017-12-30T17:21:33.473" LastActivityDate="2021-03-13T23:29:21.720" Title="Why doesn't the sun pull the moon away from earth?" Tags="&lt;the-sun&gt;&lt;orbit&gt;&lt;the-moon&gt;&lt;gravity&gt;&lt;earth&gt;" AnswerCount="6" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10947" PostTypeId="2" ParentId="10946" CreationDate="2015-06-08T19:31:58.407" Score="4" Body="&lt;p&gt;The Moon is in orbit about the Sun, much as the Earth is. Although this is not the usual perspective from the Earth, a plot of the Moon's trajectory shows the Moon in an elliptic orbit about the Sun.&#xA;Essentially the Earth, Moon, Sun system is (meta) stable, like that of other planets orbiting the Sun.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2015-06-08T19:31:58.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10948" PostTypeId="1" CreationDate="2015-06-09T09:09:10.930" Score="5" ViewCount="753" Body="&lt;p&gt;Umm al-Qura calendar of Saudi Arabia determine the new moon if &lt;a href=&quot;http://www.staff.science.uu.nl/~gent0113/islam/ummalqura.htm&quot;&gt;these&lt;/a&gt; criteria meet at 29th day:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The geocentric conjunction occurs before sunset.&lt;/li&gt;&#xA;&lt;li&gt;The moon sets after the sun.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I've got the formula to calculate when the sun (seen by the observer) sets, considering observer's latitude, longitude, height (altitude from sea level), and the date (Julian Day). I would like to ask the formula to calculate:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;When (the time) geocentric conjunction happens.&lt;/li&gt;&#xA;&lt;li&gt;How much (the angle) the altitude of the moon (seen by the observer) at a given time is.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="7432" LastEditorUserId="7432" LastEditDate="2015-06-09T09:15:06.910" LastActivityDate="2016-09-16T20:34:01.480" Title="How to calculate geocentric conjunction time and moon altitude at a given time" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;the-sun&gt;&lt;time&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10949" PostTypeId="2" ParentId="10946" CreationDate="2015-06-09T09:20:07.690" Score="1" Body="&lt;p&gt;I agree with Adrian's answer.   If you look at the moons orbit, in a very real sense it orbits the sun maybe more than it orbits the earth.  The Earth/Moon system orbits the sun at 30 KM/s, the Moon orbits the earth at about 1 KM per second.    Both orbits are reasonably eliptical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The entire solarsystem orbits around the center of the Milky-way, so orbiting more than one center of mass isn't unusual.   Orbits can exist within other orbits, within limits.   The orbital limit is sometimes referred to as the Sphere of Influence   &lt;a href=&quot;http://en.wikipedia.org/wiki/Sphere_of_influence_%28astrodynamics%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sphere_of_influence_%28astrodynamics%29&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the moon was a bit more than twice as far as it is from the Earth as it is now, the Earth might lose it.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-06-09T09:20:07.690" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10950" PostTypeId="2" ParentId="10946" CreationDate="2015-06-09T13:02:07.240" Score="20" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why doesn't the sun pull the moon away from earth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Short answer: Because the Moon is much closer to the Earth than it is to the Sun. This means the gravitational acceleration of the Earth toward the Sun is almost the same as is the gravitational acceleration of the Moon toward the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon's acceleration toward the Sun, $-GM_\odot\frac{\boldsymbol R+\boldsymbol r}{||\boldsymbol R+\boldsymbol r||^3}$ is indeed about twice that of the Moon toward the Earth, $-GM_\oplus\frac{\boldsymbol r}{||\boldsymbol r||^3}$. This is irrelevant. What is relevant is the Moon's earthward acceleration due to gravitation compared to the difference between the Moon's and Earth's sunward gravitational acceleration,&#xA;$$\boldsymbol a_{\odot,\text{rel}} = -GM_{\odot}\left(\frac{\boldsymbol R + \boldsymbol r}{||\boldsymbol R + \boldsymbol r||^3} - \frac{\boldsymbol R}{||\boldsymbol R||^3}\right)$$&#xA;This relative acceleration toward the Sun is a small perturbation (less than 1/87&lt;sup&gt;th&lt;/sup&gt; in magnitude) on the Moon's gravitational acceleration toward the Earth. Given the current circumstances, the Sun can't pull the Moon away from the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longer answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational force exerted by the Sun on the Moon is more twice that exerted by the Earth on the Moon. So why do we say the Moon orbits the Earth? This has two answers. One is that &quot;orbit&quot; is not a mutually exclusive term. Just because Moon orbits the Earth (and it does) does not mean that it doesn't also orbit the Sun (or the Milky Way, for that matter). It does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other answer is that gravitational force as-is is not a good metric. The gravitational force from the Sun and Earth are equal at a distance of about 260000 km from the Earth. The short-term and long-term behaviors of an object orbiting the Earth at 270000 km are essentially the same as those of an object orbiting the Earth at 250000 km. That 260000 km where the gravitational forces from the Sun and Earth are equal in magnitude is effectively meaningless. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better metric is the distance at which an orbit remain stable for a long, long, long time. In the two body problem, orbits at any distance are stable so long as the total mechanical energy is negative. This is no longer the case in the multi-body problem. The Hill sphere is a somewhat reasonable metric in the three body problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hill sphere is an approximation of a much more complex shape, and this complex shape doesn't capture long-term dynamics. An object that is orbiting circularly at (for example) 2/3 of the Hill sphere radius won't remain in a circular orbit for long. Its orbit will instead become rather convoluted, sometimes dipping as close to 1/3 of the Hill sphere radius from the planet, other times moving slightly outside the Hill sphere. The object escapes the gravitational clutches of the planet if one of those excursions beyond the Hill sphere occurs near the L1 or L2 Lagrange point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the N-body problem (for example, the Sun plus the Earth plus Venus, Jupiter, and all of the other planets), the Hill sphere remains a reasonably good metric, but it needs to be scaled down a bit. For an object in a prograde orbit such as the Moon, the object's orbit remains stable for a very long period of time so long as the orbital radius is less than 1/2 (and maybe 1/3) of the Hill sphere radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon's orbit about the Earth is currently about 1/4 of the Earth's Hill sphere radius. That's well within even the most conservative bound. The Moon has been orbiting the Earth for 4.5 billion years, and will continue to do so for a few more billions of years into the future.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2017-12-30T22:15:30.167" LastActivityDate="2017-12-30T22:15:30.167" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10951" PostTypeId="1" CreationDate="2015-06-10T06:25:06.823" Score="4" ViewCount="215" Body="&lt;p&gt;I am doing research on SSN-like parameters. As I can see, effective SSN is a parameter that is derived from SSN or other parameters. Do anyone know how to calculate effective SSN?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;ftp://ftp.swpc.noaa.gov/pub/latest/DSD.txt&quot; rel=&quot;nofollow&quot;&gt;this file&lt;/a&gt; has some parameters. See also &lt;a href=&quot;http://spawx.nwra.com/spawx/ssne.html&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7442" LastEditorUserId="2153" LastEditDate="2016-10-19T02:00:55.030" LastActivityDate="2016-10-19T15:51:56.137" Title="Calculating effective SSN (sunspot number)" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10952" PostTypeId="1" AcceptedAnswerId="10955" CreationDate="2015-06-10T08:34:00.970" Score="4" ViewCount="631" Body="&lt;p&gt;The Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Photometry_%28astronomy%29&quot; rel=&quot;nofollow&quot;&gt;photometry&lt;/a&gt; says that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;These have largely been replaced with CCD cameras that can simultaneously image multiple objects, although photoelectric photometers are still used in special situations, such as where fine time resolution is required.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How much the research community is using photometers today? I looked up the prices of some photometers from &lt;a href=&quot;http://telescopes.net/store/telescope-accessories/photometers.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. The device, filters, etc. can cost nearly $2500. One could get a good CCD camera with that price.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some applications where the good time resolution (apparently around 1 ms) would be needed? Are photometers equally useful with small telescopes (e.g. 8-inch) as CCDs?&lt;/p&gt;&#xA;" OwnerUserId="6991" LastEditorUserId="1288" LastEditDate="2015-06-10T11:34:55.887" LastActivityDate="2015-06-10T20:29:01.847" Title="Photometer vs. CCD-camera" Tags="&lt;telescope&gt;&lt;photography&gt;&lt;photometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10953" PostTypeId="1" AcceptedAnswerId="11008" CreationDate="2015-06-10T09:53:50.343" Score="3" ViewCount="107" Body="&lt;p&gt;Noctilucent clouds are at around 85km altitude and are formed of ice crystals. The ISS recently posted a number of photographs of &lt;a href=&quot;https://www.facebook.com/ISS/posts/998323033533961&quot; rel=&quot;nofollow noreferrer&quot;&gt;these clouds&lt;/a&gt; and pointed out the key feature, which is the bright blue glow. &lt;img src=&quot;https://i.stack.imgur.com/WwrWD.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these known to have an impact on ground-based or orbital telescopes?&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2015-06-20T02:13:47.750" Title="Do Noctilucent clouds cause problems with telescopes?" Tags="&lt;telescope&gt;&lt;light-pollution&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10954" PostTypeId="1" AcceptedAnswerId="10961" CreationDate="2015-06-10T11:36:48.640" Score="0" ViewCount="284" Body="&lt;p&gt;If I understand correctly, the date of the Big Bang is an extrapolation of acceleration of the universe's expansion through time based on the erroneous assumption that the universe is approximately 13 billion years old. The age of the universe, as far as I can tell is based on the speed of light and the fact that the furthest we could possibly see into the depths of the universe is approximately 13 billion light-years. Isn't that sort of a cat chasing his tail?  It seems that there is no valid reason for believing that the universe is really of a radius of ~13 billion years. That is just as far as we can see. If this is wrong and there is a valid reason for believing that the universe is ~ 13 billion years old, then that fact makes for a very interesting possibility.  That the expansion rate of the universe is equal to the speed of light, which leads to a &lt;strong&gt;&lt;em&gt;very&lt;/em&gt;&lt;/strong&gt; interesting suggestion that I pursued for quite awhile (until it became apparent that the suggested size of the universe is not what it seems), that the expansion rate of the universe is equal to the speed of light. Or, put another way, is it possible that the speed of light is based on the expansion rate of the universe?&lt;/p&gt;&#xA;" OwnerUserId="7443" LastEditorUserId="2153" LastEditDate="2015-06-10T17:42:26.330" LastActivityDate="2015-08-16T14:13:21.820" Title="Isn't the date of the Big Bang a bit bogus?" Tags="&lt;big-bang-theory&gt;&lt;expansion&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10955" PostTypeId="2" ParentId="10952" CreationDate="2015-06-10T13:32:25.413" Score="2" Body="&lt;p&gt;You have correctly identified a niche for photometers. Another point in their favour used to be that they were much more sensitive in the U-band than CCDs, but I think that newer CCDs can almost match or surpass the U-band response of photometers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CCDs take quite a time to readout. The faster you read them out, generally speaking, the higher the readout noise. By comparison, the output from photometers can be very fast indeed with little impact on the noise characteristics. One of the beauties of photometers is also that they give you an instant readout of how many photons have been detected, requiring little analysis to provide real-time monitoring of signals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, things move on. So now there &lt;em&gt;are&lt;/em&gt; CCD instruments that can work on extremely fast timescales. The primary example of this is a UK-built and operated instrument called &lt;a href=&quot;http://www.vikdhillon.staff.shef.ac.uk/ultracam/index.html&quot; rel=&quot;nofollow&quot;&gt;ULTRACAM&lt;/a&gt;, which allows data-taking at 100 Hz and which uses a special &quot;drift&quot; mode of readout to achieve this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to your main question - what are the scientific objectives of such instruments? On the link I provided above you will see a series of press releases that describe some of the science. These range from measuring the transits of Trans-Neptunian-Objects in front of background stars; examining detailed time series of flares on magnetic stars; investigating the eclipses caused by white dwarfs passing in front of companion stars; looking at the flickering output from accreting black holes; attempting to get phase resolved light curves of the optical emission from rapidly rotating pulsars and much more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rule of thumb is, that if you are able to obtain a time resolution $\Delta t$, then you are probing size scales of $\leq c \Delta t$, where $c$ is the speed of light. Thus if you can get a resolution of 0.01s, this corresponds to size scales of $\leq 3000$ km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to whether a photometer would be good on an 8-inch telescope, I'm not sure. When I was a student I used a photoelectric photometer on a small telescope to (i) study the scintillation of stars as a probe of conditions in the upper atmosphere - e.g. &lt;a href=&quot;http://adsabs.harvard.edu/full/1985AN....306..145S&quot; rel=&quot;nofollow&quot;&gt;Stecklum (1985)&lt;/a&gt; - which has timescales of order $0.01-1$ seconds. (ii) To monitor &quot;flare&quot; stars with a time resolution of about a second in an effort to get real-time information so that spectra could be obtained at various points during a flare.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-06-10T20:29:01.847" LastActivityDate="2015-06-10T20:29:01.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10957" PostTypeId="1" CreationDate="2015-06-10T14:21:00.490" Score="7" ViewCount="22900" Body="&lt;p&gt;I've been having trouble getting really sharp focus on my telescope. Using Saturn as a reference, with a 20mm eye piece it is very sharp. But when I go to 10mm or 6mm, it gets bigger and blurrier. I re-adjust focus each time I change eye pieces, but the best I can get at the stronger magnifications is still very blurry compared to 20mm. The 20mm will have nice sharp edges on the planet and rings, whereas 10mm and 6mm have fuzzy edges on the rings, and the planet itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using the Celestron NexStar Evolution 8 Schmidt-Cassegrain Telescope. I'm in a suburban area so I'm sure there is light pollution, could that be a factor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last night I was observing for about an hour, which I believe would be enough time for the temperature of the telescope to match the temperature outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also tried combining the 20mm with a Celestron 2x Barlow lens, just with an idea that maybe it would make a difference, but it was pretty much the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there something I'm doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="7361" LastEditorUserId="1378" LastEditDate="2015-06-11T19:19:06.233" LastActivityDate="2021-04-30T17:37:26.023" Title="I'm having trouble achieving sharp telescope focus" Tags="&lt;telescope&gt;&lt;collimation&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10958" PostTypeId="1" AcceptedAnswerId="10959" CreationDate="2015-06-10T15:29:07.890" Score="3" ViewCount="1207" Body="&lt;p&gt;Suppose a telescope has aperture $D = 20$ cm. The angular resolution of such telescope, according to the Rayleigh criterion (if I have understood it correctly), is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta = 1.22\cdot\frac{\lambda}{D} \text{ rad}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\lambda$ is the wavelength of light and $D$ is the aperture. Suppose $\lambda = 550$ nm (mid-point of visible light). Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta = 1.22\cdot\frac{\lambda}{D} \text{ rad}&#xA;= 1.22\cdot\frac{550 \text{ nm}}{20\text{ cm}} \text{ rad}&#xA;= 1.22\cdot\frac{5.5\cdot10^{-5}\text{ cm}}{20\text{ cm}}\cdot 206264.8''&#xA;\approx 0.69''&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the angular resolution of the telescope is $0.69$ seconds of arc for $\lambda=550$ nm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, is it really so that the eye-piece of the telescope has no play in this at all? &lt;a href=&quot;http://en.wikipedia.org/wiki/Angular_resolution&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; says that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and D is the diameter of the lens' aperture.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which I would understand to be taken as the size of the primary mirror alone in a Newtonian telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if there is for example a double (binary) star, I will see them as one no matter what kind of magnification/eye-piece I am using if the angle between the stars (from my point of view) is less than $0.69''$? And vice versa, if the angle between the stars is greater than $0.69''$ I will see them as separate stars with &lt;em&gt;any&lt;/em&gt; eye-piece or magnification?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In theory, I know there's a lot of limitations from optics, seeing, etc.)&lt;/p&gt;&#xA;" OwnerUserId="6991" LastEditorUserId="6991" LastEditDate="2015-06-10T16:07:42.857" LastActivityDate="2015-06-10T18:03:57.710" Title="Is the angular resolution of a telescope irrespective of used eye-piece?" Tags="&lt;telescope&gt;&lt;binary-star&gt;&lt;newtonian-telescope&gt;&lt;angular-resolution&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10959" PostTypeId="2" ParentId="10958" CreationDate="2015-06-10T16:40:57.273" Score="3" Body="&lt;p&gt;The resolution of a telescope is the resolution of the image created by the primary mirror at the focal plane.  It provides the minimum separation between two equal brightness stars that appear separate in that image.  Often, astronomers put a photographic plate or a CCD at the focal plane, create long exposure photos and these can then be examined at arbitrary resolution (with a magnifying glass, microscope, or zoom software on a screen).  But, no matter how much you magnify the image, stars separated by less than the resolution will appear merged.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that if you use, say, a wide-angle lens when observing directly with a telescope the resolution that you get can be limited by the resolution of your eye.  What is happening is the telescope is forming an image on your retina that has the full resolution, but your eye cannot make use of the finest details.  On the other hand, a hawk looking through the same eyepiece may get full resolution benefits since hawks' eyes have better resolution than humans'. &lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2015-06-10T16:40:57.273" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10960" PostTypeId="2" ParentId="10769" CreationDate="2015-06-10T17:02:06.227" Score="2" Body="&lt;p&gt;It is not unusual to have pairs of pairs or even more complex stellar systems.  Two stars make a stable pair.  A pair of these make a stable system and a pair of these would also make a stable system.  It is, however, necessary for stability for each pair to be separated by roughly 10 or more times the separation of the previous level.  This could continue ad infinitum were it not for the tidal field of neighboring systems or the Galactic potential limiting the overall size. &lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2015-06-15T19:18:07.093" LastActivityDate="2015-06-15T19:18:07.093" CommentCount="2" CommunityOwnedDate="2015-06-10T17:12:07.410" ContentLicense="CC BY-SA 3.0" />
  <row Id="10961" PostTypeId="2" ParentId="10954" CreationDate="2015-06-10T17:51:56.040" Score="3" Body="&lt;p&gt;I'd like to correct a couple misonceptions that might be confusing you.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;If I understand correctly, the date of the Big Bang is an extrapolation of acceleration of the universe's expansion through time based on the erroneous assumption that the universe is approximately 13 billion years old.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a false assumption. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Observable_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;observable universe&lt;/a&gt; is much, much larger than a sphere with a radius of 13 billion light-years. The radius is actually ~46 billion light-years (see, for example, &lt;a href=&quot;http://iopscience.iop.org/0004-637X/624/2/463/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gott et al. (2005)&lt;/a&gt;).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;The age of the universe, as far as I can tell is based on the speed of light and the fact that the furthest we could possibly see into the depths of the universe is approximately 13 billion light-years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Look at &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10677/calculating-the-age-of-the-universe/10678#10678&quot;&gt;pela's answer here&lt;/a&gt;, if you don't want to go off Stack Exchange (or see &lt;a href=&quot;http://en.wikipedia.org/wiki/Age_of_the_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; for another starting point). See how the age of the universe is calculated. As an approximation,$$t_\mathrm{Uni} \sim \frac{1}{H_0} = 14\,\mathrm{billion\,years}.$$This only relies upon observations of the movements of stars.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;That the expansion rate of the universe is equal to the speed of light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The universe is not expanding at the same rate at all distances from us. Objects further from us recede at greater speeds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-08-16T14:13:21.820" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10962" PostTypeId="2" ParentId="10958" CreationDate="2015-06-10T18:03:57.710" Score="2" Body="&lt;p&gt;The Raleigh criterion is the maximum theoretical limit that ignores the architecture, quality, and state of maintenance of optics. It basically says &quot;assuming the optics in this instrument are PERFECT, this is the resolution you could get out of it&quot;. It's a calculation that looks only at the diameter and ignores everything else. In other words, no matter how good the instrument, you cannot beat Raleigh - but you could make things worse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, of course things are worse. Take a parabolic mirror, as used in many telescope architectures, such as the newtonian. All parabolic mirrors generate perfect images only in the center of the field of view. Anything off-center is subject to coma, an aberration that even &quot;perfect&quot; paraboloids will exhibit. So the real resolving power gets worse as you move towards the edge of the field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On top of that, you have to add real-world manufacturing imperfections that any mirror will have. Also add distortions caused by temperature differences, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these contribute to distort the image formed in the focal plane of the primary mirror. The eyepiece's role is to examine and magnify that image, for you to see. That's how telescopes work - primary optics form an image in the primary focal plane, which is then examined with the eyepiece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the quality of the eyepiece will contribute to further degradation of what you actually see. Even with &quot;magic&quot; primary optics, if you had a &quot;perfect&quot; image in prime focus, a bad eyepiece will blur it. In real life, most eyepieces are at least half-decent in the center of the field (some are not), but quality degrades towards the edge. High quality eyepieces will not introduce visible degradation in the center and over most of the field. Top of the line eyepieces will not degrade the image visibly anywhere.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2015-06-10T18:03:57.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10963" PostTypeId="2" ParentId="10931" CreationDate="2015-06-10T21:56:13.877" Score="3" Body="&lt;p&gt;The radial velocity curve of a star in a binary system (with another star or a planet) is defined through 6 free parameters&#xA;$$V_r(t) = K\left(\cos(\omega + \nu) +e \cos \omega \right) + \gamma,$$&#xA;where $K$ is the semi-amplitude, $\gamma$ is the centre of mass radial velocity, $\omega$ is the usual angle defining the argument of the pericentre measured from the ascending node and $\nu$ is the true anomlay, which is a function of time, the fiducial time of pericentre passage $\tau$, the orbital period $p$ and the eccentricity $e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To proceed you estimate what all these parameters are - i.e. an initial guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, for each time $t_i$ of a data point in your RV curve you:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Calculate the mean anomaly&#xA;$$M(t) = \frac{2\pi}{p}(t - \tau),$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Solve &quot;Kepler's equation&quot;&#xA;$$M(t) = E(t) - e \sin E(t)$$&#xA;numerically (its a transcendental equation, you could use Newton-Raphson or similar) to give $M(t_i)$, the eccentric anomaly.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use&#xA;$$\tan \frac{E(t)}{2} = \left(\frac{1+e}{1-e}\right)^{-1/2} \tan \frac{\nu(t)}{2}$$&#xA;to calculate the true anomaly $\nu(t_i)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Calculate $V_r(t_i)$ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You then calculate some figure of merit (e.g. chi-squared) for how closely the model and data agree and go through an iterative process to adjust the parameters and optimise the fit of model to data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more sophisticated discussion can be found in this paper by &lt;a href=&quot;http://www.gehirn-und-geist.de/sixcms/media.php/370/Leseprobe.406372.pdf&quot; rel=&quot;nofollow&quot;&gt;Beauge et al.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have the RV curves of both stars, then you can fit them both simultaneously. Obviously, they have $p$, $e$, $\gamma$ and $\omega$ in common, but their RV amplitudes $K_1$ and $K_2$ will be different. The ratio of $K_1/K_2$ gives you the ratio of the two stellar masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only have one RV curve you are limited to estimating the &lt;a href=&quot;http://cmi2.yale.edu/bh/week4/pages/massfunction.html&quot; rel=&quot;nofollow&quot;&gt;mass function&lt;/a&gt; of the binary system.&#xA;$$ \frac{M_2^{3} \sin^{3} i}{(M_1 + M_2)^2} = \frac{p K_1^{3}}{2\pi G},$$&#xA;where $i$ is the inclination of the orbit with respect to the line of sight.&#xA;This can only give you a lower limit to $M_2$ unless $i$ is known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking your specific case study. If you know $M_1$ and $i$ (this could be the case for a transiting exoplanet, or maybe a binary featuring an eclipsed black hole candidate), then the primary radial velocity curve gives you $K_1$ and hence $M_2$. If the masses and $p$ are known then Kepler's laws give the orbital separation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of options if you want an off-the-shelf solution to fitting RV curves. Perhaps the best free one is &lt;a href=&quot;http://oklo.org/downloadable-console/&quot; rel=&quot;nofollow&quot;&gt;Systemic Console&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no fundamental difference between analysing the RV curves of stars with exoplanets and stars with unseen (stellar) companions.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-06-10T21:56:13.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10964" PostTypeId="2" ParentId="10957" CreationDate="2015-06-11T05:43:59.930" Score="10" Body="&lt;p&gt;Your telescope has a 2000 mm focal length, and 200 mm aperture. With the 20 mm eyepiece, you get 100x magnification. With the 10 mm lens, you get 200x. Several things could cause what you describe:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Astronomical_seeing&quot; rel=&quot;noreferrer&quot;&gt;Seeing&lt;/a&gt; (turbulence) might be bad. It is common in many places that less than great seeing means everything at 200x and over might start looking blurry. But seeing changes with the seasons, or day to day, hour to hour, or indeed from one second to another. It's random. If you get the exact same results all the time, and the image never improves even after many weeks of trying, then perhaps it's not seeing that's the culprit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your telescope might not be collimated. Look in the user's manual. Is there a collimation procedure you need to follow? (EDIT: yes, there is - page 20) A miscollimated scope indeed behaves as you described - seems &quot;sharp&quot; at low-ish magnification, but quickly becomes blurry if you push magnification up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point the scope at a bright star and defocus. Are the rings perfectly circular? If not, it might be miscollimated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astrophoto.fr/collim.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.astrophoto.fr/collim.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another factor is optics quality. Optics with lots of aberrations cause &quot;mushy&quot; images and a difficulty to find the true focus. High quality optics produce a scope that is very &quot;snappy&quot; when achieving perfect focus, and make very sharp images when seeing is good. There's nothing you can do about this, but the good news is - this problem is not very common nowadays; optics quality, while not always great, tends to be at least passable in many cases. Quality could be evaluated either on the test bench (which requires equipment), or via direct observation of stars (which requires a very experienced observer and takes time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, have you compared views with another instrument of similar aperture? The image at higher magnification will always seem a bit more soft, compared to lower magnification - even in perfect seeing with a collimated instrument. Also, the contrast always decreases as you increase magnification. This is why there is always an optimal magnification for each view - sometimes higher, sometimes lower, depending on many factors. The best scope in the world will look mushy and blurry if you push magnification up too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you easily see the Cassini division in Saturn's rings? If yes, then you're probably in okay shape (collimation and seeing), you're just not used to the softness at higher magnification. I have a very high quality reflector (self-made), similar aperture class to yours (a bit smaller), and Cassini is immediately visible at 180x and looks sharp; it's also visible at 255x but the whole thing starts to look mushy and faded. It looks super-sharp and high-contrast at 136x but it's kind of small. This is in perfect collimation, perfect thermal equilibrium, high quality optics throughout the stack (primary and secondary mirrors, eyepieces), and good seeing typical of N. California.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When seeing is bad, Cassini starts to fade no matter what I do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iWuKh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does Saturn look equally soft in the center of the image, as well as near the edge? Good optics (mirrors, eyepieces) make a good image everywhere, cheap optics make an okay image in the center and a blurry image at the edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My bet is that it's either bad collimation, or you haven't compared the view at 200x with another scope, or both. But it's hard to diagnose things over the Internet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should probably get a 15mm eyepiece as well. You'll use it a lot, probably more than the 10mm.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Light pollution is never a factor for: planets, the Moon, the Sun, most double stars. These things are just too bright to care about it, so observing them from the city is fine. What does matter is seeing (turbulence). Also, it's important that the scope is in perfect collimation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light pollution only matters for the &quot;faint fuzzies&quot;: nebulae and galaxies. These are low-brightness, low-contrast objects, that are difficult to see from the city. Being low contrast, your eye will not perceive a lot of detail anyway, so for these objects seeing doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Congratulations for being aware of, and following, the rule that says you need to acclimate the scope to ambient temperature. A lot of people are not aware of it. At least you're removing one unknown from the equation. It's always a good idea to let the scope &quot;breathe&quot; outside for an hour before you even begin to observe.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One way to be 100% sure that your scope is focused perfectly is to use a &lt;a href=&quot;http://en.wikipedia.org/wiki/Bahtinov_mask&quot; rel=&quot;noreferrer&quot;&gt;Bahtinov mask&lt;/a&gt;. Just make (or buy) one to match the scope's diameter, put it over the scope's mouth, point scope at a bright star, and tweak the focuser until all spikes intersect exactly in the center of the star. Then remove the mask - the scope is in perfect focus. This will probably not help in your case, but it's just one way to remove another uncertainty. It also helps with regular observations - I use it almost every time I observe, it makes focusing so much easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mask could be made from a piece of cardboard and a sharp knife in like 20 minutes. It doesn't need to be a marvel of engineering precision to work well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.deepskywatch.com/Articles/make-bahtinov-mask.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.deepskywatch.com/Articles/make-bahtinov-mask.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One thing that is certainly NOT a factor is the central obstruction. You'll often hear on the Internet how Cassegrain instruments, or other scopes with a large central obstruction (secondary mirror) are supposedly less &quot;sharp&quot; than scopes without a central obstruction, or ones with a very small obstruction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, the obstruction does matter a little, but not to the extent that Internet mythology would have you believe. A Cassegrain instrument, with good optics, in perfect collimation, can be a superb telescope even at a 40% obstruction. It simply performs like a somewhat smaller instrument, that's all (the physics of the central obstruction are very complex, and would require an entirely different discussion on this forum to fully elucidate).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(Added in reply to a comment below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Miscollimation is quite likely the #1 performance killer of all amateur telescopes. Some well-made refractors are more or less immune to it, but most reflectors need periodic collimation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a good idea to google around for various collimation techniques, there's so many of them. Some require the observation of stars, some require special devices known as collimators. The technique is also different for different types of telescopes (newtonian, cass, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you own a reflector, it's a good idea to collimate it once in a while, just in case, like changing the oil on a car. Dobsonian telescopes probably require most frequent collimation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check Howie Glatter's site - he makes some of the most precise collimators currently available:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://collimator.com/&quot; rel=&quot;noreferrer&quot;&gt;http://collimator.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless, just learn any method you can and apply it. If the telescope is massively miscollimated, you'll get an improvement immediately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not being able to see Cassini is not a good sign. Maybe seeing is persistently bad (it can happen), or maybe the scope is off-collimation. Can anyone else in your area see Cassini these days in a similar aperture, in a scope known to be collimated well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surface detail on Saturn is not easy to see, but should be doable with apertures over 100mm. It's like zones of latitude similar to Jupiter's equatorial belts, but very, very pale and faded. You have to keep looking for a while to see it. Don't worry if you don't see that stuff for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter's Great Red Spot is harder to see nowadays. It's been shrinking and fading for decades now. Go on Wolfram Alpha and type in Great Red Spot. It will tell you if it's visible at the moment. Usually it looks more like a dent in the equatorial belts, rather than an actual visible spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=great+red+spot&quot; rel=&quot;noreferrer&quot;&gt;http://www.wolframalpha.com/input/?i=great+red+spot&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter works best at not so high magnification, otherwise the contrast gets too low. Around 140x ... 180x should give you decent contrast in your aperture.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-06-11T19:20:47.327" LastActivityDate="2015-06-11T19:20:47.327" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10965" PostTypeId="2" ParentId="10939" CreationDate="2015-06-11T07:01:10.973" Score="1" Body="&lt;p&gt;If you want protection from radiation on Mars, you'd have to either live in caves, or else build some kind of thick dome overhead, or live behind thick walls/ceiling. Cosmic radiation will reach you anyway as long as the sky is visible, so deep valleys can only help you so much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way to stop hard radiation is to place a lot of &quot;stuff&quot; (thick mass) between you and the source. A pile of dirt would suffice in many cases. The geologic layers on top of a cave would work well. I guess a dome could also work if it's thick enough. Habitats with thick ceilings (and walls) could also be an option. Have a large water reservoir? Put it on the roof of the habitat: voila, instant radiation shield.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main concern is the stuff coming in from directly overhead; near horizon even Mars' atmosphere is thick enough to reduce radiation a little. So prioritize the ceiling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess at the base of a hill with some overhang above your head you'd get some shielding, but again - is the sky visible? That's the main concern. Also, areas under an overhang may or may not be safe (stable) long term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find a very large boulder (the size of a house or bigger), much taller than your habitat, and place the habitat right next to it - you've cut 50% of radiation right there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a shorter term stay, a simple habitat with thin panels would be enough - not very different from the interplanetary vehicle after all.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-06-11T07:18:05.410" LastActivityDate="2015-06-11T07:18:05.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10967" PostTypeId="2" ParentId="10220" CreationDate="2015-06-11T13:27:54.020" Score="13" Body="&lt;p&gt;I'll add to Wayfaring Stranger's comments. In fact most of the time you would be able to see &lt;em&gt;fewer&lt;/em&gt; stars in the night sky of Mars, than in a good dark night sky on Earth, because of dust obscuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in favourable conditions, the optical depth of the Martian atmosphere is usally somewhere between 0.5 and 1 per airmass. (&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0032063311002960&quot;&gt;Petrova et al. 2012&lt;/a&gt;; &lt;a href=&quot;http://arxiv.org/abs/1403.4234&quot;&gt;Lemmon et al. 2014&lt;/a&gt;) and is nearly wavelength independent. This corresponds to a reduction in flux to between 37% and 60% of its value above the atmosphere. This compares to a typical optical (V-band) extinction of around 0.1 magnitude at a good site on Earth, which allows 90% of the flux through.  There are times when it can be much worse than this on Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the limiting magnitude on Mars would be somewhere between 0.44 and 0.96 magnitudes brighter than it is on a dark site on Earth and that all the stars would be fainter by these values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a rough guide to the effect this has I took the Hipparcos catalogue and made a frequency histogram of V-magnitude of the stars. If we define a Hipparcos magnitude of 6 as the limit you can see from a good dark site on Earth, then you would see 4559 stars (that's over the whole of the sky in both hemispheres). If that limit was reduced to 5.56 or 5.04, then this number decreases to 2745 or 1560 respectively. Thus there would be a &lt;em&gt;reduction&lt;/em&gt; in the number of stars you could see by typically between a factor of 1.66 and 2.92.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-06-11T13:27:54.020" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10968" PostTypeId="1" AcceptedAnswerId="10974" CreationDate="2015-06-11T18:49:39.613" Score="3" ViewCount="3164" Body="&lt;p&gt;I was thinking about the fact that all the largest Solar System moons are tidally locked to its primary and this question arose.&lt;/p&gt;&#xA;" OwnerUserId="7453" LastActivityDate="2020-06-28T17:39:15.763" Title="What would be the practical consequences (on earth) if the Moon was not tidally locked?" Tags="&lt;the-moon&gt;&lt;solar-system&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10969" PostTypeId="1" AcceptedAnswerId="10971" CreationDate="2015-06-11T19:13:14.700" Score="2" ViewCount="98" Body="&lt;p&gt;I'm learning about LOSVDs (Line Of Sight Velocity Distributions) and I'm having a bit of trouble understanding the used terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand, the LOSVD of a given (elliptical) galaxy is the density distribution of the LOS-velocities. The full LOSVD is difficult to find and it's easier to find 2 parameters of the distribution: $\bar v_{_{LOS}}$ and $\sigma_{_{LOS}}$ by fitting a (Gaussian) model to the spectrum of the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a limited understanding of statistics, so I'm having trouble intuitively understanding what these 2 parameters mean later on.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that $\bar v_{_{LOS}}$ is simply the average value of the LOS-velocity for the whole galaxy while $\sigma_{_{LOS}}$ is the equivalent of a standard deviation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now later on in my coursebook, there's an explanation of how to get a LOSVDs for every point (/pixel) in the (projection onto the celestial sphere of the) galaxy by using 3D spectrography.&lt;br&gt;&#xA;From this we can get a 3D dynamic model from our 2D kinematic model. In this newly found 3D distribution there's also 3 Sigmas, one for every dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here comes the part I don't understand:&lt;br&gt;&#xA;The book is looking at the movement of stars in a spiral galaxy, the Milky way in particular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're trying to find a correlation between the age of MS-stars and their dispersion and so there's a dataset of a few stars in the neighbourhood of the sun with their dispersion in every dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the meaning of dispersion in this context? How can a single object have a dispersion if it's a parameter of a density distribution?&lt;/p&gt;&#xA;" OwnerUserId="7318" LastActivityDate="2015-06-11T22:18:57.447" Title="Trouble understanding speed-dispersion in (elliptical) galaxies" Tags="&lt;galaxy&gt;&lt;stellar-dynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10970" PostTypeId="2" ParentId="10968" CreationDate="2015-06-11T22:18:18.357" Score="4" Body="&lt;p&gt;If the Moon was not tidally locked, it would mean that the Moon is in the wrong orbital distance from Earth. The Moon is tidally locked because it is close to the Earth. If the moon were closer, it would approach the Earth's roche limit, be torn apart, and its debris would become a ring for ~100-200 million years. If it were too far, it would continue to drift outward until it was free of Earths' gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By being tidally locked, the Moon has been extending the Earth day by slowing down the Earth's spin, to about ~6 hours from an 18 hour day to a 24 hour day. This process will continue into the future until the Earth is also tidally locked with the Moon. The Moon, while in this state, has also stabilized the Earth's seasons by stabilizing the Earths' orbital axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, a possible indirect consequence would be upon human history. Mankind would have probably abandoned the celestial sphere model in favor of the spinning Earth model much sooner had they seen proof that a celestial body could spin and rotate, much like when Galileo saw that moons were orbiting Jupiter.&lt;/p&gt;&#xA;" OwnerUserId="7430" LastActivityDate="2015-06-11T22:18:18.357" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10971" PostTypeId="2" ParentId="10969" CreationDate="2015-06-11T22:18:57.447" Score="1" Body="&lt;p&gt;The total light of a galaxy is the sum of perhaps billions of stars, each with their own line of sight velocity. You can characterise this distribution with a mean and a standard deviation (dispersion), and this will be reflected in the absorption line profiles of that galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally you might choose to measure this distribution as a function of position in a galaxy, if you are able to spatially resolve different regions. For instance one could put a spectrograph slit across a galaxy and get the velocity dispersion as a function of position along the slit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you are talking about nearby stars, you can measure the 3D velocities individually, providing you can measure the radial velocity, proper motion and distance of each star. You can then find the mean and standard deviation for this group of measurements in each velocity coordinate.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-06-11T22:18:57.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10972" PostTypeId="2" ParentId="10934" CreationDate="2015-06-12T01:01:50.053" Score="1" Body="&lt;p&gt;I can only speculate but here goes nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;67P/Churyumov–Gerasimenko apparently came from the &lt;a href=&quot;http://en.wikipedia.org/wiki/67P/Churyumov%E2%80%93Gerasimenko&quot; rel=&quot;nofollow&quot;&gt;Kuiper Belt&lt;/a&gt;. There are a lot of KBO (Kuiper Belt Objects) in that region, so it is not unreasonable to expect some of these objects to have crashed into other objects of about the same mass, resulting in irregular shapes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small moons (or other moons in the Solar System) have been isolated from other objects of the same size. This allows gravity work work undisturbed for a longer period of time than those in the Kuiper Belt, molding the object into more or less a spherical shape. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is my best stab at the question, though I can't imagine some other cause that would result in such a drastic irregular shape in something as large as a comet (relatively speaking).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I just found something interesting. So the smallest moon in the solar system is &lt;a href=&quot;http://www.ciclops.org/media/ma/2011/6956_16708_0.pdf&quot; rel=&quot;nofollow&quot;&gt;Aegaeon&lt;/a&gt;. It happens to be a couple of orders of magnitude smaller than the comet mentioned above. In addition, this moon appears to have an oblate spheroidal shape to it. It may just be the case that we have only seen spherical moons, but there are many more moons out in the cosmos that have odd shapes to them.&lt;/p&gt;&#xA;" OwnerUserId="7455" LastEditorUserId="7455" LastEditDate="2015-06-12T04:15:15.533" LastActivityDate="2015-06-12T04:15:15.533" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10974" PostTypeId="2" ParentId="10968" CreationDate="2015-06-12T10:25:36.603" Score="4" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;What would be the practical consequences (on earth) if the Moon was&#xA;not tidally locked?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Honestly, I think the consequences would be pretty small, except we'd see the dark side of the moon from time to time.   On the moon, the consequences would be bigger.&lt;/p&gt;&#xA;&lt;p&gt;All tidally locked means is that the moon's rotation matches the moon's orbit, so that the same side of the moon always faces the earth.  If the moon wasn't tidally locked, it would spin from our point of view.  The moon spinning wouldn't affect the earth hardly at all - at least, in no way I can see.&lt;/p&gt;&#xA;&lt;p&gt;The reason most moons are tidally locked to their planets is because the planets gravitation on their moons is quite large.  Strong gravitation, or, strong tidal effects is perhaps more correct, slows down an orbiting objects rotation, so tidal locking of moons is common.  Tidal locking of planets - less so.   Both Pluto and it's moon Charon are tidally locked to each other cause they're pretty close to each other.    Mercury is also, nearly tidally locked to our sun.&lt;/p&gt;&#xA;&lt;p&gt;Now, Phuc's answer&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;By being tidally locked, the Moon has been extending the Earth day by&#xA;slowing down the Earth's spin, to about ~6 hours from an 18 hour day&#xA;to a 24 hour day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;As HDE pointed out, this isn't so.   It's the earth's rotation being ahead of the moon's orbit that's caused the earth to slow down.   The moon's tidal effect on the earth plays a role in that, but the Moon being tidally locked to the earth is irrelevant.&lt;/p&gt;&#xA;&lt;p&gt;Also, the earth's spin was much much faster than an 18 hour day when the moon was young.   By this article, a day on earth was only a few hours long.   &lt;a href=&quot;http://sservi.nasa.gov/articles/nasa-scientist-jen-heldmann-describes-how-the-earths-moon-was-formed/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://sservi.nasa.gov/articles/nasa-scientist-jen-heldmann-describes-how-the-earths-moon-was-formed/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The earth, 4 billion years ago was spinning unusually fast for an object in our solar system.  It might help to consider what makes planets spin.  When they form, it's conservation of angular momentum, but according to the giant impact hypothesis, the earth was hit, not dead center but at an angle.   The giant impact that formed the moon also set the earth spinning very fast.   The moon was also very close when it formed - maybe just twice the Roche limit, so, that close, that the moon slowed the earth's spin and the (at the time) much larger tidal effects pulled on the moon, causing it to move farther away.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2020-06-28T17:39:15.763" LastActivityDate="2020-06-28T17:39:15.763" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10975" PostTypeId="1" CreationDate="2015-06-12T17:18:38.013" Score="7" ViewCount="199" Body="&lt;p&gt;I'm trying to avoid an opinion-based question, so before I outline the comparison I'm proposing, I will qualify the specific facts that yield this comparison. By focusing answers on the relative accuracy of those facts, hopefully we can avoid primarily opinion-based answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Facts&lt;/strong&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Over the course of millions of years, planetary bodies, especially those closer to the sun, &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_escape&quot;&gt;tend to lose&lt;/a&gt; their water and atmosphere as it is blown away by solar wind, and radiated or boiled off by other forces, unless they have a strong magnetosphere to repel the bulk of that solar wind, or enough gravity to hang on to their matter. On a much faster time scale, this also happens to comets as they approach the sun; thus, the tail.&lt;/li&gt;&#xA;&lt;li&gt;Over the course of millions of years, bombardment of planets &lt;a href=&quot;http://earthsky.org/space/did-comets-bring-water-to-earth&quot;&gt;by comets&lt;/a&gt; can increase the water and gas on a planet, or at least counter-balance the gradual loss of the same (depending on other factors like proximity to the sun, size, and magnetic protection).&lt;/li&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;http://space-facts.com/oort-cloud/&quot;&gt;Oort cloud&lt;/a&gt; is composed mostly of icy bodies (including solids that would be a gas at Earth temperatures). And it is from this region that comets &quot;fall&quot; toward the inner solar system.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Also Fact?&lt;/strong&gt;&lt;br&gt;&#xA;So I started wondering what ultimately happens to the water and gas that gets blown off of planetary bodies. I would guess that given the outward pressure of solar wind and solar radiation, the direction of such material is generally outward, away from the sun. Well, where to? Media coverage of the departure of Voyager 1 from our solar system into the realm of interstellar space alerted me to the idea that there is an outer edge to solar influence, and I would guess that this is where the lighter molecules, like water and gases would tend to land. And indeed there is a collection of similar material out there at that solar radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given these facts, is it accurate to view the Oort cloud as behaving similarly to a terrestrial cloudscape? Our clouds form at the altitude where water tends to deposit after it evaporates from the heated surface of the Earth. It gathers there and coalesces to a form which is then overcome by the opposing force of gravity. Then it returns from whence it came. Similarly, I imagine water and gases ultimately being driven from the heat of the sun to an “altitude above its surface”, where ultimately, over a much longer time scale, that material coalesces until other forces eventually drive them back toward the hot sun. Is this an accurate understanding? Is this comparison justified?&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2016-02-08T23:41:48.333" Title="Is it accurate to compare comets to clouds and rain?" Tags="&lt;comets&gt;&lt;planetary-atmosphere&gt;&lt;oort-cloud&gt;&lt;solar-wind&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10976" PostTypeId="1" CreationDate="2015-06-13T02:02:16.563" Score="2" ViewCount="288" Body="&lt;p&gt;Another way of asking this is, &lt;strong&gt;what is the difference between dwarf planets and asteroids? Aren't they the same things?&lt;/strong&gt; It's just that some asteroids are spherical and others aren't...or is there something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can understand why Pluto was demoted. Same thing happened to Ceres (after 50 years of being called a planet?) when we found out it was in an asteroid belt. It would be more helpful to laymen, I think, if the Kuiper Belt was renamed to the Second Asteroid Belt or maybe the Icy Asteroid Belt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why not stick to the original classes where we only had planet and asteroid? (and moon i guess.) Why add a new class called dwarf planets? Was it just to temper the protests?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I forgot to add, we never needed a dwarf planet class before, despite knowing that Ceres was spherical for a long time. So where has this need, if it really exists, come from?&lt;/p&gt;&#xA;" OwnerUserId="7449" LastActivityDate="2020-05-21T12:05:22.057" Title="Is the Dwarf Planet class really necessary?" Tags="&lt;asteroids&gt;&lt;dwarf-planets&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10977" PostTypeId="2" ParentId="10976" CreationDate="2015-06-13T02:36:17.303" Score="9" Body="&lt;p&gt;A lot of the naming conventions were originally &quot;because they remind us of things we already called this&quot;, or simply &quot;tradition&quot;.  How we name things has slowly but surely adjusted with time as more objects were found, and a more robust classification system was needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine it like having bins to sort your toys into.  If you have a small number of balls, perhaps you'd stick them all into a single bin called &quot;balls&quot;.  But as you get more and more balls, you find they no longer fit into that bin.  Looking at them, you notice they are all either quite small, or quite large.  So you make two bins: small balls and large balls.  Or maybe you notice they are all either very soft (stuffed or Nerf versions, perhaps), or very hard (baseballs, for example).  So you could have a &quot;soft balls&quot; and &quot;hard balls&quot; bins.  Keep collecting ever more balls, and you may need three, four, etc. boxes to fit them all in.  And each time you will naturally try to order them so that each box contains balls that are similar to each other: these are all soft and green, these are all hard and white, these are all big and bouncy, etc.  This applies equally well to all sorts of collections: baseball cards may start all together, but then get sorted into teams and years or even individual players as a collection gets very large, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is basically what's happened as we've classified and reclassified objects we've found in our solar system.  We started with a small number of objects, but as we found more and more of them it became too unwieldy to stick them all together, so we started to separate them more and more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ceres was labeled a planet because it seemed like a tiny version of, but otherwise very similar object to, existing planets.  Then we eventually discovered there are &lt;em&gt;a lot&lt;/em&gt; of things in the same general orbital region as Ceres, and if those were to also be planets then we'd soon have a gigantic list of planets.  Since all these objects seemed pretty similar, and came from the same general region, they were given their own class of objects: asteroids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pluto was labeled a planet due in part to some initial errors in the data that suggested it was larger than it actually is.  Despite a growing number of oddities about it that made it look increasingly un-planet-like, tradition left it as a &quot;planet&quot;; it wasn't particularly problematic for this one little odd-ball to be lumped with the others.  Finding a host of new, similar, and sometimes&amp;mdash;in the case of Eris, at least&amp;mdash;bigger objects forced a reconsideration.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, there are also comets, which are not the same as asteroids.  Asteroids are rocky objects in the inner regions of the solar system.  Comets are icy objects from the outskirts of the solar system: the Kuiper belt and Oort cloud, in particular.  They are vaguely similar to each other: generally very small, lumpy, potato-like objects, but we noticed that asteroids were mostly rocks and metals between Jupiter and Mars, and comets were mostly ices beyond Neptune.  So we separated them into their own groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pluto is more comet than asteroid, and likely originated from either the Kuiper belt or Oort cloud.  In this case, it likely got knocked into a closer orbit via gravitational interactions with the gas giants, or possibly a star that got close enough to the Oort cloud.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2015-06-13T02:36:17.303" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10978" PostTypeId="1" AcceptedAnswerId="10982" CreationDate="2015-06-13T06:57:50.770" Score="9" ViewCount="659" Body="&lt;p&gt;Here's what I just read from Wikipedia's page on the Hubble Space Telescope:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;While Hubble helped to refine estimates of the age of the universe, it also cast doubt on theories about its future. Astronomers from the High-z Supernova Search Team and the Supernova Cosmology Project used ground-based telescopes and HST to observe distant supernovae and uncovered evidence that, far from decelerating under the influence of gravity, &lt;strong&gt;the expansion of the universe may in fact be accelerating&lt;/strong&gt;. The cause of this acceleration remains poorly understood; the most common cause attributed is dark energy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't take this at face value because we should expect more distant objects to have higher &lt;strong&gt;observed&lt;/strong&gt; speeds and therefore higher &lt;strong&gt;observed&lt;/strong&gt; red-shifts. Here's why.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Let's start with a model where the Universe expanded very fast early on, but has been slowing down ever since due to gravity, as one would normally expect.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Remember that, the farther away a cosmic object is, &lt;strong&gt;the farther back in the past&lt;/strong&gt; we are observing it. An object 1,000 light years away, if it's light is reaching us now, is being observed in its state that existed 1,000 years ago. We are effectively looking through a time machine.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;So if we observe a more distant object, we're observing an older state of that object. Therefore, we are observing it at a time when the Universe was expanding faster than it is now, so it has higher red-shifts.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And isn't that what we observe today? The more distant the galaxy, the higher its red-shift? This is not inconsistent with a &quot;normal&quot; model where the expansion is slowing down due to gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing? Why are scientists trying to explain such things with weird dark matter and dark energy that otherwise have never been detected or found evidence of and aren't needed for any other model, and in fact get in the way of our models of physics and quantum dynamics?&lt;/p&gt;&#xA;" OwnerUserId="7449" LastActivityDate="2021-02-09T14:51:29.937" Title="Is the Universe really expanding at an increasing rate?" Tags="&lt;cosmology&gt;&lt;cosmological-inflation&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10979" PostTypeId="1" AcceptedAnswerId="10984" CreationDate="2015-06-13T15:10:42.493" Score="16" ViewCount="6712" Body="&lt;p&gt;The Earth revolves around the Sun and the Moon revolves around the Earth.&#xA;Out of curiosity I started thinking about the orbit of the Moon around the Sun and expected (assumed) it to be as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PWZVl.gif&quot; alt=&quot;Assumed Path&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But on &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbit_of_the_Moon&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; and some other sites I found out that the orbit is actually like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sQTE7.gif&quot; alt=&quot;Actual Path&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 3 questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the reason for this difference between assumed and actual path variation?&lt;/li&gt;&#xA;&lt;li&gt;Has this path been like this since the formation of the Moon?&lt;/li&gt;&#xA;&lt;li&gt;Do natural satellites of other planets also follow the same orbit around the Sun?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;After further searching I found out a better, easier explanation regarding the orbit path on &lt;a href=&quot;https://www.youtube.com/watch?v=3cJ3AemeUFM&quot; rel=&quot;noreferrer&quot;&gt;YouTube&lt;/a&gt; for those interested, be sure to check it out.&lt;/p&gt;&#xA;" OwnerUserId="7466" LastEditorUserId="9186" LastEditDate="2015-11-09T16:14:54.007" LastActivityDate="2020-05-21T03:11:34.990" Title="Moon's orbit around the Sun" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;the-sun&gt;&lt;earth&gt;&lt;natural-satellites&gt;" AnswerCount="8" CommentCount="6" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10980" PostTypeId="2" ParentId="10979" CreationDate="2015-06-13T15:28:39.667" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the reason for this difference between assumed and actual path variation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The orbit of a moon around the sun depends on the time to orbit the planet and the planet's time to orbit the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case where the moon takes a long time to orbit the planet (like the Earth-Moon), the orbit just wiggles along the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case where the moon has a short orbital period compared to the planets year (like Jupiter-Io), the path is like you drew in the first figure.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Has this path been like this since the formation of the Moon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the Earth-Moon system, ...&lt;br&gt;&#xA;Yes, it has always been that way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do Natural Satellites of other planets also follow the same orbit around the Sun?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="1322" LastEditDate="2015-06-13T16:55:18.987" LastActivityDate="2015-06-13T16:55:18.987" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10981" PostTypeId="1" AcceptedAnswerId="10983" CreationDate="2015-06-13T20:37:06.013" Score="2" ViewCount="319" Body="&lt;p&gt;So I was wondering how astronomers are able to get the texture/detail of an exoplanet or how they know a planet has water or not.  I know how they discover planets by using the light of a star and seeing if it gets faint or by seeing if the star is getting gravitationaly pulled back and forth,  to determine the planet size. But I just don't understand how they are able to get the texture of a planet. Thanks for the help! &lt;/p&gt;&#xA;" OwnerUserId="7467" LastActivityDate="2015-06-14T00:45:54.917" Title="How do astronomers determine the texture of an exoplanet?" Tags="&lt;star&gt;&lt;orbit&gt;&lt;exoplanet&gt;&lt;kepler&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10982" PostTypeId="2" ParentId="10978" CreationDate="2015-06-13T22:13:45.293" Score="11" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I don't take this at face value because we should expect more distant&#xA;  objects to have higher observed speeds and therefore higher observed&#xA;  red-shifts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's true.   That was the original Hubble discovery - the farther away things were, the faster they were moving away from us.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Here's why.  Let's start with a model where the Universe expanded very&#xA;  fast early on, but has been slowing down ever since due to gravity, as&#xA;  one would normally expect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes - that's what everybody thought following Hubble's discovery.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Remember that, the farther away a cosmic object is, the farther back&#xA;  in the past we are observing it. An object 1,000 light years away, if&#xA;  it's light is reaching us now, is being observed in its state that&#xA;  existed 1,000 years ago. We are effectively looking through a time&#xA;  machine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is not lost on Astrophysicists.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So if we observe a more distant object, we're observing an older state&#xA;  of that object. Therefore, we are observing it at a time when the&#xA;  Universe was expanding faster than it is now, so it has higher&#xA;  red-shifts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;OK, 2 points.   2nd point first.   The red-shift has to do with relative velocity, not speeding up or slowing down.    Something can be more red-shifted and slowing down and something can be less red-shifted and speeding up, especially since the acceleration/deceleration is comparatively slow compared to the relative velocity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and other point - lets keep in mind, we don't know what a galaxy 3 billion light years away is doing now.  We can guess and we can run models, but we can only see what it's doing 3 billion years ago.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And isn't that what we observe today? The more distant the galaxy, the&#xA;  higher its red-shift? This is not inconsistent with a &quot;normal&quot; model&#xA;  where the expansion is slowing down due to gravity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, the more distant the galaxy the higher it's red-shift.    But no, that's not inconsistent with expansion.  That's what you'd see, expansion or contraction, because red-shift is just relative velocity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What am I missing? Why are scientists trying to explain such things&#xA;  with weird dark matter and dark energy that otherwise have never been&#xA;  detected or found evidence of and aren't needed for any other model,&#xA;  and in fact get in the way of our models of physics and quantum&#xA;  dynamics?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A lot of these ideas are confusing.   They're confusing to scientists too, especially when they were first discovered - so you're not alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark matter was observed because galaxies were behaving strangely.   The stars in the outer arms of the galaxy were observed to be moving much too fast and faster than the stars more towards the middle of the galaxy and that made no sense.   The galaxies also weighed too much and the only way to explain this was extra mass in kind of a halo around the galaxy, but this extra mass, also, didn't interact with electromagnetic waves like the mass here on earth does - so they called this extra mass (and there's a lot of it, more than there is regular mass), but since it's invisible, they called it &quot;dark matter&quot; and it's not dark like dirt or coal, it's dark as in - invisible.   It's completely transparent to light, but it has mass and they still don't know what it is.  They have some OK theories, but nothing definite.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, dark energy - think about the big bang and all matter flying apart - the galaxies twice as far are moving away twice as fast, BUT, as you said, because of gravity, we should see the galaxies that are twice as far moving away more than twice as fast, cause the nearer the galaxy, the more time it's had to slow down - aha, they thought, if we can compare the speed of the galaxies 4 billion light years away to the speed of the galaxies 2 billion light years away to the speed 1 billion light years - etc, etc and measure it all carefully, we can measure the rate at which gravity is slowing down the universe. - that makes sense right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And with careful measurement of Type 1A supernovas, which temporarily outshine entire galaxies - with remarkable consistency (what they call a standard candle - a very bright standard candle, but a standard candle all the same) - with that, they thought they could measure the gravitational slow down of expansion - exactly what you're talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem was, the measurements told them the opposite of what they expected to find.   The measurements told them that the galaxies 2 billion light years away were traveling slightly more than half as fast as the galaxies 4 billion light years away, and so on.  They checked this, cause it had to be wrong, then they re-checked it, and re-checked again and the only conclusion was, stuff out there is speeding up, not slowing down - cause that's what the telescopes tell us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark energy wasn't a hair-brained scheme that mad scientists thunk up.  It was an observed reality that nobody expected (well, cept just maybe for Einstein and his cosmological constant, but that's another story). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark energy's just a name anyway.   They have to call it something, even if they're not sure what it is or how it works.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-06-13T22:13:45.293" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10983" PostTypeId="2" ParentId="10981" CreationDate="2015-06-14T00:45:54.917" Score="1" Body="&lt;p&gt;This is known as spectroscopy.  Every molecule and atom in the universe emits and absorbs light at specific frequencies.  This is a result of the quantization of the energy levels (for electrons) in an atom.  Although there are lots of complicating factors, such as redshift, to account for, the patterns are usually so distinctive that the complications can be accounted for, and do not prevent scientists from figuring out chemical compositions (and even, to some extent, how abundant each chemical is).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, here are the emission spectra of several common atoms at rest (with respect to the observer):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/QaQCR.gif&quot; alt=&quot;emission spectra of common elements&quot;&gt;&#xA;Absorption spectra, on the other hand, appear as darker regions in the band.  Here's an example of an absorption spectra, which identities several elements by knowing that it is distinguished by absorbing light at specific wavelengths that others do not:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/CdJaJ.jpg&quot; alt=&quot;sample absorption spectra&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For exoplanets with significant atmospheres, we can only expect to see the spectra for its upper atmosphere.  All other signals will be muted out by the rest of the atmosphere.  We will not be able to tell what the surface looks like, or what it is made of.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2015-06-14T00:45:54.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10984" PostTypeId="2" ParentId="10979" CreationDate="2015-06-14T01:18:26.367" Score="11" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the reason for this difference between assumed and actual path variation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even your second image isn't correct. Imagine zooming in on a small portion of the Moon's orbit about the Sun, for example, one full moon to the next, with the Sun zoomed out of the picture. Now imagine drawing a line segment from one outer cusp (full moon) to the next. In both of your images, that line segment crosses outside of the curve. In other words, both of your curves are concave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare that to the he Moon's orbit about the Sun. This is a convex curve. If you pick any two points on that curve and draw a line segment between them, the entirety of that segment will be on or inside the curve. The reason the Moon's orbit about the Sun is convex is because the gravitational force exerted by the Sun on the Moon is more than twice than exerted by the Earth on the Moon. The orbit would be concave if the Moon was closer to the Earth than 259000 km (about 40.6 Earth radii). Since the Moon orbits at about 385000 km (about 60.4 Earth radii), the Moon's orbit about the Sun is convex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether the orbit of a moon about the Sun is non-simple (first image in the question), simple/concave (second image in the question), or simple/convex (Moon's orbit about the Sun), the deviations from an ellipse are tiny. With regard to the Earth-Moon system, the deviations are so very small that at the plotted resolution (288x288 pixels), the orbits of the Earth, the Earth-Moon barycenter, and the Moon about the Sun will be right on top of one another. The reason the variations are so small (less than one pixel at 288x288 pixels) is because of the huge ratio of the size of Earth/Moon orbit about the Sun compared to the size of the Moon's orbit about the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those backward loops in your first image don't happen for any object orbiting the Earth. That would require an orbital velocity about the Earth greater than the Earth's orbital velocity about the Sun. The Earth's orbital velocity about the Sun is about 30 km/sec, considerably more than the orbital velocity of an object in low Earth orbit is about 7.8 km/sec.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Has this path been like this since the formation of the Moon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. The Moon formed at four to six Earth radii, far less than the 40.6 Earth radii figure cited above. The Moon's orbit initially looked like your second image.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do Natural Satellites of other planets also follow the same orbit around the Sun?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The massive planets are much further from the Sun than is the Earth and are much more massive than is the Earth. The orbits of most of the moons of Jupiter about the Sun are concave rather than convex. Only the outermost moons of Jupiter have convex orbits about the Sun. A few of Jupiter's innermost moons (Metis, Adrastea, Amalthea, Thebe, Io, and Europa) exhibit the retrograde motion depicted in your first image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regard to moons whose orbit about the Sun is convex, the distances that correspond to the 259000 km value for the Earth are 129000 km for Mars, 24.1 million kilometers for Jupiter, 24.2 million kilometers for Saturn, 19.0 million kilometers for Uranus, and 32.3 million kilometers for Neptune. Both of Mars' moons orbit close-in. However, all four of the giant planets have moons whose semi-major axis orbit fall outside the corresponding limit.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2015-06-14T15:30:49.690" LastActivityDate="2015-06-14T15:30:49.690" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10985" PostTypeId="1" CreationDate="2015-06-14T02:04:06.467" Score="5" ViewCount="291" Body="&lt;p&gt;Let's limit this to optical telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the higher you are, the less atmosphere is above you to get in the way of observations. From reading around, there seem to be a few more but I'm not sure:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Not near an active or dormant volcano (but Mauna Kea seems to disprove this?). This is kinda a bummer because a lot of tall mountains seem to be volcanic.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Somewhere with clear and/or dry weather for as much of the year as possible&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Not near major light pollution like cities&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Cold weather is better than hot weather? Not sure I understand that; if it's uniformly hot or cold then I don't see how it makes a difference.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Am I missing any? Or are these wrong in some way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a concrete example, would the Tibetian plateau make a really good observatory site? (Not considering construction or transportation costs to a remote area, of course.)&lt;/p&gt;&#xA;" OwnerUserId="7449" LastEditorUserId="7472" LastEditDate="2017-07-15T22:26:44.053" LastActivityDate="2017-07-15T22:26:44.053" Title="What makes a really good observatory site, besides altitude?" Tags="&lt;weather&gt;&lt;observatory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10986" PostTypeId="2" ParentId="10979" CreationDate="2015-06-14T05:09:13.217" Score="15" Body="&lt;p&gt;Not an answer, but I thought this was a good slice of a picture of the Moon's orbit around the sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gz2cl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source:  &lt;a href=&quot;http://www.wired.com/2012/12/does-the-moon-orbit-the-sun-or-the-earth/&quot; rel=&quot;noreferrer&quot;&gt;http://www.wired.com/2012/12/does-the-moon-orbit-the-sun-or-the-earth/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-06-14T05:09:13.217" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10987" PostTypeId="1" AcceptedAnswerId="11006" CreationDate="2015-06-14T07:48:26.523" Score="4" ViewCount="163" Body="&lt;p&gt;Recent stars (and standard candle supernova), afaik, contain more heavy elements than older stars. As consequence, I guess, recent stars and supernova should be slightly dimmer than older stars? If previous sentence is true, slightly dimmer Ia supernova will appear as if it is more distant leading to the conclusion that universe is expanding faster. Does measuring of universe expansion rate and &quot;dark energy&quot; takes such possibility in to the computation?&lt;/p&gt;&#xA;" OwnerUserId="7469" LastEditorUserId="7469" LastEditDate="2015-06-14T14:43:23.993" LastActivityDate="2015-06-16T15:33:27.610" Title="Accelerating universe expansion and standard candle" Tags="&lt;universe&gt;&lt;expansion&gt;&lt;supernova&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10988" PostTypeId="1" AcceptedAnswerId="10995" CreationDate="2015-06-14T07:55:39.893" Score="1" ViewCount="11650" Body="&lt;p&gt;I previously asked that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/10958/is-the-angular-resolution-of-a-telescope-irrespective-of-used-eye-piece=&quot;&gt;Is the angular resolution of a telescope irrespective of used eye-piece?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I learned that the resolution is fixed when the light enters the telescope, and that the eye-piece is used just for magnifying the obtained image at the focal plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the magnification of a telescope is calculated as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\text{magnification} = \frac{\text{focal length of telescope}}{\text{focal length of eye-piece}} \qquad (1)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the focal length of my telescope is $900$ mm and I am using a $25$ mm eye-piece, the magnification I get is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\text{magnification} = \frac{900\ \text{mm}}{25\ \text{mm}} = 36.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, where does the formula $(1)$ come from? Why is the magnification affected by the focal length of the &lt;em&gt;telescope and eye piece&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am using a Newtonian telescope if it matters.)&lt;/p&gt;&#xA;" OwnerUserId="6991" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2017-03-12T03:11:45.663" Title="Why does the focal length of a telescope have an effect on the magnification?" Tags="&lt;telescope&gt;&lt;newtonian-telescope&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10989" PostTypeId="1" CreationDate="2015-06-14T10:24:47.233" Score="5" ViewCount="690" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Andromeda-Milky_Way_collision#The_fate_of_the_Solar_System&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Such an event would have no adverse effect on the system and the chances of any sort of disturbance to the Sun or planets themselves may be remote.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm imagining that this is because the distance between stars is so big, it'd be highly unlikely for any star to actually collide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the Oort Cloud is 2 light years in diameter. Given that the average distance between stars in the Milky Way is about 5 light years, that's almost half way. If Alpha Centauri had an Oort Cloud of the same size, we'd already almost be overlapping each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that we're not the only star with an Oort Cloud, isn't there a high probability that the Sun will actually be inside the Oort Cloud of a passing star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would imagine this would because the Sun and our planet would be bombarded with lots of comets which have been thrown off their orbits by either the gravity of the Sun or the passing star. &lt;/p&gt;&#xA;" OwnerUserId="7470" LastEditorUserId="2153" LastEditDate="2015-06-14T14:23:09.140" LastActivityDate="2015-06-17T02:13:17.133" Title="Why won't the Milky Way colliding with Andromeda affect the Solar System?" Tags="&lt;solar-system&gt;&lt;milky-way&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10990" PostTypeId="2" ParentId="10989" CreationDate="2015-06-14T12:50:03.810" Score="4" Body="&lt;p&gt;Outside my field, but collecting information from the Wiki article, I can make an attempt nevertheless:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're right that we will probably pass through Oort-like clouds of other stars. But the largest of the planetesimals of the Oort Cloud is measured in kilometres, whereas the distance between these objects is measured in AU, so the total cross section of object compared to that of the cloud is still miniscule. Smaller objects are larger in number, but still it's a very low probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitationally, the clouds also won't affect us. The total mass of the Oort Clouds is probably of the order of 10 Earth masses, i.e. much smaller than the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outer parts of the Oort Cloud is only loosely bound to the Sun. I suppose the best evidence that we won't have any major collisions is that then we would already have similar objects bombarding us from our neighboring stars' clouds.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2015-06-14T12:50:03.810" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10993" PostTypeId="2" ParentId="10989" CreationDate="2015-06-14T17:27:14.323" Score="0" Body="&lt;p&gt;I've always found that somewhat strange when they say no stars will hit out of billions, but I also trust the scientists that they know their stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets look closer at your numbers. If the Oort Cloud has a diameter of 2 light years (I think it's a bit less than that, but numbers vary), that's a radius of 1 light year, and the Solar System to the Oort Cloud has a volume of $\frac{4}{3} \pi R^3$, or $\frac{4}{3} \pi$ cubic light-years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 5 light-years average distance from stars, that's an area 125 times larger. So our Sun fills up (to its Oort Cloud) 1/125th of its region of space, and 5 light-years might be low too. The average closest star might be a bit more than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But not all stars are as big as our Sun; many are smaller and if two Oort Clouds pass through each other, not much happens. A star has to pass within our Oort Cloud for the gravitational changes to be interesting, and that doesn't happen very often.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to consider is that the Earth is in a spiral arm of our galaxy, and both the Milky Way and Andromeda are flat disk-shaped galaxies. Are we crashing head on or at an angle where our spiral arms won't be directly hit? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think we'll care much 4 billion years from now. I like to think we'll have traveled beyond the Solar System by then, but the danger to the Sun from Andromeda probably isn't huge. That said, I'd guess that some stars will pass near other stars, even if the experts say direct hits are unlikely (from the article)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the Sun were a ping-pong ball, Proxima Centauri would be a pea about 1,100 km (680 mi) away, and the Milky Way would be about 30 million km (19 million mi) wide. Although stars are more common near the centres of each galaxy, the average distance between stars is still 160 billion (1.6×1011) km (100 billion mi). That is analogous to one ping-pong ball every 3.2 km (2.0 mi). Thus, it is extremely unlikely that any two stars from the merging galaxies would collide&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If stars are ping pong balls every two miles and we use our Solar System as an example (it's really not, because the ping-pong-ball-every-two-miles example is in the center of the Solar System), but if the Sun is a ping pong ball, the Earth orbits at a radius of 4.3 meters and Neptune at a bit over 120 meters and the outer edge of the Kuiper Belt about 200 meters. That's still small compared to the average distance between stars, even in the galactic center, but not so small that interaction would never happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ping pong balls might miss each other but I think it's a safe bet that there will be the occasional star-to-outer planet close encounter. From a local perspective these will be rare, but from an entire galaxy perspective, I'd think there would be quite a few of what could be called, planetary fly-bys/near misses.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="2153" LastEditDate="2015-06-14T18:09:43.243" LastActivityDate="2015-06-14T18:09:43.243" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10994" PostTypeId="2" ParentId="10979" CreationDate="2015-06-14T18:12:29.453" Score="3" Body="&lt;p&gt;[NOTE: The animated GIFs are too large to copy into this post, but the &#xA;URLs should work]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: all 3 animated GIFs are now available on youtube:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://youtu.be/7xN5hwcxspg&quot; rel=&quot;nofollow&quot;&gt;https://youtu.be/7xN5hwcxspg&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://youtu.be/G5YfPQdNw5g&quot; rel=&quot;nofollow&quot;&gt;https://youtu.be/G5YfPQdNw5g&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://youtu.be/qSbozfxAfDQ&quot; rel=&quot;nofollow&quot;&gt;https://youtu.be/qSbozfxAfDQ&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;END EDIT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://test.bcinfo3.barrycarter.info/farmoon.gif&quot; rel=&quot;nofollow&quot;&gt;http://test.bcinfo3.barrycarter.info/farmoon.gif&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image above would apply if: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the Earth and Moon both had circular orbits (approximately true) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the Moon's siderial period was exactly 1/12th year (approximately true) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the Earth's distance from the Sun was 150 (million km) (approximately true) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the Moon's distance from the Earth was 30 (million km) (completely false) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here, you see the loop-de-loop orbit you originally expected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what if we reduce the distance to 10 million km (still very large): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://test.bcinfo3.barrycarter.info/nearmoon.gif&quot; rel=&quot;nofollow&quot;&gt;http://test.bcinfo3.barrycarter.info/nearmoon.gif&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the loop-de-loops are gone, although the orbit still &#xA;has some &quot;sharp points&quot;, which we don't see in the Moon's real orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we reduce the distance to 3 million km, we get something closer to &#xA;what expect: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://test.bcinfo3.barrycarter.info/closemoon.gif&quot; rel=&quot;nofollow&quot;&gt;http://test.bcinfo3.barrycarter.info/closemoon.gif&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, we have a wobbly circle, closer to what actually happens. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the Moon's actual distance from the Earth is only 0.35 &#xA;million km, so the actual wobbles are much smaller. I tried to do a &#xA;graphic of those, but the two orbits ended up on top of each other. &lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2015-11-08T18:18:40.327" LastActivityDate="2015-11-08T18:18:40.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10995" PostTypeId="2" ParentId="10988" CreationDate="2015-06-14T18:37:51.990" Score="3" Body="&lt;p&gt;The objective lens of a telescope forms an real image of the night sky, the size of that image is in proportion to the focal length of the objective lens. The reason for this is simple geometry: If two stars are 1 arcminute apart, and the lens is forming an image of them, then the further the image is from the lens, the further apart the images of the two stars will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The eyepiece then magnifies the image. When using a lens to magnify you put the object at (or near) the focal distance from the lens. The shorter the focal length of the eyepiece, the closer you can get to the object and so the larger it appears. All this is a round about way of saying the the magnification of a lens is inversely proportional to the focal length. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a telescope, the &quot;object&quot; being magnified is the image formed by the objective lens. Putting this together: the apparent size of the image is directly proportional to the focal length of the objective lens, and inversely proportional to the focal length of the eyepiece, and the constant of proportionality is 1 (just imagine that both the objective and eyepiece have the same focal length). So &#xA;$$\mathrm{Mag} = \frac{L_o}{L_e}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might be a useful point to mention that magnification is usually the least useful function of an astronomical telescope. Their main purpose is to gather light.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-06-14T18:37:51.990" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10996" PostTypeId="1" AcceptedAnswerId="10999" CreationDate="2015-06-14T23:34:23.427" Score="3" ViewCount="1164" Body="&lt;p&gt;Since Jupiter is the biggest planet in our solar system and is made up of mostly hydrogen and helium, (the gases the Sun uses to create energy), how come it didn't form as far out as say Uranus or Neptune since when our star formed it took most of the gases around it to create? Just a little confused on this, thanks for the help! &lt;/p&gt;&#xA;" OwnerUserId="7467" LastActivityDate="2015-06-15T00:14:42.057" Title="How did Jupiter form where it is?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;jupiter&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10997" PostTypeId="2" ParentId="10996" CreationDate="2015-06-15T00:04:26.587" Score="2" Body="&lt;p&gt;Here's something from &lt;a href=&quot;http://en.wikipedia.org/wiki/Formation_and_evolution_of_the_Solar_System#Formation_of_the_planets&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; (emphasis mine):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The giant planets (Jupiter, Saturn, Uranus, and Neptune) formed further out, beyond the frost line, the point between the orbits of Mars and Jupiter where the material is cool enough for volatile icy compounds to remain solid. The ices that formed the Jovian planets were more abundant than the metals and silicates that formed the terrestrial planets, allowing the giant planets to grow massive enough to capture hydrogen and helium, the lightest and most abundant elements. Planetesimals beyond the frost line accumulated up to 4 M⊕ within about 3 million years. Today, the four giant planets comprise just under 99% of all the mass orbiting the Sun. &lt;strong&gt;Theorists believe it is no accident that Jupiter lies just beyond the frost line. Because the frost line accumulated large amounts of water via evaporation from infalling icy material, it created a region of lower pressure that increased the speed of orbiting dust particles and halted their motion toward the Sun. In effect, the frost line acted as a barrier that caused material to accumulate rapidly at ~5 AU from the Sun.&lt;/strong&gt; This excess material coalesced into a large embryo (or core) on the order of 10 M$_⊕$, which began to accumulate an envelope via accretion of gas from the surrounding disc at an ever increasing rate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Quoting from the actual Wikipedia article on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Frost_line_(astrophysics)&quot; rel=&quot;nofollow&quot;&gt;frost line&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The lower temperature in the nebula beyond the frost line makes many more solid grains available for accretion into planetesimals and eventually planets. The frost line therefore separates terrestrial planets from giant planets in the Solar System.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any giants that lie inside the frost line in stellar systems most likely &lt;a href=&quot;http://en.wikipedia.org/wiki/Planetary_migration&quot; rel=&quot;nofollow&quot;&gt;migrated inwards&lt;/a&gt; after interactions with the protoplanetary disk or, as happened in our Solar System (but didn't cause Jupiter to move that far inwards), with other giant planets. The Nice model describes such incidents&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2015-06-15T00:04:26.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10998" PostTypeId="1" AcceptedAnswerId="11001" CreationDate="2015-06-15T00:07:52.730" Score="5" ViewCount="1223" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Grand_Tack_Hypothesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Grand Tack Hypothesis&lt;/a&gt; states that Jupiter formed at roughly 3.5 AU from the Sun, then interacted with a large portion of objects in the Solar System before settling into its current orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet as I found from researching for an answer to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10996/how-did-jupiter-form-where-it-is/10997#10997&quot;&gt;this question&lt;/a&gt;, giant planets generally form beyond the frost line - in fact, this explains why the giant planets are cut off from the terrestrial planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the Grand Tack Hypothesis get around this?&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-06-15T07:54:21.473" Title="How does the Grand Tack Hypothesis explain how Jupiter formed inside the frost line?" Tags="&lt;solar-system&gt;&lt;jupiter&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10999" PostTypeId="2" ParentId="10996" CreationDate="2015-06-15T00:10:42.407" Score="3" Body="&lt;p&gt;It's at Jupiter's distance and beyond that ices were able to form out of the disk of material surrounding the early sun.  Go much further in and there is too much energy from the sun for them to stay as solids (and will sublimate into gases); this is why asteroids are principally rocks and metals.  So at this distance more of the materials of the planetary disk can form the base planetesimals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, your question is answered largely by a geometric consideration (or two) and one of Kepler's laws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the geometric consideration.  A circle of radius $r$ has area $\pi r^2$.  The bigger the radius, the more area.  The material for Jupiter (or any other planet) came from an annulus: stuff outside one circle, but inside a slightly bigger circle.  This annulus had a lot more area for the out planets than the inner planets, and so could contain a lot more mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, that could makes us think that Jupiter shouldn't be the largest of the gas giants: it's the closest of them all to the sun, after all.   The density of the disk needn't have been approximately constant throughout these regions, though.  Quite possibly the density was such that Jupiter's region had more mass than the areas for the other planets.  As HDE's answer (posted as I was finishing this) points out, these ices probably also helped stop materials from passing into the inner solar system, maintaining a higher density than you might otherwise expect in the inner solar system, as well as causing materials to sort of &quot;dam up&quot; right around Jupiter's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the Kepler's law.  The further you get from the sun, the slower your orbital period.  Picking the correct units, we have $P^2=a^3$, where $P$ is the period measured in years, and $a$ is the semi-major axis of the orbit measured in AU.  The further out you go, the slower you go around the sun; indeed, it's not just that it takes you a longer total time, but your actual velocity goes down.  We can also see this as a Newton's law of gravity consequence.  At Jupiter's furthest point from the Sun, the escape velocity is a little more than 18 km/s.  At Saturn's maximum distance, the escape velocity drops to around 13.25 km/s.  So things can go roughly 35% faster within Jupiter's orbit than they can closer to Saturn's, and they have less distance to travel to make a complete orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means is that it takes longer for planetesimals to get close enough to each other to accrete together the further out you go, and there is a longer mean time between collisions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, eventually, the Sun &quot;turned on&quot; and started blasting space with it's solar wind (before that, the heat came primarily from thermal radiation from the gravitational contraction of the sun).  This ended up clearing out most of the unaccreted particles out of the solar system, stopping planetary growth (and removed portions of existing atmospheres; a very young Earth probably had a lot of H and He in its atmosphere, until the sun hit it with enough energy to knock any of it not locked up in rocks away).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Jupiter was probably in a bit of a goldilocks situation.  The average density of the region it formed in was probably higher than where the other giants formed in, it was at the perfect spot for lots of materials to start accreting early, and the accretion process would have been faster.  So Jupiter is growing faster, and this gives it a competitive advantage: the bigger the growing planetesimals get the further their influence extends and the faster they can pull in more materials, and subsequently interfere with the growth of other planets (or planetesimals).  Somewhere around a mass of 10-15 earth masses, the giants can start pulling in large quantities of the hydrogen and helium gasses.  And, again, Jupiter likely hit this mass well before the other giants, and had more material to pull from, so it became much larger than the others could before the solar wind stopped the process.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastEditorUserId="2188" LastEditDate="2015-06-15T00:14:42.057" LastActivityDate="2015-06-15T00:14:42.057" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11000" PostTypeId="2" ParentId="525" CreationDate="2015-06-15T00:29:39.753" Score="2" Body="&lt;p&gt;This is just an amusing addition to the existing answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that a metallic hydrogen layer (which lets electrons move freely, and moving electrons means a magnetic field can form) is not enough to account for the size of Jupiter's magnetosphere.  It's off by a factor of approximately 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Io_(moon)#Interaction_with_Jupiter.27s_magnetosphere&quot; rel=&quot;nofollow&quot;&gt;The rest of it is mostly thanks to Io&lt;/a&gt;.  The wiki page will give a more complete description (of a rather complex system) and references, but here's the short of the matter.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Io is on an eccentric orbit, thanks to a resonance with the other Galilean moons.  This gives it significant tidal heating (and as it is the closest of the Galilean moons, it has the most significant heating effect of them all).  This gives it volcanic activity, which puts new gaseous materials (sulfur, oxygen, and chlorine mostly) into its atmosphere.  Jupiter strips material from Io's upper atmosphere at about 1 metric ton per second.  This material ultimately forms ionized bands that generate substantial electric current, and significantly increases the magnetosphere around Jupiter as a result.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2015-06-15T00:29:39.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11001" PostTypeId="2" ParentId="10998" CreationDate="2015-06-15T02:56:42.037" Score="5" Body="&lt;p&gt;A direct cribbing of the wiki's page for the &lt;a href=&quot;https://en.wikipedia.org/wiki/Frost_line_(astrophysics)&quot; rel=&quot;noreferrer&quot;&gt;frost line&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In astronomy or planetary science, the frost line, also known as the snow line or ice line, is the particular distance in the solar nebula from the central protostar where it is cold enough for volatile compounds such as water, ammonia, methane, carbon dioxide, carbon monoxide to condense into solid ice grains. This condensation temperature depends on the volatile substance and the partial pressure of vapor in the protostar nebula. The actual temperature and distance for the snow line of water ice depend on the physical model used to calculate it:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;170 K at 2.7 AU (Hayashi, 1981)&lt;/li&gt;&#xA;  &lt;li&gt;143 K at 3.2 AU to 150 K at 3 AU (Podolak and Zucker, 2010)&lt;/li&gt;&#xA;  &lt;li&gt;3.1 AU (Martin and Livio, 2012)&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;Occasionally, the term snow line is also used to represent the present distance at which water ice can be stable (even under direct sunlight). This current snow line distance is different from the formation snow line distance during the formation of Solar System, and approximately equals 5 AU. The reason for the difference is that during the formation of Solar System, the solar nebula was an opaque cloud where temperature were lower close to the Sun,[citation needed] and the Sun itself was less energetic. After formation, the ice got buried by infalling dust and it has remained stable a few meters below the surface. If ice within 5 AU is exposed, e.g. by a crater, then it sublimates on short timescales. However, out of direct sunlight ice can remain stable on the surface of asteroids (and the Moon) if it is located in permanently shadowed craters, where temperature may remain very low over the age of the Solar System (e.g. 30–40 K on the Moon).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Observations of the asteroid belt, located between Mars and Jupiter, suggest that the water snow line during formation of Solar System was located within this region. The outer asteroids are icy C-class objects (e.g. Abe et al. 2000; Morbidelli et al. 2000) whereas the inner asteroid belt is largely devoid of water. This implies that when planetesimal formation occurred the snow line was located at around $R_{snow} = 2.7$ AU from the Sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So in all of the three listed models, as well as the snow line inferred from asteroid properties, it seems that the 3.5 AU starting distance for Jupiter in the Grand Tack hypothesis remains comfortably behind the snow line (for water, at least, which is still an advantage over the inner planets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it ultimately fell well within the snow line for a time, one would expect any exposed ices to rapidly sublimate.  I would guess that it made up for this by accreting the rockier planetesimals found in the inner parts of the Solar System, which could then shield the remaining ices.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.boulder.swri.edu/~kwalsh/GrandTack.html&quot; rel=&quot;noreferrer&quot;&gt;This site&lt;/a&gt; offers a pretty good detailing of the basic structure of the Grand Tack Hypothesis.  The most digestible piece of information is at the very bottom, in the form of an AVI file that models the first few million years of the solar system.  The short answer to your question is &quot;Jupiter is assumed to have already formed some fundamental core at the start of the hypothesis, then moves around accreting and scattering things through the inner solar system, until returning to the outer solar system to finish growing&quot;.  The initial inward migration is caused by interactions with the relatively thick gaseous disc (the majority of it will never become part of any planet), which creates drag on the planets/planetesimals and causes them to lose velocity and fall inward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the video's pretty cool, so I'm going to link it and talk about it just for its own sake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.boulder.swri.edu/~kwalsh/Fred/GT-all.AVI&quot; rel=&quot;noreferrer&quot;&gt;Get the video, as produced by the paper Walsh et. al., 2011&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may find it best to run this at half speed or slower to really see what's going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how it starts out.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ISkJW.jpg&quot; alt=&quot;starting configuration&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solid black orbits are the gas giant orbits, with Jupiter being the innermost one, and Saturn the one after that.  The dashed circle represents a 2 AU distance from the sun (modern day Mars is currently at about 1.5 AU, and the modern asteroid belt stretches roughly 2.2-3 AU).  This is noteworthy since the model is trying to explain why Mars is so small, and why there is so little material (relatively speaking) in the asteroid belt.  The numerous small blue dots are the icy comets/planetesimals that form right around the point Jupiter starts out, and points beyond.  The numerous red dots are the rocky asteroids/planetesimals characteristic of the inner solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first several seconds we see Jupiter creep in towards the inner solar system.  You'll notice a lot of rocky planetesimals get scattered out into the outer solar system, while some of the icy ones are scattered to the inner solar system.  Indeed, it seems to be a general feature of this model that the total amount of material within approximately 1 AU undergoes a net increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn is also slowly moving in, but not nearly as quicky as Jupiter is (Neptune/Uranus seem pretty stable, but do &quot;pulse&quot; in and out a bit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wZFQ6.jpg&quot; alt=&quot;jupiter hits the 2 AU mark&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this image we see that Jupiter hits the 2 AU mark at ~3 seconds in, corresponding to a model time of about 68,000 years.  At this point you can see a dashed circle at about the 3.5 AU distance, corresponding to Jupiter's starting distance.  Saturn is about 4 AU out now, having started at over 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third through fourth second are fairly stable-looking, with Jupiter moving in to about 1.5 AU but the other outer planets staying mostly put.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the 5th second (~94,000 years model time)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/25OF2.jpg&quot; alt=&quot;Saturn makes its move&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/N2gYH.jpg&quot; alt=&quot;Saturn still moving fast&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see Saturn make a very rapid move inwards.  This is the main reason I suggest playing it slowly, as you can very easily miss it and lose track of where each solid black orbit has come from.  By model time 98,000 years (still in the 5th second of the video time), Saturn has already come to within about 2 AU of the Sun; Jupiter is still about 1.5 AU, and there is a thick concentration of mostly rocky planetesimals within 1 AU.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/trda6.jpg&quot; alt=&quot;Saturn and Jupiter rebound back out&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By model time of about 120,000 years, Saturn and Jupiter are rebounding back out into the outer solar system, to much the same distances we see today, scattering most of the planetesimals that remained in the areas that would become Mars and the asteroid belt.  This leaves a fairly dense disc of material for Earth, Venus, and Mercury to form and grow from, while leaving significantly less for Mars, and lesser still for the asteroid belt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole 10 second movie covers a model time of 600,000 years.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/xhbGl.jpg&quot; alt=&quot;The End&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Walsh, K. J., Morbidelli, A., Raymond, S. N., O’Brien, D. P., Mandell, A. M. (2011). A low mass for Mars from Jupiter’s early gas-driven migration. Nature 475, 206-209.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Walsh, K. J., Morbidelli, A., Raymond, S. N., O’Brien, D. P., Mandell, A. M. (2012). Populating the asteroid belt from two parent source regions due to the migration of giant planets – ”The Grand Tack”. Meteoritics &amp;amp; Planetary Science 47(12), 1941-1947.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastEditorUserId="2188" LastEditDate="2015-06-15T07:54:21.473" LastActivityDate="2015-06-15T07:54:21.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11003" PostTypeId="2" ParentId="10989" CreationDate="2015-06-15T08:34:22.457" Score="0" Body="&lt;p&gt;I've seen some articles, &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10871/how-much-overlap-will-the-andromeda-galaxy-and-the-milky-way-have-when-they-coll/10873#10873&quot;&gt;which I've referenced in a recent answer of mine concerning this collision&lt;/a&gt;, which make a claim to the tune of &quot;the Earth and Sun will probably not be significantly affected&quot; (other than that the Sun is dead by this point, and Earth possibly consumed by it).  This was based on N-body simulations of the merger, themselves based in part on recently collected data on Andromeda's tangential velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, this is likely true for the sun because it is true for most of the stars in both galaxies: &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_density&quot; rel=&quot;nofollow noreferrer&quot;&gt;the distances between stars are so vast that the galaxies are almost entirely empty space&lt;/a&gt; (well, okay, there's gas, and there's going to be all sorts of stuff happening with the gas, in the form of a lot of compression leading to new star formation; but stellar density is basically nothing), and as the galaxies merge the stars will mostly just harmlessly pass through the vast open spaces between each other.  Indeed, in the neighborhood of our sun stars give an average density of about $4\times 10^{−24} \ g/cm^3.$  The densest parts of our galaxy are around 500 times as dense as this.  This gives about 70 solar masses in a cubic parsec.  So while the dense regions may have several stars within a cubic light year, that's still a very small fraction of space occupied by stars within that volume.  Our sun has a volume of about $1.41\times 10^{18} \  km^3$, whereas a cubic light year is $8.46732407 \times 10^{38} \ km^3$: over 20 orders of magnitude larger!  Even the largest stars only have volumes about $5\times 10^{10}$ that of sun, which still makes the cubic light year ten orders of magnitude larger.  That's a very tiny fraction of 1% for even the biggest stars, and much, much, much tinier for stars more like our Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbital paths of stars themselves may see significant changes, but the disturbing influence generally won't extend to structures on the scale of a solar system.  The existing star(s) will still have the overwhelming influence over their own systems, since other stars are simply too far away.  Some stars' orbital paths will undergo drastic alterations over the course of the merger (which is expected to take a few billion years), and may be flung into the halo or completely ejected from the merging system.  Still, if any of these stars have planetary systems of their own then it is unlikely that those systems will be significantly disrupted.  There may be a few exceptions, the most notable being the expected merger of the supermassive blackholes, but in the overwhelming number of cases the structure of a star system won't be altered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is all fairly model-dependent, but the general expectation for any model is that very little will change locally for any given star system (or even star cluster), due to the low density of stars.  It's just the global picture that changes a lot.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-06-15T08:41:29.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11005" PostTypeId="1" CreationDate="2015-06-15T14:36:36.303" Score="0" ViewCount="113" Body="&lt;p&gt;Supposedly how accurate is it when you see a movies showing something entering Earth's atmosphere traveling 1 mile to 10 miles an hour or barely at all but skinning the atmosphere they show it heating up and burning just by entering the atmosphere no speed or barely just entering?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a second question, something coming in - say, a vehicle with rockets - to ensure that the speed does not exceed that would cause  any heating at all. What speed would that be?&lt;/p&gt;&#xA;" OwnerUserId="7477" LastEditorUserId="7982" LastEditDate="2021-02-28T11:52:59.677" LastActivityDate="2021-02-28T11:52:59.677" Title="How accurate are rederings of something entering Earth's atmosphere?" Tags="&lt;planetary-atmosphere&gt;&lt;atmospheric-effects&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11006" PostTypeId="2" ParentId="10987" CreationDate="2015-06-15T15:12:14.607" Score="1" Body="&lt;p&gt;Most type Ia supernovae are thought to arise from the thermonuclear detonation of white dwarfs that are composed almost entirely out of carbon and oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These white dwarfs are the cores of relatively low-mass stars that have lived their lives, gone through stages of core hydrogen and helium burning, leaving behind degenerate carbon/oxygen cores that become cooling white dwarfs after the outer envelope has been shed during the asymptotic giant branch and planetary nebula phases. As such, their composition, at least to first order, is almost independent of the initial composition of the star from which they were formed. That is, even if the progenitor star had a very low initial metal content, the white dwarf produced would still be almost exclusively a carbon/oxygen mixture, which had a similar Chandrasekhar mass and a similar explosive potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well known however that not all type Ia supernovae are the same. It has long been known that their light curves are subtley different and there is a so-called &lt;a href=&quot;http://ned.ipac.caltech.edu/level5/Sept05/Leibundgut/Leibundgut3.html&quot; rel=&quot;nofollow&quot;&gt;stretch factor&lt;/a&gt; that can be applied to get a &quot;corrected&quot; peak magnitude. a.k.a The width-luminosity relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More recently there has been a realisation that type Ia supernovae could arise from both accretion or mergers and there is clear evidence that the amount of radioactive Ni varies from explosion to explosion. A very recent paper by &lt;a href=&quot;http://arxiv.org/abs/1408.1706&quot; rel=&quot;nofollow&quot;&gt;Milne et al. (2015)&lt;/a&gt; has however challenged the view of metallicity independence. They claim there are two populations of type Ia SNe, connected with progenitor metallicity, and that these populations become more apparent at high redshift when looking at rest-frame ultraviolet emission. The gist of their conclusions is indeed, as your question supposes, that this may go some way to ameliorating (but not eliminating) the need for dark energy.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-06-16T15:33:27.610" LastActivityDate="2015-06-16T15:33:27.610" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11007" PostTypeId="2" ParentId="10921" CreationDate="2015-06-15T15:18:03.380" Score="0" Body="&lt;p&gt;No, since any matter inside a black hole will not be in a stable solid state. This happens because the gravitational force at the inside of a black hole is huge and powerful, because it's formed by the collapse of a star under its own mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So say for your question there might have been a immensely huge supernova, which became a black hole. In turn, the Big Bang happened to form our universe. Would not that be reverting the black hole laws?&lt;/p&gt;&#xA;" OwnerUserId="7478" LastEditorUserId="2153" LastEditDate="2015-06-15T21:22:18.277" LastActivityDate="2015-06-15T21:22:18.277" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11008" PostTypeId="2" ParentId="10953" CreationDate="2015-06-15T19:37:34.743" Score="4" Body="&lt;p&gt;Noctilucent clouds are not a problem for space telescopes because their orbits are always more than 85 km. The Hubble Telescope orbits at about 570 km.  Noctilucent clouds are a problem for ground-based telescopes (although only at high latitude sites,&gt; 50$^{\circ}$), especially if you are trying to get accurate photometric brightnesses.  However, typically you can see them reflecting city lights or moonlight, or you can tell from the large variance in your photometry counts that they are there and you should try again another night.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2015-06-20T02:13:47.750" LastActivityDate="2015-06-20T02:13:47.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11009" PostTypeId="1" AcceptedAnswerId="11014" CreationDate="2015-06-15T20:03:11.070" Score="4" ViewCount="682" Body="&lt;p&gt;I've been to some talks that mention how stable the period of a millisecond pulsar is over long periods of time. Recently, it was mentioned that astronomers have calculated the change in period over time to be less than 10^-12 seconds per year for several pulsars. No one I've talked to seems to know any details of this calculation.&#xA;How do we calculate such small differences in period? How much data must be collected and what are the exposure times for imaging such rapid phenomena?&#xA;A source/paper would be excellent. I apologize that I don't have a citation for the 10^-12s figure, but the lack of citation is mostly my reason for posting this question.&lt;/p&gt;&#xA;" OwnerUserId="7480" LastActivityDate="2015-06-16T10:00:10.997" Title="Pulsars: How do astronomers measure minute changes in period (~picoseconds per year)?" Tags="&lt;astrophysics&gt;&lt;pulsar&gt;&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11010" PostTypeId="2" ParentId="10904" CreationDate="2015-06-15T20:11:42.333" Score="0" Body="&lt;p&gt;The answer is there on the &lt;a href=&quot;http://www.stjarnhimlen.se/comp/riset.html&quot; rel=&quot;nofollow&quot;&gt;stjarnhimlen.se site&lt;/a&gt; and also &lt;a href=&quot;http://www.stargazing.net/kepler/altaz.html&quot; rel=&quot;nofollow&quot;&gt;stargazing.net&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now we can compute the Sun's altitude above the horizon:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;sin(h) = sin(lat) * sin(Decl) + cos(lat) * cos(Decl) * cos(LHA)&#xA;LHA = LST - RA&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;h=Altitude=30$^\circ$,&#xA;LHA = Local Hour Angle,&#xA;lat= Your latitude on Earth,&#xA;Decl = Object's declination,&#xA;Ra = Object's right ascension, and&#xA;LST = Local Standard Time. &#xA;Just need to solve for LST. &lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2015-06-15T20:11:42.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11011" PostTypeId="2" ParentId="11005" CreationDate="2015-06-15T22:54:32.837" Score="1" Body="&lt;p&gt;In the movies you reference, I doubt that it was ever the intention of a movie maker to suggest that the craft was moving at a crawl when it began to enter the atmosphere. If the craft doesn't look like it is moving very fast, the idea is probably that the point of reference (the &quot;camera&quot;) is moving at nearly the same speed as the craft, or perhaps that the craft is so huge, that the reference point is miles away. Everything is relative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is essentially no realistic scenario where any object would enter the atmosphere at such a slow speed. Imagine, for example that you have an object, like &lt;a href=&quot;https://en.wikipedia.org/wiki/Felix_Baumgartner&quot; rel=&quot;nofollow&quot;&gt;Felix Baumgartner&lt;/a&gt; or something, which is standing approximately stationary in space above the Earth's atmosphere. Gravity will immediately begin pulling on that object and because there is no atmosphere to slow the object down yet, it just keeps accelerating as it falls. Even as it enters the edge of the atmosphere, it doesn't slow down much, because there isn't &lt;em&gt;much&lt;/em&gt; atmosphere yet. Felix was moving more than 800 mph before he started to slow down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The International Space Station is another object that's sort of &quot;stationary&quot;, in a manner of speaking. It's not going anywhere because it's in a stable orbit (sort of). But in order to stay up there, it's actually moving more than 17,000 mph, relative to the atmosphere. If something is &quot;stationary&quot; with respect to the sun, just sitting there in Earth's orbit as Earth comes at it, the Earth's atmosphere would smack it at 67,000 mph. That's how fast Earth orbits the sun. So there's really isn't anything that would just gradually drift into the atmosphere at 1 to 10 mph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a craft were to use rockets to halt itself before entry, as you suggest, remember Felix again; it can become stationary with respect to the Earth, but then gravity begins its work, so if the craft wants to remain at a very slow speed through the entire decent, it has to keep firing those rockets all the way down. And certainly, nothing would burn up at that pace, other than a lot of rocket fuel. What causes the burning is not something special and hot about the atmosphere, but the fact that the object is smashing into the air (thin as it may be at that altitude) at thousands of mile per hour. It is even theoretically possible for meteors to enter at such speeds that the atoms are literally forced to hit each other's nuclei, and a part of the fire generated would actually a brief &lt;em&gt;nuclear&lt;/em&gt; reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Response to Question Edit:&lt;/strong&gt;&lt;br&gt;&#xA;(Thank you HDE. The initial question was very unclear.)&lt;br&gt;&#xA;The speed at which friction completely stops is 0. Any movement at all, even in that 1 - 10 mph range, produces a small amount of heat from friction, even if the medium you are moving through is very thin air (though certainly nothing one would characterize as &quot;burning&quot;). You seem to be looking for the speed at which that friction would be low enough to not cause a problem, but there's no magic answer. It depends on the situation. How delecate is the craft? How much wear is acceptable? Essentially, this question becomes something very similar to asking &quot;How fast can I drive with something sticking out the window?&quot; You are trying to keep the entry speed low enough that the wind is not a problem for you as the density of that wind gradually increases from practically nothing to surface air density. If your craft has a heavy, durable heat sheild, properly oriented, then extraordinary speeds are still not a real problem. But if you are trying to create a sand mandala on the roof of your craft as you descend, then you are going to want to control that wind level very carefully.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditDate="2015-06-16T22:01:16.523" LastActivityDate="2015-06-16T22:01:16.523" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11013" PostTypeId="2" ParentId="11009" CreationDate="2015-06-15T23:43:21.720" Score="3" Body="&lt;p&gt;My comment notwithstanding, &lt;a href=&quot;http://www.astro.ucla.edu/~ianc/courses/08SA082/week6_homework4_pulsars.pdf&quot; rel=&quot;nofollow&quot;&gt;here's a solution to a homework problem that does the calculation&lt;/a&gt;.  It doesn't specify the exact mechanism that converts angular momentum (aka rotational energy) into electromagnetic radiation.  In this case, it is just an assertion of the problem (partially justified with what I said in my comment: the energy must come from somewhere, and if there doesn't seem to be any sources other than angular momentum, then it must be coming from the angular momentum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slightly rephrasing that link's content for the sake of future accessibility:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The pulsar is radiating energy (which we observe as radio waves). Since the total&#xA;  energy in the universe must be conserved, this radio energy must come from&#xA;  somewhere. In this case, it is taken out of the rotational kinetic energy of the pulsar:&#xA;  thus, it gradually slows down.&#xA;  We're interested in a relation between the pulsar luminosity and its rotational period.&#xA;  In general, the kinetic energy of a rotating body is given by&#xA;  $$E=\frac{1}{2} I \omega^2 = 2\pi^2 I P^2.$$&#xA;  Since Luminosity is the time derivative of energy, we are&#xA;  now in a position to relate the quantities we are interested in:&#xA;  $$L= \frac{\partial E}{\partial t} = -4\pi^2 I P^{-3} \frac{\partial P}{\partial t}.$$&#xA;  Rearranging this in terms of the quantity we want – the rate of&#xA;  change of the period – gives:&#xA;  $$\frac{\partial P}{\partial t} = \frac{-L P^3}{4\pi^2 I}.$$&#xA;  If we assume that this neutron star is a homogeneous sphere (not really true, but a&#xA;  simple approximation), then its moment of inertia is just:&#xA;  $$I_{\text{sphere}}=\frac{2}{5}M R^2,$$&#xA;  and so the final rate of change of period we get is:&#xA;  $$\frac{\partial P}{\partial t} = -\frac{5}{8\pi ^2} \frac{L P^3}{MR^2}.$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So as long as we have measurements of the quantities on the right-hand side, we can just plug them in to get a value for the rate of change in the period.  The hardest to measure is usually the moment of inertia (where the mass, $M$, and radius, $R$, terms come from).  These are easier to get from eclipsing binary systems, since then there are nice relations between their orbital paths and masses.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2015-06-15T23:43:21.720" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11014" PostTypeId="2" ParentId="11009" CreationDate="2015-06-16T09:54:47.517" Score="5" Body="&lt;p&gt;Let us suppose that the pulsar is spinning down at a uniform rate. So it has a period $P$ and a rate of change of period $dP/dt$ that is positive and constant (in practice there are also second, third, fourth etc. derivatives to worry about, but this doesn't change the principle of my answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's assume you can measure the period very accurately - say you look at the pulsar today and measure its radio signals for a few hours, do a Fourier transform of the signal and get a nice big peak with a period of 0.1 seconds (for example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that period, you can &quot;fold&quot; the data to create an average pulse profile. This pulse profile can then be cross-correlated with subsequent measurements of the pulse to determine an offset between the predicted time of &quot;phase zero&quot; in the profile, calculated using the 0.1 s period, and the actual time of phase zero. This is often called an &quot;O-C&quot; curve or a residuals curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have the correct period and $dP/dt=0$, then the residuals will scatter randomly around zero with no trend as you perform later and later observations (see plot (a) from Lorimer &amp;amp; Kramer 2005, The Handbook of Pulsar Astronomy). If the initial period was in error, then the residuals would immediately begin to depart from zero on a linear trend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If however, you have the period correct, but $dP/dt$ is positive, then the residuals curve will be in the form of a parabola (see plot (b)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have second, third etc. derivatives in the period, then this will affect the shape of the residuals curve correspondingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The residuals curve is modelled to estimate the size of the derivatives of $P$. The reason that $dP/dt$ can be measured so precisely is that pulsars spin fast and have repeatable pulse shapes, so changes in the phase of the pulse quickly become apparent and can be tracked over many years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0BMzj.png&quot; alt=&quot;Pulsar residual timing curves&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically it works something like this. The phase $\phi(t)$ is given by&#xA;$$\phi(t) \simeq \phi_0 + 2\pi \frac{\Delta t}{nP} - \frac{2\pi}{2}\frac{(\Delta t)^2}{nP^2} \frac{dP}{dt} + ...,$$&#xA;where $\phi_0$ is an arbitrary phase zero, $\Delta t$ is the time between the first and last observation and $n$ is the integer number of full turns the pulsar has made during that time. If the period is approximately correct, then $n = int(\Delta t/P)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;residual curve&quot; would be given by&#xA;$$\phi_0 - 2\pi\frac{\Delta t}{nP} -\phi(t) \simeq  \frac{2\pi}{2}\frac{(\Delta t)^2}{nP^2} \frac{dP}{dt} + ...,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the period of a $P \sim 0.01$ second pulsar changed by a picosecond in a year, then there would be an accumulated residual of almost $10^{-4}$ seconds after 1 year of observation. Depending on how &quot;sharp&quot; the pulse is, then this shift of about 1% in the phase of the pulse might be detectable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps needless to say, but there are a host of small effects and corrections to make in order to get this very high precision timing. You need to know exactly how the Earth is moving in its orbit. The proper motion of the pulsar on the sky also has an effect. These and more can be found in Lorimer and Kramer's book, but there is also a summary &lt;a href=&quot;http://www.cv.nrao.edu/course/astr534/PulsarTiming.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-06-16T10:00:10.997" LastActivityDate="2015-06-16T10:00:10.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11015" PostTypeId="2" ParentId="10988" CreationDate="2015-06-16T11:32:17.960" Score="1" Body="&lt;p&gt;Sometimes this can be difficult to wrap your head around in Astronomy, as telescopes generally have a fixed aperture and focal distance, and simply use an eyepiece at the end to make a difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you, instead, look at a camera you can get the concept quite quickly.  DSLR cameras have swappable lenses and many lenses include non-fixed focal distances (zoom lenses). So as you zoom with the lens, you are effectively changing the focal distance of the camera, which results in an image zooming in and out. If you've been in photography for a while, you probably also realize that when you do this, it also effects your lighting as the greater your focal distance, the less total light you are actually utilizing. Here's a pretty rough (and simplified) diagram of what is happening:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/n9WSb.jpg&quot; alt=&quot;Diagram of focal distance and resulting image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crop size is defined by the sensor size. Any light that spreads beyond the sensor is not captured. By increasing the area of the image, the portion that is within the sensor decreases, resulting in a cropped image that appears to be zoomed in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A telescope works in a very similar fashion, but the telescope has a fixed focal length and fixed aperture. As these are all fixed, the telescope is generally engineered to utilize all light, where as a DSLR level camera will be built to allow diverse lens selection, which results in a lot of light wasted (light that is cropped by not reaching the viewfinder or sensor). The light is finally adjusted by the eyepiece. As everything before the eyepiece is more or less fixed, the type of eyepiece is a direct trade between total light and image size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should probably also be noted, that 'More distance = More zoom' isn't true everywhere in optics.  Microscopes, for instance, tend to be designed in the opposite fashion.&lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="1288" LastEditDate="2015-06-16T11:48:01.550" LastActivityDate="2015-06-16T11:48:01.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11016" PostTypeId="1" AcceptedAnswerId="11020" CreationDate="2015-06-16T15:50:15.690" Score="2" ViewCount="332" Body="&lt;p&gt;I have seen this in a couple of places, but when the authors in the paper - &lt;a href=&quot;http://chandra.as.utexas.edu/~kormendy/kfcb-accepted.pdf&quot; rel=&quot;nofollow&quot;&gt;http://chandra.as.utexas.edu/~kormendy/kfcb-accepted.pdf&lt;/a&gt; -&#xA;wrote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;All 10 of our ellipticals with total absolute magnitudes M_VT ≤ −21.66 have cuspy cores – “missing light” – at small radii. Cores are well known and naturally scoured by binary black holes formed in dissipationless (“dry”) mergers. All 17 ellipticals with −21.54 ≤ M_VT ≤ −15.53 do not have cores.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since absolute magnitude is the magnitude of that object at 10 pc away from Earth, then this is why the galaxies have such negative (meaning, very bright) values for total absolute magnitude, correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in this paper - &lt;a href=&quot;http://arxiv.org/pdf/astro-ph/9602044v1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/astro-ph/9602044v1.pdf&lt;/a&gt; - the author writes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The morphological properties of galaxies between 21 mag &amp;lt; I &amp;lt; 25 mag in the Hubble Deep Field are investigated using a quantitative classification system based on mea- surements of the central concentration and asymmetry of galaxian light.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The galaxies here are between 21 to 25 absolute magnitudes (?) in the infrared (I?) band, but why are these magnitudes not negative? &lt;/p&gt;&#xA;" OwnerUserId="2913" LastEditorUserId="2913" LastEditDate="2015-06-16T23:24:29.107" LastActivityDate="2015-06-26T17:32:38.713" Title="Question about total absolute magnitudes of galaxies - negative or not?" Tags="&lt;galaxy&gt;&lt;magnitude&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11017" PostTypeId="1" AcceptedAnswerId="11023" CreationDate="2015-06-16T16:20:07.787" Score="7" ViewCount="1139" Body="&lt;p&gt;We can observe many astronomical objects and structures beyond our solar system,  however our observations are as the objects appeared in the past due to the speed of light. My question: &quot;&lt;strong&gt;Is it possible to use Hubble Telescope to observe Earth as it was in the past&lt;/strong&gt;&quot; by turning Hubble's lens to a previous location of Earth in space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UQp4v.png&quot; alt=&quot;Observing the Earth by Hubble.&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7483" LastEditorUserId="1980" LastEditDate="2015-06-17T02:58:43.440" LastActivityDate="2019-09-14T09:32:43.153" Title="Is it possible to use Hubble Telescope to observe Earth in the past?" Tags="&lt;hubble-telescope&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11018" PostTypeId="2" ParentId="11017" CreationDate="2015-06-16T16:26:54.080" Score="4" Body="&lt;p&gt;Not without some hypothetical faster-than-light travel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hubble is seeing distant galaxies as they were in the past because they're so far away that the light from those galaxies takes a very long time to get here. If the galaxy is 1,000 light-years away, you're seeing it as it was 1,000 years ago. The light reflected from Earth moves much more quickly than the Earth itself moves. Pointing a telescope to where the Earth was won't show you what the Earth looked like at that time because the light coming from Earth at that time is already gone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to be in a place that was x light-years away from Earth, you could see the Earth as it was x years ago. For instance, if you wanted to see the dinosaurs you would need to find a way to get more than 65 million light-years away from where the Earth was at that time. To see the Earth as it was in the past from where the Earth is now isn't possible.&lt;/p&gt;&#xA;" OwnerUserId="1980" LastEditorUserId="1980" LastEditDate="2015-06-16T17:17:43.250" LastActivityDate="2015-06-16T17:17:43.250" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11019" PostTypeId="2" ParentId="11017" CreationDate="2015-06-16T16:46:07.027" Score="4" Body="&lt;p&gt;Technically yes you would be able to.  Picture this. If you were to position a gigantic mirror 10 light years from earth (let's say you are 10 at the time) and the mirror was pointing right at you. When you are 30 (so in 20 years because it would take light 10 years to get there and 10 years to get back) you would be able to see yourself when you were 10. Now picture that was the earth and yes you would be able to. Another way is if you were able to move faster than the speed of light then you could get to a point in space where earth's light hasn't reached yet and you could observe earth from that light that it enjoyed a long time ago.  I hope these 2 answers helped and answered your question. &lt;/p&gt;&#xA;" OwnerUserId="7467" LastActivityDate="2015-06-16T16:46:07.027" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11020" PostTypeId="2" ParentId="11016" CreationDate="2015-06-16T20:49:46.317" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The morphological properties of galaxies between 21 mag &amp;lt; I &amp;lt; 25 mag&#xA;  in the Hubble Deep Field are investigated using a quantitative&#xA;  classification system based on mea- surements of the central&#xA;  concentration and asymmetry of galaxian light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those are apparent magnitudes, not absolute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The convention is to use $M_X$ for absolute magnitude and just $X$ for apparent mag (or m$_X$), where X is the bandpass symbol (U, B, V, R, I, J, L, H, K, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, the $V_T$ and $M_{V_T}$ (note the correct subscripting, which the authors failed to do) means the V band radial profile is extrapolated to infinite radius to give a total for the whole galaxy.  &lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2015-06-26T17:32:38.713" LastActivityDate="2015-06-26T17:32:38.713" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11022" PostTypeId="2" ParentId="10989" CreationDate="2015-06-17T02:13:17.133" Score="0" Body="&lt;p&gt;In a simpler way than the preceding answers, think probability. The probability of a galaxy colliding with another is small, but it affecting a single star greatly is very tiny, especially 3/4 from the center. The most affect we would get is a wonderful show...&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=qnYCpQyRp-4&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=qnYCpQyRp-4&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stars in our (and all other) galaxies are extremely far apart, plus, our star would probably have exploded most likely by then anyway, so goodbye Earth way before the collision :)&lt;/p&gt;&#xA;" OwnerUserId="7484" LastActivityDate="2015-06-17T02:13:17.133" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11023" PostTypeId="2" ParentId="11017" CreationDate="2015-06-17T09:45:35.970" Score="10" Body="&lt;p&gt;The only way you can use Hubble to view the Earth &quot;in the past&quot; is to swivel it and point it at the Earth now. Assuming that this didn't break the telescope (it would) you would collect images of the Earth's surface that were approximately $500\times10^{3}/3\times10^{8} = 0.0017$ seconds old, since that is how long it takes light to travel from the Earth's surface directly below HST to its $\sim 500$ km orbital height above the Earth's surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with what you suggest, is that although we might be able to pinpoint where the Earth was in the past with respect to its position now and the Galactic centre (actually that would be very hard in practice), the light that came from the Earth &lt;em&gt;then&lt;/em&gt; has long since travelled beyond where the Earth is &lt;em&gt;now&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, let's say the Earth/Sun travel around the Galaxy at approximately 230 km every second and say we want to look at the spot where the Earth was 1000 years ago. Well, it turns out then that the Earth was at some location in space that is currently about $7\times 10^{12}$ km away (7 trillion). But the light that came from the Earth then, has already been travelling outwards at the speed of light for 1000 years, covering a distance of 1000 light-years, which is about $10^{16}$ km. So that light passed by our &lt;em&gt;current&lt;/em&gt; location in space more than 999 years ago.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-06-17T09:45:35.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11024" PostTypeId="2" ParentId="11017" CreationDate="2015-06-17T12:56:02.467" Score="2" Body="&lt;p&gt;Hubble &lt;a href=&quot;http://hubblesite.org/hubble_discoveries/10th/photos/slide05.shtml&quot; rel=&quot;nofollow&quot;&gt;orbits at about 569 km&lt;/a&gt;. If you point it at the earth and take a photo, the speed of light (300,000 km/sec) will result in you obtaining a photo of the earth as it appeared 1.9 milliseconds before you clicked the shutter.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-06-17T12:56:02.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11025" PostTypeId="1" CreationDate="2015-06-17T21:44:23.623" Score="0" ViewCount="131" Body="&lt;p&gt;Can there be any truth to the rumour that the Apollo Moon landing in 1969 was staged ? I dismissed this rumour as being nonsense but today on the news I was surprised to hear that the head of the Police in Russia is calling for an investigation concerning this . Why does this rumour that it never happened persist even today ?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastActivityDate="2015-06-17T22:34:54.097" Title="Apollo Moon Landing" Tags="&lt;moonlanding&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2015-06-18T15:39:52.253" ContentLicense="CC BY-SA 3.0" />
  <row Id="11026" PostTypeId="1" AcceptedAnswerId="11030" CreationDate="2015-06-17T21:57:00.440" Score="3" ViewCount="79" Body="&lt;p&gt;So I just had a quick question and I was wondering how we convert other waves of the electromagnetic spectrum to visible light waves. For example we use X-ray's or radio waves to see things in space like nebulae that we wouldn't normally be able to see with just visible light waves. But how do we convert those waves to visible light so we can actually see them.  I don't know if I'm staying this right but another example that just came to me is pictures of the Sun we see that was taken with radio waves, we see them,  which must mean that they were covered into visible light some how, so I was just wondering how, thanks in return =)&lt;/p&gt;&#xA;" OwnerUserId="7467" LastActivityDate="2015-06-18T06:41:36.760" Title="How do we convert other waves of the EMS to visible light?" Tags="&lt;telescope&gt;&lt;radio-astronomy&gt;&lt;wavelength&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11027" PostTypeId="2" ParentId="10985" CreationDate="2015-06-17T22:13:17.570" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;• Not near an active or dormant volcano (but Mauna Kea seems to disprove this?). This is kinda a bummer because a lot of tall mountains seem to be volcanic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's nothing intrinsically wrong with &lt;em&gt;dormant&lt;/em&gt; volcanos, as Mauna Kea and the observatories in the Canary Islands demonstrate. Not sure where you would have gotten that idea.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;• Somewhere with clear and/or dry weather for as much of the year as possible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clear weather is essential -- you can't observe when it's cloudy!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dry weather is very good, especially for infrared observations (water vapor blocks a lot of infrared light).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;• Not near major light pollution like cities&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;• Cold weather is better than hot weather? Not sure I understand that; if it's uniformly hot or cold then I dont see how it makes a difference.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cold climates tend to have drier air. (Antarctica is an extreme case.) Of course, being on top of a mountain means colder air, which is why Mauna Kea is good despite it's being in the tropics.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Am I missing any? Or are these wrong in some way?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You also want &lt;em&gt;stable&lt;/em&gt; air with good &quot;seeing&quot;, which rules out places with turbulent air and lots of wind (and is another reason mountaintops are good: they're usually above the most turbulent layers of the atmosphere).&#xA;Quoting from &lt;a href=&quot;http://www.astr.ua.edu/keel/techniques/sites.html&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt; (which has a good discussion of the general topic):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Seeing ... requires at minimum a lack of extra turbulence at all atmospheric levels, and seems to be best satisfied in the convergence zones just outside the tropics, at latitudes about ± 30º. Also, minimal local turbulence is often associated with mountain peaks that reach into the otherwise undisturbed oceanic airflow, as on islands or coastal ranges (and given the direction of the planet's rotation, this generally favors western coast ranges).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2015-06-17T22:13:17.570" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11028" PostTypeId="2" ParentId="11025" CreationDate="2015-06-17T22:27:01.250" Score="5" Body="&lt;p&gt;No, there's not an ounce (gram?) of truth that the Apollo moon landings were faked.  The guys on Mythbusters did a good job of busting several of the so-called problems.  The light on the astronaut descending the ladder on the lunar lander was light reflected from the white space suit of the astronaut snapping the picture; the shadows that appear to indicate different light sources is actually a matter of topography; missing stars is because of basic photography (the foreground is so bright, the lens has to be stopped down to take the picture, thus not allowing the light from the stars to register on the film); and it is not actually possible to duplicate the 1/6th gravity by simply slowing down the playback speed.  Oh...and let's not forget the 100s of pounds of lunar rocks returned from several missions.  Here's a link to Mythbusters: &lt;a href=&quot;http://www.discovery.com/tv-shows/mythbusters/mythbusters-database/apollo-moon-landing-pictures-fake/&quot; rel=&quot;noreferrer&quot;&gt;http://www.discovery.com/tv-shows/mythbusters/mythbusters-database/apollo-moon-landing-pictures-fake/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7108" LastEditorUserId="7108" LastEditDate="2015-06-17T22:34:54.097" LastActivityDate="2015-06-17T22:34:54.097" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11030" PostTypeId="2" ParentId="11026" CreationDate="2015-06-18T06:41:36.760" Score="4" Body="&lt;p&gt;It's just as simple as taking the flux at some wavelength (just a number) and using this number to represent a visible intensity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only have one wavelength then you can only get a monochrome picture. However, if you have flux information at more than one wavelength, let's say three, you can use the flux at the longest wavelength to represent red (r), the middle wavelength to represent green (g) and the shortest wavelength to represent blue (b).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put this together and you have 3 numbers representing a visible rgb signal that can be used to create a picture. Of course there may be lots of fiddling that goes on with the colour balance and the contrast to produce an effective picture, but this is the basic process.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-06-18T06:41:36.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11031" PostTypeId="1" AcceptedAnswerId="11033" CreationDate="2015-06-18T17:13:22.230" Score="4" ViewCount="368" Body="&lt;p&gt;I've been reading a little bit about coronal mass ejections lately and they seem really interesting, and they cause a lot of amazing things to happen. However, one thing I haven't seen is the physical force they can have on objects. CMEs seem to have a lot of mass and they're moving at a high rate of speed so I imagine they must have &lt;em&gt;some&lt;/em&gt; effect on objects in space, right? What kind of effect can they have on smaller objects like meteoroids, comets, or satellites (as far as orbit, rotation, anything else)?&lt;/p&gt;&#xA;" OwnerUserId="1980" LastActivityDate="2015-06-19T11:46:49.473" Title="What is the force of a CME on objects in space?" Tags="&lt;coronal-mass-ejection&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11032" PostTypeId="2" ParentId="687" CreationDate="2015-06-18T18:45:54.677" Score="5" Body="&lt;p&gt;We are &lt;a href=&quot;http://www.nature.com/nature/journal/v522/n7556/full/522290a.html&quot; rel=&quot;noreferrer&quot;&gt;now&lt;/a&gt; able to find smaller exoplanets than the Earth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Analysis of Kepler data has yielded the smallest known mass for an exoplanet orbiting a normal star. Its mass and size are similar to those of Mars, setting a benchmark for the properties of exoplanets smaller than Earth. See Letter p.321&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Read the article published in &lt;em&gt;Nature&lt;/em&gt; from &lt;a href=&quot;http://www.nature.com/nature/journal/v522/n7556/full/nature14494.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Earth is twice bigger than Mars.)&lt;/p&gt;&#xA;" OwnerUserId="6991" LastActivityDate="2015-06-18T18:45:54.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11033" PostTypeId="2" ParentId="11031" CreationDate="2015-06-18T20:27:15.263" Score="4" Body="&lt;p&gt;This isn't a complete answer, but it's a partial one:   &lt;a href=&quot;http://pwg.gsfc.nasa.gov/istp/outreach/sunearthmiscons.html&quot; rel=&quot;nofollow&quot;&gt;http://pwg.gsfc.nasa.gov/istp/outreach/sunearthmiscons.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Despite the electromagnetic havoc they play with satellite electronics&#xA;  and with Earth's magnetic field, the solar wind and CMEs barely exert&#xA;  a pressure or force that you could measure. In fact, they couldn't&#xA;  ruffle the hair on your head. The solar wind has fewer particles per&#xA;  cubic centimeter than the best vacuums scientists have ever created on&#xA;  Earth. Our own air is billions of times denser than the solar wind,&#xA;  such that a cubic centimeter of air has as many particles as a cube of&#xA;  solar wind measuring 10 kilometers on each side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, for your question on changing orbits - not very much.   Granted everything moves a little under any pressure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the lighter the object is to it's surface area, the more a CME can move it, that's why a solar sail, very light weight and large area, might work.    CMEs can also move gases, like the tail of a comet can be pushed visibly away from the sun (I read that, but misplaced the article), or lighter, high atmospheric gas can be blown away from smaller planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, no 2 CMEs are exactly alike.   A smaller and/or a younger star will emit much stronger CMEs, so around a red dwarf for example, the effect might be more noticeable for tiny asteroids and the like.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-06-19T07:22:51.203" LastActivityDate="2015-06-19T07:22:51.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11034" PostTypeId="1" CreationDate="2015-06-19T04:20:23.927" Score="1" ViewCount="126" Body="&lt;p&gt;On 2013 September 11th, a meteor impacted the Moon leaving a 34-meter crater. So far the only info I've found was from youtube (&lt;a href=&quot;https://www.youtube.com/watch?v=zCFDkj2JtyA&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=zCFDkj2JtyA&lt;/a&gt;) and this site: &lt;a href=&quot;http://lroc.sese.asu.edu/posts/810&quot; rel=&quot;nofollow&quot;&gt;http://lroc.sese.asu.edu/posts/810&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm surprised this doesn't have its own wikipedia article (none that I could find, anyway), and google wasn't helpful either. Is the astronomical world more or less oblivious to this event?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the data I'm most interested in is the meteor's orbital trajectory. And also, why would such a small meteor leave a &quot;fireball&quot; for 8 seconds according to the youtube video.&lt;/p&gt;&#xA;" OwnerUserId="7449" LastEditorUserId="7449" LastEditDate="2015-06-19T20:49:39.000" LastActivityDate="2015-08-25T08:30:04.870" Title="What was the orbit of the meteor that impacted the Moon on 2013 September 11?" Tags="&lt;the-moon&gt;&lt;meteor&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11035" PostTypeId="2" ParentId="11034" CreationDate="2015-06-19T09:23:39.310" Score="2" Body="&lt;p&gt;Out of curiosity I looked into this a bit.    North/South on the moon is pretty irrelevant for the objects orbit, but it struck at East 339 degrees (or West 21 degrees) near the Lubiniezky craters in the northwest part of Mare Nubium - which can be seen here - towards the center but a little bit in the south eastern part of the map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1wTro.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1wTro.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you can see the phase of the moon here. It changes slowly enough that a 24 hour range doesn't make a huge difference.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://lunaf.com/lunar-calendar/2013/09/11/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://lunaf.com/lunar-calendar/2013/09/11/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but apart from the crater looking roughly straight on, it's probably impossible to say the orbit it came from because the angle of approach is too big an unknown.   It hit from kind of both the earth side and sun side direction, so it could have been a temporarily captured asteroid around the earth or a small comet or asteroid coming back from around the sun.  Either could have made that impact I would think.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorDisplayName="user8024" LastEditDate="2015-08-25T08:30:04.870" LastActivityDate="2015-08-25T08:30:04.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11036" PostTypeId="2" ParentId="11031" CreationDate="2015-06-19T11:46:49.473" Score="4" Body="&lt;p&gt;&lt;a href=&quot;http://www.spaceweather.com/&quot; rel=&quot;nofollow&quot;&gt;A large CME&lt;/a&gt; will run 700km/sec at earth's distance from the sun.&#xA;Proton flux can get up to 20 protons per cubic cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a 1 square cm object facing the sun in earth orbit that amounts to being hit by a mass of 2.3e-18 kg traveling 700000 m/sec every second.&#xA;That works out to a force of 0.000001127 &lt;a href=&quot;https://en.wikipedia.org/wiki/Newton_%28unit%29&quot; rel=&quot;nofollow&quot;&gt;Newtons&lt;/a&gt; per second.&#xA;Not much push. (barring math errors of course, they &lt;em&gt;love&lt;/em&gt; to creep in)&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-06-19T11:46:49.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11037" PostTypeId="1" AcceptedAnswerId="11043" CreationDate="2015-06-19T16:09:47.593" Score="14" ViewCount="2923" Body="&lt;p&gt;The new &lt;a href=&quot;https://www.google.com/maps/&quot; rel=&quot;noreferrer&quot;&gt;Google Maps&lt;/a&gt;&lt;sup&gt;1&lt;/sup&gt; presents an actual view of Earth, with the current position of Sun illuminating half part of Earth in real time. It is quite an exquisite view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is based on the following image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qAayT.png&quot; alt=&quot;Earth&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, Sun's reflection from southern Atlantic Ocean looks very charming. But does it really happen like that? If we travel far&lt;sup&gt;2&lt;/sup&gt; from Earth, can we actually see Sun's reflection or it is just something Google added for aesthetics?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;: Yes, I don't like most of its new features too, especially when they &lt;em&gt;broke&lt;/em&gt; several of the old ones. But that's a separate discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2&lt;/sup&gt;: Although, the camera's supposed position would be close to geostationary orbit, I guess the reflection can also been seen as close as ISS or Hubble.&lt;/p&gt;&#xA;" OwnerUserId="698" LastActivityDate="2020-10-01T04:33:42.440" Title="Does Sun have a reflection on Earth?" Tags="&lt;orbit&gt;&lt;the-sun&gt;&lt;earth&gt;&lt;light&gt;" AnswerCount="7" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11038" PostTypeId="2" ParentId="891" CreationDate="2015-06-19T17:09:36.790" Score="1" Body="&lt;p&gt;5/2013 I was thinking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one googles the question - you get back earth answers: cheetah, cars, airplanes.  I wanted to know in the universe.  I was thinking about the &quot;thought experiment&quot; of a ladder moving at light speed that contracts enough to fit into a too small garage.  I reasoned if there is NO macro mass (like a ladder)  in the universe that approaches light velocity, what is the point of the ladder thought experiment?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I read about how fast NGC 1365 is spinning:  &quot;spinning so fast that its surface is traveling at nearly the speed of light.&quot; Press Release: 2013-07, 2/27/13 01:00:00 PM EST&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One generally reads it would take infinite energy to move mass to the speed of light.  I've reasoned this is why one generally hears about massless particles moving at the speed of light (photons and ?).  But now we have NGC 1365 spinning at nearly the speed of light, with its two numbers of mass, and spin.  I am not sure what &quot;nearly&quot; is - say 90% or ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though we are speaking about the spin speed, nonetheless at 2 million miles across, this NGC 1365 black hole with mass is assuredly the fastest velocity mass we know about in the universe, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I reason: article says &quot;Imagine a sphere more than 2 million miles across&quot; - this description is its diameter, D  = 2,000,000 miles or 3,218,688 km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The circumference of this object is Pi x D = 3.14 x 3,281,688 km = 10,106,680.32 km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interesting question is what is it like for an object that sits on the tangent to the circumference [&quot;circumference&quot; means &quot;Innermost Stable Circular Orbit&quot;, at the point in common with the line and the Innermost Stable Circular Orbit].  I lose track if this raises accuracy vs. precision?  With an object 2,000,000 miles across, is a &quot;point&quot; on its Innermost Stable Circular Orbit equal to a semi truck, a small car, a refrigerator, a book, marble, molecule, or atom???  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatever the size of the mass at this point, the tangent to the Innermost Stable Circular Orbit does describe an asymptote with a &quot;macro&quot; length.  The mass's movement along this asymptote does explain its velocity, straight line.  Hence this has got to be the highest velocity, straight line, not angular, mass we know of in the universe.  RIGHT???&#xA;Does speed angular vs. straight line matter (have effect) in a large object.  We are on a spinning planet and don't notice its speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks, JMc&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-06-19T17:09:36.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11042" PostTypeId="2" ParentId="11037" CreationDate="2015-06-20T05:45:42.390" Score="10" Body="&lt;p&gt;I think what this photo is actually capturing is the reflection of the sun off of the ocean's surface.  Were the sun over a landmass, I don't think this &quot;reflection&quot; would be seen.&lt;/p&gt;&#xA;" OwnerUserId="7108" LastActivityDate="2015-06-20T05:45:42.390" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11043" PostTypeId="2" ParentId="11037" CreationDate="2015-06-20T09:19:29.767" Score="16" Body="&lt;p&gt;It's called &lt;a href=&quot;https://earthobservatory.nasa.gov/glossary/s&quot; rel=&quot;nofollow noreferrer&quot;&gt;sunglint&lt;/a&gt;. This can be problematic for Earth-observing satellites in low Earth orbit. Such satellites typically don't take a &quot;picture&quot;. They instead continuously scan the Earth a line at a time. This means the sunglint moves with the satellite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see this effect yourself while flying in an airplane. Little ribbons of rivers and lakes can appear to be on fire. Here's a nice short &lt;a href=&quot;https://www.youtube.com/watch?v=VXtwkqTx-2o&quot; rel=&quot;nofollow noreferrer&quot;&gt;YouTube video&lt;/a&gt; that shows this perfectly.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="23114" LastEditDate="2019-08-20T01:12:12.577" LastActivityDate="2019-08-20T01:12:12.577" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11046" PostTypeId="2" ParentId="10886" CreationDate="2015-06-20T16:22:58.123" Score="2" Body="&lt;p&gt;In addition to the previous answer about Romily's explanation, the black hole is just a black hole at the center of the star system they're exploring. It's the remnants of the large star, not a wormhole. When they first come through the wormhole in the movie they're fairly far from the &quot;Gargantua&quot;. That's why it takes them some time to get to Miller's planet that is orbiting the black hole. Hope this helps. :)&lt;/p&gt;&#xA;" OwnerUserId="7505" LastActivityDate="2015-06-20T16:22:58.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11047" PostTypeId="1" AcceptedAnswerId="11062" CreationDate="2015-06-20T17:00:44.310" Score="2" ViewCount="413" Body="&lt;p&gt;I have a six inch telescope on an equatorial mount and I've been struggling to use the mount properly. For a couple months now, the only strategy I've used is just 'point and look.' I've watched some &lt;a href=&quot;https://www.youtube.com/watch?v=plx6XXDgf2E&quot; rel=&quot;nofollow&quot;&gt;youtube videos&lt;/a&gt; on polar aligning a telescope, but none of the videos ever tell me how I can use polar alignment to find other objects after I've aligned. Is polar alignment even the right method of finding objects? If so, how can I use the degrees and minutes on my RA and DEC knobs to point directly at an object?&lt;/p&gt;&#xA;" OwnerUserId="7132" LastActivityDate="2017-05-15T08:25:43.190" Title="How can I use Polar Alignment to find objects in the sky?" Tags="&lt;telescope&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11048" PostTypeId="1" CreationDate="2015-06-20T18:38:10.763" Score="3" ViewCount="62" Body="&lt;p&gt;In a book by Alexander Moszkowski, there is an Einstein quote about a hypothetical size of the universe (100 million Light years). Moszkowski claims that Einstein had deduced that from the gravitational constant and the mass distribution of the Milky Way, as known around 1920. My question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was the data and the best estimate of the mass density of the Milky Way, &lt;strong&gt;back then in 1920?&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="7507" LastActivityDate="2015-06-20T18:38:10.763" Title="Available data on the Milky way around 1920" Tags="&lt;cosmology&gt;&lt;milky-way&gt;&lt;history&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11049" PostTypeId="1" AcceptedAnswerId="11060" CreationDate="2015-06-20T18:50:07.960" Score="1" ViewCount="69" Body="&lt;p&gt;As I understand it, the distribution of matter is roughly homogeneous across space -- anywhere I stand in the universe, there will about the same amount of matter in every direction around me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the distribution of matter across &lt;em&gt;velocities&lt;/em&gt; roughly homogeneous?  If I flew away from the center of the Milky Way at .999c, would all directions look roughly the same, or would I notice a wind of galaxies and neutrinos traveling at a preferred velocity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tags I cannot add: cosmological principle, homogeneity, isotropy&lt;/p&gt;&#xA;" OwnerUserId="7506" LastActivityDate="2015-06-21T07:13:47.197" Title="Does the universe prefer a velocity?" Tags="&lt;cosmology&gt;&lt;velocity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11050" PostTypeId="1" AcceptedAnswerId="11066" CreationDate="2015-06-20T18:55:19.427" Score="8" ViewCount="350" Body="&lt;p&gt;A couple of decades ago I read an article (could have been either Astronomy, Space, or Science News) that suggested the contents of the universe may be older than the universe itself.  Of course, that result was so preposterous that it was summarily dismissed.  However, when I read that article, I thought it made sense.  Surely the universe is much more massive than its contents, so as General Relatively predicts, should experience time more slowly.  This conclusion seemed so obvious that a) a bunch of other folks much brighter than I would have thought of it, and b) therefore my conclusion is completely erroneous.  So my question is: why wouldn't the mass of the universe cause it to experience time at a rate slower than its contents as GR predicts?  (There are a couple of questions on SE that are similar, but I do not believe this is a duplicate.)&lt;/p&gt;&#xA;" OwnerUserId="7108" LastActivityDate="2018-01-16T20:19:59.900" Title="Age of the universe vs. its contents" Tags="&lt;universe&gt;&lt;time-dilation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11051" PostTypeId="1" AcceptedAnswerId="11052" CreationDate="2015-06-20T18:59:04.023" Score="2" ViewCount="279" Body="&lt;p&gt;I've been working on a programming project involving Peter Duffett-Smith's &lt;em&gt;Practical Astronomy With Your Calculator (3rd Ed.)&lt;/em&gt;, which has a large number of mathematical/astronomical formulas in it. There's a part at the end of the book for symbols and abbreviations, which contains this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QWUdk.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;After doing some Google-fu, I was unable to find the name of the symbol. Would anyone smarter than me (which should be quite a few of you... heck, quite a &lt;em&gt;lot&lt;/em&gt;) know what this is? Hopefully there's Unicode for it...&lt;/p&gt;&#xA;" OwnerUserId="5462" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-06-20T19:34:37.587" Title="Name and language of symbol in astronomy book" Tags="&lt;the-sun&gt;&lt;fundamental-astronomy&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11052" PostTypeId="2" ParentId="11051" CreationDate="2015-06-20T19:04:30.693" Score="2" Body="&lt;p&gt;This is &lt;a href=&quot;http://en.wikipedia.org/wiki/Pi_(letter)#Variant_pi&quot; rel=&quot;nofollow noreferrer&quot;&gt;a variant of pi&lt;/a&gt;, or &amp;quot;pomega&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;In LaTeX, you can get it by using&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;                                          \varpi&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\varpi$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://tex.stackexchange.com/questions/216358/using-mathematical-symbols-alpha-beta-implies-in-latex/216362#216362&quot;&gt;TeX&lt;/a&gt; has a question on the 'var' prefix, and &lt;a href=&quot;http://en.wikibooks.org/wiki/LaTeX/Mathematics#List_of_Mathematical_Symbols&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikibooks&lt;/a&gt; has something relevant.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The various forms of pi are present in Unicode as:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;U+03A0 Π greek capital letter pi (HTML Π · Π)&lt;/li&gt;&#xA;&lt;li&gt;U+03C0 π greek small letter pi (HTML π · π)&lt;/li&gt;&#xA;&lt;li&gt;U+03D6 ϖ greek pi symbol (HTML ϖ · ϖ)&lt;/li&gt;&#xA;&lt;li&gt;U+220F ∏ n-ary product (HTML ∏ · ∏)&lt;/li&gt;&#xA;&lt;li&gt;U+1D28 ᴨ greek letter small capital pi (HTML ᴨ)&lt;/li&gt;&#xA;&lt;li&gt;U+1D70B  mathematical italic small pi (HTML )&lt;/li&gt;&#xA;&lt;li&gt;U+1D6D1  mathematical bold small pi (HTML ).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It is used to denote the &lt;a href=&quot;http://en.wikipedia.org/wiki/Longitude_of_the_periapsis&quot; rel=&quot;nofollow noreferrer&quot;&gt;longitude of the periapsis&lt;/a&gt;, as in your case.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-06-20T19:09:55.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11053" PostTypeId="1" CreationDate="2015-06-20T20:26:43.070" Score="5" ViewCount="239" Body="&lt;p&gt;Is metallicity low at the central region or nucleus of the Milky Way? Does metallicity decrease or increase as we move from the center out to the edges of our galaxy?&lt;/p&gt;&#xA;" OwnerUserId="7509" LastEditorUserId="2153" LastEditDate="2015-06-20T21:18:30.167" LastActivityDate="2015-06-21T20:25:00.397" Title="Is metallicity low at the central region or nucleus of the Milky Way?" Tags="&lt;star&gt;&lt;galaxy&gt;&lt;milky-way&gt;&lt;galaxy-center&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11054" PostTypeId="1" AcceptedAnswerId="11064" CreationDate="2015-06-20T21:18:42.383" Score="4" ViewCount="297" Body="&lt;p&gt;How can you calculate the length of day for a planet (given a particular latitude and longitude on that planet) orbiting multiple stars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about a planet that is orbiting one star, but that star has another star orbiting it?&lt;/p&gt;&#xA;" OwnerUserId="7511" LastEditorUserId="7511" LastEditDate="2015-06-20T21:24:17.063" LastActivityDate="2015-06-22T15:08:35.900" Title="How do you calculate length of day in a multiple-star system?" Tags="&lt;solar-system&gt;&lt;positional-astronomy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11057" PostTypeId="1" CreationDate="2015-06-21T02:20:22.770" Score="2" ViewCount="665" Body="&lt;p&gt;What does the image in the link mean?  Are the stars in the halo high in metallicity or low?  How about at the bottom by the disk?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://science.psu.edu/alert/images/SDSSmetals.jpg/image_view_fullscreen&quot; rel=&quot;nofollow&quot;&gt;http://science.psu.edu/alert/images/SDSSmetals.jpg/image_view_fullscreen&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7509" LastActivityDate="2015-06-21T03:49:24.707" Title="Metal rich or metal poor?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;galaxy-center&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11058" PostTypeId="2" ParentId="11057" CreationDate="2015-06-21T03:38:19.687" Score="0" Body="&lt;p&gt;Red is higher blue lower metalicity, so the halo has low metalicity and the disk higer.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-06-21T03:38:19.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11059" PostTypeId="2" ParentId="11057" CreationDate="2015-06-21T03:42:11.423" Score="2" Body="&lt;p&gt;The ones in the halo  have blue colors, which is the &quot;old&quot; part of the spectrum.  Each type of &lt;a href=&quot;http://en.wikipedia.org/wiki/Metallicity&quot; rel=&quot;nofollow&quot;&gt;metallicity&lt;/a&gt; is usually measured using a form of logarithmic scale, since the flat percentage of metals (especially iron peak elements) of any star will be very small.  In this case, we have&#xA;$$[Fe/H] = \log(N_{\text{Fe}}/N_H)_{\text{star}} - \log(N_{\text{Fe}}/N_H)_{\text{sun}},$$&#xA;So we take the ratio of iron to hydrogen in the target star, measure it as well in our own sun, then take the ratio of those two and take the logarithm of that value.  Since the $\log$ function is monotonic and has $\log(1)=0$, a negative value of $[Fe/H]$ means our Sun has more iron in it (relative to the amount of hydrogen).  A value of $[Fe/H]=-1$ means our sun has an iron to hydrogen ratio that is 10 times greater than the given star's, for example.  The more negative this value, the more metal-poor the star is (or, at least, iron poor) than our own sun, which is considered of a fairly recent generation.  Lower metallicities are typical of earlier generations of stars, and so may be described as &quot;old&quot;.  Note, however, that they need not necessarily be old stars.  In principle it is possible for a recently formed star to have formed in an unusually metal-poor region, or otherwise been unable to accrete the normal quantity of metals.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastEditorUserId="2188" LastEditDate="2015-06-21T03:49:24.707" LastActivityDate="2015-06-21T03:49:24.707" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11060" PostTypeId="2" ParentId="11049" CreationDate="2015-06-21T07:01:12.487" Score="2" Body="&lt;p&gt;There is a preferred standard of rest, that of the Cosmic Microwave Background (CMB). You can detect motion with respect to the microwave background. So move fast with respect to the background and the CMB in the direction of motion will be blue-shifted and in the opposite direction will be red-shifted compared to a direction orthogonal to the direction of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth motion has been measured with respect to the CMB, &lt;a href=&quot;http://www.astro.ucla.edu/~wright/CMB-dipole-history.html&quot; rel=&quot;nofollow&quot;&gt;see here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could/can also detect your motion from the anisotopy of galactic red-shift in the same way.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-06-21T07:13:47.197" LastActivityDate="2015-06-21T07:13:47.197" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11061" PostTypeId="1" CreationDate="2015-06-21T07:27:49.063" Score="5" ViewCount="120" Body="&lt;p&gt;Comets are mostly icy but they should have some rocky components too, right? I wonder if there's any rocky piece of a former comet among collected meteorites. Is it possible today to identify a meteorite as having cometary origin? Or has surviving pieces from comet impacts been too rare (or violent vapourizing) events?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-07-01T12:06:10.343" LastActivityDate="2015-07-02T13:24:10.137" Title="Is any meteorite known to come from a comet?" Tags="&lt;comets&gt;&lt;meteorite&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11062" PostTypeId="2" ParentId="11047" CreationDate="2015-06-21T15:31:34.787" Score="4" Body="&lt;p&gt;If a mount is polar aligned, it becomes very easy to find objects that have the same Right Ascension or Declination as a bright star. It works like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) you align the mount to the pole&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) using a star map, or an app, you try to find an object that has the same R.A. or declination as a bright star. As an example, use Messier 4, which has almost the same declination as Antares. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) you point the telescope at Antares. Lock the declination axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Look through the finder, or through the scope using the eyepiece that gives you the biggest field of view. This most often this is the eyepiece that gives you the lowest magnification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) move the scope to the west, it rotates around the RA. axis. Messier 4 should now enter the field. Move slowly, and move back to Antares when you get lost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) When you see Messier 4, lock the RA. axis too, and center Messier 4 in the field using the fine adjustment controls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7) There are a couple of other bright objects you can try. I'm not familiar with the Southern sky, but for the Northern sky, Messier 13 can be found starting at eta Her(cules), and moving in declination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8) a bit harder is Messier 57 in Lyra, you need to start a bit to the north of gamma Lyr, or a bit to the south of beta. Another easy target is IC 4665, which is very close to Beta Ophiuchi (Cebalrai), almost but not quite to the north.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9) Other objects to try are Messier 5 starting with Beta Librae (Zubeneschamali), and Messier 10 starting at epsilon Ophiuch (Yed Posterior).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10) The next step is to try and find objects by using multiple steps, alternating between moving in RA and in declination. You would move to a star or better, to a recognizable star pattern, and continue from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11) This way of star hopping works best if the angles moved are not huge. Think 10, 15 degrees. As you get more experienced you will start using the stars in between as waypoints so you can move bigger distances.&lt;/p&gt;&#xA;" OwnerUserId="7514" LastActivityDate="2015-06-21T15:31:34.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11063" PostTypeId="2" ParentId="2521" CreationDate="2015-06-21T18:20:58.467" Score="5" Body="&lt;p&gt;Just to clarify a concept here: there are no pictures of the Milky Way other than the ones taken from &lt;em&gt;within the Solar system&lt;/em&gt;, and they all look somewhat like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/W1Jgf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything else you see are artists' depictions of the Milky Way based on the latest research. This is the currently most accepted shape:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gFwNY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.: a spiral galaxy with four main arms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an ongoing discussion about whether our Galaxy is made up of four arms or just two. See for example: &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014MNRAS.442.2993V&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vallée (2014)&lt;/a&gt;, which sadly is not available online :(&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2015-06-21T18:20:58.467" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11064" PostTypeId="2" ParentId="11054" CreationDate="2015-06-21T18:48:51.223" Score="3" Body="&lt;p&gt;Are you asking how to measure a day for an exoplanet?  Because that is pretty hard given that they are far away and hard to image their surfaces.  But, if you are asking how to define the day, say for simulations, then that is pretty easy, actually.&#xA;From Wikipedi's entry on Day.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;For a given planet, there are three types of day defined in astronomy:&lt;/p&gt;&#xA;&lt;p&gt;stellar day - an entire rotation of a planet with respect to the distant stars&lt;/p&gt;&#xA;&lt;p&gt;sidereal day - a single rotation of a planet with respect to the vernal equinox&lt;/p&gt;&#xA;&lt;p&gt;mean solar day - average time of a single rotation of a planet with respect to the sun as the central star&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The first two have to do with the spin of the planet with respect to the remote stars and therefore the complexity of the stellar system does not enter, to first approximation.  The 'stellar day' depends on how much time it takes for some remote galaxy, for instance, to come back to the same point in the sky as seen by an observer on the planet.  The 'sidereal day' depends on how much time it takes for the direction that is at the extension of the equator of the planet with the plane of the orbit (pick 1 of the 2 points) to come back to the same point in the sky. So presuming that the spin axis and orbital plane are not changing much (and conservation of angular momentum law helps here), this is all well defined.  If the planet's orbital plane is changing rapidly then there simply is no well defined sidereal day, but still there would be a valid stellar day.&lt;/p&gt;&#xA;&lt;p&gt;The 'mean solar day' is the average time it takes for the nearest star to return to the same spot in the sky.  Generally, even in a binary or triple system, the planet is close enough to one that this is well defined.  This is because, if  not, the planet is probably not in a stable orbit and will quickly leave the system.  However, if the stars are in a tight configuration  and the planet's orbit is wide and far enough away from them, it could be stable.  Then the simplest definition for 'mean solar day' would use the center of mass of the stellar system instead of the nearest star (which would change over time).&lt;/p&gt;&#xA;&lt;p&gt;There are a few stable complicated orbits, like a figure 8 orbit, where the planet can switch back and forth between two stars.  Unlikely, but theoretically possible.  In such a case, the mean solar day is ill-defined, but the stellar and sidereal day are well defined.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-06-22T15:08:35.900" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11065" PostTypeId="2" ParentId="11053" CreationDate="2015-06-21T20:25:00.397" Score="5" Body="&lt;p&gt;The stars in the Galactic bulge are predominantly metal-rich (by that I mean have a metallicity similar to the Sun or even a little higher).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though these stars are predominantly old, the bulge is thought to have formed extremely quickly and the interstellar medium from which the stars were formed would have been enriched with metals very quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a plot from &lt;a href=&quot;https://www.eso.org/sci/publications/messenger/archive/no.136-jun09/messenger-no136-48-53.pdf&quot; rel=&quot;noreferrer&quot;&gt;Zoccali et al. (2009)&lt;/a&gt; (which I recommend reading). It shows the metallicity distribution of many stars in the Galactic bulge, measured using high-resolution spectroscopy. It shows that the highest metallicity stars are towards the middle (it is very hard to get samples right in the middle because of extinction) and the averge metallicity falls as you mover further from the centre (the samples with more negative Galactic latitude $b$.). The percentages in the plots are the estimated contamination from the disk population in each sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LdoPw.png&quot; alt=&quot;Metallicity of the Galactic bulge from Zoccali et al. (2009).&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dependence of metallicity on height and radial coordinate in the Galactic disk is still keenly debated. There is general consensus that the metallicity falls with radial distance from the Galactic centre and with height above the Galactic plane. The gradient is of order a few hundredths of a dex per kpc between a few kpc and 10-12 kpc from the Galactic centre. It may flatten beyond this. You could have a look at the discussion in &lt;a href=&quot;http://iopscience.iop.org/0004-637X/746/2/149&quot; rel=&quot;noreferrer&quot;&gt;Cheng et al. (2012)&lt;/a&gt;, but there are many other attempts at parameterising the gradient and it is a technically difficult thing to do.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-06-21T20:25:00.397" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11066" PostTypeId="2" ParentId="11050" CreationDate="2015-06-22T06:55:17.087" Score="6" Body="&lt;p&gt;&lt;strong&gt;Models are uncertain&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing that you are referring to age estimates of very old stars. The age of a star can be determined from certain observables, among one is its absolute luminosity, which in turn requires a precise measurement of its distance. Even today, the uncertainties involved yield estimates with mean values longer than the age of the Universe, which of course is impossible. For instance, the age of the star &lt;a href=&quot;https://en.wikipedia.org/wiki/HD_140283&quot; rel=&quot;noreferrer&quot;&gt;HD 140283&lt;/a&gt; was found as late as in 2013 to be $14.46\pm0.8$ billion years (Gyr). This is not really in conflict with the estimated age of the Universe, &lt;a href=&quot;https://en.wikipedia.org/?title=Age_of_the_universe&quot; rel=&quot;noreferrer&quot;&gt;$13.798\pm0.037$ Gyr&lt;/a&gt;; all this tells us is that our methods are still not perfect, but that HD 140283 was formed shortly after the Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Time dilation is small&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General relativity predicts that time evolves more slowly in a graviational field. This is with respect to an observer outside the gravitational field, and since there is no external observer wrt. the Universe, the global time of the Universe can be said to be the &quot;right&quot; time. You're right that in principle the total mass of the Universe makes time run slower than in a hypothetical, empty universe, but almost anywhere in the Universe, the gravitational field is way too weak to cause any noticeable dilation. Close to massive objects, time &lt;em&gt;does&lt;/em&gt; run slow. On the surface of a star, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_time_dilation&quot; rel=&quot;noreferrer&quot;&gt;time dilation factor&lt;/a&gt; is roughly $(1 - GM/rc^2)^{-1/2} = 1.000002$. Inside the star, the factor is even closer to unity, because only the mass between the center and you contributes, so the farther you dive into the star (kids, don't try this at home), the &lt;em&gt;less&lt;/em&gt; the time dilation. Only if you compress the star, so as to squezze it into a neutron star of a black hole, will you get a larger time dilation. For instance, at the &quot;surface&quot; of a black hole, time stops (but again, only wrt. an external observer; for the person at the surface, time evolves as expected)&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2015-06-22T06:55:17.087" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11067" PostTypeId="1" CreationDate="2015-06-22T07:27:32.657" Score="1" ViewCount="128" Body="&lt;p&gt;I was looking at Arcturus with a pair of binoculars, and noticed that when they vibrate due to my heartbeat, Arcturus seemed to move differently than the other dimmer stars.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started playing with that visual effect and noticed that it happens both vertically and horizontally, and that I could even make it &quot;overlap&quot; with a nearby star.  It is like it either lags or is ahead of the others in terms of how it moves in the field of view as the binoculars do their slight shake.  Why is that?&lt;/p&gt;&#xA;" OwnerUserId="7515" LastEditorUserId="2365" LastEditDate="2015-11-15T13:16:10.073" LastActivityDate="2015-11-15T13:16:10.073" Title="Why some stars appear to move differently when viewing device shakes" Tags="&lt;amateur-observing&gt;&lt;binoculars&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11069" PostTypeId="2" ParentId="11037" CreationDate="2015-06-22T08:06:27.660" Score="2" Body="&lt;p&gt;Most probably not! But that being said, it also depends on &quot;&lt;em&gt;how far&lt;/em&gt;&quot; you have travelled above the surface. The earth is surrounded by a multitude of different things at different levels from the ground, like clouds, smoke, satellites and satellite-debris etc. which we come across progressively on the outward journey. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At lower levels we may get a somewhat clearer view of the reflection, which typically would not be a complete spherical reflection but a bunch of rays/ glints, or a blurred bright light cloud. But as we progress further, most probably this view would be blocked out by the above mentioned factors. And in case of clear-skies too, a spherical reflection is highly unlikely considering the distance between the sun and the earth. &lt;/p&gt;&#xA;" OwnerUserId="7517" LastActivityDate="2015-06-22T08:06:27.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11070" PostTypeId="1" CreationDate="2015-06-22T13:06:54.533" Score="14" ViewCount="33616" Body="&lt;p&gt;what is our Solar System called rather than &quot;the Solar System&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found &quot;Sol System&quot; and &quot;Monmatia&quot;, too, but is there more to it?&lt;/p&gt;&#xA;" OwnerUserId="7518" LastEditorUserId="2153" LastEditDate="2015-06-22T15:02:40.333" LastActivityDate="2015-12-27T19:03:49.983" Title="What is the name of our Solar System?" Tags="&lt;solar-system&gt;&lt;naming&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11071" PostTypeId="2" ParentId="11070" CreationDate="2015-06-23T00:21:26.110" Score="12" Body="&lt;p&gt;It's just called &quot;the Solar System&quot;. (Plenty of places and objects have names like that; it's no different from &quot;the Arctic&quot; or &quot;the Moon&quot; or &quot;the Sun&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&quot;Sol system&quot; is an invention of science fiction writers; it has no general use outside some science fiction contexts. Anything else is going to be something similar, or general crackpottery of one kind or another.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2015-06-23T08:59:20.943" LastActivityDate="2015-06-23T08:59:20.943" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11072" PostTypeId="2" ParentId="11037" CreationDate="2015-06-23T01:05:43.287" Score="15" Body="&lt;p&gt;Yes.  A web search for &quot;photo sun glint space&quot; turns up a number of images, including this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ujSzR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I really wanted to just say &quot;Yes.&quot; with the photo, but it wouldn't accept it.)&lt;/p&gt;&#xA;" OwnerUserId="53" LastActivityDate="2015-06-23T01:05:43.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11073" PostTypeId="2" ParentId="842" CreationDate="2015-06-23T05:35:18.730" Score="4" Body="&lt;p&gt;You can absolutely see it in rural areas. I have a gorgeous view of it on clear nights, living thirty miles west of Lincoln.&lt;/p&gt;&#xA;" OwnerUserId="7523" LastActivityDate="2015-06-23T05:35:18.730" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11074" PostTypeId="2" ParentId="11070" CreationDate="2015-06-23T14:00:19.543" Score="2" Body="&lt;p&gt;To answer your question succinctly, the Solar System also goes by the names: The Copernican System, The Heliocentric System, and The Planetary System, in addition to the ones you have mentioned. There aren't too many other names, actually, so just stick to Solar System since it's the most widely accepted.&lt;/p&gt;&#xA;" OwnerUserId="7332" LastActivityDate="2015-06-23T14:00:19.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11075" PostTypeId="2" ParentId="11070" CreationDate="2015-06-23T16:30:46.920" Score="7" Body="&lt;p&gt;The solar system was named long before we knew that other solar systems existed. Just like the Sun is not the only Sun in the universe but we still call it the Sun.&lt;/p&gt;&#xA;" OwnerUserId="7528" LastActivityDate="2015-06-23T16:30:46.920" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11076" PostTypeId="1" AcceptedAnswerId="11078" CreationDate="2015-06-23T16:41:17.217" Score="7" ViewCount="3141" Body="&lt;p&gt;If photon is mass-less and gravity can interact only with matter, then how  does gravity alter the trajectory of light?&lt;/p&gt;&#xA;" OwnerUserId="7528" LastEditorUserId="38" LastEditDate="2015-06-24T14:55:49.347" LastActivityDate="2015-09-25T19:32:07.457" Title="How does gravity interact with a photon?" Tags="&lt;gravity&gt;&lt;light&gt;&lt;gravitational-lensing&gt;&lt;matter&gt;&lt;photons&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11077" PostTypeId="2" ParentId="11076" CreationDate="2015-06-23T17:14:23.907" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If photon is mass-less and gravity can interact only with matter, then how gravity can alter the trajectory of light?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second part of your &quot;if&quot; clause is incorrect. It's mass-energy, not just mass, that gravitates.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-06-23T17:14:23.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11078" PostTypeId="2" ParentId="11076" CreationDate="2015-06-23T18:05:55.797" Score="8" Body="&lt;p&gt;It is simply not true that gravity can only interact with mass. Rather, any long-range spin-2 force interacts with all energy-momentum equally, and it source is the stress-energy-momentum tensor. That is one way to state the equivalence principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that a massive object in its own rest frame has an associated energy $E = mc^2$, which under ordinary conditions is usually much higher than any stress terms (including pressure) or momentum (outside the rest frame). Thus one can usually pretend that gravity couples to mass, but it isn't so--rather, the gravitational charge is energy, and the entire stress-energy-momentum tensor couples to the gravitational field. This is analogous to how for electromagnetism, there is electric charge, but the electric currents also make a difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the weak-field limit of general relativity, one can consider the fact that light is gravitationally deflected twice as much as a naive Newtonian prediction would be (for an object at the speed of light under Newtonian mechanics, that is) as the fault of light putting a pressure in the direction of its propagation equal to its energy density. But in general, the motion of a test particle is determined by its four-velocity and the geometry of spacetime &lt;em&gt;only&lt;/em&gt;, and the light would just be a special case of having a 'lightlike' four-velocity.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2015-06-23T18:05:55.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11079" PostTypeId="1" CreationDate="2015-06-23T19:20:21.243" Score="3" ViewCount="926" Body="&lt;p&gt;If I have understood correctly, stars form in big clouds of gas and dust that are pulled together by their gravity. And the stars are often ignited when something disturbs the cloud, such as a passing massive body, or a wave like the once which create spirals in galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any phenomena on earth that behave like these clouds, which could be used to illustrate and explain star formation? Something people might be familiar with and which you could show a video clip of?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking about some gas that is disturbed then takes solid form? Or something completely different which looks similar?&lt;/p&gt;&#xA;" OwnerUserId="7529" LastEditorUserId="7529" LastEditDate="2015-06-23T19:39:25.520" LastActivityDate="2016-11-14T14:36:23.993" Title="Star formation analogy" Tags="&lt;star-formation&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11080" PostTypeId="2" ParentId="11079" CreationDate="2015-06-23T20:19:20.083" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If I have understood correctly, stars form in big clouds of gas and dust that are pulled together by their gravity. And the stars are often ignited when something disturbs the cloud, such as a passing massive body, or a wave like the once which create spirals in galaxies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You have the sequence backwards. Protostars ignite by themselves once their cores reach a critical temperature and pressure. The gravitational collapse that eventually results in protostars (and then a stars) oftentimes needs a little help from an external disturbance such as a passing star or a supernova shock.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there any phenomena on earth that behave like these clouds, which could be used to illustrate and explain star formation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not really. The interstellar gas clouds that eventually collapse to form stars are very large, very massive, very cold, and very tenuous. The best way to understand them is to understand them directly rather than looking for a poor analogy. There's nothing on Earth like them.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2015-06-23T20:26:56.837" LastActivityDate="2015-06-23T20:26:56.837" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11082" PostTypeId="1" AcceptedAnswerId="24398" CreationDate="2015-06-24T13:59:54.900" Score="14" ViewCount="2405" Body="&lt;p&gt;Stars plotted by luminosity and surface temperature fit into patterns in a &lt;a href=&quot;https://en.wikipedia.org/wiki/Hertzsprung%E2%80%93Russell_diagram&quot; rel=&quot;noreferrer&quot;&gt;Hertzsprung–Russell&lt;/a&gt; diagram.  A rough diagonal subset of this plot is called the main sequence.  Is this in any sense a temporal sequence?  There's a clue in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hertzsprung%E2%80%93Russell_diagram#Diagram.27s_role_in_the_development_of_stellar_physics&quot; rel=&quot;noreferrer&quot;&gt;stellar physics&lt;/a&gt; section of the wikipedia article that the answer is no, but that it once was thought so:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Contemplation of the diagram led astronomers to speculate that it might demonstrate stellar evolution, the main suggestion being that stars collapsed from red giants to dwarf stars, then moving down along the line of the main sequence in the course of their lifetimes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So does the word &quot;sequence&quot; in this case now mean just a particular ordering, and not a progression over time that any one star makes?  Is the main sequence just some kind of plateau in stellar evolution where stars spend a significant duration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CGzow.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CGzowl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="71" LastEditDate="2018-01-20T17:53:51.210" LastActivityDate="2018-01-20T17:53:51.210" Title="Is &quot;Main Sequence&quot; a temporal sequence?" Tags="&lt;stellar-evolution&gt;&lt;main-sequence&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11084" PostTypeId="1" CreationDate="2015-06-24T14:27:23.810" Score="3" ViewCount="495" Body="&lt;p&gt;Pretty much everything in our solar system has collisions with smaller objects that lead to an increase in the mass of the larger.  With Pluto being one of the largest things in the area of which it resides, and being surrounded by small stuff. Assuming that it does not &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1518&quot;&gt;hit Neptune, which seems unlikely&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How long will it take Pluto to grow to &lt;a href=&quot;https://en.wikipedia.org/wiki/Definition_of_planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;planet size?&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clarify&lt;/strong&gt; or what ever the name of the largest object in the Kuiper belt turns out to be, at the time it reaches the &lt;a href=&quot;https://en.wikipedia.org/wiki/IAU_definition_of_planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;IAU definition of planet&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="513" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-11-15T22:33:11.833" Title="How long will it take Pluto to grow to planet size?" Tags="&lt;planet&gt;&lt;pluto&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" ClosedDate="2016-11-15T22:33:22.733" ContentLicense="CC BY-SA 3.0" />
  <row Id="11085" PostTypeId="2" ParentId="19069" CreationDate="2015-06-24T14:39:59.743" Score="14" Body="&lt;p&gt;Pluto will never be a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of technical papers that give more precise meaning to the concept of &quot;clearing the neighborhood&quot;. It's not just now, it's can the object in question clear the neighborhood of its path while the Sun is still a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of Pluto, Ceres, Eris, and a host of other not-quite-planet objects, that will not happen. Ever. Or at least so long as the Sun shines. After that, does it matter?&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="6" LastEditDate="2016-11-15T22:38:52.830" LastActivityDate="2016-11-15T22:38:52.830" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11086" PostTypeId="2" ParentId="11082" CreationDate="2015-06-24T15:50:09.073" Score="8" Body="&lt;p&gt;The main sequence is &lt;em&gt;mostly&lt;/em&gt; a plateau that a star reaches after it is fully formed but before it begins to run low on hydrogen to fuel normal fusion reactions. And yes, the sequence is &lt;em&gt;mostly&lt;/em&gt; an ordering - by mass, not age. I say mostly because age does have some effect (see the section from the Wikipedia article on the main sequence regarding &lt;a href=&quot;https://en.wikipedia.org/wiki/Main_sequence#Luminosity-color&quot; rel=&quot;nofollow noreferrer&quot;&gt;temperature-luminosity variation&lt;/a&gt;). The result is that older stars a slightly hotter and brighter than younger stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For most stars, much of the light it emits is &lt;a href=&quot;https://en.wikipedia.org/wiki/Black-body_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;black-body radiation&lt;/a&gt;. The amount of energy that a star produces is complicated (as explained on the page for the &lt;a href=&quot;https://en.wikipedia.org/wiki/Mass%E2%80%93luminosity_relation&quot; rel=&quot;nofollow noreferrer&quot;&gt;mass-luminosity relation&lt;/a&gt;) but the bottom line is that for stars with a larger mass, the energy output increases significantly relative to its surface area and thus is hotter. The page on black-body radiation has a nice &lt;a href=&quot;https://en.wikipedia.org/wiki/Black-body_radiation#Explanation&quot; rel=&quot;nofollow noreferrer&quot;&gt;explanation&lt;/a&gt; including a temperature graphic showing how the surface temperature of smaller stars red and as mass increases, orange, yellow, green, and blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The higher fusion rate (relative to size) explains why larger stars run out of hydrogen faster than smaller stars.&lt;/p&gt;&#xA;" OwnerUserId="38" LastEditorUserId="38" LastEditDate="2018-01-08T14:28:12.797" LastActivityDate="2018-01-08T14:28:12.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11087" PostTypeId="2" ParentId="11082" CreationDate="2015-06-24T16:07:26.630" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this in any sense a temporal sequence?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not really. At least not in the sense of a star sliding along the main sequence. That doesn't happen. Instead, a star remains more or less parked at one spot on the main sequence during it's lifetime as a main sequence star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A protostar is more luminous and cooler than the zero age main sequence star it will become. Once a star &quot;ignites&quot; (starts fusing hydrogen (not deuterium)) is when a star enters the main sequence. This is where the star spends the majority of its life. In the case of small stars, stars whose mass is less than about 40% of the Sun's mass, this is where the star will spend the entirety of its life as a star. Small stars get dimmer and dimmer as they age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Larger stars don't thoroughly mix from the innermost core to the outermost regions.  These larger stars build up an ash of helium as they age. This hydrogen fusion eventually comes to an end when all of the hydrogen in the core has been fused into helium. That's when the star departs the main sequence. Unlike small stars, larger stars get brighter (more luminous) as they age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Larger stars (stars larger than 40% solar masses) might double or triple in luminosity as they age. That's a third to a half order of magnitude increase, and that's tiny compared to the eleven or twelve orders of magnitude difference in luminosity shown on an HR diagram between the smallest red dwarf and the largest blue giant. What this means is that once a non-tiny star enters the main sequence, it more or less stays at that spot on the main sequence until it leaves the main sequence.  &lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-06-24T16:07:26.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11088" PostTypeId="2" ParentId="10631" CreationDate="2015-06-24T17:43:55.803" Score="2" Body="&lt;p&gt;You've put in the moon filter, right? The moon is awfully bright without it. In any case, take the scope out in the daytime, and try focusing on distant objects; the tree at the end of the block, or the radio tower on top of the hill at the edge of town. A scope is easier to learn how to use when you can see what you're doing. &lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-06-24T17:43:55.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11089" PostTypeId="2" ParentId="4794" CreationDate="2015-06-24T18:21:18.857" Score="3" Body="&lt;p&gt;The answer to your question for brane cosmology is simply that gravity follows the inverse-square law. We don't feel the gravitational pull of, say, Mars because it's so far away. Any other universes are so unimaginably (all right, that's hyperbole) far away in comparison that any gravitational effects by the matter in them would be very, very, very, tiny.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar argument might be to say that an infinite series always goes to infinity because you keep adding on terms. But this is often not the case, as with the series&#xA;$$\sum_{n=0}^{\infty}2^{-n}=2$$&#xA;It is infinite, but it converges.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-06-25T20:32:20.340" LastActivityDate="2015-06-25T20:32:20.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11090" PostTypeId="1" AcceptedAnswerId="11091" CreationDate="2015-06-25T07:25:56.793" Score="12" ViewCount="342" Body="&lt;p&gt;We know that the Andromeda Galaxy is moving towards the Milky Way and vice versa, so there will be a collision. Because of the vast space between them, the  stars as they are defined will not themselves come into contact until much later if at any time at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is whether Andromeda has had a collision prior to this 4 billion year inevitability.&lt;/p&gt;&#xA;" OwnerUserId="2811" LastEditorUserId="2811" LastEditDate="2016-03-05T21:34:28.933" LastActivityDate="2016-03-05T21:34:28.933" Title="Has the great Andromeda Galaxy ever collided with any galaxies?" Tags="&lt;galaxy&gt;&lt;galactic-dynamics&gt;&lt;m31&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11091" PostTypeId="2" ParentId="11090" CreationDate="2015-06-25T09:13:58.930" Score="14" Body="&lt;p&gt;Galaxies grow through cosmic time by accretion of the surrounding matter. Some of its mass increase happens through smooth accretion of gas, but much also happens through merging with small clumps of dark matter, gas, and stars, called satellite galaxies. This is called &quot;minor merging&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If merging galaxies are similar in size, it's called major merging. Most large galaxies have gone through of the order of one major merger through its life (&lt;a href=&quot;http://arxiv.org/abs/1410.3479&quot;&gt;Man et la. 2014&lt;/a&gt;), but whether or not it happened for Andromeda, we cannot know for certain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you take a look at Andromeda, you'll see that it has quite a large &lt;em&gt;bulge&lt;/em&gt;, i.e. the central, reddish part of the galaxy. This region is dominated by stars with a more &quot;chaotic&quot; velocity patterns. That is, their paths aren't &quot;rotation-dominated&quot; (lie in the disk), but more &quot;dispersion-dominated&quot; (leave the disk). To achieve such velocities, the system has to be disturbed by infalling matter, so a large bulge is probably a signature of a major merger. Investigating the velocity patterns of Andromeda, &lt;a href=&quot;http://arxiv.org/abs/1502.03820&quot;&gt;Dorman et al. (2015)&lt;/a&gt; find that it has had a much more violent history than the Milky Way, and in fact models by &lt;a href=&quot;http://arxiv.org/abs/1203.6081&quot;&gt;Davidge et al. (2012)&lt;/a&gt; and simulations by &lt;a href=&quot;http://arxiv.org/abs/1010.0679&quot;&gt;Hammer et al. (2010)&lt;/a&gt; of the history of Andromeda suggest that &lt;strong&gt;8-9 billion years ago, Andromeda went through a major merger&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note however that a large bulge &lt;em&gt;can&lt;/em&gt; also be explained, at least in cosmological simulations, by a more smooth and &quot;cold&quot; accretion, i.e. gas that doesn't get shock-heated (Dekel et al.)&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-06-26T11:25:28.253" LastActivityDate="2015-06-26T11:25:28.253" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11093" PostTypeId="1" CreationDate="2015-06-25T19:16:01.367" Score="6" ViewCount="147" Body="&lt;p&gt;From our point of view, no object can fall into a black hole in finite time. The object will seem to be more and more red-shifted to us, but it will never quite pass the schwarzschild-radius / event horizon. (Of course the time frame of the falling object is another thing, but we are well away from any direct effects of black holes on us)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, how can black holes come to existence? As soon as there would be a schwarzschild radius, nothing can cross it. To me this seems like a chicken-and-egg problem, so no black holes can possibly exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously the black holes in our universe do not seem to be aware of this, so what am I missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance :-)&lt;/p&gt;&#xA;" OwnerUserId="7548" LastActivityDate="2015-06-25T19:16:01.367" Title="If nothing can pass the schwarzschild radius, how can black holes even start?" Tags="&lt;black-hole&gt;" AnswerCount="0" CommentCount="4" FavoriteCount="3" ClosedDate="2015-06-26T17:00:03.750" ContentLicense="CC BY-SA 3.0" />
  <row Id="11094" PostTypeId="2" ParentId="10631" CreationDate="2015-06-26T00:55:28.403" Score="2" Body="&lt;p&gt;Your telescope is not focused (most likely), or is having some major collimation issues (less likely).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try and move the eyepiece back and forth a little. Go through the whole range of the focuser. You must catch the primary focal plane in order for the image to become clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that doesn't work, pull the eyepiece out a few mm and try to move through the whole range again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that doesn't work, try a different eyepiece.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The &quot;advice&quot; regarding the Moon filter is bogus. Be careful regarding what you hear on the Internet. The eye is certainly capable of adapting to various levels of brightness. Even with very large telescopes, the Moon filter is useless. All it does is reduce the brightness and contrast - the former is no big deal, but the latter is a major loss. You can never have enough contrast in a telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the Moon is visible, there's no point in going through deep dark adaptation anyway, because the &quot;faint fuzzies&quot; are obscured by Moon's glare. When I watch the Moon, I do it from a fully illumined backyard, or even on the street under the street lights. In fact, this is the best way to observe this object - no filters, but have some ambient light around you. Your eyes will function at optimal parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deep dark adaptation is only needed when observing faint nebulae and galaxies. But such objects can be observed in good conditions only when the Moon is below horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me make it clear: &lt;strong&gt;a filter, any filter, will not fix the problem you're having&lt;/strong&gt;, because it does not address the root cause.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beware of most cases &lt;strong&gt;when the advice you receive is &quot;use a filter&quot;. In 99% of cases, it's bogus.&lt;/strong&gt; Many people own filters, but only a very small fraction know how to use them. In the vast, vast majority of cases you don't need any filters. Please steer clear of this superstition. It's a fad that vendors are more than happy to feed, because it makes them money.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are legitimate ways to use filters (in the rare cases when they're justified), but that would be the subject of a different discussion - those rare cases are related to certain techniques of increasing the apparent contrast in the image. A neutral density filter (a.k.a. &quot;moon filter&quot;) always decreases apparent contrast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this hobby, like in most other hobbies, a lot of people fixate on purchasing all sorts of accessories (in this case: filters), hoping to get extra performance, when all they need to do in reality is learn how to use the device / machine / etc (in this case: telescope) correctly. I see this tendency in all my other hobbies. It's unfortunate, and a huge money sink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Save the money and use it instead towards purchasing a better telescope, or better eyepieces, or a good sky atlas, or an observing chair, etc.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-06-26T01:12:41.817" LastActivityDate="2015-06-26T01:12:41.817" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11095" PostTypeId="2" ParentId="4794" CreationDate="2015-06-26T05:29:14.027" Score="3" Body="&lt;p&gt;An exact answer would require a rather specific and mathematical formulation of the multiverse in consideration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a simple approximating example, suppose we have a countably infinite number of (observable) universes of the same mass $M$.  Suppose the dimension of the full multiverse is one higher than each individual universe, and suppose the universes are all separated by the same minimum distance $\epsilon&amp;gt;0$ from each other.  In a 2-D picture, this would just look like a bunch of parallel lines all separated by the same distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick your home universe and put an observer.  Another universe of distance $n\epsilon$ away (meaning they're $n$ universes up or down from you in the 2-D picture) exerts a gravitational force on the observer in its direction approximately proportional to $\displaystyle{\frac{M}{n^2\epsilon^2}}.$  With the right units, we can just say &quot;approximately&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum gravitational force occurs when the observer is at the bottom (or top) of the picture, and all over universes are above it (or below): universes on each side will pull in opposing directions and so lead to cancellations.  So the net gravitational force from the other universes (in the right units) is at most &#xA;$$\sum_{n=1}^\infty \frac{M}{n^2\epsilon^2} = \frac{M}{\epsilon^2}\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2 M}{6\epsilon^2}&amp;lt;\infty.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the observer was &quot;in the middle&quot;&amp;mdash;infinitely many universes above and below, with the distribution identical in either direction&amp;mdash; the net gravitational force from the other universes is exactly 0.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2015-06-26T05:29:14.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11096" PostTypeId="1" CreationDate="2015-06-26T06:34:13.587" Score="7" ViewCount="722" Body="&lt;p&gt;I would like to check in which constellation the Sun was in at some time in the past, any recommendations how to do that?&lt;/p&gt;&#xA;" OwnerUserId="7549" LastActivityDate="2016-05-02T05:59:02.207" Title="How/where to check where the Sun is (constellation)?" Tags="&lt;positional-astronomy&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11097" PostTypeId="2" ParentId="11096" CreationDate="2015-06-26T08:31:42.470" Score="1" Body="&lt;p&gt;There are plenty of online planetarium sites which should be able to do this for you. &lt;a href=&quot;http://neave.com/planetarium/&quot; rel=&quot;nofollow&quot;&gt;This one&lt;/a&gt; is the first hit for a Google search for &quot;online planetarium software&quot;&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-06-26T08:31:42.470" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11099" PostTypeId="1" AcceptedAnswerId="11101" CreationDate="2015-06-26T19:35:18.383" Score="6" ViewCount="622" Body="&lt;p&gt;Is the Hubble Space Telescope, and every other space telescope for that matter, completely outside the Earths atmosphere? &lt;/p&gt;&#xA;" OwnerUserId="7467" LastEditorUserId="2531" LastEditDate="2015-07-02T11:00:54.697" LastActivityDate="2015-07-02T11:00:54.697" Title="Are space telescopes completely out of the earth's atmosphere?" Tags="&lt;hubble-telescope&gt;&lt;space-telescope&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11100" PostTypeId="2" ParentId="11099" CreationDate="2015-06-26T20:16:12.250" Score="8" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_space_telescopes&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt; contains a list of space telescopes. It's likely to be nearly complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extent of the Earth's atmosphere is not very well defined. The altitude at which Hubble orbits (about 550 kilometers above the surface) is above &lt;em&gt;almost&lt;/em&gt; all of the atmosphere, but there's still enough residual air to cause some slight drag. It's not higher because it was deployed from the Space Shuttle, which couldn't carry it much higher that its current orbit. It was also useful to be able to fly servicing missions.&lt;/p&gt;&#xA;" OwnerUserId="53" LastEditorUserId="1437" LastEditDate="2015-07-02T09:52:26.587" LastActivityDate="2015-07-02T09:52:26.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11101" PostTypeId="2" ParentId="11099" CreationDate="2015-06-26T21:47:49.940" Score="10" Body="&lt;p&gt;A handful of space telescopes are located in Langrange point L2, 1.5 million km from Earth. This is much farther away than the Moon, and far outside Earth's atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/WMAP&quot; rel=&quot;noreferrer&quot;&gt;WMAP&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Planck_(spacecraft)&quot; rel=&quot;noreferrer&quot;&gt;Planck&lt;/a&gt;, which measure the cosmic microwave background (CMB), are located here because Earth is a hundred times brighter than the CMB in this wavelength region. &lt;a href=&quot;https://en.wikipedia.org/wiki/Herschel_Space_Observatory&quot; rel=&quot;noreferrer&quot;&gt;Herschel&lt;/a&gt; observes in the infrared, which is heat radiation, so it was also put in L2 to avoid Earth's heat. Others can be found in the list provided in Keith's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another advantage of putting a telescope in L2 is that the Earth and Sun are always in the same direction, making it easier to observe most of the sky. The cost of putting it here is probably more than an order of magnitude higher, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether or not low Earth orbit telescopes like Hubble at 550 km are outside Earth's atmosphere is a bit subjective. The atmosphere doesn't stop at some particular point, instead gradually thinning. These space telescope are definitely far enough away that they are not heated up. Neither does the atmosphere affect their observations (by &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_seeing&quot; rel=&quot;noreferrer&quot;&gt;seeing&lt;/a&gt;). They &lt;em&gt;are&lt;/em&gt;, however, slowly decelerated and thus need a boost once in a while not to fall down.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="6980" LastEditDate="2015-06-27T20:11:22.980" LastActivityDate="2015-06-27T20:11:22.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11102" PostTypeId="1" CreationDate="2015-06-27T08:04:55.997" Score="2" ViewCount="965" Body="&lt;p&gt;Is there a set of standard abbreviations for the days of the week for use in astronomy? I couldn't find any reference to it in the &lt;a href=&quot;http://www.iau.org/static/publications/stylemanual1989.pdf&quot; rel=&quot;nofollow&quot;&gt;The IAU Style Manual (1989)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5388" LastActivityDate="2017-04-17T00:01:30.460" Title="Abbreviations for days of the week" Tags="&lt;standards&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11103" PostTypeId="1" CreationDate="2015-06-27T13:36:42.350" Score="3" ViewCount="103" Body="&lt;p&gt;I have practiced astronomy as a hobby for a while now. I have an entry-level telescope (6-inch Newtonian). In addition to observing the night sky, I have been studying physics related to optics, astronomical distance &amp;amp; temperature measurement, gravity, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My plan is to next get a CCD camera, and do some astrophotography. It should keep me busy the next winter. However, I feel I should try to find some kind of project where I could gradually make progress. Otherwise, after learning all the basics, and seeing everything that can be seen with the smallish telescope I have, I might gradually start to lose interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would you recommend as the &quot;next stage&quot;?&lt;/p&gt;&#xA;" OwnerUserId="7557" LastActivityDate="2015-06-28T09:48:58.950" Title="What to do after first year of amateur astronomy?" Tags="&lt;amateur-observing&gt;&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-06-29T16:49:54.123" ContentLicense="CC BY-SA 3.0" />
  <row Id="11104" PostTypeId="1" AcceptedAnswerId="11105" CreationDate="2015-06-27T15:52:47.233" Score="3" ViewCount="126" Body="&lt;p&gt;A paper appeared at the arXiv.org (in April or May) claimed that the writers have found the gamma rays emitted by annihilation of dark matter in a satellite galaxy of the Milky Way by examining the Fermi data.  I cannot find this article anymore.  (It is possible that the writers withdrew the article.) Has someone actually detected gamma rays emitted by neutralino (or any other dark matter) annihilation?&lt;/p&gt;&#xA;" OwnerUserId="7558" LastActivityDate="2015-06-28T01:19:35.903" Title="Dark Matter Detection" Tags="&lt;dark-matter&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11105" PostTypeId="2" ParentId="11104" CreationDate="2015-06-27T22:17:23.803" Score="4" Body="&lt;p&gt;Is it &lt;a href=&quot;http://arxiv.org/abs/1503.02320&quot; rel=&quot;nofollow noreferrer&quot;&gt;the paper&lt;/a&gt; &lt;a href=&quot;https://astronomy.stackexchange.com/questions/11104/dark-matter-detection#comment15027_11104&quot;&gt;David H pointed out&lt;/a&gt;, by Geringer-Sameth et al. (2015)? If so, then I should point out that they're not the first. Just using &lt;a href=&quot;http://en.wikipedia.org/wiki/Dark_matter#Indirect_detection_experiments&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, I came across &lt;a href=&quot;http://arxiv.org/abs/1210.3013&quot; rel=&quot;nofollow noreferrer&quot;&gt;Weniger (2012)&lt;/a&gt; and &lt;a href=&quot;http://iopscience.iop.org/0004-637X/679/1/428/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Albert et al. (2008)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geringer-Sameth et al.'s abstract reads (in part)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We present a search for gamma-ray emission from the direction of the newly discovered dwarf galaxy Reticulum 2. . . . Reticulum 2 has the most significant gamma-ray signal of any known dwarf galaxy. If Reticulum 2 has a dark matter halo that is similar to those inferred for other nearby dwarfs, the signal is consistent with the s-wave relic abundance cross section for annihilation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Albert et al. also studied a dwarf galaxy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The nearby dwarf spheroidal galaxy Draco, with its high mass to light ratio, is one of the most auspicious targets for indirect dark matter (DM) searches. Annihilation of hypothetical DM particles can result in high-energy γ-rays, e.g., from neutralino annihilation in the supersymmetric framework. A search for a possible DM signal originating from Draco was performed with the MAGIC telescope during 2007.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The point is, this has been done before. These folks aren't the first.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Has someone actually detected gamma rays emitted by neutralino (or any other dark matter) annihilation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More analysis is probably needed. I'll quote the paper itself:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;While Ret2's X-ray signal is tantalizing, it would be premature to conclude it has a dark matter origin. Among alternative explanations, perhaps the most&#xA;  mundane is the possibility that an extragalactic source lies in the same direction. Searching the BZCAT [54] and CRATES [55] catalogs reveals a CRATES quasar (J033553-543026) that is 0:46 from Ret2. Further work must be done to determine whether this particular source contributes to the emission, though we note that at spectrum radio quasars rarely have a spectral index less than 2 [56]. One of the much-discussed astrophysical explanations for the apparent Galactic Center excess is millisecond pulsars [24, 26, 57{61]. In the case of Ret2, it is the high-energy behavior which disfavors a pulsar model, as millisecond pulsars exhibit an exponential cut-offat around 2.5 to 4 GeV [26, 30, 61{64]. Alternatively, high-energy cosmic ray production could potentially arise in the vicinity of young massive stars. Upcoming photo-metric and spectroscopic analysis of Ret2 will check this possibility.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't jump to conclusions. They didn't. It stinks, but the answer to your question is, &quot;Nobody's sure.&quot;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-06-28T01:19:35.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11106" PostTypeId="1" CreationDate="2015-06-28T09:31:42.807" Score="3" ViewCount="124" Body="&lt;p&gt;Of the exoplanets found so far, is there one that can exchange material with its host? How could we prove that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Io might be a good example, but it is not a planet.&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="2153" LastEditDate="2016-10-04T12:52:35.953" LastActivityDate="2016-11-03T17:44:40.387" Title="Material exchange between a planet and its host" Tags="&lt;planet&gt;&lt;exoplanet&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11107" PostTypeId="2" ParentId="11103" CreationDate="2015-06-28T09:48:58.950" Score="1" Body="&lt;p&gt;Astrophotography is always a good step to take after just observation. That way, you can see things through your telescope that you cannot see with the naked eye,  and also share your findings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing you could do is start on a different part of astronomy, maybe getting a hydrogen alpha filter or a solar telescope to observe the Sun, which again could be combined with the astrophotography.&lt;/p&gt;&#xA;" OwnerUserId="1494" LastActivityDate="2015-06-28T09:48:58.950" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11109" PostTypeId="1" AcceptedAnswerId="11111" CreationDate="2015-06-28T15:48:21.510" Score="4" ViewCount="762" Body="&lt;p&gt;I'm a software developer and I'm developing a planetarium and the first thing that I don't know is: What is the shape of the sky?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I though that it is a semi-sphere, but searching on Internet I have found a lot of youtube tutorials, and all of them draw a semi-sphere but flatten (you can search skydome on Google and find a lot of tutorials).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how can I represent sky on a virtual 3D world? How much do I need to flat a sphere to get a real sky representation?&lt;/p&gt;&#xA;" OwnerUserId="7566" LastEditorUserId="7449" LastEditDate="2015-06-29T10:01:07.767" LastActivityDate="2017-02-18T22:33:55.267" Title="Which is the shape of the sky?" Tags="&lt;fundamental-astronomy&gt;&lt;coordinate&gt;&lt;software&gt;&lt;declination&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11110" PostTypeId="1" CreationDate="2015-06-29T01:10:33.690" Score="6" ViewCount="2512" Body="&lt;p&gt;I've found various sources on the web stating that the limiting apparent magnitude of the Hubble Space Telescope is about 31 or 31.5. Wikipedia lists that &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_magnitude#Table_of_notable_celestial_objects&quot;&gt;here&lt;/a&gt; and it has a cite to a pdf about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to know is &lt;strong&gt;how&lt;/strong&gt; is that calculated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware of two methods found at the bottom of a table &lt;a href=&quot;http://www.twcac.org/Tutorials/limiting_magnitude_table.htm&quot;&gt;here&lt;/a&gt;, but they seems to be general calculations for ground-based telescopes, because if I plug in Hubble's numbers, I get something around 21. The difference between an apparent magnitude of 21 and 31 is actually something like ten thousands times (in terms of brightness).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, one key factor missing in their formulas is &lt;strong&gt;exposure time.&lt;/strong&gt; I'm pretty sure the limiting magnitude has something to do with exposure time. Hubble Ultra Deep Field, for example, had an exposure time of about 1 million seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to sum up, how would we calculate the limiting magnitude of a space-based telescope, in visible light?&lt;/p&gt;&#xA;" OwnerUserId="7449" LastActivityDate="2015-07-02T12:07:52.470" Title="How to calculate the limiting magnitude of Hubble?" Tags="&lt;light&gt;&lt;space-telescope&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11111" PostTypeId="2" ParentId="11109" CreationDate="2015-06-29T04:18:08.683" Score="5" Body="&lt;p&gt;For computer software, the easiest way to take a sphere (and/or hemisphere) and flatten it into a flat shape (usually a rectangle) is the &lt;strong&gt;equi-rectangular projection&lt;/strong&gt; (also known as the plate carrée), because it has the simplest formula relating pixels and coordinates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = w \times \frac\lambda{360} + \frac w2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y = -h \times \frac\phi{180} + \frac h2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x$ and $y$ is the pixel point, $w$ and $h$ are the width and height (in pixels) of your map's rectangle, $\lambda$ is the longitude in degrees and $\phi$ is the latitude in degrees&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This all assumes that your map puts the 0 longitude, 0 latitude point in the center. In other words, the prime meridian runs down the map's middle vertically and the equator runs down the map's middle horizontally. It also assumes that north and east are positive, while west and south are negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that you don't even have to take the sine or cosine of anything. That's why the equi-rectangular projection is so simple. It basically just takes longitude and latitude and makes them $x$/$y$ coordinate with only an offset. The offset comes because, usually in computer programming, the &lt;strong&gt;origin is at the top left&lt;/strong&gt; instead of the center of the screen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice also that I negated the latitude in the $y$ formula, because most computer systems in 2D coordinates have the &lt;strong&gt;positive $\bf y$ direction pointing down.&lt;/strong&gt; This is opposite of a standard graph in math, where positive $y$ goes up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, longitude and latitude will actually be right ascension and declination respectively. They are basically stellar coordinates for the celestial sphere as opposed to ground coordinates for the geode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many, many other &lt;a href=&quot;https://en.wikipedia.org/wiki/Map_projection&quot; rel=&quot;nofollow noreferrer&quot;&gt;map projections&lt;/a&gt; possible, but equi-rectangular is the simplest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt;, your question is a little vague because you mentioned a &quot;virtual 3D world&quot;, yet also asked how much you need to flatten it. If you already have a 3D model of a sphere in your computer graphics, you &lt;strong&gt;don't have to&lt;/strong&gt; flatten it at all. You just draw the 3D model (sphere) with a texture on it. If you want to represent it as a map, then of course you have to flatten it, and equi-rectangular projection is the simplest way.&lt;/p&gt;&#xA;" OwnerUserId="7449" LastEditorUserId="9301" LastEditDate="2017-02-18T22:33:55.267" LastActivityDate="2017-02-18T22:33:55.267" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11112" PostTypeId="1" CreationDate="2015-06-29T04:45:24.280" Score="2" ViewCount="2396" Body="&lt;p&gt;What day will the summer solstice and full moon coincide? As far as I know the next time will be in 2016. I would like to know what day. &lt;/p&gt;&#xA;" OwnerUserId="7567" LastActivityDate="2020-06-20T07:02:50.327" Title="When will the next summer solstice and a full moon coincide? What day?" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;the-sun&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11113" PostTypeId="2" ParentId="8300" CreationDate="2015-06-29T07:35:07.490" Score="0" Body="&lt;p&gt;The chances of what you are suggesting are almost nil. Though we might be constantly scanning our surrounding atmosphere to detect fragments of, or actual meteorites, other than the debris and artificial satellites that we ourselves have put in orbit around the earth, there is no second moon currently in orbit around us to the best of our knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is, however, a theory that proposes the presence of another moon in the past. As per this theory, there was a planet &quot;Theia&quot; immediately close to earth that underwent a close collision with Earth to form two moons, which ultimately coalesced to form the single moon we see today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.space.com/12529-earth-2-moons-collision-moon-formation.html&quot; rel=&quot;nofollow&quot;&gt;http://www.space.com/12529-earth-2-moons-collision-moon-formation.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://news.discovery.com/space/two-moons-earth-110804.htm&quot; rel=&quot;nofollow&quot;&gt;http://news.discovery.com/space/two-moons-earth-110804.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.telegraph.co.uk/news/science/space/10165036/Earth-had-two-moons-scientists-claim.html&quot; rel=&quot;nofollow&quot;&gt;http://www.telegraph.co.uk/news/science/space/10165036/Earth-had-two-moons-scientists-claim.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7517" LastActivityDate="2015-06-29T07:35:07.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11114" PostTypeId="2" ParentId="10017" CreationDate="2015-06-29T08:06:42.203" Score="5" Body="&lt;p&gt;No. The sun does not revolve around another big star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It revolves around the center of our galaxy along with the whole solar system, including comets, asteroids, and a large amount of other stars and stellar systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As per some theories, however, at the center of our Milky Way galaxy lies a Super Massive black hole, which was essentially once a huge star, at least more than 10 times the mass of our sun, that had collapsed into itself due to its enormous gravity forming the black hole. So based on these theories, it wouldn't be technically incorrect to say that the sun revolves around a 'dead' star.&lt;/p&gt;&#xA;" OwnerUserId="7517" LastActivityDate="2015-06-29T08:06:42.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11115" PostTypeId="2" ParentId="11079" CreationDate="2015-06-29T10:23:54.293" Score="3" Body="&lt;p&gt;To begin with, star-formation occurs when relatively stable and huge clouds of gases and dust are disturbed by external disturbances such as shock waves from a supernova explosion. This brings some material closer when gravity kicks in, bringing together more of the gases, which accumulate and grow dense to pull in more material.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process continues till an enormous amount of material has been accumulated, making it dense at the core i.e. the &quot;center&quot; of the gigantic gas cloud, as well as raising the temperatures and pressures there. This is when the star &quot;ignites&quot;, meaning that a fusion reaction kicks in, releases the outward-energy required to balance the inner-crushing gravity of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the magnitude of the pressures, temperatures, size of the gas clouds etc. involved here are huge, it's impossible to find such phenomena on the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can view the episode &quot;Extreme Stars&quot; from the series How the Universe Works (Season 1 Episode 4) that were aired on The Discovery Channel for a detailed visual understanding of the Star formation process.  &lt;/p&gt;&#xA;" OwnerUserId="7517" LastActivityDate="2015-06-29T10:23:54.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11116" PostTypeId="1" CreationDate="2015-06-29T01:40:33.550" Score="1" ViewCount="200" Body="&lt;p&gt;Seems like it is the most convenient SE community for this question.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine the sphere, witch center is the Sun and radius is, say, ~1.5 distance from Sun to earth (in arbitrary time, precise number doesn't matter much).&#xA;&lt;code&gt;EDIT&lt;/code&gt;: the center of sphere is 'attached' to the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, imagine, the Earth's axis projection on that sphere in time. It leaves some trace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What will be the picture? Will it repeat itself through some period? What will be the &quot;scansion&quot; for different time periods?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;EDIT&lt;/code&gt;: as reply to comments I'll add, that my interest is purely out of curiosity, and degree of required accuracy is low. I don't want high precision here, just rough pattern. As I understand this could be done, using mentioned &lt;code&gt;Milankovitch cycles&lt;/code&gt;, but I'm not much into mathematics. Though, if I'll not find answer here, I will be forced (by myself) to figure this out, when i'll get some spare time heh.&lt;/p&gt;&#xA;" OwnerUserId="7568" OwnerDisplayName="zloidooraque" LastActivityDate="2015-07-29T16:30:49.393" Title="What is the projection of Earth's axis on the sphere with the Sun in center (see explanation)" Tags="&lt;earth&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11117" PostTypeId="2" ParentId="11112" CreationDate="2015-06-29T10:45:20.457" Score="1" Body="&lt;p&gt;The Summer Solstice in 2016 will occur at 22:34 UT on the 20th June, and the full moon will occur at 11:02 UT on that day (but since they are 12 hours apart that won't be on the same day everywhere).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solstice data from &lt;a href=&quot;http://wwp.greenwichmeantime.com/longest-day/equinox-solstice-2010-2019.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; (GMT- Grenwich Mean Time =UT)&#xA;&lt;img src=&quot;https://i.stack.imgur.com/7095m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;New and full moon data from &lt;a href=&quot;http://www.moongiant.com/Full_Moon_New_Moon_Calendar.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; (UTC- Coordinated Universal Time =GMT=UT)&#xA;&lt;img src=&quot;https://i.stack.imgur.com/yDFey.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-06-29T10:45:20.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11118" PostTypeId="1" CreationDate="2015-06-29T11:45:52.370" Score="8" ViewCount="387" Body="&lt;p&gt;The diameter of my observable universe is 90 billion ly measured in proper distance. But isn't lenght a relative quantity in the theory of Relativity? Could an observer moving at a different velocity than me measure a different diameter for my observable Universe?&lt;/p&gt;&#xA;" OwnerUserId="6920" LastActivityDate="2015-08-05T15:28:26.800" Title="Is the diameter of the observable universe a relative quantity?" Tags="&lt;universe&gt;&lt;cosmology&gt;&lt;distances&gt;&lt;general-relativity&gt;&lt;observable-universe&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11119" PostTypeId="2" ParentId="11116" CreationDate="2015-06-29T15:54:59.000" Score="1" Body="&lt;p&gt;It will be two images of the nearly-circular ellipse of the Earth's orbit (one for the north pole, one for the south), distorted by being projected onto the interior of a sphere, which means that the side of each orbit that has the axis tilted &quot;towards&quot; it will appear enlarged and closer to the ecliptic, and the side of each orbit that has the axis tilted &quot;away&quot; from it will be smaller and closer to the pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To a first approximation the pattern will retrace itself every year. The Earth's orbit is stable with respect to the Sun and the fixed stars. Over a period of tens of thousands to hundreds of thousands of years, the Earth's orbit will &lt;a href=&quot;https://en.wikipedia.org/wiki/Apsidal_precession&quot; rel=&quot;nofollow&quot;&gt;precess&lt;/a&gt; (causing the figures to rotate around the sphere), and &lt;a href=&quot;https://en.wikipedia.org/wiki/Nutation&quot; rel=&quot;nofollow&quot;&gt;nutation&lt;/a&gt; will also cause small wiggles at a much higher frequency.&lt;/p&gt;&#xA;" OwnerUserId="4198" LastActivityDate="2015-06-29T15:54:59.000" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11120" PostTypeId="1" CreationDate="2015-06-29T18:26:06.860" Score="2" ViewCount="330" Body="&lt;p&gt;If an instrument like LSST quickly spots something as soon as it reaches magnitude 24&amp;half;, how far away would a 60km first-time visitor from the Oort cloud be when it is spotted? I'm thinking that a huge iceball will make a visible coma sooner, but have no idea what actual numbers would be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, given that distance and a parabolic orbit, how much time would that give before it crosses Earth's orbit?&lt;/p&gt;&#xA;" OwnerUserId="7142" LastActivityDate="2015-07-09T09:13:32.793" Title="How far away would a supernucleus comet be visible?" Tags="&lt;orbit&gt;&lt;comets&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11121" PostTypeId="1" AcceptedAnswerId="11122" CreationDate="2015-06-29T19:14:19.423" Score="3" ViewCount="848" Body="&lt;p&gt;I know that a &lt;strong&gt;day&lt;/strong&gt; represent the time for the earth to make a full rotation around its own axis and a &lt;strong&gt;year&lt;/strong&gt; is the time for the earth to make a full rotation around the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;strong&gt;does a week represent something?&lt;/strong&gt; And a month?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ps: Not sure what tags I need to use, feel free to add/remove them&lt;/p&gt;&#xA;" OwnerUserId="7571" LastActivityDate="2015-06-30T13:16:17.757" Title="Does a week represent something in astronomy?" Tags="&lt;earth&gt;&lt;fundamental-astronomy&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2015-06-30T11:14:27.390" ContentLicense="CC BY-SA 3.0" />
  <row Id="11122" PostTypeId="2" ParentId="11121" CreationDate="2015-06-29T20:35:39.120" Score="5" Body="&lt;p&gt;The concept of a month derived from the amount of time it takes for the moon to cycle from new moon to new moon (which is roughly 29.5 days). The modern month has experienced changes from this original concept due to trying to fit a standard number of months within a solar year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you divide the lunar month into quarters, each quarter is approximately 7 days long. This is believed to be the origin of the seven-day week.&lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorUserId="6" LastEditDate="2015-06-30T13:16:17.757" LastActivityDate="2015-06-30T13:16:17.757" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11123" PostTypeId="2" ParentId="11121" CreationDate="2015-06-30T10:21:48.947" Score="0" Body="&lt;p&gt;I'm pretty sure that weeks are seven days long because in ancient times people worshipped the sun, moon and the five then-known planets.  The days are even named after these in various languages...&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2015-06-30T10:21:48.947" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11124" PostTypeId="2" ParentId="11118" CreationDate="2015-06-30T14:49:13.217" Score="2" Body="&lt;p&gt;If the observer has always been moving at a high speed, then yes, s/he will measure a different value than you. But if s/he accelerated to a high speed just now, you would obtain the same result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is that the radius of the observable Universe is defined as the distance light has had the time to travel since the Big Bang, and that does not depend on your instantaneous velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if an observer started moving at, say, 0.866 times the speed of light when the Universe was born, then in the reference frame of that observer, the Universe would only be $\sqrt{1 - 0.866^2} = $ 1/2 the age measured by you, i.e. 6.9 Gyr. At this age, the radius of the observable Universe was only ~39 Gly, i.e. ~0.8 times the present value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in order for this result to be exact, the observer would have to slowly accelerate. The reason is that in an expanding Universe, a given &quot;peculiar&quot; velocity wrt. the surrounding matter will slowly decrease, because the comoving coordinate system expands with the Universe. For instance, an observer starting out with $v = 0.866c$ at the time the cosmic microwave background was released 380,000 yr after the Big Bang, would only be moving at 80 km/s today if s/he didn't keep accelerating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the observer were clever, s/he would probably realize that s/he did't live in an asymmetric universe where all matter seemed to be whirling by at close to the speed of light, but rather that the Universe were quite isotropic, but s/he had somehow acquired a high speed with repect to everything else. It would then be possible to deduce the &quot;true&quot; age of the Universe.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-07-30T21:25:43.343" LastActivityDate="2015-07-30T21:25:43.343" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11125" PostTypeId="2" ParentId="11096" CreationDate="2015-06-30T15:35:40.090" Score="1" Body="&lt;ul&gt;&#xA;&lt;li&gt;If you're talking about short time periods and don't need a lot of detail, a lookup table such as this one: &lt;a href=&quot;http://spaceplace.nasa.gov/starfinder3/en/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://spaceplace.nasa.gov/starfinder3/en/&lt;/a&gt; will work just fine.&lt;/li&gt;&#xA;&lt;li&gt;For more detail, you'll have to calculate the position of the Sun on the background stars and determine that moment when the position crosses the (somewhat arbitrarily defined) constellation boundaries. Wikipedia has good introductory equations to determine the Sun's position (&lt;a href=&quot;https://en.wikipedia.org/wiki/Position_of_the_Sun&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Position_of_the_Sun&lt;/a&gt;) though David Colarusso's site has much more information: &lt;a href=&quot;http://www.davidcolarusso.com/astro/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.davidcolarusso.com/astro/&lt;/a&gt;. There are other equations you can use to determine which side of the line the Sun's position lies: &lt;a href=&quot;https://stackoverflow.com/questions/1560492/how-to-tell-whether-a-point-is-to-the-right-or-left-side-of-a-line&quot;&gt;https://stackoverflow.com/questions/1560492/how-to-tell-whether-a-point-is-to-the-right-or-left-side-of-a-line&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;For long time periods, you'll have to do what's in the previous point and also determine how the Earth's position changes in relation to the Sun and where in the galaxy the Sun lies. I'll leave that as an exercise for you since I'm not sure you want that level of detail.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7154" LastEditorUserId="-1" LastEditDate="2017-05-23T12:37:21.700" LastActivityDate="2015-06-30T15:35:40.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11126" PostTypeId="1" AcceptedAnswerId="11894" CreationDate="2015-06-30T19:00:29.543" Score="3" ViewCount="689" Body="&lt;p&gt;I need a stellar mass database that also has the distance of the star/system from the sun. Can anybody tell me the name of such of database and where I can find it. I need to do a project on this. Thank you for the answers.&lt;/p&gt;&#xA;" OwnerUserId="7580" LastActivityDate="2015-10-27T09:43:29.170" Title="Stellar mass database" Tags="&lt;star&gt;&lt;mass&gt;&lt;star-systems&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11127" PostTypeId="2" ParentId="10374" CreationDate="2015-06-30T19:32:52.180" Score="1" Body="&lt;p&gt;The mass of the observable universe can be derived from its density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Dr. Jagadheep D. Pandian:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The density of matter in the universe can be measured by various means, which are too technical to go into at this point: people measure the density by studying the fluctuations in the Cosmic Microwave Background, superclusters, Big Bang nucleosynthesis, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using the density and the size of the observable universe, the mass can be derived to be 3 x 10&lt;sup&gt;55&lt;/sup&gt; g. This figure includes both dark matter and traditional matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine that a historical deviation in mass might be detectable if it were significant, but I can't imagine what might cause a deviation that would be detectable on the scale we're talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://curious.astro.cornell.edu/our-solar-system/101-the-universe/cosmology-and-the-big-bang/general-questions/579-what-is-the-mass-of-the-universe-intermediate&quot; rel=&quot;nofollow&quot;&gt;What is the mass of the Universe? (Intermediate) - Cornell University, Ask an Astronomer&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6" LastActivityDate="2015-06-30T19:32:52.180" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11128" PostTypeId="1" AcceptedAnswerId="11133" CreationDate="2015-06-30T20:49:22.153" Score="10" ViewCount="300" Body="&lt;p&gt;How would we be able to detect Exomoons? We detect exoplanets by seeing if the light gets faint from a star, but could we really use the same method for determining if there is an exomoon around that exoplanet? Thanks for your thoughts and answers! &lt;/p&gt;&#xA;" OwnerUserId="7467" LastActivityDate="2016-10-06T03:56:56.757" Title="How to find Exomoons?" Tags="&lt;the-moon&gt;&lt;exoplanet&gt;&lt;kepler&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11129" PostTypeId="1" AcceptedAnswerId="11184" CreationDate="2015-06-30T21:14:04.990" Score="0" ViewCount="96" Body="&lt;p&gt;I've heard that we have a great shot at finding life on Mars if we just drill deep enough down to the surface to the water and detect if its sterile or not. Are there planned missions to do this? If not, what is the next (or existing) mission where finding life is likely? Is it launching something to Mars, Europa, Titan, or some other moon? Why does the mission have a good shot at finding life? &lt;/p&gt;&#xA;" OwnerUserId="36" LastEditorUserId="36" LastEditDate="2015-12-17T11:28:36.340" LastActivityDate="2015-12-17T11:28:36.340" Title="What is the next planned mission that can discover life on another planet?" Tags="&lt;mars&gt;&lt;astrobiology&gt;" AnswerCount="3" CommentCount="5" ClosedDate="2015-12-17T15:24:06.803" ContentLicense="CC BY-SA 3.0" />
  <row Id="11130" PostTypeId="1" AcceptedAnswerId="11131" CreationDate="2015-06-30T21:43:01.190" Score="1" ViewCount="2091" Body="&lt;p&gt;Is it &lt;a href=&quot;https://en.wikipedia.org/wiki/Rho_Cassiopeiae&quot; rel=&quot;nofollow&quot;&gt;Rho Cassiopeiae&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/Mu_Sagittarii&quot; rel=&quot;nofollow&quot;&gt;Mu Sagittarii&lt;/a&gt;?&lt;br&gt;&#xA;I see in the &lt;a href=&quot;https://launchpad.net/stellarium/+milestone/0.13.0&quot; rel=&quot;nofollow&quot;&gt;Stellarium 0.13.0&lt;/a&gt; that Mu Sagittarii has absolute magnitude -11.43!!&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2016-04-08T21:41:36.490" LastActivityDate="2016-04-08T21:41:36.490" Title="What is the brightest star (by absolute magnitude) that we can see by naked eye?" Tags="&lt;star&gt;&lt;magnitude&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11131" PostTypeId="2" ParentId="11130" CreationDate="2015-06-30T21:58:01.737" Score="5" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Mu_Sagittarii&quot; rel=&quot;noreferrer&quot;&gt;Mu Sagittarii&lt;/a&gt; is a star system, not a single star. If that can be included, then &lt;a href=&quot;https://en.wikipedia.org/wiki/Eta_Carinae&quot; rel=&quot;noreferrer&quot;&gt;Eta Carinae&lt;/a&gt; should be included, and it has an absolute magnitude of -12.0. It's a star system about 7,500 light-years from Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like the brightest (absolute magnitude) single star visible to the unaided eye is WR 24 (in Carina Nebula). Its absolute magnitude is −11.1 and apparent magnitude is 6.48, so just barely visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;source: &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_most_luminous_known_stars&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia - List of most luminous known stars&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: &lt;a href=&quot;https://en.wikipedia.org/wiki/Rho_Cassiopeiae#cite_note-gorlova-6&quot; rel=&quot;noreferrer&quot;&gt;Rho Cassiopeiae&lt;/a&gt;'s absolute magnitude is -9.5.&lt;/p&gt;&#xA;" OwnerUserId="1980" LastEditorUserId="1980" LastEditDate="2015-06-30T22:09:06.787" LastActivityDate="2015-06-30T22:09:06.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11133" PostTypeId="2" ParentId="11128" CreationDate="2015-07-01T00:18:26.300" Score="6" Body="&lt;p&gt;I'm going to try to take a stab at answering this. With our current technologies, detecting exomoons can prove hard however there are various techniques being used today such as:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Analyzing data from the Kepler Spacecraft&lt;/li&gt;&#xA;&lt;li&gt;Dynamic effects – the exomoon tugs the planet, which causes deviations in the times and durations of the host planet’s transits. This is similar to the radial velocity technique for detecting exoplanets. Source: &lt;a href=&quot;http://www.universetoday.com/103238/the-hunt-for-exomoons-begins/&quot;&gt;UniverseToday&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Transit effects – the exomoon may transit the star immediately before or just after the planet does. This will cause an added dip in the observed light. See this video for a great demonstration. This is similar to the light curve technique for detecting exoplanets. Source: &lt;a href=&quot;http://www.universetoday.com/103238/the-hunt-for-exomoons-begins/&quot;&gt;UniverseToday&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Gravitational Microlensing - which is a technique used to detect exoplanets like you stated above however it may also reveal signs of an exomoon. Read this &lt;a href=&quot;http://www.space.com/25438-exomoon-around-alien-planet-discovery.html&quot;&gt;source&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I found this information doing some simple googling, feel free to edit or add to.&lt;/p&gt;&#xA;" OwnerUserId="7258" LastActivityDate="2015-07-01T00:18:26.300" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11135" PostTypeId="2" ParentId="10374" CreationDate="2015-07-01T09:00:02.143" Score="4" Body="&lt;p&gt;Even if you're only referring the &quot;ordinary&quot; matter (such as stars, gas, and bicycles) and &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_matter&quot; rel=&quot;nofollow&quot;&gt;dark matter&lt;/a&gt;, the mass of the observable Universe does increase, not because mass is being created, but because the size of the observable Universe increases. In a billion years from now, we can see stuff that today is too far away for the light to have reached us, so its radius has increased. Since the mass $M$ equals density $\rho_\mathrm{M}$ times volume $V$, $M$ increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As called2voyage mentions, we have several ways of measuring the density, and we know it's close to $3\times10^{-30}\,\mathrm{g}\,\mathrm{cm}^{-3}$. The radius is $R = 4.6\times10^{28}\,\mathrm{cm}$, so the mass is&#xA;$$M = \rho_\mathrm{M} \frac{4\pi}{3}R^3 \simeq 10^{57}\,\mathrm{g},$$&#xA;or $5\times10^{23}M_\odot$ (Solar masses).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, another factor contributes to the mass increase, namely the so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_energy&quot; rel=&quot;nofollow&quot;&gt;dark energy&lt;/a&gt;, which is a form of energy attributed to empty space. Since the Universe expands, dark energy is being created all the time.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2015-07-01T09:00:02.143" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11136" PostTypeId="1" CreationDate="2015-07-01T12:13:18.480" Score="2" ViewCount="101" Body="&lt;p&gt;Meteor showers happen when Earth passes through the debris left by a comet in its orbit. Meteors are generally tiny, like snow flings or pebbles and don't make it to the ground as meteorites. But periodic comets sublimate and disintegrate. Don't they shed kilometer sized fragments too? Maybe it is not only the comet which is the threat, but its entire orbit is a minefield too? Are bolides and observed meteorite falls more frequent during meteor showers?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-02-14T11:32:32.373" Title="Are mass extinction events more likely during meteor showers / passing through comet debris?" Tags="&lt;comets&gt;&lt;meteor&gt;&lt;impact-probability&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11137" PostTypeId="1" AcceptedAnswerId="11138" CreationDate="2015-06-30T22:48:27.373" Score="11" ViewCount="956" Body="&lt;p&gt;How can we know that Halley's Comet is going to come again every 76 years? I understand that the comet has an orbit around the sun but how can scientists know that it won't be thrown off course for example by passing Jupiter and Jupiters gravity giving it a tug in a different direction? Its not a very big object in space so it seems to me that its orbit could be changed very easily.&lt;/p&gt;&#xA;" OwnerUserId="7586" OwnerDisplayName="Chad Watkins" LastEditorUserId="13" LastEditDate="2015-07-01T15:09:18.847" LastActivityDate="2019-04-23T15:28:33.653" Title="How can Halley's Comet be predictable?" Tags="&lt;gravity&gt;&lt;comets&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11138" PostTypeId="2" ParentId="11137" CreationDate="2015-06-30T23:27:11.757" Score="23" Body="&lt;p&gt;Each time Halley's comet passes us, we can make a pretty good estimate of its current orbit, and determine how close it will get to the massive bodies of the solar system like Jupiter, Saturn, Uranus, or Neptune on its next orbit. We can make good estimates of gravitational perturbation effects, and thereby know where to look for it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, &lt;a href=&quot;https://en.wikipedia.org/wiki/Halley&amp;#39;s_Comet#Computation_of_orbit&quot;&gt;astronomers have been doing exactly that&lt;/a&gt; for centuries with remarkable accuracy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Halley thus concluded that all three comets were, in fact, the same object returning every 76 years, a period that has since been amended to every 75–76 years. After a rough estimate of the perturbations the comet would sustain from the gravitational attraction of the planets, he predicted its return for 1758.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Halley's prediction of the comet's return proved to be correct, although it was not seen until 25 December 1758, by Johann Georg Palitzsch, a German farmer and amateur astronomer. It did not pass through its perihelion until 13 March 1759, the attraction of Jupiter and Saturn having caused a retardation of 618 days. This effect was computed prior to its return (with a one-month error to 13 April) by a team of three French mathematicians, Alexis Clairaut, Joseph Lalande, and Nicole-Reine Lepaute. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, Halley's position turns out to be hard to predict for time scales much beyond that of a single orbit; over the long term, &lt;a href=&quot;http://arxiv.org/abs/1409.7762&quot;&gt;the orbit is chaotically unstable&lt;/a&gt; and it's possible that it will collide with a planet, or be ejected from the solar system, within some hundreds of thousands of years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mention: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Its not a very big object in space so it seems to me that its orbit could be changed very easily.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As it turns out, the mass of an object in space doesn't affect how the gravity of other bodies changes its trajectory. A feather, a baseball, a comet, and a small moon making a close flyby of Jupiter along the same trajectory, one after another, will all exit Jupiter's neighborhood in the same direction and at the same speed. The small moon will have more effect on &lt;em&gt;Jupiter&lt;/em&gt; than the feather will, but aside from that there's no difference.&lt;/p&gt;&#xA;" OwnerUserId="2573" OwnerDisplayName="Russell Borogove" LastActivityDate="2015-07-01T02:26:04.847" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11139" PostTypeId="2" ParentId="11061" CreationDate="2015-07-01T16:20:58.230" Score="1" Body="&lt;p&gt;Well to quote the &lt;a href=&quot;https://en.wikipedia.org/wiki/Leonids&quot; rel=&quot;nofollow&quot;&gt;wikipedia page&lt;/a&gt; on the Leonids: &quot;The Leonids are a prolific meteor shower associated with the comet Tempel-Tuttle.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if a meteorite falls to Earth during the Leonids, it may have originated in Comet Tempel-Tuttle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, there appears to be no conclusive evidence that any particular meteorite came from any particular Comet. The problem is that we do not have direct samples of cometary material to compare against. See for example &lt;a href=&quot;https://www.google.co.uk/url?sa=t&amp;amp;source=web&amp;amp;rct=j&amp;amp;q=leonid%20meteorite&amp;amp;ei=4RCUVfKLIKPB7gafko7YAg&amp;amp;url=https://journals.uair.arizona.edu/index.php/maps/article/download/15038/15009&amp;amp;ved=0CDMQFjAHOCg&amp;amp;usg=AFQjCNHMnMQjYZLE3aB1CViyw7wpPmzuiw&amp;amp;sig2=2DArP_2AEJB9DBDjH0ZYAA&quot; rel=&quot;nofollow&quot;&gt;Swindles &amp;amp; Campin 2004&lt;/a&gt;, though somebody may know of something more recent?&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-07-01T16:40:08.053" LastActivityDate="2015-07-01T16:40:08.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11141" PostTypeId="1" AcceptedAnswerId="11142" CreationDate="2015-07-01T19:17:39.247" Score="9" ViewCount="6029" Body="&lt;p&gt;So, the recent conjunction of Jupiter and Venus seems to have spawned lots of excitement over this &quot;rare&quot; event. But what I can't figure out, is exactly how rare it is. And I've seen such conflicting claims and calculations that I figured I'd better calculate it myself. The only problem is I'm no astronomer. So, does anyone know of a program that can calculate all the conjunctions between a date range that are within a degree of each other? Or does smart person know how to calculate it manually? :) &#xA;    I have a program called Stellerium which I've been using to view the various conjunctions that people have mentioned. I don't know if I can use it in the way I need it though. Any help? :)&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7588" LastActivityDate="2020-12-23T01:26:59.337" Title="How to calculate conjunctions of 2 planets" Tags="&lt;jupiter&gt;&lt;venus&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11142" PostTypeId="2" ParentId="11141" CreationDate="2015-07-01T19:57:49.050" Score="5" Body="&lt;p&gt;The PyEphem library allows you to create Python scripts that could calculate any conjunction you want (and many other things besides that).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rhodesmill.org/pyephem/&quot; rel=&quot;nofollow&quot;&gt;http://rhodesmill.org/pyephem/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a script that already does that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://shallowsky.com/blog/science/astro/predicting-conjunctions.html&quot; rel=&quot;nofollow&quot;&gt;http://shallowsky.com/blog/science/astro/predicting-conjunctions.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the result of running that script now:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pastebin.com/ehjxV66m&quot; rel=&quot;nofollow&quot;&gt;http://pastebin.com/ehjxV66m&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notable event: extremely close conjunction between Jupiter and Venus in 2016:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Conjunction of Venus and Jupiter lasts from 2016/8/25 to 2016/8/31.&#xA;  Venus and Jupiter are closest on 2016/8/28 (0.1 deg).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-07-02T17:46:36.733" LastActivityDate="2015-07-02T17:46:36.733" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11143" PostTypeId="1" AcceptedAnswerId="11144" CreationDate="2015-07-01T20:12:03.997" Score="3" ViewCount="149" Body="&lt;p&gt;I assume it would have to be considerably larger since rock isn't as flammable as gases are? But it's also more dense to begin with..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we know of any suns that are made like this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it even possible to have a thermonuclear reaction with solid rock?&lt;/p&gt;&#xA;" OwnerUserId="7589" LastActivityDate="2015-08-12T03:06:44.850" Title="How much larger would a star have to be to cause thermonuclear reactions if it was made out of mostly rock like Earth, instead of gases?" Tags="&lt;star&gt;&lt;matter&gt;&lt;size&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11144" PostTypeId="2" ParentId="11143" CreationDate="2015-07-01T22:00:59.370" Score="6" Body="&lt;p&gt;It really depends what you mean by &quot;rock&quot;. At the temperatures and pressures at the cores of stars (and at which nuclear fusion reactions are possible), &quot;rocks&quot; as I suspect you are thinking of, do not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermonuclear reactions do not occur because the gas is &quot;flammable&quot;, they occur because the kinetic energies of the nuclei in the gas (at these temperatures, atoms are fully ionised) are sufficient to get them close enough together for nuclear fusion to take place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Rocks&quot; are made of atoms of silicon and oxygen (for example) in the form if silicates. But these are dissociated at fairly low temperatures compared to the centres of stars. Oxygen and Silicon thermonuclear fusion ignition requires temperatures in excess of $10^9$ K, and these temperatures are only reached late in the lives of stars of mass $&amp;gt;8 M_{\odot}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It &lt;em&gt;is&lt;/em&gt; possible to have stars with solid cores. This is thought to be the fate of white dwarf stars as they cool. Most white dwarfs are made of a mixture of carbon and oxygen and this &quot;crystallises&quot; once the core of the white dwarf cools below about a few million degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ordinarily, the cores of white dwarfs are inert as far as nuclear reactions are concerned, because the temperatures are too cool. However, if the white dwarf is massive enough (or mass is added to it), then the central densities climb, and for white dwarfs of mass $\sim 1.38M_{\odot}$ it is thought that the densities become high enough ($\sim 10^{13}$ kg/m$^3$) to start nuclear fusion of carbon via the zero point energy oscillations in the crystalline material (so-called &lt;a href=&quot;https://books.google.co.uk/books?id=eez38xjCYGkC&amp;amp;pg=PA380&amp;amp;lpg=PA380&amp;amp;dq=pyconuclear%20reactions&amp;amp;source=bl&amp;amp;ots=OIBlhAD7En&amp;amp;sig=qC64JFfOOSPA-i8E8Y_5pdrYuks&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=KmKUVdKpIsqisAHl35HgDw&amp;amp;redir_esc=y#v=onepage&amp;amp;q=pyconuclear%20reactions&amp;amp;f=false&quot;&gt;pyconuclear reactions&lt;/a&gt;). Such reactions in degenerate matter are highly explosive and might result in the detonation of the whole star in a type Ia supernova.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-07-01T22:00:59.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11146" PostTypeId="1" CreationDate="2015-07-02T03:34:28.507" Score="1" ViewCount="98" Body="&lt;p&gt;What kinds of eclipse is seen by the person on the moon, when there is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Solar eclipse on the earth, and&lt;/li&gt;&#xA;&lt;li&gt;Lunar eclipse on the earth?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2990" LastEditorUserId="9805" LastEditDate="2016-03-08T05:49:33.627" LastActivityDate="2016-03-08T05:49:33.627" Title="Eclipse on the moon" Tags="&lt;the-moon&gt;&lt;solar-eclipse&gt;&lt;lunar-eclipse&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11147" PostTypeId="2" ParentId="11146" CreationDate="2015-07-02T05:37:00.363" Score="2" Body="&lt;p&gt;This kinda depends on where you are on the moon.  If you're on the far side during a solar eclipse, you won't see a darn thing but the sun.  If you're on the near side, you will see the moon's shadow pass across the earth.  If you're on the far side during a lunar eclipse, you won't see anything but the stars. If you're on the near side, you will see the earth completely blot out the sun.  You should also see all the Earth's sunrises and sunsets at the same time (kinda cool!).&lt;/p&gt;&#xA;" OwnerUserId="7108" LastActivityDate="2015-07-02T05:37:00.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11148" PostTypeId="1" CreationDate="2015-07-02T12:06:50.287" Score="0" ViewCount="178" Body="&lt;p&gt;We know that the visual size of the Sun and Moon from the Earth in our lifetime (!) is nearly equivalent.&lt;br&gt;&#xA;Now my question: Is this event is random? Or have a scientific reason?&lt;br&gt;&#xA;For example, maybe there is a connection between this and the formation of life on the Earth...&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2016-04-21T07:06:02.117" LastActivityDate="2016-04-21T07:06:02.117" Title="Accidental or deliberate?" Tags="&lt;the-moon&gt;&lt;the-sun&gt;&lt;life&gt;&lt;formation&gt;&lt;origin-of-life&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2015-07-03T00:31:15.983" ContentLicense="CC BY-SA 3.0" />
  <row Id="11149" PostTypeId="2" ParentId="11110" CreationDate="2015-07-02T12:07:52.470" Score="4" Body="&lt;p&gt;The table you linked to gives limiting magnitudes for direct observations through a telescope &lt;em&gt;with the human eye&lt;/em&gt;, so it's definitely not what you want to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quoted number for &lt;em&gt;HST&lt;/em&gt; is an empirical one, determined from the actual &quot;Extreme Deep Field&quot; data (total exposure time ~ 2 million seconds) after the fact; the Illingworth et al. PDF you linked to explains how it was done. Briefly, they decided the limit would be defined by a signal-to-noise ratio of 5, and then figured out what the noise level in the final combined images was by integrating the total flux in many circular apertures of diameter 0.35 arc seconds placed in regions of blank sky background, and then computing the standard deviation ($\sigma$) of these measurements. Five times $\sigma$ was then the limiting flux. Knowing the photometric calibration (the conversion between observed counts and apparent magnitude; see the paper for specific details), they converted this $S/N = 5$ limit into an apparent magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, instead, you want try to estimate in advance what the limiting magnitude &lt;em&gt;should&lt;/em&gt; be, then you have to consider several factors. Simplifying a bit, the equation for long-exposure, background-limited $S/N$ is&#xA;$$&#xA;S/N \approx \frac{s t}{ \sqrt{ s t + n_{\rm pix}(b t)} }&#xA;$$&#xA;where $s$ = detected flux from the source (star, galaxy, etc.) in electrons (i.e., detected photons) per second, $t$ is the integration time in seconds , $b$ is the flux from the background in electrons per pixel, and $n_{\rm pix}$ is the number of pixels you integrate over for the measurement. $s$ can be computed from the combination of the object's apparent magnitude, the diameter of the telescope's mirror and the overall efficiency of the telescope + detector system. $b$ can be determined in a similar fashion, using the estimated background brightness (which, for a telescope like &lt;em&gt;HST&lt;/em&gt;, is mostly sunlight reflected from dust in the inner solar system). $n_{\rm pix}$ is a number chosen as a compromise between maximizing the amount of light from the source (higher $n_{\rm pix}$) and minimizing the noise from the background (lower $n_{\rm pix}$). This depends on the telescope's resolution (better resolution = smaller images of stars and other point-like sources = fewer pixels needed); the empirical calculations by Illingworth et al. used an aperture size amounting to about 100 pixels.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2015-07-02T12:07:52.470" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11150" PostTypeId="2" ParentId="11061" CreationDate="2015-07-02T13:24:10.137" Score="3" Body="&lt;p&gt;This page -- &lt;a href=&quot;http://www.amsmeteors.org/fireballs/faqf/&quot; rel=&quot;nofollow&quot;&gt;http://www.amsmeteors.org/fireballs/faqf/&lt;/a&gt; -- suggests that it's &lt;em&gt;very&lt;/em&gt; unlikely to find cometary-origin meteorites, even though the majority of observed &lt;em&gt;meteors&lt;/em&gt; are cometary in origin, because the latter are too fragile to survive all the way to the ground:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Based upon photographic fireball studies, cometary meteoroids have&#xA;  extremely low densities, about 0.8 grams/cc for class IIIA fireballs,&#xA;  and 0.3 grams/cc for class IIIB fireballs. This composition is very&#xA;  fragile and vaporizes so readily when entering the atmosphere, that it&#xA;  is called “friable” material. These meteoroids have virtually no&#xA;  chance of making it to the ground unless an extremely large piece of&#xA;  the comet enters the atmosphere, in which case it would very likely&#xA;  explode at some point in its flight, due to mechanical and thermal&#xA;  stresses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It goes on to claim that cometary meteoroids make up about 95% of observed meteors, 38% of observed fireballs, and 0% of fresh meteorites. (Since a hypothetical cometary-origin meteorite would weather faster than meteorites of asteroidal origin, they would be even less likely to be found later, and so it's not surprising that the page says 0% of all known meteorites are of cometary origin.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2015-07-02T13:24:10.137" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11151" PostTypeId="2" ParentId="11136" CreationDate="2015-07-02T13:37:36.447" Score="1" Body="&lt;p&gt;The &lt;a href=&quot;http://www.amsmeteors.org/fireballs/faqf/&quot; rel=&quot;nofollow&quot; title=&quot;American Meteor Society FAQ page&quot;&gt;American Meteor Society FAQ page&lt;/a&gt; notes that about 95% of observed meteors are probably of cometary origin. But they're apparently so fragile that they're only about 1/3 of observed fireballs, and &lt;em&gt;0%&lt;/em&gt; of known meteorites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, since the Earth passes through streams of cometary debris several times a year, and yet there's been only one plausible mass extinction due to cometary or asteroid impact in the last few hundred million years (the one that did in the dinosaurs), it appears to be &lt;em&gt;extremely&lt;/em&gt; unlikely for cometary debris to be a meaningful danger.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2015-07-02T13:37:36.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11153" PostTypeId="2" ParentId="11148" CreationDate="2015-07-02T17:26:54.383" Score="1" Body="&lt;p&gt;As the front image on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_eclipse&quot; rel=&quot;nofollow&quot;&gt;wikipedia-page&lt;/a&gt; already indicates, a total solar eclipse is not always total.&#xA;Earth's orbit is slightly elliptic, and so is the Moons orbit around the Earth.&#xA;Now take the Moon's slight orbital inclination into account and far from all total eclipses are really total.&lt;br&gt;&#xA;Unlike stated usually.&#xA;In fact wiki states &quot;On average, the Moon appears to be slightly smaller than the Sun as seen from the Earth, so the majority (about 60%) of central eclipses are annular.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a sidenote: It's important to understand that if the inclination and eccentricity of the orbit would be zero, then we'd have a perfect total solar eclipse on earth every month!&#xA;But yeah you can look it up, that this is not the case. There are, after all, not every month news reports from ppl staring into the sun, are there ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On other note, we know that the Moon is increasing its distance to earth. This happens because Orbital angular momentum is converted into internal heat for both bodies through &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_force&quot; rel=&quot;nofollow&quot;&gt;tidal forces&lt;/a&gt;.&lt;br&gt;&#xA;This situation will evolve over time, so yes, it's a coincidence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And no, I'm not aware of the Moon's distance having any strong implications for life.&lt;br&gt;&#xA;Tell me, if this is not clarified enough from the above.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2015-07-02T17:26:54.383" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11155" PostTypeId="2" ParentId="11148" CreationDate="2015-07-02T19:39:58.193" Score="1" Body="&lt;p&gt;That the Moon and Sun both have about the same angular size is just a coincidence; we (humans) just happen to be around at a time when this is so. Tidal torque is causing the Moon to speed up (which in turn is causing the Earth to slow down and the Moon's distance to increase as it gains orbital velocity), so eventually the Moon's angular size will always be less than the Sun's. Then no eclipses will be total, with the Moon only covering up a portion of the Sun's disk.&lt;br&gt;&lt;br&gt;The Moon raises tides on the earth, but Earth's more rapid rotation than the Moon's orbit causes the Earth to pull the tidal bulge ahead of the Moon.&lt;br&gt;&lt;br&gt;There are some hypothesis that life may have begun in tidal pools where the Sun's ultraviolet radiation may have been the energy that caused the chemical reactions that created the precursors to life. Because the Moon was so much closer billions of years ago, the tides would have been much larger than today, so tidal pools would have been undisturbed for weeks at a time. This is, in my opinion, hypothesis, not theory, though it certainly seems plausible.&lt;/p&gt;&#xA;" OwnerUserId="7108" LastEditorUserId="7108" LastEditDate="2015-07-02T22:29:17.390" LastActivityDate="2015-07-02T22:29:17.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11156" PostTypeId="1" AcceptedAnswerId="11159" CreationDate="2015-07-02T21:38:23.750" Score="5" ViewCount="3329" Body="&lt;p&gt;What causes the horizontal and vertical lines coming out of pictures of stars, instead of them simply appearing as circles? For example, this picture from Wikipedia: &lt;img src=&quot;https://i.stack.imgur.com/tMp7E.jpg&quot; alt=&quot;this picture&quot;&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7603" LastActivityDate="2015-07-08T04:47:36.487" Title="What causes horizontal and vertical lines coming out of pictures of stars?" Tags="&lt;star&gt;&lt;photography&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11157" PostTypeId="2" ParentId="11156" CreationDate="2015-07-02T21:51:19.583" Score="11" Body="&lt;p&gt;They are called &lt;a href=&quot;https://en.wikipedia.org/wiki/Diffraction_spike&quot;&gt;diffraction spikes&lt;/a&gt;, and they're artifacts from a supporting structure inside a reflector-type telescope.&lt;/p&gt;&#xA;" OwnerUserId="2573" LastActivityDate="2015-07-02T21:51:19.583" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11159" PostTypeId="2" ParentId="11156" CreationDate="2015-07-02T22:48:09.960" Score="-1" Body="&lt;p&gt;(I'd make this a comment, but no reputation yet.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/VVAKFJ8VVp4&quot; rel=&quot;nofollow&quot;&gt;Minute Physics&lt;/a&gt; has a good video on this! Filtering light through any lens causes spikes that identify what kind of aperture was used. That looks like a Hubble picture because of the diamond spikes.&lt;/p&gt;&#xA;" OwnerUserId="7606" LastActivityDate="2015-07-02T22:48:09.960" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11160" PostTypeId="1" AcceptedAnswerId="11166" CreationDate="2015-07-03T04:04:33.107" Score="1" ViewCount="75" Body="&lt;p&gt;i was retrieving a some personalities &lt;code&gt;birthday&lt;/code&gt; date through computer programming &lt;/p&gt;&#xA;&#xA;&lt;p&gt;when i retrieve a person's birthday the result was&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;1962-07-03+02:00&lt;/code&gt;   [this is tom cuise's birthday for information]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above result i could understand his birthday is on 1962-07-03 , &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what is &lt;code&gt;+02:00&lt;/code&gt; , is that the location of birth or GMT ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for more information the data is retrieved from the URL&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dbpedia.org/page/Tom_Cruise&quot; rel=&quot;nofollow&quot;&gt;http://dbpedia.org/page/Tom_Cruise&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sorry to ask this question here , I asked this question because it deals with time and i believe astronomers can help with this &lt;/p&gt;&#xA;" OwnerUserId="7607" LastActivityDate="2015-07-03T17:20:12.327" Title="what is the time mentioned after date (1962-07-03+02:00) the +02:00?" Tags="&lt;earth&gt;&lt;rotation&gt;&lt;time&gt;&lt;clock&gt;" AnswerCount="1" CommentCount="7" ClosedDate="2015-07-03T19:16:30.203" ContentLicense="CC BY-SA 3.0" />
  <row Id="11161" PostTypeId="1" AcceptedAnswerId="11163" CreationDate="2015-07-03T09:56:15.600" Score="-1" ViewCount="2152" Body="&lt;p&gt;I'm curious; does the Sun actually move? The concept behind the question is that if you wear a shirt and you move, does the shirt actually move? The Solar System orbits around the Milky Way, but does the Sun move?&lt;/p&gt;&#xA;" OwnerUserId="7610" LastEditorUserId="2153" LastEditDate="2015-07-08T19:03:17.547" LastActivityDate="2015-07-08T19:03:17.547" Title="Does the Sun move?" Tags="&lt;the-sun&gt;&lt;movement&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2015-07-08T19:49:14.270" ContentLicense="CC BY-SA 3.0" />
  <row Id="11163" PostTypeId="2" ParentId="11161" CreationDate="2015-07-03T13:03:01.127" Score="3" Body="&lt;p&gt;The Sun orbits in the Galactic potential. The motion is quite complex; it takes about 230 million years to make a circuit (meaning an orbital speed of around 220 km/s), but at the same time it oscillates up and down with respect to the Galactic plane every $\sim 70$ million years and also wobbles in and out every $\sim 150$ million years (this called epicyclic motion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can also measure how the Sun moves with respect to the average motion of the stars in the solar vicinity - the so-called local standard of rest. Estimates of the Sun's motion with respect to the LSR vary a little. According to &lt;a href=&quot;http://arxiv.org/abs/astro-ph/9710077&quot; rel=&quot;nofollow&quot;&gt;Dehnen (Walter!) &amp;amp; Binney 1998&lt;/a&gt; the Sun moves at 10 km/s inwards, at about 5 km/s faster than the average star tangentially to the Galactic centre and at about 7 km/s upwards out of the Galactic plane. Though as I mentioned previously, these motions are of an oscillatory nature around the mean value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could cast our net wider and ask how fast Sun moves with respect to the standard of rest defined by the cosmic microwave background - the local cosmological rest frame. The Sun moves at about 370 km/s relative to the CMB in the direction of the constellation Leo (as revealed by the dipole-like anisotropy in the CMB temperatures).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-07-03T13:03:01.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11164" PostTypeId="2" ParentId="11161" CreationDate="2015-07-03T13:12:47.087" Score="1" Body="&lt;p&gt;Yes. No. It depends on your perspective. Suppose you're in a plane, flying from New York to Los Angeles. If you stay in your seat all the way, you haven't moved (from your seat). On the other hand, you clearly have moved, since you were in New York and now you're in Los Angeles. Motion is relative to some reference point, and you can always pick a reference point that makes something stationary. However, some reference points are easier to use than others; that's why the geocentric model of the universe isn't used any more: cycles and epicycles and so on are just too complicated, and using the sun or the the center of the galaxy makes it much easier to visualize and calculate what's going on.&lt;/p&gt;&#xA;" OwnerUserId="1867" LastActivityDate="2015-07-03T13:12:47.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11166" PostTypeId="2" ParentId="11160" CreationDate="2015-07-03T17:20:12.327" Score="0" Body="&lt;p&gt;As indicated in the comments, the &lt;code&gt;+02:00&lt;/code&gt; is the &lt;a href=&quot;https://en.wikipedia.org/wiki/ISO_8601&quot; rel=&quot;nofollow&quot;&gt;ISO 8601&lt;/a&gt; syntax for the time zone. Specifically, it indicates a time zone 2 hours east of UTC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;whois dbpedia.org&lt;/code&gt; indicates that the site is hosted in Berlin, Germany, which is consistent with the &lt;code&gt;+02:00&lt;/code&gt; time zone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the data includes the date and time zone but not the time is probably an error. This particular datum &lt;code&gt;1962-07-03+02:00&lt;/code&gt; is Tom Cruise's birthday; in fact he was born on that day in Syracuse, New York.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A data representation that includes a date and time zone but no time is peculiar, and I'm not sure that it conforms to the ISO 8601 standard. The description in the Wikipedia article seems to imply that a time zone can only follow a time. I haven't checked the standard itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A representation of just the date: &lt;code&gt;1962-07-03&lt;/code&gt; would make more sense, and would be more useful. This is probably a bug in the software that runs dbpedia.org.&lt;/p&gt;&#xA;" OwnerUserId="53" LastActivityDate="2015-07-03T17:20:12.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11168" PostTypeId="1" AcceptedAnswerId="11173" CreationDate="2015-07-03T23:00:45.973" Score="4" ViewCount="159" Body="&lt;p&gt;Specifically, I am wondering what the pressure is at a distance of around 1 AU. Does it it decrease quadratically with distance?&lt;/p&gt;&#xA;" OwnerUserId="7615" LastEditorUserId="7982" LastEditDate="2021-02-28T11:29:34.927" LastActivityDate="2021-02-28T11:29:34.927" Title="What is the air pressure in the heliosphere (Sun's atmosphere)?" Tags="&lt;the-sun&gt;&lt;stellar-atmospheres&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11169" PostTypeId="1" AcceptedAnswerId="11172" CreationDate="2015-07-03T23:54:41.800" Score="5" ViewCount="166" Body="&lt;p&gt;My question is simple:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Is the Solar system stable?&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can see &lt;a href=&quot;https://en.wikipedia.org/wiki/Stability_of_the_Solar_System&quot;&gt;this&lt;/a&gt; Wikipedia page.&lt;br&gt;&#xA;&lt;strong&gt;Edit&lt;/strong&gt;: Sorry, because i think my question is more about mathematics and classical mechanics of planets in billion years scale, than astronomy. But i think this is our fortuity that Earth didn't doomed in past billion years ago and it's possible that we will have less than 8 planet's before Sun will be ruin.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2015-07-04T16:40:59.763" LastActivityDate="2015-07-04T16:40:59.763" Title="Stability of solar system" Tags="&lt;solar-system&gt;&lt;gravity&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11171" PostTypeId="1" CreationDate="2015-07-04T01:49:32.283" Score="4" ViewCount="45" Body="&lt;p&gt;Inflation seems to have occurred when the symmetry breaking of the electroweak occurred.   But do we know of any reason why we did not have inflation when gravity separated from the other forces or the strong and electroweak separated?  Are all these other symmetry breakings associated with the Higgs field and a mexican hat energy distribution.  Sorry it is all a bit confusing.  &lt;/p&gt;&#xA;" OwnerUserId="7616" LastActivityDate="2015-07-04T01:49:32.283" Title="Why didn't we have inflation when the theory of everything and GUTS broke symmetry?" Tags="&lt;cosmology&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2015-07-05T13:53:41.463" ContentLicense="CC BY-SA 3.0" />
  <row Id="11172" PostTypeId="2" ParentId="11169" CreationDate="2015-07-04T03:08:55.650" Score="6" Body="&lt;p&gt;The wikipedia page you linked to tells you that the solar system is gravitationally &quot;chaotic&quot;, in part because the mass of the sun is not fixed over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even more simply than that, focusing just on the gravity (ignoring loss of stellar mass, etc.), the solar system is an N-body problem.  We have 8 planets, a sun, and millions of asteroids, comets, and who knows how many individual particles gravitationally bound to our sun (plus ones that aren't and are just passing through the neighborhood, so to speak).  When you have more than 2 bodies, the solutions to the N-body problem are unstable.  What this means is that, say we describe the N-body problem with data $D$ (the &quot;initial conditions&quot;, or a perfect description of the state of the system at some specific point in time).  With a given complete data set the (Newtonian) gravitational evolution of the system is completely determined (but so difficult to do we can only approximate it).  What instability means here is that if we have some other data set $D'$ that is only a little bit different from $D$, then the differences between the evolution from $D$ and $D'$ will become exponentially large over long enough time scales.  So what may seem like minor differences now will result in radically different looking solar systems in the long run.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since all of our observations can never give exact values, but only a range of values, there is necessarily a bit of uncertainty in what the exact gravitational state of our solar system is.  We have very poor data on the exact asteroid and comet content of our solar system, and even planetary data has significant error margins.  All of this means that there are lots of justifiable picks for the data $D$, each differing by a small amount from each.  But due to the instability, eventually these data will produce radically different futures from each other.  Currently we can only predict the solar system's evolution up to a few million years or so (the exact value stated can vary wildly depending on how you opt to define and compute the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lyapunov_time&quot; rel=&quot;noreferrer&quot;&gt;Lyapunov time&lt;/a&gt;).  After that the evolutionary tracks become so disparate we can't really say we're predicting anything other than &quot;it'll definitely do something&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way or another, it is currently impossible for us to make any clear assertions about what the solar system will look like on a timescale of billions of years.  Maybe all 8 planets will still be there; maybe their orbits will be very similar, but maybe they'll have much different orbits; maybe several planets will have been ejected from the solar system.  At best we can observe a few things that lead certain objects to be most likely to undergo significant alteration.  For example, Jupiter and Mercury appear to have a certain orbital resonance right now which could ultimately lead Mercury to undergo a significant orbit change.  This may ultimately cause it to collide with another planet, or the sun, or be ejected from the solar system entirely.  But maybe it won't.  It's hard to say.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastEditorUserId="2188" LastEditDate="2015-07-04T04:10:14.467" LastActivityDate="2015-07-04T04:10:14.467" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11173" PostTypeId="2" ParentId="11168" CreationDate="2015-07-04T08:24:10.790" Score="7" Body="&lt;p&gt;I am going to assume that what you mean here is &quot;what is the pressure&quot; in the solar wind? There is no &quot;air&quot;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solar wind is pretty complex, consisting of a &quot;fast wind&quot; observed predominantly at high solar latitudes and a slower more variable wind a low latitudes. Both components essentially consist of an expanding stream of protons, electrons plus a small fraction of helium nuclei.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think all the information you want is contained in this paper by &lt;a href=&quot;http://heliophysics.nasa.gov/SolarMinimum24/papers/Ebert2009.pdf&quot; rel=&quot;noreferrer&quot;&gt;Ebert et al. (2009)&lt;/a&gt;, but I will attempt a broad-brush summary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fast wind has a typical proton number density of $n_p=3$ cm$^{-3}$, a temperature of $T=2\times10^{5}$ K and a speed of $V=700$ km/s. Note that &quot;temperature&quot; is a slippery concept here since the velocity distribution of the particles is non-Maxwellian. However, ignoring this and assuming the protons are matched by numbers of electrons, then the &quot;thermal pressure&quot; $P_{th} = (n_p + n_e)kT \simeq 10^{-11}$ Pa. However, this is negligible compared with the dynamic pressure $P_d = \rho V^2$, where $\rho$ is the mass density given roughly by $\rho = n_p m_p$. Thus $P_d \simeq 2\times 10^{-9}$ Pa (i.e. about $2 \times 10^{-14}$ bar).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;slow wind&quot; is quite variable, but mean values could be $n_p = 6$ cm$^{-3}$, $T=8\times 10^{4}$ K and $V=400$ km/s. Here again, the thermal pressure $P_{th} \simeq 10^{-11}$ Pa is dwarfed by the dynamic pressure $P_d = 1.6 \times 10^{-9}$ Pa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thermal pressure component falls as distance from the Sun &lt;em&gt;cubed&lt;/em&gt;, ie. as $P_{th} \propto R^{-3}$. This is because for mass conservation, then a sherically expanding wind must have a density that falls as $R^{-2}$, but it is experimentally found that the proton temperature also falls as $\simeq R^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dynamic pressure does fall almost as $R^{-2}$. This is because there is very little radius dependence for the proton velocity (i.e. the protons do not slow down, right out to the heliopause - as measured for instance by &lt;a href=&quot;http://web.mit.edu/space/www/voyager_science.html&quot; rel=&quot;noreferrer&quot;&gt;Voyager 2&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-07-04T08:24:10.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11176" PostTypeId="1" CreationDate="2015-07-05T10:33:58.207" Score="9" ViewCount="510" Body="&lt;p&gt;Water is very useful for rocket fuel and life support in space. But does water, in abundance and form which conceivably could be usefully mined this century, really exist in NEA's? Or do we have to go all the way out to the frost line, the Asteroid Belt, to find asteroids with available water? (Lunar polar water ice is not topic here, I think, but low delta-v asteroidal water).&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-07-05T12:50:05.733" LastActivityDate="2015-07-07T11:44:21.707" Title="Do NEA (Near Earth Asteroids) have minable water ice?" Tags="&lt;asteroids&gt;&lt;water&gt;&lt;near-earth-object&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11177" PostTypeId="2" ParentId="11176" CreationDate="2015-07-05T13:07:28.390" Score="5" Body="&lt;p&gt;Unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging numbers into the Stefan–Boltzmann law &lt;a href=&quot;https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law#Temperature_of_the_Earth&quot; rel=&quot;nofollow&quot;&gt;gives us a temperature&lt;/a&gt; near 273°K (0°centigrade) for bodies near earth's orbit. The exact answer for atmospherless bodies depends on albedo.&#xA;Any water on nearby asteroids will thus boil until it freezes, and then sublimate.&#xA;That's why the search for nearby ice is focused on lightless, cold regions of &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_south_pole&quot; rel=&quot;nofollow&quot;&gt;craters near the moon's south pole&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space probe Rosetta and its attendant comet are still &lt;a href=&quot;http://sci.esa.int/where_is_rosetta/&quot; rel=&quot;nofollow&quot;&gt;200 million km from the sun&lt;/a&gt;, and already outgassing water. IIRC, that started back in January, when the comet was 390 million km from the sun, well beyond Mars. Black body temp. out there would be around -100°C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looks like I was wrong about when jets first appeared: &lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/2014/0919-more-jets-from-rosettas-comet.html&quot; rel=&quot;nofollow&quot;&gt;More jets from Rosetta's comet!&lt;/a&gt; September 19, 2014. On that day, the comet was &lt;strong&gt;500 million km&lt;/strong&gt; from the sun. &lt;a href=&quot;https://en.wikipedia.org/wiki/Asteroid_belt&quot; rel=&quot;nofollow&quot;&gt;That's &lt;em&gt;outer&lt;/em&gt; belt&lt;/a&gt;. No spectra that I know of, so &lt;em&gt;possibly not water&lt;/em&gt;. Water seems most likely though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likely we'll have to go out &lt;em&gt;at least&lt;/em&gt; that far to find ice on small bodies.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="2031" LastEditDate="2015-07-07T11:44:21.707" LastActivityDate="2015-07-07T11:44:21.707" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11178" PostTypeId="1" AcceptedAnswerId="11235" CreationDate="2015-07-05T18:59:06.580" Score="5" ViewCount="325" Body="&lt;p&gt;In images of other galaxies they look very bright and vivid against the blackness of space (&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/9/98/Andromeda_Galaxy_%28with_h-alpha%29.jpg&quot; rel=&quot;noreferrer&quot;&gt;example&lt;/a&gt;). If I was located a few tens of thousands of lightyears away from a galaxy, would I see something similar or would it be more like a slightly brighter smudge in space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering since it seems we can barely see the features of our own galaxy with the naked eye from earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I see that someone has downvoted the question, it would have been good to have some feedback! I realise the question might be a bit vague, but I thought it was interesting to know if galaxies would look as bright as we see them in the pictures.&lt;/p&gt;&#xA;" OwnerUserId="7624" LastEditorUserId="7624" LastEditDate="2015-07-06T06:54:05.223" LastActivityDate="2015-07-09T13:08:29.787" Title="How clearly would I be able to see a galaxy with the naked eye if viewed from a &quot;close&quot; distance?" Tags="&lt;galaxy&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11179" PostTypeId="1" CreationDate="2015-07-06T00:54:41.617" Score="8" ViewCount="221" Body="&lt;p&gt;Over &lt;a href=&quot;https://en.wikipedia.org/wiki/Stability_of_the_Solar_System#Neptune.E2.80.93Pluto_resonance&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;, I was shocked to learn:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Neptune–Pluto system lies in a 3:2 orbital resonance. C.J. Cohen and E.C. Hubbard at the Naval Surface Warfare Center Dahlgren Division discovered this in 1965.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: there is no cite for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure the orbit of Pluto, including it's period, was calculated many years before 1965, to say nothing of Neptune. So how is it possible that astronomers missed the 2:3 resonance? Was orbital resonance just not a conceived concept until the sixties?&lt;/p&gt;&#xA;" OwnerUserId="7449" LastActivityDate="2015-07-07T03:03:04.317" Title="Why was Pluto's resonance not discovered until 1965?" Tags="&lt;history&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11180" PostTypeId="1" CreationDate="2015-07-06T01:02:59.590" Score="1" ViewCount="183" Body="&lt;p&gt;After searching through many videos, I finally found &lt;a href=&quot;https://www.youtube.com/watch?v=LQpO1TpcSH0&quot; rel=&quot;nofollow&quot;&gt;one&lt;/a&gt; that shows, &lt;strong&gt;from the perspective of low earth orbit&lt;/strong&gt;, the Moon transiting the horizon. It is actually a Moonset, but it shouldn't make a difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon does not appear to change its color, neither gold nor red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This contrasts with my understanding of atmospheric effects. Here on Earth, the Moon does appear more yellow (and when halfway or less above the horizon, even red) during Moonrise. The atmosphere tends to scatter blue light, leaving only yellow and red behind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why isn't this happening in the video? From orbit, we should be seeing the Moon through &lt;strong&gt;even more&lt;/strong&gt; atmosphere. But I can't detect a trace of red or yellow in the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Granted it's not the best quality video that it could be. It could be more zoomed in. But it was the &lt;strong&gt;only&lt;/strong&gt; video I could find that showed the Moon transiting the horizon from space.&lt;/p&gt;&#xA;" OwnerUserId="7449" LastEditorUserId="7982" LastEditDate="2021-02-28T11:52:32.030" LastActivityDate="2021-02-28T11:52:32.030" Title="Why doesn't the Moonrise appear red when viewed from orbit?" Tags="&lt;the-moon&gt;&lt;atmospheric-effects&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11181" PostTypeId="2" ParentId="11180" CreationDate="2015-07-06T03:42:35.380" Score="2" Body="&lt;p&gt;In this video there is blue diffused sunlight present in the atmosphere.  At any point where you might expect blue light to be diffused from the Moon, there is also blue light diffused from the Sun being added back into what you see. To know for sure, you would need to have a recording from night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The density of air could also be a consideration. The perspective in the video shows the moon from above cloud clover passing below cloud cover. The further the light passes from the Earth's surface doesn't only result in less distance spent in the atmosphere, but also the density of air throughout the entire passage.&lt;/p&gt;&#xA;" OwnerUserId="1288" LastActivityDate="2015-07-06T03:42:35.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11182" PostTypeId="1" AcceptedAnswerId="11188" CreationDate="2015-07-06T10:28:26.733" Score="4" ViewCount="821" Body="&lt;p&gt;I am very much a beginner on the astronomy front but I understand about planets having different axial tilts, hence why Venus turns the opposite direction from the Earth and Uranus turns sideways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am confused as to what the axial tilt bearing is measured from. For example, if all planets were in a perfect line from the sun, would they spin perfect in alignment on their own various axial tilts? Would some still be 'off-set' by a few degrees to one side or the other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we actually know please?&lt;/p&gt;&#xA;" OwnerUserId="7629" LastActivityDate="2015-07-07T10:10:00.067" Title="What is the axial tilt of a planet measured relative to?" Tags="&lt;planet&gt;&lt;rotation&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11183" PostTypeId="1" AcceptedAnswerId="11189" CreationDate="2015-07-06T11:49:23.693" Score="2" ViewCount="264" Body="&lt;p&gt;By what process can a user running the latest version of OS X Yosemite execute MOOG with &lt;code&gt;$MOOG&lt;/code&gt; in bash? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://bathompso.com/blog/moog/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Due to the removal of 32bit libraries in OS X 10.7+, the Fortran77 code for the MOOG installer cannot be compiled on OS X 10.7+&lt;/a&gt;. As a result, the &lt;a href=&quot;https://github.com/andycasey/moog&quot; rel=&quot;nofollow noreferrer&quot;&gt;pip installer&lt;/a&gt; will not work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have followed the instructions in the above guide, except that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I cannot install scisoft because the link in the page does not work&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://bi-jmyles.blogspot.com/2015/07/moog-vald-sm-marcs-ares-cont.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;I have supermongo installed independently of MOOG&lt;/a&gt;, so &lt;code&gt;$sm&lt;/code&gt; works for me&lt;/li&gt;&#xA;&lt;li&gt;I have made my computer compile Fortran code with the &lt;a href=&quot;https://apple.stackexchange.com/questions/99077/how-to-set-gcc-4-8-as-default-gcc-compiler?rq=1&quot;&gt;true version of gcc, not Apple gcc&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, in addition to being unable to compile the code, the &lt;a href=&quot;https://github.com/bathompso/moog/tree/master/bin&quot; rel=&quot;nofollow noreferrer&quot;&gt;pre-compiled binaries&lt;/a&gt; aren't working for me right now on Yosemite 10.10.4. The error I get when I &lt;code&gt;$MOOG&lt;/code&gt; is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;open: No such file or directory&#xA;apparent state: unit 35 named /.moog/Barklem.dat&#xA;lately writing sequential formatted external IO&#xA;Abort trap: 6&#xA;&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I do have .moog in my home folder, and it has the Barklem files inside, and both the directory and files have been made readable/writable with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;$sudo chmod +rw .moog&lt;/code&gt; and &lt;code&gt;$sudo chmod +rw Bark*&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Motivation: &lt;a href=&quot;https://github.com/andycasey/moog/issues/6&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomers try to solve the problem independently&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Justification: AstronomySE topics include: &quot;Setting up, using and maintaining your astronomy related equipment&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user5341" LastEditorUserId="-1" LastEditorDisplayName="user5341" LastEditDate="2017-04-13T12:45:31.967" LastActivityDate="2015-07-29T14:31:18.853" Title="How to run MOOG on OS X Yosemite" Tags="&lt;software&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11184" PostTypeId="2" ParentId="11129" CreationDate="2015-07-06T11:57:25.183" Score="2" Body="&lt;p&gt;The &lt;a href=&quot;http://exploration.esa.int/mars/&quot; rel=&quot;nofollow&quot;&gt;Exomars rover&lt;/a&gt; of ESA and Roscosmos would be the obvious answer. To be launched in 2018. It will drill 2 meters deep and as far I know is the only mission since the Vikings in the 1970's, to explicitly be equipped to find biosignatures, signs of life. But the Russians, who will land the rover, have had a very poor Mars mission success rate, ESA cooperates with them since NASA abandoned the project a few years ago, and ESA's only own landing attempt on Mars failed too, &lt;a href=&quot;https://en.wikipedia.org/wiki/Beagle_2&quot; rel=&quot;nofollow&quot;&gt;Beagle 2&lt;/a&gt;. And some biologists think it is more challenging to detect sparse exotic microbial life than what that rover is capable of. It weights about 1/3 of MSL Curiosity, and underground life can maybe be very local.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to recommend the blogger &lt;a href=&quot;http://www.science20.com/robertinventor&quot; rel=&quot;nofollow&quot;&gt;Robert Walker&lt;/a&gt; who writes at great length and well informed, still interestingly speculative, about possibilities for life on Mars, what one maybe should be looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should add that since 1960 &lt;a href=&quot;http://www.seti.org/talks&quot; rel=&quot;nofollow&quot;&gt;SETI&lt;/a&gt; uses telescopes to pick up evidence of interstellar life which is powerful enough to somehow change its environment to make itself astronomically detectable. That's the other main potential except for probes in the Solar system. But not even SETI people sit up waiting for it any more.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-07-06T12:19:31.037" LastActivityDate="2015-07-06T12:19:31.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11185" PostTypeId="2" ParentId="11176" CreationDate="2015-07-06T11:59:57.090" Score="6" Body="&lt;p&gt;No and yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll address the &quot;no&quot; answer first. The answer is no if you truly mean &quot;water ice&quot;. Ceres is close to the limit of where water can exist as water, as opposed to in the form of hydrated rock. Those intriguing bright spots on Ceres may well be water ice that has been exposed to vacuum (but they might just be salts). Ceres most likely is inside the frost line, but just slightly. That's part of what makes Ceres so intriguing. Ceres orbits at 2.7+ AU. It is not a Near Earth Asteroid. Contrary to mainstream thinking, perhaps the current frost line is at 2.7 AU (as opposed to 5 AU). There's no way that it is at 1 AU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is yes if you look beyond ice as a source of water. Some carbonaceous asteroids contain water in the form of hydrated minerals. Some of them contain quite a bit of water, up to 22%. We'll find out more when &lt;a href=&quot;https://en.wikipedia.org/wiki/OSIRIS-REx&quot; rel=&quot;noreferrer&quot;&gt;OSIRIS-REx&lt;/a&gt; launches next year and returns a sample from &lt;a href=&quot;https://en.wikipedia.org/wiki/101955_Bennu&quot; rel=&quot;noreferrer&quot;&gt;101955 Bennu&lt;/a&gt; in 2023.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2015-07-06T12:06:20.460" LastActivityDate="2015-07-06T12:06:20.460" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11186" PostTypeId="1" AcceptedAnswerId="11796" CreationDate="2015-07-06T13:42:50.567" Score="10" ViewCount="174" Body="&lt;p&gt;The James Webb Space Telescope (JWST) which is slated to launch in 2018 will bring us a better view of exoplanets, but will it be sufficient to detect signs of life on at least some of those worlds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good answer will provide an estimate of the James Webb Space Telescope's capabilities to detect atmospheric data relevant to the presence of life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The preferred answer would also include an analysis of what it would take to say with a decent level of confidence that life was present on an exoplanet and whether or not JWST will be capable of this.&lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorUserId="37191" LastEditDate="2021-03-11T13:22:03.480" LastActivityDate="2021-03-11T13:22:03.480" Title="Could the James Webb Space Telescope detect biosignals on exoplanets?" Tags="&lt;exoplanet&gt;&lt;astrobiology&gt;&lt;james-webb-space-telescope&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11187" PostTypeId="2" ParentId="11182" CreationDate="2015-07-06T14:22:01.593" Score="1" Body="&lt;p&gt;Draw a line perpendicular to the planet's orbit.  Measure the angle between this line and the planet's axis.  That's the axial tilt, and it doesn't matter what other planets are doing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_tilt&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page on axial tilt&lt;/a&gt; has some diagrams that help clarify this.&lt;/p&gt;&#xA;" OwnerUserId="1595" LastActivityDate="2015-07-06T14:22:01.593" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11188" PostTypeId="2" ParentId="11182" CreationDate="2015-07-06T14:43:38.113" Score="3" Body="&lt;p&gt;If you look at the &lt;em&gt;Astronomical Almanac for the year 2011&lt;/em&gt; as an example, in the table at the top of page E3, you find two measures of axial tilt. The third column is the declination of the planet's north pole for the mean equinox and equator of date 2011 January 0, 0 hours terrestrial time. So this is measured with respect to the Earth's orbit. The last column is the inclination of the planet's equator to the planet's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the right ascension column in the same table, you see that on that particular date the right ascensions are all different.&lt;/p&gt;&#xA;" OwnerUserId="7385" LastEditorUserId="7385" LastEditDate="2015-07-07T10:10:00.067" LastActivityDate="2015-07-07T10:10:00.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11189" PostTypeId="2" ParentId="11183" CreationDate="2015-07-06T15:02:49.280" Score="1" Body="&lt;p&gt;In order to execute &lt;a href=&quot;https://github.com/bathompso/moog/tree/master/bin&quot; rel=&quot;nofollow noreferrer&quot;&gt;the compiled binaries&lt;/a&gt;, all libraries on which the executable depends must be in the correct location. In particular, the error specified in the question is due to the fact that my .moog folder was in ~/, my user home folder, rather than /, the root of my hard drive, as explained &lt;a href=&quot;https://github.com/andycasey/moog/issues/6#issuecomment-118881869&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;cd&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;mv .moog /.moog&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://astronomy.stackexchange.com/questions/11433/how-to-specify-sm-libraries-for-moog/11434#11434&quot;&gt;this link&lt;/a&gt; for answer on how to get MOOG to find SM libraries for plotting.&lt;/p&gt;&#xA;" OwnerDisplayName="user5341" LastEditorUserId="-1" LastEditorDisplayName="user5341" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-07-29T14:31:18.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11190" PostTypeId="2" ParentId="8525" CreationDate="2015-07-06T15:14:14.913" Score="2" Body="&lt;p&gt;If observed Black Hole magnetic fields are generated in part by moving charges within the Black Hole, then curvature of space-time as the means of intermediation of action at a distance for gravity would be undermined as an explanatory conceptual construct.  Magnetars are Neutron Stars with extreme magnetic fields.  Like Black Holes, their matter becomes so dense that their gravitational fields become more forceful than the electro-magnetic fields that characterize ordinary matter, causing the electro-magnetic structure of atoms to collapse.  They're just smaller in total mass than Black Holes.  If Neutron stars can generate extreme magnetic fields, presumably from un-matched protons and/or electrons within the dense mass and moving with star-rotation, then Black Holes should be capable of doing so, as well.  Magnetic fields exist in space surrounding their source - moving charged particles - and if space is warped extremely, per space-time-curvature concepts, then the field lines should be warped with the space.  If they extend beyond the event horizon, then we'd have to find a different means of action-at-a-distance for gravitation, consistent with the proven formulas, if not the under-lying concept &quot;explaining&quot; action at a distance, of General Relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The observational challenge is that accretion disks also contain moving charges that generate strong magnetic fields, and it is not clear how one might distinguish between fields generated within the event horizon from those generated without. The enormous power of Magnetars, however, without apparent accretion disks, suggests that the possibility of internal magnetic fields may be difficult to ignore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is some other common means of intermediating action at a distance (other than curvature of space-time and exchange of virtual particles), that serves to explain both gravitational and electro-magnetic phenomena, very large, fast-spinning BHs would also be expected to generate strong gravito-magnetic fields with angular momentum vectors similar to magnetic poles.  It has never been clear why curved space-time would generate a gravito-magnetic effect, nor does exchange of virtual particles provide a very satisfying explanation for electro-magnetic induction and fields, for that matter.  A new common explanation for action at a distance might be expected to offer a better explanation for these phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is an excellent question that deserves careful analysis, including from the perspective that, perhaps, our explanations for action at a distance are flawed.&lt;/p&gt;&#xA;" OwnerUserId="7631" LastActivityDate="2015-07-06T15:14:14.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11191" PostTypeId="1" CreationDate="2015-07-06T15:16:14.587" Score="0" ViewCount="1520" Body="&lt;p&gt;To calculate for example the polaris hour angle I use this formula to calculate polaris hour angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;τ = θ - α&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;τ= hour angle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;θ= sideral time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;α= polaris RA (~ 2h32m)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But where does the α=2h32m come from? Is there a list? Does this change somehow over time?&#xA;Is my understanding correct that this is the time where polaris crosses the upper meridian at Greenwich?&lt;/p&gt;&#xA;" OwnerUserId="7630" LastActivityDate="2016-07-01T15:58:34.117" Title="Where to get Polaris Right ascension value from" Tags="&lt;star&gt;&lt;positional-astronomy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11192" PostTypeId="2" ParentId="11191" CreationDate="2015-07-06T15:25:48.037" Score="1" Body="&lt;p&gt;There are several published star catalogs that list right ascensions for stars, along with a lot of other almanac-type data.  Wikipedia's a good resource for common stars (e.g. &lt;a href=&quot;https://en.wikipedia.org/wiki/Polaris&quot; rel=&quot;nofollow&quot;&gt;Polaris&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right ascensions change over time, as stars move through space.  Star catalogs will have a year associated with them, which is the year that the measurement was taken.&lt;/p&gt;&#xA;" OwnerUserId="1595" LastActivityDate="2015-07-06T15:25:48.037" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11193" PostTypeId="1" CreationDate="2015-07-07T01:26:42.480" Score="2" ViewCount="641" Body="&lt;p&gt;Generally there may be some line ratio between the two lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If  H$\alpha$ is strong, it is quite possible we can see H$_\beta$ too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible we can see H$_\beta$, but can not see H$_\alpha$? In what kind of environments can it happen?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="609" LastEditDate="2015-07-20T01:26:54.500" LastActivityDate="2015-07-20T01:26:54.500" Title="about H$_\alpha$ and H$_\beta$ lines in astronomical objects' spectra" Tags="&lt;spectra&gt;&lt;star&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11194" PostTypeId="2" ParentId="11179" CreationDate="2015-07-07T03:03:04.317" Score="5" Body="&lt;p&gt;This much, I can answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Was orbital resonance just not a conceived concept until the sixties?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Laplace recognized orbital resonance long before then, and he died in 1827.  I looked, couldn't find a precise year of publication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_resonance#Laplace_resonance&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Orbital_resonance#Laplace_resonance&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, 3/2 resonances were perhaps suggested/discovered quite a bit later, I'm not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some mathematics on the subject in the following link, a little bit over my skill-set, but it makes the point of stable resonance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.applet-magic.com/solaresonance.htm&quot; rel=&quot;noreferrer&quot;&gt;http://www.applet-magic.com/solaresonance.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the link points out that the the first resonance discovered, Io, Europa, Ganymede the 1-2-4 resonance, have ratios of 2.0076 and 2.0143, or, 0.4% and 0.7% off resonance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm tempted to say that small variations like that fall within a stable zone, but I'd basically just be guessing.   I can't say for sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pluto was discovered in 1930 and Charon, it's rather large moon, not discovered till 1977.   Perhaps in the 1930s, Pluto was no longer cutting edge physics and all the big brains were looking at Quantum physics, Hubble's discovery and other galaxies, implications from Einstein's discoveries, such as black holes and Nuclear bombs . . . but again, I'm only speculating.   It does seem that a discovery of the 3/2 resonance should have come before 1965 and I think it's a good question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not to take this too off subject, but I love the story of how Neptune was discovered.   Short summary here:   &lt;a href=&quot;http://coolcosmos.ipac.caltech.edu/ask/146--When-was-Neptune-discovered-&quot; rel=&quot;noreferrer&quot;&gt;http://coolcosmos.ipac.caltech.edu/ask/146--When-was-Neptune-discovered-&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longer one here: &lt;a href=&quot;https://en.wikipedia.org/wiki/Discovery_of_Neptune&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Discovery_of_Neptune&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My favorite potential orbital resonance is the Jupiter/Mercury one, which could, eventually, pull Mercury away from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Stability_of_the_Solar_System#Mercury.E2.80.93Jupiter_1:1_perihelion-precession_resonance&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Stability_of_the_Solar_System#Mercury.E2.80.93Jupiter_1:1_perihelion-precession_resonance&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Jupiter/Mercury resonance is described as &quot;coincidence&quot; in the article, where as, Pluto's 3/2 with Neptune and Jupiter's 3 inner moons are (I think), probobly not coincidental resonance but some form of stable orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope you don't mind my amateur/hobbyist answer.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-07-07T03:03:04.317" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11195" PostTypeId="2" ParentId="11193" CreationDate="2015-07-07T09:48:07.483" Score="2" Body="&lt;p&gt;In a normal stellar atmosphere, where the temperature decreases with height, I think this is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H-alpha absorption arises from the $n=2$ to $n=3$ transition; H-beta from the $n=2$ to $n=4$ transition. Thus both transitions are governed by the number of atoms in the $n=2$ lower level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absorption coefficient for a transition can be written in proportionality terms by (in local thermodynamic equilibrium)&#xA;$$ \alpha_\nu \propto  \nu B_{lu} n_l ( 1 - \exp[-h\nu/kT]),$$&#xA;where $B_{ul}$ is the Einstein absorption coefficient, $n_l$ is the population of the lower energy level and $\nu$ is the photon frequency corresponding to the transition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the well-known relationship between the Einstein emission and absorption coefficients and values for the emission coefficients found &lt;a href=&quot;http://www.physics.byu.edu/faculty/christensen/Physics%20612/FTI/H%20Einstein%20Coefficients.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, we can estimate a ratio of absorption coefficients for a given temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\alpha_{H\beta}}{\alpha_{H\alpha}} = \frac{g_{n=4}}{g_{n=3}} \frac{A_{H\beta}}{A_{H\alpha}} \left(\frac{\nu_{\alpha}}{\nu_{\beta}}\right)^2 \left(\frac{1 - \exp(-h\nu_{\beta}/kT)}{1-\exp(-h\nu_{\alpha}/kT}\right),$$&#xA;where $A_X$ are the Einstein emission coefficients and the statistical weights $g_n$ are given by $2n^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if we take $\nu_\alpha = 4.57\times10^{14}$ Hs, $\nu_{\beta}=6.17\times10^{14}$ Hz, $A_{H\alpha} \simeq 10^{8}$ s$^{-1}$, $A_{H\beta} \simeq 3\times 10^{7}$ s$^{-1}$, then&#xA;$$\frac{\alpha_{H\beta}}{\alpha_{H\alpha}} = 0.29 \left( \frac{1 - \exp(-29642/T)}{1 - \exp(-21956/T)}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when $T$ (in Kelvin) is small, the ratio is about 0.3. When $T$ becomes large then the ratio increases, but above $T \sim 12,000$ K, all the Hydrogen is ionised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus in thermodynamic equilibrium is it very difficult to manufacture a circumstance where the optical depth in the H$\beta$ line is larger than that in the H$\alpha$ line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[I'd be grateful if someone can tell me whether the ratio I calculated above is correct, since I could not find it in any reference and need to work it out from scratch].&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-07-07T09:48:07.483" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11196" PostTypeId="1" CreationDate="2015-07-07T12:24:20.013" Score="4" ViewCount="3173" Body="&lt;p&gt;Based on &lt;a href=&quot;http://en.wikipedia.org/wiki/Crater_counting&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, we can estimate the age of a planet's surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to tell &lt;strong&gt;a particular crater&lt;/strong&gt; is newly formed or not? It may have sharp edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different astronomical objects may have different kinds of environments which may make this more complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any peer-reviewed papers?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditDate="2015-07-08T01:35:11.633" LastActivityDate="2017-02-08T02:14:12.583" Title="How to tell a crater is old or not?" Tags="&lt;crater&gt;&lt;age&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11197" PostTypeId="2" ParentId="11196" CreationDate="2015-07-07T17:28:44.827" Score="4" Body="&lt;p&gt;One criterion that works well is the presence or absence of smaller craters inside. If the crater is clean, it's probably new. If it's full of smaller craters, it's old. This applies to planets and satellites with no atmosphere, or with a rarefied atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2015-07-07T17:28:44.827" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11198" PostTypeId="1" CreationDate="2015-07-07T17:50:38.720" Score="4" ViewCount="200" Body="&lt;p&gt;Could Olympus Mons have thrown ejecta out of Mars' orbit when it was active? Does it fit with the composition and dating of meteorites with martian origin?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="9527" LastEditDate="2020-09-24T07:28:23.863" LastActivityDate="2020-09-24T22:08:05.243" Title="Could martian meteorites come from Olympus Mons?" Tags="&lt;mars&gt;&lt;meteorite&gt;&lt;volcanism&gt;&lt;olympus-mons&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11199" PostTypeId="1" AcceptedAnswerId="11871" CreationDate="2015-07-07T19:54:42.107" Score="10" ViewCount="1317" Body="&lt;p&gt;Are the four stars mentioned in the title stars that are going to go supernova, or are they stars that are in mid-life, like our Sun, but just happen to be so much larger than our Sun?&lt;/p&gt;&#xA;" OwnerUserId="5401" LastEditorUserId="2153" LastEditDate="2015-09-20T16:00:47.343" LastActivityDate="2016-12-29T18:39:01.183" Title="Are stars NML Cygni, UY Scuti, VY Canis Majoris and VV Cephei near the ends of their lives?" Tags="&lt;star&gt;&lt;stellar-evolution&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11200" PostTypeId="1" CreationDate="2015-07-07T19:59:10.883" Score="3" ViewCount="965" Body="&lt;p&gt;The sun is a main-sequence star; as I understand it the sun will expand into a red giant in a few billion years.  I have heard conflicting reports of what this entails - some sources I found say the sun will expand until the current orbit of Earth is contained within it, others say it will stop expanding short of Earth's current orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every source has agreed that the planet will have long-since stopped being able to support life as we know it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming humanity takes no action to significantly alter the natural course of events, such as dramatic alterations of our atmosphere, what will the last few decades of Earth's life be like?  Will the planet be heated to the point where its surface is like Mercury's sun-side?  Will the sun's output drop low enough that it instead cools, like Mars?  Will the sun expand and engulf the planet, or will the planet continue to orbit an expanded sun as a lifeless ball?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the sun's expansion be gradual or rapid?  How will this expansion affect the rocky planets nearest to it?&lt;/p&gt;&#xA;" OwnerUserId="7638" LastActivityDate="2015-07-11T10:10:14.353" Title="What will the last days of Earth be like?" Tags="&lt;the-sun&gt;&lt;planet&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11201" PostTypeId="2" ParentId="11193" CreationDate="2015-07-07T20:57:03.010" Score="2" Body="&lt;p&gt;As a small addendum to Rob Jeffries' excellent answer, I'll note that you probably can't get a scenario of &quot;H-beta, but no H-alpha&quot; for emission lines, either. The emission lines are the reverse of the absorption lines, so to have H-beta emission without H-alpha, you'd have to have $n = 4$ to $n = 2$ transitions &lt;em&gt;without&lt;/em&gt; also having $n = 3$ to $n = 2$ transitions. But even if you could somehow magically arrange to have all your H atoms in the $n = 4$ level, some of them would transition to $n = 3$, and then to $n = 2$, so you really can't avoid H-alpha emission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, dust extinction is worse at shorter wavelengths, so dust extinction could in principle give you the appearance of H-alpha emission without H-beta emission, but not the reverse.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2015-07-07T20:57:03.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11202" PostTypeId="1" AcceptedAnswerId="11208" CreationDate="2015-07-07T22:32:38.680" Score="3" ViewCount="4436" Body="&lt;p&gt;I was wondering how much you had to crush an object for it to become a black hole. Recently I learned that anything could become a black hole (even you) if it were crushed down small enough, for example our sun would have to be crushed to the size of Manhattan to become a black hole (According to a video I saw). I did a little research and found out that something would have to be crushed down to Schwarzchild radius meaning all of that objects mass would have to be crushed into that objects radius. But if you say crush the sun down to the size of Manhattan, that is much smaller than the sun's actual radius. So I haven't really gotten a clear answer and I'm a little confused, thanks for clarifying =)&lt;/p&gt;&#xA;" OwnerUserId="7467" LastActivityDate="2015-07-08T19:38:42.977" Title="How small would you have to crush an object for it to become a black hole?" Tags="&lt;gravity&gt;&lt;black-hole&gt;&lt;mass&gt;" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11204" PostTypeId="2" ParentId="11156" CreationDate="2015-07-08T04:47:36.487" Score="1" Body="&lt;p&gt;sometimes you can get these spike artifacts from the microlenses over the CCD's sensor array as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;also if your CCD does not have antiblooming logic, a very bright star can cause neighboring pixels to saturate as they are being read out, leading to a &quot;spike&quot; only along the readout axis. here's a good document with some common artifacts, from the hubble team: &#xA;&lt;a href=&quot;http://www.stsci.edu/~levay/color/HandoutIIIc.pdf&quot; rel=&quot;nofollow&quot;&gt;HandoutIIIc&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the spikes in the OP's image are likely caused by the spider vanes which hold the secondary mirror in place, as Russell says.&lt;/p&gt;&#xA;" OwnerUserId="7642" LastActivityDate="2015-07-08T04:47:36.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11205" PostTypeId="2" ParentId="11202" CreationDate="2015-07-08T09:36:24.553" Score="2" Body="&lt;p&gt;Well, it has to be crushed (compressed would be the better word to use here) below the Schwarzschild radius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Schwarzschild radius is the radius of the object in which the escape velocity would be the speed of light from that object. When this radius becomes even smaller, even light cannot escape it and voilà, there's your black hole!&lt;/p&gt;&#xA;" OwnerUserId="7643" LastEditorUserId="2153" LastEditDate="2015-07-08T19:38:42.977" LastActivityDate="2015-07-08T19:38:42.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11206" PostTypeId="1" AcceptedAnswerId="11209" CreationDate="2015-07-08T12:30:43.590" Score="3" ViewCount="984" Body="&lt;p&gt;The $\log{gf}$ value of an element in a star, where $f$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Oscillator_strength&quot; rel=&quot;nofollow&quot;&gt;oscillator strength&lt;/a&gt;, or probability of transition in an atom, is &lt;a href=&quot;http://arxiv.org/pdf/0712.1185v1.pdf&quot; rel=&quot;nofollow&quot;&gt;a scientifically pursued quantity&lt;/a&gt; with relation to the model atmosphere of the star. Why is the oscillator strength tied to $g$? And what is the motivation for including this quantity in a stellar model?&lt;/p&gt;&#xA;" OwnerDisplayName="user5341" LastActivityDate="2015-07-08T18:08:16.123" Title="Why is $g$ tied to the oscillator strength $f$ in $\log{gf}_{\odot}$" Tags="&lt;spectroscopy&gt;&lt;stellar-astrophysics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11207" PostTypeId="2" ParentId="11202" CreationDate="2015-07-08T14:17:32.657" Score="2" Body="&lt;p&gt;$$r_s = \frac{2GM}{c^2}$$&#xA;--&lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_radius#Formula&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$r_s$ is the Schwarzschild radius;&lt;br&gt;&#xA;  $G$ is the gravitational constant;&lt;br&gt;&#xA;  $M$ is the mass of the object;&lt;br&gt;&#xA;  $c$ is the speed of light in vacuum.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The proportionality constant, $2G/c^2$, is approximately 1.48×10&lt;sup&gt;27&lt;/sup&gt; m/kg&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="2153" LastEditDate="2015-07-08T17:04:42.050" LastActivityDate="2015-07-08T17:04:42.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11208" PostTypeId="2" ParentId="11202" CreationDate="2015-07-08T17:40:12.227" Score="10" Body="&lt;p&gt;Well, I wasn't going to answer but the other two answers are wrong, or at least incomplete. If you wish to make a black hole from a stellar-sized object, then there is no need to compress it to as small as the Schwarzschild radius (though that would certainly work and would certainly be the answer for smaller objects with negligible self-gravity). Instead, you just need to compress it to a size at which it cannot be possibly supported by any plausible equation of state against further gravitational collapse. It turns out that this is somewhat bigger than the Schwarzschild radius and hence the density required is considerably lower. More details below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a radius, larger than the Schwarzschild radius at which a neutron star, quark matter, whatever its equation of state, cannot be supported against collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are limits imposed by causality and General Relativity on the structure of compact stars. In &quot;Black Holes, White Dwarfs and Neutron Stars&quot; by Shapiro &amp;amp; Teukolsky, pp.260-261, it is shown, approximately, that even if the equation of state hardens to the point where the speed of sound equals the speed of light, that $(GM/Rc^2)&amp;lt;0.405$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Schwarzschild radius is $R_s=2GM/c^2$ and therefore $R &amp;gt; 1.23 R_s$ for stability. This limit is reached for a neutron star with $M \simeq 3.5 M_{\odot}$. A more accurate treatment in &lt;a href=&quot;http://arxiv.org/abs/1305.3510&quot; rel=&quot;noreferrer&quot;&gt;Lattimer (2013)&lt;/a&gt; suggests that a maximally compact neutron star has $R\geq 1.41R_s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the equation of state is softer, then collapse will occur at smaller masses, and higher densities but at a similar multiple of $R_s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus it is not necessary to compress matter within $R_s$ to form a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture below (from &lt;a href=&quot;http://adsabs.harvard.edu/abs/2010Natur.467.1081D&quot; rel=&quot;noreferrer&quot;&gt;Demorest et al. 2010&lt;/a&gt;) shows the mass-radius relations for a wide variety of equations of state. The limits in the top-left of the diagram indicate the limits imposed by (most stringently) the speed of sound being the speed of light (labelled &quot;causality&quot; and which gives radii slightly larger than Shapiro &amp;amp; Teukolsky's approximate result) and then in the very top left, the border marked by &quot;GR&quot; coincides with the Schwarzschild radius. Neutron stars become unstable where their mass-radius curves peak, so stable neutron stars are always significantly larger than $R_s$ at all masses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/c4Iry.jpg&quot; alt=&quot;Neutron star mass-radius relations&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-07-08T17:40:12.227" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11209" PostTypeId="2" ParentId="11206" CreationDate="2015-07-08T18:08:16.123" Score="2" Body="&lt;p&gt;You don't make it clear, but you &lt;em&gt;may&lt;/em&gt; be confused about what $g$ is. It is the statistical weight of an atomic energy level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $gf$ value, which refers to a particular transition between two energy levels in an atom/ion, is used because there is symmetry in terms of emission/absorption processes once the statistical weight is taken into account.&#xA;$$g_1 f_{12} = - g_2 f_{21}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just quoted $f$, then you would also need to know what the appropriate statistical weight was in order to calculate transition probabilities.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-07-08T18:08:16.123" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11210" PostTypeId="5" CreationDate="2015-07-08T21:07:34.353" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-07-08T21:07:34.353" LastActivityDate="2015-07-08T21:07:34.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11211" PostTypeId="4" CreationDate="2015-07-08T21:07:34.353" Score="0" Body="Questions about the parameters needed to specify the characteristics of an orbit." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-07-08T21:07:34.353" LastActivityDate="2015-07-08T21:07:34.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11212" PostTypeId="5" CreationDate="2015-07-08T21:10:24.150" Score="0" Body="&lt;p&gt;For questions about Johannes Kepler, use the &lt;a href=&quot;/questions/tagged/johannes-kepler&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;johannes-kepler&amp;#39;&quot; rel=&quot;tag&quot;&gt;johannes-kepler&lt;/a&gt; tag.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-07-08T21:10:24.150" LastActivityDate="2015-07-08T21:10:24.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11213" PostTypeId="4" CreationDate="2015-07-08T21:10:24.150" Score="0" Body="Questions about the NASA space observatory famous for finding potentially habitable exoplanets. Do not use this tag for questions about Johannes Kepler." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-07-08T21:10:24.150" LastActivityDate="2015-07-08T21:10:24.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11214" PostTypeId="5" CreationDate="2015-07-08T21:11:37.257" Score="0" Body="&lt;p&gt;For questions about the Kepler spacecraft, use the &lt;a href=&quot;/questions/tagged/kepler&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;kepler&amp;#39;&quot; rel=&quot;tag&quot;&gt;kepler&lt;/a&gt; tag.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-07-08T21:11:37.257" LastActivityDate="2015-07-08T21:11:37.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11215" PostTypeId="4" CreationDate="2015-07-08T21:11:37.257" Score="0" Body="Questions about the German astronomer famous for his three laws of planetary motion. Do not use this tag for questions about the Kepler spacecraft." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-07-08T21:11:37.257" LastActivityDate="2015-07-08T21:11:37.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11216" PostTypeId="5" CreationDate="2015-07-08T21:14:44.040" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-07-08T21:14:44.040" LastActivityDate="2015-07-08T21:14:44.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11217" PostTypeId="4" CreationDate="2015-07-08T21:14:44.040" Score="0" Body="Questions about the large ring of rocky bodies between Mars and Jupiter." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-07-08T21:14:44.040" LastActivityDate="2015-07-08T21:14:44.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11218" PostTypeId="5" CreationDate="2015-07-08T21:15:48.887" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-07-08T21:15:48.887" LastActivityDate="2015-07-08T21:15:48.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11219" PostTypeId="4" CreationDate="2015-07-08T21:15:48.887" Score="0" Body="Questions about large, dense, massive collections of stars orbiting galaxies, thought to have formed in the early universe." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-07-08T21:15:48.887" LastActivityDate="2015-07-08T21:15:48.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11220" PostTypeId="5" CreationDate="2015-07-08T21:17:38.210" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-07-08T21:17:38.210" LastActivityDate="2015-07-08T21:17:38.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11221" PostTypeId="4" CreationDate="2015-07-08T21:17:38.210" Score="0" Body="Questions about the chances of collisions between two objects or groups of objects, based on factors such as velocity, mass and number density." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-07-08T21:17:38.210" LastActivityDate="2015-07-08T21:17:38.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11222" PostTypeId="5" CreationDate="2015-07-08T21:20:19.947" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-07-08T21:20:19.947" LastActivityDate="2015-07-08T21:20:19.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11223" PostTypeId="4" CreationDate="2015-07-08T21:20:19.947" Score="0" Body="Questions about orbits with high eccentricities, which deviate greatly from perfect circles." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-07-08T21:20:19.947" LastActivityDate="2015-07-08T21:20:19.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11224" PostTypeId="5" CreationDate="2015-07-08T21:21:13.083" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-07-08T21:21:13.083" LastActivityDate="2015-07-08T21:21:13.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11225" PostTypeId="4" CreationDate="2015-07-08T21:21:13.083" Score="0" Body="Questions about quasi-stellar radio sources, active galactic nuclei that are far away and quite old." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-07-08T21:21:13.083" LastActivityDate="2015-07-08T21:21:13.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11226" PostTypeId="5" CreationDate="2015-07-08T21:23:16.143" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-07-08T21:23:16.143" LastActivityDate="2015-07-08T21:23:16.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11227" PostTypeId="4" CreationDate="2015-07-08T21:23:16.143" Score="0" Body="Questions about a mathematical distribution explaining how common different masses of stars are." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-07-08T21:23:16.143" LastActivityDate="2015-07-08T21:23:16.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11228" PostTypeId="5" CreationDate="2015-07-08T21:25:32.850" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-07-08T21:25:32.850" LastActivityDate="2015-07-08T21:25:32.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11229" PostTypeId="4" CreationDate="2015-07-08T21:25:32.850" Score="0" Body="Questions about the speed at which an object can go to a point at infinity while ending at rest. " OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-07-08T21:25:32.850" LastActivityDate="2015-07-08T21:25:32.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11230" PostTypeId="5" CreationDate="2015-07-08T21:28:09.523" Score="0" Body="" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-07-08T21:28:09.523" LastActivityDate="2015-07-08T21:28:09.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11231" PostTypeId="4" CreationDate="2015-07-08T21:28:09.523" Score="0" Body="Questions about the region of space around a star where a planet can sustain life." OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-07-08T21:28:09.523" LastActivityDate="2015-07-08T21:28:09.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11232" PostTypeId="2" ParentId="11198" CreationDate="2015-07-09T06:22:10.537" Score="4" Body="&lt;p&gt;Since this has gone unanswered, I'm going to give it a shot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the answer is no.   We know that on some moons, volcanic eruptions do send material into space.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Enceladus&quot; rel=&quot;nofollow&quot;&gt;Enceladus&lt;/a&gt; for example, feeds Saturn's E-ring.   It regularly shoots about 200 KG per second into space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other highly volcanic moon in our solar system is &lt;a href=&quot;https://en.wikipedia.org/wiki/Io_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Io&lt;/a&gt; which shoots material high into it's orbit, and it has some of the largest if not the largest observed volcanic eruptions.   Source:   &lt;a href=&quot;http://www.space.com/26732-jupiter-moon-io-volcano-eruption-photos.html&quot; rel=&quot;nofollow&quot;&gt;http://www.space.com/26732-jupiter-moon-io-volcano-eruption-photos.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I couldn't find any references that Io's volcanoes shot material into orbit.   &lt;a href=&quot;http://www.whillyard.com/science-pages/our-solar-system/jupiters-rings.html&quot; rel=&quot;nofollow&quot;&gt;short article on Jupiter's rings&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between Enceladus and Io is size.   Enceladus's escape velocity is 0.239 km/s (860.4 km/h) Source in link above, and it's gravitation is just over 1% of Earth's.    Io's escape velocity is 2.558 km/s and it's gravity is similar to the Moon's, about 18% of earth's.  If Io has a hard time sending volcanic material into orbit, on Mars, it would be nearly impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Io can shoot material up to 300 miles into the air, Source:  &lt;a href=&quot;https://en.wikipedia.org/wiki/Volcanology_of_Io&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Volcanology_of_Io&lt;/a&gt;  but that's still only about 1/7th to 1/8th of it's radius, probably well within it's &lt;a href=&quot;https://en.wikipedia.org/wiki/Sphere_of_influence_%28astrodynamics%29&quot; rel=&quot;nofollow&quot;&gt;Sphere of Influence&lt;/a&gt;.  On Mars, it's Sphere of influence is significantly larger and it's gravity twice as much.   If a super large Volcano shot material 300 miles into the air above Mars, that's still well below escape velocity.  (I could run the Maths if you like, but in a nutshell, escape velocity changes with the square of the distance, so, ignoring air resistance, Io's enormous eruption reaches 14% of it's diameter or 28% of the radius is about 40% of escape velocity - assuming I got my math right.   On the Mars, an equivalent enormous eruption would be what, 20% of necessary escape velocity?  Less?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, while Olympus Mons is enormous, it didn't all erupt at once.   Source:  &lt;a href=&quot;http://www.space.com/20133-olympus-mons-giant-mountain-of-mars.html&quot; rel=&quot;nofollow&quot;&gt;http://www.space.com/20133-olympus-mons-giant-mountain-of-mars.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Olympus Mons is over a hot spot on Mars, subject to regular volcanic eruptions (regular meaning, perhaps every few million years), but because Mars doesn't have plate movement, the eruptions happen in the roughly the same location and that, in combination with virtually no erosion and low gravity, is why Mons is as large as it is.   It probably does erupt impressively when it goes, but I'd be pretty skeptical that it erupts violently enough to send magma into space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, most of the Martian meteors on Earth are from a recent impact, about 5 million years ago:    &lt;a href=&quot;http://www.space.com/24962-mars-meteorites-single-impact.html&quot; rel=&quot;nofollow&quot;&gt;http://www.space.com/24962-mars-meteorites-single-impact.html&lt;/a&gt;   Olympus Mons may have erupted about 5 million years ago, so the timing works, but I still think it's unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I could add some links on how far Earth volcanoes can send magma/pumice if interested, I looked that up as well).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get that kind of velocity, you need significant amounts of concentrated energy.   A gun works by creating a controlled explosion with only one place for the energy to go.   Rockets work by sustained acceleration over time.  While it seems quite different, a meteor strike is similar in some ways to an enormous gun.   When a meteor strikes the surface of a planet, it hits with so much force in a localized area that the ground get's crushed and then rebounds with enormous force in a localized area.   A volcano simply can't (I think) generate that kind of concentrated energy unless it's very low gravity like Enceladus, 1.1% of the gravity of Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That probably needs tidying up a little.   I'll try to clean it up later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A somewhat related conversations on this subject if interested:   &lt;a href=&quot;http://www.reddit.com/r/askscience/comments/1gkiz3/could_a_volcano_eruption_theoretically_be/&quot; rel=&quot;nofollow&quot;&gt;http://www.reddit.com/r/askscience/comments/1gkiz3/could_a_volcano_eruption_theoretically_be/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-07-09T06:22:10.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11233" PostTypeId="2" ParentId="11120" CreationDate="2015-07-09T08:55:06.383" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Even 100 meter NEOs could cause significant damage, and there are far&#xA;  more of them. LSST will be able to detect objects as faint as 24.5 in&#xA;  magnitude in a 30s visit, enabling it to detect 140m NEOs &lt;strong&gt;&lt;a href=&quot;http://www.lsst.org/lsst/science/scientist_solar_system&quot; rel=&quot;nofollow&quot;&gt;as far away&#xA;  as the Main Belt asteroids.&lt;/a&gt;&lt;/strong&gt;&#xA;  [...] During its survey of the sky, LSST can find 90% of the PHAs over&#xA;  140 meters in diameter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Earth is 1 AU from the Sun, so &lt;a href=&quot;http://space-facts.com/asteroid-belt/&quot; rel=&quot;nofollow&quot;&gt;that would be&lt;/a&gt; &lt;strong&gt;1.2 to 2.2 AU away from us:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Asteroid Belt is located in an area of space between the orbits of&#xA;  Mars and Jupiter. That places it between 2.2 and 3.2 astronomical&#xA;  units (AU) from the Sun. The belt is about 1 AU thick. The average&#xA;  distance between objects in the Asteroid Belt is quite large. If you&#xA;  could stand on an asteroid and look around, the next one would be too&#xA;  far away to see very well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Timewise, it would be &lt;a href=&quot;http://www.lsst.org/lsst/sites/default/files/NEOwarning-1.pdf&quot; rel=&quot;nofollow&quot;&gt;about 8 years away:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The bottom right panel shows an example of a very different hazardous object - the 3 km large comet C/1996 B2 Hyakutake, which passed within 0.10 AU from Earth&#xA;  in 1996. The apparent magnitude vs. distance diagram indicates that the current&#xA;  generation of wide-angle imaging surveys, such as SDSS with V &amp;lt; 22,&#xA;  could have provided an early warning about 3 years before the closest&#xA;  approach (if they had sufficient cadence and sky coverage). &lt;strong&gt;With LSST,&#xA;  the warning time would be ∼8 years&lt;/strong&gt;, with over 500 observations over&#xA;  that period.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fortunately, your comet is 60 km, making the &lt;a href=&quot;https://en.wikipedia.org/wiki/Comet_tail&quot; rel=&quot;nofollow&quot;&gt;comet tail&lt;/a&gt; easy to spot:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;While the solid nucleus of comets is generally less than 50 km across,&#xA;  the coma may be larger than the Sun, and &lt;strong&gt;ion tails have been observed&#xA;  to extend 3.8 astronomical units&lt;/strong&gt; (570 Gm; 350×106 mi).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://neo.jpl.nasa.gov/risks/&quot; rel=&quot;nofollow&quot;&gt;This table of impact risks&lt;/a&gt; shows that the most likely (5.8e-02) impact is between 2095 and 2113.&lt;/p&gt;&#xA;" OwnerUserId="4068" LastEditorUserId="4068" LastEditDate="2015-07-09T09:13:32.793" LastActivityDate="2015-07-09T09:13:32.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11234" PostTypeId="2" ParentId="11178" CreationDate="2015-07-09T10:58:08.220" Score="2" Body="&lt;p&gt;The surface brightness of an extended source in independent of distance. So for the surface brightness of a large spiral go look at M31 (The Great Galaxy in Andromeda) . What will change with distance is the apparent size, but the surface brightness will remain the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface brightness of a near edge-on galaxy will be greater than one face-on due to the greater number of stars close to any sight-line, and the integrated brightness will of course vary with distance.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-07-09T13:08:29.787" LastActivityDate="2015-07-09T13:08:29.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11235" PostTypeId="2" ParentId="11178" CreationDate="2015-07-09T12:56:48.523" Score="2" Body="&lt;p&gt;Well, I didn't vote you down but it's a strange question in that there are some uncertainties, but it's also not hard to do the math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, which Galaxy? - they vary quite a bit, and from what angle?  10,000 light years from the side or from the &quot;top&quot; for lack of a better word, above the brighter galactic core or 10,000 light years from the tip of a spiral arm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Andromeda is 10 times brighter than the Milky way, and galaxies get a good deal larger than Andromeda and a good deal smaller than the Milky way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, the short answer to your question is, from 10,000 light years, a galaxy wouldn't appear that bright . . . er, mostly but it would appear quite large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see the Milky way as kind of a white smudge across the southern hemisphere, but much of that bright smudge that we see is stars that are closer than 10,000 light years, so that's not really apples to apples, as much of that smudge is alot closer to us than your example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets take our sun, from 10,000 light years away, lets, just for fun, say you had 10 billion of our suns, each 10,000 light years away.   10 billion suns is a lot, but light decreases by the square of the distance and there's 63,000 AUs in a light year so each sun at that distance would be 1/(397 million billion) less bright than the sun is from the earth.   10 billion of them, it's still 1/40 million times less bright than the sun is form the earth, roughly 1/100th the brightness of the full moon - that's what 10 billion of our suns would look like from that distance.   All together, that's still 10 times brighter than Venus, so if they were close together they'd shine clear as day, but spread out, like a galaxy would be at 10,000 light years away - not so much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a galaxy is more complicated cause some of the stars light would be blocked by dust and some stars would be much brighter than the sun, others much dimmer and a single supernova can outshine an entire galaxy for a couple of weeks.   There's no precise answer, but in general, a galaxy would look very large but not very bright from 10,000 light years away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another, and perhaps better way to look at this is to take a look at Andromeda, which is about 10 times the brightness of the Milky way and we have a pretty clear view of it, but it's also about 2.5 million light years away from us.    If it was, in your scenario, 250 times closer it would be 62,500 times brighter and much larger.   It would spread out over most of the sky, though, like the Milky way, we'd see a smudge of brightness, or we might be able to make out the spiral arms depending on the angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Andromeda's apparent brightness or apparent magnitude is 3.4, and each number in the brightness scale is 2.5 times brighter, so 62,500 times as bright, would be between 12 and 12.1 scales brighter, or -8.6 to -8.7, which for an object in the sky, is fairly bright, (yeah, negative numbers = brighter), but it's still quite a bit less bright than the full moon, and at the same time, spread out over a much larger area - so, if Andromeda was that close, the galactic center would probably be a fair bit more visible than Venus, brighter in total but also more spread out, and much less bright than the full moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short answer - the photos are probably brighter by a healthy amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a very cool and loosely related article.   &lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2014/01/01/moon_and_andromeda_relative_size_in_the_sky.html&quot; rel=&quot;nofollow&quot;&gt;http://www.slate.com/blogs/bad_astronomy/2014/01/01/moon_and_andromeda_relative_size_in_the_sky.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-07-09T12:56:48.523" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11236" PostTypeId="2" ParentId="9920" CreationDate="2015-07-09T20:14:51.240" Score="3" Body="&lt;p&gt;There are tons a amateur radio astronomy clubs and groups.  Radio Astronomy was basically started by an amateur named Grote Reber, where he basically did the experiment that you are pondering to do in 1937 in his backyard in Wheaton, Illinois.  You can visit that exact instrument at the NRAO facility in Green Bank, West Virginia.  It was named a National Historic Landmark in 1989.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just Google Amateur Radio Astronomy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Reber_Radio_Telescope&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Reber_Radio_Telescope&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nrao.edu/whatisra/hist_reber.shtml&quot; rel=&quot;nofollow&quot;&gt;http://www.nrao.edu/whatisra/hist_reber.shtml&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7653" LastActivityDate="2015-07-09T20:14:51.240" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11237" PostTypeId="1" CreationDate="2015-07-10T00:40:26.633" Score="6" ViewCount="124" Body="&lt;p&gt;I've been playing with simulations of co-orbital bodies similar to Saturn's moons Janus &amp;amp; Epimetheus- horseshoe orbits where the two bodies are of comparable mass- and I'm seeing some very odd patterns that I can't explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What &quot;should&quot; happen is that the body with the slightly smaller orbit will eventually overtake the outer body; when their mutual interaction becomes significant compared to their interactions with the primary, the lower body is accelerated forward into a higher orbit, while the higher body is accelerated backward into a lower orbit, they switch positions around their average orbital radius, and then separate again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm seeing in simulation, however, is a little more complicated. When I simulate the Saturn+Janus+Epimetheus system, the moons' semimajor axis very slowly migrate farther away from their average up until they swap; right around the swap, the difference in semimajor axes is maximized, and they drift closer together for half a cycle and then farther apart again before the next swap. This creates &quot;u&quot; and &quot;n&quot; shapes on the plot of semimajor axis over time, with the plots for each moon in opposite phases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Janus and Epimetheus, the effect is very small, but increasing the masses of the moons amplifies it, to the point that the maximum separation in orbital radii that occurs during the moons' closest approach to each other is ten or more times larger than the &quot;normal&quot; separation during the middle of a cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is of course the possibility that this an artifact of errors in my simulation software, but I think that is unlikely given the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;All other aspects of the Janus+Epimetheus system are reproduced perfectly- individual orbital periods are correct, average semimajor axes are correct, the frequency of the swaps is correct, and my simulation conserves energy to within less than 1 part in 1 trillion over a petasecond of simulation time (approx. 31,689 years).&lt;/li&gt;&#xA;&lt;li&gt;The effect is perfectly regular, and simulated systems remain stable. If the effect was a result of simulation errors, I wouldn't expect them to behave so nicely.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My first thought was that perhaps the semimajor axis of each moon was being affected by the added gravity of the other moon when they are on opposite sides of the primary from each other, and when they approach more closely they won't be pulling each towards the primary as strongly. That, however, should result in a uniform expansion and contraction of both moons' orbits as they get respectively closer and farther away from each other. What I actually see is that the inner moon's orbit gets &lt;em&gt;smaller&lt;/em&gt; while the outer moons orbit gets &lt;em&gt;larger&lt;/em&gt;, and vice-versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is this likely to just be a simulation error after all (if it helps, I am using the 4th-order Hermite integration algorithm)? If not, what is the explanation for it? Has such an effect been documented before, and if so, where?&lt;/p&gt;&#xA;" OwnerUserId="7655" LastEditorUserId="7655" LastEditDate="2015-07-10T00:55:30.357" LastActivityDate="2015-07-10T00:55:30.357" Title="Semimajor axis variations in co-orbital moons" Tags="&lt;orbit&gt;&lt;gravity&gt;&lt;saturn&gt;&lt;n-body-simulations&gt;" AnswerCount="0" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11238" PostTypeId="1" AcceptedAnswerId="11239" CreationDate="2015-07-10T02:11:39.130" Score="1" ViewCount="159" Body="&lt;p&gt;This is a homework question and I don't want an answer but just want to understand what its asking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are given that a star with a mass of &lt;code&gt;x solar masses&lt;/code&gt; in our galaxy turns to a supernova and releases &lt;code&gt;y&lt;/code&gt; amount of energy. Using this we are asked to calculate the average speed of the ejected mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;v = sqrt(2*E_k/M)&lt;/code&gt; (Kinetic energy equation) gives the average speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the questions asks the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Compare  your  result with our Sun’s orbital speed around the centre of the Milky Way galaxy. Where would you expect this material to end up?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it asking to solve and check that the average speed &gt; escape velocity if the star was at the same distance from the sun? Or how far from the center of the Galaxy would it end up?&lt;/p&gt;&#xA;" OwnerUserId="7656" LastActivityDate="2015-07-10T02:27:38.920" Title="Suns Orbital speed and escape velocity question" Tags="&lt;velocity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11239" PostTypeId="2" ParentId="11238" CreationDate="2015-07-10T02:27:38.920" Score="1" Body="&lt;p&gt;It's asking exactly what it says- &lt;em&gt;where&lt;/em&gt; would you expect this material to end up? Giving a distance from the center of the galaxy may be relevant, or it may not, depending on exactly what speed you end up with, and how it compares to galactic orbital / escape velocity. If the velocity is well above galactic escape velocity, giving a specific distance from the galactic core doesn't make much sense- there will be no fixed distance, the stuff will just keep going, achieving greater and greater distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the first step is indeed to solve and check if average speed &gt; orbital velocity or not (because the question says to do so), and a good second step would indeed be to check if it's less or greater than escape velocity. And from that information, you can guess generally where the stuff will go- stay reasonably close to the original orbit of the star? or stay in galactic orbit, but spread out in some wildly different range of orbits than what the original star had? or completely ejected from the galaxy?&lt;/p&gt;&#xA;" OwnerUserId="7655" LastActivityDate="2015-07-10T02:27:38.920" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11240" PostTypeId="1" CreationDate="2015-07-10T13:07:57.567" Score="1" ViewCount="85" Body="&lt;p&gt;Is China fully integrated in the international astronomical community, or does military secrecy and language problems and sanctions (ITAR) or something else stand in the way? Does China have valuable assets for astronomical observations, and a vibrant national astronomical academical community?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;China has an ambitious and successful HSF and Lunar program with orbiters and a lander. They have also announced a very ambitious program for &lt;a href=&quot;http://www.planetary.org/blogs/guest-blogs/van-kane/20141125-a-potpourri-of-future-mission.html?referrer=https://www.google.se/&quot; rel=&quot;nofollow&quot;&gt;inter-planetary probes&lt;/a&gt; to practically the entire Solar system, to date they've flewn by the NEO Toutatis. The are also building a &lt;a href=&quot;https://en.wikipedia.org/wiki/Five_hundred_meter_Aperture_Spherical_Telescope&quot; rel=&quot;nofollow&quot;&gt;500 meter radio telescope&lt;/a&gt; super-version of Arecibo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will for example US astronomers be able to use that telescope? Will they be offered samples of the upcoming Chinese lunar sample return, maybe from the lunar far side? Have lunar ground penetrating radar data from Chang'e 3 been presented publicly and led to papers in international journals? Is there any prominent Chinese astronomical journal?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="6" LastEditDate="2019-01-29T14:49:09.610" LastActivityDate="2019-01-29T14:49:09.610" Title="How significant and accessible is modern Chinese astronomy?" Tags="&lt;the-moon&gt;&lt;radio-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11242" PostTypeId="1" AcceptedAnswerId="11254" CreationDate="2015-07-10T21:59:35.323" Score="3" ViewCount="330" Body="&lt;p&gt;The Wikipedia page for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Blink_comparator&quot; rel=&quot;nofollow&quot;&gt;blink comparator&lt;/a&gt; fails to mention who invented it. Many other pages extol the importance of the device, but we cannot find any mention of an inventor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia does mention that Ben Mayer invented the The Projection Blink Comparator (PROBLICOM), which is a low-cost improvement on the blink comparator. However, he did not invent the original device.&lt;/p&gt;&#xA;" OwnerUserId="7662" LastEditorUserId="2153" LastEditDate="2015-07-11T15:18:58.153" LastActivityDate="2017-01-28T05:51:23.007" Title="Who invented the blink comparator?" Tags="&lt;telescope&gt;&lt;history&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11243" PostTypeId="1" AcceptedAnswerId="19857" CreationDate="2015-07-10T22:51:28.247" Score="19" ViewCount="2563" Body="&lt;p&gt;First, I know that modeling orbital mechanics of 8 planets is hard, but there are some theories out there, for example, Jupiter is thought to have moved in towards the sun then started moving away.   &lt;a href=&quot;http://www.nasa.gov/topics/solarsystem/features/young-jupiter.html&quot; rel=&quot;noreferrer&quot;&gt;Article&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and Uranus and Neptune may have switched spots &lt;a href=&quot;http://news.nationalgeographic.com/news/2007/12/071219-planet-swap.html&quot; rel=&quot;noreferrer&quot;&gt;Article&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any pretty good evidence on how the Earth's orbit has changed over time.   I remember reading some geological evidence that a year used to be longer, implying that the Earth used to be farther form the Sun, but I've since been unable to find that article, and for purposes of this question, lets count a day as 24 hours even though a day used to be quite a bit shorter hundreds of millions or billions of years ago.   - footnote, I've still not been able to find that article but it occurs to me, it could have been counting shorter days, not longer years - so, take that part with a grain of salt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any good studies out there on how many 24 hour days were in a year, 100, 300, 500, 800 million years ago?  or 1 or 2 billion years ago?  Either geological or orbital modeling?   Preferably something a layman can read, not something written by and for PHDs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or any good summaries, also encouraged.    Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also found this article, but it seems more theoretical than evidence based.   &lt;a href=&quot;http://www.futurity.org/did-orbit-mishap-save-earth-from-freezing/&quot; rel=&quot;noreferrer&quot;&gt;http://www.futurity.org/did-orbit-mishap-save-earth-from-freezing/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="2531" LastEditDate="2017-01-25T19:38:35.777" LastActivityDate="2017-01-25T19:38:35.777" Title="How has the Earth's orbit changed over hundreds of millions or billion of years?" Tags="&lt;orbital-migration&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11244" PostTypeId="1" CreationDate="2015-07-10T23:06:33.063" Score="6" ViewCount="132" Body="&lt;p&gt;Mars has much more trojans in L5. Does it reflect a common pattern?&lt;/p&gt;&#xA;" OwnerUserId="2541" LastActivityDate="2016-03-11T21:04:38.533" Title="Are trojans in L5 more likely than in L4?" Tags="&lt;orbit&gt;&lt;trojan-asteroids&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11245" PostTypeId="2" ParentId="11200" CreationDate="2015-07-10T23:29:46.173" Score="2" Body="&lt;p&gt;A lot of the answer depends on Orbital mechanics and calculating that isn't easy.   If no action is taken, Earth's orbit is still likely to change, but how much and in which direction is hard to say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long before the sun goes red giant it will get too hot for the Earth.  If the earth doesn't move, the sun's increased temperature will likely make the earth uncomfortable in a few hundred million years and perhaps scorching in as little as a billion years.   Warmer air and more water vapor in the air will trap heat and at a certain point, you'd get a run-away green house effect, long before the red-giant phase.    Short term, changes in CO2 in the atmosphere might slow that.   We could see ice ages before we see run-away heating, but over time, with no human action, run-away heating is likely inevitable - on a very large timescale, like a billion years or so.   Longer if the Earth drifts farther from the sun.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;what will the last few decades of Earth's life be like? Will the&#xA;  planet be heated to the point where its surface is like Mercury's&#xA;  sun-side? Will the sun's output drop low enough that it instead cools,&#xA;  like Mars? Will the sun expand and engulf the planet, or will the&#xA;  planet continue to orbit an expanded sun as a lifeless ball?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Will the sun's expansion be gradual or rapid? How will this expansion&#xA;  affect the rocky planets nearest to it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm going to answer the 2nd part first.   The sun's expansion will first be gradual, but the red giant stage will be fairly rapid.   Over the next 4 billion years or so, the sun will get some 60% more luminous, so significantly much more heat at the same orbit.   A very rough rule of thumb is 10% more luminous every billion years. (that's not for every star, just for 1 solar mass) - and that's a pretty ballpark estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But just 10% more heat from the sun would raise the Earth's temperature uncomfortably, and with the added water vapor in the air and greenhouse effect, the earth might be a permanent steam room, even at the poles, at just 10% more heat from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;asking about the last few decades is the wrong time frame as changes would probably happen much more slowly, though atmospheric/feedback mechanisms in the climate might make changes happen more quickly - ice age speed for example, a few thousand years or so.   Also, as Antarctica drifts away from the south pole, we might see an end to ice ages - there's lots of moving parts to long term climate predictions, but in general, you'd see gradual heating to the point where complex life might have a hard time adapting.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The suns output won't drop until after it goes red giant, it's on a steady but gradual increase, followed by a relatively fast increase at the red giant stage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if we terraform Titan, that might be a safe distance to watch the sun go red giant from and it would probably be quite a spectacular show, well, over enough time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to how fast it will go red-giant, fast for a star means, about 200 million years, so, from the perspective of a human lifetime, watching a star go red giant would be like watching grass grow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A much more intelligent discussion on that than I can give, here:   &lt;a href=&quot;https://physics.stackexchange.com/questions/25622/how-fast-will-the-sun-become-a-red-giant&quot;&gt;https://physics.stackexchange.com/questions/25622/how-fast-will-the-sun-become-a-red-giant&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-07-11T10:10:14.353" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11246" PostTypeId="1" AcceptedAnswerId="11248" CreationDate="2015-07-10T23:49:52.357" Score="5" ViewCount="1115" Body="&lt;p&gt;With the rate at which the moon is further receding into a higher orbit, how long until the barycentre between us and the moon leaves the earth, and going by the IAUs 2006 definition update, we become a double planet?&lt;/p&gt;&#xA;" OwnerUserId="7663" LastActivityDate="2015-07-11T04:58:42.010" Title="How long until the Earth and Moon become a binary planet?" Tags="&lt;the-moon&gt;&lt;gravity&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11247" PostTypeId="1" CreationDate="2015-07-11T00:31:13.947" Score="9" ViewCount="1386" Body="&lt;p&gt;Why are radio telescopes typically only a dish with a receiver above it, while optical telescopes have a primary, plus secondary and sometimes even a tertiary mirror?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, why do radio telescopes have only one reflector, while optical telescopes have up to three or more?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same wave phenomena, such as focusing, should apply in both cases. So I don't understand why the geometry would be radically different. You could just replace actual mirrors with a convenient solid, such as plastic, that reflects radio waves just as wells as mirrors reflect optical waves.&lt;/p&gt;&#xA;" OwnerUserId="7449" LastActivityDate="2015-07-17T01:25:29.773" Title="Why are radio telescopes shaped so differently than optical telescopes?" Tags="&lt;telescope&gt;&lt;radio-astronomy&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11248" PostTypeId="2" ParentId="11246" CreationDate="2015-07-11T00:52:29.733" Score="7" Body="&lt;p&gt;I'm not sure I agree with the double planet POV, but the calculation is pretty simple.   The earth weighs 81 moons, so for the Barycenter to be outside the earth, the distance (center of Moon to surface of earth), = 81 earth radii.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XRg2K.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or about 515,000 KM.   It's current farthest distance is 405,000 KM, average distance 384,000 KM and closest 363,000 KM  &lt;a href=&quot;http://www.space.com/18145-how-far-is-the-moon.html&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;, so, it depends on whether you mean, temporarily outside, outside more than 50% of the time or always outside - each would provide different answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But moving away 4 CM per year (same source) or 1 KM every 25,000 years, it would need 2.75 billion years to move away the necessary 110,000 KM necessary to have the barycenter move outside of earth at the moon's furthest point.  (less if the orbit gains eccentricity which is possible).   But probably more than that, cause as the moon moves away from the earth, the tidal forces that continue to push it away grow weaker and the earth slows down a little.    If the earth ever gets tidally locked with the moon, the reverse will happen and the sun's tidal forces will draw the Earth and Moon slowly towards each other, so a precise answer is too mathematically difficult for me, but longer than 2.75 billion years seems a pretty good guestimate.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-07-11T04:58:42.010" LastActivityDate="2015-07-11T04:58:42.010" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11249" PostTypeId="2" ParentId="11247" CreationDate="2015-07-11T04:40:53.673" Score="3" Body="&lt;p&gt;Radio telescopes are shaped differently primarily because we can't see microwaves or radio waves. Optical telescopes are designed so that there is a focal point where you can look and see the image. However, radio telescopes and optical telescopes actually work very similarly, and sometimes radio telescopes do have secondary reflectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an optical telescope the secondary mirrors are generally meant to redirect the light and focus the image for your eye. The primary mirror is what is gathering the light, so it's what is doing the magnification. You can see a great example of this with the image of the Newtonian telescope below (thank you, Wikipedia!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zOOmW.png&quot; alt=&quot;Newtonian telescope&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Radio Telescopes&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Radio telescopes actually work in a very similar manner. The &quot;dish&quot; portion of the telescope is reflecting the waves, the same as the primary mirror in the optical scope. It is then received at the LNB/LNA/receiver part. You can think of that as the focal point where the secondary mirror is positioned in the optical telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to that, sometimes radio telescopes actually do have a secondary reflector. I'll use an image of a radio telescope at NASA Jet Propulsion Laboratory's Goldstone Deep Space Communications Complex to show this (also from Wikipedia). The &quot;dish&quot; is the primary reflector, then it's being reflected again at the secondary reflector being held up by the metal arms. After the second reflection the signal goes into the receiver attached to the primary reflector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DCy5r.jpg&quot; alt=&quot;Cassegrain radio antenna&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1980" LastEditorUserId="1980" LastEditDate="2015-07-11T04:48:28.410" LastActivityDate="2015-07-11T04:48:28.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11251" PostTypeId="1" AcceptedAnswerId="11252" CreationDate="2015-07-11T08:14:32.840" Score="10" ViewCount="562" Body="&lt;p&gt;When the Sun runs out of hydrogen to fuse it will grow bigger in size and my question is does it rotate more slowly, like a spinning ice skater extending their arms to reduce angular velocity? What consequences will accompany with a slower dynamo (big ball of plasmas), such as more frequent solar flares or perhaps more dark spots?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="2531" LastEditDate="2015-07-11T10:00:12.667" LastActivityDate="2015-07-11T10:00:12.667" Title="Will the Sun rotate slower in red giant form?" Tags="&lt;star&gt;&lt;stellar-evolution&gt;&lt;magnetic-field&gt;&lt;red-giant&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11252" PostTypeId="2" ParentId="11251" CreationDate="2015-07-11T09:56:46.840" Score="10" Body="&lt;p&gt;Indeed conservation of angular momentum dictates that in a single star like the the Sun, rotation should be much slower when it becomes a red giant. This is because at the present time the Sun does not rotate at vastly different rates with depth, thus when it expands, the moment of inertia increases drastically and convection in the outer envelope will ensure that slow rotation is enforced throughout most of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The consequence for magnetic activity will be that the Sun will become magnetically &lt;em&gt;inactive&lt;/em&gt; (in relative terms - magnetic activity will not be absent, but its signatures will be much reduced when expressed as a fraction of the stellar luminosity) as a red giant, because dynamo-generated magnetic activity is strongly correlated with rotation rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general it is found that giant stars are magnetically inactive and slowly rotating, but there are exceptions. If a star considerably more massive ($\sim 3-5M_{\odot}$) than the Sun evolves to become a giant it may have preserved a considerable amount of angular momentum, since such stars are not magnetically active on the main sequence and are unable to lose their angular momentum through a magnetised wind in the same way that a solar-type star does. Secondly, it is possible that some rapidly rotating giants (e.g. the FK Com stars) could be the result of mergers in binary systems or possibly even giant planet engulfment. In which case the angular momentum is from the binary system and these stars can be very magnetically active. Finally it is also found that giants in close binary systems (the RS CVn stars) can be fast-rotating and magnetically active because tidal locking in the binary system enforces rapid rotation. The FK Com and RS CVn stars manifest many signs of extreme magnetic activity - hot X-ray coronae, chromospheres, coverage of large fractions of the surface by cool starspots. We thus deduce that the difference between these and the vast majority of magnetically-inactive giants is their rapid rotation.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-07-11T09:56:46.840" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11254" PostTypeId="2" ParentId="11242" CreationDate="2015-07-11T14:30:34.547" Score="5" Body="&lt;p&gt;It would appear to have been developed by &lt;a href=&quot;http://www.city-of-light.com/en/topics/topics-detailpage/?tx_ttnews%5Btt_news%5D=275&amp;amp;cHash=2e1a29de74aaa837a979293e8e926538&quot;&gt;Carl Pulfrich working for Zeiss&lt;/a&gt; in 1904.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively &lt;a href=&quot;http://www.math.harvard.edu/~knill/history/wolf/bea_proofs_wolf.pdf&quot;&gt;Max Wolf&lt;/a&gt; in 1900 again working with the Zeiss company.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looks like the idea was Wolf's and the realisation Pulfrich's. From the second link we have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Wolf was a codeveloper of the stereo comparator together with Carl&#xA;  Pulfrich from the Zeiss company. The  stereo comparator consists of a&#xA;  pair of microscopes arranged so that one can see simultaneously two&#xA;  photographic plates of the same region taken at diff erent times. Wolf&#xA;  seems to have experimented with such techniques as early as 1892, but&#xA;  without success. When Pulfrich approached him to adapt the technique&#xA;  from geodesy to astronomy, Wolf was delighted. A steady exchange of&#xA;  letters followed. Wolf and Pulfrich then worked together to analyze&#xA;  the rapidly growing accumulation of photographic plates. Tragically,&#xA;  Pulfrich lost one eye in 1906, preventing him from using the&#xA;  stereographic tool from then on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-07-11T14:59:56.753" LastActivityDate="2015-07-11T14:59:56.753" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11255" PostTypeId="1" AcceptedAnswerId="11264" CreationDate="2015-07-11T15:43:32.460" Score="4" ViewCount="7964" Body="&lt;p&gt;So as I understand it the source of most geothermal energy on Earth is due to radioactive decay. Jupiter is the most massive planet but if the bulk of it's mass comes from gaseous elements instead of heavier ones like uranium or whatever than it being the most massive might not necessarily mean it has the most radioactive material.&lt;/p&gt;&#xA;" OwnerUserId="7668" LastActivityDate="2019-05-20T12:44:22.773" Title="what planet is the most radioactive?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;radiation&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11256" PostTypeId="1" AcceptedAnswerId="11263" CreationDate="2015-07-11T21:48:19.213" Score="1" ViewCount="421" Body="&lt;p&gt;I'm just getting started with very primitive analysis of FITS data, and I have a 'raw' FITS file which I don't know how to get different wavebands of data from (if this is even possible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Filename: casa_raw.fits&#xA;No.    Name         Type      Cards   Dimensions   Format&#xA;0    PRIMARY     PrimaryHDU      33   ()&#xA;1    EVENTS      BinTableHDU    793   14621280R x 19C   [1D, 1I, 1I, 1J, 1I, 1I, 1I, 1I, 1E, 1E, 1E, 1E, 1J, 1J, 1E, 1J, 1I, 1I, 32X]&#xA;2    GTI         BinTableHDU     28   2R x 2C      [1D, 1D]&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone explain what exactly this information conveys?&lt;/p&gt;&#xA;" OwnerUserId="7671" LastActivityDate="2015-10-10T12:13:45.797" Title="What does the BinTableHDU store?" Tags="&lt;photography&gt;&lt;astropy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11257" PostTypeId="1" AcceptedAnswerId="11268" CreationDate="2015-07-11T22:39:38.490" Score="3" ViewCount="4539" Body="&lt;p&gt;I'm making an application that needs to calculate a planet's current position within the solar system - I don't want to calculate its position in the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, what I want to do is exactly what this website does:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.theplanetstoday.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.theplanetstoday.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to create an overhead view of all the planet's positions in the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where do I start? Is there a latitude/longitude coordinate system for the solar system? Everything I've found seems to calculate a planet's position in the sky from the observer's point on earth.&lt;/p&gt;&#xA;" OwnerUserId="7672" LastEditorUserId="7672" LastEditDate="2015-07-12T01:27:43.733" LastActivityDate="2016-02-08T13:51:06.623" Title="How to calculate a planet's current position within the solar system" Tags="&lt;solar-system&gt;&lt;planet&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11260" PostTypeId="1" AcceptedAnswerId="11262" CreationDate="2015-07-12T03:35:09.263" Score="11" ViewCount="1972" Body="&lt;p&gt;I think that &lt;em&gt;size&lt;/em&gt; and &lt;em&gt;mass&lt;/em&gt; do not correlate to &lt;em&gt;temperature&lt;/em&gt;, but then again these factors contribute to the internal pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know if there is a limit to &lt;strong&gt;how hot a star can get and what mechanism(s) could drive a star to get unusually hot&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know that negative temperature occurs in laser is hotter than a positive temperature, and can a star produce negative temperature?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorDisplayName="user2216" LastEditDate="2015-07-13T01:39:21.300" LastActivityDate="2015-07-29T03:55:19.393" Title="Is there a limit to how hot a star can be?" Tags="&lt;star&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11262" PostTypeId="2" ParentId="11260" CreationDate="2015-07-12T07:57:44.740" Score="9" Body="&lt;p&gt;Yes, there is a limit. If the radiation pressure gradient exceeds the local density multiplied by the local gravity, then no equilibrium is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radiation pressure depends on the fourth power of temperature. Radiation pressure gradient therefore depends on the third power of temperature multiplied by the temperature gradient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence for stability&#xA;$$ T^3 \frac{dT}{dr} \leq \alpha \rho g,$$&#xA;where $\rho$ is the density, $g$ is the local gravity and $\alpha$ is some collection of physical constants, including how opaque the material is to radiation. Because there must be a temperature gradient in stars (they are hotter on the inside than the outside) this effectively puts an upper limit to the temperature. It is this that sets an upper limit of around 60,000-70,000 K to the surface temperature of the most massive stars, which are dominated by radiation pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In regions of higher density or higher gravity, radiation pressure is not such an issue and temperatures can be much higher. The surface temperatures of White dwarf stars (high density and gravity) can be 100,000 K, the surfaces of neutron stars may exceed a million K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course stellar interiors are much denser and consequently can be much hotter. The maximum temperatures there are controlled by how quickly heat can be moved outwards by radiation or convection. The very highest temperatures of $\sim 10^{11}$ K are reached at the centres of core-collapse supernovae. Ordinarily, these temperatures are unattainable in a star because cooling by neutrinos can carry energy away highly effectively. In the final seconds of a CCSn, the density gets high enough that neutrinos become trapped and so the gravitational potential energy released by the collapse cannot escape freely - hence the high temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the last part of your question, yes there are &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Astrophysical_maser&quot; rel=&quot;nofollow&quot;&gt;astrophysical masers&lt;/a&gt; found in the envelopes of some evolved stars. The pumping mechanism is still debated. The brightness temperatures of such masers can be much higher than anything discussed above.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-07-12T19:21:11.737" LastActivityDate="2015-07-12T19:21:11.737" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11263" PostTypeId="2" ParentId="11256" CreationDate="2015-07-12T09:21:49.903" Score="0" Body="&lt;p&gt;This shows that the file contains 3 Header-Data Units (HDUs; you can read more about this &lt;a href=&quot;http://fits.gsfc.nasa.gov/fits_primer.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;). The headers contain meta-data. The primary HDU (0) has some 'cards' (basically keyword/value pairs most of the time, or comments) but no data. The next HDU (1, called 'EVENTS' in your file) is a binary table which contains 19 columns. Finally, the last HDU (2, also called GTI) contains another table with 2 columns. You can access the data in one of the columns of the EVENTS table using for example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from astropy.io import fits&#xA;hdulist = fits.open('myfile.fits')&#xA;hdulist['EVENTS'].data['colname']&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where &lt;code&gt;colname&lt;/code&gt; is the name of one of the columns. You can find out the column names by looking at:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;hdulist['EVENTS'].data.names&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7675" LastActivityDate="2015-07-12T09:21:49.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11264" PostTypeId="2" ParentId="11255" CreationDate="2015-07-12T11:07:50.913" Score="4" Body="&lt;p&gt;There's a few parts to this question, I'll give it a shot.  Also, since you're asking about the geothermal energy of the earth, I'm going to ignore radiation from space and radioactive van allen belts (Jupiter has a very radioactive van Allen belt).   Also, as voyager discovered when it left the solar system, there's more radiation in the form of cosmic rays out there, that the sun effectively shields most of the planets from.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, because you mention the Earth's geothermal heat from radiation, so I'll focus on that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a very well answered similar question about radioactive elements inside the earth.  &lt;a href=&quot;https://earthscience.stackexchange.com/questions/4798/what-percent-of-the-earths-core-is-uranium&quot;&gt;what percentage of the Earth's core is Uranium&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So as I understand it the source of most geothermal energy on Earth is&#xA;  due to radioactive decay.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A healthy share of Earth's core heat is from the planets formation.  This is a pretty good article on that subject - &lt;a href=&quot;http://physicsworld.com/cws/article/news/2011/jul/19/radioactive-decay-accounts-for-half-of-earths-heat&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://physicsworld.com/cws/article/news/2011/jul/19/radioactive-decay-accounts-for-half-of-earths-heat&lt;/a&gt;, it says that of the current heat, about 50% is from the energy left from the formation of the planet and about 50% is from radiation - so the radiation is very important, but it's probably not the cause of &quot;most&quot; of the heat in the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, no similar studies have been done on any of the other planets.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Jupiter is the most massive planet but if the bulk of it's mass comes&#xA;  from gaseous elements instead of heavier ones like uranium or whatever&#xA;  than it being the most massive might not necessarily mean it has the&#xA;  most radioactive material.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That I pretty much agree with.  The two primary differences between the 4 gas giant planets (or, two gas giant, two ice giants if you like) and the 4 inner rocky planets is that the gas giants formed outside the &lt;a href=&quot;https://en.wikipedia.org/wiki/Frost_line_%28astrophysics%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;frost line&lt;/a&gt; and key difference #2, the 4 outer planets are much larger, so they were more able to hold onto more hydrogen and helium, which are the 2 most abundant elements in the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, the 4 outer planets have a lower percentage of heavy elements overall.    As to their percentage of Uranium - well, your guess is as good as mine, but, probably a lower percentage of radioactive elements than the inner planets seems reasonable to assume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, any radioactive elements in the 4 outer planets would likely be well below their thick gaseous surfaces, so it would be difficult to detect, outside of Neutrino detection I suppose, and more trouble than it's worth to try to collect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 4 inner planets (and lets throw the moon in there as well, and asteroids if you like) are more interesting I'm assuming David Hammen's answer is correct in the linked question above, and that most of the Uranium on earth is in the crust or mantle, the highest percentage in the crust, and so that's probably true for any other rocky planet that formed and was, at one time, molten enough to circulate, so all the inner planets probably have similarly small amounts of uranium and other radioactive elements as Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury may have had much of it's crust and some of it's mantle blown off in a giant collision, and if that's the case, Mercury might have lost a portion of it's radioactive elements.  Brief summary of that here, point #6:  &lt;a href=&quot;http://messenger.jhuapl.edu/faq/faq_science.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://messenger.jhuapl.edu/faq/faq_science.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;unrelated footnote, but if Mercury's crust was blown off, it might be a good planet to mine for rare earth elements, or maybe asteroids will be easier, who knows, but I've always been curious of the mining possibilities on Mercury.  Abundant solar energy too, but it's a hard planet to fly to because of the sun's gravity, so, maybe asteroids are the way to go for space mining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon was also formed by a giant impact, so, the moon was formed by material from Earth's but it has a much higher percentage of mantle than the Earth, so it could have a higher percentage of Uranium and other radioactive elements than the earth.  How well the Moon's material was circulated in it's cooling process is another question.   Being smaller it probably cooled more quickly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus and Mars, I see no reason why they shouldn't have similar percentages of radioactive elements than Earth.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RQN7S.png&quot; alt=&quot;enter image description here&quot;&gt; Source:  &lt;a href=&quot;https://en.wikipedia.org/wiki/Internal_structure_of_the_Moon&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Internal_structure_of_the_Moon&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to look at this question is to investigate how quickly the other inner planets cooled.  Far from exact, but that probably tells us something, since heat generated by radioactive decay inside the planet would probably lead to volcanism at some point.  Even the Moon appears to have had volcanic eruptions as recently as 33 million years ago, so some ongoing internal heating seems likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://news.discovery.com/space/imps-reveal-volcanoes-erupted-recently-on-the-moon-141014.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Volcanoes on the moon&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Size is also a key factor in what keeps the inside of a planet hot, and in some cases, tidal heating (Io, Enceladus), but that's not really relevant to this question.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither Mars nor Venus are volcanically active like the earth, but Olympus Mons may have erupted as recently as 25 million years ago, and there is evidence of large lava flows on Venus.   As to how big a role radioactivity plays in heating up the cores of those 2 planets, I don't know precisely how much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, I think this is a good question on planet cores and radioactivity, as the inside of the planet is just as interesting as the surface, well, to me, at least.   I know this doesn't answer your question specifically as to which is the most radioactive, cause I don't know, but I like the question.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:29.267" LastActivityDate="2015-07-12T13:22:36.203" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11265" PostTypeId="2" ParentId="10934" CreationDate="2015-07-12T13:37:01.343" Score="1" Body="&lt;p&gt;Seems to me that moons are regularly subjected to tidal forces that comets, asteroids and the like are not.&#xA;Repeated stress, such as rocks in a rock tumbler, or two circles of glass on a &lt;a href=&quot;https://en.wikipedia.org/wiki/Amateur_telescope_making#Mirror_making&quot; rel=&quot;nofollow&quot;&gt;mirror maker's table&lt;/a&gt;, tends to smooth things out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure there's a pile of academic papers on sub-disruptive levels of strain tending to increase the symmetry of many types of systems, likely even some formal proofs, but I can't search at that level this morning.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-07-12T13:37:01.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11266" PostTypeId="2" ParentId="11240" CreationDate="2015-07-12T13:55:03.420" Score="1" Body="&lt;p&gt;China is pretty well integrated into modern astronomy -- there are regular international conferences held in China (the most recent General Assembly of the International Astronomical Union was held in Beijing in 2012), lots of Chinese astronomers at other international conferences, Chinese participation in international observatory projects (the Joint Institute for VLBI, the Thirty Meter Telescope, the Square Kilometer Array, etc.), and so forth. I'm not sure I'd say they have any really significant facilities &lt;em&gt;yet&lt;/em&gt; (with the possible exception of LAMOST), but they have very ambitious plans, as you've noted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ITAR has very little to do with astronomy except when it comes to a few things like nuclear physics codes produced by places like Los Alamos.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how much international participation is anticipated for the FAST project, though I gather the instrument development has some involvement from British and Australian institutes, so they'd likely get some time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, results from the Chang’E-3 mission have been published in scientific papers, e.g.&#xA;&lt;a href=&quot;http://www.sciencemag.org/content/347/6227/1226.abstract&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencemag.org/content/347/6227/1226.abstract&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several Chinese astronomy journals (e.g., Acta Astronomica Sinica; Research in Astronomy and Astrophysics), but like most &quot;national&quot; journals they aren't that important internationally.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2015-07-12T13:55:03.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11267" PostTypeId="2" ParentId="8105" CreationDate="2015-07-12T15:02:39.100" Score="7" Body="&lt;p&gt;Even if we assume a dark-adjusted human eye could in principle discern artificial lights on the night side of the Earth, there's the problem that your eyes won't be dark-adjusted because the day side will be much, &lt;em&gt;much&lt;/em&gt; brighter than the night side. There's not enough dynamic range in the eye to sense that sort of a brightness difference at once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if you'd be looking when the Earth is new, so there's no day side visible, you'd have an even worse problem: the Sun would be in your field of view. A solar eclipse (a lunar eclipse as seen from the Earth) would be your best bet - except that the atmosphere would form a brilliant ring around the dark Earth...&lt;/p&gt;&#xA;" OwnerUserId="2134" LastEditorUserId="2134" LastEditDate="2015-07-12T21:33:54.807" LastActivityDate="2015-07-12T21:33:54.807" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11268" PostTypeId="2" ParentId="11257" CreationDate="2015-07-12T15:28:43.673" Score="2" Body="&lt;p&gt;Buy &lt;em&gt;Astronomical Algorithms&lt;/em&gt; 2nd ed. by Jean Meeus. It is published by Willmann-Bell Inc. If you buy it direct from the publisher, you will get a corrected printing with all errata up to August 10, 2009. There are a number of systems for describing the location of a planet. This book provides a method to find the heliocentric longitude, latitude, and distance from the Sun. It also explains the different coordinate systems and how to transform from one to another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method you use depends on how accurate you need to be. If you state your needed accuracy some of us might be able to suggest other alternatives. &lt;/p&gt;&#xA;" OwnerUserId="7385" LastActivityDate="2015-07-12T15:28:43.673" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11269" PostTypeId="2" ParentId="11247" CreationDate="2015-07-12T21:27:49.443" Score="3" Body="&lt;p&gt;One of the reasons for the difference is the sheer number of different optical (and near-infrared) instruments available. Most professional optical telescopes have two or more standard instruments (e.g., an imager and a spectrograph), with the possibility of adding guest instruments from time to time; some have as many as five standard instruments at the same time. Having the instruments mounted at the base of the telescope makes it much easier to switch between them (sometimes, as with a Nasmyth mount, by simply rotating the tertiary mirror 90 or 180 degrees) than it would be if the instruments were mounted at prime focus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See, for example, the picture at this web page for the SOAR Telescope, which has ports for five different instruments:&#xA;&lt;a href=&quot;http://www.lna.br/soar/telescope_e.html&quot; rel=&quot;nofollow&quot;&gt;http://www.lna.br/soar/telescope_e.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2015-07-12T21:27:49.443" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11271" PostTypeId="1" CreationDate="2015-07-13T15:20:39.263" Score="1" ViewCount="123" Body="&lt;p&gt;I hope this is the right place to ask this question, and I apologise if not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've come across a date written in an old book, but it is written using what seems to be a lunar/zodiac system.&#xA;Below is a picture of the date (ignore the text above it):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eW4gz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to work out what this date would have been in our modern calendar, but I can't seem to find any tables or almanacs that go back that far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any publicly available tables that do go back to 1587? If not, how exactly could I go about calculating the answer for myself? I am a mathematician, so a bit of wacky maths should be within my capabilities.&#xA;This is the answer I'm probably expecting from this site: some kind of program/algorithm/formula to be able to work out on what date the moon would have been in Taurus in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, any ideas as to where best for me to ask/look next?&lt;/p&gt;&#xA;" OwnerUserId="7680" LastActivityDate="2016-01-05T22:48:24.997" Title="Historical astronomical lunar tables" Tags="&lt;the-moon&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11272" PostTypeId="1" AcceptedAnswerId="11281" CreationDate="2015-07-13T16:17:43.923" Score="4" ViewCount="1487" Body="&lt;p&gt;Is there freely available and reliable source, where one can find positions of planets and moons of the Solar system at given time in cartesian coordinates? (I am writing a simple gravity simulator and it would save me a lot of work with conversion.)&lt;/p&gt;&#xA;" OwnerUserId="2529" LastActivityDate="2015-07-14T05:11:32.330" Title="Planets and moons positions in cartesian coordinates?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;data-analysis&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11273" PostTypeId="1" CreationDate="2015-07-13T16:26:27.900" Score="5" ViewCount="8482" Body="&lt;p&gt;I need a way of to calculate the moon phase. Do you know resources for research the theory from behind this? I basically have searched in Google, but the results were not very satisfactory. Exists a mathematical formula for do this calculation?&lt;/p&gt;&#xA;" OwnerUserId="7681" LastActivityDate="2015-09-03T21:29:36.793" Title="How to calculate the moon phase" Tags="&lt;the-moon&gt;&lt;moon-phases&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" ClosedDate="2015-09-08T10:27:41.897" ContentLicense="CC BY-SA 3.0" />
  <row Id="11274" PostTypeId="2" ParentId="6376" CreationDate="2015-07-13T18:19:51.693" Score="1" Body="&lt;p&gt;Cosmic ray &quot;hits&quot; are artifacts caused when high energy particles (cosmic rays, often muons) slam into atoms in the CCD itself and liberate large numbers of electrons, which then show up as bright spots and streaks when the CCD data is read out. They can also occur as a result of radioactive decay processes in the material in the detector and instrument itself and also the immediate surroundings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are a tremendous nuisance that can be removed to a certain extent by median stacking images - though this also fails if you get into a situation where cosmic ray hits are found in the same pixels of $&amp;gt;1$ images in your stack. This limits the lengths of exposures that can safely be made with astronomical CCDs. Depending on exact usage, this limit is usually between 30 minutes and an hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flux of muons and the distribution of angles at which they reach the detector is actually very well known already. It is largely unaffected by local shielding (unless your telescope is buried far underground!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to distinguish different types of events (muons, locally produced gamma rays etc.) from looking at the pattern made on the detector (e.g. &lt;a href=&quot;http://www.astronomy.ohio-state.edu/MDM/OSMOS/CCD_CosmicRays_groom.pdf&quot; rel=&quot;nofollow&quot;&gt;Groom 2004&lt;/a&gt;), but I don't think you can tell much about their energies, unless it could be by looking at the flux as a function of how much shielding (lead?) or how far underground the detector was, since often only a small fraction of the cosmic ray energy ends up deposited in the CCD.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-07-13T18:46:36.090" LastActivityDate="2015-07-13T18:46:36.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11275" PostTypeId="2" ParentId="1812" CreationDate="2015-07-13T19:26:10.280" Score="6" Body="&lt;p&gt;If, Venus had started out at Mars' position, it is possible it would've ended up evolving into an ice giant. At the greater distance of Mars the Sun's solar wind is weaker and would've stripped off much less of Venus' atmosphere, even in comparison to the amount Earth lost, as Earth is considerably closer to the Sun than Mars. So, Venus would perhaps have been left with a thick helium envelope and become a mini-Neptune/super-Earth rather than a second Earth. &#xA;And also, if Mars had evolved at the distance of Venus, closer in, it would've been stripped of any appreciable atmosphere and possibly would've evolved into another Mercury, a grey and airless world.&lt;/p&gt;&#xA;" OwnerUserId="7682" LastActivityDate="2015-07-13T19:26:10.280" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11276" PostTypeId="2" ParentId="11271" CreationDate="2015-07-13T23:21:50.400" Score="2" Body="&lt;p&gt;You can calculate this using the program &quot;KAIROS&quot;, See here: &lt;a href=&quot;http://www.raymondm.co.uk/&quot; rel=&quot;nofollow&quot;&gt;http://www.raymondm.co.uk/&lt;/a&gt;&#xA;It will let you play through the variables: Is this a Julian or a Gregorian date? Is it using true or mean positions? And so forth.&lt;/p&gt;&#xA;" OwnerUserId="5442" LastActivityDate="2015-07-13T23:21:50.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11277" PostTypeId="2" ParentId="11247" CreationDate="2015-07-14T00:01:47.693" Score="10" Body="&lt;p&gt;They're not different. Same principles do apply. You could have secondary, tertiary, quaternary, and so on, mirrors with instruments at any wavelength, either optical, or radio, or infrared, etc. You could also have instrumentation placed directly in prime focus (so no mirrors other than the primary) with any kind of instrument - radio or infrared or visible or whatever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See this image of the 5 meter Hale telescope on Mt. Palomar - there is no secondary mirror in this case, the observer is sitting in a little cage at prime focus, using the primary mirror directly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i9Ane.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, for other scenarios, the Hale telescope employs secondary and tertiary mirrors - it depends on the particulars of the telescope, the instrumentation, the experiment or research you're doing, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One reason many of the large optical telescopes very often have at least a secondary mirror is that the architecture preferred in most of these cases is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ritchey%E2%80%93Chr%C3%A9tien_telescope&quot; rel=&quot;noreferrer&quot;&gt;Ritchey–Chrétien&lt;/a&gt; - chosen often for the largest professional telescopes because it eliminates &lt;a href=&quot;https://en.wikipedia.org/wiki/Coma_(optics)&quot; rel=&quot;noreferrer&quot;&gt;coma&lt;/a&gt;, an aberration that is detrimental to astrometry (with coma, images of stars are not round, so it's hard to measure angular distances between them). You could use the primary mirror of such a telescope directly, sure, but being a concave hyperbolic mirror, it has strong aberrations of its own, and so requires the convex hyperbolic secondary (often a strong hyperbola, with a large eccentricity) to correct the aberrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hale telescope pictured above has a parabolic primary, so using it directly is not a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, all of the above are not strict rules, just statistical observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some radio telescopes have instrumentation at prime focus simply because it's convenient for that particular case. Other radio telescopes have secondary mirrors. Again, it all depends on what you try to achieve. E.g., the Arecibo radio telescope could be used either in prime focus, or with a secondary mirror in a Gregorian configuration - here's the image with the prime focus instrumentation and the Gregorian mirror to the left:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G8ie5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the Arecibo scope, the N-ary mirrors are sometimes used to correct the aberrationa of the spherical primary reflector, but that's not the only reason why they're used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astro.uni.torun.pl/~kb/Telescopes/RT90/RTH.htm&quot; rel=&quot;noreferrer&quot;&gt;Here's a discussion&lt;/a&gt; comparing various architectures (classic Cassegrain versus Ritchey-Chrétien versus anastigmatic aplanat) for a large radio telescope, highlighting various design, performance and operational issues for each. TLDR: classic Cassegrain is traditional for radio telescopes, but the R-C architecture performs better and is not significantly more difficult to build; OTOH, with R-C you must always use the secondary.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-07-17T01:25:29.773" LastActivityDate="2015-07-17T01:25:29.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11278" PostTypeId="1" AcceptedAnswerId="11293" CreationDate="2015-07-14T02:00:32.807" Score="3" ViewCount="868" Body="&lt;p&gt;I'm not sure whether I should ask this question here but I'll give it a shot. I have a XT8 Dobsonian Reflector telescope and it works fine. But recently when I use the focuser to focus the image when I change eyepieces it focuses fine but it makes a scratching noise that is really loud and it doesn't sound right. Now I don't know what this screw that I am about to explain is neccesarly for because I am no where near a telescope expert. Next to the eyepiece there is a screw that you can turn loose and if you make it as tight as possible the focus knob won't focus the image but when it is a little loose it starts to focus it, but it keeps making a scratching sound and a high pitched rubbing sound that doesn't sound right.  I know I explained this pretty vaguly but any help would be helpful! So far I could not find anything on the web for this weird problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ljZxZ.jpg&quot; alt=&quot;the eyepiece,screw is on the left&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="7467" LastEditorUserId="7467" LastEditDate="2015-07-15T00:49:07.137" LastActivityDate="2015-08-15T06:29:08.913" Title="Problem with Dobsonian reflector telescope" Tags="&lt;dobsonian-telescope&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11279" PostTypeId="1" CreationDate="2015-07-14T02:47:25.453" Score="8" ViewCount="739" Body="&lt;p&gt;So I had an argument with a friend, which was about whether the Moon is a planet or a satellite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.iau.org/static/resolutions/Resolution_GA26-5-6.pdf&quot; rel=&quot;noreferrer&quot; title=&quot;IAU 2006 Resolution B5&quot;&gt;IAU 2006 Resolution B5&lt;/a&gt; gives definitions on what it is to be a planet, but there is a vagueness, as I perceived, is that it does not define satellites, while using it to define dwarf planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a &lt;a href=&quot;http://www.iau.org/static/archives/releases/doc/iau0601_q_answers.doc&quot; rel=&quot;noreferrer&quot;&gt;supplementary Q&amp;amp;A sheet&lt;/a&gt;, and it uses the &lt;strong&gt;barycenter&lt;/strong&gt; to define a satellite, but it is &lt;em&gt;implied&lt;/em&gt;, not explicitly &lt;em&gt;defined&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions follow:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Should the supplementary Q&amp;amp;A sheet considered as legitimate official definitions, or should it be considered as a one way of interpreting the Resolution B5?&lt;/li&gt;&#xA;&lt;li&gt;If Q&amp;amp;A is legitimate, then why is the definition of the satellite is not included in the Resolution B5? If not, is the official definition of the satellite nonexistent currently?&lt;/li&gt;&#xA;&lt;li&gt;Are there any source or archive of the &lt;em&gt;official&lt;/em&gt; or widely accepted definitions of glossaries in the fields of astronomy?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7141" LastEditorUserId="2618" LastEditDate="2015-07-15T11:51:10.823" LastActivityDate="2015-07-15T18:26:25.763" Title="Official Definition of Satellite?" Tags="&lt;the-moon&gt;&lt;solar-system&gt;&lt;planet&gt;&lt;natural-satellites&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11280" PostTypeId="1" AcceptedAnswerId="11282" CreationDate="2015-07-14T04:41:21.853" Score="4" ViewCount="753" Body="&lt;p&gt;Hypothetical question: when a pair of identical stellar black holes are very close to one another wouldn't the side facing each other experience lower gravitational force? How is this logic un/sound?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastActivityDate="2015-07-14T08:32:49.983" Title="Can stellar black holes cancel each other?" Tags="&lt;gravity&gt;&lt;black-hole&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11281" PostTypeId="2" ParentId="11272" CreationDate="2015-07-14T05:11:32.330" Score="5" Body="&lt;p&gt;&lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;noreferrer&quot;&gt;Horizons Ephemeris Generator&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a screen capture where vector option is chosen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/03hkS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1232" LastActivityDate="2015-07-14T05:11:32.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11282" PostTypeId="2" ParentId="11280" CreationDate="2015-07-14T06:45:51.417" Score="5" Body="&lt;p&gt;Your logic isn't unsound. By symmetry there would be a gravitational neutral point between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the black holes could not then be described by the Schwarzschild metric, since that only applies to a spherically symmetric situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you brought the black holes closer together, the isopotentials would become distinctly aspherical, the event horizons would be complicated shapes and ultimately would merge - although as the black holes would presumably be in orbit and rotating, the spacetime geometry would be even more complex than that.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-07-14T06:45:51.417" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11283" PostTypeId="2" ParentId="11280" CreationDate="2015-07-14T07:54:48.477" Score="2" Body="&lt;p&gt;What would really happen is that the 2 black holes would spiral around and into each other so fast that any theoretical safe point wouldn't be safe for long, like a few millionths of a second at most.    But if you imagine a snapshot in time, where 2 black holes are near each other and your equidistant from both of them, that would be a gravitationally neutral point - as Rob Jeffries says.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The experience of zero gravity happens in orbit anyway, so perhaps a better way to look at is is, is whether you'd experience gravitational curvature of space, where your direction is pulled off in the direction of the massive object, and if you're between 2 massive objects and you experience an equal tug from both, you would in a sense, fly in a straight line between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would still experience tidal effects though and the tidal effects near a stellar mass black hole would be sufficient to tear you apart.   The Neutral point is exactly that - a point, so a 3D ship would inevitably feel more tug from one on one side and more tug from the other on the other side, as well as being drawn towards the point on approach and experience a tug back to the point on exit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's 2 super-massive black holes then the tidal effects are less of a problem and that might be a more interesting and longer lasting journey, but the gravitationally Neutral part is still just a point, but the &quot;drop off&quot; away from the point would be less intense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider if the Schwarzchild radii of the 2 black holes were touching - in our thought experiment, that's a different story and the gravitational sweet spot is now inside a bigger black hole, not in a safe spot between 2 black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about what happens when you have a 3 stellar mass black hole, radius 8.8 KM and another 3 stellar mass black hole, 8.8 KM, and the Schwarzchild radii are just touching, the singularities are 17.6 KM apart in that instant, but you now have enough mass for a 17.6 KM Schwarzchild radius black hole, so in effect, it's now 1 bigger, not 2 smaller black holes and the zero gravity sweet spot, is right in the middle of the 17.6 KM radius black hole, Because as you double the mass, you double the size.   If the black holes aren't touching, then it's theoretically possible to fly through there, but the instant they touch, both black holes and everything in between them is inside a bigger black hole.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, of-course, an oversimplified model assuming stationary black holes.  In reality, I think you'd get a much more complex gravitational dance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the answer to your question &quot;can they cancel each other&quot; - no.  They can only add to each other.   Gravity between 2 objects can have a balancing out effect, but never a cancelling out effect.   Think of the 2D representation of gravity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZMFpi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as you push 2 black holes closer together, there's never a point where the holes cancel each other out and create a zero gravity between them, but there is a ledge of sorts between them, below the near zero gravity of empty space.  As the holes approach, they make each other bigger.   The balancing point between them, is still inside the gravity well and as they get closer, the sweet spot between gets deeper and deeper in the 2-d representation.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-07-14T08:32:49.983" LastActivityDate="2015-07-14T08:32:49.983" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11284" PostTypeId="1" CreationDate="2015-07-14T08:35:36.800" Score="11" ViewCount="461" Body="&lt;p&gt;Regarding solar system dynamics, i.e. planets in stellar systems and moons in planetary systems, this is often mentioned in the literature, but it is difficult to find a good analysis/explanation of this phenomenon.&lt;/p&gt;&#xA;" OwnerUserId="7685" LastEditorUserId="7685" LastEditDate="2015-07-14T14:59:19.113" LastActivityDate="2016-07-03T12:35:18.837" Title="Why are retrograde orbits more stable than prograde ones?" Tags="&lt;orbit&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11285" PostTypeId="2" ParentId="11279" CreationDate="2015-07-14T14:14:14.083" Score="4" Body="&lt;p&gt;&lt;strong&gt;In fact there is no official definition of a satellite.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Issac Asimov makes &lt;a href=&quot;https://www.physicsforums.com/threads/issac-asimov-moon-planet-theory.6562/&quot; rel=&quot;nofollow noreferrer&quot;&gt;a good argument that the Moon is a planet&lt;/a&gt; since its orbit is convex around the Sun for its entire orbit, unlike the other moons of the solar system which are concave when in opposition and convex (to the Sun) when in conjunction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, the idea that a body is a satellite of another body based on the barycenter being inside one body is flawed. Consider moving the Moon further or closer from/to the Earth. In which direction would you move it to make the two a &quot;binary system&quot; as opposed to a primary/moon system. In fact, though naively one would consider two &lt;em&gt;closer&lt;/em&gt; bodies to be a binary system, moving the Moon closer puts the barycenter closer to the center of the Earth. Yet &lt;em&gt;moving the moon further away&lt;/em&gt; moves the barycenter outside the Earth. Would you consider the Earth/Moon system as a binary system if the Moon were 30% further from the Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we see that all proposed definitions for a satellite are flawed and have corner cases which make a precise definition contentious. &lt;strong&gt;This is the same problem as we had when trying to define a planet.&lt;/strong&gt; No definition will nicely fit all the corner cases that Nature will give us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note that this answer was translated into English by my father, who is the one typing it. I'm eight years old and &lt;a href=&quot;https://aviation.meta.stackexchange.com/questions/1693/move-a-question-to-a-new-user&quot;&gt;don't use SE unsupervised&lt;/a&gt;. A good portion of this information I learned just now because of this question, so thank you to the OP.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="7662" LastEditorUserId="-1" LastEditDate="2017-03-16T15:52:28.417" LastActivityDate="2015-07-15T18:26:25.763" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11286" PostTypeId="1" AcceptedAnswerId="11287" CreationDate="2015-07-14T22:02:59.480" Score="6" ViewCount="259" Body="&lt;p&gt;Articles written about fast radio bursts describe the signal's short-wavelength components arriving before its longer-wavelength components, suggesting energy-dependent time delays in the interstellar medium (ex. &lt;a href=&quot;http://www.scientificamerican.com/article/a-brilliant-flash-then-nothing-new-fast-radio-bursts-mystify-astronomers/&quot; rel=&quot;noreferrer&quot;&gt;http://www.scientificamerican.com/article/a-brilliant-flash-then-nothing-new-fast-radio-bursts-mystify-astronomers/&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why are lower-energy signals slowed slightly more than higher-energy signals when passing through clouds of electrons?&lt;/strong&gt; Also, &lt;strong&gt;how do we know about this slowing effect&lt;/strong&gt; if it appears to be so small (what objects or events have exhibited the time difference)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first-order logic guessed that higher energy radiation is slowed more, remembering the mnemonic &quot;&lt;em&gt;Blue bends best&lt;/em&gt;&quot; from optics. I assume these intergalactic clouds behave quite differently from a solid prism.&lt;/p&gt;&#xA;" OwnerUserId="7480" LastEditorUserId="616" LastEditDate="2019-12-14T22:24:19.463" LastActivityDate="2019-12-14T22:24:19.463" Title="Speed of light through the ISM and Wavelength" Tags="&lt;light&gt;&lt;interstellar-medium&gt;&lt;extinction&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11287" PostTypeId="2" ParentId="11286" CreationDate="2015-07-15T10:18:06.483" Score="6" Body="&lt;p&gt;You can find a neat description and some examples of the effect &lt;a href=&quot;http://www.jb.man.ac.uk/distance/frontiers/pulsars/section4.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. This is known as the &lt;a href=&quot;http://astronomy.swin.edu.au/cms/astro/cosmos/p/Pulsar+Dispersion+Measure&quot; rel=&quot;nofollow noreferrer&quot;&gt;pulsar dispersion measure&lt;/a&gt;. As you correctly say, waves with longer wavelength (lower energy photons) are delayed with respect to shorter wavelength radiation from the same phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When electromagnetic waves travel through a plasma, they excite currents in the free charged particles. In such cases it can be shown (using Maxwell's equations) that the waves propagate with a relationship between their frequency &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; and &quot;wavenumber&quot; &lt;span class=&quot;math-container&quot;&gt;$k = 2\pi/\lambda$&lt;/span&gt; given by&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\omega^2 = \omega_p^{2} + c^2k^2,$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\omega_p$&lt;/span&gt; is known as the &quot;plasma&quot; frequency and equals &lt;span class=&quot;math-container&quot;&gt;$(n_e e^2/\epsilon_0 m_e)^{1/2}$&lt;/span&gt; for the electrons in the plasma (i.e. it depends on the electron number density &lt;span class=&quot;math-container&quot;&gt;$n_e$&lt;/span&gt;.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you have a bunch of photons emitted as a pulse, the relevant velocity is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Group_velocity&quot; rel=&quot;nofollow noreferrer&quot;&gt;group velocity&lt;/a&gt; given by &lt;span class=&quot;math-container&quot;&gt;$v_g=d\omega/dk$&lt;/span&gt;. So&#xA;&lt;span class=&quot;math-container&quot;&gt;$$v_g = \frac{c^2 k}{(\omega_p^{2} +c^2 k^2)^{1/2}} = c\left(1 - \frac{\omega_p^2}{\omega^2}\right)^{1/2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This converges to &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; when the frequencies are high (wavelengths are short), but is slower when frequencies are low (wavelengths are long).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of an intuitive physical picture, yes you could think that the refractive index is frequency dependent, but the difference here is that the reason for this dispersion is that waves travelling through a conductive medium are &quot;lossy&quot; - that is the currents induced also encounter resistivity and therefore the waves heat the plasma.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-12-14T10:37:06.223" LastActivityDate="2019-12-14T10:37:06.223" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11288" PostTypeId="2" ParentId="11279" CreationDate="2015-07-15T12:24:39.320" Score="2" Body="&lt;p&gt;Your final three questions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Should the supplementary Q&amp;amp;A sheet considered as legimate official definitions, or should it be considered as a one way of interpreting the Resolution B5?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That supplementary Q&amp;amp;A absolutely should not be considered as legitimate official definitions. That Q&amp;amp;A pertains to the original draft of the resolution. The final, voted-upon version represents a significant departure from the original. The final version added the dynamicists' concept of &quot;clearing the neighborhood&quot; to the definition of a planet, and the concept of double planets is gone. The IAU currently deems Pluto to be a dwarf planet, but not Charon (but that may change). From &lt;a href=&quot;https://www.iau.org/public/themes/pluto/&quot; rel=&quot;nofollow&quot;&gt;https://www.iau.org/public/themes/pluto/&lt;/a&gt;, &quot;For now, Charon is considered just to be Pluto's satellite. The idea that Charon might qualify to be called a dwarf planet in its own right may be considered later.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the Q&amp;amp;A is legitimate, then why is the definition of the satellite is not included in the Resolution B5? If not, is the official definition of the satellite nonexistent currently?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This question is a bit moot as that Q&amp;amp;A is not legitimate. There is no official definition of what constitutes a satellite. By any sane definition of the concept, Pluto is not a satellite of Charon, so Pluto's status as a dwarf planet is secure. Whether or not Charon is a satellite of Pluto is a matter of debate, and I suspect the IAU has had it's fill with rhetorical debates given the heated objections to their definition of the term &quot;planet&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there any source or archive of the official or widely accepted definitions of glossaries in the fields of astronomy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is one word that does have an official meaning, and that is &quot;planet&quot;. &quot;Satellite&quot;? No. &quot;Star&quot;? No. &quot;Galaxy&quot;? No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the debate with your friend that made you ask this question, the only way in which one could view the Moon as a planet is by the fact that its orbit about the Sun is convex. That is a consequence of the gravitational force exerted on the Moon by the Sun is always greater in magnitude than that exerted by the Earth on the Moon. That's not a good definition; the IAU certainly doesn't use it. Along with the Earth's Moon, there are several outer moons of the giant planets that have this same characteristic, and they're all listed as satellites.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-07-15T12:24:39.320" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11289" PostTypeId="1" CreationDate="2015-07-15T13:47:10.447" Score="5" ViewCount="374" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Vela_X-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vela X-1&lt;/a&gt; has a 283s spin period, while &lt;a href=&quot;http://adsabs.harvard.edu/abs/1994ApJ...429..356F&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt; may have a 2.7hr spin period.&lt;br/&gt;&#xA;However neither of them is included in ATNF catalog.&lt;br/&gt;&#xA;They are not included in $p-\dot{p}$ diagram either&lt;img src=&quot;https://i.stack.imgur.com/MPOAw.gif&quot; alt=&quot;$p-\dot{p}$&quot;&gt;. The longest period is just about 10s in this diagram.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question is since Vela X-1 is definitely a pulsar, why is it not accepted? &lt;br/&gt;&#xA;Which one has the longest spin period so far?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;When this diagram is mentioned in textbooks, we do not say which kind of pulsars are included at all. Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p-\dot{p}$ diagram only shows radio pulsars?  &lt;strong&gt;no!&lt;/strong&gt;&#xA;you can see magnetars,radio quiet pulsars and pulsars which are only active in $\gamma-ray$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Except magnetars, there are no single pulsars only pulse in X-ray band?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a pulsar pulses in X-ray band, that does not necessarily mean it is an X-ray pulsar?&lt;br/&gt;&#xA;Only pulsars in binaries and accretion-powered pulsars could be called X-ray pulsars? They should be radio-quiet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In binaries, there are no pulsars which pulse in radio and x-ray band simultaneously?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditDate="2015-07-16T11:04:05.470" LastActivityDate="2015-07-16T20:22:38.327" Title="which pulsar has the longest spin period so far?" Tags="&lt;pulsar&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11290" PostTypeId="1" CreationDate="2015-07-15T19:29:54.670" Score="3" ViewCount="89" Body="&lt;p&gt;If we look at our solar system we have the sun which has planet orbits and planets have moon orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at the possible size difference from hypergiants to small asteroids why shouldn't it be possible to nest orbits to an astounding degree: supergiant orbit a hypergiant, star orbits supergiant and so on ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first guess is that the size and distance must be so small in contrast to the primary that the gravitational gradient of the primary does not cause too much disturbance, but it should be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone tried to find it out with celestial mechanics ? While it may be completely theoretical, it is IMHO a solid question and it would be nice to find out how deep celestial bodies can be nested and how it limits stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some conditions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The orbits should be (very probably) stable for one million years.&#xA;Hypergiants are burning fast, so I limit the stability for a relative short timeframe.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The barycenter of both satellite and its primary are inside the primary.&#xA;No double system.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The first primary should be a star, not a black hole or a neutron star.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7695" LastActivityDate="2015-07-16T17:04:23.437" Title="How many nested stable (1 Mio years) orbits are theoretically possible?" Tags="&lt;star-systems&gt;&lt;orbital-migration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11291" PostTypeId="2" ParentId="11290" CreationDate="2015-07-15T19:40:28.093" Score="4" Body="&lt;p&gt;Yes, it's possible to have many nested levels, all stable. Each nested orbit needs to be inside the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;nofollow&quot;&gt;Hill sphere&lt;/a&gt; of its central body in order to be stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't nest any further when the Hill sphere gets too close to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Roche_limit&quot; rel=&quot;nofollow&quot;&gt;Roche limit&lt;/a&gt;. Below the Roche limit the satellite disintegrates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact number of nested levels depends very greatly on the characteristics of all bodies involved.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2015-07-15T19:40:28.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11292" PostTypeId="2" ParentId="11289" CreationDate="2015-07-15T19:54:52.803" Score="3" Body="&lt;p&gt;Vela X-1 is an example of an accretion-powered pulsar. These emit pulsed emission because they are accreting material onto their magnetic poles. If the magnetic poles and rotation axis are misaligned this results in pulsed X-ray emission. The power for the pulsar comes from the infalling material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classic P-Pdot diagram you show is for radio pulsars, or rather rotationally-powered pulsars. These are objects that are pulsators by virtue of accelerating charged particles from their magnetic poles out along the field lines. These accelerated relativistic particles emit synchrotron and curvature radiation that is beamed and intensified. The energy ultimately comes from the rotational kinetic energy of the pulsar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, other than the fact that they both involve neutron stars, these are completely different phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of radio pulsars, your diagram looks reasonably up to date. I think the longest period object on your diagram is &lt;a href=&quot;https://en.wikipedia.org/wiki/PSR_J2144-3933&quot; rel=&quot;nofollow&quot;&gt;PSR J2144-3933&lt;/a&gt;, which has a period of 8.51 seconds (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1999Natur.400..848Y&quot; rel=&quot;nofollow&quot;&gt;Young, Manchester &amp;amp; Johnston 1999&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your diagram has a line marked as &quot;graveyard&quot;. I believe this is a locus defined by &lt;a href=&quot;http://adsabs.harvard.edu/abs/1992A%26A...254..198B&quot; rel=&quot;nofollow&quot;&gt;Bhattacharya et al. (1992)&lt;/a&gt; and has the form&#xA;$$P = 2.42\times 10^{-6} B^{1/2}\ s,$$&lt;br&gt;&#xA;where the pulsar magnetic field $B$ is in units of Gauss (typical pulsar values would be $B=10^{10}-10^{13}$ Gauss (as also marked on your diagram). The theory behind this &quot;death line&quot; is discussed by &lt;a href=&quot;http://adsabs.harvard.edu/abs/1975ApJ...196...51R&quot; rel=&quot;nofollow&quot;&gt;Ruderman &amp;amp; Sutherland (1975)&lt;/a&gt;. Briefly it arises from the requirement of a minimum potential difference to be generated such that accelerated particles produce energetic enough radiation to stimulate the production of further electron/positron pairs. If the magnetic field strength falls or the rotation period gets too long, then this mechanism fails and the pulsar is quenched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PSR J2144-3933 appears to lie beyond this death line, but there are other ideas and models of how this death line may arise (of which I am not very familiar, but see for example &lt;a href=&quot;http://iopscience.iop.org/1538-4357/531/2/L135&quot; rel=&quot;nofollow&quot;&gt;Zhang et al. 2000&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that millisecond pulsars are thought to recycled accretion-powered pulsars. That is that they gain sufficient angular momentum from an accretion process that they spin up to become radio pulsars again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radio-quiet pulsars are thought to be rotationally powered but where the radio beam is narrower than say the beaming of gamma rays in which they are detected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnetars are something different again. Their power comes from the decay of extraordinarily strong magnetic fields.&#xA;The soft gamma ray repeaters and anomalous X-ray pulsars fall in this category.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can see that the word &quot;pulsar&quot; might include all sorts of different types of objects and physics. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-07-16T20:22:38.327" LastActivityDate="2015-07-16T20:22:38.327" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11293" PostTypeId="2" ParentId="11278" CreationDate="2015-07-16T06:02:23.060" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Next to the eyepiece there is a screw that you can turn loose and if&#xA;  you make it as tight as possible the focus knob won't focus the image&#xA;  but when it is a little loose it starts to focus it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sounds like the lock screw. If that's what it is (your description is a bit vague), it's designed to lock the focuser in place, like when you're doing prime focus photography with a heavy camera, and you don't want the focuser to move around and lose focus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's the lock screw, don't mess with it. Don't tighten it, you don't need that for regular observations. Keep it comfortably loose. If it's a bit tight, it could be rubbing against the drawtube (or whatever moving part it's designed to stop), and then it would make that noise - and you don't want that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a very heavy &lt;a href=&quot;https://i.imgur.com/vRfiW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;30mm 82deg Explore Scientific eyepiece&lt;/a&gt;, and even with that thing (bigger than a soda can, and weighing about 1 kilogram) I don't need the lock screw - and this is on a regular mass produced two-speed focuser, not some fancy precision hand-crafted MoonLite. Visual observations are all about moving the focuser around, hunting for the best focus, and changing eyepieces for the best view. Let the focuser move easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless it's moving too easily, but then you have a problem with the focuser.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-07-16T06:13:38.960" LastActivityDate="2015-07-16T06:13:38.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11294" PostTypeId="2" ParentId="11290" CreationDate="2015-07-16T07:07:28.570" Score="2" Body="&lt;p&gt;Tidal forces limit the stability.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: &lt;a href=&quot;http://www.universetoday.com/109666/can-moons-have-moons/&quot; rel=&quot;nofollow&quot;&gt;http://www.universetoday.com/109666/can-moons-have-moons/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easier if you set every orbiter far away from the orbited, but by setting large distances, that limits how many levels of nesting you can do.   There are complicated star systems with like 6 or 7 suns, but they aren't nested in the way you suggest.  Pluto is a 5 body system but it's also quite far from the sun and it stays far from all the outer planets, and it's got 2 inner bodies (Pluto and Charon) and 3 outer, smaller bodies.   Pluto is also quite far from any large gravitational bodies.   If Pluto orbited one of the planets, it's 5 body system wouldn't be stable.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-07-16T17:04:23.437" LastActivityDate="2015-07-16T17:04:23.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11295" PostTypeId="1" AcceptedAnswerId="11299" CreationDate="2015-07-16T14:05:37.777" Score="1" ViewCount="355" Body="&lt;p&gt;I have been thinking about this question :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can they tell no asteroids (big enough to create damage) will hit earth in the next few hundred years?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because, so far, we were unable to get a clear and nice view of a planet of the solar system planet which is not that big, but still bigger than an asteroid. So how can they tell that nothing from the immensity of the Galaxy and Universe, bigger than a few yards, will cross our path and hit us within the next 100 years? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We all remember Chelyabinsk meteor in 2013, or whatever how they called that thing. They never saw that coming but it still caused damage and it was only few meters (± 20 meters according to that page &lt;a href=&quot;https://en.wikipedia.org/wiki/Chelyabinsk_meteor&quot; rel=&quot;nofollow&quot;&gt;Wiki link&lt;/a&gt;), Arround 1,500 injuries and ± 7200 damaged buildings... It's hard to believe that they know how to see something not so much wider than 20 metres wide outside of our solar system.&lt;/p&gt;&#xA;" OwnerUserId="777" LastEditorUserId="2365" LastEditDate="2016-03-06T11:09:51.967" LastActivityDate="2016-03-06T11:17:51.313" Title="How can they tell no asteriods will hit earth in the next hundreds years?" Tags="&lt;earth&gt;&lt;asteroids&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11296" PostTypeId="1" AcceptedAnswerId="11395" CreationDate="2015-07-16T14:09:14.243" Score="2" ViewCount="473" Body="&lt;p&gt;I am an amateur observer and Olympus 10x50 binocular is my tool. As I keep locating celestial objects (Planets, Stars etc), I would like to maintain a log of the observed objects. The purpose of the log is to list down the number of celestial objects I have observed/located. I need suggestion on what data points should I capture which helps in making some sense out of it and also encourage my friends to get into star gazing.&lt;/p&gt;&#xA;" OwnerUserId="7698" LastActivityDate="2018-07-16T13:18:56.127" Title="Any tip on maintaining a log of celestial objects?" Tags="&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11298" PostTypeId="2" ParentId="10904" CreationDate="2015-07-16T17:43:38.277" Score="0" Body="&lt;p&gt;You're not saying what programming language you're using. If it's Python, or if you could link Python libraries from it, then PyEphem would provide everything you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rhodesmill.org/pyephem/&quot; rel=&quot;nofollow&quot;&gt;http://rhodesmill.org/pyephem/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2015-07-16T17:43:38.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11299" PostTypeId="2" ParentId="11295" CreationDate="2015-07-16T18:10:59.690" Score="2" Body="&lt;p&gt;A couple of points based on some basic orbital mechanics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They don't need to get a &quot;good&quot; view, like the clear, crisp photos of Pluto to see one coming.   They only need to get a picture over time to calculate trajectory.   The unclear snapshots work just fine to calculate if it'll hit us or miss us.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, an object as far as Mars at it's closest pass to earth, a bit over 1/2 AU, would still take a few months to reach earth if it's in solar orbit.   Mostly we don't need to track anything as far out as Pluto, they can look much closer to the Earth and still have sufficient warming time.   The hard part, is tracking things that approach from the Sun side, cause those are harder to see.  That's why the Chelyabinsk meteor wasn't spotted.   it was also on the small side, smaller than NASA is currently looking for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The good news is that, we don't get struck by things that size very often.   The Solar system is pretty enormous and pretty empty and pretty big strikes like that one are rare, like, maybe once a century.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, virtually all of the injuries from the Chelyabinsk meteor were from people who didn't know what to do.   If you see a big fireball in the sky, it's human nature to watch it, but use some common sense.  A space rock of that size will make a shock-wave that travels at roughly the speed of sound and the shock-wave can break windows, even knock over trees and buildings if it's big enough.   You don't want to be standing in-front of a window when the shock wave its.  Lay down next to a couch or under a table in case your building gets shaken and cover your ears.   If everyone had done that, there would have been very few injuries.   You only need to wait maybe 2 minutes or so to be on the safe side.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're in a car, stop, cause the shock-wave could knock down trees or debris in-front of you and stay in the car, cause that's safer than being outside.     All told, the damage to buildings was tiny compared to natural disasters like Earthquakes, floods or volcanoes which happen to us several times a year.    It's good that NASA is watching for this kind of thing, but it's also a pretty rare event.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-07-16T18:10:59.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11300" PostTypeId="1" AcceptedAnswerId="11303" CreationDate="2015-07-16T18:23:09.857" Score="3" ViewCount="163" Body="&lt;p&gt;Does NASA make publicly available real images of missions in an organized manner? What I mean is: Say I'm looking for the images taken by New Horizons - a simple list of the images in chronological order. Even if it's not a comprehensive list, at least all of those made publicly available. Does that exist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be clear - I'm not interested in &quot;artist concept&quot;. On the other hand - &quot;false color&quot; pictures of infrared etc. are fine.&lt;/p&gt;&#xA;" OwnerUserId="7706" LastActivityDate="2015-07-16T21:08:19.550" Title="Real images (not &quot;artist concept&quot;) by NASA" Tags="&lt;star&gt;&lt;solar-system&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11301" PostTypeId="1" CreationDate="2015-07-16T18:24:00.920" Score="3" ViewCount="143" Body="&lt;p&gt;For example, if star 1 is 4x bigger than star 2 but both are the exact same temperature how do you determine which is more luminous?&lt;/p&gt;&#xA;" OwnerUserId="7707" LastActivityDate="2015-09-15T15:54:12.383" Title="How do you determine which star, between two, is more luminous?" Tags="&lt;amateur-observing&gt;&lt;star-systems&gt;&lt;luminosity&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11303" PostTypeId="2" ParentId="11300" CreationDate="2015-07-16T21:08:19.550" Score="3" Body="&lt;p&gt;NASA mission data is archived at a handful of NASA archives.  If you do not know which archive a data set is in, then you can search the Master Catalog at the &lt;a href=&quot;http://nssdc.gsfc.nasa.gov/&quot; rel=&quot;nofollow&quot;&gt;NASA Space Science Data Coordinated Archive (NSSDC)&lt;/a&gt;.&#xA;The main astrophysics archives are: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;HEASARC (High Energy Astrophysics Science Archive Research Center)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IRSA (Infrared Science Archives)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LAMBDA (Legacy Archive for Microwave Background Data Analysis)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;a href=&quot;https://archive.stsci.edu/&quot; rel=&quot;nofollow&quot;&gt;MAST&lt;/a&gt; (Mikulski Archive for Space Telescopes)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planetary/Solar System Data are at the various Nodes of the &lt;a href=&quot;https://pds.nasa.gov/&quot; rel=&quot;nofollow&quot;&gt;PDS&lt;/a&gt;.&#xA;NSSDC holds data from many of the older missions that predate these archives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;New Horizons data is being archived at the &lt;a href=&quot;http://pds-smallbodies.astro.umd.edu/data_sb/missions/newhorizons/index.shtml&quot; rel=&quot;nofollow&quot;&gt;Small Bodies Node&lt;/a&gt; of PDS. But, it usually takes several months for data to get into these archives.  So, if you are looking for very current images, you have to rely on what the mission team has put up on its web site.  That is at their discretion.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2015-07-16T21:08:19.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11304" PostTypeId="1" AcceptedAnswerId="11308" CreationDate="2015-07-17T04:08:43.117" Score="4" ViewCount="151" Body="&lt;p&gt;Now that we have seen what Pluto has to offer and New Horizons is headed to the Kuiper Belt, is there anything we can gain from the mission in interstellar space after it travels beyond the Kuiper Belt?&lt;/p&gt;&#xA;" OwnerUserId="5283" LastEditorUserId="2153" LastEditDate="2015-08-30T17:43:52.323" LastActivityDate="2015-09-06T23:15:03.483" Title="What's to be gained from the New Horizons mission once it's beyond the Kuiper Belt?" Tags="&lt;pluto&gt;&lt;interstellar-travel&gt;&lt;kuiper-belt&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11305" PostTypeId="1" AcceptedAnswerId="11321" CreationDate="2015-07-17T06:24:40.163" Score="11" ViewCount="18475" Body="&lt;p&gt;I wonder, how did our ancestors discover the Solar System? They did not have any telescopes to see distant objects, right? Even a planet looks like a star from a distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They discovered the rotations of different planets without having much technology.&lt;/p&gt;&#xA;" OwnerUserId="1598" LastEditorUserId="2153" LastEditDate="2015-07-17T20:16:37.937" LastActivityDate="2018-05-04T17:29:42.057" Title="How did our ancestors discover the Solar System?" Tags="&lt;solar-system&gt;&lt;history&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11306" PostTypeId="2" ParentId="11305" CreationDate="2015-07-17T06:52:52.380" Score="8" Body="&lt;p&gt;In the absence of the lighting levels in cities, there are a number of objects in the night sky that are easily seen to move against the background of stars: the Sun, the Moon, Venus, Jupiter, Saturn, Mars and Mercury. These are obvious to anyone who looks at the sky regularly from any dark sky site on clear nights, like most places in the pre-industrial world. These objects are the Planets or Wanderers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term Solar System is anachronistic when used of ancient astronomy. You have the Fixed Stars which maintain their relative positions but appear to rotate as if fixed to a sphere centred on the observer about a celestial axis about once per day. then you have the Planets which move relative to the Fixed Stars. Together with the Earth itself these constitute the World, Universe or what-have-you.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-07-17T17:31:12.807" LastActivityDate="2015-07-17T17:31:12.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11308" PostTypeId="2" ParentId="11304" CreationDate="2015-07-17T12:41:33.910" Score="7" Body="&lt;p&gt;&lt;a href=&quot;http://www.smithsonianmag.com/science-nature/where-will-new-horizons-probe-go-after-pluto-180955929/?no-ist&quot; rel=&quot;nofollow&quot;&gt;Possible rendezvous with a Kuiper belt object and Voyager type data on the heliosphere&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;We should have power until the 2030s, so we can get into the outer part of the heliosphere,&quot; says Spencer. &quot;As long as we can continue to get good data—and persuade NASA to pay for it—we will keep getting the data, because we will be in a unique environment that we've never been in before.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Added:&#xA;&lt;a href=&quot;http://www.vox.com/2015/8/29/9223011/new-horizons-nasa&quot; rel=&quot;nofollow&quot;&gt;NASA just picked the object for flyby&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a 30-mile-wide chunk of ice known as 2014 MU69 that orbits nearly a billion miles farther out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/2014_MU69&quot; rel=&quot;nofollow&quot;&gt;Not much is currently known about the object&lt;/a&gt;. We'll no doubt learn more from the January 2019 encounter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More info from Emily Lakdawalla and Casey Dreier at the Planetary Society:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/2015/09011608-new-horizons-extended-mission-pt1.html&quot; rel=&quot;nofollow&quot;&gt;New Horizons extended mission target selected&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="2031" LastEditDate="2015-09-06T23:15:03.483" LastActivityDate="2015-09-06T23:15:03.483" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11309" PostTypeId="1" CreationDate="2015-07-17T15:23:00.553" Score="2" ViewCount="222" Body="&lt;p&gt;I'm doing some writing set on a fictional earthlike planet that for reasons needs to be substantially larger than Earth but can't have substantially higher gravity. I've been able to compromise with a planet that's 6x the circumference, having a surface gravity of 1.5G, yielding a density of about 4700kg/m^3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is quite a bit (~15%) less than Earth and Venus, but (20%) more than Mars. It would proportionally have less heavy metal, meaning less iron for a protective magnetic field. I would like this planet to have some basis in reality - could such a planet be earth-like, and is there a definitive range of planet densities where life as we know it could exist? &lt;/p&gt;&#xA;" OwnerUserId="7712" LastEditorUserId="7712" LastEditDate="2015-07-20T16:05:22.963" LastActivityDate="2015-07-20T16:05:22.963" Title="How significant is a planet's density to the formation of life?" Tags="&lt;planet&gt;&lt;earth&gt;&lt;earth-like-planet&gt;&lt;geology&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11310" PostTypeId="2" ParentId="11301" CreationDate="2015-07-17T15:34:38.770" Score="1" Body="&lt;p&gt;One way to determine the luminosity of a star is to first determine its distance, d.  Then one can use the formula, Luminosity = Flux * 4 * Pi * d$^2$.  Flux is related to the apparent brightness that is measured with a telescope (and CCD or photometer).  mag = -2.5 * Log(Flux) + c.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the star is relatively nearby, one can use &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1314/how-does-a-telescope-measure-parallax-angle&quot;&gt;parallax&lt;/a&gt; to get the distance.  If the star is in a binary system, one can solve for the distance from the period, orbital velocities and apparent size of the orbit.  If you know the velocities throughout one orbit, then one knows how far it has physically traveled.  Comparing this with the apparent size gives an approximate distance.  However, since one usually does not know the inclination, this causes uncertainty.  If it is an eclipsing binary, then you know that it is not very inclined and this works well.  On the other hand, if it is eclipsing, one can usually determine the radii of the binary pair and then get distances from Stefan Boltzmann Law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method for estimating luminosity is to locate the star in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hertzsprung%E2%80%93Russell_diagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;HR Diagram&lt;/a&gt; (Luminosity vs Color).   The color (difference between the magnitudes in two different filters) locates it horizontally, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_classification&quot; rel=&quot;nofollow noreferrer&quot;&gt;stellar class&lt;/a&gt; locates it vertically in the diagram.  One needs a spectrum of the star to determine the class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we have one more trick up our sleeves.  If the star is a periodic variable such as a Cepheid or an RR Lyra, then there are relations between period of variability and luminosity.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-07-17T15:34:38.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11312" PostTypeId="2" ParentId="11305" CreationDate="2015-07-17T18:11:41.553" Score="0" Body="&lt;p&gt;By very carefully tracking the movements of the planets across the sky. And then coming up with the counterintuitive idea that they move just like apples move as they fall from trees.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-07-17T18:11:41.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11313" PostTypeId="2" ParentId="11309" CreationDate="2015-07-18T00:23:34.463" Score="2" Body="&lt;p&gt;You could probably get away with replacing molten iron, 7.87 g/cc, in the core with aluminum, 2.70 g/cc, and still generate a substantial magnetic field: &lt;a href=&quot;http://physics.aps.org/story/v19/st3&quot; rel=&quot;nofollow&quot;&gt;Molten Metal Magnet&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It’s easy to create a magnetic field by using a battery to force an electric current through a loop of wire. But Earth’s core, a rotating mix of iron and nickel with internal flows driven by the passage of heat, has no battery and no wires. Instead, it creates magnetism by means of self-sustaining feedback. Liquid metal moving through a magnetic field generates a current, similar to that induced in the moving coil of an electric generator. That current in turn generates the magnetic field. This “self-generation” mechanism can dramatically amplify the small, random fields that always exist in magnetic materials. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sodium was used in the linked article, however any molten metal or conductor, should do, and sodium, at 0.97 g/cc is just not dense enough to sink to a planet's core.&#xA;You'd have to do some hand waving over &quot;an iron poor/aluminum rich &lt;a href=&quot;http://Bok%20globule&quot; rel=&quot;nofollow&quot;&gt;Bok globule&lt;/a&gt;&quot;, but I think you could get your planet, with a sufficient magnetic field to fend off solar winds.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-07-18T00:23:34.463" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11314" PostTypeId="1" AcceptedAnswerId="11315" CreationDate="2015-07-18T06:40:34.167" Score="7" ViewCount="1131" Body="&lt;p&gt;I've heard from a lot of sources that Pluto doesn't seem to have &lt;em&gt;any&lt;/em&gt; impact craters, but I see a few things that look like impact craters, highlighted in this picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uL9HL.png&quot; alt=&quot;Pluto impact craters&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparison, here's something I think looks like an impact crater on Pluto next to an impact crater on Mercury:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YwTvJ.png&quot; alt=&quot;Pluto Mercury comparison&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are they impact craters on Pluto?  If not, what are they?  If they are, why are people saying it has &lt;em&gt;no&lt;/em&gt; impact craters?&lt;/p&gt;&#xA;" OwnerUserId="7717" LastActivityDate="2020-05-19T15:09:42.273" Title="Does Pluto have impact craters?" Tags="&lt;pluto&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11315" PostTypeId="2" ParentId="11314" CreationDate="2015-07-18T11:26:35.510" Score="5" Body="&lt;p&gt;It was predicted by some, before the images arrived, that Pluto's craters would be much shallower than perhaps on other rocky objects/satellites. The reason is the icy composition and the relatively low impact velocities in the outer solar system (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2015Icar..246..156B&quot; rel=&quot;nofollow&quot;&gt;Bray &amp;amp; Schenk 2015&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are indeed reported quotes in recent days that there are no impact craters on Pluto (e.g. &lt;a href=&quot;http://uk.businessinsider.com/pluto-surface-photos-no-craters-geologically-active-2015-7?r=US&amp;amp;IR=T&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;https://www.youtube.com/watch?v=ZAUMDv0XTb4&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;), based on the highest resolution images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm with you - the things you have marked look like very large impact craters, or the remains of them, and so what we are seeing is that there are no &lt;em&gt;small&lt;/em&gt; ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gist of what I have heard/read is that this indicates geological activity on Pluto, which is effectively removing the evidence of relatively small impacts and that the surface we see is mostly younger than 100 million years old.&#xA;Watch this space, a lot of people are just flapping their lips at the moment (including me).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility - raised by Wayfaring Stranger - could be that the expected number of impacts was (hugely) overestimated. This is discussed by &lt;a href=&quot;http://www.hou.usra.edu/meetings/craterstats2015/pdf/9031.pdf&quot; rel=&quot;nofollow&quot;&gt;Singer &amp;amp; Stern (2015)&lt;/a&gt;. There are indeed uncertainties, and it is hoped that cratering on Pluto &amp;amp; Charon would help constrain the number densities and size distribution of small impactors. My impression is that the uncertainties could be at levels of up to factors of 10 - and this seems to be discussed quite carefully in &lt;a href=&quot;http://adsabs.harvard.edu/abs/2015Icar..258..267G&quot; rel=&quot;nofollow&quot;&gt;Greenstreet et al. (2015)&lt;/a&gt; (behind a paywall, sorry), who also comment in the abstract that surface ages judged from cratering are &quot; entirely dependent on the extrapolation to small sizes [of the impactor distribution]&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further Edit: LDC3 suggests all the craters could be too small to be seen. The high resolution images that have been released could resolve craters of a diameter of a couple of km or more and cover a few percent of the surface. According to &lt;a href=&quot;http://www.hou.usra.edu/meetings/craterstats2015/pdf/9031.pdf&quot; rel=&quot;nofollow&quot;&gt;Singer &amp;amp; Stern (2015)&lt;/a&gt;, the number of craters at this size and above on Pluto &lt;em&gt;they expected&lt;/em&gt; in such an area is $\sim 1000$. So I think that either the planet has been resurfaced recently, the craters have been eroded in some other way or their estimates for the number of impacts/size of the impactors is wrong.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-07-18T15:13:28.783" LastActivityDate="2015-07-18T15:13:28.783" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="11316" PostTypeId="1" CreationDate="2015-07-18T11:31:21.767" Score="1" ViewCount="588" Body="&lt;p&gt;I was googling Neptune's trojan companions and found on Wiki this claim about  ~200 km diameter &lt;a href=&quot;https://en.wikipedia.org/wiki/(316179)_2010_EN65&quot; rel=&quot;nofollow noreferrer&quot;&gt;(316179) 2010 EN65&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot; the object is actually a jumping trojan, is jumping from the&#xA;  Lagrangian point L4 into L5 via L3.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So some asteroids not only oscillate between L4 and L5, between being trojans or greeks, but even go via L3, on the opposite side of the Sun? Are their orbits circular as Neptune's (in this case) or highly eccentrical? How could an object in the same orbit as Neptune have another orbital period than Neptune? Or how else can it migrate between L3, L4 and L5?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And are &quot;trojan&quot; objects expected to gather in L3 too?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &#xA;I'd like to add &lt;a href=&quot;https://en.wikipedia.org/wiki/Hilda_family&quot; rel=&quot;nofollow noreferrer&quot;&gt;these illustrations&lt;/a&gt; because I think they illustrate well how a regularly orbiting object seems to &quot;stay&quot; longer at L3, L4, L5. (I'm not sure whether &quot;jumping trojans&quot; have some significant other characteristics, though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eOq9X.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3Xj6t.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-07-20T20:00:44.487" LastActivityDate="2015-08-23T17:58:01.663" Title="What is a &quot;jumping trojan&quot;? And what do their orbits look like?" Tags="&lt;lagrange-point&gt;&lt;orbital-migration&gt;&lt;neptune&gt;&lt;trojan-asteroids&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11317" PostTypeId="1" CreationDate="2015-07-18T13:13:44.600" Score="2" ViewCount="219" Body="&lt;p&gt;The Moon &lt;a href=&quot;https://en.wikipedia.org/wiki/Occultation#Occultations_by_the_Moon&quot; rel=&quot;nofollow noreferrer&quot;&gt;occults&lt;/a&gt; a lot of stars. But are these events visible to the naked eye? Won't it be blinded out by the crescent even at its thinnest just before sunrise? Did ancient astronomers actually observe them, or did they deduce them theoretically according to orbital calculations? (I suppose occultation during an eclipse is both invisible and unlikely, right?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus, the brightest planet, like the new Moon always stays near the horizon. But could one actually see even Venus disappear behind the lunar anti-crust? A popular medieval islamic symbol suggests to me that they were well aware of this phenomena. Both the Moon and Venus are still symbolically important in their culture. But was occultation directly visible or inferred?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6QKzs.jpg&quot; alt=&quot;Photo (non-naked eye) of the Moon occulting Aldebaran&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OTVy9.png&quot; alt=&quot;Ancient symbol for a lunar occultation (please disregard whatever it is interpreted to stand for today)&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-07-18T13:27:47.873" LastActivityDate="2015-09-11T18:39:41.247" Title="Are lunar occultations visible to the naked eye?" Tags="&lt;the-moon&gt;&lt;venus&gt;&lt;history&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11318" PostTypeId="1" CreationDate="2015-07-18T14:05:59.750" Score="3" ViewCount="150" Body="&lt;p&gt;Is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Pluto%27s_heart&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tombaugh Regio&lt;/a&gt; on Pluto discernably visible from Earth ground based or Earth orbiting telescopes? And more generally, can surface features seen in Hubble images be matched with images from New Horizons? I can imagine a match from the image compilation I attach, with TR being the bright triangle, but is that intuition valid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/P9qNK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-04-09T16:25:33.977" Title="Is the Tombaugh Regio on Pluto visible from Earth?" Tags="&lt;hubble-telescope&gt;&lt;pluto&gt;&lt;space-probe&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11319" PostTypeId="2" ParentId="10382" CreationDate="2015-07-18T17:02:22.727" Score="-2" Body="&lt;p&gt;Brilliant question and great guess. The farther we see in space, the farther we see back in time; and the farther we see back in time, the faster &lt;strong&gt;was&lt;/strong&gt; acceleration. And vice versa.&#xA;In fact, our local universe--Laniakea--is contracting. Laniakea is 500 million light years across. It suggests that the expansion stopped 500 million years ago and contraction began.&#xA;Galaxies farther than 500 light years away we still see redshifted because more than 500 million years ago they still were running away. Today they are blueshifted but humans will see this blueshift only in very distant future.  &lt;/p&gt;&#xA;" OwnerUserId="7721" LastActivityDate="2015-07-18T17:02:22.727" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11320" PostTypeId="2" ParentId="11316" CreationDate="2015-07-19T00:27:12.230" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Are their orbits circular as Neptune's (in this case) or highly&#xA;  eccentrical?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;L4 and L5 stability is rather narrow&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eJs1e.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://ccar.colorado.edu/asen5050/projects/projects_2010/singh/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ccar.colorado.edu/asen5050/projects/projects_2010/singh/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more elliptical or eccentric orbit in resonance with a planet with a circular orbit would reach well outside of L4 and L5 and wouldn't be an L4 or L5 object.   (though I have to admit, a planet's eccentricity and the effect that has on L4 and L5 stable zones has gotten me curious), but they would likely need to be pretty close to the same eccentricity or it wouldn't be L4 or L5.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth's curious orbital partner &lt;a href=&quot;https://en.wikipedia.org/wiki/3753_Cruithne&quot; rel=&quot;nofollow noreferrer&quot;&gt;3753 Cruithne&lt;/a&gt; probobly isn't stable, long term anyway.   It's elongated as you describe and it could be stable for a long time, perhaps millions of years but it's not an L4 or L5.  3753 Cruithne has an orbit around the sun of 364 days, so it, in effect, laps the earth every 364 years, rarely coming close enough to hit or significantly change it's orbit cause it's on a different orbital plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an object go from L4, past L3 into L5 it would need to be orbiting slightly faster than the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, that object oscillates between L4 and L5, that would be more exotic and much harder to explain.   But passing though L3 is possible if the object is orbiting slightly slower or slightly faster than the orbit of the planet.   It could be a Jupiter or other planet's resonance that's causing it, but that's just a guess.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How could an object in the same orbit as Neptune have another orbital&#xA;  period than Neptune? Or how else can it migrate between L3, L4 and L5?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Saying an L4 or L5 object has exactly the same orbit as the planet is probobly inaccurate.   The planet tugs on the L4 or L5 object anyway, so there's kind of a dance with Trojan objects.   It might average out to being the same orbit, but it's never precisely the same on any given pass.   Tugs from other planets affect the dance too.   Jupiter, Saturn and Uranus all probobly tug on Neptune's Trojans, throwing off the clockwork a bit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And are &quot;trojan&quot; objects expected to gather in L3 too?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gather, no.   Pass through, yes.    L3 isn't stable, but that doesn't mean an orbit couldn't pass through L3 and be in a near L3 Orbit for several rotations around the sun, but it's not stable.  An L3 orbit would tend to move away from L3.    L1, L2 and L3 aren't stable, but they are still good points to put a satellite because less energy is needed to keep a satelite in those regions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source:  &lt;a href=&quot;http://map.gsfc.nasa.gov/mission/observatory_l2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://map.gsfc.nasa.gov/mission/observatory_l2.html&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that's not too clumsy an answer.  My layman's attempt.    I can try to clean up if needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit / Update&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/%28316179%29_2010_EN65&quot; rel=&quot;nofollow noreferrer&quot;&gt;2010 EN65&lt;/a&gt;, per your original link is a co-orbital with Neptune, so it's not a 3/2 or 2/3 or any other fraction it's 1 orbit to 1 orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also a &lt;a href=&quot;https://en.wikipedia.org/wiki/Centaur_%28minor_planet%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Centaur&lt;/a&gt;, which means it's long term orbit is not stable.  A true Trojan asteroid is stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found an article on Jumping Mechanisms of Trojan Asteroids here - it gives a brief summary but it's a pay article:  &lt;a href=&quot;http://link.springer.com/article/10.1007%2Fs10569-015-9609-4&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://link.springer.com/article/10.1007%2Fs10569-015-9609-4&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and a map of EN65's orbit (has to be calculated, it's not been observed for nearly that long), but map below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xrIqV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://inspirehep.net/record/1190480/plots#2&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://inspirehep.net/record/1190480/plots#2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's pattern looks similar to some of Earth's co-orbital asteroids, including 3753 Cruithne that I mentioned above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;diagrams here:  &lt;a href=&quot;http://223.252.18.210/petercaspari/bdi/coorb.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://223.252.18.210/petercaspari/bdi/coorb.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, that's probobly a more accurate comparison than Hildas.  Look at 3753 Cruithne, or other planet's co-orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What confuses me - and I hope you forgive me for only enough to be dangerous, as they say, but what confuses me is that your original link says it's an elliptical orbit but it also seems to be drawn to both L4 and L5, spending more time there, so I might have been incorrect in suggesting the elliptical orbit wouldn't be possible. - sorry bout that.   At least, some additional eccentricity seems likely, but I'm guessing, probobly not a huge amount.   If I was to guess, I'd guess that EN65 is less eccentric compared to Neptune than 3753 Cruithne is compared to Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a cool picture of Jupiter's Trojans (Jupiter's the King of Trojans, it's got like 50,000 in both L4 and L5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/om9i5.png&quot; alt=&quot;http://www.minorplanets.org/ExpeditionLaSagra/img/Jupiter_Trojans.png&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you'll probobly notice, there's one point at (or, perhaps more accurately, passing through) L3.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-07-20T21:03:08.560" LastActivityDate="2015-07-20T21:03:08.560" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11321" PostTypeId="2" ParentId="11305" CreationDate="2015-07-19T03:57:43.253" Score="20" Body="&lt;h1&gt;1. Ancient cultures observed the sky&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Night skies are naturally dark and there was no light-pollution in ancient times. So if weather permits, you can easily see a lot of stars. No need to tell about the Sun and the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ancient people had good reasons to study the night skies. In many cultures and civilizations, stars (and also the Sun and the Moon) where perceived to have religious, legendary, premonitory or magical significance (astrology), so a lot of people were interested in them. It did not took long to someone (in reality a lot of different people independently in many parts of the world) to see some useful patterns in the stars that would be useful to navigation, localization, counting hours, counting days and relating days to seasons, etc. And of course, those patterns in the stars were also related to the Sun and the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, surely all ancient cultures had people who dedicated many nights of their lifes to study the stars in detail right from the stone age. They would also perceive meteorites (falling stars) and eclipses. And sometimes a very rare and spetacular comet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there are the planets Mercury, Venus, Mars, Jupiter and Saturn. They are quite easily to notice to be distinct from the stars because all the stars seems to be fixed in the celestial sphere, but the planets don't. They are very easily to notice to be wandering around in the sky with the passage of the days, specially for Venus, which is the brightest &quot;star&quot; in the sky and is also a formidable wanderer. Given all of that, the ancient people surely become very aware of those five planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About Mercury, initially the Greeks thought that Mercury were two bodies, one that showed up only in the morning a few hours before the sunrise and another only a few hours after the sunset. However, soon they figured out that in fact it was only one body, because either one or the other (or neither) could be seen in a given day and the computed position of the unseen body always matched the position of the seen body.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;2. The Earth seems to be round&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Now, out of the stone age, already into ancient times, navigators and merchants who travelled great distances perceived that the Sun rising and setting points could variate not only due to the seasonal variation, but also accordingly to the location. Also, the distance from the polar star to the horizon line also variates accordingly with the location. This fact denounces the existence of the concept nowadays known as latitude, and this was perceived by ancient astronomers in places like Greece, Egypt, Mesopotamia and China.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomers and people who dependend on astronomy (like navigators) would wonder why the distance from the polar star to the horizon varied, and one possibility was that it is because the Earth would be round. Also, registering different Sun angles in different locations of the world on a given same day and at a given same hour, also gives a hint that the Earth is round. The shadow on the Moon during a lunar eclipse also gives a hint that the Earth is round. However, this by itself is not a proof that the Earth is round, so most people would bet on some other simpler thing, or simply don't care about this phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most cultures in ancient times presumed that the world was flat. However the idea of the world being round exists since the ancient Greece. Contrary to the popular modern misconception, &lt;a href=&quot;http://www.asa3.org/ASA/topics/history/1997Russell.html&quot; rel=&quot;noreferrer&quot;&gt;in the Middle Ages, almost no educated person on the western world thought that the world was flat&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the Earth's size, by observing different Sun positions and shadows angles in different parts of the world, &lt;a href=&quot;https://en.wikipedia.org/wiki/Eratosthenes&quot; rel=&quot;noreferrer&quot;&gt;Erasthotenes&lt;/a&gt; in ancient Greece calculated the size of Earth and the distance between the Earth and the Sun correctly for the first time as back as the third century B.C. However, due to the confusion about all the different and inconsistent unit measures existent back then and the difficulty to precisely estimate long land and sea distances, confusion and imprecision persisted until the modern times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ancient cultures also figured out that the shiny part of the Moon was illuminated by the Sun. Since the Full Moon is easily seen even at midnight, this implies that the Earth is not infinite. The fact that the Moon enters in a rounded shadow when exactly in the opposite side of the sky as the Sun also implies that it is the Earth's shadow on the Moon. This also implies that Earth is significantly larger than Moon.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;3. Geocentrism&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;So, people observed the Sun, Moon, Mercury, Venus, Mars, Jupiter, Saturn and the fixed sphere of stars all revolving around the sky. They naturally thought that the Earth would be the center of the universe and that all of those bodies revolved around the Earth. This culminated with the work of the phylosopher Claudius Ptolemaeus about &lt;a href=&quot;https://en.wikipedia.org/wiki/Geocentric_model#Ptolemaic_model&quot; rel=&quot;noreferrer&quot;&gt;geocentrism&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Altough we now know that the ptolomaic geocentric model is fundamentally wrong, it could be used to compute the position of the planets, the Sun, the Moon and the celestial sphere of stars, with a somewhat acceptable precision at the time. It accounted to include the observation of planets velocity variations, retrograde motions and also for coupling Mercury and Venus to the Sun, so they would never go very far from it. Further, based on the velocity of the motion of those bodies in the sky, then the universe should be something like:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Earth at the center.&lt;/li&gt;&#xA;&lt;li&gt;Moon orbiting the Earth.&lt;/li&gt;&#xA;&lt;li&gt;Mercury orbiting the Earth farther than the Moon.&lt;/li&gt;&#xA;&lt;li&gt;Venus orbiting the Earth farther than Mercury.&lt;/li&gt;&#xA;&lt;li&gt;Sun orbiting the Earth farther than Venus.&lt;/li&gt;&#xA;&lt;li&gt;Mars orbiting the Earth farther than Sun.&lt;/li&gt;&#xA;&lt;li&gt;Jupiter orbiting the Earth farther than Mars.&lt;/li&gt;&#xA;&lt;li&gt;Saturn orbiting the Earth farther than Jupiter.&lt;/li&gt;&#xA;&lt;li&gt;The celestial sphere of stars rotating around the Earth, being the outermost sphere.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In fact, the ptolomaic model is a very complicated model, way more complicated than the copernic, keplerian and newtonian models. Particularly, this could be compared to softwares that are based on severely flawed concepts but that are still working due to a lot of complex, tangled and unexplainable hacks and kludges that are there just for the sake of making the thing work.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;4. The discovery of the Americas&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Marco_Polo&quot; rel=&quot;noreferrer&quot;&gt;Marco Polo&lt;/a&gt;, in the last years of the 1200's, was the first European to travel to China and back and leave a detailed chronicle of his experience. So, he could bring a lot of knowledge about what existed in the central Asia, the East of Asia, the Indies, China, Mongolia and even Japan to the Europeans. Before Marco Polo, very few was known to the Europeans about what existed there. This greatly inspired European cartographers, philosophers, politicians and navigators in the years to come.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Portugal and Spain fough &lt;a href=&quot;https://en.wikipedia.org/wiki/Reconquista&quot; rel=&quot;noreferrer&quot;&gt;a centuries-long war against the invading Moors on the Iberian Peninsula&lt;/a&gt;. The Moors were finally expelled in 1492. The two states, were looking for something profitable after so many years of war. Since Portugal ended its part of the war first, it had a head start and went to explore the seas first. Both Portugal and Spain were trying to find a navigation route to reach the Indias and the China in order to trade highly profitable spices and silk. Those couldn't be traded by land efficiently anymore due to the fact that the lands on West Asia and North Africa were dominated by Muslim cultures unfriendly to Christian Europeans, a situation that were just made worse after the fall of Constantinople in 1453.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Portugal, were colonizing the Atlantic borders of Africa and eventually they managed to reach the Cape of Good Hope in 1488 (with &lt;a href=&quot;https://en.wikipedia.org/wiki/Bartolomeu_Dias&quot; rel=&quot;noreferrer&quot;&gt;Bartolomeu Dias&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Genovese navigator called &lt;a href=&quot;https://en.wikipedia.org/wiki/Christopher_Columbus&quot; rel=&quot;noreferrer&quot;&gt;Cristoforo Colombo&lt;/a&gt; believed that if he sailed west from the Europe, he could eventually reach the Indies from the east side. Inspired by Marco Polo and subestimating the size of Earth, he estimated that the distance between the Canary Islands and the Japan to be 3700 km (in fact it is 12500 km). Most navigators would not venture in such voyage because they (rightly) tought that Earth was larger than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Colombo tried to convice the king of Portugal to finance his journey in 1485, but after submitting the proposal to experts, the king rejected it because the estimated journey distance was too low. Spain, however, after finally expelling the Moors in 1492, were convinced by him. Colombo's idea was far-fetched, but, after centuries of wars with the Muslims, if that worked, then Spain could profit quickly. So, the Spanish king approved the idea. And just a few months after expelling the Moors, Spain sent Colombo to sail west towards the Atlantic and then he reach the Hispaniola island in Central America. After coming back, the news about the discovery of lands in the other side of the Atlantic spread quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Portugal and Spain then divided the world by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Treaty_of_Tordesillas&quot; rel=&quot;noreferrer&quot;&gt;Treaty of Tordesillas&lt;/a&gt; in 1494. In 1497, &lt;a href=&quot;https://en.wikipedia.org/wiki/Amerigo_Vespucci&quot; rel=&quot;noreferrer&quot;&gt;Amerigo Vespucci&lt;/a&gt; reached the mainland America.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Portugal would not be left behind, they managed to navigate around Africa to reach the Indies in 1498 (with &lt;a href=&quot;https://en.wikipedia.org/wiki/Vasco_da_Gama&quot; rel=&quot;noreferrer&quot;&gt;Vasco da Gama&lt;/a&gt;). And they sent &lt;a href=&quot;https://en.wikipedia.org/wiki/Pedro_%C3%81lvares_Cabral&quot; rel=&quot;noreferrer&quot;&gt;Pedro Álvares Cabral&lt;/a&gt;, who reached the Brazil in 1500 before crossing the Atlantic back in order to go for the Indies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, Portugal and Spain quickly started to explore the Americas and eventually colonize them. France, England and Netherlands also came to the Americas some time later.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;5. The Earth IS round&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;After, the Spanish discovered and settled into the Americas (and Colombo's plan in fact didn't worked). The question that if it was possible to sail around the globe to reach the Indies from the east side remained open and the Spanish were still interested on it. They eventually discovered the Pacific Ocean after crossing the Panama Ishtums by land in 1513.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eager to find a maritime route around the globe, the Spanish crown funded an expedition leadered by the Portuguese &lt;a href=&quot;https://en.wikipedia.org/wiki/Ferdinand_Magellan&quot; rel=&quot;noreferrer&quot;&gt;Fernão de Magalhães&lt;/a&gt; (or Magellan as his name was translated to English) to try to circle the globe. Magellan was an experienced navigator, and had reached what is present day Malaysia traveling through the Indian Ocean before. They departed from Spain in September 20th, 1519. It was a long and though journey that costed the lives of most of the crew. Magellan himself did not survived, having died in a battle in the Phillipines on 1521. At least, he lived enough to be aware that they in fact reached East Asia by traveling around the globe to the west, which also proves that the Earth is round.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The journey was eventually completed by the leadership of &lt;a href=&quot;https://en.wikipedia.org/wiki/Juan_Sebasti%C3%A1n_Elcano&quot; rel=&quot;noreferrer&quot;&gt;Juan Sebatián Elcano&lt;/a&gt;, one of the crewmen of Magellan. They reached Spain back through the Indian and Atlantic Oceans on September 6th, 1522 after traveling for almost three years a distance of 81449 km.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;6. Heliocentrism&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;There were some heliocentric or hybrid geo-heliocentric theories in ancient times. Notably by the Greek philosopher &lt;a href=&quot;https://en.wikipedia.org/wiki/Philolaus&quot; rel=&quot;noreferrer&quot;&gt;Philolaus&lt;/a&gt; in the 5th century BC. By &lt;a href=&quot;https://en.wikipedia.org/wiki/Martianus_Capella&quot; rel=&quot;noreferrer&quot;&gt;Martianus Capella&lt;/a&gt; around the years 410 to 420. And by &lt;a href=&quot;https://en.wikipedia.org/wiki/Aristarchus_of_Samos&quot; rel=&quot;noreferrer&quot;&gt;Aristarchus of Samos&lt;/a&gt; around 370 BC. Those models tried to explain the motion of the stars as rotation of the Earth and the position of the planets, specially Mercury and Venus as translation around the Sun. However those early models were too imprecise and flawed to work appropriately, and the ptolomaic model still was the model with the better prediction of the positions of the heavenly bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea that the Earth rotates was much less revolutionary than heliocentrism, but &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth%27s_rotation#History&quot; rel=&quot;noreferrer&quot;&gt;was already more-or-less accepted with reluctancy in the middle ages&lt;/a&gt;. This happens because if the stars rotated around Earth, they would need do so at an astonishing velocity, dragging the Sun, the Moon and the planets with it, so it would be easier if Earth itself rotated. People were uncomfortable with this idea, but they still accepted it, and this became easier to be accepted after the Earth sphericity was an established concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first years of the 1500's, while the Portuguese and Spanish were sailing around the globe, a polish and very skilled matemathical and astronomer called &lt;a href=&quot;https://en.wikipedia.org/wiki/Nicolaus_Copernicus&quot; rel=&quot;noreferrer&quot;&gt;Nikolaus Kopernikus&lt;/a&gt; took some years thinking about the mechanics of the heavenly bodies. After some years making calculations and observations, he created a model of circular orbits of the planets around the Sun and perceived that his model were much more simpler than the ptolomaic geocentric model and was at least as precise. His model also features a rotating Earth and fixed stars. Further, his model implied that the Sun was much larger than the Earth, something that was already strongly suspected at the time due to calculations and measurements and also implied that Jupiter and Saturn were several times larger than Earth, so Earth would definitively be a planet just like the other five then-known planets were. This could be seen as the borning of the model today know as Solar System.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fearing persecution and harsh criticism, he avoided to publish many of his works, sending manuscripts to only his closest acquaintances, however his works eventually leaked out and he was convinced to allow its full publication anyway. Legend says that he was presented to his finally fully published work in the very day that he died in 1543, so he could die in peace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was a heated debate between supporters and oppositors of Copernic's heliocentric theory in the middle 1500's. One argument for the opposition was that star parallaxes could not be observed, which implied that either the heliocentric model was wrong or that the stars were very very far and many of them would be even larger than the Sun, which seemed to be a crazy idea at the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Tycho_Brahe&quot; rel=&quot;noreferrer&quot;&gt;Tycho Brache&lt;/a&gt;, which did not accepted heliocentrism, in the latest years of the 1500's tried to save geocentrism with an hybrid geo-heliocentric model that featured the five heavenly planets orbiting the Sun while the Sun and the Moon orbited Earth. However, he also published a theory which better predicted the position of the Moon. Also, by this time, the observation of some supernovas showed that the celestial sphere of the stars were not exactly immutable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1600, the astronomer &lt;a href=&quot;https://en.wikipedia.org/wiki/William_Gilbert_(astronomer)&quot; rel=&quot;noreferrer&quot;&gt;William Gilbert&lt;/a&gt; provided strong argument for the rotation of Earth, by studing magnets and compasses, he could demonstrate that the Earth was magnetic, which could be explained by the presence of enourmous quantities of iron in its core.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;7. With telescopes&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;All of what I wrote above happened without telescopes, only by using naked eye observations and measurements around the globe. Now, add even some small telescopes and things change quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The earliest telescopes &lt;a href=&quot;https://en.wikipedia.org/wiki/Telescope&quot; rel=&quot;noreferrer&quot;&gt;were invented in 1608&lt;/a&gt;. In 1609, the astronomer &lt;a href=&quot;https://en.wikipedia.org/wiki/Galileo_Galilei&quot; rel=&quot;noreferrer&quot;&gt;Galieu Galilei&lt;/a&gt; heard about that, and constructed his own telescope. In January of 1610, &lt;a href=&quot;https://en.wikipedia.org/wiki/Galileo_Galilei&quot; rel=&quot;noreferrer&quot;&gt;Galieu Galilei&lt;/a&gt;, using a small telescope, observed four small bodies orbiting Jupiter at different distances, figuring out that they were Jupiter's &quot;moons&quot;, he also could predict and calculate its positions along their orbits. Some months later, he also observed that Venus had phases as seen from the Earth. He also observed Saturn's rings, but his telescope was not powerful enough to resolve them as rings, and he tought that they were two moons. These observations were incompatible with the geocentric model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A contemporary of Galilei, &lt;a href=&quot;https://en.wikipedia.org/wiki/Johannes_Kepler&quot; rel=&quot;noreferrer&quot;&gt;Johannes Kepler&lt;/a&gt;, working on Copernicus' heliocentric model and making a lot of calculations, in order to explain the differing orbital velocities, created an heliocentric model where the planets orbits the Sun in elliptic orbits with one of the ellipse's focus in the Sun. His works were published in 1609 and 1619. He also suggested that tides were caused by the motion of the Moon, though Galilei was skeptical to that. His laws predicted a transit of Mercury in 1631 and of Venus in 1639, and such transit were in fact observed. However, a predicted transit of Venus in 1631 could not be seen due to imprecision in calculations and the fact that it was not visible in much of the Europe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1650 the &lt;a href=&quot;https://en.wikipedia.org/wiki/Double_star&quot; rel=&quot;noreferrer&quot;&gt;first double star&lt;/a&gt; were observed. Further in the 1600's, the Saturn rings were resolved by the use of better telescopes by &lt;a href=&quot;https://en.wikipedia.org/wiki/Robert_Hooke&quot; rel=&quot;noreferrer&quot;&gt;Robert Hooke&lt;/a&gt;, who also observed a double star in 1664 and developed microscopes to observe cellular structures. From them on, many stars were discovered to be double. In 1655, Titan were discovered orbiting Saturn, putting more confidence on the heliocentric model. More four Saturnian moons were discovered between 1671 and 1684.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;8. Gravitation&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Heliocentrism was reasonably well-accepted in the middle 1600's, but people was not confortable with it. Why the planets orbits the Sun afterall? Why the Moon orbits Earth? Why Jupiter and Saturn had moons? Although Keplerian mechanics could predict their moviment, it was still unclear what was the reason that makes them move that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1687, &lt;a href=&quot;https://en.wikipedia.org/wiki/Isaac_Newton&quot; rel=&quot;noreferrer&quot;&gt;Isaac Newton&lt;/a&gt; who was one of the most brilliant physic and mathematic that ever lived (although he was also an implacable persecutor of his opponents), provided the gravitational theory (based on prior work by Robert Hooke). Ideas for the gravitation theory and the inverse square law already were developed in the 1670's, but he could publish a very simple and clear theory for gravitation, very well-fundamented in physics and mathematics and it explained the motions of the celestial bodies with a great precision, including comets. It also explained why the planets, the Moon and the Sun are spherical, explained tides and it also served to explain why things falls to the ground. This made heliocentrism to be definitely widely accepted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, Newton gravitational law predicted that Earth rotation would make it not exactly spherical, but a bit ellipsoidal by a factor of 1:230. Something that agreed with measures done using pendulums in 1673.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;9. What are the stars and the Solar System afterall?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In the early 1700's, &lt;a href=&quot;https://en.wikipedia.org/wiki/Edmond_Halley&quot; rel=&quot;noreferrer&quot;&gt;Edmund Halley&lt;/a&gt;, already knowing about newtonian laws (he was a contemporary of Newton) perceived that comets who passed near Earth would eventually return, and he found that there was a particular instance of sightings every 76 years, so he could note that those comets in reality were all the same comet, which is called after him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only remaining problem with the heliocentric model was the lack of observation of parallax to the stars. And nobody knew for sure what the stars were. However, if they in fact are very distant bodies, most of them would be much larger than the Sun. At the first half of the 1700's, trying to observe parallax, &lt;a href=&quot;https://en.wikipedia.org/wiki/James_Bradley&quot; rel=&quot;noreferrer&quot;&gt;James Bradley&lt;/a&gt; perceived phenomena like the aberration of light and the Earth's nutation, and those phenomena also provides a way to calculate the speed of light. But the observation of parallax remained a challenge during the 1700's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1781, Uranus were discovered orbiting the Sun beyond Saturn. Although barely visible to the naked eye in the darkest skies, it was so dim that it escaped observation from astronomers until then, and so were discovered with a telescope. The first asteroids were also discovered in the early 1800's. Investigation on pertubations on Uranus' orbit due to the predicted newtonian and keplerian movement eventually leaded to the discovery of Neptune in 1846.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1838, the astronomer &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedrich_Wilhelm_Bessel&quot; rel=&quot;noreferrer&quot;&gt;Friedrich Wilhelm Bessel&lt;/a&gt; who measured the position of more than 50000 stars with the greatest precision as possible, could finally measure the parallax of the star 61 Cygni successfully, which proved that stars were in fact very distant bodies and that many of them were in fact larger than the Sun. This also demonstrates that the Sun is a star. Vega and Alpha Centauri also had their parallaxes measured successfully in 1838. Further, those measurements permitted to estimate the distance between those stars and the Solar System to be on the order of many trillions of kilometers, or several light-years.&lt;/p&gt;&#xA;" OwnerUserId="609" LastEditorUserId="609" LastEditDate="2018-05-04T17:29:42.057" LastActivityDate="2018-05-04T17:29:42.057" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="11322" PostTypeId="1" AcceptedAnswerId="11323" CreationDate="2015-07-19T13:16:59.800" Score="2" ViewCount="142" Body="&lt;p&gt;The basis of my confusion is that atomic orbitals, even when described accurately by quantum numbers, have definite energies, meaning they represent energy eigenstates of the wave function. They're stationary states, and thus separable solutions to the Schrödinger equation. The general wave function of an electron, however, is a linear combination of such states, so that general solution does not have a definite energy. Still, we observe discrete spectral lines at specific wavelengths (with some broadening).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why do we observe spectral lines of specific energy when the wave function of the electrons in an atom are linear combinations of energy eigenstates&lt;/strong&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like the spectral lines that are observed are just the energy eigenstates, while, according to quantum mechanics, the wave representation of electrons is a linear combination of those states.&lt;/p&gt;&#xA;" OwnerDisplayName="user5341" LastEditorUserId="2365" LastEditDate="2018-07-27T22:35:04.927" LastActivityDate="2018-07-27T22:35:04.927" Title="Why do we observe spectral lines of specific energy?" Tags="&lt;spectroscopy&gt;&lt;spectra&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11323" PostTypeId="2" ParentId="11322" CreationDate="2015-07-19T17:38:11.373" Score="1" Body="&lt;p&gt;When an interaction occurs that results in the absorption or emission of a photon, we are not dealing with stationary states. Stationary states are solutions of the time-independent Schrödinger equation, but the absorption and emission of photons is a time-dependent phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spectral lines in fact do not have a definite energy. They have a natural width associated with the time taken for the transition to occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability that a transition might occur resulting in the absorption or emission of a photon of a given frequency can only be tackled by considering a time-dependent Hamiltonian. What one finds is that transition probabilities become large for frequencies close to the those corresponding to energy differences between stationary states.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-07-19T17:38:11.373" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11326" PostTypeId="1" CreationDate="2015-07-20T17:04:38.337" Score="2" ViewCount="1162" Body="&lt;p&gt;&lt;a href=&quot;http://pluto.jhuapl.edu/News-Center/News-Article.php?page=20150717-3&quot; rel=&quot;nofollow&quot;&gt;http://pluto.jhuapl.edu/News-Center/News-Article.php?page=20150717-3&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Scientists working with NASA’s New Horizons spacecraft have observed Pluto’s atmosphere as far as 1,000 miles (1,600 kilometers) above the surface of the planet, demonstrating that Pluto’s nitrogen-rich atmosphere is quite extended.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I note that conventional descriptions of the gas giants' sizes include mostly atmosphere. &lt;/p&gt;&#xA;" OwnerUserId="7728" LastActivityDate="2015-07-20T17:51:51.973" Title="Shall we say now that Pluto is &quot;larger than Mercury&quot; ?" Tags="&lt;planetary-atmosphere&gt;&lt;pluto&gt;&lt;mercury&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11327" PostTypeId="1" AcceptedAnswerId="11361" CreationDate="2015-07-20T17:39:33.673" Score="4" ViewCount="247" Body="&lt;p&gt;It is known that &lt;a href=&quot;https://en.wikipedia.org/wiki/Spacetime&quot; rel=&quot;nofollow&quot;&gt;spacetime&lt;/a&gt; is curved,so I am wandering to know if its possible to look into space and we see earth (light from earth long long ago). Did I understood spacetime correctly and if its possible then can some one after million years can see dinosaurs by looking into space?&lt;/p&gt;&#xA;" OwnerUserId="379" OwnerDisplayName="SpringLearner" LastActivityDate="2015-07-23T15:34:39.333" Title="Can we see Earth by looking into space?" Tags="&lt;space&gt;&lt;space-time&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11328" PostTypeId="2" ParentId="11326" CreationDate="2015-07-20T17:46:05.703" Score="7" Body="&lt;p&gt;The implication of the question is that this extra 1000 miles should be added to Pluto's radius. The answer is no. For all of the solid planets, it's that solid surface (or solid+liquid surface in the case of the Earth) that counts, not the outer reaches of the atmosphere. The surface is a clear-cut, non-arbitrary boundary. The atmosphere? They can extend a long way out. A non-arbitrary boundary is always going to be preferred over an arbitrary one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's not possible in the case of the Sun and the giant planets. A somewhat arbitrary boundary is needed for those bodies. That somewhat arbitrary boundary explicitly excludes the upper reaches of the atmosphere, which extends out for many thousands of kilometers. For the giant planets, some use a tenth of the Earth's atmosphere as defining the arbitrary &quot;surface&quot;, others, one atmosphere, yet others, ten atmospheres. That factor of 100 variance in pressure amounts to about a hundred kilometers. That's not that much considering how big the gas giants are.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2015-07-20T17:51:51.973" LastActivityDate="2015-07-20T17:51:51.973" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11329" PostTypeId="1" CreationDate="2015-07-20T23:06:54.300" Score="6" ViewCount="587" Body="&lt;p&gt;BACKGROUND&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hot Jupiters are thought to have migrated inwards, implying that another giant planet has been ejected in order to conserve the orbital momentum of those planetary systems. The number of interstellar &quot;vagabond planets&quot; has been estimated to be as many as the number of stars. And &lt;a href=&quot;http://arxiv.org/ftp/arxiv/papers/1507/1507.04654.pdf&quot; rel=&quot;noreferrer&quot;&gt;simulations show&lt;/a&gt; that if the Solar System started out with an extra gas giant next to Jupiter which was ejected, then one can end up with a very good fit with today's Solar System. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;QUESTIONS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Where would such a hypothetical ejected ice giant be today?&lt;/strong&gt; Should it be closer or further away than hot Jupiter-ejected planets because it is much smaller? Could it stay in an Oort cloud type orbit during several billion years (maybe ~15 orbits around the Milky Way) without being pulled off by passing stars? At what distance could we today observe a Uranus sized planet? Should it stay in the ecliptical plane or could its orbit have become greatly inclined when it was ejected?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-07-22T01:42:26.923" Title="Could an ejected &quot;extra ice-giant&quot; still be lurking in distant solar orbit?" Tags="&lt;solar-system&gt;&lt;gas-giants&gt;&lt;celestial-mechanics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11330" PostTypeId="1" CreationDate="2015-07-21T03:21:23.670" Score="2" ViewCount="360" Body="&lt;p&gt;Is there any naturally occurring event powerful enough that can cause a galaxy regardless of shape and size to divide itself into two or more smaller galaxies?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastActivityDate="2015-07-21T10:14:28.383" Title="Can a galaxy splits into multiple galaxies?" Tags="&lt;galaxy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11331" PostTypeId="2" ParentId="11330" CreationDate="2015-07-21T10:14:28.383" Score="6" Body="&lt;p&gt;Small galaxies (say of masses $\sim10^7M_\odot$) can be torn apart by stellar feedback, i.e. the energy provided by exploding stars in a starburst. Whether you want to call the debris galaxies is a matter of opinion, I suppose, but since these guys typically are satellites of larger galaxies, it will probably be torn completely apart by tidal forces, eventually accreting onto the larger one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Large galaxies (say of masses $\sim10^{11}M_\odot$) are too massive to be split by feedback, but may lose matter (mostly gas, rarely stars) through large-scale outflows created by stellar feedback (or a quasar jet). However, when two large galaxies collide and merge, sometimes large chunk of gas, stars, and dark matter are slung out, which arguably could be called individual galaxies. Most of the matter of the two original galaxies become one, though.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2015-07-21T10:14:28.383" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11332" PostTypeId="1" CreationDate="2015-07-21T11:23:27.213" Score="2" ViewCount="96" Body="&lt;p&gt;BACKGROUND&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.breakthroughinitiatives.org/&quot; rel=&quot;nofollow&quot;&gt;The Breakthrough Initiatives&lt;/a&gt; is in the news because of a $100 million private donation for a SETI project. It doesn't seem to be for building any new observatory (but maybe for buying computing power for handling the data for SETI purposes?). It will use two of the greatest radio telescopes in the world, Parks and Green Banks, and optical Automated Planet Finder Telescope at Lick, some part of the time during 10 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QUESTIONS&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Does this mean a zero sum redistribution of observatory time from other astronomical purposes to SETI?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How does it work when someone can buy the usage of these telescopes? Isn't there a danger in letting popular scifi ideas decide how the great telescopes should be used? Do recruited big names like Hawkins and Drake decide how they should be used anyway??&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="6" LastEditDate="2019-01-29T14:45:59.607" LastActivityDate="2019-01-29T14:45:59.607" Title="How will the Breakthrough Initiatives affect other astronomy observations?" Tags="&lt;radio-astronomy&gt;&lt;observatory&gt;&lt;extra-terrestrial&gt;&lt;interstellar-travel&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11333" PostTypeId="1" CreationDate="2015-07-21T16:15:02.737" Score="2" ViewCount="121" Body="&lt;p&gt;&lt;a href=&quot;http://www.seti.org/&quot; rel=&quot;nofollow&quot;&gt;SETI&lt;/a&gt; collects alot of data from space, mostly radio. I wonder if this data has any astrophysics value other than SETI?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-08-18T23:28:54.040" Title="Has SETI data been used for astrophysics? How, or why not?" Tags="&lt;radio-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11334" PostTypeId="1" CreationDate="2015-07-21T17:11:49.180" Score="3" ViewCount="267" Body="&lt;p&gt;For a non-astronomy personal project I would like to have a large set of optical stellar spectra combined with absolute magnitude when available. I search for a dataset that contains 10k - 10M objects, with total compressed size below 100Gb (ideally below 10Gb), is freely available via ftp/http/rsync and machine-readable. Presence of most 'naked eye visible' stars and their individual names in the set is desired, but not required. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that some digital survey data are freely available via FTP, so maybe some reasonable spectral data catalog exists as well? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, I'm planning to download the data for local processing, so retrieving them through on-line forms (or even automated http interfaces) is undesirable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best I was able to find is this &lt;a href=&quot;http://www.as.utexas.edu/~hebe/s4n/data.html&quot; rel=&quot;nofollow&quot;&gt;catalog&lt;/a&gt;  . Dropping aside that FITS is not something I'm used to (this apparentrly can be rectified) it is too small. I'd like at least 1k stars, 10k preferable. There is apparently a lot of surveys in 0.1-10k objects range that are focused on specific type of stars (say, nearby M-class dwarfs) and stellar libraries containing spectral data on representative objects from various classes. However I'd like a 'representative set' here, which contains stars of various nature in proportion similar to their 'natural abundance' in some region. An example would be the set of stars of solar neighborhood above, but again it's too small. &lt;/p&gt;&#xA;" OwnerUserId="7735" LastEditorUserId="7735" LastEditDate="2015-07-23T07:33:52.440" LastActivityDate="2015-07-23T07:33:52.440" Title="Any freely available large stellar spectra catalog?" Tags="&lt;data-analysis&gt;" AnswerCount="0" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11335" PostTypeId="2" ParentId="11333" CreationDate="2015-07-21T19:27:43.467" Score="3" Body="&lt;p&gt;Well, that might depend on exactly what definitions you use.  SETI itself is a broad program.  In this answer I'll focus solely on what I consider the most iconic part of modern SETI, which is SETI@home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have the following quote from &lt;a href=&quot;http://setiathome.berkeley.edu/sah_about.php&quot; rel=&quot;nofollow&quot;&gt;SETI@home's webpage&lt;/a&gt; (emphasis their own):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;SETI&lt;/strong&gt; (Search for Extraterrestrial Intelligence) is a scientific area whose goal is to detect intelligent life outside Earth. One approach, known as &lt;strong&gt;radio SETI&lt;/strong&gt;, uses radio telescopes to listen for narrow-bandwidth radio signals from space. Such signals are not known to occur naturally, so a detection would provide evidence of extraterrestrial technology.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;SETI@home's specific purpose deals with the radio SETI data.  So they are implicitly asserting that the data they collect is expected to be useless outside of their namesake: they're specifically looking for things that are not known to occur naturally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, it turns out that the SETI@home data collection is almost never active (see &lt;a href=&quot;http://setiathome.berkeley.edu/sci_status.html&quot; rel=&quot;nofollow&quot;&gt;here on their site&lt;/a&gt;).  It is done passively, as a piggyback collection when the Arecibo radio telescope is being used for other science projects.  So if your question is &quot;is any data useful outside of SETI actually collected while SETI(@home) is gathering data?&quot;, then the answer is &quot;yes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we can also note that one of the original goals of the SETI@home project was to prove the viability and practicality of distributed computing for scientific projects.  This we can say was a resounding success, as it spawned &lt;a href=&quot;https://en.wikipedia.org/wiki/BOINC&quot; rel=&quot;nofollow&quot;&gt;BOINC&lt;/a&gt;, which serves as a distributed computing platform for a wide range of non-SETI science projects: Folding@home and Rosetta@home deal with protein folding, Einstein@home is for gravity wave detection, the now-defunct Mesenne@home was for finding Mersenne primes, and many others (my copy of BOINC specifically states that there are over 30 currently active).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you are willing to consider the advances they spurred in distributed computing in scientific projects, and how this has impacted scientific projects inside astrophysics (or outside), then the answer is again &quot;yes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT:  Einstein@Home recently made &lt;a href=&quot;http://einstein.phys.uwm.edu/forum_thread.php?id=11375&quot; rel=&quot;nofollow&quot;&gt;the following announcement&lt;/a&gt; about a gamma ray pulsar that was detected via the program; it includes links to a published article on the detection, including an arxiv version link if you find the official publication is behind a paywall.  So that's a concrete, published, and current discovery in astrophysics.  If you're willing to give SETI some of the credit for having spawned the @Home style of distributed computing projects in science, as suggested above, then there you go.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastEditorUserId="2188" LastEditDate="2015-08-18T23:28:54.040" LastActivityDate="2015-08-18T23:28:54.040" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11336" PostTypeId="1" AcceptedAnswerId="11337" CreationDate="2015-07-21T21:25:55.413" Score="2" ViewCount="3773" Body="&lt;p&gt;Today I was wondering with a friend about the birth of a star. So we know that stars are born from a nebula, but the residue of a dead star (like a supernova) is itself a nebula. So can a star be born from this nebula? &lt;/p&gt;&#xA;" OwnerUserId="7674" LastEditorUserId="6980" LastEditDate="2015-07-22T19:42:44.377" LastActivityDate="2015-07-22T19:42:44.377" Title="Can a supernova make a new star?" Tags="&lt;star&gt;&lt;supernova&gt;&lt;star-formation&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ClosedDate="2015-07-24T10:53:00.407" ContentLicense="CC BY-SA 3.0" />
  <row Id="11337" PostTypeId="2" ParentId="11336" CreationDate="2015-07-21T22:14:56.727" Score="3" Body="&lt;p&gt;This idea has been around for decades, so I'm not sure who first came up with it. Here's a reasonably sourced article on the involvement of supernova in solar system formation:&lt;a href=&quot;http://www.space.com/16943-supernova-explosion-solar-system-formation.html&quot; rel=&quot;nofollow&quot;&gt;Exploding Star May Have Sparked Formation of Our Solar System&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The shock wave from an exploding star likely helped trigger the formation of our solar system, according to a new 3D computer model, researchers say.&#xA;  The solar system is thought to have coalesced from a giant rotating cloud of gas and dust known as the solar nebula about 4.6 billion years ago. For decades, scientists have suspected a star explosion called a supernova helped trigger our solar system's formation. In particular, the shock wave from the explosion is thought to have compressed parts of the nebula, causing these regions to collapse.&#xA;  According to this theory, the shock wave would have injected material from the exploding star into the solar nebula.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So yes, material from supernova can end up triggering the formation of new suns, and material from supernova do end up mixed in with the nebular material that makes the new stars.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-07-21T22:14:56.727" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11339" PostTypeId="1" AcceptedAnswerId="11341" CreationDate="2015-07-22T01:08:24.970" Score="4" ViewCount="477" Body="&lt;p&gt;How fast does the blast front of a supernova expand at? Is it close to the speed of light or is it less than a quarter of the speed of light?&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2015-07-22T04:10:39.327" Title="Speed of blast from supernova" Tags="&lt;supernova&gt;&lt;explosion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11340" PostTypeId="2" ParentId="11329" CreationDate="2015-07-22T01:42:26.923" Score="4" Body="&lt;p&gt;Since I'm a hobbyist, I usually wait to see if someone a bit smarter wants to answer first, but I can give a couple thoughts on this.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hot Jupiters are thought to have migrated inwards, implying that&#xA;  another giant planet has been ejected in order to conserve the orbital&#xA;  momentum of those planetary systems.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the article you posted (I'll pull the quote)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The conservation of orbital momentum dictates, as planetesimals move&#xA;  from the outer disk inward, that Saturn, Uranus and Neptune must move&#xA;  outward. This process is known as the planetesimal driven migration&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The conservation of orbital momentum he's talking about happens when objects cross each others orbits and interact, one gains orbital energy, the other loses energy.   That's not the only way planetary orbits can change.   Over time, there's tidal effects, orbital resonance and a sun can lose mass leading to gradual drifting or perhaps a sun can gain mass if the solar system passes through a dense dust/gas cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But mostly what we're talking about here is a form of gravity assist as 2 planets get close enough that they kind of swing past each other, one moves in the other out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More on that here:   &lt;a href=&quot;http://science.howstuffworks.com/nomad-planet2.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://science.howstuffworks.com/nomad-planet2.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, were gas giants ejected from our solar system, according to this question, somebody said no:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/747/is-there-any-evidence-that-the-gas-giant-planets-in-our-solar-system-are-experie?rq=1&quot;&gt;Is there any evidence that the Gas Giant planets in our solar system are experiencing orbital migration?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, this article, and the article you posted that suggests otherwise, that it probobly did happen based on models, but when the solar system was young, like about 4 billion years ago, this model suggests the Solar system could have lost 1 or 2 ice giants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.space.com/13584-extra-giant-planet-solar-system.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.space.com/13584-extra-giant-planet-solar-system.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we could still lose Mercury (though we probobly won't).   &lt;a href=&quot;http://www.universetoday.com/14032/could-jupiter-wreck-the-solar-system/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.universetoday.com/14032/could-jupiter-wreck-the-solar-system/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where would such a hypothetical ejected ice giant be today? Should it&#xA;  be closer or further away than hot Jupiter-ejected planets because it&#xA;  is much smaller?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tough call.  Using basic gravity assist mathematics, the closer to the sun the faster an object could be ejected from the solar system so a hot Jupiter probobly eject inner planets at a faster velocity on average than more distant Jupiter, but the bigger factor is the size of the star system, not the size of the planet.   The more massive the star, the faster the orbital velocities, the faster the Ejection is likely to be, even after accounting for escape velocity.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;it's also important that the object being ejected is smaller than the object doing the ejecting, for example, Jupiter could pull Mercury away from the sun but Mercury couldn't pull Jupiter into the sun.   More massive wins the tug of war, but how much smaller the ejected object needs to be isn't that relevant.  Jupiter could throw a Neptune sized planet out of a solar system at almost the same speed as it could throw a Mercury sized planet out of a solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as to where it would be, planets aren't likely to escape a solar system at too fast a speed.   Jupiter orbits the sun at 47,000 KM/Hour and an optimal gravity assist could eject something way from Jupiter at 94,000 KM/Hour, (plus any velocity of the object's own, which would depend largely on eccentricity, but since it would cross Jupiter's path, probably wouldn't be hugely different than Jupiter's), and if you subtract escape velocity from the sun at that distance (47,000 * square root of 2), or 66.4 KM/Hour, your left with a maximum exit the solar system velocity of, roughly 28,000 KM/Hour, plus a bit - and that's maximum.  Average escape velocity would likely be less, but lets say 20,000 KM/Hour, which works out to be 1 light year every 50,000 years.  considering it left the solar system an estimated 4 billion years ago, that's 80,000 light years, which from where we are, is pretty much the far end of the Milky way and who can say how it's orbital velocity might have changed over that time as it passed by other stars, but basically, it could be anywhere in the Milky way, and, if you consider, our sun might have come from the Sagittarius dwarf galaxy, not the Milky way, it could be practically anywhere, even ejected from the Milky way.  4 billion years is a long time, even for a slow ejection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or, it could be orbiting another sun - article on that here:  &lt;a href=&quot;http://www.universetoday.com/94656/rogue-planets-can-find-homes-around-other-stars/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.universetoday.com/94656/rogue-planets-can-find-homes-around-other-stars/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Could it stay in an Oort cloud type orbit during several billion years&#xA;  (maybe ~15 orbits around the Milky Way) without being pulled off by passing stars? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Entirely possible though billions of years is a long time.   Passing stars usually don't pass very close, but I'm sure it's happened at least a few times in the past 4 billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;close enough to disrupt the inner planets - that's different and more rare.   Close enough to disrupt the Oort cloud, probobly happens every so often - but I wouldn't want to guess how often.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At what distance could we today observe a Uranus sized planet? Should&#xA;  it stay in the ecliptical plane or could its orbit have become greatly&#xA;  inclined when it was ejected?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;it depends if we know where to look.    If we saw it pass in-front of a star and we could plot it's path, then a number of light-years.   Looking into the blackness of space, much less distance, and it would depend on the planet's reflectivity too.   I wouldn't even know what to guess, but Eris, for example is 1/20th the diameter of Uranus and it's one of the larger Kuiper belt objects.   With relative ease, they could likely spot a Uranus sized object at well over 20 times the distance of Eris.   Now, if the question is, what's the furthest a Uranus sized object could be and not have been seen yet - that's hard for me to say but an object that size and at that distance would likely have it's own Trojans and I find it hard to believe it would have gone unnoticed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as to &quot;would it stay in an elliptical plane&quot; - not necessarily.   A gravitational assist around a planet can alter the elliptical plane depending on the angle of approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hope that's not too wordy.  I can try to tidy up later.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-07-22T01:42:26.923" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11341" PostTypeId="2" ParentId="11339" CreationDate="2015-07-22T04:10:39.327" Score="5" Body="&lt;p&gt;The speed of the blast front depends on the initial energy release and the density of the medium into which it is expanding, see &lt;a href=&quot;http://www.astronomy.ohio-state.edu/~ryden/ast825/ch5-6.pdf&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theory suggests and &lt;a href=&quot;http://jila.colorado.edu/~dick/Racusin09.pdf&quot;&gt;measurements confirm expansion&lt;/a&gt; rates of the order of thousands of km/s or a few $\times 10^6\ \mbox{m/s}$ or  $\sim 1\% \mbox{c}$.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-07-22T04:10:39.327" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11342" PostTypeId="1" AcceptedAnswerId="11345" CreationDate="2015-07-22T07:59:51.647" Score="2" ViewCount="347" Body="&lt;p&gt;I am doing a project which requires that I have data for the mass of several stars. However, the data for mass of stars is obviously not readily available as it's really hard to calculate. In such a scenario, would it be alright to use the mass-luminosity relationship with different values of p(3.5 for a main sequence star) without getting major errors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what value of p should i use for other stars like white and brown dwarfs?&lt;/p&gt;&#xA;" OwnerUserId="7580" LastActivityDate="2015-07-22T12:54:00.610" Title="Calculating the mass of star" Tags="&lt;star&gt;&lt;astrophysics&gt;&lt;mass&gt;&lt;data-analysis&gt;&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11343" PostTypeId="2" ParentId="11314" CreationDate="2015-07-22T11:26:54.800" Score="3" Body="&lt;p&gt;Further imagery has confirmed billion year old surface with impact craters on one part and younger which is very recent and smooth on another part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jeff Moore, of the New Horizons team has said. “We now know Pluto has some surfaces that are a billion years old,” &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit... new estimate quote from NASA:&#xA;Based on the lack of impact craters, scientists suspect the surface of Pluto is less than a few hundred million years old, &#xA;Read more at &lt;a href=&quot;https://web.archive.org/web/20150807092534/http://newsdaily.com/2015/07/nasa-spacecraft-shows-pluto-wrapped-in-haze-ice-flows/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://newsdaily.com/2015/07/nasa-spacecraft-shows-pluto-wrapped-in-haze-ice-flows/#ytDT3emJZSKRbL7b.99&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He expects that the bulk of the data that he wants to study wont have been beamed back before early 2016.&#xA;Large scale image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/u6bGY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/u6bGY.png&quot; alt=&quot;http://pluto.jhuapl.edu/images/featuredImages/nh-pluto-mountain-range.png&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/6GZb8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6GZb8.jpg&quot; alt=&quot;boundary of old and new, old dateable by number of craters&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/WH6AI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WH6AI.jpg&quot; alt=&quot;recent zone similar to a convection surface&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="7472" LastEditDate="2020-05-19T15:09:42.273" LastActivityDate="2020-05-19T15:09:42.273" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="11344" PostTypeId="1" CreationDate="2015-07-22T11:35:25.780" Score="10" ViewCount="278" Body="&lt;p&gt;The images of New Horizons from Pluto will take more a year to arrive from space, for the moment with have some compressed images as a preview. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nasa has advanced some information regarding the geologically inactive zone, but there has been no information regarding the possible geologically active zones, on the formation of the plateaus and the mountains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What forces could be driving geothermal activity of what nature on Pluto? What can be learnt from theories of the moons of Jupiter and Saturn to understand activity inside of Pluto? What is the theoretical temperature of the center of the dwarf planet? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uKYLz.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uKYLz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/FfZWF.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FfZWF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2015-07-22T17:36:53.917" LastActivityDate="2015-07-25T17:23:35.350" Title="Pluto- Current theories regarding its geological nature?" Tags="&lt;planet&gt;&lt;pluto&gt;&lt;dwarf-planets&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11345" PostTypeId="2" ParentId="11342" CreationDate="2015-07-22T11:46:12.550" Score="1" Body="&lt;p&gt;You don't say what other information you have for the &quot;several stars&quot;. Yes, you can use a mass-luminosity relationship if the stars are on the main sequence. In terms of mass uncertainties I would estimate that you might be at the level of 20% unless you can absolutely pinpoint them on a Hertzsprung-Russell diagram, because the luminosity of a (fixed mass) main sequence star changes, even whilst it is on the main sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the stars are in any other phase of their evolution then your problem is much harder, even if you can place them on the HR diagram. Not least because there is considerable disagreement between different evolutionary models as regards the positions of stars of a given mass at a given age, on both the pre- and post-main sequence. Furthermore, their luminosity and temperature evolution can be influenced by factors such as rotation, metallicity, magnetic fields and mass-loss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compact stellar remnants and brown dwarfs do not follow a mass-luminosity relation. In general you can only estimate their masses if they are in binary systems, or for white dwarfs, you can estimate their radii and either their surface gravity or gravitational redshift. Estimating a brown dwarf's mass needs both its luminosity and age.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-07-22T12:54:00.610" LastActivityDate="2015-07-22T12:54:00.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11346" PostTypeId="2" ParentId="11342" CreationDate="2015-07-22T11:47:49.400" Score="1" Body="&lt;p&gt;Using a mass-luminosity relation will give you a decent estimate on the mass (probably within a magnitude), but as usual, the more data you have on the stars, the better your estimate will be. There are a lot of sources of error (&lt;a href=&quot;http://arxiv.org/abs/1205.5484&quot; rel=&quot;nofollow&quot;&gt;for example, age and metallicity&lt;/a&gt;), and whatever errors on your luminosity measurement will be propagated to your mass measurement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A compilation of various relations like this is given in &lt;a href=&quot;http://adslabs.org/adsabs/abs/2000MNRAS.315..543H/&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;, for various evolution stages. If you don't know various parameters of your stars (like age), it might be good to try out a variety of possible values for these models, and see how it impacts your estimates.&lt;/p&gt;&#xA;" OwnerUserId="608" LastActivityDate="2015-07-22T11:47:49.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11347" PostTypeId="1" AcceptedAnswerId="11350" CreationDate="2015-07-22T19:56:49.730" Score="4" ViewCount="125" Body="&lt;p&gt;Is there a possibility that I will find records of stars with masses around 25-30 solar masses within 30 parsecs of solar system. If not, how far do I need to look?&lt;/p&gt;&#xA;" OwnerUserId="7580" LastActivityDate="2015-07-23T04:08:24.733" Title="Massive Nearby Stars" Tags="&lt;star&gt;&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11348" PostTypeId="1" CreationDate="2015-07-22T21:03:50.620" Score="9" ViewCount="171" Body="&lt;p&gt;I want to calculate the distance between two quasars of which I know the angular position and the red shift. Let $Q_1=(\alpha_1,\delta_1, z_1)$ and $Q_2=(\alpha 2,\delta 2, z_2)$ and suppose $z_2 &amp;gt; z_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to find the angular separation $\theta$ between them by means of the angular coordinates. But how to find the comoving distance (at epoch $z=0$) from them ? I know how to find the comoving distance from the Earth of the two quasars, can we find the distance between them using these two distances?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A related question is to find the redshift of $Q_2$ as seen by $Q_1$ at the epoch when $Q_1$ received the light emitted from $Q_2$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is some standard method to solve this problem ?&lt;/p&gt;&#xA;" OwnerUserId="7741" LastActivityDate="2016-04-27T18:03:47.627" Title="How to find the distance between two quasars" Tags="&lt;cosmology&gt;&lt;distances&gt;&lt;quasars&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11349" PostTypeId="2" ParentId="10229" CreationDate="2015-07-22T22:04:21.237" Score="2" Body="&lt;p&gt;Simple answer: the material is just too spread out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forming a planetary-size object takes LOTS of collisions to build up enough matter to start gravitationally attracting the surrounding stuff.  The Oort cloud is very far from the sun (starting at ~1000 AU, compared to a maximum of ~50 AU for Pluto), meaning that the icy fragments are moving very slow in their orbits.  Furthermore, the material is spread out over a huge volume, especially since the Oort cloud is a spherical shell (the matter is spread in 3 dimensions), as opposed to the 2-d disk that makes up the vast majority of matter in the Solar system (Sun, Planets, Asteroid &amp;amp; Kuiper Belts).  The timescale for these objects to accrete to any graviationally significant size is far beyond the lifetime of the Solar system (or even the universe).&lt;/p&gt;&#xA;" OwnerUserId="2259" LastActivityDate="2015-07-22T22:04:21.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11350" PostTypeId="2" ParentId="11347" CreationDate="2015-07-22T22:12:20.347" Score="7" Body="&lt;p&gt;There aren't any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 25-30 $M_{\odot}$ main sequence star would have a spectral type of $\sim$ O7 and an absolute $V$ magnitude $M_V \simeq -5$ (see &lt;a href=&quot;http://ads.harvard.edu/cgi-bin/bbrowse?book=hsaa&amp;amp;page=72&quot;&gt;Zombeck 1982&lt;/a&gt;). A giant star with this mass would be even more luminous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a distance of 30pc, the apparent magnitude of such a star would be $V=-2.6$. Closer examples would of course be brighter. Sirius is the brightest star in the sky with $V=-1.46$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nearest O-star may be zeta Ophiuchus which is at about 450 light years (or about 140 parsecs). This might be a bit lower than $25M_{\odot}$.  Betelgeuse, a red supergiant at a Hipparcos-measured distance of $152\pm 20$ pc, probably has  a mass of about $15 M_{\odot}$, but I think this is quite uncertain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The secondary of $\gamma^{2}$ Velorum has an O7.5V spectral type and reasonably well-determined mass of about $30M_{\odot}$ (&lt;a href=&quot;http://arxiv.org/abs/0909.0504&quot;&gt;Eldridge 2009&lt;/a&gt;) and a distance of $336 \pm 8$ pc (&lt;a href=&quot;http://arxiv.org/abs/astro-ph/0702375&quot;&gt;North et al. 2007&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-07-22T22:29:41.583" LastActivityDate="2015-07-22T22:29:41.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11351" PostTypeId="1" AcceptedAnswerId="11353" CreationDate="2015-07-23T01:36:09.653" Score="4" ViewCount="251" Body="&lt;p&gt;Complete newbie here. I have an opportunity to visit one of the few designated dark sky sites here in Pennsylvania, &lt;a href=&quot;http://www.dcnr.state.pa.us/stateparks/findapark/cherrysprings/&quot; rel=&quot;nofollow&quot;&gt;Cherry Springs State Park.&lt;/a&gt; I plan on camping out over night, but I have never gone stargazing before, don't use a telescope, and I don't know what the protocols are as far as binoculars? glasses? Etc. for viewing the night skies. The park allows no lights, save for flashlights covered with red tint. Is there anything else that you more experienced astronomers would suggst that I &lt;em&gt;need&lt;/em&gt; to have with me? What should I bring? What should I expect for my first observation tour ever? Thank you.&lt;/p&gt;&#xA;" OwnerUserId="7738" LastActivityDate="2015-07-23T04:40:26.740" Title="First time visit to dark sky stargazing site: what do I need?" Tags="&lt;amateur-observing&gt;&lt;star-gazing&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11352" PostTypeId="2" ParentId="11347" CreationDate="2015-07-23T04:08:24.733" Score="1" Body="&lt;p&gt;Rob Jeffries nailed it, but I'll add a few points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;25-30 stellar mass stars are quite rare.   &lt;a href=&quot;https://en.wikipedia.org/wiki/O-type_star&quot; rel=&quot;nofollow&quot;&gt;O-type stars&lt;/a&gt; begin at a mass of about 16 suns and they're about 0.00003% of the main sequence  (that's 1 in 3 million).   They're also short lived, a few million to maybe 10 million years in main sequence.   That's part of the reason there are so few of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at what stars are near us - lets just go 20 light years, there's about 150 stars and star like objects - source:  &lt;a href=&quot;http://www.solstation.com/stars/s20ly.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.solstation.com/stars/s20ly.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each time you double the distance, you have 8 times the volume, so within 100 light years (and this is a very rough estimate), 5 to the 3rd power * 150, roughly 20,000 stars within 100 light years of earth.   So if a star of the size you're looking for is 1 in 3 million (or less common than that at 25-30 solar masses), and there's 20,000 stars within 100 light-years, it's not surprising that there's none that close to earth.   That doesn't mean there's never been one that close, cause star orbits in the Milky way move around in relation to each other quite a bit over time, but mostly stars of that size don't get that close cause they're very rare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Loosely related footnote, but &lt;a href=&quot;https://en.wikipedia.org/wiki/SN_1054&quot; rel=&quot;nofollow&quot;&gt;SN 1054&lt;/a&gt;, which is now the Crab Nebula was so bright when it went super nova that it could be seen during the day (for about 23 days) and that's 6,500 light years away.  We can see stars that large well over 100 light years.   &lt;a href=&quot;https://en.wikipedia.org/wiki/Betelgeuse&quot; rel=&quot;nofollow&quot;&gt;Betelgeuse&lt;/a&gt; is 640 light years away and it's the 8th brightest star in the sky.   They don't have to be that close to be impressive.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-07-23T04:08:24.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11353" PostTypeId="2" ParentId="11351" CreationDate="2015-07-23T04:40:26.740" Score="6" Body="&lt;p&gt;If you have yet to plan your adventure and the moon is not you main target, be sure to set your date nearest a &quot;New Moon&quot;. During this phase the moon will not be visible at night, reducing the light pollution that much more. This will increase the amount of stars you should be able to see. Peak of any meteor shower event would be a bonus aswell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Red light is also important as you suggested, but keep in mind a cellphone is just as bad as a flashlight. There are free apps available for many devices to reduce your brightness levels even further. Your eyes can take roughly 20 to 30 minutes to adjust to the darkness after exposure to light. This will significantly impact your stargazing experience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wether its your first or hundredth time, naked eye observation is always an enjoyable experience. Being that you shall be going to a region with rather dark skies, you should see a suprisingly bright milky way. There should be so many stars in the sky, one could get &quot;lost&quot; looking for often easily  recognized constellations. You may even see meteors, satellites or even the ISS. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, binoculars are not neccesary but are a great tool for the avid or begginer gazer. With a steady hand and a decent pair of binoculars it is possible to see the rings on saturn and moons of jupiter. Many galaxies and starclusters can also be enhanced with the aid of the binoculars. Be sure to lean on something though or it will be way to shaky to see clearly. If you have the space and a half decent pair, I would bring them. If not, no worries the pea soup of stars in the sky should keep you busy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many free resources online that can help you find what your looking for if you have any celestial targets in mind. If you have a mobile device, an astronomy/planetarium app is a very useful tool for any stargazer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your first time, the most important thing I can suggest to bring is a comfortable reclining chair or blanket and an open mind. There are so many amazing things to learn about the universe and there is no better place to do it than under the stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also be sure not to forget the bug repellent. Can't see much of the stars if your hiding in your tent. Best of luck and hope you get great weather.&lt;/p&gt;&#xA;" OwnerUserId="7745" LastActivityDate="2015-07-23T04:40:26.740" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11354" PostTypeId="1" AcceptedAnswerId="11359" CreationDate="2015-07-23T06:32:31.220" Score="4" ViewCount="1319" Body="&lt;p&gt;I realise that night time on the Moon is two weeks, however I am curious if during the night there, whether or not the Earth reflects light in similar manner to the Moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the one hand, I know it must (as seen in photographs from astronauts), but what I'm really wanting to know is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the Earth have phases like the Moon and do they work in a similar manner for an observer on the Moon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume the Earth isn't as reflective as the Moon since the Moon is much lighter in colour, however the Earth is also fairly larger, does this make any difference in the amount of light reflected?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if the Earth does appear in phases, does that mean that the two week long nights on the Moon are sometimes pitch black (new Earth) and sometimes less than pitch black (full Earth)?&lt;/p&gt;&#xA;" OwnerUserId="1768" LastActivityDate="2016-05-01T17:22:45.610" Title="During night on the Moon is there Earth light and Earth phases?" Tags="&lt;the-moon&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11355" PostTypeId="1" CreationDate="2015-07-23T07:07:15.847" Score="3" ViewCount="58" Body="&lt;p&gt;I would like to calculate how the RA/Dec changes when an object has an apparent linear motionin the equatorial sphere. For example, more or less what an artificial satellite would do in a circular orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, what I'm looking for is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the RA/dec of the object at two times, &lt;em&gt;t1&lt;/em&gt;, and &lt;em&gt;t2&lt;/em&gt;. Assuming it has a linear motion at constant speed, what RA/dec will it have a &lt;em&gt;t3&lt;/em&gt;? So it will have an angular separation proportional to the one it had between &lt;em&gt;t1&lt;/em&gt; and &lt;em&gt;t2&lt;/em&gt;, and the three points will form a straight line.&lt;/p&gt;&#xA;" OwnerUserId="4048" LastActivityDate="2015-07-23T07:07:15.847" Title="Linear motion in the equatorial frame" Tags="&lt;orbit&gt;&lt;observational-astronomy&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11356" PostTypeId="2" ParentId="10710" CreationDate="2015-07-23T09:23:51.647" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Those sites where one views a light transit curve or tries to classify a galaxy type are not very interesting to me. I feel I could investigate things more in depth than that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is a lot more to many citizen science projects than that! I'd recommend having a look through all the projects available as there may be one in an area you are more interested in : &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_citizen_science_projects&quot; rel=&quot;nofollow&quot;&gt;Wikipedia: list of citizen science projects&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.zooniverse.org/#/&quot; rel=&quot;nofollow&quot;&gt;Zooniverse&lt;/a&gt; is a great collection of Astronomy related projects. They are not just toys or made for school children, they are real science projects where you will be contributing to important and current research. Members who show they have advanced skills often get given more credit in the results weighting algorithms, or upgraded to more difficult tasks. People have even spotted things professional scientists had missed, and been able to bring new things to the attention of the scientific community and be credited on published papers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amateur observational astronomers are also useful members of the scientific community. Some beautiful and intriguing images of object have been taken by skilled amateurs, who don't have the pressures of short observing times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you focused on research or on generally aiding science?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As someone with a lifetime of experience, you would be invaluable as a volunteer at outreach and engagement activities. You could ask your local University or even city council if they are running any science outreach projects you could collaborate with. Again, this does not just mean school children, although that can be a part of it. Lobbying for science literacy in politics, organising science fairs, giving talks to interested groups, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will try to think of more research related examples later today. It would be helpful to know what country you are based in, as my experience may be very different to yours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later edit: If you are interested in exoplanets but want more autonomy of research than the citizen science projects afford, you may be interested to know that &lt;a href=&quot;http://exoplanetarchive.ipac.caltech.edu/docs/intro.html&quot; rel=&quot;nofollow&quot;&gt;NASAs Exoplanet Archive&lt;/a&gt; is a publicly available archive of data from exoplanet missions along with the tools to work with the data. Happy planet hunting!&lt;/p&gt;&#xA;" OwnerUserId="2270" LastEditorUserId="2270" LastEditDate="2015-07-23T16:23:11.080" LastActivityDate="2015-07-23T16:23:11.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11358" PostTypeId="2" ParentId="11354" CreationDate="2015-07-23T10:57:02.323" Score="3" Body="&lt;p&gt;Yes, there are Earth phases, viewing from the Moon.  Full earths, half earths, quarter earths, waning and waxing earths.   The easiest way to visualize this is, imagine the earth is still, one half of the Earth facing the sun, the other half away from the sun, so you have half the Earth is light, half is dark, now, imagine you're on the moon orbiting the Earth every 28 days.    When you're over the sunny half of the Earth (night on your part of the Moon) the Earth is full.   When you're over the dark side of the Earth (day on your part of the Moon), the Earth is new.   As the moon takes 28 days to orbit the earth, like the moon in our sky, every 28 days would complete one cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's different is the Earth wouldn't move in the night sky.  It would actually go back and forth a bit, but it would stay in the same general area, every day, every year, every century, because the Moon is tidally locked to the Earth, but apart from not moving, it would be similar to the Lunar cycles.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another difference is that you could observe the Earth's rotation.  Here's a pretty good video on what it would look like.  28 days squeezed into about 1 minute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=-HgHEO0DUig&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=-HgHEO0DUig&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would imagine the Earth looks quite bright from the point of view of the Moon, and I'd guess the pictures don't really do it justice, but that's just a guess.   I've never seen it for myself.    I'm also not sure it would be pitch black and not visible as a &quot;New Earth&quot; either.   I remember reading that you can see stars from the moon even during the day, that's because there's no atmosphere to diffract the light so you could probably see the Earth even at new earth too.   We can see the new moon from Earth sometimes, so I would think a &quot;new earth&quot; would be visible but dark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a discussion on being able to see the new moon.   I would think, seeing a new earth from the moon would be even easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short discussion:  &lt;a href=&quot;http://scienceline.ucsb.edu/getkey.php?key=26&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://scienceline.ucsb.edu/getkey.php?key=26&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long discussion: &lt;a href=&quot;https://physics.stackexchange.com/questions/1907/why-can-we-see-the-new-moon-at-night&quot;&gt;https://physics.stackexchange.com/questions/1907/why-can-we-see-the-new-moon-at-night&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-07-23T10:57:02.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11359" PostTypeId="2" ParentId="11354" CreationDate="2015-07-23T11:00:26.777" Score="6" Body="&lt;p&gt;From the perspective of the Moon, the Earth exhibits phases just as does the Moon when viewed from the perspective of the Earth. There are however some differences. One difference is that the phase of the Earth at a fixed local lunar time will be pretty much the same from one day to the next. This is a result of the fact that the Moon is tidally locked. Another difference is that earthlight is considerably brighter than moonlight. This is a consequence of the Earth's significantly larger albedo and significantly larger diameter compared to the Moon. A third difference is that except for locations close to ±90° lunar longitude, the Earth doesn't rise and set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I assume the Earth isn't as reflective as the Moon since the Moon is much lighter in colour, however the Earth is also fairly larger, does this make any difference in the amount of light reflected?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You have this backwards. The Moon is rather dark. It's reflectivity is about that of old asphalt. The Earth is significantly more reflective than is the Moon.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Finally, if the Earth does appear in phases, does that mean that the two week long nights on the Moon are sometimes pitch black (new Earth) and sometimes less than pitch black (full Earth)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. This is the first difference between Moon phases and Earth phases. Earth phase on the Moon is a function of lunar longitude and time of lunar day. You won't see a full Earth one night and a new Earth some other night.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-07-23T11:00:26.777" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11360" PostTypeId="1" CreationDate="2015-07-23T11:52:31.333" Score="8" ViewCount="3078" Body="&lt;p&gt;Pluto has a high orbital inclination compared to the planets in our solar system.  What has caused Pluto to have such a different orbit -- was it always like this, or did something happen to make it so?&lt;/p&gt;&#xA;" OwnerUserId="7750" LastActivityDate="2019-07-30T08:02:17.347" Title="why does pluto have a high orbital inclination" Tags="&lt;pluto&gt;&lt;inclination&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11361" PostTypeId="2" ParentId="11327" CreationDate="2015-07-23T15:34:39.333" Score="5" Body="&lt;p&gt;The answer, for the universe we live in, is almost certainly No. The best guess is that the universe, on the largest scale, is very nearly flat and probably extends infinitely in all directions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physics of general relativity does permit a hypothetical universe that curves back on itself in the way you describe, and if it were small enough, we could indeed see ourselves. But I don't know whether stars and planets could evolve in such a universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, it doesn't have to even have curvature to wrap around as you describe. The universe could be like the game Asteroids, where you wrap off one edge and come back on the other. That has the topology of a torus, but is geometrically flat -- for instance, the angles inside a triangle still add up to 180 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it doesn't look like our universe is like that.&lt;/p&gt;&#xA;" OwnerUserId="7752" LastActivityDate="2015-07-23T15:34:39.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11362" PostTypeId="2" ParentId="10710" CreationDate="2015-07-23T17:23:56.780" Score="0" Body="&lt;p&gt;If you have a telescope and use it often, there are things that you could definitely contribute. Discovering new comets is still something that amateurs do well. Tracking variable stars is also worthwhile.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2015-07-23T17:23:56.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11363" PostTypeId="2" ParentId="11360" CreationDate="2015-07-23T17:34:44.257" Score="2" Body="&lt;p&gt;I think the best answer is that there is no particular reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any planetary system, most orbits tend to stay close to a common plane - close, but not exactly there. Even the big planets are slightly out-of-plane with each other, but the differences are tiny.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you move away from the bulk of the planetary system's mass, smaller bodies tend to be more random. They are more easily deviated from encounters and near misses. Their origins might be different from the rest. If they get close to the outskirts of the system, that's a more unruly place, and orbit-changing events are more likely - especially for small bodies. Finally, the whole distribution of orbital planes has a bit of a stochastic (random) component anyway, and there will always be outliers that are quite different from the bulk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory, even some larger planets could be significantly &quot;out of plane&quot; (much more than they are right now), but that's unlikely and it tends to be smoothed out over large periods of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, Pluto is different because it's smaller and it's not really a planet, but don't attach too much significance to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to install &lt;a href=&quot;http://universesandbox.com/&quot; rel=&quot;nofollow&quot;&gt;Universe Sandbox&lt;/a&gt; and play with some of the included scenarios. Big systems bound by gravity can be pretty complex, and completely orderly configurations are rare.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2015-07-23T17:34:44.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11364" PostTypeId="2" ParentId="11354" CreationDate="2015-07-24T05:34:48.180" Score="3" Body="&lt;p&gt;The moon only seems light in color against the backdrop of the void. It's actually fairly dark. Moon's albedo (scientific term for how reflective something is) is .12 and Earth's .3. So earth is two and a half times more reflective. Moreover the moon subtends about half a degree as seen from earth while the earth subtends nearly two degrees as seen from the moon. A full earth is about 34 times as bright as a full moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the point of view of someone on the moon, the earth seems to stay in one part of the sky. This is because the moon is tidelocked to the earth. That is it always presents the same face to us. But since the moon's orbit isn't perfectly circular and it's equator is tilted with regard to it's orbital plane, the earth moves a little.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we're standing in the lunar crater Mösting A. This is the region of the moon closest to the earth. From this vantage point the earth is always straight up. At high noon the sun seems to be near the earth and we see either a thin crescent earth or a new earth. At midnight the sun's moved 180º and we're seeing a bright, full earth. So in this location earth is brightest at midnight and darkest at noon. This area always enjoys either earthlight and/or sunlight except doing an eclipse&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we're standing in a region near what we see as the edge of the moon, the earth is low in the sky near the horizon. Again, the earth is a thin crescent or a new earth when the sun is close to the earth in the sky. So there can be a new earth near the horizon while the sun is still behind the horizon (i.e. just before sunrise or just after sunset). In this case you'd have neither earthlight nor sunlight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the near side enjoys earthlight all or nearly all of the night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the far side of the moon, you never see the earth. The nights are always dark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the &lt;a href=&quot;http://hopsblog-hop.blogspot.com/2013/01/the-dark-side-of-moon.html&quot; rel=&quot;nofollow&quot;&gt;Pink Floyd lyrics make some sense after all&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1232" LastEditorUserId="1232" LastEditDate="2015-07-25T04:52:03.100" LastActivityDate="2015-07-25T04:52:03.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11365" PostTypeId="1" AcceptedAnswerId="11366" CreationDate="2015-07-24T10:55:55.393" Score="8" ViewCount="1343" Body="&lt;p&gt;We have been looking for planets around other stars for a while (see related &lt;a href=&quot;https://space.stackexchange.com/questions/1484&quot;&gt;https://space.stackexchange.com/questions/1484&lt;/a&gt; ) and we are finding lots of planets, some of them are &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler-452b&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth like&lt;/a&gt; According to an the answer at &lt;a href=&quot;https://space.stackexchange.com/questions/4062&quot;&gt;https://space.stackexchange.com/questions/4062&lt;/a&gt; there are over 14,000 stars within 100 light years of Earth, the answer goes on to say there are &quot;1,500 potentially habitable planets within 100 light-years&quot; of Earth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on current trends it looks like there are lots of stars with lots of planets.  I am beginning to assume that our solar system is not unique and that  every star has several planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any Stars that we know are without any planets?&lt;/p&gt;&#xA;" OwnerUserId="513" OwnerDisplayName="James Jenkins" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2015-07-24T13:57:56.740" Title="Are there any Stars we know don't have planets?" Tags="&lt;exoplanet&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11366" PostTypeId="2" ParentId="11365" CreationDate="2015-07-24T11:53:52.857" Score="14" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am beginning to assume that our solar system is not unique and that every star has several planets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not quite, but indeed a &lt;a href=&quot;http://www.nature.com/nature/journal/v481/n7380/full/nature10684.html&quot;&gt;study published in Nature in 2012&lt;/a&gt; found that, based on our observations so far, roughly 17% of stars host Jupiter-mass planets, 52% host &quot;Cool Neptunes&quot; and 62% host Super-Earths. (Note that these percentages do not add up to 100%, because they are not mutually exclusive possibilities). This was particularly surprising, because half of all visible stars are believe to be in &lt;a href=&quot;https://en.wikipedia.org/wiki/Binary_star&quot;&gt;binary systems&lt;/a&gt;, which would make planetary systems very unstable, but some binary systems have been &lt;a href=&quot;http://physicsworld.com/cws/article/news/2014/oct/31/planet-forming-stream-found-in-binary-star-system&quot;&gt;found to have planets&lt;/a&gt; too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So indeed it seems &lt;em&gt;the majority&lt;/em&gt; of stars have planets, but it's very unlikely that all of them do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the exact answer to your question &quot;do we &lt;strong&gt;know&lt;/strong&gt; of any stars with no planets&quot; has got to be &quot;no&quot;, because there remains a possibility that they have planets that we simply haven't been able to detect, because of &lt;a href=&quot;http://blog.thingswedontknow.com/2013/03/detection-and-discovery-of-exoplanets.html&quot;&gt;limitations in our techniques&lt;/a&gt; to detect them.&lt;/p&gt;&#xA;" OwnerUserId="7759" OwnerDisplayName="Eddy" LastActivityDate="2015-07-24T13:54:05.403" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11367" PostTypeId="1" AcceptedAnswerId="11370" CreationDate="2015-07-24T14:38:30.097" Score="4" ViewCount="484" Body="&lt;p&gt;I've read once in &lt;a href=&quot;https://en.wikipedia.org/wiki/Intracluster_medium&quot; rel=&quot;nofollow noreferrer&quot;&gt;an article&lt;/a&gt; that the temperature of the medium within galaxy clusters (the intracluster medium or ICM) is extremely high. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What does that temperature really mean? Particle velocity? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why is that so?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is that medium the &quot;Void&quot;? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is it in conflict with the CMB temperature (that's about 3 K)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7761" LastEditorUserId="14094" LastEditDate="2019-07-04T10:05:52.860" LastActivityDate="2019-07-04T10:20:26.547" Title="Temperature of the intracluster medium (ICM)" Tags="&lt;temperature&gt;&lt;cmb&gt;&lt;galaxy-cluster&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="11368" PostTypeId="2" ParentId="11367" CreationDate="2015-07-24T20:29:47.487" Score="5" Body="&lt;p&gt;The answer to your first question is &quot;Yes, the temperature referred to is the 'normal' temperature, reflecting the average kinetic energy of the gas particles&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to your second question is a bit more complex:&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Cooling function&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Gas cools by various processes, with an efficiency depending on the temperature, the density, and the composition of the gas. At &quot;low&quot; temperatures, most cooling happens because particles collide and their kinetic energy is used to excite or even ionize atoms. The atoms subsequently recombine or de-excite, emitting radiation which can carry energy away from the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the gas in the Universe is hydrogen, which cools most efficiently around &lt;span class=&quot;math-container&quot;&gt;$10^4\,\mathrm{K}$&lt;/span&gt;. At higher temperatures, helium becomes the major contributor. Additionally, various metals contribute at various temperatures depending on their density and ionization state. At very high temperatures (&lt;span class=&quot;math-container&quot;&gt;$\gtrsim10^7\,\mathrm{K}$&lt;/span&gt;) where the gas is highly ionized, cooling mainly happens through bremsstrahlung, i.e. deceleration of charged particles by other charged particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure below (from &lt;a href=&quot;http://staff.ustc.edu.cn/~wzhao7/c_index_files/main.files/HJMo.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mo, Bosch &amp;amp; White 2010&lt;/a&gt;) shows the cooling function for various metallicities (with my own annotations):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/m1c85.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/m1c85.png&quot; alt=&quot;Cooling function&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Collapse of gasous halos&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;So what makes the intracluster medium (ICM) different from the interstellar medium (ISM)? In the expanding Universe, overdensities try to contract. These overdensities can reach hydrostatic equilibrium only if radiative cooling is small. If the cooling timescale &lt;span class=&quot;math-container&quot;&gt;$t_\mathrm{cool}$&lt;/span&gt; is much smaller than the free-fall timescale &lt;span class=&quot;math-container&quot;&gt;$t_\mathrm{ff}$&lt;/span&gt;, it can collapse and form stars. From the virial theorem, which gives the relation between potential and kinetic energy of the system, you can calculate the relation between the two timescales. It's a bit too much math for this post, but not difficult. I recommend reading chapter 8.4 in Mo, Bosch &amp;amp; White (2010). Their Fig. 8.6 shows a cooling diagram with the locus of &lt;span class=&quot;math-container&quot;&gt;$t_\mathrm{cool} = t_\mathrm{ff}$&lt;/span&gt; in the density-temperature plane:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qt8Jq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qt8Jq.png&quot; alt=&quot;Cooling diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above the locus, cooling is effective, and the cloud can collapse. The tilted dashed lines are lines of constant gas mass. Now you see that even for solar metallicity (the &lt;span class=&quot;math-container&quot;&gt;$Z = Z_\odot$&lt;/span&gt; line), clouds of gas masses larger that &lt;span class=&quot;math-container&quot;&gt;$\sim10^{13}\,M_\odot$&lt;/span&gt; are unable to cool. &lt;strong&gt;This is the fundamental difference between a galaxy and a cluster. They are both overdensities that resisted the expansion, but the gas in the cluster is not able to cool and form stars&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Voids&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;When people talk about &quot;voids&quot;, they usually refer &lt;em&gt;not&lt;/em&gt; to the ICM — which is the hot and dilute gas between the galaxies of a cluster — but to the immense regions of even more dilute gas which are… well, void of galaxies. Take a look at this image (from &lt;a href=&quot;http://ganymede.nmsu.edu/tharriso/ast110/class25.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;). Here, I marked clusters with green and voids with purple. I also marked some filaments with cyan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/t0TnP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t0TnP.jpg&quot; alt=&quot;Cosmological simulation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The cosmic microwave background&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;As for your last question, the ICM doesn't interact much with the CMB, so they're not in thermodynamic equilibrium, and there's no conflict. A small fraction (&amp;lt;10%) of the CMB photons &lt;em&gt;does&lt;/em&gt; interact. This doesn't change the state of the gas, but it tends to polarize a fraction of the CMB, and this we can use to study the ionization history of the Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I forgot about the &lt;em&gt;Sunyaev-Zel'dovich effect&lt;/em&gt;, which is the interaction of hot electrons with CMB photons. Rob Jeffries discusses this in &lt;a href=&quot;https://astronomy.stackexchange.com/a/11370/14094&quot;&gt;his answer&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="14094" LastEditDate="2019-07-04T10:20:26.547" LastActivityDate="2019-07-04T10:20:26.547" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11369" PostTypeId="1" AcceptedAnswerId="11371" CreationDate="2015-07-24T21:13:39.637" Score="0" ViewCount="175" Body="&lt;p&gt;Let's say that Person A is on a planet orbiting a black hole (like the one from Interstellar) and time is slowed for Person A. Person B is back home on Earth. Now let's say a supernova happens and it's possible for both parties to see it happen at the same time (ignoring the time it would take for the light to get there). Would Person A see the supernova at the same time as Person B? And if so, if Person A is in 1,000 B.C. and Person B is in 2,000 A.D, would the supernova then happen at two different times at the same time?&lt;/p&gt;&#xA;" OwnerUserId="7764" LastEditorUserId="1435" LastEditDate="2018-12-10T07:40:07.487" LastActivityDate="2018-12-10T07:40:07.487" Title="If you're near a black hole and your time is slowed down, would a supernova be observable to both you and someone outside of the blackhole's pull?" Tags="&lt;space&gt;&lt;supernova&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="11370" PostTypeId="2" ParentId="11367" CreationDate="2015-07-24T21:26:05.043" Score="3" Body="&lt;p&gt;I second Pela's answer, apart from one addition. The CMB and the intracluster gas do interact with each other via the &lt;a href=&quot;https://en.wikipedia.org/wiki/Compton_scattering#Inverse_Compton_scattering&quot; rel=&quot;nofollow&quot;&gt;inverse Compton effect&lt;/a&gt;, whereby electrons in the gas cool by giving their energy to CMB photons, increasing their energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process, when applied to hot gas in the intracluster medium, is also known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Sunyaev%E2%80%93Zel%27dovich_effect&quot; rel=&quot;nofollow&quot;&gt;Sunyaev-Zel'dovich&lt;/a&gt; effect. This causes the brightness of the CMB to be different when looking towards a galaxy cluster. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The importance of this effect can be found in these notes by &lt;a href=&quot;http://ned.ipac.caltech.edu/level5/Birkinshaw/Birk4_1.html#4.1&quot; rel=&quot;nofollow&quot;&gt;Mark Birkinshaw&lt;/a&gt;. Essentially, the S-Z effect when combined with X-ray images of the hot gas can provide vital information on the density structure and thermal structure of the hot gas. It also turns out that the comparison of the X-ray and S-Z properties of a cluster can yield a value for the Hubble parameter and provide other cosmological constraints (for example, see Planck results presented in &lt;a href=&quot;http://arxiv.org/abs/1502.01597&quot; rel=&quot;nofollow&quot;&gt;Ade et al. 2015&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-07-24T21:26:05.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11371" PostTypeId="2" ParentId="11369" CreationDate="2015-07-24T22:08:08.133" Score="1" Body="&lt;p&gt;Time is relative therefore this is a bit of a nonsense question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But yes, if the gravitational effect near the black hole is strong enough (depends on how close you are) it would affect 'when' the observer would notice it.  This is because the light must travel a 'further' distance through the spacetime to get to you since gravity is distorting it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may find this interesting: &lt;a href=&quot;https://web.archive.org/web/20130424041627/http://www.thebigview.com/spacetime/spacetime.html&quot; rel=&quot;nofollow&quot;&gt;https://web.archive.org/web/20130424041627/http://www.thebigview.com/spacetime/spacetime.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7765" LastActivityDate="2015-07-24T22:08:08.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11373" PostTypeId="1" CreationDate="2015-07-25T01:26:29.733" Score="-2" ViewCount="106" Body="&lt;p&gt;Large is bigger than Earth. Assuming that such planets exist (though I guess if such planets don't exist that would explain much of the Fermi paradox).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe some amount of iron is needed though, to make a magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess you could also have a planet like that with very good climate and gravity for Earthlings but toxic air and a lot more carbon or ammonia or methane or hydrogen or formaldehyde or acetone or alcohol or hydrocarbons than here. It wouldn't be Earthlike but it'd look like a lush land where the rocks are all graphite or diamond and the sky is brown or something. That planet could be like the Garden of Eden for a sentient that's mostly acetone and would get drunk from an ounce of H2O (or whatever chemicals are appropriate).&lt;/p&gt;&#xA;" OwnerUserId="7767" LastActivityDate="2015-07-25T22:38:31.337" Title="What would Earthlike planets bigger than Earth with &lt;10 m/s^2 gravity and complex life be like?" Tags="&lt;gravity&gt;&lt;exoplanet&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-08-05T16:15:25.887" ContentLicense="CC BY-SA 3.0" />
  <row Id="11374" PostTypeId="1" AcceptedAnswerId="11378" CreationDate="2015-07-25T01:32:58.653" Score="3" ViewCount="80" Body="&lt;p&gt;Well, I am a post graduate studying mathematics. Current I am working on the mathematical theory of star(not an expert thought). When I am making some assumption on the equation of state (for a star or sun of course), I simply sum up the radiation pressure, idea gas pressure and degenerate pressure together. In some sense I believe this choice is an appropriate approximate to some model. But when someone ask me why, I am kind of stuck, even thought some of the lecture note posted online supports my guess(for example, &lt;a href=&quot;http://www.jb.man.ac.uk/~smao/starHtml/equationState.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.jb.man.ac.uk/~smao/starHtml/equationState.pdf&lt;/a&gt;). But I am still wondering, is there any standard text book introducing something like this? I know Chandrasekhar's book has already told something, but it looks like it did not really state that I can simply sum all the pressure together linearly. Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="7766" LastActivityDate="2015-07-26T19:39:56.800" Title="About the reference for equation of state" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11375" PostTypeId="1" CreationDate="2015-07-25T01:47:22.633" Score="7" ViewCount="1455" Body="&lt;p&gt;I just observed a satellite almost directly above me travel for a few minutes while constantly dimming and then getting brighter. It did this for about 3 minutes until it faded out. I also observed a satellite with similar behavior on July 4th. I was thinking the satellite might be rotating, but I do not know of any satellites that rotate. Then I thought it could be a piece of space junk, but I don't think they get big enough to be visible from the ground. Does anyone know why the brightness of some satellites change while others remain constant? I am in the Northwest of the US. (Northern hemisphere) &lt;/p&gt;&#xA;" OwnerUserId="7221" LastEditorUserId="1030" LastEditDate="2015-07-25T11:01:06.363" LastActivityDate="2018-08-16T18:12:01.433" Title="Why do some artificial satellites decrease and increase in brightness multiple times?" Tags="&lt;earth&gt;&lt;observational-astronomy&gt;&lt;light&gt;&lt;satellite&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11376" PostTypeId="1" CreationDate="2015-07-25T05:06:03.017" Score="1" ViewCount="19" Body="&lt;p&gt;How can exoplanet moons be detected? Are there current methods to differentiate exoplanets with moons and exoplanets without moons? Or, are only the combined masses of the exoplanet and its moons detectable?&lt;/p&gt;&#xA;" OwnerUserId="7768" LastActivityDate="2015-07-25T05:06:03.017" Title="How can exoplanet moons be detected?" Tags="&lt;exoplanet&gt;&lt;natural-satellites&gt;&lt;satellite&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2015-07-26T06:18:26.810" ContentLicense="CC BY-SA 3.0" />
  <row Id="11377" PostTypeId="2" ParentId="11375" CreationDate="2015-07-25T07:46:34.117" Score="4" Body="&lt;p&gt;I do not know for sure, but you are viewing the satellites by the reflection of sunlight. As the satellite moves across the sky, the Sun-satellite-you angle changes, therefore there is no reason to expect that the brightness should stay the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for why it should fade and brighten again - if you imagine the satellite to consist of a number of surfaces, these surfaces may essentially cast a beam of light onto the Earth, which move across the Earth's surface. The satellite will brighten as one of these beams crosses you. I suppose this could happen several times for the same satellite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extreme examples of this are the Iridium satellite that &quot;flare up&quot; to magnitude -8, before fading in a matter of seconds.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-07-25T07:46:34.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11378" PostTypeId="2" ParentId="11374" CreationDate="2015-07-25T07:55:57.777" Score="2" Body="&lt;p&gt;You cannot just add the ideal gas pressure and degeneracy pressure together for any particular species of particle. (Well, you can, but it would be a poor approximation to the actual pressure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trouble is that they are not separate things. In general, the electrons in the gas are partially degenerate and always obey Fermi-Dirac statistics. Unfortunately there is no analytic solution that gives you pressure as a function of temperature and density; you either have to use Tables of pre-calculated results, or there are analytic approximations that work well in certain regimes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clayton's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0226109534&quot; rel=&quot;nofollow&quot;&gt;Principles of Stellar Evolution and Nucleosynthesis&lt;/a&gt; contains some of these tables and also some better analytical approximations that work in certain regimes of (partial) degeneracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is true is that you are fine to add to this pressure, the radiation pressure due to photons and the ideal gas pressure of the &lt;em&gt;ions&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I realise that what may be troubling you is not the above, but why you can add the separate pressures together at all (eqn 122 in the notes you reference). The reason can be traced to eqn 31 in the same notes. This says that the pressure is an integral over the number density and momenta of individual particles. Note that this equation does not contain anything that &lt;em&gt;identifies&lt;/em&gt; the particles (i.e. their mass). The integral can therefore be treated as a summation over all particles, which in turn can be separated into three integrals that cover the ions, electrons and photons separately.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-07-26T19:39:56.800" LastActivityDate="2015-07-26T19:39:56.800" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11379" PostTypeId="2" ParentId="11344" CreationDate="2015-07-25T13:18:33.370" Score="5" Body="&lt;p&gt;Geothermal activity can originate from a combination of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Residual heat from planet formation:&lt;/strong&gt; During the formation of a planet, half of the potential energy of&#xA;the parent region of dust and gas can be theoretically converted to&#xA;kinetic energy---this is known as the &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/astro/gravc.html&quot; rel=&quot;nofollow&quot;&gt;Virial Theorem&lt;/a&gt;.&#xA;The kinetic energy of the particles translate to a temperature.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Internal radioactivity:&lt;/strong&gt; Around half of the $4.5 \times 10^{13}$W of heat that come from the Earth's interior is driven by &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth%27s_internal_heat_budget#Radiogenic_heat&quot; rel=&quot;nofollow&quot;&gt;radioactive decay&lt;/a&gt;.  Should Pluto have the same proportion of radioactive materials in&#xA;its composition? Some would assume so, but others would say that the&#xA;heavier radioactive materials should have fallen towards the centre&#xA;of the solar system during its formation and would thus exist mostly in the inner planets. Note that geothermal activity on&#xA;Mars has now ceased despite it being much larger (Mars has a radius of $3400$km while that of Pluto is $1200$), so a combination of residual heat and radioactive decay is&#xA;perhaps unlikely in the case of Pluto.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Tidal effects:&lt;/strong&gt; Tidal forces basically apply whenever a part of a planetary body experiences a changing gravitational field with time (see &lt;a href=&quot;http://www.astro.washington.edu/users/smith/Astro150/Tutorials/TidalHeat/&quot; rel=&quot;nofollow&quot;&gt;tidal heating&lt;/a&gt;)---a number of processes can cause this:&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the moon having an elliptical orbit;&lt;/li&gt;&#xA;&lt;li&gt;extra moons in the vicinity (relevant to &lt;a href=&quot;http://www.planetaryexploration.net/jupiter/io/tidal_heating.html&quot; rel=&quot;nofollow&quot;&gt;Io&lt;/a&gt;);&lt;/li&gt;&#xA;&lt;li&gt;moon's orbital period and planet's rotational period not being synchronised;&lt;/li&gt;&#xA;&lt;li&gt;non-uniform density in planet.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;With time, the New Horizons mission should shed more light on these $3$ possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the geothermal activity on Pluto likely refers explicitly to substances such as Nitrogen, Methane, etc., where far less energy is required to drive activity; that is to say, the ability to melt silicates has long since passed. &lt;/p&gt;&#xA;" OwnerUserId="7770" LastEditorUserId="7770" LastEditDate="2015-07-25T17:23:35.350" LastActivityDate="2015-07-25T17:23:35.350" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11380" PostTypeId="2" ParentId="10065" CreationDate="2015-07-25T18:12:03.613" Score="3" Body="&lt;p&gt;&lt;strong&gt;Orange&lt;/strong&gt;: the orange represents &lt;a href=&quot;http://www.esa.int/var/esa/storage/images/esa_multimedia/images/2012/04/centaurus_a_far-infrared/10147086-2-eng-GB/Centaurus_A_Far-infrared_medium.jpg&quot; rel=&quot;nofollow&quot;&gt;far infrared&lt;/a&gt;. The flurry of star formation activity in the galactic disk has created large quantities of dust, which when heated by the stars themselves, glows in infrared wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Blue&lt;/strong&gt;: the blue false colour represents &lt;a href=&quot;http://chandra.harvard.edu/photo/2014/cena/cena_red.jpg&quot; rel=&quot;nofollow&quot;&gt;X-ray&lt;/a&gt; wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;White&lt;/strong&gt;: the bright white across the galaxy, and the fuzzy haze surrounding it is the optical light from the billions of stars that make up the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the far-infrared is quite close to the &lt;a href=&quot;http://www.mpe.mpg.de/~hcs/Cen-A/Pictures/SuW200801.jpg&quot; rel=&quot;nofollow&quot;&gt;radio continuum&lt;/a&gt;. The two lobes above and below the galactic disk can also be viewed in the radio continuum and is Synchrotron emission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also this &lt;a href=&quot;http://apod.nasa.gov/apod/image/0208/cenA_chandra_big.gif&quot; rel=&quot;nofollow&quot;&gt;false colour image of Centaurus A&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7770" LastEditorUserId="7770" LastEditDate="2015-07-25T18:20:05.130" LastActivityDate="2015-07-25T18:20:05.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11381" PostTypeId="1" AcceptedAnswerId="11385" CreationDate="2015-07-25T19:25:08.713" Score="1" ViewCount="183" Body="&lt;p&gt;Where can I find a catalog for stars within 450 light-years of Earth? I'd prefer one that has approximated masses as well.&lt;/p&gt;&#xA;" OwnerUserId="7580" LastEditorUserId="2153" LastEditDate="2015-07-25T21:49:04.093" LastActivityDate="2015-07-27T21:35:41.647" Title="Where can I find a catalog for stars within 450 light-years of Earth?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11382" PostTypeId="1" AcceptedAnswerId="11383" CreationDate="2015-07-25T22:20:34.773" Score="10" ViewCount="975" Body="&lt;p&gt;If two celestial bodies are in orbit, will they always eventually collide if not acted upon by outside forces?&lt;/p&gt;&#xA;" OwnerUserId="7119" LastEditorUserId="2153" LastEditDate="2015-07-25T22:44:42.850" LastActivityDate="2015-07-28T17:47:47.107" Title="Do all orbiting bodies eventually collide?" Tags="&lt;orbit&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11383" PostTypeId="2" ParentId="11382" CreationDate="2015-07-25T22:37:22.380" Score="13" Body="&lt;p&gt;&lt;strong&gt;Yes.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two bodies in orbit around each other will inevitably collide. The reason for this is that the system will give off energy in the form of &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_wave&quot; rel=&quot;nofollow&quot;&gt;gravitational waves&lt;/a&gt;. This effect is commonly cited in binary neutron star systems, where the two stars are isolated and close together. One of the most famous of these systems is the &lt;a href=&quot;http://en.wikipedia.org/wiki/PSR_B1913%2B16&quot; rel=&quot;nofollow&quot;&gt;Hulse-Taylor binary&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time it will take for the objects to collide &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_wave#Orbital_lifetime_limits_from_gravitational_radiation&quot; rel=&quot;nofollow&quot;&gt;can be calculated&lt;/a&gt;:&#xA;$$t=\frac{5}{256}\frac{c^5}{G^3}\frac{r^4}{(m_1m_2)(m_1+m_2)}$$&#xA;where $r$ is the initial radius, $m_1$ and $m_2$ are the masses of the bodies, and $c$ and $G$ are the familiar constants, the speed of light in a vacuum and Newton's universal gravitational constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;However&lt;/em&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Tidal_acceleration&quot; rel=&quot;nofollow&quot;&gt;tidal acceleration&lt;/a&gt; could offset some of the effects.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-07-28T17:47:47.107" LastActivityDate="2015-07-28T17:47:47.107" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="11384" PostTypeId="2" ParentId="11373" CreationDate="2015-07-25T22:38:31.337" Score="1" Body="&lt;p&gt;It's not really clear what you're asking and most of these questions can be researched individually, but I'll give it a shot:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What would Earthlike planets bigger than Earth with &amp;lt;10 m/s^2 gravity&#xA;  and complex life be like?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Large is bigger than Earth. Assuming that such planets exist (though I&#xA;  guess if such planets don't exist that would explain much of the Fermi&#xA;  paradox).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure I follow the Fermi Paradox argument cause life probably could develop on small earths just as easily as large earths, though tectonic plate movement, a core maintaining a magnetic field and atmosphere retention are all factors, so there are sets of circumstances that would be more likely than others but there's no direct relation with one specific type &quot;big lower G earths&quot; and the Fermi Paradox.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, to answer your other question, there's no reason why a lower iron, larger but lower gravity earth couldn't exist.   How common they would be depends on how common Iron is in that the gas cloud that forms solar systems but lower Iron content is certainly possible.   There's other ways too, for example, a giant impact, like the one that formed our moon could form a low Iron core high mantle planet.   How common a larger but lower G earth would be - hard to say, but I'm sure there's at least some of them out there.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Maybe some amount of iron is needed though, to make a magnetic field.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some iron is likely inevitable as it's such a common element released in supernovas.   Of the 4 inner planets, only Earth has a really good magnetic field, and all 4 inner planets have iron cores, so clearly Iron isn't the only factor in a magnetic field.  You also need the right temperature and pressure and, I think, inner core rotation and probobly some measure of radioactive elements to keep it hot.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I guess you could also have a planet like that with very good climate&#xA;  and gravity for Earthlings but toxic air &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Toxic to whom?    If life evolved on a planet, they would likely adapt to whatever elements were in the air.   Oxygen used to be toxic to early life when it started to collect in the oceans once the dissolved Iron has been used up.  Early one-celled life first protected themselves from Oxygen, then, later, began to use it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and a lot more carbon or ammonia or methane or hydrogen or&#xA;  formaldehyde or acetone or alcohol or hydrocarbons than here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's too much to do here as each element could be written up separately, but one thing to consider, while complex elements like acetone or alcohol do appear to exist in giant gas clouds in space, and they likely exist in comets too, when a planet is formed, there's enormous heat in the process.  Anything complex like that is likely to be destroyed in the planets formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, photosynthesis, if that's going to be the basis for lush life, that creates Oxygen and Oxygen is very reactive, it would bind with any hydrogen and free Iron and likely free Sulfur and other elements and it would in time, react with and remove any CH4 from the air.  If your &quot;lush&quot; planet doesn't operate on photosynthesis, then you can set up a different set of rules, but Oxygen is very reactive and certain other elements aren't possible once Oxygen becomes present in the atmosphere.  Oxygen from photosynthesis would react with and remove any fairly large amounts of acetone or alcohol if it was present on a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water is also very common in the universe, so, unless the planet is below freezing, or very high pressure, or very hot, it's hard to imagine oceans of anything other than water.   On titan there are oceans of Methane, but that's cause it's really cold.   Oceans of alcohol is a fun sounding idea, but I think, unlikely.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It wouldn't be Earthlike but it'd look like a lush land where the&#xA;  rocks are all graphite or diamond and the sky is brown or something.&#xA;  That planet could be like the Garden of Eden for a sentient that's&#xA;  mostly acetone and would get drunk from an ounce of H2O (or whatever&#xA;  chemicals are appropriate).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Graphite and Diamond are formed by high pressure inside a plant, so all the rocks being graphite seems unlikely.   More present - sure.   All rocks, I don't think so.   A Brown sky - sure, why not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;a garden of eden that's mostly acetone&quot;, because acetone is both somewhat complex and fairly reactive and it contains oxygen and because most primordial oxygen on any planet would probably be bound more simply to hydrogen or Iron or Carbon, I have a really hard time seeing that as a likely outcome.   Unexpected combinations could still happen, so I like where you're going with this, but that specific example seems unlikely.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take the assumption that life forms in water / needs liquid water, it's hard to figure that water would get anything drunk.  It's too essential to life.   Our bodies are 80% water.   I don't think it's possible for life to develop where the body is 80% alcohol for example.   An Alcohol water-mix, kind of an anti-freeze of sorts on a colder planet. . . . who knows, maybe.   On a cold enough planet, cold blooded animals resist freezing by limiting the amount of water in their bodies, so some exotic outcomes are possible, but no water, I have a hard time seeing that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've only got 1 example of the evolution of life to study so far, so how it might work on another planet with a different conditions is far from exact science.  There's always going to be certain parameters defined by the chemistry, like carbon based life is probably the only one that works, other molecules simply aren't able to do what Carbon does and liquid water may be essential too, but there might be a lot of exotic and unexpected ways life can happen within those parameters.  Too many unknowns to say for sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps some.   At least, that's my 2 cents on this question.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-07-25T22:38:31.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11385" PostTypeId="2" ParentId="11381" CreationDate="2015-07-26T00:37:48.230" Score="4" Body="&lt;p&gt;The Hipparcos catalogue (use the revision by &lt;a href=&quot;http://cdsarc.u-strasbg.fr/viz-bin/Cat?I/311&quot; rel=&quot;nofollow&quot;&gt;van Leeuwen 2007&lt;/a&gt;) contains trigonometric parallaxes and is complete down to about 9th magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means is that you will have a complete set of measurements for F-type main sequence stars and hotter, and all giant stars within 450 light years. However, the sample will not be complete for G dwarfs or cooler, since these are fainter than 9th magnitude at 450 light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The catalogue does not contain mass, since mass is not a measurable quantity for single stars. If you want masses you will have to use relationships between absolute magnitude and mass for main sequence stars (e.g. &lt;a href=&quot;http://arxiv.org/abs/astro-ph/9701213&quot; rel=&quot;nofollow&quot;&gt;Kroupa 2007&lt;/a&gt;), but note that these depend on metallicity and a little bit on stellar age. Masses are extremely difficult to estimate for giants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can wait a couple of years, the Gaia mission will report parallax distances for all stars to magnitudes of 19 or 20. This will be almost complete at 450 light years, even for M dwarfs.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-07-27T21:35:41.647" LastActivityDate="2015-07-27T21:35:41.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11386" PostTypeId="1" AcceptedAnswerId="11390" CreationDate="2015-07-26T02:04:48.543" Score="1" ViewCount="211" Body="&lt;p&gt;I suspect this may be an open problem in astronomy, but I have to ask anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There doesn't seem to be any large-scale difference in the formation of pulsars from the formation of &quot;normal&quot; neutron stars. Yet pulsars end up emitting electromagnetic radiation, while others - &quot;radio quiet&quot; neutron stars - do not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do are some neutron stars formed as pulsars, while others are not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is partly based on &lt;a href=&quot;https://astronomy.stackexchange.com/questions/42/what-causes-a-star-to-become-a-pulsar&quot;&gt;What causes a star to become a pulsar?&lt;/a&gt;, but the answer does not cover this part.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-07-26T11:11:55.517" Title="Why do some neutron stars become pulsars?" Tags="&lt;neutron-star&gt;&lt;pulsar&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-07-27T21:04:23.957" ContentLicense="CC BY-SA 3.0" />
  <row Id="11387" PostTypeId="2" ParentId="2115" CreationDate="2015-07-26T02:47:17.083" Score="2" Body="&lt;p&gt;The 'stuff' comprising the portion of a neutron star theoretically comprised mostly of neutrons has been referred to as &quot;neutronium&quot; in various sources, one of which is science fiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have always found imagining what this 'neutronium' would actually look like to be an amusing exercise.  However, attempting to define this 'stuff' as an element in and of itself is no different than performing the same attempt at categorization for an exclusive mass of protons or elections.  None of these components define atoms or elements singly.  &lt;/p&gt;&#xA;" OwnerUserId="7776" LastActivityDate="2015-07-26T02:47:17.083" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11388" PostTypeId="1" CreationDate="2015-07-25T08:39:09.267" Score="4" ViewCount="2311" Body="&lt;p&gt;In the recent $100m search for extra terrestrial life project project, &lt;a href=&quot;http://www.washingtonpost.com/news/speaking-of-science/wp/2015/07/20/stephen-hawking-announces-100-million-hunt-for-alien-life/&quot; rel=&quot;nofollow&quot;&gt;Stephen Hawking is quoted&lt;/a&gt; in the following way:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;We believe that life arose spontaneously on Earth,&quot; Hawking said at Monday's news conference, &quot;So in an infinite universe, there must be other occurrences of life.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My understanding is that scientists believe that the universe &lt;em&gt;isn't&lt;/em&gt; infinite - there there is a finite number of stars, it's continuously expanding but has a finite size at a given time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's Stephen Hawking refering to here?&lt;/p&gt;&#xA;" OwnerDisplayName="dwjohnston" LastEditorUserId="2153" LastEditDate="2015-07-26T13:43:36.020" LastActivityDate="2019-04-08T08:23:48.753" Title="What does Stephen Hawking mean by 'an infinite universe'?" Tags="&lt;universe&gt;&lt;cosmology&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11389" PostTypeId="1" AcceptedAnswerId="12741" CreationDate="2015-07-26T08:12:07.857" Score="11" ViewCount="1733" Body="&lt;p&gt;I know of three sets of stable orbits in a binary star system: orbiting closely around star A, orbiting closely around star B, or orbiting distantly around both stars (and their mutual center of gravity) at once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a fourth set of stable orbits, around the mutual center of gravity, but &lt;em&gt;inside&lt;/em&gt; both stars' orbits?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jcFfG.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jcFfG.png&quot; alt=&quot;Two stars orbiting their mutual center of gravity, with a planet orbiting that same point&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1559" LastEditorUserId="1559" LastEditDate="2015-07-26T19:44:45.520" LastActivityDate="2016-03-08T18:35:48.223" Title="Orbits in a binary star system" Tags="&lt;orbit&gt;&lt;binary-star&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11390" PostTypeId="2" ParentId="11386" CreationDate="2015-07-26T09:13:45.053" Score="3" Body="&lt;p&gt;My understanding is that it would be quite usual for the majority of neutron stars to be pulsars shortly after they formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The requirements are that they should have strong magnetic fields and rapid rotation (faster than a rotation period of a few seconds). The latter seems an almost inevitable consequence of conservation of angular momentum and the former must surely be assisted (if not fully explained) by conservation of magnetic flux as the core of a massive star collapses (see the discussion in &lt;a href=&quot;http://arxiv.org/abs/0710.2839&quot; rel=&quot;nofollow&quot;&gt;Reisenegger 2007&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to be &lt;em&gt;observed&lt;/em&gt; as a pulsar, the spin and magnetic field axes have to be oriented in a specific way, so that Doppler-boosted and beamed radiation sweeps over the Earth. This requirement means that we can observe only a fraction of the pulsars out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact there are a category of &quot;radio quiet pulsars&quot; - these are detected by their high energy radiation pulsations but not seen as radio pulsars. The idea here is that the beaming is less extreme for the high energy radiation, allowing it to be seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pulsar phenomenon is also transient. Once the rotational kinetic energy of the pulsar has been extracted and it spins down beyond the &quot;pulsar death line&quot;, or if magnetic field decay occurs rapidly, then the pulsar should switch off. Most of the billion (or so) neutron stars in our Galaxy are no longer pulsars. The slope of the death line is such (e.g. see &lt;a href=&quot;http://arxiv.org/abs/astro-ph/9911478&quot; rel=&quot;nofollow&quot;&gt;Arons 1999&lt;/a&gt;) that even if the pulsar is born with a relatively weak field ($10^{6}$ T !) it should still be born with a rotation period short enough ($&amp;lt;1$ s) to produce the pulsar phenomenon initially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might find the discussion/simulations in &lt;a href=&quot;http://www.slac.stanford.edu/econf/C041213/papers/2311.PDF&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; interesting.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-07-26T11:11:55.517" LastActivityDate="2015-07-26T11:11:55.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11391" PostTypeId="2" ParentId="42" CreationDate="2015-07-26T09:32:12.263" Score="5" Body="&lt;p&gt;The endpoint in the lives of massive stars between about 10 and 25 solar masses is thought to be a core-collapse supernova that produces a condensed remnant called a neutron star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lower mass limit for neutron star progenitors is reasonably well known and due to the evolutionary paths taken by stars of differing masses. Below 10 solar masses it is probable that the core of the star reaches an electron-degenerate state &lt;em&gt;before&lt;/em&gt; it is able to fuse elements like Magnesium and Silicon to form Iron. An electron degenerate core can support the star and the remnant will cool forever as a white dwarf. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above 10 solar masses, nuclear fusion will proceed all the way to iron-peak elements, beyond which fusion reactions would be endothermic. Electron degeneracy is insufficient to support the core of the star and it collapses. If the core is not too massive, or so long as not too much material falls onto the collapsed core afterwards, then it is possible that a combination of neutron degeneracy pressure and the repulsive nature of short-range strong nuclear forces can support the remnant as a neutron star. The upper limit to the progenitor mass is uncertain. Whilst progenitor mass is very important, the rotational state and magnetic field of the progenitor are also thought to determine the outcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A neutron star is a 10km radius ball made mostly of neutrons, but it has a crust of exotic nuclear material and a fluid interior that also contains some protons and neutrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conservation of angular momentum dictates that whatever spin the core of the massive star had before it collapsed is magnified for a neutron star; so they should be born as extremely rapidly rotating objects the 1000 year old Crab pulsar spins 33 times per second).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conservation of magnetic flux also amplifies whatever magnetic field is around, and the rapidly rotating, superconducting protons enhance it even more, so that neutron stars are born with surface magnetic fields of 100 million to 100 trillion Teslas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rapid rotation generates a huge electric field at the neutron star surface that can rip off charged particles and hurl them along the magnetic field lines. These particles lose energy by radiating synchrotron and curvature radiation that is boosted and beamed in the forward direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the magnetic and rotation poles are misaligned, this can in favourable orientations lead to a beam of radiation sweeping over the Earth like that from a lighthouse. This is a pulsar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pulsars are not eternal. The energy of the radiation is ultimately powered from the spin of the pulsar. The pulsar spins down and for, as yet poorly understood reasons, the phenomenon turns off when the spin period slows beyond a few to 10 seconds.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-07-26T10:48:41.157" LastActivityDate="2015-07-26T10:48:41.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11392" PostTypeId="1" AcceptedAnswerId="11396" CreationDate="2015-07-26T12:49:25.243" Score="5" ViewCount="788" Body="&lt;p&gt;Planetary systems as we understand it exist from a disk of mass when the parent star is young.  That's why all our planets are in the same plane, or close to it.  See also &lt;em&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/q/2224/1462&quot;&gt;Why is the solar system often shown as a 2D plane?&lt;/a&gt;&lt;/em&gt;,  &lt;em&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/q/6001/1462&quot;&gt;Inclination of planets&lt;/a&gt;&lt;/em&gt;, and answers there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet the newly announced &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler-452b&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kepler-452b&lt;/a&gt; apparently has an inclination of close to 90° (89.806+0.134−0.049°).  Do we have any model of solar system formation that explains how this can form?&lt;/p&gt;&#xA;" OwnerUserId="1462" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-07-26T18:42:01.003" Title="How can a planet have a 90° inclination?" Tags="&lt;solar-system&gt;&lt;exoplanet&gt;&lt;inclination&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11393" PostTypeId="1" CreationDate="2015-07-26T13:16:34.320" Score="8" ViewCount="957" Body="&lt;p&gt;A system with two massive bodies has potential energy proportional to their separation. Since the universe is expanding, is the potential (and total?) energy of such a system slowly increasing? What is the scale of this effect if it exists?&lt;/p&gt;&#xA;" OwnerUserId="7779" LastActivityDate="2021-03-18T18:58:32.650" Title="Is expanding universe adding potential energy?" Tags="&lt;gravity&gt;&lt;expansion&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11394" PostTypeId="2" ParentId="11388" CreationDate="2015-07-26T13:32:43.137" Score="4" Body="&lt;p&gt;Due to the finite speed of light, and the finite age of the Universe, only a portion of it is observable. When people talk about &quot;the size of the Universe&quot;, &quot;the number of stars in the Universe&quot;, etc., they usually refer to the &lt;em&gt;observable Universe&lt;/em&gt;, i.e. the sphere in which we are centered, and which has a radius given by the distance light has been able to travel in the 13.8 billion years since the Big Bang. Note that since the Universe is expanding, this radius is &lt;em&gt;more&lt;/em&gt; than 13.8 billion light-years. In fact it's 46.3 billion light-years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmological_principle#Observations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Observations&lt;/a&gt; indicate that, on large scales (i.e. above roughly half a billion light-years), the Universe is &lt;em&gt;homogeneous&lt;/em&gt; (the same everywhere) and &lt;em&gt;isotropic&lt;/em&gt; (the same in all directions). Assuming that this is indeed true is known as the &lt;em&gt;cosmological principle&lt;/em&gt;. If the rest of the Universe follows this principle, then there are three possible overall &quot;versions&quot; of universes that we can live in. We call these versions &quot;flat&quot;, &quot;closed&quot;, and &quot;open&quot;. Whereas a globally closed universe would have a finite extent, globally flat or open universes must be infinitely large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The observable Universe is, within measuring uncertainties but too a very high precision, flat (e.g. &lt;a href=&quot;https://arxiv.org/abs/1502.01589&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck Collaboration et al. 2016&lt;/a&gt;). Hence, we might think that the whole Universe is, in fact, infinite. But sort of like standing in a large forest with limited visibility doesn't tell you whether the forest is just larger than you can see, or if it's infinitely large, we can't with our current theories and observations know whether the Universe is finite or infinite.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2019-04-08T08:23:48.753" LastActivityDate="2019-04-08T08:23:48.753" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="11395" PostTypeId="2" ParentId="11296" CreationDate="2015-07-26T13:59:07.773" Score="2" Body="&lt;p&gt;I suggest you to take a look at the &lt;a href=&quot;http://www.lies.com/aaol/&quot; rel=&quot;nofollow&quot;&gt;Amateur Astronomy Observers Log Web Site&lt;/a&gt;, where everybody can share their astronomical logs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logs contain:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Instrumentation used&lt;/li&gt;&#xA;&lt;li&gt;Sky condition (seeing, light pollution, ...)&lt;/li&gt;&#xA;&lt;li&gt;Accurate date and time of the observation&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Specific informations that you could add depend by the kind of object that you are watching. For example you could try to estimate the magnitude of a variable star, or you could describe colour variation of Jupiter bands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some software can give you more support. There are specific astronomical log software for smartphone (for example &lt;em&gt;Stargazing Log&lt;/em&gt; for Android).&#xA;Morover in the Linux planetarium software &lt;em&gt;KStars&lt;/em&gt; when you open the detail of an object you have a specific tab where you can register logs in form of simple text.&lt;/p&gt;&#xA;" OwnerUserId="7780" LastActivityDate="2015-07-26T13:59:07.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11396" PostTypeId="2" ParentId="11392" CreationDate="2015-07-26T15:04:14.090" Score="5" Body="&lt;p&gt;As &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_inclination&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; notes,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since the word 'inclination' is used in exoplanet studies for this line-of-sight inclination then the angle between the planet's orbit and the star's rotation must use a different word and is termed the spin-orbit angle or spin-orbit alignment. In most cases the orientation of the star's rotational axis is unknown.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As of now (7/26/2015), no other planets, smaller objects, or debris disks have been found around Kepler-452. This means that the situation is not as unusual as might be thought. If there were other planets in the system with orbital inclinations near 0°, then things would appear exceedingly odd. Why would this one planet be different than the others?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible scenario is that the star's rotation axis has been changed. Think about how Uranus's axial tilt is skewed. Nobody knows for sure what happened, though one popular theory is that it was hit at some point in time by another object. Granted, it would take a rather large object to have any effect whatsoever on the axial tilt of Kepler-452, but it is still possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point of this idea is that perhaps it is the star, not the planet, which is unusual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several papers have been written by those much more knowledgeable than I, notably &lt;a href=&quot;http://arxiv.org/abs/1405.0960&quot; rel=&quot;nofollow&quot;&gt;Crida &amp;amp; Batygin (2014)&lt;/a&gt; and &lt;a href=&quot;http://iopscience.iop.org/0004-637X/784/1/66/pdf/apj_784_1_66.pdf&quot; rel=&quot;nofollow&quot;&gt;Xue et al. (2014)&lt;/a&gt;. Both refer to Hot Jupiter models, but the ideas behind them are still possible for other planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The former suggest&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Contemporary observational surveys suggest that a considerable fraction of solar-type stars are born as binary or multiple systems (Ghez et al. 1993; Kraus et al. 2011). Moreover, most stars form in embedded cluster environments (Lada &amp;amp; Lada 2003) where dynamical evolution can lead to the acquisition of transient companions (Malmberg et al. 2007). Recently, Batygin (2012) showed that the presence of a companion to a young star can force the proto-planetary disk to precess around the binary’s axis, so that the plane in which planets eventually form and the equatorial plane of the central star can differ.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They also suggest that encounters with other planets could be responsible for some cases of this (although an angle of 90° makes this seem rather improbable); a related idea that has been applied to Kuiper Belt Objects is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kozai_mechanism&quot; rel=&quot;nofollow&quot;&gt;Kozai mechanism&lt;/a&gt;. The companion to this theory is that the star was perturbed by an encounter, possibly with a companion (as I discussed earlier).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Xue et al. discuss similar scenarios, focusing in part on &quot;dynamic&quot; interactions between planets. Direct tidal interactions between the star and the planet are also a possible scenario. &lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-07-26T18:42:01.003" LastActivityDate="2015-07-26T18:42:01.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11397" PostTypeId="2" ParentId="11393" CreationDate="2015-07-26T18:22:25.477" Score="1" Body="&lt;p&gt;Universe is &lt;a href=&quot;http://darkmatterdarkenergy.com/tag/integrated-sachs-wolfe-effect/&quot; rel=&quot;nofollow&quot;&gt;picking up kinetic energy due to the expansion of space&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A photon will gain energy (blueshift) when it heads into a supercluster on its way to the Earth. This is an effect of general relativity. And as it leaves the other side of the supercluster as it continues its journey, it will lose energy (redshift) as it climbs out of the gravitational potential well. But while it is passing through the supercluster, that structure is spreading out due to the Big Bang overall expansion, and its gravitational potential is weakening. So the redshift or energy loss is smaller than the original energy gain or blueshift. So net-net, photons gain energy passing through a supercluster.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Potential energy on that scale, not so much.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-07-26T18:22:25.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11398" PostTypeId="2" ParentId="11388" CreationDate="2015-07-26T19:08:16.070" Score="1" Body="&lt;p&gt;A relativistic universe which is expanding faster than light (like ours) is effectively infinite for all practical purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also due to its relativistic nature and faster than light expansion, even if you assume it's not infinite at some given moment, it still doesn't have any edges or borders - for you. You're in some random place in it, the maximum speed of any possible interaction is speed of light, and the universe is expanding faster than light - then anything that happens at some (real or imaginary) &quot;edge&quot; is outside your realm of existence. It is effectively infinite - for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give it enough time and it would grow as big as you want. Give it asymptotically infinite time and it will grow asymptotically infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to what the &quot;edge&quot; might be, see &lt;a href=&quot;https://en.wikipedia.org/wiki/Eternal_inflation&quot; rel=&quot;nofollow&quot;&gt;Eternal Inflation&lt;/a&gt;. This is a model in modern cosmology where local bubbles like ours have stopped inflating (they still expand, but not at the tremendous rate of the initial inflationary phase); however, inflation continues forever outside the bubbles and at the edges. Therefore the bubbles keep growing indefinitely at faster than light speeds. Because speed of light is a limit for any interaction inside the bubbles, for any internal observer each bubble is effectively infinite for any practical purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be aware that there is no proof that this is actually the case, but this is a model that fits well what we now observe.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: TBH, I'm not even sure this is a question for StackExchange. It's very open-ended, and we don't really have the conclusive answers. All we can say for now is that the universe appears to be infinite for all practical purposes, but we can't know for sure. So often scientists like Hawking just simplify the language and refer to the universe as &quot;infinite&quot; without any of the qualifiers that would be required in a strict context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think there's any single, final answer here.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-07-26T22:15:57.203" LastActivityDate="2015-07-26T22:15:57.203" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11399" PostTypeId="1" AcceptedAnswerId="11411" CreationDate="2015-07-26T21:57:03.040" Score="4" ViewCount="298" Body="&lt;p&gt;How much more massive would Pluto have to be to clear its neighborhood?&lt;/p&gt;&#xA;" OwnerUserId="7785" LastEditorUserId="2153" LastEditDate="2015-07-26T22:04:39.690" LastActivityDate="2015-07-28T10:48:47.147" Title="How much more massive would Pluto have to be to clear its neighborhood?" Tags="&lt;mass&gt;&lt;pluto&gt;&lt;tno&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11401" PostTypeId="2" ParentId="2115" CreationDate="2015-07-26T22:23:30.743" Score="11" Body="&lt;p&gt;To be considered an element, they would have to have a positively charged nucleus. They don't. The neutron star is largely neutral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They would have to have a cloud of surrounding electrons. These electrons would have to share &quot;orbitals&quot; with other electrons around other, nearby neutron stars. That doesn't happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, at these scales, gravitational interaction predominates. Even if neutron stars did have some positive charge, and did have electrons orbiting them, interaction with other neutron stars would still depend a lot on gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, even if a neutron star was charged, it's not clear whether electrons around them would actually fall into some kind of orbitals governed by quantum mechanics. QM tends to not happen at such a scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, no, they are completely unlike atoms and their nuclei.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2015-07-26T22:23:30.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11402" PostTypeId="2" ParentId="11399" CreationDate="2015-07-27T01:23:51.503" Score="1" Body="&lt;p&gt;As we know, Pluto is within the Kuiper belt, and it only makes up about $0.077\%$ times the mass of all of the object within its orbit. Now, in comparison, Earth has objects in its orbit, but it is $1.7$ million times the mass of all of the objects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's been debated in the past (by supporters of Pluto being a planet) that if Earth were in Pluto's place, it wouldn't be able to clear the neighborhood. So, with Pluto being even smaller than Earth's moon, it would have to gain a significant amount of mass to be able to clear its orbit. &lt;/p&gt;&#xA;" OwnerUserId="6783" LastActivityDate="2015-07-27T01:23:51.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11403" PostTypeId="1" AcceptedAnswerId="11407" CreationDate="2015-07-27T03:47:47.003" Score="18" ViewCount="1302" Body="&lt;p&gt;The photos of Pluto from New Horizons are truly beautiful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But considering that Pluto is so far away from its nearest start - our Sun - how is it so well lit up? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did the New Horizons have a massive flash mounted on its camera to illuminate the planet or is there just a lot of ambient light in space? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really curious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qSUR1.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qSUR1.jpg&quot; alt=&quot;Pluto&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7787" LastEditorUserId="2153" LastEditDate="2015-07-27T14:29:36.830" LastActivityDate="2019-12-09T14:54:53.423" Title="How did New Horizons take such well-lit pictures of Pluto?" Tags="&lt;light&gt;&lt;pluto&gt;&lt;artificial-satellite&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11407" PostTypeId="2" ParentId="11403" CreationDate="2015-07-27T11:00:51.387" Score="30" Body="&lt;p&gt;It's brighter on Pluto than you think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NASA developed a tool called &lt;a href=&quot;https://www.nasa.gov/feature/nasa-lets-you-experience-pluto-time-with-new-custom-tool/&quot; rel=&quot;noreferrer&quot;&gt;Pluto time&lt;/a&gt;, which tells you when at your place the ambient light conditions are similar to the ones on Pluto.  This occurs when the Sun is only 2° below the horizon!  That's quite shortly after sunset, and considerably before the end of civil twilight, which is when it's 6° below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these photos were taken at local &quot;Pluto time&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VuUnu.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VuUnu.jpg&quot; alt=&quot;Pluto time&quot;&gt;&lt;/a&gt;&#xA;&lt;br /&gt;Pluto time, according to NASA.  &lt;a href=&quot;https://www.nasa.gov/feature/nasa-lets-you-experience-pluto-time-with-new-custom-tool/&quot; rel=&quot;noreferrer&quot;&gt;Source: NASA&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question: all it takes is a slightly longer exposure time / larger aperture / higher gain (what photographers may call ISO) than taking photos closer to the Sun.  It's easily bright enough for outdoor activities (except perhaps on a cloudy evening and/or in a dense forest, but neither should be common beyond the orbit of Neptune), so have fun &lt;a href=&quot;http://www.nature.com/news/nitrogen-glaciers-flow-on-pluto-1.18062&quot; rel=&quot;noreferrer&quot;&gt;glacier hiking&lt;/a&gt; on Pluto!&lt;/p&gt;&#xA;" OwnerUserId="1462" LastEditorUserId="1462" LastEditDate="2019-12-09T14:54:53.423" LastActivityDate="2019-12-09T14:54:53.423" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="11408" PostTypeId="1" CreationDate="2015-07-27T19:41:19.283" Score="1" ViewCount="606" Body="&lt;p&gt;I have got confused with some cosmology formulas which calculate distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most famous formula is indeed the Hubble law which says (nowadays) velocity is proportional to Hubble's parameter at current time. &#xA;On the other hand we can find the velocity from redshift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Hubble law it is possible to evaluate distance but it's not clear for me that what does this distance mean...! As the photons, which are now observed, are belong to another redshift (larger) and by traveling toward Earth the have become &quot;redder&quot; because of expansion of universe, what does the observed redshift stand for?&#xA;The actual redshift of the galaxy at observation time (although we can't recieve the photons now), the redshift at emission time (that means we do not calculate current distance of the galaxy) or something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot.&lt;/p&gt;&#xA;" OwnerUserId="7761" LastActivityDate="2015-07-27T20:59:25.183" Title="Hubble law, cosmological redshift and distance" Tags="&lt;cosmology&gt;&lt;redshift&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11409" PostTypeId="1" AcceptedAnswerId="11412" CreationDate="2015-07-27T20:45:38.020" Score="1" ViewCount="241" Body="&lt;p&gt;I've studied this in a few optics classes before, but I'm drawing a blank at the moment.  If I know the aperture size and focal length of a telescope, as well as how far away the object I want to view is, can I calculate how large/wide/big that object must be in order to be seen?&lt;/p&gt;&#xA;" OwnerUserId="7800" LastEditorUserId="2153" LastEditDate="2015-07-28T19:43:03.063" LastActivityDate="2015-07-28T19:43:03.063" Title="How large must an object be to be seen through a telescope?" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11410" PostTypeId="2" ParentId="11408" CreationDate="2015-07-27T20:59:25.183" Score="3" Body="&lt;p&gt;I think that the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia page on Hubble's law&lt;/a&gt; is reasonably clear. The distance in Hubble's law is the proper distance. This is the separation between two objects measured by observers at the same cosmic time. That is, if you imagine (instantaneously) stretching a load of metre rulers end-to-end from us to a distance galaxy, it is how many metre rules you would need. The velocity in Hubble's law is the rate of change of proper distance with cosmic time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On large enough scales we find that that the ratio of velocity to proper distance is the Hubble parameter, which itself changes with cosmic time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v(t) = H(t) D(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice we cannot measure the velocity or proper distance in Hubble's law - we can generally only estimate these quantities for a galaxy as we observe it some time in the past, when it emitted the light we detect. This is why Hubble's law in terms of measured redshifts and estimated distances is only applicable over relatively small (cosmologically speaking) distances and for recession velocities much less than the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more distant galaxies the distance cannot be calculated from a redshift without a cosmological model for how the Hubble parameter changes with cosmic time. This in turn depends on the adopted cosmological parameters. See &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10585/evolution-of-the-hubble-parameter&quot;&gt;Evolution of the Hubble parameter&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proper distance is also discussed &lt;a href=&quot;https://en.wikipedia.org/wiki/Comoving_distance#Uses_of_the_proper_distance&quot; rel=&quot;nofollow noreferrer&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-07-27T20:59:25.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11411" PostTypeId="2" ParentId="11399" CreationDate="2015-07-27T20:59:49.867" Score="7" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Clearing_the_neighbourhood&quot;&gt;This Wikipedia page&lt;/a&gt; does a decent job of describing the orbit-clearing criterion, based on the original paper by Stern &amp;amp; Levison (2002), which can be found &lt;a href=&quot;http://www.boulder.swri.edu/~hal/PDF/planet_def.pdf&quot;&gt;here (PDF)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to have cleared its orbit over a period of billions of years, an object needs a &quot;Stern-Levison parameter&quot; $\Lambda$ which is $&amp;gt; 1$; Pluto has $\Lambda \approx 3$-$4 \times 10^{-3}$. Since $\Lambda$ depends on the square of the object's mass (see the Wikipedia page, or Equation 4 of Stern &amp;amp; Levison), Pluto would need to be about fifteen to twenty times its current mass in order to cross the threshold.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2015-07-27T20:59:49.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11412" PostTypeId="2" ParentId="11409" CreationDate="2015-07-28T01:08:13.967" Score="3" Body="&lt;p&gt;A few clarifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Telescopes in general operate at very large distances - &quot;at infinity&quot; is the term used in optics parlance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bright enough object can be seen from any distance, no matter what its size is. All that matters is that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It's bright enough to produce an impression on whatever sensor&#xA;you're using (or your eye)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The background is dark enough to produce sufficient contrast&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But then it would be just a bright but tiny spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe what you're really asking for is: what is the combination of factors that shows the object as bigger than a simple dot? In that case, it's two factors: aperture of the telescope, and angular size of the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a flawless telescope, its aperture is what determines its resolving power. The resolving power is the angle at which two dots can be separated by the telescope. The formula is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;resolving power = 1 / (10 * aperture)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where resolving power is in arcseconds, and aperture is in meters. Examples:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;aperture         resolving power&#xA;10 cm            1 arcsec&#xA;20 cm            0.5 arcsec&#xA;1 m              0.1 arcsec&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As long as the object's angular size is bigger than the resolving power, it will appear bigger than a dot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's all. In astronomy, we don't speak of an object's absolute size, we only speak of the angular size. But that should be enough. As soon as you have the angular size, and say the distance, then you could deduce the absolute size, it's a simple matter of trigonometry.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;absolute size = distance * tangent(angular size)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;E.g., this is the size of an object of 1 arcsec angular size, situated at 384,000 km (the orbit of the Moon):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=384000+km+*+tangent%281+arcsec%29&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=384000+km+*+tangent%281+arcsec%29&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's 1.8 km (in case the link above is unavailable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, that's the minimum absolute distance resolved by a telescope 10 cm in aperture, for objects on the Moon. Any two dots closer together than 1.8 km, placed on the Moon, are seen as one dot in a 10 cm telescope.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-07-28T04:27:03.110" LastActivityDate="2015-07-28T04:27:03.110" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11414" PostTypeId="1" CreationDate="2015-07-28T07:29:23.793" Score="4" ViewCount="226" Body="&lt;p&gt;I'm in search of historical knowledge of the outer ice giants, which seems so hard to come by. Right now I would be happy with just a photo of each, the best photo we had of them &lt;strong&gt;before Voyager 2&lt;/strong&gt;. Such a photo from the 60's to early 80's would be great. This will almost surely be from the Hale Telescope at the Palomar Observatory, because it's 5 meter primary mirror was the biggest for a long time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I'm not entirely sure that Neptune would even be a resolvable disk, but nevertheless I still want to see what our best photo of it (from that time) would be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After googling this for so long, I've come to the conclusion that the best bet would be old astronomy books, which sadly, I do not have. Maybe someone here can help me out.&lt;/p&gt;&#xA;" OwnerUserId="7449" LastActivityDate="2017-01-15T20:40:54.767" Title="Before Voyager 2, what were the highest quality photos of Uranus and Neptune?" Tags="&lt;history&gt;&lt;neptune&gt;&lt;uranus&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11415" PostTypeId="1" CreationDate="2015-07-28T09:34:24.350" Score="8" ViewCount="933" Body="&lt;p&gt;I followed the New Horizons Mission a little, and saw among others this image of Pluto:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1fVnh.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1fVnh.jpg&quot; alt=&quot;Pluto&amp;#39;s eclipse&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder, why you can't see any stars on it. As far as my very basic knowledge in astronomy goes, I think you can even see stars during a &quot;Sun-Moon-Eclipse&quot; here on earth. So besides having here a spectacular &quot;Sun-Pluto-Eclipse&quot;, I further more ask myself if that eclipse is even necessary to see other stars, as those pictures are already taken quit far away from sun, and outside an atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that not enough to see stars? Or does NASA remove the stars from the images, before publishing them? (If so, why? Not to distract?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for answers! &lt;/p&gt;&#xA;" OwnerUserId="7807" LastEditorUserId="7807" LastEditDate="2015-07-28T11:57:28.327" LastActivityDate="2015-10-17T04:10:22.110" Title="Why are there no stars on New Horizons images of Pluto" Tags="&lt;star&gt;&lt;pluto&gt;&lt;solar-eclipse&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11416" PostTypeId="1" AcceptedAnswerId="11424" CreationDate="2015-07-28T16:11:21.397" Score="4" ViewCount="160" Body="&lt;p&gt;Are there statistics on the expected number of asteroids (of various sizes) to orbit planets and moons in our (or other) solar systems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we know of other natural bodies orbiting the Earth (besides the Moon) or asteroids orbiting the Moon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, why not? Though the likelihood of natural captures are low, it is my understanding that there are lots of asteroids (though, of course, all orbits will eventually decay).&lt;/p&gt;&#xA;" OwnerUserId="4019" LastActivityDate="2015-07-28T22:53:57.797" Title="Do celestial bodies have many asteroids orbiting them?" Tags="&lt;asteroids&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11418" PostTypeId="1" CreationDate="2015-07-28T17:48:29.043" Score="2" ViewCount="137" Body="&lt;p&gt;It appears that there is already a calculated ballpark for meteor fatalities, but what about other factors such as space trash, comets, and other miscellaneous items in space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is it possible to obtain a ballpark estimate?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7095" LastActivityDate="2017-01-15T20:36:17.967" Title="What is the probability of an astronomical body hitting you?" Tags="&lt;comets&gt;&lt;meteor&gt;&lt;impact-probability&gt;&lt;space-debris&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11419" PostTypeId="1" AcceptedAnswerId="11423" CreationDate="2015-07-28T18:00:55.510" Score="4" ViewCount="736" Body="&lt;p&gt;&lt;em&gt;Our Earth, being a beautiful blue orb, has an overall perceived wavelength of somewhere near 470 nm, that being the wavelength of light blue, because the oceans play an enormous role in the weighted average of all of the Earth's colors.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But what about the &lt;strong&gt;whole universe&lt;/strong&gt;? The Earth is nothing in comparison to the host of multicolored stars, ranging from deep red to fiery blue.&lt;br&gt; Taking into consideration the number of stars and each separate color's weighted average, &lt;br&gt; could we calculate the average color of the universe?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="7095" LastActivityDate="2015-07-28T20:05:59.167" Title="What is the average color of the universe?" Tags="&lt;light&gt;&lt;wavelength&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11420" PostTypeId="2" ParentId="11419" CreationDate="2015-07-28T18:32:38.250" Score="0" Body="&lt;p&gt;The wave lengths emitted by hydrogen are a good candidate.&#xA;Limiting the idea of color to the narrow band that earth's current atmosphere does not filter out and is not emitted in great quantiles &#xA;to our sun out will probably miss the mark.  &lt;/p&gt;&#xA;" OwnerUserId="7708" LastActivityDate="2015-07-28T18:32:38.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11421" PostTypeId="1" AcceptedAnswerId="11427" CreationDate="2015-07-28T19:42:16.477" Score="5" ViewCount="143" Body="&lt;p&gt;Is it possible to create the following system of satellites?:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Place a satellite that orbits the some object X, once per day. Call it D.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Place a satellite that orbits D, once per hour. Call it H&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Place a satellite that orbits H, once per minute. Call it M.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Place a satellite that orbits M, once per second. Call it S.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If possible, what are some values for the weights, rotational speed, and radii that would work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, does the universe have a limit on how many layers of satellite systems it permits?&lt;/p&gt;&#xA;" OwnerUserId="7814" LastActivityDate="2015-07-30T09:49:10.510" Title="Is it possible to create a traditional clock solar system?" Tags="&lt;orbit&gt;&lt;artificial-satellite&gt;&lt;clock&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11422" PostTypeId="2" ParentId="11419" CreationDate="2015-07-28T20:05:27.497" Score="1" Body="&lt;p&gt;According to astronomers at Johns Hopkins University, it's beige. Note, they previously announced it was turquoise, but they identified a bug in their calculations, redid the numbers, and now it was determined the average color is beige. Kind of boring, yes, but kind of what you'd expect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, they collected &quot;detailed light measurements from more than 200,000 galaxies. They then constructed a &quot;cosmic spectrum,&quot; which represents all the energy in the local universe emitted at different optical wavelengths of light&quot; and averaged it. More details at link below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.npr.org/programs/atc/features/2002/jan/aas/020110.aas.html&quot; rel=&quot;nofollow&quot;&gt;NPR : The Color of the Universe Is...&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2015-07-28T20:05:27.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11423" PostTypeId="2" ParentId="11419" CreationDate="2015-07-28T20:05:59.167" Score="1" Body="&lt;p&gt;In RGB Hex notation, it's #FFF8E7, a light beige color called &quot;Cosmic Latte&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The color was calculated by a team of astronomers at Johns Hopkins University in 2002 from a sample of over 200,000 galaxies as a side-effect of a more &quot;serious&quot; study on the spectral features of different galaxies in relation to star formation.&lt;/p&gt;&#xA;" OwnerUserId="7655" LastActivityDate="2015-07-28T20:05:59.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11424" PostTypeId="2" ParentId="11416" CreationDate="2015-07-28T22:53:57.797" Score="2" Body="&lt;p&gt;The short answer is not very often.   Article on that here:  &lt;a href=&quot;http://www.universetoday.com/109666/can-moons-have-moons/&quot; rel=&quot;nofollow noreferrer&quot;&gt;can moons have moons?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing can orbit the Moon long term cause the Earth affects a moon/moon-satellite orbit and it draws tidal energy from it so an object would crash into the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 2nd moon around earth might be more stable than one around the moon, but I suspect, not long term stable.   If it fell into orbital resonance with our moon, like the inner 3 moons of Jupiter, I don't know, maybe, but I suspect not, cause the Moon's tidal force on a 2nd moon would (I suspect) be too great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From time to time, the Earth captures an asteroid, but they usually don't stay for long:  &lt;a href=&quot;http://www.scientificamerican.com/podcast/episode/earth-usually-has-second-tiny-tempo-11-12-27/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.scientificamerican.com/podcast/episode/earth-usually-has-second-tiny-tempo-11-12-27/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars' 2 moons and many of the gas giant planet's moons are captured asteroids or Kupier belt objects and if the earth didn't have our large moon, we might have a few captured asteroids in our orbit as well, so planets absolutely can capture asteroids, but the closer to the sun (Mercury/Venus), the smaller the planet's gravitational sphere's of influence and the less stable long term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Curiously, some asteroids even have asteroids, and Pluto in particular is a rather strange system with a comparatively enormous and very close moon, Charon and 4 outer moons that orbit around Pluto and Charon.   you can't get objects orbiting the Earth-Moon system, cause that would extend past the stable part of the Earth's Hill Sphere.   Pluto's Hill Sphere is 'about 3 times the radius of Earth's because it's so far away from any large bodies.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, stable orbits only exist in the inner 1/2 to 1/3rd of a body's hill sphere, so those are the 2 factors to look at, tidal effects and hill sphere (or Sphere of influence, which is related to Hill Sphere).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sphere or influence numbers for the 8 planets and Pluto:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sphere_of_influence_%28astrodynamics%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Sphere_of_influence_%28astrodynamics%29&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related:  &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6348/what-is-the-difference-between-sphere-of-influence-and-hill-sphere&quot;&gt;What is the difference between Sphere of Influence and Hill sphere?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-07-28T22:53:57.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11427" PostTypeId="2" ParentId="11421" CreationDate="2015-07-29T08:16:42.317" Score="7" Body="&lt;p&gt;&lt;em&gt;Is it possible to create the proposed system of satellites?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Yes&lt;/strong&gt; (&lt;em&gt;in theorey&lt;/em&gt;). Hierarchical multiple systems (like the one proposed in the OP) tend to be stable if the periods differ by $\sim 5$ or more and the orbits are near-circular. So, such a system could be stable. A slight problem may be the commensurability of the periods (the fact that their ratios are rational numbers), which implies orbital resonances. However, these resonances are rather weak (the lowest ratio is 1:24), but they may mess up the system in the very long term (a detailed analysis would be required to find out). &lt;em&gt;In practice&lt;/em&gt;, it will be extremely hard to set-up such a system, even with advanced space engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What are some values for the weights, rotational speed, and radii that would work?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is a strongly hierarchical system, the masses $m$ and semi-major axes $r$ (=radius if the orbit is circular) can be worked out from Kepler's third law&#xA;$$&#xA;    \frac{2\pi}{T} = \Omega = \sqrt{\frac{G(m_1+m_2)}{r^3_{1-2}}}&#xA;$$&#xA;for each sub-system. Giving&#xA;\begin{align}&#xA;   \frac{4\pi^2}{1\mathrm{sec}^2} &amp;amp;= \frac{G(m_S+m_M)}{r^3_{S-M}} \\&#xA;   \frac{4\pi^2}{1\mathrm{min}^2} &amp;amp;= \frac{G(m_S+m_M+m_H)}{r^3_{SM-H}} \\&#xA;   \frac{4\pi^2}{1\mathrm{hour}^2} &amp;amp;= \frac{G(m_S+m_M+m_H+m_D)}{r^3_{SMH-D}} \\&#xA;   \frac{4\pi^2}{1\mathrm{day}^2} &amp;amp;= \frac{G(m_S+m_M+m_H+m_D+m_X)}{r^3_{SMHD-X}},&#xA;\end{align}&#xA;where $r_{ABC-D}$ denotes the distance (more precisely: the orbital semi-major axis) between object $D$ and the centre of mass of objects $A$, $B$, and $C$. Since these are 4 equations for 8 unknowns (4 masses and 4 distances), there is some freedom in the design of your 'clock'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Does the universe have a limit on how many layers of satellite systems it permits?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, our formula only considered gravitational forces and neglected all else. Moreover, post-Newtonian effects (general relativity) have been ignored. As long as these assumptions remain valid (orbital speeds $\ll c$, orbital distances $\gtrsim\,$cm, solid objects with sizes $\ll$ distances so that tides are neglible, no electric or magnetic fields, vacuum), there is no limit. However, in practice, there is always some elctro-magnetic field and never perfect vacuum...&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2015-07-30T09:49:10.510" LastActivityDate="2015-07-30T09:49:10.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11429" PostTypeId="1" CreationDate="2015-07-29T10:25:11.097" Score="4" ViewCount="423" Body="&lt;p&gt;Are there many single transits in the Kepler data, which might be exoplanets with a longer orbital period than the time they have been observed for transits? Can any conclusions be drawn about how common they are? Is there reason to believe that the outer solar system is a common configuration? I suppose that slow moving planets are hard to detect also with doppler measurements. Will the upcoming telescopes aimed at nearby stars be able to find 10+ AU orbital radius planets?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-07-29T20:09:59.730" Title="How common are exoplanets in distant orbits?" Tags="&lt;orbit&gt;&lt;exoplanet&gt;&lt;kepler&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11430" PostTypeId="1" CreationDate="2015-07-29T10:44:12.360" Score="2" ViewCount="83" Body="&lt;p&gt;Jupiter is believed to have metallic hydrogen in its core. And gas giants that migrate to become hot Jupiters are believed to evaporate, have their atmospheres blown away by their nearby star. Can metallic hydrogen (or some other exotic materia) be detected on the surface of such a planet, or does it require high pressure to exist? What happens to the metallic hydrogen? Does it release much energy if and when it is transformed?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-07-29T12:24:39.057" LastActivityDate="2015-11-06T19:36:02.160" Title="Could evaporating hot Jupiters have metallic hydrogen on their surfaces?" Tags="&lt;core&gt;&lt;hydrogen&gt;&lt;hot-jupiter&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11432" PostTypeId="2" ParentId="11429" CreationDate="2015-07-29T12:13:40.477" Score="5" Body="&lt;p&gt;There are many planets known which have orbits longer than the longest exoplanet orbital periods found by Kepler. These planets were discovered using the &quot;doppler wobble&quot; or radial velocity technique. The plot below (a few months out of date now) shows many planets orbiting with similar periods to Mars and Jupiter. The red points we discovered by transits (including Kepler planets), whilst the green were discovered by the doppler technique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sensitivity of current observational doppler techniques (shown by the blue line) means that so long as we observe for a sufficiently long time (decades, in order to see an oscillation in the radial velocity), we will be able to detect Jupiter-like (or even a bit lower-mass) planets out to much longer periods.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/96gTU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/96gTU.png&quot; alt=&quot;Period distribution for exoplanets&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The frequency of planets in wider orbits is also probed well by surveys for gravitational microlensing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Single transit events cannot be securely identified as being caused by planets. I think there are a number of other possible causes, both astronomical and instrumental, that might also be to blame. Nevertheless I'm sure somebody will analyse these in some statistical way at some point to draw some inferences, but at the moment, the securest data are from the radial velocity technique. These show that the solar system is not necessarily unusual at all - there are plenty of stars that have Jupiter-sized objects in Jupiter-sized orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the review by &lt;a href=&quot;http://arxiv.org/abs/1305.0542&quot; rel=&quot;nofollow noreferrer&quot;&gt;Howard et al. (2013)&lt;/a&gt;, from radial velocity surveys, about 10 per cent of G/K stars have a 0.3-10 Jupiter-mass planet in orbit between 0.03 and 3 au from the parent star and around 7% have a Jupiter-sized planet orbiting between 3 and 20 au. This latter number probably has a significant error bar at the moment, but it is consistent with the information from microlensing surveys  (e.g. &lt;a href=&quot;http://arxiv.org/abs/1001.0572&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gould et al. 2010&lt;/a&gt;). This latter paper has an interesting statistic in the abstract. they say that if all stars had a solar system like ours, they would have detected 18.2 &quot;events&quot; - 11.4 due to Jupiters,  6.4 due to Saturns and 0.5 Uranus/Neptunes; they would also have detected 6 &quot;two-planet&quot; events. In actual fact they detected 6 events, one of which was a two-planet event, and thus conclude that a broad-brush figure for the frequency of &quot;solar system occurrence&quot; is 1/6. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-07-29T20:09:59.730" LastActivityDate="2015-07-29T20:09:59.730" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11433" PostTypeId="1" AcceptedAnswerId="11434" CreationDate="2015-07-29T14:30:59.730" Score="3" ViewCount="481" Body="&lt;p&gt;I am running MOOG on OS X Yosemite and with &lt;code&gt;$MOOG&lt;/code&gt; running the abfind driver I can force fit elemental abundances with an appropriate input line list and model atmosphere. MOOG crashes, however, when I try to use it to generate plots, even though I know I have SuperMONGO (sm) installed on my computer because &lt;code&gt;$sm&lt;/code&gt; works fine. In particular, MOOG will crash with the following errors:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;*******************************************************************************&#xA;                  MOOG IS CONTROLLED BY DRIVER abfind                      &#xA;*******************************************************************************&#xA;&#xA;wav. correl.:  slope =  -7.791E-07  intercept =   7.505  corr. coeff. =  -0.122&#xA;Fonts file not found&#xA;File graphcap is not defined&#xA;Can't get graphcap entry for nodevice&#xA;No such device nodevice&#xA;File graphcap is not defined&#xA;Can't get graphcap entry for x11&#xA;No such device x11 -bg black -title MOOGplot -geom 700x800+650+000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How can I get MOOG to find these libraries?&lt;/p&gt;&#xA;" OwnerDisplayName="user5341" LastActivityDate="2016-09-29T01:31:39.733" Title="How to specify SM libraries for MOOG" Tags="&lt;spectroscopy&gt;&lt;software&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11434" PostTypeId="2" ParentId="11433" CreationDate="2015-07-29T14:30:59.730" Score="1" Body="&lt;p&gt;MOOG will search for the libraries through a .sm file. To solve this error, download an example sm file like &lt;a href=&quot;https://gist.github.com/jtmyles/4085763ec32cdfbbf843&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;, and specify where your libraries are in this file. Then use &lt;code&gt;mv .sm ~/.sm&lt;/code&gt; to place this in your home folder, then MOOG should be able to create plots with SM.&lt;/p&gt;&#xA;" OwnerDisplayName="user5341" LastActivityDate="2015-07-29T14:30:59.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11435" PostTypeId="2" ParentId="11429" CreationDate="2015-07-29T19:03:30.210" Score="0" Body="&lt;p&gt;I know of two groups working on counting the number of long period Jupiter's in Kepler data (based on finding one transit only), but I haven't seen any results yet. Radial Velocities are the best way to answer your question because of the much longer timebase of observations.&lt;/p&gt;&#xA;" OwnerUserId="7827" LastActivityDate="2015-07-29T19:03:30.210" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11436" PostTypeId="1" CreationDate="2015-07-29T20:47:57.927" Score="3" ViewCount="65" Body="&lt;p&gt;If we orbit a non-spherical asteroid or moon at a sufficient distance I believe that we can consider it to be a point mass. Therefore we can take up a conventional orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming our lander craft is of insignificant mass compared with the body we wish to orbit and land on, how do we cope when we get near to the surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably an orbit will become more and more chaotic the nearer we get. As we try to land, unless we somehow synchronise with the body's rotation things will be equally tricky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What implications did this have for Rosetta's Philae and what implications will it have for much larger and maybe more irregular bodies?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some rule-of-thumb method to get an approximate answer or do spacecraft simply have to make minute by minute adjustments?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand mathematics to a reasonable level but really I'm more interested in the practicalities of such landings. To what extent can they be pre-calculated and to what extent must they be adjusted on the fly.&lt;/p&gt;&#xA;" OwnerUserId="7828" LastActivityDate="2015-07-29T22:27:55.340" Title="Orbiting and landing on non-sperical objects" Tags="&lt;orbit&gt;&lt;asteroids&gt;&lt;eccentric-orbit&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11437" PostTypeId="2" ParentId="10710" CreationDate="2015-07-29T21:18:31.610" Score="0" Body="&lt;p&gt;I haven't managed to formulate a coherent single answer, but here are several suggestions for where to find some hints of current topical research where you might be able to contribute.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Open source projects&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Astronomy is increasingly using large open source projects, many written in Python, which itself free. For example, the &lt;a href=&quot;http://www.astropy.org/&quot; rel=&quot;nofollow&quot;&gt;Astropy&lt;/a&gt; project is trying to create an extensive Python library for astronomical data manipulation, so you could try to contribute some of the desired features in a project like that. (There are other, more substantial, projects, especially in various kinds of modelling, but they require an understanding of the physics and are often big chunks of code, written in Fortran.)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Review articles&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If possible, I'd suggest looking at review articles (e.g. in &lt;a href=&quot;http://www.annualreviews.org/journal/astro&quot; rel=&quot;nofollow&quot;&gt;Annual Reviews in Astronomy &amp;amp; Astrophysics&lt;/a&gt;). Though many of the articles are probably behind paywalls, most of the recent ones should also be available on the &lt;a href=&quot;http://arxiv.org/list/astro-ph/recent&quot; rel=&quot;nofollow&quot;&gt;arXiv&lt;/a&gt;. Similarly, you can also try searching arXiv for lecture notes from summer/winter schools.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Departmental PhD project listings&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In a similar vein to contacting people, you might find that potential projects are listed online, in which case you'll see what kinds of things people would like to do.  For example, so quick Googling netted me information at &lt;a href=&quot;http://www.physics.manchester.ac.uk/media/eps/schoolofphysicsandastronomy/postgraduate/Project_Booklet_2014-15.pdf&quot; rel=&quot;nofollow&quot;&gt;Manchester&lt;/a&gt;, &lt;a href=&quot;http://www.st-andrews.ac.uk/physics/prosp_pg/phd/current_phd_projects.php?theme=6&amp;amp;filter=3&quot; rel=&quot;nofollow&quot;&gt;St Andrews&lt;/a&gt;, &lt;a href=&quot;http://astro.qmul.ac.uk/phd/available-projects&quot; rel=&quot;nofollow&quot;&gt;QMUL&lt;/a&gt;, and &lt;a href=&quot;https://www.ucl.ac.uk/star/phd/projects&quot; rel=&quot;nofollow&quot;&gt;UCL&lt;/a&gt;. (Clearly Google thinks I'm in the UK...)  While it probably doesn't make sense to actually try to carry out these projects, they might give you a better idea of the sorts of things that need doing.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Observation projects with public data&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I'd particularly watch out for anything that involves data-mining, since that mostly involves spending time crunching some of the huge datasets available.  I'm mostly aware of projects in the time domain (e.g. &lt;a href=&quot;http://ogle.astrouw.edu.pl/&quot; rel=&quot;nofollow&quot;&gt;OGLE&lt;/a&gt; and &lt;a href=&quot;http://www.superwasp.org/&quot; rel=&quot;nofollow&quot;&gt;WASP&lt;/a&gt;) but there are also larger projects like &lt;a href=&quot;http://www.sdss.org/&quot; rel=&quot;nofollow&quot;&gt;SDSS&lt;/a&gt; that I think have more data than people to sift (intelligently!) through it all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd note here the special cases of Kepler and it's continuation K2.  In these cases the actual analysis of the cameras' pixel data is still an open question, especially for K2. Any clever progress on automatically reducing the data better would be a boon in that field, although several active research groups are also working on it full time.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2015-07-29T21:18:31.610" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11438" PostTypeId="2" ParentId="11436" CreationDate="2015-07-29T22:27:55.340" Score="3" Body="&lt;p&gt;For bodies small enough to be non-spherical, the mass is relatively low (as an example, Rosetta's bounces took a long time - it touched the surface at 15:34, 17:25 and 17:32 GMT comet time) but the centre of mass is still the point the lander will orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So a chaotic orbit is not a problem here&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What will be problems are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;trying to arrange for the lander's velocity to match that of the landing site. For example if the body is spinning rapidly and an outcrop spins round into the path of the lander that could end badly&lt;/li&gt;&#xA;&lt;li&gt;staying on the surface - Rosetta tried to use harpoons. Another proposed solution was a low powered rocket pointing upwards&lt;/li&gt;&#xA;&lt;li&gt;while small bodies won't have an atmosphere as such, outgassing and dust can cause problems&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm assuming 1 is the one you are most interested in here. The solution Rosetta favoured was to pick a site that was on one of the outer surfaces, rather than on the 'neck' of the comet, as this offered the cleanest route to landing. A landing on the neck would have necessitated flying into a channel on a spinning body, and that would make the calculations much harder.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2015-07-29T22:27:55.340" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11440" PostTypeId="1" AcceptedAnswerId="11448" CreationDate="2015-07-30T13:17:46.337" Score="3" ViewCount="286" Body="&lt;p&gt;There is an interesting discussion about whether there could be life on Saturn's moon Titan. (For example &lt;a href=&quot;https://www.youtube.com/watch?v=bbkTJeHoOKY&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.) The life could use reaction of hydrogen and acetylene which are being produced by photolysis in Titan's atmosphere. I would like to know what total energy is available for such life per square meter per day, or at least how much hydrogen is produced in Titan's atmosphere per day. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some estimates in article &lt;a href=&quot;http://astrobiology.jhu.edu/wp-content/uploads/2010/06/Icarus-2005-McKay.pdf&quot; rel=&quot;nofollow&quot;&gt;McKay and Smith, Icarus, 2005&lt;/a&gt;, but I have problem translating the quantities they use for the estimate to total amount of hydrogen produced per unit of time or energy produced per unit of time.&lt;/p&gt;&#xA;" OwnerUserId="2529" LastEditorUserId="2529" LastEditDate="2015-08-01T16:20:23.013" LastActivityDate="2015-08-01T16:20:23.013" Title="Available energy for potential life on Titan" Tags="&lt;life&gt;&lt;titan&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11441" PostTypeId="1" CreationDate="2015-07-30T14:01:28.390" Score="14" ViewCount="565" Body="&lt;p&gt;This arose from &lt;a href=&quot;https://worldbuilding.stackexchange.com/questions/21508/equatorial-icecaps-and-polar-jungles-a-fantasy-or-reality#comment55064_21509&quot;&gt;a comment on Worldbuilding&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have data from four planets in the Solar System with rings, which doesn't make for a very good sample size. Observations of exoplanets could change that, but I expect that for now, any answer will simply be theoretical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How thick can planetary rings be? I assume it would depend on the amount of material, but I don't know.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:52:24.020" LastActivityDate="2016-10-17T16:44:41.260" Title="How thick can planetary rings be?" Tags="&lt;size&gt;&lt;planetary-ring&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11442" PostTypeId="1" AcceptedAnswerId="19767" CreationDate="2015-07-30T17:08:34.940" Score="3" ViewCount="228" Body="&lt;p&gt;It has been suggested by some futuristic or sci fi leaning thinkers, that Ceres' surface might be mined for water to support human exploration and settlement of space. But NASA's Dawn mission and other observations show that it is very dark and seems to consist of &quot;hydrated minerals&quot; in an environment warm enough to make water ice sublimate. Does this mean that mining water on Ceres is really hard? Requiring deep drilling or excavating through thick hard surface layers before useful water ice is encountered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could a lander on Ceres just heat up the surface and collect water sublimating from it?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2017-01-15T20:28:07.930" Title="How easy is it to mine water on Ceres?" Tags="&lt;water&gt;&lt;surface&gt;&lt;ceres&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11445" PostTypeId="1" CreationDate="2015-07-31T03:46:19.020" Score="12" ViewCount="211" Body="&lt;p&gt;Planetary rings seem to generally form on the ecliptic plane because of the rotation of the planet, but I was wondering what kind of effect the magnetic field of the planet might have on the ring system. Would it help keep the ring system together, disrupt it, or have little to no effect? What if the magnetic poles were on or near the ecliptic?&lt;/p&gt;&#xA;" OwnerUserId="1980" LastActivityDate="2021-02-11T20:56:14.977" Title="Do magnetic fields affect planetary rings?" Tags="&lt;magnetic-field&gt;&lt;planetary-ring&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11446" PostTypeId="2" ParentId="11393" CreationDate="2015-07-31T04:53:49.783" Score="3" Body="&lt;p&gt;Yes, in an expanding universe there is an increase in potential energy (with same caveats).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The (Friedmann) equations of a homogeneous and isotropic universe with no spatial curvature or cosmological constant can, in fact, be derived very easily from nonrelativistic Newtonian gravity. (This derivation is shown, e.g., in Mukhanov's &lt;em&gt;Physical Foundations of Cosmology&lt;/em&gt;.) So it is actually valid to think of such a cosmos in Newtonian terms, using randomly but approximately homogeneously distributed point masses. At early times, the point masses are flying apart at high speed but the average distance between them is small. At late times, they will have slowed down, but the average distance between them will have increased. All that kinetic energy is converted into gravitational potential energy. This also remains true if we take into account Newtonian perturbation theory (the growth of overdense and underdense regions from initial perturbations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the caveats: First, as I mentioned above, when spatial curvature is present, the Newtonian solution is no longer valid. Second, there is a century of literature with a variety of attempts to define a meaningful stress-energy tensor for the gravitational field in general relativity, none completely satisfying. Last but not least, a cosmological constant/dark energy term adds its own interesting twist to the story, as its energy density remains constant in an expanding universe, causing the expansion to accelerate: in this case, both the kinetic and the potential energy increase, thanks to the role played by dark energy's negative pressure.&lt;/p&gt;&#xA;" OwnerUserId="7838" LastActivityDate="2015-07-31T04:53:49.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11447" PostTypeId="1" CreationDate="2015-07-31T08:32:40.070" Score="6" ViewCount="195" Body="&lt;p&gt;The magnetic field activity in a star can be very chaotic and my question is are they somehow related?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="2531" LastEditDate="2015-08-01T08:53:32.977" LastActivityDate="2016-07-17T11:50:14.153" Title="Does the shape of a supernova remnant depend on the progenitor star's magnetic field?" Tags="&lt;star&gt;&lt;supernova&gt;&lt;magnetic-field&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11448" PostTypeId="2" ParentId="11440" CreationDate="2015-07-31T09:35:08.003" Score="2" Body="&lt;p&gt;I can give a rough answer to your question and some brief thoughts on the paper, and I invite correction from anyone smarter than me.   Your article talks about energy per mole, but energy of production on a daily basis would be a product of solar energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To produce hydrogen in Titan's atmosphere requires UV light, from this article, UV light of 1600 Angstroms (160 NM) &lt;a href=&quot;http://www.astrobio.net/topic/solar-system/saturn/titan/the-missing-methane/&quot; rel=&quot;nofollow&quot;&gt;(Source)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quote: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Methane is a carbon atom surrounded by four hydrogen atoms, and it can&#xA;  be broken apart by ultraviolet light at wavelengths of about 1600&#xA;  Angstroms. The fragments, or radicals, that are produced from this&#xA;  process are very chemically reactive. They’re things like CH, CH2, and&#xA;  in some cases CH3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only a small percentage of the sun's light is in that range.   8%-10% of the sun's light is in the UV range, but maybe only 1%-2% in the upper UV3 range of 160 NM.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Sunlight#Composition_and_power&quot; rel=&quot;nofollow&quot;&gt;(Source)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same Wiki article, using Saturn as a guide, Saturn receives between 13.4 and 16.7 Watts per square meter &lt;a href=&quot;https://en.wikipedia.org/wiki/Sunlight#Intensity_in_the_Solar_System&quot; rel=&quot;nofollow&quot;&gt;(Source)&lt;/a&gt;, so figure an estimate of 15 watts per square meter total solar energy, 1%-2% of that strong enough to pull a hydrogen atom off Methane, so 0.15 - 0.3 watts per square meter.   That might not sound like much, but on a moon the size of titan, that's a lot of square meters so the low wattage isn't a big deal.   That's a peak estimate though as some of the hydrogen would recombine with the CH3 it split from, some energy would likely be converted to heat, and perhaps the biggest problem, mentioned in the article above, a share of the hydrogen, perhaps the lion's share would simply be lost off the planet.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the atmospheres of the giant planets, the hydrogen stays around&#xA;  because of the high gravity of those planets. Their atmospheres are&#xA;  primarily hydrogen anyway, and after the fragmentation occurs, the&#xA;  products sink into the deeper atmosphere, and methane is&#xA;  reconstituted. It’s a complete chemical cycle.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;On Titan, as far as we understand, that does not occur. Because the&#xA;  gravity is low, hydrogen should escape. The Voyager ultraviolet&#xA;  spectrometer saw a corona of hydrogen around Titan, which is a good&#xA;  indication that hydrogen is escaping. Now, if hydrogen is going away,&#xA;  the products that can be made from the methane are going to have a&#xA;  higher carbon to hydrogen ratio than methane itself, and you’re not&#xA;  going to be able to remake methane. So as far as we understand the&#xA;  photochemistry, Titan should be destroying methane and making more&#xA;  carbon-rich products, the simplest of which are acetylene – C2H2 – and&#xA;  ethane – C2H6. These are made directly from methane&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is my problem with the article.  I don't see any way the hydrogen finds it's way down to the planet's surface to be used in the way the article suggests, life in the lakes of Titan.    It's an interesting idea and the people who wrote it are probably PHD educated and I'm not, but I suspect the majority of hydrogen produced high in Titan's atmosphere would remain high in it's atmosphere and over time, lost to the solar wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does raise the question, because Titan should lose it's Methane over time, why does it still have so much Methane - a very good question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets look at the timing.   &lt;a href=&quot;https://en.wikipedia.org/wiki/Huygens_%28spacecraft%29&quot; rel=&quot;nofollow&quot;&gt;Huygens&lt;/a&gt; landed on Titan January 14th, 2005.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your article was written  14 January 2005; revised 18 April 2005&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and here's an article from a few months later, November 2005.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.spaceref.com/news/viewpr.html?pid=18410&quot; rel=&quot;nofollow&quot;&gt;http://www.spaceref.com/news/viewpr.html?pid=18410&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The origin of methane in Titan's atmosphere is a mystery because it&#xA;  gets broken down by sunlight and particle radiation from space in the&#xA;  upper atmosphere. If surface lakes and pools were the only source, all&#xA;  of Titan's methane would be lost by this mechanism in less than a&#xA;  hundred million years, a short time for a moon that's been around&#xA;  since the formation of the solar system 4.5 billion years ago.�&#xA;  Components of the methane molecules react with each other and&#xA;  atmospheric nitrogen. As they descend, they form larger and heavier&#xA;  molecules that comprise the orange haze that blankets the moon.&#xA;  Because Titan is very cold (292 degrees below zero F, or minus 180&#xA;  degrees Celsius) these heavy compounds condense and rain out on the&#xA;  surface.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;We have determined that Titan's methane is not of biological origin,&#xA;  so it must be replenished by geologic processes on Titan, perhaps&#xA;  venting from a supply in the interior that could have been trapped&#xA;  there as the moon formed,&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, I'd chalk this one up to timing.   January-April 2005, life on Titan was a viable theory, which has since been de-theoried, or, whatever the correct term is.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-07-31T09:35:08.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11449" PostTypeId="1" AcceptedAnswerId="11452" CreationDate="2015-07-31T15:56:38.520" Score="2" ViewCount="66" Body="&lt;p&gt;I know this is not supposed to be precise since it is just a movie but I would suppose it is based on real things. In the Contact movie intro&#xA;&lt;a href=&quot;https://youtu.be/EWwhQB3TKXA?t=1m&quot; rel=&quot;nofollow&quot;&gt;https://youtu.be/EWwhQB3TKXA?t=1m&lt;/a&gt; I think the Milk Way is shown at around 2:12, I am wondering what is the structure shown around 2:32. Does that a representation of anything real? Virgo maybe?&lt;/p&gt;&#xA;" OwnerUserId="7842" LastEditorUserId="7842" LastEditDate="2015-07-31T16:13:04.167" LastActivityDate="2015-07-31T18:12:13.693" Title="Which structure is represented in this video (if any)?" Tags="&lt;galaxy&gt;&lt;universe&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11450" PostTypeId="2" ParentId="11118" CreationDate="2015-07-31T16:49:41.953" Score="2" Body="&lt;p&gt;What would an observer see if they were moving at nearly the speed of light?  They would see distances &lt;strong&gt;along the direction of motion&lt;/strong&gt; appear shorter, but perpendicular to that direction distances would appear normal (ie. the same as observers in the non-moving frame).  &quot;&lt;strong&gt;Non-moving&lt;/strong&gt;&quot;, you say angrily.  &quot;All motion is relative. How can there be a non-moving frame?&quot;  Well, there is a special frame relative to the mean of all of the mass in the visible universe and in this frame the Microwave Background Radiation is uniform over all directions in the sky.  This is a natural non-moving (and non-rotating) frame for doing cosmology called the comoving frame.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now back to the moving frame.  There, one observes all nearby galaxies in the direction of motion as blueshifted and all nearby galaxies behind as redshifted.  Therefore, Hubble's Law would be a function of direction and would be pretty complicated unless one made all the appropriate relativistic corrections.  If the observers did not know relativity, and just go ahead and simply correct Hubble's Law according to measured velocities and velocity independent distance measures, they would conclude that the universe is shaped like an oblate ellipsoid (too short in 1 direction and correct in the other 2 directions, from our point of view).  But, in fact, to them it actually is that shape and they actually can travel more quickly between stars and galaxies in that 1 direction than we can (at least in their forward direction).  Of course, if they land anywhere in a typical galaxy then they will be switching to the comoving frame and see a full sized spherical universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They will also observe a Microwave Background that is quite a bit warmer in one direction and colder in the opposite direction and 2.7K in the circle between these.  If they know relativity they will understand why this is and why the density of galaxies is higher in one dimension and why Hubble's Law is so complicated.  But yes, their universe will be, in a real sense, smaller (in one dimension) than ours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxies do all have some motion in the comoving frame and this is called peculiar velocity.  The Milky Way is being pulled toward Andromeda galaxy at roughly 50 km/s and the whole Local Group is pulled toward the Virgo Cluster (the center of the Local Supercluster) at 100 - 200 km/s.  Galaxies in the Virgo Cluster have velocity dispersion of 700 km/s, so some move at about twice that.  And a  large region around is moving at ~ 500 km/s due to superclusters at large distances.  But, since these peculiar velocities tend to add up to less than 2,000 km/s, no alien race is going to be at relativistic speeds unless they are in a spaceship of their own devise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more issue.  On the time scale of a billion years or several, this moving frame will move to a location in space that is moving at a similar velocity and so they will merge with the non-moving (but expanding) frame.  So, slowly their oblate universe is transforming to a spherical one, unless they have a means to continuously accelerate.  &lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2015-08-05T15:28:26.800" LastActivityDate="2015-08-05T15:28:26.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11452" PostTypeId="2" ParentId="11449" CreationDate="2015-07-31T18:12:13.693" Score="1" Body="&lt;p&gt;I think it is supposed to be a representation of the Centaurus A/M83 Galaxy group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Centaurus_A/M83_Group&quot; rel=&quot;nofollow&quot;&gt;https://en.m.wikipedia.org/wiki/Centaurus_A/M83_Group&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are a bunch of galaxies between 12 million and 15 million light years away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big central dust lane in Cen A is very recognisable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the sequence then continues through the Virgo cluster and supercluster which are roughly in the same direction.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-07-31T18:12:13.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11454" PostTypeId="1" CreationDate="2015-08-01T10:55:30.530" Score="5" ViewCount="176" Body="&lt;p&gt;Are sodium lasers useful for Earth observing space telescopes/spy satellites?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="7982" LastEditDate="2021-02-28T11:29:09.970" LastActivityDate="2021-02-28T11:29:09.970" Title="Could spy satellites use laser guide stars (for adaptive optics)?" Tags="&lt;optics&gt;&lt;space-telescope&gt;&lt;atmospheric-effects&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11455" PostTypeId="2" ParentId="11393" CreationDate="2015-08-01T14:58:07.890" Score="3" Body="&lt;p&gt;Not within general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Except in special circumstances, such as an asymptotically flat spacetime, there is no definable total energy in general relativity. The Friedmann–Roberston–Walker family of spacetimes that describe a homogeneous and isotropic universe are not an exception, although there's a separate argument that can be made for the case of spatially compact universes towards their total energy being exactly zero, no matter how they expand or contract.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the matter distributions, gravitational potential can be useful in the weak-field limit of general relativity when the matter covers a finite region of space, but in general it's not present in general relativity. Or rather, the metric itself can be thought of as taking the role of a 'gravitational potential', but the way it does so is very different from the Newtonian theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Newtonian cosmology having formally analogous behavior to the pressureless FRW cosmology with kinetic and potential terms can be done:&#xA;$$E = \frac{1}{2}\sum_k m_k\dot{r}_k^2 - G\sum_{j\neq k}\frac{m_jm_k}{|\mathbf{r}_j-\mathbf{r}_k|} - \frac{1}{6}\Lambda\sum_k m_kr_k^2\text{.}$$&#xA;However, this is incompatible with an infinite universe, and is neither homogeneous nor isotropic (being rather spherically symmetric with a distinguished center), and therefore violates the cosmological principle. That issue &lt;em&gt;can&lt;/em&gt; be fixed by introducing a potential for the scale factor itself, but in that case interpreting it as the gravitational potential of the individual gravitating masses is problematic.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2015-08-01T14:58:07.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11456" PostTypeId="1" AcceptedAnswerId="11631" CreationDate="2015-08-01T19:51:28.517" Score="9" ViewCount="5105" Body="&lt;p&gt;I recently heard the claim that mid-July, the &quot;Star of Bethlehem&quot; formed for the first time in 2,000 years, where the &lt;a href=&quot;https://en.wikipedia.org/wiki/Star_of_Bethlehem#Planetary_conjunction&quot;&gt;Star of Bethlehem&lt;/a&gt; is a three-way conjunction between Venus, Jupiter, and Regulus. The Wikipedia page I just linked mentions one such conjunction occurring in 2 BC, and other sources (like &lt;a href=&quot;http://earthsky.org/tonight/see-venusjupiter-conjunction-in-western-sky-after-sunset&quot;&gt;EarthSky&lt;/a&gt;) mention the conjunction that happened last month, in 2015 AD. Has this three-way conjunction never happened in the intervening 2016 years?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For the purpose of this question, I'll say that it counts as a conjunction when each pair of bodies is separated by no more than 10 degrees, as suggested by this &lt;a href=&quot;http://cosmoquest.org/forum/showthread.php?127683-Venus-Jupiter-Regulus-Conjuction&quot;&gt;CosmoQuest forum thread&lt;/a&gt; on the subject.)&lt;/p&gt;&#xA;" OwnerUserId="129" LastActivityDate="2021-04-05T06:31:05.233" Title="Has the conjunction between Venus, Jupiter, and Regulus only occurred twice in 2,000 years?" Tags="&lt;jupiter&gt;&lt;venus&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11457" PostTypeId="1" CreationDate="2015-08-02T10:43:10.203" Score="2" ViewCount="314" Body="&lt;p&gt;Caught a sight of moon tonight (by chance) and came in surprise that moon looks thrice in size to its usual size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mostly called as Blue moon. Didn't succeed in getting the details and explanations of the phenomenon on Internet. Wikipedia just tells about the origin of the word &lt;em&gt;blue moon&lt;/em&gt; and particular dates at which super moon is more obvious but doesn't shed the light on the true physical causes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just wondering what could be the causes that makes the appearance of moon (near the horizon) so big. &lt;/p&gt;&#xA;" OwnerUserId="1842" LastEditorUserId="1842" LastEditDate="2015-08-23T22:38:28.903" LastActivityDate="2015-08-23T22:38:28.903" Title="What makes the moon, a super moon" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11458" PostTypeId="2" ParentId="11454" CreationDate="2015-08-02T13:49:00.380" Score="4" Body="&lt;p&gt;The guide star laser is used to adapt the optics at a specific time and location, for specific atmospheric conditions. It is not, as your question implies, used to calibrate the optics for long-term use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefor, if order to calibrate the satellite with a laser fired from Earth, the reconnaissance target would have to provide the laser. This is useless for non-cooperative targets, which I would imagine make up the bulk of the reconnaissance targets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose that one could fire a laser from the satellite itself, but this would have the obvious disadvantage of making it even more obvious when a spy satellite was overhead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, the guide star image does not penetrate the entire atmosphere, I believe that it produces an image just a few tens of kilometers up. This is useful because it is the thicker, denser and more turbulent lower air of which is the most concern to optics. Firing the laser from above would thus see the laser energy absorbed in the thinner, stiller, more predictable upper atmosphere, thus it would have must less advantage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One final reason not to use a guide star laser on a reconnaissance satellite would be power budget. These satellites are as small as possible, with as small a heat and reflectivity index as possible. The added power source, be it RTG, conventional batteries, or solar, would likely increase the reflectivity of the satellite.&lt;/p&gt;&#xA;" OwnerUserId="209" LastActivityDate="2015-08-02T13:49:00.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11459" PostTypeId="1" AcceptedAnswerId="11462" CreationDate="2015-08-02T13:56:01.730" Score="3" ViewCount="3497" Body="&lt;p&gt;I use a Celestron NexStar 130SLT Azimuthal Newtonian Reflector. I noticed that recently, due to the wet season, some tiny spots of what appears to be fungus have appeared. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having taking appropriate precautions to prevent the same in future, how can I presently get rid of these fungi colonies on my primary reflector?&lt;/p&gt;&#xA;" OwnerUserId="7854" LastEditorUserId="1378" LastEditDate="2015-08-02T18:42:07.717" LastActivityDate="2015-08-02T21:44:51.027" Title="How do I remove fungus from a telescope mirror?" Tags="&lt;telescope&gt;&lt;newtonian-telescope&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11462" PostTypeId="2" ParentId="11459" CreationDate="2015-08-02T18:32:33.360" Score="7" Body="&lt;p&gt;Several points need to be made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. Performance&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The performance of the whole optical stack is incredibly resilient w.r.t. small spots on the primary mirror. The mirror might look visually very, very dirty, but the performance of the whole instrument will remain essentially the same. Even if you chip the mirror at the edge, it shouldn't matter. Take a marker and scribble a dozen lines on the mirror - it doesn't make a large difference for performance, not greater anyway than what the 4 vanes of the secondary mirror already do (we can start a separate discussion about diffraction artifacts from spots and lines on the mirror - but the bottom line is, don't worry about it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that in mind, you should be conservative when deciding to clean the mirror. Let it accumulate some amount of dirt, the scope will be okay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Permanent damage&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, mold or other living things may chemically corrode the mirror. Again, a few spots of corrosion do not impact performance, but you don't want to allow this process to continue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dust will generally not corrode the mirror. Salt (from ocean breeze) will usually corrode it in time. With mold it's hard to say, but it's best if it's removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3. Prevention&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mold grows because the mirror is wet. The mirror fogs up usually when you bring the scope back into a warm room from outside; the mirror is cold, and water in the warm air condenses on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Procure a cheap, small fan, and point it at the rear end of the scope while it's warming up after you bring it back in. The air current will reduce water condensation on the mirror. Turn the fan off after 30 ... 60 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, keep the OTA (the big tube of the telescope) capped off while not in use. This will reduce dust deposits, and therefore mold will find less food to grow. At least cap off the top end and store the scope in a vertical position; that will reduce dust deposits greatly. The bottom end is less important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Avoid storing the scope in a place with large day/night temperature variations. When the mirror is colder than air, condensation may form. This can happen spontaneously with scopes stored in a shed outside. The mirror fogs up in the morning, when air is getting warmer but the scope is still cold. This is much less of an issue with small scopes, but it's something to keep in mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could cap off both ends while in storage, and place a bag of silica gel inside the OTA (don't drop it on the mirror). That will suck up all humidity inside the OTA, but you have to bake the silica gel in the oven every so often to keep it working effectively. You could buy silica gel online, it's very cheap. Avoid temperature variations and the gel is not necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4. Clean up procedure&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the procedure recommended by Optic Wave Labs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=9Y8xFnXFVGQ&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=9Y8xFnXFVGQ&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I make mirrors but I don't coat them myself (yet) - however, I've found that procedure to work very well on both coated and non-coated glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few minutes of initial soaking in water makes everything easier later on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After soaking it, add a few drops of detergent and rub very gently with your clean fingertips or with a soft, clean, lint-free cloth. Don't put pressure on the mirror, just try to detach the dirt from mirror in the horizontal direction and let it wash off. Don't drag dust particles across the glass, just detach them and let them go; this is why fingers are better than cloth - because you can feel large particles coming off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When done, wash off the detergent with lots and lots and lots of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using tap water for most of the cleaning procedure is absolutely fine. If you can drink the water, it's fine for the mirror too. Don't use hot water if the mirror is cold, there's a very small chance of cracking it, especially with cheap, mass produced glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you're done cleaning, rinse the mirror with distilled water. Again, a few water spots on the mirror don't matter to performance, but the mirror looks better if it's spot-free. Tap water tends to leave water spots if you don't follow up with distilled water rinsing, and most people find the spots not pleasing from an esthetic perspective (but performance is same). So just splash a bottle of distilled water on it at the end and you'll be fine. No touching after rinsing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some dirt or mold may not be removed by the usual water-based procedure. In that case, use isopropyl alcohol, the highest concentration you can find (over 90% is fine). Rub gently with your clean, non-greasy fingertips or a lint-free cloth. If it doesn't come off quickly, leave it alone. Alcohol has probably killed the mold anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acetone might be used for things that resist even alcohol. But again, you have to use your judgement: a few little spots won't affect performance, so is it worth the trouble?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, rinse with distilled water at the end, then let it dry off in an inclined position (carefully to not tip the mirror over on its face). Put a towel on the table in front of the mirror, so if it does topple over, at least it falls on something soft. Do not let it dry in a horizontal position, you'll get water spots and dust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there's some permanent damage in a few little spots, don't worry about it. The mirror will work essentially the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is recommended that you do the water-based procedure every 1 or 2 years, depending on how dirty the mirror is. If dust contamination is moderate, leave it alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5. Post-cleanup&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After everything is done, re-collimate the telescope carefully. But this should go without saying, since collimation is part of the regular maintenance of newtonian telescopes.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-08-02T21:44:51.027" LastActivityDate="2015-08-02T21:44:51.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11463" PostTypeId="1" CreationDate="2015-08-02T19:18:54.357" Score="3" ViewCount="518" Body="&lt;p&gt;Is the moon the most full only for one moment &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As stated  &lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20080721155747AAsraEM&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://m.space.com/24511-moon-myths-lunar-new-year.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or does if stay the same full for a longer period of time (maybe because, the sun is grater then it, and light spreads out)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/q/8380&quot;&gt;Full Moon Position of Sun&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7858" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-08-03T17:51:06.320" Title="How long is the moon full" Tags="&lt;the-moon&gt;&lt;moon-phases&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11465" PostTypeId="2" ParentId="11457" CreationDate="2015-08-02T20:28:16.383" Score="5" Body="&lt;p&gt;A Supermoon is when the point in the Moons orbit when it is at its closest to the Earth coincides with full or new Moon (&lt;a href=&quot;https://en.wikipedia.org/wiki/Supermoon&quot; rel=&quot;noreferrer&quot;&gt;see Wikipedia entry&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A blue Moon is (in this instance) the occurrence of a second full Moon in the same calendar month (&lt;a href=&quot;https://en.wikipedia.org/wiki/Blue_moon&quot; rel=&quot;noreferrer&quot;&gt;see Wikipedia entry&lt;/a&gt;), and is unrelated to the coincident Supermoon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the apparent size difference between a full Supermoon and a more usual full Moon is inlikely to be detectable by a casual observer. As can be seen in the side-by-side comparison graphic on the Wikipedia page reproduced below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/z3ZVg.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z3ZVg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The apparent giant Moon you report is the well known &lt;a href=&quot;https://en.wikipedia.org/wiki/Moon_illusion&quot; rel=&quot;noreferrer&quot;&gt;Moon illusion&lt;/a&gt; where the Moon appears larger when observed near the horizon compared to when seen higher in the sky. The Moon illusion is startling if you are unfamiliar with it and very much more noticeable than a Supermoon to the casual observer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While out in the morning of the 31st, walking the dog in the local recreation ground, between ~04:00 and 05:00 BST (03:00-04:00 UT) I observed the apparent size of the Moon and commented to the dog that we will probably see someone (in the media I thought) mistake the apparent size illusion I was seeing for the effect of the Supermoon. (I had also noted the same effect the day before at a similar time, but not on the 1st as we go out later at the weekend)&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-08-02T21:06:54.487" LastActivityDate="2015-08-02T21:06:54.487" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11466" PostTypeId="1" AcceptedAnswerId="12210" CreationDate="2015-08-02T22:06:54.497" Score="2" ViewCount="512" Body="&lt;p&gt;What is the equilibrium temperature that a black body will reach at the Earth's distance from Sun?&lt;/p&gt;&#xA;" OwnerUserId="2541" LastActivityDate="2015-10-21T14:26:09.840" Title="What is space temperature around Earth?" Tags="&lt;solar-system&gt;&lt;temperature&gt;&lt;radiation&gt;&lt;habitable-zone&gt;&lt;radiative-transfer&gt;" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="11467" PostTypeId="1" CreationDate="2015-08-03T00:14:26.237" Score="8" ViewCount="22141" Body="&lt;p&gt;I am new to telescopes and I'm planning to buy an astronomical telescope like the &quot;Celestron AstroMaster 114 EQ Reflector&quot;. Just curious to know whether it can be used to view objects on Earth (e.g. traffic and people miles away from tall buildings). Telescopes are used to observe stars; they must have higher power than binoculars. Right?&lt;/p&gt;&#xA;" OwnerUserId="7859" LastEditorUserId="23940" LastEditDate="2018-12-31T09:13:44.883" LastActivityDate="2021-05-24T21:14:18.057" Title="Can an astronomical telescope view objects on Earth?" Tags="&lt;telescope&gt;" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="11468" PostTypeId="1" CreationDate="2015-08-03T01:12:16.360" Score="13" ViewCount="1282" Body="&lt;p&gt;Venus is somewhat lighter then Earth, yet has a much thicker atmosphere. One would imagine that the following should be true:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;During the formation phase, all inner planets had captured as much gas as they could hold per gravitational/thermodynamic equilibrium. After all, even puny Mars managed to capture a sizable atmosphere.&lt;/li&gt;&#xA;&lt;li&gt;The rate of atmospheric escape should be much higher for Venus:&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Venus receives more heat from the sun, thus higher Jeans escape rate&lt;/li&gt;&#xA;&lt;li&gt;Venus has negligible magnetic field, so some of its atmosphere must be lost to direct &quot;blow-off&quot; by solar wind&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Yet, it is Earth that seemingly misses a great deal of atmospheric volume. So the question is: what are the current theories regarding the &quot;thinning&quot; of the Earthian atmosphere? When and why the atmospheric gases had left the planet?&lt;/p&gt;&#xA;" OwnerUserId="412" LastEditorUserId="9527" LastEditDate="2020-10-07T13:43:27.087" LastActivityDate="2020-10-07T13:43:27.087" Title="Why is the Earthian atmosphere so thin?" Tags="&lt;earth&gt;&lt;earth-like-planet&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="11469" PostTypeId="2" ParentId="11467" CreationDate="2015-08-03T02:47:25.887" Score="3" Body="&lt;p&gt;Yes.&#xA;Most consumer telescopes can be used for terrestrial viewing (or at least the ones I have used) and are often utilized for wildlife viewing or other uses for extreme telephoto requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The manual doesn't list the minimum focus distance (usually the information provided for photographic lens to describe how near they can focus), but the &lt;a href=&quot;http://www.celestron.com/c3/images/files/downloads/1218147775_astromasterinst.pdf&quot; rel=&quot;nofollow&quot;&gt;manual&lt;/a&gt; does include a short passage (pg 25) on terrestrial viewing.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Terrestrial Photography&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Your telescope makes an excellent telephoto lens for terrestrial&#xA;  (land) photography. You can take images of various scenic views,&#xA;  wildlife, nature, and just about anything. You will have to experiment&#xA;  with focusing, speeds, etc. to get the best image desired. You can&#xA;  adapt your camera per the instructions at the top of this page. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="1288" LastEditDate="2015-08-03T08:07:21.133" LastActivityDate="2015-08-03T08:07:21.133" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11471" PostTypeId="2" ParentId="11467" CreationDate="2015-08-03T09:03:56.843" Score="9" Body="&lt;p&gt;Yes, but many (not all) telescopes made for astronomy will invert the image. Presenting the image upright is important for terrestrial instruments, but it doesn't matter for astronomy. Also, it usually requires fewer optical components to make an instrument that inverts the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, terrestrial instruments are made to operate in an environment where there's plenty of ambient light (daylight), whereas many telescope designs (such as truss dobsonians) are built to operate in a dark environment and are usually hampered by excess ambient light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in a nutshell, yes, you could turn a telescope towards objects here on Earth, and it would work, but for best results and easiest operation just get some good binoculars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, &quot;power&quot; is a much misunderstood characteristic of optical instruments. It is technically true that a large dobsonian could in theory outperform any pair of binoculars you could care to try, but in practice a large optical stack is difficult to swing around and is fickle with regard to collimation, thermal equilibrium, etc. If you're willing and able to put up with the operating difficulties, with the necessity to flip the image around, etc, then I guess that indeed a large dobsonian could surpass anyone's little binoculars in a terrestrial context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But very large magnification always reduces the field of view. Go ahead and crank magnification all the way up, and it's like looking through a straw. It's hard to locate anything. Astronomers have specific methods to deal with this difficulty, but for a terrestrial observation the narrow field of view is a liability, not an advantage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a reason why various optical instruments have evolved into different shapes and designs to operate in different environments. &quot;Horses for courses&quot;, as they say.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2019-01-02T19:10:42.653" LastActivityDate="2019-01-02T19:10:42.653" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11473" PostTypeId="2" ParentId="11463" CreationDate="2015-08-03T17:38:55.407" Score="5" Body="&lt;p&gt;tl;dr: Perfectly full moon lasts, never more than, approximately 2 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Estimating an answer from the point of view of Plain Geometry:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we approximate, that the sun is at an infinite distance, then exactly half of the moon is illuminated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, how much of the Moon do I see when I look at it from Earth? (This has nothing to do with illumination of the Moon.) Estimate: Radius of moon (Rm), 1740km. Distance to the moon (Dm), 384,000km. Doodle this into a diagram that is VERY much not to scale...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HIPGI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HIPGI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;arctan(Rm / Dm) gives us 0.260 degrees. So as long as the observer is within that angle from the Moon-Sun-centerline, the portion of the Moon which is &lt;em&gt;visible&lt;/em&gt;, would be a circular disc fully illuminated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, I suspect that the variation in the Moon's distance is the biggest thing I've estimated away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question now is: combining the Moon's motion of revolution (aka orbit about the Earth) with the Earth's rotation, how long does it take an observer to perceive the moon moving half a degree (1/4 degree either side of the Moon-Sun-centerline)? Rotation of the Earth gives 15 degrees-per-hour, minus the Moon's 1/2 degree per hour... 14.5 degrees per hour apparent motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get approximately 2 minutes duration when the Moon's shifting of apparent position would only hide/show illuminated portions of it's disc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But the lunar orbit is inclined&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's only 2 minutes if the Moon is very close to the plane of the Equator. As the Moon's orbit carries it out of the plane, the time would be reduced. Beyond half a degree out of the plane, you would actually never (pure geometry) seen a fully illuminated disc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think the best answer is probably, &quot;Never more than 2 minutes.&quot;&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="7" LastEditDate="2015-08-03T17:51:06.320" LastActivityDate="2015-08-03T17:51:06.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11474" PostTypeId="1" CreationDate="2015-08-03T17:59:53.790" Score="6" ViewCount="119" Body="&lt;p&gt;I'm trying to find equations that would help me determine the amount of solar radiation hitting a certain latitude on a certain planet given the following inputs:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the degrees of latitude of the location in question&lt;/li&gt;&#xA;&lt;li&gt;this hemisphere's current season (winter or summer)&lt;/li&gt;&#xA;&lt;li&gt;size of the planet&lt;/li&gt;&#xA;&lt;li&gt;luminosity of the star(s) the planet orbits&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Without taking into consideration wind, air pressure, or any atmospheric effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally I would like to determine the average solar radiation of a given location in both the winter and summer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My end goal is to determine the average surface temperature of a given latitude on a planet using the base solar radiation and the effects of wind, air pressure, and surface ocean currents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know it is possible to do this for the entire planet generally, but I would like some way of doing it for a particular latitude and season.&lt;/p&gt;&#xA;" OwnerUserId="7511" LastActivityDate="2015-12-24T11:56:28.580" Title="Equation for solar radiation at a given latitude on a given exoplanet?" Tags="&lt;exoplanet&gt;&lt;planetary-atmosphere&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11475" PostTypeId="2" ParentId="11445" CreationDate="2015-08-03T20:09:38.623" Score="3" Body="&lt;p&gt;(I will assume that these planets orbit a star.) Planetary rings are formed by the gravitational capture of small objects, mostly ice and dust. These particles can be magnetic and in such cases would be affected by the planetary magnetic field. It is also to be pointed out that the magnetic field is not really uniform, but let us assume that it is. The presence of a magnetic field would be important for the sustenance of a ring, but the stability of the ring is an entirely different issue. If the ring does consist of magnetically sensitive materials, then large enough fluctuations in the magnetic field can in principle cause the ring to become unstable. So depending on the answer of stability, I choose to say yes, it does depend on the magnetic field. To calculate the stability would be a complex n-body problem but can be averaged over. &lt;/p&gt;&#xA;" OwnerUserId="7867" LastEditorUserId="7867" LastEditDate="2015-08-04T08:29:47.610" LastActivityDate="2015-08-04T08:29:47.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11476" PostTypeId="1" AcceptedAnswerId="11478" CreationDate="2015-08-03T21:00:21.037" Score="3" ViewCount="88" Body="&lt;p&gt;When we say that 'accuracy of gaia satellite is 10 $\mu$as at $V = 10$ magnitude in position and annual proper motion'. What does this mean? Does it change over time?&lt;/p&gt;&#xA;" OwnerDisplayName="user7330" LastEditorUserId="15415" LastEditorDisplayName="user7330" LastEditDate="2018-07-15T11:57:27.190" LastActivityDate="2018-07-15T11:57:27.190" Title="What is the accuracy of an artificial satellite at a certain magnitude?" Tags="&lt;astrophysics&gt;&lt;artificial-satellite&gt;&lt;mathematics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="11477" PostTypeId="1" AcceptedAnswerId="11485" CreationDate="2015-08-03T21:07:29.393" Score="2" ViewCount="1101" Body="&lt;p&gt;Some moon phase calculation algorithms (apparently derived from Duffet-Smith's book, &lt;a href=&quot;http://bazaar.launchpad.net/~keturn/py-moon-phase/trunk/view/head:/moon.py#L141&quot; rel=&quot;nofollow&quot;&gt;example here&lt;/a&gt;) seem to use a parameter called &quot;longitude of perigee at epoch&quot;. What exactly is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I assume this is the same as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Longitude_of_the_periapsis&quot; rel=&quot;nofollow&quot;&gt;longitude of the periapsis&lt;/a&gt;? Can I just calculate it adding the longitude of the ascending node and the argument of periapsis together?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, all the data that I find on this parameter is either related to Duffet-Smith's book or algorithms based on it. So, I have no idea on how this can be calculated for a specific epoch.&lt;/p&gt;&#xA;" OwnerUserId="7869" LastActivityDate="2020-08-24T15:52:21.013" Title="What is exactly the &quot;longitude of the perigee&quot;" Tags="&lt;orbit&gt;&lt;orbital-elements&gt;&lt;algorithm&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11478" PostTypeId="2" ParentId="11476" CreationDate="2015-08-03T21:58:11.810" Score="1" Body="&lt;p&gt;What you are referring to is the precision with which the astrometry (position) of stars will be measured with the &lt;a href=&quot;http://sci.esa.int/gaia/&quot; rel=&quot;nofollow&quot;&gt;Gaia&lt;/a&gt; satellite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sentences you refer to in &lt;a href=&quot;http://arxiv.org/abs/1406.0129&quot; rel=&quot;nofollow&quot;&gt;Perryman et al. (2014)&lt;/a&gt;  are &quot;Gaia will achieve accuracies of some 10μas (microarcsec) in positions and annual&#xA;proper motions for bright stars (V ∼ 10), degrading to around 25μas at V = 15, and to around 0.3 mas (300μas) at V = 20 (Lindegren et al. 2008).&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means is that after the 5-year Gaia mission is complete, the angular position (or coordinates) of 10th magnitude stars on the sky will be determined with a relative precision of 10 millionths of an arcsecond.  This will also be the accuracy with which the annual tangential motion (the so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Proper_motion&quot; rel=&quot;nofollow&quot;&gt;proper motion&lt;/a&gt;) of stars can be measured. i.e. 10th magnitude stars will have an uncertainty in their proper motion of $\pm 10 \mu$as/year.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-08-04T11:09:40.877" LastActivityDate="2015-08-04T11:09:40.877" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11480" PostTypeId="1" AcceptedAnswerId="11509" CreationDate="2015-08-04T06:05:14.330" Score="7" ViewCount="2875" Body="&lt;p&gt;Earth has the van Allen radiation belts. Which other planets have one? Can it be determined if distant planets in other solar systems have one? &lt;/p&gt;&#xA;" OwnerUserId="7872" LastActivityDate="2018-08-30T12:16:07.787" Title="Do all planets have a van Allen radiation belt?" Tags="&lt;planet&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11482" PostTypeId="1" CreationDate="2015-08-04T14:51:15.613" Score="5" ViewCount="385" Body="&lt;p&gt;This seemed to be the closest group for this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every few months there are a few articles asking &quot;Are we alone in the universe&quot; which focus on the Fermi Paradox, the Drake Equation, and end by talking about SETI and its lack of progress. As I understand it SETI is looking for signs of technological civilizations beyond our solar system, both in our galaxy and beyond through radio astronomy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is about what would count as a &quot;sign&quot; of a technological civilization, and is there a maximum distance beyond which such a signal would be indistinguishable from other radio emissions from stars and galaxies?&lt;/p&gt;&#xA;" OwnerUserId="7874" LastEditorUserId="7982" LastEditDate="2021-02-19T01:29:07.247" LastActivityDate="2021-02-19T01:29:07.247" Title="What types of signals is SETI (and other such efforts) looking for?" Tags="&lt;radio-astronomy&gt;&lt;drake-equation&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11485" PostTypeId="2" ParentId="11477" CreationDate="2015-08-04T17:45:51.010" Score="5" Body="&lt;p&gt;Perigee is the Earth-specific name for periapsis. People use longitude (which is a composite angle rather than an angle) because this solves the problems of circular and equatorial orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason you need to use an epoch time to specify the Moon's argument of perigee and longitude of ascending node is because the Moon's orbit about the Earth precesses. Neither the argument of perigee nor the longitude of ascending node is constant. Instead, they are functions of time (and hence, so is longitude of perigee).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of an inclined, non-circular orbit, &quot;longitude&quot; is the sum of the right ascension of the ascending node (which is measured on the fundamental plane of the reference system) and some angle (or angle-like measure such as mean anomaly) on the orbital plane. Thus longitude of periapsis (or longitude of perigee in the case of an object orbiting the Earth) is the sum of the object's right ascension of ascending node and it's argument of periapsis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The use of &quot;longitude&quot; as orbital elements are for two reasons: To be able take advantage of Hamiltonian mechanics (this was done by Delaunay, resulting in the Delaunay orbital elements), and to address problems related to orbits with very small inclinations and/or very small eccentricities (this was done by Poincaré, resulting in the Poincaré orbital elements). Anomaly and longitude are synonymous in the case of an orbit with zero inclination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it is invalid to add two angles on different planes and treat the sum as if it were an angle, it is not necessarily invalid to treat that sum as what it is, a composite angle or dogleg angle, which is exactly what the Poincaré orbital elements do. (&quot;Dogleg angles&quot; aren't angles. Analogous terms include &quot;dwarf planets&quot;, which aren't planets, and &quot;red herrings&quot;, which oftentimes are neither red nor herrings.)&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2015-08-06T17:08:11.907" LastActivityDate="2015-08-06T17:08:11.907" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11486" PostTypeId="1" AcceptedAnswerId="11488" CreationDate="2015-08-04T19:34:32.890" Score="1" ViewCount="830" Body="&lt;p&gt;I'm making a code where I calculate the sun position in a given time. But I also want to learn astronomy with it. I'm following a .pdf to make the program, and in this .pdf it says &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Julian Day (JD) is calculated using UT and the Julian Ephemeris&#xA;  Day (JDE) is calculated using TT.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it the only difference?&lt;/p&gt;&#xA;" OwnerUserId="7878" LastActivityDate="2015-08-05T06:11:37.200" Title="What's the difference between the Julian and the Julian Ephemeris Day?" Tags="&lt;time&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11488" PostTypeId="2" ParentId="11486" CreationDate="2015-08-05T06:11:37.200" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it the only difference?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By definition, yes. It's important to keep in mind that the difference between TT and UT (UT1) is not constant.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-08-05T06:11:37.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11489" PostTypeId="2" ParentId="11482" CreationDate="2015-08-05T16:49:55.970" Score="3" Body="&lt;p&gt;&lt;strong&gt;What signals do SETI receive?&lt;/strong&gt;&#xA;SETI uses &lt;a href=&quot;http://www.seti.org/faq#ata1&quot; rel=&quot;nofollow&quot;&gt;radio telescopes&lt;/a&gt; to look at large portions of the sky across a range of wavelengths. They believe that we are most likely to find alien civilisation buy looking for &lt;a href=&quot;http://www.seti.org/faq#obs8&quot; rel=&quot;nofollow&quot;&gt;microwaves.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How would we know if they were from aliens?&lt;/strong&gt;&#xA;There are radio signals from many different sources bouncing around the universe, many of which we don't quite understand yet. However, natural sources of energy tend to produce a range of wavelengths. SETI are hoping to find &quot;&lt;a href=&quot;http://www.seti.org/faq#obs3&quot; rel=&quot;nofollow&quot;&gt;narrow-band signals&lt;/a&gt;&quot;, which  are strong signals at a single wavelength. These (as far as we know) only come from devices made to produce them, so only a technologically advanced species would emit these signals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there a maximum distance for detecting these?&lt;/strong&gt;&#xA;Yes, these signals would decrease with distance. It is not so much that they become indistinguishable form natural sources, as they are a different type of signal, but as the waves travel they spread out, like dropping a pebble in a pond, so the signal is much weaker as you get further away. Also, space is not empty. If the waves are emitted from a planet in a galaxy, there will probably be lots of dust and gas (interstellar medium) in the galaxy which will absorb the radio waves, so it might be like looking for car headlights on a foggy day - we will see something vague and fuzzy but it will be much weaker if there is more matter between us and them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to find out more about SETI they have a &lt;a href=&quot;http://www.seti.org/&quot; rel=&quot;nofollow&quot;&gt;website&lt;/a&gt; with FAQs and news. If you want to get involved there is a project called &lt;a href=&quot;http://setiathome.berkeley.edu/&quot; rel=&quot;nofollow&quot;&gt;SETI@home&lt;/a&gt; which uses spare processing power on your home computer to analyse the data they are receiving and search for these narrow-band signals.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2015-08-05T16:49:55.970" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11491" PostTypeId="1" CreationDate="2015-08-06T01:33:07.370" Score="8" ViewCount="125" Body="&lt;p&gt;For years I've been fascinated with the mutually tide-locked bodies Pluto and Charon. In July 2012, The Astronomical Journal published an article &lt;a href=&quot;http://iopscience.iop.org/1538-3881/144/1/15/pdf/1538-3881_144_1_15.pdf&quot;&gt;The Orbit of Charon Is Circular&lt;/a&gt; by Buie, Tholen and Grundy. The authors thought that the eccentricity of Charon's orbit is very close to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me their opinion is somewhat speculative given the low quality of images from Hubble. Has New Horizons data verified Charon's circular orbit? Is there an online source giving more precise orbital elements from New Horizons data?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also interested in the obliquity of Pluto and Charon.&lt;/p&gt;&#xA;" OwnerUserId="1232" LastActivityDate="2016-04-09T20:27:23.897" Title="Has New Horizon's data updated Charon's orbital elements?" Tags="&lt;pluto&gt;&lt;charon&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11492" PostTypeId="2" ParentId="11482" CreationDate="2015-08-06T08:54:15.783" Score="2" Body="&lt;p&gt;Besides radio, SETI is also searching for signs of artificial technology in data from infrared (space) telescopes. The idea is that civilizations make good use of energy, and solar energy is the most abundant and easy to access. Fusion power for free. Fissile elements are mostly inaccessibly hidden deep inside planets. If a civilization absorbs a substantial part of a star's light for its industry, the natural star light would be replaced by redshifted waste heat radiation from that industry. Something we should be able to detect with today's technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jason T Wright seems to be the one man driving force behind this kind of actual search today, within the SETI and wider science community. Here are some links to a &lt;a href=&quot;http://www.seti.org/weeky-lecture/wise-search-large-extraterrestrial-civilizations-complementary-approach-traditional&quot; rel=&quot;nofollow&quot;&gt;talk&lt;/a&gt;, &lt;a href=&quot;http://astro.fnal.gov/events/Seminars/Slides/JWright_61614.pdf&quot; rel=&quot;nofollow&quot;&gt;slides&lt;/a&gt;, a &lt;a href=&quot;http://arxiv.org/pdf/1408.1134.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; and a &lt;a href=&quot;http://sites.psu.edu/astrowright/&quot; rel=&quot;nofollow&quot;&gt;blog&lt;/a&gt; by Jason. Lots of galaxies have been studied, so at least it isn't common that civilizations extract most of the energy from entire galaxies. The universe might be too young for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I fear that our focus on life and intelligence as we know it, expressed in the in my opinion naive &lt;a href=&quot;https://en.wikipedia.org/wiki/Kardashev_scale&quot; rel=&quot;nofollow&quot;&gt;Kardashev scale&lt;/a&gt; may miss that there might exist completely different kinds of phenomena out there. Life on Earth is all just one single instance. We have no diversity, no statistics, but we do know that the combinatorics of chemistry is mindboggling. But we can only do what we can do, and it would be a bit strange if there's nothing like us out there, considering how extremely ancient and successful DNA-life is on Earth, which doesn't seem to have unique circumstances. A problem might be that biology seems to use all tricks, &lt;em&gt;except&lt;/em&gt; radio.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-08-06T10:18:07.373" LastActivityDate="2015-08-06T10:18:07.373" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11493" PostTypeId="1" CreationDate="2015-08-06T17:13:21.000" Score="5" ViewCount="188" Body="&lt;p&gt;It has been reported that perhaps half of nearby sun-like stars are found in multiple systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this so common, and how do planets form in such a situation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kaitlin M. Kratter, “Sibling Rivalry Begins at Birth,” Nature, Vol. 518, 12 February 2015, p. 173&lt;/p&gt;&#xA;" OwnerUserId="7889" LastEditorUserId="7889" LastEditDate="2015-08-06T21:19:16.167" LastActivityDate="2015-09-12T08:52:41.853" Title="Why are so many sun-like stars found in multiple systems? How do planets form in such a situation?" Tags="&lt;exoplanet&gt;&lt;binary-star&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11494" PostTypeId="1" AcceptedAnswerId="11498" CreationDate="2015-08-06T17:53:28.653" Score="6" ViewCount="1448" Body="&lt;p&gt;I'm trying to get a better understanding of what things look like in &quot;outer space.&quot; The main problem I'm trying to solve right now, is how much illumination is provided by the background stars. Is it even perceivable? If we were to ignore the light contributions from our sun and local planets, how difficult would it be to see an object (within arm's reach) that was lit only by the remaining ambient light?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way this might be thought of is: how dark is the dark side of Pluto? If a person were able to stand on the surface of Pluto, would a moonless/Charonless night be totally pitch black, or would there be enough ambient light to see? How bright might that be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The example about Pluto is a little deceptive because Pluto will block a large portion of starlight. I'm thinking of the starlight that is neither blocked by nor supplemented by any of the bodies in our solar system.&lt;/p&gt;&#xA;" OwnerUserId="7890" LastEditorUserId="7890" LastEditDate="2015-08-10T15:10:29.357" LastActivityDate="2020-03-09T07:10:13.227" Title="How much illumination do the background stars provide?" Tags="&lt;star&gt;&lt;light&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11496" PostTypeId="1" AcceptedAnswerId="11499" CreationDate="2015-08-07T04:19:32.950" Score="4" ViewCount="930" Body="&lt;p&gt;I was doing some calculations to see how hard it would be to observe the speed of light and discovered an interesting correlation between the speed of light and the average distance from the earth to the sun. In my example, an object would be placed in orbit around the sun so that its  shadow would pass over the earth at the speed of light.  After doing some math, I found that an object roughly 1 km above the sun, orbiting at 2 m/s would achieve this.  I though that these numbers were a bit odd, so I took a closer look at the speed of light (299,792,458 m/s) and AU (149,597,871 km.)  I found that 299,792,458 / 149,597,871 = 2 within a .2% degree of inaccuracy.  To me, this seems too close to be chance.  Is there an established relationship here, or is this just luck?&lt;/p&gt;&#xA;" OwnerUserId="7895" LastActivityDate="2015-08-07T17:01:12.723" Title="Is the ratio between earth's distance from the sun and the speed of light just a coincidence?" Tags="&lt;orbit&gt;&lt;earth&gt;&lt;light&gt;&lt;speed&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11497" PostTypeId="1" AcceptedAnswerId="11502" CreationDate="2015-08-07T05:27:47.677" Score="1" ViewCount="220" Body="&lt;p&gt;I am trying to code up the following article &lt;a href=&quot;http://dl.acm.org/citation.cfm?id=2349512&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://dl.acm.org/citation.cfm?id=2349512&lt;/a&gt; (&lt;a href=&quot;http://www.cs.bilkent.edu.tr/~cagatay/Modeling_Daytime_and_Night_Lighting_V6.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;alternative free access version&lt;/a&gt;) to calculate the light reflected from the moon to a given point on earth at a given time. However I occasionally get a gross error when calculating the distance of the light path through the atmosphere (Figure 2. of the article.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9GxqA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9GxqA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The approximation given is to take the inverse sine of right ascension in radians, however whenever RA approaches pi, the number blows out to 8e15. Can you suggest a better method?&lt;/p&gt;&#xA;" OwnerUserId="7897" LastEditorUserId="4085" LastEditDate="2015-08-09T04:52:39.840" LastActivityDate="2015-08-09T07:58:18.583" Title="Calculate Atmospheric Extinction Using Source Altitude Angle" Tags="&lt;the-moon&gt;&lt;distances&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11498" PostTypeId="2" ParentId="11494" CreationDate="2015-08-07T06:18:18.080" Score="5" Body="&lt;p&gt;Although &lt;a href=&quot;http://adsabs.harvard.edu/full/1961ApJ...133..228R&quot; rel=&quot;noreferrer&quot;&gt;sources go way back&lt;/a&gt;, the current best estimate seems to be that of &lt;a href=&quot;http://www.oocities.org/aa_institute1/ahads_constant_refereed_paper.pdf&quot; rel=&quot;noreferrer&quot;&gt;Abdul Ahad from March 2004&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For an observer located anywhere within the Solar System - excluding the contribution of light from &#xA;  the Sun (Solar constant) - provisional integrations using stellar magnitude data sourced from astronomical catalogs places an an approximation for this constant in the range:  -6.0 &amp;lt;= &lt;strong&gt;m&lt;/strong&gt;Ahad &amp;lt;= -7.0&#xA;  having a mid-point value of &lt;strong&gt;-6.5 magnitudes&lt;/strong&gt; as a consensus within the astronomical community. This &#xA;  is the total integrated brightness of the night sky that we visually experience from Earth. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other &lt;a href=&quot;http://firstarktoalphacentauri.wikia.com/wiki/Ahad%27s_Constant&quot; rel=&quot;noreferrer&quot;&gt;sources&lt;/a&gt; for same &lt;a href=&quot;http://www.oocities.org/aa_spaceagent/ahads_sphere.htm&quot; rel=&quot;noreferrer&quot;&gt;value&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;-6.5. Brighter than Venus, but dimmer than the brightest Iridium satellite flares.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-08-07T06:18:18.080" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11499" PostTypeId="2" ParentId="11496" CreationDate="2015-08-07T07:58:27.927" Score="10" Body="&lt;p&gt;It is a co-incidence. The units are what you need to look at, in both the speed of light and the distance of the Sun the units of length are the same (m) (your ratio is $\approx 0.002 s^{-1}$ when you use metres as the unit for the distance to the Sun). The definition of the second is in a sense arbitrary and is a consequence of how we have chosen to sub-divide the day into smaller units of time (there is nothing cosmic in dividing the day into 24 units each subdivided into sixtieths which are again subdivided into sixtieths).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we had chosen to take as our basic unit of time $1/100000$ of a day rather than $1/86400$ the nice round answer vanishes (if my calculations are right the $\approx 0.002$ becomes $\approx0.0023$)&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-08-07T16:03:27.803" LastActivityDate="2015-08-07T16:03:27.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11500" PostTypeId="2" ParentId="11482" CreationDate="2015-08-07T12:49:45.913" Score="3" Body="&lt;p&gt;Other answers covered most of it but I would like to bring a couple of points up:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;As the distance between the radio emitter and the radio receiver increases, the signal will be received from a smaller patch of sky, requiring better telescope resolution (see: radio telescope effective area and solid angle), and also requiring more time to scan the sky, i.e. making everyone's life more difficult. &lt;/li&gt;&#xA;&lt;li&gt;Natural light is unpolarised and is, basically, white noise (even though there are few natural sources of polarised light). SETI hopes to find signals in small bandwidths (distinguishing them from the broad bandwidth noise) near natural lines that are familiar to intelligent life, such as between the 21 cm band that is a characteristic of water. This is purely an assumption that aliens will be smart enough to emit a radio frequency at 21 cm and we will be there in the right place to detect it.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7809" LastEditorUserId="7809" LastEditDate="2016-01-30T19:46:42.873" LastActivityDate="2016-01-30T19:46:42.873" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11501" PostTypeId="2" ParentId="11496" CreationDate="2015-08-07T17:01:12.723" Score="3" Body="&lt;p&gt;Pure coincidence. Your result of 2 is better written as 2 meters/kilometer/second. If we had six fingers on each hand we would probably use base twelve rather than base ten. A base 12 metric system would have 1728 (in our base 10) meters in a kilometer. Your coincidence would vanish in a twelve-fingered world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our Earth takes 86400 seconds to make one rotation with respect to the Sun. Your coincidence would vanish if our day was longer or shorter (but we still divided the day into 86400 seconds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our division of the day into 86400 seconds is a historical oddity that results from&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The ancient Egyptians and their infatuation with base 12. They divided day and night into 12 parts each, hence the origin of our 24 hour day.&lt;/li&gt;&#xA;&lt;li&gt;The ancient Babylonians and their infatuation with base 60. They divided the circle into 360 parts (degrees), that into 60 parts each (minutes), and that into 60 parts each again (seconds), hence the origin of our division of an hour into 60 minutes, and a minute into 60 seconds.&lt;/li&gt;&#xA;&lt;li&gt;The not-so-ancient French revolutionaries and their infatuation with base 10 failing to catch on with regard to time. They tried to metricize time and angle, but unlike the length, area, volume, and mass, metric time and angle didn't take hold.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Your coincidence would vanish if the French metric time had caught on, or if those ancients hadn't been so numerologically inspired, or if they had been inspired by a different base.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-08-07T17:01:12.723" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11502" PostTypeId="2" ParentId="11497" CreationDate="2015-08-08T15:13:42.240" Score="1" Body="&lt;p&gt;You seem to be referring to the computation of $m$ in equation (15) for the extinction as the light from the source passes through the atmosphere. The author intends to say that $m\approx \frac{1}{\sin(\alpha_s)}$, where $\alpha_s$ is the altitude of the source (in radian), so restricted to the range $0$ to $ \pi/2$. (Note this is not the inverse of the sine function as the author says but the reciprocal of the sine function)  The altitude is the angle measured from the horizon to the source, which is $=\pi/2-\theta$ where $\theta$ is as in figure 2. Note this blows up when $\alpha_s=0$ ($\theta=\pi/2$) because in the simple approximation being used the path length through the atmosphere in infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The approximation being used here is that of a flat Earth, and $m$ is the number of equivalent atmosphere heights that the light path travellels through&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-08-09T07:58:18.583" LastActivityDate="2015-08-09T07:58:18.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11503" PostTypeId="1" AcceptedAnswerId="11508" CreationDate="2015-08-08T21:45:53.483" Score="2" ViewCount="4454" Body="&lt;p&gt;Let Line A be the projection of the axis of Earth's rotation onto the plane of the eliptic.  Let Line B the line connecting the center of the Earth and center of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At what times during the year is Line A perpendicular to Line B?&lt;/p&gt;&#xA;" OwnerUserId="7907" LastActivityDate="2015-08-09T04:08:55.283" Title="Earth Rotation and Orbit Geometry" Tags="&lt;orbit&gt;&lt;the-sun&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11505" PostTypeId="1" AcceptedAnswerId="11506" CreationDate="2015-08-09T00:30:00.690" Score="17" ViewCount="68569" Body="&lt;p&gt;I thought dark and anti matter were kinda the same, but after saw a video, they mention that dark matter is not antimatter but their explanation is a little fast so I got doubts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the differences between matter, dark matter and antimatter? Are they related? How they interact each other? Where can I see an example of this interaction? and also, in terms of percentage, how much of matter, dark matter and antimatter exist in the universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7832" LastActivityDate="2016-01-28T14:21:48.973" Title="What are the differences between matter, dark matter and antimatter?" Tags="&lt;dark-matter&gt;&lt;matter&gt;&lt;antimatter&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11506" PostTypeId="2" ParentId="11505" CreationDate="2015-08-09T01:40:45.447" Score="13" Body="&lt;p&gt;Matter is the stuff you are made of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Antimatter is the same as matter in every way, looks the same, behaves the same, except its particles have electrical charges opposite to matter. E.g., our electrons are negatively charged, whereas a positron (an antimatter &quot;electron&quot;) is positively charged. The positron is the &quot;anti-particle&quot; of the electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a particle meets its anti-particle, they &quot;annihilate&quot;: the two particles disappear, and gamma photons are released carrying off their energy. For this reason, should a lump of matter touch a lump of antimatter, they would annihilate, and a giant explosion would result because of the huge energy released (E=mc^2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matter and antimatter are definitely related: same thing, but with opposite signs. Twins, but opposites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not clear why, but it seems like there isn't that much antimatter out there, more like trace amounts. Definitely not as much as regular matter as far as we can tell. This is puzzling to physicists and cosmologists, because you'd expect the Big Bang to make roughly equal amounts of matter and antimatter. Scientists agree that the paradox of &quot;excess matter&quot; will advance physics even further once it's solved.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Dark matter - we don't really know what it is. It's not even sure it's &quot;matter&quot; in a conventional sense, or related to it in any way. We just know that galaxies are rotating in such a way that indicates there's a lot more mass out there, but it is mass that we cannot see and cannot be accounted for in the usual ways. Hence the name &quot;dark&quot; (as in invisible) matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark matter doesn't seem to interact much with regular matter, except gravitationally. Right now dark matter could be passing through you and you wouldn't notice. Dark matter also does not interact with light, so you can't see it. It doesn't seem to interact much with itself either, so for this reason dark matter cannot form &quot;clumps&quot; such as planets or stars. Instead, it probably exists in a diffuse form. Bottom line, dark matter interacts pretty much only via gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape of galaxies is a proof of the existence of dark matter, and is a result of the interaction between matter and dark matter. Without dark matter, galaxies would be much less massive, and the outer parts would rotate much more slowly compared to the center. Due to dark matter, galaxies are quite massive, and they rotate almost as solid objects - the outer parts rotate approximately as fast as the central parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Estimates vary, but it seems like there's something like 5x to 6x more dark matter out there compared to regular matter.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-08-09T01:46:24.297" LastActivityDate="2015-08-09T01:46:24.297" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11507" PostTypeId="2" ParentId="11505" CreationDate="2015-08-09T02:56:14.993" Score="3" Body="&lt;p&gt;A pretty good site for quick explanations is the &lt;a href=&quot;http://www.particleadventure.org/index.html&quot; rel=&quot;nofollow&quot;&gt;Particle Adventure&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.particleadventure.org/antipreface.html&quot; rel=&quot;nofollow&quot;&gt;Anti Matter&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.particleadventure.org/dark.html&quot; rel=&quot;nofollow&quot;&gt;Dark Matter&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anti Matter is really quite simple and very similar to regular matter.  It just happens to explode violently when it touches regular matter - like a positron (Positive Electron) and an electron (negative) will touch and evaporate into a pair of gamma rays.  The Proton/anti Proton or Neutron/anti Neutron or Proton/antineutron or Neutron/antiProton (they interact because they have some quarks/anti-quarks in common).  Those reactions are more complicated, but the gist is the same.  They explode violently when they touch.  So, because matter and anti matter tend to evaporate each other, there's really no primordial anti-matter left in the universe, cause there was slightly more matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, other than the explosive interaction, Antimatter is almost exactly the same as matter, You could, in theory build a star, a planet, trees and life out of anti-matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark Matter is a lot more different.   We don't really know what it is, but it's a different kind of matter, it's transparent and it's non binding but it has mass.   Regular matter can bind together gravitationally, that's why it forms into things like stars, planets, comets, asteroids, etc.   Dark matter doesn't do that.  It loosely collects around galaxies, or, perhaps more accurately, galaxies collect inside large pockets of dark matter.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-08-09T03:47:09.990" LastActivityDate="2015-08-09T03:47:09.990" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11508" PostTypeId="2" ParentId="11503" CreationDate="2015-08-09T04:08:55.283" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qCwR7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qCwR7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The axis points in pretty much the same direction all the time, though it does wobble slightly, ranging between 22.1 degrees and 24.5 degrees every 41,000 years.   &lt;a href=&quot;https://en.wikipedia.org/wiki/Milankovitch_cycles#Axial_tilt_.28obliquity.29&quot; rel=&quot;nofollow noreferrer&quot;&gt;(Source)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perpendicular between two lines is a 2 dimensional property, so, while the axis is never &quot;straight up&quot; so to speak, there is a 90 degree angle between the axis and the line between the center of the sun and the center of the earth during  the spring and fall equinox.  Those are the only times.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-08-09T04:08:55.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11509" PostTypeId="2" ParentId="11480" CreationDate="2015-08-09T17:24:53.823" Score="9" Body="&lt;p&gt;Mars does not have a magnetic field of any strength and so does not have a van Allen belt (and this is a serious problem for potential human exploration of the planet). Similarly, neither Venus nor Mercury (nor the Moon, which is arguably part of a two-planet system with the Earth) have a van Allen belt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gaseous giants do, however, have radiation belts which are analogous to Earth's van Allen belt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spotting aurorae on distant exoplanets would be an indication of a strong magnetic field and thus of the likelihood of a radiation belt. I am not aware that such aurorae have been detected but distant aurorae have been seen around a brown dwarf - a substellar object somewhat bigger than a planet (see &lt;a href=&quot;http://www.theguardian.com/science/2015/jul/29/astronomers-find-aurora-a-million-times-brighter-than-the-northern-lights&quot; rel=&quot;noreferrer&quot;&gt;http://www.theguardian.com/science/2015/jul/29/astronomers-find-aurora-a-million-times-brighter-than-the-northern-lights&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, the radio emissions from high energy electrons trapped in planetary radiation belts might be detectable - this was how Jupiter's radiation belt was discovered around 60 years ago. Work seems to be underway to detect such radiation: &lt;a href=&quot;https://skaoffice.atlassian.net/wiki/download/attachments/22183971/PoS-exopla-AASKA14.pdf?version=1&amp;amp;modificationDate=1403003959989&amp;amp;api=v2&quot; rel=&quot;noreferrer&quot;&gt;https://skaoffice.atlassian.net/wiki/download/attachments/22183971/PoS-exopla-AASKA14.pdf?version=1&amp;amp;modificationDate=1403003959989&amp;amp;api=v2&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="823" LastEditorUserId="823" LastEditDate="2015-08-12T14:09:40.357" LastActivityDate="2015-08-12T14:09:40.357" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11510" PostTypeId="2" ParentId="914" CreationDate="2015-08-10T01:51:08.703" Score="2" Body="&lt;p&gt;This really isn't as simple as you may think, the moon does not orbit the Earth at a constant speed just like how the Earth doesn't orbit the sun at a constant speed. The phases repeat an average of every 29.53 days. The amount of time between the new moon can very.&lt;/p&gt;&#xA;" OwnerUserId="7914" LastActivityDate="2015-08-10T01:51:08.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11512" PostTypeId="1" AcceptedAnswerId="11514" CreationDate="2015-08-10T04:14:30.793" Score="2" ViewCount="357" Body="&lt;p&gt;Due to Hawking Radiation, after a very long time, a black hole will eventually disappear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But where does this &quot;evaporation&quot; go? What is this made of?&lt;/p&gt;&#xA;" OwnerUserId="7832" LastEditorUserId="2270" LastEditDate="2015-08-10T14:53:32.733" LastActivityDate="2015-08-10T15:43:25.097" Title="What particles does a black hole emit when it evaporates itself?" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11513" PostTypeId="1" AcceptedAnswerId="11515" CreationDate="2015-08-10T05:17:56.090" Score="0" ViewCount="129" Body="&lt;h1&gt;Take me to the moon!&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I know this is &lt;em&gt;literally&lt;/em&gt; rocket science but, I would like any &lt;strong&gt;sources&lt;/strong&gt; or &lt;strong&gt;formulas&lt;/strong&gt; that would help me &lt;strong&gt;plot a trip to the Moon&lt;/strong&gt;. &lt;strong&gt;I'm not sending anything, nor do I ever plan to&lt;/strong&gt;, I just want to do the next best thing by plotting the trip. So if anyone could give me any info as to how I can plot the trip, that'd be great! :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I'm perfectly aware that the moon isn't always the same distance away, that it's like hoping onto a taxi in New York City, and how it's got &lt;em&gt;a lot&lt;/em&gt; of work (I wouldn't mind doing it really), I'd just like to plot the trip. And please don't hate on me for asking, let's just be &lt;strong&gt;civil adults&lt;/strong&gt; here. :)&lt;/p&gt;&#xA;" OwnerUserId="7914" LastEditorUserId="6" LastEditDate="2019-01-29T14:48:45.177" LastActivityDate="2019-01-29T14:48:45.177" Title="Help plotting a trip to the Moon!" Tags="&lt;the-moon&gt;&lt;space-travel&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ClosedDate="2015-08-18T13:12:15.783" ContentLicense="CC BY-SA 3.0" />
  <row Id="11514" PostTypeId="2" ParentId="11512" CreationDate="2015-08-10T08:01:12.483" Score="5" Body="&lt;p&gt;Although black holes are widely believed to emit Hawking radiation, it should be stressed that it has not actually been observed (yet?). The Hawking radiation should consist of electromagnetic radiation/waves that have a near-perfect &lt;a href=&quot;https://en.wikipedia.org/wiki/Black_body&quot; rel=&quot;nofollow&quot;&gt;black body&lt;/a&gt; spectrum, which is at a temperature that is inversely proportional to the mass of the black hole - the smaller the black hole, the higher the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radiation is caused because &quot;vacuum&quot; is not empty when considered in quantum mechanics. Particle/anti=particle pairs are created for brief moments of time, before they annihilate again. Close to the event horizon of a black hole, it is possible that one particles from a &lt;a href=&quot;https://en.wikipedia.org/wiki/Virtual_particle&quot; rel=&quot;nofollow&quot;&gt;virtual particle&lt;/a&gt; pair travels within the event horizon of the black hole and cannot escape to annihilate with its anti-particle. In principle, these particle/anti-particle pairs could be any types of particle, but in practice they are more likely to be the lightest particles. The lightest charged particles are electron/positron pairs, since these need to &quot;borrow&quot; less energy from the vacuum to be created. I do not think it is essential for one of the pair to disappear, since &quot;Unruh radiation&quot;, a close relation of Hawking radiation should also be seen whenever there is acceleration with respect to the vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To stay out of the black hole event horizon the particles must be accelerating. Accelerating charged particles locally emit electromagnetic radiation, which is then gravitationally redshifted when seen by an observer a long way from the black hole. It turns out that to be in a thermal equilibrium, the radiation must have a blackbody spectrum form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperatures and amount of radiation emitted are very small for stellar-sized black holes. The temperature of a non-rotating Schwarzschild black hole is given by&#xA;$$ T = \frac{6.2\times 10^{-8}}{M}\ K,$$&#xA;where the mass $M$ is in solar maases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power emitted by a black hole as Hawking radiation is&#xA;$$ P = \frac{9\times 10^{-29}}{M^2}\ W$$&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-08-10T15:43:25.097" LastActivityDate="2015-08-10T15:43:25.097" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11515" PostTypeId="2" ParentId="11513" CreationDate="2015-08-10T14:09:12.603" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://braeunig.us/apollo/apollo11-TLI.htm&quot; rel=&quot;nofollow&quot;&gt;This website&lt;/a&gt; has a mathematical in depth analysis of the Apollo 11 translunar trajectory. It looks to be a reasonably reliable resource but I've only had a quick skim through so you should check on the sources cited.&#xA;There should be enough information there to make a computer program to calculate the orbits.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2015-08-10T14:09:12.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11517" PostTypeId="2" ParentId="11456" CreationDate="2015-08-10T20:05:33.603" Score="0" Body="&lt;p&gt;Though my computer is still crunching the numbers (and subsequently slowing everything else down... ) I can tell you the answer is no. It's up to the 1700s already (starting from 2 BC) and I've already seen multiple conjunctions in quite decent proximity of all three. I set the max to 5 degrees. I'm using a modified form of this python script.&#xA;&lt;a href=&quot;http://shallowsky.com/blog/science/astro/predicting-conjunctions.html&quot; rel=&quot;nofollow&quot;&gt;http://shallowsky.com/blog/science/astro/predicting-conjunctions.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, the script has finished. I don't have it nicely filtered down to ONLY the conjunctions of all three, so I have 3000+ lines of results. I can copy a few instances in real quick though.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Conjunction of Venus, Jupiter and Regulus lasts from 2015/6/23 to 2015/8/4.&#xA;Venus and Jupiter are closest on 2015/7/2 (0.4 deg).&#xA;Venus and Regulus are closest on 2015/7/16 (2.4 deg).&#xA;Jupiter and Regulus are closest on 2015/8/4 (1.8 deg).&#xA;&#xA;Conjunction of Venus, Regulus and Jupiter lasts from 1885/7/25 to 1885/8/11.&#xA;Venus and Regulus are closest on 1885/7/29 (1.1 deg).&#xA;Venus and Jupiter are closest on 1885/8/7 (0.4 deg).&#xA;&#xA;Conjunction of Jupiter, Regulus and Venus lasts from 1861/6/19 to 1861/8/6.&#xA;Jupiter and Regulus are closest on 1861/7/16 (0.5 deg).&#xA;Regulus and Venus are closest on 1861/7/30 (1.1 deg).&#xA;Jupiter and Venus are closest on 1861/8/2 (0.6 deg).&#xA;&#xA;Conjunction of Venus, Jupiter and Regulus lasts from 1837/7/24 to 1837/8/3.&#xA;Venus and Jupiter are closest on 1837/7/28 (0.7 deg).&#xA;Venus and Regulus are closest on 1837/7/31 (1.2 deg).&#xA;Jupiter and Regulus are closest on 1837/7/31 (3.6 deg).&#xA;&#xA;Conjunction of Venus, Regulus and Jupiter lasts from 1802/7/16 to 1802/8/1.&#xA;Venus and Regulus are closest on 1802/7/19 (1.1 deg).&#xA;Venus and Jupiter are closest on 1802/7/27 (0.5 deg).&#xA;&#xA;Conjunction of Jupiter, Regulus and Venus lasts from 1778/6/12 to 1778/8/2.&#xA;Jupiter and Regulus are closest on 1778/7/11 (0.5 deg).&#xA;Regulus and Venus are closest on 1778/7/19 (1.2 deg).&#xA;Jupiter and Venus are closest on 1778/7/21 (0.6 deg).&#xA;&#xA;Conjunction of Venus, Jupiter and Regulus lasts from 1754/7/12 to 1754/7/31.&#xA;Venus and Jupiter are closest on 1754/7/16 (0.8 deg).&#xA;Venus and Regulus are closest on 1754/7/21 (1.2 deg).&#xA;Jupiter and Regulus are closest on 1754/7/31 (2.5 deg).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You should probably verify these finds with a program like Stellerium.&#xA;Note that I set the view point to be in the middle east. Roughly where Persia used to be.&lt;/p&gt;&#xA;" OwnerUserId="7588" LastActivityDate="2015-08-10T20:05:33.603" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11518" PostTypeId="1" AcceptedAnswerId="11521" CreationDate="2015-08-10T23:40:03.900" Score="2" ViewCount="153" Body="&lt;p&gt;How do we know that other planetary systems aren't composed of antimatter? If they never touch matter, then they don't annihilate. All their observable physics (spectra) are the same as matter, right?&lt;/p&gt;&#xA;" OwnerUserId="7768" LastEditorUserId="6" LastEditDate="2016-05-23T17:54:34.987" LastActivityDate="2016-05-23T17:54:34.987" Title="How do we know that other planetary systems aren't composed of antimatter?" Tags="&lt;star&gt;&lt;planet&gt;&lt;antimatter&gt;&lt;planetary-systems&gt;" AnswerCount="3" CommentCount="0" ClosedDate="2015-08-15T22:42:52.743" ContentLicense="CC BY-SA 3.0" />
  <row Id="11519" PostTypeId="2" ParentId="11518" CreationDate="2015-08-11T05:23:44.273" Score="0" Body="&lt;p&gt;Assuming the parent star of such a system emits a solar wind similar to our sun's, then those antimatter particles would collide at that system's heliopause with regular-matter cosmic ray particles. I would think such a matter-antimatter reaction would be visible across much of the galaxy.&lt;/p&gt;&#xA;" OwnerUserId="7108" LastActivityDate="2015-08-11T05:23:44.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11520" PostTypeId="1" CreationDate="2015-08-11T05:28:04.000" Score="4" ViewCount="488" Body="&lt;p&gt;I do understand that galaxies were formed in the famous Big Bang. However, I also understand that if such an explosion occurs in empty space it will never form distinct galaxies, it should more or less evenly fill the whole empty space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of these might clot together to form galaxies but I cannot agree that between the void of galaxies there are no other bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="6950" LastEditorUserId="2153" LastEditDate="2015-08-15T23:59:57.893" LastActivityDate="2015-08-15T23:59:57.893" Title="Are there heavenly bodies between galaxies?" Tags="&lt;galaxy&gt;&lt;big-bang-theory&gt;&lt;star-formation&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11521" PostTypeId="2" ParentId="11518" CreationDate="2015-08-11T07:13:28.493" Score="2" Body="&lt;p&gt;There are two major points about this, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, and ultimately most importantly, is that if we have matter here, and they have antimatter over there, then somewhere in between we must have a region that transitions from matter to antimatter.  Even if this region is located in the intergalactic medium, where densities are typically very low, collisions between matter and antimatter would happen often enough that there would be a tell-tale radiation signature from this region.  To date, no confirmed detections of such a region have been made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the second point, it is not actually true that matter and antimatter have the same observable physics, though the observable differences are hard to detect even within our own laboratories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electromagnetism (classical or quantum), gravity, and the strong force all obey &lt;a href=&quot;https://en.wikipedia.org/wiki/C-symmetry&quot; rel=&quot;nofollow noreferrer&quot;&gt;C-symmetry&lt;/a&gt;, and so you would not see any observable differences between matter and antimatter with these forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, C-symmetry is not preserved in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Weak_interaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;weak interaction&lt;/a&gt;/weak force.  It is also known that &lt;a href=&quot;https://en.wikipedia.org/wiki/Parity_(physics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;parity symmetry&lt;/a&gt; can be violated in weak interactions.  Parity symmetry, loosely, means reflecting all particles in a mirror, and violation of it means that something has &quot;handedness&quot;: you can tell your left hand from a right hand, and the mirror makes your left look like a right.  Remarkably, there is reason to believe that the weak force also violates the combined &lt;a href=&quot;https://en.wikipedia.org/wiki/CP_violation&quot; rel=&quot;nofollow noreferrer&quot;&gt;CP symmetry&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the weak force introduces this sort of handedness into particle physics.  It does so with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Neutrino&quot; rel=&quot;nofollow noreferrer&quot;&gt;neutrinos and anti-neutrinos&lt;/a&gt; in particular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this handedness of neutrinos is something that could, in principle, let us detect that a mass is composed of antimatter.  Beta decay is a result of the weak force and emits neutrinos or anti-neutrinos, depending on whether the original atom is matter or anti-matter.  So theoretically we might tell a planet is antimatter if it has some radioactive material undergoing beta decays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Except that will actually be pretty impossible.  The scale of even an entire Jupiter's worth of radioactive mass would just be too small a source of neutrino emissions.  Current neutrino detectors confirm on the order of a handful of neutrinos that probably originated from outside of our solar system &lt;em&gt;every decade&lt;/em&gt;.  And those aren't really looking for neutrinos from any particular area; they'll look for any neutrino coming in from anywhere in the universe (well, where you are on the earth matters a bit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So our current technology basically isn't good enough to detect these weak sector asymmetries from sources outside of our solar system.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, assuming some improvements, a lot of time, and/or one heck of a lot of luck, perhaps our best shot at confirming a large source of antimatter comes from supernovae.  Supernovae are very, very bright (they will outshine the entire galaxy they occur in), and therefore very noticeable.  They also emit tremendous numbers of neutrinos very quickly.  So if we chance to notice a supernova, and if we luck into our neutrino detectors having enough neutrino detections at the right time to correlate to that supernova, then if somehow that was an anti-matter star we might be able to tell due to detecting the difference in the neutrinos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that's really unlikely.  We're substantially more likely to detect the energy signature of annihilation events at the matter/anti-matter interface, as mentioned in the first point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10756/how-can-we-tell-the-difference-between-matter-and-antimatter-by-observation-in-s&quot;&gt;this Q&amp;amp;A&lt;/a&gt;, which proposes using polarized light as a means of detection.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-08-11T07:18:51.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11522" PostTypeId="2" ParentId="11518" CreationDate="2015-08-11T07:21:10.197" Score="2" Body="&lt;p&gt;You would/should be able to detect the annihilation of electron-positron pairs at the boundary between normal-matter and anti-matter space. This has a distinctive energy and so could be identified unambiguously. For an example of the detection of electron-positron annihilation see &lt;a href=&quot;http://www.nasa.gov/centers/goddard/news/topstory/2007/antimatter_binary.html&quot; rel=&quot;nofollow&quot;&gt;this NASA news item&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-08-11T07:21:10.197" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11523" PostTypeId="2" ParentId="11520" CreationDate="2015-08-11T07:34:20.803" Score="9" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The Big-Bang was not an explosion in empty space.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Inter-galactic space is not empty, there is an &lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/I/Intergalactic+Medium&quot;&gt;inter-galactic medium&lt;/a&gt;, &lt;a href=&quot;http://www.natureworldnews.com/articles/1833/20130509/blobs-intergalactic-hydrogen-gas-observed-first-time-hold-secrets-universe.htm&quot;&gt;gas clouds&lt;/a&gt; and material ejected from galaxies, including &lt;a href=&quot;https://en.wikipedia.org/wiki/Intergalactic_star&quot;&gt;stars&lt;/a&gt; and possibly &lt;a href=&quot;http://www.skyandtelescope.com/astronomy-news/intergalactic-globulars/&quot;&gt;globular clusters&lt;/a&gt;, by various mechanisms...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-08-11T07:34:20.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11524" PostTypeId="2" ParentId="11480" CreationDate="2015-08-11T07:59:08.800" Score="1" Body="&lt;p&gt;Yes, according to this &lt;a href=&quot;http://news.discovery.com/space/alien-life-exoplanets/monstrous-aurora-detected-beyond-our-solar-system-150729.htm&quot; rel=&quot;nofollow&quot;&gt;recent news&lt;/a&gt; an aurora has been found on the &lt;a href=&quot;https://en.wikipedia.org/wiki/LSR_J1835%2B3259&quot; rel=&quot;nofollow&quot;&gt;brown dwarf&lt;/a&gt; LSR J1835+3259 which is 18½ light years from here. I suppose that an aurora which is a million times more powerful than those on Earth comes together with some kind of van Allen Belts, but real astronomers maybe have more imagination. An M class brown dwarf is maybe not the typical &quot;planet&quot;, but it's more like Jupiter than the Sun. (All planets certainly don't have it, the headline is confusing).&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-08-11T07:59:08.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11525" PostTypeId="1" AcceptedAnswerId="11526" CreationDate="2015-08-11T09:17:07.217" Score="2" ViewCount="245" Body="&lt;p&gt;I have recently read &lt;a href=&quot;http://www.space.com/4271-huge-hole-universe.html&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; and I am finding it a little hard to believe that there would be a place that is without matter. So, if there is a galaxy at the other side of the void, we would be unable to detect it. I am still very much a rookie at this, so please correct me if I'm wrong.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;So could this article be telling the truth? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If so, what is the science/reasoning behind this?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What if say, a black hole were to appear in such a void, what would happen?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7926" LastEditorUserId="7926" LastEditDate="2015-08-11T11:33:19.753" LastActivityDate="2015-08-11T12:22:16.600" Title="Is it possible to have a part of space devoid of matter?" Tags="&lt;black-hole&gt;&lt;intergalactic-space&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11526" PostTypeId="2" ParentId="11525" CreationDate="2015-08-11T12:03:28.977" Score="3" Body="&lt;p&gt;OK, so I don't think the article is claiming that it is a place &quot;without matter&quot;, but that it is a large region of space where the density of matter is lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://arxiv.org/abs/0704.0908&quot; rel=&quot;nofollow&quot;&gt;original research&lt;/a&gt;, published in 2007, suggests that a lack of radio sources seen in a particular direction coincides with a &quot;cold-spot&quot; that is seen in the cosmic microwave background by WMAP (which has since been confirmed by Planck). They calculate that &lt;em&gt;if&lt;/em&gt; the void were completely empty it would need to be about 280 Mpc across and at redshifts $z\leq 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxies certainly could be seen on the other size of this void. Many of the galaxies in their survey must have considerably higher redshifts and are more distant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cold spot &lt;em&gt;could&lt;/em&gt; be caused by such a void, because light is redshifted as it travels through gravitational fields. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Sachs%E2%80%93Wolfe_effect&quot; rel=&quot;nofollow&quot;&gt;Sachs-Wolfe&lt;/a&gt; effect, as it is known, causes large scale perturbations in the cosmic microwave background as light travels from the distant past, when the CMD was generated, to us in the present day. It is argued that a large void that is comparatively empty of both visible and dark matter, but which has dark energy. The way it works is that photons works against gravity to enter a void, loses energy, but then gain back some of that energy as it leaves the void. However, the expansion of the universe means that in the time taken to cross the void the amount that is regained is not as much as was lost in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What if a black hole were to appear&quot;? Not clear what you mean by this. Black holes are black, you wouldn't necessarily see them, however if the &quot;void&quot; or lack of radio galaxies was compensated for by an equivalent amount of mass that was for some reason invisible then it would not explain the CMB cold spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea of a &quot;supervoid&quot; causing the CMB cold spot has also been supported by other surveys of extragalactic sources (&lt;a href=&quot;http://arxiv.org/abs/1405.1566&quot; rel=&quot;nofollow&quot;&gt;Szapudi et al. 2015&lt;/a&gt;), but you are correct that the size of this void is difficult to understand in terms of current cosmological assumptions.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-08-11T12:22:16.600" LastActivityDate="2015-08-11T12:22:16.600" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11527" PostTypeId="1" AcceptedAnswerId="11536" CreationDate="2015-08-11T14:36:55.470" Score="2" ViewCount="104" Body="&lt;p&gt;Could it be possible for Earth-like rogue planet to maintain large bodies of liquid water &lt;em&gt;on the surface&lt;/em&gt; without a sun by relying mostly on tidal forces of a moon (or multiple moons).&#xA;How big/far/fast would the moon have to be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about a binary planet instead of planet and a moon?&lt;/p&gt;&#xA;" OwnerUserId="7929" LastEditorUserId="7929" LastEditDate="2015-08-12T05:46:13.290" LastActivityDate="2015-08-12T08:56:18.733" Title="Configuration required to tidally heat Earth-like rogue planet" Tags="&lt;tidal-forces&gt;&lt;heat&gt;&lt;rogue-planet&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11528" PostTypeId="1" AcceptedAnswerId="11529" CreationDate="2015-08-11T22:49:14.253" Score="3" ViewCount="166" Body="&lt;p&gt;How does the idea that the universe is dying (trumpeted as some big new revelation the past few days on several new sites I keep seeing pop up on google news) square with the idea that the universe is ever expanding?&lt;/p&gt;&#xA;" OwnerUserId="7932" LastActivityDate="2015-08-12T22:02:29.153" Title="The universe is dying vs the universe is ever expanding" Tags="&lt;universe&gt;&lt;space&gt;&lt;cosmological-inflation&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11529" PostTypeId="2" ParentId="11528" CreationDate="2015-08-11T23:48:14.027" Score="4" Body="&lt;p&gt;Both are true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe is &quot;dying&quot; in the sense that stars eventually run out of hydrogen, and there aren't infinite amounts of hydrogen in galaxies to replace old stars. This will take a very long time, but eventually all stars will burn out and new stars will stop forming. This is what the study recently in the news is saying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe is ever expanding in the sense that distances between galaxies which are not bound together gravitationally is expanding all the time. If you look at very distant galaxies, they are all receding into the distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combine these two facts and the result is that the universe will become more and more &quot;rarefied&quot; (distances between galaxies will keep expanding), and also all existing stars in galaxies will burn out over trillions of years.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2015-08-11T23:48:14.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11530" PostTypeId="1" CreationDate="2015-08-12T00:43:12.237" Score="6" ViewCount="513" Body="&lt;p&gt;How did we get 13.2 billion light years from the faraway galaxies being discovered with Hubble?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we all started near each other at the Big Bang, and we all travelled slower than the speed of light, and the universe is about 13.9 billion years old, how could we possibly be 13.2 billion light years away from recently discovered galaxies?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know &quot;inflation&quot; was supposedly superluminal, but I bet no one thinks that matter dispersed several billion light years apart during that operation.&lt;/p&gt;&#xA;" OwnerUserId="7933" LastEditorUserId="2153" LastEditDate="2015-08-15T23:58:00.713" LastActivityDate="2015-12-09T16:52:55.283" Title="How can we be 13.2 billion light years from another galaxy?" Tags="&lt;fundamental-astronomy&gt;&lt;big-bang-theory&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11531" PostTypeId="1" CreationDate="2015-08-12T01:54:53.353" Score="-6" ViewCount="277" Body="&lt;p&gt;In northern hemisphere, we have GBT, Arecibo, Parkes, Jodrell and VLA, while&#xA;in southern hemisphere, only Australia has big radio telescopes.&lt;br/&gt;&#xA;ALMA is in Chile, but it is in sub-mm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is not appropriate to put big radio telescopes on the plateaus of Chile?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="2365" LastEditDate="2016-05-15T06:34:38.073" LastActivityDate="2016-05-15T06:34:38.073" Title="Why does the Southern hemisphere seem to be short of radio telescopes?" Tags="&lt;telescope&gt;&lt;radio&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2015-08-13T13:35:38.430" ContentLicense="CC BY-SA 3.0" />
  <row Id="11532" PostTypeId="2" ParentId="11531" CreationDate="2015-08-12T02:40:03.927" Score="2" Body="&lt;p&gt;There are radio telescopes at many sites in the southern hemisphere.&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_radio_telescopes#South_America&quot; rel=&quot;nofollow&quot;&gt;Here's a list&lt;/a&gt;.&#xA;Chile has a big one going in at &lt;a href=&quot;http://at%20Llano%20de%20Chajnantor&quot;&gt;Llano de Chajnantor&lt;/a&gt; and several already operational. See second link for details.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-08-12T02:40:03.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11533" PostTypeId="2" ParentId="11143" CreationDate="2015-08-12T03:06:44.850" Score="0" Body="&lt;p&gt;If you put a large enough mass of earth-like rock together, you might get lithium fusion.  Stars fuse what little lithium they have at masses just under the mass needed to become a red dwarf (&quot;stars&quot; that only burn lithium and not hydrogen are large mass brown dwarf stars).  But the lithium concentration in a &quot;rock&quot; star (hur, hur.  Rock star) is certainly much higher than in a regular brown dwarf.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Lithium_burning&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Lithium_burning&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7934" LastActivityDate="2015-08-12T03:06:44.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11534" PostTypeId="2" ParentId="11493" CreationDate="2015-08-12T03:33:06.850" Score="0" Body="&lt;p&gt;Alpha Centauri, which is a binary star, is likely to have planets (they may have been detected, but not everyone is convinced).  The most likely scenario for planets to form stably would be if the two stars were widely separated.  For instance, a pair consisting of a sun-like star and a small red dwarf at a distance of 50 AU apart would allow plenty of room around the sun-like star for planet formation.  In fact, if there were a 90 Jupiter-mass red dwarf orbiting our sun at a distance of 50 AU, it would have less gravitation pull on us than Jupiter does.&lt;/p&gt;&#xA;" OwnerUserId="7935" LastActivityDate="2015-08-12T03:33:06.850" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11535" PostTypeId="2" ParentId="8482" CreationDate="2015-08-12T03:49:40.450" Score="-1" Body="&lt;p&gt;If you are orbiting a 1-sun-mass black hole at 1 AU, gravity-wise it would be just like it is for us orbiting the sun.  Gravity is gravity and mass is mass.  The planets would have to have been captured after formation of the black hole or moved in from distant orbits.  However, the accretion disk would be likely to be emitting lots of X-rays and not anywhere near enough visible light, so life on that planet would be pretty grim.&lt;/p&gt;&#xA;" OwnerUserId="7936" LastActivityDate="2015-08-12T03:49:40.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11536" PostTypeId="2" ParentId="11527" CreationDate="2015-08-12T05:43:16.717" Score="1" Body="&lt;p&gt;Do you need tidal heat?  What about core heat?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ice is a fair insulator.  Put enough ice on top and the heat flux from the core can keep the bottom of the oceans liquid, no tides required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need the water on the &lt;em&gt;surface&lt;/em&gt;, I don't think it's possible. Even a huge amount of interior heat (whether from tides, radiation, or primordial) wouldn't be sufficient to hold the surface liquid.  The Earth possibly had periods where there was no surface water (snowball earth), and that was with the advantages of an atmosphere and solar input. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lack of a sun is going to make having an Earth-like atmosphere difficult.  In an interstellar environment, any thin atmosphere will freeze out.  Without an atmospheric blanket, the surface temperature plummets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think you could have an ocean, but not with surface water.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more thing.  Assuming the current heat production of the earth, and its size, I calculate that something around 5.6km of ice is sufficient insulation.  Water below that depth with average heat flow would be liquid (ignoring pressure effects and only assuming temperature and heat flow issues).  That's a lot of ice, but not implausible.  &lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-08-12T08:56:18.733" LastActivityDate="2015-08-12T08:56:18.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11538" PostTypeId="2" ParentId="11520" CreationDate="2015-08-12T07:28:17.523" Score="0" Body="&lt;p&gt;Certainly &quot;empty&quot; space between galaxies has some things in it.   Space has properties in and of itself, such as dark energy and virtual particles appearing in pairs and then disappearing, but ignoring the virtual and the not very well understood, empty space is full of photons, more specifically, &lt;a href=&quot;http://Cosmic%20background%20radiation&quot; rel=&quot;nofollow&quot;&gt;Cosmic Background Radiation&lt;/a&gt; and all kinds of wavelengths of light from galaxies as well as cosmic rays and a boatload of neutrinos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the matter that came out of the big bang was gravitationally drawn into galaxies, but likely not all of it, so there likely is some primordial matter between galaxies too, just not very much of it.   (precisely how much is a bit over my pay grade).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as Conrad Turner points out, 3 or more body gravitational interactions can from time to time, kick a planet or star out of the galaxy but gaining sufficient velocity to escape the dark matter halo completely is probably quite rare, outside of galaxy on galaxy collisions, but rogue stars/planets outside of galaxies probably happens from time to time.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-08-12T07:28:17.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11539" PostTypeId="1" CreationDate="2015-08-12T11:01:11.547" Score="1" ViewCount="2741" Body="&lt;p&gt;The planets range hugely in sizes, Jupiter is more than 300 times as massive as Earth. But the sizes of the large moons seem to be uncorrelated with the sizes of the planets they orbit. Why doesn't Jupiter have a giant moon with a mass hundreds times larger than the Earth's Moon's?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among gas giants, Saturn at a third of Jupiter's mass has the largest of all moons and Triton way out at 30 AU Neptune is the 7th largest of all moons. The theories of the origin of the moons seem to vary much too, without explaining any upper size limit. The Moon was formed by an impact. Triton and maybe Titan were captured. Other large moons were formed along with their planet. Is this upper size limit on moons imaginary, coincidental or is there some reason for it to be this way? I of course mean a hypothetical expected upper size limit for exomoons in general, not only statistics about the Solar System moons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wonderful illustration here is from Emily Lakdawalla at &lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/2015/07141338-the-not-planets.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Planetary Society&lt;/a&gt;, it is now my desktop background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XOWdJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XOWdJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="2618" LastEditDate="2017-03-07T01:06:29.570" LastActivityDate="2017-03-07T01:06:29.570" Title="Is there some upper limit in the moon size distribution?" Tags="&lt;natural-satellites&gt;&lt;formation&gt;&lt;size&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11540" PostTypeId="2" ParentId="11531" CreationDate="2015-08-12T11:55:50.947" Score="5" Body="&lt;p&gt;ALMA is the main example of radio-telescope located in Chile. And, what an example! the most powerful radio-telescope in the world working at millimetre and sub-millimetre frequencies. Although there are bigger single dish radio-telescopes in other places, ALMA outperforms them in general. A bigger radio-telescope has a higher sensitivity because it collects more &quot;light&quot; (radio waves) but its field of view is very limited. A telescope like ALMA, composed of smaller dishes, can survey a bigger area of the sky than a single dish telescope. A good resolution is achieved using &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_interferometer&quot;&gt;interferometry&lt;/a&gt; and the sensitivity problem is compensated by the use of many dishes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed out in other answers there are also other interesting radio-telescopes in Chile like, for example, APEX.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why there are mainly millimetre and sub-millimetre radio-telescopes in Chile? Because mm and submm wavelengths are highly absorbed by the water in the atmosphere. To get rid of this problem you have to go to very high altitudes with very low content of water vapour. This make the high plains of Chile an exceptional place to build this kind of radio-telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why there are not many mid and low-frequency radio telescopes in Chile? I can only guess, but these are some reasons that I can think about:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mid and low-frequency radio-telescopes do not require exceptional atmospheric conditions. In most cases it is possible to observe through the clouds. There is no point in locating these telescopes in a remote place in Chile when you can do that closer to your place for a lower price. I think this is the main reason.&lt;/li&gt;&#xA;&lt;li&gt;The astronomical community in Chile has a strong expertise in other wavelengths like optical, infra-red or mm and submm as mentioned. There is so much exceptional science to do in those areas that there was no need to push or focus in low and mid radio frequencies (at least not as much as to build their own radio-telescopes).&lt;/li&gt;&#xA;&lt;li&gt;Many current radio-telescopes are improvements made on old ones like the VLA. You do not need to build new dishes but focus in improving the receivers, the data processing, etc.  &lt;/li&gt;&#xA;&lt;li&gt;Future radio-telescopes like &lt;a href=&quot;https://www.skatelescope.org/&quot;&gt;SKA&lt;/a&gt; will extend over a vast amount of area. They need antennas located at very big distances from West to East which made Chile alone an unsuitable option. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7924" LastActivityDate="2015-08-12T11:55:50.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11541" PostTypeId="1" AcceptedAnswerId="11543" CreationDate="2015-08-12T14:16:26.250" Score="1" ViewCount="105" Body="&lt;p&gt;According to the eclipse calculator I use (Alcyone), a partial eclipse 0.314 magnitude occurred August 1, 45 A.D. in Rome starting at 8:35am local time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would seem to correspond to an eclipse which was predicted to fall on the birthday of the emperor Claudius in the consulship of Marcus Vinicius and Statilius Corvinus (45 AD), as described in Dio Cassius Book 6, Section 26. The birthday of Claudius was Kalends Augustus (the first of August).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the case, it would seem that there is an exact correspondence between the Gregorian and Julian calendar going back to 45 AD. However, I would expect that small differences in leap years and other irregularities would make this exact match on the same day unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is the eclipse of 1140 at London which the Saxon Chronicle says occurred on 13 Kalends April (March 20th) and Alcyone says occurred March 20, again a perfect match. Yet another example is the eclipse of 809 which passed over the Faroe islands and the Saxon Chronicle says occurred 17 Kalends Augustus (16 July) on the &lt;em&gt;second day of the week&lt;/em&gt;. Now, of course, 16 July 809 is a Thursday in the Gregorian calendar, not the second day of the week, so something would seem to be fishy somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is Alcyone/NASA making some kind of adjustment so that these pre-Gregorian dates match up?&lt;/p&gt;&#xA;" OwnerUserId="2800" LastActivityDate="2015-08-12T15:54:09.290" Title="How could eclipse happen on the same day in a different calendar? (45 AD)" Tags="&lt;solar-eclipse&gt;&lt;time&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11542" PostTypeId="1" AcceptedAnswerId="11547" CreationDate="2015-08-12T14:36:55.790" Score="8" ViewCount="608" Body="&lt;p&gt;So there is this &lt;a href=&quot;http://cdn.spacetelescope.org/archives/images/screen/heic1502a.jpg&quot; rel=&quot;noreferrer&quot;&gt;huge picture&lt;/a&gt; by Hubble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Better yet, with the &lt;a href=&quot;http://www.spacetelescope.org/images/heic1502a/zoomable/&quot; rel=&quot;noreferrer&quot;&gt;zoom tool&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if you zoom to the maximum level, you can see a seemingly uniformly distributed dots of orange and blue. Are those actually stars and maybe galaxies behind them and not just camera noise?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore I'm trying to get a grasp on how large portion of the sky this photo represents.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At 3.4, the apparent magnitude of the Andromeda Galaxy is one of the brightest of any Messier objects, making it visible to the naked eye on moonless nights even when viewed from areas with moderate light pollution. Although it appears more than six times as wide as the full Moon when photographed through a larger telescope, only the brighter central region is visible to the naked eye or when viewed using binoculars or a small telescope.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's say this is about the half of the diameter of the galaxy so this photo is about ~3 Moons wide, is that correct?&lt;/p&gt;&#xA;" OwnerUserId="7942" LastEditorUserId="4085" LastEditDate="2015-08-13T10:17:42.617" LastActivityDate="2015-08-13T20:43:28.860" Title="Are those noise or stars?" Tags="&lt;galaxy&gt;&lt;photography&gt;&lt;hubble-telescope&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11543" PostTypeId="2" ParentId="11541" CreationDate="2015-08-12T15:54:09.290" Score="2" Body="&lt;p&gt;From the documentation for your program, &lt;a href=&quot;http://www.alcyone.de/plsv/documentation/overview.html&quot; rel=&quot;nofollow&quot;&gt;http://www.alcyone.de/plsv/documentation/overview.html&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dates are in the Julian Calendar through 4 October 1582, and then in the Gregorian Calendar, which begins on 15 October 1582. (Compute for 1582, and you will see that 1. oct to 1. nov. is short.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-08-12T15:54:09.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11544" PostTypeId="2" ParentId="11520" CreationDate="2015-08-12T18:33:28.310" Score="1" Body="&lt;p&gt;There are heavenly bodies between the galaxies. In a simple way a group of stars is known as star cluster. But there are stars which are not in the star clusters. Similarly a group of planets and around a star makes up a star system(solar system for our sun) but there are planets (can't call them planets but do are small heavenly bodies) not  a part of any solar system. A galaxy is a group of heavenly bodies. But there may be heavenly bodies not a part of galaxy.&lt;/p&gt;&#xA;" OwnerUserId="7946" LastActivityDate="2015-08-12T18:33:28.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11547" PostTypeId="2" ParentId="11542" CreationDate="2015-08-13T01:04:42.883" Score="5" Body="&lt;p&gt;This is most likely not noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photographic noise is generally going to be limited to roughly single pixels. This is because much of it is generated at the individual pixel level (often a single color, so not even an entire pixel) on the sensor. There is also some relationship to this and overheating - I'm not familiar with the cooling systems of the Hubble, but presumably it's sophisticated enough to prevent serious noise from overheating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When examining the photo, you can see that these are not individual pixels but rounded seemingly point-source light. It's hard to say whether this is an accurate representation of their actual angular diameter, as they may have acquired some of that rounding by diffraction. Regardless, this suggests that these are not artifacts from the sensor but are actual light sources. For a long exposure photograph, this also means that they are a persistent light source. The colors that they are composed of also fits the colors we would expect to see from stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless there has been some very heavy editing done with the intention of representing noise as stars, I think it's fairly safe to say that these are all stars or other star-like light sources.&lt;/p&gt;&#xA;" OwnerUserId="1288" LastActivityDate="2015-08-13T01:04:42.883" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11548" PostTypeId="2" ParentId="11539" CreationDate="2015-08-13T06:33:44.760" Score="2" Body="&lt;p&gt;Small point, but Ganymede is slightly larger than Titan as moons are measured by their solid surface, though with atmosphere, Titan appears slightly bigger.   The drawing is probably accurate, but slightly misleading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth adding that Titan is out-gassing so it may have, quite some time ago, been larger than Ganymede.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://purelyfacts.com/question/1/which-is-bigger-titan-or-ganymede?DDA=10&amp;amp;DDB=9&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to wikipedia, the larger, close to Jupiter and Saturn moons are unlikely to have been captured moons.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Formation_and_evolution_of_the_Solar_System#Moons&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Formation_and_evolution_of_the_Solar_System#Moons&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Jupiter and Saturn have several large moons, such as Io, Europa,&#xA;  Ganymede and Titan, which may have originated from discs around each&#xA;  giant planet in much the same way that the planets formed from the&#xA;  disc around the Sun.[76] This origin is indicated by the large sizes&#xA;  of the moons and their proximity to the planet. These attributes are&#xA;  impossible to achieve via capture, while the gaseous nature of the&#xA;  primaries also make formation from collision debris unlikely. The&#xA;  outer moons of the giant planets tend to be small and have eccentric&#xA;  orbits with arbitrary inclinations. These are the characteristics&#xA;  expected of captured bodies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the 3 listed methods of moon formation are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Co-formation&#xA;Impact&#xA;Capture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest co-formation ratio is Titan to Saturn, about 1 to 4,000 in terms of mass.   Ganymede to Jupiter, less than 1 to 12,000.   The wiki article points out that ratios of 100 to 1 are likely giant impacts and you're unlikely to see that kind of ratio in co formation, so a conservative estimate, the largest co-formation you mgiht see might be around a few hundred to 1.   If you figure the largest planets are about 12-12.5 Jupiters (more than that, they become brown dwarfs, though a cool brown dwarf orbiting a sun might get called a planet too at greater than 13 Jupiter masses), but lets say, 12 Jupiters.  That's about 4,000 earth masses.    If we look at a co-formation ratio of a few hundred to perhaps over a thousand to 1 as about the maxiumum, there's a possible ballpark answer of a handul of earth masses might be about as big as a co-formation moon is likely to get in a perfect formation situation, And those would probably be pretty rare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Giant impacts are different.  Pluto-Charon could be an impact generated dwarf planet/moon system and Charon is 1/8th the mass of Pluto.    How likely that kind of impact is to happen on a super earth is hard to say, but theoretically a super-earth with a giant impact could have a large moon too - a bigger earth with a bigger impact and bigger moon, but there's problems here.  Giant planets are likely to form close to the sun and the orbital sphere of influence gets smaller closer to the sun.   Also, giant planets might tend to form thick gaseous atmospheres, retaining all their hydrogen and helium, so too big, it might begin to resemble a gas giant, which aren't as good at creating debris when impacted, so there's a few limiting factors here and we might be limited to, certainly several times larger than our moon, but perhaps not as large as co-formation moons, further from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, moon capture.    While there's no examples of this in our solar system, I see no reason why a large planet couldn't capture another planet, though you then have to determine whether it's a co-planet orbit, which I think a captured planet should be considered, or whether the captured object is a moon. - personally I think a captured planet that orbits a planet would be a co-planet system, so I don't like the capture as a method for getting as large a moon as possible, though I suppose it could be looked at either way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know almost nothing about moons in other solar systems (exomoons).  The observations are just barely possible with current technology, but very difficult.  That's likely going to change in the next few years, but for now we know virtually nothing in other solar systems, so, any present predictions on the largest possible moons would have to be modeled or estimated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few articles on exomoons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astrobio.net/news-exclusive/new-exomoon-hunting-technique-could-find-solar-system-like-moons/&quot; rel=&quot;nofollow&quot;&gt;http://www.astrobio.net/news-exclusive/new-exomoon-hunting-technique-could-find-solar-system-like-moons/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://discovermagazine.com/2012/jul-aug/06-hunting-moons-outside-the-solar-system&quot; rel=&quot;nofollow&quot;&gt;http://discovermagazine.com/2012/jul-aug/06-hunting-moons-outside-the-solar-system&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.space.com/25438-exomoon-around-alien-planet-discovery.html&quot; rel=&quot;nofollow&quot;&gt;http://www.space.com/25438-exomoon-around-alien-planet-discovery.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://phys.org/news/2014-10-exomoons-abundant-sources-habitability.html&quot; rel=&quot;nofollow&quot;&gt;http://phys.org/news/2014-10-exomoons-abundant-sources-habitability.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://blogs.scientificamerican.com/life-unbounded/has-an-exomoon-been-found2/&quot; rel=&quot;nofollow&quot;&gt;http://blogs.scientificamerican.com/life-unbounded/has-an-exomoon-been-found2/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-08-13T06:33:44.760" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11549" PostTypeId="2" ParentId="2581" CreationDate="2015-08-13T07:54:06.303" Score="3" Body="&lt;p&gt;One of the absolutely fundamental differences between the LMC and SMC, and why they have become among the most studied of astronomical objects, is their metallicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average metallicity of interstellar gas in the LMC is about half that of the Sun, whereas the average metallicity in the SMC is only a fifth that of the Sun.&#xA;Hence the clouds act as two different, relatively nearby, analogues with which to study what happens to lower metallicity (usually much more distant) galaxies in the early universe, or how metallicity affects things like star formation.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-08-13T15:10:34.287" LastActivityDate="2015-08-13T15:10:34.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11550" PostTypeId="2" ParentId="11542" CreationDate="2015-08-13T13:50:03.430" Score="3" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_magnitude&quot; rel=&quot;nofollow&quot;&gt;this table&lt;/a&gt;, Hubble is good down to around magnitude 31.5. What with digital stacking of exposures, that number has got to reflect a point at which detector noise is equal to incoming signal.&#xA;At the distance of M31, 778000pc, that brightness corresponds to an &lt;a href=&quot;https://en.wikipedia.org/wiki/Absolute_magnitude&quot; rel=&quot;nofollow&quot;&gt;absolute magnitude&lt;/a&gt; of M = 31.5 -5((log778000)-1) = 7.&#xA;That's about as bright as &lt;a href=&quot;http://www.astro.wisc.edu/~dolan/constellations/extra/nearest.html&quot; rel=&quot;nofollow&quot;&gt;Epsilon Indus&lt;/a&gt;, a K3 class star, &lt;a href=&quot;https://en.wikipedia.org/wiki/K-type_main-sequence_star&quot; rel=&quot;nofollow&quot;&gt;0.45 to 0.8 times the mass of the Sun&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Hubble is seeing that deep, it's a little surprising how &lt;em&gt;few&lt;/em&gt; stars it picked up in that image of our neighboring galaxy.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="2031" LastEditDate="2015-08-13T20:43:28.860" LastActivityDate="2015-08-13T20:43:28.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11552" PostTypeId="1" CreationDate="2015-08-14T00:59:45.827" Score="4" ViewCount="311" Body="&lt;p&gt;The Moon's orbit is more eccentric, 0.0549, than most &lt;a href=&quot;http://www.astronomynotes.com/tables/tablesb.htm&quot; rel=&quot;nofollow&quot;&gt;planets&lt;/a&gt;. I can understand that planets get eccentric by disturbing each other like under the late heavy bombardment. And likewise for multiple moon systems. But shouldn't the debris after the &lt;a href=&quot;https://www.youtube.com/watch?v=Fwl_JBQtH9o&quot; rel=&quot;nofollow&quot;&gt;collision&lt;/a&gt; with &lt;a href=&quot;https://en.wikipedia.org/wiki/Theia_(planet)&quot; rel=&quot;nofollow&quot;&gt;Theia&lt;/a&gt; have been circulirized? Is its orbit being changed by the planets, at every opposition with Venus?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-09-18T00:36:16.727" Title="How did the moon's orbit become eccentric?" Tags="&lt;the-moon&gt;&lt;formation&gt;&lt;eccentric-orbit&gt;&lt;theia&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11553" PostTypeId="1" AcceptedAnswerId="11557" CreationDate="2015-08-14T04:58:29.570" Score="4" ViewCount="717" Body="&lt;p&gt;In a solar system or a planet that has many moons, why do their orbiting objects all orbit in the same direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible for two orbiting masses to orbit in the opposite direction of each other?&lt;/p&gt;&#xA;" OwnerUserId="7954" LastActivityDate="2015-08-16T19:33:44.293" Title="Why do objects all orbit in the same direction?" Tags="&lt;orbit&gt;&lt;gravity&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11554" PostTypeId="1" AcceptedAnswerId="11556" CreationDate="2015-08-14T05:15:25.910" Score="2" ViewCount="8994" Body="&lt;p&gt;If an astronaut went into space with a sealed container of oxygen and opened it up in space and let all the oxygen go, what would happen to it? would it all just dissipate into some other particle or matter or would it just stay there until something consumed it&lt;/p&gt;&#xA;" OwnerUserId="7954" LastActivityDate="2017-07-25T00:35:14.677" Title="If oxygen were pumped into space where would it go?" Tags="&lt;matter&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2017-07-28T15:03:06.720" ContentLicense="CC BY-SA 3.0" />
  <row Id="11555" PostTypeId="2" ParentId="11243" CreationDate="2015-08-14T05:18:44.543" Score="5" Body="&lt;p&gt;Earth's orbital eccentricity varies over time from being nearly circular (low eccentricity of 0.0034) and mildly elliptical (high eccentricity of 0.058). It takes roughly 100,000 years for Earth to undergo a full cycle. In periods of high eccentricity, radiation exposure on Earth can accordingly fluctuate more wildly between periods of perihelion and aphelion. Those fluctuations are likewise far milder in times of low eccentricity. Currently, the Earth's orbital eccentricity is at about 0.0167, which means its orbit is closer to being at its most circular.&lt;/p&gt;&#xA;" OwnerUserId="7954" LastActivityDate="2015-08-14T05:18:44.543" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11556" PostTypeId="2" ParentId="11554" CreationDate="2015-08-14T06:04:19.017" Score="3" Body="&lt;p&gt;Gases tries to occupy all the volume of the container in which they are encased in equilibrated pressure. This happens because the gas molecules are always colliding and kicking each other in a 3D high velocity brownian motion. The result is that the gas molecules occupies all the container's volume homogeneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if the sealed container is opened up in space, the oxygen would quickly leak out, because the molecules near the vacuum edge would be instantaneosuly kicked out by the gas innermost molecules. This would happen rather quickly and in a chained manner, but is dependent on the size of the open hole. Macroscopically, the result would be that all the gas would be running out in a form of a wind/jet out of the open hole into outer space. This would only stop when all the molecules becomes too far apart from each other to have any further intermolecular interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the oxygen (or any other gas) would be dissipated into outer space. The gas molecules would initially be spread out by the intermolecular collisions and then would go outwardly due to the momentum acquired from each individual gas molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When freed in outer space, each molecule would eventually reach something else to interact, so they could be attracted and retained by some gravity well (most likely from a planet or moon), or interact with some magnetosphere (if near a planet) or interact with some other molecule wandering around, or with a cosmic ray or be catched by the solar wind particles (or any of the equivalent if this happens near some other star).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this happens in intergalactic space where there is no body with appreciable gravity or magnetic field nearby, no stellar wind and the cosmic rays are very sparse and rare and the very few that exists are randomly directed, then the molecules would just spread out in virtually linear trajectories and travel the intergalactic space lonely, quietly and virtually undisturbed for some billions or trillions of years. In their journey, each molecule would only very occasionally interact with some other molecule (likely to be hydrogen) or some other lost particle wandering around.&lt;/p&gt;&#xA;" OwnerUserId="609" LastEditorUserId="609" LastEditDate="2015-08-14T06:50:55.447" LastActivityDate="2015-08-14T06:50:55.447" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11557" PostTypeId="2" ParentId="11553" CreationDate="2015-08-14T07:02:30.437" Score="3" Body="&lt;p&gt;This video provides a visual answer: &lt;a href=&quot;https://youtu.be/MTY1Kje0yLg?t=3m24s&quot; rel=&quot;nofollow&quot;&gt;https://youtu.be/MTY1Kje0yLg?t=3m24s&lt;/a&gt;&#xA;objects orbiting in different directions are more likely to collide with each other. By the process of elimination, eventually one direction of orbit dominates.&lt;/p&gt;&#xA;" OwnerUserId="7955" LastActivityDate="2015-08-14T07:02:30.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11558" PostTypeId="1" AcceptedAnswerId="11559" CreationDate="2015-08-14T10:27:15.573" Score="1" ViewCount="1087" Body="&lt;p&gt;As the title states, If earth were the size of a marble would it be smoother than a marble?&lt;/p&gt;&#xA;" OwnerUserId="7954" LastActivityDate="2015-08-14T10:56:18.500" Title="If earth were the size of a marble would it be smoother than a marble?" Tags="&lt;earth&gt;&lt;size&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2015-08-18T00:59:23.333" ContentLicense="CC BY-SA 3.0" />
  <row Id="11559" PostTypeId="2" ParentId="11558" CreationDate="2015-08-14T10:56:18.500" Score="3" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://what-if.xkcd.com/46/&quot; rel=&quot;nofollow&quot;&gt;This What-If&lt;/a&gt; page describes the case for 'smoother than a bowling ball&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;These scans (along with various measurements of ball roughness&lt;a href=&quot;https://what-if.xkcd.com/46/&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt; tell us that a high-end bowling ball is quite smooth. If blown up to the scale of the Earth, the ridges and bumps[2] would be between 10 and 200 meters high, and the peaks would be between one and three kilometers apart:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;By Earth standards, this is quite smooth; our highest mountains are 40 times higher.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And it is simple to continue the scaling process to show that the same holds true for marbles. (Although I have not found a good source of data from people who scan marbles...) &lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2015-08-14T10:56:18.500" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11560" PostTypeId="2" ParentId="11552" CreationDate="2015-08-14T12:52:16.983" Score="1" Body="&lt;p&gt;First, your claim that the moon is more eccentric isn't entirely true (lets ignore Pluto):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://nssdc.gsfc.nasa.gov/planetary/factsheet/&quot; rel=&quot;nofollow&quot;&gt;(Source)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury .205&lt;br&gt;&#xA;Mars .094&lt;br&gt;&#xA;Saturn .057&lt;br&gt;&#xA;Moon .055&lt;br&gt;&#xA;Jupiter .049&lt;br&gt;&#xA;Uranus .046&lt;br&gt;&#xA;Earth .017&lt;br&gt;&#xA;Neptune .011&lt;br&gt;&#xA;Venus .007&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it's really not abnormal, though it might be unusually eccentric for a non captured moon, but that's may be due to it being as close to the sun as it is.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, our solar system might not be exactly normal.  It's fairly common for solar systems to have a gas giant close to the sun &lt;a href=&quot;http://www.universetoday.com/113153/is-our-solar-system-weird/&quot; rel=&quot;nofollow&quot;&gt;(explained here)&lt;/a&gt;.  And there are some very odd solar systems - same link.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="2502" LastEditDate="2015-08-19T00:35:37.580" LastActivityDate="2015-08-19T00:35:37.580" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11562" PostTypeId="2" ParentId="2392" CreationDate="2015-08-15T03:12:02.573" Score="2" Body="&lt;p&gt;Before about 1966, Mercury was thought to be tide-locked, almost half always sunlit and another near-half always dark – as most moons, including ours, are tide-locked to their primaries, and for the same reason.  The difference in the strength of the primary's gravity between the inner and outer ‘poles’ creates a force tending to pull those points away from the center of the satellite, along the line joining it to the primary.  If the satellite is ellipsoidal rather than spherical, the tide will tend to align the ellipsoid's long axis to the primary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But Mercury's orbit is so eccentric that the strength of the solar tide varies by a ratio of 4:7 (if I've computed correctly).  The rotation rate nearly matches the revolution rate at perihelion, when tide is strongest and Mercury is moving fastest; if the match were perfect (if the orbital eccentricity were a bit less), the sun's apparent path would have cusps rather than little loops.  Presumably the imperfection is because the tidal effect does not vanish away from perihelion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The loops have nothing to do with axial tilt; Envite was probably thinking of the analemma.&lt;/p&gt;&#xA;" OwnerUserId="7959" LastEditorUserId="7959" LastEditDate="2015-08-19T20:43:20.200" LastActivityDate="2015-08-19T20:43:20.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11563" PostTypeId="1" CreationDate="2015-08-15T03:25:57.203" Score="1" ViewCount="74" Body="&lt;p&gt;Mercury rotates at such a rate that the Sun appears to stand nearly still at perihelion, when the tide is strongest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a mechanism that tends to adjust the orbital eccentricity to improve the match – minimize the tidal torque (integrated over time), or some such?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, might there be other stable combinations of eccentricity and rotation coupling, other than the obvious 1:1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I'll rephrase, since most of you seem to think my question was something elementary like “how does Mercury rotate?” or “what the heck is tidal locking anyway?”.  I suppose it's not a bad heuristic to assume that the newcomer is completely ignorant, but I wish that assumption didn't override a literal reading of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each planet's orbital eccentricity varies over time.  Mercury's is now suspiciously near to what it would need to be for the sun to stand exactly still at perihelion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's conceivable that in the dawn of time Mercury orbited at some arbitrary eccentricity and rotated at roughly 3/2, then resonance brought the rotation rate to exactly 3/2, and then something adjusted the eccentricity so that Mercury's long axis – the line of the lumps on which the tidal torque acts – tracks the sun more closely around perihelion.  (I wouldn't expect a perfect match at perihelion, because though tide is strongest then it doesn't vanish at other times!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that adjustment effect does happen, how does it work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if that happened for 3/2, could it also happen for other resonances, say 2/1 or 5/4?  For an odd denominator (other than 1) does the effect, whatever it is, behave differently (unless Mercury has a weird shape)?&lt;/p&gt;&#xA;" OwnerUserId="7959" LastEditorUserId="7959" LastEditDate="2015-08-16T01:16:56.470" LastActivityDate="2016-11-03T00:32:40.940" Title="attractors of Mercury's spin-orbit relation" Tags="&lt;mercury&gt;&lt;eccentric-orbit&gt;&lt;tidal-forces&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11565" PostTypeId="1" CreationDate="2015-08-15T11:59:42.530" Score="3" ViewCount="54" Body="&lt;p&gt;The E-mode polarization power spectrum of the Cosmic Microwave Background displays the same acoustic peaks that can be seen in the (more famed) temperature power spectrum. However, they are out of phase with respect to each other. A number of sources give answers that all come down to the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The main &lt;code&gt;EE' power spectrum has peaks that are out of phase with those in the&lt;/code&gt;TT' spectrum, because the polarization anisotropies are sourced by the fluid velocity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I fail to understand this. Can someone elaborate on this?&lt;/p&gt;&#xA;" OwnerUserId="6965" LastActivityDate="2015-08-15T11:59:42.530" Title="Why is the E-mode polarization spectrum out of phase with the Temperature spectrum?" Tags="&lt;cosmology&gt;&lt;cmb&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11566" PostTypeId="2" ParentId="11563" CreationDate="2015-08-15T18:27:08.117" Score="1" Body="&lt;p&gt;Over enough time there is a tendency of all orbiting objects to fall into resonance with the object they're orbiting.   1:1 being is most common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1:1 usually called Tidal Locking, is quite common. The Moon to the Earth.   Pluto and Charon to each other.  All four of the Galilean moons are tidally locked to Jupiter, likely others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury's 3:2 ratio is more unusual but not necessarily uncommon, especially for a planet with an eccentric orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_locking#Rotation.E2.80.93orbit_resonance&quot; rel=&quot;nofollow&quot;&gt;See Here&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An eccentric orbit likely helps create a spin-orbit resonance other than 1:1 and resonance other than 3:2 should certainly be possible (the Wiki article implies that) though there are no known examples that I know of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a really crazy orbital spin - check this out:  &lt;a href=&quot;http://www.businessinsider.com/pluto-moons-weird-behavior-2015-6&quot; rel=&quot;nofollow&quot;&gt;http://www.businessinsider.com/pluto-moons-weird-behavior-2015-6&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-08-15T18:27:08.117" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11567" PostTypeId="1" CreationDate="2015-08-16T03:28:05.430" Score="1" ViewCount="236" Body="&lt;p&gt;We know that two celestial bodies rotate around a center of mass. How does this concept work in case of solar systems where there are several planets versus one/two stars? (Even if the center of mass lies within a star, the center of mass for every planet would be different.) How is it that the rotation of celestial bodies does not become unsteady?&lt;a href=&quot;https://i.stack.imgur.com/FNKxP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FNKxP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7964" LastEditorUserId="2270" LastEditDate="2015-08-17T21:29:33.647" LastActivityDate="2015-08-17T21:29:33.647" Title="How does 'centre of mass' concept work?" Tags="&lt;eccentric-orbit&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11568" PostTypeId="1" AcceptedAnswerId="11569" CreationDate="2015-08-16T01:17:53.327" Score="4" ViewCount="127" Body="&lt;p&gt;I know that meteor showers have a &quot;peak&quot; observation time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does the distribution in time look like? Is one side of the peak steeper/shallower or longer/shorter than the other side, and if so, &lt;em&gt;why&lt;/em&gt;? &lt;/p&gt;&#xA;" OwnerUserId="7965" OwnerDisplayName="jvriesem" LastEditorUserId="2153" LastEditDate="2015-08-16T17:45:54.403" LastActivityDate="2015-08-29T22:37:35.780" Title="What's the distribution curve for meteor sightings?" Tags="&lt;observational-astronomy&gt;&lt;meteor&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11569" PostTypeId="2" ParentId="11568" CreationDate="2015-08-16T10:31:05.717" Score="4" Body="&lt;p&gt;The meteor sightings in a meteor shower can be approximated to a normal distribution. In fact, &lt;a href=&quot;https://books.google.co.in/books?id=Fj_wCAAAQBAJ&amp;amp;pg=PA307&amp;amp;lpg=PA307&amp;amp;dq=normal+distribution+meteor+shower&amp;amp;source=bl&amp;amp;ots=B34RlVR7ff&amp;amp;sig=TziF9nStVVvORDYDqfV9fWLyEUM&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0CDIQ6AEwA2oVChMI_ui08K6txwIVFE6OCh23yACT#v=onepage&amp;amp;q=normal%20distribution%20meteor%20shower&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;research papers&lt;/a&gt; use a normal distribution to model meteor showers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact characteristics of the shower will depend on the location, local weather conditions etc. The number of sightings and the number of actual meteors impacting the earth's atmospheric conditions are different as it is the local conditions that determine the number of sightings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The International Meteor Organisation maintains data on the &lt;a href=&quot;http://www.imo.net/zhr&quot; rel=&quot;nofollow&quot;&gt;number of meteor sightings&lt;/a&gt; which shows that the sightings can be approximated to a normal distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this data has to be used carefully because these refer to the reported sightings and the errors are significant.&lt;/p&gt;&#xA;" OwnerUserId="7970" OwnerDisplayName="aeroalias" LastEditorUserId="2674" LastEditDate="2015-08-29T22:37:35.780" LastActivityDate="2015-08-29T22:37:35.780" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11570" PostTypeId="2" ParentId="11553" CreationDate="2015-08-16T19:33:44.293" Score="1" Body="&lt;p&gt;In different systems, the cloud that creates it starts with one direction of angular momentum (spin). This is because of the initial conditions as the gas cloud collapsed in on itself to create a star and planets around it. &#xA;However, some orbits will move in the opposite direction but will normally be destroyed due to collisions with other orbits in the dominant orbit direction. In Galaxies though there will always be a few exceptions&lt;/p&gt;&#xA;" OwnerUserId="7967" LastActivityDate="2015-08-16T19:33:44.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11572" PostTypeId="1" AcceptedAnswerId="11576" CreationDate="2015-08-17T07:03:04.560" Score="1" ViewCount="313" Body="&lt;p&gt;I became curious about the maximum mass in a star's accretion disc while watching an episode of &lt;a href=&quot;https://en.wikipedia.org/wiki/Relics_(Star_Trek:_The_Next_Generation)&quot; rel=&quot;nofollow&quot;&gt;Star Trek&lt;/a&gt; involving a &lt;a href=&quot;https://en.wikipedia.org/wiki/Dyson_sphere&quot; rel=&quot;nofollow&quot;&gt;Dyson Sphere&lt;/a&gt;. I wondered if some maximum amount of stellar material would limit natural and artificial structures around any system of one or more stars. What affects these limits? Is there a logarithmic/linear/exponential correlation between stellar mass and maximum accretion mass? &lt;/p&gt;&#xA;" OwnerUserId="7973" LastActivityDate="2015-08-17T15:44:09.787" Title="What is the mass limit in a stellar accretion disc?" Tags="&lt;solar-system&gt;&lt;star-formation&gt;&lt;accretion-discs&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11573" PostTypeId="1" CreationDate="2015-08-17T08:46:47.660" Score="1" ViewCount="1288" Body="&lt;p&gt;In &lt;a href=&quot;https://en.wikipedia.org/wiki/Hindu_astrology&quot; rel=&quot;nofollow&quot;&gt;Hindu astrology&lt;/a&gt;, for creating horoscopes (birth charts), I am looking for calculations of &lt;a href=&quot;https://en.wikipedia.org/wiki/Hindu_astrology#Grahas_.E2.80.93_planets&quot; rel=&quot;nofollow&quot;&gt;planetary&lt;/a&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Hindu_astrology#R.C4.81.C5.9Bi_.E2.80.93_zodiacal_signs&quot; rel=&quot;nofollow&quot;&gt;positions&lt;/a&gt;. So, how could I calculate planetary positions according to given date, time and place? &lt;/p&gt;&#xA;" OwnerUserId="1769" LastEditorUserId="2270" LastEditDate="2015-08-17T21:29:14.707" LastActivityDate="2015-08-18T08:50:33.853" Title="Looking for planetary position calculation" Tags="&lt;solar-system&gt;&lt;planet&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11574" PostTypeId="2" ParentId="11567" CreationDate="2015-08-17T13:18:16.227" Score="0" Body="&lt;p&gt;You are missing a step in the understanding of this concept. &quot;Two celestial bodies rotate around a centre of mass&quot; in a system which only has two bodies. In larger systems all objects will rotate about the centre of mass of the whole system. &#xA;So each planet in our solar system and the sun will all be rotating around this common centre of mass, called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Barycenter&quot; rel=&quot;nofollow&quot;&gt;barycentre&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2015-08-17T13:18:16.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11575" PostTypeId="2" ParentId="11567" CreationDate="2015-08-17T13:26:45.480" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;We know that two celestial bodies rotate around a center of mass.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;We&quot; do not know that. Rhetorical question: Pick up an apple and drop it at high noon. Does it fall towards the solar system barycenter? The answer is of course &quot;No.&quot; It falls away from the barycenter, towards the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll steal a graph from an answer I wrote to a question at the sister site, &lt;a href=&quot;http://physics.stackexchange.com&quot;&gt;physics.stackexhange.com&lt;/a&gt;, &lt;a href=&quot;https://physics.stackexchange.com/a/188817/52112&quot;&gt;Do the planets really orbit the Sun?&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BjKVY.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the above graph, it appears that it's much better to say that Venus orbits the Sun (red curve) rather than the solar system barycenter (black curve). So why do people consistently use a barycentric frame rather than a heliocentric frame? The answer is simple: The barycentric frame is the frame in which the equations of motion undertake their simplest form. This is particularly so when one wants to incorporate some of general relativity into the equations of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our solar system is well-behaved. Other star systems aren't quite as well-behaved. For an example of a hypothetical system that is extremely far from well-behaved, I suggest you read &lt;a href=&quot;http://www.ams.org/notices/199505/saari-2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Saari and Xia, &quot;Off to infinity in finite time,&quot; Notices of the AMS&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-08-17T13:26:45.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11576" PostTypeId="2" ParentId="11572" CreationDate="2015-08-17T15:32:34.037" Score="3" Body="&lt;p&gt;The accretion disc is formed by material in orbital motion against a central body, which can be a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size, mass and other characteristics are usually &lt;a href=&quot;http://www.scholarpedia.org/article/Accretion_discs&quot; rel=&quot;nofollow&quot;&gt;determined&lt;/a&gt; by the central object, in this case the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the protoplanetary accretion discs are the largest ones (with the largest mass) and as the age on the central star increases, the average size decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because as the age of the system increases, more mass is drawn into the system and less is left out in the disc itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the central star crosses the &lt;a href=&quot;https://en.wikipedia.org/wiki/Chandrasekhar_limit&quot; rel=&quot;nofollow&quot;&gt;Chandrasekhar limit&lt;/a&gt;, it will become unstable and become a supernova and once it passes the &lt;a href=&quot;https://en.wikipedia.org/wiki/Tolman%E2%80%93Oppenheimer%E2%80%93Volkoff_limit&quot; rel=&quot;nofollow&quot;&gt;Tolman–Oppenheimer–Volkoff limit&lt;/a&gt; it will become a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physics of black hole accretion disc is entirely different and is possible to derive a &lt;a href=&quot;http://iopscience.iop.org/0004-637X/712/2/1129/&quot; rel=&quot;nofollow&quot;&gt;relation&lt;/a&gt; between the black hole mass and size of the accretion disc. The relation appears logarithmic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any natural or artificial structure around a star would feel the gravity of the star and consequently, it is the mass of the star that determines amount of the counteracting force (rotation, radiation pressure etc.) required to maintain stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, as the stellar mass increases, the rotation speed must increase if the orbit is maintained at the present position or the orbit must become larger. In case of stellar radiation pressure is used to counteract gravity, as the mass (and gravity) increases, the system becomes unstable (as the ratio of radiation pressure to gravity is constant).  &lt;/p&gt;&#xA;" OwnerUserId="7970" LastEditorUserId="7970" LastEditDate="2015-08-17T15:44:09.787" LastActivityDate="2015-08-17T15:44:09.787" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11577" PostTypeId="2" ParentId="3663" CreationDate="2015-08-17T19:00:10.423" Score="7" Body="&lt;p&gt;In astronomy, there is no formal definition of the threshold between gas and dust. Gas can be monoatomic, diatomic, or molecular (or &lt;a href=&quot;https://en.wikipedia.org/wiki/Photon_gas&quot; rel=&quot;nofollow noreferrer&quot;&gt;made of photons&lt;/a&gt;, in principle). Molecules can be very large, and in principle, dust particles are just very large molecules. I've seen various authors use various definitions, ranging from $\sim100$ to $\sim1000$ atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not to say that there isn't a distinct difference between molecules and dust. They have very different properties, but the transition between them is just not perfectly well-defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADDED 13.6.2018: I'm currently attending a conference on cosmic dust, and 200 astronomers weren't able to answer the question of the threshold between large molecules and small dust grains. One way to define a distinction, however, is to look at the associated spectral lines: Molecules emit/absorb and specific wavelengths, whereas dust can emit/absorb over a broader region. But there isn't a specific number of atoms — for instance, the fullerene C&lt;sub&gt;540&lt;/sub&gt; is a very large carbon molecule, but if you reshuffle its 540 C atoms into amorphous carbon, it is considered a dust grain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gas, molecules, and dust can all be hot or cold, but if it becomes too hot, larger particles are destroyed in collisions. So while a &lt;a href=&quot;https://en.wikipedia.org/wiki/Molecular_cloud&quot; rel=&quot;nofollow noreferrer&quot;&gt;molecular cloud&lt;/a&gt; typically is very cold and consists of both gas and dust, dust tends to be destroyed (though not completely) in the &lt;a href=&quot;https://en.wikipedia.org/wiki/H_II_region&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\mathrm{H\,II}$ regions&lt;/a&gt; around hot stars through collisions with other grains, sputtering due to collisions with ions, sublimation or evaporation, or even explosions due to ultraviolet radiation (see e.g. &lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/bib_query?1976Ap&amp;amp;SS..39....9G&quot; rel=&quot;nofollow noreferrer&quot;&gt;Greenberg 1976&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your final question, I haven't heard the term &quot;gas&quot; used for dust particles, but metals$^\dagger$ and molecules can both be referred to as gas. For instance, &lt;a href=&quot;http://astronomy.nmsu.edu/cwc/Research/MgIIreview/mgii-over.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\mathrm{Mg\,II}$ gas&lt;/a&gt; is routinely used to detect distant galaxies, and molecular clouds contain $\mathrm{H}_2$ and $\mathrm{CO}$ gas.&#xA;In the interstellar medium, roughly 2/3 of the metals are in the gas phase, while 1/3 is in dust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^{^\dagger}$&lt;sup&gt;&quot;Metals&quot; in the astronomical sense, i.e. all other elements than hydrogen and helium.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-06-13T07:59:05.503" LastActivityDate="2018-06-13T07:59:05.503" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="11578" PostTypeId="1" AcceptedAnswerId="11590" CreationDate="2015-08-17T20:09:54.767" Score="-1" ViewCount="394" Body="&lt;p&gt;As we all know, There is no sound in space, because sound waves don't travel in a vacuum.&#xA;But even if Big Bang did sound, what was the main source of it's cause?&lt;/p&gt;&#xA;" OwnerUserId="7978" LastActivityDate="2015-08-19T10:29:14.503" Title="Did Big Bang sound as loud as we think?" Tags="&lt;big-bang-theory&gt;&lt;space&gt;&lt;wavelength&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11579" PostTypeId="1" CreationDate="2015-08-17T20:25:41.610" Score="2" ViewCount="178" Body="&lt;p&gt;Is it possible that it is just our observable part of the universe that is expanding, in the time that we exist, and other parts are both expanding and contracting at different rates and times?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would light from a shrinking or slower/faster-expanding part of the universe reach us?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could that be an alternative to the big-bang theory, in that the red-shift is a temporary situation local to our time and part of the universe? A &quot;throbbing&quot; energy that leads to the non-uniformity that we see in galaxy clusters; something like foam awash on the ocean surface.&lt;/p&gt;&#xA;" OwnerUserId="7979" LastActivityDate="2015-09-05T18:06:20.520" Title="Is it just the observable universe that is expanding?" Tags="&lt;big-bang-theory&gt;&lt;expansion&gt;&lt;observable-universe&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11580" PostTypeId="1" CreationDate="2015-08-17T20:32:35.740" Score="3" ViewCount="56" Body="&lt;p&gt;I imagine there are several possible conditions for determining the time of a helical rising of a star, the elevation of the star at sunrise, the latitude of the observer, and the azimuth of the star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any kind of equation (or plot/graphic) that relates these variables and can be used to predict whether a star will be visible? (Ignoring weather, ie assuming some standard atmospheric condition).&lt;/p&gt;&#xA;" OwnerUserId="2800" LastActivityDate="2015-11-28T13:57:35.347" Title="Standard conditions for a heliacal rising" Tags="&lt;star&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11582" PostTypeId="2" ParentId="11579" CreationDate="2015-08-17T20:55:47.747" Score="1" Body="&lt;p&gt;There is no evidence to support the idea that some part of the universe (at a cosmological scale) is contracting. Obviously one could construct such a theory but there is currently no observational data to support it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until relatively recently there were three broad ideas about the Universe's expansion - that gravitational pull would slow the expansion but by never enough to halt it, that (a boundary case) there was exactly enough matter in the universe for expansion to halt at an infinite point in the future, or that eventually gravitational attraction would first slow and then reverse the expansion, leading to a big crunch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, though, the observational evidence suggests that the expansion of the Universe is accelerating - that there is some &quot;dark energy&quot; that is speeding up the expansion. This is now the generally accepted view, though arguments continue as to what this &quot;dark energy&quot; is.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2015-08-17T20:55:47.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11583" PostTypeId="1" CreationDate="2015-08-17T21:36:14.907" Score="1" ViewCount="85" Body="&lt;p&gt;Now that NASA's LADEE spacecraft has confirmed the presence of Neon in the moon's atmosphere could the official explanation of the moon's light end up being change from reflecting the sun's light to some sort of electromagnetic interaction with the neon caused by solar winds, or something like that?&lt;/p&gt;&#xA;" OwnerUserId="7932" LastActivityDate="2015-08-17T22:20:48.137" Title="What's the significance of finding neon on lunar atmospehere? new explanation of moon light?" Tags="&lt;the-moon&gt;&lt;moon-phases&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11584" PostTypeId="1" AcceptedAnswerId="11586" CreationDate="2015-08-17T21:44:44.373" Score="2" ViewCount="163" Body="&lt;p&gt;Suppose that we could measure the current acceleration with respect to us of a distant galaxy chosen at random in our observable universe (Ignore the peculiar motion of the galaxy). Is the acceleration constant, is it increasing or is it decreasing?&lt;/p&gt;&#xA;" OwnerUserId="6920" LastEditorUserId="6920" LastEditDate="2015-08-17T21:50:54.480" LastActivityDate="2015-08-19T20:39:01.457" Title="Are galaxies moving away with constant acceleration?" Tags="&lt;cosmology&gt;&lt;expansion&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11585" PostTypeId="2" ParentId="11583" CreationDate="2015-08-17T22:20:48.137" Score="1" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_the_Moon&quot; rel=&quot;nofollow noreferrer&quot;&gt;moon's &amp;quot;atmosphere&amp;quot;&lt;/a&gt; isn't much of an atmosphere, and is sometimes called an &lt;a href=&quot;https://en.wikipedia.org/wiki/Exosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;exosphere&lt;/a&gt;.  The particle density in the moon's atmosphere is low enough to be considered a very good vacuum for Earth experiments.&lt;/p&gt;&#xA;&lt;p&gt;The discovery is mentioned in &lt;a href=&quot;http://phys.org/news/2015-08-ladee-spacecraft-neon-lunar-atmosphere.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article on phys.org&lt;/a&gt;, which specifically says&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;There's not enough neon to make the moon visibly glow because the moon's atmosphere is extremely tenuous, about 100 trillion times less dense than Earth's atmosphere at sea level.&lt;/p&gt;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&lt;p&gt;The behavior of a dense atmosphere is driven by collisions between its atoms and molecules. However, the moon's atmosphere is technically referred to as an exosphere because it's so thin, its atoms rarely collide.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;As far as the question &amp;quot;what's the importance&amp;quot; in your title, it seems mainly to be a confirmation of something that has been predicted for a while, as well as providing more data concerning exospheres: the most common type of &amp;quot;atmosphere&amp;quot; in the solar system.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-08-17T22:20:48.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11586" PostTypeId="2" ParentId="11584" CreationDate="2015-08-17T22:41:03.697" Score="3" Body="&lt;p&gt;I think the correct answer is that nobody knows - it all depends on the behaviour, or equation of state, of the dark energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the dark energy takes the form of a cosmological constant, then the energy density due to dark energy will end up completely dominant as the universe expands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The acceleration of the scale factor $a$ is given by&#xA;$$\frac{\ddot{a}}{a} = -\frac{4\pi G}{3}\left(\rho + \frac{3P}{c^2}\right) $$&#xA;where $\rho$ is the energy density, and $P$ is the pressure. A cosmological constant has $P = -\rho c^2$, which leads to&#xA;$$ \ddot{a} = \frac{8\pi G \rho a}{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if dark energy dominates, then $\rho a$ increases with time, so the acceleration is positive and increasing with $a$ and hence time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If matter dominates, which it did in the past, then $P \simeq 0$ and $\rho a \propto a^{-2}$, so the acceleration is negative, with a magnitude that becomes smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transition from deceleration to acceleration happens when $\rho_{\Lambda} + \rho_m + 3P/c^2= 0$. If we assume a cosmological constant with $P = -\rho_{\Lambda} c^2$, where $\rho_{\Lambda}$ is the dark energy density (which is constant), and $\rho_m$ is the matter density, where $\rho_m = \rho_0 a_0^{3}/a(t)^3$ and $\rho_0$ and $a_0$ are the present-day matter density and scale factor respectively. Thus&#xA;$$ \rho_{\Lambda} + \rho_m -3\rho_{\Lambda} = 0$$&#xA;and the transition point occurs when $\rho_m = 2 \rho_{\Lambda}$. But because $\rho_{\Lambda}/\rho_0 = 0.7/0.3$, then&#xA;$$ \rho_{\Lambda} = \frac{7\rho_0}{3} = \frac{7}{3}\left(\frac{a}{a_0}\right)^3 \rho_m$$&#xA;and so the scale factor at the time of transition from decleration to acceleration is found from&#xA;$$ \rho_m =  \frac{14}{3}\left(\frac{a}{a_0}\right)^3 \rho_m,$$&#xA;$$ a = \left(\frac{3}{14}\right)^{1/3} a_0$$&#xA;i.e. when the scale factor was 60% of the present day value. According to this &lt;a href=&quot;http://home.fnal.gov/~gnedin/cc/&quot; rel=&quot;nofollow&quot;&gt;cosmology calculator&lt;/a&gt;, this happened roughly 6 billion years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the universe was decelerating then accelerated. So clearly the acceleration now is &lt;em&gt;increasing&lt;/em&gt; with time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if for whatever reason, the dark energy started to &quot;decay&quot; then it might be that the acceleration would decrease.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-08-19T20:39:01.457" LastActivityDate="2015-08-19T20:39:01.457" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11587" PostTypeId="2" ParentId="11573" CreationDate="2015-08-18T08:50:33.853" Score="2" Body="&lt;p&gt;If you're only interested in the positions, you'll probably be best off just using a program instead of diving into celestial mechanics. Depending on what you exactly want with &quot;planetary position&quot;, here are three different approaches to finding a planet's position at a given time:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;Stellarium&lt;/em&gt; is a free program to visually solve for the position of a planet, and it gives some precise data for the positions of the planets too, but it needs to be installed and run on your computer so it won't be the quickest way to find the position.&lt;/li&gt;&#xA;&lt;li&gt;If you're looking to get a planet's coordinates on ecliptic and/or equatorial planes, you can let &lt;a href=&quot;http://keisan.casio.com/exec/system/1224748262&quot; rel=&quot;nofollow&quot; title=&quot;Planet&amp;#39;s position on ecliptic and equatorial planes Calculator&quot;&gt;this&lt;/a&gt; web app do the calculations for you for a given time.&lt;/li&gt;&#xA;&lt;li&gt;Also, a quick Google search came up with &lt;a href=&quot;http://www.currentplanetarypositions.com/cgi-bin/calculate_planets.cgi&quot; rel=&quot;nofollow&quot; title=&quot;Custom Planet Position Calculations&quot;&gt;this&lt;/a&gt; web app meant for working with zodiacs. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7015" LastActivityDate="2015-08-18T08:50:33.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11589" PostTypeId="1" AcceptedAnswerId="11591" CreationDate="2015-08-18T19:06:41.927" Score="1" ViewCount="2430" Body="&lt;p&gt;It is said that when our solar system passes through the dark rift of galaxy (what is it exactly?), a phenomenon named '&lt;strong&gt;Galactic Eclipse&lt;/strong&gt;' happens, when our solar system witnesses 3 days of darkness. Is it possible even in this case that there would be no light in the night sky, as there are many stars in the night sky. (don't consider city lights and other artificial human interference)  &lt;/p&gt;&#xA;" OwnerUserId="7964" LastActivityDate="2015-08-19T10:45:36.020" Title="What is a Galactic Eclipse?" Tags="&lt;galaxy&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11590" PostTypeId="2" ParentId="11578" CreationDate="2015-08-19T09:35:29.270" Score="4" Body="&lt;p&gt;&quot;Was it as loud as we think&quot; is difficult to answer, since it's opinion-based. But since sound is nothing but longitudal oscillations in a gaseous medium, &lt;strong&gt;Big Bang was not at all silent&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Universe were completely homogeneous, it would stay like that. But primordial quantum fluctuations ensured that space was a tiny bit more dense in some places, and a tiny bit more dilute in other places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitation then ensured that the overdensities attracted matter and grew in size, until the pressure thus built up resisted further compression, and waves traveled outward from these overdensities. They then contracted and expanded again a few times, until 380,000 years after Big Bang when the photons &lt;a href=&quot;https://en.wikipedia.org/wiki/Decoupling_%28cosmology%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;decoupled&lt;/a&gt; from the gas, relieving the gas of its pressure, and freezing the waves in (comoving) space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This phenomenon is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Baryon_acoustic_oscillations&quot; rel=&quot;nofollow noreferrer&quot;&gt;baryonic acoustic oscillations&lt;/a&gt; (BAOs). You may also be interested in &lt;a href=&quot;https://physics.stackexchange.com/questions/162184/what-is-the-speed-of-sound-in-space/162210#162210&quot;&gt;my answer to the question &quot;What is the speed of sound in space?&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, humans wouldn't be able to hear it, since the wavelength (at decoupling) were roughly half a million lightyears, and the corresponding frequency thus orders of magnitues below the human threshold of ~20 Hz. Due to the expansion of space, the frequency increases as we go back in time, and scaling by a factor of $\sim10^{26}$, it is possible to get a notion of what the early Universe sounded like. This has been done e.g. by &lt;a href=&quot;http://faculty.washington.edu/jcramer/BBSound_2013.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;John Cramer from U. of Washington&lt;/a&gt; on the basis of &lt;a href=&quot;http://WMAP&quot; rel=&quot;nofollow noreferrer&quot;&gt;WMAP&lt;/a&gt;'s observations of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow noreferrer&quot;&gt;cosmic microwave background&lt;/a&gt; which hold information about the BAOs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can hear it &lt;a href=&quot;http://faculty.washington.edu/jcramer/BigBang/Planck_2013/BBSnd100.wav&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. It doesn't really sound nice, though.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-08-19T10:29:14.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11591" PostTypeId="2" ParentId="11589" CreationDate="2015-08-19T10:45:36.020" Score="3" Body="&lt;p&gt;Definitely a hoax as other commenters have said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark rifts exist in the galaxy - large clouds of dust, several can be made out by eye, overlapping the milky way, in ideal conditions.  See for example the &lt;a href=&quot;https://en.wikipedia.org/wiki/Great_Rift_(astronomy)&quot; rel=&quot;nofollow&quot;&gt;Great Rift&lt;/a&gt;.  However these are so large it would take far longer than three days for a solar system to pass through them.  Probably thousands or millions of years.  So this is definitely incorrect information.&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2015-08-19T10:45:36.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11592" PostTypeId="1" AcceptedAnswerId="11593" CreationDate="2015-08-19T15:24:15.607" Score="5" ViewCount="1268" Body="&lt;p&gt;Famously, when Eddington attempted to measure gravitational lensing during the Eclipse, it was the measured &lt;em&gt;magnitude&lt;/em&gt; of the lensing that gave gravity [pun, obviously, intended] to General Relativity - not the measurement of lensing itself. That is, Newtonian physics also predicted a lensing deflection, but only half of the deflection predicted by GR. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question is: Why? I've read a lot about this, and I can see how when one integrates the Newtonian acceleration along the photon path with constant |v|=c this gives the 'correct' Newtonian value - but intuitively, I can't wrap my head around it. Why should a massless photon be affected by gravity at all in Newtonian gravity?&lt;/p&gt;&#xA;" OwnerUserId="6965" LastEditorUserId="6965" LastEditDate="2015-08-20T09:33:50.857" LastActivityDate="2015-08-27T22:21:14.817" Title="Gravitational lensing in Newtonian physics" Tags="&lt;general-relativity&gt;&lt;gravitational-lensing&gt;&lt;newtonian-gravity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11593" PostTypeId="2" ParentId="11592" CreationDate="2015-08-19T17:32:39.783" Score="4" Body="&lt;p&gt;A photon is an entity defined in the context of a relativistic field theory, and so it doesn't really make sense to talk about the Newtonian bending of a photon. Necessarily, we need to substitute an analogous question that's sensible in the Newtonian framework. To do so, we can imagine a classical &lt;em&gt;corpuscle&lt;/em&gt; of light--appropriately enough, a theory of light advanced by Newton himself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many problems with the Newtonian conception of light, but that's an issue for electromagnetism rather than gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Critically, for a test particle, the trajectory depends on the initial velocity only, and not the mass. So we don't even have to address the mass of the light corpuscle at all to talk about its trajectory, as the acceleration is the gradient of the gravitational potential and if considering force (or gravitational potential &lt;em&gt;energy&lt;/em&gt;), the mass cancels out anyway. If you wish to bring in the mass explicitly, we can still think of the trajectory of a massless particle as a limit of trajectories of particles with masses tending to zero but having the same velocity &amp;mdash; a trivial limit because they're all the same trajectory in the Newtonian theory. On the other hand, if we recognize that light carries momentum, a Newtonian light corpuscle &lt;em&gt;shouldn't&lt;/em&gt; have zero mass, so the question of what to do with a genuinely massless particle evaporates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When people talk about the Newtonian deflection of light, they are typically considering a hyperbolic trajectory of test particle at the speed of light under Newtonian gravity. If the angle between the asymptotes is $\theta$, then $\theta = \pi$ represents a completely straight trajectory unaffected by gravity, and the eccentricity is&#xA;$$e = 1 + \frac{v_\infty^2R_p}{GM} = \frac{1}{\cos\left(\frac{\theta}{2}\right)} = \frac{1}{\sin\psi}\text{,}$$&#xA;where $R_p$ is the periapsis distance and $2\psi = \pi-\theta$ is the measure of deflection. On the scale of the solar system, it doesn't matter whether we set $v_\infty = c$ or anywhere else along the trajectory. For example, if the velocity at peripasis is $c$ instead, then $e\mapsto e-2$, and so is negligible for light deflection due to the Sun, $e &amp;gt; 10^{5}$. The total deflection is approximately&#xA;$$2\psi \approx e^{-1} \approx \frac{2GM}{c^2R_p}\text{,}$$&#xA;which is half the correct general-relativistic prediction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that here we did &lt;em&gt;not&lt;/em&gt; assume that the speed is constant along the hyperbolic orbit. That would not be consistent with Newtonian gravity. Rather, what we have is a situation where if $v = c$ &lt;em&gt;anywhere&lt;/em&gt; along the trajectory, then the speed along any other point is so close to $c$ that it doesn't practically matter for considering light deflection.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="6980" LastEditDate="2015-08-27T22:21:14.817" LastActivityDate="2015-08-27T22:21:14.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11594" PostTypeId="1" AcceptedAnswerId="11595" CreationDate="2015-08-20T01:27:38.807" Score="4" ViewCount="1773" Body="&lt;p&gt;I am wondering what is the largest known core hydrogen-burning star?  A look at the &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_largest_known_stars&quot; rel=&quot;nofollow&quot;&gt;list of largest known stars on Wikipedia&lt;/a&gt; seems to indicate VV Cephei B (at the bottom of the list), but I would like to know for sure if it is the largest known.  In addition to knowing which star it is, I would also like to know its temperature, size, and expected lifetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also curious to know if the largest &lt;em&gt;known&lt;/em&gt; core hydrogen-burning star is similar to what astrophysicists theorize is the largest possible core hydrogen-burning star (given current metallicity conditions in the universe; I know star-formation timescales have a metallicity dependence) and the expected temperature, size, and lifetime of such an object.&lt;/p&gt;&#xA;" OwnerUserId="128" LastEditorUserId="128" LastEditDate="2015-08-20T11:14:53.913" LastActivityDate="2015-08-20T12:18:08.027" Title="What is the largest hydrogen-burning star?" Tags="&lt;star&gt;&lt;star-formation&gt;&lt;main-sequence&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11595" PostTypeId="2" ParentId="11594" CreationDate="2015-08-20T12:11:43.247" Score="6" Body="&lt;p&gt;I assume by largest, you mean largest radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well it won't be VV Cep B since this is merely a B-type main sequence star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;O-type main sequence stars are known and these have both larger masses and larger radii on the main sequence (when they are burning hydrogen in their cores).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A selection of the most massive objects can be found in the R136 star forming region in the Large Magellanic Clouds. If you look at &lt;a href=&quot;https://en.wikipedia.org/wiki/R136#Components&quot; rel=&quot;noreferrer&quot;&gt;this list&lt;/a&gt; (though I recommend having a look at the primary literature), you will see that O3V stars are listed. Such objects are also present in our Galaxy, for instance in the supercluster NGC 3603 (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1998MNRAS.296..622C&quot; rel=&quot;noreferrer&quot;&gt;Crowther &amp;amp; Dessart 1998&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such stars have masses of maybe $100 M_{\odot}$, luminosities of $2\times 10^{6} L_{\odot}$ and temperatures of 50,000 K. Using Stefan's law, we can deduce radii of $\sim 20 R_{\odot}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are suggestions that even more massive main sequence stars have existed in R136 and NGC 3603 (see &lt;a href=&quot;http://arxiv.org/abs/1007.3284&quot; rel=&quot;noreferrer&quot;&gt;Crowther et al. 2010&lt;/a&gt;), which are now seen as evolved Wolf-Rayet objects, possibly up to $300 M_{\odot}$ on the main sequence (though this is a model-dependent extrapolation), and these would have had radii $&amp;gt;20 R_{\odot}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the very early universe, population III main sequence stars without metals could have been much more massive and larger. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-08-20T12:18:08.027" LastActivityDate="2015-08-20T12:18:08.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11596" PostTypeId="1" AcceptedAnswerId="11598" CreationDate="2015-08-20T16:47:58.170" Score="1" ViewCount="41" Body="&lt;p&gt;Suppose I am orbiting a rocky planet (density = 4.4 g/cm3) two-thirds the mass of Mars in the Habitable Zone of Gliese 251. This planet has a thin atmosphere, presumably from out-gassing, still has internal heating, has a weak magnetic field, but has no sign of tectonic activity. I need to mine precious and radioactive elements to replenish my ship. Gliese 251 only has about 8% of the metallicity of the sun. Can I assume that the planet (theoretically) would have about 8% of the elements (per volume of crustal material) as the Earth? Would lack of tectonic activity make them much harder to get to?&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2015-08-20T22:43:09.590" Title="Elemental abundances in the crust of exoplanets and accessibility to them" Tags="&lt;planet&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11598" PostTypeId="2" ParentId="11596" CreationDate="2015-08-20T20:17:51.987" Score="2" Body="&lt;p&gt;No, it doesn't work like that. A rocky planet, to first order, will probably contain a similar mix of elements to the Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most likely reason that the star has a low metallicity is because it is old, and formed when the interstellar medium was less metal rich. But because rocky planets do not incorporate (much) hydrogen and helium, what you are really asking is how do the abundances of other elements scale to the iron abundance (which is often what is meant by metallicity)? The r-process events, such as type II supernovae that produce the heaviest radioactive elements (beyond lead), enrich the interstellar medium &quot;promptly&quot;, because the progenitor lifetimes are short compared to the age of the Galaxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, because a lot of the iron and nickel in the interstellar medium is not produced/disseminated by type II supernovae, but by type Ia supernovae with longer-lived, lower mass progenitors, then if anything I would expect a &lt;em&gt;larger&lt;/em&gt; ratio of elements produced/disseminated by type II supernovae to iron/nickel. This will also be true of the &quot;alpha&quot; elements like Silicon, Oxygen and Magnesium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The planet might be a bit short in those elements predominantly produced by the s-process in lower mass giant stars - so elements like carbon (!), nitrogen, barium, rubidium, strontium, zinc and many others may be scarce.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a star of this age/metallicity I would expect these variations to be plus or minus a factor of two or so compared to the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, I am discussing bulk abundances. The accessibility of the material, where it is etc., is more of a speculative geology question, and I don't know the answer. A further complication is where exactly the planet forms compared with the condensation temperature of various elements and materials. The above discussion implicitly assumes that the planet forms in a protoplanetary disc material at a similar temperature and density to the conditions in which the Earth formed. However, a change in these conditions &lt;em&gt;will&lt;/em&gt; alter the detailed compositional mix.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-08-20T22:43:09.590" LastActivityDate="2015-08-20T22:43:09.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11599" PostTypeId="1" CreationDate="2015-08-20T22:02:55.057" Score="1" ViewCount="58" Body="&lt;p&gt;If an object enters Earth's atmosphere, what determines whether it burns up or not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A feather is released from a space-ship 30,000 km away from Earth and falls under the influence of Earth's gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will it burn up on re-entry or will it float down to the surface unharmed. If it would be unharmed, where is the cutoff point for which objects survive?&lt;/p&gt;&#xA;" OwnerUserId="7828" LastActivityDate="2015-08-21T23:30:14.927" Title="Meteors made of differing substances. Will they all burn up?" Tags="&lt;meteor&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11601" PostTypeId="1" AcceptedAnswerId="11602" CreationDate="2015-08-21T00:58:07.037" Score="-2" ViewCount="829" Body="&lt;p&gt;The Virgo Galaxy Cluster has a mass of $10^{14} M_{\odot}$ and its centre is $16Mpc$ from Earth. The large elliptical galaxy $M87$ lies at the centre of the Virgo cluster. $M87$ has a supermassive black hole at it's centre with an estimated mass of $6 \times 10^9 M_{\odot}$. Take Hubble's constant to be $H_0=70 km s^{-1} Mpc^{-1}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the Virgo cluster to be spherically symmetric with a radial density profile given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho(r)=\rho_0 (\frac{r}{1Mpc})^{-2}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine the value of the constant $\rho_0$ is S.I units assuming the radius of the Virgo cluster is 1Mpc. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am confused with how to approach this question, &#xA;I know that density $\rho=\frac{M}{\frac{4}{3} \pi r^3}$, when I substitute it into the given radial density profile, the $r$ variable doesn't cancel, should I substitute the radius of the cluster into $r$? Is it really that straight forward?&lt;/p&gt;&#xA;" OwnerUserId="7998" LastActivityDate="2015-08-21T02:15:10.433" Title="Radial Density Profile Equation" Tags="&lt;galaxy&gt;&lt;galaxy-cluster&gt;&lt;density&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2015-09-01T20:59:22.393" ContentLicense="CC BY-SA 3.0" />
  <row Id="11602" PostTypeId="2" ParentId="11601" CreationDate="2015-08-21T02:15:10.433" Score="0" Body="&lt;p&gt;If the density is radially symmetric and given by $\rho = \rho(r)$, then a shell of radius $r$ and infinitesimal thickness $\mathrm{d}r$ has volume $\mathrm{d}V = 4\pi r^2\, \mathrm{d}r$, and therefore a mass of $\mathrm{d}M = \rho\,\mathrm{d}V$. The total mass enclosed by some radius $R$ is of course the sum of the masses of all shells up to that radius. I think this should enable you to finish the problem yourself.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2015-08-21T02:15:10.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11603" PostTypeId="1" AcceptedAnswerId="11604" CreationDate="2015-08-21T02:35:45.397" Score="2" ViewCount="69" Body="&lt;p&gt;I have started a &lt;a href=&quot;http://nanosatrevolution.com&quot; rel=&quot;nofollow&quot;&gt;project&lt;/a&gt; in Melbourne, Australia called nanosat eye in the sky to put a telescope in a cubesat and put it in LEO. The idea is then to add the telescope to &lt;a href=&quot;http://itelescope.net&quot; rel=&quot;nofollow&quot;&gt;itelescope&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to learn a heap about designing telescopes. I have limited space and want to understand how to measure the theoretical performance of our telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is- how do I measure the performance of the telescope? &lt;/p&gt;&#xA;" OwnerUserId="7999" LastActivityDate="2015-08-21T05:38:44.497" Title="What measure should I use to help optimise the design of a telescope in a cubesat?" Tags="&lt;space-telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11604" PostTypeId="2" ParentId="11603" CreationDate="2015-08-21T05:08:51.530" Score="2" Body="&lt;p&gt;I make telescopes and telescope mirrors. You're asking a hard question, since it depends on so many things. But it's an interesting project, so I'll try to help as much as I can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hard limit for performance is aperture, or diameter of the primary optics. Bigger aperture = higher hard limit for performance. On a nanosat you only have room for so much aperture, so performance will necessarily be limited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, you'd need some kind of compact design. A reflector for sure, but not newtonian. Some kind of Cassegrain variant, you can make those pretty short and stubby. Unless you make it so that it &quot;unfolds&quot; in space, but that's going to be tricky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the whole optical stack has to be very rigid and must hold its shape through launch and while in space. The optical system must remain collimated at all times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you probably have some weight limits. That will make it harder to create a rigid system.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One measure of performance is the resolving power, which indicates the angular size of the smallest details resolved by the telescope. The relationship is linear. An aperture of 100 mm provides a resolving power of 1 arcsec. An aperture of 200 mm provides a resolving power of 0.5 arcsec. And so on.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Telescopes operating at ground level are limited by atmosphere in several ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is the so-called &quot;seeing&quot;, or air turbulence. This blurs the image and reduces the effective resolving power. Since small telescopes have less resolving power to begin with, they are less affected. A 100 mm aperture is unaffected most of the time. A 200 mm aperture is affected most of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your telescope will be above the atmosphere, so the effects of seeing will be zero. It will operate at full resolving power all the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way the air influences telescope is via light pollution and other forms of luminescence and reflected light from the air. This reduces contrast and &quot;erases&quot; the faintest objects from the image; bright objects such as planets are not affected at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your telescope will not be affected by light pollution, so its performance on faint objects (such as galaxies or nebulae) will always be 100% of the maximum for that aperture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line: you are going to have a pretty small telescope, but operating completely free of the effects of the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Since it's operating in microgravity, the optics will not deform under their own weight, so you can make them quite thin and lightweight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The instrument will operate under extremes of temperature. You will need some kind of dynamic focusing to compensate for expansion / contraction. Be careful, with some catoptric designs such as Cassegrain you also need to pay attention to the distance from primary mirror to secondary elements, etc. Using carbon fiber for the skeleton would reduce expansion / contraction a great deal, and would also provide excellent stiffness per unit of weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quartz mirrors could deal very well with huge temperature variations, but quartz is expensive to process to optical precision. Borosilicate glass (the material used for most amateur telescopes) should also work well; Supremax 33 for example is cheap and fairly good. It depends on the optician who will make the mirrors. If you find someone who could give a discount on quartz, go for it; otherwise use borosilicate and don't worry about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will also need some kind of inertial elements to change the orientation of the telescope in space. That's not a trivial task.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Bottom line: you begin with a performance budget limited by aperture. That's the ideal performance. Then real world issues will push real performance below that level: collimation errors, temperature shifts, optics quality, focusing errors, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Performance of a real instrument is affected by &lt;em&gt;everything&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-08-21T05:38:44.497" LastActivityDate="2015-08-21T05:38:44.497" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11605" PostTypeId="1" AcceptedAnswerId="11606" CreationDate="2015-08-21T06:50:21.967" Score="5" ViewCount="403" Body="&lt;p&gt;From my understanding, it is believed that almost every big galaxy and especially spiral galaxies have supermassive black holes (SMBH's) at the center. Also, from what I've read, a SMBH isn't required for a galaxy to exist since &lt;em&gt;and in layman's terms&lt;/em&gt;, the matter inside a galaxy such it gas clouds and formation of stars will keep it gravitational-ly in check. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am right or even wrong, then would there be significant difference if our galaxy did not have a SMBH at the center? Significant enough for noticeable differences even here on Earth?&lt;/p&gt;&#xA;" OwnerUserId="7258" LastEditorDisplayName="user1569" LastEditDate="2015-08-21T17:12:26.657" LastActivityDate="2015-08-21T22:38:27.423" Title="What if our galaxy didn't have a SMBH?" Tags="&lt;galaxy&gt;&lt;supermassive-black-hole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11606" PostTypeId="2" ParentId="11605" CreationDate="2015-08-21T07:31:29.810" Score="5" Body="&lt;p&gt;The supermassive black hole (SMBH) in the center of the Milky Way (MW) — called Sgr A* [Sagittarius A-star] — has no direct impact on our galaxy. Its mass is only a few million Solar masses, and if you remove it$^\dagger$, it will only affect the most central stars, which would suddenly continue in straight paths out through the MW. These stars would almost surely not hit any other stars or something like that (since stars are really, really far apart), but &lt;a href=&quot;https://en.wikipedia.org/wiki/S2_%28star%29&quot; rel=&quot;nofollow&quot;&gt;some of them&lt;/a&gt; have velocities high enough that they may escape the MW.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Sgr A* weren't there to begin with, things might look a little different. There seems to be a relation between the mass of a galaxy's SMBH and the velocity dispersion of the stars in its central bulge; the so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/M%E2%80%93sigma_relation&quot; rel=&quot;nofollow&quot;&gt;M-sigma relation&lt;/a&gt;. so MW without Sgr A* would mean a more ordered center. Our Solar system is located in the disk, far from the center, and their is evidence that SMBHs have little impact on the disk (&lt;a href=&quot;http://arxiv.org/abs/astro-ph/0107135&quot; rel=&quot;nofollow&quot;&gt;Gebhardt et al. 2001&lt;/a&gt;). However, in their early phase (as an &lt;a href=&quot;https://en.wikipedia.org/wiki/Active_galactic_nucleus&quot; rel=&quot;nofollow&quot;&gt;active galactic nucleus&lt;/a&gt;), their extreme luminosities cause galactic superwinds which blow out gas and may quench star formation (&lt;a href=&quot;http://arxiv.org/abs/1501.07664&quot; rel=&quot;nofollow&quot;&gt;Tombesi et al. 2015&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^{^\dagger}$&lt;sup&gt;Removing Milky Way's SMBH is left as an exercise for the reader.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-08-21T22:38:27.423" LastActivityDate="2015-08-21T22:38:27.423" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11607" PostTypeId="1" CreationDate="2015-08-21T07:47:14.680" Score="0" ViewCount="86" Body="&lt;p&gt;We know that planets revolve around their parent star (or specifically Barycenter), they maintain a specific speed of revolution so that they don't end up smashing into their planet star. Which force or energy do that, so that they maintain their speed? If we have infinite amount of time, would planets smash into their parent star ?&lt;/p&gt;&#xA;" OwnerUserId="7964" LastEditorUserId="2531" LastEditDate="2015-08-21T13:12:14.980" LastActivityDate="2015-08-21T16:10:15.707" Title="Which force makes planets to continuously revolve around their parent star?" Tags="&lt;fundamental-astronomy&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11608" PostTypeId="2" ParentId="11607" CreationDate="2015-08-21T13:39:11.000" Score="1" Body="&lt;p&gt;Newton's First Law of Motion states that &quot;an object in motion stays in motion with the same speed ... unless acted upon by an unbalanced force&quot;. On earth, an object in motion will slow down and stop due to air resistance (the unbalanced force), but in space, there is no atmosphere to cause air resistance, so the object will maintain its speed.&lt;/p&gt;&#xA;" OwnerUserId="8001" LastEditorUserId="8001" LastEditDate="2015-08-21T16:10:15.707" LastActivityDate="2015-08-21T16:10:15.707" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11609" PostTypeId="1" AcceptedAnswerId="11612" CreationDate="2015-08-21T15:11:13.767" Score="1" ViewCount="145" Body="&lt;p&gt;A Nova episode said that during stages of some moons' elliptical orbits, the moon is stretched, and friction causes the inside of the moon to heat up. I do not understand how this is compliant with the Law of Conservation of Energy - what energy would be lost to make the inside of the moon heat up?&lt;/p&gt;&#xA;" OwnerUserId="8001" LastEditorUserId="8001" LastEditDate="2015-08-21T16:00:15.110" LastActivityDate="2015-08-22T00:41:54.837" Title="How is the heat production inside moons compliant with Conservation of Energy?" Tags="&lt;orbit&gt;&lt;astrophysics&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11611" PostTypeId="2" ParentId="11599" CreationDate="2015-08-21T18:18:09.013" Score="1" Body="&lt;p&gt;I can give a partial answer to this.   In general, larger objects are more likely to break up than smaller ones, and comets are more likely to break up than meteors.   A feather would likely burn up too, here's why:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;30,000 KM above the surface, or roughly 4.7 earth radii, the escape velocity at that height would be the square root of that or roughly 2.17 times less or 46% of the escape velocity from the surface, so a feather dropped from that height would be traveling at nearly 54% of earth's escape velocity when it hit the atmosphere, or about 6 KM per second.  That's about 7 times as fast as a bullet and I'd wager, that would burn up a feather pretty quick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speed of space collisions is faster than free-fall velocity can reach cause anything that hits the earth from space is also in free fall and you can add to that any relative velocity between the objects.   The slowest meteors hit the earth at about 11 KM per second, not surprisingly, the Earth's escape velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.amsmeteors.org/fireballs/faqf/#12&quot; rel=&quot;nofollow&quot;&gt;http://www.amsmeteors.org/fireballs/faqf/#12&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you drop a feather from the highest hot air balloon, then it probably would float down to the earth, accelerating faster at first in the thinner atmosphere, then more slowly as the air got thicker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small point to add, angle of approach matters too.   A glancing blow and the object can effectively bounce off the atmosphere, not burn up in it.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-08-21T23:30:14.927" LastActivityDate="2015-08-21T23:30:14.927" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11612" PostTypeId="2" ParentId="11609" CreationDate="2015-08-22T00:41:54.837" Score="3" Body="&lt;p&gt;The heat production inside moons is due to &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_heating&quot; rel=&quot;nofollow noreferrer&quot;&gt;tidal heating&lt;/a&gt;. Due to this, the orbit of the moon would change.&lt;/p&gt;&#xA;&lt;p&gt;The orbital and rotational energy of the moon is dissipated as heat either in the surface ocean or in the interior of the moon. This is especially seen in the moons of large planets, for example Io.&lt;/p&gt;&#xA;&lt;p&gt;Io, being Jupiter's innermost moon, is pulled by Jupiter on one side and the other moons on the opposite side. Due to this, Io's orbit is highly elliptic. During this orbit, the shape of the Io changes continuously (the surface flexes by ~100m). This causes a large amount of friction in the moon.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9AHRs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9AHRs.png&quot; alt=&quot;Tidal Flexing of Io&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Tidal_heating_on_Io.png#/media/File:Tidal_heating_on_Io.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tidal heating on Io&lt;/a&gt;&amp;quot; by Lsuanli - Own work. Licensed under CC BY-SA 3.0 via &lt;a href=&quot;https://commons.wikimedia.org/wiki/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Commons&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Usually, the tidal heating leads to tidal dissipation and circularizes the orbit. However, it hasn't happened in Io because it is in orbital resonance with other moons of Jupiter.&lt;/p&gt;&#xA;&lt;p&gt;If the orbital period of the moon is less than the rotational period of the planet, the moon is decelerated if the moons are out of the synchronous orbit, the opposite thing happens, conserving energy and angular momentum.&lt;/p&gt;&#xA;&lt;p&gt;On the other hand, Europa, another of Jupiter's moons is tidally locked (it has same orbital period as the rotational period of the larger body) to the Jupiter, which causes tidal flexing on the surface.&lt;/p&gt;&#xA;" OwnerUserId="7970" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-08-22T00:41:54.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11613" PostTypeId="1" AcceptedAnswerId="11614" CreationDate="2015-08-22T09:32:54.257" Score="1" ViewCount="687" Body="&lt;p&gt;For what I know, L4 and L5 are high points. Although, they are stable points. Why? I thought it was the same being a high point and being a unstable point. If I'm off-topic, please let me know where could I get an answer.&#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="8006" LastActivityDate="2015-08-24T16:13:10.473" Title="L4 and L5 stability" Tags="&lt;lagrange-point&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11614" PostTypeId="2" ParentId="11613" CreationDate="2015-08-22T13:54:08.497" Score="4" Body="&lt;p&gt;It does indeed seem counterintuitive that $L_4$ and $L_5$ would be at the same time both high points of potential as well as stable points in the system. In fact, a quick look at an example contour plot with all five Lagrangian points demonstrated would also suggest that $L_4$ and $L_5$ would be unstable: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/i4tw0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i4tw0.jpg&quot; alt=&quot;From Wikipedia: Lagrangian point&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the picture you can see that $L_1 - L_3$ are placed in saddle nodes, and intuitively one concludes that the smallest perturbation would set them &quot;falling&quot; in either direction, i.e. they need station keeping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, $L_4$ and $L_5$ would still seem to be unstable, so what gives? What makes $L_4$ and $L_5$ stable is that each of them is located equidistant from both of the masses. This leads to the gravitational forces from each of the bodies towards $L_4$ and $L_5$ to be in the same ratio as the two bodies' masses, hence the resultant acceleration points towards the barycentre of the system (which is also the centre of rotation for a three-body system).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This resultant acceleration, and by extension force, leads to the object in either $L_4$ or $L_5$ to in fact have just the correct amount of acceleration / correct force towards the system's barycentre so that it will not fall out of orbit (it &quot;falls&quot; towards the barycentre as the system rotates, like Earth similarly falls towards Sun throughout its orbit), as counterintuitive as it sounds at first. (Furthermore, this effect is not only limited to bodies of negligible mass, but is independent of mass, which is why you can have more massive bodies orbiting in $L_4$ and $L_5$ like Jupiter's trojans, as well as smaller, artificial satellites.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Lagrangian_point#L4_and_L5&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lagrangian point - Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7015" LastEditorUserId="7015" LastEditDate="2015-08-22T20:55:15.293" LastActivityDate="2015-08-22T20:55:15.293" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11615" PostTypeId="1" CreationDate="2015-08-23T06:02:50.980" Score="5" ViewCount="12335" Body="&lt;p&gt;I am a 9th class student so I'm a beginner at using a telescope, hence please help me.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How many bodies can I see using a 90x telescope?&lt;/li&gt;&#xA;&lt;li&gt;How will I know when and where I should point a telescope to see planets?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8015" LastEditorUserId="-1" LastEditDate="2018-09-20T13:05:43.613" LastActivityDate="2019-08-11T11:18:50.193" Title="How many celestial bodies can I see with a 90x telescope?" Tags="&lt;telescope&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="11616" PostTypeId="1" AcceptedAnswerId="11636" CreationDate="2015-08-23T06:29:02.833" Score="8" ViewCount="2107" Body="&lt;p&gt;Since I learned about Newton's gravity law, I wonder if I can measure the moon's gravity using (almost) no special equipment. The best I could come up with is a pendulum, which is in a resting position. The moon (at the horizon) will pull on the weight at the bottom of the pendulum and cause it to shift. 6 hours later, with the moon above or below the pendulum, the weight should be back. Maybe we could detect the shift, just like we see the tides. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rQZ5q.jpg&quot; alt=&quot;moon puling on a pendulum&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Running the math for a 5 meter long pendulum and 1 kg mass, I get an amplitude of 0,017 mm. I fear that external forces (wind, unstable surface, ...) will be bigger than this effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: &lt;em&gt;is it possible (again, without specific equipment) to measure or detect the moon's gravitational force? What would be the best way to measure/detect it?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2189" LastEditorUserId="2189" LastEditDate="2015-08-23T12:53:10.473" LastActivityDate="2019-01-17T22:02:59.077" Title="Can I measure the moon's gravity?" Tags="&lt;the-moon&gt;&lt;newtonian-gravity&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11617" PostTypeId="1" CreationDate="2015-08-23T13:57:03.260" Score="1" ViewCount="258" Body="&lt;p&gt;If I draw the orbit of Mars in stellarium 0.10.4 for 23 Sep 2015, it passes south of Chi Leonis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nuOJZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nuOJZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I advance the date to 17 Oct 2015, however, the orbit of Mars correctly goes right through Chi Leonis (Mars will occult Chi Leonis on 18 Oct 2015):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wWGgm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wWGgm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a known bug in 0.10.4 that's been fixed? (I unfortunately can't upgrade)&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2015-11-10T01:46:36.143" Title="Stellarium 0.10.4: planet orbits change over time?" Tags="&lt;orbit&gt;&lt;observational-astronomy&gt;&lt;software&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11619" PostTypeId="2" ParentId="11579" CreationDate="2015-08-23T17:47:27.673" Score="0" Body="&lt;p&gt;We do not know what the dark energy is , which is responsible for the expanding universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if it's a property of space-time, the following two points will result:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The universe expansion is isotropic, meaning it expands the same way in all directions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The expansion is homogeneous, meaning it expands the same way in all its parts, not only what we have observed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The homogeneity and isotropy of the universe are fundamental principles, from which the laws of conservation of impulse and angular momentum follow. All the observational results support them. If these laws would be wrong, all the physical theory should be revised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in case dark energy is a property of the universe, all the universe should expand the same way as the observable universe.&lt;/p&gt;&#xA;" OwnerUserId="4135" LastActivityDate="2015-08-23T17:47:27.673" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11620" PostTypeId="2" ParentId="11316" CreationDate="2015-08-23T17:58:01.663" Score="3" Body="&lt;p&gt;In short, most of the trojans stay orbiting arround L4 or L5. These can be called tad-pole orbit asteroids. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some trojan asteroids, however, that their orbit never get's too close to L4 nor L5 to get trapped and have a tad-pole orbit, or they do but they have too much energy to get trapped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These trojan asteroids have larger orbits, following a path like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;L4 --&gt; L3 --&gt; L5 --&gt; L3 --&gt; L4 and so on. This kind of orbit is called horseshoe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is, however, a third kind of trojan asteroid. This one appears to follow a tad-pole orbit (I.E. arround L4), but numerical analysis proves they will eventually adopt a horseshoe-like orbit. When they reach the other stable point (L5) they  will get trapped again in a tad-pole orbit. This kind of orbit is very rare (rarity decreases as planet mass increases) and could be considered (personal opinion) as an hybrid tadpole - horseshoe orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information, see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Jumping mechanisms of Trojan asteroids in the planar&#xA;restricted three- and four-body problems&quot; from: Kenta Oshima and Tomohiro Yanao.&lt;/p&gt;&#xA;" OwnerUserId="8006" LastActivityDate="2015-08-23T17:58:01.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11622" PostTypeId="1" CreationDate="2015-08-24T00:16:04.087" Score="2" ViewCount="565" Body="&lt;p&gt;Is a supernova explosion really fast or is it just fast, metaphorically speaking? Can you cite the fastest known phenomenon in astronomy (neutron stars, pulsars) besides supernovae in terms of explosions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: yes, i mean the shockwave from supernova&lt;/p&gt;&#xA;" OwnerUserId="8016" LastEditorUserId="8016" LastEditDate="2015-08-28T08:32:14.477" LastActivityDate="2015-08-28T08:32:14.477" Title="What is faster than a supernova explosion?" Tags="&lt;supernova&gt;&lt;speed&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11623" PostTypeId="1" AcceptedAnswerId="11624" CreationDate="2015-08-24T02:33:19.663" Score="3" ViewCount="230" Body="&lt;p&gt;What is the highest recorded temperature in space? I'm not asking about Planck temperature that is a fraction of a second before the Big Bang. I'm asking about the temperature that has been recorded by an instrument.&lt;/p&gt;&#xA;" OwnerUserId="7926" LastActivityDate="2015-08-24T03:42:39.583" Title="What is the highest recorded temperature in space" Tags="&lt;temperature&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11624" PostTypeId="2" ParentId="11623" CreationDate="2015-08-24T03:42:39.583" Score="3" Body="&lt;p&gt;Well &lt;a href=&quot;http://www.fromquarkstoquasars.com/coldest-hottest-temperatures-known-universe-infographic/&quot; rel=&quot;nofollow&quot;&gt;this infograhic&lt;/a&gt; puts gas heated by a supernova at about 55 million Celsius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that's not even the highest temperature we can just &quot;find&quot; out there, even if we discard supernovae and stellar cores.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Intracluster_medium&quot; rel=&quot;nofollow&quot;&gt;The gas in a cluster of galaxies (the ICM)&lt;/a&gt; can have temperatures on the orders of 10 million to 100 million Kelvin.  Such a high temperature gas emits high energy photons such as X-rays via bremsstrahlung radiation, and these are observable (and have been observed) by X-ray telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's &lt;a href=&quot;http://arxiv.org/pdf/astro-ph/0310557v1.pdf&quot; rel=&quot;nofollow&quot;&gt;this review article&lt;/a&gt; from 2003 that covers the basic methodologies and measurements made on ICMs, including (but decidedly not limited to) temperature. &lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2015-08-24T03:42:39.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11625" PostTypeId="2" ParentId="609" CreationDate="2015-08-24T08:08:19.533" Score="4" Body="&lt;p&gt;The rest frame for measuring (astronomical) speeds and velocities depends on the context and purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A geocentric frame, based on the Earth's centre of mass might be appropriate for objects in orbit around the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A heliocentric reference frame, centred on the Sun, is often used when describing the line of sight velocities of astronomical objects, but more usually, a barycentric frame at the solar system centre of mass is used. For example, this is how the radial velocities measurements of exoplanet host stars or components of eclipsing binaries would be quoted. It would also be appropriate for the motion of objects in the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motions in the Galaxy are commonly defined in two ways. One is the Local Standard of Rest (LSR), which measures velocities with respect to the average motion of stars near the Sun. The second attempts to measure velocities with respect to the Galactic plane and radially and tangentially in the plane with respect to the Galactic centre. These are the so-called $U,V,W$ velocities. The position and motion of the LSR with respect to the Galactic centre are uncertain at the level of a few km/s, whereas the solar motion with respect to the LSR is more precisely known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Galactic centre can also be used as a reference frame for the motion of galaxies in our local group. e.g. when talking about the motion of the Andromeda Galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we can determine a cosmic standard of rest - the co-moving stationary frame of the Hubble flow - using precise measurements of the cosmic microwave background (CMB). In other words, we can determine our peculiar motion with respect to the Hubble flow by observing the dipole doppler signature in the CMB.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-08-24T08:15:29.533" LastActivityDate="2015-08-24T08:15:29.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11626" PostTypeId="1" AcceptedAnswerId="11627" CreationDate="2015-08-24T12:48:15.597" Score="10" ViewCount="4441" Body="&lt;ol&gt;&#xA;&lt;li&gt;How are non-rotating black holes formed?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I guess that non-rotating black holes are formed from non-rotating stars but I couldn't find any proof that such a star exists on the internet. If this is true can you also put up an example of such a star?&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Does the matter absorption of a non-rotating black hole differ from a rotating black hole?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7926" LastActivityDate="2017-12-08T09:23:13.890" Title="How is a non-rotating black hole created" Tags="&lt;black-hole&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11627" PostTypeId="2" ParentId="11626" CreationDate="2015-08-24T13:28:35.663" Score="5" Body="&lt;p&gt;All stars have &lt;em&gt;some&lt;/em&gt; angular momentum. Conservation of angular momentum when the core of a star collapses means that it is very likely that most black holes are rapidly rotating, just as newly-born neutron stars are observed to be extremely fast-rotating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if one were to have a star with a non-rotating core as it collapsed, the subsequent accretion of material with angular momentum would give the black hole that angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-spinning black holes are a theoretical ideal that is unlikely to be found in nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The details of the accretion of material onto a black hole are complex and still the subject of theoretical and observational research. But yes, the details of accretion in spacetime governed by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kerr_metric&quot; rel=&quot;noreferrer&quot;&gt;Kerr metric&lt;/a&gt; appropriate for rotating black holes will differ from those of a non-rotating &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_metric&quot; rel=&quot;noreferrer&quot;&gt;Schwarzschild&lt;/a&gt; black hole. Kerr black holes have an &quot;ergosphere&quot;, outside their event horizons, where material is force to co-rotate with the black hole.  &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-08-24T13:28:35.663" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11628" PostTypeId="1" CreationDate="2015-08-24T15:05:10.327" Score="2" ViewCount="54" Body="&lt;p&gt;After folding a light curve, we may  get some kind of shape.&lt;br/&gt;&#xA;How to tell whether there is a real periodical signal quantifiably?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fluctuation may be small, and the folded shape may not be perfect. &lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2015-08-25T07:25:04.307" Title="how to assess a periodical signal?" Tags="&lt;light&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11629" PostTypeId="2" ParentId="11628" CreationDate="2015-08-24T16:07:16.673" Score="1" Body="&lt;p&gt;You have it the wrong way around. You have to establish a periodicity before doing the folding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various ways to do this, but the most frequently (pun intended) used is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Least-squares_spectral_analysis#The_Lomb.E2.80.93Scargle_periodogram&quot; rel=&quot;nofollow&quot;&gt;Lomb-Scargle&lt;/a&gt; periodogram. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with a datset consisting of a set of points $(t_i, m_i\pm \sigma_i )$, where $t_i$ are times of observation and $m_i$ and $\sigma_i$ are the fluxes and their uncertainties at those times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then calculate an array of chi-squared statistics, by fitting a mean level $\bar{m}$ and sinusoidal functions over a range of discrete frequencies $\omega$. The aim is to &lt;em&gt;maximise&lt;/em&gt; the statistic&#xA;$$p(\omega) = \frac{\chi^2_{0} - \chi^2(\omega)}{\chi^2_{0}},$$&#xA;where &#xA;$$\chi^{2}_{0} = \sum (m_i - \bar{m})^2/\sigma_i^{2}$$ and &#xA;$$\chi^{2}(\omega) = \sum  (y_i - \bar{m})^2/\sigma_i^{2},$$ with $y_i = A\sin(\omega t_i + \phi) + A_0$, where $A$, $A_0$ and $\phi$ are free parameters that are optimised for each frequency (i.e. you find the minimum possible $\chi^2(\omega)$ at each frequency).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once a period is obtained from a peak in the power spectrum, $p_{best}$, &lt;em&gt;then&lt;/em&gt; it must be established whether this peak is &quot;real&quot; or significant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically, the probability that $p&amp;gt;p_{best}$ is $(1-p_{best})^{(N-3)/2}$ for $N$ independent measurements (see &lt;a href=&quot;http://arxiv.org/abs/astro-ph/9906466&quot; rel=&quot;nofollow&quot;&gt;Cumming, Marcy &amp;amp; Butler 1999&lt;/a&gt;). This is the &quot;false alarm probability&quot; - the chance that you would get $p&amp;gt;p_{best}$ from random, aperiodic data with those uncertainties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, real data has correlated noise and the points are not independent. So what you do instead (see &lt;a href=&quot;http://arxiv.org/abs/0908.0189&quot; rel=&quot;nofollow&quot;&gt;Collier Cameron et al. 2009&lt;/a&gt;) is shuffle all the times of the data points (i.e. randomly reallocate the times of each observation) and go through the process again. Do this a squillion times with different random reallocations and see how many times you end up with $p&amp;gt;p_{best}$. This gives you some idea about how confident you can be that your detected periodic signal is real. This is called a &quot;bootstrap Monte-Carlo method&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other ways to do this; my answer is not complete; but it's the one I'm familiar with.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-08-25T07:25:04.307" LastActivityDate="2015-08-25T07:25:04.307" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11630" PostTypeId="2" ParentId="11613" CreationDate="2015-08-24T16:13:10.473" Score="0" Body="&lt;p&gt;Strictly speaking, the premise of your question is false.  The L4 and L5 points in a &lt;em&gt;general&lt;/em&gt; system are not necessarily stable.  Whether they are stable or not depends on the ratio of masses in the system.  It does turn out that in many systems of interest, the ratios are such that the these points are stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When they are stable, it's basically because the Coriolis force balances the gravitational force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See, for example,&#xA;&lt;a href=&quot;http://www.physics.montana.edu/faculty/cornish/lagrange.html&quot; rel=&quot;nofollow&quot;&gt;http://www.physics.montana.edu/faculty/cornish/lagrange.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8019" LastActivityDate="2015-08-24T16:13:10.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11631" PostTypeId="2" ParentId="11456" CreationDate="2015-08-24T16:55:15.323" Score="7" Body="&lt;p&gt;The 2 BCE conjunction had Regulus, Jupiter, and Venus within 5.346&#xA;degres, and the Sun was 19.371 degrees distant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 2015 conjunction had the three within 5.488 degrees, with the Sun&#xA;30.930 degrees away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 81 conjunctions between those two dates where Regulus,&#xA;Jupiter, and Venus are less than 5.488 degrees apart, although, in&#xA;some cases, the Sun would be too close to see these conjunctions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I list all of these conjunctions below, and also include conjunctions&#xA;as far back as 999 BCE and as far forward as 2999 CE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first two columns are the date and time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third column is the maximum pairwise separation (in degrees) of&#xA;Regulus, Jupiter, and Venus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fourth column is the smallest angular distance from the Sun (in&#xA;degrees) of the three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional notes follow the list:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&#xA;-998-08-13 17:47:22 4.460 17.341&#xA;-986-06-14 19:33:17 5.078 31.064&#xA;-962-06-13 20:42:13 1.599 37.154&#xA;-939-08-24 16:05:37 0.960 32.337&#xA;-903-06-23 23:23:31 4.166 23.499&#xA;-879-06-22 18:41:22 2.388 28.906&#xA;-856-09-02 02:38:31 0.900 40.426&#xA;-820-07-03 18:56:04 3.059 15.483&#xA;-796-07-02 11:36:42 3.326 19.880&#xA;-773-09-09 05:10:53 2.068 46.122&#xA;-737-07-14 22:02:17 1.914  7.202&#xA;-713-07-13 13:16:57 4.432 10.473&#xA;-654-07-25 04:03:45 0.919  0.676&#xA;-618-06-16 16:24:53 2.111 36.170&#xA;-571-08-04 08:00:55 0.890  9.482&#xA;-535-06-08 09:43:32 1.519 43.497&#xA;-512-08-15 20:18:35 4.304 21.310&#xA;-488-08-14 11:02:13 2.216 17.658&#xA;-452-06-14 20:40:52 1.507 38.039&#xA;-429-08-26 12:36:08 3.022 30.516&#xA;-405-08-25 05:36:22 3.500 25.655&#xA;-369-06-24 14:26:28 1.383 30.683&#xA;-346-09-04 07:24:03 2.055 38.915&#xA;-322-09-03 07:20:58 4.678 33.128&#xA;-310-07-05 12:23:59 4.755 15.643&#xA;-286-07-04 05:57:12 1.723 21.719&#xA;-263-09-10 01:35:27 1.743 44.339&#xA;-239-09-10 16:58:48 5.467 39.382&#xA;-227-07-15 14:17:13 3.299  7.717&#xA;-203-07-14 05:55:31 3.137 12.390&#xA;-156-09-12 06:11:05 5.008 41.107&#xA;-144-07-25 20:01:44 1.779  0.850&#xA;-132-06-30 22:00:27 4.887 21.625&#xA;-120-07-24 10:48:43 4.594  2.846&#xA; -61-08-06 01:53:32 0.781  8.312&#xA; -25-06-11 09:01:53 2.012 43.414&#xA;  -2-08-17 13:43:06 5.346 19.371 (conjunction in 2 BCE)&#xA;&#xA;  -1-06-11 09:00:23 4.978 45.464&#xA;  22-08-16 03:46:55 1.066 16.885&#xA;  58-06-16 21:31:59 1.924 38.499&#xA;  81-08-27 08:28:28 4.344 28.668&#xA;  82-06-16 01:28:35 4.809 41.268&#xA; 105-08-26 00:47:42 2.203 25.078&#xA; 141-06-25 09:01:41 1.347 31.319&#xA; 164-09-05 10:04:04 3.548 37.319&#xA; 165-06-24 06:31:05 5.379 33.701&#xA; 188-09-04 07:11:50 3.049 33.055&#xA; 224-07-04 23:17:54 1.226 23.421&#xA; 247-09-13 15:10:43 3.233 44.173&#xA; 271-09-13 05:17:58 3.587 40.061&#xA; 283-07-17 06:44:26 5.353  7.568&#xA; 307-07-15 23:33:14 1.106 14.268&#xA; 330-09-10 21:05:11 4.715 41.177&#xA; 354-09-17 08:06:60 3.418 44.034&#xA; 366-07-27 11:51:60 3.985  0.992&#xA; 390-07-26 03:05:29 2.339  4.782&#xA; 449-08-06 18:09:31 2.584  6.358&#xA; 473-08-05 08:52:09 3.774  1.408&#xA; 485-06-13 23:13:34 3.793 41.669&#xA; 509-06-14 10:03:10 3.248 45.079&#xA; 532-08-16 22:56:25 1.136 15.985&#xA; 556-08-15 13:18:21 5.306  9.232&#xA; 568-06-18 04:59:43 3.729 37.949&#xA; 592-06-17 10:00:36 3.097 42.534&#xA; 615-08-27 17:07:18 0.513 24.909&#xA; 651-06-27 07:39:52 2.799 31.415&#xA; 674-09-07 10:08:57 5.021 35.627&#xA; 675-06-26 05:23:47 3.875 35.349&#xA; 698-09-06 05:27:18 1.613 32.746&#xA; 734-07-06 15:41:56 1.585 24.008&#xA; 757-09-15 06:00:16 4.344 43.078&#xA; 758-07-05 10:40:10 4.927 26.804&#xA; 781-09-14 13:11:04 2.606 39.663&#xA; 817-07-16 15:15:47 1.096 16.042&#xA; 840-09-17 11:49:26 4.635 45.023&#xA; 864-09-19 23:02:46 2.770 44.576&#xA; 876-07-28 03:47:10 5.109  0.757&#xA; 900-07-26 19:38:38 1.354  6.699&#xA; 959-08-08 09:59:38 3.685  4.399&#xA; 983-08-07 01:00:51 2.596  0.922&#xA; 995-06-18 07:09:17 4.556 40.276&#xA;1019-06-19 20:30:29 2.698 43.446&#xA;1042-08-18 14:56:17 2.469 14.017&#xA;1066-08-17 05:54:20 3.837  8.756&#xA;1078-06-20 14:43:27 5.059 37.822&#xA;1102-06-19 21:49:33 1.707 43.667&#xA;1125-08-28 15:18:36 1.383 23.486&#xA;1149-08-27 06:37:45 4.987 17.075&#xA;1161-06-28 07:08:46 4.549 31.267&#xA;1185-06-27 05:43:18 2.039 36.937&#xA;1208-09-07 05:59:08 0.502 32.528&#xA;1244-07-07 11:31:14 3.647 23.669&#xA;1268-07-06 06:22:18 2.826 28.583&#xA;1291-09-17 00:48:31 0.553 40.580&#xA;1327-07-18 07:04:40 2.460 15.742&#xA;1351-07-17 00:10:09 3.975 19.529&#xA;1374-09-23 07:58:08 1.568 46.055&#xA;1410-07-28 09:20:39 1.073  7.770&#xA;1434-07-27 02:11:36 5.317 10.122&#xA;1493-08-07 16:51:42 0.882  0.938&#xA;1529-06-29 06:47:15 2.990 37.035&#xA;1552-08-19 07:13:29 4.307 12.057&#xA;1576-08-17 22:28:03 2.198  8.457&#xA;1612-07-01 16:15:13 0.936 44.541&#xA;1635-09-09 08:38:22 2.760 21.568&#xA;1659-09-08 00:14:25 3.635 16.511&#xA;1695-07-09 07:11:03 1.123 38.488&#xA;1718-09-20 00:58:55 1.552 30.735&#xA;1742-09-18 17:58:01 4.959 24.413&#xA;1754-07-20 07:11:04 4.686 24.403&#xA;1778-07-19 02:46:21 1.812 30.338&#xA;1801-09-29 19:39:12 0.682 39.105&#xA;1837-07-31 00:36:14 3.503 16.558&#xA;1861-07-29 17:54:30 2.897 21.392&#xA;1884-10-06 07:55:46 1.204 45.253&#xA;1920-08-11 02:21:34 2.259  8.435&#xA;1944-08-09 18:40:08 4.058 12.042&#xA;2003-08-22 07:15:20 1.041  0.825&#xA;&#xA;2015-07-16 11:48:33 5.488 30.930 (conjunction in 2015 CE)&#xA;2027-08-20 23:44:27 5.261  2.523&#xA;2086-09-01 14:32:37 0.790  8.374&#xA;2122-07-08 16:51:26 1.516 43.744&#xA;2145-09-14 01:53:17 4.836 19.637&#xA;2146-07-08 12:23:30 5.417 45.470&#xA;2169-09-12 17:38:23 1.559 16.679&#xA;2205-07-15 09:10:00 1.372 38.734&#xA;2228-09-24 20:25:03 3.776 28.900&#xA;2229-07-14 10:39:27 5.357 41.055&#xA;2252-09-23 13:21:48 2.792 24.725&#xA;2288-07-23 23:00:59 1.054 31.616&#xA;2311-10-05 21:49:18 2.753 37.499&#xA;2335-10-04 18:19:54 3.841 32.416&#xA;2371-08-04 11:28:25 1.029 23.258&#xA;2394-10-13 02:17:26 2.210 44.307&#xA;2418-10-12 12:48:16 4.632 38.993&#xA;2430-08-15 18:58:06 4.175  8.408&#xA;2454-08-14 11:37:59 2.188 13.939&#xA;2477-10-10 03:40:59 3.897 41.142&#xA;2501-10-16 12:47:51 4.549 41.762&#xA;2513-08-26 23:54:34 2.667  0.875&#xA;2537-08-25 15:50:58 3.690  4.447&#xA;2596-09-06 05:26:08 1.151  6.635&#xA;2620-09-05 21:40:43 5.240  0.897&#xA;2632-07-15 02:33:17 2.386 43.108&#xA;2656-07-15 11:20:55 4.650 45.236&#xA;2679-09-18 10:25:21 0.705 15.875&#xA;2715-07-21 15:26:25 2.319 39.102&#xA;2738-09-30 15:29:27 4.810 27.036&#xA;2739-07-20 18:15:52 4.425 42.356&#xA;2762-09-29 08:12:17 1.644 24.014&#xA;2798-07-29 19:09:49 1.599 32.315&#xA;2821-10-09 21:38:10 3.932 35.818&#xA;2822-07-28 16:03:44 4.946 35.075&#xA;2845-10-08 17:02:38 2.697 31.836&#xA;2881-08-08 05:45:37 0.984 24.514&#xA;2904-10-18 16:53:19 3.435 43.217&#xA;2928-10-17 21:25:11 3.325 38.965&#xA;2964-08-19 03:36:41 0.956 15.872&#xA;2987-10-21 20:20:21 4.041 45.058&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You can (and should) check these numbers against a reliable source, such as Stellarium or HORIZONS (&lt;a href=&quot;http://ssd.jpl.nasa.gov/?horizons&quot; rel=&quot;noreferrer&quot;&gt;http://ssd.jpl.nasa.gov/?horizons&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;These numbers are imperfect for several reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Like most planetarium programs, I neglect light travel&#xA;time (Stellarium neglects light travel time by default, but you&#xA;can change this in the settings). This is probably the largest&#xA;error in the numbers above.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;NASA solves differential equations from known constants to&#xA;publish planetary positions. The constants aren't necessarily&#xA;accurate (they are updated occasionally), and NASA publishes only&#xA;approximations to the differential equations solutions. The&#xA;approximations are usually good within a few meters, but if the&#xA;constants were/are drastically different in the past/future&#xA;(and/or are erroneous), these results would not apply.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;My methodology:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I used the SPICE kernels&#xA;(&lt;a href=&quot;http://naif.jpl.nasa.gov/pub/naif/toolkit_docs/C/req/kernel.html&quot; rel=&quot;noreferrer&quot;&gt;http://naif.jpl.nasa.gov/pub/naif/toolkit_docs/C/req/kernel.html&lt;/a&gt;)&#xA;to find the positions of Jupiter and Venus (in the ICRF J2000&#xA;frame) daily from 999 BCE to 2999 CE.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I assumed the position of Regulus was constant in the ICRF J2000&#xA;frame. Since ICRF J2000 is a non-precessing frame, this is&#xA;essentially accurate, but neglects Regulus' small proper motion.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I computed the daily maximal angular separation between Jupiter,&#xA;Venus, and Regulus.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I found local minima among the daily separations, and used the&#xA;ternary method to find the instant of the actual local minimum.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I looked at the separations for 2 BCE and 2015 CE, and filtered&#xA;the results to only show conjunctions with separations less than&#xA;the maximum of these two separations.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I computed the position of the Sun on the filtered list, and&#xA;included the separation between the Sun and the closest of the&#xA;Jupiter, Regulus, and Venus.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I did most of the work in Mathematica, but used the Unix program&#xA;j2d to convert Julian dates to calendar dates, because Mathematica&#xA;uses the proleptic Gregorian calendar, which most people do not&#xA;use.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can see what I did (in extremely cluttered form) at:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/barrycarter/bcapps/tree/master/ASTRO&quot; rel=&quot;noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/tree/master/ASTRO&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I wanted to present the results in a sortable HTML table, but&#xA;stackexchange doesn't allow tables of any sort.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Here are Stellarium screenshots of some of these&#xA;conjunctions. Regulus is the light blue star, Venus is bright&#xA;yellow, and Jupiter is the one with the moons. The object I've&#xA;selected (if any) is not necessarily relevant to the conjunction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dJts6.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dJts6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/htK2g.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/htK2g.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/mp19i.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mp19i.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/t18gY.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t18gY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2015-08-24T16:55:15.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11632" PostTypeId="2" ParentId="11626" CreationDate="2015-08-24T19:27:01.247" Score="3" Body="&lt;p&gt;For the first part of your question: A real black hole is formed from a collapsing start is likely to have angular momentum.  In theory a non-spinning black hole could be formed by collapse of a non-rotating star.  In theory, you could potentially also realize a non-rotating black hole if angular momentum were extracted from a rotating black hole via the Penrose process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Penrose_process&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Penrose_process&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part of your question, I think it's not quite clear what you mean.  The best that I can understand your question, I would say that matter is &quot;absorbed&quot; the same, although the path taken into the black hole may depend on its spin.&lt;/p&gt;&#xA;" OwnerUserId="8019" LastActivityDate="2015-08-24T19:27:01.247" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11633" PostTypeId="1" CreationDate="2015-08-25T00:28:54.840" Score="2" ViewCount="109" Body="&lt;p&gt;Since a few days after the flyby there are no more images published on the NASA page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is NASA waiting to publish any more images for some reason?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is there something happened to New Horizon?&lt;/p&gt;&#xA;" OwnerUserId="5501" LastEditorUserId="5501" LastEditDate="2015-08-25T03:31:28.583" LastActivityDate="2015-08-25T12:53:21.657" Title="Why are there no more images revealed by New Horizon?" Tags="&lt;pluto&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11634" PostTypeId="1" AcceptedAnswerId="11938" CreationDate="2015-08-25T00:56:27.563" Score="8" ViewCount="234592" Body="&lt;p&gt;I was camping this weekend away from the city, so my friends and I decided to spend one of the nights watching the sky, since it's impossible to see anything around where we live. After a little while, of staring up, I saw what looked like a star moving. It moved fast, but not a long distance, and various directions and back. It was about the same size of the other stars near by. It was not as bright as other stars though. I stopped looking up for a little bit, to make sure my eyes were seeing things correctly. But then I looked up again I continued to see the light moving. I thought I was going crazy, so I pointed it out to my friend. It took him a little while, but after a few minutes he also saw it. His girlfriend also saw it a few minutes later. My friend brought up something about satellite reflections and other theories, but we have no idea what we saw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what could it be?&lt;/p&gt;&#xA;" OwnerUserId="8022" LastEditorUserId="8022" LastEditDate="2017-07-16T18:17:47.397" LastActivityDate="2017-10-30T13:22:46.647" Title="Star like light moving in the sky, what could it be?" Tags="&lt;star&gt;&lt;light&gt;" AnswerCount="3" CommentCount="14" FavoriteCount="9" ClosedDate="2017-10-30T13:22:53.447" ContentLicense="CC BY-SA 3.0" />
  <row Id="11635" PostTypeId="1" AcceptedAnswerId="11637" CreationDate="2015-08-25T03:11:15.817" Score="2" ViewCount="636" Body="&lt;p&gt;What is the largest possible size for a brown dwarf before it has enough mass, capable of sustained fusion, thus becoming a star? &lt;/p&gt;&#xA;" OwnerUserId="7926" LastEditorUserId="7926" LastEditDate="2015-08-25T03:17:50.033" LastActivityDate="2015-08-25T09:00:28.930" Title="What is the largest possible size for a brown dwarf?" Tags="&lt;brown-dwarf&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11636" PostTypeId="2" ParentId="11616" CreationDate="2015-08-25T05:17:27.877" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Running the math for a 5 meter long pendulum and 1 kg mass, I get an amplitude of 0,017 mm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are off by quite a bit. There is essentially no horizontal deflection when the Moon is at the horizon. The maximum horizontal deflection occurs when the Moon is about 45 degrees above or below the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tidal acceleration at some point on the surface of the Earth due to some external body (e.g., the Moon) is the difference between the gravitational acceleration toward that external body at the point in question and the gravitational acceleration of the Earth as a whole toward that body:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec a_{rel} = GM_\text{body} \left(\frac {\vec R - \vec r} {||\vec R - \vec r||^3} - \frac {\vec R} {||\vec R||^3} \right)$$&#xA;where&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$M_{body}$ is the mass of the external body,&lt;/li&gt;&#xA;&lt;li&gt;$\vec R$ is the displacement vector from the center of the Earth to the center of the external body, and&lt;/li&gt;&#xA;&lt;li&gt;$\vec r$ is the displacement vector from the center of the Earth to the point in question on the surface of the Earth.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The resultant force looks like this:&lt;br&gt;&#xA;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/d/d8/Field_tidal.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the tidal force is away from the center of the Earth when the Moon is directly overhead or directly underfoot, toward the center of the Earth when the Moon is on the horizon, and horizontal when the Moon is halfway between directly overhead/underfoot and on the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tidal force is maximum when the Moon is directly overhead, and even then it's only about 10&lt;sup&gt;-7&lt;/sup&gt; &lt;em&gt;g&lt;/em&gt;. You need a sensitive instrument to read that. A simple pendulum or a simple spring will not do the trick.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-08-25T05:17:27.877" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11637" PostTypeId="2" ParentId="11635" CreationDate="2015-08-25T07:43:47.633" Score="2" Body="&lt;p&gt;I &lt;em&gt;think&lt;/em&gt; what you are asking is what is the radius of an object that has the maximum mass to be classed as a brown dwarf?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass threshold for this is about $0.075 M_{\odot}$. The boundary is a bit fuzzy - there is still &lt;em&gt;some&lt;/em&gt; hydrogen fusion, but not enough to produce the luminosity of the object), it also depends slightly on the chemical composition of the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the fact that you can still see (with a telescope!) brown dwarfs raises a more fundamental problem with your question. Their luminosity comes from gravitational potential energy; they are contracting all the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can only ask what is the radius of the most massive brown dwarf at a given age?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oldest (about 10 billion years, though there is no good way to estimate the age of a brown dwarf in the field, or measure its mass), most massive brown dwarf's should be about the size of Jupiter. A similar brown dwarf in a star forming region, born a few million years ago, should be several times bigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plot below (from &lt;a href=&quot;http://adsabs.harvard.edu/abs/1986ApJ...311..226N&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nelson, Rappaport &amp;amp; Joss 1986&lt;/a&gt; - newer models are available, but show the same behaviour) shows the radius of substellar objects as a function of time. The plateau at early times is caused by Deuterium fusion in object more massive than about $0.012M_{\odot}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4ruoD.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4ruoD.jpg&quot; alt=&quot;Brown dwarf radius versus time from Nelson et al. (1986).&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-08-25T09:00:28.930" LastActivityDate="2015-08-25T09:00:28.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11638" PostTypeId="2" ParentId="11622" CreationDate="2015-08-25T11:51:09.080" Score="2" Body="&lt;p&gt;I am assuming you are talking about how long the event lasts. See question on this site &quot;How long does a supernova explosion last?&quot;. The answer is that if you are talking about how long it brightens then dims (its light curve), then it typically takes a couple of days to get to full brightness then weeks to months to fade. Core collapse can be a fraction of a second (keep in mind there are different kinds of supernovae).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;fastest&quot; pulsar found so far spins 716 times a second. If you really want fast, how about the standard for atomic clocks? The used transition is between two ground states of the element caesium that are very close together and lasts about $1\times 10^{-10}$ seconds. There are even better atomic clocks now.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="7595" LastEditDate="2015-08-27T22:21:48.930" LastActivityDate="2015-08-27T22:21:48.930" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11639" PostTypeId="1" AcceptedAnswerId="11641" CreationDate="2015-08-25T12:11:40.660" Score="3" ViewCount="126" Body="&lt;p&gt;I am a bit confused with point 2.d of the IAU's resolution on dwarf planets.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;(2) A &amp;quot;dwarf planet&amp;quot; is a celestial body that&lt;/p&gt;&#xA;&lt;p&gt;(a) is in orbit around the Sun&lt;/p&gt;&#xA;&lt;p&gt;(b) has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape&lt;/p&gt;&#xA;&lt;p&gt;(c) has not cleared the neighbourhood around its orbit&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;(d) is not a satellite.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.iau.org/static/resolutions/Resolution_GA26-5-6.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.iau.org/static/resolutions/Resolution_GA26-5-6.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;I notice that a lot of websites exclude this point all together? Some of&#xA;them even rather combine it with point 2.a ? Like over &lt;a href=&quot;http://lasp.colorado.edu/education/outerplanets/kbos_dwarfplanets.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Feels a bit redundant, since orbits the sun implies not orbiting a planet?&lt;/p&gt;&#xA;&lt;p&gt;Or am I missing something here?&lt;/p&gt;&#xA;" OwnerUserId="8026" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-08-25T15:37:38.093" Title="IAU Dwarf planet definition Point 2.b - is not a satellite" Tags="&lt;dwarf-planets&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11640" PostTypeId="2" ParentId="11633" CreationDate="2015-08-25T12:53:21.657" Score="4" Body="&lt;p&gt;&lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/2015/06240556-what-to-expect-new-horizons-pluto.html&quot; rel=&quot;nofollow&quot;&gt;http://www.planetary.org/blogs/emily-lakdawalla/2015/06240556-what-to-expect-new-horizons-pluto.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The transmission of the High Priority data set will be complete on July 20, and then image transmission will pause. For nearly two months, until September 14, New Horizons will switch to near-real-time downlinking of data from other, so-called &quot;low-speed&quot; instruments while it transmits just housekeeping information for all of the rest of the data. No new images will arrive on the ground during this time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5501" LastActivityDate="2015-08-25T12:53:21.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11641" PostTypeId="2" ParentId="11639" CreationDate="2015-08-25T13:19:58.550" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Or am I missing something here?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. You are missing the word &quot;Sun&quot;. The IAU did not want to tackle the problem of how to categorize exoplanets. There are only eight planets in the entire universe, plus a good number of (five and counting) dwarf planets. An expoplanet orbiting Alpha Centauri is not a planet or a dwarf planet, per the IAU definition. Some hypothesize that the early solar system contained five giant planets, but one of them was expelled. That rogue giant planet, if it exists, is not a planet per the IAU definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are treating the word &quot;orbit&quot; as if it's a mutually exclusive term. It's not. The Earth's Moon, for example, orbits the Earth and it orbits the Sun. But since the Moon is a satellite of the Earth, it's not a planet or a dwarf planet. Side note: the concept of what distinguishes a &quot;satellite&quot; from a non-satellite is not defined.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2015-08-25T13:30:59.903" LastActivityDate="2015-08-25T13:30:59.903" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11642" PostTypeId="1" AcceptedAnswerId="11643" CreationDate="2015-08-25T14:37:56.073" Score="2" ViewCount="113" Body="&lt;p&gt;Using the engines today how long will take humanity to reach earth like planets?&lt;/p&gt;&#xA;" OwnerUserId="8028" LastActivityDate="2015-08-25T15:23:45.640" Title="How long will it take to reach the closest earth like planet" Tags="&lt;space&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11643" PostTypeId="2" ParentId="11642" CreationDate="2015-08-25T15:23:45.640" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://www.washingtonpost.com/news/speaking-of-science/wp/2015/07/31/nasa-just-found-the-closest-super-earth-ever/&quot; rel=&quot;nofollow&quot;&gt;HD 219134b is apparently closest &quot;super earth&quot;&lt;/a&gt;; 21 light years away.&#xA;New Horizons, our speediest yet, is currently running at &lt;a href=&quot;http://www.livecometdata.com/new-horizons-spacecraft/&quot; rel=&quot;nofollow&quot;&gt;about 15 km/sec relative to the sun&lt;/a&gt;.&#xA;At 9.46e12 km/lightyear HD 219134b is 2e14 km away &#xA;At 3.16e7 seconds per year, New Horizons will travel about 4.74e8 km/year.&#xA;2e14/4.74e8 = &lt;strong&gt;422000 years&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HD 219134b isn't even a very good earth; gravity is too high.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-08-25T15:23:45.640" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11647" PostTypeId="2" ParentId="6519" CreationDate="2015-08-26T06:57:27.417" Score="1" Body="&lt;p&gt;I believe NASA is stating it is not only when solar wind but gravitational pull shifts... That's not to say the sun has no pull or solar wind but that the influence of the sun is now less than the surrounding environment. To put it simply, when the sun is no longer winning the tug of war.&lt;/p&gt;&#xA;" OwnerUserId="8035" LastActivityDate="2015-08-26T06:57:27.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11649" PostTypeId="2" ParentId="11592" CreationDate="2015-08-26T09:39:52.877" Score="2" Body="&lt;p&gt;Newtonian treatments of the bending of light go back to Laplace who, in 1798, wrote about light escaping from massive bodies, ie: black holes!  See Appendix A of Hawking and Ellis &quot;Large Scale Structure of Space-Time&quot;  where there is a nice translation of Laplace's paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newtonian treatments cannot properly deal with all aspects of light bending.  Notably, the important difference between the 'Luminosity Distance' and the 'Angular Diameter Distance' of a cosmological object is only a feature of Einstein-style theories of gravitation.  It is that difference which, for example, enables us to test different cosmological models (as in the Alcock-Paczynski test).  See for example the paper of Anderson etal. in arXiv:1303.4666 for technical details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This should probably have been a comment - but I don't have the rep for that).&lt;/p&gt;&#xA;" OwnerUserId="5301" LastActivityDate="2015-08-26T09:39:52.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11652" PostTypeId="2" ParentId="11622" CreationDate="2015-08-26T12:50:13.450" Score="2" Body="&lt;p&gt;If you count unique events as well, the fastest known event would be the exponential expansion of the Universe during &lt;a href=&quot;https://en.wikipedia.org/wiki/Inflation_%28cosmology%29&quot; rel=&quot;nofollow&quot;&gt;inflation&lt;/a&gt; which lasts from about $10^{-36}\,\rm s$ to $10^{-32}\,\rm s$ right after the Big Bang. In this time, the Universe grew by a factor of about $10^{26}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to note, however, that space itself expands, so you are not limited to the speed of light which would be the case for phenomena such as stellar explosions or pulsars.&lt;/p&gt;&#xA;" OwnerUserId="7595" LastActivityDate="2015-08-26T12:50:13.450" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11655" PostTypeId="1" AcceptedAnswerId="11656" CreationDate="2015-08-26T20:28:42.443" Score="3" ViewCount="111" Body="&lt;p&gt;A few years or so ago there was a project that distributed small very low cost telescopes (perhaps around US$20). What was the name of (and links to) that project?&lt;/p&gt;&#xA;" OwnerUserId="8043" LastActivityDate="2015-08-26T21:22:26.217" Title="Ultra Low Cost Telescope Project" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11656" PostTypeId="2" ParentId="11655" CreationDate="2015-08-26T21:16:30.730" Score="4" Body="&lt;p&gt;If you're interested in a low cost, but usable telescope that provides decent performance, look into the Galileoscope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://galileoscope.org/&quot; rel=&quot;nofollow&quot;&gt;http://galileoscope.org/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It provides 50 mm (2 in) of aperture, at a focal length of 500 mm - so it's an f/10 instrument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kit is very easy to assemble, and instructions are provided on the Internet in great detail. Even kids can assemble it, with some supervision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objective lens (the primary optics) is an achromat doublet - it's made of two actual lenses combined to reduce chromatic aberration. I was impressed by the quality of that lens, at least when compared to its price - it seems to work as well as optics in instruments that cost several times more. This is one of the main reasons why I recommend this instrument to beginners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It uses standard 1.25&quot; eyepieces. That means you can either use the eyepieces that come with the kit, or you could borrow any 1.25&quot; eyepiece from another telescope and just use it. This is rare in an instrument this size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kit gives you the option to build either a galilean eyepiece, or a 2x barlow. My advice: don't bother with the galilean eyepiece; its field is far too narrow. Instead, build the 2x barlow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's another eyepiece in the kit, use it either alone or combined with the barlow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory, the maximum magnification is 100x, but in practice things will get a bit difficult when you push magnification much over 50% of the max value. This is typical for many telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Important: Like any telescope used for astronomy, this scope will not work well if you just hold it with your hands. You must use some kind of mount. The scope comes with a standard mounting nut that fits any tripod used for photography or astronomy. Use the sturdiest, most rigid tripod you can find. Even a very little amount of wobble will make it very difficult for you to use the scope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, make and use a DIY mount such as this one (results will vary depending on skill and materials used):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://aquillam.wordpress.com/2010/01/04/a-cheap-tripod-alternative/&quot; rel=&quot;nofollow&quot;&gt;https://aquillam.wordpress.com/2010/01/04/a-cheap-tripod-alternative/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the very least, support one end of the telescope on a fixed, tall object such as a fence; this will make things a little better, but it's not a replacement for a true mount.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-08-26T21:22:26.217" LastActivityDate="2015-08-26T21:22:26.217" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11657" PostTypeId="2" ParentId="11615" CreationDate="2015-08-26T22:11:49.607" Score="9" Body="&lt;p&gt;The key issue for a telescope is not magnification but light gathering capability - which is (crudely) the size of the main aperture. Thus a 90x magification on a very large (wide) telescope would let you see a very large number of things (if you are in an area where the sky is dark), but 90x on a small telescope would let you see a number of interesting things (the Moon, planets, some nebulae and star clusters) but not relatively faint objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small, cheaper, telescopes are still worth buying if you want to dabble, seeing the Moon through a telescope is always amazing and a small telescope can show lots, for instance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, for a small telescope don't go for big magnifications - 90x would be fine for a low cost 50mm refractor for instance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for where the planets are - Venus, Mars, Jupiter, Saturn are generally easy to spot if you look up their positions online and know your way round the constellations. Because the planets are not, as stars effectively are, points at infinite distance, but small disks they (on most nights) do not twinkle as the stars do.&lt;/p&gt;&#xA;" OwnerUserId="823" LastEditorUserId="823" LastEditDate="2015-08-26T22:24:48.857" LastActivityDate="2015-08-26T22:24:48.857" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11659" PostTypeId="1" AcceptedAnswerId="11668" CreationDate="2015-08-27T05:14:20.190" Score="1" ViewCount="60" Body="&lt;p&gt;I'm researching high altitude photography and want to know a schedule for taking pictures of the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Preferably I want a picture of the (full) moon on the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go about scheduling such an event from a high altitude location?&lt;/p&gt;&#xA;" OwnerUserId="8047" LastActivityDate="2015-08-27T14:27:46.093" Title="Simulating location of the moon" Tags="&lt;the-moon&gt;&lt;photography&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11660" PostTypeId="1" AcceptedAnswerId="11664" CreationDate="2015-08-27T06:09:32.117" Score="3" ViewCount="174" Body="&lt;p&gt;What compensate the energy loss in tidal heating? Is it orbital decay?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="1678" LastEditDate="2015-08-27T09:27:50.633" LastActivityDate="2015-08-27T10:20:33.653" Title="Does tidal heating imply orbit degradation?" Tags="&lt;orbit&gt;&lt;planet&gt;&lt;tidal-forces&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11661" PostTypeId="1" CreationDate="2015-08-27T08:25:55.407" Score="0" ViewCount="44" Body="&lt;p&gt;New Horizons, speediest yet, is currently running at about 15 km/sec relative to the sun.  What is the roadmap for future ? How is it possible to reach much faster speeds ?&lt;/p&gt;&#xA;" OwnerUserId="8028" LastActivityDate="2015-08-28T22:54:58.983" Title="Road map for a faster spaceship" Tags="&lt;interstellar-travel&gt;&lt;space-travel&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-08-31T21:38:34.843" ContentLicense="CC BY-SA 3.0" />
  <row Id="11662" PostTypeId="2" ParentId="11661" CreationDate="2015-08-27T09:02:53.303" Score="2" Body="&lt;p&gt;Here's &lt;a href=&quot;http://en.wikipedia.org/wiki/Interstellar_travel&quot; rel=&quot;nofollow&quot;&gt;a Wikipedia page&lt;/a&gt; discussing various theoretical methods of propulsion. It has some very interesting ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one that piqued my interest was the Alcubierre drive. For all intents and purposes it is a warp drive - well, as close as science can get to the fictional engine of &lt;em&gt;The Enterprise&lt;/em&gt; anyway. It has been nicknamed the warp drive, I believe, by NASA scientists.&lt;/p&gt;&#xA;" OwnerUserId="7791" LastEditorUserId="2153" LastEditDate="2015-08-28T22:54:58.983" LastActivityDate="2015-08-28T22:54:58.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11663" PostTypeId="1" CreationDate="2015-08-27T09:38:24.103" Score="3" ViewCount="110" Body="&lt;p&gt;The unique characteristics of &lt;a href=&quot;http://en.wikipedia.org/wiki/B%28e%29_star&quot; rel=&quot;nofollow&quot;&gt;Be stars&lt;/a&gt; are mainly because of their &lt;strong&gt;circumstellar discs&lt;/strong&gt;.&lt;br/&gt;&#xA;I wonder what makes Be stars special that only they have unique characteristics?&lt;br/&gt;&#xA;Which kind of stars have discs generally?&lt;br/&gt;&#xA;Why are there no &lt;strong&gt;Oe and Ae&lt;/strong&gt; stars?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="2153" LastEditDate="2015-08-27T16:16:06.240" LastActivityDate="2015-08-27T16:54:49.057" Title="Why are there no Oe or Ae stars?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11664" PostTypeId="2" ParentId="11660" CreationDate="2015-08-27T10:08:31.963" Score="3" Body="&lt;p&gt;There is a wonderful &lt;a href=&quot;https://physics.stackexchange.com/questions/9290/why-does-the-moon-drift-away-from-earth&quot;&gt;post&lt;/a&gt; at Physics Stackexchange:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Gravitational coupling between the Moon and the tidal bulge nearest the Moon acts as a torque on the Earth's rotation, draining angular momentum and rotational kinetic energy from the Earth's spin. In turn, angular momentum is added to the Moon's orbit, accelerating it, which lifts the Moon into a higher orbit with a longer period. As a result, the distance between the Earth and Moon is increasing, and the Earth's spin slowing down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find more reading material and information in the original thread.&lt;/p&gt;&#xA;" OwnerUserId="7595" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-08-27T10:20:33.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11666" PostTypeId="1" AcceptedAnswerId="11667" CreationDate="2015-08-27T11:15:42.390" Score="12" ViewCount="5586" Body="&lt;p&gt;Given that the stars' distances to Earth are measured in light-years (for example, Sirius is 8.6 light-years away from Earth), what we are seeing as Sirius now is actually its state 8.6 years ago, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is possible that a star (maybe not Sirius, I don't know, it's just an example) somehow explodes and creates a supernova, and if this is the case, we will see this event 8.6 years later (I assume everything is right up to this point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is, is it possible for me while looking at the sky on a lucky day, suddenly see the explosion of a star that happened x years ago and be the first eye witness of this event? In other words, is there a technology on Earth (emphasis on &quot;on Earth&quot; here, the satellites or space shuttles do not count since they might be slightly closer to the star than the Earth is) that can see this before me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My logic is that even the greatest telescope &quot;sees&quot; whatever light it receives. So since a telescope cannot increase the speed of light it receives, it shouldn't be more fast than me. And since light is the fastest way of transferring information, I assume that I am as possible as NASA to see such an event. Is there any way this assumption is wrong?&lt;/p&gt;&#xA;" OwnerUserId="8051" LastEditorUserId="2153" LastEditDate="2015-08-27T16:57:23.080" LastActivityDate="2015-08-29T19:44:26.067" Title="Is it possible to witness a star's death?" Tags="&lt;star&gt;&lt;supernova&gt;&lt;stellar-evolution&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11667" PostTypeId="2" ParentId="11666" CreationDate="2015-08-27T11:59:38.203" Score="13" Body="&lt;p&gt;Naked eye nova are fairly common, several per year. &lt;a href=&quot;http://apod.nasa.gov/apod/ap150325.html&quot;&gt;Here's one&lt;/a&gt;.&#xA;Naked eye supernova are far rarer. SN1987a in the large Magellanic cloud &lt;a href=&quot;https://www.youtube.com/watch?v=2gOBBUUY3zk&quot;&gt;was naked eye visible&lt;/a&gt; (vid). &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_supernovae&quot;&gt;From this list&lt;/a&gt;, it appears the supernova in 1987 was the most recent naked eye supernova. &#xA;There was a &lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2008/21mar_nakedeye/&quot;&gt;naked eye gamma ray burst in 2008&lt;/a&gt;, but I don't think anyone actually got outside in time to see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have 50 years to look at the stars, you &lt;em&gt;might&lt;/em&gt; see a supernova. &#xA;If you have a small telescope, you can pick them up pretty regularly in nearby galaxies.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-08-27T11:59:38.203" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11668" PostTypeId="2" ParentId="11659" CreationDate="2015-08-27T13:38:17.020" Score="1" Body="&lt;p&gt;The US Naval Observatory has a &lt;a href=&quot;http://aa.usno.navy.mil/data/docs/RS_OneYear.php&quot; rel=&quot;nofollow&quot;&gt;web page&lt;/a&gt; which lets you print a table of moonrise/moonset times for any location, together with twilight times. If you use this approach, you would print the data and then examine it for a suitable date. I don't know of a computing resource where you would put in your requirements it would report the appropriate date and time.&lt;/p&gt;&#xA;" OwnerUserId="7385" LastActivityDate="2015-08-27T13:38:17.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11669" PostTypeId="2" ParentId="11659" CreationDate="2015-08-27T14:27:46.093" Score="1" Body="&lt;p&gt;The Isaac Newton Group of Telescopes has a web resource for plotting the altitude of celestial bodies, given a date and your latitude, longitude, altitude, etc. It is used for general RA, Dec. angles, but it automatically plots the altitude of the moon as well, for any plot that you generate. It's called &lt;a href=&quot;http://catserver.ing.iac.es/staralt/&quot; rel=&quot;nofollow&quot;&gt;Staralt&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8054" LastActivityDate="2015-08-27T14:27:46.093" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11670" PostTypeId="1" AcceptedAnswerId="11696" CreationDate="2015-08-27T15:39:26.420" Score="6" ViewCount="99" Body="&lt;p&gt;A decade or so ago, when I was still a science undergrad, one of the open questions in astrophysics was to explain the uneven distribution of galaxies in the observable universe.  That is, why did the universe look lumpy instead of evenly distributed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this still an open question, and if so who is working on it?  &lt;/p&gt;&#xA;" OwnerUserId="8055" LastActivityDate="2015-09-02T00:19:47.337" Title="Is the Astronomy community still concerned about the lumpyness of matter distribution in the universe?" Tags="&lt;cosmology&gt;&lt;galactic-dynamics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11671" PostTypeId="2" ParentId="11663" CreationDate="2015-08-27T16:54:49.057" Score="2" Body="&lt;p&gt;Well, there might be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I strongly disagree with your statement that circumstellar discs are what differentiate B(e) stars from other related stars. In fact, there are four primary criteria for stars that satisfy &quot;the B(e) phenomenon&quot;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Strong emission lines in &lt;a href=&quot;http://en.wikipedia.org/wiki/Balmer_series&quot; rel=&quot;nofollow&quot;&gt;the Balmer series&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Emission of lines of certain (low ionization) metals such as Fe II&lt;/li&gt;&#xA;&lt;li&gt;&quot;Forbidden emission lines&quot; in the visible part of the electromagnetic spectrum such as O I&lt;/li&gt;&#xA;&lt;li&gt;Hot circumstellar dust&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The dust is not the key to the B(e) phenomenon, though it can change the spectrum slightly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HAeB(e) stars are a class of stars (and a subclass of B(e) stars) related to &lt;a href=&quot;http://en.wikipedia.org/wiki/Herbig_Ae/Be_star&quot; rel=&quot;nofollow&quot;&gt;Herbig Ae/Be stars&lt;/a&gt;, which have not yet left the main sequence. There are additional criteria that a star must satisfy to be considered an HAeB(e) star, too. These stars can be of spectral type A, meaning that you can have B(e) stars that are not of spectral type B, if you use a certain definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer is based largely on &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1998A%26A...340..117L&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow&quot;&gt;Lamers et al. (1998)&lt;/a&gt;, a fascinating paper on the classification of B(e) stars.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2015-08-27T16:54:49.057" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11672" PostTypeId="2" ParentId="11666" CreationDate="2015-08-27T18:07:56.557" Score="10" Body="&lt;p&gt;Supernova create huge spikes in neutrino emissions.  Since neutrinos pass through a stellar mass mostly unimpeded, they're visible up to 3 hours before the shockwave even starts to affect the star's surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since neutrinos travel at the speed of light, they will always keep their 3 hour head start.  Thus, unless you have a neutrino detector buried a few miles below your house, you're unlikely to be the first to observe a naked eye supernova, even with a telescope pointed directly at the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first supernova definitively detected by a neutrino spike before it was visible was &lt;a href=&quot;http://en.wikipedia.org/wiki/SN_1987A#Neutrino_emissions&quot; rel=&quot;nofollow&quot;&gt;SN 1987A&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As more neutrino detectors come online, and as their ability to pinpoint the exact direction that neutrinos come from is improved, it's almost certain that the next naked eye supernova will have dozens of observatories and thousands of amature telescopes pointed at it before it's even possible for these telescopes to detect the event.&lt;/p&gt;&#xA;" OwnerUserId="8059" LastEditorUserId="2674" LastEditDate="2015-08-29T19:44:26.067" LastActivityDate="2015-08-29T19:44:26.067" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11673" PostTypeId="1" AcceptedAnswerId="11675" CreationDate="2015-08-27T18:32:30.090" Score="6" ViewCount="6284" Body="&lt;p&gt;Black Holes are regions of space where things get weird &lt;sup&gt;[Citation Needed]&lt;/sup&gt;. Past the event horizon of a black hole, any moving particle instantaneously experiences a gravitational acceleration towards the black hole that will cancel out it's current velocity, even light. That means that the gravity well of the black hole must be able to accelerate from -C¹ to 0 instantly². Given that fact, we can assume the gravitational acceleration of black holes is C/instant³. Given this, it stands to reason that in successive instants, the particle will be moving at speeds greater than C, because it is experiencing greater gravitational forces and continuous gravitational acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this actually make sense? Is there something I'm missing here? By this logic, it seems like anything inside of the event horizon of a black hole could and should move faster than C due to gravitational acceleration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I showed this question to a friend and he questioned if the hypothetical particles that were radiating from the singularity (The photon traveling exactly away from the black hole) might be hawking radiation; that is, the gravitation acceleration of a black hole is only strong enough to curve the path of light around a non-zero radius (thus not actually stopping it, but altering it's course), and not powerful enough to decelerate light. Is this actually what hawking radiation is, or is he as confused as I am?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;¹ Where movement towards the singularity would be considered a positive value, movement away from the singularity is a negative value, that is, anything moving at the speed of light away from the singularity would be moving with a velocity of -C relative to the singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;² If it couldn't accelerate from -C to 0 instantly, any photon traveling exactly away from the black hole would be able to escape the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;³ An instant is an arbitrary amount of time, it could be a fraction of a second, a second, a minute....&lt;/p&gt;&#xA;" OwnerUserId="4026" LastEditorUserId="16673" LastEditDate="2019-10-16T07:36:29.110" LastActivityDate="2019-10-23T20:05:11.530" Title="Can things move faster than light inside the event horizon of a black hole?" Tags="&lt;black-hole&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="11674" PostTypeId="1" CreationDate="2015-08-27T20:56:11.450" Score="6" ViewCount="126" Body="&lt;p&gt;Recently, I have been going out to observe the ISS when it is within sight of my location.  I have noticed that the ISS, based on the time that I have to observe the space station during any given transit, moves at a fairly constant speed across the sky, depending on the degree of elevation and maybe some other factors related to the seeing conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question relates to the different course the space station takes when traversing over my location. There may be as long as several weeks when the ISS is not within viewing parameters for my location. Why does the course across the sky vary from time to time?&lt;/p&gt;&#xA;" OwnerUserId="8061" LastEditorUserId="7258" LastEditDate="2015-08-28T06:49:01.043" LastActivityDate="2015-10-05T18:36:15.083" Title="International Space Station" Tags="&lt;amateur-observing&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11675" PostTypeId="2" ParentId="11673" CreationDate="2015-08-27T21:51:31.727" Score="20" Body="&lt;p&gt;I think your initial question is a good one, but the text gets a bit more jumbled and covers a few different points.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can things move faster than light inside the event horizon of a black&#xA;  hole?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nice question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Black Holes are regions of space where things get weird.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm 100% OK with this statement.  I think it's a true enough summary and I'm sure I've heard physicists say this too.   Even if &quot;Weird&quot; isn't a clearly defined scientific term, I'm 100% fine with this (even without a citation).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Past the event horizon of a black hole, any moving particle&#xA;  instantaneously experiences a gravitational acceleration towards the&#xA;  black hole that will cancel out it's current velocity, even light.&#xA;  &lt;strong&gt;That means that the gravity well of the black hole must be able to&#xA;  accelerate from -C* to 0 instantly&lt;/strong&gt;✝.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Are you quoting somebody here?   Anyway, this isn't quite true.   Black holes don't accelerate things from -c (which I'm guessing would be a light beam trying to fly away from the singularity but inside the event horizon), to 0 &quot;instantaneously&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps a better way to look at it is to consider curvature of space, and inside a black hole, space curves so much that all directions point to the singularity.   It's the &quot;all roads go to Rome&quot; scenario, even if you do a complete 180, you're still on a road that leads to the singularity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the temptation to look at that as deceleration, but I think that's a bad way to think about it.  Light doesn't decelerate, it follows the curvature of space.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given that fact, we can assume the gravitational acceleration of black&#xA;  holes is C/instant**. Given this, it stands to reason that in&#xA;  successive instants, the particle will be moving at speeds greater&#xA;  than C, because it is experiencing greater gravitational forces and&#xA;  continuous gravitational acceleration.&#xA;  Does this actually make sense? Is there something I'm missing here? By&#xA;  this logic, it seems like anything inside of the event horizon of a&#xA;  black hole could and should move faster than C due to gravitational&#xA;  acceleration.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;outside of a black hole, continuous acceleration would never lead to a speed greater than C.  You can accelerate for billions and trillions of years and all you'd do is just add more 9s to the right of the decimal point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You seem to be assuming that inside a black hole this can happen, but I'm not sure why you'd assume that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;continuous gravitational acceleration&quot; - no matter how strong, is no guarantee for faster than light travel.   That's logically inconsistent with the laws of relativity.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Edit: I showed this question to a friend and he questioned if the&#xA;  hypothetical particles that were radiating from the singularity (The&#xA;  photon traveling exactly away from the black hole) might be hawking&#xA;  radiation; that is, the gravitation acceleration of a black hole is&#xA;  only strong enough to curve the path of light around a non-zero radius&#xA;  (thus not actually stopping it, but altering it's course), and not&#xA;  powerful enough to decelerate light. Is this actually what hawking&#xA;  radiation is, or is he as confused as I am?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think, a more correct way to look at hawking radiation is to see it as something that forms just outside of the black hole, a particle/anti particle pair and one escapes and the other falls inside, and that's probably not 100% correct either, but the singularity itself doesn't send out particles.   Hawking radiation has to do with quantum properties of space.  It's not a property of black holes.  The black hole just happens to be unique in that it can capture one half of a virtual particle pair and the other half can escape.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also is a pretty different topic than your original question.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;*Where movement towards the singularity would be considered a positive value, movement away from the singularity is a negative value, that&#xA;  is, anything moving at the speed of light away from the singularity&#xA;  would be moving with a velocity of -C relative to the singularity.&#xA;  ✝If it couldn't accelerate from -C to 0 instantly, any photon&#xA;  traveling exactly away from the black hole would be able to escape the&#xA;  event horizon.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;**An instant is an arbitrary amount of time, it could be a fraction of a second, a second, a minute....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think it's a good idea to differentiate mass-less pure energy particles and particles with mass.   You seem to be saying that a ray of light can be traveling away from a black hole at the speed of light, get caught in the gravity, slow down and then fall back into the black hole like a ball that's tossed straight up into the air from the surface of the Earth.   That's probably not what happens.   The ray of light follows the path of space time ahead of it, which happens to be curved so much that it points into the black hole, even if, in the classical sense, the light begins by pointing away.   All space curves into the singularity once you're inside the event horizon, so there is no &quot;away from&quot; anymore.   At least, that's how I think it works.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-08-27T23:31:38.920" LastActivityDate="2015-08-27T23:31:38.920" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11676" PostTypeId="1" CreationDate="2015-08-27T22:29:11.417" Score="2" ViewCount="194" Body="&lt;p&gt;I'm having difficulty with an apparent contradiction about the sun's motion during the summer solstice above the arctic circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The YouTube video &lt;a href=&quot;https://youtube.com/watch?v=ZZcafg-meJA&quot; rel=&quot;nofollow&quot;&gt;The North Pole Entire Summer of never ending sun Summer Solstice High speed camera&lt;/a&gt; and solar motion models show the sun to move clockwise around the horizon for the full 360 degrees during the summer solstice.&lt;br /&gt;&#xA;However, when I walk in a circle counterclockwise, to represent the earth's rotation, and focus upon a fixed object outside the circle to represent the sun, I get a 180 degree motion, from &quot;east to west&quot;, and then the sun's apparent motion returns again from west to east, not the full 360 degrees which everyone reports is the actual phenomenon.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone help me to understand my dilemma?  &lt;/p&gt;&#xA;" OwnerUserId="8063" LastEditorDisplayName="user1569" LastEditDate="2015-08-28T22:11:56.957" LastActivityDate="2015-08-30T04:52:00.923" Title="Sun's apparent motion above the arctic circle during summer solstice" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11678" PostTypeId="1" AcceptedAnswerId="11681" CreationDate="2015-08-28T02:25:19.997" Score="7" ViewCount="313" Body="&lt;p&gt;It is my impression that most planetary systems have, at most, three to four planets. Is that real? Or is it just that we lack the ability to detect most planets?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="2531" LastEditDate="2015-08-28T05:31:15.483" LastActivityDate="2015-08-28T05:31:15.483" Title="Do most planetary systems have fewer planets than the solar system?" Tags="&lt;solar-system&gt;&lt;exoplanet&gt;&lt;planet&gt;&lt;planetary-formation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11680" PostTypeId="2" ParentId="11678" CreationDate="2015-08-28T05:01:56.297" Score="0" Body="&lt;p&gt;There are two things to consider when thinking about this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is that by virtue of living within our solar system, we are much closer to our planets, have a much easier time seeing and documenting them. We can point ground based telescopes at them and get reasonably detailed pictures of them. The point is, since we're so close, we're less likely to miss planets; &lt;a href=&quot;http://www.sciencedaily.com/releases/2015/01/150115083036.htm&quot; rel=&quot;nofollow&quot;&gt;But it should be noted there may still be planets in our Solar System we have not documented&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly is that we're really not all that great at identifying extra solar planets. Most techniques for doing this require the exo-planet to be rather large, on the scale or our gas giants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of these two points, it's hard to definitively answer this question. We have identified multiple solar systems containing more than one planet, but that doesn't mean we have identified ALL planets in these systems.&lt;/p&gt;&#xA;" OwnerUserId="4026" LastActivityDate="2015-08-28T05:01:56.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11681" PostTypeId="2" ParentId="11678" CreationDate="2015-08-28T05:19:36.623" Score="7" Body="&lt;p&gt;Perhaps the way to answer this is ask - could we detect the planets in our solar system if we were looking at the Sun, using current technology, from distances of many light years?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short answer is that we could detect Jupiter using the Doppler radial velocity technique, if we observed for more than 10 years (at least one orbit is required). If we were lucky, and the orientation is right, we might then also be able to detect a transit of Venus &lt;em&gt;or&lt;/em&gt; the Earth, using a satellite observatory like Kepler. Kepler could detect Earthlike planets by the transit technique, but the solar system is not &quot;flat&quot; enough that you would observe multiple transiting planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is that we would currently have seen Jupiter and maybe one other planet. Therefore we cannot at the moment conclude that 8 planets is an unusually high number; it may be quite typical. Although we do know that solar systems can be much more &lt;em&gt;densely&lt;/em&gt; populated with planets (in their &quot;terrestrial planet zones&quot;) than our own.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-08-28T05:19:36.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11682" PostTypeId="2" ParentId="11674" CreationDate="2015-08-28T07:08:46.287" Score="1" Body="&lt;p&gt;I recommend you to take a look at Wikipedia about the ISS's orbit: &lt;a href=&quot;https://en.wikipedia.org/wiki/International_Space_Station#Orbit&quot; rel=&quot;nofollow&quot;&gt;Wikiepdia - ISS: Orbit&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a very interesting video that I think will answer to your question: &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Iss.ogg&quot; rel=&quot;nofollow&quot;&gt;ISS Orbit animation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1678" LastActivityDate="2015-08-28T07:08:46.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11683" PostTypeId="1" AcceptedAnswerId="11684" CreationDate="2015-08-28T07:55:04.910" Score="2" ViewCount="360" Body="&lt;p&gt;I encountered this statement that the dark side of the moon is hidden from our sights.&#xA;Could someone please explain? &lt;/p&gt;&#xA;" OwnerUserId="551" LastEditorUserId="551" LastEditDate="2015-08-28T22:05:46.827" LastActivityDate="2015-08-28T22:05:46.827" Title="The dark side of the moon" Tags="&lt;the-moon&gt;&lt;natural-satellites&gt;&lt;satellite&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ClosedDate="2015-09-01T20:57:36.157" ContentLicense="CC BY-SA 3.0" />
  <row Id="11684" PostTypeId="2" ParentId="11683" CreationDate="2015-08-28T09:38:27.197" Score="7" Body="&lt;p&gt;Yes, the dark side of the moon is mostly hidden from earthbound observers. This is due to &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_locking&quot; rel=&quot;nofollow noreferrer&quot;&gt;tidal locking&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A tidally locked body takes just as long to rotate around its own axis as it does to revolve around its partner. This causes one hemisphere constantly to face the partner body.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, one moon-year is just as long as one moon-day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left-hand side shows the actual configuration of earth and moon: The orbit is locked and we can only observe one side of the moon. On the (fictional) right-hand side, the moon does not rotate around itself and we can observe different areas of the moon at different times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JclGv.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JclGv.gif&quot; alt=&quot;Left: Tidally locked orbit. Right: Orbit without any spin.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should note that we can actually observe not 50%, but around 59% of the moon's surface because of &lt;a href=&quot;https://en.wikipedia.org/wiki/Libration&quot; rel=&quot;nofollow noreferrer&quot;&gt;libration&lt;/a&gt; and parallax effects that cause the moon to slightly &quot;wobble&quot;:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/wRTb2.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wRTb2.gif&quot; alt=&quot;Simulated view of the moon, demonstrating librations.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find all those images and further explanation in the linked wikipedia articles on &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_locking&quot; rel=&quot;nofollow noreferrer&quot;&gt;tidal locking&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Libration&quot; rel=&quot;nofollow noreferrer&quot;&gt;libration&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7595" LastEditorUserId="7595" LastEditDate="2015-08-28T11:19:18.230" LastActivityDate="2015-08-28T11:19:18.230" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11686" PostTypeId="1" CreationDate="2015-08-28T16:54:57.523" Score="7" ViewCount="1036" Body="&lt;p&gt;The atoms inside the core of sun are continuously fusing and forming other atoms. Can this continuous fusing make gold atoms?&lt;/p&gt;&#xA;" OwnerUserId="8033" LastEditorUserId="2188" LastEditDate="2015-08-29T22:40:22.993" LastActivityDate="2019-03-27T00:29:37.160" Title="Can the Sun become a big ball of gold atoms?" Tags="&lt;the-sun&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11687" PostTypeId="1" AcceptedAnswerId="11698" CreationDate="2015-08-28T18:57:11.253" Score="2" ViewCount="96" Body="&lt;p&gt;I saw a talk recently mentioning that another factor that may determine a star's metallicity (besides age) may be how far from the center of its &quot;birth cluster&quot; it is when the short lived hot central stars of the cluster go supernova. I wonder if someone could comment on this. I also wonder if, since now we have better data, have we found more of a continuum of metallicities or is the &quot;Population I, Population II, Population III&quot; idea still going strong??&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2015-08-29T22:37:05.537" Title="stellar metallicity" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11689" PostTypeId="1" CreationDate="2015-08-28T19:35:43.743" Score="6" ViewCount="477" Body="&lt;p&gt;All I could come up with is it depends on the atmospheric composition of the planet as they will interact with the charged solar particles which might be true but is there is there a definite reason behind the different colors of aurorae on other worlds (Saturn's blue aurora for example)?&lt;/p&gt;&#xA;" OwnerUserId="8073" LastActivityDate="2018-12-26T20:01:00.713" Title="What causes the aurora of other planets to have different colors?" Tags="&lt;planetary-atmosphere&gt;&lt;planet&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11690" PostTypeId="1" AcceptedAnswerId="11693" CreationDate="2015-08-28T20:53:59.257" Score="3" ViewCount="431" Body="&lt;p&gt;I know Seyfert galaxies are types of active galaxies, but I do not understand what separates them from other types of active galaxies.  Are they brighter/not as bright as other types of active galaxies?  Do they emit in different wavelengths relative to other active galaxies?  I've read that 10% of all galaxies are Seyfert galaxies, so it seems to me that understanding what these galaxies are is very important to understanding galaxies overall.&lt;/p&gt;&#xA;" OwnerUserId="128" LastEditorUserId="37191" LastEditDate="2021-04-15T10:54:19.640" LastActivityDate="2021-04-15T10:54:19.640" Title="What is a Seyfert galaxy?" Tags="&lt;galaxy&gt;&lt;active-galaxy&gt;&lt;seyfert-galaxy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11691" PostTypeId="2" ParentId="11686" CreationDate="2015-08-28T20:58:41.663" Score="13" Body="&lt;p&gt;Gold is heavier (further down the periodic table of elements, not necessarily more dense, although this is also the case) than iron. Iron's elemental number is 26. Gold is 79.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any star that attempts to fuse iron will go into a supernova phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At that point, gold and all of the elements heavier than iron will be created, but only in the brief moments while the star is tearing itself apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No star can ever create gold in any circumstances other than during a supernova.  Our own sun is not nearly large enough to go supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any supernova will create gold, not just the type of supernova described above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More reading: &lt;a href=&quot;https://en.wikipedia.org/wiki/Type_II_supernova&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Type_II_supernova&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8059" LastEditorUserId="8059" LastEditDate="2015-08-31T04:05:58.100" LastActivityDate="2015-08-31T04:05:58.100" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11693" PostTypeId="2" ParentId="11690" CreationDate="2015-08-28T22:09:51.493" Score="4" Body="&lt;p&gt;Seyfert galaxies differ from other active galaxies (most notably quasars) in that their galactic nuclei are lower in luminosity compared to the rest of the galaxy. Quasars have nuclei that easily outshine the rest of the galaxy. Seyfert galaxies, on the other hand, host active nuclei that do not outshine the rest of the galaxy by the same amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly enough, in his original analysis, Seyfert focused on emission lines, noting that in this class of galaxies, there are strong high-ionization emission lines present in certain parts of the spectrum. This alternate definition is also used today. LINERS (Low Ionization Nuclear Emission line RegionS) are very similar to Seyfert galaxies if this definition is used; they are differentiated because LINERS also contain low-ionization emission lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1943ApJ....97...28S&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow&quot;&gt;Seyfert (1943)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.astro.rug.nl/~gpopp/projects/agnhostgalaxies.pdf&quot; rel=&quot;nofollow&quot;&gt;Popping (2008)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ned.ipac.caltech.edu/level5/Glossary/Essay_seyfert.html&quot; rel=&quot;nofollow&quot;&gt;A Knowledgebase for Extragalactic Astronomy and Cosmology: Seyfert Galaxies&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-08-29T20:37:42.893" LastActivityDate="2015-08-29T20:37:42.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11695" PostTypeId="1" CreationDate="2015-08-29T03:01:58.770" Score="-4" ViewCount="145" Body="&lt;p&gt;What will happen to us are we going to move in a tangential direction with earth axis of rotation ?&lt;/p&gt;&#xA;" OwnerUserId="8033" LastActivityDate="2015-09-03T07:27:43.280" Title="What if the earth stopped rotating and revolving at the same time?" Tags="&lt;earth&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2015-09-09T03:27:28.693" ContentLicense="CC BY-SA 3.0" />
  <row Id="11696" PostTypeId="2" ParentId="11670" CreationDate="2015-08-29T03:39:24.357" Score="2" Body="&lt;p&gt;The &quot;lumpy&quot; distribution of galaxies in the universe is now well understood.  In the current cosmological paradigm, the very early Universe was seeded with fluctuations in the distribution of matter and energy.  Some regions of the universe happened to be slightly more dense than average, and some happened to be slightly under-dense.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over time, these perturbations evolve.  Gravity, for instance, will cause the initially over-dense regions to become more over-dense.  When galaxy formation occurs, it usually happens in the dense regions where there is more material with which to form galaxies.  Therefore, galaxies will tend to trace (to some degree) the underly fluctuations in the matter density.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, the lumpy distribution of galaxies results from (a) the initial density fluctuations set down in the early universe, and (b) the evolution of these fluctuations according to well known laws of physics.  Cosmologists frequently model the lumpy distribution of galaxies using, for instance, numerical simulations or theoretical models; we find that these models agree extremely well with the observed distribution of galaxies in our Universe.  In fact, measurement of the galaxy distribution is used as a powerful tool to test and constrain our cosmological models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the above answer doesn't explain the process responsible for laying down the initial density perturbations.  One theory that attempts to address this question is called Inflation.&lt;/p&gt;&#xA;" OwnerUserId="8075" LastActivityDate="2015-08-29T03:39:24.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11697" PostTypeId="1" AcceptedAnswerId="11708" CreationDate="2015-08-29T12:01:07.970" Score="1" ViewCount="483" Body="&lt;p&gt;How might it be possible to determine (via some specific online empemeris or via some well-understood algorithm) the observed position of some distant celestial body (e.g. Betelgeuse)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be concrete: suppose I wished to do what any number of mobile phone applications do, i.e. computationally model the observed position of stars and project them onto the screen using GPS coordinates and compass direction, so that an `overlay' of the night sky is produced. Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Where could I source the relevant data from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) What coordinate system would it be best to use?&lt;/p&gt;&#xA;" OwnerUserId="8078" LastEditorUserId="8078" LastEditDate="2015-09-01T18:44:40.777" LastActivityDate="2015-09-01T19:48:14.953" Title="Real-time position of a distant celestial body" Tags="&lt;positional-astronomy&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11698" PostTypeId="2" ParentId="11687" CreationDate="2015-08-29T16:54:23.667" Score="1" Body="&lt;p&gt;The idea of self-pollution in a cluster may be viable. When you talk about metallicity, there are two things - one is what you see, the photospheric metallicity; the second is the bulk metallicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To alter the former only requires that adjacent stars accrete enriched material and then that they are unable to mix and dilute this. The second is much harder. If you want to form a star with high metallicity, the material it forms from must be enriched first, e.g. a supernova progenitor must have lived and died first. As the earliest age that a supernova can explode at is about 5 million years, this means having an extended period of star formation in a single cluster. This is controversial. There is a lot of evidence that young star clusters form very quickly, in which case the supernova mechanism is unviable. Nearby open clusters also show no evidence of star-to-star composition variations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, ancient globular clusters may be different. There is evidence for multiple populations with differing compositions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your final question is too broad. Yes, there are many nuances and gradations in populations, though the catch-all terms of Populations I and II are still used. But for example we also talk about thin and thick disc populations or bulge populations.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2270" LastEditDate="2015-08-29T22:37:05.537" LastActivityDate="2015-08-29T22:37:05.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11699" PostTypeId="2" ParentId="11697" CreationDate="2015-08-29T18:08:48.577" Score="0" Body="&lt;p&gt;The most robust, but generally not computationally feasible way would be a full N-body simulation of all the relevant interacting neighbours of the object in question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a more crude note, one could observe the eigen motion of the object (its velocity vector), its distance to us, and from that information (assuming the eigen motion is constant over the period between now and when the light was emitted from the object) one may infer the current position. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can of course make a hybrid of the above methods: use the observed motion of the object, and use simulations to include the interaction with other objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that 'current position' is not unambiguously defined. Astronomers would often consider the observed position to be its current position, even though we are looking back in time. And, needless to say, on cosmological scales these concepts pose more problems. &lt;/p&gt;&#xA;" OwnerUserId="6965" LastActivityDate="2015-08-29T18:08:48.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11700" PostTypeId="1" CreationDate="2015-08-29T18:43:05.630" Score="5" ViewCount="137" Body="&lt;p&gt;I was reading about the history of the early universe in Carroll &amp;amp; Ostlie and specifically looking at Inflation, which, as I understand, occurred because our [observable] part of the universe entered a false vacuum state containing negative pressures, thus fueling expansion. (Carroll &amp;amp; Ostlie state the opposite, but as per the discussion below, that appears to be incorrect). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that mean that there are other parts of the Universe (outside of our horizon) that experienced deflation, as inflationary parts pressed against them? Would there be boundaries between separate parts of the larger universe where inflationary parts encounter one another? Any thoughts on what those parts of the Universe are like, such as containing their own CMB that's NOT close to uniform, or where there is a much higher density of matter?&lt;/p&gt;&#xA;" OwnerUserId="8080" LastEditorUserId="8080" LastEditDate="2015-09-03T22:37:41.967" LastActivityDate="2015-10-03T23:03:10.613" Title="Were parts of the Universe deflating during Inflation?" Tags="&lt;cosmological-inflation&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11701" PostTypeId="2" ParentId="11676" CreationDate="2015-08-30T04:52:00.923" Score="2" Body="&lt;p&gt;Instead of walking in a circle, spin. Take a tiny step forward with your right foot, backward with your left. Then repeat. You are spinning counterclockwise when viewed from above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, imagine holding a landscape photograph at arms length, and you initially facing a floor lamp. Now take that tiny step forward with your right foot, backward with your left. From your body fixed perspective, the landscape photograph doesn't move. It's the lamp that appears to be moving, and it moves to the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, imagine a helmet cam on your head that rotates to keep the floor lamp in the center of the camera image. From this perspective, it's the landscape photograph that appears to be moving, and it moves to the left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's what you are seeing in that youtube video.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-08-30T04:52:00.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11703" PostTypeId="2" ParentId="11689" CreationDate="2015-08-31T02:23:34.583" Score="5" Body="&lt;p&gt;Ultimately, you are correct. However, there may also be some clarification in order: what blue aurora of Saturn are you referring to? The most well-known auroral images of Saturn &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Saturn_with_auroras.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;like this one&lt;/a&gt;, are not &lt;em&gt;true color&lt;/em&gt; images. That is, what we see in the photo isn't what you'd see with the naked eye; instead, the photo has been enhanced to emphasize something interesting, in this case the aurora. In reality, that beautiful blue you see in the picture isn't even of light that's visible with our eyes. It's UV light which would be completely invisible to us. People working with data taken by a specialized camera have to give the UV light some visible color or we wouldn't see it at all. The Saturn aurora images I am aware of all have been modified in this manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, aurora themselves are the result of charged particles funneling down a planet's magnetic field and crashing into the planet's atmosphere. On Earth, the aurora occurs because of the atoms in the atmosphere getting energized by these collisions, and then emitting light when they de-energize. Here, it's the oxygen and nitrogen in the atmosphere that are responsible for aurora light (primarily greens from oxygen, but also reds at higher altitudes from oxygen and occasionally blues from nitrogen). In other planets' atmospheres dominated by different atoms and molecules, the reactions will be different, and the wavelengths of light emitted will be different as well. For example, Saturn's aurora are due to energized H2 molecules. On Jupiter, it can be &lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2007/29mar_bigauroras/&quot; rel=&quot;nofollow noreferrer&quot;&gt;the infalling particles themselves&lt;/a&gt; (sulfur and oxygen) that are responsible for auroral emission. &lt;/p&gt;&#xA;" OwnerUserId="8080" LastEditorUserId="16746" LastEditDate="2018-12-26T20:01:00.713" LastActivityDate="2018-12-26T20:01:00.713" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="11704" PostTypeId="1" CreationDate="2015-08-31T12:16:40.060" Score="2" ViewCount="190" Body="&lt;p&gt;What am I looking for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from any point Latitude, Longitude and date D &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know where the sun is in the sky at any given time anywhere on our planet. I even don't know what the outcome will be for this calculation. I Imagine an angle and a height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone send me on the right path for doing this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not looking for a quick formula that let's me calculate this bur rather understand the components :)&lt;/p&gt;&#xA;" OwnerUserId="8090" LastActivityDate="2015-08-31T16:33:56.430" Title="Finding the suns position in the sky anywhere at any time" Tags="&lt;the-sun&gt;&lt;positional-astronomy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11705" PostTypeId="2" ParentId="11704" CreationDate="2015-08-31T16:33:56.430" Score="1" Body="&lt;p&gt;There are many resources that will let you compute the position of the Sun in a coordinate system with its center at the center of the Earth. The answer will be an angle above or below the equator, an angle eastward from the vernal equinox (first point of Aries, and a distance. Books such as &lt;em&gt;Astronomical Algorithms&lt;/em&gt; 2nd ed. by Jean Meeus have information on how to transform such a position to a horizon system centered at an observer on the surface of the Earth. The answer will be an angle from the north, an angle above the horizon, and the distance. Wikipedia has articles about software libraries that will do this kind of calculation, such as &quot;SOFA (astronomy)&quot; and &quot;Naval Observatory Vector Astrometry Subroutines&quot;. These subroutines are quite complicated because they are meant to be suitable for the most critical work, with angles accurate to a few milliarcseconds. The Meeus formulas are much simpler.&lt;/p&gt;&#xA;" OwnerUserId="7385" LastActivityDate="2015-08-31T16:33:56.430" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11706" PostTypeId="1" CreationDate="2015-08-31T18:41:07.937" Score="2" ViewCount="804" Body="&lt;p&gt;I have been reading about celestial mechanics and particularly about planetary orbits. I understand that a planet's orbit can be tilted (pitched) with respect to the Earth's ecliptic and that it might precess (yaw) over time. But I have not come across any indication that another planet's orbital plane might be rolled with respect to the Earth's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also found references to Keplerian orbital elements, but I don't see that any of them indicate roll. Am I misunderstanding the situation? If not, would someone point me at an elementary resource where I can learn about this?&lt;/p&gt;&#xA;" OwnerUserId="8092" LastEditorUserId="2153" LastEditDate="2015-09-01T12:03:16.657" LastActivityDate="2015-09-03T10:39:08.487" Title="Roll, Pitch and Yaw of Orbital Planes" Tags="&lt;orbital-elements&gt;&lt;celestial-mechanics&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11707" PostTypeId="2" ParentId="11706" CreationDate="2015-09-01T01:56:04.010" Score="1" Body="&lt;p&gt;You're using the wrong terms. Engineers use yaw, pitch, and roll to describe the orientation of a vehicle. Some erroneously call these rotations Euler angles. Astronomers and physicists use true Euler angles, a rotation about the Z axis of some reference plane, followed by a second rotation about the once-rotated X axis, followed by a third rotation about the twice-rotated Z axis. Note that the Tait-Bryan angles (aka Cardano angles) use a sequence of rotations about three distinct axes. Euler angles use only two axes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first rotation is the planet's &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_precession&quot; rel=&quot;nofollow&quot;&gt;axial precession&lt;/a&gt; angle. The second rotation is the planet's &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_tilt&quot; rel=&quot;nofollow&quot;&gt;axial tilt&lt;/a&gt;, or obliquity. The third rotation represents the planet's daily rotation. The rates at which the precession and obliquity change are much smaller than the quickly-changing third angle.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2015-09-03T10:39:08.487" LastActivityDate="2015-09-03T10:39:08.487" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11708" PostTypeId="2" ParentId="11697" CreationDate="2015-09-01T19:48:14.953" Score="0" Body="&lt;p&gt;A fine place for reliable information on the location of the vast majority of cataloged stars is the Simbad Astronomical Database. For example, here is the data for &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-basic?Ident=betelgeuse&amp;amp;submit=SIMBAD%20search&quot; rel=&quot;nofollow&quot;&gt;Betelgeuse&lt;/a&gt;. As you will see, the first line after &quot;other object types&quot; is the position given in the ICRS coordinate system (the first set of numbers is Right Ascension, in degrees minutes seconds and the second set of number is Declination in degrees minutes seconds). For all but the highest level research-grade astrometry or proper-motion purposes, this position will be more than sufficient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, given you have the 'global' position of the star, you can calculate its relative position as seen from wherever you are on Earth. Doing so is a function of your latitude, longitude and current time. There are online calculators that can do this &lt;a href=&quot;http://koti.mbnet.fi/jukaukor/star_altitude.html&quot; rel=&quot;nofollow&quot;&gt;such as this one&lt;/a&gt;, or you can make a program to crunch through the math yourself (it's not exactly trivial, but if there are lots of stars you want to do it for, or you want to do it on the fly, this is a better bet). If you &lt;em&gt;really&lt;/em&gt; want to be clever about it and you have lots of stars, you only have to do the calculation once, since all the stars are essentially fixed relative to one another. When all is said and done, the most intuitive local coordinate system to use for your projecting would be altitude &amp;amp; azimuth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are concerned about also including the relative motions of stars over time, I would say don't be. Nearly all stars have proper motions of less than 1 arcsecond a year, which means it would be half a century before most stars have moved in even the most remotely perceptible way to the naked eye (the human eye has an angular resolution of roughly one arcminute). Other apparent motions, such as parallax, are also negligibly small for objects outside of our solar system (again, unless you are doing research-grade work).&lt;/p&gt;&#xA;" OwnerUserId="8080" LastActivityDate="2015-09-01T19:48:14.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11709" PostTypeId="1" AcceptedAnswerId="11710" CreationDate="2015-09-01T22:17:10.140" Score="10" ViewCount="16919" Body="&lt;p&gt;I am new to astrophotography and am looking for guidance. I have the following equipment:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Orion StarBlast 9814 4.5&quot; Altazimuth Reflector Tabletop telescope&lt;/li&gt;&#xA;&lt;li&gt;1.25in Telescope Adapter Extension Tube T Ring for Canon DSLR SLR Camera DC618&lt;/li&gt;&#xA;&lt;li&gt;17mm and 6mm lenses&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When I use the lenses in the Telescope Adapter Extension, focus is not achievable. My theory is that the focal plane of most DSLR are 55mm back focus plane. Reflectors are positioned farther back than where the human eye would normally be placed, and hence it is usually difficult if not impossible to bring the camera into focus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true? Also, would a Barlow lens help achieve focus or be a waste of money? Any other suggestions?&lt;/p&gt;&#xA;" OwnerUserId="8100" LastEditorUserId="735" LastEditDate="2018-03-18T14:28:56.213" LastActivityDate="2018-03-18T14:28:56.213" Title="Astrophotography - Unable to achieve focus with current setup" Tags="&lt;photography&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11710" PostTypeId="2" ParentId="11709" CreationDate="2015-09-02T00:05:18.203" Score="12" Body="&lt;p&gt;Yes, the vast majority of small telescopes made for visual observations (like most dobsonians and other newtonian-based architectures) have very limited back focus. In other words, with the focuser retracted all the way in, the focal plane of the primary mirror is sticking out only a short distance. A large back focus is not typically seen, nor is it desirable, in a visual scope. (However, most astrographs have a large back focus.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is compounded by the fact that most DSLR have a flange distance (focus-flange distance) which is pretty large - over 45...50 mm in many cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, an additional distance is wasted by the classic T-ring adapters, which tend to push the camera way back from the focuser's shoulder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these factors conspire in your case to prevent you from achieving focus.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Some ideas and solutions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Adapters&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tell everyone who is trying to do prime focus photography with visual scopes (dobsonians, regular newtonians) to ditch the old T-ring system and just buy dedicated adapters. All you need is a piece of metal that goes in the bayonet mount on the camera on one side, and into the focuser on the other side. No need for T- components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some such adapters are made by telescopeadapters.com:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.telescopeadapters.com/&quot; rel=&quot;noreferrer&quot;&gt;http://www.telescopeadapters.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two adapters like this, they save a very large amount of back focus, compared to the classic T-ring / T-adapter system. In other words, they allow the camera to get much closer to the focuser, and therefore catch the focal plane on the sensor. You just need to choose the kind of adapter that works with your camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of these adapters are 2&quot;, so in a 1.25&quot; focuser a size adapter is needed - but that would waste some back focus distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Camera&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classic DSLR cameras have a very large flange distance, so a large back focus is needed to overcome it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other removable-lens, large-sensor cameras have a much smaller flange distance, if they are mirrorless. Examples: Micro Four Thirds cameras (Panasonic G series), Sony NEX, Samsung NX, etc. All these systems have large sensors (comparable to DSLR) which work well for astrophotography, but don't have moving mirrors, so their flange distances are tiny.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Mirrorless_interchangeable-lens_camera&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Mirrorless_interchangeable-lens_camera&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a Panasonic G1, used with the adapter mentioned above. I only need 25 mm of back focus for this to work, which is an extremely small value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Use a barlow&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A barlow would increase the amount of back focus available. Try and see what happens. Keep in mind it cannot work miracles - if you need a very large amount of back focus, it may not work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Use eyepiece projection&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With some eyepieces, you can thread adapters on the user side, and then put the camera body (without lens) on the adapter. Move the eyepiece WAY out of focus, and it will generate a real image which can be projected on the sensor. This is not prime focus photography anymore, but it works and it's better than nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, such adapters are hard to find, and many eyepieces won't support them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Use a dedicated USB &quot;eyepiece camera&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find relatively cheap USB cameras made for imaging, some are less than $100. With one of those, you could achieve focus relatively easy. Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.telescope.com/Astrophotography/Astrophotography-Cameras/Orion-StarShoot-USB-Eyepiece-Camera-II/pc/-1/c/4/sc/58/p/102083.uts&quot; rel=&quot;noreferrer&quot;&gt;http://www.telescope.com/Astrophotography/Astrophotography-Cameras/Orion-StarShoot-USB-Eyepiece-Camera-II/pc/-1/c/4/sc/58/p/102083.uts&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would not offer the same performance like a DSLR, but it's better than nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Use a telescope made for imaging&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Visual scopes, like a small newtonian, are not really that great for photography, for a variety of reasons. A dedicated astrograph would always work better. There are many Cassegrain systems out there made specifically for imaging, some are relatively inexpensive (not much more expensive than a DSLR).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many cases, you need very long exposure, and you need to track the sky while the shutter is open. Many such scopes come with a tracking mount, or are made to be installed on a tracking mount. That would make it very easy to do tracking well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be very hard to do tracking with other kinds of scopes (tracking platforms for dobsonians do exist, but are not very common).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Move the primary mirror up&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about it - if you move the primary mirror up in its cell, the focal plane will stick out. This is a way to gain back focus only if you know what you're doing. Additional spacers in the primary mirror cell could do it. Modifying the cell is also doable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beware, you could ruin the scope if you make a mistake. Also, if the secondary mirror is too small to begin with, you'll increase the amount of clipping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Afocal photography&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put the eyepiece in the scope, just like for a normal visual observation. Now put the camera at the eyepiece and move it around until you capture the image. Click the shutter. This is called afocal photography.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll need a camera mount, that will attach to the focuser, holding the camera at the eyepiece. Something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.telescope.com/Astrophotography/Astrophotography-Accessories/Orion-SteadyPix-Deluxe-Camera-Mount/pc/-1/c/4/sc/61/p/5338.uts&quot; rel=&quot;noreferrer&quot;&gt;http://www.telescope.com/Astrophotography/Astrophotography-Accessories/Orion-SteadyPix-Deluxe-Camera-Mount/pc/-1/c/4/sc/61/p/5338.uts&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that your camera is very likely too heavy for this trick to work. Also, afocal is usually the lowest quality of all techniques, and rarely works well enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There are special adapters that go on the eyepiece on one end, and thread into the camera lens like a filter on the other end. These would work better, but are hard to find, only a few eyepieces support them, and your camera is too heavy anyway.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that your scope has a 1.25&quot; focuser. Even if you do get the sensor in focus, the small size of the focuser will clip the image; you will only be able to use the central part of the sensor. It is recommended to use 2&quot; focusers for prime focus imaging with large sensors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, your f/4 scope would exhibit strong aberrations at the edge anyway, which would blur the image. For prime focus imaging, longer focal ratios are usually desirable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all, your scope and your camera are mismatched. You have a pretty good camera, but a very small scope that is not made for imaging. Either use a dedicated astrograph with that camera, or keep the scope and use a cheap USB &quot;eyepiece camera&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-09-02T02:40:15.960" LastActivityDate="2015-09-02T02:40:15.960" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11711" PostTypeId="2" ParentId="11670" CreationDate="2015-09-02T00:19:47.337" Score="1" Body="&lt;p&gt;The issue is twofold: (a) whether the formation of cosmological structure is an open question and (b) if that is the case who is working on it?  Structure formation always was and still is an important issue for cosmology since we need to understand that structure in order to say that we understand our Universe.  As with all aspects of science, the nature of the study of structure formation has changed considerably over the past decades: today's problems are quite different from yesterday's.  For one thing, we now realise that the baryonic component of the Universe is but a few percent of the total mass density.  Over 97% of our Universe is made up of dark matter and dark energy.  The game has changed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) The issue of cosmological structure formation goes back to the work of Gamow in the 1940's, but we now have very strong, I would say almost incontrovertible, evidence that the formation of the structure on large scales was driven by gravity.  We understand the formation of what is referred to as the &quot;Cosmic web&quot;, the large scale filamentary structures surrounding great voids that dominate the observed structure.  This is the environment for the formation of the galaxies that we see.  We have models for the process of galaxy formation.  These are based around numerical gravitatinal N-Body models dressed up with semi-analytic descriptions of the star formation process.  The results can look rather good:&#xA;&lt;a href=&quot;http://www.illustris-project.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.illustris-project.org/&lt;/a&gt;&#xA;and we can choose parameters that bring such models into some kind of accord with what is observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The models tell us, for example, how we might study the dark matter component through observations of satellite galaxies of larger galaxies.  Since we can observe the galaxy formation process out to redshifts of 2 or more (the most distant galaxy observed has a redshift in excess of 8) we can check the evolution seen in the models.  If you are a cynic you might say we can &quot;re-tune&quot; the models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) We are now measuring the 6-7 parameters that describe our universe by observing various aspects of that large scale structure.  We can for example look at the clustering on scale &gt; 100Mpc and detect an excess of clustering at what is called the Baryonic Acoustic Oscillation (BAO) scale, a feature that emerges from the primordial Universe.  We can also look at the voids in the cosmic web to do the same thing in  an entirely different way.  All that can be calibrated by N-Body models and compared with the answers obtained from the cosmic microwave background.   So we find groups with nice names like WiggleZ looking at the BAOs:&#xA;&lt;a href=&quot;http://wigglez.swin.edu.au/site/&quot; rel=&quot;nofollow&quot;&gt;http://wigglez.swin.edu.au/site/&lt;/a&gt;&#xA;who use redshift surveys containing millions of galaxies.  It's quite amazing, even to one who works in this field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the importance of all this?  We find that the Universe can be described by just 6 parameters.  We can measure these parameters with any of these methods with an accuracy of a few percent or better.  What is convincing us that we have a good model is that these quite independent ways of measuring the 6 parameters all come up with the same answer to within the experimental accuracy.  This is &quot;Precision Cosmology&quot;.  We can feel confident that the Universe is made up of ~70% of dark energy because we detect its influence in different ways and all those ways provide the same values.  What is it?  That's the 21st century question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short - the important questions are still open and hundreds, if not thousands, of astronomers are working on it. &lt;/p&gt;&#xA;" OwnerUserId="5301" LastActivityDate="2015-09-02T00:19:47.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11712" PostTypeId="1" CreationDate="2015-09-02T01:59:13.517" Score="2" ViewCount="838" Body="&lt;p&gt;What is the difference between spicules and prominences/filaments. I know that they are both in the chromosphere, and I know the difference between provinces and filaments (depends on the view), but how are spicules different? &lt;/p&gt;&#xA;" OwnerUserId="8102" LastActivityDate="2015-09-02T08:55:49.413" Title="Spicules vs Prominences/filaments" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11713" PostTypeId="2" ParentId="11686" CreationDate="2015-09-02T05:57:16.780" Score="3" Body="&lt;p&gt;Carl Sagan answers this in the Cosmos series. Here is an excerpt.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The matter in the known Universe is made up of 74% Hydrogen. In most&#xA;  of the stars we see, hydrogen nuclei are being jammed together to form&#xA;  helium nuclei. Every time a nucleus of helium is made, a photon of&#xA;  light is generated. This is why the stars shine. Helium accounts for&#xA;  24% of matter in known Universe. In fact, helium was detected on the&#xA;  sun before it was ever found on the Earth. These two elements have&#xA;  accounted for 98% of matter. Might the other chemical elements have&#xA;  somehow evolved from hydrogen and helium? &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Three units, put together in different patterns make, essentially,&#xA;  everything – “Neutron, Proton and Electron”. If you’re an atom and you&#xA;  have just one proton you’re hydrogen. Two protons, helium. And so on.&#xA;  Protons have positive electrical charges. But since like charges repel&#xA;  each other, why does the nucleus hold together? Why don’t the&#xA;  electrical repulsion of the protons make the nucleus fly to pieces?&#xA;  Because there’s another force in nature. Not electricity, not gravity.&#xA;  The nuclear force!! We can think of it as short-range hooks which&#xA;  start working when protons or neutrons are brought very close&#xA;  together.The nuclear force can overcome the electrical repulsion of&#xA;  the protons. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A lump of two protons and two neutrons is the nucleus of a helium atom&#xA;  and is very stable. Three helium nuclei, stuck together by nuclear&#xA;  forces makes carbon. Four helium nuclei makes oxygen. There’s no&#xA;  difference between four helium nuclei stuck together by nuclear forces&#xA;  and the oxygen nucleus. They’re the same thing.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;How easy is that to fuse nuclei? To avoid the&#xA;  electrical repulsion protons and neutrons must be brought very close&#xA;  together so the hooks which represent nuclear forces are engaged. This&#xA;  happens only at very high temperatures, where particles move so fast&#xA;  that there’s no time for electrical repulsion to act. Temperatures of&#xA;  tens of millions of degrees. Such high temperatures are common in&#xA;  nature. Where? In the insides of the stars. Atoms are made in the&#xA;  insides of stars.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;It is possible to make atoms with up to 26 Protons [Iron] in a Star.&#xA;  Above that, We need a supernova to create atoms with 30 protons, 40&#xA;  protons, 50 protons or even 60 protons. Nature prefers ‘even’ numbers&#xA;  for stability. But ‘Gold’ is an odd-numbered atom. It has 79 Protons.&#xA;  It needs more than a super nova for its creation. It needs 2 Neutron&#xA;  stars to collide directly. This collision alone can favor the creation&#xA;  of stable atoms of Gold.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Since neutron star collisions are also suggested as the origin of&#xA;  short duration gamma-ray bursts, it is possible that you already own a&#xA;  souvenir from one of the most powerful explosions in the universe.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Except for hydrogen and helium every atom in the sun and the Earth was&#xA;  synthesized in other stars. The silicon in the rocks, the oxygen in&#xA;  the air, the carbon in our DNA, the gold in our banks, the uranium in&#xA;  our arsenals were all made thousands of light-years away and billions&#xA;  of years ago. Our planet, our society and we ourselves are built of&#xA;  “star stuff“.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8103" LastActivityDate="2015-09-02T05:57:16.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11714" PostTypeId="2" ParentId="11712" CreationDate="2015-09-02T08:55:49.413" Score="1" Body="&lt;p&gt;Spicules are much smaller, more dynamic, and possibly hotter. They are also much more abundant. While the term spicules is strictly defined for objects off the solar limb, their disk counterparts (let's call them fibrils for simplicity) cover the whole Sun, all the time. Spicules have lifetimes of a few minutes and extend up to ~15 Mm, with the most energetic of them being ejected with speeds of around 100 km/s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While prominences/filaments have a relatively stable temperature around 10,000 K, spicules/filaments are also seen in filters sensitive to temperatures around 100,000 K, meaning that some of them will undergo a thermal evolution to at least the transition region. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spicules/fibrils are a candidate for energy and mass deposition in the corona.&lt;/p&gt;&#xA;" OwnerUserId="8104" LastActivityDate="2015-09-02T08:55:49.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11715" PostTypeId="2" ParentId="11700" CreationDate="2015-09-02T09:53:49.827" Score="1" Body="&lt;p&gt;That sounds... confused, or else just wrong. Negative pressure is what &lt;em&gt;causes&lt;/em&gt; inflation: something (inflaton field, cosmological constant, or &quot;dark energy&quot; in general) with an energy density that stays approximately constant as space expands has negative pressure and will drive the exponential expansion of space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transition to &quot;true&quot; vacuum is what &lt;em&gt;shuts off&lt;/em&gt; inflation, after which you get  the standard Big Bang, during which the expansion continues but slows down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The cosmological constant -- or &quot;dark energy&quot; -- also has negative pressure, and so the universe is gradually moving towards a state where the space between galaxy groups and galaxy clusters will again be expanding exponentially.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Eternal_inflation&quot; rel=&quot;nofollow&quot;&gt;&quot;eternal inflation&quot; model&lt;/a&gt; has our visible universe -- almost certainly a tiny part of a much larger region of true vacuum -- as one of many such pockets of true vacuum embedded within a vastly larger false vacuum that's still expanding exponentially.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2015-09-02T09:53:49.827" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11716" PostTypeId="1" CreationDate="2015-09-02T11:27:17.173" Score="2" ViewCount="4561" Body="&lt;p&gt;I have a question about iron stars. I was fascinated about the theory about the inability of protons to decay, and it led me to a point where I read something about some special kind of star called an &quot;iron star&quot;. Can anybody enlighten me a bit, if possible, on what exactly it is and why it may be significant? And how do iron stars form? &lt;/p&gt;&#xA;" OwnerUserId="7332" LastEditorUserId="2153" LastEditDate="2015-09-02T14:35:09.163" LastActivityDate="2019-09-27T14:51:07.503" Title="What are iron stars?" Tags="&lt;star&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11717" PostTypeId="1" CreationDate="2015-09-02T15:40:59.957" Score="5" ViewCount="299" Body="&lt;p&gt;When a nova gets matter, the outcome is that it may lose more matter because of eruptions. Only some kind of nova's masses can increase stably. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A white dwarf may become a supernova after it exceeds the Chandrasekhar limit, and we study that much. A neutron star can get mass too, why do not we talk about their end much?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about a neutron star when it gets matter from accretion? It is necessary that its mass will increase? &lt;br/&gt;&#xA;When will it collapse?&lt;br/&gt;&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditDate="2015-09-03T07:11:35.677" LastActivityDate="2015-10-03T07:57:43.340" Title="When a neutron star accretes matter, will its mass increase?" Tags="&lt;neutron-star&gt;&lt;accretion-discs&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11718" PostTypeId="2" ParentId="11673" CreationDate="2015-09-02T19:40:58.903" Score="5" Body="&lt;p&gt;The short answer to your top-level question &quot;Can things move faster than light inside the event horizon of a black hole?&quot; is no.  The text of your question gets confusing and contains a lot of assumptions that are not correct.  There's no &quot;special&quot; acceleration happening at or near the event horizon.  If things &lt;em&gt;could&lt;/em&gt; travel faster than the speed of light there, then they would potentially be able to escape the black hole.  The hole is &quot;black&quot; because things are bounded by the speed of light and because of the spacetime curvature.&lt;/p&gt;&#xA;" OwnerUserId="8019" LastActivityDate="2015-09-02T19:40:58.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11719" PostTypeId="1" CreationDate="2015-09-03T01:06:43.580" Score="4" ViewCount="255" Body="&lt;p&gt;I recently read that if you're in the northern hemisphere and have an analog watch, then you can point the hour hand at the sun and know that a south line lies between (bisection) the hour hand and the 12 o'clock position:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2dvlD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2dvlD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently the trick works in the southern hemisphere as well and needs to be adjusted for day-light saving (image taken from Wikipedia):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hq3tV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hq3tV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know the reasoning behind this trick, and would like to know if the trick can be applied to tell the approximate time provided I only have a Brunton compass.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I made an illustration if it helps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TvR0z.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TvR0z.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8110" LastEditorDisplayName="user1569" LastEditDate="2015-09-03T12:52:11.243" LastActivityDate="2015-09-05T13:41:43.000" Title="Determining North-South Line Via Watch Method: Theory &amp; Reason" Tags="&lt;solar-system&gt;&lt;time&gt;&lt;geology&gt;&lt;celestial-mechanics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11720" PostTypeId="2" ParentId="11717" CreationDate="2015-09-03T07:04:53.053" Score="-1" Body="&lt;p&gt;A nova is generally understood as an explosion of a star.  Thus comparing a nova to a stable neutron star is kind of a nonsensical thing to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're asking if a star goes nova and the core that remains is a neutron star after the resulting explosion, will it's mass increase overall?  The answer is no, since it loses a lot of mass in the process of going nova.  However, it is now more dense than before and the concentration of matter is in a much smaller space (the neutron star).  (The explosion compresses part of the star and pushes the rest of it outward off into space).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're asking if a neutron star's mass will increase if you throw matter at it, the answer is yes.  Just like any other form of matter.  Is there a limit to how bit you can increase it?  No; however, past a certain density it will compress into a black hole (which do not have a size limit as far as we know).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may want to edit your question to clarify what you mean.&lt;/p&gt;&#xA;" OwnerUserId="7765" LastActivityDate="2015-09-03T07:04:53.053" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11721" PostTypeId="2" ParentId="11717" CreationDate="2015-09-03T07:11:18.077" Score="1" Body="&lt;p&gt;The neutron stars have about 1.1 to 3 solar masses. If the neutron star accretes mass, its mass will increase and once it crosses the &lt;a href=&quot;https://en.wikipedia.org/wiki/Tolman%E2%80%93Oppenheimer%E2%80%93Volkoff_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tolman–Oppenheimer–Volkoff limit&lt;/a&gt;, the star will collapse to form a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On way of this happening is in the case where one of the binary stars becomes a neutron star as in &lt;a href=&quot;https://en.wikipedia.org/wiki/X-ray_pulsar&quot; rel=&quot;nofollow noreferrer&quot;&gt;X-ray pulsars&lt;/a&gt;. The neutron star receives gas supply from its companion, which increases its mass,for example &lt;a href=&quot;https://en.wikipedia.org/wiki/Centaurus_X-3&quot; rel=&quot;nofollow noreferrer&quot;&gt;CenX3&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EtmJX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EtmJX.jpg&quot; alt=&quot;Swift J1749&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: nasa.gov&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though this method could lead tothe formation of black hole, no observational evidence exists as of today.&lt;/p&gt;&#xA;" OwnerUserId="7970" LastActivityDate="2015-09-03T07:11:18.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11722" PostTypeId="2" ParentId="11716" CreationDate="2015-09-03T07:18:58.740" Score="1" Body="&lt;p&gt;The basic premise of an iron star is a 'dead' star whose mass is made primarily of iron.  It's a hypothetical type of compact star.  Wikipedia has a bit of a blurg here: &lt;a href=&quot;https://en.wikipedia.org/wiki/Iron_star&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Iron_star&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process of forming an iron star would be tricky, because normally iron does not fuse and so the star would normally collapse under it's own weight - this is why the Wiki article describes a cold-fusion process via quantum tunneling processes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This article describes the process of fusion in stars and might help you understand why this is more significant compared to regular stars: &lt;a href=&quot;http://www.eso.org/public/usa/news/eso0129/&quot; rel=&quot;nofollow&quot;&gt;http://www.eso.org/public/usa/news/eso0129/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7765" LastActivityDate="2015-09-03T07:18:58.740" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11723" PostTypeId="2" ParentId="11695" CreationDate="2015-09-03T07:27:43.280" Score="1" Body="&lt;p&gt;This is more of a physics question/off topic question.  Due to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Momentum#Conservation&quot; rel=&quot;nofollow&quot;&gt;law of conservation of momentum&lt;/a&gt; things that are in motion stay in motion.  If the earth was magically seized upon by some cosmic tractor beam, it would basically be squished flat (imagine dropping an egg on the floor, where the floor represents the opposing force of the tractor beam).  If it was a gradual controlled reduction of speed (like an elevator's motion), I guess nothing would happen.  But that doesn't really make any sense either, because that doesn't explain how you got rid of the gravitational effects of the sun and everything else.&lt;/p&gt;&#xA;" OwnerUserId="7765" LastActivityDate="2015-09-03T07:27:43.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11724" PostTypeId="2" ParentId="914" CreationDate="2015-09-03T21:29:36.793" Score="2" Body="&lt;p&gt;I highly recommend picking up a copy of &lt;em&gt;Astronomical Algorithms&lt;/em&gt; by Jean Meeus. Chapter 49 covers the calculation methods that are used by many (if not most) online calculators.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The somewhat dated page at &lt;a href=&quot;http://www.aphayes.pwp.blueyonder.co.uk/sun_moon.html&quot; rel=&quot;nofollow&quot;&gt;http://www.aphayes.pwp.blueyonder.co.uk/sun_moon.html&lt;/a&gt; has a link to &quot;moon.js&quot;, the javascript that does its lunar calculations.  The calculations you need are in that file.&lt;/p&gt;&#xA;" OwnerUserId="8117" LastActivityDate="2015-09-03T21:29:36.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11726" PostTypeId="1" AcceptedAnswerId="11727" CreationDate="2015-09-04T16:42:26.153" Score="5" ViewCount="2077" Body="&lt;p&gt;Specifically, I am wondering if some areas of the universe expand faster than other areas and whether the faster expanding areas diffuse the expansion through the slower expanding areas or does the expansion occur at a uniform rate throughout the entire universe.&lt;/p&gt;&#xA;" OwnerUserId="9124" LastActivityDate="2016-09-11T09:48:46.077" Title="Does the universe expand at the same rate everywhere in the universe?" Tags="&lt;expansion&gt;&lt;cosmological-inflation&gt;&lt;redshift&gt;&lt;dark-energy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11727" PostTypeId="2" ParentId="11726" CreationDate="2015-09-04T21:01:41.663" Score="6" Body="&lt;p&gt;What's outside the observable Universe, we can't say anything about, but averaged over large enough scales ($\gtrsim$ a billion lightyears), it does indeed seem to be expanding uniformly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the presence of mass, or more generally energy, retards the expansion. This means that on the scale of clusters of galaxies, the Universe expands more slowly, and on the scale of galaxy groups, the galaxies' mutual gravitational attraction will prevent them from receding from each other. This is also why our galaxy, Solar system, planet, and bicycles will never get torn apart (unless the cosmological constant is not a constant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, in mass underdensities, i.e. the huge voids between clusters and filaments of gas and galaxies, expansion is increased (relative to denser regions). In fact, it has been hypothesized that that the observed accelerated expansion of the Universe is not due to dark energy, but could be an &quot;illusion&quot; from accidentally living in the center of a huge underdensity (e.g. &lt;a href=&quot;http://arxiv.org/abs/0809.3761&quot;&gt;Zibin et al. 2008&lt;/a&gt;). &lt;a href=&quot;http://hubblesite.org/newscenter/archive/releases/2011/08/text/&quot;&gt;More recent observations&lt;/a&gt; seem to rule out this possibility, though&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-09-05T11:35:18.550" LastActivityDate="2015-09-05T11:35:18.550" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11728" PostTypeId="1" AcceptedAnswerId="11736" CreationDate="2015-09-05T00:02:37.173" Score="3" ViewCount="717" Body="&lt;p&gt;I recently read this article: &lt;a href=&quot;http://www.cbsnews.com/news/supermoon-lunar-eclipse-coming-this-month/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.cbsnews.com/news/supermoon-lunar-eclipse-coming-this-month/&lt;/a&gt; I'm having some trouble understanding how a &quot;supermoon&quot; lunar eclipse is possible. &lt;a href=&quot;https://i.stack.imgur.com/wJ7zK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wJ7zK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Judging from this diagram, it would seem that the moon always reaches its apsides above or below the orbital plane of the earth around the sun, making any kind of eclipse at these points impossible. Can someone clarify for me? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-09-06T17:40:23.783" Title="Supermoon Lunar Eclipse?" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;lunar-eclipse&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11729" PostTypeId="1" AcceptedAnswerId="11730" CreationDate="2015-09-05T06:37:45.477" Score="1" ViewCount="68" Body="&lt;p&gt;Can it be (or has it been already?) measured how much and what kind of EM radiation (apart from visible light)/particles Earth &quot;receives&quot; from a specific planet in our solar system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can other planets (giants like Jupiter/Saturn) also emit neutrinos or some other high energetic particles (like Sun)?&lt;/p&gt;&#xA;" OwnerUserId="9127" LastActivityDate="2015-09-05T07:08:15.430" Title="Amount of EM radiation (and particles) Earth &quot;receives&quot; from other planets in our solar system?" Tags="&lt;the-sun&gt;&lt;planet&gt;&lt;radiation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11730" PostTypeId="2" ParentId="11729" CreationDate="2015-09-05T07:08:15.430" Score="2" Body="&lt;p&gt;Of course all the planets have been observed by many telescopes at different wavelengths. Yes, they do emit energy at wavelengths other than in the visible part of the spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Broadly speaking, the peak of their output depends on their temperatures. Most of the light from the outer giant planets emerges in the infrared for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, some planets also are strong radio emitters. For example Jupiter has a strong magnetic field that accelerates charged particles which then emit radio waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutrinos are hard to pinpoint - telescopes currently struggle to say exactly where they come from in the sky. The planets are not expected to be strong neutrino emitters.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-09-05T07:08:15.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11731" PostTypeId="2" ParentId="11719" CreationDate="2015-09-05T13:41:43.000" Score="2" Body="&lt;p&gt;When the Sun is on the meridian and is due South (local noon in Northern Hemisphere) your hour hand will point to (or close to 12) - assuming your local noon and civil noon are similar! (Daylight saving means 1pm fills this role).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun crosses the sky in 24 hours, the hour hand travels round the watch in 12 hours. Hence the watch hour hand is twice as fast, so when the hour hand has reached 2pm, the Sun has only gone half as far (ie &quot;to 1pm&quot;), so if you point the hour hand towards the Sun in this case, the meridian is an hour further behind rather than 2 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the time we get to 4pm the Sun will have travelled &quot;two watch hours&quot; away from the meridian and so on.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2015-09-05T13:41:43.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11732" PostTypeId="2" ParentId="11579" CreationDate="2015-09-05T18:06:20.520" Score="0" Body="&lt;p&gt;Yes, we don't know what dark energy is. It's better to say that whatever is responsible for expanding, we call it dark energy.&lt;/p&gt;&#xA;" OwnerUserId="4135" LastActivityDate="2015-09-05T18:06:20.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11733" PostTypeId="1" AcceptedAnswerId="11735" CreationDate="2015-09-05T20:09:25.083" Score="8" ViewCount="2888" Body="&lt;p&gt;Why does the Earth not have rings like Saturn or Uranus?&lt;/p&gt;&#xA;" OwnerUserId="8033" LastEditorUserId="2153" LastEditDate="2015-09-05T22:15:37.920" LastActivityDate="2018-02-24T17:48:46.240" Title="Why Earth does not have rings?" Tags="&lt;earth&gt;&lt;planetary-ring&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11735" PostTypeId="2" ParentId="11733" CreationDate="2015-09-05T22:19:30.757" Score="18" Body="&lt;p&gt;A moon is held together by its own gravity, and pulled apart by the tidal action of a planet. If a moon comes too close to a planet it will be ripped apart by the planet's gravity and become a ring. The closest a moon can come to a planet is known as the Roche limit, and it is dependent on the mass and density of the planet and moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A large planet, such as as Saturn, has a large Roche limit, and has collected many moons over its history, some of which have been ripped apart to make rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Roche limit for the Earth-moon system is at a radius of about 10000 km, which is very close to the surface of the earth (about 6400km). And unlike the gas giants, terrestrial planets don't have many moons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So with few moons, and small Roche limits, the opportunities for ring formation are rare. That said, Phobos, one of the satellites of Mars, is now so close to Mars, that it is held together as much by its rigidity as by gravity, and is slowly getting closer to Mars. Within 50 million years it will be ripped apart and form a (small) ring around Mars &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-09-05T22:19:30.757" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11736" PostTypeId="2" ParentId="11728" CreationDate="2015-09-06T10:09:30.260" Score="2" Body="&lt;p&gt;Lunar Eclipses occur when a full moon occurs when the the moon crosses the ecliptic. The ecliptic is the plane in which the Earth orbits the sun, and so also the track which the sun appears to trace in the sky relative to the stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears from the diagram that there are two nodes each month (ascending and descending), but this is only approximately correct. The influence of the sun causes the nodes to move.It takes 29.5 days between two full moons, but only 27.2 days between two ascending and descending nodes). This means that a node can occur on any day of the year, and any day of the month. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;supermoon&quot; occurs when the moon when it is closest to the earth, the perigee. This point of perigee also moves, so the perigee can also occur at any time during the month, and the time of perigee is not related to the time of the node. Unlike terms like &quot;node&quot; which astro(nom|log)ers have used for many years, Supermoon is a neologism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus is is possible, but rare for a full moon, a lunar node, and a perigee to occur at about the same time, and if it does, a supermoon eclipse will occur.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2015-09-06T17:40:23.783" LastActivityDate="2015-09-06T17:40:23.783" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11737" PostTypeId="1" CreationDate="2015-09-06T10:37:03.013" Score="2" ViewCount="223" Body="&lt;p&gt;If I point in the sky and draw a line from the tip of my finger to the end of the universe, how many galaxies would this line cross? I had a discussion with my dad and he said due to the universe being so empty the line wouldn't cross any solid matter. I said that the universe is so enormous in size that there would be many galaxies and stars crossing the line.&#xA;Is this calculable? If yes which one is correct?&lt;/p&gt;&#xA;" OwnerDisplayName="user9132" LastActivityDate="2016-01-06T18:29:14.660" Title="If I were to point into the sky, how many galaxies would be in this line?" Tags="&lt;universe&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11738" PostTypeId="2" ParentId="11737" CreationDate="2015-09-06T11:11:04.257" Score="1" Body="&lt;p&gt;Take a look at the &quot;Deep field&quot; image from Hubble: &lt;a href=&quot;http://www.spacetelescope.org/science/deep_fields/&quot; rel=&quot;nofollow&quot;&gt;http://www.spacetelescope.org/science/deep_fields/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see in this tiny part of the sky, hundreds of galaxies, but between there is lots of dark sky. Most lines of sight don't end on a galaxy. I'd say dad was right. I would estimate, from the deep field, that you have about a 1/10 chance of pointing at a galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if your finger does point at a galaxy, the distances between stars are vast. Think of two footballs separated by the distance between Europe and China as a model for the distances between stars. A finger pointing at a galaxy would be unlikely to point exactly at one of the stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is probably a Good Thing. Olber's paradox supposed that if each line of sight ended on a star, the night sky would be a uniform blaze of light.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-09-06T11:11:04.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11740" PostTypeId="1" AcceptedAnswerId="11750" CreationDate="2015-09-06T13:01:53.577" Score="3" ViewCount="118" Body="&lt;p&gt;To receive light from objects like &lt;a href=&quot;https://en.wikipedia.org/wiki/MACS0647-JD&quot; rel=&quot;nofollow&quot;&gt;MACS0647-JD&lt;/a&gt;, I suppose, it has to has extraordinary dimensions and intensity of electromagnetic radiation. What are the mass, the dimensions and other characteristics for this objects?&lt;/p&gt;&#xA;" OwnerUserId="4242" LastEditorUserId="4242" LastEditDate="2015-09-06T17:44:04.800" LastActivityDate="2015-09-07T23:48:58.903" Title="What do we know about the characteristics of stellar objects, 13,3 billion light years away from us?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11741" PostTypeId="2" ParentId="11737" CreationDate="2015-09-06T16:37:57.517" Score="1" Body="&lt;p&gt;How thick is the line? An extended finger is about 30 arcminutes across, the size of the moon, but a line is a line, 0 arcminutes.&#xA;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble_Deep_Field&quot; rel=&quot;nofollow&quot;&gt;Hubble Deepfield, as mentioned by Kilfiger, is only 2.5 arcminutes across&lt;/a&gt;, 1/12th of a finger-width, yet contains hundreds of galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to be a stickler for true lines, the image does contain dark space too. At a guess, there's not easily accessible software for counting such things, it looks like 70-80% blackness (ignoring obvious stars), so 1 chance in 5 your line is pointing at a galaxy.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="2031" LastEditDate="2015-09-06T22:00:30.063" LastActivityDate="2015-09-06T22:00:30.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11742" PostTypeId="1" AcceptedAnswerId="11746" CreationDate="2015-09-06T17:04:15.237" Score="2" ViewCount="143" Body="&lt;p&gt;It is a said fact that 299792458 m/s is the speed limit,nothing can go faster than this speed, or it will take an infinite amount of energy for propulsion.What we perforce in vacuum is one thing.Can someone explain what will be observed(from the particle's frame and an observer frame at rest) if a particle of a non zero test mass moves at light speed inside a non vacuum medium(say water) as this velocity is definitely lesser than the limit.&#xA;THANKS IN ADVANCE &lt;/p&gt;&#xA;" OwnerUserId="551" LastActivityDate="2015-09-06T22:10:37.073" Title="Travelling at light speed(inside water)" Tags="&lt;light&gt;&lt;general-relativity&gt;&lt;speed&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2015-09-09T15:04:48.330" ContentLicense="CC BY-SA 3.0" />
  <row Id="11743" PostTypeId="1" CreationDate="2015-09-06T17:17:36.393" Score="7" ViewCount="9249" Body="&lt;p&gt;There are many large reflecting/optical telescopes around the world, just to name a few: the Very Large Telescope, Large Zenith Telescope, and European Extremely Large Telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is does actually building larger telescopes improve visibility of the universe, quality of an image, and various other attributes of a telescope? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well it does seem kind of obvious that a larger telescope may have more magnification and better imaging, however if that is true than why don't we just build a super large one (think really really big, unlike the size of the current large ones) so we can see the universe in HD - so to speak?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we haven't done this yet (I think) is it because of financial issues or is it because simply building a bigger telescope won't improve much to visibility and image quality?&lt;/p&gt;&#xA;" OwnerUserId="9134" LastActivityDate="2020-09-28T21:25:57.717" Title="Do bigger telescopes equal better results?" Tags="&lt;telescope&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11744" PostTypeId="2" ParentId="11743" CreationDate="2015-09-06T18:14:00.590" Score="12" Body="&lt;p&gt;Magnification is not the main purpose of a telescope. A very small telescope will have a lower resolution than a larger one, but for larger ones, the atmosphere limits the amount of detail visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Large telescopes capture more light than small telescopes. More light means that dimmer objects are visible, and there is more light to be analysed. A very dim object, such as a planet orbiting another star, needs a very large mirror to collect enough light to analyse spectroscopically. And this is the reason why larger telescopes are made. Of course larger telescopes cost a lot, and cost is the reason that telescopes like the &quot;Overwhelmingly large telescope&quot; was not built.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much of the development in telescopes in the last 25 years has been not in ever larger mirrors, but in adaptive optics, or using two telescopes together as an interferometric telescope.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-09-06T18:14:00.590" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11745" PostTypeId="2" ParentId="11737" CreationDate="2015-09-06T19:15:15.817" Score="0" Body="&lt;p&gt;&lt;strong&gt;On average, you will point at one galaxy.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument goes as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the spectrum of &lt;a href=&quot;https://en.wikipedia.org/wiki/Quasar&quot; rel=&quot;nofollow noreferrer&quot;&gt;quasars&lt;/a&gt;, which are effectively point sources, and which lie at distances of the order of the size of the observable Universe, it turns out that typically, roughly one &lt;a href=&quot;https://en.wikipedia.org/wiki/Damped_Lyman-alpha_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;damped Lyman $\alpha$ system&lt;/a&gt; (DLA) is detected. DLAs are huge reservoirs of neutral gas, which are early stages of galaxy formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least in a handful of cases, the galaxy counterpart of the DLA has been confirmed. These observation are very challenging, since they are a faint sources at very small projected distances from the very bright quasar. Here's the setup: &lt;a href=&quot;https://i.stack.imgur.com/3joeF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3joeF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, accepting a DLA as a galaxy, I'd say that you will typically point at one galaxy.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-01-06T11:34:06.687" LastActivityDate="2016-01-06T11:34:06.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11746" PostTypeId="2" ParentId="11742" CreationDate="2015-09-06T22:10:37.073" Score="5" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Cherenkov_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cherenkov radiation&lt;/a&gt; (Wikipedia)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cherenkov radiation, also known as Vavilov–Cherenkov radiation,[a] is electromagnetic radiation emitted when a charged particle (such as an electron) passes through a dielectric medium at a speed greater than the phase velocity of light in that medium. The characteristic blue glow of an underwater nuclear reactor is due to Cherenkov radiation. It is named after Soviet scientist Pavel Alekseyevich Cherenkov, the 1958 Nobel Prize winner who was the first to detect it experimentally.&lt;a href=&quot;https://en.wikipedia.org/wiki/Cherenkov_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt; A theory of this effect was later developed within the framework of Einstein's special relativity theory by Igor Tamm and Ilya Frank, who also shared the Nobel Prize. Cherenkov radiation had been theoretically predicted by the English polymath Oliver Heaviside in papers published in 1888–89.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YbAjT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YbAjT.jpg&quot; alt=&quot;Eerie blue glow&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-09-06T22:10:37.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11747" PostTypeId="1" CreationDate="2015-09-06T23:05:15.977" Score="4" ViewCount="286" Body="&lt;p&gt;Are micrometeorites just dust grains at high relative speed, or is there some difference between them and cosmic or interstellar or solar nebula dust? Dust as in the clouds which obscure parts of the Milky Way in our night sky.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-09-08T07:15:15.890" Title="Are micrometeorites cosmic dust?" Tags="&lt;dust&gt;&lt;meteorite&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11748" PostTypeId="2" ParentId="11747" CreationDate="2015-09-06T23:26:47.727" Score="2" Body="&lt;p&gt;It might seem like the way to figure this out would be to test the composition of micrometeorites and see what objects this matches up with. This, however, is not a viable technique, as &lt;a href=&quot;http://lmcm.mnhn.fr/cos/doc/2008%20Genge%20et%20al.%20MAPS.pdf&quot; rel=&quot;nofollow&quot;&gt;Genge et al. (2008)&lt;/a&gt; explains:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In interplanetary space dust particles are inherently transported sunwards by P-R light drag (Dohnanyi 1967) and once captured by the Earth their deceleration occurs at high altitudes allowing fragile micrometeoroids to survive atmospheric entry (Love and Brownlee 1991). The parent bodies of MMs, therefore, are likely to differ from those of meteorites.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is, however, possible to analyze micrometeorites and group them in various categories, and determine the parent bodies. The two main types of parent bodies are thought to be comets and asteroids. Distinguishing between the micrometeorites of these two groups, however, can be difficult. Many large micrometeorites are from main-belt asteroids (see &lt;a href=&quot;http://oro.open.ac.uk/8727/1/the_distribution_of_asteroids_-_evidence_from_Antarctic_micrometeorites.pdf&quot; rel=&quot;nofollow&quot;&gt;Genge &amp;amp; Grady (2002)&lt;/a&gt;). However, &lt;a href=&quot;http://adsabs.harvard.edu/abs/2010ApJ...713..816N&quot; rel=&quot;nofollow&quot;&gt;Nesvorny et al. (2010)&lt;/a&gt; suggests that many micrometeorites (and large amounts of cosmic dust in general) are from comets.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2015-09-06T23:26:47.727" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11750" PostTypeId="2" ParentId="11740" CreationDate="2015-09-07T23:48:58.903" Score="2" Body="&lt;p&gt;Holger, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I highly recommend you check out the NASA story on the discovery MACS0647-JD, which includes many dimensional facts (mass, distance, age, etc.) and comparisons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hubblesite.org/newscenter/archive/releases/2012/36/full/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://hubblesite.org/newscenter/archive/releases/2012/36/full/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the article: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The estimated total mass of the stars in this baby galaxy is roughly&#xA;  equal to 100 million or a billion suns, or about 0.1 percent to 1&#xA;  percent the mass of our Milky Way's stars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Measuring the mass of a galaxy is very tricky indeed (especially one as far as 13.3 billion light-years from earth!), but there are a couple different methods that can one can use, see here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www2.astro.psu.edu/~caryl/a480/lecture12_10.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;PSU Astro Measuring Mass of Clusters &lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this Physics Stackexchange thread as well:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/123/how-does-one-measure-the-mass-of-a-galaxy-and-other-such-large-quantities&quot;&gt;https://physics.stackexchange.com/questions/123/how-does-one-measure-the-mass-of-a-galaxy-and-other-such-large-quantities&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9142" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-09-07T23:48:58.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11751" PostTypeId="1" AcceptedAnswerId="11756" CreationDate="2015-09-08T05:34:05.053" Score="2" ViewCount="254" Body="&lt;p&gt;Is reference 6 in &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Tidal_locking#Planets&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia article&lt;/a&gt; &quot;dubious&quot; and this passage in need of excising or clarifying?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Venus's 583.92-day interval between successive close approaches to Earth is equal to 5.001444 Venusian [sic] solar days, making approximately the same face visible from Earth at each close approach. Whether this relationship arose by chance or is the result of some kind of tidal locking with Earth is unknown.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gold T., Soter S. (1969), Atmospheric tides and the resonant rotation of Venus, Icarus, v. 11, p 356-366&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it still an &quot;unknown&quot; or &quot;crazy idea that didn't develop with further investigation&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea that Venus is tidally locked with &lt;em&gt;Earth&lt;/em&gt; sounds far-fetched. We understand the slowing of Earth's rotation, and I don't recall ever reading of Venus's resistance to the change!&lt;/p&gt;&#xA;" OwnerUserId="7142" LastEditorUserId="332" LastEditDate="2015-09-08T19:39:26.510" LastActivityDate="2015-09-08T22:18:05.007" Title="Venus presents same face toward Earth at closest approach. What do you make of that?" Tags="&lt;orbit&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11752" PostTypeId="2" ParentId="11747" CreationDate="2015-09-08T07:15:15.890" Score="2" Body="&lt;p&gt;Micrometeorites are just tinier versions of regular meteorites. As such, they can have the same composition as any other meteor class. There's nothing that we know of that says they have a chemistry more like one kind of dust source, although this is controlled entirely by observational bias. Certain types of particles are more likely to make it to our planetary neighborhood. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that interstellar dust grains are found in meteorites, as are all kinds of other grains. Most (or many) meteorites we have contain grains from all over our cosmic neighborhood, including those formed long before our Earth, and they're conglomerated together in the meteorite matrix. Micrometeorites are smaller and will have less types of grains and likely more weathering due to their higher surface area to volume ratio. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To answer your question: are micrometeorites cosmic dust?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some probably &lt;em&gt;contain&lt;/em&gt; microscopic bits of cosmic dust, but some won't, and those that do aren't necessarily entirely made of cosmic dust. &lt;/p&gt;&#xA;" OwnerUserId="7965" LastActivityDate="2015-09-08T07:15:15.890" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11753" PostTypeId="1" AcceptedAnswerId="11758" CreationDate="2015-09-08T15:27:46.953" Score="3" ViewCount="130" Body="&lt;p&gt;I'm trying to follow the calculations in &lt;a href=&quot;http://www.siamese-heritage.org/jsspdf/1991/JSS_088_0r_Eade_RulesForInterpolationInThaiCalendar.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt; on page 5 (see image below). The paper is describing traditional methods of calculation for a luni-solar calendar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The math is not hard, but I can't make sense of the notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured that the &lt;code&gt;... = 150 : 1780&lt;/code&gt; (as in step 3) means that 150 is the integer result, and 1780 is the remainder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what is &lt;code&gt;... = 2; 19:28&lt;/code&gt; (step 8)? Degree; minute:second? Okay, but then I'm lost at step 10 with &lt;code&gt;134 * sin(0:32) = 0:01&lt;/code&gt;. I haven't managed to get that result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bMGRW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bMGRW.png&quot; alt=&quot;calculations&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9153" LastActivityDate="2015-09-09T02:29:08.933" Title="How to interpret this old degree notation?" Tags="&lt;notation&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11754" PostTypeId="1" CreationDate="2015-09-08T19:08:28.367" Score="3" ViewCount="1525" Body="&lt;p&gt;I have started to watch the skies in order to learn stars, constellations planets etc. The few references I checked state that Saturn presents itself to the naked eye as a pale yellow tiny little ball, easy to notice if one knows beforehand its position. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, using Stellarium I have managed to learn that Saturn is near the Scorpio constellation, just below its claws, which is easy to spot because Antares has distinct red-orange colour. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I first spotted the planet I noticed that it was like a blue ball. I should say it looked like a pale, near white blue or a pale blue, near white -- I can not be more precise than this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am spotting it correctly? I was taught that planets do not scintillate like stars. The ball I am talking about certainly did not do it. If I did spot it correctly, I would like to ask if and how often Saturn changes its colours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;p.s. I thought it might be a case of city pollution, however I should say I live in a small mountain village with very few if any air or light pollution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="9156" LastActivityDate="2015-09-08T20:54:23.980" Title="The colour of saturn as it appears to the naked eye" Tags="&lt;planet&gt;&lt;saturn&gt;&lt;naked-eye&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11755" PostTypeId="2" ParentId="11754" CreationDate="2015-09-08T20:45:46.300" Score="2" Body="&lt;p&gt;Planets do scintillate, just much less often - it takes much greater turbulence to make them twinkle. I've seen Venus twinkle quite a few times, especially when it was close to horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light pollution makes no difference when observing the planets in the Solar System. They are far too bright to be affected by it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn should be an off-white when observed with the naked eye. That being said however, if the planet is close to horizon and atmosphere is turbulent, it can cause some weird color effects. But then the color would appear to change rapidly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, as someone else said in comments, no planet looks like a &quot;ball&quot; to the naked eye. They are all points of light just like the stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check this guide, it should tell you the best method to locate Saturn these days (use the image at the top of the article):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.skyandtelescope.com/observing/sky-at-a-glance/this-weeks-sky-at-a-glance-september-4-12/&quot; rel=&quot;nofollow&quot;&gt;http://www.skyandtelescope.com/observing/sky-at-a-glance/this-weeks-sky-at-a-glance-september-4-12/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the object you're looking at is in the right place with regard to Antares and the other things in Scorpio, then it's probably Saturn, no matter what color it seems to be. Color is very subjective and can be modified by the atmosphere in some cases. Position is not subjective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't worry too much about color. For comparison, Jupiter is a more pronounced shade of butter-yellow than Saturn, so that's less controversial, you don't see people argue about the color of Jupiter. But a lot of people see Saturn as plain white; I see it as off-white, the color of old PC cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can borrow some good binoculars, you should see the ring. Lean the instrument (or your elbows) against something solid (like a table or wall), try and keep the binoculars very steady, and watch the object. Be patient and hold steady. You should see the ring as a tiny streak across the planet, kind of like the letter phi: Φ. Just regular binoculars should be enough to make this observation, and it's pretty awesome when you see it for the first time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a bigger instrument the real shape of the ring becomes evident.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-09-08T20:54:23.980" LastActivityDate="2015-09-08T20:54:23.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11756" PostTypeId="2" ParentId="11751" CreationDate="2015-09-08T22:18:05.007" Score="4" Body="&lt;p&gt;I wouldn't call the reference dubious so much as very much out-of-date. The paper in question was written in 1969, which was still pretty early in our understanding of the dynamics of the Solar System, including Venus. There would have been a lot of uncertainty in many of the measurements, such as rotation speed, which would have been determined by radar. Consequently, the &lt;em&gt;possibility&lt;/em&gt; of resonance was there, at least until better measurements were available. However, &lt;a href=&quot;https://en.wikipedia.org/wiki/Venus#Orbit_and_rotation&quot; rel=&quot;nofollow&quot;&gt;the wikipedia article on Venus&lt;/a&gt;, and the &lt;a href=&quot;http://adsabs.harvard.edu/abs/1979ApJ...230L.123S&quot; rel=&quot;nofollow&quot;&gt;attendant 1979 paper&lt;/a&gt; citation (#90) indicate the idea of a resonant Venus was rejected decades ago once sufficiently accurate radar observations of the planet had been made. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, clarifying of the wikipedia article you mention is probably in order.&lt;/p&gt;&#xA;" OwnerUserId="8080" LastActivityDate="2015-09-08T22:18:05.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11757" PostTypeId="2" ParentId="11753" CreationDate="2015-09-08T22:30:29.600" Score="3" Body="&lt;p&gt;This doesn't really answer your question, but I read &lt;a href=&quot;https://github.com/profound-labs/calculating-the-uposatha-moondays/blob/master/references/Rules%20For%20Interpolation%20in%20the%20Thai%20Calendar.pdf&quot; rel=&quot;nofollow&quot;&gt;https://github.com/profound-labs/calculating-the-uposatha-moondays/blob/master/references/Rules%20For%20Interpolation%20in%20the%20Thai%20Calendar.pdf&lt;/a&gt; and noted that footnote 13 (which is referenced by equation 8, although it sort of looks like 3^13) reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;The routine subtraction of 3 arcmins is a causing Ashadha to be month &#xA;10; but by this Wat geographical longitude correction for the sun, as &#xA;is the Tapotharam reckoning Vaisakha is month 7 (Keng subtraction of &#xA;40 arcmins for the moon (sec. C13). Tung style), causing Ashadha to be &#xA;month 9. &#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which suggests the calculation is being done in arcminutes, not degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, this doesn't help, since the result is 76.5283 arcminutes, which is nowhere near 2 degrees 19 minutes 28 seconds (which is 139.467 arcminutes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;134*sin(0:32) may refer to 32 seconds of arc, and the result is 0.02 which is 1 minute or arc (0:01). Terrible use of notation here if that actually is the case.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2015-09-08T22:30:29.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11758" PostTypeId="2" ParentId="11753" CreationDate="2015-09-09T01:13:30.747" Score="3" Body="&lt;p&gt;The notation $(x;\;y:z)$ seems to be $(30\times 60)x+60y+z$ in minutes of arc. The calculation seems to take just the integer part rather than rounding. Thus, for example, step 8 is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}((64552 / 292207) \times 360) - 3 &amp;amp;=&amp;amp; 79^\circ \lfloor 31.7\rfloor' - 3' \\ &amp;amp;=&amp;amp; (2\times 30 + 19)^\circ 28' \\ &amp;amp;=&amp;amp; (2;\;19:28)\text{,}\end{eqnarray}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and step 10 is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$134\,\sin(0:32) = \lfloor 1.247\rfloor = 1 = (0:01)\text{.}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some passages in the paper that point towards this interpretation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(2) How does one interpret what the numerical values of the &lt;em&gt;rӕk&lt;/em&gt; are taken to represent? It is universally agreed that the counting of the &lt;em&gt;rasi&lt;/em&gt; (ราศี, signs of the zodiac) begins with Aries (Mesa) = 0; but the &quot;r&quot; at 0. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$^{12}$ Where desirable, values in arcmins are here converted to signs, degrees, and arcmins in order to make them compatible with following operations. Thus at stage C12, the value 258 arcmins becomes 0; 4, 18 to make it compatible with 2; 19, 28.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since the zodiac partitions the celestial longitude into twelve $30^\circ$ divisions, this makes sense, and seems to work out fairly straightforwardly though a bit cryptically, because the calculation mixes degrees and minutes: e.g., the minuend (first part) of Step 8 is in degrees, which is only &lt;em&gt;implicitly&lt;/em&gt; converted to minutes of arc to be compatible with the subtrahend ($3'$).&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditDate="2015-09-09T02:29:08.933" LastActivityDate="2015-09-09T02:29:08.933" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11760" PostTypeId="1" AcceptedAnswerId="11761" CreationDate="2015-09-09T08:20:22.573" Score="1" ViewCount="39" Body="&lt;p&gt;High mass x-ray binaries necessarily accrete via wind, no disc accretion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes the difference between a hmxb and a lmxb?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In what kind of circumstances does an accretion disc form?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does it form?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When do we think there is a disc?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a clear cut between disc and wind accretion?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditDate="2015-09-09T12:33:56.453" LastActivityDate="2015-09-09T12:33:56.453" Title="accretion discs in h/lmxbs" Tags="&lt;accretion&gt;&lt;x-ray&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11761" PostTypeId="2" ParentId="11760" CreationDate="2015-09-09T08:45:53.210" Score="2" Body="&lt;p&gt;The physical difference between high- and low-mass X-ray binaries is that the latter has a donor star that fills its &lt;a href=&quot;https://en.wikipedia.org/wiki/Roche_lobe&quot; rel=&quot;nofollow&quot;&gt;Roche lobe&lt;/a&gt;. Usually, the compact object is &lt;em&gt;more&lt;/em&gt; massive than the donor star. Mass transfer occurs via Roche lobe overflow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In high-mass X-ray binaries usually the donor star is more massive than the compact object and mass transfer is via a stellar wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that a disc may form (in either case actually, but almost always for a LMXB) is that there is too much specific angular momentum (angular momentum per unit mass) in the accreted material to allow it to fall directly onto the compact object (angular momentum must be conserved). Broadly speaking, a condition for the formation of an accretion disc would be that the specific angular momentum of the accreted material exceeds the specific angular momentum of material in a Keplerian orbit at the surface of the compact object. The formation of an accretion disc would be favoured by short orbital periods or slower wind speeds. Material can then only be transferred onto the compact object as angular momentum is transferred outward by (possibly) viscous torques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This simple picture is immensely complicated by the presence of any strong magnetic fields.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-09-09T08:45:53.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11762" PostTypeId="2" ParentId="11617" CreationDate="2015-09-09T15:04:47.030" Score="1" Body="&lt;p&gt;Probably this is due to the fact that orbits of the Earth and Mars are inclined to the ecliptic differently so our viewing angle of the orbit of Mars changes slightly over the year.  The program is probably showing the orbit of Mars projected onto the Celestial Plane as seen by us on the one specific date. Note, this is different from the ephemeris orbital data which gives the apparent position as seen by us at each date/time.   Just a guess though.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2015-09-10T18:07:30.677" LastActivityDate="2015-09-10T18:07:30.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11764" PostTypeId="1" AcceptedAnswerId="11766" CreationDate="2015-09-09T17:45:44.820" Score="3" ViewCount="479" Body="&lt;p&gt;How can I find simulations of trinary, 4-nary.. etc. star systems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do they look like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example the Alpha Centauri system has 3 stars and possible planets. What does that orbit look like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot find any simulation on &lt;a href=&quot;http://www.solstation.com/orbits/ac-absys.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, it includes 2 Centauri A and B does not include the third dwarf star Proxima.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have been observing the system for many years, I would think a lot of simulations would come out of each new star system we find, or are they all very similar to one another.&lt;/p&gt;&#xA;" OwnerUserId="9163" LastEditorUserId="9163" LastEditDate="2015-09-10T21:25:17.290" LastActivityDate="2015-09-10T21:25:17.290" Title="What does the orbit of multiple star systems look like?" Tags="&lt;orbit&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11766" PostTypeId="2" ParentId="11764" CreationDate="2015-09-10T06:58:42.240" Score="7" Body="&lt;p&gt;In the case of Alpha Centauri A&amp;amp;B and Proxima Centauri,   A&amp;amp;B orbit each other at elliptical orbits, getting almost as close as the Sun is to Saturn and about as far as the Sun is to Neptune.   Proxima Centauri is about 1,000 times further out and it's a tiny compared to the inner 2, so for simulation purposes, it's essentially a 2 star system with a far distant 3rd orbiting the inner 2.    The 3rd star is so far and comparatively small that it has essentially zero effect on the simulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/n5k8G.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/n5k8G.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://blogs.agu.org/martianchronicles/2012/10/20/earth-sized-planet-discovered-around-alpha-centauri/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source of Photo&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in general, most 3 or more star systems operate in similar ways where you can get a 1 or 2 stars operating as the center of mass and then you need to extend a fair bit of distance for the 3rd star.  3 stars all about the same distance and about the same size can be set up on a computer, but it's unlikely to happen in reality cause instabilities tend to grow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun-Earth-Moon is effectively a stable 3 body system and not subject to 3-body chaos because the Moon is safely within the stable region of the Earth's &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hill Sphere&lt;/a&gt; and those 3 objects have clear cut hierarchy in terms of gravitational influence.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pluto, Neptune&amp;amp; the Sun, for example, is probably long term unstable.   &lt;a href=&quot;https://en.wikipedia.org/wiki/Stability_of_the_Solar_System#Neptune.E2.80.93Pluto_resonance&quot; rel=&quot;nofollow noreferrer&quot;&gt;See here&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is Wikipedia's 3 body simulation, but this is probably not something you'd see in solar-systems because of the inherent instability.   In such a 3 body orbital system, one of the 3 bodies would likely either be ejected or 2 of the 3 bodies would collide, creating a more stable 2 body system.   It wouldn't have long term stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AcSxe.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AcSxe.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/N-body_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/N-body_problem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could potentially get a 3 body system where the 2 outer bodies are in Trojan points to each other (L4 or L5 only).   I'm not sure how common that is but I'd think it happens occasionally, provided the central star had significantly more mass than the 2 outer stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Lagrangian_point&quot; rel=&quot;nofollow noreferrer&quot;&gt;L4 and L5 Trojans&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-09-10T07:54:39.230" LastActivityDate="2015-09-10T07:54:39.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11767" PostTypeId="1" AcceptedAnswerId="11772" CreationDate="2015-09-10T18:54:45.360" Score="11" ViewCount="562" Body="&lt;p&gt;Are there any stars visible to the naked eye whose position, over a normal human lifetime, can be seen to have changed (relative to other stars in the vicinity)?&lt;/p&gt;&#xA;" OwnerUserId="9170" LastEditorUserId="7982" LastEditDate="2020-04-14T02:24:20.033" LastActivityDate="2020-04-14T02:24:20.033" Title="Perceiving movement of stars" Tags="&lt;star&gt;&lt;amateur-observing&gt;&lt;night-sky&gt;&lt;apparent-motion&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11768" PostTypeId="2" ParentId="11767" CreationDate="2015-09-10T19:36:56.443" Score="0" Body="&lt;p&gt;Yes, there are plenty.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun, being the closest star, will appear to move the most quickly, covering the expanse of the sky from the East(ish) to the West(ish) in a single day.  Especially around sunset and sunrise, the motion is very visible and you can watch it moving position over the course of a few minutes (don't look directly at it, though; it's very bad for your eyes).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Northern hemisphere, other stars will appear to rotate around the North Star every night, and if you look at the stars a few hours apart they will have rotated into different positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a photo resulting from someone leaving their camera shutter open for several hours: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3Jfng.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3Jfng.jpg&quot; alt=&quot;Star Trails&quot;&gt;&lt;/a&gt;&#xA;(&lt;a href=&quot;https://en.wikipedia.org/wiki/Star_trail#/media/File:Star_trails_over_the_ESO_3.6-metre_telescope.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image source&lt;/a&gt;: Wikipedia article on &lt;a href=&quot;https://en.wikipedia.org/wiki/Star_trail&quot; rel=&quot;nofollow noreferrer&quot;&gt;Star Trails&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="7849" LastEditorUserId="7849" LastEditDate="2015-09-10T19:43:22.323" LastActivityDate="2015-09-10T19:43:22.323" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11769" PostTypeId="1" AcceptedAnswerId="11775" CreationDate="2015-09-10T20:00:11.060" Score="1" ViewCount="293" Body="&lt;p&gt;How do I get the error value out of an astropy.constants quantity?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;In [87]: from astropy import constants as c&#xA;&#xA;In [88]: c.M_sun&#xA;Out[88]: &amp;lt;Constant name='Solar mass' value=1.9891e+30 error=5e+25 units='kg' reference=&quot;Allen's Astrophysical Quantities 4th Ed.&quot;&amp;gt;&#xA;&#xA;In [89]: c.M_sun.value&#xA;Out[89]: 1.9891e+30&#xA;&#xA;In [90]: c.M_sun.error&#xA;AttributeError: 'Constant' object has no 'error' member&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="113" LastActivityDate="2015-09-11T05:17:00.010" Title="How to get error out of astropy constants" Tags="&lt;astropy&gt;&lt;python&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11770" PostTypeId="2" ParentId="11767" CreationDate="2015-09-10T20:32:38.497" Score="2" Body="&lt;p&gt;Really difficult.  While the motion of several stars is significant on that time frame, most of them are simply not close enough to other stars such that the motion can be determined.  Barnard's star moves about a third of the width of the moon over 60 years.  If it were next to another bright star, that could be seen.  But without some image enhancement to bring up fainter stars or a particularly close conjunction, I think it's unlikely that any single naked-eye observer could be certain of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speculation on my part is that Tycho Brahe could have just managed it if he had been very lucky.  Although he didn't use lenses, he had an observatory, excellent vision, and detailed records.  The observatory and the ability to measure accurate angles removes the need for a nearby star.  But as far as I know, he did not suggest any such changes.  Only later astronomers (often working with Brahe's data) were able to tease out the motions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry to suggest Barnard's in a question for naked eye observations.  As it is the fastest known proper motion, I tend to think of it for an order of magnitude check.  If we were looking for the best-case scenario, it's probably 61 cygni.  Tycho Brahe's accuracy limit was probably around 1 arcminute.  61 cygni will move that much in around 12 years.  So it's theoretically possible, but terribly unlikely given the rarity of high proper motion stars.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-09-10T22:06:25.973" LastActivityDate="2015-09-10T22:06:25.973" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11772" PostTypeId="2" ParentId="11767" CreationDate="2015-09-10T21:11:28.333" Score="11" Body="&lt;p&gt;To answer this, I interrogated the revised Hipparcos catalogue produced by van Leeuwen (2007), accessed at the &lt;a href=&quot;http://cdsarc.u-strasbg.fr/viz-bin/Cat?cat=I%2F311&amp;amp;target=brief&amp;amp;&quot;&gt;CDS, Strasbourg&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 4559 stars in the catalogue with a Hipparcos magnitude $&amp;lt;6$ and which you might class as &quot;naked eye&quot; objects, though you'd need good eyes, a good site and plenty of patience to be dealing with stars fainter than about 5th magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorting these objects by total proper motion, we find there are three objects with proper motions between  4.0 and 5.3 arcseconds per year:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Star           Mag       Total PM (arcsecs/yr)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HIP19849       4.56       4.1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HIP108870      4.83       4.7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HIP104214      5.37       5.3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say a human lifetime is 80 years. This means that these stars move an angular distance of between 328 and 424 arcseconds on the sky with respect to the average position of all the other stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angular resolution of the human eye is about &lt;a href=&quot;https://en.wikipedia.org/wiki/Naked_eye#Naked_eye_in_astronomy&quot;&gt;30-60 arcsecs&lt;/a&gt; (for really good vision and a completely dark adapted eye). So the separation between two stars could be estimated to some fraction of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this would be very difficult though unless these stars were very close to another naked eye star, so that you could see the separation changing with time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I did another search. I looked for naked eye stars that were within 10 arcminutes of another naked-eye star &lt;em&gt;and&lt;/em&gt; which had a high relative proper motion that would make them move significantly with respect to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best candidate pair is HIP71500 (which is basically stationary) and HIP71502 (which is a high proper motion star), magnitude 5.58 and 5.85, which are separated currently by about 15 arcseconds, but after 80 years, this will have changed by 52 arcseconds. What this means is that for someone with acute naked-eye vision (is it possible to have that over your lifetime), a star that was apparently single to the naked eye will gradually become a (just) resolved visual binary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nice bit of trivia!&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-09-10T22:52:02.443" LastActivityDate="2015-09-10T22:52:02.443" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11773" PostTypeId="2" ParentId="11317" CreationDate="2015-09-10T21:35:11.507" Score="2" Body="&lt;p&gt;Yes - especially as the occultation does not have to occur on the bright side. Planets may also be occulted (see &lt;a href=&quot;http://www.lunar-occultations.com/iota/planets/planets.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.lunar-occultations.com/iota/planets/planets.htm&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally the symbol is not originally Islamic. It was the flag of Constantinople when conquered by the Ottomans and then adopted by them. The Ottomans later became the Caliphs - the nominal leaders of the world's Sunni Muslims and so the flag was widely adopted across Islam.&lt;/p&gt;&#xA;" OwnerUserId="823" LastEditorUserId="823" LastEditDate="2015-09-11T18:39:41.247" LastActivityDate="2015-09-11T18:39:41.247" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11775" PostTypeId="2" ParentId="11769" CreationDate="2015-09-11T05:17:00.010" Score="5" Body="&lt;p&gt;While I'm not familiar with the package, a &lt;a href=&quot;http://docs.astropy.org/en/stable/api/astropy.constants.Constant.html#astropy.constants.Constant&quot;&gt;very quick look at the documentation&lt;/a&gt; suggests that you want&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;In [90]: c.M_sun.uncertainty&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;instead.  I've just checked and this appears to be correct.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt; python -c &quot;from astropy import constants as c ; print c.M_sun.uncertainty&quot;&#xA;&#xA;5e+25&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2015-09-11T05:17:00.010" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11776" PostTypeId="2" ParentId="10901" CreationDate="2015-09-11T11:42:50.217" Score="1" Body="&lt;p&gt;The mean instant of an equinox JDE_{0} (or solstice) is the statistical mean calculated over a sample of equinox's instants (since they do not happen exactly in the same instant/day every year). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first eight equations are derived by interpolation and the corrections (the 24 terms) are derived as a truncated iteration, I suppose from the JDE{N}-JDE{0,N}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The truth is that Meeus does not explain anything at all (though he mentions interpolating the apparent longitude of the sun in 3 dates for higher accuracy). Without an explanation in the book, it is pretty difficult to guess how he did it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find some more details about the iterative method here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.spacebanter.com/showthread.php?t=171014&quot; rel=&quot;nofollow&quot;&gt;http://www.spacebanter.com/showthread.php?t=171014&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps :)&lt;/p&gt;&#xA;" OwnerUserId="9174" LastActivityDate="2015-09-11T11:42:50.217" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11777" PostTypeId="1" CreationDate="2015-09-11T11:47:19.490" Score="2" ViewCount="171" Body="&lt;p&gt;I was watching a documentary and they said that Cosmologists have determined that the Universe is infinite. I find this to contradict the expansion theory. If the Universe had a starting point then how could it be infinite ? It can only reach out as far as it has expanded in the age of the Universe, 13.8 billion years . If it was infinite then it would not be expanding because it would already reach out to infinity. I hope it is clear what it is that I am trying to say ?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastActivityDate="2015-11-23T11:11:58.673" Title="infinite Universe" Tags="&lt;universe&gt;" AnswerCount="3" CommentCount="3" ClosedDate="2015-09-12T23:31:25.600" ContentLicense="CC BY-SA 3.0" />
  <row Id="11778" PostTypeId="2" ParentId="11777" CreationDate="2015-09-11T16:49:48.697" Score="0" Body="&lt;p&gt;In geometry one may have a line segment, of finite length, with defined start and end points. One may also have a line, continuing without bound in both directions. Then again you may have a ray, which has one end, continuing without bound in one direction. You could even have a circle, with no end, but finite length. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what is the time dimension of the universe like? That depends on how space-time is curved by the gravity of the mass and energy in the universe. Measurements suggest that time in the universe has one end, at the &quot;big bang&quot;, but continues without bound: an &quot;infinite universe&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without knowing the exact context it is hard to be certain, but my guess is this is what is intended. The universe has start time, but no end time.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-09-11T16:49:48.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11779" PostTypeId="2" ParentId="11777" CreationDate="2015-09-11T19:36:13.430" Score="1" Body="&lt;p&gt;There are infinite number of points between 0cm and 1cm on your ruler, there are also an infinite number of points between 0cm and 2cm but it is plain that the 2cm gap is twice as big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not an exact parallel (or even a very good inexact one) but is offered to show you that infinity may mean quite a lot of things: for instance what if space were infinite before the big bang but what happened then was that &quot;things&quot; (ie the spacetime metric) became ever further distant from one another - if space was already infinite then this expanded space could not reach a limit.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2015-09-11T19:36:13.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11780" PostTypeId="1" CreationDate="2015-09-12T01:26:09.490" Score="3" ViewCount="35" Body="&lt;p&gt;Where can I find a large list of online star cameras with wide fields of view?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.itelescope.net/all-sky-cameras&quot; rel=&quot;nofollow&quot;&gt;http://www.itelescope.net/all-sky-cameras&lt;/a&gt; is one example, but the images are fairly low resolution and there aren't many of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a better site or sites to see wide-field star images?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize there are several online telescopes, but they all have a small limited field of view (which makes sense), so you can only see small portions of the sky at once, not the entire sky as above.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2015-09-12T01:26:09.490" Title="Online wide-field star cameras?" Tags="&lt;amateur-observing&gt;&lt;photography&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11781" PostTypeId="2" ParentId="11777" CreationDate="2015-09-12T01:36:43.193" Score="0" Body="&lt;p&gt;I think you're operating under a few false presumptions, the first being that that all infinities are &lt;em&gt;instantaneous&lt;/em&gt;, which they are not!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine I told you to start counting at 1:00 PM tomorrow, starting at the number one, and then to keep counting, once a second, until I say stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at 1:00 PM you'd start: 1...2...3...4...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 1:05 PM you'd be at: 300...301...302...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And at 1:10 PM you'd be at: 600...601..602...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok now I tell you to keep counting &lt;em&gt;to infinity&lt;/em&gt;, so that you never stop counting! Eventually you'll reach every number you can think of (and even more, since you can always count another second after the one you just got to), but it might take you weeks, or months, or centuries to get to there!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in that example we had a start, 1, and we're going all the way to infinity (i.e. we'll never stop). In the same way, the universe could start at one time and keep expanding forever (the only difference is that the universe &quot;starts counting&quot; at 0 instead of 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why 0 and not 1? Or 2? Or -356? Or even &lt;em&gt;-infinity&lt;/em&gt;? Why did the universe start 13.8 billion years ago? What about the time &lt;em&gt;before&lt;/em&gt; the big bang?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very, very tricky question (and one which may never be answered), but most cosmologists will tell you that there was no time before the Big Bang (since there was no space!), and even if there was, we probably wouldn't be able to probe it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will suggest you check out Sean Carroll's blog with a few great explanations to these kind of questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://preposterousuniverse.com/writings/cosmologyprimer/faq.html&quot; rel=&quot;nofollow&quot;&gt;http://preposterousuniverse.com/writings/cosmologyprimer/faq.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9142" LastEditorUserId="9142" LastEditDate="2015-11-23T11:11:58.673" LastActivityDate="2015-11-23T11:11:58.673" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11782" PostTypeId="1" AcceptedAnswerId="11783" CreationDate="2015-09-12T05:45:38.367" Score="2" ViewCount="1811" Body="&lt;p&gt;We know that, the orbit of pluto ambiguously cuts the orbit of Neptune. I want to know, whether the revolving speed of Neptune is equal to the revolving speed of pluto. If not, is there a possibility in future,  that the two planets, or rather two celestial bodies will collide in future? &lt;/p&gt;&#xA;" OwnerUserId="9176" LastActivityDate="2015-09-12T08:22:28.447" Title="Can Pluto and Neptune collide anytime in future?" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;pluto&gt;&lt;neptune&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-09-13T13:47:55.010" ContentLicense="CC BY-SA 3.0" />
  <row Id="11783" PostTypeId="2" ParentId="11782" CreationDate="2015-09-12T07:45:58.100" Score="5" Body="&lt;p&gt;It's hard to find a good 2 dimensional representation, but Pluto and Neptune's orbits don't actually cross.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lxgCE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lxgCE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the orbits (not the planets but the orbits) never get within 100 million miles (I remember reading that, but can't find a link right now).  This is because Pluto's orbit is on a different plane.   It's easy to represent in 3D, hard to draw in 2D, but when Pluto is effectively the same distance from the sun as Neptune it's &quot;below&quot; Neptune, looking at the solar system from a sideways point of view.   When it's on the same plane as Neptune's orbit, which it only crosses twice in one of it's years, it's further away from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sciencefocus.com/qa/could-neptune-and-pluto-ever-collide-their-orbits-intersect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source_1&lt;/a&gt; and &lt;a href=&quot;http://www.livescience.com/32505-will-pluto-ever-hit-neptune.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source_2&lt;/a&gt; and &lt;a href=&quot;http://www.spaceanswers.com/solar-system/is-it-possible-for-neptune-and-pluto-to-collide-given-that-their-orbits-cross/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source_3&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An astronomer diagrammed the 2 objects orbits for the next 4 billion years and they don't intersect during that time.  &lt;a href=&quot;http://www.astronomycafe.net/qadir/q364.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is a quirk in the calculations.   Pluto's resonance with Neptune is probably long term unstable and because of this, it's orbit could be subject to variations that leave open a Jim Carrey's chance of hitting Neptune.   &lt;a href=&quot;https://en.wikipedia.org/wiki/Stability_of_the_Solar_System#Neptune.E2.80.93Pluto_resonance&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very unlikely, because they're not on the same plane and they're not orbiting at the same period and they're currently in orbital resonance which they would need to fall out of to even have a chance, so it's a lottery ticket's chance of happening but it could theoretically happen, in a billion or couple billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's more likely that Pluto would fly past Neptune and have it's orbit changed, and even that's still a long shot.  The most likely scenario is that they never get close, at least over the lifespan of the sun, but if they do get close, Pluto could fly further out into the solar system or closer in due to gravity assist.   It might even hit Earth.   But, I repeat, this is crazy-long-shot territory, but I think it's worth mentioning that it's about as likely to hit one of the inner planets as it is to hit Neptune.  Jupiter being the largest, might have the best chance of being hit by Pluto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is also answered &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1518/plutos-orbit-overlaps-neptunes-does-this-mean-pluto-will-hit-neptune-sometime&quot;&gt;here&lt;/a&gt;, but because everyone else was answering this question no, I wanted to point out that it's just barely possible but extremely unlikely.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-09-12T08:22:28.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11784" PostTypeId="2" ParentId="11493" CreationDate="2015-09-12T08:52:41.853" Score="2" Body="&lt;p&gt;You are correct - when we look at stars in the sky, about half of them are actually binary systems, with a range of separations. The frequency of binarity is strongly mass dependent; the 50% figure is appropriate for stars like the Sun, but it appears to be closer to 100% for high-mass O-stars and as small as 20% for the lowest mass M-dwarf stars.&#xA;There is also a change in the separation (and corresponding orbital period) distribution. The distributions of separations looks &quot;log-normal&quot; (like a Gaussian when you plot log separation on the x-axis). For solar type stars the peak is at around 50-100 au, but binaries become more compact (maybe 10-30 au) for low-mass stars. There is a tail of binaries with much smaller and larger separations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These observational data are explained (or are in the process of being explained) as a combination of nature and nurture. The basic process of star formation results from the collapse and fragmentation of molecular clouds. If a collapsing fragment has significant angular momentum then the most energetically favourable way it can continue to collapse whilst conserving angular momentum is to sub-fragment into two pieces orbiting each other. This is thought to be the most basic binary formation process. Other processes include the formation of low mass companions through instabilities in cicumstellar disks or interactions between young stars in dense protostellar environments. Exactly how this leads to the mass dependence of the binary frequency and separation distribution is a topic of contemporary research; there is no definitive answer I can give.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that I think there are two basic ways, or rather two basic locations, in which planets could form in a binary stem. Either around one or other of the stars or as circumbinary objects. The likelihood of each will depend how and when the binary formed, the binary component separation and mass ratio. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The planets would likely form in the same way that they do around single stars. If the binary star components are originally formed close together, there could be a circumbinary disk from which a stable planetary system formed. Both circumbinary disks and circumbinary planets have been observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disks around individual components are truncated by tidal forces at a radius &lt;em&gt;roughly&lt;/em&gt; corresponding to a third of the binary separation. Inside this radius it may be possible to form planets. Thus widely separated binaries could have planets orbiting the individual stars. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-09-12T08:52:41.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11786" PostTypeId="1" CreationDate="2015-09-13T02:19:50.403" Score="3" ViewCount="782" Body="&lt;p&gt;Or has it been gravitationally steered that way because of how it was formed?&lt;/p&gt;&#xA;" OwnerUserId="9179" LastEditorUserId="9179" LastEditDate="2015-09-14T16:51:57.387" LastActivityDate="2015-09-14T19:55:00.847" Title="Is it just coincidence that the Moon follows the ecliptic?" Tags="&lt;the-moon&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11788" PostTypeId="2" ParentId="11786" CreationDate="2015-09-13T16:25:59.097" Score="1" Body="&lt;p&gt;From what I can find out on-line (Wikipedia, etc.) most natural moons tend to be in the equatorial plane of the planet (formed in a circumplanetary disc). The Earth's moon is commonly thought to have formed in a collision with a Mars size object. It probably settled initially into an orbit inclined about 10 degrees to the equatorial plane of the Earth due to accretion disc processes. Since then, tidal forces from the sun (due to Earth's proximity) have brought the Moon more inline with the ecliptic.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2015-09-13T16:25:59.097" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11789" PostTypeId="2" ParentId="11786" CreationDate="2015-09-13T16:41:04.840" Score="1" Body="&lt;p&gt;The Earth and Moon are rather better thought of as a two planetoid system both orbiting the Sun, than a planet and a satellite (the Moon does indeed orbit the Sun as well as the Earth). The Moon orbits Earth at a point where Earth's gravitational pull is of the same order as the Sun's. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also believe that the Moon was formed as a result of a catastrophic event involving the proto-Earth - ie one planet split into two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such it is not a surprise that it has an orbit close to the plane of the ecliptic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other planets' satellites are not the same - the gas giants dominate their satellites and in Mars's case it seems likely that Phobos and Demos are captured planetoids from the asteroid belt rather than anything much to do with Mars as such.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2015-09-13T16:41:04.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11790" PostTypeId="1" AcceptedAnswerId="11803" CreationDate="2015-09-13T19:21:12.370" Score="0" ViewCount="195" Body="&lt;p&gt;The Big Bang scattered planets and stars everywhere in three dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But after billion years of moving and interacting with each others through gravity, planets moved on the same plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the very small but non zero friction in space, could we give a good estimate of the age of the universe?&lt;/p&gt;&#xA;" OwnerUserId="9182" LastEditorUserId="2153" LastEditDate="2015-09-13T19:24:15.937" LastActivityDate="2015-09-14T09:57:24.143" Title="Could we estimate the age of the universe based on the planar property of the Solar System?" Tags="&lt;universe&gt;&lt;big-bang-theory&gt;&lt;age&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11791" PostTypeId="2" ParentId="11790" CreationDate="2015-09-13T19:28:56.040" Score="1" Body="&lt;p&gt;First, &lt;strong&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/136860/did-the-big-bang-happen-at-a-point&quot;&gt;the Big Bang did not happen at a point.&lt;/a&gt;&lt;/strong&gt; I cannot emphasize this enough. It happened everywhere in space at the same time. You can't think of it as an explosion from one location pushing everything else away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, planets and stars didn't exist at the time of the Big Bang. The Solar System, for example, formed about 4.5 billion years ago, while the oldest stars - Population III stars - formed &lt;a href=&quot;http://www2.astro.psu.edu/users/rbc/a534/redman.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;within a few million years&lt;/a&gt; after the Big Bang. At the beginning of the universe, everything was in an exotic soup-like state: a &lt;a href=&quot;http://en.wikipedia.org/wiki/Quark%E2%80%93gluon_plasma&quot; rel=&quot;nofollow noreferrer&quot;&gt;quark-gluon plasma&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, your hypothesis is flawed in two places.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-09-13T19:28:56.040" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11793" PostTypeId="2" ParentId="11786" CreationDate="2015-09-13T19:40:34.917" Score="1" Body="&lt;p&gt;It is not a coincidence at all! (There was, however, some statistical chances involved.) A key question to answer before answering your question, though, is Why is there an ecliptic at all? It is fair to see all the planets and a majority of the smaller bodies existing in a nice flat plane (the ecliptic) and think that it didn't have to be that way. Why not have different planets orbiting in different planes, or even backwards? To get the answer, we have to go back to the origins of the Solar System, when it was still a collapsing cloud of interstellar gas and dust. As gravity collapsed the cloud, the small amount of angular momentum it had became more significant, much as an ice skater pulling her arms in while spinning will make her spin faster. So, with the help of friction, the early Solar System flattened and had a 'spin' in a preferential direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the Solar System evolved, planets formed, and all of them had ingrained in them this spin. Today, we see that in not only the existence of the ecliptic and the fact that planets &lt;em&gt;orbit&lt;/em&gt; in the same direction, but in the fact that most planets &lt;em&gt;rotate&lt;/em&gt; in the same direction as the sun does (Uranus and Venus being notable exceptions) and most large moons orbit their planets in the same direction (Triton around Neptune being a notable exception).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the moon? A leading theory for its formation is that it formed from the debris that resulted in a collision between a still-forming Earth and another planet-sized object. Statistically, both these objects would be most likely to have similar angular momenta to that of the overall Solar System (though it was by no means guaranteed), so their resulting debris would have it as well. When that debris coalesced into our moon, it too would also keep the original angular momentum 'signature,' and its orbit would thus be comparable to that of the ecliptic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is all a very chaotic and statistical process. While angular momentum is always conserved, there are ways of exchanging it between bodies (via gravitational interactions &amp;amp; collisions). Consequently, there are variations away from the ecliptic: the moon is inclined 5 degrees from the ecliptic; Earth's rotation is inclined 23 degrees from it; smaller bodies like comets will sometimes orbit well outside the ecliptic; etc. But the general trend holds, and is a marker you can see present throughout the solar system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an interesting aside, while all major planets in our Solar System lie roughly along the ecliptic, that is not always the case for other planets in other star systems. &lt;a href=&quot;https://en.wikipedia.org/wiki/WASP-17b&quot; rel=&quot;nofollow&quot;&gt;Wasp 17-b&lt;/a&gt;, for example, is observed to have a retrograde orbit around its host star, while others have been observed to orbit outside of their own system's ecliptics entirely. &lt;/p&gt;&#xA;" OwnerUserId="8080" LastActivityDate="2015-09-13T19:40:34.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11794" PostTypeId="1" CreationDate="2015-09-13T20:40:30.993" Score="2" ViewCount="109" Body="&lt;p&gt;I watched a cosmology lecture given by Leonard Susskind that convinced me to put my money on dark energy being a pressure caused by energy that is evenly distributed throughout the &quot;vacuum&quot; of space. However, quantum physics predicts that there should be a much larger energy density (i.e. many magnitudes larger). Have there been any more recent findings or theories that could shed light on this discrepancy?&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="2153" LastEditDate="2015-11-04T00:01:24.417" LastActivityDate="2015-11-04T00:01:24.417" Title="The largest discrepancy in the history of science" Tags="&lt;cosmology&gt;&lt;dark-energy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11796" PostTypeId="2" ParentId="11186" CreationDate="2015-09-13T21:35:50.073" Score="7" Body="&lt;p&gt;From what I understand, James Webb, if used in conjunction with a successful starshade (being developed at MIT), should be able to detect close in planets orbiting nearby stars. However, getting good atmospheric spectra of these planets directly (from planet's blackbody IR emission) is unlikely. What we must hope for is that TESS, which should be going up in 2017, will find a few nearby stars with transiting planets. Then, James Webb will be able to look for atmospheric absorption lines from stellar light passing through a planet's atmosphere during transit. This method may still be limited to large (Jupiter size) planets.&#xA;In an ideal situation (say looking at absorption lines of a super-earth) there are many &quot;bio-signatures&quot;, but one of the easiest to detect would be an ozone line in the infrared. By itself, this would not be proof, even though there would need to be a constant replenishing source of O2 in the atmosphere to maintain the O3. If methane could also be found we could rightly get VERY excited since methane and oxygen don't co-exist very well.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2015-09-13T21:35:50.073" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11798" PostTypeId="1" AcceptedAnswerId="18749" CreationDate="2015-09-13T23:35:27.527" Score="6" ViewCount="129" Body="&lt;p&gt;In May, I did my own amateur analysis of &lt;em&gt;Kepler&lt;/em&gt; data that was supplied &lt;a href=&quot;https://kepler.nasa.gov/Mission/discoveries/index.cfm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. This was a table of all confirmed &quot;Kepler planets&quot; to date (April 1, 2015). This table had some unbelievably high masses/densities listed in the first column after the planet designation (ie: planets of Kepler- 23, -24, -25, -27, -28, -32, -39, -48, -52, -54, -57, -58, -59, -60 and others). I didn't take these seriously, but would like to know if there was a &quot;units&quot; error or something I am missing. To give you an idea, Kepler-23b is given a radius of $1.9 R_{\text{Earth}}$ and a mass of $254.3 M_{\text{Earth}}$ - and that's nothing compared to some others. I checked just now and the data is the same.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="2153" LastEditDate="2016-10-16T19:34:38.037" LastActivityDate="2016-10-16T19:43:41.333" Title="Unbelievably high masses/ densities in NASA website data" Tags="&lt;exoplanet&gt;&lt;mass&gt;&lt;data-analysis&gt;&lt;kepler&gt;&lt;density&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11799" PostTypeId="1" AcceptedAnswerId="11804" CreationDate="2015-09-14T00:10:01.287" Score="4" ViewCount="106" Body="&lt;p&gt;This question is from an amateur data analysis I did in May 2015 of data from &lt;a href=&quot;https://web.archive.org/web/20150423011745/http://kepler.nasa.gov/Mission/discoveries/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kepler.nasa.gov/Mission/discoveries&lt;/a&gt;. There were 80 Jupiter size planets in the Kepler &amp;quot;confirmed&amp;quot; planet table on this website (through 4/1/15) that had host stars with metallicity data. Of these, 24 stars have [Fe/H}&amp;gt;0.2 and 40 have -0.1&amp;lt;[Fe/H]&amp;lt;+0.2. This is a much higher percentage than the sample as a whole. Would this be strong evidence that high metallicity measurements are due to accretion? Or, do the metallicity measurements given reflect abundance deeper into the star than accretion can count for (ie. higher metallicity stars are more likely to have Jupiters in general)?&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="26112" LastEditDate="2021-04-06T01:46:16.313" LastActivityDate="2021-04-06T01:46:16.313" Title="Reason for a correlation between Hot Jupiters and higher metallicity in Kepler data" Tags="&lt;data-analysis&gt;&lt;kepler&gt;&lt;metallicity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11800" PostTypeId="1" CreationDate="2015-09-14T00:43:30.577" Score="2" ViewCount="71" Body="&lt;p&gt;The data from Kepler has taught us that there is a much larger number of &quot;Neptune-class&quot; planets out there than we previously thought. I wonder, however, if this &quot;dominance&quot; of Neptunes is because of the wide range of radii we use to define this class. If I'm not mistaken, we classify a planet as &quot;Earth-like&quot; when it is in the range 0.75 to 1.25 R(Earth), &quot;Super-earth&quot; from about 1.25 to 2.0 R(Earth) and &quot;Neptune class&quot; from 2.0 to 6.0 R(Earth) (a much larger range which would stand out even more if we knew masses). Are we maintaining the nomenclature to classify planets of different characteristics or are we just &quot;solar system biased&quot;?&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2015-09-14T00:43:30.577" Title="Classification of Planets?" Tags="&lt;data-analysis&gt;&lt;kepler&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11801" PostTypeId="1" AcceptedAnswerId="11812" CreationDate="2015-09-14T01:22:20.627" Score="0" ViewCount="185" Body="&lt;p&gt;I did my own amateur Kepler data analysis in May of this year of data from Kepler.nasa.gov/Mission/ discoveries. I found a strong correlation between star size and planet size (percentages). 37 of the 44 planets confirmed (84.1%) orbiting M dwarfs were less than 2.0 R(Earth) with only one hot Jupiter. 129 of 241 (53.6%) orbiting K dwarfs were less than 2.0 R(Earth) with 9 hot Jupiters (3.7%). 196 of 517 (37.9%) orbiting G dwarfs were less than 2.0 R(Earth) with 49 hot Jupiters (9.5%). 92 of 213 (43.2%) orbiting F dwarfs were less than 2.0 R(Earth) with 23 hot Jupiters (10.8%). I would like to know how much of this correlation is probably real and how much is biased by the limits of the study? Biases I am aware of are the fact that the Kepler study is primarily of systems with &quot;close in&quot; planets, (only 90 of 1015 confirmed as of 4/1/15 are over 0.3 AU) and that only 4.3% of the &quot;Kepler stars&quot; are M dwarfs. I also see biases that might arise from the methods of confirming the planets (Kepler stars from 110 to 405 had planets confirmed by the &quot;multiple candidate&quot; method). It also appears that there may be a greater difference between K and G star systems than between G and F. Where can I find a more professional analysis of this type?&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2015-09-14T20:12:02.870" Title="Correlation of planet sizes with star sizes?" Tags="&lt;data-analysis&gt;&lt;kepler&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11803" PostTypeId="2" ParentId="11790" CreationDate="2015-09-14T04:02:37.703" Score="0" Body="&lt;p&gt;No, the formation of planetary systems is not the result of the interaction of stars and planets which form at random in space post Big Bang. Stars form from density fluctuations in the gas in galaxies and planetary systems usually form (to the best of our knowledge) from the disks of gas and dust around stars that are a side effect of the formation of stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is planar planetary systems should be usual because they form from planar systems which are largely the result of dynamical processes in star formation.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-09-14T09:57:24.143" LastActivityDate="2015-09-14T09:57:24.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11804" PostTypeId="2" ParentId="11799" CreationDate="2015-09-14T07:02:58.560" Score="4" Body="&lt;p&gt;This is a well known, well researched phenomena. Yes, there certainly is a correlation between metallicity and the likelihood of observing a hot Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two classes of explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) The correlation is real and due to the fact that it is easier to form planetary embryos from metal-rich material in the core-accretion model of giant planet formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) The observed exoplanets are those that weren't swallowed. The hosts are metal rich because they did swallow others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, it was thought possible that there could be an observational bias. Planets are easier to find and measure around high metallicity stars. This applied mainly to the Doppler technique, because high metallicity stars have stronger spectral lines. Transit detection should be much less affected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most think that some form of (1) is going on. (2) is argued against because the correlation still appears in stars with a wide variety of convection zone depths into which planetary material would have mixed. I'm not sure it is dead though, there have been a number of claims for abundance enhancements that look like the product of planet engulfment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll add some references later. Nice work though to discover the correlation if you knew nothing of it previously.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-09-14T07:02:58.560" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11805" PostTypeId="2" ParentId="11786" CreationDate="2015-09-14T09:30:32.230" Score="2" Body="&lt;p&gt;&lt;strong&gt;No&lt;/strong&gt;, this is not entirely a coincidence, though perhaps not in the way you expect (and other answers have indicated): it may have been coincidal at formation, but is required for long-term stability of the Moon-Earth system itself and, possibly, Earth climate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;formation mechanism&lt;/strong&gt; of the Moon (or the Moon-Earth system) is still debated. The favoured model involves a &lt;strong&gt;collision&lt;/strong&gt; (between a pre-Earth and a Mars-sized object), when the resulting orientation of the Moon-Earth orbit depends critically on the impact parameters, which are essentially random, i.e. &lt;strong&gt;coincidal&lt;/strong&gt;. Conversely, if the Moon-Earth system formed by some form of &lt;strong&gt;capture&lt;/strong&gt; (possibly involving some surrounding gas to absorb the excess energy), a orientation in the eccliptic is more &lt;strong&gt;likely&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Howerver, the formation mechanism is not all there is to this. It is also required that the &lt;strong&gt;Moon-Earth system remains stable&lt;/strong&gt; over 4.5 Gyr. This excludes any orbit initially circular and inclined by more than $i=\arccos\sqrt{3/5}\approx39.2^\circ$, when the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kozai_mechanism&quot; rel=&quot;nofollow&quot;&gt;Kozai mechanism&lt;/a&gt; would result in a collision of the Moon with Earth within a few years. The critical incliniation for a non-circular orbit is even smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even an inclination $i\approx30^\circ$ may not be viable, since it may imply that the Earth's spin axis (which is after all tidally coupled to the Moon's orbit) would be/have been tilted considerably stronger than by merely $\approx23^\circ$. A strong tilt of the Earth spin axis would have significant consequences for the Earth's climate and may well have prevented the formation of higher live forms. This is essentially a weak form of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Anthropic_principle&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;antropic principle&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2015-09-14T19:55:00.847" LastActivityDate="2015-09-14T19:55:00.847" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11806" PostTypeId="1" AcceptedAnswerId="11808" CreationDate="2015-09-14T11:36:02.910" Score="1" ViewCount="45" Body="&lt;p&gt;Back in 2002, me and my friends observed a green glow in the sky at night in the form of giant slow moving spurts of flame, looking more dense near the horizon and more blurred at height. It was surrounding the area, not just on one side, as I see on most of the imagery available. It was bright enough to see for several people sitting around a bonfire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible for the natural phenomena, 'Aurora Borealis' take place as far south as 56° N, 37° E; in a Moscow suburb, Russia?&lt;/p&gt;&#xA;" OwnerUserId="9189" LastEditorUserId="2365" LastEditDate="2015-09-14T21:37:01.833" LastActivityDate="2015-09-14T21:37:01.833" Title="Is Polar Lights observation possible at the latitude of 56° N?" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11807" PostTypeId="1" CreationDate="2015-09-14T13:20:59.213" Score="4" ViewCount="58" Body="&lt;p&gt;According to the suggested tags, this must be the right forum:&lt;/p&gt;&#xA;&lt;p&gt;I &lt;a href=&quot;https://landsat.gsfc.nasa.gov/landsat-8/operational-land-imager&quot; rel=&quot;nofollow noreferrer&quot;&gt;found&lt;/a&gt;  that for each band of Landsat 8 over 7000 detectors are on the array. Does anyone know how big each of the sensors is?&lt;/p&gt;&#xA;" OwnerUserId="9190" LastEditorUserId="37191" LastEditDate="2021-03-19T11:59:25.317" LastActivityDate="2021-03-21T12:16:22.740" Title="What is the detector size of LANDSAT 8's detector array?" Tags="&lt;satellite&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="11808" PostTypeId="2" ParentId="11806" CreationDate="2015-09-14T16:44:23.220" Score="2" Body="&lt;p&gt;Yes. The Aurora form in a ring, roughly centred on the Magnetic pole. The pole does wander slowly, but it has been located in the Canadian Arctic for some time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the pole is on the American side of the Arctic ocean, aurora are seen at lower latitudes in America than in Russia, nevertheless, when a powerful solar storm occurs, aurora can be seen at surprisingly low latitudes, even into the subtropics, as Wayfaring stranger notes in her comment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aurora in Moscow is a notable, but not exceptional occurrence, as this news report indicates &lt;a href=&quot;http://www.themoscowtimes.com/article/507096.html&quot; rel=&quot;nofollow&quot;&gt;http://www.themoscowtimes.com/article/507096.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-09-14T16:44:23.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11809" PostTypeId="1" CreationDate="2015-09-14T17:12:17.717" Score="6" ViewCount="543" Body="&lt;p&gt;The popular, but still disputable, Moon formation theory is that a pre-Earth was hit by a Mars sized planet. The mass ratio of Earth:Mars:Moon is roughly 100:10:1.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Does this mean that pre-Earth was roughly 9 Moon masses (~10%) lighter&#xA;than today's Earth? Or could much mass have been ejected from the&#xA;Earth-Moon system?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Did the impact substantially and immediately change Earth's orbit,&#xA;such as its average distance from the Sun?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there a better planet name for &quot;pre-Earth&quot;, like there is for&#xA;Theia? It was after all another planet than Earth is.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-09-15T11:32:23.283" Title="How did pre-Earth's mass and orbit change at the Moon formation event?" Tags="&lt;the-moon&gt;&lt;earth-like-planet&gt;&lt;impact&gt;&lt;orbital-migration&gt;&lt;theia&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11810" PostTypeId="2" ParentId="1966" CreationDate="2015-09-14T17:35:43.967" Score="3" Body="&lt;p&gt;I'll be honest. I looked up Galactic Habitable Zone on Wikipedia. The idea was introduced in 1986 and expanded upon in the book &quot;Rare Earth&quot; by Brownlee and Ward. Essentially the idea is that in the bulge there is too much activity (supernovae, etc.) and in the halo of the galaxy there is not enough heavy elements (carbon, etc.) for life to exist. This was in reference to complex life &quot;as we know it&quot; (Check out another Peter Ward book entitled &quot;Life As We Do Not Know It&quot;). Stellar habitable zones are based on the presence of liquid water. This definition would greatly increase the theoretical &quot;Galactic Habitable Zone&quot;. I think we need to look at a region and calculate all of the pluses and minuses of that specific region then keep our minds open.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2015-09-14T17:35:43.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11811" PostTypeId="1" AcceptedAnswerId="18565" CreationDate="2015-09-14T19:46:46.990" Score="3" ViewCount="210" Body="&lt;p&gt;I was extremely stoked when I read an article in January of this year that the Kepler2 mission found three Earth-sized objects (one possibly in the habitable zone) transiting a star only 150LY away. I haven't heard anything since, not even if this star and planets (confirmed) have been given a Kepler designation. I guess I would like to know if JWST or some other upcoming mission will be able to give us more info on this system. I can't imagine how excited I will be if TESS gives us something even closer!!&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2016-10-10T13:32:52.607" Title="EPIC 201367065?" Tags="&lt;exoplanet&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11812" PostTypeId="2" ParentId="11801" CreationDate="2015-09-14T20:12:02.870" Score="1" Body="&lt;p&gt;I think there are a number of studies that look at these statistics; I'll try to dig some out. In the meantime I can give you some further things to ponder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let's assume a null hypothesis that the statistics and mass distribution of planets were independent of stellar mass - what would we observe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well one selection bias you don't mention is that the likelihood of transit detection depends on the ratio of planet to star size. This means you are &lt;em&gt;less&lt;/em&gt; likely to see small planets around F-stars than M-stars. Or to put it another way, you expect the fraction of large planets to increase with stellar mass - as you have found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is quite likely that the material available to form planets is correlated with the stellar mass - in other words there is a correlation between stellar mass and protoplanetary disc mass. In which case it could be quite difficult for smaller stars to have the requisite protoplanetary disc density to form giant planets before the disc disperses.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-09-14T20:12:02.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11814" PostTypeId="2" ParentId="1961" CreationDate="2015-09-15T03:26:16.350" Score="1" Body="&lt;p&gt;I think you're like me and need more of a layman's answer. If you want a good, easy to understand explanation of what happens look at &quot;Formation and Evolution of the Solar System&quot; in Wikipedia then click on 5.3 (The Sun and planetary environments). The sun will actually expand twice: Once when the core gets so hot from accelerated hydrogen fusion (as the sun's core gets hotter the hydrogen burns faster) that the hydrogen in the shell around the core starts to fuse (this hydrogen fusion in the shell is what pushes the outer layers out to about 1AU). Then after as long as about 2 billion years. The core reaches a critical density/temperature (due to the increased amount of helium) that the helium starts to fuse into carbon. At this point, there is a helium &quot;flash&quot; and the sun shrinks back down to about 11 times its original size. The helium in the core fuses into carbon for about 100 million years until the same sort of thing happens (except this time hydrogen and helium in the shell around the core start to fuse causing outer layers to expand again. It's after the helium starts getting used up (or &quot;polluted&quot; with carbon enough to stop the fusion process) and there is not enough mass to start carbon fusion that a planetary nebula is ejected and the star starts to &quot;die&quot;.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="7997" LastEditDate="2015-09-15T03:45:15.447" LastActivityDate="2015-09-15T03:45:15.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11815" PostTypeId="1" AcceptedAnswerId="11818" CreationDate="2015-09-15T04:24:51.510" Score="0" ViewCount="162" Body="&lt;p&gt;Delta Pavonis has about the same mass as the sun so I would think that the evolutionary path would be about the same (the mass is given as O.991 M(Sun)/ not sure how we get this so accurately). It has been estimated to be about 6.6 to 6.9 Ga with a luminosity of 1.22 L(Sun). This all makes sense if you put it on the main sequence where we will be in a couple of billion years (after T(max) in about a billion years). The problem is that the surface temperature is only 5550K (spectral) and for this reason is thought to be in subgiant phase which isn't supposed to happen to us for about another 5.5 billion years. Delta Pav also has a very high metallicity [Fe/H] of +0.33 which also screams &quot;subgiant&quot;. What am I missing?? (data from Wikipedia sources)&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2015-09-15T08:49:33.807" Title="Conflicting information about the star Delta Pavonis?" Tags="&lt;stellar-evolution&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11816" PostTypeId="2" ParentId="11243" CreationDate="2015-09-15T05:17:04.640" Score="3" Body="&lt;p&gt;Since the late heavy bombardment, the Earth's orbit couldn't have changed much. After all, life has been here ever since. Most scientists think that the answer to the faint young sun paradox is a thicker CO2 atmosphere and not the Earth being closer in. In earlier solar system history, the Earth may have been &quot;shepherded&quot; in by Jupiter when it migrated closer.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2015-09-15T05:17:04.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11817" PostTypeId="2" ParentId="11815" CreationDate="2015-09-15T08:44:20.220" Score="1" Body="&lt;p&gt;I haven't looked in too much detail, but the first point is to note that there are two estimates of the age: 6.6–6.9 Gyr and 9.3 Gyr.  The latter is given without uncertainties, but I looked it up on Vizier and the range is 5.8–10.7 Gyr, basically from isochrones. The former is computed using gyrochronology or activity-age relations, which are usually only calibrated up to about the age of the Sun: 4.6 Gyr.  So this star is probably older than the oldest calibration point. In addition, recent results from asteroseismology (some presented at a conference just a few months ago) suggest that the activity-rotation-age indicators are unreliable in older stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this is basically pointing to the fact that measuring a star's age is difficult and model/calibration dependent.  I wouldn't take those results too seriously.  The 5.8–10.7 Gyr range is the more realistic range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, higher metallicity stars tend to be redder, so I'm not surprised that it would be a bit redder than an older version of the Sun.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2015-09-15T08:44:20.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11818" PostTypeId="2" ParentId="11815" CreationDate="2015-09-15T08:49:33.807" Score="3" Body="&lt;p&gt;To answer a question, we &lt;em&gt;do&lt;/em&gt; need the source of information (wikipedia just isn't good enough).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=delta%20Pavonis&amp;amp;NbIdent=1&amp;amp;Radius=2&amp;amp;Radius.unit=arcmin&amp;amp;submit=submit%20id&quot; rel=&quot;nofollow&quot;&gt;SIMBAD&lt;/a&gt; database, delta Pavonis is a G8 subgiant, with a temperature of 5512K and a $\log g$ of 4.23 (in cgs units) (&lt;a href=&quot;http://cdsbib.u-strasbg.fr/cgi-bin/cdsbib?2006AJ....132..161G&quot; rel=&quot;nofollow&quot;&gt;Gray et al. 2006&lt;/a&gt;), which looks correct for a subgiant classification. The metallicity is given as $+0.13$ in the same source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a large number of other spectroscopic determinations of the star's parameters listed in the SIMBAD measurements. All seem to agree on the subgiant nature, but for example &lt;a href=&quot;http://cdsbib.u-strasbg.fr/cgi-bin/cdsbib?2014A%26A...562A..71B&quot; rel=&quot;nofollow&quot;&gt;Bensby et al. (2014)&lt;/a&gt; makes the star a little warmer (5635 K) and more metal rich (+0.37).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what appears to be at odds with this is the age/mass combination that you quote. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wikipedia entry and information you quote appears to be taking information from &lt;a href=&quot;http://adsabs.harvard.edu/abs/2007ApJS..168..297T&quot; rel=&quot;nofollow&quot;&gt;Takeda et al. (2007)&lt;/a&gt;. They estimate the age and mass of the star from the temperature and gravity using a Bayesian fitting method and the YREC stellar evolution models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is really important to you, then you need to read the Takeda et al. paper. They show that the posterior probability distribution of age and mass can have &lt;em&gt;multiple&lt;/em&gt; peaks. In the case of delta Pavonis it appears that wikipedia has got things wrong (what a surprise). The age that is quoted is actually for a &lt;em&gt;secondary&lt;/em&gt; peak in the posterior probability distribution. The secondary peak in the mass distribution is for $M= 1.101M_{\odot}$. I think that age, mass pair makes much more sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They do not provide detailed information/discussion on individual cases. Like you, I cannot see how they arrive at 1 solar mass for a star of any age, unless they have picked up a &lt;em&gt;pre main sequence&lt;/em&gt; solution in the HR diagram - for which the age corresponding to 1 solar mass would be very small. However, they claim that their models start at the ZAMS. The fact that they list no primary peak in the age posterior pdf suggest to me that the pdf look like a declining function from zero (the ZAMS), with a secondary peak at the 7Gyr (quoted by wikipedia). This then suggests that this secondary solution is also not a brilliant fit and that the fitting process would really have liked to make this younger than ZAMS - i.e. a PMS star (which it isn't). In turn then, this may suggests that the gravity used by Takeda et al. might be too low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Takeda et al. used spectroscopically derived parameters of temperature 5590 K, $\log g= 4.31$ and $[M/H]=+0.26$ from &lt;a href=&quot;http://adsabs.harvard.edu/abs/2005ApJS..159..141V&quot; rel=&quot;nofollow&quot;&gt;Valenti &amp;amp; Fischer (2005)&lt;/a&gt;. These authors also did their own isochrone fits to these parameters finding $M=1.045 \pm 0.058 M_{\odot}$ and an age of 5.3-6.8 Gyr. They say that the mass at the best-fitting isochronal age is $1.07M_{\odot}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think part of what is puzzling you is that you have to remember that having a metal content that is twice the Sun's &lt;em&gt;does&lt;/em&gt; affect the evolutionary tracks considerably. It is not that the evolutionary timescales are massively different, but the effective temperature of the star is changed considerably for a given mass and age.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-09-15T08:49:33.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11819" PostTypeId="1" AcceptedAnswerId="11821" CreationDate="2015-09-15T10:07:51.453" Score="1" ViewCount="1250" Body="&lt;p&gt;I have 9DOF sensor (accelerometer, gyroscope, and magnetometer) that gives orientation via Euler angles (yaw, pitch, roll). Along with GPS data (latitude, longitude, elevation, time) this is passed to a PC application that has to calculate Dec and RA of a target (e.g. star). Sensor and GPS receiver are mounted on a telescope, so it should give the celestial coordinates of a star. The problem is that I don't know how to get Dec and RA from those data. I would very much appreciate a detail explanation of how I could get this.&lt;/p&gt;&#xA;" OwnerUserId="9197" LastActivityDate="2015-09-15T14:35:37.083" Title="Calculate Dec and RA of a star from Euler angles and GPS data" Tags="&lt;star&gt;&lt;telescope&gt;&lt;observational-astronomy&gt;&lt;declination&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11820" PostTypeId="2" ParentId="11809" CreationDate="2015-09-15T11:32:23.283" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does this mean that pre-Earth was roughly 9 Moon masses (~10%) lighter&#xA;  than today's Earth? Or could much mass have been ejected from the&#xA;  Earth-Moon system?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think that's likely about right.  In a nutshell, there's two basic questions here.   How much mass was lost as a result of the giant impact and how much mass was gained after the giant impact during the late heavy bombardment and following collisions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that smaller bombardments on Mars has resulted in debris landing on Earth, so it seems highly likely that some debris was lost when Theia and Earth collided, but estimating how much is tricky and you'd really need to be well versed in the models to have a pretty good guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One theory is that Theia formed in an Earth Trojan point so it shared the same orbit around the sun that the Earth did.   &lt;a href=&quot;https://en.wikipedia.org/wiki/Giant_impact_hypothesis#Possible_origin_of_Theia&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;   This would cause a slower impact than a more oblique approach where the velocity of impact wouldn't be much greater than the Escape Velocity of Earth.   If the impact velocity is only a bit above the necessary escape velocity and if a fair bit of energy is transferred into orbital momentum and heat, it's unlikely that a high percentage of debris would reach escape.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if it was a faster impact from a more oblique orbital approach or a head on collision that would create greater compession and rebound, those two scenarios would likely lead to more material escaping.   I can't give you precise numbers, this is more of general thinking about what might happen.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, according to &lt;a href=&quot;https://en.wikipedia.org/wiki/Giant_impact_hypothesis#Basic_model&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; about 20% of the mass of Theia entered the earth's orbit after the impact.    Orbital Velocity has to be at least Escape Velocity over the square root of 2 and less than escape velocity.   &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/vesc.html#c2&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we use 11 KM/s as an escape velocity estimate, anything that flew off the impact at less than 7.5 KM/s would have returned to Earth pretty quickly.   A bit above 7.5 and maybe about 9 or so KM/s would be low Earth orbit and about 10 KM/s high earth, with 11 KM/s, likely escape and enter in a near earth orbit around the sun.     You'd need a bit faster than that to have a truly outside of earth orbit.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What percentage of Theia flew off at those velocities is hard to say, but I think it's likely that much of the planet slowed down upon impact and not too much escaped the Earth's Hill sphere, but I'd be guessing if I tried to put percentages to that.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Did the impact substantially and immediately change Earth's orbit,&#xA;  such as its average distance from the Sun?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because it was a (I assume) a Trojan object, the 2 young planets would have shared the same orbit and so there probably wouldn't have been a big change.  Orbital momentum should be largely be conserved though some orbital velocity might be converted to rotational momentum, I don't think there would be a huge change.   Similarly, if any debris from the impact flew off in one particular direction, the remaining material would react with similar momentum in the opposite direction, but having much greater mass, the effect would be small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly some orbital change probably happened, but compared to the entire Earth, it was probably small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the impact was different than theorized, that might change the answers quite a bit.    (My 2 cents anyway)&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-09-15T11:32:23.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11821" PostTypeId="2" ParentId="11819" CreationDate="2015-09-15T14:35:37.083" Score="0" Body="&lt;p&gt;Obtain from the publisher Willmann-Bell the book ''Astronomical Algorithms'' by Jean Meeus. If obtaining elsewhere, be sure to obtain the 2nd ed. with corrections as of August 10, 2009. The equations you want are in Chapter 13, &quot;Transformations of Coordinates&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some variables must be defined:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha$ = right ascension, if obtained from formula it is in radians&lt;br&gt;&#xA;$\delta$ = declination, positive north, negative south&lt;br&gt;&#xA;$h$ = altitude, positive above the horizon, negative below horizon&lt;br&gt;&#xA;$A$ = azimuth, measured westward from the South, other sources often measure from the North&lt;br&gt;&#xA;$\psi$ = observer's latitude&lt;br&gt;&#xA;$H$ = local hour angle&lt;br&gt;&#xA;$\theta$ = local sidereal time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step is to transform horizon coordinates (azimuth and altitude) to equatorial coordinates (local hour angle and declination).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\tan H  = \frac{\sin A}{\cos A \sin \psi + \tan h  \cos \psi}\\&#xA;\sin \delta  = \sin \psi  \sin h  - \cos \psi  \cos h  \cos A&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the local hour angle H is transformed to right ascension $\alpha$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\alpha = \theta - H&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="7385" LastActivityDate="2015-09-15T14:35:37.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11822" PostTypeId="1" CreationDate="2015-09-15T14:59:15.393" Score="3" ViewCount="218" Body="&lt;p&gt;Is this true? And if so, why?&#xA;&lt;a href=&quot;http://www.snapple.com/real-facts/cap-view/1003&quot; rel=&quot;nofollow&quot;&gt;http://www.snapple.com/real-facts/cap-view/1003&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9200" LastActivityDate="2015-09-15T17:38:04.207" Title="If two pieces of the same type of metal touch in outer space, they will bond together permanently?" Tags="&lt;space&gt;&lt;metal&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2015-09-23T16:10:12.507" ContentLicense="CC BY-SA 3.0" />
  <row Id="11823" PostTypeId="1" AcceptedAnswerId="11831" CreationDate="2015-09-15T15:04:43.450" Score="8" ViewCount="431" Body="&lt;p&gt;In turn, would that increase our gravitational pull?&lt;/p&gt;&#xA;" OwnerUserId="9200" LastActivityDate="2015-09-16T05:58:59.103" Title="If 50 tons or more of debris falls to earth everyday, is Earth getting heavier?" Tags="&lt;gravity&gt;&lt;earth&gt;&lt;space&gt;&lt;space-debris&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11824" PostTypeId="2" ParentId="1961" CreationDate="2015-09-15T16:01:09.450" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My question is this: Why does fusion cease in the core?! It seems to&#xA;  me that as gravity crushes the star, stellar fusion would reignite in&#xA;  the core itself—not in a sphere around the core. Why does the stellar&#xA;  core remain “dead” while its “shell” begins fusion???&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our sun is about halfway through its &quot;main sequence&quot; or the hydrogen fusing stage. Fusion in a star's core is part of its &lt;em&gt;dynamic equilibrium&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The gravitational field of the star (produced by its mass) tends to compress its mass toward the core.  The more compressed the matter, the hotter it becomes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The release of energy produced by fusion of elements at the core tends to disperse matter away from the core.  Dispersion of matter from the core tends to reduce its temperature.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The size of a star is due then, at least in part, by the dynamic equilibrium formed at which the gravitational compressive forces are equal to the fusion-produced expansive forces.  This is called a star's &lt;em&gt;hydrostatic equilibrium&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of energy that is released on a per-mass basis declines as heavier elements are fused.  The most energy is released for fusing hydrogen, less is released by fusing helium, and so on.  Eventually, a point is reached (fusion of iron) at which the amount of energy needed to fuse the elements is greater than the energy released by the fusion reaction.  The iron core of such stars is thought to be &quot;non-fusing&quot; because if the core were heated to a temperature to enable iron fusion, insufficient energy would be released from the reaction to maintain the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, the star becomes increasingly unable to maintain its hydrostatic equilibrium, even as its mass condenses.  What happens next depends on how massive the star is and whether its gravitational field is strong enough to exceed the electron degeneracy pressure of its mass.&lt;/p&gt;&#xA;" OwnerUserId="9201" LastEditorUserId="9201" LastEditDate="2015-09-15T16:07:37.847" LastActivityDate="2015-09-15T16:07:37.847" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11825" PostTypeId="2" ParentId="1961" CreationDate="2015-09-15T16:54:49.940" Score="1" Body="&lt;p&gt;I suggest you to read &lt;a href=&quot;http://www.space.com/22471-red-giant-stars.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; article on &lt;a href=&quot;http://www.space.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.space.com/&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Quoting from it:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Most of the stars in the universe are main sequence stars — those converting hydrogen into helium via nuclear fusion. A main sequence star may have a mass between a third to eight times that of the sun and eventually burn through the hydrogen in its core. Over its life, the outward pressure of fusion has balanced against the inward pressure of gravity. Once the fusion stops, gravity takes the lead and compresses the star smaller and tighter.&lt;/p&gt;&#xA;&lt;p&gt;Temperatures increase with the contraction, eventually reaching levels where helium is able to fuse into carbon. Depending on the mass of the star, the helium burning might be gradual or might begin with an explosive flash. The energy produced by the helium fusion causes the star to expand outward to many times its original size.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;Wikipedia provides some more insight :&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;When the star exhausts the hydrogen fuel in its core, nuclear reactions can no longer continue and so the core begins to contract due to its own gravity. This brings additional hydrogen into a zone where the temperature and pressure are adequate to cause fusion to resume in a shell around the core. The higher temperatures lead to increasing reaction rates, enough to increase the star's luminosity by a factor of 1,000–10,000. The outer layers of the star then expand greatly, thus beginning the red-giant phase of the star's life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1156" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-09-16T04:44:30.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11827" PostTypeId="2" ParentId="11823" CreationDate="2015-09-15T19:44:49.657" Score="-5" Body="&lt;p&gt;Most of it burns up in the atmosphere and never actually makes it the surface. So no, as far as I know or have ever heard/read the Earth is not getting more massive. Any increase in mass due to space debris (or other things like more humans being born) is negligible compared to the mass of the Earth anyways so it has a null effect in terms of the gravitational attraction. &lt;/p&gt;&#xA;" OwnerUserId="9204" LastActivityDate="2015-09-15T19:44:49.657" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11828" PostTypeId="2" ParentId="1961" CreationDate="2015-09-15T19:46:07.313" Score="11" Body="&lt;p&gt;(This is somewhat simplified but I hope it gets the idea across.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reactions stop in the core because it runs out of fuel. During the main sequence, the star is supported by the fusion of hydrogen into helium. Eventually, the hydrogen runs out at the centre, so hydrogen fusion is no longer possible there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why doesn't it start fusing helium into carbon right away? That's because the core isn't hot or dense enough yet. Different reactions broadly rely on the presence of different resonant states in the nuclei and, in the case of helium, such a state cannot be reached often enough until the core temperature is about $10^8$ kelvin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to get that hot, the core has to contract and heat up. It eventually does (if the star is massive enough) but it doesn't happen instantaneously. Remember that the gas is still hot and at high pressure, which it exerts on itself and its surroundings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, at the edge of the core, the star (partly as a result of said contraction) &lt;em&gt;is&lt;/em&gt; hot enough to turn hydrogen into helium, so it does so. This is exactly the nuclear-burning shell that distinguishes the internal structure of a red giant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So maybe think of it this way. Imagine a star at the end of the main sequence. Where is it hot enough to fuse hydrogen into helium? Everywhere up to the edge of the core! Does it fuse in the core? No, because it's out of fuel. So where does it fuse? At the edge of the core, which we recognize as the shell.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2015-09-15T19:46:07.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11829" PostTypeId="1" AcceptedAnswerId="11832" CreationDate="2015-09-15T23:40:42.037" Score="10" ViewCount="251" Body="&lt;p&gt;The &quot;Lithium Problem&quot; relates to the fact very-low-metallicity stars appear to have a Li/H ratio approximately one third of what would be expected. The ratio should be the same as the prediction from Big Bang nucleosynthesis theory. A possibly related problem is that the ⁶Li/⁷Li ratio is several orders of magnitude too high.&lt;/p&gt;&#xA;" OwnerUserId="9186" LastEditorUserId="9186" LastEditDate="2015-09-16T00:14:32.637" LastActivityDate="2020-04-15T14:46:31.097" Title="Is any consensus forming on the solution to the &quot;Lithium Problem&quot;?" Tags="&lt;big-bang-theory&gt;&lt;astrophysics&gt;&lt;spectroscopy&gt;&lt;stellar-astrophysics&gt;&lt;elemental-abundances&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11830" PostTypeId="2" ParentId="11823" CreationDate="2015-09-16T03:40:06.443" Score="7" Body="&lt;p&gt;The Earth has a net loss of mass each year. The infall of debris from space is more than matched mainly by the loss of hydrogen from the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the BBC radio show &lt;a href=&quot;http://www.bbc.co.uk/news/magazine-16787636&quot;&gt;More or Less&lt;/a&gt; the net annual loss of mass is ~50,000 tons/year.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-09-16T03:40:06.443" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11831" PostTypeId="2" ParentId="11823" CreationDate="2015-09-16T05:58:59.103" Score="8" Body="&lt;p&gt;The Earth loses mass because hydrogen and helium (plus other elements in trace amounts compared to hydrogen and helium) escapes the Earth's atmosphere. The Earth gains mass because incoming asteroids (most of them very small) impact the Earth's atmosphere; a few make it all the way to the surface of the Earth. Whether those incoming asteroids burn up in the atmosphere or make it all the way to the surface is irrelevant; the Earth gains mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether the net result is a mass gain or mass loss is a bit up in the air; the uncertainties on both are rather large, and they overlap. The Earth might be gaining or losing a tiny bit of mass every year. That said, most research leans toward atmospheric losses being greater than mass gain from impacting asteroids, comets, and dust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whichever is the case, it's a bit irrelevant. Even the most extreme upper estimates on mass loss or mass accumulation are incredibly tiny compared to the mass of the Earth itself.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-09-16T05:58:59.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11832" PostTypeId="2" ParentId="11829" CreationDate="2015-09-16T11:36:14.030" Score="14" Body="&lt;p&gt;No. There is no consensus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discrepancy between the predicted big bang nucleosynthetic abundance of Lithium 7 and the measured value can be summarised as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take what we know about the the baryonic mass density of the universe and the Hubble constant, we get a self-consistent picture between the cosmic microwave background, observations of galaxy recession etc. and the estimated primordial abundances of Helium and Deuterium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem arises because these same cosmological parameters predict a primordial lithium abundance of &lt;span class=&quot;math-container&quot;&gt;$3\times10^{-10}$&lt;/span&gt;, when expressed as a ratio to the hydrogen abundance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, measurements of the Li abundance present in the photospheres of the &lt;em&gt;oldest&lt;/em&gt; stars (a.k.a. &quot;halo stars&quot;) in our Galaxy suggest that the primordial abundance was about &lt;span class=&quot;math-container&quot;&gt;$1.2\times10^{-10}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The factor of 2-3 difference between these numbers is about 4-5 times the measurement &lt;em&gt;precision&lt;/em&gt;. This is the so-called &quot;Lithium problem&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The potential solutions are reviewed by &lt;a href=&quot;http://arxiv.org/abs/1203.3551&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fields (2012)&lt;/a&gt;. They fall into the following categories.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Astrophysical solutions - that we don't understand our measurements of the Li abundances because of an imperfect understanding of the atmospheres of low metallicity stars; or that we don't understand interior mixing mechanisms that mean at the photosphere, we see material that has been mixed upwards from the interior where the Li has been depleted in nuclear reactions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Nuclear physics - maybe the details of the reaction rates and cross-sections in the big bang model are awry? There are still some sizeable uncertainties here which remain to be nailed down, but are seen as rather unlikely solutions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Additions to the standard big bang model. This includes things like inhomogeneous nucleosynthesis in the early universe - i.e. that it was clumpy even at this early stage. Other possibilities include the possibility that the equilibrium reactions in big bang nucleosynthesis were upset by the decay of massive dark matter particles.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thus there are lots of ideas to solve this problem, and other ideas which suggest it is not so much a problem, but that we can't do the measurements properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT: Update 18/11/19 from the &quot;Lithium in the Universe&quot; conference in Frascati, Rome.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having sat through some review talks on the problem, I can summarise progress as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explanation 2 (Nuclear Physics) is dead. All avenues appar to be explored and the likely remaining uncertainties are at the level of 10%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the following possibilities were also discussed.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;Perhaps the fundamental constants have changed with time resulting in somewhat different binding energy differences between different nuclei? &lt;a href=&quot;https://arxiv.org/abs/1707.03852&quot; rel=&quot;nofollow noreferrer&quot;&gt;Scherrer &amp;amp; Scherrer (2017)&lt;/a&gt; for example discuss a scenario where the mass difference between 2 He nuclei and a 8Be nucleus changes with time in such a way that lots of 8Be is produced during big bang nucleosynthesis, but decays back into He later on. This does not alter the primordial He abundance inferred from present-day observations, but the removal of He at early epochs results in a lower production of 7Li.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Inhomogeneous magnetic fields, contributing significant energy density in the early universe may have led to temperature fluctuations at the epoch of nucleosynthesis. Such fluctuations could reduce the predicted Li abundances a bit, but they also change the predicted deuterium abundance and this wouldn't fit with the now quite precise measurements of the primordial D abundance and the Planck value of &lt;span class=&quot;math-container&quot;&gt;$\Omega_b h^2$&lt;/span&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It has been hypothesised that magnetic fields can play a role in filtering how much primordial Li makes it into the structures that form galaxies. i.e. There is a sort of chemical differentiation caused when there is a gradient of the magnetic field strength at right angles to the magnetic field. This accelerates ions down the gradient and possibly away from forming structures. Since Li is easily ionised in the early universe (compared with H and He), this might lead to the gas that formed the first stars being Li-poor.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;At the end of the meeting, participants were asked (for fun) to vote on what they thought to most likely solution was. #1 Was first, with #3 a distant second.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2020-04-15T14:46:31.097" LastActivityDate="2020-04-15T14:46:31.097" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="11833" PostTypeId="1" AcceptedAnswerId="11834" CreationDate="2015-09-16T13:22:38.610" Score="10" ViewCount="2514" Body="&lt;p&gt;If an Asteroid was to strike the Earth, would it affect noticeably the Earth's rotation, and if so, how large would this Asteroid have to be?&lt;/p&gt;&#xA;" OwnerUserId="9200" LastEditorUserId="2365" LastEditDate="2015-11-06T20:31:57.220" LastActivityDate="2020-10-30T19:14:14.933" Title="If an Asteroid was to strike the Earth, would it affect the Earth's rotation?" Tags="&lt;earth&gt;&lt;asteroids&gt;&lt;impact&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11834" PostTypeId="2" ParentId="11833" CreationDate="2015-09-16T18:31:48.807" Score="8" Body="&lt;p&gt;To have a noticeable effect the impactor needs to be BIG.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most questions about &quot;what would happen if ... hits&quot; can be answered by the &quot;Earth impact effects program&quot; (&lt;a href=&quot;http://impact.ese.ic.ac.uk/&quot; rel=&quot;noreferrer&quot;&gt;http://impact.ese.ic.ac.uk/&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are calculations for a &lt;a href=&quot;http://impact.ese.ic.ac.uk/cgi-bin/crater.cgi?dist=1&amp;amp;diam=100000&amp;amp;pdens=3000&amp;amp;pdens_select=0&amp;amp;vel=20&amp;amp;theta=10&amp;amp;tdens=2500&amp;amp;tdens_select=0&quot; rel=&quot;noreferrer&quot;&gt;100km stony asteroid...&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A brute like this would have a good chance of wiping out most complex life on the planet. There has been nothing like this in the last 4 billion years (or so)... It could cause the length of the day to change by &quot;up to 2.42 seconds&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As gerrit said, it would be the last of our worries.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2015-09-17T17:29:45.217" LastActivityDate="2015-09-17T17:29:45.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11835" PostTypeId="1" AcceptedAnswerId="28599" CreationDate="2015-09-17T03:17:57.760" Score="12" ViewCount="491" Body="&lt;p&gt;In 2014, Robertson et al. made a case that stellar activity is masquerading as planets (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2014Sci...345..440R&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014ApJ...793L..24R&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;) for these well-known systems. In both cases, they claim that the only real planets discovered so far are planets &quot;b&quot; and &quot;c&quot;. In 2015, Anglada-Escude et al. questioned the methods of Robertson et al. (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2015Sci...347.1080A&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;). Also in 2015, Brewer et al. suggested that there is evidence for multiple planets at Gliese 581 &lt;a href=&quot;http://adsabs.harvard.edu/abs/2015MNRAS.448.3206B&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; (note, I only have access to the abstract - fulltext is &lt;a href=&quot;https://arxiv.org/pdf/1501.06952.pdf&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;). Is there a direction that a majority of the professional community is leaning?&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="17348" LastEditDate="2017-07-06T23:15:21.980" LastActivityDate="2019-06-13T19:27:26.060" Title="The battle over Gliese 581 and Gliese 667C?" Tags="&lt;exoplanet&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11836" PostTypeId="1" CreationDate="2015-09-17T03:39:32.433" Score="8" ViewCount="2135" Body="&lt;p&gt;For example, I saw the next expression for matallicity in a paper: $\log(O/H) + 12$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand O/H is the O(Oxygen) to H(Hydrogen) ratio but why is there the number 12? And why the logarithm?&lt;/p&gt;&#xA;" OwnerUserId="9215" LastEditorUserId="4085" LastEditDate="2015-09-17T10:07:03.110" LastActivityDate="2015-09-21T07:20:50.450" Title="Why you use log to measure metallicity in galaxies?" Tags="&lt;fundamental-astronomy&gt;&lt;metal&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11837" PostTypeId="2" ParentId="11836" CreationDate="2015-09-17T07:05:57.947" Score="2" Body="&lt;p&gt;The logarithm is there because the ratio $O/H$ is really tiny.  The log converts essentially points out the order of magnitude.  If we have $O/H= 10^{-14}$, then $\log(O/H)=-14$.  Taking logarithms is a standard practice in science and mathematics when the numbers range over several orders of magnitude (especially so when they are either very large or very small).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the +12, that is most likely because they want their preferred example to have value 0.  It is then easy to tell at a glance if other stars/galaxy have more ($&amp;gt;0$) or less ($&amp;lt;0$) oxygen-to-hydrogen metallicity than the preferred example.  In most cases this will be our own Sun, but without knowing where you saw this it's impossible to say for certain.  We have far more information about our own sun than any other star, so it is the most reliable measuring stick we have for comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, the value tells you approximately how many of orders of magnitude more or less metallicity it has.  For, supposing $\log(O_\mbox{sun}/H_\mbox{sun}))=-12$, then $$\log(O/H)+12 = \log(O/H)-\log(O_\mbox{sun}/H_\mbox{sun})) = \log\left(\frac{O/H}{O_\mbox{sun}/H_\mbox{sun}}\right).$$ &#xA;So if the value of this is 1, then you know the star in question has an $O/H$ ratio $10^1=$ten times that of our sun.  Similarly, if it were $-3$, then the star in question would have an $O/H$ ratio that is $10^{-3}=1/1000$-th that of our sun's (or, equivalently, that our sun's ratio is $1000$ times larger).  A value of $0$ means the ratios are the same.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2015-09-17T07:05:57.947" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11840" PostTypeId="1" AcceptedAnswerId="11846" CreationDate="2015-09-17T15:47:00.263" Score="13" ViewCount="4365" Body="&lt;p&gt;Given the Earth's current speed around the sun and current rate &amp;amp; axis of rotation, what is the best way to keep time to avoid a &lt;a href=&quot;https://en.wikipedia.org/wiki/Leap_year&quot;&gt;leap year&lt;/a&gt;? How many hours should we have in the day and days in a year would keep things balanced to not need to add or remove days from the year? Further, how many minutes per hour and seconds per minute should we have to avoid a &lt;a href=&quot;https://en.wikipedia.org/wiki/Leap_second&quot;&gt;leap second&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2015-09-18T14:54:48.650" Title="How can we avoid needing a leap year/second?" Tags="&lt;orbit&gt;&lt;rotation&gt;&lt;time&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11841" PostTypeId="1" AcceptedAnswerId="11844" CreationDate="2015-09-17T16:03:51.160" Score="3" ViewCount="8045" Body="&lt;p&gt;I was looking at a full moon the other night with my telescope and noticed a black dot move across the moon quite rapidly. This happened a few times and others who were with me were able to confirm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no way the ISS would make its orbit around the Earth that quickly for us to be able to see it a few times. Therefore there must have been others. How many are there, and how long does the average satellite take to make a full orbit around the earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dots we saw looked much like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/BDgZ2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BDgZ2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9200" LastEditorUserId="2365" LastEditDate="2015-09-17T17:06:36.893" LastActivityDate="2015-09-17T17:25:17.810" Title="How many satellites orbit the Earth? At what rate do they orbit the earth?" Tags="&lt;orbit&gt;&lt;earth&gt;&lt;satellite&gt;&lt;the-moon&gt;" AnswerCount="2" CommentCount="8" ClosedDate="2015-09-18T14:47:44.373" ContentLicense="CC BY-SA 3.0" />
  <row Id="11842" PostTypeId="2" ParentId="11841" CreationDate="2015-09-17T16:57:30.573" Score="3" Body="&lt;p&gt;Im not sure of the number, upwards of 2k probably, Godards lists it at 2271, but who knows how accurate that is as some are inactive, or not listed due to government privacy. &#xA;The period would depend on the satellite, its orbutal distance from the earth and its velocity. &lt;/p&gt;&#xA;" OwnerUserId="9219" LastActivityDate="2015-09-17T16:57:30.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11843" PostTypeId="2" ParentId="10455" CreationDate="2015-09-17T17:18:38.033" Score="6" Body="&lt;p&gt;Because planets actually do twinkle. Most people were told that the major difference between stars and planets is that only the former twinkle - but that's an oversimplification. Given the right conditions, planets will twinkle too, it just happens more rarely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several factors that contribute to it:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;lots of air turbulence; or, as astronomers call it, &quot;bad seeing&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;closeness to horizon; if the planets are high in the sky, the air column is shorter so there's less chance they will twinkle; but when they are low, their light goes through more air and so it is perturbed to a larger degree&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The observation you've made, Venus twinkling, is not very unusual. Many stargazers are used to seeing that once in a while. I've seen Venus scintillate several times in the past, always at sunset when it was about to drop below horizon; I would presume you could see the same behavior very early in the morning as Venus has just risen.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2015-09-17T17:18:38.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11844" PostTypeId="2" ParentId="11841" CreationDate="2015-09-17T17:25:17.810" Score="4" Body="&lt;p&gt;There are over &lt;a href=&quot;http://www.universetoday.com/42198/how-many-satellites-in-space/&quot; rel=&quot;nofollow&quot;&gt;1000 operational satellites&lt;/a&gt;, and probably as many again defunct or non-operational satellites in orbit, plus a great deal more small pieces of space junk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About half of these are in a low earth orbit (LEO). Satellites in LEO have a period of between &lt;a href=&quot;https://en.wikipedia.org/wiki/Low_Earth_orbit&quot; rel=&quot;nofollow&quot;&gt;90 minutes and 130 minutes&lt;/a&gt;. But you would not expect a satellite to follow the same path in the sky in two orbits, as the earth will have moved between orbits of the satellite. So it is very unlikely that you will see the same satellite pass in front of the moon twice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Satellites in LEO are moving fast! They will cross the moon's disc in less than a second, and transits are rare, as each is a path only 7km wide. at any one time, a transit is visible from less than 100th of 1% of the Earth's surface. To obtain the awesome photo that you have linked, the astrophotographer would have had to use some orbital prediction software to calculate the exact time of a solar transit by the ISS and travel to a carefully calculated position (and prayed for clear skies). You could be observing the moon all year and never see a transit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the remaining satellites are in Geostationary orbits, They orbit at a great distance over the equator. As their orbital period is exactly one day, they appear to hang in space above a point on the Earth. Such satellites are too distant to be seen as &quot;dots&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-09-17T17:25:17.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11845" PostTypeId="2" ParentId="11840" CreationDate="2015-09-17T18:06:49.583" Score="4" Body="&lt;p&gt;This doesn’t really work the way that you are thinking, at least not in a way that is practical for society at all.  They problem is that we define a day to be based on Earth rotations relative to the sun, and a year as a full orbit around the sun, and if you find the number of rotations of the earth in a single orbit, it is not an integer (~365.24 rotations (days) in a year).  To avoid a leap year, you would need to define the day such that there are an integer number of days in a year (i.e 365 days exactly).  The problem with this is that day and night will drift relative to our clocks, and after 2 years, day and night will be switched.  The length of the year is also variable and not fundamental, so in order to keep this exact relationship, you'd have to constantly redefine the length of the day, which is not a practical improvement over having leap years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The leap second has the same type of problem.  We want to define the number of seconds in a day as 86400 seconds/day, but the Earth’s rotation is not constant.  So, in order to keep clocks from drifting, you have to add leap seconds.&lt;/p&gt;&#xA;" OwnerUserId="9210" LastEditorUserId="7142" LastEditDate="2015-09-17T23:42:32.197" LastActivityDate="2015-09-17T23:42:32.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11846" PostTypeId="2" ParentId="11840" CreationDate="2015-09-17T18:13:51.630" Score="18" Body="&lt;p&gt;Leap years exist for two reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are not an integer number of days in a year.&lt;/li&gt;&#xA;&lt;li&gt;People perceive a need to keep the seasons where they are on the calendar.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Given the above, there is no way to avoid leap years, or something similar. Defining the calendar year as being a fixed number of days (e.g., 365 days) would result in the seasons shifting by one day per four years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leap seconds exist for two reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The length of a day as measured by an atomic clock is not constant.&lt;/li&gt;&#xA;&lt;li&gt;People perceive a need to keep midnight at midnight, noon at noon.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Given the above, there is no way to avoid leap seconds, or something similar. Defining the day as being a fixed number of atomic clock seconds (e.g., 86400) would result in your clock and the Sun disagreeing on mean local noon, but by a very small amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, there are serious proposals to eliminate leap seconds. Some people such as those who use UTC to timestamp financial transactions do not like them. So far, those proposals have been rejected. The standard response is that it's not UTC that's broken; it's using of UTC in a context where it shouldn't be used that is broken. If you need a monotonically increasing time scale, use TAI or GPS time instead.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2015-09-18T14:54:48.650" LastActivityDate="2015-09-18T14:54:48.650" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="11847" PostTypeId="2" ParentId="11840" CreationDate="2015-09-17T23:09:18.850" Score="2" Body="&lt;p&gt;I'm a software engineer, and I can speak about the issue with leap seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are unpredictable.  You don't know far in advance whether you will have one.  Code that cares about accurate number of seconds will need some kind of update or feed to continue working correctly.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also a step that adds complexity.  You have to allow for a minute that contains 61 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first issue, a compromise that keeps reasonable tracking between the Earth's rotation and the time of day would be to allow looser tolerance.  Rather than being within one second, correct it on schedule every 10 years.  Software doesn't have to worry about year-by-year issues, and the clock stays 7 seconds (or ±4 if you jump ahead) to true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that we already have time zones, the sun will not be exactly at the midnight position at midnight &lt;em&gt;anyway&lt;/em&gt; but will be half an hour ahead or behind.  Astronomers already need a special offset clock.&lt;/p&gt;&#xA;" OwnerUserId="7142" LastActivityDate="2015-09-17T23:09:18.850" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11848" PostTypeId="2" ParentId="11840" CreationDate="2015-09-18T00:03:51.447" Score="4" Body="&lt;p&gt;We not only can avoid leap seconds, that's how it used to work in fact.  And there is a common newer system which avoids leap seconds as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before 1960, seconds were defined as 1/86400 of a mean solar day.  Then when variations in the earth's rotation caused it to get out of sync, a new mean solar day could be computed and divided by 86400 - changing the length of the second in absolute terms, stretching or shrinking it very slightly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That was a mess, as you can imagine.  So the second was defined in terms of a specific number of atomic oscillations which could be made extremely precise.  Instead of shrinking and stretching the second to keep an exact number of them in a day, we keep the second fixed and add or subtract one from the (integer) count when we need to adjust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those are pretty much the ways to keep earth rotation timing in sync with our clock time - you need some give somewhere, either by changing the length of the second and keeping the count fixed, or you keep the length fixed and change the count.  For somebody just writing a simple program to, say, compute the civil seconds between two UTC timestamps, the old way was easier (a fixed count of seconds between two times is trivial).  But if you are doing scientific or engineering calculations or experiments to great precision, it's WAY better to have a very firmly fixed length of a second, not changing it from time to time - much worse than the inconvenience of taking leap seconds into account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the way, another approach is to just ignore leap seconds and keep your clocks running continuously.  That's how GPS time works - it started in sync with UTC, but has not been adjusted for the leap seconds since then, so they are out of sync by a quarter minute or so (I haven't check in some while).  That's nice for GPS orbital calculations that cross leap second adjustment boundaries.  In the GPS data packet there is information about the current delta between UTC and GPS time so you can calculate civil time from GPS time, as well as a few months advanced warning when a new leap second is  going to be added or omitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another answer suggested queuing up leap seconds and making a multi-second leap every decade.  That doesn't really simplify your software much tho - now you have to allow minutes with, say, 67 seconds, every decade.  Easier to just deal with leap seconds using a table and meanwhile never be off by even 1 second.  (The standard allows for them to added or omitted by the way - you could have a 59 second minute or a 61 second minute when you need an adjustment.  It's generally the latter tho.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, one other solution.  The organization which really tracked all this was called the &lt;a href=&quot;https://en.wikipedia.org/wiki/International_Earth_Rotation_and_Reference_Systems_Service&quot; rel=&quot;nofollow&quot;&gt;International Earth Rotation Service, later renamed to  International Earth Rotation and Reference Systems Service (IERS)&lt;/a&gt;.  Imagine the chaos if they stopped being funded and the Earth stopped rotating.  Anyway, I suppose you could just ask them to rotate it more consistently.  :-)&lt;/p&gt;&#xA;" OwnerUserId="9225" LastEditorUserId="525" LastEditDate="2015-09-18T14:53:19.757" LastActivityDate="2015-09-18T14:53:19.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11849" PostTypeId="1" AcceptedAnswerId="11852" CreationDate="2015-09-18T10:17:36.740" Score="7" ViewCount="4647" Body="&lt;p&gt;I want to know the angle between the plane that passes through the centre of sun and is perpendicular to its spin axis, and the plane that contains the orbit of earth around sun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a fixed value or it is variable over time (and if so, how does it vary) ! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5LdTe.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5LdTe.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="862" LastEditorUserId="4085" LastEditDate="2015-09-18T18:12:50.443" LastActivityDate="2021-01-01T00:02:55.680" Title="What is the angle between the equator of the sun, and the plane of the Earth's orbit" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;distances&gt;&lt;angular-diameter&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11850" PostTypeId="2" ParentId="11836" CreationDate="2015-09-18T11:02:01.863" Score="5" Body="&lt;p&gt;The distribution of metallicities appear to be more evenly spread out in logspace than in linear space. The reason for this can be ascribed to there being no preferred scale for the abundance of a given element; rather they span several orders of magnitude. The same can be said for instance about the distribution of dust grain sizes, the distribution of the mass of dark matter halos, and the distribution of the area of lakes on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you measure the number of oxygen atoms (or ions) in ten stars, and divide by the number of hydrogen atoms, you may get values like&#xA;$$\mathrm{O}/\mathrm{H} = \{0.03,3.5,25,0.003,0.9,0.4,0.09,0.01,8,0.02\}\times10^{-4}.$$&#xA;Plotting this in a linear and a logarithmic plot, you see that on a log scale, the values are easier to tell apart:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Zptdv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zptdv.png&quot; alt=&quot;lin&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/siLRc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/siLRc.png&quot; alt=&quot;log&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Another reason, as zibadawa timmy writes in his/her answer, is that taking the log you don't need to write all those $10^x$ factors).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now why add 12? This factor corresponds to measuring the number of a given atom per $10^{12}$ hydrogen atoms. I've been asking colleagues from students to professor level, and the best answer we can come up with, is that the least abundant elements in the Sun (uranium, rhenium, thorium, …) have abundances of the order of one per $10^{12}$ hydrogen atoms, so adding 12 to $\log(X/\mathrm{H})$ makes all values positive. This angst of negative values, I don't understand, though. Usually, we're okay with minuses. And measuring metallicities in stars or gas of sub-solar metallicity yield negative values anyway. Moreover, the scale was in use even when we had only lower thresholds for uranium and bismuth of $\log(\mathrm{U}/\mathrm{H})+12&amp;lt;0.8$ and $\log(\mathrm{Bi}/\mathrm{H})+12&amp;lt;0.6$, respectively (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1969SoPh....6..381G&quot; rel=&quot;nofollow noreferrer&quot;&gt;Grevesse 1969&lt;/a&gt;) and didn't know if they were actually smaller.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-09-21T07:20:50.450" LastActivityDate="2015-09-21T07:20:50.450" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11852" PostTypeId="2" ParentId="11849" CreationDate="2015-09-18T17:44:52.297" Score="8" Body="&lt;p&gt;The plane that contains the orbit of the Earth is known as the &quot;ecliptic&quot;. The rotation of the sun is tilted by 7.25 degrees to the ecliptic, and this value does not vary over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rotation of the Earth is also tilted, by 23.45 degrees to the ecliptic, it is this angle that causes seasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An old paper &lt;a href=&quot;http://adsbit.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1921PA.....29..325H&amp;amp;db_key=AST&amp;amp;page_ind=0&amp;amp;data_type=GIF&amp;amp;type=SCREEN_VIEW&amp;amp;classic=YES&quot; rel=&quot;nofollow noreferrer&quot;&gt;Position of the Sun's axis&lt;/a&gt; describes how the Earth crosses the solar equator on June 4th and Dec. 6th each year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nasa has a &lt;a href=&quot;http://nssdc.gsfc.nasa.gov/planetary/factsheet/sunfact.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;factsheet&lt;/a&gt;, that has this and other information about our nearest star.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-04-16T11:24:43.937" LastActivityDate="2018-04-16T11:24:43.937" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11853" PostTypeId="1" CreationDate="2015-09-18T19:43:31.873" Score="12" ViewCount="1158" Body="&lt;p&gt;Is there any argument against this situation: two black holes, one within the event horizon of the other, and the system is stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is interesting to me because that if this works, we could have the system's event horizon not in the shape of a perfect sphere. (The event horizon would probably be moving with the orbit)&lt;/p&gt;&#xA;" OwnerDisplayName="Köksal" LastActivityDate="2015-09-19T10:24:36.167" Title="two black holes orbiting each other within their event horizon" Tags="&lt;black-hole&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11854" PostTypeId="2" ParentId="11853" CreationDate="2015-09-18T20:23:29.780" Score="0" Body="&lt;p&gt;It is an unlikely scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The event horizon is the point where orbital velocity exceeds C (the speed of light).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matter is incapable of equalling or exceeding C within its frame. Therefore, a stable orbit is implausible, since the second black hole would need to be moving faster than C. Thus, the orbit is inherently unstable, and decaying. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We should be able to detect a velocity shift in the orbits of objects outside the event horizon in the accretion disk, and the event horizon should also have a bulge, but depending upon the current spot in orbit, that may become undetectably small as the smaller black hole spirals in closer and closer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I'm not up on all the details of frame dragging and other black hole related weirdnesses, so this is only a first order examination of the principles, hence implausible rather than impossible.&lt;/p&gt;&#xA;" OwnerUserId="9242" LastActivityDate="2015-09-18T20:23:29.780" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11855" PostTypeId="2" ParentId="11853" CreationDate="2015-09-18T20:44:49.607" Score="10" Body="&lt;p&gt;Inside the event horizon is a weird place to be. Inside a black-hole space-time is flowing towards the singularity faster than the speed of light (as measured from outside the black-hole. The effect is that any thing inside a black hole will, in a finite (and usually short) amount of time end up at the singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Models of merging black holes exist such as &lt;a href=&quot;https://www.youtube.com/watch?v=7y9AtdN4BlM&quot; rel=&quot;noreferrer&quot;&gt;video from nasa&lt;/a&gt;. It requires super-computers to solve Einstein's equations numerically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I noted in a comment, no stable orbit can exist within 1.5 times the radius of the event horizon. The orbital velocity is c at that distance (1.5 Schwarzschild); it's also called the &quot;photon sphere&quot;. Within that radius all orbits are unstable. And the radiation of gravitation waves means that as two black holes orbit each other, they will lose energy, and so their orbits will decay. As the black holes approach, the event horizons are distorted, and merge into an &lt;a href=&quot;https://www.youtube.com/watch?v=Tr1zDVbSjTM&quot; rel=&quot;noreferrer&quot;&gt;droplet shape&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="-1" LastEditDate="2015-09-19T10:24:36.167" LastActivityDate="2015-09-19T10:24:36.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11856" PostTypeId="1" CreationDate="2015-09-18T20:45:05.310" Score="5" ViewCount="252" Body="&lt;p&gt;I am trying to match up a detail of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Astronomical_ceiling_of_Senemut_Tomb&quot; rel=&quot;nofollow noreferrer&quot;&gt;Senemut ceiling&lt;/a&gt; to the time of the year it might involve. In the detail (shown below), the belt of Orion is slanted upwards from right to left:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/M2Y4z.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/M2Y4z.png&quot; alt=&quot;senmut&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Orion_Nebula&quot; rel=&quot;nofollow noreferrer&quot;&gt;Orion Nebula&lt;/a&gt; is to the right. When I try to duplicate this in Stellarium, I can only get the belt to slant left to right, going up as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KKCVp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KKCVp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some particular time of the year when the belt of Orion is visible from Egypt (Banî Suwayf) and slants upward going right to left, as in the Egyptian ceiling?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One answer below suggested that Orion's belt is like this when it rises, but I cannot duplicate it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TPqv1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TPqv1.jpg&quot; alt=&quot;orion rising&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Stellarium, I found by experimentation that Orion seems to rise on the morning of July 18th at the latitude in question and date range in question. However, as can be seen in the screen shot the belt still goes left to right.&lt;/p&gt;&#xA;" OwnerUserId="2800" LastEditorUserId="2153" LastEditDate="2015-09-20T00:19:24.267" LastActivityDate="2015-09-20T18:08:48.147" Title="When is the belt of Orion slanted from right to left, going up? (Egypt)" Tags="&lt;constellations&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11857" PostTypeId="2" ParentId="11856" CreationDate="2015-09-18T21:24:33.283" Score="3" Body="&lt;p&gt;EDIT: entirely new answer based on comments&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may actually be on to something here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most obvious (and thus least interesting) answer is that they were &#xA;depicting Orion shortly after sunset, not shortly before sunrise (all &#xA;times UTC): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mdEKD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mdEKD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the helical rising of Orion today, and it does indeed go right &#xA;to left: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BUDfL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BUDfL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we go back 1000 years, it's much closer to being a straight line: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qdYn6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qdYn6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if we go back to 1473 BCE (which is -1472), the rising is clearly &#xA;left to right: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2KYn3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2KYn3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, if we go far enough south, the rising is right to left &#xA;even in -1472: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ItWZz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ItWZz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that position in which Orion rises doesn't change much from day &#xA;to day. In other words, the position of Orion's stars at helical &#xA;rising are the same as the position of Orion's stars at rising any &#xA;other day of the year. Because of precession, the position &lt;em&gt;does&lt;/em&gt; &#xA;change over long periods of time, but not day to day. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only problem with the &quot;most obvious&quot; answer above is that we know &#xA;the Egyptians were very interested in helical risings, so it would be &#xA;odd to see a depiction of a &quot;helical setting&quot;. &lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2015-09-20T18:08:48.147" LastActivityDate="2015-09-20T18:08:48.147" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11858" PostTypeId="2" ParentId="11853" CreationDate="2015-09-19T00:13:40.577" Score="1" Body="&lt;p&gt;The event horizon is a &lt;em&gt;global&lt;/em&gt; property of the spacetime.  There is only &lt;em&gt;one&lt;/em&gt; event horizon in this case.  For there to be &quot;two&quot; black holes, we usually mean that their &lt;em&gt;apparent horizons&lt;/em&gt; are separate at some time, and the ultimately merge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This technicality aside, there's a ton of research in this are because black hole mergers are a likely source of gravitational radiation, and there are multiple experiments around the world attempting to directly detect such radiation, including LIGO in the US.  There are also several space-based experiments proposed but not currently funded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of the theoretical work in this are uses large numerical simulations on supercomputers to compute important quantities, including event and apparent horizons.  Due to the emission of gravitational radiation, the black holes that you describe will slowly in-spiral and will eventual merge completely. (The gravitational radiation carries away energy, so their orbital radius will shrink.) Eventually they will settle (asymptotically) into either a Kerr or a Schwarzschild solution.&lt;/p&gt;&#xA;" OwnerUserId="8019" LastActivityDate="2015-09-19T00:13:40.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11859" PostTypeId="2" ParentId="11853" CreationDate="2015-09-19T00:16:46.513" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;two black holes, one within the event horizon of the other&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The premise is wrong. There can be no such thing. What you and me, external observers, understand to be a &quot;black hole&quot; is actually the whole volume inside the event horizon. It is a chunk of spacetime causally disconnected from where we are now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When two black holes get close enough together, the event horizons bulge towards each other. When they touch and become a single event horizon, the merger process has begun in force (sometimes this is called &quot;collision&quot;). The only possible outcome at that point is that the two BHs will merge and become a single, bigger black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See this video for an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=p647WrQd684&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=p647WrQd684&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(note: the two red spheres at the end of the simulation have no physical reality, please ignore them)&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-09-19T00:21:47.373" LastActivityDate="2015-09-19T00:21:47.373" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11860" PostTypeId="1" AcceptedAnswerId="11861" CreationDate="2015-09-19T02:36:36.627" Score="3" ViewCount="225" Body="&lt;p&gt;I know that models of ocean planets typically have deep liquid water that transitions into exotic forms of ice. What I am wondering is if it makes sense to have an &quot;ocean floor&quot; composed of debris from bombarding asteroids during the early formation stages?&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2015-09-22T02:59:34.273" Title="Ocean floors on ocean planets?" Tags="&lt;exoplanet&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11861" PostTypeId="2" ParentId="11860" CreationDate="2015-09-19T10:24:09.450" Score="3" Body="&lt;p&gt;Super-pressurized ice probably wouldn't be static.  It would (almost certainly) move around and shift with the seasons and planetary climate changes and all that good stuff, so I'm extremely doubtful that anything would stay &quot;littered&quot; on the floor in this scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other factors would be whether the planet had life, which can drop shells or scales and other stuff, which falls to the ocean floor, and volcanic activity which likely happens on most planets and would eject some matter to cover whatever was on the surface over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plate tectonics, obviously another factor though I'm not sure how plate tectonics would affect a thick layer of ice.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://news.nationalgeographic.com/news/2007/05/070517-hot-planet_2.html&quot; rel=&quot;nofollow&quot;&gt;This planet&lt;/a&gt; is too hot for life (probably) and likely too large for plate tectonics, but I'm sure the hot ice still shifts and moves around and the interior heat has to escape from time to time, either cracks in the surface or volcanoes.   I think you'd need a nearly dead surface like the moon or mars for things to &quot;collect&quot; like that.   I don't think an exotic ice planet would cooperate in that way.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;update, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;what I am trying to do is come up with a good scenario where life &quot;as we know it&quot; could arise and flourish on an ocean planet. To me that meant needing some sort of ocean floor that wasn't just some exotic form of ice. I've seen standard models of ocean planet interiors, but hadn't really considered that there will be a wide variety (ie. the interior of Ganymede is probably quite different than Europa). So, in the right conditions I suppose volcanism and plate tectonics may be the answer (super-earths may be more tectonically active than Earth)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lots to cover here, I'll give it a shot briefly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Life may have arisen on earth near the ocean floor but not on the ocean floor.  it was the hot vents of gas that &lt;a href=&quot;https://en.wikipedia.org/wiki/Extremophile&quot; rel=&quot;nofollow&quot;&gt;Extremophiles&lt;/a&gt; lived in.   I don't think they require the ocean surface, just the volcanic jets with methane and other gasses that they could use.  (I think they use Methane, not 100% sure).   I think they require liquid water, but not ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the deep below the surface of planets tends to be hot, exotic ice would usually be quite hot, I would think, just solid under very high pressure, I suspect that only Extremophiles could live near exotic ice in those high temperatures.   High pressure is less of an obstacle but high temperature might limit the options for life to the very primitive/tiny (I think).    The good news is, nearer the surface, you might get lower temperatures and that might open the door to more advanced life.   I think what's on the surface is generally more interesting than what's on the floor.    Ganymede and Europa both have frozen surfaces with occasional breaches from tidal forces.   A larger planet might have exotic ice deep down but more earth like oceans towards the surface.   I think that would be a lot more interesting to us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how many varieties of water worlds there are.   Neptune is a water world of sorts though it's usually categorized as an ice giant and it probably doesn't have what we would consider oceans but it has a lot of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On, Plate Tectonics, it might not be essential to life on water worlds.   Volcanism can exist without plate tectonics and Volcanism is probably essential to life on water worlds.   Don't get me wrong.  Nothing beats plate tectonics for circulation and recycling of materials and preventing unwanted buildup of waste, so plate tectonics is really cool and better than just volcanism, but I think active Volcanism would be sufficient for life in oceans.   It's really hard to say anything definite about this subject though, as it's based on theory.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's probably possible to have volcanic eruptions even though dozens or maybe even hundreds of miles of exotic ice because the heat has to go somewhere, eventually, assing it's likely to build up over time, so either by circulation of eruption, the heat has push through at some point.  This even happens on so called &quot;dead&quot; planets like Mars or even the Moon.   Mars still has the occasional volcanic eruption, just not very often.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But water worlds certainly can have plate tectonics.  There's nothing in the water that would prevent it from happening.  Plate Tectonics is, as I understand it, primarily a factor of the size of the planet.   Gas planets - different story, but planets with a hard surface, Earth sized, a tiny bit smaller to a fair bit but not much bigger are good candidates for plate tectonics (I think).  There's some debate on how large, I think, still going on.   But I remember reading that ocean/water worlds might even be more likely to have plate tectonics.   Plate tectonics is definitely something we'd look for if we ever get a close enough look at other planets in different solar-systems (exoplanets).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just my thoughts on this.  Not meant to be complete or definitive.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-09-22T02:59:34.273" LastActivityDate="2015-09-22T02:59:34.273" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11862" PostTypeId="2" ParentId="6555" CreationDate="2015-09-19T19:37:25.240" Score="1" Body="&lt;p&gt;The precession of the perihelion of the planets depends mostly on the effects of the other planets perturbations, but also to a lesser extent on relativistic effects, the aspherical shape of the sun and solar tides. See also &lt;a href=&quot;https://en.wikipedia.org/wiki/Apsidal_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia's Apsidal precession&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node115.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This calculation of the effects of pertubations&lt;/a&gt; includes a comparison of the actual and observed rates of precession of perihelion. It gives rates in arcseconds per year:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Mercury  5.75&#xA;Venus    2.04&#xA;Earth   11.45&#xA;Mars    16.28&#xA;Jupiter  6.55&#xA;Saturn  19.50&#xA;Uranus   3.34&#xA;Neptune  0.36&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For Mercury the observed rate of precession is rather more than would be predicted from perturbations. The reason for this is general relativity, and was one of the first pieces of observational evidence for relativity.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="6593" LastEditDate="2019-08-12T16:21:22.063" LastActivityDate="2019-08-12T16:21:22.063" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11863" PostTypeId="2" ParentId="6180" CreationDate="2015-09-19T20:01:29.860" Score="3" Body="&lt;p&gt;This was clouds.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Eclipses occur at full moon, not last quarter.&lt;/li&gt;&#xA;&lt;li&gt;Eclipses take several hours as the moon traverses the Earth's shadow, not a few minutes.&lt;/li&gt;&#xA;&lt;li&gt;Eclipses don't look like the moon becoming a gleaming point of light. In a lunar eclipse the moon becomes a copper colour, as it is illuminated by red light that has passed round the Earth, through our atmosphere.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;On the other hand, clouds can pass in front of the moon at any time, and when low in the sky would take a few minutes to pass in front of the moon (depending on wind), and could entirely cover the moon, giving a the impression of an eclipse, if you've not seen one before. But lunar eclipses look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0fMMq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0fMMq.jpg&quot; alt=&quot;Lunar eclipse look like this.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photo Credit: Jimmy Westlake &lt;a href=&quot;http://www.space.com/15689-lunar-eclipses.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;space.com&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-09-19T20:01:29.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11864" PostTypeId="2" ParentId="6084" CreationDate="2015-09-19T23:08:14.070" Score="1" Body="&lt;p&gt;You simply cannot obtain &lt;em&gt;any&lt;/em&gt; information on the half-light radius from the quadrupole moment only of a galaxy, without knowing the galaxy profile.  Consider for example two (extreme) light profiles:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A galaxy consisting of a uniform disk of radius $R$ – that is, its surface brightness would be something like $I(r) \propto H(R_1 - r)$, where $H$ is the Heaviside function.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A galaxy consisting of a point source + a uniform disk – that is $I(r) \propto \delta(r) + H(R_2 - r)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can easily find a combination of radii $R_1$ and $R_2$ such that the quadrupole moments of the two galaxies are identical.  However, clearly their half-light radii cannot be (for the second galaxy, in particular, the half-light radius vanishes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I can see, the relation you wrote holds for a Gaussian profile, and even then it is not correct: your $r$ is not the half-light radius but the standard deviation of the Gaussian profile.  For different profiles there is no guarantee that your relation will work (in general it will not): in general, it will only give you something &lt;em&gt;proportional&lt;/em&gt; to the square of the half-light radius, but the proportionality constant will depend on the specific light profile.&lt;/p&gt;&#xA;" OwnerUserId="9249" LastActivityDate="2015-09-19T23:08:14.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11865" PostTypeId="1" CreationDate="2015-09-19T23:57:00.283" Score="1" ViewCount="51" Body="&lt;p&gt;Could this pulse potentially shield spacecraft from harmful solar radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was assuming an EMP would have little effect as a radiation shield because the pulse would not be rapid enough to counter act the suns energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f we pumped an EMP into high earth orbit and it detinated would it create a pocket of radiation-free space&lt;/p&gt;&#xA;" OwnerUserId="4175" LastActivityDate="2015-09-20T13:35:45.220" Title="If we pumped an large EMP into High Earth orbit and it detinated would it act as a shield against solar radiation?" Tags="&lt;radiation&gt;&lt;the-sun&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11866" PostTypeId="2" ParentId="2362" CreationDate="2015-09-20T02:41:52.030" Score="3" Body="&lt;p&gt;If it works for Arecibo it could be made to work on the Moon, Ceres, and maybe even Mars. The point about sending a prefab is a good one. And using two radio telescopes in tandem separated in space would allow for astronomers to take advantage of the improved angular resolution this affords provided they were arranged so as to allow them to simultaneously focus on the same point. That would be a neat problem in astrophysics!  Radio telescopes aimed at the same point simultaneously from the Earth and Mars while they are on opposite sides of their orbits would have incredible resolution.  &lt;/p&gt;&#xA;" OwnerUserId="9250" LastEditorUserId="7472" LastEditDate="2018-12-17T10:39:06.723" LastActivityDate="2018-12-17T10:39:06.723" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11867" PostTypeId="1" AcceptedAnswerId="11873" CreationDate="2015-09-20T03:54:02.660" Score="7" ViewCount="5706" Body="&lt;p&gt;I've read in multiple sources that the Moon appears the other way up in Northern Hemisphere vs Southern Hemisphere. Here is one that explains why: &lt;a href=&quot;http://curious.astro.cornell.edu/our-solar-system/46-our-solar-system/the-moon/observing-the-moon/135-does-the-moon-look-different-in-the-northern-and-southern-hemispheres-beginner&quot;&gt;Does the Moon look different in the northern and southern hemispheres?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, where is the line drawn in this case where the moon 'flips'? Is it exactly the equator? I'm not sure if it is, the simple 'thought experiment' in the above link assumes (for simplicity's sake) that the orbit of the Moon is exactly above the equator. According to Wikipedia, &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbit_of_the_Moon&quot;&gt;the orbit of the Moon&lt;/a&gt; is slightly tilted, and our own earth also tilts from time to time. Where does the moon flip exactly? Does it change throughout the year as our earth change its tilt? If so, how? Which are the areas that always see the northern look or the southern one throughout the year?&lt;/p&gt;&#xA;" OwnerUserId="9251" LastEditorUserId="9251" LastEditDate="2015-09-20T04:32:22.927" LastActivityDate="2017-05-08T20:07:20.060" Title="Where exactly does the Moon flip, given that it appears the other way up in the other hemisphere?" Tags="&lt;the-moon&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11868" PostTypeId="2" ParentId="11867" CreationDate="2015-09-20T07:14:55.523" Score="9" Body="&lt;p&gt;The moon doesn't flip. Is us who flip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those in the antipodes have their heads pointing in the opposite direction. So the moon appears inverted. But there is no sudden point where it &quot;flips&quot;. We don't live on a discworld.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see this yourself. Because as the Earth spins you will change the direction in which your head is pointing. Look at the full moon over the course of one night. As it appears to travel from East to West in the sky, it's orientation will slowly change. There is no mystery here. You are being rotated by the Earth. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-09-20T07:14:55.523" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11869" PostTypeId="2" ParentId="11860" CreationDate="2015-09-20T13:19:43.327" Score="3" Body="&lt;p&gt;You get an ocean planet because in the early phase of formation it was bombarded with ice. But everything is relative including the composition of asteroids and planets which will accrete more or less based on the composition of the primordial disk from which they are formed. Some ocean planets will have more ocean and less to no rock while others will be mostly rock with a few miles of ocean overlaying the hard parts like Earth. Your proposed scenario is a good one except keep in mind that what we are dealing with with that much water is a whole bunch of a substance famous for being the &quot;universal solvent&quot;!  So whatever solid material crashes into the ocean of that planet that isn't vaporized by the impact would still probably end up in solution or suspension. Have you ever microwaved a cup of water and then put in a spoonful of sugar?  &lt;/p&gt;&#xA;" OwnerUserId="9250" LastActivityDate="2015-09-20T13:19:43.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11870" PostTypeId="2" ParentId="11865" CreationDate="2015-09-20T13:35:45.220" Score="1" Body="&lt;p&gt;I can think of 2 ways to create an EMP and while both involve explosives in both cases the EMP is an effect of the blast and not a detonation per se. Would it create a radiation free &quot;bubble &quot;? Yes but only briefly and hardly worth the trouble. In fact the first way an EMP is created is as a side effect of a nuclear explosion. Hardly worth the trouble then if the goal is to clear out a patch of space to then fill it full of fallout. Rapidly collapsing a strong magnetic field essentially by crushing the magnet in an explosion may also produce an EMP although less strong and your space cleared of radiation would be but for a fleeting moment and only potentially cleared of radiation susceptible to electromagnetic influence. If you think of the charged particles that you are trying to deflect as bullets and the electromagnetic lines of force as walls then your walls have to be strong enough to withstand the bullets or the particles  will blithely rip through the EMP like it wasn't even there. &lt;/p&gt;&#xA;" OwnerUserId="9250" LastActivityDate="2015-09-20T13:35:45.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11871" PostTypeId="2" ParentId="11199" CreationDate="2015-09-20T16:00:27.613" Score="2" Body="&lt;h3&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/NML_Cygni&quot; rel=&quot;nofollow noreferrer&quot;&gt;NML Cygni&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/VY_Canis_Majoris&quot; rel=&quot;nofollow noreferrer&quot;&gt;VY Canis Majoris&lt;/a&gt;&lt;/h3&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1207.1850&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zhang et al. (2012)&lt;/a&gt; considered several values for &lt;span class=&quot;math-container&quot;&gt;$T_{\text{eff}}$&lt;/span&gt; (effective temperature) and &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; (luminosity) and found that NML Cygni matches up with the evolutionary track of a star of mass &lt;span class=&quot;math-container&quot;&gt;$\sim25M_\odot$&lt;/span&gt;, near the HR-diagram location of a similar star, VY Canis Majoris (denoted as VY CMa). Using these data points, the model estimates that NML Cygni is approximately 8 million years old and in the post-main sequence phase of its life.&lt;/p&gt;&#xA;&lt;p&gt;However, NML Cygni is thought to be related to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cygnus_OB2&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cygnus OB2 association&lt;/a&gt;, which has an age of 2-3 million years. This is strange because all the stars in the association should be nearly the same age  (see &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0007442&quot; rel=&quot;nofollow noreferrer&quot;&gt;Knodlseder (2008)&lt;/a&gt;). That age may be questionable, however, as evidence for older stars nearby has been presented (see &lt;a href=&quot;http://arxiv.org/abs/1003.2463&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wright et al. (2010)&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;p&gt;The exact location of VY Canis Majoris on the evolutionary tracks is not certain. &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0604253&quot; rel=&quot;nofollow noreferrer&quot;&gt;Massey et al. (2006)&lt;/a&gt; created models that placed it on a lower-mass track to help it avoid the &amp;quot;forbidden zone&amp;quot; beyond the Hayashi limit.&lt;/p&gt;&#xA;&lt;h3&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/VV_Cephei&quot; rel=&quot;nofollow noreferrer&quot;&gt;VV Cephei&lt;/a&gt;&lt;/h3&gt;&#xA;&lt;p&gt;Based on standard parameters, stellar wind measurements and &lt;a href=&quot;http://obswww.unige.ch/Recherche/evol/Geneva-grids-of-stellar-evolution&quot; rel=&quot;nofollow noreferrer&quot;&gt;Geneva grids&lt;/a&gt;, &lt;a href=&quot;http://arxiv.org/abs/1004.1853&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bennett (2010)&lt;/a&gt; suggested that the red supergiant component of VV Cephei could be near the end of its life, assuming a mass of &lt;span class=&quot;math-container&quot;&gt;$\sim20$&lt;/span&gt;-&lt;span class=&quot;math-container&quot;&gt;$25M_\odot$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h3&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/UY_Scuti&quot; rel=&quot;nofollow noreferrer&quot;&gt;UY Scuti&lt;/a&gt;&lt;/h3&gt;&#xA;&lt;p&gt;Wikipedia claims&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;UY Scuti should fuse lithium, carbon, oxygen, neon, and silicon in its core within the next million years. After this, its core will begin to produce iron, disrupting the balance of gravity and radiation in its core and resulting in a core collapse supernova.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;As far as I know, that claim is made only through rough estimates of evolutionary tracks, not precise calculations. We do know that UY Scuti is certainly in the later period of its hydrogen-burning phase, if it is indeed at that point in its life. &lt;a href=&quot;https://arxiv.org/abs/1305.6179&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arroyo-Torres et al. (2013)&lt;/a&gt; believe observations place it near evolutionary tracks of stars originating with masses of &lt;span class=&quot;math-container&quot;&gt;$\sim25$&lt;/span&gt;-&lt;span class=&quot;math-container&quot;&gt;$40M_{\odot}$&lt;/span&gt;. Therefore, it may have already lost a substantial fraction of its original mass - not surprising for a red hypergiant.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-12-29T18:39:01.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11872" PostTypeId="2" ParentId="11141" CreationDate="2015-09-20T16:42:04.457" Score="7" Body="&lt;p&gt;EDIT: &lt;a href=&quot;http://wgc.jpl.nasa.gov:8080/webgeocalc/#AngularSeparationFinder&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://wgc.jpl.nasa.gov:8080/webgeocalc/#AngularSeparationFinder&lt;/a&gt; lets you find planetary conjunctions online using NASA's data. It's still iterative, but fairly fast (since NASA uses fairly powerful servers, even for their websites).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summary: I'm still researching, but there appears to be no well-known,&#xA;reliable non-iterative method to find conjunctions. Using the&#xA;iterative method and the C SPICE libraries, I created a table of&#xA;conjunctions for the visible planets (Mercury, Venus, Mars, Jupiter,&#xA;Saturn, Uranus) here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://search.astro.barrycarter.info/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://search.astro.barrycarter.info/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Full &quot;answer&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am still researching a general answer to this question (&quot;How to&#xA;calculate conjunctions of 2 planets&quot;), but here's what I have so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The iterative method:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Compute the positions of the planets at regular intervals (eg,&#xA;daily). The &quot;daily&quot; works for planets (but not some asteroids and&#xA;definitely not the Moon) because the planets move through the sky&#xA;relatively slowly.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Find local minima in the daily lists.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For efficiency, carefully discard local minima that are too&#xA;large. For example, Mercury and Venus may approach each other, reach&#xA;a minimal distance of 20 degrees, and then drift apart. The 20&#xA;degrees is a local minima, but not a conjunction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;However, be careful when discarding minima. If you are searching&#xA;for 5-degree conjunctions, two planets may be 5.1 degrees apart one&#xA;day, and 5.2 degrees apart the next day, but less than 5 degrees&#xA;apart sometime in the interim.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For 5-degree conjunctions, you only need daily minima less than 8&#xA;degrees, and even &lt;em&gt;that&lt;/em&gt; is overkill. The fastest a planet can move in&#xA;the sky is 1.32 degrees per day (Mercury), and the second fastest is&#xA;1.19 degrees per day (Venus). In theory, these movements could be in&#xA;opposite directions, so the fastest two planets can separate is 2.51&#xA;degrees per day. So, if two planets are more than 8 degrees apart&#xA;two days in a row, there is no way they could be closer than 5&#xA;degrees between the days.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In reality, planets maximum retrograde angular speed is slower&#xA;than the prograde maximum speed, so the 2.51 degree limit above is&#xA;never actually reached.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;After finding local minima, use minimization techniques (eg, the&#xA;ternary method) to find the instant of closest approach.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I ended up using the C SPICE libraries, and found 32,962&#xA;six-degrees-or-less conjunctions between -13201 and 17190, averaging&#xA;about 1 conjunction per year. Of these, 2,185 occur between the &quot;star&#xA;of Bethlehem&quot; and the 2015 conjunctions:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://12d4dc067e0d836f1541e50125c24a03.astro.db.mysql.94y.info/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://12d4dc067e0d836f1541e50125c24a03.astro.db.mysql.94y.info/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This iterative process works, but can be tedious. Since planetary&#xA;positions are semi-well-behaved, you'd think there would be a faster,&#xA;non-iterative method. Well...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;As &lt;a href=&quot;https://math.stackexchange.com/questions/95250/&quot;&gt;https://math.stackexchange.com/questions/95250/&lt;/a&gt; notes, people&#xA;have been interested in finding planetary conjunctions for at least&#xA;2000 years, probably longer. The first such known tool is the&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Antikythera_mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Antikythera_mechanism&lt;/a&gt;. The &quot;Operation&quot;&#xA;section of the wikipedia page suggests this device was hand-cranked,&#xA;and thus iterative.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;SPICE&#xA;(&lt;a href=&quot;https://en.wikipedia.org/wiki/Spacecraft_Planet_Instrument_C-matrix_Events&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Spacecraft_Planet_Instrument_C-matrix_Events&lt;/a&gt;)&#xA;is one of the oldest digital computer attempts to compute planetary&#xA;positions, and it includes a routine to compute conjunctions:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://emfisis.physics.uiowa.edu/Software/C/cspice/doc/html/cspice/gfsep_c.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://emfisis.physics.uiowa.edu/Software/C/cspice/doc/html/cspice/gfsep_c.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this also uses an iterative method, as the long description&#xA;of the &quot;step&quot; parameter indicates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;step must be short enough for a search using step to locate the time&#xA;intervals where the specified angular separation function is monotone&#xA;increasing or decreasing. However, step must not be &lt;em&gt;too&lt;/em&gt; short, or&#xA;the search will take an unreasonable amount of time&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By experimentation, I found that a step size of 6 days does not find&#xA;all Mercury/Venus minimal separations, although a step size of 1 day&#xA;does find these. In other words, reducing the step size from 6 days to&#xA;1 day found additional conjunctions, but reducing the step size to&#xA;well below 1 day did not produce any additional conjunctions.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Akkana Peck (akkana@shallowsky.com), creator of a well known script to find&#xA;conjunctions (&lt;a href=&quot;https://github.com/akkana/scripts/blob/master/conjunctions.py&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/akkana/scripts/blob/master/conjunctions.py&lt;/a&gt;)&#xA;tells me:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;M[y script] iterates through dates.&lt;p&gt;&#xA;&#xA;[...]&lt;p&gt;&#xA;&#xA;In Astronomical Algorithms, Meeus has a chapter (17) on Planetary&#xA;Conjunctions, but the chapter is less than three pages and lacks&#xA;detail except that he's pretty clearly iterating and looking for&#xA;the smallest separation, same as my Python program does.&lt;p&gt;&#xA;&#xA;He does a little better in Astronomical Formulae for Calculators:&#xA;sounds like he's advocating iterating over big steps to find the&#xA;place where the separation stops decreasing and starts increasing,&#xA;then using &quot;inverse interpolation&quot; to find the exact time.&lt;p&gt;&#xA;&#xA;&quot;Mathematical Astronomy Morsels&quot; (also Meeus) talks starting on p.&#xA;246 about triple conjunctions (defined as two planets having several&#xA;conjunctions in a short time, due to retrograde motion, not three&#xA;planets all in conjunction at once) and gives some examples for very&#xA;specific pairs, like Jupiter and Saturn, with ways you could look for&#xA;those conjunctions.&lt;p&gt;&#xA;&#xA;It doesn't look like any of these books would be very helpful in&#xA;finding a non-iterative solution.&#xA;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I haven't had a chance to read the books above, but did find:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1981JRASC..75...94M&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/1981JRASC..75...94M&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where Meeus confirms the standard iterative method, but also provides&#xA;a different, less accurate method. Unfortunately, Meeus only uses this&#xA;method to compute solar conjunctions, elongations and oppositions, not&#xA;interplanet conjunctions.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Jon Giorgini of NASA (jdg@tycho.jpl.nasa.gov) tells me:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&#xA;As far as NASA computed planetary/stellar conjunctions/occultations, I&#xA;don't know of anyone within NASA that does that routinely. There is an&#xA;external network of volunteers that does that under the umbrella of&#xA;the International Occultation Timing Association (IOTA), and they have&#xA;developed pretty refined internal software for that purpose.&#xA;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I contacted IOTA (&lt;a href=&quot;http://occultations.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://occultations.org/&lt;/a&gt;) and they confirmed&#xA;they use an iterative method as well:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&#xA;[...] the software package Occult does generate planetary conjunction&#xA;predictions - based on a two-body solution. The approach used is a&#xA;crude brute-force method of generating the planetary ephemerides on a&#xA;daily basis.  It is not particularly efficient - but it is sufficient&#xA;for its intended purpose. [...]&#xA;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;As a note, IOTA focuses on asteroid occultations, so computing&#xA;positions daily doesn't always work. Especially for near-Earth&#xA;asteroids, IOTA must iterate considerably more frequently.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I also tried contacting Fred Espenak, the creator of&#xA;&lt;a href=&quot;http://eclipse.gsfc.nasa.gov/SKYCAL/SKYCAL.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://eclipse.gsfc.nasa.gov/SKYCAL/SKYCAL.html&lt;/a&gt;, but was unable to&#xA;do so. Jon Giorgini tells me that Fred has retired.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I'm still looking, but my current conclusion is that there is no&#xA;good well-known non-iterative way to find planetary conjunctions. As&#xA;the image in my&#xA;&lt;a href=&quot;https://mathematica.stackexchange.com/questions/92774&quot;&gt;https://mathematica.stackexchange.com/questions/92774&lt;/a&gt; shows,&#xA;planetary separations aren't really as well-behaved as I had hoped.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I just got a reply from Arnold Barmettler, who runs calsky.com:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;I'm using a very time consuming iterative approach to pre-calculate&#xA;Bessel Elements for conjunctions. This allows fast online calculation&#xA;for any place on earth.&#xA;Initial calculations are only done every few years, so CPU time does not&#xA;matter. This would change if I'd enter the business to calculate&#xA;asteroidal occultations.&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;re-iterating (pun intended) the same theme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Miscellaneous:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I used planetary system barycenters (the center of mass of a&#xA;planet and its moons) for an entirely different reason. If you ask&#xA;HORIZONS (&lt;a href=&quot;http://ssd.jpl.nasa.gov/?horizons&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ssd.jpl.nasa.gov/?horizons&lt;/a&gt;) for the position of&#xA;Mars and set the date, you'll see this notice:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Available time span for currently selected target body:&#xA;1900-Jan-04 to 2500-Jan-04 CT.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you use Mars' barycenter, this becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Available time span for currently selected target body:&#xA;BC 9998-Mar-20 to AD 9999-Dec-31 CT.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, NASA computes the position of Mars' planetary system&#xA;barycenter for a much longer interval than they compute Mars' actual&#xA;position. Since I wanted to compute conjunctions for a long period of&#xA;time, I went with the barycenters (DE431 computes barycenters even&#xA;beyond 9998 BC and 9999 AD).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've complained that this is silly, especially for Mars, since the&#xA;distance between Mars' center and Mars' planetary system barycenter is&#xA;only about 20cm (yes, centimeters, since Phobos and Deimos have very&#xA;little mass) per&#xA;&lt;a href=&quot;http://naif.jpl.nasa.gov/pub/naif/generic_kernels/spk/planets/aareadme_de430-de431.txt&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://naif.jpl.nasa.gov/pub/naif/generic_kernels/spk/planets/aareadme_de430-de431.txt&lt;/a&gt;. However,&#xA;NASA apparently plans to keep it this way.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I ignore light travel time, which introduces a small error. Most&#xA;planetarium programs, like Stellarium, ignore light travel time by&#xA;default, although Stellarium has an option to turn it on.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I also ignore refraction, aberration, and similar minor effects.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;More details on how I generated these tables (in highly fragmented form):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/barrycarter/bcapps/tree/master/ASTRO&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/tree/master/ASTRO&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(README.conjuncts in the above is a good starting point)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the &quot;cooler&quot; conjunctions I found are at:&#xA;&lt;a href=&quot;http://search.astro.barrycarter.info/table.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://search.astro.barrycarter.info/table.html&lt;/a&gt;. Here's a screenshot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WnzUo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WnzUo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2019-10-07T23:56:47.523" LastActivityDate="2019-10-07T23:56:47.523" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11873" PostTypeId="2" ParentId="11867" CreationDate="2015-09-20T18:48:52.600" Score="7" Body="&lt;p&gt;This doesn't really answer your question, but the orientation of the moon is changing constantly, because we tend to compare the moon to the horizon, and think of the horizon as fixed. Here's a video demonstrating the effect (simulated) for the full moon over London on August 28-29 (all times GMT):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=QXVpRfaJXmY&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=QXVpRfaJXmY&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were changing location, you would see a similar gradual effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only real place you would see a true &quot;flip&quot; is near where the moon is at zenith (overhead), because, as you passed the zenith point, the moon may go from being 89.999 degrees high in the western sky to 89.999 degrees high in the eastern sky (and flip directions because you're now comparing it to a different horizon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Realistically, when the moon is that high, you probably wouldn't be comparing it to the nearest horizon.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2015-09-20T18:48:52.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11874" PostTypeId="1" CreationDate="2015-09-21T05:08:05.637" Score="4" ViewCount="2504" Body="&lt;p&gt;Okay so scientists use the red shift to know that galaxies are moving away from us faster than light. How do we know that its not just red light from red stars. i mean most stars are in the infrared spectrum.&lt;/p&gt;&#xA;" OwnerUserId="9259" LastActivityDate="2019-07-08T20:31:46.390" Title="How Do we know about redshift?" Tags="&lt;redshift&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11875" PostTypeId="2" ParentId="11874" CreationDate="2015-09-21T06:38:57.050" Score="7" Body="&lt;p&gt;Stars consist of similar atoms and molecules that we can find here on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radiation from those atoms and molecules often occurs at discrete, identifiable wavelengths that form a unique pattern. This is a consequence of quantum physics - the energies of atomic and molecular states are quantised, and radiation arises from transitions between these. These wavelengths can be measured (in a laboratory) so that we know the &lt;em&gt;rest wavelengths&lt;/em&gt; of light emitted by atoms and molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we look at the spectrum of light from a star (or a Galaxy made up of stars) we identify atoms and molecules from the pattern of absorption (and sometimes emission) lines and can work out whether the &lt;em&gt;whole pattern&lt;/em&gt; has been redshifted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Occasionally, the difficulty you propose does occur. Galaxies which emit a high proportion of their light in the form of a smooth continuum (e.g. some types of active galaxies, where the light is dominated by the surroundings of the black hole at the centre) can exhibit a featureless spectrum or with perhaps a single identifiable emission line. In those circumstances it can be impossible to estimate a redshift.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-09-21T06:38:57.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11876" PostTypeId="1" AcceptedAnswerId="11877" CreationDate="2015-09-21T10:16:29.387" Score="3" ViewCount="613" Body="&lt;p&gt;I'm in Brisbane right now. I took this picture 15 minutes ago from my balcony. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CqayI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CqayI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you help me with identification of marked Lunar maria? I cannot compare any of these to &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/6/61/Moon_names.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think 1 is &lt;strong&gt;Mare Serenitatis&lt;/strong&gt; and 2 is &lt;strong&gt;Mare Tranquillitatis&lt;/strong&gt;. Am I right?&lt;/p&gt;&#xA;" OwnerUserId="9260" LastEditorUserId="9260" LastEditDate="2015-09-21T10:23:19.627" LastActivityDate="2015-09-21T11:42:31.447" Title="Identification of Lunar mare (maria) from a taken picture" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11877" PostTypeId="2" ParentId="11876" CreationDate="2015-09-21T11:42:31.447" Score="4" Body="&lt;p&gt;So your image looks more like this, with the red line being the terminator. The illuminated side in your image is the left side of the reference image, turned upside down.  I think you've gotten it backwards, the mare that you've labeled &quot;1&quot; is the Sea of Tranquility, and the one you've labeled &quot;2&quot; is the Sea of Serenity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iZWWC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iZWWC.png&quot; alt=&quot;Your image is upside down&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9262" LastActivityDate="2015-09-21T11:42:31.447" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11878" PostTypeId="1" CreationDate="2015-09-21T15:03:13.033" Score="3" ViewCount="106" Body="&lt;p&gt;Is there a place on Earth, where polar axis lies on the horizont and heaven equator is in zenit? If it exists, can I, in theory, see all the constelations available worldwide this night?&lt;/p&gt;&#xA;" OwnerUserId="9264" LastActivityDate="2016-07-21T04:14:30.753" Title="Universal point for sky observation" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11879" PostTypeId="1" CreationDate="2015-09-22T00:34:41.153" Score="5" ViewCount="676" Body="&lt;p&gt;Okay, I am seriously ashamed for asking this especialy when I geniualy study on physics but there is something that bugs me with the simulation I am working on.&lt;/p&gt;&#xA;&lt;p&gt;I am re-creating the solar system in an n-body simulation that i programmed before. And I've a problem with scaling down the solar data. So even if I use kg and km for the metric units, the values are far bigger than the variables can hold in programming. Also as some of you know, bigger the value is, bigger the floating point error it makes. (error noise in data) Also it makes it slower to process.&lt;/p&gt;&#xA;&lt;p&gt;I decided to scale down the data with a reference point, and for that, I took the earth's radius as 1 unit. And scaled down every other distance and radius according to it. (So a unit is 6371 km just to be clear)&lt;/p&gt;&#xA;&lt;p&gt;But I am not sure whether if I should scale down the mass or not. My common sense says that I should scale down the mass so density of each body should remain same. So I took the density, and calculated a new mass value for each body, with the new scaled down radius.&#xA;But I am somehow couldn't conviced myself about If it's true or not. So here I am, asking to you :) Should I also scale down the mass?&lt;/p&gt;&#xA;&lt;p&gt;PS.1: I used using &lt;span class=&quot;math-container&quot;&gt;$F = GMm/r^2$&lt;/span&gt; equation for the calculactions as usual. (Iterating it through each body pairs)&lt;/p&gt;&#xA;&lt;p&gt;If there are other programmers like me interested in making a simulation like this, how did you accomplish this data size problem? Are there any better solutions than scaling down the values?&lt;/p&gt;&#xA;&lt;p&gt;PS. I have created an excel file that does the scale conversion. So I am sharing  the sheet in OneDrive. (&lt;a href=&quot;http://1drv.ms/1NIekGo&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://1drv.ms/1NIekGo&lt;/a&gt;) If you can check my calculations and values, that I'd also be really helpful to me. Thanks for any help.&lt;/p&gt;&#xA;" OwnerUserId="9265" LastEditorUserId="37191" LastEditDate="2021-02-11T09:26:02.847" LastActivityDate="2021-02-11T09:26:02.847" Title="How to scale down solar system data to simulatable values" Tags="&lt;solar-system&gt;&lt;n-body-simulations&gt;&lt;units&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="11880" PostTypeId="1" AcceptedAnswerId="11881" CreationDate="2015-09-22T01:53:51.200" Score="17" ViewCount="2780" Body="&lt;p&gt;I understand that iron and all heavier elements consume more energy to produce than they make, and that is what eventually leads to a supernova. I also understand that a lot of the heavier elements are produced during that supernova. However, what I'm wondering is, before the star goes supernova, does any of the iron fuse with other elements? Yes, there would be a net energy loss, but if there is only small amount of iron in the star, it would probably be able to handle that.&lt;/p&gt;&#xA;" OwnerUserId="9266" LastActivityDate="2020-03-26T09:07:34.230" Title="Does any iron fuse in stars before they go supernova?" Tags="&lt;star&gt;&lt;supernova&gt;&lt;element&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11881" PostTypeId="2" ParentId="11880" CreationDate="2015-09-22T03:32:58.537" Score="10" Body="&lt;p&gt;Yes, but it's slow.   (I'm not an expert, so feel free to correct if I miss something important), but once the star is into the later stages, past the helium stage, up to iron, fusion mostly takes place by fusing a helium onto a heavier element, raising each atomic number by 2.  That's not the only method but it's the most common.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron can also fuse into nickel in this way inside a star and it does in small amounts, but mostly beyond iron, and certainly beyond nickel, heavier elements are created through the &lt;a href=&quot;https://en.wikipedia.org/wiki/S-process&quot; rel=&quot;nofollow noreferrer&quot;&gt;S-Process&lt;/a&gt;.   (short for slow neutron capture process).   This happens when a free neutron binds to the atomic nucleus and over time, the addition of neutrons can lead to beta decay, where an electron is ejected and a proton remains - adding to the atomic number.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;but if there is only small amount of iron in the star, it would&#xA;  probably be able to handle that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is undoubtedly true.  The stars that go super-nova are incredibly large and the iron doesn't exactly sink to the core right away.  It takes some time.    For a star to go kablooie (supernova), it needs an iron core of both enough purity where it's no longer undergoing expansion from nearby fusion, and enough size for it to undergo rapid collapse in a way that effects the star around it almost instantaneously.    I'm not clear on the exact process, but it requires way more than just a little iron.    As a layman's guess, it might require a Jupiter sized ball of iron.  Perhaps a fair bit more than that. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="9527" LastEditDate="2020-03-26T09:07:34.230" LastActivityDate="2020-03-26T09:07:34.230" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11882" PostTypeId="1" CreationDate="2015-09-22T11:24:41.843" Score="14" ViewCount="5236" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Tide&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article on tides&lt;/a&gt; shows a diagram with a tidal bulge both towards the Moon and on the antipodal point (the Sun isn't included in the diagram). Surely that's wrong - what would cause the antipodal bulge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/e/eb/Tide_overview.svg/871px-Tide_overview.svg.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/e/eb/Tide_overview.svg/871px-Tide_overview.svg.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;Image courtesy of Wikipedia user Jhbdel under the &lt;a href=&quot;http://creativecommons.org/licenses/by-sa/3.0/deed.en&quot; rel=&quot;noreferrer&quot;&gt;Creative Commons Attribution-Share Alike 3.0 Unported license&lt;/a&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="9270" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:47.413" LastActivityDate="2018-11-08T21:49:17.273" Title="What causes the antipodal bulge?" Tags="&lt;tidal-forces&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11883" PostTypeId="2" ParentId="11882" CreationDate="2015-09-22T12:12:43.680" Score="22" Body="&lt;p&gt;That is because the moon attracts both the water &lt;em&gt;and&lt;/em&gt; the earth. The gravity of the moon reduces with distance (by the inverse square law). So, the moon's gravity is &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Greatest at the point nearest to the moon (called sub-moon point)&lt;/li&gt;&#xA;&lt;li&gt;Is lesser on the earth's center, and'&lt;/li&gt;&#xA;&lt;li&gt;Is the least in the opposite side of the earth.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Assume that earth is falling to wards the moon. Now, as the force on the point near the moon is the maximum and it has 'fallen' the maximum distance, while the earth 'falls' a lesser distance and the farthest of all, the antipodal point 'falls the least'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CFfae.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CFfae.png&quot; alt=&quot;Tidal forces&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Tidal-forces.png#/media/File:Tidal-forces.png&quot; rel=&quot;noreferrer&quot;&gt;Tidal-forces&lt;/a&gt;&quot;. Licensed under &lt;a href=&quot;http://creativecommons.org/licenses/by-sa/3.0/&quot; rel=&quot;noreferrer&quot;&gt;CC BY-SA 3.0&lt;/a&gt; via &lt;a href=&quot;//commons.wikimedia.org/wiki/&quot; rel=&quot;noreferrer&quot;&gt;Wikimedia Commons&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above figure shows the forces acting on earth due to moon, The top figure shows the actual gravitational forces acting, while the bottom one shows the forces after subtracting the force on the earth itself. That should explain the antipodal bulge. See the wikipedia ariticle in &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_force&quot; rel=&quot;noreferrer&quot;&gt;Tidal forces&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7970" LastEditorUserId="7472" LastEditDate="2018-11-06T07:08:30.660" LastActivityDate="2018-11-06T07:08:30.660" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11884" PostTypeId="2" ParentId="11879" CreationDate="2015-09-22T13:26:33.583" Score="4" Body="&lt;p&gt;As &lt;a href=&quot;https://astronomy.stackexchange.com/users/21/barrycarter&quot;&gt;barrycarter&lt;/a&gt; mentioned in his comment, you should be more concerned with units and less with scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, it's best to stick to conventional units that people recognize. (It will keep your head on straight and make it easier for others to review your work.) In astronomy, these are a little different than the standard SI units, since things are—shall we say—&lt;em&gt;big&lt;/em&gt;, and numbers get out of hand quickly (as you have noted). Here are some units suggested in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_system_of_units&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomical system of units Wikipedia page&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Time&lt;/em&gt;: Day. This is probably not too important, so feel free to use seconds here. (After all, it's only a 5 order of magnitude difference.)&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Mass&lt;/em&gt;: There are several conventions, but the one I've personally seen most is solar masses. Since you're modeling our solar system, it makes a convenient reference.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Length&lt;/em&gt;: I would definitely use astronomical units. (What an appropriate name!)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you stick to these units, you should be able to limit error due to rounding and reduce the computational load of large numbers.&lt;/p&gt;&#xA;" OwnerUserId="9271" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-09-22T13:26:33.583" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11885" PostTypeId="2" ParentId="11874" CreationDate="2015-09-22T14:55:40.610" Score="16" Body="&lt;p&gt;Let's start with a quick clarification: &lt;em&gt;Red Shift&lt;/em&gt; is not the same as &lt;em&gt;red light&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Red light&lt;/strong&gt; is just electromagnetic radiation with a 400–484 THz frequency range, the lowest our eyes can see - highest being violet light, with a 668–789 THz frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Red Shift&lt;/strong&gt; is an observable effect when you analyze the spectroscopic signature of an electromagnetic radiation source that is so far from us that the expansion of the universe (whose unit of measurement is called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble Constant&lt;/a&gt;) is noticeable because of the frequency drop, called &lt;a href=&quot;https://en.wikipedia.org/wiki/Doppler_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Doppler effect.&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, it works like this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Pick one of the elements the universe is made of - hydrogen (75%), helium (24%) and trace amounts of everything else (1%). For this example, let's stick with hydrogen.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Determine the &lt;em&gt;spectrum signature&lt;/em&gt; of hydrogen by running the light generated via &lt;a href=&quot;https://en.wikipedia.org/wiki/Glow_discharge&quot; rel=&quot;nofollow noreferrer&quot;&gt;electric discharge on the gas&lt;/a&gt; through a spectrometer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f6ZMV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f6ZMV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;source: &lt;a href=&quot;http://images.flatworldknowledge.com/averillfwk/averillfwk-fig06_009.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;flatworldknowledge.com&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll notice some markings as you examine the light dispersed by the prism. Those markings are called &lt;a href=&quot;https://en.wikipedia.org/wiki/Narrowband&quot; rel=&quot;nofollow noreferrer&quot;&gt;narrow bands&lt;/a&gt;, and they appear as a consequence of &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/Bohr.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;quantized energy states&lt;/a&gt;. Hydrogen narrow bands are well known : &lt;strong&gt;656.2&lt;/strong&gt;, &lt;strong&gt;486.1&lt;/strong&gt;, &lt;strong&gt;434.0&lt;/strong&gt; and &lt;strong&gt;410.1&lt;/strong&gt; nanometers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now look at the sky. Pick any bright dot, run its light through a prism, and knowing that it is also made of roughly 75% hydrogen and 24% helium compare its spectroscopic signature to the bands you obtained earlier. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Because of the Doppler effect caused by the expansion of the universe the light waves are stretched, making the frequency drop - &lt;strong&gt;&lt;em&gt;shifting the whole spectroscopic signature towards lower frequencies&lt;/em&gt;&lt;/strong&gt;. The further the object is, more noticeable the shift towards red is.&lt;a href=&quot;https://i.stack.imgur.com/lFmgt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lFmgt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the term &lt;em&gt;Red Shift&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, answering your question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do we know that its not just red light from red stars?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because even regular light from red stars will shift towards the infrared side of the spectrum as a consequence of the &lt;em&gt;Doppler&lt;/em&gt; effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image Sources:&lt;br&gt;&#xA;&lt;a href=&quot;http://catalog.flatworldknowledge.com/bookhub/4309?e=averill_1.0-ch06_s03&quot; rel=&quot;nofollow noreferrer&quot;&gt;General Chemistry: Principles, Patterns, and Applications, v. 1.0&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://startswithabang.com/?p=1543&quot; rel=&quot;nofollow noreferrer&quot;&gt;Starts With A Bang!&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More information:&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Spectral_line#Spectral_lines_of_chemical_elements&quot; rel=&quot;nofollow noreferrer&quot;&gt;Spectral Lines of chemical elements&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="893" LastEditorUserId="893" LastEditDate="2019-07-08T20:31:46.390" LastActivityDate="2019-07-08T20:31:46.390" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11886" PostTypeId="1" AcceptedAnswerId="11892" CreationDate="2015-09-22T14:56:25.293" Score="3" ViewCount="128" Body="&lt;p&gt;Apparently, a few nights from now (more specifically on the night between the 27th and 28th of September this year, nights being Belgian nights), we will have a lunar eclipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's great, and I wish I could find a nice spot to go see that with friends!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sadly I never took a lot of time watching the moon with a compass in my hands, so I'm wondering:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we know at what orientation (as in North, South etc) will the moon show up from Brussels between 12PM and 4AM GMT in the night of 27th to 28th September?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the calculations and/or data source involved?&lt;/p&gt;&#xA;" OwnerUserId="9273" LastActivityDate="2015-09-23T17:15:31.877" Title="Can we know where (South, East, West...) the moon will be at a specific time?" Tags="&lt;the-moon&gt;&lt;lunar-eclipse&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11887" PostTypeId="2" ParentId="11468" CreationDate="2015-09-22T15:06:24.873" Score="12" Body="&lt;p&gt;&lt;em&gt;The short answer:&lt;/em&gt; Atmospheric gases never &lt;em&gt;left&lt;/em&gt; Earth, they're &lt;em&gt;in&lt;/em&gt; it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The long answer to this question isn't just about the planets' current states, but rather the processes that led them there. Let's start at the very beginning (a very good place to start).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The very early years&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When our solar system began forming 4.6 billion years ago, most of the mass from the collapsed portion of a molecular cloud (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Nebular_hypothesis&quot;&gt;Nebular hypothesis&lt;/a&gt;) collected in the center to form the Sun. The mass that didn't collapse into the Sun left a protoplanetary disk—a cloud of dust and gas—surrounding the new star. Gradually, dust particles began to come together through accretion, attracting more and more particles into fledgling planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Close to the Sun, where Venus and Earth both reside, it was too hot for many particles to condense, so the planets in this region formed with metals and silicates, which have high melting points. This is why the four planets in the inner Solar System are called the &quot;rocky&quot; or &quot;terrestrial&quot; planets. The earliest atmospheres on these planets began forming with the gradual collection of gases from the solar nebula, mostly hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Goldilocks and the two planets&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point in the evolution of the two planets, they looked pretty similar, but there is one major difference: the distance to the Sun. Earth, it seems, got lucky to be in the &quot;Goldilocks zone&quot;, where the temperature is &lt;em&gt;just right&lt;/em&gt; to support life. Being in this zone has two major implications: liquid water and, as a consequence, active plate tectonics. (See &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/524012/pdf&quot;&gt;this article&lt;/a&gt; for an in-depth view into why the two are related.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Carbon sinks&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Earth, there is considerable water contained in liquid oceans. On Venus, this is not the case. It is simply too hot that close to the Sun, so all the water evaporated into the atmosphere. (Venus likely contained liquid water in its early stages, but it all evaporated after a billion years or so.) It is also likely that the young Earth once had a dense, punishing atmosphere like Venus's today. However, surface oceans and plate tectonics both provided ample pathways for gases to be absorbed into the surface of the Earth. Oceans and plate tectonics offer huge amounts of carbonate storage, allowing the transfer and balance of carbon compounds to and from the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we have two things reinforcing the differences between Earth's and Venus's atmospheres:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Evaporation of liquid water&lt;/em&gt;: It is too hot on Venus for liquid water to exist. All water has evaporated, resulting in a denser atmosphere. On Earth, water can reside on the surface, lessening the amount in the atmosphere.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Carbon sinks&lt;/em&gt;: Liquid water and plate tectonics allow Earth to absorb considerable amounts of gas, allowing the atmosphere to be thinned of certain compounds like carbon dioxide. There is no such pathway on Venus, forcing all gas to remain in the atmosphere.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With no major mechanisms for gases to be absorbed by the planet, Venus is experiencing a runaway greenhouse effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Atmospheric escape&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mention Jeans escape. It is true that this effect is greater at higher temperatures; however, it is much easier for smaller molecules to escape than for larger ones. Hydrogen and helium, being the two smallest elements, are the most greatly effected by this phenomenon. In comparison, carbon dioxide, which makes up the majority of Venus's atmosphere, is not much affected by Jeans escape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also mention solar winds. While these play an effect, especially on planets without a magnetic field, this phenomenon is not as powerful as you might think. Ultraviolet light (i.e. photoionizing radiation) causes ionization in the uppermost region of the atmosphere. These charged particles now form a shell (called the ionosphere) that deflect solar winds, much like a magnetic field would. On Venus, the thick atmosphere provides more particles for ionization, resulting in a more powerful deflection. (Compare this to Mars, where solar wind is the primary non-thermal escape mechanism due to the thin atmosphere with few ionized particles.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary atmospheric escape mechanism for Venus is actually a bit more complicated. In the absence of a magnetic field, it is easier for charged particles to escape. In particular, electrons are most susceptible due to their small mass. As electrons escape, the net charge of the ionosphere leans positive, causing the ejection of positive ions, mostly H&lt;sup&gt;+&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While Earth and Venus formed similarly, Earth got lucky. It has pathways to remove gases from the atmosphere, while Venus does not. Additionally, the two planets do not experience significantly different rates of atmospheric escape. This results in the atmospheric densities where know today: 66 kg/m&lt;sup&gt;3&lt;/sup&gt; for Venus and just 1.2 kg/m&lt;sup&gt;3&lt;/sup&gt; for Earth.&lt;/p&gt;&#xA;" OwnerUserId="9271" LastActivityDate="2015-09-22T15:06:24.873" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11888" PostTypeId="2" ParentId="11880" CreationDate="2015-09-22T16:00:32.957" Score="14" Body="&lt;p&gt;The &quot;iron core&quot; in a supernova is actually the end product of a nuclear statistical equilibrium that begins when the silicon core begins to fuse with alpha particles (helium nuclei). Exothermic reactions are possible right up to Nickel-62 (which is actually the nucleus with the highest binding energy per nucleon). In fact, successive, rapid alpha captures produce nuclei with the same number of protons and neutrons, but at the same time, the competing processes of photodisintegration and radioactive decay work in the other direction. The process is thought to mostly stop at Nickel-56 which, because heavier nuclei are more stable with $n/p&amp;gt;1$, then undergoes a couple of $\beta^{+}$ decays via Cobalt-56 to Iron-56. However, the core of a supernova just before it explodes is likely to contain a bit of a mixture of iron-peak isotopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Before&lt;/em&gt; all this happens it &lt;em&gt;is&lt;/em&gt; possible for iron and nickel to undergo nuclear reactions &lt;em&gt;if&lt;/em&gt; there is an appropriate source of free neutrons. The elements beyond iron in our universe are predominantly created by neutron-capture in either the &lt;a href=&quot;https://en.wikipedia.org/wiki/R-process&quot; rel=&quot;noreferrer&quot;&gt;r-process&lt;/a&gt; or the &lt;a href=&quot;https://en.wikipedia.org/wiki/S-process&quot; rel=&quot;noreferrer&quot;&gt;s-process&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The r-process is thought to occur &lt;em&gt;after&lt;/em&gt; a core-collapse supernova (or a type Ia supernova) has been initiated. The neutron flux is created by the neutronisation of protons by a dense, degenerate electron gas in the collapsing core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the s-process can occur outside the core of a massive star before it explodes. It is a &lt;em&gt;secondary process&lt;/em&gt; because it needs iron nuclei to be present already - that is, the iron that is used for the seed nuclei is not produced inside the star, it was already present in the gas from which the star formed. The s-process in massive stars uses free neutrons produced during neon burning (so at advanced nuclear burning stages beyond helium, carbon and oxygen burning) and results in the addition of neutrons to iron nuclei. This builds up heavy isotopes, which may either be stable or undergo $\beta$ decay and/or further neutron captures to build up a chain of &quot;s-process elements&quot; (e.g.  Sr, Y, Ba) all the way up to lead. The overall process is &lt;em&gt;endothermic&lt;/em&gt;, but the yields and reaction rates are so small that it has no major influence on the overall energetics of the star. The newly-minted s-process elements are easily blasted into the interstellar medium shortly afterwards when the supernova explodes.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-10-02T06:05:11.153" LastActivityDate="2015-10-02T06:05:11.153" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11889" PostTypeId="2" ParentId="8592" CreationDate="2015-09-23T01:36:05.963" Score="1" Body="&lt;p&gt;I read this as &quot;Benenaim&quot; (another name for Alkaid) too (not Etamin, which is another name for Gamma Draconis), and that star does indeed transit the meridian near midnight on April 4th (12:44am to be more precise). Note that the hour angle is 0 in the Stellarium screenshot below, and the star is only 35 minutes from the zenith position:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8KkpP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8KkpP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2015-09-23T01:36:05.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11890" PostTypeId="2" ParentId="11468" CreationDate="2015-09-23T02:03:42.407" Score="4" Body="&lt;p&gt;I think dpwilson's answer is excellent and I voted him up, but I wanted to post this chart with the old picture is worth a thousand words point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jK9t8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jK9t8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Venus is somewhat lighter then Earth, yet has a much thicker&#xA;  atmosphere. One would imagine that the following should be true:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;During the formation phase, all inner planets had captured as much gas&#xA;  as they could hold per gravitational/thermodynamic equilibrium. After&#xA;  all, even puny Mars managed to capture a sizable atmosphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe.   But during the early solar system, once the sun has formed and starts pumping out light and solar flares, (and the early sun was likely much more active in shooting out solar flares in part due to a faster rotation), a key factor to consider is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Frost_line_%28astrophysics%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Frost Line&lt;/a&gt; - which is well past earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/B43QY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B43QY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, 1 of 2 things can happen in the early formation of the Solar-system.   One, the planets form and collect available ices and gas they can before the sun starts melting/pushing away any ice and gas inside the frost line, or 2, the sun forms first and the inner planets have very little gas and water as they form.   They do get bombarded by hydrogen ejected from the sun, but mostly the inner planets aren't good at holding onto this hydrogen.  In the 2nd scenario, any atmosphere and water they get would have to come from comet impacts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The early atmosphere of the inner planets were mostly CO2, CH4, NH3 maybe some N2.    If Venus was struck by a couple extra comets, that alone would explain it and it's not statistically unreasonable.   Now, I'm not saying this is what happened, only that it's possible.  Venus retains most of it's CO2, but it could lose over time most of it's H20, CH3, NH3, maybe N2 if it was present, leading to the mostly CO2 atmosphere it has today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also theoretically possible that the giant impact that formed the moon also blew away a lot of Earth's early atmosphere.  (not certain of that but the enormous addition of heat and rotation, it's possible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the chart above, it suggests Venus won't lose much H20, but other charts have Venus closer to the H20 line.   (google gas escape velocity planets for more charts)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The rate of atmospheric escape should be much higher for Venus: Venus&#xA;  receives more heat from the sun, thus higher Jeans escape rate Venus&#xA;  has negligible magnetic field, so some of its atmosphere must be lost&#xA;  to direct &quot;blow-off&quot; by solar wind&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is true.  It might well explain why Venus has so little water which is common in the solar-system.    But on your last point, Venus has an induced magnetic field - see &lt;a href=&quot;http://www.astrobio.net/topic/solar-system/venus/a-magnetic-surprise-from-venus/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.   dpwilson explained this in more detail.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Yet, it is Earth that seemingly misses a great deal of atmospheric&#xA;  volume. So the question is: what are the current theories regarding&#xA;  the &quot;thinning&quot; of the Earthian atmosphere? When and why the&#xA;  atmospheric gases had left the planet?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I gather there's still come uncertainty on precisely what Earth's atmosphere was like billions of years ago.  It might well have started out with an even denser atmosphere than Venus currently has, but it's hard to know with any certainty (at least, nothing I've read suggests certainty on the subject).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth pointing out that coal, oil and natural gas don't form naturally, but they are the product of dead plants and sea life buried over hundreds of millions of years.   Also, many of the rocks we see all around us have Oxygen in them.   Granite has Oxygen, for example.  (There's no, or at least, very little Granite on Venus).   The absorption of atmosphere by life on earth and by Oxygen binding to surface and dissolved ocean minerals likely played an enormous role in thinning out Earth's atmosphere.   Life on earth, all by itself, could be enough explain the difference in Earth's to Venus' atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comets: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Comets used to be larger.   Every pass near the sun, comets shrink.   Besides, it's not just comets it's icey moon like objects and asteroids, which, as Jupiter migrated and the late heavy bombardment took place, some of those could have been quite large.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See:  &lt;a href=&quot;http://viewzone2.com/alienatmospherex.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; and &lt;a href=&quot;http://www.technologyreview.com/view/414215/comets-seeded-earths-early-atmosphere/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Origin_of_water_on_Earth#Extraplanetary_sources&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides, I didn't say it as definitive, I said it was possible a large part of Venus' atmosphere came from a large comet strike.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-09-23T05:34:30.803" LastActivityDate="2015-09-23T05:34:30.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11891" PostTypeId="2" ParentId="11468" CreationDate="2015-09-23T04:02:06.577" Score="0" Body="&lt;p&gt;It so appears, there exist surprisingly little solid theories as to why Earth and Mars managed to lose the majority of their atmospheric gases, while Venus had managed to retain a most magnificent atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One plausible theory was put forward by a prominent chemist Octave Levenspiel et alii, based on the old Soviet research into Earthian crust composition conducted in the 1950s (I could not find any substantial updates to the crust composition model developed since that time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a glance, the theory works as following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Earth had formed with an atmosphere similar or denser than Venusian. Its most abundant ingredient should have been CO2 (analogously to Venus and Mars).&lt;/li&gt;&#xA;&lt;li&gt;Earthian atmosphere managed to cool down enough for water to start condensing into liquid phase. The exact mechanism for this is not quite clear to me (Goldilocks zone notwithstanding) because hot and dense CO2/H2O atmosphere should have caused a prominent &quot;green house&quot; effect, preventing the planetary surface from cooling down (unless &quot;green house&quot; effect models are too exaggerated).&lt;/li&gt;&#xA;&lt;li&gt;Atmospheric CO2 started to dissolve in liquid water (this alone would account for about 50% reduction in partial CO2 pressure). Strongly acidic water started to erode calcium from the crust, kick-starting the limestone formation process.&lt;/li&gt;&#xA;&lt;li&gt;Emerging life had accelerated the process, sequestering the remaining atmospheric CO2 into the gigantic limestone and somewhat smaller coal deposits.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The detailed outline of the theory can be seen here:&#xA;&lt;a href=&quot;http://pubs.acs.org/subscribe/archive/ci/30/i12/html/12learn.html&quot; rel=&quot;nofollow&quot;&gt;http://pubs.acs.org/subscribe/archive/ci/30/i12/html/12learn.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hoped that some answer's here can suggest plausible alternative theories. In particular:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&quot;Plate tectonics&quot; have probably nothing to do with the composition and parameters of the present day atmosphere. As far as I know nobody had ever suggested that mantle can re-absorb gases from the atmosphere - on a contrary, gases released from a cooling mantle through a volcanic activity should contribute to a more denser atmosphere (clearly, this process never helped Mars and is not helping Earth that much either). Volcanic gases are mostly composed of CO2 and water vapor (up to 90% by mass), while those two substances are hardly present in the modern day atmosphere (CO2 - ~350ppm, vapor - 0.4%, mostly from evaporation unrelated to volcanic recycling).&lt;/li&gt;&#xA;&lt;li&gt;Comets are relatively lite objects (a decent comet weighs 10000-100000 times less that even a thinnish, present day Earthian atmosphere) with low density. High kinetic energy impact of a comet with a planet will most probably result in escape of the majority of comet contained gases back into space (and impact heating will also add some of the planetary gases to the escaping mix - a process known as &quot;impact erosion&quot;). It is long believed that no substantial matter transfer is possible between comets and planets (&lt;a href=&quot;http://adsabs.harvard.edu/full/1998ASPC..148..364Z&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/full/1998ASPC..148..364Z&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;li&gt;&quot;Early thin atmosphere&quot; - atmosphere being eroded or lost shortly after Earth (Moon) formation is not plausible for the obvious reason: where does the limestone/coal comes from? If Earthian atmosphere was lost and then replenished through tectonic activity this brings us to back to an original question.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="412" LastEditorUserId="412" LastEditDate="2015-09-23T05:26:01.443" LastActivityDate="2015-09-23T05:26:01.443" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11892" PostTypeId="2" ParentId="11886" CreationDate="2015-09-23T17:15:31.877" Score="1" Body="&lt;p&gt;A planetarium programme will tell you exactly where to look, and can be invaluable when looking for the location of planets, but the Moon is very easy to find in the sky, even after several glasses of Flanders Red Ale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what you really want to know is a) is the eclipse visible from Belgium and b) roughly where in the sky will the moon be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://eclipse.gsfc.nasa.gov/LEplot/LEplot2001/LE2015Sep28T.pdf&quot; rel=&quot;nofollow&quot;&gt;This map&lt;/a&gt; shows that the Eclipse will be fully visible from Belgium. So that's good. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The the Moon will be full (a lunar eclipse always happens at full moon) so the moon will rise in the East as the sun sets, and travel into the West. At Midnight (12 AM, but many people make that mistake) the Moon will be in the South, and at 4AM it will be South West, and quite low in the sky. But even eclipsed the moon hard to miss!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't go out too early. Nothing much will be visible until the &quot;Umbral eclipse&quot; starts at 1:07 UTC (=3:07 am in Belgium) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish you clear skies.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-09-23T17:15:31.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11893" PostTypeId="1" AcceptedAnswerId="11922" CreationDate="2015-09-23T18:28:26.857" Score="9" ViewCount="5816" Body="&lt;p&gt;Sirius is currently approaching the solar system, at a rate of 5.5 km/s, consequently in the future it will be closer and therefore brighter. When will Sirius pass closest to the solar system? How far will it then be from the sun, and what magnitude will it reach?&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-02-07T06:02:19.177" Title="When will Sirius be closest to the solar system?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11894" PostTypeId="2" ParentId="11126" CreationDate="2015-09-24T00:24:24.570" Score="3" Body="&lt;p&gt;Check out the on-line SIMBAD astronomical data base. You can even make separate lists based on criteria (like: list all of the stars between 10 and 20 parsecs (have to convert parallax to light years/ parsecs = 1/parallax (in arcsec/ 1 parsec = 3.262 LY) with class between G0 and G9 from 0 to +30 deg declination). It's pretty cool and shows numerous listings of distance values, brightness, metallicity, etc. complete with date of study and source. I just looked again and masses aren't listed (sorry). Masses aren't directly measured anyway (except in the case of binaries where we know the distance). Masses are essentially determined from the Hertzprung -Russell (mass/ luminosity) diagram and can be refined if we can determine age by looking at spectra. Back to Sol-Station or Wikipedia or I should say their references.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="7997" LastEditDate="2015-09-26T23:55:12.247" LastActivityDate="2015-09-26T23:55:12.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11895" PostTypeId="1" AcceptedAnswerId="11912" CreationDate="2015-09-24T02:58:00.017" Score="4" ViewCount="36" Body="&lt;p&gt;What are some possible atmospheric bio-signatures for a planet with anaerobic life? or even microorganisms before oxygen builds up in the atmosphere? Would there be a way of telling if methane, etc. is possibly not from abiotic causes?&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2015-09-25T20:21:45.613" Title="Biosignatures for anaerobic or other early microbiology?" Tags="&lt;astrobiology&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11899" PostTypeId="1" AcceptedAnswerId="11901" CreationDate="2015-09-24T19:21:44.990" Score="7" ViewCount="293" Body="&lt;p&gt;I'm looking for accurate positions and velocities of the planets in the solar system over several decades. I want to simulate their trajectories using Newtonian laws of motion, and compare the precession of Mercury's perihelion with observed data. I used the &lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;noreferrer&quot;&gt;JPL's HORIZONS system&lt;/a&gt; but I can't manage to find these positions and velocities in the output data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you explain how the data is formatted? (The HORIZONS web-interface Tutorial didn't help.) &lt;/p&gt;&#xA;" OwnerUserId="9288" LastEditorUserId="2365" LastEditDate="2015-09-25T21:38:36.923" LastActivityDate="2015-10-05T18:37:01.667" Title="What is the format of the data from the JPL's HORIZONS system?" Tags="&lt;solar-system&gt;&lt;mercury&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11900" PostTypeId="1" CreationDate="2015-09-24T19:43:41.867" Score="3" ViewCount="52" Body="&lt;p&gt;Some colleagues and I were planning to launch a stratospheric balloon in December. A thought occurred to me: what if we were to put some sort of xray camera in the payload? We are expecting a maximum altitude of about 80,000 feet. Would it work any better at that altitude? &lt;/p&gt;&#xA;" OwnerUserId="9287" LastActivityDate="2016-04-27T11:42:07.847" Title="Stratospheric Balloon with an Xray camera" Tags="&lt;x-ray&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11901" PostTypeId="2" ParentId="11899" CreationDate="2015-09-25T01:25:17.287" Score="1" Body="&lt;p&gt;WebGeocalc (&lt;a href=&quot;http://wgc.jpl.nasa.gov:8080/webgeocalc/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://wgc.jpl.nasa.gov:8080/webgeocalc/&lt;/a&gt;) is actually a &#xA;fairly useful tool, and the next best thing to downloading the CSPICE &#xA;libraries, so let me provide a slightly more detailed answer to your &#xA;question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you first visit WebGeocalc, you will see something like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/53p1m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/53p1m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because you are looking for perihelions, which involve distance, &#xA;scroll down and choose &quot;distance finder&quot;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kLQsv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kLQsv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you want to find the perihelion of Mercury, fill out the top &#xA;half of the form like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Pk3S0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pk3S0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this example, I'll find all perihelions from 1900 to 2201. A &#xA;perihelion is a minimum of distance, and since you're looking for all &#xA;perihelions, choose &quot;is local minimum&quot; on the lower half of the form, &#xA;so it looks as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VQMo1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VQMo1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scroll down once again (you can leave the remaining form values as is) &#xA;and click the &quot;Calculate&quot; button: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/q4Y1t.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q4Y1t.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you get your results, click &quot;Save All Intervals&quot; to save the &#xA;times of Mercury's perihelions. Note that each result is an interval, &#xA;but the interval is 0 seconds long, so it's actually a point in time: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/D2YES.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D2YES.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now click on &quot;Calculation Menu&quot; to get back to the main menu: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yLQZZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yLQZZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll now find Mercury's position at all of these perihelions so you &#xA;can see how the perihelion position changes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find Mercury's position at these times, first click on &quot;State &#xA;Vector&quot;, which will take you to a form. Fill out the top half of the &#xA;form as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tYROf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tYROf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fill out the lower half of the form as follows, dragging the results &#xA;window (for the times of Mercury's perihelions) into the input times &#xA;window: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H1rcX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H1rcX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click &quot;Calculate&quot; to receive your results: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jSWSX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jSWSX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://naif.jpl.nasa.gov/pub/naif/toolkit_docs/Tutorials/pdf/individual_docs/36_webgeocalc.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://naif.jpl.nasa.gov/pub/naif/toolkit_docs/Tutorials/pdf/individual_docs/36_webgeocalc.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: if you're going to do extensive calculations, you'll probably &#xA;want to download the C SPICE library and kernels: &#xA;&lt;a href=&quot;http://naif.jpl.nasa.gov/naif/tutorials.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://naif.jpl.nasa.gov/naif/tutorials.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I emailed Jon D. Giorgini (jdg@tycho.jpl.nasa.gov) re the open-sourceness of HORIZONS, and got this answer (which I'm not sure is helpful, but...):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;&#xA;Horizons output is copyrighted in the sense one can't take a&#xA;table output by Horizons and re-publish it as one's own.&#xA;JPL is a contractor organization and so the output is not prepared&#xA;by an officer or employee of the US government, and it is not&#xA;transferred to the US government, so copyright remains relevant.&#xA;&#xA;The data values themselves, say the RA/DEC of Pluto at&#xA;some instant, are not legally copyrighted, though it would &#xA;be unethical to publish it without citation.&#xA;&#xA;Legal copyright protection is for fixed representations&#xA;analogous to a book; the book has copyright protection, but &#xA;an individual word extracted from the book (i.e., into a &#xA;different representation) is not itself protected.  Up to &#xA;a point&#xA;&#xA;Note that Horizons reads SPICE files for planets, natural&#xA;satellites, and spacecraft, then derives results it outputs &#xA;from scratch from that basis. But for comets and asteroids, &#xA;it creates all information on demand from scratch.&#xA;&#xA;Further, we (JPL Solar System Dynamics) put the planet/satellite/&#xA;spacecraft information into the SPICE format to begin with. So &#xA;a SPICE file might be like the blank pages of a book; not &#xA;copyrighted until the original work is written into it, but not&#xA;itself copyrighted (beyond the source code, of course).&#xA;&#xA;But if you need a more trustworthy legal opinion for a specific &#xA;situation, I can put you in touch with legal staff. If this is an &#xA;actual issue and not just a point of curiosity, that's the way to&#xA;since perhaps other rules would apply to your actual situation. &#xA;&lt;/pre&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2015-10-05T18:37:01.667" LastActivityDate="2015-10-05T18:37:01.667" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11903" PostTypeId="1" CreationDate="2015-09-25T05:41:11.300" Score="10" ViewCount="528" Body="&lt;p&gt;Venus surface isn't older than about 300 million years. The only explanation I've come across is that some kind of global volcanism resurfaced the planet. But couldn't it have been an impact event? A huge comet hitting it by chance, maybe making Venus tip over and stop spinning too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or could Venus have had a moon like Phobos which spiraled inwards, disintegrated and rained down during a few million years? Phobos will do so in only 50 million years, maybe it is a common feature for terrestrial planets to form with such moons that hit them billions of years later. Maybe Earth is an exception because of the large Moon and how it formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would an impact event leave visible traces like impact basins, or could the entire surface melt and reform as it is today, as I suppose Earth did when the Moon formed? Could Venus have been a very different planet up until 0.3 billion years ago? How could one find out, what kind of investigation would be needed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pluto's mass is about 1/50th of Mars (and Theia), so I suppose that it would be enough for a Pluto sized comet to have had about 7 times the impact speed compared to the pre-Earth/Theia collision. Comets could have up to 70+35 km/s speed relative to Venus. Someone maybe can make a rough estimate of what's needed to resurface Venus and how likely such an impact event would be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And could reformation and redifferentiation after an impact have eliminated a previous magnetic field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diversity of the planets and moons, I suggest, depend more on their vast history of random events, than on where and how they once formed. I suspect that planetary scientists generally want to discount a unique planet reforming event only 0.3 billion years ago, just to make their job a bit easier ;-)&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-11-07T06:19:45.680" LastActivityDate="2020-09-19T08:12:19.463" Title="Could an impact have resurfaced Venus 300 million years ago?" Tags="&lt;venus&gt;&lt;surface&gt;&lt;impact&gt;&lt;volcanism&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11904" PostTypeId="2" ParentId="11903" CreationDate="2015-09-25T08:33:44.043" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Would an impact event leave visible traces like impact basins, or&#xA;  could the entire surface melt and reform as it is today, as I suppose&#xA;  Earth did when the Moon formed? Could Venus have been a very different&#xA;  planet up until 0.3 billion years ago? How could one find out, what&#xA;  kind of investigation would be needed?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Certainly giant impacts were fairly common, though I don't think they were common any more as recent as 300-500 million years ago.   That would have been a kind of unusual event.   Mostly they occurred when the solar-system was young.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.space.com/26447-mercury-composition-giant-impact.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mercury probably had one&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.space.com/5025-venus-mysteries-blamed-colossal-collision.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Venus may have too&lt;/a&gt; - and I recommend this article as very loosely related.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth had one - as you know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and Mars too, but not quite as &quot;giant&quot;, though still very big and it left a visible mark that spans nearly 1/2 of Mars' surface.  See &lt;a href=&quot;http://www.science20.com/news_articles/mars_dichotomy_did_a_giant_impact_shape_the_southern_and_northern_hemispheres-152691&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; and &lt;a href=&quot;http://www.space.com/5558-huge-impact-created-mars-split-personality.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon even had one and there is evidence in the thickness of the Moon's crust and location of it's volcanic magma pools/dark spots.   See &lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2014/07/01/the_moon_s_two_faces_why_are_they_so_different.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; and &lt;a href=&quot;http://www.universetoday.com/87934/second-moon-may-have-orbited-earth-billions-of-years-ago/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point I'm trying to make with all these examples is that unless the Impact is very very large, it would leave a impact footprint of sorts.   The impact has to be large enough to not only liquefy the surface but to also allow the surface enough time to reset in mostly uniform layers so you don't get 1/2 of the planet with a thicker crust than the other half.   Impacts of that size are statistically improbable, within the last 300 million years.  Most of the big impacts happened long before then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia suggests that there's other reasons for believing the global resurfacing though interior heat idea.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Geology_of_Venus#Global_resurfacing_event&quot; rel=&quot;nofollow noreferrer&quot;&gt;Venus Global resurfacing event - Wikipedia&lt;/a&gt; - Venus lack of a magnetic field, it's influx and loss of water and D to H ratio and it's Sulfur content.     Do these have had to happened through internal heat resurfacing?   Now, the article says there's no proof, only some pretty good evidence, so it it possible Venus resurfacing was due to an impact?    I see no reason why it wouldn't be possible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given Venus' Heat trapping atmosphere, higher temperature and greater gravity, which also means, greater impact velocity, at least escape velocity of about 10 km/s where as the Mars impact that covers a hemisphere was thought to be about 5 km/s and the impacting object into Mars, at least 1,600 km in radius which is nearly the size of the Moon.   You might not need that big to impact Venus and create a complete resurfacing, but I wouldn't think you could go too much smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I love the Venus had a moon that crashed into it idea cause that seems to work on a few levels.  A giant impact that created Venus could have happened 4 and change billion years ago and created a Venusian Moon, which over time, could both slow Venus' rotation and work it's way towards Venus in the process, eventually crashing into it.   It seems at least, possible, though I'm not sure what evidence could be gathered to prove it.    I like your idea a lot and I think it's much more likely than this one:  &lt;a href=&quot;http://www.space.com/22966-earths-moon-from-venus.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth stole the Moon from Venus&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to Venus being very different prior to it's resurfacing.    The sun was only about 3%-5% less luminous when Venus resurfaced and given Venus' proximity to the sun, I have a very hard time seeing Venus as not already in a run-away greenhouse effect even then.   I love the Venus once had oceans and maybe life hypothesis, but with a CO2 rich atmosphere, it's hard for me to see it as likely.   My guess is that Venus was hot and virtually without water 300 million years ago too.   Unless it somehow stored a huge amount of it's CO2 underground, I have a hard time not seeing it as too hot for oceans 300-500 million years ago.   Maybe, but I'm skeptical.    It would be very neat to have a picture of what Venus was like prior to it's resurfacing and over the hundreds of millions and billions of years before that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this &lt;a href=&quot;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20140017487.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; it mentions that it's improbable that Venus once had a Moon based on angular momentum.   A planet-Moon system maintains it's angular momentum and if the moon spirals in towards the planet, it should carry significant angular momentum with it.   Such an impact should give Venus a pretty rapid rotation which would be difficult to explain given that it rotates so slowly now.  Unless perhaps, the Moon was in a retrograde orbit, Moon going in one direction the planet in another and the combined angular momentum largely cancelled out, but adding more specific conditions tends to decrease the probability.   I still like the hypothesis, but the article above says it's unlikely.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-10-25T10:39:33.490" LastActivityDate="2017-10-25T10:39:33.490" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="11905" PostTypeId="1" CreationDate="2015-09-25T11:57:14.933" Score="4" ViewCount="105" Body="&lt;p&gt;I've read that photon's have zero mass, hence no force is needed to accelerate them. But then the second question. How can gravity affect something which has 0 mass?. Yes, gravity(black hole) can bend space and time but how can 0 mass particle be affected by this?. What am i missing.?&lt;/p&gt;&#xA;" OwnerUserId="9295" LastActivityDate="2015-09-25T12:30:15.047" Title="How does light so travel so fast?." Tags="&lt;gravity&gt;&lt;black-hole&gt;&lt;light&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2015-09-27T11:01:04.103" ContentLicense="CC BY-SA 3.0" />
  <row Id="11908" PostTypeId="2" ParentId="6555" CreationDate="2015-09-25T13:45:48.643" Score="2" Body="&lt;p&gt;I might be misunderstanding your question, but using HORIZONS (&lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ssd.jpl.nasa.gov/horizons.cgi&lt;/a&gt;) with these settings should give you the data you need:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jgcMX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jgcMX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2015-09-25T13:45:48.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11909" PostTypeId="2" ParentId="10138" CreationDate="2015-09-25T13:54:38.327" Score="1" Body="&lt;p&gt;This is wrong, but may help someone find the right answer: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can convert right ascension and declination to a 3 dimensional &#xA;unit vector in the J2000.0 ICRF reference frame using the standard &#xA;formula for converting spherical coordinates to rectangular &#xA;coordinates and using a radius of 1: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;{Cos[dec] Cos[ra], Cos[dec] Sin[ra], Sin[dec]}&lt;/code&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ra and dec of the north ecliptic pole is ra,dec of &#xA;&lt;code&gt;{270, 66.5607083333}&lt;/code&gt; per Wikipedia, so the unit vector representing &#xA;the north ecliptic pole is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;{0, -0.3977771648286046, 0.9174820582119942}&lt;/code&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it turns out, there is conflicting data for Ceres, perhaps because &#xA;of the semi-recent DAWN flyby. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, I'll use Vesta, where wikipedia explicitly states (&lt;a href=&quot;https://en.wikipedia.org/wiki/4_Vesta#Rotation&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/4_Vesta#Rotation&lt;/a&gt;): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;north pole pointing in the direction of right ascension 20 h 32 min, &#xA;declination +48 [... which] gives an axial tilt of 29 [degrees]&lt;/code&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting to degrees, Vesta's north pole's ra,dec is &#xA;&lt;code&gt;{308,48}&lt;/code&gt;. Using the formula above to find the unit vector, we get: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;{0.411957936296447, -0.527282113378167, 0.743144825477394}&lt;/code&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take the dot product of two vectors and divide by the product of &#xA;their lengths, we get the cosine of the angle between them. In this &#xA;case, both vectors have length 1, and the dot product is &lt;code&gt;0.891563&lt;/code&gt; &#xA;whose arc-cosine is right around 27 degrees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &lt;em&gt;if&lt;/em&gt; Vesta's orbit were the same plane as Earth's orbit, the axial &#xA;tilt would be 27 degrees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since Vesta's orbit is inclined 7.14043 degrees to the &#xA;ecliptic, this answer is incorrect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think you can find the correct answer without using Vesta's &#xA;inclination and the longitude of its ascending node. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of these values are known, but I can't figure out how to use them &#xA;to get the correct asnwer. &lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2015-09-25T13:54:38.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11910" PostTypeId="1" AcceptedAnswerId="11913" CreationDate="2015-09-25T14:56:43.010" Score="5" ViewCount="394" Body="&lt;p&gt;I've been trying to figure out how to calculate the latitude of my location with just the shadow of a stick cast by the sun. I am doing this under the assumption that I am stranded in some place with no almanac or any other data. (not even day or month).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find the latitude roughly. But I have problems picturing the globe along with it's tilted axis and the angles formed by the shadow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you help me understanding the concept and give me a definitive picture of angles that can be calculated. A perspective like a side-view of the earth and sun (orbiting plane - perpendicular to the screen) would help me understand much better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my efforts to understand so far - &lt;a href=&quot;https://i.stack.imgur.com/4ylx7.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4ylx7.png&quot; alt=&quot;My confusions&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not an expert nor have a good background in maths (my maths teacher ruined my life). So please explain me in simple words.&lt;/p&gt;&#xA;" OwnerUserId="9296" LastActivityDate="2015-09-26T15:16:53.443" Title="Understanding Earth Tilt, Sun's Position and Lattitude Calculation" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;navigation&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11911" PostTypeId="2" ParentId="11076" CreationDate="2015-09-25T18:14:29.463" Score="4" Body="&lt;p&gt;What we call &quot;gravity&quot; is really just the distortion of spacetime. If spacetime was completely straight, there would be zero gravity. But the existence of a massive body is one of the things that can distort spacetime, so that &quot;straight&quot; lines are not straight anymore. We perceive that distortion as &quot;gravity&quot;. It is the distortion that actually generates the gravitational force of attraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a very crude analogy, imagine a lemon. As long as you're not squeezing it, the juice stays put, because no force is exerted on it. But when you squeeze the lemon, distorting its structure, juice starts moving around inside. The lemon is spacetime. The juice is anything existing in spacetime. Gravity is the distortion of the shape of the lemon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photons may not have any rest mass, but they move in a straight line. When there's gravity, straight lines are not straight anymore, and therefore photons curve around massive bodies.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-09-25T19:32:07.457" LastActivityDate="2015-09-25T19:32:07.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11912" PostTypeId="2" ParentId="11895" CreationDate="2015-09-25T20:21:45.613" Score="3" Body="&lt;p&gt;Anaerobic life has much less impact on the atmosphere than aerobic life. Anaerobic life could exist in the deep sea and have no gas exchange with the atmosphere at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methane would get everyone excited. Biotic methane might be expected to have a different isotope mix compared with abiotic methane. But it is hard enough to distinguish isotope mixes when you have a rover on a planet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More complex compounds would be very exciting, but a very sensitive detector would be needed to find the very low concentrations required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be that the marker of abiotic life is the detection of anomalously low CO2 levels. Most abiotic atmospheres have very high levels of Carbon dioxide, but life will progressively reduce this level. A low level of carbon dioxide, in an atmosphere containing water, especially if coupled with methane would look bery interesting.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-09-25T20:21:45.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11913" PostTypeId="2" ParentId="11910" CreationDate="2015-09-25T21:26:50.457" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I've been trying to figure out how to calculate the latitude of my&#xA;  location with just the shadow of a stick cast by the sun. I am doing&#xA;  this under the assumption that I am stranded in some place with no&#xA;  almanac or any other data. (not even day or month).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You really cannot find your latitude with only the sun and sticks at your disposal and no declination data or time data. Maybe if you could record your shadow over a period of a day, you could tell if you are in the northern or southern hemisphere by observing which way the shadow moves. If it moves clockwise, you are in northern hemisphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if you could repeat and observe for 365 days (from the day you are stranded, assuming you are very accurately measuring the angles) you could find if you are within the tropic of cancer or above it, as you might not have the sun right on top of your stick over the year if you are above the 23.5 N parallel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that's as far as you can get. To determine the precise location, you will need the sun's &lt;code&gt;declination&lt;/code&gt; at any given time. May be you will get lucky, if the day you are measuring the angle and the equinox matches, but you wouldn't know it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Could you help me understanding the concept and give me a definitive&#xA;  picture of angles that can be calculated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A picture is worth thousand words. And &lt;a href=&quot;http://astro.unl.edu/naap/motion3/animations/sunmotions.swf&quot; rel=&quot;nofollow&quot;&gt;an interactive demonstration&lt;/a&gt;... It's worth at least a hundred 2D diagrams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional Interactive media - &lt;a href=&quot;http://astro.unl.edu/naap/motion1/animations/seasons_ecliptic.html&quot; rel=&quot;nofollow&quot;&gt;http://astro.unl.edu/naap/motion1/animations/seasons_ecliptic.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="557" LastActivityDate="2015-09-25T21:26:50.457" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11914" PostTypeId="2" ParentId="11882" CreationDate="2015-09-25T23:15:09.703" Score="7" Body="&lt;p&gt;I am probably going to get slammed for this, because it violates everything we were taught about tidal forces, but the antipodal tide is caused by the centrifugal force created by the Earth's rotation about the earth/moon barycenter, not differential gravitational forces.  While the moon's gravity is less on the side of the earth furthest from it, that force is still towards the moon, not away from it.&lt;br&gt;&lt;br&gt;There is a good explanation of this centrifugal force explanation on NOAA's website &lt;a href=&quot;https://tidesandcurrents.noaa.gov/restles3.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. I'll cite some pictures and pertinent text, but this site is a good read for anyone trying to understand the forces responsible for the Earth's tides.&lt;br&gt;&lt;br&gt;Here is a diagram showing the earth and moon's movement around the system's barycenter:&lt;br&gt;&lt;a href=&quot;https://i.stack.imgur.com/1TxAd.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1TxAd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;And some of the pertinent text:&lt;br&gt;&lt;br&gt;&lt;blockquote&gt;The center of revolution of this motion of the earth and moon around their common center-of-mass lies at a point approximately 1,068 miles beneath the earth's surface, on the side toward the moon, and along a line connecting the individual centers-of-mass of the earth and moon. (see G, Fig. 1) The center-of-mass of the earth describes an orbit (E1, E2, E3..) around the center-of-mass of the earth-moon system (G) just as the center-of-mass of the moon describes its own monthly orbit (M1, M2, M3..) around this same point. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The Effect of Centrifugal Force. It is this little known aspect of the moon's orbital motion which is responsible for one of the two force components creating the tides. As the earth and moon whirl around this common center-of-mass, the centrifugal force produced is always directed away from the center of revolution. All points in or on the surface of the earth acting as a coherent body acquire this component of centrifugal force. And, since the center-of-mass of the earth is always on the opposite side of this common center of revolution from the position of the moon, the centrifugal force produced at any point in or on the earth will always be directed away from the moon. This fact is indicated by the common direction of the arrows (representing the centrifugal force Fc) at points A, C, and B in Fig. 1, and the thin arrows at these same points in Fig. 2.&lt;/blockquote&gt;&lt;br&gt;And finally another diagram that the blockquote cites:&lt;br&gt;&lt;br&gt;&lt;a href=&quot;https://i.stack.imgur.com/zVOzk.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zVOzk.jpg&quot; alt=&quot;Moon gravitational diagram&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;br&gt;Yes, one can probably find lots and lots of quotations declaring that the antipodal tide is caused because the moon's gravitational force is much less on the far side than on the near side, but that doesn't make it true. And I believe NOAA ought to be a pretty authoritative source. If they can't get it right...&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7108" LastActivityDate="2015-09-25T23:15:09.703" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11915" PostTypeId="1" CreationDate="2015-09-26T06:44:26.160" Score="5" ViewCount="81" Body="&lt;p&gt;What visual artifacts should we expect from the images that will be released from the James Webb Space Telescope? Specifically, do we expect 3-pronged diffraction spikes (any preview how they would look?) and do we expect any artifacts due to the honeycomb-like structure of the main mirror?&lt;/p&gt;&#xA;" OwnerUserId="9302" LastActivityDate="2016-02-19T19:32:10.200" Title="What visual artifacts are expected from the JWST?" Tags="&lt;telescope&gt;&lt;space-telescope&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11916" PostTypeId="1" AcceptedAnswerId="11919" CreationDate="2015-09-26T10:29:19.470" Score="1" ViewCount="365" Body="&lt;p&gt;I know that gravity will turn a mass into a sphere, which is why planets and stars are that shape. But then we have asteroids and small moons which are not spherical. Such as a couple of Pluto's moons. Is there a size at which they will become spheres? Is it dependent on their composition? Does it need a certain mass? Is it only because of the heat of collisions melted the rocky planets that they became this shape while they were semi liquid?&lt;/p&gt;&#xA;" OwnerUserId="9304" LastActivityDate="2015-09-26T13:16:09.277" Title="Spheres in space" Tags="&lt;planet&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2015-10-04T13:54:46.033" ContentLicense="CC BY-SA 3.0" />
  <row Id="11917" PostTypeId="1" AcceptedAnswerId="11923" CreationDate="2015-09-26T11:59:50.277" Score="4" ViewCount="1277" Body="&lt;p&gt;I've watched the Venus transit 2004 with a small telescope, a relatively rare event, considering most people can only observe two in their life time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If E.T. were sitting in a sunny place on a moon of the outer planets,&#xA;sipping on a liquid nitrogen drink, would there ever be (has there ever been) a transit of an inner gas giant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess if E.T. were in &lt;a href=&quot;https://en.wikipedia.org/wiki/Sputnik_Planum&quot; rel=&quot;nofollow&quot;&gt;Sputnik Planum&lt;/a&gt;, there would be no transits due to Pluto's orbital plane being inclined too much wrt. the Ecliptic.&lt;/p&gt;&#xA;" OwnerUserId="6593" LastActivityDate="2015-09-27T14:05:56.173" Title="Was there ever a Jupiter transit or Saturn transit?" Tags="&lt;solar-system&gt;&lt;jupiter&gt;&lt;saturn&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11919" PostTypeId="2" ParentId="11916" CreationDate="2015-09-26T13:16:09.277" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I know that gravity will turn a mass into a sphere, which is why&#xA;  planets and stars are that shape. But then we have asteroids and small&#xA;  moons which are not spherical. Such as a couple of Pluto's moons. Is&#xA;  there a size at which they will become spheres?  Is it dependent on&#xA;  their composition?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Google searches provide a wide variety of answers to this and while the other question is answered, it's only answered briefly, so I thought I'd give this a try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mikebrownsplanets.com/2008/09/what-is-dwarf-planet.html&quot; rel=&quot;nofollow&quot;&gt;Mike Brown's Planets&lt;/a&gt; was the best link I could find on the subject, and I can't swear by his correctness, but his numbers are in the range you find in other articles.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;While we can't see most of the objects in the Kuiper belt well enough&#xA;  to determine whether they are round or not, we can estimate how big an&#xA;  object has to be before it becomes round and therefore how many&#xA;  objects in the Kuiper belt are likely round. In the asteroid belt&#xA;  Ceres, with a diameter of 900 km, is the only object large enough to&#xA;  be round, so somewhere around 900 km is a good cutoff for rocky bodies&#xA;  like asteroids. Kuiper belt objects have a lot of ice in their&#xA;  interiors, though. Ice is not as hard as rock, so it less easily&#xA;  withstands the force of gravity, and it takes less force to make an&#xA;  ice ball round.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The best estimate for how big an icy body needs to be to become round&#xA;  comes from looking at icy satellites of the giant planets. The&#xA;  smallest body that is generally round is Saturn's satellite Mimas,&#xA;  which has a diameter of about 400 km. Several satellites which have&#xA;  diameters around 200 km are not round. So somewhere between 200 and&#xA;  400 km an icy body becomes round. Objects with more ice will become&#xA;  round at smaller sizes while those with less rock might be bigger. We&#xA;  will take 400 km as a reasonable lower limit and assume that anything&#xA;  larger than 400 km in the Kuiper belt is round, and thus a dwarf&#xA;  planet. We might be a bit off in one direction or another, but 400 km&#xA;  seems like a good estimate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other estimates I've read have 600 KM diameter for rocky bodies and Ceres is ice and rock, not really a rocky body, so I think there's some uncertainty in there.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does it need a certain mass?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I personally don't like talking about planets by their mass cause it gets rather unwieldy in size.  Take Ceres, mentioned above.  It's 8.958 × 10^20 kg.  A 400 KM diameter ice world would have a mass of, figure a water-ice average density, about 3 x 10^19 and 200 KM diameter, 1/8th of that, about 3.75 x 10^18.   The minimum mass is somewhere in  those ranges, but I think radius and composition are easier to work with.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it only because of the heat of collisions melted the rocky planets&#xA;  that they became this shape while they were semi liquid?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can only give an intuitive answer here, but collisions that liquefy a planet are rare and, especially with smaller planets, would be more likely to blow the planet to bits as soon as melt it.  Take the giant impact on Mars.  Article &lt;a href=&quot;http://www.space.com/5558-huge-impact-created-mars-split-personality.html&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; and &lt;a href=&quot;http://www.caltech.edu/news/giant-impact-explains-mars-dichotomy-1442&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;.   The 2nd article says that the object that hit Mars was thought to be traveling at about &quot;&lt;em&gt;6 to 10 kilometers per second&lt;/em&gt;.&quot; and size &quot;&lt;em&gt;roughly 1,600 to 2,700 kilometers across&lt;/em&gt;&quot; - so, bigger than Ceres, Smaller than the Moon and this may have melted half of Mars' surface but it didn't melt all of it cause it left a measurably difference in Mars crust one side to the other.  If a collision of that enormous size didn't melt Mars all the way around - I'd wager that such melting is pretty rare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd point, as a general principal, the Solar system objects orbit in the same direction and the further out you get from the sun, the slower the orbital speed (though the more likely you get eccentric orbits).  The speed of collisions in space are roughly the vector addition of the 2 objects orbital speeds, which are often somewhat in line with each other, plus the escape velocity of the more massive object (or a bit more than that if both objects are massive), so once objects get to be a pretty good size, about the size where they become spheres, collisions that would melt the object grow less and less likely, and if it is large enough to do that, you should also expect a fair bit of debris from impact.   My guess, based on just barely enough knowledge to think about this stuff, is that planetoids and large enough to maybe get spherical asteroids are unlikely to form into spheres due to melting from heat from collisions because due to the relatively weak gravity of smaller objects like that (Ceres has an escape velocity of 0.51 km/s, about 1140 MPH), an impact to generate enough heat to melt it would blow half the dwarf planet away, give or take.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you have a very hot planet that's very very close to the sun and at near liquid temperature anyway or if you have a coalescing and a multitude of successive impacts and lots of heat collecting in the object, then, sure.   It's certainly possible but my guess is that a planetoid or asteroid forming into a sphere due to melting it's rare.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a layman's answer.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-09-26T13:16:09.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11920" PostTypeId="2" ParentId="11910" CreationDate="2015-09-26T15:16:53.443" Score="0" Body="&lt;p&gt;Assuming you aren't north of the Arctic Circle or south of the Antarctic Circle, you can determine your latitude my making observations throughout the course of a day, and over the course of a year. You'll need&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A rather straight stick,&lt;/li&gt;&#xA;&lt;li&gt;A fairly flat piece of ground,&lt;/li&gt;&#xA;&lt;li&gt;A plumb bob (which you can make out of string and a rock),&lt;/li&gt;&#xA;&lt;li&gt;Some small pebbles to mark the tip of the shadow of the stick over the course of a year at solar noon, and&lt;/li&gt;&#xA;&lt;li&gt;A trigonometry table or a calculator.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Use your plumb bob to ensure your stick is as close to vertical as you can make it. You'll want to place the stick in exactly the same place every day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing you'll want to find is the north-south line that passes through the base of the stick. To do this, place a pebble at the tip of the stick's shadow when the shadow is at its shortest. If you do this perfectly, you'll have the north-south line on day number one. You almost certainly won't do this perfectly, so you'll need to repeat this for a few days. Because the Sun rises more or less in the east and sets in the west, you also know which way is north and which is south. You have a compass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll also have a very rough idea of your latitude. If the Sun sets to the left of the north-south line, you know you are somewhere north of 23.44 south latitude. If it sets to the right, you are south south of 23.44 north latitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a better idea of your latitude, keep doing this until the day you see the Sun rise exactly in the East. This happens twice a year, typically on March 20 and September 23. Use a string to measure the length of the stick's shadow when the tip of the shadow crosses the north-south line. You do not need a ruler; all you need to know is the ratio of the shadow's length to the stick's length. Now it's a matter of trigonometry:&#xA;$$\phi = \arctan\left(\frac {\text{shadow length}}{\text{stick length}}\right)$$&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-09-26T15:16:53.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11921" PostTypeId="2" ParentId="2176" CreationDate="2015-09-26T15:31:28.000" Score="1" Body="&lt;p&gt;The Earth would not appear fully dark because we have an atmosphere to scatter and adsorb the Sunlight. The edge of the Earth would be a ring of orange /red light... this is the limb or &quot;terminator&quot; that separates day from night and the ring is a view of all the sunsets and sunrises seen simultaneously. You may also (with magnification) see the artificial lights from major cities in darkness. &#xA;You may be able to see the Sun&quot;s corona as a pearly light spreading out from behind the disk of the Earth.&#xA;Around you, the lunar surface would be dramatically darkened and the ash grey lunar regolith would be lit red. There would be no shadows as the Sun/Earth/Moon are lined up and any light would be straight on&lt;/p&gt;&#xA;" OwnerUserId="9306" LastActivityDate="2015-09-26T15:31:28.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11922" PostTypeId="2" ParentId="11893" CreationDate="2015-09-26T19:05:17.833" Score="7" Body="&lt;p&gt;Sirius is moving towards the sun at 5.5 km/s, and according to Sky and Telescope, will reach its closest distance in 60000 years, at which time it will be &lt;a href=&quot;https://en.wikipedia.org/wiki/Historical_brightest_stars&quot; rel=&quot;noreferrer&quot;&gt;7.8 light years distant, and be at magnitude -1.64&lt;/a&gt;, only slightly brighter than it's current magnitude -1.46.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculations using the latest Hippacos results give a slightly different value: a closest approach of 8.18 ly in 46000 years, and consequently a slightly dimmer peak. However Sirius is not moving rapidly relative to the sun, and will not change its brightness greatly in the next 100000 years or so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will be the brightest star in the sky for another 90000 years, and no star will match its current brightness for another million years (when delta Scuti will reach magnitude -1.84.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2015-09-27T20:22:15.627" LastActivityDate="2015-09-27T20:22:15.627" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11923" PostTypeId="2" ParentId="11917" CreationDate="2015-09-26T19:25:54.183" Score="4" Body="&lt;p&gt;Wikipedia actually has &lt;a href=&quot;https://en.wikipedia.org/wiki/Transit_of_Jupiter_from_outer_planets&quot; rel=&quot;nofollow&quot;&gt;many pages on this subject&lt;/a&gt;. Transits of Jupiter, viewed from Saturn are rare. The calculations are done from the centre of the planet, rather than a moon, but the principle is clear: Transits of Jupiter are rare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next, non-grazing transit of Jupiter visible from Saturn is in 7541. From Uranus in 2714, and from Neptune in 2188.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But probably the most photogenic transit will be the &lt;a href=&quot;https://en.wikipedia.org/wiki/Transit_of_Saturn_from_outer_planets&quot; rel=&quot;nofollow&quot;&gt;transit of Saturn from Uranus&lt;/a&gt; in 2669. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn will also transit from Neptune in 2061, the only transit which we could live to see. Uranus will not transit from Neptune until 38172, and transits from Pluto or other KBO are even more rare, due to the high orbital inclinations.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2015-09-27T14:05:56.173" LastActivityDate="2015-09-27T14:05:56.173" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11924" PostTypeId="2" ParentId="2092" CreationDate="2015-09-26T19:41:49.213" Score="7" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrostatic_equilibrium#Planetary_geology&quot; rel=&quot;noreferrer&quot;&gt;Icy objects, such as most in the Kuiper belt can reach an equilibrium if they are about 400km across&lt;/a&gt;, whereas the rocky asteroid Pallas, at 572km clearly has an irregular, non spherical shape. All rocky objects larger than Pallas (and there aren't many) are spherical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rock tends to be stronger than ice. Rocky objects are able to withstand their own gravity for longer than icy ones. Pallas is a reasonable cut of point. The next smaller asteroids (Vesta, Hygiea etc) are round-ish, but not in hydrostatic equilibrium. On the other hand Small, icy moons such as Miranda and Mimas are in, or close to equilibrium. Mimas has a diameter of just under 400km.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-07-17T09:30:54.227" LastActivityDate="2016-07-17T09:30:54.227" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11925" PostTypeId="1" AcceptedAnswerId="11926" CreationDate="2015-09-26T19:57:58.270" Score="3" ViewCount="181" Body="&lt;p&gt;If we put a station in orbit around the Sun, and wanted to send transports as regularly as possible without the travel costs being prohibiting; should we have it rotate around the Sun clockwise, or anti-clockwise like the other planets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'll be checking the page frequently so feel free to post comments and questions should you have them. I'll respond as quickly as possible.)&lt;/p&gt;&#xA;" OwnerUserId="9309" LastEditorUserId="9309" LastEditDate="2015-09-26T21:33:39.180" LastActivityDate="2015-09-26T21:33:39.180" Title="Should a station in orbit around the Sun travel in the same or the opposite direction to the planets?" Tags="&lt;orbit&gt;&lt;space-travel&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11926" PostTypeId="2" ParentId="11925" CreationDate="2015-09-26T20:44:37.843" Score="3" Body="&lt;p&gt;It must orbit the sun in the same direction as the Earth and other planets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main difficulty in sending a transport to the station will be the the velocity change needed to put the transport into the station's orbit. If the orbit is retrograde with respect to the Earth, then the transport needs to make a massive change of velocity to reach the station, and this needs a massive amount of fuel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, on the other hand, the station is prograde with respect to the earth, then the transfer can be made with an economical Hohmann transfer. This is much more important than any apparent benefit of a retrograde orbit passing close to the Earth more often.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-09-26T20:44:37.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11927" PostTypeId="1" AcceptedAnswerId="11928" CreationDate="2015-09-26T22:42:04.680" Score="8" ViewCount="4055" Body="&lt;p&gt;I am not sure if my question is true or not. But this is what I observe regularly. When I was young and when I look at the sky, there were too many many stars were there. But now, from few months I'm observing the sky (by eyes, not any telescope), I see very few stars. Perhaps, countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Stars are dead?&lt;/strong&gt; (Not all) As I read somewhere, by the time you see the star, it would have been dead. Because, the light would've traveled so many years to reach earth. By the time, star would have completed 'X' light years and it is dead by now?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Current position of the earth in galaxy&lt;/strong&gt;: As I read, the earth will not always traverse in the same path around the sun. So the current view from the earth to the sky, may not have any star? (This is my guess/imagination, I'm not sure)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is it because of the city light/ pollution?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PS:&lt;/strong&gt; Please be easy. May be the question is broad or simple, just want to know the brief answer and if its complete explanatory, I'm happy for that.&lt;/p&gt;&#xA;" OwnerUserId="9284" LastActivityDate="2019-06-12T04:36:34.267" Title="Less stars in the night, compared to 15-20 years back" Tags="&lt;star&gt;" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11928" PostTypeId="2" ParentId="11927" CreationDate="2015-09-27T01:07:09.387" Score="10" Body="&lt;p&gt;In my own case, if I see fewer stars than I did when I was younger, it's partly because my eyes have deteriorated with age. I don't know whether that applies to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most likely explanation is increased light pollution. If you can manage to get to an area far enough away from city lights, on a cloudless and moonless night, you should see just as many stars as you did when you were younger.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Stars are dead?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No.  All the stars you can see in the night sky are within a few hundred light-years of Earth. That means that you're seeing them as they were no more than a few hundred years ago. Most stars live for &lt;em&gt;billions&lt;/em&gt; of years; the Sun, for example, is about 5 billion years old and is expected to live for another 5 billion or so years in essentially its current form. Some of the brighter stars have shorter lifespans (because they're larger and &quot;burn&quot; their nuclear fuel more quickly), but even they have lifespans of at least millions of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No naked-eye visible stars have &quot;died&quot; in the last several centuries. We do sometimes see stars explode (as novas or, more rarely, as supernovas), but I don't believe any of the stars that have done this were naked-eye visible before they exploded.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Current position of the earth in galaxy&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No.  The stars do move relative to each other, but not quickly enough for the motion to be visible over a human lifetime. Barnard's Star has the fastest proper motion of any star in the sky, but it only moves at about 10 arcseconds per year; it would take it nearly 2 centuries to move the width of a full Moon. And Barnard's star isn't even naked-eye visible. The stars you see in the night sky are in very nearly the same apparent positions as they were thousands of years ago.&lt;/p&gt;&#xA;" OwnerUserId="53" LastActivityDate="2015-09-27T01:07:09.387" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11929" PostTypeId="1" AcceptedAnswerId="11932" CreationDate="2015-09-27T01:27:57.163" Score="7" ViewCount="168" Body="&lt;p&gt;I noticed in today's newspaper (Boston Globe, September 26) that sunrise and sunset were both at 6:25. That's the twelve hour day I'd have expected at the equinox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Please retag if there's a better one)&lt;/p&gt;&#xA;" OwnerUserId="9311" LastActivityDate="2015-09-27T14:04:09.963" Title="September 26 twelve hours sunrise to sunset?" Tags="&lt;time&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11930" PostTypeId="2" ParentId="11927" CreationDate="2015-09-27T02:15:36.217" Score="3" Body="&lt;p&gt;If you've moved to an area with major aircraft traffic, you may be seeing fewer stars due to &lt;a href=&quot;http://www.atmos-chem-phys.org/5/2155/2005/acp-5-2155-2005.pdf&quot; rel=&quot;nofollow&quot;&gt;cirrus cloud formation&lt;/a&gt;. Such clouds are often subtle enough not to be visible as clouds but still block out starlight. For example: &lt;a href=&quot;http://www.nature.com/nature/journal/v403/n6765/full/403068a0.html&quot; rel=&quot;nofollow&quot;&gt;Forecasting Andean rainfall and crop yield from the influence of El Niño on Pleiades visibility&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We find that poor visibility of the Pleiades in June—caused by an increase in subvisual high cirrus clouds—is indicative of an El Niño year, which is usually linked to reduced rainfall during the growing season several months later. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the above, climate is obviously also a variable in star visibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh yes, aging is also a factor. The &lt;a href=&quot;https://books.google.com/books?id=alfNBQAAQBAJ&amp;amp;pg=PA8&amp;amp;lpg=PA8&amp;amp;dq=aging%20eyes%20star%20magnitude%20limit&amp;amp;source=bl&amp;amp;ots=zkRasRUhqw&amp;amp;sig=HXprgs7jYgcSPbB4EYrJ762Fnw8&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0CE8Q6AEwCWoVChMI3_OZuYaayAIVCAqSCh3U7QhG#v=onepage&amp;amp;q=aging%20eyes%20star%20magnitude%20limit&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;eyes opacify with age&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The normal aging processes in the eye mean that the retina of a 60 year old person recieves only about 30% of the amount of light seen by a person half that age.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="2031" LastEditDate="2015-09-28T15:39:11.180" LastActivityDate="2015-09-28T15:39:11.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11931" PostTypeId="2" ParentId="2318" CreationDate="2015-09-27T09:34:07.940" Score="-1" Body="&lt;p&gt;I noticed an important error in Gerald's otherwise excellent calculation above. A layer of water 0.142 m thick per meter squared has a mass of 142 kg, not 14.2 kg. This means Gerald's results should be multiplied by 10:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;14.2 kg/m2 x 3.711 m/s2 = 528 N/m2 = 5.28 hPa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is comparable to the current average Martian surface pressure of ~6.3 hPa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A side note: The orbit at impact could also be an ellipse with current orbit as apoapsis and the Martian surface as periapsis. This hypothetical change in orbit would increase the impact speed to ~4.3 km/s (first equation at &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_speed#Precise_orbital_speed&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Orbital_speed#Precise_orbital_speed&lt;/a&gt; ). However, this should probably be achieved artificially and has nothing to do with the natural decay of the moon.&lt;/p&gt;&#xA;" OwnerUserId="9314" LastEditorUserId="9314" LastEditDate="2015-09-27T11:52:47.330" LastActivityDate="2015-09-27T11:52:47.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11932" PostTypeId="2" ParentId="11929" CreationDate="2015-09-27T13:06:50.407" Score="4" Body="&lt;p&gt;A day would be twelve hours long at the equinoxes if the Earth had no atmosphere and if the Sun was a point rather than a sphere. Sunrise and sunset are defined as the time at which the upper limb of the Sun appears to rise above or set below an ideal oblate spheroid Earth and assuming average atmospheric conditions. While this calculation ignores terrain and variations in atmospheric condition, it does not ignore that the basic facts that the Earth does have an atmosphere and that Sun is a sphere. The standard atmosphere conditions means that sunrise (sunset) occur when the upper limb of the Sun first reaches (first falls below) 34 arc minutes below the horizon. That the Sun is not a point adds another 16 arc minutes to this, making sunrise/sunset occur when the center of the Sun is 50 arc minutes below the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that at the equinoxes, a &quot;day&quot; is at least six minutes longer than the twelve hours we are naively taught (and incorrectly reported twice a year by naive weathermen). As an aside, days are always longer than nights at the equator.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-09-27T13:06:50.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11933" PostTypeId="2" ParentId="11929" CreationDate="2015-09-27T14:04:09.963" Score="2" Body="&lt;p&gt;The day and night are not equal on the equinox, though they are about equal. Equinox is when the Earth's axis is not tilted towards the sun, or equivalently the path of the sun passes from the Northern to the Southern hemisphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sunrise and sunset are measured from the &quot;top&quot; of the sun, not the centre, and the refraction of the atmosphere means that the sun is visible from the surface, even after it would have set were there no air to bend its light. Finally the &lt;a href=&quot;http://www.analemma.com/Pages/framesPage.html&quot; rel=&quot;nofollow&quot;&gt;Earth's elliptical orbit&lt;/a&gt; also varies the length of the day. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The day on which the day and night are equal varies with latitude. In Boston, at 42 degrees North (it's as far south as the French Riviera - always surprises me) &lt;a href=&quot;http://www.timeanddate.com/astronomy/equinox-not-equal.html&quot; rel=&quot;nofollow&quot;&gt;the day is 12 hours long on Sept. 26&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-09-27T14:04:09.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11934" PostTypeId="1" CreationDate="2015-09-27T14:36:42.647" Score="3" ViewCount="177" Body="&lt;p&gt;The data precision now can exclude a &gt;&gt;$1M_{\odot}$ black hole companion, which has a, say 10000 year orbital period with our sun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we draw a two dimensional diagram(Mass VS. distance) to show the region that is still a grey area? Within 1 pc, there is no stellar mass black hole? If there is one, the orbit change of objects in solar system may be detectable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to calculate its proper motion or its planets' proper motion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am serious to ask. Please give some reference/technical method to calculate PM. For example, is there some reference paper about the distance limitation of the black hole based on the motion of asteroids/comets?  &lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditDate="2016-12-13T12:26:20.833" LastActivityDate="2016-12-13T12:26:20.833" Title="Can we draw a min(distance) and max(mass) relation based on current observation data of nearby astronomical objects within several parcsecs?" Tags="&lt;solar-system&gt;&lt;black-hole&gt;&lt;positional-astronomy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11935" PostTypeId="2" ParentId="11934" CreationDate="2015-09-27T15:09:45.083" Score="3" Body="&lt;p&gt;First lets make sure we know the scale of the solar system. Neptune, the most distant major planet, has an orbit of 165 years. Sedna, the object with the longest known period, has an orbit of 11400 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes come in different sizes. A black hole that has formed from the collapse of a giant star would have a mass more than three times the mass of the sun. If a black hole of this size were inside the orbit of Neptune, it would not be orbiting the sun, the sun would be orbiting it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about further out? Blackholes can be spotted in X-rays when matter falls into them forming a disc of matter orbiting at very high speeds, and heating up to enormous temperatures.  A black hole without an accretion disc would be very dark and hard to detect from a distance. But we can be reasonably confident that there are no in our immediate neighbourhood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motion of a blackhole in orbit with the sun would be described by Newtons laws of gravity. The sun and the black hole would orbit on elliptical orbits around their common centre of gravity. Planets of the sun or the black hole would be stable within their Hall radii. Close to the blackhole one would need General relativity to describe the gravitational field, but from further out, its gravity would be almost exactly as predicted by Newton. You could model it in any gravity simulator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that micro black-holes exist in the universe. These could pass through the earth without doing much damage, as the are sub-atomic size and since they interact only with matter by gravity, and don't weigh much, they don't have much effect. They might be &quot;hot&quot; with Hawking radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of these are much of a risk to the survival of the solar system. Black holes are rare. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2015-12-08T16:48:33.240" LastActivityDate="2015-12-08T16:48:33.240" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11936" PostTypeId="2" ParentId="11927" CreationDate="2015-09-27T15:26:19.830" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it because:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Stars are dead?&lt;/li&gt;&#xA;  &lt;li&gt;Current position of the earth in galaxy?&lt;/li&gt;&#xA;  &lt;li&gt;Is it because of the city light/ pollution?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not #1 or #2. It's #3, plus other factors.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;Light pollution.&lt;br&gt;&#xA;Anyone who lives in or near a large city (that probably describes most of the the stackexchange network members) won't be able to see as many stars as someone who lives a hundred kilometers away from that city. Light pollution is a growing problem to astronomical observatories, even those very far removed from cities.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Loss of visual acuity.&lt;br&gt;&#xA;If you're an adult, your visual acuity isn't as good as it was when you were a child. Visual acuity peaks between 8 and 15 years of age, depending on who you read. After 15, it's all downhill.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Different levels of humidity, particulates, and air pollution.&lt;br&gt;&#xA;You will see fewer stars than nominal if you live in an area with high humidity, high levels of particulates, or high levels of air pollution. If you made such a change in locale recently, you will remember seeing a lot more stars than you can see now.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-09-27T15:26:19.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11937" PostTypeId="1" AcceptedAnswerId="12027" CreationDate="2015-09-27T16:32:49.763" Score="4" ViewCount="160" Body="&lt;p&gt;My knowledge of CCDs is that these are sensors which collect photoelectrons. That's about it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a difference between CCDs used for astrometry, spectroscopy, and photometry? As an example, each type of CCD is currently used by Gaia. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recommendations of where else to read about this at a detailed engineering level is appreciated. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="9171" LastActivityDate="2015-10-06T13:23:53.557" Title="Gaia: What is the difference between CCDs used for astrometry, photometry, and spectroscopy?" Tags="&lt;spectroscopy&gt;&lt;astrometry&gt;&lt;photometry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11938" PostTypeId="2" ParentId="11634" CreationDate="2015-09-27T17:15:17.570" Score="19" Body="&lt;p&gt;There was a object, apparently flying above you, that you couldn't identify. By definition this is an unidentified flying object. However this does not imply that it was an extra-terrestrial spacecraft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UFO reports can be explained by a combination of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Not recognising a known natural object, such as Venus, or unusual clouds.&lt;/li&gt;&#xA;&lt;li&gt;Planes, drones, Chinese lanterns or satellites.&lt;/li&gt;&#xA;&lt;li&gt;Optical illusion, dreams or the effects of drugs.&lt;/li&gt;&#xA;&lt;li&gt;Deliberate falsification.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There is no real way to decide what it was you saw. There are a couple of points in your report that are odd. You describe it as the same size as stars, but less bright. Stars have no visible size,  and have many magnitudes of brightness. Your report suggests that you are someone who doesn't regularly observe the sky, and so is likely to be more susceptible to optical illusions and tricks that eyes can play on us.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-09-27T17:15:17.570" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11939" PostTypeId="1" CreationDate="2015-09-27T19:54:15.873" Score="1" ViewCount="283" Body="&lt;p&gt;I watched a documentary where I heard some very strange theories that are difficult for me to comprehend. It is even difficult for me to explain what I heard. &quot;Objects that fall into a black hole leave a copy of themselves outside the black hole.&quot; &quot;We are not living in reality but in a hologram .&quot; Does any of this make any sense to anyone ? O.K. so I went to check the title of this documentary because how can anyone explain it without knowing what I saw. &quot;Fingerprints of God -the Dark Universe&quot; this is the title. I do admit that I did sleep through a good part of it because it was almost 2:00 a.m. Also if anyone wants to watch it then look for it on youtube because that is where I saw it.&lt;/p&gt;&#xA;" OwnerUserId="2562" LastActivityDate="2015-09-29T03:11:07.390" Title="Is our Universe a hologram?" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ClosedDate="2015-10-14T08:54:58.020" ContentLicense="CC BY-SA 3.0" />
  <row Id="11941" PostTypeId="1" AcceptedAnswerId="12361" CreationDate="2015-09-27T21:41:46.140" Score="4" ViewCount="4684" Body="&lt;p&gt;My task is to calculate the heliocentric velocity of an object to correct for wavelength in a spectrum. I have the RA, Dec and observation date, and I know that the  NED velocity correction calculator(&lt;a href=&quot;https://ned.ipac.caltech.edu/forms/vel_correction.html&quot; rel=&quot;nofollow&quot;&gt;https://ned.ipac.caltech.edu/forms/vel_correction.html&lt;/a&gt;) will do the job, and I found out the equation here (&lt;a href=&quot;https://ned.ipac.caltech.edu/help/velc_help.html#notes&quot; rel=&quot;nofollow&quot;&gt;https://ned.ipac.caltech.edu/help/velc_help.html#notes&lt;/a&gt;), but honestly I don't have a clue where these parameters come from. Could some one give me some hints on the reasoning behind it? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="9317" LastEditorUserId="9317" LastEditDate="2015-11-02T07:32:33.177" LastActivityDate="2016-01-31T18:03:39.633" Title="How to calculate the heliocentric velocity of an object?" Tags="&lt;velocity&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11943" PostTypeId="2" ParentId="11939" CreationDate="2015-09-28T09:26:29.907" Score="6" Body="&lt;p&gt;I can't give a good answer but I did find the documentary and out of curiosity, watched some of it, skipped forward and found the hologram section you're talking about, it's at the end about 1:18:40 in.  (I'm not going to post a link to the video cause I'm not sure it's correctly copyrighted by those who are showing it)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The video itself isn't bad, but it's basically a science channel level program, which is written to appeal to viewers and explanations are often over-simplified.   This video in particular seemed to try to cover so much ground that it explains very little.   But on the positive side, they have Brian Greene and Leonard Susskind discussing the hologram idea and those guys are both real physicists who've talked about this before.  I don't think it's a main stream idea, just an idea that a few people like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://discovermagazine.com/2011/jun/03-our-universe-may-be-a-giant-hologram&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's&lt;/a&gt; a magazine article on the subject and an excerpt from Brian Greene's book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;a href=&quot;https://www.youtube.com/watch?v=2DIl3Hfh9tY&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's&lt;/a&gt; Leonard Susskind giving a talk on the subject at Stanford, not at a class, more of a lecture cause he mentions that not everyone in attendance might be a physicist, but I've since watched his lecture and he makes a few of these points more clear, at least to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And &lt;a href=&quot;https://physics.stackexchange.com/questions/79198/what-are-the-implications-of-the-holographic-principle&quot;&gt;Here's&lt;/a&gt; a stack exchange question and answer on a similar subject.    There's also no shortage of articles that can be found on the Holographic principal or the &quot;universe is a hologram&quot;, though I think the first is a better set of words to search and the 2nd might lead you to some sub-par articles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/79198/what-are-the-implications-of-the-holographic-principle/79246#79246&quot;&gt;Philip Gibbs - inactive's&lt;/a&gt; answer (also in the link above) on what the Holographic Principal implies is better than anything I could say about it, so I'd start there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a rule, it's a good idea to post links and if the video is 2 hours long, the specific time as well.   YouTube links can also be removed so they're not ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit to answer questions.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Had to think about this one a bit, and sometimes trying to answer &quot;why&quot; is opening a can of worms, but . . . here I go.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why do they think that it is our Universe that is the Holograph?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not a fan of the theory myself and I'd guess that not all physicists agree with it either.   One big problem with the theory is that it can't be proven, which really makes it more of a guess based off a mathematical model and not a good theory.  One definition of a good theory is one that can be tested.   If it can't be tested true or false, it's really more of an idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for why Brian Greene likes the theory, you might need to ask him or search his interviews, but maybe he likes it so can sell books - I don't know.  Maybe he likes the idea of the Universe being a Hologram.   He's obviously a pretty smart guy.  You don't get a PHD in physics and get recognition and invited to ted talks and all that by being dumb, but smart guys are wrong all the time when it comes to new theories.  There's no evidence that this theory is true.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I should point out that the Holographic principal as it applies to black holes and the information paradox, all that stuff is good and valid theoretical physics rooted in ideas we understand such as entropy and time dilation and red-shift.   The Holographic principal seems a very reasonably deduced feature of black holes that also, cleverly answers the information paradox, but whether it can be extrapolated to apply to the entire universe is a whole different ball game and to me, it's basically a guess.  Not much more.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why couldn't our Universe be reality and a Holograph exists of it&#xA;  somewhere else?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, I could say &quot;you're right&quot;, but the truth is, you're just flipping around the question.  The more relevant questions are whether the theory is true at all, and as I said above (and Susskind said in his video), whether it can be tested.   Susskind's video is really quite good though he starts slowly, but if you're really interested in this it's worth watching.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What makes them think that this is the Holograph instead of the&#xA;  reality?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One thing about Physics to keep in mind.  It's often surprised us.  Physics has given us answers that physicists didn't expect many many times, perhaps more than any other field of study.  Physics is probably more likely to yank the rug of reality out from under us than any other field and with that in mind, it's not a subject to study if you want to feel secure in what you believe and know.    In the days after Newton, Physicists were very confident in what they knew and felt that nothing very new was out there, though the constant speed of light was a puzzle they'd not expected.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then Einstein came along and pretty much threw everybody's understanding out the window when he started to figure things out and not long after that, the quantum physicists came up with ideas that made Einstein's head spin and to which he directly disagreed.   After that the big bang was unexpected, background radiation was unexpected.  Dark matter was unexpected (except for maybe Fritz Zwicky) and dark energy was quite the surprise when it was discovered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being regularly surprised, of-course, doesn't make the hologram theory any more true or false, cause nobody knows, but good physicists these days try to keep an open mind to mathematically valid new ideas.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;my 2 cents as a hobbyist.  Hope that wasn't too long and off the point, and please, take what I say with a grain of salt.   I'm not an expert.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-09-29T03:11:07.390" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11946" PostTypeId="1" AcceptedAnswerId="11949" CreationDate="2015-09-28T11:55:17.920" Score="7" ViewCount="1118" Body="&lt;p&gt;A solar eclipse only puts part of Earth at a time in shadow, so I'm not talking about a solar eclipse that casts a shadow over all of Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I mean is: are there any areas on Earth where, no matter how Sol, Earth and Luna are aligned, a total solar eclipse is impossible? Can the penumbra cast a shade on any part of Earth, or are there certain areas where a penumbra due to a solar eclipse cannot occur?&lt;/p&gt;&#xA;" OwnerUserId="9319" LastActivityDate="2017-11-12T15:40:51.047" Title="Can all areas on Earth experience a total solar eclipse?" Tags="&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11949" PostTypeId="2" ParentId="11946" CreationDate="2015-09-28T12:32:55.227" Score="6" Body="&lt;p&gt;Yes, total Solar eclipses can occur anywhere. On any given place, they occur roughly every 400 years, on average. If you take a look on the picture below (from &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_eclipse#Occurrence_and_cycles&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;), you'll see that within the last one thousand year, almost every place on Earth has experienced a total Solar eclipse. The places that &lt;em&gt;haven't&lt;/em&gt; experienced one have no special latitude or longitude, so it's just a matter of coincidence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5UOwZ.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5UOwZ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2015-09-28T12:32:55.227" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11953" PostTypeId="1" CreationDate="2015-09-28T16:04:17.047" Score="9" ViewCount="1708" Body="&lt;p&gt;From a horizontal frame of reference, the path of the Sun looks like a part of a circle in a tilted plane, as shown in this picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/uc0u6.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you go to a heliocentric system, one would say that this is because the Earth is rotating around it's axis. Is there any more detailed explanation of this change of reference? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how can one explain in detail the following facts (seen from a geocentric point of view - horizontal reference system) starting from a heliocentric system and the fact that the Earth is rotating around it's axis:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why is the path of the Sun a part of a circle? &lt;/li&gt;&#xA;&lt;li&gt;Why is this path in a plane?&lt;/li&gt;&#xA;&lt;li&gt;Why is this plane tilted?&lt;/li&gt;&#xA;&lt;li&gt;How to get this &quot;tilt-angle&quot;?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Why does this motion seem to be uniform from an equatorial point of view, and can I neglect the ecliptic to understand the points above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to model it using a ball as Earth, attached a camera to it and used another ball as the Sun, but I didn't succeed in reproducing the effect of the apparent motion in this model. Perhaps there is any animation which makes the points above clear.&lt;/p&gt;&#xA;" OwnerUserId="9328" LastEditorUserId="13" LastEditDate="2015-09-28T19:04:18.423" LastActivityDate="2016-05-26T01:17:18.553" Title="How can one explain the apparent motion of the Sun from a heliocentric point of view?" Tags="&lt;orbit&gt;&lt;the-sun&gt;&lt;earth&gt;&lt;apparent-motion&gt;&lt;frame-of-reference&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11954" PostTypeId="2" ParentId="11953" CreationDate="2015-09-28T18:09:31.997" Score="3" Body="&lt;p&gt;The picture shows 3 paths of the sun over a single day, plotted on three particular days: Midsummer, midwinter, and an equinox. The apparent motion of the sun is due to the rotation of the Earth, not the motion of the Earth around the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;&lt;b&gt;Why is the path of the sun a part of a circle?&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the Earth is spinning, so you are moving in a circle around an axis that passes through the poles of the Earth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;&lt;b&gt;Why is this path in a plane?&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The local surface of the Earth is modelled as a plane tangent to the sphere of the Earth. As the Earth spins, this plane rotates, and so relative to this plane, the sun appears to move. Because the plane is rotating in a single axis and at a uniform rate, the sun appears to move in a circle, across the sky. As it is a circle, it is in a plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;&lt;b&gt;Why is this plane tilted?&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is tilted because you are not at the poles, the surface of the Earth on which you stand is tilted with respect to the axis of the Earth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;&lt;b&gt;How to get this &quot;tilt-angle&quot;?&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is your latitude, or more exactly 90 - latitude. If you are at the poles the sun would travel in a horizontal plane. If you are at the equator the sun travels in a vertical plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The different paths (red green and blue in the picture) are due to the tilt of the Earth at 23.4 degrees, and so in Summer (in the north) the sun is higher at midday and further North at rise and set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;&lt;b&gt;Why does this motion seem to be uniformly from a equatorial point of view.&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image appears to show the path of the sun from a location about 45 degrees North, for example Northern USA. Again, I may be misunderstanding the question.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2015-09-28T18:37:23.637" LastActivityDate="2015-09-28T18:37:23.637" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11955" PostTypeId="1" CreationDate="2015-09-28T19:09:13.160" Score="8" ViewCount="838" Body="&lt;p&gt;There were some suggestions that asteroids could one day be mined for precious metals. I have found an outdated study that says that some siderites contain as much as 0.001% of gold: &#xA;&lt;a href=&quot;http://pubs.usgs.gov/circ/1968/0603/report.pdf&quot; rel=&quot;nofollow&quot;&gt;http://pubs.usgs.gov/circ/1968/0603/report.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there recent studies on precious metals in space rocks, i.e. platinum, rhodium, iridium, rhenium, osmium, ruthenium, palladium, germanium and gold, and is it genuinely a viable mining research subject?&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="2618" LastEditDate="2015-10-02T03:54:41.497" LastActivityDate="2021-04-04T11:29:00.543" Title="What are the most valuable meteorites and asteroids in terms of precious metals?" Tags="&lt;asteroids&gt;&lt;meteor&gt;&lt;meteorite&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11956" PostTypeId="1" CreationDate="2015-09-28T21:16:25.780" Score="3" ViewCount="141" Body="&lt;p&gt;There is a lot of information on different star types and compositions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we make estimates of the total mass of rock, ice, and non stellar matter that orbits stars based on star evolution models?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take, for example, the sun - are there studies of the total mass of non-solar material associated with the sun and it's composition regardless of accretion. What about nearby stars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we estimate how much non-incandescent material the galaxy contains aside from dark matter?&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="2270" LastEditDate="2015-09-30T09:55:53.577" LastActivityDate="2015-09-30T09:55:53.577" Title="Estimation of average rock and asteroid mass associated with different stars" Tags="&lt;planet&gt;&lt;exoplanet&gt;&lt;asteroids&gt;&lt;oort-cloud&gt;&lt;star-systems&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11957" PostTypeId="1" AcceptedAnswerId="11971" CreationDate="2015-09-28T22:06:29.270" Score="8" ViewCount="352" Body="&lt;p&gt;I am wondering what the brightest individual star is in M31, the Andromeda Galaxy.  Specifically, brightest as seen from Earth (so relative magnitude).&lt;/p&gt;&#xA;" OwnerUserId="128" LastEditorUserId="9186" LastEditDate="2015-09-29T06:25:06.980" LastActivityDate="2015-09-29T21:41:13.003" Title="What is the brightest star (relative magnitude) in M31?" Tags="&lt;star&gt;&lt;apparent-magnitude&gt;&lt;magnitude&gt;&lt;hypergiants&gt;&lt;m31&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11958" PostTypeId="1" CreationDate="2015-09-29T01:26:32.420" Score="9" ViewCount="3465" Body="&lt;p&gt;I saw on the tv show &lt;a href=&quot;http://www.sciencechannel.com/tv-shows/through-the-wormhole/&quot; rel=&quot;noreferrer&quot;&gt;Through the Wormhole (hosted by Morgan Freeman)&lt;/a&gt; that the planets in our solar system have been continuously moving away from the sun for millions of years. however, when I try to Google about it, there isn't much written about it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a proven fact that the planets are moving away from the sun?&lt;/p&gt;&#xA;" OwnerUserId="9332" LastEditorUserId="15415" LastEditDate="2018-01-22T23:16:32.653" LastActivityDate="2018-09-08T07:43:40.727" Title="Are planets moving away from the sun?" Tags="&lt;the-sun&gt;&lt;orbit&gt;&lt;distances&gt;&lt;mass&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11960" PostTypeId="1" AcceptedAnswerId="11965" CreationDate="2015-09-29T02:00:52.040" Score="21" ViewCount="1022" Body="&lt;p&gt;Could dark matter compress and form black holes? Since dark matter is even more abundant than normal matter, a dark matter black hole should not be rare...right?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9186" LastEditDate="2015-10-09T22:04:19.940" LastActivityDate="2019-06-17T08:43:36.490" Title="Could there be dark matter black holes?" Tags="&lt;gravity&gt;&lt;black-hole&gt;&lt;dark-matter&gt;&lt;supermassive-black-hole&gt;&lt;matter&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11961" PostTypeId="1" CreationDate="2015-09-29T02:02:44.447" Score="2" ViewCount="84" Body="&lt;p&gt;I thought we have found bacteria or other kinds of life on Mars.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is liquid water.  Is there such a big difference between liquid water and iced water? &lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="9186" LastEditDate="2015-09-29T06:32:09.180" LastActivityDate="2015-09-29T06:32:09.180" Title="Is liquid water on Mars such a big leap compared with ice?" Tags="&lt;mars&gt;&lt;life&gt;&lt;water&gt;&lt;astrobiology&gt;&lt;origin-of-life&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11963" PostTypeId="2" ParentId="11961" CreationDate="2015-09-29T03:27:40.893" Score="3" Body="&lt;p&gt;We have not found life, bacterial or otherwise, anywhere other than Earth ... yet. I think you mean the difference between liquid water and water ice. This difference relates to the hospitality to life. Life of the type we are used to needs liquid water, even if this is just trapped moisture, unless the life is temporarily in suspended animation. If there was no liquid water on Mars, there would &lt;strong&gt;&lt;em&gt;seem&lt;/em&gt;&lt;/strong&gt; to be no possibility of ongoing life there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to distinguish hospitable conditions from evidence for life. While having a hospitable environment is &lt;strong&gt;&lt;em&gt;necessary&lt;/em&gt;&lt;/strong&gt; for life, it is far from proof of life. The big hurdle is starting life, or bringing it from somewhere else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a variety of hospitable environments in the solar system and on exoplanets, such that if certain life from Earth (such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Extremophile&quot; rel=&quot;nofollow&quot;&gt;extremophile&lt;/a&gt; bacteria) were somehow placed there they could live. But that does not mean there is in fact life there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another important point, Mars was way more hospitable early in it's history, with bodies of liquid water. Which is important because even though starting life on Mars under current conditions is unimaginable, early Mars was rather Earthlike and could possibly start, or receive, life.&lt;/p&gt;&#xA;" OwnerUserId="9186" LastActivityDate="2015-09-29T03:27:40.893" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11964" PostTypeId="2" ParentId="11960" CreationDate="2015-09-29T03:48:11.407" Score="5" Body="&lt;p&gt;Black holes are the result of mass so concentrated that gravity does not let anything out, including light. Just about the only things we know about dark matter is that it has mass and &lt;em&gt;seems&lt;/em&gt; to only interact with ordinary matter through gravity. Since we do not know the physics of dark matter at all, it is impossible to say what processes might concentrate dark matter enough to form a black hole, but theoretically a black hole could be formed from dark matter, or even both ordinary matter and dark matter. Some theorists [&lt;em&gt;footnote&lt;/em&gt;] even think that is exactly what super-massive black holes are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;footnote&lt;/strong&gt;:&lt;/em&gt; Jeremiah P. Ostriker of Princeton University, &lt;a href=&quot;http://arxiv.org/pdf/astro-ph/9912548.pdf&quot; rel=&quot;nofollow&quot;&gt;Collisional Dark Matter and the Origin of Massive Black Holes&lt;/a&gt;, and Zoltán Haiman of Columbia University, &lt;a href=&quot;http://arxiv.org/pdf/1203.6075.pdf&quot; rel=&quot;nofollow&quot;&gt;The Formation of the First Massive Black Holes&lt;/a&gt;. Also interesting, &lt;a href=&quot;https://scholar.google.com/scholar?as_ylo=2011&amp;amp;hl=en&amp;amp;as_sdt=5,24&amp;amp;sciodt=1,24&amp;amp;cites=12977079148967989561&quot; rel=&quot;nofollow&quot;&gt;these papers&lt;/a&gt;, from 2011 to now, citing Ostriker's seminal paper.&lt;/p&gt;&#xA;" OwnerUserId="9186" LastEditorUserId="9186" LastEditDate="2015-10-09T04:26:47.737" LastActivityDate="2015-10-09T04:26:47.737" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11965" PostTypeId="2" ParentId="11960" CreationDate="2015-09-29T06:42:51.170" Score="24" Body="&lt;p&gt;The problem with trying to form a black hole with dark matter is that dark matter can only weakly interact (if at all) with normal matter and itself, other than by gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This poses a problem. To get dark matter concentrated enough to form a black hole requires it to increase its (negative) gravitational binding energy &lt;em&gt;without&lt;/em&gt; at the same time increasing its internal kinetic energy by the same amount. This requires some sort of dissipative interaction between dark matter and normal matter (or itself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following scenario should make this clear. Suppose we have a lump of dark matter that gravitationally attracts another lump of dark matter. As the two approach each other, they accelerate and gain kinetic energy. The kinetic energy gained will be exactly enough to then separate them to a similar degree to which they started, &lt;em&gt;unless&lt;/em&gt; some dissipative process takes place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example is to suppose that dark matter is weakly interacting massive particles (WIMPs). WIMPs are gravitationally drawn towards the centres of stars. If the weak interactions happen sufficiently frequently then it &lt;em&gt;might&lt;/em&gt; be possible for them to accumulate in stars, rather than shoot through and out the other side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has been hypothesised that black holes could be made like this near the centre of a Galaxy, seeded by dense neutron stars. The density of neutron star matter, combined with the enhanced density of dark matter near galaxy centres &lt;em&gt;could&lt;/em&gt; result in dark matter accumulation in the neutron stars, leading to the formation of black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once a black hole is formed then any dark matter that enters the event horizon cannot emerge regardless of what kinetic energy it gains in the process. However, there is still a problem. Material in orbit around a black hole has &lt;em&gt;less&lt;/em&gt; angular momentum the closer it orbits. To pass inside the event horizon requires the dark matter to lose angular momentum. Normal matter does this via an accretion disc that can transport angular momentum outwards by viscous torques, allowing matter to accrete. Dark matter has almost zero viscosity so this can't happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So building a supermassive black hole from a smaller seed would be difficult, but forming small black holes out of neutron stars might be easier. It has been proposed that a relative lack of pulsars observed towards our own Galactic centre &lt;em&gt;could&lt;/em&gt; be due to this process. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-09-29T22:20:16.677" LastActivityDate="2015-09-29T22:20:16.677" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11966" PostTypeId="2" ParentId="11956" CreationDate="2015-09-29T06:56:22.810" Score="2" Body="&lt;p&gt;The composition of the gas from which stars and their planetary systems form is reasonably well known. About 1-2% of this gas is in the form of chemical elements heavier than Helium (the so-called metallicity of the gas).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fraction of these &quot;metals&quot; - the iron, silicon, oxygen etc. is capable of forming dust and then accumulating to form &quot;rocky&quot; material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer your last question first, given the available reservoir it seems most unlikely that even 1% of the baryonic (normal) matter in the universe could be rocks, even if none of it were gathered into luminous stars and galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forming stars are surrounded by circumstellar material from which planets and other rocks form. Observations of young stars suggest these discs can be as massive as 10% of the stellar mass, but more usually 1% or less. This means that as a fraction of the whole formed star system the rocks will be at most 1% of 10% (ie 0.1%) of the stellar mass. For a Sun-like star that means there's less than 330 Earth masses of rocky material around it.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-09-30T06:22:03.593" LastActivityDate="2015-09-30T06:22:03.593" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11967" PostTypeId="2" ParentId="11958" CreationDate="2015-09-29T11:13:25.813" Score="9" Body="&lt;p&gt;I asked a somewhat similar question but just about the Earth.  &lt;a href=&quot;https://astronomy.stackexchange.com/questions/11243/how-has-the-earths-orbit-changed-over-hundreds-of-millions-or-billion-of-years&quot;&gt;Here&lt;/a&gt; and in my question there's some links that you might be interested in.   &lt;a href=&quot;http://www.space.com/28901-wandering-jupiter-oddball-solar-system.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jupiter&lt;/a&gt; for example is thought to have moved closer to the sun during the late heavy bombardment, then back outwards.   So I suspect that through the wormhole didn't get that point quite right.   Observations by &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler_%28spacecraft%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Kepler Spacecraft&lt;/a&gt; in the hunt for exoplanets suggests that a number of planets in other solar systems have moved closer to their suns over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, there's a few parts to this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;On the sun losing mass:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, it is, but only a little bit.  &lt;a href=&quot;http://solar-center.stanford.edu/FAQ/Qshrink.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;See Here&lt;/a&gt;.  Their math isn't complete as they don't factor in coronal mass ejections or take into account the sun's expected increase in luminosity, but the gist of their point remains true.   The sun is only losing a tiny percentage of it's mass.  The effect on planetary orbits is pretty small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sunlight also creates a pressure against the planets, but this effect, isn't very much due to the comparatively enormous mass of the planets compared to the light and coronal mass ejection material that hits the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tides&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tidal effects can work either way depending on the rate of rotation compared to the planet's orbital rotation.  If the star turns ahead of the planet, the tidal force should push the planet slowly outwards, like the Earth is currently doing to the Moon.   If the planet moves around the star faster than the star turns, the tidal effect is reversed and it slowly pulls the planet in.  At our Sun's current rate of rotation it's moving ahead of all the planets, so there should be a small tidal force that pushes the planets outwards, but very very slowly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Orbital debris and effective loss of angular momentum during planet formation.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the orbit is littered with dust and comets, asteroids and such, the sun can gain mass faster than it loses it and the planets can effectively slow down as they move through the dust or orbital debris with more oblique orbits.   This may be the explanation for hot Jupiters and so many observed planets very close to their suns in other solar systems.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Planets can affect each other.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/1551/does-the-gravity-of-the-planets-affect-the-orbit-of-other-planets-in-our-solar-s&quot;&gt;See here&lt;/a&gt; and a curious possibility &lt;a href=&quot;https://astronomy.stackexchange.com/questions/2383/can-mercury-hit-earth-or-mars-in-the-next-5-billion-years&quot;&gt;here&lt;/a&gt;.   Planet to planet effects are tiny unless they are in resonance where the effect can be amplified and become significant over time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that's clear enough.  I can tidy up and provide additional links for reference if needed.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-09-29T11:13:25.813" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11969" PostTypeId="1" AcceptedAnswerId="11970" CreationDate="2015-09-29T20:42:16.920" Score="7" ViewCount="7353" Body="&lt;p&gt;I mean in intrinsic detail. What would happen to our bodies besides &quot;burst into flames&quot;?&lt;/p&gt;&#xA;" OwnerUserId="9342" LastActivityDate="2015-09-30T15:09:55.403" Title="What would happen if we stepped on the Sun?" Tags="&lt;solar-system&gt;&lt;the-sun&gt;&lt;heat&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11970" PostTypeId="2" ParentId="11969" CreationDate="2015-09-29T20:44:41.683" Score="11" Body="&lt;p&gt;Well first thing's first: You would disintegrate. At the temperature of the Sun, most of the molecules that make up our bodies could not even survive, that is why we would not only fry and die, we would really disintegrate (all the molecules breaking apart, leaving only loose atoms). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's now pretend heat doesn't exist. This is what would happen. First, we must remember that there is no solid surface to the Sun just like there is no surface to Uranus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As soon as you reached the Sun itself, you would sink. The sun's density is less than 1 — and 1 is the density of water. So you would be sucked inside kind of. Except that there are convection currents. If we happen to be just above one, we might be kept close to the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, eventually, the eddies would entrain us sideways until we came to a &quot;downdraft&quot; that would take us downwards. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, let's talk about gamma rays now. Gamma rays are a form of light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even though heat is out of the way, all of these light rays are now trying to tear up whatever remaining molecules we have. This could trigger nuclear reactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, you (not me, I'm out of there) would cycle between two layers, in a convection cell moving you up and down, above and below the depth where the density of the gas is the same as the density of your body. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I guess there's one sure thing about all of this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're going to need some sunscreen.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2015-09-29T20:44:41.683" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11971" PostTypeId="2" ParentId="11957" CreationDate="2015-09-29T21:41:13.003" Score="1" Body="&lt;p&gt;The Variable stars that Hubble studied in M31 when he showed that it was a galaxy are among the most luminous. They include Var-A1, a luminous blue variable, and &lt;a href=&quot;http://arxiv.org/pdf/1403.6859.pdf&quot; rel=&quot;nofollow&quot;&gt;magnitude about 16.5&lt;/a&gt;. Var A-1 is one of the most luminous stars known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nasa has a &lt;a href=&quot;http://heasarc.gsfc.nasa.gov/W3Browse/star-catalog/m31stars.html&quot; rel=&quot;nofollow&quot;&gt;catalog of stars&lt;/a&gt; in the m31 field, The brightest star in this catalog is an &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=%401560657&amp;amp;Name=TYC%202788-276-1&amp;amp;submit=submit&quot; rel=&quot;nofollow&quot;&gt;11.4 magnitude&lt;/a&gt; star. But I think that database needs careful interpretation.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-09-29T21:41:13.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11972" PostTypeId="1" CreationDate="2015-09-29T23:52:59.803" Score="4" ViewCount="381" Body="&lt;p&gt;We need to figure if liquid water is present throughout a Mars year, and how long the planet has been at its current distance from Sun. That will give us the length of time that liquid water has been present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can then compare this with Earth's data to figure out how long life would have had to evolve and what state it may have reached. In particular would there have been enough time for humanoids to evolve?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the different composition of the atmosphere and surface gravity produce life that looks different to that on Earth today?&lt;/p&gt;&#xA;" OwnerUserId="9326" LastEditorUserId="4085" LastEditDate="2015-09-30T10:11:33.180" LastActivityDate="2015-09-30T19:09:24.547" Title="Does liquid water on Mars means life?" Tags="&lt;mars&gt;&lt;life&gt;&lt;water&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11973" PostTypeId="2" ParentId="11972" CreationDate="2015-09-30T05:26:26.197" Score="4" Body="&lt;p&gt;I don't think applying Earth data to Mars makes much sense, well, except perhaps at the very beginning.  The reason I say that is that life on Earth took a long time to get what we might consider &quot;interesting&quot;, for lack of a better word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1wQfT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1wQfT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ck12.org/book/CK-12-Biology/r23/section/1.2/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt; - and, OK, it's a children's book, but I like the picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 billion years ago - simple cells.&#xA;3.5 billion years ago - photosynthesis.&#xA;2 billion years ago - cells with Nuclei&#xA;1 billion years ago - bacteria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the Earth took 3 billion years to make - bacteria.   Granted it's only one planet as a sample, but if that's standard, then, arguably, most of the life in the solar system might be 1 celled and, you know, kind of boring to you and me, though I'm sure scientists would love to study 1 celled organisms from other planets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If planets need a few billion years before things like plants, fish, insects, etc, evolve, then many planets won't stay in the Goldilocks zone that long.  . . . or, option B, maybe things happened slower on earth than other planets, but for now, we just don't know.   Finding life on Mars or records of past life, if there was any, could help answer some of those questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars may have lost it's oceans fairly quickly, in less than a billion years.   That's not a long time for life to evolve, at least based on the Earth Model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, low gravity, probably not a big deal.   Animals and trees on a low-G planet grow grow taller and thinner/weaker - thinner bones, smaller muscles for example.   High-G, just the opposite.  Shorter and stronger with thicker bones.    Prior to lungs, Insect size is determined by Oxygen content in the atmosphere much more than gravity.   &lt;em&gt;(Fish-Fin size point removed as it probably wasn't correct).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably none of this happened on Mars cause I doubt life evolved that far but in theory on a low G planet you might get very large animals, perhaps the size of 10 story buildings, because size is a good evolutionary defense mechanism.  (Fun to think about, right?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thin Air, that's a bigger problem cause thin air leads to evaporating oceans and once the oceans evaporate, life would have a hard time continuing the evolutionary process   Life needs water, at least here on Earth.   Life can adapt to living with small amounts of water but life needs water.  In ice, primitive Life hibernates, but it doesn't continue to evolve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it's possible that Mars' thinning atmosphere coincided with it cooling off and it could have had oceans frozen at the top but liquid below that for some time, while heat from inside the planet maintained liquid water and life was able to survive in that.   This happened to the Earth, when the Earth was covered by ice during the snowball earth period and when Snowball Earth ended, the Cambrian explosion started.   The problem on Mars was, if Mars had a snowball with liquid water below the ice period, it never ended.   It never went back to liquid oceans and that's a problem for evolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's possible that snowball earth (limiting the sunlight and atmospheric gases available to life inside the ocean) forced bacteria to evolve, hunting for food rather than simply absorbing it via sunlight and air, but that's just speculation on my part but the timing was pretty consistent.  The Cambrian explosion followed snowball earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Humanoids on other planets (assuming we ever find any) would probably have characteristics adapted to that planet.  We stand upright because our ancestors found it helpful to run.   That might not be the case on other planets, perhaps that have thicker brush and no open planes.  That's just one example, there's probably others.   Our height likely developed out of the optimal height for hunting.  If we just were gatherers and trappers, not spear throwers we might be quite a bit smaller with shorter arms cause there's be no evolutionary advantage to being bigger or longer arms if we climbed trees to get away from predators.  You can clime trees higher if you're small for example and be safe that way.   Humanoids might have tails on other planets if they had a use for them, like, impressing the opposite sex with them or for balance for tree swinging/jumping.   There's a variety of possibilities that go way beyond gravity (which might just have an effect on body size and shape) or air pressure, which might affect things like wings for gliding, running (as thick air might reduce running speed) and lung size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On red dwarf planets for example, plants and people might be significantly different colors.  Plants might be blue or black, to absorb more red light, Animals, for example, might have colors into the Infra-red spectrum that we don't see.   One blue sun planets, plants could be red and animals might have colors into the UV spectrum.   (Some birds on earth have UV spectrum colors that we don't see but they do).  Thicker air pressure might affect sound as well, they might talk much quieter perhaps higher pitch and have smaller vocal cords and smaller ears, where as, thinner air, bigger animals might make thunderous yells when they communicate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a variety of possible factors. . . . hope I haven't gone too deep into speculation here, but speculation is fun sometimes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I don't believe Mars ever came close to any of the ideas I mentioned above.  I don't see how it could.  it wasn't habitable for long enough.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-09-30T19:09:24.547" LastActivityDate="2015-09-30T19:09:24.547" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11974" PostTypeId="1" CreationDate="2015-09-30T09:41:21.897" Score="3" ViewCount="79" Body="&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most planets have circular orbit&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What else?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="633" LastEditDate="2015-10-05T08:05:38.310" LastActivityDate="2015-10-05T08:05:38.310" Title="How special is the Solar system compared with other planetary systems?" Tags="&lt;solar-system&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2015-10-02T14:12:34.820" ContentLicense="CC BY-SA 3.0" />
  <row Id="11975" PostTypeId="2" ParentId="2572" CreationDate="2015-09-30T10:23:22.030" Score="0" Body="&lt;p&gt;The asteroid that destroyed the dinosaurs was equivalent to about a billion hiroshima bombs or a million &quot;Tsar Bombs&quot; (biggest atom bombs). The counter force of energy involved to defuse that kinetic energy is a complex dynamic, scientifically designed and emulated involving: deflection, disintegration and reducing speed of impact.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most dangerous thing for the earth is a concentrated short blast on the earth's crust which causes nuclear winters and runaway volcanic activity and sending mineral clouds into the sky as opposed to the ice that arrived through the atmosphere, and it's difficult to say beyond which point it's better to scorch the earth surface than to shock it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a very technical study. The energy is either released as heat or a shock wave or both, so if you wanted to reduce damage to the earth you would have to measure weather it would withstand a massive shock wave better than massive heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately the current SENTRY asteroid monitoring system and future asteroid safeguards will give enough time so that it's more a choice of deflecting the asteroid entirely than breaking it in such a way that it would all be still on course with the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A grain of dust is enough to deflect a billiard ball by a large span, given enough distance, so the most efficient use of energy is one of deflection.&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2015-09-30T10:30:41.093" LastActivityDate="2015-09-30T10:30:41.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11976" PostTypeId="2" ParentId="11960" CreationDate="2015-09-30T12:01:31.873" Score="10" Body="&lt;p&gt;As pointed out by Rob Jeffries, &lt;strong&gt;forming&lt;/strong&gt; a black hole (BH) from dark matter (DM) is impossible (&lt;em&gt;unless&lt;/em&gt; there is a [hypothetical] interaction by which dark-matter can lose energy that evades all detection). &lt;strong&gt;Accreting DM into an existing BH&lt;/strong&gt; is still unlikely (since DM cannot lose its excess energy and angular momentum as easily as gas), but not impossible and a small fraction of matter accreted into the supermassive BHs (SMBHs) at galactic centres was likely dark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, once matter has accreted into a BH, the information about its origins (dark matter or baryonic) is lost. Therefore, &lt;strong&gt;it makes no sense to talk about &lt;em&gt;dark matter black holes&lt;/em&gt;&lt;/strong&gt;: black holes don't have any property beyond their mass, spin, and charge (&lt;a href=&quot;https://en.wikipedia.org/wiki/No-hair_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;no-hair theorem&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This very property makes &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Primordial_black_hole&quot; rel=&quot;nofollow noreferrer&quot;&gt;primordial black holes&lt;/a&gt;&lt;/strong&gt; interesting as candidates for dark matter, since they would not violate constraints (for example from &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Bang_nucleosynthesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;big-bang nucleosynthesis&lt;/a&gt;) about the amount of baryonic matter. However, AFAIK, primordial black holes are out of favour as DM candidates (for reasons beyond the scope of this answer).&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2019-06-17T08:43:36.490" LastActivityDate="2019-06-17T08:43:36.490" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="11977" PostTypeId="2" ParentId="11969" CreationDate="2015-09-30T12:22:23.410" Score="3" Body="&lt;p&gt;Even if you could withstand the conditions (pressure, temperature, radiation, charged particles, magnetic field, gravitational crunch) on the 'surface' of the Sun, you &lt;strong&gt;cannot step onto&lt;/strong&gt; the Sun: it has &lt;strong&gt;no surface&lt;/strong&gt;. But why does it appear to have a sharp edge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun is a ball of hot plasma permeated with magnetic fields and photons (=electro-magnetic radiation). The photons are permanently absorbed and re-emitted, whereby transporting energy from the Solar depths (another transport process is convection). To very good approximation, the photons follow a &lt;a href=&quot;https://en.wikipedia.org/wiki/Black-body_radiation&quot; rel=&quot;nofollow&quot;&gt;black-body spectrum&lt;/a&gt; with the temperature equal to that of the ambient gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the image of the Sun that we perceive is made from those photons that managed to escape unscathed. The likelihood for a photon coming from a certain depth and escaping the Sun decreases exponentially with depth $d$ as $\mathrm{e}^{-d/\tau}$ with &lt;em&gt;optical depth&lt;/em&gt; $\tau$. Since $\tau\ll R_{\odot}$, the radius of the Sun, the Sun appears to have a sharp edge. The same also applies to gas planets: you cannot step onto them, but would sink.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2015-09-30T12:22:23.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11978" PostTypeId="1" AcceptedAnswerId="11980" CreationDate="2015-09-30T14:09:46.130" Score="1" ViewCount="8676" Body="&lt;p&gt;Let's assume it is possible for a spacecraft to travel at the speed of light (I've read the interstellar book by Kip Thorne, apparently this is theoretically possible if you swing around two black holes)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have watched the video where the traveling of a light particle is simulated and while planets are certainly big enough so you can adjust your route, what would happen if a rocket (or spaceship in general) would collide with an earth-like planet (i.e. a planet having a massive surface). Would we burst through the whole planet, make it only to a core, or just a few km? I know what happens when a meteor lands, but obviously they aren't traveling at the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And one similar another question - when a spaceship is traveling at the speed of light, is there a problem with collision with smaller objects? I think you can't avoid millions of rockets in outer space, it would surely damage the spaceship, wouldn't it?&lt;/p&gt;&#xA;" OwnerUserId="9349" LastActivityDate="2016-09-02T17:37:01.310" Title="What would happen if a rocket traveling at speed of light would collide with a planet (like earth)?" Tags="&lt;space&gt;&lt;speed&gt;&lt;interstellar-travel&gt;&lt;space-travel&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2015-10-09T17:37:40.973" ContentLicense="CC BY-SA 3.0" />
  <row Id="11980" PostTypeId="2" ParentId="11978" CreationDate="2015-09-30T15:08:41.067" Score="1" Body="&lt;p&gt;I don't know about a spaceship, but the XKCD guy wrote an interesting article about what would happen if Earth was hit by a solid asteroid, travelling at various different speeds:&#xA;&lt;a href=&quot;https://what-if.xkcd.com/20/&quot; rel=&quot;nofollow&quot;&gt;what-if question: Diamond&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ship would be a lot smaller than an asteroid, so I think the damage would be a lesser version of these descriptions.  (The most extreme case described is the whole planet being vapourised and even neighbouring planets being affected by the radiation.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As for the second part of the question - protecting the craft from minor collisions - I admit I have no idea how to work out the kinetic energy from collisions at such high speeds. But even a grain of sand would at least damage the front of the craft, and I guess some sort of thick shield would be needed to protect against erosion from interstellar dust...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even at normal orbital speeds, Space Shuttle windows have had minor damage as &lt;a href=&quot;http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19950019959.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; paper describes...&lt;/p&gt;&#xA;" OwnerUserId="7491" LastEditorUserId="7491" LastEditDate="2015-09-30T16:49:27.867" LastActivityDate="2015-09-30T16:49:27.867" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11982" PostTypeId="2" ParentId="11978" CreationDate="2015-09-30T18:35:53.000" Score="0" Body="&lt;p&gt;Given Earth's atmosphere, the added drag on the ship would make it nearly impossible to collide with the planet without losing momentum (if we're traveling at 99% the speed of light here). Assuming somehow the ship were to maintain speed, the heat and drag cause by the atmosphere at this speed would tear any known design to pieces, causing it to burn up and be destroyed before colliding with the surface (any pieces that fall to earth would be traveling slower than the speed of light). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this aside, estimating gamma at ~70.2 for the speed of light, per kilogram, a collision should release ~5.47 x 10^17 joules (or around 132 megatonnes). To give some perspective, Meteor Crater in Arizona hit with an estimated 10 megatonnes. This sounds like a huge amount of energy (it is) but wouldn't transfer into a spaceship having the mass to vaporize or even melt the earth. The blast that led to the extinction of the dinosaurs was estimated to be ~100 million megatonnes. So if the rocket were around 1 x 10^6 kg then a collision could hypothetically (in this unrealistic scenario) have the same climactic impacts.&lt;/p&gt;&#xA;" OwnerUserId="9343" LastEditorUserId="9343" LastEditDate="2016-09-02T17:37:01.310" LastActivityDate="2016-09-02T17:37:01.310" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11983" PostTypeId="1" CreationDate="2015-10-01T11:29:45.863" Score="6" ViewCount="125" Body="&lt;p&gt;I'm trying to do a brief literature review to find out what the best current models for population III stellar evolution are. I was hoping that someone with more expertise in the area could perhaps summarise or point me in the right direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/26271/how-can-a-population-iii-star-be-so-massive&quot;&gt;This post&lt;/a&gt; on physics.SE is a great intro, but it's now over 4 years old. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recent observations (see &lt;a href=&quot;http://arxiv.org/pdf/1504.01734.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.nytimes.com/2015/06/18/science/space/astronomers-report-finding-earliest-stars-that-enriched-cosmos.html?_r=0&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;) suggest hints of these kinds of stars at relatively low redshift (z ~ 6) - does this provide significant tension on current models? Have there been other observational hints?&lt;/p&gt;&#xA;" OwnerUserId="941" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-04-27T07:13:33.507" Title="What's the bleeding-edge model for how Population III stars are born and evolve?" Tags="&lt;star&gt;&lt;stellar-evolution&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11985" PostTypeId="1" CreationDate="2015-10-01T22:35:31.713" Score="1" ViewCount="80" Body="&lt;p&gt;You've seen it in all kinds of science fiction, be it books, games, movies, and such. Space is depicted all sorts of ways, with asteroids far too close together, nebula thick enough to hide in and much more. Yet for a writer that want's to describe space accurately, this is a problem. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;So I need stellar features and reasons why they would form/exist&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here's a rough list of what I'm talking about:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Thick asteroid style fields, averaging around 100 meters apart&lt;/li&gt;&#xA;&lt;li&gt;Nebula regions that are actually thick enough to obscure light, like clouds&lt;/li&gt;&#xA;&lt;li&gt;Dust cloud regions thick enough to obscure light&lt;/li&gt;&#xA;&lt;li&gt;Any other kind of cloud of material that's similar to the two above&lt;/li&gt;&#xA;&lt;li&gt;Naturally occurring fields of explosive gas or a physical material that does the same&lt;/li&gt;&#xA;&lt;li&gt;Radiation levels high enough to eat through ship plating&lt;/li&gt;&#xA;&lt;li&gt;Naturally occurring anomalies of any sort, such as black holes, rogue planets and stars&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;An example would be a protoplanetary disc. Could this allow for material in close proximity so you have rocks in orbit that are close together? What about clouds of material? Would a forming star would have a thick cloud of material around it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Anything naturally occurring or even unnaturally occurring is fine, so long as space magic or whatever isn't involved.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If I missed anything you can think of, please elaborate. This topic has been difficult for me to find information on.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extra question: Could a cloud or clouds of opaque gas exist, even for a short time in space? With or without other material present works. &lt;/p&gt;&#xA;" OwnerUserId="9360" LastEditorUserId="9360" LastEditDate="2015-10-02T16:37:01.413" LastActivityDate="2015-10-02T16:37:01.413" Title="Realistic space environments" Tags="&lt;space&gt;&lt;star-systems&gt;&lt;stellar-dynamics&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2015-10-04T13:49:58.033" ContentLicense="CC BY-SA 3.0" />
  <row Id="11986" PostTypeId="2" ParentId="11985" CreationDate="2015-10-02T00:06:29.307" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Thick asteroid style fields, averaging around 100 meters apart&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's a very high density. It's not any &quot;asteroid&quot; field that would occur naturally and would be stable for a long time. I can only think of a few environments that would be close to that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the ring around a big planet, like Saturn&lt;/li&gt;&#xA;&lt;li&gt;the debris field shortly after a major planetary collision, like the one that created the Moon. It would be a very dangerous place to be in, because not all fragments would move at the same speed; there would be quite an ongoing amount of collisions.&lt;/li&gt;&#xA;&lt;li&gt;maybe, and this is a very cautious maybe, the pile of material coalescing into a protoplanet&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All of the above would be short-lived, on a cosmic scale. It would be a coincidence that you're there to see it, and you'd probably have to explain the coincidence somehow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Asteroid fields&quot;, the way most people imagine them nowadays (very high density), are largely fiction, originating in the iconic space chase sequences in the Star Wars movies, and the like.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Nebula regions that are actually thick enough to obscure light, like&#xA;  clouds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On what scale? Quite a few nebulae obscure light, but we're talking on the scale of distance between two neighboring stars. There are some nebulae that glow from within but we don't quite see the star - that's an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're talking about blocking light on the scale of mere kilometers, that's not possible &quot;in outer space&quot;. Not for nebulae, not for dust clouds or anything else. That would be a density approaching the atmosphere of a planet. Consider the Earth - you can see just fine kilometers away through its atmosphere, unless fog gets in the way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Naturally occurring fields of explosive gas or a physical material&#xA;  that does the same&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even if you could find something like that, which is pretty much impossible, it would be highly unstable, and would eventually decay into the lower energy state - i.e. it would tend to explode. Such things might exist for the duration of a cosmic blink of an eye. It would be quite a coincidence that you could find something like that out there. A coincidence that would require some explanation, and very likely would not be a &quot;mere coincidence&quot;. I think that's called a plot point. ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such things just don't occur naturally.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Radiation levels high enough to eat through ship plating&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It would melt off, rather than being eaten through. Radiation tends to be absorbed; it doesn't tend to chip material away. And when it's absorbed, it yields its energy as heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there is no material that could shield nuclear radiation so well that it would melt off on one side, while blocking it to non-lethal levels on the other side. At least not within any reasonable thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, such conditions might be found near extreme places such as the polar jets of massive black holes - but only while the accretion disk is very active.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Astrophysical_jet&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Astrophysical_jet&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Naturally occurring anomalies of any sort, such as black holes, rogue&#xA;  planets and stars&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Black holes are pretty ordinary, not anomalous in any way. There are huge black holes in the center of galaxies, and regular-size blackholes sprinkled throughout.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure what you mean by &quot;rogue planets and stars&quot;. Planets not belonging to any star system are actually thought to be pretty common. Stars not belonging to any galaxy are also not extraordinary, but of course the distances between them would be gigantic, since any such star, by definition, would be alone in intergalactic space.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An example would be a protoplanetary disc. Could this allow for&#xA;  material in close proximity so you have rocks in orbit that are close&#xA;  together? What about clouds of material?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes and yes. But at that point it would evolve rapidly, coalescing into a planet. Not rapidly on the human scale, but rapidly on cosmic scale. To you, as a human being, it would seem quite stable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Would a forming star would have a thick cloud of material around it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The transition between a disperse, rarefied cloud of gas, and an actively burning star with a clearly defined &quot;surface&quot; is very rapid. However, for a very brief time (maybe thousands of years? not sure), you might find something with a dense center, surrounded by a thin cloud; but even then, density near the central area would increase rapidly. Perhaps you need to ask an actual expert in stellar evolution; there are models that show exactly how each phase looks like and how long it would last.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're looking for large expanses of gas or dust at densities comparable to Earth atmosphere, such a thing is simply not stable on a cosmic scale. It either dissipates if there's not enough gravity to hold it together, or it's part of an active process of collapse as a protostar or protoplanet.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-10-02T00:17:04.083" LastActivityDate="2015-10-02T00:17:04.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11987" PostTypeId="2" ParentId="11955" CreationDate="2015-10-02T04:16:48.797" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I have found an outdated study that says that some siderites contain as much as .001% of gold. Is it genuinely a viable mining research subject?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not yet, and not for quite a long time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gold is precious because humans only mine about 2.5 million kilograms of the stuff per year. Compare this to the huge amount of infrastructure (and huge cost) needed to retrieve the few kilograms of stuff that has been brought back to Earth from space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We haven't the foggiest idea how to mine minerals in space, let alone minerals that at best might have a concentration of ten parts per million (another way to look at your 0.001% figure). On the other hand, there are many ideas regarding taking advantage of untapped resources right here at home. For example, there are manganese nodules all over the sea floor. These nodules contain not just manganese , but also iron, nickel, and precious metals. Space mining cannot compete with those untapped local resources until they run out. (And then we'll find yet another untapped local resource.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mining stuff in space will be profitable only if there's a need for that stuff in space. If space mining becomes reality, the first stuff to be mined will be rather mundane volatiles such as water and methane rather than metals. Those volatiles are the low hanging fruit in space. How to extract and make use of those mundane volatiles is a good problem for our current generation of young adults. How to go beyond that is a problem for their grandchildren.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-10-02T04:16:48.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11988" PostTypeId="1" AcceptedAnswerId="11991" CreationDate="2015-10-02T21:08:42.817" Score="4" ViewCount="67" Body="&lt;p&gt;currently I am curious about proto-planetary nebulae. At first I was curious about how planets spin but I read in a couple of articles that they got their spin from when they were in a proto-planetary nebula and the momentum stayed because of conservation of momentum. But now a new question arises, where exactly do proto-planetary nebula's get their momentum from? I've searched this up on the internet but I haven't really gotten a clear response.&lt;/p&gt;&#xA;" OwnerUserId="7467" LastEditorUserId="4085" LastEditDate="2015-10-03T06:47:23.173" LastActivityDate="2015-10-03T06:59:26.033" Title="How do proto-planetary nebulae gain momentum?" Tags="&lt;planet&gt;&lt;planetary-formation&gt;&lt;nebula&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11989" PostTypeId="1" AcceptedAnswerId="11993" CreationDate="2015-10-02T22:19:44.937" Score="4" ViewCount="534" Body="&lt;p&gt;According to my understanding of &lt;a href=&quot;http://arxiv.org/abs/1311.7024&quot; rel=&quot;nofollow&quot;&gt;Soderblom et al. (2014)&lt;/a&gt;, lithium ages of stars are determined as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Determine lithium abundance from equivalent width measurement of Li$_{\mathrm{I}}$ transitions at 6708 angstroms for all stars in a coeval group.&lt;/li&gt;&#xA;&lt;li&gt;Plot all stars in group on color-magnitude diagram, with each star labeled as Li-rich or Li-poor.&lt;/li&gt;&#xA;&lt;li&gt;On plot, look for the boundary where the stellar population transitions from depleted to undepleted. That is the Lithium Depletion Boundary (LDB).&lt;/li&gt;&#xA;&lt;li&gt;Use the luminosity of the LDB and the age-LDB luminosity relationship to determine the age of the stars in the group.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What is the age-LDB luminosity relationship, and what are its physical origins? &lt;/p&gt;&#xA;" OwnerDisplayName="user5341" LastEditorUserId="15415" LastEditDate="2018-07-15T04:51:05.110" LastActivityDate="2018-07-15T04:51:05.110" Title="How is the Lithium Depletion Boundary used to determine the age of a stellar cluster?" Tags="&lt;star&gt;&lt;star-formation&gt;&lt;luminosity&gt;&lt;age&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11990" PostTypeId="2" ParentId="6333" CreationDate="2015-10-03T06:24:40.723" Score="7" Body="&lt;p&gt;The Sun currently accounts for more than 99.86% of the mass of the Solar System. Based on spectrographic estimates of the composition of the sun and it's centrifugal position and the mass of metals, you can deduce it also contains the most of all kinds of metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example to illustrate:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The milky way contains roughly 0.00011% of $\mathrm{Fe}$ (1.1ppm).&lt;/li&gt;&#xA;&lt;li&gt;The sun contains roughly 0.1% of $\mathrm{Fe}$, it contains about 333 earth masses of $\mathrm{Fe}$.&lt;/li&gt;&#xA;&lt;li&gt;The planets combined weigh about 500 earth masses.&lt;/li&gt;&#xA;&lt;li&gt;The sun contains only about 3% of our planet's weight of gold.&lt;/li&gt;&#xA;&lt;li&gt;The sun contains about 30% of our planet's weight of platinum.&lt;/li&gt;&#xA;&lt;li&gt;If you have time to do the maths, I think you will find &lt;a href=&quot;http://www.periodictable.com/Properties/A/SolarAbundance.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; the same is true for other metals.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The best thing to do is probably to compare graphs of elemental abundance of the earth, and of the sun, and multiply by weight, as the earth contains alot of heavy elements compared to further away planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is estimated that the mass of average new born star is between 1 and 10% of baryonic elements held in a surrounding dust cloud that later forms an accretion disc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's my understanding that 90+ percent of the accretion disk falls into the star, depending on the star's mass compared to that of the cloud, and the rest of the metals and other elements have time to condensate into ice and asteroids and planets. I am told that about 1 percent of our solar system's composition was originally held in the dust cloud, and 99% was in the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, the elements in our solar system measure no more than 500 earth masses in all the planets and the Oort cloud. And the sun is 330 000 earth masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that less than 0.15 of the solar system currently lies outside of the sun. The other 85 or 850% of the original accretion disk with all it's metals must have fallen into the sun, if we follow the rule that, 1 to 10% of a new born star is found in it's surrounding dust cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements#Abundance_of_elements_in_the_Earth&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is a list of elemental abundance for our galaxy and our solar system that is fairly interesting, I didn't find more precise figures.&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1891" LastEditDate="2017-12-30T22:07:27.037" LastActivityDate="2017-12-30T22:07:27.037" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11991" PostTypeId="2" ParentId="11988" CreationDate="2015-10-03T06:59:26.033" Score="2" Body="&lt;p&gt;The angular momentum comes from the chaotic/turbulent collapse of the parent molecular cloud. The collapse is turbulent because the initial conditions in the parent cloud are not spherically symmetric. This leave the proto-stellar clumps with non-zero angular momentum even if the parent cloud started with zero angular momentum. The sum of the angular momenta of all the material of the original cloud after fragmentation is still that of the cloud before fragmentation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some other/external triggering events for molecular cloud collapse can impart non-zero angular momentum to the cloud as a whole, but this is not necessary for the the individual fragments to end up with angular momentum.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-10-03T06:59:26.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11992" PostTypeId="2" ParentId="1911" CreationDate="2015-10-03T07:17:31.333" Score="4" Body="&lt;p&gt;If the sun was 10 cm wide, Alpha Centauri would be 3200 kilometers away. &lt;a href=&quot;http://www.wolframalpha.com/input/i=%281mm%2Fearth+diameter+%29+*alpha+centauri+distance&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sun was an orange, the termination shock, the extreme boundary of the solar system, where the interstellar winds are more consequent than solar wind, is about 1 km diameter. The different stars wouldnt tend to interfere with each other's termination shock boundary. you wouldn't even see a degree of change in earth temperature from another star passing near to the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the most dense parts of our galaxy are 3000 times more star dense than our neighborhood, we are talking about a cloud of oranges, each of which is separated by a kilometer, in the most dense scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most probably it's the dramatic collision of a cloud of oranges each separated by 3000 km, the Milky Way in total would be 10 million kilometers across, if the sun was an orange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wiki sais that it would be negligable probability of even two stars colliding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The collision of the galaxies doesn't change the challenge of transporting life away from our planet, compared to what it already is. we already have to cross the 32000km to a dozen nearby stars, if we humans measured 1.4 Angstroms tall... alpha centaur would be 3200 km walk away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Andromeda%E2%80%93Milky_Way_collision#Stellar_collisions&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Andromeda%E2%80%93Milky_Way_collision#Stellar_collisions&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- &lt;a href=&quot;http://www.wolframalpha.com/input/i=%281mm%2Fearth+diameter+%29+*alpha+centauri+distance&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/i=%281mm%2Fearth+diameter+%29+*alpha+centauri+distance&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2015-10-03T07:56:10.130" LastActivityDate="2015-10-03T07:56:10.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11993" PostTypeId="2" ParentId="11989" CreationDate="2015-10-03T16:38:16.623" Score="2" Body="&lt;p&gt;When low mass stars are very young, they are termed &lt;em&gt;pre main sequence&lt;/em&gt; (PMS) stars. These PMS stars have &lt;em&gt;larger&lt;/em&gt; radii than main sequence stars of the same mass, and energy transport in their interiors occurs primarily through convection. The convection ensures that the star is thoroughly mixed and chemically uniform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the PMS star radiates away its gravitational potential energy, it contracts. The virial theorem tells us that as it does so, its interior becomes hotter. Roughly speaking, the core temperature is proportional to $M/R$, where $M$ is the mass and $R$ the radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nuclear fusion of hydrogen will not commence until the core temperature reaches more than 10 million K, however there are other fusion reactions that become possible at lower temperatures - namely deuterium burning at around 1 million K and then lithium burning at around 3 million K. The latter reaction is not energetically important in the star's life because there is not much lithium in the star to begin with (about 1 part in a billion), however this lithium can be observed in the photosphere of the PMS star (via the 670.8 nm Li I resonant absorption line) and the convective mixing means whatever we see at the photosphere also represents the abundance at the core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Li burning reaction is extremely temperature dependent (like $T^{20}$ or thereabouts), so it turns on like a switch once the core reaches the appropriate temperature (e.g. see &lt;a href=&quot;http://arxiv.org/abs/astro-ph/9612155&quot; rel=&quot;nofollow&quot;&gt;Bildsten et al. 1997&lt;/a&gt;). The time it takes for a PMS star to reach this core temperature basically depends on its mass. More massive, and hence more luminous stars contract faster and reach the Li burning temperature quicker. Once they do so then the Li in the star is rapidly and thoroughly consumed by fusion. The relationship between the age at Li destruction and the luminosity of the PMS star at that time is the age-LDB luminosity relationship that you refer to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is that if you look at a bunch of stars in a cluster (assuming they all have the same age), then the more massive, more luminous PMS stars will have destroyed their Li, whilst the lower mass, lower luminosity stars will still contain their original Li content. The luminosity at the reasonably sharp transition between these two regimes is known as the LDB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LDB ages are arguably the most &lt;em&gt;accurate&lt;/em&gt; way to find the ages of stars in clusters. All age determinations are to some extent dependent on what physical ingredients are in stellar evolution models, but the sensitivity of the age-LDB luminosity relationship to various uncertainties is quite weak (e.g. &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0309461&quot; rel=&quot;nofollow&quot;&gt;Burke et al. 2004&lt;/a&gt;) - we basically understand the physics of a contracting, fully convective ball of gas quite well. The LDB ages can also be precise, because the very sharp turn-on of Li burning and its rapidity should lead to a sharp transition between stars with low luminosities that have Li and stars with only slightly higher luminosities that don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, in the last couple of years it has become apparent to us (and others) that there are some model ingredients that are not completely understood, namely the effects of dynamo generated magnetic fields and dark starspots. Both of these may lead to the suppression of heat transport, either throughout the star or just at the surface, slowing the PMS contraction so that at a given age, the PMS star has a cooler core temperature. This might delay the onset of Li burning and mean that the currently determined LDB ages are underestimates by 10-20% or so (e.g. &lt;a href=&quot;http://arxiv.org/abs/1410.0022&quot; rel=&quot;nofollow&quot;&gt;Jackson &amp;amp; Jeffries 2014&lt;/a&gt;; &lt;a href=&quot;http://arxiv.org/abs/1410.4238&quot; rel=&quot;nofollow&quot;&gt;Somers &amp;amp; Pinsonneault 2015&lt;/a&gt; ).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-10-03T16:38:16.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11994" PostTypeId="1" CreationDate="2015-10-03T19:06:54.357" Score="4" ViewCount="1172" Body="&lt;p&gt;I am trying to follow this guide and move to a programming language&#xA;but I have doubts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What means the strings A, B and C in this manual? &#xA;&lt;a href=&quot;http://aastro.org/wp-content/uploads/2010/11/Calculation-of-Local-Sidereal-Time.pdf&quot; rel=&quot;nofollow&quot;&gt;http://aastro.org/wp-content/uploads/2010/11/Calculation-of-Local-Sidereal-Time.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9372" LastActivityDate="2015-10-04T16:23:13.603" Title="Calculate Local Sidereal Time" Tags="&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11995" PostTypeId="2" ParentId="11994" CreationDate="2015-10-03T20:55:02.000" Score="0" Body="&lt;p&gt;Sidereal time is the time reckoned with respect to the stars rather than the sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We normally take the length of the day to be the 24 hours it takes the sun to go from mid-day to mid-day. But as the Earth is spinning on its axis and moving around the sun, the time it takes for a star to appear to move in a full circle once is slightly less: 23.9344 hours. This is called the sidereal day. Sidereal time is time reckoned by this &quot;day&quot; length. Consult &lt;a href=&quot;https://en.wikipedia.org/wiki/Sidereal_time&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, for more details&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three constants A, B and C in the linked document are chosen to convert a day, and and local time to sidereal time. They have no special meaning, but note that $C=1+\frac{1}{\rm number\ of\ days\ in\ year}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Implementation is straightforward. You will however need to know, or calcualate&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the day number and the decimal hour. But these can be easily found from the calendar date and time in hours minutes and seconds&lt;/li&gt;&#xA;&lt;li&gt;The difference between the the time zone and local mean time. This depends on location, and whether daylight saving time is in operation.&lt;/li&gt;&#xA;&lt;li&gt;that the algorithm returns a decimal hour. You may want to convert to hours minutes and seconds.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-10-03T20:55:02.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11997" PostTypeId="1" AcceptedAnswerId="11998" CreationDate="2015-10-03T22:51:05.817" Score="2" ViewCount="6751" Body="&lt;p&gt;The astronomical zodiac contains a bunch of constellations along the ecliptic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some sources say there are 13 constellations in the astronomical zodiac. Other sources claim there are 12. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to NASA (&lt;a href=&quot;http://spaceplace.nasa.gov/starfinder3/en/&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If we count all the traditional constellations, there are 13 in the&#xA;  zodiac, not 12. So here are the dates (give or take one day each year)&#xA;  when the Sun is between Earth and each of these 13 constellations.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There seems to be some confusion out there between zodiac &lt;em&gt;constellations&lt;/em&gt; (in astronomy) and zodiac &lt;em&gt;signs&lt;/em&gt; (in astrology). From the research I've done so far (including the quote above), it seems as though one of them (Ophiuchus) is a &quot;recent addition&quot;, possibly due to the Earth's precession since ancient times (&lt;a href=&quot;http://io9.com/5733004/your-zodiac-sign-may-have-changed-this-week&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What (officially?) are the &lt;em&gt;constellations&lt;/em&gt; of the astronomical zodiac? &#xA;Why do some sources list 12 and others list 13? &lt;/p&gt;&#xA;" OwnerUserId="7965" LastEditorUserId="7965" LastEditDate="2015-10-04T20:53:40.607" LastActivityDate="2017-09-09T19:57:21.527" Title="How many constellations in the Zodiac?" Tags="&lt;history&gt;&lt;constellations&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11998" PostTypeId="2" ParentId="11997" CreationDate="2015-10-04T00:26:30.777" Score="6" Body="&lt;p&gt;There are thirteen &lt;em&gt;modern&lt;/em&gt; constellations in the Zodiac. In modern astronomy, a constellation is a specific area of the celestial sphere as defined by the &lt;a href=&quot;https://en.wikipedia.org/wiki/International_Astronomical_Union&quot; rel=&quot;noreferrer&quot;&gt;International Astronomical Union&lt;/a&gt;. In total, there are 88 constellations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomy and Astrology are not the same thing. Astronomy is a science while Astrology is not. As such, I'll restrict myself to the historical and modern constellations of the Zodiac.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.britannica.com/topic/zodiac&quot; rel=&quot;noreferrer&quot;&gt;Encylopedia Brittanica&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Zodiac (is) a belt around the heavens extending 9° on either side of the ecliptic, the plane of the earth’s orbit and of the sun’s apparent annual path. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In historical astronomy, the zodiac is a circle of twelve 30° divisions of celestial longitude that are centered upon the ecliptic, the apparent path of the Sun across the celestial sphere over the course of the year. Historically, each of these divisions were called signs and named after a constellation: Sagittarius, Capricornus, Aquarius, Pisces, Aries, Taurus, Gemini, Cancer, Leo, Virgo, Libra, and Scorpius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1930, the International Astronomical Union &lt;a href=&quot;http://www.southastrodel.com/Page207.htm&quot; rel=&quot;noreferrer&quot;&gt;defined the boundaries between the various constellations&lt;/a&gt;, under &lt;a href=&quot;https://en.wikipedia.org/wiki/Eug%C3%A8ne_Joseph_Delporte&quot; rel=&quot;noreferrer&quot;&gt;Eugène Delporte &lt;/a&gt;, who,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... drew his boundaries along vertical lines of right ascension and horizontal parallels of declination. One governing principle was that all variable stars with an established designation would remain in that constellation, as requested by the IAU’s Variable Stars committee.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/T28AQ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/T28AQ.png&quot; alt=&quot;Constellations&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Constellations_ecliptic_equirectangular_plot.svg#/media/File:Constellations_ecliptic_equirectangular_plot.svg&quot; rel=&quot;noreferrer&quot;&gt;Constellations ecliptic equirectangular plot&lt;/a&gt;&quot; by &lt;a href=&quot;//commons.wikimedia.org/wiki/User:Cmglee&quot; rel=&quot;noreferrer&quot; title=&quot;User:Cmglee&quot;&gt;Cmglee&lt;/a&gt;, &lt;a href=&quot;//commons.wikimedia.org/wiki/User:Timwi&quot; rel=&quot;noreferrer&quot; title=&quot;User:Timwi&quot;&gt;Timwi&lt;/a&gt;, NASA - Own work, &lt;a href=&quot;http://svs.gsfc.nasa.gov/vis/a000000/a003500/a003572&quot; rel=&quot;noreferrer&quot;&gt;http://svs.gsfc.nasa.gov/vis/a000000/a003500/a003572&lt;/a&gt;. Licensed under Public Domain via &lt;a href=&quot;https://commons.wikimedia.org/wiki/&quot; rel=&quot;noreferrer&quot;&gt;Commons&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, the path of the ecliptic now officially passes through thirteen constellations: the twelve traditional 'zodiac constellations' and Ophiuchus (which was one of the 48 constellations listed by the 2nd-century astronomer Ptolemy), the bottom part of which interjects between Scorpio and Sagittarius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JR9l8.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JR9l8.gif&quot; alt=&quot;Zodiacal Constellations&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: journeytothestars.wordpress.com&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the 13 constellations of the Zodiac are Capricornus, Aquarius, Pisces, Aries, Taurus, Gemini, Cancer, Leo, Virgo, Libra, Scorpius, Sagittarius and Ophiuchus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As seen from the (first) figure, the ecliptic also touches the edge of the constellation Cetus, though it in not usually included in the Zodiacal constellations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note about Precession: Because the Earth in inclined (by $23.45^{\circ}$), it rotates like a top. This is called precession, which results in a shift in the position of the constellations relative to us on Earth. The result is that the 'Signs of the Zodiac' are off by about one month.&lt;/p&gt;&#xA;" OwnerUserId="7970" LastActivityDate="2015-10-04T00:26:30.777" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11999" PostTypeId="1" CreationDate="2015-10-04T03:27:40.777" Score="3" ViewCount="99" Body="&lt;p&gt;A few years ago I purchased a Celestron Nexstar 130 SLT computerised telescope thinking that it would be a perfect to get me into astronomy. The issue is the computer doesn't work at all and manually finding objects was a pain so I haven't used it in years. Now I want to get back into the hobby the 'proper' way by tracking objects with star maps and I want to get it right since photography and spectroscopy really excite me and that's what I want to end up doing. Despite all of this I can't seem to find a dovetail Equatorial mount that will fit my optical tube so now I feel I am stuck with a motorised mount that doesn't allow me to point my tube precisely and is really no fun. Is there any way for me to get into the hobby with my current optical tube using a different mount, use my computerised mount accurately enough for decent observation, or do I really have to buy a whole new optical tube just for the mount?&lt;/p&gt;&#xA;" OwnerUserId="9376" LastActivityDate="2016-04-27T14:04:15.187" Title="Moving from computerised to normal amateur telescope?" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12000" PostTypeId="1" AcceptedAnswerId="12006" CreationDate="2015-10-04T05:34:39.253" Score="2" ViewCount="127" Body="&lt;p&gt;According to &lt;a href=&quot;http://www.theaustralian.com.au/news/miranda-devine-perth-electrical-engineers-discovery-will-change-climate-change-debate/story-e6frg6n6-1227555674611&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;, the greatest cause of global warming would be from the waxing and waning of reflected radiation from the Sun. (Placing carbon emissions at 10-20%) So really, how much heat is generated from the process that stays on Earth? And would it really affect it that much?&lt;/p&gt;&#xA;" OwnerUserId="7926" LastActivityDate="2015-10-05T04:36:54.217" Title="How much heat is generated from waxing and waning of reflected radiation from the Sun?" Tags="&lt;radiation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12001" PostTypeId="1" AcceptedAnswerId="12032" CreationDate="2015-10-04T06:04:41.937" Score="6" ViewCount="883" Body="&lt;p&gt;Recently I purchase Chinese 8 inch reflector telescope from local store. When I observing starts and planets I've been having trouble getting really sharp image from this telescope. I observe Venus and Jupiter with several eyepieces but I got some &quot;&lt;em&gt;coma like&lt;/em&gt;&quot; effect all the time with this telescope. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To improve the image quality I decided to collimate this telescope with my laser collimator and cheshire collimators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before attaching the collimation device(s) I check primary and secondary mirror alignment through draw tube and it looks like this:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;a href=&quot;https://i.stack.imgur.com/mTgHs.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mTgHs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;But when I test this setup with laser collimator it show vast miss-alignment in both primary and secondary mirrors. After align both secondary and primary mirrors I've got view similar to this through the draw tube:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/urzZk.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/urzZk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also check above setup with Venus and few starts and I didn't notice any improvements of the image. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I try to collimate this telescope with both cheshire and laser collimator but still I didn't get any clear image with this telescope. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was the &quot;most likely&quot; to have issue in this telescope?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is problem in draw-tube assembly alignment can cause problem like this &lt;em&gt;or&lt;/em&gt; is this happens because of the problem in primary / secondary mirror of this telescope? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I test this telescope with several &lt;em&gt;branded&lt;/em&gt; and &lt;em&gt;unbranded&lt;/em&gt; eyepieces and I didn't notice any improvement. &lt;/p&gt;&#xA;" OwnerUserId="9377" LastEditorUserId="1378" LastEditDate="2015-10-06T20:24:28.460" LastActivityDate="2015-10-07T04:35:23.900" Title="Telescope collimation issue" Tags="&lt;telescope&gt;&lt;newtonian-telescope&gt;&lt;collimation&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12002" PostTypeId="1" CreationDate="2015-10-04T07:27:58.857" Score="1" ViewCount="9127" Body="&lt;p&gt;Are falling stars as we know only just space debris entering and burning up in the atmosphere and not a star at all? &lt;/p&gt;&#xA;" OwnerUserId="9326" LastActivityDate="2018-10-24T01:07:46.863" Title="Why can falling stars only be seen moving down towards horizon not up away from horizon?" Tags="&lt;space&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12003" PostTypeId="1" CreationDate="2015-10-04T10:37:47.960" Score="0" ViewCount="319" Body="&lt;p&gt;I have a problem explaining for my students how to get from the earth to the moon? Is there someone who can give a good explanation?&#xA;I Think these are words that might be included. Escape velocity, gravity, force, free fall, weightless, earth orbit, lunar orbit.(How do the spaceship get into the lunars orbit how do the spaceship get from lunars orbit to the surface of the moon?)&lt;/p&gt;&#xA;" OwnerUserId="9378" LastActivityDate="2015-10-04T19:28:47.667" Title="How to get from the earth to another planet/ the moon? What happens on the way?" Tags="&lt;orbit&gt;&lt;near-earth-object&gt;&lt;space-travel&gt;&lt;the-moon&gt;&lt;escape-velocity&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-10-05T11:47:23.843" ContentLicense="CC BY-SA 3.0" />
  <row Id="12004" PostTypeId="2" ParentId="12002" CreationDate="2015-10-04T12:35:52.067" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why can falling stars only be seen moving down towards horizon not up&#xA;  away from horizon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Had to think about this one a bit, cause, I'm not really sure what you mean.  I've watched meteors and they appear to streak across the sky, not exactly towards the horizon and only the occasional larger one will actually approach the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But meteors can approach the earth from pretty much any angle, but generally towards the earth.   Some, at the right angle, can skip off the atmosphere but in order to see them, they need to fall into the earth.   What this means is, at the birth of the visual travels across the sky, the meteor is usually further from the Earth's surface than it is at the end of it's visual travels across the sky.   Being closer to the Earth it's more likely to be nearer the horizon, so it's more likely a meteor will end closer to the horizon than it will begin, but it's not automatic.   A meteor could enter the sky 60-80 miles east of you for example and burn up above you and give the opposite impression, traveling horizon to center of the sky where it burns up.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are falling stars as we know only just space debris entering and&#xA;  burning up in the atmosphere and not a star at all?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This one is kind of silly, and could have been answered with a 2 second google search, but yes, falling stars aren't really falling stars or even pieces of stars.   They're mostly space dust, the majority being (I read once) remains of tails of comets though they can be debris from asteroid collisions or even older space dust that's been around for a long time.  Usually tiny grains like sand, sometimes larger, but because they travel so fast they glow white hot and can make the air ahead of them glow white hot and from our perspective they look a lot like stars - tiny white points of light.    Needless to say, they're not star like at all from any point of view other than how they appear to our eyes.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-10-04T12:35:52.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12005" PostTypeId="2" ParentId="12002" CreationDate="2015-10-04T14:08:10.360" Score="5" Body="&lt;p&gt;The falling star or shooting star has nothing to do with a star. If shooting stars were actual stars, we'll be burning in their atmosphere, rather than the other way around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a meteoroid, comet or (small) asteroid (or space debris in some cases) enters the Earth's atmosphere at high speed (typically in excess of 20 km/s), aerodynamic heating produces a streak of light, both from the glowing object and the trail of glowing particles that it leaves in its wake. This is called a meteor or shooting star (or a falling star). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the meteors travel in a (approximately) straight path. We usually see only the large ones travel down towards the horizon while the small ones are burnt up long before that (giving the impression of coming &lt;em&gt;from&lt;/em&gt; the horizon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vbnOf.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vbnOf.gif&quot; alt=&quot;Meteor path&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image from &lt;em&gt;Meteors? Even More Remarkable&lt;/em&gt; by Bruce Maccabee at brumac.8k.com&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the meteor path in the image above. it can be seen that,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The observer will see the meteor only after it passes the initial altitude $H_{i}$. If he sees the meteor at this point, and it burns up in the atmosphere, the impression is that the meteor comes from the horizon.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the observer notices the meteor after it passes the initial height and tracks it through its travel in the atmosphere, till is exits at the final altitude $H_{f}$, it will appear as if the meteor is moving down towards the horizon.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Another factor is the thickness of the atmosphere itself. The thickness of the atmosphere increases in path of the meteor from $H_{i}$ to $H_{m}$, and decreases from there to $H_{f}$. As the 'glowing' of the shooting star is directly dependent on the density of the atmosphere, its visibility trails of near the observer's horizon. This is another reason for the meteor looking like its falling towards the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be seen clearly in case of meteor showers which appear to radiate from a point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/29k4W.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/29k4W.jpg&quot; alt=&quot;Meteor shower&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image from &lt;em&gt;Lyrids Meteor Shower 2014&lt;/em&gt; by Steve Owens at darkskydiary.wordpress.com&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the meteors appear to come away from the (radiant, which is above the) horizon.&lt;/p&gt;&#xA;" OwnerUserId="7970" LastActivityDate="2015-10-04T14:08:10.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12006" PostTypeId="2" ParentId="12000" CreationDate="2015-10-04T14:43:16.167" Score="3" Body="&lt;p&gt;The article doesn't go into specifics and appears, not to be written by Dr. Evans at all, but I'll pull some quotes. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When it is completed his work will be published as two scientific&#xA;  papers. Both papers are undergoing peer review&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;He has been summarising his results in a series of blog posts on his&#xA;  wife Jo Nova’s blog for climate sceptics.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;He is about half way through his series, with blog post 8, “Applying&#xA;  the Stefan-Boltzmann Law to Earth”, published on Friday.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(footnote, I'm guessing in Australia, summarizing and skeptics are spelled differently than in the US, cause that's a copy-paste)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've only read one of his summaries, and in Mathematics, you have to look at the details, which he's not provided, so, he's basically saying little more than &quot;this is true, here's generally why, I've done the math, please take my word for it, I'll publish the numbers later&quot;.   Now Michael Mann is also presenting non peer reviewed work to Paris, so, what's good for the goose is good for the gander I suppose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm just a guy who likes science, but my understanding is that ideas have been presented before they are finished or peer reviewed fairly often.   Einstein did this in fact regarding general relativity and another scientist actually beat him to publishing the theory (though, the other scientist was gracious about it and gave Einstein full credit), so, I don't think it's necessarily bad to present a summary prior to peer review.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is, however, unusual to post summaries of an idea on a blog saying &quot;I figured it out, the majority of research on this subject is wrong&quot;.   Dr. Evans is saying he has proof, but he's acting like a junk science blogger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dr. Evans &quot;trust me I did the math&quot; claim kind of requires that we look &#xA;at his track record, and his track record isn't very strong, though he will say, he's being attacked by the establishment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to this &lt;a href=&quot;https://denierlist.wordpress.com/2013/02/26/dr-david-evans/&quot; rel=&quot;nofollow&quot;&gt;site&lt;/a&gt;, he hasn't published anything peer-reviewed since the 1980s and he's on a their &lt;a href=&quot;https://watchingthedeniers.wordpress.com/climate-sceptic-conspiracies/profile-dr-david-evans/&quot; rel=&quot;nofollow&quot;&gt;Climate Denier List&lt;/a&gt;.  He's also on Skeptical Science's &lt;a href=&quot;https://www.skepticalscience.com/misinformers.php&quot; rel=&quot;nofollow&quot;&gt;Climate Misinformers List&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here's a &lt;a href=&quot;http://www.skepticalscience.com/David_Evans_arg.htm&quot; rel=&quot;nofollow&quot;&gt;list&lt;/a&gt; of debunked claims he's made (and if this list is accurate, he's not a scientist at all, just a guy on a fishing expedition.   No good scientist would agree with pretty much every counter argument against climate change cause that's not how the scientific method works.  You can disagree with something, that's fine, but to agree with every counter argument - that's silly.  &lt;a href=&quot;http://www.skepticalscience.com/david-evans-understanding-goes-cold.html&quot; rel=&quot;nofollow&quot;&gt;Here's&lt;/a&gt; another more detailed explanation of what he's gotten wrong, from 2011.    Evans seems to be in this debate to disagree with it much more than he's in it to do scientific research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's pretty much impossible to make a true scientific argument against Evans &quot;proof&quot; without his specifics, and, that kind of proof/disproof can get a little long and complicated, but for now, disproving him is impossible.   But should we listen to him?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says above he's published 8 blogs related to his recent research.   I'm not going to dig up all 8, but here's the most recent one. &lt;a href=&quot;http://joannenova.com.au/2015/10/new-science-8-applying-the-stefan-boltzmann-law-to-earth/&quot; rel=&quot;nofollow&quot;&gt;Applying the Stefan-Boltzmann Law to Earth&lt;/a&gt;.   Now, I'm just a layman, but even I can see problems with his argument here (and he should too given that he's an engineer with a PhD).   The Stefan-Boltzmann law is an approximation.   It's a physical model to calculate radiation into space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with his approach is, the best way to measure how much heat/energy leaves the earth by radiation is to measure it directly, by satellite.   The amount of energy that radiates from the Earth into space varies with temperature, snow cover, cloud cover, even humidity, and probably 1 or 2 other things I'm overlooking.   If you try to calculate this energy leaving the earth into space by playing with with the Stefan-Boltzmann Law instead of relying on direct measurements, you're allowing yourself a lot of fudge factors and inviting a far greater error than direct measurements would give you.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the quoted article, let me pull out an example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dr Evans has a theory: solar activity. What he calls “albedo&#xA;  modulation”, the waxing and waning of reflected radiation from the&#xA;  Sun, is the likely cause of global warming.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;OK, so, which is it, solar activity or albedo modulation, cause they're not the same thing.  The first takes place on the sun, the 2nd, the earth.   This paragraph makes no sense to me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;He predicts global temperatures, which have plateaued, will begin to&#xA;  cool significantly, beginning between 2017 and 2021. The cooling will&#xA;  be about 0.3C in the 2020s. Some scientists have even forecast a mini&#xA;  ice age in the 2030s.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, this paragraph is particularly devious.   El Nino's tend to warm the earth, La Nina's cool it.  The effect is temporary and not huge, but enough to cause yearly variation.   A strong El Nino drove the big spike in global temperature for 1998 and we're in an &lt;a href=&quot;http://www.weather.com/news/climate/news/strong-el-nino-noaa-update-september2015&quot; rel=&quot;nofollow&quot;&gt;El Nino now&lt;/a&gt; (edited my answer, since 2014 they've been talking about entering an El Nino, I gather it's officially started now).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We had more La Nina years than El Nino 2006-2013 with the only small El Nino coinciding with 2010, which set records for temperature.  A lot of the hiatus in warming that is often talked about is related to there being only 1 small El Nino over 7 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Predicting 2017 as the time when the cooling will &quot;begin&quot; is devious because that could be around the time the El Nino has ended and the oceans could switch back to a La Nina (which usually follows El Nino).   This will create a temporary cooling for a year or two, which he, no doubt, will take credit for if it happens.   Now, he also predicts 2021 which could go either way and he gives an amount, but that doesn't change the fact that he's making a prediction and hoping the El Nino of 2015 will end and make it prediction look good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Real global warming or cooling can't be measured in 1 year anyway, unless, maybe, if it's ocean current and the occasional mega-volcano adjusted - then, maybe you can get some measure of warming/cooling based on one year, but it's still only one year.  That's a really really short period to make any predictions on and not something I'd trust very far at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and on the &quot;scientists have predicted a mini ice age in 2030&quot;, that's not actually true.   There was a study on sun-spots and they predicted that we could see a sun-spot low period around 2030, perhaps similar to the Maunder Minimum that may have caused the mini ice age, but the scientists who predicted that were very clear that they were not predicting a new mini ice age, they said the effect would be smaller than the effect of CO2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wired.co.uk/news/archive/2015-07/13/mini-ice-age-earth-sunspots&quot; rel=&quot;nofollow&quot;&gt;Here's&lt;/a&gt; the mini ice age prediction, which a few people made (but not the scientists who did the research).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.slate.com/blogs/future_tense/2015/07/13/sunspot_cycles_won_t_cause_a_mini_ice_age_by_2030.html&quot; rel=&quot;nofollow&quot;&gt;Here's&lt;/a&gt; an article that explains why it isn't true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there's a lot of bad and a handful of false statements in that article you quoted, which, granted, wasn't written by Dr. Evans himself, but still, it's hard for me to take it seriously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until he publishes his results, he can't be proved or disproved but based on what I've read, I find it hard to take him seriously.   My hunch is, he's not trying to reach scientists at all, but he's trying to reach his target audience.  Those who question climate change and he gives them a name and an alternate argument that they can stand on.   An argument doesn't have to be correct, it only needs to sound correct and with that, you can usually convince a percentage of people to agree with you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure how much that helps, but that's my take and I went through and tried to clean up my long answer a bit.   If you'll forgive me, it reminds me of the the old joke.   How can you tell Dr. David Evans is lying?   He's talking or writing.  :-)&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-10-05T04:36:54.217" LastActivityDate="2015-10-05T04:36:54.217" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12007" PostTypeId="2" ParentId="12003" CreationDate="2015-10-04T15:43:16.553" Score="2" Body="&lt;p&gt;Short answer is easy.   Long answer, there's lots of maths involved, and I'm out of practice with my maths, but there's a few basic parts that your students should be able to follow.   It's a little confusing, not too bad. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can ignore the Earth's motion around the sun for the most part but not the Earth's escape velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with &lt;a href=&quot;https://en.wikipedia.org/wiki/Newton&amp;#39;s_cannonball&quot; rel=&quot;nofollow noreferrer&quot;&gt;Newton's cannonball thought experiment&lt;/a&gt;.  That will get you into space, and that's an example of orbital velocity.  With a fast enough launch, the cannonball enters into and stays in Earth's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd, explain the difference between &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/vesc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;escape velocity and orbital velocity&lt;/a&gt;.   Escape velocity is greater than orbital velocity and the cannonball that orbits the earth hasn't really escaped the earth, it's still in the Earth's gravitational sphere of influence.   You need escape velocity (or very close to it) to get to the moon, about 11.2 KM/s.  You can get away with a bit less cause the Moon hasn't fully escaped the Earth either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronauts can't get shot out of a cannon at high speeds cause it's too many G forces, so what rocket launches do is accelerate over a few minutes.  (you could probably look up the precise amount of G forces), but within G forces that humans can tolerate.   If they experience 4Gs, they're accelerating the equivalent of 3Gs (or 29.4 meters per second), at least at first, as they move away from the Earth, the gravitational acceleration towards the earth drops and after only a few minutes they can shut down the launch and drift towards the moon.   Most of the trip to the moon is done in this way, just drifting through space.   You don't want to go too fast cause you'd just need to slow down as you approach the moon again and the moon has no atmosphere to assist the deceleration like the Earth does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ships momentum traves through the combined gravitational field of, primarily, the Earth, Sun and Moon, and the Maths get pretty complicated, so the the launch needs to be quite precise.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon also orbits the Earth at about 2,300 miles per hour and that orbital speed has to be matched as well as, since the rocket will often enter a partial orbit around the moon, that orbital velocity around the Earth needs to be matched too.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once this is done, landing on the moon requires a simple enough deceleration to have a gentle enough touchdown to not harm any equipment.  (this probably isn't as easy as it sounds).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An orbit around the moon isn't strictly necessary but a partial orbit was used to assist landing on the desired spot.   A partial orbit around the Earth was also used in the launch to the moon.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They accelerate away from the Earth in 2 different times, first to leave the earth and get into Earth orbit.  That's the big one, then from a different spot (and not even an hour later) to accelerate out of Earth's orbit towards the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See diagram.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/3j01n.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3j01n.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://taylorspectrum.wordpress.com/2011/12/19/apollo-11/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least, that's a pretty basic summary.   I welcome correction if I got anything wrong or missed anything important.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-10-04T15:43:16.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12008" PostTypeId="2" ParentId="11994" CreationDate="2015-10-04T16:23:13.603" Score="1" Body="&lt;p&gt;A highly authoritative source is &lt;em&gt;Explanatory Supplement to the Astronomical Almanac&lt;/em&gt; 3rd ed., Urban &amp;amp; Seidelmann, 2013.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chapter 1 gives the introductory explanations, without all the refinements if you want millisecond accuracy. Page 12 explains that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;local sidreal time = Greenwich sidereal time + east longitude&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This book assumes the reader is able to convert between angle notation and time notation whenever necessary.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If accuracy of 0.1 second is sufficient, the formulas given by &lt;a href=&quot;http://aa.usno.navy.mil/faq/docs/GAST.php&quot; rel=&quot;nofollow&quot; title=&quot;the US Naval Observatory&quot;&gt;the US Naval Observatory&lt;/a&gt; should do the job. Note that for subsecond accuracy you will have to determine UT1, which can be up to 0.9 second different from the commonly available UTC. You will have to look up the difference each time you do the calculation because the difference cannot be predicted more than about 6 months into the future.&lt;/p&gt;&#xA;" OwnerUserId="7385" LastActivityDate="2015-10-04T16:23:13.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12009" PostTypeId="1" AcceptedAnswerId="12010" CreationDate="2015-10-04T18:03:37.277" Score="3" ViewCount="237" Body="&lt;p&gt;This is related to another question I asked. Might we actually be observing dark energy when we do experiments that verify the Casimir effect?&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="2153" LastEditDate="2015-10-04T18:58:18.703" LastActivityDate="2015-10-04T19:32:47.920" Title="Relationship between the Casimir force and dark energy" Tags="&lt;dark-energy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12010" PostTypeId="2" ParentId="12009" CreationDate="2015-10-04T18:55:50.210" Score="5" Body="&lt;p&gt;There are two problems that arise here. One obvious problem is that &lt;a href=&quot;http://en.wikipedia.org/wiki/Casimir_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Casimir effect&lt;/a&gt; is attractive&lt;sup&gt;1&lt;/sup&gt;, while &lt;a href=&quot;http://en.wikipedia.org/wiki/Dark_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;dark energy&lt;/a&gt; is repulsive. The other problem is one of scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.dwc.knaw.nl/DL/publications/PU00018547.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Casimir (1948)&lt;/a&gt; shows that, between two particles (instead of the oft-cited case of two plates)&#xA;$$\delta E \propto R^{-7}$$&#xA;That's an &lt;em&gt;enormous&lt;/em&gt; drop-off. On large scales, this should be negligible. Dark energy, however, is clearly non-negligible on these scales, while its effects are unobservable on small scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Casimir effect is certainly one manifestation of &lt;a href=&quot;http://en.wikipedia.org/wiki/Vacuum_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;vacuum energy&lt;/a&gt;, which may also be the cause of dark energy (in some theories), but that doesn't mean that dark energy is an example of the Casimir effect, or vice versa.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1 There &lt;em&gt;are&lt;/em&gt; exceptions, as &lt;a href=&quot;https://astronomy.stackexchange.com/questions/12009/relationship-between-the-casimir-force-and-dark-energy#comment16695_12010&quot;&gt;pointed out by Stan Liou&lt;/a&gt;, which arise under certain conditions, but they wouldn't play a part here.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-10-04T19:32:47.920" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12012" PostTypeId="2" ParentId="2427" CreationDate="2015-10-04T19:31:15.747" Score="-1" Body="&lt;p&gt;Technically the true way to find the period between alignment of all 8 planets is to find the LCM of all 8 of their year lengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LCM (88, 225, 365, 687, 4333, 10759, 30685, 60189) = 814252949520007202031000.&#xA;I understand that this is a rough estimate since these are rounded to the nearest integer, but it gives a good idea of the number of days it would take.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;814252949520007202031000/365 = 2230829998684951238441. That's how many years.&lt;/p&gt;&#xA;" OwnerUserId="9384" LastActivityDate="2015-10-04T19:31:15.747" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12013" PostTypeId="2" ParentId="11674" CreationDate="2015-10-04T20:25:52.080" Score="1" Body="&lt;p&gt;The iss is orbiting around the earth, but even as it does so, the earth is rotating, and the orbital rate of the iss is not aligned to the  rotation of the earth. The effect is that the track that the iss takes changes with each orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly the Iss move back about 10 degrees with each orbit, though this amount varies, as the Iss is not in a stable orbit wrt the earth, over long periods of tim. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-10-04T20:25:52.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12014" PostTypeId="1" CreationDate="2015-10-04T21:26:18.120" Score="4" ViewCount="390" Body="&lt;p&gt;What places on Pluto and Charon are facing each other? Are Pluto and Charon perfectly tidally locked, or is there a slow rotation over time? And how long after formation should they have become tidally locked? Before or after the late heavy bombardment?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT to add the image series of Pluto below, found at &lt;a href=&quot;http://www.universetoday.com/123533/see-daytime-views-of-pluto-and-charons-rotation/#comments&quot; rel=&quot;nofollow noreferrer&quot;&gt;University Today&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nFDmv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nFDmv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-11-21T13:55:18.470" LastActivityDate="2015-11-21T13:55:18.470" Title="What places on Pluto and Charon are facing each other?" Tags="&lt;pluto&gt;&lt;tidal-forces&gt;&lt;charon&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12015" PostTypeId="1" CreationDate="2015-10-04T21:40:44.017" Score="8" ViewCount="279" Body="&lt;p&gt;Pluto and Charon seem to have surprisingly young surfaces, considering that resurfacing events were expected to be rare for them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could the explanation be that they actually &lt;em&gt;are&lt;/em&gt; young? Formed later than the Solar system at some other star and then ejected and captured? Or that they are ancient but have extra-Solar origin and have little cratering because they have spent billions of years as a vagabond planet system in empty interstellar space? &lt;a href=&quot;http://www.universetoday.com/121637/stealing-sedna/&quot;&gt;Sedna&lt;/a&gt; is believed to be &lt;a href=&quot;http://arxiv.org/abs/1506.03105&quot;&gt;captured&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="2365" LastEditDate="2015-11-06T20:30:23.850" LastActivityDate="2015-11-06T20:30:23.850" Title="Could Pluto and Charon have extra-Solar origin?" Tags="&lt;pluto&gt;&lt;surface&gt;&lt;rogue-planet&gt;&lt;charon&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12016" PostTypeId="2" ParentId="12015" CreationDate="2015-10-05T04:50:12.907" Score="1" Body="&lt;p&gt;(Edit, I think my original conclusion here was wrong, having read up about it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it seems probable that Pluto and Charon were formed by collision, I gather it's unlikely that this collision was so recent as to explain their lack of craters and young surface.   The Collision, by most articles I've read, happened when the solar system was young, not in the last 100 million years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it's possible that none of this is certain, but I retract that part of my original post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, the creation of Charon by giant collision remains likely  &lt;a href=&quot;http://www.space.com/23136-pluto-moons-orbits-giant-impact.html&quot; rel=&quot;nofollow&quot;&gt;See here&lt;/a&gt;, and &lt;a href=&quot;http://www.researchgate.net/publication/7281381_A_giant_impact_origin_for_Pluto&amp;#39;s_small_moons_and_satellite_multiplicity_in_the_Kuiper_belt&quot; rel=&quot;nofollow&quot;&gt;here.&lt;/a&gt;   That would suggest formation within our solar-system, not from outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pluto/Charon are also among a fairly common orbital region called &lt;a href=&quot;https://en.wikipedia.org/wiki/Trans-Neptunian_object&quot; rel=&quot;nofollow&quot;&gt;Trans-Neptunian objects&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Charon_%28moon%29#Orbit&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; they suggest the collision that formed Charon happened 4.5 billion years ago.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sedna (love the story, by the way), has a very elongated and distant orbit which seems more likely with capture of an object.   I agree with everything James Kilfinger said.   Rogue planet/Rogue object capture may happen from time to time, but the Rogue would need to get quite close to the sun for the sun to have sufficient gravitation to achieve the capture.   Highly elongated orbits would seem likely for solar-system orbital capture.  Pluto is only slightly elongated and not a good candidate.   More massive suns should have significantly greater capture numbers, but that's kinda obvious I suppose.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-10-06T00:38:44.380" LastActivityDate="2015-10-06T00:38:44.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12017" PostTypeId="2" ParentId="2427" CreationDate="2015-10-05T08:37:04.473" Score="3" Body="&lt;p&gt;The correct answer is '&lt;strong&gt;never&lt;/strong&gt;', for several reasons. &lt;strong&gt;First&lt;/strong&gt;, as pointed out in Florin's comment, the planet's orbits are not co-planar and hence cannot possibly align, even if each planet could be placed arbitrarily in its orbital plane. &lt;strong&gt;Second&lt;/strong&gt;, even pure &lt;em&gt;radial alignment&lt;/em&gt; never happens because the planet's periods are incommensurable -- their ratios are not rational numbers. &lt;strong&gt;Finally&lt;/strong&gt;, the planets' orbits evolve over timescales of millions of years, mainly due to their mutual gravitational pull. This evolution is (weakly) chaotic and thus unpredictable for very long times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://astronomy.stackexchange.com/a/2428/633&quot;&gt;wrong answer by &#xA;harogaston&lt;/a&gt; essentially approximates the orbital periods by the nearest commensurable numbers, yielding a very long time (though he got that wrong by a factor of merely $10^{16}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A much more interesting question (and perhaps the one you were actually interested in) is how often do the 8 planets &lt;strong&gt;nearly align radially&lt;/strong&gt;. Here, '&lt;em&gt;nearly&lt;/em&gt;' could simply mean '&lt;em&gt;to within $10^\circ$ as seen from the Sun&lt;/em&gt;'. &#xA;At such an occasion, the mutual gravitational pull of the planets will align and hence result in stronger orbital changes than the average.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-10-05T08:37:04.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12018" PostTypeId="1" AcceptedAnswerId="12020" CreationDate="2015-10-05T09:31:22.373" Score="2" ViewCount="697" Body="&lt;p&gt;What is the exact definition of a &quot;continuum-subtracted spectrum&quot;, such that one finds in SDSS spectra? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://classic.sdss.org/dr7/dm/flatFiles/spSpec.html#specmask&quot; rel=&quot;nofollow&quot;&gt;http://classic.sdss.org/dr7/dm/flatFiles/spSpec.html#specmask&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a definition somewhere? Thanks&lt;/p&gt;&#xA;" OwnerUserId="9171" LastEditorUserId="2365" LastEditDate="2016-02-14T20:44:09.890" LastActivityDate="2016-02-14T20:44:09.890" Title="Definition question: continuum-subtracted spectrum" Tags="&lt;spectra&gt;&lt;sky-survey&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12019" PostTypeId="1" AcceptedAnswerId="12023" CreationDate="2015-10-05T11:45:48.887" Score="4" ViewCount="79" Body="&lt;p&gt;K2 seems to become an even better mission than what a continuation of Kepler would have been, at least to some astronomers. By what means is differential star rotation being dealt with? The rotational period of the Sun is 45% longer at its poles than at its equator. Together with the unknown orientation of the star's rotational axis, can this really be sorted out from Doppler broadening of spectral lines? Is a star's rotational period defined as that of its equator or as some average across its surface?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="2531" LastEditDate="2015-10-05T21:20:33.863" LastActivityDate="2015-10-05T21:30:30.937" Title="How does gyrochronology deal with differential rotation and axial tilt?" Tags="&lt;rotation&gt;&lt;stellar-astrophysics&gt;&lt;kepler&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12020" PostTypeId="2" ParentId="12018" CreationDate="2015-10-05T11:51:35.333" Score="2" Body="&lt;p&gt;A spectrum of, say, a star or a galaxy, consists of wavelengths emitted from various physical processes. In the case of the star, there is the blackbody spectrum that reflects the temperature of all of its gas, whether it is hydrogen, helium, or metals. On top of this, there are &lt;em&gt;lines&lt;/em&gt;, i.e. emission from certain atomic transitions that are excited and then de-excite emitting photons in a very narrow wavelength interval. The blackbody spectrum is said to comprise the continuum of the total spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of galaxies, the continuum consists of the combined spectrum of many different stars' different blackbody spectra, infrared light from dust, etc. On top of this, there are lines, e.g. from nebular processes such as my favorite line Lyman $\alpha$, which is emitted when hard UV radiation from hot O and B stars ionize their surrounding neutral hydrogen, which subsequently recombines emitting mostly Lyman $\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want to measure the total flux in a given line, we are usually only interested in the flux that comes from a particular physical process, and not the part of the flux that comes from the &quot;underlying&quot; sources which contribute to the continuum. Hence, we need to &lt;em&gt;subtract the continuum&lt;/em&gt;. This is usually done by fitting some functional form to the continuum (e.g. a straight line or a second order polynomial), substract this fit from the spectrum, and finally integrate from one side of the line to the other in the resulting &quot;continuum-subtracted spectrum&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the top figure below, the total spectrum's flux density$^\dagger$ is in black, and the fit is the red lines. Underneath is drawn the continuum-subtracted spectrum, and a few dashed green lines show some lines that rise above the continuum. The integrated line flux is indicated by light red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/z04nX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z04nX.png&quot; alt=&quot;specs&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;Note that many authors use the term &quot;flux&quot; where it should actually be &quot;flux density&quot;. Integrating flux &lt;em&gt;density&lt;/em&gt; in $\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}\,\mathrm{Å}^{-1}$ over wavelength in Å$^{-1}$ gives &lt;em&gt;flux&lt;/em&gt; in $\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}$.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-10-05T15:08:52.557" LastActivityDate="2015-10-05T15:08:52.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12021" PostTypeId="2" ParentId="12015" CreationDate="2015-10-05T17:30:37.373" Score="6" Body="&lt;p&gt;The surface of Pluto is not just young, but &lt;em&gt;very&lt;/em&gt; young, and also differentiated. The 'heart', Sputnik Planum, may be currently active, with flowing glaciers, and nitrogen snow. This area may have be resurfaced during Pluto's perihelion. Your theory cannot account for the very young age of the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next capture of a rogue planet is not easy. It would need Pluto to interact with the outer planets, to slow it down to enter solar orbit (otherwise a rogue planet would just pass through the solar system and come out the other side) A captured rogue would be unlikely to have the relatively low eccentricty of pluto. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, while we don't know how common rogue planets are, they are probably not common, and given the massive distances between stars, the chances of any rogue just happening to have passed anywhere near the solar system in the last billion years in the right direction to be captured to Pluto's orbit is very low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conclusion. Pluto formed around the sun. A capture theory does not explain the observed young surface of Pluto, and modelling suggests a capture would be very unusual.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-10-05T17:30:37.373" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12022" PostTypeId="2" ParentId="11674" CreationDate="2015-10-05T18:36:15.083" Score="0" Body="&lt;p&gt;To a first-order approximation, the ISS (as all satellites do), orbits in a fixed plane around the planet.  As the orbital period is about 90 minutes, and observational conditions that allow viewing usually last longer than that, you might expect that if you could see it at one time, you would be able to in the future.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two major effects that change the relationship of the orbit with respect to twilight locations on the earth.  The first is the orbit of the earth around the sun.  If the plane of the orbit were fixed, this would change where the orbit encountered twilight with a yearly cycle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stronger effect though is precession.  At the altitude of the ISS, the non-spherical mass of the earth causes the plane of rotation to move (about 5 degrees a day).  So it takes a bit more than 2 months to precess all the way back around.  Depending on your latitude, you should have two good periods of viewing during the full precession.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that for periods of weeks or so, the ISS will be passing overhead during unobservable times (middle of the day/middle of the night).  As the orbit continues to precess, it will begin to pass overhead closer to twilight and you'll have opportunities for viewing based on the specifics of the orbit.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-10-05T18:36:15.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12023" PostTypeId="2" ParentId="12019" CreationDate="2015-10-05T19:34:27.037" Score="3" Body="&lt;p&gt;Gyrochronology uses the rotation &lt;em&gt;periods&lt;/em&gt; of stars, caused by rotational modulation by starspots to estimate a stellar age. In the absence of differential rotation, the rotation axis inclination has &lt;em&gt;no effect&lt;/em&gt; on this measurement. The doppler broadening of spectral lines plays no role in gyrochronology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rotation period of a star is just that. It is the period that is measured from the light curve. So it is some sort of emission-weighted average rotation period that depends on the distribution of starspot latitudes, the stellar inclination and the limb darkening. Sometimes there is evidence of differential rotation because the rotation period of the star changes from epoch to epoch. This can be used to calibrate the uncertainty in an individual rotation period measurement. Differential rotation &lt;em&gt;is&lt;/em&gt; a source of uncertainty in gyrochronology, however its effect is limited because (i) at least on the Sun, and maybe other stars too, spots are confined to relatively low latitudes; (ii) spots occur at a range of latitudes over this latitude range. It is likely that younger, faster rotating stars do have spots at higher latitudes, but these stars also appear to have much weaker differential rotation than the Sun ($\Delta P/P =0.05 P^{0.3}$ according to &lt;a href=&quot;http://adsabs.harvard.edu/abs/1996ApJ...466..384D&quot; rel=&quot;nofollow&quot;&gt;Donahue et al. 1996&lt;/a&gt;, where $\Delta P$ characterises the range of measured periods for a typical star of period $P$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect of differential rotation on gyrochronology is discussed at length by &lt;a href=&quot;http://arxiv.org/abs/1203.1618&quot; rel=&quot;nofollow&quot;&gt;Epstein &amp;amp; Pinsonneault (2014)&lt;/a&gt; and rather more concisely by &lt;a href=&quot;http://arxiv.org/abs/1404.7156&quot; rel=&quot;nofollow&quot;&gt;Jeffries (2014)&lt;/a&gt;. It appears that differential rotation probably does set the precision limit of a measured period to be around 10 per cent (though I would argue that it is smaller for younger stars). Because the rotation rate of a star roughly follows the Skumanich-type spin-down law $\Omega \propto t^{-1/2}$, then this leads to an age uncertainty of around 20% in older stars. For younger stars it is the initial spreads in rotation rates which are more important than differential rotation as a source of error.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-10-05T21:30:30.937" LastActivityDate="2015-10-05T21:30:30.937" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12024" PostTypeId="1" AcceptedAnswerId="12025" CreationDate="2015-10-06T08:41:00.343" Score="4" ViewCount="301" Body="&lt;p&gt;Just a quick (and hopefully simple) question on the distribution of stars in the Milky Way and in globular clusters. I was wondering &lt;strong&gt;why&lt;/strong&gt; it &lt;em&gt;IS NOT&lt;/em&gt; possible to describe the distributions of stars in both these systems by analogy to the distribution of air molecules in a room. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are we talking of the more random motion of air molecules as well as the relatively large number of collisions between air molecules...something which we wouldn't see in a galaxy. I'm assuming also that air molecules do not have the same central gravtational potential...and therefore do not move relative to some fixed point. That is to say, that it is much more random? Any advise on this point would be greatly appreciated. &lt;/p&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2015-10-06T11:50:02.053" Title="Distribution of Stars in Milky Way and globular cluster analogy" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;galactic-dynamics&gt;&lt;globular-clusters&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12025" PostTypeId="2" ParentId="12024" CreationDate="2015-10-06T11:50:02.053" Score="5" Body="&lt;p&gt;Particles in a gas approximate to point-like objects that interact roughly elastically through short-range forces when they collide, but are otherwise non-interacting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars interact gravitationally over long ranges, occasionally with each other, but always with the overall gravitational potential of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes people do talk thermodynamically about star clusters. You can discuss the &quot;temperature&quot; of stars when you are referring to the velocity dispersion in a cluster. The concept of heating or cooling a cluster also has some merit.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-10-06T11:50:02.053" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12026" PostTypeId="1" AcceptedAnswerId="12029" CreationDate="2015-10-06T12:50:28.843" Score="4" ViewCount="387" Body="&lt;p&gt;According to the website &lt;a href=&quot;http://aa.usno.navy.mil/faq/docs/GAST.php&quot; rel=&quot;nofollow&quot;&gt;http://aa.usno.navy.mil/faq/docs/GAST.php&lt;/a&gt;, a low-precision way of calculating the GMST is via the following formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GMST = 18.697374558 + 24.06570982441908 D &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The website doesn't actually state where the numbers come from. Another website tries to do this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;To compute of sidereal time with low precision use of the low precision formula: GMST = 18.697374558 + 24.06570982441908 D, where 18.697… is sidereal time of the reference time 2000-01-01 at 0 UT, 24.0657… is a ratio of synodic and sidereal periods of Earth and D is days (and its fractions) since the reference time.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I don't understand what this 'ratio of synodic and sidereal periods' means. I know what the sidereal and synodic days are and their values, but how do we use those to end up with the number 24.0657...? If anyone could help me out that'd be much appreciated!&lt;/p&gt;&#xA;" OwnerUserId="9392" LastActivityDate="2015-10-07T22:12:29.733" Title="Low precision GMST formula clarification" Tags="&lt;declination&gt;&lt;ascension&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12027" PostTypeId="2" ParentId="11937" CreationDate="2015-10-06T13:23:53.557" Score="1" Body="&lt;p&gt;CCDs are optimized for a certain wavelength range, and for a certain expected signal level. In astronomy, we tend to be short of light, so here we almost always want them to be as sensitive as possible (an exception may be observations of the Sun, which I don't know much about). But for instance, the Nordic Optical Telescope has a CCD which is optimized for blue wavelengths, but has quite a lot &lt;a href=&quot;https://www.astro.uni-bonn.de/theli/gui/aboutbackground.html#fringing&quot; rel=&quot;nofollow&quot;&gt;fringing&lt;/a&gt; in the near-infrared. And further out in the IR, CCDs aren't even used, instead using something which are just called &quot;detectors&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, whether the CCD is used for imaging (photometry and astrometry) or spectroscopy does not have anything to do with the CCD; it's just a matter of inserting a grism or not. I'm not really into the instruments of Gaia, but I assume that differences in the CCDs are due to different wavelength regions being probed. There may be a difference in how its sub-parts (it's actually an array of CCDs) are positioned (for instance, for spectroscopy in principle you don't need a large field of view, but can do with a long array rather than a more square one), but the design of the individual CCDs are the same.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2015-10-06T13:23:53.557" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12029" PostTypeId="2" ParentId="12026" CreationDate="2015-10-06T17:36:49.957" Score="2" Body="&lt;p&gt;The number 18.697374558 was the value of GMST (in hours) at noon UT on 1 January, 2000. The number 24.06570982441908 reflects the fact that it takes slightly less than 24 hours (23.934469591898 hours to be precise) for a star to the appear to rotate by 360 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a consequence of the fact that the Earth orbits the Sun. Our 24 hour day is based on the time between two consecutive solar noons. Because the Earth moves a bit along its orbit, the Earth has to rotate by more than 360 degrees (with respect to the stars) to see the Sun rotate by 360 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to look at it: The 23.934469591898 hour long sidereal day means that a solar day is 24.06570982441908 hours long expressed in sidereal hours.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2015-10-07T22:12:29.733" LastActivityDate="2015-10-07T22:12:29.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12030" PostTypeId="2" ParentId="12014" CreationDate="2015-10-06T19:43:31.860" Score="3" Body="&lt;p&gt;As &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/0/07/New_Horizons_Pluto_Encounter_%28Eyes_on_the_Solar_System%29.webm&quot; rel=&quot;nofollow&quot;&gt;New Horizons past through the Pluto system&lt;/a&gt;, it travelled so that the spacecraft was on the far side of Pluto, as seen from Charon. It did this so both Pluto and Charon were roughly in the same direction, so the space craft would not have to rotate 180 degrees between imaging Pluto to imaging Charon.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It means that in images of Charon, the face that you can see is roughly the face that faces Pluto, However the it is the far side of Pluto (the side with the &quot;whale&quot; and the &quot;dots&quot; that faces Charon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tidal locking is pretty stable. See our own moon: it is locked, and has been for a long time. There is likely to be some wobble, just like the libations of the moon, but no overall rotation of either Pluto or Charon.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-10-06T19:43:31.860" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12031" PostTypeId="2" ParentId="10072" CreationDate="2015-10-06T19:50:12.880" Score="0" Body="&lt;p&gt;The orbit of the ISS does vary. Drag from the very thin atmosphere slowly causes the orbit to lower, and the speed to increase. So the ISS needs to be boosted about a couple of times a month.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its position can be predicted, at least in the short term, and websites such as &lt;a href=&quot;http://www.heavens-above.com/&quot; rel=&quot;nofollow&quot;&gt;Heavens Above&lt;/a&gt; publish current position, predictions, and orbital information.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-10-06T19:50:12.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12032" PostTypeId="2" ParentId="12001" CreationDate="2015-10-06T20:05:16.150" Score="2" Body="&lt;p&gt;First off - good job for paying attention to collimation. For a newtonian telescope, it is as important as changing the oil on your car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would help to know a little bit more about the telescope. What make and model? What is the focal length?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saw your comment below. It's an f/4 parabolic mirror, you're always going to have coma at the outer edge. But the center should be aberration-free if the scope is collimated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beware - it's an inexpensive scope, so depending on your luck some residual aberrations may remain even after perfect collimation. But that doesn't mean you should stop trying to collimate it. On the contrary, the better the collimation the better the performance, always.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I observe Venus and Jupiter with several eyepieces but I got some &quot;coma like&quot; effect all the time with this telescope.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the image in the center is fairly good, but you get &quot;coma like&quot; effects at the edge, then that is indeed coma (perhaps mixed with other aberrations). It is normal at focal ratios of f/5 ... f/4 or faster. If the focal ratio is around f/6 ... f/8 or slower, coma is not so easily visible, or not visible at all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But when I test this setup with laser collimator it show vast miss-alignment in both primary and secondary mirrors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is a very common issue with cheap laser &quot;collimators&quot; that the collimator itself is miscollimated. The laser only helps if it is very precisely centered, otherwise it is misleading. Unfortunately, many, many commercial lasers are woefully off-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plug the laser into the focuser. Tighten the screw only a little bit, so that the laser is sitting there semi-snugly, but you can still turn it around its own axis (rotate it in place in the focuser). Fire up the laser. Now gently turn the laser around, while watching the laser spot on the primary. What does the spot do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the laser is centered, the spot should never move as you rotate the laser in place. If the spot moves even a little bit (try to avoid rocking the laser tube sideways, as that would falsify the test), if the spot describes a circle on the primary as you rotate the laser in place, throw that laser in the garbage can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Ok, some lasers can be adjusted, provided they have the 2 or 3 adjustment screws on the collar, on the lateral part of the laser tube. Sometimes the screws are hidden under a plastic or metal ring. If that's the case, try and rescue the laser by carefully adjusting those screws. Otherwise dispose of the laser, or destroy it, since it's misleading.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other things to try&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See if you can get better results with the Cheshire only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try and make a very simple collimation cap. It's basically a plastic cap with a tiny hole in the exact center, that's all. The procedure for using it is explained here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=YAVGcGEBmCE&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=YAVGcGEBmCE&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more detailed procedure is explained here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.garyseronik.com/?q=node/169&quot; rel=&quot;nofollow&quot;&gt;http://www.garyseronik.com/?q=node/169&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're patient, you could collimate by the stars. It's complex and time-consuming, but it's guaranteed to be 100% accurate. It is always a good idea to use a quick star test at the end of whatever other collimation procedure you use - think of it as a reality check. The laser may or may not work properly, but the stars never lie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astrophoto.fr/collim.html&quot; rel=&quot;nofollow&quot;&gt;http://www.astrophoto.fr/collim.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Laser collimators can be very good and very quick - provided they are centered very precisely from factory. I use Howie Glatter's collimators, but they are not cheap. If you could borrow a Glatter collimator it would show you what's really going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://collimator.com/&quot; rel=&quot;nofollow&quot;&gt;http://collimator.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is your primary center-dotted with good precision? In many cases, mass-produced telescopes come from factory with an off-center dot on the primary, which is as misleading as a bad laser. Even the best laser will not work properly if the primary is not precisely center-dotted. But you can always re-center the dot with good precision. See here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.garyseronik.com/?q=node/168&quot; rel=&quot;nofollow&quot;&gt;http://www.garyseronik.com/?q=node/168&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Always use a donut shape to center the primary, instead of a solid dot. The donut allows you to use laser collimators, unlike the solid dot markers.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More thoughts on collimation below. These articles are written by very experienced amateur astronomers and telescope makers. Highly recommended to read these before you begin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.garyseronik.com/?q=node/165&quot; rel=&quot;nofollow&quot;&gt;http://www.garyseronik.com/?q=node/165&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://web.telia.com/~u41105032/myths/myths.htm&quot; rel=&quot;nofollow&quot;&gt;http://web.telia.com/~u41105032/myths/myths.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try something simple (like the collimation cap) before you open up the scope and do &quot;major surgery&quot; on it. Then do a star test. Does it look pretty good on the stars? Then you're probably doing something right. The stars are looking pretty bad no matter what? Then perhaps you need to open up the scope, re-center the primary, check if the mirrors are sitting properly in their cells, etc.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Final comments&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use too much magnification, the image will not look very sharp, no matter what. Stay within 100x ... 200x while testing. It is pretty rare that atmospheric conditions and/or the quality of mass-produced optics allow good performance above 250x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Always check the results in the center of the image. With most mass-produced optics, performance degrades at the edge, and there's not much you can do about it (other than replacing them with better optics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before you do anything, take the scope outside and let it &quot;breathe&quot; for 30 ... 60 minutes. After it reaches ambient temperature, its performance might get better. More details here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.garyseronik.com/?q=node/55&quot; rel=&quot;nofollow&quot;&gt;http://www.garyseronik.com/?q=node/55&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-10-07T04:35:23.900" LastActivityDate="2015-10-07T04:35:23.900" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12033" PostTypeId="1" AcceptedAnswerId="12046" CreationDate="2015-10-06T20:52:51.123" Score="4" ViewCount="604" Body="&lt;p&gt;Just a question I am having trouble understanding. I have the Schechter luminosity function for galaxies, given as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Phi(L)dL=\Phi_{0}\left({\frac{L}{L_{\star}}}\right)^{\alpha}e^{-\frac{L}{L_{\star}}}\frac{dL}{L_{\star}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to consider the case when $\alpha=-1$. And then show that the average luminosity of a galaxy is exactly $L_{\star}$. Could somebody explain how I could go about doing this and perhaps a hint or some part of a setup would be excellent. I really need to understand this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another part, which is related to the above question, is asking me to explain why the total luminosity is a finite number, whereas the total number of galaxies diverges. By this does it mean that the total number is infinite? Any extra comments on this would also be really appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2091" LastEditorUserId="9301" LastEditDate="2016-06-29T04:39:17.103" LastActivityDate="2016-06-29T04:39:17.103" Title="Luminosity Schechter function for galaxies" Tags="&lt;galaxy&gt;&lt;luminosity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12034" PostTypeId="1" AcceptedAnswerId="12774" CreationDate="2015-10-07T01:24:08.463" Score="8" ViewCount="516" Body="&lt;p&gt;Some cartoon simulations (&lt;a href=&quot;https://youtu.be/r_m7VgVSBoM?t=1863&quot;&gt;an example&lt;/a&gt;) of an object (assume here a large moon with Earth-like density) which crosses the Roche limit of a much more massive giant planet, illustrate a circle suddenly disintegrating and forming a ring. But such a process occurs gradually over geological time scales. I doubt that there would be a single bad day when the whole moon disintegrates (as in &lt;em&gt;-Oh, there Mt. Everest broke off and flew away!&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wouldn't the tidal tugging cause volcanism and gradually melt the moon as its (somewhat eccentric) orbit spirals inwards across the limit during millions of years?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would the melting moon deform? Would it really assume an oval shape, elongated towards the primary planet? Since the near side wants to orbit faster than the far side, wouldn't that cause the moon to spin at an accelerated rate, even if it initially was tidally locked? Would the melting, deforming and spinning save it from disintegrating further to within the Roche limit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some examples of Roche limit crossings I know of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Comet Shoemaker-Levy 9 may have broken up suddenly because it had&#xA;such a high velocity relative to Jupiter.&lt;/li&gt;&#xA;&lt;li&gt;Phobos will cross its Roche limit to Mars in ~50 million years. It&#xA;will hardly be a sudden process. I suppose its very low density and&#xA;mass won't allow any volcanism and melting, though.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/1307.4080&quot;&gt;KOI1843.03&lt;/a&gt;, an exoplanet candidate with a density of ~7 g/cm³,&#xA;is already inside the Roche limits for its less dense compounds which&#xA;it has shed (as I interpret it).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-12-06T04:59:42.147" Title="How does gradual crossing over of the Roche limit transform a planet or moon?" Tags="&lt;exoplanet&gt;&lt;tidal-forces&gt;&lt;roche-limit&gt;&lt;phobos&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12035" PostTypeId="1" CreationDate="2015-10-07T08:15:05.857" Score="4" ViewCount="79" Body="&lt;p&gt;Does anyone have any idea were I can get a dataset of stars in the milky way galaxy, their mass and their planets and distance of planets from their star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried various catalogs of NASA. I need something specific. I am planning to run some data analytics on them. So it needs to be in a format I can use preferably excel or data.&lt;/p&gt;&#xA;" OwnerUserId="9399" LastEditorUserId="7991" LastEditDate="2015-10-07T12:29:11.680" LastActivityDate="2015-10-07T12:29:11.680" Title="dataset of stars in the milky way galaxy" Tags="&lt;star&gt;&lt;planet&gt;&lt;exoplanet&gt;&lt;milky-way&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2015-10-07T12:29:34.363" ContentLicense="CC BY-SA 3.0" />
  <row Id="12036" PostTypeId="2" ParentId="2427" CreationDate="2015-10-07T13:18:47.410" Score="1" Body="&lt;p&gt;Any estimate of the common period of more than two planets (i.e., after how much time do they approximately align in heliocentric longitude again?) depends very strongly on how much deviation from perfect alignment is acceptable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the period of planet $i$ is $P_i$, and if the acceptable deviation &lt;em&gt;in time&lt;/em&gt; is $b$ (in the same units as $P_i$), then the combined period $P$ of all $n$ planets is approximately $$P \approx \frac{\prod_i P_i}{b^{n-1}}$$ so reducing the acceptable deviation by a factor of 10 means increasing the common period by a factor of $10^{n-1}$, which for 8 planets is a factor of 10,000,000.  So, it is meaningless to quote a common period if you don't also specify how much deviation was acceptable.  When the acceptable deviation declines to 0 (to achieve &quot;perfect alignment&quot;), then the common period increases to infinity.  This corresponds to several commenters' statements that there is no common period because the periods are not commensurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the planets' periods listed by harogaston, $\prod_i P_i \approx 1.35\times10^6$ when the $P_i$ are measured in Julian years of 365.25 days each, so the common period in years is approximately $$P \approx \frac{1.35\times10^6}{b^7}$$ if $b$ is measured in years as well.  If the periods are approximated to the nearest day, then $b \approx 0.00274$ years and $P \approx 1.2\times10^{24}$ years.  If the periods are approximated to the nearest 0.01 day, then $b \approx 2.74\times10^{-5}$ and $P \approx 1.2\times10^{38}$ years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivation of the above formula is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Approximate the planets' periods by multiples of a base unit $b$: $P_i \approx p_i b$ where $p_i$ is a whole number.  Then the common period is at most equal to the product of all $p_i$.  That product is still measured in units of $b$; we must multiply by $b$ to go back to the original units.  So, the common period is approximately $$P \approx b \prod_i p_i \approx b \prod_i \frac{P_i}{b} = b \frac{\prod_i P_i}{b^n} = \frac{\prod_i P_i}{b^{n-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above derivation doesn't take into account that the $p_i$ might have common factors so that the alignment occurs sooner than $\prod_i p_i$ suggests.  However, whether or not any two $p_i$ have common factors depends strongly on the chosen base period $b$, so it is effectively a random variable and does not affect the global dependence of $P$ on $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you express the acceptable deviation in terms of &lt;em&gt;angle&lt;/em&gt; rather than &lt;em&gt;time&lt;/em&gt;, then I expect you'll get answers that depend on the size of the acceptable deviation as strongly as for the above formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://aa.quae.nl/en/reken/periode.html&quot; rel=&quot;nofollow&quot;&gt;http://aa.quae.nl/en/reken/periode.html&lt;/a&gt; for a graph of $P$ as a function of $b$ for all planets including Pluto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an estimate with acceptable deviation in terms of &lt;em&gt;angle&lt;/em&gt;.  We&#xA;want all planets to be within a range of longitude of width $δ$&#xA;centered on the longitude of the first planet; the longitude of the&#xA;first planet is free.  We assume that all planets move in the same&#xA;direction in coplanar circular orbits around the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the planets' periods are not commensurate, all combinations of&#xA;longitudes of the planets occur with the same probability.  The&#xA;probability $q_i$ that at some specific moment of time the longitude&#xA;of planet $i &amp;gt; 1$ is within the segment of width $δ$ centered on the&#xA;longitude of planet 1 is equal to $$q_i = \frac{δ}{360°}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability $q$ that planets 2 through $n$ are all within that&#xA;same segment of longitude centered on planet 1 is then $$q =&#xA;\prod_{i=2}^n q_i = \left( \frac{δ}{360°} \right)^{n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To translate that probability to an average period, we need to&#xA;estimate for how much time all planets are aligned (to within $δ$)&#xA;each time they are all aligned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first two planets to lose their mutual alignment are the fastest&#xA;and slowest of the planets.  If their synodic period is $P_*$, then&#xA;they'll be in alignment for an interval $$A = P_* \frac{δ}{360°}$$ and&#xA;then out of alignment for some time before coming into alignment&#xA;again.  So, each alignment of all planets lasts about an interval $A$,&#xA;and all of those alignments together cover a fraction $q$ of all time.&#xA;If the average period after which another alignment of all planets&#xA;occurs is $P$, then we must have $qP = A$, so $$P = \frac{A}{q} = P_*&#xA;\left( \frac{360°}{δ} \right)^{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are only two planets, then $P = P_*$ regardless of $δ$, which is as expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are many planets, then the fastest planet is a lot faster&#xA;than the slowest one, so then $P_*$ is very nearly equal to the&#xA;orbital period of the fastest planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, too, the estimate for the average time between successive&#xA;alignments is &lt;em&gt;very&lt;/em&gt; sensitive to the chosen deviation limit (if there&#xA;are more than two planets involved), so it is meaningless to quote&#xA;such a combined period if you don't also mention what deviation was&#xA;allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also important to remember that (if there are more than two&#xA;planets) these (near-)alignments of all of them do not occur at&#xA;regular intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's plug in some numbers.  If you want all 8 planets to be&#xA;aligned to within 1 degree of longitude, then the average time between&#xA;two such alignments is roughly equal to $P = 360^6 = 2.2×10^{15}$&#xA;orbits of the fastest planet.  For the Solar System, Mercury is the&#xA;fastest planet, with a period of about 0.241 years, so then the&#xA;average time between two alignments of all 8 planets to within 1&#xA;degree of longitude is about $5×10^{14}$ years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are satisfied already with an alignment to within 10 degrees&#xA;of longitude, then the average period between two such alignments is&#xA;roughly equal to $P = 36^6 = 2.2×10^9$ orbits of Mercury, which is&#xA;about 500 million years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the best alignment that we can expect during the coming 1000&#xA;years?  1000 years are about 4150 orbits of Mercury, so $(360°/δ)^6&#xA;\approx 4150$, so $δ \approx 90°$.  In an interval of 1000 years&#xA;chosen at random, there is on average one alignment of all 8 planets&#xA;to within a segment of 90°.&lt;/p&gt;&#xA;" OwnerUserId="9401" LastEditorUserId="9401" LastEditDate="2015-10-08T13:16:02.457" LastActivityDate="2015-10-08T13:16:02.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12038" PostTypeId="2" ParentId="12034" CreationDate="2015-10-07T19:00:45.443" Score="2" Body="&lt;p&gt;Bodies typically progress outward rather than inward. (See &lt;a href=&quot;https://astronomy.stackexchange.com/questions/420/why-is-the-moon-receding-from-the-earth-due-to-tides-is-this-typical-for-other&quot;&gt;Why is the Moon receding from the Earth due to tides? Is this typical for other moons?&lt;/a&gt; .) The only orbiting bodies that might approach are ones that orbit faster than the main object spins, IOW, closer than synchronous orbit. Even then they could recede if locked into resonance w/ other bodies, &lt;em&gt;eg&lt;/em&gt;, moons, further out. (See &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1080/does-anyone-know-why-three-of-jupiters-largest-moons-orbit-in-124-resonance&quot;&gt;Does anyone know why three of Jupiter&amp;#39;s largest moons orbit in 1:2:4 resonance?&lt;/a&gt; .) Deimos and Phobos are approaching Mars though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supposing you do have an approaching body, eccentricity and inclination will be damped into a slowly degenerating circular orbit. As the body approaches, the acceleration of the body, through the main body's two tides' net pull, increases roughly as the &lt;strong&gt;&lt;em&gt;6th&lt;/em&gt;&lt;/strong&gt; power of distance. (See &lt;a href=&quot;http://www.exo.net/~pauld/physics/tides/tidalevolution.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tidal Evolution of a Planet and its Moon&lt;/a&gt;.) &lt;strong&gt;&lt;em&gt;And&lt;/em&gt;&lt;/strong&gt; tidal heating will soften the body, allowing even more deformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a runaway process at some point, and that point may be quite far into the Roche limit. How far would depend on body size, material tensile strength, structure of the body, thermal conductivity, changes with temperature, &lt;em&gt;etc&lt;/em&gt;. It would take detailed modelling to describe the process. The catastrophe could start in the most vulnerable locality of the body and spread from there (Ka-Boom!), or involve the whole body concurrently (Squish!). It may not go &quot;boom&quot;, but at the end you might be able to observe it in real time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is that the body will start disintegrating at the near end, where the forces are strongest. By conservation of momentum (or would it be energy?), every time a piece leaves, the rest of the body is pushed slightly the other way, sending the remainder slightly higher, delaying the process. This could take quite a while, but there is always a chance that things will destabilize at some point and go into catastrophe.&lt;/p&gt;&#xA;" OwnerUserId="9186" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-10-07T19:27:58.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12039" PostTypeId="1" AcceptedAnswerId="12398" CreationDate="2015-10-08T02:06:39.503" Score="4" ViewCount="706" Body="&lt;p&gt;I always thought that mass was the sole determinant of a star's fate. Then I saw the table &lt;a href=&quot;http://en.wikipedia.org/wiki/Supernova#Core_collapse&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&#xA;So why does metallicity influence a star's ability to become a black hole or neutron star? Does it have as much influence as mass?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="2153" LastEditDate="2015-10-08T02:16:06.113" LastActivityDate="2015-11-04T23:49:33.533" Title="Why is metallicity important in the death of stars?" Tags="&lt;star&gt;&lt;black-hole&gt;&lt;mass&gt;&lt;stellar-evolution&gt;&lt;neutron-star&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12040" PostTypeId="1" AcceptedAnswerId="12054" CreationDate="2015-10-08T04:05:57.603" Score="6" ViewCount="5125" Body="&lt;p&gt;There wouldn't be enough oxygen for any bacteria to decompose the body, right? Not to mention, the radiation of space might kill off most organisms on it. So would it decompose, given millions of years?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2018-04-04T18:01:11.867" Title="If a person were to die on the Moon or Mars, would the body decompose?" Tags="&lt;the-moon&gt;&lt;solar-system&gt;&lt;planet&gt;&lt;space&gt;&lt;mars&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12041" PostTypeId="1" AcceptedAnswerId="12087" CreationDate="2015-10-08T04:06:45.320" Score="5" ViewCount="292" Body="&lt;p&gt;Right now I'm writing a sci-fi story that develops in a double planet. But that's not all, the double planet (along with a couple more planets) orbit a big star and the whole system is orbited by another star (about 100 AU from the central star.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to know how implausible this is. I don't want my story to drift from speculation into impossible worlds. I'm guessing that the occurrence of such a planetary arrangement is very low likely, but is such a system even possible at all? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if it's possible, how messed up the days and seasons would be? I'm bending my brain to figure out what would be the length of day and night in such a system (the planets are very close to each other.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help with this will be really appreciated.&lt;/p&gt;&#xA;" OwnerUserId="9411" LastActivityDate="2015-10-11T03:52:29.570" Title="Double Planet Orbiting Wide Binary Star?" Tags="&lt;binary-star&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12042" PostTypeId="1" CreationDate="2015-10-08T05:34:19.690" Score="3" ViewCount="208" Body="&lt;p&gt;When Jupiter is undergoing retrograde motion as seen from the Earth, would you expect the eclipses of Jupiter’s moons to occur several minutes early, several minutes late, or neither? &lt;br&gt;&#xA;Can someone please explain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a novice in astronomy, and I read a couple of articles on the retrograde motion. I understand it is an illusion. But how would I expect the eclipses of Jupiter's moon?&lt;/p&gt;&#xA;" OwnerUserId="9412" LastActivityDate="2015-10-08T08:31:18.087" Title="Eclipses of Jupiter's Moon during Retrograde motion" Tags="&lt;the-moon&gt;&lt;jupiter&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12044" PostTypeId="1" AcceptedAnswerId="12065" CreationDate="2015-10-08T08:25:36.293" Score="3" ViewCount="338" Body="&lt;p&gt;There are many hypothesis saying that there will a collision between the milky way and Andromeda galaxy...so what happens when two black holes will collide?? And how will it effect the other objects revolving around them??&lt;/p&gt;&#xA;" OwnerUserId="9403" LastActivityDate="2015-10-09T10:45:05.957" Title="What happens when two black holes collide?" Tags="&lt;black-hole&gt;&lt;milky-way&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2015-10-19T16:32:40.227" ContentLicense="CC BY-SA 3.0" />
  <row Id="12045" PostTypeId="2" ParentId="12042" CreationDate="2015-10-08T08:31:18.087" Score="2" Body="&lt;p&gt;The fact that eclipses are either &quot;early&quot;, &quot;on time&quot; or &quot;late&quot; only depend on the distance between Jupiter and the Earth. Since the speed of light is not infinite, a longer distance means a longer delay between the actual eclipse and the time it is observed from Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For superior planets, retrograde motion occurs around opposition, at a time when the distance between the planet and the Earth is minimal. So you should expect eclipses to appear early during retrograde motion. But note that the early eclipses are not a consequence of retrograde motion: they are both consequences of keplerian motion.&lt;/p&gt;&#xA;" OwnerUserId="9415" LastActivityDate="2015-10-08T08:31:18.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12046" PostTypeId="2" ParentId="12033" CreationDate="2015-10-08T08:36:10.280" Score="2" Body="&lt;p&gt;Total Luminosity with $\alpha=-1$:&#xA;$$&#xA;\begin{aligned}&#xA;L_{tot}&amp;amp;=\int_0^{\infty}L \Phi_0 \frac{L^*}{L}e^{-\frac{L}{L^*}}\frac{dL}{L^*}\\&#xA;&amp;amp;=\int_0^{\infty}\Phi_0 e^{-\frac{L}{L^*}}dL&#xA;\end{aligned}&#xA;$$&#xA;Now put $L'=L/L^*$ and you get:&#xA;$$&#xA;L_{tot}=\Phi_0L^* \int_0^{\infty}e^{-L'}dL'=\Phi_0L^*&#xA;$$&#xA;Total number:&#xA;$$&#xA;\begin{aligned}&#xA;N_{tot}&amp;amp;=\int_0^{\infty} \Phi_0 \frac{L^*}{L}e^{-\frac{L}{L^*}}\frac{dL}{L^*}\\&#xA;&amp;amp;=\int_0^{\infty} \Phi_0 \frac{1}{L}e^{-\frac{L}{L^*}}dL&#xA;\end{aligned}&#xA;$$&#xA;which diverges at the lower limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observe with $\alpha=-1$ we have no gamma functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also IIRC with $\alpha \in (-1,0)$ when computing $L_{tot}/N_{tot}$ the gamma functions cancel leaving $(1+\alpha)L^*$&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="2091" LastEditDate="2015-10-08T10:12:17.060" LastActivityDate="2015-10-08T10:12:17.060" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12049" PostTypeId="1" AcceptedAnswerId="12055" CreationDate="2015-10-08T11:52:28.020" Score="0" ViewCount="1751" Body="&lt;p&gt;I would like to know if you can see the Pleiades and Sirius in the sky at the same time, from Greece?&lt;br /&gt;&#xA;What time of the year would that be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I would be very greatful if someone knows an interactive starmap where I could see how that would look like.&lt;/p&gt;&#xA;" OwnerUserId="9419" LastEditorDisplayName="user1569" LastEditDate="2017-01-24T16:35:13.077" LastActivityDate="2017-01-24T16:35:13.077" Title="Can you see Pleiades and Sirius at the same time on the sky, from Greece?" Tags="&lt;amateur-observing&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12050" PostTypeId="2" ParentId="12040" CreationDate="2015-10-08T12:53:07.570" Score="2" Body="&lt;p&gt;Mars is hot enough to keep a body unfrozen oxygen and moisture also speed up the decaying process. A dry environment and constant heat like a desert would mummify a person like beef jerky.If the moon is always under freezing and no air the body would be freeze dried and look like a wax version of themselves forever.    &lt;/p&gt;&#xA;" OwnerUserId="9326" LastActivityDate="2015-10-08T12:53:07.570" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12051" PostTypeId="2" ParentId="2513" CreationDate="2015-10-08T13:56:34.243" Score="16" Body="&lt;p&gt;To first order, the relative abundances of the heavier elements to iron (for instance) are constant. So the metal content of a star is shorthand for the content of any element heavier than He. (NB: we now know this is &lt;em&gt;not&lt;/em&gt; true in many circumstances and elements can be grouped by synthesis process - for example we can talk about &amp;quot;alpha elements&amp;quot; - O, Mg, Si etc. produced by alpha particle capture; or s-process elements - Ba, Sr etc. produced by the s-process. We know that the ratio of say O/Fe gets larger in more &amp;quot;metal-poor&amp;quot; stars, but Ba/Fe gets smaller. So talking about a single &amp;quot;metallicity&amp;quot; parameter only gets you so far, and the truth is more complex (and interesting).&lt;/p&gt;&#xA;&lt;p&gt;The next point is why they are referred to as &amp;quot;metals&amp;quot; rather than another term like &amp;quot;heavies&amp;quot; or something. I would guess this is down to a bit of history and the fact that initial abundance analyses in stars were (and still are on the whole) done in the visible part of the spectrum (e.g. in the early part of the 19th century by Hyde Woolaston and &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Fraunhofer_lines&quot; rel=&quot;noreferrer&quot;&gt;Fraunhofer&lt;/a&gt;). The most abundant elements heavier than He are in fact &lt;em&gt;not&lt;/em&gt; metals; they are Oxygen, Carbon, Nitrogen and Neon. However, the signatures of &lt;em&gt;these&lt;/em&gt; elements are not at all obvious in the visible spectra of (most) stars. Whereas the signatures (absorption lines) of elements like Fe, Na, Mg, Ni etc., which decidedly are metals, are often very prominent.&lt;/p&gt;&#xA;&lt;p&gt;Thus, there &lt;em&gt;is&lt;/em&gt; a reason and some history behind the name &amp;quot;metals&amp;quot;. It is that apart from hydrogen and helium, the metallic elements have the most prominent features in the optical spectra of stars, whereas in most stars the signatures of the more abundant non-metals are hard to see.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2020-01-28T16:39:50.893" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12052" PostTypeId="1" AcceptedAnswerId="12053" CreationDate="2015-10-08T14:01:20.607" Score="4" ViewCount="77" Body="&lt;p&gt;This is a follow-up to &lt;a href=&quot;https://astronomy.stackexchange.com/q/11753/9153&quot;&gt;How to interpret this old degree notation?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is about an old calendar system which generates a luni-solar calendar with a procedural method, using a handful of calculations and some constants derived from astronomical observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method originates from ancient India. Today it is used in Thailand to determine public holidays and observance days (&lt;em&gt;uposatha&lt;/em&gt; days) for Buddhist monks, which fall on New- and Full Moon days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually it's a neat system that had allowed village &quot;astronomers / astrologers&quot; construct a calendar with a method they could memorize by rote learning, without astronomical tools to make precise measurements. It is dreadfully underdocumented, and I've been collating the method and it's practices in &lt;a href=&quot;https://github.com/profound-labs/calculating-the-uposatha-moondays&quot; rel=&quot;nofollow noreferrer&quot;&gt;Calculating the Uposatha Moondays&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most useful have been the actual formulas in &lt;a href=&quot;http://www.siamese-heritage.org/jsspdf/1991/JSS_088_0r_Eade_RulesForInterpolationInThaiCalendar.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rules for Interpolation in the Thai Calendar&lt;/a&gt; by J.C. Eade, it is just some basic arithmetic, but following his notation is a puzzle in itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Effectively I'm trying to implement these formulas (see image below), in a &lt;a href=&quot;https://github.com/splendidmoons/suriya-go/blob/master/suriya.go&quot; rel=&quot;nofollow noreferrer&quot;&gt;golang package&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the earlier help, I arrived at &lt;strong&gt;step 13 (Mean Moon)&lt;/strong&gt;. Hooray :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step &lt;strong&gt;14.&lt;/strong&gt; and &lt;strong&gt;15.&lt;/strong&gt; is a puzzle again.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Step 14.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How would you interpret this?&lt;/strong&gt; &quot;(1780 + 80) * 3 on base 808, and add 2&quot;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Step 15.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Eade has &lt;code&gt;(8; 11 : 7) - (6; 27 : 12) = (1; 3 : 55)&lt;/code&gt;, but that doesn't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That specific substr. gives me &lt;code&gt;(1; 13 : 55)&lt;/code&gt;, which could be a typo in the paper, but that value doesn't carry things forward either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things straighten up at step &lt;strong&gt;16., 17., 18.&lt;/strong&gt;, which would produce the paper's results, to get there I faked the values for step 14. and 15, to work at least in the example case given in the paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;https://github.com/splendidmoons/suriya-go/blob/master/suriya.go&quot; rel=&quot;nofollow noreferrer&quot;&gt;suriya.go&lt;/a&gt;&#xA;this business is happening in:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;func (suDay *SuriyaDay) Init(ce_year int, lunar_year_day int)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bMGRW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bMGRW.png&quot; alt=&quot;JC Eade, Rules for Interpretation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-10-09T07:36:33.940" Title="From Mean Moon to True Moon in an old procedural calendar" Tags="&lt;notation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12053" PostTypeId="2" ParentId="12052" CreationDate="2015-10-08T15:33:12.357" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;How would you interpret this?&lt;/strong&gt; &quot;(1780 + 80) * 3 on base 808, and add 2&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As before, there is a confusing avoidance of specifying the unit under consideration, but the pattern is still that the last number is in minutes of arc, the middle in degrees, and the first in $30^\circ$ for the Zodiac divisions, so:&#xA;$$\begin{eqnarray*}&#xA;\frac{(1780+80)\times 3}{808} &amp;amp;=&amp;amp; 6+\frac{27}{30} + \frac{10}{30\times 60} + \frac{41}{30\times 60^2}+\ldots \\&#xA;&amp;amp;=&amp;amp; (6;\;27\!:\,10)\text{.}\end{eqnarray*}$$&#xA;The $+2$ makes this $(6;\;27\!:\,12)$, although this makes the second addend in minutes of arc while the first one is not. As before, the calculation is not rounded.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Eade has &lt;code&gt;(8; 11 : 7) - (6; 27 : 12) = (1; 3 : 55)&lt;/code&gt;, but that doesn't work.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;That specific substr. gives me &lt;code&gt;(1; 13 : 55)&lt;/code&gt;, which could be a typo in the paper, but that value doesn't carry things forward either.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think that you are correct that this is a typographical error, but the corrected value probably &lt;em&gt;is&lt;/em&gt; carried forward--the error is just repeated again in Step 16. Note that $(1;\;13\!:\,55) = 43^\circ 55'$, which makes the corrected Step 16&#xA;$$296\sin(1;\;13\!:\,55) = 205.309 = (0;\;3\!:25)\text{,}$$&#xA;following the convention that the numerical value is re-interpreted as minutes of arc. On the other hand, if we follow what Step 16 actually says, we get a very different value:&#xA;$$296\sin(1;\;3\!:\,55) = 165.164 = (0;\;2\!:45)\text{.}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Granted, the value actually carried forward to Step 17 is $(0;\;3\!:24)$ instead, but the former is definitely much closer. This further discrepancy could also be an undocumented geographical correction (the procedure does, after all, have things like $+3$ and $+2$ at times), but I'm more inclined to guess that this is due to imprecision in the calculation of sine, as the paper indicates that a table was used.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditDate="2015-10-08T20:27:11.567" LastActivityDate="2015-10-08T20:27:11.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12054" PostTypeId="2" ParentId="12040" CreationDate="2015-10-08T17:06:31.277" Score="5" Body="&lt;p&gt;Space, as &lt;a href=&quot;https://what-if.xkcd.com/134/&quot;&gt;Randall notes&lt;/a&gt;, is really dry. Mars, (&lt;a href=&quot;https://www.nasa.gov/press-release/nasa-confirms-evidence-that-liquid-water-flows-on-today-s-mars/&quot;&gt;recent discoveries&lt;/a&gt; notwithstanding) is not much moister.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In these conditions, bodies mummify. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The microbes that live in you, wouldn't survive the freezing, dessication and radiation. There is no real upper limit on how long a mummified body could exist in space. There would be nothing to cause the body to change, and so it would remain. There would be a slow breaking down of surface proteins, due to UV light, and eventually micrometeorites would erode the body, but these processes would take many millions of years.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2015-10-08T17:22:46.377" LastActivityDate="2015-10-08T17:22:46.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12055" PostTypeId="2" ParentId="12049" CreationDate="2015-10-08T17:50:33.813" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://www.stellarium.org/&quot; rel=&quot;nofollow&quot;&gt;Stellarium&lt;/a&gt; is a free planetarium software, and it allows you view the position of stars in the sky at any date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sirius and the Pleiades are actually rather close in the sky. On the line of Orion's belt, on opposite sides. So for much of the the winter months, both are visible in the sky. For example in mid-December, they will be visible at midnight, in south-western skies. At the time of writing (8th October) they will be visible in the early morning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a well known starscape, with Orion central, the Pleiades and the V of the Hyades on one side, and Sirus, scorching on the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the Pleiades and Sirius are &quot;fixed stars&quot;, so the separation between them doesn't change. There is no time when they approach each other.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2015-10-08T18:55:41.853" LastActivityDate="2015-10-08T18:55:41.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12059" PostTypeId="1" CreationDate="2015-10-08T21:39:13.577" Score="2" ViewCount="172" Body="&lt;p&gt;In this &lt;a href=&quot;https://www.youtube.com/watch?v=kYAdwS5MFjQ&quot; rel=&quot;nofollow&quot;&gt;youtube video&lt;/a&gt;, Prof. Michio Kaku mentions there could be multiple universes and &lt;strong&gt;if&lt;/strong&gt; there are, then there could be a &lt;strong&gt;worm hole&lt;/strong&gt; to some other universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: We are not sure if the Multiverse exists, so how can we believe or have imagined that worm holes could exist and create a path to another universe?&lt;/p&gt;&#xA;" OwnerUserId="9284" LastEditorUserId="4085" LastEditDate="2015-10-09T05:35:38.223" LastActivityDate="2015-10-10T11:55:23.393" Title="Existence of multiple universe?" Tags="&lt;multiverse&gt;&lt;wormhole&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12060" PostTypeId="1" CreationDate="2015-10-09T00:51:58.070" Score="2" ViewCount="2429" Body="&lt;p&gt;How can i find the location of the sun and which constellation it would occupy based on the location and date of year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for example: an observer is in Des Moines, Iowa on April 10th, which constellation does the sun appear to occupy?&lt;/p&gt;&#xA;" OwnerUserId="9427" LastActivityDate="2015-10-10T06:33:28.337" Title="What constellation does sun occupy by location and date?" Tags="&lt;the-sun&gt;&lt;constellations&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12061" PostTypeId="2" ParentId="12059" CreationDate="2015-10-09T05:58:07.793" Score="3" Body="&lt;p&gt;We don't know. The Multiverse Theory is really just a hypothethis; sure, it makes sense, but it cannot be scientifically proven. We really have no reason to doubt that this is our only universe. Wormholes are just as hypothetical, since they would require infinite energy or negative energy to be created, neither of which we can get. Also, keep in mind that String Theory is, for the most part, &lt;strong&gt;unprovable&lt;/strong&gt;. It cannot be confirmed through scientific experiments and is mostly guesswork.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it will take trillions of years before the Universe gets too cold for life, and we will be long gone by then. However way you put it, it's unlikely that humans will ever reach other habitable planets before we go extinct, should our planet be destroyed. The closest potentially-habitable planet is nearly 480 lightyears from us – as in, it would take us 480 years to get there, if we travelled at the fastest speed in the Universe. We surely won't live to see the end of the Universe.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2015-10-10T11:55:23.393" LastActivityDate="2015-10-10T11:55:23.393" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12062" PostTypeId="1" AcceptedAnswerId="12067" CreationDate="2015-10-09T06:37:19.610" Score="3" ViewCount="132" Body="&lt;p&gt;I know massive objects can curve light, can we capture the image of past Earth by lights from Earth that curved by massive objects?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastActivityDate="2015-10-12T10:53:12.350" Title="Can we see the past image of Earth because of curving light by massive objects?" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12063" PostTypeId="2" ParentId="12044" CreationDate="2015-10-09T06:50:56.343" Score="0" Body="&lt;p&gt;When two black holes come close, they will orbit each other extremely quickly, greatly increasing the strength of their gravity. Over the course of billions of years, they will get closer and closer. Once they come so close that they cannot escape each other's gravity, they will merge to become one bigger black hole.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2015-10-09T06:50:56.343" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12064" PostTypeId="1" CreationDate="2015-10-09T08:39:34.233" Score="2" ViewCount="567" Body="&lt;p&gt;I read in many astronomy sites that Sirius is the brightest star of the night sky and that Orion’s Belt points to Sirius. For some days I have been regularly watching Orion (just before dawn) but the star that the Orion's belt points towards is of very average brightness; I don't think it is Sirius. On the other hand I found a very bright star near Orion , albeit not along the line of its belt. On the morning of October 9 (around 5:30 am) this star was adjacent to the crescent moon and shining slightly more brightly ! Even in the early morning glow when all the stars faded away it could still be seen along with the moon. Is this Sirius?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I live in Delhi so I am talking of the Delhi night sky.&lt;/p&gt;&#xA;" OwnerUserId="9430" LastEditorUserId="2365" LastEditDate="2015-10-10T21:57:20.113" LastActivityDate="2015-10-11T17:01:20.410" Title="Brightest star of the night sky" Tags="&lt;star&gt;&lt;amateur-observing&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12065" PostTypeId="2" ParentId="12044" CreationDate="2015-10-09T10:45:05.957" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;There are many hypothesis saying that there will a collision between&#xA;  the milky way and Andromeda galaxy...so what happens when two black&#xA;  holes will collide??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's a few answers in the links in my comment above copied, &lt;a href=&quot;https://physics.stackexchange.com/questions/4881/black-holes-in-a-head-on-collision&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://astronomy.stackexchange.com/questions/2111/collision-of-2-black-holes&quot;&gt;here&lt;/a&gt;, but it's such a fun question I thought I'd answer it anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With Andromeda and the Milky way's super-massive black holes in their respective centers, those objects are so massive, they are likely to be largely unaffected by any stars in their path and they will just fly towards each other at whatever rate and direction gravity dictates and by the time they get fairly close to each other, their respective gravity should have them moving quite fast towards one another.   I don't know if anyone knows if they will orbit around each other for an extended period of time or spiral in fairly quickly.   That will depend on the angle of approach.  It's possible they could pass by each other, miss and each could get sent far out away from the other, part of an enormous orbit around each other, taking perhaps billions of years to lead to an eventual collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a few simulated videos out there on what happens when 2 super massive black holes spiral into each other.   &lt;a href=&quot;https://www.youtube.com/watch?v=GVds0q0y5RM&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's one.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Super-Massive black holes approach, each will pass through roughly half of the other galaxy, disrupting the orbits of any stars they pass hear-by, though they'd likely need to pass within less than a light year to have significant effects on the star's direction, which wouldn't happen that often, but it would happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sagittarius A is huge for a black hole, but quite small compared to the space between stars.  It's estimated to be about 44 million KM in diameter, which means it would fit (Just barely) inside the orbit of Mercury and our Sun.   Mostly it will fly past stars.   Andromeda's super massive black hole may be several times larger, but still fairly small compared to the distance between stars it's likely to pass on the way towards their mutual collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's possible that one or both stars will pass through each other's galaxy relatively collision free but it's possible that one or both of them will get close enough to a star to generate a large accretion disk.   While black holes don't have charge, their accretion disks do, and if that happens, that could be an interesting and not very well understood interaction between the 2 super-massive objects and their accretion disks.   Also, as they spiral in towards each other, several of their near-by stars will be cast any-which way, some of them, inevitably inside towards one of the 2 black holes.   It could be a very impressive show.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, as they merge, which, might take quite a bit of time if they end up orbiting each other, perhaps millions, even billions of years.   If/when they do merge, there could be some curious and not very well understood gravitational wave effects. bending and stretching of space like ripples.  (Gravity already bends space, but not in measurable ripples.   Our observation's of gravity warped space is a smooth curve.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll re-post this &lt;a href=&quot;http://www.universetoday.com/13002/what-happens-when-supermassive-black-holes-collide/&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; here from comments that says it's possible for 2 black holes to repel each other if they bend space in opposite spin-directions and approach each other on a level plane.   Less close and they could still easily disrupt outer planet orbits, casting some planets every which way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And how will it effect the other objects revolving around them??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Imagine jellybeans in a salad spinner going as fast as you can spin it, and you remove the top.  That's basically what will happen to any near-by stars.   Both galaxy centers are quite crowded with stars (and perhaps several black holes) orbiting their centers.   The gravity assists from 2 super-massive objects moving towards each other will be significant and, basically stars will be flying all over the place.   Stars have tiny masses compared to those objects and they could be sent in any direction.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-10-09T10:45:05.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12066" PostTypeId="2" ParentId="12064" CreationDate="2015-10-09T12:21:47.043" Score="2" Body="&lt;p&gt;The best way to find a particular star/planet/etc. is almost certainly to use a star map of some sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, doing a Google Image search for &quot;star map sirius&quot; would get you quite a few pictures such as the one in &lt;a href=&quot;http://beforeitsnews.com/science-and-technology/2013/09/worlds-oldest-temple-probably-built-to-worship-the-dog-star-sirius-2634354.html&quot; rel=&quot;nofollow&quot;&gt;this article about Sirius&lt;/a&gt; that give a better picture of how Orion's belt &quot;points&quot; to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately, there are a number of computer programs that offer a real time views/maps of the night sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also check out &lt;a href=&quot;http://www.google.com/sky/&quot; rel=&quot;nofollow&quot;&gt;Google Sky&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="38" LastActivityDate="2015-10-09T12:21:47.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12067" PostTypeId="2" ParentId="12062" CreationDate="2015-10-09T13:05:24.527" Score="1" Body="&lt;p&gt;Light can be made to do this around &lt;a href=&quot;https://en.wikipedia.org/wiki/Black_hole&quot; rel=&quot;nofollow&quot;&gt;black holes&lt;/a&gt;, sort of. At a certain distance from a black hole you have what's called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Photon_sphere&quot; rel=&quot;nofollow&quot;&gt;photon sphere&lt;/a&gt;. At this distance, determined by the mass of the black hole, photons travel in orbits because of the space-time curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer your question, &lt;strong&gt;maybe&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, a massive entity would need to be located at some point in space in order to influence the photons into an orbit - okay yes, possible. Secondly, this massive object would need to have a mechanism of instantaneously disappearing from space-time in order for the photos to tangentially come out of orbit and back towards earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read on a previous question on Physics.SE that there is no such solution to the Einstein Field Equations which allows for a massive object to just disappear - the closest thing would be for the density distribution of the object to change but I can't say if that would yield the result you are asking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rob Jeffries mentioned that images do exist which suggest that photons experience such a change in trajectory - this could happen on paths which are very close to a photon sphere-like one, but deviate enough from it to allow photos to escape.&lt;/p&gt;&#xA;" OwnerUserId="5472" LastEditorUserId="5472" LastEditDate="2015-10-12T10:53:12.350" LastActivityDate="2015-10-12T10:53:12.350" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12068" PostTypeId="1" AcceptedAnswerId="12075" CreationDate="2015-10-09T16:50:58.443" Score="15" ViewCount="1016" Body="&lt;p&gt;Consider this scenario.  One of the probes sent to Mars either very soon, or within the next century, finds conclusive micro-fossil evidence that life once existed on Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One theory today that is quite common is that life might that have arisen on Mars was then ejected into space, floated to Earth and then seeded life on our planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm tracking with you and that could make sense, assuming we found the conclusive evidence for life on Mars.  However, why would we assume that the early martian life originated on Mars, rather than Earth?  Is it not likely that somewhere between 2.5-3 billion years ago, a major volcanic eruption or meteor impact could have hit an earth teeming with simple-life and sent it hurdling through space to Mars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that my theory has the gravity of the sun working against it, but I also see it not being entirely impossible.  I mean, 100 years ago, one could assume that life originating on Mars and then being sent this was was ridiculous and impossible but now it's a rational, viable theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering such a scenario, how would scientist verify that these life-forms didn't originate from Earth, rather than Mars?  It seems to me that this question is the first question that would need to be asked and answered after the fossil-records were found and verified.&lt;/p&gt;&#xA;" OwnerUserId="9434" LastActivityDate="2017-09-13T07:40:02.460" Title="If we found evidence of life on Mars, how would we know that it originated on Mars rather than Earth?" Tags="&lt;earth&gt;&lt;mars&gt;&lt;life&gt;&lt;origin-of-life&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12069" PostTypeId="2" ParentId="12068" CreationDate="2015-10-09T18:04:40.227" Score="2" Body="&lt;p&gt;If any amino acid chain on Mars is the same or similar to any amino acid chain on Earth, then we know with final certainty that they have the same origin. Simple combinatorics proves it within as many sigmas you wish. Do the math. A median protein on Earth is about 300 amino acids long. That combination has never occurred by chance anywhere in the visible universe ever. Evolution has no chance to converge them, because evolution can only test that which occurs by chance.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-10-09T18:04:40.227" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="12070" PostTypeId="1" AcceptedAnswerId="12071" CreationDate="2015-10-09T18:44:37.397" Score="8" ViewCount="418" Body="&lt;p&gt;For those that haven't seen &lt;a href=&quot;http://media.skysurvey.org/interactive360/index.html&quot;&gt;this link&lt;/a&gt; - please have a look as it's a an awe-inspiring 360 interactive panormic image of the milky way (may need a bit of time to load though!). I've found myself idly browsing it several times and as an astronomy layman I wondered if anyone could help me navigate what I'm looking at. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I can orientate myself and visualise where we are in the milky way, what is the distinctive black band? I'm assuming it is caused by the absence of stars in these areas. However, as I understood it, the disc (as well as the center) are the areas of highest concentration of stars. Are these 'gaps' caused by dark matter?&lt;/p&gt;&#xA;" OwnerUserId="9436" LastActivityDate="2016-09-12T11:11:09.113" Title="Disc of the milky way" Tags="&lt;milky-way&gt;&lt;photography&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12071" PostTypeId="2" ParentId="12070" CreationDate="2015-10-09T19:41:37.680" Score="10" Body="&lt;p&gt;&lt;strong&gt;It's dust!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black band is not absence of stars, but rather &lt;em&gt;clouds of gas and dust&lt;/em&gt; — a significant component of almost all spiral galaxies$^\dagger$ — which block the light of background stars and luminous gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the image, you see both individual stars, scattered all over, and the distinctive bright band of the Milky Way, most of the stars of which are so far away that they blend together. Most of the dust is in the plane of the Milky Way. The individual stars are closer to us than the bright band and the dust clouds, so what you see is the billions of stars in the Milky Way band, some of the light of which is blocked by dust, and then on top of this you see a few 1000s individual stars that are closer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The components of galaxies&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxies consist of roughly 85% dark matter and 15% normal (&quot;baryonic&quot;) matter. By far, most of the normal matter is hydrogen and helium, some of which is locked up in stars, and some of it in huge gas clouds, sometimes glowing (the pink clouds you see in the Milky Way image are probably hydrogen clouds being excited by the hard UV radiation from hot and massive stars, subsequently emitting &lt;a href=&quot;https://en.wikipedia.org/wiki/H-alpha&quot; rel=&quot;noreferrer&quot;&gt;H$\alpha$ light&lt;/a&gt;). A small fraction (1–2%) of the normal matter are heavier elements, by astronomers lazily referred to as &quot;metals&quot;. Roughly 2/3 of the metals are in the gas phase, but the remaining 1/3 has depleted into dust grains, e.g. silicates and soot. This dust is mixed with the gas clouds and often become dense enough that they block light of stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A strong wavelength dependence of the dust's extinction properties&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, dust scatters and absorbs light with short wavelengths (such as blue and UV) much more efficiently than long wavelengths (such as red and infrared). So while it may effectively block visual and UV light, infrared light passes more or less unhindered. For instance, if you look at the molecular cloud &lt;a href=&quot;https://en.wikipedia.org/wiki/Barnard_68&quot; rel=&quot;noreferrer&quot;&gt;Barnard 68&lt;/a&gt; in red and near-infrared, it looks black (left in the image below), but if you look farther out in the infrared, you can actually see the background stars (right below):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/54UBd.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/54UBd.jpg&quot; alt=&quot;B68&quot;&gt;&lt;/a&gt;&#xA;&lt;sup&gt;Image taken from &lt;a href=&quot;http://www.eso.org/public/images/eso0102c/&quot; rel=&quot;noreferrer&quot;&gt;ESO&lt;/a&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dark matter&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark matter cannot be seen. It's… well, dark. It interacts with normal matter and light &lt;em&gt;only&lt;/em&gt; through gravity. That means that if you could place a lump of dark matter in front of a star (you can't, really), you wouldn't block its light. It would pass right through. If the lump were big enough, it might gravitationally deflect the light, so the background star would look distorted to you, as would a lump of normal matter (e.g. a black hole).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;In contrast, the interstellar medium of &lt;em&gt;elliptical&lt;/em&gt; galaxies tend to be much more gas- and dust-depleted.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-09-12T11:11:09.113" LastActivityDate="2016-09-12T11:11:09.113" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12072" PostTypeId="1" CreationDate="2015-10-09T20:08:30.740" Score="7" ViewCount="484" Body="&lt;p&gt;My understanding is that a pulsar is a neutron star whose magnetic poles don't coincide with the rotational poles and whose magnetic poles sweep through Earth (allowing us to hear radio noise).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we keep hearing pulsars, the polar offset must be very stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is such an alignment offset so stable?&lt;/p&gt;&#xA;" OwnerUserId="9433" LastEditorUserId="6980" LastEditDate="2015-10-11T23:49:44.463" LastActivityDate="2020-09-02T12:44:06.063" Title="Why are the magnetic poles of a pulsar so far off the rotational axis, yet stable?" Tags="&lt;astrophysics&gt;&lt;magnetic-field&gt;&lt;pulsar&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12073" PostTypeId="2" ParentId="12070" CreationDate="2015-10-09T20:25:58.573" Score="2" Body="&lt;p&gt;Pela has a nice answer, I'll just add my 2 cents.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;While I can orientate myself and visualise where we are in the milky&#xA;  way, what is the distinctive black band? I'm assuming it is caused by&#xA;  the absence of stars in these areas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The black band probably blocks some of the more dense areas of stars, not a lack of stars at all, with the exception, I think, for the larger dark area on the far left, which might be a lack of Milky-way stars.   I believe that window of low star density is where Hubble spends most of it's time looking into deep space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, consider which is brighter a flashlight 6 inches from your eye or a car head-light 60 feet away.   The center of our galaxy may be a lot brighter, but we're pretty far away from it, so the relative brightness we see of the Milky way is more evened out.   A lot of the light in that white streak we see is coming from relatively close stars.  It's a very different perspective than our view of Andromeda where the center stands as much brighter.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, as I understood it, the disc (as well as the center) are the&#xA;  areas of highest concentration of stars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can see it gets a bit thicker in the center and the brightest part is the lower-center.   The bright center is there in the picture, it's just largely blocked by gas and dust.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are these 'gaps' caused by dark matter?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Dark matter is essentially invisible to all light, unless there's enough of it to bend light, then it can be observed indirectly.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-10-09T20:25:58.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12074" PostTypeId="2" ParentId="12060" CreationDate="2015-10-09T20:55:51.590" Score="0" Body="&lt;p&gt;There are a few ways to do this, but I would recommend a decent star chart or software that allows you to set a specific date/time and location. There are a couple of really good software programs that offer WAY more than a Zodiacal reference, and that are free.&lt;/p&gt;&#xA;" OwnerUserId="2394" LastActivityDate="2015-10-09T20:55:51.590" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12075" PostTypeId="2" ParentId="12068" CreationDate="2015-10-09T21:40:50.017" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why would we assume that the early martian life originated on Mars,&#xA;  rather than Earth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's still a whole lot we don't know.  As Wayfaring stranger points out in the comments, Origin is a whole different question.  It's possible that life originated outside our solar system and came to either Mars and/or Earth from outside the solar-system.   I don't think anyone who studies this idea is &quot;Assuming&quot; life originated on Mars, only that the idea has a chance of being true.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it not likely that somewhere between 2.5-3 billion years ago, a&#xA;  major volcanic eruption or meteor impact could have hit an earth&#xA;  teeming with simple-life and sent it hurdling through space to Mars?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As I understand it, volcanic eruptions are unlikely to send anything into space unless it's a smaller sized moon.   The escape velocity (young earth, maybe 10 km/s, young mars, maybe 3-4), volcanic eruptions, as far as I know, don't shoot things out at 10,000 - 20,000 MPH.   But meteor impacts of sufficient size can do that.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars is a better meteor debris making target than the Earth cause it's smaller, so the gravity is lower and presumably it's mostly had a thinner atmosphere too.  We've found martial meteors on Earth.   We might not find any Earth meteors on mars cause it takes a much bigger impact to knock bits of rock off Earth and because the Atmosphere slows objects down, both coming in and going out.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I understand that my theory has the gravity of the sun working against&#xA;  it, but I also see it not being entirely impossible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sun isn't as big a factor as you might think.  Once something is knocked off a planet and it gets into orbit around the sun, gravitational assists can move it around further out or further inside the solar system.  What presumably happens is that, with a big enough impact, many thousands if not millions of bits of debris get into solar-system orbit and from there, some of them land on other planets - probably much less than 1% of those hit Earth, but if it carries life that can survive the trip, all you need is one rock.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I mean, 100 years ago, one could assume that life originating on Mars&#xA;  and then being sent this was was ridiculous and impossible but now&#xA;  it's a rational, viable theory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While that's true, the &quot;People used to think this was impossible&quot; isn't a scientific approach for what might be true.   We should determine what is possible and/or thought to be likely, based on physical evidence, not what wasn't understood 100 years ago.   Your example is a good point on why it's important to keep an open mind about the unknown.   You can still make theories based on evidence, and keep an open mind on the unknown.  There's really no conflict between the two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Considering such a scenario, how would scientist verify that these&#xA;  life-forms didn't originate from Earth, rather than Mars? It seems to&#xA;  me that this question is the first question that would need to be&#xA;  asked and answered after the fossil-records were found and verified.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's a good question.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simple answer is that Mars cooled first and Mars (likely) had oceans first, so it's a better candidate to have developed life first though extremophiles can live in hot oceans, so . . . time will tell.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-10-09T21:40:50.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12076" PostTypeId="2" ParentId="12072" CreationDate="2015-10-09T22:22:02.007" Score="5" Body="&lt;p&gt;The magnetic field of a star is not entirely a result of the global spin of the star. The global spin is part of it, but there are other mechanisms as well. Within the star, there are convection zones, meridional flow, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://solarscience.msfc.nasa.gov/dynamo.shtml&quot; rel=&quot;noreferrer&quot;&gt;http://solarscience.msfc.nasa.gov/dynamo.shtml&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these flows generate their own field components. The overall field is simply the sum of all little fields. Its general orientation might be close to the global spin, if the strongest components are aligned to it, but there are many smaller components with different orientations. Therefore, the total field of the star can be somewhat slanted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then the star collapses into a neutron star, and its field is compressed. The collapse itself may be slightly asymmetrical, and may further deviate the magnetic field axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result of all of the above, it's by no means unusual that the magnetic field of the neutron star is not aligned with the spin.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2015-10-09T22:22:02.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12077" PostTypeId="2" ParentId="3611" CreationDate="2015-10-10T01:21:40.140" Score="0" Body="&lt;p&gt;No, there are not. Try as I might, I cannot without a dynamic screen construct the actual effect. I know something of log vision scales and light effects and the eye's own internal zoom (this is the origin of the large moon error). In order to make it look right it is necessary to render it wrong, and at a very high DPI (&gt; 400 I believe).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see at once the foreground stars, and behind two sheets of light (I know it's really only one sheet with a dust band in front of it but that's not what it looks like), and at the same time I see the stars that make up the light sources of the sheets of light. Not that I can really resolve them, but as if I were to pick up sand from the beach where the boundaries are obvious to existence but counting is not possible. That is, it's just at the point of switching between a texture and individual entities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot make out any color at all because the total light level is just too low, but the night sky provides enough light to walk around by (neither Venus nor the Moon were visible when I tested the walking around by).&lt;/p&gt;&#xA;" OwnerUserId="8056" LastActivityDate="2015-10-10T01:21:40.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12078" PostTypeId="1" CreationDate="2015-10-10T04:49:47.660" Score="7" ViewCount="195" Body="&lt;p&gt;Are there any plans for the 'Hubble2'? Maybe with an even bigger mirror capable of even finer resolution, to look even deeper into the origins of the universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, when can we expect to see it in action?&lt;/p&gt;&#xA;" OwnerUserId="7069" LastEditorUserId="2365" LastEditDate="2015-10-10T13:55:52.387" LastActivityDate="2016-03-17T16:33:08.640" Title="Successor to the Hubble Telescope" Tags="&lt;hubble-telescope&gt;&lt;photography&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12079" PostTypeId="2" ParentId="12060" CreationDate="2015-10-10T06:33:28.337" Score="1" Body="&lt;p&gt;You can see for yourself for any date and location using the free Stellarium program (&lt;a href=&quot;http://stellarium.org/&quot; rel=&quot;nofollow&quot;&gt;http://stellarium.org/&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constellation that the Sun is in as seen from Earth does not depend noticeably on the precise location of the observer on Earth.  It depends mostly on the time of year.  Here are the approximate days of the year at which the Sun enters a new constellation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Aries: April 18&lt;/li&gt;&#xA;&lt;li&gt;Taurus: May 14&lt;/li&gt;&#xA;&lt;li&gt;Gemini: June 21&lt;/li&gt;&#xA;&lt;li&gt;Cancer: July 20&lt;/li&gt;&#xA;&lt;li&gt;Leo: August 10&lt;/li&gt;&#xA;&lt;li&gt;Virgo: September 16&lt;/li&gt;&#xA;&lt;li&gt;Libra: October 31&lt;/li&gt;&#xA;&lt;li&gt;Scorpius: November 23&lt;/li&gt;&#xA;&lt;li&gt;Ophiuchus: November 29&lt;/li&gt;&#xA;&lt;li&gt;Sagittarius: December 17&lt;/li&gt;&#xA;&lt;li&gt;Capricornus: January 19&lt;/li&gt;&#xA;&lt;li&gt;Aquarius: February 16&lt;/li&gt;&#xA;&lt;li&gt;Pisces: March 12&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The &quot;approximate&quot; part is that the Sun may be up to one day earlier or later than the quoted dates, just like the start of the seasons may be one day earlier or later than the average.  (Actually, it is the calendar that is early or late with respect to the phenomenon.)  Also, because of the precession of the equinoxes, these dates shift by about 1 day every 70 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://aa.quae.nl/en/antwoorden/sterrenbeelden.html#6&quot; rel=&quot;nofollow&quot;&gt;http://aa.quae.nl/en/antwoorden/sterrenbeelden.html#6&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9401" LastActivityDate="2015-10-10T06:33:28.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12080" PostTypeId="2" ParentId="12078" CreationDate="2015-10-10T06:53:04.437" Score="7" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/James_Webb_Space_Telescope&quot;&gt;James Webb Space Telescope&lt;/a&gt; (JWST) will be Hubble 2.0, and, my personal favorite, it could get a better look at planets in other solar-systems than we've ever had.   Scheduled for launch in 2018.  (Footnote, as Rob Jeffries points out, it's infra-red through long-wave visible light only).   Hence, no Blue/Indigo/Violet or Green, but some yellow.   So, it's not quite like Hubble, but it's still often called Hubble's replacement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Chandra_X-ray_Observatory&quot;&gt;Chandra&lt;/a&gt; observes X-rays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Spitzer_Space_Telescope&quot;&gt;Spitzer&lt;/a&gt; observes infra-red.    They both followed Hubble, but they're not the same type of telescope. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-10-10T09:19:26.407" LastActivityDate="2015-10-10T09:19:26.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12081" PostTypeId="2" ParentId="12041" CreationDate="2015-10-10T14:50:55.843" Score="0" Body="&lt;p&gt;This might not be a complete answer but in fact a planet does not exactly orbit a star. A star and a planet (or any celestial body) orbit the mass center of both. In example the Sun-Earth system is orbiting a point located inside the sun, that is not it's exact center of the sun but the mass center of both.&#xA;In multi-body systems, the mass center might not be a single point in space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this and assuming the mass of a star is much bigger than the mass of a planet, it can be assumed that a star will hardly be orbiting both star and planets. What could happen is that the planets would orbit arround both stars (maybe one, maybe the other and maybe both). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to have a planet and a star be orbiting another star, the &quot;main&quot; star should be much bigger than the &quot;secondary&quot; star and the planet should be much closer to the main star than the secondary. So it would be something similar to having the Sun, Mercury and Jupiter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to have a binary planet orbiting the main star, I guess they should be very close one to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must say I'm not an expert in this subject.&lt;/p&gt;&#xA;" OwnerUserId="8006" LastActivityDate="2015-10-10T14:50:55.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12082" PostTypeId="1" CreationDate="2015-10-10T19:27:33.837" Score="6" ViewCount="250" Body="&lt;p&gt;I see &lt;a href=&quot;https://en.wikipedia.org/wiki/Timeline_of_the_far_future&quot;&gt;here&lt;/a&gt; that after a very long time into the future, &quot;Estimated time for random quantum fluctuations to generate a new Big Bang&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This begs the question, do we speculate that our big bang was the first?  Or are we one of many in a never ending chain?  Or do I miss the point entirely (being ignorant in the subject)?&lt;/p&gt;&#xA;" OwnerUserId="7312" LastActivityDate="2015-10-15T22:11:43.097" Title="Is ours the first universe?" Tags="&lt;universe&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12083" PostTypeId="2" ParentId="12082" CreationDate="2015-10-10T21:13:14.260" Score="2" Body="&lt;p&gt;It is very hard to tell whether we are in a unique Universe. The difficulty is the closer we attempt to probe in the fractions of a nano second after the Big Bang our current forumlations of Gravity and the three our fundamental forces do not work. So it is very hard to tell whether our Universe is unique. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, with all that said, my own personal belief is that our Universe is not unique. The estimated time for random quantum fluctuations to generate a new Big Bang as quoted in the page that you have referred to is an &lt;em&gt;UNIMAGINELY&lt;/em&gt; large time scale. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question I will pose back to you is, how could we ever tell we are a unique Universe, or part of a collection of Universes?  &lt;/p&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2015-10-10T21:13:14.260" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12084" PostTypeId="2" ParentId="12064" CreationDate="2015-10-10T22:10:39.663" Score="1" Body="&lt;p&gt;Sirius is the brightest &quot;star&quot; in the night sky, but not the brightest object in the sky in the early morning, at the time of writing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In autumn of 2015, Venus is a very bright in the early morning sky, and it outshines any star by several magnitudes of brightness. It is, perhaps, in comparison to Venus that you find that Sirius is &quot;of very average brightness&quot; &#xA;On the night you mention, Venus would have appeared as a bright &quot;star&quot; near to the crescent moon, and visible even after sunrise. No other star or planet would have been visible after sunrise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compared to the other stars, Sirius is exceptionally bright. However, at this time of the year it is close to the horizon (as seen from Delhi) which diminishes its apparent magnitude. Sirius will also appear to &quot;twinkle&quot; much more than Venus, and as of the time of writing,  Venus will appear as clearly a &quot;half disc&quot; in a small telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Learn to follow the sky, with either a map, or &lt;a href=&quot;http://www.stellarium.org/&quot; rel=&quot;nofollow&quot;&gt;stellarium&lt;/a&gt; so you can recognise the patterns made by the stars of the sky, then you can find a particular star, such as Sirius, much more easily and confidently.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-10-10T22:10:39.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12085" PostTypeId="2" ParentId="11617" CreationDate="2015-10-11T00:16:03.507" Score="0" Body="&lt;p&gt;Edit (I looked, unfortunately most of the sources are astrology, not astronomy), but Mars last went in Retrograde March 2014 and it's not due to go again till May 2016 (every 2 years, 2 months).    My guess below was incorrect (though it's kind of cool, I'll leave the picture up).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars does have a tilted orbit compared to earth and that could be the reason.  That and the angle from the earth changes too. &#xA;&lt;a href=&quot;http://mars.nasa.gov/allaboutmars/nightsky/retrograde/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6k8MY.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6k8MY.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Mars, from our point of view, slows down, temporarily reverses direction then returns to it's original direction and continues on.   The entire &quot;S&quot; motion takes maybe 3-4 months and your 2 photos are 24 days apart.&lt;/em&gt;  (but this won't happen till May 2016)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6cAt2.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6cAt2.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source of photo and explanation &lt;a href=&quot;http://cmb.physics.wisc.edu/pub/tutorial/briefhist.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;:   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-10-11T00:49:53.423" LastActivityDate="2015-10-11T00:49:53.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12086" PostTypeId="1" CreationDate="2015-10-11T03:07:21.110" Score="4" ViewCount="529" Body="&lt;p&gt;Does Ceres belong to the terrestrial-type celestial bodies?&lt;/p&gt;&#xA;" OwnerUserId="2541" LastActivityDate="2015-10-11T19:53:21.137" Title="Is Ceres a terrestrial-type (dwarf) planet?" Tags="&lt;dwarf-planets&gt;&lt;terrestrial-planets&gt;&lt;asteroid-belt&gt;&lt;ceres&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12087" PostTypeId="2" ParentId="12041" CreationDate="2015-10-11T03:52:29.570" Score="2" Body="&lt;p&gt;Generally speaking, there's 2 types of planet systems in a binary - see pretty picture.   &lt;a href=&quot;http://jenomarz.com/designing-a-planetary-system-extension/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.  The writer is a sci-fi writer so the entire article might be of interest to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PWDRw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PWDRw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to binary star tidal forces, there are some setups that are unlikely but in your example with a 100 AU distance between the stars, an orbit around the larger star is reasonable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A double planet system is less likely.   It's unlikely to form on it's own during planet formation as that requires too much planetary angular momentum during formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's possible, but also unlikely to form by giant impact, as that's more likely to leave 1 planet and 1 moon.  I've read (but can't find an article right now) that there's a giant impact size ratio and it's in the planet-moon range, not planet planet.   Much less than 1 to 1.  Pluto-Charon is 9-1 and Earth-Moon 81-1.   A giant impact is also unlikely to create an Iron rich core for both objects.   It's not a good way to create planet-planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That leaves a 3rd possibility, also unlikely, but perhaps the most likely of the bunch is planet capture.    Planets can form in Trojan points in the same orbit (&lt;a href=&quot;https://en.wikipedia.org/wiki/Theia_%28planet%29#Orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Theia&lt;/a&gt;).  The difficulty with orbital capture is that the velocity needs to be just right and capture's are likely to be significantly elongated orbits, which, maybe, over time, perhaps with a 3rd asteroid, could even out to slightly more circular.   This is very improbable, but it might be the most likely way to form a double planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My 2 cents, as a layman.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-10-11T03:52:29.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12088" PostTypeId="2" ParentId="12086" CreationDate="2015-10-11T06:22:03.377" Score="2" Body="&lt;p&gt;I'm inclined to say no (and footnote, I realize Wikipedia isn't a good source for scientific proof as it's not always right, but I'm using it more to demonstrate a point than than use it as an authoritative definition).     &lt;a href=&quot;https://en.wikipedia.org/wiki/Terrestrial_planet&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A terrestrial planet, telluric planet or rocky planet is a planet that&#xA;  is composed primarily of silicate rocks or metals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;All terrestrial planets have approximately the same type of structure:&#xA;  a central metallic core, mostly iron, with a surrounding silicate&#xA;  mantle. The Moon is similar, but has a much smaller iron core. Io and&#xA;  Europa are also satellites that have internal structures similar to&#xA;  that of terrestrial planets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dwarf planets, such as Ceres and Pluto, and large small Solar System&#xA;  bodies are similar to terrestrial planets in the fact that they do&#xA;  have a solid surface, but are, on average, composed of more icy&#xA;  materials (Ceres and Pluto have densities 2.17 and 1.87 g cm−3,&#xA;  respectively, and Haumea's density is similar to Pallas's 2.8 g cm−3).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, in my opinion, half ice/half rock if a different category and not terrestrial.   There probably are planets that formed outside the frost line that are half rock-half ice in other solar systems, but I don't think I'd call them terrestrial.   We probably need a new word for them.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Ceres was to drift closer to an Earth's orbit, it's ice would melt and it would have oceans (and an H20 rich atmosphere), at least, for a little while anyway until it lost it's atmosphere.   Planets with Ceres composition would likely be water worlds if they were warm enough.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I can't speak for any the official answer, but that's my opinion.  If Ceres is Terrestrial and Haumea is more rocky than Ceres, then Ceres probably shouldn't be the only terrestrial dwarf planet in the solar-system.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;edit.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Geology_of_solar_terrestrial_planets&quot; rel=&quot;nofollow&quot;&gt;Link&lt;/a&gt; provided that says Ceres is a terrestrial dwarf planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Terrestrial planets have numerous similarities to plutoids (objects&#xA;  like Pluto), which also have a solid surface, but are composed of more&#xA;  icy materials.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some of this is just semantics, but lets look at density, which is a pretty good measure of Water-Ice and other ices to Rock content.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ceres 2.08 g/cm^3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the other dwarfs, by size (not all of them are well measured so density isn't certain).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eris 2.3&#xA;Pluto 1.88&#xA;Makemake ~2&#xA;Haumea  2.6-3.3&#xA;Quaoar ~2.2&#xA;Sedna (2.0?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I could go on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets do a few moons, just for fun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Io:  3.55  (extreme volcanism)&#xA;Europa:  3.0&#xA;Ganymede:  1.93&#xA;Calisto:  1.83&#xA;Titan:  1.88&#xA;Triton:  2.06&#xA;Enceladus:   1.61&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moons are easy, we don't call them dwarf planets even if they are larger than dwarf planets cause we call them moons. - definition averted.   :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have a hard time seeing why Ceres gets a different classification than other very similar objects which just happen to be further away.   Now, granted, Ceres, while it formed outside the frost line, it's now inside the frost line so surface ice on Ceres doesn't last.  It's got a rocky surface while most Kuiper belt objects have an icy surface.    I can see that, defining an object based on it's surface, but I still think a half rock/half ice object (of significant size, large enough to be round and meet dwarf planet criteria), should be called half rock-half ice or whatever that definition is.   I don't think it should be rocky/terrestrial if it's Ceres and icy if it's Pluto, even if that's how they look the surface, but that's just my opinion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;/// hope that wasn't too much of a rant.  :-)&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-10-11T19:53:21.137" LastActivityDate="2015-10-11T19:53:21.137" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12089" PostTypeId="2" ParentId="12064" CreationDate="2015-10-11T10:03:38.517" Score="1" Body="&lt;p&gt;Sirius is in Canis Major, and if you take the belt of Orion to be the reference point then we can proceed as follows. If we define Alnitak, Alnilam and Mintaka as going left to right as we perceive it to be in the Northern Hemisphere, then we need to draw a line going from Mintaka in a SW-ish (left and down) orientation through Alnitak to Canis Major. You should see it. It's coming up to the best time of year to see Sirius and the Orion constellation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;N.B. I'm pretty certain what you are referring to as the object you're seeing at this present moment is Venus. &lt;/p&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2015-10-11T10:03:38.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12090" PostTypeId="1" CreationDate="2015-10-11T12:22:59.913" Score="2" ViewCount="1603" Body="&lt;p&gt;Does anyone know how to find a location by using :&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the direction of shadow relative to the normal (for example azimuth&#xA;angle) &lt;/li&gt;&#xA;&lt;li&gt;the Universal time (UT) and date&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I have a satellite image and I know north, time and date. One of the cues that I can use is, for example, in the northern hemisphere the direction of shadow is north.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would anyone know if it were possible, and how, to calculate the location based on these parameters?&lt;/p&gt;&#xA;" OwnerUserId="9450" LastEditorUserId="2091" LastEditDate="2015-10-11T16:58:47.630" LastActivityDate="2016-09-05T21:28:18.240" Title="Finding a location using the direction of shadow, UT and date approximately" Tags="&lt;coordinate&gt;&lt;positional-astronomy&gt;&lt;azimuth&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12093" PostTypeId="2" ParentId="12090" CreationDate="2015-10-11T17:12:25.363" Score="1" Body="&lt;p&gt;Planetarium software, such as &lt;a href=&quot;http://www.stellarium.org/&quot; rel=&quot;nofollow&quot;&gt;Stellarium&lt;/a&gt;, or astronomical computation software such as &lt;a href=&quot;http://rhodesmill.org/pyephem/&quot; rel=&quot;nofollow&quot;&gt;pyephem&lt;/a&gt; can compute the exact position of the sun at any date, time and location. It is then simple trigonometry to calculate the length of a shadow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if the sun is $\theta$ degrees above the horizon, a vertical building of height $h$ will cast a shadow on horizontal ground of length $\frac{h}{\tan(\theta)}$.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-10-11T17:12:25.363" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12095" PostTypeId="1" CreationDate="2015-10-12T08:49:30.817" Score="6" ViewCount="1203" Body="&lt;p&gt;I would like to be able to connect to a web-service and retrieve JSON data containing information on current (as of today or so) orbits of planets, comets, asteroids, and dwarf planets in our solar system. (Even just planets would be fine but better if I can get info about asteroids, comets, and dwarf planets as well).&#xA;I would like to draw planets and their orbits accurately in a 2D animation of our solar system in an Android app in the X-Y coordinate plane as observed from the top of the Z axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where can I find such public web service and how can I access it and get that data?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;&#xA;I tried using HORIZONS Web-Interface but out of all variables I need, I can see only &quot;Semi-Annual Axis&quot; information.  I need following information:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;orbital ellipse angle in X-Y coordinate system (i.e 1.5rad or degrees or so)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;motion direction (pro-grade or retrograde)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;current position of planet/comet/asteroids on that ellipsis in degrees or radians, say 60deg from x (horizontal) axis&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;time it takes plant to go around the sun (I believe this is called orbital period)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;eccentricity of ellipse (orbit)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;semi-annual axis&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These are the values that give me enough information to draw orbit (ellipsis) and correctly position planet on that ellipsis, and to calculate speed at which it moves around the sun in my 2D model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I am not interested in inclination or any parameters that would be useful for a 3D model.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can find this data online but it takes lots of searching since I have not been able to find a central place for it.  HORIZONS seems to be such place but out of variables I need (as described above), I can find there only semi-annual axis in HORIZONS.  I am surprised that I cannot fine a public web-service offering this info (REST or SOAP, or whatever, does not matter as long as I can fetch that data and process it for my needs from my application).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I apologize if my explanation is somewhat unclear, I am not an astronomer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that explains.&lt;/p&gt;&#xA;" OwnerUserId="9457" LastEditorUserId="9457" LastEditDate="2015-10-14T18:18:34.540" LastActivityDate="2019-03-04T08:28:02.053" Title="Is there a web service providing orbital data of solar bodies (planets, comets, asteroids, dwarf planets)?" Tags="&lt;orbit&gt;&lt;planet&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12096" PostTypeId="2" ParentId="8605" CreationDate="2015-10-12T10:07:50.187" Score="6" Body="&lt;p&gt;First, I think Rob Jeffries answer is brilliant.  I'll just add a minor points that might be worth mentioning.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What would we be able to detect and with what telescope(s)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alpha Centauri A is a binary star with Alpha Centauri B and they are close enough in size to have no stable L4 or L5, so anything that orbits either one of them would need to be either very close (Mercury distance perhaps Venus) or very far and very cold, much greater than Pluto distance, orbiting both stars like Proxima Centauri does.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VUb9d.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VUb9d.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you put a Jupiter in it's solar orbit around A or B, the 3 body effect would almost certainly create a wildly unstable orbit for the planet that likely wouldn't last long, so, one answer to this question is that our solar-system type of orbit around A or B is impossible.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Could we detect radio transmissions and/or any atmospheres?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For now, our detection of atmosphere is very limited and only to large planets very close to their stars, but, the article says they're working on that with bigger telescopes on the way, so maybe in a few years we'll get something on that for habitable zone planets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.universetoday.com/92962/how-well-can-astronomers-study-exoplanet-atmospheres/&quot; rel=&quot;noreferrer&quot;&gt;exoplanet atmosphere detection&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On radio-waves and worth mentioning, visible light, I couldn't find a good article, but if an alien planet shoots a message towards us in a tight beam - then, I'm sure we could detect that, provided they shoot a big enough beam, But could we detect another earth with our current output?  I don't think we're close to that kind of detection technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(and if I got any of that wrong, I welcome correction).  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(I asked something like this during a von Karman lecture by Neil&#xA;  Turner)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Did he answer you?   Did he say anything good?&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-10-12T10:07:50.187" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12098" PostTypeId="1" AcceptedAnswerId="12112" CreationDate="2015-10-13T00:31:28.720" Score="1" ViewCount="244" Body="&lt;p&gt;Using the standard formula for effective temperature of a planet, Venus has T(eff)=220K, Earth has T(eff)=255K and Mars has T(eff)=212K (using albedo for Venus a=0.7, Earth a=0.367, Mars a=0.25). Average surface temperatures of 730K, 288K and 218K respectively are easily explained by differences in greenhouse effect. The moon however with an albedo of a=0.1 should have a T(eff) of about 270K, but the average surface temperature is about 250K. &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Effective_temperature&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is where I got the T(eff) formula and &lt;a href=&quot;http://www.asi.org/adb/02/05/01/surface-temperature.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; is where i got the average temperature of the moon. (the latter is a new citing and I have edited the mean temperature value) I have revised my planet temperatures using &lt;a href=&quot;http://www.atmos.washington.edu/2002Q4/211/notes_greenhouse.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; reference.&#xA;What explains this negative difference?&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="7997" LastEditDate="2015-10-15T01:24:59.287" LastActivityDate="2015-10-15T01:24:59.287" Title="Moon's negative greenhouse effect?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12099" PostTypeId="1" AcceptedAnswerId="12117" CreationDate="2015-10-13T00:43:59.067" Score="4" ViewCount="456" Body="&lt;p&gt;Take for example an ocean planet slightly smaller than the Earth. Let's say it has a planet-wide ocean that is on the average 10 miles deep. Say for argument sake, that there is a molten, rotating outer core producing a magnetic field and that there are plate tectonics (although not to the same degree as Earth). Also, imagine that there are volcanoes that reach as high as 5 miles below sea level. The planet's effective temperature is similar to Earth (not counting greenhouse). Would such a world without land always be covered with clouds? Would it necessarily have a runaway greenhouse? How about such a world with no vulcanism?&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="7997" LastEditDate="2015-10-15T01:30:00.063" LastActivityDate="2020-05-15T16:27:16.817" Title="Are all models of ocean planets theoretically cloud covered?" Tags="&lt;exoplanet&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12100" PostTypeId="1" CreationDate="2015-10-13T04:25:53.967" Score="2" ViewCount="1708" Body="&lt;p&gt;If a black hole can swallow everything, even light, then how can a black hole power anything, particularly something as bright as a quasar. Why doesn't the black hole swallow the light of the quasar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are quasars formed when two black holes combine? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assumed it might be like an orange when we press it from middle then its juice comes out from both ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my assumption right...or is there another reason?&lt;/p&gt;&#xA;" OwnerUserId="9458" LastEditorUserId="2365" LastEditDate="2015-10-13T21:04:14.787" LastActivityDate="2015-10-14T18:24:32.873" Title="If quasars are powered by black holes, why are they so bright?" Tags="&lt;black-hole&gt;&lt;quasars&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12101" PostTypeId="2" ParentId="12100" CreationDate="2015-10-13T06:30:28.333" Score="5" Body="&lt;p&gt;The light emitted isn't actually from the black hole itself but the swirling matter that surrounds it called the accretion disk. The light emitted from the accretion disk is far enough to escape from the pull of the quasar. A link from wikipedia sufficiently explains this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Quasars are believed to be powered by accretion of material into supermassive black holes in the nuclei of distant galaxies, making these luminous versions of the general class of objects known as active galaxies. &lt;strong&gt;Since light cannot escape the black holes, the escaping energy is actually generated outside the event horizon by gravitational stresses and immense friction on the incoming material.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am supposing that part of the question, you are asking why quasars are brighter than black holes. That's because of the larger amount of in falling matter into the black hole creates greater friction thus more light.&lt;/p&gt;&#xA;" OwnerUserId="7926" LastEditorUserId="7926" LastEditDate="2015-10-13T12:19:18.890" LastActivityDate="2015-10-13T12:19:18.890" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12102" PostTypeId="1" AcceptedAnswerId="12103" CreationDate="2015-10-13T06:36:22.507" Score="1" ViewCount="161" Body="&lt;p&gt;2 black holes are about to merge to become a larger black hole. Does this mean it'll become a quasar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I know, quasars are supermassive black holes or a collective amount of them. So, if it merges and becomes a supermassive black hole, does this mean that it is a quasar? If the size is smaller than what would be considered as a supermassive black hole would it not be a quasar?&lt;/p&gt;&#xA;" OwnerUserId="7926" LastActivityDate="2015-10-13T22:52:43.907" Title="Does 2 merging black holes necessarily make a quasar?" Tags="&lt;black-hole&gt;&lt;quasars&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12103" PostTypeId="2" ParentId="12102" CreationDate="2015-10-13T08:33:29.127" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does 2 merging black holes necessarily make a quasar?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Basically no.  While the merging of 2 black holes is a very interesting event, a quasar is what you get when 1 very large black hole eats a whole bunch of matter and the light from the quasar comes from the intense heat and interactions from that tightly bound, rapidly spiraling and very excited matter.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quasar's were probably most common when galaxies are young but there are a few more recent ones (see examples in comments).  &lt;a href=&quot;http://www.universetoday.com/73222/what-is-a-quasar/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Quasar&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/8623/how-often-does-the-supermassive-black-hole-in-the-milky-way-become-a-quasar?rq=1&quot;&gt;Related Stack Exchange Question&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the size is smaller than what would be considered as a supermassive&#xA;  black hole would it not be a quasar?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Almost all large galaxies have a super-massive black hole in their center.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Supermassive_black_hole&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.  The sizes vary with the size of the galaxy.   Quasars are much more rare, only in a few galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the specific size of black hole that can form a quasar, credit to Rob Jeffries below.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-10-13T22:52:43.907" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12104" PostTypeId="1" CreationDate="2015-10-13T10:12:49.147" Score="-2" ViewCount="362" Body="&lt;p&gt;The pictures show exactly what my naked eye visualized.What is the cause of this?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yFNCc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yFNCc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/4E2Sa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4E2Sa.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/m7pBx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/m7pBx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/BEXAv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BEXAv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9326" LastEditorUserId="7982" LastEditDate="2021-02-28T11:51:56.707" LastActivityDate="2021-02-28T11:51:56.707" Title="Why don't these Clouds show no depth perception in correspondence to the Sun" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;planetary-atmosphere&gt;&lt;atmospheric-effects&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2015-12-18T10:23:51.700" ContentLicense="CC BY-SA 3.0" />
  <row Id="12105" PostTypeId="1" CreationDate="2015-10-13T10:28:11.573" Score="-3" ViewCount="219" Body="&lt;p&gt;On Sep 25 2015 in California, at 8pm the moon was observed between 11 o'clock and 12 o'clock position in the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Sep 26 at 8pm at the same location the moon appeared roughly between 11 o' clock and 12 o'clock position as night before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Sept 27 at 8pm same location the super moon was not between the 11 o'clock and 12 o'clock but seen at 9:30pm at the 9 o'clock way behind trajectory. What is the cause of this? &lt;/p&gt;&#xA;" OwnerUserId="9326" LastEditorUserId="2365" LastEditDate="2015-10-13T21:27:20.660" LastActivityDate="2015-10-17T20:29:36.397" Title="Super moon or Super lag?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ClosedDate="2015-10-15T19:53:02.943" ContentLicense="CC BY-SA 3.0" />
  <row Id="12106" PostTypeId="2" ParentId="12070" CreationDate="2015-10-13T12:45:19.933" Score="0" Body="&lt;p&gt;The most basic reason is that the human naked eye cannot see all parts of the electromagnetic spectrum. There are a lot more stars than what we see on a regular basis, probably lying in the infrared region. &lt;/p&gt;&#xA;" OwnerUserId="9394" LastActivityDate="2015-10-13T12:45:19.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12107" PostTypeId="1" AcceptedAnswerId="12110" CreationDate="2015-10-13T15:53:10.463" Score="1" ViewCount="114" Body="&lt;p&gt;I am having difficulty with some terminology:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;sub-Martian hemisphere of Deimos ... what place on Deimos does it refer to?&lt;/li&gt;&#xA;&lt;li&gt;leading/trailing apex of Phobos ... what place are those?&lt;/li&gt;&#xA;&lt;li&gt;Apex ... in general it means a top of something, or an end of something pointed; but what does it mean on a relatively spherical body?&lt;/li&gt;&#xA;&lt;li&gt;Anti-Mars meridian ... is this a meridian on the side of the moon facing away from Mars?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="9469" LastEditorDisplayName="user1569" LastEditDate="2015-10-14T10:12:47.517" LastActivityDate="2015-10-14T17:11:42.733" Title="Questions about terminology used for Mars and its moons" Tags="&lt;planet&gt;&lt;fundamental-astronomy&gt;&lt;terminology&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12109" PostTypeId="2" ParentId="12104" CreationDate="2015-10-13T18:39:45.683" Score="1" Body="&lt;p&gt;Since the sun is opaque, there is no such thing as seeing clouds behind it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture seems to show clouds of different densities passing in front of the sun. &lt;/p&gt;&#xA;" OwnerUserId="420" LastActivityDate="2015-10-13T18:39:45.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12110" PostTypeId="2" ParentId="12107" CreationDate="2015-10-13T21:24:09.177" Score="3" Body="&lt;p&gt;Deimos is, like our own moon, tidally locked to Mars, so one hemisphere of Deimos faces the planet, and the faces away. The Sub-Martian hemisphere faces Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Deimos is locked, there is a point on its surface that always faces in the direction of orbital motion, that point is the leading apex. It is an apex in the sense of a &quot;leading point&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A meridian is a line of longitude, running from the North to the South pole. One of Deimos's meridians goes through the centre of its hemisphere that faces away from Mars, this is the anti-Mars Meridian. As you say, the &quot;meridian on the side of Deimos facing away from Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image here shows a map of Deimos:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XJdkj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XJdkj.jpg&quot; alt=&quot;The leading face is the left side of the image, and the trailing face is on the right. The central 50% is the sub-martian hemisphere and the anti mars meridian is the line on the left and right edge, where the 3D image of Deimos has been cut to make a 2D map&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The leading face is the left side of the image, and the trailing face is on the right. The central 50% is the sub-martian hemisphere and the anti mars meridian is the line on the left and right edge, where the 3D image of Deimos has been cut to make a 2D map&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2015-10-14T17:11:42.733" LastActivityDate="2015-10-14T17:11:42.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12111" PostTypeId="1" AcceptedAnswerId="12116" CreationDate="2015-10-13T23:03:42.087" Score="6" ViewCount="658" Body="&lt;p&gt;Previously, I was told that in order for any bodies to orbit each other for over a long period of time, the orbital period, distance and masses have to be precisely matched such that the bodies won't eventually collide and forming larger bodies or being swung out of orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is it then, that all 8 planets (and countless dwarf planets), and their moons, and the Sun are all in such equilibrium for over thousands of years (if not millions)? Could it be possible what we see today are what's left, i.e. countless planets or moons have been swung out of the Solar System and collided, until a stable (relatively speaking) system is reached as we observe today?&lt;/p&gt;&#xA;" OwnerUserId="9472" LastEditorUserId="2153" LastEditDate="2015-10-14T00:21:30.167" LastActivityDate="2015-10-14T09:45:01.617" Title="How is it that all planets (and moons) in our solar system are in equilibrium orbits?" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;planet&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12112" PostTypeId="2" ParentId="12098" CreationDate="2015-10-13T23:05:30.120" Score="2" Body="&lt;p&gt;You don't say how you got your average temperature. If it is a straightforward average, then I think I have your answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature of the moon is highly non uniform, ranging from around 100 K on the night side to 370 K on the sunlit side. This gives a crude average of 235 K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://space.stackexchange.com/questions/3290/what-is-the-temperature-variation-between-the-lunar-day-and-night-on-the-moon&quot;&gt;https://space.stackexchange.com/questions/3290/what-is-the-temperature-variation-between-the-lunar-day-and-night-on-the-moon&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in terms of blackbody radiation, if we treat both hemispheres as separate blackbodies, then together they radiate as much energy as a hotter blackbody because of the $T^4$ factor in Stefan's law. The &quot;effective temperature&quot; will be &#xA;$$ T \simeq (0.5\times 100^4 + 0.5\times 370^4)^{0.25}= 311\ K$$&#xA;So this is more like your equilibrium temperature and a more appropriate area weighting and/or tweaks to the exact range of temperatures could bring it into close agreement.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2015-10-13T23:05:30.120" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12113" PostTypeId="2" ParentId="12104" CreationDate="2015-10-13T23:12:01.377" Score="0" Body="&lt;p&gt;It may not seem related, but here's a detailed answer:  &lt;a href=&quot;https://photo.stackexchange.com/questions/27643/why-dont-cameras-capture-dynamic-range-as-our-eyes-do&quot;&gt;https://photo.stackexchange.com/questions/27643/why-dont-cameras-capture-dynamic-range-as-our-eyes-do&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The camera doesn't differentiate between different colors and brightness nearly as well as our eyes do so in the photos you linked, both the clouds in-front of the sun and sky that's kind of around but looks behind the sun appear the same color and to our brains, it appears the sun is rising in the middle of the clouds - like a tall building or mountain peak might appear in the middle of some clouds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were there in person when the photo was taken, this illusion probably wouldn't be there.   (nice photos though).&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:44:06.413" LastActivityDate="2015-10-13T23:12:01.377" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12114" PostTypeId="2" ParentId="12105" CreationDate="2015-10-14T02:56:22.307" Score="5" Body="&lt;p&gt;This is the animation of the Moon's position over Los Angeles, CA skies at 8 p.m. local (PDT) time from September 25 to September 27, 2015 (made with &lt;a href=&quot;http://www.skymaponline.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sky Map Online&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&lt;a href=&quot;https://i.stack.imgur.com/rfnKo.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rfnKo.gif&quot; alt=&quot;Sky map&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, there isn't any way to describe its apparent movement with a single angular distance, whichever method you use. In astronomy, position of objects on the night skies is most commonly given by RA/Dec (&lt;a href=&quot;https://en.wikipedia.org/wiki/Right_ascension&quot; rel=&quot;nofollow noreferrer&quot;&gt;Right Ascension&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Declination&quot; rel=&quot;nofollow noreferrer&quot;&gt;Declination&lt;/a&gt;), which places an object on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Celestial_sphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;celestial sphere&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless we count the &lt;a href=&quot;https://en.wikipedia.org/wiki/September_2015_lunar_eclipse&quot; rel=&quot;nofollow noreferrer&quot;&gt;Supermoon Eclipse&lt;/a&gt; on September 27, 2015 (your time, September 28 UTC), one that won't occur again until &lt;a href=&quot;https://en.wikipedia.org/wiki/October_2033_lunar_eclipse&quot; rel=&quot;nofollow noreferrer&quot;&gt;October 8, 2033&lt;/a&gt;, or exactly one &lt;a href=&quot;https://en.wikipedia.org/wiki/Saros_%28astronomy%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;saros period&lt;/a&gt; since the last one, there wasn't anything else as extraordinarily dramatic about the Moon's movement on those nights as you describe. So what happened is most likely that measurements by your observations weren't taken precisely enough; Be it you didn't time your observations precisely, didn't measure Moon's position relative to some fixed direction from some fixed vantage point, didn't project its position onto your scale taking the curvature of the celestial sphere into account, or a bit of any or all of these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon, if observed at the same solar time from the same location, moves in apparent eastward direction (when it's visible and above the horizon, westwards when it's below the horizon, regardless if that's during day or night) roughly 13° per day, or, saying it otherwise, each day, at the same time, it will appear to lag for about an hour relative to its position a day before. Why? Simply because it completes one orbit around the Earth every 27.321582 days, so at exact same solar time the next day, it will be 360°/27.321582 or 13.18° more East. Its apparent movement during the same day is of course still East to West, because the Earth rotates on its axis towards East much faster than the Moon rotates around the Earth.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditDate="2015-10-17T20:29:36.397" LastActivityDate="2015-10-17T20:29:36.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12115" PostTypeId="1" AcceptedAnswerId="12139" CreationDate="2015-10-14T09:35:51.577" Score="1" ViewCount="57" Body="&lt;p&gt;Just a quick question relating to the thermal Blackbody temperature of a celestial object. In the c.g.s regime of astrophysics, is it more 'sophisticated' to quote the thermal Blackbody temperature in units of Kelvin ($\mathrm{K}$) or in units of $\mathrm{keV}$? That is to say to use $k_{B}T_{\mathrm{bb}}$ and convert to an energy?&lt;/p&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2016-10-25T18:13:59.963" Title="In what units to quote the thermal Blackbody temperature" Tags="&lt;astrophysics&gt;&lt;high-energy-astrophysics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12116" PostTypeId="2" ParentId="12111" CreationDate="2015-10-14T09:45:01.617" Score="5" Body="&lt;p&gt;I think at our current stage of solar system evolution, because of the fact that we have been able to evolve to our current level of sophistication, it could be considered to be very stable and in a very calm period in its evolutionary history. Unstable objects usually will be flung out very early on in the formation of such a system, hence why we do not see them today. However, just look at the Earth-Moon evolutionary history. It is theorised that our current Moon was captured due to a collision between Earth and perhaps a Mars (or similar size) object. This is anything but stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why, is of course, related to gravity. Or more correctly, the gravitational potential of the system. All systems want to thermalise (in the same sense that a room of air molecules want to reach a thermal equilbirium). Although our solar system is anything but thermalised, it is constantly working to achieve this. Hence, why, at our present state of the solar system's evolution, we seem like we are in a fairly calm state. At this state, to go into more depth, we can employ Bertrand's theorem which tells us that for a central potential with an $r^{-1}$ dependance on radial distance the orbits will be stable. The stability of orbits in three spatial dimensions is due to the fact that the gravitational potential decreases with distance $r$ as $r^{-1}$.      &lt;/p&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2015-10-14T09:45:01.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12117" PostTypeId="2" ParentId="12099" CreationDate="2015-10-14T10:12:06.380" Score="2" Body="&lt;p&gt;Having given this some thought, it's a bit of a monster question with a few variables that you don't mention.  Planetary tilt, which causes seasons is one.   Length of day, for example, a rotation like Venus' 116 days would be quite different than an earth rotation of 24 hours and a tidally locked planet would also be different.   Atmospheric pressure is another.   I'm assuming Earth like tilt, 24 hour day since you didn't mention, and 1 ATM, but I think, if and when we finally get a good look at earth-like exo-planets, which, granted, might be several decades away, I think we'll see a few unexpected things.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, moving on.   Earth likely got much of it's atmosphere from comets which suggests healthy amounts of CO2, CH4 and NH3, maybe N2, some from the NO family, maybe some Helium, Argon and Neon, those 3 I think we can mostly ignore, some hydrogen perhaps from solar flares, which might get stripped along with the helium from the planet over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total amount of comet and asteroid impacts and ratio of gasses add some variability and a really important factor is whether or not the planet has life and undergoes photosynthesis.   Photosynthesis would put O2 into the atmosphere and reduce CO2 and would also have the effect of, over time, reacting with and eliminating the virtually all of the CH4 from the air and any dissolved Iron in the oceans, turning the oceans from a muddy brownish/red to clear blue.  Photosynthesis likely played a key role in turning the Earth from a hot planet to a &lt;a href=&quot;http://www.snowballearth.org/cause.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;snowball&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, lots of variables, but it's a fun speculative question, so I'll give it a shot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much smaller than Earth?   I'm going to go with Venus size but if you have different in mind, let me know.   .815 Earth masses, 6,052 km in Radius and gravity of .905 of Earth.  &lt;a href=&quot;http://nssdc.gsfc.nasa.gov/planetary/factsheet/venusfact.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10 miles deep oceans (16.09 km), surface area, 4 Pi R^2 about 7.4 billion cubic km of water.  (Earth, by comparison, about 1.37 billion cubic km)  &lt;a href=&quot;http://hypertextbook.com/facts/2001/SyedQadri.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.   Your planet has about 5.4 times as much water as earth (not counting water in the crust, but you didn't mention that, so lets not go there) - and planet size doesn't affect the predictions much.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I wonder how much outgassed CO2 etc. would stay dissolved in the ocean&#xA;  and how much would accumulate in the atmosphere? Nitrogen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm going to assume that the theory is correct that Earth (and your water world) get most of it's atmosphere and water from asteroids and comets &lt;a href=&quot;http://news.discovery.com/earth/how-early-earth-got-its-nitrogen-150120.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt; and &lt;a href=&quot;http://www.livescience.com/10229-earth-owe-life-comets.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;, though the 2nd source suggests some could have come from the primordial material of the Earth.   If your water-world has a lot of gas trapped below it's crust in it's primordial material, then out-gassing from volcanism becomes a bigger factor and the volcanoes, not the atmosphere above would keep the oceans saturated with gas.   That's possible, but I'm going to assume that most of the gas is already in the atmosphere following a late heavy bombardment period.   I'll touch a bit more on tectonics later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Titan Probably got it's thick atmosphere from outgassing &lt;a href=&quot;http://www.nasa.gov/mission_pages/cassini/media/methane20060302.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt; and perhaps Venus from a recent large outgassing event too - just throwing that out there.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also assuming that your planet wouldn't have permanent ice over it's oceanic poles due to the warmish temperature you implied and likely oceanic circulation not blocked by land-masses would prevent ice formation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Would it necessarily have a runaway greenhouse?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Greenhouse Gases, CO2, CH4, NO family, (H20, more indirectly)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your planet has a lot of CO2/CH4, comparable amounts to the amount of H20, then it's unlikely the oceans could begin to dissolve enough gas and this would lead to a run-away Greenhouse, almost without question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your planet has more earth like CO2/CH4 levels, perhaps driven by photosynthesis capturing carbon and released oxygen chemically reacting with the CH4, then the runaway greenhouse can be avoided and CH4 be very low concentration.   CH4 on Earth is currently about 1.8 PPM and prior to farming and livestock and oil drilling &amp;amp; Fracking which can release some CH4, it was probably less than half that much, less than 1 PPM.  In a water world CH4 would probably be even less as there's no biodegrading or digestion of plant-mass releasing CH4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CO2 is more complicated and the amount dissolved in the ocean depends on the total amount available.  If there's not too much CO2 to saturate the oceans, then you have an ocean/air equilibrium which is also affected by temperature.    For a more detailed answer on this, look into &lt;a href=&quot;http://butane.chem.uiuc.edu/pshapley/GenChem1/L23/web-L23.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Henry's law&lt;/a&gt;, but I'm going to cheat and do a quick and dirty calculation rather than use Henry's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;on Earth, the oceans contain about 50 times the CO2 as the Atmosphere:   &lt;a href=&quot;http://www.waterencyclopedia.com/Bi-Ca/Carbon-Dioxide-in-the-Ocean-and-Atmosphere.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.  And the oceans weigh about 265-270 times more than the atmosphere.   (ocean mass given above, atmosphere mass &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Earth&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we estimate this as parts per million (ppm), the CO2 concentration in Earth's atmosphere is about 5 times the concentration in Earth's oceans, and at higher temperatures, that ratio goes up, lower temperatures it goes down, but it never gets to the point that there's no CO2 in the air, cause not possible in an equilibrium.   With 10 mile deep oceans, we can assume a more CO2 in the oceans, maybe 99% vs about 98% ocean to air ratio currently on earth but the 1% in the air would depend on the total mass of CO2 in the water world's ecosystem, ignoring anything permanently trapped deep on the ocean floor, such as sea-shells.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Would such a world without land always be covered with clouds?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lets say you have an Earth like Oxygen/Nitrogen atmosphere, 1 bar, a planet very similar to earth but just oceans.  Similar weather (as a starting point), this planet should have more clouds as clouds are more common over the oceans, &lt;a href=&quot;http://www.giss.nasa.gov/research/briefs/rossow_01/distrib.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;, and you would have no low humidity/dry pockets of land, but I would think, probably not 100% clouds, just more clouds.   The cloud effect on temperature has some uncertainty to it, as clouds both both reflect sunlight making the planet colder during the day and they trap heat at night, so they play for both the warming team and the cooling team, but what I've read, the overall effect is pretty small.   Days could be colder, nights warmer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, evaporation tends to cool the air at the surface, which is why islands surrounded by ocean don't ever get the scorching hot temperatures you get in death valley for example, even if they're closer to the equator, so you wouldn't get heatwaves, but you wouldn't get freezing cold either.  (I would think).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, oceans have lower albedo and a water-world earth would likely have no ice-caps, so, Overall, I think it's likely a water-world earth would be warmer primarily due to lower albedo and this would increase atmospheric water-vapor (not clouds but transparent water vapor, which is a greenhouse gas), so a water-earth would probably several degrees warmer on average with no ice-caps.    How much warmer - I have no idea but I don't think a water-earth would be run-away greenhouse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A water world (Earth) could also have less CO2 in the atmosphere due to more dissolved in water and probably less CH4 from decaying matter on land, so, it might actually be colder, but this would more likely depend on the life cycle's O2 to CO2 ratio rather than oceanic absorption of CO2.   Too many unknowns to say for sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another curious effect to a water world earth is you might get some hurricanes that last for weeks, maybe even months or years in the right conditions, kind of like mini versions of Jupiter's great red spot.   All you need for a hurricane to gain strength is warm water and cold air and with no land for the Hurricane to lose strength on, You could get some real jumbos.   Maybe category 7s, perhaps 8s.  That would be awesome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if (and I think this is the gist of your question), what happens if you decrease the atmospheric pressure by significantly lowering the amount of O2/N2 in the atmosphere (or raising it).  Lower atmospheric pressure lowers the boiling point of water and if you lower the atmospheric pressure enough, then water starts to boil and you get a partial water vapor atmosphere as a result of low atmospheric pressure and lots of water.   I suspect this is unlikely to actually happen as a planet should always have enough other types of gas to prevent this unlikely outcome but a water vapor atmosphere can be approached logically, even if it's not going to really happen.  Now, this isn't just high humidity, this is an actual water vapor atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine a water-vapor rich atmosphere would form droplets and rain fairly regularly and be permanent or nearly permanent cloud cover but that's just a guess, though it would probably appear more like low to the ground smog than cloud cover, the clouds might hover much closer to the earth.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you increase atmospheric pressure, then water has a higher boiling point and takes more energy to vaporize.   A denser atmosphere might hold heat better and be a bit warmer.  It might also have fewer clouds due to more air and perhaps less variation in temperature and circulation, as it's circulation and warm air cooling off that's the primary driver in cloud formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water vapor would still form under a higher pressure atmosphere by wind and by photons.  Sunlight plays an important role in evaporation, perhaps even more than temperature based on &lt;a href=&quot;https://en.wikipedia.org/wiki/Pan_evaporation&quot; rel=&quot;nofollow noreferrer&quot;&gt;pan evaporation&lt;/a&gt; studies.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a very thin atmosphere made almost entirely of water (getting back to our improbable example).  You'd drown if you tried to breath it, but as far as temperature of such a planet, at least with earth temperatures, water stays liquid even at quite quite low atmospheric pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 1/2 PSI (1/29th of 1 atm) the boiling temperature of water is 79.6 degrees C.  &lt;a href=&quot;http://www.engineeringtoolbox.com/boiling-point-water-d_926.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.   I'm not sure how much heat 1/29th of an atmosphere could trap even if it was mostly water which is a greenhouse gas, so this one is hard to predict.  You'd likely see wild night to daytime temperature swings, frequent boiling of the oceans during the day and very heavy rainfall at night, much more rain than we ever see on earth, as 1/29th of an ATM of water vapor is several times more water than is in the earth's atmosphere at any given time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a water-vapor atmosphere planet (unlikely), got enough heat from it's sun, it could certainly turn into a run-away greenhouse, but I'm not sure if that happens at the radiance earth gets.   But a planet like this would probably be subject to bigger temperature swings than we see on Earth and very fast wind.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How about such a world with no vulcanism?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Vulcanism - as in, spock, . . . sorry).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We think of plate tectonics and volcanism as a process by which gas is given to the atmosphere, and that's true to an extent.  Certainly any gas trapped inside a planet at formation can be outgassed through Volcanism, but perhaps a more important aspect of plate tectonics isn't the release of gas but the absorption of gas.   Oxygen is very reactive and it combines with basaltic rock to make lighter/stronger rock like granite and bedrock which leads to continents and mountain ranges and all that good stuff.  Nitrogen can bind with basalt too, but Oxygen I think, binds more readily.   Without photosynthesis and the creation of oxygen, not just life on earth but the continents would look very different and they'd probably be much less permanent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if all the volcanoes are under water, that limits the atmospheric absorption of oxygen and other gases to just what's disolved in the oceans which could be a lot of NH3, but relatively lower amounts of other gasses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another effect of all tectonic activity and volcanoes being under water is that volcanic gas like sulfates and tiny dust particles which can cool the planet after a large eruption, are unlikely to reach the atmosphere in any significant volume, so you're likely to avoid any volcanic cooling like you get on occasion on earth.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you might think that volcanoes under oceans would warm the oceans and they would locally, but not much overall.   A significant percentage of Earth's volcanoes are under water, but because ocean water circulates, most of the deep oceans stay at a chilly 4 degrees C, as opposed to land which gets warmer as you dig into the ground.  Convection circulates heat much faster than conduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Undersea volcanoes in your water world could be very helpful in the continuation and maintenance of extremophile life due to repleneshment of nutrients but not much in the way of temperature change or cloud formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally - a fun bit worth mentioning.   NH3, which is a very common gas/ice in the solar system.  It's in all the outer planets and comets and outer moons.   The neat thing about NH3 is it's water soluble and it's an energy source to primitive life, so simply by having oceans, stinky but useful NH3 is fairly quickly, mostly removed from the atmosphere and dissolved in liquid oceans and if there's life, it can be consumed, some released as N2 or NO-family and some can be built into amino acids.  Some nitrogen can bind with basaltic rock and magma from undersea volcanoes forming different types of rock, but I'm no geologist, so I'm not sure how much that would happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no one simple answer to this, but that's my best guess as a layman.   I enjoy thinking about stuff like this.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="7472" LastEditDate="2020-05-15T16:27:16.817" LastActivityDate="2020-05-15T16:27:16.817" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="12118" PostTypeId="1" AcceptedAnswerId="12166" CreationDate="2015-10-14T12:42:50.650" Score="3" ViewCount="6072" Body="&lt;p&gt;I was reading a book about the sun which suggested that a solar flare can destroy every electronic item on Earth. Is this really true and if so, how would it happen? Could we do anything to to prevent it? &lt;/p&gt;&#xA;" OwnerUserId="9458" LastEditorUserId="2270" LastEditDate="2015-10-14T15:59:51.383" LastActivityDate="2016-08-18T13:28:25.583" Title="Can a solar flare destroy every electronic item on earth?" Tags="&lt;the-sun&gt;&lt;flare&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12119" PostTypeId="2" ParentId="12100" CreationDate="2015-10-14T18:24:32.873" Score="2" Body="&lt;p&gt;A black hole, in deep space is basically black, and very hard to detect. But if a black hole is surrounded by material, that material will fall towards the black hole and enter into orbit about it. (&lt;em&gt;Black holes don't suck, they gravitate&lt;/em&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The material may come from, for example another star, or in the case of the giant black holes at the centre of many galaxies, from the gas, dust and stars that are found in the cores of galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As objects orbit the black hole they will tend to collide with each other, releasing energy in the form of heat, and causing them to fall to lower orbits. This process tends to cause the gas orbiting a black hole to form into a disk, called an accretion disc. Pretty soon any larger objects will be broken apart, and the accretion disk will be composed of gas, and as it heats up, plasma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now as objects fall to lower orbits they speed up. And for a black hole, this speed up is extreme. The gas will be orbiting at speeds that approach the speed of light. This makes friction and collisions between the particles that are orbiting the black extreme as well. The accretion disk heats up, to &lt;strong&gt;immense&lt;/strong&gt; temperatures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now something weird happens, and the physics of it is not really sorted out. Magnetic fields get tangled up in the hot matter and cause a portion of it to be ejected away from the black hole in a jet, perpendicular to the accretion disk. The speeds of the particles in the jet is close to the speed of light. &lt;strong&gt;Massive&lt;/strong&gt; amounts of electromagnetic energy is also released along this jet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quasars are active galaxies that happen have their jets pointing towards us. The large amounts of gas need to supply a massive black hole with the energy to make a quasar were more common in the early universe, so many quasars are very distant and very old, but the youngest is only about 700 million light years distant, and there is every reason to suppose that quasars still exist today. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't need a binary black hole to make a quasar, but the merging of two black holes could also release massive energy, and may be a type of gamma ray burst, and should also release gravitation waves.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-10-14T18:24:32.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12121" PostTypeId="1" AcceptedAnswerId="12125" CreationDate="2015-10-15T15:50:53.583" Score="1" ViewCount="343" Body="&lt;p&gt;What is the relation between azimuth angle of the sun the bearing of an vertical object's shadow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if the bearing of the shadow is 75° can we conclude that azimuth angle of the sun is 195°? Is the diagram below correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x1A0I.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x1A0I.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9487" LastEditorUserId="2365" LastEditDate="2015-10-15T18:38:38.833" LastActivityDate="2015-10-15T18:44:42.933" Title="relation between azimuth angle of the sun and angle of the shodow" Tags="&lt;the-sun&gt;&lt;positional-astronomy&gt;&lt;azimuth&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12122" PostTypeId="1" AcceptedAnswerId="12124" CreationDate="2015-10-15T17:09:37.577" Score="2" ViewCount="808" Body="&lt;p&gt;I have been reading &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0001141&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; paper on the separation of a binary quasar in which it states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At a projected separation of 8.3/h kpc (Omega_matter = 1), CTQ 839 would be the smallest projected separation binary quasar currently known.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is meant by projected separation and where does this unit come from?&lt;/p&gt;&#xA;" OwnerUserId="9488" LastEditorUserId="9186" LastEditDate="2015-10-15T19:54:31.427" LastActivityDate="2019-07-07T16:17:00.507" Title="What is projected separation and how can I make sense of its unit (h^-1 kpc)?" Tags="&lt;units&gt;&lt;quasars&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12123" PostTypeId="1" CreationDate="2015-10-15T17:27:54.890" Score="4" ViewCount="604" Body="&lt;p&gt;Is it possible that the intense gravity of a black hole converts matter into antimatter? Does this mean that matter or energy that goes inside the black hole is converted to antimatter?&lt;/p&gt;&#xA;" OwnerUserId="9231" LastEditorUserId="2153" LastEditDate="2015-10-16T00:19:36.760" LastActivityDate="2019-09-21T17:18:19.587" Title="Conversion of matter into antimatter" Tags="&lt;black-hole&gt;&lt;matter&gt;&lt;antimatter&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12124" PostTypeId="2" ParentId="12122" CreationDate="2015-10-15T17:42:47.013" Score="5" Body="&lt;p&gt;It means the separation between the two objects &lt;em&gt;if&lt;/em&gt; they were both at the same distance away. This separation is found by multiplying the angular separation (in radians) by the distance to the objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often, the distance to extragalactic objects depends on the assumed value of the Hubble parameter. In this case, $h$ is a dimensionless number corresponding to $h=H_0/100$, where $H_0$ is the Hubble parameter in km/s per Mpc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current best value of $H_0$ is about 70 km/s per Mpc, so $h\simeq 0.7$.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-10-15T17:42:47.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12125" PostTypeId="2" ParentId="12121" CreationDate="2015-10-15T18:44:42.933" Score="2" Body="&lt;p&gt;Your diagram is correct, your maths isn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bearing of the object's shadow is, as your diagram shows, away from the sun. If the bearing of the shadow is 75°, then the sun must have an azimuth of of 75°+180°, or 255°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly if the bearing of the shadow was 300° (pointing towards the northwest) the sun would be at an azimuth of 300 - 180°, or 120° (in the southeast)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-10-15T18:44:42.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12126" PostTypeId="1" CreationDate="2015-10-15T18:58:36.997" Score="1" ViewCount="48" Body="&lt;p&gt;In this &lt;a href=&quot;https://en.wikipedia.org/wiki/Nebula#Planetary_nebulae&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt;, it describes planetary nebula as &quot;short-lived,&quot; but I'm not sure in relation to what?  This could mean billions of years in some contexts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How long is short-lived?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus if there are any useful rules of thumb in astronomy terms for future uses (how stars timelines are described, versus galaxies, etc.)&lt;/p&gt;&#xA;" OwnerUserId="7370" LastEditorUserId="4085" LastEditDate="2015-10-15T19:16:39.837" LastActivityDate="2015-10-15T19:30:34.980" Title="What is meant by &quot;short lived&quot; in the duration of a planetary nebula?" Tags="&lt;nebula&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12127" PostTypeId="2" ParentId="12126" CreationDate="2015-10-15T19:30:34.980" Score="2" Body="&lt;p&gt;In the case of planetary nebulae we are talking of ~10,000 years (follow the link to the main article on &lt;a href=&quot;https://en.wikipedia.org/wiki/Planetary_nebula#Lifetime&quot; rel=&quot;nofollow&quot;&gt;Planetary Nebulae&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general a short lifetime for a phenomena means a time short compared to the total life of the entity it is related to. It is an informal term and does not have a fixed maximum value, but 10,000 years is a short time on most astronomical scales.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-10-15T19:30:34.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12128" PostTypeId="2" ParentId="12123" CreationDate="2015-10-15T21:56:26.613" Score="3" Body="&lt;p&gt;I'm assuming that by &quot;inside the black hole&quot; you mean inside the event horizon.  In that case, the answer is no.  The &lt;em&gt;local&lt;/em&gt; space of a particle inside the event horizon is the same as outside - The difference is just that the spacetime points inside and out are causally disconnected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right at the singularity, no one knows for sure what happens, but I think it's not likely that matter would turn to antimatter.&lt;/p&gt;&#xA;" OwnerUserId="8019" LastActivityDate="2015-10-15T21:56:26.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12129" PostTypeId="2" ParentId="12082" CreationDate="2015-10-15T22:11:43.097" Score="-2" Body="&lt;p&gt;With regards to first Universe I must say that it is first in many ways from within the Universe and within our own perspectives. I will try to stay close to the idea of first because you are unique. In the whole Universe there cannot be an exact, (exact containing the same molecules atoms and energy), copy of you out in this Universe so you, the rest of us and the planet are unique. Extrapolating from this, an argument could be made for the Universe being uniquely the first based on many assumptions but the crucial fact is, with the arrow of time the way it is, we are able to formulate such questions within our minds. Having an answer to such a question may not be as important as the question itself. Going forward the true realization maybe understanding how far our minds may reach. Since this question speaks of first it really is a temporal one and the passage of time can be understood as the progression of events in only one way. The singularity at the beginning would be where there is only that event {with no events?). Be that as it may we are living in side the result of that singularity at this event and as far as we can understand it is singular.      &lt;/p&gt;&#xA;" OwnerUserId="9489" LastActivityDate="2015-10-15T22:11:43.097" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12130" PostTypeId="2" ParentId="12123" CreationDate="2015-10-15T22:56:20.827" Score="4" Body="&lt;p&gt;It would be more accurate to say that a black hole eventually turns whatever falls in it into equal quantities of matter and antimatter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between matter and antimatter is defined by some conserved (or approximately conserved) quantum numbers. For example, matter baryons and quarks have positive baryon number, while antibaryons and antiquarks have negative baryon number. Leptons (e.g., electrons, neutrinos) and their antiparticles have a corresponding lepton number instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Semiclassically, a black hole evaporates through Hawking radiation. For astrophysically large black holes, this will be overwhelmingly be in terms of photons, which are their own antiparticles, but once the black hole evaporates to a small enough size, it will also radiate massive particles as well. Because there are no long-range forces that couple to baryon number or lepton number, a black hole is under no obligation to conserve them, and thus should radiate equal amounts of matter and antimatter, regardless of what fell into it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is unlike, say, energy-momentum, electric charge, and angular momentum, the presence of which makes a difference in the long-range gravitational and/or electromagnetic fields. Thus the total amount of charge the black hole will radiate should match the total amount that fell into it, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless, of course, quantum gravity gives us more surprises, which is definitely possible.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2015-10-15T22:56:20.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12131" PostTypeId="1" CreationDate="2015-10-15T23:07:51.273" Score="-1" ViewCount="160" Body="&lt;p&gt;What are the orbits and suspected objects or planets that have been noted around this star? KIC 8462852&lt;/p&gt;&#xA;" OwnerUserId="9489" LastActivityDate="2017-07-16T22:32:45.693" Title="My question relates to KIC 8462852. What are the orbits and suspected objects or planets that have been noted around this star?" Tags="&lt;kepler&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2017-07-24T17:07:23.627" ContentLicense="CC BY-SA 3.0" />
  <row Id="12132" PostTypeId="1" CreationDate="2015-10-16T04:42:56.037" Score="1" ViewCount="772" Body="&lt;p&gt;What would be the size of Sun as seen from surface of Pluto? It obviously depends on the distance and one would expect it to be very small. However, due to different optical phenomenon, the size of Sun seen on surface of Pluto could be larger. Can it can be estimated from the annular eclipse &lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2015/07/24/pluto_ice_flows_and_a_ring_of_light.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sun-Pluto eclipse&lt;/a&gt; picture or other pictures taken by New Horizons probe ? Thanks for your insight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture is also posted on &lt;a href=&quot;https://astronomy.stackexchange.com/questions/11415/why-are-there-no-stars-on-new-horizons-images-of-pluto&quot;&gt;this forum question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: From the answer given by @Conrad Turner below, Sun should be just pin point at that distance. However, when seen by New Horizons probe from behind Pluto, its size is seen to be almost as large as Pluto. The atmosphere of Pluto must be bending the light so much to make this happen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: This picture was taken from a distance of &lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2015/07/24/pluto_ice_flows_and_a_ring_of_light.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;2 million km&lt;/a&gt; from Pluto. At this distance Sun will have an angular size of 0'49&quot; (distance 5804550000 km, diameter 1392000 km), while Pluto has the angular size of 4'5&quot; (distance 2000000 km, diameter 2372 km). However, the picture is showing Sun to have an angular size slightly larger than Pluto (or nearly the same size). Hence, high refraction or scattering or some other phenomenon are at work here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering sizes and distances, if atmosphere of Pluto does nothing to light, the ring of Sun should have been visible around Pluto once probe was 10 million km beyond Pluto. However, the ring of Sun is seen in this picture when probe is  only 2 million km beyond Pluto. The scientists must have known the optical characteristics of Pluto's atmosphere and hence they got the probe early in proper position to take this picture. &lt;/p&gt;&#xA;" OwnerUserId="9491" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-10-18T11:22:24.870" Title="Size of Sun as seen on Pluto" Tags="&lt;pluto&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12133" PostTypeId="2" ParentId="12132" CreationDate="2015-10-16T04:58:37.127" Score="1" Body="&lt;p&gt;The atmoshere on Pluto will have very little effect on the apparent size of the Sun, It is negligable compared to that on Earth ($\sim 0.3-1.0 \mbox{ Pa}$ compared to $\sim 10^5 \mbox{ Pa}$, approximately the air pressure at 100 km on the Earth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now angular size is proportional to the ratio of the objects diameter to its range. At the Earth's distance ($1 \mbox{ AU}$) the Sun subtends about $30'$ of arc. So at Pluto (as it is at a distance from the Sun of between $26.7 \mbox{AU}$ and $49.3 \mbox{AU}$ depending on where in its orbit it is) the Sun will subtend between $30/26.7\approx 1.1'$ and $30/49.3\approx 0.6'$ of arc. Which is close to the resolution of the human eye corresponding to 20/20 vision.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-10-16T05:07:40.510" LastActivityDate="2015-10-16T05:07:40.510" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12134" PostTypeId="1" CreationDate="2015-10-16T06:17:07.220" Score="0" ViewCount="127" Body="&lt;p&gt;New Horizons probe has found astonishing features on Pluto, while the likelihood of making even greater discoveries is remote in further space. Is it possible for New Horizons probe to turn back and start orbiting Pluto? It may save much time and resources as compared to sending another probe to Pluto. Thanks for your responses. &lt;/p&gt;&#xA;" OwnerUserId="9491" LastEditorUserId="9491" LastEditDate="2015-10-16T06:30:28.713" LastActivityDate="2017-09-01T09:00:39.667" Title="Can New Horizons probe turn back and start orbiting Pluto" Tags="&lt;pluto&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12135" PostTypeId="1" CreationDate="2015-10-16T06:24:11.230" Score="3" ViewCount="149" Body="&lt;p&gt;Is it possible for a man-made probe to orbit Pluto given its closeness to Charon? The gravitational field interaction of Pluto and Charon is likely to make simple orbit structure difficult for a man-made probe. Thanks for your insight.&lt;/p&gt;&#xA;" OwnerUserId="9491" LastActivityDate="2015-10-21T06:08:15.290" Title="Can a probe orbit Pluto given Charon's gravity?" Tags="&lt;orbit&gt;&lt;pluto&gt;&lt;charon&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12137" PostTypeId="2" ParentId="12134" CreationDate="2015-10-16T08:56:06.633" Score="1" Body="&lt;p&gt;No, unfortunately this is not possible. To get back to Pluto, it has to reduce its speed. Even more then it is currently flying (because it has passed Pluto already). Because New Horizon is flying at a speed of approximately 13,8 km/s, it has to use massive amounts of fuel/energy to get a full stop and fly back. Let alone having enough fuel to maneuver into a stable orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://io9.com/heres-why-the-new-horizons-spacecraft-wont-be-stopping-1710069685&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9494" LastActivityDate="2015-10-16T08:56:06.633" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12138" PostTypeId="1" CreationDate="2015-10-16T09:12:16.223" Score="4" ViewCount="201" Body="&lt;p&gt;As we all know matter is attracted by the gravitational field of black hole. So, my question is: Is there any possibility of a black-hole attracting antimatter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you explain how/why please?&lt;/p&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="4085" LastEditDate="2015-10-16T10:19:45.543" LastActivityDate="2015-10-16T20:32:56.340" Title="Is antimatter also attracted by gravitational field?" Tags="&lt;black-hole&gt;&lt;antimatter&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12139" PostTypeId="2" ParentId="12115" CreationDate="2015-10-16T09:19:55.497" Score="2" Body="&lt;p&gt;I'm not entirely sure what you mean, but the (Planck's) formula for blackbody radiation is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_{\lambda} = \frac{8 \pi h c}{\lambda^5} \frac{1}{e^{hc/\lambda kT} - 1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $h$ is in $\mathrm{J \cdot s}$, $c$ in $\mathrm{m/s}$, $\lambda$ in $\mathrm{m}$, $k$ in $\mathrm{J/K}$ and $T$ in $\mathrm{K}$.&#xA;So, the temperature is just in kelvin, not in energy.&#xA;This formula, with these units, gives $S$ in $\mathrm{W/m^2/m}$, which describes the amount of energy per temperature and wavelength. There is no need to convert a given temperature to energy.&lt;/p&gt;&#xA;" OwnerUserId="9494" LastEditorUserId="9301" LastEditDate="2016-10-25T18:13:59.963" LastActivityDate="2016-10-25T18:13:59.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12141" PostTypeId="2" ParentId="12138" CreationDate="2015-10-16T10:20:53.400" Score="1" Body="&lt;p&gt;Yes, common believes are because gravitation has effect on particles with mass. Antimatter are equivalent to there counterpart, sharing the same mass but opposite charges. Therefore, antimatter is attracted by gravitation.&#xA;But, this is still not experimentally proven, therefore a more theoretical explanation of the gravitational interaction.&#xA;Some researches indeed suggest anti-gravitational interaction between matter and antimatter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_interaction_of_antimatter&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9494" LastActivityDate="2015-10-16T10:20:53.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12142" PostTypeId="1" CreationDate="2015-10-16T10:46:59.527" Score="1" ViewCount="122" Body="&lt;p&gt;What datasets containing KIC 8462852 are available online ? I've found only light curve from Kepler survey.&lt;/p&gt;&#xA;" OwnerUserId="9492" LastActivityDate="2015-10-16T10:46:59.527" Title="Datasets containing KIC 8462852" Tags="&lt;exoplanet&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12143" PostTypeId="2" ParentId="12132" CreationDate="2015-10-16T12:10:30.333" Score="3" Body="&lt;p&gt;New Horizons encountered Pluto when it was about 33.5 au from the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solar diameter is $1.39\times 10^{9}$ m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An astronomical unit is $1.496\times 10^{11}$ m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle subtended by the Sun at Pluto (in radians) is the ratio of the diameter of the Sun to its distance from Pluto. Converting to degrees, minutes, seconds, gives 0.01589 degrees, or 0.95 arcminutes.&#xA;The atmosphere of Pluto is way too thin (much thinner than the Earth's) to make any significant difference to this at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diameter of Pluto is $2.37\times 10^6$ m. In order for Pluto and the Sun to subtend a similar angle in an image, you just need to move far enough away from Pluto to make the same angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&#xA;$$ D_P/R_P  \simeq (D_P + D_{PS})/R_S,$$&#xA;where $D_P$ would be the distance from New Horizons to Pluto, $D_{PS}$ the distance from Pluto to the Sun, $R_P$ the radius of Pluto and $R_S$ the &lt;em&gt;radius&lt;/em&gt; of the Sun. Because I will show that $D_P \ll D_{PS}$ we can say that&#xA;$$ D_P \simeq \frac{R_P}{R_S} D_{PS} \simeq 0.057\ au$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;New Horizons is travelling at about 14.5 km/s. So a simple bit of Maths tells us it would takes about 6.5 days to get into this position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture I think you are talking about was taken on July 15th 2015 only 7 hours after the flyby. The Sun and Pluto &lt;em&gt;do not&lt;/em&gt; share the same angular size in this photo. You are not seeing the Sun at all in this picture. It is eclipsed by Pluto completely. All that is being seen is sunlight refracted (or more likely scattered) through Pluto's atmosphere. The Sun acts more-or-less like a point source of light at more-or-less infinite distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle that the light needs to deviate in order to be seen through the atmosphere is about $0.19$ degrees (the half-angle subtended by Pluto at the Sun [negligible] plus the angle subtended by Pluto at New Horizons after 7 hours). If we were to model the atmosphere as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Prism#Deviation_angle_and_dispersion&quot; rel=&quot;nofollow&quot;&gt;glass prism&lt;/a&gt;. The deviation angle is $\delta \simeq (n-1) \alpha$, where $n$ is the refractive index and $\alpha$ is the opening angle of the prism. If you let $\alpha$ be quite large (let's say 1 radian) to simulate rays coming through the atmosphere, then $n = 1 + \delta$, where $\delta$ is in radians. So $n\simeq 1.003$. NB: This assumes there is no &lt;em&gt;scattering&lt;/em&gt; in Pluto's atmosphere, but I suspect that is not true and that my estimate of the refractive index is a large overestimate. In fact I'm sure it is an overestimate because this is larger than the refractive index of air on Earth. i.e. I am sure that what we are seeing is scattered light, not refracted light.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-10-17T20:55:18.057" LastActivityDate="2015-10-17T20:55:18.057" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12144" PostTypeId="2" ParentId="115" CreationDate="2015-10-16T16:54:02.067" Score="2" Body="&lt;p&gt;As other answers have said, the definition of a &quot;star&quot; is generally taken to be an object that is undergoing sufficient hydrogen fusion to reach an equilibrium between energy produced by fusion and the energy it is radiating. The exact definition varies, but does not affect this answer much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When &quot;stars&quot; are young, they are large, their cores are too cool to initiate hydrogen fusion. They then contract and hydrogen fusion is initiated when their cores reach about &lt;strong&gt;3 million K&lt;/strong&gt; (e.g. see &lt;a href=&quot;http://arxiv.org/abs/astro-ph/9705201&quot; rel=&quot;nofollow&quot;&gt;Burrows et al. 1997&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why so hot? Because the Coulombic repulsion between positively charged protons prevents fusion. The fusion reaction proceeds by quantum mechanical tunneling, but even then requires that the protons have sufficient kinetic energy to at least partially overcome their Coulomb repulsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of their &lt;em&gt;surface&lt;/em&gt; temperatures, the lowest mass objects that begin hydrogen fusion are about $0.075 M_{\odot}$. Their surface temperatures when fusion commences are about 2800 K, but then their surfaces continue to cool, so that the oldest in our Galaxy might now be around 2300 K and &quot;L dwarfs&quot; (for example see &lt;a href=&quot;http://arxiv.org/abs/astro-ph/9704118&quot; rel=&quot;nofollow&quot;&gt;Chabrier &amp;amp; Baraffe 1997&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, red giants are also stars - either burning hydrogen or helium, or both in shells around an inert core. Their interior temperatures are much hotter than the low-mass objects described above, but because they are very large, their surfaces can be very cool. The coolest red giants also have temperatures of around 2600-2800 K.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-10-16T16:54:02.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12146" PostTypeId="1" CreationDate="2015-10-16T16:57:27.903" Score="2" ViewCount="67" Body="&lt;p&gt;I suppose that (especially space) telescope costs is dominated by design and development, not by material and manufacturing. So once a telescope has been developed and designed, how much extra would it cost to build a second copy of it? It seems as if spy satellites are built in batches of similar designs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the science value of a second similar telescope be modest, compared to a unique instrument, or improvements on the single telescope, for the same cost?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose that using two telescopes as an interferometer would be quite a different instrument design to begin with than just two of the same.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="6" LastEditDate="2019-01-29T14:48:27.880" LastActivityDate="2019-01-29T14:48:27.880" Title="What is the cost and benefit of building two identical telescopes?" Tags="&lt;telescope&gt;&lt;space-telescope&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12147" PostTypeId="1" CreationDate="2015-10-16T16:59:40.127" Score="4" ViewCount="89" Body="&lt;p&gt;The &lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2015/07/24/pluto_ice_flows_and_a_ring_of_light.html&quot; rel=&quot;nofollow&quot;&gt;Sun-Pluto eclipse picture&lt;/a&gt; was taken from 2 million km from Pluto. At that distance, Pluto will have angular size of just 4'5&quot;. Hence this picture must be a zoomed up image. Approximately how much will be the zoom of this image? Also, is it possible to give the power of its camera in megapixel terms? Thanks for your responses.&lt;/p&gt;&#xA;" OwnerUserId="9491" LastActivityDate="2015-12-03T22:58:31.817" Title="How much is the zoom of Sun-Pluto eclipse image" Tags="&lt;pluto&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12148" PostTypeId="1" AcceptedAnswerId="12149" CreationDate="2015-10-16T17:23:32.790" Score="0" ViewCount="267" Body="&lt;p&gt;Nuclear fusion is the main way for stars to generate energy. But we know that nuclear fission and fusion are not controlled, then it causes chain reaction and sudden explosion. But why in the case of stars it does not happen? I mean, that nuclear fusion happen in stars but it take few billion years to complete the amount of hydrogen. Is there any controlling mechanism? If no, what happens here?&lt;/p&gt;&#xA;" OwnerUserId="9267" LastActivityDate="2015-10-21T03:09:12.850" Title="Why nuclear fusion is a controlled process in stars?" Tags="&lt;star-systems&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12149" PostTypeId="2" ParentId="12148" CreationDate="2015-10-16T18:00:11.947" Score="3" Body="&lt;p&gt;Because in most stars, the pressure where nuclear burning takes place is proportional to temperature. In order for the nuclear reaction rate to increase, the temperature and hence pressure must increase. This would cause the gas to push out the layers above it, then to expand and cool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if the nuclear reaction rate fell, so would the temperature and pressure. The smaller pressure would allow the outer layers to move inwards, compressing the gas, heating it up and speeding up the nuclear reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way the nuclear reaction rate (for hydrogen burning) can be kept almost constant. The small changes (increase) that occur during a star's main sequence life are driven by the increasing mean atomic mass as hydrogen turns to helium. This means you need a gradually higher temperature to maintain the same pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the temperature and pressure are decoupled - for instance in the degenerate helium core of a low(ish) mass star near the red giant tip, where electron degeneracy pressure is independent of temperature, then runaway nuclear reactions &lt;em&gt;can&lt;/em&gt; be initiated - the &quot;helium flash&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-10-16T18:00:11.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12150" PostTypeId="2" ParentId="11415" CreationDate="2015-10-16T18:01:01.043" Score="4" Body="&lt;p&gt;This picture was taken when the probe was 2 million km further away from Pluto (&lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2015/07/24/pluto_ice_flows_and_a_ring_of_light.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.slate.com/blogs/bad_astronomy/2015/07/24/pluto_ice_flows_and_a_ring_of_light.html&lt;/a&gt;). At that distance the angular size of Pluto would be only 4'5&quot;. The picture itself must be a zoomed up image. It is quite possible that there was no bright star present in and around this small angle. The unzoomed image could be like following manually created graphic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LOWZX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LOWZX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is remarkable how a spacecraft so far away was guided to come on that line passing through Sun &amp;amp; Pluto with camera properly aligned to take &lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2015/07/24/pluto_ice_flows_and_a_ring_of_light.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this great picture&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9491" LastEditorUserId="9491" LastEditDate="2015-10-17T04:10:22.110" LastActivityDate="2015-10-17T04:10:22.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12151" PostTypeId="1" AcceptedAnswerId="12573" CreationDate="2015-10-16T18:58:02.137" Score="2" ViewCount="941" Body="&lt;p&gt;When we are talking about relativistic motions, say from a relativistic jet, what is meant by the term &quot;bulk Lorentz factor&quot;, and the bulk speed, $\beta$?&lt;/p&gt;&#xA;" OwnerUserId="2091" LastEditorUserId="2091" LastEditDate="2015-10-17T10:22:38.420" LastActivityDate="2015-11-21T10:15:39.890" Title="What is the bulk Lorentz factor?" Tags="&lt;special-relativity&gt;&lt;high-energy-astrophysics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12152" PostTypeId="2" ParentId="12138" CreationDate="2015-10-16T20:32:56.340" Score="1" Body="&lt;p&gt;The consensus opinion of physicists is that antimatter has mass, and it is attracted to other massive objects by gravity in exactly the same way as matter: Antimatter isn't anti-gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proving this is difficult. It is hard to obtain enough antimatter in one place to observe any gravitational interactions. The best observations aren't even able to conclusively show that antimatter &quot;falls&quot; in a gravitational field. However for theoretical reasons it is considered extremely likely. If antimatter were repulsed by matter, it would allow for violations of the conservation of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A black hole is a region of extreme gravity, and a black-hole would attract matter in just the same way as it attracts antimatter. It would even be possible for antimatter to form a black hole. In fact there are only 11 numbers that define a black hole: mass (made of either matter, antimatter, or energy), position, velocity, spin rate and direction, and electrical charge.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-10-16T20:32:56.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12153" PostTypeId="2" ParentId="11415" CreationDate="2015-10-16T20:39:48.503" Score="5" Body="&lt;p&gt;The answer by @mso is very good. Another contributing factor is the exposure time of the camera on New Horizons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of light reflecting from Pluto is greater than that from the background stars. As a result the exposure time is set to achieve a good image for the desired target and detail required. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over the integration time the amount of light from background stars is not enough to register onto the focal plane and subsequent image. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to get good pictures of stars one needs to integrate over long period of time while maintaining accurate pointing. &lt;/p&gt;&#xA;" OwnerUserId="9502" LastActivityDate="2015-10-16T20:39:48.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12154" PostTypeId="1" AcceptedAnswerId="12167" CreationDate="2015-10-16T21:18:21.223" Score="5" ViewCount="4558" Body="&lt;p&gt;I'm trying to write a gravity simulation (suns planets etc), and was hoping tidal locking could be one feature demonstrated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a simple equation for gravity has produced some interesting results, but (unless its emergent behaviour) I see nothing that would encourage tidal locking. But, after some reading it appears tidal locking is quite common, planets and their satellites, planets and suns, suns and other suns (binary stars).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a result of the formation stage of these objects, or is it somehow a function of the equation of gravity?&lt;/p&gt;&#xA;" OwnerUserId="9503" LastEditorUserId="9186" LastEditDate="2015-10-17T03:07:46.623" LastActivityDate="2015-10-18T20:53:00.667" Title="What causes objects to become tidally locked?" Tags="&lt;orbit&gt;&lt;gravity&gt;&lt;rotation&gt;&lt;tidal-forces&gt;&lt;celestial-mechanics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12157" PostTypeId="1" AcceptedAnswerId="12160" CreationDate="2015-10-17T10:01:50.867" Score="2" ViewCount="209" Body="&lt;p&gt;As we all know, antimatter is present across all of space. Is it also present on Earth? If it is present on Earth then by the interaction of matter and antimatter, why don't we feel the energy around us?&lt;/p&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="53" LastEditDate="2015-10-19T14:51:12.283" LastActivityDate="2015-10-19T14:51:12.283" Title="Is antimatter present on Earth?" Tags="&lt;antimatter&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12158" PostTypeId="1" AcceptedAnswerId="12161" CreationDate="2015-10-17T10:39:50.337" Score="0" ViewCount="174" Body="&lt;p&gt;The Big Bang theory is the approved theory of the &quot;birth&quot; of Universe. Are there any other theories which can explain how our Universe came into existence?  &lt;/p&gt;&#xA;" OwnerUserId="9403" LastEditorUserId="2091" LastEditDate="2015-10-18T15:25:58.907" LastActivityDate="2015-10-18T15:25:58.907" Title="Is there any other theory, apart from the Big Bang paradigm, which describes the birth of universe?" Tags="&lt;universe&gt;&lt;cosmology&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-10-19T16:34:42.467" ContentLicense="CC BY-SA 3.0" />
  <row Id="12160" PostTypeId="2" ParentId="12157" CreationDate="2015-10-17T12:49:19.940" Score="3" Body="&lt;p&gt;Antimatter is present on Earth and is being naturally created all the time (by &lt;a href=&quot;https://en.wikipedia.org/wiki/Beta_decay&quot; rel=&quot;nofollow&quot;&gt;Beta decay&lt;/a&gt;) as well as being created as product of cosmic ray collisions and in particle accelerators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the universe appears to be principally made of matter and so the antimatter thus created is annihilated: and although the production of antimatter through nuclear decay is essentially continuous, it is not of a high volume so the energy created in this way is not significant in the normal course of events. The real question is why is matter so much more abundant than antimatter: which is an open question though various theories have been proposed about why we might live in such a universe. &lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2015-10-17T12:49:19.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12161" PostTypeId="2" ParentId="12158" CreationDate="2015-10-17T14:11:14.633" Score="0" Body="&lt;p&gt;This is not an answer but a comment which is too long to post as such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no such thing as a scientifically approved theory, there are theories consistent with the evidence some of which are simpler in some sense than others, and philosophy of science tells us that we should prefer those that are simpler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I doubt that the big-bang theory is what you think it is, which is simply that at a certain time in the past the Universe was in a hot dense state which then expanded, forming the lightest chemical elements... The first part of this is a simple extrapolation of the observation of the way that galactic Doppler shift increases with distance and the observed cosmic microwave background. The second part is the consequence of the application of well know nuclear properties of matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there are plenty of more-or-less widely accepted additions to this, mainly to explain the uniformity and fluctuations in the microwave background, and the emergence of first stars and galaxies etc. But all of these are to some extent speculative since we have no generally accepted unification of Quantum Mechanics and General Relativity or require physics that has no predicted consequence (so far) beyond the phenomena it has been proposed in order to explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make this approximate an answer I will point you to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Non-standard_cosmology&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page on Non-Standard Cosmologies&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-10-18T14:17:00.193" LastActivityDate="2015-10-18T14:17:00.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12162" PostTypeId="1" AcceptedAnswerId="12219" CreationDate="2015-10-17T14:31:55.497" Score="9" ViewCount="2604" Body="&lt;p&gt;Recently, I read yet another news about &lt;a href=&quot;https://en.wikipedia.org/wiki/European_Extremely_Large_Telescope&quot;&gt;E-ELT&lt;/a&gt;. It will have 39.3-metre-diameter segmented primary mirror. And I was interested in the next question:&#xA;Theoretically, what size of the primary mirror (single/multiple/segmented) can have a telescope on Earth for observing at optical wavelengths? And why? I mean what are physical limitations exists?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the same question about space (not on Earth)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the advice of @TildalWave, to make this question answerable let's make a few adjustments:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Primary mirror should be segmented (or it variations) like on E-ELT.&lt;/li&gt;&#xA;&lt;li&gt;Suppose we have a large (several square kilometers), flat surface high above sea level.&lt;/li&gt;&#xA;&lt;li&gt;We have to build telescope for observing at optical wavelengths.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I know that, there is concept of &lt;a href=&quot;https://en.wikipedia.org/wiki/Overwhelmingly_Large_Telescope&quot;&gt;OWL&lt;/a&gt; with 100 metre-diameter segmented primary mirror.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about 500 metre-diameter or 1000? Is it possible in theory?&lt;/p&gt;&#xA;" OwnerUserId="9508" LastEditorUserId="9508" LastEditDate="2015-10-21T09:58:17.710" LastActivityDate="2015-10-22T19:19:37.363" Title="Theoretically, what is the biggest optical telescope that may exist?" Tags="&lt;telescope&gt;&lt;space-telescope&gt;&lt;optics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12164" PostTypeId="1" CreationDate="2015-10-18T07:55:02.630" Score="1" ViewCount="39" Body="&lt;p&gt;I have an exposure of a star field with a small galaxy located near the centre. It is of course a greyscale image. I would like to add colours (realistic) to this image for aesthetic purposes. What software can I use to play around with colours that is Linux compatible and relatively easy to get the hang of? I've been using DS9 which has some colour options but to my knowledge doesn't quite produce a realistic colour image.&lt;/p&gt;&#xA;" OwnerUserId="9511" LastActivityDate="2015-10-18T07:55:02.630" Title="How does one add false colour to a fits image" Tags="&lt;exposure&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12165" PostTypeId="2" ParentId="997" CreationDate="2015-10-18T15:43:53.480" Score="2" Body="&lt;p&gt;The question is compromised by saying that you allow arbitrarily perfect measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have a bolometer that can measure the amount of flux from a star, at a distance that is known to arbitrary accuracy, with arbitrarily good spatial resolution, then what we do is measure the bolometric luminosity from a 1 m$^2$ area at the centre of the stellar disk. This flux is $\sigma T_{eff}^4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now of course, stars do not have homogeneous atmospheres (spots, granulation, meridional flows, non-sphericity due to rotation...), so the result you would get would depend on exactly what 1 m$^2$ bit of atmosphere you were looking at. So, with my arbitrarily accurate instruments I would have to measure the luminosity from every 1 m$^2$ patch over the entire surface of the star. Each one would give me another estimate of $T_{eff}^4$; each would be somewhat different. That would be difficult, but the form of your question allows me to ignore those problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this level of precision, the utility of a single $T_{eff}$ for the whole star is questionable, but if you wanted one then it would be the flux-weighted mean of all the above measurements, and as far as I can see one can &lt;em&gt;instantaneously&lt;/em&gt; determine it to whatever accuracy you desire. Of course it will then vary if you have a variable star, and it will vary from point to point with time due to granulation; so the accuracy of the $T_{eff}$ could depend on how quickly and by how much it varies compared with how long it takes you to do your arbitrarily accurate measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think to get a better answer, you do need to specify some realistic observational constraints - such as (a) you cannot resolve the star at all, or (b) that you can resolve it, but observations can only take place from an earth-bound observatory (thus not allowing you to take flux measurements from the whole surface at once).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing occurs, is that in unresolved observations, even with an absolutely accurately measured luminosity (assuming isotropic radiation) there is still the issue of what radius to use. The radius at which the radiation escapes the star (at optical depth $\sim 2/3$) is ill-defined and wavelength dependent. An error bar of maybe tens of km is appropriate here, since atmospheres are 100-200 km &quot;thick&quot;. For a solar-type star this would limit $T_{eff}$ accuracy to $\sim 0.1 K$ !&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-10-18T18:11:15.657" LastActivityDate="2015-10-18T18:11:15.657" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12166" PostTypeId="2" ParentId="12118" CreationDate="2015-10-18T19:06:59.973" Score="1" Body="&lt;p&gt;Here's the how: very energetic charged particles interact with the Earth's magnetic field, and when they do so they emit electromagnetic radiation. If the energy of this radiation was high enough then when it reached wires/conductors at the Earth's surface the opposite effect would take place: ie the EM radiation would result in the flow of charged particles (a current). Such currents, if strong enough, could damage very many electronic and electric devices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the possibility: it seems unlikely that such a high energy event would take place (ie to destroy all/nearly all electronic equipment), but it could be possible. We don't have sufficient data.&lt;/p&gt;&#xA;" OwnerUserId="823" LastEditorUserId="823" LastEditDate="2015-10-18T19:33:45.017" LastActivityDate="2015-10-18T19:33:45.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12167" PostTypeId="2" ParentId="12154" CreationDate="2015-10-18T20:53:00.667" Score="7" Body="&lt;p&gt;Tidal locking occurs because the planet deforms the satellite into an oval, with long axis pointing towards the planet. If the satellite is rotating the long axis will move away from being pointing towards the planet, and the gravity of the planet will tend to pull it back, slowing the rotation until one face is permanently facing the planet. Tidal locking isn't a result of the formation processes, but a consequence of satellites not being perfectly rigid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to model the effects of tides on the orbits and rotation periods of satellites you need to know several important pieces of information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First you obviously need to know the size of the planet and the satellite (both in terms of mass and radius) the shape of the orbit and the rotation rate of both planet and satellite. For many objects, these values are well known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, and this is the tricky bit, you need to know how the satellite and planet will be deformed by the other's gravity, and how much tidal heating will occur. These are the so-called &quot;love number&quot; (after Augustus Love) and the dissipation function, Q. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is hard to estimate these. For the Earth Moon system the ratio k/Q is known to be 0.0011. (but the Earth is a poor model for other planets, which don't have a substantial ocean, or a liquid core) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For other planets &lt;a href=&quot;http://www.es.ucsc.edu/~pkoch/EART_206/09-0127/Goldreich%20&amp;amp;%20Soter%2066%20Icarus%205-375.pdf&quot; rel=&quot;noreferrer&quot;&gt;the value of Q varies between 10 and 10000&lt;/a&gt;, with larger values for the gas giants, and k can be estimated from the rigidity of the bodies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple gravitation model is not able to capture the subtleties of the gravitational interaction between two mutually deforming bodies, indeed for most simulations, the planets are modelled as points, or at most as spheres, and this is good enough for all but the highest precision calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tidal locking takes a long time (by human standards) but a relatively short time compared with the age of the solar system. The time taken is very strongly dependent (order 6) on the radius of the orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Direct simulation would be more or less impossible: the deformations are too small, and the time scale of locking is too large. It would be possible (though difficult) to model tidal locking in a simulation with unrealistic values for the rigidity of the satellite, and the size of the planet (think jelly world, orbiting a (Newtonian) black hole) so the deformation is greater and the locking time shorter. However modelling the  elastic deformation of a body under gravity is far from trivial. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-10-18T20:53:00.667" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12168" PostTypeId="1" CreationDate="2015-10-18T22:07:52.613" Score="4" ViewCount="97" Body="&lt;p&gt;I read a lot about the amount of radiation bodies like the Moon and Mars receive, but it's hard to find information about where the radiation is coming from. It seems to me that it's cosmic radiation plus whatever is being radiating from the Sun. Would it be useful to build a habitat somewhere that's in perpetual shadow (e.g. deep crater), or to seek shelter in the shadow of a crater wall or mountain during a solar storm. Mostly I'm thinking about a habitat design for Mars, where the habitat will always be out of view of the sun. Sunlight could light the habitat via something like huge Mylar mirrors, that would not reflect much harmful radiation. But I'd like to know if the applies to the Moon.&lt;/p&gt;&#xA;" OwnerUserId="9516" LastEditorUserId="9516" LastEditDate="2015-10-19T05:17:28.923" LastActivityDate="2016-01-03T00:44:26.837" Title="How much of the ambient radiation on Mars is arriving line of sight from the Sun?" Tags="&lt;the-sun&gt;&lt;planet&gt;&lt;radiation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12169" PostTypeId="1" CreationDate="2015-10-19T02:07:29.187" Score="2" ViewCount="507" Body="&lt;p&gt;As the Moon is tidally locked to Earth, that means, from Earth we can only ever see one side of the moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we see all the sides of the Earth, from the moon?  &lt;/p&gt;&#xA;" OwnerUserId="9519" LastEditorUserId="2365" LastEditDate="2015-10-19T19:28:38.413" LastActivityDate="2015-10-21T13:16:57.177" Title="Tidal lock of earth and moon" Tags="&lt;the-moon&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12170" PostTypeId="1" AcceptedAnswerId="12183" CreationDate="2015-10-19T02:18:02.460" Score="3" ViewCount="401" Body="&lt;p&gt;How astronomers really distinguishes between the natural and artificial signals? From time to time astronomers detects many signals from outer space and SETI is always trying to find signals that carries the sign of intelligent beings. The Wow signals was such a much hyped signal that we received. So what properties a signal should carry so that we can conclude it is from some alien civilizations?  &lt;/p&gt;&#xA;" OwnerUserId="9213" LastEditorUserId="9213" LastEditDate="2015-10-20T01:06:45.030" LastActivityDate="2015-10-22T01:33:26.120" Title="How astronomers distinguishes between natural and artificial signals coming from outer space?" Tags="&lt;extra-terrestrial&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12173" PostTypeId="2" ParentId="12169" CreationDate="2015-10-19T05:58:40.197" Score="5" Body="&lt;p&gt;The Earth and the Moon &lt;b&gt;do&lt;/b&gt; interact, but because they have different sizes and masses the effect on each is different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon rotates once in its orbit around the Earth, causing it to constantly show the same face. This was the Earth's effect on the Moon due to tidal dragging by gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's pretty stable, the Moon will continue to show the same face to the Earth even as the tidal forces cause it to orbit farther and farther away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We actually see slightly more than 50% of the Moon. The Moon's orbit causes its aspect to nod because its orbit is slightly elliptical and shimmy because it's not orbiting directly over the equator. Over time we actually see almost 60 percent of the surface of the Moon (an effect called &lt;a href=&quot;https://en.wikipedia.org/wiki/Libration&quot; rel=&quot;nofollow&quot;&gt;libration&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon's effect on the Earth has been to slow its daily rotation to 24 hours. In the past, the Moon was much closer to the Earth, and the day was much shorter. As time passes and the Moon moves into a wider orbit, the length of the Earth's day will increase (see e.g. &lt;a href=&quot;http://www.scientificamerican.com/article/earth-rotation-summer-solstice/&quot; rel=&quot;nofollow&quot;&gt;this Scientific american post&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read that the Moon would recede to a certain point, then begin approaching the Earth again, but unfortunately the Sun will swell into a red giant before then and swallow the Earth and the Moon.&lt;/p&gt;&#xA;" OwnerUserId="9516" LastEditorDisplayName="user1569" LastEditDate="2015-10-21T13:16:57.177" LastActivityDate="2015-10-21T13:16:57.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12174" PostTypeId="1" AcceptedAnswerId="12180" CreationDate="2015-10-19T13:03:15.180" Score="4" ViewCount="187" Body="&lt;p&gt;This morning at 6:03am Israel time (3:03 UTC) I saw, simultaneously, what appeared to be 2 satellites streaking across the sky directly overhead. If I remember correctly, they were both coming from NW (but it was early).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know what they were, or how I can find out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I live in Mitzpe Netofa, lon/lat about 35.39/32.80. &lt;/p&gt;&#xA;" OwnerUserId="9526" LastActivityDate="2018-01-06T17:51:21.427" Title="What were the 2 satellites I saw this morning?" Tags="&lt;amateur-observing&gt;&lt;artificial-satellite&gt;&lt;satellite&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12175" PostTypeId="1" AcceptedAnswerId="12179" CreationDate="2015-10-19T16:42:38.867" Score="1" ViewCount="118" Body="&lt;p&gt;We say that our universe is 13.7 billion years ago. During the big bang, it doubled at least 90 times in trillionth of a second (as given &lt;a href=&quot;http://www.space.com/52-the-expanding-universe-from-the-big-bang-to-today.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;), and other topological statements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is which time are we talking about. Is the time-lapse considered? If yes, how?&lt;/p&gt;&#xA;" OwnerUserId="9528" LastActivityDate="2015-10-19T21:11:48.430" Title="Is the time lapse considered when estimating the age of the universe?" Tags="&lt;big-bang-theory&gt;&lt;space-time&gt;&lt;time-dilation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12176" PostTypeId="1" CreationDate="2015-10-19T17:21:59.803" Score="0" ViewCount="132" Body="&lt;p&gt;Let's assume that Earth in its current state is suddenly replaced with the Earth completely identical, but made entirely of antimatter. Ignoring meteorites and asteroids which would cause mass destruction, extinction events and whatnot, what would the night sky look like as the anti-Earth swept up the space dust?&lt;/p&gt;&#xA;" OwnerUserId="7929" LastActivityDate="2015-10-22T07:21:56.200" Title="What would night sky look like if Earth was made of antimatter" Tags="&lt;antimatter&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12177" PostTypeId="2" ParentId="12176" CreationDate="2015-10-19T17:59:58.050" Score="0" Body="&lt;p&gt;I think night sky will be the same as it look today. But if only earth is the only antimatter present in the solar system then it will destroyed after some time and converted into energy. As we know antimatter + matter = energy. But the night sky will be as same as today.&lt;/p&gt;&#xA;" OwnerUserId="9231" LastActivityDate="2015-10-19T17:59:58.050" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12178" PostTypeId="1" AcceptedAnswerId="14323" CreationDate="2015-10-19T19:01:54.057" Score="9" ViewCount="2361" Body="&lt;p&gt;Is the reason Mars is red because its surface contains a lot of iron? (When dirt is red on Earth, it is sometimes caused by a high amount of iron).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, does Mars contain more iron than the earth? &lt;/p&gt;&#xA;" OwnerUserId="9530" LastActivityDate="2018-09-09T18:49:20.360" Title="Does Mars contain more iron than the Earth?" Tags="&lt;earth&gt;&lt;mars&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12179" PostTypeId="2" ParentId="12175" CreationDate="2015-10-19T21:11:48.430" Score="1" Body="&lt;p&gt;As relativistic effects will cause clocks to run slower, a frame of reference must be chosen when considering the time of the &quot;big bang&quot;. There is a natural and convenient choice of reference frame, based on the cosmic microwave background. The cosmic background appears to be extremely red-shifted light, indicating it is receding from us very fast, due to the expansion of the universe. If we choose a frame in which the CMB is receding equally fast in all directions, we have a convenient frame of reference. It is called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Comoving_distance&quot; rel=&quot;nofollow&quot;&gt;Comoving frame&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we have a frame of reference, we can talk about time and distance in a way that all observers that share this frame can agree. In the comoving frame, the &quot;big bang&quot; occurred about 13.8 billion years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To directly answer the question: The time measured is the time in the comoving frame and relativistic time dialations (time-lapse) are considered.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-10-19T21:11:48.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12180" PostTypeId="2" ParentId="12174" CreationDate="2015-10-19T21:28:12.523" Score="4" Body="&lt;p&gt;The best way to figure it out would be to use the site &quot;In the Sky&quot; (&lt;a href=&quot;https://in-the-sky.org/&quot; rel=&quot;nofollow&quot;&gt;https://in-the-sky.org/&lt;/a&gt;). You can enter time, date and location to identify satellites. This would at least be able to tell you if what you saw was a satellite.&lt;/p&gt;&#xA;" OwnerUserId="9537" LastActivityDate="2015-10-19T21:28:12.523" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12181" PostTypeId="1" AcceptedAnswerId="41397" CreationDate="2015-10-19T21:36:39.530" Score="7" ViewCount="578" Body="&lt;p&gt;Is it theoretically possible that planets exist with mountains so high that their peaks overtop the planets atmosphere? And which physical laws are relevant for this question? I'm just curious.&lt;/p&gt;&#xA;&lt;p&gt;The question came while I imagined the atmosphere like a second ocean above our ocean of water. And I thought it would be nice if such a gaseous ocean could have islands as well.&lt;/p&gt;&#xA;" OwnerUserId="9535" LastEditorUserId="37191" LastEditDate="2021-02-12T21:33:17.097" LastActivityDate="2021-02-13T05:55:58.507" Title="Mountains are higher than the atmosphere?" Tags="&lt;planet&gt;&lt;exoplanet&gt;&lt;planetary-atmosphere&gt;" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="12182" PostTypeId="2" ParentId="12181" CreationDate="2015-10-19T22:27:25.190" Score="8" Body="&lt;p&gt;This is a bit of a gray area, as an atmosphere doesn't have a clear boundary. That being said, Olympus Mons on Mars is so tall, the atmospheric pressure on top of it is only 12% the average pressure on the surface of Mars. That's near vacuum by terrestrial standards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Olympus_Mons#Description&quot;&gt;https://en.wikipedia.org/wiki/Olympus_Mons#Description&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, for this to happen you need:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a pretty thin atmosphere to begin with&lt;/li&gt;&#xA;&lt;li&gt;some exceptional geology that ends up producing very tall anomalies like Olympus&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It's not a very likely combination, but it can happen, as seen on Mars.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2015-10-19T22:27:25.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12183" PostTypeId="2" ParentId="12170" CreationDate="2015-10-20T01:38:20.633" Score="2" Body="&lt;p&gt;We could begin to conclude if a signal is from some Extra Terrestrial Intelligence (ETI) if the signals are very difficult or impossible to explain by non-life origins or terrestrial origins. One commonly cited example is with a very narrowband radio signal. In natural radio emission, the frequencies are spread out due to processes like Doppler shift which raise and lower the frequency or radio emission. The Wow signal was so narrowband that it behaved just like one would expect for ETI, but having only one detection that disappeared makes it hard to be sure it wasn't interference or a problem with that particular antenna.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest that ETI detection should have some of these properties, but feel free to add more:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;No natural abiotic terrestrial phenomenon that can produce it&lt;/li&gt;&#xA;&lt;li&gt;Verifiability - multiple detection methods, instruments and teams that can confirm the signal is not due to errors in processing, analysis or local interference&lt;/li&gt;&#xA;&lt;li&gt;Evidence for non-random patterns such as prime numbers&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;KIC 8462852 currently has a feasible solution that dust escaping from a comet or multiple comets can obscure light from its host star. This natural explanation has ways of being confirmed (such as wavelength dependence to the stellar blockage), which will help with the first criterion. Hopefully, the flux dips observed in Kepler's days ~800 and ~1500 will repeat in the future but if they don't, it fails the second criterion. As with definition life, it is hard to come up with a rule book. However, if we see a lot of the properties we associate with ETI, we can start the classification process. As more evidence comes in, like polarization, multi-wavelength and Doppler shift data for KIC 8462852, we can evaluate which criteria are satisfied.&lt;/p&gt;&#xA;" OwnerUserId="9440" LastActivityDate="2015-10-20T01:38:20.633" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12184" PostTypeId="1" CreationDate="2015-10-20T02:03:47.677" Score="4" ViewCount="1823" Body="&lt;p&gt;As there are millions of rogue planets in the Milky Way, what would happen if one hit a planet in the Solar System. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the rogue planet hit Pluto, would Pluto fall in to the Sun due to to its gravity? What kind of catastrophic damage could it do in the Solar System?&lt;/p&gt;&#xA;" OwnerUserId="9519" LastEditorUserId="11264" LastEditDate="2019-03-18T08:23:51.920" LastActivityDate="2019-03-18T08:23:51.920" Title="What would happen if a rogue planet hit one of the planets in our Solar System?" Tags="&lt;solar-system&gt;&lt;rogue-planet&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="12185" PostTypeId="2" ParentId="12184" CreationDate="2015-10-20T09:54:10.250" Score="3" Body="&lt;p&gt;It would depend entirely on the size, composition and trajectory of the two objects coming into the collision. They could annihilate each other leaving a mess of debris. Or one or both could survive the collisions in some form. And each of the surviving objects could be flung out of the solar system, flung into a spiral into the sun, flung into an elliptical orbit, or thrown into other objects. The space between planets is big so odds of direct and immediate collision of the &quot;planets like billiard balls&quot; variety you see on some sci-fi aren't likely. But anything the size of a planet would be seen well before any impact.&lt;/p&gt;&#xA;" OwnerUserId="9540" LastActivityDate="2015-10-20T09:54:10.250" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12186" PostTypeId="2" ParentId="12184" CreationDate="2015-10-20T09:55:48.300" Score="1" Body="&lt;p&gt;This really depends on the size, density and velocity the rogue planet was traveling before the collision and which planet it hit. As a side note, the possibility of this happening is quite rare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that rogue planet have sufficient velocity and under perfect conditions (such as the angle of collision), yes there is a possibility that it could reach the sun (bringing along with the 'planet' it collided say Pluto). The effect of that rogue planet would be enough to slightly disrupt other planets' orbit if it passes close and could disrupt the entire solar system. Pluto may not just fall to the sun based on the the pull of gravity, but also the velocity from the rouge planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly addressing the 'sun' problem, yes if a sizable planet punches in to the sun, it'll create something like a mass coronal ejection which in return would bombard the solar system with harmful rays. Already this has affected the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a sizable rouge planet hit one of the rocky planets such as Mars, large chunks of debris would come shooting out of the collision, a very high chance that a few pieces would land on Earth potentially creating another mass extinction; similarly to the other rocky planets such as Venus and Mercury. Similar to one of the solar system's gas giants if say, another gas giant hit it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summaries this, even a rogue planet entering our system would have already changed a few things not to mention a collision.&lt;/p&gt;&#xA;" OwnerUserId="7926" LastEditorUserId="7926" LastEditDate="2015-10-21T00:38:30.697" LastActivityDate="2015-10-21T00:38:30.697" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12187" PostTypeId="1" AcceptedAnswerId="12199" CreationDate="2015-10-20T13:10:04.287" Score="4" ViewCount="311" Body="&lt;p&gt;In the far future the Milky Way and Andromeda will collide.&#xA;As far I know direct collision between large dense objects (black holes, stars, planets, etc.) is unlikely due to the relative &quot;low density&quot; of both galaxies.&#xA;My naive intuition is it's like two big clouds of dust merging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know at the collision time the Sun will be most like a red star and Earth a scorched dead planet.&#xA;For the sake of the next questions let's hypothesise another system with a star resembling the Sun and another small beautiful third rocky blue planet holding some advanced live form.&#xA;Also, for acceleration measures, let's use a frame of reference like if someone is observing from far away both galaxies getting close one to another and interacting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some questions pop up in my mind.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Can the gravitational interaction sling shot an entire stellar system? How much Km/s2 we can expect these system to accelerate? (In a most extreme plausible case).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How this acceleration can affect the system? Can it de-stabilise planet orbits for example?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How it can affect life? Can it affect magnetosphere? change it's shape? Can solar flares/wind get stronger in some path due particles acceleration?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If another star passes really near (let's say 1 light year at closest approach) but at great speed can its light reach that planet at high energy levels, let's say x-ray and affect that planet life?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2951" LastEditorUserId="2365" LastEditDate="2015-10-20T19:33:34.913" LastActivityDate="2015-10-21T08:18:35.810" Title="Acceleration in Galaxies Collision" Tags="&lt;galaxy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12188" PostTypeId="2" ParentId="12187" CreationDate="2015-10-20T20:00:07.670" Score="4" Body="&lt;p&gt;The short answer is no. There is no huge accelerations of star systems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at a pair of interacting galaxies &lt;a href=&quot;https://i.stack.imgur.com/4bJvq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4bJvq.jpg&quot; alt=&quot;Antenna Galaxies&quot;&gt;&lt;/a&gt; These are the &lt;a href=&quot;http://apod.nasa.gov/apod/ap100507.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Antenna galaxies&lt;/a&gt; (image copyright Star Shadows Remote Observatory and PROMPT/CTIO, from apod)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will see the long &quot;antennae&quot;, which are formed of stars which have been ejected from their galaxies by the interaction. But the accelerations needed are not large, because the interaction takes a long time. The acceleration nothing like $km/s^2$, The force required to accelerate a planet like that would smash it. remember 1$km/s^2$ is about 100g. But small accelerations over long periods can have big effects. The only way to cause objects to accelerate is to apply a force, and the only force that can move planets is gravity. A galaxy has a great big gravitational field, but it is very spread out, and as you note, most of the galaxy is empty space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planet orbits should be stable, certainly inner planets would remain bound to their stars. There could be disruption to oort clouds, but the gravitational field around a star would not change much. Similarly the star's magnetosphere would not be changed much. As above there are no large forces applied to stars or planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A close-ish approach, to 1 light year, would cause disruption of the oort cloud, but nothing more. Stars may have larger relative motions, but no where near enough to cause blue shifting of light into x-ray, in fact it would be hard for observers on the planet to tell which stars came from which galaxy. Andromeda is approching at 120km/s. But proper motions of stars in the milkyway can be greater than this, Barnards star has a relative velocity of 142km/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of the interaction, the star and its planet &lt;a href=&quot;http://mnras.oxfordjournals.org/content/386/1/461.full&quot; rel=&quot;nofollow noreferrer&quot;&gt;is likely to be scattered to the halo of an elliptical galaxy&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-10-20T20:00:07.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12189" PostTypeId="2" ParentId="12181" CreationDate="2015-10-20T23:55:26.473" Score="3" Body="&lt;p&gt;Some planets have no atmosphere at all, so every bump, hill, and mountain would meet your requirement.  There's no fundamental physics that has to do with the size of mountains and the thickness of the atmosphere.  They are controlled by completely different processes.&lt;/p&gt;&#xA;" OwnerUserId="8019" LastActivityDate="2015-10-20T23:55:26.473" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12191" PostTypeId="2" ParentId="12184" CreationDate="2015-10-21T02:41:15.450" Score="1" Body="&lt;p&gt;There are lot of ifs, buts, sizes, etc to answer:&#xA;- Any sizable objects entering our solar system will disturb our orbital 'equilibrium'.&#xA;- First the empty space between the planets is huge, So very unlikely a rogue planet will collide with a solar system planet but once it has entered in to the solar system gravity well it sure headed towards Sun. No 'Deep Impact' or 'Armageddon' please.&#xA;- Second the rogue planet will have a high probability to alter the Oort Cloud objects sending them our way with multiple results.&#xA;- Third it may get captured by the gas giants(Triton/Neptune).&#xA;- Fourth if the rogue planets escapes the gas giants it will encounter Asteroid Belt altering the orbits there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally any interaction direct or indirect with the inner planets is going to be catastrophic. Just think about the magnitude of tsunamis if a earth size planet comes anywhere close to us.&lt;/p&gt;&#xA;" OwnerUserId="9546" LastActivityDate="2015-10-21T02:41:15.450" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12192" PostTypeId="2" ParentId="12181" CreationDate="2015-10-21T02:51:30.007" Score="-1" Body="&lt;p&gt;I think the planet's size and it's geological activity will dictate the mountains height. Mars is smaller than Earth and no active core, So has high probability to have mountain like Olympus Mons. &lt;/p&gt;&#xA;" OwnerUserId="9546" LastActivityDate="2015-10-21T02:51:30.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12193" PostTypeId="2" ParentId="12170" CreationDate="2015-10-21T02:54:33.947" Score="2" Body="&lt;p&gt;A signal artificial is a product from no natural radio frequency composition, so ,we principally change Amplitude (AM) or frequency(FM) and combinations, but required a pattern like a time or position, and a common &quot;simple structure of communication&quot;, this algorithms are used to detect a signal from Intelligence&lt;/p&gt;&#xA;" OwnerUserId="775" LastEditorUserId="6" LastEditDate="2015-10-21T13:02:24.130" LastActivityDate="2015-10-21T13:02:24.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12194" PostTypeId="2" ParentId="12148" CreationDate="2015-10-21T03:09:12.850" Score="0" Body="&lt;p&gt;Adding to Rob explanation. Once that equilibrium stops the bang happens known as Supernova and for smaller stars(our Sun) they will swell in to Red Giants.&lt;/p&gt;&#xA;" OwnerUserId="9546" LastActivityDate="2015-10-21T03:09:12.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12195" PostTypeId="2" ParentId="12135" CreationDate="2015-10-21T03:11:57.887" Score="1" Body="&lt;p&gt;Yes it is possible if you have the probe orbiting the Pluto-Charon system at one their Lagrange points.&lt;/p&gt;&#xA;" OwnerUserId="9546" LastActivityDate="2015-10-21T03:11:57.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12196" PostTypeId="1" CreationDate="2015-10-21T03:19:42.903" Score="3" ViewCount="354" Body="&lt;p&gt;I got some interesting answers for &lt;a href=&quot;https://astronomy.stackexchange.com/questions/12184/what-would-happen-if-a-rogue-planet-hit-one-of-the-planets-in-our-solar-system&quot;&gt;What would happen if a rogue planet hit one of the planets in our Solar System?&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have seen some documentaries that state that rogue planets from other planetary systems (which are ejected from the planetary systems in which they formed) have become part of the other planetary systems and orbit around those new stars as planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, please check this &lt;a href=&quot;http://www.universetoday.com/94656/rogue-planets-can-find-homes-around-other-stars/&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The latest research suggests that sometimes, these rogue, nomadic worlds can find a new home by being captured into orbit around other stars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How is this possible? Stars' gravity can pull the rogue planet into them. How do rogue planets escape this and orbit around the new stars. This seems to be a bit mysterious.&lt;/p&gt;&#xA;" OwnerUserId="9519" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-11-21T19:57:26.037" Title="How do rogue planets orbit around stars in other planetary systems?" Tags="&lt;rogue-planet&gt;&lt;celestial-mechanics&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12197" PostTypeId="2" ParentId="12135" CreationDate="2015-10-21T06:08:15.290" Score="4" Body="&lt;p&gt;Avoiding hard math, which I'm not very good at, the simple answer is yes, provided there's enough distance.   Pluto/Charon have 4 moons orbiting them a bit further away, all in relatively stable orbits.   &lt;a href=&quot;https://en.wikipedia.org/wiki/Moons_of_Pluto#Small_moons&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's distances to &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/d/d8/Pluto_and_Charon_system_new.png/3600px-Pluto_and_Charon_system_new.png&quot; rel=&quot;nofollow&quot;&gt;scale&lt;/a&gt;.  - same source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the ratio of gravitational field strength to size is exponential (Power of 1.5), the apparent closeness of Pluto and Charon to the 4 small moons looks unusual, but it's perfectly acceptable for objects of that size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lagrange point stability requires a mass ratio of about 26 to 1.  (.0385 to 1 per &lt;a href=&quot;http://www.spaceacademy.net.au/library/notes/lagrangp.htm&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;).  The Earth has satellites in unstable Lagrange points, so it's certainly doable, it's just not technically &quot;stable&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hardest part about getting a satellite to orbit Pluto is that Pluto's sufficiently small that any ship that approaches it would need to slow down significantly on it's own to get captured into a Pluto Orbit.   That's why New Horizon was a flyby, not an orbit. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-10-21T06:08:15.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12198" PostTypeId="1" CreationDate="2015-10-21T07:03:40.167" Score="3" ViewCount="262" Body="&lt;p&gt;Couldn't it be just a relatively large object, anywhere in the interstellar space between the star and Earth? I mean, look out the window at a far away car or a tree, then stretch out your hand and you can block it entirely with your thumb. What if the &lt;em&gt;thumb&lt;/em&gt; is that object (a planet, a star, a comet, some space debris, etc.) that's between Earth and the star, and it's just the right size to only block that star, and no other stars. Why has this very simple explanation been ruled out (if it has been)?&lt;/p&gt;&#xA;&lt;p&gt;I know that the debris next to the star explanation has been ruled out because of the lack of IR that would come from that debris being heated up if it was close to the star. But this is not a problem here, if the blocking object is far away from the star.&lt;/p&gt;&#xA;" OwnerUserId="9548" LastEditorUserId="6593" LastEditDate="2021-01-25T14:05:37.347" LastActivityDate="2021-01-25T14:05:37.347" Title="Why can't this be the simple (and obvious) explanation for the dimming of KIC 8462852?" Tags="&lt;star&gt;&lt;universe&gt;&lt;exoplanet&gt;&lt;extra-terrestrial&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="12199" PostTypeId="2" ParentId="12187" CreationDate="2015-10-21T08:18:35.810" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Some questions pop up in my mind.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Can the gravitational interaction sling shot an entire stellar system?&#xA;  How much Km/s2 we can expect these system to accelerate? (In a most&#xA;  extreme plausible case).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Basically this comes down to gravity assists.   A gravity assist is essentially a 3 body interaction.   You need a point of reference and 2 velocities.   When we send a spacecraft around a planet for a gravity assist, the point of reference is the sun.   The maximum gravity assist velocity is (see picture), 2U+v.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/r3L8r.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/r3L8r.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Gravity_assist&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree with James Kiflinger.  Reaching galactic escape velocity is very unlikely because galaxies have large dark matter halos where most of the mass is contained and to actually escape all that requires a very high velocity.  The vast vast vast, even with spiral arms extending outwards, still, most of the stars would stay in the gravitaitonal control of the merged galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this &lt;a href=&quot;https://space.stackexchange.com/questions/3948/escape-velocity-for-our-galaxy&quot;&gt;question&lt;/a&gt; 0ur Sun orbits the center of the Milky way at 220 km/s and it's escape velocity 537 km/s so it would need an acceleration of 317 km/s to escape the Milky way.   That kind of acceleration is possible given that Andromeda and Milky way will fly towards each other at something like escape velocity speed (537 km/s) - perhaps a bit more as Andromeda is larger, and there's orbital speed (similarish to our sun's 220 km/s), so there's enough velocity to kick a star out if the gravity assist is positioned just right, but you'd need a very heavy star, as a star the mass of our sun simply couldn't effect another star the mass of our sun if they fly past each other at several hundred km/s - the gravitational bend would be much less than 180 degrees, so you'd really need a very large sun, maybe a few dozen solar masses traveling at high speed to (maybe) kick another star out of the combined galaxy.   Such an even would be hugely rare but theoretically possible.    Now, an individual star could get multiple gravity assists, but 2 stars getting close enough for a good gravity assist would be rare enough, so that, a star actually being ejected, while it could happen, would be very rare.  Certainly billions of stars per galaxy it could happen to a significant number of stars, but it would still be a 1 in a million (so to speak) event.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How this acceleration can affect the system? Can it de-stabilise&#xA;  planet orbits for example?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First things first, accelerations are rare.  For a star to accelerate around another star they need to get quite close and in such a scenario, the first thing you'd notice is there would be a 2nd sun in the sky, growing gradually brighter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even when Andromeda &amp;amp; the Milky Way collide, form the point of view of your average star, it wouldn't be very interesting unless you actually saw Andromeda's galactic core move towards you - then it might look cool but mostly it would just look like a bunch of stars - like we see in the sky today.   The average closest distance between stars in our neck of the woods is maybe about 5-6 light years.   In the center of the galaxy, less, but there's still a lot lot lot of space between stars, so even as Andromeda and Milky-way collide into each other, from one stars perspective, most of the time, all the stars would be quite distant and another star passing as close as 1 lightyear would be very rare - like maybe every million years or so, rare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If our sun was to pass another similar star about 1 light year away, it would be bright in the sky but otherwise the effect would be negligible unless the star was very large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a sun like star to have the gravitational effect on Earth that Jupiter has (Jupiter is 4 AU from Earth at closest pass, our sun 1047 times the mass of Jupiter), 32 times further, about 128 AU, which is a teeny/tiny fraction of 1 light year, about 1/470.   Even 100 solar masses, such a star would have to pass within 1/47th of a light-year to have the gravitational tug on Earth that Jupiter has.   That's crazy-close and would be enormously rare.  It would happen to some stars, but your average individual solar-system with life on it would on average not be effected in that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luminosity could be more likely a factor than gravity, as a sun with 100 solar masses would be about a million times as luminous as our sun, and 1 million solar luminosity, to equal the solar output from our sun it would only need to pass within 1,000 AU or 1/60th of a light-year, and a sun that bright would put out lots of UV light.   But even so, fly-bys that close would be very rare, but we would still be more likely to be cooked by a large star than orbitally perturbed by one.   Smaller stars would need to fly much closer to have an effect, but smaller stars are also much more common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Outer planets would be more vulnerable and oort cloud objects even more so, to near fly-bys and orbital perturbations, but again, that kind of near fly-by wouldn't happen often.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How it can affect life? Can it affect magnetosphere? change it's&#xA;  shape? Can solar flares/wind get stronger in some path due particles&#xA;  acceleration?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As stated, 2 stars would have to get unusually close for there to be any effect at all.  The most likely any effect would be interactions between the two stars oort cloud, possibly leading to a big increase in comet collisions, which in and of itself could be pretty devastating when a good sized comet hits a planet.  Such events would still be rare, but rather than every few million years, it might be every few thousand (ballpark guess).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnetosphere, doubtful.  If a star gets close enough to affect a planet's magnetosphere, the planet would be vaporized. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solar wind/solar flares - also, no measurable effect.   A sun's solar flares are internally driven and not changed by objects flying past, especially at the distance of 2 stars even in a close flyby.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If another star passes really near (let's say 1 light year at closest&#xA;  approach) but at great speed can its light reach that planet at high&#xA;  energy levels, let's say x-ray and affect that planet life?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, no measurable effect.  With the colliding of 2 galaxies you might see stars fly by each other at about 1000 km/s but that's nowhere close to enough velocity to create any kind of blue-shifted x-ray effect.   Other than the night sky changing more rapidly over centuries than usual, there's no effect to stars flying past each other at 1000 km/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mostly, there would be little to worry about.   Star on star dangerously close fly-bys would be rare, even in a galaxy on galaxy collision.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2015-10-21T08:18:35.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12200" PostTypeId="2" ParentId="12196" CreationDate="2015-10-21T08:43:31.960" Score="-1" Body="&lt;p&gt;There are a few ways in which this can happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of them is by a supernova explosion. A star explodes pushing out the planets who were once orbiting this into free space, going too fast for recapture by other stars. Some planets closer to the star would obviously be decimated but the outer ones could survive the blast(as said &lt;a href=&quot;https://astronomy.stackexchange.com/questions/12201/when-the-sun-explodes-will-some-of-the-planets-in-our-solar-system-survive-and/12204#12204&quot;&gt;here&lt;/a&gt; by UserLTK).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another is that a space object with enough velocity and mass to knock out a planet from its orbit into space. Such as if the originally solar system has too much planets, planet to planet interaction could send one if not both out of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rare occasion for this way of separating planets by their host star is by a black hole. There is current 16 (as far as I could remember) stars travelling at hyper speeds through the galaxy because it came too close to the black hole at the center of the milky way, then spinning outwards at incredible speeds and then the velocity finally comes over the blackhole's gravitational pull and shoots out. The speeds in which they come out is about 7-10 million miles per hour and the top speed achievable is at 30 million miles per hour. That's about crossing earth's diameter in about 10 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As LocalFluff said and UserLTK's contribution to it in the comment above, that's one of the ways too for a planet to be ejected out of the system. Link for that: &lt;a href=&quot;http://blogs.discovermagazine.com/badastronomy/2011/11/16/did-jupiter-toss-a-giant-planet-out-of-the-solar-system/#.Vicx5aQx5mo&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://blogs.discovermagazine.com/badastronomy/2011/11/16/did-jupiter-toss-a-giant-planet-out-of-the-solar-system/#.Vicx5aQx5mo&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's some good sites for answering your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.space.com/15308-rogue-alien-planets-billions-stars.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.space.com/15308-rogue-alien-planets-billions-stars.html&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.space.com/15023-warp-speed-planets-light-speed.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.space.com/15023-warp-speed-planets-light-speed.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7926" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-10-21T11:42:48.167" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12201" PostTypeId="1" AcceptedAnswerId="12203" CreationDate="2015-10-21T08:54:41.757" Score="2" ViewCount="4508" Body="&lt;p&gt;Say if a large enough star, that has a number of planets, were to explode in a supernova:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Would any of the planets survive?&lt;/li&gt;&#xA;&lt;li&gt;Would they become rogue planets?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7926" LastEditorUserId="2365" LastEditDate="2015-10-22T17:57:49.657" LastActivityDate="2019-02-12T07:48:25.117" Title="When the sun explodes, will some of the planets in the Solar System survive and become rogue planets?" Tags="&lt;supernova&gt;&lt;rogue-planet&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12202" PostTypeId="2" ParentId="12169" CreationDate="2015-10-21T09:10:43.480" Score="2" Body="&lt;p&gt;Yes. The tidal locking of the moon with respect to the earth means that the moon rotates once and it axis in exactly the same time it requires to orbit the earth. To always see the same side of earth you have to orbit it in the same time as the earth needs to rotate about its axis, i.e. in 24 hours. The moon orbit around the earth takes about 27 days, so from the moon you'll see the earth rotate below you, and in time you get to see all sides of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbit period is dictated by the distance from the earth; objects farther from the earth will orbit it more slowly, objects closer by more rapidly. In order to always see the the same side of earth an object needs to be at a particular place, a &lt;strong&gt;geostationary orbit&lt;/strong&gt;, which is in the equatorial plane. This is at 42000 km from the center of the earth, or about 36000 km above its surface.&lt;br&gt;&#xA;Television and other communication satellites are in a geostationary orbit, and this orbit is so crowded that satellites sit at just a few tens of kilometers from their neighbors. Everybody wants her satellite at the right longitude to service North America or Europe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kepler's 3rd Law says that there's a fixed relationship between the orbital period squared and the distance cubed. For instance the moon's period in days squared is about 750. If you divide the moon's distance cubed by the geostationary orbit's distance cubed you get this same number.&lt;/p&gt;&#xA;" OwnerUserId="1050" LastActivityDate="2015-10-21T09:10:43.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12203" PostTypeId="2" ParentId="12201" CreationDate="2015-10-21T09:16:00.120" Score="9" Body="&lt;p&gt;&lt;a href=&quot;http://www.universetoday.com/18847/life-of-the-sun/&quot; rel=&quot;noreferrer&quot;&gt;The Sun will not become a supernova&lt;/a&gt;, it will never explode. A star must have about 8 times more mass than the Sun in order to cause a supernova explosion. When it starts to fuse helium, the Sun will become a red giant and expand out to about Earth's orbit and throw out gas to form a beautiful so called planetary nebula. Mercury, Venus and maybe Earth will be inside of its photosphere when it is at the largest. I don't know if they will evaporate completely inside the thin hot gas of the red giant Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining planets will keep orbiting the white dwarf remnant, which is not a star (no fusion) but a compact planet sized object. It will shine like a star because of its stored heat. Orbits will change because the Sun will have only 40% of its mass left, the rest having been thrown out to form a nebula, over very long time. Orbits should become larger, and planets maybe will be pushed out a bit by the ejections from the red giant too. Gas planets maybe get their atmospheres a bit stripped off, icy dwarf planets like Ceres and Pluto could melt and evaporate a bit of their mass since the helium burning red giant is hotter than the Sun today. But I'm kind of guessing here, I don't know if those are significant effects.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-10-22T06:05:17.447" LastActivityDate="2015-10-22T06:05:17.447" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12204" PostTypeId="2" ParentId="12201" CreationDate="2015-10-21T09:32:23.287" Score="3" Body="&lt;p&gt;The short answer to your question is that the Sun's expansion will likely be insufficient to blow any planets out of orbit.   It's possible that the Sun gradually losing mass could over long periods of time cause an outer planet to spiral out of orbit and become a rogue planet but that would be a gradual effect over a billion or more years, and, then, only maybe.   No planets would be ejected from the rather rapid effect of the Sun expanding into a red giant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I think (I couldn't find a good link, will keep looking), but as our sun begins to burn Helium it will form a kind of stellar nebula.  It won't be close to a supernova but it will cast off a fair bit of material and losing an outer planet or two during that stage might be possible - but far from a guarantee.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Would any of the 8 planets currently known, survive?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mercury and Venus are not expected to survive the Sun going red giant but they would be eaten, not ejected.   Earth is 50/50.  (lots of articles on this, google &quot;what will happen to the Earth when the sun goes red giant&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what happens to planets when a star goes SuperNova (which our sun won't as LocalFluff pointed out) - that's a good question.   Some answers here:   &lt;a href=&quot;http://news.nationalgeographic.com/news/2011/08/110805-planets-survive-supernovas-ejected-rogues-space-science/&quot; rel=&quot;nofollow&quot;&gt;http://news.nationalgeographic.com/news/2011/08/110805-planets-survive-supernovas-ejected-rogues-space-science/&lt;/a&gt;    A supernova is likely capable of ejecting outer planets (and eating inner ones).&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-10-21T09:40:18.097" LastActivityDate="2015-10-21T09:40:18.097" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12205" PostTypeId="1" AcceptedAnswerId="12208" CreationDate="2015-10-21T10:32:18.603" Score="4" ViewCount="986" Body="&lt;p&gt;We know the mass a white dwarf needs to be. That's well defined by the Chandrasekhar limit, but before a main sequence star turns into a white dwarf it tends to lose a fair bit of its matter in a stellar nebula.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.uni.edu/morgans/astro/course/Notes/section2/new8.html&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt;, the white dwarf that remains is about half the mass of the main sequence star, with larger stars losing a bit more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the question: Is it correct to say that a star with a mass of about three solar masses will eventually go supernova, similar to a type 1 supernova, even when it's not part of a binary system? Has that kind of supernova ever been observed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or does something else happen like in the final stages of that star? Does it keeps going though collapse and expand cycles, losing enough mass that when it finally becomes a white dwarf it's below the Chandrasekhar limit in mass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mostly, what I've read on supernovae says that type 1 supernovae happen when a white dwarf accretes extra matter and reaches the limit and type 2 supernovae are much larger and require about 8-11 solar masses to generate the iron core which triggers the supernova. What happens with the death of the star between three solar masses and eight solar masses?&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="2365" LastEditDate="2015-11-04T20:18:32.273" LastActivityDate="2015-11-04T20:18:32.273" Title="How massive does a main sequence star need to be to go type 1 supernova?" Tags="&lt;supernova&gt;&lt;stellar-evolution&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12206" PostTypeId="1" CreationDate="2015-10-21T10:45:47.323" Score="1" ViewCount="376" Body="&lt;p&gt;I thought a supernova's core temperature was absolute zero just before it collapsed, as it has run out of all nuclear and thermal energy, but I haven't been able to confirm this online anywhere and I was wondering if anyone can confirm this for me.&lt;/p&gt;&#xA;" OwnerUserId="9549" LastEditorUserId="2153" LastEditDate="2015-10-21T23:23:47.733" LastActivityDate="2015-10-21T23:23:47.733" Title="Is a supernova's core temperature absolute zero just before collapse?" Tags="&lt;supernova&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12207" PostTypeId="2" ParentId="12206" CreationDate="2015-10-21T11:29:59.080" Score="7" Body="&lt;p&gt;No, absolutely not. The core of a core-collapse supernova is one of the hottest places in the present-day universe. The temperature as the star runs out of nuclear fuel in its core is around 6-10 billion Kelvin. As it collapses, the core gets even hotter, perhaps as high as 100 billion Kelvin for a few seconds, before neutrino cooling starts to become effective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the temperatures are getting as high as this because at least two neutrino instruments on Earth detected roughly &quot;thermal neutrinos&quot; with energies of $\sim 10$ MeV coinciding with SN1987A in the Large Magellanic Cloud (see for instance &lt;a href=&quot;http://arxiv.org/abs/1508.00785&quot;&gt;Mirizzi et al. 2015&lt;/a&gt;). &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-10-21T11:29:59.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12208" PostTypeId="2" ParentId="12205" CreationDate="2015-10-21T11:50:33.290" Score="3" Body="&lt;p&gt;This is ground that is probably duplicated in a variety of questions here and on Physics SE, so I'll keep it brief. You have also mixed in several different questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Chandrasekhar mass has &lt;em&gt;very little&lt;/em&gt; to do with determining what initial mass of object will end up as what particular type of stellar remnant (or black hole).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether a star will end up exploding as a supernova depends primarily on its initial mass, but also whether it has a binary companions. There are (basically) two routes to supernovahood.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If the star is more massive than about $8M_{\odot}$ it will progress through several nuclear burning stages. The core of the star does not become degenerate and continues to get more dense and hot through each burning stage. It ends up as iron. Once the core mass of iron exceeds about $1.2M_{\odot}$ (which &lt;em&gt;is&lt;/em&gt; the Chandrasekhar mass for an iron composition), then it collapses and we get a type II (core-collapse) supernova.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In this route a $3 M_{\odot}$ star gets nowhere near being able to go supernova. It will burn hydrogen and helium, produce a &lt;em&gt;degenerate&lt;/em&gt; core of carbon and oxygen. This degenerate core can cool whilst maintaining the same pressure. The outer layers are shed through thermal pulsations and a dense stellar wind in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Asymptotic_giant_branch&quot; rel=&quot;nofollow&quot;&gt;asymptotic giant branch phase&lt;/a&gt;, leaving behind a white dwarf. The relationship between the initial mass of the progenitor and the final mass of the white dwarf is not a straightforward fraction. It probably is about 50% for a star like the Sun, but the fraction is more like 15% for a $7M_{\odot}$ initial mass. The maximum mass of a white dwarf formed in this way is probably about $1.1-1.2M_{\odot}$ and some way below the Chandrasekhar limit for a C/O white dwarf ($\simeq 1.39M_{\odot}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The preceding paragraph is more-or-less what should happen for all stars between about $0.6 M_{\odot}$ (except they haven't had time to do so yet) and $8M_{\odot}$, except that there is a small &quot;grey area&quot; at the upper mass end ($7-9M_{\odot}$) where you might produce &lt;em&gt;slightly&lt;/em&gt; more massive O/Ne white dwarfs.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Once a white dwarf has formed and &lt;em&gt;if&lt;/em&gt; it is in a binary system, then the white dwarf could merge or accrete more mass. At some point close to the Chandrasekhar limit, it ignites. This causes a type Ia (detonation or deflagration) supernova explosion (or at least this is the leading model for how this works). This is really the only route whereby a star with initial mass $&amp;lt;8M_{\odot}$ could end up going supernova.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-10-21T11:50:33.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12209" PostTypeId="2" ParentId="11466" CreationDate="2015-10-21T13:26:54.320" Score="0" Body="&lt;p&gt;This is not a simple question to answer. Outside the influence of the earth's magnetic field then &quot;space&quot; is actually filled with the solar wind, which has a high nominal temperature but low energy density - in other words the particles in the solar wind are moving very fast but there are not that many of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mean what temperature a black body in constant sunlight would reach then probably not that dis-similar from the temperature of the Earth, but you are not measuring the temperature of space then, but rather the intensity of solar radiation.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2015-10-21T13:26:54.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12210" PostTypeId="2" ParentId="11466" CreationDate="2015-10-21T14:26:09.840" Score="4" Body="&lt;p&gt;Assume you have a spherical blackbody.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solar flux at the radius of the Earth is given to a good approximation by $L/4\pi d^2$, where $d = 1$ au. This is $f=1367.5$ W/m$^2$ (though note the distance between the Earth and the Sun has an &lt;em&gt;average&lt;/em&gt; of 1 au).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is a &lt;em&gt;blackbody&lt;/em&gt; sphere it absorbs all radiation incident upon it. Assuming this is &lt;em&gt;just&lt;/em&gt; the radiation from the Sun (starlight being negligible), then an easy bit of integration in spherical polar coordinates tells us that the body absorbs $\pi r^2 f$ W, where $r$ is its radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is then able to reach thermal equilibrium and it entire surface is at the same temperature, then it will re-radiate all this absorbed power. Hence&#xA;$$\pi r^2 f = 4 \pi r^2 \sigma T^4,$$&#xA;where $T$ is the &quot;blackbody equilibrium temperature&quot;. Hence&#xA;$$ T = \left( \frac{f}{4\sigma}\right)^{1/4} = 278.6\ K$$&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-10-21T14:26:09.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12211" PostTypeId="1" CreationDate="2015-10-21T16:28:35.900" Score="2" ViewCount="244" Body="&lt;p&gt;I'm interested, is there any database of stars which I could browse on the internet?&#xA;It would be good, because I could read their parameters, features and maybe find a picture about it.&#xA;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="9554" LastActivityDate="2015-10-21T18:23:58.760" Title="Is there a database for stars?" Tags="&lt;star&gt;&lt;photography&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-10-21T19:59:30.417" ContentLicense="CC BY-SA 3.0" />
  <row Id="12212" PostTypeId="1" CreationDate="2015-10-21T16:31:36.123" Score="2" ViewCount="166" Body="&lt;p&gt;I measured the FWHM (Full Width at Half Maximum) of several emission lines using SPLOT in IRAF, and now need to calculate the uncertainty. How can I do that? &lt;/p&gt;&#xA;" OwnerUserId="9553" LastEditorUserId="13" LastEditDate="2015-10-21T21:34:36.180" LastActivityDate="2015-10-21T21:34:36.180" Title="Uncertainty in the FWHM (IRAF)" Tags="&lt;software&gt;&lt;observational-astronomy&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12213" PostTypeId="2" ParentId="12211" CreationDate="2015-10-21T18:23:58.760" Score="2" Body="&lt;p&gt;HD? Perhaps start with: &lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR&quot; rel=&quot;nofollow&quot;&gt;VizeR&lt;/a&gt;. You can search for any star you like, regardless of catalogue. &lt;/p&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2015-10-21T18:23:58.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12214" PostTypeId="1" CreationDate="2015-10-21T18:29:18.057" Score="3" ViewCount="266" Body="&lt;p&gt;Frustratingly, I find this could be a very simplistic question and may leave me feeling a little stupid for asking it. But I have to ask this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My lecturer keeps refering to &quot;matter distributions&quot; for various rotation cases, e.g., Keplerian rotation, solid body rotation, flat rotation etc etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if anyone knew exactly what this term means, or the specifics for each case?&lt;/p&gt;&#xA;" OwnerUserId="2091" LastEditorUserId="9301" LastEditDate="2016-06-29T04:40:16.530" LastActivityDate="2016-07-03T19:16:23.483" Title="What is meant by matter distribution?" Tags="&lt;orbit&gt;&lt;galaxy&gt;&lt;rotation&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12216" PostTypeId="2" ParentId="12198" CreationDate="2015-10-21T20:12:17.423" Score="3" Body="&lt;p&gt;The simplest reason why your proposed explanation is not really that obvious, and thus was not considered is that the dimming of KIC 8462852 was not an isolated event. The dimming is a pattern that has been observed over years. There is no known phenomenon in interstellar space that would cause a repeated obstruction of that star, so it doesn't make any sense to consider that possibility when there are other phenomena (such as those captured in Boyajian's paper) that are more likely to be the source of the dimming.&lt;/p&gt;&#xA;" OwnerUserId="6" LastActivityDate="2015-10-21T20:12:17.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12217" PostTypeId="2" ParentId="12198" CreationDate="2015-10-21T21:20:35.513" Score="2" Body="&lt;p&gt;As others have said, there is repeated dimming of the star, that suggests something in orbit around the star. But in addition to that, you are having the perfectly reasonable and common failure to comprehend the vastness and emptiness of space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, comets, planets and the like are quite common going around stars, but in interstellar space they are very rare and to line up exactly with a distant star is incredibly unlikely. To get an idea of the scale, think of the shadow cast by a flea in London, from a light bulb in Beijing, being seen in New York. It just doesn't happen. Because the distances are too vast (even this example is wrong, by a factor of about 1000). Space is just too big and too empty for random debris to exactly line up with a star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if the debris is near the Sun, as is passes in front of a star, the light from that star is abruptly cut out. This is called an occulation, and they happen &lt;a href=&quot;http://www.asteroidoccultation.com/&quot; rel=&quot;nofollow&quot;&gt;very often&lt;/a&gt;. The conclusion is that whatever is causing the dimming of this star must be close to it. Probably a cloud of dust in orbit around the star, but this doesn't explain how that dust got there, and why it doesn't appear in the infrared.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-10-21T21:20:35.513" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12218" PostTypeId="2" ParentId="12214" CreationDate="2015-10-21T22:11:00.287" Score="5" Body="&lt;p&gt;I assume you are referring to the rotation curve of galaxies, i.e. the speed of matter (stars, gas clouds, etc.) as a function of its distance from the center. This speed will depend on the &lt;em&gt;matter distribution&lt;/em&gt;, that is, how the galaxies' constituents are distributed throughout the galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Keplerian rotation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If (almost) all of the mass $M$ were located in the center, then a star at a distance $R$ would have a speed $V = \sqrt{GM/R}$. This is called &lt;em&gt;Keplerian rotation&lt;/em&gt;, because this is how planets in the Solar system move, which were described by Johannes Kepler. In this case, the mass inside a radius $R$ is (almost) constant, i.e. $M'(&amp;lt;R) = M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Flat rotation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most disk galaxies have a large concentration of visible mass in the center (the &quot;bulge&quot;), so one might think that Keplerian rotation would be a good description of the stars' speeds. But this is not what is observed. In fact, the speed is seen to rise steeply the first few kpc, and then stay constant out to distances where no light is detected. This will happen if the mass $M'(&amp;lt;R)$ inside $R$ increases with $R$, i.e. if the matter is distributed throughout the galaxy. This is one of several pieces of evidence that 1) there exists some matter that we cannot see, in addition to what we &lt;em&gt;can&lt;/em&gt; see, 2) that this matter dominates the kinematics of the galaxies, and 3) that it resides in a halo that extends beyond the visible matter. We call it dark matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solid body rotation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If in addition to a given matter distribution the distances between the individual parts are fixed, the object we are discussing would describe a solid body. Of course this can't be the case for a galaxy, but it will be the case for a rocky planet. In this case the velocity of a given point will depend linearly on its perpendicular distance to the axis of rotation, i.e. $V \propto R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Density profile of the matter distribution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Excactly &lt;em&gt;how&lt;/em&gt; the matter is distributed in the galaxy is not easily deducted, and varies from galaxy type to galaxy type. An early description of the matter density as a function of distance from the center was the so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Navarro%E2%80%93Frenk%E2%80%93White_profile&quot; rel=&quot;nofollow noreferrer&quot;&gt;NFW profile&lt;/a&gt;, but other models may be better, e.g. a &lt;a href=&quot;https://en.wikipedia.org/wiki/Sersic_profile&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sérsic profile&lt;/a&gt; or an &lt;a href=&quot;https://en.wikipedia.org/wiki/Einasto_profile&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einasto profile&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to summarize, &lt;strong&gt;the term &quot;matter distribution&quot; refers to density as a function of location, i.e. $\rho(\mathbf{r})$, for some object&lt;/strong&gt;. This can be for any object — solid, liquid, gaseous, or consisting of several components that are held together by gravity — but in any case it is true that its form determines its rotation, and the three different rotation forms you mention are often discussed in the context of galaxies.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The three different cases are illustratred below. Keplerian, solid, and flat rotation are drawn in green, red, and blue, respectively. For comparison, a typical spiral galaxy velocity curve is drawn in magenta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/m7UyF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/m7UyF.png&quot; alt=&quot;rotation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-10-25T09:22:16.393" LastActivityDate="2015-10-25T09:22:16.393" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12219" PostTypeId="2" ParentId="12162" CreationDate="2015-10-21T22:11:25.540" Score="12" Body="&lt;p&gt;It's complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until late-20th century, we've tried to make bigger and bigger monolithic telescopes. That worked pretty well up to the 5 meter parabolic mirror on Mount Palomar in California in the 1940s. It kind of worked, but just barely, for the 6 meter mirror on Caucasus in Russia in the 1970s. It did work, but that was a major achievement, for the twin 8.4 meter mirrors for the LBT in Arizona in the 2000s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've learned eventually that the way to go is not by pouring larger and larger slabs of low-expansion glass. It is generally accepted that somewhere just below 10 meters diameter is about as large as possible for monolithic mirrors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way to go is by choosing to make smaller mirror segments (1 meter to a few meters in diameter each) and combining those into a tiled mirror. It's somewhat harder to carve the asymmetrical parabolic (or hyperbolic, or elliptic, or spherical) reflecting curved surface in a segment like that, but it's far easier to manage thermal and cooling issues when you have to deal with smaller solid objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each segment is mounted in an active mirror cell, with piezo actuators that very precisely control its position. All segments must combine into a single smooth surface with a precision better than 100 microns (much better than that in reality). So now you have a large array of massive objects, dynamically controlled via computer, each with its own vibration modes, each with its own source of mechanical noise, each with its own thermal expansion motions, all of them &quot;dancing&quot; up and down a few microns on piezo elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to orchestrate a very large system like that? Yes. The 100 meter OWL was considered feasible technically. From the perspective of keeping the mirrors aligned, an even larger structure should be doable; the computer-controlled actuators should overcome most vibrations and shifts up to quite large distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like you said, the real limits are financial. The complexity of such a system increases with the square of the diameter, and with complexity comes cost.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The entire discussion above was about &quot;filled aperture&quot; telescopes: given a round shape of a certain diameter, it is filled with mirror segments. For a given aperture, this design captures the largest amount of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the aperture does not have to be filled. It can be mostly empty. You could have a few reflecting segments on the periphery, and the center would be mostly void. You'd have the same resolving power (you would see the same small details), it's just that the brightness of the image would decrease, because you're capturing less light total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the principle of the interferometer. The twin 10 meter segmented Keck mirrors in Hawaii can work as an interferometer with a baseline of 85 meters. This is effectively equivalent to a single 85 meter aperture in terms of resolving power, but obviously not in terms of image brightness (amount of light captured).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The US Navy has an interferometer in Arizona with mirrors placed on 3 arms in a Y shape, each arm 250 meters long. That gives the instrument a baseline (equivalent aperture) of several hundreds of meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;U of Sydney has a 640 meter baseline interferometer in the Australian desert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interferometers cannot be used to study very faint objects, because they can't capture enough light. But they can produce very high resolution data from bright objects - e.g. they are used to measure the diameter of stars, such as Betelgeuse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The baseline of an interferometer can be made extremely large. For terrestrial instruments, a kilometers-wide baseline is very doable now. Larger will be doable in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are talks about building interferometers in outer space, in orbit around Earth or even bigger. That would provide a baseline at least in the thousands of kilometers. That's not doable now, but seems feasible in the future.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-10-22T19:19:37.363" LastActivityDate="2015-10-22T19:19:37.363" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12220" PostTypeId="1" AcceptedAnswerId="12221" CreationDate="2015-10-21T23:23:56.180" Score="4" ViewCount="1363" Body="&lt;p&gt;As describe in &lt;a href=&quot;http://www.cnn.com/2015/10/21/us/asteroid-earth-nasa-halloween-feat/index.html&quot; rel=&quot;nofollow&quot;&gt;this CNN article&lt;/a&gt;, an asteroid is going to pass near the Earth on Halloween.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article says that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As it misses Earth by about 300,000 miles (slightly farther away than the moon), the asteroid, named 2015 TB145, will be visible to those with good telescopes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It's a good thing it will miss, though. The asteroid is estimated to be 300 to 600 meters wide and traveling at 78,000 mph.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If, hypothetically, this asteroid actually hit the moon, could it knock it out of orbit? How about another, contrived asteroid with whatever properties are needed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be clear, when I say &quot;out of its orbit&quot; I'm including both the orbit decaying and the moon crashing into Earth, and the moon being knocked out into space away from the Earth's influence. Either would do.&lt;/p&gt;&#xA;" OwnerUserId="9559" LastEditorUserId="9559" LastEditDate="2015-10-22T00:39:30.853" LastActivityDate="2015-10-23T00:45:38.800" Title="Could an asteroid knock the moon out of its orbit?" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;asteroids&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12221" PostTypeId="2" ParentId="12220" CreationDate="2015-10-22T00:25:05.003" Score="5" Body="&lt;p&gt;Basically no and not by a long shot.   The asteroid's measurements are pretty vague, but lets give it a nice clean 600 meters in diameter and the same density of the moon.   The moon is 3,474 km in diameter or 3,474,000 meters.   So the moon is nearly 5,790 times the diameter and that gives is 194 billion times the mass as that asteroid.    Conservation of momentum, if the asteroid transfers all of it's 78,000 mph into the moon, with no energy going into angular momentum, 78,000 / 194,000,000,000 = .0000004 mph.   That's roughly the speed your fingernails grow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To knock the moon out of orbit, it's velocity would need to be increased by oh, very rough ballpark, maybe 400-500 mph.   The additional velocity needed for the moon to reach escape velocity from the Earth is .414 x 2288 or about 947 mph, but given that it has an elongated orbit and that it would escape the Earth's sphere of influence well below escape velocity, maybe as little 400 extra MPH could knock the moon into a far enough orbit that it could escape the Earth.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figuring 78,000 MPH, and you want to move the moon some 400 or more MPH with ah it just right, you'd need an object about a 5th or 6th the diameter of the moon or 600 KM in diameter give or take, assuming it was straight on impact as a glancing blow would lose a share of energy into angular momentum and assuming it hits the moon away from the Earth, not towards as towards would require even more energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This space pebble doesn't even come close.   You'd need something the size of a small dwarf planet to have a chance at knocking the moon out of the Earth's orbit and that speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;300-600 meters is huge, if it hits the Earth.  That's probably a few square miles of earth practically leveled to the ground and broken windows and knocked down trees a good deal further than that.   If it hits the moon it would leave a nice crater, maybe a miles or two across but that's about it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It seems like you're only talking about the moon being kicked out into&#xA;  space. Would the same amount of energy be needed to knock the moon out&#xA;  of its orbit in towards Earth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The mathematics of crashing is a bit more complicated.   Escape Velocity vs Orbital Velocity is easy, just straight multiplication.   &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/vesc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;More details here&lt;/a&gt;, and because the Moon wouldn't need to reach full escaple velocity, I cut the number in half - and it might be a bit less than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moon's orbital Speed (2,288 MPH) x (Square root of 2 minus 1, or 0.414).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crashing the Moon into the Earth - a bit harder.  There's probably a simple enough formula, but I'm not sure what it is, but I believe that would take even more energy.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's discussed &lt;a href=&quot;https://physics.stackexchange.com/questions/212494/what-happens-if-you-slow-down-a-moon&quot;&gt;here&lt;/a&gt; without math and &lt;a href=&quot;https://www.quora.com/How-much-energy-would-it-take-to-crash-the-moon-into-the-Earth&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, with math.  But it would take a huge amount of energy, probably 3 or 4 times more energy to push the Moon into the Earth than to push it out of orbit if I was to guess.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-10-22T01:16:03.483" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12222" PostTypeId="1" AcceptedAnswerId="12626" CreationDate="2015-10-22T00:25:46.423" Score="3" ViewCount="392" Body="&lt;p&gt;I've often come across sun path charts &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Sun-path-polar-chart.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;like this (Wikimedia Commons)&lt;/a&gt; that show the position of the sun at any point of the year, at any time of day, for a specific location (or a set of locations), on an azimuth-elevation polar plot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WaTMS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WaTMS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to get the exact same chart (or data, to create my own chart) for &lt;strong&gt;any given star,&lt;/strong&gt; such as Aldebaran or Sirius, for a given point on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking into Stellarium scripting, as well as the HYG stellar database (I am not familiar with the mathematics of right ascension and declination, or how to convert this to azimuth-elevation like in the example). Is this a very unusual data product to ask for?&lt;/p&gt;&#xA;" OwnerUserId="9560" LastActivityDate="2015-11-24T16:46:25.000" Title="Annual path polar chart for a star" Tags="&lt;amateur-observing&gt;&lt;data-analysis&gt;&lt;constellations&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12224" PostTypeId="2" ParentId="12170" CreationDate="2015-10-22T01:33:26.120" Score="1" Body="&lt;p&gt;Considering the vastness of space, an alien(artificial) signal will have to be pointed in our direction and when it reaches us the strength will be very very small. Natural signals are more pronounced and 'appear' all the time. To duplicate a natural signal the artificial source have to have lot of energy resources at disposal. &lt;/p&gt;&#xA;" OwnerUserId="9546" LastActivityDate="2015-10-22T01:33:26.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12225" PostTypeId="2" ParentId="138" CreationDate="2015-10-22T01:40:12.877" Score="7" Body="&lt;p&gt;Yes, Sun has differential rotation. Movement of Sun spots is one of the proofs that Sun rotates. The differential rotation causes the weird twisted magnetic fields which shows in the Sun's prominence. &lt;/p&gt;&#xA;" OwnerUserId="9546" LastActivityDate="2015-10-22T01:40:12.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12227" PostTypeId="2" ParentId="12176" CreationDate="2015-10-22T07:21:56.200" Score="1" Body="&lt;p&gt;It's kind of a strange question but I'll give it a shot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It depend, kind of obviously, on how much of the &quot;space dust&quot; hits the earth.   Estimates vary pretty widely on how much dust from space hits the Earth anyway, by this &lt;a href=&quot;http://www.universetoday.com/94392/getting-a-handle-on-how-much-cosmic-dust-hits-earth/&quot; rel=&quot;nofollow&quot;&gt;site&lt;/a&gt;, between 5 and 300 metric tons per day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets start with the low end estimate, 5 metric tons (5,000 KG) per day.   Antimatter will meet with and turn into energy an equal amount of matter, so 10,000 KG per day turned into energy, presumably almost all gamma-rays, but a lot of that energy would reflect off the upper atmosphere, lets estimate 40% of it reaches the earth, so 4,000 KG of gamma ray energy hits Earth from Space every day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4,000 KG, * C^2 (300,000,000^2) = 3.6*10^20 joules of energy per day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun (I'll just borrow from this &lt;a href=&quot;http://www.jiskha.com/display.cgi?id=1186705458&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; rather than calculate).  1.6x10^22, so, the heat from the matter-anti matter interactions would be 2% of sunlight.   That would be enough to warm the earth measurably.  A 2% increase in energy form space could warm the Earth several degrees, maybe 10 degrees, maybe more.   The Earth would be measurably warmer, but that would be the least of our problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think we can bring the number down a bit, cause you said no meteors, or asteroids, and even if you hadn't, larger objects like that in a matter-antimatter interaction, the contact would be so hot that most of the meteor wouldn't have a chance to evaporate but would simply be exploded away from the Earth, so we can round the number down quite a bit, like maybe a few hundred KG per day gets converted to gamma-rays and travels through the Earth's atmosphere and hits the Earth.   At this stage, the total heat the Earth gets is negligible enough at those levels, but would the sky glow visibly at night?   Honestly, I don't know, but maybe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gamma rays aren't visible, but as the rays pass through the atmosphere you might get enough visibly hot atmosphere/visible thermal radiation that you might see it.   I don't know how to begin to calculate that, but it seems possible.  The gamma ray effect would be similar to what would happen to the Earth if it was hit by a gamma ray burst (from a non lethal, but perhaps uncomfortably close distance).  &lt;a href=&quot;http://space.about.com/od/deepspace/a/Could-A-Gamma-Ray-Burst-Destroy-Life-On-Earth.htm&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;'s a description of what that would be like.   We'd lose much of our ozone layer and the chemical composition of our atmosphere would change.   over time, it could end much of the life on the Earth's land.  Life in the oceans could survive, but life on land would have a hard time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's not the only effect though.   As a few hundred tons of antimatter hits the upper atmosphere, the upper atmosphere would heat up and the Earth would lose it's atmosphere due to higher temperature much faster.   Over thousands, perhaps millions of years the Earth would lose much of it's atmosphere, which wouldn't be fun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A final effect is that antimatter would change chemistry.   If an Oxygen atom is hit it becomes a Nitrogen, if a Nitrogen is hit it becomes a carbon (I think, unless the gamma ray energy of the evaporation causes further photo-disintigration), but you'd see chemical changes as a result, perhaps some of them toxic and some radioactive isotopes.   So, no ozone layer, gamma ray radiation and toxic chemicals forming in the upper atmosphere - not exactly sunshine and puppies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Teeny-Tiny amounts of antimatter hitting the Earth isn't an issue, but in the amount you propose it would probably make the earth pretty inhospitable pretty fast. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-10-22T07:21:56.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12228" PostTypeId="1" AcceptedAnswerId="22262" CreationDate="2015-10-22T09:47:59.380" Score="9" ViewCount="911" Body="&lt;p&gt;As &lt;a href=&quot;http://en.wikipedia.org/wiki/Kozai_mechanism&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; says, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In celestial mechanics, the Kozai mechanism, or the Lidov–Kozai mechanism, is a perturbation of the orbit of a satellite by the gravity of another body orbiting farther out, causing libration (oscillation about a constant value) of the orbit's argument of pericenter. As the orbit librates, &lt;strong&gt;there is a periodic exchange between its inclination and its eccentricity.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question A&lt;br/&gt;&#xA;Which is the least massive object? The tertiary object, which is the furthest object, or the satellite in the inner binary? It seems it is not necessary that the tertiary object be the least massive one, which violates what Wikipedia says.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question B&lt;br/&gt;&#xA;How does the three-body system evolve?&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a periodic exchange between its inclination and its eccentricity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whose inclination and whose eccentricity? &lt;strong&gt;Please specify them using m0, m1 or m2 in the figure below.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbit of the inner binary should become and more circular. Can it become circular, eccentric, circular, eccentric?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question C&lt;br/&gt;&#xA;The inner binary will lose energy during the whole process, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BmTh3.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BmTh3.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditDate="2019-01-18T12:40:20.770" LastActivityDate="2019-01-18T12:40:20.770" Title="clarification of the Kozai mechanism" Tags="&lt;celestial-mechanics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="12230" PostTypeId="1" CreationDate="2015-10-22T15:44:56.490" Score="3" ViewCount="231" Body="&lt;p&gt;I would like to know which are the best books to learn about stellar astrophysics at (just) graduate level. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a basic formation in general astrophysics but I'm interested in learning about stars, specially their evolution and constitution. I also have a good mathematical formation so it should not be a problem (but I don't care if it focus on the ideas as long as it is a good book and make itself clear).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found several books at the university library, but I don't know which of them are good (in my situation at least).&lt;/p&gt;&#xA;" OwnerUserId="8011" LastActivityDate="2016-03-30T12:44:38.727" Title="Recommendation for learning about stellar astrophysics" Tags="&lt;star&gt;&lt;stellar-evolution&gt;&lt;stellar-astrophysics&gt;&lt;resource-request&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12231" PostTypeId="2" ParentId="12201" CreationDate="2015-10-22T18:18:06.737" Score="2" Body="&lt;p&gt;A supernova would vapourise all the planets in the stars planetary system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a rule of thumb when estimating the effects of supernovae: however powerful you think they are, they are more powerful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A supernova releases about $10^{44}$ joules in a matter of a few seconds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, a planet is a pretty small target, only about 1 billionth of the sun's heat hits a planet like (say) Jupiter, but 1 billionth of $10^{44}$ is still  $10^{35}$ Joules. This is the same order of magnitude as the gravitational binding energy of Jupiter. In conclusion, a supernova has enough power to completely destroy Jupiter. It is possible that there might be a small core that survives, and would be ejected as part of the supernova remnant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, a smaller earth-like planet would be toast (although steam is a closer description, the planet would be vapourised.) &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-10-22T18:18:06.737" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12232" PostTypeId="2" ParentId="12230" CreationDate="2015-10-22T18:57:04.840" Score="2" Body="&lt;p&gt;Best book I have used for this topic is: An Introduction to Stellar Astrophysics by Le Blanc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Excellent book, very good introductory start into this topic.&lt;/p&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2015-10-22T18:57:04.840" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12233" PostTypeId="2" ParentId="12228" CreationDate="2015-10-22T23:45:19.613" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which is the least massive object?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Quoting Wikipedia,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the hierarchical, restricted three-body problem, it is assumed that the satellite has negligible mass compared with the other two bodies (the &quot;primary&quot; and the &quot;perturber&quot;), . . .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the case studied in &lt;a href=&quot;http://adsabs.harvard.edu/abs/1962AJ.....67..591K&quot; rel=&quot;nofollow&quot;&gt;Kozai (1962)&lt;/a&gt;, specifically, the case of asteroids being perturbed by Jupiter. While not massless, the difference in mass is large enough that the asteroid's mass is negligible.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How does the three-body system evolve? . . . Whose inclination and whose eccentricity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wikipedia is again rather direct in this, stating that the conserved quantity $L_z$ depends on the satellite's orbit's eccentricity and inclination:&#xA;$$L_z=\sqrt{1-e^2}\cos i$$&#xA;As the satellite's mass is negligible, it will not have any significant effect on its perturber.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can it become circular, eccentric, circular, eccentric?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is basically asking if the eccentricity (and therefore inclination) can be described by some periodic function. Again, this is given in the Wikipedia article. A slightly different but just as simple xpresison is given in &lt;a href=&quot;http://faculty.wcas.northwestern.edu/rasio/Papers/106.pdf&quot; rel=&quot;nofollow&quot;&gt;Takeda &amp;amp; Rasio (2005)&lt;/a&gt;:&#xA;$$\text{Kozai Period}=P_{\text{perturbed}}\left(\frac{m_{\text{star}}+m_{\text{perturbed}}}{m_{\text{perturber}}}\right)\left(\frac{a_{\text{perturber}}}{a_{\text{perturbed}}}\right)^3(1-e_{\text{perturber}}^2)^{3/2}$$&#xA;In the approximation discussed above, in cases of extreme mass difference, $m_{\text{perturbed}}\to0$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The inner binary will lose energy during the whole process, right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The whole thing is periodic, so no energy is lost.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-10-23T01:42:09.940" LastActivityDate="2015-10-23T01:42:09.940" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12234" PostTypeId="2" ParentId="10017" CreationDate="2015-10-23T00:31:28.573" Score="2" Body="&lt;p&gt;Yes and no....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is now thought that all galaxies have a super massive black hole at their center. In that sense each galaxy thus orbits around its central black hole. See: &lt;a href=&quot;http://www.astro.ucla.edu/~ghezgroup/gc/journey/smbh.html&quot; rel=&quot;nofollow&quot;&gt;UCLA Article on galatic center&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a black hole isn't a &quot;star&quot; in the sense of emitting sunlight. A black hole is a &quot;sun&quot; which is so massive that it has collapsed into a singularity. &lt;/p&gt;&#xA;" OwnerUserId="9570" LastActivityDate="2015-10-23T00:31:28.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12235" PostTypeId="2" ParentId="12220" CreationDate="2015-10-23T00:45:38.800" Score="1" Body="&lt;p&gt;You misunderstand the very nature of such an interaction. An elastic collision with enough energy to knock the moon out of orbit would need a dwarf planet sized object to impact the moon. But such a collision wouldn't be elastic but disintegrate the two bodies.  &lt;/p&gt;&#xA;" OwnerUserId="9570" LastActivityDate="2015-10-23T00:45:38.800" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12236" PostTypeId="1" AcceptedAnswerId="12237" CreationDate="2015-10-23T04:50:19.093" Score="12" ViewCount="156" Body="&lt;p&gt;Recently I have been looking into planetoids in our asteroid belt and I have found one that caught my interest, Ceres. One of the main points that was said about it was that it had a subterranean ocean. But I'm puzzled as to how astronomers can come to this conclusion. Any explanations would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="9573" LastEditorUserId="9573" LastEditDate="2015-10-23T06:47:43.197" LastActivityDate="2015-10-31T14:17:15.413" Title="Subterranean Oceans On Other Planets/Planetoids: How Do Astronomers Deduce This" Tags="&lt;planet&gt;&lt;ceres&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12237" PostTypeId="2" ParentId="12236" CreationDate="2015-10-23T08:01:03.550" Score="1" Body="&lt;p&gt;How they have concluded that there could b.e a subterranean ocean on Ceres is by &lt;a href=&quot;http://www.enchantedlearning.com/subjects/astronomy/glossary/spectroscopy.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;Spectralscopy&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Spectral Signature can be summarised to like this:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Different elements emit different emission spectra when they are&#xA;excited because each type of element has a unique energy shell or&#xA;energy level system. Each element has a different set of emission&#xA;colors because they have different energy level spacings. We will see&#xA;the emission spectra or pattern of wavelengths (atomic spectra)&#xA;emitted by six different elements in this lab. We will then identify&#xA;an unknown element by comparing the color of the unknown with the&#xA;flame color of our knowns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And &lt;a href=&quot;http://www.800mainstreet.com/spect/emission-flame-exp.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;another&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;When something is hot enough to glow (like a star), it gives you information about what it is made of, because different substances give off a different spectrum of light when they vaporize. Each substance produces a unique spectrum, almost like a fingerprint.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So how the scientists would have concluded that there was water because of the water vapours in the atmosphere. It gives off a certain wavelength that could be compared the closest element that gives off the same wavelength which is water.&lt;/p&gt;&#xA;&lt;p&gt;Then the scientists would have used Galileo's magnetometer(an instrument which measures the strength and direction of magnetic fields) to conclude that there could be an ocean on the dwarf planet. The strength and response of the induced field would tell the scientists a rough estimate of the dwarf planet's sub surface which in return they could deduce the fact the there is a large amount of water present which equals an ocean.&lt;/p&gt;&#xA;" OwnerUserId="7926" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-10-23T09:21:07.243" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12238" PostTypeId="1" AcceptedAnswerId="12245" CreationDate="2015-10-23T05:41:00.403" Score="6" ViewCount="348" Body="&lt;p&gt;Recently the Kepler telescope in its study of white dwarfs detected the first planetary object transiting a white dwarf in the data from the K2 mission. It was consistent with earlier theories' prediction that a planetary object orbiting a white dwarf would slowly disintegrate. Why would a planetary object that is orbiting a white dwarf disintegrate? I read it &lt;a href=&quot;http://www.thehindubusinessline.com/news/science/nasa-probe-spots-dead-star-killing-a-planet/article7792728.ece&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The original paper published in Nature (preprint): &lt;a href=&quot;http://cfa.harvard.edu/~avanderb/wd1145_017.pdf&quot; rel=&quot;nofollow&quot;&gt;A Disintegrating Minor Planet Transiting a White Dwarf&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="9574" OwnerDisplayName="r2_d2" LastEditorUserId="9186" LastEditDate="2015-10-24T03:35:28.047" LastActivityDate="2015-10-26T03:31:48.040" Title="White dwarf's impact on orbiting bodies" Tags="&lt;planet&gt;&lt;exoplanet&gt;&lt;white-dwarf&gt;" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12240" PostTypeId="1" AcceptedAnswerId="12242" CreationDate="2015-10-23T16:14:16.423" Score="2" ViewCount="140" Body="&lt;p&gt;Most apparent stars have ancient names. But some of the newer (since 200 or so years) discoveries have been given the name of some prominent astronomer. &lt;a href=&quot;https://en.wikipedia.org/wiki/Kapteyn%27s_Star&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kapteyn's star&lt;/a&gt; is one example, and there are several more in that Wikipedia article. Edward Barnard et cetera. But there aren't too many, are there? It seems hard to get ones name attached to a star, or am I missing a bunch?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VaOGG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VaOGG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/tkzdd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tkzdd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Kapteyn and his tie knot matching star&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RhGUJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RhGUJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/1/10/George_Van_Biesbroeck.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomy was never easy&lt;/a&gt;, there's always more to see. Star namer van Biesbroeck in action with his flaring tiny red dwarf.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-10-23T23:03:22.583" LastActivityDate="2015-10-23T23:03:22.583" Title="Which stars have been named after astronomers?" Tags="&lt;star&gt;&lt;history&gt;&lt;naming&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12241" PostTypeId="2" ParentId="12240" CreationDate="2015-10-23T16:45:57.677" Score="0" Body="&lt;p&gt;I know of a Barnard's Star, and also Herschel's star...&lt;/p&gt;&#xA;" OwnerUserId="5505" LastActivityDate="2015-10-23T16:45:57.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12242" PostTypeId="2" ParentId="12240" CreationDate="2015-10-23T20:12:26.897" Score="2" Body="&lt;p&gt;You probably need someone better versed in the history of astronomy than me, but I'll give it a stab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you've already noticed, the Wikipedia article on &lt;a href=&quot;https://en.wikipedia.org/wiki/Stars_named_after_people&quot; rel=&quot;nofollow&quot;&gt;stars named after people&lt;/a&gt; has some entries, and that's probably about it.  In terms of modern naming procedures, we tend to use catalogue names. Unlike, say, asteroids, there's no systematic way of naming stars after people or anything else, so stars usually continue to be referred to only by catalogue number. Those rare stars that are named after people (or &lt;a href=&quot;https://en.wikipedia.org/wiki/Pistol_Star&quot; rel=&quot;nofollow&quot;&gt;something else&lt;/a&gt;) are usually because they are now somehow historically associated with that system.  Unless, of course, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Henry_Draper_Catalogue&quot; rel=&quot;nofollow&quot;&gt;catalogue is named after someone&lt;/a&gt;!  Another exception are those very bright stars with pre-telescope names, but those aren't named after people (e.g. many scientific papers refer to Betelgeuse as such).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess in short, there's just no standard modern naming procedure by which a star can end up with a person's name, so few exist, and probably most are listed in the Wikipedia article.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2015-10-23T20:12:26.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12243" PostTypeId="1" CreationDate="2015-10-24T02:02:16.740" Score="4" ViewCount="434" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Pluto#Surface&quot; rel=&quot;nofollow noreferrer&quot;&gt;wiki page&lt;/a&gt;, the mean temperature of Pluto is 44 degree K (-229 degree C). However, given that Pluto is tilted greatly and there is daylight for long periods on part of Pluto facing the Sun and also given that atmosphere of Pluto &lt;a href=&quot;https://astronomy.stackexchange.com/questions/12132/size-of-sun-as-seen-on-pluto&quot;&gt;scatters light to a high degree&lt;/a&gt;, the surface temperature on that part of Pluto may be higher. The atmosphere of Pluto contains methane which is a powerful greenhouse gas, and that also could push up the temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can be the maximum temperature on part of Pluto facing the Sun? Does available temperature estimate take into account findings of New Horizons probe which flyby Pluto earlier this year? Thanks for your insight.&lt;/p&gt;&#xA;" OwnerUserId="9491" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-10-24T19:54:34.040" Title="What is the temperature on surface of Pluto facing Sun?" Tags="&lt;pluto&gt;&lt;temperature&gt;&lt;surface&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12244" PostTypeId="2" ParentId="12243" CreationDate="2015-10-24T04:09:33.433" Score="2" Body="&lt;p&gt;According to &lt;a href=&quot;http://www.universetoday.com/13893/temperature-of-pluto/&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt;, Pluto may experience temperatures of up to 55K, which I assume happens during the daytime. This low temperature is due to Pluto's distance from the sun, its high albedo (about 0.5), and its extremely thin atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="9580" LastEditorUserId="9580" LastEditDate="2015-10-24T19:54:34.040" LastActivityDate="2015-10-24T19:54:34.040" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12245" PostTypeId="2" ParentId="12238" CreationDate="2015-10-24T08:16:19.350" Score="3" Body="&lt;p&gt;I think Aabaakawad's link gives a complete answer, but to give an astronomy for dummies answer, there's nothing about a white dwarf that causes a planet's orbit to decay at least, not directly.  Your article (I've pulled quote below the caption):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Slowly the object will disintegrate, leaving a dusting of metals on&#xA;  the surface of the star.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's only talking about this particular situation and there's a difference between disintegrate and decay.    This planetoid is enormously close to the white dwarf.  So close, that what we think of as normal white dwarf/planet dynamics (very cold) is no longer true.  This planetoid is slowly being vaporized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the orbital period of 4.5 hours (about 1948 orbital periods in 365.25 days).   The orbital distance to orbital ratio relation is exponential to the power of 2/3 (this varies a bit due to eccentricity, but it's generally correct), so an orbital period 1948 times faster means about 156 times closer, and giving the white dwarf equal mass to our sun, that puts the planetoid at a bit under 1 million KM.  If this white dwarf is lighter than the sun, the planetoid would need to be even closer.  That's close to the Roche limit and would be inside it if the planetoid wasn't dense and rocky/metallic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we estimate the white dwarf to be about the size of the Earth, which is a common size given for white dwarfs, An Earth sized object from 1 million KM would be larger in the sky than the sun appears from Earth, and presumably quite a bit hotter than the surface of our sun too, so this isn't a tiny white dwarf in the sky from the perspective of the planetoid.   It's a blazing furnace of a sun, so hot, it vaporizes metallic gas and dust off the surface of the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article mentions this (end of page 3), that Poynting-Robertson drag see &lt;a href=&quot;https://en.wikipedia.org/wiki/Poynting%E2%80%93Robertson_effect&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://planetfacts.org/poynting-robertson-effect/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, and that may be a factor in any orbital decay in this scenario.  The article is clear that there's a good deal of uncertainty on with that effect, and that only affects tiny particles, but enough tiny particles could create a drag over time. . . . (maybe).   The general scenario with this orbit is a planet scorched and as a result, is losing material.  It's likely the very high heat that's driving any orbital decay, not gravity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational decay / &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_decay&quot; rel=&quot;nofollow&quot;&gt;orbital decay&lt;/a&gt; does happen, usually much more slowly.   That's probably not what's happening here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some interesting orbital effects that can happen when a main sequence star goes red dwarf and later when it creates a planetary nebula, significant increases in tidal forces due to the star's greater size in the first case and increased drag in the 2nd, but at the white dwarf stage, there's no significant orbital decay effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why not Poynting Robertson drag and Orbital decay effect the planetoid&#xA;  when white dwarf was a star or even red giant? Is there any&#xA;  &quot;interesting orbital effects&quot; when a star undergoes red giant?.Can you&#xA;  update your answer to summarize the forces and their effect on the&#xA;  planetoid in each phase of the star. &#xA;  and also what do you mean by orbital decay? Does it have something to deal with the Roche limit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;OK, I think, having read more about it, Poynting-Robertson effect only matters when the orbiting objects are very small.   I've linked it twice above, but the simple explanation is that objects in orbit move and so any light or debris from the sun hits the moving object at an angle, not direct on.    If the object is small enough, this over time drives the dust and maybe grain of sand sized particles into the sun.   This doesn't affect larger objects, so it's not really relevant to any planets or planetoids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as &quot;interesting red dwarf&quot; effects.   That really has to do with tides.   Using the Moon/Earth example, the Moon creates tides on the Earth, a tidal bulge towards the Moon, but because the Earth Rotates faster than the Moon orbits, this tidal bulge is always ahead of the Moon and this creates a gravitational tug on the moon that pulls it away from the Earth - very slowly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same thing happens with planets around stars, but even more slowly,   lets pretend it's just the Earth and the Sun - a 2 body system (in reality, with several planets it's much more complicated), but just Earth and sun, teh Earth creates a tidal bulge on the sun, the sun rotates ahead of the earth, this causes the earth to very slowly spiral away from the sun - so slowly that it might take a trillion years for the Earth to spiral away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now when the sun goes Red Giant, the sun is essentially the same mass but much more spread out and parts of it, much closer to the Earth and less gravitationally bound to the sun.   This creates a far larger tidal tug.   Also, as the sun expands it's orbital velocity drops, because orbital momentum is concerved, so when the Sun is Red Giant, the tidal bulge will be behind the Earth which drags it in towards the Sun.    Due to the size and proximity of the Red Giant star, this draws the remaining near-by planets towards the sun fairly quickly, at least compared to main sequence stages which, provided the sun rotates faster than the planets orbit, has a much smaller outward tidal pressure on the planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And when the sun goes planetary nebula, any debris in the planet's path can also cause the planets to slow down slightly - the precise process there I'm less clear on, but in general, any orbital debris creates drag and can slow down a planet's orbit.   This may be a key factor in the formation of hot jupiters, cause they can't form close to their suns but enough orbital debris can drive them in closer to their suns.  (or planet to planet gravitational interactions can too).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the gist of the Sun-Planet orbital relation.  When the sun is young, planets are mostly driven outwards, and young suns can have far greater solar flares and stronger solar wind.   How much that effects the planets, I'm not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the Main sequence stage, stars tend to push planets outwards (unless they rotate very slowly, in which case the tidal effect is reversed), but this tidal effect is very small and very gradual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the Red Giant stage, stars tend to drag planets in wards, and I assume, during the planetary nebula stage as well.   This effect is larger for closer planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also asked about Orbital Decay - if you click on the link, there are examples of that.  That probably gives a better explanation than I could.   In general, Orbital decay happens very slowly unless you're talking Neutron Star or Black hole in which case the relativistic effects can cause orbital decay to happen quite fast.  There's nothing about a white dwarf star that would cause faster than normal orbital decay but a white dwarf star would lose any tidal bulge tugging that a main sequence star has, so there would be essentially no tidal outwards pressure either which could in theory speed up decay cause you've lost a small outwards pressure but you would still have any debris or space dust clouds causing a small inwards pressure.  (if that makes sense?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's my layman's explanation anyway.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-10-26T03:31:48.040" LastActivityDate="2015-10-26T03:31:48.040" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12246" PostTypeId="1" AcceptedAnswerId="12257" CreationDate="2015-10-24T09:57:25.597" Score="1" ViewCount="98" Body="&lt;p&gt;Would an exoplanet that has more ocean than Earth rotate at a different speed as a result of this? Would the amount of water impact the weight, gravitational pull, and/or tidal forces and cause a difference in the exoplanet's rotation period?  (For simplicity's sake, say, if it was earth-like in every respect other than having more ocean than Earth does at about 70% of the surface area.)  &lt;/p&gt;&#xA;" OwnerUserId="2301" LastEditorUserId="2301" LastEditDate="2015-10-25T00:24:25.107" LastActivityDate="2015-10-25T11:42:34.887" Title="Does more ocean on an Earth analog produce a different rotation period?" Tags="&lt;gravity&gt;&lt;exoplanet&gt;&lt;rotation&gt;&lt;earth-like-planet&gt;&lt;tidal-forces&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12247" PostTypeId="1" AcceptedAnswerId="12254" CreationDate="2015-10-24T10:22:21.670" Score="6" ViewCount="883" Body="&lt;p&gt;According to &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/astro/orbtilt.html&quot;&gt;this explanation&lt;/a&gt; at the Georgia State University website, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Axis Tilt is Critical for Life&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;. . .&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The Earth's spin axis is tilted 23.5° with respect to the ecliptic, giving moderate seasons and preventing temperature extremes anywhere on the planet. . . . Not only is the Earth's angle of tilt close to the optimum value, it also seems to have been essentially constant. That is crucial for the development of advanced life. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is this claim correct?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Why does it matter if there are some areas of a planet with extreme temperatures, as long as there are other spots on the planet that are not extreme?&lt;/li&gt;&#xA;&lt;li&gt;Why are &quot;moderate&quot; seasons required for life to exist?&lt;/li&gt;&#xA;&lt;li&gt;If humans can live at the equator on Earth where there is the least amount of tilt, why would an exoplanet with less tilt or no tilt be necessarily non-inhabitable?&lt;/li&gt;&#xA;&lt;li&gt;Even if humans could not live on a planet without axial tilt, are there no other forms of known &quot;advanced&quot; life that can? We know that extremophiles exist, such as &lt;a href=&quot;http://www.nytimes.com/2015/09/08/science/the-tardigrade-water-bear.html&quot;&gt;tardigrades' ability to survive in the vacuum of space&lt;/a&gt;.  What is the most &quot;advanced life&quot; that could live on a planet without axial tilt?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2301" LastActivityDate="2015-10-25T05:00:19.787" Title="Is axial tilt critical for life?" Tags="&lt;rotation&gt;&lt;habitable-zone&gt;&lt;weather&gt;&lt;astrobiology&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12248" PostTypeId="2" ParentId="12247" CreationDate="2015-10-24T15:50:25.017" Score="2" Body="&lt;p&gt;This is a case where the Southern US phrase &lt;a href=&quot;https://en.wiktionary.org/wiki/might_could&quot; rel=&quot;nofollow&quot;&gt;might could&lt;/a&gt; is appropriate. Intelligent life&lt;sup&gt;1&lt;/sup&gt; &lt;strong&gt;might could&lt;/strong&gt; arise only on terrestrial planets with just the right axial tilt. Then again, intelligent life might could arise on all kinds of other planets as well. Extrapolating from a sample size of one is always a risky endeavor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hyperphysics page referenced in the question implicitly poses the &lt;a href=&quot;https://en.wikipedia.org/wiki/Rare_Earth_hypothesis&quot; rel=&quot;nofollow&quot;&gt;rare Earth hypothesis&lt;/a&gt; proposed by Peter Ward and Donald Brownlee  as if it is fact. It is not fact. It's a hypothesis that attempts to answer the &lt;a href=&quot;https://en.wikipedia.org/wiki/Fermi_paradox&quot; rel=&quot;nofollow&quot;&gt;Fermi paradox&lt;/a&gt;. There's a lot to be said both for and against the rare Earth hypothesis. Stating it as if it is fact is bad form, particular for &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu&quot; rel=&quot;nofollow&quot;&gt;http://hyperphysics.phy-astr.gsu.edu&lt;/a&gt;, which overall is a very good site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hyperphysics page goes even further than do Ward and Brownlee, forgetting that Ward and Brownlee addressed the issue of intelligent life. Could life arise on a Venus-like or Mars-like planet that orbits at just the right distance from it's star? That's a good question, the correct answer to which is &quot;we don't know (yet).&quot; Could intelligent life arise on such a planet? Once again, the correct answer is &quot;we don't know (yet).&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extrapolating from a sample size of one (the solar system, which contains one planet that harbors intelligent life), life in general and intelligent life in particular might should be quite common. Extrapolating from a sample size of one (the solar system, which contains a very large number of objects that don't harbor life), life in general and intelligent life in particular might should be quite rare. Which is the case is a good question, the answer to which as of now can  only be rightfully stated as &quot;we don't know (yet).&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; In the context of the Fermi paradox, &quot;intelligent life&quot;  is defined as life that is capable of communicating with life on planets orbiting other stars.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-10-24T15:50:25.017" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12249" PostTypeId="2" ParentId="12238" CreationDate="2015-10-24T18:57:24.720" Score="2" Body="&lt;p&gt;Let's assume the white dwarf has a mass of $0.6 M_{\odot}$ (there's probably a more accurate value, but most white dwarfs are close to this...). With a period of 4.5 hours we can use &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law&quot; rel=&quot;nofollow&quot;&gt;Kepler's third law&lt;/a&gt;, assuming the planetary mass is negligible compared to the white dwarf, to infer an orbital radius of 0.0054 au ($8.1\times 10^{8}$ m).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tidal forces this close to a white dwarf are very large. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Roche_limit&quot; rel=&quot;nofollow&quot;&gt;Roche limit&lt;/a&gt; for the total tidal disintegration of a satellite, in synchronous rotation, held together only by its own gravity is roughly&#xA;$$ d = 1.44 R_{WD} \left( \frac{\rho_{WD}}{\rho_p}\right)^{1/3},$$&#xA;where $R_{WD}$ is the radius of the white dwarf (similar to the radius of the Earth), $\rho_{WD}$ is the average density of the white dwarf (a few times $10^{9}$ kg/m$^3$) and $\rho_p$ the density of the planet (let's assume 5000 kg/m$^3$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $d \simeq 6 \times 10^{8}$ m and is very similar to the actual orbital radius of the planet. i.e. It will be tidally disintegrating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess it will be an observational selection effect that such objects &lt;em&gt;will&lt;/em&gt; be detected at the tidal breakup radius, since if they were further way they would not be disintegrating and would not be detected, and if they were closer they would have already disintegrated and wouldn't be seen!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: On reading the paper - the authors claim that these objects are not tidally disintegrating. In fact they argue that this must be debris from a rocky planet precisely because the density must be large enough to avoid tidal disintegration according to the formula above. However I find the whole discussion rather incoherent. They specifically talk about &quot;disintegrating planetesimals&quot; (note the tense) which are being evaporated in a Parker-type wind due to heating by the radiation from the white dwarf. I cannot see where they explain then how the planetesimals disintegrate.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-10-24T21:11:11.650" LastActivityDate="2015-10-24T21:11:11.650" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12250" PostTypeId="2" ParentId="12049" CreationDate="2015-10-24T22:19:46.820" Score="0" Body="&lt;p&gt;The simple answer is yes. Look around midnight in the South Easterly direction and you should be able to see them both at around this time of year. As James K says above, Orion is your friend.&lt;/p&gt;&#xA;" OwnerUserId="823" LastEditorUserId="2365" LastEditDate="2017-01-17T21:47:07.283" LastActivityDate="2017-01-17T21:47:07.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12251" PostTypeId="2" ParentId="12246" CreationDate="2015-10-25T01:25:17.350" Score="-1" Body="&lt;p&gt;I would assume so seeing as how it would change the density&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is exactly like earth but has a crap ton more of an ocean- it is perceivable that it's (average) density would be lower, &#xA;Also, minutely, it would compress this water more and heat it and make something of an atmosphere given that it doesn't escape. Trapping more and more energetic particles into a thick foggy atmosphere  can change the rotation so slightly it will probably never be noticed.....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if it has a moon, tides/ waves due to gravity can also minutely push the planet in a different state of rotation. Very slightly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read it wrong, but I'm keeping that in case it helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It really depends on how much more water there is. If it was 100% coverage, it would be more &quot;tidy&quot; and the ocean would certainly be more compact (same mass as earth, but add A HUGE amount of water to that to get from 70% to 100% coverage of the area)-- tides may not be as wild. I suppose it would effect the rotation period, but not much. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, it might be easier to post this on physics stack exchange, you get better answers, much sooner. This seems like a good question to get up votes off of...&lt;/p&gt;&#xA;" OwnerUserId="6769" LastEditorUserId="6769" LastEditDate="2015-10-25T01:31:22.193" LastActivityDate="2015-10-25T01:31:22.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12252" PostTypeId="1" CreationDate="2015-10-25T01:51:59.150" Score="5" ViewCount="580" Body="&lt;p&gt;So, you can determine the location of an object in an orbit at any given time via Kepler's laws, but how do you determine the orbital inclination at any given point in time?&lt;/p&gt;&#xA;" OwnerUserId="6769" LastEditorUserId="9186" LastEditDate="2015-10-27T13:29:05.323" LastActivityDate="2016-04-29T19:04:45.897" Title="Inclination in Kepler's laws" Tags="&lt;orbit&gt;&lt;orbital-elements&gt;&lt;inclination&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12253" PostTypeId="1" AcceptedAnswerId="12261" CreationDate="2015-10-25T03:08:59.250" Score="1" ViewCount="648" Body="&lt;p&gt;The light pollution fades the starlight coming from the Sky and this dont let people to see the Milky Way disc. Is there any study about this?&lt;/p&gt;&#xA;" OwnerUserId="2150" LastEditorUserId="2150" LastEditDate="2015-10-27T16:31:34.427" LastActivityDate="2015-10-27T16:31:34.427" Title="What is the minimum distance from city that allow to see Milky Way plane?" Tags="&lt;milky-way&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2015-10-26T14:30:36.690" ContentLicense="CC BY-SA 3.0" />
  <row Id="12254" PostTypeId="2" ParentId="12247" CreationDate="2015-10-25T05:00:19.787" Score="3" Body="&lt;p&gt;I agree with David Hammen.   Hyperphysics is mostly a very good site but they dropped the ball on that page IMHO.   Hope you don't mind a partially speculative answer, but here goes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does it matter if there are some areas of a planet with extreme&#xA;  temperatures, as long as there are other spots on the planet that are&#xA;  not extreme?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It shouldn't matter if part of the planet is uninhabitable.   There are deserts on Earth which are all but uninhabitable but that doesn't effect life elsewhere.   Prior to 5.3 million years ago the Mediterranean sea evaporated and that entire region could have had a salty basin and been hugely hot, but I've not read of life on Earth having any problem with that.   &lt;a href=&quot;http://phys.org/news/2009-12-mediterranean-sea-years.html&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why are &quot;moderate&quot; seasons required for life to exist?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's a boatload we don't know about the evolution of life on other planets but seems perhaps universally true that life adapts, so I find it difficult to believe that moderate seasons are necessary.    Very extreme changes could be difficult, but change can force adaptation.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If humans can live at the equator on Earth where there is the least&#xA;  amount of tilt, why would an exoplanet with less tilt or no tilt be&#xA;  necessarily non-inhabitable?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The tilt is planet wide but the lowest variation happens near the equator, but animals that thrive near the poles adapt by hibernation or migration and smaller stuff can be frozen and then come back to life, so, I don't agree with the article on this point.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Even if humans could not live on a planet without axial tilt, are&#xA;  there no other forms of known &quot;advanced&quot; life that can? We know that&#xA;  extremophiles exist, such as tardigrades' ability to survive in the&#xA;  vacuum of space. What is the most &quot;advanced life&quot; that could live on a&#xA;  planet without axial tilt?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One of the interesting historical facts of life on Earth, at least to me, is how long it took what we might consider advanced life to develop.   One celled life in various forms was around for over 3 billion years but the first fossils are about 650 million years old.   It took life a very long time on earth to get from too small to see to large enough to leave a footprint . . . but, I digress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree 100%, one celled life or Tardegrades could live on a planet with no tilt or 90 degree tilt.  Easy.   Ocean life in general should be fine cause oceans are more adaptive.  Evaporation keeps ocean surfaces colder than land gets during peak heat and while a completely frozen over ocean isn't great for life, cold oceans hold more oxygen and CO2 which can be good for life.   Oceans also circulate as an effective means of temperature moderation and fish don't really care how windy it is or how much or little it rains.   The tilt question, I think, is really just about life on land.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Land life could be more vulnerable to high wind, extreme temperature shifts, droughts or floods, which could be driven by greater axial tilt, but I find it hard to believe that Axial Tilt is the be-all and end all.   Day length and year length are key factors too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One point I agree with the article on, is that a close to 90 degree tilt might not be ideal with one part of the planet always facing the sun and the other part never facing it but outside of extreme tilts, I don't see why it would be a big deal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A thick cloud cover, for example, reduces seasonal changes.   There's a number of factors.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-10-25T05:00:19.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12255" PostTypeId="1" AcceptedAnswerId="12278" CreationDate="2015-10-25T06:16:30.950" Score="2" ViewCount="212" Body="&lt;p&gt;Are the dark grey areas on the moon formed from the last lava flows from volcanoes? And that is why there are hardly any craters in those dark grey areas because the lava has covered the old craters?And If so are there still active volcanoes that are still flowing to this day?&lt;/p&gt;&#xA;" OwnerUserId="9326" LastEditorUserId="2365" LastEditDate="2015-10-25T07:21:23.443" LastActivityDate="2015-11-25T12:56:05.763" Title="Volcano activity on moon" Tags="&lt;the-moon&gt;&lt;volcanism&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12256" PostTypeId="1" CreationDate="2015-10-25T06:30:54.333" Score="1" ViewCount="688" Body="&lt;p&gt;I seen with my naked eye in mid 2014 from Ca. right after sunrise a black spot in front of the Sun.I observed it for about a hour with my naked eye. Here is a picture that closely resembles what I observed, except the object was a third the size of the black dot in the posted picture and not in the middle of the sun but off to a angle.Since Venus did not transcend with the Sun in 2014 my speculation would have to say it might have been a Brown Dwarf.The link provided below of civilizations from the past documenting Venus transcending with the Sun throughout history could be observed with the naked eye,So a Brown Dwarf would be easier to see with the naked eye.The picture is from google and is the closest image I can give to what I seen.I have no knowledge of the picture and can not determine if it is real or fake.     &lt;img src=&quot;https://farm4.staticflickr.com/3110/3147227521_396e16bcfc.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof of venus or a sunspot seen with naked eye transending past the sun&#xA;&lt;a href=&quot;http://sunearthday.nasa.gov/2012/articles/ttt_73.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://sunearthday.nasa.gov/2012/articles/ttt_73.php&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9326" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:47.420" LastActivityDate="2015-12-16T10:54:31.057" Title="Brown Dwarf transcending past the sun with the naked eye" Tags="&lt;solar-system&gt;&lt;the-sun&gt;&lt;planet&gt;&lt;venus&gt;&lt;dwarf-planets&gt;" AnswerCount="3" CommentCount="15" ClosedDate="2015-12-17T15:22:59.527" ContentLicense="CC BY-SA 3.0" />
  <row Id="12257" PostTypeId="2" ParentId="12246" CreationDate="2015-10-25T07:17:44.047" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Would an exoplanet that has more ocean than Earth rotate at a&#xA;  different speed as a result of this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Basically no.   Rotation speed, or &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_velocity&quot; rel=&quot;nofollow noreferrer&quot;&gt;angular velocity&lt;/a&gt; doesn't measurably change with your proposed example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain this in more detail, there's two important concepts.  First, is Angular Momentum, (&lt;a href=&quot;http://csep10.phys.utk.edu/astr161/lect/solarsys/angmom.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Short&lt;/a&gt; explanation or &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_momentum&quot; rel=&quot;nofollow noreferrer&quot;&gt;longer&lt;/a&gt;) and the 2nd concept is &lt;a href=&quot;https://en.wikipedia.org/wiki/Moment_of_inertia&quot; rel=&quot;nofollow noreferrer&quot;&gt;Moment of Inertia&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic formula is that Angular velocity (time for 1 full rotation) = Angular Momentum divided by Moment of Inertia.  Formulas and explanations are in the link(s) above, and if I was to explain it, it would get wordy, but your question seemed to be more general, less about doing the mathematical calculations, so I'll skip the formulas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In nutshell, talking about a planet, the state of matter doesn't affect the angular momentum.   Angular momentum is conserved and that, divided by moment of inertia determines rotation speed.  Now if you get something spinning fast enough the angular rotation can overwhelm the gravity and when this happens, the planet can begin to fly apart, which happens more easily with water than a rocky surface which has some cohesion, but ignoring crazy super fast rotations, a water world, a deep ocean world, a shallow ocean world and a rocky world all obey the the angular velocity law and composition doesn't matter.  The angular momentum is conserved.  The moment of inertia of a planet can change but for the most part, doesn't change much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth's moment of Inertia, for example, changes as glaciers grow or shrink, or when there's an earthquake.   Even, every time we a tall building is built the Earth's moment of inertia increases a teeny tiny bit, similar to a skater extending their arms to slow down coming out of a spiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When there's an &lt;a href=&quot;http://www.nasa.gov/topics/earth/features/japanquake/earth20110314.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;earthquake&lt;/a&gt; which, for the most part, settles the Earth, there's a small increases in the Earth's rotational speed.   The total angular momentum and total mass remaining the same but shifting of material changes the moment of inertia.   (Granted space dust and tidal effects change the Earth's moment of Inertia, but quite slowly).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Would the amount of water impact the weight, gravitational pull,&#xA;  and/or tidal forces and cause a difference in the exoplanet's rotation&#xA;  period?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is harder question to answer precisely because adding water changes the mass of the planet and changing the mass changes the moment of inertia, but sticking to the principal of your question, there's no measurable effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets take a somewhat simpler example without changing the planet's mass.  Ice ages.  When the Earth is in an ice age there's less liquid oceans and more ice at the poles but the total mass is unchanged.   More mass at the poles and less mass in the oceans decreases the Earth's moment of inertia because the bulk of the Earth's moment of inertia is around the equator,  so, as a result, the Earth rotates slightly faster during an ice age and slightly slower after an ice age.   Over time, the Earth's crust has a tendency adjust for this effect but that takes tens of thousands of years.  Parts of the Earth's crust is still rebounding from the last ice age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational pull isn't relevant.  Neutron Stars with enormous gravitational pull can rotate very fast and the planet with the fastest rotation in our solar-system is Jupiter and the one with the slowest rotation is Mercury.   Angular velocity has no direct correlation to mass or gravity though there is an indirect correlation.   As a star, for example condenses it's rotation speeds up, because the angular momentum is conserved but as it settles the moment of inertia decreases.   That's why young stars, White Dwarfs and Neutron stars can spin very fast.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tidal forces can create drag on rotation but the effect is slow, taking millions or hundreds of millions of years.  With enough time, tidal forces cause a planet or moon to stop rotating and become tidally locked but there's no short term affect.  (I'll say a bit more on this later).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, a planet with large oceans wouldn't rotate any slower than a planet with no oceans because liquid or solid can have equal angular velocity, but over time, tides will slow a planet with oceans more quickly than a planet without them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the Earth has oceans, the Moon's gravitation on the Earth's tidal bulge does slow down the Earth's rotation, but this has been happening for 4 billion years and the Earth still rotates every 24 hours - one of the faster planets.   If the Earth had more water the Moon's tidal tug would slow the earth down a bit faster, but it would still be very gradual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter, which is basically a ball of gas, is the fastest rotating planet and Mercury, basically a rock, the slowest, so those are 2 examples of composition not being a factor, though Mercury's slow rotation is in large part due to the strong tidal forces it receives from the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I Understand the logical approach to your question, as there's something apparent about water resisting rotation - touched on in this &lt;a href=&quot;https://physics.stackexchange.com/questions/71925/why-does-an-ice-cube-not-twirl-along-when-twirling-a-drink-glass&quot;&gt;question&lt;/a&gt;, but the fact that water doesn't spin with a glass when you spin a glass is an example of conservation of angular momentum, not an argument against it.  On a planet, the oceans are rotating with the planet and the angular momentum is already there.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that wasn't too long, but that's the gist of it.   I can try to clean up or clarify if needed.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-10-25T07:34:34.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12258" PostTypeId="2" ParentId="12255" CreationDate="2015-10-25T07:49:27.307" Score="1" Body="&lt;p&gt;The grey areas are known as &quot;maria&quot; (singular:&quot;mare&quot;) and they were formed from flood  basalts. These are a type of volcano and they have formed on earth too, examples include the lake eruption on Iceland, and the Deccan traps, that played a role in the mass extinction 65 million years ago. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the moon, the flood basalt is old. About 3-4 billion years old, with some regions probably younger, but nothing less than 1 billion. Volcanism is driven by the heat of radioactive decay. On the moon there are no longer enough radioactive elements to melt rock, and so the moon is no longer a volcanic active body. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no active volcanoes on the moon, and there is no prospect of any in the future. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-10-25T07:49:27.307" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12259" PostTypeId="2" ParentId="12253" CreationDate="2015-10-25T08:33:17.930" Score="2" Body="&lt;p&gt;The premise is certainly true. An exact distance isn't possible to determine since it would depend on how much light pollution is coming from the city, and how much haze (water droplets and particulates) is in the air. You could also be a &quot;short&quot; distance away but on a mountain with much greater elevation than the city. &lt;/p&gt;&#xA;" OwnerUserId="9570" LastActivityDate="2015-10-25T08:33:17.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12261" PostTypeId="2" ParentId="12253" CreationDate="2015-10-25T10:45:58.663" Score="2" Body="&lt;p&gt;The biggest factor defining the minimum distance is light pollution from the particular city you are talking about. You can find light pollution maps &lt;a href=&quot;http://www.lightpollutionmap.info/#zoom=4&amp;amp;lat=5759860&amp;amp;lon=1619364&amp;amp;layers=B0TFFFFTT&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://darksitefinder.com/maps/world.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. You can also find charts &lt;a href=&quot;http://www.cleardarksky.com/csk/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; that describe the quality of viewing based on a combination of cloud cover, haze, turbulence and wind, temperature, light pollution, and other factors. They look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lhfaW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lhfaW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9575" LastActivityDate="2015-10-25T10:45:58.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12262" PostTypeId="2" ParentId="12246" CreationDate="2015-10-25T11:42:34.887" Score="1" Body="&lt;p&gt;If the mass, radius and total angular momentum of the planet is the same as Earth, then I can put forward an answer. (Of course if the angular momentum is allowed to be different you could have any rotation rate...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, surface gravity is unchanged since this only depends on the mass and radius of a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, if the mass is constant then the &lt;em&gt;average&lt;/em&gt; density must be the same as Earth. But water is less dense than the average density of the Earth, so the interior of the planet would need to be denser than that of the Earth in order for the average to be similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the density is higher than Earth on the inside, but lower on the outside, then the &lt;em&gt;moment of inertia&lt;/em&gt;, which depends on how far mass is from the axis of rotation &lt;em&gt;squared&lt;/em&gt;, will be lower. Then, because angular momentum (assumed constant) is the product of moment of inertia and rotation rate, we deduce that &lt;strong&gt;the rotation rate would need to be faster.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB: Simply covering a larger area with Earthlike ocean would make hardly any difference whatsoever. You need a &lt;em&gt;lot&lt;/em&gt; of surface water to make a difference.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-10-25T11:42:34.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12263" PostTypeId="2" ParentId="12256" CreationDate="2015-10-25T11:44:02.630" Score="2" Body="&lt;p&gt;&lt;em&gt;Possibly&lt;/em&gt; an aircraft or flock of birds, or maybe an insect really close to your eye, but more likely an illusion or misperception. It's not very &quot;bright&quot; to look directly at the Sun with your naked eye!&lt;/p&gt;&#xA;" OwnerUserId="9575" LastEditorUserId="9575" LastEditDate="2015-10-25T12:17:15.150" LastActivityDate="2015-10-25T12:17:15.150" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12264" PostTypeId="1" AcceptedAnswerId="12265" CreationDate="2015-10-25T09:52:27.427" Score="3" ViewCount="2106" Body="&lt;p&gt;If a star/galaxy is moving towards us, then its light is blueshifted, and if it is moving away from us, then its light is redshifted. How do we exactly know that the color we are seeing is not part of the fuel burnt by the star/galaxy, and is instead caused by the Doppler effect?&lt;/p&gt;&#xA;" OwnerUserId="142" OwnerDisplayName="Madeyedexter" LastEditorUserId="2153" LastEditDate="2015-10-25T13:40:25.773" LastActivityDate="2015-10-25T13:47:49.527" Title="How do we know that light is redshifted/blueshifted and not the original light of a star/galaxy?" Tags="&lt;redshift&gt;&lt;doppler-effect&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-10-26T14:27:09.970" ContentLicense="CC BY-SA 3.0" />
  <row Id="12265" PostTypeId="2" ParentId="12264" CreationDate="2015-10-25T13:47:49.527" Score="9" Body="&lt;p&gt;If you had a simple slit spectroscope, and looked at an incandescent light, you'd see a smear of light with red on one end and blue on the other. This is because the filament is producing light by glowing from being heated.&lt;/p&gt;&lt;p&gt;If you looked at one of those orange colored sodium vapor street lamps, instead of a smear of color, you'd see a &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/c/ca/Spectrum-hp-sodium.jpg&quot;&gt;group of lines&lt;/a&gt;. This light is produced by ionizing the gas.&lt;/p&gt;&lt;p&gt;The lines represent specific frequencies of light coming from the lamp. You could add a horizontal scale and find that the lines represent a specific frequency in the light spectrum.&lt;/p&gt;&lt;p&gt; If the street lamp was coming at you at a significantly high speed, the individual lines would be shifted in frequency towards blue, but would still have the same pattern. Conversely, if the light was moving away from you, you'd again see the same pattern of light lines, but their frequencies would be shifted toward red.&lt;/p&gt;&lt;p&gt;This is what's being measured when the spectra of stars and galaxies are measured: not just what the color looks like, but whether the spectra of things like hydrogen, helium and iron are shifted in frequency towards red or blue.&lt;/p&gt;&lt;p&gt;So, it's the specific line patterns produced by these elements when they are ionized in stars that helps us identify them. Comparing the light frequencies of locally ionized elements to the frequencies coming from distant stars that tell us if the stars themselves are approaching or receding.&lt;/p&gt; &lt;/p&gt;&#xA;" OwnerUserId="9516" LastActivityDate="2015-10-25T13:47:49.527" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12266" PostTypeId="2" ParentId="12256" CreationDate="2015-10-25T14:00:30.157" Score="1" Body="&lt;p&gt;Although the picture is too low rez to be certain, it looks like the object is very nearly circular. Pixels along the edge vary from pink to dark blue, while pixels in the center seem to be gray to black.&lt;/p&gt;&lt;p&gt;Without seeing the original image, I don't think one can draw any conclusion from the image, but the fact that it's so circular and seems to be exactly centered on the sun makes me suspect that this is a photoshopped hoax.&lt;/p&gt;&lt;p&gt;FWIW, the bright feature towards the left side of the picture is a sun dog, a refraction of sunlight through ice crystals, and shows no indication of the black spot, which means that if it was a real object, it was below that cloud layer, rather than above it.&lt;/p&gt;&lt;/p&gt;&#xA;" OwnerUserId="9516" LastActivityDate="2015-10-25T14:00:30.157" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12267" PostTypeId="1" AcceptedAnswerId="20797" CreationDate="2015-10-25T17:00:35.727" Score="8" ViewCount="962" Body="&lt;p&gt;I read recently about &lt;a href=&quot;https://en.wikipedia.org/wiki/VFTS_352&quot; rel=&quot;noreferrer&quot;&gt;VFTS 352&lt;/a&gt;, an overcontact binary star system where both stars have roughly equal mass. All of the reports I've read (in mass-media type publications) have said that the system has one of two fates: either the two stars will merge, or they'll supernova. But when will this happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wikipedia page for &lt;a href=&quot;https://en.wikipedia.org/wiki/Contact_binary&quot; rel=&quot;noreferrer&quot;&gt;contact binaries&lt;/a&gt; says that they have a lifespan of millions to billions of years, but doesn't say if that's different for overcontact binaries. It also says that they're often confused with &lt;a href=&quot;https://en.wikipedia.org/wiki/Common_envelope&quot; rel=&quot;noreferrer&quot;&gt;common envelopes&lt;/a&gt;, which have a lifespan of months to years, and I'm not sure where in that spectrum an overcontact lies (or really what the distinction is, since the page for contact binaries says they share an envelope, which sounds the definition of a common envelope). I'm also not sure whether the fact that both stars have roughly equal mass affects the lifespan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass-media articles I've read have implied that the merger-or-supernova is happening soon, but I don't know if this is on a human scale (months) or galactic scale (millions of years).&lt;/p&gt;&#xA;" OwnerUserId="9590" LastEditorUserId="9590" LastEditDate="2015-10-26T06:35:01.360" LastActivityDate="2019-04-20T19:12:23.167" Title="How long does an over contact binary star system last?" Tags="&lt;stellar-evolution&gt;&lt;binary-star&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12268" PostTypeId="2" ParentId="12256" CreationDate="2015-10-25T19:24:55.673" Score="1" Body="&lt;p&gt;Might've been a large sunspot. Quite a few have been naked eye visible this solar cycle. Apparently there were &lt;a href=&quot;http://apod.nasa.gov/apod/ap130710.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;some big ones in July 2013&lt;/a&gt; (APOD)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a white light shot of the &lt;a href=&quot;http://topics181.rssing.com/browser.php?indx=7886493&amp;amp;item=6984&quot; rel=&quot;nofollow noreferrer&quot;&gt;full sun on July 7 2013&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7Kea6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Kea6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="9186" LastEditDate="2015-10-26T18:04:28.307" LastActivityDate="2015-10-26T18:04:28.307" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12269" PostTypeId="1" AcceptedAnswerId="12270" CreationDate="2015-10-25T21:07:54.400" Score="21" ViewCount="4367" Body="&lt;p&gt;Mercury rotates three times for every two revolutions around the Sun, apparently due to a gravitational resonance with the Sun.&lt;/p&gt;&lt;p&gt;Venus takes about 225 days to rotate, and rotates in the opposite direction of any of the inner planets. Maybe because its extreme nature makes it ornery.&lt;/p&gt;&lt;p&gt;Earth rotates once every 24 hours, a condition caused by the tidal interaction between Earth and its Moon. It's believed that the Earth was rotating about once every 5 hours before the theorized collision with a Mars sized coorbiting object referred to as Theia.&lt;/p&gt;&lt;p&gt;Mars shows no signs of a similar collision. Its two moons appear to be asteroids that were captured from the asteroid belt. So how did Mars come to have a day so close to the length of an Earth day?&lt;/p&gt;&lt;/p&gt;&#xA;" OwnerUserId="9516" LastEditorUserId="9186" LastEditDate="2015-10-27T02:08:20.123" LastActivityDate="2019-01-16T20:15:44.100" Title="How did Mars come to have a 24 hour 39 minute day?" Tags="&lt;planet&gt;&lt;mars&gt;&lt;rotation&gt;&lt;tidal-forces&gt;&lt;planetary-formation&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12270" PostTypeId="2" ParentId="12269" CreationDate="2015-10-25T22:12:23.133" Score="23" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;It's believed that the Earth was rotating about once every 5 hours&#xA;  before the theorized collision with a Mars sized coorbiting object&#xA;  referred to as Theia.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Almost. Theia did not have to be co-orbiting, just an intersecting orbit. We have no idea what the Earth's spin was before the collision, but it is theorized that the Earth rotation had a 5 hour period &lt;strong&gt;&lt;em&gt;after&lt;/em&gt;&lt;/strong&gt; the collision with Theia, at the time of the Moon's formation from the debris. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that Mars and Earth have such a similar period is a coincidence, perhaps you are asking why Mars is spinning so fast? Well actually Mars is not the odd man out, Mercury and Venus are. Most planets spin fast. exactly which spin orientation is somewhat arbitrarily determined by the vagaries of the ways the planetesimals collided to form them. The fact that Venus and Uranus have unusual spin &lt;em&gt;orientations&lt;/em&gt; is just the way things turned out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both Mercury and Venus used to spin much faster. Mercury's spin was tidally slowed down by the Sun &lt;em&gt;and&lt;/em&gt; Mercury's orbit was (and still is being) driven further away by the Sun (just like the Moon and Earth: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/420/why-is-the-moon-receding-from-the-earth-due-to-tides-is-this-typical-for-other&quot;&gt;Why is the Moon receding from the Earth due to tides? Is this typical for other moons?&lt;/a&gt;). Eventually Mercury was held in that 2:3 resonance. Which, by the way had a certain amount of luck involved (see: &lt;a href=&quot;http://www.researchgate.net/publication/8492094_Mercurys_capture_into_the_32_spin-orbit_resonance_as_a_result_of_its_chaotic_dynamics._Nature&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mercury’s capture into the 3:2 spin-orbit resonance as a result of its chaotic dynamics&lt;/a&gt; ). Venus, we are not so sure of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tidal &lt;em&gt;force&lt;/em&gt; from the sun is much much less for Venus than for Mercury, but much more than for Earth. &lt;em&gt;However&lt;/em&gt; Venus has a dense hot &lt;em&gt;massive&lt;/em&gt; atmosphere, which can be forced into both gravitational bi-modal (two peaks) tides and thermal uni-modal (one peak) tides. The bulge lags behind the tidal forcing peak, which creates a torque by the sun to slow it down. This is fiendishly complex (See: &lt;a href=&quot;http://www.imcce.fr/Equipes/ASD/preprints/prep.2002/venus1.2002.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Long term evolution of the spin of Venus - I. Theory&lt;/a&gt; )&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;P.S.&lt;/em&gt; Actually Phobos, and probably Deimos, are thought to be constructed fairly recently (millions of years) from debris from a collision of Mars with a large asteroid. There is no way to capture a whole asteroid into orbits that close.&lt;/p&gt;&#xA;" OwnerUserId="9186" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-10-27T15:51:24.267" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12272" PostTypeId="1" AcceptedAnswerId="12309" CreationDate="2015-10-26T06:52:54.920" Score="3" ViewCount="181" Body="&lt;p&gt;For example, if I throw two planets to orbit a star at random direction, would they form an orbital resonance?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="2153" LastEditDate="2015-10-26T15:14:14.000" LastActivityDate="2015-10-28T17:10:53.717" Title="Do orbital resonances always form naturally?" Tags="&lt;orbit&gt;&lt;celestial-mechanics&gt;&lt;orbital-resonance&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12273" PostTypeId="1" AcceptedAnswerId="18570" CreationDate="2015-10-26T07:48:48.037" Score="5" ViewCount="313" Body="&lt;p&gt;Mars's moons &lt;a href=&quot;http://marsmobile.jpl.nasa.gov/allaboutmars/extreme/moons/&quot; rel=&quot;nofollow noreferrer&quot;&gt;have the appearance of captured asteroids&lt;/a&gt;.  How could this capture have taken place.  An asteroid in solar orbit would need to loose some momentum to be captured.  The gas Giants can capture new moons through 3 body interactions between the asteroid and their existing moons,  but Mars doesn't have this issue. Airbreaking would lead to impact, not capture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/12269/how-did-mars-come-to-have-a-24-hour-39-minute-day&quot;&gt;An answer here&lt;/a&gt; suggests that the moons could have formed recently (millions of years, not billions) after an impact ejected a substantial amount of material in Martian orbit. What is the evidence for this?  Is there an impact crater of the right size, and age? Were both moons formed together? &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-10-04T13:23:47.053" Title="Formation and capture of Mars' moons, Phobos and Deimos" Tags="&lt;mars&gt;&lt;phobos&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12275" PostTypeId="1" CreationDate="2015-10-26T08:36:22.157" Score="3" ViewCount="741" Body="&lt;p&gt;I was watching a documentary on the big bang, one of the astronomers said that initially the four fundamental forces were combined as one, they then emerged to become: The strong and weak nuclear forces, the electromagnetic force and the gravitational force. He claimed that the first force to emerge was the gravitational force, he went on to say that had the gravitational force been weaker than it is everything would fly apart so no galaxies would form and if gravity was too large than we would end up with black holes everywhere, so the force of gravity needed to be just right, as it is now. I couldn't find anything backing up this information online or the reasoning behind this information, any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="9599" LastActivityDate="2017-06-24T19:49:07.297" Title="Strength of gravity during the big bang?" Tags="&lt;gravity&gt;&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12276" PostTypeId="2" ParentId="12275" CreationDate="2015-10-26T10:21:07.610" Score="1" Body="&lt;p&gt;The truth is we do not have a working or widely accepted &quot;theory of everything&quot; that unifies gravitation with the other fundamental, and quantum, field theories. What we do have is strong evidence that the other field theories - electromagnetic, strong nuclear and weak nuclear - are unified at high energies (indeed the evidence is so strong that I doubt it is widely disputed at all).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do have a theory of gravitation - general relativity - that appears to work well in &quot;most&quot; circumstances, though it also has a singularity (black holes) and there is a fundamental difference between it and the quantum theories about vacuum energy (the so-called vacuum catastrophe).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we also know is that gravitation appears to much weaker than the other forces, though it can act over very large distances and so is fundamental in shaping the nature of our universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various arguments about how/why our universe is able to support life/is the way it is. Perhaps it is just one of many/an infinite number of universes that have a range of physical properties and we just happen to be in one that &quot;works&quot; for life - obviously there could be no life in a universe that had physical properties that didn't support life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At another extreme perhaps we are all inside a computer simulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd suggest hunting down works by &lt;a href=&quot;http://space.mit.edu/home/tegmark/&quot; rel=&quot;nofollow&quot;&gt;Max Tegmark&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Hidden_Reality&quot; rel=&quot;nofollow&quot;&gt;Brian Greene&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2015-10-26T10:21:07.610" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12278" PostTypeId="2" ParentId="12255" CreationDate="2015-10-26T12:33:58.133" Score="1" Body="&lt;p&gt;There was once a very active argument over vulcanism on the Moon - with the distinguished amateur astronomers &lt;a href=&quot;https://en.wikipedia.org/wiki/Patrick_Moore&quot; rel=&quot;nofollow&quot;&gt;Patrick Moore&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Axel_Firsoff&quot; rel=&quot;nofollow&quot;&gt;V. A. Firsoff&lt;/a&gt; the two leading advocates of vulcanism as having had a decisive impact on the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Apollo programme essentially settled this argument as far as craters were concerned - no serious figure seems to doubt these are the result of impacts (though Moore was still forcefully arguing, if knowingly from a losing position, for vulcanism long after the end of Apollo).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the lunar mare it is not quite so clear cut - it is not disputed that these are flood basalt plains and so are the result of some form of vulcanism. However, they also appear to lie inside impact craters and so there is a suggestion of a connection, at least, with impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no volcanoes on the Moon today, though there are some signs that there may be very limited vulcanism or similar activity - see &lt;a href=&quot;http://www.armaghplanet.com/blog/whatever-happened-to-transient-lunar-phenomena.html&quot; rel=&quot;nofollow&quot;&gt;http://www.armaghplanet.com/blog/whatever-happened-to-transient-lunar-phenomena.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2015-10-26T12:33:58.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12279" PostTypeId="1" CreationDate="2015-10-26T16:53:25.027" Score="3" ViewCount="109" Body="&lt;p&gt;Many years ago (early 80s or late 70s), I bought a cardboard &quot;device&quot; which I think was called &quot;Planet Timer&quot; and I think was sold by Edmund Scientific. Does anyone have any information on this device? Google was unhelpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The device had concentric revolving circles around a fixed Sun at the center. By revolving the circles to the current time (via tables provided with the device), you could presumably see what constellation each planet was in. You could also use it to approximate planetary conjunctions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The circles were equal distance from each other, so I'm pretty sure the device couldn't actually have worked, but I'm curious to see if I can find out more about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: this is not a &quot;planisphere&quot; device, which did work, and for which I can still find plenty of information.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2015-10-29T20:08:24.710" Title="Planet-timer &quot;device&quot; from 1970s or 1980s possibly by Edmund Scientific" Tags="&lt;observational-astronomy&gt;&lt;planet&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12280" PostTypeId="1" CreationDate="2015-10-26T17:27:46.873" Score="5" ViewCount="225" Body="&lt;p&gt;The difference between &lt;em&gt;meteorites&lt;/em&gt;, &lt;em&gt;meteors&lt;/em&gt;, and &lt;em&gt;meteoroids&lt;/em&gt; is one of altitude relative to a celestial surface: in space, it's a meteoroid; in the atmosphere, it's a meteor; and on the surface, it's a meteorite. (See &lt;a href=&quot;http://www.universetoday.com/100075/infographic-whats-the-difference-between-a-comet-asteroid-and-meteor/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.universetoday.com/36398/what-is-the-difference-between-asteroids-and-meteorites/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for more information.)&lt;/p&gt;&#xA;&lt;p&gt;The International Astronomical Union (IAU) defined the term &lt;em&gt;Small Solar System Body&lt;/em&gt; (SSSB) in 2006 with &lt;a href=&quot;https://www.iau.org/static/resolutions/Resolution_GA26-5-6.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Resolution B5&lt;/a&gt; as&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;(3) All other objects³,except satellites, orbiting the Sun shall be referred to collectively as &amp;quot;Small Solar System Bodies&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;code&gt;[Footnote 3]&lt;/code&gt; &amp;quot;These currently include most of the Solar System asteroids, most Trans-Neptunian Objects (TNOs),comets, and other small bodies.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And, although this official definition clearly states, &amp;quot;All other objects ... orbiting the Sun&amp;quot;, I'm not sure how verbatim this definition is meant to be treated.&lt;/p&gt;&#xA;&lt;p&gt;What I mean by this is, well, for example, in Law, all definitions are, by default, unless previously stated, treated in an context of exactness. But although science is oftentimes based on specificity, it is not always, and the degree thereof in this context is thus not wholly clear to me.&lt;/p&gt;&#xA;" OwnerUserId="9611" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-12-26T23:43:00.263" Title="Are meteors and meteorites considered &quot;Small Solar System Bodies&quot;?" Tags="&lt;meteor&gt;&lt;terminology&gt;&lt;meteorite&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12281" PostTypeId="1" CreationDate="2015-10-26T19:19:23.590" Score="6" ViewCount="1984" Body="&lt;p&gt;When I look up in the night sky, I assume the brightest area, commonly known as &quot;The Milky Way&quot;, is the Galactic Center of our home galaxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/s1nzo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/s1nzo.jpg&quot; alt=&quot;The Milky Way&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then I stumbled upon this illustration, which seems to suggest what I'm seeing is (and by far) not the Galactic Center. In fact, from this picture it looks like the brightest area might be the closest spiral arm (on the opposite direction of the galactic center).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lttRT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lttRT.jpg&quot; alt=&quot;what can be seen in our night sky&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this illustration is not meant to be precise, so it's probably not the closest spiral arm either. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what am I really looking at when I stare at the brightest spot of the night sky? What percentage of it would be stars, planets, and nebulae? And what's the farthest object I can see in that direction with my unaided eyes?&lt;/p&gt;&#xA;" OwnerUserId="9612" LastEditorUserId="9186" LastEditDate="2015-10-26T22:21:03.060" LastActivityDate="2015-10-26T22:21:03.060" Title="What is in the brightest area of the night sky?" Tags="&lt;observational-astronomy&gt;&lt;milky-way&gt;&lt;galaxy-center&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12282" PostTypeId="2" ParentId="12281" CreationDate="2015-10-26T20:49:49.520" Score="4" Body="&lt;p&gt;All quoted text in this answer is from image captions in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Milky_Way&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article on the Milky Way&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;360-degree panorama view of the Milky Way (an assembled mosaic of&#xA;  photographs) by &lt;a href=&quot;http://en.wikipedia.org/wiki/European_Southern_Observatory&quot; rel=&quot;nofollow noreferrer&quot;&gt;ESO&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Milky_Way#/media/File:ESO_-_The_Milky_Way_panorama_(by).jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CJEDn.jpg&quot; alt=&quot;From ESO&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This magnificent 360-degree panoramic image, covering the entire&#xA;  southern and northern celestial sphere, reveals the cosmic landscape&#xA;  that surrounds our tiny blue planet. This gorgeous starscape serves as&#xA;  the first of three extremely high-resolution images featured in the&#xA;  &lt;a href=&quot;http://www.gigagalaxyzoom.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;GigaGalaxy Zoom project&lt;/a&gt;, launched by ESO within the framework of the&#xA;  International Year of Astronomy 2009 (IYA2009). The plane of our Milky&#xA;  Way Galaxy, which we see edge-on from our perspective on Earth, cuts a&#xA;  luminous swath across the image. The projection used in GigaGalaxy&#xA;  Zoom place the viewer in front of our Galaxy with the Galactic Plane&#xA;  running horizontally through the image — almost as if we were looking&#xA;  at the Milky Way from the outside. From this vantage point, the&#xA;  general components of our spiral galaxy come clearly into view,&#xA;  including its disc, marbled with both dark and glowing nebulae, which&#xA;  harbours bright, young stars, as well as the Galaxy’s central bulge&#xA;  and its satellite galaxies. As filming extended over several months,&#xA;  objects from the Solar System came and went through the star fields,&#xA;  with bright planets such as Venus and Jupiter. For copyright reasons,&#xA;  we cannot provide here the full 800-million-pixel original image,&#xA;  which can be requested from Serge Brunier. The high resolution image&#xA;  provided here contains 18 million pixels.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Here is a schematic map of our POV in the Milky Way galaxy.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Observed (normal lines) and extrapolated (dotted lines) structure of&#xA;  the spiral arms. The gray lines radiating from the Sun's position&#xA;  (upper center) list the three-letter abbreviations of the&#xA;  corresponding constellations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Milky_Way#/media/File:Milky_Way_Arms.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/n5Ebj.png&quot; alt=&quot;From Wikipedia&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A &quot;God's view&quot; map of Milky Way as seen from far Galactic North (in&#xA;  Coma Berenices). The star-like lines center in a yellow dot&#xA;  representing the position of Sun. The spokes of that &quot;star&quot; are marked&#xA;  with constellation abbreviations, &quot;Cas&quot; for &quot;Cassiopeia&quot;, etc. The&#xA;  spiral arms are colored differently in order to highlight what&#xA;  structure belongs to which arm. H II regions are marked as dots&#xA;  colored in the same color as their spiral arm. They come in three&#xA;  sizes, measured by the excitation parameter U: small - U &gt; 200 pc cm-2&#xA;  medium - 200 &gt; U &gt; 110 pc cm-2 large - 110 &gt; U &gt; 70 pc cm-2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It turns out we are &lt;strong&gt;&lt;em&gt;in&lt;/em&gt;&lt;/strong&gt; an arm -- the Orion-Cygnus Arm. The much brighter part of the Milky Way from our POV is in the direction of the galactic center, but the actual nucleus around the &lt;a href=&quot;http://en.wikipedia.org/wiki/Galactic_Center#Supermassive_black_hole&quot; rel=&quot;nofollow noreferrer&quot;&gt;supermassive black hole&lt;/a&gt; is obscured by dust. If it was visible, it would be quite bright. What we are seeing that is bright is mostly the pseudobulge, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Spiral_galaxy#Milky_Way&quot; rel=&quot;nofollow noreferrer&quot;&gt;galactic bar&lt;/a&gt; formation, in the middle of the galaxy. We are looking at the galactic bar almost end-on, so it resembles a sphere from our POV.&lt;/p&gt;&#xA;" OwnerUserId="9186" LastActivityDate="2015-10-26T20:49:49.520" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12283" PostTypeId="1" AcceptedAnswerId="16652" CreationDate="2015-10-26T23:05:01.370" Score="2" ViewCount="6137" Body="&lt;p&gt;Now, I was looking the sky when I saw the Moon showing up in a sky completely cloudy, like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oX5de.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oX5de.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a little different. The clouds were doing an eye shape. What does explain that phenomenon? Looks as if moonbeams drilled through the clouds! &lt;/p&gt;&#xA;" OwnerUserId="9615" LastEditorUserId="9186" LastEditDate="2015-10-27T13:29:35.273" LastActivityDate="2017-11-10T13:48:12.503" Title="Why does the moon appear on cloudy night sky?" Tags="&lt;the-moon&gt;&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12284" PostTypeId="2" ParentId="12252" CreationDate="2015-10-27T01:00:05.477" Score="0" Body="&lt;p&gt;The position of a planet is determined by 6 numbers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First there are the eccentricity and semi-major-axis.  These give the shape and size of the ellipse. By Kepler Law,  the sun is at one focus of this ellipse, &#xA;Then there is the plane in which the orbit is placed.  This can be conveniently determined by the angle between the planet's or it and the elliptical, and the position of the point where the orbit crosses the plane: the longitude of the ascending node.  Finally there is the position of the planet on this path.  The angular position the ginger to me hard  at some reference time(the mean anomaly at epoch) and the amount of time passing since the epoch. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-10-27T01:00:05.477" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12285" PostTypeId="2" ParentId="12252" CreationDate="2015-10-27T02:07:48.340" Score="1" Body="&lt;p&gt;The orbital inclination is constant -- does not change with time. Except over extremely long timescales due to interaction with other bodies, and that interaction would be considered non-Keplerian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, measure orbital inclination by observation, then expect it not to change over time.&lt;/p&gt;&#xA;" OwnerUserId="9186" LastActivityDate="2015-10-27T02:07:48.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12286" PostTypeId="1" AcceptedAnswerId="12287" CreationDate="2015-10-27T06:05:22.013" Score="1" ViewCount="1979" Body="&lt;p&gt;Sorry I forgot where this statements come from, but I also remember the reason behind it is due to young Jupiter moves inwards and destroys the original super earth in inner solar system, and the current terrestrial planets are formed after that, is that true?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="9186" LastEditDate="2015-10-27T22:20:06.230" LastActivityDate="2015-10-27T22:20:06.230" Title="Do the terrestrial planets form later than gas giants in our solar system?" Tags="&lt;solar-system&gt;&lt;gas-giants&gt;&lt;planetary-formation&gt;&lt;accretion-discs&gt;&lt;terrestrial-planets&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12287" PostTypeId="2" ParentId="12286" CreationDate="2015-10-27T09:41:55.363" Score="2" Body="&lt;p&gt;The current ideas are that both terrestrial planets and giant planets start their formation in a similar manner. Dust settles towards the mid-plane of a predominantly gaseous disk, starts to stick together and eventually small (km-sized) planetesimals are formed. This process may be quicker in the outer parts of the solar systems where the gas is colder and the condensed material (ices) is probably &quot;stickier&quot;. The planetesimals then interact with each other gravitationally and can grow by merger. The rate of growth is controlled by their spatial densities and their relative velocities, but is thought to occur quite quickly in both the inner and outer parts of the solar system ($\sim$ 1-3 million years, e.g. &lt;a href=&quot;http://www.pnas.org/content/108/48/19165.full.pdf&quot; rel=&quot;nofollow&quot;&gt;Righter &amp;amp; O'Brien 2011&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thereafter, the inner and outer parts of the solar system differ. The gas is cold enough in the outer parts of the solar system to accrete onto rocky cores and build giant planets on timescales of another few million years.&#xA;In the inner solar system the gas is too hot to be accreted and instread the next tens of millions of years are characterised by high velocity collisions between planetary embryos and planetesimals. This indeed may be sculpted and influenced by the early migration inwards of Jupiter to about 1.5 au, followed by migration outwards - the so-called &quot;Grand Tack model&quot; (&lt;a href=&quot;http://arxiv.org/abs/1409.6340&quot; rel=&quot;nofollow&quot;&gt;Raymond &amp;amp; Morbidelli 2014&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall it probably takes the inner solar system and terrestrial planets of order 100 million years to settle down into its final configuration. The collision that formed the moon may have been several tens of millions of years after the formation of the Sun and certainly long after the giant planets formed.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-10-27T09:41:55.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12288" PostTypeId="2" ParentId="12252" CreationDate="2015-10-27T10:18:25.247" Score="4" Body="&lt;p&gt;So, basically you're trying to merge &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_inclination&quot; rel=&quot;nofollow noreferrer&quot;&gt;orbital inclination&lt;/a&gt; to Kepler's laws.  The simplest way is to take the measured orbital inclination of the planet, which is constant, and apply the Pythagorean theorem to any given location and that gives you 3 dimensions of distance from the 2 dimensions defined in Kepler's laws.  That's probably what Kepler did.  (see picture from the Wikipedia link above)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QLNxU.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QLNxU.gif&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A much too simple explanation of orbital inclination &lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/O/orbital+inclination&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And a much more complex one, factoring in orbital movement &lt;a href=&quot;http://www.braeunig.us/space/orbmech.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kepler's laws only work in 2 dimensions because one object orbiting another is a 2 dimensional calculation and bringing a 3rd dimension into his law complicates it unnecessarily.   3 dimensions is essential for accurate planetary tracking, but it's unnecessarily in a 2 body gravitational system that takes into account distance, eccentricity and velocity.  Kepler was likely aware of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kepler, in fact, discovered orbital inclination by looking at Copernicus' observations, and that was something he was very proud of and it was a very important discovery.  Source &lt;a href=&quot;https://books.google.com/books?id=OgRAAQAAIAAJ&amp;amp;pg=PA383&amp;amp;lpg=PA383&amp;amp;dq=when%20was%20angle%20of%20inclination%20of%20the%20planets%20first%20observed&amp;amp;source=bl&amp;amp;ots=PWoP0sTkHb&amp;amp;sig=xFsoFQtZB1yIlWGNaXjSLKV3kUk&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0CD0Q6AEwBWoVChMIyIC27qziyAIVgTs-Ch0jqwC7#v=onepage&amp;amp;q=when%20was%20angle%20of%20inclination%20of%20the%20planets%20first%20observed&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.   Kepler deduced that orbital inclination explained away previously not understood and overly complicated orbital oscillations.   He didn't just undo Copernicus' circles, he worked things out in 3 dimensions and he might never have figured out his laws without understanding and measuring orbital inclination.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nice question.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;footnote, my answer is more history of science and mathematics, but I think your question, the way you asked it is more astronomy, so, I think the question should stay here, but the powers that be can decide otherwise if they like&lt;/em&gt;).&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-01-28T04:32:16.130" LastActivityDate="2016-01-28T04:32:16.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12289" PostTypeId="1" CreationDate="2015-10-27T13:13:10.327" Score="2" ViewCount="47" Body="&lt;p&gt;even if they did how will time pass in that ship. Also that ship would require a infinite amount of energy to fight the drag.but what would provide the drag since in space there is nothing to resist?&lt;/p&gt;&#xA;" OwnerUserId="9576" LastActivityDate="2015-10-27T13:13:10.327" Title="nasaworking on a way to travel faster than speed of light" Tags="&lt;space-time&gt;&lt;nasa&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="0" ClosedDate="2015-10-27T13:30:07.730" ContentLicense="CC BY-SA 3.0" />
  <row Id="12290" PostTypeId="1" CreationDate="2015-10-27T14:59:13.177" Score="3" ViewCount="70" Body="&lt;p&gt;I am doing photometry on a number of objects in a number of different galaxies. I need the extinction coefficients for these galaxies for various filters passes. Is there some database that displays extinction values for different galaxies at different filter passes?&lt;/p&gt;&#xA;" OwnerUserId="9624" LastEditorUserId="9186" LastEditDate="2015-10-27T16:08:57.343" LastActivityDate="2015-10-27T16:08:57.343" Title="How does one find the extinction coefficient of a particular galaxy?" Tags="&lt;galaxy&gt;&lt;deep-sky-observing&gt;&lt;resource-request&gt;&lt;extinction&gt;&lt;photometry&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2016-10-04T17:42:59.583" ContentLicense="CC BY-SA 3.0" />
  <row Id="12291" PostTypeId="2" ParentId="12283" CreationDate="2015-10-27T20:43:01.107" Score="3" Body="&lt;p&gt;Clouds are obviously not completely opaque, and what you are seeing is a break in the clouds. It is of course possible for the clouds to cover the moon, however, often the moon can shine brightly through the clouds and illuminate the clouds in some nice way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This video might help in a visual way: &lt;a href=&quot;https://www.youtube.com/watch?v=aLd_yhIWE48&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=aLd_yhIWE48&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2091" LastEditorUserId="957" LastEditDate="2017-11-10T13:48:12.503" LastActivityDate="2017-11-10T13:48:12.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12293" PostTypeId="1" AcceptedAnswerId="12294" CreationDate="2015-10-27T21:56:04.003" Score="8" ViewCount="5042" Body="&lt;p&gt;In my textbook, the H–R diagram’s y-axis is $\log(L/L_{\odot})$ and values are higher as we go higher on the scale, but the x-axis is $\log(T_\text{eff})$ and gets smaller as we proceed to the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This confuses me. Why not let the x-axis go from lower values to higher values? This will also help visualising the “linear” relation between the luminosity and the temperature of a star in the main sequence.&lt;/p&gt;&#xA;" OwnerUserId="7554" LastEditorUserId="9630" LastEditDate="2015-10-28T21:44:52.717" LastActivityDate="2016-03-27T09:40:41.857" Title="Why does the Hertzsprung–Russell diagram's x-axis go from large temperatures to lower?" Tags="&lt;history&gt;&lt;luminosity&gt;&lt;classification&gt;&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12294" PostTypeId="2" ParentId="12293" CreationDate="2015-10-27T23:17:18.683" Score="7" Body="&lt;p&gt;Originally, what was plotted was &lt;em&gt;luminosity against colour&lt;/em&gt;, and by colour I mean the wavelength of the peak intensity. Short wavelengths on the left, and long on the right, as you would expect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.stars-planets-galaxies-qa.com/stars&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eHK0i.png&quot; alt=&quot;stars-planets-galaxies-qa.com&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since stars emit (nearly) black body radiation, there is a close relationship between colour and temperature. I suspect that the reason that the x axis isn't inverted when temperature is used is just inertia:- We have always plotted out H-Z diagrams like this and we will keep doing so.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="9186" LastEditDate="2015-10-28T09:20:51.467" LastActivityDate="2015-10-28T09:20:51.467" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12295" PostTypeId="2" ParentId="12280" CreationDate="2015-10-27T23:29:34.593" Score="1" Body="&lt;p&gt;Your question is really well answered by &lt;a href=&quot;http://www.britannica.com/topic/small-body&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; at the Encyclopedia Britannica website, written by &lt;a href=&quot;http://www.psi.edu/about/staff/eft/eft.html&quot; rel=&quot;nofollow&quot;&gt;Prof. Ed Tedesco&lt;/a&gt; of the Planetary Science Institute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two quotes that resonate well are...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The small bodies populate the solar system in vast numbers and include ... the fragments of these bodies—commonly called meteoroids—over a continuum of sizes down to microscopic grains known as interplanetary dust particles or micrometeoroids.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The term small body is sometimes restricted to objects that can be observed telescopically while still outside Earth’s atmosphere. In practice, this results in a lower limit of a few metres on the diameter of a small body.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in general you can assume that anything other than the Sun, planets, minor planets, and their moons is a small solar system body. But in practice it makes sense to limit the minimum size for a SSSB, and the &quot;observable telescopically&quot; caveat provides that limit so that we don't have to include single molecules or one grain of sand in the list.&lt;/p&gt;&#xA;" OwnerUserId="9575" LastActivityDate="2015-10-27T23:29:34.593" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12296" PostTypeId="1" AcceptedAnswerId="12303" CreationDate="2015-10-28T04:14:59.293" Score="1" ViewCount="1901" Body="&lt;p&gt;We know that black hole possesses immense power and can destroy anything coming in front of it's path. If such a black hole appears near our solar system, are we going to survive? Is there any way not to fall in a black hole? &lt;/p&gt;&#xA;" OwnerUserId="9213" LastActivityDate="2015-11-04T12:44:13.013" Title="What will happen if a black hole appears near our solar system?" Tags="&lt;solar-system&gt;&lt;black-hole&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12297" PostTypeId="1" CreationDate="2015-10-28T04:49:31.770" Score="2" ViewCount="255" Body="&lt;p&gt;I would like to use a system of co-ordinates (x,y,z,t) based on a specific well known point. My original choice (the center of the universe in the big bang theory, Bigbang0), was wrong as the universe did not explode everything moved apart from everything else and is still doing so.&lt;/p&gt;&#xA;&lt;p&gt;So I would like a reference position useful to an observer within at least 100,000 light years, and valid for 250,000 years. I will call it Bigref0 to remind me of my error.&lt;/p&gt;&#xA;&lt;p&gt;From Bigref0 I would have a position - x y z.&lt;/p&gt;&#xA;&lt;p&gt;The time value I would like to use as an origin is based on the following criteria.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The centers of the Sun and Earth are most closely aligned to the Bigref0.&lt;/li&gt;&#xA;&lt;li&gt;The time point t0 should be the closest point within 256 tera (2 to the power 48) seconds or about 8 million years which provides a reasonable accuracy. I doubt we can do better than 1 part in 25,000 or +/- 2 to the power 15.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;The questions are&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;When was/will be t0 ?&lt;/li&gt;&#xA;&lt;li&gt;What is the distance (r0) to Bigref0 from the center of the Sun at time t0?&lt;/li&gt;&#xA;&lt;li&gt;The x y z position of the Earth relative to the center of the Sun at time t0 ?&lt;/li&gt;&#xA;&lt;li&gt;The accuracy/stability of the 5 values (t0,r0,x,y,z) ?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;em&gt;Example&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;The point in space time where the line through Earth and Sun is the closest to a line running from the Our Galactic center to the center of the local cluster.&lt;/p&gt;&#xA;" OwnerUserId="9627" LastEditorUserId="9627" LastEditDate="2021-03-22T13:39:09.090" LastActivityDate="2021-03-22T13:39:09.090" Title="A universal reference point" Tags="&lt;solar-system&gt;&lt;universe&gt;&lt;big-bang-theory&gt;&lt;space-time&gt;&lt;observable-universe&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2015-10-28T21:44:15.403" ContentLicense="CC BY-SA 4.0" />
  <row Id="12298" PostTypeId="2" ParentId="12296" CreationDate="2015-10-28T05:06:23.197" Score="7" Body="&lt;p&gt;The way not to fall into a black hole is essentially the same as the way not to fall into the Sun.  It's no more dangerous than any other body with the same mass, which is typically a few times the Sun's mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The danger of a passing black hole (or other star) is not that it could swallow us but that it could perturb our orbit, with unpredictable effects on climate; as well as stir up the comets of the (hypothetical) Oort Cloud, sending some of them toward the inner planets, with the possibility of an impact that could ruin your whole day.&lt;/p&gt;&#xA;" OwnerUserId="7959" LastActivityDate="2015-10-28T05:06:23.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12299" PostTypeId="1" CreationDate="2015-10-28T05:07:46.807" Score="2" ViewCount="100" Body="&lt;p&gt;A little over half a year ago, I think it was mid March 2015, I glanced out of my window at about 4 am, to see what appeared to be a very bright total lunar eclipse, as in, the moon was full and bright orange-red. As I watched, it disappeared before my eyes, looking as if someone slid a paper down in front of it. The moon was not full that night (and by extension that means that there couldn't have been an eclipse that night) and the next few nights, the moon was on the other side of the sky at 4 am.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The room I was in was dark, so it couldn't have been a reflection on the window of something inside the house. It might have been a reflection on the window of something outside the house, but that's not very likely, because it was very bright and opaque (something would have to be really bright to look so bright and opaque when reflected off a window, and I can't think of anything that would have been so bright. Plus, I probably would have noticed something something so bright), it didn't move when I moved my head around, and it looked exactly like the moon – it had the same dark and light spots. I would have taken a picture of it, it would be a lot easier to understand what I mean if you would see it, but it disappeared really fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might have been a reflection on the clouds. I don't recall if there were clouds in the sky, I wasn't paying that much attention to the sky around it, because while it was happening I just thought it was an eclipse. But if it was a reflection on the clouds, there's still the question of what was being reflected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked someone who knows a lot about a bunch of topics, but is not good at explaining what he means, what it could have been, and he said something about solar flares, I didn't really get what he was saying. I would ask him again, but I don't know him personally and I don't know how to contact him, and in any case, I don't think he would explain it any better the second time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I searched everywhere I could possibly think of (on the web, on Q&amp;amp;A sites, on SE) for a cause for this, but couldn't find anything like it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm asking is this: Could it have been a reflection on the clouds, and if so, what could it have been of? Could it have been caused by a solar flare or something like it, and if so, how? Is there an astronomical event that occurred around that time that could have caused something like that, and if so, what and how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question was put on hold as unclear what you're asking, so if anyone has any suggestions that can make the question more clear, I would love to hear it. I wish I could be more precise about when it happened, or what exactly it looked like, but I didn't write it down or note it in any way at the time, so this is pretty much all I remember.&lt;/p&gt;&#xA;" OwnerUserId="9346" LastEditorUserId="9346" LastEditDate="2015-11-01T10:05:18.920" LastActivityDate="2015-11-01T10:05:18.920" Title="Illusion of a lunar eclipse" Tags="&lt;the-moon&gt;&lt;amateur-observing&gt;&lt;solar-flare&gt;&lt;lunar-eclipse&gt;" AnswerCount="0" CommentCount="10" ClosedDate="2015-10-30T13:32:00.810" ContentLicense="CC BY-SA 3.0" />
  <row Id="12300" PostTypeId="1" AcceptedAnswerId="12304" CreationDate="2015-10-28T06:19:38.673" Score="3" ViewCount="260" Body="&lt;p&gt;Often I see books which state that Jupiter has a rocky core, if it is true, does it originate from super earth with excessive collection of hydrogen?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="9485" LastEditDate="2015-10-28T09:20:36.060" LastActivityDate="2015-10-28T10:09:17.533" Title="Is Jupiter just a super earth with hydrogen atmosphere?" Tags="&lt;jupiter&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12301" PostTypeId="2" ParentId="12300" CreationDate="2015-10-28T08:52:11.803" Score="2" Body="&lt;p&gt;Jupiter &lt;em&gt;probably&lt;/em&gt; originated with a dense core, and then accreted a gaseous envelope. To call the core &quot;rocky&quot; could be misnomer, since the planet formed much further out from the Sun where it may have been possible to form ices. The exact location of Jupiter's birth and the composition of its core are not known, but the composition was unlikely to have been the same as Earth. It is also probable (again, theory not fact), that Jupiter's core may have reached multiple Earth masses before accreting the gaseous envelope.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-10-28T08:52:11.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12302" PostTypeId="2" ParentId="12297" CreationDate="2015-10-28T09:18:44.340" Score="1" Body="&lt;p&gt;You are at the centre of the Big Bang. It happened here and here and here. In fact everywhere in the universe was the centre of the Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of it this way - if there was somewhere that exists now that wasn't at the centre of the Big Bang then that point existed outside space (a nonsense) or the Big Bang was not the beginning of space (and so not the Big Bang).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the strongest evidence in support of the theory is the cosmic microwave background which comes from all directions precisely because the Big Bang took place everywhere in the Universe.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2015-10-28T09:18:44.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12303" PostTypeId="2" ParentId="12296" CreationDate="2015-10-28T09:21:41.207" Score="5" Body="&lt;p&gt;When you say &quot;destroy anything in it's path&quot;, that's true, but it's also true for a star, even a red-dwarf or white-dwarf star would effectively eat or destroy almost anything in their path.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also worth mentioning that your scenario is very unlikely.  According to this &lt;a href=&quot;http://hubblesite.org/explore_astronomy/black_holes/encyc_mod3_q7.html&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;, stars outnumber black holes 1,000 to 1 in the Milky way and most of those likely near the center of the Milky way.   There are probably a few stellar black holes within 75 or maybe 100 light years, based on probability and since they would be hard to detect (see Rob jeffries' comment).   But even so, one passing near enough to affect the Earth is very rare.  Story about a star passing within 1 light year of earth &lt;a href=&quot;http://www.universetoday.com/119038/a-star-passed-through-the-solar-system-just-70000-years-ago/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, onto your question: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If such a black hole appears near our solar system, are we going to&#xA;  survive? Is there any way not to fall in a black hole?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It depends how near and how big.   A &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_black_hole&quot; rel=&quot;nofollow&quot;&gt;stellar mass&lt;/a&gt; black hole, with a mass range of about 3.8 solar masses to maybe 16 solar masses would need to get quite close to cause real problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our Solar system has a very sparse &lt;a href=&quot;https://en.wikipedia.org/wiki/Oort_cloud&quot; rel=&quot;nofollow&quot;&gt;Oort Cloud&lt;/a&gt;, extending perhaps almost 2 light-years and a much more densely packed, but still quite diffuse, &lt;a href=&quot;https://en.wikipedia.org/wiki/Kuiper_belt&quot; rel=&quot;nofollow&quot;&gt;Kuiper-belt&lt;/a&gt;, mostly about 30-50 AU (less than a thousandth of a light year).   If we assume that a black hole, as much as 16 times the mass of our sun has similar orbital debris at larger orbits, an oort cloud around a 16 stellar mass black hole it could extend several light-years though still likely very diffuse, as well as perhaps a more densely packed asteroid belt and maybe even a few planets in the I dunno, 5-300 AU range or something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect of a close pass gets pretty speculative and depends on the amount of objects in distant orbit around the black hole, but a pass as close as 1 or 2 light-years could increase the chance of a good sized comet or other objects striking Earth.   Not a guarantee at that distance, but an increased chance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you get a pass-by of a few hundred AU to maybe 1,000 AU (about 1/50th to 1/100th of a light-year), you begin to get measurable Kuiper belt stirring up.  If the black hole has it's own Kuiper belt equivalent, this could create a significant increase in asteroids and meteors flying around the solar system.   A significant meteor strike on earth at this point becomes a distinct possibility, perhaps even likely and it could be a life killer and ocean evaporator.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With proper technology, it might be possible to deflect any asteroids on an impact trajectory.   We'd probably observe multiple impacts on other planets too and at this distance, our solar system might even pick up a planet from the black hole if there was one in distant orbit.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At about 100-200 AU, the black hole has a chance of pulling some of the outer planets out of our solar system.  It could also visibly alter Saturn's or maybe Jupiter's orbits and stir things up in our solar-system pretty good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the most likely effect of a very near pass of either a star or black hole is a stirring up of more distant orbiting objects and a significant increase in the chance of planet killing or at least, dinosaur killing impact, but it needs to be very close.  The article I linked above says that the star that passed within 0.8 light years (about 50,000 AU), some 70,000 years ago didn't measurably increase comet impacts.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To measurably change the Earth's orbit, maybe 30-50 AU.  About the distance of Uranus or Pluto.   This wouldn't pull the Earth away from the Sun, but it could change the Earth's orbit enough to alter the seasons and change climate on Earth.  An elongation of our orbit could make the seasonal changes bigger and change the length of a year.  The gas giants would all be moved.   If the Earth survives any bombardment from a pass that close, it might quickly enter a new ice age, as one possible outcome, but we might be able to fix that by burning lots of fossil fuels.  :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At somewhere around Maybe 12-20 AU, the Earth might get moved too much for us to adapt, perhaps outside of the Goldilocks Zone.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At somewhere around 5-10 AU (estimates are pretty rough), but this is where things get really interesting.   We begin to see a measurable increase in tides and at this point and we could lose the moon.   The earth could also get pulled outside the solar system, thrown into the sun or captured by and end up in an orbit around the black hole.   We might begin to see a small tidal bulge on the sun from the black hole and a corresponding small increase in solar output, but that might require the black hole to be closer. Not sure.  If the black hole gets close enough to the sun it could stretch the sun somewhat oblong and start to draw some solar gases towards it, but that might be pretty close, maybe 1 or 2 AU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At maybe 3-6 AU, depending on the relative velocities, the sun might get captured into an fairly close orbit around the black hole, turning our solar-system into a binary and the planets could end up pretty much anywhere in that scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 1 or maybe 2 AU from Earth, we could see some cracking of the Earth's crust and a big increase in volcanoes and earthquakes.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even at 1 AU, the black hole still wouldn't be visible unless it's acquired enough matter to form a bright accretion disk, in which case it might be very bright, but still visibly very small like a super bright star in the sky.   Without an accretion disk the gravitational lensing should be visible by telescope but not to the naked eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much closer than that and it hardly matters.   The Earth's would be largely be re-surfaced with magma and the oceans would probably boil.   All that said, the Earth still isn't being eaten.  To eat the Earth a black hole would need to pass somewhere around maybe 5 or 10 million miles.  Super-close for another stellar object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I welcome correction if any of my math is broken, but that's my roughly calculated estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.askamathematician.com/2012/02/q-what-would-happen-if-a-black-hole-passed-through-our-solar-system-2/&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; with some of the same conclusions.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit - due to comment above&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I mean a supermassive black hole, like the one we have in the center&#xA;  of Milkyway&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My long answer above is basically gravitational effects when the black hole is in the general vicinity of the solar system.   That kind of very close pass happens very rarely, but the odds of a black hole hitting earth is very slim.  The odds of a black hole passing close enough to our solar-system and stirring things up in an undesirable way, also unlikely but more likely than a black hole actually hitting earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most likely effect is some gravitational swirling (if you will) due to the massive object passing kinda close to the solar system, and as a result, possible impacts, possible orbital changes - stuff that wouldn't be fun, even if it passes outside the orbit of Pluto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a super-massive black hole, it's basically the same answers but the distance can be much further out.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Sagittarius_A*&quot; rel=&quot;nofollow&quot;&gt;Sagittarius A&lt;/a&gt; is about 4 million solar masses, compared to a stellar mass black hole which is about 4 to 16 solar masses.   That's 250,000 to a million times bigger, so equivalent gravitational occurs at the square root of that, about 500 to 1,000 times further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tidal effects are actually much smaller.   A 4 to 16 stellar mass black hole could fly by at roughly Pluto's orbital distance (30-50 AU) and as it passes, it could pull Jupiter into a very different orbit, perhaps even away from the sun, but a super-massive black hole fling past some 500 times the distance of pluto (about 1/3 to 1/2 light year) would exert the same gravitational tug, but that tug would be much the same across the entire solar system, so the entire solar system could be pulled effectively into orbit around the super-massive black hole without changing shape too much.  It still be visibly very tiny though the gravitational lensing might be visible to the naked eye at that point, but not much bigger than a star.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real danger with a super-massive black hole passing within a few light-years of our solar system is the stuff it carries with it.   Super-massive black holes are orbited by stars and presumably all the stuff that generally orbits stars, like planets, moons, asteroids, comets.   All that stuff would approach our solar-system long before the black hole came close and that's the mostly likely answer.  Our solar-system would likely be pelted by orbital debris when the super-massive black hole was still a few light-years away.   I don't think I should try to guess how many but a super-massive black hole would carry with it orbiting debris for several light years.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My quick and dirty answer to the super-massive question.   You mgiht think a black hole of that size would strip the atmosphere off the earth and stuff like that, but with a super-massive black hole, the tidal forces are much smaller at the event horizon.  The atmospheric striping and earthquakes and other effects don't happen until it's enormously close, at which point the earth would be orbiting around it quite fast, perhaps fast enough that the night sky might visibly appear to spin to the naked eye.  We could get pretty close to it before life on earth became unlivable, provided we're not bombarded by orbiting debris first.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There could be other effects.  Gamma rays from any matter falling into the black hole or perhaps strong magnetic effects.  More precise predictions gets a bit difficult for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's my layman's attempt at an answer anyway.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-11-04T12:44:13.013" LastActivityDate="2015-11-04T12:44:13.013" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12304" PostTypeId="2" ParentId="12300" CreationDate="2015-10-28T10:09:17.533" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does it originate from super earth with excessive collection of&#xA;  hydrogen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Basically no, and this is why:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eq4si.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eq4si.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;em&gt;Source from the water article linked below.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter likely formed outside the &lt;a href=&quot;https://en.wikipedia.org/wiki/Frost_line_%28astrophysics%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;frost-line&lt;/a&gt;, so it never had a rocky or even a molten magma surface.  It formed with much too high a percentage of ices, which, under pressure, became hot gas and perhaps, at one time, a very deep ocean around the whole planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Rob Jeffries said, the formation of Jupiter isn't well understood, so there's some uncertainty there, but I don't see how a planet that forms with as much water, CO2, NH3 and other gases as Jupiter did could ever form what we would consider a solid surface.  It's large size and heat of formation probably kept it highly gaseous during it's entire formation.   Perhaps it was once a water world of sorts, but my guess is it always resembled a gas giant once it was recognizable as a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth is about &lt;a href=&quot;http://www.universetoday.com/65588/what-percent-of-earth-is-water/&quot; rel=&quot;nofollow noreferrer&quot;&gt;0.02% water&lt;/a&gt; (some websites say 0.05% but whichever percentage, the Earth is still almost entirely Rocky Mantle &amp;amp; Metallic core.   I don't think a planet that forms outside the frost line would ever become Earth like or a super-earth.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside the frost line, A super earth could in time become a gas giant, but outside, I don't think so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, Jupiter might well have an earth like ratio of elements at it's core, but I don't believe that means it was ever a super-earth.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-10-28T10:09:17.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12305" PostTypeId="2" ParentId="12293" CreationDate="2015-10-28T10:27:55.783" Score="7" Body="&lt;p&gt;The original Hertzsprung-Russell diagrams constructed by Henry Russell and Eijnar Hertzsprung consisted of absolute magnitude on the y-axis and a spectral type or an indicator of spectral type on the x-axis. Below you can see an original HR diagram produced by Russell in 1913.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CzNH1.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CzNH1.gif&quot; alt=&quot;Original HR diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the diagrams were constructed, it was &lt;em&gt;not at all clear&lt;/em&gt; what the sequence of spectral types or spectral type indicators actually meant. It turned out of course that the sequence (in modern day parlance O,B,A,F,G,K,M) actually corresponds to &lt;em&gt;decreasing&lt;/em&gt; temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomers have simply stuck with this convention to the present day, there is no particular reason for that. Most HR diagrams are now plotted with temperature (decreasing) along the x-axis, although that is not what the original HR diagram was. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-03-27T09:40:41.857" LastActivityDate="2016-03-27T09:40:41.857" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12306" PostTypeId="1" CreationDate="2015-10-28T13:05:05.430" Score="3" ViewCount="75" Body="&lt;p&gt;If I understand it correctly, boulders on the Moon are only found near fresh craters, because micrometeorites erode them over time. Asteroids are believed to have formed sometimes even earlier than the Moon, but some images show asteroids covered by boulders. And NASA is planning the ARM mission to go pick up a boulder from an asteroid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are boulders more frequent on asteroids than on the Moon, and if so, by what kind of mechanism? For example, it is not primarily micrometeorites, but temperature changes that erodes them on the Moon. Or don't microgravity objects attract as many micrometeorites as the Moon does?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intact boulders on asteroid 25143 Itokawa:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/3bOkY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3bOkY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eroded boulders on the Moon:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/x9XPq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x9XPq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-10-30T22:32:56.273" Title="Do boulders erode differently on asteroids than on the Moon?" Tags="&lt;the-moon&gt;&lt;asteroids&gt;&lt;surface&gt;&lt;meteorite&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12308" PostTypeId="2" ParentId="12296" CreationDate="2015-10-28T15:25:03.803" Score="0" Body="&lt;p&gt;Massive and thus stable black holes are rare and as dangerous and predictable as any equally massive object. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small black hole does not last very long (uS) see &lt;a href=&quot;https://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Hawking_radiation&lt;/a&gt;, 1 second for a 100 ton black hole. 1us for a 1Kg black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 1 million ton (1e9 kg) black hole should last about 32 thousand million years. &#xA;Its event horizon (see &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/astro/blkhol.html&quot; rel=&quot;nofollow&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/hbase/astro/blkhol.html&lt;/a&gt;) is about 100 million times smaller than an atomic radius.&#xA;At 1 meter it exerts an attraction of less that 100 th of a g.&#xA;It is the equivalent of a cubic kilometer of water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing stops a moving black hole.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is attracted to the earth from outer space it would pass through and leave within 1200 seconds perhaps taking a few grams of earth with it.&lt;/p&gt;&#xA;" OwnerUserId="9627" LastActivityDate="2015-10-28T15:25:03.803" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12309" PostTypeId="2" ParentId="12272" CreationDate="2015-10-28T17:10:53.717" Score="1" Body="&lt;p&gt;If the question is &quot;if I throw two planets to orbit a star at random direction, would they form an orbital resonance?&quot; -- then in general, no. A resonance is an integral ratio (1/1, 2/1, 3/5, etc.) between the periods of motion of objects -- i.e., the ratio of their periods forms a rational number. Formally speaking the odds of getting a integral ratio (let alone a strong, low-order ratio, since those are the dynamically interesting ones) if you set the system up &quot;randomly&quot; should be infinitesimal, because irrational numbers are (infinitely) more abundant that rationals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the orbits of one or both of the planets can &lt;em&gt;change&lt;/em&gt; over time, then the ratio between their periods changes, and they &lt;em&gt;can&lt;/em&gt; end up in a resonance. (Which is maybe answering the title question.) How often this happens depends on whether the planets happen to start near a strong resonance, and on how rapidly the orbits change. (If the orbit of a planet changes slowly, then it won't encounter new resonances very often; on the other hand, rapid orbital change can overwhelm the effect of weak resonances, so that the planet passes through the resonance without being caught.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, it's thought that Neptune and Pluto were originally not in resonance; but the gradual outward migrations of Neptune (due to various gravitational encounters between planetesimals and the giant planets) changed its orbital period and meant that eventually it reached 2/3 resonance with Pluto, and Pluto was &quot;captured&quot; by the resonance, after which it stayed in resonance with Neptune.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vast majority of objects in the Solar System are &lt;em&gt;not&lt;/em&gt; in resonance with anything else, which is perhaps another way of answering your question. (I.e., in practice it doesn't happen very often.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2015-10-28T17:10:53.717" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12310" PostTypeId="1" AcceptedAnswerId="12311" CreationDate="2015-10-28T19:39:09.033" Score="10" ViewCount="1058" Body="&lt;p&gt;What is the current best approximation for the path the Sun takes around the center of the Milky Way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found some information on the approximate position of the Milky Way's center, the speed of the Sun's orbit, and the inclination/declination of the Sun's orbit. I have also seen models suggesting the Solar System oscillates up and down as it revolves around the galactic center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering if there is any agreed upon approximation (a 3d plot of position vs time) that includes all of these factors.&lt;/p&gt;&#xA;" OwnerUserId="9635" LastEditorUserId="9186" LastEditDate="2015-10-28T21:15:56.187" LastActivityDate="2016-09-15T13:41:38.193" Title="Best approximation for Sun's trajectory around galactic center?" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;the-sun&gt;&lt;milky-way&gt;&lt;galaxy-center&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12311" PostTypeId="2" ParentId="12310" CreationDate="2015-10-28T22:10:20.473" Score="4" Body="&lt;p&gt;The Sun orbits in the Galactic potential. The motion is complex; it takes about 230 million years to make a circuit with an orbital speed of around 220 km/s, but at the same time it oscillates up and down with respect to the Galactic plane every $\sim 70$ million years and also wobbles in and out every $\sim 150$ million years (this is called epicyclic motion). The spatial amplitudes of these oscillations are around 100 pc vertically and 300 pc in the radial direction inwards and outwards around an average orbital radius (I am unable to locate a precise figure for the latter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is established by measuring how the Sun moves with respect to the average motion of the stars in the solar vicinity - the so-called local standard of rest. Estimates of the Sun's motion with respect to the LSR vary a little. According to &lt;a href=&quot;http://arxiv.org/abs/astro-ph/9710077&quot; rel=&quot;nofollow&quot;&gt;Dehnen &amp;amp; Binney (1998)&lt;/a&gt; the Sun &lt;em&gt;currently&lt;/em&gt; moves at 10 km/s inwards, at about 5 km/s faster than the average star tangentially to the Galactic centre and at about 7 km/s upwards out of the Galactic plane. Though as I mentioned previously, these motions are of an oscillatory nature around the mean value. The Sun is currently about 8.5 kpc from the Galactic centre and about 20 pc &quot;north&quot; of the Galactic plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I leave it as an exercise for the reader to visualise this orbit. From a side-view it would look like a pseudo-sinusoidal oscillation. From above the disk it would look like a slightly eccentric ellipse with a very large precession.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the above assumes that the Sun moves in a smooth Galactic potential. In fact the presence of spiral arms, a Galactic bar and the clumpy presence of giant molecular clouds can and will perturb the Sun's Galactic orbit around the regular behaviour described above, so predicting the Galactic location of the Sun in a billion years or so is not necessarily going to be very accurate.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-09-15T13:41:38.193" LastActivityDate="2016-09-15T13:41:38.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12312" PostTypeId="1" AcceptedAnswerId="12313" CreationDate="2015-10-28T23:32:19.770" Score="2" ViewCount="137" Body="&lt;p&gt;Given an equivalent width measurement $W$ of a spectral line of element $X$ and the effective temperature $T_{eff}$ of a star, how can you determine the atomic number density of $X$ in that star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://ac.els-cdn.com/0022407365900555/1-s2.0-0022407365900555-main.pdf?_tid=7ee852e4-7dcf-11e5-842f-00000aacb361&amp;amp;acdnat=1446076764_280f65bab859d5a489c834444d50b957&quot; rel=&quot;nofollow&quot;&gt;Warner 1965&lt;/a&gt;, for the 6707.8 Lithium line we can apply the approximation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log{\frac{N^*(X)}{N^{\odot}(X)}} = \log{\frac{W^*(X)}{W^{\odot}(X)}} - 4.93 (\gamma -1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\gamma = \frac{T^{\odot}}{T^*}$ and $\log{W^{\odot}(X)}$ must be measured. The same paper says that $\log{W^{\odot}}(Li)=-6.15$ (my goal is to solve for the abundance of Lithium given a measured W of the 6707.8 $\overset{\circ}A$ line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under what regime is this approximation valid? Also, I am unable to solve this equation because I don't know what $N^{\odot}(Li)$ is. How can I determine a reputable value for $N^{\odot}(Li)$?&lt;/p&gt;&#xA;" OwnerDisplayName="user5341" LastEditorDisplayName="user5341" LastEditDate="2015-10-28T23:57:39.673" LastActivityDate="2015-10-28T23:57:39.673" Title="How to determine atomic number density of an element in a star based on equivalent width measurements" Tags="&lt;star&gt;&lt;spectroscopy&gt;&lt;elemental-abundances&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12313" PostTypeId="2" ParentId="12312" CreationDate="2015-10-28T23:48:13.717" Score="2" Body="&lt;p&gt;The measurement of a chemical abundance is not a question of using a simple equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest it gets is using a &quot;curve of growth&quot;, which relates equivalent width to abundance and assumes you already know the temperature of the star and its surface gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Li 6708A line, the relevant tables, that can be interpolated, are found in Soderblom et al. (1993). &lt;a href=&quot;http://adsabs.harvard.edu/abs/1993AJ....106.1059SS&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/1993AJ....106.1059SS&lt;/a&gt; . These are LTE curves of growth based on Atlas atmospheric models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, the equivalent width of this line in the Sun is 3 mA. The formula you quote looks like some sort of linear approximation, but this resonance line rapidly enters the saturated (non-linear) portion of the curve of growth for equivalent widths of 50 mA or more. As the equation is couched in terms of ratios, then the units you use are entirely up to you.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-10-28T23:48:13.717" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12314" PostTypeId="1" CreationDate="2015-10-29T01:44:08.577" Score="4" ViewCount="202" Body="&lt;p&gt;We measure a star's relative velocity towards or away from us via its Doppler-shifted spectrum. This is also how we measure the distance of very distant stars: measuring the shifts in the spectrum tells us its relative velocity, which we interpret as a distance via Hubble's Law. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we knew the star's relative velocity and distance exactly. It seems to me that if the star were moving away from us slightly less than would be expected from expansion alone, it would be because it was moving towards us more than the matter surrounding it: it would have some additional velocity component towards us. If we were to try to infer that star's distance from expansion alone, we would measure its relative velocity via the Doppler shift and would conclude that it was closer than it actually is. Conversely, if it were moving away from us faster than it should based on expansion alone, we would estimate its distance further than it actually is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to disentangle this degeneracy? &lt;/p&gt;&#xA;" OwnerUserId="7965" LastActivityDate="2015-10-29T07:34:29.037" Title="How to disentangle a very distant star's relative velocity vs. redshift distance" Tags="&lt;cosmology&gt;&lt;expansion&gt;&lt;redshift&gt;&lt;spectra&gt;&lt;doppler-effect&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12315" PostTypeId="2" ParentId="12314" CreationDate="2015-10-29T05:06:21.807" Score="2" Body="&lt;p&gt;We measure the aggregated red-shift of galaxies not single stars, which averages out the motion of the stars within the galaxy. This leaves us with the proper motion of the galaxy to deal with, which can be estimated from the red-shifts of other galaxies in the same group/cluster/super-cluster and their distribution. But this is all moot at high red-shift since proper motions become small compared to the Hubble flow.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-10-29T05:06:21.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12316" PostTypeId="1" AcceptedAnswerId="12323" CreationDate="2015-10-29T05:50:32.503" Score="2" ViewCount="1349" Body="&lt;p&gt;Is there meaningful reason to distinguish 90 and 270 degree rotational axis?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="53" LastEditDate="2015-10-29T20:30:01.013" LastActivityDate="2015-10-29T21:42:35.800" Title="Is there any difference between rotation axis 90 degrees and 270 degrees?" Tags="&lt;rotation&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2015-11-12T12:39:00.793" ContentLicense="CC BY-SA 3.0" />
  <row Id="12317" PostTypeId="2" ParentId="12314" CreationDate="2015-10-29T07:34:29.037" Score="5" Body="&lt;p&gt;Conrad is almost right. It is true generally that if a Galaxy is close enough to take spectra of individual stars (e.g. luminous supergiants) then it is not far enough away to be regarded as part of the &quot;Hubble flow&quot; and so applying Hubble's law to this star, or its host galaxy, would not yield a reliable distance in any case, but would reflect the &quot;peculiar motion&quot; of that galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put some numbers on this. Galaxy peculiar motions tend to be a few 100 km/s, as do the individual velocities of stars with respect to their galaxies. Taking a Hubble constant of 70 km/s per Mpc, we see that we need to be at distances of 15 Mpc before Hubble recession velocities ($v = H_0 d$) become large compared with peculiar motions. At these distances we cannot observe individual stars - they are too faint and unresolved from the bulk of the Galactic light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;exceptions&lt;/strong&gt; are supernovae. The redshifts of individual supernovae, that briefly outshine their galaxies, can be measured right across the universe. Here you are correct that the measured redshift is a combination of cosmological redshift due to the expansion of the universe &lt;em&gt;and&lt;/em&gt; a velocity of the star relative to the Hubble flow at that distance. There is no way to distinguish between these two unless velocity measurements could be obtained for other objects in the same galaxy. Given the rarity of supernovae, we might wait a long time for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But does it matter? Even if we look at a &quot;low redshift&quot; supernova at $z=0.1$, its Hubble recession velocity is 30,000 km/s and far in excess of any peculiar velocity contribution at the level of $\sim 1$%.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-10-29T07:34:29.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12318" PostTypeId="1" CreationDate="2015-10-29T10:54:55.047" Score="4" ViewCount="307" Body="&lt;p&gt;In radio/microwave observations, I understand the beamsize is the response of the telescope to a point source - effectively, the telescope's resolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the Planck satellite (par exemple) combines different frequency channels in a variance-weighted map. These different channels in general will have different sensitivities, and different beamsizes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you coadd two (or any number really) of these channels. What will be the effective resolution of the resulting map?&#xA;For instance, suppose we take channel 1 with a 5 arcminute beamsize (FWHM), and channel 2 with a 12 arcminute beamsize (and let's suppose they have the same sensitivity). &#xA;&lt;strong&gt;What will be the resolution of the inverse-variance weighted map? How will this change if the sensitivities differ?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="6965" LastActivityDate="2015-10-29T12:14:41.753" Title="Beamsize in radio/microwave observations" Tags="&lt;radio-astronomy&gt;&lt;angular-resolution&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12319" PostTypeId="2" ParentId="12318" CreationDate="2015-10-29T12:14:41.753" Score="1" Body="&lt;p&gt;This is a fundamental problem in modern multi-color/frequency astronomy. You can not simply add two spectrally significant different channels because they contain information about different solid angles on the sky. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have two maps with different inherent beam sizes one common approach is to resample the map with the higher spatial resolution to the lower resolution of the other map by convolving with a Gaussian beam of size:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta_{Gauss-kernel}=\sqrt{\theta_{low}^2-\theta_{high}^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you have two maps with resolution (FWHM) 5&quot; and 15&quot; you need to smooth the 5&quot; map with a Guassian kernel of size&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta_{Gauss-kernel}=\sqrt{15^2-5^2}=14.14&quot;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to obtain a comparable map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the beams are never true Gaussians and it depends whether you observe point sources or extended (w.r.t. the beam) source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example you can take a look at the &lt;a href=&quot;http://herschel.esac.esa.int/Docs/SPIRE/html/spire_om.html&quot; rel=&quot;nofollow&quot;&gt;SPIRE Observers manual&lt;/a&gt;. SPIRE was an instrument onboard the Herschel Space Observatory. They discuss the involved problems in somewhat more detail.&lt;/p&gt;&#xA;" OwnerUserId="1988" LastActivityDate="2015-10-29T12:14:41.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12320" PostTypeId="1" CreationDate="2015-10-29T13:02:00.783" Score="1" ViewCount="169" Body="&lt;p&gt;In astronomy class, we have to state how gravity work in the black hole? I know for one fact is that black hole is strong enough to stretch you thin enough like a noodle. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="9641" LastActivityDate="2015-10-29T15:07:58.630" Title="How gravity works in a black hole?" Tags="&lt;gravity&gt;&lt;black-hole&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2015-11-03T17:07:58.957" ContentLicense="CC BY-SA 3.0" />
  <row Id="12321" PostTypeId="2" ParentId="12320" CreationDate="2015-10-29T15:07:58.630" Score="3" Body="&lt;p&gt;Gravity works &lt;strong&gt;exactly the same&lt;/strong&gt; around a black hole as it does everywhere else in the universe: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the black hole distorts spacetime, and then objects follow the shortest route in the curved spacetime. The only difference between the gravity of a black-hole and the gravity of Earth is that the Black hole has a much larger mass in a much smaller volume, so the gravitational field is much stronger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strength of gravity has several consequences I'll mention three: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First close to the black hole spacetime is so curved that light can't escape. Now all gravity bends light a bit, but you usually don't notice it. A black hole bends light a lot.  The edge of the black hole is called an event horizon and it is the closest distance that light can go to a black hole and not fall in. Nothing can ever pass from within the event horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next there are tidal forces. All gravity can cause tides: the force of the earth gravity is a bit more on your feet than on your head. Usually you don't notice. However a solar mass black hole has such extreme tides that you would be pulled apart by the difference in force (this is the spaghettification you mention) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, inside the black hole, space works like time. You can't change your direction in time (unless you have a Delorean) Inside a black hole you can't change direction in space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But remember the only difference between the gravity around a black hole and the Earth is a matter of degree.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-10-29T15:07:58.630" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12322" PostTypeId="2" ParentId="12279" CreationDate="2015-10-29T20:08:24.710" Score="2" Body="&lt;p&gt;I ended up finding this on page 14 of the Edmund Scientific Holiday Sale catalog, with sale ending on December 31, 1983:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tStEM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tStEM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;except that it's &quot;Planetimer&quot; (and not &quot;Planet-timer&quot;), catalog number H9454.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 1200dpi image somewhat shows how it's supposed to work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/L9cM4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L9cM4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My original question was going to be: could this device have worked?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also appreciate any other information on this product (including a better scan, ideally of the product itself, if anyone still has one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference, here is the page on which the ad appears and the cover of the catalog in which I found it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3IbF8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3IbF8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Zo9fs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zo9fs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2015-10-29T20:08:24.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12323" PostTypeId="2" ParentId="12316" CreationDate="2015-10-29T20:29:31.320" Score="1" Body="&lt;p&gt;No, I don't believe there's any meaningful distinction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A planet's axial tilt can be thought of as a 2-dimensional quantity: its angle relative to the ecliptic (or to whatever baseline you're using), and the direction in which it's tilted. When we talk about a planet's axial tilt as a number of degrees, we're only talking about the first component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A planet whose rotation axis is perpendicular to the ecliptic would have a tilt of 0°. If its rotation axis is parallel to the ecliptic (rotating &quot;on its side&quot;), its tilt is 90°. If its axis is &quot;vertical&quot; but it's spinning backwards, its tilt is 180°. The range from 0° to 180° is enough to express all possible tilts, both prograde and retrograde.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider taking a planet with a 0° tilt and tilting its axis by 270°. The result is exactly the same as tilting it 90° &lt;em&gt;in the opposite direction&lt;/em&gt;. Since the axial tilt is usually expressed as just the magnitude of the tilt, and not its direction, the distinction between 90° and 270° can be ignored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if we want to express the direction as well, we'll just say that it's tilted 90° in some specified direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We &lt;em&gt;could&lt;/em&gt; have had a convention where the tilt ranges from 0° to just under 360°, with the direction specified more narrowly, but that would be less useful; the particular direction of a planet's axial tilt is less interesting than its magnitude. Also, the direction varies over time due to precession.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could also have had a convention where the tilt ranges from 0° to 90°, and we also specify whether it's retrograde or not; then Venus would have a tilt of 3° rather than 177°.&lt;/p&gt;&#xA;" OwnerUserId="53" LastActivityDate="2015-10-29T20:29:31.320" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12325" PostTypeId="2" ParentId="12267" CreationDate="2015-10-29T23:22:48.730" Score="2" Body="&lt;p&gt;The articles indicate one of two possible outcomes: merger, followed ultimately by gamma ray burst, or permanent separation, separate supernovae leading to binary black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second case the supernovae will be in a few million years, the typical life span for massive stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first case the merger could happen sooner, perhaps hundreds of thousands of years, so &quot;soon&quot; in astronomical terms, but long in comparison to a human life.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-10-29T23:22:48.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12327" PostTypeId="1" AcceptedAnswerId="24182" CreationDate="2015-10-30T10:49:08.087" Score="7" ViewCount="233" Body="&lt;p&gt;&lt;a href=&quot;http://www.stsci.edu/jwst/instruments/niriss/ObservationModes/ami&quot; rel=&quot;noreferrer&quot;&gt;NIRISS&lt;/a&gt; is an instrument on the James Webb Space Telescope. It has a &lt;em&gt;&quot;non-redundant aperture mask&quot;&lt;/em&gt; which obviously covers most of the area of the sensor. It seems to be advantageous for high contrast imaging (like finding an exoplanet next to a star) and an alternative to &lt;a href=&quot;https://en.wikipedia.org/wiki/Coronagraphhttps://en.wikipedia.org/wiki/Coronagraph&quot; rel=&quot;noreferrer&quot;&gt;coronagraphs&lt;/a&gt;. But however does that work? Why is it good to cover most of a sensor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have associated interferometers with creating as large as possible baselines for higher resolution, like the Very Large Baseline Array and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Spektr-R&quot; rel=&quot;noreferrer&quot;&gt;Spectr-R&lt;/a&gt; radio space telescope which gives up to a 390,000 km long baseline. So what is the magic with sacrificing sensor area to turn a single small telescope into an interferometer? Aren't all photons welcome? Would such an instrument do as well with a correspondingly smaller main mirror (maybe in separate fragments)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/muKIp.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/muKIp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2017-12-26T08:40:03.643" Title="What makes small interferometers useful? Like NIRISS on JWST" Tags="&lt;telescope&gt;&lt;optics&gt;&lt;infrared&gt;&lt;interferometry&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12329" PostTypeId="1" AcceptedAnswerId="12333" CreationDate="2015-10-30T12:51:10.690" Score="10" ViewCount="407" Body="&lt;p&gt;What raw data can I possibly acquire from an 8&quot; Classical Dobsonian Telescope, and a DSLR? Could anything eye-opening to amateur astronomers be computed or calculated first-hand with such equipment? I'm sure scientists must've considered this equipment &quot;advanced technology&quot; at some point in history not too far back...Could I rediscover or calculate some Laws (like Kepler's laws) or some other things amateur astronomers would be amazed to calculate themselves (like the distance to a planet) using this equipment?&lt;/p&gt;&#xA;" OwnerUserId="9649" LastEditorUserId="9186" LastEditDate="2015-10-30T16:11:05.060" LastActivityDate="2018-02-06T18:57:12.320" Title="Acquirable Raw Data in Amateur Astrophotography" Tags="&lt;amateur-observing&gt;&lt;history&gt;&lt;data-analysis&gt;&lt;photography&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12330" PostTypeId="2" ParentId="12329" CreationDate="2015-10-30T13:43:48.210" Score="7" Body="&lt;p&gt;You are absolutely right: amateurs can do a lot of science with the apparatus you own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book &quot;&lt;a href=&quot;http://www.springer.com/us/book/9783319156590&quot; rel=&quot;noreferrer&quot;&gt;Astronomical Discoveries You Can Make, Too!&lt;/a&gt;&quot;, by Robert Buchheim, lists famous historical observations that can be replicated by amateurs.&lt;/p&gt;&#xA;" OwnerUserId="9651" LastEditorUserId="9186" LastEditDate="2015-10-30T17:21:41.213" LastActivityDate="2015-10-30T17:21:41.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12331" PostTypeId="1" CreationDate="2015-10-30T15:58:02.503" Score="5" ViewCount="316" Body="&lt;p&gt;I was wondering if anyone knew of any resources to understand the linear limb darkening coefficient, $u$. That is to say, how the $\theta$-dependent coefficient $u$ (or sometimes $b_{\nu}$) varies with wavelength and with stellar effective temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've looked at a number of sources, such as Schwarzschild (1906), Milne (1921) and my own lecture notes and others, but nothing on this coefficient? Does anyone know where to even start looking for information of the properties of $u$?  &lt;/p&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2015-10-30T15:58:02.503" Title="Linear limb darkening coefficient, u" Tags="&lt;the-sun&gt;&lt;radiative-transfer&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12332" PostTypeId="1" AcceptedAnswerId="12336" CreationDate="2015-10-30T17:47:08.083" Score="1" ViewCount="220" Body="&lt;p&gt;I'm a French student and I'm planning to go to the United States for a graduate program in astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I will arrive in the U.S. I will begin by the first year of graduate. Then I will pursue with a Ph.D. and some postdoctoral research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know which are the best American universities/colleges for what I want to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning what division of astronomy I am interested in:&#xA;I'm not interested in the conception of telescopes, instruments and other systems. I'm not interested in engineering. And I'm not interested in extremely theoretical astronomy either.&#xA;I'm interested in observational astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding what subfields of astronomy I am interested in:&#xA;I'm not interested in extragalactic astronomy or cosmology.&#xA;I'm interested in planetary astronomy, exoplanetology, exobiology, SETI, stellar astronomy in the solar neighborhood.&#xA;I'm not interested in things that are less than 2 Astronomical Units or more than 70 light years away from us.&lt;/p&gt;&#xA;" OwnerUserId="6786" LastActivityDate="2015-10-31T09:44:06.440" Title="Which are the best American universities/colleges for observational astronomy?" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ClosedDate="2015-11-05T10:03:22.393" ContentLicense="CC BY-SA 3.0" />
  <row Id="12333" PostTypeId="2" ParentId="12329" CreationDate="2015-10-30T18:22:00.873" Score="10" Body="&lt;p&gt;First off, pairing a classic dob with a DSLR is a bit like a shotgun marriage. A dobsonian is fundamentally a visual telescope. Most manufacturers don't even consider the possibility that these instruments could be used for data collection via a sensor. There are 2 issues here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. The dobsonian is not tracking&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sky is moving, the dob stays still. You have to push the dob to keep up with the sky. Any long-exposure photo would be smeared. To remedy this, you'll need an &lt;a href=&quot;http://www.equatorialplatforms.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;equatorial platform&lt;/a&gt;, which will move the dob in sync with the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that only the best platforms allow reasonably long exposure times. Then &lt;a href=&quot;http://www.equatorialplatforms.com/image.of.month.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;the results&lt;/a&gt; can be fairly good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. There isn't enough back-focus&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best photos are taken when you remove the lens from the camera, &lt;a href=&quot;http://www.telescopeadapters.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;plug it into the telescope&lt;/a&gt; directly, and allow the primary mirror to focus the image directly on the sensor. This is called prime focus photography. But most dobs can't reach the sensor within the camera, because their prime focus doesn't stick out far enough. There are several remedies for this, like using a barlow, moving the primary mirror up in its cell, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line is that it takes some effort to make a dob and a DSLR play nice together. Is it doable? Yes. Is it simple and immediate? No. So the literal answer to your question is that there isn't much you can do with just a dob and a DSLR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can take photos of the Moon and the Sun, because the short exposure there does not require tracking, but that's pretty much it. Here is an image of the Moon I took with a home-made 6&quot; dob (with home-made optics) and a mirrorless camera (prime focus, about 1/320 sec exposure):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vg3kE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vg3kE.jpg&quot; alt=&quot;the Moon&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Makes a cute little desktop background, I guess, but it's definitely not research-grade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now add a tracking platform and things become more interesting, and the possibilities open up quite a lot.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In a more general sense:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are telescopes that are specifically made for astrophotography. They have lots of back-focus, they are short and lightweight and therefore can easily be installed on tracking mounts. More importantly, there are tracking mounts made specifically for imaging - very precise, delicate mechanisms that follow the sky motion with great accuracy. In fact, the mount is more important than the scope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A typical example would be a &lt;a href=&quot;http://www.celestron.com/browse-shop/astronomy/optical-tubes/edgehd-8-optical-tube-assembly&quot; rel=&quot;nofollow noreferrer&quot;&gt;C8 telescope&lt;/a&gt; installed on a &lt;a href=&quot;http://www.celestron.com/browse-shop/astronomy/mounts-and-tripods/cgem-computerized-mount&quot; rel=&quot;nofollow noreferrer&quot;&gt;CGEM mount&lt;/a&gt;, or anything equivalent. Barring that, a dob with lots of back-focus sitting on a very smooth tracking platform (probably not as accurate as a GEM, but good enough for many purposes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure you don't exceed the load capacity of the mount. If the mount claims it can carry X amount of weight, it's best if the telescope weight doesn't exceed 1/2 of that amount. Close to the weight load limit, all mounts become imprecise. The exceptions are high end (the most expensive) mounts which cost many thousands of dollars and usually honor their promises in terms of load capacity 100%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have: a tracking mount, a good camera, and a telescope (listed here from most important to least important), you can start imaging various portions of the sky for research. There are 2 main classes of objects that you could image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. Solar system objects&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They're called &quot;solar system objects&quot; but the class includes anything that's pretty bright, not very big, and it's high resolution. Tracking is important but not that crucial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need a sensitive, high speed camera that can take thousands of images quickly (a movie, basically). These are called &lt;a href=&quot;https://astronomy-imaging-camera.com/products/usb-3-0/asi224mc/&quot; rel=&quot;nofollow noreferrer&quot;&gt;planetary cameras&lt;/a&gt;. They generally have small sensors, are high sensitivity, and can operate at high frame rates (hundreds of frames per second).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a cheap alternative in the beginning you could use a webcam, there are tutorials on the Internet about that. A DSLR in video mode in prime focus might work, but it's going to do a lot of pixel binning, so resolution would be greatly reduced unless you use a very powerful barlow (or a stack of barlows).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll load all those images in &lt;a href=&quot;http://www.autostakkert.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;a software that will perform &quot;stacking&quot;&lt;/a&gt; to reduce them all to one single, much clearer image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scope needs to operate at a long focal length, f/20 being typical, so a barlow is usually required. The bigger the aperture, the better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Deep space objects (DSO)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are anything that's pretty faint and fuzzy, like galaxies, but some comets are also DSO-like in their appearance. You need to take extremely long exposures; usually a dozen or a few dozen images, each one between 30 sec and 20 min of exposure, sometimes even longer. Extremely precise tracking is paramount, so you need the best tracking mount you could buy. &lt;a href=&quot;https://en.wikipedia.org/wiki/Autoguider&quot; rel=&quot;nofollow noreferrer&quot;&gt;Autoguiding&lt;/a&gt; is also needed to correct tracking errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scope needs to operate at short focal ratios, f/4 is pretty good, but as low as f/2 is also used; focal reducers (opposite of barlows) are used with some telescopes, &lt;a href=&quot;http://www.celestron.com/browse-shop/astronomy/astroimaging-accessories/reducers/reducer-lens-7x-edgehd-800&quot; rel=&quot;nofollow noreferrer&quot;&gt;like this&lt;/a&gt; or &lt;a href=&quot;http://starizona.com/acb/HyperStar/&quot; rel=&quot;nofollow noreferrer&quot;&gt;like this&lt;/a&gt;. Aperture doesn't mean much; small refractors are used with good results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The camera needs to be very low noise; &lt;a href=&quot;https://astronomy-imaging-camera.com/products/asi-cooled-cameras/asi071mc-pro/&quot; rel=&quot;nofollow noreferrer&quot;&gt;DSO cameras&lt;/a&gt; use active cooling that lowers their temperature 20 ... 40 C below ambient. Typically they have large sensors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DSLRs can also provide decent results, but their noise is typically higher than dedicated cameras, so you need to work harder for the same results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://deepskystacker.free.fr/english/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Specific software&lt;/a&gt; is used for processing, stacking, noise reduction, etc.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So what can you do with such a setup?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comet- or asteroid-hunting works pretty well. Terry Lovejoy has discovered &lt;a href=&quot;https://en.wikipedia.org/wiki/Comet_Lovejoy&quot; rel=&quot;nofollow noreferrer&quot;&gt;several comets&lt;/a&gt; recently using equipment and techniques as described above. &lt;a href=&quot;http://www.scientificamerican.com/article/ask-the-experts-how-do-amateur-astronomers-discover-comets/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's Terry talking about his work&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.aavso.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tracking variable stars&lt;/a&gt; is also open to amateurs. This could also be done visually, without any camera, just a dob, meticulous note-taking, and lots of patience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a bit of luck, you could also be the person who discovers a new supernova in a nearby galaxy. You don't need professional instruments, you just need to happen to point the scope in the right direction at the right time and be &lt;a href=&quot;http://www.cbat.eps.harvard.edu/HowToReportDiscovery.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;the first to report it&lt;/a&gt;. This also could be done purely visually, no camera, just a dob.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2018-02-06T18:57:12.320" LastActivityDate="2018-02-06T18:57:12.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12334" PostTypeId="2" ParentId="12306" CreationDate="2015-10-30T22:32:56.273" Score="1" Body="&lt;p&gt;The moon does attract more particles.  But more importantly, those particles hit the moon with velocities of the order of the Moon's escape velocity, 2.4 km/s.  That's a &lt;strong&gt;&lt;em&gt;very&lt;/em&gt;&lt;/strong&gt; powerful sandblaster...&lt;/p&gt;&#xA;" OwnerUserId="420" LastActivityDate="2015-10-30T22:32:56.273" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12335" PostTypeId="1" AcceptedAnswerId="12339" CreationDate="2015-10-31T04:44:17.560" Score="2" ViewCount="61" Body="&lt;p&gt;As of now the New Horizons probe is well beyond the orbit of Pluto, moving towards the Kuiper belt. In an article I read that New Horizons will eventually follow the path of Voyager probes to the interstellar space. But how long it will remain functional and as further it will travel away from us it will also required very long time to receive data from the probe. Is there any pre-planned fate of the probe ? &lt;/p&gt;&#xA;" OwnerUserId="9213" LastActivityDate="2015-11-02T04:01:58.847" Title="How long New Horizons will remain operational?" Tags="&lt;space-probe&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-11-02T16:08:58.340" ContentLicense="CC BY-SA 3.0" />
  <row Id="12336" PostTypeId="2" ParentId="12332" CreationDate="2015-10-31T09:44:06.440" Score="1" Body="&lt;p&gt;Out of curiosity: where does this limit of 70 lightyears come from? Exoplanets can be detected to way beyond that. This strikes me as a strangely restrictive limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be that as it may. There are several great astronomy departments in the US, and it really depends on what you want to do exactly. For exoplanets, Harvard would be a great choice. Professor David Charbonneau is a heavy weight in the field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, many of the other renowned Universities have good astronomy departments. Yale, for instance, or Berkeley (although since the &lt;a href=&quot;http://www.nature.com/news/berkeley-astronomer-in-sexual-harassment-case-to-resign-1.18582&quot; rel=&quot;nofollow&quot;&gt;Geoff Marcy&lt;/a&gt; incident I imagine the exoplanet department has taken a hit). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at this point it is not really a matter of choosing 'the best' University. I suggest you have a look at some papers by the professors at different Universities, to get a sense of what you'd like to work on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And don't forget: for your research to be fruitful the environment is also very relevant. Just have a look and see what city you'd like to live in. &lt;/p&gt;&#xA;" OwnerUserId="6965" LastActivityDate="2015-10-31T09:44:06.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12337" PostTypeId="1" CreationDate="2015-10-31T10:47:15.310" Score="2" ViewCount="50" Body="&lt;p&gt;Recently, the Rosetta mission found oxygen on the comet 67P/Churyumov-Gerasimenko. According to theories, oxygen should not be in the state in which it is on the comet. What are the previous theories that didn't support this, and what is the explanation for the oxygen-rich environment? How will this change our understanding of the formation of the Solar System?&lt;/p&gt;&#xA;" OwnerUserId="9574" LastEditorUserId="2153" LastEditDate="2015-10-31T17:36:27.663" LastActivityDate="2015-12-01T10:49:17.623" Title="What does oxygen on comet 67P/Churyumov-Gerasimenko mean?" Tags="&lt;comets&gt;&lt;element&gt;&lt;67p&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12338" PostTypeId="2" ParentId="12236" CreationDate="2015-10-31T14:17:15.413" Score="0" Body="&lt;p&gt;From what I have read including &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014Natur.505..525K&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2014/22jan_ceres&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1029/2004JE002244/full&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, &lt;em&gt;some&lt;/em&gt; models predict a layer of liquid water. The presence of water (liquid or ice) is predicted based on the density measurement of 2.1 g/cm3 which is a little more than Ganymede, Titan, and Callisto and considerably less than Europa. Early in Ceres history it's interior would have been hotter due to the residual heat from formation and decay of radioactive elements and there would have been a large amount of liquid water. As Ceres cooled with age by way of convection of heat to the surface and lessening radioactive decay, most or all of the subsurface liquid water would have frozen.There &lt;em&gt;may&lt;/em&gt; still be a layer of liquid water due to some remaining interior heat and the mixing of materials such as ammonia in the water that would reduce the freezing temperature.&#xA;As far as surface water vapor detection goes, there is a chance that there is some &quot;cryovolcanism&quot; (water volcanoes) on Ceres (like Enceladus and Europa), which would indicate subsurface water, but it is thought to be from ice sublimation on the surface similar to what happens on comets. According to one of the articles I read above, scientists believe that when Ceres is closest to the sun this sublimation occurs and when farther away it doesn't.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2015-10-31T14:17:15.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12339" PostTypeId="2" ParentId="12335" CreationDate="2015-11-01T01:30:44.590" Score="1" Body="&lt;p&gt;The answer is in the other section:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://space.stackexchange.com/questions/4827/how-long-can-probes-function-e-g-new-horizons-can-and-should-we-build-centur&quot;&gt;New Horizons Mission Info&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I can say off the top of my head is that the mission has been extended and they have identified another object that the spacecraft can reach. However, after the next rendezvous it will be out of fuel and will not be able to maneuver. &#xA;BUT -- the probe is powered by a Plutonium-238 radioisotope thermal generator (RTG), and will likely to continue functioning for the foreseeable future. The relevant question is: how long will the government fund the NASA operations team that is 'talking' to the probe. That is anyone's guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pluto.jhuapl.edu/Mission/Spacecraft/Systems-and-Components.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;New Horizons Spacecraft Components&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS - I'll update this answer later once I talk with some colleagues. I started, and worked for the company (KinetX, Inc) that did the navigation for New Horizons. I no nothing of the science, but my friends did the navigation and know the near-term details for the mission.&lt;/p&gt;&#xA;" OwnerUserId="9648" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2015-11-02T04:01:58.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12341" PostTypeId="1" AcceptedAnswerId="12343" CreationDate="2015-11-01T06:08:21.843" Score="10" ViewCount="2973" Body="&lt;p&gt;How is a &lt;strong&gt;Dead Comet&lt;/strong&gt; different from the normal comet? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are they formed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And why is the Halloween asteroid &lt;strong&gt;2015 TB145&lt;/strong&gt; called a &lt;strong&gt;&lt;em&gt;dead&lt;/em&gt;&lt;/strong&gt; comet?.&lt;/p&gt;&#xA;" OwnerUserId="9574" LastEditorUserId="2091" LastEditDate="2015-11-01T10:42:05.937" LastActivityDate="2016-01-06T03:25:34.010" Title="What is a dead comet?" Tags="&lt;asteroids&gt;&lt;comets&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12342" PostTypeId="2" ParentId="12337" CreationDate="2015-11-01T09:52:08.173" Score="1" Body="&lt;p&gt;It's not that oxygen &lt;em&gt;should&lt;/em&gt; not be in this state, but rather we had not discovered it (e.g. on other comets). One of the reasons for this is that interstellar O2 is very reactive and quickly reacts to form e.g. H20 and O3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current models also favoured O2-less comets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, the paper you're referring to also mentions the O2 abundance was rather constant over the half year of observation, which hints to the O2 actually being primordial, i.e. from when the solar system condensed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To reconcile these  observations with models, a couple of possible solutions are offered: a slightly warmer primordial cloud (from which the solar system formed), or reactions with ices. &lt;/p&gt;&#xA;" OwnerUserId="6965" LastActivityDate="2015-11-01T09:52:08.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12343" PostTypeId="2" ParentId="12341" CreationDate="2015-11-01T09:58:41.703" Score="8" Body="&lt;p&gt;A comet is usually characterized by its tail. A dead comet has lost all its ices and gases (responsible for producing this tail), leaving just a rocky core. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Halloween comet is such a dead comet, in that it has no tail, but furthermore it resembles a skull, making it particularly relevant for Halloween.&lt;/p&gt;&#xA;" OwnerUserId="6965" LastEditorUserId="6965" LastEditDate="2015-11-01T15:02:22.980" LastActivityDate="2015-11-01T15:02:22.980" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12344" PostTypeId="1" AcceptedAnswerId="12345" CreationDate="2015-11-01T10:26:43.997" Score="3" ViewCount="115" Body="&lt;p&gt;This question relates formalising the gravitational constant, $G$, (which in S.I. units is usually quoted as $\sim6.67\times10^{-11}\,\mathrm{m}^{3}\,\mathrm{kg}^{-1}\,\mathrm{s}^{-2}$) in units which are more appropriate to astrophysics, such as in the field of galaxy dynamics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would $G$ be in more appropriate units to astrophysics? &lt;/p&gt;&#xA;" OwnerUserId="2091" LastEditorUserId="2153" LastEditDate="2015-11-01T20:01:58.330" LastActivityDate="2015-11-01T20:01:58.330" Title="Astrophysical &quot;unitised&quot; version of the Gravitational constant" Tags="&lt;units&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12345" PostTypeId="2" ParentId="12344" CreationDate="2015-11-01T10:30:17.327" Score="3" Body="&lt;p&gt;Useful formulations which have taken some time to dig out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G\sim{4.3\times{10^{-6}}}\,\mathrm{kpc}\,\mathrm{M}_{\odot}^{-1}\,\mathrm{km}^{2}\,\mathrm{s}^{-2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G\sim{4.3\times{10^{-3}}}\,\mathrm{pc}\,\mathrm{M}_{\odot}^{-1}\,\mathrm{km}^{2}\,\mathrm{s}^{-2}$&lt;/p&gt;&#xA;" OwnerUserId="2091" LastEditorUserId="2091" LastEditDate="2015-11-01T18:53:58.033" LastActivityDate="2015-11-01T18:53:58.033" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12346" PostTypeId="2" ParentId="12341" CreationDate="2015-11-01T10:35:31.403" Score="4" Body="&lt;p&gt;It is actually an asteroid. I think it just helps to promote astronomy near Hallowe'en by calling it a dead comet and releasing some pretty nice photos in which it can be likened to a skull. But yes, a dead comet has been shed of all of it's ice and gas, and is left as the rocky core. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the quote relating to the comet you are refering to and perhaps a theory on its origins:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;We found that the object reflects about six percent of the light it&#xA;  receives from the sun,&quot; said Vishnu Reddy, a research scientist at the&#xA;  Planetary Science Institute, Tucson, Arizona. &quot;That is similar to&#xA;  fresh asphalt, and while here on Earth we think that is pretty dark,&#xA;  it is brighter than a typical comet which reflects only 3 to 5 percent&#xA;  of the light. That suggests it could be cometary in origin - but as&#xA;  there is no coma evident, the conclusion is it is a dead comet.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2015-11-01T10:35:31.403" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12347" PostTypeId="1" AcceptedAnswerId="12351" CreationDate="2015-11-01T12:13:44.987" Score="3" ViewCount="223" Body="&lt;p&gt;I'm trying to develop a planetarium using &lt;a href=&quot;https://en.wikipedia.org/wiki/Tycho-2_Catalogue&quot; rel=&quot;nofollow&quot;&gt;Tycho 2 catalogue&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To read it I use the &lt;a href=&quot;http://tdc-www.harvard.edu/catalogs/tycho2.html&quot; rel=&quot;nofollow&quot;&gt;WCSTools&lt;/a&gt;. But I have a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Wikipedia I have found that Rigel star has RA: 05h 14min 32,3s and Dec: -08º 12’ 06’’. I have translate them to decimal degrees (78.6375, -8.2017). I've used that to find it on Tycho 2 catalogue (file catalog.dat) but I can't find it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To search into the catalogue I have used the function TY2READ from &lt;a href=&quot;https://github.com/kapadia/PinpointWCS/blob/master/libwcs/ty2read.c&quot; rel=&quot;nofollow&quot;&gt;ty2read.c&lt;/a&gt; file.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is how I search the star:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;void ReadTy2Catalog()&#xA;{&#xA;    double raCenter = 84.5;&#xA;    double decCenter = -1.2;&#xA;    double raDistance = 10.;&#xA;    double decDistance = 10.;&#xA;&#xA;    int nMaxStars = 20000;&#xA;&#xA;    char* filePath = &quot;D:\\Fuentes\\Repos\\Planetarium\\StarsCataloguesLib\\Resources\\&quot;;&#xA;    Ty2Read read;&#xA;&#xA;    read.Read(&#xA;        filePath,&#xA;        TYCHO2,         /* Catalog code from wcscat.h */&#xA;        raCenter,       /* Search center J2000 right ascension in degrees */&#xA;        decCenter,      /* Search center J2000 declination in degrees */&#xA;        raDistance,     /* Search half width in right ascension in degrees */&#xA;        decDistance,    /* Search half-width in declination in degrees */&#xA;        0,              /* Limiting separation in degrees (ignore if 0) */&#xA;        0,              /* Inner edge of annulus in degrees (ignore if 0) */&#xA;        1,              /* 1 to sort stars by distance from center */&#xA;        WCS_J2000,      /* Search coordinate system */&#xA;        2000.0,         /* Search coordinate equinox */&#xA;        0.0,            /* Proper motion epoch (0.0 for no proper motion) */&#xA;        -20.0,          /* Limiting magnitudes (none if equal) */&#xA;        30.0,           /* Limiting magnitudes (none if equal) */&#xA;        1,              /* Magnitude by which to sort (1 or 2) */&#xA;        nMaxStars,      /* Maximum number of stars to be returned */&#xA;        1);             /* Verbose*/&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I have checked all the stars returned and none has Rigel's RA and DEC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="7566" LastEditorUserId="2365" LastEditDate="2015-11-01T16:38:52.380" LastActivityDate="2015-11-01T16:38:52.380" Title="I can't find Rigel on Tycho 2 catalogue" Tags="&lt;software&gt;&lt;astrometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12348" PostTypeId="2" ParentId="12341" CreationDate="2015-11-01T13:25:09.367" Score="2" Body="&lt;p&gt;Comets are bodies that formed in the outer solar system, and are composed largely of ices (water, CO2 and others). The Rosetta mission is discovering lots of new science about the compostion of comets right now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The asteroids are more varied. Some are rocky, some metallic and some have a lot of ice. The asteroids are have various origins, but most orbit between Mars and Jupiter and in the plane of the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, comets tend to have highly elliptical orbits and often their orbits are highly inclined, relative to the rest of the solar system. This is because they are falling towards the sun having been disturbed from their birthplace in the outer solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chief characteristic of a comet is that, as it comes near the sun, the ices sublimate off the comet and form an a coma: a giant sphere of gas and dust that surrounds the comet's icy nucleus. The solar wind pushes this into a tail that point away from the sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After many orbits of the sun a comet will eventually run out of ice, at least on its surface, and will no longer form a coma. How long this takes depends on the orbit of the comet, but about 1/2 million years seems to be an estimate of the average life span of a comet. As it is no longer active, it is a &quot;dead comet&quot;. Although having a &quot;dead comet&quot; on Halloween is a bit of spin, the term is real and has been used prior to 2015-TB145, for example in &lt;a href=&quot;http://spaceguard.rm.iasf.cnr.it/NScience/neo/neo-what/com-life.htm&quot; rel=&quot;nofollow&quot;&gt;this page about near earth objects&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, as noted there, many near earth &quot;asteroids&quot; may actually be dead comets.  Evidence that 2015-TB145 is one is that it has a highly inclined elliptical orbit, and it is very dark. Other comets, such as Halley, are blacker than soot.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-11-01T13:25:09.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12351" PostTypeId="2" ParentId="12347" CreationDate="2015-11-01T16:33:51.160" Score="2" Body="&lt;p&gt;This frustrated me as well when I first used the Tycho-2 catalog. Not finding a well known bright star can be disconcerting. Quoting from &lt;a href=&quot;http://heasarc.nasa.gov/W3Browse/all/tycho2.html&quot; rel=&quot;nofollow&quot;&gt;http://heasarc.nasa.gov/W3Browse/all/tycho2.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Supplement-1 (not part of this HEASARC database but available at&#xA;  &lt;a href=&quot;ftp://cdsarc.u-strasbg.fr/cats/I/259/suppl_1.dat.gz&quot; rel=&quot;nofollow&quot;&gt;ftp://cdsarc.u-strasbg.fr/cats/I/259/suppl_1.dat.gz&lt;/a&gt;) lists stars from&#xA;  the Hipparcos and Tycho-1 Catalogues which are not in Tycho-2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I only skimmed that file, but I think this might be the line you're looking for:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5331 01752 1|H|078.63446353|-08.20163919|    1.9|   -0.6|  0.6|  0.5|  0.7|  0.5| | 0.312|0.012| 0.283|0.010|999|T| 24436 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(the &quot;-0.6&quot; is the U-B color index, not the magntiude)&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2015-11-01T16:33:51.160" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12353" PostTypeId="1" CreationDate="2015-11-02T00:45:45.720" Score="2" ViewCount="311" Body="&lt;h2&gt;The problem:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I want to simulate a travel from Earth to the closest 15-25 stars. Which is the fastest way to accomplish this?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;What i have:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I currently have the distance between Earth and the nearby stars, according to this link: &lt;a href=&quot;http://www.space.com/18964-the-nearest-stars-to-earth-infographic.html&quot; rel=&quot;nofollow&quot;&gt;http://www.space.com/18964-the-nearest-stars-to-earth-infographic.html&lt;/a&gt;. But from it I can't know if &lt;em&gt;Alpha Centauri&lt;/em&gt; (4.3 light-years from Earth) is closest to &lt;em&gt;Sirius&lt;/em&gt; (8.58 light-years from Earth) or to &lt;em&gt;Wolf 359&lt;/em&gt; (7.8 light-years from Earth). &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;What I need:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I need the distance between any two stars, at least the closest ones to Earth. Then it would be possible to run a &lt;em&gt;Traveling Salesman problem&lt;/em&gt;, in order to visit all the nearby stars (or a subset of 15-25 of them) in the fastest time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Where can I get this information?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9662" LastEditorUserId="9186" LastEditDate="2015-11-02T02:42:25.670" LastActivityDate="2015-11-02T05:16:16.070" Title="Where can I look up the 3D positions of the closest stars?" Tags="&lt;star&gt;&lt;distances&gt;&lt;positional-astronomy&gt;&lt;interstellar-travel&gt;&lt;resource-request&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-11-02T16:16:31.677" ContentLicense="CC BY-SA 3.0" />
  <row Id="12354" PostTypeId="2" ParentId="12353" CreationDate="2015-11-02T05:16:16.070" Score="0" Body="&lt;p&gt;You can determine the 3D position for a star, $p$, by using the distance from Earth, $R$, as well as the right ascension, $\Omega$, and declination, $\delta$, within the following formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p = R \begin{bmatrix} \cos \Omega \cos \delta \\ \sin \Omega \cos \delta \\ \sin \delta \\ \end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note this will yield positions within the Earth-centered inertial reference frame.&lt;/p&gt;&#xA;" OwnerUserId="9575" LastActivityDate="2015-11-02T05:16:16.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12355" PostTypeId="1" AcceptedAnswerId="12357" CreationDate="2015-11-02T05:26:09.070" Score="2" ViewCount="357" Body="&lt;p&gt;Tonight around 12:20 AM I realized that the top part (instead of the side) of the moon was missing. I live at a country in the tropics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took this picture with my phone so it isn't very good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FRx6w.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After googling I found out about the wet moon phase in Wikipedia and I wonder if that is what I was looking at. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FRx6w.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wet Moon&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9665" LastActivityDate="2015-11-02T06:52:03.470" Title="Is this a wet moon?" Tags="&lt;moon-phases&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12356" PostTypeId="1" AcceptedAnswerId="12360" CreationDate="2015-11-02T06:44:24.420" Score="12" ViewCount="816" Body="&lt;p&gt;After Pluto's demotion as a planet, we have currently eight planets in our solar system. But Sun's gravitational pull can be felt well beyond Pluto, so is it possible to have a ninth planet beyond Pluto? I asked this question because we have Kuiper belt beyond Pluto, so there might be some Kuiper belt object which may qualified to be a ninth planet, isn't it possible? &lt;/p&gt;&#xA;" OwnerUserId="9213" LastEditorUserId="182" LastEditDate="2016-01-21T07:33:47.327" LastActivityDate="2016-01-21T07:33:47.327" Title="Is there a ninth planet?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;9th-planet&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12357" PostTypeId="2" ParentId="12355" CreationDate="2015-11-02T06:52:03.470" Score="1" Body="&lt;p&gt;No it is not a wet moon, as wet moon occurs when the horn of the crescent moon pointed at an angle away from the horizon. In your picture, that is not a crescent Moon. It is may be in the first quarter or last quarter, so Gibbous moon is more suitable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_phase&quot; rel=&quot;nofollow&quot;&gt;Moon phases.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9213" LastActivityDate="2015-11-02T06:52:03.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12358" PostTypeId="1" AcceptedAnswerId="12359" CreationDate="2015-11-02T08:48:17.643" Score="8" ViewCount="851" Body="&lt;p&gt;I am curious that there are lots of celestial objects (planets, comets, stars, galaxies, etc.) in space. How do we name them?&lt;/p&gt;&#xA;" OwnerUserId="9666" LastEditorUserId="2091" LastEditDate="2015-11-03T18:24:55.257" LastActivityDate="2020-12-17T16:13:06.903" Title="How do scientists name space objects?" Tags="&lt;naming&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12359" PostTypeId="2" ParentId="12358" CreationDate="2015-11-02T10:17:37.640" Score="5" Body="&lt;p&gt;Millennia ago, when only a few thousand stars, a handful of planets and nebulae, and some transient objects like supernovae and comets, were known, people usually named these objects after gods and heros, but also after everyday objects. Stellar constellations that vaguely resembled something known, was named after this. Mars is red, so it was named after the god of war; Mercury, being the fastest to complete one cycle around the Sun, was named the god of messaging and traveling.&lt;/p&gt;&#xA;&lt;p&gt;With the advent of telescopes, so many objects are now known that we had to come up with a more pragmatic convention, so usually an object is named according to its place in some catalogue, its type, and/or it coordianates on the sky. For instance, a quasar detected at the &amp;quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Epoch_%28astronomy%29#Julian_years_and_J2000&quot; rel=&quot;nofollow noreferrer&quot;&gt;J2000&lt;/a&gt;&amp;quot; coordinates &lt;span class=&quot;math-container&quot;&gt;$\{\mathrm{dec.},\mathrm{RA}\} = \{22^\circ\, 22'\, 56'', -09^\mathrm{h}\, 46^\mathrm{m}\, 36^\mathrm{s}\}$&lt;/span&gt; may be called J2222–0946. An x-ray source observed in the constellation Cygnus may be called Cyg X-1, while its binary counterpart is number 226,868 in the Henry Draper Extension Catalogue and is thus named &amp;quot;HDE 226868&amp;quot; (not to be confused with one of our site moderators).&lt;/p&gt;&#xA;&lt;p&gt;Oftentimes, an object appears in multiple catalogues, and will thus have multiple names. For instance, a Lyman &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;-emitting galaxy appears in a catalogue of Ly&lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; emitters, but due to its infrared properties, it will also appear in some catalogue over dusty galaxies (though Ly&lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; emitters tend to contain little dust).&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="31410" LastEditDate="2020-12-17T16:13:06.903" LastActivityDate="2020-12-17T16:13:06.903" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="12360" PostTypeId="2" ParentId="12356" CreationDate="2015-11-02T15:39:29.143" Score="4" Body="&lt;p&gt;There could well be, since when we had nine planets we were looking for planet X (planet number 10). However, the modern day designation of a planet is: &quot;is massive enough for its own gravity to make it roughly spherical (slight &quot;oblate-ness&quot;), and has &quot;cleared its neighbourhood&quot; of smaller objects around its orbit&quot;. Hence, we would more than likely discard anything far out such as Pluto because it would belong to either the Kuiper belt or the Oort cloud. Anything further out than Neptune would likely be considered a Dwarf planet or proto-planet.&lt;/p&gt;&#xA;" OwnerUserId="2091" LastEditorUserId="2091" LastEditDate="2015-11-02T20:23:33.177" LastActivityDate="2015-11-02T20:23:33.177" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12361" PostTypeId="2" ParentId="11941" CreationDate="2015-11-02T16:31:38.710" Score="1" Body="&lt;p&gt;The heliocentric velocity $V_\mathrm{H}$ of an object is its velocity wrt. the Sun. When you measure an object's velocity, you measure it in the reference frame of Earth, which revolves around the Sun with ~30 km/s (varying a bit from aphelion to perihelion), so convert to $V_\mathrm{H}$ you need to know the time of the year of the observation (unless the line of sight toward your object is exactly perpendicular to the ecliptic plane), as well as the angle between the line of sight, and the line of sight toward the Sun. This involves a number of sines and cosines that you can find in e.g. &lt;a href=&quot;https://books.google.dk/books?id=JhU6V1yDRCgC&amp;amp;pg=PA153&amp;amp;lpg=PA153&amp;amp;dq=geocentric%20to%20heliocentric%20velocity&amp;amp;source=bl&amp;amp;ots=NBLRwDXsOW&amp;amp;sig=NyR5Kz4lechA7DqrWrnFaBRsydY&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0CB4Q6AEwADgKahUKEwiw8PiOlPLIAhVLcRQKHVtsAkE#v=onepage&amp;amp;q=geocentric%20to%20heliocentric%20velocity&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Barbieri (2006)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you further want to convert from $V_\mathrm{H}$ to the reference frame in which the Milky Way's center is at rest, the Local Group is at rest, or the Cosmic Microwave Background is isotropic (the &quot;cosmic&quot; frame), &lt;em&gt;then&lt;/em&gt; you use the formula you link to, adding a similar term as described above, but instead using the velocity (i.e. speed and direction) of the Sun wrt. to the given frame.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2015-11-02T16:31:38.710" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12362" PostTypeId="2" ParentId="12358" CreationDate="2015-11-02T18:53:33.390" Score="1" Body="&lt;p&gt;Names are only what we agree to call things. Astronomers have a practical need for names, and a romantic sense of discovery being confirmed by the naming. So Astronomers have cooperated to agree on names for astronomical objects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1919 several astronomical associations merged to form the International Astronomers Union (IAU), and it is now the umbrella body to which professional astronomers and national scientific academies belong. As such what the IAU calls things is as close to &quot;official&quot; names as we can get. It is the IAU who set standards and guidelines for the naming of new asteroids, Trans Neptunian Objects, craters on Pluto, exoplanets, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The procedure for naming is roughly: Following discovery the IAU checks that the object really exists, and which astronomer actually discovered it first (sometimes several people report a new object at about the same time, the IAU sorts out priority). The IAU then asks the discoverer to propose a name. The name must be unique, and some objects are named systematically: for example moons of Pluto are named after gods from underworld mythology. The IAU checks this and then confirms the name, if everything is okay. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars aren't named by the IAU, with the exception of those that already had names (mostly arabic) from ancient times. There are so many stars that naming them individually would be impractical, and they just get catalogue numbers. Most stars have many catalogue designations each from a different catalogue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some objects do get named by general adoption by the media, for example &quot;Tabby's star&quot; has not been designated as such by the IAU, but if people continue to call KIC 8462852 by that name, it may enter general use.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2015-11-02T19:00:34.410" LastActivityDate="2015-11-02T19:00:34.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12363" PostTypeId="1" CreationDate="2015-11-02T20:34:57.093" Score="4" ViewCount="222" Body="&lt;p&gt;Our universe is supposedly 13+ billion years old and our Sun is a third generation star. It seems to me that we are now in a relatively young stage of the universe. How many generations of stars will there be before the end of the universe? When is 'now', relative to the all possible timepoints in the universe, that life could appear? &lt;/p&gt;&#xA;" OwnerUserId="9670" LastEditorUserId="2153" LastEditDate="2015-11-02T22:23:28.010" LastActivityDate="2015-11-04T10:16:30.520" Title="Where are we in an approximate timeline of the possibly habitable universe?" Tags="&lt;star&gt;&lt;universe&gt;&lt;age&gt;" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="12367" PostTypeId="1" CreationDate="2015-11-03T04:33:47.437" Score="5" ViewCount="1152" Body="&lt;p&gt;I'm a little bit confused about reference frames, and I was wondering if someone could help clarify a few things?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's say I have equatorial coordinates that refer to the mean equator and equinox of an non standard arbitrary epoch. Now, if those coordinates are referred to the mean equator and equinox that epoch, then is the mean equator/equinox of the specific epoch considered the reference frame for those coordinates? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;…and then how would you go about converting those coordinates to another reference frame like say the IRCF or FK5? I assume that I would need to precess them to J2000 accounting for any applicable proper motion (and if they referred to the true equator nutation). What else do you need to put them in either the ICRF or FK5 frames?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, something else I'm confused about is the epoch of a reference frame. So, let's say I have the IRCF astrometric coordinates of a planet at JD 2456963.5 TDB (or any other arbitrary date). Since they're IRCF coordinates, then the epoch of the reference frame is still J2000, but the coordinates refer to the instant of 2456963.5 TDB correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any clarification would be greatly appreciated, and thanks in advance! &lt;/p&gt;&#xA;" OwnerUserId="9673" LastActivityDate="2015-12-03T17:27:02.730" Title="Conversion between Astronomical Frames, ex. IRCF, FK5, FK4, etc…" Tags="&lt;coordinate&gt;&lt;frame-of-reference&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12368" PostTypeId="1" CreationDate="2015-11-03T07:14:49.997" Score="7" ViewCount="373" Body="&lt;p&gt;During the super-moon total lunar eclipse on September 27th a meteor-like streak passed very close within view of the event. It appeared to happen at the very middle, the maximum eclipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://petapixel.com/assets/uploads/2015/09/shooting.jpg&quot; alt=&quot;supermoon&quot;&gt;&#xA;&lt;em&gt;[CafferyPhoto][1]&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;At the time, I thought it must have been some organization shedding an unused satellite. This would be a pretty epic time to do so as so many eyes around the world are looking at the same thing. Maybe even NASA marking the halftime of the event. Can anyone confirm or shed any amount of light on what this was?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1][&lt;a href=&quot;https://www.facebook.com/cafferyphoto/photos/a.279223605465113.81307.120967987957343/915070541880413/?type=3&amp;amp;theater]&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.facebook.com/cafferyphoto/photos/a.279223605465113.81307.120967987957343/915070541880413/?type=3&amp;amp;theater]&lt;/a&gt;&#xA;&lt;em&gt;note - photo for proof only - shutter cut off leading edge&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;QUESTIONS:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;People on the Caffery's Facebook page seem to have seen in in California and other locations as well, I happened to be in St Thomas, Virgin Island at the time of the event.  CafferyPhoto quotes. &quot;I was only exposing for half a second so the shutter cut off the latter half of the trail, darn it!&quot; - Seems in was taken in the NorthWestern US sometime around 7:30pm.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="9675" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:47.433" LastActivityDate="2019-01-23T02:46:54.167" Title="Was that actually a shooting star during the supermoon eclipse on September 27th 2015?" Tags="&lt;amateur-observing&gt;&lt;meteor&gt;&lt;meteorite&gt;&lt;near-earth-object&gt;&lt;the-moon&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12370" PostTypeId="2" ParentId="11530" CreationDate="2015-11-03T14:15:07.133" Score="4" Body="&lt;p&gt;The distance you are referring to is the &quot;light travel distance&quot;. It is the length of time that light has been travelling from the distant galaxy to us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the light was emitted, our Galaxy was much closer to that distant galaxy. What has happened in the meantime is that the space between the galaxies has expanded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore we can ask where is that galaxy right &lt;em&gt;now&lt;/em&gt; (where &lt;em&gt;now&lt;/em&gt; is a time measure since the big bang) and it turns out that the galaxy would be much further away than 13.2 billion light years - more like 40 billion light years if you were to somehow measure that distance with a very long ruler.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-11-03T14:15:07.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12371" PostTypeId="2" ParentId="12358" CreationDate="2015-11-03T16:47:56.570" Score="1" Body="&lt;p&gt;One of the caveats to this question is that often one object, say a star, will have multiple names or &quot;designations&quot;&quot; for example, one of the most famous stars I can think of (and, as it happens, one of my favourite) is Vega. Vega is also known as $\alpha$-Lyrae, being the brightest star in the constellation of Lyra. This is known as a Bayer designation: whereby the most of the brighter stars were assigned their first systematic names such as alpha ($\alpha$), beta ($\beta$), gamma ($\gamma$), etc. in order of brightness. The Vega name is the Latin name of the star’s parent constellation in genitive (possessive) form. The entirity of Vega's other designations and names are: Wega, Lucida Lyrae, Alpha Lyrae, α Lyrae, 3 Lyr, GJ 721, HR 7001, BD +38°3238, HD 172167, GCTP 4293.00, LTT 15486, SAO 67174, HIP 91262, 织女一. So you can see, there is really multiple names for one object on the sky, depending on which catalogue you use!&lt;/p&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2015-11-03T16:47:56.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12372" PostTypeId="2" ParentId="10668" CreationDate="2015-11-03T17:19:35.527" Score="0" Body="&lt;p&gt;&lt;strong&gt;There is no transient part&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In solving a linear ode y'' + ay' + by = f(x), one finds that the solution is the superimposition of a general solution of y''+ay'+by=0, and a particular solution of the ode. In the situation in which the ode represents damped harmonic motion, the general solution part is said to be transient. This is a particular solution to this ODE, and not a general technique for solving all differential equations, and does not apply to the ode that describes newtonian gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there is another way of solving the newtonian equations, in the case of a two body problem, and that is &lt;em&gt;Kepler's laws of planetary motion&lt;/em&gt;. These give a complete solution, and there is not transient part. Given a body's initial state, you can immediately find it's orbit using Kepler's laws. The body will be in an elliptical orbit, and return to its starting point, having made one complete orbit.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-11-03T17:19:35.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12373" PostTypeId="2" ParentId="12367" CreationDate="2015-11-03T17:23:02.153" Score="2" Body="&lt;p&gt;I'm not sure this answers your question, but ICRF is a non-precessing &#xA;reference frame. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On J2000 (2000-01-01 12:00:00 UTC), the z axis of the ICRF frame was &#xA;exactly parallel to a line connecting the Earth's center to the &#xA;Earth's north pole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since then, the Earth's north pole has precessed, so the ICRF z axis &#xA;and the geocenter-north pole line are no longer parallel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, you can visit &lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ssd.jpl.nasa.gov/horizons.cgi&lt;/a&gt; and &#xA;use these settings: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3j8zM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3j8zM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(this is actually looking from the north pole &quot;down&quot; to the Earth's &#xA;center, so we would expect a negative z value). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interesting part of the results: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;$$SOE&#xA;2445701.000000000, A.D. 1984-Jan-01 12:00:00.0000,  1.007696415090285E+01, -3.712508526111991E-02, -6.356744326939433E+03, -2.994380889331686E-07, -2.570088696418019E-07, -4.731801778858530E-10,  2.120384334099952E-02,  6.356752314245179E+03,  5.329972950848752E-25,&#xA;2451545.000000000, A.D. 2000-Jan-01 12:00:00.0000,  1.601081935638483E-01,  1.746805655485580E-01, -6.356752309828780E+03,  2.230286001476848E-07, -8.078442785945724E-07, -1.658173608778433E-11,  2.120384334099953E-02,  6.356752314245181E+03, -5.549834335071262E-23,&#xA;2457389.000000000, A.D. 2016-Jan-01 12:00:00.0000, -9.877267476902714E+00,  3.153970319756332E-01, -6.356744632653835E+03,  1.794887917287106E-07, -5.409368308792473E-07, -3.057333255094828E-10,  2.120384334099952E-02,  6.356752314245179E+03, -5.736383388350969E-23,&#xA;$$EOE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It's a little hard to notice because of the scientific notation (and &#xA;the embarrassing fact that x and y aren't exactly zero at J2000), but &#xA;the x and y values at J2000 (the middle entry) are nearly zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The y value is nearly zero for all three entries, but the x value &#xA;changes. In 1984, the x value is about +10m, and, in 2016, the x value &#xA;is almost -10m, showing the effects of precession. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One benefit of a non-precessing frame like ICRF is that star positions &#xA;are relatively constant (excluding proper motion and the solar system's own &#xA;motion). For example, the z axis will always point (very roughly) to &#xA;Polaris, even after precession has changed the north star to, for &#xA;example, Deneb. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By default, NASA now uses ICRF coordinates to indicate where planets &#xA;are, so, if you're looking to map planets on a J2000 starchart, you &#xA;should use the ICRF coordinates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only problem you'll have using ICRF/J2000 coordinates (instead of &#xA;J2016 coordinates) is that any calculations you make for &#xA;stellar/planetary azimuth and altitude will be off slightly, not &#xA;enough to really notice unless you are doing precision work. &lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2015-11-03T17:23:02.153" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12374" PostTypeId="1" AcceptedAnswerId="12391" CreationDate="2015-11-03T18:24:16.283" Score="2" ViewCount="210" Body="&lt;p&gt;I have a homework assignment question on accretions discs (essentially an estimation of the number of electron scatterings, but this is just for background). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few parameters, one of them being $L$, which is the linear size of the medium (the medium in this case being an accretion disc around a blackhole)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I have been given the mass of the black hole. Other than that, nothing else which could give me the linear size of the accretion disc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could I assume that the linear extent of the accretion disc is perhaps of the order of a few Schwarzschild radii? Which could be calculated from the mass, which is given. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone could shed some light on this I would be very appreciative. I need a nudge in the right direction on this.&lt;/p&gt;&#xA;" OwnerUserId="2091" LastEditorUserId="2365" LastEditDate="2015-11-05T20:58:26.353" LastActivityDate="2015-11-06T08:38:47.600" Title="Linear extent of an accretion disk" Tags="&lt;black-hole&gt;&lt;accretion-discs&gt;&lt;accretion&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12376" PostTypeId="2" ParentId="12368" CreationDate="2015-11-03T20:36:51.193" Score="1" Body="&lt;p&gt;There are about 5 sporadic meteors visible per hour most nights, and since thousands of people were taking photographs of the moon that night, it is not impossible that someone captured a meteor in the same shot as the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of things it was &lt;strong&gt;not&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It was not an unused satellite. When they re-enter you see a fireball as it breaks up, not a single streak, as here.&lt;/li&gt;&#xA;&lt;li&gt;It was not &quot;Nasa marking half-time&quot;. The object would have been visible from quite wide area (but not world wide, not even countrywide) But the association with the eclipsed moon would only have been visible from a particular location. Nasa isn't going to be interested in entertaining the  good folks of Yakima, at the expense of everyone else.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What it could be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It might be a meteor. Normally you would see a brighter head as the meteor changes brightness as it falls, but if this was a 1/2 second exposure, that could have been cut off.&lt;/li&gt;&#xA;&lt;li&gt;It could be a satellite, which appear to be a steadily moving star, and would look like a streak in a timed exposure. However, looking at Heavens above doesn't show any satellites near the moon at that time.&lt;/li&gt;&#xA;&lt;li&gt;It could be a deliberate forgery.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I reserve judgement. One would need exact locations and timings.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2015-11-24T19:44:53.113" LastActivityDate="2015-11-24T19:44:53.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12377" PostTypeId="2" ParentId="11794" CreationDate="2015-11-03T21:43:08.150" Score="2" Body="&lt;p&gt;There are roughly two possibilities: either there isn't a large vacuum energy, this would imply that there is something missing from quantum field theory. We don't know what it could be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or the zero point energy really is as large as QFT predicts, but there is something else that prevents it from having a large cosmological effect. We have no idea what that could be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We might get more of an idea if we find out what Dark Matter consists of, or a good quantum theory of gravity we might know more.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-11-03T21:43:08.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12378" PostTypeId="5" CreationDate="2015-11-03T21:52:31.930" Score="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma-ray_burst&quot; rel=&quot;nofollow&quot;&gt;Gamma ray burst&lt;/a&gt; are formed in various ways, including hypernovae and the mergers of binary neutron stars.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2153" LastEditDate="2015-11-04T01:27:41.847" LastActivityDate="2015-11-04T01:27:41.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12379" PostTypeId="4" CreationDate="2015-11-03T21:52:31.930" Score="0" Body="Gamma ray bursts are flashes of gamma rays, produced by extremely powerful explosions in distant galaxies. They are among the most powerful events in the universe" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2015-11-04T01:27:26.573" LastActivityDate="2015-11-04T01:27:26.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12381" PostTypeId="1" AcceptedAnswerId="12382" CreationDate="2015-11-04T02:20:34.273" Score="9" ViewCount="4670" Body="&lt;p&gt;Through nuclear fusion, the Sun can (or at the very least, someday will) &lt;a href=&quot;https://en.wikipedia.org/wiki/Triple-alpha_process&quot;&gt;produce atoms of all elements up to and including oxygen&lt;/a&gt;.  And in terrestrial chemistry at least, when you combine oxygen, hydrogen, and a small amount of heat you get water.  Or lithium and oxygen and heat will produce lithium oxide.  Those individual components are all readily available in a star like the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my questions is, does the oxygen produced by the Sun (or any other star, now, in the future, or in the past) react chemically (for instance, through oxidation or combustion processes) with the other elements present in the star's atmosphere?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or I suppose more generally, do the elements formed through nuclear fusion in a star chemically interact to produce more complex molecules (and if not, then why not)?&lt;/p&gt;&#xA;" OwnerUserId="9687" LastEditorUserId="9687" LastEditDate="2015-11-04T05:56:06.093" LastActivityDate="2020-12-19T12:06:29.243" Title="What happens to oxygen produced on the Sun (or other stars)?" Tags="&lt;the-sun&gt;&lt;stellar-evolution&gt;&lt;astrochemistry&gt;&lt;stellar-atmospheres&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12382" PostTypeId="2" ParentId="12381" CreationDate="2015-11-04T04:39:40.213" Score="8" Body="&lt;p&gt;The Sun is a small main sequence star. It does not produce oxygen via fusion. It can't. The temperature and pressure in the Sun's core are too low. Fusion in the Sun is currently limited to production of helium. This will remain the case for several billion years.&lt;/p&gt;&#xA;&lt;p&gt;That said, there is oxygen in the Sun, about 1% by mass. This oxygen was produced long ago by other stars at the end stages of their lives. Our Sun is a third generation (or more) star. Most of the Sun is far too hot for those oxygen atoms to combine chemically. One exception is sunspots, relatively cool areas on the Sun's photosphere. (Relatively cool means less than 4500 kelvins, so still quite hot.) Molecules can form at these lowish temperatures, and scientists do see signatures of many different molecules in the light coming from the Sun.&lt;/p&gt;&#xA;&lt;br&gt;&#xA;**Update, in response to edits to the question**&#xA;&lt;p&gt;Molecules cannot form inside of a star. The temperatures are just too high. Molecules decompose (split apart) into their constituent parts at high temperatures. The Sun's photosphere is about 5800 kelvins, which is already too hot to sustain very many molecules. Temperature rises rapidly with increasing depth below the photosphere. The Sun's core temperature is about 15 million kelvins (27 million Fahrenheit), and the Sun is a small star. Larger stars have even higher core temperatures. At 15 million kelvins, there aren't even atoms, let alone molecules. There are instead atomic nuclei and electrons. Atoms are stripped of their electrons at those extreme temperatures.&lt;/p&gt;&#xA;&lt;p&gt;In five to seven billion years, our Sun will have fused all of the hydrogen in the core into helium. That's when our Sun will become a red giant. Even then, it still will not produce oxygen. The first stage a one solar mass star experiences after leaving the main sequence is the red giant phase, where the core is an inert mass of helium surrounded by a shell of fusing hydrogen.&lt;/p&gt;&#xA;&lt;p&gt;Eventually (after another billion years or so), the temperature of that helium core will rise to the point where the helium starts fusing into carbon, plus a little bit of oxygen via the first step on the alpha ladder. At this point, the Sun will leave the red giant phase and join the horizontal branch of the Hertzsprung–Russell diagram. This is a rather short-lived phase of a star's life. The carbon and oxygen produced by helium fusion quickly (in stellar timeframes) form an inert core. At that point, our sun will become an asymptotic red giant.&lt;/p&gt;&#xA;&lt;p&gt;The red giant and asymptotic red giant phases are rather messy affairs, wracked by convulsions where the star expels lots of gas. Our Sun will lose about half its mass to such convulsions. Molecules do form when this expelled gas cools. This results in some of the prettiest pictures in astronomy, shown below.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BZGtW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BZGtW.jpg&quot; alt=&quot;&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;https://inspirehep.net/record/1083878/files/figs_f03.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;inspirehep.net&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="7472" LastEditDate="2020-12-19T12:06:29.243" LastActivityDate="2020-12-19T12:06:29.243" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="12383" PostTypeId="2" ParentId="12381" CreationDate="2015-11-04T07:54:43.713" Score="7" Body="&lt;p&gt;The answer is yes. Molecule formation is common in the outer photospheres of cool stars, and those molecules frequently contain oxygen. Obvious and common examples are TiO, VO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This chemistry almost entirely happens when temperatures fall below 5000K, because otherwise the molecules are dissociated. Therefore it never occurs in stellar interiors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun will form molecules (involving O) in the later stages of its life, first in its atmosphere when it becomes a red giant (in small quantities). The oxygen in these molecules was not &lt;em&gt;produced&lt;/em&gt; in the Sun, it was already there. More extensive molecule formation happens when it ascends the Asymptotic Giant Branch, prior to losing about half its envelope to thermal pulsations and a strong stellar wind. Here, some of the oxygen was made in a helium burning shell inside the Sun and transported to the surface by convective mixing.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-11-04T08:08:29.233" LastActivityDate="2015-11-04T08:08:29.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12384" PostTypeId="1" AcceptedAnswerId="25255" CreationDate="2015-11-04T07:54:56.693" Score="9" ViewCount="297" Body="&lt;p&gt;Just because we observed that our milky way galaxy is spinning in a certain direction therefore we assume it is applicable to all other galaxies, I am curious to find out if hypothetically most of the celestial objects such as natural satellites, planets, stars even galaxies within our observable universe are spinning/rotating in the same direction as the milky way what kind of implications can we say about the condition in the early universe? or is it just an coincident? &lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="9301" LastEditDate="2016-06-29T04:37:42.923" LastActivityDate="2018-02-23T20:03:41.603" Title="Are there equal number of planets, stars, galaxies etc in observable universe spinning in both directions?" Tags="&lt;planet&gt;&lt;galaxy&gt;&lt;star-systems&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12385" PostTypeId="1" AcceptedAnswerId="13263" CreationDate="2015-11-04T09:06:12.437" Score="6" ViewCount="361" Body="&lt;p&gt;&lt;a href=&quot;http://www.telegraph.co.uk/news/science/space/5711129/Uranium-could-be-mined-on-the-Moon.html&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt; talks about the possibility of mining Uranium on the moon. Since the Moon lacks the geological forces that have created veins of concentrated minerals on Earth, would extraterrestrial mining be more difficult or impractical? How does Moon, Martian, or asteroid mining overcome the lack of mineral veins otherwise formed by Earth's geology? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like extraterrestrial mining would be at least as difficult as trying to refine desert sand, which isn't a thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/HOjWRGgBpyU&quot; rel=&quot;nofollow&quot;&gt;A short video on how gold veins form&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wayfaring Stranger added an interesting link on &lt;a href=&quot;https://en.wikipedia.org/wiki/Ore_genesis#Uranium&quot; rel=&quot;nofollow&quot;&gt;ore genesis.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7973" LastEditorUserId="37191" LastEditDate="2021-03-07T03:37:58.143" LastActivityDate="2021-03-07T03:37:58.143" Title="Is extraterrestrial mining more difficult or impractical for bodies without plate tectonics?" Tags="&lt;the-moon&gt;&lt;planet&gt;&lt;asteroids&gt;&lt;geology&gt;&lt;terrestrial-planets&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12386" PostTypeId="2" ParentId="12363" CreationDate="2015-11-04T10:16:30.520" Score="1" Body="&lt;p&gt;Here's a very roughly calculated partial answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A first generation star and solar-system would obviously not be a good candidate for life or even planets cause you can't do much with mostly hydrogen and helium.   You can't even build an ocean with mostly those 2 elements.  Stars do just fine, but planets don't.  Gas Giants and Suns only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://scienceblogs.com/startswithabang/2013/07/05/why-did-the-universe-start-off-with-hydrogen-helium-and-not-much-else/&quot; rel=&quot;nofollow noreferrer&quot;&gt;After the big bang:&lt;/a&gt; 75%-76% Hydrogen by mass, 24-25% Helium (with a teeny-tiny trace amount of Lithium).   I'm going to assume 76% / 24% for rough calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using our solar-system and the Milky-way as a model, elements by mass:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements#Abundance_of_elements_in_the_Universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;Milky-Way&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;73.9% Hydrogen&lt;/p&gt;&#xA;&#xA;&lt;p&gt;24.0% Helium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.04% Oxygen&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.46% Carbon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.13% Neon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.11% Iron&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.10% Nitrogen&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.07% Silicon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.06% Magnesium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.04% Sulfur&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smaller amounts of Argon, Calcium, Nickel, Sodium, Phosphorus, Chlorine, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/tables/suncomp.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;composition of the Sun&lt;/a&gt; by mass (pretty similar except for more Helium)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;71.0% Hydrogen&lt;/p&gt;&#xA;&#xA;&lt;p&gt;27.1% Helium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.97% Oxygen&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.40% Carbon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.06% Neon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.14% Iron&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.10% Nitrogen&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.10% Silicon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.08% Magnesium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.04% Sulfur&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these numbers as a rough guideline, if 1.9% to 2.1% of the baryonic mass of the milky-way converts from Hydrogen to Carbon or heavier elements after (about) 2 generations, and we extrapolate a 1% increase in all Carbon or heavier elements to each stellar generation or every 6 billion years, then we can begin to make an estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Helium percentage could stay somewhat constant at about 24% cause it's both consumed and created stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your original question, how many generations, if we figure 1% increase in heavier elements (Carbon or higher) per stellar generation, 10 generations we're still looking at, very roughly, (24% helium), 66% hydrogen, 5% oxygen, 2%-3% carbon, 2%-3% other elements.  It still looks like a workable scenario to me though the stars, being made of some denser elements might have denser cores and as a result, could burn hydrogen a bit faster.    Past a certain point the ratios probably stop being ideal but I see no reason why quite a few generations couldn't still work create life-friendly solar-systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our sun is expected to cast off about half it's matter into a planetary nebula before it settles down into a white dwarf and most of that cast off matter will be hydrogen and some helium.   Some time later if our sun as a white dwarf accretes enough to meet the Chandrasekhar mass and goes Type-1 Nova, it will again cast off a significant percentage of it's matter, so, to put it simply, Stars recycle a pretty good percentage of their hydrogen and helium, which is kinda cool.   If you see Rob Jeffries' brilliant answer &lt;a href=&quot;https://astronomy.stackexchange.com/questions/12205/how-massive-does-a-main-sequence-star-need-to-be-to-go-type-1-nova&quot;&gt;here&lt;/a&gt;, stars of 7-9 solar masses will eject an even higher percentage of their elements before they go white dwarf, as much as 85% and most of what gets ejected is the lighter elements, hydrogen and helium from the outer layers.   Even smaller red-dwarf stars expel a percentage of their elements due to more active solar flares and lower gravity throughout their lives.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add to this, the possibility of continued merger of the galaxies and dwarf galaxies in our &lt;a href=&quot;https://en.wikipedia.org/wiki/Local_Group&quot; rel=&quot;nofollow noreferrer&quot;&gt;local group&lt;/a&gt;, and life capable solar-systems could continue to form for quite some time, perhaps several tens of billions of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer isn't meant to be definitive, but only a rough estimate.  Corrections welcome.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-11-04T10:16:30.520" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12390" PostTypeId="1" CreationDate="2015-11-04T12:07:25.027" Score="3" ViewCount="41" Body="&lt;p&gt;Molecular oxygen O2 has been found on comet 67P/C-G in a ratio of 3.8% to water, which is much higher than expected. An explanation proposed is that the Solar System formed from a molecular cloud warmer at 45 K, than typical molecular clouds at 10 K. I speculate this could have had many kinds of consequences for the Solar System and wonder what the most important or easiest to detect would be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, my uneducated guess would be that more hydrogen escapes from warmer clouds. Exoplanets with low density could then be the result of higher hydrogen abundance, more frequently than here giving them large hydrogen atmospheres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nature.com/news/rosetta-sniffs-oxygen-around-comet-67p-1.18658&quot; rel=&quot;nofollow noreferrer&quot;&gt;News article&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.abc.net.au/science/audio/2015/11/04/4343809.htm?topic=space&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ten minutes radio interview&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Haven't found any paper)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/T7K2B.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/T7K2B.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A star forming molecular cloud, for decoration.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-11-04T12:07:25.027" Title="What are the implications if the Sun was formed in a warm nebula?" Tags="&lt;temperature&gt;&lt;formation&gt;&lt;nebula&gt;&lt;hydrogen&gt;&lt;67p&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12391" PostTypeId="2" ParentId="12374" CreationDate="2015-11-04T13:29:22.597" Score="1" Body="&lt;p&gt;I think the outer edge of an accretion disk is not well-defined, and observationally the radius will depend on which wavelength you consider, since the farther you get from the BH, the softer the radiation will be. But if you look in the UV, then &lt;a href=&quot;http://arxiv.org/abs/0707.0305&quot; rel=&quot;nofollow&quot;&gt;Morgan et al. (2010)&lt;/a&gt; find the following relation between $R_{2500}$ (the radius when observed at $\lambda = 2500$ Å) and the mass $M_\mathrm{BH}$ of the black hole:&#xA;$$&#xA;\log\left( \frac{R_{2500}}{\mathrm{cm}}\right) = 15.78 + 0.80&#xA;\log\left( \frac{M_\mathrm{BH}}{10^9M_\odot}\right),&#xA;$$&#xA;(modulo some uncertainties that you can look up in the paper).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, if your BH has a mass of $10^8 M_\odot$, its radius will be $R_{2500}\sim64\,\mathrm{AU}$, or roughly 1/3 lightdays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparison, its Schwarzschild radius is $\sim2\,\mathrm{AU}$, so your estimate was actually pretty good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This result is consistent with &lt;a href=&quot;http://arxiv.org/pdf/1501.05951v2.pdf&quot; rel=&quot;nofollow&quot;&gt;Edelson et al. (2015)&lt;/a&gt;, who find 0.35 lightdays, also in the UV. However, in you look in longer wavelengths, the disk is much, much larger. If you're interested beyond your homework assignment, take a look at the accretion disk theory review by &lt;a href=&quot;http://arxiv.org/abs/1203.6851&quot; rel=&quot;nofollow&quot;&gt;Armijo (2013)&lt;/a&gt;, who shows that in the radio regime, the disk is thousands of AU, and even up to ~100 pc.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-11-06T08:38:47.600" LastActivityDate="2015-11-06T08:38:47.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12392" PostTypeId="1" CreationDate="2015-11-04T16:50:35.517" Score="3" ViewCount="341" Body="&lt;p&gt;I'm trying to calculate the longitude on the earth where it's noon at some time. (That is, the longitude which is coplanar with the plane defined by the sun and the earth's axis.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my python code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from math import sin, cos, tan, atan, pi&#xA;def sun_longitude(when):&#xA;    &quot;&quot;&quot;Given time in ms since 1/1/1970, return the longitude the sun is over at that moment&quot;&quot;&quot;&#xA;    # https://en.wikipedia.org/wiki/Position_of_the_Sun&#xA;    jdn = 2440587.5 + when / (1000.0 * 3600 * 24)&#xA;    n = jdn - 2451545.0 # 1/1/2000&#xA;    L = (280.460 + 0.9856474 * n) % 360.0&#xA;    g = (357.528 + 0.9856004 * n) % 360.0&#xA;    degtorad = 2.0 * pi / 360.0&#xA;    lambda_ = L + 1.915 * sin(g * degtorad) + 0.020 * sin(2 * g * degtorad)&#xA;&#xA;    # https://en.wikipedia.org/wiki/Axial_tilt#Short_term&#xA;    T = n / (365 * 100.0) # julian centuries from 2000&#xA;    # ε = 23° 26′ 21.406″ − 46.836769″ T − 0.0001831″ T2&#xA;    epsilon = 23.4392794 - 0.780612817 * T - 5.0861e-8 * T * T;&#xA;    alpha = atan(cos(epsilon * degtorad) * tan(lambda_ * degtorad)) / degtorad&#xA;&#xA;    # https://en.wikipedia.org/wiki/Sidereal_time&#xA;    # https://en.wikipedia.org/wiki/Hour_angle&#xA;    GMST = (18.697374558 + 24.06570982441908 * n) % 24.0&#xA;    print &quot;jdn = {jdn}, n = {n}, L = {L}, g = {g}, lambda_ = {lambda_}, T = {T}, epsilon = {epsilon}, alpha = {alpha}, GMST = {GMST}&quot;.format(**locals())&#xA;    LHA = GMST * 360/24 + alpha;&#xA;&#xA;    return LHA&#xA;&#xA;from datetime import datetime&#xA;def sun_long_from_str(whenstr):&#xA;    when = datetime.strptime(whenstr, &quot;%Y-%m-%d %H:%M&quot;)&#xA;    secs = (when - datetime(1970, 1, 1)).total_seconds()&#xA;    return sun_longitude(secs * 1000.0)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm aware that I'm not correcting for the correct quandrant in figuring alpha, but I have errors elsewhere; for example, for noon GMT Jan 1, 2000, I'm getting 279° rather than close to 0, which is what I'd expect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't giving me correct answers, and I'm at a bit of a loss for how to debug it. Can anyone find my mistakes or point me to some reasonable sample code or worked-through example for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because, once I get the algorithm right, I'll be reimplementing for an embedded device, I can't just use an implementing package, and I haven't found a library which is straightforward enough for me to understand how I can implement this specific question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for a couple of error corrections. Now, when I run it, I get the following for these examples:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2000-01-01 12:00:&#xA;  jdn = 2451545.0, n = 0.0, L = 280.46, g = 357.528, lambda_ = 280.375680197, T = 0.0, epsilon = 23.4392794, alpha = -78.7141369122, GMST = 18.697374558&#xA;  result: 201.74648145775257&#xA;&#xA;2000-03-20 07:35:&#xA;  jdn = 2451623.81597, n = 78.815972222, L = 358.144758099, g = 75.2090537484, lambda_ = 360.006175505, T = 0.00215934170471, epsilon = 23.4375937902, alpha = 0.00566598789588, GMST = 19.4596915825&#xA;  291.9010397253072&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As you can see, the first query (noon on Jan 1, 2000) now has a correct Julian day number. At noon GMT, I would expect the sun to be above a longitude near 0, so 201 is incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second query is the time of the spring equinox in 2000, which I expected to result in zeroing out the sidereal part of the calculation, but it does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I attempted to consult the NASA HORIZONS web interface for ephemerides for the sun on 1/1/2000 at noon GMT, for both &quot;Geocentric&quot; and Greenwich locations, but I don't know how to evaluate the result and compare it to my work above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geocentric:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; Date__(UT)__HR:MN     R.A._(ICRF/J2000.0)_DEC  APmag  S-brt            delta      deldot    S-O-T /r    S-T-O&#xA;**************************************************************************************************************&#xA; 2000-Jan-01 12:00     18 45 09.36 -23 01 59.7 -26.78 -10.59 0.98332762653520  -0.0127281   0.0000 /?   0.0000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Greenwich:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; Date__(UT)__HR:MN     R.A._(ICRF/J2000.0)_DEC  APmag  S-brt            delta      deldot    S-O-T /r    S-T-O&#xA;**************************************************************************************************************&#xA; 2000-Jan-01 12:00 *m  18 45 09.36 -23 02 08.3 -26.78 -10.59 0.98331613178086  -0.0166655   0.0000 /?   0.0000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thanks again for any help you can provide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edited to correct Julian day calculation and use of degrees vs. radians for alpha and to add examples]&lt;/p&gt;&#xA;" OwnerUserId="9669" LastEditorUserId="9669" LastEditDate="2015-11-05T15:48:54.230" LastActivityDate="2015-11-05T18:51:06.643" Title="Math for calculating the terrestrial longitude directly under the sun with time" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;coordinate&gt;&lt;software&gt;&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12393" PostTypeId="2" ParentId="12392" CreationDate="2015-11-04T19:47:53.567" Score="2" Body="&lt;p&gt;There are two things I can see in this code:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the Julian date is measured from Midday, whereas the unix epoch is measured from midnight. &lt;code&gt;jdn = 2440587.5 + when / (1000.0 * 3600 * 24)&lt;/code&gt; should be the correct expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, &lt;code&gt;alpha = atan(cos(epsilon * degtorad) * tan(lambda_ * degtorad))&lt;/code&gt;&#xA;calculates a right ascension in radians, you should convert this to degrees for the last part of the calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;&#xA;You do need to fix the quadrant, for example at when n==0, you should get a right ascension of 281.285. However the last error is in the calcualtion of the longitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relevant equation is $LHA = GMST -\lambda - \alpha$. When the sun is on the meridian, LHA = 0, and (rearranging) $\lambda = GMST -\alpha$. Thus the line &lt;code&gt;LHA = GMST * 360/24 + alpha&lt;/code&gt; should be &lt;code&gt;longitude = GMST*360/24 - alpha&lt;/code&gt;. (and return &lt;code&gt;longitude&lt;/code&gt;) If you do this with n=0, you get a longitude of -1.5 degrees. The sun is at meridian at noon on Jan 1 2000, if you are 1.5 degrees west of Greenwich.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You seem to be attempting a very accurate ephemerides for the sun. You can check you accuracy against &lt;a href=&quot;http://rhodesmill.org/pyephem/index.html&quot; rel=&quot;nofollow&quot;&gt;PyEphem&lt;/a&gt;, or the &lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow&quot;&gt;Nasa Horizon ephemerides&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2015-11-05T18:51:06.643" LastActivityDate="2015-11-05T18:51:06.643" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12394" PostTypeId="5" CreationDate="2015-11-04T20:16:58.493" Score="0" Body="&lt;p&gt;A white dwarf in a binary system can pull hydrogen from its binary partner onto its surface. When a certain amount has built up, nuclear fusion can start in this Hydrogen layer, which causes the white dwarf to suddenly increase in brightness, and can give the appearance of a &quot;New star&quot; (hence the name)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The white dwarf is not destroyed in the nova, and one white dwarf can give rise to recurrent novae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Novae are distinct from supernovae, which can occur when a white dwarf has built up sufficient mass for the entire star to restart fusion. Supernovae are many magnitudes more powerful than novae, and result in the destruction of the white dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most novae that occur in the Milky way are too distant and faint to be seen with the naked eye, but occasionally a nearby nova can be seen. The last bright nova was in 1975 in the constellation of cygnus, and was as of the second magnitude at peak brightness.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2015-11-04T21:50:43.523" LastActivityDate="2015-11-04T21:50:43.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12395" PostTypeId="4" CreationDate="2015-11-04T20:16:58.493" Score="0" Body="A nova is a type of explosive variable star. A white dwarf which undergoes sudden and explosive brightening. It is distinct from a supernova, which is much more powerful." OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2015-11-04T21:50:58.473" LastActivityDate="2015-11-04T21:50:58.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12396" PostTypeId="5" CreationDate="2015-11-04T20:34:12.500" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-11-04T20:34:12.500" LastActivityDate="2015-11-04T20:34:12.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12397" PostTypeId="4" CreationDate="2015-11-04T20:34:12.500" Score="0" Body="M31 is the Andromeda Galaxy. A spiral galaxy and the largest member of the local group of galaxies. M31 is currently about 2.2 million light years from the Milky Way, but is approaching and is expected to collide and merge in about 4 billion years." OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2015-11-04T21:51:01.550" LastActivityDate="2015-11-04T21:51:01.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12398" PostTypeId="2" ParentId="12039" CreationDate="2015-11-04T23:49:33.533" Score="5" Body="&lt;p&gt;I can't give a detailed answer; the details are buried in the depths of numerical stellar evolution models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing that changes most with the metallicity of a newborn star is the radiative opacity of the gas. Higher metallicity leads to more opacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has two immediate effects - it makes energy harder to get out of the stellar interior and makes it more likely that convection will take over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Convection has the property of mixing up all the material within the convective zone. This can have knock on effects as to how long each nuclear burning phase lasts and how much material is consumed. It also mixes synthesised material from the interior outwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A further important effect is that mass loss from massive stars is very extensive and is due to radiatively accelerated winds. For a given luminosity, high metallicity gas is more opaque and easier to accelerate. Hence mass loss is very sensitive to metallicity and determines how massive the star is as it reaches the end of its life. This in turn has a large bearing on what the remnant will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a further feedback in that the wind metallicity is the metallicity at the surface, but this in turn can be affected by interior mixing that is in turn metallicity-dependent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that sounds complicated, that's because it is, and detailed numerical models are required to see how these things play out.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-11-04T23:49:33.533" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12399" PostTypeId="1" AcceptedAnswerId="12403" CreationDate="2015-11-05T04:26:45.733" Score="1" ViewCount="190" Body="&lt;p&gt;Hot stars like &lt;a href=&quot;http://en.wikipedia.org/wiki/O-type_star&quot; rel=&quot;nofollow&quot;&gt;O-type stars&lt;/a&gt; show no hydrogen in their &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_classification&quot; rel=&quot;nofollow&quot;&gt;spectra&lt;/a&gt;. Does this mean they are made entirely of helium?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any explanation would be really helpful.&lt;/p&gt;&#xA;" OwnerUserId="9696" LastEditorUserId="2153" LastEditDate="2015-11-05T16:38:46.870" LastActivityDate="2015-11-05T16:38:46.870" Title="Are hot stars like O-type stars entirely composed of helium?" Tags="&lt;star&gt;&lt;spectra&gt;&lt;hydrogen&gt;&lt;helium&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12401" PostTypeId="1" CreationDate="2015-11-05T08:23:04.000" Score="5" ViewCount="6679" Body="&lt;p&gt;If the universe is expanding at an accelerating rate, such that the galaxies' moving away from each other is accelerated, then time should also slow down.&#xA;And when universe will accelerate to the speed of light, then time should stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does time slow down because universe is expanding at an accelerating rate?&#xA;Can any physicist explains if this is true or not?&lt;/p&gt;&#xA;" OwnerUserId="9698" LastEditorUserId="9186" LastEditDate="2015-11-05T19:10:51.100" LastActivityDate="2019-10-20T15:23:09.120" Title="Does time slow down because the universe is expanding at an accelerating rate?" Tags="&lt;universe&gt;&lt;space-time&gt;&lt;cosmological-inflation&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12403" PostTypeId="2" ParentId="12399" CreationDate="2015-11-05T09:58:35.260" Score="4" Body="&lt;p&gt;The lines that appear in a stars spectrum mainly reflects its temperature not its composition, &lt;a href=&quot;http://www.star.ucl.ac.uk/~pac/spectral_classification.html&quot; rel=&quot;nofollow&quot;&gt;see here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;O-type stars start out with the same sort of &lt;a href=&quot;https://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements#Abundance_of_elements_in_the_Universe&quot; rel=&quot;nofollow&quot;&gt;composition as other stars&lt;/a&gt;, that is they are mainly H and He (approximately 75% and 25% by mass) with traces heavier elements.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-11-05T09:58:35.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12404" PostTypeId="1" AcceptedAnswerId="12508" CreationDate="2015-11-05T11:11:57.613" Score="4" ViewCount="181" Body="&lt;p&gt;I was just wondering what are some of the stars, constellations or any other things that one can just identify by looking up at the night sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, when ever i look up, I immediately see Orion and can identify Betelgeuse, Rigel, Bellatrix. Apart from that i can identify some of the bright stars liike sirius , vega.&#xA;So, what other stars are there that one can easily identify and point?&lt;/p&gt;&#xA;" OwnerUserId="7757" LastActivityDate="2015-11-20T17:35:34.587" Title="What are the stars/constellations a beginner/enthusiast can easily identify?" Tags="&lt;star&gt;&lt;amateur-observing&gt;&lt;constellations&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12405" PostTypeId="2" ParentId="12401" CreationDate="2015-11-05T11:13:26.120" Score="3" Body="&lt;p&gt;In general, better refrain from using 'inflation' unless you're talking about the early Universe-inflation. Rather, the Universe is &lt;em&gt;expanding&lt;/em&gt; at an accelerated rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crux of this question seems to be in what you call 'the' time. There is no such thing as 'the' time, and I'm not sure what definition of time you're expecting to stop.&#xA;Time is perfectly well defined in any restframe and will not be affected by the overall expansion of the Universe, at whatever rate.&lt;/p&gt;&#xA;" OwnerUserId="6965" LastActivityDate="2015-11-05T11:13:26.120" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12407" PostTypeId="1" AcceptedAnswerId="12442" CreationDate="2015-11-05T16:04:11.147" Score="1" ViewCount="434" Body="&lt;p&gt;Could a planet's main use for natural satellites be that of a photon sail?&lt;a href=&quot;https://i.stack.imgur.com/ehRuF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ehRuF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9326" LastEditorUserId="9326" LastEditDate="2015-12-16T11:26:14.763" LastActivityDate="2015-12-16T11:26:14.763" Title="Does the Moon capture radiation pressure from the sun causing momentum from photon propulsion?" Tags="&lt;the-moon&gt;&lt;the-sun&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="13" ClosedDate="2015-11-09T16:46:47.507" ContentLicense="CC BY-SA 3.0" />
  <row Id="12408" PostTypeId="1" AcceptedAnswerId="12411" CreationDate="2015-11-05T16:07:41.000" Score="1" ViewCount="19509" Body="&lt;p&gt;According to Newton, gravity is the pulling - or in fact the attracting - force of any heavenly body towards any object to its center. But, to the contrary, Einstein once said that the four dimensions of space and time push the object downwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can which one of them is correct?&lt;/p&gt;&#xA;" OwnerUserId="9701" LastEditorUserId="2153" LastEditDate="2015-11-05T21:03:46.263" LastActivityDate="2019-10-02T22:46:38.503" Title="Gravitation - Pulling or Pushing force?" Tags="&lt;gravity&gt;&lt;general-relativity&gt;&lt;newtonian-gravity&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12409" PostTypeId="2" ParentId="12408" CreationDate="2015-11-05T16:36:12.357" Score="3" Body="&lt;p&gt;I'm guessing that this misunderstanding is a result of the oft-used &lt;a href=&quot;http://theory.uwinnipeg.ca/users/gabor/black_holes/slide5.html&quot; rel=&quot;nofollow&quot;&gt;rubber sheet analogy&lt;/a&gt;. The rubber sheet analogy says that, according to general relativity, mass curves space-time like a heavy bowling ball on a near-taut blanket (or rubber sheet) curves the blanket/sheet. This resulting curve makes &lt;em&gt;other&lt;/em&gt; bits of matter/energy move in different ways. I'm guessing that this is your confusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rubber sheet analogy fails massively in one area, any demonstration of it involves gravity on Earth. If I use a bowling ball to deform a sheet, and then role a golf ball along the sheet nearby, the golf ball will move a bit towards the bowling ball because of the force of gravity around me - not &quot;gravity&quot; in the simulation. It thus makes it seem like gravity pulls the golf ball &quot;down&quot; because the bowling ball pulls the rubber sheet &quot;down&quot;. This is the result of using a two-dimensional analogy of a three-dimensional universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is, there is no &quot;pushing&quot; going on in the general relativistic model of gravity. Gravity is attractive (so long as the strong energy conditions holds for the object in question), just like Newton postulated.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-11-05T21:05:17.553" LastActivityDate="2015-11-05T21:05:17.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12410" PostTypeId="1" AcceptedAnswerId="14526" CreationDate="2015-11-05T18:51:05.437" Score="5" ViewCount="93" Body="&lt;p&gt;Colliding galaxies sometimes merge and sometimes pass through each other. Either way, there are huge changes as a result. What are the parameters that matter in determining whether a collision will result in a merger or a hit and run, or something in between? What will be the result of the Milky Way-Andromeda collision?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking about this some more, I suppose the most clearcut distinctions for the result would be whether the two supermassive black holes end up unbound, bound, or merged. Would it be possible for the bulk of the two galaxies to merge, but one or two of the SMBH's go on their merry way?&lt;/p&gt;&#xA;" OwnerUserId="9186" LastEditorUserId="9186" LastEditDate="2015-11-06T01:16:52.980" LastActivityDate="2016-04-13T09:57:28.130" Title="What parameters determine whether galaxies colliding will result in a merger or a hit and run?" Tags="&lt;milky-way&gt;&lt;supermassive-black-hole&gt;&lt;galactic-dynamics&gt;&lt;impact&gt;&lt;m31&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12411" PostTypeId="2" ParentId="12408" CreationDate="2015-11-05T19:08:00.887" Score="3" Body="&lt;p&gt;In general relativity, gravity neither pushes nor pulls. To explain why ball travels in an arc you note the start and end points of the throw in 4d space time (3 space co-ordinates and 1 time coordinate) You then find the shortest path between these two 4d points in the curved spacetime surrounding the Earth. This shortest path is the path in spacetime that the ball travels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in General relativity, gravity is not seen as being a force, instead it is the result objects travelling in the most direct way in a region of curved spacetime. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However for nearly all practical purposes, the effect of gravity in relativity is amost identical to that produced by an attractive force, as supposed by Newton.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-11-05T19:08:00.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12412" PostTypeId="2" ParentId="12401" CreationDate="2015-11-05T20:00:51.323" Score="6" Body="&lt;p&gt;Yes, time does run slower for far-away objects, as observed from our point of view; this is a prediction of general relativity. And yes, because expansion accelerates, this time dilation slowly, very slowly, becomes more pronounced (this would happen even if the expansion didn't &lt;em&gt;accelerate&lt;/em&gt;, but just continued at the same rate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This time dilation  is a well-known effect, and is always taken into account when doing observations. For instance, when observing distant supernovae, one is often interested in how their luminosities decrease as a function of time. This is called their &lt;em&gt;lightcurve&lt;/em&gt;. In order to compare lightcurves at different redshifts, they are usually converted to their restframe, i.e. how they would look if you were &quot;standing next to the supernova&quot; (e.g. &lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2001ApJ...558..359G&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goldhaber et al 2001&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, time dilation does not work exactly as you seem to think. At galaxy at a redshift of &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; has its time dilated by a factor of &lt;span class=&quot;math-container&quot;&gt;$1+z$&lt;/span&gt;, so time runs twice as slow for a galaxy at, say, &lt;span class=&quot;math-container&quot;&gt;$z=3$&lt;/span&gt; than for a galaxy at &lt;span class=&quot;math-container&quot;&gt;$z=1$&lt;/span&gt;. Galaxies with redshifts larger than &lt;span class=&quot;math-container&quot;&gt;$z\sim1.5$&lt;/span&gt; recede faster than the speed of light, and time does not at all stop here. Only for &lt;span class=&quot;math-container&quot;&gt;$z\rightarrow\infty$&lt;/span&gt;, i.e. at the beginning of time at Big Bang, does the time dilation approach infinity.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2019-10-20T15:23:09.120" LastActivityDate="2019-10-20T15:23:09.120" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="12413" PostTypeId="2" ParentId="12374" CreationDate="2015-11-05T21:35:23.447" Score="1" Body="&lt;p&gt;$10^{8} M_{\odot}$ SMBH.  Eddington Luminosity (dodgy estimate, since assumes &lt;em&gt;spherical&lt;/em&gt; accretion) is $ L/L_{\odot} \simeq 3\times 10^{4} (M/M_{\odot}) = 3\times 10^{12} L_{\odot}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume we are seeing all this emerge from the face of a flat disc of total area (front and back) $\pi R^2$ and temperature $T \simeq 10^{5}$ K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume black body emission, so $L = 2\pi R^2 \sigma T^4$. And thus&#xA;$$ R \simeq \left[\frac{3\times 10^{12} L_{\odot}}{2 \pi \sigma T^4}\right]^{1/2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting in the numbers I get $R \simeq  40$ au. However it is very sensitive to the assumed temperature (&lt;em&gt;much smaller&lt;/em&gt; for $T = 10^{6}$ K).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-11-05T21:35:23.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12414" PostTypeId="1" CreationDate="2015-11-05T21:44:32.293" Score="2" ViewCount="412" Body="&lt;p&gt;I know the luminosity classes are: Ia-0 ( Hypergiants ), Ia ( bright supergiants ), ... , VII ( white dwarf ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also have learned that you can use the presence of absorption lines ( ie. use spectral profiles ) to determine which spectral type a star belongs to: presence of helium lines imply O-type, and Hydrogen lines in A-type, getting fainter as the star gets cooler, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was told that one can use spectral profiles to determine luminosity class ( Ia-0, Ia, Ib, ... ), that the width of the spectral lines indicate the luminosity class: Ia stars have sharp lines, and V stars have broad lines...  How about Ib, II, III, IV, VI, VII?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean... how can I distinguish between Ib and II?  Given examples of Ia lines and V lines, you can tell whether an additional line is broader or sharper, but then guess that this additional line is Ib, II, III, or IV?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there a more concrete method of determining luminosity class from spectral profiles?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9703" LastActivityDate="2015-11-05T21:44:32.293" Title="How to use spectral profiles to determine luminosity class?" Tags="&lt;spectroscopy&gt;&lt;luminosity&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12416" PostTypeId="1" AcceptedAnswerId="12438" CreationDate="2015-11-05T23:36:33.030" Score="4" ViewCount="4182" Body="&lt;p&gt;I know all about how black holes form and why their gravity is so strong. However, is the gravity equally powerful in all directions? Will the event horizon appear as a &lt;strong&gt;perfect&lt;/strong&gt; sphere?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="1378" LastEditDate="2015-11-06T22:37:51.703" LastActivityDate="2019-01-15T16:57:58.203" Title="Is a black hole a perfect sphere?" Tags="&lt;gravity&gt;&lt;black-hole&gt;&lt;light&gt;&lt;stellar-evolution&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12417" PostTypeId="1" AcceptedAnswerId="12420" CreationDate="2015-11-06T01:26:54.183" Score="3" ViewCount="2405" Body="&lt;p&gt;When a star ignites ( ie. fusion starts ), the star maintains its form by balancing gravity's inward pressure, and radiation's outward pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get that the fusion of hydrogen atoms releases energy... fine...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does gravity keep it together if the mass is lessening as a result of fusion( mass being converted into energy from fusion) while gravity is weakening( as mass lessens )?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wouldn't the radiation overpower the force of gravity and tear the star apart?&lt;/p&gt;&#xA;" OwnerUserId="9703" LastActivityDate="2015-11-07T20:18:32.210" Title="Why does gravity increase in star formation?" Tags="&lt;gravity&gt;&lt;star-formation&gt;&lt;radiation&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12418" PostTypeId="1" AcceptedAnswerId="12434" CreationDate="2015-11-06T01:40:04.483" Score="6" ViewCount="1585" Body="&lt;p&gt;I am thinking about making a telescope. I have a 100cm focal length lens which I can use as objective lens in my telescope. So which is the best focal length I can use as eyepiece in my astronomical telescope?&lt;/p&gt;&#xA;" OwnerUserId="9705" LastEditorUserId="1378" LastEditDate="2015-11-06T21:44:57.170" LastActivityDate="2020-03-29T12:22:06.880" Title="Astronomical telescope making" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;optics&gt;&lt;refractor-telescope&gt;&lt;telescope-making&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12419" PostTypeId="1" AcceptedAnswerId="12428" CreationDate="2015-11-06T02:59:34.427" Score="1" ViewCount="434" Body="&lt;p&gt;Specifically, I have often wondered whether gravitational lensing might be able to (theoretically) be used in order to see light reflecting off the Earth that has zoomed off into space and been bent by one object after another until it comes full circle back to Earth again. Given the distance the light would have travelled, could we not use this technique to peer back through time at our own planet?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is it reasonable that this could happen given the size of the universe and the mass and distribution of objects within it?&lt;/li&gt;&#xA;&lt;li&gt;Given the restraints above, would be be able to estimate how far back in time we are looking?&lt;/li&gt;&#xA;&lt;li&gt;Given the above, how bright would something have to be in order for its light to travel this distance and still be detectable?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9707" LastEditorUserId="2365" LastEditDate="2015-11-06T20:37:49.117" LastActivityDate="2018-08-02T08:27:20.440" Title="Could the Earth use gravitational lensing / bending of light to see it's own bottom?" Tags="&lt;space-time&gt;&lt;gravitational-lensing&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12420" PostTypeId="2" ParentId="12417" CreationDate="2015-11-06T03:25:25.913" Score="3" Body="&lt;h2&gt;I am going to start with &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_fusion#Nuclear_fusion_in_stars&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paragraph from Wikipedia&lt;/a&gt; &lt;em&gt;(emphasis mine)&lt;/em&gt;:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The most important fusion process in nature is the one that powers&#xA;  stars. In the 20th century, it was realized that the energy released&#xA;  from nuclear fusion reactions accounted for the longevity of the Sun&#xA;  and other stars as a source of heat and light. The fusion of nuclei in&#xA;  a star, starting from its initial hydrogen and helium abundance,&#xA;  provides that energy and synthesizes new nuclei as a byproduct of that&#xA;  fusion process. &lt;strong&gt;&lt;em&gt;The prime energy producer in the Sun is the fusion of&#xA;  hydrogen to form helium&lt;/em&gt;&lt;/strong&gt;, which occurs at a solar-core temperature of&#xA;  14 million kelvin. The net result is the fusion of four protons into&#xA;  one alpha particle, with the release of two positrons, two neutrinos&#xA;  (which changes two of the protons into neutrons), and energy.&#xA;  Different reaction chains are involved, depending on the mass of the&#xA;  star. For stars the size of the sun or smaller, the &lt;strong&gt;&lt;em&gt;proton-proton&#xA;  chain&lt;/em&gt;&lt;/strong&gt; dominates. In heavier stars, the &lt;strong&gt;&lt;em&gt;CNO cycle&lt;/em&gt;&lt;/strong&gt; is more important.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;The &lt;strong&gt;&lt;em&gt;proton-proton chain&lt;/em&gt;&lt;/strong&gt; set of reactions look like this:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PLExf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PLExf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;The &lt;strong&gt;&lt;em&gt;CNO&lt;/em&gt;&lt;/strong&gt; cycle looks like this:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VVxm1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VVxm1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Net Result&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Either way, the net result is 4 protons ($^1\!$H nuclei) are turned into 1 alpha particle ($^4\!$He nucleus) plus 2 positrons (e$^+$). The 2 positrons go on to annihilate 2 electrons, so altogether we have a mass change of $$ \Delta M = M_{\mathsf \alpha} - 2M_{\mathsf e} - 4M_{\mathsf P}\,. $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's find out the fractional change in mass: $$ f_\Delta = \frac{\Delta M}{4M_{\mathsf P}} = \frac{M_{\mathsf \alpha} - 2M_{\mathsf e} - 4M_{\mathsf P}}{4M_{\mathsf P}}\,. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now &lt;a href=&quot;http://physics.nist.gov/cgi-bin/cuu/Value?malsmp&quot; rel=&quot;nofollow noreferrer&quot;&gt;the ratio of the mass of an alpha particle to a proton is $3.9726$&lt;/a&gt;, or $$ M_{\mathsf \alpha} = 3.9726\times M_{\mathsf P}\,. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.nist.gov/cgi-bin/cuu/Value?mpsme&quot; rel=&quot;nofollow noreferrer&quot;&gt;The ratio of the mass of a proton to an electron is $1836.1$&lt;/a&gt;, or $$ M_{\mathsf e} = \frac{M_{\mathsf P}}{1836.1} = 0.0005446\times M_{\mathsf P}\,. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting into the $f_\Delta$ equation, $$ f_\Delta = \frac{3.9726\times M_{\mathsf P} - 0.0011\times M_{\mathsf P} - 4\times M_{\mathsf P}}{4\times M_{\mathsf P}} = \frac{-0.0285}4 = -0.007125 = -0.7125\%\,\,.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So obviously, Even if &lt;em&gt;all&lt;/em&gt; of the hydrogen were converted (only a fraction actually is) the loss of mass to the star would be too negligible to matter.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A more important mass loss for large stars is that from their &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_wind&quot; rel=&quot;nofollow noreferrer&quot;&gt;stellar wind&lt;/a&gt;, which for very large main sequence stars (types O or B) removes a sizable &lt;strong&gt;&lt;em&gt;fraction&lt;/em&gt;&lt;/strong&gt; of the very large star's mass over it's lifetime.&lt;/p&gt;&#xA;" OwnerUserId="9186" LastEditorUserId="9186" LastEditDate="2015-11-07T20:18:32.210" LastActivityDate="2015-11-07T20:18:32.210" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12421" PostTypeId="2" ParentId="12417" CreationDate="2015-11-06T03:39:24.443" Score="-1" Body="&lt;p&gt;As a star runs out of hydrogen fuel, the fusion slows, causing the gravity to overpower the outward force of pressure, thus contracting. Contraction of the star causes high temperature and pressure, to the extent that it is enough to fuse helium into carbon, then the energy released is stronger than the gravity, increasing the size of the star into a red giant. The following paragraph from an article explains this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Over its life, the outward pressure of fusion has balanced against the&#xA;  inward pressure of gravity. Once the fusion stops, gravity takes the&#xA;  lead and compresses the star smaller and tighter. Temperatures&#xA;  increase with the contraction, eventually reaching levels where helium&#xA;  is able to fuse into carbon. Depending on the mass of the star, the&#xA;  helium burning might be gradual or might begin with an explosive&#xA;  flash. The energy produced by the helium fusion causes the star to&#xA;  expand outward to many times its original size.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The amount of mass lost is more due to stellar wind, rather than to fusion. To answer your second question, the pressure will never overcome completely the force of gravity. When a star reaches the iron nickel stage of fusing, it'll stop, unable to go further. This causes a rapid contraction for either a star to go supernova (which is really tearing most of the star apart except for its core), or cool down to a black dwarf.&lt;/p&gt;&#xA;" OwnerUserId="7926" LastActivityDate="2015-11-06T03:39:24.443" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12422" PostTypeId="1" AcceptedAnswerId="12440" CreationDate="2015-11-06T05:21:05.527" Score="3" ViewCount="694" Body="&lt;p&gt;Mars looks reddish and Mercury looks gray because they lack thick atmospheres and we can see their &quot;real color&quot; easily, how about Venus? Is the surface of Venus really yellowish in color?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="2365" LastEditDate="2015-11-06T17:11:03.383" LastActivityDate="2015-11-07T04:35:34.913" Title="What is the color of Venus if it has no atmosphere?" Tags="&lt;venus&gt;&lt;surface&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12423" PostTypeId="2" ParentId="12418" CreationDate="2015-11-06T05:40:43.370" Score="1" Body="&lt;p&gt;There are a few &lt;a href=&quot;http://www.telescopes.com/blogs/helpful-information/18940868-magnification-and-using-eyepieces&quot; rel=&quot;nofollow&quot;&gt;rules of thumb&lt;/a&gt; you will need: Maximum useful magnification $M_{max}\approx D\times 50$ where $D$ is the diameter of the objective  is in inches, and magnification $M=l_{obj}/l_{eye}$, where $l_{obj}$ and $l_{eye}$ are the focal lengths of the objective and eye piece respectively.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-11-06T05:40:43.370" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12424" PostTypeId="2" ParentId="12417" CreationDate="2015-11-06T06:53:35.187" Score="3" Body="&lt;p&gt;Some good answers, I'm going to give kind of summary, cause you touched on a few points.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does gravity increase in star formation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gravitation is a product of a few forces.   Mass, density and, not to be ignored, rotation speed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not actually the fusion process that keeps the sun from contracting, at least, not directly.   It's heat that keeps the star expanded.   That's the balancing act.  High temperature wants to expand, gravity wants to contract.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fusion process is actually pretty slow, which is why stars like our sun have a main sequence of about 10 billion years, and a lot of the heat that a star starts out with is from the heat of formation.  Potential energy gets converted to heat due to the coalescing and condensing of all that matter so stars start out hot, even before fusion begins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, a star in formation can be many times brighter than the star is during it's main sequence due to the high heat of formation.   Here's an &lt;a href=&quot;http://cseligman.com/text/stars/mldiagram.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; that says the forming sun was 200 times brighter than it is now.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Young proto-stars, as a result of conservation of angular momentum, tend to &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_rotation&quot; rel=&quot;nofollow noreferrer&quot;&gt;rotate very fast&lt;/a&gt; and that fast rotation can create a bulge and increases &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_rotation#During_formation&quot; rel=&quot;nofollow noreferrer&quot;&gt;ejection of matter&lt;/a&gt;.   The formation process is pretty chaotic compared to the main sequence stage.  Lots of ejected matter, much bigger solar storms, lots of lheat from formation, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the main sequence stage is underway and rotation is slowed down, then there's more of a balance between heat and gravity mentioned above.  The fusion process continues to add heat to the core of star which the star, convects or conduct heat away from the core into the outer layers and then, radiates from it's surface, but during the main sequence, in general, the core of the star gradually heats up and in most cases, the energy added from fusion isn't nearly strong enough to blow apart the star, unless the star is enormously large like over 150 or 200 solar masses, then the star doesn't really work without blowing off a bunch of matter.  See: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/328/is-there-a-theoretical-maximum-size-limit-for-a-star&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I get that the fusion of hydrogen atoms releases energy... fine...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;How does gravity keep it together if the mass is lessening as a result&#xA;  of fusion( mass being converted into energy from fusion) while gravity&#xA;  is weakening( as mass lessens )?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As others have said, mass loss by solar wind is a bigger factor especially for young and smaller stars, but there's a few factors at play.   The short answer to this question is that the mass loss, at least by fusion, is quite very compared to the total mass of the star.   Another factor, as hydrogen becomes helium, the core of the star becomes denser and greater density tends to be smaller and that increases gravity, but there are competing factors.   The inner core grows denser as it becomes more hydrogen rich and the fusion tends to expand outwards on the outside of the helium core, so a star like our sun gets a denser inner core over time, but the layers around the core can grow hotter and larger, even as they lose mass.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Wouldn't the radiation overpower the force of gravity and tear the&#xA;  star apart?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As mentioned above, this happens if you have 150 or 200 solar masses.   lower mass stars, the fusion isn't nearly powerful enough to blow the star apart.   Stars and white dwarfs blow apart when they go supernova, but that's different than the main sequence fusion process.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our sun will blow off some of it's matter when it has it's helium flash, so there are examples of what you're describing happening, but not during the main sequence for stars like our sun when material is expelled primarily by magnetic storms causing coronal mass ejections.  Fusion is, generally speaking, more like a slow burn, than a big explosion when it's up against the enormous gravitational binding energy of a star.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-11-06T10:55:36.887" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12425" PostTypeId="2" ParentId="12416" CreationDate="2015-11-06T07:16:05.330" Score="0" Body="&lt;p&gt;As I understand it, general relativity says that there's gravitation everywhere in the universe, and this gravitation creates dips in space, so to speak, often represented in 2D as a weight on a rubber sheet, like the picture below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mmPP2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mmPP2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.daviddarling.info/encyclopedia/G/genrel.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so, a black hole might generate a perfectly spherical event horizon, but it generates it on a not perfectly flat 3 dimensional surface, and I think the gravitation from other objects makes it not quite a perfect sphere.  For example, a star or planet that orbits a black hole would drag around a ripple on the event horizon as it orbits the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precisely what shape that ripple would be . . . I'm not sure.    That said, if you had a black hole as the only object in a universe, then I think the event horizon might be a perfect sphere, or as perfect as possible, given quantum fluctuation, hawking radiation and the impossibility of precise observation and all that good stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non black holes tend to have lumpy/inconsistent gravity - see &lt;a href=&quot;http://www.universetoday.com/116801/the-potsdam-gravity-potato-shows-earths-gravity-variations/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.  Even Neutron stars have some inconsistency, but Black holes probably avoid that because the matter is condensed to a point, so there's equal gravitational pull from all directions from the singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a Kerr black hole,  that's a whole different question.   I'm not smart enough to try to answer that one.   That might not be a sphere at all.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-11-06T07:16:05.330" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12426" PostTypeId="2" ParentId="12417" CreationDate="2015-11-06T07:42:10.243" Score="1" Body="&lt;p&gt;Here's the bare bones reason for stars like our sun.  The full story is much more...full.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expansion means cooling.  Cooling means less fusion.  Less fusion means less energy driving expansion, meaning the outward pressure is going down.  Eventually gravity is pulling inward more strongly than radiation is pushing outward.  So the material collapses again.  Collapsing means heating.  Heating means more fusion.  More fusion means more radiation pushing outward on the star.  Produce enough energy, and you'll overcome gravity and the star expands.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rinse and repeat.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The star naturally sits at an equilibrium where gravity and radiation balance each other. Deviations from this are self-correcting.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2015-11-06T07:42:10.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12427" PostTypeId="2" ParentId="12417" CreationDate="2015-11-06T07:58:27.557" Score="4" Body="&lt;p&gt;The Sun's luminosity is $3.8\times 10^{26}$ W. Application of mass energy equivalence tells you it loses mass at a rate of 4.25 million tonnes per second as hydrogen turns into helium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is practically nothing as far as the structure of the star goes. Over its lifetime, the Sun has lost about 0.03% of its mass in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radiation pressure &lt;em&gt;is&lt;/em&gt; a feature in stellar evolution calculations. It is almost negligible in the solar interior (at the 1% level compared with thermal pressure). However, it does become more important in more massive stars with hotter interiors and higher luminosities.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-11-06T07:58:27.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12428" PostTypeId="2" ParentId="12419" CreationDate="2015-11-06T09:12:11.787" Score="1" Body="&lt;p&gt;Whimsical question. Similarly, you could reason that someone on a planet far, far away would be holding up a mirror, such that we could see our own reflected past again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, both options are practically infeasible, even if the light would be undistorted and redirected perfectly towards us. Remember that (1) Earth is small, (2) the reflected signal is weak as it is, (3) any object capable of reflecting this back towards us would be a black hole, the closest of which is further away than most known exoplanets, and the signal falls of with $r^{2}$ with $r$ twice (!) the distance to the object doing this lensing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, &lt;strong&gt;if&lt;/strong&gt; we somehow were able to observe it anyway, it would be straightforward enough to estimate the look back time - the distance to the lens would presumably be fairly well known. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How bright it would have to be is not really a sensible question - there are too many variables in play for that (which (hypothetical future) telescope are we using, what are the observing conditions, how far away is the source, what kind of source are we talking about...). And at the end of the day, the brightness of your source is not even the biggest problem; finding a suitable lens is. &lt;/p&gt;&#xA;" OwnerUserId="6965" LastActivityDate="2015-11-06T09:12:11.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12430" PostTypeId="1" AcceptedAnswerId="12436" CreationDate="2015-11-06T03:05:05.697" Score="19" ViewCount="619" Body="&lt;p&gt;I was browsing NASA featured items and came across this - &lt;a href=&quot;https://www.nasa.gov/feature/goddard/out-with-the-old-in-with-the-new-telescope-mirrors-get-new-shape&quot; rel=&quot;noreferrer&quot;&gt;Out With the Old, In With the New: Telescope Mirrors Get New Shape&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Called freeform optics, this emerging mirror technology, brought about&#xA;  by advances in computer-controlled fabrication and testing, has&#xA;  triggered a sea change in optical engineering... the technology holds&#xA;  great promise for scientists who want to develop compact telescopes&#xA;  for CubeSat and other small satellites — an increasingly popular and&#xA;  cost-effective alternative to more traditional missions that are more&#xA;  expensive to build and launch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Telescopes for &lt;em&gt;cubesats&lt;/em&gt;?!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article had this illustration, and I'm trying to figure out how that is converted into a clear image without distortions, and why they say it allows the telescope to be much more compact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ibh9Q.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ibh9Q.jpg&quot; alt=&quot;freeform optics mirror design&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was a short article that didn't even try to explain this, and when I went looking for other articles on the topic for non-experts I found zip. Perhaps this is very challenging to explain, but I thought I'd ask anyhow. I told an enthusiastic youth just a few weeks ago that it wasn't possible to launch a telescope capable of anything useful on a cubesat. Oops...&lt;/p&gt;&#xA;" OwnerUserId="2810" OwnerDisplayName="kim holder" LastEditorUserId="2365" LastEditDate="2015-11-06T20:35:23.207" LastActivityDate="2015-11-06T20:49:39.033" Title="Why does a mirror bent 'like a potato chip' allow space telescopes to be smaller and have a wider field of view?" Tags="&lt;space-telescope&gt;&lt;optics&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12434" PostTypeId="2" ParentId="12418" CreationDate="2015-11-06T19:20:48.310" Score="9" Body="&lt;p&gt;The first question anyone asks about a telescope is &quot;what is the magnification?&quot; It is almost always not the most important thing. Any telescope can magnify a million times, given a short enough eyepiece - the problem is, how good the image is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For observing planets, the main thing is resolving power - the ability of the telescope to discern tiny details. The resolving power is limited by aperture, or the diameter of the primary lens or mirror. If the primary lens diameter is measured in mm, and the resolving power in arcseconds, then the limit is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;resolving power = 100 / aperture&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your 50 mm diameter lens would have a resolving power limited to 2 arcseconds in ideal conditions. Any details smaller than that would be blurry, no matter how much magnification you put into that thing. For comparison, Jupiter's apparent diameter is between 30 and 50 arcsec, Mars' between 3.5 and 25 arcsec, Saturn's between 14 and 20 (without the rings).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that aperture you would see the rings of Saturn, and a couple equatorial belts on Jupiter. Venus shows a crescent when it's far from the Sun. Mars is a little orange disk when it's closest to Earth. I've looked at planets through 50 mm of aperture, and you can see quite a few things that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon also looks interesting, you can see craters and mountains, plus it's available every month.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Because resolving power is limited by aperture, as you give it more and more magnification, at some point the image is just huge and bloated; big, but blurry. So there's a limit to the useful magnification - again, this depends on aperture. If the aperture is measured in mm, the formula is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;maximum magnification = 2 * aperture&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your 50 mm aperture would support up to 100x magnification before things get too blurry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do you calculate magnification for a given instrument? It is the ratio between the focal length of the primary lens (or mirror) and the focal length of the eyepiece (the lens close to your eye):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M = F / f&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your 1000 mm focal length primary, you achieve 100x magnification with an eyepiece with 10 mm focal length. Anything shorter than that would just bloat the image uselessly. Eyepiece focal lengths that would make sense in your instrument would be between 10 mm and 120 mm.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Words of advice:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure the lenses are perfectly aligned. When you build the instrument, have some way to adjust the direction of the primary lens, so as to align it with the main axis of the instrument. Some tweaking will be required - tilt the primary a fraction of mm, check the results, tilt again, repeat until the image looks best. This is called collimation and it's very important for the overall performance of the instrument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have some way for the eyepiece to move back and forth until you achieve perfect focus. This will have to be re-adjusted every time you observe. The adjustment is quite sensitive. A real focuser would help a lot, but this could be achieved also with two tubes sliding into each other, held in place by friction, like in an old pirate captain's telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For astronomy, when the scope is focused at infinity, the distance between primary lens and eyepiece is the sum of their focal lengths. E.g. with a 1000 mm primary and a 10 mm eyepiece, the distance would be 1010 mm. This assumes a convergent eyepiece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The telescope will not work well hand-held. The image will be jumping around too much. You will need some kind of mount. At the very least put the top of the tube on a fence or something. But astronomical instruments work best when they are on a rock solid mount. Even a broomstick hammered into the ground, with a loop on top for the tube, is better than nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whichever way you mount the primary lens, don't squeeze it too hard in the lens cell or tube. It will deform the lens and make the image bad. Some pressure on the edge is okay; lots and lots of pressure not okay. You could glue the edge of the lens on some kind of ring. Just use common sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is totally doable. I've built a telescope just like that, many years ago. It's great to see what you can achieve on your own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-11-06T21:54:20.220" LastActivityDate="2015-11-06T21:54:20.220" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12435" PostTypeId="2" ParentId="11430" CreationDate="2015-11-06T19:36:02.160" Score="4" Body="&lt;p&gt;Metallic hydrogen is an odd substance. When you push hydrogen atoms very close together, their electrons can come free, and move around, instead of being tightly bound to the atomic nuclei. As this form of hydrogen would conduct electricity, it behaves like a metal. At least this is the theory. Nobody has been able to produce enough pressure to actually make any metallic hydrogen in the lab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the pressure inside Jupiter should be high enough to form metallic hydrogen. In extrasolar planets there could be large amounts of metallic hydrogen too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However as soon as you release the pressure from metallic hydrogen, it turns back to normal molecular hydrogen. So it could not exist on the surface of a &quot;hot jupiter&quot;, even one from which the outer layers had been stripped away by the solar wind. The metallic hydrogen that had been in the interior would change back into molecular hydrogen as it approached the surface.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-11-06T19:36:02.160" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12436" PostTypeId="2" ParentId="12430" CreationDate="2015-11-06T19:46:29.150" Score="9" Body="&lt;p&gt;Freeform optics are a response to the specific challenge of cramming a telescope in a very limited space. A traditional instrument would have all optics symmetrical and aligned on the same axis. It would waste a lot of space within the cubesat. Also, traditional designs tend to be much longer than they are wider; they don't fit well in a cube; it is very hard to make classic instruments that are as short as they are wide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But with freeform optics you could bounce light in a few directions within the cube. You'd still achieve a decent focal length, and you would use all the volume available to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/h2Vxw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h2Vxw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since light is reflected from mirrors at angles different from normal, you cannot use the traditional symmetric shapes such as parabolic, spherical, etc. You need to basically take a paraboloid and squish it in one direction so that it works about the same like a parabolic mirror (I'm simplifying), but at an angle of reflection of, say, 45 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In such an instrument you could have multiple &quot;potato chip&quot; mirrors, as in the diagram above. You have to design the instrument as a whole; computer simulations will adjust the shape of each mirror until the performance of the whole instrument is close to a classic straight design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can tell, the manufacturing precision is such that freeform optics are only usable at long wavelengths such as infrared, where less precise optics can be used. But technology improves all the time. It also depends on how much aberration you can tolerate in your image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For usage from ground level this is less useful, unless you absolutely need a telescope in a very small form factor for some reason. Classic optics are still preferred when space and shape are not restricted.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-11-06T20:49:39.033" LastActivityDate="2015-11-06T20:49:39.033" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12438" PostTypeId="2" ParentId="12416" CreationDate="2015-11-06T22:23:38.807" Score="7" Body="&lt;p&gt;You cannot see the event horizon. That being said:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A non-rotating black hole, free of external influences, has perfect spherical symmetry. All its properties are exactly the same in any direction, period. This is the Schwarzschild metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_metric&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Schwarzschild_metric&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if it is electrically charged, if it's non-rotating, and free of external influences, it is still perfectly spherically symmetrical - this is the Reissner–Nordström metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Reissner%E2%80%93Nordstr%C3%B6m_metric&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Reissner%E2%80%93Nordstr%C3%B6m_metric&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As soon as it begins rotating, the black hole is no longer in perfect spherical symmetry. It acquires an ergosphere, which is shaped like a flattened ball. This is the Kerr metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Kerr_metric&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Kerr_metric&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RxP11.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RxP11.png&quot; alt=&quot;Kerr metric with ergosphere&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Significant external factors can distort the shape of the event horizon. Two black holes merging will go through a process where spherical symmetry is lost temporarily:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=JOQMvyLYmd4&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=JOQMvyLYmd4&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically, anything near a BH would slightly distort the metric of spacetime, but in practice this would not be easily measurable unless the external factor is very large (another very massive object). So for most practical purposes, the metrics described above would apply.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-11-06T22:30:46.130" LastActivityDate="2015-11-06T22:30:46.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12439" PostTypeId="2" ParentId="12416" CreationDate="2015-11-07T01:37:12.833" Score="4" Body="&lt;p&gt;The Schwarzschild metric by definition can be written in a spherically symmetric form, the Kerr metric is not spherically symmetric, but in  Boyer–Lindquist coordinates the event horizon has a constant radial coordinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For most people this will be good enough to say the event horizon of Kerr black holes (including Schwarzschild black holes) are perfect spheres. A physical black hole has a quasi-neutral charge, but the more general case of a charged Kerr-Newman black hole still has a perfectly spherical event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now of course no physical situation actually possess the exact symmetries of Kerr black holes. However the outer Kerr metric is stable, which means that the event horizon mantains a near-perfect spherical shape, except in extreme situations such as when two black holes are undergoing a merger.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastActivityDate="2015-11-07T01:37:12.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12440" PostTypeId="2" ParentId="12422" CreationDate="2015-11-07T04:35:34.913" Score="1" Body="&lt;p&gt;According to the webpage from the Planetary Habitability Laboratory of the University of Puerto Rico: &lt;a href=&quot;http://phl.upr.edu/library/media/therealcolorsofvenuswithouttheclouds&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Real Colors of Venus without the Clouds&lt;/a&gt;, did the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We combined radar topography and emissivity to generate the surface texture assuming a generalized basalt composition with gray to light brown tones, depending on the iron and sulfur content. In the images we also included the effect of the 'orange sky' of Venus thus given its surface a more dark brown appearance &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As in the image below. &#xA;Additionally, according to the model the scientists made, they state that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Under white light the surface should look grayish and more similar to the Moon. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lDv2j.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lDv2j.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;As an aside, the website also shows a fictional map of Venus with 70% surface water.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user9654" LastActivityDate="2015-11-07T04:35:34.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12441" PostTypeId="1" CreationDate="2015-11-07T05:56:05.810" Score="4" ViewCount="75" Body="&lt;p&gt;A large fraction (a third?) of all comets found (with observational bias thanks to SOHO) have been sungrazers, breaking up with perihelion of a few of Sun's radius. Is there reason to believe that this replenishes some kind of &lt;a href=&quot;https://en.wikipedia.org/wiki/Vulcanoid&quot; rel=&quot;nofollow noreferrer&quot;&gt;vulcanoid&lt;/a&gt; population of small objects? Or would they evaporate and be blown away by the Solar wind, or continue in their comet's orbit, or be ejected by Mercury's eccentricity? Would the constant replenishment, even now, not billions of years ago, help keep a population there in spite of all above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BghYT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BghYT.jpg&quot; alt=&quot;Comet ISON&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-11-07T07:07:24.533" LastActivityDate="2015-12-03T18:36:39.907" Title="Do sungrazing comets leave a field of meteoroids near the Sun?" Tags="&lt;the-sun&gt;&lt;asteroids&gt;&lt;comets&gt;&lt;mercury&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12442" PostTypeId="2" ParentId="12407" CreationDate="2015-11-07T06:17:49.287" Score="3" Body="&lt;p&gt;If we assume the 15 CM per year is accurate (which I'm not 100% sure it is), then it's possible to answer this.  In a loose sense, yes, the pressure should move the moon away from the sun faster, but more accurately and the real answer is that doesn't happen because there are 3 bodies in play and when you have 3 body gravitation, things get a lot more complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, just for fun, a &lt;a href=&quot;https://www.newscientist.com/article/dn17228-why-is-the-earth-moving-away-from-the-sun/&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt; for the 15 CM.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The moon away each year by photon momentum.The moon is the perfect&#xA;  color,size and The moon has a mass of 7.35 x 10²² kilograms. It is&#xA;  only about 60 percent as dense as Earth so it should be effected by&#xA;  the photons from the sun causing movement.And that movement should be&#xA;  combined with earths 15cm giving a new total distance a year right&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we only look at equal force per surface area, which is what you're describing, it's not density, it's the ratio surface area to mass.  The Moon's surface area is about 7.4% of Earth's (about 1/13.5) and it's mass about 1/81st of Earths.  &lt;a href=&quot;http://www.universetoday.com/20489/moon-compared-to-earth/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.  81/13.5 = about 6.  That means given equal pressure, the moon should accelerate 6 times as much which corresponds over small distances to 6 times the distance or 90 CM per year - BUT, that's if you ONLY take into account the pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect of the sun losing mass has equal effect on both the Earth and the smaller Moon.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And a 3rd factor to consider is the tidal bulge on the sun caused by the Earth-Moon system, which is very small but all these effects are small.   The Earth-Moon system creates a tiny bulge on the surface of the Sun and because the Sun rotates ahead of the Earth-Moon, that tiny tidal bulge has a tiny pull on the Earth-Moon system that slowly accelerates them and pushes them slowly away from the sun.  That effect is equal for both the Moon and Earth too.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All 3 of those are factors in the 15 CM per year estimate and only one of them has 6 times the effect on the Moon than the Earth.   The planets Venus and Jupiter might also be factors in that 15 CM per year estimate too, but lets leave that alone for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider the 3 body problem it gets very mathy, but I'll just talk about how it applies to solar pressure.   Lets start with a picture.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OAdJP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OAdJP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the Moon is waning its moving towards the sun and any pressure from the sun slows the moon down, (a tiny bit) and that slow-down moves the Moon closer to the Earth which speeds it up even more - funny how that works, pushing to slow something down in orbit and it goes faster - but that's how it works, cause potential energy converts to kinetic as the orbit drops, kinda like how falling makes things go faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the Moon is waxing, it's moving away from the sun and any pressure from the sun speeds it up which moves it away from the earth and that in turn, slows it down.  So that's your answer in a nutshell.  The solar pressure doesn't push the moon away from the sun so much as it pushes the moon into a lower and then higher orbit around the Earth depending on where the Moon is in it's orbit around the Earth and the various positions in both the Moon's and Earth's elliptical orbits.  The overall effect is very difficult to calculate and I suspect it could go either way depending in part on the timing of large coronal mass ejections hitting the moon and in part on any possible resonance between the eccentricity of the Moon around the Earth and the Earth around the sun, er, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is safe to say is that the Earth-Moon system obeys the same surface area to mass ratio the Earth alone because that's simple momentum which needs to be conserved.    The Earth-Moon system has a combined surface area of 1.074 Earths and a combined mass of 1.012 Earths, so the Moon being in the Earth-Moon system makes the Earth move away due to solar pressure about 6% faster.  15 CM to 15.9 CM.  Not a very big change, and it's possible that the 15 CM estimate is based on the Earth-Moon barycenter not the Earth itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the effect is more of a wobble in the Moon's orbit around the earth than an actual force that pushes the Moon away from the sun faster than it pushes the Earth away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this, is, of-course, unimportant compared to the much bigger solar tidal force on the Earth-Moon system.   As the Moon gets closer to the sun (new moon in the picture above) The sun effectively pulls the Moon a little bit away from the Earth and when it's further from the sun (full moon), The Moon is effectively pulled towards the earth.  The net effect of this solar tide is a measurable wobble in the Moon's orbit around the earth.  The effect on the Moon's orbit around the earth due to photon pressure and coronal mass ejection pressure - basically insignificant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that's clear.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-11-08T12:00:48.933" LastActivityDate="2015-11-08T12:00:48.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12445" PostTypeId="1" AcceptedAnswerId="12450" CreationDate="2015-11-07T12:43:33.690" Score="3" ViewCount="1322" Body="&lt;p&gt;Is it possible that the redshift observed by Edwin Hubble is really from a gravitational redshift and the universe isn't expanding as he has predicted?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I think I know thus far is this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Redshift due to the Doppler effect of a moving object is happening due to universe expansion. Galaxies further away are speeding away from us faster than those close to us, hence having a stronger redshift effect. &lt;/li&gt;&#xA;&lt;li&gt;Gravitational redshift is happening due to light &quot;escaping&quot; the gravitational pull of the source (star) and it has nothing to do with the movement of such source.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="692" LastEditorUserId="7982" LastEditDate="2019-05-06T12:43:54.907" LastActivityDate="2019-05-06T12:43:54.907" Title="Gravitational red shift vs Doppler redshift: Is the universe really expanding?" Tags="&lt;gravity&gt;&lt;expansion&gt;&lt;redshift&gt;&lt;doppler-effect&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="12446" PostTypeId="1" AcceptedAnswerId="12448" CreationDate="2015-11-07T13:17:26.030" Score="1" ViewCount="301" Body="&lt;p&gt;What exactly is a &quot;stellar association&quot; in the strict kinematic sense? &lt;/p&gt;&#xA;" OwnerUserId="2091" LastEditorUserId="2365" LastEditDate="2015-11-07T15:26:06.397" LastActivityDate="2015-11-07T15:36:14.953" Title="What exactly is a stellar association?" Tags="&lt;star-cluster&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12447" PostTypeId="2" ParentId="12445" CreationDate="2015-11-07T13:21:02.773" Score="1" Body="&lt;p&gt;I would hazard a guess at this point, but effects of gravitational red-shift would perhaps average out, so approximately 50% of light would orginate from objects with stronger gravitational fields and 50% from weaker gravitational fields. Hence, we would see blue-shifted and red-shifted objects if gravitational redshift is an prominent as assumed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, with the exception of the local group (where due to being gravitationally bound M33, Andromeda and ourselves are all moving towards each other), most of the extragalactic objects we observe have a redshift. Hence, the Doppler effect must be dominating. &lt;/p&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2015-11-07T13:21:02.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12448" PostTypeId="2" ParentId="12446" CreationDate="2015-11-07T15:36:14.953" Score="2" Body="&lt;p&gt;A stellar association is a loose cluster of stars, that formed at the same time from the same molecular cloud, and so have the same proper motion. Unlike open clusters, they are not gravitationally bound together, so the stars in a stellar association will gradually separate, forming a moving group of stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Scorpius%E2%80%93Centaurus_Association&quot; rel=&quot;nofollow&quot;&gt;Scorpius–Centaurus Association&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-11-07T15:36:14.953" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12450" PostTypeId="2" ParentId="12445" CreationDate="2015-11-07T21:46:43.373" Score="5" Body="&lt;p&gt;Ultimately this is an application of Occam's razor&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for the light to be gravitationally redshifted, it would have to be coming out of a &lt;strong&gt;deep&lt;/strong&gt; gravitational well. For the red shift observed in galaxies to be gravitational, you would have to suppose several things. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, that the stars in distant galaxies are somehow much denser: more than neutron star dense. (or possibly that entire galaxies are as dense as neutron stars.) &lt;a href=&quot;http://www.nature.com/nature/journal/v420/n6911/abs/nature01159.html&quot; rel=&quot;noreferrer&quot;&gt;A neutron star has a redshift of about z=0.35&lt;/a&gt; distant galaxies have redshift of more than 7. No known object has a gravitational redshift like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly that the density is proportional to distance from us. Placing us at a special position in the universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is much simpler to interpret the redshift as a doppler redshift, and therefore an expanding universe&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-11-07T21:46:43.373" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12452" PostTypeId="1" CreationDate="2015-11-08T01:30:44.620" Score="0" ViewCount="344" Body="&lt;p&gt;Is it posssible for us to time travel?&#xA;Is light essential factor or the dark matter and dark energy?&lt;/p&gt;&#xA;" OwnerUserId="9701" LastActivityDate="2015-11-08T03:28:21.530" Title="TIME TRAVEL-Can it be really done?" Tags="&lt;light&gt;&lt;dark-matter&gt;&lt;dark-energy&gt;&lt;matter&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-11-09T16:46:11.000" ContentLicense="CC BY-SA 3.0" />
  <row Id="12453" PostTypeId="2" ParentId="12452" CreationDate="2015-11-08T03:28:21.530" Score="2" Body="&lt;p&gt;In theory it is possible to do time travel. There are several possibilities from theoretical point of view. Two aspects are there, travelling to the future and travel back to past. Although many scientists including Stephen Hawking said that travelling to the past is not possible (as there are several paradoxes including &quot;the grand father paradox&quot;) that results in some weird phenomena (like in grandfather paradox, if you are allowed to travel back in time and you went to the time of your grandfather's youth and kill him then you are not going to born in the first place in future, thus a contradiction). But travelling to the future is seems possible. There are several ways to do that, thanks to &lt;strong&gt;The Theory of Relativity&lt;/strong&gt;. As Einstein said, time flows at different speeds in different part of the universe, faster in Earth as compared to the near vicinity of a black hole, due to the stronger gravitational pull of the black hole then the Earth. So if you travel near to a black hole and keeping a safe distance and starts revolving around it for few hours and return back to Earth, then you will find the many decades has been passed here, due to the different speed rate of time in the two places, so essentially you are in future (but alas you cannot go back!). Another way is to travel at a very high speed (almost near the speed of light) in a spaceship for few years also take you to the future back on Earth, this is because time slows down in case you speed up. And many more concepts of time travel exists involving &lt;a href=&quot;https://en.wikipedia.org/wiki/Wormhole&quot; rel=&quot;nofollow&quot;&gt;Wormhole&lt;/a&gt; and others. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Grandfather_paradox&quot; rel=&quot;nofollow&quot;&gt;Grandfather paradox&lt;/a&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Time_travel&quot; rel=&quot;nofollow&quot;&gt;Time travel&lt;/a&gt;&#xA;&lt;a href=&quot;http://spaceplace.nasa.gov/review/dr-marc-space/time-travel.html&quot; rel=&quot;nofollow&quot;&gt;See this&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9213" LastActivityDate="2015-11-08T03:28:21.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12455" PostTypeId="1" CreationDate="2015-11-08T04:15:15.880" Score="2" ViewCount="228" Body="&lt;p&gt;I've recently learned about the theory of relativity and time.  I'm trying to understand it more.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My (basic) understanding is that there is no distinction between past present and future and that all of time already exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is true, what prevents us from seeing the past and the future? And/Or what causes us to only be able to see the &quot;present&quot;?&lt;/p&gt;&#xA;" OwnerUserId="9722" LastActivityDate="2015-11-26T12:02:01.787" Title="Does all of time exist and if so where is it?" Tags="&lt;space-time&gt;&lt;time-dilation&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12456" PostTypeId="1" AcceptedAnswerId="12462" CreationDate="2015-11-08T09:33:25.097" Score="3" ViewCount="413" Body="&lt;p&gt;I am interested in current estimate of distribution of planets of various radii in various distances from their parent star. There is many sources, where one can find database of presently known exoplanets, for example &lt;a href=&quot;http://exoplanets.org/&quot; rel=&quot;nofollow&quot;&gt;exoplanets.org&lt;/a&gt;, &lt;a href=&quot;http://www.openexoplanetcatalogue.com/&quot; rel=&quot;nofollow&quot;&gt;www.openexoplanetcatalogue.com&lt;/a&gt; and others. However, the known exoplanets are, of course, subject to conditional probability of detection by various methods. One cannot take the database as representative distribution, because, for example, hot Jupiters or big planets in general would be greatly overestimated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the best resources to obtain current unbiased estimates of exoplanet radii and distances?&lt;/p&gt;&#xA;" OwnerUserId="2529" LastActivityDate="2015-11-09T14:58:16.877" Title="Estimates of exoplanets distribution consistent with current data" Tags="&lt;exoplanet&gt;&lt;star-systems&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12457" PostTypeId="2" ParentId="12455" CreationDate="2015-11-08T10:13:49.420" Score="3" Body="&lt;p&gt;Your question is difficult to answer exactly, because it touches the question of consciousness, which is problematic in general. But the answer is basically this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theory of relativity describes the space-time as four-dimensional space that exists as a whole. You are described by a &lt;a href=&quot;https://en.wikipedia.org/wiki/World_line&quot; rel=&quot;nofollow&quot;&gt;world-line&lt;/a&gt;, which contains all states you go through from birth to death. There is many slices, which describe &quot;you&quot; in different times. What you can see is given by what information reached your sensory inputs (your senses) and how they are processed by your brain. The brain creates a model of reality based on your sensory inputs, which is stored in space-like slice of your world-line in the space-time, which we call present. This is what &quot;you&quot; can see. So the reason basically is in how the brain processes the data it is getting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Along you world-line, there is many versions of you at different ages and each version feels its own present and the versions more in the future also have memories of the past (because the brain stores the past sensory inputs). There is no single &quot;present time&quot; in the theory of relativity, only the space-time as a whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you should't forget that the theory of relativity is still only a very well working model. Even within the theory of relativity, the question whether space-time exists as a whole, or whether there is &quot;present state&quot;, which evolves in time, is rather philosophical than scientific, because there is no observable difference between the two.&lt;/p&gt;&#xA;" OwnerUserId="2529" LastActivityDate="2015-11-08T10:13:49.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12458" PostTypeId="2" ParentId="12455" CreationDate="2015-11-08T11:17:00.943" Score="2" Body="&lt;p&gt;Time isn't quite like space. &lt;a href=&quot;https://i.stack.imgur.com/x1ETQ.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x1ETQ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first difference is that we can't stop moving in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this picture time is on the vertical axis, and space is horizontal. You have a history and a future. if you don't move you will travel straight up the vertical axis. Even if you do move you can never escape the cone, which shows the path that light takes in spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can receive information from anything in the past light cone, and you can send information to something in the future light cone. Points in spacetime outside the cone can never give or get information of you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, there is Entropy: Entropy increases with time. This gives a direction to time. This prevents us from remembering the future.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-11-08T11:17:00.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12459" PostTypeId="5" CreationDate="2015-11-08T12:44:39.097" Score="0" Body="&lt;p&gt;Phobos is the larger of Mars's two moons, being 11km in diameter. It is also the nearer to the planet's surface, and orbits the planet every 7 hours 39 minutes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is predicted that within 50 million years, Phobos will break up or collide with the planet's surface.&lt;/p&gt;&#xA;" OwnerUserId="9527" LastEditorUserId="9527" LastEditDate="2019-12-11T13:39:48.380" LastActivityDate="2019-12-11T13:39:48.380" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12460" PostTypeId="4" CreationDate="2015-11-08T12:44:39.097" Score="0" Body="Questions about Phobos, the larger of Mars's two moons." OwnerUserId="2365" LastEditorUserId="9527" LastEditDate="2019-12-11T13:39:03.937" LastActivityDate="2019-12-11T13:39:03.937" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12461" PostTypeId="2" ParentId="12401" CreationDate="2015-11-08T22:28:55.637" Score="2" Body="&lt;p&gt;&lt;strong&gt;Our time does not slow down due to the expansion of the universe. And time will not stop.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However &lt;a href=&quot;https://en.wikipedia.org/wiki/Time_dilation&quot; rel=&quot;nofollow&quot;&gt;time dilation&lt;/a&gt; does occur, and this is something different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When something is moving at a very high velocity, relative to us, we see time slowing down for it. As more distant galaxies are moving away from us at great speeds, events in those galaxies would (from our point of view) appear to run slower. This is observed in the at which supernovae brighten and fade &lt;a href=&quot;http://arxiv.org/abs/0804.3595&quot; rel=&quot;nofollow&quot;&gt;described in this paper&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inhabitants of those galaxies would not perceive any change in the rate of time. (rather they would see our clocks slow down, as we are the ones moving fast, from their perspective)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate of expansion is increasing (though why this should be is mysterious) but no part of the universe will accelerate to the speed of light, as nothing may travel at the speed of light. So time dilation will remain finite.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-11-08T22:28:55.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12462" PostTypeId="2" ParentId="12456" CreationDate="2015-11-08T23:53:43.760" Score="7" Body="&lt;p&gt;As you've correctly mentioned observation biases play a huge role in current understanding of planetary populations. Let's have a look at a slightly outdated plot (age 2 months) that shows us the semimajor-axis vs mass distribution of currently known planets including KOI's (unconfirmed planets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2avQl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2avQl.png&quot; alt=&quot;Current planetary population&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we see two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hot Jupiters seem to make up a significant part of the confirmed planets (confirmed = RV + Transit)&lt;/li&gt;&#xA;&lt;li&gt;The farther out we go, the less likely it is to have a measured planetary radius (black dots)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now knowing how our methods and instrumentation work, we can try and debias this data.&lt;br&gt;&#xA;Transit measurments pose the easiest ones to do this: as for transits the detection probability is a simple geometric function of planetary alignment, radius and stellar radius one can overplot the found planets with this probability. So this is usually of not very massive interest to the planetary community.&lt;br&gt;&#xA;&lt;strong&gt;edit&lt;/strong&gt;&lt;br&gt;&#xA;The probability usually is given as $\sim r_{stellar}/2 a$, with $r_{stellar}$ being the stellar radius and $a$ the planet's semi-major axis. A simple derivation is &lt;a href=&quot;http://kepler.nasa.gov/Science/about/characteristicsOfTransits/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. Discussion how everything gets more complicated when combining with RV measurments &lt;a href=&quot;http://astrobites.org/2013/05/23/transit-probabilities-not-as-simple-as-they-seem/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;br&gt;&#xA;&lt;strong&gt;/edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any instrument one uses however, will have their own quirks and biases and starspots or eclipsing binaries can lead to false positives. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013PNAS..11019273P&quot; rel=&quot;nofollow noreferrer&quot;&gt;Petigura+ 2013, fig. 1&lt;/a&gt; try to asses those effects for Kepler transits and come up with the following detection probability map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fXkJW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fXkJW.png&quot; alt=&quot;Rough correction for transit incompleteness&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corrected population then would be found by dividing $\rm\frac{found population}{completeness}$. Mathematical issues like dividing by zero are easily interpreted as issues stemming from the incomplete data. As where the detectability is zero, we can't know if there are 0 or 1 or 20 planets hidden in a bin.&#xA;When projected on one axis or the other those corrected data-plots then result in corrected Radius- or Period occurence rates.&lt;br&gt;&#xA;I won't show those for the transit method, as I want to encourage to check the original reference.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I show one of those plots for the radial velocity method however. From &lt;a href=&quot;http://adsabs.harvard.edu/abs/2011arXiv1109.2497M&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mayor+ 2011, fig. 12&lt;/a&gt; that analyse their HARPS-spectrograph data from the preceeding decade:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/L0ph7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L0ph7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In black&lt;/strong&gt;: Found planets, &lt;strong&gt;red&lt;/strong&gt;: debiased radial-velocity population&#xA;Note: A Jupiter-mass is roughly 320 Earth-masses, so in this plot the Hot Jupiters are on the far right. This now closes the thematic circle to my initial note about Hot Jupiters: While in black Jupiter-like planets seem very common, in the debiased version this is not the case anymore. Mayor et al. estimate the Hot Jupiters to make up only 0.5% of the total planetary population with this method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that although the debiased number of terrestrial planets increases significantly, their error bars explode in size. This stems simply from doing statistics with low numbers. Also for reasons of simplicity and clarity the error bars on the first two plots are not shown.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Concluding&lt;/strong&gt;:&lt;br&gt;&#xA;It is important to debias the data we have about exoplanets by method.&lt;br&gt;&#xA;When doing this, we get rough ideas about how the true planetary populations look like.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recent results and non-stop public announcments about &quot;hundreds of newly found planets&quot; can lead to a certain, totally understandable euphoria that we now know everything about planets.&lt;br&gt;&#xA;This is however very far from true. The Number distribution in full mass-radius-semimajor axis space is still very, very thinly spread. So in the end we do low-number statistics when debiasing, which is scientifically simply unsatisfying.&lt;br&gt;&#xA;Or said differently: There is still a lot of exciting work to be done!&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditDate="2015-11-09T14:58:16.877" LastActivityDate="2015-11-09T14:58:16.877" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12463" PostTypeId="1" CreationDate="2015-11-09T12:14:19.537" Score="0" ViewCount="454" Body="&lt;p&gt;It seems to me that the expansion of the universe is only being looked at one way. a repelling force, basically being the opposite of gravity.  After pondering this for many nights, I began to wonder if its possible that the changing pressure outside the universe could cause the expansion thst we observe.&lt;/p&gt;&#xA;" OwnerUserId="9730" LastEditorUserId="2365" LastEditDate="2015-11-09T18:19:59.877" LastActivityDate="2015-11-09T18:54:39.563" Title="could dark energy be a force coming from the hyperspace outside the universe?" Tags="&lt;cosmology&gt;&lt;dark-energy&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12464" PostTypeId="2" ParentId="12463" CreationDate="2015-11-09T18:54:39.563" Score="3" Body="&lt;p&gt;Nobody really has any idea what &quot;dark energy&quot; is. But there is a principle known as Occam's razor. It can be expressed in various ways, for example &quot;When an observation can be explained in many ways, the one with fewest assumptions should be used.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain the properties of dark energy as being due to changing pressure of hyperspace, requires the assumption that &quot;hyperspace&quot; exists, that it has a property that pressurizes the universe, that this &quot;pressure&quot; is changing, and that affects the rate of expansion of the universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_energy#Theories_of_explanation&quot; rel=&quot;nofollow&quot;&gt;other hypothesis&lt;/a&gt;: a cosmological constant, or a as yet undiscovered force field. None have much observational evidence, but have the benefit of requiring fewer assumptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Rob mentioned what you describe is similar to &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Brane_cosmology&quot; rel=&quot;nofollow&quot;&gt;brane cosmology&lt;/a&gt;&quot;, in which the universe is embedded in a higher dimensional space. These could influence our space, &lt;a href=&quot;http://arxiv.org/abs/hep-th/0304256&quot; rel=&quot;nofollow&quot;&gt;by controlling the scale of a cosmological constant, making it small, but non-zero&lt;/a&gt;. As with much of string theory, the experimenters are well behind the theorists.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-11-09T18:54:39.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12466" PostTypeId="1" CreationDate="2015-11-10T03:50:28.930" Score="7" ViewCount="6531" Body="&lt;p&gt;In general, when a star runs out of nuclear fuel, gravity gets the upper hand and the material in the core is compressed even further and creates black holes. I am clear till  here.&lt;/p&gt;&#xA;&lt;p&gt;Now the question is: &lt;strong&gt;Does that depend on the mass of the star (or object)?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Naturally, the more massive the core of the star, the greater the force of gravity that compresses the material and easy to create Black Hole.&#xA;For smaller stars, when the nuclear fuel is exhausted and there are no more nuclear reactions to fight gravity, the repulsive forces among electrons within the star eventually create enough pressure to halt further gravitational collapse. The star then cools and dies peacefully.&lt;/p&gt;&#xA;&lt;p&gt;In science videos and discussions is said that we can create black holes from any object. For example, &lt;strong&gt;if we are able to compress Earth to the size of the tennis ball, can we then create a black hole?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Isn't the nuclear fusion / reaction necessary to create a black hole? Of course, to fight against gravity we may require some kind of energy in this case, Nuclear reactions. But I am generally concerned like, does Nuclear Reaction is only to fight gravity, it does not play any other role in creating Black Hole.&lt;/p&gt;&#xA;" OwnerUserId="9519" LastEditorUserId="37191" LastEditDate="2021-02-15T14:13:33.960" LastActivityDate="2021-02-15T14:13:33.960" Title="Can we compress any object to create black Holes?" Tags="&lt;star&gt;&lt;black-hole&gt;&lt;nucleosynthesis&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="12467" PostTypeId="2" ParentId="12466" CreationDate="2015-11-10T07:49:05.163" Score="14" Body="&lt;p&gt;Do you mean, &quot;Can we &lt;em&gt;in principle&lt;/em&gt;&quot;, or just, &quot;Can we...&quot; ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to the latter is no. We have no technology to create black holes. It was a remote possibility that was considered when creating the high energy densities in the LHC, but in practice the energies aren't high enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we in principle do it? The answer is yes. If you can squash enough mass (or energy) into a very small volume then the inevitable consequence is a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most would say that the volume in question is governed by the Schwarzschild radius, $r_s$  which is roughly 3km multiplied by the mass in solar units. Thus for an Earth mass $r_s$ is about 1 cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually the radius within which collapse to a black hole is inevitable is a little bit larger because even the &quot;hardest&quot; equation of state, where the speed of sound equals the speed of light, cannot resist collapse at radii below about $1.5 r_s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that to compress materials this much would be incredibly difficult. For the example of an Earth mass, we are talking densities of order $10^{30}$ kg/m$^3$. But (long) before you got to that stage, yes you would have the problem of overcoming the nuclear fusion reactions would be initiated that would oppose your compression (ie you would end up making a mini-star before you could get to a black hole). Because the Schwarzschild radius depends linearly on mass, then less massive black holes need to be even denser. However, we are again talking about &quot;in principle&quot;. So assuming compression could overcome the various opposing forces, then a black hole could be formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It really doesn't matter what the black hole is made of. In a star, the sequence of nuclear fusion events is a consequence of the physical conditions in the stellar interior. It is not anything that is &lt;em&gt;required&lt;/em&gt; for black hole formation.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-11-13T00:45:28.470" LastActivityDate="2015-11-13T00:45:28.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12468" PostTypeId="1" CreationDate="2015-11-10T08:11:35.887" Score="3" ViewCount="1235" Body="&lt;p&gt;I have studied some basic astronomy, but I have difficulty conceptualizing the physics of luminosity and optics.  We use electromagnetic spectra to detect existence and properties of distant planets, but I am curious about the extent of what could theoretically be optically captured.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am unsure if optical is the correct term, but I'm referring to what could be directly captured vs. computed.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the use of technology far beyond what we currently are capable of, what is the maximum possible optical angular resolution we could theoretically obtain of a distant sky object?  Or, for example, of the surface topography of an outer planet/moon/dwarf planet?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the limiting factor at the extreme end of possible angular resolution?&lt;/p&gt;&#xA;" OwnerUserId="1398" LastEditorUserId="1398" LastEditDate="2015-11-10T08:20:49.953" LastActivityDate="2018-01-03T03:15:45.837" Title="What would be the maximum theoretically possible angular resolution?" Tags="&lt;telescope&gt;&lt;optics&gt;&lt;angular-resolution&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12469" PostTypeId="1" CreationDate="2015-11-10T10:11:48.677" Score="4" ViewCount="294" Body="&lt;p&gt;What is the explanation of the color similarity that the majority of natural satellites obtain?   &lt;/p&gt;&#xA;" OwnerUserId="9326" LastEditorUserId="53" LastEditDate="2015-11-26T19:43:31.370" LastActivityDate="2016-02-24T22:03:35.487" Title="Why is the Color scheme of natural satellites in our solar system based of light shades of grey?" Tags="&lt;the-moon&gt;&lt;solar-system&gt;&lt;planet&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12470" PostTypeId="1" CreationDate="2015-11-10T11:45:51.230" Score="3" ViewCount="241" Body="&lt;p&gt;A raw iron meteor or asteroid in outer space should be in its raw form having no signs of fusion crust from the friction of atmospheric entry and should therefore show distinctive signs like the posted image. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/Di3Pu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Di3Pu.jpg&quot; alt=&quot;Iron Meteor Original state &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9326" LastEditorUserId="2270" LastEditDate="2016-02-04T12:56:57.230" LastActivityDate="2016-02-04T12:56:57.230" Title="Is there a photo that shows a iron meteor or asteroid in space in raw form having no layers of fusion crust?" Tags="&lt;solar-system&gt;&lt;space&gt;&lt;meteorite&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12471" PostTypeId="1" CreationDate="2015-11-10T12:29:51.677" Score="2" ViewCount="286" Body="&lt;p&gt;The observable universe has a radius of about 46.5 billion light years. That’s big but I just wonder how that can be big enough to fit in everything we know exists in the universe. There are hundreds of billions of galaxies and each galaxy has billions of stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plus there are vast spaces between galaxies. And galaxies tend to cluster up into local groups, so there would be even bigger gulfs of empty space between these local groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the observable universe was only 46.8 billion light years, and there were this many and galaxies stars that need to fit inside, wouldn’t everything be much closer together?&lt;/p&gt;&#xA;" OwnerUserId="9734" LastActivityDate="2015-11-11T08:14:50.437" Title="How can the observable universe be so small if there are so many stars in it?" Tags="&lt;observable-universe&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12472" PostTypeId="1" CreationDate="2015-11-10T12:49:56.893" Score="2" ViewCount="120" Body="&lt;p&gt;Earth gets its radiation input primarily from Sun, then from reflected sunlight from Moon, and stars. Among these, what is the share of stars (or sources outside solar system) to this total radiation input of Earth. Is it negligible compared to solar radiation? (I am referring to top-of-atmosphere situation.)&lt;/p&gt;&#xA;" OwnerUserId="9735" LastEditorUserId="2365" LastEditDate="2015-11-10T21:34:46.773" LastActivityDate="2020-04-22T12:31:53.077" Title="What is the share of stars in total radiation input of Earth?" Tags="&lt;apparent-magnitude&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12474" PostTypeId="2" ParentId="12471" CreationDate="2015-11-10T16:30:25.577" Score="2" Body="&lt;p&gt;If you know the total number $N$ of galaxies, and you know the size $V$ of the (observable) Universe, then you can calculate the number density $n = N/V$ and mean distance (of the order $1/n^{1/3}$) between galaxies, and convince yourself that it actually works out fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the reason we know how many galaxies are in the observable Universe in the first place is the other way round:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Observe&lt;/em&gt; an average number density of galaxies (which of course varies quite a lot from dense clusters to voids).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Calculate&lt;/em&gt; the size of the Universe by integrating the &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;nofollow&quot;&gt;Friedmann equation&lt;/a&gt; (using as input observed values of the densities of the constituents of the Universe; dark energy, dark matter, gas, stars, radiation, etc.).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, multiply the two numbers to get the total number.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note that when observing the number density, we must not look too far away from the local Universe, since this means looking back in time to a period where the number density was different from present-day's value.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-11-11T08:14:50.437" LastActivityDate="2015-11-11T08:14:50.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12475" PostTypeId="2" ParentId="12472" CreationDate="2015-11-10T21:32:05.207" Score="1" Body="&lt;p&gt;It is negligible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of energy per square metre per second is called the &quot;flux&quot;. The flux of the brightest star in the sky, Sirius, is &lt;a href=&quot;http://www.astro.wisc.edu/~hill/ast103/a103_notes_oct23.pdf&quot; rel=&quot;nofollow&quot;&gt;one ten billionth of the sun&lt;/a&gt;. That mean the energy the earth receives from the sun in &lt;em&gt;one second&lt;/em&gt; is the same as the energy that it gets from Sirius in about 300 years. And Sirius (at magnitude -1.47) is by far the brightest star. All the others add next to nothing.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-11-10T21:32:05.207" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12476" PostTypeId="2" ParentId="10975" CreationDate="2015-11-10T21:53:21.970" Score="1" Body="&lt;p&gt;Dating cometary material is tricky, just because you have to go to space to get it. But the dates from the &quot;stardust&quot; mission show that comets are very very old. As old as the oldest asteroids, and date back to the origin of the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now while there is evidence that material that formed in the inner solar system when the sun was young, and the planets still forming, was carried to the outer solar system. There is no suggestion that this process is ongoing. There is no astrononomical water cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the planets might lose a little water to solar wind. It is very very little. Earth is protected by a magnetic field. Mars has lost all its  water already, as has Venus, and Mercury probably never had much.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-11-10T21:53:21.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12477" PostTypeId="1" CreationDate="2015-11-11T02:10:05.340" Score="1" ViewCount="966" Body="&lt;p&gt;This is a follow up question to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/11726/does-the-universe-expand-at-the-same-rate-everywhere-in-the-universe&quot;&gt;Does the universe expand at the same rate everywhere in the universe?&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the universe is expanding, then it would seem necessary for each object (ie galaxy) to be accelerating away from every other object at the same rate of acceleration which is constant (or not?). If that is true, what is the acceleration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, relative to the Milky Way, does it seem that all the other galaxies are moving directly away from us at the same rate of acceleration? If so, is it possible for us to prove that for any other galaxy that we know, the same observation would be true, were we able to view the universe from that galaxy?&lt;/p&gt;&#xA;" OwnerUserId="2800" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-11-20T07:16:50.460" Title="Are objects in the universe moving away from each other at the same acceleration?" Tags="&lt;universe&gt;&lt;cosmology&gt;&lt;apparent-motion&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12478" PostTypeId="1" AcceptedAnswerId="12486" CreationDate="2015-11-11T04:14:21.823" Score="10" ViewCount="739" Body="&lt;p&gt;I know that our galaxy is spiral in shape, but I'm wondering how the scientists found out that our galaxy has a spiral shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think we can see the entire galaxy from telescopes on Earth, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it makes sense that they say that Andromeda has a spiral shape because we can see the whole galaxy, but how do scientist know about our galaxy?&lt;/p&gt;&#xA;" OwnerUserId="9739" LastEditorDisplayName="user1569" LastEditDate="2015-11-12T15:01:43.040" LastActivityDate="2015-11-12T15:01:43.040" Title="How do we know that our galaxy is a spiral galaxy?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12480" PostTypeId="2" ParentId="12466" CreationDate="2015-11-11T04:38:33.350" Score="5" Body="&lt;p&gt;Kind of a quick answer, if you don't mind.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can we compress any object to create black Holes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The pressure at the bottom of the ocean is enormous, it would kill a person in a fraction of a second, but it's tiny compared to the pressure in the center of the earth, and that's tiny compared to the pressure in the center of Jupiter, and that's tiny compared to the pressure inside the sun, and the pressure inside of the sun is also, tiny-tiny-tiny compared to the pressure required to create a black hole.    &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the science videos and discussions, they say that, we can create&#xA;  Black hole from any object, i mean for example, if we are able to&#xA;  compress EARTH to the size of the Tennis Ball, then we can create&#xA;  Black Hole, Is it True?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think it's closer to the size of a golf-ball, but yes, that's true.  But it's also impossible.  There's no known way to do that.   We probably don't even have the means to crush a single car to the size of a golf-ball or tennis-ball and if we did, it would just rebound once we removed the compression force.   Matter is enormously difficult to crush anywhere close to that much.   That's why it only happens in the core of large stars.   That's the only place where there's enough pressure.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;does Nuclear Reaction is only to fight gravity, it does not play any&#xA;  other role in creating Black Hole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nuclear reaction doesn't play a role in creating a black hole.  A black hole by definition, is gravity strong enough that the escape velocity is greater than the speed of light.  Nuclear energy doesn't come close to that.   Nuclear fusion prevents black holes form forming when there's enough matter for a black hole to form, but that's about it.   Black holes can't happen without an enormous amount of mass, about 3 solar masses and no sustained nuclear fusion. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-11-13T18:34:57.957" LastActivityDate="2015-11-13T18:34:57.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12481" PostTypeId="2" ParentId="12477" CreationDate="2015-11-11T05:16:23.033" Score="3" Body="&lt;p&gt;I assume you mean is $\frac{dv}{dt}$ the same for all objects and/or is it constant for all objects, where $v$ is recessional velocity (and $t$ is cosmic time)? The answer is no it is not the same for all objects and it is not constant for all objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v=\frac{dD}{dt}=H(t)D$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $D$ is proper distance and $H(t)$ is the Hubble parameter. If we take the Hubble parameter as a constant, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D = D_0e^{Ht}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $D_0$ is the proper distance at the present time. So,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D''(t) = \frac{dv}{dt} = D_0H^2e^{Ht}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the &quot;recessional acceleration&quot; of an object depends on its present proper distance and is exponentially increasing with time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the Hubble parameter isn't a constant and in LCDM cosmology is currently asymptotically decreasing to a constant. In the current epoch though it is fair to say that the &quot;recessional acceleration&quot; of an object depends on its  proper distance and is increasing with time.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastActivityDate="2015-11-11T05:16:23.033" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12482" PostTypeId="1" CreationDate="2015-11-11T05:47:32.747" Score="4" ViewCount="419" Body="&lt;p&gt;I realize this is kind of general as it would depend on the size of the white dwarf and the rate of accretion. The general idea I got thinking about is what would happen if a white dwarf star - lets say 0.8 solar mass, well below Chandrasekhar was to accrete a bunch of hydrogen rather quickly.  I thought it might be get very hot and bright, perhaps similar to a red giant for a relatively brief period of time.&lt;/p&gt;&#xA;&lt;p&gt;Mostly White dwarfs add mass by siphoning from the outer layer from a near by co-orbiting star, but I imagine, that mostly happens quite slowly, not fast enough to make the white dwarf shine like a main sequence star.&lt;/p&gt;&#xA;&lt;p&gt;Are there studies on how quickly mass can be added in this way and how bright it makes the white dwarf?  Does Hydrogen quickly fuse on the surface of a white dwarf or does it need to build up to a certain thickness first, perhaps creating a mini flash? or does the fusion happen rather quickly.&lt;/p&gt;&#xA;&lt;p&gt;A very rough calculation, the escape velocity of a white dwarf, which would be similar to the terminal velocity of matter (mostly hydrogen) falling into it is about 5,500,000 meters/second for a 1 solar mass white dwarf.  &lt;a href=&quot;http://www.astronomynotes.com/evolutn/s13.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The velocity needed for hydrogen fusion is about 20,000,000 meters per second.  &lt;a href=&quot;http://physics.ucsd.edu/do-the-math/2012/01/nuclear-fusion/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt; (under the what makes fusion hard section).   So, white dwarf's aren't massive enough to create fusion upon impact, but the high density and high heat of impact seems like it could create fusion relatively quickly as matter builds up around them, perhaps creating something that perhaps glows like a star and expands rather rapidly - assuming there's a sustained burn and not a kind of very quick explosion of sorts.   I think it could glow very hot and perhaps expand rather large, but I'm just kind of guessing.&lt;/p&gt;&#xA;&lt;p&gt;If a white dwarf was to crash into, say, a brown dwarf or very low mass star, hydrogen rich, say about .1 solar mass?   Such a brown dwarf would be much larger than the white dwarf but also quite a bit lighter, probably light enough to get penetrated completely though the impact would also likely create fusion and it's difficult for me to imagine what would happen with so violent a crash.   It seems to me that the density of the white dwarf, if it could retain most of the matter from the brown dwarf would be sufficient that you'd be left with a fast burning light weight fusion star, with a very dense core.   This might be an impossible question, but how quickly would a .8 solar mass white dwarf consume the hydrogen from a .1 solar mass brown dwarf, should this rare kind of collision happen.   Would it actually form a star-like object for a period of time, say a million years or so, or would the enormous energy of impact basically blow it apart no new star?  Perhaps a near miss and an inside the Roche limit approach would be better than a direct impact for new star formation.&lt;/p&gt;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="31410" LastEditDate="2020-11-17T18:32:51.777" LastActivityDate="2020-11-17T18:32:51.777" Title="How bright can white dwarf stars glow as they accrete matter?" Tags="&lt;star&gt;&lt;white-dwarf&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="12483" PostTypeId="2" ParentId="12478" CreationDate="2015-11-11T07:54:17.143" Score="8" Body="&lt;ol&gt;&#xA;&lt;li&gt;It is clearly a disk galaxy since the Milky-Way is a stream across the sky and so we conclude that the Galaxy is a highly flattened structure.&lt;/li&gt;&#xA;&lt;li&gt;We observe that external galaxies which are highly flattened are almost all spirals of some form. (Also the stellar populations and arrangement of globular clusters are similar to those of external spirals)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://astrobites.org/2015/07/03/mapping-the-milky-way/&quot;&gt;Radio surveys/mapping radial velocities of molecular clouds&lt;/a&gt; etc.&lt;/li&gt;&#xA;&lt;li&gt;Other (which I'm sure other posters will expand on).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-11-11T07:54:17.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12485" PostTypeId="2" ParentId="12478" CreationDate="2015-11-11T13:42:25.660" Score="-2" Body="&lt;p&gt;You can see that the milky way appears as a stream across the night sky. We can calculate the distance between us and the stars in this stream, therefor we also know the distance between those stars (by using the Pythagorean algorithm for example). Because we now know the distance between them, we know their relative position of each other (by triangulation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you would then construct a 3D model by using these relative positions it would look like a spiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Here's a Kahn Academy video explaining how distance from earth to a star is calculated: &lt;a href=&quot;https://www.khanacademy.org/science/cosmology-and-astronomy/stellar-life-topic/stellar-parallax-tutorial/v/stellar-distance-using-parallax?v=lVadjWOjvV8&quot; rel=&quot;nofollow&quot;&gt;Stellar distance using parallax&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9741" LastEditorUserId="9741" LastEditDate="2015-11-11T13:49:49.913" LastActivityDate="2015-11-11T13:49:49.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12486" PostTypeId="2" ParentId="12478" CreationDate="2015-11-11T14:12:25.443" Score="9" Body="&lt;p&gt;To elaborate on Conrad Turner's excellent list:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;The disk structure is even more easily seen in the infrared, where dust extinction is much less than in the optical. See for instance this image from the &lt;a href=&quot;http://www.ipac.caltech.edu/2mass/gallery/images_misc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;2MASS infrared survey&lt;/a&gt;:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/SV9Av.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SV9Av.jpg&quot; alt=&quot;2MASS&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Stars are not really (as Milan van Dijck suggests) used for mapping the spiral arms. Rather the gas clouds of gas in between stars are used, more specifically neutral hydrogen observed through its  21 cm observation. The figure below (from &lt;a href=&quot;http://adsabs.harvard.edu/abs/1959IAUS....9..409O&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oort 1959&lt;/a&gt;) shows the resulting map. As you can see, its not really as well-defined as seen in many artist's-conception pictures.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Bb6TH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bb6TH.png&quot; alt=&quot;HI map&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Neutral hydrogen is fuel for stars. Massive stars ionize the gas around them, and since star formation mainly takes place in the spiral arms, the gas clouds are accompanied by pockets of ionized hydrogen, as seen in this map from &lt;a href=&quot;http://adsabs.harvard.edu/abs/1976A%26A....49...57G&quot; rel=&quot;nofollow noreferrer&quot;&gt;Georgelin &amp;amp; Georgelin (1976)&lt;/a&gt;:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/m5AJK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/m5AJK.png&quot; alt=&quot;HII map&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The physical properties of the Milky Way (stellar population, gas fraction, color, kinematics, etc.) are also similar to other spirals.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-11-11T14:27:39.280" LastActivityDate="2015-11-11T14:27:39.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12491" PostTypeId="2" ParentId="12470" CreationDate="2015-11-12T20:19:51.637" Score="3" Body="&lt;p&gt;In short, no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is that while finding meteorites on Earth is hard, finding them in space is a lot harder. A small asteroid, weighing only a few kg can't be spotted while it is still in space. The first we know of them is when they make their fiery descent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, there are some large asteroids that are metallic, such as &lt;a href=&quot;https://en.wikipedia.org/wiki/16_Psyche&quot; rel=&quot;nofollow&quot;&gt;16 Psyche&lt;/a&gt;, and we have &lt;a href=&quot;http://www.indiana.edu/~geol105/images/gaia_chapter_2/comets_and_asteroids.htm&quot; rel=&quot;nofollow&quot;&gt;direct images of other large asteroids&lt;/a&gt;. Metallic asteroids are relatively rare, and have a higher albedo than stony asteroids, and much more than comets. However they are not shiny balls of metal. After a few billion years in space they absorb 80% of the light that falls on them.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2015-11-29T09:26:49.730" LastActivityDate="2015-11-29T09:26:49.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12493" PostTypeId="1" AcceptedAnswerId="12498" CreationDate="2015-11-13T13:40:26.963" Score="8" ViewCount="1541" Body="&lt;p&gt;In &lt;a href=&quot;http://youtu.be/UHU226kWZIA&quot;&gt;this video&lt;/a&gt; the host of Test Tube Plus states that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;you can go out a buy a laser, point it at the moon, and see the retroreflector left by the astronauts.  When you point at the right spot you'll see a reflection, elsewhere not.  He reiterates that this is not just a hypothetical use of 2nd person for effect, but that &lt;em&gt;you&lt;/em&gt; the viewer can go out and try this yourself, and anyone paying any attention to moonlanding conspiracy theories ought to try it&lt;/li&gt;&#xA;&lt;li&gt;tracks and artefacts at the landing site can be seen with amateur telescopes, and again this is approachable to anyone who cares to look&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Neither of these seem correct to me.  Could someone with real numbers weigh in on this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I seem to recall that seeing artefacts is only possible in the latest high-res images taken from lunar orbit, and the telescope that uses the reflector gets back single photons on some trials and is a much more powerful laser than you could buy!  But, a continuous beam may be different? How big of a laser would you need to see a reflection with a viewing instrument that doesn't cost more than the laser?&lt;/p&gt;&#xA;" OwnerUserId="7142" LastActivityDate="2016-07-18T15:37:19.240" Title="Visibility of human activity on the moon" Tags="&lt;amateur-observing&gt;&lt;moonlanding&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12494" PostTypeId="2" ParentId="12493" CreationDate="2015-11-13T16:01:59.830" Score="5" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You can point a laser at the moon, but you won't be able to perceive any photons reflecting back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The beam spreads out due to diffraction. The rest of the illuminated side of the moon would far overwhelm your eyes, even at a slim crescent, to perceive any reflected photons from your laser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://what-if.xkcd.com/13/&quot; rel=&quot;nofollow&quot;&gt;xkcd's What If? #13&lt;/a&gt; shows what would happen if you got more people and more lasers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The retroreflectors left by Apollo astronauts won't help. You need a lot of &lt;a href=&quot;http://physics.ucsd.edu/~tmurphy/apollo/basics.html&quot; rel=&quot;nofollow&quot;&gt;complicated equipment&lt;/a&gt; to get even one photon back that you can distinguish.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Phil Plait, author of the Bad Astronomy blog, &lt;a href=&quot;http://physics.ucsd.edu/~tmurphy/apollo/basics.html&quot; rel=&quot;nofollow&quot;&gt;covered&lt;/a&gt; why we can't use telescopes, even the Hubble, to resolve artifacts on the moon. He walks through the calculations, and manmade objects on the moon are far smaller than the resolving power of any telescope.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="7142" LastEditDate="2015-11-14T18:23:34.497" LastActivityDate="2015-11-14T18:23:34.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12495" PostTypeId="2" ParentId="12493" CreationDate="2015-11-13T16:06:49.693" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Neither of these seem correct to me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are correct. The Apollo lunar lander was 4 meters across, or about 2 milliarcseconds at a distance of 385000 kilometers. The Hubble has a resolution of 50 milliarcseconds. Even the Hubble can't see the tracks and artifacts that the Apollo astronauts left on the Moon, let alone a backyard astronomer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding a backyard astronomer &quot;seeing&quot; the retroreflectors left on the Moon, that too is incorrect. &quot;Seeing&quot; those retroreflectors requires an extremely powerful pulsed laser, a rather large telescope, and rather expensive photodetectors. The final item is why I wrote &quot;seeing&quot; (in quotes) rather than seeing (without quotes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The professional astronomers who use those retroreflectors to measure the distance between the Earth and the Moon receive one reflected photon for every 10&lt;sup&gt;17&lt;/sup&gt; photons sent toward the Moon. They are looking for individual photons. Using pulsed lasers achieves two ends. First and foremost, it lets the astronomers measure distance. A continuous laser wouldn't let them do that. Secondly, it enables them to distinguish those individual photons from background noise. The lasers used in the lunar laser ranging experiments have a peak power in the gigawatts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An amateur backyard astronomer who a few million dollar or so to spare might be able to build a system that &quot;sees&quot; the retroreflectors left on the Moon. An amateur backyard astronomer cannot see the artifacts left on the Moon.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-11-13T16:06:49.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12496" PostTypeId="1" CreationDate="2015-11-13T17:08:57.740" Score="3" ViewCount="710" Body="&lt;p&gt;I did some googling around and couldn't get answer to a simple question of black hole mass compared to the star from which it was created. It is well known that giant stars burn very bright and relatively very quickly. They end up their lives as red giants and eventually collapse into a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that the star is mid-life and has some mass M. Over the time it burns its fuel, turns into a red giant and after supernova, turns into a black hole. This black hole has some mass B. Why B &gt;&gt; M?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to some simple physics laws, mass could not be created nor destroyed.&lt;/p&gt;&#xA;" OwnerUserId="9761" OwnerDisplayName="xeon" LastActivityDate="2018-02-13T22:45:34.350" Title="Is a black hole heavier than the star from which it was created?" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12497" PostTypeId="2" ParentId="12496" CreationDate="2015-11-13T17:45:04.143" Score="6" Body="&lt;p&gt;The mass of a black hole is always much less than the star from which it formed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very large star such as Eta Carinae (which is expected to form a black-hole some day) is about 100 times that of the sun, Having already lost as much as 50 solar masses, blown off the star by powerful outbursts in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The future of the star is not fully understood. It is possible that it will collapse in a supernova, in which most of the rest of the star will be blasted into space, forming a supernova remnant, and the core of the star will collapse. The black hole formed may have a mass of more than 3 times the sun, but much less than the mass of the original star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while the black hole is not heavier, it is much denser. A 3 solar mass black hole would have a diameter of about 20km. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-11-13T17:45:04.143" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12498" PostTypeId="2" ParentId="12493" CreationDate="2015-11-13T19:33:56.473" Score="17" Body="&lt;p&gt;The video is hilariously wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the principle of laser ranging is more or less right, and it does require the reflectors left behind by the astronauts on the Moon. It's just that the physics and technology involved are far beyond just pointing a toy laser at the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Project APOLLO (Apache Point Observatory Lunar Laser-ranging Operation) is actually doing this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.ucsd.edu/~tmurphy/apollo/basics.html&quot; rel=&quot;noreferrer&quot;&gt;http://physics.ucsd.edu/~tmurphy/apollo/basics.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need a fairly large telescope to start with - both for collimating the outgoing light pulse, and for receiving as much of the reflection as possible. APOLLO uses a 3.5 meter telescope at the &lt;a href=&quot;http://www.apo.nmsu.edu/&quot; rel=&quot;noreferrer&quot;&gt;Apache Point observatory&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need a laser that can generate a high-energy light pulse that is very short. The pulse is injected into the telescope's optical train and sent to the Moon. This is not a laser pointer; it's a high power Q-switched laser for research, &lt;a href=&quot;http://physics.ucsd.edu/~tmurphy/apollo/onscope.html&quot; rel=&quot;noreferrer&quot;&gt;a device the size of a refrigerator&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the receiving end, you need a very good detector also plugged into the telescope. Of the many, many photons sent to the Moon in the pulse, only between 1 and 5 photons make it back down to the detector. You need a detector that can tell the extra 1 to 5 photons from the background noise of light coming from the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jHEhu.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jHEhu.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this technique, the distance from Earth to Moon can be measured with very high precision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the APOLLO system in action:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LYFQv.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LYFQv.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Regarding observing human artifacts on the Moon with terrestrial telescopes, I wish it was doable, but it's not. Again the YouTube video is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest telescope accessible to amateurs has an aperture (diameter) in the range of 1 meter or a little bit larger (the aperture of the largest amateur-owned telescopes currently). The resolving power of a telescope (the size of finest details discernible) depends on aperture - if aperture is measured in mm and the resolving power in measured in arcseconds, then the formula is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;resolving power = 100 / aperture&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a 1 meter telescope has a resolving power of 0.1 arcseconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance from Earth to Moon is 384000 km (3.84 * 10^8 meters). With a resolving power of 0.1 arcseconds, the size of the smallest detail discernible on the Moon is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;detail size = distance_from_Earth_to_Moon * arctan(resolving power)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;detail size = &lt;a href=&quot;http://www.wolframalpha.com/input/?i=3.84%20*%2010%5E8%20*%20arctan%280.1%20arcsec%29&quot; rel=&quot;noreferrer&quot;&gt;3.84 * 10^8 * arctan(0.1 arcsec)&lt;/a&gt; = 186 meters&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything smaller than 186 meters would be blurred into a single dot. There's nothing we've done on the Moon that's as big as that. It's not possible to see traces of human activity with amateur-level telescopes, even with extremely large meter-class dobsonian telescopes. Even with professional telescopes, we just don't have the aperture yet to resolve that kind of detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, satellites in orbit around the Moon, such as the LRO, were able to image traces of the Apollo missions. That's because they are a lot closer to the site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nasa.gov/mission_pages/LRO/news/apollo-sites.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.nasa.gov/mission_pages/LRO/news/apollo-sites.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/S5Eiz.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S5Eiz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2015-11-16T03:20:07.113" LastActivityDate="2015-11-16T03:20:07.113" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12499" PostTypeId="5" CreationDate="2015-11-13T19:56:57.803" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-11-13T19:56:57.803" LastActivityDate="2015-11-13T19:56:57.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12500" PostTypeId="4" CreationDate="2015-11-13T19:56:57.803" Score="0" Body="Questions about the dwarf planet, and the largest asteroid n the Solar System, comprising 1/3 of the mass of the asteroid belt. Ceres orbits between Mars and Jupiter. " OwnerUserId="2365" LastEditorUserId="2153" LastEditDate="2015-11-14T00:59:41.777" LastActivityDate="2015-11-14T00:59:41.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12501" PostTypeId="1" AcceptedAnswerId="12502" CreationDate="2015-11-13T21:51:07.310" Score="6" ViewCount="2021" Body="&lt;p&gt;Could a gaseous moon exist in the same way as a giant gas planet? All the moons in the solar system are rocky, or icy. Why shouldn't gas planets  have gas moons? &lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="211" LastEditDate="2015-11-15T14:48:10.173" LastActivityDate="2015-11-15T14:48:10.173" Title="Can a gas moon exist?" Tags="&lt;natural-satellites&gt;&lt;gas-giants&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2020-02-06T02:29:39.420" ContentLicense="CC BY-SA 3.0" />
  <row Id="12502" PostTypeId="2" ParentId="12501" CreationDate="2015-11-14T00:57:47.273" Score="6" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/12501/can-a-gas-moon-exist#comment17661_12501&quot;&gt;LocalFluff's comment&lt;/a&gt; is spot-on. You need mass to have gas, and moons just don't have enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is thought that gas giants gathered the gas they have today by accreting large amounts of it from the protoplanetary disk in the early days of the Solar System. At first, they were only gas-less cores (not rocky, exactly, but not gaseous), but they quickly became the most massive objects in their immediate vicinities, and thus gobbled up all the gas nearby them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, less massive bodies, like Earth, can still accrete gas. However, they have difficulty retaining it. &lt;a href=&quot;http://mnras.oxfordjournals.org/content/early/2014/02/17/mnras.stu085.full.pdf&quot;&gt;Lammer et al. (2014)&lt;/a&gt; calculated that under conditions like those experienced in the young Solar System, planets of one Earth mass or less could retain captured hydrogen envelopes for no more than ~100 million years - a long time for us humans, but shorter for astronomical objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydrogen escapes via &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_escape&quot;&gt;atmospheric escape&lt;/a&gt;, which happens when particles in the atmosphere have velocities higher than escape velocity. This phenomenon is known as Jeans escape. Atmospheric escape happens more often with lighter gases, such as hydrogen. It is also influenced by stellar winds, which can cause even gas giants (generally close to the star, such as hot Jupiters) to lose some or all of their atmospheres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't call gas moons impossible, but certainly very unlikely.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2015-11-14T00:57:47.273" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12503" PostTypeId="1" AcceptedAnswerId="12504" CreationDate="2015-11-14T16:15:13.153" Score="2" ViewCount="521" Body="&lt;p&gt;I am not an astronomy expert, but as of today, all other stars than the Sun have little to no effect on life on Earth. So I was thinking, would there be any differences for the Earth &lt;strong&gt;at all&lt;/strong&gt; if all stars would disappear except the Sun?&lt;/p&gt;&#xA;" OwnerUserId="9765" LastEditorUserId="9186" LastEditDate="2015-11-17T12:28:41.193" LastActivityDate="2015-11-17T12:28:41.193" Title="Notable differences if all stars would disappear except the Sun" Tags="&lt;star&gt;&lt;life&gt;&lt;habitable-zone&gt;" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12504" PostTypeId="2" ParentId="12503" CreationDate="2015-11-14T16:36:17.127" Score="1" Body="&lt;p&gt;There are two main things we should look at here, because there are two main ways these distant stars can affect us: light and gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Light&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all the stars in the universe disappeared (besides the Sun), the most of the luminous objects in the universe would simply vanish. There would still be objects emitting radiation - stellar remnants, accretion disks, etc. - but the notable sources of light would be gone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this be an issue? I can't give you any numbers, but I would assume that there wouldn't be much of a difference. Night/day cycles would continue as normal, because the rotation of the Earth also would not be affected. The behavior of &lt;a href=&quot;https://en.wikipedia.org/wiki/Stoma&quot; rel=&quot;nofollow&quot;&gt;stoma&lt;/a&gt; in plants should remain the same, as should all the other natural rituals of life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Gravity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where I'm slightly more concerned. Stars have mass, and so they influence other objects with gravity. There are varying estimates of how much of the Milky Way is made up of stars. For example, &lt;a href=&quot;http://mnras.oxfordjournals.org/content/414/3/2446.full.pdf&quot; rel=&quot;nofollow&quot;&gt;McMillan (2011)&lt;/a&gt; estimated the stellar mass of the Milky Way to be about 6.43$\times$10&lt;sup&gt;10&lt;/sup&gt; solar masses, while the virial (total) mass was estimated to be about 1.26$\times$10&lt;sup&gt;12&lt;/sup&gt; solar masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extra matter consists of gas, dust, planets, stellar remnants, and, of course, dark matter (a lot of it). The mass of the Milky Way would be reduced by about 5% - not a terrible amount, but certainly non-negligible. I don't know how this would impact the motion of the Solar System; detailed simulations based on density models would be needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, any effects from stars wouldn't reach us any time soon. Light and gravity both travel at the speed of light, so we'd have at least four years before we started seeing the effects (4 light-years is the distance to the nearest star, Proxima Centauri). We wouldn't feel the effects from the rest of the galaxy for many more years.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-11-14T18:14:54.507" LastActivityDate="2015-11-14T18:14:54.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12505" PostTypeId="2" ParentId="12503" CreationDate="2015-11-14T21:32:20.713" Score="1" Body="&lt;p&gt;Do you mean if they disappeared now, or disappeared at some arbitrary time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the latter, then I would say that if they had disappeared more than a few thousand years ago, we might still be living in mud huts. The movement of stars and planets in the heavens motivated many of the ancient scholars and led directly to things like Newton's laws and a theory of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the former, no immediate physical consequences, but - I would say a worldwide panic and financial meltdown for starters.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-11-14T21:32:20.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12506" PostTypeId="1" AcceptedAnswerId="12507" CreationDate="2015-11-15T00:50:50.370" Score="4" ViewCount="1941" Body="&lt;p&gt;I know it can't be a perfect ellipse or smooth orbit due to all the matter in every direction, and our Sun could be affected by a gravity assist of sorts if it passes close enough to another massive star. I also know that galaxy orbits are different than Solar System orbits because most of the mass is in the dark matter halo and as a result, the outer stars orbit faster than in a standard orbit, so there's a number of variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, if we ignore gravity assists, is the sun mostly in a standard, or lightly inclined orbit around the center of the Milky way or more sinusoidal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reason I'm asking - see (attached and in my opinion, bad) &lt;a href=&quot;http://finance.yahoo.com/news/leading-harvard-physicist-radical-theory-160000092.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt;, and picture included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/v5Th9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/v5Th9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen pictures like this before that suggests that the Solar System moves in a kind of sine wave motion. I've seen estimates between 26 and 32 million years with the solar-system, rising above then below the plane of the Milky way with consistency (and perhaps, mass extinctions tied to this movement).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can kind of see that happening, as the sun rises above it might have more gravitational pull below it drawing it back down and you kind of have a horizontal velocity to displacement cycle creating a sine wave in the orbit. That makes logical sense, but I'm wondering if that 26-32 million year sine wave type of movement is generally accepted or more somebody's imagination. Is anything definitive known about stars' orbits within galaxies?&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-11-15T01:15:50.690" LastActivityDate="2015-11-15T10:01:36.373" Title="Does the Sun orbit the Milky way in a (kind of) flat or inclined orbit or more of a sine wave?" Tags="&lt;orbit&gt;&lt;galaxy&gt;&lt;stellar-dynamics&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2015-11-16T08:13:28.300" ContentLicense="CC BY-SA 3.0" />
  <row Id="12507" PostTypeId="2" ParentId="12506" CreationDate="2015-11-15T10:01:36.373" Score="4" Body="&lt;p&gt;It is correct that the sun bobs &quot;up and down&quot; (relative to the plane of the galaxy as it orbits the galaxy, and it takes about 64 million years  to complete a full cycle, so it does pass through the plane of the galaxy every 32 million years or so (there are quite large error bars on those numbers.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This idea is uncontroversial, but it gets caught up in other theories that are less well established (for example the idea that there are dense clumps of dark matter in the galactic plane, that can have disrupt the Oort cloud, or otherwise cause mass extinction) or are simply nonsense (the idea that the solar system is a &quot;vortex&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2013/03/04/vortex_motion_viral_video_showing_sun_s_motion_through_galaxy_is_wrong.html&quot; rel=&quot;nofollow&quot;&gt;Phil Plait rubbishes this second idea on his blog&lt;/a&gt;, and includes some further discussion of the &quot;bobbing&quot; path of the sun. He notes that the disk is about 1000 ly thick, and we bob up and down by about 200 light years every 64 million years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lisa Randall's is a serious scientist, who would have no truck with &quot;solar system vortices&quot;, and the ideas in her book are reasonable, but not uncontroversial. We have no evidence of dense clumps of dark matter, nor do we know exactly where the Earth was 65 million years ago. &lt;a href=&quot;http://earthsky.org/human-world/will-our-solar-system-cross-the-galactic-plane-on-december-21-2012&quot; rel=&quot;nofollow&quot;&gt;We last passed through the galactic plane about 3 million years ago&lt;/a&gt;, without serious consequences.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-11-15T10:01:36.373" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12508" PostTypeId="2" ParentId="12404" CreationDate="2015-11-15T13:12:54.860" Score="1" Body="&lt;p&gt;It varies throughout the year. A book or &lt;a href=&quot;http://starryskies.com/The_sky/constellations/index.html&quot; rel=&quot;nofollow&quot;&gt;a website with maps&lt;/a&gt; will help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Spring, Leo and Boötes dominate the southern skies, with Coma Berencies between them&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Summer, the summer triangle of Deneb, Altair and Vega is seen, with Cygnus flying along the Milky Way, and the rich star fields of Saggitarius closer to the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Autumn there is the square of Pegasus, with Andomeda riding, past Perseus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in Winter, Orion dominates the southern skies, with his dogs following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the North, Cassiopea and the Ursa Major, are on opposite sides of the Pole star.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-11-15T13:12:54.860" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12509" PostTypeId="1" AcceptedAnswerId="12511" CreationDate="2015-11-15T19:53:37.357" Score="2" ViewCount="397" Body="&lt;p&gt;I found this &lt;a href=&quot;https://en.wikipedia.org/wiki/Color_index&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt; on this property called color index. Unfortunately, I could not find a source that explains this property well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My interpretation is that color index is the brightness difference of an object through some 2 different wavelengths. Is my understanding correct? If not, could someone explain it to me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if my understanding is correct, why would there need to be a calibration table, as shown in the above link?&lt;/p&gt;&#xA;" OwnerUserId="9580" LastActivityDate="2015-11-16T23:01:16.207" Title="What is Color Index?" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12510" PostTypeId="1" CreationDate="2015-11-16T04:59:27.343" Score="2" ViewCount="169" Body="&lt;p&gt;Would more questions regarding the planet be answered if this was the case? &lt;/p&gt;&#xA;" OwnerUserId="9326" LastEditorUserId="2365" LastEditDate="2015-11-16T19:04:49.280" LastActivityDate="2015-11-19T18:43:19.850" Title="Is it plausible that Mars could have been one of Jupiter's natural satellites at one time?" Tags="&lt;mars&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12511" PostTypeId="2" ParentId="12509" CreationDate="2015-11-16T07:36:33.283" Score="3" Body="&lt;p&gt;A colour index measures the &lt;em&gt;ratio&lt;/em&gt; of brightnesses in two wavelength intervals, not the difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason an index is written as the difference between two numbers is due to the logarithmic astronomical magnitude system - the difference between two logarithms is the logarithm of the ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that a red colour is bigger than a blue colour is also because of the perverse way that objects with smaller magnitudes are brighter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calibration table you refer to appears to just show that stars of different spectral class and temperature exhibit different colours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You do however need a calibration to relate real brightness ratios and real properties of stars to the colours. That is because each logarithmic magnitude scale has its own zeropoint (the flux associated with a zeroth magnitude star in that band). The baseline colours are defined using the star Vega as a standard, such that all the colours of an A0 main sequence star are approximately zero (as shown in that table).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-11-16T23:01:16.207" LastActivityDate="2015-11-16T23:01:16.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12512" PostTypeId="2" ParentId="12510" CreationDate="2015-11-16T08:31:37.143" Score="5" Body="&lt;p&gt;I'm going to say no, for a few reasons, but if anyone wants to give a more detailed answer, feel free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Planets and moons that form inside the &lt;a href=&quot;https://en.wikipedia.org/wiki/Frost_line_%28astrophysics%29&quot;&gt;frost-line&lt;/a&gt; have much less water and other ices than those that form outside.   Mars likely once had water and oceans but if it had formed outside the frost-line it should have a lot more water and a lower density.   The frost line moves away from the sun as the sun grows more luminous and planets move, so there is a degree of imprecision with this argument, but I believe this on it's own is a pretty good reason why it's not likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) A moon that escapes Jupiter's orbit should enter a near Jupiter orbit initially, and at some point later, it could receive a gravity assist and be propelled either in towards the sun or away, but it's unlikely that Mars could have obtained it's relatively circular orbit had it originated as one of Jupiter's moons, unless Jupiter was much closer to the sun when Mars escaped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) it's difficult to see how the asteroid belt could have stayed around if Mars originated outside the asteroid belt and moved inside it.   Mars passing through the asteroid belt would have caused significant disruptions to it.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-11-16T08:31:37.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12513" PostTypeId="2" ParentId="12503" CreationDate="2015-11-16T12:10:28.897" Score="1" Body="&lt;p&gt;One example: there might be some effects on animal migration, as migrating species have been shown to respond to star patterns in a planetarium in various experiments over the years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even dung beetles were found to respond to the &quot;stripe&quot; of the milky way across the sky, in an experiment publicised a few years ago.  This was not for long distance migration by the way, but for moving in a straight line for a short distance when a food source is found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.bbc.co.uk/news/science-environment-21150721&quot; rel=&quot;nofollow&quot;&gt;BBC news story about dung beetles&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://web.mnstate.edu/wisenden/AnBeh13%20Paper%205%20Dacke%20et%20al%202013.pdf&quot; rel=&quot;nofollow&quot;&gt;Paper about dung beetles&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2015-11-16T12:10:28.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12514" PostTypeId="1" AcceptedAnswerId="12519" CreationDate="2015-11-16T17:10:41.957" Score="2" ViewCount="276" Body="&lt;p&gt;This question sounds like silly and rather fantasy filled. But can our earth's gravity capture a sizable asteroid and make it rotate around our earth like our moon which become visible from ground. Is there any theoretical possibility or practical probability? &lt;/p&gt;&#xA;" OwnerUserId="9213" LastActivityDate="2015-11-17T18:25:52.397" Title="Can we have a second moon?" Tags="&lt;the-moon&gt;&lt;asteroids&gt;&lt;natural-satellites&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12515" PostTypeId="1" CreationDate="2015-11-16T17:51:14.837" Score="4" ViewCount="9075" Body="&lt;p&gt;I have a home made astronomical telescope with 100 cm objective lens and 5 cm eyepiece.when viewing an object in infinity the distance between eyepiece and objective is 105 cm. So how could I calculate distance between objective and eyepiece for and far away object.For example a planet 1 AU from earth how much should be the distance between objective and eyepiece ?&lt;/p&gt;&#xA;" OwnerUserId="9705" LastEditorUserId="2365" LastEditDate="2015-11-16T19:01:53.390" LastActivityDate="2018-05-19T09:56:13.477" Title="Equation to find distance between objective and eyepiece" Tags="&lt;telescope&gt;&lt;telescope-making&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12516" PostTypeId="2" ParentId="12515" CreationDate="2015-11-16T19:01:14.493" Score="1" Body="&lt;p&gt;Astronomical objects are so far away that they focus the same point as an object at infinite distance. In fact any object more than a hundred metres away or so can be treated as being an infinite distance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To quantify this consider the &lt;a href=&quot;http://www.physicsclassroom.com/class/refln/Lesson-3/The-Mirror-Equation&quot; rel=&quot;nofollow&quot;&gt;focal distance equation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the focal distance = 1m, and the object distance $d_o$ = 1AU = $150\times 10^9$m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The the image distance ($d_i$) will be 100.000000001cm (less than one atom different)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-11-16T19:01:14.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12517" PostTypeId="2" ParentId="12503" CreationDate="2015-11-16T19:02:52.330" Score="1" Body="&lt;p&gt;I think that you mean to say: if the stars would disappear &lt;em&gt;right now&lt;/em&gt;, but in case you meant to say &lt;em&gt;if there had never been stars&lt;/em&gt;, here is an additional answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without stars human development would have followed completely different paths, and there's no way to know where we would have ended by now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably as long as human beings developed some form of intelligence (whatever that may be), and especially its components curiosity and inquisitiveness, the movements of the stars will have puzzled them: what are these scintillating, unreachable pinpricks of light that follow regular (or sort-of-regular) patterns?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has influenced religion, philosophy, and science in no small part: stories about deities in the sky, measurement of time during the year, navigation directions, all astronomical knowledge (probably including relativity), etc...&lt;/p&gt;&#xA;" OwnerDisplayName="user1569" LastActivityDate="2015-11-16T19:02:52.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12518" PostTypeId="2" ParentId="12515" CreationDate="2015-11-16T19:06:07.450" Score="3" Body="&lt;p&gt;1 A.U. is same as infinity. The difference in terms of eyepiece position is infinitesimal, you can't measure it. Anything beyond a few kilometers away is pretty much &quot;at infinity&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless of that - from the practice of designing and building telescopes, calculations only offer you a starting point. You do the math, and the distance is 105 cm. But in practice lenses will deviate from the ideal focal length. Even if they didn't deviate at some temperature, put them in a cold environment, and the focal length will change a fraction of mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So take the calculations as a starting point, and build the instrument in such a way as to allow fine adjustments of the position of the eyepiece. There's a device called focuser that allows such fine adjustments. Or simply rely on friction to move the eyepiece back and forth until the image looks best, and hold it there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When using the instrument in practice, you'll forget the ideal distance. What you will do is adjust the position of the eyepiece until the image looks best. You will do that every time you observe, and often multiple times during the same observation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you want some math, take a look at the thin lens equation, and apply it to the objective lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f = focal length of the lens&lt;/p&gt;&#xA;&#xA;&lt;p&gt;o = distance from lens to object&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i = distance from lens to image&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the thin lens equation is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1/f = 1/i + 1/o&#xA;&#xA;i = 1/(1/f - 1/o)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If o = infinity, then i = f.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what happens if o = 1 A.U.?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i = 1/(1/1 - 1/(1.5 * 10^11)) = 1.0000000000067 meters&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference is something like 6.7 * 10^-12 meters. It's smaller than an atom.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2015-11-16T19:06:07.450" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12519" PostTypeId="2" ParentId="12514" CreationDate="2015-11-16T19:54:03.677" Score="5" Body="&lt;p&gt;At this stage of our solar system it is impossible for something with size of moon to fall in inner orbit. But smaller things like asteroids do fall in, and become captured by earths gravity. but they aren't big enough to see with naked eye. For example 2006 RH120 is a near earth asteroid which orbited Earth from September 2006 to June 2007.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OlZaL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OlZaL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9786" LastActivityDate="2015-11-16T19:54:03.677" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12520" PostTypeId="1" AcceptedAnswerId="12524" CreationDate="2015-11-16T20:16:45.327" Score="4" ViewCount="114" Body="&lt;p&gt;How do the size and heat of stars affect development of solar systems and what kinds of planets can develop? Is it possible to orbit supermassive stars? Furthermore, would star's unleashing violent winds such as wolf-rayet stars be candidates for solar system formations?&lt;/p&gt;&#xA;" OwnerUserId="9645" LastActivityDate="2015-11-17T07:10:21.700" Title="Is the size and heat of a star important in the development of a solar system?" Tags="&lt;star&gt;&lt;orbit&gt;&lt;solar-system&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12524" PostTypeId="2" ParentId="12520" CreationDate="2015-11-17T04:59:05.390" Score="4" Body="&lt;p&gt;I think the quick answer to this is no, at least, any solar-system around a Wolf-Rayet or similarly large star wouldn't have time to develop.   I would think that large stars have solar-systems cause I see no reason why they wouldn't, but the stars don't last long enough for their solar-systems to develop much.   The formation period, including bombardment period(s) and time for the planets to cool down from the heat of formation takes a significant amount of time, perhaps a couple hundred million years minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our solar system's &lt;a href=&quot;https://en.wikipedia.org/wiki/Late_Heavy_Bombardment&quot; rel=&quot;nofollow&quot;&gt;late heavy bombardment&lt;/a&gt; occurred between about 4.1 and 3.8 billion years ago, ending some 700 million years after our solar-system formed and Earth didn't cool down enough to have a magnetic field until about &lt;a href=&quot;http://www.wired.com/2010/03/earths-magnetic-field-is-35-billion-years-old/&quot; rel=&quot;nofollow&quot;&gt;3.5 billion years ago&lt;/a&gt;, roughly 1 billion years after formation.   Granted, that's a sample size of one, so numbers could vary quite a bit, but I have a hard time imagining that planet's could cool down and complete most of their early bombardment in the lifespan of a very large star.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For developed solar-systems where the planets had time to cool and the solar-systems had time to mostly coalesce and get past early-bombardment, I don't think there's much of a window beyond maybe 2 solar masses  (a 2 solar mass star would have roughly 1/16th the lifespan of our sun, or a bit less than a billion years in main sequence), based on an exponential ratio to the 4th power of the star's mass.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Mass%E2%80%93luminosity_relation&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as to whether a solar-system could survive a Wolf-Rayet star's enormous ejections, that's another question.   I'm not 100% sure but large planets are pretty resilient, so my guess is yes, but it's just a guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another interesting question is gravity vs luminosity.   Our solar-system the furthest planet Neptune, about 30 astronomical units from the Sun, and, presumably a planet could orbit a fair bit further out than that and be in a stable orbit, but lets go with 30 AU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A star with 10 times the mass of our sun during it's main sequence has roughly 3,000 times the luminosity (using the exponential relationship of 3.5 Mass to Luminosity - see &lt;a href=&quot;http://cseligman.com/text/stars/mldiagram.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravity for 10 solar masses, gravity reduces by the square of the distance, so square root of 10, a Neptune equivalent orbital velocity would be about 95 Astronomical Units, so, at 95 astronomical units from a star with 3,000 times the luminosity of our sun, the energy received from that star at that distance would be roughly 1/3rd the energy the Earth gets from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Too much larger than that, somewhere around 20 solar masses or, not much more than that, the star gets too hot to have much of a solar-system within it's theoretical Goldilocks zone.   Not that that matters much, cause the short lifespan of the star kind of answers the question anyway, but my guess would be, if planets survive around a Wolf-Rayet star, even the outer parts of such a solar-system should be hot unless they were enormously distant, like a couple hundred AUs.   There might be no ice or water planets around a star of that size.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-11-17T07:10:21.700" LastActivityDate="2015-11-17T07:10:21.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12527" PostTypeId="1" AcceptedAnswerId="12528" CreationDate="2015-11-17T16:16:32.103" Score="2" ViewCount="76" Body="&lt;p&gt;I'm continue working on my Planetarium software and I have a doubt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this moment my star catalogue is very small, but I'm planning to use a bigger one. To optimize the search inside this catalogue I think I have to divide it in regions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These regions will be delimited by right ascension and declination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using stars' right ascension and declination, how can I divide the sky to optimize my search?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to retrieve all stars visible at this moment. Maybe not all of them, only the visible by the user. For example, if user is looking to the North, he/she won't see the stars at the South. The region will be the area visible by the user at this moment, and I want to get all stars for that given region using also a limit magnitude.&lt;/p&gt;&#xA;" OwnerUserId="7566" LastEditorUserId="7566" LastEditDate="2015-11-19T10:29:47.093" LastActivityDate="2015-11-19T10:29:47.093" Title="Divide a star catalogue to optimize star search" Tags="&lt;star&gt;&lt;software&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12528" PostTypeId="2" ParentId="12527" CreationDate="2015-11-17T16:42:40.250" Score="2" Body="&lt;p&gt;This is really a computing problem, but I suppose the only point from an astronomical perspective is what the RA, Dec distributions of your catalogues look like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not that familiar with optimal search techniques but I guess that you want roughly similar numbers of stars in each region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your catalogue is just of the brightest stars, then these are fairly uniformly distributed around the sky, so splitting them up in equal bands of right ascension would probably do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you are dealing with much larger catalogues then there is likely to be a significant concentration towards the galactic plane. In these circumstance you might be better off splitting your regions in terms of their galactic latitude. So you would calculate galactic coordinates by transforming the RA, Dec and then plot the distribution of number of stars versus galactic latitude to see where you would split your regions.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-11-17T16:42:40.250" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12529" PostTypeId="2" ParentId="2594" CreationDate="2015-11-17T17:23:47.977" Score="4" Body="&lt;p&gt;You'd probably be most interested in the results of the Lunar Dust EXperiment (LDEX).  A &lt;a href=&quot;http://www.hou.usra.edu/meetings/leag2015/pdf/2083.pdf&quot; rel=&quot;nofollow&quot;&gt;2015 paper&lt;/a&gt; states&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;LDEX data show no evidence for an electrostaticallylofted&#xA;  dust component at densities greater than a few&#xA;  per m3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am assuming the solar flare stuff didn't pan out, otherwise it would be mentioned in the various &lt;a href=&quot;http://www.sciencedaily.com/releases/2015/06/150617135035.htm&quot; rel=&quot;nofollow&quot;&gt;LDEX summary papers&lt;/a&gt;.  In addition, the NASA DREAM group site doesn't seem to have anything relevant LADEE results, and they're the ones that brought the subject up in the first place.&lt;/p&gt;&#xA;" OwnerUserId="9749" LastActivityDate="2015-11-17T17:23:47.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12530" PostTypeId="2" ParentId="12514" CreationDate="2015-11-17T18:25:52.397" Score="0" Body="&lt;p&gt;Here's a second and third moon: &lt;a href=&quot;https://en.wikipedia.org/wiki/Telstar&quot; rel=&quot;nofollow&quot;&gt;Telstar 1 and 2&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Telstar 1 and 2—though no longer functional—still orbit the Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=R0jUvufQgQo&quot; rel=&quot;nofollow&quot;&gt;Here's a song about Telstar 1&lt;/a&gt;. It was a hit for The Tornados back in 1962.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-11-17T18:25:52.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12531" PostTypeId="1" AcceptedAnswerId="12532" CreationDate="2015-11-17T18:36:37.307" Score="13" ViewCount="2766" Body="&lt;p&gt;So, we have surface pictures from two alien planets, Venus (captured by the Venera 13), and Mars (captured by the rover).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of these pictures appear to be very dusty. For Venus we see strong storms; that makes sense. However, the Martian air also appears very dusty. Is Earth relatively unique in its clear atmosphere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BYQw2.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BYQw2.jpg&quot; alt=&quot;Martian Landscape&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&#xA;Mars&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dIjXZ.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dIjXZ.jpg&quot; alt=&quot;Venusian Landscape&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&#xA;Venus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kllDJ.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kllDJ.jpg&quot; alt=&quot;Titan Landscape&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&#xA;Titan (Thanks LocalFluff!)&lt;/p&gt;&#xA;" OwnerUserId="4026" LastEditorUserId="7982" LastEditDate="2021-02-28T11:51:10.873" LastActivityDate="2021-02-28T11:51:10.873" Title="Is Earth unique in its fairly clear atmosphere?" Tags="&lt;earth&gt;&lt;planetary-atmosphere&gt;&lt;atmospheric-effects&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12532" PostTypeId="2" ParentId="12531" CreationDate="2015-11-17T20:34:05.060" Score="21" Body="&lt;p&gt;Our atmosphere is only transparent to visible light, In most other wavelengths, some or all of the light is absorbed &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wvref.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wvref.png&quot; alt=&quot;Image from wikipedia&quot;&gt;&lt;/a&gt; &#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Optical_window#/media/File:Atmospheric_electromagnetic_opacity.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image from Wikipedia&lt;/a&gt;, adapted from image by NASA&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our eyes have evolved to take advantage of the transparency at these wavelengths. If we had evolved in an atmosphere with a very different mix of gases. One in which visible light was absorbed, we would have evolved eyes that see different wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a notable &quot;window&quot; at about $10\mu m$ in the diagram above. And you might wonder if any animal has evolved to see in this window. However, &lt;a href=&quot;https://en.wikipedia.org/wiki/Black-body_radiation#Human%20body%20emission&quot; rel=&quot;nofollow noreferrer&quot;&gt;our own bodies emit thermal radiation at about $10\mu m$&lt;/a&gt;, eyes wouldn't work as they would be swamped by their own glow. However this window is used by thermal imaging devices.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2015-11-18T17:51:06.160" LastActivityDate="2015-11-18T17:51:06.160" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="12534" PostTypeId="1" CreationDate="2015-11-17T23:13:41.210" Score="3" ViewCount="1042" Body="&lt;p&gt;I know that each element has it's own distinct spectra (this is a good site that lists them: &lt;a href=&quot;http://chemistry.bd.psu.edu/jircitano/periodic4.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Periodic Table&lt;/a&gt;) but when looking at a spectra of a galaxy (or even a star) how do we match those elements to their distinct spectra? Looking at this image from &lt;a href=&quot;http://www.ifa.hawaii.edu/~barnes/ASTR110L_F05/spectralab.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Spectra in the Lab&lt;/a&gt; : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e54tm.png&quot; alt=&quot;spectra&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the lines in the spectra overlap. The galaxies are also moving away, so it makes it more difficult to figure out the elements producing the emission/absprtion lines, right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, each element usually has more than one absorption/emission line, so how can we say that just one line is the H-alpha line, or the K and H calcium lines, for example, as in this spectra: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wFjtd.gif&quot; alt=&quot;spectra&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(image from: &lt;a href=&quot;http://spiff.rit.edu/classes/phys301/lectures/doppler/doppler.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://spiff.rit.edu/classes/phys301/lectures/doppler/doppler.html&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="2913" LastEditorUserId="7472" LastEditDate="2018-12-28T23:15:21.103" LastActivityDate="2019-03-28T19:24:28.660" Title="How do we know which elements are in a galaxy spectra?" Tags="&lt;galaxy&gt;&lt;spectra&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="12535" PostTypeId="2" ParentId="12534" CreationDate="2015-11-18T01:03:52.110" Score="6" Body="&lt;p&gt;Because the wavelength ratio of the lines remains constant despite any cosmological red shift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the redshift is &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;, &lt;em&gt;all&lt;/em&gt; the lines are shifted redward in wavelength by a factor &lt;span class=&quot;math-container&quot;&gt;$(1+z)$&lt;/span&gt;. This means that a pattern of lines can still be recognisable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also have a pretty good idea of what the spectra should look like, which chemical elements will produce visible absorption features with what relative strengths and so on (see below). This usually makes identification of line features straightforward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course if there were just a single line visible in the spectrum (it does happen, usually in high redshift quasars) it can be difficult to pinpoint the redshift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of analysing what's in a Galaxy, well usually the light is dominated by the mixed spectrum of billions of stars. This spectrum is interpreted and modelled using galaxy evolution models and population synthesis models that predict a spectrum from a given ensemble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a Galaxy is near enough, the chemical abundances of its interstellar medium can be estimated from resolved spectra of emission nebulae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In comparison, interpreting spectra from an individual star is trivial. Hundreds if not thousands of absorption lines can easily be identified and matched with the predictions of very detailed stellar atmosphere models to estimate chemical abundances. These models contain up to millions of possible radiative transitions as well as the various line strengths and broadening processes that affect the spectrum.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2019-03-28T19:24:28.660" LastActivityDate="2019-03-28T19:24:28.660" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12536" PostTypeId="1" AcceptedAnswerId="12538" CreationDate="2015-11-18T09:31:05.460" Score="8" ViewCount="772" Body="&lt;p&gt;A rogue planet (or a rogue anything, for that matter: a celestial body other than a star) is something that's drifting through space without being attached by gravity to any star. They're just out there in the big black void. Is there any evidence that they actually exist? Or are they purely hypothetical?&lt;/p&gt;&#xA;" OwnerUserId="9510" LastActivityDate="2015-11-19T00:02:44.990" Title="Is there any hard evidence that rogue planets exist?" Tags="&lt;rogue-planet&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12537" PostTypeId="2" ParentId="12531" CreationDate="2015-11-18T09:49:05.883" Score="8" Body="&lt;p&gt;No, the clarity of the Earth's atmosphere cannot be considered unique. We don't have to speculate about exoplanets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could argue the answer is no, because both the Moon and Mercury have (very, very) thin atmospheres, and these are obviously &quot;clear&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you regard that argument as tricksy, then we can turn to Mars. Yes Mars has occasional dust storms. In normal conditions, the optical depth of the Martian atmosphere is usually somewhere between 0.5 and 1 per airmass. (&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0032063311002960&quot;&gt;Petrova et al. 2012&lt;/a&gt;; &lt;a href=&quot;http://arxiv.org/abs/1403.4234&quot;&gt;Lemmon et al. 2014&lt;/a&gt;). Most of this extinction is caused by dust and is nearly wavelength independent. i.e. between 60% and 37% of light would travel through it's atmosphere from outside. This compares with typical extinctions of about  0.2-0.4 &lt;em&gt;magnitudes&lt;/em&gt; of visual extinction per airmass on Earth (0.1 mag at the best astronomical sites in the world), corresponding to 80% to 69% of light passing through the Earth's atmosphere from outside (to sea level). Most of this extinction is due to dust, though there is some absorption by water and other aerosols).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, though Mars is dustier than Earth on average, it is not outrageously so. It would be stretching the use of the word unique to say that the clarity of the Earth's atmosphere was &quot;unique&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-11-18T09:49:05.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12538" PostTypeId="2" ParentId="12536" CreationDate="2015-11-18T10:29:47.410" Score="11" Body="&lt;p&gt;Rogue planets have been discovered by infra-red imaging because planets are hot when they form.  Here's a &lt;a href=&quot;https://en.wikipedia.org/wiki/Rogue_planet#Known_or_possible_rogue_planets&quot;&gt;list&lt;/a&gt; of a few.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's 2 types of Rogue planets.  One is failed stars.  condensing pockets of gas and dust that form similar to how our solar-system formed, but that are too small to form stars.   &lt;a href=&quot;http://www.universetoday.com/104210/rogue-planets-could-form-on-their-own-in-interstellar-space/&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 2nd type is planets that escape from a star's orbit.   This can happen by the star ejecting material and losing mass which causes the planets to expand in more distant orbits, some eventually escaping, or by gravitational assist either planet on planet or by two stars passing quite close to each other.    It's statistically impossible for Rogue Planets not to, from time to time, get ejected from solar systems, so they have to exist at least in reasonably large numbers throughout the galaxy, though I don't think it's well known how common there are.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our solar-system might have ejected a planet - see &lt;a href=&quot;https://en.wikipedia.org/wiki/Hypothetical_fifth_giant_planet&quot;&gt;here&lt;/a&gt;.   In general, larger planets can eject smaller ones but mostly not the other way around, but two stars that pass too close to each other can eject any planets, mostly ones with more distant orbits.  A star that loses a lot of it's mass or a star that goes nova can push planets out of it's orbit too.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-11-18T10:29:47.410" CommentCount="25" ContentLicense="CC BY-SA 3.0" />
  <row Id="12539" PostTypeId="1" CreationDate="2015-11-18T13:49:48.900" Score="9" ViewCount="99" Body="&lt;p&gt;If we were able to send a probe to 550AU from the Sun, which exoplanets could be examined, to what resolution, and what form would the examination take?  It is radio telescopy only, or would it be possible to optically see what the surface of an exoplanet looked like?&lt;/p&gt;&#xA;" OwnerUserId="4146" LastActivityDate="2016-04-17T15:19:20.967" Title="What are the best candidates for solar lensing of exoplanets, and what would the results be?" Tags="&lt;gravitational-lensing&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12541" PostTypeId="1" CreationDate="2015-11-18T15:37:48.250" Score="4" ViewCount="1427" Body="&lt;p&gt;I'm a bit puzzled and not an observer, so please bear with me if I'm being stupid here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The code &lt;a href=&quot;http://www.iap.fr/magphys/magphys/MAGPHYS.html&quot; rel=&quot;nofollow&quot;&gt;MAGPHYS&lt;/a&gt; specifies Jansky (&lt;code&gt;Jy&lt;/code&gt;) as the input unit for flux through a filter (see &lt;code&gt;Section 3.2.3&lt;/code&gt; in the &lt;a href=&quot;http://www.iap.fr/magphys/magphys/MAGPHYS_files/readme.pdf&quot; rel=&quot;nofollow&quot;&gt;documentation&lt;/a&gt;). Meaning MAGPHYS wants me to use &lt;code&gt;Jy&lt;/code&gt; as unit, when specify the total light of a whole galaxy in e.g. the SDSS-g band. But &lt;code&gt;Jy&lt;/code&gt; is a unit of spectral &lt;em&gt;density&lt;/em&gt;, i.e. &lt;code&gt;W/m²/Hz&lt;/code&gt;. I would expect a flux as input unit when talking about the amont of light coming through a filter, i.e. in &lt;code&gt;W/m²&lt;/code&gt; or in $L_\odot$ (solar luminosity), so integrated over all frequencies that go through the filter and weighted by the filter response function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have the absolute AB magnitude of a galaxy in the SDSS-g band, how do I convert it to &lt;code&gt;Jy&lt;/code&gt; so that MAGPHYS will be happy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_{\text{AB}\nu} = -2.5 \log_{10} \left(\frac{f_{\nu}}{3631\text{Jy}}\right) \qquad \text{or} \qquad \left(\frac{f_{\nu}}{\text{Jy}}\right) = 10^{-0.4\  (m_{\text{AB}\nu} - 8.9)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that is in one frequency, not a whole band. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, I have found the formula below in &lt;a href=&quot;http://www.star.bris.ac.uk/~mbt/topcat/&quot; rel=&quot;nofollow&quot;&gt;TOPCAT&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(\frac{F}{\text{Jy}}\right) = 10^{\left(23-0.4 \ (m_\text{AB}+48.6)\right)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does the &lt;code&gt;23&lt;/code&gt; in the exponent come from?&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2015-11-18T17:20:04.630" LastActivityDate="2021-05-12T11:19:03.300" Title="Conversion of magnitudes to Jansky and MAGPHYS?" Tags="&lt;luminosity&gt;&lt;magnitude&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12543" PostTypeId="1" AcceptedAnswerId="12544" CreationDate="2015-11-18T19:06:29.027" Score="6" ViewCount="416" Body="&lt;p&gt;...or how small can a solar system be, to sustain an Earth-like planet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sun is not really small, and 1AU is pretty far when you look at orbital radii of exoplanets and size of their stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How small a star can be, to still provide Earth's equivalent of 1.36 kW/m&lt;sup&gt;2&lt;/sup&gt; above its own Roche Limit (of Earth-sized planet)? What would the star's diameter and the orbit providing such irradiance be in such a case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess a Brown Dwarf wouldn't be enough to give that much within its Roche Limit, meaning any planet receiving enough light would be torn apart by tidal forces, and likely stepping up to smallest Main Sequence stars would step up the energy output enough that a planet at the Roche Limit would be scorched like Mercury, so its orbit would need to be considerably larger... but I don't quite know how to perform the needed calculations.&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2017-10-08T11:23:37.503" Title="How small a star can provide Sun-level illumination to its planets?" Tags="&lt;star&gt;&lt;earth-like-planet&gt;&lt;roche-limit&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12544" PostTypeId="2" ParentId="12543" CreationDate="2015-11-18T22:00:37.897" Score="10" Body="&lt;p&gt;I consider small stars to mean main sequence stars (or brown dwarfs) with mass less than the Sun, and then also consider compact stellar remnants - white dwarfs and neutron stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Main sequence stars and brown dwarfs&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you need is a mass-luminosity relation combined with an expression for the tidal radius in terms of the stellar mass. The latter also depends on the mass of the planet, so when you say earth-sized, I'll assume that means mass and radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So going through the calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flux at the planet is $L/4\pi r^2$, where $L$ is the luminosity and $r$ the orbital radius (assumed circular).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's next use an approximate relation that&#xA;$$\frac{L}{L_{\odot}} = \left(\frac{M}{M_{\odot}}\right)^3$$&#xA;A more accurate numerical relationship could be obtained from evolutionary model calculations. There is a complication that the luminosity of very low mass stars and brown dwarfs does depend on age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a rigid body the tidal radius (Roche limit) is&#xA;$$ r_t\simeq 1.4 R_E\left(\frac{M_{\odot}}{M_E}\right)^{1/3},$$&#xA;where $R_E$ and $M_E$ are the radius and mass of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So setting $r=r_t$, the flux to $1400$ W m$^{-2}$ and replacing $L$ in terms of mass, we find&#xA;$$\frac{M}{M_{\odot}} = \left[\frac{4\pi \times 1400}{L_{\odot}} (1.4 R_E)^2 \left(\frac{M_{\odot}}{M_E}\right)^{2/3}\right]^{1/3}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that remains is to put the numbers in and I get $M = 0.026 M_{\odot}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, a small brown dwarf - but there are caveats. First, and most importantly, as I said, the luminosity-mass relation is a bit rough and ready, and is certainly age dependent for something as low mass as this. Second, the expression for the Roche limit depends a bit on structural properties of the planet, but I think this introduces relatively little uncertainty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use that mass and the tidal radius formula, you find that the orbital radius is only $5R_E$. As the minimum size of a brown dwarf is about a Jupiter radius, it appears that tidal breakup would not be the limiting factor &lt;em&gt;for an Earth-like planet&lt;/em&gt;. Also note that the spectrum of radiation from a brown dwarf might contain the same power per unit area as sunlight, but it has a very different spectrum; mostly infrared with almost nothing in the visible band.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Compact stellar remnants&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smaller &quot;stars&quot; are possible in the form of the collapsed remnants known as white dwarfs or neutron stars. Planets &lt;em&gt;have&lt;/em&gt; been detected around neutron stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;White dwarf stars&lt;/strong&gt; are &quot;born&quot; very hot, cool rapidly and then more slowly. To answer your question requires an addendum, which is a minimum time you want this illumination for. A newborn WD has about a solar luminosity and has a similar diameter to the Earth. So you planet could be 1 au away from this and still receive a sun-like power per unit area. BUT, it would be short-lived and the spectrum of the light would be nothing like the Sun. The WD would have a temperature in excess of 50,000K and most of the light would be in the UV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cooling curves (and observations) suggest they cool to $10^{-4}$ solar luminosities in less than a billion years and have more sun-like temperatures, but to receive sun-like illumination at this point, your planet would need to be only 0.01 au from the white dwarf. White dwarfs have a mass a little less than the Sun. Using the same tidal radius formula as above we get $r_t \sim 0.004$au. So I conclude that yes, it might be possible to have a long-lived planet with Earth-like illumination around a white dwarf. But note (a) conditions change rapidly (compared with the Sun) and (b) you have to arrange to get the planet that close, since an orbit of that size would have been &lt;em&gt;inside&lt;/em&gt; the red giant progenitor of the white dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Neutron stars&lt;/strong&gt; are born in the cores of supernovae explosions and have a radius of about 10 km and mass about 1.5 times that of the Sun. &lt;em&gt;If&lt;/em&gt; your planet can survive the supernova (or be constructed afterwards) it faces a similar set of problems to a planet around a white dwarf. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutron stars are born with surface temperatures of billions of degrees, emitting copious gamma ray radiation. They cool very rapidly - within a million years the surface temperature is down to an X-ray emitting million degrees and perhaps to UV-emitting temperatures of tens of thousands within 100 million years. It may be that they never cool much below this, since they are heated by their rotational spin-down, the decay of their strong magnetic fields and accretion of gas from the interstellar material. Even at 50,000K, because of its small size, the neutron star has a luminosity of only a millionth that of the Sun. That means your planet would have to be moved within 0.001 au of the neutron star to receive the same power and this would be well &lt;em&gt;inside&lt;/em&gt; the tidal breakup radius of $\sim 0.005$ au.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I conclude that there really is no long-term possibility for sun-like illumination from a neutron star. Note also, that I have ignored magnetospheric and pulsar-like emission from the neutron star. These are also short-lived phenomena only seen in the first million years or so of a neutron stars existence.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-10-08T11:23:37.503" LastActivityDate="2017-10-08T11:23:37.503" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12545" PostTypeId="1" CreationDate="2015-11-18T23:02:57.237" Score="2" ViewCount="109" Body="&lt;p&gt;This isn't a question so much as a request. I'm looking for an image of the Orion constalletion I could use, but I would like to have a very high resolution HD image, whereby a lot of the deep sky objects can be seen to high detail other than just the 7 hourglass stars of the belt and four corners. Would love to have perhaps the Orion nebula (Messier 42), the Horsehead Nebula (near ζ Orionis), the emission nebula NGC 2174, Messier 43 and Barnard's Loop.  &lt;/p&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2015-11-24T20:21:04.267" Title="Orion Constellation request" Tags="&lt;amateur-observing&gt;&lt;observational-astronomy&gt;&lt;photography&gt;&lt;orion&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12546" PostTypeId="2" ParentId="12536" CreationDate="2015-11-19T00:02:44.990" Score="5" Body="&lt;p&gt;Giant planets when first formed are big and hot. They radiate their own light, mostly in the infrared. So &lt;em&gt;young&lt;/em&gt; isolated planets can be seen directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been various claims in the literature that objects as small as a few Jupiter masses have been identified in young star forming regions. See various papers by the IAC brown dwarf research group&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2000Sci...290..103Z&quot; rel=&quot;noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2000Sci...290..103Z&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2002ApJ...578..536Z&quot; rel=&quot;noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2002ApJ...578..536Z&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2014A%26A...568A..77Z&quot; rel=&quot;noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2014A%26A...568A..77Z&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2013MmSAI..84..926Z&quot; rel=&quot;noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2013MmSAI..84..926Z&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another object that is part of the beta Pic moving group, recently discovered by Liu et al. (2013), has an estimated mass of about 8 Jupiter masses (Biller et al. 2015).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1510.07625&quot; rel=&quot;noreferrer&quot;&gt;http://arxiv.org/abs/1510.07625&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These claims are open to criticism - sometimes it is hard to tell whether a faint object really belongs to the star forming region observed, rather than being an unassociated background object. The claimed masses also depend heavily on models for the luminosity-mass relation as a function of age, and the ages of these objects are not easily constrained. The likelihood is that at least some of these objects are below 10 Jupiter masses and would rank as planets by some definitions; though none of the individual objects could be said to be proven beyond any doubt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless it would not be surprising if, in the maelstrom of the formation of a cluster of stars, some planetary systems were stripped from their parent stars by close encounters with other objects and indeed numerical simulations of planetary systems in dense star clusters show that this process occurs (e.g. Davies 2011).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2011IAUS..276..304DD&quot; rel=&quot;noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2011IAUS..276..304DD&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chances of seeing older, isolated, planetary mass objects are slim, but microlensing appears to be the only technique presently available. The microlensing signature of a free-floating planet is of course unrepeatable so a discovered planet could not be followed up in any way. However, surveys of microlensing events could be a way of saying something statistically about how common such objects are. See for example &lt;a href=&quot;http://astrobites.org/2011/05/24/free-floating-planets-might-outnumber-stars/&quot; rel=&quot;noreferrer&quot;&gt;http://astrobites.org/2011/05/24/free-floating-planets-might-outnumber-stars/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also worth noting that the whether these things really are &quot;planets&quot; at all is disputed. They could either be genuine planets, formed in the same way that is hypothesised for most giant planets - that is by accretion onto a rocky core that formed around a star. They could then have been displaced from their parent star by dynamical interactions with other bodies in their system or with a third body. As I said above, N-body simulations do predict that this will happen (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013ApJ...772..142L&quot; rel=&quot;noreferrer&quot;&gt;Liu et al. 2013&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand they could represent the very lowest mass gas fragments that are able to form during the collapse and fragmentation of a molecular cloud and that for some reason were unable to accrete further gas (i.e. they are really more like low-mass brown dwarfs). This so-called &quot;fragmentation limit&quot; is of order 10 Jupiter masses, but if it were a little lower it might explain the free-floating &quot;planets&quot; that have been seen so far.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-11-19T00:02:44.990" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12547" PostTypeId="1" CreationDate="2015-11-19T01:54:39.273" Score="2" ViewCount="142" Body="&lt;p&gt;This &lt;a href=&quot;https://astronomy.stackexchange.com/questions/12543/how-small-a-star-can-provide-sun-level-illumination-to-its-planets&quot;&gt;question&lt;/a&gt; got me thinking about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter has, and presumably super-Jupiters can have a strong magnetospheres.  A solid metallic core and rotating material around the core, creates a magnetic field and all 4 gas planets in our solar-system have magnetic fields, though Jupiter's is by far the strongest.   &lt;a href=&quot;http://www.astronomynotes.com/solarsys/plantblb.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, a planetary magnetic field requires a solid core and a slight variation in rotation where the liquid outer core rotates in relation to the solid inner core.  The planet's speed of rotation might not be necessary, but it could be a factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the hot period of formation, gas giants might not form magnetic fields but when they cool down enough to have a solid core, they probably would have them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the question is, is there a rough mass for which planet-wide magnetic field formation becomes unstable, where the core of the heavy Jupiter or small or old brown dwarf takes a very very long time to solidify, say, many billions of years?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd think the high temperature of formation and additional heat from fusion in a Brown Dwarf wouldn't be a good situation for that kind of star-wide magnetic field to form due too inevitable convection and probably no solid core.   That kind of star would probably form multiple magnetic dipoles like our sun has, but perhaps in an older brown dwarf a single magnetic field might be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/White_dwarf#Magnetic_field&quot; rel=&quot;nofollow noreferrer&quot;&gt;White dwarfs&lt;/a&gt; have very strong magnetic fields, so pressure obviously isn't an issue, in fact, I think high pressure can create a stronger magnetic field.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-11-19T07:54:48.027" Title="Formation of a magnetosphere for gas giants / small stars" Tags="&lt;magnetic-field&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12548" PostTypeId="1" AcceptedAnswerId="39942" CreationDate="2015-11-19T02:05:02.097" Score="5" ViewCount="171" Body="&lt;p&gt;Has any official information been published regarding the estimated age of the Occator crater on Ceres?  I ran some quick searches but couldn't find anything putting a definitive (or even speculative) date on the moment of impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask because based upon my own very amateur analysis of the imagery:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Sd3Uu.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sd3Uu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...there seems to be a marked decrease in the amount of secondary cratering visible within a fairly uniform radius of the main crater.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My assumption is that when the impact occurred, the surrounding area was likely blanketed by material kicked up from the impact site, obscuring most pre-existing craters (smaller ones, in particular).  Kind of like fresh snowfall, except made of rock.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for a (small) secondary crater to be visible near Occator, the impact would have to have occurred &lt;em&gt;after&lt;/em&gt; the impact that created Occator crater.  Meaning that the relative lack of small craters near Occator would seem to imply a fairly recent impact (&quot;recent&quot; in terms of Ceres geological timescales), I believe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any official data that would confirm or refute this?&lt;/p&gt;&#xA;" OwnerUserId="9687" LastActivityDate="2020-11-19T09:53:06.263" Title="Age of Occator Crater" Tags="&lt;ceres&gt;&lt;crater&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12549" PostTypeId="2" ParentId="12545" CreationDate="2015-11-19T02:42:04.273" Score="0" Body="&lt;p&gt;Install either Stellarium or Cartes du Ciel, drag the sky chart around, zoom in or out, frame the constellation any way you like, then take a screenshot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.stellarium.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.stellarium.org/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ap-i.net/skychart/en/start&quot; rel=&quot;nofollow&quot;&gt;http://www.ap-i.net/skychart/en/start&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2015-11-19T02:42:04.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12550" PostTypeId="2" ParentId="12541" CreationDate="2015-11-19T04:04:55.033" Score="4" Body="&lt;p&gt;Caveat this with the fact that I am not familiar with this code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The input asks you for the flux through the filter, but a filter has a bandpass, so if you have a 10 nm bandpass filter and you measure $1 W/m^2$, then you measured $0.1 W/m^2/nm$ (sorry, I'm a physicist so I never got used to measuring wavelengths in Hz).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your second question, the factor of 23 is because in cgs units a Jansky is $10^{23} erg/s/Hz/cm^{-1}$ and that TOPCAT equation is in cgs (the equation is $m_\text{AB} = -\frac{5}{2}\,log_{10}\,f_v - 48.6$) and if you are normalizing it to Janskys, you need to divide by $10^{-23}$.&lt;/p&gt;&#xA;" OwnerUserId="9749" LastEditorUserId="2246" LastEditDate="2015-11-20T11:57:56.813" LastActivityDate="2015-11-20T11:57:56.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12551" PostTypeId="2" ParentId="12548" CreationDate="2015-11-19T04:22:55.283" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://gsa.confex.com/gsa/2015AM/webprogram/Paper260706.html&quot; rel=&quot;nofollow&quot;&gt;It seems&lt;/a&gt; not yet.  &lt;a href=&quot;https://gsa.confex.com/gsa/2015AM/webprogram/Paper260352.html&quot; rel=&quot;nofollow&quot;&gt;The non-uniform density of craters&lt;/a&gt; seen on Ceres is still under investigation.  From the abstracts of those papers, they're still working on trying to date the ejecta material from that crater, and trying to understand the resurfacing that appears to be going on.&lt;/p&gt;&#xA;" OwnerUserId="9749" LastActivityDate="2015-11-19T04:22:55.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12552" PostTypeId="2" ParentId="12545" CreationDate="2015-11-19T07:33:34.187" Score="0" Body="&lt;p&gt;I think what you want it for is all important?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use World Wide Telescope (&lt;a href=&quot;http://www.worldwidetelescope.org&quot; rel=&quot;nofollow&quot;&gt;http://www.worldwidetelescope.org&lt;/a&gt;) as a demonstration tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use it to look at any part of the sky using a variety of catalogues and imagery. The default &quot;background&quot; imagery is from the digitised sky survey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way it works is that if you zoom into an area it downloads higher and higher resolution images, but you can cache these. Thus you can scan and cache the whole of Orion so that you can pan and zoom (particularly effective with a mouse wheel) rapidly and at will at a later date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also supplement the DSS images with (e.g. HST images of the Orion Nebula cluster), which are overlaid and rotated correctly onto the DSS image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have an HD display, then of course you can take a screenshot of any view you arrange.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-11-19T07:33:34.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12553" PostTypeId="2" ParentId="12547" CreationDate="2015-11-19T07:48:28.230" Score="3" Body="&lt;p&gt;We do not know whether Jupiter has a solid core. That is one of the primary goals of the Juno mission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general it is clear that a solid core is not required to generate magnetic fields. The Sun does so and magnetic fields have been measured (with average surface field strengths of around 3 kG) on very low mass stars. Incidentally, the Sun's magnetic field does have a large scale dipole component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More recently, there have been observations of cyclotron and auroral emission suggesting that even old, cold brown dwarfs have strong, organised magnetic fields. See for example Kuznetsov et al. (2012); Hallinan et al.(2015) and references therein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1111.7019&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1111.7019&lt;/a&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/1507.08739&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1507.08739&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment, nobody can say for sure how exactly these fields are generated, sustained and organised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact you could do worse than look at a paper that popped up at the top of today's astro-ph listings by Williams et al.  The abstract begins:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The well-studied M9 dwarf TVLM 513-46546 is a rapid rotator (P_rot ~ 2 hr) hosting a stable, dipolar magnetic field of ~3 kG surface strength.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1511.05559&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1511.05559&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-11-19T07:54:48.027" LastActivityDate="2015-11-19T07:54:48.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12554" PostTypeId="1" AcceptedAnswerId="12555" CreationDate="2015-11-19T10:23:10.917" Score="1" ViewCount="612" Body="&lt;p&gt;They actually do, but let's set Relativity aside for a moment. And the whole barycenter thing too. Here's the scenario:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somewhere in the backwaters of a galaxy there is this huge planet. It is NOT a failed star. It could be considered a gas giant, except most of the gas is nitrogen, with traces of oxygen. The planet has a nucleus with a rocky surface covered in part by water. There's life on the planet, maybe even intelligent life. Orbiting this planet is a dwarf star. It is sufficiently close to the planet to provide energy and warmth, and not close enough to incinerate it. Its orbital speed is sufficient to warm the equatorial parts of the planet without singing them when it passes over them and to come back to them before they freeze over. Or maybe there are two stars of slightly different sizes, traveling along different orbits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this at all possible?&lt;/p&gt;&#xA;" OwnerUserId="9510" LastActivityDate="2016-03-17T23:07:49.363" Title="Could a star revolve around a planet?" Tags="&lt;planet&gt;&lt;star-systems&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2015-11-20T18:22:47.530" ContentLicense="CC BY-SA 3.0" />
  <row Id="12555" PostTypeId="2" ParentId="12554" CreationDate="2015-11-19T10:54:34.937" Score="2" Body="&lt;p&gt;I think the answer is no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A star is much more massive than the most massive planet you could imagine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take Jupiter as an example : it is a massive planet, with a gravitational pull so intense it makes the Sun move. Not that much, but it actually does. However, the Sun is not in the slightest orbiting any planet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a star to orbit a planet, it would require a wandering star to be caught into a planet's orbit. But it's pretty sure that only the opposite can work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... No, no star orbiting around planets.&lt;/p&gt;&#xA;" OwnerUserId="9679" LastEditorUserId="10172" LastEditDate="2016-03-17T23:07:49.363" LastActivityDate="2016-03-17T23:07:49.363" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="12556" PostTypeId="1" CreationDate="2015-11-19T16:05:53.033" Score="2" ViewCount="130" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SONX0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SONX0.png&quot; alt=&quot;STAR CLASSIFICATION DIAGRAM&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this diagram the $x$-axis is spectral type for stars and the $y$-axis is B-V colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking &lt;strong&gt;Series 1&lt;/strong&gt; as the Main Sequence, &lt;strong&gt;Series 2&lt;/strong&gt; as Giants and &lt;strong&gt;Series 3&lt;/strong&gt; as Super-giants. Why the spectral class for all these stars differ from the other ones?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were to say differences, for instance, how would one express that information?&lt;/p&gt;&#xA;" OwnerUserId="9823" LastEditorUserId="9186" LastEditDate="2015-11-23T08:04:50.127" LastActivityDate="2015-11-23T08:04:50.127" Title="Why does spectral class and U-V colour correlate slightly differently for main sequence, giant and supergiant stars?" Tags="&lt;spectra&gt;&lt;observational-astronomy&gt;&lt;classification&gt;&lt;main-sequence&gt;&lt;magnitude&gt;" AnswerCount="0" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12557" PostTypeId="2" ParentId="12510" CreationDate="2015-11-19T18:43:19.850" Score="2" Body="&lt;p&gt;Yet another indication that Earth and Mars formed in the inner Solar System is that they have the same deuterium to hydrogen ratios. Primordial water trapped in 4.55 billion years old (0.017 bn years after Earth formation) lava from the deep mantle has been compared with Martian meteorites. The ratio should increase further out in the Solar System.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.abc.net.au/science/audio/2015/11/18/4354018.htm?topic=space&quot; rel=&quot;nofollow&quot;&gt;10 minutes radio interview&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.abc.net.au/news/2015-11-13/origin-of-earths-water-found-in-ancient-rocks/6935928&quot; rel=&quot;nofollow&quot;&gt;News article&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-11-19T18:43:19.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12558" PostTypeId="1" AcceptedAnswerId="18564" CreationDate="2015-11-19T19:43:39.493" Score="8" ViewCount="1154" Body="&lt;p&gt;The exoplanet &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler-10c&quot; rel=&quot;nofollow&quot;&gt;Kepler-10c&lt;/a&gt; has a mass between 15 and 19 times the mass of the Earth (making it comparable in mass to Neptune), and yet is thought to have a density of about 7g/cm&lt;sup&gt;3&lt;/sup&gt; and to be a terrestrial planet, with a substantial proportion of &quot;hot ice&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an upper limit to the mass of terrestrial planets, or can rocky planets form that are larger than Kepler-10c?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, older, article in Universe Today suggests that &lt;a href=&quot;http://www.universetoday.com/13757/how-big-do-planets-get/&quot; rel=&quot;nofollow&quot;&gt;terrestrial planets can't form more than 5-10 Earth Masses&lt;/a&gt;, substantially smaller than Kepler 10c.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2153" LastEditDate="2016-10-03T20:09:12.310" LastActivityDate="2016-10-03T20:09:12.310" Title="Is there an upper limit to the mass of terrestrial planets?" Tags="&lt;exoplanet&gt;&lt;mass&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12559" PostTypeId="2" ParentId="12554" CreationDate="2015-11-19T23:11:45.633" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;And the whole barycenter thing too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're throwing out the definition of orbit at this point, and moving into a different realm. If you throw out barycenter, you have to leave me with my frames of reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A star can definitely orbit a planet, if one such as this did in fact exist. You are assuming it is simply a planet-star system, which makes things easier. If you're on the planet, the star is orbiting you, if you're on the star (ouch), the planet is orbiting you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once they are of comparable size, the barycenter thing has to be taken into account, you can't just throw out this crucial part of the physics, even in a thought experiment. If you want that part out, the planet must be much bigger than the star.&lt;/p&gt;&#xA;" OwnerUserId="9826" LastActivityDate="2015-11-19T23:11:45.633" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12560" PostTypeId="1" AcceptedAnswerId="12570" CreationDate="2015-11-20T00:27:04.573" Score="0" ViewCount="1069" Body="&lt;p&gt;If the Universe is infinite, one would imagine that if we kept moving through space, we'd constantly find newer and newer galaxies and objects forever. This would imply that there is an infinite amount of matter in the Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat death theory states that, eventually, there will be no free energy left in the Universe. Stars would no longer form, black holes would eventually evaporate, and the Universe would be effectively dead. But if there's an infinite amount of matter, won't there always be more gas clouds that haven't formed stars, more stars that have yet to become black holes, etc.? If there is an infinite amount matter and energy, shouldn't more activity be possible no matter how many stars go out or how many black holes evaporate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One might say that with enough time, all matter in the Universe would still run dry. But if there is an &lt;em&gt;infinite&lt;/em&gt; amount, there should always be areas in which activity is still going on...right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is what I'm saying correct? Does the heat death of the Universe contradict the theory that it is infinite?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2015-11-20T21:04:56.610" LastActivityDate="2015-11-20T21:04:56.610" Title="If the Universe is infinite, would heat death be impossible?" Tags="&lt;gravity&gt;&lt;universe&gt;&lt;dark-energy&gt;&lt;fate-of-universe&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12561" PostTypeId="5" CreationDate="2015-11-20T00:33:38.497" Score="0" Body="&lt;p&gt;The fate of the Universe is a topic in cosmology, regarding if and how the Universe will come to an end. Numerous theories have been proposed for both infinite and finite futures.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2015-11-20T17:33:21.590" LastActivityDate="2015-11-20T17:33:21.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12562" PostTypeId="4" CreationDate="2015-11-20T00:33:38.497" Score="0" Body="Questions regarding the future and ultimate fate of the Universe, including the heat death, Big Freeze, Big Rip, or Big Crunch." OwnerUserId="9301" LastEditorUserId="6" LastEditDate="2015-11-20T18:28:20.723" LastActivityDate="2015-11-20T18:28:20.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12564" PostTypeId="2" ParentId="12477" CreationDate="2015-11-20T07:16:50.460" Score="0" Body="&lt;p&gt;Here's a non-mathematical answer. Something you can think about in your head. The further away a galaxy is from us the faster it will move away from us. There is no exact way to measure the exact speed because the distance is phenomenal, however, we can use red-shift to work out that this is actually happening. A way to prove that this happens is to get a balloon and draw dots on it. Blow it up a little bit, maybe half way. Focus on one dot and blow it up even more. You will notice that the other dots move away from the one you focus on faster than the dot you are looking at. This is a good way of explaining what is actually happening in the universe - and why things further away from us move away faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This applies to whatever galaxy we are in. If we were in the andromeda galaxy, we would still experience every other galaxy moving away from us faster.&lt;/p&gt;&#xA;" OwnerUserId="9645" LastActivityDate="2015-11-20T07:16:50.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12565" PostTypeId="2" ParentId="12404" CreationDate="2015-11-20T14:49:07.027" Score="3" Body="&lt;p&gt;In the Fall (early morning) and the Winter (early-late evening), Orion is a large an easily identifiable constellation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Spring (early morning) and the Summer (early-late evening), Cygnus and Pegasus are two large and easily identifiable constellations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out an application called &lt;a href=&quot;http://www.stellarium.org&quot; rel=&quot;nofollow&quot;&gt;Stellarium&lt;/a&gt;.  It's free and available for every current platform and has a large database.&lt;/p&gt;&#xA;" OwnerUserId="9832" LastEditorUserId="2365" LastEditDate="2015-11-20T17:35:34.587" LastActivityDate="2015-11-20T17:35:34.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12567" PostTypeId="1" AcceptedAnswerId="12569" CreationDate="2015-11-20T19:58:03.997" Score="3" ViewCount="679" Body="&lt;p&gt;The &lt;a href=&quot;http://www.astronomy.ohio-state.edu/~dhw/A142/notes10.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;luminosity of a black hole accretion disc&lt;/a&gt; gaining mass at a rate $\mathrm{d}M\over \mathrm{d}t$ can be estimated as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${1\over 12}{\mathrm{d}M\over \mathrm{d}t}c^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is a substantial proportion of the rest mass of the in-falling matter. The linked document explains that the factor 1/12 is because at less than 3 times the event horizon radius the matter &quot;spirals in without&#xA;radiating more energy&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What causes matter in the accretion disk not to radiate beyond this limit? It is twice the radius of the photon sphere, is the reason general relativity?&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="1203" LastEditDate="2016-11-02T10:05:25.723" LastActivityDate="2016-11-02T21:40:23.767" Title="Luminosity of black hole accretion disc" Tags="&lt;black-hole&gt;&lt;accretion-discs&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12568" PostTypeId="2" ParentId="12567" CreationDate="2015-11-20T20:16:27.857" Score="1" Body="&lt;p&gt;Yes it is to do with general relativity. For a non-rotating Schwarzschild black hole the last &lt;strong&gt;stable&lt;/strong&gt; orbit is at 3 times the Schwarzschild radius and hence the inner edge of the accretion disk.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastActivityDate="2015-11-20T20:16:27.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12569" PostTypeId="2" ParentId="12567" CreationDate="2015-11-20T20:19:46.193" Score="4" Body="&lt;p&gt;Three times the Schwarzschild radius corresponds to the closest stable circular orbit around a black hole. The general idea is that as matters moves in towards the black hole it gets stuck in an accretion disc where angular momentum has to be moved outwards in order to allow the matter to move inwards. The generic mechanism is some sort of viscosity, which heats the gas and hence you get radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, once the matter gets inside $3r_s$, that problem disappears. There are no stable orbits, no angular momentum loss or viscosity is needed and the material is able to flow (rapidly) straight into the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus when we observe black hole accretion discs we expect them to be truncated at $3r_s$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think the argument then is along the lines of - the gravitational potential energy of unit mass falling to $3r_s$ is converted into an orbital kinetic energy of $0.5v^2 = GM/6r_s$ per unit mass and the rest is converted to radiation. Thus&#xA;$$L = \left[\frac{GM}{3r_s} - \frac{GM}{6r_s}\right] \frac{dM}{dt}$$&#xA;$$ L = \frac{GM}{6r_s} \frac{dM}{dt} = \frac{1}{12}c^2 \frac{dM}{dt} .$$&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-11-02T21:40:23.767" LastActivityDate="2016-11-02T21:40:23.767" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12570" PostTypeId="2" ParentId="12560" CreationDate="2015-11-20T20:25:57.343" Score="3" Body="&lt;p&gt;Heat death of the Universe doesn't mean that there is no energy in it, it means it has reached thermodynamic equilibrium and hence no useful energy in it. This could in theory occur in an infinite Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course thermodynamics is statistical in nature, so an argument could be made that some extremely unlikely event in a region of the Universe could destroy thermal equilibrium in that region and that the larger the Universe the more likely that some region like that exists.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastActivityDate="2015-11-20T20:25:57.343" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12573" PostTypeId="2" ParentId="12151" CreationDate="2015-11-21T10:15:39.890" Score="4" Body="&lt;p&gt;I think it is referring to the speed and Lorentz factor $(\beta = v/c$ and $\gamma = [1-\beta^2]^{-1/2})$ of the gas as a whole. Within the gas, there could be particles moving with a variety of velocities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you pick up a ball of gas at 10,000 K (ouch) and throw it at 100 m/s then the bulk speed is 100 m/s, but obviously the particles in the gas have their own individual velocities.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-11-21T10:15:39.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12574" PostTypeId="1" AcceptedAnswerId="12577" CreationDate="2015-11-21T16:02:25.170" Score="1" ViewCount="157" Body="&lt;p&gt;We all know that we need a lot of fuel to escape from earth. Even 90% fuel of an spaceship is wasted in escaping the earth which definitely affect our power to explore the universe. we have to do a lot of calculations and even a small failure can waste a lot of money. for ex- the Manglayan Indian mission to Mars due to under performance of the engine was almost lost. As it a cheapest mission to Mars. They have a very limited fuel. So why can't we connect our space station with a long pipe and Make our space ship refuel at that point. I know a lot of energy will be required to pump the fuel up. But I was thinking about a better option. why can't we use vacuum to transfer liquid. Like if we have a pipe and we suck one end and the water automatically travels. Is it possible or what are the technical difficulty here.&lt;/p&gt;&#xA;" OwnerUserId="9839" LastActivityDate="2015-11-21T18:22:52.240" Title="can we use connect a long pipe with space station and use vacuum for fuel transfer?" Tags="&lt;space-probe&gt;&lt;space-travel&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2015-11-23T15:30:44.857" ContentLicense="CC BY-SA 3.0" />
  <row Id="12575" PostTypeId="2" ParentId="12574" CreationDate="2015-11-21T16:13:34.653" Score="0" Body="&lt;p&gt;The ISS travels over 7 kilometers per second, so the challenge of refueling would require bringing the pipe from Earth to the same speed as the ISS in order to make the connection. However, the pipe could be used to supply a stationary supply space station that has intermediate orbiters that connect the stationary supply station to the ISS.  That is the idea of the space elevator. &lt;/p&gt;&#xA;" OwnerUserId="9124" LastActivityDate="2015-11-21T16:13:34.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12576" PostTypeId="1" AcceptedAnswerId="12578" CreationDate="2015-11-21T16:41:33.400" Score="7" ViewCount="3626" Body="&lt;p&gt;I understand the expansion of the universe as actually an increase in the ratio of space to matter. Is this a correct understanding?  Otherwise, I don't understand how an infinite structure can expand.&lt;/p&gt;&#xA;" OwnerUserId="9124" LastEditorUserId="9124" LastEditDate="2015-11-24T13:52:32.743" LastActivityDate="2018-07-05T17:18:50.597" Title="How can an infinite universe expand?" Tags="&lt;expansion&gt;" AnswerCount="7" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12577" PostTypeId="2" ParentId="12574" CreationDate="2015-11-21T16:44:41.253" Score="1" Body="&lt;p&gt;A pump can lift water about thirty feet. That's the height where the column of water in the pipe weighs the same as an equivalent column of air (and that's why deep wells use submersible pumps: they push the water up from below, instead of pulling from above and relying on atmospheric pressure to provide the lift). While fuel would be less dense than water, it's still much more dense than air, and the pressure of the atmosphere simply doesn't provide enough force to raise that column of fuel very far.&lt;/p&gt;&#xA;" OwnerUserId="1867" LastEditorUserId="1867" LastEditDate="2015-11-21T18:22:52.240" LastActivityDate="2015-11-21T18:22:52.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12578" PostTypeId="2" ParentId="12576" CreationDate="2015-11-21T17:14:59.683" Score="7" Body="&lt;p&gt;Expansion means that distances are increasing as a function of time. Say if the distance between two galactic clusters is $D$, then in an expanding Universe the distance is governed by some strictly increasing function of time $a(t)$ called the scale factor where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D=a(t)D_0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $D_0$ is the distance at the present time and by definition $a(t_{0})=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cosmology assumes that the Universe is on large scales the same everywhere (homogeneous) and the same in all directions (isotropic) so the above applies to all distances above a certain scale. The scale factor $a(t)$ can be found from the Friedmann equations and initial conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expansion is possible in Universes of both finite and infinite spatial extent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the volume of a (large enough) region of space increases in proportion to $[a(t)]^3$, but the amount of matter remains constant, the matter density changes in proportion to $[a(t)]]^{-3}$. Expansion however also decreases the kinetic energy of of its contents, so the energy density decreases by a greater factor if the contents has kinetic energy.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastEditorUserId="2092" LastEditDate="2015-11-21T22:49:41.193" LastActivityDate="2015-11-21T22:49:41.193" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12580" PostTypeId="2" ParentId="12576" CreationDate="2015-11-21T18:01:24.683" Score="-4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I understand the expansion of the universe as actually an increase in the ratio of space to matter. Is this a correct understanding?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It isn't wrong. The ratio &lt;em&gt;is&lt;/em&gt; increasing. But it isn't a &quot;correct understanding&quot;. It's merely an observation of one of the results of the expansion of space. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the universe is infinite how can it expand?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know. Nor do I know how big bang cosmology can be reconciled to an infinite universe. If you look around on the internet, you can find articles like &lt;a href=&quot;http://www.physicsoftheuniverse.com/topics_bigbang_timeline.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; which say this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;The linear dimensions of the early universe increases during this period of a tiny fraction of a second by a factor of at least 10$^{26}$ to around 10 centimetres (about the size of a grapefruit)&quot;.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in 2013 results from the WMAP mission appeared to confirm that space is flat. Then a non-sequitur crept in. See &lt;a href=&quot;http://map.gsfc.nasa.gov/universe/uni_shape.html&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; and pay careful attention to this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;We now know (as of 2013) that the universe is flat with only a 0.4% margin of error. This suggests that the Universe is infinite in extent; however, since the Universe has a finite age, we can only observe a finite volume of the Universe. All we can truly conclude is that the Universe is much larger than the volume we can directly observe.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a massive error. It absolutely doesn't suggest that the universe is infinite in extent. Or that the Universe is much larger than the volume we can directly observe. But this myth has legs, and people repeat it ad-infinitum, even though they can't explain how it fits in with Big Bang cosmology. What you tend to hear is that the &lt;em&gt;observable&lt;/em&gt; universe was the size of a grapefruit, but it absolutely doesn't satisfy. Moreover there's a dreadful flaw lurking in the shadows. Take a look at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor&quot; rel=&quot;nofollow&quot;&gt;stress-energy-momentum tensor&lt;/a&gt;, and note the energy-pressure diagonal. A gravitational field is something like a spatial pressure gradient, and you can think of space as having an innate &quot;pressure&quot;. So you can reason that the universe must expand. As to why Einstein didn't, I just don't know. But anyway, for an analogy, squeeze a stress-ball down in your fist, and let go. It expands because of the pressure. However if that material was infinite in extent, the pressure is counter-balanced at all locations. So it can't expand. In similar vein, in my opinion, &lt;em&gt;an infinite universe can't expand&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;People claim the universe must be infinite because of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmological_principle&quot; rel=&quot;nofollow&quot;&gt;cosmological principle&lt;/a&gt;. But this is merely an assumption. There's an &lt;em&gt;assumption&lt;/em&gt; that the universe is homogeneous and isotropic, but this isn't fact. You cannot use it to make sweeping claims about an infinite universe that was always infinite. For all we know some observer 50 billion light years away might be looking up at the night sky wondering why half of it is black. Or a mirror-image of the other. Or some kind of &lt;em&gt;edge&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is said that in days gone by, people could not conceive of a world that was curved. They could only conceive of a world with an edge. Nowadays I rather fancy that there are some people who cannot conceive of a world that is &lt;em&gt;not&lt;/em&gt; curved. They cannot conceive of a world with an edge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://einsteinpapers.press.princeton.edu/vol6-trans/197?&quot; rel=&quot;nofollow&quot;&gt;The Foundation of the General Theory of Relativity&lt;/a&gt;: &lt;em&gt;&quot;the energy of the gravitational field shall act gravitatively in the same way as any other kind of energy&quot;&lt;/em&gt;. Energy is the source of the stress-energy tensor. Matter is only a source because of the energy-content. Also see &lt;a href=&quot;http://arxiv.org/abs/1209.0563&quot; rel=&quot;nofollow&quot;&gt;Inhomogeneous and interacting vacuum energy&lt;/a&gt; which refers to spatial energy. An interesting read is the article &lt;a href=&quot;http://edition.cnn.com/2004/TECH/space/05/24/universe.wide/&quot; rel=&quot;nofollow&quot;&gt;Universe 156 billion light-years wide&#xA;&lt;/a&gt; featuring Neil Cornish. This isn't entirely accurate, but the compound interest and the hall of mirrors concepts are of interest. As for the non-sequitur, see &lt;a href=&quot;http://www.esa.int/Our_Activities/Space_Science/People/Is_the_Universe_finite_or_infinite_An_interview_with_Joseph_Silk&quot; rel=&quot;nofollow&quot;&gt;this interview&lt;/a&gt; with Joseph Silk:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;We do not know whether the Universe is finite or not.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope nobody will argue with that. Reading on:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;To give you an example, imagine the geometry of the Universe in two dimensions as a plane. It is flat, and a plane is normally infinite. But you can take a sheet of paper [an 'infinite' sheet of paper] and you can roll it up and make a cylinder, and you can roll the cylinder again and make a torus [like the shape of a doughnut]. The surface of the torus is also spatially flat, but it is finite&quot;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is akin to the old &lt;a href=&quot;https://en.wikipedia.org/wiki/Asteroids_(video_game)&quot; rel=&quot;nofollow&quot;&gt;Asteroids game&lt;/a&gt;. But the Planck mission &lt;a href=&quot;http://xxx.lanl.gov/abs/1502.01593&quot; rel=&quot;nofollow&quot;&gt;found no evidence of any torus&lt;/a&gt;. Reading on further: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;So you have two possibilities for a flat Universe: one infinite, like a plane, and one finite, like a torus, which is also flat.&quot;&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I dispute that. There is a third possibility. A flat finite universe with no intrinsic curvature. If anybody can cite some reliable sources that support the assertion that a flat universe must be infinite, I'd like to see them. &lt;/p&gt;&#xA;" OwnerUserId="9772" LastEditorUserId="9772" LastEditDate="2015-11-23T17:57:12.373" LastActivityDate="2015-11-23T17:57:12.373" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="12581" PostTypeId="2" ParentId="12196" CreationDate="2015-11-21T19:57:26.037" Score="1" Body="&lt;p&gt;Capture of any object is never common. A rogue planet that passes a star will be accelerated by the star's gravity, and provided that it doesn't hit anything will pass the star in a hyperbolic path. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For capture to occur, the rogue planet has to lose momentum, and there are a few ways in which this can happen. The most general way is for the rogue planet to interact with one of the stars existing planets. If the rogue planet passes close one of the star's orbiting planets, it can interact gravitationally, and transfer momentum to the planet, and slow down in the process. This is rather like the reverse of a gravitational slingshot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also possible that a double planet can be captured, if one of the bodies transfers momentum to the other. One of the double planet will escape, the other is captured by the star's gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Capture of rogue planets is very rare. There is no evidence that it has occurred in our solar system (never in 4.7 billion years is a big Never) though some have speculated that Sedna might be a captured object. But we know that captures like this are possible, as Jupiter and Saturn have a collection of moons, some of which seem to have been captured from the asteroid belt.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-11-21T19:57:26.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12582" PostTypeId="1" CreationDate="2015-11-22T03:43:02.473" Score="5" ViewCount="179" Body="&lt;p&gt;I took a picture of Jupiter through my telescope, using an iPhone. I know the magnification of the telescope lens I used, and can Google to get the magnification of the iPhone lens (the magnification without zooming). Using a standard Photo Editor, I also know the pixels across of Jupiter in my image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Can I, using all of this information above, calculate how many kilometres one pixel represents, and consequently the diameter of Jupiter?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="9650" LastEditorUserId="9186" LastEditDate="2015-11-22T22:42:09.837" LastActivityDate="2015-11-22T22:42:09.837" Title="Diameter of Astronomical Object Using Magnification" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;jupiter&gt;&lt;photography&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12583" PostTypeId="2" ParentId="12582" CreationDate="2015-11-22T09:17:32.363" Score="3" Body="&lt;p&gt;You could use your image to estimate the angular size of Jupiter that night. To find the actual diameter of Jupiter you would also need to know the distance that Jupiter is from Earth, and since Jupiter and Earth are both orbiting the sun, this distance changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find the distance with software (&lt;a href=&quot;http://www.stellarium.com&quot; rel=&quot;nofollow&quot;&gt;Eg Stellarium&lt;/a&gt;) or using Nasa's &lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow&quot;&gt;Horizons&lt;/a&gt; web interface (You will need to use the column labeled &quot;delta&quot; which gives distance in &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_unit&quot; rel=&quot;nofollow&quot;&gt;astronomical units&lt;/a&gt;). To find the diameter you will need to do some trigonometry. (Diameter = distance $\times$ angular size in radians)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively you could use the published Diameter of Jupiter (70000km) and use your observations to calculate the distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the distance to Jupiter without knowing it's diameter requires more than a single observation. You will need to have:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Tycho_Brahe&quot; rel=&quot;nofollow&quot;&gt;A long enough series of observations&lt;/a&gt; to establish the orbital period and elements of Jupiter.&lt;/li&gt;&#xA;&lt;li&gt;Knowledge of &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion&quot; rel=&quot;nofollow&quot;&gt;Kepler's laws of planetary motion&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Knowledge of the distance of Earth from the sun, and to get this you can &lt;a href=&quot;https://en.wikipedia.org/wiki/First_voyage_of_James_Cook&quot; rel=&quot;nofollow&quot;&gt;fund a expedition to the South Pacific during a transit of Venus&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;A sharp pencil to do all the maths.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm being a little glib... The point is that it took more than 150 years of research between the broad acceptance of heliocentric solar system, and our knowledge of its scale, or the actual sizes of planets.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2015-11-22T10:07:03.467" LastActivityDate="2015-11-22T10:07:03.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12585" PostTypeId="2" ParentId="8618" CreationDate="2015-11-22T13:33:07.040" Score="4" Body="&lt;p&gt;Looking at the primary literature  (and the source of the wikipedia information), it appears that &lt;a href=&quot;http://arxiv.org/abs/1305.6179&quot; rel=&quot;nofollow&quot;&gt;Arroyo-Torres et al. (2013)&lt;/a&gt; have estimated a radius of $1708\pm 192R_{\odot}$, using an interferometric angular diameter and a highly uncertain distance (where I'm not sure the distance uncertainty has been adequately included). Using similar methods &lt;a href=&quot;http://arxiv.org/abs/1203.5194&quot; rel=&quot;nofollow&quot;&gt;Wittkowski et al. (2012)&lt;/a&gt; estimated a radius of $1420\pm 120 R_{\odot}$ for VY CMa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking these numbers at face value then UY Scuti is &lt;em&gt;probably&lt;/em&gt; larger than VY CMa. However, I wouldn't take these numbers at face value.  I don't believe the distances to these stars are known to better than 10% as claimed in these papers (they soon will be with Gaia trigonometric parallaxes). You also have to bear in mind that these are pulsational variables (hence the variable star designations), so their radii are variable.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-11-22T17:07:33.813" LastActivityDate="2015-11-22T17:07:33.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12586" PostTypeId="1" AcceptedAnswerId="12588" CreationDate="2015-11-22T18:21:55.773" Score="3" ViewCount="206" Body="&lt;p&gt;Science fiction seems to depict the experience of traveling through wormholes (if possible of course) as seeing psychedelic colors or lines of light where stars once were (ie. 2001 or Contact). I wonder if theorists have any idea of what, if anything, the traveler would see.&#xA;This question may be subjective, but I know that there are theories about what an observer would see if traveling near the speed of light.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="2365" LastEditDate="2015-11-22T18:37:55.767" LastActivityDate="2015-11-23T04:08:49.860" Title="What you would see in a wormhole?" Tags="&lt;general-relativity&gt;&lt;wormhole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12587" PostTypeId="2" ParentId="12576" CreationDate="2015-11-22T18:22:07.457" Score="3" Body="&lt;p&gt;There is absolutely not contradiction between being infinite and able to expand (in contrast to what your question seems to suggest). This simple fact is not confined to the actual universe we are living in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an illustraction, take the infinite 'universe' of the natural numbers $i=0\dots\infty$. Now consider the sets $2i$ and $2i+1$, each equally infinite as the natural numbers, but stretched. Now combine those two sets to get an expanded 'universe' and you obtain the natural numbers again.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2015-11-22T18:22:07.457" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12588" PostTypeId="2" ParentId="12586" CreationDate="2015-11-22T18:36:18.697" Score="2" Body="&lt;p&gt;Around the worm hole would be a region with an intense gravitational field, and there would be gravitational lensing occurring, similar to that around a black hole. The difference would be that light that has passed through the worm hole would would be visible. So, a distorted view of the &quot;other side&quot; could be seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://physicsbuzz.physicscentral.com/2014/11/what-does-journey-through-wormhole.html&quot; rel=&quot;nofollow&quot;&gt;physicsbuzz article&lt;/a&gt; has some images and discussion. Note that the wormhole would be a spherical region in space, and not a 2d &quot;stargate&quot;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="7110" LastEditDate="2015-11-23T04:08:49.860" LastActivityDate="2015-11-23T04:08:49.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12589" PostTypeId="1" AcceptedAnswerId="12591" CreationDate="2015-11-22T19:05:06.503" Score="7" ViewCount="601" Body="&lt;p&gt;The Sun has solar flares that are caused when there is a magnetic reconnection in the Sun's atmosphere, causing a loop of magnetic field to be ejected at high energy, along with a large number of protons, giving a intense burst of radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other stars - notably some red dwarfs - have much more intense flares than the Sun does. Is the mechanism the same in these stars as in the Sun, and in larger stars? What causes the flares in red dwarfs, and even brown dwarfs? Why do less powerful stars have more powerful flares?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter also has a magnetic field, and it's magnetosphere is composed of plasmas. Why aren't &quot;Jovian Flares&quot; seen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/624/how-do-large-solar-flares-compare-to-flares-on-other-stars&quot;&gt;How do large solar flares compare to flares on other stars?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2018-10-23T07:32:31.760" Title="Is the mechanism of solar flares on red dwarfs and brown dwarfs the same as that on the Sun?" Tags="&lt;solar-flare&gt;&lt;brown-dwarf&gt;&lt;red-dwarf&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12590" PostTypeId="1" AcceptedAnswerId="12606" CreationDate="2015-11-22T19:25:42.927" Score="6" ViewCount="6715" Body="&lt;p&gt;How does the appearance of an analemma vary with latitude?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This what I take to be a definition of an analemma: If the position of the sun is plotted or photographed at noon clock time (say UTC) every day from a position facing due south (in the northern hemisphere) it forms a figure of eight shape as in the picture. &lt;a href=&quot;https://i.stack.imgur.com/UTxlq.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UTxlq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photographs usually show it with the long axis at an angle. My question is, assuming that the angle is related to the latitude of the camera, how does this angle vary as the camera is moved from the equator to the pole, and why?&lt;/p&gt;&#xA;" OwnerUserId="2022" LastEditorUserId="2022" LastEditDate="2015-11-23T11:29:49.227" LastActivityDate="2018-05-12T13:41:28.107" Title="How does the appearance of the analemma vary with latitude" Tags="&lt;the-sun&gt;&lt;time&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12591" PostTypeId="2" ParentId="12589" CreationDate="2015-11-22T21:29:54.467" Score="9" Body="&lt;p&gt;Yes, the basic mechanism is thought to be the same on red dwarfs and at least the hotter brown dwarfs, but the details can be different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you say, magnetic reconnection in the corona is the starting point. Well, actually it is fluid motions at the magnetic loop footpoints that is the starting point. The B-field and partially ionised plasma are coupled and photospheric motions put (magnetic) potential energy into the B-field structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This potential energy can be released suddenly in reconnection events. These can drive coronal mass ejections or accelerate charged particles along the loop field lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A flare occurs when a significant amount of energy goes into accelerating charged particles down the field lines towards the loop footpoints. These charge particles emit radio waves and then non-thermal hard X-rays as they impact the thicker chromosphere/photosphere. Their energy is then thermalised, heating the chromosphere and possibly causing hot ($&amp;gt;10^{6}$ K) material to evaporate into the coronal loops. Here it can cool by radiation and conduction before falling back, or perhaps forming cool prominences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar things must be going on in low-mass stars. The X-ray and optical light curves of their flares do resemble what is seen on the Sun as do the relationships between hard- and soft X-rays and the evolution of plasma temperatures. The details may be different because the temperature and density structure of their photospheres, chromospheres and coronae are a bit different to the Sun's, and there are some indications that their coronae can be denser or sometimes that flares occur in much larger structures than are seen on the Sun. &quot;White light&quot; flares are also more common in M-dwarfs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are flares on red dwarfs so powerful? Partly it is contrast - you are comparing the flare emission with something that is intrinsically less luminous. In absolute terms the flares on the Sun and flares on red dwarfs are not hugely different. What is different is that the flare power as a fraction of the bolometric luminosity and the frequency of large flares can be much higher on M-dwarfs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The underlying reasons are likely to do with the magnetic field strengths and structures on red dwarfs. Magnetic activity is empirically connected to rotation and convection. Magnetic activity is higher on rapidly rotating stars and those with deep convection zones. M-dwarfs have very deep convection zones (or are even fully convective). They also tend to rotate much more rapidly than the Sun, since their spin-down timescales appear to be much longer than G and K dwarfs. They are thus more magnetically active relative to their bolometric luminosities. It appears that active M dwarfs have very strong magnetic fields (like those in sunspots) covering a very large fraction of their surfaces and this along with the convective turbulence where the magnetic loop footpoints are anchored likely leads to their strong flaring activity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brown dwarfs are a bit more tricky. The younger, hotter ones probably behave much like low-mass M-dwarfs (in fact they are M-type objects). The magnetic activity on cooler/older L- and T-dwarfs is much more mysterious. I guess a few flares have been seen, but it is not clear this is related to similar mechanisms as in higher mass and hotter stars. These cool brown dwarfs have neutral atmospheres and the magnetic field is not frozen into the plasma like it is in the partially ionised photospheres of hotter stars. This means that the loop footpoints may not be stressed by photospheric motions in the same way. It is not even clear that brown dwarfs generate a magnetic field in the same way as more massive stars, though it is clear that at least some of them do have magnetic fields.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-11-22T21:29:54.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12592" PostTypeId="5" CreationDate="2015-11-22T21:46:10.460" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-11-22T21:46:10.460" LastActivityDate="2015-11-22T21:46:10.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12593" PostTypeId="4" CreationDate="2015-11-22T21:46:10.460" Score="0" Body="Questions regarding the possible existence of other universes." OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2015-11-22T22:42:20.377" LastActivityDate="2015-11-22T22:42:20.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12595" PostTypeId="1" CreationDate="2015-11-22T22:00:55.377" Score="2" ViewCount="283" Body="&lt;p&gt;It's hard to search through the entire (observable) universe, but the most luminous thing there is must on the other hand be easy to find. Can we now say that the most luminous quasar we know, is the (intrinsically) most luminous one that exists, or will we soon?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-11-23T05:30:37.683" LastActivityDate="2015-11-23T05:56:07.740" Title="Has the most luminous object in the universe been found?" Tags="&lt;observable-universe&gt;&lt;deep-sky-observing&gt;&lt;quasars&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12596" PostTypeId="2" ParentId="11634" CreationDate="2015-11-22T22:34:34.433" Score="12" Body="&lt;p&gt;Most likely a satellite. They look exactly like stars, but they glide across the sky smoothly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Airplanes may have multiple light sources, some blinking lights, and you can definitely perceive how low it seems to be: An airplane somewhat seems to come from the horizon and disappear the same way, while a satellite &quot;seems to always be at the same distance from you&quot;, like gliding on an imaginary half-sphere or dome of a night sky covering you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last summer I went for a night bike ride for 6 hours and during this time, I spotted 12 satellites just by observing. They're a very common thing. This time I got very lucky though, because one of the 12 satellites unexpectedly was something called an Iridium flare and I had never seen one of those in real life before. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a lot of man-made space junk orbiting the Earth, some satellites still operated, some not. There's a bunch of satellites known as Iridium, which happen to have a design that includes large, reflective solar panels.&lt;br /&gt;I live in the North so in the summer, the sun only barely goes below the horizon and the summer nights are short. When I'm just in the dark side but an Iridium satellite happens to fly approximately above me, if everything is in a specific angle, the sun from behind the horizon hits the satellite's solar panels, and like a mirror, reflects those rays to the dark side, into a viewer's eyes.&lt;br /&gt;It was my first time seeing that and I must say, it's extremely impressive when you're merely looking at a &quot;moving star&quot;, then unexpectedly it grows very bright, brighter than any star in the night sky, and finally reverts back to the normal looking gliding satellite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because people are very much aware of all of the space junk that's up there, there are plenty of websites and mobile apps that you can use to see where the most common satellites are at the moment, and, because we know the route of each satellite, as well as the year and location specific data regarding the Earth and the Sun, Iridium flares can also be predicted, like you would predict things like Solar eclipses. &lt;/p&gt;&#xA;" OwnerUserId="9848" LastEditorDisplayName="user1569" LastEditDate="2017-07-16T14:28:37.800" LastActivityDate="2017-07-16T14:28:37.800" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12597" PostTypeId="2" ParentId="12595" CreationDate="2015-11-22T22:36:18.220" Score="2" Body="&lt;p&gt;As it stands I think the answer is no. The most luminous object in the (observable) universe is probably hidden behind many magnitudes of dust extinction. There are lots of faint, infrared-detected objects that may be extremely luminous galaxies in the early universe (and indeed some have turned out to be) but it is difficult to get their redshifts, so it is difficult to be sure we have a complete luminosity function. We now know that objects with central black holes $&amp;gt;10^{9} M_{\odot}$ already exist in the universe after a billion years or so. To get to that size there must have been an awful lot of accretion and an awful lot of accretion luminosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, &lt;em&gt;the&lt;/em&gt; most luminous galaxy/quasar is going to be a rare object and only ridiculously small areas of the sky are covered by things like the HST ultra-deep field. i.e. There is almost bound to be a more luminous one in some other patch of sky that has not (yet) been surveyed. The logistical problems of getting a UDF over the whole sky (or even a significant fraction of the sky) means that isn't going to happen any time soon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the brightest quasar is certainly not the most luminous. 3C273 is quite a modest object, in terms of luminosity, but it is very close to us as quasars go. But perhaps you didn't mean bright in that sense? Brightness is normally reserved to refer to the apparent brightness. Luminosity is the extinction-free, distance-corrected power.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-11-23T05:56:07.740" LastActivityDate="2015-11-23T05:56:07.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12598" PostTypeId="1" AcceptedAnswerId="12628" CreationDate="2015-11-23T02:47:22.793" Score="3" ViewCount="326" Body="&lt;p&gt;A few days ago I decided to take on the little project of converting a Qbasic program into Python (as a side project to my Radio Jove project), and I've managed to get it to run, but the math is definitely off. I was hoping for a fellow astronomer or astrophysicist to assist in the mathematical side of my Python code. Obviously something is wrong with the math aspect of the code since most of the values in the python program are very skewed. My apologies in advance for the single character variables. I would make them better if I knew everything going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Qbasic code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;'Program to flag Jovian Decametric windows&#xA; m$ = &quot;JANFEBMARAPRMAYJUNJULAUGSEPOCTNOVDEC&quot;&#xA;     wi = 42.46 / 360&#xA;     pi = 3.141593&#xA;     kr = pi / 180&#xA;     f$ = &quot;###  \ \ ##   ##.#     ###      ###    ##.##     \  \&quot;&#xA; OPEN &quot;jovrad.txt&quot; FOR OUTPUT AS #1&#xA; INPUT &quot;Year for which predictions are required&quot;; yy&#xA; e = INT((yy - 1) / 100)&#xA; f = 2 - e + INT(e / 4)&#xA; jd = INT(365.25 * (yy - 1)) + 1721423 + f + .5&#xA; d0 = jd - 2435108&#xA; incr = 0&#xA; IF yy / 400 - INT(yy / 400) = 0 THEN incr = 1&#xA; yyly = yy / 4 - INT(yy / 4)&#xA; yylc = yy / 100 - INT(yy / 100)&#xA; IF yyly = 0 AND yylc &amp;lt;&amp;gt; 0 THEN incr = 1&#xA; ty = 59 + incr&#xA; dmax = 365 + incr&#xA; tx = ty + .5&#xA; PRINT #1, &quot;******************************************************&quot;&#xA; PRINT #1, &quot;  JOVIAN IO-DECAMETRIC EMISSION PREDICTIONS FOR&quot;; yy&#xA; PRINT #1, &quot;******************************************************&quot;&#xA; PRINT #1,&#xA; PRINT #1, &quot;Day   Date   Hr(UT)  Io_Phase   CML   Dist(AU)  Source&quot;&#xA; PRINT #1,&#xA; th = 0&#xA; DO&#xA;   GOSUB Compute&#xA;   s$ = &quot;&quot;&#xA;       IF L3 &amp;lt; 255 AND L3 &amp;gt; 200 AND U1 &amp;lt; 250 AND U1 &amp;gt; 220 THEN s$ = &quot;Io-A&quot;&#xA;   IF L3 &amp;lt; 180 AND L3 &amp;gt; 105 AND U1 &amp;lt; 100 AND U1 &amp;gt; 80 THEN s$ = &quot;Io-B&quot;&#xA;       IF L3 &amp;lt; 350 AND L3 &amp;gt; 300 AND U1 &amp;lt; 250 AND U1 &amp;gt; 230 THEN s$ = &quot;Io-C&quot;&#xA;   IF s$ &amp;lt;&amp;gt; &quot;&quot; THEN GOSUB Outdat&#xA;   th = th + .5&#xA; LOOP UNTIL INT(th / 24) + 1 &amp;gt; dmax&#xA; PRINT &quot;Program completed - results in file JOVRAD.TXT&quot;&#xA; END&#xA;&#xA;Compute:&#xA; d = d0 + th / 24&#xA; v = (157.0456 + .0011159# * d) MOD 360&#xA; m = (357.2148 + .9856003# * d) MOD 360&#xA; n = (94.3455 + .0830853# * d + .33 * SIN(kr * v)) MOD 360&#xA; j = (351.4266 + .9025179# * d - .33 * SIN(kr * v)) MOD 360&#xA; a = 1.916 * SIN(kr * m) + .02 * SIN(kr * 2 * m)&#xA; b = 5.552 * SIN(kr * n) + .167 * SIN(kr * 2 * n)&#xA; k = j + a - b&#xA; r = 1.00014 - .01672 * COS(kr * m) - .00014 * COS(kr * 2 * m)&#xA; re = 5.20867 - .25192 * COS(kr * n) - .0061 * COS(kr * 2 * n)&#xA; dt = SQR(re * re + r * r - 2 * re * r * COS(kr * k))&#xA; sp = r * SIN(kr * k) / dt&#xA; ps = sp / .017452&#xA; dl = d - dt / 173&#xA; pb = ps - b&#xA; xi = 150.4529 * INT(dl) + 870.4529 * (dl - INT(dl))&#xA; L3 = (274.319 + pb + xi + .01016 * 51) MOD 360&#xA; U1 = 101.5265 + 203.405863# * dl + pb&#xA; U2 = 67.81114 + 101.291632# * dl + pb&#xA; z = (2 * (U1 - U2)) MOD 360&#xA; U1 = U1 + .472 * SIN(kr * z)&#xA; U1 = (U1 + 180) MOD 360&#xA;RETURN&#xA;&#xA;Outdat:&#xA; dy = INT(th / 24) + 1&#xA; h = th - (dy - 1) * 24&#xA; IF dy &amp;gt; ty THEN&#xA;   m = INT((dy - tx) / 30.6) + 3&#xA;   da = dy - ty - INT((m - 3) * 30.6 + .5)&#xA; ELSE&#xA;   m = INT((dy - 1) / 31) + 1&#xA;   da = dy - (m - 1) * 31&#xA; END IF&#xA; mn$ = MID$(m$, (m - 1) * 3 + 1, 3)&#xA;     PRINT #1, USING f$; dy; mn$; da; h; U1; L3; dt; s$&#xA;RETURN&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Sample output:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;******************************************************&#xA;  JOVIAN IO-DECAMETRIC EMISSION PREDICTIONS FOR 1994 &#xA;******************************************************&#xA;&#xA;Day   Date   Hr(UT)  Io_Phase   CML   Dist(AU)  Source&#xA;&#xA;  4  JAN  4   11.0     228      205     5.81     Io-A&#xA;  4  JAN  4   11.5     233      224     5.81     Io-A&#xA;  4  JAN  4   12.0     237      242     5.81     Io-A&#xA;  6  JAN  6    6.0     233      325     5.78     Io-C&#xA;  6  JAN  6    6.5     237      343     5.78     Io-C&#xA;  7  JAN  7    6.5      81      133     5.76     Io-B&#xA;  7  JAN  7    7.0      86      152     5.76     Io-B&#xA;  7  JAN  7    7.5      90      170     5.76     Io-B&#xA;  9  JAN  9   20.0     241      203     5.73     Io-A&#xA;  9  JAN  9   20.5     246      222     5.73     Io-A&#xA; 11  JAN 11   12.5     225      232     5.70     Io-A&#xA; 11  JAN 11   13.0     229      251     5.70     Io-A&#xA; 11  JAN 11   14.5     242      305     5.70     Io-C&#xA; 11  JAN 11   15.0     246      323     5.70     Io-C&#xA; 12  JAN 12   15.0      90      113     5.68     Io-B&#xA; 12  JAN 12   15.5      94      131     5.68     Io-B&#xA; 12  JAN 12   16.0      98      150     5.68     Io-B&#xA; 14  JAN 14    8.5      82      179     5.66     Io-B&#xA; 16  JAN 16   21.0     234      214     5.61     Io-A&#xA; 16  JAN 16   21.5     238      232     5.61     Io-A&#xA; 16  JAN 16   22.0     242      250     5.61     Io-A&#xA; 18  JAN 18   15.5     234      315     5.59     Io-C&#xA; 18  JAN 18   16.0     238      333     5.59     Io-C&#xA; 19  JAN 19   16.0      82      124     5.57     Io-B&#xA; 19  JAN 19   16.5      86      142     5.57     Io-B&#xA; 19  JAN 19   17.0      91      160     5.57     Io-B&#xA; 19  JAN 19   17.5      95      178     5.57     Io-B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Python code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#!/usr/bin/env python&#xA;from __future__ import print_function, division&#xA;import math&#xA;print('Program to flag Jovian Decametric windows')&#xA;month = ['JAN','FEB','MAR','APR','MAY','JUN','JUL','AUG','SEP','OCT','NOV','DEC']&#xA;week = 42.46/360&#xA;pi = 3.141593&#xA;kr = pi / 180&#xA;form = &quot;###  \ \ ##   ##.#     ###      ###    ##.##     \  \\&quot;&#xA;num1 = open('jovrad.txt', 'w')&#xA;yy = int(raw_input((&quot;Year for which predictions are required &quot;)))&#xA;e = math.trunc(((yy-1) / 100))&#xA;f = 2 - e + math.trunc(e/4)&#xA;jd = math.trunc(365.25 * (yy - 1)) + 1721423 + f + .5&#xA;d0 = jd - 2435108&#xA;incr = 0&#xA;dmax = 0&#xA;tx = 0&#xA;ty = 0&#xA;yyly = 0&#xA;yylc = 0&#xA;if yy / 400 - math.trunc((yy / 400)) == 0:&#xA;    incr = 1&#xA;    yyly = yy / 4 - math.trunc((yy / 4))&#xA;    yylc = yy / 100 - math.trunc((yy / 100))&#xA;if yyly == 0 and yylc != 0:&#xA;    incr = 1&#xA;    ty = 59 + incr&#xA;    dmax = 365 + incr&#xA;    tx = ty + .5&#xA;print(&quot;******************************************************&quot;, file=num1)&#xA;print(&quot;  JOVIAN IO-DECAMETRIC EMISSION PREDICTIONS FOR &quot;,yy, file=num1)&#xA;print(&quot;******************************************************&quot;, file=num1)&#xA;print(&quot;\n&quot;, file=num1)&#xA;print(&quot;Day   Date   Hr(UT)  Io_Phase   CML   Dist(AU)  Source&quot;, file=num1)&#xA;print(&quot;\n&quot;, file=num1)&#xA;th = 0&#xA;def compute(d0, th, dmax):&#xA;    global U1, L3, dt, s&#xA;    d = d0 + math.trunc(th / 24)&#xA;    v = (157.0456 + .0011159 * d) % 360&#xA;    m = (357.2148 + .9856003 * d) % 360&#xA;    n = (94.3455 + .0830853 * d + .33 * math.sin(kr * v)) % 360&#xA;    j = (351.4266 + .9025179 * d - .33 * math.sin(kr * v)) % 360&#xA;    a = 1.916 * math.sin(kr * m) + .02 * math.sin(kr * 2 * m)&#xA;    b = 5.552 * math.sin(kr * n) + .167 * math.sin(kr * 2 * n)&#xA;    k = j + a - b&#xA;    r = 1.00014 - .01672 * math.cos(kr * m) - .00014 * math.cos(kr * 2 * m)&#xA;    re = 5.20867 - .25192 * math.cos(kr * n) - .0061 * math.cos(kr * 2 * n)&#xA;    dt = math.sqrt(re * re + r * r - 2 * re * r * math.cos(kr * k))&#xA;    sp = r * math.sin(kr * k) / dt&#xA;    ps = sp / .017452&#xA;    dl = d - dt / 173&#xA;    pb = ps - b&#xA;    xi = 150.4529 * math.trunc((dl)) + 870.4529 * (dl - math.trunc((dl)))&#xA;    L3 = (274.319 + pb + xi + .01016 * 51) % 360&#xA;    U1 = 101.5265 + 203.405863 * dl + pb&#xA;    U2 = 67.81114 + 101.291632 * dl + pb&#xA;    z = (2 * (U1 - U2)) % 360&#xA;    U1 = U1 + .472 * math.sin(kr * z)&#xA;    U1 = (U1 + 180) % 360&#xA;    s = &quot;&quot;&#xA;    while math.trunc(th / 24) + 1 &amp;lt; dmax and math.trunc(th / 24) + 1 == dmax:&#xA;        if L3 &amp;lt; 255 and L3 &amp;gt; 200 and U1 &amp;lt; 250 and U1 &amp;gt; 220:&#xA;            s = &quot;Io-A&quot;&#xA;        if L3 &amp;lt; 180 and L3 &amp;gt; 105 and U1 &amp;lt; 100 and U1 &amp;gt; 80:&#xA;            s = &quot;Io-B&quot;&#xA;        if L3 &amp;lt; 350 and L3 &amp;gt; 300 and U1 &amp;lt; 250 and U1 &amp;gt; 230:&#xA;            s = &quot;Io-C&quot;&#xA;        if s != &quot;&quot;:&#xA;            outdat()&#xA;        th = th + .5&#xA;    print(&quot;Program completed - results in file JOVRAD.TXT&quot;, file=num1)&#xA;compute(d0, th, dmax)&#xA;&#xA;def outdat(th,tx,ty):&#xA;    dy = math.trunc((th / 24)) + 1&#xA;    h = th - (dy - 1) * 24&#xA;    if(dy &amp;gt; th):&#xA;        m = math.trunc((dy - tx) / 30.6) + 3&#xA;        da = dy - ty - math.trunc((m - 3) * 30.6 + .5)&#xA;    else:&#xA;        m = math.trunc((dy - 1) / 31) + 1&#xA;        da = dy - (m - 1) * 31&#xA;    mn = month[(m-1)*3+1:(m-1)*3-1+3]&#xA;    print(dy, mn, da, h, U1, L3, dt, s, file=num1)&#xA;outdat(th,tx,ty)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;My current output:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;******************************************************&#xA;  JOVIAN IO-DECAMETRIC EMISSION PREDICTIONS FOR  1998&#xA;******************************************************&#xA;&#xA;&#xA;Day   Date   Hr(UT)  Io_Phase   CML   Dist(AU)  Source&#xA;&#xA;&#xA;-12.9775 1.0 ['AUG'] -0.5 0.0 135.325782651 10.6054462674 5.7071322535 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9849" LastEditorUserId="9849" LastEditDate="2015-11-24T03:17:03.650" LastActivityDate="2015-12-26T11:55:40.920" Title="Need help with the math in Python program to flag Jovian radio emissions" Tags="&lt;jupiter&gt;&lt;radio-astronomy&gt;&lt;python&gt;&lt;radio&gt;" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="12599" PostTypeId="1" AcceptedAnswerId="12602" CreationDate="2015-11-23T04:19:58.680" Score="4" ViewCount="1426" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_horizon&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia page&lt;/a&gt; for apparent horizon is pretty sketchy and requires some GR knowledge. Is there any simple definition?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2018-08-07T10:53:47.573" Title="What's the difference between the apparent horizon and event horizon of a black hole?" Tags="&lt;gravity&gt;&lt;black-hole&gt;&lt;light&gt;&lt;general-relativity&gt;&lt;supermassive-black-hole&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12600" PostTypeId="1" CreationDate="2015-11-23T06:42:02.963" Score="3" ViewCount="547" Body="&lt;p&gt;Time moves more slowly near a mass than in relatively empty space.  For example, I get that an observer would see someone falling into a black hole to appear to move more slowly and get &quot;stuck&quot; on the event horizon, while the falling body would see the observer as aging more rapidly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, an observer watching a passenger craft moving near the speed of light (high mass equivalent) might see a passenger who tripped falling in slow motion, because the forward momentum of the passenger + the speed of the craft must never exceed the speed of light, while the passenger would experience no effect of this time dilation, and would simply trip and fall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which brings me to my question...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes, even supermassive black holes, are nowhere near the mass of the universe as a whole; so when the universe itself was much smaller and far more dense, then at a universal scale time itself would have progressed more &quot;slowly&quot; than it does today, and since the expansion of the universe is accelerating, doesn't it mean that time is also accelerating?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Assuming, of course, there was some &quot;outside&quot; to observe this from).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a duplicate of &lt;a href=&quot;https://astronomy.stackexchange.com/questions/12401/does-time-slow-down-because-the-universe-is-expanding-at-an-accelerating-rate&quot;&gt;Does time slow down because the universe is expanding at an accelerating rate?&lt;/a&gt;.  That question and its answers seem more concerned with local, relativistic time (which is as malleable as space itself).  My question asks more about the effect across a universal scale as it relates to density of mass across that same universal scale.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-11-23T20:02:45.353" Title="Is time itself speeding up universally?" Tags="&lt;universe&gt;&lt;space-time&gt;&lt;early-universe&gt;&lt;time-dilation&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12601" PostTypeId="1" CreationDate="2015-11-23T08:31:50.603" Score="20" ViewCount="3915" Body="&lt;p&gt;Reading about the Star KIC 8462852, it has been said that the SETI project turned its radio telescopes towards the star to search for extra terrestrial radio signals as the star had strange fluctuations in light. How can we point from earth a radio telescope towards a star which is 1480 light years away whilst the earth is spinning at 1675 km/h and keep it focused or, in the case of a radio telescope, aligned in order to try and receive radio waves??&lt;/p&gt;&#xA;" OwnerUserId="1478" LastActivityDate="2015-11-27T09:39:28.617" Title="How can we focus radio telescopes on a star when the earth is spinning?" Tags="&lt;telescope&gt;" AnswerCount="7" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12602" PostTypeId="2" ParentId="12599" CreationDate="2015-11-23T09:30:55.330" Score="4" Body="&lt;p&gt;The black hole region of a spacetime is defined as a region where nothing can escape to infinity and an event horizon at a given time is the boundary of a connected region of space which is part of the black hole region. As you're after a simple answer I won't give a formal definition of a black hole or an event horizon, but they can be found in Wald.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem though is to know whether anything in a region of space can escape to infinity depends on precise knowledge of the future, also not all spacetimes have a suitable notion of infinity. Yet clearly even when we don't know the entire history of the spacetime or when there isn't a suitable notion of infinity, we can still identify objects that are functionally equivalent to black holes. The apparent horizon if you like is the spatial boundary of what we might consider to functionally be a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As nothing can escape a black hole, even light directed away from it is pulled back towards the singularity so we know that a black hole can cause even light directed outwards, relative to a point in space, to move inwards to that point. The apparent horizon is the boundary between where outwardly directed light moves outwards and where it moves inwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Schwarzschild spacetime, or more generally Kerr-Newman spacetime (in standard coordinates) the apparent horizon and the event horizon coincide. However, more generally, the location of the apparent horizon depends on how you 'slice' spacetime (it is observer-dependent if you like - unlike the event horizon). Apparent horizons needn't be associated with (formally-defined) black holes, however as long as the spacetime has certain properties, they will indeed be associated with black holes and will lie at or inside the event horizon.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastEditorUserId="2092" LastEditDate="2015-11-23T15:35:11.863" LastActivityDate="2015-11-23T15:35:11.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12604" PostTypeId="2" ParentId="12601" CreationDate="2015-11-23T11:32:29.950" Score="11" Body="&lt;p&gt;First you are talking about pointing the telescope at the source not focusing it on the source. Telescopes are generally focused at infinity, and there is no need to compensate for the Earth's rotation in the focusing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed of motion of the telescopes location on the Earth is also not directly relevant, what is relevant is the apparent rotation of the sky around the projection of the Earth's axis onto the sky. That is (in the Northern hemisphere) the rotation of the sky about the pole star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of ways of dealing with the earth rotation.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Actually use it to scan over the sources&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Drive the telescope to keep it pointing in the direction of interest&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Track the source (use multiple channels to measure the source error from boresight and drive the telescope to to null the error).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This really has nothing to do with radio telescopes per se, but is common to all telescopes including optical.  &lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-11-23T12:00:12.840" LastActivityDate="2015-11-23T12:00:12.840" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12605" PostTypeId="2" ParentId="12590" CreationDate="2015-11-23T11:37:35.843" Score="0" Body="&lt;p&gt;I think that the answer is that the analemma would always be vertical, at right angles to the equator, if photographed at midday. The apparent slope in photographs is due to perspective, and being taken at a different time of day, not at noon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the equator the analemma will be directly overhead at the equinoxes, and off this by the angle of the ecliptic at the solstices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the poles the analemma will be centred on the horizon, the top half giving the long period without night in the summer, and the obscured half the long winters without daylight.&lt;/p&gt;&#xA;" OwnerUserId="2022" LastActivityDate="2015-11-23T11:37:35.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12606" PostTypeId="2" ParentId="12590" CreationDate="2015-11-23T11:45:58.307" Score="6" Body="&lt;p&gt;Analemma is a diagram showing the deviation of the Sun from its mean motion in the sky, as viewed from a fixed location on the Earth. Note that it says nothing about time- it can be any fixed time. Analemmas created at different times of the day have &lt;a href=&quot;http://solar-center.stanford.edu/art/analemma.html&quot; rel=&quot;noreferrer&quot;&gt;slightly different shapes&lt;/a&gt;. For example, see the solar analemmas taken at the same place at different times.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aL04l.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aL04l.jpg&quot; alt=&quot;700 UTC&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Solar analemma at 0700 UTC by Anthony Ayiomamitis, from solar-center.stanford.edu&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/P0tvN.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P0tvN.jpg&quot; alt=&quot;1400 UTC&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Solar analemma at 1400 UTC by Anthony Ayiomamitis, from solar-center.stanford.edu&lt;/p&gt;&#xA;&lt;p&gt;The best way to understand the effect of latitude on the shape of the analemma is to consider the &lt;a href=&quot;http://www.timeanddate.com/astronomy/solar-analemma.html?BHT-0bc8ea03-716f-4f06-85be-7cfb4cbb27e9.7&quot; rel=&quot;noreferrer&quot;&gt;effect of earth's axial tilt on the shape of the analemma&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If the Earth’s orbital path was elliptical, but its axis was not tilted, the analemma curve would be oval shaped. At the Equator, this line would be a straight line spanning from left to right or West to East.&lt;/p&gt;&#xA;&lt;p&gt;The 23.5-degree axial tilt of the Earth affects the Sun’s apparent position in the sky – as the year progresses and the Earth continues to spin at an tilted axis and orbit around the Sun, the Sun seems to move up and down (North-South) in the sky. This has the effect of generating the two loops of the figure 8.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You're correct that the tilt of the analemma varies with the latitude. Observers in the Northern hemisphere will obtain an analemma curve with the broader loop at the bottom. This reverses for observers in the Southern hemisphere, where the broader loop is on the top of the curve. At equator, the anlaemma lies on its side. At north pole, only the top of the analemma would be visible, while it is the opposite at the south pole.&lt;/p&gt;&#xA;" OwnerUserId="7970" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-11-23T11:45:58.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12607" PostTypeId="2" ParentId="12601" CreationDate="2015-11-23T13:02:40.270" Score="6" Body="&lt;p&gt;Because the speed of light is so much faster than the speed of the telescope, the star looks like it is standing still in the sky so the telescope only needs to track it as it moves across the sky at 15-degrees per hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the speed of light isn't infinite, and there is a measurable effect there.  When you're riding in a car while it is raining and the rain preferentially hits your windshield, to you it looks like the rain is coming from some location in front of you even though it is falling straight down, and because of this when you try to look directly at the source of the rain, you look tilt your head forward rather than look straight up.  The same thing happens with light from stars.  Because the Earth is going around in its orbit and spinning on its axis, light falling &quot;straight down&quot; on us looks like it is coming from a position a little ahead.  This is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Aberration_of_light&quot;&gt;stellar aberration&lt;/a&gt;.  It isn't a large effect, but it is large enough that if you're trying to figure out very precisely where the stars are, then you need to correct for it.&lt;/p&gt;&#xA;" OwnerUserId="9749" LastActivityDate="2015-11-23T13:02:40.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12608" PostTypeId="2" ParentId="12601" CreationDate="2015-11-23T13:33:14.990" Score="3" Body="&lt;p&gt;Let's say you go out on a warm summer day, lie down and look up at the stars. For some reason you manage to not fall asleep, while you only look at one star the whole night. You will have no problems pointing your eyes at this star (except for falling eyelids), like it is no problem to point a telescope at one star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;curiousdannii is right, I didn't explain how. I will do this now: There is a machine called motor, or motor drive, or engine, which converts electrical energy into motion energy. With a little engineering, you can use this motion energy to turn the telescope. &lt;/p&gt;&#xA;" OwnerUserId="9854" LastEditorUserId="9854" LastEditDate="2015-11-24T13:28:36.123" LastActivityDate="2015-11-24T13:28:36.123" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12609" PostTypeId="2" ParentId="12601" CreationDate="2015-11-23T14:05:19.350" Score="5" Body="&lt;p&gt;There are two processes to manage this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the telescopes (really, big antennas) are aimed mechanically and move so they can maintain their reception of a specific star/source/sky location over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, except for stars immediately in the vicinity of the pole stars, the star will eventually go below the horizon.  Once this happens the telescope/antenna cannot receive anything further until the source appears above the horizon again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens at this point is we have many telescopes/antennas around the world that are collectively controlled.  Long before a star/source/etc falls below the horizon for one telescope, another telescope further west has already pointed at it, and is receiving the same signal.  Once this switchover has occurred, the previous telescope is free to select another target - something else on the other side of the planet that will be falling below the horizon for telescope further east.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The telescopes are under constant use pointing at interesting things&lt;/li&gt;&#xA;&lt;li&gt;Things which need continuous monitoring can be monitored without interruption despite the world turning&lt;/li&gt;&#xA;&lt;li&gt;We can observe anything at any time, as long as there's available time on the radio telescope network&lt;/li&gt;&#xA;&lt;li&gt;Sharing resources allows scientists to conduct science more completely and inexpensively&lt;/li&gt;&#xA;&lt;li&gt;By having 2 or more telescopes pointing at the same object at once, we can effectively increase the signal to noise ratio and get better data - it's technically very similar to having one earth sized single antenna rather than two tiny (relatively) antennas.&lt;/li&gt;&#xA;&lt;li&gt;With central control of a whole participating worldwide network, scientists can react very quickly to sudden phenomena, like bursts, at any time, regardless of the position of the earth&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="559" LastActivityDate="2015-11-23T14:05:19.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12610" PostTypeId="2" ParentId="12601" CreationDate="2015-11-23T15:14:41.003" Score="16" Body="&lt;p&gt;Part of the answer that I suspect the original questioner needs is that although the Earth is indeed spinning very fast, the amount the surface of the Earth moves relative to an astronomical object is tiny.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you put motors in the base of the telescope so that it slowly turns to look at the same patch of sky.   You don't need to refocus because telescopes are looking at objects so far away that focus doesn't matter.  You don't need to do anything else because the Earth's movement is smooth and continuous, and it's not about how fast you're moving, it's about how quickly you're turning.  In our case, one complete circle every 24 hours which is pretty slow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Focus at infinity just means that you set the focus of the telescope so that an object which is infinitely far away would be in perfect focus.  It depends on the quality of the telescope, but the practical difference between focus at infinity and focus at the actual distance disappears after a few miles or so.  At the distance of stars, there is essentially no difference at all.&lt;/p&gt;&#xA;" OwnerUserId="9745" LastActivityDate="2015-11-23T15:14:41.003" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12611" PostTypeId="2" ParentId="12576" CreationDate="2015-11-23T15:29:31.550" Score="1" Body="&lt;p&gt;How do you describe how far away two points are?  You have to have some way of describing the concept of distance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we say that the universe is expanding, what we really mean is that distances inside it are increasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea that the expanding universe is some sort of 3D bubble or balloon that can be seen to expand from outside isn't meaningful, as there is no outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps a more helpful way to think of it is to say that the concept of distance is a property of the universe, and that property is changing over time.&lt;/p&gt;&#xA;" OwnerUserId="9745" LastActivityDate="2015-11-23T15:29:31.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12612" PostTypeId="2" ParentId="12601" CreationDate="2015-11-23T16:09:54.433" Score="3" Body="&lt;p&gt;The way telescopes work is pretty much the same in the optical and radio wavelengths- the telescopes collect electromagnetic radiation, rather than focusing at a point. There are multiple reasons for this, the main one being the amount of photons reaching the telescope from the region of interest is pretty low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to collect more photons, the telescope (or telescope array) has to 'look' into the area of interest for a long time- this is achieved in case of earth based telescopes by mechanically steering the antennae, so that they are pointed in the same direction for a long amount of time. The principle is pretty much &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble_Ultra-Deep_Field&quot; rel=&quot;nofollow&quot;&gt;the same in space&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For looking at KIC 8462852, SETI used the &lt;a href=&quot;https://en.wikipedia.org/wiki/Allen_Telescope_Array&quot; rel=&quot;nofollow&quot;&gt;Allen Telescope Arry&lt;/a&gt;, which is basically a set of 42 antennae scanning the skies in radio wavelengths. The problem of earth's rotation is basically solved in two steps by (radio) telescopes.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;By steering the antenna(e) as decided by the software so that the antenna points at the same position of the sky. For a star at ~1500 light years, the angular speed required is pretty small and can be easily supplied by the modern telescopes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Even if the star (or any other object of interest) passes below the horizon, the telescope can simply continue its work on the next day, gathering more photons. Of course, other telescopes can take over from this one, but the end result is the same- collecting more photons.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7970" LastActivityDate="2015-11-23T16:09:54.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12613" PostTypeId="1" CreationDate="2015-11-23T17:19:52.643" Score="2" ViewCount="819" Body="&lt;p&gt;why has not Nasa gone back to our Moon since the first landing on the moon&lt;/p&gt;&#xA;" OwnerUserId="9857" LastActivityDate="2015-11-23T18:42:24.757" Title="why hasn't Nasa gone back to our Moon?" Tags="&lt;the-moon&gt;&lt;nasa&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ClosedDate="2015-11-23T19:15:55.587" ContentLicense="CC BY-SA 3.0" />
  <row Id="12614" PostTypeId="2" ParentId="12613" CreationDate="2015-11-23T17:41:12.983" Score="3" Body="&lt;p&gt;Nasa has not gone back to the moon for a few reasons. Nasa was no longer being funded for the apollo missions, but also don't want to risk the lives of astronauts. The political aim of the apollo missions was to make sure every astronaut returned alive, and the scientific aim was to learn as much as possible about the moon - they would love to have someone up there for weeks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something NASA has been working on is a virtual simulation environment of the moon. They believe this can help save the risks of any lunar accidents and allow them to study the moon safely from earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did a few google searches to find this answer, so information might not be exact.&lt;/p&gt;&#xA;" OwnerUserId="9645" LastActivityDate="2015-11-23T17:41:12.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12615" PostTypeId="2" ParentId="12613" CreationDate="2015-11-23T18:42:24.757" Score="5" Body="&lt;p&gt;We didn't only visit the Moon once! Neil Armstrong, Buzz Aldrin and Michael Collins were the first manned mission to land on the Moon with the famous Apollo 11 landing, but 5 later Apollo missions also landed safely on the Moon and did various exciting things. &#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Moon_landing&quot;&gt;This Wikipedia article&lt;/a&gt; gives a brief list of all the manned and unmanned missions that have investigated the Moon, and &lt;a href=&quot;http://www.nasa.gov/mission_pages/apollo/missions/index.html&quot;&gt;this NASA article&lt;/a&gt; gives a nice history of the Apollo program with all the manned missions. (So far the US are the only country to land astronauts on the Moon.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for why there haven't been more visits since Apollo, there are a couple of points: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) We have continued investigating the Moon, through remote observation, close encounters and landings, just not with humans on board. The future of human space flight is quite a debated question, since humans are very fragile, have lots of requirements in terms of food, space, health etc. and are not really very efficient at many tasks. Automated investigation of space is by far cheaper and easier. The main advantages humans have are intuition and adaptability. We can change plans, make snap decisions and respond much quicker than a robot controlled by people far away on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Turn the question around: why would we go back to the Moon? Why spend money and time investigating this rather boring lump of cold rock when we could go to places we have absolutely no data on, or very shaky theories, or places that have environments hugely different to our own? If we want to test scientific theories, we often want extreme environments, not just something  a little bit different to home. And if, as your comment says, we are looking for other life or future homes, these things are very rare so we will have to look &lt;a href=&quot;http://www.joshworth.com/dev/pixelspace/pixelspace_solarsystem.html&quot;&gt;a lot further away&lt;/a&gt; than the Moon to find something that interesting!&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2015-11-23T18:42:24.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12616" PostTypeId="2" ParentId="12600" CreationDate="2015-11-23T19:07:11.020" Score="2" Body="&lt;p&gt;The rate at which a clock ticks is a local property. There is no universal rate of time which could speed up, so the answer is no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A clock stationary in the same frame of reference, and the same gravitational field as me will tick at the same rate, (one second every second) It is only clocks that are moving, or in different gravitational fields that that tick at a different rate. If I am falling into a black hole, and I stop the check the time, I wouldn't see my watch slow down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is no standard clock rate. Clocks in the early universe (or at least things that depend on time, such as nuclear decay) ran at the same rate, in their local gravitational field (at one second per second)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were to observe a clock from the early universe, it would be receding at great velocity (and so it would be redshifted and time dilated) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the gravitational time dilation does not depend on the amount of mass directly, but on the intensity of the gravitational field. A galaxy has huge mass, but the only place in which the gravitational field is significant (from a GR perspective) is in the neighbourhood of Neutron stars and black holes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get gravitational time dialtion you need an intense gravitational field. Just having a lot of mass is not enough. After the period of inflation, the universe was homogeneous, and there were no significant &quot;lumps&quot; to give a net gravitational field. As the gas in the universe collapsed into galaxies and stars, and eventually black holes regions of intense gravity formed, in which clocks would run slowly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there is no general speeding up of clocks in the universe.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-11-23T19:07:11.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12617" PostTypeId="2" ParentId="12600" CreationDate="2015-11-23T20:02:45.353" Score="1" Body="&lt;p&gt;Gravitational time dilation applies to asymptotically flat static spacetimes, that is spacetime describing an isolated system in space, which is independent of time (does not change with time). The reason for this is that gravitational time dilation needs a stationary clock that is faraway from the gravitational influence of the system and the field must be static so that clocks can be stationary relative to the field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spacetime describing an expanding Universe is neither static nor asymptotically flat, so gravitational time dilation does not apply. Unfortunately we can't invoke an 'outside observer', the nearest thing to an &quot;outside observer&quot; in GR is a stationary observer faraway from an isolated system, which clearly cannot be invoked in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted in the other question, when we look faraway we are also looking back in the past and the furthest, and therefore also furthest back in time, objects we see are red-shifted. When a clock is red-shifted it &lt;em&gt;appears&lt;/em&gt; to run slower. We could define a new kind of time dilation based on this observation to say that clocks ran slower in the past, however this is not the standard way of looking at things (where we attribute red-shift as due to the expansion of space) and not a particularly useful way of looking at things either.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastActivityDate="2015-11-23T20:02:45.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12618" PostTypeId="1" CreationDate="2015-11-24T02:11:51.410" Score="1" ViewCount="478" Body="&lt;p&gt;My question is about the &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0004-637X/812/1/8/pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;journal paper&lt;/a&gt; mentioned in &lt;a href=&quot;https://academia.stackexchange.com/questions/58502/journal-paper-and-conference-proceedings-note-appear-to-be-identical-is-this-pl&quot;&gt;an Academia Stack Exchange post&lt;/a&gt;. Please understand that this paper has never been posted on arXiv, and I can provide only a link whose content is behind a paywall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary of my question:&lt;/strong&gt; it boils down to &quot;whether a spatial coordinate of a fiducial observer can have a nonzero partial derivative with respect to the coordinate time.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in the validity of the central result of this paper. It is Eq. 4.24, which reads&#xA;\begin{equation}&#xA;\begin{split}&#xA;&amp;amp;\nabla\cdot\left[\frac{\alpha}{\varpi^{2}}\left\{1-\left(\frac{\omega-\Omega^{F}}{\alpha}\varpi\right)^{2}\right\}\nabla\Psi\right]&#xA;- \frac{\omega-\Omega_{F}}{\alpha}\nabla\Omega^{F}\cdot\nabla\Psi\\&#xA;&amp;amp;+ \frac{4\pi\dot{\varpi}}{\alpha^{2}\varpi}\left(1-\frac{\dot{\Phi}}{4\pi}\right)\frac{\partial \Omega^{F}}{\partial z} + 4\pi \frac{\partial}{\partial z}\left[\frac{\dot{\varpi}}{\alpha\varpi}\frac{\omega-\Omega^{F}}{\alpha}\left(1-\frac{\dot{\Phi}}{4\pi}\right)\right]\\&#xA;&amp;amp;+\frac{1}{\alpha\varpi^{2}}\left[\left(\frac{\dot{\alpha}}{\alpha}+\frac{\dot{\varpi}}{\varpi}\right)\dot{\Psi}-\ddot{\Psi}\right] + \frac{\dot{\varphi}}{\varpi}\frac{\omega-\Omega^{F}}{\alpha}\frac{\partial\Psi}{\partial\varpi}\\&#xA;&amp;amp;-\frac{16\pi^{2}\xi}{\varpi^{2}}\left(1-\frac{\dot{\Phi}}{4\pi}\right) = 0,&#xA;\end{split}&#xA;\end{equation}&#xA;where a dot on top of a symbol denotes a &lt;strong&gt;partial&lt;/strong&gt; derivative with respect to the coordinate time $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above partial differential equation is supposed to describe the magnetosphere of a Kerr black hole. The authors use the spherical coordinates $(r,\theta,\varphi)$ and define $\varpi$ as follows:&#xA;\begin{equation}&#xA;\varpi \equiv \frac{\Sigma}{\rho}\sin\theta,&#xA;\end{equation}&#xA;where&#xA;\begin{equation}&#xA;\rho^{2} \equiv r^{2} + a^{2}\cos^{2}\theta,&#xA;\end{equation}&#xA;\begin{equation}&#xA;\Sigma^{2} \equiv (r^{2}+a^{2})^{2}- a^{2}\Delta\sin^{2}\theta,&#xA;\end{equation}&#xA;and&#xA;\begin{equation}&#xA;\Delta \equiv r^{2} + a^{2} - 2Mr.&#xA;\end{equation}&#xA;Note also that $\alpha$ is the lapse function defined as&#xA;\begin{equation}&#xA;\alpha\equiv \frac{\rho}{\Sigma}\sqrt{\Delta}.&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The functions $\Psi(t,\textbf{r})$ and $\Phi(t,\textbf{r})$ denote the magnetic and electric fluxes through an $\textbf{m}$-loop passing through $\textbf{r}$, where $\textbf{m} \equiv \varpi\hat{e}_{\varphi}$ is the Killing vector associated with axisymmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What confuses me is the following: $\varpi$, $\varphi$, and $\alpha$ are simply spatial coordinates or combinations thereof, and their (partial) time derivatives should all be identically equal to zero because space and time coordinates are independent variables. This would render many parts of Eq. 4.24 nothing but convoluted ways to express the number zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also tried to follow the derivation of Eq. 4.24, and figured out that the authors implicitly assumed the following relations:&#xA;\begin{equation}&#xA;\dot{\Phi} = \dot{\varpi}\frac{\partial\Phi}{\partial \varpi}&#xA;\end{equation}&#xA;and&#xA;\begin{equation}&#xA;\dot{\Psi} = \dot{\varpi}\frac{\partial\Psi}{\partial \varpi}.&#xA;\end{equation}&#xA;Recall that a dot means a partial derivative with respect to time. As $\dot{\varpi}$ is identically zero, the above relations seem to be wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what makes me somewhat unsure about my conclusion is that this paper is published in &lt;em&gt;The Astrophysical Journal&lt;/em&gt;, a renowned peer-reviewed journal in astrophysics. (I have little expertise in astrophysics.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone verify whether my suspicion is well founded or correct me where I am wrong? Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="9862" LastEditorUserId="-1" LastEditDate="2017-04-13T12:49:59.040" LastActivityDate="2015-11-25T10:17:48.323" Title="Does this black hole magnetohydrodynamics equation even superficially make sense?" Tags="&lt;black-hole&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12619" PostTypeId="2" ParentId="12618" CreationDate="2015-11-24T03:59:32.673" Score="2" Body="&lt;p&gt;I believe your assumption of the partial derivatives with respect to time being identically zero is incorrect.  When you are talking about working in phase space, &lt;em&gt;i.e.&lt;/em&gt;, Liouville's theorem with functions of positions and velocities, then you are correct that the partial derivatives of one phase space variable with respect to the others is zero.  But here (I presume), they're looking at the temporal dynamics of the black hole magnetosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a little dense (but if you're working through Equation 4.24 above then you'll have no trouble with this), but you might find the stuff &lt;a href=&quot;http://www3.kis.uni-freiburg.de/~peter/teach/mhd/mhd.pdf&quot; rel=&quot;nofollow&quot;&gt;in this paper&lt;/a&gt; as filling in a lot of the background equations.&lt;/p&gt;&#xA;" OwnerUserId="9749" LastActivityDate="2015-11-24T03:59:32.673" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12620" PostTypeId="1" AcceptedAnswerId="12622" CreationDate="2015-11-24T08:48:56.713" Score="6" ViewCount="1015" Body="&lt;p&gt;Wikipedia coyly suggests that some notions and views kept circulating (apparently since Adam and Eve) until finally Kepler, of all people,  ... etc. I find this hard to believe. Please elucidate.&lt;/p&gt;&#xA;" OwnerUserId="9510" LastEditorUserId="2153" LastEditDate="2015-11-25T23:10:01.307" LastActivityDate="2015-11-25T23:10:01.307" Title="When was the term &quot;orbit&quot; (in the modern sense) first used and by whom?" Tags="&lt;orbit&gt;&lt;history&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12621" PostTypeId="2" ParentId="12620" CreationDate="2015-11-24T08:56:28.567" Score="6" Body="&lt;p&gt;The word orbit could refer to three latin words : orbis, which means &quot;ring&quot;, orbitus, which describes et circular shape, and orbita, which describes the track of a wheel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the meanings of this word are quite old. It is therefore hard to tell the first time it was used to describe a celestial body's round trajectory. But the word itself comes from latin.&lt;/p&gt;&#xA;" OwnerUserId="9679" LastActivityDate="2015-11-24T08:56:28.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12622" PostTypeId="2" ParentId="12620" CreationDate="2015-11-24T09:17:14.263" Score="10" Body="&lt;p&gt;Searching for the &lt;a href=&quot;http://dictionary.reference.com/browse/orbit&quot;&gt;definition&lt;/a&gt; in dictionary.com you can find this under the word origin and history section:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Late 14c., &quot;the eye socket,&quot; from Old French orbite or directly from&#xA;  Medieval Latin orbita, transferred use of Latin orbita &quot;wheel track,&#xA;  beaten path, rut, course, orbit Astronomical sense first&#xA;  recorded 1690s in English; it was in classical Latin, revived in&#xA;  Gerard of Cremona's translation of Avicenna.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe that this is what you are looking for.&lt;/p&gt;&#xA;" OwnerUserId="7926" LastActivityDate="2015-11-24T09:17:14.263" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12623" PostTypeId="1" AcceptedAnswerId="12625" CreationDate="2015-11-24T14:14:09.440" Score="1" ViewCount="130" Body="&lt;p&gt;I have been pondering the question of how we can look back to the state of the early universe for some time. So I came here and found, to my delight, that several such questions have been asked, such as &lt;a href=&quot;https://astronomy.stackexchange.com/questions/3621/why-can-we-still-see-10-billion-year-old-galaxies&quot;&gt;this one&lt;/a&gt;. Fortunately the answers to that (and other questions) has helped immeasurably, but I have one lingering area of confusion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I often hear, in cosmological TV documentaries (such as Horizon in the UK) words such as &quot;The further away we look, the further back in time we are looking. Our technology has now advanced to the point we can see objects so far away they began emitting their light just [n] after the big bang&quot; [this is not a quote from a programme but my synthesis of what I have heard].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this: I understand the limitation of c and light taking finite time to reach us from distant objects. Having now understood that inflation caused the early universe to &quot;move&quot; so far away we are only now receiving its light; How can today's technology evolve into tomorrow's technology and enable us to see &quot;older&quot; light as we can only see what is arriving at the Earth right now. I.e. if a time-traveller turned up today bearing a better telescope from the future, I should not be able to use it to see further away and therefore back in time, because it is dependent on the age of the light arriving at my location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only solution I can think of is that due to tricksy inflation rate effects, the light of successively further objects is arriving at, essentially, the same time (when looked at over a universal age). But that doesn't seem to make sense and even if it did how would we sort the older light from the newer light from the same object if it is all arriving at once (notwithstanding red shift, which I do (hopefully) understand)...&lt;/p&gt;&#xA;" OwnerUserId="9869" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-11-25T09:01:03.953" Title="Looking back in time by looking further away" Tags="&lt;cosmological-inflation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12624" PostTypeId="2" ParentId="387" CreationDate="2015-11-24T14:19:38.140" Score="2" Body="&lt;p&gt;The disk flattens from a 3d cloud to a 2d disk. A ring forms from a disk over-density band, reducing to a nominally 1d ring from a 2d band. The ring particles fall towards each other around the ring, and such attractions cause the ring to contract its circumference. The ring thus migrates inwards from the disk to the sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ring particles collide and the average particle size increases while the number of particles decreases. The angular momentum of the ring increases as particles falling against the direction of spin of the ring fall inward from the circumference while the particles falling with the direction of spin fall outwards, so that mutual gravitational attraction of particles causes them to orbit one another rather than collide. This effect converts spin of the ring to spin of mutually bound particles within the ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally the ring collapses to one remaining particle which is the planet hatched by the ring. Moons are remaining large particles in the ring distant from the planet. Once the planet has formed and any remaining moons have bound themselves the planet, the ring no longer exists, leaving a planet and its moons if any in a stable orbit around the sun. Later other planets form from other over-dense bands of the disk. &lt;/p&gt;&#xA;" OwnerUserId="9870" LastEditorUserId="2365" LastEditDate="2015-11-24T20:27:20.660" LastActivityDate="2015-11-24T20:27:20.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12625" PostTypeId="2" ParentId="12623" CreationDate="2015-11-24T15:09:04.050" Score="4" Body="&lt;p&gt;Before the advent of telescopes, we could only look back in time from a few years (for nearby stars) to a few thousand years (for the most distant stars visible to the unaided eye). In addition to this, a handful of galaxies are visible without binoculars, letting us look back a few million years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first telescopes allowed us to see much farther, like hundreds of megaparsec, so we could look back hundreds of millions of years. As technology improves and our telescopes become better, we are able to observe more and more distant objects, and thus look further and further back in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The farther away an object is, the less light we receive, and hence the bigger telescope and better detector we need in order to see it. So, being able to look further back in time than previously is not about the light from more distant objects eventually reaching us, but simply a matter of developing the tools necessary to see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, even the most distant galaxy we will ever see (which are at earliest a few hundred million years after the Big Bang, i.e. roughly 13.5 billion years (&quot;Gyr&quot;) ago), are not farther away than the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow&quot;&gt;cosmic microwave background&lt;/a&gt;, which was emitted 380,000 years after Big Bang, i.e. ~13.8 Gyr ago, and which we have been able to detect since the mid-1950's. If we ever manage to build an efficient neutrino detector, we may be able to look back to seconds after the Big Bang, when these particles decoupled from matter and began traveling freely through space.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-11-24T15:16:06.877" LastActivityDate="2015-11-24T15:16:06.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12626" PostTypeId="2" ParentId="12222" CreationDate="2015-11-24T16:46:25.000" Score="1" Body="&lt;p&gt;I wrote some JavaScript to help answer my question. It's in terrible shape right now, but it plots the &lt;em&gt;hourly&lt;/em&gt; position of Aldebaran in azimuth &amp;amp; elevation when viewed from Dayton, Ohio, USA, every &lt;em&gt;fortnight&lt;/em&gt; for a &lt;em&gt;year&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://fasiha.github.io/starpath/&quot; rel=&quot;nofollow&quot;&gt;http://fasiha.github.io/starpath/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The source code is at &lt;a href=&quot;https://github.com/fasiha/starpath&quot; rel=&quot;nofollow&quot;&gt;https://github.com/fasiha/starpath&lt;/a&gt; but everything in it is hardcoded. However, when I have a bit of time, I can make it into a nice little webapp where you can input your position and select a celestial object (or inputs its RA/dec) and see the same plot—possibly along with information about when the sun obscures it.&lt;/p&gt;&#xA;" OwnerUserId="9560" LastActivityDate="2015-11-24T16:46:25.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12627" PostTypeId="2" ParentId="12545" CreationDate="2015-11-24T20:21:04.267" Score="2" Body="&lt;p&gt;The &lt;a href=&quot;http://apod.nasa.gov/apod/ap151123.html&quot; rel=&quot;nofollow&quot;&gt;Astronomy Picture of the day for the 23rd November 2015&lt;/a&gt; has all the neblosity around Orion, in fantatic detail. The result of 212 hour exposure.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-11-24T20:21:04.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12628" PostTypeId="2" ParentId="12598" CreationDate="2015-11-24T23:17:12.950" Score="2" Body="&lt;p&gt;There were a few other small errors in your translation (such as indenting too much stuff in two of the &lt;code&gt;if&lt;/code&gt; statements), and I think you might have been trying to be too clever with changing &lt;code&gt;SUBROUTINE&lt;/code&gt; into defined functions when it was a lot easier to just drop the code in-line since it was only used once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There were minor tweaks and as far as I can tell this seems to work (you'll want to play with the formatted print to clean it up, and this runs under Python 3; if you need it to run under Python 2, you'll probably need to change the &lt;code&gt;print&lt;/code&gt; and &lt;code&gt;num1.write&lt;/code&gt; statements):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#!/usr/bin/env python&#xA;&#xA;import math&#xA;&#xA;print('Program to flag Jovian Decametric windows')&#xA;month = ['JAN','FEB','MAR','APR','MAY','JUN','JUL','AUG','SEP','OCT','NOV','DEC']&#xA;week = 42.46/360&#xA;pi = 3.141593&#xA;kr = pi / 180&#xA;num1 = open('jovrad.txt', 'w')&#xA;yy = int(input((&quot;Year for which predictions are required &quot;)))&#xA;e = math.trunc(((yy-1) / 100))&#xA;print(e)&#xA;f = 2 - e + math.trunc(e/4)&#xA;print(f)&#xA;jd = math.trunc(365.25 * (yy - 1)) + 1721423 + f + .5&#xA;print(jd)&#xA;d0 = jd - 2435108&#xA;print(d0)&#xA;incr = 0&#xA;dmax = 0&#xA;tx = 0&#xA;ty = 0&#xA;yyly = 0&#xA;yylc = 0&#xA;if yy / 400 - math.trunc((yy / 400)) == 0:&#xA;    incr = 1&#xA;    print(&quot;in if-1&quot;)&#xA;yyly = yy / 4 - math.trunc((yy / 4))&#xA;yylc = yy / 100 - math.trunc((yy / 100))&#xA;if yyly == 0 and yylc != 0:&#xA;    print(&quot;in if-2&quot;)&#xA;    incr = 1&#xA;ty = 59 + incr&#xA;dmax = 365 + incr&#xA;tx = ty + .5&#xA;num1.write(&quot;******************************************************\n&quot;)&#xA;pout = &quot;  JOVIAN IO-DECAMETRIC EMISSION PREDICTIONS FOR &quot; + str(yy) + &quot;\n&quot;&#xA;num1.write(pout)&#xA;num1.write(&quot;******************************************************\n&quot;)&#xA;num1.write(&quot;\n&quot;)&#xA;num1.write(&quot;Day   Date   Hr(UT)  Io_Phase   CML   Dist(AU)  Source&quot;)&#xA;num1.write(&quot;\n&quot;)&#xA;th = 0&#xA;&#xA;while int(th / 24) + 1 &amp;lt;= dmax:&#xA;    d = d0 + (th / 24)&#xA;    v = (157.0456 + .0011159 * d) % 360&#xA;    m = (357.2148 + .9856003 * d) % 360&#xA;    n = (94.3455 + .0830853 * d + .33 * math.sin(kr * v)) % 360&#xA;    j = (351.4266 + .9025179 * d - .33 * math.sin(kr * v)) % 360&#xA;    a = 1.916 * math.sin(kr * m) + .02 * math.sin(kr * 2 * m)&#xA;    b = 5.552 * math.sin(kr * n) + .167 * math.sin(kr * 2 * n)&#xA;    k = j + a - b&#xA;    r = 1.00014 - .01672 * math.cos(kr * m) - .00014 * math.cos(kr * 2 * m)&#xA;    re = 5.20867 - .25192 * math.cos(kr * n) - .0061 * math.cos(kr * 2 * n)&#xA;    dt = math.sqrt(re * re + r * r - 2 * re * r * math.cos(kr * k))&#xA;    sp = r * math.sin(kr * k) / dt&#xA;    ps = sp / .017452&#xA;    dl = d - dt / 173&#xA;    pb = ps - b&#xA;    xi = 150.4529 * math.trunc((dl)) + 870.4529 * (dl - math.trunc((dl)))&#xA;    L3 = (274.319 + pb + xi + .01016 * 51) % 360&#xA;    U1 = 101.5265 + 203.405863 * dl + pb&#xA;    U2 = 67.81114 + 101.291632 * dl + pb&#xA;    z = (2 * (U1 - U2)) % 360&#xA;    U1 = U1 + .472 * math.sin(kr * z)&#xA;    U1 = (U1 + 180) % 360&#xA;    s = &quot;&quot;&#xA;    if L3 &amp;lt; 255 and L3 &amp;gt; 200 and U1 &amp;lt; 250 and U1 &amp;gt; 220:&#xA;        s = &quot;Io-A&quot;&#xA;    if L3 &amp;lt; 180 and L3 &amp;gt; 105 and U1 &amp;lt; 100 and U1 &amp;gt; 80:&#xA;        s = &quot;Io-B&quot;&#xA;    if L3 &amp;lt; 350 and L3 &amp;gt; 300 and U1 &amp;lt; 250 and U1 &amp;gt; 230:&#xA;        s = &quot;Io-C&quot;&#xA;    if s != &quot;&quot;:&#xA;        dy = math.trunc((th / 24)) + 1&#xA;        h = th - (dy - 1) * 24&#xA;        if(dy &amp;gt; th):&#xA;            m = math.trunc((dy - tx) / 30.6) + 3&#xA;            da = dy - ty - math.trunc((m - 3) * 30.6 + .5)&#xA;        else:&#xA;            m = math.trunc((dy - 1) / 31) + 1&#xA;            da = dy - (m - 1) * 31&#xA;        mn = month[(m-1)]&#xA;#        mn = month[(m-1)*3+1:(m-1)*3-1+3]&#xA;        outstring = &quot;%i  %s  %i  %5.3f  %5.3f  %5.3f  %5.3f  %s\n&quot; % (dy, mn, da, h, U1, L3, dt, s)&#xA;        num1.write(outstring)&#xA;    th = th + .5&#xA;&#xA;num1.close()&#xA;print(&quot;Program Complete  - results in file JOVRAD.TXT&quot;)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9749" LastActivityDate="2015-11-24T23:17:12.950" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12629" PostTypeId="1" CreationDate="2015-11-25T02:04:40.837" Score="-1" ViewCount="826" Body="&lt;p&gt;Every day, babies are born and people grow, which makes their respective masses greater. However, this change in mass (should) come out of the food that they consume - it is used as energy and thus converted into this growth. Likewise, building new structures is just a redistribution of mass already on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the only way I can see the actual mass of the Earth changing is by meteors that have landed here (increase in mass) and things that leave Earth such as space shuttles and rockets which would be  a decrease in mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But have I missed anything out? If no meteors crashed into Earth, and we had not yet figured out how to make machines that could fly, would the mass of the Earth remain constant? Or is it somehow loosing or gaining mass? Am I wrong in assuming that the growth of animals doesn't affect the mass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, Toastrackenigma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Mass includes atmosphere :)&lt;/p&gt;&#xA;" OwnerUserId="9876" LastEditorUserId="4085" LastEditDate="2015-11-25T07:38:27.107" LastActivityDate="2016-04-10T16:29:10.847" Title="Does the mass of the Earth change?" Tags="&lt;earth&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12630" PostTypeId="1" AcceptedAnswerId="12635" CreationDate="2015-11-25T02:47:39.220" Score="3" ViewCount="169" Body="&lt;p&gt;Would dark matter have a similar effect on the Solar System as it does on the Milky Way? I'd imagine it could help hold together some of the Solar System bodies or at least affect their motion, right? Maybe to a lesser extent, but would it be significant?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2015-11-25T07:47:23.267" Title="Does Dark Matter affect the motion of the Solar System?" Tags="&lt;solar-system&gt;&lt;the-sun&gt;&lt;gravity&gt;&lt;galaxy&gt;&lt;dark-matter&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12633" PostTypeId="1" AcceptedAnswerId="12634" CreationDate="2015-11-25T07:14:21.170" Score="5" ViewCount="303" Body="&lt;p&gt;Why does hydrogen ionization happen in HII regions? Why is the hydrogen there ionized? &lt;/p&gt;&#xA;" OwnerUserId="9645" LastEditorUserId="2365" LastEditDate="2015-11-25T18:23:31.500" LastActivityDate="2015-11-25T18:23:31.500" Title="Why does hydrogen ionization happen in HII regions?" Tags="&lt;nebula&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12634" PostTypeId="2" ParentId="12633" CreationDate="2015-11-25T07:33:29.080" Score="8" Body="&lt;p&gt;Stars are responsible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HII regions$^\dagger$ can refer to several things, but usually I guess one thinks of the volumes around star-forming regions. The more massive a star is, the faster it burns its fuel, and at a higher temperature, meaning that the peak of their spectra are more toward the high frequencies. The most massive stars of a stellar population — the so-called O and B stars — produce enough photons with wavelengths below the hydrogen ionization threshold of $\lambda = 912$ Å that they carve out bubbles in their surrounding HI clouds, giving rise to HII regions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tj74l.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tj74l.png&quot; alt=&quot;pic&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;em&gt;Right&lt;/em&gt;: The HII region NGC 604 (from &lt;a href=&quot;https://en.wikipedia.org/wiki/H_II_region&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;). &lt;em&gt;Left&lt;/em&gt;: The spectra of three different stars. Only the B star has a significant portion of its spectrum above the hydrogen ionization threshold. Note the logarithmic scale on the intensity (from &lt;a href=&quot;http://boojum.as.arizona.edu/~jill/NS102_2006/Lectures/Lecture19/StarProperties/starproperties.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;).&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the high densities, the HII quickly recombines to HI. If the recombination goes directly to the ground state, a new ionizing photon is emitted, which again is absorbed by a hydrogen atom, but if it goes to one of the higher states, the emitted radiation is no longer capable of ionizing the gas. In this way the ionizing radiation is converted into photons of specific wavelengths, corresponding the energy differences between the excited level of hydrogen, most notably the Lyman $\alpha$ emission line with $\lambda = 1216$ Å.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because hydrogen is the most abundant element in the Universe, and because Lyman $\alpha$ is the most common transition, the Lyman $\alpha$ line is an excellent probe of the most distant galaxies where other wavelengths are not observable. This is especially so because the most distant galaxies are also the earliest and hence still in the process of forming, meaning a high star formation rate which, in turn, means that the shortlived OB stars are still present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to this distinct regions of HII, ionized hydrogen also exist in a more diffuse component between the stars of a galaxy, in huge bubbles caused by stellar feedback and supernovae, and in the intergalactic medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;$^\dagger$The terms HI and HII refers to neutral and ionized hydrogen, respectively.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-11-25T10:26:16.447" LastActivityDate="2015-11-25T10:26:16.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12635" PostTypeId="2" ParentId="12630" CreationDate="2015-11-25T07:47:23.267" Score="5" Body="&lt;p&gt;The dark matter model that is used to explain the &quot;missing mass&quot; problem relating to our Galactic rotation curve, consists of a pseudo-spherical distribution that is much more extended than the visible stars and gas. Even though this &quot;halo&quot; contains more than ten times the mass of the visible matter, when you work out what it's density should be in the solar neighbourhood, it turns out to be around 100 times &lt;em&gt;less&lt;/em&gt; than the &quot;normal&quot; interplanetary medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means, that inside the Earth's orbit, there could be as much as $10^{10}$ kg of dark matter. This sounds a lot and in principle, yes, this would speed up the orbits of planets in a way that depended on their distance from the Sun. However, the predicted effects of this are still about 6 orders of magnitude below the current precision of measurement.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-11-25T07:47:23.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12636" PostTypeId="1" CreationDate="2015-11-25T10:26:56.647" Score="1" ViewCount="428" Body="&lt;p&gt;I have a set of equivalent width measurements for different emission lines. How can I use them in order to determine the corresponding line luminosities?&lt;/p&gt;&#xA;" OwnerUserId="9879" LastEditorUserId="2365" LastEditDate="2015-11-25T18:19:26.203" LastActivityDate="2015-11-25T18:48:18.720" Title="How can I determine line luminosities from equivalent width measurements?" Tags="&lt;star&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12637" PostTypeId="1" CreationDate="2015-11-25T13:36:11.740" Score="6" ViewCount="374" Body="&lt;p&gt;I'd like to know how the orbits of planets were first 'mapped out' and to understand the maths behind it. e.g. How do we know what position in space a planet will be at certain time so that spacecraft trips to planets could be planned etc.?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd also like to know the most primitive way that this could be done and if there is anyway this could be done using only maths and simple measuring tools assuming no previous knowledge about the planets positions.&lt;/p&gt;&#xA;" OwnerUserId="9515" LastEditorUserId="2153" LastEditDate="2015-11-25T23:11:37.023" LastActivityDate="2016-01-25T07:51:44.697" Title="How were the orbits of planets first mapped out?" Tags="&lt;orbit&gt;&lt;amateur-observing&gt;&lt;fundamental-astronomy&gt;&lt;history&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12638" PostTypeId="1" CreationDate="2015-11-25T16:08:32.830" Score="2" ViewCount="107" Body="&lt;p&gt;I'm trying to think through what the cost would be of an unmanned mission to a nearby asteroid.  To me it seems like the high-level costs would be the fixed cost of the &quot;spaceship&quot; itself (including any scientific or mining equipment required to carry out the mission), the cost of getting into orbit which would depend on the ship mass, and the fuel to get there and return (which would be a function of fuel cost, ship mass, how fast you want to get there, and the mass of anything collected at the asteroid site that would be returned).  Is there anywhere that this is discussed or has anyone thought this through further?&lt;/p&gt;&#xA;" OwnerUserId="9214" LastActivityDate="2015-11-25T18:37:35.727" Title="What would the cost be of visiting an asteroid?" Tags="&lt;asteroids&gt;&lt;space-travel&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12639" PostTypeId="1" CreationDate="2015-11-25T16:32:47.277" Score="2" ViewCount="874" Body="&lt;p&gt;I have a homework question in which I need to estimate a parameter known as $\beta_{P}$ and also the core temperature, both for the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, prior to doing this I need to know the mean molecular weight of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First question is, how would I go about doing this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second question is, can I assume that it is about 70% hydrogen and 30% Helium? &lt;/p&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2015-11-25T17:07:19.460" Title="How to calculate the mean molecular weight of the Sun" Tags="&lt;astrophysics&gt;&lt;the-sun&gt;&lt;hydrostatic-equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12640" PostTypeId="2" ParentId="12639" CreationDate="2015-11-25T17:07:19.460" Score="3" Body="&lt;p&gt;For your second question, it depends upon whether you are talking about composition by number, or composition by mass (the numbers you give are roughly the composition by mass).  If you want to get really specific, you can &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/tables/suncomp.html&quot; rel=&quot;nofollow&quot;&gt;add in the other elements too&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the first, you add them up &lt;a href=&quot;http://facstaff.cbu.edu/rprice/lectures/compos.html&quot; rel=&quot;nofollow&quot;&gt;proportionally&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9749" LastActivityDate="2015-11-25T17:07:19.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12641" PostTypeId="2" ParentId="12638" CreationDate="2015-11-25T18:37:35.727" Score="4" Body="&lt;p&gt;This is something that has already been done, so we already have a point of reference. &lt;a href=&quot;https://en.wikipedia.org/wiki/Hayabusa&quot; rel=&quot;nofollow&quot;&gt;Hayabusa&lt;/a&gt; landed on the asteroid &lt;a href=&quot;https://en.wikipedia.org/wiki/25143_Itokawa&quot; rel=&quot;nofollow&quot;&gt;25143 Itokawa&lt;/a&gt; in 2005. The mission was considered a success and a sample was returned to Earth. According to &lt;a href=&quot;http://www.astrobio.net/topic/solar-system/meteoritescomets-and-asteroids/the-little-spacecraft-that-could/&quot; rel=&quot;nofollow&quot;&gt;sources from the time&lt;/a&gt;, the total mission cost was around 170 million dollars and &lt;a href=&quot;http://phys.org/news/2010-06-japanese-space-probe-hayabusa-home.html&quot; rel=&quot;nofollow&quot;&gt;other sources&lt;/a&gt; put the development cost of the spacecraft at around 138 million dollars. There is also a planned &lt;a href=&quot;https://en.wikipedia.org/wiki/Hayabusa_2&quot; rel=&quot;nofollow&quot;&gt;Hayabusa 2&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="2092" LastActivityDate="2015-11-25T18:37:35.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12642" PostTypeId="2" ParentId="12636" CreationDate="2015-11-25T18:48:18.720" Score="1" Body="&lt;p&gt;If you &lt;em&gt;also&lt;/em&gt; have photometry, then there is a reasonably accurate conversion between the apparent magnitude and the continuum flux (per unit wavelength interval) at the wavelength of the photometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have this conversion factor, you multiply your equivalent width by it to get a flux.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-11-25T18:48:18.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12643" PostTypeId="1" AcceptedAnswerId="12644" CreationDate="2015-11-25T20:22:41.827" Score="4" ViewCount="132" Body="&lt;p&gt;So for an assignment I have to answer the question what I can conclude concerning the shape of the orbit, knowing that it is both an eclipsing and spectroscopic binary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I think the answer I supposed to be that, since we know it is a spectroscopic binary, the orbits will most likely be circular. And I've read on various places that this is true, because of the tidal forces that cause for circular orbits. These tidal forces are quite strong, because apparently spectroscopic binaries are quite close to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here is where I lose it, I don't understand why spectroscopic binaries should per se be close to each other. I have considered that this is because they are not visual binaries; if they were visual binaries we would be able to tell them apart. But we can't tell them apart, so they are not visual binaries. From this you could conclude that this means that they are close to each other, but I think it could just as well mean that they are very very far away, and that that is the reason we cannot tell them apart. So I can't find a proper explanation for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks if you can help me out!!&lt;/p&gt;&#xA;" OwnerUserId="9713" LastEditorUserId="9186" LastEditDate="2015-11-26T08:49:08.390" LastActivityDate="2015-11-26T08:49:08.390" Title="Why do spectroscopic binaries have approximate circular orbits?" Tags="&lt;orbit&gt;&lt;observational-astronomy&gt;&lt;binary-star&gt;&lt;tidal-forces&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12644" PostTypeId="2" ParentId="12643" CreationDate="2015-11-26T00:37:42.473" Score="2" Body="&lt;p&gt;Spectroscopic binaries are not necessarily close and not necessarily in circular orbits. If the orbit is circular, then a time series of radial velocity measurements for one or both stars would be well modelled with a simple sine wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, &lt;em&gt;eclipsing&lt;/em&gt; binaries are almost certainly close, since eclipses in a system with widely separated components are improbable (but not impossible). If you have the eclipsing light curve, then the primary and secondary eclipses will be equi-spaced in time if the orbit is circular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key factor for circularisation is the orbital period. If it is less than about 7 days (for solar-type stars), then circularisation due to tidal forces is almost certain.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-11-26T00:37:42.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12645" PostTypeId="1" AcceptedAnswerId="12648" CreationDate="2015-11-26T02:53:58.307" Score="1" ViewCount="215" Body="&lt;p&gt;For example, the Sun is a giant sphere of positively charged plasma. About 72% of the elements in it are protium. This means that these protium atoms would have had to lose their electrons, right? If so, the majority of the Sun would be protons. So just as there are neutron stars, would stars like our Sun be considered &quot;proton stars&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may be a stupid question, but what I'm really getting at is: are stars mostly protons? Is what I described how it all works?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2018-04-11T21:09:12.927" Title="If there are neutron stars, would most stars be considered &quot;proton stars&quot;?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;neutron-star&gt;&lt;hydrogen&gt;&lt;astrochemistry&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12646" PostTypeId="1" AcceptedAnswerId="12650" CreationDate="2015-11-26T02:55:54.040" Score="4" ViewCount="432" Body="&lt;p&gt;A black hole doesn't necessarily need to form from a star; theoretically, it could form from any extremely dense object. In fact, many astronomers differentiate certain black holes, like &lt;a href=&quot;http://en.wikipedia.com/wiki/Supermassive_black_hole&quot; rel=&quot;nofollow&quot;&gt;supermassive ones&lt;/a&gt;, from &lt;a href=&quot;http://en.wikipedia.com/wiki/Stellar_black_hole&quot; rel=&quot;nofollow&quot;&gt;stellar ones&lt;/a&gt; (ones that form from stars).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, could the same apply for neutron stars? Neutron stars only form because of the intense gravity during a star's collapse: electron capture is forced to happen, and the majority of the star becomes neutrons. Could this potentially happen to non-stellar objects, if the gravity forces electron capture?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, why don't we see as many of these &quot;neutron objects&quot; as we do non-stellar black holes?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="2365" LastEditDate="2018-07-27T22:35:43.573" LastActivityDate="2020-10-20T06:39:24.383" Title="If any object could become a black hole, could any object become a neutron star?" Tags="&lt;star&gt;&lt;black-hole&gt;&lt;gravity&gt;&lt;neutron-star&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12647" PostTypeId="1" AcceptedAnswerId="12651" CreationDate="2015-11-26T03:01:34.707" Score="5" ViewCount="81" Body="&lt;p&gt;This question relates to a diffuse hot gas halo of our Milky Galaxy. I've read that there is a hot diffuse halo of gas surrounding our Galaxy (NED, Caltech)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering why such a halo can exist? Why doesn't it collapse to a disk shape? Is it because the gas itself is still hot and so remains largely unaffected by the Galaxy potential?&lt;/p&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2015-11-26T07:32:53.413" Title="Gas halo of our Milky Way Galaxy" Tags="&lt;gas&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12648" PostTypeId="2" ParentId="12645" CreationDate="2015-11-26T05:36:13.423" Score="5" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Isotopes_of_hydrogen#Hydrogen-1_.28protium.29&quot; rel=&quot;noreferrer&quot;&gt;Protium&lt;/a&gt; is a proton + an electron.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under enormously high pressure, it's energetically favorable for electrons to merge with protons and become neutrons - see &lt;a href=&quot;https://en.wikipedia.org/wiki/Degenerate_matter#Neutron_degeneracy&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;are stars mostly protons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By mass, yes, at least before they get too old.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the universe is more complicated, but anything solid that we think of as matter is made of atoms, which are by mass, mostly protons and neutrons (you can break it up further than that if you like, but that's best for another question).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most hydrogen has no neutrons so any hydrogen rich object (the sun, most young stars, gas giant planets) are by mass, mostly protons.  That's no longer true when a star gets close to the end of it's life and has burned much of it's hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter, by mass is roughly about 80% protons.  The sun, because it's been turning hydrogen into helium for about 4.5 billion years, is roughly about 67% protons by mass.   The Earth, mostly other elements, Oxygen, Silicon, Iron, etc, is a about 50% proton by mass.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Would stars like our Sun be considered &quot;proton stars&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I suppose you could use that term, but I don't see any benefit to it.    It's not too different than calling the sun a &quot;hydrogen&quot; star.  All stars start out as hydrogen stars.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-11-26T05:36:13.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12649" PostTypeId="2" ParentId="12637" CreationDate="2015-11-26T07:09:56.947" Score="4" Body="&lt;p&gt;I think your question covers a lot of ground cause you're asking it from 2 different perspectives.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How were the orbits of planets &lt;strong&gt;first&lt;/strong&gt; mapped out?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do we know what position in space a planet will be at certain time&#xA;  so that spacecraft trips to planets could be planned etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These should probably be separate questions, as they're quite different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First one first, here's a bit of &lt;a href=&quot;http://www.universetoday.com/117843/how-did-we-find-the-distance-to-the-sun/&quot; rel=&quot;nofollow noreferrer&quot;&gt;history&lt;/a&gt;:  Using parallax Hipparchus was able to get a pretty good estimate of the distance to the Moon about 200 BC.  The distance to the sun was much more difficult but Aristarchus was the first to come up with a method and others were able to improve upon it to get more accurate measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relative orbits of the planets wasn't hard to calculate once you had one of them, more on that &lt;a href=&quot;https://physics.stackexchange.com/questions/36005/when-and-how-were-relative-distances-to-the-planets-first-measured&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can measure the length of a year on Mercury or Venus quite easily, as well as their relative angle to the sun (though Mercury's orbit is quite eccentric, you'd want to take several measurements).  Even the wacky Copernican model with just circles on top of other circles made pretty good distance calculations with the sun in the center and the planets orbiting the sun.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kepler was the first to recognize that the planets orbited at slightly different orbital planes, and from what I've read, Kepler was prouder of that discovery than the discovery of the elliptical orbits that he's most often remembered for.  The combination of slightly different orbital planes and elliptical orbits and Brache's improved accuracy to his measurements was a huge step forward in planet mapping.  Kepler still didn't have the distance to the sun well plotted out, but he figured out the model for planetary orbits.   After that, it was just a matter of improving the distance measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to how NASA does it now.   I'm not an expert, so somebody else might be able to answer that one better.   It gets pretty mathematical, 3 position vectors, 3 direction vectors and velocity changes based on position, but we have nearly pinpoint laser measurements now.     &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'd also like to know the most primitive way that this could be done&#xA;  and if there is anyway this could be done using only maths and simple&#xA;  measuring tools assuming no previous knowledge about the planets&#xA;  positions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It depends what you mean by primitive.    Galileo's telescope enabled him to see phases of Venus, and that isn't possible without a telescope.  The telescope also made it possible for Galileo to see 4 of Jupiter's moons, which orbited Jupiter and not Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aristarchus tried to eyeball the difference in shadow from the sun between the earth and the moon and he estimated the Sun was some 18-20 times more distant than the moon.  Details &lt;a href=&quot;https://en.wikipedia.org/wiki/On_the_Sizes_and_Distances_%28Aristarchus%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.  To his credit, he successfully deduced that the Sun is much further than the moon, and he estimated the angle ratio to be 87 degrees when it was 89.83 degrees.  Despite his efforts, his estimate was off by a factor of about 20, in distance though he was only off by 2.83 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brahe used the best equipment available to him at the time, much of it made specifically for his observations.  More on that &lt;a href=&quot;http://ircamera.as.arizona.edu/NatSci102/NatSci102/lectures/tycho.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.  Brahe also had the advantage of having accurate time pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brahe believed in the Ptolemaic model but he was unable to get it to match his data and he urged Kepler to figure it out when he was old and knew he wasn't going to figure it out himself.  (There are rumors that Kepler poisoned Brahe to get more credit for himself, but I don't think that actually happened, Brahe's body was even exhumed to test that theory some time later but no traces of poison were found).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kepler, unlike Brahe, liked the Copernican model but he couldn't get that to work because he believed in Copernican circles.   He tried ellipses, more out of desperation because he couldn't get circles to work and the ellipse model matched the data.   Kepler had not only Brahe's but years of Copernicus' records to work with as well as observations preceding them, some going all the way back to ancient Greece.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure it would be possible to figure out that planets have elliptical orbits and how far they were from Earth without good equipment.  Kepler had a lot to work with and a determination to find a simple explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that's not too wordy.  I can try to clean it up a bit later if you like.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-11-26T07:09:56.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12650" PostTypeId="2" ParentId="12646" CreationDate="2015-11-26T07:11:22.137" Score="5" Body="&lt;p&gt;There can be no such thing as a &amp;quot;supermassive neutron star&amp;quot;. The theoretical upper mass limit for a neutron star is somewhere between 2.2 and 3 solar masses. Any more massive and they inevitably form black holes. So I am not clear what kind of &amp;quot;neutron objects&amp;quot; you were thinking of?&lt;/p&gt;&#xA;&lt;p&gt;Nor is it clear what you mean by &amp;quot;non-stellar&amp;quot; objects that will have the densities required to make neutron-degenerate matter? There aren't any apart from (I) the cores of massive stars at the ends of their lives. (II) Massive white dwarfs if they accrete matter over and above their Chandrasekhar limit. Furthermore, there is a &lt;em&gt;minimum&lt;/em&gt; mass for a neutron star. Although the observed lower limit to those seen in nature is (so far) around &lt;span class=&quot;math-container&quot;&gt;$1.15 M_{\odot}$&lt;/span&gt;, there is a theoretical lower limit of about &lt;span class=&quot;math-container&quot;&gt;$0.1$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$0.2M_{\odot}$&lt;/span&gt; set by the stability of the neutron against decay in self-gravitating material ( See &lt;a href=&quot;https://physics.stackexchange.com/questions/143166/what-is-the-theoretical-lower-mass-limit-for-a-gravitationally-stable-neutron-st/143174#143174&quot;&gt;https://physics.stackexchange.com/questions/143166/what-is-the-theoretical-lower-mass-limit-for-a-gravitationally-stable-neutron-st/143174#143174&lt;/a&gt; ).&lt;/p&gt;&#xA;&lt;p&gt;It is true if you could arrange to compress any matter to densities above &lt;span class=&quot;math-container&quot;&gt;$\sim 10^{15}$&lt;/span&gt; kg/m&lt;span class=&quot;math-container&quot;&gt;$^3$&lt;/span&gt; it would form neutron-degenerate material. But this requires (as far as we know) the conditions I listed above.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2020-10-20T06:39:24.383" LastActivityDate="2020-10-20T06:39:24.383" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12651" PostTypeId="2" ParentId="12647" CreationDate="2015-11-26T07:32:53.413" Score="9" Body="&lt;p&gt;The scale height of gas in a disk (if it were in equilibrium) is roughly $kT/mg$, where $T$ is the temperature, $g$ is the gravitational field, $m$ the mean mass of agas particle, and $k$ the Boltzmann constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume most of the mass is in a thin disk, then Gauss's law for gravitation tells us that that $g = 2\pi G \sigma$, where $\sigma$ is the mass per unit area in the disk. According to Rix &amp;amp; Bovy, $\sigma \simeq 70 M_{\odot}$ pc$^{-2}$ at the location of the Sun (&lt;a href=&quot;http://arxiv.org/abs/1309.0809&quot;&gt;http://arxiv.org/abs/1309.0809&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume hydrogen gas, then the effective particle mass is that of a proton, and this means the gas scale height is&#xA;$$ H = 4300 \left(\frac{T}{10^6\ K}\right)\ pc$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus gas hotter than a million degrees will have a very substantial scale height and is not expected to be confined to the Milky Way disk.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-11-26T07:32:53.413" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12652" PostTypeId="1" AcceptedAnswerId="12653" CreationDate="2015-11-26T09:04:39.750" Score="3" ViewCount="150" Body="&lt;p&gt;Observing from Earth what are the odds that Venus does not line up in conjunction with Earth’s moon and does not transit behind the moon but slightly below or above with a complete visual of Venus during the entire conjunction of the two? What are the chances of a miscalculation in this event?&lt;/p&gt;&#xA;" OwnerUserId="9326" LastEditorUserId="2287" LastEditDate="2016-09-10T20:37:56.250" LastActivityDate="2016-09-10T20:37:56.250" Title="Venus transiting behind the Earth’s moon December 7, 2015" Tags="&lt;the-moon&gt;&lt;venus&gt;&lt;planetary-transits&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12653" PostTypeId="2" ParentId="12652" CreationDate="2015-11-26T10:06:04.530" Score="5" Body="&lt;p&gt;Well, depending on where you live, the probability is either 0 or 1, modulo cloudy weather.&lt;/p&gt;&#xA;&lt;p&gt;From the picture below (from &lt;a href=&quot;http://www.lunar-occultations.com/iota/planets/1207venus.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this site&lt;/a&gt;), you see that you'll need to be in the US to see the occultation. However, only in the regions outlined in cyan (West Alaska, East Siberia, East Canada, and in Caribbean islands east of the Dominican Republic) will it be during darkness (just before sunrise / after sunset).&lt;/p&gt;&#xA;&lt;p&gt;But with binoculars (or without, if you have very sharp eyes), you should be able to see it over all of the US.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iS2wG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iS2wG.jpg&quot; alt=&quot;occultation map&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;(cyan=occultation at moonrise/moonset; red dotted=daytime occultation; blue=twilight occultation; white=nighttime occultation)&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;strong&gt;Miscalculation&lt;/strong&gt;&#xA;&lt;p&gt;After your edit, I see that you are not referring to the probability of seeing the occultation, but the probability of a mis-calculation. I'd say that those odds cannot really be calculated, but it can be said that they are extremely small. No physical theory can ever be proved, but after a sufficient amount of verifications, we usually accept a theory as &amp;quot;true for all practical purposes, until disproved&amp;quot;. Many, many factors go into a calculation like this (all the way down to mathematical axioms).&lt;/p&gt;&#xA;&lt;p&gt;But a calculation like this has to do with celestial mechanics, which is very well understood, and which continuously makes accurate predictions and hence is continuously empirically verified. So the odds you request are definitely much, much smaller than, say, the odds of calculating the weather tomorrow, or the mass of a galaxy cluster.&lt;/p&gt;&#xA;&lt;strong&gt;Uncertainties&lt;/strong&gt;&#xA;&lt;p&gt;Of course there are always &lt;em&gt;uncertainties&lt;/em&gt; associated with such calculations, which gives an uncertainty in the position of celestial bodies. But if you take a look at &lt;a href=&quot;https://www.youtube.com/watch?v=yox5Isqwk4Q&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video&lt;/a&gt;, you'll see that the predicted path of Venus is more or less through the middle of the Moon. If the actual path should not cross the disk of the Moon, the calculation would be off by &lt;span class=&quot;math-container&quot;&gt;$\sim15$&lt;/span&gt; arcmin, the probability of which is virtually zero.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-11-26T11:59:20.557" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12654" PostTypeId="1" AcceptedAnswerId="18863" CreationDate="2015-11-26T11:14:56.210" Score="3" ViewCount="440" Body="&lt;p&gt;This is similar to a previously asked question, but I am just asking about theory rather than observational evidence. Assuming there were a much larger number of protoplanets in the early Solar System, what percentage of these would theoretically have been ejected (orbital velocity increased beyond escape velocity due to interaction) and what percentage would have been just moved out to a more distant orbit (say, in the Kuiper Belt or the Oort Cloud)?&#xA;In other words, should we expect to find rocky worlds in the far outer Solar System that were formed inside the current radius of the asteroid belt?&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="2153" LastEditDate="2016-10-26T15:06:09.527" LastActivityDate="2016-10-26T15:20:08.327" Title="Ejected planets during the early stages of the formation of the Solar System" Tags="&lt;solar-system&gt;&lt;planetary-formation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12655" PostTypeId="1" CreationDate="2015-11-26T11:22:22.350" Score="-1" ViewCount="167" Body="&lt;p&gt;My theory of why Brown Dwarfs are the rogue celestial bodies that travel through galaxies without a set course. My first theory would be their matte black in color due to a layer of soot.Light does not reflect off a thick layer of black powdered soot,The light is absorbed fully making it impossible to view in a color light spectrum.Second these giants are iron all the way through making its mass weight extremely dense in proportion to its volume.Third They are extremely   cold having no energy to burn,some might be completely dead giving off no heat and would be almost impossible to capture but by chance with FIR technology heating a spot up and captured with IR.These dead cold Dwarfs might even be completely stationary in a solar system having no momentum from photon light scattering.Some might have collided with a plant causing velocity momentum and set to a linear path.Some might be smoldering out still emitting some FIR radiation making a small gap to be captured by IR technology.They can travel through solar systems or galaxies as pleased with little interference of gravity from planets due to its extreme mass weight,until it transcends extremely close to an active star and slingshotted around the star,Now hot and glowing the Brown Dwarf can be seen through color spectrum and will radiate FIR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could this theory be plausible? If not describe with links to creditable data no pictures or diagrams.&lt;/p&gt;&#xA;" OwnerUserId="9326" LastEditorUserId="9326" LastEditDate="2015-11-27T07:12:28.927" LastActivityDate="2015-11-27T07:12:28.927" Title="Plausible? Brown Dwarfs are rogue celestial bodies because they absorb all light due to their thick layer of matte black soot" Tags="&lt;light&gt;&lt;brown-dwarf&gt;&lt;photons&gt;" AnswerCount="2" CommentCount="10" ClosedDate="2015-11-27T19:33:06.950" ContentLicense="CC BY-SA 3.0" />
  <row Id="12656" PostTypeId="1" AcceptedAnswerId="12667" CreationDate="2015-11-26T11:54:29.753" Score="2" ViewCount="86" Body="&lt;p&gt;Large telescopes don't use a single CCD; they use &lt;a href=&quot;http://www.dailytech.com/Update+Kepler+Mission+Sets+Out+to+Find+Planets+Using+CCD+Cameras/article14421.htm&quot; rel=&quot;nofollow&quot;&gt;arrays of them&lt;/a&gt;. Since CCDs can't be seamlessly joined, this means that every captured image must have &lt;a href=&quot;https://www.eso.org/sci/facilities/paranal/instruments/omegacam/inst.html&quot; rel=&quot;nofollow&quot;&gt;gaps&lt;/a&gt;. But, the resulting images almost never have gaps when they are presented to the public.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are these gaps filled? Does the telescope capture a number of overlapping images and then merge the results? Wouldn't this result in differing exposure levels for differing areas of the image, which, even though it could be compensated for in the processing, would give different noise characteristics to the different areas?&lt;/p&gt;&#xA;" OwnerUserId="9889" LastEditorUserId="9301" LastEditDate="2015-11-26T18:57:17.847" LastActivityDate="2015-11-27T01:07:34.970" Title="How are the gaps filled in images captured by CCD arrays?" Tags="&lt;telescope&gt;&lt;photography&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12658" PostTypeId="1" AcceptedAnswerId="34636" CreationDate="2015-11-26T12:17:19.297" Score="11" ViewCount="713" Body="&lt;p&gt;Serious modern telescopes use CCDs to capture the results. This makes it convenient to process or display the results &lt;a href=&quot;http://www.brighthub.com/science/space/articles/29608.aspx&quot; rel=&quot;noreferrer&quot;&gt;halfway around the world&lt;/a&gt;. However, there is something visceral about looking through an eyepiece and seeing the results live.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My telescopic experience has been with smaller (&amp;lt;=8&quot;) devices in not-great-seeing conditions, and I've generally found the results dim and fuzzy. I've imagined what it would be like to look through an eyepiece on a serious telescope, showing a nebula bright and clear. Imagine stereo eyepieces on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Large_Binocular_Telescope&quot; rel=&quot;noreferrer&quot;&gt;Large Binocular Telescope&lt;/a&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence my question: what is the largest aperture telescope that was ever fitted with an eyepiece so that someone could look through it in real-time? This should include situations where a normally-digital telescope was temporarily fitted with an eyepiece.&lt;/p&gt;&#xA;" OwnerUserId="9889" LastActivityDate="2020-05-15T16:42:31.700" Title="What was the largest telescope ever fitted with an eyepiece" Tags="&lt;telescope&gt;&lt;telescope-lens&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12659" PostTypeId="2" ParentId="12655" CreationDate="2015-11-26T14:24:11.877" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;They are so dark they absorb almost every bit of light from any star and receive no photon velocity momentum or electromagnetic radiation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The momentum that a planet receives or does not receive from photons is negligible for its orbit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;These dark masses are extremely high in mass and do not interact with light like other planets and moons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A dark object very much interacts with light: it absorbs light.  An object that does not interact with light would be transparent, not dark.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I believe different levels of light interaction on a mass causes planets to move away from stars or stay at bay.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Planetary orbits are governed by gravitational effects.  The effect of light interaction on orbits is negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, none of this has anything to do with planets in orbit or not.  Planetary orbits are due to gravitation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Only another collision from a celestial body can interfer with the direction they are heading they outweigh most of the competition so distraction is rare.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Indeed, the only force massive enough to alter the orbit of a planet would be another planet, either hitting it full force, or passing closely thus altering its angular momentum.  You'd need an awful lot of this to cause a planet to reach a solar system escape trajectory, however!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it has nothing to do with colour or photon interaction.  It's all about mass.&lt;/p&gt;&#xA;" OwnerUserId="1462" LastEditorUserId="1462" LastEditDate="2015-11-26T15:22:12.077" LastActivityDate="2015-11-26T15:22:12.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12660" PostTypeId="2" ParentId="12658" CreationDate="2015-11-26T17:00:24.833" Score="8" Body="&lt;p&gt;&lt;a href=&quot;https://www.quora.com/What-is-the-largest-telescope-you-can-use-with-the-naked-eye&quot; rel=&quot;noreferrer&quot;&gt;From this answer&lt;/a&gt;, it seems that the &lt;a href=&quot;https://en.wikipedia.org/wiki/Harlan_J._Smith_Telescope&quot; rel=&quot;noreferrer&quot;&gt;Harlan J Smith telescope&lt;/a&gt; located at the McDonald Observatory, in Texas, with a 107-inch (2.7 m) aperture is the largest aperture telescope that was ever fitted with an eyepiece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some &lt;a href=&quot;https://books.google.co.in/books?id=XM0DAAAAMBAJ&amp;amp;pg=PA58&amp;amp;lpg=PA58&amp;amp;dq=harlan+smith+telescope+eyepiece&amp;amp;source=bl&amp;amp;ots=olMalB6quF&amp;amp;sig=9B7yskuudQh-NtxqrrY9d35IzA0&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwiikNXypa7JAhVQW44KHfkqBYEQ6AEIOjAF#v=onepage&amp;amp;q=harlan&quot; rel=&quot;noreferrer&quot;&gt;independent evidence&lt;/a&gt; is there to indicate that this indeed had an eyepiece.&lt;/p&gt;&#xA;" OwnerUserId="7970" LastActivityDate="2015-11-26T17:00:24.833" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12661" PostTypeId="1" CreationDate="2015-11-26T18:42:03.213" Score="4" ViewCount="249" Body="&lt;p&gt;On Nov. 25 at 2:30a.m. from southern California while viewing the moon with a 70mm D x 400mm F telescope a point of light was seen at about the 7 o'clock position below the moon. As viewed the point of light was clear and seemed to be tracking with the moon. As I tracked the moon the point of light stayed within the same distance and in view. If it was a planet or star wouldn't they track slightly different than the moon ?&lt;/p&gt;&#xA;" OwnerUserId="9890" LastActivityDate="2019-01-05T14:47:18.150" Title="Would a satellite in geosynchronous orbit between the earth and moon track across the sky together?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12662" PostTypeId="1" AcceptedAnswerId="12669" CreationDate="2015-11-26T19:16:52.633" Score="3" ViewCount="244" Body="&lt;p&gt;Imagine we have a black hole and a star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are very close to each other so black hole slowly consumes the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to have planets in this system, and for them not to be consumed by this black hole but just follow their unique orbits. Could there even have life there?&lt;/p&gt;&#xA;" OwnerUserId="9891" LastEditorUserId="2365" LastEditDate="2015-11-28T15:47:38.987" LastActivityDate="2015-11-28T15:47:38.987" Title="A star a black hole and planets around them" Tags="&lt;black-hole&gt;&lt;exoplanet&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12663" PostTypeId="2" ParentId="12469" CreationDate="2015-11-26T19:24:40.990" Score="1" Body="&lt;p&gt;The colour of a natural satellite is almost entirely based off the atmosphere and surface of the satellite. The moon Io, of Jupiter is not grey and is coloured because of sulfur on its surface that create a colour. Io has a very thin atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that brings us to your question, why is the &lt;strong&gt;majority&lt;/strong&gt; of natural satellites grey? We know that moon's colours are dependent on what they are made up of and if they have an atmosphere, so the basic answer is that the majority of moons are grey because they do not hold an atmosphere, and are made up of rock. Our moon is made out of basalts which are a grey rock. You can probably go out and research what certain moons are made of, and find out more about the rocks on them and how they affect the colour of the planet.&lt;/p&gt;&#xA;" OwnerUserId="9645" LastEditorUserId="9645" LastEditDate="2015-11-26T19:44:03.463" LastActivityDate="2015-11-26T19:44:03.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12664" PostTypeId="2" ParentId="12662" CreationDate="2015-11-26T19:36:46.773" Score="1" Body="&lt;p&gt;Planet's that are not part of a solar system are called rogue planets. These planets are planets which are starless, and support themselves with gravity in space. It would most likely be very hard for life to form on these types of planets because there is no star to provide heat for the life. I'm not an expert on astrobiology though, so I can't answer much about life on these planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a black hole consumes a star an accretion disk is formed around it. An accretion disk is a large disk of the stellar material that forms around the black hole as it draws mass from the star. If the black hole is big enough, it could swallow the star whole. If a black hole were to come into a range where it was drawing mass from the star, then all of the planets would most likely be destroyed due to the immense gravitational pull which they possess. If a planet would be orbiting the star, it would be extremely unlikely that it would not be destroyed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is almost certain that tidal forces would of destroyed the planet if it was orbiting the star, and unless the planet is large enough to support itself without a star it cannot become a rogue planet. I have no idea if a rogue planet can support life.&lt;/p&gt;&#xA;" OwnerUserId="9645" LastActivityDate="2015-11-26T19:36:46.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12665" PostTypeId="2" ParentId="12655" CreationDate="2015-11-26T20:00:10.017" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My first theory would be there dark color or no color because light does not reflect back off these giants but is absorbed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's nothing more than a long way of saying that they're dark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A brown dwarf is a body that's more massive than a planet but less massive than a star. It's dark simply because it's not massive enough to sustain the same kind of nuclear fusion as our Sun, the process by which stars generate light and heat. (Some may be able to fuse deuterium or lithium.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That just tells us that they don't glow; it doesn't say anything about how they would look in reflected light. But we don't see brown dwarfs in reflected light, because there don't happen to be any close enough to the Sun or to another nearby star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trajectory of any body that massive is not going to be significantly affected by light pressure or by magnetic fields, whether it's hot enough to generate its own light or not. The same is true for &lt;em&gt;much&lt;/em&gt; smaller bodies, down to the size of moons or asteroids. (Some smaller bodies in our Solar System have their rotation influenced by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Yarkovsky_effect&quot; rel=&quot;nofollow&quot;&gt;Yarkovsky effect&lt;/a&gt;, but it's not significant for anything the size of a planet or star.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If&lt;/em&gt; it were close enough to a star, a brown dwarf might look similar to Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://en.wikipedia.org/wiki/Brown_dwarf&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Brown_dwarf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="53" LastEditorUserId="53" LastEditDate="2015-11-27T03:05:21.163" LastActivityDate="2015-11-27T03:05:21.163" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="12666" PostTypeId="1" CreationDate="2015-11-26T23:36:27.173" Score="1" ViewCount="1450" Body="&lt;p&gt;I need to determine Right Ascension and Declination from Azimuth and Altitude, working in C#. The problem is that the formula for calculating hour angle, for some reason, doesn't work. Here's the code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        az = az * DEG_TO_RAD;&#xA;        alt = alt * DEG_TO_RAD;&#xA;&#xA;        lati = latitude * DEG_TO_RAD;&#xA;&#xA;        // Julian day&#xA;        JD = CalculateJDN(year, month, day, h, m, s);&#xA;&#xA;        // Greenwich mean sidereal time&#xA;        GMST = CalculateGMST(JD);&#xA;&#xA;        LST = GMST + longitude / 15;&#xA;&#xA;        dec = Math.Asin((Math.Sin(lati) * Math.Sin(alt)) + (Math.Cos(lati) * Math.Cos(alt) * Math.Cos(az)));&#xA;&#xA;        ha = Math.Atan2(Math.Sin(az), (Math.Cos(az) * Math.Sin(lati) + Math.Tan(alt) * Math.Cos(lati))); &#xA;&#xA;        ha = ha * RAD_TO_DEGREE / 15;&#xA;        dec = dec * RAD_TO_DEGREE;&#xA;&#xA;        ra = LST - ha;&#xA;&#xA;        // Input data for Mintaka (delta Ori):&#xA;        // az = 47.5, alt = -35.3 on 13:57 UTC, 1 Dec 2015&#xA;        // latitude = 43.897, longitude = 20.344&#xA;        // Required output: dec = -0.19, ra = 5.5&#xA;        // Given output: &#xA;        // dec = -0.19, ra = 17.5, ha = 2.5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Az and alt are given in degrees, so they are first converted into radians. Functions for calculating Julian day number and GMST are correct, since I've already tested them. Formula for declination is good, but for some reasons formula for hour angle (ha) doesn't work. I don't know where's the error.&lt;/p&gt;&#xA;" OwnerUserId="9197" LastEditorUserId="9197" LastEditDate="2015-12-01T14:06:47.520" LastActivityDate="2015-12-01T14:06:47.520" Title="Calculation of hour angle" Tags="&lt;fundamental-astronomy&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12667" PostTypeId="2" ParentId="12656" CreationDate="2015-11-27T01:07:34.970" Score="4" Body="&lt;p&gt;In general, the common practice is to &lt;a href=&quot;http://www.noao.edu/kpno/manuals/dim/&quot; rel=&quot;nofollow noreferrer&quot;&gt;dither exposures and fill in the gaps&lt;/a&gt;. The dither serves multiple purposes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;provides some exposure in the gaps between the CCDs&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;smooths out pixel-to-pixel variations in the response.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;prevents one bad pixel from ruining an entire observation&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The dither pattern varies from telescope to telescope. For example, the following image shows the dither pattern of Chandra telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NZEi7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NZEi7.png&quot; alt=&quot;Dither patter&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image from &lt;a href=&quot;http://cxc.harvard.edu/ciao/why/dither.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;havard.edu&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect of dither is removed during high-level ground processing of the data.&lt;/p&gt;&#xA;" OwnerUserId="7970" LastActivityDate="2015-11-27T01:07:34.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12668" PostTypeId="1" AcceptedAnswerId="12670" CreationDate="2015-11-27T03:30:14.760" Score="1" ViewCount="131" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_common_misconceptions#cite_note-184&quot; rel=&quot;nofollow&quot;&gt;List of Common Misconceptions &lt;/a&gt;by Wikipedia, under the heading of astronomy, this line can be found: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A black hole &lt;strong&gt;can&lt;/strong&gt; act like a &quot;cosmic vacuum cleaner&quot; and pull a&#xA;  substantial inflow of matter, but only if the star it forms from is&#xA;  already having a similar effect on surrounding matter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is referenced to a paper written in Yale university:  &quot;Frontiers And Controversies In Astrophysics Transcript 9&quot;, although that's a dead link. By linking so, I believe that it is a reliable source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This got me thinking that is there a black hole that doesn't suck in surrounding matter? And is there a star that don't take in surrounding matter?&lt;/p&gt;&#xA;" OwnerUserId="7926" LastActivityDate="2015-11-27T04:51:40.157" Title="A black hole that doesn't take in matter?" Tags="&lt;star&gt;&lt;black-hole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12669" PostTypeId="2" ParentId="12662" CreationDate="2015-11-27T04:24:23.570" Score="5" Body="&lt;p&gt;It's possible to have planets orbiting a binary pair of stars, your scenario of a close orbit, sometimes called &quot;short orbit binaries&quot;.    See &lt;a href=&quot;https://en.wikipedia.org/wiki/Circumbinary_planet&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, also posted above in comments.   In such a binary-system, nothing can orbit an individual star, but at some distance, plants can orbit and some systems like this have even been observed, listed in the link.  The orbital dynamics is the same for a star-black hole short-orbit binary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there are problems.   Black holes form out of very large stars and the formation is one of the biggest explosions in the universe, a &lt;a href=&quot;https://en.wikipedia.org/wiki/Type_II_supernova&quot; rel=&quot;nofollow&quot;&gt;Type II supernovas&lt;/a&gt;, and that's not very friendly to any planets in orbit.  A star might survive it, planets would be harder, though it might be possible for new planets to form from nebula material remaining after the nova (I'm just guessing there).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A black hole could also form from a Neutron star accreting matter, but you still have the problem that the formation of a Neutron Star also only happens out of a Type-II supernova, so such a system has a difficult beginning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  While some planets have been observed around Neutron stars, these appear to be quite rare.   2 Neutron Stars have been observed &lt;a href=&quot;https://en.wikipedia.org/wiki/Pulsar_planet#Confirmed_planets&quot; rel=&quot;nofollow&quot;&gt;with planets&lt;/a&gt;, out of over &lt;a href=&quot;http://www.universetoday.com/25376/pulsars/&quot; rel=&quot;nofollow&quot;&gt;1,600 Neutron stars&lt;/a&gt; observed.    A type II nova is very planet unfriendly.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically a close gravitational capture is possible, but those are very rare, as stars rarely get that close.   There's many stars that are known to orbit the super-massive black holes at the center of our galaxy (Andromeda galaxy too), but stars and stellar mass black holes are much more rare.   A few have been observed, but they don't appear to be common.   Such a system would be easy to observe, so the fact that there are only a few that have been noticed is evidence to them being rare.  Here's a few mentions of them.  &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/astro/blkbin.html&quot; rel=&quot;nofollow&quot;&gt;One&lt;/a&gt;, &lt;a href=&quot;https://www-xray.ast.cam.ac.uk/xray_introduction/Blackholebinary.html&quot; rel=&quot;nofollow&quot;&gt;Two&lt;/a&gt;, &lt;a href=&quot;http://www.astronomy.com/news/2014/01/spanish-researchers-discover-the-first-black-hole-orbiting-a-spinning-star&quot; rel=&quot;nofollow&quot;&gt;Three&lt;/a&gt;, &lt;a href=&quot;http://www.sciencedaily.com/releases/2011/03/110325082725.htm&quot; rel=&quot;nofollow&quot;&gt;Four&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the 4th article, which is from 2011, so more may be known now, but it says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Only about 20 binary stellar systems are known to contain a black&#xA;  hole, out of an estimated population of around 5,000 in the Milky Way&#xA;  Galaxy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A 2nd problem is that a star feeding a black hole would create an &lt;a href=&quot;https://en.wikipedia.org/wiki/Accretion_disk&quot; rel=&quot;nofollow&quot;&gt;accretion disk&lt;/a&gt; which would be very radioactive and not ideal for life on an orbiting planet.   Maybe the planet could have a very thick atmosphere that might protect it, but that would also likely reduce sunlight reaching the surface.   The star feeding the black hole would also be losing mass, and over time, grow smaller and provide less light and heat to the planet.   Ideally, you'd want it to be a very slow feed.   It's pretty far from an optimal life on planet situation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So it is possible to have planets in this system not to be consumed by&#xA;  this black hole but just follow their unique orbits&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This part is certainly possible.   Things can orbit a black hole at a safe distance without any problem.  As for life, we don't know how common life is in other solar-systems so nobody can say how likely it might be, but it's theoretically possible, but, in my opinion, pretty far from ideal.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2015-11-28T11:13:52.797" LastActivityDate="2015-11-28T11:13:52.797" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12670" PostTypeId="2" ParentId="12668" CreationDate="2015-11-27T04:51:40.157" Score="2" Body="&lt;p&gt;This is the full &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_common_misconceptions#Astronomy&quot; rel=&quot;nofollow noreferrer&quot;&gt;quote&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If, for example, the Sun were replaced by a black hole of equal mass,&#xA;  the orbits of the planets would be essentially unaffected. A black&#xA;  hole can act like a &quot;cosmic vacuum cleaner&quot; and pull a substantial&#xA;  inflow of matter, but only if the star it forms from is already having&#xA;  a similar effect on surrounding matter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's badly written, but what it's basically saying is that it doesn't matter the form of the massive object that's being orbited by a planet or other object.  A stable orbit would remain stable, so, if our sun was to collapse into a black hole and nothing else changes, Earth's orbit would remain the same.  It would get dark and cold, but the orbit would be unchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The misconception is, if the sun became a black hole than Earth would be sucked into the sun.   That's 100% false and that's all they're saying.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity is a function of mass and distance.   Black holes have very high gravity, but a big part of the reason for that is because when they form out of dying stars, they become very small, a few solar masses squeezed into only about 10 miles across so the distance to the center of mass gets very small.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a planet to be in any danger from a black hole, it would probably need to be inside the Roche Limit, perhaps only a few million miles away, which is several times closer than Mercury is to the sun, for example.   The safe distance, of-course, varies with the density and solidity of the orbiting object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a Q on &lt;a href=&quot;https://astronomy.stackexchange.com/questions/8503/do-roche-limits-apply-to-black-holes&quot;&gt;Roche Limits and Black holes&lt;/a&gt; if interested.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-11-27T04:51:40.157" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12671" PostTypeId="2" ParentId="2441" CreationDate="2015-11-27T07:23:04.533" Score="11" Body="&lt;p&gt;We need to think about just &lt;em&gt;where&lt;/em&gt; the time dilation effect occurs. By then thinking about the observations from each point of view, that is the free falling object and the external observer, we can come to terms with just what is &lt;em&gt;happening&lt;/em&gt; as opposed to what &lt;em&gt;appears&lt;/em&gt; to be happening.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The experience of time&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;We must remember that an object moving at a certain speed will travel through time (or the 4th dimension) at a slower rate. This does not mean that it moves slower, otherwise it would obviously not be travelling &quot;at a certain speed&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where time slows is in the ticking of the physical processes of the object itself. In other words, my clock would tick twice as slow according to you as I flew past you at 87% the speed of light. I would be waving my arms normally, but according to you, I would appear to be waving my arms twice as slow and would also appear to be squeezed in size (not really relevant to this).&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The falling object's point of view&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;If you were the object falling into the black hole, you would accelerate as you approached the event horizon, but you would take longer and longer to &lt;em&gt;react&lt;/em&gt; to the approach, to the point where you would fall into the black hole in no time at all. From your perspective, your approach to the event horizon would become exponentially faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, you would fall incredibly fast into the black hole, but you would have barely registered it in your mind because there just wasn't enough &lt;em&gt;time&lt;/em&gt; for you due to relativity.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The stationary observer's point of view&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Now, the stationary observer outside the black hole's influence would &lt;em&gt;observe&lt;/em&gt; something very different. The light (or rather, information) about your descent would become more and more redshifted, but &lt;em&gt;also&lt;/em&gt; take longer and longer to actually reach their eyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that according to &lt;em&gt;the observer&lt;/em&gt;, the falling object would slow down to a halt at the event horizon and have disappeared.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;So what really &quot;happened&quot;?&lt;/h1&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The falling object fell in very quickly, but hardly realised it happening&lt;/li&gt;&#xA;&lt;li&gt;The stationary observer would think that the object disappeared and never reached the event horizon.&lt;/li&gt;&#xA;&lt;li&gt;Cooper taps on some gravity books and saves the human race.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9895" LastEditorUserId="9895" LastEditDate="2016-12-28T23:13:01.503" LastActivityDate="2016-12-28T23:13:01.503" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12672" PostTypeId="2" ParentId="12601" CreationDate="2015-11-27T07:56:19.163" Score="2" Body="&lt;p&gt;The smart antennas are already in work and software controlled beam forming is also pretty much in use now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, at even this very high spinning speed of the earth, the tracking of the stars at large distances is not that difficult. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also high speed data acquisition and compression algorithms are there to help. So it's been with help of control engineering been possible to point at a specific celestial object.&lt;/p&gt;&#xA;" OwnerUserId="9896" LastActivityDate="2015-11-27T07:56:19.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12674" PostTypeId="2" ParentId="12666" CreationDate="2015-11-27T10:14:04.567" Score="2" Body="&lt;p&gt;I would suggest that this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;LST = GMST + longitude;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;line is wrong. 360 degrees of longitude correspond to 24 hours, so it should be something like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;LST = GMST + longitude/15.0 ;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming these are all in floating point and you are subsequently handling the potential overflow of time into the next day correctly...&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-11-27T10:14:04.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12675" PostTypeId="1" CreationDate="2015-11-27T14:13:42.383" Score="4" ViewCount="91" Body="&lt;p&gt;I just had a conceptual question, hopefully it makes for some really interesting answers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering, as most galaxies have an SMBH at their centre (see Kormandy &amp;amp; Richstone 1995; Richstone et al. 1998), could dark matter accrete onto the SMBH over time (perhaps perturbed by a galaxy collision, interaction or merger) and then recycle this matter as energy? Can dark matter help an SMBH grow?   &lt;/p&gt;&#xA;" OwnerUserId="2091" LastEditorUserId="2365" LastEditDate="2015-11-27T23:35:10.847" LastActivityDate="2015-11-27T23:35:10.847" Title="Can an SMBH recycle dark matter into energy?" Tags="&lt;dark-matter&gt;&lt;supermassive-black-hole&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12676" PostTypeId="1" AcceptedAnswerId="12679" CreationDate="2015-11-27T18:28:25.153" Score="9" ViewCount="256" Body="&lt;p&gt;I'm trying to understand the examples of gravity lensing (using the general relativity property of large masses to bend light like a lens). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the examples I see are of some galaxy (presumably a large mass) between us and a star (or are at least vague enough to not specify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet my intuitive understanding of the sky is that all the stars we see are relatively close, all entirely within the Milky Way galaxy (and many of the stars in our galaxy provide a general glow), and that other galaxies are far enough away that it is difficult to image individual stars. The only single objects large enough comparable to a galaxy would be a quasar. Is that right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what then is going on usually with examples of gravity lensing? I find it hard to believe the captions that say a galaxy is enabling seeing more distant stars. I would think one could only apply that concept to a star or galaxy to see something much further away and as big or much bigger. Could one really use a galaxy as a lens to see a star? I wouldn't expect a star to be &lt;em&gt;behind&lt;/em&gt; a galaxy from us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, whatever the objects themselves, what is the scale? If the lensing is of a galaxy done by a star, I'd expect the star to galaxy distance ratio to be well below 1:1000 (~width of Milky Way to distance to Andromeda). But for galaxy to galaxy or galaxy cluster to galaxy or quasar lensing what are the likely relative distances?&lt;/p&gt;&#xA;" OwnerUserId="673" LastEditorUserId="673" LastEditDate="2015-11-29T16:03:43.660" LastActivityDate="2017-01-02T20:57:55.480" Title="What is the scale of things you can see with gravitational lensing?" Tags="&lt;star&gt;&lt;galaxy&gt;&lt;distances&gt;&lt;gravitational-lensing&gt;&lt;quasars&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12677" PostTypeId="1" AcceptedAnswerId="12691" CreationDate="2015-11-27T19:16:36.407" Score="5" ViewCount="1930" Body="&lt;p&gt;I have implemented the formulas to convert AR and Dec into altitude and Azimuth in C++ following the book &lt;a href=&quot;http://www.amazon.es/Practical-Astronomy-Calculator-Spreadsheet-Spiral/dp/0521146542/ref=sr_1_2?ie=UTF8&amp;amp;qid=1448651526&amp;amp;sr=8-2&amp;amp;keywords=practical%20astronomy%20with%20your%20calculator&quot; rel=&quot;nofollow noreferrer&quot;&gt;Practical Astronomy with your Calculator or Spreadsheet 4th Edition&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the book there is an example to check if I have do it fine. This example works perfectly but I want to check if my formulas are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have use this &lt;a href=&quot;http://www.stargazing.net/mas/al_az.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;page&lt;/a&gt; to check it but I get different minutes values in both altitude and azimuth values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have checked on that page that I get the same LST value, but ALT and AZ have different values. This data I have tested:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Altitude: -15° 27' 40&quot;&#xA;Azimuth:  273° 12' 34&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is the data get it from that page:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/TD44D.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TD44D.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know if there is another web page or program to check if the result of my formulas are correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt;&lt;br/&gt;&#xA;I have tried to get Azimuth and Altitude using a date in year 2000 in web page and my program and this is the result (you can find my program's result at the bottom of the image):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ccf6s.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ccf6s.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the same problem like the previous one (I'm not taking account for precession)?&lt;/p&gt;&#xA;" OwnerUserId="7566" LastEditorUserId="7566" LastEditDate="2015-11-29T07:54:15.687" LastActivityDate="2015-11-29T07:54:15.687" Title="Software to convert RA and DEC into ALT and AZ" Tags="&lt;coordinate&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12678" PostTypeId="2" ParentId="12675" CreationDate="2015-11-27T19:21:33.260" Score="2" Body="&lt;p&gt;The accretion of dark matter by a black hole is possible, but difficult. Unless its trajectory takes it into the black hole, then it will simply approach the black hole, speed up and then escape again, since non-interacting dark matter cannot dissipate energy in any way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presuming a black hole did accrete the dark matter then the energy possessed by the dark matter would become part of the rest mass energy of the black hole. So yes, the black hole would grow. Not sure what you mean by recycle the energy. Energy is always recycled (conserved). If you mean can the black hole lose its rest mass energy? I suppose it may do via Hawking radiation, but this is vanishingly small for a SMBH. The dark matter will also carry angular momentum into the BH and there are ways that this rotational energy can be extracted (e.g. the Penrose process &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Penrose_process&quot; rel=&quot;nofollow&quot;&gt;https://en.m.wikipedia.org/wiki/Penrose_process&lt;/a&gt; ).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-11-27T19:21:33.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12679" PostTypeId="2" ParentId="12676" CreationDate="2015-11-27T21:11:23.670" Score="6" Body="&lt;p&gt;You are right that the stars seen on the sky are within the Milky Way. Only with a large telescope is it possible to resolve individual stars in other galaxies, and only for the nearest ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know which sources you refer to, by I think perhaps you are confusing the different types of gravitational lensing. I cannot explain them better than &lt;a href=&quot;https://astronomy.stackexchange.com/a/242/5264&quot;&gt;the excellent review by astromax&lt;/a&gt;, but briefly, there are three types:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Strong lensing&lt;/strong&gt;, where are foreground galaxy cluster (i.e. a group of $\sim$100–1000 galaxies) magnify and severely distort background galaxies,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Weak lensing&lt;/strong&gt;, where clusters or individual galaxies distort the shapes of many background galaxies on the percent scale, which can only be seen statistically, and&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Microlensing&lt;/strong&gt;, where a single object inside the Milky Way happen to pass in front of another single object, also inside the Milky Way. These objects are usually stars or planets, and do not distort the images of the background objects, but merely increase the flux for a while. This effect has been used to find exoplanets.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Whereas type 1 and 2 are more or less static in a human lifetime, type 3 is an event that happens once only for a given set of stars (as seen from Earth).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Scales&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Strong and weak lensing happen on very large scales, from a few hundred million lightyears, up to the order of the size of the observable Universe (e.g. &lt;a href=&quot;http://arxiv.org/abs/1405.3661&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wong et al. 2014&lt;/a&gt;). While the lenses themselves are cluster of galaxies, and thus a few to $\sim10$ &lt;a href=&quot;https://en.wikipedia.org/wiki/Parsec#Megaparsecs_and_gigaparsecs&quot; rel=&quot;nofollow noreferrer&quot;&gt;megaparsec&lt;/a&gt; across, the lenses object are typically individual galaxies. Gravitational lensing is most efficient when the lens is halfway between us and the background source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Occurring in the Milky Way, microlensing, on the other hand, happen on the scale of a few kiloparsec, again with us-lens distance being of the same order as the lens-background object distance (see e.g. &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_exoplanets_detected_by_microlensing&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Wikipedia list&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2017-01-02T20:57:55.480" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12680" PostTypeId="2" ParentId="12677" CreationDate="2015-11-27T23:58:13.867" Score="1" Body="&lt;p&gt;Minutes of a degree are &quot;minúte&quot;. There are a few things that could cause errors. One is atmospheric refraction: it can cause an error of about 1 minute at the an alt of 45degrees, and over 30minutes on the horizon. When observing objects close to the horizon, this is a sigificant source of error&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its pretty much impossible to check without access to the source code. However There is plenty of astronomical software that can calulate Az-Alt positions. Pyephem, xephem, stellarium come to mind, and you can ask them to calculate with or without atmospheric distortion. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-11-27T23:58:13.867" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12681" PostTypeId="2" ParentId="12658" CreationDate="2015-11-28T00:29:38.523" Score="8" Body="&lt;p&gt;Realize that a large telescope would have a large prime-focus image, that would in turn necessitate  a very extraordinary eyepiece to fully utilize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This said, if you look at this image from the 200 inch  Palomar telescope (apparently published in Life magazine):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eMrHZ.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eMrHZ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see an astronomer inserting a plate in the camera at the 200 inch prime focus.  Close to his right thumb you see a vertical tube.  This is an eyepiece that is used to examine a star at the edge of the prime focus image outside of the plate dimensions.  The astronomer will watch this star through the eyepiece and manually adjust the tracking of the telescope to keep the star stationary. For hours.  In the dark.  With no source of heat. And no plumbing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have an eyepiece being used in parallel with a photographic plate.  It may be a stretch, but...&lt;/p&gt;&#xA;" OwnerUserId="420" LastActivityDate="2015-11-28T00:29:38.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12682" PostTypeId="1" CreationDate="2015-11-28T03:11:15.713" Score="4" ViewCount="328" Body="&lt;p&gt;A couple questions, please: I know that the Lagrangian points 1, 2 and 3 are unstable and special Lissajous orbits plus some station-keeping are required to place a spacecraft around them. But I was wondering if they are so totally unstable that they can't temporarily &quot;capture&quot; a passing natural object (let's say an asteroid or a cloud of cosmic dust, in a stationary way or entering some kind of not-very-stable orbit around them)? Or could they?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if they were able to, how much time would (approximately) be required for this/these object(s) to go &quot;off-orbit&quot; and abandon the L(1,2,3) area? In the case of the cloud of cosmic dust, would it just slowly (or quickly) drift away, or would they be &quot;launched&quot; towards another direction or orbit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance! &lt;/p&gt;&#xA;" OwnerUserId="9902" LastActivityDate="2015-11-28T16:37:24.557" Title="How (un)stable are the Lagrangian points 1, 2 and 3?" Tags="&lt;orbit&gt;&lt;lagrange-point&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12683" PostTypeId="1" AcceptedAnswerId="12699" CreationDate="2015-11-28T04:02:29.663" Score="4" ViewCount="183" Body="&lt;p&gt;This is a hypothetical question, but it really is meant as a vivid way of asking about the affects of high-speed galactic collisions. &lt;a href=&quot;http://apod.nasa.gov/apod/ap151127.html&quot; rel=&quot;nofollow&quot;&gt;The November 27 APOD&lt;/a&gt; showed the Cheshire Cat galaxy group, and the caption says that the &quot;eyes&quot; are two large elliptical galaxies, closing on one another at over 1000 km/s, and bathed in a glow of X-rays from heating of the associated interstellar media.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How intense is this glow? I suspect it is very faint, but I don't know how to quantify it. Would it hinder X-ray astronomy for residents of the galaxies? Would it be strong enough to affect life? Does anybody have quantitative information and know how to interpret it? (I do realize that elliptical galaxies are thought to be made up of old, low-metallicity stars that aren't good candidates for habitable planets, but let's ignore that.)&lt;/p&gt;&#xA;" OwnerUserId="7752" LastActivityDate="2015-11-29T13:47:06.520" Title="Are high-speed galactic collisions survivable?" Tags="&lt;galaxy-cluster&gt;&lt;x-ray&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12686" PostTypeId="2" ParentId="11580" CreationDate="2015-11-28T13:57:35.347" Score="1" Body="&lt;p&gt;What I do is compute the limiting magnitude for the position of the object I am interested in, and subtract it from the magnitude of the object (I call this the magnitude contrast).  If negative, it should be visible with the naked eye.  I use the function limmag_jd() in the Fortran library &lt;a href=&quot;http://libthesky.sourceforge.net&quot; rel=&quot;nofollow&quot;&gt;libTheSky&lt;/a&gt; for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to find the best moment in a given night, I compute the contrast for say every 15 minutes, and find the minimum.  You can do this for a number of nights, until the best contrast found is negative, which would be the heliacal rising.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure whether there is an easy way using a simple equation - the limiting magnitude will depend on things like the positions of Sun, Moon and the object in question, the phase of the Moon, the magnitude of the object, etc.&lt;/p&gt;&#xA;" OwnerUserId="286" LastActivityDate="2015-11-28T13:57:35.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12687" PostTypeId="2" ParentId="6535" CreationDate="2015-11-28T14:30:42.820" Score="1" Body="&lt;p&gt;You can take a look at &lt;a href=&quot;http://hemel.waarnemen.com/mercurius/mercurius_elongaties_2001-2025.html&quot; rel=&quot;nofollow&quot;&gt;this table&lt;/a&gt; - it's in Dutch, but the dates and numbers in the table shouldn't be too hard to understand.  The first date is that of the greatest elongation (GE), the second (halfway the columns) that of the best visibility (BV).  You should ignore the lines with red dates - they're unfavourable appearances, and the code may find a date far from that of the GE.  The table shows that the difference between the moment of GE and that of the BV is typically a few days, although it can be about a week. (You can click on the first date for a page which includes a sky map of the appearance at constant Sun altitude (-6&amp;deg; I believe)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hemel.waarnemen.com/venus/venus_elongaties_2001-2050.html&quot; rel=&quot;nofollow&quot;&gt;Venus&lt;/a&gt; is a different story, because it is usually visible for months.  As you may know, in the northern hemisphere morning appearences of the inner planets are unfavourable in spring and favourable in autumn, and for evening appearances this is the other way around.  The table shows that Venus had its GE (west/morning) on 8 June 2001, but was best visible on 22 August.  In the mean time, the planet had moved somewhat closer to the Sun, but the effect of moving from spring towards autumn was clearly stronger (click the date link for the weird map that goes with this).  Hence, for Venus the difference can be several months.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, these are just examples for one location (52&amp;deg;N).  I think the process may be too complex to be described by a single, simple equation.&lt;/p&gt;&#xA;" OwnerUserId="286" LastActivityDate="2015-11-28T14:30:42.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12688" PostTypeId="1" CreationDate="2015-11-28T15:54:38.717" Score="2" ViewCount="435" Body="&lt;p&gt;So I'd like to ask whether the semi-major axis of planetary orbits are aligned? Logically, they shouldn't be, but many websites act as if they are. If they're not aligned, what's the angle between the reference xy axis and the semi-major axis of the orbits of planets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="9905" LastActivityDate="2015-11-28T17:15:13.083" Title="Alignment of Semi-Major Axis of Orbits" Tags="&lt;orbit&gt;&lt;planet&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12689" PostTypeId="2" ParentId="12688" CreationDate="2015-11-28T16:12:41.217" Score="3" Body="&lt;p&gt;The major axes of the planets are not aligned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The major axis passes through the perihelion, the solar system barycentre and the aphelion. So by &lt;a href=&quot;http://ssd.jpl.nasa.gov/txt/p_elem_t1.txt&quot; rel=&quot;nofollow&quot;&gt;looking up&lt;/a&gt; the &lt;a href=&quot;https://en.wikipedia.org/wiki/Longitude_of_the_periapsis&quot; rel=&quot;nofollow&quot;&gt;Longitude of perihelion&lt;/a&gt; we can find the direction of the major axis:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;planet  Longitude of perihelion &#xA;M       77&#xA;V       132&#xA;E       103&#xA;M       -24&#xA;J       15&#xA;S       93&#xA;U       171&#xA;N       45&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There is no evidence of alignment.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-11-28T16:12:41.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12690" PostTypeId="2" ParentId="12682" CreationDate="2015-11-28T16:37:24.557" Score="2" Body="&lt;p&gt;All five Lagrange points are unstable &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/z2fw3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z2fw3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;L1, 2, and 3 are &quot;saddle points&quot; in the effective potential formed from the combination of gravity and the centrifugal force of a rotating frame of reference. An object which is in front or behind in the orbit would tend to approach the Lagrange point, and then move away, either towards or away from the sun, following the lines in the picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However note that in the picture, the lines around the Lagrange points are not close together, this means that the motion towards, or away from the point is very slow (in the rotating frame) An object near the Lagrangian point would not be &quot;ejected&quot; at high speed. Any cosmic dust that was near the Lagrangian point would drift slowly away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However although the L4 and L5 points are unstable, there are quite stable orbits that (in the rotating frame) orbit around the L4 and L5 point, allowing for &lt;a href=&quot;https://i.stack.imgur.com/z2fw3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Trojan&lt;/a&gt; satellites.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-11-28T16:37:24.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12691" PostTypeId="2" ParentId="12677" CreationDate="2015-11-28T16:37:27.460" Score="2" Body="&lt;p&gt;Based on stellarium, I think you're not accounting for precession (more after the picture):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bGbVv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bGbVv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, Rigel's J2000 coordinates and Rigel's current coordinates are different by about the same amount as your calculation.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2015-11-28T16:37:27.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12692" PostTypeId="2" ParentId="12688" CreationDate="2015-11-28T17:15:13.083" Score="3" Body="&lt;p&gt;The current longitudes of perihelion show no alignment. Furthermore, they change because all the planets have &lt;a href=&quot;http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node115.html&quot; rel=&quot;nofollow&quot;&gt;orbital precession&lt;/a&gt; - that is, the longitude of perihelion changes by between 0.36 arcseconds/year for Neptune (i.e. the longitude of perhelion goes through the full range of angles in 10,000 years) to 19.5 arcseconds/year for Saturn. So not only is there no alignment, but the relative directions of the semi-major axes change comparatively quickly.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-11-28T17:15:13.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12694" PostTypeId="1" AcceptedAnswerId="12695" CreationDate="2015-11-29T01:00:56.460" Score="6" ViewCount="670" Body="&lt;p&gt;Earth gravity is 9.8 $ms^{-2}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity on ISS(international space station) is 8.2 $ms^{-2}$ approx. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As per this &lt;a href=&quot;https://physics.stackexchange.com/questions/29929/gravity-on-the-international-space-station&quot;&gt;https://physics.stackexchange.com/questions/29929/gravity-on-the-international-space-station&lt;/a&gt; ISS never hit the ground because of it's vertical velocity.(free fall)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we create artificial gravity on ISS by using centrifugal force?&#xA;[Have anyone watched &quot;The martian&quot;? they have used this technique]&lt;/p&gt;&#xA;" OwnerUserId="9908" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-09-17T15:51:15.313" Title="Can we simulate Earth's gravity in space?" Tags="&lt;gravity&gt;&lt;space&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12695" PostTypeId="2" ParentId="12694" CreationDate="2015-11-29T07:25:57.790" Score="6" Body="&lt;p&gt;Simulating gravity in space basically means simulating weight, which requires acceleration. So basically, the question is, how do we create acceleration in space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest method for simulating gravity in space is by spinning the space station. In this case, the reaction force to centripetal force substitutes the force of gravity, which pushes the inner contents to be pushed against the outer edge, giving a sensation of weight. This method is theoretically sound and can be used to simulate gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with doing this in ISS boils down to two things- size and shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best shape for a space station having a simulated gravity is something like a doughnut spinning about its axis, so that the occupants are pushed against the outer wall. The shape of ISS is nowhere near this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one excludes the solar panels and the truss structure, the ISS is around 45m along the axis of the habitable modules, if we spin along the axis of the truss structure, this gives a radial distance of ~22.5m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simulating earth's gravity in ISS, we will require,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega^{2} r = g$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega = \sqrt{\frac{g}{r}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For g = 9.8 $ms^{-2}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega = \sqrt{\frac{9.8}{22.5}} = 0.66 rad/sec = 6.3 rpm$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that at this rotational speed, the head and foot of the astronomer will have different linear velocities. The head of a 1.5m tall astronaut will be moving at 13.86 $ms^{-1}$, while the foot will be moving at 14.85 $ms^{-1}$, never mind they are standing at the bottom of a barrel. This will lead to Coriolis effect, which will be uncomfortable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, while earth's gravity cannot be simulated in ISS due to various constraints, we can settle for reduced gravity (with reduced, ~2 rpm) in an ISS sized spacecraft. &lt;/p&gt;&#xA;" OwnerUserId="7970" LastEditorUserId="9301" LastEditDate="2016-09-17T15:51:15.313" LastActivityDate="2016-09-17T15:51:15.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12696" PostTypeId="1" AcceptedAnswerId="12703" CreationDate="2015-11-29T09:41:33.987" Score="0" ViewCount="158" Body="&lt;p&gt;Even if there isn't enough heat and energy in the star to cause nucleosynthesis, could atoms quantum entangle to create a new atomic nuclei? Or would it be quantum tunneling - if this is possible?&lt;/p&gt;&#xA;" OwnerUserId="9645" LastEditorUserId="9645" LastEditDate="2015-11-29T14:10:23.227" LastActivityDate="2015-11-29T22:20:30.033" Title="Can quantum entanglement cause nucleosynthesis to happen in stars?" Tags="&lt;star&gt;&lt;nucleosynthesis&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12697" PostTypeId="5" CreationDate="2015-11-29T10:39:29.900" Score="0" Body="&lt;p&gt;Questions about nucleosynthesis may encompass at least one of the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Stellar nucleosynthesis&lt;/li&gt;&#xA;&lt;li&gt;Supernova nucleosynthesis&lt;/li&gt;&#xA;&lt;li&gt;Cosmic ray spallation&lt;/li&gt;&#xA;&lt;li&gt;Big bang nucleosynthesis&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9645" LastEditorUserId="2153" LastEditDate="2015-11-29T16:14:12.340" LastActivityDate="2015-11-29T16:14:12.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12698" PostTypeId="4" CreationDate="2015-11-29T10:39:29.900" Score="0" Body="Questions to do with how new atomic nuclei are created inside the universe." OwnerUserId="9645" LastEditorUserId="9645" LastEditDate="2015-11-29T16:14:17.677" LastActivityDate="2015-11-29T16:14:17.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12699" PostTypeId="2" ParentId="12683" CreationDate="2015-11-29T13:47:06.520" Score="2" Body="&lt;p&gt;Interesting question.   I did a bit of reading and gave it some thought and and I can sort of give an answer, though I invite corrections and input from anyone more knowledgeable than me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, those 2 galaxies are enormous.  I didn't see specific sizes listed, but per this &lt;a href=&quot;http://arxiv.org/abs/1505.05501&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1.2-1.5 x 10^14 solar mass fossil group&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's about 100 Andromeda galaxies.  Also, since what we're seeing is 4.6 billion years old, it's reasonable to assume that those 2 galaxies probably have more free dust and gas and, being larger, the gravitational collision is happening at higher speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's really the question with how much background x-ray radiation you get when 2 galaxies collide.   how much free gas and dust and how fast the galaxies move into each other.   My guess is that being larger, and younger, those 2 galaxies are glowing much more brightly than the Milky-way and Andromeda will 4 billion years from now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to how intense the glow is, my hunch agrees with yours, it's probably not very bright if you're there.   The faint glow and the 2 bright eyes you see in the picture are long exposures.   This &lt;a href=&quot;http://chandra.si.edu/photo/2015/cheshirecat/&quot; rel=&quot;nofollow&quot;&gt;site&lt;/a&gt; (scroll to the bottom) says it's 19 hours and 30 minutes of exposure.  Andromeda, for example is barely visible to the naked eye, but when photographed it's made many times brighter.   I'm not smart enough to say for 100% sure, but my guess is that the faint purple glow would be barely visible, if visible at all, assuming it even reached into the visible spectrum, which X-ray's don't.  Chandra might be setting the color to a visible spectrum in the picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How dangerous is it?   Again, I'm guessing, but I think in the Milky-way-Andromeda collision, we'd probably get more UV from the sun than we would from high speed gas and dust collisions, but that's just a guess.   It would be kind of fun if the entire night sky glowed a little bit though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the Milky-way-Andromeda collision, Wikipedia says that there won't be much free gas in the collision, &lt;a href=&quot;https://en.wikipedia.org/wiki/Andromeda%E2%80%93Milky_Way_collision#Possible_triggered_stellar_events&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, so it's a safe bet there won't be any visible glow in the sky and not even that much of an acceleration of new star formation, though there should be some.   The Milky-way currently forms about 7 new stars every year, and that I imagine will gradually slow down over the next 4 billion years, then speed up significantly as the collision gets underway, but from the point of view of Earth, I don't think there's any guarantee we'd get a close look at any new star formations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.npr.org/2012/06/08/154587994/what-happens-when-two-galaxies-collide&quot; rel=&quot;nofollow&quot;&gt;Here's a fun article/interview&lt;/a&gt; on the collision between the Milky Way and Andromeda, though I think Roeland van der Marel might be exaggerating the new star formation a bit if we're to believe Wikipedia where it says there won't be much free gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the speed of our collision, Andromeda is currently heading towards us (or, we're heading towards it, whichever you prefer) at about 110 km/s &lt;a href=&quot;https://en.wikipedia.org/wiki/Andromeda%E2%80%93Milky_Way_collision#Certainty&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;.  and it's currently about 2.5 million light years away.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Andromeda_Galaxy&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Covering 2.5 million light years (about 24 million trillion km) in 4 billion years (126 million billion seconds), works out to an average speed of 190 km/s, so we can roughly estimate that the speed of impact between our two galaxies will reach a peak somewhere around 270 or so km/s.   But (see &lt;a href=&quot;http://hubblesite.org/newscenter/archive/releases/2012/20/video/a/&quot; rel=&quot;nofollow&quot;&gt;video&lt;/a&gt; you also have to take into account the rotation speed and it appears (video above) the rotations will be going in opposite directions when the galaxies collide.    We orbit the Milky way at about 250 km/s and Andromeda, being larger, probably a bit faster orbit speed.   Adding the 2 velocities together, in our neck of the woods, we might see some relative velocity and some gas &amp;amp; dust collisions at (guessing here, cause I don't know if the rotations will slow down as the galaxies approach), but lets say 500 to 600 km/s or so.   For any close bright stars from Andromeda, that could be fast enough for visible changes to a few near-by Andromeda stars in constellations over a single human lifetime.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps, we'll pass through an oort cloud of another star every few thousand years or so, perhaps even the occasional kuiper belt equivalent, every ten or 50 million years - er, maybe.   We could see some very impressive meteor showers and perhaps an occasional, slightly more frequent, dinosaur killing level comet or meteor impact - but I'm just speculating.  There might be lots of interesting things to see between 4 and 5 billion years from now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When stars form they can be (per this &lt;a href=&quot;http://cseligman.com/text/stars/mldiagram.htm&quot; rel=&quot;nofollow&quot;&gt;website&lt;/a&gt;) 100 to 200 times brighter than they are during their main sequence, so if our solar-system finds itself nearish to a colliding gas cloud and some new star formation, that could be interesting.  Likewise if we pass near the core of Andromeda we might, for a while, have a very bright night sky.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-11-29T13:47:06.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12700" PostTypeId="2" ParentId="12696" CreationDate="2015-11-29T19:09:23.540" Score="0" Body="&lt;p&gt;I think you're referring to quantum tunneling. Well, quantum tunneling is actually essential in nucleosynthesis in stars. Usually quantum tunneling is a very rare occurence — but in a body as large as the Sun (whose diameter is 4 light seconds long), it is bound to happen quite frequently. This is what mainly allows atomic nuclei to collide.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2015-11-29T19:09:23.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12701" PostTypeId="1" CreationDate="2015-11-29T19:52:11.830" Score="1" ViewCount="223" Body="&lt;p&gt;Does anyone out there have a script to replicate the star of Bethlehem.  In this case meaning a conjunction of Venus and Jupiter near Regulus.   &lt;/p&gt;&#xA;" OwnerUserId="9915" LastEditorUserId="2365" LastEditDate="2015-11-29T21:23:29.773" LastActivityDate="2021-03-19T10:35:43.000" Title="Stellarium script star of bethlehem" Tags="&lt;star&gt;&lt;planet&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12702" PostTypeId="1" CreationDate="2015-11-29T21:30:43.300" Score="4" ViewCount="471" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Sculptor_Dwarf_Galaxy&quot; rel=&quot;nofollow&quot;&gt;Sculptor Dwarf Galaxy&lt;/a&gt; is a small satellite Galaxy of the Milky Way. There are many such satellites galaxies and I'm trying to compile basic data on them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I can't find an estimate for the number of stars in the Sculptor Dwarf (and many other satellite galaxies too). I'm disappointed that Wikipedia doesn't have this basic information, and that makes me suspicious that estimating the number of stars in other galaxies is not very easy or reliable for current astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Googling these things has resulted in nothing but a bunch of pop science sites...which is no surprise except that these sites don't seem to have the number of stars either. This lack of basic data is surprising and only adds to me worries that no reliable estimates exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone find a link to this info? It would also be very helpful to point me to a database that has the number of stars for all galaxies in the local group, if such a database exists.&lt;/p&gt;&#xA;" OwnerUserId="7449" LastActivityDate="2016-03-18T09:21:28.483" Title="How many stars are in the Sculptor Dwarf Galaxy?" Tags="&lt;star&gt;&lt;galaxy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12703" PostTypeId="2" ParentId="12696" CreationDate="2015-11-29T22:20:30.033" Score="3" Body="&lt;p&gt;You are referring to &lt;a href=&quot;https://books.google.co.uk/books?id=eez38xjCYGkC&amp;amp;pg=PA380&amp;amp;lpg=PA380&amp;amp;dq=pyconuclear%20reactions&amp;amp;source=bl&amp;amp;ots=OICoivA8Kn&amp;amp;sig=VGJQ-4h-N6G813g2frBaCDUTcoY&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwix1Ozx0rbJAhUHthoKHZCTD_0Q6AEILjAD#v=onepage&amp;amp;q=pyconuclear%20reactions&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;pyconuclear reactions&lt;/a&gt; - these are reactions that are initiated even when the temperature is effectively zero. They are caused by zeropoint oscillations of particles that are trapped in a deep potential well and hence are purely a quantum mechanical effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is far from just of theoretical interest. It may be that pyconuclear reactions occur in the condensed matter of white dwarfs; possibly even initiating the carbon burning that can destroy the star in a type Ia supernova.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-11-29T22:20:30.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12704" PostTypeId="1" AcceptedAnswerId="12710" CreationDate="2015-11-30T02:54:18.007" Score="4" ViewCount="356" Body="&lt;p&gt;Note: I'm not talking about a star orbiting around a single or lonely planet :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know a star orbiting a planet is almost impossible because if a planet is more massive than a star, that &quot;planet&quot; would probably be a star. But how about if the planet is not lonely like:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A multiple planetary system which consists of free floating planets only but the planets are so massive so that their total mass is larger than a star&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A planet itself is not more massive than a star but it has so many massive moons&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is it possible for a very small star to orbit around planets in such situations?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastActivityDate="2018-09-22T18:15:06.953" Title="Can a star orbit around multiple planets or a planet with massive moons?" Tags="&lt;star&gt;&lt;orbit&gt;&lt;planet&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12705" PostTypeId="1" AcceptedAnswerId="12742" CreationDate="2015-11-30T03:20:45.387" Score="7" ViewCount="332" Body="&lt;p&gt;Currently, most explorations about new planets are pointing to area outside Neptune's orbit, but how about inside Mercury orbit? Can we say for certain there are no planets inside Mercury's orbit?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="19" LastEditDate="2015-12-03T05:05:16.420" LastActivityDate="2015-12-03T17:52:21.193" Title="Are we sure there are no planets inside Mercury's orbit?" Tags="&lt;planet&gt;&lt;mercury&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12707" PostTypeId="1" AcceptedAnswerId="12708" CreationDate="2015-11-30T06:01:33.370" Score="5" ViewCount="951" Body="&lt;p&gt;Mass and metallicity are the two main determinants for a star's fate. This is simple enough. What's more complicated is how exactly these determine the star's fate. For example, you can see in this image I got from Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Supernova&quot; rel=&quot;noreferrer&quot;&gt;supernova article&lt;/a&gt;:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/0hOyb.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0hOyb.png&quot; alt=&quot;Star remnants&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Generally, as metallicity decreases and mass increases, black holes become more likely to form. At a specific mass and metallicity range, however, there are no remnants left behind. From that point, increasing mass and decreasing metallicity allows for black holes again.&lt;/p&gt;&#xA;&lt;p&gt;How does this work? Why would there ever be an area in which no remnants are left behind? In fact, it seems almost random, since the range lies right between direct black hole ranges.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-08-23T12:46:43.010" Title="Why do certain massive stars leave no remnants?" Tags="&lt;star&gt;&lt;black-hole&gt;&lt;stellar-evolution&gt;&lt;mass&gt;&lt;metallicity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12708" PostTypeId="2" ParentId="12707" CreationDate="2015-11-30T07:47:30.193" Score="8" Body="&lt;p&gt;The gap appears because of &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Pair-instability_supernova&quot;&gt;pair instability supernovae&lt;/a&gt;&lt;/em&gt;. In short, as one looks at such massive stellar cores at increasing temperatures, an ever-larger fraction of the photons are sufficiently energetic to spontaneously form electron-positron pairs. True, they soon recombine, but there is nevertheless a loss in (radiation) pressure, which causes contraction, and thus a further increase of temperature. If unstable, as appears to be the case in such massive stars, this leads to more high-energy photons, more pairs, and a further loss of pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Models of such objects suggest that the collapse leads to a sudden, rapid ignition of oxygen- and silicon-burning reactions. In a mass range between about 150 and 250 solar masses (corresponding to core masses of very roughly half of that), the thermonuclear explosion is enough to rip the core apart, and there's nothing left to collapse. This is a bit like how the sudden onset of carbon burning destroys white dwarfs in type Ia supernovae. More massive cores are more strongly bound by gravity, and models suggest that the nuclear explosion isn't enough to unbind the core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So below about 150 solar masses (and at higher metallicities, I think), the core doesn't reach pair-unstable conditions. Above about 250 solar masses, the induced nuclear explosion isn't powerful enough to destroy the core.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2015-11-30T07:47:30.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12709" PostTypeId="1" AcceptedAnswerId="12714" CreationDate="2015-11-30T09:11:31.557" Score="6" ViewCount="384" Body="&lt;p&gt;It’s well-known that the axial tilt of the Earth (with respect to the ecliptic) is about 23.4 degrees. However, two angles is needed to specify the orientation of any rigid body, so it’s unclear to me exactly how Earth is oriented with respect to its orbital plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more precise, consider the plane that satisfies these two conditions:&lt;br&gt;&#xA;1. Perpendicular to the ecliptic plane&lt;br&gt;&#xA;2. Contains the line connecting the Sun and the Earth when Earth is at perihelion (Jan. 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Basically, this plane cuts through the orbit of the Earth such that for half the year Earth is one side and for the other half Earth is on the other side.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions:&lt;br&gt;&#xA;1. When the Earth is at perihelion, what is the angle between its rotational axis and the plane described above? I would guess this angle is very small from the images of Earth’s orbit. Is there a technical term for this angle?&lt;br&gt;&#xA;2. Does the direction of the axis remain approximately the same throughout its entire orbit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please ignore precession for this question.&lt;/p&gt;&#xA;" OwnerUserId="9917" LastActivityDate="2015-11-30T19:01:25.683" Title="Ambiguity in Earth's &quot;Tilt&quot;" Tags="&lt;orbit&gt;&lt;earth&gt;&lt;rotation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12710" PostTypeId="2" ParentId="12704" CreationDate="2015-11-30T11:50:19.670" Score="4" Body="&lt;p&gt;There's theoretical ways to do it but it's so unlikely as to probably not exist.   Assuming you want a system where the planets are in stable orbits around each other.   The basic difficulty is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Three-body_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;3 body problem&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/N-body_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;n body problem&lt;/a&gt;.  More on it &lt;a href=&quot;http://news.sciencemag.org/physics/2013/03/physicists-discover-whopping-13-new-solutions-three-body-problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.askamathematician.com/2011/10/q-what-is-the-three-body-problem/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a massive planet could (in theory) have a single super-massive moon of similar mass to the planet, though that would probably be considered a 2 planet system.  It's not possible for one planet to have two super-massive moons because that wouldn't be stable.   In general, planets are many times the mass of all their moons combined, similarly stars are many times the mass of all their planets combined and when that stops being true the system is no longer stable.   It's very difficult to generate sufficient mass by adding moons to a system, or by adding enough planets to get more massive than the sun, unless the planets crash into each other but past a certain mass, they'd stop being planets and become a kind of star when enough planets combined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at gravitational systems, like our solar System, something like 99.7% of the mass is in the sun, so the sun dominates and everything orbits the sun.   A few of the larger objects have moons, and curiously, but only cause they're in relatively calm orbits far away from other planets, a few asteroids orbit each other, but the system is very structured around the sun with much smaller stable orbital zones around the planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pluto also has a rather curious orbital system of it's own, likely caused by an impact, where Pluto and Charon are comparatively large and 4 tiny moons orbit around them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WGGKJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WGGKJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.space.com/29559-pluto-moons-weird-orbit-chaos.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But these kind of structured systems are only possible if you have a significant mass differential.   When you have 3 or more bodies of similar mass and similar distance you get a high degree of mathematical chaos and instability.   There are creative mathematical tricks to make it work, but none of them are stable or likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what 3 similar mass bodies look like, and in such a system, with constant changes, the most likely scenario is that one of the bodies eventually gets ejected.  (source, N-body problem above)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/n91VT.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/n91VT.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are star systems with several stars, but they are either unstable or contain significant differences in mass.   The gravitational structure applies for large planets the same as it does for stars, and there's an article about that &lt;a href=&quot;https://en.wikipedia.org/wiki/Star_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can create stability by having 2 objects orbiting each other and a 3rd massive object quite distant.  (Alpha Centauri is that kind of set-up though Proxima Centauri is quite a bit smaller, but it's the same system).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can even create the heirarchy where you have 2 objects orbiting each other and then 2 more, orbiting each other, but distant so the 2 co-orbitals orbit around each other, and if you do this enough times, you can kind of create enough planetary mass where a more massive star could distantly orbit the entire thing, but it gets very structured and very distant.   It's not what I'd consider a normal orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also cheat and have several planets in a wildly unstable general proximity orbit and have a star some distance away orbiting the chaotic mess in the middle, but it wouldn't be stable for long.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-11-30T11:50:19.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12713" PostTypeId="2" ParentId="12702" CreationDate="2015-11-30T18:46:49.670" Score="5" Body="&lt;p&gt;Estimates of star numbers are &lt;a href=&quot;http://www.space.com/25959-how-many-stars-are-in-the-milky-way.html&quot;&gt;difficult to obtain&lt;/a&gt;. Red dwarfs are too dim to be seen as individual points so star counts cannot be obtained. The best guesses come from obtaining the mass of the galaxy and then guessing how much of that mass is made of stars. It is usually a low proportion, between 1 and 3%. You then guess how big the stars are on average to obtain a rough estimated of the star count.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/pdf/0802.4220v2.pdf&quot;&gt;Sources&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dwarf galaxies tend to have a high mass to light ratio meaning that more of their mass is in the form of dust, gas and &quot;dark matter&quot;. Sculptor has particularly high ratio, of about 160&lt;a href=&quot;https://en.wikipedia.org/wiki/Mass-to-light_ratio&quot;&gt;$\Upsilon_\odot$&lt;/a&gt;, and a mass of about about 200 million times the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 1% of that mass is in the form of stars, but the stars are on average half the mass of the sun then Sculptor would have about &lt;strong&gt;4 million stars&lt;/strong&gt;.  The error bars on that number are very large: The mass could be as much as 320 million suns, and the other values are guessed by me (a&lt;a href=&quot;http://www.space.com/25959-how-many-stars-are-in-the-milky-way.html&quot;&gt;bout 3% of the milky way is in stars&lt;/a&gt;, but the milky way has a much lower mass to light ratio) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the final reason that this number is not easily available is that it is of relatively little use. While the size of a galaxy can be measured spatially or by mass, and the nature of the stars is of interest, the number of them is of less use.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-11-30T18:46:49.670" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12714" PostTypeId="2" ParentId="12709" CreationDate="2015-11-30T19:01:25.683" Score="5" Body="&lt;p&gt;The fact that the perihelion is close to the the winter solstice is purely coincidental. The perihelion might occur in any season. It wanders around in a somewhat chaotic fashion because of the gravitational &lt;a href=&quot;http://www.timeanddate.com/astronomy/perihelion-aphelion-solstice.html&quot; rel=&quot;noreferrer&quot;&gt;influence of the other planets&lt;/a&gt;. At the moment the perihelion moves a day every 58 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let A = Earths axis, and E = the line perpendicular to the orbital plane. The angle between A and E is called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_tilt&quot; rel=&quot;noreferrer&quot;&gt;axial tilt&lt;/a&gt;. Let S = the line between the centers of the Earth and the Sun. Let P = the plane containing S and E. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q1 is (maybe) asking what is the angle between A and P on Jan 2. It will be about 23.4 times sine(2*pi*12/365).  Because there are 12 days from Dec 21 to Jan 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q2. yes.&lt;/p&gt;&#xA;" OwnerUserId="9926" LastActivityDate="2015-11-30T19:01:25.683" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12715" PostTypeId="1" AcceptedAnswerId="12770" CreationDate="2015-11-30T19:11:41.477" Score="5" ViewCount="237" Body="&lt;p&gt;Is it possible for a small, close-to-star planet to have a stable ring system--icy or otherwise? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A smaller planet means the rings would have to be closer to the planet, which means the radial shear (difference in orbital speed vs. radius) in the rings would be higher. A higher radial sheer would be more likely to cause turbulence...or so I would think. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously one would not expect icy rings around a planet that's inside the frost line....&lt;/p&gt;&#xA;" OwnerUserId="7965" LastActivityDate="2015-12-05T10:28:26.977" Title="Rings around a smaller, close-in planet?" Tags="&lt;planetary-ring&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12716" PostTypeId="1" AcceptedAnswerId="12725" CreationDate="2015-11-30T20:14:10.183" Score="5" ViewCount="937" Body="&lt;p&gt;I want to simulate the orbits of the planets from our solar system. I want to use orbital elements to calculate the current position(xyz) at a time t. The simulation doesn't have to be too exact, but the initial position of the bodies should be somewhat realistic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing some research on the calculation I stumbled upon the following formula:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R,X,Y,Z-Heliocentric Distances&#xA;TA - True Anomaly&#xA;N - Longitude of the Ascending Node&#xA;w - Argument of the Perihelion&#xA;&#xA;R = a * (1 - e ^ 2) / (1 + e * Cos(TA))&#xA;X = R * (Cos(N) * Cos(TA + w) - Sin(N) * Sin(TA+w)*Cos(i)&#xA;Y = R * (Sin(N) * Cos(TA+w) + Cos(N) * Sin(TA+w)) * Cos(i))&#xA;Z = R * Sin(TA+w) * Sin(i)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://www.physicsforums.com/threads/calculating-elliptic-orbits-in-cartesian-coordinates.712979/&quot; rel=&quot;noreferrer&quot;&gt;https://www.physicsforums.com/threads/calculating-elliptic-orbits-in-cartesian-coordinates.712979/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For testing purposes I calculated the true anomaly with the help of the following calculator: &lt;a href=&quot;http://www.jgiesen.de/kepler/kepler.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.jgiesen.de/kepler/kepler.html&lt;/a&gt;&lt;br&gt;&#xA;It needs the mean anomaly and eccentricity which I took from Wikipedia. Doing some more research on the mean anomaly, I now believe that the time at which I want the position will be somehow fed into the calculation of the mean anomaly along with the time of the initial position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody clarify for me on how to correctly calculate the position at time t with the above or a different formula. With the above formula I also think I need the initial values at a specific time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also want to say that I am not an astronomer and I don't have a clue on how to handle the calculation or which formula to use. The result of the calculations should mimic our solar system as closely as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Rene Hollander&lt;/p&gt;&#xA;" OwnerUserId="9927" LastActivityDate="2015-12-01T16:57:07.557" Title="Simulate an orbit with orbital elements" Tags="&lt;orbit&gt;&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12717" PostTypeId="1" AcceptedAnswerId="12719" CreationDate="2015-11-30T22:53:20.147" Score="2" ViewCount="96" Body="&lt;p&gt;I know an Einstein-De Sitter universe is a &quot;flat&quot; universe, i.e. with a K (&quot;bendiness&quot;, sorry, don't really know what to call it in English) factor of 0, but does that entail anything as to its contents? The Wikipedia page is quite lackluster on this aspect, and my astronomy course textbook merely slightly less so (it says it's made up mainly of &quot;non-relativistic, dust-like matter&quot;).&lt;/p&gt;&#xA;" OwnerUserId="9928" LastActivityDate="2015-12-01T09:34:14.433" Title="What does an Einstein-De Sitter universe look like?" Tags="&lt;universe&gt;&lt;matter&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12719" PostTypeId="2" ParentId="12717" CreationDate="2015-11-30T23:31:17.360" Score="1" Body="&lt;p&gt;Bendiness = space-time curvature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, it is flat (k=0) for the EdS model. Essentially it is the epoch at which only matter is thought to dominate (no contributions from radiative pressure or dark energy, $\Lambda$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is only a model. The EdS universe is thought of as a static Universe which is expanding indefinitely. It just makes solving certain equations in Cosmology a little easier, but it was perhaps due to Einstein's stubbornness in accepting a dynamic Universe.&lt;/p&gt;&#xA;" OwnerUserId="2091" LastEditorUserId="2091" LastEditDate="2015-12-01T09:34:14.433" LastActivityDate="2015-12-01T09:34:14.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12720" PostTypeId="1" CreationDate="2015-12-01T03:40:07.597" Score="2" ViewCount="136" Body="&lt;p&gt;In several equations of orbital elements (such as the determination of true anomaly from mean anomaly), the terms 1-&lt;em&gt;e&lt;/em&gt; and 1+&lt;em&gt;e&lt;/em&gt; appear. These are the ratios of the orbital periapsis and apoapsis to the semi-major axis, but do the ratios themselves have names?&lt;/p&gt;&#xA;" OwnerUserId="2573" LastEditorUserId="2573" LastEditDate="2015-12-01T16:46:11.477" LastActivityDate="2016-01-30T18:41:52.490" Title="Name for 1-e and 1+e terms?" Tags="&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12721" PostTypeId="2" ParentId="12720" CreationDate="2015-12-01T10:14:32.263" Score="4" Body="&lt;p&gt;As the commenter states, $e$ is indeed called the orbital eccentricity. If you add a radial scale length (e.g., semi-major axis) to both of those values the $1-e$ describes the periapsis (closest approach of orbit) and the $1+e$ apoapsis (furthest) of an elliptical orbit. They don't have a specific special name, as they are dimensionless measures, but can be quite useful in determining the orbit of planets and other Keplerian systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Periapsis:&#xA;$$&#xA;r_{p}=a(1-e)&#xA;$$&#xA;Apoapsis:&#xA;$$&#xA;r_{a}=a(1+e)&#xA;$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the could be named the maximum radial eccentricity and minimum radial eccentricity, if you needed to describe these parameters in a report or class homework etc.&lt;/p&gt;&#xA;" OwnerUserId="2091" LastEditorUserId="2091" LastEditDate="2015-12-01T17:08:06.520" LastActivityDate="2015-12-01T17:08:06.520" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12722" PostTypeId="1" CreationDate="2015-12-01T11:13:55.127" Score="4" ViewCount="742" Body="&lt;p&gt;I sometimes hear about astronomers using an arc spectrum to calibrate observations. For example a &quot;He-Ar arc spectrum&quot;. What is an &quot;arc&quot; in this context? I assume it's nothing got to do with angles (arcmin, arcsec, etc.). &lt;/p&gt;&#xA;" OwnerUserId="9932" LastActivityDate="2015-12-01T16:01:14.043" Title="What is an &quot;arc&quot; spectrum ?" Tags="&lt;observational-astronomy&gt;&lt;amateur-observing&gt;&lt;spectroscopy&gt;&lt;spectra&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12723" PostTypeId="1" AcceptedAnswerId="12727" CreationDate="2015-12-01T12:00:38.667" Score="9" ViewCount="586" Body="&lt;p&gt;From my understanding, a ring can form around a planet when a moon gets too close to its Roche limit, and gets ripped appart by the planet's gravity pull. That makes sense to me, but I don't understand why Saturn has both moons and rings at the same place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the F ring is supposedly created by Enceladus' rejections, but I don't get why Pandora and Prometheus are not ripped appart as well as the ancient objects that formed the ring they are in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't find specific explanations for this. Does it have any link to the moon's density? &lt;/p&gt;&#xA;" OwnerUserId="9679" LastEditorUserId="2153" LastEditDate="2015-12-01T23:06:38.817" LastActivityDate="2016-07-30T10:03:38.203" Title="Why does Saturn have both moons and rings?" Tags="&lt;natural-satellites&gt;&lt;tidal-forces&gt;&lt;saturn&gt;&lt;planetary-ring&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12724" PostTypeId="2" ParentId="12722" CreationDate="2015-12-01T16:01:14.043" Score="5" Body="&lt;p&gt;An arc spectrum is one produced by a discharge lamp where the discharge is through ionised gas, in the case of He-Ar a mixture of Helium and Argon, which produces a predictable line emission spectrum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are often used to provide a &lt;a href=&quot;http://www.ing.iac.es/astronomy/observing/manuals/ps/tech_notes/tn070_1.pdf&quot;&gt;calibration spectrum for spectrometers&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-12-01T16:01:14.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12725" PostTypeId="2" ParentId="12716" CreationDate="2015-12-01T16:57:07.557" Score="2" Body="&lt;p&gt;As it happens, I was writing code for the exact same transformation last night using &lt;a href=&quot;https://downloads.rene-schwarz.com/download/M001-Keplerian_Orbit_Elements_to_Cartesian_State_Vectors.pdf&quot; rel=&quot;nofollow&quot;&gt;this memo&lt;/a&gt; as a reference. I haven't yet tested my implementation but the document is quite clear about what each step is doing, which pleased me greatly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll have to understand the terminology to some degree; &quot;epoch&quot; means a reference time for which you have the orbital elements, and &quot;considered epoch&quot; means the time at which you're evaluating the position, for example. Other points requiring caution: the equations inconsistently drop subscripts; the use of the Newton-Raphson solver isn't particularly clear (i.e. how many times should it run)? Make sure you get the units right also.&lt;/p&gt;&#xA;" OwnerUserId="2573" LastActivityDate="2015-12-01T16:57:07.557" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12726" PostTypeId="1" CreationDate="2015-12-01T21:39:47.383" Score="6" ViewCount="859" Body="&lt;p&gt;I'm trying to fit Gaussians to several lines in a spectrum that I have. Some of them overlap with one another, causing the fitting program that I'm using to not be able to give reasonable estimates for the errors on the measurements. For example, sometimes it will give the width of the spectral line to be 0.5 GHz but give the uncertainty at 1,000 GHz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To fix this, I've been doing manual estimates on the ones that it can't get. I read that you can estimate the error on the peak of the Gaussian with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Peak Error = (1/2)*(Width / Signal To Noise Ratio)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I can't find anything for the height or width. Are there similar ways to estimate the uncertainty on these measurements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="9935" LastActivityDate="2016-09-24T13:49:21.330" Title="How do you estimate the error on the height/width of a Gaussian?" Tags="&lt;spectroscopy&gt;&lt;spectra&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12727" PostTypeId="2" ParentId="12723" CreationDate="2015-12-01T23:05:19.243" Score="5" Body="&lt;p&gt;You're right that density is the important thing here. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Roche_limit&quot; rel=&quot;noreferrer&quot;&gt;Roche limit&lt;/a&gt; is the distance from the main body $d$ such that&#xA;$$d=1.26R_M\left(\frac{\rho_M}{\rho_m}\right)^{\frac{1}{3}}$$&#xA;where $_M$ denotes the main body and $_m$ denotes the satellite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see from the chart on the Wikipedia page, Pandora and Prometheus are both at least one and a half times the Roche limit from Saturn. Therefore, they're in no danger of being ripped apart any time soon.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2015-12-01T23:05:19.243" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12729" PostTypeId="1" CreationDate="2015-12-01T02:30:57.950" Score="11" ViewCount="665" Body="&lt;p&gt;I am developing a RPG game (or so I like to tell myself) within outer space. Within the game, players would be able to mine small asteroids and collect resources, in order to make money. I'm having some difficulty determining exactly &lt;em&gt;how much&lt;/em&gt; of the resources there would be in relation to everything else gotten from the asteroid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://i0.wp.com/www.universetoday.com/wp-content/uploads/2009/08/infog-high-value-asteroids1.jpg&quot;&gt;This&lt;/a&gt; image has provided me with a good idea of what asteroids generally contain. It does not, however, show me the quantities. For example, I'm fairly certain that an asteroid would have more oxygen than palladium, but I don't know by how much. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Can someone tell me (or better yet show me) the percentage of materials within an asteroid? I realize not all asteroids are the same, so categorizing or averaging is fine. The more detailed the better. The ideal information would be a list of values that I can scale with asteroid size. &lt;/p&gt;&#xA;" OwnerUserId="9937" OwnerDisplayName="Tommy Myron" LastEditorUserId="9937" LastEditDate="2015-12-02T00:21:05.583" LastActivityDate="2015-12-02T03:59:17.023" Title="What is the composition of an asteroid in percentages?" Tags="&lt;space&gt;&lt;asteroids&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12730" PostTypeId="2" ParentId="10279" CreationDate="2015-12-02T03:42:46.947" Score="2" Body="&lt;p&gt;In the Q on Earth Sciences SE &lt;a href=&quot;https://earthscience.stackexchange.com/questions/6528/what-is-the-mass-of-our-planets-atmosphere-hydrosphere-and-cryosphere-combine&quot;&gt;What is the mass of our planet's atmosphere, hydrosphere, and cryosphere combined?&lt;/a&gt; this formula was derived: $$ M_{\mathrm{atm}_\bigoplus} \approx \frac{4\pi R^2_{\scriptscriptstyle \bigoplus} P_{\mathrm{sea\,level}}}{g_{_\bigoplus}} $$ which can be rearranged to derive pressure from the total mass of the atmosphere. $$  P_{\mathrm{sea\,level}} \approx \frac{g_{_\bigoplus}M_{\mathrm{atm}_\bigoplus}}{4\pi R^2_{\scriptscriptstyle \bigoplus}} $$&lt;/p&gt;&#xA;" OwnerUserId="9186" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:29.267" LastActivityDate="2015-12-02T03:42:46.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12731" PostTypeId="2" ParentId="12729" CreationDate="2015-12-02T03:59:17.023" Score="3" Body="&lt;p&gt;You might find the information you seek in &lt;a href=&quot;http://www.nature.com/nature/journal/v505/n7485/fig_tab/nature12908_F4.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article by Nature on the compositional mass distribution of asteroids.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NSO3P.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NSO3P.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Wikipedia &lt;a href=&quot;https://en.wikipedia.org/wiki/Asteroid_spectral_types&quot; rel=&quot;nofollow noreferrer&quot;&gt;Asteroid Spectral Types&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The majority of bodies fall again into the three broad C, S, and X&#xA;  categories, with a few unusual bodies categorized into several smaller&#xA;  types:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C-group of carbonaceous objects including:[citation needed]&#xA;    B-type largely overlapping with the Tholen B and F types.&#xA;    C-type the most 'standard' of the non-B carbonaceous objects&#xA;    Cg Ch Cgh somewhat related to the Tholen G type&#xA;    Cb transition objects between plain C and B types.&#xA;S-group of silicaceous (stony) objects including:&#xA;    A-type&#xA;    Q-type&#xA;    R-type&#xA;    K-type a new category (181 Eucharis, 221 Eos)&#xA;    L-type a new category (83 Beatrix)&#xA;    S-type the most 'standard' of the S group&#xA;    Sa, Sq, Sr, Sk, and Sl transition objects between plain S and the other types in the group.&#xA;X-group of mostly metallic objects including:&#xA;    X-type the most 'standard' of the X group including objects classified by Tholen as M, E, or P-type.&#xA;    Xe, Xc, and Xk transition types between plain X and the appropriately lettered types.&#xA;T-type&#xA;D-type&#xA;Ld-type: a new type with more extreme spectral features than the L-type&#xA;O-type a small category (3628 Božněmcová)&#xA;V-type (4 Vesta)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You will have to go to the Wikipedia article to drill down for more detail on the probable minerals contained in these asteroid classes.  We really only have good evidence on those classes that we think we have meteor samples of.  The rest would be guesses.&lt;/p&gt;&#xA;" OwnerUserId="7168" LastActivityDate="2015-12-02T03:59:17.023" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12732" PostTypeId="1" CreationDate="2015-12-02T09:11:12.450" Score="1" ViewCount="176" Body="&lt;p&gt;As part of my physics project, I investigated the relationship between a light bulb's illuminance and the distance from the measuring device to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Illuminance was measured in flux, and distance in metres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I derived an inverse relationship. First of all, can I check if this is the relationship I was supposed to get? (I understand that there is an inverse square relationship between luminosity and distance, but since illuminance is measured in lumens/sq metre, then I think an inverse relationship is correct...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, using this relationship, I need to extend my project so that I can find distances to stars using this relationship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as 2 stars have a similar mass/size/density, then I can use this relationship to find the distance to the 2nd star as long as the illuminance of the stars are known, and the distance to the 1st star is known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I tried to do this with an example - the Sun and Tau Ceti, as they are reasonably similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have values for the Sun's distance/illuminance - $1.5\!\times\!10^8 $km, $1.2\!\times\!10^5$lux.&#xA;I now am trying to calculate the distance to Tau Ceti using this method.... But nowhere can I find a value for its illuminance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please can someone help me with this? Is there a way to convert from apparent magnitude or luminosity to illuminance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The method I want to use from the inverse relationship is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_2=d_1l_1/l_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $d_1$ and $l_1$ are the known distance and illuminance of the 1st star, whilst $d_2$ and $l_2$ is that of the second, and $d_2$ is unknown.&lt;/p&gt;&#xA;" OwnerUserId="9940" LastEditorUserId="9186" LastEditDate="2015-12-03T11:23:27.377" LastActivityDate="2015-12-03T11:23:27.377" Title="What is the illuminance of Tau Ceti?" Tags="&lt;star&gt;&lt;luminosity&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12733" PostTypeId="2" ParentId="12732" CreationDate="2015-12-02T12:56:53.517" Score="3" Body="&lt;p&gt;You can easily convert flux to luminosity and vice versa using the spherical nature of the propagation of light. The equation you need is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;L=4\pi D^{2}f&#xA;$$&#xA;Where $f$ will be your flux, and $D$ is the distance to the object, Tau Ceti in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Pijpers (2003) Tau Ceti's luminosity is $0.52\pm0.03L_{\odot}$. You should have everything you need there. $L_{\odot}$ is of course the luminosity of the Sun. &lt;/p&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2015-12-02T12:56:53.517" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12735" PostTypeId="1" AcceptedAnswerId="12763" CreationDate="2015-12-02T22:40:34.737" Score="4" ViewCount="986" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Analemma&quot; rel=&quot;nofollow&quot;&gt;The Earth's solar analemma is a diagram showing the deviation of the Sun from its mean motion in the sky, as viewed from a fixed location on the Earth. ... For [a planet] with a &lt;strong&gt;&lt;em&gt;circular orbit but significant axial tilt, the analemma would be a figure of eight&lt;/em&gt;&lt;/strong&gt; with northern and southern lobes equal in size.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This statement seems to be true, but it doesn't make any sense to me. If the eight is not due to the equation of time, why is there an eight? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="9949" LastEditorUserId="9186" LastEditDate="2015-12-03T11:23:01.030" LastActivityDate="2015-12-04T21:36:22.267" Title="Why would the Earth's solar analemma would be still a figue eight even if Earth's orbit was circular?" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;observational-astronomy&gt;&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12736" PostTypeId="1" CreationDate="2015-12-03T01:27:15.870" Score="2" ViewCount="405" Body="&lt;p&gt;I am currently taking a first course on stellar astrophysics, and I noticed that in some cases we use the ideal gas equation of state for stars, so we also use $\gamma =5/3$. Of course it can only be applied where there is no nuclear reaction, so there is a limit in the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if the radiation pressure if relevant, you have to consider the parameter $\beta$ to calculate the total pressure and the adiabatic coefficient $\gamma$. It is also incorrect (I think) if there are degeneration or relativistic considerations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if I have to consider other factors before I can use this equation of state and value for $\gamma$, maybe it is important if it is a convective or radiative area, or other factors I didn't consider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: can anybody tell me the limits of the ideal gas approximation? (quantitative better than qualitative but any help will be well received)&lt;/p&gt;&#xA;" OwnerUserId="8011" LastActivityDate="2016-02-01T15:27:00.593" Title="When is it a good aproximation to consider a star to be an ideal gas?" Tags="&lt;star&gt;&lt;astrophysics&gt;&lt;gas&gt;&lt;stellar-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12737" PostTypeId="2" ParentId="12736" CreationDate="2015-12-03T06:04:29.363" Score="1" Body="&lt;p&gt;There is no hard and fast answer. To be treated as an idea gas (your title question), the particles in your gas should be point-like and they should be non-interacting if you are to use the ideal gas approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means you can take the mean separation ($\sim n^{-1/3}$) and compare that with the size of the particles - it should be much bigger. You can also compare the particle kinetic energies with the potential energy of their interactions at typical separations. eg The ratio of $kT$ to the Coulomb potential energy should be large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest of your question is confused. You are asking about an adiabatic index, but that is a completely separate question from whether the gas is ideal. It depends on the equation of state, on interactions, on internal degrees of freedom, on the way energy is transported.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relativistically degenerate gases &lt;em&gt;can&lt;/em&gt; be ideal gases and are often assumed to be so. Do not confuse an ideal gas with a perfect gas (even though many do so).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-12-03T13:44:05.857" LastActivityDate="2015-12-03T13:44:05.857" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12741" PostTypeId="2" ParentId="11389" CreationDate="2015-12-03T17:32:49.957" Score="5" Body="&lt;p&gt;The point you appear to refer to, is called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lagrangian_point&quot; rel=&quot;noreferrer&quot;&gt;Lagrangian point&lt;/a&gt; $L_1$.&#xA;This point is a saddle in the field of gravity, hence not to be considered to be stable in the strict sense.&#xA;Two other Lagrangian points, called $L_4$ and $L_5$, can be stable, provided the considered orbiting objects are of small mass in comparison to the two main bodies of the system, and if the masses of the binary components are sufficiently different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to theorem 4.1 of &lt;a href=&quot;http://www.math.cornell.edu/~templier/junior/final_paper/Thomas_Greenspan-Stability_of_Lagrange_points.pdf&quot; rel=&quot;noreferrer&quot;&gt;this paper&lt;/a&gt;, $L_4$ and $L_5$ are stable in all directions, if and only if the mass ratio of the two main binary components $\frac{m_1}{m_2}\geq\frac{25+3\sqrt{69}}{2}\approx 24.9599$.&#xA;According to theorem 3.1 of the same paper all Lagrangian points are stable in z-direction, which is the direction perpendicular to the orbital plane of the binary system.&#xA;(Credits for this corrected version go to user DylanSp.)&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2016-03-08T18:35:48.223" LastActivityDate="2016-03-08T18:35:48.223" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12742" PostTypeId="2" ParentId="12705" CreationDate="2015-12-03T17:52:21.193" Score="4" Body="&lt;p&gt;Yes, Vulcanoids larger than 5.7 km diameter are considered to be ruled out:&#xA;&lt;a href=&quot;http://arxiv.org/abs/1301.3804&quot; rel=&quot;nofollow&quot;&gt;A Search for Vulcanoids with the STEREO Heliospheric Imager&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An object sufficiently large to be called &quot;planet Vulcan&quot; would change Mercury's orbit in a detectable way by gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there are two lines of evidence, that such a planet doesn't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More to read in Wikipedia:&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Vulcan_(hypothetical_planet)&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Vulcan_(hypothetical_planet)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2015-12-03T17:52:21.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12743" PostTypeId="2" ParentId="12441" CreationDate="2015-12-03T18:36:39.907" Score="2" Body="&lt;p&gt;The SOHO image you attached, provides some evidence, that most of the debris continues the trajectory of the original body.&#xA;Most of the debris hence won't replenish the Vulcanoid population.&#xA;But due to the disruptive event you get an additional delta-v, which may change the orbit a bit, however insufficient to slow down to a Vulcanoid orbit; compare the velocity of the comet near perihelion with the speed of sound as an order of magnitude estimate of the upper limit of the delta-v for debris.&#xA;Dust may either be slowed down in this or in future orbits in the solar corona to end up in the Sun, or blown away by solar radiation pressure, again no Vulcanoids.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2015-12-03T18:36:39.907" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12744" PostTypeId="2" ParentId="10463" CreationDate="2015-12-03T21:42:34.223" Score="3" Body="&lt;p&gt;The space itself is falling into the black hole. Below the event horizon - in the timelike zone - an observer is falling towards the singularity, together with the space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seen from this falling observer, space looks like vacuum outside the black hole. Close to the singularity tidal forces become an increasing problem. But for a sufficiently large and &quot;benign&quot; black hole, an observation can't locally distinguish between spacetime inside and spacetime outside the black hole.&#xA;Otherwise it would violate the &lt;a href=&quot;https://en.wikipedia.org/wiki/Principle_of_relativity&quot; rel=&quot;nofollow noreferrer&quot;&gt;principle of relativity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So - seen from the falling observer - we get short-lived &lt;a href=&quot;https://en.wikipedia.org/wiki/Virtual_particle&quot; rel=&quot;nofollow noreferrer&quot;&gt;virtual particles&lt;/a&gt;, as in usual vacuum. Due to space-time curvature, the segment of vacuum in the black hole isn't perfectly symmetric, including &lt;a href=&quot;https://en.wikipedia.org/wiki/Quantum_fluctuation&quot; rel=&quot;nofollow noreferrer&quot;&gt;quantum fluctuations&lt;/a&gt;.&#xA;This tiny asymmetry results in (few) real particles, when seen from outside the black hole, when applied to the (light-like) zone near the event horizon.&#xA;It should be fundamentally possible to probe the tiny asymmetry of the &lt;a href=&quot;https://physics.stackexchange.com/questions/64862/resonances-in-high-energy-physics&quot;&gt;resonances&lt;/a&gt; with particle experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An observer hovering just below the event horizon would be tachyonic in the in-falling spacetime.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-12-03T21:42:34.223" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12745" PostTypeId="1" CreationDate="2015-12-03T22:15:16.470" Score="1" ViewCount="186" Body="&lt;p&gt;Using the centripetal force to create artificial gravity has been proposed for space exploration. From our perspective, the force is an outward force, pushing objects away. Could it be inverted to attract them instead? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could we, instead of having a spacecraft, or a &quot;dish,&quot; rotate to create the force that would push the travelers to the inside of the hull - could we instead have a cluster of dishes rotating in such a manner that the travelers are pressed to the outside of a sphere? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, could we construct an artificial mini-planet with a gravity exceeding by far the gravity of a natural celestial body with the same mass?&lt;/p&gt;&#xA;" OwnerUserId="9510" LastEditorUserId="9510" LastEditDate="2015-12-04T11:26:55.557" LastActivityDate="2015-12-04T13:21:35.360" Title="Can the centripetal force be inverted?" Tags="&lt;newtonian-gravity&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2015-12-07T15:32:20.037" ContentLicense="CC BY-SA 3.0" />
  <row Id="12746" PostTypeId="2" ParentId="12147" CreationDate="2015-12-03T22:52:24.510" Score="4" Body="&lt;p&gt;The &lt;a href=&quot;http://pluto.jhuapl.edu/soc/Pluto-Encounter/view_obs.php?image=029932/lor_0299323899_0x630_sci_2.jpg&amp;amp;utc_time=2015-07-16%3Cbr%3E03:39:40%20UTC&amp;amp;description=OpNav%20Campaign%204,%20LORRI%201X1&amp;amp;target=PLUTO&amp;amp;range=2.0M%20km&amp;amp;exposure=100%20msec&amp;amp;imgType=approved&quot; rel=&quot;nofollow&quot;&gt;&quot;raw&quot; LORRI image&lt;/a&gt; is 1024x1024 pixels. The field of view is about 0.29°, see section 3.6 of &lt;a href=&quot;http://sbn.pds.nasa.gov/holdings/nh-x-lorri-2-plutocruise-v1.0/document/morgan_spie.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;, or abstract of &lt;a href=&quot;http://arxiv.org/ftp/arxiv/papers/0709/0709.4278.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;.&#xA;Pluto is a little less than 250 pixels in the raw image, corresponding to 0.29° x 250/1024 = 4'15'', well consistent with your estimate.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2015-12-03T22:58:31.817" LastActivityDate="2015-12-03T22:58:31.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12747" PostTypeId="1" AcceptedAnswerId="12772" CreationDate="2015-12-03T23:30:11.840" Score="7" ViewCount="285" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Wow!_signal&quot; rel=&quot;noreferrer&quot;&gt;The Wow! signal&lt;/a&gt; is still an enigma. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the incredibly high signal strength it entails. This during some time-period because the sweep of the alignment (or rather the receiving station). That could explain the rise and falloff. This because the horns tracking of that part of sky, once, because of earths movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The signal never returned. Everybody understands the narrow band of the signal, and the frequency of the hydrogen line. This incredible signal strength was logged, but only that, the spectacular strength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the contents of said signal was not logged, it was not even capable of recording that. Is this right? Is there a possibility that the burst could have carried information, however unlikely? It could have carried a one-time &quot;message&quot;? Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="9883" LastEditorDisplayName="user1569" LastEditDate="2015-12-04T15:37:04.210" LastActivityDate="2020-06-19T05:51:06.630" Title="The contents of potential &quot;message&quot; in the Wow! signal lost?" Tags="&lt;radio-astronomy&gt;&lt;extra-terrestrial&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12748" PostTypeId="2" ParentId="12168" CreationDate="2015-12-03T23:50:59.010" Score="2" Body="&lt;p&gt;Besides solar energetic particle (SEP) events, it's mostly (almost isotropic) galactic cosmic rays (GCR). GCRs may cause secondary radiation by interaction with the atmosphere or with rock. The smaller the &lt;a href=&quot;https://en.wikipedia.org/wiki/Solid_angle&quot; rel=&quot;nofollow&quot;&gt;solid angle&lt;/a&gt; of the visible sky, the smaller the mean galactic radiation dose. On Mars you get additional shielding by the atmosphere, hence dependence of the atmospheric pressure. More detail in &lt;a href=&quot;http://arxiv.org/pdf/1507.03473.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;.&#xA;Shieldings from SEP events should be rich in protons, e.g. water or food, or you need to go deeper into underground, as soon as a SEP event is detected.&#xA;GCRs are partially shielded by the heliosphere. Hence solar maxima protect partially from GCRs, but SEP events get more likely during solar maxima.&#xA;Since GCRs are unpredictable, and SEP events can be detected, for solar maxima, protection strategies resultng in an overall reduced radiation dose can be elaborated.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2015-12-03T23:50:59.010" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12749" PostTypeId="1" CreationDate="2015-12-04T00:41:08.607" Score="1" ViewCount="694" Body="&lt;p&gt;When the big bang happened and all that exist today came into existence from singularity, how do you determine which part of the universe is the oldest?  Its natural to think that looking back at the center of the universe, we would find the oldest galaxies(not sure if this is true); but isn't there a certain logic that the farthest from the center should have the oldest galaxies?  Is there a proven theory behind where we would find the oldest galaxies? or is this one of those space-time enigmas?&lt;/p&gt;&#xA;" OwnerUserId="9958" LastEditorUserId="9301" LastEditDate="2016-06-29T04:37:31.890" LastActivityDate="2016-07-07T11:56:28.013" Title="Oldest in the universe" Tags="&lt;galaxy&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12750" PostTypeId="2" ParentId="8864" CreationDate="2015-12-04T01:36:24.433" Score="2" Body="&lt;ol&gt;&#xA;&lt;li&gt;May, or may not.&lt;/li&gt;&#xA;&lt;li&gt;No.&lt;/li&gt;&#xA;&lt;li&gt;Q1: A formula: yes, a simple formula: no.&lt;/li&gt;&#xA;&lt;li&gt;Q2: Not for each point on Earth. This is e.g. due to different distance quotients.&#xA;Q3: Depends on distances, sizes, angular velocities, position on Earth.&lt;/li&gt;&#xA;&lt;li&gt;Q4: Yes, but it's different for each point, not just by latitude. Even at the equator, there may exist points with, and points without totality. That's since the shadow of the Moon isn't cylindrical, but a cone. And because the Earth isn't a plane perpendicular to the Sun - Moon axis. And the distance Moon - Sun is varying due to its orbit around Earth.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tYpd4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tYpd4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2015-12-04T01:36:24.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12753" PostTypeId="2" ParentId="12745" CreationDate="2015-12-04T02:27:34.980" Score="2" Body="&lt;p&gt;Close to a black hole, but still outside the event horizon, orbiting arbitrarily fast, wouldn't result in a circular orbit, but in a trajectory towards the black hole.&#xA;In this extreme scenario, centrifugal force could be interpreted as attractive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more usual settings, centrifugal forces point outward.&#xA;Since the &lt;a href=&quot;https://en.wikipedia.org/wiki/Centrifugal_force#Force&quot; rel=&quot;nofollow&quot;&gt;centrifugal force&lt;/a&gt; (in usual settings) $m \omega^2r$ is proportional to square of the angular velocity $\omega$, you would need an &lt;a href=&quot;https://en.wikipedia.org/wiki/Imaginary_number&quot; rel=&quot;nofollow&quot;&gt;imgaginary&lt;/a&gt; angular velocity, or alternatively a negative mass to get an inward pointing centrifugal force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not aware of a physical realization of imaginary angular velocities, at least not for circular orbits,&#xA;although &lt;a href=&quot;http://arxiv.org/ftp/arxiv/papers/1312/1312.0967.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; introduces imaginary velocities for hyperbolic orbits. This may be derived from &lt;a href=&quot;https://en.wikipedia.org/wiki/Hyperbolic_angle#Imaginary_circular_angle&quot; rel=&quot;nofollow&quot;&gt;imaginary circular angles&lt;/a&gt;.&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Negative_mass&quot; rel=&quot;nofollow&quot;&gt;Negative masses&lt;/a&gt; aren't entirely impossible, but those potentially feasible are too tiny to be of any practical value in your sense.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2015-12-04T02:27:34.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12754" PostTypeId="2" ParentId="12726" CreationDate="2015-12-04T02:46:24.753" Score="1" Body="&lt;p&gt;If you know the number of overlapping Gauss functions, you may write them as $\sum_{i=1}^n a_ie^{-(\frac{x-\mu_i}{\sigma_i})^2}$, then subtract your time series, with more than $3i$ items, to reduce the risk of an ill-posed problem.&#xA;Then apply a root mean square (RMS) minimization / &lt;a href=&quot;https://en.wikipedia.org/wiki/Least_squares&quot; rel=&quot;nofollow&quot;&gt;least squares&lt;/a&gt; method to infer the $3i$ parameters.&#xA;If your FWHM, or $\sigma$ is known to be the same for all peaks, your number of parameters reduces to $2i+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;https://en.wikipedia.org/wiki/Rietveld_refinement&quot; rel=&quot;nofollow&quot;&gt;Rietveld refinement&lt;/a&gt; as a related approach.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2015-12-04T02:46:24.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12757" PostTypeId="2" ParentId="12745" CreationDate="2015-12-04T13:21:35.360" Score="0" Body="&lt;p&gt;The simple answer is no.&lt;br&gt;&#xA;The centripetal force is what we call an 'inertial force', as contrary to 'fundamental force'. This means there are no charges attached to it, and no field (-analogue) that propagates this force - unlike e.e. for electromagnetism and gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A inertial forces originate purely in changes of your frame of reference (or 'viewpoint') and how this frame moves relative to your originating frame.&lt;br&gt;&#xA;The name &lt;a href=&quot;https://en.wikipedia.org/wiki/Fictitious_force&quot; rel=&quot;nofollow noreferrer&quot;&gt;fictious force&lt;/a&gt; for this concept is also very common. To this the corresponding wikipedia-page discusses an example with a deccelerated bus, which I can only very strongly recommend to study in order to understand the concept of inertial forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The direction of the centripetal force is tied to the acceleration at any given moment. Any path that a massive object has through space has a curvature at every point of this path. This local curvature defines a so-called osculating circle (see below) towards which the centripetal force points. Thus, there is no way to reverse this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Y9LfM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y9LfM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;(c) Wikipedia, Wiki commons license&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, if this answer seems a bit vague, but I think this is the best I can do without going deeper into the math of classical mechanics that covers those topics.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2015-12-04T13:21:35.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12758" PostTypeId="1" CreationDate="2015-12-04T15:59:39.087" Score="1" ViewCount="1100" Body="&lt;p&gt;In &lt;a href=&quot;https://www.youtube.com/watch?v=s1E1ZiiFSec&quot; rel=&quot;nofollow&quot;&gt;this video&lt;/a&gt;, between 4 and 6 minutes, the lecturer describes M type stars as being less massive, smaller, and redder than G type stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He then gives Betelgeuse as an example of an M type star and mentions that the sun is a G type star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, Betelgeuse is more massive than our sun, and larger than our sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I reconcile these facts?&lt;/p&gt;&#xA;" OwnerUserId="9963" LastEditorUserId="9963" LastEditDate="2015-12-04T17:47:17.460" LastActivityDate="2015-12-04T18:00:34.380" Title="Betelgeuse and sun classification" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;classification&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12759" PostTypeId="2" ParentId="12758" CreationDate="2015-12-04T17:12:06.793" Score="3" Body="&lt;p&gt;If you take a look at the Hertzsprung-Russell Diagram that helps us to classify stars &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/goffK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/goffK.png&quot; alt=&quot;HR-Diagram&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;(c) Wikimedia commons&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(while this one uses data gathered from the Hipparcos satellice, roughly 100.000 stars) we see, that there are 2 main types of red stars: Giants and M-dwarves. The latter are stars on the main sqeuence (they have stable hydrogen-burning in their cores), while the giants are stars that have left the main sequence due to aging. They appear red as they bloat up to bigger radii compared to their main-sequence life which also makes them cooler, and thus redder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Betelgeuse is now one of those giants, and he's actually more massive than the sun (somewhere between 7 and 20 solar masses, according to wikipedia), so there is no contradiction with M-Dwarves also being red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please let me know if you desire to know more details on this.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2015-12-04T17:12:06.793" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12760" PostTypeId="2" ParentId="12758" CreationDate="2015-12-04T17:31:49.987" Score="1" Body="&lt;p&gt;Type M is a spectral class: that is a star colour. Type M stars are red, and relatively cool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are, roughly, two types of stars that are Type M:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Main sequence stars that are small, have relatively low rates of hydrogen fusion, and cool exteriors. These are called &quot;red dwarfs&quot; and account for the majority of stars. All red dwarfs are too dim to be visible with the naked eye, but the closest star, Proxima centuri, is an example.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Stars near the end of their life swell up, and the outer layers cool and become cool and red. Very massive stars like Betelgeuse have a similar colour to red dwarfs but are much larger. Betelgeuse is a Red Giant; it is not on the Main Sequence.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The video shows the Hertzsprung-Russell diagram which shows the two groups. It correctly describes Main Sequence type M stars as smaller, redder and dimmer than the sun. It then gives Betelgeuse as an example of a Red Giant. Which is larger and brighter than the sun, yet still cooler on the outside.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2015-12-04T18:00:34.380" LastActivityDate="2015-12-04T18:00:34.380" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12762" PostTypeId="2" ParentId="8864" CreationDate="2015-12-04T19:10:53.780" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Every new moon would be a total solar eclipse.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Incorrect.  Even if the Earth had no axial tilt and the Moon had an equatorial orbit, that doesn't mean that every new moon would be a total solar eclipse.  Yes, the moon would always generate some kind of an eclipse at every new moon, but it's orbit shape hasn't changed -- it's still an ellipse.  It's further at some points in its orbit and closer at others.  When it's further away at new moon, it appears smaller than the sun in the sky, making an &lt;em&gt;annular&lt;/em&gt; (ring) eclipse, not a total eclipse.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;The path of totality would always cover the same area: a 250km wide corridor with its centre at the equator (so 125km over the northern hemisphere and 125km over the southern hemisphere).&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the same reason as (1), the path of totality would vary from a maximum amount down to nothing.  For annular eclipses there would be a similar path of annularity instead.  But whichever path it is would always be centered on the equator.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;I think the path of totality wouldn't always cover the same area, e.g. always over Africa, so Q1: Is there a simple formula to predict which areas would be in the path of totality?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm sure that there are formulas that would predict which areas would be in the path of totality/annularity, but they would not be simple.  They would depend on exactly when new moon occurs and the exact distance of the moon from the earth.  That would control which strip along the equator would see the eclipse and how wide it would be.  Because the moon slows down in its orbit when it's further away from the earth, and it speeds up in its orbit when it's closer to the earth, the time between new moons is not constant.  The fact that Earth's own orbit around the sun is also elliptical, and it also slows down/speeds up when it is far/close from/to the Sun complicates this as well.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;4&quot;&gt;&#xA;  &lt;li&gt;In real life, the moment of totality can vary from seconds to seven minutes. But in my study example, Q2: the period of totality should always be the same, right? Because the crossing path of moon and sun is always the same (while in real life the Moon can overlap the Sun at different angles and going in different directions) Q3: How can I calculate this period of time?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, because of the elliptical orbit of the Moon, the length of totality or annularity still can vary.  Even during a single eclipse, the length of totality depends on your exact location on Earth.  Your exact location determines your exact distance from the moon during the eclipse, which will vary depending on whether the Moon is getting closer/farther and how fast.  There are formulas for calculating this, but I'm sure they are quite complicated even with the moon orbiting over the equator and with Earth having no axial tilt.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;5&quot;&gt;&#xA;  &lt;li&gt;What about the latitudes for whom the solar eclipse would only be seen as partial? I understand that the nearest to the path of totality, the more covered the sun will be, and vice-versa. Q4: Is there a set value that says e.g. for each km, another degree is visible?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Generally, the closer you are to the equator, the more the sun would be covered.  But this would vary slightly depending on the distance of the Moon from the Earth, and it wouldn't depend solely on your latitude.  As Gerald points out, the moon's shadow, in which you would see a total solar eclipse, is cone-shaped.  Different points on the Earth close to the equator would experience a different cross-section of the cone, which may be narrower or wider depending on your exact location.  Where the eclipse is partial, the apparent size of the moon would control how much of the Sun is covered.  A further-away moon would appear smaller in the sky and thus cover less of the Sun.  Close to the borderline where the Moon appears to graze the Sun, the Moon may miss the Sun entirely if it's far enough and it appears small enough.&lt;/p&gt;&#xA;" OwnerUserId="1082" LastActivityDate="2015-12-04T19:10:53.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12763" PostTypeId="2" ParentId="12735" CreationDate="2015-12-04T21:36:22.267" Score="2" Body="&lt;p&gt;It is correct that the Earth's axial tilt alone would result in an analemma of a figure of eight with equally sized lobes.  The axial tilt contributes to the equation of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Circular Orbit, no Axial Tilt&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at an example case with a perfectly circular orbit and no axial tilt, to observe what happens with no equation of time.  In this example, it's noon at the equator and prime meridian.  The Sun is directly overhead.  Next, the Earth rotates exactly once.  This isn't a normal solar day, but a sidereal day, about 23 hours, 56 minutes.  The Earth has rotated exactly once with respect to the stars.  Where is the Sun directly overhead now?  The Earth has moved a little bit in it's orbit around the Sun, so that our observer at the equator and 0 degrees longitude has observed that the Sun has moved in the sky with respect to the stars.  The Sun is now directly overhead at a location to the east, a little less than 1 degree longitude away (360 degrees in a circle, usually 365 days in a year).  The Earth must rotate a little more so that the Sun is directly overhead again (which completes the 24 hours in a day).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After each rotation, the Earth must rotate a little more each day to bring the Sun back overhead at 0 degrees latitude, 0 degrees longitude.  After a quarter of a year, the Earth must rotate another quarter of a rotation (90 degrees) for the Sun to be overhead at that spot.  Without that extra quarter of a rotation, the Sun is overhead at 90 degrees east longitude.  That makes sense; the Earth has moved 90 degrees in its orbit.  With a perfectly circular orbit, the rate at which this point moves eastward per day is constant, a little under a degree per day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These points define a great circle around the Earth that coincides with the equator.  A great circle is a circle on the surface of a sphere whose center is also the center of the sphere; it's the biggest possible circle that can exist on the surface of a sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A Large Axial Tilt&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's keep Earth's perfectly circular orbit, but let's give it a very large axial tilt, 80 degrees, for the purpose of this explanation.  This Earth now begins at noon on the northward equinox at 0 degrees latitude, 0 degrees longitude.  After one Earth rotation, where on Earth is the Sun overhead now?  It's still a little less than one degree of angle away, but the direction of the displacement has changed.  Instead of this point moving due east, it is mostly northward and only a little bit east.  The Earth needs to rotate far less than 4 more minutes for the Sun to reach local noon for our observer at 0 degrees latitude, 0 degrees longitude.  These points are west of where they would be without axial tilt.  Solar noon has occurred before 24 hours have passed.  As days continue to pass after the equinox, solar noons continue to occur earlier each solar day, as these points continue mostly northward and only a little bit eastward.  The analemma shape as seen in the Northern hemisphere sky is moving up and to the right (west is to the right when looking south towards the Sun in the Northern hemisphere).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quarter of a year later, it's the northern solstice, so let's see what has happened to our sun-overhead plots as the sidereal days have piled up.  These points have continued to move mostly northward and a little eastward, but now they are moving purely eastward, because at the solstice the sun has stopped moving northward in the sky.  It's still moving at a little less than 1 degree per day along that great circle, but because longitude lines are spaced much more closely together at 80 degrees latitude, one degree along a great circle covers many degrees of longitude.  In other words, the eastward movement of these points has been &quot;catching up&quot; with the longitude of the corresponding no-axial tilt points.  At the solstice, both the no-axial tilt point and the axial tilt point have the same longitude, 90 degrees east.  The analemma shape as seen in the Northern hemisphere sky is moving up and to the left (eastward), where it reaches its highest point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These points will continue to move through longitude lines rapidly while the Sun's overhead point remains far in the north.  Now, the sun's overhead point is further east than it would be with no axial tilt.  The analemma shape as seen in the Northern hemisphere sky is moving down and to the left (eastward).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the southward equinox approaches, the sun-overhead points are moving mostly southward and only a little eastward.  The analemma shape as seen in the Northern hemisphere sky is moving down and to the right (westward).  This allows the non-axial tilt points to &quot;catch up&quot;, and the longitude differences begin to decrease again, until the time of the southward equinox comes, when the two points coincide again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a crude ASCII drawing of these points so far, with longitude along the horizontal direction, and latitude along the vertical direction.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                                   (2)&#xA;                   K       L        M        N      O&#xA;            J                                              P&#xA;         I                                                    Q&#xA;      H                                                          R&#xA;    G                                                               S&#xA;(1)F                                                                 T(3)&#xA;  E                                                                   U&#xA; D                                                                     V&#xA; C                                                                     W&#xA;B                                                                       X&#xA;A  B  C  D  E  F  G  H  I  J  K  L  M  N  O  P  Q  R  S  T  U  V  W  X  Y&#xA;&#xA;A = northward equinox&#xA;M = northern solstice&#xA;Y = southward equinox&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The points around (1) are further west than their corresponding points on the equator. At (2), the points have caught up in longitude.  At (3), they are further east than their corresponding points on the equator, but the equator points are catching up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the southward equinox, the longitudes of the sun-overhead point diverge once again, with the same mechanism as described above for the northward equinox.  Just reverse north and south, and the eastward movements are the same.  The analemma shape as seen in the Northern hemisphere sky first moves down and to the right, then down and to the left where it reaches its lowest point at the southern solstice, then up and to the left, then up and to the right where it reaches its starting point at the northward equinox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Back to Reality&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With an axial tilt of 80 degrees, the equation of time would show some extreme values, approaching almost 6 hours of divergence from the no-axial tilt case.  With the true axial tilt of about 23.5 degrees, our equation of time difference values are far less substantial, but the effect is real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The true analemma shape we see on Earth is the combined effect of the axial tilt that we have seen plus the fact that Earth's orbit is elliptical and it slows down during the parts of its orbit when it's farther from the Sun and it speeds up during the parts of its orbit when it's closer to the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The site &lt;a href=&quot;http://www.analemma.com/Pages/framesPage.html&quot; rel=&quot;nofollow&quot;&gt;Analemma&lt;/a&gt; has good explanations about the individual effects of the elliptical orbit and the axial tilt and how they combine to create our analemma.&lt;/p&gt;&#xA;" OwnerUserId="1082" LastActivityDate="2015-12-04T21:36:22.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12764" PostTypeId="1" AcceptedAnswerId="12838" CreationDate="2015-12-04T21:54:58.530" Score="1" ViewCount="140" Body="&lt;p&gt;I was reading an elementary book on dark matter (in fact, a historical perspective) and there were mentioned how the scientific community react to the idea of dark matter proposed as a solution to observed discrepancy between the actual mass of astronomical systems and the predicted mass from Newton's theory. I was wondering where Einstein theory stands in relation to dark matter, did it somehow predict it, or does dark matter prove the incompleteness of Einstein's theory? And what about dark energy? &lt;/p&gt;&#xA;" OwnerUserId="9952" LastActivityDate="2015-12-11T22:30:27.587" Title="What's the relation between Einstein's Gravitational Theory and Dark Matter?" Tags="&lt;general-relativity&gt;&lt;dark-matter&gt;&lt;dark-energy&gt;&lt;newtonian-gravity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12765" PostTypeId="1" CreationDate="2015-12-04T23:11:12.100" Score="3" ViewCount="618" Body="&lt;p&gt;Is there any &lt;em&gt;real&lt;/em&gt; proof that alien civilizations exist in outer space? What if someone were to say that Earth is the only inhabited planet in the whole universe? How would you respond?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to write a five-paragraph argumentative essay to prove that alien civilizations exist. I need to write an introduction, three paragraphs supporting my statement that aliens exist, and a conclusion. In each of the three &quot;body&quot; paragraphs I need to describe in detail a piece of evidence supporting my claim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't even know where to start. I mean, I want to find out the opinion of some scientists who think that we are alone, and then disprove their claim in my essay. I need to know the latest scientific evidence on the existence of alien civilizations. I don't mean like alien abduction stories or any of that science fiction stuff. I need some real proof that will be convincing and not undermine my argument. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you know for sure if there are alien civilizations in outer space? What do they look like? On which planets do they live? What kind of technology do they use? Why are we not alone in the universe? What evidence disproves the claims of Earth chauvinists?&lt;/p&gt;&#xA;" OwnerUserId="9966" LastEditorUserId="2153" LastEditDate="2015-12-05T01:11:16.403" LastActivityDate="2015-12-09T06:44:17.000" Title="Is there any real evidence to prove or disprove the existence of alien civilizations?" Tags="&lt;planet&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12766" PostTypeId="2" ParentId="12765" CreationDate="2015-12-04T23:29:27.087" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there any REAL proof that alien civilizations exist in outer space?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What if someone would say that Earth is the only inhabited planet in the whole Universe, how would you respond?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Depends on what their goal is. If their goal is to convince me of that, I'd just try to avoid the conversation. If their goal is to come to an understanding of the current body of scientific knowledge on the matter, I'd talk about how we don't know that that statement is true.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I need to write a 5 paragraph argumentative essay to prove that alien civilizations exist. I need to write an introduction, 3 paragraphs supporting my statement that aliens exist, and a conclusion. In each of the three &quot;body&quot; paragraphs I need to describe in detail a piece of evidence supporting my claim.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That will be difficult, given that there is no such proof. Best you could do is argue that alien civilizations &lt;em&gt;likely&lt;/em&gt; exist, given a certain set of &lt;em&gt;assumptions&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I don't even know where to start. I mean, I want to find out the opinion of some scientists who think that we are alone, and then disprove their claim in my essay.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is not a good goal for an essay. Don't start with the goal of disproof. Start with the goal of understanding.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I need to know the latest scientific evidence on the existence of alien civilizations. I don't mean like alien abduction stories or any of that science fiction stuff. I need some real proof that will be convincing and not undermine my argument.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Look at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Drake_equation&quot;&gt;Drake Equation&lt;/a&gt; and &lt;a href=&quot;http://www.popsci.com/science/article/2012-01/new-exoplanet-analysis-determines-planets-are-more-common-stars-milky-way&quot;&gt;recent research that concludes that there are many planets in the galaxy&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can you know for sure if there are alien civilizations in outer space?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You would need observational evidence.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What do they look like? On which planets do they live? What kind of technology do they use?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All unknown.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why are we not alone in the Universe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This question presumes the conclusion that we are not alone. We might be alone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What evidence disproves the claims of Earth chauvinists?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;None.&lt;/p&gt;&#xA;" OwnerUserId="9963" LastActivityDate="2015-12-04T23:29:27.087" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12768" PostTypeId="1" AcceptedAnswerId="12769" CreationDate="2015-12-05T07:37:37.290" Score="6" ViewCount="2257" Body="&lt;p&gt;Has there been any study to suggest that either quasars of gamma ray bursts are more detectable from further distances in the universe? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read that quasars are the most powerful energy in the universe but gamma ray bursts are detectable from further distances. Logically one would think that if some entity's energy were detectable from further distances than another entity's energy, then it'd be the entity with more energy than the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just looking to see if anyone has any definitive study references that suggest why either quasar or gamma ray bursts have more energy than the other and to explain why if true the one with less energy is visible from further distances than the one with more energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not really an astrophysicist type guy but I do some reading and watch different scientific shows that interest my since a lot is still unknown, so I'm hopeful someone here more familiar with this can answer or clarify this for me or give me some references to anything that gives clue to any of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I apologize for not having direct reference to any readings I've read that bring about this question in my mind but I know I read something that said gamma ray burst are detectable from further distances than any other energy entity and I've also read quasars are the most powerful energy sources in the universe, so I'm just looking for some clarify on this topic whether right or wrong with anything further you can provide.&lt;/p&gt;&#xA;" OwnerUserId="9969" LastEditorUserId="9969" LastEditDate="2015-12-05T14:25:29.117" LastActivityDate="2018-10-31T08:07:06.363" Title="Quasars and Gamma Ray Bursts" Tags="&lt;quasars&gt;&lt;gamma-ray-bursts&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12769" PostTypeId="2" ParentId="12768" CreationDate="2015-12-05T09:06:16.420" Score="9" Body="&lt;p&gt;I think you may be mixing up &quot;Energy&quot; and &quot;Power&quot; a little here. Power is the rate of Energy output per time. If Power is the speed, then Energy is the distance travelled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quasars look like stars, from Earth, they give out a light, and while they do vary a little (or a lot) you would not expect to see new ones forming or existing ones fading away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gamma Ray Bursts are a short lived. While a quasar will emit more energy over it's long life. Gamma Ray Bursts are more powerful. They are formed mostly as a result of energy released in gravitational collapse during a hypernova. Moreover this energy is released in a narrow beam. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine comparing a laser beam with a equivalent power light bulb.  If the laser is pointing right at you, it appears much brighter than the bulb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;GRB are more powerful than quasars, but only last a short time.&lt;/li&gt;&#xA;&lt;li&gt;GRB are form a beam of radiation.&lt;/li&gt;&#xA;&lt;li&gt;So if a GRB and a quasar are equally distant, the GRB would appear brighter.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-12-05T09:06:16.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12770" PostTypeId="2" ParentId="12715" CreationDate="2015-12-05T10:28:26.977" Score="3" Body="&lt;p&gt;I think the related answer that James Kilginger provided is very good and worth a read.  I'll touch on a few points related to your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's 2 parts to this, how do ring systems form and where are they more likely to be stable and long lasting.   I doubt that any ring system would be permanent, but some should be much longer lasting than others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, lets talk stability first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you pointed out, as ices melt inside the frost line (somewhere around Ceres, though different ices have different melting points, but in general, around Jupiter distance or greater ice particles are largely stable and can be part of a ring system (I'm not sure about Jupiter's gonzo magnetosphere though).   Closer to the sun, certainly by the time you get to Mars, probably a bit further out than that, solar radiation melts ices and it turns into gas which the solar wind blows away from ring systems pretty quickly.   What's more, any smaller rocky debris mixed with ice could experience spin or acceleration as ice is melted and escapes from one side as gas.   Those local accelerations could blow some material from a ring system too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smaller dust particles in a ring system near a star would get blown away too by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Poynting%E2%80%93Robertson_effect&quot; rel=&quot;nofollow&quot;&gt;Poynting-Robertson effect&lt;/a&gt; which is much more significant closer to a star.   Some of these problems would be avoided if the star is in it's white dwarf stage and quite cold, but a main sequence star is much more friendly to distant ring systems than closer ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ring systems are only stable inside the &lt;a href=&quot;https://en.wikipedia.org/wiki/Roche_limit&quot; rel=&quot;nofollow&quot;&gt;Roche Limit&lt;/a&gt;.  That's covered in the other question and outside the Roche Limit, the material would tend to reform into a moon.   Solid objects however can orbit inside a liquid or fine particle Roche limit and maintain their shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets close to the sun still have Roche limits usually well inside their Hill Spheres, but long term stability is still a problem for planets close to their stars.   Here's an &lt;a href=&quot;https://planetpailly.wordpress.com/2015/02/13/why-doesnt-mercury-have-a-moon-sciency-words-hill-sphere/&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; that touches on this regarding Mercury and &lt;a href=&quot;http://www.universetoday.com/14250/does-venus-have-moons/&quot; rel=&quot;nofollow&quot;&gt;one on Venus&lt;/a&gt;.  The same principal would apply to ring systems where the significantly larger solar tide would have a destabilizing effect on all objects in orbit around a planet close to it's star.   It wouldn't destabilize immediately but over many orbits, objects could either spiral into the planet or outside the Roche limit, so Ring systems would be significantly shorter lived as planets get closer to their star.  The tidal force grows with the square of the distance.  Earth, for example experiences 90 times the tidal force from the Sun that Saturn experiences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn is also about 95 times the mass of the Earth, so it's not hard to see why Saturn is a much better candidate for a long term ring system than one of the inner planets.   Much more mass and much further form the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formation of ring systems can happen by 3 methods, (that I can think of).  Large collision, break-up of a moon that travels inside the Roche limit or accretion.   The collision can be on a planet's surface where debris is blown into orbit or similar high speed impact on a moon in orbit around the planet.    Collisions of this size are rare, but likely more common in young solar systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A moon that moves too close to a planet can get broken up and form into a ring system if it travels inside the &lt;a href=&quot;https://en.wikipedia.org/wiki/Roche_limit&quot; rel=&quot;nofollow&quot;&gt;Roche limit&lt;/a&gt; see pictures top right.   There's no reason why that couldn't happen from time to time on a number of planets.   It might one day happen to Phobos as it spirals in closer to Mars.   Moons tend to move towards or away from planets quite slowly so this doesn't happen often, but it can happen from time to time, even when solar-systems are billions of years old and mostly settled into stable orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, accretion, Enceladus see &lt;a href=&quot;http://www.universetoday.com/48796/enceladus/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; is feeding Saturn's e-ring and may have fed other rings in the past.   Enceladus is unique among all the moons in our solar-system in that it's able to steadily feed a ring system.   This is possible because it's quite small, so volcanoes on it's surface are able to eject material at greater than it's escape velocity and because it probably has a liquid water-like layer under it's surface, maintained by tidal heating because it's close enough to Saturn to experience enormous tidal forces.   Enceladus is likely responsible for only a tiny part of Saturn's rings, but this is still a viable method for ring formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Europa on Jupiter experiences similar consistent volcanism due to large tidal forces that also keep some of it's inner layers liquid, but Europa's a much larger moon, so far less material gets ejected at or above escape velocity than with Enceladus.   Only small moons close to large planets can supply material to a ring system in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, because moons may play a role in ring formation and Large planets are more likely to have more moons due to greater gravity and Planets far away from their star are also more likely to capture more moons because they have larger Hill Spheres and objects don't orbit the sun quite as fast, that's probably the gist of it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Large planets are more likely to have moons and rings than small planets and further away from the sun, a planet is more likely to maintain a ring system then close to the sun.    It's not impossible that planets close to a star or smaller planets could have a ring system, but quite a bit less likely.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A smaller planet means the rings would have to be closer to the&#xA;  planet, which means the radial shear (difference in orbital speed vs.&#xA;  radius) in the rings would be higher. A higher radial sheer would be&#xA;  more likely to cause turbulence...or so I would think.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;High orbital speed, I suspect, makes little difference, in fact, a larger planet would have higher orbital speed in it's rings than a smaller one.  A heavy Jupiter, 8 or 10 Jupiter masses and 2 or 3 times denser than the Earth could have a large orbital region inside it's Roche limit and potentially, very large, very fast orbiting rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that wasn't too long.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-12-05T10:28:26.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12771" PostTypeId="1" AcceptedAnswerId="12914" CreationDate="2015-12-05T14:46:08.853" Score="6" ViewCount="295" Body="&lt;p&gt;We need to calculate P/Pdot?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pulsar may switch between theses two states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a rotation powered pulsar, it can pulse and, the pulsed Lx can be much larger than the luminosity given by its rotation?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If Lx is very small, smaller than the luminosity given by its rotation, how to tell it is rotation-powered or accretion-powered? There are not necessarily X-ray pulsations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;You can see I am asking a very tricky scenario.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditDate="2015-12-18T16:26:35.000" LastActivityDate="2015-12-19T11:38:36.267" Title="How to tell a pulsar is rotation-powered or accretion-powered?" Tags="&lt;pulsar&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12772" PostTypeId="2" ParentId="12747" CreationDate="2015-12-05T15:26:32.967" Score="7" Body="&lt;p&gt;I'm not an expert in this, but it's a fun little blip in the history of SETI.  Pretty much the only blip I think. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I understand the incredibly high signal strength it entails.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wouldn't call it &quot;incredibly high&quot;.  It peaked at 30 times normal. &lt;a href=&quot;http://www.space.com/17151-alien-wow-signal-response.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;, and that's inside the &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Water_hole_%28radio%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;waterhole&lt;/a&gt;&quot; a frequency range where the background radiation is the lowest in the cosmos.  30 times a normally very low signal is still pretty small, I think.  It was unusual, no question, but looking at the sheet of paper most often reprinted in articles about this, each space is 12 seconds, there are a handful of visible spikes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4wFyb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4wFyb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just on that one printout there's a 3,3,2,space,7 and a 3,6,1,3 and a 4,3,4,4&#xA;The Wow, 6,E,Q,U,J,5 or 6,15,26,30,20,5 compared to those numbers is maybe 6 or 7 times larger than some fairly common spikes.   Statistically speaking, that's high, but not enormous.  It's still an unexplained oddity, but it wasn't an &quot;incredibly loud&quot; signal, just unusually loud.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1977 they couldn't turn the dishes, so they had to wait 24 hours to scan that part of the sky again and 24 hours later, and every day since, and no similar signals have ever been found, so yeah, it's a mystery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way they had it set up, they couldn't tell which dish either, so there's 2 tiny slices of the sky that it might have come from, or it could have (theoretically) come from a ship of some sort transmitting it's location back to it's home planet, and the earth just passed through (or it could have been a mouse chewing on a wire, or a Russian prank) . . . it could have been lots of things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1980 they set up the ability to have the dishes track a specific part of space, which would have come in handy in 1977, but new new signals of that strength were ever heard.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, you're right, the big ear wasn't designed to record.  As I understand it, the system was designed to eliminate background signals, so if both dish's caught similar wavelengths while pointing at slightly different parts of the sky the duplication could be eliminated.  It was what one received that the other didn't that was printed on paper, but no way to tell which dish did the receiving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some cool pictures and write-ups about the Big ear &lt;a href=&quot;http://www.bigear.org/about.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.bigear.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.  It was a monster of engineering for the 1950s when it was designed but it's primitive by today's standards.   The Wow signal has no triangulation to verify if it was from outer-space vs a closer event, and they couldn't turn their equipment to track it.   It was initially designed to look deep into space, by the way, not to record alien signals.   Seti began using it in 1973 after it was somewhat obsolete for more traditional astronomical pursuits, but throughout the 60s it was one of the better observatories.  Another problem, it didn't detect a wide range of radio waves, just a tight band, which is good for looking for a signal, but less good for eliminating rare stellar events that might have spanned a broader range of radio-frequencies.   Some possibilities &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_radio_source#The_galactic_center&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for stellar explanations and without triangulation, the possibility of rare atmospheric reflection of some kind seems possible too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 6 data points looks like a bell curve and that's exactly what we should expect if we pass through a tight signal.  It's not what I'd expect if a signal was directed at us, but passing through a tight beam would explain the bell curve of the spike.  A black hole in that part of space might explain it too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kA431.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kA431.gif&quot; alt=&quot;http://www.bigear.org/Wow30th/WowSigL.gif&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a possibility that the burst could have carried information,&#xA;  however unlikely? It could have carried a one-time &quot;message&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &quot;one time message&quot; idea strikes me as very unlikely because I don't see any point to it.   I also don't think a lot of information could be contained in 72 seconds, but some could, but there's no point in sending information for just 72 seconds and just once.  The point of sending information is so the person on the other end receives it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if it was Technology based and not natural in origin, Radio-waves can carry a lot of information, otherwise radios and antenna televisions wouldn't work, but the information is contained in signal variation usually in either amplitude or frequency, like these images below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cgZn3.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cgZn3.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BLXDW.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BLXDW.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.qrg.northwestern.edu/projects/vss/docs/communications/1-how-is-data-put-on-radio-waves.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big ear measured total signal variation between the disks every 12 seconds, so it wouldn't detect Amplitude or Frequency variations that would happen in much shorter intervals than 12 seconds.  We simply have no way of knowing.   Another possible method to transmit a information could perhaps be polarization variation in a consistent signal, but big ear couldn't pick that up either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, we simply don't have enough information to know.  It's like somebody snapped a terrible picture of something that might have happened.   You can look at the picture forever, but it will never give you any new information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More on it &lt;a href=&quot;http://www.bigear.org/wowmenu.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, and I liked this article &lt;a href=&quot;https://skeptoid.com/episodes/4342&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.  Hope that wasn't too wordy or too much already known information.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="7472" LastEditDate="2020-05-21T09:26:40.153" LastActivityDate="2020-05-21T09:26:40.153" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12774" PostTypeId="2" ParentId="12034" CreationDate="2015-12-06T04:59:42.147" Score="3" Body="&lt;p&gt;A large moon (the size of a planet) would be modeled by the fluid model, meaning friction and tensile strength are too weak to modify the shape of the moon significantly, shape is determined by rotation, vortices, self-gravitation, and tidal forces.&#xA;Smaller moons, like Phobos would likely follow a &lt;a href=&quot;https://www.boulder.swri.edu/~bottke/rubble/node2.html&quot; rel=&quot;nofollow&quot;&gt;rubble pile model&lt;/a&gt;.&#xA;On circular orbits tidal heating doesn't play a major role.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/ftp/arxiv/papers/1409/1409.3858.pdf&quot; rel=&quot;nofollow&quot;&gt;The fluid model has been studied extensively&lt;/a&gt;. The occuring shapes are known as &lt;a href=&quot;http://keith.aa.washington.edu/papers/stability.pdf&quot; rel=&quot;nofollow&quot;&gt;equilibrium figures&lt;/a&gt;.&#xA;For a full understanding, the parent body should be considered together with the moons, as one system. (Fluid) moons getting too close to the Roche limit gradually deform (tidal heating depends on inner friction), and start to shed mass at some point. There exist solutions with vortices, and vortex-free solutions, depending on the properties of the fluid. The lost mass may either &lt;a href=&quot;http://yly-mac.gps.caltech.edu/A_DPS/dps%202011%20/a_dps%202011%20program%20+%20abstracts/pdf/EPSC-DPS2011-1559.pdf&quot; rel=&quot;nofollow&quot;&gt;form an n-body system&lt;/a&gt; outside the Roche limit, or form a ring system, mostly inside the Roche limit.&#xA;Closer to the planet, it may either impact, and fuse with the planet to a the overall spheroid of the planet, or - if the planet is already rotating fast, and the mass of the moon was sufficiently high - increase the rotation of the planet, such that it may undergo e.g. the &lt;a href=&quot;http://farside.ph.utexas.edu/teaching/336L/Fluidhtml/node39.html&quot; rel=&quot;nofollow&quot;&gt;Maclaurin&lt;/a&gt; - &lt;a href=&quot;http://farside.ph.utexas.edu/teaching/336L/Fluidhtml/node40.html&quot; rel=&quot;nofollow&quot;&gt;Jacobi&lt;/a&gt; series, up to the point, that it cannot accrete more mass. &lt;a href=&quot;https://hal.archives-ouvertes.fr/jpa-00208894/document&quot; rel=&quot;nofollow&quot;&gt;There are several more possibilities in detail&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rubble piles with tensile strength are more likely to undergo sudden shape changes or disruptions than rubble piles without tensile strength. Behaviour of rubble piles without tensile strength depends on friction. The less friction, the closer to the fluid model.&#xA;The more tensile strength of the rubble pile, the more it behaves like the ridgid model. Ridgid objects may survive well below the Roche limit for fluid objects, but disrupt suddenly when tidal forces get too strong.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2015-12-06T04:59:42.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12776" PostTypeId="1" AcceptedAnswerId="12779" CreationDate="2015-12-06T08:23:29.913" Score="4" ViewCount="240" Body="&lt;p&gt;Please see this image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://astrobites.org/wp-content/uploads/2014/11/fig1.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder two things about these black holes. The first is, what are the oblong, smaller black hole-looking voids outside the perimeter of the major black holes? Also, why are these black holes not entirely circular, given that their mass is concentrated in a point?  Is this just visual distortion?&lt;/p&gt;&#xA;" OwnerUserId="7312" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:47.443" LastActivityDate="2015-12-07T22:45:08.577" Title="Two black holes photo interpretation" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12777" PostTypeId="1" AcceptedAnswerId="12778" CreationDate="2015-12-06T10:48:57.673" Score="18" ViewCount="2647" Body="&lt;p&gt;I have heard a lot of buzz about distant planets that could potentially be second homes for human existence, but what is that approximate number?&lt;/p&gt;&#xA;" OwnerUserId="9975" LastEditorUserId="9975" LastEditDate="2015-12-07T01:25:44.973" LastActivityDate="2016-03-04T17:40:58.923" Title="How many planets have we discovered that can support human life?" Tags="&lt;earth&gt;&lt;life&gt;&lt;earth-like-planet&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12778" PostTypeId="2" ParentId="12777" CreationDate="2015-12-06T11:37:13.207" Score="26" Body="&lt;p&gt;There is currently only one planet known to be capable of supporting human life, and you're on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several planets have been found in the region in which we expect water to be liquid on much of the planet. Of these, only one fits the criteria of being Earth-sized and well placed in the habitable zone: &lt;a href=&quot;http://www.nasa.gov/ames/kepler/nasas-kepler-discovers-first-earth-size-planet-in-the-habitable-zone-of-another-star/&quot;&gt;Kepler 186-f&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However we know nothing about it's atmosphere (or lack of one). The star is a red dwarf, so it could be subject to dramatic solar flares. The planet is rather colder than earth, so could be in a perpetual &quot;snowball world&quot; state, depending on the composition of the atmosphere and the strength of the greenhouse effect. The atmosphere would be very unlikely to be even close to breathable, and it is nearly 500 light-years from Earth, so could not be reached in a reasonable amount of time, even with much more advanced propulsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment we can't usually detect most Earth-like planets in the habitable zone of brighter stars like the sun, though the probably do exist and may be common.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-12-06T11:37:13.207" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12779" PostTypeId="2" ParentId="12776" CreationDate="2015-12-06T19:50:33.563" Score="4" Body="&lt;p&gt;The image is about a merger of two black holes, before the actual merger takes place.&#xA;The two largest of the dark regions are slightly distorted images of two the black holes. Distortion is due to light being bended asymmetrically by the combined gravity of the two black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two smaller dark areas show the respective other black hole in a strongly distorted version, since the light - or missing light - of the other black hole is bended around the respective less distorted appearing black hole by about 180 degrees, somewhat similar to an arrangement of uneven mirrors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may interprete the phenomenon of the smaller dark areas as an extreme version of &lt;a href=&quot;http://hubblesite.org/hubble_discoveries/science_year_in_review/pdf/2006/einstein_rings_natures_gravitational_lenses.pdf&quot; rel=&quot;nofollow&quot;&gt;Einstein rings&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="2365" LastEditDate="2015-12-07T22:45:08.577" LastActivityDate="2015-12-07T22:45:08.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12780" PostTypeId="2" ParentId="12777" CreationDate="2015-12-06T20:39:17.003" Score="11" Body="&lt;p&gt;You may be interested in the number of habitable zone planets. These are planetary-mass objects in a sort of &quot;Goldilocks&quot; region from their respective stars: This region is just right, given adequate atmospheric conditions, to possibly permit the existence of liquid water on the surface. That's not a surefire sign that the planet is capable of supporting humans, but it's certainly a good start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scientists Kane and Gellino have &lt;a href=&quot;http://www.hzgallery.org/&quot;&gt;tabulated such information at the HZ Gallery&lt;/a&gt;. As of Sep 2015, the site assesses nearly 1600 exoplanets of which 62 spend their time entirely within the habitable zone.&lt;/p&gt;&#xA;" OwnerUserId="9983" LastEditorUserId="9983" LastEditDate="2015-12-06T21:58:37.533" LastActivityDate="2015-12-06T21:58:37.533" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12781" PostTypeId="1" CreationDate="2015-12-06T21:26:07.637" Score="-1" ViewCount="1168" Body="&lt;p&gt;I want to buy some meteorite and put stones made from it into gold or platinum ring. Or maybe I can use some metal meteorite for making ring? What kinds of meteorites can I use? It must be strong, non radioactive, it will be good if it pretty look. What minerals do fit?&lt;/p&gt;&#xA;" OwnerUserId="9985" LastActivityDate="2015-12-07T02:47:45.837" Title="What types of meteorites can I use for making a wedding ring?" Tags="&lt;meteorite&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2015-12-07T15:30:26.560" ContentLicense="CC BY-SA 3.0" />
  <row Id="12782" PostTypeId="2" ParentId="12781" CreationDate="2015-12-06T21:50:10.857" Score="2" Body="&lt;p&gt;Iron-nickel meteorites are made of metal alloys, either &lt;a href=&quot;https://en.wikipedia.org/wiki/Kamacite&quot; rel=&quot;nofollow&quot;&gt;kamacite&lt;/a&gt;, or &lt;a href=&quot;https://en.wikipedia.org/wiki/Taenite&quot; rel=&quot;nofollow&quot;&gt;taenite&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm a little sceptical about whether it's a good idea to make a wedding ring of it, since it may oxidize over time, and it may contain toxic elements like cobalt, or it may cause nickel allergy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's probably better not to use it for direct contact with skin, or humidity.&#xA;However, etching it to reveal &lt;a href=&quot;https://en.wikipedia.org/wiki/Widmanst%C3%A4tten_pattern&quot; rel=&quot;nofollow&quot;&gt;Widmanstätten patterns&lt;/a&gt; may look good.&#xA;Better protect it from skin and air, e.g. by melting it into something like a glass containment. Account for different &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermal_expansion#Coefficient_of_thermal_expansion&quot; rel=&quot;nofollow&quot;&gt;coefficients of thermal expansion&lt;/a&gt; between meteoric material and containment.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2015-12-06T21:50:10.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12783" PostTypeId="1" AcceptedAnswerId="12787" CreationDate="2015-12-07T00:47:08.383" Score="2" ViewCount="483" Body="&lt;p&gt;If I have a binary object system of equal masses, why don't they always have the same circular orbit around their center of mass, like in the photo? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/D06Ke.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D06Ke.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9988" LastEditorUserId="9186" LastEditDate="2015-12-07T11:14:37.127" LastActivityDate="2015-12-07T11:14:37.127" Title="Why don't the stars in a binary star system of equal masses always orbit their center of mass in a circular orbit?" Tags="&lt;binary-star&gt;&lt;orbital-elements&gt;&lt;celestial-mechanics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12784" PostTypeId="1" AcceptedAnswerId="12797" CreationDate="2015-12-07T00:49:14.757" Score="2" ViewCount="374" Body="&lt;p&gt;In a binary star system, why does $$e_1 = e_2 = e$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $e_1$,$e_2$, and $e$ are the eccentricities of the three orbits of $m_1$,$m_2$, and the reduced mass, respectively.&lt;/p&gt;&#xA;" OwnerUserId="9988" LastEditorDisplayName="user1569" LastEditDate="2015-12-08T14:35:35.603" LastActivityDate="2015-12-08T14:35:35.603" Title="In a binary star system, what relation determines the eccentricity of the three orbits (for $m_1$, $m_2$, and the reduced mass)?" Tags="&lt;celestial-mechanics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12785" PostTypeId="2" ParentId="12781" CreationDate="2015-12-07T02:47:45.837" Score="3" Body="&lt;p&gt;I agree with @Gerold, &quot;meteorite jewelry&quot; usually means a band made of iron meteorite. The good thing about the way you are doing it with &quot;stone&quot; settings, is that it practically prevents &lt;a href=&quot;http://en.wikipedia.org/wiki/Nickel#Toxicity&quot; rel=&quot;nofollow&quot;&gt;allergic reaction&lt;/a&gt; plus less rusting. Frequent oiling also helps. But even so, get ONLY &lt;a href=&quot;http://en.wikipedia.org/wiki/Gibeon_(meteorite)&quot; rel=&quot;nofollow&quot;&gt;Gibeon&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Seymchan_(meteorite)&quot; rel=&quot;nofollow&quot;&gt;Seymchan&lt;/a&gt; meteorite material for the best patterns and rust-resistance. You MUST find a reputable dealer, some will LIE. Good Luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh there is also &lt;a href=&quot;https://en.wikipedia.org/wiki/Pallasite&quot; rel=&quot;nofollow&quot;&gt;Pallasite&lt;/a&gt; material, very pricey, which ranges in content from silicate inclusions in an Fe-Ni matrix to Fe-Ni inclusions in a silicate matrix. Pallasites are only used as stones or inlays. Far too fragile for band material. &lt;/p&gt;&#xA;" OwnerUserId="9186" LastActivityDate="2015-12-07T02:47:45.837" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12787" PostTypeId="2" ParentId="12783" CreationDate="2015-12-07T06:22:23.667" Score="4" Body="&lt;p&gt;There's no reason why 2 bodies of equal mass couldn't have elliptical orbits around each other.   There's an example of that &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_two-body_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.    &lt;a href=&quot;https://i.stack.imgur.com/ylmhG.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ylmhG.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simple way to think about this is, if two bodies of similar mass approach each other, one of two things can happen, they either have sufficient velocity to pass each other with some hyperbolic curving of both objects or there's enough gravitational attraction that they capture each other in orbit, but the shape or eccentricity of the orbit depends on the ratio between the tangential velocity and the orbital speed at the closest pass.   For a circular capture the ratio needs to be exactly 1, and that kind of exactness is rare, so we nearly always get an elliptical orbit with an orbital capture.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solar-system planet or moon formation (not moon capture) tends to be much more circular, because elliptical orbits of objects tends to cancel out as large bodies coalesce, but you're never going to get 2 similar mass objects orbiting each other in a condensing gas cloud.   Most of the mass inevitably collects in the gravitational center.  There can be, however, a tidal influence that circularizes orbits over time, never reaching a full circle, cause that's impossible, but slowly becoming more circular.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-12-07T06:22:23.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12788" PostTypeId="1" CreationDate="2015-12-07T11:47:12.650" Score="14" ViewCount="5923" Body="&lt;p&gt;What would be different for us if Earth and Moon revolved around each other like Pluto and Charon do? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YM3TW.gif&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YM3TW.gif&quot; alt=&quot;Pluto and Charon&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9997" LastActivityDate="2015-12-09T19:29:49.203" Title="What if Earth and Moon revolved around each other like Pluto and Charon?" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;earth&gt;&lt;pluto&gt;&lt;charon&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12789" PostTypeId="2" ParentId="12788" CreationDate="2015-12-07T13:26:43.617" Score="37" Body="&lt;p&gt;&lt;strong&gt;They do&lt;/strong&gt;, but due to the ratio of masses being vastly different, they seem like they would not to do so as moon seems to rotate just around (the centre of) Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio of Earth and Moon's masses is &lt;a href=&quot;http://www.wolframalpha.com/input/?i=earth%20moon%20mass%20ratio&quot;&gt;$\frac{M_{Earth}}{M_{Moon}} = 81.3$&lt;/a&gt; whereas for Pluto and Charon the same ratio is &lt;a href=&quot;http://www.wolframalpha.com/input/?i=pluto%20charon%20mass%20ratio&quot;&gt;$\frac{M_{Pluto}}{M_{Charon}} = 8.09$&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the ratio for Pluto and Charon is relatively small, the center of the system – barycenter, around which the two bodies orbit – is somewhere on a line drawn between the two celestial bodies's mass centers. But for Earth and Moon, because Earth is proportionally much heavier, the system's barycenter does not reach outside of Earth, but instead is about 4,500 kilometers from the center of Earth (see the photo below, too):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In cases where one of the two objects is considerably more massive than the other (and relatively close), the barycenter will typically be located within the more massive object. Rather than appearing to orbit a common center of mass with the smaller body, the larger will simply be seen to &quot;wobble&quot; slightly. &lt;strong&gt;This is the case for the Earth–Moon system, where &lt;em&gt;the barycenter is located on average 4,671 km from the Earth's center&lt;/em&gt;, well within the planet's radius of 6,378 km.&lt;/strong&gt; &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Barycenter&quot;&gt;Source: Wikipedia - Barycenter&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SPYK4.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SPYK4.jpg&quot; alt=&quot;Earth-Moon System Barycenter: http://astronomy.stackexchange.com/questions/11246/how-long-until-the-earth-and-moon-become-a-binary-planet&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The major effect of this co-rotational system is that Earth seems to &quot;wobble&quot; on its orbit, as mentioned in the quote from Wikipedia above.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;@JeppeStigNielsen makes a good point about differences in tidal locking in the comments below. In the Earth-Moon system, only Moon is tidally locked (which causes us to see only one face of it, so only roughly a half of it, from Earth), whereas in Pluto-Charon both of the bodies are tidally locked. Earth is not tidally locked because of the higher mass ratio between it and Moon, but the lower mass ratio Pluto-Charon system is, as the lower mass Charon has slowly changed Pluto's rotation to match its orbital movement.&lt;/p&gt;&#xA;" OwnerUserId="7015" LastEditorUserId="7015" LastEditDate="2015-12-07T14:40:26.777" LastActivityDate="2015-12-07T14:40:26.777" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="12792" PostTypeId="2" ParentId="12749" CreationDate="2015-12-07T18:08:31.347" Score="1" Body="&lt;p&gt;The key point is that the Big Bang happened everywhere at the same time, hence there is no centre (or alternatively, the centre is any point). This was not an explosion into space, it created space. (Think of a balloon, when you blow it up points on the surface all move further apart without there being a centre of the expansion on the surface itself.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A related idea is the &quot;cosmological principle&quot; (CP) which, simply put, states that, at a large scale, the universe looks the same everywhere. That means the &quot;oldest&quot; parts (as measured by how long it look radiation to get here) of the universe can be seen in any direction - and the CMB is just that and it does come from all directions. You can think of it as the burst of light that was emitted when the universe, through expansion, cooled just enough to let atomic nuclei recapture electrons. The electrons emitted photons and those have now been redshifted by the universe's expansion to match the spectrum of a body 'glowing' at about 3 degrees Kelvin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In fact there is some evidence to suggest that the CP is not valid or at least that there is a question about what scale constitutes &quot;large scale&quot;: there is more here &lt;a href=&quot;https://en.wikipedia.org/wiki/Huge-LQG#Cosmological_principle&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Huge-LQG#Cosmological_principle&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2015-12-07T18:08:31.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12793" PostTypeId="2" ParentId="12788" CreationDate="2015-12-07T18:30:54.733" Score="14" Body="&lt;h2&gt;On barycentres&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The Pluto-Charon couple is not &lt;em&gt;qualitatively&lt;/em&gt; different to the Earth-Moon couple with regards to orbits. As was pointed out in other answers, in both cases, the two bodies revolve around each other, i.e. they are best described as orbiting around their barycentre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more physical terms, the referential centred on the barycentre of the Earth-Moon system is &lt;a href=&quot;https://en.wikipedia.org/wiki/Inertial_frame_of_reference&quot;&gt;&quot;more Galilean&quot;&lt;/a&gt; than the referential centred on the Earth geometrical centre: if you do high-precision measures of physical systems on Earth, you will see some &quot;jitter&quot; that shows that the Earth is not really Galilean. Most of which is due to the Earth rotation (the jitter being most famously demonstrated by &lt;a href=&quot;https://en.wikipedia.org/wiki/Foucault_pendulum&quot;&gt;Foucault's pendulum&lt;/a&gt;) but even if you account for the rotation, you still get some residual perturbations coming from the revolution of the Earth around the Earth-Moon barycentre. (And if you fix these ones, you still get some due to the Earth revolution around the Sun -- really, the revolution of the Earth around the Solar system barycentre -- and then some because of the rotation of the Galaxy, and so on, but they are increasingly difficult to detect.)&lt;/p&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;h2&gt;About tides&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;When two round bodies orbit each other, they have a tendency to go into &quot;tidal locking&quot;: their individual rotation speed will synchronize with the revolution, so that, &lt;em&gt;in fine&lt;/em&gt;, the two bodies always keep the same hemisphere toward each other. Pluto and Charon are at that step. The Moon is also tidally locked with the Earth: we always see the same hemisphere (in fact we see a tiny bit more than half of the Moon, because it wobbles a bit). The Earth is not tidally locked... yet. But it will ultimately be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, the Earth and the Moon exert tidal forces on each other. This is most easily explained by considering orbital speed: when a very small satellite orbits around a big planet, it must go at a speed that depends on the satellite's altitude: the further the satellite is, the slowest it goes (e.g. low-orbit satellites zoom at around 8 km/s, while the Moon goes at a leisurely 1 km/s or so). But the Moon is quite bulky: its radius is a bit more than 1700 km. This means that if the Moon's centre goes at the right speed for its orbit, the rocks on the far side of the Moon are 1700 km further from the Earth, and thus are a wee bit too fast for that orbit, so they want to leave. Similarly, the rocks on the near side of the moon are 1700 km closer to the Earth, and are thus going too slowly: they tend to &quot;fall&quot; toward the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phenomenon is symmetrical: the Earth also experiences tidal forces from the Moon. In fact, both Earth and Moon experience tidal forces from the gravitational Earth-Moon couple. This engenders tides, where water moves around in response to the forces; rocks do not because they're rocks, i.e. not very fluid under normal conditions -- they would like to move, but are too rigid to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tidal forces are somehow opposed to the Earth rotating faster than the 27 days for an Earth-Moon revolution, and the rotational energy is slowly dissipated: some of it happens to be injected in the Earth-Moon gravitational coupling, which drives them apart of each other (it has been measured thanks to reflectors from space probes and Apollo missions: the Moon is fleeing us at a rate of about 38 mm per year); the rest is lost in the friction from moving water, thus ultimately converted to heat radiated into space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom-line: the Earth rotation is slowing down. For instance, a day would have lasted about 22 hours at the time of the dinosaurs (the big ones, not the birds). The slowing-down is known around time-keeping circles as the &lt;a href=&quot;https://en.wikipedia.org/wiki/%CE%94T&quot;&gt;ΔT&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even when the Earth becomes tidally locked with the Moon, there will still be tides (at least, if there is still liquid water at that time, which is not a given, since the Sun energy output is predicted to go down sharply 5 billion years from now). Indeed, the Sun-Earth couple also produces tidal forces. The Earth-Moon tidal forces are about twice stronger, so the Moon-induced tides are bigger, but in a tidally locked situation, we should still witness tides induced by the Sun -- but on a smaller scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Without the Moon, the Earth would ultimately tidally lock with the Sun and a 365-days rotation -- using today's length for a day, of course. I am not entirely clear about what should become of the Sun-Earth-Moon system in the very long term; but it appears that this is still widely open research, notably because there are other planets in the mix, leading to a very complex situation.)&lt;/p&gt;&#xA;" OwnerUserId="10006" LastActivityDate="2015-12-07T18:30:54.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12794" PostTypeId="1" CreationDate="2015-12-07T19:17:06.433" Score="2" ViewCount="50" Body="&lt;p&gt;I understand that coordinates of stars are in RA,DEC. I am looking for a mathematical expression that describes how the positions of stars change due to the expanding universe.&lt;/p&gt;&#xA;" OwnerUserId="10008" LastEditorUserId="2365" LastEditDate="2015-12-07T21:48:59.737" LastActivityDate="2015-12-07T22:02:11.687" Title="Looking for an equation that describes how star (x,y,z) positions are affected by expanding universe" Tags="&lt;expansion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12795" PostTypeId="1" CreationDate="2015-12-07T20:32:56.273" Score="3" ViewCount="88" Body="&lt;p&gt;So I got this information from a magazine , Scientific American issue Physics at the Limits. The theory is that our Universe came about, when in the &quot;Bulk Universe&quot;, which is four dimensional, a four dimensional Star collapsed and formed a black hole which became our 3 dimensional &quot;Brane Universe&quot;, in which we live. My question is, how could this one Star have contained all the material that is in our Universe ?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastActivityDate="2015-12-08T15:28:57.030" Title="Formation of our Universe" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12796" PostTypeId="2" ParentId="12794" CreationDate="2015-12-07T22:02:11.687" Score="3" Body="&lt;p&gt;The position of stars change very slowly for a couple of reasons, but not due to an expanding universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxies distant from our own, are all &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/astro/hubble.html&quot; rel=&quot;nofollow&quot;&gt;moving away from our galaxy at a rate proportional to their distance.&lt;/a&gt; The equation is simple: &#xA;$$\mathrm{speed}=H_0 \times \mathrm{distance}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $H_0$ is a constant of proportionality, with a value of about 70 km/s per kiloparsec. Since the galaxies are moving away, they don't change their position in the sky at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The nearest galaxies are exceptions to this rule, they may be moving towards us)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://spiff.rit.edu/classes/phys301/lectures/parallax/parallax.html&quot; rel=&quot;nofollow&quot;&gt;Stars in our galaxy are also moving&lt;/a&gt;. They have their own proper motion, but the motions of the stars relative to our own are essentially random. You can look up the proper motion of stars on the &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/&quot; rel=&quot;nofollow&quot;&gt;SIMBAD&lt;/a&gt; database. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars also appear to wobble due to the annual rotation of the Earth about the Sun and our consequentially changing perspective. The amount of wobble is in inverse proportion to their distance (the nearest stars appear to wobble most)   &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-12-07T22:02:11.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12797" PostTypeId="2" ParentId="12784" CreationDate="2015-12-08T07:49:03.770" Score="0" Body="&lt;p&gt;Isn't it just conservation of momentum? Without any external forces, the centre of mass of the binary system must stay in the same place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, imagine that you had a binary system, where one star had a circular orbit and the other, with equal mass, was eccentric. Clearly, as the orbits proceeded, the centre of mass, which is half way between the stars, would have a position which oscillates with time. But with no external forces applied to the binary system, this is forbidden by the conservation of linear momentum.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-12-08T07:49:03.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12798" PostTypeId="2" ParentId="12795" CreationDate="2015-12-08T15:28:57.030" Score="1" Body="&lt;p&gt;I didn't read the SciAm article, but: Any cosmological theory compatible with &quot;the&quot; mainstream big bang theory doesn't assume material to have been contained in &quot;Our Universe&quot; (needs to be defined) from the very beginning.&#xA;Instead a very hot, unified state, is assumed, which then underwent several symmetry breakings (&lt;a href=&quot;https://en.wikipedia.org/wiki/Symmetry_breaking&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Symmetry_breaking&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Higgs_mechanism&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Higgs_mechanism&lt;/a&gt;), before &quot;material&quot; came into existence, as we observe it by now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the very hypothetical, presumed four-dimensional star just needs to trigger an initial state causing the big bang. Whether &quot;Our Universe&quot; is a brane universe embedded in a higher-dimensional universe, is another mere hypothesis. Whatever &quot;star&quot; or &quot;black hole&quot; in a higher-dimensional universe would mean, the physics would be very different from the physics we can observe. For instance, inverse &lt;a href=&quot;https://en.wikipedia.org/wiki/Inverse-square_law&quot; rel=&quot;nofollow&quot;&gt;square laws&lt;/a&gt; wouldn't hold due to the additional dimension(s), hence orbital dynamics, electromagnetism needed for atoms, etc., all would either fail, or work rather different. Presuming the existence of stars in such a hypothetical universe is very daring, already.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2015-12-08T15:28:57.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12799" PostTypeId="1" AcceptedAnswerId="12800" CreationDate="2015-12-08T15:29:22.300" Score="16" ViewCount="4440" Body="&lt;p&gt;Rainbow is a property of light splitting due to the water molecule present in the clouds and is related to the light emitted by our Sun. Our sun is a yellow star. Now if we replace sun by a blue giant star or red giant star, or any other star of different color, are we still going to have rainbows? If yes is it still going to compose of seven colors? What kind of possible change will occurs, if any? &lt;/p&gt;&#xA;" OwnerUserId="9213" LastActivityDate="2015-12-18T00:01:21.100" Title="Are we still going to have rainbows if Sun is replaced by another star?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;light&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12800" PostTypeId="2" ParentId="12799" CreationDate="2015-12-08T16:01:49.503" Score="18" Body="&lt;p&gt;Rainbows would lack most blue, and some green for red stars.&#xA;For a blue star, the blue part of the rainbow would be more intense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more complex colors, the rainbow may show some gaps. A rainbow is essentially a spectrum of that star light portion, which is visible to our eyes, and to which the atmosphere is transparent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars vary in brightness. A blue giant would be large and glaring, a red dwarf faint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Colors in the rainbow would be blurred, hence closer to white, for large stars, and sharper, more distinct, for small stars, according to the angular size of the respective light source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... This all assumes, that there is still rain. With small, red stars, it would get too cold for rain. With large blue stars, Earth would heat up too much.&#xA;To adjust for these effects, the distance to the star would need to be modified.&#xA;And of course, the length of a year, and the orbital velocity may change.&#xA;This could then cause different tides, changes in volcanism, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exchanging the star could cause various other effects, too, other polar lights, effects to the ionosphere, the ozone layer, atmospheric erosion, more...&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2015-12-08T17:27:32.133" LastActivityDate="2015-12-08T17:27:32.133" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12801" PostTypeId="2" ParentId="12799" CreationDate="2015-12-08T16:30:21.067" Score="4" Body="&lt;p&gt;A proxy would be Moonbows, Lunar rainbows. I created them as a child by simply spraying a mist of water from a garden hose during a full Moon. It was much fainter and bluer in the night than on many of the long exposure professional photos &lt;a href=&quot;https://www.google.se/search?q=lunar+rainbow&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ved=0ahUKEwjo3su61MzJAhWDDSwKHSwVAsYQ_AUIBygB&amp;amp;biw=1299&amp;amp;bih=761&quot; rel=&quot;nofollow noreferrer&quot;&gt;one can find&lt;/a&gt; of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon reflects only parts of the Sun's spectrum and it generates a different kind of rainbow. I bet that this is true for different stars too. There will still be some kind of bow, for sure, as long as you can see the star or whatever reflects its light brightly enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image below is said to be natural as seen through the water spray of Victoria Falls. (I hope it is not caused by the background city lights! That they look bright only because of the long exposure needed to capture the colors out clearly). Some moonbows and starbows can be caused by atmospheric ice crystals or dust storms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6TtUa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6TtUa.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-12-08T16:36:47.627" LastActivityDate="2015-12-08T16:36:47.627" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12802" PostTypeId="2" ParentId="12799" CreationDate="2015-12-08T16:36:32.373" Score="12" Body="&lt;p&gt;The colors that we see in rainbows has to do with a few things.  First, its the variability of the index of refraction of water as a function of wavelength.  Since the index of refraction is different at different wavelengths, as light propagates through the water, the different colors split up.  This happens over a wide range of wavelengths outside of the visible range as seen below (data from &lt;a href=&quot;https://en.wikipedia.org/wiki/Optical_properties_of_water_and_ice&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, I edited the plot because excel did something strange stretching out the x-scale)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OHVf6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OHVf6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that in the visible range ($\sim 0.4\mu\text{m}-0.7\mu\text{m}$, dashed red lines), the slope of the index is not very large compared to other parts of the spectrum, so for stars of different colors, you might see a larger spread in the colors in the rainbow, allowing you do distinguish between more or less wavelengths of light depending on what part of the spectrum you are in (if you could perceive this light, I'll discuss this later).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the rainbow that we see is dependent on the sensitivity of our eyes at certain wavelengths.  There are more colors in the rainbow than we can perceive beyond red and violent (hence the names infrared and ultraviolet).  In within the visible spectrum, the rods and cones in our eyes can only distinguish between so many wavelengths.  Also, the concept of &quot;different&quot; colors outside of our visible range doesn't have much meaning either, since we don't know how they would be distinguished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, for different stars, the colors that you would see would really depend on what light is coming from the stars and where they fall on this curve.  Our visible range is fixed, so if we confine ourselves to that, the colors won't change much, but the relative brightness between them will.  And as the stars light changes color range, some colors will appear brighter than the would from the sun.  Others may be to faint to perceive.  If we had a wider range of sensitivity, it would be more colorful than we could possibly imagine in some cases.&lt;/p&gt;&#xA;" OwnerUserId="9210" LastEditorUserId="9210" LastEditDate="2015-12-09T15:01:18.283" LastActivityDate="2015-12-09T15:01:18.283" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12803" PostTypeId="2" ParentId="12661" CreationDate="2015-12-08T16:57:45.370" Score="2" Body="&lt;p&gt;The answer to the headline question is clearly: No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The full moon has an apparent diameter of about 0.5°. Its apparent motion is roughly 360°/24h, neglecting the orbital motion of Earth and Moon. Hence it apparently moves about 24h*0.5°/360° = 0.033h = 2 min.&#xA;Hence the moon moves one moon diameter in about 2 min.&#xA;(The neglected orbital motion induces an error of about 3.65% (0.997d/27.3d), divide the &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth#Rotation&quot; rel=&quot;nofollow&quot;&gt;siderial Earth day&lt;/a&gt; by a &lt;a href=&quot;https://en.wikipedia.org/wiki/Moon#Orbit&quot; rel=&quot;nofollow&quot;&gt;siderial moon orbit&lt;/a&gt;, hence an error of just a few seconds for the apparent motion of 0.5°.)&#xA;In 35 minutes, Moon's apparent motion would be about 17 moon diameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The geosynchronous satellite, in contrast, remains staying at the same apparent position, by definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets or stars move apparently about the above error estimate different from the moon, hence roughly 27.3-times slower relative to the apparent moon than a geosynchronous satellite.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2015-12-08T16:57:45.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12805" PostTypeId="2" ParentId="12799" CreationDate="2015-12-08T19:35:35.967" Score="9" Body="&lt;p&gt;Just for completeness, here's another angle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, as Joseph von Fraunhofer and many other people have demonstrated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a whole scientific discipline that, putting things somewhat simplistically, involves &lt;em&gt;making rainbows&lt;/em&gt; out of the starlight from non-Sol stars and seeing what they look like.  It is optical stellar spectroscopy.  So yes, there already are such rainbows, and people do know how they differ from the ones made by light from Sol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Get into that in detail, of course, and one learns that &quot;Sol is a yellow star&quot; and &quot;There are seven colours.&quot; are (gross) simplifications.  You've already seen from other answers here that there is a wealth of complexities, including the effect of the local star upon weather and the existence of liquid water rain, questions about what atmosphere one has and what time of day it is, and how much atmosphere (and indeed what purity of water) one is looking through.  The study of rainbows is actually quite complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that, it's quite difficult to answer such questions, because Sol is &lt;em&gt;not really&lt;/em&gt; yellow, that largely being an artefact of looking at a star &lt;em&gt;that emits all colours&lt;/em&gt; through a certain atmosphere in a certain way; and there are only &quot;seven colours&quot; if (a) one adopts certain cultural and linguistic biases and (b) one ignores the rest of the electromagnetic spectrum, which astronomers most definitely &lt;em&gt;do not&lt;/em&gt; (radio and x-ray stellar spectroscopy &lt;em&gt;also&lt;/em&gt; being scientific disciplines).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sticking to visible light for the sake of simplicity, therefore:  Here is a set of &quot;rainbows&quot; made from other stars &lt;a href=&quot;http://www.noao.edu/image_gallery/html/im0649.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;by the Kitt Peak National Observatory 0.9m Telescope on the 30th of May, 2000&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://apod.nasa.gov/apod/image/0105/obafgkm_noao_big.jpg&quot; alt=&quot;Image from NOAO/AURA/NSF&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The codes down the right hand side are astronomical stellar catalogue numbers of the individual stars whose light was used to make each one.  This should give you the beginnings of an idea of what rainbows made by different stars would look like.  But it's quite a complex thing.  Just look at what a &lt;em&gt;solar&lt;/em&gt; rainbow &lt;em&gt;really&lt;/em&gt; looks like.  Here's what a solar rainbow looks like when viewed in great detail (&lt;a href=&quot;http://www.noao.edu/image_gallery/html/im0600.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;by the McMath-Pierce Solar Observatory on the 30th of August, 2000&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://apod.nasa.gov/apod/image/0008/solarspectrum_noao.jpg&quot; alt=&quot;Image from FTS/NSO/KPNO/AURA/NSF&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not yellow, and there are a lot more than seven colours.  &amp;#9786;&lt;/p&gt;&#xA;" OwnerUserId="10024" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:47.453" LastActivityDate="2015-12-08T19:35:35.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12806" PostTypeId="2" ParentId="12799" CreationDate="2015-12-08T19:39:46.020" Score="4" Body="&lt;p&gt;For most stars (probably all stars) out there the answer is yes.  Stars are blackbody emitters, which means that the light they emit is based on their surface temperature.  A blackbody emitter emits all frequencies of light lower than a certain cutoff.  Any star with a higher surface temperature than our own Sun would have a higher frequency (bluer) peak emission.  However, they also emit more light at every frequency below that than a cooler star would.  So hotter stars will emit more red light, more blue light, more green light, more infrared light, more radio light, etc.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So our only worry are stars that are cooler.  It turns out that it's not until you reach an effective surface temperature of around 2500K that you really start to have a huge drop-off in the blue and green part of the spectrum.  There aren't any stars that I know of that are this cool.  The lower end of the temperature spectrum tends to be around 3500K for very red stars.  All the colours of the rainbow would be fainter than they would with our star, but they should still be detectable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The different compositions of stars result in certain black lines (absence of light at that frequency) in the rainbow if you look closely enough.  However, this happens even with our own star, and we can't see it until we use a spectrometer - the lines simply aren't wide enough and the rainbow is too compressed for us to notice with the naked eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this is all assuming you could hold everything else on Earth constant and switch out stars like you would a light bulb.  The biggest factor in creating rainbows are the composition of the atmosphere, the size of raindrops, and the ability to actually create rain.  All of these would begin to change once our star is replaced.  Sometimes very rapidly, as in the case of an O class star (or a star that goes supernova from being shaken too hard during transport to our solar system).&lt;/p&gt;&#xA;" OwnerUserId="10025" LastActivityDate="2015-12-08T19:39:46.020" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12807" PostTypeId="1" AcceptedAnswerId="12808" CreationDate="2015-12-08T21:10:40.470" Score="1" ViewCount="157" Body="&lt;p&gt;Forgive me if my question has an obvious answer, but I need to know the answer. I always thought that more massive/energetic objects had a stronger force of gravity than less massive ones; that is, the Sun would create a stronger gravitational pull than the Earth does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon looking into Newton's law of universal gravitation, I began to wonder whether I misunderstood this. Does the gravity between the Sun and Earth have an equal force on both? As in, does the Earth pull the Sun with as much force as the Sun does on Earth? If so, would the Sun just resist accelerating to Earth because it is extremely massive? Mass is just an object's resistance toward accelerating from a force, right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the case, would it apply to GR? Sorry for hypothesizing, I'm honestly trying to get a grasp on it.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2015-12-08T21:50:15.743" Title="Understanding gravity" Tags="&lt;gravity&gt;&lt;general-relativity&gt;&lt;newtonian-gravity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12808" PostTypeId="2" ParentId="12807" CreationDate="2015-12-08T21:50:15.743" Score="4" Body="&lt;p&gt;The magnitude of the force of gravity between two bodies is proportional to the product of their masses:&#xA;$$F=G\frac{m_1m_2}{r^2}$$&#xA;This doesn't change depending on which body you're applying the force to, i.e. if you interchange the masses. The magnitude is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What &lt;em&gt;does&lt;/em&gt; change is the direction of the force. Force is a vector quantity, denoted as $\vec{F}$ or $\mathbf{F}$. If we write the equation for gravity using proper vector notation, we have&#xA;$$\mathbf{F}=G\frac{m_1m_2}{|\mathbf{r}_1-\mathbf{r_2}|^2}\frac{\mathbf{r_1}-\mathbf{r_2}}{|\mathbf{r_1}-\mathbf{r_2}|}$$&#xA;Here, the positions of the objects are represented by vectors, $\mathbf{r}_1$ and $\mathbf{r_2}$. Additionally, $|\mathbf{x}|$ denotes the norm of a vector $\mathbf{x}$ - its magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Now&lt;/em&gt;, if you interchange the masses, the direction of the force changes, although $|\mathbf{r_1}-\mathbf{r_2}|=|\mathbf{r_2}-\mathbf{r_1}|$, because this refers to the magnitude of the vectors. So the force applied on one object is the opposite of the force applied on the other object. This is Newton's third law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The acceleration is more interesting. The force on object $1$ due to gravity is&#xA;$$F_1=m_1g_1$$&#xA;Here,&#xA;$$g_1=\frac{Gm_2}{r^2}$$&#xA;where $m_2$ is the other mass. This should tell you that $g_1\neq g_2$, &lt;em&gt;except&lt;/em&gt; when $m_1=m_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not an expert in general relativity, but I do know that it describes how spacetime curves due to the presence of one body. The solution to the Einstein Field Equations, the metric, is different for different bodies, because one piece of it, the stress-energy tensor, is different for objects of different mass/energy/etc.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2015-12-08T21:50:15.743" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12809" PostTypeId="2" ParentId="12749" CreationDate="2015-12-08T23:50:49.103" Score="3" Body="&lt;p&gt;So, I agree with the other answers, that as we look further away from our Galaxy, we see galaxies as they were in the past. Therefore, by and large, these galaxies are &lt;em&gt;younger&lt;/em&gt; than ours. That is, we see them as they were when they had an age that could be &lt;em&gt;at most&lt;/em&gt; 13.7 billion (the age of the universe) minus their distance in light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore to find the oldest objects in the universe, we must look locally. Probably the most ancient galaxies that we can study, are those that are thought to be the building blocks of larger galaxies. These are the &quot;dwarf spheroidal galaxies&quot;, several of which have been identified as companions to the Milky Way and M31, and are probably remnants of a larger population of such objects. Their stellar populations appear to consist of mostly intermediate age stars through to extremely ancient stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other ancients objects include the extremely metal-poor stars of the halo population of our Galaxy and a number of very metal poor globular clusters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Estimates for the ages of such objects are uncertain, but are likely in the range 12-13 billion years old and so only a short time after the big bang.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-12-09T12:12:05.697" LastActivityDate="2015-12-09T12:12:05.697" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12811" PostTypeId="2" ParentId="11530" CreationDate="2015-12-09T16:52:55.283" Score="0" Body="&lt;p&gt;Don't think of inflation as super luminal movement of our galaxy from others. Think about the 'stuff' that makes space-time between the galaxies simply increasing (this is inflation). This causes the light to travel through more and more 'stuff' as time is passing by. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a clump of space-time A, and another B, next to each other at the start of the Big Bang. Then add some more clumps of space-time between them. You wouldn't think that the objects within space-time A and B have actually moved. A really basic analogy would be two pool balls next to each other with an ant on each. The ants can only travel to the next ball by walking on the the balls (ignore the ground). Then imagine the same scenario, but you add another ball between them. The ants haven't moved on the balls, but they would have to travel further now to get to the ball the other ant is on. The balls in the example would represent space-time, and the ants any galaxy/star. &lt;/p&gt;&#xA;" OwnerUserId="10003" LastActivityDate="2015-12-09T16:52:55.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12812" PostTypeId="2" ParentId="12788" CreationDate="2015-12-09T19:29:49.203" Score="4" Body="&lt;p&gt;TLDr answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both answers are very good.   There's a few more details to consider if we want to look at all the what-ifs in this amusing but wacky scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Already mentioned, the ratio of size is 8 to 1, not 81 to 1, so for starters, the Charon like Moon would be much larger in the sky.   The Moon, with roughly 10 times the mass, figuring slightly greater density due to some minor compacting, would still be 2.1 times as big across, assuming the same distance, that would make it 4 times as bright in the night sky.  A full moon would be quite impressive.  Perhaps (just barely) bright enough to read by if it was a large text book.   (some people claim to be able to read by Moonlight now, most people can't, but 4 times brighter, a full moon might just be bright enough.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solar eclipses would become more frequent and last about twice as long and you might think the Earth would be slightly colder due to the Moon blocking some sunlight, but the Moon, believe it or not, radiates more heat on the earth than it blocks because the lit up moon that faces us is nearly 400 degrees F in peak daytime and it's not hard to see that a surface that temperature radiates some heat.   Not a lot, but some.  A question on that &lt;a href=&quot;https://physics.stackexchange.com/questions/89181/how-is-the-earth-heated-by-a-full-moon&quot;&gt;here&lt;/a&gt;, so a little over 4 times the energy (ignoring solar eclipse losses), about 1/2,500th the heat from the sun, might work out to 1/10th of a degree at night during a full moon.  Not a lot, certainly, but measurable to anyone with sensitive enough instruments.   The brightness and size of the Moon would obviously be more noticeable than about 1/10th of 1 degree in temperature (C not F).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A moon of that mass would slow down the Earth's rotation significantly faster, already mentioned, but this one, we have to give some thought to.    When the Moon formed it was much closer to the Earth, about 3-5 times the radius of the Earth away.  &lt;a href=&quot;http://curious.astro.cornell.edu/our-solar-system/37-our-solar-system/the-moon/the-moon-and-the-earth/31-how-close-was-the-moon-to-the-earth-when-it-formed-intermediate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.    That's outside the Hill Sphere, and the formation of the Moon left the Earth rotating very rapidly so the effects (rapidly rotating earth, very powerful lunar tides) would still be there but the Lunar tides would be 10 times greater, so we're looking at earth-quake level tides when the moon, 10 times the mass, was 3-5 earth radius away.   The Moon, because during formation it wouldn't have much angular momentum, would quickly settle into a tidally locked rotation around the Earth.   The Tidal effect on the Earth, being 10 times greater would cause (roughly) 10 times the tidal bulge on the Earth would would push the Moon away from the Earth about 10 times faster, but, at the same time, the tidal drag slowing down the Earth, would be 10 times as great too (I assume that corresponds to about 10 times as fast).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, basically, the Moon and Earth would follow the system they're in now, but it would proceed about 10 times as fast with a Moon with 10 times the mass.   The estimate (here) is that it will take about 50 billion years for the Moon to slow down the Earth enough to enter into tidal locking, so divide that by 10, we would be very close to a tidal lock today.  The Earth would rotate very slowly.  The moon would also (likely) be a bit further from the Earth and probably have a more wobbly orbit due to solar perturbations, and perhaps, escaped completely.   This is a complicated bit of mathematics that I'd prefer not to attempt (at the Moon's current mass, the Sun will go Red Giant long before either the Moon escapes or the Earth gets tidally locked but with a Moon 10 times as massive, that's probably no longer the case and either the Moon is gone or the Moon is more distant, has a more elongated orbit, and the Earth is at or close to tidally locked.   If the Moon escapes, we'd have a near earth orbit object of enormous size that could later crash into us or swing past the earth and move our orbit - either effect and simply the effect of having no moon, would be enormous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Discussion on the Moon/earth escape vs tidal locking &lt;a href=&quot;https://www.quora.com/Which-will-happen-first-the-Earth-becoming-tidally-locked-with-the-Moon-or-the-Moon-leaving-Earths-orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume full tidal locking, &lt;a href=&quot;http://community.dur.ac.uk/john.lucey/users/lunar_sid_syn.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;29.5 days&lt;/a&gt; (synodic, not sidreal) and a moon a bit further out so we might be looking at 30 something to maybe 40 days for 1 earth rotation, that's 20 days of sunshine, 20 days of night.  That would play absolute havoc on the weather systems and seasons.   Day to night would have a bigger effect than Summer vs winter, and the summer days would be scorching, though some regions might due just fine because of rainfall.   Evolution could probably adapt to that, but it doesn't sound fun to me.  The further distance might make the Moon just 3 times as bright in the night sky instead of 4.  still pretty bright though.  You'd still get 6 months of sun and 6 months of night at the poles, but for most of the earth, this would be a radical change having days and nights that long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other possible effects, Obliquity (no moon, perhaps greater, a larger ice age driver), see &lt;a href=&quot;https://en.wikipedia.org/wiki/Milankovitch_cycles#Axial_tilt_.28obliquity.29&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.  Also, if the Earth still had the Moon but the Moon was in a more elongated orbit, we'd still have tides as the Moon moved in and out to appogee and perogee.  see picture&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HwPOl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HwPOl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://jerseyastronomyclub.weebly.com/breaking-news-and-cool-facts/the-moon-illusion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line, while we might not give it much thought, a different sized moon would actually change quite a lot.  A smaller moon would move away from the Earth more slowly and earth might be able to have a 2nd moons perhaps by capture, if the moon was smaller, also we might have more aggressive ice ages and climate changes due to a greater obliquity variation and, assuming the giant impact still happens in a similar way but a smaller amount of debris (which doesn't make sense, but lets pretend), then a smaller moon wouldn't have slowed the Earth's rotation as much and the Earth might be rotating quite a bit faster, 10 or 15 hour days instead of 24.  The effects would be pretty significant.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-12-09T19:29:49.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12813" PostTypeId="1" CreationDate="2015-12-09T20:08:43.370" Score="1" ViewCount="118" Body="&lt;p&gt;So I read that if an astronaut travels at the speed of light and goes to a nearby star system , when he or she returns, he or she will have aged 10 years but 1000 years will have gone by on Earth. Travel at the speed of light is not achievable ,so I was wondering if the spacecraft was traveling at an achievable speed of 10% the speed of light , how much will the astronaut have aged and how much time has elapsed here on Earth upon the return of the astronaut ?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastActivityDate="2015-12-11T14:44:35.597" Title="Space Travel and length of time" Tags="&lt;space-travel&gt;" AnswerCount="1" CommentCount="8" ClosedDate="2015-12-25T22:01:49.580" ContentLicense="CC BY-SA 3.0" />
  <row Id="12816" PostTypeId="2" ParentId="12799" CreationDate="2015-12-10T00:49:05.250" Score="0" Body="&lt;p&gt;Yes, of course.  The passage of light through raindrops (or hosepipe spray, waterfalls etc) would be the same.  And your shadow would still point to the rainbow.&lt;/p&gt;&#xA;" OwnerUserId="10046" LastActivityDate="2015-12-10T00:49:05.250" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12817" PostTypeId="1" CreationDate="2015-12-10T03:20:27.363" Score="5" ViewCount="131" Body="&lt;p&gt;I'm doing an Astronomy project and need to terraform a moon. It is about the size of Mars and orbits a gas giant which is at a distance of 1.28 AU of its star. I though about putting a sunshade at the Lagrange point L1 between the planet and the star, but I don't think that would be consistently cooling the moon (by about 30K). For the purpose of this project, it's surface is mostly water, and the atmosphere is mostly water vapor. Any ideas how to tackle this problem? Any way to raise the albedo, or maybe getting a sunshade to work? Be it at a Lagrange point or in orbit around the moon? I still want to do this project by myself, I could just use an idea or two, maybe some initial help with the calculations. Thanks in advance :)&lt;/p&gt;&#xA;" OwnerUserId="10047" LastEditorUserId="2618" LastEditDate="2017-03-07T00:49:40.510" LastActivityDate="2017-03-07T00:49:40.510" Title="How to cool down a moon?" Tags="&lt;star&gt;&lt;orbit&gt;&lt;natural-satellites&gt;&lt;albedo&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12818" PostTypeId="1" CreationDate="2015-12-10T04:07:59.827" Score="2" ViewCount="402" Body="&lt;p&gt;This is very confusing to me, so the question might be same. Please go through it, if required i will add more text.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Einstein's theory of general relativity, photons are not technically affected by large gravitational fields; instead space and time become distorted around massive objects and the light simply follows this distorted curvature of space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&#xA;1) For example: 1 hour near a black hole, is equivalent to 100 Years on Earth. If the light travels through this distorted space and time around a black hole, how much earth time does it take for light to cross the black hole's gravity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) We say the Universe is 14 Billion Years old, and its diameter is 91 Billion Light Years approximatly. My assumption is, these age and diameter of universe is based on the measurement of the light travel (Light is one of the factor in many).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) So, if first (1) and second (2) points are right, how can we say the Universe is created 14 Billion Years ago (if light takes 100 Earth Years to cross over the black hole's gravity)?&lt;/p&gt;&#xA;" OwnerUserId="9519" LastEditorUserId="6965" LastEditDate="2015-12-10T12:21:22.777" LastActivityDate="2015-12-10T12:21:22.777" Title="Does gravity bend light, and how much time does it take for light to cross gravity of a Black Hole?" Tags="&lt;gravity&gt;&lt;black-hole&gt;&lt;universe&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12819" PostTypeId="1" CreationDate="2015-12-10T04:23:06.963" Score="3" ViewCount="273" Body="&lt;p&gt;I wrote an experiment in HTML5 and Javascript to try to show what is happening to the curvature of spacetime as a body moves through a region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As time progresses, I'm using the &lt;code&gt;g=Gm/r2&lt;/code&gt; equation to move the spacetime grid toward the object. When a grid point is within the body, I use &lt;code&gt;g(1-h/R)&lt;/code&gt; to modify the bending according to the acceleration at the surface. This assumes uniform density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is that it takes a very large mass and very long time for any sort of noticeable curvature to occur (as you'd expect). By decreasing the Earth's radius by a factor of 3, I was able to see the curvature very well as in the screenshot below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What occurred to me then is that the the warping of space would remain long after the body of mass had moved on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this warping in space ever return to truly &quot;flat&quot; or is the fact that we almost never orbit through the same absolute region of space mean we never experience this existing warping?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If light went through this &quot;wake&quot; would it bend according to the spacetime curvature created by the long distant planet? Or have I gone wrong in my model?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VBRld.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VBRld.png&quot; alt=&quot;Spacetime warping demo&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This demo uses real values except that the radius is shrunk by 3x to increase its gravitation effect.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;earth.radius = 6.371e6 / 3;&#xA;earth.mass = 5.972e24;  &#xA;spacetimeGrid.extent = 4.5e7;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You can run the demo here: &lt;a href=&quot;https://dl.dropboxusercontent.com/u/2236585/spacetime/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://dl.dropboxusercontent.com/u/2236585/spacetime/index.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9895" LastEditorUserId="9895" LastEditDate="2015-12-10T04:28:07.550" LastActivityDate="2015-12-11T22:38:31.870" Title="Does spacetime return to being flat?" Tags="&lt;gravity&gt;&lt;space-time&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12820" PostTypeId="1" CreationDate="2015-12-10T07:32:04.307" Score="1" ViewCount="903" Body="&lt;p&gt;To escape the Sun's gravity, the Earth would need sufficient amount of energy. Let's say a stellar black hole has entered our solar system and at a certain distance from earth, the black hole exerts same and equal gravitational force on the Earth as the sun is exerting. Earth feels a zero net force. Will it help earth to fly away?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the black hole had an empty stomach before eating our solar system.&lt;/p&gt;&#xA;" OwnerUserId="10049" LastEditorUserId="2365" LastEditDate="2015-12-10T22:00:08.503" LastActivityDate="2015-12-11T02:17:22.910" Title="Can earth escape sun's gravity with the help of a black hole heading towards our solar system?" Tags="&lt;solar-system&gt;&lt;earth&gt;&lt;black-hole&gt;&lt;escape-velocity&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12821" PostTypeId="2" ParentId="12817" CreationDate="2015-12-10T07:49:47.273" Score="3" Body="&lt;p&gt;Assuming the major heat input is from the star (the gas giant may also radiate significant IR) then you could seed clouds or freeze the water in order to increase the albedo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you could increase the radiative efficiency by somehow painting the side of the moon facing away from the star black. Tricky, as no doubt this hemisphere changes with time quite rapidly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If tidal heating from the gas giant is significant then you need to get the moon into a wider orbit!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Space_sunshade&quot; rel=&quot;nofollow&quot;&gt;sunshade&lt;/a&gt; at the L1 point idea is widely discussed. Usually it is lenses that are considered, to avoid big problems with radiation pressure. I'm sure you could calculate some appropriate numbers to block a certain percentage of light in your system.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2153" LastEditDate="2015-12-10T22:20:22.263" LastActivityDate="2015-12-10T22:20:22.263" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12822" PostTypeId="2" ParentId="12818" CreationDate="2015-12-10T11:15:13.357" Score="2" Body="&lt;p&gt;You are quite right: Einstein's theory says the curvature of space is locally deformed. The essence of this is captured in the spacetime 'metric', a mathematical tool that tells us what space looks like and, derived from this, what is meant by 'a straight path', which photons take. If there is no source of gravitation present, the path of a photon will be what you know intuitively as a straight line. However, for some mass concentration (e.g. a black hole, as you say), this path will be bended such that the mass concentration acts as a lens. This is immediately clear from this image from the CFHTLenS survey:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ze2nN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ze2nN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crucially to your question I think, you must remember that photons do not experience time and their speed is equal to $c$, &lt;em&gt;the&lt;/em&gt; speed of light. Photons are not unaffected by their movement through a gravitational field, mind you: but this shows up as a &lt;em&gt;gravitational redshift&lt;/em&gt; (for a time-varying potential), rather than a time delay (other than the slightly elongated path, perhaps). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides that, the age of the Universe is typically not quite measured as you say, but rather through parameter estimation in e.g. the cosmic microwave background. The effect of gravitational lensing needs to be taken into account for that, but not in the way you presume. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, remember that an extreme gravitational field such as that of a black hole is relatively rare, and even if it did delay the photon for a 1000 years that is still a tiny fraction of the age we would, according to your way of thinking, infer; the photon would have to encounter a LOT of such black holes for this 'effect' (which does not occur) to have a huge impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a general point, you seem to have some (Interstellar-induced?) misconceptions about the universe, but I think that's not for me to address here.&lt;/p&gt;&#xA;" OwnerUserId="6965" LastEditorUserId="6965" LastEditDate="2015-12-10T11:21:26.510" LastActivityDate="2015-12-10T11:21:26.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12824" PostTypeId="1" AcceptedAnswerId="12826" CreationDate="2015-12-10T13:48:33.520" Score="15" ViewCount="95750" Body="&lt;p&gt;From the time that the sun appears on the horizon, or meets it on its setting, to the time that it is fully visible, or no longer visible on its setting, how much time passes?&#xA;Secondly, is there a place in the world where a sunrise/sunset occurs over a period of a few days? Meaning, that from the time it begins to appear over the horizon until it is fully visible, a period of a few days pass without night intervening (and the same for the opposite with sunset)?&lt;/p&gt;&#xA;" OwnerUserId="10056" LastActivityDate="2018-07-02T03:24:43.453" Title="How long does a sunrise or sunset take?" Tags="&lt;the-sun&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12825" PostTypeId="2" ParentId="12824" CreationDate="2015-12-10T14:35:10.613" Score="-1" Body="&lt;p&gt;The Sun's diameter being ½ degree out of 360, I figure it is 2 minutes.&#xA;Very even precisely two, because the division of time as into minutes, very very long ago, was designed with the movement of the Sun as its base.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2015-12-10T14:35:10.613" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12826" PostTypeId="2" ParentId="12824" CreationDate="2015-12-10T21:47:36.540" Score="11" Body="&lt;p&gt;The time it takes depends on various factors: the angle that the path of the sun makes with the horizon is the main one, though there are also optical effects caused by the atmosphere have an effect too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally the closer to the equator you live, the steeper is the angle, and so the faster is the sunset. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Stellarium I did a couple of tests:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In the UK (50 degrees North) on 10th December, it took the sun 4min 47s seconds to sink below a simulated horizon.&lt;/li&gt;&#xA;&lt;li&gt;In Angloa (10 degrees south), on the same day it took 2min 26s for the sun to set.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It seems that in most populated regions, a sunset takes between 2 and 5 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are locations, close to the Antarctic circle at this time of year, at which the sun merely partially sets, and then rises again. And at the Pole, the sun moves around in horizontal circles in the sky each day. During summer there is a permanent sun, as winter approaches the sun gets closer to the horizon, and then sets over several days. (Randall calculates 38 to 40 hours in the blog that Barry links) &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-08-21T06:50:32.707" LastActivityDate="2017-08-21T06:50:32.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12827" PostTypeId="2" ParentId="12820" CreationDate="2015-12-10T21:59:04.063" Score="1" Body="&lt;p&gt;There is no need to hypothesise a black hole. A black hole's gravity is indistinguishable from that of a similarly sized star, except in the region closely surrounding the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes can't be only 1 solar mass, there is no known mechanism that could cause a single solar mass object to undergo complete collapse. Black holes have at least three times the mass of the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So could a massive star disrupt the orbit of Earth? &lt;strong&gt;Yes it could&lt;/strong&gt;. If the star were to pass by the Earth, it could cause a slingshot effect, transferring some of it's momentum to the Earth, which could cause Earth to leave solar orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the precise situation that you describe, Earth would initially be travelling in a straight line, but then would follow a complex and possibly chaotic path, as the sun and the black hole pass each other. It is quite possible that the Earth could be ejected, or transferred to an orbit of the black hole in that case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately the gaps between stars are so vast (and black holes even more widely spaced) that in the nearly 5 billion years of the Earth, not one has come close enough to cause major disruption to Earth (we know because we are still here).&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-12-10T21:59:04.063" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12828" PostTypeId="2" ParentId="1083" CreationDate="2015-12-11T00:55:34.817" Score="8" Body="&lt;p&gt;The way it works is as follows. We do detailed studies of stars in the solar neighbourhood. This establishes the local density of stars and the mix of masses they possess (called the stellar mass function). We compare that with the mass function of clusters of stars and note that to first order it appears invariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can then triangulate the problem in various ways: we can make a model for the stellar density of the Galaxy, assume it all has the same mass function and hence get a number of stars. The model may be based on crude light-to-mass conversions, but more often would be based on deep surveys of the sky - either narrow pencil beam surveys from HST, or broader surveys like SDSS, The key is to be able to count stars but also estimate how far away they are. This is highly uncertain and relies on some assumptions about symmetry to cover regions of our Galaxy we cannot probe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method is to count up bright objects that might act as tracers of the underlying stellar population (eg red giants), compare that with the number of giants in our well-studied locale, and from this extrapolate to a total number of stars, again relying on symmetry arguments for those bits of the Galaxy that are distant or obscured by dust. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A third way is to ask, how many stars have lived and died in order to enrich the interstellar medium with heavy elements (a.k.a. metals). For example, it turns out there must have been about a billion core-collapse supernovae to create all the oxygen we see. If we assume the mass function is invariant with time and that supernovae arise from stars above 8 solar masses, then we also know how many long-lived low-mass stars were born with their high-mass siblings and hence estimate how many stars exist today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number, whether it be 100 billion or 300 billion is no more accurate than a factor of a few, but probably more accurate than an order of magnitude. The main issue is that the most common stars in the Galaxy are faint M dwarfs,that contribute very little light or mass to the Galaxy, so we really are relying on an extrapolation of our local knowledge of these objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of galaxies problem is easier, though the number is less well defined. We assume that on large scales the universe is homogeneous and isotropic. We count up how many galaxies we can see in a particular area, multiply it up to cover the whole sky. The number must then be corrected for distant faint galaxies that cannot be seen. The difficult here is that we are looking into the past and the number of galaxies may not be conserved, either through evolution or mergers. So we have to try and come up with a statement like &quot;there are n galaxies in the observable universe today that are more luminous than L&quot;. I think this number is certainly only an order of magnitude estimate.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-12-11T01:10:23.547" LastActivityDate="2015-12-11T01:10:23.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12829" PostTypeId="2" ParentId="12820" CreationDate="2015-12-11T02:17:22.910" Score="1" Body="&lt;p&gt;I'm thinking, this is the gist of your question&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Earth feels a zero net force. Will it help earth to fly away?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First, zero net gravitational force between two large objects is certainly possible, well, not exactly zero, not for more than an instant anyway, but close to zero, absolutely possible, but whether it leads to an object flying away is more complicated.   It depends on the relative motion of the 3 objects.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon, for example, orbits the Earth, but from the Moon's point of view, the Sun is about 333,000 times more massive than the Earth and about 388 times further away, (on average) when the Moon is between the Earth and the Sun (390 times when the Moon is on the opposite side of the Earth, again on average.  There's some variation in there).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because gravitation drops with the square of the distance, 388 times more distant means about 151,000 times less G force at that relative distance, but with 333,000 times more mass, the Moon actually experiences over twice the gravitational tug from the Sun than it gets from the Earth, so, even though, from the Moon's surface, the Earth is much larger than the Sun, the sun's mass is sufficient to exert the greater gravitational pull.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if, by some magical power, you were to grab a hold of the Earth and stop it from moving and grab a hold of the Moon and stop it too, then let the Moon go, the Moon would fall more towards the Sun than the Earth cause the gravitational pull in that direction is over twice as much.   (Ask this great magical being not to let go of the Earth, because if he does, the Earth would fall into the Sun too).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's not quite the same as your scenario but it points out that zero net gravitation doesn't govern where an object ends up.   The Moon orbits both the Earth and the Sun, and it's in a stable orbit around the earth even though it's feeling more gravitation from the Sun.  That's because the Moon is inside the stable part of the Earth's &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hill Sphere&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your scenario, however, a passing object the mass of another star could certainly pull the Earth away from the Sun.   It wouldn't even need to achieve a net zero gravitation to accomplish that, nor would it need to be nearly so massive.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture below covers the Earth orbiting around the Sun.  If you bring the net Gravity to zero, in theory  the &quot;F&quot; in the diagram shrinks to zero and the Earth continues straight in direction V for that time period, increasing it's distance from the Sun.  &lt;a href=&quot;http://physics.bu.edu/~duffy/py105/Gravity.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YklmL.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YklmL.gif&quot; alt=&quot;http://buphy.bu.edu/~duffy/PY105/Earthsun.GIF&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth's tangential velocity relative to the sun is 30 km/s and it's escape velocity is just the square root of 2 times that, about 42.5 km/s, so an acceleration of the Earth of 12.5 km/s or moving the Earth to an orbit a bit outside Mars' orbit and keeping the velocity the same would both work (or some combination of the two).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model is a bit more complicated because a gravitational assist, which would also happen in your scenario and a gravity assist can work both ways, increasing or decreasing the orbital velocity.  It's possible, depending on direction of the pass, that a passing star could push the Earth closer to the Sun, even passing outside, if it slows the earth's velocity by gravity assist.  Drawing it away isn't the only possible outcome.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More on gravity assists here, &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/doubal.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Short&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravity_assist&quot; rel=&quot;nofollow noreferrer&quot;&gt;Longer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As James Kilfinger points out, stars passing that close is extremely extremely rare so this kind of thing, for all practical purposes, virtually never happens.   It's much more rare than a dinosaur killing meteor for example.  It's hugely unlikely.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-12-11T02:17:22.910" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12830" PostTypeId="1" CreationDate="2015-12-11T02:28:34.470" Score="3" ViewCount="679" Body="&lt;p&gt;I'm writing a program to randomly generate star systems and I have been having a difficult time figuring this out. I've found the equations I need to get the heliocentric equatorial coordinates, but I haven't been able to figure out how to determine the longitude of perihelion. Is it just the opposite of the longitude of the ascending node or is it a completely different variable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might just me, but the names for this stuff is definitely making it more confusing than it should be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is all my notes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;** = power&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P = planet's period&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a = semi-major axis&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b = semi-minor axis&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M = mean anomaly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E = Eccentric anomaly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e = eccentricity of orbit&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v = true anomaly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;r = radial distance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A = longitude of asending node&lt;/p&gt;&#xA;&#xA;&lt;p&gt;w = longitude of perihelion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i = inclination&lt;/p&gt;&#xA;&#xA;&lt;p&gt;t = time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(r, v) = polar coordinates&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(x, y, z) = Heliocentric Ecliptic coordinates&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(X, Y, Z) = Heliocentric Equalorial coordinates&lt;/p&gt;&#xA;&#xA;&lt;p&gt;eccentricity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e = sqrt(1 - (b ** 2 / a ** 2))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Period&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P = sqrt(a ** 3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kepler's Equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M = E - e * sin(E)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mean Anomaly&#xA;use as starting point for eccentric anomaly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M = (2 * pi * t) / P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;True Anomaly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v = 2 * atan(sqrt((1 + e) / (1 - e)) * tan(E / 2))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radial Distance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;r = (a(1 - e ** 2)) / (1 + e * cos(v))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heliocentric Ecliptic coordinates&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x = r(cos(A) * cos(w + v) - sin(A) * sin(w + v) * cos(i)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y = r(sin(A) * cos(w + v) + cos(A) * sin(w + v) * cos(i)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z = r * sin(w + v) * sin(i)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heliocentric Equalorial coordinates&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X = x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Y = y * cos(i) - z * sin(i)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Z = y * sin(i) + z * cos(i)&lt;/p&gt;&#xA;" OwnerUserId="6755" LastEditorUserId="286" LastEditDate="2015-12-22T20:32:27.903" LastActivityDate="2020-11-24T01:06:09.590" Title="How to get the longitude of the perihelion to find the Heliocentric Equatorial coordinates of a planet" Tags="&lt;orbit&gt;&lt;planet&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12831" PostTypeId="2" ParentId="12819" CreationDate="2015-12-11T09:36:07.190" Score="3" Body="&lt;p&gt;To answer the question&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does spacetime return to being flat?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes it does. The picture you have calculated is not correct. It is not like that in gravity a solid body plow through spacetime and creates this valley like dragging a bowling ball throw snow would. The curvature of each point is affected by the mass and changes as the mass moves. So if the mass moves away far enough from a given point in space time the curvature becomes flat again.&lt;/p&gt;&#xA;" OwnerUserId="6712" LastEditorUserId="6712" LastEditDate="2015-12-11T14:17:12.267" LastActivityDate="2015-12-11T14:17:12.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12832" PostTypeId="1" AcceptedAnswerId="12833" CreationDate="2015-12-11T10:47:05.403" Score="15" ViewCount="415" Body="&lt;p&gt;Contaminating other planets is &lt;a href=&quot;http://planetaryprotection.nasa.gov/overview&quot;&gt;a concern in space exploration.&lt;/a&gt; My question is what organisms, or small collections of organisms, from earth would be able form a self-sustaining population on Mars? I'm not worried about how they might get there, so deep sea microbes and the like which would have to be purposely transported are in scope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not the same as &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1068/could-any-known-living-organisms-on-earth-survive-on-mars&quot;&gt;this earlier question&lt;/a&gt;, as that one does not consider sustainability.&lt;/p&gt;&#xA;" OwnerUserId="9152" LastEditorUserId="9152" LastEditDate="2015-12-14T09:24:36.633" LastActivityDate="2015-12-14T09:24:36.633" Title="What earth organisms would survive if they arrived on Mars?" Tags="&lt;mars&gt;&lt;life&gt;&lt;astrobiology&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12833" PostTypeId="2" ParentId="12832" CreationDate="2015-12-11T14:02:22.983" Score="8" Body="&lt;p&gt;On the surface of Mars probably none, since it's too dry or too cold, or both, to stay active.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spores or other &lt;a href=&quot;https://en.wikipedia.org/wiki/Dormancy&quot;&gt;dormant&lt;/a&gt; forms probably could survive for centuries, until radiation will gradually destroy the organic molecules necessary to get back into an active state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are &quot;&lt;a href=&quot;http://mepag.jpl.nasa.gov/reports/Rummel_et_al_Astrobiology_14-SR-SAG2.pdf&quot;&gt;Mars Special Regions&lt;/a&gt;&quot;, where either Earth microbes or potential Martian microbes - if some exist - cannot be ruled out to be able to spread.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things might look quite different in the underground, especially in warm and wet zones, which may exist by geothermal heat and ground water.&#xA;There, at least &lt;a href=&quot;https://en.wikipedia.org/wiki/Lithotroph&quot;&gt;lithotrophic bacteria&lt;/a&gt;/&quot;chemolithoautotrophs&quot; could survive and spread.&#xA;Those could then form a basis for a &lt;a href=&quot;https://en.wikipedia.org/wiki/Food_chain&quot;&gt;food chain&lt;/a&gt; of other organisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plants dependent on sunlight, or animals dependent on oxygen would hardly survive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a list of species, see table 2 on page 895 of the &lt;a href=&quot;http://mepag.jpl.nasa.gov/reports/Rummel_et_al_Astrobiology_14-SR-SAG2.pdf&quot;&gt;above reference paper&lt;/a&gt;. The following species and &lt;a href=&quot;https://en.wikipedia.org/wiki/Genus&quot;&gt;genera&lt;/a&gt; are mentioned in &lt;a href=&quot;http://mepag.nasa.gov/reports/sr-sag/SR-SAG2%20Report_Tables_v18.pdf&quot;&gt;the table&lt;/a&gt;:&#xA;&lt;a href=&quot;https://microbewiki.kenyon.edu/index.php/Psychromonas_ingrahamii&quot;&gt;Psychromonas ingrahamii&lt;/a&gt;, &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/23389107&quot;&gt;Planococcus halocryophilus strain Or1&lt;/a&gt;, Paenisporoarcina sp. and &lt;a href=&quot;http://www.bacterio.net/chryseobacterium.html&quot;&gt;Chryseobacterium&lt;/a&gt;, Rhodotorula glutinis (yeast), &lt;a href=&quot;https://microbewiki.kenyon.edu/index.php/Colwellia_psychrerythraea&quot;&gt;Colwellia psychroerythraea&lt;/a&gt;, &lt;a href=&quot;http://www.nrcresearchpress.com/doi/abs/10.1139/m88-198#.VmrWRtLhCt8&quot;&gt;Nitrosomonas cryotolerans&lt;/a&gt;, (lichen) &lt;a href=&quot;http://bomax.botany.pl/pubs/data/article_pdf?id=2331&quot;&gt;Pleopsidium chlorophanum&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2015-12-11T14:02:22.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12834" PostTypeId="1" CreationDate="2015-12-11T14:09:00.910" Score="3" ViewCount="538" Body="&lt;p&gt;I am trying to calculate the position of a point on the Earth's surface with respect to the Sun's center at a certain date and time.&#xA;For what I am trying to do it is OK to ignore precession of Earth's orbit and precession, nutation and polar motion of the rotating Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say that I have coordinate system centered at the Sun $X,Y,Z$, where the $X$ axis points at the point of vernal equinox, the $XY$ plane is the plane of Earth's orbit around the Sun, the $Z$ axis pointing toward the ecliptic north pole and then $Y$ completes the right-handed system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I've been able to find the following orbital elements at the epoch J2000:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a=1 \ \mathrm{AU}$ — major radius &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bar{\lambda_0}=100.47^\circ$ — mean longitude at epoch&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e=0.01673$ — eccentricity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I=0^\circ$ — inclination to the ecliptic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bar{\omega}=102.93^\circ$ — longitude of perihelion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T=1 \ \text{year}$ — orbital period&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{m}{M}=3.039\times 10^{-6}$ — planetary-solar mass ratio&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now knowing this I could determine the orbit of the Earth at a certain date, however I am unable to determine Earth's orientation.&#xA;Lets say that we have ECEF coordinate $x,y,z$ system centered at Earth's center of mass with the $z$ axis pointing toward the north pole (we can assume that it coincides with Earth's rotational axis), the $x$ axis pointing toward the intersection of the prime meridian and the equator and then $y$ completing the right-handed system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the point of vernal equinox by definition the equatorial plane crosses the center of the sun.Knowing this and the axial tilt of $~23.44^\circ$ of the Earth's rotational axis I can determine the $z$ axis and $xy$ plane with respect to $X,Y,Z$ coordinates but I cannot determine the exact positions of $x$ and $y$ axes which I need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I searched a lot for data showing me the orbital elements at certain epoch AND the exact orientation of the Earth ($x,y,z$ coordinates expressed as $X,Y,Z$ at the vernal equinox) so that I can use it for my calculations but I didn't succeed. From all the things that I found I am still unable to tell what the orientation of the Earth is at certain point so I can't really determine the position of certain point on its surface at certain time with respect to the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that I am missing something or that I have some fault in my approach since I am complete noob at astronomy and I've never dealt with this before I had to find a way to perform this calculation. I will take any suggestions if there is a better way to solve the initial problem that I am trying to solve or if there is a place where I can find the data that I need.&lt;/p&gt;&#xA;" OwnerUserId="10059" LastEditorUserId="9301" LastEditDate="2016-10-17T15:01:00.863" LastActivityDate="2016-10-17T15:01:00.863" Title="Earth-Centered Earth-Fixed coordinates with respect to heliocentric coordinate system at vernal equinox" Tags="&lt;earth&gt;&lt;rotation&gt;&lt;coordinate&gt;&lt;orbital-elements&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12835" PostTypeId="2" ParentId="12832" CreationDate="2015-12-11T14:44:03.933" Score="3" Body="&lt;p&gt;I don't think there is necessarily a problem with it being dry or cold. An excellent similacrum of the Martian surface conditions is found in Antarctica. There are plenty of &lt;em&gt;anaerobic&lt;/em&gt; microbes that exist here. A brief search reveals&#xA;&lt;a href=&quot;https://www.google.co.uk/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=2&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0ahUKEwj_8q_Mg9TJAhUEdA8KHXGcAMkQFggjMAE&amp;amp;url=http%3A%2F%2Flink.springer.com%2Fcontent%2Fpdf%2F10.1007%2FBF02703452.pdf&amp;amp;usg=AFQjCNEKgzxFX_a8I5SKFEzG-28pWXhsAw&amp;amp;sig2=J3pD4naDJOoWxsfgRMSMnA&quot; rel=&quot;nofollow&quot;&gt;Shivaji (1988)&lt;/a&gt;; &lt;a href=&quot;http://journals.cambridge.org/action/displayFulltext?type=1&amp;amp;fid=222320&amp;amp;jid=ANS&amp;amp;volumeId=4&amp;amp;issueId=02&amp;amp;aid=222319&quot; rel=&quot;nofollow&quot;&gt;Franzmann &amp;amp; Rhode (1992)&lt;/a&gt;; &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/11118605&quot; rel=&quot;nofollow&quot;&gt;Dube et al. (2001)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To quote the wikipedia entry on the &lt;a href=&quot;https://en.wikipedia.org/wiki/McMurdo_Dry_Valleys#Life&quot; rel=&quot;nofollow&quot;&gt;&quot;McMurdo dry valleys&quot;&lt;/a&gt; - &quot;&lt;em&gt;Endolithic photosynthetic bacteria have been found living in the Dry Valleys, sheltered from the dry air in the relatively moist interior of rocks. Summer meltwater from the glaciers provides the primary source of soil nutrients. Scientists consider the Dry Valleys perhaps the closest of any terrestrial environment to the planet Mars, and thus an important source of insights into possible extraterrestrial life.&lt;/em&gt;&quot; I think that's quite direct.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-12-11T14:44:03.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12836" PostTypeId="2" ParentId="12813" CreationDate="2015-12-11T14:44:35.597" Score="2" Body="&lt;p&gt;You're actually describing the &lt;a href=&quot;https://en.wikipedia.org/wiki/Twin_paradox&quot; rel=&quot;nofollow&quot;&gt;twin paradox&lt;/a&gt;.&#xA;That's a little beyond &lt;a href=&quot;https://en.wikipedia.org/wiki/Special_relativity&quot; rel=&quot;nofollow&quot;&gt;Special Relativity&lt;/a&gt;, since the astronaut accelerates, and deaccelerates. Since Special Relativity treats only &lt;a href=&quot;https://en.wikipedia.org/wiki/Inertial_frame_of_reference&quot; rel=&quot;nofollow&quot;&gt;inertial frames of reference&lt;/a&gt;, accurate treatment would require some extension towards &lt;a href=&quot;https://en.wikipedia.org/wiki/General_relativity&quot; rel=&quot;nofollow&quot;&gt;General Relativity&lt;/a&gt;.&#xA;But you may consider two astronauts, one travelling with constant velocity towards Earth, the other one with the same (by amount) velocity away from the Earth, and then apply the formula for &lt;a href=&quot;https://en.wikipedia.org/wiki/Time_dilation&quot; rel=&quot;nofollow&quot;&gt;time dilation&lt;/a&gt; of Special Relativity to both astronauts, as an approximation, and to avoid the twin paradox.&#xA;Then you get the almost ubiquitous factor $\sqrt{1-\beta^2}$ of Special Relativity, with $b=v/c$ the ratio of the relative velocity to the speed of light. For $\beta=0.1$, hence 10% the speed of light, we get a factor of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{1-\beta^2}=\sqrt{1-0.1^2}=\sqrt{1-0.01}=\sqrt{0.99}\approx 0.9949874371$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide 10 years by 0.9949874371, and you get 10 years plus 18.4 days on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\beta=1$ this division would fail, implying a break-down of the formulas of Special Relativity. That's why travelling with the speed of light doesn't work for astronauts respecting Special Relativity.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2015-12-11T14:44:35.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12838" PostTypeId="2" ParentId="12764" CreationDate="2015-12-11T22:30:27.587" Score="0" Body="&lt;p&gt;Dark matter was originally hypothesised because there is a greater degree of rotational energy in galaxies than visible matter would allow for - crudely they rotate so fast that they ought to spin apart and therefore it was hypothesised that there is an additional source of gravitational force in some form of invisible matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not as a result of a discrepancy in Einstein's General Relativity. Newtonian gravitation is a very good approximation of Einstein's GR at most ordinary scales and energies and this is true in this case also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure, exactly, what your &quot;what about dark energy&quot; question means, but assuming you mean, does this imply an incompleteness in GR, then again the answer is no, or at least not necessarily, but it's a more subtle point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark energy is the hypothesised source of the energy that causes the universe to expand at an accelerating rate. It can be inserted as a term into the field equations of GR (other explanations of DE exist though) - but that is just a mathematical term, rather than an explanation of the physicality of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Einstein originally inserted such a term into his solutions - to predict an essentially static universe. When it was shown that the universe was expanding he described this as his &quot;greatest mistake&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The evidence for the accelerating rate of expansion is relatively recent and came long after Einstein's death so he never lived to see the essential idea - of a &quot;cosmological constant&quot; in his equations - revived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This explains more - &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmological_constant&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Cosmological_constant&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2015-12-11T22:30:27.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12839" PostTypeId="2" ParentId="12819" CreationDate="2015-12-11T22:38:31.870" Score="3" Body="&lt;p&gt;You model the massive object moving through space time. Remember Relativity. In this case I mean just Galilean relativity: The laws of motion are the same in any inertial frame. If I choose a frame that moves with your planet it would mean that the planet does not move at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That would make it clear that your model is wrong, a motionless planet should not create a wake in spacetime. Your planet is motionless (wrt to an inertial frame) but is leaving a wake. Your model is incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Glancing over the code you seem to be dynamically distorting your grid. But really this is a static situation. The grid distortions can be determined for any position of the planet, where the planet has been before should not need to be calculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this is not spacetime you are describing here, rather this is just Newtonian gravity dressed to look like spacetime. For the sort of mass you discribe here there is no great difference between GR and Newtonian gravity. GR is asymptotically approximated by Newtonian gravity. It is possible to model real spacetime, but it is very computationally intensive. This doesn't make your simulation uninteresting but take care in what is claimed for it. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-12-11T22:38:31.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12843" PostTypeId="1" CreationDate="2015-12-12T23:33:36.323" Score="6" ViewCount="201" Body="&lt;p&gt;I am wondering if there are any models of interstellar cloud collapse out there which take dark matter into account. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If dark matter has local perturbations or density fluctuations, then that may be a large enough effect to cause collapse of an interstellar cloud. This might increase the mass and density of a cloud enough to overcome the kinetic energy of hydrogen atoms leading to collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there dark matter density variations at the scale of interstellar clouds?&lt;/p&gt;&#xA;" OwnerUserId="7786" LastEditorUserId="9186" LastEditDate="2015-12-13T15:12:21.390" LastActivityDate="2016-04-26T10:05:26.200" Title="Can dark matter decrease the Jeans length?" Tags="&lt;dark-matter&gt;&lt;star-formation&gt;&lt;interstellar-medium&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12845" PostTypeId="1" AcceptedAnswerId="12848" CreationDate="2015-12-13T17:15:21.257" Score="12" ViewCount="206" Body="&lt;p&gt;Ok, I read about the Supernova of type Ia and I found out that there are two detonations happening. First one is in helium shell around the white dwarf and second one seems to be triggered by the first one and it is the main part of whole supernova. I did not find how this helium shell explodes at all so that`s what I am looking for.&#xA;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="9677" LastActivityDate="2015-12-13T22:15:50.493" Title="How is the first detonation in Supernove type Ia triggered?" Tags="&lt;supernova&gt;&lt;white-dwarf&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12846" PostTypeId="1" AcceptedAnswerId="12851" CreationDate="2015-12-13T17:38:16.750" Score="1" ViewCount="45" Body="&lt;p&gt;Unlike all other meteor showers, which originate from comets, Geminids originate from asteroid 3200 Phaethon. Seeing a large number of meteors last night, I noticed they seemed to be much more green than other meteors, but is there anything else I should expect to see different to other meteor showers? Brightness, size, length of trail etc?&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2015-12-14T19:03:57.907" Title="What visible differences do Geminids have to other meteors" Tags="&lt;meteor&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12847" PostTypeId="1" AcceptedAnswerId="12849" CreationDate="2015-12-13T19:44:06.340" Score="9" ViewCount="834" Body="&lt;p&gt;I've read Wikipedia and found a &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_stars_in_Canis_Minor&quot;&gt;list of stars in Canis Minor&lt;/a&gt;, and counted the number of stars in the list. I came up with 56 stars, a surprising number. I did some more &lt;a href=&quot;http://stars.astro.illinois.edu/sow/cmi-t.html&quot;&gt;research&lt;/a&gt; and found that Canis Minor was just the two binary stars orbiting together: Procyon and Gomeisa. So I came to the conclusion that the answer could be found here somewhere. Please help and tell me how many stars the constellation Canis Minor is made of. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="10079" LastEditorUserId="9186" LastEditDate="2015-12-14T08:34:42.673" LastActivityDate="2015-12-14T23:59:25.397" Title="How many stars are in the constellation Canis Minor?" Tags="&lt;star&gt;&lt;amateur-observing&gt;&lt;constellations&gt;&lt;magnitude&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12848" PostTypeId="2" ParentId="12845" CreationDate="2015-12-13T22:15:50.493" Score="10" Body="&lt;p&gt;Nobody really knows how type Ia supernovae detonate (or deflagrate) - there are a number of possibilities. The &quot;vanilla&quot; possibility is not what you state in your question, it is that the white dwarf accretes sufficient mass that it approaches the Chandrasekhar limit and becomes dense enough in its core to commence carbon burning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the emerging &lt;em&gt;diversity&lt;/em&gt; that is seen among type Ia supernovae, once thought to be a single population, suggests there may be other possibilities. There is some evidence that white dwarfs may explode at masses well below the Chandrasekhar limit. If a white dwarf in a binary accretes enough He-rich matter, this can become compressed enough to ignite He burning near the surface (this happens at a lower density threshold than Carbon burning). This then drives a shock wave into the white dwarf and the compression caused by this can ignite the carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the He &quot;explode&quot;? Well, the accreted He will form an electron-degenerate layer at the surface. A fundamental property of this degenerate gas is that the pressure is independent of the temperature. Thus, if the He ignites then at least initially, the temperature goes up but the pressure does not. Since the He fusion rate depends on something like $T^{40}$ this allows a runaway reaction to develop that could be characterised as an &quot;explosion&quot; in the surface layers.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-12-13T22:15:50.493" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12849" PostTypeId="2" ParentId="12847" CreationDate="2015-12-13T23:45:28.820" Score="11" Body="&lt;p&gt;You possibly are confused by these two entries in Wikipedia (click on the quotations to go to the original distinct entries:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Canis_Minor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Canis Minor contains only two stars &lt;strong&gt;&lt;em&gt;brighter than the fourth magnitude&lt;/em&gt;&lt;/strong&gt;, Procyon (Alpha Canis Minoris), with a magnitude of 0.34, and Gomeisa (Beta Canis Minoris), with a magnitude of 2.9.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_stars_in_Canis_Minor&quot; rel=&quot;nofollow noreferrer&quot;&gt;This is the list of [56] &lt;strong&gt;&lt;em&gt;notable stars&lt;/em&gt;&lt;/strong&gt; in the constellation Canis Minor, sorted by decreasing brightness.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first statement is about stars brighter than the 4th magnitude, of which there are two, and not about all the others. Those two stars are NOT a binary. Procyon is 11 lightyears away and Gomeisa is 170 lightyears away. Procyon itself is a binary, but for the purpose of general observation is considered to be one star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second statement is about &quot;notable stars&quot;, which is rather arbitrary. The list is 56 stars long, but in fact is one were to map every star visible to telescopes, there would be thousands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another quibble: What perchance is the exact area in the sky that is assigned to Canis Minor? This is not precisely defined. See below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ybmo5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ybmo5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;~~~~~~~~~~~~~~~~~~~~~~ and ~~~~~~~~~~~~~~~~~~~~~~&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zVbfr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zVbfr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the completely different boundaries.&lt;/p&gt;&#xA;" OwnerUserId="9186" LastEditorUserId="9186" LastEditDate="2015-12-14T00:05:31.267" LastActivityDate="2015-12-14T00:05:31.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12850" PostTypeId="2" ParentId="12847" CreationDate="2015-12-14T08:22:16.813" Score="7" Body="&lt;p&gt;Aabaakawad's answer highlights an important aspect of any constellation: They are defined by a surface area on the sky, behind which many, many stars hide. In order to answer just &lt;em&gt;how&lt;/em&gt; many stars, you'll have to ask &quot;&lt;em&gt;above which brightness threshold?&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you should accept his answer, but I have two comments:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;&lt;em&gt;…if one were to map every star visible to telescopes, there would be thousands.&lt;/em&gt;&quot; Even inside the Milky Way, 'thousands' is a severe understatement. Canis Minor spans an area of 183 deg$^2$. Although this is one of the smallest constellations, it still corresponds to roughly 0.5% of the sky. And although the direction to Canis Minor is more or less away from the center of the Milky Way, with 100 billion stars in the Milky Way this is still millions of stars. But why stop in the Milky Way? If you continue your sightline, roughly $10^{20}$ stars lie within the observable Universe, of probably infinitely (!) many more outside.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The region, and hence the area, of Canis Minor and any other constellation is exactly defined by its coordinates on the sky, as shown by Aabaakawad's first figure. His second figure is from Johann Bode's &lt;em&gt;Uranographia&lt;/em&gt; from 1801, at which time the boundaries weren't well-defined, but in 1930 the Belgian astronomer Eugène Delporte devised the modern boundaries (which at that time were along vertical and horizontal lines of right ascension and declination, but due to Earth's precession are becoming more and more skewed).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-12-14T08:45:38.377" LastActivityDate="2015-12-14T08:45:38.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12851" PostTypeId="2" ParentId="12846" CreationDate="2015-12-14T19:03:57.907" Score="2" Body="&lt;p&gt;Well done on your observations, I've been completely clouded out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are quite right, the Geminids are noted for a &lt;a href=&quot;http://www.theskyscrapers.org/meteor-showers/&quot; rel=&quot;nofollow&quot;&gt;variety of colours&lt;/a&gt;, and a good number of fireballs, though most observers note a &lt;a href=&quot;http://solarsystem.nasa.gov/planets/geminids.cfm&quot; rel=&quot;nofollow&quot;&gt;predominiance of yellow meteors&lt;/a&gt; This is partly due to their composition and density (coming from a peculiar &quot;rock comet&quot;) and partly to their velocity (35km/s) which is fairly fast. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a good chance of seeing some good trails with this shower. Brighter meteors and most fireballs will leave a trail, and the Geminids have more of these than most other showers.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-12-14T19:03:57.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12854" PostTypeId="1" AcceptedAnswerId="12856" CreationDate="2015-12-15T00:51:33.153" Score="124" ViewCount="19134" Body="&lt;p&gt;Sound can't travel through outer space. But if it could, how loud would the Sun be? Would the sound be dangerous to life on Earth, or would we barely hear it from this distance?&lt;/p&gt;&#xA;" OwnerUserId="10093" LastEditorUserId="37191" LastEditDate="2021-03-26T09:48:34.880" LastActivityDate="2021-03-26T09:48:34.880" Title="How loud would the Sun be?" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;asteroseismology&gt;&lt;sound&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="30" ContentLicense="CC BY-SA 3.0" />
  <row Id="12856" PostTypeId="2" ParentId="12854" CreationDate="2015-12-15T01:41:42.277" Score="102" Body="&lt;p&gt;The Sun is immensely loud. The surface generates thousands to tens of thousands of watts of sound power for every square meter. That's something like 10x to 100x the power flux through the speakers at a rock concert, or out the front of a police siren. Except the &quot;speaker surface&quot; in this case is the entire surface of the Sun, some 10,000 times larger than the surface area of Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite what &quot;user10094&quot; said, we do in fact know what the Sun &quot;sounds&quot; like -- instruments like SDO's HMI or SOHO's MDI or the ground-based GONG observatory measure the Doppler shift everywhere on the visible surface of the Sun, and we can actually see sound waves (well, infrasound waves) resonating in the Sun as a whole! Pretty cool, eh? Since the Sun is large, the sound waves resonate at very deep frequencies -- typical resonant modes have 5 minute periods, and there are about a million of them going all at once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resonant modes in the Sun are excited by something. That something is the tremendous broadband rushing of convective turbulence. Heat gets brought to the surface of the Sun by convection -- hot material rises through the outer layers, reaches the surface, cools off (by radiating sunlight), and sinks. The &quot;typical&quot; convection cell is about the size of Texas, and is called a &quot;granule&quot; because they look like little grains when viewed through a telescope. Each one (the size of Texas, remember) rises, disperses its light, and sinks in five minutes. That produces a heck of a racket. There are something like 10 million of those all over the surface of the Sun at any one time. Most of that sound energy just gets reflected right back down into the Sun, but some of it gets out into the solar chromosphere and corona. No one can be sure, yet, just how much of that sound energy gets out, but it's most likely between about 30 and about 300 watts per square meter of surface, on average. The uncertainty comes because the surface dynamics of the Sun are tricky. In the deep interior, we can pretend the solar magnetic field doesn't affect the physics much and use hydrodynamics, and in the exterior (corona) we can pretend the gas itself doesn't affect the physics much. At the boundary layers above the visible surface, neither approximation applies and the physics gets too tricky to be tractable (yet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of dBA, if all that leaked sound could somehow propagate to Earth, well let's see... Sunlight at Earth is attenuated about 10,000 times by distance (i.e. it's 10,000 times brighter at the surface of the Sun), so if 200 W/m2 of sound at the Sun could somehow propagate out to Earth it would yield a sound intensity of about 20 mW/m2. 0dB is about 1pW/m2 , so that's about 100dB. At Earth, some 150,000,000 kilometers from the sound source. Good thing sound doesn't travel through space, eh?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The good folks at the SOHO/MDI project created some sound files of resonant solar oscillations by speeding up the data from their instrument by 43,000 times. You can hear those here, at the &lt;a href=&quot;http://solar-center.stanford.edu/singing/#sing&quot;&gt;Solar Center website&lt;/a&gt;. Someone else did the same thing with the SDO/HMI instrument, and &lt;a href=&quot;https://www.youtube.com/watch?v=GWCJkG31h0c&quot;&gt;superposed the sounds on first-light videos from SDO&lt;/a&gt;. Both of those sounds, which sound sort of like rubber bands twanging, are heavily filtered from the data -- a particular resonant spatial mode (shape of a resonant sound) is being extracted from the data, and so you hear mainly that particular resonant mode. The actual unfiltered sound is far more cacophonous, and to the ear would sound less like a resonant sound and more like noise.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2015-12-15T06:56:11.097" LastActivityDate="2015-12-15T06:56:11.097" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="12857" PostTypeId="5" CreationDate="2015-12-15T01:58:10.520" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-12-15T01:58:10.520" LastActivityDate="2015-12-15T01:58:10.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12858" PostTypeId="4" CreationDate="2015-12-15T01:58:10.520" Score="0" Body="Questions about the properties of vacuums or perfect vacuums" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2015-12-17T14:43:58.190" LastActivityDate="2015-12-17T14:43:58.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12860" PostTypeId="1" CreationDate="2015-12-15T02:58:30.533" Score="3" ViewCount="392" Body="&lt;p&gt;Trying to create a periodic table that shows &quot;where&quot; elements are made. For example big bang, cosmic rays, stars, supernovae etc...but reading a lot about gold,and other heavy neutron rich elements, being formed in colliding neutron stars. Is there a definitive list of elements that may have been formed in neutron stars?&lt;/p&gt;&#xA;" OwnerUserId="10096" LastActivityDate="2015-12-15T10:18:18.870" Title="trying to understand stellar nucleosynthesis" Tags="&lt;nucleosynthesis&gt;&lt;star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12861" PostTypeId="1" AcceptedAnswerId="12870" CreationDate="2015-12-15T03:32:04.123" Score="3" ViewCount="1177" Body="&lt;p&gt;Gravity is the result of the curvature of space, so could the curvature actually send objects away from the source?&lt;/p&gt;&#xA;" OwnerUserId="10093" LastActivityDate="2015-12-17T23:59:24.910" Title="Why can't gravity repel things?" Tags="&lt;gravity&gt;&lt;space&gt;&lt;general-relativity&gt;&lt;space-time&gt;&lt;newtonian-gravity&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12863" PostTypeId="1" AcceptedAnswerId="12866" CreationDate="2015-12-15T05:44:44.650" Score="2" ViewCount="280" Body="&lt;p&gt;I heard that most supermassive blackholes are not formed by stars – in fact, we aren't even sure how they're formed. But could a star, with enough mass and low enough metallicity, form a supermassive black hole? Rather, would there be any stars that fit this category?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2015-12-15T09:12:13.350" Title="Can any stars ever form supermassive black holes?" Tags="&lt;star&gt;&lt;black-hole&gt;&lt;supermassive-black-hole&gt;&lt;stellar-evolution&gt;&lt;mass&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12864" PostTypeId="2" ParentId="12863" CreationDate="2015-12-15T08:20:55.153" Score="2" Body="&lt;p&gt;Very early stars were constructed out of material produced in the big bang. This would be almost totally hydrogen and helium with no &quot;metals&quot;. It is easier to form more massive stars from such material as it is more transparent to radiation. Nevertheless, the largest of these &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_population#Population_III_stars&quot; rel=&quot;nofollow&quot;&gt;population III&lt;/a&gt; stars are only mooted to be about 1000 solar masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supermassive black holes therefore need to grow (quickly), either by mergers in clusters or between galaxies, or by rapid accretion of gas onto a much lower mass &quot;seed&quot; black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exactly how is a major unsolved astrophysics problem.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-12-15T09:09:19.437" LastActivityDate="2015-12-15T09:09:19.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12866" PostTypeId="2" ParentId="12863" CreationDate="2015-12-15T09:12:13.350" Score="6" Body="&lt;p&gt;As you say, making black holes quickly in the early Universe is a major unsolved problem in astrophysics. There are various hypotheses, of which two roughly correspond to supermassive stars.  All basically involve trying to give the black hole a headstart in mass.  There isn't really enough time to grow a $100\,M_\odot$ black hole to $10^9\,M_\odot$, so the idea is to rather get something more massive than a few $\times1000\,M_\odot$. The most recent review I know of offhand is probably &lt;a href=&quot;http://arxiv.org/abs/1003.4404&quot;&gt;Volonteri (2010)&lt;/a&gt;, but I'm not up to date on the literature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, imagine a few hundred thousand solar masses of gas collapsing into a primordial galaxy.  As the gas collapses, it potentially fragments, depending on whether or not it can cool efficiently.  If so, the fragments can presumably form stars, but in a very dense cluster.  Ultimately, either (a) the massive stars in the cluster collapse into black holes that then merge into a larger black hole, which can subsequently accrete its way to supermassiveness (supermass?); or (b) the individual stars first merge in the centre, creating a star of perhaps some thousands of solar masses, which would collapse into a massive black hole that could grow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the primordial galactic cloud doesn't fragment, we expect a sort of monolithic collapse.  Somewhere in the middle, gas will start to reach hydrostatic equilibrium: a protostar forms.  But it's a protostar that's potentially accreting several solar masses of material per year (or even faster).  So what happens next depends on whether that rapidly infall of material has to time to reach local thermal equilibrium with the protostar, or if it just piles up on the outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the former, then the protostar can become very large: thousands or tens of thousands of solar masses, which, like the supermerger product above, presumably leaves a massive black hole in the end. If the small protostar evolves independently of the infalling gas, it's probably also big enough to leave a black hole, just a much smaller one: tens, maybe hundreds of solar masses.  But it's a black hole embedded in this enormous cloud of infalling gas, which potentially settles into an envelope around the black hole. This structure has been dubbed a &quot;quasi-star&quot;, and the black hole inside can grow very rapidly in this cocoon. Eventually, the envelope will evaporate/disperse, leaving the now massive black hole to continue to accrete its way to supermassiveness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that these formation mechanisms are expected to be particular to the early Universe.  Once you add even a small amount of metals to the gas, then star formation is expected to look much more like the &quot;modern&quot; Universe.  In fact, star formation in these scenarios is still far from settled.  The main reason is that you need to follow how the gas evolves from the scale of the protogalaxy all the way down to the protostars inside.  This is a range of scales something like $1\,\mathrm{AU}/10,000\,\mathrm{ly}\approx10^{-9}$, which is numerically very difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, finally, to answer the question directly, in neither of the supermassive star options do the stars really collapse in supermassive black holes.  SMBHs are so big that they can only have grown so large through accretion.  The supermassive stars would collapse into their progenitors (or &quot;seeds&quot;), which would subsequently grow to such large masses.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2015-12-15T09:12:13.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12867" PostTypeId="2" ParentId="12860" CreationDate="2015-12-15T09:59:19.043" Score="3" Body="&lt;p&gt;The elements you are talking about are created by the &lt;a href=&quot;https://en.wikipedia.org/wiki/R-process&quot; rel=&quot;nofollow noreferrer&quot;&gt;r-process&lt;/a&gt; - the rapid neutron capture onto heavy elements that can occur in neutron-rich environments. The r-process is probably responsible &lt;em&gt;to some extent&lt;/em&gt; for all the most neutron rich (i.e. more neutrons than protons) nuclei with atomic masses above that of iron and is probably responsible for all elements heavier than lead and bismuth. However, there are three broad peaks in the solar-system abundances, centred at Germanium, Xenon and Platinum (the first, second and third r-process peaks), where the r-process is thought to dominate the production over competing processes such as the s- and p-processes (plot taken from &lt;a href=&quot;http://zuserver2.star.ucl.ac.uk/~idh/PHAS2112/Lectures/2009/Nuclear/topic6.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9M5j6.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9M5j6.gif&quot; alt=&quot;Origin of the elements&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one subtracts the expected contributions from other sources one can estimate the fraction of a given element that comes from the r-process (plot taken from &lt;a href=&quot;http://zuserver2.star.ucl.ac.uk/~idh/PHAS2112/Lectures/2009/Nuclear/topic6.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ny8Vp.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ny8Vp.gif&quot; alt=&quot;Fractional contribution of the r-process&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a contemporary astrophysical problem (i.e. there is no definitive answer) for the source of the r-process. In other words, it is an unsolved mystery as to what fraction of each r-process element comes from which astrophysical location and it may well vary as a function of cosmic time or location. The two main candidates for the sites of creation for r-process elements are core-collapse supernovae, the neutrino driven winds from supernovae and neutron star mergers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main problems to be solved are the detailed physics of supernova explosions and neutron star mergers, the relative populations and numbers of these events as a function of cosmic time and location and even the nuclear physics ingredients that go into calculating the yields of such elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A recent paper by &lt;a href=&quot;http://arxiv.org/abs/1407.7039&quot; rel=&quot;nofollow noreferrer&quot;&gt;van de Voort (2015)&lt;/a&gt; performs simulations that suggests &lt;em&gt;most&lt;/em&gt; of the r-process elements in the universe could be created by neutron star mergers, and they point out that it is especially difficult for core-collapse supernovae to produce what are known as the &quot;third-peak&quot; r-process nuclei. These claims are all contentious, although the latter point about the heavier r-process elements is echoed in a number of previous works. (e.g. &lt;a href=&quot;http://arxiv.org/abs/1305.0371&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wanajo 2013&lt;/a&gt;). Nevertheless, even this is disputed; for example &lt;a href=&quot;http://arxiv.org/abs/1501.06567&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nishimura et al. (2015)&lt;/a&gt; perform simulations of rapidly rotating supernovae progenitors with strong magnetic fields and find that they &lt;em&gt;can&lt;/em&gt; reproduce the abundance pattern of all the r-process elements seen in the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if I wanted a vague stab in the dark at directly answering your question, I would choose the group of elements in the third r-process peak, with mass numbers $191 &amp;lt; A&amp;lt; 197$ as being most likely to have been (mostly) produced by neutron star mergers. This includes Osmium (192), Iridium (191, 193), Platinum (194, 195, 196) and possibly gold (197).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-12-15T10:18:18.870" LastActivityDate="2015-12-15T10:18:18.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12868" PostTypeId="1" AcceptedAnswerId="12880" CreationDate="2015-12-15T13:30:25.177" Score="19" ViewCount="9404" Body="&lt;p&gt;Having seen many pictures produced by artists of neutron stars and planets that orbit some of them, I was wondering how a pulsar would appear to a human being, in visible light (assuming the intense radiation etc. doesn't kill us in the process).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand, the pulsar's beam is projected from the star's magnetic poles rather than rotational poles, which are not necessarily in line with each other. Given that pulsars rotate extremely quickly and the beam could be visible across vast distances - such as if it were shining through the pulsar's nebula - would it appear as a straight line, curved line or perhaps a cone? This is assuming the beam can be seen in visible light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the incredible density of neutron stars and their small physical sizes, would the night sky be visibly distorted to the point where (for example) just after sunset on a hypothetical planet, one could possibly observe other planets near or behind the star that would otherwise be blocked by it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given their small surface areas, would a neutron star still appear as luminous as say, the Sun, at a similar distance? How close would you have to get to a neutron star for its apparent magnitude to match the Sun's from Earth?&lt;/p&gt;&#xA;" OwnerDisplayName="user10106" LastActivityDate="2018-06-01T02:23:17.967" Title="How Would a Neutron Star Actually Appear?" Tags="&lt;star&gt;&lt;observational-astronomy&gt;&lt;pulsar&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12869" PostTypeId="2" ParentId="12861" CreationDate="2015-12-15T13:47:26.223" Score="1" Body="&lt;p&gt;As far as our current understanding of gravity, no. The common analogy is a rubber sheet with marbles. The sheet can only be pulled downwards, so you cannot have a repelling force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To pull the sheet upwards would require something like inverse gravity, which is not yet known to current physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course dark energy seems to look like this, but as the origin of dark energy is currently completely unknown, it is not useful to speculate on this at this point in time.&lt;/p&gt;&#xA;" OwnerUserId="5308" LastActivityDate="2015-12-15T13:47:26.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12870" PostTypeId="2" ParentId="12861" CreationDate="2015-12-15T15:21:50.537" Score="4" Body="&lt;p&gt;Apart from the field-theoretical standpoint presented by Stan, one can repel objects in a sense, when taking orbital mechanics into account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravity_assist&quot; rel=&quot;nofollow&quot;&gt;slingshot maneuver&lt;/a&gt; extracts angular momentum and energy of an orbiting mass by the use of gravity. The trick here is that the probe's velocity just gets redirected in the planet's reference frame. But this results in an acceleration in the solar rest-frame that the probe essentially tries to leave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you were asking about curvature, I don't know how and if centrifugal potentials that play a role here are described by general relativity.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2015-12-15T15:21:50.537" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12871" PostTypeId="1" AcceptedAnswerId="12878" CreationDate="2015-12-15T16:07:18.000" Score="3" ViewCount="932" Body="&lt;p&gt;In our solar system, there are 8 planets orbiting a star, the Sun.&#xA;And I understand that there are about 500 confirmed solar systems out there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why is it always planets orbiting stars? Why can't it be several stars orbiting a planet, or a star orbiting a star? Why is a star by definition stationary as opposed to planets which are moving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, in that case, it wouldn't make much sense to call it a solar system, but still.&lt;/p&gt;&#xA;" OwnerDisplayName="user10113" LastEditorUserId="823" LastEditDate="2015-12-15T19:54:41.393" LastActivityDate="2021-03-01T00:31:56.323" Title="Why is it always planets orbiting stars?" Tags="&lt;star&gt;&lt;planet&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12872" PostTypeId="2" ParentId="12871" CreationDate="2015-12-15T16:11:30.927" Score="0" Body="&lt;p&gt;The gravitational pull from these stars. There is so much matter compacted into these stars that the gravitational pull is strong enough to hold all these planets together. &lt;/p&gt;&#xA;" OwnerUserId="10114" LastActivityDate="2015-12-15T16:11:30.927" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12873" PostTypeId="2" ParentId="12871" CreationDate="2015-12-15T16:38:12.480" Score="0" Body="&lt;p&gt;Planets often circle stars because they have a very strong gravitational pull. Infact, our moon orbits the earth and we can look at a solar system in the same way. The earth has enough matter inside it to keep the moon in orbit, and the sun has enough matter inside it to create a gravitational pull strong enough to keep a entire solar system in its orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't have several stars circling a planet, the pull of a single planet wouldn't be strong enough (the planet's center would be so dense nuclear reactions would start occurring and it would be a star anyway) . You can have binary star systems though, which are where 2 stars orbit around their common barycenter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically, things orbit other things with a stronger gravitational pull.&lt;/p&gt;&#xA;" OwnerUserId="9645" LastActivityDate="2015-12-15T16:38:12.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12875" PostTypeId="1" CreationDate="2015-12-15T16:52:33.293" Score="4" ViewCount="398" Body="&lt;p&gt;In other words, does everything spin counterclockwise according to their axes (in relation to each other)? I suppose my question goes for stars, moons and other objects too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then again, all you have to do is look at any rotation upside down and it goes clockwise. So depending how you're looking at it, one could say everything rotates in the same direction whether they do or not. It's all a matter of perspective, right? And in space, there is no &quot;right side up&quot;. Maybe my question is moot now that I think about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we re-drew all of Earth's maps and globes upside down, New York or Sydney would be on the &quot;west coast&quot; and the Sun would rise in the &quot;west&quot; and set in the &quot;east&quot;. Is it just by luck or maybe by the prominent early mapmakers, or that the majority of human population lives in the Northern Hemisphere, that it's given &quot;dominance&quot; on &quot;top&quot;? (I've heard that theory before)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ugh. This is all so confusing.&lt;/p&gt;&#xA;" OwnerUserId="10117" LastEditorUserId="2153" LastEditDate="2015-12-15T23:12:48.457" LastActivityDate="2016-01-14T23:38:48.447" Title="Do all planets rotate in the same direction in relation to each other?" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;planet&gt;&lt;rotation&gt;&lt;history&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12876" PostTypeId="1" CreationDate="2015-12-15T17:46:29.157" Score="3" ViewCount="265" Body="&lt;p&gt;How fast stars accelerating away from a central point? Are they increasing in acceleration or decreasing? If possible could you please provide a plot of the acceleration of stars away from the central fixed point.&lt;/p&gt;&#xA;" OwnerUserId="5533" LastEditorUserId="2531" LastEditDate="2015-12-15T19:55:30.890" LastActivityDate="2015-12-16T17:35:29.807" Title="At what rate are stars accelerating?" Tags="&lt;cosmology&gt;&lt;astrophysics&gt;&lt;speed&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12877" PostTypeId="2" ParentId="12875" CreationDate="2015-12-15T19:30:59.853" Score="1" Body="&lt;p&gt;This Q can be split into two:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Do all the planets in a stellar system always rotate around the star[s] in the same direction as the star[s] itself/themselves?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Question #1 is a duplicate of &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6183/why-do-most-of-the-planets-rotate-counterclockwise-i-e-the-same-way-the-sun&quot;&gt;Why do (most of) the planets rotate counterclockwise, i.e. the same way the Sun does?&lt;/a&gt;. As you mentioned, calling it counter-clockwise is simply due to the convention of North on top.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;How did the convention of North being the top of maps and globes get established?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Question #2 is about earth science history or the history of cartography. I encourage you to ask that as a separate Q either on the &lt;a href=&quot;https://earthscience.stackexchange.com/&quot;&gt;Earth Science Stack Exchange&lt;/a&gt; or on the &lt;a href=&quot;https://gis.stackexchange.com/&quot;&gt;Geographic Information Systems Stack Exchange&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many cultures, East was/is put on top, as a nod to the rising sun. I imagine if the convention had come from the Southern hemisphere, South or East would have ended up on top.&lt;/p&gt;&#xA;" OwnerUserId="9186" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-12-15T19:37:00.427" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12878" PostTypeId="2" ParentId="12871" CreationDate="2015-12-15T19:50:50.910" Score="12" Body="&lt;p&gt;&quot;Why is a star by definition stationary as opposed to planets which are moving.&quot; This just isn't true. Both star and planets orbit a point in the system known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Barycenter&quot; rel=&quot;noreferrer&quot;&gt;barycenter&lt;/a&gt; (a.k.a. the center of mass of the system).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because stars are much more massive, this barycenter is much closer to the center of the star than it is to the planets. Hence it &lt;em&gt;appears&lt;/em&gt; to the casual observer that the less massive object orbits around the more massive object. In fact both objects orbit the barycenter. Like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bJQ8D.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bJQ8D.gif&quot; alt=&quot;Orbit around a barycenter&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following diagram (seen on the wikipedia page above) shows the motion of the solar system barycenter with respect to the center of the Sun. Note that the barycenter is close to the center of the Sun, but often spends time outside the visible surface of the Sun. It follows this complicated path, mainly because of Jupiter but all the other planets make smaller contributions too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/enbLf.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/enbLf.png&quot; alt=&quot;Solar system barycenter&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-12-15T19:50:50.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12879" PostTypeId="2" ParentId="12876" CreationDate="2015-12-15T19:59:19.503" Score="1" Body="&lt;p&gt;No part of the universe, at least on a large scale, is believed to be accelerating away from a central point. And, in any case, the movement of stars in galaxies is dominated by the effect of galactic gravitation rather than the expansion of spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe as a whole is believed to be expanding (the observational evidence for this is generally not questioned these days), but this is also believed to be happening everywhere at once. The big bang - the moment at which this expansion started also occurred everywhere at once and so there is no central point of the expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My favourite analogy - which I have used on this site a few times now - is of the surface of a balloon as it is blown up - all the points on the surface move away from each other but there is no central point on the surface.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2015-12-15T19:59:19.503" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12880" PostTypeId="2" ParentId="12868" CreationDate="2015-12-15T20:51:49.313" Score="13" Body="&lt;p&gt;Your question is too general, you need to get to specific examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, very few neutron stars are pulsars. Pulsars are either a brief phase during a pulsar's spin-down at the start of a neutron star's life, or they are the product of the spin-&lt;em&gt;up&lt;/em&gt; of a neutron star in a binary system. Most neutron stars fall in neither of these categories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A standard neutron star will look like any other star at a similar temperature. Most of them will be very hot indeed - 100,000 K or more, though the cooling histories of neutron stars are still uncertain and depend on some exotic physics. Such an object is &quot;white hot&quot; - it emits black body radiation at all frequencies visible to the eye (as well as lots more at UV wavelengths).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How close would you have to get for it's apparent luminosity/magnitude to match the Sun? Well that depends on the size and temperature of the neutron star. Most are thought to have a diameter of 20 km. The way you would do the calculation is equate the blackbody radiative flux per unit area at a given distance to the solar radiation constant of about 1300 W per square metre. However, there are two wrinkles for a neutron star: First, the radiation is gravitationally redshifted, so the temperature we measure is lower than the temperature at the surface. Second, General Relativity tells us that we can see more than just a hemisphere of the neutron star - i.e. we can see around the back - and this increases the flux we observe. These are roughly factor of two effects, so just to get an order of magnitude estimate, ignore GR and assume a 10 km radius NS with $T=10^{5}$ K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Stefan's law for a blackbody, then at a distance $d$, we have that&#xA;$$\frac{4 \pi r^2}{4\pi d^2} \sigma T^4 = 1300\ W\ m^{-2},$$&#xA;where $\sigma$ is the Stefan-Boltzmann constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $r=10$ km, then $d=7 \times 10^{8}$ m, which is coincidentally about a solar radius. Of course this distance depends on the square of the temperature, so a younger NS with $T=10^6$ K, then $d \sim 1$ au. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the distances where the total flux at all wavelengths would be similar to that from the Sun. To do the calculation just for the visible range we need to account for the bolometric correction, which converts a visual magnitude to a bolometric magnitude. The bolometric correction for the Sun is $\sim 0$, whereas the bolometric correction for a very hot star could be -5 mag. This means that only 1% as much flux from the hot neutron star emerges in the visible band compared with sunlight. This means that the distances calculated above, if we require the &lt;em&gt;visual&lt;/em&gt; brightness of the neutron star be similar to the Sun, must be reduced by a factor of 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To turn to pulsars. Note that the pulsed radiation &lt;em&gt;does&lt;/em&gt; have an optical component and pulsed optical radiation has been seen from a number of pulsars. Optical synchrotron emission would just appear to be a periodic, intense brightening of the pulsar, as the beam sweeps across the line of sight. If you were not in the line of sight, then you would not see the pulsed optical emission. If you could observe the beam passing through nebulosity or some other medium around the pulsar then yes there may well be some effects you could see in terms of ionisation or scattered light coming from along the beam path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, the gravitational lensing effect. Yes, this should be strong close to a neutron star. The deflection angle (in radians) is given by&#xA;$$ \alpha  = \frac{4GM}{c^2 b},$$&#xA;where $b$ is how close the light passes to the neutron star and $M$ is the neutron star mass. Expressing $b$ in terms of the 10km radius of the neutron star:&#xA;$$ \alpha \simeq 0.83 \left(\frac{M}{1.4M_{\odot}}\right) \left(\frac{b}{10 km}\right)^{-1},$$&#xA;where strictly speaking this formula is only valid for $\alpha \ll 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So consider a planet directly behind the neutron star at a distance of 1 au. The light from this would only need to be bent through an angle of $\sim 2 \times 10\ km/1\ au \sim 10^{-7}$ radians in order to be seen from a planet diametrically opposite at a distance of 1 au. So this is easily possible. However, the image would likely be highly distorted, especially if the neutron star was spinning. It would not look dissimilar to this simulated black hole image, but with a bright neutron star in the middle rather than a black disc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/v65Lg.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/v65Lg.jpg&quot; alt=&quot;distorted images&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-12-15T23:28:16.987" LastActivityDate="2015-12-15T23:28:16.987" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12881" PostTypeId="2" ParentId="12868" CreationDate="2015-12-15T21:07:59.520" Score="2" Body="&lt;p&gt;I can kind of give an answer, but I welcome correction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I was wondering how a pulsar would appear to a human being, in visible&#xA;  light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It wouldn't look like much in the visible light spectrum unless there was a significant nebula, then we might see the effect of the pulsar on the nebula, but not the pulsar itself.  X-rays and radio waves aren't visible, and if the pulsar wasn't directed at us, we wouldn't see it pass through empty space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutron Stars are generally too hot for us to see.  If one was to cool down significantly, to maybe 10 or 20 thousand degrees on the surface, then it might glow visibly blue and look like the brightest star in the sky, still just a point in the sky, but the brighest point in the sky at 1 AU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But mostly they're too hot to glow in visible light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you might see from 1 AU from a Neutron Star could be the accretion disk.   Matter that falls into a Neutron Star gets very hot and the energy if impact is far greater than the energy of fission, so as matter gets close to the Neutron star and spirals in, you're probably talking x-rays and gamma rays, but you might see a visibly glowing accretion disk at some distance out, perhaps in a gradually decaying orbit.   In effect, what you could see would depend on what's around the Neutron star than it would depend on the star itself.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As I understand, the pulsar's beam is projected from the star's&#xA;  magnetic poles rather than rotational poles, which are not necessarily&#xA;  in line with each other. Given that pulsars rotate extremely quickly&#xA;  and the beam could be visible across vast distances - such as if it&#xA;  were shining through the pulsar's nebula - would it appear as a&#xA;  straight line, curved line or perhaps a cone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The problem here is, you can't see the beam.   You see light as it's pointed towards you, you can't see a light beam in space (even if it's visible light).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see a beam not pointed at you in the atmosphere because of reflection off dust and water molecules in the air.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(see little picture)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/I161O.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I161O.jpg&quot; alt=&quot;http://cache1.asset-cache.net/xt/516070391.jpg?v=1&amp;amp;g=fs1|0|FLF|70|391&amp;amp;s=1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In space, matter is far more spread out.   It's true that a pulsar can light up part of a nebula, though the nebula may also glow on it's own anyway (I'm not 100% sure on that), but a Nebula is very large and very spread out.   To see it from the naked eye, I don't think you'd see much other than perhaps a large glow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could see a pulsar beam, it takes light 8 minutes to for light to travel 1 AU, and a pulsar can rotate hundreds of times, perhaps thousands of times in 8 minutes, so if you could actually see the beam, it would be enormously curved, like a spiral.  The light itself would travel in a straight line but since the source of the light was rapidly rotating it would appear like this (picture below), if there was sufficient material for the light to reflect off of (which there probably wouldn't be, not within 1 AU).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H5AtL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H5AtL.png&quot; alt=&quot;http://orig10.deviantart.net/193f/f/2011/095/d/9/spiral_by_10binary-d3dbvut.png&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, it would look nothing like that, but if you could see the the beam, that's what it would look like.   What that spiral looks like from a single point is a pulsar, off, on, off, on, off, on, etc.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the light never travels in a spiral, it travels in a direct line away from the Pulsar, but like the water spiral &lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2013/03/23/water_spiral_amazing_video_of_water_spiraling_out_of_a_hose.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, which falls down in a straight line, but it looks like it falls in a spiral (if that makes sense).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given the incredible density of neutron stars and their small physical&#xA;  sizes, would the night sky be visibly distorted to the point where&#xA;  (for example) just after sunset on a hypothetical planet, one could&#xA;  possibly observe other planets near or behind the star that would&#xA;  otherwise be blocked by it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, for starters, without a sun there, planets would probably not be visible.   If the Neutron Star glowed brightly due to a hot accretion disk you couldn't see anything behind it cause the brightness of it would make seeing light bent around it pale by comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if the Neutron star was dark, to our eyes, then we could see gravity lensing around it, but stars, not planets cause planets would be dark.  (The moon would be very dark too, visible more by what it blocks than what it shines).   The lensing would be quite small however.   Visible lensing would only be a few times the diameter of the Neutron star, maybe 100 miles across, which, 93 million miles away is really tiny.  You might see some odd warping of a star here or there when properly lined up, but to see any interesting visible lensing you'd need a pretty powerful telescope.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given their small surface areas, would a neutron star still appear as&#xA;  luminous as say, the Sun, at a similar distance? How close would you&#xA;  have to get to a neutron star for its apparent magnitude to match the&#xA;  Sun's from Earth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Kind of touched on this above.   The Neutron Star can give off a lot of energy in it's pulsar beam, but it's mostly x-rays, not visible light.  How bright it is would depend on how much material is falling into it at the time, so there's no right answer to how close the Earth would need to be to have equal brightness.   It's a different kind of brightness too, mostly not visible light.   But there's no way to answer that question cause it depends on too many things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a Neutron star is just formed (which usually happens after a supernova so there's enormous energy released), but when the star just forms, it's maybe 12-15 miles in diameter but it's surface temperature can be (guessing) perhaps a billion degrees, though it cools very quickly.    A very young Neutron Star might emit more energy to our sun, though much of it would be in Neutrinos that would largely pass through the Earth.   But that level of energy output wouldn't last long.   It would cool down to about a million degrees within a few years.    &lt;a href=&quot;https://en.wikipedia.org/wiki/Neutron_star#Properties&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2015-12-15T21:07:59.520" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12884" PostTypeId="1" CreationDate="2015-12-16T02:42:57.943" Score="5" ViewCount="931" Body="&lt;p&gt;I am reading a book on spherical astronomy. Surprisingly, my problem is not with the mathematics. The author mentions several times words like 'eastward movement' and 'westward movement'. For example, the Earth revolves around the Sun in an eastward direction. When I look at the diagram from the perspective of someone above the north pole of the Sun, the Earth seems to go around the Sun in an anticlockwise manner. My problem is I don't understand what is meant by eastward movement ? Can someone explain what this means without ambiguity ? Is there some sort of a right-hand rule that one can follow such as Fleming's right-hand rule in electricity ?&lt;/p&gt;&#xA;" OwnerUserId="10137" LastActivityDate="2016-02-14T14:36:57.913" Title="What exactly is 'eastward movement'" Tags="&lt;fundamental-astronomy&gt;&lt;movement&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12885" PostTypeId="2" ParentId="12884" CreationDate="2015-12-16T07:38:09.423" Score="1" Body="&lt;p&gt;From the point of view of an observer hovering above the Earth's surface, the rotation of the Earth is eastward. That is, points on the surface appear to move in an easterly direction as the Earth spins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extending this to the orbital motion, there are at least two possible interpretations. One is that the spin angular momentum and orbital angular momentum are in the same direction. This is roughly true (and certainly closer to true than the opposite). Another interpretation would be that the Earth moves around in its orbit such that it heads in an easterly direction when referred to a longitude on the Sun. This is also approximately true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are asking what is a general definition of the north pole, then I suppose it could be the direction of the angular momentum vector. Then yes indeed, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Right-hand_rule&quot; rel=&quot;nofollow&quot;&gt;right hand grip rule&lt;/a&gt; tells you the spin is eastward as it relates to north.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-12-16T11:35:47.993" LastActivityDate="2015-12-16T11:35:47.993" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12886" PostTypeId="1" AcceptedAnswerId="12948" CreationDate="2015-12-16T09:48:13.130" Score="5" ViewCount="1902" Body="&lt;p&gt;I am working on an astrology project (don't panic, my question is about astronomy or math), where I need to figure out some values by the longnitude of the planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a program for this, the &lt;a href=&quot;http://www.astro.com/swisseph/swephinfo_e.htm&quot; rel=&quot;noreferrer&quot;&gt;Swiss ephemeris&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this, I can get the position of the planets by the time and the geolocation of the observer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I need to do is to do it backward, so I need to get the exact date and time by the given position of the sun and the position of the observer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, today (2015.12.16 00:00:00 UT) the longitude of the sun is 263.6893755 from postion (lng,lat,alt) 19.2,47.29,0 on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need a formula, what tells me, what was the UT time from this longitude of sun -88 degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody give me a clue, how can I calculate this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made some research of course on the net, and I found the related &lt;a href=&quot;https://en.wikipedia.org/wiki/Position_of_the_Sun&quot; rel=&quot;noreferrer&quot;&gt;wiki&lt;/a&gt; article, but if I am honest, there are too many unknown terms and phrase here, what I really do not understand, and I do not want to dig in too deep. &lt;/p&gt;&#xA;" OwnerUserId="10055" LastActivityDate="2021-05-27T11:08:41.810" Title="Get date and time by position of the sun and the observer position" Tags="&lt;positional-astronomy&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12888" PostTypeId="2" ParentId="12854" CreationDate="2015-12-16T12:16:13.527" Score="29" Body="&lt;p&gt;While Sir Cumference's post is a very intriguing answer, but I'm afraid it's wrong. The sun's surface is clearly in motion, but that does not necessarily result in the radiation of audible sound, even if the sun and earth where in a fluid medium (such as a air) that would allow sound transfer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain why, we can actually apply the same line of analysis to the earth's ocean. The surface moves a lot, so sound should be radiated. However, we hear nothing unless you are really close by and have breaking waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's run the math with rough numbers: The ocean has a surface area of about 510 million square kilometers. $150 \cdot 10^{12} m^2$. Let's say the average wave height is 1m and the average wave frequency is 0.1 Hz (1 wave every 10 s). If the ocean were a spherical source this would create a sound power of $5 \cdot 10^{24} W$ and the sound pressure at 1000 km away would be 240 dB SPL. That's obviously not the case, otherwise we'd all be  dead. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why not? In order for sound to actually radiate, the surface must move uniformly. For every ocean wave that moves air up there is a wave nearby that moves air down and so the contributions simply cancel. Technically speaking, we need to calculate the power by integrating the normal intensity over the entire surface, the intensity has equal amounts of positive and negative components and the sum over those is zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the same reason why you put a loudspeaker in a box: in open air, the air motion from the front of the cone and from the rear of the cone will simply cancel out, so you put it in a box to get rid of the sound from the rear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think the real answer here is: you would hear absolutely nothing since the sound contributions from different parts of the sun's surface would cancel each other out. Sound radiation over that distance would only occur if the sun's surface moves uniformly, i.e. the whole sun expands or contracts. That does happen to some degree but only at very, very low frequencies which are inaudible and where sound radiation is a lot less efficient. &lt;/p&gt;&#xA;" OwnerUserId="10127" LastEditorUserId="11622" LastEditDate="2018-08-24T05:34:25.457" LastActivityDate="2018-08-24T05:34:25.457" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="12890" PostTypeId="1" CreationDate="2015-12-16T16:06:22.617" Score="7" ViewCount="845" Body="&lt;p&gt;It is known that a physicist must have, at least in general, a good level in mathematics in order to research and understand other people's works. Linear algebra is a good example, as ODEs and calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose that some astrophysicists have deep knowledge of abstract mathematics and there are some applications of it (for example, I've been told that there is a connection between topology and cosmology). I wondered if there are some field in Astrophysics in which abstract algebra (i.e. fields, rings, category theory...) has been applied.&lt;/p&gt;&#xA;" OwnerUserId="8011" LastEditorUserId="6" LastEditDate="2016-05-25T15:37:51.520" LastActivityDate="2016-05-28T19:27:25.870" Title="Are there real uses of abstract algebra in astrophysics?" Tags="&lt;astrophysics&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12891" PostTypeId="2" ParentId="12876" CreationDate="2015-12-16T17:35:29.807" Score="2" Body="&lt;p&gt;As has already been mentioned, stars within a galaxy do not generally expand. The stars in a galaxy are gravitationally bound together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as you observe more and more distant galaxies, you see a general trend that distant galaxies are moving away from us at a speed that's proportional to their distance.  (There are some small variations due to the random motion of individual galaxies; for example, the Andromeda galaxy is approaching us, and will collide with the Milky Way in about 4 billion years.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does not indicate that everything is expanding from some central point.  An observer in a galaxy a billion light-years away would see the same thing, all other galaxies receding at a speed proportional to the distance from the observer.  This can be interpreted as space itself expanding.  Going into more detail would require General Relativity, which I'm not competent to explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question is about the rate of expansion.  The answer to that is what's called the Hubble Constant, which describes the ratio between a galaxy's speed of recession and its distance from us.  The current best estimate of this ratio is 67 kilometers per second per megaparsec.  A parsec is about 3.26 light-years; a galaxy one million parsecs away will be moving away from us at about 67 kilometers per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble%27s_law#Determining_the_Hubble_constant&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Hubble%27s_law#Determining_the_Hubble_constant&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="53" LastActivityDate="2015-12-16T17:35:29.807" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12892" PostTypeId="1" AcceptedAnswerId="12893" CreationDate="2015-12-16T17:46:22.980" Score="4" ViewCount="109" Body="&lt;p&gt;So I had some problem and I think I solved it, then I began to dig out things around it and I found some equations using red shift etc. to get speed of the object in Universe relative to us and then I found this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;distance=speed of light/Hubble constant*red shift&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It took some time before I found the thing which seems to be wrong with this thing. &#xA;How can I get the distance traveled if there is no time mentioned?&#xA;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="9677" LastActivityDate="2015-12-16T17:54:25.750" Title="Why is there no time in the distance equation using Hubble constant, red shift and speed of light?" Tags="&lt;redshift&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12893" PostTypeId="2" ParentId="12892" CreationDate="2015-12-16T17:54:25.750" Score="2" Body="&lt;p&gt;Are you familiar with Hubble's Law? Because this is just it, in a slightly less familiar form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the distance in this formula is the proper distance, the distance between us and some source, not the distance travelled. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than that, the speed of light in a sense defines our conception of time and space, so it is mentioned in the formula. ;)&lt;/p&gt;&#xA;" OwnerUserId="6965" LastActivityDate="2015-12-16T17:54:25.750" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12894" PostTypeId="2" ParentId="11129" CreationDate="2015-12-16T19:00:04.367" Score="1" Body="&lt;p&gt;I think it would be The Europa Jupiter System Mission – Laplace (EJSM/Laplace) was a proposed joint NASA/ESA unmanned space mission slated to launch around 2020 for the in-depth exploration of Jupiter's moons.&lt;/p&gt;&#xA;" OwnerUserId="10161" LastActivityDate="2015-12-16T19:00:04.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12895" PostTypeId="2" ParentId="11129" CreationDate="2015-12-16T19:29:19.700" Score="1" Body="&lt;p&gt;In 2026 or thereabouts it is entirely possible that the Square Kilometre Array could pick up radio communications, at the level at which we leak stuff into space, from any intelligent civilisations that are within around 150 light years - see the last paragraph of my answer to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/8146/how-far-away-could-we-detect-that-earth-has-life&quot;&gt;how far away could we detect that Earth has life?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-12-16T19:29:19.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12896" PostTypeId="1" CreationDate="2015-12-16T23:20:42.090" Score="8" ViewCount="559" Body="&lt;p&gt;Is there graphic of typical temporal paths that stars take through the Hertzsprung-Russell diagram?&lt;/p&gt;&#xA;" OwnerUserId="10165" LastActivityDate="2015-12-20T08:45:54.750" Title="Temporal path through Hertzsprung-Russell diagram?" Tags="&lt;stellar-evolution&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12897" PostTypeId="1" AcceptedAnswerId="12912" CreationDate="2015-12-16T23:26:09.507" Score="9" ViewCount="14913" Body="&lt;p&gt;I know that asteroids are huge chunks of rock, orbiting a solar system. Do asteroids have a gravitational field and do they gravitationally attract each other to form planets?&lt;/p&gt;&#xA;" OwnerUserId="10164" LastActivityDate="2015-12-23T12:03:12.980" Title="Do asteroids have a gravitational field?" Tags="&lt;gravity&gt;&lt;planet&gt;&lt;asteroids&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12898" PostTypeId="2" ParentId="12897" CreationDate="2015-12-16T23:28:48.917" Score="6" Body="&lt;p&gt;Sure. &lt;em&gt;Any&lt;/em&gt; mass has its gravitational field. However, its size is proportional to the mass, so as most asteroids have little mass, they have little gravitational field, and therefore pull only very slightly at each other, resulting in not enough effect to get them to lump together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, their difference in momentum/speed is too large to be removed by the little pull of the gravitation between them.&lt;/p&gt;&#xA;" OwnerUserId="10084" LastActivityDate="2015-12-16T23:28:48.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12899" PostTypeId="1" AcceptedAnswerId="12900" CreationDate="2015-12-17T00:09:23.547" Score="7" ViewCount="118" Body="&lt;p&gt;I've heard of the magnetic field of the earth reversing poles over the course of thousands of years but do stars and other planets (specifically gas giants) also experience this? If so how does it differ and if not why?&lt;/p&gt;&#xA;" OwnerUserId="10166" LastActivityDate="2015-12-17T00:41:31.313" Title="Do the magnetic fields of stars and/or other planets reverse?" Tags="&lt;star&gt;&lt;planet&gt;&lt;magnetic-field&gt;&lt;gas-giants&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12900" PostTypeId="2" ParentId="12899" CreationDate="2015-12-17T00:41:31.313" Score="3" Body="&lt;p&gt;Sure. &lt;a href=&quot;http://www.nasa.gov/content/goddard/the-suns-magnetic-field-is-about-to-flip/&quot; rel=&quot;nofollow&quot;&gt;It even happens with the Sun&lt;/a&gt;, as part of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_cycle&quot; rel=&quot;nofollow&quot;&gt;solar cycle&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth's magnetic field is described by something called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Dynamo_theory&quot; rel=&quot;nofollow&quot;&gt;dynamo theory&lt;/a&gt;, which states that the movement of fluids in the core generates the magnetic field. A modification of this is used to model the magnetic fields of stars. Given that the cause of &lt;a href=&quot;https://en.wikipedia.org/wiki/Geomagnetic_reversal&quot; rel=&quot;nofollow&quot;&gt;geomagnetic reversals&lt;/a&gt; is related to the movement of currents deep on Earth's core, it makes sense that the same would be the case in other bodies.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2015-12-17T00:41:31.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12901" PostTypeId="1" AcceptedAnswerId="12902" CreationDate="2015-12-16T22:06:07.907" Score="1" ViewCount="2840" Body="&lt;p&gt;What is the difference between a neutron star and a white dwarf? I know that both are very dense even if they go through different phases.&lt;/p&gt;&#xA;" OwnerUserId="10164" OwnerDisplayName="Austin Phillips" LastEditorUserId="2153" LastEditDate="2016-01-10T13:17:41.337" LastActivityDate="2020-11-26T17:06:32.677" Title="What is the difference between a neutron star and a white dwarf?" Tags="&lt;neutron-star&gt;&lt;white-dwarf&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12902" PostTypeId="2" ParentId="12901" CreationDate="2015-12-16T22:10:04.537" Score="2" Body="&lt;p&gt;In a neutron star, the force of gravity is strong enough to press the protons and electrons together to form neutrons &lt;a href=&quot;http://www.space.com/22180-neutron-stars.html&quot; rel=&quot;nofollow&quot;&gt;{1}&lt;/a&gt;, White dwarfs are only very compact. With even more mass, you get a black hole.&lt;br&gt;&#xA;All three types are outcomes of star death, when the failing fusion in the middle of a star is no longer able to counteract gravity. What a star become when it collapses is depending on its mass.&lt;/p&gt;&#xA;" OwnerUserId="10172" OwnerDisplayName="Hohmanfan" LastActivityDate="2015-12-17T05:34:11.817" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12903" PostTypeId="1" AcceptedAnswerId="12910" CreationDate="2015-12-17T07:13:15.143" Score="9" ViewCount="18110" Body="&lt;p&gt;As we know, according to &lt;a href=&quot;https://en.wikipedia.org/wiki/Inner_core&quot;&gt;Wikipedia on &lt;strong&gt;Earth's inner core&lt;/strong&gt;&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Earth's inner core is the Earth's innermost part and according to seismological studies, it is primarily a solid ball with a radius of about 1220 kilometers, or 760 miles (about 70% of the Moon's radius). It is believed to consist primarily of an iron–nickel alloy and to be approximately the same temperature as the surface of the Sun: approximately 5700 K (5400 °C).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now the question is, do all planets have molten inner cores?&lt;/p&gt;&#xA;" OwnerUserId="10164" LastActivityDate="2019-01-04T08:56:15.017" Title="Do all planets have a molten core?" Tags="&lt;planet&gt;&lt;core&gt;" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12904" PostTypeId="2" ParentId="12896" CreationDate="2015-12-17T07:34:01.410" Score="6" Body="&lt;p&gt;This is typically a bit tricky to show because (1) some phases (e.g. the main sequence) are much, much longer lived than other phases and (2) a number of phases are spaced very close together in HR-space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be that as it may: any elementary text book on stellar structure will guide you through the path of a star through the HR diagram, usually accompanied by pictures. The book by Prialnik on this subject matter, or Ostlie &amp;amp; Carroll, are both excellent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a very crude version of the temporal path through the HR diagram I made some time ago. Numbers show the different phases (2 is MS), colorcoding shows the ages.&lt;a href=&quot;https://i.stack.imgur.com/MXumb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MXumb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6965" LastActivityDate="2015-12-17T07:34:01.410" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12906" PostTypeId="2" ParentId="12861" CreationDate="2015-12-17T09:31:35.543" Score="2" Body="&lt;p&gt;I'm not sure how accepted this is in the physics community, but a book by Andrew Thomas called &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B008ABSSIW&quot; rel=&quot;nofollow&quot;&gt;Hidden in Plain Sight&lt;/a&gt; posits a theory that gravity may be repulsive within black holes (i.e. on scales smaller than the Schwartzchild radius for a given mass).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's an interesting theory, and it avoids the need for a singularity, but to my knowledge it's completely untestable and therefore not really science.&lt;/p&gt;&#xA;" OwnerUserId="10174" LastActivityDate="2015-12-17T09:31:35.543" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12907" PostTypeId="2" ParentId="12890" CreationDate="2015-12-17T13:18:17.480" Score="4" Body="&lt;p&gt;There are certainly applications of number theory (as opposed to group theory/algebra - but you do also mention abstract mathematics in general) to astrophysics, though many would suggest these applications are on the boundary between real science and metaphysics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One such application is the  &lt;a href=&quot;https://en.wikipedia.org/wiki/Measure_problem_(cosmology)&quot; rel=&quot;nofollow&quot;&gt;measure problem&lt;/a&gt; - if we assume our universe is one of an infinite number of multiverses how can we draw conclusions about the fundamental physics of the many multiverses, for instance to say that intelligent life is possible in a given proportion.&lt;/p&gt;&#xA;" OwnerUserId="823" LastEditorUserId="823" LastEditDate="2015-12-17T13:26:12.393" LastActivityDate="2015-12-17T13:26:12.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12909" PostTypeId="2" ParentId="12903" CreationDate="2015-12-17T15:33:58.363" Score="1" Body="&lt;p&gt;A lot of the data we have on the inner structure of the Earth comes from seismology - the study of earthquake waves travelling through the ground. This requires sensors placed at different points around the planet to pick up the waves and infer the structures they passed through. I haven't looked into it but I would think that this would be very difficult to do on other planets in terms of getting lots of sensors there and positioning them correctly, so maybe we just don't know this sort of thing about other planets yet!&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2015-12-17T15:33:58.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12910" PostTypeId="2" ParentId="12903" CreationDate="2015-12-17T17:46:24.123" Score="4" Body="&lt;p&gt;The short answer is no. Take Mercury for example in this &lt;a href=&quot;http://www.solstation.com/solsys/mer3core.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;comparison of Earth Mercury core&lt;/a&gt;. Mercury is thought to have a liquid outer core and solid inner core. The gas giants like &lt;a href=&quot;http://lasp.colorado.edu/education/outerplanets/images_giants/thumbs/pie.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jupiter&lt;/a&gt; are thought to have a relatively tiny rocky core but the convective motion in the metallic hydrogen is what gives them the strong magnetic fields. &lt;br/&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10894/is-mercurys-core-liquid?rq=1&quot;&gt;Is Mercury&amp;#39;s core liquid?&lt;/a&gt; for more on Mercury's core.&lt;/p&gt;&#xA;" OwnerUserId="10166" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2015-12-17T17:46:24.123" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12912" PostTypeId="2" ParentId="12897" CreationDate="2015-12-17T19:10:55.460" Score="9" Body="&lt;p&gt;By definition, gravity is a result of mass. Any body with a non-zero mass (even atoms) will have a gravitational field associated with it. The higher the mass the stronger will be the field. This is basic of classical mechanics. Until we reach quantum scale where the gravitational force is dominated by other 3 forces and the gravitational field becomes irrelevent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to graitational field of asteroids, it exists, but is very weak. However over a course of few million years these small asteroids combine together to form large masses of bodies that we now call planets. That is one of the prominent theory of Solar system formation, where the gravity of small dust particles from first our generation disintegrated star over a course of time accumulated to give us what we now know as our Solar system. &#xA;Think of it like this, every planet that you see now would once have been an asteroid at some point during its evolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another proof to support this is the presence of numerous binary asteroids that orbit each other around a common center of mass, which requires gravitational attraction.&lt;/p&gt;&#xA;" OwnerUserId="10177" LastActivityDate="2015-12-17T19:10:55.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12914" PostTypeId="2" ParentId="12771" CreationDate="2015-12-17T21:46:32.223" Score="1" Body="&lt;p&gt;The spin behaviour of the two types of pulsar would usually be quite different. The $\dot{P}$ for a rotationally powered pulsar is always positive, and higher order time derivatives of $P$ are quite small. This is because the rotational kinetic energy is powering the pulsar emission and the neutron star continually spins down as it loses rotational energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accretion powered pulsars can have very variable spin down or spin up characteristics, because they are powered by mass transfer and accretion in binary systems, and are influenced by a variety of factors affecting the accretion flow and how it couples to the pulsar magnetic field. The rates of spin down can be much higher than can be plausibly be accounted for by magnetic spin down as in a rotationally powered pulsar. Spin up cannot be accounted for in a rotationally powered model and neither can sign reversals or dramatic variability in $\dot{P}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, an accretion-powered pulsar would necessarily need to be in a short period binary system - therefore there would be a very obvious periodic modulation in the pulse period caused by the doppler shift as it travels in ts orbit.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-12-19T11:38:36.267" LastActivityDate="2015-12-19T11:38:36.267" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12915" PostTypeId="1" AcceptedAnswerId="12916" CreationDate="2015-12-18T12:15:44.677" Score="15" ViewCount="317" Body="&lt;p&gt;Suppose there's a star out there that's a lot more massive than the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose further that orbiting this star is a planet not unlike Earth. Water, oxygen, civilization, and all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the star decides to go supernova. How quick, or slow, is the process?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How long will it take it to heat up to make life on that planet impossible? A month? A year? A hundred years? A thousand years?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And how long will it take for the supernova to engulf the planet? A day? A month? A year? A million years?&lt;/p&gt;&#xA;" OwnerUserId="9510" LastActivityDate="2015-12-18T21:16:56.957" Title="How quickly does a supernova heat up/expand?" Tags="&lt;supernova&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12916" PostTypeId="2" ParentId="12915" CreationDate="2015-12-18T21:16:56.957" Score="12" Body="&lt;p&gt;Ricky, it is &lt;em&gt;very&lt;/em&gt; rapid. The core collapse and initial neutrino burst takes seconds to tens of seconds. We don't normally think about neutrino interactions, but so many are released that even this might be a problem for a nearby habitable planet. It then takes a few hours for the shockwave from the core collapse and bounce to make it out to the surface, accompanied by an intense flash of UV light that would likely sterilise anything in its planetary system. The outer layers of the exploding star are hurled out at around 1000 km/s, so could travel an astronomical unit in a day or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The supernova continues to become more luminous for about a week thereafter, increasing in luminosity, from what must have already been enormous (thousands of solar luminosities) by another factor of $10^5$. The equilibrium temperature of any planet scales roughly as $L^{1/4}$, so temperatures would rise by more than an order of magnitude in a week.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-12-18T21:16:56.957" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12917" PostTypeId="2" ParentId="12886" CreationDate="2015-12-19T00:13:02.117" Score="3" Body="&lt;p&gt;I think you can very easily use the transformations available in the wiki link to obtain the Julian day, then exploit the position of observer to get local sidereal time.&#xA;Lets start by computing the Julian day&#xA;you have a given longitude $L$ which gives you $n$ number of days since Greenwich noon as&lt;br&gt;&#xA;$$n = \dfrac{L-280.4860}{0.9856474}$$&#xA;giving the corresponding Julian day as&lt;br&gt;&#xA;$$JD = n+2451545$$&#xA;once you have Julian day you can use the following link to obtain UT&lt;br&gt;&#xA;&lt;a href=&quot;http://ssd.jpl.nasa.gov/tc.cgi&quot; rel=&quot;nofollow&quot;&gt;http://ssd.jpl.nasa.gov/tc.cgi&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="10177" LastEditorUserId="9301" LastEditDate="2016-08-25T22:24:35.273" LastActivityDate="2016-08-25T22:24:35.273" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12918" PostTypeId="2" ParentId="12903" CreationDate="2015-12-19T06:11:36.123" Score="0" Body="&lt;p&gt;Depends on what you call &quot;planet.&quot; For example, in comparison to Earth, the Earth and Moon is a &quot;binary planet system.&quot; And the moon doesn't have any molten core. Also, you said inner. Inner core of Earth (and many other planets) are solid because of (what I think is pressure). Finally, if it is outer core you meant, and you didn't count the moon as a &quot;planet&quot;, would you count extrasolar planets as &quot;planets&quot;. Please clarify, but to your current question, no.   &lt;/p&gt;&#xA;" OwnerUserId="10161" LastActivityDate="2015-12-19T06:11:36.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12919" PostTypeId="1" AcceptedAnswerId="12926" CreationDate="2015-12-19T10:53:57.220" Score="3" ViewCount="387" Body="&lt;p&gt;The materials inside a neutron star are densely packed (extremely close to each other) and as time goes by say about an &lt;strong&gt;&lt;em&gt;eon&lt;/em&gt;&lt;/strong&gt; later, most of the neutrons would eventually tunnel into each other and voila! the neutron star becomes black hole? sound suspicious to me but can it happen?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastActivityDate="2015-12-20T04:53:35.020" Title="Would all neutron stars inevitably collapse into black holes due to quantum tunneling?" Tags="&lt;black-hole&gt;&lt;neutron-star&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12920" PostTypeId="2" ParentId="12903" CreationDate="2015-12-19T11:11:34.497" Score="0" Body="&lt;p&gt;Saturn has a rock or ice core thirty times the mass of earth while jupiter has ice core ten-thirty times the mass of earth.Mars has a liquid core,moon however doesn't have a core .Composition of venus core varies significantly (Iron-88%,nickel-5%,sulphur-5%).Mercury Core occupies 84% of planets radius making it the largest core relative to the size of the planet in our solar system .While Gilese 1214 b is an exoplanet 42 light years from the sun has ice core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this answers your question.&lt;/p&gt;&#xA;" OwnerUserId="10191" LastEditorUserId="10191" LastEditDate="2015-12-26T10:45:33.633" LastActivityDate="2015-12-26T10:45:33.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12921" PostTypeId="1" AcceptedAnswerId="12928" CreationDate="2015-12-19T14:59:43.400" Score="4" ViewCount="233" Body="&lt;p&gt;That is the question. I know the concepts of luminosity &lt;span class=&quot;math-container&quot;&gt;$(L=4\pi R^2F)$&lt;/span&gt; and the flow &lt;span class=&quot;math-container&quot;&gt;$F=\sigma· T^4$&lt;/span&gt;, with &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; the temperature in its surface. But how I use that to know if the Earth would get warmer or not if &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; increases and &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; decreases?&lt;/p&gt;&#xA;" OwnerUserId="9949" LastEditorUserId="9949" LastEditDate="2020-09-24T15:28:09.707" LastActivityDate="2020-09-24T15:28:09.707" Title="If the Sun were bigger but colder, Earth would be hotter or colder?" Tags="&lt;astrophysics&gt;&lt;luminosity&gt;&lt;spectrometry&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="12923" PostTypeId="2" ParentId="12919" CreationDate="2015-12-19T18:36:29.467" Score="-1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Would all neutron stars inevitably collapse into black holes due to quantum tunnelling?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not to my knowledge. Where did you hear that? I don't think it's true, because if it was you could make the same claim about any massive body. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The materials inside a neutron star are densely packed (extremely close to each other) and as time goes by say about an eon later, most of the neutrons would eventually tunnel into each other and voila! the neutron star becomes black hole? sound suspicious to me but can it happen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't say &lt;em&gt;no it can't happen&lt;/em&gt; with confidence. But nor can I say with confidence that you won't receive a visit tomorrow from a fairy riding on the back of a unicorn. The thing to focus on here is the &quot;quantum tunnelling&quot;. Do you know of any actual experiments where two particles do some &quot;quantum tunnelling&quot; and morph into something else. Because I don't.  &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2015-12-19T18:36:29.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12924" PostTypeId="2" ParentId="12897" CreationDate="2015-12-19T19:11:12.163" Score="2" Body="&lt;p&gt;Certainly! &#xA;Anything that you see around you that has mass, your dog, your house, your car or yourself, they all have gravitational field and they exert gravitational pull to everything around them. And everything around them exerts that gravitational pull back. This pull however is so weak, that we can't perceive it with our senses. Gravity is a direct result of mass and the bigger the mass that an object has, the bigger its gravitational pull. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can extrapolate this paradigm to everything that exists in space! From the smallest particles of dusts and comets to the biggest of stars and galaxies. An asteroid that crashes on a planet, is attracted by the planet's gravitational pull but at the same time, the asteroid attracts the planet. Eventually, this is &lt;a href=&quot;http://www.stellarwise.com/how-big-can-rocky-planet-get/&quot; rel=&quot;nofollow&quot;&gt;how planets get to grow&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All celestial bodies in our sky would not exist if they hadn't any gravitational field. (1) Small dust particles collide with each other, forming bigger rocks. (2) Bigger rocks collide further to each other (or if they are big enough -several dozen of meters- might attract each other) to form comets and asteroids. (3) Comets and asteroids on their turn, will coalesce with other asteroids and rocks and will form dwarf planets and other terrestrial planets. (4) If those planets gain more mass they will be able to attract gas and they will form gas giants. (5) And if gas giants gain even more mass they will turn into smaller or bigger stars.&lt;/p&gt;&#xA;" OwnerUserId="10193" LastEditorUserId="10193" LastEditDate="2015-12-19T23:35:05.037" LastActivityDate="2015-12-19T23:35:05.037" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12925" PostTypeId="1" AcceptedAnswerId="12927" CreationDate="2015-12-20T04:28:55.367" Score="7" ViewCount="2032" Body="&lt;p&gt;I've been looking around, and can't find a solid answer. Many websites say darkness is the speed of light due to the fact it is the absence of light, but other sources say otherwise, that in some cases darkness is actually faster in some cases. Could someone clarify this for me?&lt;/p&gt;&#xA;" OwnerUserId="10197" LastEditorUserId="2365" LastEditDate="2015-12-20T08:22:08.507" LastActivityDate="2017-01-13T18:01:29.363" Title="Is Darkness Faster Than Light?" Tags="&lt;light&gt;&lt;speed&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2017-01-14T06:09:44.780" ContentLicense="CC BY-SA 3.0" />
  <row Id="12926" PostTypeId="2" ParentId="12919" CreationDate="2015-12-20T04:53:35.020" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The materials inside a neutron star are densely packed (extremely close to each other) and as time goes by say about an eon later, most of the neutrons would eventually tunnel into each other&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on what ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The material's collapse was &lt;em&gt;halted&lt;/em&gt; from an astonishingly high speed collapse by the neutron degeneracy pressure that exists precisely because the universe doesn't want to let two neutrons fuse.  The enormous forces involved in the collapse and the enormous forces keeping the mass compact are not high enough to overcome this.  That's why it exists in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over time &quot;tunnelling&quot; works in many, many ways. If two neutrons were to fuse in this manner the list of possible by products is enormous, and a black hole is most unlikely, IMO.  We've a lot of experience of whacking particles into other particles in extreme conditions ( think CERN ) and we haven't seen any sign of a black hole yet ( and believe me, they're looking for them ).  So I think the odds of a fuse into a black hole are extremely low.  Over time it's more likely the neutron star would &quot;evaporate&quot; into other particles.  But over those time scales you're looking at the cold death of the universe and protons decaying and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if, for argument sake, a mini black hole formed from two neutron's &quot;fusing&quot; it's life expectancy would be minuscule as tunnelling effects ( predicted by Hawking ) suggest it would &quot;evaporate&quot; in the blink of an eye.  So isolated fusing events, even f they happened, would be unlikely to produce a black hole that grew, but instead one that just blew up as fast as it was created.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and voila! the neutron star becomes black hole? sound suspicious to me but can it happen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No evidence I am aware of suggests neutrons will fuse in the manner you're describing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Neutron Star&quot; is probably something of a misnomer.  Apart from neutrons, current models ( &quot;educated guesses&quot; ) suggest that the core of such a body might be a lump of something called quark-gluon plasma, which is essentially mush made from neutrons.  This is what you get when you fuse neutrons together -  a form of mush made up of the stuff that neutrons and protons are made from - quarks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So theory and evidence suggests that you don't get a black hole from just waiting for a neutron star to change into one.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2015-12-20T04:53:35.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12927" PostTypeId="2" ParentId="12925" CreationDate="2015-12-20T04:58:07.443" Score="7" Body="&lt;p&gt;The darkness does not have a speed, as it's just the space the light travels through.  It's always there until the brief instant a photon whizzes through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be analogous to asking what speed the road is travelling at, rather than the car.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2015-12-20T04:58:07.443" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12928" PostTypeId="2" ParentId="12921" CreationDate="2015-12-20T08:17:06.087" Score="3" Body="&lt;p&gt;The equilibrium temperature of the Earth, $T_E$, scales roughly as $L^{1/4}$, which is proportional to $R^{1/2} T$, where $L$, $R$ and $T$ are the solar luminosity, radius and temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual approximate relationship is derived by equating the power received by the Earth, which is proportional to the solar luminosity $L$, with the power radiated by the Earth, which is proportional to $T_E^4$ for a blackbody. Hence $T_E \propto L^{1/4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to your question depends on by how much you increase the radius compared with the decrease in temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There will be second order effects that do depend on the spectrum of radiation from the Sun (and therefore its temperature) compared with the wavelength dependence of the albedo and emissivity of the Earth. So I will post a better question...&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-12-20T23:08:51.217" LastActivityDate="2015-12-20T23:08:51.217" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12929" PostTypeId="1" CreationDate="2015-12-20T08:24:35.740" Score="9" ViewCount="1284" Body="&lt;p&gt;A recent question &lt;a href=&quot;https://astronomy.stackexchange.com/questions/12921/if-the-sun-were-bigger-but-colder-earth-would-be-hotter-or-colder/12928#12928&quot;&gt;If the Sun were bigger but colder, Earth would be hotter or colder?&lt;/a&gt; asked - if the Sun got bigger and cooler, would the Earth heat up or cool down. I think the answer to that is mainly that it depends on the final luminosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what I want to know here (hypothetically), is if the Sun got larger and it's effective temperature decreased &lt;em&gt;such that it's luminosity was unchanged&lt;/em&gt;; how would that affect the equilibrium temperature of the Earth? I suspect the answer may involve the wavelength dependence of the albedo, emissivity and atmospheric absorption of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another, less hypothetical, way of asking this is, if you put an Earth-like planet at different distances from stars with a variety of temperatures, such that the total flux incident at the top of the atmosphere was identical, how would the temperatures of those planets compare?&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2018-07-07T17:15:23.860" Title="If the Sun got larger, but maintained its luminosity, would the Earth get hotter or colder?" Tags="&lt;the-sun&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12930" PostTypeId="2" ParentId="12896" CreationDate="2015-12-20T08:45:54.750" Score="0" Body="&lt;p&gt;There are a number of illustrations of this on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_evolution&quot; rel=&quot;nofollow&quot;&gt;Stellar evolution&lt;/a&gt; page of wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly a star begins as a large but cool ball of contracting gas, to the right of the main sequence until fusion starts in its core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once fusion starts it has reached the main sequence. It gradually moves up as it slowly brightens during its life, and then moves to the right and up as it expands into a red giant. There are then some significant movements as heavier elements begin to fuse. Sun-like stars undergo a significant change when the helium fuses in a matter of days (or less according to some models), called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Helium_flash&quot; rel=&quot;nofollow&quot;&gt;Helium flash&lt;/a&gt;. As this occurs, the star moves significantly down and left, back towards the main sequence, before expanding again into an even larger red giant. The expelling its outer layers in a planetary nebula, and rapidly moving to the left and down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Larger stars undergo other fluctuations, changing from red supergiants, to Luminous Blue Variables, and Wolf-Rayet stars, and ending in a supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several illustrations of this on the wikipedia page, but noting the sometimes significant differences between the diagrams suggests that this is a topic in which the details are not certain, in part due to the difficulty in getting observational evidence of changes in stars, which occur over very long time scales.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2015-12-20T08:45:54.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12931" PostTypeId="2" ParentId="12929" CreationDate="2015-12-20T11:41:22.973" Score="6" Body="&lt;p&gt;To a first approximation if the Sun got bigger (by a relatively small factor, say a few times) but maintained its present luminosity, as the Earth would still be intercepting the same total energy per unit time the temperature would stay the same. Constant luminosity gives a constant intensity at the Earth's distance from the Sun (intensity is the energy crossing unit area in unit time) which would mean that the total solar energy intercepted by the Earth would remain the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in this scenario the colour of the sun would change and we would get secondary effects due to the reflectivity of the Earth being frequency dependent so the fraction of the incident energy reflected rather than absorbed would change, which would result in a change in temperature. Which direction this would go in is difficult to say as an increase in temperature should result in more cloud which would increase reflectivity ...&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2015-12-20T11:41:22.973" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12932" PostTypeId="2" ParentId="12576" CreationDate="2015-12-20T18:37:14.560" Score="0" Body="&lt;p&gt;If you he universe is finite in size, something(matter) would have to be present to to fill in the empty space surrounding the entire entire universe or else it would be just more empty space.  Let me explain what I mean.  A room is finite in size because when you get to a certain point that the walls, floor, and ceiling stop you.  So if a universe is finite, it would have to have some equivalent to walls, to keep it the same size and finite and/or to constitute an end.  If space ends at some point because nothing can go any further, like if we were inside of an incomprehensibley large hollow sphere and traveled to the end of space and found the end of the universe,  something would still be on the other side of whatever the end of the was.  Infinite is not a hard concept at all and is really the only possibility if you think a about.  Infinite can't expand.  It's like a child saying  infinite plus one.  Only now its college educated grown men now and all the logic in the world wont stop them from trying to make it make sense.  The universe is definitely infinite, anyone can observe that by looking up at the night sky.  We dont live inside of a cosmic domicile with limited space.  We live in infinite empty space on a finite in size spherical shaped mass of a finite amount of matter. an  Infinite universe of empty space that has stars and planets, moons and whatever else.  But the universe itself is already infinite and therfore can't get any bigger.  Its empty space, in every direction, forever(with stars and planets here and there).  Why is that so hard for science to accept?  &lt;/p&gt;&#xA;" OwnerUserId="10198" LastActivityDate="2015-12-20T18:37:14.560" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12933" PostTypeId="1" CreationDate="2015-12-21T01:41:21.787" Score="1" ViewCount="329" Body="&lt;p&gt;Can planets have orbits other than elliptical or circular orbits? All of the planets in our solar system have either a circular orbit or an elliptical orbit (perhaps like Pluto).&lt;/p&gt;&#xA;" OwnerUserId="10164" LastEditorUserId="10164" LastEditDate="2015-12-22T05:17:03.500" LastActivityDate="2015-12-22T05:17:03.500" Title="Can planets have orbits other than elliptical or circular orbits?" Tags="&lt;orbit&gt;&lt;planet&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12934" PostTypeId="2" ParentId="12843" CreationDate="2015-12-21T01:41:23.580" Score="-2" Body="&lt;p&gt;Dark Matter is a kind of &quot;fill in the blank&quot; at the moment.  No one really has any definitive theory that has wide acceptance on what &quot;dark matter&quot; may be, never mind how it interacts with other matter.  There's still a little resistance to even saying it exists at all, although it's pretty much the mainstream now that it does.  But what it is, that's still up for grabs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I did find this from Google :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://iopscience.iop.org/article/10.1086/306334/fulltext&quot; rel=&quot;nofollow&quot;&gt;http://iopscience.iop.org/article/10.1086/306334/fulltext&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It presumes a certain type of model for dark matter, but may be what you're looking for.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2015-12-21T01:41:23.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12935" PostTypeId="1" AcceptedAnswerId="12937" CreationDate="2015-12-21T09:53:45.560" Score="12" ViewCount="3051" Body="&lt;p&gt;I often hear about planetary ring systems, and even some moons might have them, but how about stars? Can a star also have rings?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="13" LastEditDate="2015-12-22T06:21:13.063" LastActivityDate="2018-08-22T15:09:59.600" Title="Can a star have a ring system?" Tags="&lt;star&gt;&lt;planetary-ring&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12936" PostTypeId="2" ParentId="12933" CreationDate="2015-12-21T10:40:53.183" Score="5" Body="&lt;p&gt;Orbits are conic sections therefore can be either circular, elliptical, parabolic or hyperbolic. &#xA;Of these 4, only first two form a closed curve under 2 body hypothesis, while the later two extend to infinity.&lt;br&gt;&#xA;If you talk about planet,  by definition it has to orbit a star which would require it to have a closed orbit hence circular or elliptical, with respect to the star. For any other kind of orbit the body will just fly away to infinity never to return back.  Infact comets are considered to be parabolic, but in theory they have highly elliptical orbits with their aphelion lying near the edge of the solar system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it is possible for a planet to have other kind of orbits if we  consider their motion from a different reference frame such as with respect to another planet. &#xA;So for an inertial frame of reference a planet will have a circular or elliptical orbit, even Pluto. &lt;/p&gt;&#xA;" OwnerUserId="10177" LastActivityDate="2015-12-21T10:40:53.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12937" PostTypeId="2" ParentId="12935" CreationDate="2015-12-21T10:40:59.147" Score="11" Body="&lt;p&gt;They certainly can. A ring is often formed around a celestial body when its gravity rips apart another smaller celestial body. The Sun is really massive, so it could destroy any object that is not dense enough. Just Google about the Roche Limit for more informations (and better explanations). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, take a look at our solar system : You have the asteroid belt between Mars and Jupiter, and you also have the Kuiper belt beyond Neptune. Would you consider these being rings ? They sure are not as smooth as Saturn's, but to my mind, they still are rings.&lt;/p&gt;&#xA;" OwnerUserId="9679" LastEditorUserId="9679" LastEditDate="2015-12-21T16:21:42.007" LastActivityDate="2015-12-21T16:21:42.007" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12938" PostTypeId="2" ParentId="12933" CreationDate="2015-12-21T10:45:10.333" Score="-2" Body="&lt;p&gt;Certainly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you'd have to think in Einsteinian, rather than Newtonian, terms. Inertial frames of reference and all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you view our planet as motionless, then the orbit of Venus, for instance, would resemble a telephone coil cord.&lt;/p&gt;&#xA;" OwnerUserId="9510" LastActivityDate="2015-12-21T10:45:10.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12939" PostTypeId="2" ParentId="12897" CreationDate="2015-12-21T13:22:34.397" Score="4" Body="&lt;p&gt;You asked two questions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do asteroids have a gravitational field.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course. Even a microscopic grain of dust has a gravitational field.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do they gravitationally attract each other to form planets?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not any more. During the formation of the solar system, asteroid-like and comet-like objects collided to build larger objects, which in turn collided to form even larger objects, and so on, eventually building the cores of giant planets and later, the terrestrial planets. But that stage ended long ago, shortly after the solar system formed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asteroids do of course gravitationally attract other objects, but this attraction is so weak due to the small masses of asteroids that it is easily overwhelmed by other perturbing forces. The vast majority of the asteroids lie between Mars and Jupiter, and Jupiter is the primary culprit in explaining why no planet exists in that gap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When two astronomical bodies collide, one of the outcomes is a purely inelastic collision that makes two bodies form a single body. This only happens with a rather mild collision. A more energetic collision will result in some mass being expelled. An even more energetic collision will result in lots of mass being expelled; the colliding bodies become many smaller bodies. With a few exceptions, the latter is what is what is happening amongst the asteroids today, and for the last four-plus billion years or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter is such a huge perturbing body that collisions in the asteroid belt are generally very energetic. Instead of forming ever larger bodies, the asteroid belt is gradually being broken up into smaller and smaller bodies. Some of these collisional bodies are ejected from the solar system thanks to interactions with Jupiter. The smallest results of these collisions migrates sunward thanks to the Poynting-Robertson effect.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-12-21T13:22:34.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12940" PostTypeId="1" AcceptedAnswerId="13064" CreationDate="2015-12-21T17:10:30.147" Score="6" ViewCount="420" Body="&lt;p&gt;The Winter Solstice in the Northern hemisphere in 2015 is December 22 at 04:49 UTC.  Where I'm located in Nova Scotia, Canada, that's December 22, 00:49 AST.  And so I would say the shortest day of the year is tomorrow, the 22nd.   But for my friends in Ontario, that's December 21, 23:49 EST.  And so, for them, the shortest day is today, the 21st.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this make sense?&lt;/p&gt;&#xA;" OwnerUserId="6830" LastEditorUserId="6830" LastEditDate="2015-12-28T12:32:31.053" LastActivityDate="2016-01-07T21:26:19.473" Title="22nd is shortest day in some places, but the 21st is shortest in other places. Can this be true?" Tags="&lt;earth&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12941" PostTypeId="2" ParentId="12940" CreationDate="2015-12-21T20:23:13.100" Score="5" Body="&lt;p&gt;Sure it can be (and is) true. Everyone is going through the Earth's &quot;zone of lightness&quot; at different times. Presumably whichever of those times is &quot;nearest to&quot; the actual moment of solstice will be the &quot;shortest period between sunrise and sunset&quot; for any given person (longitude) (ignoring natural and time-zone variations as noted in other comments). As you note, for your friends and everyone west of them up to the Date Line, that will likely be 12/21. And for you and everyone east of you around to the Date Line, that will likely be 12/22.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe it helps (or maybe not) when you realize that people very close to the Date Line itself but on opposite sides experience 12/21 and 12/22, resp., as &quot;the same day&quot;.&lt;/p&gt;&#xA;" OwnerUserId="10005" LastEditorUserId="10005" LastEditDate="2015-12-21T20:35:47.470" LastActivityDate="2015-12-21T20:35:47.470" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12942" PostTypeId="1" CreationDate="2015-12-22T00:42:36.617" Score="2" ViewCount="692" Body="&lt;p&gt;As a data visualization task, I'd like to plot some celestial bodies in 3D coordinate space. I assumed there would be a lot of databases where celestial bodies are recorded along with an XYZ coordinate relative to our sun. As I ponder more on it I begin to realize other features, for example how is the axis of alignment achieved and agreed upon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the below data set, is it clear what system is in use and how I may interpret it in a way that gives me an XYZ I can begin plotting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.com/fusiontables/DataSource?docid=1TZ3eoWstpR8d0O2-eqoz3VFDsLwSSzNA5wX4LxE&quot; rel=&quot;nofollow&quot;&gt;https://www.google.com/fusiontables/DataSource?docid=1TZ3eoWstpR8d0O2-eqoz3VFDsLwSSzNA5wX4LxE&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10210" LastActivityDate="2016-04-20T20:45:58.837" Title="A beginner project in plotting celestial bodies in 3D coordinate space" Tags="&lt;coordinate&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12943" PostTypeId="2" ParentId="10859" CreationDate="2015-12-22T02:14:33.650" Score="0" Body="&lt;p&gt;Given the the discovery of Pluto was made based on the search for a body to explain orbits being different from that predicted by known bodies, it seems very unlikely with the even more precise and powerful calculations available today that we've missed a large body of any kind in that region, let alone a whole planet.  I think we have simply amassed too much high precision data for that to be the case.  Although I'd like to be wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also suggest that the &quot;&lt;em&gt;clear cut line between planets and dwarf planets&lt;/em&gt;&quot; is not at all clear cut, but a rather arbitrary line drawn by a committee.  I personally consider it artificial and any decision of that sort that can demote a body the size of Pluto from planet to dwarf-planet is one I can only view with derision.  The measure chosen to justify the decision seems more to do with the need to have a definition for the &quot;rule book&quot;, rather than any scientific need.  I don't think I'm the only one with that view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a chunk of rock ( or ice ) in that region large enough to be a planet by any reasonable common sense view, much less the IAU definition, seems unlikely to have gone undiscovered to date.  It's not as if we're not looking.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2015-12-22T02:14:33.650" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12944" PostTypeId="2" ParentId="12385" CreationDate="2015-12-22T02:27:29.853" Score="1" Body="&lt;p&gt;Quakes and other large scale geological activity ( and indeed running water ! ) cause no end of trouble for mining operations.  Much better to be able to sit on a nice quiet rock in space without all that craziness going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have, at this time, no real basis for saying one way or another, that there are not mineral concentrations.  If all I let you do was dig a few meter deep boreholes in the surface of the Earth at random locations, what are the odds you would know anything about mineral concentrations for mining ?&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2015-12-22T02:27:29.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12947" PostTypeId="1" AcceptedAnswerId="12965" CreationDate="2015-12-22T10:53:52.060" Score="8" ViewCount="333" Body="&lt;p&gt;I read that a black hole can sometime &quot;choke&quot; on a star:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://news.utexas.edu/2015/01/27/black-hole-swallowed-star&quot;&gt;&quot;...the disrupted stellar matter was generating so much radiation&#xA;  that it pushed back on the infall. The black hole was choking on the&#xA;  rapidly infalling&#xA;  matter.&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why does a black hole choke? How do astronomers manage to observe this rare event?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="13" LastEditDate="2015-12-23T12:42:22.280" LastActivityDate="2017-05-01T17:06:26.213" Title="Why do black holes choke?" Tags="&lt;black-hole&gt;&lt;radiation&gt;&lt;accretion&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12948" PostTypeId="2" ParentId="12886" CreationDate="2015-12-22T13:03:56.093" Score="2" Body="&lt;p&gt;Ok guys, I've found the solution. Sadly, if you really want to know the really  exact datetime, there is no exact and universal formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found several formulas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/25336/reverse-sun-position-algorithm&quot;&gt;https://physics.stackexchange.com/questions/25336/reverse-sun-position-algorithm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.pveducation.org/pvcdrom/properties-of-sunlight/declination-angle&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.pveducation.org/pvcdrom/properties-of-sunlight/declination-angle&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.itacanet.org/the-sun-as-a-source-of-energy/part-3-calculating-solar-angles/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.itacanet.org/the-sun-as-a-source-of-energy/part-3-calculating-solar-angles/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several problem with these formulas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What I'd like to know?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I mentioned in my original post, I have a datetime, and a geolocation, and with that Swiss ephemeris program, I can &quot;query&quot; the angle of the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this point, I substract 88 degree with this angle, and I want to know the exact date time by the angle and the geolocation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The problems&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first problem is the declination. There are several problems with this. The declination is not the same at every day in a year as you know. Ok, let's say, I want to figure it out. I can not. Since, I want to know the exact date first, I do not know, the ordinal date. Sigh, at this point the whole formula has a missing value. But, let's assume, I know. Even, if I know the ordinal date, the formula is not right, because the declination is not the same at 2015-01-01 and 2016-01-01. Let's see:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;swetest.exe -edirD:/Apache/htdocs/astro/vendors/swiss/ -b01.01.2010 -geopos19.2,47.29,0 -topo  -p0 -eswe -fPlTd -g, -head -n5 -s365.4&#xA;Sun            , 280.4507684,01.01.2010,-23.0283899&#xA;Sun            , 280.6104798,01.01.2011 9:36:00 ET,-23.0160751&#xA;Sun            , 280.7713793,01.01.2012 19:12:00 ET,-23.0008866&#xA;Sun            , 280.9345061,01.01.2013 4:48:00 ET,-22.9876598&#xA;Sun            , 281.0890407,01.01.2014 14:24:00 ET,-22.9741190&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;-22...., -23... is the declination. Declination is the same in a day. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So these formulas can gives you approximate results, but not the exact, even if you know the ordinal date. And I think, the formula is does not calculate with the leap years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The solution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my case, the solution was to track back 95 days, and get 20 recors for every day from that time. 95 because the sun is moving 1 degree about 1 day and 2-3 hours. The sun has different speed at different dates, 95days seems enought for me for 88 degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With my program, I am checking, where the integer values are equals. For example, if I need to get 69.217465 degree I am just checking, where the sun was in 69.xxxxx position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I have this date, I am track back 3 days, and list the values by hours, 72 hours. I get the  proximate date and time. And after this, I track back 3 hours, and listing the results by seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will be the exact (most proximate) date and time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a Yahoo group, a guy told me, this is an interpolation. This is how you can get the exact date time for a given angle, I do not know more precise method. If you know, please write me.&lt;/p&gt;&#xA;" OwnerUserId="10055" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2015-12-22T13:03:56.093" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12949" PostTypeId="1" CreationDate="2015-12-22T13:22:14.140" Score="3" ViewCount="519" Body="&lt;p&gt;The sphere of influence is a region around a supermassive black hole in which the gravitational potential of the black hole dominates the gravitational potential of the host galaxy(found it in wiki). So is it the same? If not, what holds the milky way together?&lt;/p&gt;&#xA;" OwnerUserId="10049" LastActivityDate="2015-12-22T14:19:29.870" Title="Is the sphere of influence of our milky way's center super massive black hole same as the Milky way's radius?" Tags="&lt;gravity&gt;&lt;black-hole&gt;&lt;milky-way&gt;&lt;supermassive-black-hole&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12950" PostTypeId="2" ParentId="12949" CreationDate="2015-12-22T14:19:29.870" Score="3" Body="&lt;p&gt;No. The black hole at the centre of the Milky way is only a few million solar masses. The same amount of mass, made up mostly of stars, would be enclosed within $\sim 10$ parsecs of the Milky Way centre. The bulge of the Milky Way, which is a few thousand parsecs across contains more than 20 billion solar masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the dynamics of our Galaxy, outside of the central few parsecs are hardly influenced by the black hole at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What holds the Milky Way together? The gravity of everything else - stars, gas, dust, dark matter (the latter probably being the most important once you get to galactocentric radii beyond the Sun).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2015-12-22T14:19:29.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12951" PostTypeId="2" ParentId="12942" CreationDate="2015-12-22T15:13:41.170" Score="0" Body="&lt;p&gt;I think the answer to your question is the game Universe Sandbox 2. With such a software I think it would be pretty easy to obtain what you want.&lt;/p&gt;&#xA;" OwnerUserId="9679" LastActivityDate="2015-12-22T15:13:41.170" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12952" PostTypeId="1" AcceptedAnswerId="12960" CreationDate="2015-12-22T16:17:45.567" Score="8" ViewCount="1214" Body="&lt;p&gt;If we suppose we and everything in the universe is perfectly flat.  That is, we have width and depth, but no height.  And we are situated on the surface of a balloon blowing up with air and expanding &quot;outwards&quot;.  Us flat people can move around on the surface of our balloon but not through it, in much the same way we navigate around the earth, in a plane say. No matter where something is situated on the surface of the balloon, one point is no more central than any other point.  In other words, the surface of a balloon (the &quot;universe&quot;) has no centre.  Is this why we say it makes no sense to say the universe has a centre, or edge for that matter, or is this analogy stupid?&lt;/p&gt;&#xA;" OwnerUserId="5306" LastEditorUserId="5306" LastEditDate="2015-12-22T17:59:19.997" LastActivityDate="2016-02-01T04:44:22.863" Title="Why does it make sense to say the universe has no centre?" Tags="&lt;universe&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12953" PostTypeId="1" AcceptedAnswerId="12956" CreationDate="2015-12-22T17:13:16.583" Score="7" ViewCount="1388" Body="&lt;p&gt;Are there natural satellites in the Solar System that are geologically active? This includes volcanism, existence and motion of tectonic plates, et cetera. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a common or a rather rare feature among such bodies?&lt;/p&gt;&#xA;" OwnerUserId="62" LastActivityDate="2020-11-11T14:06:39.627" Title="Are moons geologically active?" Tags="&lt;natural-satellites&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12954" PostTypeId="2" ParentId="12953" CreationDate="2015-12-22T18:01:45.710" Score="3" Body="&lt;p&gt;Yes there are. However in some moons it is caused by tidal friction not due to the natural structure of the moon. Take &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/5/55/PIA01081-Color_Mosaic_and_Active_Volcanic_Plumes_on_Io.jpg&quot; rel=&quot;nofollow&quot;&gt;This volcano on Io&lt;/a&gt; it is thought to be 1 of 150+ volcanoes on the satellite. However the lava these volcanoes spew is mostly in forms of molten sulfur. Enceladus is a moon of Saturn which experiences volcanic activity due to tidal friction resulting in geysers bursting from poles. Titan is also thought to have volcanism but instead of hot lava they would spout liquidy methane. As far as I know there is no analogue to Earth's plates on any satellites. There are moon quakes.&lt;/p&gt;&#xA;" OwnerUserId="10166" LastActivityDate="2015-12-22T18:01:45.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12955" PostTypeId="2" ParentId="12952" CreationDate="2015-12-22T18:52:03.913" Score="-1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does it make sense to say the universe has no centre?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can't be sure it does. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If we suppose we and everything in the universe is perfectly flat. That is, we have width and depth, but no height. And we are situated on the surface of a balloon blowing up with air and expanding &quot;outwards&quot;. Us flat people can move around on the surface of our balloon but not through it, in much the same way we navigate around the earth, in a plane say. No matter where something is situated on the surface of the balloon, one point is no more central than any other point. In other words, the surface of a balloon (the &quot;universe&quot;) has no centre. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This analogy is often used to justify that the claim that the universe has no centre. But there's no evidence of any higher dimensions. We simply don't know that there's any kind of &quot;asteroids&quot; curvature such that if you keep going one way you come back the other. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this why we say it makes no sense to say the universe has a centre, or edge for that matter, or is this analogy stupid?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wouldn't say it's stupid. But I would say it's presenting hypothesis as some kind of fact. For all we know our universe does have a centre and an edge. &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2015-12-22T18:52:03.913" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12956" PostTypeId="2" ParentId="12953" CreationDate="2015-12-22T20:14:56.380" Score="6" Body="&lt;p&gt;Yes.  Moons around Jupiter (Io, Europa and Ganymede), Saturn (Titan and Enceladus) and Neptune (Triton) all have some form of geological activity.  Charon also may have geological activity, being in a binary system with Pluto.  However, while Earth's geological activity is caused by internal heating and tectonic plates, the geological activity of the moons around Jovian planets comes in the form of tidal forces.  Io is the most iconic instance of tidal stress, because Io's plumes are frequent, volatile and make the world look extremely chaotic, with its surface frequently being altered and renewed by its non stop volcanic activity.  (Because it is chaotic)&lt;/p&gt;&#xA;&lt;p&gt;As for tectonic plates, &lt;a href=&quot;http://www.jpl.nasa.gov/news/news.php?feature=4285&quot; rel=&quot;nofollow noreferrer&quot;&gt;Europa&lt;/a&gt; is the closest you get to tectonic plates with moons in our star system.  Water replaces lava when it comes to ice worlds.  Ice worlds being worlds that have ice instead of rock for their crust.  This means that water mantles are a frequent occurrence, with the core of ice worlds being mineral rich stone.  This is the case for Triton as well, which has cyro-volcanism from the sheer tidal stress Neptune exerts on the captured dwarf planet.&lt;/p&gt;&#xA;&lt;p&gt;Enceladus and Titan have water mantles, Enceladus being the world notable for its massive plumes, high albedo and tiger stripe surface fractures.  Titan may also have tectonic activity for similar reasons to Europa.&lt;/p&gt;&#xA;" OwnerUserId="8017" LastEditorUserId="9527" LastEditDate="2020-11-11T14:06:39.627" LastActivityDate="2020-11-11T14:06:39.627" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12957" PostTypeId="1" AcceptedAnswerId="12964" CreationDate="2015-12-23T01:21:58.427" Score="5" ViewCount="681" Body="&lt;p&gt;In this photo taken by SpaceX during one of their Falcon 9 launches, I noticed a familiar group of stars but I'm not sure of what it really is. I encircled it in red, is this group the Cancer constellation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xFouy.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xFouy.jpg&quot; alt=&quot;Encircled group of stars&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://flic.kr/p/rokQXY&quot; rel=&quot;noreferrer&quot;&gt;Original Image&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10216" LastActivityDate="2019-02-19T15:14:11.517" Title="Is this the Cancer constellation?" Tags="&lt;constellations&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12958" PostTypeId="2" ParentId="12952" CreationDate="2015-12-23T02:59:33.243" Score="1" Body="&lt;p&gt;The centre of something ( anything ) has to be defined in some way by the properties of the something and a coordinate system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can in principle define a centre to the universe, but it only has a meaning in the context of the chosen definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could easily define a different measure that produces a different centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you can define a centre even on the surface of your balloon, simply by arbitrary selection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An edge to the universe requires a context.  How do we define it ?  How do we detect it ?  The only meaningful measure of this ( in a human sense ) that I know of is &quot;how far can I see ?&quot;.  That keeps changing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at the moment I would say &quot;centre is arbitrary&quot; and &quot;edge is unknown&quot;.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2015-12-23T02:59:33.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12960" PostTypeId="2" ParentId="12952" CreationDate="2015-12-23T05:43:25.007" Score="3" Body="&lt;p&gt;The answer to this question lies in cosmology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's because of relativity in positions.  Every galaxy appears to be moving away from us and the universe we see is only the observable universe.  Further points are more in the distant past, when you observe the deep field you observe the past.  This may seem irrelevant to your question, but the idea of a &quot;center&quot; to a universe where any position in the universe, be it from our galaxy or from a galaxy in the Sloan Great Wall is the true center is not consistent with observation.  That's because galaxies/clusters of galaxies are moving away from each other, relative to each other.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every galaxy is slowly being pulled apart from other galaxies, save for those in gravitationally bounded clusters.  (The whole architecture of the universe, with massive voids of absolute nothing and bunched up clusters of galaxies reflects this quite a bit)  If the universe had a central point, this would be easily observed because there would be variation in the color shift of galaxies that could be analyzed to deduce a central point where the galaxies are moving away from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is not what is observed.  Instead, pretty much most galaxies (save for exceptions like Andromeda or the Triangulum, who prove the rule since those galaxies are gravitationally connected to our galaxy via the local group) are moving away from us.  The Great attractor being not a path to the center of the universe, but the center of the recently discovered Laniakea super cluster.  So if there's attraction between galaxies, it's because of gravitational interactions.  The universe is also without any real edge unless you count the edge of the observable universe.  The observable universe being more a historical record than what the contemporary universe around us looks like and is a byproduct of light speed being so limited.  The speed of light in a vacuum is also the speed of information.  The whole universe is far more massive by comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to speak about a center of the universe would require the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A region of space that galaxy is moving away from or towards. (Which does not exist, instead relative trajectories and gravitational clustering is all regional at most.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Edges that are defined enough to pinpoint a center (If such edges exist, they are impossible to observe due to the limited speed information travels in the universe)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Absolute positioning to pinpoint where this center would be. (All positions are relative, the galaxy, the sun and our planet all are moving at extremely fast velocities we don't notice thanks to relativity.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8017" LastEditorUserId="8017" LastEditDate="2015-12-23T05:48:33.240" LastActivityDate="2015-12-23T05:48:33.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12962" PostTypeId="1" CreationDate="2015-12-23T07:23:51.967" Score="3" ViewCount="65" Body="&lt;p&gt;For a video game I'm looking for footage of a collision of meteoroids or maybe asteroids, not planets. Even though I have searched for a while I couldn't find anything. So has anyone ever captured footage?&lt;/p&gt;&#xA;" OwnerDisplayName="user10219" LastEditorDisplayName="user10219" LastEditDate="2015-12-23T17:17:11.297" LastActivityDate="2015-12-28T00:58:21.927" Title="Has someone ever captured footage of the collision of a meteoroid and a meteoroid?" Tags="&lt;asteroids&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12963" PostTypeId="2" ParentId="12962" CreationDate="2015-12-23T08:13:25.510" Score="1" Body="&lt;p&gt;I'm fairly certain it has already happened a lot. But the closest footage I found from what you are looking for is the aftermath of a collision, right &lt;a href=&quot;http://www.space.com/11537-nuclear-force-asteroid-collisions-aftermath.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. However, I don't think we'll ever get to record an actual collision. Space is big, you know. Chances of a collision are quite rare. But you can find a lot of simulations on Youtube.&lt;/p&gt;&#xA;" OwnerUserId="9679" LastActivityDate="2015-12-23T08:13:25.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12964" PostTypeId="2" ParentId="12957" CreationDate="2015-12-23T08:36:27.040" Score="8" Body="&lt;p&gt;Nope, that constellation is Perseus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BBWjK.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BBWjK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1678" LastActivityDate="2015-12-23T08:36:27.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12965" PostTypeId="2" ParentId="12947" CreationDate="2015-12-23T10:02:57.003" Score="9" Body="&lt;p&gt;There is a limit to how quickly a black hole can accrete matter. As a star gets close to a black hole it may be tidally disrupted and pulled apart. The material from the star will then fall toward the black hole and as it does so, the gravitational potential energy will go into heating it up. Now the stellar material would not normally be able to fall directly into the black hole - angular momentum must be conserved, so first it will form an &quot;accretion disk&quot;. Viscosity in the disk can transport angular momentum outwards, allowing material to then fall through the event horizon. The ultimate rate at which the black hole can swallow mass will be controlled by the rate at which material can reach the inner parts of the accretion disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the accretion disk is very hot and luminous, there could be a limit to how fast material can be added to it. This is known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Eddington_luminosity&quot;&gt;Eddington limit&lt;/a&gt;. Basically what happens is that the radiation from the inner parts of the accretion disk is so intense that it can &quot;blow away&quot; any new material falling towards the black hole. This is then a self-regulating phenomenon because the luminosity of the disk is derived from the material falling onto it - and so an equilibrium will be set up that limits the accretion rate to that which produces only enough luminosity that allows the accretion to proceed. This limit on the accretion rate is what the article you refer to means by &quot;choking&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An estimate of the maximum accretion rate is set by equating the accretion luminosity and the Eddington luminosity (the maximum luminosity that an object may have before radiation pressure will halt accretion). This in turn depends on the efficiency $\epsilon \sim 0.1$ with which gravitational potential energy is turned into radiation as material falls into the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The journal paper that reports this discovery - &lt;a href=&quot;http://arxiv.org/abs/1410.6014&quot;&gt;Vinko et al. (2015)&lt;/a&gt; - analyse the light curve of the event and suggest this is black hole of $10^{5}$ to $10^{6}$ solar masses, tidally disrupting and swallowing a star of about a solar mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Eddington luminosity (for spherical accretion) would be $\leq 3\times 10^{10}$ solar luminosities , but this thing had a peak luminosity (even in X-rays) of $10^{11}$ solar luminosities indicating that it must be in the regime where radiation pressure would reduce the accretion rate. To supply this luminosity would require $\epsilon \dot{M} c^2 \sim 10^{11} L_{\odot}$. &#xA;For $\epsilon \sim 0.1$ this gives a peak mass accretion rate $\dot{M} \sim 2\times 10^{-4} M_{\odot}$ per day. Thus it would take more than 10 years to &quot;swallow&quot; the star even at this peak rate.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-12-23T16:30:49.143" LastActivityDate="2015-12-23T16:30:49.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12967" PostTypeId="1" AcceptedAnswerId="13002" CreationDate="2015-12-23T11:28:13.460" Score="2" ViewCount="240" Body="&lt;p&gt;While cleaning my desk, I found an out of date star map (puts the cleaning into perspective, huh?) where the &lt;em&gt;Scutum&lt;/em&gt; constellation is still referred to as &lt;em&gt;Scutum Sobiescianum&lt;/em&gt;).  This is the original name Johannes Hevelius gave it in 1684&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My map is (unfortunately) modern and comes from the 2nd half of the 20th century. It's a Polish one so it's not odd for it to still use the traditional name. The main illustration of the &lt;a href=&quot;https://pl.wikipedia.org/wiki/Gwiazdozbi%C3%B3r_Tarczy&quot; rel=&quot;nofollow&quot;&gt;Scutum article in Polish Wikipedia&lt;/a&gt; happens to do so as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/9/90/Scutum_constelation_PP3_map_PL.jpg&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/9/90/Scutum_constelation_PP3_map_PL.jpg&quot; alt=&quot;Scutum in Polish Wikipedia&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the name of the constellation has been changed but I can't find any information on who did it and when. I suppose it must have been the IAU (International Astronomical Union) but I have no idea regarding the time of the change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When was the English name of the constellation officially changed and who did it?&lt;/p&gt;&#xA;" OwnerUserId="211" LastActivityDate="2019-02-06T10:30:44.623" Title="When did the Scutum constellation receive its current name?" Tags="&lt;history&gt;&lt;constellations&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12972" PostTypeId="2" ParentId="12947" CreationDate="2015-12-23T16:43:44.797" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I read that a black hole can sometime &quot;choke&quot; on a star: &lt;em&gt;&quot;...the disrupted stellar matter was generating so much radiation that it pushed back on the infall. The black hole was choking on the rapidly infalling matter.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I read the report and thought it was reasonable, and in line with Rob's answer. But note that there's no certainty that this is what actually happened. They saw a very bright &quot;optical transient&quot; event, circa ten times brighter than a normal supernova. See the paper on the arXiv: &lt;a href=&quot;http://arxiv.org/abs/1410.6014&quot; rel=&quot;nofollow&quot;&gt;A Luminous, Fast Rising UV-Transient Discovered by ROTSE: a Tidal Disruption Event?&lt;/a&gt; There's a question mark on the end of the title, they don't know for sure. And see this from the news article:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;To narrow it down from four possibilities, they studied Dougie with the orbiting Swift telescope and the giant Hobby-Eberly Telescope at McDonald, and they made computer models. These models showed how Dougie's light would behave if created by different physical processes. The astronomers then compared the different theoretical Dougies to their telescope observations of the real thing.&quot;&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They came up with what they thought was the best fit. But we don't know for sure that this was actually a black hole &quot;choking&quot; on a star. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does a black hole choke? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We don't know for sure. Note in the paper on page 8 and 9 they talk about an off-axis GRB interpretation. It's possible that this is what happened. Or something else altogether. The tidal disruption idea looks like a good fit, but they don't know for certain. Anyway, have a look at the summary on page 12 for a nice  round-up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;The tidal disruption scenario was explored by fitting the event to an amended version of the model presented in Guillochon et al. (2014). The TDE model yielded a good fit to the photometric and spectral evolution of the flare, with the highest-likelihood models suggesting a disruption of a solar-mass star by a black hole.&quot;&lt;/em&gt;     &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do astronomers manage to observe this rare event?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They didn't actually observe a black hole choking on a star. What they saw was an optical transient. Something that looked like a supernova, but didn't fit the supernova pattern. So they think it was the disruption of a star by a black hole.    &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2015-12-23T16:43:44.797" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12973" PostTypeId="1" CreationDate="2015-12-23T16:44:30.647" Score="3" ViewCount="276" Body="&lt;p&gt;Wolf 1061 (only 13.8 ly away) was recently found to have three rocky planets one of which is in the habitable zone. It was stated on the english wikipedia site that it has a very stable light curve and was not found to have &quot;any significant activity such as sunspots or flares.&quot; &#xA;Would this imply that this star is very old? The expected lifetime of this M dwarf with a mass estimated to be 0.25 solar mass could potentially reach a trillion years.&lt;/p&gt;&#xA;" OwnerUserId="7786" LastEditorUserId="7786" LastEditDate="2015-12-23T21:36:36.920" LastActivityDate="2015-12-24T10:43:40.933" Title="Is there a way to estimate the age of M dwarf stars?" Tags="&lt;red-dwarf&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12974" PostTypeId="1" CreationDate="2015-12-23T18:13:33.063" Score="3" ViewCount="132" Body="&lt;p&gt;My question is, during the evolution of a solar system is it possible to end up with a massive planet around a star with two small planets around their $L_4$ and $L_5$.&lt;br&gt;&#xA;Since we know it is not possible to have any bodies around $L_1, L_2, L_3$ due to their low stability, hence any planet trying to evolve their can be easily nudges out of its position.&lt;/p&gt;&#xA;" OwnerUserId="10177" LastActivityDate="2016-03-08T21:20:03.427" Title="Theoretically, can a solar system evolve around lagrange points" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;lagrange-point&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12976" PostTypeId="2" ParentId="12962" CreationDate="2015-12-23T19:41:19.683" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Deep_Impact_%28spacecraft%29&quot; rel=&quot;nofollow&quot;&gt;Deep Impact&lt;/a&gt;, the NASA comet probe, not the movie, may be as close as you can get to what you want now.&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=TA5Y9FBR2DM&quot; rel=&quot;nofollow&quot;&gt;Here's a video of the action&lt;/a&gt; when &lt;a href=&quot;http://www.nasa.gov/mission_pages/deepimpact/spacecraft/impactor.html&quot; rel=&quot;nofollow&quot;&gt;370 kg of copper&lt;/a&gt; hits a comet at 10.2 km/s.&#xA;There may be better video out there, but the movie trailers really mess up the search.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="2031" LastEditDate="2015-12-24T13:34:56.260" LastActivityDate="2015-12-24T13:34:56.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12977" PostTypeId="2" ParentId="12973" CreationDate="2015-12-23T21:09:29.347" Score="1" Body="&lt;p&gt;Really difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very young ($&amp;lt;50$ Myr) M dwarfs may still exhibit lithium in their photospheres. Any older and it will have been depleted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For older dwarfs you then move on to looking at rotation and activity. Both decline with age, but in an M dwarf as cool as this, the decay timescale is many Gyr, so it is only weakly constraining and not very well calibrated for such low mass stars. A very slow rotation rate or very low level of magnetic activity might indicate it was older than 5 Gyr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lack of flares and the lack of light curve modulation by starspots &lt;em&gt;may&lt;/em&gt; indicate that it has spun down, is magnetically inactive and is therefore older than 5 Gyr ( I cannot immediately locate the source of this Wikipedia claim). The ratio of X-ray to bolometric luminosity would be more definitive and could probably be derived from the ROSAT all-sky survey (or a constraining upper limit would be found).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A useful paper is by Stelzer et al. (2013). &lt;a href=&quot;http://arxiv.org/abs/1302.1061&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1302.1061&lt;/a&gt;&#xA;She finds various activity diagnostics for nearby stars, including this one. Its projected rotation velocity is 1.5 km/s, which suggests a rotation period of $\sim 10$ days for a star of this size. This could not be less constraining on the age! The X-ray to bolometric flux ratio is $10^{-4.45}$ which suggests it is neither very young or very old - say between 1 and 5 Gyr for a star of this mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is to look at the Galactic space motion and position. Older stars &lt;em&gt;tend&lt;/em&gt; to have higher velocities, especially perpendicular to the Galactic plane, and also &lt;em&gt;tend&lt;/em&gt; to be found further from the Galactic plane. The velocities for this object do not look large, suggesting it is not very old, ie younger than perhaps 10 Gyr and likely a young disk object with age of $&amp;lt;5$ Gyr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also look at the metallicity. This is difficult to measure in M-dwarfs. Maldonado et al. (2015) give [Fe/H]$=-0.05 \pm 0.03$. This is very close to the solar value and does not constrain the age beyond suggesting the star is younger than 10Gyr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond this, it is basically impossible. Even if you had a very precise distance and could place the star on a HR diagram, low mass M dwarfs evolve so little in luminosity and temperature over many Gyr that this offers no constraint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my conclusion - the age is between 1 and 5 Gyr, with the probability peaking somewhere in the middle.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-12-24T10:43:40.933" LastActivityDate="2015-12-24T10:43:40.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12979" PostTypeId="2" ParentId="11474" CreationDate="2015-12-24T11:56:28.580" Score="1" Body="&lt;p&gt;The rotational period ( or whether a body is tidal locked ) is a factor here.  A tidal locked body has a hot spot on one side and a cold spot on another.  Rotation may also affects average temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Season depends on orbital factors.  Simply saying it's one season or another does not tell you the distance you are from the star, which is the single most important factor ( combined with the power of the star ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in terms of making a very basic calculation you can use a black body radiation calculation and thermal equilibrium to get something.  You can see more on this at these links :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://claesjohnson.blogspot.ie/2010/04/black-body-temperatures-of-planets.html&quot; rel=&quot;nofollow&quot;&gt;http://claesjohnson.blogspot.ie/2010/04/black-body-temperatures-of-planets.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.acs.org/content/acs/en/climatescience/energybalance/predictedplanetarytemperatures.html&quot; rel=&quot;nofollow&quot;&gt;http://www.acs.org/content/acs/en/climatescience/energybalance/predictedplanetarytemperatures.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a pretty good predictor considering it's simplicity.  Note the impact of atmosphere and ocean.  Note also that significant energy is being radiated by some bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This slide show has several equations that may help you.  It also has an interesting model on the effect of an ice cap, which I'd not seen before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.umn.edu/~mcgehee/Seminars/ClimateChange/presentations/2011-2Fall/20111102Handouts.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.umn.edu/~mcgehee/Seminars/ClimateChange/presentations/2011-2Fall/20111102Handouts.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2015-12-24T11:56:28.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12980" PostTypeId="1" AcceptedAnswerId="12990" CreationDate="2015-12-24T18:39:48.530" Score="5" ViewCount="281" Body="&lt;p&gt;Detecting planets orbiting around stars can be made with several &lt;a href=&quot;https://en.wikipedia.org/wiki/Methods_of_detecting_exoplanets&quot; rel=&quot;noreferrer&quot;&gt;methods&lt;/a&gt;, transit photometry seems to be by far the most popular method nowadays. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder how we could detect planets orbiting black holes. Which methods are we left with? Has it been done before?  &lt;/p&gt;&#xA;" OwnerUserId="10232" LastActivityDate="2015-12-26T15:00:54.387" Title="How could we detect planets orbiting black holes?" Tags="&lt;planet&gt;&lt;black-hole&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12981" PostTypeId="1" CreationDate="2015-12-24T20:29:20.930" Score="1" ViewCount="273" Body="&lt;p&gt;What it says on the tin: using the most efficient launch opportunity afforded by orbital mechanics, what delta-v would you have to impart to a payload to send it from the surface of Mercury to crash on the surface of Venus?&lt;/p&gt;&#xA;" OwnerDisplayName="user10233" LastActivityDate="2016-02-25T22:15:42.340" Title="Delta-v from Mercury surface to Venus surface" Tags="&lt;venus&gt;&lt;mercury&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12984" PostTypeId="1" CreationDate="2015-12-25T00:56:14.747" Score="2" ViewCount="80" Body="&lt;p&gt;How pristine and ancient are asteroids really? I suppose they melt and reform when they collide with each other. How frequently do they collide &lt;strong&gt;with each other&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-12-25T11:34:50.787" LastActivityDate="2016-01-05T01:09:34.833" Title="How long time since a typical asteroid collided with another asteroid?" Tags="&lt;asteroids&gt;&lt;impact-probability&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12985" PostTypeId="2" ParentId="12984" CreationDate="2015-12-25T02:00:27.803" Score="3" Body="&lt;p&gt;How ancient they are, I'd have to say as old as the solar system.&#xA;Collision frequency:&lt;br&gt;&#xA;&lt;strong&gt;Collision amongst Asteroid Belts&lt;/strong&gt;&#xA;From ref 1  below: Based on the particle in a box model which assumes a homogeneous distribution in the interactive volume, the intrinsic collision probability is computed to be around $5 \times 10^{-18} km^{-2}yr^{-1}$.&lt;br&gt;&#xA;However Wetherill(ref below) have some modifications to Opik's formulation for collision between two elliptical orbits, suggests that the impact is nearly independent of the orbital elements of the test body and is nearly half of the value obatined from PIAB model. It also seems to suggest that intrafamily collisions are higher than collision amongst non family objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Collision with the Earth&lt;/strong&gt; Here it characterized in various classes:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;bolides or upper atmospheric explosion,  &lt;/li&gt;&#xA;&lt;li&gt;land impacts leading to local or regional devastation,  &lt;/li&gt;&#xA;&lt;li&gt;ocean impacts leading to regional disasters,  &lt;/li&gt;&#xA;&lt;li&gt;global ecological catastrophe,  &lt;/li&gt;&#xA;&lt;li&gt;mass extinctions  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Based on the scale, the impact events are categorized as harmless, Class-1, to devastating, Class-5.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 440-KT Chelyabinsk event of 2013 belongs to impact hazard class 2, causing regional or local devastation. This was caused by a mere 20 meters wide meteor entering the atmosphere of Earth. For a cumulative number of impacts described for a given reference time interval of $T_{ref}$ = 1Ga ($10^9$ years)&#xA;which is long enough to statistically include a meaningful number of impacts from all the classesabove, the class 2 impacts cumulative will be $2\times 10^6$ , with small bodies falling in the size range of 20-100 m. Where the characteristic time scale is of $10^2$ a.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For class 3-5 events leading to regional disasters to global extinction events, the cumulative number of impacts within the reference interval is $10^4$ with characteristic time scale being $10^5$ a. These events can lead to large devastation ranging from an entire country or region (class 3, with an explosive energy of approximately 104 MT TNT), to mass extinctions (class 5 releasing an approximate energy 108 MT of TNT). An example from the history for such event is the classic K-T extinction events 65 Ma ago that wiped out more than 99% of the living species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;br&gt;&#xA;1. &lt;a href=&quot;http://articles.adsabs.harvard.edu//full/1992LPI....23..283D/0000283.000.html&quot; rel=&quot;nofollow&quot;&gt;http://articles.adsabs.harvard.edu//full/1992LPI....23..283D/0000283.000.html&lt;/a&gt;&lt;br&gt;&#xA;2. &lt;a href=&quot;http://www.springer.com/fr/book/9780792354666&quot; rel=&quot;nofollow&quot;&gt;http://www.springer.com/fr/book/9780792354666&lt;/a&gt;&lt;br&gt;&#xA;3. &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1029/JZ072i009p02429/abstract&quot; rel=&quot;nofollow&quot;&gt;http://onlinelibrary.wiley.com/doi/10.1029/JZ072i009p02429/abstract&lt;/a&gt;&lt;br&gt;&#xA;(PS: If anyone has a link to the open access for final paper, or a pdf; I would really appreciate a copy.)&lt;/p&gt;&#xA;" OwnerUserId="10177" LastEditorUserId="10177" LastEditDate="2016-01-05T01:09:34.833" LastActivityDate="2016-01-05T01:09:34.833" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12986" PostTypeId="2" ParentId="12980" CreationDate="2015-12-25T15:59:24.047" Score="2" Body="&lt;p&gt;Theoretically we could use radial velocity using X ray or gamma ray spectra, but it would only work with relatively active black holes. If a black hole doesn't have an accretion disk detecting planets around it is impossible, because all methods require at least partial visibility of an object and black holes are invisible unless you can see their influence in their surroundings (stars orbiting invisible objects, accretion disks etc.). Besides, chances that a planet could exist near a black hole are very slim, &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10373/life-planets-orbiting-black-holes-can-do-they-really-exist&quot;&gt;detalled in this post (1st answer)&lt;/a&gt;, because black holes are formed in supernova explosions which aren't very planet friendly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way a planet could exist near a black hole is if it's a captured planet far enough away to not get torn by tidial forces or if one of the stars orbiting a black hole has planets, while we could detect the latter kind of planets using one of the known methods they wouldn't count as planets orbiting a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far no planets orbiting a black holes were found due to reasons stated above.&lt;/p&gt;&#xA;" OwnerUserId="10228" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2015-12-25T15:59:24.047" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12989" PostTypeId="1" AcceptedAnswerId="12993" CreationDate="2015-12-26T06:59:49.367" Score="5" ViewCount="153" Body="&lt;p&gt;After reading &lt;a href=&quot;http://www.sciencealert.com/this-timeline-shows-the-entire-history-of-the-universe-and-where-it-s-headed&quot; rel=&quot;nofollow&quot;&gt;this space timeline&lt;/a&gt;, I have wondered how they managed to make these assumptions about various events like when the first black holes formed. This is obviously far beyond the good ol' carbon-14 dating procedure.&lt;/p&gt;&#xA;" OwnerUserId="10246" OwnerDisplayName="Silviu Burcea" LastEditorUserId="2365" LastEditDate="2016-01-12T20:50:45.650" LastActivityDate="2016-01-12T20:50:45.650" Title="Dating very old objects/events" Tags="&lt;age&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12990" PostTypeId="2" ParentId="12980" CreationDate="2015-12-26T10:58:07.027" Score="4" Body="&lt;p&gt;Gravitational microlensing is a way of finding planets that does not care how luminous or otherwise the hosts of the exoplanets are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way it would work is that you stare at a dense background field of stars; then when a foreground black hole passes in front of a background star, the light from the star is magnified by gravitational lensing. Typically, the lensing event takes a few weeks for the magnification to develop and subside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the black hole has a planet, it may be possible to &quot;see&quot; its graviational potential, which will manifest itself as an asymmetry in the lens light curve or possibly even a little extra spike in the light curve lasting a few hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This technique is well established and is already used to detect planets around unseen objects. It is more sensitive than the transit and Doppler methods to planets orbiting a fair distance from their parent star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty here is not finding planets around a black hole, it is proving that the planets you found were actually orbiting a stellar sized black hole. The microlensing event would only be seen once and the black hole system would likely be invisible. There is a possibility I suppose that it may undergo some sort of accretion activity after having been found by lensing, and this accretion might then be picked up by telescopes trying to identify the lensing source. I guess a good place to start would be events where a stellar-mass lens is inferred, but no star can be seen after sufficient time has elapsed that the lens and source star ought to be resolved.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2015-12-26T15:00:54.387" LastActivityDate="2015-12-26T15:00:54.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12992" PostTypeId="1" CreationDate="2015-12-26T12:26:25.527" Score="9" ViewCount="680" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/67P/Churyumov%E2%80%93Gerasimenko&quot;&gt;Comet 67P&lt;/a&gt;/Churyumov–Gerasimenko flew by Jupiter at only 0.05 AU distance in 1959. This encounter lowered its perihelion from 2.7 to 1.3 AU (almost becoming an Earth crossing comet). Since it happens to be the target of the Rosetta mission (and was picked only because of a launch delay), I wonder if it is as common as it seems that a 4+ billion year old comet shifted its orbit drastically less than 60 years ago?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How often does Jupiter do something like this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many known comets are known to have had their orbits substantially changed by a single flyby?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-12-26T17:20:29.640" LastActivityDate="2016-03-09T06:10:24.317" Title="How frequently does a comet drastically change its orbit in the Solar System (like 67P did in 1959)?" Tags="&lt;comets&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12993" PostTypeId="2" ParentId="12989" CreationDate="2015-12-26T14:21:13.020" Score="2" Body="&lt;p&gt;One property of distant objects, that follows from the limited speed of light, is that we are observing them &lt;em&gt;in the past&lt;/em&gt;; the light that they emit takes time to reach us. For very distant objects, we are effectively observing the universe at a very early time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objects that are very far away, such as distant galaxies, are receding away from us, which causes the light that they emit to be doppler shifted. The recession velocity is proportional to the distance away, and in turn so is the magnitude of the redshift. This redshift allows us to put a number to the distance away; combining this with the speed of light limit allows us to determine how old the light we are observing from them is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two effects, the limited speed of light and redshift, effectively allow us to determine an evolutionary timeline for the universe, and place constraints on when certain objects appear.&lt;/p&gt;&#xA;" OwnerUserId="941" LastEditorUserId="941" LastEditDate="2015-12-26T17:44:26.103" LastActivityDate="2015-12-26T17:44:26.103" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12994" PostTypeId="2" ParentId="12989" CreationDate="2015-12-26T18:16:28.283" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I have wondered how they managed to make these assumptions about various events like when the first black holes form.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When the first black holes formed ( and many other things ) are things we predict from theoretical models of the development of the universe.  They are no proven facts, but better classed as speculative predictions based on theories.  Different theories give different predictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way it works ( broadly ) is this : We try and develop theoretical models of the universe's creation and early development that will develop into a universe that matches what ours looks like.  Doing this has led to such controversial concepts as string theory, which are not proven facts in a scientific sense, but theories subject to ongoing investigation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We try and verify ( or disprove ) theories by comparing what they predicts with what we see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So these &quot;histories&quot; should be considered as &quot;best guesses&quot; to some extent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The approximate date of the Sun's formation, for example, is based on our current best models for the evolution of stellar bodies like the sun, which of course are based on our current best measurements of whatever we can measure in relation to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For history related to objects we can examine in more detail, like stuff on Earth, some of which came from space, we can perform detailed analysis and look for patterns in related objects and areas.  This makes estimates dating from the start of life on Earth on reasonably good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the far distant history beyond that is concerned we're relying primarily on models driven by astronomical observations and the law of physics we regard as &quot;safe&quot;.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2015-12-26T18:16:28.283" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12996" PostTypeId="1" CreationDate="2015-12-27T09:12:46.893" Score="4" ViewCount="2582" Body="&lt;p&gt;We know that it is a commonly heard claim that in every black hole there is a universe. It is also believed that the universe is expanding. If we use both of these, can we conclude that every black hole is expanding?&lt;/p&gt;&#xA;" OwnerUserId="9231" LastEditorUserId="4085" LastEditDate="2015-12-30T05:39:25.887" LastActivityDate="2018-05-23T00:10:04.043" Title="Are black holes expanding?" Tags="&lt;black-hole&gt;&lt;universe&gt;&lt;expansion&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12998" PostTypeId="2" ParentId="12996" CreationDate="2015-12-27T17:44:16.500" Score="7" Body="&lt;p&gt;Most$^\dagger$ black holes &lt;em&gt;are&lt;/em&gt; indeed expanding, but not because the Universe is expanding. Rather, their size (more precisely their &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;noreferrer&quot;&gt;Schwarzschild radius&lt;/a&gt;) increases proportionally to their mass, so they grow as they accrete more matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a common misconception that everything expands along with the expanding Universe. It doesn't. On small scales, i.e. black holes, planets, stars, galaxies, and even galaxy groups, gravity ensures that these things don't increase in size. Only on large scales, i.e. galaxy clusters and beyond, does the expanding Universe pull everything apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the theory about a universe inside, I… I don't even wanna go there. I'll just refer to HDE 226868's comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;Very small black holes will decrease in size due to &lt;a href=&quot;https://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;noreferrer&quot;&gt;Hawking evaporation&lt;/a&gt;.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-12-27T22:35:09.200" LastActivityDate="2015-12-27T22:35:09.200" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12999" PostTypeId="2" ParentId="12981" CreationDate="2015-12-27T19:03:35.617" Score="0" Body="&lt;p&gt;If you launch from the side of Mercury furthest from the Sun, &#xA;parallel to the horizon in the direction of Mercury's orbit at &#xA;13 km/s when the Mercury-Sun-Venus angle is 53 degrees, your &#xA;payload will reach Venus in 35 days, and its path would look &#xA;something like this (each black dot = 1 day). IMPORTANT CAVEATS BELOW IMAGE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MlWxO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MlWxO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't really answer your question, since I'm not sure &#xA;13 km/s is the minimal required speed. I also made several &#xA;simplifying assumptions (perhaps too many), so the answer &#xA;above isn't exact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the heavily commented Mathematica script I wrote, which &#xA;also explains the simplifications I made. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&#xA;(* cleaned up and well-commented version for SE *)&#xA;&#xA;(*&#xA;&#xA;I assume that Mercury's orbit is planar, and that Venus' orbit lies in&#xA;the same plane, or at least intersects it at the appropriate time.&#xA;&#xA;This allows me to use a two-dimensional equation for acceleration due&#xA;to gravity, instead of a 3 dimensional one.&#xA;&#xA;The equation (below) is the acceleration imparted to an object of mass&#xA;m1 at {x1,y1} by an object of mass m2 at {x2,y2}, given that the&#xA;gravitational constant is g.&#xA;&#xA;Note that the mass of the object being accelerated (m1) is actually&#xA;irrelevant; however, I include it as a parameter for symmetry&#xA;&#xA;*)&#xA;&#xA;accel[{x1_,y1_},{x2_,y2_},m1_,m2_,g_]=g*m2/Norm[{x2-x1,y2-y1}]^3*{x2-x1,y2-y1}&#xA;&#xA;(* &#xA;&#xA;The mass, semimajor axis, period, and radius of Mercury, in kg, m (not&#xA;km), and s&#xA;&#xA;*)&#xA;&#xA;mercsma = 57909050000;&#xA;mercper = 87.9691*86400&#xA;mercrad = 2439700&#xA;mercmass = 3.3011*10^23&#xA;&#xA;(* solar mass, in kg *)&#xA;&#xA;sunmass = 1.98855*10^30&#xA;&#xA;(* gravitational constant of universe, in kg-m-s system *)&#xA;&#xA;g = 6.6740*10^-11&#xA;&#xA;(* &#xA;&#xA;Heliocentric, so Sun is always at origin. In theory, the positions of&#xA;the other planets (eg, Jupiter) could help boost your payload, so you&#xA;might be able to launch with a lower speed than I find below&#xA;&#xA;*)&#xA;&#xA;sun[t_] = {0,0}&#xA;&#xA;(*&#xA;&#xA;I assume Mercury's orbit is circular. Since the actual orbit is&#xA;elliptical, you could get a boost for your payload by launching it&#xA;when Mercury's distance from the Sun is increasing the fastest (in&#xA;other words, solar radial velocity is greatest)&#xA;&#xA;I've chosen the x axis to be the line connecting the Sun to Mercury at time 0.&#xA;&#xA;*)&#xA;&#xA;merc[t_] = {Cos[t*2*Pi/mercper],Sin[t*2*Pi/mercper]}*mercsma&#xA;&#xA;(*&#xA;&#xA;I also ignore Venus' own gravity: you can do slightly better by noting&#xA;that Venus will pull the payload towards itself once the payload gets&#xA;close enough.&#xA;&#xA;I do want to plot Venus' orbit, so I use the semi-major axis and&#xA;period values below.&#xA;&#xA;Venus' starting angle (vsa below) was found by trial and error to make&#xA;sure Venus was at the right place when the payload crossed its orbit.&#xA;&#xA;*)&#xA;&#xA;vensma = 108208000000&#xA;venper = 224.701*86400;&#xA;vsa = 53*Degree;&#xA;ven[t_] = {Cos[t*2*Pi/venper+vsa],Sin[t*2*Pi/venper+vsa]}*vensma&#xA;&#xA;(*&#xA;&#xA;If we launch from side of Mercury furthest from the Sun, the payload's&#xA;starting position will be Mercury's position plus Mercury's radius in&#xA;the x direction&#xA;&#xA;NOTE: This start position is completely arbitrary; you may get better&#xA;results by starting at different positions on Mercury's surface.&#xA;&#xA;*)&#xA;&#xA;s0 = {mercsma+mercrad,0}&#xA;&#xA;(*&#xA;&#xA;The initial velocity of the payload (with respect to the Sun) will be&#xA;Mercury's velocity + whatever velocity (delta v) we impart to the&#xA;payload.&#xA;&#xA;Note that both the direction I choose for initial velocity (in the&#xA;same direction as Mercury's orbit) and the magnitude are&#xA;arbitrary. You may get better results by aiming the payload at a 45&#xA;degree angle or straight up or something.&#xA;&#xA;NOTE: If I change 13000 to 12500 below, Mathematica will refuse to&#xA;solve the differential equation. This doesn't necessarily mean 13000&#xA;is a minimal velocity, but there is apparently some sort of important&#xA;change between 12500 m/s and 13000 m/s&#xA;&#xA;*)&#xA;&#xA;v0 = merc'[0] + {0,13000}&#xA;&#xA;(*&#xA;&#xA;Mathematica won't close-form integrate this problem, so we integrate&#xA;numerically, which requires a start time (0) and an end time (below).&#xA;&#xA;I chose 35 days after confirming that's how long it takes the payload&#xA;to reach Venus.&#xA;&#xA;*)&#xA;&#xA;timelimit = 86400*35;&#xA;&#xA;nds = NDSolve[{s[0]==s0, s'[0] == v0,&#xA; s''[t] == accel[s[t],sun[t],1,sunmass,g] + accel[s[t],merc[t],1,mercmass,g]&#xA;},s,{t,0,timelimit}]&#xA;&#xA;(* The use of [[1,1,2]] below is just Mathematica nesting weirdness *)&#xA;&#xA;g= ParametricPlot[{nds[[1,1,2]][t],merc[t],ven[t]},{t,0,timelimit}, &#xA; Mesh -&gt; timelimit/86400, AxesOrigin-&gt;{0,0}, PlotStyle -&gt; {Blue,Red,Green},&#xA; MeshStyle -&gt; {Black}&#xA;]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2015-12-27T19:03:35.617" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13001" PostTypeId="2" ParentId="12996" CreationDate="2015-12-27T20:02:08.037" Score="6" Body="&lt;p&gt;In the standard ΛCDM cosmology, the expansion of the universe is modelled by a cosmological constant $\Lambda$, usually interpreted as a constant density of dark energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nonzero cosmological constant somewhat modifies the usual black hole geometries. For example, the simplest Schwarzschild black hole (isolated, uncharged, nonrotating) is described by&#xA;$$\mathrm{d}s^2 = -\left(1-\frac{2GM}{r}\right)\mathrm{d}t^2 + \left(1-\frac{2GM}{r}\right)^{-1}\mathrm{d}r^2 + r^2\mathrm{d}\Omega^2\text{,}$$&#xA;which with a nonzero $\Lambda$ would be the Schwarzschild-de Sitter spacetime described by the replacement&#xA;$$\left(1-\frac{2GM}{r}\right)\mapsto\left(1-\frac{2GM}{r}-\frac{\Lambda r^2}{3}\right)\text{.}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, a cosmological constant would not cause a black hole to expand or contract. Rather, its size (i.e. of the horizon) is slightly modified by the cosmological constant. The same kind of thing happens to other bound systems, such as star systems and galaxies: in principle the expansion of the universe modifies their size by a slight amount, but the forces that bind them simply find an equilibrium against cosmic expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would take some extended model of dark energy, in which it is not a constant, to expand of shrink black holes. But there is currently no evidence of such a thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side note: since the Hawking temperature of any astrophysical black hole would be much lower than the surrounding space (e.g., cosmic microwave background), we can expect black holes to slightly expand due to absorbing more energy from their environment than their lose by Hawking radiation. But this effect is both tiny and has a completely different mechanism than the expansion of the universe.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2015-12-27T20:02:08.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13002" PostTypeId="2" ParentId="12967" CreationDate="2015-12-27T21:42:06.587" Score="2" Body="&lt;p&gt;IAU's &lt;a href=&quot;http://www.iau.org/public/themes/naming/&quot; rel=&quot;nofollow&quot;&gt;official website&lt;/a&gt; mentions that the names of the 88 current constellations (Scutum included) were standardised during the organizations inaugural meeting in 1922 and apparrently haven't been changed ever since.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Celestial nomenclature has long been a controversial topic. At its inaugural meeting in 1922 in Rome, the IAU standardized the &lt;a href=&quot;http://www.iau.org/public/themes/constellations/&quot; rel=&quot;nofollow&quot;&gt;constellation names and abbreviations&lt;/a&gt;. More recently IAU Committees or Working Groups have certified the names of astronomical objects and features. In the following links you can find further information on how different objects and features are named.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="211" LastActivityDate="2015-12-27T21:42:06.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13003" PostTypeId="2" ParentId="12996" CreationDate="2015-12-28T00:10:56.453" Score="3" Body="&lt;p&gt;&lt;strong&gt;Simple Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Large black holes are usually expanding by an incredibly small amount as they suck in more stuff (gases, planets, stars, etc.) through gravity. So they are expanding but not because of our expanding universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exceptions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia, small black holes might shrink. Stephen Hawking predicts that all black holes have radiation. Small black holes, that suck in less, might emit more energy than they pull in, so they theoretically shrink and close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Big Asterisk&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many additional details that could be included in this thread. I think the most important to point out is that fifty years ago black holes were still largely science fiction. So it's a relatively new science. And even if we had been studying them for 200-300 years, they're hard to observe and practically impossible to experiment with. Point being, most black hole knowledge is actually black hole theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Observing Black Holes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a relevant &lt;a href=&quot;https://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow&quot;&gt;Wiki&lt;/a&gt; excerpt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In June 2008, NASA launched the Fermi space telescope, which is&#xA;  searching for the terminal gamma-ray flashes expected from evaporating&#xA;  primordial black holes. In the event that speculative large extra&#xA;  dimension theories are correct, CERN's Large Hadron Collider may be&#xA;  able to create micro black holes and observe their&#xA;  evaporation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Relevant Detail on Black Holes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes were once massive stars. Stars have massive gravity but they don't collapse until they run out of fuel. When they do run out of fuel, they expand then collapse. Big stars have so much gravity that they collapse into a small sphere with gravity so intense that light can't escape it. That is when a black &quot;hole&quot; is born. Really, it's more like a black sphere. It appears to be a hole only because no light escapes. Inside the sphere there could be a hole, but no one knows. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Further Explanation on Black Hole Expansion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bigger stars have more mass so when they collapse, they have more gravity and the perceived &quot;hole&quot; is bigger. Typically, large galaxies have large black holes at their center and small galaxies have small black holes. Over time, the black hole will pull more matter (gases, planets, asteroids, etc.) into its sphere of blackness. This adds to its mass and slowly increases its gravity. More gravity means a wider radius of black where from which not even light escapes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Unanswered / Theorized&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When does a black hole stop growing and why? That's hard to answer because we have no data about the inside of a black hole. Some people theorize the immense gravity bends space/time to create a wormhole. Many questions about space are being answered in our lifetime by observing and experimenting. Black holes are hard to observe and even harder to experiment with. Most &quot;answers&quot; in our lifetime will be more theory than proven physics. &lt;/p&gt;&#xA;" OwnerUserId="10247" LastEditorUserId="10247" LastEditDate="2015-12-28T00:24:32.760" LastActivityDate="2015-12-28T00:24:32.760" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13004" PostTypeId="2" ParentId="12962" CreationDate="2015-12-28T00:58:21.927" Score="0" Body="&lt;p&gt;&lt;strong&gt;It's a Needle in a Haysta... In The Pacific Ocean&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes galaxies' gravities catch each other and they slowly merge. Even with billions of stars and planets merging with billions of other stars and planets, its theorized that almost no collisions take place. Point being that although it does happen:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Space is mostly &quot;empty&quot; and statistical odds of collision are close to nil&lt;/li&gt;&#xA;&lt;li&gt;Unlike planets, astroids often have complex trajectories that aren't shared with other solid bodies (two asteroids very rarely even have a similar route, like planets and moons do)&lt;/li&gt;&#xA;&lt;li&gt;Humanity's ability to monitor the night sky is limited. We see very little of what happens out there.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Sorry to say but I think you're unlikely to find what you're looking for!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Next Best Bet:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try Googling asteroid belt collisions? Or collisions in planet's rings? Those areas are rocks and ice, so a collision might be similar to an asteroid/asteroid collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No two asteroid collisions would be alike anyhow. They would vary widely. If one is huge and one relatively small, that collision would be very different from a collision where they were the same size. Or head on versus rear-ending, etc. So, you've got some artistic freedom for your video game. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, you could Google collisions of asteroids that hurdled into terrestrial  planets or moons without atmospheres. Watching a collision with Earth or Titan does you no good because the atmosphere changes the dynamic of the impact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just remember to think of the physics. If the two sides colliding are well lit, there shouldn't be ice. Well-lit means the sun likely melted the ice. The ice is on the dark side and in the crevasses. &lt;/p&gt;&#xA;" OwnerUserId="10247" LastActivityDate="2015-12-28T00:58:21.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13005" PostTypeId="2" ParentId="12953" CreationDate="2015-12-28T01:11:12.323" Score="2" Body="&lt;p&gt;It's important to point out that geological activity on moons is &lt;strong&gt;not only common, but its at the core of modern space exploration&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because moon geological activity is often caused by tidal forces, the activity correlates with heat. Tidal forces create friction, which in turn creates heat. Moderate heat is an important part of creating an atmosphere and nurturing life. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moons Europa and Titan both are geologically active present the best chances for finding organic life in our solar system.&lt;/p&gt;&#xA;" OwnerUserId="10247" LastActivityDate="2015-12-28T01:11:12.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13006" PostTypeId="2" ParentId="12576" CreationDate="2015-12-28T01:26:09.497" Score="0" Body="&lt;p&gt;&lt;strong&gt;Don't Forget This Part!!&lt;/strong&gt; The other answers simply explain that an infinite universe can continuously expand because the increase in size is filled with space, not new matter. They confirm your explanation. However, the answer to &quot;how can an infinite universe expand?&quot; requires an explanation of &quot;empty space&quot; and mention of the fact that the universe's expansion is also &lt;em&gt;accelerating&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Empty Space&lt;/strong&gt; gives physicists headaches. Is there even such a thing as empty space? Or just space filled with matter we cannot yet detect? More on this below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Accelerating Expansion:&lt;/strong&gt; The universe is not only expanding, &lt;em&gt;it's speeding up&lt;/em&gt;. That keeps some very smart people awake at night. &quot;How can an infinite universe expand?&quot; Well the big bang would explain it... if it were slowing down. It's not. So... it has to be pushed or pulled, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why Does That Happen?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Einstein's theories first predicted that the gravity of the universe would cause the universe to collapse in on itself so he introduced something called the &quot;cosmological constant&quot; in his theory of relativity. That constant enabled the equations to predict a static universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmological_constant&quot; rel=&quot;nofollow&quot;&gt;Wiki&lt;/a&gt; excerpt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[The cosmological constant] was originally introduced by Albert&#xA;  Einstein in 1917 as an addition to his theory of general relativity to&#xA;  &quot;hold back gravity&quot; and achieve a static universe, which was the&#xA;  accepted view at the time. Einstein abandoned the concept after&#xA;  Hubble's 1929 discovery that all galaxies... are moving away from each&#xA;  other, implying an overall expanding universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Modern Understanding&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although Einstein would later call that &quot;cosmological constant&quot; his &quot;greatest blunder&quot;, it turned out to be correct. However, he also thought the value would be used to explain why things are static. It turned out the value would explain why things are continually expanding. But what is that constant, besides a number in an equation? Today most physicists seem to think it's dark matter or dark energy. It's a very new science but it is highly researched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Accelerating_universe#Dark_energy&quot; rel=&quot;nofollow&quot;&gt;Wiki&lt;/a&gt; excerpt from 'Accelerating Universe': &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Different theories of dark energy suggest different values of w, with&#xA;  w &amp;lt; -1/3 for cosmic acceleration... The simplest explanation for dark&#xA;  energy is that it is a cosmological constant... this leads to the&#xA;  Lambda-CDM model, which has generally been known as the Standard Model&#xA;  of Cosmology from 2003 through the present...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmological_constant&quot; rel=&quot;nofollow&quot;&gt;Wiki&lt;/a&gt; excerpt from Dark Energy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;around 70% of the mass–energy density of the universe can be&#xA;  attributed to dark energy. While dark energy is poorly understood at a&#xA;  fundamental level, the main required properties of dark energy are&#xA;  that it functions as a type of anti-gravity, it dilutes much more&#xA;  slowly than matter as the universe expands... The cosmological constant&#xA;  is the simplest possible form of dark energy since it is constant in&#xA;  both space and time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answering Your Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think the real answer to your question isn't that an infinite universe is possible simply because the amount of empty space increases. Actually, that might not even be possible. There may be no such thing as empty space. Empty space might just be a type of matter we can't see or detect yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_matter&quot; rel=&quot;nofollow&quot;&gt;Wiki&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dark matter is a hypothetical kind of matter that cannot be seen with&#xA;  telescopes but accounts for most of the matter in the universe. The&#xA;  existence and properties of dark matter are inferred from its&#xA;  gravitational effects on visible matter, on radiation, and on the&#xA;  large-scale structure of the universe. Dark matter has not been&#xA;  detected directly, making it one of the greatest mysteries in modern&#xA;  astrophysics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To understand the &lt;em&gt;infinite&lt;/em&gt; expansion of the universe, ask first how such a thing could be possible. Ask why it expands infinitely instead of the mutual gravity of all the galaxies slowing each other down and causing a collapse. &lt;/p&gt;&#xA;" OwnerUserId="10247" LastEditorUserId="10247" LastEditDate="2015-12-31T00:07:06.963" LastActivityDate="2015-12-31T00:07:06.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13007" PostTypeId="2" ParentId="969" CreationDate="2015-12-28T05:03:22.000" Score="3" Body="&lt;p&gt;I'm probably misunderstanding the question, but &#xA;&lt;a href=&quot;http://nssdc.gsfc.nasa.gov/planetary/factsheet/neptunefact.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://nssdc.gsfc.nasa.gov/planetary/factsheet/neptunefact.html&lt;/a&gt; &#xA;notes that Neptune's magnitude from Earth is 7.8 at opposition, when it's &#xA;4347.31 million km away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you get 3 times closer, 1449.1 million km away, Neptune &#xA;will appear 9 times brighter, bringing its magnitude to 5.4, &#xA;well within our visible range. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Stellarium notes below, Neptune is almost visible &#xA;(magnitude 5.66) from Uranus, an entire planet away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gNn9i.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gNn9i.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My version of Stellarium can't simulate the view from Naiad &#xA;(Neptune's closest moon), but at a distance of only 48,227 km &#xA;(Naiad's semi major axis), Neptune's magnitude would be almost &#xA;-2, much fainter than our own moon, but brighter than Sirius &#xA;appears to us. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This brightness would be spread out across Neptune's disk, but &#xA;the brightness at any point would be about magnitude 3, still &#xA;quite visible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, Naiad probably has a much thinner atmosphere (and no &#xA;light pollution), making Neptune even easier to see. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other outer planets are visible from Earth (Uranus just &#xA;barely), and so would also be visible from their own moons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't done the calculations for Pluto, which is &#xA;technically no longer a planet. &lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2015-12-28T05:10:08.650" LastActivityDate="2015-12-28T05:10:08.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13008" PostTypeId="1" AcceptedAnswerId="13009" CreationDate="2015-12-28T14:59:28.420" Score="8" ViewCount="2443" Body="&lt;p&gt;I am trying to find information on the equinox/solstice dates before 1600, more exactly referring up to 10,000 BCE so as to create a sort of calculator on an Excel spreadsheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read about the SOLEX, but it is not available anymore. Is there any other software that can accurately (as possible, of course) predict the equinoxes and solstices keeping in mind the perihelion/aphelion effects on the period between equinoxes/solstices?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you believe it is possible to try a formula on excel to calculate these dates for me (instead of just copying down the dates into a database), please keep in mind that my mathematical/programming knowledge is limited. &lt;/p&gt;&#xA;" OwnerUserId="5504" LastActivityDate="2020-06-13T17:44:10.280" Title="Are there accurate equinox and solstice predictions for the distant past?" Tags="&lt;orbit&gt;&lt;the-sun&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13009" PostTypeId="2" ParentId="13008" CreationDate="2015-12-28T16:39:48.050" Score="8" Body="&lt;p&gt;I wrote&#xA;&lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/STACK/bc-solve-astro-13008.c&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/STACK/bc-solve-astro-13008.c&lt;/a&gt;&#xA;to find historical solstices and equinoxes. The full results are at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/solstices-and-equinoxes.txt.bz2&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/solstices-and-equinoxes.txt.bz2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accuracy:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I'm using the EARTH_IAU_1976 precession model and the&#xA;EARTH_IAU_1980 nutation and obliquity models, which are&#xA;the only models supported by the CSPICE library I'm&#xA;using:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hesperia.gsfc.nasa.gov/ssw/stereo/gen/exe/icy64/doc/frames.req&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://hesperia.gsfc.nasa.gov/ssw/stereo/gen/exe/icy64/doc/frames.req&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the same library and the same models NASA uses to&#xA;do its own calculations, but the models are known to have&#xA;limited accuracy. Quoting &quot;&lt;a href=&quot;http://aa.usno.navy.mil/faq/docs/SpringPhenom.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://aa.usno.navy.mil/faq/docs/SpringPhenom.php&lt;/a&gt;&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;The times given in the tables are accurate to within two or&#xA;three hours for 25 to 5 BCE, and one or two hours for 4 BCE&#xA;to 38 CE. The uncertainty in these times arises from the&#xA;stochastic, that is unpredictable, change in the Earth's&#xA;rotation rate.&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The accuracy gets worse the further back OR the further&#xA;forward you go: not only don't we know precisely about the&#xA;Earth's rotation in the past, but we can't even predict the&#xA;Earth's rotation precisely for the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sample output:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&#xA;EQU 511720269.432607 A.D.   2016-03-20 04:30:01&#xA;SOL 519734120.174820 A.D.   2016-06-20 22:34:11&#xA;EQU 527826138.004142 A.D.   2016-09-22 14:21:09&#xA;SOL 535589116.137776 A.D.   2016-12-21 10:44:07&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Format:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The first column indicates whether this is a solstice or an equinox.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The second column is the ephemeris time of the&#xA;solstice/equinox. If you're doing serious astronomical&#xA;work, this is the column you should use. The approximate definition is the number of seconds since the J2000 epoch. The precise definition as used in CSPICE is: &lt;a href=&quot;https://naif.jpl.nasa.gov/pub/naif/toolkit_docs/FORTRAN/req/time.html#Ephemeris%20Time%20(ET)&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://naif.jpl.nasa.gov/pub/naif/toolkit_docs/FORTRAN/req/time.html#Ephemeris%20Time%20(ET)&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The remaining columns indicate the UTC time of the&#xA;solstice/equinox in a more human-readable format:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The CSPICE libraries assume that the Gregorian&#xA;calendar reformation occurred on 4 October 1582&#xA;(meaning the day after 4 October 1582 was 15 October&#xA;1582). Looking at these lines:&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&#xA;SOL -13191695511.794357 A.D.   1581-12-11 20:07:27&#xA;EQU -13183992131.003845 A.D.   1582-03-10 23:57:07&#xA;SOL -13175951250.170158 A.D.   1582-06-12 01:31:48&#xA;EQU -13167875920.223862 A.D.   1582-09-13 12:40:38&#xA;SOL -13160138634.917915 A.D.   1582-12-22 01:55:23&#xA;EQU -13152434815.793312 A.D.   1583-03-21 05:52:23&#xA;SOL -13144394485.035870 A.D.   1583-06-22 07:17:53&#xA;EQU -13136319216.460808 A.D.   1583-09-23 18:25:42&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;you can see that the dates of the solstices/equinoxes jump&#xA;ahead 10 days per the reformation.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The CSPICE libraries use the Julian calendar prior to&#xA;4 October 1582. In reality, the Julian calendar was&#xA;introduced in 46 BCE:&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Julian_calendar&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Julian_calendar&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Prior to 46 BCE, there were other calendar systems in&#xA;use, but the CSPICE libraries assume the Julian&#xA;calendar goes back indefinitely:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Proleptic_Julian_calendar&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Proleptic_Julian_calendar&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My calculations go back to 13201 BCE (the limits of DE431,&#xA;the ephemeris I'm using), and it's possible human beings&#xA;weren't even using calendars regularly at the time: quoting&#xA;&quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/History_of_calendars#Prehistory&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/History_of_calendars#Prehistory&lt;/a&gt;&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;A mesolithic arrangement of twelve pits and an arc found in&#xA;Warren Field, Aberdeenshire, Scotland, dated to roughly&#xA;10,000 years ago, has been described as a lunar calendar&#xA;and dubbed the &quot;world's oldest known calendar&quot; in 2013.&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;It takes the Earth 365.256363004 days to revolve around&#xA;the Sun with respect to the fixed stars, but the time&#xA;between vernal equinoxes is slightly less (365.242190402&#xA;days) because the position of the vernal equinox moves&#xA;(precesses) with respect to the stars. Source:&#xA;&lt;a href=&quot;http://hpiers.obspm.fr/eop-pc/models/constants.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://hpiers.obspm.fr/eop-pc/models/constants.html&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Gregorian calendar's average day length of 365.2425&#xA;is much closer to 365.242190402 days than the Julian&#xA;calendar's 365.25 average day length, but it's still not&#xA;perfect. As noted in&#xA;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/1918/do-solstices-and-equinoxes-shift-over-time&quot;&gt;Do solstices and equinoxes shift over time?&lt;/a&gt;&#xA;if we continue using the Gregorian calendar in the far&#xA;future, the equinoxes and solstices will drift&#xA;backwards. By 17090 (the limit of DE431), they will look&#xA;like this:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&#xA;EQU 476198945887.238159 A.D.  17090-02-22 08:56:59&#xA;SOL 476207018540.040894 A.D.  17090-05-26 19:21:11&#xA;EQU 476214808218.146362 A.D.  17090-08-24 23:09:09&#xA;SOL 476222655067.609985 A.D.  17090-11-23 18:49:59&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;about a month behind their &quot;regular&quot; times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MY EARLIER PARTIAL ANSWER FOR REFERENCE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since HORIZONS (&lt;a href=&quot;http://ssd.jpl.nasa.gov/?horizons&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ssd.jpl.nasa.gov/?horizons&lt;/a&gt;) and SPICE (&lt;a href=&quot;http://naif.jpl.nasa.gov/naif/tutorials.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://naif.jpl.nasa.gov/naif/tutorials.html&lt;/a&gt;) can compute the ecliptic and solar position back that far, it should be possible to compute equinoxes and solstices with reasonable accuracy. However, I haven't been able to find a site that actually lists these dates (I'm pretty sure USNO did this at one point, but I can't find their list). Other possibly helpful sources/questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/10901/how-did-meeus-calculate-equinox-and-solstice-dates&quot;&gt;How did Meeus calculate equinox and solstice dates?&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://ns1763.ca/equinox/eqindex.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ns1763.ca/equinox/eqindex.html&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://data.giss.nasa.gov/ar5/srvernal.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://data.giss.nasa.gov/ar5/srvernal.html&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.usno.navy.mil/USNO/astronomical-applications/data-services/spring-phenom&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.usno.navy.mil/USNO/astronomical-applications/data-services/spring-phenom&lt;/a&gt; (appears to be down at the moment)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2020-06-13T17:44:10.280" LastActivityDate="2020-06-13T17:44:10.280" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="13012" PostTypeId="2" ParentId="12940" CreationDate="2015-12-28T18:37:32.050" Score="0" Body="&lt;p&gt;Here's a simple python3 script that calculates when the shortest day of the year occurs at various longitudes. If you don't already have PyEphem, it's a cinch to install it: &lt;code&gt;pip3 install PyEphem&lt;/code&gt; (or just &lt;code&gt;pip install PyEphem&lt;/code&gt; if you've moved beyond python2).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import ephem&#xA;&#xA;sun = ephem.Sun()&#xA;obs = ephem.Observer()&#xA;obs.lat = '49'&#xA;&#xA;start_year = 2015&#xA;end_year   = 2017&#xA;for iyear in range(start_year,end_year+1) :&#xA;    solstice = ephem.next_solstice(42357+(iyear-2015)*365.25)&#xA;    print(&quot;December solstice in&quot;, iyear, &quot;is at&quot;, solstice, &quot;UTC&quot;)&#xA;&#xA;    print(&quot;Shortest day of the year, by longitude:&quot;)&#xA;    print(&quot;Date  Longitude&quot;)&#xA;    for lon in range(-180, 180+10, 10) :&#xA;        tzoffset = lon/15/24;&#xA;        obs.lon = str(lon)&#xA;&#xA;        # Start with a date that makes the next transit of the Sun two or three&#xA;        # days prior to the solstice.&#xA;        obs.date = solstice-3&#xA;        sun.compute (obs)&#xA;&#xA;        data = []&#xA;        for day in range(6) :&#xA;            obs.date = obs.next_transit(sun)&#xA;            sun.compute (obs)&#xA;            day_length = obs.next_setting(sun) - obs.previous_rising(sun)&#xA;            date = ephem.Date(obs.date+tzoffset)&#xA;            data.append ((day, date, day_length))&#xA;&#xA;        shortest = min(data, key=lambda entry: entry[2])&#xA;        if (shortest[0] == 0) or (shortest[0] == 5) :&#xA;            raise RuntimeError(&quot;Can't find shortest day&quot;)&#xA;        print(&quot;12/&quot;+str(shortest[1].datetime().day), lon)&#xA;    print()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that if you change the start and end years so that you can see up to 2196, you'll see that the shortest day of the year in that year for much of the northern hemisphere will occur on the 20th of December.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2016-01-04T05:10:58.457" LastActivityDate="2016-01-04T05:10:58.457" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13013" PostTypeId="1" AcceptedAnswerId="13018" CreationDate="2015-12-29T01:15:28.083" Score="22" ViewCount="851" Body="&lt;p&gt;When listening to podcasts or watching youtube videos of astronomers discussing galaxy mergers, I often hear talk about how the super massive black holes at their centers will themselves merge during or shortly after the collision. Why do we believe this to be the case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A priori, I would expect SMBH's to behave the same as all other galactic objects. They might be extraordinarily massive, but physically they are still minuscule compared to the vast empty space between stars in a galaxy. Collisions between objects (not counting gigantic clouds of gas and dust) would be extraordinarily rare, so why do we make an exception for the SMBH's?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could see a case for them merging in the (rare?) instance where the host galaxies hit each other in such a way that the mutual center of mass happens to coincide with their individual centers of mass. In that case, the SMBH's could be close enough to orbit about one another, losing energy to gravitational waves and eventually merging. I would think that this scenario is rather rare, however. I find it more plausible that an average galaxy merger would leave the SMBH's independently orbiting the center of the combined galaxy, too far apart to lose any significant kinetic energy to gravitational waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The astronomers who talk about galactic mergers know a heck of a lot more than I do about the subject, so I presume there are flaws in my assumptions or my understanding of the physics. What am I missing?&lt;/p&gt;&#xA;" OwnerUserId="10252" LastEditorUserId="9301" LastEditDate="2016-06-29T04:38:49.180" LastActivityDate="2016-07-11T17:31:43.377" Title="Why do we believe that the super massive black holes at the centers of two merging galaxies would themselves merge?" Tags="&lt;galaxy&gt;&lt;supermassive-black-hole&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13015" PostTypeId="1" CreationDate="2015-12-29T03:56:41.123" Score="1" ViewCount="108" Body="&lt;p&gt;My husband and I are trying to figure out what type of telescope to get our young daughter. We would like to be able to see planets but nothing too extreme. Just something to get her started. She has shown a big interest in astronomy and we would like see if it sticks then we will get her a much better one. Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="10255" LastActivityDate="2015-12-30T08:16:16.780" Title="Beginner telescope for my daughter" Tags="&lt;planet&gt;&lt;telescope&gt;" AnswerCount="1" CommentCount="7" ClosedDate="2016-01-06T16:10:47.363" ContentLicense="CC BY-SA 3.0" />
  <row Id="13017" PostTypeId="1" AcceptedAnswerId="13020" CreationDate="2015-12-29T05:32:14.147" Score="2" ViewCount="326" Body="&lt;p&gt;If I travelled near a black hole, my time would progress slower relative to someone on Earth. This is clear enough. However, what if we sent a probe with a camera to a black hole? When we watch the screen, would we see time through the camera's perspective — that is, would the Universe appear to progress faster as the probe got closer and closer to the black hole?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2020-10-13T15:00:03.373" Title="A camera and time dilation?" Tags="&lt;gravity&gt;&lt;black-hole&gt;&lt;general-relativity&gt;&lt;space-time&gt;&lt;time-dilation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13018" PostTypeId="2" ParentId="13013" CreationDate="2015-12-29T13:33:34.993" Score="10" Body="&lt;p&gt;The SMBHs reside in the bottom of the galactic potentials, which are dominated by the galaxies' dark matter halos. But although dark matter dominates gravity, collisions between gas and dust particles in the interstellar medium causes enough friction that the baryonic component of the galaxies is decelerated. This will cause the other components of the galaxies to decelerate as well, through gravitational attraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, despite dark matter (and, in practice, stars and black holes since they're so small) being collisionless, there are several ways of &quot;relaxing&quot;, i.e. to evolve towards an equilibrium. In the context of galaxy merging, the most important mechanism (I think) is &quot;violent relaxation&quot;, where the rapid change of the gravitational potential causes particles to relax, e.g. more massive particles tend to transfer more energy to their lighter neighbors and so become more tightly bound, sinking towards the center of the gravitational potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although SMBHs are…, well, supermassive, the potential will (usually) be dominated by dark matter, gas, and stars, so the new gravitational potential will also cause the SMBHs to seek towards the bottom in the same fashion, and eventually merge.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2015-12-29T14:44:44.913" LastActivityDate="2015-12-29T14:44:44.913" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13019" PostTypeId="1" AcceptedAnswerId="13026" CreationDate="2015-12-29T15:56:19.067" Score="1" ViewCount="111" Body="&lt;p&gt;I recently became aware of the work of the &lt;a href=&quot;https://www.google.com/search?q=+Kepler+Cluster+Study&quot; rel=&quot;nofollow&quot;&gt;Kepler Cluster Study&lt;/a&gt;. In January of this year it was announced that a long awaited method of finding ages of &quot;cool&quot; stars is now available. This is huge and I'm surprised that I just found out about it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if there has been a little too much &quot;extrapolation&quot; and that more data is needed. I also wonder if they have gotten around the fact that it is vsini that is being measured and not the actual rotation velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line is that I want to see the catalogs of updated ages based on this new information. Does anybody know of any such publications in the works? It will be interesting to compare results to ages based on metallicity.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorDisplayName="user1569" LastEditDate="2016-01-02T11:29:04.603" LastActivityDate="2016-01-02T11:29:04.603" Title="Gyrochronology?" Tags="&lt;stellar-evolution&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13020" PostTypeId="2" ParentId="13017" CreationDate="2015-12-29T16:13:50.013" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If I travelled near a black hole, my time would progress slower relative to someone on Earth. This is clear enough. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, no problem with the gravitational time dilation. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, what if we sent a probe with a camera to a black hole? When we watch the screen, would we see time through the camera's perspective — that is, would the Universe appear to progress faster as the probe got closer and closer to the black hole?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. We'd see the universe progressing normally, because we aren't subject to that gravitational time dilation. (I presume we're at a safe distance). Let's suppose it was a TV camera that took a picture 25 times a second and sent it back to us, adequately catering for redshift. The camera starts off taking 25 pictures a second as measured by us. But after a while we notice we're only getting 24 pictures a second, then 23, and so on. We see events in the wider universe progressing at their normal rate, but eventually the movie starts getting jerky as the frame-rate reduces. In the end the frame-rate reduces to zero, and that's the end of the show. &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2015-12-29T16:13:50.013" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="13021" PostTypeId="2" ParentId="13017" CreationDate="2015-12-29T20:35:29.260" Score="4" Body="&lt;p&gt;For simplicity, let's say that the black hole is isolated and non-rotating (and uncharged), so that the situation is described by the comparatively simple Schwarzschild spacetime. Let's also suppose that the camera free-falls radially into the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the camera looking at? Suppose it is looking at some stationary object that does something with a known frequency. Your question is basically how at what frequency it will be observed on the video feed emitted by the camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without loss of generality, we can suppose that the camera is looking at &lt;em&gt;us&lt;/em&gt;, and that we're shining a laser beam at it: the 'doing something at a known frequency' would be the oscillations in the electromagnetic wave of the laser beam. We can do this because time dilation affects every physical process, so we might as well pick one that is more convenient to think about. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, it is straightforward why the camera feed will not show any time dilation: being equivalent to a reflected laser beam, the gravitational blueshift when going inward will be cancelled by the gravitational redshift going outward.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2015-12-29T20:35:29.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13022" PostTypeId="1" CreationDate="2015-12-30T02:45:25.920" Score="6" ViewCount="8595" Body="&lt;p&gt;How much gravitational force is felt on earth from the other planets in the solar system? The sun exerts the strongest g-force, holding us in it's orbit, followed by the moon which affects the tides on earth, but how much force do we feel from Jupiter, Saturn, Venus, etc?&lt;/p&gt;&#xA;" OwnerUserId="10262" LastEditorUserId="6870" LastEditDate="2015-12-30T21:17:24.480" LastActivityDate="2021-04-22T07:59:46.473" Title="What is the gravitational force felt on Earth from the other planets in our solar system?" Tags="&lt;solar-system&gt;&lt;gravity&gt;&lt;planet&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13023" PostTypeId="1" AcceptedAnswerId="13024" CreationDate="2015-12-30T03:55:23.250" Score="3" ViewCount="96" Body="&lt;p&gt;As we know, if a planet is near a star and is tidally locked, then at near side it is very hot and at far side it is very cold. But between 2 sides there should be a gradient of temperature change and should have an area which the temperature is suitable for lives. Can we also say it has a habitable zone?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastActivityDate="2015-12-30T07:48:50.167" Title="Can a tidally locked planet have their own habitable zone?" Tags="&lt;extra-terrestrial&gt;&lt;habitable-zone&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13024" PostTypeId="2" ParentId="13023" CreationDate="2015-12-30T07:42:59.283" Score="2" Body="&lt;p&gt;As you suggest, it might be possible for a habitable corridor to exist along the stationary terminator. But there are ideas around for more than that. The planet might be exposed to tidal forces the volcanism of which warms the far side. It might have a thick stormy ocean or atmosphere which evens out the surface temperature (All but the smallest planets have atmospheres). It might have habitable moons which when tidally locked to the planet rotate regularly versus the star. And if it, like Mercury, has an eccentric enough orbit it might rotate relative to its star although it is tidally locked. Planetary diversity is huge.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2015-12-30T07:48:50.167" LastActivityDate="2015-12-30T07:48:50.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13025" PostTypeId="2" ParentId="13015" CreationDate="2015-12-30T08:16:16.780" Score="0" Body="&lt;p&gt;I have been looking to get into astronomy myself and I have been impressed with the reviews on the Celestron Firstscope, which is around $50. It has a decent optic size and is small enough to pack up and get out to a clear area if you live in a high light pollution area. The only drawback I find with the scope (without actually using it) is it doesnt have a viewfinder but you can find one of those easily.&lt;/p&gt;&#xA;" OwnerUserId="10264" LastActivityDate="2015-12-30T08:16:16.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13026" PostTypeId="2" ParentId="13019" CreationDate="2015-12-30T10:49:11.847" Score="1" Body="&lt;p&gt;Please point us to the peer-reviewed reference! As far as I'm aware the problems that plague gyrochronology still persist. These are (in no particular order).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A lack of calibrating clusters for stars with ages older than the Sun and for stars older than about 1 Gyr and mass much less than a solar mass.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It just doesn't work for stars younger than about 200 Myr at a solar mass and younger than 1 Gyr at half a solar mass, because rotation rates have not converged to a single-valued function of age.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You need rotation &lt;em&gt;periods&lt;/em&gt; to use it. You cannot get around the unknown inclination angle so if you only have a projected equatorial velocity $v \sin i$, then this is considerably less useful. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Measuring the rotation periods for stars takes a lot of effort. For young stars they potentially have a large photometric modulation caused by starspots that can be measured through monitoring over timescales of days to weeks. For older stars like the Sun, the spot modulation is so small that you need something like Kepler in order to be able to measure it. It is not something that could be routinely done over huge swathes of the sky.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The precision is not that great. Aside from the calibration issue there is the problem that there is a dispersion in rotation rates at a given age that leads to an uncertainty in the estimated age for a given rotation rate. This amounts to at least 10-20% for stars younger than a Gyr but probably decreases for older stars. However, this is counterbalanced by the fact that older stars appear to suffer more differential rotation with latitude. That means if you measure the rotation period on one occasion it is quite possible it will be different the next time you measure it, if spots are at a different latitude. This would give rise to errors approaching 20% in a star like the Sun.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Work to overcome these issue will be slow and incremental. Asteroseismology is certainly helping for solar-type stars because for these stars, you can estimate the age in an independent (though model-dependent) way, which offers an alternative calibration route. The original Kepler project has also observed a number of older clusters and there are a few more which might prove to be decent calibrators in the K2 fields. However, my bet would be that the problems of calibration for K and M dwarfs will still be there even after all this data has come in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find a discussion of the above and a comparison with other age estimation techniques in &lt;a href=&quot;http://arxiv.org/abs/1404.7156&quot; rel=&quot;nofollow&quot;&gt;Jeffries (2014)&lt;/a&gt;. Estimates of age based on metallicity would be &lt;em&gt;very&lt;/em&gt; crude indeed, and subject to catastrophic errors. For instance, the metallicity of young, star-forming clusters in the solar vicinity is almost identical to the Sun. What you can &lt;em&gt;probably&lt;/em&gt; say, is that if a star has a metallicity less than about 0.5 solar then it is &lt;em&gt;likely&lt;/em&gt; to be older than 5 Gyr; anything less than 0.1 solar is &lt;em&gt;likely&lt;/em&gt; to be older than 10 Gyr. But that's about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: So, the paper that triggered your question is &lt;a href=&quot;http://arxiv.org/abs/1501.05651&quot; rel=&quot;nofollow&quot;&gt;Meibom et al. (2015)&lt;/a&gt; &quot;A spin-down clock for cool stars from observations of a 2.5-billion-year-old cluster&quot;, which talks about Kepler observations of the open cluster NGC 6819, in which the rotation periods have been found for 30 cool stars with masses between 0.85 and 1.4 solar masses. These data are indeed an extremely useful calibration set which fills in the gap for solar type stars between the ages of the Sun and the well-studies clusters like the Hyades and Praesepe below 1 Gyr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, all the problems I mentioned above still exist. The range of masses probed here is rather small, but what they are able to establish is that the 20 or so stars below about 1.2 solar masses have a dispersion around a mean period-colour relationship with a standard deviation of only 5%. This means that the &lt;em&gt;solar mass&lt;/em&gt; stars have converged to a fairly tight relationship between rotation period, mass and age and that errors in measuring the periods are also relatively small. Because $t \propto P^{2}$, this leads to the&#xA; claim that ages could be determined with a precision of 10% &lt;em&gt;at this age, if&lt;/em&gt; the age-rotation-mass space were fully calibrated. Note though that these rotation periods were measured by Kepler; that the median &lt;em&gt;peak-to-peak&lt;/em&gt; amplitude was only 4 milli-mag and that in most cases the period is a mean from several independent measurements of the same star. Such excellent data will not be forthcoming for very large samples of field stars anytime soon.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-01-01T18:46:21.833" LastActivityDate="2016-01-01T18:46:21.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13027" PostTypeId="2" ParentId="13022" CreationDate="2015-12-30T17:03:48.047" Score="9" Body="&lt;p&gt;Because of the inverse square law for Newtonian gravity we have the acceleration due the gravity $g_b$ at the surface of the Earth due to a body of mass $m_b$ at a distance $d_b \gg r_e$ (where $r_e\approx 6371 \mbox{km}$ denotes the radius of the Earth, note all distances will need be in $\mbox{km}$ in what follows) is:&#xA;$$&#xA;g_b=g\times \frac{m_b}{m_e}\times \left(\frac{r_e}{d_b}\right)^2&#xA;$$&#xA;where $g$ is the usual accelleration due to gravity (from the Earth at the Earth's surface $\approx 10 \mbox{m/s}^2$, and $m_e\approx 6.0 \times 10^{24} \mbox{kg}$. We get the maximum acceleration due to a body when that body is at its closest to the Earth, which is what we do from now on (except for the Sun and Moon where the mean distance is used).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the Moon $r_b\approx 0.384 \times 10^6 \mbox{km}$, and $m_b\approx 7.3 \times 10^{22} \mbox{kg}$, so the accelleration at the Earth's surface due to the Moon $g_b\approx 3.3 \times 10^{-5} \mbox{m/s}^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then putting this relation and Solar-System data into a spread sheet we get:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Nf3Bi.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nf3Bi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-12-30T17:10:46.427" LastActivityDate="2015-12-30T17:10:46.427" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13028" PostTypeId="1" AcceptedAnswerId="13029" CreationDate="2015-12-30T17:06:40.107" Score="-4" ViewCount="186" Body="&lt;p&gt;my question is regarding  space travel and also will some countries have colonies on Mars etc cause the clear reason NASA etc are trying to send manned Mars mission is to get raw materials etc or maybe send a large part of human population to Mars because of limited resources on Earth?&lt;/p&gt;&#xA;" OwnerUserId="10191" LastEditorUserId="14021" LastEditDate="2016-09-14T19:55:36.570" LastActivityDate="2016-09-14T19:55:36.570" Title="humans leave Earth forever?" Tags="&lt;space&gt;" AnswerCount="2" CommentCount="5" ClosedDate="2016-04-14T16:48:27.227" ContentLicense="CC BY-SA 3.0" />
  <row Id="13029" PostTypeId="2" ParentId="13028" CreationDate="2015-12-30T18:54:09.413" Score="1" Body="&lt;p&gt;Nasa are sending humans to mars because it has a background of evolution very similar to earth, meaning we can discover more about the past and future of our planet. Mars had the capability to have life in the past, and if we can work this out we can answer the fundamental question: Does life exist anywhere else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, about humans leaving earth. Humanity will have to leave earth at some point - due to issues like overpopulation, resource shortages and NASA as well as many other scientists are planning things like human missions to asteroids, and eventually developing technology for a human mission into deep space. Astronauts on the ISS are helping us understand how the body changes in space (for example, fluids inside your body aren't pulled down by gravity). Furthermore, these astronauts are helping us prove the technologies needed for exploration into deep space. I found a site about stephen hawking talking about how we need to leave earth &lt;a href=&quot;http://www.space.com/8924-stephen-hawking-humanity-won-survive-leaving-earth.html&quot; rel=&quot;nofollow&quot;&gt;http://www.space.com/8924-stephen-hawking-humanity-won-survive-leaving-earth.html&lt;/a&gt;. &lt;a href=&quot;http://www.theregister.co.uk/2015/02/20/hawking_alert_leave_planet_earth_stupid_humans/&quot; rel=&quot;nofollow&quot;&gt;http://www.theregister.co.uk/2015/02/20/hawking_alert_leave_planet_earth_stupid_humans/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, another thing about the matter of exploration further into space, and why we haven't encountered aliens is that some scientists believe that there a many, many, many sentient civilisations out there, but they wipe themselves out so quickly they never cross paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, there are so many things we need to think about if we will travel and colonise other planets. Like how the planet's gravity will most likely be different, the atmosphere will be at a different density. This is why we need to expand our presence in the solar system, and one day send a long term manned mission to mars.&lt;/p&gt;&#xA;" OwnerUserId="9645" LastActivityDate="2015-12-30T18:54:09.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13030" PostTypeId="1" CreationDate="2015-12-30T19:21:15.127" Score="1" ViewCount="26" Body="&lt;p&gt;Capella is a system with four stars in two binary pairs (&lt;a href=&quot;https://en.wikipedia.org/wiki/Capella&quot; rel=&quot;nofollow&quot;&gt;source: Wikipedia&lt;/a&gt;). If an observer were located at the L1 LaGrange point between the larger pair, how many degrees of the sky would each of the stars - both the first and second pairs - occupy?&lt;/p&gt;&#xA;" OwnerUserId="10267" LastEditorUserId="4085" LastEditDate="2015-12-31T08:55:01.053" LastActivityDate="2015-12-31T08:55:01.053" Title="What would the apparent size of the four stars in the Capella system be from the L1 point between the first pair?" Tags="&lt;lagrange-point&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13032" PostTypeId="1" AcceptedAnswerId="13044" CreationDate="2015-12-31T02:24:15.750" Score="2" ViewCount="599" Body="&lt;p&gt;I often hear the hypothesis that Jupiter decreases the chances of an asteroid strike on the Earth, but my question is, why does it only decrease the chances ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While Jupiter can deflect some asteroids that were originally going to hit the Earth, it can also deflect some asteroids towards the Earth. How can we be sure that Jupiter only decreases the chances of an asteroid impact on Earth ? &lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="4085" LastEditDate="2015-12-31T08:32:28.997" LastActivityDate="2016-01-01T07:18:03.107" Title="Why doesn't Jupiter increase the chances of an asteroid to strike the Earth?" Tags="&lt;orbit&gt;&lt;asteroids&gt;&lt;jupiter&gt;&lt;impact&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13033" PostTypeId="1" CreationDate="2015-12-31T05:09:38.603" Score="8" ViewCount="347" Body="&lt;p&gt;You all probably know the famous &lt;a href=&quot;http://www.jpl.nasa.gov/spaceimages/details.php?id=pia17172&quot; rel=&quot;nofollow noreferrer&quot;&gt;The day the Earth Smiled&lt;/a&gt; image. For the life of me I can't understand what's going on with the rings there. Seems like the &quot;bottom&quot; half of the rings is the half that goes &quot;behind&quot; Saturn (relative to the camera), but then what's going on with the distortion of the rings on the upper half when they (presumably) cross in front of Saturn? I can't make sense of of this image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IVdFL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Higher resolution and annotated versions &lt;a href=&quot;http://www.jpl.nasa.gov/spaceimages/details.php?id=pia17172&quot; rel=&quot;nofollow noreferrer&quot;&gt;available&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10268" LastEditorUserId="7982" LastEditDate="2019-08-15T23:16:53.457" LastActivityDate="2019-08-15T23:16:53.457" Title="Trying to understand the way Saturn's ring look in this famous Cassini image" Tags="&lt;saturn&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="13034" PostTypeId="2" ParentId="13033" CreationDate="2015-12-31T05:35:58.253" Score="6" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1YHF2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1YHF2.jpg&quot; alt=&quot;Saturn&amp;#39;s ring&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, Saturn is eclipsing the sun in this image. This eclipsing causes the white outlining of the planet in this photo. The bottom ring is indeed passing on the other side of Saturn which would explain why it is not visible in this image. However, the ring passing in front of Saturn is being cast in the shadow of the planet. The light cast on the planet, which could be the cause of this &quot;distortion&quot; is the light reflected from the Rings of Saturn. Since the Sun is being eclipsed, the ring passing in front is not receiving the light from the Sun causing it to look like a black shadow band across the planet.&lt;/p&gt;&#xA;" OwnerUserId="10083" LastEditorUserId="10083" LastEditDate="2016-01-01T04:26:32.013" LastActivityDate="2016-01-01T04:26:32.013" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13035" PostTypeId="1" AcceptedAnswerId="13037" CreationDate="2015-12-31T12:38:39.063" Score="3" ViewCount="389" Body="&lt;p&gt;This question came to my mind after reading &lt;a href=&quot;https://astronomy.stackexchange.com/questions/13017/a-camera-and-time-dilation/13020?noredirect=1#comment18789_13020&quot;&gt;this.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is safe to assume that the supposedly thrown camera would not reach the event horizon in one piece. So my question is : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I dropped a TV camera into a black hole, will it be destroyed before it reaches the event horizon, and if so, why?&lt;/p&gt;&#xA;" OwnerUserId="9679" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-01-01T13:49:20.357" Title="Camera reaching the event horizon" Tags="&lt;gravity&gt;&lt;black-hole&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13037" PostTypeId="2" ParentId="13035" CreationDate="2015-12-31T15:28:15.313" Score="8" Body="&lt;p&gt;If the black hole has no surrounding matter, so that there is no violent radiation generated by accretion or the like, then it still depends on the mass of the black hole. If it is in the thousands of solar masses or more, it is possible for a realistic camera to survive free-fall to the horizon. This is essentially a non-issue for supermassive black holes. On the other hand, smaller black holes are quite a bit more punishing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Newtonian gravity with potential $\Phi$, in a free-falling frame, a particle at $x^k$ near the origin of the frame will be accelerated at&#xA;$$\frac{\mathrm{d}^2x^j}{\mathrm{d}t^2} = -\frac{\partial\Phi}{\partial x^j} = -\frac{\partial^2\Phi}{\partial x^j\partial x^k}x^k\text{,}$$&#xA;where the the second derivatives of the potential, $\Phi_{,jk}$, form the so-called &lt;em&gt;tidal gravitational field&lt;/em&gt;. Since $\Phi = -GM/r$ for a point-source, you should expect the tidal forces on a free-falling object to be proportional to $GM/r^3$ times the size of the object. Thus, at the Schwarzschild radius, this is on the order of $c^6/(GM)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, black holes are not Newtonian. However, it turns out that for the non-rotating, uncharged (Schwarzschild) black hole, radial free-fall of a test particle has the same form as in Newtonian theory, except in Schwarzschild radial coordinate (not radial distance) and proper time of the particle (not universal time), so the above is essentially correct even for Schwarzschild black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be relativistically correct, the tidal forces on an free-falling object are described by the equation of &lt;a href=&quot;https://en.wikipedia.org/wiki/Geodesic_deviation&quot; rel=&quot;noreferrer&quot;&gt;geodesic deviation&lt;/a&gt;, in which the gravitoelectric part of the Riemann curvature provides forms the tidal tensor:&#xA;$$\frac{\mathrm{D}^2x^\alpha}{\mathrm{d}\tau^2} = -R^\alpha{}_{\mu\beta\nu} u^\mu u^\nu x^\beta\text{.}$$&#xA;In Schwarzschild spacetime, this turns out to be $+2GM/r^3$ in the radial direction, stretching the free-falling object, and $-GM/r^3$ in the orthogonal directions, squeezing it. This stretching-and-squeezing due to gravitational tidal forces is sometimes called &lt;a href=&quot;https://en.wikipedia.org/wiki/Spaghettification&quot; rel=&quot;noreferrer&quot;&gt;spaghettification&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some example numbers: say the camera size is on the order of $0.1\,\mathrm{m}$. The the following are the approximate tidal accelerations near the horizon for black holes of different multiples of solar masses:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$M\sim 10\,\mathrm{M_\odot}$: $\sim 10^6$ Earth gravities;&lt;/li&gt;&#xA;&lt;li&gt;$M\sim 10^4\,\mathrm{M_\odot}$: $\sim 1$ Earth gravity;&lt;/li&gt;&#xA;&lt;li&gt;$M\sim 10^6\,\mathrm{M_\odot}$: $\sim 10^{-4}$ Earth gravities.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The curvature around a rotating black hole is more complicated, but the moral of the story is basically the same.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditDate="2015-12-31T19:04:09.913" LastActivityDate="2015-12-31T19:04:09.913" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13038" PostTypeId="2" ParentId="13032" CreationDate="2015-12-31T16:05:17.910" Score="2" Body="&lt;p&gt;With regard to collisions with earth, or inner planet collisions in general, you're right that Jupiter has both a positive and negative effect on asteroid and comet trajectories. On balance though, the effect is mainly positive.&lt;/p&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;http://earthsky.org/space/is-it-true-that-jupiter-protects-earth&quot; rel=&quot;nofollow noreferrer&quot;&gt;earthsky.org&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Some astronomers believe that one reason Earth is habitable is that the gravity of Jupiter does help protect us from some comets. Long-period comets, in particular, enter the solar system from its outer reaches. Jupiter’s gravity is thought to sling most of these fast-moving ice balls out of the solar system before they can get close to Earth. So long-period comets are thought to strike Earth only on very long timescales of millions or tens of millions of years. Without Jupiter nearby, long-period comets would collide with our planet much more frequently.&lt;/p&gt;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&lt;p&gt;But Jupiter creates both good and bad conditions for earthly life. Consider that its powerful gravity prevented space rocks orbiting near it from coalescing into a planet, and that’s why our solar system today has an asteroid belt, consisting of hundreds of thousands of small flying chunks of debris... today, Jupiter’s gravity continues to affect the asteroids – only now it nudges some asteroids toward the sun, where they have the possibility of colliding with Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="941" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2015-12-31T16:05:17.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13039" PostTypeId="2" ParentId="13033" CreationDate="2015-12-31T17:53:09.027" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What's going on with the distortion of the rings on the upper half when they (presumably) cross in front of Saturn?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The brownish areas you see on Saturn are ring light, analogous to seeing the Earth by moonlight. Saturn's rings light up Saturn's night sky, particularly just after sunrise and just before sunset. The two dark bands across the face of Saturn are the A ring (upper dark band) and B ring (lower dark band). Of all of Saturn's rings, these two are the most opaque. The parts of Saturn underneath those two dark bands are brilliantly lit by ring light, but Cassini can't see those parts of Saturn because the opaque A and B rings block that light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a much better and more thorough explanation of all of the features in this incredible image, I suggestion you spend eleven minutes &lt;a href=&quot;https://www.youtube.com/watch?v=aIVFbbIlnLQ&quot;&gt;watching this youTube video&lt;/a&gt; by &lt;a href=&quot;https://en.wikipedia.org/wiki/Emily_Lakdawalla&quot;&gt;Emily Lakdawalla&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;sub&gt;This is one of the very few times I will post a link to a youTube video. It is worth every eleven minutes.&lt;/sub&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2015-12-31T17:53:09.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13040" PostTypeId="2" ParentId="5894" CreationDate="2015-12-31T21:12:23.157" Score="6" Body="&lt;p&gt;At 87 degrees south latitude on the prime meridian (0 degrees east&#xA;longitude), the moon set at 0121 UTC on Nov 22 2012, and didn't rise&#xA;again until 2356 on Dec 6 2012, for a total moonless period of 14&#xA;days, 22 hours, and 35 minutes. See caveats below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can &quot;verify&quot; this calculation by visiting&#xA;&lt;a href=&quot;http://aa.usno.navy.mil/data/docs/RS_OneYear.php&quot; rel=&quot;noreferrer&quot;&gt;http://aa.usno.navy.mil/data/docs/RS_OneYear.php&lt;/a&gt; and entering these&#xA;parameters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6Wbo7.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6Wbo7.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to get this result:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&#xA;Sun or Moon Rise/Set Table for One Year&#xA;             o  ,    o  ,                                                                              Astronomical Applications Dept.&#xA;Location:  000 00, S87 00                         Rise and Set for the Moon for 2012                   U. S. Naval Observatory        &#xA;                                                                                                       Washington, DC  20392-5420     &#xA;                                                            Universal Time                                                            &#xA;&#xA;&#xA;       Jan.       Feb.       Mar.       Apr.       May        June       July       Aug.       Sept.      Oct.       Nov.       Dec.  &#xA;Day Rise  Set  Rise  Set  Rise  Set  Rise  Set  Rise  Set  Rise  Set  Rise  Set  Rise  Set  Rise  Set  Rise  Set  Rise  Set  Rise  Set&#xA;     h m  h m   h m  h m   h m  h m   h m  h m   h m  h m   h m  h m   h m  h m   h m  h m   h m  h m   h m  h m   h m  h m   h m  h m&#xA;01  ---- ----  ---- ----  ---- ----  ---- ----  1821       **** ****  **** ****  **** ****       0649  ---- ----  ---- ----  ---- ----&#xA;02  ---- ----  ---- ----  ---- ----  ---- ----  1255 0021  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----&#xA;03  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----&#xA;04  ---- ----  ---- ----  ---- ----  1718       **** ****  **** ****  **** ****  1900 1121  ---- ----  ---- ----  ---- ----  ---- ----&#xA;05  ---- ----  ---- ----  ---- ----  1030 0834  **** ****  **** ****  **** ****       0634  ---- ----  ---- ----  ---- ----  ---- ----&#xA;06  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  2356     &#xA;07  ---- ----  ---- ----  2222       **** ****  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  **** ****&#xA;08  ---- ----  ---- ----  1634 0339  **** ****  **** ****  **** ****       1038  ---- ----  ---- ----  ---- ----  ---- ----  **** ****&#xA;09  ---- ----  2057       **** ****  **** ****  **** ****  **** ****  0200 0451  ---- ----  ---- ----  ---- ----  0427 1338  **** ****&#xA;09                                                                                                                2341                &#xA;10  ---- ----  1425 1149  **** ****  **** ****  **** ****  2154 1518  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****&#xA;11  ---- ----  **** ****  **** ****  **** ****  **** ****       1029  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****&#xA;12  ---- ----  **** ****  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****&#xA;13  2049       **** ****  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  0357       **** ****  **** ****&#xA;14  **** ****  **** ****  **** ****  **** ****       1502  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****&#xA;15  **** ****  **** ****  **** ****  **** ****  0304 1024  ---- ----  ---- ----  ---- ----  0824 1654  **** ****  **** ****  **** ****&#xA;16  **** ****  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  0325       **** ****  **** ****  **** ****&#xA;17  **** ****  **** ****  **** ****       1533  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****&#xA;18  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****&#xA;19  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  0754       **** ****  **** ****  **** ****  1143 0118&#xA;19                                                                                                                                2046&#xA;20  **** ****  **** ****       2102  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****  ---- ----&#xA;21  **** ****  **** ****  0252 1604  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****  ---- ----&#xA;22  **** ****       2040  ---- ----  ---- ----  ---- ----  ---- ----  1324 1819  **** ****  **** ****  **** ****       0121  ---- ----&#xA;23  **** ****  0810 1556  ---- ----  ---- ----  ---- ----  ---- ----  0758       **** ****  **** ****  **** ****  ---- ----  ---- ----&#xA;24  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****  ---- ----  ---- ----&#xA;25  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  1253 2338  **** ****  **** ****  **** ****  1214 0649  ---- ----  ---- ----&#xA;26       2026  ---- ----  ---- ----  ---- ----  ---- ----  0757       **** ****  **** ****  **** ****       0154  ---- ----  ---- ----&#xA;27  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----&#xA;28  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  1659 0645  ---- ----  ---- ----  ---- ----&#xA;29  ---- ----  ---- ----  ---- ----  ---- ----  1303       **** ****  **** ****  **** ****       0213  ---- ----  ---- ----  ---- ----&#xA;30  ---- ----             ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----&#xA;31  ---- ----             ---- ----             **** ****             **** ****  **** ****             ---- ----             ---- ----&#xA;&#xA;(**** object continuously above horizon)                                                      (---- object continuously below horizon)&#xA;&#xA;NOTE: BLANK SPACES IN THE TABLE INDICATE THAT A RISING OR A SETTING DID NOT OCCUR DURING THAT 24 HR INTERVAL.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Notice that from Nov 22 through Dec 06 (non-inclusive) the rise and&#xA;set are both &quot;----&quot;, indicating &quot;object continuously below horizon&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Caveats/notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Like the table above, I assume 34 arcminutes of refraction at the&#xA;horizon. That's reasonable for most locations, but may be&#xA;unreasonable at 87 degrees south latitude. If refraction was higher&#xA;during the calculated moonless period above, the moon may have risen&#xA;briefly.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This is the longest moonless period I could fine between 2 Jan&#xA;2000 and 12 Mar 2016, the time limits of the program I'm&#xA;using. There might be longer moonless periods outside this interval.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The process I used isn't guarenteed to find the longest moonless&#xA;period even &lt;em&gt;within&lt;/em&gt; the interval above. In other words, there might&#xA;be a longer moonless period that I missed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It's somewhat surprising that the longest moonless period occurs&#xA;so far from the poles (ie, 3 degrees from the south pole). This is&#xA;due in part to refraction and the fact that the moon has an angular&#xA;diameter and isn't just a geometric point. Also, the moon's&#xA;elevation can be 3 degrees lower at 87 degrees than it is at the&#xA;pole, allowing the moon to remain below the horizon just a little&#xA;longer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The longest moonless period at the south pole (with caveats above)&#xA;was from 1 Jun 2013 at 1104 to 16 Jun 2013 at 0719, about two and a&#xA;quarter hours less than the shortest overall moonless period I&#xA;found.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The longest moonless period at the north pole was from 18 Dec 2008&#xA;at 2311 to 2 Jan 2009 at 1155, about half an hour shorter than the south&#xA;pole time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The program I wrote to help me find this:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-solve-astro-5894.c&quot; rel=&quot;noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-solve-astro-5894.c&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2015-12-31T21:12:23.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13041" PostTypeId="1" CreationDate="2016-01-01T02:02:32.840" Score="2" ViewCount="586" Body="&lt;p&gt;According to &lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2014/01/01/new_year_2014_how_astronomers_define_the_year.html&quot; rel=&quot;nofollow&quot;&gt;this essay by Phil Plait&lt;/a&gt;, the duration of Earth's sidereal year is 31,558,149 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this figure stable? If I were to meaure Earth's orbit several million (or billions) of years in the past or future, should I expect to see roughly the same duration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, what would be the cause of any change observed?&lt;/p&gt;&#xA;" OwnerUserId="7007" LastActivityDate="2016-01-01T08:15:21.807" Title="Is the duration of a sidereal year stable?" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;earth&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13042" PostTypeId="2" ParentId="13041" CreationDate="2016-01-01T06:34:16.630" Score="2" Body="&lt;p&gt;It changes some.   The solar energy pressing against the Earth from the sun, the kinetic energy from coronal mass ejections hitting the Earth, tidal forces from the Earth-Sun relation and even gravitational tugs form other planets and meteors and space dust that hit the earth all change the Earth's orbit a teeny tiny bit, also, changes in the Sun's mass either by the sun accumulating mass from space dust or comets or losing mass, which most suns do over time.   Each tiny orbital change changes the sidereal year (and all the other years) a tiny little bit.    Perhaps this only amounts to millionths of a second per year but some of these forces are consistent and can lead to predictable changes over time.  Others variable, leading to uncertainty.   The long term changes to the Earth's orbital distance are currently &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth&amp;#39;s_orbit#Future&quot; rel=&quot;nofollow noreferrer&quot;&gt;unpredictable&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's generally believed that the Earth is slowly moving away from the sun, but I don't believe there's much agreement on how fast this is currently happening.   Articles &lt;a href=&quot;https://www.newscientist.com/article/dn17228-why-is-the-earth-moving-away-from-the-sun/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://curious.astro.cornell.edu/about-us/41-our-solar-system/the-earth/orbit/83-is-the-distance-from-the-earth-to-the-sun-changing-advanced&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.   The 2nd one estimates 1.5 CM per year or 15 KM per million years which even over a billion years (15,000 KM) is quite a small distance compared to 150 million KM.   But estimates are difficult for a variety of reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked a related question &lt;a href=&quot;https://astronomy.stackexchange.com/questions/11243/how-has-the-earths-orbit-changed-over-hundreds-of-millions-or-billion-of-years&quot;&gt;here&lt;/a&gt;, with a bounty and didn't get any definite answers.   It's possible to mathematically model the movement of some planets (Jupiter to an extent and Neptune/Uranus and perhaps Saturn based on the Jupiter model).  It's thought Jupiter moved both towards then away from the sun over the first few billion years of the solar system.  The Moon's distance from the Earth is also mostly predictible and can be modeled back into the past based on the Oceanic tidal bulge, but modeling the Earth;s distance from the sun over the past is much harder.  It's a tiny object compared to the Sun and Jupiter and N body systems have a lot of chaos and unpredictability to them.   4 billion years ago it's possible that the Earth was quite a bit closer to the sun, but I don't think anyone has a good model for how much closer.   The &lt;a href=&quot;https://en.wikipedia.org/wiki/Faint_young_Sun_paradox&quot; rel=&quot;nofollow noreferrer&quot;&gt;faint young sun paradox&lt;/a&gt; and the theory that young suns eject much more matter and have more active coronal mass ejections suggests to me that the Earth was likely closer to the Sun billions of years ago and not further away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The earth's orbital eccentricity also changes with time, on a cycle of roughly every 100,000 years but according to &lt;a href=&quot;https://en.wikipedia.org/wiki/Milankovitch_cycles#Orbital_shape_.28eccentricity.29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, that doesn't change the length of the year, only the shape of the orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, and just for fun, it makes logical sense to me that the Moon orbiting the Earth creates small apparent fluctuations at least from the point of view of the Earth, because the Earth in effect orbits the Earth-Moon barycenter in an ellipse roughly 1/81st the size of the distance to the moon, or about 4,480 KM in radius every 29 days.     Depending on whether the Moon is ahead of the earth or behind it, the Earth year can change by as much as 5 minutes.    I suspect that this is largely ignored when measuring Earth years though and the Earth-Moon barycenter is used.   If we measured the Earth's center we should get much larger variations year to year.  (see picture).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ta97l.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ta97l.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.education.com/science-fair/article/barycenter-balancing-point/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(if anyone can improve on this, feel free, I have the feeling it's only a partial answer, cause logically it makes sense that the Earth-Jupiter positions or Earth-Venus should change, if only slightly, the Sidereal year).*&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-01-01T06:34:16.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13043" PostTypeId="2" ParentId="13041" CreationDate="2016-01-01T07:15:31.687" Score="-1" Body="&lt;p&gt;It fluctuates a little now and then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon's gravity is responsible for tides. This means that the oceans are not constantly in synch with the planet's rotation. The Moon pulls the oceans back a little and then releases them. This creates drag - a form of friction. However minimal, it is still telling over the millennia. However, certain types of volcano eruptions, along with the tsunami it creates, speeds Earth's rotation up a little bit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the mantle and core are in fact liquid, they also create a degree of drag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This changes (a little bit) the duration of each second. Thus, ancient Roman seconds were a trillionth of a fraction faster than ours, give or take.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, the orbit itself fluctuates a little bit. It is believed today that our planet is moving slowly away from the Sun. This would make the orbit longer without changing Earth's velocity. Thus, yes, each year must be getting longer, although the difference is ridiculously small. &lt;/p&gt;&#xA;" OwnerUserId="9510" LastActivityDate="2016-01-01T07:15:31.687" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13044" PostTypeId="2" ParentId="13032" CreationDate="2016-01-01T07:18:03.107" Score="1" Body="&lt;p&gt;Polyphant is correct, it can operate either way.   It's hypothesized that Jupiter in 2:1 resonance with Saturn was the cause of the late heavy bombardment.   &lt;a href=&quot;https://en.wikipedia.org/wiki/Late_Heavy_Bombardment#Gas_giant_migration&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 2 largest objects in the solar system operating in resonance tended to stir everything up.  Many small objects including Neptune and Uranus were thrown around, some into the sun, some in towards the inner solar-system where some crashed into other planets and some out of the solar system or into the Kuiper belt, but all that ended 3.8 billion years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dummies answer is that most of the stuff in the solar system that had orbits that were likely to crash into a planet have already done so, and Jupiter sped that process up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slightly longer answer follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the time being, the asteroid belt is mostly stable, as are Jupiter's two Trojan regions, L4 and L5 and Jupiter's hildas are also stable.   So that's basically why we're in a low bombardment period now.   The inner solar-system objects are mostly in stable orbits where they aren't in position to crash into planets.   To collide, orbits need to cross and mostly the inner solar system objects don't cross orbits with planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.universetoday.com/32856/asteroid-belt/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Computer simulations suggest that the original asteroid belt may have&#xA;  contained mass equivalent to the Earth. Primarily because of&#xA;  gravitational perturbations, most of the material was ejected from the&#xA;  belt a million years after its formation, leaving behind less than&#xA;  0.1% of the original mass. Since then, the size distribution of the asteroid belt is believed to have remained relatively stable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Similarly, Jupiter Trojan objects, L4 and L5 and Jupiter Hildas are in mostly stable orbits due to the high mass of Jupiter, so it's a two fold answer.   First is that late heavy bombardment removed most of the inner solar system objects that were bombardment candidates, and 2nd, the huge mass of Jupiter compared to the inner planets and the significant relative distance between Jupiter and Mars and the small mass of Mars), together creates a healthy amount of relatively stable orbital regions where asteroids can stay in stable orbit and not be in position to crash into the 4 inner planets or be subject to n-body instability of multiple body orbital systems.   (I can add various articles on the stability of Jupiter's L4 and L5 and Hildas, but they're not hard to google).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Kuiper belt objects, I don't believe Jupiter has much of an effect pro or con, for earth impacts, though it blocks a few of those, but it should only be a small percentage, but I suspect Kuiper belt object collisions with inner planets are less frequent.  For inner solar system objects (asteroid belt, L4, L5 &amp;amp; Hildas), that's where Jupiter significantly reduces impacts with inner planets, for most of the last 3.8 billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A curious sidebar is that Jupiter may, at some point in the future, stir things up in the solar-system again and create another much smaller heavy bombardment period.  This would depend on the Jupiter-Mercury resonance and if Jupiter was able to pull Mercury away from the sun, as some models suggest it might.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is low probability and if it happens it won't happen for billions of years, but if Jupiter does cause Mercury to migrate away from the sun, Mercury passing through the asteroid belt, hildas and L4 or L5 would likely cause a significant increase in bombardment.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://www.universetoday.com/14032/could-jupiter-wreck-the-solar-system/&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; doesn't mention the asteroid factor, but Mercury passing through any of the asteroid rich regions of the inner solar-system would stir things up quite a bit and significantly increase the chance of bombardment.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-01-01T07:18:03.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13045" PostTypeId="2" ParentId="13041" CreationDate="2016-01-01T08:15:21.807" Score="2" Body="&lt;p&gt;Unfortunately, I don't have any precise information on the time scales you want (and I doubt that it's possible to make any quantitative statements with any degree of certainty). But vaguely, yes, it will almost certainly be different from today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, even on shorter time scales, the duration of the sidereal year is not constant, as a variety of physical effects, primarily perturbations from other objects in the solar system, introduce a certain level of messiness into Earth's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one thinks of the duration of the sidereal year as a continuous parameter defined at an instant of time, then it expresses the instantaneous rate of change of the mean longitude. Essentially, we ask how long it would take for the body to make a full $360^\circ$ motion if its mean longitude changed at the current rate. Specifically, if $t$ is measured in thousands of Julian years from the J2000 epoch, then we can express the mean longitude as a power series&#xA;$$\lambda = \lambda_0 + \lambda_1t + \lambda_2t^2+\lambda_3t^3 + \ldots$$&#xA;with the coefficients&lt;sup&gt;1&lt;/sup&gt;&#xA;$$\begin{eqnarray*}\lambda_0 &amp;amp;=&amp;amp; +\!100^\circ\!.46645683\text{,} \\&#xA;\lambda_1 &amp;amp;=&amp;amp; +\!1295977422''\!.83429\text{,} \\&#xA;\lambda_2 &amp;amp;=&amp;amp; -\!2''\!.04411\text{,} \\&#xA;\lambda_3 &amp;amp;=&amp;amp; -\!0''\!.00523\text{.}\end{eqnarray*}$$&#xA;The rate of change is therefore&#xA;$$\dot{\lambda} = \lambda_1 + 2\lambda_2t + 3\lambda_3t^2 + \ldots\;\text{.}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can check what this gives for J2000: at $t = 0$, the rate of change is simply $\lambda_1$, and a Julian millennium is $3.15576\times 10^{10}$ seconds, so:&#xA;$$\frac{360\times 3600\times 3.15576\times 10^{10}}{1.29597742283429\times 10^9} = 31558149.76\,\mathrm{s}\text{,}$$&#xA;which is within a second of the figure Phil Plait gave.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Simon, J. L.; Bretagnon, P.; Chapront, J.; Chapront-Touze, M.; Francou, G.; Laskar, J. &quot;Numerical expressions for precession formulae and mean elements for the Moon and planets.&quot; &lt;em&gt;Astronomy and Astrophysics&lt;/em&gt; &lt;strong&gt;282&lt;/strong&gt; (2): &lt;a href=&quot;http://adsabs.harvard.edu/abs/1994A&amp;amp;A...282..663S&quot; rel=&quot;nofollow&quot;&gt;663-683&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="661" LastActivityDate="2016-01-01T08:15:21.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13047" PostTypeId="2" ParentId="13035" CreationDate="2016-01-01T13:49:20.357" Score="-2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If I dropped a TV camera into a black hole, will it be destroyed before it reaches the event horizon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think so. I also think it will be destroyed before tidal forces have any effect.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If so, why?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because it would otherwise end up falling faster than the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sounds strange I know, but take a look at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Shapiro_delay&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shapiro delay&lt;/a&gt;: &lt;em&gt;&quot;the speed of a light wave depends on the strength of the gravitational potential along its path&quot;&lt;/em&gt;. Or see Professor Ned Wright's &lt;a href=&quot;http://www.astro.ucla.edu/~wright/deflection-delay.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Deflection and Delay of Light&lt;/a&gt;:  &lt;em&gt;&quot;in a very real sense, the delay experienced by light passing a massive object is responsible for the deflection of the light&quot;&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jiogu.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively see &lt;a href=&quot;http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/speed_of_light.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this PhysicsFAQ article&lt;/a&gt; by Don Koks: &lt;em&gt;&quot;this difference in speeds is precisely that referred to above by ceiling and floor observers&quot;.&lt;/em&gt; He's referring to the way Einstein said light curves because the speed of light varies with position. See the &lt;a href=&quot;http://einsteinpapers.press.princeton.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein digital papers&lt;/a&gt; for examples of that. &lt;a href=&quot;http://einsteinpapers.press.princeton.edu/vol7-trans/156?highlightText=%22speed%20of%20light%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's&lt;/a&gt; one from 1920, see the second paragraph: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8KXbI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light doesn't curve because spacetime is curved. Einstein never actually said that. It curves because the speed of light is lower at a lower altitude, rather like the way &lt;a href=&quot;http://fas.org/man/dod-101/navy/docs/es310/SNR_PROP/snr_prop.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;sonar&lt;/a&gt; waves curve: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sNuHt.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;As to why this isn't common knowledge, I don't know. There's this myth that Einstein gave up on a varying speed of light in 1911, but he didn't, see &lt;a href=&quot;https://en.wikipedia.org/wiki/Variable_speed_of_light#Einstein.27s_VSL_attempt_in_1911&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Wikipedia article&lt;/a&gt; and &lt;a href=&quot;http://einsteinpapers.press.princeton.edu/vol4-trans/271?highlightText=%22writer%20of%20these%20lines%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;this example from 1914&lt;/a&gt;. I don't know why the reason matter falls down isn't common knowledge either. You know about pair production and electron diffraction and the wave nature of matter, just think of an electron as a wave going round a closed path, then simplify it to a square path, like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ATcJA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horizontals bend downwards, so the electron falls down. Could it &lt;em&gt;be&lt;/em&gt; any simpler? Anyway, your TV camera falls down because the speed of light is reducing with altitude. We tend to call this the &quot;coordinate&quot; speed of light nowadays, even though Einstein just called it the speed of light. But regardless of what we call it, you don't have to be the Brain of Britain, or the Brain of France, to work out that half down, there's some kind of problem. At this point the camera will be falling as fast as the &quot;coordinate&quot; speed of light at that location. And it isn't going to slow down. Things always fall faster, not slower. But matter can't go faster than light, because of the wave nature of matter. When matter is made of waves, there is no way that matter can move faster than the speed of those waves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what's going to happen? I can see no other option: that wave has to &lt;em&gt;break&lt;/em&gt;. Again it sounds strange, but take a look at an old version of the firewall article on &lt;a href=&quot;https://en.wikipedia.org/w/index.php?title=Firewall_(physics)&amp;amp;oldid=587309921&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;. Follow the reference 7 link to Friedwardt Winterberg's &lt;a href=&quot;http://www.znaturforsch.com/aa/v56a/56a0889.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gamma-Ray Bursters and Lorentzian Relativity&lt;/a&gt;: &lt;em&gt;&quot;If the balance of forces holding together elementary particles is destroyed near the event horizon, all matter would be converted into zero rest mass particles which could explain the large energy release of gamma ray bursters&quot;.&lt;/em&gt; I think this is what happens to your TV camera. &lt;em&gt;Flash!&lt;/em&gt; It turns into a gamma-ray burst. It would be like an atom bomb, but a lot more efficient. So make sure you drop it from a safe distance. &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2016-01-01T13:49:20.357" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13049" PostTypeId="2" ParentId="13028" CreationDate="2016-01-01T18:57:46.967" Score="0" Body="&lt;p&gt;I think that the need to leave Earth won't be because of &quot;limited resources&quot; but rather that it will be too hot to stay here, when the Sun expands outward in size billions of years from now.&lt;/p&gt;&#xA;" OwnerUserId="2562" LastActivityDate="2016-01-01T18:57:46.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13051" PostTypeId="1" CreationDate="2016-01-02T03:26:50.013" Score="6" ViewCount="321" Body="&lt;p&gt;Okay, bear with me. In layman's terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The estimates vary, but on a clear night, away from big cities, one can see up to 5000 stars in the sky. All those are in the Milky Way, but they're hardly equidistant from Earth, which would mean that they revolve around the galactic center at different orbits. According to some sources, some are positioned very closely to us while others hang out as far as 1000 light year away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now by astronomical standards, 5000 stars is not a great number. It's still pretty impressive, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some stars are larger, and others are smaller. It stands to reason that some of them might be &quot;hiding&quot; behind others. Because they all orbit the galactic center at different speeds, using different orbits, wouldn't it follow that from time to time, over the centuries, decades, or even years (months?) a &quot;new&quot; star would emerge from behind another one? Shouldn't this be happening often; or are they so sparsely distributed around the galaxy that no &quot;new&quot; stars have appeared out there over the past three thousand years or so?&lt;/p&gt;&#xA;" OwnerUserId="9510" LastEditorUserId="9510" LastEditDate="2016-01-03T00:39:38.510" LastActivityDate="2016-01-05T03:45:11.047" Title="Why aren't new stars in Earth's relative proximity constantly discovered?" Tags="&lt;galaxy&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13052" PostTypeId="1" CreationDate="2016-01-02T04:45:47.833" Score="0" ViewCount="135" Body="&lt;p&gt;I am going to ask an odd question, It is: In most of the images of our solar system we see that the planets are arranged in  a straight line. As the paths of the planets are elliptical, could it be possible for planets be arranged other than in straight line (a bit up and down) ?&lt;/p&gt;&#xA;" OwnerUserId="10278" LastEditorUserId="4085" LastEditDate="2016-01-02T11:28:36.957" LastActivityDate="2016-01-02T11:28:36.957" Title="Alignments of planets" Tags="&lt;solar-system&gt;&lt;planet&gt;" AnswerCount="1" CommentCount="9" ClosedDate="2016-05-13T01:56:14.877" ContentLicense="CC BY-SA 3.0" />
  <row Id="13053" PostTypeId="2" ParentId="12824" CreationDate="2016-01-02T06:02:09.247" Score="16" Body="&lt;p&gt;As noted in &lt;a href=&quot;http://aa.quae.nl/en/antwoorden/zonpositie.html#14&quot; rel=&quot;noreferrer&quot;&gt;http://aa.quae.nl/en/antwoorden/zonpositie.html#14&lt;/a&gt; the&#xA;length of sunrise/sunset varies from approximately 128/cos(latitude)&#xA;seconds at the equinoxes to approximately 142/cos(1.14*latitude) at&#xA;the solstices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, here's the length of sunrise/sunset at various latitudes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Rvetn.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rvetn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond 65 degrees north or south latitude, the sun does not rise or&#xA;set daily, and the length of sunrise/sunset increases significantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data plotted above is the length of sunrise, but the length of&#xA;sunset is very similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All calculations for this program were made with this program:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-solve-astro-12824.c&quot; rel=&quot;noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-solve-astro-12824.c&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The raw output of sunrise/sunset times:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/sun-rise-set-multiple-latitudes.txt.bz2&quot; rel=&quot;noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/sun-rise-set-multiple-latitudes.txt.bz2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can verify these results at:&#xA;&lt;a href=&quot;http://aa.usno.navy.mil/data/docs/RS_OneYear.php&quot; rel=&quot;noreferrer&quot;&gt;http://aa.usno.navy.mil/data/docs/RS_OneYear.php&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The longest sunrise I found for 2015 was at 89 degrees 51 minutes&#xA;south latitude, 125 degrees east longitude. There, the sun starts&#xA;rising 20 Sep 2015 at 2352, bobbles up and down a bit (but never quite&#xA;sets), and finally finishes rising 43 hours and 21 minutes later, at&#xA;22 Sep 2015 at 1913, but see caveat at the end of this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can &quot;verify&quot; this by first visiting&#xA;&lt;a href=&quot;http://aa.usno.navy.mil/data/docs/RS_OneYear.php&quot; rel=&quot;noreferrer&quot;&gt;http://aa.usno.navy.mil/data/docs/RS_OneYear.php&lt;/a&gt; with these&#xA;parameters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gjEOk.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gjEOk.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to get: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&#xA;Sun or Moon Rise/Set Table for One Year&#xA;             o  ,    o  ,                                                                              Astronomical Applications Dept.&#xA;Location: E125 00, S89 51                          Rise and Set for the Sun for 2015                   U. S. Naval Observatory        &#xA;                                                                                                       Washington, DC  20392-5420     &#xA;                                                            Universal Time                                                            &#xA;&#xA;&#xA;       Jan.       Feb.       Mar.       Apr.       May        June       July       Aug.       Sept.      Oct.       Nov.       Dec.  &#xA;Day Rise  Set  Rise  Set  Rise  Set  Rise  Set  Rise  Set  Rise  Set  Rise  Set  Rise  Set  Rise  Set  Rise  Set  Rise  Set  Rise  Set&#xA;     h m  h m   h m  h m   h m  h m   h m  h m   h m  h m   h m  h m   h m  h m   h m  h m   h m  h m   h m  h m   h m  h m   h m  h m&#xA;01  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****&#xA;02  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****&#xA;03  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****&#xA;04  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****&#xA;05  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****&#xA;06  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****&#xA;07  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****&#xA;08  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****&#xA;09  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****&#xA;10  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****&#xA;11  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****&#xA;12  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****&#xA;13  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****&#xA;14  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****&#xA;15  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****&#xA;16  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****&#xA;17  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****&#xA;18  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****&#xA;19  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****&#xA;20  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  2352       **** ****  **** ****  **** ****&#xA;21  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****&#xA;22  **** ****  **** ****  1842 1614  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****&#xA;23  **** ****  **** ****       0708  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****&#xA;24  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****&#xA;25  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****&#xA;26  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****&#xA;27  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****&#xA;28  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****&#xA;29  **** ****             ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****&#xA;30  **** ****             ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****&#xA;31  **** ****             ---- ----             ---- ----             ---- ----  ---- ----             **** ****             **** ****&#xA;&#xA;(**** object continuously above horizon)                                                      (---- object continuously below horizon)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that the sun rises at 2352 on September 20th, and doesn't set for&#xA;the rest of the year, verifying the sunrise start time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Verifying the end time is a little tricker. To do this, visit&#xA;&lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;noreferrer&quot;&gt;http://ssd.jpl.nasa.gov/horizons.cgi&lt;/a&gt; with the following parameters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NQAtX.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NQAtX.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&#xA;Revised : Jul 31, 2013                  Sun                                 10&#xA;&#xA; PHYSICAL PROPERTIES (revised Jan 16, 2014):&#xA;  GM (10^11 km^3/s^2)   = 1.3271244004193938  Mass (10^30 kg)   ~ 1.988544&#xA;  Radius (photosphere)  = 6.963(10^5) km  Angular diam at 1 AU  = 1919.3&quot;&#xA;  Solar Radius (IAU)    = 6.955(10^5) km  Mean density          = 1.408 g/cm^3&#xA;  Surface gravity       =  274.0 m/s^2    Moment of inertia     = 0.059&#xA;  Escape velocity       =  617.7 km/s     Adopted sidereal per  = 25.38 d&#xA;  Pole (RA,DEC in deg.) =  286.13,63.87   Obliquity to ecliptic = 7 deg 15'        &#xA;  Solar constant (1 AU) = 1367.6 W/m^2    Solar lumin.(erg/s)   =  3.846(10^33)&#xA;  Mass-energy conv rate = 4.3(10^12 gm/s) Effective temp (K)    =  5778&#xA;  Surf. temp (photosphr)= 6600 K (bottom) Surf. temp (photosphr)=  4400 K (top)&#xA;  Photospheric depth    = ~400 km         Chromospheric depth   = ~2500 km&#xA;  Sunspot cycle         = 11.4 yr         Cycle 22 sunspot min. =  1991 A.D.&#xA;&#xA;  Motn. rel to nrby strs= apex : RA=271 deg; DEC=+30 deg&#xA;                          speed: 19.4 km/s = 0.0112 AU/day&#xA;  Motn. rel to 2.73K BB = apex : l=264.7+-0.8; b=48.2+-0.5&#xA;                          speed: 369 +-11 km/s&#xA;&#xA;Results&#xA;&#xA;*******************************************************************************&#xA;Ephemeris / WWW_USER Fri Jan  1 21:49:19 2016 Pasadena, USA      / Horizons    &#xA;*******************************************************************************&#xA;Target body name: Sun (10)                        {source: DE431mx}&#xA;Center body name: Earth (399)                     {source: DE431mx}&#xA;Center-site name: (user defined site below)&#xA;*******************************************************************************&#xA;Start time      : A.D. 2015-Sep-22 19:00:00.0000 UT      &#xA;Stop  time      : A.D. 2015-Sep-22 20:00:00.0000 UT      &#xA;Step-size       : 1 minutes&#xA;*******************************************************************************&#xA;Target pole/equ : IAU_SUN                         {East-longitude +}&#xA;Target radii    : 696000.0 x 696000.0 x 696000.0 k{Equator, meridian, pole}    &#xA;Center geodetic : 125.000000,-89.850000,7.057E-13 {E-lon(deg),Lat(deg),Alt(km)}&#xA;Center cylindric: 125.000000,16.7540774,-6356.730 {E-lon(deg),Dxy(km),Dz(km)}&#xA;Center pole/equ : High-precision EOP model        {East-longitude +}&#xA;Center radii    : 6378.1 x 6378.1 x 6356.8 km     {Equator, meridian, pole}    &#xA;Target primary  : Sun&#xA;Vis. interferer : MOON (R_eq= 1737.400) km        {source: DE431mx}&#xA;Rel. light bend : Sun, EARTH                      {source: DE431mx}&#xA;Rel. lght bnd GM: 1.3271E+11, 3.9860E+05 km^3/s^2                              &#xA;Atmos refraction: NO (AIRLESS)&#xA;RA format       : HMS&#xA;Time format     : CAL &#xA;RTS-only print  : NO       &#xA;EOP file        : eop.160101.p160324                                           &#xA;EOP coverage    : DATA-BASED 1962-JAN-20 TO 2016-JAN-01. PREDICTS-&amp;gt; 2016-MAR-23&#xA;Units conversion: 1 au= 149597870.700 km, c= 299792.458 km/s, 1 day= 86400.0 s &#xA;Table cut-offs 1: Elevation (-90.0deg=NO ),Airmass (&amp;gt;38.000=NO), Daylight (NO )&#xA;Table cut-offs 2: Solar Elongation (  0.0,180.0=NO ),Local Hour Angle( 0.0=NO )&#xA;*******************************************************************************&#xA; Date__(UT)__HR:MN     Azi_(a-appr)_Elev&#xA;****************************************&#xA;$$SOE&#xA; 2015-Sep-22 19:00 *m  128.1772  -0.3117&#xA; 2015-Sep-22 19:01 *m  127.9272  -0.3109&#xA; 2015-Sep-22 19:02 *m  127.6771  -0.3101&#xA; 2015-Sep-22 19:03 *m  127.4270  -0.3093&#xA; 2015-Sep-22 19:04 *m  127.1770  -0.3085&#xA; 2015-Sep-22 19:05 *m  126.9269  -0.3077&#xA; 2015-Sep-22 19:06 *m  126.6769  -0.3069&#xA; 2015-Sep-22 19:07 *m  126.4268  -0.3061&#xA; 2015-Sep-22 19:08 *m  126.1767  -0.3053&#xA; 2015-Sep-22 19:09 *m  125.9267  -0.3045&#xA; 2015-Sep-22 19:10 *m  125.6766  -0.3037&#xA; 2015-Sep-22 19:11 *m  125.4266  -0.3029&#xA; 2015-Sep-22 19:12 *m  125.1765  -0.3021&#xA; 2015-Sep-22 19:13 *m  124.9264  -0.3013&#xA; 2015-Sep-22 19:14 *m  124.6764  -0.3005&#xA; 2015-Sep-22 19:15 *m  124.4263  -0.2997&#xA; 2015-Sep-22 19:16 *m  124.1762  -0.2989&#xA; 2015-Sep-22 19:17 *m  123.9262  -0.2981&#xA; 2015-Sep-22 19:18 *m  123.6761  -0.2973&#xA; 2015-Sep-22 19:19 *m  123.4261  -0.2964&#xA; 2015-Sep-22 19:20 *m  123.1760  -0.2956&#xA; 2015-Sep-22 19:21 *m  122.9259  -0.2948&#xA; 2015-Sep-22 19:22 *m  122.6759  -0.2940&#xA; 2015-Sep-22 19:23 *m  122.4258  -0.2932&#xA; 2015-Sep-22 19:24 *m  122.1757  -0.2923&#xA; 2015-Sep-22 19:25 *m  121.9257  -0.2915&#xA; 2015-Sep-22 19:26 *m  121.6756  -0.2907&#xA; 2015-Sep-22 19:27 *m  121.4256  -0.2899&#xA; 2015-Sep-22 19:28 *m  121.1755  -0.2890&#xA; 2015-Sep-22 19:29 *m  120.9254  -0.2882&#xA; 2015-Sep-22 19:30 *m  120.6754  -0.2874&#xA; 2015-Sep-22 19:31 *m  120.4253  -0.2865&#xA; 2015-Sep-22 19:32 *m  120.1753  -0.2857&#xA; 2015-Sep-22 19:33 *m  119.9252  -0.2849&#xA; 2015-Sep-22 19:34 *m  119.6751  -0.2840&#xA; 2015-Sep-22 19:35 *m  119.4251  -0.2832&#xA; 2015-Sep-22 19:36 *m  119.1750  -0.2823&#xA; 2015-Sep-22 19:37 *m  118.9250  -0.2815&#xA; 2015-Sep-22 19:38 *m  118.6749  -0.2807&#xA; 2015-Sep-22 19:39 *m  118.4248  -0.2798&#xA; 2015-Sep-22 19:40 *m  118.1748  -0.2790&#xA; 2015-Sep-22 19:41 *m  117.9247  -0.2781&#xA; 2015-Sep-22 19:42 *m  117.6746  -0.2773&#xA; 2015-Sep-22 19:43 *m  117.4246  -0.2764&#xA; 2015-Sep-22 19:44 *m  117.1745  -0.2756&#xA; 2015-Sep-22 19:45 *m  116.9245  -0.2747&#xA; 2015-Sep-22 19:46 *m  116.6744  -0.2739&#xA; 2015-Sep-22 19:47 *m  116.4243  -0.2730&#xA; 2015-Sep-22 19:48 *m  116.1743  -0.2721&#xA; 2015-Sep-22 19:49 *m  115.9242  -0.2713&#xA; 2015-Sep-22 19:50 *m  115.6742  -0.2704&#xA; 2015-Sep-22 19:51 *m  115.4241  -0.2696&#xA; 2015-Sep-22 19:52 *m  115.1740  -0.2687&#xA; 2015-Sep-22 19:53 *m  114.9240  -0.2678&#xA; 2015-Sep-22 19:54 *m  114.6739  -0.2670&#xA; 2015-Sep-22 19:55 *m  114.4239  -0.2661&#xA; 2015-Sep-22 19:56 *m  114.1738  -0.2652&#xA; 2015-Sep-22 19:57 *m  113.9237  -0.2644&#xA; 2015-Sep-22 19:58 *m  113.6737  -0.2635&#xA; 2015-Sep-22 19:59 *m  113.4236  -0.2626&#xA; 2015-Sep-22 20:00 *m  113.1735  -0.2618&#xA;$$EOE&#xA;*******************************************************************************&#xA;Column meaning:&#xA;&#xA;TIME&#xA;&#xA;  Prior to 1962, times are UT1. Dates thereafter are UTC. Any 'b' symbol in&#xA;the 1st-column denotes a B.C. date. First-column blank (&quot; &quot;) denotes an A.D.&#xA;date. Calendar dates prior to 1582-Oct-15 are in the Julian calendar system.&#xA;Later calendar dates are in the Gregorian system.&#xA;&#xA;  Time tags refer to the same instant throughout the universe, regardless of&#xA;where the observer is located.&#xA;&#xA;  The dynamical Coordinate Time scale is used internally. It is equivalent to&#xA;the current IAU definition of &quot;TDB&quot;. Conversion between CT and the selected&#xA;non-uniform UT output scale has not been determined for UTC times after the&#xA;next July or January 1st.  The last known leap-second is used over any future&#xA;interval.&#xA;&#xA;  NOTE: &quot;n.a.&quot; in output means quantity &quot;not available&quot; at the print-time.&#xA;&#xA;SOLAR PRESENCE (OBSERVING SITE)&#xA;  Time tag is followed by a blank, then a solar-presence symbol:&#xA;&#xA;        '*'  Daylight (refracted solar upper-limb on or above apparent horizon)&#xA;        'C'  Civil twilight/dawn&#xA;        'N'  Nautical twilight/dawn&#xA;        'A'  Astronomical twilight/dawn&#xA;        ' '  Night OR geocentric ephemeris&#xA;&#xA;LUNAR PRESENCE WITH TARGET RISE/TRANSIT/SET MARKER (OBSERVING SITE)&#xA;  The solar-presence symbol is immediately followed by another marker symbol:&#xA;&#xA;        'm'  Refracted upper-limb of Moon on or above apparent horizon&#xA;        ' '  Refracted upper-limb of Moon below apparent horizon OR geocentric&#xA;        'r'  Rise    (target body on or above cut-off RTS elevation)&#xA;        't'  Transit (target body at or past local maximum RTS elevation)&#xA;        's'  Set     (target body on or below cut-off RTS elevation)&#xA;&#xA;RTS MARKERS (TVH)&#xA;  Rise and set are with respect to the reference ellipsoid true visual horizon&#xA;defined by the elevation cut-off angle. Horizon dip and yellow-light refraction&#xA;(Earth only) are considered. Accuracy is &amp;lt; or = to twice the requested search&#xA;step-size.&#xA;&#xA; Azi_(a-appr)_Elev =&#xA;   Airless apparent azimuth and elevation of target center. Adjusted for&#xA;light-time, the gravitational deflection of light, stellar aberration,&#xA;precession and nutation. Azimuth measured North(0) -&amp;gt; East(90) -&amp;gt; South(180) -&amp;gt;&#xA;West(270) -&amp;gt; North (360). Elevation is with respect to plane perpendicular&#xA;to local zenith direction.  TOPOCENTRIC ONLY. Units: DEGREES&#xA;&#xA;&#xA; Computations by ...&#xA;     Solar System Dynamics Group, Horizons On-Line Ephemeris System&#xA;     4800 Oak Grove Drive, Jet Propulsion Laboratory&#xA;     Pasadena, CA  91109   USA&#xA;     Information: http://ssd.jpl.nasa.gov/&#xA;     Connect    : telnet://ssd.jpl.nasa.gov:6775  (via browser)&#xA;                  telnet ssd.jpl.nasa.gov 6775    (via command-line)&#xA;     Author     : Jon.Giorgini@jpl.nasa.gov&#xA;&#xA;*******************************************************************************&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The sun's angular diameter is about 32 arcminutes, so the sun's lower&#xA;limb is 16 arcminutes below the sun's center. When the center of the&#xA;sun has geometric elevation -18 arcminutes (-0.3 degrees), the lower&#xA;limb has geometric elevation -34 arcminutes. Since refraction near the&#xA;horizon is also 34 arcminutes, the sun's lower limb rises when the&#xA;sun's geometric elevation is -0.3 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the table above, this occurs between 1914 and 1915, but my program&#xA;uses slightly more accurate data for the sun's angular diameter, and&#xA;the sun actually finishes rising between 1913 and 1914 (and closer to&#xA;1913).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then fly almost halfway across the world to latitude 89&#xA;degrees 51 minutes and longitude -19 degrees to see the&#xA;one-minute-shorter longest sunset, which starts at 23 Sep 2015 at 2128&#xA;and ends at 25 Sep 2015 at 1648, a length of 43 hours and 20 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, you would use&#xA;&lt;a href=&quot;http://aa.usno.navy.mil/data/docs/RS_OneYear.php&quot; rel=&quot;noreferrer&quot;&gt;http://aa.usno.navy.mil/data/docs/RS_OneYear.php&lt;/a&gt; to verify the ending&#xA;time of the sunset, and HORIZONS to verify the start time of the&#xA;sunset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polar sunrises and sunsets are considerably shorter:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;At the North Pole, the sun starts rising at 18 Mar 2015 at 2015,&#xA;and finishes rising at 20 Mar 2015 at 0441, a length of 32 hours and&#xA;26 minutes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At the South Pole, the sun starts setting at 21 Mar 2015 at 1650,&#xA;and finishes setting at 23 Mar 2015 at 0117, a length of 32 hours&#xA;and 27 minutes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At the South Pole, the sun starts rising at 21 Sep 2015 at 0508,&#xA;and finishes rising at 22 Sep 2015 at 1400, a length of 32 hours and&#xA;52 minutes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At the North Pole, the sun starts setting at 24 Sep 2015 at 0243,&#xA;and finishes setting at 25 Sep 2015 at 1131, a length of 32 hours&#xA;and 48 minutes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Main caveat: Like HORIZONS and the sunrise/sunset tables above, I&#xA;assume 34 arcminutes of refraction at the horizon. That's reasonable&#xA;for most locations, but may be unreasonable close the pole, where the&#xA;longest sunrises and sunsets occur. In particular, refraction can&#xA;change rapidly at these latitudes, allowing for potentially much&#xA;longer sunrises and sunsets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I now believe that &lt;a href=&quot;http://what-if.xkcd.com/42/&quot; rel=&quot;noreferrer&quot;&gt;http://what-if.xkcd.com/42/&lt;/a&gt; is inaccurate, and will&#xA;ping the author to let him know.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2016-01-02T06:02:09.247" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13054" PostTypeId="2" ParentId="13052" CreationDate="2016-01-02T06:43:06.730" Score="1" Body="&lt;p&gt;The orbits of the planets are coplanar (in the same plane) because supposedly during the Solar System's formation, the planets formed out of a disk of dust (ha, ha)  which surrounded the Sun. Because it was a &lt;strong&gt;&lt;em&gt;disk&lt;/em&gt;&lt;/strong&gt;, all in one plane, all of the planets formed in that one lousy plane as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Single rings and disks are common in astronomy. Jupiter's moons are coplanar too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The common explanation is that orbits are unstable and they all eventually tend to get into the same plane and stay there. Wouldn't it be fun to have a planet orbiting the Sun at a right angle to the other planet's orbits? You'd never know when the damn thing's going to hit what. &lt;/p&gt;&#xA;" OwnerUserId="9510" LastActivityDate="2016-01-02T06:43:06.730" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13055" PostTypeId="1" AcceptedAnswerId="13066" CreationDate="2016-01-02T10:37:49.730" Score="1" ViewCount="307" Body="&lt;p&gt;Yes, I know that most galaxies have a red shift and that means they are moving away from us. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that the farthest galaxies are 13.8 billion light years away. That means that the info is 13.8 billion years old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What has been puzzling me for years is that the red shift observation shows that the galaxies were receding 13.8 billion years ago. We don't have direct data on what they are doing now. It is natural that the universe was expanding with a tremendous speed at that time because it was near to the big bang, but how do we know that it still is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the furthest galaxies are moving with higher speed than the nearer, that means to me that their speed is reducing with the time.&lt;/p&gt;&#xA;" OwnerUserId="1848" LastEditorUserId="1848" LastEditDate="2016-01-02T12:45:46.413" LastActivityDate="2016-01-04T10:30:19.320" Title="How do we know that the Universe is still expanding now?" Tags="&lt;universe&gt;&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13057" PostTypeId="2" ParentId="1273" CreationDate="2016-01-02T15:36:16.260" Score="0" Body="&lt;p&gt;Start off with binoculars like &lt;a href=&quot;http://www.celestron.com/browse-shop/sport-optics/binoculars/upclose-g2-10x50-porro-binocular&quot; rel=&quot;nofollow&quot;&gt;Celestron Upclose 10x50&lt;/a&gt; available for around ₹2500 on &lt;a href=&quot;http://spacearcade.in/&quot; rel=&quot;nofollow&quot;&gt;spacearcade.in&lt;/a&gt;. You will also benefit from a tripod adaptor and tripod.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then study astronomy on line or from a book .... In my opinion don't buy a scope ,instead make one yourself.&lt;/p&gt;&#xA;" OwnerUserId="10279" LastEditorUserId="4085" LastEditDate="2016-01-02T15:56:06.737" LastActivityDate="2016-01-02T15:56:06.737" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13059" PostTypeId="2" ParentId="13051" CreationDate="2016-01-03T02:12:57.277" Score="10" Body="&lt;p&gt;To the naked eye, the answer is almost certainly no because of the enormously slow movement of stars across the sky and because 5,000 stars may be a lot but only a tiny percentage of the sky is covered by visible stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To Hubble, which can see perhaps tens of millions of stars, maybe more, the link &lt;a href=&quot;https://en.wikipedia.org/wiki/Proper_motion#History&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;, has a picture of two stars that are approaching crossing each other's path from our point of view.  With a big enough telescope it probably happens from time to time, though I wouldn't want to try to calculate how often, but to the naked eye, I'm comfortable saying no, in fact, it was often assumed that stars didn't move and were fixed in the sky (contrary to what Macrobius said).  That was the popular point of view prior to Halley's observation.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was also Tycho Brahe's &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/SN_1572&quot; rel=&quot;noreferrer&quot;&gt;De Nova Stella&lt;/a&gt;&quot; or &quot;new star&quot; which we now know to be a super-nova, and that was quite the surprise at the time.   Nobody thought a new star could appear because they thought the stars were fixed and permanent, but that appearance wasn't by the method you suggest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider how small stars are from our point of view.    Alpha Centauri A, the larger one, it's about 1.7 million KM across and it's about 4.3 light years away, or, 41 trillion KM.   It's diameter is 23 million times smaller than it's distance from us.  That's the equivalent of looking at a golf ball from nearly 200 miles away.    Now if you scatter 5000 golf balls each 200 miles away across the sky and you let them move around very very slowly, how often do you think one golf ball passes in-front of another?   Not very often.   Granted, that's not quite right as the atmosphere spreads stars out a bit so each golf ball is smudged to maybe the size of a basketball, but they almost never pass infront of one another, at least, not if we only take into account 5,000 visible stars.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, binary stars, it happens more often if they are lined up right, then they can pass infront of each other and this has certainly been observed by telescope but not to the Naked Eye, we can't visibly tell that Alpha Centauri is 2 stars (3 with the more distant Proxima but that can't be seen by the eye).  They are on average about a billion miles apart but that can't be seen by the naked eye.   It was observed by telescope in &lt;a href=&quot;https://en.wikipedia.org/wiki/Alpha_Centauri#Observational_history&quot; rel=&quot;noreferrer&quot;&gt;1689&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There simply aren't enough visible stars (and taking HDE's point that most of the 5,000 visible stars weren't cataloged until recently), there's essentially zero chance that it was ever observed that a star appeared &quot;new&quot; by passing from behind another star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Hubble, it can happen, but not to human sight.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-01-05T03:45:11.047" LastActivityDate="2016-01-05T03:45:11.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13060" PostTypeId="1" CreationDate="2016-01-03T19:08:20.443" Score="4" ViewCount="142" Body="&lt;p&gt;Big bang is not an explosion, but an expansion of space and time. Universe had almost infinite density... wait a moment. If it had ALMOST infinite density, it had a certain volume, and thus, space already existed. &#xA;Am I commiting a mistake?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/669/what-is-in-the-center-of-the-universe&quot;&gt;I had this question after reading this post.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8006" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-01-11T08:14:34.473" Title="Center of the universe" Tags="&lt;big-bang-theory&gt;&lt;space-time&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13061" PostTypeId="1" AcceptedAnswerId="13078" CreationDate="2016-01-03T21:18:27.797" Score="1" ViewCount="991" Body="&lt;p&gt;Looking into the history of how the speed of light came to be determined, to what we know it to be today, James Bradley is often mentioned. He was credited with the discovery of the aberration of light and having used the aberration of light to calculate the speed of light. There are many online sources that confirm this, though the year of discovery and calculated speed of light differs here and there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read through Bradley's &quot;&lt;em&gt;A Letter from the Reverend Mr. James Bradley Savilian Professor of Astronomy at Oxford, and F.R.S. to Dr.Edmond Halley Astronom. Reg. &amp;amp;c. Giving an Account of a New Discovered Motion of the Fix'd Stars&lt;/em&gt;&quot; published in &quot;&lt;em&gt;Philisophical Transaction vol 35&lt;/em&gt;&quot; which explains his findings of the aberration of light but makes no mention of calculating the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did James Bradley calculate the speed of light and what publication of his shows these calculation?&lt;/p&gt;&#xA;" OwnerUserId="10288" LastActivityDate="2016-01-04T22:48:45.447" Title="What was James Bradley's calculation to calculate the speed of light?" Tags="&lt;light&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13062" PostTypeId="2" ParentId="13060" CreationDate="2016-01-04T01:07:02.983" Score="-1" Body="&lt;p&gt;You're right; space did exist. In fact, the universe was infinite in size at the moment of the Big Bang, and it still is. This is because a common misconception is wrong: &lt;a href=&quot;https://physics.stackexchange.com/questions/136860/did-the-big-bang-happen-at-a-point&quot;&gt;The Big Bang did not happen at a point&lt;/a&gt;. Thinking of it like an explosion isn't correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You asked in a comment about the multiverse. This is speculative, to say the least. There are some theories involving objects called branes that posit that we are on the surface of some object floating through a higher dimensional &quot;bulk&quot;. However, these theories of so-called large extra dimensions have yet to be supported by experimental evidence.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2016-01-04T01:07:02.983" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13064" PostTypeId="2" ParentId="12940" CreationDate="2016-01-04T03:58:24.913" Score="1" Body="&lt;p&gt;@DavidHammen is correct in noticing that &#xA;&lt;a href=&quot;http://aa.usno.navy.mil/data/docs/RS_OneYear.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://aa.usno.navy.mil/data/docs/RS_OneYear.php&lt;/a&gt; rounds to the nearest &#xA;minute (and it turns out they do this inconsistently), so I wrote &#xA;&lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-solve-astro-12940.c&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-solve-astro-12940.c&lt;/a&gt; &#xA;to solve this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it was pretty much a waste of time, since it just verifies &#xA;@JeffY's comment that the 21st and 22nd are equally long at longitude &#xA;72 degrees and 15 minutes west. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equal day lines actually runs northwest/southeast, starting at &#xA;about &lt;code&gt;22N, 71.25W&lt;/code&gt; and ending at &lt;code&gt;67N, 73.05W&lt;/code&gt;, looking something &#xA;like this on an equiangular map (longitude scale greatly exaggerated): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7V7Ox.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7V7Ox.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;West of this line, the 21st is the shortest day; east of this line, &#xA;the 22nd is the shortest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;South of 22N, we cross the Tropic of Cancer, and neither day would be &#xA;the shortest. North of 67N, we enter the land of 24-hour sunlessness, &#xA;where there are multiple days with zero sunlight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADDENDUM: Per @JeffY's observation, the shortest/longest day will occur on&#xA;the same day worldwide (roughly speaking) when the solstice occurs at&#xA;noon GMT. Here are the times this century when the solstice&#xA;occurs within 1 hour of noon GMT:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2004-12-21 12:41:32 &#xA;2006-06-21 12:25:54 &#xA;2008-12-21 12:03:48 &#xA;2010-06-21 11:28:21 &#xA;2012-12-21 11:11:34 &#xA;2035-06-21 12:33:08 &#xA;2039-06-21 11:57:22 &#xA;2041-12-21 12:18:22 &#xA;2045-12-21 11:35:09 &#xA;2064-06-20 12:46:15 &#xA;2068-06-20 11:54:24 &#xA;2072-06-20 11:14:32 &#xA;2074-12-21 12:36:04 &#xA;2078-12-21 11:58:57 &#xA;2082-12-21 11:05:49 &#xA;2097-06-20 12:15:11 &#xA;&lt;/code&gt;&lt;/pre&gt; &#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2016-01-07T21:26:19.473" LastActivityDate="2016-01-07T21:26:19.473" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13065" PostTypeId="1" CreationDate="2016-01-04T07:29:23.917" Score="2" ViewCount="82" Body="&lt;p&gt;The diffraction limit deals with the ability to determine if two things are separate.  I am interested in the ability to find the the centroid of a single object.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a star with no near neighbors would it be possible to determine the centroid of that star at a resolution higher than the diffraction limit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The star is essentially a point source and I wonder if looking at the edges of the airy disk would allow a centroid to be determined if I had a magical camera with infinite pixels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that I should be able to see the disk and then calculate a centroid that is smaller than the airy disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help you can provide =)&lt;/p&gt;&#xA;" OwnerUserId="10291" LastEditorUserId="616" LastEditDate="2021-02-22T21:17:25.987" LastActivityDate="2021-02-22T21:17:25.987" Title="Can angular information be known more precisely than the diffraction limit?" Tags="&lt;telescope&gt;&lt;optics&gt;&lt;astrometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13066" PostTypeId="2" ParentId="13055" CreationDate="2016-01-04T10:30:19.320" Score="1" Body="&lt;p&gt;You're right that in principle we cannot know if the distant galaxies are still receding from us. When, however, we think that they indeed are, it hinges on the &lt;em&gt;cosmological principle&lt;/em&gt;, viz. that the Universe is homogeneous and isotropic, i.e. &quot;looks the same everywhere and in all directions&quot;. This is a philosophically motivated statement, which is continuously tested observationally. So far, there has been no significant experimental evidence that this is not correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we observe the velocities of galaxies throughout the Universe, we see a smooth velocity as a function of time, or distance, with some intrinsic scatter which is due to the gaalxies' &lt;em&gt;peculiar motion&lt;/em&gt;, i.e. their velocities &lt;em&gt;through&lt;/em&gt; space, which is usually of the order of 100–1000 km/s. That is, distant galaxies move away from us &lt;em&gt;fast&lt;/em&gt;, and nearby galaxies move away from us &lt;em&gt;slowly&lt;/em&gt;. As you say, the expansion rate decreased with time, but only until a certain point, after which it started increasing again (due to dark energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that we see that nearby galaxies are &lt;em&gt;still&lt;/em&gt; receding, together with the cosmological principle, implies that the distant galaxies should &lt;em&gt;also&lt;/em&gt; still be receding. We know of no reason that they shouldn't. Of course this is no proof, but in physics, there &lt;em&gt;are&lt;/em&gt; no proofs, only verifications and falsifications. To postulate that the distant galaxies no longer recede, and to be taken seriously, you would need to come up with an experiment that could verify this hypothesis, preferably along with a mechanism that would be able to cause this phenomenon (this looks a bit condescending when I type it; this is not my intention).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expansion rate is currently — and hence locally — around $70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$. Due to the aforementioned velocity scatter, we cannot probe expansion below ~1 Mpc, but at this scale, space doesn't really expand anyway due to the mutual gravitational attraction of the Local Group. This means that in principle, the expansion of the Universe &lt;em&gt;could&lt;/em&gt; have come to a halt within the last few million years, and still be in accord with the cosmological priciple. However, this would mean that the evolution of expansion velocity as a function of time would have a sharp kink. Again, this is not impossible, but we know of no physical mechanism which would be able to cause this.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2016-01-04T10:30:19.320" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13067" PostTypeId="1" AcceptedAnswerId="13068" CreationDate="2016-01-04T13:18:47.223" Score="1" ViewCount="1859" Body="&lt;p&gt;I'm developing my own Planetarium software using C++ following the book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B010WFOATY&quot; rel=&quot;nofollow&quot;&gt;Practical Astronomy with your Calculator or Spreadsheet 4th Edition&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if I my formulas are correct or not, but when I check my results (Azimuth and Altitude) with Stellarium, they are different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also checked my values with this &lt;a href=&quot;http://www.stargazing.net/mas/al_az.htm&quot; rel=&quot;nofollow&quot;&gt;page&lt;/a&gt;, but they are different. By the way, the azimuth and altitude given in that page are also different from the values given in Stellarium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know where can I find the formulas to covert celestial coordinate to horizontal coordinates?&lt;/p&gt;&#xA;" OwnerUserId="7566" LastEditorUserId="7566" LastEditDate="2016-01-04T15:07:19.510" LastActivityDate="2016-01-04T15:07:19.510" Title="Conversion from Equatorial Coordinate to Horizon Coordinates" Tags="&lt;star&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13068" PostTypeId="2" ParentId="13067" CreationDate="2016-01-04T14:23:41.270" Score="1" Body="&lt;p&gt;It depends from which celestial coordinates you are converting. I suppose you are working in equatorial coordinates. Then, you can convert as following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan A = {\sin h \over \cos h \sin\phi_o - \tan\delta \cos\phi_o} $$ $$ \begin{cases}&#xA; \cos a \sin A = \cos\delta \sin h \\&#xA; \cos a \cos A =  \cos\delta \cos h \sin\phi_o - \sin\delta \cos\phi_o&#xA;\end{cases}$$ $$\sin a = \sin\phi_o \sin\delta + \cos\phi_o \cos\delta \cos h$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $A$ is azimuth, $a$ altitude, $\alpha$ right ascension, $\delta$ declination, $h$ hour angle and $\phi_0$ the observer's latitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Celestial_coordinate_system#Equatorial_.E2.86.90.E2.86.92_horizontal&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8011" LastActivityDate="2016-01-04T14:23:41.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13069" PostTypeId="1" CreationDate="2016-01-04T14:38:56.330" Score="3" ViewCount="124" Body="&lt;p&gt;Recently I had to implement Laplace method and apply it to 3 observations of Mars (10 days in between two observations). The results were pretty good, with discrepancies with real data well below 5% in most of the orbital elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I supposed that Laplace method was very accurate, so I tried to apply it to an asteroid (CASLEO, 5387). The 3 observations were also 10-days spaced but now the results widely diverge from the real orbital elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expected it to be less accurate than in the Mars case since the observations covered less fraction of the orbit, but some of the elements were quite far from expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: What is the accuracy of the Laplace Method? Is it really that sensitive to the distance?&lt;/p&gt;&#xA;" OwnerUserId="8011" LastActivityDate="2017-02-12T23:29:52.490" Title="Accuracy of Laplace Method for determining orbital elements" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;asteroids&gt;&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13070" PostTypeId="2" ParentId="12940" CreationDate="2016-01-04T17:14:19.060" Score="0" Body="&lt;p&gt;EDIT: Ignore below, see comments for correction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@DavidHammen, I ran your script with start and stop year both set to 2096, and got this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;('December solstice in', 2096, 'is at', 2096/12/20 20:45:17, 'UTC')&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Shortest day of the year, by longitude:&#xA;Date  Longitude&#xA;('12/19', -180)&#xA;('12/19', -170)&#xA;('12/19', -160)&#xA;('12/19', -150)&#xA;('12/19', -140)&#xA;('12/19', -130)&#xA;('12/19', -120)&#xA;('12/19', -110)&#xA;('12/19', -100)&#xA;('12/19', -90)&#xA;('12/19', -80)&#xA;('12/19', -70)&#xA;('12/19', -60)&#xA;('12/19', -50)&#xA;('12/19', -40)&#xA;('12/19', -30)&#xA;('12/19', -20)&#xA;('12/19', -10)&#xA;('12/20', 0)&#xA;('12/20', 10)&#xA;('12/20', 20)&#xA;('12/20', 30)&#xA;('12/20', 40)&#xA;('12/20', 50)&#xA;('12/21', 60)&#xA;('12/21', 70)&#xA;('12/21', 80)&#xA;('12/21', 90)&#xA;('12/21', 100)&#xA;('12/21', 110)&#xA;('12/21', 120)&#xA;('12/21', 130)&#xA;('12/21', 140)&#xA;('12/21', 150)&#xA;('12/21', 160)&#xA;('12/21', 170)&#xA;('12/20', 180)&#xA;()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I take it you're getting different results?&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2016-01-05T16:23:51.403" LastActivityDate="2016-01-05T16:23:51.403" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13072" PostTypeId="1" AcceptedAnswerId="13077" CreationDate="2016-01-04T18:08:35.893" Score="2" ViewCount="710" Body="&lt;p&gt;Does exist something like a distribution of number of galaxies per redshift interval? I'm studying galaxy formation and evolution, and I'm curious to know how many todays' galaxies (with redshift &amp;lt;0.01 I would say) we see.&lt;/p&gt;&#xA;" OwnerUserId="7919" LastEditorUserId="9301" LastEditDate="2016-06-29T04:36:11.350" LastActivityDate="2017-09-06T07:23:16.597" Title="Number of galaxies per redshift" Tags="&lt;galaxy&gt;&lt;redshift&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13073" PostTypeId="1" CreationDate="2016-01-04T20:23:10.350" Score="3" ViewCount="2763" Body="&lt;p&gt;I am confused with nucleosynthesis inside supernovae. I have read that the heavier elements are made through fusion of lighter element's namely hydrogen and helium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the star &quot;store&quot; all the elements they fuse inside themselves or are they radiated out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example, A star fused hydrogen into helium and 2 helium atoms into carbon. Would it keep fusing until it hits the limit and store it inside only to be released as an explosion.&lt;/p&gt;&#xA;" OwnerUserId="10297" LastActivityDate="2016-01-11T10:44:20.830" Title="Fusion of elements inside heavy stars" Tags="&lt;supernova&gt;&lt;star-formation&gt;&lt;nebula&gt;" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13074" PostTypeId="2" ParentId="13073" CreationDate="2016-01-04T21:08:37.937" Score="2" Body="&lt;p&gt;You are correct to say that all the heavier elements in the universe were formed in stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars like the sun fuse hydrogen into helium. When they get older they can fuse the helium into carbon (it actually takes 3 helium to make one carbon). Larger stars can fuse carbon into oxygen, and neon and elements in the first half of the periodic table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the star runs out of fuel, the outer layers are gently ejected in what is called a planetary nebula (though it has directly to do with actual planets). The ejected gas is enriched with the heavier elements that the star has fused. The heavier elements in the planetary nebula mix with the hydrogen and helium gas in outer space and can later go on to be found in new stars. Most of the carbon and oxygen and nitrogen on earth was formed by this process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very large stars will fuse all the way up to iron, and then collapse in a supernova. This releases a very large amount of energy, some of which is used to form elements heavier than iron. All the heavier elements (copper gold, uranium for example) are formed in supernovae. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The supernovae explosion distributes these elements back into space, mixed in with the debris from the star, as time goes on, and stars are born and die, the gas in space gets enriched with more and more heavier elements&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the heavier elements are formed in the cores of stars, and are stored there until the star dies, when some of the star's atoms are released into space. The heavier elements are not &quot;radiated out&quot; of the star, until the star dies. Some stars die in a supernova explosian, but most have a more peaceful death in a planetary nebula.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-01-04T21:08:37.937" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13075" PostTypeId="2" ParentId="1918" CreationDate="2016-01-04T21:32:35.787" Score="0" Body="&lt;p&gt;Heres my take on whats going on. As per precession, in 13,000 years from now, june should be winter for us in the northern hemisphere. But, being on spaceship earth, our season is moving along with its month. So in june which is currently aperihelion will actually still be dec 20 or 21 in 13,000 years but at aperihelion. Currently the winter solstice falls around perihelion. Actually, and I forgot where i read it from, but the last time the winter solstice and perihelion fell on the same day was back in the year 1246. But due to precession, perihelion is now occurring around Jan 3. So using the perihelion point as a reference,and according to my calculations, perihelion is going to go through every day of the month in approx. 2150 years time. The perihelion will then jump into Feb...&lt;/p&gt;&#xA;" OwnerUserId="10300" LastActivityDate="2016-01-04T21:32:35.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13077" PostTypeId="2" ParentId="13072" CreationDate="2016-01-04T22:22:05.460" Score="5" Body="&lt;strong&gt;Yes, for a given galaxy type…&lt;/strong&gt;&#xA;&lt;p&gt;The term &amp;quot;a distribution of number of galaxies per redshift interval&amp;quot; &lt;em&gt;does&lt;/em&gt; make sense, but usally we restrict ourselves to a certain type of galaxies, e.g. &lt;a href=&quot;https://en.wikipedia.org/wiki/Lyman-alpha_emitter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lyman &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; emitters&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Lyman-break_galaxy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lyman break galaxies&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Submillimetre_astronomy&quot; rel=&quot;nofollow noreferrer&quot;&gt;sub-mm galaxies&lt;/a&gt;, etc. The reason is that galaxies are detected in a number of different ways, and no single selection technique captures all types of galaxies.&lt;/p&gt;&#xA;&lt;strong&gt;…and &lt;em&gt;comoving&lt;/em&gt; volume…&lt;/strong&gt;&#xA;&lt;p&gt;Because the Universe is expanding with time, the &amp;quot;true&amp;quot; (physical) number density of galaxies — and everything else (except dark energy) — decreases with time. To be able to compare densities at different times, it is useful to measure them in &lt;a href=&quot;https://en.wikipedia.org/wiki/Comoving_distance&quot; rel=&quot;nofollow noreferrer&quot;&gt;comoving coordinates&lt;/a&gt;, which is the coordinates that expand along with the Universe. In these coordinates, densities only change if some physical process happen. For instance, the  physical &lt;em&gt;energy&lt;/em&gt; density of photons decreases both because the volume of the Universe increases &lt;em&gt;and&lt;/em&gt; because the photons redshift. In comoving coordinates it &lt;em&gt;also&lt;/em&gt; decreases, but only due to the redshifting since the volume change is factored out.&lt;/p&gt;&#xA;&lt;strong&gt;…and luminosity.&lt;/strong&gt;&#xA;&lt;p&gt;Moreover, since galaxies have no well-defined lower size/mass/luminosity threshold, we have to define a threshold above which we count them.&#xA;For this reason, rather than talking about &amp;quot;the number density of galaxies per redshift&amp;quot;, we usually use the very useful observable &lt;a href=&quot;https://en.wikipedia.org/wiki/Luminosity_function_%28astronomy%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;luminosity function&lt;/a&gt; (LF) which measures the number of (a certain type of) galaxies &lt;em&gt;per luminosity bin&lt;/em&gt;, and per comoving volume, at a given redshift.&lt;/p&gt;&#xA;&lt;p&gt;LFs have been probed out to very high redshifts, at the times when galaxies were just beginning to form at &lt;span class=&quot;math-container&quot;&gt;$\sim$&lt;/span&gt;half a gigayear after Big Bang (e.g. &lt;a href=&quot;http://arxiv.org/abs/1212.4819&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schenker et al. 2013&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;p&gt;You end your question by asking about the number of galaxies at &lt;span class=&quot;math-container&quot;&gt;$z\le0.01$&lt;/span&gt;, which would be considered very low redshift in the context of galaxies. You can find the answer by considering all kinds of selection criteria, checking for overlaps between surveys (most local galaxies will be detected by multiple techniques), and integrating over volume the LFs down to your chosen threshold (e.g. Small Magellanic Cloud-sized). Doable, but not trivial. An alternative is to integrate numerically calculated dark matter halo mass functions, using e.g. &lt;a href=&quot;http://hmf.icrar.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this online tool&lt;/a&gt;, and assuming the every halo hosts a galaxy. I get a number density of &lt;span class=&quot;math-container&quot;&gt;$n\sim0.4$&lt;/span&gt; halos per &lt;span class=&quot;math-container&quot;&gt;$h^{-3}\,\mathrm{Mpc}^{3}$&lt;/span&gt;, or &lt;span class=&quot;math-container&quot;&gt;$1.2\,\mathrm{Mpc}^{-1}$&lt;/span&gt;. The comoving volume out to &lt;span class=&quot;math-container&quot;&gt;$z=0.01$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$V=3.3\times10^5$&lt;/span&gt; Mpc, so the total number of (super-SMC-sized) galaxies is &lt;span class=&quot;math-container&quot;&gt;$N = n V \simeq 3.8\times10^5$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-09-06T07:23:16.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13078" PostTypeId="2" ParentId="13061" CreationDate="2016-01-04T22:48:45.447" Score="0" Body="&lt;p&gt;Despite many online sources crediting James Bradley with having calculated the speed of light to a 295.000 km/sec (britannica) or 301.000 km/sec (wikipedia), a small nuance ought to be added. Like &lt;a href=&quot;https://en.wikipedia.org/wiki/R%C3%B8mer%27s_determination_of_the_speed_of_light#Did_R.C3.B8mer_measure_the_speed_of_light.3F&quot; rel=&quot;nofollow&quot;&gt;Ole Romer&lt;/a&gt; before him in 1676, James Bradley calculated the speed of light relative to something else but both never provided a value in Earth-based units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He wrote,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;This therefore being 20’’,2, AC will be to AB that is the Velocity of Light to the Velocity of the Eye (which in&#xA;  this Case may be supposed the same as the Velocity of the Earth's&#xA;  annual Motion in its Orbit) as 10210 to One from whence it would follow&#xA;  that Light moves or is propagated as far as from the Sun to the Earth&#xA;  in 8’ 12’’.&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(&lt;a href=&quot;https://books.google.nl/books?id=5V5EAAAAcAAJ&amp;amp;pg=PA159#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The calculation to get to the ratio of 10210 is,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;size of a radian (206265’’) / angle of aberration (20’’,2)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This ratio also holds for the speed of light to the speed of the earth. Not sure though exactly how britannica and wikipedia arrive at their respective values.&lt;/p&gt;&#xA;" OwnerUserId="10288" LastActivityDate="2016-01-04T22:48:45.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13079" PostTypeId="2" ParentId="13073" CreationDate="2016-01-04T23:14:26.320" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the star &quot;store&quot; all the elements they fuse inside themselves or are they radiated out?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It depends on the star.  While the heaviest elements can only be produced (and released) by massive stars that undergo supernova, many other elements are created within relatively smaller stars that become &lt;a href=&quot;https://en.wikipedia.org/wiki/Asymptotic_giant_branch&quot; rel=&quot;nofollow&quot;&gt;asymptotic giant branch stars&lt;/a&gt;.  These can undergo cycles of convection that bring the heavy elements in the core to the surface, and then release them in the stellar wind.  &lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2016-01-04T23:14:26.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13081" PostTypeId="1" CreationDate="2016-01-05T00:27:33.310" Score="6" ViewCount="2729" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Is it insanely hot? Cause of so much gravity?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is it close to absolute zero? Cause the matter is so closely packed, there is hardly any space for particles to move?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is it room temperature? Cause Cooper didn't die in &lt;em&gt;Interstellar&lt;/em&gt; when he went into a Black Hole?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10298" LastEditorUserId="9186" LastEditDate="2016-01-05T15:03:41.463" LastActivityDate="2018-05-31T19:30:50.650" Title="What is the temperature inside a Black Hole?" Tags="&lt;black-hole&gt;&lt;temperature&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13082" PostTypeId="2" ParentId="13081" CreationDate="2016-01-05T04:12:10.440" Score="0" Body="&lt;p&gt;Temperature of a black hole is determined by 'black body radiation temperature' of radiation which comes from it (if something is hot enough to give off bright blue light,it is hotter than something that is merely a dim red hot.) For black holes the mass of Sun, radiation emitted from it is so weak and so cool that temperature is only one-millionth of a degree above absolute zero. some black holes are thought to weigh a billion times as much as the Sun and they would be a billion times colder, far colder than what scientists have achieved on Earth. &#xA;Even though these things are very cold, they can be surrounded by very hot material. As they pull gas and stars down into their gravity wells, material rubs against itself at a good fraction of speed of light. this heats it up to hundreds of millions of degrees. Radiation from this hot, in-falling material is what high-energy astronomers study.&lt;br&gt;&#xA;A black hole behaves as though its horizon has a temperature and that temperature is inversely proportional to the hole's mass: T = (6 x 10^-8)M. Here M is in units of solar mass (2 x 1033 grams). Temperature is in degrees Kelvin. &#xA;This means that a hole recently formed by gravitational collapse of a star (which has to have a mass larger than about 2 suns) has a temperature less than 3 x 10^-8 centigrade above absolute zero which is very cold.&#xA;Mass with a finite temperature radiated energy. Anything which radiates energy is also losing mass.As black hole loses mass, emission of energy from black hole increases and its temperature increases and thus rate of mass loss increases.As mass of black hole gets small,one has unstable 'runaway effect'. Black hole gets hotter and hotter which causes M to decrease rapidly.When the hole is reduced to a fraction of the size of an atomic nucleus, it will be trillions of degrees. Hole will burn up and disappear. Lifetime of a black hole is greater than age of the universe.&lt;/p&gt;&#xA;" OwnerUserId="9639" LastActivityDate="2016-01-05T04:12:10.440" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13083" PostTypeId="1" AcceptedAnswerId="14655" CreationDate="2016-01-05T06:20:08.310" Score="6" ViewCount="619" Body="&lt;p&gt;I've been reading an article today on how the European Space Agency (ESA) has an interest in building a space village on the moon which will be 3D printed and assembled by robots. Eventually people will start to go there for long periods of time (presumably).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I always think to myself when hearing such news or watching a sci-fi movie, isn't there a real chance that it will get hit by a meteorite?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resource:&#xA;&lt;a href=&quot;http://www.iflscience.com/space/esa-build-moon-village-2030&quot; rel=&quot;nofollow&quot;&gt;http://www.iflscience.com/space/esa-build-moon-village-2030&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a video of their plan to build a moon village as they call it:&#xA;&lt;a href=&quot;http://www.iflscience.com/space/esa-reveal-their-plans-build-moon-village-20-years-time&quot; rel=&quot;nofollow&quot;&gt;http://www.iflscience.com/space/esa-reveal-their-plans-build-moon-village-20-years-time&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE #2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something interesting to see :) &lt;a href=&quot;http://www.iflscience.com/space/space-debris-has-chipped-one-isss-windows&quot; rel=&quot;nofollow&quot;&gt;http://www.iflscience.com/space/space-debris-has-chipped-one-isss-windows&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10304" LastEditorUserId="37191" LastEditDate="2021-02-19T10:18:01.633" LastActivityDate="2021-02-19T10:18:01.633" Title="Is there any real chance of getting hit with a meteorite if we build a base on the moon?" Tags="&lt;the-moon&gt;&lt;space&gt;&lt;meteor&gt;&lt;esa&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13084" PostTypeId="1" AcceptedAnswerId="14496" CreationDate="2016-01-05T08:09:51.107" Score="2" ViewCount="276" Body="&lt;p&gt;Wikipedia describes the five following &lt;a href=&quot;https://en.wikipedia.org/wiki/Celestial_coordinate_system#Coordinate_systems&quot; rel=&quot;nofollow&quot;&gt;celestial coordinate systems&lt;/a&gt; each with rectangular/spherical variants:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;horizontal&lt;/li&gt;&#xA;&lt;li&gt;equatorial&lt;/li&gt;&#xA;&lt;li&gt;ecliptic&lt;/li&gt;&#xA;&lt;li&gt;galactic&lt;/li&gt;&#xA;&lt;li&gt;supergalactic&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What are the practical preferred uses of each, in relation to the others. If you were navigating within the solar system, which system would you use, or has been used by existing craft?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example wikipedia describes the ecliptic system as &quot;still useful for computing the apparent motions of the [celestial bodies]&quot;, which seems pretty bloody obsolete. Even though it seems to me as the most intuitively useful system.&lt;/p&gt;&#xA;" OwnerUserId="10307" LastEditorUserId="13359" LastEditDate="2016-08-12T16:41:42.173" LastActivityDate="2016-08-12T16:41:42.173" Title="What are the key coordinate systems used in Astronomy, and where are they used?" Tags="&lt;coordinate&gt;&lt;navigation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13085" PostTypeId="2" ParentId="13073" CreationDate="2016-01-05T15:50:52.170" Score="0" Body="&lt;p&gt;In addition to nucleosynthesis as the star ages, the action of core collapse into a supernova produces heavier elements too - &lt;a href=&quot;https://en.wikipedia.org/wiki/Supernova_nucleosynthesis&quot; rel=&quot;nofollow&quot;&gt;supernova nucleosynthesis&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the intense energy of the collapse causes fusion of some of the material surrounding the core.  This material will escape, not be trapped in the core itself.&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2016-01-05T15:50:52.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13086" PostTypeId="1" AcceptedAnswerId="13087" CreationDate="2016-01-05T20:28:00.127" Score="4" ViewCount="431" Body="&lt;p&gt;Say Barry, could you abuse the answer-your-own-question feature of&#xA;this site, and answer the following question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted in &lt;a href=&quot;https://astronomy.stackexchange.com/questions/11997/&quot;&gt;How many constellations in the Zodiac?&lt;/a&gt; the&#xA;ecliptic itself touches 13 constellations, but the Encylopedia&#xA;Brittanica defines the Zodiac as &quot;a belt around the heavens extending&#xA;9 [degrees] on either side of the ecliptic&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.britannica.com/topic/zodiac&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.britannica.com/topic/zodiac&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Old Farmer's Almanac for 2014 (and previous years) notes (page&#xA;114) that Moon occasionally crosses into 5 &quot;non-Zodiac&quot; constellations&#xA;at times:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EuI9P.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EuI9P.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, exactly which constellations are &quot;in the Zodiac&quot; if the Zodiac&#xA;extends 9 degrees on either side of the ecliptic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And why 9 degrees? I know Venus can be as much as 8.25 degrees from&#xA;the ecliptic,&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-01-05T20:28:51.663" Title="What constellations touch the 9-degree wide Zodiac?" Tags="&lt;constellations&gt;&lt;ecliptic&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13087" PostTypeId="2" ParentId="13086" CreationDate="2016-01-05T20:28:51.663" Score="2" Body="&lt;p&gt;While I can't answer the 9 degrees part, &#xA;&lt;a href=&quot;http://www.space.com/5417-ecliptic-zodiac-work.html&quot; rel=&quot;nofollow&quot;&gt;http://www.space.com/5417-ecliptic-zodiac-work.html&lt;/a&gt; notes, the band of &#xA;the zodiac actually passes through 22 constellations: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt; &#xA;In fact, as pointed out by the well-known astronomical calculator, &#xA;Jean Meeus, along with Ophiuchus, there are nine other constellations &#xA;that occasionally can be visited by the Moon and planets: Auriga, the &#xA;Charioteer; Cetus, the Whale; Corvus, the Crow; Crater, the Cup; &#xA;Hydra, the Water Snake; Orion, the Hunter; Pegasus, the Flying Horse; &#xA;Scutum, the Shield; and Sextans, the Sextant.&#xA;&#xA;&lt;p&gt;&lt;p&gt;&#xA;&#xA;So in truth, there really aren't twelve zodiacal constellations, but &#xA;twenty-two! &#xA;&#xA;&lt;/blockquote&gt; &#xA;" OwnerDisplayName="user21" LastActivityDate="2016-01-05T20:28:51.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13090" PostTypeId="2" ParentId="11271" CreationDate="2016-01-05T22:48:24.997" Score="1" Body="&lt;p&gt;You probably got this far yourself, but here is my absolutely wild &#xA;speculative guess on why this could be 24 May 1587 on the Gregorian &#xA;calendar: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;It seems fairly clear that this refers to the 24th of some month &#xA;in 1587, but it's not as clear which month this is, and whether this &#xA;is in the Gregorian or Julian calendar. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The moon symbol followed by the symbol for Taurus may indicate &#xA;&quot;the month where the new moon was in Taurus&quot;. This is slightly &#xA;ambigious, since the new moon is was Taurus on both June 6th and May &#xA;7th, but I believe May 7th slightly is more likely for the reasons &#xA;below. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Wikipedia notes that &quot;cerasus&quot; is a a dated synonym of the genus &#xA;Prunus (&lt;a href=&quot;https://en.wikipedia.org/wiki/Cerasus&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Cerasus&lt;/a&gt;), which includes &#xA;cherries. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Historically, there is no &quot;Cherry Moon&quot; (although there are songs, &#xA;movies, and possibly more that mention it), but cherry trees are &#xA;known for their flowers, so this may be a reference to the Full &#xA;Flower Moon of May: &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://farmersalmanac.com/full-moon-names/&quot; rel=&quot;nofollow&quot;&gt;http://farmersalmanac.com/full-moon-names/&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The word &quot;gexia&quot; translates to &quot;Your Mightiness&quot; in Chinese: &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://translate.google.com/#auto/en/gexia&quot; rel=&quot;nofollow&quot;&gt;https://translate.google.com/#auto/en/gexia&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I'm pretty sure this is red herring, since the paper doesn't &#xA;appear to be in Chinese, and I'm not sure Chinese transliteration even &#xA;existed at the time. However, it would fit in nicely. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;So this could be saying &quot;the full cherry (flower) moon occurs on &#xA;24 May 1587&quot;. Unfortunately, the actual full moon occurs late on the &#xA;22nd GMT, so the 24th is a bit of a stretch even allowing for the &#xA;16th century equivalent of &quot;time zones&quot; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Gregorian calendar reform occurred in 1582. Many countries &#xA;didn't adopt it until much later, but the text appears to be in a &#xA;Latin-based language (ie, a Romance language), and countries like &#xA;this were quick to switch to the Gregorian calendar: &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Adoption_of_the_Gregorian_calendar#Adoption_in_Catholic_countries&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Adoption_of_the_Gregorian_calendar#Adoption_in_Catholic_countries&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If this is 24 May 1587 in the Julian calendar, it would be 10 days &#xA;later, 3 Jun 1587, in the Gregorian calendar. This date seems less &#xA;likely, since there is no full moon on that date, though there is &#xA;almost a new moon in Taurus, so maybe. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So why would the author write &quot;month of moon in Taurus&quot; instead of the &#xA;actual month name? It's possible he was trying to be careful, since &#xA;the months would be different in the many countries still using the &#xA;Julian calendar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it may just be a reference to the lunar eclipse of 24 Mar 1587: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://eclipsewise.com/lunar/LEprime/1501-1600/LE1587Mar24Tprime.html&quot; rel=&quot;nofollow&quot;&gt;http://eclipsewise.com/lunar/LEprime/1501-1600/LE1587Mar24Tprime.html&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2016-01-05T22:48:24.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13091" PostTypeId="1" AcceptedAnswerId="13094" CreationDate="2016-01-06T00:14:29.660" Score="3" ViewCount="130" Body="&lt;p&gt;I read an &lt;a href=&quot;http://space.com/31526-black-hole-burp-solves-star-mystery.html&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; which suggests black holes could influence the production of stars in the galaxy as the black hole can burst out powerful X-rays to &quot;snowplow&quot; the surrounding dust and gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the article didn't suggest that spiral galaxies are made this way, I like to think that a supermassive black hole that &quot;burps&quot; lies in the core of every spiral galaxies. Am I mistaken?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complementary question: does the supermassive black holes inside the Milky Way still &quot;burp&quot; or it is already cured?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="9301" LastEditDate="2016-06-29T04:37:23.907" LastActivityDate="2016-06-29T04:37:23.907" Title="Could a &quot;burping&quot; supermassive black hole be responsible for a spiral galaxy's look?" Tags="&lt;galaxy&gt;&lt;supermassive-black-hole&gt;&lt;star-formation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13092" PostTypeId="2" ParentId="13073" CreationDate="2016-01-06T01:31:39.017" Score="2" Body="&lt;p&gt;It's worth noting that elements with atomic mass greater than about 140 were likely created in the collision of neutron stars as opposed to supernovae. There is an interesting article &lt;a href=&quot;http://phys.org/news/2011-09-cosmic-forging-gold.html&quot; rel=&quot;nofollow&quot;&gt;on Physics.org&lt;/a&gt; and another on the &lt;a href=&quot;https://www.washingtonpost.com/news/achenblog/wp/2013/07/17/blingnova-the-origin-of-gold/&quot; rel=&quot;nofollow&quot;&gt;Washington Post&lt;/a&gt;. There is also another &lt;a href=&quot;http://beyondearthlyskies.blogspot.com/2013/10/nucleosynthesis-of-gold-in-neutron-star.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;br&gt;&lt;br&gt; I too thought that all elements heavier than iron were created in supernova explosions, and when I first heard the theory that elements heavier than gold were likely created in neutron star collisions, I rejected that idea. But after reading these and other articles, I've come to accept it. &lt;strike&gt;Apparently, the energy generated in a supernova, while considerable, is not sufficient enough to synthesis the elements heavier than gold.&lt;/strike&gt;&lt;br&gt;&lt;br&gt;&lt;b&gt;Edit:&lt;/b&gt;I found this &lt;a href=&quot;http://www.smithsonianmag.com/science-nature/all-the-gold-in-the-universe-could-come-from-the-collisions-of-neutron-stars-13474145/?page=1&quot; rel=&quot;nofollow&quot;&gt;interesting article&lt;/a&gt; on the Smithsonian Magazine's website. The theory that NS-NS collisions create elements heavier than gold came from data observed from such a collision, which only happen, on average, every 100,000 years. Apparently, this theory better answers questions that the supernova creation of heavy elements cannot. The observed NS-NS collision was observed on June 3, 2013&lt;/p&gt;&#xA;" OwnerUserId="7108" LastEditorUserId="7108" LastEditDate="2016-01-06T19:59:14.307" LastActivityDate="2016-01-06T19:59:14.307" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13094" PostTypeId="2" ParentId="13091" CreationDate="2016-01-06T08:09:19.690" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the supermassive black holes inside the Milky Way still &quot;burp&quot; or&#xA;  it is already cured&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Technically, black holes don't burp it's the material that falls into a black hole, as it gets tightly squeezed and under enormous spiral and magnetic forces it shoots out gamma rays at the poles perpendicular to the accretion disk.   More details &lt;a href=&quot;https://en.wikipedia.org/wiki/Astrophysical_jet&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The super-massive black hole (not holes as there's just one) in the center of the milky way isn't currently &quot;burping&quot; but the next time it eats a star it's expected to &quot;burp&quot; again.   I remember reading that this may happen sometime in the next 10,000 years or so, but I couldn't find the article just now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Burping&quot; or Jets of material are more active when galaxies are young.  The quasars observed by large telescope operate in that way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I like to think that a supermassive black hole that &quot;burps&quot; lies in&#xA;  the core of every spiral galaxies. Am I mistaken?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is essentially correct.   Quoting this &lt;a href=&quot;http://science.nasa.gov/astrophysics/focus-areas/black-holes/&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Astronomers believe that supermassive black holes lie at the center of&#xA;  virtually all large galaxies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Black holes are surprisingly good at shooting material back into space, strange as that may sound.  &lt;a href=&quot;http://www.space.com/22586-milky-way-giant-black-hole-food.html&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-01-06T08:09:19.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13095" PostTypeId="1" CreationDate="2016-01-06T11:45:04.163" Score="4" ViewCount="284" Body="&lt;p&gt;I understand that - although appearing yellow - the sun is white. And that this is because the other light colours are scattered by the atmosphere. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.google.com/search?q=the%20sun%20from%20space&amp;amp;espv=2&amp;amp;biw=1024&amp;amp;bih=631&amp;amp;tbm=isch&quot; rel=&quot;nofollow&quot;&gt;Google image search for &quot;The Sun from Space&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That was difficult for me to come to terms with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if yellow light is what's getting through the atmosphere - why doesn't everything have a yellow tinge to it?&lt;/p&gt;&#xA;" OwnerUserId="10318" LastEditorUserId="9186" LastEditDate="2016-01-08T09:11:59.953" LastActivityDate="2016-01-08T09:11:59.953" Title="Why isn't everything yellow(ish)?" Tags="&lt;the-sun&gt;&lt;observational-astronomy&gt;&lt;light&gt;&lt;spectra&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13096" PostTypeId="2" ParentId="13095" CreationDate="2016-01-06T12:22:39.613" Score="0" Body="&lt;p&gt;The solar spectrum peaks somewhere near green, but that is not directly relevant to its apparent colour which is a result of the actual spectrum and the sensor response. If there is enough intensity in the pass bands of the three colour sensors for them to saturate the sensor will apparently record &quot;white&quot;. This is true of camera sensors, film and the eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The light from the Sun at the bottom of the atmosphere peaking in the yellow is only a secondary factor in your perception of the colour of objects around you. If Sun light were mono-chromatic, that is concentrated in a very narrow band, say at yellow, then everything would appear yellow, like under sodium street lighting. But it is not mono-chromatic and the eyes perception of the colour of an object is effectively the result of the different reflectivity at different frequencies as sensed through three relatively wide bandwidth colour filters. the visual system is also doing a lot of other things to make the perceived colour of objects more or less constant under differing lighting conditions.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2016-01-06T12:22:39.613" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13097" PostTypeId="1" AcceptedAnswerId="13100" CreationDate="2016-01-06T12:26:30.013" Score="4" ViewCount="5596" Body="&lt;p&gt;When looking up at the night sky, we can sometimes observe a bright band of stars stretching across it. I know this is our galaxy but exactly what are we looking at? &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Could we look at the centre of the galaxy or is there another spiral arm blocking our view?&lt;/li&gt;&#xA;&lt;li&gt;How does our view change as earth rotates around the sun? (E.g. What part of the milky way do we see in a winter night sky)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10321" LastActivityDate="2016-02-05T17:53:16.967" Title="What part of the milky way do we see from earth?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13098" PostTypeId="1" AcceptedAnswerId="13099" CreationDate="2016-01-06T12:50:32.087" Score="10" ViewCount="929" Body="&lt;p&gt;If I'm told correctly (eg. &lt;a href=&quot;http://solar-center.stanford.edu/SID/activities/GreenSun.html&quot; rel=&quot;noreferrer&quot;&gt;What Color is the Sun?&lt;/a&gt;) that the sun is actually white, should the photo on that page actually be white too? Is it just doctored to meet peoples expectations?:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wiwkW.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wiwkW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise for all the red photo's in a &lt;a href=&quot;http://www.google.com/search?q=the+sun%27s+surface&amp;amp;espv=2&amp;amp;biw=1024&amp;amp;bih=631&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;gws_rd=ssl&quot; rel=&quot;noreferrer&quot;&gt;Google image search for &quot;The Sun's surface&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10319" LastEditorDisplayName="user1569" LastEditDate="2016-01-08T10:28:43.637" LastActivityDate="2016-01-09T15:24:02.427" Title="Should this photo of the sun's surface actually be white?" Tags="&lt;the-sun&gt;&lt;observational-astronomy&gt;&lt;amateur-observing&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13099" PostTypeId="2" ParentId="13098" CreationDate="2016-01-06T15:36:58.053" Score="14" Body="&lt;p&gt;The term &quot;color&quot; is a label that humans have assigned to denote the &lt;em&gt;ratio&lt;/em&gt; between the intensity at various wavelengths in the three different wavelength &lt;em&gt;bands&lt;/em&gt;, or regions, that the human eye is able to perceive. These bands are centered roughly at 430, 545, and 570 nm, but are quite broad and even overlap:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UnH59.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UnH59.png&quot; alt=&quot;vision&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Human cone response, normalized to the same height. In reality, the response of the blue cones is significantly smaller, and the green is somewhat larger (from &lt;a href=&quot;https://en.wikipedia.org/wiki/Color_vision&quot;&gt;Wikipedia&lt;/a&gt;).&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an object emits light &lt;em&gt;only&lt;/em&gt; at, say, 450 nm, the ratio is roughly 0.1:0.2:1 (in the order R:G:B); it then looks a special way to us, and we call it &quot;blue&quot;, or maybe &quot;violet&quot;. If it emits at 550 nm, or 650 nm, we call it &quot;green&quot; or &quot;red&quot;. An object that emits light in a more continuous spectrum that covers the region 500–600 nm, we'd name something like orange-/brown-/olive-ish, depending on the exact spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun emits photons at all wavelengths, but not in an equal amount at all wavelengths. The particular ratios between the three bands that we can see, we have labeled &quot;white&quot;. However, when the Sun's light enters our atmosphere, some of the light is absorbed, especially at the blue wavelength. Filtering out the blue results in a spectrum that looks more orange to us. The figure below shows the Sun's &quot;true&quot; spectrum (in yellow), and the spectrum seen from the surface of Earth (in red):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Pzw4Y.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pzw4Y.png&quot; alt=&quot;sun&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;The Sun's spectrum measured outside our atmosphere (&lt;em&gt;yellow&lt;/em&gt;) and at sea level (&lt;em&gt;red&lt;/em&gt;) (modified image from &lt;a href=&quot;https://en.wikipedia.org/wiki/Sunlight&quot;&gt;Wikipedia&lt;/a&gt;, with data from &lt;a href=&quot;http://www.globalwarmingart.com/&quot;&gt;Global Warming Art&lt;/a&gt;).&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes we want to observe the Sun in a wavelength region that is invisible to humans, for instance in UV or X-rays. This can be done with a telescope and a detector that is sensitive to light in that particular region, but in order for us to see it, we represent the image with a color that we &lt;em&gt;can&lt;/em&gt; see. The image in the top of the link you provide is taken with the European spacecraft &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_and_Heliospheric_Observatory&quot;&gt;SOHO&lt;/a&gt;'s instrument EIT at 19.5 nm, which we call &quot;extreme UV&quot;, bordering on soft X-rays. Since this is invisible to humans, they arbitrarily chose to represent it using green. They might as well have chosen pink or brown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZOEIR.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZOEIR.png&quot; alt=&quot;EUV&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;The Sun in Extreme UV, during a particularly violent solar flare (from the &lt;a href=&quot;http://sohowww.nascom.nasa.gov/gallery/images/eitbigflare.html&quot;&gt;SOHO gallery&lt;/a&gt;).&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several of the photos in your second link are images taken by the Japanese space telescope &lt;a href=&quot;https://en.wikipedia.org/wiki/Hinode&quot;&gt;Hinode&lt;/a&gt;, which observes both in the optical (i.e. visible by humans), X-rays, and far UV. If these are shown in orange, again it's just to make them visible to us, and you may say that they have been &quot;doctored to meet our expectation&quot;. In this way, I like better when they choose a color such as all green, so we know it's &quot;false color&quot;.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-01-09T15:24:02.427" LastActivityDate="2016-01-09T15:24:02.427" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13100" PostTypeId="2" ParentId="13097" CreationDate="2016-01-06T17:27:53.107" Score="2" Body="&lt;p&gt;&lt;strong&gt;Fraser Cain&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We’re seeing the galaxy edge on, from the inside, and so we see the&#xA;  galactic disk as a band that forms a complete circle around the sky.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Which parts you can see depend on your location on Earth and the time&#xA;  of year, but you can always see some part of the disk.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The galactic core of the Milky Way is located in the constellation&#xA;  Sagittarius, which is located to the South of me in Canada, and only&#xA;  really visible during the Summer. In really faint skies, the Milky Way&#xA;  is clearly thicker and brighter in that region.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want to know more, watch this video from &lt;strong&gt;Fraser Cain&lt;/strong&gt; which explains it in details:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=pdFWbEwsOmA&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=pdFWbEwsOmA&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find your answer in the first minute&lt;/p&gt;&#xA;" OwnerUserId="10322" LastEditorUserId="10322" LastEditDate="2016-01-06T17:52:09.220" LastActivityDate="2016-01-06T17:52:09.220" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13101" PostTypeId="2" ParentId="11737" CreationDate="2016-01-06T18:29:14.660" Score="0" Body="&lt;p&gt;Actually we can't have an specific answer to that. Because we don't have enough knowledge about the size of the universe and the density of galaxies in other parts of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You both might be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens if we are in a low density of matters but far from us we have a high density of galaxies and stars but we can't see them because of being too much far or being hidden from us behind something like Dark Matter or anything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Human are just guessing and are not too sure about anything.&lt;/p&gt;&#xA;" OwnerUserId="10322" LastActivityDate="2016-01-06T18:29:14.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13102" PostTypeId="2" ParentId="13098" CreationDate="2016-01-06T20:39:26.390" Score="3" Body="&lt;p&gt;The credit for that green image includes &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_and_Heliospheric_Observatory&quot; rel=&quot;nofollow&quot;&gt;SOHO&lt;/a&gt;.&#xA;SOHO routinely &lt;a href=&quot;http://www.n3kl.org/sun/&quot; rel=&quot;nofollow&quot;&gt;uses that green for images of the sun in &lt;strong&gt;hard&lt;/strong&gt; X-Rays at 19.5 nanometers.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Visible light only goes down to about 390nm for extreme violet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture &lt;em&gt;looks&lt;/em&gt; like it's taken in hard X-ray wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;•••&#xA;Later: Sorry, I meant hard UV/soft X-Rays. Don't know where my head was when I wrote this answer. SOHO's still the green pix in hard UV source though.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="2031" LastEditDate="2016-01-07T16:23:24.177" LastActivityDate="2016-01-07T16:23:24.177" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13103" PostTypeId="2" ParentId="13098" CreationDate="2016-01-06T21:22:33.197" Score="2" Body="&lt;p&gt;Unobscured sunlight is white by definition.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You may find this article interesting, &lt;a href=&quot;http://www.osa-opn.org/Content/ViewFile.aspx?id=11147&quot; rel=&quot;nofollow&quot;&gt;The Yellow&#xA;Sun Paradox&lt;/a&gt;, as to why we consider the sun yellow in casual observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would add another possible physiological explanation to the three suggested in this article: The retina's blue cones saturate at lower illumination than the red and green ones, which creates a deficit of blue signal to the optical nerve under heavy stimulation, resulting in a perception of red + green = yellow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely the sensitivity of the blue cones is why very dim objects such as faint stars in the night sky usually seem to be blue.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, as to the images. There are three possible reasons for an image to be yellow or orange. Perhaps, as a nod to expectations, it has been colorized. Possibly it is a false color image that maps a real but non-visible spectrum (&lt;em&gt;radio, microwave, IR, UV&lt;/em&gt;, or &lt;em&gt;X-ray&lt;/em&gt;) onto the visible spectrum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But most of the images you present us with in your google search are using a technique sometimes called flame gradation, where a gray-scale (&lt;em&gt;black→gray→white&lt;/em&gt;) is mapped on to a flame-scale:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;black → [purple?] → brown → red → orange → gold → yellow → white → [cyan?]&lt;/em&gt;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to allow a much greater &lt;strong&gt;&lt;em&gt;range&lt;/em&gt;&lt;/strong&gt; of gradation in intensity to be perceived. It also suggests rather naturally a &lt;em&gt;cold→cool→warm→hot&lt;/em&gt; gradation.&lt;/p&gt;&#xA;" OwnerUserId="9186" LastEditorUserId="9186" LastEditDate="2016-01-07T23:14:32.313" LastActivityDate="2016-01-07T23:14:32.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13104" PostTypeId="1" AcceptedAnswerId="13125" CreationDate="2016-01-06T23:07:16.987" Score="0" ViewCount="19694" Body="&lt;p&gt;I need a way to calculate the effective temperature (surface temperature) of a star for a stellar model. I need something in the form Te=....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Radius in m&lt;/li&gt;&#xA;&lt;li&gt;mass in kg&lt;/li&gt;&#xA;&lt;li&gt;the composition of particles (eg H 90%, He 8% etc)&lt;/li&gt;&#xA;&lt;li&gt;the combined stored thermal energy of the body in J &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Constants (any really but I'm using these for now):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;G=gravity constant=6.67408E-011&lt;/li&gt;&#xA;&lt;li&gt;k=kbolzmann=1.3806485279E-023&lt;/li&gt;&#xA;&lt;li&gt;s=sbolzmann=5,67036713E-008&lt;/li&gt;&#xA;&lt;li&gt;PI=pi ~3.14...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Example of the sun:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;mp=average mass of a particle=1,7E-027&lt;/li&gt;&#xA;&lt;li&gt;M=total mass of the body=2E30&lt;/li&gt;&#xA;&lt;li&gt;r=radius of the body=700000000&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm using this equation to estimate the core temperature :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;(G*mp*M)/(r*(3/2)*k)&lt;/code&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which nets 15653011 for the sun which is close enough given that that is the only star core temperature known (afaik).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using this to estimate the luminosity L:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;4*PI*(r^2)*s*(Te^4)&lt;/code&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which results in an error of ~1-5%  with 90% of my sample stars which is close enough. For the sun this results in &lt;code&gt;3,95120075975041E+026 W&lt;/code&gt; which is only &lt;code&gt;2,7%&lt;/code&gt; off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is I need Te for the 2nd formula which I don't have in my scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the formula for L being dependent on the surface temperature to the power of 4 this value has to be relatively precise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumptions of my model:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;uniform distribution of particles: so every slice of the body has the same composition as the entire body.&lt;/li&gt;&#xA;&lt;li&gt;perfect sphere: every body is a perfect sphere, no handling for elliptic bodies needed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My sample values (first line is the sun with a core temp of 15000000): &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     emitted energy Surface temp    radius       mass &#xA;     (in Lsun)      (in K)          (in m)       (in Msun)&#xA;     1              5800            700000000    1&#xA;     8700000        53000           25200000000  265&#xA;     6300000        50100           23100000000  110&#xA;     2900000        42000           23660000000  132&#xA;     2000000        44000           16800000000  80&#xA;     1260000        13500           140000000000 45&#xA;     57500          3600            618100000000 12.4&#xA;     78             5700            6440000000   2.56&#xA;     78.5           4940            8540000000   2.69&#xA;     15100          7350            51100000000  9.7&#xA;     1.519          5790            858900000    1.1&#xA;     0.5            5260            605500000    0.907&#xA;     370000         3690            994000000000 19.2&#xA;     123000         33000           7560000000   56&#xA;     2200000        52500           12600000000  130&#xA;     200000         10000           151900000000 22&#xA;     446000         19000           43330000000  42.3&#xA;     25.4           9940            1197700000   2.02&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Errors in luminosity to actual value (the maximum error is about 100% which I can live with since it might just be inaccurate measurements for the sample stars)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2.74%&#xA;6.71%&#xA;-1.13%&#xA;11.29%&#xA;-2.00%&#xA;-4.27%&#xA;106.76%&#xA;3.99%&#xA;2.51%&#xA;-6.50%&#xA;1.12%&#xA;4.00%&#xA;-8.27%&#xA;2.10%&#xA;1.57%&#xA;113.75%&#xA;1.64%&#xA;2.15%&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="10326" LastEditorUserId="2270" LastEditDate="2016-02-07T18:17:00.487" LastActivityDate="2016-06-07T05:44:20.523" Title="How to calculate the temperature of a star" Tags="&lt;star&gt;&lt;temperature&gt;&lt;surface&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13105" PostTypeId="1" CreationDate="2016-01-07T00:20:25.210" Score="0" ViewCount="358" Body="&lt;p&gt;In the Big Bang model of the Universe, every observable thing is thought to have expanded very abruptly from a point of infinite density and zero volume. However, the problem with this assertion is that for something to expand it first needs a locality in which to do such, which brings us to my question. How can scientists believe that &quot;everything&quot; was condensed and then suddenly expanded, since such is automatically an oxymoron?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expansion is a physical property of the universe and yet if &quot;everything&quot; had no where to expand, how could it have expanded? There was &quot;nothing,&quot; was there not? Even if we drop the big bang model and talk about &quot;everything's&quot; being in a singularity point, the question still stands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where was the singularity point? Asked differently, into what is everything expanding (which is a rhetorical question I ask to implicate space-time as necessarily existing irrespective of any theoretical expansion and likely so vastly that one might as well assume that space-time is infinitely large).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As well, it seems counter-intuitive, as little impressive a word as that is given its context here, that something like... well, &quot;everything&quot; could be condensed in the first place, let alone infinitely condensed and with zero volume. When thinking about the big-bang model, I can't help but conclude that a mere analogy was originally used for stellar birth, evolution and finally explosion. The issue though is still extension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understood it, the big bang model states that space itself was also condensed along with everything else in the observable universe to a point of singularity. So, if that's the case, then [where] was the singularity point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It ends up becoming an infinite regress, of course. Or, if you don't like the issue of &quot;time&quot; presupposed by my question, then look at the question after the universe started expanding. All the matter had to expand somewhere and &quot;somewhere&quot; is not &quot;nothing.&quot; - It just seems that the model is somehow overlooking said feature of the actual universe and trying to account for it by mere assertion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to call it space-time, extension, space, where everything supposedly expanded to and is and is still expanding to, then that's my question. How does the theory allege where everything is yet at the same time assert that prior to the singularity point, there was literally nothing.&lt;/p&gt;&#xA;" OwnerUserId="7886" LastEditorUserId="9301" LastEditDate="2016-03-08T05:36:34.077" LastActivityDate="2018-02-28T00:44:24.157" Title="Question about space-time" Tags="&lt;big-bang-theory&gt;&lt;space-time&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13106" PostTypeId="1" CreationDate="2016-01-07T01:49:31.407" Score="3" ViewCount="137" Body="&lt;p&gt;Well, I just read &lt;a href=&quot;http://news.yahoo.com/rare-galaxy-2-black-holes-1-starved-stars-180552639.html&quot; rel=&quot;nofollow&quot;&gt;an article&lt;/a&gt; about a galaxy with two black holes, and it stated that the galaxy is 1 billion light-years away. Why is there no direction given? If the Big Bang happened, and things are moving away from each other, we can still have at least two directions - away from the center, and to it. Is that a right way to think about space and our drift in space?&lt;/p&gt;&#xA;" OwnerUserId="1305" LastEditorUserId="2153" LastEditDate="2016-01-07T02:45:06.203" LastActivityDate="2016-01-07T03:47:46.010" Title="When I read that a system is located x light years away, why is there no direction?" Tags="&lt;space&gt;&lt;big-bang-theory&gt;&lt;space-time&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13107" PostTypeId="2" ParentId="12868" CreationDate="2016-01-07T01:49:46.677" Score="0" Body="&lt;p&gt;If we assume that the pulsar's surface is like that of other neutron stars, unless the beam is pointed at you, it will look like other neutron stars. RX J1856.5-3754 (&lt;a href=&quot;https://en.wikipedia.org/wiki/RX_J1856.5-3754&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/RX_J1856.5-3754&lt;/a&gt;) is one of very few neutron stars we can see at optical wavelengths.  It has a visual magnitude of 25.6 at ≈61 parsecs (the Sun's apparent visual magnitude at that distance would be about 8.75). &#xA;   Turning the cranks I get an absolute visual magnitude MV of 21.67 and a  visual luminosity of ≈.00000018.  Taking the square root, I'd need to be about .00043 AU away, or about a tenth the diameter of the Sun for it to be as bright as the Sun is from Earth, visually. At only 14 km or so in diameter, it would very small, about 4.7% the apparent diameter of the Sun--not much more than a point. &#xA;   But as noted above, the actual, bolometric, luminosity of the neutron star would be much, much higher.  A person looking at it (unprotected) from that distance it would be blinded and fried in short order.  One might also be far enough down the gravity well at that distance that the relativistic effects that dim the star would be less and the star would appear even brighter.  And one might note some tidal effects as well.  This situation calls for the &quot;General Products Hull&quot; Larry Niven used for his story, &quot;Neutron Star!&quot;&lt;/p&gt;&#xA;" OwnerUserId="10327" LastActivityDate="2016-01-07T01:49:46.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13108" PostTypeId="2" ParentId="13106" CreationDate="2016-01-07T03:47:46.010" Score="5" Body="&lt;p&gt;The galaxy's catalog name, SDSS J1126+2944, also gives its direction. This means the galaxy is at approximately 11h26m right ascension and 29 degrees 44 minutes north declination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;full name&quot; of the galaxy, given here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sci-news.com/astronomy/pair-black-holes-distant-galaxy-03546.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.sci-news.com/astronomy/pair-black-holes-distant-galaxy-03546.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is &quot;SDSS J112659.54+294442.8&quot;, which gives its position even more precisely. A finder chart is here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=SDSS+J112659.54%2B294442.8&quot; rel=&quot;noreferrer&quot;&gt;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=SDSS+J112659.54%2B294442.8&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2016-01-07T03:47:46.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13109" PostTypeId="2" ParentId="13095" CreationDate="2016-01-07T04:22:33.220" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But if yellow light is what's getting through the atmosphere - why doesn't everything have a yellow tinge to it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everything does have a yellowish tinge to it, but only for roughly the hour after sunrise and the hour before sunset. These a photographer's &lt;a href=&quot;https://en.wikipedia.org/wiki/Golden_hour_(photography)&quot; rel=&quot;nofollow noreferrer&quot;&gt;golden hours&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vHcIC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vHcIC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the Sun rises, its color changes from reddish to orangish to yellowish -- and then to nearly white, even when the Sun never does get very high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HRHb4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HRHb4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must not (and cannot) look at the Sun, even for the most fleeting of moments, once the Sun gets above 10 to 15 or so degrees above the horizon. The only time you can look at the Sun, even briefly, is when the Sun is low on the horizon and &lt;a href=&quot;https://en.wikipedia.org/wiki/Air_mass_(astronomy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;air mass&lt;/a&gt; is high. We think the Sun is yellow (or in some civilizations, orange or even red) because we cannot see the Sun for what it truly is, which is nearly white.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What bit of off-white color sunlight does have when the Sun is 10 to 15 or so degrees above the horizon is offset by the blue light from the sky, aka &quot;diffuse sky radiation&quot;. The color of the light from the Sun and the sky when the Sun is high in the sky is about as white as light can get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a bit of physiology going on here as well. To those who have lived their most of their lives inside buildings, bright sunlight is a bit yellowish and a bit too harsh. To those who spend much of their time outdoors, the fluorescent lighting that now dominates the urban landscape is a bit bluish and a bit too harsh. What city dwellers who rarely ventures outdoors thinks of as white light is not very white to those who spends most of their waking hours outdoors, and vice versa.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2016-01-07T05:10:21.297" LastActivityDate="2016-01-07T05:10:21.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13111" PostTypeId="1" AcceptedAnswerId="13113" CreationDate="2016-01-07T04:54:16.173" Score="1" ViewCount="259" Body="&lt;p&gt;Say hypothetically an UMBH with 50 billion solar masses (supposedly upper limit) is beginning to merge with another (Super)MBH similar to Sagittarius A* that sits at the center of the Milky Way Galaxy, what would happen next in this clash of the titans?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastActivityDate="2016-01-07T06:17:27.893" Title="What happens when an Ultra Massive Black Hole cannibalize another?" Tags="&lt;supermassive-black-hole&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13112" PostTypeId="1" CreationDate="2016-01-07T06:07:41.200" Score="4" ViewCount="415" Body="&lt;p&gt;Solar wind may contain some hydrogen and oxygen atoms which is the composition of water, some of them may move to the poles of earth, my question is, is solar wind also a source of water in earth?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastActivityDate="2016-05-28T15:12:09.483" Title="Is solar wind a source of water?" Tags="&lt;solar-wind&gt;&lt;the-sun&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13113" PostTypeId="2" ParentId="13111" CreationDate="2016-01-07T06:17:27.893" Score="0" Body="&lt;p&gt;From this &lt;a href=&quot;http://scitechdaily.com/black-holes-could-grow-as-large-as-50-billion-suns-before-their-food-crumbles-into-stars-research-shows/&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;“Bigger black hole masses are in principle possible – for example, a&#xA;  hole near the maximum mass could merge with another black hole, and&#xA;  the result would be bigger still. But no light would be produced in&#xA;  this merger, and the bigger merged black hole could not have a disc of&#xA;  gas that would make light.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The theoretical size limit has to do with the ability to form an accretion disk, not the hole itself.   While it's probably an extremely rare occurrence due to space expansion, if a 50 billion solar mass black hole was to merge with another 50 billion solar mass black hole, you'd have a 100 billion solar mass black hole.    There might be some interesting spiraling in towards each other and some &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_wave&quot; rel=&quot;nofollow&quot;&gt;gravitational waves&lt;/a&gt; made in the process.   It would be cool to study, but in such a scenario, there's no force that would prevent the two black holes from merging into a bigger black hole.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-01-07T06:17:27.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13115" PostTypeId="1" CreationDate="2016-01-07T11:58:16.803" Score="2" ViewCount="362" Body="&lt;p&gt;The brightest star on the night sky is Sirius, here in the Earth. But are there any exoplanets, where the brightest star on the night sky is our sun?&lt;/p&gt;&#xA;" OwnerUserId="23135" OwnerDisplayName="Iter Ator" LastActivityDate="2016-01-08T06:06:31.893" Title="From which exoplanets is our sun the brightest star on the night sky?" Tags="&lt;exoplanet&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13116" PostTypeId="2" ParentId="13115" CreationDate="2016-01-07T13:05:01.043" Score="0" Body="&lt;p&gt;Dependents. Unless blocked or outshone by a neighboring star Sirius A would continue to be the brightest star visible, Sirius A is bright due to its &lt;a href=&quot;http://hubblesite.org/newscenter/archive/releases/2005/36/full/&quot; rel=&quot;nofollow&quot;&gt;size&lt;/a&gt; and our atmosphere, not because of distance, if the planet had a different atmosphere that broke up wavelengths of light differently than earth other stars may show brighter.&lt;/p&gt;&#xA;" OwnerUserId="37919" OwnerDisplayName="Zacharia Samuel Grierson" LastActivityDate="2016-01-07T13:05:01.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13117" PostTypeId="2" ParentId="13105" CreationDate="2016-01-07T14:03:50.080" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In the big bang model of the Universe, every observable thing is thought to have expanded very abruptly from a point of infinite density and zero volume.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thats not quite true. Have a look at &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Bang#Singularity&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; where you can read this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Extrapolation of the expansion of the universe backwards in time using general relativity yields an infinite density and temperature at a finite time in the past.[18] This singularity signals the breakdown of general relativity and thus, all the laws of physics. How closely this can be extrapolated toward the singularity is debated—certainly no closer than the end of the Planck epoch. This singularity is sometimes called &quot;the Big Bang&quot;,[19] but the term can also refer to the early hot, dense phase itself,[20][notes 1] which can be considered the &quot;birth&quot; of our universe.&quot;&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't necessarily extrapolate back all the way to a point singularity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, the problem with this assertion is that for something to expand it first needs a locality in which to do such, which brings us to my question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's no problem with &lt;em&gt;space&lt;/em&gt; expanding. Don't think of space as nothing, Einstein didn't, and nor do modern physicists such as Nobel prizewinner &lt;a href=&quot;https://en.wikipedia.org/wiki/Aether_theories#Quantum_vacuum&quot; rel=&quot;nofollow&quot;&gt;Robert B Laughlin&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can scientists believe that &quot;everything&quot; was condensed and then suddenly expanded, since such is automatically an oxymoron?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because there's good scientific &lt;a href=&quot;https://www.google.co.uk/search?hl=en-GB&amp;amp;source=hp&amp;amp;biw=&amp;amp;bih=&amp;amp;q=evidence+expanding+universe&amp;amp;gbv=2&amp;amp;oq=evidence+expanding+universe&amp;amp;gs_l=heirloom-hp.3..0j0i22i30l9.1186.6803.0.6927.27.26.0.1.1.0.140.2182.24j2.26.0....0...1ac.1.34.heirloom-hp..0.27.2197.Qzx_JQ-rxXY&quot; rel=&quot;nofollow&quot;&gt;evidence that the universe is expanding&lt;/a&gt;. And when you &quot;wind back the clock&quot; you find that all the galaxies were in the same place 13.8 billion years ago.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Expansion is a physical property of the universe and yet if &quot;everything&quot; had no where to expand, how could it have expanded? There was &quot;nothing,&quot; was there not?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. Space expanded, and space isn't nothing. As to what caused the big bang to occur, I don't know. Nor do I know what the universe was like 13.8 billion years ago. But I know of no infinities in nature, and I would challenge the alleged point singularity at the centre of a black hole and refer instead to the &lt;a href=&quot;https://www.google.co.uk/search?q=frozen+star+black+hole&amp;amp;hl=en-GB&amp;amp;biw=&amp;amp;bih=&amp;amp;gbv=2&amp;amp;oq=frozen+star+black+hole&amp;amp;gs_l=heirloom-serp.3..0.236701.245814.0.246219.22.13.0.9.9.0.140.953.12j1.13.0....0...1ac.1.34.heirloom-serp..0.22.984.Q0uo6M303wE&quot; rel=&quot;nofollow&quot;&gt;frozen star interpretation&lt;/a&gt;. In similar vein I would challenge the alleged point-singularity 13.8 billion years ago. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Even if we drop the big bang model and talk about &quot;everything's&quot; being in a singularity point, the question still stands. Where was the singularity point?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think you shouldn't drop the big bang model, but instead you should drop the singularity. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Asked differently, into what is everything expanding?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nothing. I know this is hard to imagine, but try to. Space is expanding, and there is no space beyond it. So there &lt;em&gt;is&lt;/em&gt; no beyond it. As to whether this means you end up coming back round full circle or the universe is some &lt;a href=&quot;http://edition.cnn.com/2004/TECH/space/05/24/universe.wide/&quot; rel=&quot;nofollow&quot;&gt;&quot;hall of mirrors&quot;&lt;/a&gt; or something else, I don't know.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(which is a rhetorical question I ask to implicate space-time as necessarily existing irrespective of any theoretical expansion and likely so vastly that one might as well assume that space-time is infinitely large).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, one might not. Because the universe doesn't expand for nothing. It expands because space has a kind of innate &quot;pressure&quot;, as per the energy-pressure diagonal in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor&quot; rel=&quot;nofollow&quot;&gt;stress-energy-momentum tensor&lt;/a&gt;. And if the universe was infinite, the pressure would be counterbalanced at all locations. An infinite universe could not expand. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When thinking about the big-bang model, I can't help but conclude that a mere analogy was originally used &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For an analogy, get a &lt;a href=&quot;https://www.google.co.uk/search?q=stress+ball&amp;amp;hl=en-GB&amp;amp;gbv=2&amp;amp;prmd=ivnsp&amp;amp;tbm=isch&amp;amp;tbo=u&amp;amp;source=univ&amp;amp;sa=X&amp;amp;ved=0ahUKEwiiuuuW7ZfKAhUG1xQKHZRNCEwQsAQIMw&quot; rel=&quot;nofollow&quot;&gt;stress ball&lt;/a&gt; and squeeze it down in your fist. Then let go. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The issue though is still extension. As far as I understood it, the big bang model states that space itself was also condensed along with everything else in the observable universe to a point of singularity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See what I said above. Big bang cosmology doesn't actually say everything was condensed to point singularity.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;All the matter had to expand somewhere&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There wasn't any matter initially. Remember the wave nature of matter, and pair production. We can quite literally make matter out of electromagnetic waves. These wave propagate through space, and are effectively wave sin space. So we make matter out of waves in space. So start with space, and do somehting to it such that it rings like a bell and is full of waves and starts expanding. As to what, I don't know.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How does the theory allege where everything is yet at the same time assert that prior to the singularity point, there was literally nothing?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It doesn't. It goes back as far as a very dense universe. The stuff you hear about a point singularity and creation ex-nihilo is popscience, not big bang cosmology. &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2016-01-07T14:03:50.080" CommentCount="22" ContentLicense="CC BY-SA 3.0" />
  <row Id="13118" PostTypeId="2" ParentId="13115" CreationDate="2016-01-07T16:14:20.397" Score="1" Body="&lt;p&gt;You can use Celestia to get a view of the sun in the night sky from other solar systems, &lt;a href=&quot;http://astroblogger.blogspot.ie/2010/03/tour-of-exoplanets-in-celestia.html&quot; rel=&quot;nofollow&quot;&gt;just like this guy did&lt;/a&gt;, although the post is a little old.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2016-01-07T16:14:20.397" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13119" PostTypeId="2" ParentId="13115" CreationDate="2016-01-07T16:32:04.653" Score="11" Body="&lt;p&gt;I wrote &lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-solve-astro-13115.m&quot; rel=&quot;noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-solve-astro-13115.m&lt;/a&gt; to solve this with the full results at: &lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/brightest-stars-from-other-stars.txt.bz2&quot; rel=&quot;noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/brightest-stars-from-other-stars.txt.bz2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As viewed from RigelKentaurusB, the Sun has magnitude 0.48,&#xA;and is the 10th brightest star in the sky. This is the only&#xA;star system with known exoplanets where the Sun even makes&#xA;the top 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From EpsilonEridani, the sun shines at magnitude 2.37,&#xA;making it the 79th brightest star in their sky. Things go&#xA;rapidly downhill from there:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&#xA;79 EpsilonEridani Sun 2.36646&#xA;207 HIP113020 Sun 3.18865&#xA;208 HIP85523 Sun 3.11216&#xA;247 HIP106440 Sun 3.29606&#xA;396 HIP15510 Sun 3.74083&#xA;423 HIP74995 Sun 3.81444&#xA;685 Fomalhaut Sun 4.25746&#xA;878 HIP64924 Sun 4.48202&#xA;949 HIP109388 Sun 4.54456&#xA;972 HIP99825 Sun 4.55679&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is based on Mathematica's list of 552 exoplanets&#xA;orbiting 464 stars, and a total list of 88,637 stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the naked eye visibility limit is magnitude 5.5, the Sun&#xA;would only be visible from 28 of these stars:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&#xA;10 RigelKentaurusB Sun 0.476137&#xA;79 EpsilonEridani Sun 2.36646&#xA;208 HIP85523 Sun 3.11216&#xA;207 HIP113020 Sun 3.18865&#xA;247 HIP106440 Sun 3.29606&#xA;396 HIP15510 Sun 3.74083&#xA;423 HIP74995 Sun 3.81444&#xA;685 Fomalhaut Sun 4.25746&#xA;878 HIP64924 Sun 4.48202&#xA;949 HIP109388 Sun 4.54456&#xA;972 HIP99825 Sun 4.55679&#xA;1086 HIP57443 Sun 4.65677&#xA;1126 HIP21932 Sun 4.69871&#xA;1381 HIP57087 Sun 4.87838&#xA;1426 HIP83043 Sun 4.89689&#xA;1409 Pollux Sun 4.90049&#xA;1617 HIP10138 Sun 5.01833&#xA;1649 HIP57050 Sun 5.04143&#xA;1676 HIP3093 Sun 5.05658&#xA;1716 HIP48331 Sun 5.06528&#xA;1831 Gl317 Sun 5.13093&#xA;2104 HIP22627 Sun 5.24572&#xA;2264 Rho1Cancri Sun 5.31849&#xA;2263 HIP40693 Sun 5.32723&#xA;2397 HIP27887 Sun 5.36943&#xA;2355 HIP80337 Sun 5.37669&#xA;2382 GJ1214 Sun 5.3879&#xA;2689 UpsilonAndromedae Sun 5.47503&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If the limit is magnitude 6.5, 33 more star systems can see the Sun:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&#xA;2838 Alrai Sun 5.52682&#xA;2926 HIP53721 Sun 5.571&#xA;3399 HIP79431 Sun 5.69393&#xA;3637 MuArae Sun 5.74863&#xA;3679 HIP113357 Sun 5.76061&#xA;3783 TauBootis Sun 5.7949&#xA;3941 HIP98767 Sun 5.83457&#xA;4071 HIP85647 Sun 5.86725&#xA;4318 HIP71395 Sun 5.92513&#xA;4580 HIP6379 Sun 5.95739&#xA;4783 IotaHorologii Sun 6.01138&#xA;4835 HIP7978 Sun 6.02527&#xA;4867 RhoCoronaeBorealis Sun 6.03887&#xA;5016 HIP1292 Sun 6.0583&#xA;4949 NN3634 Sun 6.08757&#xA;4993 HIP55848 Sun 6.10352&#xA;5252 HIP49699 Sun 6.11214&#xA;5300 HIP83389 Sun 6.11214&#xA;5031 HIP65721 Sun 6.11803&#xA;5343 HIP79248 Sun 6.12236&#xA;5385 PiMensae Sun 6.12986&#xA;5453 EpsilonReticulii Sun 6.13303&#xA;6158 HIP98505 Sun 6.251&#xA;6351 HIP113421 Sun 6.30305&#xA;6436 Nu2CanisMajoris Sun 6.31593&#xA;6343 HIP99711 Sun 6.32197&#xA;6737 Hamal Sun 6.35637&#xA;6516 HIP64457 Sun 6.37975&#xA;6867 HIP58451 Sun 6.4205&#xA;7614 HIP25110 Sun 6.43775&#xA;7525 HIP109378 Sun 6.46941&#xA;7538 HIP54906 Sun 6.47034&#xA;7957 HIP96901 Sun 6.48193&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;OLD ANSWER FOR REFERENCE (note that Sirius does not have known exoplanets, so the Sirius solution I propose below wouldn't work):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As others have noted, since Sirius is 25 times more luminous than our Sun (though still not luminous as Canopus or Rigel), so it would be the brightest star for most nearby star systems, and all the exoplanets we've found are fairly closeby.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only possible exception would be an exoplanet orbiting Sirius itself, since Sirius would be considered their sun, and not a star in the &lt;em&gt;night&lt;/em&gt; sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, Sirius' smaller companion, Sirius B, would be much brighter than our Sun there. I'm also pretty sure Sirius B would be in the night sky (not just in the daytime sky), so it would count.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to poke around a bit more before declaring this answer complete. I think I can show that any point where the Sun is brighter than Sirius must be less than 4.3 light years (and probably much less) away from our solar system, so, unless we discover an star system MUCH closer than Proxima Centauri, there are no exoplanets where our Sun is the brightest star.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2016-01-08T05:15:16.203" LastActivityDate="2016-01-08T05:15:16.203" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13120" PostTypeId="1" AcceptedAnswerId="13121" CreationDate="2016-01-07T02:51:54.960" Score="18" ViewCount="421" Body="&lt;p&gt;Some other SE questions about launching ICBM's into the sun got me wondering whether we have ever observed an object on a path that intersected with the Sun? How close did it get? &lt;/p&gt;&#xA;" OwnerUserId="10302" OwnerDisplayName="rolfedh" LastEditorUserId="13" LastEditDate="2016-01-07T17:47:10.473" LastActivityDate="2016-01-07T17:47:10.473" Title="Have we ever observed a body, such as a large asteroid, &quot;hitting&quot; the Sun?" Tags="&lt;solar-system&gt;&lt;the-sun&gt;&lt;observational-astronomy&gt;&lt;asteroids&gt;&lt;comets&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13121" PostTypeId="2" ParentId="13120" CreationDate="2016-01-07T05:03:41.547" Score="20" Body="&lt;h2&gt;Yes&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;and here's a video of &quot;a Giant Comet Hitting the Sun&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=Mat4dWpszoQ&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=Mat4dWpszoQ&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.space.com/11663-comet-hits-sun-solar-explosion.html&quot; rel=&quot;noreferrer&quot;&gt;The impact occurred sometime during May 10-11, 2011.  The comet was not named but believed to be a member of the Kreutz family of comets&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Many close calls&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Before this spectacular plunge we had witnessed several other comets graze (come close without hitting) the Sun.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nasa.gov/mission_pages/sunearth/news/comet-lovejoy.html&quot; rel=&quot;noreferrer&quot;&gt;I've included a nice picture of comet Lovejoy after its close approach of the Sun brought it through the Sun's corona around December 15, 2011&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dU6XE.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dU6XE.jpg&quot; alt=&quot;Comet Lovejoy grazing the Sun&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;What happens when a comet hits the sun?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;It acts like &lt;a href=&quot;https://www.newscientist.com/article/dn27954-what-would-happen-if-a-massive-comet-crashed-into-the-sun/&quot; rel=&quot;noreferrer&quot;&gt;a supersonic snowball in Hell&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If a comet is big enough and passes close enough, the steep fall into&#xA;  the sun’s gravity would accelerate it to more than 600 kilometres per&#xA;  second. At that speed, drag from the sun’s lower atmosphere would&#xA;  flatten the comet into a pancake right before it exploded in an&#xA;  airburst, releasing ultraviolet radiation and X-rays that we could see&#xA;  with modern instruments.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The crash would unleash as much energy as a magnetic flare or coronal&#xA;  mass ejection, but over a much smaller area. “It’s like a bomb being&#xA;  released in the sun’s atmosphere,” Brown says. The momentum propelled&#xA;  by the comet could even make the sun ring like a bell with subsequent&#xA;  sun-quakes echoing through the solar atmosphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7168" OwnerDisplayName="Jim2B" LastActivityDate="2016-01-07T06:06:20.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13122" PostTypeId="2" ParentId="246" CreationDate="2016-01-07T20:26:13.937" Score="2" Body="&lt;p&gt;In our solar system, the answer to this depends on how you &#xA;define &quot;annular eclipse&quot; (as opposed to transit). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote &#xA;&lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-solve-astro-246.m&quot; rel=&quot;nofollow&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-solve-astro-246.m&lt;/a&gt; &#xA;to see how big planets' moons appear compared to the sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The full results: &#xA;&lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-solve-astro-246.txt.bz2&quot; rel=&quot;nofollow&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-solve-astro-246.txt.bz2&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planet/moon combinations closest to annular eclipse (note &#xA;these numbers are averages, not ranges): &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Earth Moon 0.972 &#xA;Saturn Epimetheus 0.790 &#xA;Saturn Prometheus 0.689 &#xA;Saturn Pandora 0.588 &#xA;Uranus Perdita 0.54 &#xA;Jupiter Amalthea 0.515 &#xA;Saturn Iapetus 0.4231 &#xA;Saturn Tarqeq 0.4 &#xA;Mars Phobos 0.388 &#xA;Uranus Cupid 0.28 &#xA;Jupiter Thebe 0.249 &#xA;Uranus Mab 0.21 &#xA;Neptune Nereid 0.199 &#xA;Saturn Pan 0.197 &#xA;Jupiter Metis 0.188 &#xA;Saturn Hyperion 0.1818 &#xA;Saturn Atlas 0.15 &#xA;&lt;/code&gt;&lt;/pre&gt; &#xA;&#xA;&lt;p&gt;Note that the Earth-Moon ratio is much closer to 1 than any of the others. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're looking for total eclipses where the moon is only &#xA;slightly larger than the Sun: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Saturn Janus 1.22 &#xA;Jupiter Callisto 1.433 &#xA;Pluto S2012P1 1.5 &#xA;Uranus Ophelia 1.64 &#xA;Uranus Cordelia 1.67 &#xA;Uranus Bianca 1.79 &#xA;Pluto S2011P1 1.9 &#xA;Uranus Desdemona 2.11 &#xA;Uranus Rosalind 2.1 &#xA;Saturn Enceladus 2.174 &#xA;Uranus Belinda 2.21 &#xA;Jupiter Europa 2.603 &#xA;Uranus Cressida 2.66 &#xA;Jupiter Ganymede 2.751 &#xA;Saturn Rhea 2.975 &#xA;Uranus Juliet 3.00 &#xA;Saturn Dione 3.057 &#xA;Saturn Tethys 3.733 &#xA;Uranus Puck 3.9 &#xA;Uranus Portia 4.22 &#xA;Saturn Titan 4.324 &#xA;Neptune Naiad 4.4 &#xA;Jupiter Io 4.83 &#xA;&lt;/code&gt;&lt;/pre&gt; &#xA;&#xA;&lt;p&gt;And, just in case you're wondering, Charon is 257 times &#xA;larger in Pluto's sky than the Sun. &lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2016-01-07T20:26:13.937" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13123" PostTypeId="1" AcceptedAnswerId="13136" CreationDate="2016-01-08T02:06:07.793" Score="1" ViewCount="139" Body="&lt;p&gt;As we know, both Jupiter and Neptune have great spot like features in South hemisphere, and Saturn has dragon storm in south hemisphere, is it true that great spot like features appear in South hemisphere more frequently? If so, what is the reason?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="9527" LastEditDate="2020-10-09T09:58:27.183" LastActivityDate="2020-10-10T10:24:41.637" Title="Do great spot like features favour appear in south hemisphere?" Tags="&lt;jupiter&gt;&lt;saturn&gt;&lt;neptune&gt;&lt;great-red-spot&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13125" PostTypeId="2" ParentId="13104" CreationDate="2016-01-08T03:14:53.810" Score="2" Body="&lt;p&gt;Empirically (I fit a regression on log(mass) vs log(surface temp)), using the table of values in the article on &lt;a href=&quot;https://en.wikipedia.org/wiki/Main_sequence&quot; rel=&quot;nofollow&quot;&gt;Main Sequence stars&lt;/a&gt;, I get a fairly well-fitting formula:  $\mathrm{estTemp} = 5740*\mathrm{mass}^{0.54}$, where estTemp is in C and mass is in multiples of the sun's mass.  Seems to work very well for all but the largest and smallest main sequence stars (and not TOO bad for those).&lt;/p&gt;&#xA;" OwnerUserId="7937" LastEditorUserId="2365" LastEditDate="2016-06-07T05:44:20.523" LastActivityDate="2016-06-07T05:44:20.523" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13126" PostTypeId="1" AcceptedAnswerId="13128" CreationDate="2016-01-08T03:26:11.960" Score="2" ViewCount="117" Body="&lt;p&gt;EDIT: I am still following up, but here's the latest response from Wolfram:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Paclet servers are updated and optimized for each version of&#xA;  Mathematica, so I suspect you may be working on an older version which&#xA;  we no longer update. Many times we get reports that data from the&#xA;  paclet data for version 9 was incorrect, only to find that it has&#xA;  already been fixed for version 10.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to Mathematica and several other sources, the&#xA;known information about HIP31978 (also known as &quot;S&#xA;Monocerotis&quot; and &quot;15 Monocerotis&quot;) includes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The star system is about 101.06 light years from our own.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The star's absolute magnitude is -2.79 (some sources say even lower).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The star's visual magnitude from our solar system is 4.66.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The star is slightly variable, but not enough to explain&#xA;the following discrepancy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the star's magnitude at 32.6 light years is -2.79&#xA;(definition of absolute magnitude), it would be about 9.6&#xA;times fainter at 101.06 light years (it's actual distance&#xA;from us). However, that's only about 2.5 magnitudes&#xA;fainter, nowhere near enough to bring it down to magnitude&#xA;4.66.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's happening here, and is this a one-off special case,&#xA;or does this happen a lot?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference, my calculation of the apparent magnitude:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;-2.79 - Log[100, (32.6/101.06)^2]*5 == -0.333192&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2016-01-14T22:07:12.437" LastActivityDate="2016-01-14T22:07:12.437" Title="Absolute/apparent magnitude and distance for HIP31978 inconsistent?" Tags="&lt;apparent-magnitude&gt;&lt;absolute-magnitude&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13127" PostTypeId="2" ParentId="13115" CreationDate="2016-01-08T04:41:55.413" Score="6" Body="&lt;p&gt;Another way of thinking about the question is, &quot;how close do you need to be for a dim star (our Sun) to outshine a brighter one?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider our Sun, and another star a distance $d$ away that is $m$ times brighter than the Sun (in terms of luminosity or absolute magnitude).  Let's look a location in space that is a distance $x$ from the Sun in the direction of the second star, and a distance $y$ in the perpendicular direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/knbdP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of light seen at the location is proportional to the inverse square of the distance.  The relative amount of light from each star is therefore just:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{1}{x^2+y^2} \qquad\qquad \frac{m}{(d-x)^2+y^2}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to know when the first is greater than the second:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{1}{x^2+y^2} &amp;gt; \frac{m}{(d-x)^2+y^2} \\&#xA;(d-x)^2+y^2 &amp;gt; m x^2 + m y^2 \\&#xA;d^2 - 2dx &amp;gt; (m-1)x^2 + (m-1)y^2 \\&#xA;\left(1+\frac{1}{m-1}\right)d^2 &amp;gt; \left(\frac{1}{m-1}\right)d^2 + 2dx + (m-1)\left(x^2 + y^2\right) \\&#xA;\left(\frac{m-1+1}{m-1}\right)d^2 &amp;gt; (m-1)\left(\frac{d^2}{(m-1)^2} + 2x\frac{d}{m-1} + x^2 + y^2\right) \\&#xA;m\left(\frac{d}{m-1}\right)^2 &amp;gt; \left(\frac{d}{m-1} + x\right)^2 + y^2 \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This inequality describes a sphere of radius $r=d \sqrt{m}/(m-1)$, centered around the point $x=-d/(m-1),\ y=0$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let's take the specific case of Sirius, for which $m=25.4$ and $d=8.60~\text{ly}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the above equations, we get $r=1.77~\text{ly}$ and $x=-0.35~\text{ly}$.  Therefore, even at half the distance of the next-closest star Sirius is brighter than the Sun.  This means that the Sun is the brightest star in the sky for exactly eight planets.&lt;/p&gt;&#xA;" OwnerUserId="10335" LastEditorUserId="10335" LastEditDate="2016-01-08T06:06:31.893" LastActivityDate="2016-01-08T06:06:31.893" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13128" PostTypeId="2" ParentId="13126" CreationDate="2016-01-08T07:44:55.367" Score="0" Body="&lt;p&gt;The absolute magnitude you quote appears to correspond to the Hipparcos parallax distance estimate, which is $\sim 1000 \mbox{ ly}$, not the $\sim 100 \mbox{ ly}$ distance you have used.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2016-01-08T07:44:55.367" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13129" PostTypeId="1" AcceptedAnswerId="13130" CreationDate="2016-01-08T08:02:41.380" Score="4" ViewCount="4102" Body="&lt;p&gt;Gravitational force is the attractive force between any two masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about gaseous planets, where do they get their attractive force from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the big rocky planets don't have enough attractive force to keep enough gases in their gravitational field, Jupiter has lots of gases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this means there is a concentrated mass inside, under which gravitational field the gases are kept 'in'? And if so, then why are gaseous planets not also considered to be rocky ones?&lt;/p&gt;&#xA;" OwnerUserId="10336" LastEditorUserId="10336" LastEditDate="2016-01-09T02:56:29.613" LastActivityDate="2016-01-09T02:56:29.613" Title="Where does Jupiter's gravitational force come from? Why don't Jupiters gasses fly away?" Tags="&lt;gravity&gt;&lt;planet&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13130" PostTypeId="2" ParentId="13129" CreationDate="2016-01-08T10:09:45.057" Score="6" Body="&lt;p&gt;&lt;em&gt;if they don't have any solid mass inside them&lt;/em&gt; is an incorrect conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some have 'ordinary' solid cores, and the pressures in the centers of gas giants are so high, that they are even 'suspected' of compressing hydrogen into a 'metallic' state. Quoting from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Planetary_core&quot; rel=&quot;noreferrer&quot;&gt;Planetary core Wikipedia page&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;Gas giants also have cores, though the composition of these are still a matter of debate and range in possible composition from traditional stony/iron, to ice or to fluid metallic hydrogen.&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For Jupiter especially:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;Jupiter has an observed magnetic field generated within its core, indicating some metallic substance is present. [... ]&lt;br /&gt;&#xA;Jupiter has a rock and/or ice core ten-thirty times the mass of the Earth [...]&lt;br /&gt;&#xA;Thermal contraction/evolution models support the presence of metallic hydrogen within the core in large abundances (greater than Saturn).&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The earth's core has an average density of about 12,000 kg per cubic meter, Jupiter's core density is estimated at 25,000 kg per cubic meter (&lt;a href=&quot;http://classroom.synonym.com/jupiters-core-vs-earths-core-21848.html&quot; rel=&quot;noreferrer&quot;&gt;link&lt;/a&gt;). Together with its much larger radius than the earth (70000 km versus 6370 km), you will have a pretty massive core &lt;a href=&quot;https://en.wikipedia.org/wiki/Jupiter&quot; rel=&quot;noreferrer&quot;&gt;with a mass of from 12 to 45 times the Earth's mass&lt;/a&gt;, so there's plenty of gravitational pull there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More to read: &lt;a href=&quot;http://lasp.colorado.edu/education/outerplanets/giantplanets_interiors.php&quot; rel=&quot;noreferrer&quot;&gt;The outer planets: Giant planets: Interiors&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user1569" LastEditorDisplayName="user1569" LastEditDate="2016-01-08T13:13:40.220" LastActivityDate="2016-01-08T13:13:40.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13131" PostTypeId="1" CreationDate="2016-01-08T10:23:23.773" Score="2" ViewCount="81" Body="&lt;p&gt;I've been looking for examples at how mass-loss rates are determined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm studying a circumstellar dust shell ejected from a Wolf-Rayet star. I have some parameters like, expansion velocity of the shell (60km/s), the dust mass of the shell (0.1 M_sun), the radii of the shell (R_in=10000 AU and R_out=60000) and its age (T=26000 yrs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if there's a formula relating those parameters and if they are sufficient to determine the dust mass-loss rate or its dependent on other factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone help?&lt;/p&gt;&#xA;" OwnerUserId="10338" LastActivityDate="2016-12-29T18:12:07.113" Title="Dust mass-loss rate from a massive star given a set of parameters?" Tags="&lt;star&gt;&lt;astrophysics&gt;&lt;mass&gt;&lt;interstellar-medium&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13134" PostTypeId="2" ParentId="13123" CreationDate="2016-01-08T12:20:54.690" Score="0" Body="&lt;p&gt;Saturn's Great White Spot phenomenon often occurs in the Northern hemisphere, so it doesn't look like there is any particular preference.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2016-01-08T12:20:54.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13136" PostTypeId="2" ParentId="13123" CreationDate="2016-01-08T17:56:11.910" Score="1" Body="&lt;p&gt;Given that we have very few examples, we could not, statistically, draw any significance from any apparent bias.  It's also hard to see any theoretical basis for expecting a bias, so my gut feeling would be there probably is no bias.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also worth remembering that human observations of these things are essentially a very brief snapshot in a huge period during which we have no idea about what may have been happening.  So again the statistical significance of what happens at the moment needs to be carefully considered.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2016-01-08T17:56:11.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13137" PostTypeId="2" ParentId="13129" CreationDate="2016-01-08T18:06:06.913" Score="5" Body="&lt;p&gt;You are confusing the &quot;mass&quot; with &quot;solid&quot;. All matter has mass, and all mass produces a gravitational field. That includes gasses, liquids and plasmas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although gasses are much less dense than solids, gasses also have mass, and if you have enough gas it will have a measurable gravitational field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter is big, it is composed of lots of Hydrogen and Helium (and some other gasses), and deep within the planet, the gasses are compressed into strange states. There may even be a rocky core, but it is under such extreme pressure that it is not much like &quot;rock&quot; as we understand it. But it is not necessary for a planet to have a solid core to produce a gravitational field, because all matter has mass not just solid matter.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-01-08T18:06:06.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13138" PostTypeId="1" AcceptedAnswerId="13144" CreationDate="2016-01-09T07:39:45.753" Score="9" ViewCount="1533" Body="&lt;p&gt;Recently NASA has revealed that they have recorded the sound of Sun. They say that it produces a sound like &quot;Om&quot;. I can't understand how they can hear it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the best of my knowledge, sound can't travel in space (vacuum). Has a new technique been invented that enables them to detect sound coming from outer space? If so, what is it? How does it work?&lt;/p&gt;&#xA;" OwnerUserId="10336" LastEditorUserId="2618" LastEditDate="2016-01-09T16:13:36.687" LastActivityDate="2019-01-28T04:36:37.313" Title="How can we hear the sound of the Sun?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;space&gt;&lt;vacuum&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13143" PostTypeId="1" CreationDate="2016-01-09T17:04:14.587" Score="1" ViewCount="408" Body="&lt;p&gt;I'm developing my own planetarium software and I using Stellarium to check if the values a get from my software are correct or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All values are roughly equal, being different only in the seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are Stellarium coordiantes followed by my coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, this is a CSV txt file. I don't know if I can attached to this question so I have pasted here:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Star;RA J2000;DEC J2000;RA J2016;DEC J2016;Hour Angle;AZ;Alt.&#xA;Betelgeuse (Stellarium);5h55m10.34s;7d24m25.6s;5h56m2.38s;7d24m22.0s;23h52m25.55s;+176d38m27.3s;+55d59m52.0s&#xA;Betelgeuse (My software);5h55m10.34s;7d24m25.596s;5h56m2.32s;7d24m31.74s;23h52m25.6662s;+176d38m29.643s;+56d00m01.836s&#xA;Rigel (Stellarium);5h14m32.28s;-8d12m05.9s;5h15m18.50s;-8d11m12.8s;00h33m09.44s;+190d42m52.5s;+39d51m52.8s&#xA;Rigel (My software);5h14m32.279s;-8d12m05.9s;5h15m18.46s;-8d11m03.24s;00h33m09.5262s;+190d42m55.996s;+39d52m02.033s&#xA;Bellatrix (Stellarium);5h25m7.86s;+6d20m58.7s;5h25m59.49s;+6d21m37.1s;00h22m28.45s;+189d40m43.7s;+54d38m14.4s&#xA;Bellatrix (MySoftware);5h25m7.9s;+6d20m59s;5h25m59.42s;+6d21m46.71s;00h22m28.5662s;+189d40m48.830s;+54d38m23.685s&#xA;Mintaka (Stellarium);5h32m00.41s;-0d17m56.7s;5h32m49.55s;-0d17m27.8s;0h15m38.38s;+185d52m14.8s;+48d11m30.9s&#xA;Mintaka (MySoftware);5h32m00.42s;-0d17m56.6988s;05h32m49.51s;-0d17m18.1958s;00h15m38.4762s;+185d52m18.043s;+48d11m40.40s&#xA;Alnilam (Stellariumm);5h36m12.82s;-1d12m06.9s;5h37m01.63s;-1d11m44.00s;00h11m26.31s;+184d13m21.1s;+47d21m31.3s&#xA;Alnilam (My software);5h36m12.82s;-1d12m06.912s;5h37m01.57s;-1d11m34.2547s;00h11m26.4162s;+184d13m24.435s;+47d21m40.929s&#xA;Alnitak (Stellarium);5h40m45.54s;-1d56m33.2s;5h41m34.07s;-1d56m16.6s;00h06m53.87s;+182d30m42,7s;+46d40m01.3s&#xA;Alnitak (My software);5h40m45.55s;-1d56m33.2160s;5h41m34.03s;-1d56m06.8887;00h06m53.9562s;+182d30m45.212s;+46d40m10.931s&#xA;Saiph (Stellarium);5h47m45.40s;-9d40m10.6s;5h48m31.01s;-9d40m03.7s;23h59m56.93s;+179d59m01.5s;+38d57m56.3s&#xA;Saiph (My software);5h47m45.38s;-9d40m10.5960s;5h48m30.9500s;-9d39m54.0031s;23h59m57.0362s;+179d59m03.631s;+38d58m05.992s&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is that different a problem? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know which value is the correct one (the one from Stellarium or the one from my software). The maximum difference is about 10 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the difference between declinations (Stellarium and mine) is always greater that the difference between right ascensions.&lt;/p&gt;&#xA;" OwnerUserId="7566" LastActivityDate="2016-08-18T05:09:56.290" Title="Calculating AZ and ALT from RA and DEC: values are slightly different from Stellarium" Tags="&lt;star&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13144" PostTypeId="2" ParentId="13138" CreationDate="2016-01-09T17:24:04.280" Score="13" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Recently NASA has revealed that they have recorded the sound of Sun. They say that it produces a sound like &quot;Om&quot;. I can't understand how they can hear it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not that recent (it was 2010), it wasn't NASA (it was researchers at the University of Sheffield who used data from a NASA satellite), it wasn't sound &lt;em&gt;per se&lt;/em&gt; (it was instead sonified data), and neither NASA nor those researchers said that the sound produced was &quot;Om&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sonification&quot; rel=&quot;noreferrer&quot;&gt;Data sonification&lt;/a&gt; is a scientific tool of growing importance. It is the sound equivalent of visualization techniques that have been used for centuries. (Graphs are a rather old visualization technique.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, researchers at the University of Sheffield used sonification techniques to translate imagery of solar coronal loops into sound. A faithful representation of that generated sound would not have been audible; the frequency would have been too low for a human to hear it. Those researchers had to speed up the playback to make the generated sound audible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's not to say that those sounds are a complete fiction. The surface of the Sun is not a vacuum, nor is its interior. There truly are low frequency sounds associated with events on the surface of the Sun. The Sun rings with sound (low frequency sound). Studying the sounds produced by the Sun gives scientists insights into what is happening inside the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a small downside in making scientific data public: Non-scientists who have a penchant to read/see/hear things into that scientific data will do just that. To me, the people who say that this is proof that the universe does indeed chant the sacred &quot;Om&quot; are no different that those who see cities on the Moon or Mars in NASA photography.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2016-01-09T17:24:04.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13145" PostTypeId="1" AcceptedAnswerId="13146" CreationDate="2016-01-09T18:07:51.397" Score="16" ViewCount="5423" Body="&lt;p&gt;It is often said that the Sun is a medium size, or even dim star. Is this true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to this &lt;a href=&quot;http://www.johnstonsarchive.net/astro/nearstar.html&quot; rel=&quot;noreferrer&quot;&gt;list of stars within 21 light years&lt;/a&gt;, there are, out of the closest 121 stars, only six brighter than the Sun. This means that the Sun is in the top 6%. If you also count brown dwarfs, the Sun ranks even higher. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doesn't this indicate that the Sun is actually a relatively large star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SfyPJ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SfyPJ.png&quot; alt=&quot;7/121&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="13591" LastEditDate="2016-07-27T23:33:17.277" LastActivityDate="2016-07-27T23:33:17.277" Title="Is the Sun really a medium size star?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;size&gt;&lt;luminosity&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13146" PostTypeId="2" ParentId="13145" CreationDate="2016-01-09T18:42:43.813" Score="18" Body="&lt;p&gt;It is true that a surprisingly large number of stars are smaller (and thus less massive) than the Sun. However, the stars that &lt;em&gt;are&lt;/em&gt; bigger than the Sun are often much bigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at this chart:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/d/d0/Comparison_of_planets_and_stars_%28sheet_by_sheet%29_%28Apr_2015_update%29.png/640px-Comparison_of_planets_and_stars_%28sheet_by_sheet%29_%28Apr_2015_update%29.png&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/d/d0/Comparison_of_planets_and_stars_%28sheet_by_sheet%29_%28Apr_2015_update%29.png/640px-Comparison_of_planets_and_stars_%28sheet_by_sheet%29_%28Apr_2015_update%29.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;Image courtesy of Wikipedia user Jcpag2012 under &lt;a href=&quot;https://creativecommons.org/licenses/by-sa/3.0/deed.en&quot;&gt;the Creative Commons Attribution-Share Alike 3.0 Unported license&lt;/a&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how small the Sun is compared to some of the other stars. It's &lt;em&gt;tiny&lt;/em&gt;! It is indeed a small star - in technical terms a &lt;em&gt;main sequence dwarf&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, despite its size, it is clear that there are many more stars less massive than the Sun that there are stars more massive than the Sun. Why? There are two reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Lower-mass stars live longer.&lt;/li&gt;&#xA;&lt;li&gt;More low-mass stars can form in a given region than high-mass stars.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;a href=&quot;http://www.astro.caltech.edu/~george/ay20/eaa-stellarmasses.pdf&quot;&gt;Encyclopedia of Astronomy and Astrophysics&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distribution of masses can be quantified in an &lt;a href=&quot;https://en.wikipedia.org/wiki/Initial_mass_function&quot;&gt;initial mass function&lt;/a&gt;, typically given in the form&#xA;$$\xi(m)=km^{-\alpha}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you integrate this over a range of masses, you can find how many stars are within that range. Not surprisingly, this number gets lower and lower as you slide the endpoints to more massive stars. You can see this decrease from the fact that $\xi'(m)&amp;lt;0$, so long as $k&amp;gt;0$ and $\alpha&amp;gt;0$ - which is assumed by the model, according to empirical data.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-01-09T23:14:16.663" LastActivityDate="2016-01-09T23:14:16.663" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13147" PostTypeId="1" AcceptedAnswerId="13148" CreationDate="2016-01-10T01:23:51.707" Score="8" ViewCount="8106" Body="&lt;p&gt;So I am not a true scientist and I was just wondering if an extraterrestrial rock from space can have some dangerous unidentified virus or bacteria that can be deadly and wipe out the Earth if it not quarantined. Is this science fiction?.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: Note I said extraterrestrial &lt;em&gt;rocks&lt;/em&gt; not extraterrestrial as in aliens.&lt;/p&gt;&#xA;" OwnerUserId="10352" LastEditorUserId="10352" LastEditDate="2016-05-20T01:10:02.757" LastActivityDate="2021-05-04T19:51:23.353" Title="Is it possible for a virus to come from a meteorite?" Tags="&lt;space&gt;&lt;comets&gt;&lt;meteor&gt;&lt;extra-terrestrial&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13148" PostTypeId="2" ParentId="13147" CreationDate="2016-01-10T01:31:00.857" Score="11" Body="&lt;p&gt;There is, believe it or not, a scientific theory regarding this: &lt;a href=&quot;https://en.wikipedia.org/wiki/Panspermia&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;panspermia&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the panspermia hypothesis, microbes &quot;hitch&quot; a lift on bodies leaving a planet. They then travel through space and eventually land somewhere else.&#xA;Panspermia involves three difficult phases of travel:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Launch (as well as the impact event)&lt;/li&gt;&#xA;&lt;li&gt;Travel in the harsh environment of space&lt;/li&gt;&#xA;&lt;li&gt;Atmospheric entry and landing, with high temperatures&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It's difficult to survive these. However, materials related to organic matter have been found that might be evidence of panspermia; see &lt;a href=&quot;http://www.pnas.org/content/early/2015/10/14/1517557112.full.pdf&quot; rel=&quot;noreferrer&quot;&gt;Bell et al. (2015)&lt;/a&gt; for one example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Panspermia is traditionally applied to bacteria known as &lt;em&gt;extremophiles&lt;/em&gt;, which are accustomed to living in harsh environments. Space contains temperatures ranging from hundreds of degrees Fahrenheit to hundreds of degrees below zero, and any organism would have to survive those. This may be possible - &lt;a href=&quot;http://www.nature.com/news/high-flying-bacteria-spark-interest-in-possible-climate-effects-1.12310&quot; rel=&quot;noreferrer&quot;&gt;bacteria have survived in conditions like there&lt;/a&gt; - but difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Viruses pose an additional challenge because they cannot survive long without a host organism on which to live. This means they would have to be carried along by an extremophilic bacteria. This also means that they might not be a huge harm to creatures on Earth. Viruses only attack certain bacteria because of proteins on their surfaces; extraterrestrial biology most likely differs from that on Earth, and these viruses might not be equipped to harm terrestrial organisms.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-01-10T22:45:38.197" LastActivityDate="2016-01-10T22:45:38.197" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13149" PostTypeId="1" AcceptedAnswerId="13150" CreationDate="2016-01-10T06:06:12.530" Score="9" ViewCount="1072" Body="&lt;p&gt;In &lt;a href=&quot;https://en.wikipedia.org/wiki/Isaac_Asimov&quot; rel=&quot;noreferrer&quot;&gt;Isaac Asimov&lt;/a&gt;'s &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Foundation_Series&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Foundation&lt;/em&gt;&lt;/a&gt; series, the main characters are looking for the long-lost Earth. There are two major features of the solar system that seem to make it unusual among the millions of inhabited systems: &lt;a href=&quot;https://en.wikipedia.org/wiki/Rings_of_Saturn&quot; rel=&quot;noreferrer&quot;&gt;the rings of Saturn&lt;/a&gt; and the size of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Moon&quot; rel=&quot;noreferrer&quot;&gt;Moon&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The conversations below are mostly between &lt;em&gt;Golan Trevize&lt;/em&gt;, ex-military, well-versed in astronomy, and &lt;em&gt;Janov Pelorat&lt;/em&gt;, a naïve and recluse history teacher.&lt;/p&gt;&#xA;&lt;h2&gt;The rings of Saturn&lt;/h2&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;That's what the poem was speaking of. Three wide rings, concentric, wider than the planet itself.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Trevize said, &amp;quot;I never heard of such a thing. I don't think rings can be that wide. Compared to the planet they circle, they are always very narrow.&amp;quot;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&amp;quot;Is that sort of thing common?&amp;quot; asked Bliss, awed.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;No,&amp;quot; said Trevize. &amp;quot;Almost every gas giant has rings of debris, but they tend to be faint and narrow. I once saw one in which the rings were narrow, but quite bright. But I never saw anything like this; or heard of it, either.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Pelorat said, &amp;quot;That's clearly the ringed giant the legends speak of. If this is really unique-&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Really unique, as far as I know, or as far as the computer knows,&amp;quot; said Trevize.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;The Moon&lt;/h2&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;A giant satellite is more difficult to accept. No other inhabited world in the Galaxy has such a satellite. Large satellites are invariably associated with the uninhabited and uninhabitable gas-giants. As a Skeptic, then, I prefer not to accept the existence of the moon.&amp;quot;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&amp;quot;Yes. It's rather farther from the planet than one might expect but it's definitely revolving about it. It's only the size of a small planet; in fact, it's smaller than any of the four inner planets circling the sun. Still, it's large for a satellite. It's at least two thousand kilometers in diameter, which makes it in the size range of the large satellites that revolve about gas giants.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;No larger?&amp;quot; Pelorat seemed disappointed. &amp;quot;Then it's not a giant satellite?&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Yes, it is. A satellite with a diameter of two to three thousand kilometers that is circling an enormous gas giant is one thing. That same satellite circling a small, rocky habitable planet is quite another. That satellite has a diameter over a quarter that of Earth. Where have you heard of such near-parity involving a habitable planet?&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Pelorat said timidly, &amp;quot;I know very little of such things.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Trevize said, &amp;quot;Then take my word for it, Janov. It's unique. We're looking at something that is practically a double planet, and there are few habitable planets that have anything more than pebbles orbiting them. Janov, if you consider that gas giant with its enormous ring system in sixth place, and this planet with its enormous satellite in third—both of which your legends told you about, against all credibility, before you ever saw them-then that world you're looking at must be Earth. It cannot conceivably be anything else. We've found it, Janov; we've found it.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;sup&gt;&lt;em&gt;Quotes are from &lt;a href=&quot;https://en.wikipedia.org/wiki/Foundation%27s_Edge&quot; rel=&quot;noreferrer&quot;&gt;Foundation's Edge&lt;/a&gt; (1982) and &lt;a href=&quot;https://en.wikipedia.org/wiki/Foundation_and_Earth&quot; rel=&quot;noreferrer&quot;&gt;Foundation and Earth&lt;/a&gt; (1986).&lt;/em&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;My question is in two parts:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Are the various descriptions and facts given above generally correct?&lt;/li&gt;&#xA;&lt;li&gt;How unusual are Saturn's rings and the Moon compared to other systems we've observed?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10354" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2021-02-11T07:42:13.700" Title="How unusual is the solar system?" Tags="&lt;the-moon&gt;&lt;solar-system&gt;&lt;saturn&gt;&lt;planetary-ring&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13150" PostTypeId="2" ParentId="13149" CreationDate="2016-01-10T07:07:52.230" Score="6" Body="&lt;p&gt;First, Azimov was writing (Edge 1982, and Earth 1986) before the first extra solar planets were tentatively detected (1989), or definitely detected 1991 or 1992. So what he wrote could not be based on any real data on extra solar planets, and must have been an educated guess.&lt;/p&gt;&#xA;&lt;p&gt;However both books were written after the discovery of Charon, so he may have been ill advised to claim that large moons were unlikely around non-gas giants.&lt;/p&gt;&#xA;&lt;p&gt;Personally I think both the large moon and broad planetary ring guesses are unlikely to pan out. Uncommon maybe but not uncommon enough for Azimov's purpose.&lt;/p&gt;&#xA;&lt;p&gt;If you believe it, we have already discovered an &lt;a href=&quot;https://www.discovermagazine.com/the-sciences/this-exoplanets-ring-system-puts-saturn-to-shame&quot; rel=&quot;nofollow noreferrer&quot;&gt;exoplanet with a more impressive ring system than Saturn's&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="24184" LastEditDate="2021-02-11T07:42:13.700" LastActivityDate="2021-02-11T07:42:13.700" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="13151" PostTypeId="2" ParentId="13147" CreationDate="2016-01-10T07:10:24.157" Score="4" Body="&lt;p&gt;I think it's a fun question and I can kind of give a layman's answer.   (though reading the above answer, there is some overlap).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems likely that &quot;tiny&quot;, life or viruses could survive/hibernate for a long time if blown into space on space rocks, so it follows that life could travel to earth that way, but I think we could apply a modified version of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Drake_equation&quot; rel=&quot;nofollow&quot;&gt;drake equation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth gets hit by comets or meteors every day but most of those burn up in the atmosphere.   There is a temperature beyond which even bacteria and viruses have a hard time surviving.  So, many of those meteors or bits of comet, I suspect, would lose any life they might have.   Only a percentage could be the right size to both slow down enough while breaking apart to not impact at too high a speed and successfully deliver life to the surface or into the ocean.   Like the real drake equation, we get to plug in our own numbers, but I suspect most impacts wouldn't successfully deliver life to the planet they impact.   How many would?  1 in 10?  1 in 100?   I'm not sure.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what percentage of comets or meteors have life on them?   This is a tough one, I'm pretty sure life can only form on a planet because you need (I would think), liquid water, atmosphere to maintain surface water and reduce evaporation and some kind of heat to drive the mixing of chemicals, perhaps lightning or undersea volcanic vents.   Life likely cannot be formed in space or in a nebula.   It can be blown into space off a planet with life, but it can't form in space.   That likely reduces the number of meteors or objects in space that even have the possibility of carrying life.    I hate to guess this one, cause who knows, but 1 in 1000?   1 in 10,000?   Most of the meteors, asteroids and comets in space have probably formed without the possibility of carrying life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some simple organic molecules in space, &lt;a href=&quot;http://www.universetoday.com/8098/deep-space-alcohol/&quot; rel=&quot;nofollow&quot;&gt;alcohol&lt;/a&gt; for example, but there's an enormous chasm between organic molecules and even the simplest life.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some planets are very difficult to blow debris off of.  Earth with an atmosphere and an escape velocity of some 25,000 miles per hour needs a very big impact to lose material into space.   Mars, with a very thin atmosphere and lower escape velocity is much easier.  Of some 60,000 recognized meteorites, 124 have been identified from Mars.  (&lt;a href=&quot;http://www2.jpl.nasa.gov/snc/&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;) and Mars is the easiest one.   Venus, with it's thick atmosphere and Mercury, being as close to the sun it is, adding the suns escape velocity to it's own, are much harder.   Material being blown off the gas giants is even less likely, so Mars is the easiest plant to potentially lose material where it could land on Earth or other planets.   Even if we use the Mars Meteor ratio of 400 to 1, which is, I think, too high, it's still a very low percentage of objects in space that potentially came off a planet or moon and carry life with them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 3rd point worth bringing up is repetition.   Lets say, just for fun, that &lt;a href=&quot;https://en.wikipedia.org/wiki/Europa_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Europa&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Enceladus&quot; rel=&quot;nofollow&quot;&gt;Enceladus&lt;/a&gt; both have life in their under the surface oceans and both planets regularly eject jets of life filled water into space, that freezes and carries basic life into orbits around Jupiter / Saturn and some of those life carrying bits of ice make their way on occasion to other planets.    If this happens it probably happens more than once and any bits of life that may come to earth from Europa or Enceladus, if it happens at all, probably happens more than once.   The first introduction of invasive species, even simple ones, can be problematic, but after the first, the planet should have adapted.  That's one of the mathematical quirks.   One Andromeda strain can in theory be a huge problem, but repeated Andromeda strains from the same source, the risk drops with each impact.   What you need, for a dangerous situation to happen is a new introduction of unfamiliar that's never happened before.   That's a pretty specific requirement.   Mars may have once had life, and if it had, life on Mars probably landed on earth many times over 4 plus billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a 4th point, space is full of radiation.  Even viruses will eventually die exposed to the radiation of space.   A space rock, carrying life from another solar-system light-years from earth has a very long journey and a very small target.   Bits of frozen ice or rock probably travel from other solar systems into ours quite rarely over very long periods of time, and, as I pointed out above, most of those probably won't have life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just a kind of &quot;shoot from the hip&quot; sort of answer, but I think you would need a perfect storm of events for your scenario to actually happen.   statistically, I suspect it's a very rare occurrence and not something we need to worry about on a human time frame. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-01-10T07:10:24.157" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13152" PostTypeId="1" CreationDate="2016-01-10T09:13:36.337" Score="-7" ViewCount="618" Body="&lt;p&gt;I am wanting to learn everything there is to learn about the Sun from our solar system to the ends of the universe. &lt;/p&gt;&#xA;" OwnerUserId="10355" LastActivityDate="2016-01-10T19:50:59.590" Title="How do we know so much about the sun since we can't go near it?" Tags="&lt;the-sun&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2016-01-13T08:39:10.910" ContentLicense="CC BY-SA 3.0" />
  <row Id="13153" PostTypeId="2" ParentId="13147" CreationDate="2016-01-10T12:15:17.890" Score="4" Body="&lt;p&gt;In short, no. It is impossible for an infectious agent to come from another planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comets formed in space, and while they may have some carbon based chemicals, nothing as complex as life could form on a comet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some meteoroids have been ejected from planets, and it is at least conceivable that something could survive &quot;launch&quot;. It is not likely that something could live in space, but I suppose a virus might.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I will assume the unlikely: An asteroid carrying an extra-terrestrial virus has landed on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is it going to do? Well no infect humans - viruses are quite specific to their hosts. Most viruses infect a specific type of bacteria. An alien virus would be specific to an alien bacteria. And alien bacteria will be very alien. Most Movies assume that &quot;DNA&quot; is fundamental for life, and talk about &quot;alien DNA merging with human DNA&quot; or the like. However DNA is something that life on this planet has discovered: aliens won't have DNA. They will have something else that carries hereditary information, but it won't be DNA. An alien virus would be utterly incapable of infecting Earth cells, because it's fundamental biochemistry will be utterly dissimilar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a science fiction story and nothing more or less than than.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-01-10T12:15:17.890" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13155" PostTypeId="2" ParentId="13147" CreationDate="2016-01-10T13:42:27.197" Score="9" Body="&lt;p&gt;Even if a virus came to Earth from space, it would likely be harmless to life on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at some biology numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Life on Earth uses 20 amino acids (out of about 400 considered chemically potential alternatives). The median length of a protein on Earth is about 300 amino acids. That's 20^300 combinations. And some proteins are 30,000 amino acids long. There are about 10^31 virus particles in Earth's global biosphere. They have such a large diversity that most of the (few hundred) virus &quot;families&quot; who had their genome mapped, are lone members of their families. Still, we all get along very well! That's because all life on Earth has the same origin and is very closely related to each other. Half of the human genome can also be found also in virus DNA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a virus to be dangerous, it needs to have evolved together with its victim and made evolutionary use out of hurting it. Given the enormous combinatorics involved, it seems most unlikely that an independently evolved virus could break our genetic code. It would take longer than the history of the universe for virus to evolve a crack to our genetic code from scratch. Earthly virus can do so because they originated and evolved here together with us. All life and virus on Earth share DNA that's why we can have relations with them for good and bad (from a humanitarian perspective, while evolutionarily virus are only long-term good for us!) I think our immune system and all kind of microbes would eat any space virus and only become stronger thanks to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alien microbe, something which is in itself alive and active (as opposed to a virus which is dormant until it finds the specific host which it was evolved to infect), could be more dangerous. Instead of attacking unique Earth life organisms directly, they could maybe eat simple molecules which were present also where they once originated and evolved, such as oxygen or simple hydrocarbons, and thereby change the ecology on Earth. But they should be more complex and sensitive to radiation and other space threats, than virus. (Unless we do some how have common origin and evolution even with space virus which thus are fit to interact with us)&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="7982" LastEditDate="2018-01-21T19:35:21.917" LastActivityDate="2018-01-21T19:35:21.917" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13156" PostTypeId="1" CreationDate="2016-01-10T16:35:31.177" Score="-1" ViewCount="176" Body="&lt;p&gt;Q. Why are solar and lunar eclipses not seen from all parts of the Earth at the same time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is because of the revolution of the Earth but I am not sure. I am looking for a more reliable answer.&lt;/p&gt;&#xA;" OwnerUserId="10284" LastEditorUserId="6593" LastEditDate="2020-06-15T00:17:02.530" LastActivityDate="2020-06-15T00:17:02.530" Title="Eclipse -- Solar and Lunar" Tags="&lt;solar-eclipse&gt;&lt;lunar-eclipse&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="13157" PostTypeId="2" ParentId="13156" CreationDate="2016-01-10T17:16:26.997" Score="2" Body="&lt;p&gt;Eclipses are a temporal alignment of the 3 bodies mentioned. Because those bodies are in constant relative motion -- and you have to be in the right location relative to that alignment to experience it (at least for solar eclipses) -- not everyone gets to see them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, to view a solar eclipse, you need to be somewhere &quot;in line&quot; with &lt;strong&gt;the line&lt;/strong&gt; from 1) the center of the earth to 2) the moon to 3) the sun ... Of course, moon's shadow on the earth is not just a point on the earth but a circle, so you don't have to be precisely &quot;in line&quot;, but close (within a few dozen miles, but you can do this computation yourself if you know the diameter of the earth, moon and sun). The circular shadow that crosses the surface of the earth moves in time (as the moon revolves around the earth and as the earth moves about the sun) so different places see it at different times. Finally, when the eclipse is over, it means that the moon's shadow has moved off the earth's surface and back out into space (where it is +99% of the time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For lunar eclipses, you only have to be able to view the moon (at night and during the period of eclipse of course). That means if you can see the moon (about 1/2 the surface of the earth can see the moon at any time) you'll see the eclipse! This is because the entire earth is casting the shadow on the moon and when you can see the moon, by definition, you are on the side of the earth that is casting the shadow and hence in position to see that shadow transit the moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again: all these effects are temporal since the earth/moon system is moving constantly (relative to the sun).&lt;/p&gt;&#xA;" OwnerUserId="9648" LastActivityDate="2016-01-10T17:16:26.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13158" PostTypeId="1" AcceptedAnswerId="13160" CreationDate="2016-01-10T17:34:38.137" Score="3" ViewCount="687" Body="&lt;p&gt;What will happen if just throw two solid objects (for example, solid footballs) parallel in outer space and those two objects are floating and going far away in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is there any possibility in which those two football-size solid objects can collide with each other, or will they will float parallel for an infinite amount of time in the universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm considering gravity also, and I find that other stars's and planets's&#xA;gravitational pull can pull them into their gravitational fields and change their direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if those two objects will never get destroyed by luck? Could those will collide each other?&lt;/p&gt;&#xA;" OwnerUserId="10361" LastEditorUserId="2153" LastEditDate="2016-01-10T17:57:17.017" LastActivityDate="2016-01-10T19:42:33.480" Title="What if we throw two solid objects parallel in space? Do those two objects have any chance to collide with each other?" Tags="&lt;gravity&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13159" PostTypeId="2" ParentId="13152" CreationDate="2016-01-10T19:33:48.847" Score="0" Body="&lt;p&gt;Well we can use for example spectroscopy, this means that we can know what is the surface made up of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is radiation coming from Sun, solar wind it is called. We can measure its speed and learn more about how it was created.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course we can just observe it with normal telescope. After astronomers found out the value of astronomical unit (AU) which happened I think around 18th century they were also able to find out the mass of Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are able to measure incoming radiation. This was very important during Apollo missions, even now lot of people think that there is no way to get through Van Allen radiation belts because of particles coming to us from Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We learned that there are sun-spots in periods of 11 years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also during the measurement of radiation we were able to find out how strong events can solar flares be. Actually if one of the very strong ones hit us, we would have extreme black out (there is page which is predicting space weather but I do not remember its name).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also found out using physics and particle accelerators that elements can fuse together. We learned that there is evolution in stars life and by observation in space we found out that there is going to be white dwarf instead of Sun in several billions of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We found that there is solar corona around Sun which is very hot but nobody is sure where it gets its energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also we can see in our Milkyway how stars are created so we know that there must have been disc of matter which happened to create Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is much more to this but this is what I was able to think of right now. &lt;/p&gt;&#xA;" OwnerUserId="9677" LastActivityDate="2016-01-10T19:33:48.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13160" PostTypeId="2" ParentId="13158" CreationDate="2016-01-10T19:42:33.480" Score="4" Body="&lt;p&gt;It all depends on the direction you &quot;throw them&quot; and the space-time they find themselves in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1 - in a perfectly empty universe (other than your footballs) the two would eventually collide. It doesn't matter what initial velocity you give them. The only thing that matters is there separation and their mass. You can just use Newtonian gravity to compute the two footballs' acceleration toward one another -- and hence the time for them to collide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such an 'empty' universe as above is called a Minkowsky space-time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you put these footballs on a trajectory into our real universe, well we know that space-time is curved by the matter/energy (just energy density in general) that occupies it. Hence a trajectory that starts out parallel will invariably end up &quot;not parallel&quot;. The balls will either collide or diverge depending on the geometry of the space-time they find themselves in. The &quot;geometry of the space-time&quot; is (again) completely dependent on the distribution of &quot;stuff&quot; (matter/energy) in the universe as related to the location of the footballs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, the footballs will follow the geometry of the space-time they find themselves in. That geometry (in our real universe) means they will not remain on parallel paths.&lt;/p&gt;&#xA;" OwnerUserId="9648" LastActivityDate="2016-01-10T19:42:33.480" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13161" PostTypeId="2" ParentId="13152" CreationDate="2016-01-10T19:50:59.590" Score="2" Body="&lt;p&gt;Ok... some of Vojta's info is pointing in the right direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Via &lt;a href=&quot;https://en.wikipedia.org/wiki/Spectroscopy&quot; rel=&quot;nofollow&quot;&gt;spectroscopy&lt;/a&gt; we can get a very good estimate on the chemical composition of the sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the sun can be deduced from Newton's laws (F = G Mm / r^2) and the period of the earth's orbit. Technically, you also need the mass of the earth, but it is so insignificant as compared to the sun that it's negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can measure the amount of energy received by the earth (which is the flux received by the earth). Since we know the earth diameter and the distance to the sun, we can compute how much energy the sun is sending out (and hence generating -- assuming its in equilibrium) each second (3.8 x 10^33 ergs/second or 3.8 × 10^26 watts).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those are some of the basics, but I'm sure if you google &quot;Sun&quot; you'll uncover a wealth of information.&lt;/p&gt;&#xA;" OwnerUserId="9648" LastActivityDate="2016-01-10T19:50:59.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13162" PostTypeId="1" AcceptedAnswerId="18975" CreationDate="2016-01-11T00:13:47.123" Score="0" ViewCount="87" Body="&lt;p&gt;I am a fifth (and last) year undergraduate student in Physics with good level of Mathematics formation and basic Astrophysics formation. I recently read some theories about Solar System formation and evolution and became interested in planetary stability, but I couldn't manage to get a good introduction to the topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if there is a standard book or resource that approach this topic with precision that you could recommend. Any advice will be welcome. &lt;/p&gt;&#xA;" OwnerUserId="8011" LastEditorUserId="6" LastEditDate="2016-05-23T17:53:21.620" LastActivityDate="2016-11-07T12:33:04.650" Title="Resources on planetary stability" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;solar-system&gt;&lt;orbital-migration&gt;&lt;planetary-systems&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13164" PostTypeId="1" CreationDate="2016-01-11T06:51:24.360" Score="2" ViewCount="34" Body="&lt;p&gt;From what I understand stars form in dense clouds of hydrogen and helium molecules. Only so much hydrogen and helium were &quot;created&quot; from the Big Bang or rather formed if you may. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it conceivably possible for the universe to not have any more hydrogen and helium to condense into a molecular cloud and form new stars? Meaning will star formation cease at some point in the far far far away future. &lt;/p&gt;&#xA;" OwnerUserId="10297" LastActivityDate="2016-01-11T06:51:24.360" Title="Star Formation in the future" Tags="&lt;fundamental-astronomy&gt;&lt;star-formation&gt;&lt;observable-universe&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2016-01-12T15:16:26.897" ContentLicense="CC BY-SA 3.0" />
  <row Id="13165" PostTypeId="1" AcceptedAnswerId="16696" CreationDate="2016-01-11T07:01:42.840" Score="9" ViewCount="2320" Body="&lt;p&gt;I'm working on a game concept that does some mild simulation of realistic stellar classes and luminosities. In particular, I'd like to roughly model the general frequencies of the classes and luminosities of the stars in the Milky Way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several sources, including Wikipedia's entry on stellar classifications, show a chart that includes the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_classification#Harvard_spectral_classification&quot;&gt;frequency distribution for spectral classification&lt;/a&gt;: the OBAFGKM categorization. So that's fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm having trouble finding is any frequency distribution chart similar to that one but for the Yerkes luminosity categories: Ia+, Ia, Iab, Ib, II, III, IV, V, sub-dwarf and dwarf. I have a copy of the Hipparcos database, which contains a &quot;Spectral Types&quot; field, but it's highly incoherent text. Still, I could write some code to parse the values in that field to try to get a rough count of luminosity categories in those roughly 116,000 stars... but I'm a little perplexed that no such chart appears to exist already somewhere in Internetland. (Either that or my search-fu is weaker than usual.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone can point me to a chart of the frequency distribution for the luminosity categories noted above, or suggest a reasonably simple way for me to calculate those values myself, I'd appreciate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Out of curiosity, I went ahead and did my own simple parsing of the spectrum fields from the Hipparcos dataset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out of 116472 rows, only 56284 (fewer than half) provided luminosity class data in the Spectrum field. Those 56284 rows broke down this way:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;Ia0     16     0.03%&#xA;Ia     241     0.43%&#xA;Iab    191     0.34%&#xA;Ib     694     1.23%&#xA;I       17     0.03%&#xA;II    1627     2.89%&#xA;III  22026    39.13%&#xA;IV    6418    11.40%&#xA;V    24873    44.19%&#xA;VI      92     0.16%&#xA;VII     89     0.16%&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note: Around 1000+ rows gave an either/or value for luminosity class (e.g., &quot;M1Ib/II&quot;). In these cases, I counted only the first value provided. This probably skewed the results slightly compared to counting both luminosity classes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm still very curious to know whether anyone else has produced or located a similar table of frequencies for the luminosity classes, if only to see how my very trivial analysis compares.&lt;/p&gt;&#xA;" OwnerUserId="10363" LastEditorUserId="10363" LastEditDate="2016-01-12T06:59:42.137" LastActivityDate="2016-10-02T18:18:51.603" Title="What is the frequency distribution for luminosity classes in the Milky Way Galaxy?" Tags="&lt;star&gt;&lt;milky-way&gt;&lt;luminosity&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13166" PostTypeId="2" ParentId="13060" CreationDate="2016-01-11T08:14:34.473" Score="1" Body="&lt;p&gt;Krotanix : nobody really knows the answer to this. We have good evidence that the universe is expanding, and we extrapolate this back to the big bang. Then some people say the universe was once the size of a grapefruit. See for example &lt;a href=&quot;http://www.physicsoftheuniverse.com/topics_bigbang_timeline.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. Since WMAP evidence in 2013 suggested the universe is flat, some people have started saying the &lt;em&gt;observable&lt;/em&gt; universe was once the size of a grapefruit. And that the universe was always infinite. But this is a non-sequitur, and a non-answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes,  it looks as if 13.8 billion years space had a much higher density and started expanding. And it may have had a certain volume, so space may have already existed. But nobody knows for sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Re your follow-on question, nobody has discovered hollowness or emptiness anywhere in the centre of the universe. The expansion is thought to be like that of a &lt;a href=&quot;https://www.google.co.uk/#q=raisin+cake+analogy&quot; rel=&quot;nofollow&quot;&gt;raisin cake&lt;/a&gt; rising in an oven. There's no big hole in the middle.   &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2016-01-11T08:14:34.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13167" PostTypeId="2" ParentId="13073" CreationDate="2016-01-11T10:44:20.830" Score="1" Body="&lt;p&gt;Restricting ourselves to stars that will end their lives as supernovae ($&amp;gt;8$ solar masses) then the answer is no; not all the products of nuclear fusion are stored inside the star until the final explosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a general conceptual picture that as a massive star evolves, that heavier elements are able to fuse more readily closer to the centre where the temperature is higher. This would result in an &quot;onion-like&quot; structure where you might have hydrogen fusing to helium in the outer part of the &lt;em&gt;core&lt;/em&gt; of the star (which is round about the size of the Earth), with a layer inside this where helium is fusing into carbon and oxygen, a layer inside this where neon is being produced etc. At the very centre, just before the core-collapse and supernova explosion, there would be a phase of very rapid silicon burning which produces &quot;iron-peak&quot; elements including iron, nickel, cobalt etc.&#xA;Outside the core there would be a much larger envelope consisting mainly of hydrogen and helium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This view is very simplistic because it ignores two important facets of the evolution of high-mass stars - mixing and mass loss. Massive stars probably lose the majority of their mass during their lives. The more massive the star, the greater percentage of its mass is lost. The mass is lost through a powerful wind that is driven by radiation pressure. There are many examples that are observed to undergo this mass loss including objects like &lt;a href=&quot;https://en.wikipedia.org/wiki/Eta_Carinae&quot; rel=&quot;nofollow&quot;&gt;Eta Carina&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Wolf%E2%80%93Rayet_star&quot; rel=&quot;nofollow&quot;&gt;Wolf-Rayet&lt;/a&gt; stars and &lt;a href=&quot;https://en.wikipedia.org/wiki/Red_supergiant&quot; rel=&quot;nofollow&quot;&gt;Red Supergiants&lt;/a&gt;. Many of these objects have lost their entire outer envelopes and have essentially exposed the products of nuclear burning, or those products are mixed to the surface by convection or other more exotic mixing processes associated with their rotation. These products are then driven into space by their powerful winds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This mechanism will be an important contributor to the carbon, nitrogen and oxygen abundances in the interstellar medium. The winds may also be enhanced in much heavier elements (even heavier than iron, such as Barium and Strontium) that have been produced via the &lt;a href=&quot;https://en.wikipedia.org/wiki/S-process&quot; rel=&quot;nofollow&quot;&gt;s-process&lt;/a&gt;, where pre-existing iron-peak nuclei in the nuclear burning regions are able to capture neutrons. This process can also occur in stars with masses substantially lower than required to produce a supernova.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-01-11T10:44:20.830" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13168" PostTypeId="1" CreationDate="2016-01-11T12:37:31.450" Score="1" ViewCount="71" Body="&lt;p&gt;It is often said, in spite of sci fi often having illustrated the opposite, that stellar mass black holes do not gravitationally attract material in their environment into themselves anymore than a star with similar mass does. But how and why do blue stars of about 8-10 or more Solar masses which turn into black holes, change how frequently and observably they accrete material from their environment when they transform into BH? Maybe because the supernova ejects material, some of which doesn't get far and is re-accreted when the stellar wind ceases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BH's are visible when they have active accretion disks. Do massive O and B stars also have such accretion disks as frequently?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="2365" LastEditDate="2016-01-12T20:41:38.983" LastActivityDate="2016-01-12T20:41:38.983" Title="Do massive OB stars accrete mass the same way as the black holes they turn into do?" Tags="&lt;star&gt;&lt;black-hole&gt;&lt;accretion-discs&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13169" PostTypeId="2" ParentId="13168" CreationDate="2016-01-11T13:36:29.810" Score="3" Body="&lt;p&gt;Many, if not all, stars will, in their birth phase, have an accretion disk, or &lt;em&gt;circumstellar disk&lt;/em&gt;, around them, formed of the material from which the star forms. This disk dissipates in a few million years, both due to the material accreting onto the star, due to material being blown away by radiation pressure, possibly assisted by dust grains, and due to depletion of the material to rocks and planets. Friction in the disk causes it to heat up and be observable in the infrared. After this phase, the star usually won't have an accretion disk. That is, until they die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you suggest, a star that later on goes supernova can accrete matter from its &quot;reverse shock&quot;, i.e. matter that is falling back due to the shockwave hitting the surrounding interstellar medium. More often, though (I think), will the accretion disk be made of material from a companion star: If one star of a binary system turns into a black hole, and then the other star starts to evolve into a red giant, threspassing the Roche limit, it will transfer matter to the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you say, the black hole's mass isn't stronger than the original star's mass, and thus the gravity at a given distance is also not stronger (in fact it's smaller, since it has blown out most of its mass in the explosion). However, due to its much smaller size, the accreting matter can come much closer before being sucked in. Coming much closer means being in a much steeper gravitational potential and reaching much higher speeds, i.e. being heated to much higher temperatures. Hence, black hole accretion disks can be detected in X-rays&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2016-01-11T13:36:29.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13170" PostTypeId="2" ParentId="2470" CreationDate="2016-01-11T16:19:55.970" Score="-2" Body="&lt;p&gt;When something explodes in a concentric manner it would require two opposing planes of like polarity to &quot;sandwich&quot; the exploding matter into the shape of an expanding flat plane. So what is your theory on what these sandwiching forces could be?&lt;/p&gt;&#xA;" OwnerUserId="10369" LastActivityDate="2016-01-11T16:19:55.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13171" PostTypeId="2" ParentId="13149" CreationDate="2016-01-11T19:08:38.017" Score="4" Body="&lt;p&gt;As addition:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What Conrad said about ring systems is true, we have a possible first detection, but nowhere near detections to make any claim about such phenomena being normal. Also our knowledge on that won't change much in the coming years. With the upcoming photometric missions on transits like NASA's &lt;a href=&quot;https://en.wikipedia.org/wiki/Transiting_Exoplanet_Survey_Satellite&quot; rel=&quot;nofollow&quot;&gt;TESS&lt;/a&gt; or ESA's &lt;a href=&quot;https://en.wikipedia.org/wiki/PLATO_%28spacecraft%29&quot; rel=&quot;nofollow&quot;&gt;PLATO&lt;/a&gt; we'll probably be only able to detect this type of bigger-than-Saturn ring systems. This is simply due to the fact that small ring systems won't give a strong enough transit signal to be detected.&lt;br&gt;&#xA;So in total, we won't know about the normality of Saturns rings for quite some time.&lt;/li&gt;&#xA;&lt;li&gt;Terrestrial Moons: There are ideas out there how to detect exomoons, but those are usually also dependent on photometric measurments. Transit-timing variations may lead to clues of exomoons. Also very big moons could lead to detectable signals in the photometry, but radial velocity detection of exomoons is out of the question at the moment.&lt;br&gt;&#xA;So in total: Exomoon distributions are totally unknown at the moment and will, for the same reasons as planetary ring distributions remain unknown for some time.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As much as this sounds a bit depressive concerning the search for rings and moons around exoplanets, we have to understand that this is due to the fact that at the moment those questions are not one of the main science drivers behind currently upcoming exoplanet missions.&lt;br&gt;&#xA;We're still trying to understand exoplanet populations themselves thoroughly, before moving on to the much tougher challenges of detecting the much fainter signals of moons and rings.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, I'm convinced that the moon/ring question will some time in the future move much more into scientific focus. This is simply because, at least for gas giants, moons and rings should be 'fine-probing' theories of planet formation. This happens in the sense that when gas giants are growing, at some point they will have a &lt;em&gt;subdisc&lt;/em&gt; of gas and dust forming inside their own Hill-sphere that will eventually do something interesting. Comparing what those subdiscs do to theories of moon and ring formation will teach us how those processes work in detail.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also +1 for Asimov because of non-professional personal bias.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditDate="2016-01-14T13:47:05.333" LastActivityDate="2016-01-14T13:47:05.333" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13172" PostTypeId="1" CreationDate="2016-01-11T19:27:15.057" Score="1" ViewCount="135" Body="&lt;p&gt;We are moving like 100,000 mph. And why do we not fall off going that fast&lt;/p&gt;&#xA;" OwnerUserId="10355" LastActivityDate="2016-01-11T20:26:36.743" Title="Why is it that we can&quot;t feel how fast the Earth is moving?" Tags="&lt;earth&gt;" AnswerCount="3" CommentCount="1" ClosedDate="2016-01-14T06:33:31.933" ContentLicense="CC BY-SA 3.0" />
  <row Id="13173" PostTypeId="2" ParentId="13172" CreationDate="2016-01-11T19:31:34.237" Score="2" Body="&lt;p&gt;You 'feel' only acceleration, not velocity. Acceleration is a change in velocity per time. The forces on the surface of Earth on a human are in equilibrium, and therefore the accelerations are too, thus we don't fall off.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2016-01-11T19:31:34.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13174" PostTypeId="2" ParentId="13172" CreationDate="2016-01-11T20:16:14.860" Score="0" Body="&lt;p&gt;Not only Earth is moving but also you with the rest like atmosphere, so you are in closed system. The edge of atmosphere does not fly off because it has nothing to rub against, it just stays in same speed. When you put your hand from the car and you feel the force against it, it is the force which you must make to push the air from your way. If the air would be moving at same velocity as you, your hand would not feel this and you would be able to hold outside of car even at 300km (do not do that, it is dangerous).&lt;/p&gt;&#xA;" OwnerUserId="9677" LastActivityDate="2016-01-11T20:16:14.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13175" PostTypeId="2" ParentId="13172" CreationDate="2016-01-11T20:26:36.743" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Because the rotation of the earth is very smooth and doesn't change, the centripetal acceleration we feel is very nearly constant. This means that the (small) centrifugal force from the rotation gets added to gravity to make up the &quot;background force&quot; we don't notice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Taken from &lt;a href=&quot;https://physics.stackexchange.com/a/12489&quot;&gt;Physics SE&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10298" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-01-11T20:26:36.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13176" PostTypeId="2" ParentId="13168" CreationDate="2016-01-11T20:53:14.920" Score="0" Body="&lt;p&gt;The size of the accretion disc in a black-hole binary system is &lt;a href=&quot;http://www.scholarpedia.org/article/Accretion_discs&quot; rel=&quot;nofollow&quot;&gt;calculated&lt;/a&gt; to be less than 1000000km. This is less than to the size of large stars (eg Spica at about 10000000km diameter). In other words, the accretion disc for a large star would be smaller than the star itself. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-01-11T20:53:14.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13177" PostTypeId="1" AcceptedAnswerId="14556" CreationDate="2016-01-12T03:49:38.003" Score="0" ViewCount="7548" Body="&lt;p&gt;Consider the non-inertial frame that is at rest with respect to the Earth's revolution around the Sun.  (Ignore the Earth's rotation around its own axis.)  My question is, what is the shape of the Sun's orbit around the Earth in this reference frame?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we assume the Earth moves in a circular orbit around the Sun, then the Sun's orbit around the Earth will also be circular. (Just use geometric the definition of a circle - the set of all points a fixed distance from a given point.). Specifically it's the great circle formed by the intersection of the ecliptic plane with the celestial sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the Earth does not move in a circular orbit - Kepler's first law states that the Earth moves in an ellipse around the Sun, with the Sun at one of the foci of the ellipse.  So if we consider the Earth's elliptical orbit around the Sun, what is the shape of the Sun's orbit around the Earth.  I doubt it's an ellipse, so would be it a more complicated-looking curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I'm not interested in gravitational influences from the moon and other planets - this is pretty much a purely mathematical question: if we assume the Earth moves in an ellipse, what would be the shape produced?&lt;/p&gt;&#xA;" OwnerUserId="10376" LastActivityDate="2016-04-15T20:32:59.990" Title="What is the shape of the Sun's orbit around the Earth taking into account elliptical orbits?" Tags="&lt;orbit&gt;&lt;the-sun&gt;&lt;earth&gt;&lt;fundamental-astronomy&gt;&lt;johannes-kepler&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13178" PostTypeId="1" AcceptedAnswerId="13201" CreationDate="2016-01-12T04:05:51.820" Score="6" ViewCount="1842" Body="&lt;p&gt;Are we still debating the relationship between Jupiter and solar cycle? If not, what is the common or mainstream opinion about this topic? Or in other words, have we disproved the solar cycle is related to the &quot;year&quot; of Jupiter?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastActivityDate="2019-04-28T13:54:46.507" Title="Are we sure solar cycle is not related to the orbit of Jupiter?" Tags="&lt;the-sun&gt;&lt;jupiter&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13179" PostTypeId="1" CreationDate="2016-01-12T05:43:02.287" Score="4" ViewCount="202" Body="&lt;p&gt;Is the universe heading towards a Big Freeze, a Big Rip, a Big Crunch or a Big Bounce? Or is it part of an infinitely recurring cyclic model?&lt;/p&gt;&#xA;" OwnerUserId="1094" LastActivityDate="2016-01-12T22:10:06.757" Title="What is the future of our universe?" Tags="&lt;solar-system&gt;&lt;the-sun&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13180" PostTypeId="2" ParentId="13179" CreationDate="2016-01-12T09:45:09.413" Score="2" Body="&lt;p&gt;Your guess is as good as any cosmologist's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't know how it came into being, not with any confidence.  There are theories, but nothing definitive has emerged to explain where it all came from.  This may be hotly disputed by proponents of each theory, but we're not there yet, IMO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're not sure yet about what all the relevant forces are ( e.g. dark matter, dark energy ) so any theory about how those affect possible futures is simply a complicated guess at this point.  This alone, IMO, makes conjectures about the universe's future pointless, IMO.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2016-01-12T09:45:09.413" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13181" PostTypeId="2" ParentId="13178" CreationDate="2016-01-12T10:21:06.653" Score="8" Body="&lt;p&gt;I think for people who want to believe the Jupiter link to solar cycles it will be impossible to disprove.  They'll cling to it forever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think in a practical sense the &quot;link&quot; is little more than numerology.  The force Jupiter can exert on the Sun's atmosphere and internal structure is a minute fraction of the magnetic forces at work generated by the Sun itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple model helps.  Place two magnets a foot apart.  Now place a magnet a thousand feet away from both.  The effect of the distant magnet is one millionth that of the two close together.  That gives you a feel for the relationship between magnetic forces within the Sun's atmosphere and Jupiter's distant influence.  And gravity is an even weaker force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it the current daring of models is the Double Dynamo, but certainly the dynamo type models ( all relating to electromagnetic fields generated by the Sun ) are the main contenders.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With people who propose the Jupiter link ( often extended to include Saturn ), there is a tendency to do a lot of algebra to find ways to relate various periods without actually justifying the supposed link on the basis of forces at work ( i.e. doing a simulation or a model based on real physics ).  This is why I call it numerology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dynamo models actually are based on real physics and make predictions, both past and future, which can be tested.  This also means the details of these theories can be refined in a proper scientific manner to tease out the truth.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2016-01-12T10:21:06.653" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13183" PostTypeId="1" CreationDate="2016-01-12T11:00:54.580" Score="1" ViewCount="79" Body="&lt;p&gt;Why Is it said that if someone is being sucked inside a black hole then he/she may see that time is running slowly?&lt;/p&gt;&#xA;" OwnerUserId="10379" LastActivityDate="2016-01-14T20:12:23.863" Title="Black holes and time relation" Tags="&lt;gravity&gt;&lt;general-relativity&gt;&lt;fundamental-astronomy&gt;&lt;supermassive-black-hole&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13185" PostTypeId="1" CreationDate="2016-01-12T14:41:02.367" Score="2" ViewCount="1393" Body="&lt;p&gt;If the Earth rotates on its own axis, then what do we mean by sunrise and sunset?&lt;/p&gt;&#xA;" OwnerUserId="10379" LastActivityDate="2016-01-17T13:20:36.263" Title="What do we mean by Sunrise and Sunset?" Tags="&lt;the-sun&gt;&lt;earth&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13186" PostTypeId="2" ParentId="13185" CreationDate="2016-01-12T14:53:23.277" Score="1" Body="&lt;p&gt;Sunrise is the moment when the point you are located on, on Earth, gets to receive its part of the sunlight (due to the Earth's rotation you spoke about). And sunset is the moment when the point you are located on, on Earth, has made its turn and faces away from the Sun, then preventing you from receiving its light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the Earth will have made another half of its rotation, you will get to see another sunrise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth makes a full rotation in a 24h cycle. This is why you get a sunrise and a sunset everyday :)&lt;/p&gt;&#xA;" OwnerUserId="9679" LastActivityDate="2016-01-12T14:53:23.277" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13187" PostTypeId="2" ParentId="13185" CreationDate="2016-01-12T19:00:32.827" Score="3" Body="&lt;p&gt;Yes the earth rotates around its axis,  which gives rise to the motion of the sun as we see it.&lt;br&gt;&#xA;To understand sun rise you'd have to start with a reference frame. For us,  the earth is considered to be the inertial frame for any apparent motion in the celestial sphere, it simplifies the computations. Thus any motion of celestial bodies is defined with respect to that frame.&lt;br&gt;&#xA;So in that context Sunrise would be the time when for a given latitude, when the sun first appears above the local horizon, and sun set would become the time when the sun disappears below that local horizon.  &lt;/p&gt;&#xA;" OwnerUserId="10177" LastEditorUserId="10177" LastEditDate="2016-01-17T13:20:36.263" LastActivityDate="2016-01-17T13:20:36.263" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13189" PostTypeId="1" AcceptedAnswerId="13193" CreationDate="2016-01-12T20:45:20.813" Score="11" ViewCount="1062" Body="&lt;p&gt;Just more of a conceptual question on the mutual inclusivity of the cosmological principle. That is to say, I was wondering if it were possible to have a Universe that were isotropic but NOT homogeneous OR a Universe that were homogeneous but NOT isotropic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My spidey sense is telling me that it is possible that a Universe could be istropic but not homoegenous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely, by default, if a Universe is homogenous then it follows that it has to be isotropic?   &lt;/p&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2016-01-21T09:37:21.410" Title="On the cosmological principle" Tags="&lt;cosmology&gt;&lt;cosmological-principle&gt;&lt;homogeneity&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13190" PostTypeId="2" ParentId="13189" CreationDate="2016-01-12T21:24:50.987" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I was wondering if it were possible to have a Universe that were isotropic but NOT homogeneous OR a Universe that were homogeneous but NOT isotropic. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most people will be happy with the &lt;a href=&quot;http://www.universeadventure.org/big_bang/expand-balance.htm&quot; rel=&quot;noreferrer&quot;&gt;UC Berkeley&lt;/a&gt; definition that says homogeneous means &lt;em&gt;&quot;looks the same at every location&quot;&lt;/em&gt; and isotropic means &lt;em&gt;&quot;looks the same in every direction&quot;&lt;/em&gt;. And some will know that as per &lt;a href=&quot;http://www.astro.ucla.edu/~wright/cosmo_01.htm&quot; rel=&quot;noreferrer&quot;&gt;Ned Wright's article&lt;/a&gt;, these attributes aren't quite the same: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KnPzb.gif&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He says &lt;em&gt;&quot;the figure above shows a homogeneous but not isotropic pattern on the left and an isotropic but not homogeneous pattern on the right&lt;/em&gt;&quot;.  However as far as I know pictures like this just don't apply to our universe scattered with galaxies.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My spidey sense is telling me that it is possible that a Universe could be isotropic but not homogeneous.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My spidey sense is telling me some guy 46 billion light years away might say the universe is &lt;em&gt;neither&lt;/em&gt; isotropic nor homogeneous. Because when he looks up, half the night sky is black or something. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Surely, by default, if a Universe is homogenous then it follows that it has to be isotropic? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I agree with the gist of that. IMHO if an observer sees a homogeneous universe, he sees an isotropic universe too. Yes, one can find &lt;a href=&quot;https://www.physicsforums.com/threads/isotropic-vs-homogeneous.173731/&quot; rel=&quot;noreferrer&quot;&gt;hypothetical scenarios&lt;/a&gt; wherein the universe is homogeneous but not isotropic. But they're only hypothetical. And let's not forget that it's &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann%E2%80%93Lema%C3%AEtre%E2%80%93Robertson%E2%80%93Walker_metric#General_metric&quot; rel=&quot;noreferrer&quot;&gt;only an assumption&lt;/a&gt;. If you lived in a forest would you assume that the world was covered in trees? Which look the same at every location and in every direction? It isn't a particularly scientific  assumption. For all you know some guy lives near the edge of the forest. I think you're better off saying &lt;em&gt;we just don't know&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2016-01-12T21:24:50.987" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="13191" PostTypeId="2" ParentId="13179" CreationDate="2016-01-12T22:10:06.757" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the future of our universe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Like StephenG said, nobody knows for sure. But we do have confidence that the universe is expanding, and we also have confidence that &lt;a href=&quot;https://en.wikipedia.org/wiki/Accelerating_expansion_of_the_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;the expansion is speeding up&lt;/a&gt;. So extrapolating from that, the future looks &lt;a href=&quot;https://en.wikipedia.org/wiki/Heat_death_of_the_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;cold and lonely and bleak&lt;/a&gt;. A bit like life for the older generation! &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the universe heading towards a Big Freeze, a Big Rip, a Big Crunch or a Big Bounce? Or is it part of an infinitely recurring cyclic model?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd say the Big Crunch and the Big Bounce are out. The universe didn't contract when it was small and dense. Instead it expanded, and that expansion is increasing. So it looks like we're in for a Big Freeze. However I wouldn't rule out a Big Rip of sorts. Have a look at page 5 of &lt;a href=&quot;http://arxiv.org/abs/0912.2678&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt; where Milgrom mentions the strength of space. Then think of the balloon analogy, but make it a bubble-gum balloon, in vacuum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SXsTK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;sup&gt;Image courtesy of &lt;a href=&quot;http://oneminuteastronomer.com/6949/where-is-the-center-of-the-universe/&quot; rel=&quot;nofollow noreferrer&quot;&gt;the one-minute astronomer&lt;/a&gt;&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The skin gets thinner and the balloon expands, then the skin gets even weaker, so the balloon expands even faster, and so on. Bubble-gum bubbles sometimes end badly, and there's something about  &lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2013/06/01/prince_rupert_s_drops_glass_droplets_exploding.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; by Phil Plait that strikes a chord.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Or is it part of an infinitely recurring cyclic model?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know. I have no evidence to suggest that there's any kind of recycling going on, and I can't think of mechanism by which this might occur. I have of course read about &lt;a href=&quot;https://en.wikipedia.org/wiki/Conformal_cyclic_cosmology&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;conformal cyclic cosmology&quot;&lt;/a&gt;, but I just don't buy it, along with &lt;a href=&quot;http://casa.colorado.edu/~ajsh/bhtalk_07/penrose_rni.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;other stuff&lt;/a&gt; from Penrose.    &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2016-01-12T22:10:06.757" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13192" PostTypeId="1" CreationDate="2016-01-12T22:56:46.417" Score="4" ViewCount="423" Body="&lt;p&gt;There's a prominent &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrogen_line&quot; rel=&quot;nofollow&quot;&gt;spectral line&lt;/a&gt; at ~1420 MHz caused by hydrogen.  From what I can tell, this is the most notable electromagnetic emission from hydrogen in radio astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an equivalent &lt;strong&gt;helium line&lt;/strong&gt; -- a single, prominent radio emission of helium in space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And are there equivalents for other elements too?)&lt;/p&gt;&#xA;" OwnerUserId="10384" LastActivityDate="2016-03-14T08:17:12.433" Title="Is there a single, most prominent helium line?" Tags="&lt;radio-astronomy&gt;&lt;helium&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13193" PostTypeId="2" ParentId="13189" CreationDate="2016-01-13T09:17:50.747" Score="6" Body="&lt;p&gt;Neither of the two cases are completely inconceivable:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;A homogeneous, anisotropic universe&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A universe with galaxies spread evenly all over, but all spinning in the same direction. This universe would look the same no matter where you lived, but have a net angular momentum, so looking in one direction you'd see all galaxies spinning along your line of sight, and in another direction, you'd see them spinning perpendicular to this direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is a universe that had been permeated by density waves in one direction. In this direction, you'd see the density of galaxies alternating between high and low, and perpendicular hereto you'd see a constant density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sXHCv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sXHCv.png&quot; alt=&quot;homo-noniso&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yesterday's papers on &lt;a href=&quot;http://arxiv.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;arXiv&lt;/a&gt; included a paper (&lt;a href=&quot;http://arxiv.org/abs/1601.04965v1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schucker 2016&lt;/a&gt;) that discusses the the possibility that we might live in another type of homogeneous, anisotropic universe, namely one in which the observed expansion rate depends upon the direction in which you look. This is called a &quot;Bianchi I universe&quot;, and isn't just a hypothetical curiosity (although the results of this paper is statisically non-significant). See also &lt;a href=&quot;https://astronomy.stackexchange.com/users/5301/jonestheastronomer&quot;&gt;@JonesTheAstronomer&lt;/a&gt;'s answer.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;An inhomogeneous, isotropic universe&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As John Rennie has taught us, &lt;a href=&quot;https://physics.stackexchange.com/questions/136860/did-the-big-bang-happen-at-a-point&quot;&gt;Big Bang didn't happen at a point&lt;/a&gt;. However, &lt;em&gt;if&lt;/em&gt; it did, &lt;em&gt;and&lt;/em&gt; we happened to live in the central region, we could observe the same in all directions, but see a gradually thinnening universe, or maybe increasing to some point and then decreasing, depending on exactly how this exsplosion came about. This scenario &lt;em&gt;would&lt;/em&gt; however imply that we inhabit a special place in the universe, which would make Kopernikus sad. If a universe is isotropic from more than one location, is must also be homogeneous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sjDXh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sjDXh.png&quot; alt=&quot;inhomo-iso&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-01-21T09:37:21.410" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13194" PostTypeId="2" ParentId="13192" CreationDate="2016-01-13T09:48:37.753" Score="3" Body="&lt;p&gt;&lt;strong&gt;I am unaware of there being a single Helium line that is as prominent or useful as the 21cm line is for Hydrogen.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/h21.html&quot; rel=&quot;nofollow&quot;&gt;21 cm line&lt;/a&gt; is due to a hyperfine transition in the &lt;em&gt;1s&lt;/em&gt; ground state of hydrogen. It is notable due to its strength, and also its usefulness for mapping neutral hydrogen in the universe, which is impossible in the optical since it is absorbed by neutral Hydrogen. It was in fact the first such spectral line observed in the radio part of the spectrum, and is so fundamental that the Pioneer and Voyager probes used the length of the transition as a time and distance unit on the plaques they carry, so that potential extraterrestrials could decode the messages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many more spectral lines in the radio from heavier elements, and a whole field dedicated to observing and understanding them: &lt;a href=&quot;http://encyclopedia2.thefreedictionary.com/Radio-Frequency+Spectroscopy&quot; rel=&quot;nofollow&quot;&gt;radio frequency spectroscopy&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Helium in particular, hyperfine lines can &lt;a href=&quot;http://wwwmpa.mpa-garching.mpg.de/~reionize/Talks/Thursday/Bagla.pdf&quot; rel=&quot;nofollow&quot;&gt;also be observed&lt;/a&gt;; the hyperfine transition of the $^3$He isotope has a wavelength ~4mm. The same isotope also emits a recombination line, where an electron recombines with an ion and falls to lower energy levels, emitting a photon in the process. The wavelengths of such recombination lines for Hydrogen like atoms (singly ionised in the case of Helium, so that it has a single electron) can be estimated using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Rydberg_formula&quot; rel=&quot;nofollow&quot;&gt;Rydberg Formula&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nu = R_{m}c\left[\frac{1}{n^2}-\frac{1}{(n+\Delta n)^2}\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R_m$ is the reduced mass Rydberg constant, $n$ is the final state and $\Delta n$ is the change in state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Estimating the strengths of such lines is a little more complicated, but there are details &lt;a href=&quot;http://www.cv.nrao.edu/course/astr534/Recombination.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="941" LastEditorUserId="941" LastEditDate="2016-03-14T08:17:12.433" LastActivityDate="2016-03-14T08:17:12.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13195" PostTypeId="1" AcceptedAnswerId="13199" CreationDate="2016-01-13T10:43:02.650" Score="2" ViewCount="225" Body="&lt;p&gt;I know that Pluto and Charon have been found to circle round each other, hence the reason Pluto is no longer a planet I believe, which would mean that the centre of rotation is an arbitrary (!) point somewhere between the two of them. Where is it please? Can it be calculated?&lt;/p&gt;&#xA;" OwnerUserId="7629" LastActivityDate="2016-01-13T13:01:56.163" Title="What is the centre of rotation of Pluto?" Tags="&lt;orbit&gt;&lt;pluto&gt;&lt;charon&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13196" PostTypeId="2" ParentId="13195" CreationDate="2016-01-13T11:59:12.970" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I know that Pluto and Charon have been found to circle round each other, hence the reason Pluto is no longer a planet &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is not correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is considered a dwarf planet because of a technical definition employed by the IAU in deciding what is a planet and is not.  This technical definition relates to an object's ability to &quot;sweep&quot; it's orbit clear of asteroids and other small debris.  The decision has nothing to do with Charon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The centre of rotation is calculated using Newtonian physics.  I think what you want is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Center_of_mass&quot; rel=&quot;nofollow&quot;&gt;Center Of Mass&lt;/a&gt; and related topics.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2016-01-13T11:59:12.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13197" PostTypeId="1" CreationDate="2016-01-13T12:05:30.300" Score="1" ViewCount="148" Body="&lt;p&gt;why doesn't a solar eclipse occur in every new moon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know one reason is the inclination of Earth's orbital plane. But is there another?&lt;/p&gt;&#xA;" OwnerUserId="10284" LastEditorUserId="2365" LastEditDate="2016-01-15T17:32:19.093" LastActivityDate="2016-03-31T16:30:36.233" Title="Reasoning questions on eclipses" Tags="&lt;solar-eclipse&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13198" PostTypeId="2" ParentId="13195" CreationDate="2016-01-13T12:49:48.193" Score="0" Body="&lt;p&gt;As StephenG stated, this is not the reason why Pluto is no longer a planet.&lt;/p&gt;&#xA;&lt;p&gt;The center of rotation of the two bodies is called the barycenter. It can be calculated with accuracy, knowing the mass of the two bodies and the distance separating them. See :&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aGsjd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aGsjd.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The little ring around Pluto in the center describes its path. The barycenter of this binary system is on the opposite side of the ring, starting from Pluto.&lt;/p&gt;&#xA;" OwnerUserId="9679" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-01-13T12:49:48.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13199" PostTypeId="2" ParentId="13195" CreationDate="2016-01-13T13:01:56.163" Score="0" Body="&lt;p&gt;It is listed in &lt;a href=&quot;https://en.wikipedia.org/wiki/Barycenter#Primary.E2.80.93secondary_examples&quot; rel=&quot;nofollow&quot;&gt;this wikipedia page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pluto's radius is listed there as 1,150 km; the radius of the barycenter is 2,110 km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the barycenter is 960 km above Pluto's surface!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note the radius of Pluto quoted differs to its main &lt;a href=&quot;https://en.wikipedia.org/wiki/Pluto&quot; rel=&quot;nofollow&quot;&gt;wikipedia page&lt;/a&gt;; that one says around 1,187 km)&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2016-01-13T13:01:56.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13201" PostTypeId="2" ParentId="13178" CreationDate="2016-01-13T13:17:10.003" Score="4" Body="&lt;p&gt;The barycenter of the Solar System is the effect which should be considered physically. Although dominated by Jupiter other planets have substantial effect too, moving the Sun, with 99+% of all mass in the Solar System, around by entire Solar radii (700,000 km). &lt;a href=&quot;http://homepages.wmich.edu/~korista/solarsystem_barycenter.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Illustrated here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diagram below I happened to stumble upon &lt;a href=&quot;https://tallbloke.wordpress.com/2011/01/14/gerry-pease-barycentric-orbital-periods/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. It shows the distance of the barycenter of the Solar System from the center of the Sun over time, with years on the circle. For some reason it spans from year 1773 to 1851. The peaks in the clover curve are regularly separated by about a sunspot cycle, or rather a bit longer. I suppose that the pattern of 2 big and 1 small sequential bulbs there reflects that Jupiter's and Saturn's orbital periods are about 3:1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As StephenG says, that is just a numerological hint or coincidence, not an explanation. Another example is that the rotational period of the Sun is roughly similar to the orbital period of the Moon around Earth, with a similar error, but no one imagines any connection there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is however known that hot Jupiters, giant planets near stars, cause large starspots which rotate along with the planet's orbit. And the relatively tiny moons of Jupiter cause visible spots in the planetary aurora. Actually, the Sun causes an auroral &quot;Jupiter spot&quot;! So there are physics around which under certain conditions create sunspot-like effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jtKoA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jtKoA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diagram: &lt;em&gt;Barycenter distance from the Sun's center over time&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vzWHt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vzWHt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hubble image:&lt;em&gt;The aurora on a pole of Jupiter in UV&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-01-13T13:17:10.003" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13202" PostTypeId="1" CreationDate="2016-01-13T17:38:55.410" Score="8" ViewCount="70" Body="&lt;p&gt;Atmospheric turbulence is known to scatter photons in a quasi-random way along their path throughout the atmosphere, resulting in lower &lt;strong&gt;imaging&lt;/strong&gt; resolution than would have been anticipated by instrument-only considerations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been thinking whether the same effects can play a relevant role in limiting the sensitivites for photometry in &lt;strong&gt;transits&lt;/strong&gt; or for spectrometry in &lt;strong&gt;radial velocity&lt;/strong&gt; measurments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thoughts so far:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Transits: as I'm no observer I don't know if atmospheric turbulence is actually strong enough to scatter source photons out of the line-of-sight, rendering them undetected. This would fiddle with the signal-to-noise ratio per measurment and would let it fluctuate over time.&lt;/li&gt;&#xA;&lt;li&gt;Radial velocity: Turbulence should be able to influence a spectral measurment from the ground, if the induced turbulent broadening is significant compared to the line width that can be resolved with the instrument considered. Taking the turbulence-induced doppler shift $\Delta v/c$ as $10cm/s /c \sim 10^{-7}$ (I assumed turbulent eddy-velocities to be comparable to typical winds) as typical for Earth's atmosphere, this should be  insignificant even for a high-resolution spectrograph like HARPS that has $\lambda/\Delta \lambda \sim 10^5$.&lt;br&gt;&#xA;However smaller eddies rotate faster, they could thus reach the detectability range when $\Delta v/c \sim 10^{-5}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here my expertise in this topic ends, and I'd hope for someone from this community to illuminate the points above. Also googling usually only points to the benefits in direct imaging. &#xA;&lt;strong&gt;Bonus question&lt;/strong&gt;: Would adaptive optics always help to remedy any issues that might arise?&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditDate="2016-01-13T17:44:57.763" LastActivityDate="2018-09-05T09:37:45.827" Title="Is atmospheric turbulence irrelevant for ExoPlanetary transits and radial velocity measurments?" Tags="&lt;planet&gt;&lt;exoplanet&gt;&lt;observational-astronomy&gt;&lt;turbulence&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13203" PostTypeId="2" ParentId="13197" CreationDate="2016-01-13T19:11:47.353" Score="1" Body="&lt;p&gt;Let me put it this way.  If the Moon's orbital plane where exactly aligned with the Earth's orbital plane (which, by the way, includes the sun, not surprisingly) then there would be a solar eclipse on Earth during each New Moon.  But, in fact, the Moon's orbital plane moves about by many degrees in a very complicated fashion.  Figuring out why this was and how the plane moves kept Isaac Newton busy for a few years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both the Moon and the Sun subtend about 1/2 of a degree on the sky.  The distance to the Moon also changes considerably, so its angular extent varies around this 1/2 degree. So, to get a total solar eclipse, they need to be aligned to a very high degree and the Moon needs to not be near the far edge of its distance range.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2016-03-31T16:30:36.233" LastActivityDate="2016-03-31T16:30:36.233" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13204" PostTypeId="1" AcceptedAnswerId="13421" CreationDate="2016-01-13T19:33:28.733" Score="8" ViewCount="775" Body="&lt;p&gt;Recently I've become more interested in the topic of the stability of planetary systems. I have been reading about it and it seems that orbital resonances play an important role in the stability of Solar System (as well as in the moons of Jupiter and other planets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I first thought that the orbital resonance is somehow more stable and thus we have several cases in the Solar System.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as I kept reading, I found out that in the asteroid belt there were some gaps precisely where the resonances happen, so the resonances are actually instable for asteroids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then thought that some resonances are stable while others are instable, but some of the resonances that makes gaps in the asteroid belt are actually present in the Solar System between planets, so I am completely lost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why resonances are sometimes stable and sometimes instable? What am I missing?&#xA;Maybe I am misunderstanding something because it doesn't make sense to me. Any help will be welcome.&lt;/p&gt;&#xA;" OwnerUserId="8011" LastActivityDate="2016-01-29T16:53:48.117" Title="Stability of orbital resonance" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;n-body-simulations&gt;&lt;orbital-resonance&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13205" PostTypeId="1" CreationDate="2016-01-13T22:47:16.337" Score="1" ViewCount="423" Body="&lt;p&gt;How can someone obtain such vector, for stars in the Milky Way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Input should be star data (from a catalog, eg: Hipparcos) and Julian Date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Output should be a Vector Position (in pc) in relation to the galactic center.&lt;/p&gt;&#xA;" OwnerUserId="2790" LastActivityDate="2016-01-15T02:38:02.897" Title="Obtaining stars vector position in relation to the center of the galaxy at a given date" Tags="&lt;star&gt;&lt;galaxy&gt;&lt;coordinate&gt;&lt;positional-astronomy&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13206" PostTypeId="2" ParentId="13185" CreationDate="2016-01-14T07:59:55.417" Score="0" Body="&lt;p&gt;This is only one way Possible due to Rotation of Earth.## Rotation Of Earth  ##&#xA;The time of sunset varies throughout the year, and is determined by the viewer's position on Earth, specified by longitude and latitude, and elevation. &#xA;During winter and spring, the days get longer and sunsets occur later every day until the day of the latest sunset, which occurs after the summer solstice. &#xA;In the Northern Hemisphere, the latest sunset occurs late in June or in early July, but not on the summer solstice of June 21.&#xA; &lt;strong&gt;This date depends on the viewer's latitude&lt;/strong&gt; (connected with the Earth's slower movement around the aphelion around July 4)&lt;/p&gt;&#xA;" OwnerUserId="10398" LastActivityDate="2016-01-14T07:59:55.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13207" PostTypeId="1" CreationDate="2016-01-14T09:23:38.220" Score="3" ViewCount="1264" Body="&lt;p&gt;Whilst I understand why stars move across our skies,(Earth rotation) I struggle to understand why they do not appear to move relative to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain:  Take the best know constellation the Plough.  It is a familiar site to most people.  But it has been there apparently un-changed for years.  If we (I mean earth, moon, stars and everything else ) are hurtling through space, then surely changes should be apparent unless we are all travelling in the same direction at exactly the same speed.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="2365" LastEditDate="2016-01-14T18:26:40.037" LastActivityDate="2016-01-15T10:09:58.770" Title="Why do stars seem not to move relative to each other?" Tags="&lt;positional-astronomy&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13208" PostTypeId="2" ParentId="13207" CreationDate="2016-01-14T09:36:59.063" Score="5" Body="&lt;p&gt;They do move - just far too slowly for you to detect by eye even over several human lifetimes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Space is big. Really big. You just won't believe how vastly, hugely, mind-bogglingly big it is. I mean, you may think it's a long way down the road to the chemist, but that's just peanuts to space&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Douglas Adams, &quot;The Hitch-hikers Guide to the Galaxy&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even the closest stars are a very, very, very long way away so their apparent movement relative to each other is going to be very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see the same effect when looking out of a moving vehicle through the side windows. You see the objects closest to you rushing past, but objects on the horizon appear to move much more slowly. Now, scale that up by many orders of magnitude to interstellar distances and you'll see why the stars don't appear to move relative to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's software that simulates the night sky and you can run time backwards and forwards - if you wind it far enough you'll see the constellations change.&lt;/p&gt;&#xA;" OwnerUserId="111" LastEditorUserId="111" LastEditDate="2016-01-14T14:10:41.763" LastActivityDate="2016-01-14T14:10:41.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13209" PostTypeId="1" CreationDate="2016-01-14T11:06:40.857" Score="2" ViewCount="162" Body="&lt;p&gt;Nothing can escape a black hole , not even light. But what if anything approaches a black hole from behind..... Correct me if I'm wrong!&#xA;Do black holes have front and back?&#xA;If yes then does they suck everything from both sides??&lt;/p&gt;&#xA;" OwnerUserId="10379" LastActivityDate="2016-01-14T13:20:26.070" Title="Behind a black hole" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13210" PostTypeId="1" AcceptedAnswerId="13211" CreationDate="2016-01-14T11:19:28.670" Score="5" ViewCount="206" Body="&lt;p&gt;Let's talk about the time when our sun becomes a supergiant. We know that our solar system may get destroyed or it may get affected due the large mass of the supergiant and also due to the large temperature. But somehow let us consider nothing happened to our solar system by luck. It's obvious that life won't be possible on earth anymore(due to the extreme heat), but is there a possibility that, the events that took place during the formation of Earth, may take place in the far off planets? Can life formation take place in some other planet just like it happened in Earth??&lt;/p&gt;&#xA;" OwnerUserId="10379" LastActivityDate="2016-01-14T12:45:02.163" Title="Sun as our Supergiant" Tags="&lt;earth&gt;&lt;distances&gt;&lt;earth-like-planet&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13211" PostTypeId="2" ParentId="13210" CreationDate="2016-01-14T12:45:02.163" Score="6" Body="&lt;p&gt;When the Sun becomes a red supergiant (more accurately, an &lt;a href=&quot;https://en.wikipedia.org/wiki/Asymptotic_giant_branch&quot;&gt;asymptotic giant branch star&lt;/a&gt;), it will have &lt;em&gt;less&lt;/em&gt; mass than it does now. That is because mass is lost by an increasingly massive stellar wind as the Sun approaches the end of its life. As a result, the orbital radii of the planets would &lt;em&gt;increase&lt;/em&gt;. The surface temperature of this future AGB Sun would be &lt;em&gt;lower&lt;/em&gt; than the temperature of the Sun, but its luminosity would be much larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, even if the Earth manged to avoid being engulfed in the outer envelope of the evolved Sun (which is still uncertain), the amount of energy received per square metre at the Earth's surface would be vastly increased over the 1.4 kW per square metre it receives from the Sun now. So your conclusion that life (as we know it) would be fried is quite correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Circumstellar_habitable_zone&quot;&gt;Goldilocks zone&lt;/a&gt; for the possible occurrence of life, uses the criterion of liquid water as the basis for allowing life to evolve. As a star becomes more luminous (and an AGB Sun would probably be about 100 times as luminous as the present Sun), the Goldilocks zone moves outwards roughly proportional to $\sqrt{L}$, so about a factor of 10 further away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem then with allowing life to evolve on another planet (e.g. Mars after its orbit has expanded a bit, or Jupiter) is that (i) you need the ingredients to be there already, or some way for them to be delivered, but more importantly (ii) there is little time for life to evolve, because the Goldilocks zone position evolves really rapidly towards the end of the Sun's life as the AGB Sun's luminosity increases - the timescales would be measured in millions rather than billions of years. Furthermore, the luminosity of AGB stars can be highly variable as they undergo big pulsations, which may also be a problem. (iii) After the (comparatively brief) AGB phase has ended, you would end up with a very hot ($T \simeq 10^{5}$ K) white dwarf remnant that might effectively sterilise the whole system with ultraviolet light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think we know enough about how life gets started to say that these problems rule out your scenario; but they perhaps make it less likely. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-01-14T12:45:02.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13212" PostTypeId="2" ParentId="13209" CreationDate="2016-01-14T13:20:26.070" Score="8" Body="&lt;p&gt;A black hole is be a spherical object so it does not have a front or back.&lt;/p&gt;&#xA;" OwnerUserId="7926" LastActivityDate="2016-01-14T13:20:26.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13213" PostTypeId="1" CreationDate="2016-01-14T13:39:00.057" Score="10" ViewCount="245" Body="&lt;p&gt;I am looking for the percentage of galaxies that rotate with trailing arms. I know that only a small percentage of galaxies rotate with leading arms and would like to know whether there are any results that determine the percentage of spiral galaxies that rotate with tailing arms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in publications in this topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt;&#xA;So far I have found the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/0803.3247&quot; rel=&quot;noreferrer&quot;&gt;Land et. al (2008)&lt;/a&gt;: &quot;Determining the rotation&#xA;direction in this way is accurate for 96% of galaxies&#xA;(Pasha &amp;amp; Smirnov 1982)&quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1982Ap%26SS..86..215P&quot; rel=&quot;noreferrer&quot;&gt;Pasha &amp;amp; Smirnov (1982)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/astro-ph/0211002&quot; rel=&quot;noreferrer&quot;&gt;Buta et al. (2003)&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Do you know of any newer publication on this question? More recent than Pasha &amp;amp; Smirnov (1982) where the sample is quite small?&lt;/p&gt;&#xA;" OwnerUserId="6712" LastEditorUserId="9301" LastEditDate="2016-06-29T04:36:38.587" LastActivityDate="2018-07-06T14:33:56.340" Title="What percentage of galaxies rotate with trailing arms?" Tags="&lt;galaxy&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13214" PostTypeId="1" AcceptedAnswerId="13215" CreationDate="2016-01-14T13:45:55.517" Score="10" ViewCount="474" Body="&lt;p&gt;What does the current theory tell us about what is going on at the core of a neutron star, I'm expecting a black hole hopefully I won't be disappointed!&lt;/p&gt;&#xA;" OwnerUserId="6760" LastActivityDate="2016-01-14T14:06:50.273" Title="What lies at the center of a neutron star if any?" Tags="&lt;neutron-star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13215" PostTypeId="2" ParentId="13214" CreationDate="2016-01-14T14:06:50.273" Score="12" Body="&lt;p&gt;Certainly not a black hole! That would not be stable situation at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The content of the core of a neutron star is the subject of much speculation. The possibilities fall into a number of categories. (i) An increasingly hard neutron equation of state, such that neutrons retain their identities as they are squeezed closer, but an increasingly repulsive many-body strong nuclear force provides support. (ii) Additional hadronic degrees of freedom, such that the neutrons (and protons) transform into other heavy hadrons such as lambda or sigma particles. (iii) Some sort of quark plasma. (iv) Boson condensations involving the neutrons decaying into pions or kaons with zero momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of diagnostics of these possibilities: primarily, the maximum possible mass of a neutron star should decrease from about 3 solar masses for (i) down to about 1.5 solar masses for (iv). Secure measurements of a 2 solar mass neutron star would seem to rule out (iv), but even that seems not completely agreed. Another diagnostic is how quickly neutron stars can cool. The presence of quark matter or boson condensations should lead to much more rapid cooling by neutrino emission. Again, nothing conclusive has emerged yet.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-01-14T14:06:50.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13217" PostTypeId="2" ParentId="13205" CreationDate="2016-01-14T18:35:09.413" Score="1" Body="&lt;p&gt;EDIT: It turns out a major problem with galaxy-center-based coordinates is that we don't actually know how far from us the center of the galaxy actually is. According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_Center&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Galactic_Center&lt;/a&gt; estimates vary between 24,800 and 28,400 light years. You could arbitrarily choose a value, but the results would be off by at least 2,000 light years in some cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was surprised that I couldn't find something like this &#xA;online directly. The closest I found is the file &#xA;&quot;heasarc_tycho2.tdat.gz&quot; on &#xA;&lt;a href=&quot;http://heasarc.nasa.gov/FTP/heasarc/dbase/dump/&quot; rel=&quot;nofollow&quot;&gt;http://heasarc.nasa.gov/FTP/heasarc/dbase/dump/&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The format is explained on &#xA;&lt;a href=&quot;http://heasarc.nasa.gov/W3Browse/all/tycho2.html&quot; rel=&quot;nofollow&quot;&gt;http://heasarc.nasa.gov/W3Browse/all/tycho2.html&lt;/a&gt; and it &#xA;includes the galactic latitude and longitude which are &#xA;explained here: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Galactic_coordinate_system&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this information, you'd still have to compute the &#xA;positions yourself, but at least all of the data is there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, as others have noted, you can take any star &#xA;catalog with any set of coordinates and convert it to the &#xA;format you need. The catalog I list above just makes it a &#xA;little easier, since they're precomputed some of the steps &#xA;for you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you mention above, SIMBAD also includes galactic &#xA;coordinates, but I couldn't find an easy way to download &#xA;the entire catalog. &lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2016-01-15T02:38:02.897" LastActivityDate="2016-01-15T02:38:02.897" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13218" PostTypeId="2" ParentId="13207" CreationDate="2016-01-14T19:22:27.847" Score="2" Body="&lt;p&gt;Here's a nice image of &lt;a href=&quot;https://en.wikipedia.org/wiki/Barnard&amp;#39;s_Star&quot; rel=&quot;nofollow noreferrer&quot;&gt;Barnard's star&lt;/a&gt; moving against the stellar background:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/z0ge0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z0ge0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;From this interesting website: &lt;a href=&quot;http://sites.psu.edu/astrowright/2012/08/14/barnards-stars-planets/&quot; rel=&quot;nofollow noreferrer&quot;&gt;AstroWright&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2016-01-14T19:22:27.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13219" PostTypeId="2" ParentId="13183" CreationDate="2016-01-14T20:12:23.863" Score="2" Body="&lt;p&gt;General relativity can be used to discuss, theoretically, what happens to objects as they approach the event horizon of a black hole; it all has to do with your point of view. Outside viewers, those not affected by the gravity of the black hole, would view an object approach the event horizon slowly and pretty much never enter past the horizon. For the observers point of view time stops for the object. From the point of view of the object, its own time runs as normal, and it is the time outside of the event horizon that runs faster than normal. For more general information at a really easy to follow format you might check out &lt;a href=&quot;https://books.google.com/books?id=e9erAgAAQBAJ&amp;amp;printsec=frontcover&amp;amp;dq=what+is+relativity&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwisos-xjqrKAhXBBywKHbggD-gQ6AEIKzAA#v=onepage&amp;amp;q=what%20is%20relativity&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;What is Relativity&lt;/a&gt; by Jeffery Bennett.&lt;/p&gt;&#xA;" OwnerUserId="2968" LastActivityDate="2016-01-14T20:12:23.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13221" PostTypeId="1" CreationDate="2016-01-15T09:50:58.403" Score="2" ViewCount="67" Body="&lt;p&gt;I came up on this stanza in various websites when looking up on experimenter bias. It says &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If the signal being measured is actually smaller than the rounding error and the data are over-averaged, a positive result for the measurement can be found in the data where none exists (i.e. a more precise experimental apparatus would conclusively show no such signal). If an experiment is searching for a sidereal variation of some measurement, and if the measurement is rounded-off by a human who knows the sidereal time of the measurement, and if hundreds of measurements are averaged to extract a &quot;signal&quot; which is smaller than the apparatus' actual resolution, then it should be clear that this &quot;signal&quot; can come from the non-random round-off, and not from the apparatus itself. In such cases a single-blind experimental protocol is required; if the human observer does not know the sidereal time of the measurements, then even though the round-off is non-random it cannot introduce a spurious sidereal variation.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that sidereal time is something related to the time measurement used by astronomers as a time-keeping system, but I don't understand by &quot;someone who knows the sidereal time of measurement&quot; can then unconsciously influence the results, because say if you saw the clock as 00.56 s and you rounded it to 00.6s, I don't think it will have any effect. But moreover, by rounding that, how can you tell it will eventually have effect on your results? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please advise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for any wrong tags. I'm new here, so still learning. &lt;/p&gt;&#xA;" OwnerUserId="10411" LastActivityDate="2016-01-15T09:50:58.403" Title="About bias in sidereal time (used by astronomers)" Tags="&lt;time&gt;&lt;observational-astronomy&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13225" PostTypeId="1" CreationDate="2016-01-15T16:15:00.743" Score="2" ViewCount="100" Body="&lt;p&gt;Both the Moon and Mercury have polar cold traps in polar craters, with signs of volatiles. This is of course not as interesting on a moon in the outer Solar System which consists of volatiles anyway. But suppose that there is some use for very cold real estate shielded from the Sun out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which moons are candidates to have eternally shadowed (or eternally sunlit) regions, and wouldn't they be even colder than those on the Moon? Jupiter and Neptune have less than 2 degrees inclination so their moons maybe are good candidates.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="2618" LastEditDate="2017-03-07T00:49:21.470" LastActivityDate="2017-03-07T00:49:21.470" Title="Which moons have cold traps? (i.e. low ecliptic inclination in orbital and rotational axes)" Tags="&lt;temperature&gt;&lt;natural-satellites&gt;&lt;ecliptic&gt;&lt;inclination&gt;&lt;pole&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13226" PostTypeId="2" ParentId="13197" CreationDate="2016-01-15T17:44:30.937" Score="0" Body="&lt;p&gt;The moon's orbital plane is inclined to the ecliptic at about 5.5 degrees, and a solar eclipse can only occur when the moon crosses the ecliptic during a new moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://www.timeanddate.com/eclipse/total-solar-eclipse.html&quot; rel=&quot;nofollow&quot;&gt;total solar eclipse&lt;/a&gt; can only occur when the apparent size of the moon exceeds that of the sun, when the moon is at apogee (greatest distance from Earth) an annular eclipse, in which the edge of sun appears as a ring encircling the moon, occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next annular eclipse will be visible from &lt;a href=&quot;http://www.timeanddate.com/eclipse/solar/2016-september-1&quot; rel=&quot;nofollow&quot;&gt;central africa on the 26th of September 2016&lt;/a&gt;, A few people in Southern Congo will this, and a second eclipse in February 2017.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-01-15T17:44:30.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13227" PostTypeId="1" AcceptedAnswerId="13228" CreationDate="2016-01-15T22:29:30.330" Score="11" ViewCount="1396" Body="&lt;p&gt;Transits of Mercury happen fairly frequently due to its short period, but transits of Venus are less frequent.  I've looked over the data available to me and found that there have been transits of both planets in the same calendar-year in 1631 (Nov &amp;amp; Dec) and in 1769 (Jun &amp;amp; Nov).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would really be interesting is if the two planets transited the Sun at the same time! With the rule being, that the second transit begins before the first transit completes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we figure out if they have ever so transited, or if they ever will so transit?&lt;/p&gt;&#xA;" OwnerUserId="718" LastEditorUserId="2424" LastEditDate="2016-01-19T02:26:01.677" LastActivityDate="2016-01-19T02:26:01.677" Title="Are there ever any simultaneous transits of both Mercury and Venus as seen from the Earth?" Tags="&lt;orbit&gt;&lt;venus&gt;&lt;mercury&gt;&lt;planetary-transits&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13228" PostTypeId="2" ParentId="13227" CreationDate="2016-01-16T03:59:39.940" Score="12" Body="&lt;p&gt;EDIT: As it turns out, I'm not the first or even the second person to run calculations like this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.fourmilab.ch/documents/canon_transits/&quot;&gt;https://www.fourmilab.ch/documents/canon_transits/&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.solexorb.it/SolexOld/Simtrans.pdf&quot;&gt;http://www.solexorb.it/SolexOld/Simtrans.pdf&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Meeus' work (second link) mentions the 13425 CE event in &quot;Table 1. Simultaneous and near-simultaneous transits of Mercury and Venus, years 1 to 300,000&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Within the limits of DE431 (7 May 13201 BCE to 7 May 17091 CE), there is no &#xA;time at which both Mercury and Venus transit the Sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest we get to this: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;On 16 Sep 13425 CE at 11:57pm UTC, Venus starts transiting the Sun. This &#xA;transit ends the next morning (17 Sep 13425 CE) at 7:30am. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Less than 9 hours later, at 4:27pm, Mercury starts transiting the &#xA;Sun. This transit ends at 10:26pm. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The program I used to compute this:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-solve-astro-13227.c&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-solve-astro-13227.c&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The list of transits I computed while solving this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/mercury-transits.txt.bz2&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/mercury-transits.txt.bz2&lt;/a&gt; &#xA;&lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/venus-transits.txt.bz2&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/venus-transits.txt.bz2&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I believe this answer is correct, Stellarium does not agree with &#xA;me, and HORIZONS doesn't compute positions past 9999 CE, so don't put too &#xA;much faith in this answer, since there's no good way to confirm it. I &#xA;believe that I'm correct and Stellarium is wrong this far in the future, but &#xA;it could be the other way around. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if my calculations are correct, the uncertainty in calculating the &#xA;relevant positions (Sun, Merucry, Venus, Earth) this far in the future is &#xA;high. On their own transit pages, NASA only computes Venus transits from &#xA;2000 BCE to 4000 CE, and Mercury transits from 1601 CE to 2300 CE, even &#xA;though they could've made the same calculations I made from 13201 BCE to &#xA;17091 CE:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://eclipse.gsfc.nasa.gov/transit/catalog/MercuryCatalog.html&quot;&gt;http://eclipse.gsfc.nasa.gov/transit/catalog/MercuryCatalog.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://eclipse.gsfc.nasa.gov/transit/catalog/VenusCatalog.html&quot;&gt;http://eclipse.gsfc.nasa.gov/transit/catalog/VenusCatalog.html&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This suggests NASA isn't confident enough of Mercury/Venus (and Earth/Sun) &#xA;positions to predict that far in the past or future, so my results may be fairly inaccurate.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2016-01-16T22:03:43.997" LastActivityDate="2016-01-16T22:03:43.997" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13230" PostTypeId="2" ParentId="13227" CreationDate="2016-01-16T15:14:52.490" Score="10" Body="&lt;p&gt;As others have calculated, there are no predicted double transits. Since Venus transits for about 12 hours each hundred years (roughly), Venus is in transit for about 1/100000 of the time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus there is a (roughly) 1 in 100000 chance that a randomly chosen transit of mercury will coincide with a transit of Venus. Since Transits of Mercury occur every 10 years or so, one would expect a double transit to happen on average once every million years. This is clearly a very rough estimate, but we should not be surprised that such an event has not happened in the historical period.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-01-16T15:14:52.490" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13231" PostTypeId="1" AcceptedAnswerId="13235" CreationDate="2016-01-16T23:24:05.203" Score="1" ViewCount="235" Body="&lt;p&gt;Just say you are immortal and will live forever. Since the universe is claimed to be unstable, what would happen after it ends? Would all of the universe's mass just fly back into another compressed ball, and explode again? Would you be torn apart and flung across the new universe?&lt;/p&gt;&#xA;" OwnerUserId="10417" LastActivityDate="2016-01-17T19:16:38.697" Title="Second Big Bang" Tags="&lt;universe&gt;&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2016-01-18T13:08:31.990" ContentLicense="CC BY-SA 3.0" />
  <row Id="13233" PostTypeId="2" ParentId="13185" CreationDate="2016-01-17T09:12:47.963" Score="0" Body="&lt;p&gt;Its simply mean ,when earth rotates (as Sun don't move) ,the instant when the light of sun falls on the earth called sunrise and its just opposite mean sunset.&lt;/p&gt;&#xA;" OwnerUserId="9458" LastActivityDate="2016-01-17T09:12:47.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13234" PostTypeId="1" CreationDate="2016-01-17T18:40:37.050" Score="1" ViewCount="125" Body="&lt;p&gt;During different days when both the sun and the moon are seen I notice that the axis of the lit, illuminated, side of the moon does not pass through the sun there is some degree of diversion. Why?&lt;/p&gt;&#xA;" OwnerUserId="10423" LastEditorUserId="2365" LastEditDate="2016-01-20T19:18:40.017" LastActivityDate="2016-01-20T19:18:40.017" Title="The relation of the lit side of the moon and the sun" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13235" PostTypeId="2" ParentId="13231" CreationDate="2016-01-17T19:16:38.697" Score="0" Body="&lt;p&gt;As such the answer 'I don't know' is more appropriate than anything else.&#xA;But for an answer to give---&#xA;Fate of universe depends upon many aspects - but mainly the density of the universe. There is a critical density of the universe which is not enough to produce gravitational effect to stop expansion and nor that weak to keep the universe to expand at an increasing rate. This can be more clearly understood with help of 4-d diagrammatic representation of the expansion(or contraction). When it is critical, the diagram is flat with no net curvature. When the density is less CD(Critical Density), there is a saddle shaped diagram which represents vigorously increasing rate of expansion. When density is greater than CD, there is a contraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, apart from baryonic matter, mysterious dark matter and dark energy come into effect. All these, at this moment, is making the universe accelerating, as we all know. Till now, expectations are towards forming a 'rip', or violent expansion of universe where even atoms will be torn away in distant future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concept of second Big Bang is still far from our views, but quoting Stephen Hawking --- &quot;God not only plays dice but also casts it where no one can see&quot; --- Let nature do what she wants.&lt;/p&gt;&#xA;" OwnerUserId="10414" LastActivityDate="2016-01-17T19:16:38.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13236" PostTypeId="2" ParentId="13234" CreationDate="2016-01-17T20:05:51.883" Score="7" Body="&lt;p&gt;I'll assume you mean the &lt;em&gt;lit&lt;/em&gt; side of the Moon (this wasn't clear from the original question, but I've edited it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there's a crescent Moon visible in the sky during the day. Plot a straight line between the tips of the crescent, then plot a perpendicular straight line across its center across the sky. Intuitively, that line &lt;em&gt;should&lt;/em&gt; pass through the Sun. You're saying it doesn't, and you're asking why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, that line does pass through the Sun, and what you're seeing is an optical illusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you imagine the sky as a perfect hemisphere, with you at the center, the geometry will match up. But we we tend to imagine it as a flattened dome, probably because the horizon is &lt;em&gt;obviously&lt;/em&gt; far away, but we have no reference points for the zenith so we imagine it being closer. This is (part of) the source of the &quot;Moon illusion&quot;, where the Moon appears bigger when it's close to the horizon. In fact, it's very nearly the same angular size, but since we imagine it being farther away it must be &quot;bigger&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same mention distortion of the sky's geometry makes it look like the axis of the Moon's lit side doesn't point directly to the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next time you see this, try holding up a straight rod intersecting both the Sun and the Moon. I think you'll find that the rod passes directly across the axis of the Moon's lit side.&lt;/p&gt;&#xA;" OwnerUserId="53" LastEditorUserId="53" LastEditDate="2016-01-18T21:09:48.760" LastActivityDate="2016-01-18T21:09:48.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13238" PostTypeId="1" AcceptedAnswerId="13247" CreationDate="2016-01-18T06:13:21.580" Score="2" ViewCount="931" Body="&lt;p&gt;Planets and other bodies in the universe have different ranges of gravitational fields. What is the maximum speed that can be achieved because of acceleration due to gravitational field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words which is the strongest and largest gravitational field discovered in the universe till date. What is the maximum speed An object accelerating in these fields can gain?&lt;/p&gt;&#xA;" OwnerUserId="10430" LastActivityDate="2016-01-18T20:07:24.797" Title="What is the Maximum Speed that can be acheived Because of Acceleration Due to Gravity?" Tags="&lt;gravity&gt;&lt;speed&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13239" PostTypeId="2" ParentId="13238" CreationDate="2016-01-18T09:46:34.683" Score="7" Body="&lt;p&gt;It all depends on your frame of reference, but the biggest speeds caused by gravity will involve black holes or perhaps neutron stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the point of view of a far external observer, an infalling object (from infinity) reaches a maximum inward speed of $0.385c$ (where $c$ is the speed of light) towards the black hole at 3 times the Schwarzschild radius. Thereafter, the object will &lt;em&gt;appear&lt;/em&gt; to slow down and tend towards being stationary at the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the point of view of a &quot;shell observer&quot;, that is somehow held stationary in the gravitational field of the black hole, the infalling object would have a speed that tended towards $c$ as the shell observer tended towards the event horizon at the Schwarzschild radius. It is meaningless to try to extend this inside the event horizon since (stationary) shell observers cannot exist inside the event horizon.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-01-18T09:46:34.683" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13242" PostTypeId="1" CreationDate="2016-01-18T12:05:09.907" Score="1" ViewCount="370" Body="&lt;p&gt;If an observer is falling toward a black hole with its face away from singularity then what will he observe after crossing the event horizon? The reason that why I am asking this question because as far as I know for an outside observer, the falling observer appear to freeze at event horizon i.e. time appear to stop for falling observer. So if the falling observer is able to look outward after crossing event horizon then he will be able to see an infinite amount of time which is impossible. So what will the observer see after crossing the event horizon?&lt;/p&gt;&#xA;" OwnerUserId="10436" LastActivityDate="2017-01-30T15:36:32.043" Title="View from inside the black hole" Tags="&lt;black-hole&gt;&lt;general-relativity&gt;&lt;time-dilation&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13243" PostTypeId="2" ParentId="13242" CreationDate="2016-01-18T12:23:12.813" Score="1" Body="&lt;p&gt;Your premise is incorrect:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The external observer's view does not imply anything about an observer within the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The event horizon itself is just an indicator of the edge of the region where light can no longer escape. Light can head inwards as you would expect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So an observer inside the event horizon will be able to see light reaching them as normal (well, not exactly as normal, as there will be lensing effects) - there will not be an infinite amount of time visible.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2016-01-18T12:23:12.813" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13244" PostTypeId="2" ParentId="13242" CreationDate="2016-01-18T14:07:26.530" Score="-4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If an observer is falling toward a black hole with his face away from singularity then what will he observe after crossing the event horizon? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nothing. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The reason that why I am asking this question because as far as I know for an outside observer, the falling observer appears to freeze at the event horizon i.e. time appears to stop for the falling observer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Correct. Gravitational time dilation goes infinite, and the distant observer says the &quot;coordinate&quot; speed of light at the event horizon is zero. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So if the falling observer is able to look outward after crossing the event horizon then he will be able to see an infinite amount of time which is impossible. So what will the observer see after crossing the event horizon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nothing. The coordinate speed of light is zero at that location, which means that by our clocks, it takes forever to see anything. So the falling observer hasn't seen anything yet, and he never ever will. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;IMHO it's worth reading the mathspages &lt;a href=&quot;http://mathpages.com/rr/s7-02/7-02.htm&quot; rel=&quot;nofollow&quot;&gt;Formation and Growth of Black Holes&lt;/a&gt; and paying attention to the frozen star interpretation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Incidentally, we should perhaps qualify our dismissal of the 'frozen star'  interpretation, because it does (arguably) give a servicable account of phenomena outside the event horizon, at least for an eternal static configuration. Historically the two most common conceptual models for general relativity have been the &quot;geometric interpretation&quot; (as originally conceived by Einstein) and the &quot;field interpretation&quot; (patterned after the quantum field theories of the other fundamental interactions). These two views are operationally equivalent outside event horizons, but they tend to lead to different conceptions of the limit of gravitational collapse. According to the field interpretation, a clock runs increasingly slowly as it approaches the event horizon (due to the strength of the field), and the natural &quot;limit&quot; of this process is that the clock asymptotically approaches &quot;full stop&quot; (i.e., running at a rate of zero). It continues to exist for the rest of time, but it's &quot;frozen&quot; due to the strength of the gravitational field. Within this conceptual framework there's nothing more to be said about the clock's existence...&quot;&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author doesn't favour it, but it squares with what Einstein said about the &lt;a href=&quot;http://einsteinpapers.press.princeton.edu/vol7-trans/156?highlightText=%22speed%20of%20light%22&quot; rel=&quot;nofollow&quot;&gt;speed of light varying&lt;/a&gt; with gravitational potential. The other interpretation doesn't. And note that Einstein didn't refer to the &quot;coordinate&quot; speed of light. He simply referred to the &lt;em&gt;speed&lt;/em&gt; of light. So we can reasonably say that at the event horizon, the speed of light is zero, and that this is why a vertical light beam can't get out. The distant observer sees the infalling observer freeze at the event horizon. But the infalling observer doesn't see himself as frozen, because the speed of light at that location is zero. He sees nothing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB: SR time dilation is symmetrical, but GR time dilation isn't. If you and I passed each other in gravity-free space at some relativistic speed, we would each claim that the other's clock was slower. But when we're at different elevations, we both agree that the lower clock is going slower.   &lt;/p&gt;&#xA;" OwnerUserId="9772" LastEditorUserId="9772" LastEditDate="2016-01-18T14:22:48.553" LastActivityDate="2016-01-18T14:22:48.553" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13245" PostTypeId="2" ParentId="7811" CreationDate="2016-01-18T16:10:11.880" Score="0" Body="&lt;p&gt;During the night the wind has died and the pollution from agriculture and man made chemicals have settled leaving air quality clear and the sunlight at sunrise is less obstructed and more intense.At sunset all of the above are still lingering in the air making sunlight less intense due to obstruction caused by the pollution.As for the winter being more intense, Precipitation is higher in winter months and the water molecules filter out a large percent of pollution. &lt;/p&gt;&#xA;" OwnerUserId="9326" LastActivityDate="2016-01-18T16:10:11.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13246" PostTypeId="1" AcceptedAnswerId="13265" CreationDate="2016-01-18T17:49:00.663" Score="1" ViewCount="359" Body="&lt;p&gt;There's a &lt;a href=&quot;http://www.gizmag.com/aragoscope-lensless-telescope/35761/&quot; rel=&quot;nofollow&quot;&gt;potential design for a space-based telescope&lt;/a&gt; that uses a large opaque circular disk in place of a transparent lens to focus light for analysis. It uses the light that bends around the disk that focuses on the Poisson or Arago spot, but requires a circular object. It could theoretically produce much clearer images (1000x) than telescopes we have today, albeit with a narrower field of vision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the Earth is not perfectly round, but I believe I've also heard it's very &quot;smooth&quot; despite mountains and such, so I wonder if instead of a spot (very small circle) on some point behind it, it might cast an oval or other oblong shape of some size that, using lenses or computer reconstruction, might produce an image that is usable, and presumably even higher resolution (millions, billions, trillions, or even more times as detailed?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's not possible, what are the issues? Does the atmosphere get in the way? If we could imagine a sufficiently circular disk roughly the size of Earth, about how far would the image processing receiver have to be from the disk opposite the studied object in order to capture the convergence of bent light? Would the gravity of the Earth affect the distance at all, due to gravitational lensing?&lt;/p&gt;&#xA;" OwnerUserId="10439" LastActivityDate="2016-01-20T16:27:35.707" Title="Could the Earth be used to cast an arago/poisson spot on something" Tags="&lt;space-telescope&gt;&lt;gravitational-lensing&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13247" PostTypeId="2" ParentId="13238" CreationDate="2016-01-18T20:07:24.797" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the maximum speed that can be achieved because of acceleration due to gravity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The speed of light. But there's a catch.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In other words which is the strongest and largest gravitational field discovered in the universe till date. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That of a black hole. &lt;a href=&quot;https://en.wikipedia.org/wiki/Sagittarius_A*&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sagittarius A*&lt;/a&gt; is thought to be the location of a supermassive black hole. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the maximum speed an object accelerating in these fields can gain?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The speed of light. See &lt;a href=&quot;https://en.wikipedia.org/wiki/Escape_velocity#List_of_escape_velocities&quot; rel=&quot;nofollow noreferrer&quot;&gt;this list&lt;/a&gt;. The escape velocity for a black hole is the speed of light, and you can flip this around. If you drop an object from a great height, it's travelling at escape velocity when it reaches the gravitating body. But like I said, there's a catch. Take a look at this by &lt;a href=&quot;http://einsteinpapers.press.princeton.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8KXbI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the second paragraph. The body falls down &lt;em&gt;because&lt;/em&gt; the speed of light is spatially variable. If this continued unabated there would come a point where the body is falling faster than the speed of light at that location. But since matter can't faster than this the &quot;coordinate&quot; speed of light, the maximum speed it &lt;em&gt;can&lt;/em&gt; fall is at the speed of light at that location. Which is circa half the speed of light at our location.   &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2016-01-18T20:07:24.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13248" PostTypeId="1" AcceptedAnswerId="13401" CreationDate="2016-01-18T21:07:24.997" Score="2" ViewCount="234" Body="&lt;p&gt;I'd like to be able to find the latitude and longitude on Earth (assuming a sphere is fine) which is closest to Pluto at the current time (or where Pluto is directly overhead). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this possible with PyEphem?&lt;/p&gt;&#xA;" OwnerUserId="4166" LastActivityDate="2016-01-27T01:18:08.733" Title="Closest point on Earth to a planet?" Tags="&lt;earth&gt;&lt;pluto&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13250" PostTypeId="1" CreationDate="2016-01-19T02:05:33.283" Score="1" ViewCount="160" Body="&lt;p&gt;I was wondering what a phone call would be like entering or falling t=into a black hole. Due to the massive density of the black hole and time would feel slower, how would that appear to the person on the other side of the call?&lt;/p&gt;&#xA;" OwnerUserId="10440" LastActivityDate="2016-06-22T21:44:32.327" Title="A phonecall inside a blackhole" Tags="&lt;gravity&gt;&lt;black-hole&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13251" PostTypeId="1" AcceptedAnswerId="13252" CreationDate="2016-01-19T03:05:03.787" Score="1" ViewCount="265" Body="&lt;p&gt;I heard some theory that sun may steal materials such as comets from neighbouring stars, is it true? But if is true, what is the primary source of those comets? Especially &quot;who&quot; creates them instead of steals them? And which stars are &quot;willing&quot; to bear our sun steals materials from it?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastActivityDate="2016-01-25T22:10:33.820" Title="If sun steals comets from other stars, then what is the primary source of comets?" Tags="&lt;solar-system&gt;&lt;comets&gt;&lt;oort-cloud&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13252" PostTypeId="2" ParentId="13251" CreationDate="2016-01-19T03:47:50.263" Score="3" Body="&lt;p&gt;The primary source of comets for our solar system comes from the Oort Cloud, a smaller amount coming from the Kuiper belt. The Oort cloud is thought to have originated from the remnants of a proto-planetary disk. &lt;a href=&quot;https://en.wikipedia.org/wiki/Oort_cloud&quot; rel=&quot;nofollow&quot;&gt;This paragraph&lt;/a&gt; explains this better:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Oort cloud is thought to be a remnant of the original&#xA;  protoplanetary disc that formed around the Sun approximately 4.6&#xA;  billion years ago. The most widely accepted hypothesis is that the&#xA;  Oort cloud's objects initially coalesced much closer to the Sun as&#xA;  part of the same process that formed the planets and minor planets,&#xA;  but that gravitational interaction with young gas giants such as&#xA;  Jupiter ejected the objects into extremely long elliptic or parabolic&#xA;  orbits. Recent research has been cited by NASA hypothesizing&#xA;  that a large number of Oort cloud objects are the product of an&#xA;  exchange of materials between the Sun and its sibling stars as they&#xA;  formed and drifted apart, and it is suggested that many — possibly the&#xA;  majority of — Oort cloud objects were not formed in close proximity to&#xA;  the Sun Simulations of the evolution of the Oort cloud from the&#xA;  beginnings of the Solar System to the present suggest that the cloud's&#xA;  mass peaked around 800 million years after formation, as the pace of&#xA;  accretion and collision slowed and depletion began to overtake supply.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7926" LastActivityDate="2016-01-19T03:47:50.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13254" PostTypeId="1" CreationDate="2016-01-19T07:45:03.210" Score="1" ViewCount="115" Body="&lt;p&gt;A ship travels 99% of c from Earth (leaving at Time A, say 2016) to Destination Star (arriving at Time B, say 2020, accounting for relativistic effects) to stretch the tube, leaves a buoy at Destination Star, then uses the tube's topology to return to Earth, effectively faster-than-light, moments after it left, according to how the Krasnikov tube is supposed to work (Time A plus a few minutes). Can the ship now return to Time B (2020) from &quot;Time A (2016) plus a few minutes&quot; using the tube just as quickly? If so, would its return to Time B mean &quot;Destination Star in the ship's year 2020, the moment when it arrived / first stretched the tube to this destination?&quot; And if so, would it mean returning to the system &quot;before&quot; the ship dropped the buoy?&#xA;My understanding of a Krasnikov tube is that it's a theoretical means for FTL travel that doesn't violate causality, because a ship traveling along the return path of a tube &quot;rolls back&quot; time to just moments before the ship originally left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I understand that there is a way to construct a return tube that would violate causality (&lt;a href=&quot;http://arxiv.org/pdf/gr-qc/9702049.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/gr-qc/9702049.pdf&lt;/a&gt;), but let's assume for a moment there's some &quot;chronology protection&quot; in place that doesn't allow that to happen (the tube implodes when you try to do that), and that the question of the exotic matter needed to create such a tube isn't an issue for the ship's makers.&#xA;The questions I have then are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) Am I even thinking about this correctly? Where have I messed up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) I don't understand how a system of Krasnikov tubes, as some sources suggest, could be used to create an interstellar network for FTL travel. It seems to me that the traveler in the ship can never actually have an impact on his exit or return points, since he will always be returning to Time A (2016) moments after he left, or Time B (2020), the moment he arrives?&lt;/p&gt;&#xA;" OwnerUserId="10441" LastActivityDate="2016-01-20T17:38:37.517" Title="How does time &quot;add up&quot; for observers outside of and travelers inside of a Krasnikov tube?" Tags="&lt;time-dilation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13255" PostTypeId="1" AcceptedAnswerId="13258" CreationDate="2016-01-19T08:13:24.890" Score="5" ViewCount="423" Body="&lt;p&gt;Why do proto stars on the Hayashi track get dimmer as they contract?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My expectation is that they would get hotter and brighter the more they contact. Why is this wrong?&lt;/p&gt;&#xA;" OwnerDisplayName="user10250" LastActivityDate="2016-01-20T00:03:12.053" Title="Why do proto stars on the Hayashi track get dimmer as they contract?" Tags="&lt;stellar-evolution&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13256" PostTypeId="2" ParentId="13251" CreationDate="2016-01-19T08:55:56.327" Score="1" Body="&lt;p&gt;The prime candidate for the origin of interstellar comets, if any of those are around, are star systems which formed together with the Sun out of the same nebula collapsing into an open star cluster. Being captured from a Solar sibling might mean nothing more exotic than that. Same time of formation, same mixture of elements, same trajectory. That is more likely because they should've had similar movement as the Sun. A star capturing anything from another random star is very unlikely unless they begin with a similar movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning capturing material in general, my user name here is &lt;a href=&quot;https://en.wikipedia.org/wiki/Local_Interstellar_Cloud&quot; rel=&quot;nofollow&quot;&gt;local fluff&lt;/a&gt; because that is the puff of gas and dust through which the Solar System sails now in human times. Material ejected from active newborn stars in the so called Scorpius–Centaurus Association. But they don't throw any comet sized objects at us, just individual nucleus, atoms, molecules and dust grains drifting past us driven by the stellar winds.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2016-01-19T09:18:39.273" LastActivityDate="2016-01-19T09:18:39.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13257" PostTypeId="2" ParentId="13254" CreationDate="2016-01-19T20:48:26.420" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A ship travels 99% of c from Earth (leaving at Time A, say 2016) to Destination Star (arriving at Time B, say 2020, accounting for relativistic effects) to stretch the tube, leaves a buoy at Destination Star, then uses the tube's topology to return to Earth, effectively faster-than-light, moments after it left, according to how the Krasnikov tube is supposed to work (Time A plus a few minutes).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It just can't work like this. Think about what a clock actually does. Time is a measure of local motion, that's all. It's &quot;how much local motion has occurred&quot;. If you move very fast, the rate of local motion is of necessity reduced because the maximum rate of motion is c, because of the wave nature of matter. Hence you suffer time dilation. See the &lt;a href=&quot;https://en.wikipedia.org/wiki/Time_dilation#Simple_inference_of_time_dilation_due_to_relative_velocity&quot; rel=&quot;nofollow&quot;&gt;simple inference of time dilation&lt;/a&gt; on Wikipedia. Note that if you have two parallel-mirror light-clocks and one stays at home whilst the other goes on a fast out-and-back trip, the light-path lengths are the same. And they meet back up at the same time. They don't miss each other by a week. Time dilation is not time travel.      &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can the ship now return to Time B (2020) from &quot;Time A (2016) plus a few minutes&quot; using the tube just as quickly? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, it absolutely can not. At .99c the Lorentz factor is 7, so if your journey takes you four years by your clock, your destination star is 28 light years away. When you get there it's circa 2044 according to Earth clocks. There is no way you can get back to Earth in 2016. For that to happen all the motion that occurred has to be undone. And there is no way that that can happen. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If so, would its return to Time B mean &quot;Destination Star in the ship's year 2020, the moment when it arrived / first stretched the tube to this destination?&quot; And if so, would it mean returning to the system &quot;before&quot; the ship dropped the buoy? My understanding of a Krasnikov tube is that it's a theoretical means for FTL travel that doesn't violate causality, because a ship traveling along the return path of a tube &quot;rolls back&quot; time to just moments before the ship originally left.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's just science fiction I'm afraid, with no basis in physics. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now, I understand that there is a way to construct a return tube that would violate causality (&lt;a href=&quot;http://arxiv.org/pdf/gr-qc/9702049.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/gr-qc/9702049.pdf&lt;/a&gt;), but let's assume for a moment there's some &quot;chronology protection&quot; in place that doesn't allow that to happen (the tube implodes when you try to do that), and that the question of the exotic matter needed to create such a tube isn't an issue for the ship's makers. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The chronology projection conjecture is yet more speculation I'm afraid. It's superfluous because we don't actually move forward through time. Hence moving backwards through time is a non-starter.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Am I even thinking about this correctly? Where have I messed up?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You didn't watch carefully, and you fell for the smoke-and-mirrors and the sleight of hand.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I don't understand how a system of Krasnikov tubes, as some sources suggest, could be used to create an interstellar network for FTL travel. It seems to me that the traveller in the ship can never actually have an impact on his exit or return points, since he will always be returning to Time A (2016) moments after he left, or Time B (2020), the moment he arrives?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If he can go from 2044 to 2016, it's a time machine. Only he can't, so it isn't.  &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2016-01-19T20:48:26.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13258" PostTypeId="2" ParentId="13255" CreationDate="2016-01-20T00:03:12.053" Score="3" Body="&lt;p&gt;The Hayashi track has an almost constant effective temperature. The pre main sequence star (not protostar) is powered by the release of gravitational potential energy caused by its &lt;em&gt;contraction&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the star is getting smaller at constant effective temperature, then of course it's luminosity ($\propto R^2 T^4$) decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So perhaps your question is why do PMS stars contract at constant temperature? In other words, why is the Hayashi track almost vertical in the HR diagram? The answer to this is derived in some detail on the relevant Wikipedia page and arises from the fact that low mass PMS stars are fully convective and have atmospheric opacity dominated by H- ions.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-01-20T00:03:12.053" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13259" PostTypeId="1" AcceptedAnswerId="18584" CreationDate="2016-01-20T00:06:53.310" Score="2" ViewCount="214" Body="&lt;p&gt;I am trying to calculate at which point gravitational equilibrium sets in for various bodies (planets stars neutron stars etc.) assuming they are perfect spheres. However, the radius I get is not equal to what it should be according to Wikipedia (I tried it for the Sun, a planet and a neutron star, but all of them were off by quite a bit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is the example of the neutron star radius I'm trying to get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My result (factor) is 1 at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radius: $2228588\mathrm{\ m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Density: $1.2768744892680034 \times 10^{11} \mathrm{\ kg/m^3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia article about neutron stars says the radius of a neutron star is about 12 km and the density is about 10 times higher than mine. My result is 2228km, which is quite a bit off, so I have a second question:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Am I calculating all the forces I need?&lt;/li&gt;&#xA;&lt;li&gt;Are the formulae I'm using correct?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I took them from Wikipedia and university slides and I found multiple variations of almost all of them (most of which I actually didn't list below), so I'm quite confused as to what is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that I'm using iron as the element and I should split it up and convert the protons to neutrons, but that would only increase the degeneracy pressure and result in an even bigger radius, if I'm not mistaken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how I calculated it:&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Mass of an electron: $M_e$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard atomic weight of hydrogen: $M_h = 1.00782503223$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass of a neutron: $M_n$&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Atomic mass of ${\mathrm{Fe}^{56}_{26}}$: $55.934936 \mathrm{\ u}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of atoms: $7.938 \times 10^{31}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of neutrons per atom: $(56-26) \times \text{number of atoms}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of electrons per atom: $26 \times \text{number of atoms}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total number of particles: $\text{number of atoms} + \text{number of electrons}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass: $\text{atomic mass} \times \text{number of atoms} = 4.4400513610370694 \times 10^{30} \mathrm{\ kg} \approx 2.3M_☉$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radius: $2,228,588 \mathrm{\ m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Temperature: $9.452 \times 10^{-7}° K$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force of gravity: $\frac{GM^2}{r}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Volume: $\frac{4\pi r^3}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Particle density: $\frac{\text{number of particles}}{\text{volume}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a: $\frac{4\sigma}{c}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radiation pressure: $\frac{a T^4}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gas pressure: $\text{particle density} \times k_B T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electron degeneracy pressure: $\frac{\pi^3 \hbar^2} {15 M_e} \times \frac{3 \times \text{number of electrons}} {V \pi}^{\frac{5}{3}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electron degeneracy pressure 2: $\frac{\pi^2 \hbar^2} {5 M_e \times M_h^{\frac{5}{3}}} \times \frac{3}{\pi}^{\frac{2}{3}} \times \frac{\frac{M}{V}}{1}^ {\frac{5}{3}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutron degeneracy pressure: $\frac{\pi^3 \hbar^2 }{15 M_e} \times \frac{3 \times \text{number of neutrons}}{V \pi}^{\frac{5}{3}}$     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutron degeneracy pressure 2: $\frac{3^{\frac{10}{3}} \hbar^2 }{ 15 \pi^{\frac{1}{3}} \times M_n^{\frac{8}{3}} \times r^5} \times \frac{M}{4}^{\frac{5} {3}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total pressure: $\text{gas pressure + radiation pressure + neutron degeneracy pressure + electron degeneracy pressure}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total pressure force: $\text{total pressure} \times \text{volume}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If total pressure force = force of gravity, then the body is in equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="10326" LastEditorUserId="9301" LastEditDate="2016-10-04T19:18:39.990" LastActivityDate="2016-10-04T19:18:39.990" Title="Formulae for gravitatitional equilibrium" Tags="&lt;star&gt;&lt;gravity&gt;&lt;neutron-star&gt;&lt;planet&gt;&lt;hydrostatic-equilibrium&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13260" PostTypeId="1" CreationDate="2016-01-20T01:08:54.063" Score="3" ViewCount="333" Body="&lt;p&gt;Why do pre-main-sequence stars on the Hayashi track remain at a constant temperature while they contract?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read &lt;a href=&quot;https://en.wikipedia.org/wiki/Hayashi_track#Derivation&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article&lt;/a&gt;, so no need to repeat the derivation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I took away from that is that the extra heat generated by contraction is given off almost exactly because the whole star is convective and the atmosphere is very capable of radiating that heat away if the opacity is mainly caused by hydrogen ions, which is the case in the cooler of the pre-main-sequence stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that correct?&lt;/p&gt;&#xA;" OwnerDisplayName="user10250" LastActivityDate="2016-10-13T01:58:14.120" Title="Why do PMS stars on the Hayashi track remain at a constant temperature while they contract?" Tags="&lt;stellar-evolution&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13261" PostTypeId="1" AcceptedAnswerId="13264" CreationDate="2016-01-20T07:47:51.100" Score="1" ViewCount="1718" Body="&lt;p&gt;I loaded my fits image to saoimage ds9. Then, Analysis-&gt;Catalogs automatically brought up source locations. &lt;a href=&quot;https://i.stack.imgur.com/hLKUx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hLKUx.png&quot; alt=&quot;screenshot&quot;&gt;&lt;/a&gt; how can i get xy image coordinates of this source locations?&lt;/p&gt;&#xA;" OwnerUserId="10413" LastActivityDate="2016-01-20T14:59:31.097" Title="how to get image coordinates of source points in saoimage ds9?" Tags="&lt;star&gt;&lt;astrophysics&gt;&lt;ds9&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13262" PostTypeId="2" ParentId="13189" CreationDate="2016-01-20T08:32:50.110" Score="2" Body="&lt;p&gt;Within the framework of General Relativity there are important solutions of the Einstein equations that are (a) homogeneous but anisotropic and (b) inhomogeneous yet isotropic (about a single point).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Class (a) are the Bianchi Cosmologies which are most simply described as homogeneous fluids that have different expansion rates in different directions, or some form of rotation.  There don't appear to be any simple descriptions of these but at a technical level it's hard to beat George Ellis' Cargese lectures: &lt;a href=&quot;http://arxiv.org/pdf/gr-qc/9812046.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/gr-qc/9812046.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Class(b) solutions are the Lemaitre-Tolman-Bondi (LTB) solutions which have the same non-uniform density distribution in all directions about one point.  See &lt;a href=&quot;https://en.wikipedia.org/wiki/Lema%C3%AEtre%E2%80%93Tolman_metric&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Lema%C3%AEtre%E2%80%93Tolman_metric&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our present universe is on average both homogeneous and isotropic, but both types of solution (a) and (b) nonetheless play an important role in cosmology. &lt;/p&gt;&#xA;" OwnerUserId="5301" LastActivityDate="2016-01-20T08:32:50.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13263" PostTypeId="2" ParentId="12385" CreationDate="2016-01-20T14:36:17.100" Score="2" Body="&lt;p&gt;I vaguely remember my dad talking about this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uranium and other heavy elements are dense. When the celestial body was molten (early in it's life) all the heavy elements sank to the core of the body. Now, with tectonics, the heavy elements are brought back up to the surface. This is why we can mine Uranium on Earth near the surface: the Uranium was brought back up due to convention current on Earth.&#xA;In conclusion, it's not worth it at all to mine on a non-tectonically active planet/moon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone can extend this, please do. &lt;/p&gt;&#xA;" OwnerUserId="10447" LastActivityDate="2016-01-20T14:36:17.100" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13264" PostTypeId="2" ParentId="13261" CreationDate="2016-01-20T14:59:31.097" Score="0" Body="&lt;p&gt;OK, it seems to me that what you have is a fits image (from where?) that has some world coordinate system information attached. When you load the 2MASS catalogue, SAOimage is able to use the RA and Dec in the catalogue to calculate the x,y positions of the catalog sources in your image (and marks them as green circles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is what you are asking - how do I get the x,y coordinates of these 2MASS sources in my image? Or do you wish to know what the x,y coordinates of the &lt;em&gt;actual sources in your image&lt;/em&gt;? If the latter then you will need some other software like Sextractor or DAOphot to do source-searching and parameterisation in your image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the former, here's a way of doing it. When you load the catalog you should see a window open entitled &quot;2MASS Point sources&quot; which lists all the sources in your image. From the &quot;File&quot; dialogue, select &quot;Copy to regions&quot;. Then, in the main image window open the &quot;Regions&quot; dialogue and select &quot;List regions&quot;. Another dialogue box will open asking you to select the format and coordinate system. From this, select &quot;x,y&quot; for the format and &quot;image&quot; for the coordinate system. This will give you a list of x,y coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to match this up with the 2MASS sources then you have to &quot;list regions&quot; again, but this time select &quot;fk5&quot; as the coordinate system. This will then give you a list of the RA, Dec of the same sources. Use you favourite editor to attach these lists to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite probably there is a more elegant solution, but this works.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-01-20T14:59:31.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13265" PostTypeId="2" ParentId="13246" CreationDate="2016-01-20T16:27:35.707" Score="2" Body="&lt;p&gt;One constraint is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Arago_spot&quot; rel=&quot;nofollow&quot;&gt;recommendation&lt;/a&gt; $F=\frac{d^2}{b\lambda}\geq 1$, in this case with $d=12700 \mbox{ km}$ about the diameter of Earth, $\lambda=600 \mbox{ nm}$ some wavelength of visible light, and $b$ the distance between circular obstacle and observer.&#xA;The distance between Earth and the observer should hence be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b\leq \frac{d^2}{\lambda}=\frac{(12700\cdot10^3\mbox{ m})^2}{600\cdot 10^{-9}\mbox{ m}}=481.67\cdot 10^{18}\mbox{ m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another constraint is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Arago_spot#The_circular_object.27s_surface_roughness&quot; rel=&quot;nofollow&quot;&gt;surface roughness of the circular object&lt;/a&gt;:&#xA;$\Delta r &amp;lt; \sqrt{r^2 + \lambda\frac{gb}{g+b}}-r$, with $r=6350\mbox{ km}$ the radius of the circular obstacle (here Earth), $g$ the distance between the point light source and the circular obstacle, and $b$ the distance between the circular obstacle and the screen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To simplify calculations, say $g\gg b$. Then approximately&#xA;$\Delta r &amp;lt; \sqrt{r^2 + \lambda\frac{gb}{g}}-r = \sqrt{r^2 + \lambda b}-r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After adding $r$ and squaring you get&#xA;$(\Delta r +r)^2&amp;lt;r^2+\lambda b$.&#xA;This simplifies to&#xA;$(\Delta r)^2+2r\Delta r&amp;lt;\lambda b$.&#xA;Assume $\Delta r\ll r$, and neglect the second order $(\Delta r)^2$ to get&#xA;$2r\Delta r&amp;lt;\lambda b$. Divide by $\lambda$ to get an approximate constraint for $b$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b&amp;gt;\frac{2r\Delta r}{\lambda}$.&#xA;With $2r=12700 \mbox{ km}$ about the diameter of Earth, $\lambda=600 \mbox{ nm}$ some wavelength of visible light, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b&amp;gt;\frac{12700\cdot 10^3\mbox{ m}\cdot\Delta r}{600\cdot 10^{-9}\mbox{ m}}&#xA;= 21.1667\cdot 10^{12}\Delta r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two constraints allow for reasonable values of $\Delta r$.&#xA;Assume a surface roughness of Earth of e.g. $\Delta r = 1\mbox{ km}$.&#xA;Then a valid range of distances of observers would be between&#xA;$0.00224$ and $50912$ &lt;a href=&quot;https://en.wikipedia.org/wiki/Light-year&quot; rel=&quot;nofollow&quot;&gt;lightyears&lt;/a&gt; of $9.4607\cdot 10^{15}\mbox{ m}$ from Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_unit&quot; rel=&quot;nofollow&quot;&gt;astronomcal units of&lt;/a&gt; $149597870700\mbox{ m}$ the closest distance of an observer would be $141.49 \mbox{ au}$ from Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to Earth's oblateness, however, you would get a &lt;a href=&quot;https://en.wikipedia.org/wiki/Point_spread_function&quot; rel=&quot;nofollow&quot;&gt;point spread function&lt;/a&gt; significantly different from a dot for this &quot;short&quot; distance from Earth. It might be possible to correct this by an appropriate telescope optics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect of &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_lens#Explanation_in_terms_of_space.E2.80.93time_curvature&quot; rel=&quot;nofollow&quot;&gt;gravitational lensing&lt;/a&gt; is&#xA;$\theta=\frac{4GM}{rc^2}=2.969\cdot 10^{-27}\frac{\mbox{m}}{\mbox{kg}}\frac{M}{r}$, after applying the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_constant&quot; rel=&quot;nofollow&quot;&gt;constant of gravitation&lt;/a&gt; $G$ and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Speed_of_light&quot; rel=&quot;nofollow&quot;&gt;speed of light&lt;/a&gt; $c$. With the mass $M=5.97237\cdot 10^{24}\mbox{ kg}$ and a radius of $r=6350000\mbox{ m}$ of &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth&quot; rel=&quot;nofollow&quot;&gt;Earth&lt;/a&gt;, we get an angle of&#xA;$\theta=2.793\cdot 10^{-9}$ by gravitational lensing at the surface of Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would focus parallel rays of light to a point near a distance of&#xA;$b=\frac{r}{\tan \theta}=130.27\cdot 10^{15}\mbox{ m}$, or $13.77$ lightyears, hence well beyond the minimum distance where an Arago spot could form. But, of course, the innermost peak of the point spread function would be closer to a circular disc at this larger distance with relevant gravitational lensing.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2016-01-20T16:27:35.707" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13266" PostTypeId="2" ParentId="13254" CreationDate="2016-01-20T17:38:37.517" Score="1" Body="&lt;p&gt;I think one of the most important things you missed in this article is the bit where they say the energy requirements are &amp;quot;unphysical&amp;quot;.  This is no different from saying it's impossible from our point of view.&lt;/p&gt;&#xA;&lt;p&gt;A couple of quotes from the article :&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Here as well, it was found that the wall of the “warp bubble” surrounding a spaceship must be unphysically thin compared to the bubble radius&lt;/p&gt;&#xA;&lt;p&gt;Estimates of the total negative energy required to construct Krasnikov tubes of even modest dimensions were shown to be unphysically large.  Similar difficulties have been recently shown to plague warp bubbles and wormholes.  The Krasnikov tube suffers from some of the same drawbacks as these other pro&#xA;posed methods of faster-than-light travel, and is hence also a very unlikely&#xA;possibility.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The model you're discussing is really best thought of as a mathematical exploration, rather than a practical proposal of any kind.  I would suggest the authors are &lt;em&gt;disproving&lt;/em&gt; the viability of Krasnikov tubes rather than anything else.  What was done was to try and fix problems with the previous model, but in doing so the authors ( to their credit ) show that they cannot bypass the fundamental problems making such a system unrealisable in practice.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-01-20T17:38:37.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13267" PostTypeId="2" ParentId="12356" CreationDate="2016-01-20T18:36:20.813" Score="8" Body="&lt;p&gt;&lt;a href=&quot;http://www.caltech.edu/news/caltech-researchers-find-evidence-real-ninth-planet-49523&quot; rel=&quot;noreferrer&quot;&gt;According to Caltech&lt;/a&gt; there is possibly a planet 10 times the mass of earth in a 15-20 thousand year orbit at an odd angle. It hasn't been observed yet. There's a &lt;a href=&quot;http://www.findplanetnine.com/&quot; rel=&quot;noreferrer&quot;&gt;search for it ongoing&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6870" LastEditorUserId="6870" LastEditDate="2016-01-21T05:44:28.077" LastActivityDate="2016-01-21T05:44:28.077" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13268" PostTypeId="1" AcceptedAnswerId="13271" CreationDate="2016-01-20T18:42:58.860" Score="18" ViewCount="601" Body="&lt;p&gt;Caltech just &lt;a href=&quot;http://www.caltech.edu/news/caltech-researchers-find-evidence-real-ninth-planet-49523&quot; rel=&quot;noreferrer&quot;&gt;released a report&lt;/a&gt; that says there is possibly a large (10 earth mass) planet in a remote orbit (10 - 20 thousand year) that explains a lot of observations of Kuiper Belt objects. How long could it take for this to be confirmed or refuted observationally?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://www.youtube.com/watch?v=6poHQ2h00ZA&quot; rel=&quot;noreferrer&quot;&gt;this video&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="6870" LastEditorUserId="6870" LastEditDate="2016-01-24T17:34:38.903" LastActivityDate="2016-01-24T17:34:38.903" Title="How long should it take for us to observationally determine if Caltech's Planet Nine is really there?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;observational-astronomy&gt;&lt;9th-planet&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13269" PostTypeId="1" CreationDate="2016-01-20T19:02:43.100" Score="4" ViewCount="159" Body="&lt;p&gt;I don't quite know how to word this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How close could a brown dwarf be to us and have gone undetected until now?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or maybe, with what probability are we sure that there is no brown dwarf within X light years of the sun? (For different values of X)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Don't worry about addressing Nemesis or a hypothetical binary partner for the sun. I'm just asking about undetected brown dwarfs..., even outside of the Oort cloud but within Alpha C. I don't know how sensitive the WISE survey was.)&lt;/p&gt;&#xA;" OwnerDisplayName="user10250" LastEditorDisplayName="user10250" LastEditDate="2016-01-20T19:27:01.153" LastActivityDate="2016-01-20T20:18:51.017" Title="We are sure that there are no brown dwarfs closer than ______" Tags="&lt;brown-dwarf&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13270" PostTypeId="1" AcceptedAnswerId="13279" CreationDate="2016-01-20T19:54:14.817" Score="36" ViewCount="5319" Body="&lt;p&gt;I've seen a number of news reports indicating there is likely a &lt;a href=&quot;http://news.discovery.com/space/astronomy/9th-planet-may-lurk-in-the-outer-solar-system-160120.htm&quot;&gt;9th planet in our Solar System&lt;/a&gt;, something with an orbital period of between 10k-20k years, that is 10 times Earth's mass. I haven't seen any real indication of where this object might be. If I had access to a sufficient telescope, would I be able to find this planet, and what way would I point a telescope to find it? How far is it likely to be, or is that not well known?&lt;/p&gt;&#xA;" OwnerUserId="182" LastEditorUserId="182" LastEditDate="2016-01-21T00:53:25.713" LastActivityDate="2016-04-13T10:51:13.933" Title="9th planet location?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;9th-planet&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13271" PostTypeId="2" ParentId="13268" CreationDate="2016-01-20T20:03:07.337" Score="13" Body="&lt;p&gt;Those are model calculations, which hint to the existence of a possible body of about 10-times the mass of Earth. Calling this a discovery would clearly be premature. The confidence level is just a little above the &quot;evidence&quot; level of 3 sigma, under the assumption, that the discoveries of the KBO objects leading to the inference aren't observationally biased.&#xA;It's a long way to direct observation, since precise orbital data haven't been inferred.&#xA;Another option is, that there has been a planet. But it might have left our Solar system. Hence predicting a date for direct observation doesn't appear reasonable at the moment.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2016-01-20T20:11:10.360" LastActivityDate="2016-01-20T20:11:10.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13272" PostTypeId="2" ParentId="13269" CreationDate="2016-01-20T20:07:51.403" Score="5" Body="&lt;p&gt;I think the answer is roughly 3-5 light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The recent WISE survey in the near-infrared should have been capable of detecting even a very cool nearby brown dwarf (and indeed it has detected some very cool brown dwarfs - e.g. a 250K brown dwarf only 6 light years away &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014ApJ...786L..18L&quot;&gt;Luhman et al. 2014&lt;/a&gt;). Since WISE was an all-sky survey and there is great interest in finding these ultracool brown dwarfs, then it is reasonable to assume that any similar, object within a few light years would already have been found. The caveat to this is that the point spread function of the WISE survey s not that great an so near the galactic plane it could be that source confusion might have hidden the brown dwarf. On the other hand, nearby objects usually have high proper motion, so it would probably have been uncovered in combination with the earlier 2MASS survey, since as I show below, such an object may have been detected in both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some details: As an example let's take a very low mass brown dwarf - say 20 Jupiter masses and 5 billion years old - and place it 4 light years from the Sun. According to the evolutionary models of &lt;a href=&quot;http://adsabs.harvard.edu/abs/2008ApJ...689.1327S&quot;&gt;Saumon &amp;amp; Marley (2008)&lt;/a&gt;, such an object has an intrinsic luminosity of $10^{-7}$ times (1 ten millionth) that of the Sun and a temperature of 400 Kelvin and would appear to have a spectral type of late T or early Y. This is similar to the coolest brown dwarfs yet found. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the calibration of absolute magnitudes versus spectral type for cool brown dwarfs in &lt;a href=&quot;http://arxiv.org/abs/1211.6977&quot;&gt;Marsh et al. (2013)&lt;/a&gt; we know that at 4 light years, such a brown dwarf would have magnitudes of $H=16$ and $W2=10$. The former is just about bright enough to have been seen in the 2MASS all-sky survey and the latter is easily detected by WISE. The combination of data would also easily have revealed the large parallax of such an object. We can conclude that an object would have to have &lt;em&gt;much&lt;/em&gt; lower mass than this to remain undetected and thus would arguaby not be called a brown dwarf, but would really be a &quot;free-floating planet&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you might ask, what about a cooler brown dwarf? Well the theoretical predictions of magnitudes and colours become &lt;em&gt;much&lt;/em&gt; more uncertain, but I would simply refer to the Luhman discovery. This brown dwarf is about as cool as a brown dwarf could be given the age of the Galaxy, may only be 10 Jupiter masses, yet was still easily detected in the WISE survey. Therefore I think this sets a fairly firm empirical limit.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-01-20T20:18:51.017" LastActivityDate="2016-01-20T20:18:51.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13273" PostTypeId="2" ParentId="13270" CreationDate="2016-01-20T20:39:10.220" Score="4" Body="&lt;p&gt;Citing the &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/0004-6256/151/2/22/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;original article&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We find that the observed orbital alignment can be maintained by a distant eccentric planet with mass $\geq \approx 10$ m⊕ whose orbit lies in approximately the same plane as those of the distant KBOs, but whose perihelion is 180° away from the perihelia of the minor bodies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As already alluded to above, the precise range of perturber parameters required to satisfactorily reproduce the data is at present difficult to diagnose. Indeed, additional work is required to understand the tradeoffs between the assumed orbital elements and mass, as well as to identify regions of parameter space that are incompatible with the existing data.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, finding out likely orbital parameters is work in progress.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2016-01-20T20:39:10.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13274" PostTypeId="1" CreationDate="2016-01-20T20:39:44.823" Score="2" ViewCount="55" Body="&lt;p&gt;Dark matter was needed to give the stars in a spiral galaxy a cause of their speed. So they proposed there must be some extra dark matter in the Galaxy. The total extra mass of dark mater is quite high. As a consequence of that, the whole galaxy becomes much heavier. But in the group or a cluster, the Galaxy also has a speed relative to other galaxies. So my question is whether the mass needed to give the stars their speed is the same mass that is needed to explain its movement within its cluster or a group?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="13" LastEditDate="2016-01-22T03:45:58.933" LastActivityDate="2016-01-22T03:45:58.933" Title="Does the quantity of dark matter needed in the Milky Way to give the stars their speed compare with the speed of the Milky Way?" Tags="&lt;milky-way&gt;&lt;dark-matter&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13275" PostTypeId="2" ParentId="13270" CreationDate="2016-01-20T21:14:07.457" Score="8" Body="&lt;p&gt;The position of the hypothetical object is not known with any certainty, so it's hard to know where to point your telescope. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper proposes a wide range of orbital distances anywhere from 400 to 1500 AU semi-major axis, with a perihelion (closest approach to the sun) of 200-300AU. This is 8 times as far as Neptune. (I didn't read the article closely enough to determine whether the body would be near perihelion or not at present; it could be over 1000 AU away, 30 times Neptune's distance.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a mass of 10 Earths, we would expect the body to be something like 2-5 times Earth's radius -- somewhat smaller than Neptune. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The combination of distance and size suggests the body would be far fainter than Neptune, no brighter than magnitude 16.5 at perihelion, and likely much dimmer.&lt;/p&gt;&#xA;" OwnerUserId="2573" LastEditorUserId="2573" LastEditDate="2016-01-21T00:20:11.000" LastActivityDate="2016-01-21T00:20:11.000" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13276" PostTypeId="1" AcceptedAnswerId="13278" CreationDate="2016-01-20T21:32:10.020" Score="1" ViewCount="2442" Body="&lt;p&gt;What does the surface of Mercury look like?  What color is it (I have seen some images portraying it grey and some a light brownish color), and would the surface be dusty like the Moon, smooth, or even cracked (e.g. see image below) due to thermal expansion and contraction?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/dqaOr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dqaOr.jpg&quot; alt=&quot;mud cracks&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="2153" LastEditDate="2016-01-20T22:17:13.243" LastActivityDate="2019-02-19T20:47:34.413" Title="What does the surface of Mercury look like?" Tags="&lt;surface&gt;&lt;mercury&gt;&lt;geology&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13277" PostTypeId="1" CreationDate="2016-01-20T21:44:55.737" Score="-1" ViewCount="118" Body="&lt;p&gt;I observed this when I was playing around with an amateur (crappy) telescope. The picture was taken using a dslr attached to the telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was taken in San Francisco on 2015-01-20. I don't remember the direction but I would guess it was pointing East-South East - that's the only direction which has a clear visibility from where I took this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone identify what I am looking at here? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0URWZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0URWZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:  I'm not discounting the fact that it could have been an aberration in the telescope or the camera.&lt;/p&gt;&#xA;" OwnerUserId="10449" LastEditorDisplayName="user1569" LastEditDate="2016-01-26T23:12:36.570" LastActivityDate="2016-01-26T23:12:36.570" Title="Identify this object" Tags="&lt;near-earth-object&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2016-01-28T16:41:08.987" ContentLicense="CC BY-SA 3.0" />
  <row Id="13278" PostTypeId="2" ParentId="13276" CreationDate="2016-01-20T22:11:47.430" Score="3" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/MESSENGER&quot; rel=&quot;nofollow noreferrer&quot;&gt;MESSENGER&lt;/a&gt; probe was able to take many true-color pictures of Mercury. A full list can be found on &lt;a href=&quot;http://photojournal.jpl.nasa.gov/targetFamily/Mercury&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL's Photojournal&lt;/a&gt;. It is clear that Mercury is light grey in color.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qia2Z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qia2Z.jpg&quot; alt=&quot;&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://photojournal.jpl.nasa.gov/jpegMod/PIA19410_modest.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;nasa.gov&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pTnJr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pTnJr.jpg&quot; alt=&quot;&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://photojournal.jpl.nasa.gov/jpegMod/PIA19283_modest.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;nasa.gov&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/amPh8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/amPh8.jpg&quot; alt=&quot;&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://photojournal.jpl.nasa.gov/jpegMod/PIA19449_modest.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;nasa.gov&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;In terms of the actual surface, Mercury is very similar to the Moon. It's surface is speckled with craters, with some smaller craters inside them, as can be seen in some of the images above. The smoothness varies - note the inside of the larger crater in the first picture. It's quite smooth, save the &lt;em&gt;smaller&lt;/em&gt; crater inside it.&lt;/p&gt;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://www.lpi.usra.edu/meetings/mercury01/pdf/8029.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this pdf&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-02-19T20:47:34.413" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="13279" PostTypeId="2" ParentId="13270" CreationDate="2016-01-20T23:07:09.203" Score="31" Body="&lt;p&gt;&lt;strong&gt;It's too dim to be seen during a normal survey during the majority of its&#xA;orbit.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;  Scientists at the University of Bern have &lt;a href=&quot;http://www.unibe.ch/news/media_news/media_relations_e/media_releases/2016_e/media_releases_2016/planet_9_takes_shape/index_eng.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;modeled&lt;/a&gt; a hypothetical 10 Earth mass planet in the proposed orbit to estimate its detectability with more precision than my attempt below.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The takeaway is that NASAs WISE mission would have probably spotted a planet of at least 50 Earth masses in the proposed orbit and that none of our current surveys would have had a chance to find one below 20 earth masses in most of its orbit.  They put the planets temperature at 47K due to residual heat from formation; which would make is 1000x brighter in infrared than it is in visible light reflected from the sun.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should however be within reach of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Large_Synoptic_Survey_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;LSST&lt;/a&gt; once it is completed (first light 2019, normal operations beginning 2022); so the question should be resolved within a few more years even if its far enough from Batygin and Brown's proposed orbit that their search with the Subaru telescope comes out empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My original attempt to handwave an estimate of detectability is below.&lt;/strong&gt;&#xA;The &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/0004-6256/151/2/22/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper&lt;/a&gt; gives potential orbital parameters of $400-1500~\textrm{AU}$ for the semi major axis, and $200-300~\textrm{AU}$ for perihelion. Since the paper doesn't give a most-likely case for orbital parameters, I'm going to go with the extreme case that makes it most difficult to find. Taking the most eccentric possible values from that gives an orbit with a $1500~\textrm{AU}$ semi-major axis and a $200~\textrm{AU}$ perihelion has a $2800~\textrm{AU}$ aphelion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the brightness of an object shining with reflected light, the proper scaling factor is not a $1/r^2$ falloff as could be naively assumed.  That is correct for an object radiating its own light; but not for one shining by reflected light; for that case the same $1/r^4$ scaling as in a &lt;a href=&quot;https://en.wikipedia.org/wiki/Inverse-square_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;radar return&lt;/a&gt; is appropriate. That this is the correct scaling factor to use can be sanity checked based on the fact that despite being similar in size, Neptune is $\sim 6x$ dimmer than Uranus despite being only $50\%$ farther away: $1/r^4$ scaling gives a $5x$ dimmer factor vs $2.25$ for $1/r^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using that gives a dimming of 2400x at $210~\textrm{AU}\;.$ That puts us down $8.5$ magnitudes down from Neptune at perihelion or $16.5$ magnitude.  $500~\textrm{AU}$ gets us to $20$th magnitude, while a $2800~\textrm{AU}$ aphelion dims reflected light down by nearly $20$ magnitudes to $28$ magnitude. That's equivalent to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_magnitude&quot; rel=&quot;nofollow noreferrer&quot;&gt;faintest stars visible from an 8 meter telescope&lt;/a&gt;; making its non-discovery much less surprising.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is something of a fuzzy boundary in both directions. Residual energy from formation/radioactive material in its core will be giving it some innate luminosity; at extreme distances this might be brighter than reflected light. I don't know how to estimate this. It's also possible that the extreme cold of the Oort Cloud may have frozen its atmosphere out. If that happened, its diameter would be much smaller and the reduction in reflecting surface could dim it another order of magnitude or two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not knowing what sort of adjustment to make here, I'm going to assume the two factors cancel out completely and leave the original assumptions that it reflects as much light as Neptune and reflective light is the dominant source of illumination for the remainder of my calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference, data from NASA's &lt;a href=&quot;https://en.wikipedia.org/wiki/Nemesis_%28hypothetical_star%29#Past.2C_current.2C_and_pending_searches_for_Nemesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;WISE&lt;/a&gt; experiment has ruled out a Saturn-sized body within $10,000~\textrm{AU}$ of the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It's also likely too faint to have been detected via proper motion; although if we can pin its orbit down tightly Hubble could confirm its motion.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbital eccentricity can be calculated as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e = \frac{r_\textrm{max} - r_\textrm{min}}{2a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging in the numbers gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e = \frac{2800~\textrm{AU} - 200~\textrm{AU}}{2\cdot 1500~\textrm{AU}} = 0.867$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging $200~\textrm{AU}$ and $e = 0.867$ into a &lt;a href=&quot;http://www.phas.ubc.ca/~newhouse/p210/orbits/#bottom&quot; rel=&quot;nofollow noreferrer&quot;&gt;cometary orbit calculator&lt;/a&gt; gives a $58,000$ year orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While that gives an average proper motion of $ 22~\textrm{arc-seconds/year}\;,$ because the orbit is highly eccentric its actual proper motion varies greatly, but it spends a majority of its time far from the sun where its values are at a minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kepler's laws tell us that the &lt;a href=&quot;http://people.bridgewater.edu/~rbowman/ISAW/KeplerCalc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;velocity at aphelion&lt;/a&gt; is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_a^2 = \frac{ 8.871 \times 10^8 }{ a }  \frac{ 1 - e }{ 1 + e }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $v_a$ is the aphelion velocity in $\mathrm{m/s}\;,$ $a$ is the semi-major axis in $\mathrm{AU},$ and $e$ is orbital eccentricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_a = \sqrt{\frac{ 8.871 \times 10^8 }{ 1500 } \cdot \frac{ 1 - 0.867 }{ 1 + 0.867 }} = 205~\mathrm{m/s}\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the proper motion we first need to convert the velocity into units of $\textrm{AU/year}:$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$205 \mathrm{\frac{m}{s}}\;  \mathrm{\frac{3600 s}{1 h}} \cdot \mathrm{\frac{24 h}{1 d}} \cdot \mathrm{\frac{365 d}{1 y}} \cdot \mathrm{\frac{1\; AU}{1.5 \times 10^{11}m}} = 0.043~\mathrm{\frac{AU}{year}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get proper motion from this, create a triangle with a hypotenuse of $2800~\textrm{AU}$ and a short side of $0.043~\textrm{AU}$ and then use trigonometry to get the narrow angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin \theta = \frac{0.044}{2800}\\ \implies \theta = {8.799×10^{-4}}^\circ   = 3.17~\textrm{arc seconds}\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is well within Hubble's angular resolution of $0.05~\textrm{arc seconds};$ so if we knew exactly where to look we could confirm its orbit even if its near its maximum distance from the sun. However its extreme faintness in most of its orbit means that its unlikely to have been found in any survey. If we're lucky and it's within $\sim 500~\textrm{AU},$ it would be bright enough to be seen by the ESA's &lt;a href=&quot;https://en.wikipedia.org/wiki/Gaia_%28spacecraft%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;GAIA&lt;/a&gt; spacecraft in which case we'll located it within the next few years. Unfortunately, it's more likely that all the GAIA data will do is to constrain its minimum distance slightly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its parallax movement &lt;a href=&quot;https://astronomy.stackexchange.com/a/13283/186&quot;&gt;would be much larger&lt;/a&gt;; however the challenge of actually seeing it in the first place would remain.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-04-13T10:51:13.933" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13280" PostTypeId="1" AcceptedAnswerId="13283" CreationDate="2016-01-21T00:50:52.610" Score="12" ViewCount="400" Body="&lt;p&gt;We know from &lt;a href=&quot;https://astronomy.stackexchange.com/q/13270/182&quot;&gt;this question&lt;/a&gt; that an object is hypothesized to be at between 200-1000 AU that is large enough to detect, known now as the 9th planet. It should move about 40 arcseconds per year. It seems to me that the parallax distance would be more than this distance per year. Would it be easier to detect the 9th planet using parallax than its movement around the Sun?&lt;/p&gt;&#xA;" OwnerUserId="182" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-01-23T11:37:04.187" Title="How would parallax affect an object at 200-1000 AU (for example, the 9th planet)?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;distances&gt;&lt;parallax&gt;&lt;9th-planet&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13281" PostTypeId="2" ParentId="13268" CreationDate="2016-01-21T01:23:23.750" Score="3" Body="&lt;p&gt;Maybe the Gaia space telescope has already caught it? First data release in &lt;a href=&quot;http://sci.esa.int/gaia/56838-gaia-data-processing-consortium-meets-ahead-of-first-catalogue-release-next-year/&quot; rel=&quot;nofollow&quot;&gt;mid-2016&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-01-21T01:23:23.750" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13282" PostTypeId="1" AcceptedAnswerId="13285" CreationDate="2016-01-21T06:01:25.047" Score="21" ViewCount="2995" Body="&lt;p&gt;Theoretical &lt;a href=&quot;http://www.findplanetnine.com/&quot;&gt;Planet Nine&lt;/a&gt;, &lt;a href=&quot;http://www.caltech.edu/news/caltech-researchers-find-evidence-real-ninth-planet-49523&quot;&gt;proposed&lt;/a&gt; by Caltech's &lt;a href=&quot;https://twitter.com/plutokiller&quot;&gt;Mike Brown&lt;/a&gt; and &lt;a href=&quot;http://web.gps.caltech.edu/~kbatygin/Home.html&quot;&gt;Konstantin Batygin&lt;/a&gt;, is said to have a 15-20 thousand year orbit. Approximately how bright would the sun appear from Planet Nine aphelion and perihelion? &lt;/p&gt;&#xA;" OwnerUserId="6870" LastActivityDate="2018-05-21T21:02:24.990" Title="How bright would the sun appear from the hypothetical Planet Nine proposed by Caltech?" Tags="&lt;the-sun&gt;&lt;planet&gt;&lt;9th-planet&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13283" PostTypeId="2" ParentId="13280" CreationDate="2016-01-21T07:46:24.627" Score="10" Body="&lt;p&gt;The parallax is found from a triangle with the Earth's orbit at its base. The annual parallax here would then be between 1/200 and 1/1000 radians ($\sim$ 1000 to 200 arcseconds) which is indeed enormous and much bigger than the likely proper motion during a year.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-01-21T07:46:24.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13284" PostTypeId="2" ParentId="13282" CreationDate="2016-01-21T07:59:24.237" Score="11" Body="&lt;p&gt;If it exists, then the orbit has a perihelion of maybe 300 au and an aphelion of perhaps 2700 au.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then just scale from the brightness $m=-26.7$ of the Sun at the Earth. It would be between 12.4 and 17.1 magnitudes fainter. So still much brighter than the next brightest star in the sky.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-01-21T07:59:24.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13285" PostTypeId="2" ParentId="13282" CreationDate="2016-01-21T08:02:58.303" Score="28" Body="&lt;p&gt;Between $1/40,000$ and $1/8,000,000$ of the brightness as seen from Earth, depending on what the actual orbit would turn out to be, and where the planet is in its $15,000$ year orbit period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brightness drops as $\dfrac{1}{r^2}$ with distance from the light source. Earth is at $1~\textrm{AU}.$ The theoretical planet is at $200~\textrm{AU}$ when it's closest to the sun, and up to $2800~\textrm{AU}$ at the point furthest from the sun at the upper end of the estimated orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;So e.g. $1/200^2 = 1/40,000$ of the brightness (luminance) as seen from Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For how a human would experience it, we can convert to &lt;a href=&quot;https://en.wikipedia.org/wiki/Exposure_value#Tabulated_exposure_values&quot;&gt;exposure value&lt;/a&gt; as used in photography: The difference in exposure value (photographic 'stops') is $\log 2$ of the luminance ratio, so we would have $15$ to $23$ stops less light than on Earth. &#xA;Sunny noon on Earth is $15~\textrm{EV}.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the brightness at noon on the planet surface would be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;At $200~\textrm{AU},$ planet orbit is closest to the sun: About $0~\textrm{EV},$ roughly the same as a dimly lit interior&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$400~\textrm{AU},$ lower bound on semi-major axis: $-2 ~\textrm{EV},$ similar to a landscape lit by the full moon&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$1500~\textrm{AU},$ upper bound on semi-major axis: $-6~\textrm{EV},$ similar to landscape lit by a quarter moon&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$2800~\textrm{AU},$ upper bound on aphelion (the point on orbit most distant from the sun): $-8~\textrm{EV}.$ This would be dark, but you would probably still see enough to avoid running into things.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9682" LastEditorUserId="-1" LastEditDate="2016-01-21T13:58:58.890" LastActivityDate="2016-01-21T13:58:58.890" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13286" PostTypeId="1" AcceptedAnswerId="13384" CreationDate="2016-01-21T08:28:50.997" Score="9" ViewCount="276" Body="&lt;p&gt;Non-periodic comets are comets which have very long orbital periods (&gt;200 years or more), spending most of their time in the outer solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planet X, recently revived by researchers at Caltech, is a proposed planet orbiting in the farthest reaches of the outer solar system. The Caltech researchers infer its presence from to its effect on the orbits of Kuiper belt objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If such a planet exists, and is the cause of the Kuiper belt object alignments, could it not also knock such objects in to the inner solar system, creating non-periodic comets with huge aphelion orbital distances? Could our current observations of non-periodic comets be used to infer properties of the planet?&lt;/p&gt;&#xA;" OwnerUserId="941" LastEditorUserId="793" LastEditDate="2016-01-21T09:53:52.130" LastActivityDate="2016-01-25T23:33:27.103" Title="Could the trajectories of non-periodic comets be used to infer properties of the 'ninth planet'?" Tags="&lt;solar-system&gt;&lt;comets&gt;&lt;kuiper-belt&gt;&lt;9th-planet&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13287" PostTypeId="1" CreationDate="2016-01-21T11:38:54.807" Score="1" ViewCount="788" Body="&lt;p&gt;how is the rotation and revolution of a planet that does not have a&#xA; natural satellite different from  a planet that has say 1 natural satellite&#xA;(ex. earth and moon system).&lt;/p&gt;&#xA;" OwnerUserId="10430" LastActivityDate="2016-02-01T07:51:14.173" Title="What effect does a natural satellite have on a planet's rotation and revolution?" Tags="&lt;rotation&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13289" PostTypeId="1" CreationDate="2016-01-21T10:55:37.250" Score="0" ViewCount="431" Body="&lt;p&gt;Is there any specific numbers on stars &amp;amp; planets count on Milky Way Galaxy ?&lt;/p&gt;&#xA;" OwnerUserId="10457" OwnerDisplayName="Mani" LastEditorUserId="1386" LastEditDate="2016-01-21T17:36:01.203" LastActivityDate="2016-03-23T06:16:17.510" Title="Number of stars &amp; planets in Milky Way Galaxy?" Tags="&lt;star&gt;&lt;planet&gt;&lt;milky-way&gt;&lt;estimate&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13290" PostTypeId="2" ParentId="13289" CreationDate="2016-01-21T12:11:35.057" Score="2" Body="&lt;p&gt;A consensus number is that there roughly $10^{11}$ stars in our Galaxy (though this number is certainly uncertain by a factor of at least two, because it is based on extrapolating what we know about stellar populations in our vicinity). Most of these stars are of lower mass and are much less luminous than the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of planets is even more uncertain. It now seems probable that most stars like the Sun have at least one planet, but we really don't know that much about the diversity of planetary systems, how typical something like the solar system is, or how planetary systems might change as a function of position in the Galaxy. We also don't know that much about planetary systems around the dominant (by number) very low-mass stars. They certainly can have planetary systems, but the fraction that do is still a work in progress.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-01-21T12:11:35.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13291" PostTypeId="1" AcceptedAnswerId="13300" CreationDate="2016-01-21T12:37:55.290" Score="1" ViewCount="43" Body="&lt;p&gt;I read somewhere that counting from today, five planets will be aligned during almost two weeks IIRC. The planets are : Mars, Jupiter, Saturn, Mercury and... us. I know Jupiter and Mars can be seen with the naked eye, but I would really like to catch a glimpse of Saturn. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming one does not own a telescope, but small binoculars, how can one maximize his chances of seeing at least Mars, Jupiter and Saturn ? &lt;/p&gt;&#xA;" OwnerUserId="9679" LastActivityDate="2016-01-21T18:41:40.913" Title="How to watch at best this week's planet alignement?" Tags="&lt;planet&gt;&lt;saturn&gt;&lt;alignment&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13293" PostTypeId="2" ParentId="13270" CreationDate="2016-01-21T13:15:03.807" Score="-1" Body="&lt;p&gt;If you had access to a sufficient telescope, you could theoretically see it, if you looked in the right place (although no one knows where the right place might be). But if it's anywhere near aphelion there are only a handful of sufficient telescopes in the world (let's say an 8m mirror or larger), so I think it highly unlikely that you have access to one of them.&lt;/p&gt;&#xA;" OwnerUserId="9235" LastActivityDate="2016-01-21T13:15:03.807" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13294" PostTypeId="5" CreationDate="2016-01-21T14:17:42.193" Score="0" Body="&lt;p&gt;On Jan 20, 2016, &lt;a href=&quot;http://www.theweek.com/speedreads/600274/astronomers-find-evidence-9th-planet-lurking-edge-solar-system&quot; rel=&quot;nofollow noreferrer&quot;&gt;an announcement was made&lt;/a&gt; that using the trajectories of Kuiper Belt Objects, a 9th planet was expected to be found in the far reaches of the Solar System. Such a planet has been hypothesized several times, but this tag is specifically intended for questions relating to a hypothetical planet sized object which dominates the Outer Kuiper Belt or inner Oort Cloud region.&lt;/p&gt;&#xA;&lt;p&gt;The planet is theorized to have a mass of 5-10 Earth masses, orbiting 400-800 AU away from the Sun.&lt;/p&gt;&#xA;" OwnerUserId="182" LastEditorUserId="31410" LastEditDate="2021-02-12T09:47:08.133" LastActivityDate="2021-02-12T09:47:08.133" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13295" PostTypeId="4" CreationDate="2016-01-21T14:17:42.193" Score="0" Body="Questions relating to the discovery of a new ninth planet (not Pluto) hypothesized in the Solar System." OwnerUserId="182" LastEditorUserId="9301" LastEditDate="2016-10-04T19:44:24.400" LastActivityDate="2016-10-04T19:44:24.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13296" PostTypeId="1" AcceptedAnswerId="13299" CreationDate="2016-01-21T14:39:35.643" Score="1" ViewCount="108" Body="&lt;p&gt;What's in upper part in our galaxy ?&#xA;If space is 3 dimensional, what lies beneath solar system, or in upper part ? Why it seems from every graphical view of solar system / galaxy, everything is in one single axis ? &lt;/p&gt;&#xA;" OwnerUserId="277" LastActivityDate="2016-01-21T18:08:33.237" Title="Why there is no axis measurement when telling distance between objects in space ?" Tags="&lt;galaxy&gt;&lt;distances&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13299" PostTypeId="2" ParentId="13296" CreationDate="2016-01-21T18:08:33.237" Score="4" Body="&lt;p&gt;You might benefit from watching this Minute Physics video on &lt;a href=&quot;https://youtu.be/tmNXKqeUtJM&quot; rel=&quot;nofollow&quot;&gt;Why the Solar System is Flat&lt;/a&gt;. A summary is due to the way objects in our solar system and particular galaxy are drawn around a central massive item by gravity. Through the conservation of this angular momentum, and the loss of vertical motion though collisions, the solar system and galaxy &quot;look&quot; flat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Talking of &quot;upper&quot; and &quot;lower&quot; doesn't really make sense in space. Gravity feels like it's pulling us down on Earth, and that is why we have this concept, but in space we lack this reference. For our own purposes, humans have created an artificial means of determining points of reference for spacial coordinate systems. Due to the rotation of the Galaxy we have established a base plane called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_plane&quot; rel=&quot;nofollow&quot;&gt;Galactic plane&lt;/a&gt;, from which a coordinate system and sense of &quot;up&quot; and &quot;down&quot; might be inferred (I think our solar system tilts at a 63 deg to the galactic plane of the Milky Way). In the context of Earth and our solar system, we use the plane that the Earth &quot;makes&quot; as it circles the Sun as the point of context for the Earth's axial tilt; we call it the &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/eclip.html&quot; rel=&quot;nofollow&quot;&gt;ecliptic plane&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the rest of the Universe you may wish to check out this blog post from Cornell called &lt;a href=&quot;http://curious.astro.cornell.edu/about-us/103-the-universe/cosmology-and-the-big-bang/geometry-of-space-time/600-why-is-the-universe-flat-and-not-spherical-advanced&quot; rel=&quot;nofollow&quot;&gt;Why is the Universe flat and not spherical?&lt;/a&gt;. Basically the Universe has a slight curvature to it but since we are on the surface, and have a limited view, it is difficult for us to see anything but flatness. This is related to why graphical views appear flat. Graphical views are generally limited due to our point of view and the sheer astronomical size of distances in space. Even in &quot;dense&quot; super clusters there are vast distances between objects. Overall it's easier to linearly show things rather than as they truly might be.&lt;/p&gt;&#xA;" OwnerUserId="2968" LastActivityDate="2016-01-21T18:08:33.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13300" PostTypeId="2" ParentId="13291" CreationDate="2016-01-21T18:41:40.913" Score="2" Body="&lt;p&gt;All these planets are naked eye objects, and except for Mercury, are easy to see. Venus is also part of this grouping, Meaning that all 5 planets known to ancient astronomers are on the same side of the Earth. (and the sun is on the other)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The planets are not exactly &quot;aligned&quot;, just they are all visible in the sky, spread over a wide part of the south eastern sky, a little while before dawn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.skyandtelescope.com/astronomy-news/observing-news/get-up-early-see-five-planets-at-once-01182015/&quot; rel=&quot;nofollow&quot;&gt;Sky and Telescope&lt;/a&gt; has some nice guides to what you can expect. Mercury is always the hardest, as it is always in the twilight. Saturn is an easy sight, appearing as a slightly yellow star. Binoculars will show its moon, Titan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best time is around the first week of Feb about 45 min before dawn.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-01-21T18:41:40.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13301" PostTypeId="1" CreationDate="2016-01-21T19:26:28.250" Score="29" ViewCount="2071" Body="&lt;p&gt;The internet is abuzz with the new evidence of the possible ninth planet in our Solar System. With those nine objects, the asteroid belt, the Kuiper belt, and everything else in our Solar System - the number of objects which orbit our star is pretty massive. Is this odd for a star its size? It's known that our Sun is a comparatively small star, so does having so many planets make it an oddity? Or do larger stars tend to have dozens or planets, even hundreds?&lt;/p&gt;&#xA;" OwnerUserId="10467" LastEditorUserId="2153" LastEditDate="2016-01-23T15:47:35.223" LastActivityDate="2016-01-23T15:47:35.223" Title="Is it odd that our Sun has so many planets?" Tags="&lt;solar-system&gt;&lt;the-sun&gt;&lt;planet&gt;&lt;exoplanet&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13302" PostTypeId="1" AcceptedAnswerId="13304" CreationDate="2016-01-21T20:03:15.120" Score="-1" ViewCount="186" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Gravity_assist&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Wikipedia article&lt;/a&gt; has this terrestrial analogy:&lt;/p&gt;&#xA;&lt;p&gt;Imagine standing on a train platform, and throwing a [tennis] ball at 30 km/h toward a train approaching at 50 km/h. The driver of the train sees the ball approaching at 80 km/h and then departing at 80 km/h after the ball bounces elastically off the front of the train. Because of the train's motion, however, that departure is at 130 km/h relative to the train platform; the ball has added twice the train's velocity to its own.&lt;/p&gt;&#xA;&lt;p&gt;I have two problems with this:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The train's speed relative to the platform is 50 km/h. The ball, traveling in the opposite direction, is going at 30 km/h. Thus, the train operator will indeed see the ball at&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;50 + 30 = 80 km/h.&lt;/p&gt;&#xA;&lt;p&gt;So far, so good. However, once the ball has hit the front of the train (let's assume that no speed at all is lost in the process), it'll have the train's velocity (50 km/h) added to its own (30 km/h):&lt;/p&gt;&#xA;&lt;p&gt;50 + 30 = 80 km/h.&lt;/p&gt;&#xA;&lt;p&gt;Because the train's velocity relative to the platform is 50 km/h, and the ball's now 80 km/h, the operator will see it departing from him at&lt;/p&gt;&#xA;&lt;p&gt;80 - 50 = 30 km/h.&lt;/p&gt;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;As already mentioned above, the ball's velocity relative to the platform after the bounce will be&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;The train's velocity (50 km/h) + the ball's initial velocity (30 km/h) = 80 km/h, and &lt;em&gt;not&lt;/em&gt; twice the train's velocity.&lt;/p&gt;&#xA;&lt;p&gt;What am I missing?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Addendum:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;This is kind of comical:&lt;/p&gt;&#xA;&lt;p&gt;You're watching a tennis match. Player A serves the ball at 200 km/h. Player B returns the serve, the speed of his racket added to DOUBLE the speed of the ball. Player A then puts in a ground stroke, and the speed of the ball doubles again. If they keep this up, then, according to the above logic, the speed of the ball relative to the court will be approaching the speed of light before they know it.&lt;/p&gt;&#xA;&lt;p&gt;Again: what am I missing?&lt;/p&gt;&#xA;" OwnerUserId="9510" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-01-22T01:18:49.433" Title="Gravity assist: why is the velocity doubled?" Tags="&lt;gravity&gt;&lt;general-relativity&gt;&lt;special-relativity&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2016-01-22T03:51:05.400" ContentLicense="CC BY-SA 3.0" />
  <row Id="13303" PostTypeId="2" ParentId="13301" CreationDate="2016-01-21T20:45:01.527" Score="9" Body="&lt;p&gt;We can't detect whether there are or are not objects of similar &quot;small&quot; sizes orbiting other stars.  We simply lack that level of ability.  From a scientific point of view I'd simply say &quot;don't know&quot;.  From a practical point of view I would expect other stars to have systems with a huge variety of things in orbit - there's no reason not to expect this, but it's bad science to assume it.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2016-01-21T20:45:01.527" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13304" PostTypeId="2" ParentId="13302" CreationDate="2016-01-21T21:48:01.160" Score="3" Body="&lt;p&gt;Since the collision is perfectly elastic, the ball's velocity goes from -80 &#xA;km/h from the train's reference point (negative being towards the train) to &#xA;+80 km/h from the train's reference point, a speed increase of 160 km/h. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a stationery observer, therefore, the velocity goes from -30 km/h &#xA;(towards the train) to 130 km/h, an increase of 160 km/h. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation you describe only applies if the train isn't moving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I made an error in units below, assuming the train is moving at 50 km per second (not hour) and that the ball is moving at 30 km per second (not hour), am I'm too lazy to correct it. The general principle, however, still applies.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The confusion may occur because we're ignoring the train's loss of momentum, &#xA;which means the train is going slower after the collision, and moving &#xA;backwards in its own frame of reference. A slightly more detailed &#xA;calculation: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Suppose the the train has mass &lt;code&gt;M&lt;/code&gt; kg and the ball has mass &lt;code&gt;m&lt;/code&gt; kg. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;From the &quot;fixed&quot; observer's point of view, the initial momentum (in &#xA;Newtons) is: &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$\rho =50 M-30 m$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the initial kinetic energy (in Joules) is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e=450 m+1250 M$ &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Let $v_t$ and $v_b$ be the train's and ball's velocities in &#xA;meters/second after the collision. Since perfectly elastic &#xA;collisions preserve both momentum and kinetic energy, we have: &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$m v_b+M v_t=50 M-30 m$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{m v_b^2}{2}+\frac{M v_t^2}{2}=450 m+1250 M$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are only two solutions to the equation above, one of which is &#xA;the initial conditions. The other is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ &#xA;   \left\{{v_b}\to -\frac{10 (3 m-13 M)}{m+M},{v_t}\to -\frac{10 (11 m-5 &#xA;    M)}{m+M}\right\} &#xA;$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging in 0.0585 kg for the mass of a tennis ball and 640000 kg for &#xA;the train, this becomes: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{{v_b}\to 129.9999854,{v_t}\to 49.99998538\}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;effectively confirming the calculation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not convinced this is a good analogy, however. Gravitational boost &#xA;occurs when a planet's gravity almost captures a spacecraft, thus &#xA;nearly making it a satellite, and giving it the same revolution &#xA;velocity around the Sun as the planet itself. The wikipedia analogy &#xA;bears only a passing resemblance to this. &lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2016-01-21T21:48:01.160" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13305" PostTypeId="2" ParentId="13301" CreationDate="2016-01-21T23:25:38.237" Score="32" Body="&lt;p&gt;From an exoplanet-finding point of view, the Sun has between one and three planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The major exoplanet-finding techniques in current use involve watching for either periodic Doppler shifts as the planet's gravitational pull causes the star to wobble, or periodic brightness shifts as the planet transits the star.  Both require that the planet is large enough and close enough to generate a measurable signal and that the orbital period is short enough to let astronomers distinguish periodic variations from one-off variations; the transit method additionally requires that the planet's orbit crosses the star from the point of view of Earth (which favors close-in orbits).  Looking at the Solar System with these techniques:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Mercury: too small&lt;/li&gt;&#xA;&lt;li&gt;Venus: Maybe visible&lt;/li&gt;&#xA;&lt;li&gt;Earth: Maybe visible&lt;/li&gt;&#xA;&lt;li&gt;Mars: too small&lt;/li&gt;&#xA;&lt;li&gt;Jupiter: Highly visible&lt;/li&gt;&#xA;&lt;li&gt;Saturn: Orbital period too long&lt;/li&gt;&#xA;&lt;li&gt;Uranus: Orbital period too long&lt;/li&gt;&#xA;&lt;li&gt;Neptune: Orbital period too long&lt;/li&gt;&#xA;&lt;li&gt;&quot;Planet 9&quot;: Orbital period too long&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you look at &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/c/cc/Exoplanet_Period-Mass_Scatter.png&quot;&gt;this graph of exoplanet discoveries&lt;/a&gt;, Jupiter is solidly in the cluster of blue Doppler discoveries, Saturn is just past the &quot;we've been watching for one full orbit&quot; right-hand edge of that cluster, Earth and Venus are somewhat below the sloped minimum-period-mass line, and everything else is nowhere near the detection range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason the Sun has far more known planets than any other star is simply because we've got a better look at it.&lt;/p&gt;&#xA;" OwnerUserId="1559" LastActivityDate="2016-01-21T23:25:38.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13306" PostTypeId="2" ParentId="13065" CreationDate="2016-01-22T00:44:49.130" Score="2" Body="&lt;p&gt;Yes, &lt;a href=&quot;https://www.dlr.de/content/en/downloads/news-archive/2013/20131219_gaia-reaches-for-the-stars_9053.pdf?__blob=publicationFile&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gaia&lt;/a&gt; makes extensive use of this technique.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The measurement accuracy of star positions for the brightest stars is 10 to 20 micro- arcseconds – up to 100 times better than that the precursor mission, Hipparcos; this angular resolution corresponds to the observed diameter of a Euro coin on the Moon, viewed from Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The underlying principle is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Standard_error#Standard_error_of_the_mean&quot; rel=&quot;nofollow noreferrer&quot;&gt;standard deviation of the mean&lt;/a&gt;. It depends on the sample size, more precisely on the square root of the sample size. In this case the sample size is the number of the measured photons. Therefore it works better for bright stars.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="616" LastEditDate="2021-02-22T18:48:43.433" LastActivityDate="2021-02-22T18:48:43.433" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13307" PostTypeId="1" AcceptedAnswerId="13309" CreationDate="2016-01-22T00:59:37.207" Score="2" ViewCount="4706" Body="&lt;p&gt;I am now in my mid 20s and I remember when I was 6-10 years old, there was a comet or asteroid that flew really close to earth. I remember being able to see the object very clearly as if it were the moon. I also live in the west coast of the United States. What was this object? &lt;/p&gt;&#xA;" OwnerUserId="7258" LastActivityDate="2019-04-21T22:12:07.473" Title="What was that Asteroid/Comet in the 90s?" Tags="&lt;asteroids&gt;&lt;comets&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13308" PostTypeId="1" CreationDate="2016-01-22T01:12:13.433" Score="-1" ViewCount="187" Body="&lt;p&gt;the sun itself is not responsible for global warming, it must be that the temperature is changing due to climate changes.  Anyone?&lt;/p&gt;&#xA;" OwnerUserId="10473" LastActivityDate="2016-01-22T19:29:40.380" Title="Does the sun itself present the problem of global warming is it the main cause?" Tags="&lt;the-sun&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2016-01-22T23:58:26.303" ContentLicense="CC BY-SA 3.0" />
  <row Id="13309" PostTypeId="2" ParentId="13307" CreationDate="2016-01-22T01:32:08.377" Score="10" Body="&lt;p&gt;Was it &lt;a href=&quot;https://en.wikipedia.org/wiki/Comet_Hale%E2%80%93Bopp&quot; rel=&quot;noreferrer&quot;&gt;Comet Hale-Bopp&lt;/a&gt;? It was discovered in 1995, but made a very close approach in 1997, earning it the nickname &quot;The Great Comet of 1997&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia states&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As it passed perihelion on April 1, 1997 the comet developed into a spectacular sight. It shone brighter than any star in the sky except Sirius, and its dust tail stretched 40–45 degrees across the sky. The comet was visible well before the sky got fully dark each night, and while many great comets are very close to the Sun as they pass perihelion, comet Hale–Bopp was visible all night to northern hemisphere observers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/d/df/Comet-Hale-Bopp-29-03-1997_hires_adj.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/d/df/Comet-Hale-Bopp-29-03-1997_hires_adj.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;Image courtesy of Wikipedia user LoopZilla under the &lt;a href=&quot;https://creativecommons.org/licenses/by-sa/2.0/at/deed.en&quot; rel=&quot;noreferrer&quot;&gt;Creative Commons Attribution-Share Alike 2.0 Austria license&lt;/a&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second guess is &lt;a href=&quot;https://en.wikipedia.org/wiki/Comet_Hyakutake&quot; rel=&quot;noreferrer&quot;&gt;Comet Hyakutake&lt;/a&gt;, which made an extremely close approach to Earth in 1996 - becoming &quot;The Great Comet of 1996&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-01-23T00:32:15.923" LastActivityDate="2016-01-23T00:32:15.923" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13310" PostTypeId="1" AcceptedAnswerId="13311" CreationDate="2016-01-22T01:37:51.897" Score="9" ViewCount="335" Body="&lt;p&gt;I was reading this &lt;a href=&quot;https://www.sciencenews.org/blog/science-ticker/evidence-mounts-hidden-ninth-planet&quot;&gt;article&lt;/a&gt; on science news and ran into the following statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Such a remote super-Earth probably originated closer to the sun only to be kicked out by the other giant planets during the solar system’s formative years, Batygin and Brown suggest. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was wondering, how was this planet kicked so far out of the Solar System? &#xA;Why only one planet?&lt;/p&gt;&#xA;" OwnerUserId="10298" LastEditorUserId="2153" LastEditDate="2016-01-22T01:53:53.843" LastActivityDate="2016-01-24T15:56:17.570" Title="How was the hypothetical ninth planet kicked so far out of the Solar System?" Tags="&lt;gravity&gt;&lt;planet&gt;&lt;9th-planet&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13311" PostTypeId="2" ParentId="13310" CreationDate="2016-01-22T01:52:37.630" Score="11" Body="&lt;p&gt;It's hard to say much about this planet, given that most of its properties are unknown. It hasn't been directly observed; instead, its effects on Trans-Neptunian Objects (TNOs) have been simulated and match observations. That said, its mass can be estimated, which is why it is conjectured to be the core of a giant planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the papers that Batygin &amp;amp; Brown cite is &lt;a href=&quot;http://adsabs.harvard.edu/abs/2012Icar..219..737M&quot; rel=&quot;nofollow&quot;&gt;Morbidelli et al. (2012)&lt;/a&gt;. Morbidelli has done prior work on the evolution of the Solar System if &lt;a href=&quot;https://en.wikipedia.org/wiki/Hypothetical_fifth_giant_planet&quot; rel=&quot;nofollow&quot;&gt;a 5th gas giant&lt;/a&gt; formed early in its history, which led to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Jumping-Jupiter_scenario&quot; rel=&quot;nofollow&quot;&gt;Jumping Jupiter scenario&lt;/a&gt;. This is all described in a modification of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Nice_model&quot; rel=&quot;nofollow&quot;&gt;Nice Model&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jumping Jupiter models with the four gas giants lead to one getting ejected from its stable orbit via interactions with the others (see &lt;a href=&quot;http://adsabs.harvard.edu/abs/2011ApJ...742L..22N&quot; rel=&quot;nofollow&quot;&gt;&#xA; Nesvorný (2011)&lt;/a&gt;). This is one of the ice giants - Uranus or Neptune. &lt;em&gt;However&lt;/em&gt;, if we postulate the existence of a 5th planet - a third ice giant - then it can be ejected from its orbit, leaving the other four in a stable arrangement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Batygin &amp;amp; Brown are implying that the object perturbing the TNOs may be the core of this 5th gas giant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since writing the first draft of this answer, I've found out that Batygin has made some estimates that this planet would have been ejected long before the Late Heavy Bombardment - thought to be the result of a Jumping Jupiter scenario - occurred, meaning that it could not be that ejected ice giant, unless we accept a different cause of the LHB. This makes things more interesting, as different initial setups are possible.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-01-24T15:56:17.570" LastActivityDate="2016-01-24T15:56:17.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13313" PostTypeId="1" AcceptedAnswerId="13316" CreationDate="2016-01-22T11:43:40.690" Score="15" ViewCount="1667" Body="&lt;p&gt;We know that the reflected sunlight will make detecting the 9th planet very difficult in the visible light. Is there another band that will be more likely to detect it? What is the surface temperature of this object likely to be, and what would that mean about its optimal detection wavelength?&lt;/p&gt;&#xA;" OwnerUserId="182" LastEditorUserId="182" LastEditDate="2016-01-22T12:34:36.290" LastActivityDate="2016-01-24T22:01:00.877" Title="What wavelength to best detect the &quot;9th planet&quot;?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;observational-astronomy&gt;&lt;9th-planet&gt;&lt;wavelength&gt;" AnswerCount="3" CommentCount="11" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13314" PostTypeId="2" ParentId="13301" CreationDate="2016-01-22T13:46:38.087" Score="8" Body="&lt;p&gt;Given that Kepler has detected stars with 5 or 6 planets that exist in an extremely flattened plane and much flatter than the solar system, then we can be reasonably sure that stars with more than 4 planets are reasonably common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the many mechanisms for scrambling the orbital inclinations out of a perfectly flat configuration, and that seeing multiple transiting planets around a star becomes impossible even if their inclinations are a little different (e.g. only one planet could be seen to regularly transit our Sun if observed from another star), then it is likely that many if not most of the exoplanet systems we detect have more planets than are currently known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the fact that we see &lt;em&gt;some&lt;/em&gt; multiple transiting systems, despite them being inherently geometrically improbable, means that there must be lots of multiple planetary systems.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-01-22T13:46:38.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13315" PostTypeId="2" ParentId="13313" CreationDate="2016-01-22T14:08:00.293" Score="8" Body="&lt;p&gt;The possible planet 9 is thought to be about 10 Earth masses and is unlikely to be a gas giant (it may be the core of an &quot;interrupted&quot; gas giant). As such, it will not be generating significant luminosity itself and would be rocky, or more likely, icy in character. It would thus only be seen by reflected light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The considerations for what wavelength to search in balance the sensitivity of the instruments at hand with the likely spectrum of the object. This in turn depends on the solar spectrum and the wavelength dependence of the reflectivity (albedo).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For most icy objects, including Pluto and Trans-Neptunian objects, the reflectance increases to the red and near-infrared, whilst the solar spectrum peaks at shorter wavelengths. This suggests that searches are best carried out with wide field optical instruments in the R or r' bands at around 600 nm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A further factor in &lt;em&gt;finding&lt;/em&gt; a candidate is that you are going to have to cover a large area. This is only feasible at optical and NIR wavelengths unless the object was bright enough in the mid-IR to show up in WISE (which I'm sure is being thoroughly checked). A press release I saw said SUBARU is being used for the search. I would bet they are using the half degree field of Suprime-Cam at optical wavelengths and not pursuing COMICS mid-IR imaging with it 42x32 &lt;em&gt;arcsecond&lt;/em&gt; field!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Confirming a candidate should be easy, given the enormous parallax and proper motion expected.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-01-22T16:23:03.473" LastActivityDate="2016-01-22T16:23:03.473" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13316" PostTypeId="2" ParentId="13313" CreationDate="2016-01-22T14:55:47.000" Score="7" Body="&lt;p&gt;Direct reflection of sunlight is the most likely scenario for a ninth planet discovery, however that does not hold if the object has a very low albedo. I assume you are interested in what wavelengths the planet would radiate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the surface temperature, the rotation of the planet is important. If it is locked with one side facing the sun, or rotates very slowly, the centre of the sun facing hemisphere radiates as much energy as it gets from the Sun. At 60 AU, the solar flux is about 0.38 W/m². Using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law&quot;&gt;Stefan-Boltzmann law&lt;/a&gt;, we obtain a equilibrium surface temperature of 51 K (that is the highest possible surface temperature, assuming it does not have an atmosphere).&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Wien&amp;#39;s_displacement_law&quot;&gt;Wien's displacement law&lt;/a&gt; tells us that radiation from a 51 k object peaks at a wavelength of 57 µm (infra-red).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a rotating body, the equator temperature is 38 K, with radiation peaking at 78 µm (still infra-red).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using an albedo of 0.5, the peaks are 68 µm and 90 µm for a non-rotating and a rotating body respectively.&#xA;Note that this is for the equator region only, the actual peak-wavelength is going to be a little bit higher, belonging in the far infra-red spectrum. Also, the high uncertainty of rotation, albedo and mass (mass is important for internal heat), makes it impossible to get a higher accuracy than that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;60 au is a very optimistic perihelion distance for the ninth planet, so for a more realistic distance of say 200 au, it is not possible to observe it in the IR spectrum, if it does not have a significant internal heat source.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="10172" LastEditDate="2016-01-24T22:01:00.877" LastActivityDate="2016-01-24T22:01:00.877" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13317" PostTypeId="2" ParentId="13313" CreationDate="2016-01-22T14:58:07.957" Score="6" Body="&lt;p&gt;There are two basic ways to detect such an object. First is to detect it through reflected sunlight. Second is from the heat that it produces. We already know that the &lt;a href=&quot;https://astronomy.stackexchange.com/a/13275/182&quot;&gt;reflected light of such an object&lt;/a&gt; likely would be around a 16.5 magnitude. To determine the infrared, we have to estimate the temperature&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature very much depends on the composition. For simplicity, let's assume a composition similar to Earth, and was created about the same time as the rest of the Solar System. These assumptions may not prove to be valid, but they are among the possibilities discussed. Earth's internal heat, in fact, is at least 50% from radioactive decay, according to &lt;a href=&quot;http://blogs.scientificamerican.com/observations/nuclear-fission-confirmed-as-source-of-more-than-half-of-earths-heat/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Scientific America&lt;/a&gt;. Of course, that's the internal heat only, not all of that will make it to the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proposed planet is somewhat akin to a &lt;a href=&quot;https://en.wikipedia.org/wiki/Rogue_planet#Retention_of_heat_in_interstellar_space&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Rogue Planet&quot;&lt;/a&gt;, where a small disk of gas collapsed into a planet without a star, or were ejected from their host system. A fair bit also depends on if there is a sizable moon of the object. If so, then tidal heating would dramatically increase the temperature of the object. Any such determination can't be made without observation, but it is possible. An atmosphere would also help to keep the planet from freezing. A &lt;a href=&quot;http://arxiv.org/pdf/1102.1108v2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper for detecting rogue planets&lt;/a&gt; comes from Abbott and Switzer. They hypothesis that a 3.5 Earth Mass object could be detected if it comes within 1000 AU, specifically in the far infrared, with a surface temperature of about 50 K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line, it would probably be wise to try to detect both in the far infrared, as well as the visible, although it might be difficult to detect, even then. Given parallax as the primary means of motion, the detection should be done at several points in Earth's orbit, probably the same spot should be searched about 90 days apart to give the maximum opportunity to move, as parallax would only be visible if the motion of the Earth was perpendicular to the location of the object.&lt;/p&gt;&#xA;" OwnerUserId="182" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-01-22T16:05:42.280" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13318" PostTypeId="2" ParentId="13248" CreationDate="2016-01-22T15:25:44.830" Score="4" Body="&lt;p&gt;Your question is valid and is a common operation. When computing when an object will be visible, many observers want to be able to draw a globe or map that is marked with the position from which an object like Pluto is directly overhead. Such a position on the globe at a given instant (and you do say “at the current time”) is, all other things being equal, the best-situated place from which to observe the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the “libastro” C library that PyEphem wraps seems to only provide one instance of this concept: Earth satellites have &lt;code&gt;.sublat&lt;/code&gt; and &lt;code&gt;.sublong&lt;/code&gt; attributes, because it is so common to want to draw the path of a satellite on a globe or map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But libastro does not generalize the concept. From what I can see, there is no way to generate the point on the Earth that is directly below any other Solar System object. And since I have not been in the habit of trying to extend libastro, it is likely that PyEphem will not gain this ability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have been developing a replacement for PyEphem that is written in pure Python and that I will be free to extend, called Skyfield. I will hopefully have this concept working there soon, and when I do so I plan to make it work for all objects, and not limit it to Earth satellites!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In the meantime, as pointed out in the comments: you could try asking PyEphem for the geocentric RA and dec and, if you could adjust RA by the Earth's current hour angle, then you could turn those into an approximate latitude and longitude.)&lt;/p&gt;&#xA;" OwnerUserId="10483" LastEditorUserId="10483" LastEditDate="2016-01-27T00:03:39.380" LastActivityDate="2016-01-27T00:03:39.380" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13319" PostTypeId="2" ParentId="13308" CreationDate="2016-01-22T15:29:19.553" Score="1" Body="&lt;p&gt;The energy input of the Sun stays constant (mostly, there are some minor variations), so no, the Sun is not responsible for climate changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature of the Earth has to do with the balance between the energy input, and the energy radiated back into space. If the temperature is not changing, they are the same.&#xA;Global warming is caused by gasses in the atmosphere limiting the energy radiated into space, therefore, the temperature rises, until the energy radiated is again equal to the solar energy input.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="10172" LastEditDate="2016-01-22T16:35:37.220" LastActivityDate="2016-01-22T16:35:37.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13320" PostTypeId="2" ParentId="10103" CreationDate="2016-01-22T15:39:07.913" Score="2" Body="&lt;p&gt;Alternatively, if you were to get the Right Ascension and Declination coordinates from the company that sold you the star you could use a database like SIMBAD to find the actual designation of the star you purchased.&#xA;You could then probably use the SIMBAD designation to find the star using Stellarium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a link to the SIMBAD database: &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-fcoo&quot; rel=&quot;nofollow&quot;&gt;http://simbad.u-strasbg.fr/simbad/sim-fcoo&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="10485" LastActivityDate="2016-01-22T15:39:07.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13321" PostTypeId="1" CreationDate="2016-01-22T02:09:10.007" Score="7" ViewCount="241" Body="&lt;p&gt;So recently it was &lt;a href=&quot;http://www.caltech.edu/news/caltech-researchers-find-evidence-real-ninth-planet-49523&quot;&gt;announced&lt;/a&gt; that there might be a ninth planet out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people (from NASA) say the odds are 50% that it is a real planet, since it hasn't been discovered by telescopes yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say there is no ninth planet. Is there another explanation for those orbits that require a big mass (8 times earth mass) out there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it even be a very small black hole (a few centimeters in radius) or maybe a clump of dark matter?&lt;/p&gt;&#xA;" OwnerDisplayName="White Noise" LastEditorUserId="2153" LastEditDate="2016-01-23T15:46:56.267" LastActivityDate="2016-01-23T15:46:56.267" Title="Ninth planet - what else could it be?" Tags="&lt;solar-system&gt;&lt;black-hole&gt;&lt;telescope&gt;&lt;planet&gt;&lt;9th-planet&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13322" PostTypeId="2" ParentId="13321" CreationDate="2016-01-22T08:46:10.613" Score="14" Body="&lt;p&gt;What else could it be?  Very simply, it could all be a coincidence...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What they have done is noted that objects in the outer solar system have their orbits grouped in interesting ways. Someone suggested this might indicate there was a heavy planet (or core of a giant protoplanet) that influenced their orbits in this way; the recent announcement is the result of examining this idea in detail.  It is &lt;strong&gt;likely&lt;/strong&gt; to be caused by gravitational influence of such a body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the TV and newspapers have picked this up and said a planet HAS been discovered. But there might be nothing at all; the loose clustering could all be a coincidence - though that seems, statistically, unlikely at the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/0004-6256/151/2/22/pdf&quot;&gt;the paper&lt;/a&gt; for anyone who wants to check.  (Beware, lots of mathematical terminology.)&lt;/p&gt;&#xA;" OwnerUserId="7491" OwnerDisplayName="Andy" LastActivityDate="2016-01-22T08:46:10.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13323" PostTypeId="2" ParentId="13308" CreationDate="2016-01-22T16:11:58.110" Score="1" Body="&lt;p&gt;Actually the sun's output over time does vary, and it does cause climate changes. There is an 11 year cycle as shown in this image from www.sciencemuseum.org.uk:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/16EEh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/16EEh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in the early half of the 20th century it did increase slightly, which &lt;strong&gt;probably did contribute to global warming during that time&lt;/strong&gt; but over the last 50 years it has actually slightly cooled, so it is absolutely not responsible for the global warming we currently see. This image (from the same source) shows some nice detail. The dotted line is the moving average:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/395Dq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/395Dq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="43" LastEditDate="2016-01-22T19:29:40.380" LastActivityDate="2016-01-22T19:29:40.380" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13324" PostTypeId="2" ParentId="13248" CreationDate="2016-01-22T16:21:44.393" Score="3" Body="&lt;p&gt;The orbital speed of Earth around the Sun is 30,000 m/s, the highest mountain is about 9,000 m and the rotational speed of the surface around the polar axis is about 450 m/s. So this makes me conclude that topography such as height over the Sea has to be of little importance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The polar axis around which the Earth turns is tilted by 23 degrees relative to the ecliptic plane of the planets' average. Pluto's orbit is 17 degrees tilted so every region on Earth between +40 and -40 degrees latitude should, sequentially in time, be the one spot on Earth's surface which is closest to Pluto. Spread out over looong times. That's about between New York and Sydney. For the least inclined planets the nearest point on Earth would be somewhere between Houston and Rio de Janeiro, i.e. every place on Earth within 24 degrees north or south of the equator (except very local depressions).&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2016-01-22T16:27:38.703" LastActivityDate="2016-01-22T16:27:38.703" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13325" PostTypeId="2" ParentId="13321" CreationDate="2016-01-22T17:15:07.337" Score="3" Body="&lt;p&gt;The introduction of &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/0004-6256/151/2/22/pdf&quot; rel=&quot;nofollow&quot;&gt;the paper&lt;/a&gt; mentions some alternative interpretations put forward in earlier papers. It seems as if this &quot;problem&quot; has been noted earlier. This new solution to the problem beats earlier solutions. But maybe it is an evolving discovery process which again will invent new better explanations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to those mentioned in the paper, I suggest the following candidate alternatives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-) Too few observations in order to hold up in a soon dramatically increased discovery rate. 4 or 5 or maybe even 8 TNOs which, as they claim, give a 0.007% significance level is certainly resting on some assumptions which will be challenged if thousands of similar objects are found in the next decade. So, as Andy has answered: possibly a coincidence due to small sample. (The paper's 0.007% probability is, well, we'll see)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-) Weird unthought-of science bias which, in math more than in lens, tends to select inadvertently among the (candidate) observations made and their characteristics. Since the authors are at the top in their fields, and Michael Brown has discovered loads of distant objects, including Sedna, because of that maybe one could suspect that some one thought has created a bias somehow.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-01-22T17:15:07.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13326" PostTypeId="1" CreationDate="2016-01-22T18:31:39.197" Score="5" ViewCount="2096" Body="&lt;p&gt;According to general relativity, would gravitational waves experience the same sort of redshift that electromagnetic waves experience due to the expansion of the universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And are there astrophysical processes that would be expected to have a characteristic 'frequency signature' analogous to the lyman or balmer series of hydrogen which would enable a measurement gravitational wave redshift. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking of processes like baryon accoustic oscillations, certain types of supernovae, neutron star mergers, etc.&lt;/p&gt;&#xA;" OwnerUserId="7786" LastActivityDate="2021-01-29T01:28:01.590" Title="Redshift for gravitational waves?" Tags="&lt;expansion&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13327" PostTypeId="1" AcceptedAnswerId="13350" CreationDate="2016-01-22T19:41:59.863" Score="7" ViewCount="88" Body="&lt;p&gt;eLisa - &lt;a href=&quot;https://en.wikipedia.org/wiki/Evolved_Laser_Interferometer_Space_Antenna&quot;&gt;Evolved Laser Interferometer Space Antenna&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A LISA-like mission is designed to directly observe gravitational waves, which are distortions of space-time travelling at the speed of light. Passing gravitational waves alternately squeeze and stretch objects by a tiny amount. Gravitational waves are caused by energetic events in the universe and, unlike any other radiation, can pass unhindered by intervening mass. Launching eLISA will add a new sense to scientists' perception of the universe and enable them to listen to a world that is invisible in normal light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &quot;duh&quot; answer to my question is &quot;Gravity waves, of course.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously this is not what I'm asking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we've deployed eLISA and we spotted a gravitational wave passing. Well, there went the wave. What data do we extract from it? What discoveries, observations will we acquire? What else can we discover with it than &quot;there was a gravity wave, and it came from this direction&quot;?&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2018-02-23T15:13:03.870" Title="What will eLISA be trying to observe?" Tags="&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13328" PostTypeId="1" AcceptedAnswerId="13329" CreationDate="2016-01-22T21:38:59.190" Score="2" ViewCount="209" Body="&lt;p&gt;I have seen a number of places on the Internet where Epsilon Eridani is referred to as Sadira. (e.g. &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_stars_in_Eridanus&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/List_of_stars_in_Eridanus&lt;/a&gt; )   However, it would seem that its an unofficial name and the star got renamed Ran by the IAU in a competition.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How did Epsilon Eridani become known as Sadira by some?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What or who was Sadira?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5401" LastActivityDate="2019-06-01T10:23:18.207" Title="Epsilon Eridani and Sadira" Tags="&lt;star&gt;&lt;iau&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13329" PostTypeId="2" ParentId="13328" CreationDate="2016-01-23T00:07:43.013" Score="3" Body="&lt;p&gt;All star names are unofficial. A few stars have ancient names (such as Sirius) all other stars are referred to by their position in a catalogue or star atlas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As well as I can tell, Al Sadira means &quot;the ostrich&quot;, or perhaps in context &quot;The (riverbank) ostrich&quot; indicating a type of tree that grows by rivers (baby name sites offer &quot;lotus tree&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be a list of Arabic star names that lists this star as Al Sadira(h) but the name is not in common use and won't be recognised by most people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The IAU name, Ran, from that of a Norse river goddess, is a recent innovation, and doesn't seem to be much used. The Bayer Designation &quot;Epsilon Eridani&quot; is by far the most well known.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-01-23T00:07:43.013" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13330" PostTypeId="1" AcceptedAnswerId="13345" CreationDate="2016-01-23T01:36:35.630" Score="55" ViewCount="7041" Body="&lt;p&gt;In the comments to &lt;a href=&quot;https://astronomy.stackexchange.com/q/13313/182&quot;&gt;this question&lt;/a&gt;, there was considerable doubt placed as to the subject of if the so called &quot;9th planet&quot; really exists. That wasn't really the intent of the previous question, so I ask this. Why hasn't this &quot;9th planet&quot; been detected before now, if it even exists?&lt;/p&gt;&#xA;" OwnerUserId="182" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2020-07-15T07:08:57.227" Title="Why hasn't the &quot;9th Planet&quot; been detected already?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;observational-astronomy&gt;&lt;9th-planet&gt;" AnswerCount="6" CommentCount="4" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13331" PostTypeId="2" ParentId="13330" CreationDate="2016-01-23T01:49:17.197" Score="36" Body="&lt;p&gt;This graph from &lt;a href=&quot;http://xkcd.com/1633/&quot; rel=&quot;noreferrer&quot;&gt;XKCD&lt;/a&gt; says a lot about why that is the case&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9uiOO.png&quot; rel=&quot;noreferrer&quot; title=&quot;Superman lies near the bird/plane boundary over a range of distances, which explains the confusion.&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9uiOO.png&quot; alt=&quot;&quot; title=&quot;Superman lies near the bird/plane boundary over a range of distances, which explains the confusion.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line is, the 9th planet is too small to be detected through WISE, and too far/small to have been detected through visible observation. Most likely this hypothetical planet is a long ways away, possibly as far as 1200 AU, and not particularly large, making it difficult to see. WISE was able to rule out the existence of any Saturn sized object up to 10,000 AU. This hypothetical object is about 1/5th to 1/10th the mass of Saturn. It is also far more likely to be more compact, as Saturn is rather low density. Bottom line, if it is visible in WISE, it would be on the very edge.&lt;/p&gt;&#xA;" OwnerUserId="182" LastEditorUserId="16151" LastEditDate="2017-06-25T16:23:46.697" LastActivityDate="2017-06-25T16:23:46.697" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13332" PostTypeId="2" ParentId="13330" CreationDate="2016-01-23T01:49:34.043" Score="23" Body="&lt;p&gt;There are many normal methods that we use to detect exoplanets, but none of them work well in the case of the 9th planet. Here are some of the main ones.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Methods_of_detecting_exoplanets#Radial_velocity&quot; rel=&quot;noreferrer&quot;&gt;Radial velocity.&lt;/a&gt;&lt;/strong&gt; The Sun is not moving significantly with respect to Earth, and the hypothetical planet is too far away from the Sun to have much of an impact.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Methods_of_detecting_exoplanets#Transit_photometry&quot; rel=&quot;noreferrer&quot;&gt;Transit.&lt;/a&gt;&lt;/strong&gt; This is obviously impossible, as the planet never passes between Earth and the Sun.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_microlensing&quot; rel=&quot;noreferrer&quot;&gt;Gravitational microlensing.&lt;/a&gt;&lt;/strong&gt; Again, the planet does not pass between us and the Sun.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Methods_of_detecting_exoplanets#Direct_imaging&quot; rel=&quot;noreferrer&quot;&gt;Direct imaging.&lt;/a&gt;&lt;/strong&gt; This is difficult to begin with. The planet is also far away and would have &lt;a href=&quot;https://en.wikipedia.org/wiki/Planet_Nine#Direct_detection&quot; rel=&quot;noreferrer&quot;&gt;a low absolute magnitude&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As to why indirect evidence for the planet wasn't accumulated earlier - well, &lt;a href=&quot;https://en.wikipedia.org/wiki/Kuiper_belt#History&quot; rel=&quot;noreferrer&quot;&gt;objects in the Kuiper Belt and beyond (including the perturbed Trans-Neptunian Objects (TNOs)) were only first observed in 1992&lt;/a&gt;, and we've only built up significant data in recent years.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-01-24T15:55:18.223" LastActivityDate="2016-01-24T15:55:18.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13333" PostTypeId="1" CreationDate="2016-01-23T05:21:48.550" Score="0" ViewCount="95" Body="&lt;p&gt;The Hubble telescope can see the light from the big bang.My question is if our universe is expanding that would make everything that's in front of our solar system be in the future? And also can Hubble see ahead of our solar system and does the bang big light ends behind our solar system?&lt;/p&gt;&#xA;" OwnerUserId="10499" LastActivityDate="2016-01-24T18:59:24.163" Title="The big bang and our expanding universe" Tags="&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13334" PostTypeId="1" CreationDate="2016-01-23T06:33:03.310" Score="3" ViewCount="1400" Body="&lt;p&gt;Aleksander Wolszczan of Pennsylvania State University found three planets circling a pulsar called PSR B1257+12.Two of the planets were the size of Earth.The pulsar observed by Spitzer, named 4U 0142+61 spotted a warm glowing disk formed from debris, The disk has a orbit of about 1 million miles and contains 10 Earth mass materials. Deepto Chakrabarty of the Massachusetts Institute of Technology, principal investigator states &quot;Pulsars emit a tremendous amount of high energy radiation&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In a pulsar star system, could human life thrive on a planet with high energy radiation?&lt;/strong&gt; &lt;em&gt;If answering the question, please provide the latest references to date that are relevant with your statement.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SjzLp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SjzLp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/gkzDp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gkzDp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/f8PRR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f8PRR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9326" LastEditorUserId="2153" LastEditDate="2016-01-23T16:00:26.140" LastActivityDate="2016-01-23T18:28:41.550" Title="Could human life thrive on a planet in a pulsar star system?" Tags="&lt;life&gt;&lt;pulsar&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13335" PostTypeId="2" ParentId="13330" CreationDate="2016-01-23T08:52:09.767" Score="13" Body="&lt;p&gt;Possible reasons that the planet has not been picked up previously:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It's not there.&lt;/li&gt;&#xA;&lt;li&gt;Photographic proper motion surveys cover the whoke sky. To avoid these planet 9 would have to be fainter than about 18th or 19th magnitude. This puts a lower bound in the size/albedo/distance combination for any planet 9. The proposed planet could easily be fainter than 20th magnitude.&lt;/li&gt;&#xA;&lt;li&gt;Targeted searches for &lt;a href=&quot;https://en.wikipedia.org/wiki/Trans-Neptunian_object&quot;&gt;trans-Neptunian objects&lt;/a&gt; (TNOs) use big telescopes with wide fields of view. Nevertheless they do not cover anywhere near the whole sky and are generally limited to perhaps magnitude 22-23.&lt;/li&gt;&#xA;&lt;li&gt;Deep searches can go much fainter, but are &quot;pencil beam&quot; surveys and so have little chance of blindly picking up a particular object, or are single epoch and so cannot detect motion due to parallax and orbital motion.&lt;/li&gt;&#xA;&lt;li&gt;Infrared surveys (&lt;a href=&quot;https://en.wikipedia.org/wiki/Wide-field_Infrared_Survey_Explorer&quot;&gt;WISE&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/2MASS&quot;&gt;2MASS&lt;/a&gt;) covered the whole sky. They were sensitive enough to pick up gas giants at the proposed distance of planet 9, because giants are kept hot by gravitational contraction. However, if planet 9 were rocky/icy and had no internal source of heat then it could have been too cold to detect.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="332" LastEditDate="2016-01-23T13:20:10.407" LastActivityDate="2016-01-23T13:20:10.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13337" PostTypeId="2" ParentId="13333" CreationDate="2016-01-23T12:35:15.520" Score="2" Body="&lt;p&gt;There are a couple of misunderstandings here. Let's take one at a time:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Hubble telescope can see the light from the big bang.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Hubble Space Telescope (HST) doesn't actually see light from the Big Bang. HST has several instruments on board, both for imaging and spectroscopy, but they all operate in the infrared, optical (i.e. &quot;visible to humans&quot;), and ultraviolet wavelength range. When you mention &quot;light from the Bing Bang&quot;, I suppose you're thinking of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow&quot;&gt;cosmic microwave background radiation&lt;/a&gt; (CMB), which is light emitted 380,000 after Big Bang. This is &quot;the closest&quot; we can get to Big Bang (yet). HST is unable to detect microwaves; instead we have other telescopes for that, the most recently-launched being &lt;a href=&quot;https://en.wikipedia.org/wiki/Planck_%28spacecraft%29&quot; rel=&quot;nofollow&quot;&gt;Planck&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My question is if our universe is expanding that would make everything&#xA;  that's in front of our solar system be in the future?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The expansion of the Universe is one of the most difficult concepts to get one's head around. But firstly, Big Bang wasn't an explosion, hurling matter outwards from a central point in an otherwise empty Universe. Galaxies, on average, don't travel much &lt;em&gt;through&lt;/em&gt; space. Rather, they lie relatively still in space, but space itself expands. The distances between galaxies increase all the time, and previously, before galaxies were formed, the distance between atoms and other particles increased. An often-used analogy — which you shouldn't take too far — is a balloon with dots painted on. The dots are fixed on its surface, and when you blow it up, the distances between the dots increase despite the dots still being fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, your question seems to confuse space and time. Even if the expansion were an explosion, we would travel through space, and yes, you could say that we travel through time as well, but saying that the future is &quot;everything in front of our Solar system&quot; doesn't really make sense (to me, at least). What's in front of you when you travel through space is just more space; you can look at it before you get there, but that doesn't mean you look into the future (in fact you look at the past, since the light you detect with your eyes has spent some time traveling toward you).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And also can Hubble see ahead of our solar system&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It depends what you mean by &quot;ahead&quot;. It can definitely see things &lt;em&gt;outside&lt;/em&gt; the Solar system. The Solar system consists of one star and it satellites (planets, comets, asteroids, etc.) out of the few hundred billions that make up the galaxy we call the Milky Way. Our Galaxy is itself only one out of at least some hundred billion — and quite possibly infinitely many — galaxies that float around in our Universe. And yes, HST can see many of these. But if by &quot;ahead&quot; you mean &quot;into the future&quot;, then the answer is no.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and does the bang big light ends behind our solar system?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As mentioned above, the Big Bang light — or the CMB — was emitted shortly after Big Bang. That means that it has been traveling for 13.8 billion years through space. Light travels one lightyear per year, but since the Universe is expanding, the distance that the CMB has traveled is &lt;em&gt;more&lt;/em&gt; than 13.8 billion lightyears; in fact it's some 46 billion lighyears. This is the most distant &quot;thing&quot; we can see, way, way beyond the Solar system which for comparion is of the order of one lightyear large.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="941" LastEditDate="2016-01-24T18:59:24.163" LastActivityDate="2016-01-24T18:59:24.163" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13338" PostTypeId="2" ParentId="13330" CreationDate="2016-01-23T15:59:49.383" Score="8" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TIHE7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TIHE7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clipping above from the Albuquerque Journal doesn't really answer your question, but astronomers have been hunting for a largeish planet beyond Pluto for quite some time now. See also:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.zetatalk.com/theword/tword26b.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.zetatalk.com/theword/tword26b.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.bibliotecapleyades.net/hercolobus/esp_hercolobus_2_02.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.bibliotecapleyades.net/hercolobus/esp_hercolobus_2_02.htm&lt;/a&gt; (under &quot;United States Naval Observatory Calculations&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.bibliotecapleyades.net/hercolobus/esp_hercolobus_3d.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.bibliotecapleyades.net/hercolobus/esp_hercolobus_3d.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Just to beat this to death, Percival Lowell &lt;strong&gt;may&lt;/strong&gt; have been the first person to detect the 10th planet. Quoting &lt;a href=&quot;https://en.wikipedia.org/wiki/Percival_Lowell&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Percival_Lowell&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pluto's mass could not be determined until 1978, when its satellite&#xA;  Charon was discovered. This confirmed what had been increasingly&#xA;  suspected: Pluto's gravitational influence on Uranus and Neptune is&#xA;  negligible, certainly not nearly enough to account for the&#xA;  discrepancies in their orbits.[21] In 2006, Pluto was reclassified as&#xA;  a dwarf planet by the International Astronomical Union.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course, the same page continues to state:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In addition, it is now known that the discrepancies between the&#xA;  predicted and observed positions of Uranus and Neptune were not&#xA;  caused&#xA;  by the gravity of an unknown planet. Rather, they were due to an&#xA;  erroneous value for the mass of Neptune. Voyager 2's 1989 encounter&#xA;  with Neptune yielded a more precise value of its mass, and the&#xA;  discrepancies disappear when using this value.[22]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you're willing to selectively believe Wikipedia, it's possible that Lowell did detect a 9th planet and that it wasn't Pluto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One reason to disbelieve Wikipedia (and the given source) and to credit Lowell here is that this 9th planet &lt;em&gt;does&lt;/em&gt; have an effect on Uranus' and Neptune's orbits, so the discrepancies Lowell noted wouldn't have disappeared, they would have simply been reduced. &lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2016-01-31T16:35:47.410" LastActivityDate="2016-01-31T16:35:47.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13339" PostTypeId="1" AcceptedAnswerId="13351" CreationDate="2016-01-23T16:34:41.193" Score="4" ViewCount="164" Body="&lt;p&gt;There are asteroid belts, and there are collections of bodies in resonant orbits, such as Trojans. Aren't these separate things?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The recent evidence for the 9th planet is an alignment in various parameters of known Kuiper belt objects. Conversely, few or no objects are known in the parts of the belt that would be swept away by the hypothetical planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Kuiper belt straddles the gravitational neighborhoods of Neptune and this other planet, and its populations reflect their influence, is it really a belt at all? A lopsided belt? Or just a region with a collection of distinct populations, each associated with a dominant body?&lt;/p&gt;&#xA;" OwnerUserId="1641" LastActivityDate="2016-01-24T11:10:21.147" Title="How can the Kuiper belt coexist with the 9th planet?" Tags="&lt;9th-planet&gt;&lt;kuiper-belt&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13340" PostTypeId="2" ParentId="13334" CreationDate="2016-01-23T16:56:17.790" Score="2" Body="&lt;p&gt;Only five pulsar planets have been confirmed or have garnered enough evidence to make a strong case for their existence. None of them are like the terrestrial planets in the Solar System insofar as the way they formed and their orbital movements.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/PSR_B1620-26_b&quot; rel=&quot;nofollow&quot;&gt;PSR B1620-26 b.&lt;/a&gt;&lt;/strong&gt; This planet is in a circumbinary orbit around a pulsar and a white dwarf, with a semi-major axis of ~23 AU. It is theorized that it formed around a Sun-like star, which became a binary star system after encountering a neutron star (another partner must have been ejected from the system, as is the case with triple-star encounters). The Sun-like star promptly became a white dwarf, and the planet's orbit would have widened after the encounter with the neutron star, which is a pulsar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, this planet did not form around the pulsar. The system was captured so to speak, and so neither the Sun-like star/white dwarf or planet would have been affected by the supernova. Additionally, the planet is fairly far away from the two stars in the system, meaning that effects from the pulsar are even less negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point of this is that life on this planet is &lt;em&gt;possible&lt;/em&gt; - especially given the planet's old age - but it would be hard for it to survive. If life formed before the encounter with the original binary system, it would have been disrupted a bit by the orbital changes, reducing the amount of starlight it received. Today, it would receive little light, as the white dwarf has a low luminosity. There wouldn't be enough energy to sustain life - as is the case for most pulsar planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as Rob Jeffries pointed out, this planet is a gas giant, not a terrestrial planet, meaning that human-like life could not develop on it and any arriving humans bent on colonization would not be able to live on it normally.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/PSR_B1257%2B12&quot; rel=&quot;nofollow&quot;&gt;The planets of PSR B1257+12.&lt;/a&gt;&lt;/strong&gt; Three planets orbit this pulsar, at distances of about 0.19, 0.36, and 0.48 AU - fairly close to the pulsar and within range of a decent amount of harmful radiation, relative to PSR B1629-26 b. The major reason that these planets are not a good place for life is that they are thought to have formed &lt;em&gt;after&lt;/em&gt; the supernova that led to the evolution of the star to a pulsar, from remnants of debris from the supernova. The star itself is quite young - ~3 billion years - and the supernova would have happened relatively soon thereafter. The planets themselves are less than 1 billion years old, meaning that life has not had time to develop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My issue with the age of the planets isn't based on the short time period for life to develop so much as on the idea that conditions might be hellish, comparable to those experienced in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hadean&quot; rel=&quot;nofollow&quot;&gt;Hadean eon&lt;/a&gt;. Geologically, these planets would have had different starts than Earth did, and they most likely would not have encountered the problems of the Late Heavy Bombardment, but life might not have a good chance at starting at all for another 500 million years.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/PSR_J1719-1438_b&quot; rel=&quot;nofollow&quot;&gt;PSR J1719-1438 b.&lt;/a&gt;&lt;/strong&gt; This planet is unusual because it is thought to be the remnant of the companion star - severely damaged by the supernova. It is also composed of exotic materials, for a planet - crystallized carbon. The conditions here are not conducive to life.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Looking at this list we see three distinct types of pulsar planets emerge:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Captured planets&lt;/li&gt;&#xA;&lt;li&gt;Newly formed planets&lt;/li&gt;&#xA;&lt;li&gt;Former companion stars&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The latter two are not good places for life, in part because they may not be composed or compounds life can use. The first group may be the best chance, as their formation could have been relatively normal. The problem lies in the fact that they might have been rogue planets, floating along without good sources of energy from stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this isn't even taking into account the radiation from the pulsar itself. The only escape is to have an orbit far from the pulsar, as with PSR B1620-26b, but then there's no outside source of energy. The best chance would be to have the planet in a circumbinary orbit around the pulsar and a companion star, as in this planet's case. At least then there would be some helpful light.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-01-23T18:28:41.550" LastActivityDate="2016-01-23T18:28:41.550" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13341" PostTypeId="1" AcceptedAnswerId="13343" CreationDate="2016-01-23T19:48:13.690" Score="5" ViewCount="332" Body="&lt;p&gt;Given that Planet Nine is hypothesized to be 10-20 times the mass of the Earth orbiting the sun some 200-2500 AU out, what kind of planetary-mass object would it be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even given that it has apparently &quot;influenced&quot; several Trans-Neptunian objects, since its orbit is so large and so slow (approximately 25,000 years) it would take an awful long time to clear out its own orbit (especially given that we do not know how long it has been in that orbit), so it seems like it would not qualify as an actual planet under the rules that re-classified Pluto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it orbits our sun, it wouldn't be an exoplanet nor a rogue planet.  And it sure seems too big to be called a dwarf planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A captured planet seems closest in other attributes, but it probably was not either a capture nor a re-capture, but more likely a partial ejection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what classification would it fall into?&lt;/p&gt;&#xA;" OwnerUserId="7536" LastEditorUserId="2153" LastEditDate="2016-01-24T16:46:22.470" LastActivityDate="2020-02-07T17:07:11.760" Title="What type of planetary-mass object would Planet Nine be?" Tags="&lt;planet&gt;&lt;9th-planet&gt;&lt;kuiper-belt&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13342" PostTypeId="1" CreationDate="2016-01-23T19:48:31.590" Score="6" ViewCount="144" Body="&lt;p&gt;Current models of the configurations of the planets conclude that Neptune once migrated outwards. AFAIK this refurbishment came about by Jupiter and Saturn gradually getting into and leaving a gravitational resonance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did Neptune (and possibly a fifth giant planet) redirect itself in a sudden encounter event, like comet 67P Churyumov–Gerasimenko did in 1959, halving the perihelion of its orbit since then? Or did it happen gradually over maybe millions of years? Is planetary migration in general sudden or gradual? Is it a matter of a single encounter event, or an adaptation to a changing gravity landscape?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-01-25T15:48:16.560" Title="Do planets migrate suddenly or gradually?" Tags="&lt;solar-system&gt;&lt;gravity&gt;&lt;planetary-formation&gt;&lt;orbital-migration&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13343" PostTypeId="2" ParentId="13341" CreationDate="2016-01-23T20:01:55.913" Score="8" Body="&lt;p&gt;&lt;strong&gt;It is a planet - either a gas giant core, a mini-Neptune, or a super-Earth.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, &lt;a href=&quot;http://www.newyorker.com/tech/elements/discovering-planet-nine&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mike Brown has stated outright&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;“It is a planet—there’s virtually no doubt,” he said. “What we now call planets are objects that can gravitationally dominate their neighborhood. Pluto is a slave to the gravitational influence of Neptune. By area, Planet Nine dominates more of the solar system than any other known planet—it’s only because of this that we can infer its existence. And because of this we’re pretty sure it’s not a small object: it’s at least ten times more massive than Earth and five thousand times more massive than Pluto. In many ways, you could argue that this is more of a planet than anything else in the solar system.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wrote &lt;a href=&quot;https://astronomy.stackexchange.com/a/13311/2153&quot;&gt;an answer to How was the hypothetical ninth planet kicked so far out of the Solar System?&lt;/a&gt; that discussed the possibility that the ninth planet is the hypothetical 5th gas giant proposed by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2011ApJ...742L..22N&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nesvorný (2011)&lt;/a&gt;. The details can be found in that answer, but basically, Brown &amp;amp; Batygin's reference to previous papers by Morbidelli et al. suggest that they think that this is what remains of an ice giant, like Uranus and Neptune, that formed near Saturn and Jupiter but was ejected in a series of gravitational interactions that led it to the Kuiper Belt and beyond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, Batygin's estimates place the ejection of this planet way before previous models of a 5th gas giant predict, so either we accept models with a  6th gas giant (also an ice giant) and say that both of these were ejected independently of one another - not impossible - or we go for some other option. The bottom line is that this is a planet, if we go with the above assumptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two other guesses have been made based on the planet's mass, estimated by Brown &amp;amp; Batygin at about 10 Earth masses. One is that it is a very massive &lt;a href=&quot;https://en.wikipedia.org/wiki/Super-Earth&quot; rel=&quot;nofollow noreferrer&quot;&gt;super-Earth&lt;/a&gt;, hanging around the proposed upper mass limit for those objects. This implies that the planet is rocky, but does not imply anything else. The other guess is that it is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Mini-Neptune&quot; rel=&quot;nofollow noreferrer&quot;&gt;mini-Neptune&lt;/a&gt;, with a predominantly gaseous composition but at about the same mass. The super-Earth hypothesis arose in part because of claims a long time ago about the ejection of a 5th &lt;em&gt;terrestrial&lt;/em&gt; planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've used the term &quot;planet&quot; throughout this entire answer, and that's because this object may be one - and most likely was in the past. Let's look at the IAU's three major criteria to be a planet:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;It orbits the Sun.&lt;/strong&gt; This is definitely true, if it exists. Also note that if this is a captured rogue planet, as some have proposed, then we can still call it a &quot;planet&quot;, because it now orbits the Sun.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;It is in hydrostatic equilibrium.&lt;/strong&gt; This is most likely true, and is true if any of the three models I suggested above are correct.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;It clears its neighborhood.&lt;/strong&gt; This is a bit iffy. It clearly influences objects around it, but we don't know much about those influences besides observations of the six Trans-Neptunian Objects (TNOs) Batygin &amp;amp; Brown studied. That said, an object with a mass of 10 Earth masses will most likely clear the neighborhood around its orbit if it has a stable orbit, which should be the case here. See the quote at the start of this answer for more information.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-01-24T15:55:42.503" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13345" PostTypeId="2" ParentId="13330" CreationDate="2016-01-23T22:52:38.660" Score="51" Body="&lt;p&gt;Brown and Batygin, the authors of the paper on the possible planet, have a &lt;a href=&quot;http://www.findplanetnine.com/p/blog-page.html&quot; rel=&quot;noreferrer&quot;&gt;webpage&lt;/a&gt; addressing this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few reasons not already covered:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It moves quite slowly - the authors estimate 0.2-0.6 arc seconds per hour - so standard surveys may not notice the movement and fail to recognize it as a solar system object.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Eris, which is the most distant confirmed object still known in the&#xA;  solar system, moves at a speed of 1.5 arcseconds per hour, which is so&#xA;  slow that it was missed the first time around. Most surveys of the&#xA;  outer solar system would not be able to find Planet Nine, even if it&#xA;  were quite bright, as they would just think it is a stationary star.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If the planet is near aphelion, it might be an order of magnitude further away than any major or minor planet we've found so far (excluding exoplanets, which are found by methods that don't apply in this case). The authors suggest an aphelion between 500 and 1200 AU. For comparison, Pluto is at 30-50 AU, while Eris at around 100 AU wasn't discovered until 2005. The potential 9th planet would be far larger than Eris, but is also likely to be much further away, and thereby fainter.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The WISE survey eliminated &lt;a href=&quot;http://www.space.com/24989-planet-x-search-nasa-telescope.html&quot; rel=&quot;noreferrer&quot;&gt;Saturn-sized planets within 10,000 AU&lt;/a&gt;, and Jupiter-sized planets within 26,000 AU. But the potential 9th planet is far smaller than those. WISE has also done a more sensitive search, which would pick up Neptune-sized objects, but that search has so far covered only a limited part of the sky.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The planet will be far harder to spot if it has the Milky Way in the background - there are too many stars potentially drowning out a faint object.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here's the authors' summary:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UDCTJ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UDCTJ.png&quot; alt=&quot;Estimated 9th planet orbit&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Estimated orbit for the putative 9th planet. The horizontal axis is the right ascension. The colored segments are regions where it should have been found by existing surveys.&#xA;&lt;br&gt;Illustration by Brown and Batygin, assuming &lt;a href=&quot;https://en.wikipedia.org/wiki/Fair_use&quot; rel=&quot;noreferrer&quot;&gt;fair use&lt;/a&gt; applies.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The biggest unexplored territory is where, statistically, it is most&#xA;  likely to be: near aphelion. Sadly, aphelion is also very close to the&#xA;  Milky Way galaxy. Ugh.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;So where is it? Probably distant. 500 AU+. Probably fainter than 22nd &#xA;  magnitude. Very possibly in the middle of the Milky Way galaxy.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now go find planet nine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More details on the authors' webpage: &lt;a href=&quot;http://www.findplanetnine.com/p/blog-page.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.findplanetnine.com/p/blog-page.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Finally, the gravitational dominance of the Sun reaches halfway to the nearest star. There's still plenty of unexplored territory for planets smaller than Saturn to hide in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fRmFX.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fRmFX.jpg&quot; alt=&quot;Solar system distances out to the Oort cloud and beyond&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&lt;em&gt;Note the log axis. We have a good map for the inner 50 AU, and are starting to find objects around 100 AU, but solar system objects might exist all the way to the outer edges of the Oort cloud.&#xA;&lt;br&gt;Illustration from &lt;a href=&quot;https://en.wikipedia.org/wiki/Oort_cloud&quot; rel=&quot;noreferrer&quot;&gt;wikipedia&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="9682" LastActivityDate="2016-01-23T22:52:38.660" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13347" PostTypeId="2" ParentId="13270" CreationDate="2016-01-23T23:20:45.327" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://www.findplanetnine.com/p/blog-page.html&quot; rel=&quot;nofollow&quot;&gt;Batygin and Brown made a website&lt;/a&gt; which describes the search for the 9th planet in clear terms. They specifically note the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;perihelion (its closest approach to the sun) at around a Right Ascension in the sky of 16 hours, which means that the perihelion position is straight overhead in late May. Conversely, the orbit comes to aphelion (the furthest point from the sun) at about 4 hours, or straight overhead in late November.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So to look for it, one should look along the ecliptic, concentrating mostly on the area directly overhead in late November. Note that this is the part of the sky where the galactic center also appears. The inclination is estimated to be 30 degrees, plus or minus 20, so that distance from the ecliptic should be searched as well.&lt;/p&gt;&#xA;" OwnerUserId="182" LastEditorUserId="182" LastEditDate="2016-01-24T00:33:15.307" LastActivityDate="2016-01-24T00:33:15.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13349" PostTypeId="2" ParentId="13342" CreationDate="2016-01-24T01:02:22.847" Score="6" Body="&lt;p&gt;&lt;a href=&quot;http://arxiv.org/pdf/astro-ph/0111290v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thommes et al. (2001)&lt;/a&gt; ran simulations and found that, at optimal conditions (namely, a planet of ~ 10 Earth masses), migration can be complete with ~ 100,000 years. Note that this was done before in-depth research was done on the Nice model, which is very similar. However, the mechanisms are different, as are the planet masses. The difference in timescales is dramatic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/pdf/0712.0553v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Levison et al. (2007)&lt;/a&gt; did explore their own model - the Nice Model. They found that it took 60 million years to 1.1 billion years for Jupiter and Saturn to break their resonance. The period of encounters and scattering lasted for 878-885 million years, followed by a period of eccentricity damping lasting for 0.3 million years to 4 million years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What &lt;em&gt;was&lt;/em&gt; relatively quick was the ejection of the hypothetical 5th giant planet. The change in the other giants' orbits, however, was not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, planetary migration on this scale takes a long time. A very long time.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For some &lt;em&gt;really&lt;/em&gt; interesting results, see the graphs of semi-major axis vs. time from the various four-, five-, and six- planet models of &lt;a href=&quot;http://arxiv.org/pdf/1208.2957v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nesvorný  &amp;amp; Morbidelli (2012)&lt;/a&gt;. There are some incredible oscillation among the orbits of Uranus and Neptune in some of the simulations, which is eventually slowly damped.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.imgur.com/lmvKzet.png&quot; alt=&quot;&quot;&gt;&#xA;  &lt;br&gt;&#xA;  Fig. 14.— Orbit histories of the giant planets in a simulation with five initial planets. See the caption of Fig. 1 for the description of orbital parameters shown here. The five planets were started in the (3:2,3:2,2:1,3:2) resonant chain, M&lt;sub&gt;disk&lt;/sub&gt; = 20 M&lt;sub&gt;Earth&lt;/sub&gt; and B(1). The fifth planet was ejected at $t =$ 6.1 Myr after the start of the simulation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's another relevant passage, if you're looking at setups with six giant planets:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The instability typically occurred in two steps, corresponding to the ejection of the two planets. Sometimes, as in Fig. 18, the ejection of the two planets was nearly simultaneous, but most of the times there was a significant delay between ejections. This was useful because the first planet’s ejection partially disrupted the planetesimal disk and reduced its capability to damp e55, which was then excited by the second planet’s ejection. While this mode of&#xA;  instability can be important, we would need to increase the statistics (&gt;100 simulations for each initial condition) to be able to properly resolve the small success fractions in the six-planet case.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-01-25T15:48:16.560" LastActivityDate="2016-01-25T15:48:16.560" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13350" PostTypeId="2" ParentId="13327" CreationDate="2016-01-24T09:23:55.723" Score="6" Body="&lt;p&gt;The first observation is whether gravitation radiation exists as predicted by General Relativity. Evidence from observations of binary neutron stars says it does, but it remains a major unknown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Now Gravitational waves have been discovered, and the predictions of GR have been confirmed&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational astronomy will be more like listening than looking. Right now I can hear my kids playing upstairs. I can learn a lot about what they are doing just by noting that sound waves are passing, from a particular direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Since the first detection of gravitational waves, we have observed several black hole merges, of different sizes. We have also observed neutron star collisions and subsequently observed these at gamma ray and visible wavelengths. This confirms these events as a potential source of many heavy elements.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would expect extreme gravitational events to produce particular wave forms, for example black hole mergers should make a &quot;tone&quot; that rises in pitch as the two event horizons merge at faster rates. Again we have lots of theory on this but if we can &quot;hear&quot; these events we can check if GR does correctly model gravity in these situation, or if there is something missing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most interesting would be if we do hear Black hole mergers, but they don't sound like what we have expected. That would mean that there is more to gravity than we understand, and would lead to new science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Now after several observations, it is the combination of gravitational observations combined with light observations that are producing the most science. E-Lisa will allow much smaller to be detected.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-02-23T15:13:03.870" LastActivityDate="2018-02-23T15:13:03.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13351" PostTypeId="2" ParentId="13339" CreationDate="2016-01-24T11:10:21.147" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The recent evidence for the 9th planet is an alignment in various parameters of known Kuiper belt objects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These are not Kuiper belt objects. The Kuiper belt is taken to refer to the neighborhood of Neptune resonances. The scattered disc is another population, further out, and the objects relating to Planet 9 are further still.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Dji4J.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dji4J.png&quot; alt=&quot;A scatter plot showing orbital parameters of many bodies.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Legend: grey = Kuiper belt, green = integer Neptune resonances, blue = scattered disc. The objects in the recent announcement are all off the chart, although their perihelions are &amp;lt;100 AU.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Scattered_disc&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape of the grey splotch at the left of this chart shows that the Kuiper belt is indeed a collection of populations dominated by Neptune. This isn't news, and it's still taken to be a belt. It goes all the way around the solar system and the orbital eccentricities are low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This postulated planet might help explain the &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Kuiper_belt#.22Kuiper_cliff.22&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kuiper cliff&lt;/a&gt;&quot;, the otherwise unexpected cutoff before the 2:1 resonance. But, that's another question. Failing that, there's simply no relation between the Kuiper belt and Planet 9 except that they're both very far away.&lt;/p&gt;&#xA;" OwnerUserId="1641" LastActivityDate="2016-01-24T11:10:21.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13352" PostTypeId="1" AcceptedAnswerId="13353" CreationDate="2016-01-24T13:19:07.627" Score="3" ViewCount="432" Body="&lt;p&gt;I was wondering about the light dimming of the 'alien megastructure' star.&#xA;From what I understand, for it to be a planetary body, it would have to be some 20 times the size of Jupiter.&#xA;Is it possible that it could be caused by some huge, dense gas cloud, orbiting around the star?&lt;/p&gt;&#xA;" OwnerUserId="10518" LastActivityDate="2016-01-28T15:18:44.387" Title="Could the KIC 8462852 (Alien megastructure star) be explained by orbiting gas clouds?" Tags="&lt;star&gt;&lt;extra-terrestrial&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13353" PostTypeId="2" ParentId="13352" CreationDate="2016-01-24T14:29:53.637" Score="4" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The light curves don't look at all like anything transiting. The ingress and egress profiles and the timing are way beyond that kind of explanation. Not planets, not comets, not clouds, not alien super structures.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No infrared light from dust or gas has been detected, as they should've been if the starlight had heated small particles. And since this is an F star with about 50% larger diameter than the Sun, which contains 99% of the mass in our Solar System, you can imagine how huge any comet/dust/gas cloud has to be to cover 22% of it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No brightening of the starlight has been detected, as this allegedly huge&quot;cloud&quot; orbits behind the star and should reflect light to us. Shouldn't those huge icy comet clouds reflect some starlight?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Jupiter is about as large as a planet can get, even with 10+ times more mass, its volume would not increase by much. Gas giants differ much more in mass than in volume. An object 20 or 200 times more massive than Jupiter would be a roughly Jupiter sized red dwarf star, and that is not what has been observed at &quot;Tabby's star&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The anomalies correlate strictly with the telescope's orbital period and orientation and are obviously a man made artefact because of some unforeseen malfunction of the instrument, nothing astrophysical is involved.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2016-01-24T14:38:35.593" LastActivityDate="2016-01-24T14:38:35.593" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13354" PostTypeId="1" CreationDate="2016-01-24T14:30:48.367" Score="0" ViewCount="63" Body="&lt;p&gt;It is discovered only in 2011.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it so late? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know where to find it.&lt;/p&gt;&#xA;" OwnerUserId="10519" LastActivityDate="2016-01-24T15:27:09.153" Title="Why is it so difficult to discover the earth Trojan?" Tags="&lt;asteroids&gt;&lt;celestial-mechanics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13355" PostTypeId="2" ParentId="13354" CreationDate="2016-01-24T15:27:09.153" Score="3" Body="&lt;p&gt;2010 TK7 has a diameter of about 300 meters (1,000 ft). Its movements do not bring it any closer to Earth than 20 million kilometers (12.4 million miles), which is more than 50 times the distance to the Moon. (from the wiki entry for it)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That should make it pretty clear? Can you see a fly from 200 km away?&lt;/p&gt;&#xA;" OwnerUserId="10084" LastActivityDate="2016-01-24T15:27:09.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13356" PostTypeId="1" AcceptedAnswerId="13361" CreationDate="2016-01-24T15:45:07.223" Score="3" ViewCount="336" Body="&lt;p&gt;I had a funny thought, and would like to pose it to you:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I was a kid, reading any kind of astronomy or similar books would all say that there was no other life in the universe, and made it pretty clear that anyone thinking there could be was probably off his rocker. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays, it seems this has become a very important question for astronomers and other scientists, a new deep-seated belief that must now be proven. Some seem to think it would be the most important discovery of all time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is, if we did 'discover' life on another planet, it'll make the news for a bit, and then the average person will go back to their every day life without much more thought about it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For me, I would think this is great, only insofar as it means a possible new source of food (i.e., some delicious exotic cuisine assuming it's edible, and possibly even some place to go conquer and colonize if there was any profitability/money in doing so). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some scientists, though, I get the sense it's more of an axe to grind with the long-dreary he-said she-said argument of science vs. religion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do you think it would matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question assumes any life discovered is probably not going to be very 'interesting' life, at least nothing we're going to have a conversation with.&lt;/p&gt;&#xA;" OwnerUserId="10520" LastActivityDate="2018-05-20T18:40:01.167" Title="Why would discovering life on another planet be important/matter to us?" Tags="&lt;extra-terrestrial&gt;" AnswerCount="5" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13357" PostTypeId="2" ParentId="13356" CreationDate="2016-01-24T16:52:47.973" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My question assumes any life discovered is probably not going to be very 'interesting' life, at least nothing we're going to have a conversation with.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is &quot;interesting&quot; depends on your personal outlook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can think of any number of living breathing entities on this planet I wish I did &lt;em&gt;not&lt;/em&gt; have to speak with.  A life form that just lies in a Petri dish and does nothing else sounds quite pleasing sometimes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a scientific point of view, I don't see how any scientist could regard the acquisition of information, like a form of life not evolved on Earth, as unimportant.  To me the difference between &quot;I think this is likely to happen anywhere conditions are favourable&quot; and &quot;I actually found it&quot; is pretty much the definition of good science.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2016-01-24T16:52:47.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13358" PostTypeId="2" ParentId="1179" CreationDate="2016-01-24T17:47:10.320" Score="11" Body="&lt;p&gt;The recent possible discovery of &lt;a href=&quot;https://en.wikipedia.org/wiki/Planet_Nine&quot; rel=&quot;noreferrer&quot;&gt;Planet Nine&lt;/a&gt; by &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/0004-6256/151/2/22/meta&quot; rel=&quot;noreferrer&quot;&gt;Batygin &amp;amp; Brown (2016)&lt;/a&gt; has caused quite a stir, in the astronomy community, Astronomy Stack Exchange, and the rest of the world. This is, of course, in part because any mention of such a discovery will cause a stir, but it is also in part because of the claimed probability if the movements of the Trans-Neptunian Objects (TNOs) being purely by chance: 0.007%, or 1 in 14,000, which corresponds to a probability of ~ 3.8&lt;a href=&quot;https://en.wikipedia.org/wiki/Standard_deviation&quot; rel=&quot;noreferrer&quot;&gt;$\sigma$&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, the actual probability that this planet exists is a bit lower - 90%, &lt;a href=&quot;https://www.washingtonpost.com/news/speaking-of-science/wp/2016/01/20/new-evidence-suggests-a-ninth-planet-lurking-at-the-edge-of-the-solar-system/&quot; rel=&quot;noreferrer&quot;&gt;according to Brown&lt;/a&gt;, and 68.3%, &lt;a href=&quot;http://www.newyorker.com/tech/elements/discovering-planet-nine&quot; rel=&quot;noreferrer&quot;&gt;according to Greg Laughlin&lt;/a&gt;, an astronomy at UC-Santa Cruz who knew of the results prior to the paper's publication. The difference between these odds and the odds mentioned in the paper is due to the fact that perhaps there are other contributing factors that could have caused the movements of the TNOs - and also that these informal numbers are more like guesses than estimates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/13330/why-hasnt-the-9th-planet-been-detected-already&quot;&gt;Why hasn&amp;#39;t the &amp;quot;9th Planet&amp;quot; been detected already?&lt;/a&gt; have already given a variety of methods that &lt;em&gt;won't&lt;/em&gt; work in the case of Planet Nine, and why they haven't worked. Besides the obvious possibility that Planet Nine just doesn't exist, here are some of them:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It might be confused with a background star, blending in with the Milky Way.&lt;sup&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/13345/2153&quot;&gt;[1]&lt;/a&gt;&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;li&gt;It's far away, and dim - perhaps a 22&lt;sup&gt;nd&lt;/sup&gt; magnitude object.&lt;sup&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/13345/2153&quot;&gt;[1]&lt;/a&gt;, &lt;a href=&quot;https://astronomy.stackexchange.com/a/13331/2153&quot;&gt;[2]&lt;/a&gt;&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;li&gt;Most methods of detecting exoplanets won't work, including&lt;sup&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/13332/2153&quot;&gt;[3]&lt;/a&gt;&lt;/sup&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Radial velocity.&lt;/li&gt;&#xA;&lt;li&gt;Transit.&lt;/li&gt;&#xA;&lt;li&gt;Gravitational microlensing&lt;/li&gt;&#xA;&lt;li&gt;Direct imaging.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Prior surveys either don't look enough in-depth, in-depth but in other areas of the sky, or cover wavelengths in which this planet doesn't show up in.&lt;sup&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/13335/2153&quot;&gt;[4]&lt;/a&gt;&lt;/sup&gt; This is another problem with the use of WISE detailed in &lt;a href=&quot;https://astronomy.stackexchange.com/a/1183/2153&quot;&gt;MBR's answer here&lt;/a&gt;. Others problems are that planet is probably not massive enough for WISE to see it, and is extremely far away. It may also be cold and have a low albedo, making direct detection hard.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I bring all this up because, surprisingly, the case of Planet Nine is very relevant to the detection of these other planets that you mentioned in your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/HD_106906_b&quot; rel=&quot;noreferrer&quot;&gt;HD 106906 b&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several similarities here between this planet and Planet Nine.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;HD 106906 b has an orbit with a semi-major axis of at least 650 AU. By comparison, Planet Nine is thought to have a semi-major axis of about 700 AU, according to the estimates in the paper. That said, other orbital properties are unknown for HD 106906 b and may differ.&lt;/li&gt;&#xA;&lt;li&gt;HD 106906 b has a mass of about 11 Earth masses; Planet Nine has a mass of about 10 Earth masses.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;One difference is that HD 106906 b may have formed where it is at the moment. Scattering, like Planet Nine experienced, appears to be unlikely. This means that the two could have different compositions - although I would think that both might be ice giants or the remains thereof, given where they formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, HD 106906 b may be very similar to Planet Nine, and while we do not know much about either property, it seems safe to say that detection methods and problems would be similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Fomalhaut_b&quot; rel=&quot;noreferrer&quot;&gt;Fomalhaut b&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fomalhaut b is a bit different. Its semi-major axis is ~177 AU - much smaller than these other two planets - and may be anywhere from 10 Earth masses to several hundred Earth masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly enough, Fomalhaut b was also indirectly detected, just like Planet Nine. A gap was found in Fomalhaut's dust disk, which could only have been caused by a massive planet. Later, it was directly imaged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Direct detection &lt;em&gt;might&lt;/em&gt; be possible if Fomalhaut b was in the Solar System, especially given its greater size and mass. Additionally, it would have an enormous impact on TNOs. However, it would be a gas giant, not an ice giant, so it would be difficult to explain how it moved out so far.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I discussed those planets to take a slight detour and talk about what might (and might not) be beyond Neptune. I'll be more explicit here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What could be beyond Neptune&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;An ice giant, like Planet Nine, or the remains of one. This would have been formed closer in, near Jupiter and Saturn, which existed in a period resonance with Saturn. Other giant planet resonances would have also existed. Then a instability broke the resonances, bringing the ice giant inwards towards Saturn and then Jupiter, where it was flung outwards. This changed the orbits of the other four gas giants. I talked about this is &lt;a href=&quot;https://astronomy.stackexchange.com/a/13311/2153&quot;&gt;another answer of mine&lt;/a&gt;, which I seem to be continuously referencing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, the initial answer was incorrect, which I have since noted, and was revised. Batygin's estimates found that Planet Nine would have been ejected many, many, millions of years before the resonances were supposedly broken (according to evidence from &lt;a href=&quot;https://en.wikipedia.org/wiki/Late_Heavy_Bombardment&quot; rel=&quot;noreferrer&quot;&gt;the Late Heavy Bombardment&lt;/a&gt;). However, this does not mean that there could not have been a sixth giant. The simulations &lt;a href=&quot;http://arxiv.org/pdf/1208.2957v1.pdf&quot; rel=&quot;noreferrer&quot;&gt;Nesvorný  &amp;amp; Morbidelli (2012)&lt;/a&gt; that explored the evolution of a Solar System with four, five, and six giant planets found some good results with five and six giant planets.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Super-Earth&quot; rel=&quot;noreferrer&quot;&gt;super-Earth&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/Mini-Neptune&quot; rel=&quot;noreferrer&quot;&gt;mini-Neptune&lt;/a&gt;.&lt;sup&gt;1&lt;/sup&gt; These are planets that are, at the most, 10 Earth masses. Super-Earths would be terrestrial planets with thick atmospheres; mini-Neptunes would be gas planets. Note that these names do not say anything else about their compositions - for example, super-Earths are not necessarily habitable.&lt;/li&gt;&#xA;&lt;li&gt;Something else. The region has not yet been explored well, and there could be some other unforeseen objects. This kind of object would likely be cold, as WISE (see next section) might have detected it otherwise.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What cannot be beyond Neptune&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A gas giant.&lt;sup&gt;2&lt;/sup&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Wide-field_Infrared_Survey_Explorer&quot; rel=&quot;noreferrer&quot;&gt;The Wide-field Infrared Survey Explorer&lt;/a&gt; found &lt;a href=&quot;http://www.space.com/24989-planet-x-search-nasa-telescope.html&quot; rel=&quot;noreferrer&quot;&gt;no Saturn-mass objects up to 10,000 AU, and no Jupiter-mass objects up to 26,000 AU&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.imgur.com/oizUp6J.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/oizUp6J.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is, of course, besides &lt;a href=&quot;http://xkcd.com/1633/&quot; rel=&quot;noreferrer&quot;&gt;many other planets that cannot exist&lt;/a&gt;. Anyway, this is we would most likely have detected Fomalhaut b by now, if it was in the Solar System.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, assuming that there &lt;em&gt;is&lt;/em&gt; something beyond Neptune, how can we fine it? Well, now that its existence has been hypothesized and we know where it &lt;em&gt;should&lt;/em&gt; be, astronomers can turn telescopes towards that location and use direct imaging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's been some excitement because &lt;a href=&quot;http://www.naoj.org/Topics/2016/01/21/index.html&quot; rel=&quot;noreferrer&quot;&gt;the Subaru Telescope will be used&lt;/a&gt;. Other optical (and possibly infrared and other wavelengths) telescopes will most likely be used as well. Answers to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/13313/what-wavelength-to-best-detect-the-9th-planet&quot;&gt;What wavelength to best detect the &amp;quot;9th planet&amp;quot;?&lt;/a&gt; (especially Rob Jeffries' answer) indicate that optical and infrared/near-infrared wavelengths will be the best choice. As of today (1/24/2016), we can only speculate on what other instruments will be used, and what their chances are of finding this planet - if it exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who knows? Maybe something else unexpected will turn up.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1 Now we get into the different things Planet Nine could be. &lt;a href=&quot;https://astronomy.stackexchange.com/a/13343/2153&quot;&gt;My answer to What type of planetary-mass object would Planet Nine be?&lt;/a&gt; covers this, but there are better answers to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/13321/ninth-planet-what-else-could-it-be?rq=1&quot;&gt;Ninth planet - what else could it be?&lt;/a&gt; than mine.&lt;/sup&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;2 Be careful of the difference between an &lt;a href=&quot;https://en.wikipedia.org/wiki/Ice_giant&quot; rel=&quot;noreferrer&quot;&gt;ice giant&lt;/a&gt; and a &lt;a href=&quot;https://en.wikipedia.org/wiki/Gas_giant&quot; rel=&quot;noreferrer&quot;&gt;gas giant&lt;/a&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-01-25T21:37:26.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13359" PostTypeId="1" AcceptedAnswerId="13360" CreationDate="2016-01-24T18:17:41.540" Score="4" ViewCount="387" Body="&lt;p&gt;After reading about &lt;a href=&quot;http://www.solstation.com/stars/barnards.htm&quot; rel=&quot;noreferrer&quot;&gt;Barnard's Star&lt;/a&gt; and how it's approaching us 140 km/sec, I'm curious if there are any visualizations of how the nearest stars to Sol will be moving over the next 100k years or so. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm amazed that Barnard's Star, currently 6 light years away, will be only 3.8 light years away in about 9,800 years. It's really moving!&lt;/p&gt;&#xA;" OwnerUserId="6870" LastActivityDate="2017-12-27T23:49:57.220" Title="Are there good animations of the movement of proximate stars over long time periods?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;distances&gt;&lt;movement&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13360" PostTypeId="2" ParentId="13359" CreationDate="2016-01-24T19:24:16.240" Score="6" Body="&lt;p&gt;This image should give you a good idea of the distances of stars relative to the Sun over the next 80,000 years (and the past 20,000 years).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/e/e9/Near-stars-past-future-en.svg/600px-Near-stars-past-future-en.svg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/e/e9/Near-stars-past-future-en.svg/600px-Near-stars-past-future-en.svg.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;Image courtesy of Wikipedia user Lambiam under &lt;a href=&quot;https://creativecommons.org/licenses/by-sa/3.0/deed.en&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Creative Commons Attribution-Share Alike 3.0 Unported license&lt;/a&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've created a modified version below. The nearest star's track in time is in red, and the four blue boxes indicate the five transitions (the second box contains two):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.imgur.com/0CA9gHC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/0CA9gHC.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proxima Centauri remains the nearest star to us for about 25 thousand years, at which point Alpha Centauri passes it. ~7.5 thousand years after that, Ross 248 swings in closer. It then moves away after about 10 thousand years, when Alpha Centauri moves just ahead of it for a short period of time. Gliese 445 becomes the nearest star for about another ~7.5 thousand years, until it wings away and Alpha Centauri again becomes the closest star. 80 thousand years in our future, Ross 128 nudges ahead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few interesting things to note are that for most of this time, the nearest star is closer to the Sun than Proxima Centauri is now - indeed, there are times when quite a few stars are closer simultaneously - and that this lasts for some time as a brief spike.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is only a short-term pattern. In the past, many stars have come much closer - even inside the Oort Cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, to address the mention of Barnard's Star: It will swing close to us within ~ 10,000 years, but will just as suddenly swing away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, a cool animation shows the movement of stars that are near the Sun throughout their past trip around the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/f/f8/Sun_in_orbit_around_Galactic_Centre.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/f/f8/Sun_in_orbit_around_Galactic_Centre.gif&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;Animation courtesy of Wikipedia user Henrykus under &lt;a href=&quot;https://creativecommons.org/licenses/by-sa/3.0/deed.en&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Creative Commons Attribution-Share Alike 3.0 Unported license&lt;/a&gt;. Data and original information from the European Southern Observatory.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above animation was the result of &lt;a href=&quot;http://www.eso.org/public/news/eso0411/&quot; rel=&quot;nofollow noreferrer&quot;&gt;1,000+ nights of observation&lt;/a&gt; over 15 years by astronomers working at the ESO. They observed or calculated the ages, velocities, and compositions of 14,000+ F and G stars. They then used the information to create orbital trajectories for these stars.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-01-24T21:18:38.453" LastActivityDate="2016-01-24T21:18:38.453" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13361" PostTypeId="2" ParentId="13356" CreationDate="2016-01-24T20:17:15.823" Score="5" Body="&lt;p&gt;It will be very important in several different ways. Not as in Messiah having descended to the World, and maybe not during one single human lifetime. But it will profoundly change the world in our minds, and thereby change our collective long term actions. The next generation would grow up taking alien life in space for granted, their minds will be more different than we old guys can change our stiff brain flesh. Like Newton's physics dominate our everyday life now, even for those of us who don't understand how all of physics has been deduced.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If life survives in other places than Earth, then maybe we (as in life from Earth) can do so too in the future.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If there's life in other places than Earth, then maybe we came from it? All life on Earth is family, all living things are closely related with each other. The question of our origin has always had high priority in human minds, see all religions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If alien life is biologically similar to that on Earth (as in using DNA and proteins) then we could use it as a huge laboratory for medicine and environment control and agriculture and what not in life science. Astronomy today contributes to physics in our everyday life by studying a laboratory (the universe) with scale far beyond what humans could ever construct. If biology could do the same, we would soon notice the results in the shape of useful products.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If alien life does not use DNA and proteins but is very different from us, then I can't even guess what discoveries would come out of that. Zweistein.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If alien life is not found, even when we have searched long and hard, then we can gradually make the greatest discovery of all times: That we are unique and special. Searching for ET without finding it would be the most mind boggling thing imaginable. I don't think this hypothesis has been well addressed yet, I suppose no philosopher knows even where to start.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-01-24T20:17:15.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13363" PostTypeId="1" CreationDate="2016-01-24T21:05:36.773" Score="2" ViewCount="307" Body="&lt;p&gt;Like a ball falls on the earth due to gravity, if suppose we are able to place a ball close enough to the sun. Will the ball fall into the sun due to its gravity or will it be pushed away due to its energy?&lt;/p&gt;&#xA;" OwnerUserId="10430" LastActivityDate="2016-01-26T09:23:15.373" Title="Will a Ball placed close to Sun fall into it?" Tags="&lt;the-sun&gt;&lt;gravity&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13365" PostTypeId="2" ParentId="13352" CreationDate="2016-01-24T21:58:24.160" Score="2" Body="&lt;p&gt;There are some theories to explain that. The following explanations, I took from &lt;a href=&quot;https://en.wikipedia.org/wiki/KIC_8462852&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipeda's article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The star has a small companion red dwarf which just crossed its Oort cloud equivalent (at 885 UA). A passing star this close would surely cause havoc and serious disturbances to comets' orbits. This could result in a swarm of comets being thrown into the inner stellar system at once in highly excentric orbits. However, there are severe doubts about this.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It was also proposed that a planet with a very big ring system transits the star (or perhaps, nearly misses and just part of the rings transit the star). This is not unprecedent, and was already seen with &lt;a href=&quot;https://en.wikipedia.org/wiki/1SWASP_J140747.93-394542.6&quot; rel=&quot;nofollow noreferrer&quot;&gt;another star&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Astronomer Jason Wright suggested that the star might be younger than it seems and it is still coalescing material around it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Astronomer Bradley Schaefer presented just a few days ago (2016-01-13) a study where he concludes that the star dimmed roughly 20% from 1890 to 1989, and that this is unprecedent for a F-type star. So, there could be still more weird things going on than previously thought.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;NASA Infrared Telescope Facility found similiarities with another star, Eta Corvi, which is undergoing a Late Heavy Bombardment.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Wikipedia's article also cites another possibility (which took some headlines around the world). But this is just some (unfortunately notable) sensacionalistic (un|pseudo)jornalistic claim that nobody can take seriously:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;There is an advanced alien civilization building a giant device like a Dyson Sphere or something similarly big.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;LocalFluff posted &lt;a href=&quot;https://astronomy.stackexchange.com/a/13353/609&quot;&gt;an answer suggesting another possibility&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;The anomalies correlate strictly with the telescope's orbital period and orientation and are obviously a man made artefact because of some unforeseen malfunction of the instrument, nothing astrophysical is involved.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, the conclusion (at least my conclusion) is that we simply don't know yet what is happening and we need many follow up observations which will likely tell us a lot. Specially somewhere around May of 2017, when it is predicted that the strange megastructure transit should happen again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally would guess that the close encounter with the red dwarf triggered a Late Heavy Bombardment. But this is only a guess from mine and I have no way to provide evidence for it.&lt;/p&gt;&#xA;" OwnerUserId="609" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-01-28T15:18:44.387" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13366" PostTypeId="2" ParentId="13352" CreationDate="2016-01-24T22:17:16.570" Score="0" Body="&lt;p&gt;I've been pondering the dimming of this star for a couple months now and have come up with a plausible explanation. &#xA;The dimming is probably caused by a very dim companion dwarf star in a highly elliptical orbit where its closest approach to KIC8462852 (Tabby's Star) is on, or near, the same path as our line of sight.&#xA;The gravitational pull from the dwarf would create tidal forces on Tabby's star causing it to bulge in the same way our moon causes the oceans to bulge.&lt;br&gt;&#xA;As is commonly known, stars that spin very fast tend to bulge at their equators due to centripetal force and this results in decreased pressure within the star.. this decreased pressure slows down the nuclear fusion causing a dimming effect at the equator.   Consequently, as the bulging equator causes the north and south poles to get closer together, they get brighter due to the increased pressure because they are closer to the stars core. &#xA;So with Tabby's star, as the dim dwarf is at its closest approach, it causes Tabby's star to bulge a bit..  that bulging causes nuclear fusion to slow, causing a dimming affect..  combine this with the dwarf star blocking some of the light itself and I think we could account for the 20% reduction.   The next mystery that has to be explained is the 100 year dimming of the star..   I believe that as the two companions interact with each other, material from each is probably ejected and probably stays within a close orbit of the main star.  This would account for a pattern of continuous dimming as this material built up on each pass.  This would also account for why we don't see any infrared signature because the ejected materials orbit is too close to distinguish it from the main star.&lt;br&gt;&#xA;If I'm right, our telescopes should be able to see the main star wobble back and forth (alot) as the dwarf rapidly transits from its approach on one side of the star, then to the other side on during its departure.&lt;/p&gt;&#xA;" OwnerUserId="10530" LastActivityDate="2016-01-24T22:17:16.570" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13367" PostTypeId="1" CreationDate="2016-01-25T06:00:23.203" Score="1" ViewCount="48" Body="&lt;p&gt;In &lt;a href=&quot;https://www.youtube.com/watch?v=aRDOqiqBUQY&quot; rel=&quot;nofollow&quot;&gt;this video&lt;/a&gt; on inertial reference frames, it is mentioned that the stars are humanity's best inertial reference frame: the earth experiences a subtle acceleration relative to the sun due to the earth's own orbit and rotation; therefore the earth is only roughly an inertial frame. But, the stars are relatively fixed, they say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the stars in galaxies move relative to each other--I assume many with some sort of rotational velocity and therefore acceleration--and galaxies themselves are all slowly moving away from one another, perhaps even with centrifugal acceleration of their own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has this acceleration been great enough that humanity has needed to update navigational tools that rely on the assumption that stars are an inertial reference frame?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible that every observable object in the universe is in fact accelerating relative to an unobservable reference frame?&lt;/p&gt;&#xA;" OwnerUserId="10536" LastActivityDate="2017-01-13T22:10:02.113" Title="Has celestial navigation been materially impacted by the imperfect nature of celestial reference frames over time?" Tags="&lt;celestial-mechanics&gt;&lt;navigation&gt;&lt;frame-of-reference&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13368" PostTypeId="1" CreationDate="2016-01-25T07:08:08.933" Score="0" ViewCount="126" Body="&lt;p&gt;I am NOT an astronomy person, so this may be a dumb question, however I have recently found a tripod that appears to be for star tracking as far as I can guess. I would like to know the type of tripod and if possible the brand as it is not labeled and I do not know how to operate it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oadVw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oadVw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vl71E.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vl71E.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fBYaI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fBYaI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10537" LastEditorDisplayName="user1569" LastEditDate="2016-01-27T12:51:43.880" LastActivityDate="2016-01-27T12:53:08.607" Title="Identify this 2 axis motorized tripod" Tags="&lt;telescope&gt;&lt;star-gazing&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13369" PostTypeId="2" ParentId="13363" CreationDate="2016-01-25T11:28:43.603" Score="6" Body="&lt;p&gt;Short answer is yes. The only way radiation pressure would be a factor was if the ball was in orbit around the Sun, this would mean that the slight pressure it feels would cause it to drift away slowly (assuming its made of the most heat resistant material on Earth!). But in your question you said &quot;placed&quot; near the sun not in orbit around it, so the strength of the sun's gravity would overcome all other forces and pull it in. &lt;/p&gt;&#xA;" OwnerUserId="10437" LastEditorUserId="1678" LastEditDate="2016-01-25T11:35:38.400" LastActivityDate="2016-01-25T11:35:38.400" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13370" PostTypeId="2" ParentId="13363" CreationDate="2016-01-25T14:18:45.087" Score="6" Body="&lt;p&gt;The answer depends on the size and mass of the ball. It also depends on its ability to reflect light (albedo $A$), but let's forget that for a moment.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Pressure vs. gravity&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Solar pressure decreases with $R^2$ (the inverse square law). At Earth, which is located at a distance of $1\,\mathrm{AU}$ from the Sun, we receive an irradiance $S_0 = 1361\,\mathrm{W}\,\mathrm{m}^{-2}$. Since the momentum of a photon of energy $E$ is $p = E/c$, the pressure at a distance $R$ from the Sun is&#xA;$$&#xA;P = \frac{S_0}{c(R/\mathrm{AU})^2}.&#xA;$$&#xA;If the ball's radius is $r$, this pressure will exert a force&#xA;$$&#xA;F_\gamma = \pi r^2 P = \frac{\pi r^2 S_0}{c (R/\mathrm{AU})^2}\qquad(\mathrm{away\,from\,the\,Sun}).&#xA;$$&#xA;Meanwhile, if the ball's mass is $m$, the gravitational force exerted by the Sun on the ball is&#xA;$$&#xA;F_g = \frac{G M_\odot m}{R^2},\qquad(\mathrm{toward\,the\,Sun})&#xA;$$&#xA;where $G$ is the gravitational constant and $M_\odot$ is the mass of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Threshold for falling&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The threshold for the ball falling into the Sun is found by equating the two oppositely directed forces:&#xA;$$&#xA;\frac{\pi r^2 S_0}{c (R/\mathrm{AU})^2} = \frac{G M_\odot m}{R^2}.&#xA;$$&#xA;The first thing to notice is that $R$ cancels out; the reason being that flux density and gravity both follow the inverse square law. Second, re-arranging terms we see that the ball will fall if its mass per area is greater than this threshold (if I have calculated correctly):&#xA;$$&#xA;\frac{m}{r^2} \gtrsim \frac{\pi S_0}{c G M_\odot } \mathrm{AU}^2 \simeq 2.4\times10^{-4}\,\mathrm{g}\,\mathrm{cm}^{-2}.&#xA;$$&#xA;Now you can plug in your favorite numbers. You will find that for most macroscopic objects, such as a football, a rock, and even a sand grain, it will fall. On the other hand, small &quot;balls&quot; such as dust grains and atoms, will tend to be pushed away from the Sun. An example of a macroscopic object that won't fall is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_sail&quot; rel=&quot;nofollow&quot;&gt;solar sail&lt;/a&gt; which seeks to maximize area per mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given density, say $\rho = 2.5\,\mathrm{g}\,\mathrm{cm}^{-3}$ which is characteristic of rocky materials, you can also calculate the maximum size before it falls toward the Sun:&#xA;$$&#xA;r_\mathrm{max} = \frac{3}{4\pi\rho} \,(2.4\times10^{-4}\,\mathrm{g}\,\mathrm{cm}^{-2}) \sim 0.1\mathrm{-}1\,\mu\mathrm{m}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Albedo&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The above calculations hold of all the radiation is absorbed by the object. If a fraction of it is reflected, the radiation will transfer &lt;em&gt;more&lt;/em&gt; momentum to the object, and for a prefectly reflecting object, the radiation force will be roughly twice the amount above (the exact factor depends on the geometry of the object).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Extinction curves&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It should be noted, though, that treating small particles as rigid spheres with a geometric cross section becomes imprecise when their size is comparable to wavelength of the light; rather, their absorption/scattering cross section should be treated quantum mechanically. In practice, &lt;em&gt;extinction curves&lt;/em&gt; — i.e. the cross section as a function of the wavelength of the light for a given dust particle size distribution — are measured observationally by comparing the light from unobscured star with similar stars behind dust clouds, and subsequently fitted with various functional forms.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-01-26T09:23:15.373" LastActivityDate="2016-01-26T09:23:15.373" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13371" PostTypeId="1" AcceptedAnswerId="13379" CreationDate="2016-01-25T14:19:42.990" Score="5" ViewCount="163" Body="&lt;p&gt;Konstantin Batygin has estimated that &lt;a href=&quot;https://en.wikipedia.org/wiki/Planet_Nine#Origin#Origin&quot; rel=&quot;nofollow&quot;&gt;if Planet Nine was ejected, it would have been ejected 3-10 million years after the formation of the Solar System&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious as to how he arrived at this figure. Was it through simulations of the early Solar System, a prediction of the alignments of orbits at that time, or something completely different?&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorDisplayName="user1569" LastEditDate="2016-01-26T13:10:48.153" LastActivityDate="2016-01-26T13:10:48.153" Title="How did Batygin estimate when Planet Nine was ejected?" Tags="&lt;solar-system&gt;&lt;9th-planet&gt;&lt;estimate&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13373" PostTypeId="1" CreationDate="2016-01-25T17:34:09.223" Score="6" ViewCount="420" Body="&lt;p&gt;I'm sure Caltech has answered this in some way, but I think it's a good question have on this site.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;What good evidence exists&lt;/strong&gt; for the 9th planet as spoken of by Caltech?&lt;/p&gt;&#xA;&lt;p&gt;As I see it, Caltech has apparently looked at the orbital parameters of many Trans-Neptunian-Objects (TNOs) and noticed that the lines between their perihilion and aphelion (I don't know what the technical term might be) all run more or less in the same direction. This is evidence that some large body has gravitationally &amp;quot;synced&amp;quot; these planets over millions of years.&lt;/p&gt;&#xA;&lt;p&gt;Except &lt;strong&gt;that alone isn't good evidence.&lt;/strong&gt; Here are two stats first:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Known TNOs: about 1,750 according to &lt;a href=&quot;http://www.minorplanetcenter.org/iau/lists/TNOs.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Minor Planet Center's list&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Estimated existing Kuiper-Belt Objects: 120,000+ according to &lt;a href=&quot;http://www.britannica.com/place/Kuiper-belt&quot; rel=&quot;nofollow noreferrer&quot;&gt;Britannica&lt;/a&gt;. (That number is only estimating objects of 100-km diameter or more, but that's fine for our purposes.) Note: 120,000 is only for KBOs. There are other TNO sets such as the oort cloud (and the scattered disc if you define it separately), so I expect the number to be much much bigger than 120,000.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Given that we've only measured about 1% or less of TNO orbits, how can we be sure we have a good random sample to say that a massive body must have aligned these orbits? In other words, all the other orbits might be randomly distributed, which would be compelling evidence that some giant body is &lt;strong&gt;not&lt;/strong&gt; &amp;quot;syncing&amp;quot; orbits or else way more than 1% of them would be &amp;quot;synced&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;If I recall my statistics class correctly, 1% is not good enough for a random sample. The border-line good enough area starts at 3%, but there's still a catch: &lt;strong&gt;It has to be a random sample.&lt;/strong&gt; There are still observation-bias questions here because obviously we've only mostly discovered the &amp;quot;nearby&amp;quot; TNOs, because they are easier to discover when they're &amp;quot;nearby&amp;quot;. That certainly would not be a random sample even if we had 3% of them.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;TLDR:&lt;/strong&gt; There are at least 100,000 distant objects, of which less than 1% of them have some kind of aligned orbit as far as we know today. Is this actually good evidence that something has aligned them?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Another analogy that might help. You have a bag of 100 marbles. You reach in and pull out 5 from the top of the bag (notice I didn't say any random 5!). They turn out to be 4 red ones and 1 black one. Is this actually good evidence that about 80% &lt;strong&gt;of the whole bag&lt;/strong&gt; is red?&lt;/p&gt;&#xA;" OwnerUserId="7449" LastEditorUserId="1891" LastEditDate="2020-12-31T19:20:34.587" LastActivityDate="2021-01-01T06:22:23.533" Title="What good evidence exists for the 9th planet as spoken of by Caltech?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;observational-astronomy&gt;&lt;9th-planet&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="13375" PostTypeId="1" CreationDate="2016-01-25T18:11:02.340" Score="8" ViewCount="332" Body="&lt;p&gt;I am researching into how coordinate systems of solar systems objects are created by reading some of the reports written by the Working Group on Cartographic Coordinates and Rotational Elements (&lt;a href=&quot;http://astropedia.astrogeology.usgs.gov/download/Docs/WGCCRE/WGCCRE2009reprint.pdf&quot; rel=&quot;noreferrer&quot;&gt;e.g. 2009&lt;/a&gt;). However, I am finding it difficult to completley understand the role of time in defining reference systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When observing at a planet from Earth, for example Jupiter, there are a variety of factors that make it difficult to construct a reference system (including no solid surface and planetary precession), so we use geometry to define a reference system. However, because our perspective is dynamic, meaning Jupiter's surface changes and the planet's move, we say that at time J2000 we know the precise orientation and position of Earth and therefore can say from position defined at J2000 this is the coordinate reference system for Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, does incorporating time (&lt;a href=&quot;https://en.wikipedia.org/wiki/Epoch_(astronomy)#Julian_years_and_J2000&quot; rel=&quot;noreferrer&quot;&gt;e.g. J2000&lt;/a&gt;) mean we can say a coordinate reference system is based on the situation of an object, Jupiter in this example, at a given moment?&lt;/p&gt;&#xA;" OwnerUserId="10544" LastEditorUserId="10544" LastEditDate="2016-01-27T12:04:51.140" LastActivityDate="2018-06-28T11:29:37.300" Title="Planetary reference systems and time" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;fundamental-astronomy&gt;&lt;coordinate&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13376" PostTypeId="2" ParentId="12252" CreationDate="2016-01-25T19:01:02.220" Score="6" Body="&lt;p&gt;I am assuming that the question you want answered is how to calculate the elevation of an orbit above a reference plane given the orbital inclination with this plane. If so, please update your question to reflect this, heeding the advice given in the comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kepler's first law tells us that planets move in elliptical orbits, which we can define as follows,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r = \frac{p}{1+\epsilon \, \cos\theta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p$ is the &lt;a href=&quot;http://mathworld.wolfram.com/SemilatusRectum.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;semilatus rectum&lt;/a&gt;, $r$ is the orbital distance from the central body, $\epsilon$ is the orbital eccentricity, and $\theta$ is the angle to the orbiting bodies current position from its closest approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $r$ along with the orbital inclination with respect to a given plane, we can calculate the height above this plane at any point in the orbit using simple trigonometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EJKwO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EJKwO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image above shows the relevant variables for a given orbit when the body is both above and below the reference plane. $r$ is again the orbital distance, $i$ is the orbital inclination, and $d$ is the height above the reference plane asked for in the question. $d$ is then given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d = r \, \sin \, i$$&lt;/p&gt;&#xA;" OwnerUserId="941" LastEditorUserId="616" LastEditDate="2016-04-29T19:04:45.897" LastActivityDate="2016-04-29T19:04:45.897" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13378" PostTypeId="2" ParentId="13277" CreationDate="2016-01-25T21:59:57.120" Score="2" Body="&lt;p&gt;Most likely it's Jupiter.  According to the software I use to run my telescope and plan my observations, Jupiter rose at about 6:40 pm and set the following day at about 8:30 am.  It would have been about -2.6 magnitude ,making it the brightest natural object in the sky that night.  It, of course, rose in the east and would have traveled in an arc across the sky that dipped a little south.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It could also have been Betelgeuse, or even Sirius (for Sirius, it would probably  need to be fairly low in the sky when the picture was taken to account for the color). But my best guess is still Jupiter.&lt;/p&gt;&#xA;" OwnerUserId="10546" LastActivityDate="2016-01-25T21:59:57.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13379" PostTypeId="2" ParentId="13371" CreationDate="2016-01-25T22:02:54.097" Score="4" Body="&lt;p&gt;Page 12 of &lt;a href=&quot;http://arxiv.org/abs/1601.05438&quot; rel=&quot;nofollow&quot;&gt;Batygin &amp;amp; Brown (2016)&lt;/a&gt; says that a speculative formation scenario can be drawn from recent solar system formation simulations by Bromley &amp;amp; Kenyon and by Izidoro et al. These suggest that the core of a nascent ice giant may have been ejected very early in the solar system's history in order to explain the properties of the observed planets; the formation of Uranus and Neptune was probably accompanied by at least one other ice giant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The source of the claim you mention is actually an article by &lt;a href=&quot;http://www.sciencemag.org/news/2016/01/feature-astronomers-say-neptune-sized-planet-lurks-unseen-solar-system&quot; rel=&quot;nofollow&quot;&gt;Eric Hand in Science Magazine&lt;/a&gt;.  He points out that possibly, to explain why this planet was ejected yet still remains part of the solar system in a much wider orbit, then you need it to have been slowed down by residual gas in the protosolar disk. So I assume the 10 million year is an (uncertain) upper limit on the dispersal of the disk and I would guess the 3 million year lower limit is just how long it takes to form a 10 Earth mass ice giant core at the distance of Neptune.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-01-26T08:55:57.693" LastActivityDate="2016-01-26T08:55:57.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13380" PostTypeId="2" ParentId="13373" CreationDate="2016-01-25T22:09:01.400" Score="2" Body="&lt;p&gt;What they have at present is best described as a theory based on data that is not precise enough to make a definitive statement.  They are telling people they think something is out there and roughly what they think it might be, but they really don't have accurate enough data to make any statements beyond that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the possible range of distances they suggest is so large tells you all you need to know about the accuracy of the data being insufficient to draw reliable conclusions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My biggest problem with this claim is that we have a historical lesson that suggests caution : Pluto.  Pluto was found when looking for an entirely different object ( coincidentally about the same mass as the new planet X, if memory serves me ).  It transpired that the orbital anomalies that led to predicting a Planet X 1.0 were later evaporated by more accurate measurements and theory.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are we now seeing the same sort of error ?  Could be, might not be.  All we can hope is that the quest for an explanation generates either the result claimed or, if not that, better data, more data and perhaps more accurate simulations and theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am reminded that, in the past, quite eminent and respectable scientists have claimed any number of things, which would later prove incorrect.  Cold fusion leaps to mind.  The problem with humans looking at data is that, sometimes, they see what they want.  I think this planet X 2.0 claim needs a colder and more objective, perhaps even cynical, examination, away from the media.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2016-01-25T22:09:01.400" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13381" PostTypeId="2" ParentId="13251" CreationDate="2016-01-25T22:10:33.820" Score="2" Body="&lt;p&gt;It is likely that during the formation of most stars, comets are formed from the same gas and dust that the star and any planets it hosts were formed from.  Extrapolating what we know and strongly believe about our own star, the Sun, it is likely that many or most stars have a cloud of comets around them analogous to our Oort-cloud (which hasn't been &quot;proven,&quot; but is strongly suggested by observation of comets). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As stars orbit the galactic center, they pass by one another - sometimes extremely closely, sometimes not very close.  During these encounters, the gravitational interactions will shake things up in the Oort-like clouds.  In some cases, comets that are otherwise minding their own business, orbiting far from the star, will be kicked inward.  Others will likely be ejected.  Some of those ejected will eventually find themselves orbiting another star, and possibly falling inward for a close encounter with the star and any planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the long run, it's likely that interstellar space is riddled with rogue comets.  Many comets in our solar system may have come from there, and many of the comets formed around our sun have likely been flung off, in some cases to find new stars to orbit.&lt;/p&gt;&#xA;" OwnerUserId="10546" LastActivityDate="2016-01-25T22:10:33.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13382" PostTypeId="1" AcceptedAnswerId="13383" CreationDate="2016-01-25T22:11:47.040" Score="3" ViewCount="2179" Body="&lt;p&gt;In our Solar system, the density of major planets varies from 0.7 g/cm3 (Saturn) to 5.5 g/cm3 (Earth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For exoplanets, the density varies from very low (0.03 g/cm3 for Kepler-51x) to very high (77.7 g/cm3 for Kepler-131c, data according to &lt;a href=&quot;http://exoplanetarchive.ipac.caltech.edu/docs/program_interfaces.html&quot; rel=&quot;nofollow&quot;&gt;NASA Exoplanet Archive&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that bodies with low density are composed of gas and others are rocky planets. &lt;a href=&quot;https://en.wikipedia.org/wiki/Terrestrial_planet&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; provides some approximations about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Is there any detailed classification of planetary bodies by density?&lt;/em&gt; A classification that would say, for example, that the density below 1.5 g/cm3 corresponds to a gas planet; between 1.5 and 3 are icy; between 3 and 8 are rocky; above 8 g/cm3 are even more rocky etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Would such a classification make sense?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If it exists, which are the numeric bounds for the different types of bodies?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="10547" LastActivityDate="2016-01-25T23:30:11.290" Title="Planets classification by density" Tags="&lt;planet&gt;&lt;classification&gt;&lt;density&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13383" PostTypeId="2" ParentId="13382" CreationDate="2016-01-25T22:42:15.260" Score="6" Body="&lt;p&gt;There is no 1:1 mapping between density and composition/structure. You have to look at detailed planetary models. For example, some hot Jupiters are extremely dense ($\geq 10$ g/cm$^3$) but they are undoubtedly gas giants. The origins of this diversity are the source of much speculation and theory, but are certainly within the realms of known physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of the difficulties can be gleaned from this plot from &lt;a href=&quot;http://www.nature.com/nature/journal/v513/n7518/full/nature13781.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lissauer et al. (2014)&lt;/a&gt; that shows the mass-radius plane for small planets together with the loci of models for various compositions. Lines of constant density are drawn on the plot too, but notice that a planet of a given composition can have a range of densities depending on its mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qjP6A.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qjP6A.jpg&quot; alt=&quot;Exoplanet mass radius relation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation appears even more strange with giant planets. Generally speaking, more massive giant planets are more dense, which is as expected from basic theory of degenerate gases, but objects of a given mass can have almost an order of magnitude spread in density. Below is a plot I made from exoplanets.org. Almost certainly, the change in the nature of the relationship at around 0.1 Jupiter masses marks the transition from gas giants to ice giants and other rocky-type planets, but notice that gas giants can be just as dense as smaller planets. i.e. A classification based on just density wouldn't work, but mass and density gives more of a clue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9V1ht.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9V1ht.png&quot; alt=&quot;Planet density mass relation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-01-25T23:30:11.290" LastActivityDate="2016-01-25T23:30:11.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13384" PostTypeId="2" ParentId="13286" CreationDate="2016-01-25T23:33:27.103" Score="3" Body="&lt;p&gt;The idea is basically to look for comets coming from the same direction. Look for a pair of comets that at some time earlier in their orbit was at the same place at the same time. That is a data point, but an extremely inaccurate one. It does not have to mean anything, and you are going to collect a large amount of statistical noise. For this purpose, normal periodic comets are close to useless, as they are in the same region for decades or even centuries as they fall slowly towards the sun. &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_hyperbolic_comets&quot; rel=&quot;nofollow&quot;&gt;Hyperbolic&lt;/a&gt; (non-periodic) comets on the other hand, have a shorter stay in each region, allowing us to better pin-point a possible encounter. The problem is, there are not actually a lot of them observed, and the length of the time horizon for accurate enough observations is not that long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The properties you can obtain about the ninth planet, if it exists, from large amount of data on non-periodic comets are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Current distance within a reasonably large error margin. (+-10 AU)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Orbital period; not so accurate, but you may get an OK estimate.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Inclination; not accurate, but together with the orbital period, estimates will grow more accurate over a longer time horizon of observations.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10172" LastActivityDate="2016-01-25T23:33:27.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13385" PostTypeId="1" CreationDate="2016-01-26T03:47:56.330" Score="2" ViewCount="72" Body="&lt;p&gt;Is there a publicly available data set for mean atmospheric absorption / &lt;a href=&quot;https://en.wikipedia.org/wiki/Extinction_%28astronomy%29#Atmospheric_extinction&quot; rel=&quot;nofollow&quot;&gt;extinction&lt;/a&gt;? I would like to be able to process and plot a graph similar to &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Atmospheric_electromagnetic_opacity.svg&quot; rel=&quot;nofollow&quot;&gt;this image&lt;/a&gt; from NASA on wiki commons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://hitran.org/&quot; rel=&quot;nofollow&quot;&gt;HITRAN&lt;/a&gt; database appears to offer individual lines and might possibly be used to compute mean absorption at standard atmosphere. However, the process is unclear and the data search appears to be broken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.spectralcalc.com/atmospheric_paths/paths.php&quot; rel=&quot;nofollow&quot;&gt;Spectral Calculator by GATS&lt;/a&gt; also appears to have done the calculations for various configurations pulling from the HITRAN database and using standard atmosphere, but access to transmittance and absorption plots is limited to subscribers paying $50 per month.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://www.qsl.net/wb9ajz/laser/laser-prop.html&quot; rel=&quot;nofollow&quot;&gt;lot&lt;/a&gt; of &lt;a href=&quot;http://sites.nationalacademies.org/cs/groups/ssbsite/documents/webpage/ssb_167819.pdf&quot; rel=&quot;nofollow&quot;&gt;individuals&lt;/a&gt; seem to be generating or copying these plots, but very few citations are given. Furthermore, indications are that there may be a fair amount of variability depending on conditions so a plot based on standard or mean conditions looking upward or downward through the atmosphere would be helpful.&lt;/p&gt;&#xA;" OwnerUserId="10548" LastEditorUserId="7982" LastEditDate="2021-02-28T11:28:28.060" LastActivityDate="2021-02-28T11:28:28.060" Title="Public Data Set on Atmospheric Absorbtion / Extinction" Tags="&lt;atmospheric-effects&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13386" PostTypeId="1" AcceptedAnswerId="14352" CreationDate="2016-01-26T06:11:57.010" Score="8" ViewCount="167" Body="&lt;p&gt;I'm writing a paper on astronomical seeing. Sir Isaac Newton identified both the phenomenon and origin of &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_seeing&quot; rel=&quot;nofollow noreferrer&quot;&gt;astronomical seeing&lt;/a&gt; in his Opticks. He writes:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If the Theory of making Telescopes could at&#xA;length be fully brought into Practice, yet there would&#xA;be certain Bounds beyond which Telescopes could&#xA;not perform. For the Air through which we look&#xA;upon the Stars, is in a perpetual Tremor; as may be&#xA;seen by the tremulous Motion of Shadows cast from&#xA;high Towers, and by the twinkling of the fix'd Stars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;is anyone aware of any older reference from Gilbert, Galileo, Harvey, Ptolemy, Copernicus or Kepler?&lt;/p&gt;&#xA;" OwnerUserId="10548" LastEditorUserId="2365" LastEditDate="2021-03-01T06:16:44.897" LastActivityDate="2021-03-01T06:16:44.897" Title="Oldest Reference to Astronomical Seeing" Tags="&lt;atmospheric-effects&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="13389" PostTypeId="1" AcceptedAnswerId="13415" CreationDate="2016-01-26T16:41:43.517" Score="6" ViewCount="396" Body="&lt;p&gt;It is known that each year the Moon is moving away from the Earth due to tidal forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the Earth relative to the Sun?&lt;/p&gt;&#xA;" OwnerUserId="10553" LastEditorUserId="10554" LastEditDate="2016-01-30T16:28:14.203" LastActivityDate="2017-01-25T23:24:34.563" Title="Does the Sun impose back it's tidal forces onto the Earth (such as the Earth's to the Moon)?" Tags="&lt;the-moon&gt;&lt;the-sun&gt;&lt;earth&gt;&lt;tidal-forces&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13392" PostTypeId="1" AcceptedAnswerId="13398" CreationDate="2016-01-26T17:57:03.427" Score="4" ViewCount="122" Body="&lt;p&gt;Rosetta Comet AKA &lt;a href=&quot;https://en.wikipedia.org/wiki/67P/Churyumov%E2%80%93Gerasimenko&quot;&gt;67P/Churyumov–Gerasimenko&lt;/a&gt; has the surface temperature range of -43 to -93 degree Celsius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if we compare that with the Mercury (I chose Mercury cause it's closest to the Sun and has the thinnest atmosphere), the minimum surface temperature is -170 degrees Celsius. A little farther from that, on Earth's Moon, while at night, the temperature can drop up to -233 degree Celsius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rosetta's Comet is farther from the Sun as compared to both objects (Perihelion 1.2432 AU, Aphelion 5.6829 AU); however, a lot warmer. Why is that?&lt;/p&gt;&#xA;" OwnerUserId="10298" LastActivityDate="2016-01-27T22:14:43.267" Title="Why is Rosetta's Comet so &quot;warm&quot;?" Tags="&lt;comets&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13393" PostTypeId="2" ParentId="13389" CreationDate="2016-01-26T18:34:39.653" Score="7" Body="&lt;p&gt;Yes - the earth and sun do have tidal forces like the moon and earth. There are two main reasons this is happening.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The sun is always losing mass due to nuclear reactions, the sun is always converting a tiny amount of its mass into energy. This means the pull on the earth is weakened.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The tidal forces that happen between the earth and moon also happen between the sun and earth, except on a much smaller scale.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The sun is also losing its angular momentum at a rate of 0.00003 seconds per year&lt;/p&gt;&#xA;" OwnerUserId="9645" LastEditorUserId="10554" LastEditDate="2016-08-16T21:17:52.933" LastActivityDate="2016-08-16T21:17:52.933" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13394" PostTypeId="2" ParentId="13389" CreationDate="2016-01-26T18:37:09.130" Score="-5" Body="&lt;p&gt;The theorized reason I saw on either the NASA or Discovery Channel for the Moon's orbit spiraling away from earth is that it's formation resulted from a collision between the Earth and an asteroid, and the resulting momentum carries it away.&lt;/p&gt;&#xA;" OwnerUserId="10554" LastEditorUserId="10554" LastEditDate="2016-08-16T20:36:17.077" LastActivityDate="2016-08-16T20:36:17.077" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13395" PostTypeId="1" AcceptedAnswerId="13403" CreationDate="2016-01-26T18:52:22.133" Score="2" ViewCount="301" Body="&lt;p&gt;When I go to the wikipedia page for any planet in our solar system, the exact mass and volume are reported. How is it possible to do this with them being so far away? I suppose after hundreds of years we can observe them going around the sun, but this doesn't really give any information since we need both the mass and distance from the sun for it to be of any use.&lt;/p&gt;&#xA;" OwnerUserId="10555" LastEditorUserId="2365" LastEditDate="2016-01-27T20:26:41.537" LastActivityDate="2016-01-27T20:26:41.537" Title="How is the mass of solar system objects calculated?" Tags="&lt;planet&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13396" PostTypeId="1" CreationDate="2016-01-26T19:44:56.060" Score="1" ViewCount="730" Body="&lt;p&gt;I've been presented with this problem:    &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Say that Jupiter, with its diameter of 142,000 km, was located where Mars now orbits.  What would be the angular size of (the newly-relocated) Jupiter during a close approach, when its distance would be 79,300,000 km?  Would we be able to see Jupiter as a round object with our unaided eye, or only as a point of light?      &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The angular size is easy enough to calculate given $ \frac{Angular \space Size}{206,000} = \frac{Linear \space Size}{Distance} $.&lt;br&gt;&#xA;So $\frac{Angular \space Size}{206,000} = \frac{142000}{79300000}$ meaning Angular Size = 369 seconds of arc.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, would the unaided eye be able to see this? Knowing that the eye can detect 1 minute of arc across, surely  it can detect 6.15? Any help appreciated, thanks!&lt;/p&gt;&#xA;" OwnerUserId="10556" LastActivityDate="2016-01-27T03:01:13.733" Title="Seconds of Arc and the Unaided Eye" Tags="&lt;naked-eye&gt;&lt;angular-resolution&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13397" PostTypeId="2" ParentId="12981" CreationDate="2016-01-26T21:41:50.470" Score="2" Body="&lt;p&gt;Assuming circular, coplanar orbits, Trans Venus Insertion (TVI) is 8 km/s from Mercury's surface. That's with horizontal take off and no gravity loss. That's for the minimum energy Hohmann transfer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exiting the Hohmann transfer and soft landing on Venus would take 11.9 km/s if Venus had no atmosphere. But much, maybe all of that 11.9 km/s could be done by aerobraking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used my &lt;a href=&quot;http://clowder.net/hop/railroad/Hohmann.xls&quot; rel=&quot;nofollow&quot;&gt;Hohmann spreadsheet&lt;/a&gt; to get these numbers.&lt;/p&gt;&#xA;" OwnerUserId="1232" LastActivityDate="2016-01-26T21:41:50.470" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13398" PostTypeId="2" ParentId="13392" CreationDate="2016-01-26T22:13:47.643" Score="5" Body="&lt;p&gt;67P/Churyumov–Gerasimenko is actually not really warm when compared to these bodies.&lt;/p&gt;&#xA;&lt;p&gt;You quoted figures for the minimum temperatures for Mercury and the Moon. Those are accurate but misleading. Both are for &lt;em&gt;nighttime&lt;/em&gt; temperatures, taken from locations on the bodies that are facing away from the Sun. Daytime temperatures can skyrocket, to 427 degrees Celsius during the day on Mercury. The Moon can also get very hot during the day, although nowhere near as hot as on Mercury,&lt;/p&gt;&#xA;&lt;p&gt;Looking at it like this, 67P/Churyumov–Gerasimenko's surface temperatures are pretty moderate, and nowhere near extreme. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2015Sci...348..232R&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rubin et al. (2015)&lt;/a&gt; suggest that it formed in extremely cold conditions - albeit ones not too extreme, for a comet.&lt;/p&gt;&#xA;&lt;p&gt;The ESA has &lt;a href=&quot;http://www.esa.int/Our_Activities/Space_Science/Rosetta/Rosetta_takes_comet_s_temperature&quot; rel=&quot;nofollow noreferrer&quot;&gt;something related to say on the matter&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;At these distances, the comet covered only a few pixels in the field of view and so it was not possible to determine the temperatures of individual features. But, using the sensor to collect infrared light emitted by the whole comet, scientists determined that its average surface temperature is about –70° Celsius.&lt;/p&gt;&#xA;&lt;p&gt;The comet was roughly 555 million km from the Sun at the time — more than three times further away than Earth, meaning that sunlight is only about a tenth as bright.&lt;/p&gt;&#xA;&lt;p&gt;Although –70° C may seem rather cold, it is some 20–30° C warmer than predicted for a comet at that distance covered exclusively in ice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;That said, evidence from the mission has made scientists think that an surface with a lot of dust may be the norm for comets.&lt;/p&gt;&#xA;&lt;p&gt;67P/Churyumov–Gerasimenko isn't warmer or colder than the other bodies at this distance from the Sun. It's simply milder.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;I would wager that the reason Mercury and the Moon can get so cold is because they don't really have atmospheres. Over the course of a day, Earth's atmosphere heats up a bit, and over night, it retains that heat, meaning that the surface can stay relatively warm. Mercury's atmosphere is extremely tenuous. The same goes for the Moon.&lt;/p&gt;&#xA;&lt;p&gt;This might not be an issue, but one day on Mercury lasts 59 Earth days. One day on the Moon lasts for about one Earth month. This means that while there's time to trap some heat, there's also the same amount of time to lose it, and without an atmosphere, the heat gained is soon be lost.&lt;/p&gt;&#xA;&lt;p&gt;67P/Churyumov–Gerasimenko also doesn't have an atmosphere to speak of, but it rotates quickly - rotating once every ~12.5 hours. This isn't necessarily perpendicular to the orbital plane, so not every part of it will face the Sun during a rotation, but it does mean that there's not as much time to lose heat.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-01-27T22:14:43.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13399" PostTypeId="1" AcceptedAnswerId="13404" CreationDate="2016-01-26T22:30:23.887" Score="0" ViewCount="75" Body="&lt;p&gt;I understand that we can observe far away galaxies to get an idea how the universe looked like in the past. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the universe is 13.82b years old, would it be correct to say that if we looked at a galaxy that was 13.82b ly away, we would see whatever happened just after the big bang? If this is correct, what would we have observed had we looked at the same point 1b years ago? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this an application for special relativity?&lt;/p&gt;&#xA;" OwnerUserId="10557" LastActivityDate="2016-01-27T19:55:22.143" Title="Observing a point 13.82b ly away, 1b years ago" Tags="&lt;special-relativity&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13400" PostTypeId="2" ParentId="13396" CreationDate="2016-01-26T23:54:03.190" Score="1" Body="&lt;p&gt;Yes, you could definitely see Jupiter as a round object if it's &#xA;angular diameter was 369 arcseconds. The limit for seeing something as &#xA;round appears to be a diameter of around 1 arcminute. Notes: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The most direct comparison is M16, the Eagle Nebula. According to &#xA;&lt;a href=&quot;http://www.nightskyinfo.com/archive/eagle_nebula/&quot; rel=&quot;nofollow&quot;&gt;http://www.nightskyinfo.com/archive/eagle_nebula/&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt; &#xA;The M16 complex is visible to the naked eye on clear nights as a hazy patch &#xA;&lt;/blockquote&gt; &#xA;&#xA;&lt;p&gt;(and thus not a single point), and &lt;a href=&quot;http://messier.seds.org/m/m016.html&quot; rel=&quot;nofollow&quot;&gt;http://messier.seds.org/m/m016.html&lt;/a&gt; &#xA;notes that M16 has an angular diameter of about 7 arcminutes. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.astropix.com/doubles/&quot; rel=&quot;nofollow&quot;&gt;http://www.astropix.com/doubles/&lt;/a&gt; notes that: &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt; &#xA;In the real world, folks with excellent eyesight can resolve down to &#xA;about 2 arcminutes. Folks with average eyesight can resolve about 3 &#xA;arcminutes. &#xA;&lt;/blockquote&gt; &#xA;&#xA;&lt;p&gt;Of course, resolving a double star as two separate points of light &#xA;isn't the same as seeing something as a circle, but provides a good &#xA;upper limit. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;And &#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Phases_of_Venus#Naked_eye_observations&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Phases_of_Venus#Naked_eye_observations&lt;/a&gt; &#xA;notes that: &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt; &#xA;The extreme crescent phase of Venus can be seen without a telescope by &#xA;those with exceptionally acute eyesight, at the limit of human &#xA;perception. The angular resolution of the naked eye is about 1 minute &#xA;of arc. The apparent disk of Venus' extreme crescent measures between &#xA;60.2 and 66 seconds of arc,[4] depending on the distance from &#xA;Earth. Nevertheless it is possible for observers with extremely acute &#xA;eyesight to see a crescent Venus under ideal atmospheric &#xA;circumstances. &#xA;&lt;/blockquote&gt; &#xA;&#xA;&lt;p&gt;Note also that Venus appears circle-like (not point-like) when it transits the Sun, but the same is true of Mercury, so this may be a special situation that doesn't count.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2016-01-27T03:01:13.733" LastActivityDate="2016-01-27T03:01:13.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13401" PostTypeId="2" ParentId="13248" CreationDate="2016-01-27T01:18:08.733" Score="1" Body="&lt;pre&gt;&lt;code&gt;#!/usr/bin/python&#xA;&#xA;import ephem, math&#xA;obs = ephem.Observer();&#xA;obs.long,obs.lat=0,0;&#xA;pl = ephem.Pluto(ephem.now())&#xA;print 180.*(pl.g_ra-obs.sidereal_time())/math.pi,180.*pl.g_dec/math.pi&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To a good approximation, the code above will tell you the longitude and latitude (in that order) where Pluto is currently overhead. You can edit it for different planets/times.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2016-01-27T01:18:08.733" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13402" PostTypeId="2" ParentId="13368" CreationDate="2016-01-27T03:27:23.503" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gTFyt.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gTFyt.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears to be an older version of the &lt;a href=&quot;http://www.bhphotovideo.com/c/product/765323-REG/Bushnell_789971_Voyager_Sky_Tour_800x70.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bushnell Voyager Sky Tour 800x70mm Refractor Telescope&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Full manual &lt;a href=&quot;http://bushnell.com/getmedia/42f4f060-ea31-4b94-a77e-0bbddde87f95/voyager-skytour-2011.pdf?ext=.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user1569" LastEditDate="2016-01-27T12:53:08.607" LastActivityDate="2016-01-27T12:53:08.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13403" PostTypeId="2" ParentId="13395" CreationDate="2016-01-27T12:45:10.130" Score="6" Body="&lt;p&gt;Planets are satellites of the Sun. Most planets have satellites.&#xA;Whenever you find a parent body with satellites, you get first good results about their masses and distances by applying &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion&quot;&gt;Kepler's laws&lt;/a&gt;.&#xA;A second approach is using &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_parallax&quot;&gt;parallax measurements&lt;/a&gt; for distance estimates.&#xA;The distance to some of the planets can be measured very accurately by radar, e.g. for &lt;a href=&quot;http://www.astroevents.no/distance.html&quot;&gt;Venus&lt;/a&gt;.&#xA;The known planets of our solar system have been visited by probes. Using &lt;a href=&quot;http://blogs.esa.int/mex/2012/08/05/time-delay-between-mars-and-earth/&quot;&gt;signal travel times&lt;/a&gt;, you can calculate the distance of the probe to Earth. One of the methods to determine the distance between probe and the planet is by &lt;a href=&quot;https://en.wikipedia.org/wiki/Photogrammetry&quot;&gt;photogrammetry&lt;/a&gt;.&#xA;Apply the measured data to the according &lt;a href=&quot;https://en.wikipedia.org/wiki/Simultaneous_equations&quot;&gt;system of equations&lt;/a&gt;. Usually more data than needed are available. Then you get an &lt;a href=&quot;https://en.wikipedia.org/wiki/Overdetermined_system&quot;&gt;overdetermined system&lt;/a&gt;. A best solution can be found by &lt;a href=&quot;https://en.wikipedia.org/wiki/Ordinary_least_squares&quot;&gt;minimizing appropriate squares&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Perturbation_theory#Beginnings_in_the_study_of_planetary_motion&quot;&gt;Subtle deviations from Kepler's law&lt;/a&gt; can be used to infer masses of planets which are hard to measure in a more direct way.&#xA;Once you know the distance of a planet, use its apparent diameter(s) to &lt;a href=&quot;https://en.wikipedia.org/wiki/Trigonometry&quot;&gt;calculate the actual diameter(s)&lt;/a&gt;, and then &lt;a href=&quot;https://en.wikipedia.org/wiki/Sphere#Enclosed_volume&quot;&gt;the volume of a sphere&lt;/a&gt; or a &lt;a href=&quot;https://en.wikipedia.org/wiki/Spheroid&quot;&gt;spheroid&lt;/a&gt;.&#xA;With mass and volume you know the mean &lt;a href=&quot;https://en.wikipedia.org/wiki/Density&quot;&gt;density&lt;/a&gt; of the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Within our solar system, for very distant objects the apparent size of which cannot be determined you may estimate the surface &lt;a href=&quot;https://en.wikipedia.org/wiki/Albedo&quot;&gt;albedo&lt;/a&gt; together with distance (by Kepler after knowing the mass of the Sun) and &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_magnitude&quot;&gt;apparent magnitude&lt;/a&gt; to obtain a rough estimate of the size of the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Better size estimates for objects of small apparent size can be obtained by &lt;a href=&quot;https://en.wikipedia.org/wiki/Occultation&quot;&gt;stellar occultations&lt;/a&gt;, meaning measuring the time a star is occulted by the object. Together with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_velocity&quot;&gt;angular velocity&lt;/a&gt; of the object you get data about its angular size.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2016-01-27T12:45:10.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13404" PostTypeId="2" ParentId="13399" CreationDate="2016-01-27T17:43:39.563" Score="4" Body="&lt;p&gt;Time is kind of funny when you look at such distances. Let's imagine that you are running away from a person throwing a ball at you. The ball will travel further to hit you, based on your speed, than the distance that was present when you started to run. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe is expanding. The early Universe expanded very quickly, giving large deviations from the expected position today than what we see through a telescope. Something that we see that is 12 billion years old is actually quite a bit further away today, although we can't know exactly where it is. As I understand, this hypothetical object would be around 40 billion light years away today (See &lt;a href=&quot;https://astronomy.stackexchange.com/q/3635/182&quot;&gt;this question&lt;/a&gt;), given current theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But yes, if you looked at a galaxy that appears in today's distance to be 13 billion light years away, then the universe was 800 million year old, roughly speaking. Galaxies took about a billion years to form, however, so you probably wouldn't see such a galaxy.&lt;/p&gt;&#xA;" OwnerUserId="182" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-01-27T19:55:22.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13406" PostTypeId="1" AcceptedAnswerId="13407" CreationDate="2016-01-27T20:01:48.797" Score="3" ViewCount="1746" Body="&lt;p&gt;Even the NASA website refers to stars as &quot;big ball of gas&quot;. It's a common statement that's popularized by movies and popular culture. However, stars are technically not in a gaseous state - they're in a state of plasma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So do stars qualify as being made of &quot;gas&quot; as the proverb would suggest, or is that considered a misconception?&lt;/p&gt;&#xA;" OwnerUserId="10566" LastEditorUserId="10566" LastEditDate="2016-01-27T20:46:33.607" LastActivityDate="2016-01-28T11:13:46.880" Title="Are stars really &quot;burning balls of gas&quot;?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13407" PostTypeId="2" ParentId="13406" CreationDate="2016-01-27T22:21:00.703" Score="9" Body="&lt;p&gt;While I agree that it boils down to semantics, I actually disagree on the scientific use of the term in the comments. In astronomy, we know there's a difference between gas and plasma, but we almost always use the term &quot;gas&quot; when talking about what's in stars. E.g. &quot;the fraction of gas locked up in stars&quot; (as opposed to in the interstellar medium). We also talk about the metallicity of gas, meaning both gas and plasma. I think we only use the term plasma when specifically discussing properties that are unique to plasma, like being &quot;frozen&quot; in magnetic fields. And googling &lt;a href=&quot;https://www.google.dk/#q=%22ionized+gas%22+-plasma&quot; rel=&quot;noreferrer&quot;&gt;[&quot;ionized gas&quot; -plasma]&lt;/a&gt; (the '-plasma' is to get rid of articles explaining what plasma is) returns 243,000 hits, of which by far the most seem to be scientific papers and websites. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think the answer is: &quot;No, stars aren't made of gas, they're made of plasma, but go ahead and call it gas, you won't offend any astronomer.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I agree with Rob Jeffries that we &lt;em&gt;don't&lt;/em&gt; like when you call it &lt;em&gt;burning&lt;/em&gt; gas. It's nuclear power! It's a million$^\dagger$ times more powerful than fire!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;Why &quot;a million&quot;? Because nuclear reactions are measured in MeV, while chemical reactions are measured in eV.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-01-28T11:13:46.880" LastActivityDate="2016-01-28T11:13:46.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13414" PostTypeId="1" CreationDate="2016-01-28T22:28:23.633" Score="3" ViewCount="131" Body="&lt;p&gt;I'm aware that there have been attempts to simulate the evolution of the Solar System with &lt;em&gt;six&lt;/em&gt; giant planets&lt;sup&gt;1&lt;/sup&gt;, as opposed to the traditional four or five. The recent proposed Planet Nine would probably invalidate some of those models, although I don't know how drastic its effects would be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would Planet Nine fit into models with six giant planets? Have any simulations been done that support such a  scenario?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1 The models I'm most familiar with were created by &lt;a href=&quot;https://arxiv.org/abs/1208.2957&quot; rel=&quot;nofollow&quot;&gt;Nesvorny &amp;amp; Morbidelli (2012)&lt;/a&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2016-01-28T22:28:23.633" Title="Would the existence of Planet Nine rule out the possibility of a sixth giant planet?" Tags="&lt;solar-system&gt;&lt;gas-giants&gt;&lt;9th-planet&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13415" PostTypeId="2" ParentId="13389" CreationDate="2016-01-29T05:50:20.363" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Well, I wasn't sure if tidal forces between the Earth and Sun were strong&#xA;  enough to have any effect on the matter. Truth to be told, the article&#xA;  barrycarter linked clarifies that tidal forces have neglible effect&#xA;  relative to the Sun's nuclear fusion mass loss.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's correct.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some more details on this.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tidal &quot;tugging&quot; is a two step process.   The Moon's gravity raises a tide on the Earth and this tidal bulge, spins ahead of the moon and pulls the moon very slightly faster, which slowly pushes the Moon away from the earth.  (Note, if the Earth rotated much more slowly, the effect would be the opposite, the orbiting moon would be ahead of the tidal bump and gradually slow down).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the Moon was closer and the tidal bulge was a lot bigger and the corresponding push away was significantly faster.  Still not fast, but quite a bit faster than the current 1.5 inches per year, perhaps several feet a year after formation if I was to guess.  Even more for the brief period when the Earth had an entirely liquid-magma surface, but that was probably pretty temporary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, most of the distance the Moon moved away from the Earth after formation happened in the first billion years.   The move away speed decreases with the distance, something roughly equal to the 4th power of the distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you said, this exact same thing happens between the Earth and Sun, but the effect is even smaller, so small that it's not the primary effect in changes in the Earth's average distance from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tides depend on gravitational field variation (not field strength), but on how much the field changes near side of object to far side.   It also helps if the larger object that's being orbited has a liquid surface, oceans or molten or a large ocean under an icy surface.  If the Earth didn't have oceans the Moon would moved away from the Earth much more slowly and it would be quite a bit closer to the earth than it is now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rocky surfaces still bulge a little so the effect still happens but it's smaller.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun's is plasma which is very fluid so the Earth and all the planets create a tidal bulge on the sun, but from the sun's point of view, all the planets are quite small, just tiny dots in the sky, so the tidal effect from the planets creates only a tiny tidal bulge on the sun.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tidal effect can be roughly compared to the size of the object in the sky.   for example, the Moon and Sun from the point of view of the Earth are about the same size in the sky, so they have similar tidal effects.  The Moon's larger tides is due to the Moon having greater density than the Sun.   More on &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/tide.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tides&lt;/a&gt; here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(this doesn't apply to a tennis-ball in your hand, however, which can appear the same size as the moon or sun, because you have to consider the size from the center of the earth, not the size to your eye on the surface.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tidal force the sun receives from the Earth from 93 million miles away when the Earth is just a tiny dot, so effect and corresponding the tidal bulge is quite small, so the corresponding push is very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A curious sidebar, is that when the Sun goes Red Giant, the larger sun will have a greater distance variation to the Earth, which will probably be the closest planet at that point.    The tidal bulge will be correspondingly larger and the Earth, quite a bit closer to the bulge, so when that happens, the tidal push the Earth gets will be measurably larger.&lt;/em&gt;   Which as Rob Jeffries' correction points out, could slowly draw the Earth into the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A point to add, when the Sun goes Red giant and loses mass, as Red Giants do, then the expansion of the Earth's orbit due to mass loss could be pretty significant.   I don't have the skill to even try to run those numbers though.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tug from other planets, while, even smaller, can have an effect in moving planets too, more so when the planets are in orbital resonance, which, currently, no planets are.   The solar system passing through a dust cloud can have a slowing down effect too.   Needless to say, all the effects are quite small.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-01-25T23:24:34.563" LastActivityDate="2017-01-25T23:24:34.563" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13416" PostTypeId="1" AcceptedAnswerId="13417" CreationDate="2016-01-29T11:31:34.607" Score="7" ViewCount="614" Body="&lt;p&gt;From Earth, Jupiter's maximum brightness is -2.94 and Saturn's is -0.24. But what about from Mars? They should be brighter, but by how much?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are equations at &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_magnitude&quot;&gt;the obvious wiki entry&lt;/a&gt; but I'm not sure I understand them. Plus I'm afraid to attempt this calculation myself because I've heard that this sort of thing does &lt;strong&gt;not&lt;/strong&gt; follow the inverse square law. They're reflecting light (not generating it), and I read somewhere that it should follow an inverse 4th power law because of that. I don't see any 4th powers in the apparent magnitude equations.&lt;/p&gt;&#xA;" OwnerUserId="7449" LastActivityDate="2016-02-08T11:10:01.627" Title="As viewed from Mars, what are Jupiter's and Saturn's maximum brightness in apparent magnitude?" Tags="&lt;solar-system&gt;&lt;mars&gt;&lt;apparent-magnitude&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13417" PostTypeId="2" ParentId="13416" CreationDate="2016-01-29T12:21:08.030" Score="9" Body="&lt;p&gt;The inverse-fourth-power law you're referring to is valid for light emitted from a source, reflected &lt;em&gt;non-&lt;a href=&quot;https://en.wikipedia.org/wiki/Specular_reflection&quot; rel=&quot;nofollow noreferrer&quot;&gt;specularly&lt;/a&gt;&lt;/em&gt;  — i.e. in all directions — from a reflector, and detected by the original emitter. If the reflector is a mirror, the observed flux just follows the normal inverse-square law with the nominator equal to $(2d)^2$ instead of $d^2$, since the light has to go back and forth. But if the reflector scatters light in all directions — i.e. into a $2\pi$ hemisphere — then the detected flux is $\sim r^2/d^4$, where $r$ is the radius of the reflector (see &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1443/by-putting-a-mirror-in-space-would-we-be-able-to-see-into-the-past&quot;&gt;this answer&lt;/a&gt; for a more throrough explanation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of this is a radar. But in our case, it isn't &lt;em&gt;us&lt;/em&gt; that emit the light, it's the Sun. The amount of light reflected from Jupiter and Saturn depends on their distance to the Sun, and that distance doesn't change if you move to Mars. The relevant distances (that I got from &lt;a href=&quot;http://nssdc.gsfc.nasa.gov/planetary/factsheet/planet_table_ratio.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA's Planetary Fact Sheet&lt;/a&gt;) are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Earth semi-major axis    $d_\mathrm{E} = 1.00\,\mathrm{AU}$&lt;/li&gt;&#xA;&lt;li&gt;Mars aphelion$^\dagger$  $d_\mathrm{M} = 1.64\,\mathrm{AU}$&lt;/li&gt;&#xA;&lt;li&gt;Jupiter semi-major axis  $d_\mathrm{J} = 5.20\,\mathrm{AU}$&lt;/li&gt;&#xA;&lt;li&gt;Saturn semi-major axis   $d_\mathrm{S} = 9.58\,\mathrm{AU}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now the differences between them:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Earth to Mars    $d_\mathrm{M-E} = 0.64\,\mathrm{AU}$&lt;/li&gt;&#xA;&lt;li&gt;Earth to Jupiter $d_\mathrm{J-E} = 4.20 \,\mathrm{AU}$&lt;/li&gt;&#xA;&lt;li&gt;Earth to Saturn  $d_\mathrm{S-E} = 8.58\,\mathrm{AU}$&lt;/li&gt;&#xA;&lt;li&gt;Mars  to Jupiter $d_\mathrm{J-M} = 3.56 \,\mathrm{AU}$&lt;/li&gt;&#xA;&lt;li&gt;Mars  to Saturn  $d_\mathrm{S-M} = 7.94 \,\mathrm{AU}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hence, from Mars the distance to Jupiter is $d_\mathrm{M-J} = 0.85 d_\mathrm{J-E}$, and the received flux  is thus $1/0.85^2 = 1.4$ times that on Earth. The change in apparent magnitude is then&#xA;$$&#xA;\Delta m = -2.5 \log\left( \frac{0.85^2}{1}\right) = -0.36,&#xA;$$&#xA;i.e. Jupiter would be $m = -2.94 - 0.36 = -3.30$ as viewed from Mars (assuming the values you provide are correct; I didn't check this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following the same approach for Saturn, I get $m = -0.41$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;The reason I used Mars' aphelion instead of Mars' semi-major axis is because Mars has a rather eccentric orbit ($e\sim0.09$). Jupiter and Saturn are somewhat closer to circular orbits, though ($e\sim0.05$). This of course is still an approximation. If you wish to take into account the eccentricities of all orbits, you also need to know the angle between their semi-major axes. This also doesn't take into account the inclination of the orbit; however, these are only 1º-2º.&#xA;And of course this minimum distance will not occur every Martian year.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-02-08T11:10:01.627" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13418" PostTypeId="1" CreationDate="2016-01-29T14:32:30.100" Score="1" ViewCount="120" Body="&lt;p&gt;Sorry if this is a very lay question. I am just an armchair physicist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Higgs Boson was just recently discovered, and it has a mass which is considerably larger than a proton. If this Higgs Boson was just recently discovered and the Higgs Boson is EVERYWHERE (ie. The Higgs Field), why wouldn't this perfectly explain the missing mass that is &quot;dark matter&quot; ?&lt;/p&gt;&#xA;" OwnerUserId="4238" LastActivityDate="2016-01-30T06:14:07.570" Title="Is Higgs Boson mass equal to the missing mass of dark matter?" Tags="&lt;dark-matter&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13419" PostTypeId="1" CreationDate="2016-01-29T14:50:01.063" Score="10" ViewCount="3869" Body="&lt;p&gt;Saturn is a gas giant like Jupiter. It has everything from&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;tiny moonlets less than 1 kilometer across to the enormous Titan, which is larger than the planet Mercury. Saturn has 62 moons with confirmed orbits.&#xA;  &lt;br&gt;&#xA;  &lt;sup&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Moons_of_Saturn&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why does Saturn have more moons?&lt;/p&gt;&#xA;" OwnerUserId="10582" LastEditorUserId="2153" LastEditDate="2016-01-29T23:04:10.923" LastActivityDate="2016-01-29T23:04:10.923" Title="Why does the planet Saturn have numerous (62) moons compared to the rest of the planets in the Solar System?" Tags="&lt;the-moon&gt;&lt;planet&gt;&lt;saturn&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13420" PostTypeId="2" ParentId="13419" CreationDate="2016-01-29T15:47:12.690" Score="4" Body="&lt;p&gt;Saturn and Jupiter have many moons for quite a few reasons, one of the main ones being that they have an absolutely immense gravitational pull. During the early stages of the formation of our solar system, there would of been many planet-like objects floating around which our gas giants would have attracted. Furthermore, these planets are so far out in the solar system water would if frozen (which explains Saturn's rings of ice). Infact, we can show that the ice can form moons by looking at some of the moons of Uranus, some of them are half made of ice!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few of the outer moons of our planets are captured asteroids. Phoebe, which is a moon of Saturn, is believed to have been a captured asteroid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't heard anything about Saturn having more moons than jupiter.&lt;/p&gt;&#xA;" OwnerUserId="9645" LastEditorUserId="9645" LastEditDate="2016-01-29T16:47:53.300" LastActivityDate="2016-01-29T16:47:53.300" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13421" PostTypeId="2" ParentId="13204" CreationDate="2016-01-29T16:53:48.117" Score="6" Body="&lt;p&gt;I am not very familiar with orbital dynamics (so please correct me if I'm wrong). I was told that, for instance in the case of the mean motion resonances that cause the majority of the Kirkwood gaps in the asteroid belt, not only the ratio of the periods, but also the timing is important. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take Pluto as an example, which is in 2:3 resonance with Neptune. Although Pluto crosses Neptune's orbit, both bodies will never get closer to each other than a certain threshold. That is, because the timing between their orbits was preventing close encounters in the first place, which is then enforced by the resonance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the timing was different, close encounters would be possible, thus destabilizing the orbit of the smaller body. In the case of mean motion resonances, there is an algebraic expression that allows to investigate the stability of a resonance (see, e.g., here &lt;a href=&quot;https://en.wikipedia.org/wiki/Resonant_trans-Neptunian_object&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Resonant_trans-Neptunian_object&lt;/a&gt; [toward a formal definition]).&lt;/p&gt;&#xA;" OwnerUserId="7144" LastActivityDate="2016-01-29T16:53:48.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13422" PostTypeId="2" ParentId="13418" CreationDate="2016-01-29T17:12:42.670" Score="2" Body="&lt;p&gt;To answer this question, we need to understand what dark matter and a Higgs Boson are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Higgs Boson&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Higgs Boson is an elementary particle in the standard field of quantum physics. It was theorised by a few people, namely Peter Higgs. The Higgs Field is a field which we can now believe exists, as we can detect it through its excitations, which become Higgs Bosons. The quantum excitations of the Higgs Field create Higgs Bosons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Higgs Boson is a particle with no spin, infact, it is the only particle with no spin, electric charge or colour charge. It's also extremely unstable and decays into other particles almost instantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dark matter&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark matter is a heavily researched topic in physics at the moment. Out of all matter in the universe, dark matter makes up about 80% of it! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One example of a theorised candidate for a dark matter particle is a WIMP. These are abbreviations of 'Weakly Interacting Massive Particles'. These particles are not normal matter, therefor we can't detect them. Our efforts to detect them include trying to look at the annihilation of WIMP particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main characteristics of WIMPS are that they only interact through the weak nuclear force or gravity. They also larger than standard particles (a dark matter particle with mass less than a electronvolt is classified as light dark matter)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wimps move slowly and are therefor cold, making them one of the main candidates for cold dark matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of dark matter as a 'skeleton' for normal matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From looking at both types of particles, we can clearly tell that Higgs Bosons are completely different from dark matter. We are still trying to find out about them both, as both of them can answer fundamental questions in physics.&lt;/p&gt;&#xA;" OwnerUserId="9645" LastEditorUserId="9645" LastEditDate="2016-01-30T06:14:07.570" LastActivityDate="2016-01-30T06:14:07.570" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13423" PostTypeId="2" ParentId="13418" CreationDate="2016-01-29T21:33:26.077" Score="2" Body="&lt;p&gt;The Higgs Boson can't be a dark matter particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Higgs Boson decays into other particles about $10^{-22}$ seconds after it forms.  (So fast that the LHC doesn't directly detect the Higgs, it detects the spray of particles decaying from the Higgs). So while the Higgs field is everywhere, there aren't a lot of Higgs bosons in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark matter must be stable for at least as long as the universe has existed. If Cern or someone else did create a dark matter particle, it would only appear as a suddenly missing mass, since it would be stable, and able to pass out of the LHC without interacting with any of the experiments. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-01-29T21:33:26.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13424" PostTypeId="1" CreationDate="2016-01-29T23:41:03.070" Score="6" ViewCount="1418" Body="&lt;p&gt;As the title suggests, I am wondering the maximum size an object can be so that it will completely burn up when entering and falling through Earth's atmosphere. I'm focusing on man made objects, not asteroids or meteors. Also, how much does the make-up of the object affect its burn rate? I.e. metal vs plastic vs glass, and so on.&#xA;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="10587" LastEditorUserId="11031" LastEditDate="2016-03-03T15:45:56.473" LastActivityDate="2016-04-02T16:30:43.830" Title="At what size do objects burn up in the atmosphere when falling from orbit?" Tags="&lt;space&gt;&lt;space-debris&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13425" PostTypeId="1" AcceptedAnswerId="13426" CreationDate="2016-01-30T03:02:55.850" Score="3" ViewCount="150" Body="&lt;p&gt;NASA's put out &lt;a href=&quot;https://www.nasa.gov/feature/jpl/new-animation-takes-a-colorful-flight-over-ceres/&quot; rel=&quot;nofollow&quot;&gt;a semi-CGI video of a Ceres flyby&lt;/a&gt; using material from the Dawn mission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I realise that visibility is relative, even on Earth: I can walk into my basement and not be able to see anything until I've stood there for a few moments letting my eyes adjust. But I'm wondering, since Ceres is so far from Earth, and despite the images we see beamed back from spacecraft, &lt;strong&gt;if I were on a spacecraft in orbit of Ceres, facing the side of Ceres that faces the Sun, would I be able to see its surface with the naked eye?&lt;/strong&gt; Or is sunlight too dim at that distance to do the job on its own?&lt;/p&gt;&#xA;" OwnerUserId="4155" LastActivityDate="2016-01-30T08:28:23.553" Title="Would I actually be able to see Ceres without shining a really bright torch onto it?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;light&gt;&lt;ceres&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13426" PostTypeId="2" ParentId="13425" CreationDate="2016-01-30T08:21:09.160" Score="8" Body="&lt;p&gt;Ceres is 2.76 AU away from the sun, so the sunlight is seven times weaker. That is comparable to the levels of illumination 30 minutes before sunset on Earth, so you are not going to have any problem seeing the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The albedo of Ceres is quite low, comparable to worn asphalt. But it is still easy to see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/X1TLUt.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X1TLUt.jpg&quot; alt=&quot;worn asphalt&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="10172" LastEditDate="2016-01-30T08:28:23.553" LastActivityDate="2016-01-30T08:28:23.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13428" PostTypeId="1" CreationDate="2016-01-30T12:45:37.930" Score="3" ViewCount="93" Body="&lt;p&gt;Hopefully the title is very self-explanatory. I was wondering if there were any non-stellar objects, such as bright galaxies, which when looking with the naked eye were mistaken for stars in the time when asterisms were documented and catalogued...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, within the Big Dipper I believe it is Mizar which is in fact a binary star system. (Correct me if I am wrong!)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was hoping perhaps for distant yet bright Messier objects being mistaken for stars...&lt;/p&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2016-01-30T22:59:52.280" Title="Are there any bright non-stellar objects which make up asterisms?" Tags="&lt;constellations&gt;&lt;night-sky&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13429" PostTypeId="1" AcceptedAnswerId="13437" CreationDate="2016-01-30T13:26:07.757" Score="4" ViewCount="4309" Body="&lt;p&gt;Many papers mention emission measure, and some of them give the expression of EM, BUT there is no clear definition of EM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is what EM is on earth? Its unit is supposed to be cm-3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to calculate EM? &lt;a href=&quot;http://www.cv.nrao.edu/course/astr534/FreeFreeEmission.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;(4B5) gives an expression, but it can not be derived from a light curve directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an X-ray light curve(not from the Sun), probably I need to devide the light curve into many bins and do spectral fitting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then what should I do? Rererence papers with clear definition and expression are welcome.&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="7982" LastEditDate="2021-04-01T08:57:39.713" LastActivityDate="2021-04-01T08:57:39.713" Title="What is &quot;emission measure&quot;?" Tags="&lt;terminology&gt;&lt;x-ray&gt;&lt;flare&gt;&lt;synchrotron-radiation&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13430" PostTypeId="1" AcceptedAnswerId="13436" CreationDate="2016-01-30T14:24:59.207" Score="5" ViewCount="123" Body="&lt;p&gt;This is purely a question about terminology, one that has eluded my googling efforts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a satellite and its parent body are tidally locked to each other, there is (in an ideal case) a single point on the surface of the parent body and on the surface of the satellite body that remain the closest two points on the two bodes at all times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a specific name for those two points? If so, what is it? I think &quot;antipodal points&quot; is descriptive, but I wonder if there is a more specific term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="10591" LastActivityDate="2016-01-30T23:22:57.190" Title="Terminology: Is there a name for the points on the surfaces of tidally locked parent/satellite bodies that always face each other?" Tags="&lt;tidal-forces&gt;&lt;satellite&gt;&lt;terminology&gt;&lt;orbital-resonance&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13431" PostTypeId="1" AcceptedAnswerId="13514" CreationDate="2016-01-30T15:08:54.937" Score="2" ViewCount="702" Body="&lt;p&gt;Here is the problem: I need to calculate longitude of observing place. And I know &lt;strong&gt;time&lt;/strong&gt; (in UTC+1) of star culmination (Sirius) and star &lt;strong&gt;altitude&lt;/strong&gt;. I also know &lt;strong&gt;RA&lt;/strong&gt; (right ascension) and &lt;strong&gt;Dec&lt;/strong&gt; (declination) of the star and I also know &lt;strong&gt;RA&lt;/strong&gt; of the Sun for that day. I already calculated latitude but I can't figure out how to calculate longitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="10592" LastActivityDate="2016-02-04T09:16:43.253" Title="Calculating longitude from star culmination" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13432" PostTypeId="2" ParentId="13428" CreationDate="2016-01-30T16:20:31.463" Score="2" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Orion&amp;#39;s_Sword&quot; rel=&quot;nofollow&quot;&gt;Sword of Orion&lt;/a&gt; includes the &lt;a href=&quot;https://en.wikipedia.org/wiki/Orion_Nebula&quot; rel=&quot;nofollow&quot;&gt;Great nebula in Orion&lt;/a&gt; (M42), which is naked eye visible. Depending on the seeing, the nebula can be mistaken for a star.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2016-01-30T16:20:31.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13433" PostTypeId="1" CreationDate="2016-01-30T17:54:54.170" Score="5" ViewCount="75" Body="&lt;p&gt;I'm sure it's a function of atmospheric entry angle, meteorite mass (and possibly composition/density), and its velocity, but what function is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above all, how does it depend on velocity? Will the slower meteors impact more often (less kinetic energy to dissipate) or the fast ones (less time spent in the atmosphere = less time to melt and evaporate all the way through)?&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2016-02-01T08:29:31.620" Title="What are the prerequisites for a meteorite to reach the ground?" Tags="&lt;meteorite&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13434" PostTypeId="2" ParentId="13430" CreationDate="2016-01-30T20:23:11.237" Score="1" Body="&lt;p&gt;I have not heard &quot;antipodal points&quot; used in this way before, but it is probably fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a tidally locked moon, this point is called the inner pole. It should in some sense also apply to the parent body, as that point is also facing the barycentre. The direction &quot;down&quot; is generally called &lt;a href=&quot;https://en.wikipedia.org/wiki/Nadir&quot; rel=&quot;nofollow&quot;&gt;Nadir&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For planetary coordinate systems, the constantly parent-body facing/moon facing point is used for 0,0.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastActivityDate="2016-01-30T20:23:11.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13435" PostTypeId="2" ParentId="13428" CreationDate="2016-01-30T22:59:52.280" Score="3" Body="&lt;p&gt;Most of Messier's objects are too dim to be seen by the naked eye. The few that are visible were mostly listed as stars in older lists. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edmund Halley lists 6 &quot;luminous patches&quot; which &quot;discover themselves only by the telescope, and appear to the naked eye like small fixt stars&quot; &lt;a href=&quot;http://adsabs.harvard.edu/full/1947JRASC..41...69S&quot; rel=&quot;nofollow&quot;&gt;(Halley)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These include M42, in Orion, part of the sword (or what ever that is between the hunter's legs), and Omega Centuari, listed as a minor star in the centaur (roughly on the horse's back) in Bayer's list. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The other objects on Halley's list were either known to be &quot;cloud-like&quot; [M31] or too faint for naked eye astronomy.]&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-01-30T22:59:52.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13436" PostTypeId="2" ParentId="13430" CreationDate="2016-01-30T23:22:57.190" Score="3" Body="&lt;p&gt;For tidally locked binary &lt;em&gt;stars&lt;/em&gt;, the two points in question are known as the &lt;em&gt;substellar points&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a tidally locked exoplanet, the point closest to the star would also be known as the substellar point. If the star was also tidally locked to the planet, then there would be a &lt;em&gt;subplanetary point&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a moon locked to a planet, the point on the moon would also be the subplanetary point. If the planet was mutually locked, it would have a sublunar point (not sure whether this latter only applies to the Earth's moon).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-01-30T23:22:57.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13437" PostTypeId="2" ParentId="13429" CreationDate="2016-01-30T23:28:48.820" Score="7" Body="&lt;p&gt;Emission measure is (usually) used in X-ray and EUV astronomy, though I suppose also in cases of optically thin radio emission. It is &lt;em&gt;defined as&lt;/em&gt; the square of the number density of free electrons integrated over the volume of the plasma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\rm EM} = \int n_e^2 \ dV$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flux of optically thin emission from a plasma (e.g. thermal bremsstrahlung) is then directly proportional to the emission measure of the plasma multiplied by a temperature dependent cooling loss law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, when you measure the flux of X-rays from an unresolved optically thin emitter, there is a degeneracy between the electron number density (squared) and the overall plasma volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you fit an X-ray spectrum with an optically thin model, the emission measure (divided by $4\pi d^2$, where $d$ is the distance to the object), is a multiplicative free parameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question about calculation is extremely difficult to answer. Suppose I measure a count-rate of $N$ X-ray counts per second using some X-ray telescope (I can only assume that's what you mean by a &quot;X-ray light curve&quot;.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The count-rate received at the telescope depends on: the emission measure (as defined above) multiplied by a term that depends on the temperature (or temperatures) of the source, the chemical composition of the source and the adopted emission process (is it free-free thermal bremsstrahlung, a thermal plasma or something else). It is then attenuated by any intrinsic absorption in the source and any absorption between us and the source and by the distance to the source (assuming the radiation is isotropic). Finally what is &lt;em&gt;detected&lt;/em&gt; is determined by the response of the X-ray detector to X-ray photons as a function of energy.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-06-23T08:37:38.340" LastActivityDate="2017-06-23T08:37:38.340" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13438" PostTypeId="1" CreationDate="2016-01-31T04:36:51.590" Score="4" ViewCount="64" Body="&lt;p&gt;Which planets have abundant amounts of photoluminiscent matter? Photoluminiscent or chemoluminiscent whatever the glow-in-the-dark stuff it may be. Planets or bodies which have good enough the photoluminiscence(light absorb and then release) property.&lt;/p&gt;&#xA;" OwnerDisplayName="user10603" LastActivityDate="2016-01-31T18:24:48.313" Title="Which planets have abundant amounts of photoluminiscent matter?" Tags="&lt;planet&gt;&lt;light&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13441" PostTypeId="2" ParentId="13438" CreationDate="2016-01-31T12:40:08.050" Score="2" Body="&lt;p&gt;Sorry, but none of the planets have substantial amounts of phosphorescent chemicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the closest to what you are thinking is the aurora, which can be seen on all the planets with atmospheres (to a greater or lesser degree) notably on the Gas Giants and on Earth. However, while the mechanism of aurora is similar to florescence, it is not phosphorescence.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No planets outside of the solar system have been directly imaged beyond a dim dot, so there is no information about exoplanets.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-01-31T12:40:08.050" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13442" PostTypeId="2" ParentId="13433" CreationDate="2016-01-31T12:58:47.693" Score="4" Body="&lt;p&gt;Your first resource here can be the &lt;a href=&quot;http://impact.ese.ic.ac.uk/ImpactEffects/&quot; rel=&quot;nofollow&quot;&gt;impact effects calculator&lt;/a&gt;, It can show you when it is likely that a high velocity impact will occur. Somewhat surprisingly, faster objects are less likely to hit the ground at hyper-velocity, as they are more likely to break up and explode at altitude. They will likely form multiple meteorites, as the Chelyabinsk meteor did&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in meteorite formation, rather than craters, then most impactors will in some part survive the fall, provided they are rocky rather than icy. Objects smaller than 5cm are mostly cometary in origin. Larger than that and they are more likely to be rocky.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="941" LastEditDate="2016-02-01T08:29:31.620" LastActivityDate="2016-02-01T08:29:31.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13445" PostTypeId="2" ParentId="13438" CreationDate="2016-01-31T18:24:48.313" Score="1" Body="&lt;p&gt;&lt;strong&gt;Earth&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the only place in the universe we know there is stuff that glows in the dark. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2P0ZM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2P0ZM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More serious though, the only possible place I can think of is &lt;a href=&quot;https://en.wikipedia.org/wiki/Io_%28moon%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Io&lt;/a&gt;. There may be some reactions taking place in the sulphur material ejected from the volcanoes that produces light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YTgPPt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YTgPPt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10172" LastActivityDate="2016-01-31T18:24:48.313" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13447" PostTypeId="1" CreationDate="2016-01-31T19:15:56.207" Score="2" ViewCount="786" Body="&lt;p&gt;...or what's the 'direction of vernal equinox'?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For both Ecliptic and Equatorial coordinate system, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Celestial_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; gives &quot;Primary direction (0° longitude)&quot; - &quot;Vernal Equinox&quot;, linking to article on the event of the Equinox which never mentions any directions associated with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the direction will be associated with the axis connecting the Sun and Earth during vernal equinox, but which direction of the line? Outward, with the meridian running opposite of the Sun, or inward, the meridian crossing the Sun at vernal equinox? Or maybe my guess is entirely wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, how would one go about converting celestial (ecliptic or equatorial) latitude to Earth latitude or vice versa - e.g. finding where on Earth an object of given celestial coordinates is in zenith at given time? The linked article has several formulas of converting between various celestial coordinate systems but either no conversion to Earth coordinates or I failed to understand some of the conversions.&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2016-01-31T20:00:07.540" Title="Where's the Prime Meridian of celestial coordinates" Tags="&lt;coordinate&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13448" PostTypeId="2" ParentId="13447" CreationDate="2016-01-31T20:00:07.540" Score="3" Body="&lt;p&gt;The equatorial plane and the ecliptic plane cross each other, like shown in this illustration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yS8gHm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yS8gHm.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;Vernal Equinox&quot; is simply the direction of the line where they cross.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can not obtain the Earth latitude from the ecliptic latitude alone, as the planes are not parallel. You must know the ecliptic longitude as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From equatorial to Earth latitude it is much simpler, the latitude is equal to the declination.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastActivityDate="2016-01-31T20:00:07.540" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13450" PostTypeId="2" ParentId="13287" CreationDate="2016-01-31T23:27:19.380" Score="1" Body="&lt;p&gt;A body and its satellite in truth form a system and both rotate around their common centre of gravity. Hence both the Earth and Moon rotate around this centre - however because the Earth is significantly more massive than the Moon this can be approximated as the Moon orbiting the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon does have a measurable impact on the Earth - tides are the obvious example, and tidal friction is also slowing down the Earth's rotation by a very small amount (around 2.3ms per century).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Moon's case, though, the Earth's gravitational pull has already effectively locked its rotation period with its orbital period around the Earth - so we essentially only ever see the same face of the Moon on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth and Moon are much closer in mass than most other planet/satellite systems in the Solar System - think of the huge mass of Jupiter - and so it is difficult to show how these planets' satellites have a similar impact on the planet - but the same effects are there, albeit in much smaller scale.&lt;/p&gt;&#xA;" OwnerUserId="823" LastEditorUserId="823" LastEditDate="2016-02-01T07:51:14.173" LastActivityDate="2016-02-01T07:51:14.173" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13452" PostTypeId="1" AcceptedAnswerId="13456" CreationDate="2016-02-01T02:30:06.043" Score="10" ViewCount="432" Body="&lt;p&gt;On Earth, hurricanes usually move from the equator to the poles. But if you look at a picture of Jupiter, you'd see that the Great Red Spot isn't anywhere near the poles. So the question is: why don't storms on gas giants move to the poles like storms on Earth do?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="9301" LastEditDate="2016-02-01T08:39:47.140" LastActivityDate="2016-02-01T11:43:52.060" Title="Why don't storms on gas giants move to the poles, like hurricanes on Earth do?" Tags="&lt;planet&gt;&lt;earth&gt;&lt;gas-giants&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13453" PostTypeId="2" ParentId="12952" CreationDate="2016-02-01T04:44:22.863" Score="-1" Body="&lt;p&gt;My opinion and 20 Cents:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Humanity did confused many things in history. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to the days of the holy bonfires on the squares of Europe, nobody didn't understood that the Earth - is not a center of universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are still believing that speed of light is a maximum speed of information, while there is discovered quantum teleportation with higher speed, infinity speed. Infinity looks much better than 299 792 458 meters per second, is not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, we can not describe why does photon getting its super-speed without any boost, immediately, as a part of its existence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is why, I'm thinking that we are confused with this great theory about center of universe, which is everywhere, there and here, anywhere...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is too beauty and too hard to understand, that is why everybody simply believing in it, without tries to imagine anything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could read about &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow&quot;&gt;Cosmic microwave background&lt;/a&gt;. There is a redshift and Milky-Way is moving to the one side of this big-bang sphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a sphere. There is a redshift. We are inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you still believe that center of universe is &quot;everywhere&quot;?&lt;/p&gt;&#xA;" OwnerUserId="10621" LastActivityDate="2016-02-01T04:44:22.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13455" PostTypeId="1" AcceptedAnswerId="21425" CreationDate="2016-02-01T10:40:51.437" Score="7" ViewCount="799" Body="&lt;p&gt;Let's consider Mars' two moons for a moment. They're small, not even spherical, and most likely are captured asteroids. They weren't discovered until the late 1800's. Telescopes are much more powerful today, so you'd think we'd have found any similar moons at Venus or Mercury by now, if they exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trouble is, there are some very dark asteroids out there. And of course, there are some much smaller ones too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we know Venus or Mercury doesn't have a small dark asteroid moon? Did any probes actually try radar to find them?&lt;/p&gt;&#xA;" OwnerUserId="7449" LastActivityDate="2017-06-27T17:33:28.330" Title="Could Venus or Mercury have a moon that we haven't detected?" Tags="&lt;observational-astronomy&gt;&lt;venus&gt;&lt;mercury&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13456" PostTypeId="2" ParentId="13452" CreationDate="2016-02-01T11:43:52.060" Score="12" Body="&lt;p&gt;The answer is the Coriolis effect, on Earth this produces cells within which storms move, converging towards the cell boundaries as you can see below.&lt;a href=&quot;https://i.stack.imgur.com/XGJH8.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XGJH8.png&quot; alt=&quot;An illustration of the Coriolis effect on Earth.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter however spins much faster that the Earth which produces a stronger Coriolis effect and thus more cells. This is the reason why there are so many different coloured bands on Jupiter (see image below). So storms like the big red spot don't have as far to move laterally within these cells compared to hurricanes on Earth (in comparison with the planets size).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/k5dfh.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k5dfh.jpg&quot; alt=&quot;Image of the planet Jupiter, showing the many different colour bands.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10437" LastActivityDate="2016-02-01T11:43:52.060" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13458" PostTypeId="1" AcceptedAnswerId="13469" CreationDate="2016-02-01T13:53:20.190" Score="3" ViewCount="1127" Body="&lt;p&gt;In the total beginning of the Earth there was CO2, N2 and H2O. So there was O2 but always in bond with Carbon and hydrogen, but what is the reason for that? Is O2 always reacting with free carbon atoms?&#xA;The temperature of the Earth was is those beginning days around 1200degree Celsius. Is that also a factor of those bindings?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="14021" LastEditDate="2016-09-14T14:34:40.313" LastActivityDate="2016-09-14T14:34:40.313" Title="Why wasn't there any oxygen in the beginning of the Earth?" Tags="&lt;earth&gt;" AnswerCount="3" CommentCount="1" ClosedDate="2016-09-14T18:30:16.323" ContentLicense="CC BY-SA 3.0" />
  <row Id="13459" PostTypeId="1" AcceptedAnswerId="13460" CreationDate="2016-02-01T14:56:00.677" Score="4" ViewCount="187" Body="&lt;p&gt;Do we consider the dwarf planet Ceres to be in the inner solar system or outer solar system (or possibly neither / in between).  It is located within the asteroid belt, but is the asteroid belt part of the inner or outer solar system?&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2016-02-01T15:22:47.430" Title="Is Ceres in the inner solar system or outer solar system?" Tags="&lt;solar-system&gt;&lt;classification&gt;&lt;asteroid-belt&gt;&lt;ceres&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13460" PostTypeId="2" ParentId="13459" CreationDate="2016-02-01T15:22:47.430" Score="5" Body="&lt;p&gt;Inner. The entire asteroid belt is in the Inner Solar System (now). The definition of &quot;Inner&quot; vs. &quot;Outer&quot; is non-arbitrary, based on the current &quot;frost line&quot;, approx. 5 a.u. radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Frost_line_%28astrophysics%29&quot;&gt;https://en.wikipedia.org/wiki/Frost_line_%28astrophysics%29&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_System#Inner_Solar_System&quot;&gt;https://en.wikipedia.org/wiki/Solar_System#Inner_Solar_System&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10005" LastActivityDate="2016-02-01T15:22:47.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13461" PostTypeId="2" ParentId="13458" CreationDate="2016-02-01T15:36:57.853" Score="3" Body="&lt;p&gt;The 1200 C temperature is a big clue, given the auto-ignition temperatures of most &quot;stuff&quot;. &lt;a href=&quot;https://en.wikipedia.org/wiki/Autoignition_temperature&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Autoignition_temperature&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any &quot;free&quot; oxygen would immediately &quot;burn&quot; with carbon and just about anything else, yes.&lt;/p&gt;&#xA;" OwnerUserId="10005" LastActivityDate="2016-02-01T15:36:57.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13462" PostTypeId="1" AcceptedAnswerId="14128" CreationDate="2016-02-01T15:59:21.850" Score="3" ViewCount="110" Body="&lt;p&gt;There is a &lt;a href=&quot;https://www.sciencenews.org/article/babylonians-used-geometry-track-jupiter%E2%80%99s-movements&quot; rel=&quot;nofollow&quot;&gt;science news article&lt;/a&gt; on how Babylonians used geometry to track Jupiter’s movements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an image of cuneiform tablet which shows Jupiter's movements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://www.sciencenews.org/sites/default/files/2016/01/main/articles/012816_bb_babylonian-main_free.jpg&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a layman's point of view, it looks like someone engraved gibberish on a tablet. How do we even know it was used to track Jupiter's movements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone translate this? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="10298" LastEditorUserId="10298" LastEditDate="2016-03-12T01:54:25.323" LastActivityDate="2016-04-17T12:22:57.287" Title="How to read Babylonians' tablet that were used to track Jupiter's movements?" Tags="&lt;jupiter&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13463" PostTypeId="1" AcceptedAnswerId="13464" CreationDate="2016-02-01T16:41:59.510" Score="3" ViewCount="409" Body="&lt;p&gt;I wanted to know the distance of the 7 planets (Mercury, Mars, Venus, Jupiter, Saturn, Uranus, Neptune ) each (or at least 5 planets) from the earth and the angle that it will make with a line connecting the centers of earth and the respective planet at a &lt;em&gt;point of time&lt;/em&gt;. Also I need the same for the earth and moon system and sun and earth as they are big influences. At point of time because in the course of revolution of planets the distances with respect to earth will vary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being an engineering student I have the wish :) to calculate the mechanical forces acting on the earth at a point of time due to nearby heavenly bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically I need the data or in case someone already attempted to calculate the force components I would be delighted to look at it.&lt;/p&gt;&#xA;" OwnerUserId="10430" LastEditorUserId="1891" LastEditDate="2019-02-18T09:53:59.987" LastActivityDate="2019-02-18T09:53:59.987" Title="What is the distance of Earth from the 7 planets, moons and the Sun at a point of time?" Tags="&lt;planet&gt;&lt;earth&gt;&lt;distances&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="13464" PostTypeId="2" ParentId="13463" CreationDate="2016-02-01T16:47:26.217" Score="3" Body="&lt;p&gt;There are plenty of sky simulators where you can calculate that information yourself. e.g. Celestia, KStars&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just install one of them and it will tell you position in sky and distance to most known bodies in Solar System, for any given time.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2016-02-01T16:47:26.217" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13467" PostTypeId="1" AcceptedAnswerId="14530" CreationDate="2016-02-01T20:52:15.533" Score="6" ViewCount="214" Body="&lt;p&gt;I just read an article about the cosmological model of an inflationary universe. As you probably know it basically states that space itself is subject to an expansion over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there is no moving matter or energy involved, this might even happen at a &quot;speed&quot; faster than light. So far, so good and obscure. What strikes me is the principle that the speed of light as a fundamental constant can only be expressed as a function of space-time. Where do we know that the one is constant but the other can suddenly be variable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any reason why the point of view of an expanding space is preferred over, say, a reduction in the speed of light or an increase in the &quot;speed&quot; of time? Is there any objective difference or is it just the good old rubber metaphor being stretched (pun intended) too far?&lt;/p&gt;&#xA;" OwnerUserId="10632" LastActivityDate="2016-04-13T16:39:38.220" Title="Is there an objective difference between space expansion and reduction in speed of light" Tags="&lt;space-time&gt;&lt;expansion&gt;&lt;speed&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13468" PostTypeId="1" CreationDate="2016-02-01T23:45:28.387" Score="-6" ViewCount="720" Body="&lt;p&gt;I want to get a tattoo of the solar system. The gap between the planets should not be preserved and the differences in sizes will only be symbolic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is if all planets were to align, do we know what side of the planets would be shown? If not, where can I find a profile picture of which side of the planet it will show?&lt;/p&gt;&#xA;" OwnerUserId="10634" LastEditorUserId="11264" LastEditDate="2017-11-10T13:47:59.553" LastActivityDate="2017-11-10T13:47:59.553" Title="Best planets profile for a tattoo of the solar system" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;planet&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2016-02-02T10:20:02.043" ContentLicense="CC BY-SA 3.0" />
  <row Id="13469" PostTypeId="2" ParentId="13458" CreationDate="2016-02-02T00:19:24.777" Score="1" Body="&lt;p&gt;A complete answer gets long and above my pay-grade, but when the earth and solar system were young, it's unlikely that the Earth had much atmosphere at all.   The young sun was hot enough to burn off most gases and ices in the inner solar-system and when the earth coalesced it was largely metals and silicates.  The oceans and atmosphere came later by comet and asteroid belt impacts, and, as Jeff Y points out, after that it's just heat and chemistry.   Oxygen or O2 binds so easily with other molecules, like Methane or Hydrogen or Iron that it goes away pretty quickly until those are used up.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like O2, N2 wasn't nearly as present in the young atmosphere as it is today either.   More on that here:   &lt;a href=&quot;http://sciexplorer.blogspot.com/2012/01/earths-atmosphere-part-4-evolution-of.html&quot; rel=&quot;nofollow&quot;&gt;http://sciexplorer.blogspot.com/2012/01/earths-atmosphere-part-4-evolution-of.html&lt;/a&gt;   The four most common early ices/gases on earth were CH4, CO2, H2O and NH3.    Over time, the NH3 and CH4 get used up by bacteria or chemical processes, lightning, volcanoes, UV light from the sun, etc.   The H20 and CO2 are much more stable, though photosynthesis did slowly pull most of the CO2 from the atmosphere, returning O2.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-02-02T00:19:24.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13470" PostTypeId="1" AcceptedAnswerId="13477" CreationDate="2016-02-02T01:49:30.690" Score="2" ViewCount="107" Body="&lt;p&gt;If a comet crashes into the sun, where will the substance (eg: water,silicon) be? Will they be inside the sun? Or they will be ejected from sun again?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastActivityDate="2016-02-02T10:25:24.243" Title="Where is the materials of comet after it crashed into the sun?" Tags="&lt;the-sun&gt;&lt;comets&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13471" PostTypeId="1" AcceptedAnswerId="13535" CreationDate="2016-02-02T02:08:38.267" Score="2" ViewCount="4575" Body="&lt;p&gt;Like when is the new moon highest in the sky? Or third quarter? How do I know this for any general moon phase? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="10635" LastActivityDate="2020-08-21T13:31:37.420" Title="How to calculate when the moon is highest in the sky to the earthbound observer?" Tags="&lt;moon-phases&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13472" PostTypeId="1" AcceptedAnswerId="13478" CreationDate="2016-02-02T03:56:19.907" Score="5" ViewCount="745" Body="&lt;p&gt;From what I've read, Jupiter's 'Great Red Spot' is an unbelievably powerful storm. Do astronomers and scientists have any idea of what that might look like if it were possible to be inside/below it? Would it look like a storm on Earth (thunder, lightning, winds, etc.), or would it look completely different?&lt;/p&gt;&#xA;" OwnerDisplayName="user10636" LastActivityDate="2021-01-06T19:26:46.823" Title="What Would it Look Like Inside Jupiter's Great Red Spot?" Tags="&lt;solar-system&gt;&lt;jupiter&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13473" PostTypeId="2" ParentId="13468" CreationDate="2016-02-02T04:20:17.010" Score="3" Body="&lt;p&gt;&lt;em&gt;[Not sure if I should answer this, but I will try to answer something while trying hard to not go off-topic.]&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The planets aspects&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Mercury_%28planet%29&quot; rel=&quot;nofollow&quot;&gt;Mercury&lt;/a&gt; surface is essentially a collection of small random craters with no discernible pattern at all, so you might not consider which side is presented. The only distinguished feature is a set of dark craters in its north pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Venus&quot; rel=&quot;nofollow&quot;&gt;Venus&lt;/a&gt; features few discernible aspects in visible light to the human eyes. There are only a few and faint distinguishable cloud bands, so you might also not consider which side is presented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Earth&quot; rel=&quot;nofollow&quot;&gt;Earth&lt;/a&gt; is the most important, because it have continents and oceans with distinct designs. It also features a lot of clouds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Mars&quot; rel=&quot;nofollow&quot;&gt;Mars&lt;/a&gt; has polar caps and a system of canyons. It's northern hemisphere is also much less cratered and has a lower altitude than its southern hemisphere, except that the greatest crater is in the southern hemisphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Jupiter&quot; rel=&quot;nofollow&quot;&gt;Jupiter&lt;/a&gt; has a banded structure of clouds covering the entire planet. It features a large red spot with some nearby fainter and smaller whiter spots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Saturn&quot; rel=&quot;nofollow&quot;&gt;Saturn&lt;/a&gt; also has a banded structure, which is faintier to Jupiter's structure but still clearly visible. It also features a curious hexagon on its north pole. But it is barely noticeable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Uranus&quot; rel=&quot;nofollow&quot;&gt;Uranus&lt;/a&gt; have very homogenous atmosphere (as seen in visible colors by human eyes), so it have almost no visible features to be drawn in your tatoo. Their presented side do not matters, because it is essentially a bland featureless ball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Neptune&quot; rel=&quot;nofollow&quot;&gt;Neptune&lt;/a&gt; has also few visible features. There are no more than a few cloud bands with low variation on color or hue. However there is a dark spot.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The rings&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Saturn has &lt;a href=&quot;https://en.wikipedia.org/wiki/Rings_of_Saturn&quot; rel=&quot;nofollow&quot;&gt;an extensive system of rings&lt;/a&gt;, which also contains some defined gaps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter, Uranus and Neptune also feature faint sets of rings (&lt;a href=&quot;https://en.wikipedia.org/wiki/Rings_of_Jupiter&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Rings_of_Uranus&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Rings_of_Neptune&quot; rel=&quot;nofollow&quot;&gt;3&lt;/a&gt;). Uranus has a weird orientation, so its rings are not aligned with the orbital plane, but roughly perpendicular to it.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The rotations&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;All the planets rotates at different velocities, so any side of them would do. No planet is showing always the same face to any other planet or to the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the planets, except Uranus, are rotating roughly in the same plane. So, their north is all pointing to a side and their south to another side (lets call those north/south axis Y). Their north-south axis is perpedicular to their planet-Sun axis (which we will call X). Uranus is special because at their summer/winter, its north/south axis points roughly to the Sun (i.e., in the X axis) and in its autumn/spring it is in a direction that is perpendicular to both X and Y (so it's the Z).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, don't forget the &lt;a href=&quot;https://en.wikipedia.org/wiki/Sun&quot; rel=&quot;nofollow&quot;&gt;Sun&lt;/a&gt;. It may also feature flares and sunspots sometimes. The Sun rotates in the same way as most of the planets, with its north/south axis along the Y direction.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The moons&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;You might also be tempted to include the planets moons. Those are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Earth's &lt;a href=&quot;https://en.wikipedia.org/wiki/Moon&quot; rel=&quot;nofollow&quot;&gt;Moon&lt;/a&gt;, which always presents the same face to Earth.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mars &lt;a href=&quot;https://en.wikipedia.org/wiki/Moons_of_Mars&quot; rel=&quot;nofollow&quot;&gt;moons&lt;/a&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Phobos_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Phobos&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Deimos_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Deimos&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Jupiter &lt;a href=&quot;https://en.wikipedia.org/wiki/Moons_of_Jupiter&quot; rel=&quot;nofollow&quot;&gt;moons&lt;/a&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Io_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Io&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Europa_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Europa&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Ganymede_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Ganymede&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Callisto_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Callisto&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Amalthea_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Amalthea&lt;/a&gt; and lots of smaller moons.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Saturn &lt;a href=&quot;https://en.wikipedia.org/wiki/Moons_of_Saturn&quot; rel=&quot;nofollow&quot;&gt;moons&lt;/a&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Titan_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Titan&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Rhea_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Rhea&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Iapetus_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Iapetus&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Dione_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Dione&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Tethys_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Tethys&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Enceladus_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Enceladus&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Mimas_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Mimas&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Hyperion_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Hyperion&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Phoebe_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Phoebe&lt;/a&gt; and lots of smaller moons.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Uranus &lt;a href=&quot;https://en.wikipedia.org/wiki/Moons_of_Uranus&quot; rel=&quot;nofollow&quot;&gt;moons&lt;/a&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Ariel_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Ariel&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Umbriel_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Umbriel&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Titania_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Titania&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Oberon_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Oberon&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Miranda_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Miranda&lt;/a&gt; and several smaller moons.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Neptune &lt;a href=&quot;https://en.wikipedia.org/wiki/Moons_of_Neptune&quot; rel=&quot;nofollow&quot;&gt;moons&lt;/a&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Triton_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Triton&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Proteus_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Proteus&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Nereid_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Nereid&lt;/a&gt; and several smaller moons.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You can get more data (and also some images) about them &lt;a href=&quot;https://en.wikipedia.org/wiki/Natural_satellite&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and also &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_natural_satellites&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The dwarf planets&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Also, you might want to include the dwarf planets. The confirmed dwarf planets are &lt;a href=&quot;https://en.wikipedia.org/wiki/Ceres_%28dwarf_planet%29&quot; rel=&quot;nofollow&quot;&gt;Ceres&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Pluto&quot; rel=&quot;nofollow&quot;&gt;Pluto&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Charon_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Charon&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Eris_%28dwarf_planet%29&quot; rel=&quot;nofollow&quot;&gt;Eris&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Makemake&quot; rel=&quot;nofollow&quot;&gt;Makemake&lt;/a&gt;. Ceres is located in the main asteroid belt. Although technically being a dwarf planet, for practical purposes Charon is Pluto's main satellite and they always shows the same face one to the other. Makemake and Eris are beyond Pluto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To complete the Solar Sytem, you could also add other transneptunian objects (likely to also be dwarf planets) like &lt;a href=&quot;https://en.wikipedia.org/wiki/50000_Quaoar&quot; rel=&quot;nofollow&quot;&gt;Quaoar&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/90377_Sedna&quot; rel=&quot;nofollow&quot;&gt;Sedna&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Haumea&quot; rel=&quot;nofollow&quot;&gt;Haumea&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/90482_Orcus&quot; rel=&quot;nofollow&quot;&gt;Orcus&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/120347_Salacia&quot; rel=&quot;nofollow&quot;&gt;Salacia&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/%28307261%29_2002_MS4&quot; rel=&quot;nofollow&quot;&gt;2002 MS&lt;sub&gt;4&lt;/sub&gt;&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/%28225088%29_2007_OR10&quot; rel=&quot;nofollow&quot;&gt;2007 OR&lt;sub&gt;10&lt;/sub&gt;&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/2012_VP113&quot; rel=&quot;nofollow&quot;&gt;2012 VP&lt;sub&gt;113&lt;/sub&gt;&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/2010_GB174&quot; rel=&quot;nofollow&quot;&gt;2010 GB&lt;sub&gt;174&lt;/sub&gt;&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/2004_XR190&quot; rel=&quot;nofollow&quot;&gt;2004 XR&lt;sub&gt;190&lt;/sub&gt;&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/%28148209%29_2000_CR105&quot; rel=&quot;nofollow&quot;&gt;2000 CR&lt;sub&gt;105&lt;/sub&gt;&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/2004_VN112&quot; rel=&quot;nofollow&quot;&gt;2004 VN&lt;sub&gt;112&lt;/sub&gt;&lt;/a&gt;. Those 12 transneptunian suspected dwarf planets, along with Eris and Makemake, are no more than singular dots of light on the telescopes, their appearance is unknown and even their size is known only crudely, with very large error margins in their size estimatives (see more about that below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want also specific asteroids, you might see &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_exceptional_asteroids&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The sizes&lt;/h1&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Sun's diameter is roughly 10 times Jupiter's diameter.&lt;/li&gt;&#xA;&lt;li&gt;Jupiter's diameter is almost 11 times Earth's diameter.&lt;/li&gt;&#xA;&lt;li&gt;Saturn has almost the size of Jupiter, having 9.5 Earth's diameters. Also, it is visibly oblate (oval).&lt;/li&gt;&#xA;&lt;li&gt;Uranus and Neptune are about the same size, with 4 times the Earth's diameter.&lt;/li&gt;&#xA;&lt;li&gt;Venus is almost the same size of Earth's (95% of Earth's diameter).&lt;/li&gt;&#xA;&lt;li&gt;Mars is a bit larger than a half of Earth's diameter (53%).&lt;/li&gt;&#xA;&lt;li&gt;Mercury's diameter is 38% of Earth's diameter.&lt;/li&gt;&#xA;&lt;li&gt;Moon has 27% of Earth's diameter.&lt;/li&gt;&#xA;&lt;li&gt;Pluto and Eris were determined to be essentially of the same size, which is 18.6% of Earth's diameter.&lt;/li&gt;&#xA;&lt;li&gt;Ceres has 14.8% of Earth's diameter.&lt;/li&gt;&#xA;&lt;li&gt;Charon has 9.5% Earth's diameter.&lt;/li&gt;&#xA;&lt;li&gt;All the others known transneptunian objects, although surely smaller than Pluto, have uncertain sizes.&lt;/li&gt;&#xA;&lt;li&gt;All the dwarf planets and asteroids are significantly smaller than the Moon. Check the links for more details.&lt;/li&gt;&#xA;&lt;li&gt;Some satellites of Jupiter, Saturn, Uranus and Neptune are pretty large. Titan and Ganymede are even larger than Mercury.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For more details about sizes, &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_Solar_System_objects_by_size&quot; rel=&quot;nofollow&quot;&gt;check this page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the transeptunian objects, only Pluto and Charon have a known appearance (due to being photographed by New Horizons). This way, all that you would need to draw in your tatoo beyond Pluto and Charon is a group of little spheres smaller than Pluto (I named 14 of them above), with only the one representing Eris with the same size as Pluto.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Final note&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;You will find images and further detailed information in the previous links, all of them points to wikipedia. I hope that you can get a perfect tatoo with that information!&lt;/p&gt;&#xA;" OwnerUserId="609" LastEditorUserId="609" LastEditDate="2016-02-27T04:25:15.943" LastActivityDate="2016-02-27T04:25:15.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13474" PostTypeId="1" AcceptedAnswerId="13487" CreationDate="2016-02-02T04:33:17.560" Score="8" ViewCount="1958" Body="&lt;p&gt;If we take a look at a logarithmic scale of the Solar System, there appears to be a large gap between the heliopause and the Oort cloud:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/gMbSH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gMbSH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is that? There are &lt;em&gt;some&lt;/em&gt; objects in the gap, but why are most of the Oort cloud objects in one area? Shouldn't it also fill the large gap in the picture just as much?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2016-02-02T16:25:50.420" Title="Lack of objects between heliopause and Oort cloud?" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;gravity&gt;&lt;oort-cloud&gt;&lt;interstellar-medium&gt;" AnswerCount="2" CommentCount="10" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13475" PostTypeId="2" ParentId="13474" CreationDate="2016-02-02T05:07:17.083" Score="1" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Kuiper_belt&quot; rel=&quot;nofollow&quot;&gt;Kuiper belt&lt;/a&gt; and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Scattered_disc&quot; rel=&quot;nofollow&quot;&gt;Scattered disk&lt;/a&gt; are widely believed to lie in the space between the outer planets and the Oort cloud, but not to reach all the way out to the Oort cloud (apparently due to resonances with Neptune and a scarcity of sighted object much outside the 1:2 resonance orbit). The various dwarf planets of the outer solar system are sometimes referred to as Kuiper belt objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it the Kuiper belt is expected to be distinct from the Oort cloud in being at least somewhat planar and aligned with the ecliptic whereas the Oort cloud is essentially spherical. If Pluto can be taken as a guide then we expect considerably larger inclinations from Kuiper belt objects than we see from large bodies in the inner solar system, but still a nod toward the ecliptic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've less familiarity with the scattered disc and the Wikipedia articles indicate that the usage is not very consistent. It seems to mean highly eccentric objects beyond the orbit of Neptune.&lt;/p&gt;&#xA;" OwnerUserId="530" LastEditorUserId="530" LastEditDate="2016-02-02T06:32:07.853" LastActivityDate="2016-02-02T06:32:07.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13476" PostTypeId="1" CreationDate="2016-02-02T07:13:02.637" Score="6" ViewCount="347" Body="&lt;h2&gt;Brief tale of woe&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;My department at a small public university in the US has decided to begin offering a descriptive astronomy general education course for the first time in a couple of decades and we have no institutional memory of how it was done before nor anyone on staff who's taught such a beast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I've been tapped to teach it.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Requirements&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;It's to be a one-semester course with three hours a week of classroom instruction and a three hour lab scheduled late enough that we'll have some dark sky early in the semester and more as the year goes on (schedule has it in Fall Semester). Expected enrollment in the neighborhood of 20 to 30. I won't have a teaching assistant unless I can rope a physics major into doing it somehow (we're an all undergrad department).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a &quot;general education&quot; course mine will mostly attract students majoring in non-science subjects, biological science or human science. The students will have had enough math to know (in principle) how to isolate any variable in, say, the law of universal gravitation but many won't remember at first and some will resist; many will have chosen this class in an effort to avoid 'hard' classes like our gen ed chemistry, physics, or geology courses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking of a textbook that touches on things like&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Observational coordinate systems, familiarization with the night sky (constellations, I suppose), finding things&lt;/li&gt;&#xA;&lt;li&gt;The Earth moon system, phases, eclipses, maybe even tides&lt;/li&gt;&#xA;&lt;li&gt;Kepler's laws&lt;/li&gt;&#xA;&lt;li&gt;The planets, moons and other bits and pieces in the solar system&lt;/li&gt;&#xA;&lt;li&gt;The sun in particular and stars in general; including the birth, evolution and death of stars in a low math way&lt;/li&gt;&#xA;&lt;li&gt;Some observational stuff; parallax, standard candles, HR diagram ...&lt;/li&gt;&#xA;&lt;li&gt;A unit or integrated material on the planetary missions would be nice&lt;/li&gt;&#xA;&lt;li&gt;Larger scale structure of the universe; a cosmology unit wouldn't be out of place; how much dark matter (or worse dark energy) can actually be done?&lt;/li&gt;&#xA;&lt;li&gt;A unit on extra-solar planet hunting would be nice.&lt;/li&gt;&#xA;&lt;li&gt;A unit on other wavelengths?&lt;/li&gt;&#xA;&lt;li&gt;What else goes into a course like this?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Please include some detail on what is good about suggested texts and how well it matches with the course I've got to teach. What does it offer that's special?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Random detail&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;At this point we own a 8&quot; scope (reflector) with a axial mount and motor drive but no automated pointing and two pairs of decent binoculars (I can probably buy more binoculars, but additional scopes are probably out of budget for this). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's been about 20 years since I pointed a scope up with my own hands, and my only &quot;real&quot; astronomical experience was doing some programming for a CCD backed 14&quot; remote-operated scope in the early 90s (we even managed to track a MACHO light curve and agree with the big boys, woo hoo science!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The department head thinks that every gen. ed. course in our department should cover the scientific method and some thermodynamics which I think we can fit in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I sincerely intend to include at least one field trip out to dark sky to just look up with adapted eyes. It's been too long...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://astronomy.meta.stackexchange.com/questions/237/policy-on-asking-for-book-recommendations&quot;&gt;only question I could find on meta&lt;/a&gt; suggested that a sufficiently focused recommendation question would be accepted. If this is a no go, perhaps I'll pop into chat at some point.&lt;/p&gt;&#xA;" OwnerUserId="530" LastEditorUserId="-1" LastEditDate="2017-03-16T15:52:10.723" LastActivityDate="2016-05-25T15:37:25.610" Title="Good textbooks for a descriptive astronomy course" Tags="&lt;naked-eye&gt;&lt;binoculars&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13477" PostTypeId="2" ParentId="13470" CreationDate="2016-02-02T08:46:00.660" Score="3" Body="&lt;p&gt;It depends on the size of the comet. If it's only $0.1$–$1\,\mathrm{km}$, it will evaporate before even reaching the Sun (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2013CBET.3496....1L&quot; rel=&quot;nofollow&quot;&gt;Li. et al. 2013&lt;/a&gt;; &lt;a href=&quot;http://arxiv.org/abs/1309.2288&quot; rel=&quot;nofollow&quot;&gt;Knight &amp;amp; Walsh 2013&lt;/a&gt;), and the released gasses will be ejected by the Sun's radiation pressure. Larger comets, so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Sungrazing_comet&quot; rel=&quot;nofollow&quot;&gt;Sungrazing comets&lt;/a&gt;, can survive close passages, although they probably will be torn apart by tidal forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the comet comes so close that it crashes into the Sun (I'm not sure this has ever been observed, despite various blogs discussing this, but there's no reason it couldn't happen), it would reach the escape velocity of the Sun, i.e. $\sqrt{2 G M_\odot / R_\odot}\sim600\,\mathrm{km}\,\mathrm{s}^{-1}$. &lt;a href=&quot;https://www.newscientist.com/article/dn27954-what-would-happen-if-a-massive-comet-crashed-into-the-sun/&quot; rel=&quot;nofollow&quot;&gt;This New Scientist article&lt;/a&gt; claims that the impact would cause an explosion and possibly a &lt;a href=&quot;https://en.wikipedia.org/wiki/Coronal_mass_ejection&quot; rel=&quot;nofollow&quot;&gt;coronal mass ejection&lt;/a&gt;, but the underlying scientific article by &lt;a href=&quot;http://arxiv.org/abs/1507.05062&quot; rel=&quot;nofollow&quot;&gt;Brown et al. (2015)&lt;/a&gt; seems to indicate that the large ablation of the comet will result in less dramatic velocities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, some of the released gas would stay at the surface of the Sun, just like some of the Sun's &quot;own&quot; gas stays. And since the convection cells, or &lt;a href=&quot;https://en.wikipedia.org/wiki/Granule_%28solar_physics%29&quot; rel=&quot;nofollow&quot;&gt;granules&lt;/a&gt;, are quite shortlived (of order 10 minutes; &lt;a href=&quot;http://adsabs.harvard.edu/abs/1961ApJ...134..312B&quot; rel=&quot;nofollow&quot;&gt;Bahng &amp;amp; Schwarzschild 1961&lt;/a&gt;), the gas will soon be engulfed by the Sun, being carried deeply into the body of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The convective zone of the Sun is in the outer ~30% of the radius, and there isn't much mixing of material between that and the inner, radiative zone. In the last phases of the Sun's life, however, when core hydrogen burning stops and the Sun becomes a red giant, convection reaches so deeply into the Sun that some mixing takes place. This is called &lt;em&gt;dredge-up&lt;/em&gt;. Eventually, roughly half of the Sun will be ejected due to the large mass-loss during the red giant phase, while the other half becomes a white dwarf. If the dredge-up didn't happen, the comet material would all be a part of the ejected material, but due to the dredge-up, some of it will end up in the white dwarf.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-02-02T10:25:24.243" LastActivityDate="2016-02-02T10:25:24.243" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13478" PostTypeId="2" ParentId="13472" CreationDate="2016-02-02T08:51:11.907" Score="5" Body="&lt;p&gt;You would not actually be able to see that much, the sunlight is less than 4% of the strength at Earth's distance to the Sun. That is still enough to see though (Much brighter than Moon light at night in fact). The problem is the transparency of the atmosphere; we can not look in, you can not look out. You would only see a weak brown or red glow below the cloud cover.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is &lt;a href=&quot;http://www.nasa.gov/centers/goddard/multimedia/largest/lightning.html&quot; rel=&quot;noreferrer&quot;&gt;plenty of lightning! &lt;/a&gt;&#xA;Strong winds and thunder exist too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IhRHY.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IhRHY.jpg&quot; alt=&quot;lighting&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lightning on Jupiter.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="10172" LastEditDate="2016-02-02T09:45:05.830" LastActivityDate="2016-02-02T09:45:05.830" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13479" PostTypeId="1" AcceptedAnswerId="13480" CreationDate="2016-02-02T10:35:57.190" Score="5" ViewCount="114" Body="&lt;p&gt;From my understanding, the Hubble constant $H_0$ calculates from observed redshifts $z$ of distant galaxys against their proper distance $D$. The current value appears to be &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble%27s_law#Observed_values&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;test&quot;&gt;67.80(77) $\frac{km}{s}Mpc^{-1}$&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating the Hubble constant via the redshift, I assume one only wants those velocity contributions due to the expansion of the universe, and not those from the real movement of the galaxies within in cluster (&lt;em&gt;peculiar motion&lt;/em&gt;) or so. This, I also read in various online resources (which I can't recall right now).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other Hand, if your galaxy cluster is far enough away (eq. ~ 1 Gpc), one can neglect peculiar motion, which is in the order of 1000 $\frac{km}{s}$ (1000 $\frac{km}{s}$ / (1 Gpc $\times$ 67.80(77) $\frac{km}{s}Mpc^{-1}$) $\approx$ 1.4%)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, how would you try to correct for the peculiar motion, or is it really just neglected? Calculating all the gravitational components in each cluster? Another idea could be to assume that the galaxies within the cluster move randomly in respect to each other and the peculiar motion cancels out over the average? Any other possibilities?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt; that this question is partly a copy of &lt;a href=&quot;https://physics.stackexchange.com/questions/232929/naive-interpretation-of-hubbles-law&quot; title=&quot;another question of mine&quot;&gt;another question of mine&lt;/a&gt; at physics.stackexchange.com, which wasn't fully answered but commented to ask it here.&lt;/p&gt;&#xA;" OwnerUserId="10611" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-02-02T11:47:24.140" Title="What accounts into calculating the Hubble constant?" Tags="&lt;cosmology&gt;&lt;expansion&gt;&lt;galaxy-cluster&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13480" PostTypeId="2" ParentId="13479" CreationDate="2016-02-02T11:47:24.140" Score="4" Body="&lt;p&gt;I think &lt;a href=&quot;http://www.astronomy.ohio-state.edu/~dhw/A1143/notes4.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; might answer your question, I have quoted the important paragraph below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If peculiar velocities could have any value, then this would make Hubble’s law useless.&#xA;  However, peculiar velocities are typically only about 300 km/sec, and they very rarely exceed&#xA;  1000 km/sec.&#xA;  Hubble’s law therefore becomes accurate for galaxies that are far away, when H0d is much larger&#xA;  than 1000 km/sec.&#xA;  Furthermore, we can often estimate what a galaxy’s peculiar velocity will be by looking at the&#xA;  nearby structures that will be pulling on it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So it seems that in answer to your question, if the galaxy is far enough away then the peculiar motion can be neglected, or if its close by, a reasonable approximation found using estimates from nearby galaxies (in a cluster for example).&lt;/p&gt;&#xA;" OwnerUserId="10437" LastActivityDate="2016-02-02T11:47:24.140" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13481" PostTypeId="1" AcceptedAnswerId="13494" CreationDate="2016-02-02T13:11:43.620" Score="19" ViewCount="4419" Body="&lt;p&gt;Upon reading on this site (and many others), one can think that Mars might have supported life in a distant past (discovery of liquid water, valleys, mountains...). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, Venus is the perfect example of the consequences of the greenhouse effect. A planet where life is simply impossible because no heat can escape the planet's ultra-thick, CO2 filled atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, with a lot of exageration, one can say that both Mars and Venus could be the Earth's distant future. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Earth is actually facing global warming and greenhouse effect, I was wondering if the Earth would one day be like Venus ? Or if it is going to be like Mars, with no more magnetosphere and almost no atmosphere left ?&#xA;Or if it is not going anywhere near those two planets' fate ? &lt;/p&gt;&#xA;" OwnerUserId="9679" LastEditorUserId="9679" LastEditDate="2016-02-02T13:21:15.503" LastActivityDate="2016-02-02T21:52:52.300" Title="Is the Earth going to evolve towards Mars' fate or Venus' fate?" Tags="&lt;earth&gt;&lt;mars&gt;&lt;venus&gt;&lt;future&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13482" PostTypeId="2" ParentId="13481" CreationDate="2016-02-02T13:31:20.957" Score="10" Body="&lt;p&gt;Either or neither. It's impossible to tell from the present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If runaway climate change occurs, then yes, the conditions on Venus could be a potential analogue for the kind of environment on Earth due to the greenhouse effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars's atmosphere is assumed to have been much thicker in the past, otherwise it could not have sustained liquid water on the surface; it would have evaporated away. This water is required to explain the gorges and riverbeds we see on the Martian surface today. Researchers are still unsure where the atmosphere went, but the prime candidate is through top-loss to outer space, rather than sequestration through minerals on the surface (see &lt;a href=&quot;http://www.nasa.gov/feature/jpl/what-happened-to-early-mars-atmosphere-new-study-eliminates-one-theory&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The likelihood of such an atmospheric escape process becoming dominant on Earth is small due to the Earth's greater mass and strong relative magnetic field, which prevents ion escape. In fact, the dominant loss process on the earth is sequestration. Some estimates put the reservoirs of sequestered carbon from original CO$_2$ at 250000 times the size of the existing atmosphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alternative is that the Earth remains in its current steady state, or close to it, far in to the future. There are many mechanisms to enable such an equilibirum, such as oceanic sequestration of C0$_2$ highlighted above. There are also more contested theories such as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gaia_hypothesis&quot;&gt;Gaia hypothesis&lt;/a&gt;. This proposes that the global biological ecosystem on the Earth is self regulating, helping to maintain life enabling conditions on the Earth. It's pretty much a dead-cert that life doesn't exist on either the Martian or Venusian surfaces, so we can accept that if the Gaia hypoethesis is true, the Earth will never reach these conditions. &lt;/p&gt;&#xA;" OwnerUserId="941" LastEditorUserId="941" LastEditDate="2016-02-02T13:46:45.037" LastActivityDate="2016-02-02T13:46:45.037" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13483" PostTypeId="2" ParentId="13476" CreationDate="2016-02-02T14:07:02.590" Score="3" Body="&lt;p&gt;Kaufman &amp;amp; Freedman's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/142923153X&quot; rel=&quot;nofollow&quot;&gt;Universe&lt;/a&gt; is a very basic astronomy book aimed at non-astronomy science students or astronomy 1st year students, although for the latter it is perhaps a bit too basic. It covers introductions to all the topics you mention, except perhaps exoplanets (but my edition is from the late 90's; my guess is they have it covered too by now). It even gives suggestions for observing projects at the end of each chapter, along with exercises and extra literature. It is also very rich in pictures and explanatory diagrams.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2016-02-02T14:07:02.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13484" PostTypeId="2" ParentId="13476" CreationDate="2016-02-02T15:09:23.347" Score="2" Body="&lt;p&gt;Zeilik's &lt;a href=&quot;https://www.goodreads.com/book/show/288807.Introductory_Astronomy_and_Astrophysics&quot; rel=&quot;nofollow&quot;&gt;Introductory Astronomy and Astrophysics&lt;/a&gt; is worth investigating. It's more in depth than Kaufman &amp;amp; Freedman's Universe, mentioned in another answer, but requires a little more mathematical background such as basic calculus and trigonometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd also recommend looking at online courses from providers such as &lt;a href=&quot;https://www.coursera.org/course/introastro&quot; rel=&quot;nofollow&quot;&gt;Coursera&lt;/a&gt;, &lt;a href=&quot;https://www.khanacademy.org/science/cosmology-and-astronomy&quot; rel=&quot;nofollow&quot;&gt;Khan Academy&lt;/a&gt; and &lt;a href=&quot;https://www.edx.org/course/introduction-solar-systems-astronomy-asux-ast111x-0&quot; rel=&quot;nofollow&quot;&gt;edx&lt;/a&gt; (and &lt;a href=&quot;https://www.edx.org/course/evolving-universe-caltechx-ay1001x&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) for inspiration on how to structure such a course.&lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2016-02-02T15:09:23.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13485" PostTypeId="2" ParentId="13481" CreationDate="2016-02-02T15:23:43.327" Score="6" Body="&lt;p&gt;This is a question that cannot be accurately answered. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the closest to a honest and accurate answer would be that neither, since the Earth is going to &quot;evolve&quot; (curious choice of words) towards &lt;em&gt;the Earth's&lt;/em&gt; fate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The presence of humans means any purely physical projections need to be taken with a grain of salt. Our ability to influence the planet will only increase. With it, so will the range of possible future states.&lt;/p&gt;&#xA;" OwnerUserId="10641" LastActivityDate="2016-02-02T15:23:43.327" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13486" PostTypeId="2" ParentId="13458" CreationDate="2016-02-02T15:32:01.107" Score="1" Body="&lt;p&gt;Oxygen is an extremely eager element to bind to other elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was, in fact, a lot of oxygen. But most of it was stuck to carbon etc., and later tied up in various ores and minerals - in general, always stuck to something else, until biological processes started &quot;liberating&quot; it (and, interestingly, causing a bit of an ecological catastrophe as free oxygen was toxic to the very microorganisms that produced it. It was not a catastrophe when considered from our point of view, obviously.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TLDR, think of oxygen atoms like stray pieces of duct tape floating around. &lt;/p&gt;&#xA;" OwnerUserId="10641" LastActivityDate="2016-02-02T15:32:01.107" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13487" PostTypeId="2" ParentId="13474" CreationDate="2016-02-02T16:25:50.420" Score="4" Body="&lt;p&gt;&lt;strong&gt;There may be Sednoids there.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sednoid&quot;&gt;Sednoids&lt;/a&gt; are a hypothetical class of &quot;inner Oort Cloud objects&quot; named after their prototype, &lt;a href=&quot;https://en.wikipedia.org/wiki/90377_Sedna&quot;&gt;Sedna&lt;/a&gt;. Sedna's aphelion is ~936 AU, bringing it close to the inner boundary of the Oort Cloud. Sednoids may have aphelions ranging from about 100 AU to 1,000 AU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, only two Sednoids have beet detected to date, 90377 Sedna and &lt;a href=&quot;https://en.wikipedia.org/wiki/2012_VP113&quot;&gt;2012 VP&lt;sub&gt;113&lt;/sub&gt;&lt;/a&gt;. &lt;a href=&quot;http://arxiv.org/pdf/astro-ph/0404456v1.pdf&quot;&gt;Brown et al. 2004)&lt;/a&gt; suggested that ~500 may be detectable; surveys simply haven't tracked objects in that area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are Sendnoids where they are? Three ideas have but put forth:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A planet at ~70 AU scattered these objects into elliptical orbits.&lt;/li&gt;&#xA;&lt;li&gt;A close pass by a nearby star.&lt;/li&gt;&#xA;&lt;li&gt;Interactions with other stars in the Sun's original cluster.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These objects would fill in the space between the Kuiper Belt/scattered disc and the Oort Cloud itself.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2016-02-02T16:25:50.420" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13488" PostTypeId="1" CreationDate="2016-02-02T18:43:24.087" Score="33" ViewCount="3465" Body="&lt;p&gt;What resources are available to find the positions of planets, stars,&#xA;moons, artificial satellites, asteroids, and other heavenly bodies?&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorUserId="7982" LastEditDate="2020-12-27T13:57:15.603" LastActivityDate="2021-04-07T11:56:14.150" Title="Where can I find the positions of the planets, stars, moons, artificial satellites, etc. and visualize them?" Tags="&lt;amateur-observing&gt;&lt;positional-astronomy&gt;&lt;resource-request&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="19" ContentLicense="CC BY-SA 4.0" />
  <row Id="13489" PostTypeId="2" ParentId="13488" CreationDate="2016-02-02T18:43:24.087" Score="33" Body="&lt;p&gt;There are many resources online, so this is a community wiki&#xA;answer. Please feel free to add to it!&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If you want to visualize the stars/planets/etc (as viewed from Earth&#xA;or another location), you are looking for &lt;strong&gt;planetarium software&lt;/strong&gt;:&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Planetarium_software&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Planetarium_software&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you want accurate positions for stars/planets/etc, you are looking&#xA;for &lt;strong&gt;horizons&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/JPL_Horizons_On-Line_Ephemeris_System&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/JPL_Horizons_On-Line_Ephemeris_System&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://ssd.jpl.nasa.gov/?horizons&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ssd.jpl.nasa.gov/?horizons&lt;/a&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If you want to &lt;strong&gt;compute&lt;/strong&gt; star/planet positions yourself, you have&#xA;several options:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;SPICE (&lt;a href=&quot;http://naif.jpl.nasa.gov/naif/tutorials.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://naif.jpl.nasa.gov/naif/tutorials.html&lt;/a&gt;) will give you&#xA;results that match HORIZONS very closely. You can also use some of&#xA;SPICE's functionality online at&#xA;&lt;a href=&quot;http://wgc.jpl.nasa.gov:8080/webgeocalc/#NewCalculation&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://wgc.jpl.nasa.gov:8080/webgeocalc/#NewCalculation&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;SPK (Spice kernel) files are available at &lt;a href=&quot;https://naif.jpl.nasa.gov/pub/naif/generic_kernels/spk/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://naif.jpl.nasa.gov/pub/naif/generic_kernels/spk/&lt;/a&gt; -- while these are primarily intended for use with CSPICE, the format is documented and you can use it directly. It's described at &lt;a href=&quot;https://github.com/skyfielders/python-skyfield/issues/19&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/skyfielders/python-skyfield/issues/19&lt;/a&gt; and implemented in Python at &lt;a href=&quot;https://github.com/brandon-rhodes/python-jplephem/tree/master/jplephem&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/brandon-rhodes/python-jplephem/tree/master/jplephem&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you want to do the SPICE computations yourself (numerically solving the differential equations), see &lt;a href=&quot;https://astronomy.stackexchange.com/a/13491/21&quot;&gt;https://astronomy.stackexchange.com/a/13491/21&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You might also want to use an n-body simulator to do the computations yourself: &lt;a href=&quot;https://physics.stackexchange.com/questions/25241/what-open-source-n-body-codes-are-available-and-what-are-their-features&quot;&gt;https://physics.stackexchange.com/questions/25241/what-open-source-n-body-codes-are-available-and-what-are-their-features&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;IAU SOFA (&lt;a href=&quot;http://www.iausofa.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.iausofa.org/&lt;/a&gt;) will give you the International&#xA;Astronomical Union's &amp;quot;official&amp;quot; libraries to compute positions.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;VSOP theory (&lt;a href=&quot;https://en.wikipedia.org/wiki/VSOP_%28planets%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/VSOP_%28planets%29&lt;/a&gt;) is&#xA;another option.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;PyEphem (&lt;a href=&quot;https://en.wikipedia.org/wiki/PyEphem&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/PyEphem&lt;/a&gt;), which derives&#xA;from MIT's libastro library&#xA;(&lt;a href=&quot;https://stuff.mit.edu/afs/athena/project/xephem/src/xephem-3.5.2/libastro/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://stuff.mit.edu/afs/athena/project/xephem/src/xephem-3.5.2/libastro/&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Skyfield: &lt;a href=&quot;https://github.com/skyfielders/python-skyfield&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/skyfielders/python-skyfield&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;libnova (&lt;a href=&quot;http://libnova.sourceforge.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://libnova.sourceforge.net/&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you want to treat planetary orbits as simple ellipses and ignore perturbations, you can find orbital elements at &lt;a href=&quot;https://ssd.jpl.nasa.gov/txt/p_elem_t1.txt&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ssd.jpl.nasa.gov/txt/p_elem_t1.txt&lt;/a&gt; but you'll probably want to visit &lt;a href=&quot;https://ssd.jpl.nasa.gov/?bodies#elem&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ssd.jpl.nasa.gov/?bodies#elem&lt;/a&gt; first&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Note that these computational libraries don't always agree with each&#xA;other or with planetarium software:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://stackoverflow.com/questions/16293146&quot;&gt;https://stackoverflow.com/questions/16293146&lt;/a&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If you want a &lt;strong&gt;catalog&lt;/strong&gt; of stars/etc, NOMAD&#xA;(&lt;a href=&quot;http://www.usno.navy.mil/USNO/astrometry/optical-IR-prod/nomad&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.usno.navy.mil/USNO/astrometry/optical-IR-prod/nomad&lt;/a&gt;) and GAIA (&lt;a href=&quot;http://gea.esac.esa.int/archive/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://gea.esac.esa.int/archive/&lt;/a&gt;) are&#xA;the largest such catalogs (about 1 billion entries each, lots of overlap). The NOMAD link also provides references to smaller catalogs.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For high-resolution &lt;strong&gt;astrophotography&lt;/strong&gt; try ALADIN (&lt;a href=&quot;http://aladin.u-strasbg.fr/#AladinLite&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://aladin.u-strasbg.fr/#AladinLite&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you want to find artificial satellites and spacecraft, start at&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Two-line_element_set&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Two-line_element_set&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Transits&lt;/strong&gt; of Mercury and Venus across the Sun? &lt;a href=&quot;http://eclipse.gsfc.nasa.gov/transit/transit.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://eclipse.gsfc.nasa.gov/transit/transit.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Planetary &lt;strong&gt;conjunctions&lt;/strong&gt;?&lt;br /&gt;&#xA;&lt;a href=&quot;http://laps.noaa.gov/albers/ast/conjun/conjun.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://laps.noaa.gov/albers/ast/conjun/conjun.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Occultations&lt;/strong&gt; of stars by asteroids? The International Occultation&#xA;Timing Association (IOTA) at &lt;a href=&quot;http://occultations.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://occultations.org/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Create your own &lt;strong&gt;Sun-Earth-Moon model&lt;/strong&gt; by using a ~17 inches (~44 cm) beach ball or box as the Sun, 4-5 mm object like a pencil eraser as the Earth and a 1 mm sized drill bit or other object to represent the Moon. Then, find a 150-foot string or tape measure. The sun and earth are ~154 feet (~46m) apart. The earth and moon are ~4.7 inches (12 cm) apart.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user21" LastEditorUserId="37191" LastEditorDisplayName="user21" LastEditDate="2021-04-07T11:56:14.150" LastActivityDate="2021-04-07T11:56:14.150" CommentCount="7" CommunityOwnedDate="2016-02-02T18:43:24.087" ContentLicense="CC BY-SA 4.0" />
  <row Id="13491" PostTypeId="2" ParentId="2416" CreationDate="2016-02-02T19:12:21.263" Score="3" Body="&lt;p&gt;You've probably long since moved on, but, just for reference, the &#xA;initial conditions HORIZONS uses are mentioned (&quot;header.431_572&quot;) in &#xA;&lt;a href=&quot;ftp://ssd.jpl.nasa.gov/pub/eph/planets/Linux/README.txt&quot; rel=&quot;nofollow&quot;&gt;ftp://ssd.jpl.nasa.gov/pub/eph/planets/Linux/README.txt&lt;/a&gt; but the only &#xA;place I could find them in &quot;table form&quot; is in my own git repository: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/header.431_572&quot; rel=&quot;nofollow&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/header.431_572&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The values are explained in &#xA;&lt;a href=&quot;http://ilrs.gsfc.nasa.gov/docs/2014/196C.pdf&quot; rel=&quot;nofollow&quot;&gt;http://ilrs.gsfc.nasa.gov/docs/2014/196C.pdf&lt;/a&gt; starting on page 39, &#xA;&quot;VI. Initial Conditions and Constants&quot;, especially in the tables &#xA;starting with Table 4 on page 47 and ending with Table 13 on page 74. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've written scripts to setup the initial conditions and numerically &#xA;solve the differential equations using Mathematica, so the following &#xA;may be helpful: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/README&quot; rel=&quot;nofollow&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/README&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-header-values.pl&quot; rel=&quot;nofollow&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-header-values.pl&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-integrate.m&quot; rel=&quot;nofollow&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-integrate.m&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I reduce the step size sufficiently (Mathematica's default step &#xA;size is too large), my results closely match those of HORIZONS: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-integrate-compare.m&quot; rel=&quot;nofollow&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-integrate-compare.m&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2016-02-02T19:12:21.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13492" PostTypeId="5" CreationDate="2016-02-02T20:32:14.477" Score="0" Body="&lt;p&gt;This tag is for questions related to Astropy (&lt;a href=&quot;http://www.astropy.org&quot; rel=&quot;nofollow&quot;&gt;http://www.astropy.org&lt;/a&gt;). &quot;The Astropy Project is a community effort to develop a single core package for Astronomy in Python and foster interoperability between Python astronomy packages.&quot;&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="38" LastEditDate="2016-02-03T21:10:50.483" LastActivityDate="2016-02-03T21:10:50.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13493" PostTypeId="4" CreationDate="2016-02-02T20:32:14.477" Score="0" Body="Questions related to Astropy, a community Python package for Astronomy" OwnerUserId="354" LastEditorUserId="354" LastEditDate="2016-02-04T11:25:38.370" LastActivityDate="2016-02-04T11:25:38.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13494" PostTypeId="2" ParentId="13481" CreationDate="2016-02-02T20:45:56.723" Score="10" Body="&lt;p&gt;If you are asking about short-term effects related to human's effect on the climate, the answer is (obviously) unclear. But in the very long term, Earth is likely to evolve to a more Venus-like state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over the next billion years or so, the Sun's luminosity will slowly increase, which will heat Earth's surface. As a result, more water vapor will evaporate into the atmosphere. Since water vapor is a greenhouse gas, this will compound the heating. It's not entirely clear what the new equilibrium temperature will be after this runaway greenhouse effect runs its course. But it will clearly be more Venus-like than Mars-like.&lt;/p&gt;&#xA;" OwnerUserId="10648" LastActivityDate="2016-02-02T20:45:56.723" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13495" PostTypeId="2" ParentId="13481" CreationDate="2016-02-02T21:52:52.300" Score="3" Body="&lt;p&gt;Since the Sun's temperature will continue to increase and it is estimated that in ~2 billion years it will be so hot on Earth that life will be impossible, I'd say it's going to evolve more toward Venus.&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2016-02-02T21:52:52.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13496" PostTypeId="2" ParentId="1767" CreationDate="2016-02-02T22:37:54.573" Score="-3" Body="&lt;p&gt;I will say, two years late I suppose, (lol) that on a true scientific basis, we cannot factually say yes or no to the question. You have been given excellent theories on the matter, but in order to answer that question, one would need to document and follow the life of one single atom, how it becomes whatever it becomes, and track it all the way to it's exsistance in a human body... Which we cannot do at this time. Perhaps one day? :) But not as of yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is one I've looked into thousands of times, but you must always look to the basis of what is fact. Fact is a verifiable observation and as no one has tracked and witnessed an atom from a supernova make its way to the human body, no one can merely answer 'yes' or 'no'. Not even a 'yes, mostly'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Odds are they are the same in some form or another, but that's using the word 'odds'. You have asked an excellent question... But it cannot be answered and probably won't be in our lifetime.&lt;/p&gt;&#xA;" OwnerUserId="10652" LastActivityDate="2016-02-02T22:37:54.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13497" PostTypeId="2" ParentId="13476" CreationDate="2016-02-02T23:13:19.393" Score="1" Body="&lt;p&gt;Some years ago I did an astronomy course for non astronomers at our university where they used &lt;em&gt;Koupelis' &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/144968775X&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;In Quest Of The Universe&lt;/strong&gt;&lt;/a&gt;&lt;/em&gt;. Although it changed a bit from my edition (fourth) to the current (seventh), it still seems to cover the same subjects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It talks about the history, modern astronomy, measurement, the planets, stars, galaxies, etc. The book also has exercises with answers for self-study. Using the index I could find most of the things you wanted the book to discuss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book also has a &lt;a href=&quot;http://physicalscience.jbpub.com/starlinks/7e/&quot; rel=&quot;nofollow&quot;&gt;website&lt;/a&gt; with different resources to help the students to study the material, such as quizzes, chapter outlines, flashcards, etc.&lt;/p&gt;&#xA;" OwnerUserId="9562" LastActivityDate="2016-02-02T23:13:19.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13498" PostTypeId="1" AcceptedAnswerId="13504" CreationDate="2016-02-02T23:52:39.193" Score="12" ViewCount="4468" Body="&lt;p&gt;I noticed that the atmospheres of Earth and Mars have a little bit of argon in them (1% to 2%). I checked Venus, too, which has 0.007% argon, but that's still more than any other noble gas in the Venusian atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I checked the 4 outer gas planets, plus the moon Titan, and for the most part could not find Argon listed in their atmospheric composition. Jupiter is listed as having &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Jupiter#Chemical_composition&quot;&gt;a very tiny trace amount&lt;/a&gt; of Argon that is hard to interpret because it's listing it as relative to helium and then relative to Jupiter/Sun, whatever that means. The moon Titan is listed as having &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Titan&quot;&gt;trace argon&lt;/a&gt; without specifying any numbers. The others did not mention Argon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So why do the rocky planets tend to have argon, but the gaseous planets don't?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why argon specifically?&lt;/strong&gt; Presumably helium is too light and will float to the top and get knocked away by the solar wind. However, there is one noble gas between helium and argon, namely, neon. So why don't we have neon in our atmosphere instead of argon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Maybe the outer planets &lt;strong&gt;do&lt;/strong&gt; have Argon, but it all sank to the bottom and thus we don't detect it? I'm also curious about where Argon lies in the &quot;food chain&quot; of nuclear fusion for an average star like our Sun. I have a hard time seeing how Argon-40 would be more populous than Neon-20 though.&lt;/p&gt;&#xA;" OwnerUserId="7449" LastEditorUserId="7449" LastEditDate="2016-02-03T01:08:10.483" LastActivityDate="2018-01-25T06:07:51.733" Title="Why argon instead of another noble gas?" Tags="&lt;planetary-atmosphere&gt;&lt;gas&gt;&lt;elemental-abundances&gt;&lt;element&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13499" PostTypeId="1" AcceptedAnswerId="13502" CreationDate="2016-02-03T00:07:30.850" Score="5" ViewCount="622" Body="&lt;p&gt;As I see it, most of the stars in the galactic bulge are Population I stars. However, as one moves farther from the galactic bulge, star metallicity decreases. In fact, halo stars are almost entirely Population II stars. Why is this?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2016-02-03T21:25:24.530" Title="Why are stars more metallic closer as you move closer to the galactic bulge?" Tags="&lt;star&gt;&lt;galaxy&gt;&lt;galactic-dynamics&gt;&lt;galaxy-center&gt;&lt;metal&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13500" PostTypeId="2" ParentId="13498" CreationDate="2016-02-03T00:15:19.623" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;why Argon specifically?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Both helium and neon are pretty lightweight, tend to vaporize easily even at low temperatures, and are chemically inert. For all these reasons combined, they tend to not get trapped when planets are formed - and when they do get trapped they leak out easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Argon is just heavy enough to not escape easily into space, so some fraction of it may stick around in an atmosphere for a longer time.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2016-02-03T00:15:19.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13501" PostTypeId="2" ParentId="13455" CreationDate="2016-02-03T00:41:17.033" Score="8" Body="&lt;p&gt;It's unlikely that either Mercury or Venus could have moons to begin with. Both of these planets are pretty close to the Sun — and in general, this prevents moons from finding stable orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a moon were too close to the planets, it would fall within the &lt;a href=&quot;https://en.wikipedia.org/wiki/Roche_limit&quot; rel=&quot;noreferrer&quot;&gt;Roche limit&lt;/a&gt; and be torn apart by tidal forces. If a moon were too far from the planets, it would fall outside the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;noreferrer&quot;&gt;Hill sphere&lt;/a&gt; and be pulled into the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The zones where moons around these planets could be stable over billions of years is probably so narrow that no body was ever captured into orbit when the planets were first being accreted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it gets even more complicated in Mercury's case. Its Hill sphere extends about 3000 miles (~4828 km) — that is, any satellite more than 3000 miles from the planet will be pulled away by the Sun's gravity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, its Roche limit extends to 3600 miles (~5794 km), so any satellite within this would be ripped apart. Venus lacks this excuse, but the region between where the Sun would pull away a satellite and the Roche limit is probably too small for a satellite to happen to have formed there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, you have to realize that inner planets generally don't have moons. Solar wind pressure usually sends dust and debris flying away from the planets, so they don't have any spare material for moons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that Earth's Moon is thought to have formed from a collision with the Mars-sized planet Theia. This makes Earth an oddity in itself. Meanwhile, Mars is thought to have captured Phobos and Deimos from the asteroid belt — so Mars and its moons didn't condense from the same dust mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With all this, it'd be extremely surprising if we found moons on either Mercury or Venus.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="1678" LastEditDate="2016-02-03T07:59:33.073" LastActivityDate="2016-02-03T07:59:33.073" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13502" PostTypeId="2" ParentId="13499" CreationDate="2016-02-03T00:49:46.597" Score="3" Body="&lt;p&gt;It has to do with the formation of the Milky Way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the beginning, the Milky Way was much more spherical than it is now - perhaps closer to what an elliptical galaxy is like than a spiral galaxy. Population III stars would have formed first, then quickly died out. Next came Population II stars. They formed when the galaxy was still somewhat spherical, and so they tend to inhabit the galactic spheroid/halo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, the rotation of the Milky Way flattened out much of the remaining gas and dust, and some of the stars. When younger stars formed, they formed in the flatter disk, nearer to the center. The disk itself became smaller that the spheroid/halo. Thus, the younger Population I stars are found in the galactic disk, and are closer in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No more stars will form in globular clusters; they are relatively dust-free and contain old, Population II stars. The same is true for the galactic halo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Source: &lt;a href=&quot;http://www.ifa.hawaii.edu/~barnes/ast626_95/pcmw.html&quot; rel=&quot;nofollow&quot;&gt;Populations &amp;amp; Components of the Milky Way&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The galactic bulge itself contains &lt;a href=&quot;http://arxiv.org/pdf/1511.07438v1.pdf&quot; rel=&quot;nofollow&quot;&gt;several populations of stars&lt;/a&gt;. Some may have come from the halo and thick disk (thus being metal-poor) while others may have formed together more recently from the thin disk itself (thus being metal-rich).&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-02-03T16:49:17.653" LastActivityDate="2016-02-03T16:49:17.653" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13503" PostTypeId="1" CreationDate="2016-02-03T02:05:17.887" Score="2" ViewCount="597" Body="&lt;p&gt;We often hear our solar system has properties for life. For example, it is far away from the center of Galaxy. It has Jupiter far away to protect us from asteroids. It has optimum size of territorial planets, plate activity on Earth, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But besides that, compared with similar stars, does our sun itself also have properties that make life possible? For example, does it have extra stable brightness? Or does it emit less high energy radiation?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="38" LastEditDate="2016-02-06T08:07:25.437" LastActivityDate="2016-02-06T08:07:25.437" Title="Compared with similar stars, does the sun itself have properties that make life on Earth possible?" Tags="&lt;solar-system&gt;&lt;the-sun&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13504" PostTypeId="2" ParentId="13498" CreationDate="2016-02-03T06:23:54.290" Score="13" Body="&lt;p&gt;Doing a bit of reading up on this, I might have an answer, though credit where credit is due, the answer isn't really mine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.reddit.com/r/askscience/comments/3wsy99/why_is_neon_so_rare_on_earth/&quot; rel=&quot;noreferrer&quot;&gt;https://www.reddit.com/r/askscience/comments/3wsy99/why_is_neon_so_rare_on_earth/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the planets coalesced, it's likely that there was very little ices/gas around the inner planets when they formed and the Earth's atmosphere and water (CH4, NH3, CO2 and H20 being the 4 most common outside the &lt;a href=&quot;https://en.wikipedia.org/wiki/Frost_line_%28astrophysics%29&quot; rel=&quot;noreferrer&quot;&gt;frost line&lt;/a&gt; ices).  These likely came from asteroids and meteors that formed outside the frost line and later crashed onto earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neon is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements&quot; rel=&quot;noreferrer&quot;&gt;5th most common element&lt;/a&gt; in the milky-way but because all noble gases have very low freezing points, it's likely not be very common even on comets or meteors for the same reason that water or CO2 aren't common inside the frost line, Neon, and other noble gases likely stay free and don't collect on comets or meteors in high amounts.   (I looked, but couldn't find an article to verify that).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if comets have low noble gas content, then we have to look for an alternate source.  With that in mind, and going back to the first link, Argon is produced by radioactive decay of Potassium 40 and that would explain it's relative abundance compared to the more common noble gas, Neon.   Helium (Alpha particles) is also produced inside the earth and Radon is is too in small amounts but Radon also decays - that's not related to your question though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Argon on planets comes primarily from Potassium 40, you should expect the amount of Argon to have a roughly similar ratio to the amount of potassium on a planet and not be relative to the percentage of atmosphere.   A 2nd factor, how much gets blown off the planet over long periods of time is a factor too.   Venus in general should be able to retain much of it's Argon &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_escape&quot; rel=&quot;noreferrer&quot;&gt;based on atomic weight&lt;/a&gt; (40) similar to CO2 (44), but if it loses even a tiny percentage of it's Argon over time, that would be a factor too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to see if this is possible, I should run some numbers, but I warn you, my math can be a little rusty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Potassium is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Abundance_of_elements_in_Earth&amp;#39;s_crust&quot; rel=&quot;noreferrer&quot;&gt;7th most common&lt;/a&gt; element in the Earth's lithosphere at about 0.26% and about 0.0117% of that Potassium is Potassium 40.   Using a &lt;a href=&quot;https://www.quora.com/What-is-the-total-mass-of-the-Earths-crust&quot; rel=&quot;noreferrer&quot;&gt;very rough estimate&lt;/a&gt; of $2.3 \times 10^{19}$ tonnes for the Earth's crust, $(2.3 \times 10^{19}) \times (2.5 \times 10^{-3}) \times (1.17 \times 10^{-4}) =$ about $6.7 \times 10^{12}$ or 6.7 trillion tons of Potassium 40 currently in the Earth's crust. (There's probably a fair bit more in the mantle, so these numbers are rough)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a &lt;a href=&quot;https://www.cns-snc.ca/media/uploads/teachers/K40_4pg_10_06.pdf&quot; rel=&quot;noreferrer&quot;&gt;half life&lt;/a&gt; of about 1.248 billion years, that's sufficient time for over 3 half lives if we start after the late heavy bombardment, which suggests a bit over 7/8ths of the original Potassium 40 in the Earth's crust has decayed into Argon 40, so there should be, given the age of the Earth and abundance of Potassium 40, a bit over 7 times 6.7 trillion tons or, lets ballpark and say a bit over 50 trillion tons of Argon that formed on earth by Potassium decay.  (I'm ignoring any that might have been produced prior to the late heavy bombardment, cause I assume that could have blown some of the atmosphere off the earth or heated the atmosphere enough for the sun to blow some of it off).    Also, doing a bit of research, only 11% of the Potassium 40 decays in to Argon 40, 89% undergoes beta decay into Calcium 40, so for this to work, there would need to be a fair bit more Potassium in the earth than I estimated, but that's still likely the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the atmosphere is about 5,140 trillion tons, and 1.288% of that (By mass, not volume) = about 66 trillion tons, so the Argon we should expect from Potassium 40 decay and the amount of Argon in the atmosphere are pretty close.   Some Argon gas might have escaped and some should still be trapped inside the earth but the numbers are close enough to work and I think that's very likely the answer.  It also suggests that the Earth has lost relatively little Argon to space, which also fits with the Atmospheric Escape article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 2nd way to look at this is that Argon 40 makes up 99.6% of the Argon in the atmosphere and Stellar Nucleosis likely wouldn't account for a ratio anywhere close to that (not a typical stellar link but &lt;a href=&quot;https://en.wikipedia.org/wiki/Potassium-40#Potassium.E2.80.93argon_dating&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; says Argon 36 is the most common isotope).   The decay of Potassium 40 does explain the 99.6% Argon40 ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we apply a similar estimate to Venus, with Venus atmosphere about 94 times the mass of Earth's, and we assume a similar amount of Argon-40 being produced in Venus' crust we could roughly expect 1.28%/60 or about 0.02% Argon by mass in Venus's atmosphere or perhaps, if the Earth lost a pretty high share of it's lighter crust elements after the giant impact, we might expect a bit more than that on Venus, perhaps 0.03% or 0.04% as a rough estimate.   Using your number of 0.007%, that's lower than I calculate it should be, but Venus could have lost a higher share of it's Argon than Earth and it also might be slower to release trapped gas inside it's crust than Earth because it doesn't have plate tectonics, so the number for Venus looks &quot;about right&quot; too.    It's the Potassium 40 in the crust.   I'm convinced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interesting question.  I learned something researching it. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2018-01-25T06:07:51.733" LastActivityDate="2018-01-25T06:07:51.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13506" PostTypeId="2" ParentId="13503" CreationDate="2016-02-03T08:17:34.840" Score="3" Body="&lt;p&gt;One thing that is, if not unique, unusual about the Sun is the size of it.&#xA;It belongs to the top 5% brightest stars, meaning that 19 out of 20 stars are smaller than the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SfyPJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SfyPJ.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having a Jupiter size planet is certainly not unique, and plate tectonics we know nothing about for exoplanets.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="10172" LastEditDate="2016-02-04T07:38:38.493" LastActivityDate="2016-02-04T07:38:38.493" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13508" PostTypeId="2" ParentId="13503" CreationDate="2016-02-03T10:24:00.827" Score="0" Body="&lt;p&gt;In terms of absolute size, the sun is at the smaller end of the spectrum (the largest stars can be over 100x the mass of the Sun) but due to the skewed mass distribution in stars, the number of these massive stars is very low and the sun actually lies in the top 5% of stars by mass. The size of the sun is definitely not unique, nor are any of its properties really, but it's size is definitely the reason why it is more stable, the more massive stars burn their fuel much faster and thus have very short lifespans in comparison with our sun, something that wouldn't be helpful to conducing life.&lt;/p&gt;&#xA;" OwnerUserId="10437" LastActivityDate="2016-02-03T10:24:00.827" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13509" PostTypeId="1" CreationDate="2016-02-03T10:41:09.473" Score="10" ViewCount="698" Body="&lt;p&gt;The axial tilts of asteroids seem to vary randomly (let me know if this premise is wrong), while the planets have a strong tendency to rotate in the same way. If planets were formed by colliding asteroids, shouldn't the sum of random tilts result in random planetary rotation? Of course other factors are important, like angle and velocity of impacts, YORP-effect, centrifugal breakup and whatnot, but how could that taken together have any systematic effect on rotation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ceres does behave with 4° tilt, but the other of the first discovered asteroids have tilts like 84°, 50°, 42°. Dust particles (and gas molecules if applicable) surely rotate randomly. The Solar nebula had a net spin which gravity and friction has manifested in the orbits of the planets. But shouldn't the netting of rotation be individual for each planet, with uncorrelated tilts, as orbital orientation is for each star?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2019-05-01T01:04:18.793" Title="Why do planets tend to rotate in the same direction although they have formed from tumbling asteroids?" Tags="&lt;planet&gt;&lt;asteroids&gt;&lt;rotation&gt;&lt;planetary-formation&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13510" PostTypeId="1" AcceptedAnswerId="13513" CreationDate="2016-02-03T11:00:27.983" Score="2" ViewCount="157" Body="&lt;p&gt;I recently saw this &lt;a href=&quot;https://www.youtube.com/watch?v=LS-VPyLaJFM&amp;amp;feature=youtu.be&quot; rel=&quot;nofollow&quot;&gt;video&lt;/a&gt;. The creator of the video says that one of the reasons for non-existence of life on planets around red dwarfs might be that these stars can increase &lt;strong&gt;luminosity roughly by a factor of 2&lt;/strong&gt; in a matter of seconds which will boil off water from planets. He also said something about Helium/Hydrogen convection belts which are responsible for this. Can someone please explain this phenomenon? &lt;/p&gt;&#xA;" OwnerUserId="10657" LastEditorUserId="941" LastEditDate="2016-02-03T18:03:21.647" LastActivityDate="2019-03-06T17:28:24.027" Title="Red dwarf variation in Luminosity" Tags="&lt;life&gt;&lt;luminosity&gt;&lt;red-dwarf&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13511" PostTypeId="2" ParentId="13509" CreationDate="2016-02-03T12:25:28.500" Score="14" Body="&lt;p&gt;You are right that the tilt of the asteroids are distributed in very random way, and that the rotation of the Solar nebula is a minor contributor to that tilt, and only skews it a little.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you are not right that randomness simply adds up. The randomness does in fact cancel out more and more when you combine a large amount of asteroids, until the rotation of the nebula becomes the dominant factor. This is related to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Law_of_large_numbers&quot;&gt;Law of large numbers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, throw a dice. The outcome is random. Throw 10 dices, calculate their sum, and divide by 10. Not so far from the average any more? You can do the same thing with thousands of dices, or millions of asteroids. When the number of asteroids that form an object is really high, the tilt is not going to be far from the average value, determined by the nebula's rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same argument goes for inclination, and the fact that even though the planets' orbits are elliptical, they are not as far from circular that a random orbit would be.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastActivityDate="2016-02-03T12:25:28.500" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13513" PostTypeId="2" ParentId="13510" CreationDate="2016-02-03T16:56:49.123" Score="4" Body="&lt;p&gt;The creator seems to be referring to &lt;a href=&quot;https://en.wikipedia.org/wiki/Flare_star&quot; rel=&quot;nofollow&quot;&gt;flare stars&lt;/a&gt;. Flares may be magnetic in origin, like various manifestations of the Sun's magnetic field. These flares can be quit luminous across the electromagnetic spectrum, including in the x-ray wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some red dwarfs are flare stars, though flare activity may be largely restricted to a small part of a red dwarf's lifetime. However, during this period, life on a planet orbiting close to a red dwarf could be in serious danger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reference to convection may be that &lt;a href=&quot;http://arxiv.org/abs/0901.1659&quot; rel=&quot;nofollow&quot;&gt;red dwarfs are generally fully convective&lt;/a&gt;, unlike other stars. Magnetic fields in stars are generated via &lt;a href=&quot;https://en.wikipedia.org/wiki/Dynamo_theory&quot; rel=&quot;nofollow&quot;&gt;the dynamo effect&lt;/a&gt;; in red dwarfs, full convectivity can play a huge role in generating strong magnetic fields, and thus stellar flares.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-02-03T21:18:30.053" LastActivityDate="2016-02-03T21:18:30.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13514" PostTypeId="2" ParentId="13431" CreationDate="2016-02-03T17:42:36.013" Score="1" Body="&lt;p&gt;This is my first post, and I've accidentally deleted almost all I've written! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;a later edit with an answer&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The RA of a culminating star gives you the local sidereal time and comparison with Greenwich Sidereal Time gives your local longitude by noting the difference between the two. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If local sidereal time is less than Greenwich, your position is west of the prime meridian at Greenwich; if it is more, then your position is east of Greenwich.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Four minutes of Right Ascension = One degree of longtitude. &#xA;One minute of RA = 15 arc minutes of longitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A worked example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Greenwich Sidereal time 17:42&lt;br&gt;&#xA;          Local Sidereal time 16:36 hours &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      Greenwich is ahead (more) than local, therefore local is West.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The difference = 42 minutes plus 24 minutes &lt;/p&gt;&#xA;&#xA;&lt;p&gt;= 1 hour 6 minutes of RA to the west of Greenwich&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, 60 minutes = 15° longitude&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add  4 minutes =   1°   longitude&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add   2 minutes =   0°30' longitude&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= longitude 16°30'W &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is accurate to 0°05' of the usually given figure for my location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.abecedarical.com/javascript/script_clock.html&quot; rel=&quot;nofollow&quot;&gt;http://www.abecedarical.com/javascript/script_clock.html&lt;/a&gt; will show GST &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use a Sidereal Clock app on my Android tablet, set to show both GST and local sidereal time at my location. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;information for background&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the historical background for the search for a reliable method of calculating longitude when at sea&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.thegreenwichmeridian.org/tgm/articles.php?article=9&quot; rel=&quot;nofollow&quot;&gt;The Quest for Longitude and the Rise of Greenwich – a Brief History&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I've edited this several times because I'm finding my way around how this works and made several editing mistakes! I've also tried to add another link, and got an error message, but no explanation. Is there a limit to the number of hyperlinks permitted?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="10625" LastEditorUserId="10625" LastEditDate="2016-02-04T09:16:43.253" LastActivityDate="2016-02-04T09:16:43.253" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13515" PostTypeId="1" CreationDate="2016-02-03T18:38:39.390" Score="2" ViewCount="542" Body="&lt;p&gt;How can we have a singularity, without a black hole around it? How would one form? What does it do?&lt;/p&gt;&#xA;" OwnerUserId="9645" LastActivityDate="2016-05-19T17:44:10.847" Title="How can a naked singularity be possible?" Tags="&lt;black-hole&gt;&lt;cosmology&gt;&lt;singularity&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13516" PostTypeId="5" CreationDate="2016-02-03T19:51:46.657" Score="0" Body="&lt;p&gt;In general relativity there are two types of singularities. Coordinate singularities are associated with a specific way of ascribing space-time coordinates to each event, and refer to events where one of the coordinates becomes infinite as your approach it. For instance the event horizon of a black hole is of this kind, and lies infinitely far in the future in the natural coordinates of a distant observer, but is a perfectly normal point in the coordinates natural to an infalling observer.  Curvature singularities are points in the topological closure of space-time, such that the curvature of space-time (a scalar quantity independent of coordinates) tends to infinity as you approach them.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="21928" LastEditDate="2020-12-13T16:18:35.267" LastActivityDate="2020-12-13T16:18:35.267" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13517" PostTypeId="4" CreationDate="2016-02-03T19:51:46.657" Score="0" Body="Questions regarding a point in space-time where matter is infinitely dense." OwnerUserId="9645" LastEditorUserId="31410" LastEditDate="2020-12-12T22:52:43.580" LastActivityDate="2020-12-12T22:52:43.580" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13519" PostTypeId="2" ParentId="13499" CreationDate="2016-02-03T21:25:24.530" Score="2" Body="&lt;p&gt;The bulge population is old, older than 10 billion years. Its stars have a broad range of metallicities, but are more metal rich than population II, have an average close to solar metallicity and a significant fraction more metal rich than the Sun. The basic idea is that the bulge population is one that formed very quickly, with a high rate of gas infall and star formation. These properties lead to rapid ISM metal-enrichment (from supernovae and high mass AGB stars), a population with a broad range of, but high average, metallicity. An example paper on the subject would be Ferreras et al. (2003). &lt;a href=&quot;http://m.mnras.oxfordjournals.org/content/345/4/1381&quot; rel=&quot;nofollow&quot;&gt;http://m.mnras.oxfordjournals.org/content/345/4/1381&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The halo population is older and formed at a relatively low rate (hence with a low ISM metal enrichment rate) before the Galaxy formed a disk. Hence the stars are still in a pseudo-spherical distribution and metal-poor.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-02-03T21:25:24.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13520" PostTypeId="2" ParentId="13503" CreationDate="2016-02-04T03:51:47.547" Score="3" Body="&lt;p&gt;Other than being kind of in the &quot;Goldilocks&quot; range, meaning, not too big, not too small, (not to be confused with Goldilocks-zone, that's something else), there's nothing special about our sun.  There might be something a little unusual about Jupiter's movement over the last 4 billion years and perhaps, something unusual in the just right formation of our moon that stabilizes the Earth's wobble some, but there's nothing special about the sun beyond being a good size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as far as the &quot;center&quot; vs &quot;outside&quot; of the galaxy, I've heard that too but I'm not sure how well established that theory is.  The &quot;OK&quot; range might be fairly broad.  Certainly you wouldn't want to be too close to a large star that goes nova, and you &quot;might&quot; not want to be on the outer edge of the galaxy assuming the galaxy has a kind of protective shield like the sun's &lt;a href=&quot;https://en.wikipedia.org/wiki/Heliosphere&quot; rel=&quot;nofollow&quot;&gt;heliosphere&lt;/a&gt;, though I'm not sure how important that is and I'm not sure anyone knows.   It's possible there's an ideal ring within the milkyway, but I'm not sure anyone knows that for sure or knows how large it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A star can't be too large because there's probably a minimum &lt;a href=&quot;https://en.wikipedia.org/wiki/Main_sequence#Lifetime&quot; rel=&quot;nofollow&quot;&gt;lifetime&lt;/a&gt; for an Earth like planet to form.   With a ratio of roughly the power of 2.5 a star twice the mass of our sun would only have a lifespan of 1.5 billion years before it goes red-giant.  If you figure it takes 0.5-1.0 billion years for a planet to cool down enough to be a good place for life and for the early bombardment periods to complete and time for the planet to cool and form a magnetic field, 1.5 billion years is pretty tight to allow time for oxygen formation and evolution, so about 2 solar masses is close to the maximum ceiling for earth like planet candidate and that ignores the sun going through increasing luminosity as it approaches it's red giant stage.  Ideally you might not want to go too much over 1.5 solar masses (about 4 billion years of main sequence).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you could probably go a fair bit smaller than our sun, cause that would extend the lifespan significantly, but if you get too small, &quot;Flare stars, coronal mass ejections and tidal locking could become issues.   Still, there's a pretty good range in the Orange-Yellow dwarf/main sequence range and perhaps upper red dwarf where Earth like planets might be possible.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bigger question (I think) is stable orbits and planet formation, I suspect.   If a star has a gas giant that moves inward and becomes a hot jupiter or a gas giant with a very high eccentricity, I don't think a solar-system like that could maintain and earth like planet.   Binary stars might have a hard time providing a billion or more years of stability.  Similarly, if there are no large gas giants, There might not be enough early bombardment and water carrying comets, or, there might be too much bombardment later on, which isn't good either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other factors like a rotating core, a magnetic field, plate tectonics, and the right mix of gases, for example, too large a planet probably retains hydrogen and you can't get an oxygen atmosphere if there's too much hydrogen, so there's likely a maximum mass beyond which earth like planets couldn't be.   There's other issues like snowball earth.   Runaway Greenhouse.   Planets really need a nearly perfect situation to be earth-like.   I suspect there's not much room for error.   Fortunately with a billion or so stars of the right size in the milky way, there's likely many potentially Earth like planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, that's my &quot;guess&quot; is that earth like planets has much more to do with solar-system formation than anything else.  The star simply has to be within a range of mass, maybe 0.1 or 0.2 to 1.5 or so solar masses.   There's still so much we don't know about other solar systems that we really can't say how common earth like planets are.   That's one reason I'm looking forward to &lt;a href=&quot;http://jwst.nasa.gov/origins.html&quot; rel=&quot;nofollow&quot;&gt;James Webb Telescope&lt;/a&gt;.   That should give us much better information on exo-planets and other solar-systems.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-02-04T03:51:47.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13522" PostTypeId="1" AcceptedAnswerId="13523" CreationDate="2016-02-04T07:56:13.970" Score="0" ViewCount="213" Body="&lt;p&gt;I have watched this scishow episode where Hank says that astronomers believe there is another planet in our solar system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=I0thBlCkSdI&quot; rel=&quot;nofollow&quot;&gt;Link&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He says that it's because how asteroids and comments are being pushed towards the sun (and us) from similar angles. &lt;strong&gt;Is this true? Do astronomers really have data that confirms the 'new ninth' planet?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I did read the other related answers and articles, the way I understand this theory explained by tilt of 6 objects that could be explained by something massive enough pulling them. My question is do astronomers believe this is sufficient to confirm it that there is a planet. &lt;/p&gt;&#xA;" OwnerUserId="9477" LastEditorUserId="9477" LastEditDate="2016-02-04T10:39:13.773" LastActivityDate="2016-02-04T10:39:13.773" Title="Is there sufficient evidence to confirm additional Planet in solar system" Tags="&lt;solar-system&gt;&lt;the-sun&gt;&lt;planet&gt;&lt;comets&gt;&lt;data-analysis&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13523" PostTypeId="2" ParentId="13522" CreationDate="2016-02-04T08:09:17.897" Score="3" Body="&lt;p&gt;This question already has muliple answers on this site... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However : Yes, astronomers DO have enough data to speculate about a possible Ninth Planet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbits of six KBO are correlated, and a possible ninth planet could be the reason for those peculiar orbits. See image below for the computed results of the possible orbit of the ninth planet.: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aP5ty.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aP5ty.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since the Planet has not yet been SEEN or DETECTED, its existence is only supposed. We have nothing to prove it really exists. But if it exists, we know where it is :p&lt;/p&gt;&#xA;" OwnerUserId="9679" LastEditorUserId="9679" LastEditDate="2016-02-04T08:31:24.190" LastActivityDate="2016-02-04T08:31:24.190" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13526" PostTypeId="1" CreationDate="2016-02-04T10:01:03.520" Score="9" ViewCount="5380" Body="&lt;p&gt;General relativity is often explained as saying spacetime is curved by gravity, what does this mean? How could we perceive a curve in spacetime when there is no external &quot;straight&quot; reference frame for instance?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="11562" LastEditDate="2017-07-14T17:20:14.647" LastActivityDate="2019-08-21T08:18:19.057" Title="What is spacetime 'made' of?" Tags="&lt;gravity&gt;&lt;space-time&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13529" PostTypeId="2" ParentId="13526" CreationDate="2016-02-04T12:31:49.583" Score="7" Body="&lt;p&gt;Space-time is not &quot;made&quot; of anything, it is merely a medium or coordinate system. Think about the grid lines on a map, they aren't &quot;made&quot; of anything, they're just a representation of the geometry of the Earth. Space-time is a concept envisaged by Einstein when he wrote his theory of Special Relativity that the properties of space and time become intrinsically linked at relativistic speeds. One of the consequences of this is that you cannot go forward in space without going forward in time, this can be visualized in a light cone:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ZJoiX.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZJoiX.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is by no means a representation of what space-time looks like, only how it behaves in 2 dimensions of space and one of time. Gravitational waves are caused when two massive objects (binary stars) orbit each other, and they cause &quot;ripples&quot; in space-time. Again another 2D visualization is needed:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/o3l1H.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o3l1H.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;This shows how the gravitational potentials of two stars interact in the fabric of space-time as they orbit, causing the radiation of gravitational waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of these are only visual guides, we will never be able to &quot;see&quot; gravitational waves or space-time itself with our own eyes, that is why we need detectors such as LIGO or VIRGO to infer their existence.&lt;/p&gt;&#xA;" OwnerUserId="10437" LastActivityDate="2016-02-04T12:31:49.583" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="13530" PostTypeId="2" ParentId="13471" CreationDate="2016-02-04T14:13:59.983" Score="1" Body="&lt;p&gt;The Moon moves round the ecliptic. At full Moon it is opposite the Sun. Hence when the Sun is furthest South (in the Northern Winter), that full Moon is that which will be furthest North in that year and so on. But plainly, the Moon moves round the ecliptic every 28 days, so every 28 days or so it is furthest North etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words there is no answer based on phase or time of year alone.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2016-02-04T14:13:59.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13531" PostTypeId="2" ParentId="1333" CreationDate="2016-02-04T14:51:26.970" Score="5" Body="&lt;p&gt;Old question, but I'll address something that hasn't been brought up by the previous answers.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Photons $\simeq$ CMB photons (to first order)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As the others has already said: yes, light has energy and hence it gravitates. The bulk of photons that permeate the Universe isn't of stellar origin, though, but is in fact the cosmic microwave background, the energy density of which several orders of magnitude larger than other photons, as seen in the graph from &lt;a href=&quot;https://physics.stackexchange.com/a/196373/70207&quot;&gt;this answer to &quot;Number density of CMB photons&quot;&lt;/a&gt;. In terms of number density, there are 4-500 photons per cm$^3$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Space is big and isotropic&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Since CMB photons are isotropically distributed, the ever-so-small radiation pressure is equal in all directions, and hence cancels out. And although we're all the time bombarded by both CMB photons and stellar photons, space is so mind-bogglingly big (&lt;a href=&quot;https://en.wikiquote.org/wiki/The_Hitchhiker&amp;#39;s_Guide_to_the_Galaxy#Chapter_8&quot; rel=&quot;nofollow noreferrer&quot;&gt;D. Adams, 1978&lt;/a&gt;) that if you consider a random photon in the Universe, the probability of it hitting anything at all is negligible. Roughly 90% of the CMB photons have traveled for 13.8 billion years without hitting anything; the remaining 10% interacted with the free electrons that were released after reionization, but weren't absorbed, just polarized, and by far most of these interactions took place shortly after reionization; by now, the Universe has simply expanded too much. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Photons are redshifted&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Although there &lt;em&gt;is&lt;/em&gt; energy in photons, and hence they add to gravitation, first of all they're homogeneously distributed in the Universe (and thus pulls equally in all directions), and second their energy density is negligible compared to baryons (&quot;normal matter&quot; like gas, stars, and planets), dark matter, and dark energy. In fact, their relative densities are $\{\rho_\mathrm{bar},\rho_\mathrm{DM},\rho_\mathrm{DE},\rho_\mathrm{phot}\}/\rho_\mathrm{total} = \{0.05,0.27,0.68,10^{-4}\}$. But this was not always the case. As the Universe expands and new space is created, the density of matter decreases as $1/a^3$, where $a$ is the scale factor (&quot;size&quot;) of the Universe. The same is true for photons, but since additionally they're redshifted proportionally to $a$, their energy density decreases as $1/a^4$. That means that as you go back in time, the relative contribution of photons to the energy budget increases, and in fact until the Universe was 47,000 years old, its dynamics was dominated by radiation.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-02-04T14:51:26.970" CommentCount="28" ContentLicense="CC BY-SA 3.0" />
  <row Id="13532" PostTypeId="1" CreationDate="2016-02-04T16:42:51.037" Score="-1" ViewCount="298" Body="&lt;p&gt;Considering that the Earth orbits the Sun at 100 000 km/h and that the Sun orbits the Milky Way at almost 1 000 000 km/h, could it be possible that the galactic cluster orbits some super cluster at speeds approaching 1 000 000 000 km/h? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it did, how would that affect the mass of the cluster and how would we know or be able to measure it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you show any calculations to prove that this additional mass is not in line with what's measured for &quot;Dark Matter&quot;? &lt;/p&gt;&#xA;" OwnerUserId="10673" LastActivityDate="2016-06-04T02:09:13.480" Title="Can &quot;dark matter&quot; be the additional mass from the speed of the Galactic Cluster?" Tags="&lt;light&gt;&lt;dark-matter&gt;&lt;mass&gt;&lt;dark-energy&gt;&lt;galaxy-cluster&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13533" PostTypeId="2" ParentId="13532" CreationDate="2016-02-04T19:01:28.750" Score="1" Body="&lt;p&gt;The first part, &quot;could the galactic cluster orbit another at 10000000km/h&quot;. Is no. For simple observational reasons. If the local group were in orbit, we would be able to see and measure the supercluster which we were orbiting. We don't see it so it isn't there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I take it that you hypothesise that the 28% of the mass of the universe that is not visible, is in fact the moving mass of the galaxies. That is not the case. Dark matter is not the moving mass of visible matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First remember that velocity is a relative concept, and when specifying it, you must give a frame of reference. The relativistic mass of an object is dependent on the frame of reference. For clarity lets agree to use the frame of reference co-moving with our galaxy (which is very nearly inertial)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our own Milky way has an excess of mass. Since we are in the Milky Way, we are not moving relative to it, there is low relative velocity, low moving mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first observation of dark matter was in the rotation rates of spiral galaxies. There is a missing mass in the galaxy in its own frame of reference, and in its frame of reference. Thus the missing mass is not relativistic mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a &quot;real&quot; missing rest mass in the universe, probably supplied by a weakly interacting massive particle.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-02-04T19:01:28.750" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13535" PostTypeId="2" ParentId="13471" CreationDate="2016-02-04T19:14:27.367" Score="1" Body="&lt;p&gt;In winter, the full moon is opposite the sun, and as the sun is low, the full moon is high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summer the full moon is low (for the same reasons). The crescent moon is high in summer (and low in winter) but as the crescent moon is near the sun, it is normally not visible during the day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During spring and autumn, the sun, and the moon follow roughly equal paths, with no phase of the moon being higher in the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third quarter, being at right angles to the sun will be at an intermediate altitude, in both summer and winter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For exact calculations either use technology, or a set of astronomical tables and a sharp pencil!&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-02-04T19:14:27.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13536" PostTypeId="2" ParentId="13526" CreationDate="2016-02-04T20:16:12.247" Score="1" Body="&lt;p&gt;A mathematicial primer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A triangle on a flat surface has three angles, and these three angles add up to 180degrees - This is a well known theorem from geometry.  It is also well known that if you draw a triangle on a ball, then the angles will add up to more than 180 degrees, due to the curvature of the ball. (and if you draw on a pringles crisp, the angle sum will be less than 180).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An creature crawling on the ball could, by observing only the angles, discover that the surface of the ball was curved. So curvature can be observed by a creature inspecting only local parts of the ball, the creature doesn't need to be able to move away from the ball to observe its curvature. Curvature is an intrinsic property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly we don't need to be outside of spacetime to observe that spacetime is curved. It is actually the case that the angles in triangles don't add up to 180 due to the gravitational curvature of space (but this effect is too small to be noticeable). We can perceive a curvature in space-time without space-time having to be &quot;in&quot; anything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most obvious consequence of spacetime being curved is that things fall down towards the centre of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some level you should remember that general relativity describes a theoretical model of how gravitational mechanics functions. That model includes coordinates for time and space, and in this model &lt;strong&gt;space-time is a mathematical construct&lt;/strong&gt;, which allows us to predict with extreme accuracy the behaviour of objects under gravitational influence.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-02-04T20:16:12.247" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13537" PostTypeId="2" ParentId="13526" CreationDate="2016-02-04T20:59:12.350" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;General relativity is often explained as saying spacetime is curved by gravity, what does this mean?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This OP's comment narrowed the question down:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So [the spacetime] doesn't have to exist, like the theorema of Pythagoras is not necessary to have a triangle? – Marijn Feb 4 '16 at 20:24&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The spacetime is a map and not a territory. Spacetime is a model. In other words it is something that exists in human brain, to help that brain exactly predict the future experiences out of previous experiences. For example, the concept of spacetime (in context of the General Theory of Relativity) helps you to predict that the gravitational lensing exists. Your brain is better equipped to process on that level, rather than calculating the same picture by  taking a bunch of individual photons and applying the equations to each photon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6uOww.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6uOww.jpg&quot; alt=&quot;Gravitational lens example&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;model&quot; inside a brain is made by imagining something very close to reality and then by deliberately &lt;em&gt;removing the details&lt;/em&gt; until you got something so simple that is becomes &lt;em&gt;computable&lt;/em&gt;. You forget all the attributes of a star (color, alien civilizations around it, its prehistory, its internal dynamics) and you only think of the star as the mass, etc. This way you turn &quot;enlightening imagination&quot; into a &quot;primitive mechanism usable for predicting experiences&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Marjin, I'd like to say I love the style of your questioning, and I've signed up especially to attempt to answer this one question.&lt;/p&gt;&#xA;" OwnerUserId="10675" LastEditorUserId="10675" LastEditDate="2019-08-21T08:18:19.057" LastActivityDate="2019-08-21T08:18:19.057" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="13539" PostTypeId="2" ParentId="13532" CreationDate="2016-02-04T22:51:42.367" Score="2" Body="&lt;p&gt;The invariant mass of a multi-particle system (in a galaxy the 'particles' being stars, etc.) can come (in part, or even in whole for systems of massless particles) from the kinetic energy of its constituent particles. The relativistic picture of gravity (i.e. general relativity) is a bit more complicated, but it is clear that in a multi-particle system (in its rest frame) the kinetic energy of individual particles contributes to the gravitational &quot;force&quot;  in the form of the components representing pressure in the stress-energy tensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How the total kinetic energy of a system, which is not at rest, contributes to its gravitational attraction is a more complicated question, particularly as for a body's total kinetic energy to have a gravitational effect it must be travelling at relativistic speeds, which means we have to depart from Newtonian approximations (such as gravitational force). However it is not entirely wrong to say that the total kinetic energy of a body does contribute to its gravitational attraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite this, contributions to the gravitational attraction from the kinetic energy can be discounted as alternative explanation to dark matter for galactic rotation curves mainly because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The centre of momentum frame of the galaxy is the correct frame to consider the contribution of kinetic energy, not its possible velocity relative to some other far away galaxies. Besides which the peculiar velocities of galaxies are very small compared to c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Though the velocity of particles within the galaxy relative to the centre of momentum frame does contribute to the gravitational attraction, there just isn't enough mass moving at fast enough speed within the galaxy for the contribution to be more than negligible.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastEditorUserId="2092" LastEditDate="2016-02-05T22:09:17.853" LastActivityDate="2016-02-05T22:09:17.853" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13540" PostTypeId="2" ParentId="13526" CreationDate="2016-02-05T00:00:38.657" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;General relativity is often explained as saying spacetime is curved by gravity, what does this mean?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It means that general relativity can be formulated in a way in which its mathematics have a very direct analogue to differential geometry on a curved four-dimensional manifold. In other words, the way test particles would behave under the influence of only gravitational forces is exactly how they would behave if moving freely on a curved four-dimensional manifold. The mathematics have a direct correspondence: nothing more, nothing less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electromagnetism has a description in which the electromagnetic field strength is the curvature of a connection on a line bundle. I realize that this statement is very cryptic to someone who hasn't studied gauge theory, but it's important to realize that an essentially geometric description is not special to gravity. What's special to gravity is that it couples to all stress-energy-momentum equally, and gravitational freefall of a test particle is completely independently of composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of this universality, it is possible to interpret the properties of the gravitational field as properties of spacetime, i.e. as property of the arena on which everything else happens. We don't &lt;em&gt;have&lt;/em&gt; to do so, and indeed there are some presentations of general relativity (e.g., Weinberg's) in which the geometric interpretation is relegated to an unimportant side note, but we can--and geometry is how general relativity was originally developed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How could we perceive a curve in spacetime when there is no external &quot;straight&quot; reference frame for instance?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We could measure it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a conceptually (but not practically) simple way to do so, we could set up a small ball consisting of initially comoving test particles. With no curvature of the gravitational field, every such ball would keep the same shape and volume because they're all the test particles are moving in the same direction with the same speed. But if the gravitational field has Ricci curvature, the volume of the ball would either start shrinking or expanding. Similarly, changes in the shape of the ball would give information about Weyl curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the same kind of answer as in the case of electromagnetism: the field strength is also a kind of curvature (though not of spacetime), but how do we perceive it? Well, we could measure it by seeing how test charges behave.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2016-02-05T00:00:38.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13541" PostTypeId="1" CreationDate="2016-02-05T02:54:12.343" Score="3" ViewCount="417" Body="&lt;p&gt;I have a simple question which I think about often but have no answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If The Big Bang is true, than if the whole space was just a point 14 billion years ago, then what was around that point, some other kind of &quot;space&quot; or nothing? In latter case how do we define nothing. &lt;/p&gt;&#xA;" OwnerUserId="10665" LastActivityDate="2016-02-05T03:33:53.760" Title="What was &quot;space&quot; like before big bang?" Tags="&lt;universe&gt;&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13542" PostTypeId="2" ParentId="13541" CreationDate="2016-02-05T03:33:53.760" Score="7" Body="&lt;p&gt;This is a common question that comes up - one I've (at least tried to) answer a few times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's try to understand what the Big Bang is. The Big Bang Theory is a model, and scientists came up with it when they observed the universe expanding - and thought that since the universe is now expanding it must have been smaller in the past. Running the clock backwards eventually means all of the universe was squeezed into a &quot;singularity&quot;. Our physical models of the universe break down at this point - so we are not too sure of what the universe was like at the Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that I said the universe was squeezed into a singularity - not the stuff inside the universe. This means space-time was also compressed into a single point. The universe is not expanding out into empty space - it is space-time itself that is expanding. Hence, there was no &quot;before&quot; the Big Bang - time (as well as space) was created when the Big Bang happened. There is no outside the universe! This means there is no outside the Big Bang!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't go back in time and watch the Big Bang - since there is no universe or space (or time) outside the singularity. This can be very confusing, but I hope this helps!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a few places to look for more info:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Planck_epoch&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Planck_epoch&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Bang#Singularity&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Big_Bang#Singularity&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1263" LastActivityDate="2016-02-05T03:33:53.760" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13543" PostTypeId="1" AcceptedAnswerId="13549" CreationDate="2016-02-05T02:02:05.937" Score="3" ViewCount="303" Body="&lt;p&gt;Since light travels at the constant speed and takes more than a billion years to travel from the distant parts of the Universe, how do we know the state of the distant regions? Do those still exist or not? Is our picture of the Universe outdated?&lt;/p&gt;&#xA;" OwnerUserId="10690" OwnerDisplayName="signsgeek" LastActivityDate="2016-02-05T09:45:34.380" Title="How do we know the current state of the universe by observing it from Earth?" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13544" PostTypeId="1" CreationDate="2016-02-05T05:28:34.440" Score="1" ViewCount="3264" Body="&lt;p&gt;My science teacher told me that every object in Universe has gravity but space is a part of Universe too, so,  &lt;strong&gt;why does the space have no gravity?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10680" LastEditorUserId="1678" LastEditDate="2016-02-05T08:16:24.387" LastActivityDate="2018-03-09T20:30:19.313" Title="Does every object in the Universe have gravity? Space has no gravity, why?" Tags="&lt;gravity&gt;&lt;universe&gt;&lt;space&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13545" PostTypeId="1" AcceptedAnswerId="13547" CreationDate="2016-02-05T06:02:00.527" Score="4" ViewCount="8342" Body="&lt;p&gt;Orion constellation is a very famous constellation. It can be seen throughout the earth. Orion looks like a warrior with a weapon and skirt. I see this constellation daily and wonder, if there is a planet there.&lt;/p&gt;&#xA;" OwnerUserId="1274" LastActivityDate="2017-02-01T10:07:39.130" Title="Have any planets been discovered in Orion constellation?" Tags="&lt;exoplanet&gt;&lt;constellations&gt;&lt;orion&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13546" PostTypeId="2" ParentId="13544" CreationDate="2016-02-05T06:35:21.270" Score="3" Body="&lt;p&gt;A quick &lt;a href=&quot;https://www.google.com.au/search?q=does%20space%20have%20gravity%3F&amp;amp;ie=utf-8&amp;amp;oe=utf-8&amp;amp;gws_rd=cr&amp;amp;ei=MjS0VpmUJMm20AT0xq_YBw&quot; rel=&quot;nofollow&quot;&gt;search&lt;/a&gt; up on google you would find this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Gravity causes every object to pull every other object toward it. &lt;strong&gt;Some&#xA;  people think that there is no gravity in space. In fact, a small&#xA;  amount of gravity can be found everywhere in space.&lt;/strong&gt; Gravity is what&#xA;  holds the moon in orbit around Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why is this? That's &lt;a href=&quot;https://www.google.com.au/search?q=what%20is%20pace%20made%20of%3F&amp;amp;ie=utf-8&amp;amp;oe=utf-8&amp;amp;gws_rd=cr&amp;amp;ei=GjS0VtWXB4Kw0gSgkpagBQ#q=what%20is%20space%20made%20of%3F&quot; rel=&quot;nofollow&quot;&gt;because&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Outer space is the closest known approximation to a perfect vacuum. It&#xA;  has effectively no friction, allowing stars, planets and moons to move&#xA;  freely along their ideal orbits. However, even the deep vacuum of&#xA;  intergalactic space is not devoid of matter, as &lt;strong&gt;it contains a few&#xA;  hydrogen atoms per cubic meter.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If there's matter in space, thus this makes &lt;a href=&quot;https://www.google.com.au/search?q=does%20any%20object%20have%20gravity&amp;amp;ie=utf-8&amp;amp;oe=utf-8&amp;amp;gws_rd=cr&amp;amp;ei=FUC0VvmEOObImAW61JigCw&quot; rel=&quot;nofollow&quot;&gt;this statement&lt;/a&gt; valid. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Yes, &lt;strong&gt;everything that has mass will cause a gravitational pull on other&#xA;  objects&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7926" LastActivityDate="2016-02-05T06:35:21.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13547" PostTypeId="2" ParentId="13545" CreationDate="2016-02-05T07:07:20.327" Score="6" Body="&lt;p&gt;Yes, there are  exo-planets discovered in the Orion constellation, for instance, the stars &lt;a href=&quot;https://en.wikipedia.org/wiki/HD_37605_b&quot; rel=&quot;nofollow noreferrer&quot;&gt;HD 37605&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/HD_38529&quot; rel=&quot;nofollow noreferrer&quot;&gt;HD 38529&lt;/a&gt; both have two known planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/HD_38858&quot; rel=&quot;nofollow noreferrer&quot;&gt;HD 38858&lt;/a&gt; is also interesting, as it has a known planet, and also a disk of comets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AzQpy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AzQpy.png&quot; alt=&quot;HD 38858&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="10172" LastEditDate="2016-02-05T07:17:40.060" LastActivityDate="2016-02-05T07:17:40.060" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13548" PostTypeId="2" ParentId="13544" CreationDate="2016-02-05T09:35:43.813" Score="2" Body="&lt;p&gt;In the intergalactic medium — the most dilute regions of the Universe between the galaxies — as CipherBot writes you'll find roughly one hydrogen per cubic meter, i.e. the density is $\sim10^{-6}\,\mathrm{atoms}\,\mathrm{cm}^{-3}$, or $\sim10^{-30}\,\mathrm{g}\,\mathrm{cm}^{-3}$, or, in terms of energy (since mass and energy are equivalent through $E=mc^2$), $\sim10^{-9}\,\mathrm{erg}^\dagger$. In addition to this, you'll find 5–6 times as much dark matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space itself can't really be considered &quot;an object&quot;. Nevertheless, even ignoring the normal and dark matter, space does have energy: the so-called &lt;em&gt;dark energy&lt;/em&gt;. We don't really know much about it, but we  can measure its presence through its effect on the expansion of the Universe. But whereas normal and dark matter decelerates this expansion, dark energy has the opposite effect of accelerating the expansion. And since the energy density of dark energy is more than twice that of the other two components together, it actually dominates the dynamics of the Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in this regard you can say that space does have gravity, although it's a &quot;negative gravity&quot;. This phrasing is a bit misinterpreted, though. The dark energy is thought to be a negative &lt;em&gt;pressure&lt;/em&gt;. Pressure has a energy density associated with is, so negative pressure has a negative energy density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;$1\,\mathrm{erg}\equiv10^{-7}\,\mathrm{Joule}$.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2016-02-05T09:35:43.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13549" PostTypeId="2" ParentId="13543" CreationDate="2016-02-05T09:45:34.380" Score="2" Body="&lt;p&gt;Since light travels at a constant &lt;em&gt;and finite&lt;/em&gt; speed, looking into the Universe means looking back in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No matter in which direction we look, on average we see the same: Locally, we see evolved galaxies, far away we see young galaxies, and even farther we see the so-called cosmic microwave background, which is light from so far back in time that galaxies hadn't yet formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the Universe seems to be &lt;em&gt;isotropic&lt;/em&gt;. Unless we occupy a special place in the Universe, this implies that the Universe is also &lt;em&gt;homogeneous&lt;/em&gt;, i.e. it would look more or less the same if you were located in another part of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is turn means that the current state of the Universe can be inferred by observering the local region. The most distant observed galaxies can then be concluded be in the same state (statistically speaking; of course they don't look exactly the same, but any given property, e.g. the distribution of their sizes, will be the same as in our neighborhood).&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2016-02-05T09:45:34.380" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13550" PostTypeId="1" CreationDate="2016-02-05T10:42:01.397" Score="-1" ViewCount="67" Body="&lt;p&gt;If Alpha Centauri A our neighboring star happens to shift into a Red Giant or a Planetary Nebula, Will our star system be affected? And if so is there a ripple effect throughout the galaxy causing a disturbance or activating a nearby star to shift into a new cycle?&lt;/p&gt;&#xA;" OwnerUserId="9326" LastActivityDate="2016-02-05T23:42:50.197" Title="Can a drastic change in a stars cycle cause harm or affect a nearby star system?" Tags="&lt;star&gt;&lt;solar-system&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13551" PostTypeId="1" AcceptedAnswerId="14677" CreationDate="2016-02-05T10:42:17.913" Score="4" ViewCount="469" Body="&lt;p&gt;First: I don't think this is a duplicate question. There are related question on mergers of black holes that orbit around each other etc., so please read on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having a hard time picturing what might happen in the following scenario:&lt;br&gt;&#xA;Assume two black holes of the same size on an &lt;strong&gt;almost&lt;/strong&gt; frontal collision course with a very high relative speed of x km/h (replace x with whatever would be necessary to enable the scenarios described below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If their event horizons don't &quot;touch&quot; I'd assume their courses will change considerably but it would be a normal flyby - right ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what happens if their event horizons &quot;touch&quot;, if their event horizons &quot;overlap&quot; just a few meters ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the one hand both objects have a massive kinetic energy with opposite directional vectors. And on the other hand nothing can escape below the gravitational radius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does all the kinetic energy go ? Would the two black holes extremely deform, like creating a long string before they collapse into spherical form?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what if in a similar scenario one of the two objects is a neutron star. Would the neutron star be ripped apart when it scrapes the black hole on high-speed flyby ?&lt;/p&gt;&#xA;" OwnerUserId="10676" LastEditorUserId="10676" LastEditDate="2016-02-07T12:11:47.547" LastActivityDate="2019-05-23T08:17:00.360" Title="highspeed black holes or neutron stars on (almost) head-on collision course and kinetic energy" Tags="&lt;black-hole&gt;&lt;kinematics&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13552" PostTypeId="1" AcceptedAnswerId="13553" CreationDate="2016-02-05T10:55:08.490" Score="11" ViewCount="2103" Body="&lt;p&gt;I often read that the CMB was released from everywhere in the Universe, in every direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that statement is true, can someone elaborate what &lt;strong&gt;&quot;everywhere in the universe, in every direction&quot;&lt;/strong&gt; means ? If not, &lt;strong&gt;why can we observe it in every direction ?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10669" LastActivityDate="2020-10-31T04:12:22.280" Title="Why can we observe the Cosmic Microwave Background no matter the direction we look?" Tags="&lt;cmb&gt;&lt;observable-universe&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13553" PostTypeId="2" ParentId="13552" CreationDate="2016-02-05T12:24:01.990" Score="24" Body="&lt;p&gt;Until the Universe was 380,000 years old, it was filled with a gas of protons an electrons. There was also radiation, in thermal equilibrium with the matter, and because it was so hot, the protons and electrons couldn't form neutral hydrogen, since every time it &quot;tried&quot;, an energetic photon would knock off the electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gas was everywhere. And photons traveled and scattered in all directions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EMiWN.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EMiWN.png&quot; alt=&quot;CMB1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Photons (&lt;em&gt;purple&lt;/em&gt;) scatter on free electrons (&lt;em&gt;green&lt;/em&gt;), and both are mixed with protons (&lt;em&gt;red&lt;/em&gt;).&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;380,000 years after the Big Bang, the temperature had fallen sufficiently that neutral atoms could form (this is called &lt;em&gt;recombination&lt;/em&gt;). The radiation, which until now had scattered continously on free electrons, could now stream freely between the atoms (this is called &lt;em&gt;decoupling&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So they did. Still in all directions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VkZyW.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VkZyW.png&quot; alt=&quot;CMB2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This free streaming is still taking place. Photons travel in all directions, and are everywhere. The photons that &lt;em&gt;you&lt;/em&gt; are able to see, are the ones that started out at a particular distance from you, and in a particular direction, but other photons started out at smaller and larger distances, and in other directions. You just don't see them, because you happen to be right &lt;em&gt;here&lt;/em&gt;. But a person in another place of the Universe would see the same as you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The photons that we observe as the CMB come from a region we call the &lt;em&gt;surface of last scattering&lt;/em&gt;, because it corresponds to the surface of a shell centered at us. But there is nothing special about this &quot;surface&quot;, except that is consists of all points in the Universe that are so far from us, that it takes a photon roughly 13.8 billion years to travel. And because of the expansion, these points are now roughly 47 billion lightyears away from us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the figure below, the arrows show CMB photons. All have the same length; they start where they were emitted and end where they are today. What &lt;em&gt;we&lt;/em&gt; observe as the CMB are all arrows that end at the Milky Way (in the center). Other arrows may be observed by other observers in other galaxies that have their own surface of last scattering around them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3xhaC.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3xhaC.png&quot; alt=&quot;CMB3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2017-06-07T08:28:00.893" LastActivityDate="2017-06-07T08:28:00.893" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13554" PostTypeId="1" CreationDate="2016-02-05T14:08:44.660" Score="1" ViewCount="798" Body="&lt;p&gt;It was discovered recently that the Milky Way galaxy lies inside a supercluster named &lt;a href=&quot;https://en.wikipedia.org/wiki/Laniakea_Supercluster&quot; rel=&quot;nofollow&quot;&gt;Laniakea Supercluster&lt;/a&gt;. On the other hand, wikipedia says that galaxy filaments are &quot;the largest known structures in the universe&quot;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Galaxy_filament&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Galaxy_filament&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: is the Laniakea Supercluster part of a galaxy filament, and if so, what's the name of the galaxy filament?&lt;/p&gt;&#xA;" OwnerUserId="6920" LastEditorUserId="9533" LastEditDate="2016-08-25T02:12:53.273" LastActivityDate="2016-08-25T02:12:53.273" Title="Is our supercluster part of a galaxy filament?" Tags="&lt;milky-way&gt;&lt;galaxy-cluster&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13555" PostTypeId="2" ParentId="13554" CreationDate="2016-02-05T14:38:34.063" Score="3" Body="&lt;p&gt;The Laniakea Supercluster contains several galaxy filaments, even just reading the wikipedia articles you provided you can see the difference in size of the two:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxy Filament&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They are massive, thread-like formations, with a typical length of 50 to 80 megaparsecs, (163 to 261 million light years) that form the boundaries between large voids in the universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Laniakea Supercluster&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Laniakea Supercluster encompasses 100,000 galaxies stretched out over 160 megaparsecs (520 million light-years).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason why Laniakea isn't the largest &quot;structure&quot; in the universe is because it inst a single object, rather an arbitrarily assigned group of galaxies. From this image below you can see how the supercluster is given an imaginary edge based on the peculiar velocities of the galaxies within it, there is no observable edge to the supercluster, unlike galaxy filaments that have visible edges.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/j5SJG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j5SJG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10437" LastActivityDate="2016-02-05T14:38:34.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13556" PostTypeId="2" ParentId="13550" CreationDate="2016-02-05T14:45:13.140" Score="3" Body="&lt;p&gt;I'm not sure if this is a misunderstanding or just a matter of scientific terminology, but the &quot;star cycle&quot; is not something you can &quot;drastically change&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The life cycle of a star (as we currently understand it) is a fairly predictable set of stages of evolution from the formation of a star from a cloud of gas, along the main sequence (the stable stage out own star is currently in) and then to &quot;death&quot; of the star as it's internal fuel runs out, and material can be released back to the interstellar medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details see &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_evolution&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia page&lt;/a&gt;, &lt;a href=&quot;http://www.nasa.gov/audience/foreducators/topnav/materials/listbytype/Lifecycle_of_a_Star.html#.VrSv_T--LtQ&quot; rel=&quot;nofollow&quot;&gt;this video from NASA&lt;/a&gt;, or &lt;a href=&quot;http://www.bbc.co.uk/schools/gcsebitesize/science/add_aqa/stars/lifecyclestarsrev2.shtml&quot; rel=&quot;nofollow&quot;&gt;this very simplified diagram from the BBC&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to answer the underlying question, stars at different stages of their life cycle can definitely affect other stars, planetary systems, or interstellar matter that is nearby in many ways. Here are a few examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When young stars are born they may be quite close together and their mutual gravitational forces will disrupt circumstellar disks (where planets form), which may inhibit the formation of planets, but the specifics of this are debated in the scientific community.(&lt;a href=&quot;http://adslabs.org/adsabs/abs/2016MNRAS.457..313P/&quot; rel=&quot;nofollow&quot;&gt;see this paper&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;At the end of their life cycle, more massive stars may explode in a supernova, which releases lots of energy and throws matter out. This can change the structure of clouds of gas and can compress it, possibly triggering new stars to form in the more dense material. This is also a topic of current study so the exact details and mechanisms are not very well understood, but the theory is logical and observations have been found that seem to confirm this.(&lt;a href=&quot;http://adslabs.org/adsabs/abs/2016arXiv160101114L/&quot; rel=&quot;nofollow&quot;&gt;see this paper&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;A white dwarf star and a red giant star in a binary system can have transfer of mass from the red giant to the white dwarf. A white dwarf has a maximum stable mass (&lt;a href=&quot;https://en.wikipedia.org/wiki/Chandrasekhar_limit&quot; rel=&quot;nofollow&quot;&gt;Chandrasekhar limit&lt;/a&gt;) so when the mass exceeds this, the white dwarf will go supernova - this is , as you stated, a drastic change in one star due to interaction with another star.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2016-02-05T14:45:13.140" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13557" PostTypeId="2" ParentId="13552" CreationDate="2016-02-05T15:18:00.407" Score="3" Body="&lt;p&gt;Yes. That's &lt;a href=&quot;http://www.astro.ubc.ca/people/scott/cmb_intro.html&quot; rel=&quot;nofollow&quot;&gt;why we called the CMB isotropic&lt;/a&gt; until a decade or two ago; isotropic meaning '&lt;a href=&quot;http://dictionary.reference.com/browse/isotropic&quot; rel=&quot;nofollow&quot;&gt;the same in every direction&lt;/a&gt;'. Since that time we've had the Wilkinson Microwave Anisotropy Probe, and its successors which show slight differences in it depending on where we point our antenna. We can easily measure our velocity with respect to this nearly uniform background radiation: &lt;a href=&quot;http://apod.nasa.gov/apod/ap090906.html&quot; rel=&quot;nofollow&quot;&gt;Speeding through the universe&lt;/a&gt;.&#xA;Subtracting that, we get &lt;a href=&quot;http://map.gsfc.nasa.gov/media/121238/index.html&quot; rel=&quot;nofollow&quot;&gt;Speed corrected WMAP data showing hot and cold spots&lt;/a&gt;. That's why we now call the CMB anisotropic.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2016-02-05T15:18:00.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13558" PostTypeId="1" CreationDate="2016-02-05T18:58:09.937" Score="-10" ViewCount="494" Body="&lt;h2&gt;I will try to give you my reasoning behind my scenario.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Ok I've researched and found &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;That 95 percent of the galaxy has been formed already, And the&#xA;other 5 percent is still to come.&lt;a href=&quot;http://www.popsci.com/science/article/2012-11/almost-all-stars-are-old-and-universe-making-hardly-any-new-ones&quot; rel=&quot;nofollow&quot;&gt;http://www.popsci.com/science/article/2012-11/almost-all-stars-are-old-and-universe-making-hardly-any-new-ones&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I've also found that stars can&#xA;    live up to 10 billion years.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the Universe is around 13 billion&#xA;years old.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Much much more stars die than new stars are born and&#xA;around 9 billion years ago the new stars being born dropped to 30&#xA;percent.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If number 4 and 2 are correct almost all the stars in&#xA;the universe should of died already.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Black holes are formed either by two neutron star colliding or a supernova. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If 5 is correct there should be a huge amount of black holes left from all&#xA;the stars that supernovaed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Black holes can consume one another and form a larger mass.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If our galaxy is 4.5 billion years old&#xA;and 5 percent of the galaxy is still unformed we are almost consider&#xA;one of the last formed galaxy and should be on the outer edge of the&#xA;universe if we are 4.5 billion years old, So that could give us&#xA;about 5 billion years left.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the references I found lean towards saying new stars hardly form.And there is a estimated 100,000,000 Black holes in the galaxy as of now.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A massive Black hole has to be 80 percent of our Universe  by now and slowly making its way toward our galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Please tell me where my info was wrong.&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Will our solar system die of old age in 5.4 billion years or will we be consumed by the massive black hole if this scenario is plausible?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9326" LastEditorUserId="9326" LastEditDate="2016-02-06T17:53:25.943" LastActivityDate="2016-02-06T17:53:25.943" Title="Is this Universe scenario correct?" Tags="&lt;orbit&gt;&lt;the-sun&gt;&lt;black-hole&gt;&lt;universe&gt;&lt;solar-system&gt;" AnswerCount="1" CommentCount="9" ClosedDate="2016-02-08T20:03:55.413" ContentLicense="CC BY-SA 3.0" />
  <row Id="13559" PostTypeId="1" AcceptedAnswerId="13561" CreationDate="2016-02-05T20:29:03.207" Score="0" ViewCount="99" Body="&lt;p&gt;In the nebulaes of galaxies are stars born, but the gasses and dust they are made of comes from exploded stars (fe supernovas). So it looks like these processes can go on continuesly. But is that really the case, or does according to entropy(?) galaxies extinguish because fe the galaxie went too big so gravity has no opportunity to form other stars any more. And before colliding with other galaxies it is passed away. So is there any sign that galaxies are getting smaller or less active?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastActivityDate="2016-02-05T22:05:32.660" Title="How probably is it that galaxies will extinguish?" Tags="&lt;galaxy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13560" PostTypeId="2" ParentId="13559" CreationDate="2016-02-05T21:12:45.120" Score="-3" Body="&lt;p&gt;Far more stars die than stars are born, The number isn't exact but I would say 90 to 10.Then what occurs next is a large percentage of those dead stars form black holes from supernovas and the occasional neutron star collision will form black holes, Giving the new stars less of a chance to reach maturity.The Universe is being consumed from the point it was created.The reason why the Universe can not repair itself is because it was almost entirely created all at once so everything will decay at once leaving no chance for restructure.  &lt;/p&gt;&#xA;" OwnerUserId="9326" LastActivityDate="2016-02-05T21:12:45.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13561" PostTypeId="2" ParentId="13559" CreationDate="2016-02-05T21:45:47.413" Score="3" Body="&lt;p&gt;Galaxies are gradually being extinguished. Most star formation activity occurs near the start of a galaxy's life, or in response to merger activity with other galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The star formation rate of the universe peaked at redshifts of around 3, corresponding to a look-back time of around 9 billion years. Since then the star formation rate has declined as the universe expands; mergers are less frequent, gas is driven out of galaxies by supernovae and active galactic nuclei.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, most of the stars that have been formed are of lower mass (K- and M-dwarfs) than the Sun and will live on for tens or hundreds of billions of years. They are however much fainter than the Sun. So although high-mass, luminous O- and B stars live their short lives and are not replaced at the same rate, the low-mass stars continue to shine. This means that galaxies &lt;em&gt;will&lt;/em&gt; get fainter on average as the remaining stellar populations increase in average age and decrease in average mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a slow process though. High mass stars are still being formed in our galaxy after 12 billion years, and most spiral galaxies have ongoing star formation. However, star formation has more-or-less ceased in gas-poor elliptical galaxies.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-02-05T22:05:32.660" LastActivityDate="2016-02-05T22:05:32.660" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13562" PostTypeId="1" CreationDate="2016-02-05T22:04:24.933" Score="7" ViewCount="368" Body="&lt;p&gt;Is it possible Planet Nine is shepherding the Kuiper Belt similar to the way Jupiter shepherds the asteroid belt? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that very little is actually known about Planet Nine at this point, but has it been postulated that it has control over the Kuiper Belt? I don't exactly intend to read any theoretical papers on the existence of Planet Nine because I am sure it will be way over my head, but I understand that it is from the position of several Trans-Neptunian objects that its existence has any merit at all.  So, I am wondering if part of that theory includes the Kuiper belt as a whole?&lt;/p&gt;&#xA;" OwnerUserId="10026" LastEditorUserId="2153" LastEditDate="2016-02-06T15:44:43.723" LastActivityDate="2021-03-03T14:24:08.403" Title="Is Planet Nine shepherding the Kuiper Belt?" Tags="&lt;planet&gt;&lt;9th-planet&gt;&lt;kuiper-belt&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13563" PostTypeId="1" CreationDate="2016-02-05T23:03:52.573" Score="-1" ViewCount="222" Body="&lt;p&gt;Gravitational force is particle or wave? I know that gravitation depend of the mass and the rotation speed but what is the transmission medium of this wave the same like transmission medium of the light ?&lt;/p&gt;&#xA;" OwnerUserId="10668" LastActivityDate="2020-12-05T08:24:27.407" Title="what is gravitational force?" Tags="&lt;gravity&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13565" PostTypeId="2" ParentId="13545" CreationDate="2016-02-06T00:25:02.420" Score="2" Body="&lt;p&gt;(RV)-discovered planet around HD 38858,&#xA;and then RV limits on undiscovered planets around HD 20794,&#xA;61 Vir (HD 115617), HD 69830, and HD 38858.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ast.cam.ac.uk/~wyatt/kmmg15.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ast.cam.ac.uk/~wyatt/kmmg15.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9326" LastActivityDate="2016-02-06T00:25:02.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13567" PostTypeId="1" CreationDate="2016-02-06T01:31:14.900" Score="1" ViewCount="295" Body="&lt;p&gt;Does the CMB contain any clues to what caused the Big Bang? Does a mathematical model theorize it? &lt;/p&gt;&#xA;" OwnerUserId="10690" LastActivityDate="2016-03-13T19:53:03.857" Title="What happened just before the Big bang?" Tags="&lt;early-universe&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2016-03-21T15:34:44.857" ContentLicense="CC BY-SA 3.0" />
  <row Id="13568" PostTypeId="1" CreationDate="2016-02-06T02:01:46.717" Score="3" ViewCount="114" Body="&lt;p&gt;Isn't it better to say that the light from the galaxy took 5 billion light years to reach us and it is a very old picture of it? However, if light travelled back to the galaxy at the constant speed and an observer was stationed to record it wouldn't it take longer to reach because of the expansion of the Universe? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also if it still exists after 10 billion years or not we could not predict. So is the picture of the Universe more of a timeline of events rather than like a conventional map that shows distance between points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do current day theories tackle these issues?&lt;/p&gt;&#xA;" OwnerUserId="10690" LastEditorUserId="2270" LastEditDate="2016-02-08T18:28:03.400" LastActivityDate="2016-02-08T18:28:03.400" Title="Since the Universe is expanding, is it accurate to say that a galaxy is 5 billion light years away?" Tags="&lt;observable-universe&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13570" PostTypeId="1" AcceptedAnswerId="13580" CreationDate="2016-02-06T04:39:21.070" Score="6" ViewCount="489" Body="&lt;p&gt;There is actually a similar question on this site: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/8525/can-magnetism-escape-a-black-hole&quot;&gt;Can magnetism escape a black hole?&lt;/a&gt;, however the answers do not focus on accretion, which is my main concern, so I start a new one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Kip Thorne's book &lt;em&gt;Black Holes and Time Warps: Einstein's Outrageous Legacy&lt;/em&gt;, there is a paragraph in Chapter 9, describing how the central black hole of a quasar (or radio galaxy) acquires a magnetic field through accretion:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where do these magnetic field lines come from? From the disk itself.&#xA;  All gas in the Universe is magnetized, at least a little bit, and the [accretion] disk's gas is no exception. As, bit by bit, the disk's gas accretes into the hole,&#xA;  it carries its magnetic field lines with it. Upon nearing the hole, each&#xA;  bit of gas slides down its magnetic field lines and through the horizon,&#xA;  leaving the field lines behind, sticking out of the horizon and threading&#xA;  it [...]. These threading field lines, firmly&#xA;  confined by the surrounding disk, should then extract the hole's&#xA;  rotational energy by the Blandford—Znajek process.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately this description does not seem clear to me (although Thorne generally is excellent in explaining things throughout the book). To be specific, I cannot get a picture of the gas &quot;sliding down&quot; its magnetic field lines through the event horizon, while making the magnetic field &quot;stick out&quot; of the horizon at the same time. And here Thorne did not seem to point to any original research papers for further study.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my questions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Do today's astronomers still believe that disk accretion is the main process by which supermassive black holes acquire a magnetic field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) If the answer to (1) is &quot;yes&quot;, then how can I build the picture (if there is such a picture) that Thorne is trying to convey in his paragraph?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Is there any research paper that addresses this problem to some detail?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is appreciated!&lt;/p&gt;&#xA;" OwnerUserId="10691" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-02-06T16:12:11.367" Title="How does an accreting black hole acquire magnetic fields?" Tags="&lt;black-hole&gt;&lt;supermassive-black-hole&gt;&lt;magnetic-field&gt;&lt;active-galaxy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13571" PostTypeId="2" ParentId="13563" CreationDate="2016-02-06T04:53:21.783" Score="5" Body="&lt;p&gt;As a hobbyist, I'll give a limited answer.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Gravitational force is particle or wave?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Until we actually find out what gravity is, the safe answer here is that we don't know, but it's probably a wave and a particle.   Most (perhaps all) fundamental particles in quantum physics are waves and particles.   Light for example, Electrons too, though we might think of an electron as a particle, it's wave properties are easily demonstrated, and wave properties of protons can be demonstrated as well.  There's a ton of stuff that can be googled on the wave-particle duality.  Here's &lt;a href=&quot;http://webs.morningside.edu/slaven/Physics/uncertainty/uncertainty3.html&quot; rel=&quot;nofollow&quot;&gt;one&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's unclear exactly what fundamental particles &quot;are&quot; so saying a fundamental particle is &quot;this or that&quot; is a bit iffy since we can't see them.  We can only study how they behave.   Gravity, if it's a fundamental particle, which it probably is, it should be both a wave and a particle, and a field.   I know, that's a little hard to think about.  We don't get to have many tangible answers in quantum physics. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I know that gravitation depend of the mass and the rotation speed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gravity depends on mass.  Period.   Rotation speed can creates a centrifugal force opposing gravity and making things lighter, but that isn't gravity.   The gravity is still 100% governed by mass.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;what is the transmission medium of this wave the same like&#xA;  transmission medium of the light ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Light (as far as I know) doesn't require a transmission medium.  Gravity shouldn't either.   Sound waves require a medium but light and gravity can pass through empty space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have more specific questions, you might have better luck on Physics rather than Astronomy.  Also, my 2 cents, Stan Liou has made some intelligent comments well worth reading.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-02-07T03:00:58.260" LastActivityDate="2016-02-07T03:00:58.260" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="13572" PostTypeId="2" ParentId="13503" CreationDate="2016-02-06T06:44:30.497" Score="-2" Body="&lt;p&gt;Well our sun was not unique 11 billion years ago, Because almost every star in the Universe were all created at once so they all went through Stellar evolution all at once also.Our Sun is 4 billion years old and in a childhood stage compared to the rest of the Stars in the Universe.Our Star systems was one of the last formed from the Big Bangin so only a few stars are in a similar stage as ours as of right now. There still is a few that haven't formed yet, but right now our Sun is a rarity because of it's age and stage giving it a uniqueness in the present Universe.&lt;/p&gt;&#xA;" OwnerUserId="9326" LastActivityDate="2016-02-06T06:44:30.497" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13573" PostTypeId="2" ParentId="13568" CreationDate="2016-02-06T10:55:08.467" Score="3" Body="&lt;p&gt;I'm not 100% sure if I'm understanding what your asking, but if your question could be rephrased as &quot;how do we measure distances in an expanding universe?&quot;, then I can try to answer that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on what astronomers measure they use different &lt;em&gt;distance measurements&lt;/em&gt;. For example the &lt;strong&gt;comoving distance&lt;/strong&gt; between two objects takes into account the expansion, and so does not change with time. If you know the redshift of the galaxy, for example by measuring the spectrum, and have a cosmological model, then you can calculate the comoving distance. Here cosmological model means constraints on the amount of dark energy, matter (both dark and regular) and radiation. The current accepted model is that the universe is around 70% dark energy and 30% matter, most of which is dark (with negligible radiation). The percent here refers to the fraction of energy density. Note that these values change over time, mostly since dark energy is like a property of space and so increases as the universe expands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some more info see:&#xA;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Distance_measures_(cosmology)&quot; rel=&quot;nofollow&quot;&gt;https://en.m.wikipedia.org/wiki/Distance_measures_(cosmology)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that light years is a definite measurement, and so we can use it a valid unit for all distance measures.&lt;/p&gt;&#xA;" OwnerUserId="1263" LastActivityDate="2016-02-06T10:55:08.467" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13574" PostTypeId="2" ParentId="13558" CreationDate="2016-02-06T11:20:26.267" Score="4" Body="&lt;p&gt;Like it was already pointed out in the comments, your assertions and assumptions are way off today's well-accepted theories. Nonetheless, I'll try to answer you questions.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Will our solar system die of old age in 5.4 billion years&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Our sun is a &lt;a href=&quot;https://en.wikipedia.org/wiki/G-type_main-sequence_star&quot;&gt;G-type main-sequence star&lt;/a&gt; with an estimated lifespan of roughly 10 billion years. Like you mentioned, it is about 4.6 billion years old and will stay in the main sequence for another 5.6ish billion years. The sun's mass is not enough to end with it's life with a supernova. Instead it will become a &lt;a href=&quot;https://en.wikipedia.org/wiki/Red_giant&quot;&gt;red giant&lt;/a&gt; with a radius of about 1 AU (= astronomical unit), meaning that it will most likely devour planet earth, but won't expand further.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;or will we be consumed by the Universal black hole?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I've never heard of something like an universal black hole, but it appear's to me that you might have a big misconception about black holes in general. If (for some completly unknown and unphysical reasons) the sun would all of the sudden turn into a black hole with the same mass, nothing much would change (on a cosmological scale). Planet earth's main energy input would seep away and we all would freeze to death, but gravitational, nothing would change. The planets orbits would be exactly the same and nothing would become 'consumed'. Black holes don't accrete mass (that's what I suppose you mean by consuming) due to some magical properties, but due to gravitational attraction, which is only dependent on the masses and the distance of the objects.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;How long should it take for the 93 percent of universe to be consumed by the black hole?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Speaking of 93% of the universe might not make any sense. Today's measurements suggest that the curvature of space is flat ($\Omega_{tot} = 1.00 \pm 0.02$) and that yields in a possible infinite universe. Please see &lt;a href=&quot;https://en.wikipedia.org/wiki/Lambda-CDM_model&quot;&gt;$\lambda CDM-Model$&lt;/a&gt;&#xA;and &lt;a href=&quot;http://arxiv.org/abs/1502.01589&quot;&gt;Planck 2015 results. XIII&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10611" LastActivityDate="2016-02-06T11:20:26.267" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13575" PostTypeId="1" CreationDate="2016-02-06T11:59:53.567" Score="0" ViewCount="128" Body="&lt;p&gt;Everything which has mass has gravity. Some theories hypothesis that gravitons are responsible for that, but do they have any clue where they come from? Like gluons are interacting between quarks and photons between eg electrons. But does gravitons have, according to those theories, also have a kind of particle on their own with whom they are interacting with. Or is it possible that fe gravitons are also coming from quarks?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="37191" LastEditDate="2021-02-16T08:53:28.067" LastActivityDate="2021-02-16T08:53:28.067" Title="Is their any clue where gravitons come from?" Tags="&lt;gravity&gt;&lt;graviton&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13576" PostTypeId="1" AcceptedAnswerId="13579" CreationDate="2016-02-06T13:52:25.433" Score="4" ViewCount="866" Body="&lt;p&gt;Astronomers say that it's strange that the outer stars of a galaxy are travelling with the same speed as the inner stars. Therefore they introduced dark matter to solve this strangeness. The theory that the inner stars 'have to' go faster is that because they are closer to the center so the gravity is more, like the earth is closer to the sun and faster than Neptune. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what I wonder is, whether the fact that just the outer stars are attracted to more mass, because there is more mass of stars that are within the orbits of the outer stars, explains the equality? So in this view it is not so strange that they have the same speed because they are influenced by more gravity. Or is it that this extra influence is considered to be too little to explain the equality? &lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="2531" LastEditDate="2018-11-03T19:42:03.097" LastActivityDate="2018-11-03T19:42:03.097" Title="Why is it strange that outer stars in a galaxy are travelling with the same speed as inner while the total mass is also increasing?" Tags="&lt;gravity&gt;&lt;galaxy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13577" PostTypeId="1" CreationDate="2016-02-06T14:54:48.423" Score="3" ViewCount="41" Body="&lt;p&gt;I've seen discussion threads where amateur astronomers estimate a satellite's &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_elements&quot; rel=&quot;nofollow noreferrer&quot;&gt;orbital parameters&lt;/a&gt; from angles-only telescope observation.  Are there any community tools or algorithms to help with this math? It's not a trivial problem to solve.&lt;/p&gt;&#xA;" OwnerUserId="10693" LastEditorUserId="37191" LastEditDate="2021-04-21T08:41:17.257" LastActivityDate="2021-04-21T08:41:17.257" Title="Tools and algorithms for angles-only orbit estimation" Tags="&lt;amateur-observing&gt;&lt;celestial-mechanics&gt;&lt;satellite&gt;&lt;astrometry&gt;" AnswerCount="0" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="13578" PostTypeId="2" ParentId="13575" CreationDate="2016-02-06T15:29:52.217" Score="2" Body="&lt;p&gt;There is no fully satisfactory quantum theory of gravity. The electromagnetic field is quantized, the quantum excitations are photons. It would be reasonable to believe that the gravitational field would also be quantized, and the quanta would be particles, called gravitons. In fact it would be very hard to explain how gravity could not be quantised. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image of a particle emitting photons is rather misleading. Instead, two particles exchange a photon in an electromagnetic interaction. And the particles are not classical, but quantum mechanical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photons carry the electromagnetic force between any charged particles: not just electrons, but also muons, tauons, quarks, and the W particles. By this analogy, anything that can create a gravitational field would be exchanging gravitons: that includes anything that has mass or momentum. (ie more or less anything). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is neither a theoretical, nor an experimental basis for the existence of gravitons.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-02-14T23:35:16.517" LastActivityDate="2017-02-14T23:35:16.517" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13579" PostTypeId="2" ParentId="13576" CreationDate="2016-02-06T15:39:08.277" Score="3" Body="&lt;p&gt;This is accounted for in real models of the rotation curves of galaxies. The rotation curve of a galaxy is more complicated than the motion of planets in the solar system, for the reasons that you describe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A trivial model would consider the galaxy as spherically symmetric and would use the shell theorem to estimate the centripetal acceleration of a star at radius $r$. Thus&#xA;$$ m r \omega^2 = G\frac{m M(r)}{r^2},$$&#xA;where $m$ is the mass of the star, $\omega$ is the angular velocity and $M(r)$ would be the mass of all gravitating matter at radii $&amp;lt;r$. Note that this only applies to a spherically symmetric distribution of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make further progress demands that you know the density distribution of the gravitating matter. Let's just assume that density $\rho$ is constant for the moment. Then&#xA;$$ r^3 \omega^2 = G \int \rho\ 4\pi r^2\ dr = \frac{4\pi}{3} G \rho r^3$$&#xA;$$ \omega = \sqrt{\frac{4\pi G \rho}{3}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the angular velocity would be constant with radius and the rotation speed, $v = \omega r$ would increase with radius. This I think is the situation that your question supposes and so yes, if there was a constant (or perhaps &lt;em&gt;slowly&lt;/em&gt; declining) density of material in the Galaxy then this would produce a rotation curve that increased (or was flat).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trouble is that the density of material in our Galaxy inferred from the matter that we can see is &lt;em&gt;not&lt;/em&gt; constant with radius. It declines rapidly and exponentially, such that there is very little visible matter beyond a radius of about 15 kpc. If we take a situation where we go to radii beyond the gravitating matter, then a similar treatment to the above suggests that&#xA; $$ m r \omega^{2} = G\frac{mM}{r^2},$$&#xA;where $M$ is now the total mass of the (visible) Galaxy. In this case&#xA;$$ \omega = \sqrt{\frac{GM}{r^{3}}}$$&#xA;and the rotation velocity $v = \omega r$ should decline as $1/\sqrt{r}$ (like it does in the solar system).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the fact that the rotation velocity of stars and gas at large radii ($&amp;gt;15$ kpc) continues to be flat or even &lt;em&gt;increase&lt;/em&gt; that leads to the conclusion that the mass that we can see is not all that there is. i.e That we need &quot;dark matter&quot; to explain the high rotation speeds at radii where there is very little visible matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Realistic models of the galaxy do not make the assumption that the visible matter is spherically symmetric (it isn't). But the conclusions I qualitatively set out above hold in the same way.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-02-06T15:39:08.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13580" PostTypeId="2" ParentId="13570" CreationDate="2016-02-06T16:12:11.367" Score="2" Body="&lt;p&gt;Addressing (2) and (3) first:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a case like this, your best option is to go straight to the source. In this case, that would be &lt;a href=&quot;http://adsabs.harvard.edu/abs/1977MNRAS.179..433B&quot; rel=&quot;nofollow noreferrer&quot;&gt;Blandford &amp;amp; Znajek (1977)&lt;/a&gt;, the originators of the proposed mechanism. they have several figures regarding different magnetospheric structures depending on different magnetic fields. However, the most important one is Figure 1, the general case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/6Xrk5wM.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D$ represents the accretion disk, while $T$ is the transition zone where matter is rapidly moving radially towards the black hole. The lines represent magnetic field lines, along which charged particles. Electrons (e&lt;sup&gt;-&lt;/sup&gt;) travel in the direction of the medium arrows on each field line; positrons (e&lt;sup&gt;+&lt;/sup&gt;) travel in the other direction. The field lines are closer together in the disk area closer to the black hole, indicating the stronger magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a good overview of the process, this is the paper to look at. That said, it is far from an easy read.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding (1), I'm not an astronomer and not really qualified to tell you what the mainstream opinion among astronomers is. However, as far as I'm aware, the Blandford-Znajek process is currently accepted as a strong possibility for generating the magnetic fields observed near most black holes (including stellar-mass black holes).&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2016-02-06T16:12:11.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13581" PostTypeId="1" AcceptedAnswerId="13583" CreationDate="2016-02-06T16:44:54.160" Score="1" ViewCount="71" Body="&lt;p&gt;I have a simple question. I can see orion constellation clearly in the night sky. I read it somewhere on this website that gemini is on the left-hand side of orion constellation. I was able to spot gemini later that night. I was wondering on how to spot other constellations using orion and gemini constellation. &lt;/p&gt;&#xA;" OwnerUserId="1274" LastActivityDate="2016-02-06T18:06:10.710" Title="How to see other celestial things using stars for direction?" Tags="&lt;star&gt;&lt;constellations&gt;&lt;orion&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13582" PostTypeId="1" CreationDate="2016-02-06T17:53:15.863" Score="0" ViewCount="1039" Body="&lt;p&gt;QUESTION: &#xA;Suppose we discover a planet orbiting a nearby star. &#xA;The distance to the star is 3 pc. &#xA;We observe the angular radius of the planet’s orbit to be 0.1 arc sec. &#xA;How many AU from the star is the planet? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;**CONCERN: I am not sure what &quot;The distance to the star is 3 pc&quot; means. The earth's distance to the star? or the sun's distance to the star? or even better, the planet's distance to the star??!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any explanation of the question would be greatly appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial approach was some fixed position (say sun/earth's distance) to the star is 3pc, &#xA;so I will have    0.1 arc sec = 3pc/distance, and solve for distance as the answer.&#xA;(I reckon this is a very simple question for astrophysics, but I need to get basics right)&lt;/p&gt;&#xA;" OwnerUserId="10696" LastEditorUserId="10696" LastEditDate="2016-02-06T18:00:47.047" LastActivityDate="2016-05-10T14:16:51.533" Title="Distance of a planet to the star?" Tags="&lt;astrophysics&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13583" PostTypeId="2" ParentId="13581" CreationDate="2016-02-06T18:06:10.710" Score="2" Body="&lt;p&gt;Use a constellation map! That will provide much of the information you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are however some things you can try to spot:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;On the lower left hand side of Orion, you can spot a very bright star, &lt;a href=&quot;https://en.wikipedia.org/wiki/Sirius&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sirius&lt;/a&gt;, in fact the brightest one in the sky. It is part of the constellation &lt;a href=&quot;https://en.wikipedia.org/wiki/Canis_Major&quot; rel=&quot;nofollow noreferrer&quot;&gt;Canis Major&lt;/a&gt; (great dog).&lt;/li&gt;&#xA;&lt;li&gt;A bit further left and up you probably see a star nearly as bright as Sirius. That is &lt;a href=&quot;https://en.wikipedia.org/wiki/Procyon&quot; rel=&quot;nofollow noreferrer&quot;&gt;Procyon&lt;/a&gt; in &lt;a href=&quot;https://en.wikipedia.org/wiki/Canis_Minor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Canis minor&lt;/a&gt; (lesser dog).&lt;/li&gt;&#xA;&lt;li&gt;Up, and a little bit to the right from Orion you can see a very red star. That is &lt;a href=&quot;https://en.wikipedia.org/wiki/Aldebaran&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aldebaran&lt;/a&gt; in the constellation of &lt;a href=&quot;https://en.wikipedia.org/wiki/Taurus_%28constellation%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Taurus&lt;/a&gt; (the bull). In the same constellation you can see the &lt;a href=&quot;https://en.wikipedia.org/wiki/Pleiades&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pleiades&lt;/a&gt;, a bunch of stars in a cluster.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Taurus with Aldebaran and the Pleiades:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/k6RAZm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k6RAZm.png&quot; alt=&quot;Taurus&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10172" LastActivityDate="2016-02-06T18:06:10.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13584" PostTypeId="2" ParentId="13582" CreationDate="2016-02-06T18:15:42.607" Score="1" Body="&lt;p&gt;Use this (&lt;a href=&quot;http://www.ast.cam.ac.uk/~mjp/calc_parallax.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.ast.cam.ac.uk/~mjp/calc_parallax.html&lt;/a&gt;) but substitute the exoplanet for the Earth, you don't need to involve the Sun in your calculations, just assume the star is 3pc (parsecs) from Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eURfp.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eURfp.gif&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From Figure 2, the distance between the Sun and the star is : &#xA;  d = r / tan P &#xA;  If P is 1 second of arc: &#xA;  d = 150 000 000 / tan 1&quot; = 30 million million km &#xA;  This distance is called one parsec and is a basic unit for measuring astronomical distances. &#xA;  Distance in parsecs = 1 / P in seconds of arc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10437" LastEditorUserId="10437" LastEditDate="2016-02-06T18:20:41.077" LastActivityDate="2016-02-06T18:20:41.077" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13585" PostTypeId="1" AcceptedAnswerId="13663" CreationDate="2016-02-06T20:25:06.417" Score="3" ViewCount="298" Body="&lt;p&gt;Moon can be fully eclipsed by the virtue of being fairly close to Earth. Any body distant enough will not be eclipsed fully, Earth's disc not fully covering Sun's disc. How's that for anything at the L2 point? Will it be lit by Sun's corona or will it bask in deep shadow of Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus question: if Umbra does reach there, how large a body could remain fully shaded at L2? What's the Umbra's radius there?&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2016-02-10T20:44:48.460" Title="Does earth's Umbra reach Sun-Earth L2?" Tags="&lt;lunar-eclipse&gt;&lt;lagrange-point&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13587" PostTypeId="1" AcceptedAnswerId="13589" CreationDate="2016-02-06T21:24:02.830" Score="6" ViewCount="424" Body="&lt;p&gt;I am looking for an answer as to why the geocentric gravitational constant, μ, defined as the product of the gravitational constant, G, and the mass of a body (earth in this case), M, can be calculated to a higher degree of precision than either G or M alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From wikipedia&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...for celestial bodies such as Earth and the Sun, the value of the product GM is known much more accurately than each factor independently. Indeed, the  limited accuracy available for G limits the accuracy of scientific &#xA;   determination of such masses in the first place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can someone explain this? I can't find a good answer anywhere I look. &lt;/p&gt;&#xA;" OwnerUserId="10697" LastActivityDate="2016-06-23T16:45:26.913" Title="Precision of geocentric gravitational constant" Tags="&lt;gravity&gt;&lt;newtonian-gravity&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13588" PostTypeId="2" ParentId="13587" CreationDate="2016-02-06T23:36:53.140" Score="2" Body="&lt;p&gt;$\mu$ is a quantity that can be easily observed (semi-) directly. It can be easily derived from orbital period of orbiting bodies or acceleration of falling bodies, even if their mass is significantly smaller than the mass of the body you're measuring $\mu$ for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the only way to really obtain G is to divide $\mu$ by mass of your celestial body. And measuring this mass without knowing G is opening a can of worms. You may try to get geophysical studies of density distribution, or try to see how this body reacts to other masses (of comparable magnitude), or a number of increasingly complex methods of dubious accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measuring mass of a simple body of range of milligrams to tons without use of gravity as a reference is not a big problem. You accelerate it with a known force, measure the acceleration, and get the mass. (possibly, the force is known because it was used to accelerate a known mass - or a model mass, that defines, say, 1kg.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do you go about accelerating a planet - or any body heavy enough that it exhibits gravitational force strong enough to be measurable with any reasonable precision - with a known force, to measure its mass? About the only forces that are capable of doing so in any observable manner, are the forces of gravity - and they are all proportional to the unknown G.&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="83" LastEditDate="2016-02-07T00:02:07.180" LastActivityDate="2016-02-07T00:02:07.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13589" PostTypeId="2" ParentId="13587" CreationDate="2016-02-06T23:53:18.207" Score="3" Body="&lt;p&gt;Ignoring details such as the oblateness of the Earth, atmospheric drag, third body influences such as the Moon and the Sun, relativity, ..., the period of a satellite of negligible mass (even the International Space Station qualifies as a &quot;satellite of negligible mass&quot;) is $T=2\pi\sqrt{\frac {a^3}{\mu_\mathrm{Earth}}}$. Neither Newton's gravitational constant nor the mass of the Earth are involved in this expression. This means that, ignoring those details, calculating $\mu_\mathrm{Earth}$ is merely a matter of calculating a satellite's rotational period and its semimajor axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Humanity has lots and lots of artificial satellites in orbit, and the people who model the orbits of those satellites don't ignore those details. A few of those satellites were specially designed to enable the determination of the Earth's non-spherical gravitational field (e.g., GRACE and GOCE), and a few were specially designed to enable extremely precise orbit determination (e.g., LAGEOS). Even with all of those details, the Earth's gravitational parameter is a directly inferable quantity (i.e., knowledge of G is not required). Moreover, the value is known to a very high degree of precision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth's mass? Not so much. The most precise way to &quot;weigh the Earth&quot; is to divide the high precision Earth's gravitational parameter by the low precision universal gravitational constant G. There's a problem here, which is the notoriously low precision of the gravitational constant when expressed in SI units.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2016-02-06T23:57:06.403" LastActivityDate="2016-02-06T23:57:06.403" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13590" PostTypeId="1" CreationDate="2016-02-07T04:39:43.130" Score="1" ViewCount="139" Body="&lt;p&gt;This is, I think, somewhat intuitively obvious, but this comment got me thinking about this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/13563/what-is-gravitational-force/13571?noredirect=1#comment20082_13571&quot;&gt;what is gravitational force?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Rotation speed can create centrifugal force opposing gravity and making things lighter,&quot; Rotation also stores energy, and energy is mass, per $E=mc^2$. A body spinning sufficiently fast will exert higher gravity, e.g. a slowly-spinning neutron star will have a weaker gravitational pull than equivalent neutron star that spins very fast.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, my answer is &quot;no way&quot;, added mass by rotational velocity I would think, could never be extensive enough to counteract the &quot;flicking off&quot; or centrifugal force of very fast rotation.  Even with a Neutron Star I would think it's impossible, but I'm not 100% sure, so I thought I'd ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we ignore relativity from motion but not $E=mc^2$, as that's the crux of the question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Centrifugal force = $m v^2/r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kinetic energy of rotation = $1/2 I w^2 = 1/5 m v^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass equivalent of kinetic energy = $1/5 m v^2/c^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gravitational force $g=\frac{Gm}{r^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so if we apply the gravitational force of kinetic energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g = G(1/5 m * v^2/c^2) / r^2$, or, simplified, $G*m*v^2 / r^2 c^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we compare the two equations &#xA;Centrifugal  $m v^2/r$&#xA;additional gravitational   $G*m*v^2 / r^2 c^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can remove $m*v^2$ from both on the top and 1 $r$ from the bottom&lt;/p&gt;&#xA;&#xA;&lt;p&gt;additional gravitation ratio to centrifugal force = $G/r * c^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G$ and $c$ are numerical.    $G$ is very small, $c$ is very big and the ratio grows smaller as the radius grows larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gravitational constant:   $6.67408 × 10^{-11} m^3 kg^{-1} s^{-2}&#xA;c = 2.998 x 10^8 m s-2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the ratio, unless my math is broken, centrifugal force to additional gravity from added mass by kinetic energy of motion = $1/r * 1.35 * 10^{27}$, so you'd need a hugely small r, almost a plank length or a singularity where the added gravity from kinetic energy of rotation would overcome the &quot;flicking off the surface&quot; or centrifugal force.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I work out the units, I get meters per kilogram, which I don't think is right.   The units should cancel out with a ratio of two forces in opposite directions, so I think I made an error, but I don't see where I made it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is two fold.    1) is my math broken? and if so, where?   and 2) is the added mass from kinetic rotational energy ever relevant, say in a very rapidly rotating Neutron star?   Could it ever assist the Neutron star in collapse or add gravity?, I can see how it could add to flattening, as rotation flattens objects naturally and perhaps, add a speck of gravity on the polls where centrifugal force is zero, but logically, I think, rotational energy would end up spinning any non black hole object apart, long before the added energy of rotation had enough mass to make a measurable difference on gravity.    Is my sense right or is there a situation where added mass from energy of rotation could overcome the centrifugal force?&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-04-09T20:27:25.930" Title="Is it ever possible that the graviation from the mass of kinetic rotational energy will overcome centrifugal force?" Tags="&lt;general-relativity&gt;&lt;rotation&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13591" PostTypeId="1" CreationDate="2016-02-07T08:18:24.753" Score="7" ViewCount="258" Body="&lt;p&gt;On Earth, it's fairly well published, mostly in climate change related articles, that the Earth's orbital eccentricity operates on a 413,000 year cycle with roughly 90,000–125,000 year variation within that cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Milankovitch_cycles#Orbital_shape_.28eccentricity.29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cause of this variation is primarily the other planets, with Jupiter and Saturn usually mentioned as the primary causes (same link).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question is two fold.   How stable is the Earth's orbital eccentricity cycle?  413,000 years sounds enormously precise, but logically, I would think small changes in planetary orbits would create some variability.   Is the 413,000 years well established and repeating or is it more uncertain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, do we have a good estimate of other planets orbital eccentricity variation?  I looked, but couldn't anything at all on other planets eccentricity cycles.   The closest I came was this gravity simulation chart below.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;90,000 year eccentricity chart of the 4 inner planets &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Eccentricity_rocky_planets.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;,&#xA;and &lt;a href=&quot;http://www.orbitsimulator.com/gravity/articles/what.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;method&lt;/a&gt; used.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_eccentricity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="7142" LastEditDate="2016-11-20T20:23:28.710" LastActivityDate="2018-01-17T16:04:46.343" Title="Orbital eccentricity variation of the other planets?" Tags="&lt;orbital-elements&gt;&lt;orbital-migration&gt;&lt;eccentric-orbit&gt;&lt;orbital-resonance&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13592" PostTypeId="1" CreationDate="2016-02-07T17:24:32.567" Score="-3" ViewCount="104" Body="&lt;p&gt;Whether the collision of particle will lead to a massive WORMHOLE or if BLACKHOLE is also possible? How much I know is that it's created for einstien's E=mc^2 ,conservation of energy from mass and speed of light. I guess also for the STRING THEORY.&lt;/p&gt;&#xA;" OwnerUserId="10704" LastActivityDate="2016-02-07T20:18:37.730" Title="Large hadron collider" Tags="&lt;black-hole&gt;&lt;universe&gt;&lt;cosmology&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2016-02-08T21:31:48.207" ContentLicense="CC BY-SA 3.0" />
  <row Id="13593" PostTypeId="2" ParentId="13592" CreationDate="2016-02-07T19:25:01.973" Score="4" Body="&lt;p&gt;Even if the LHC did create a black hole, its mass would be no greater than the sum of the particles that formed it, i.e. very tiny. It would evaporate almost instantly due to Hawking radiation. Secondly, subatomic particles have been colliding with atoms high in earth's atmosphere for eons, and we're still here.&lt;/p&gt;&#xA;" OwnerUserId="7108" LastActivityDate="2016-02-07T19:25:01.973" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13594" PostTypeId="2" ParentId="13592" CreationDate="2016-02-07T20:18:37.730" Score="3" Body="&lt;p&gt;The LHC has a maximum collision energy of 13 TeV. This yields in mass of $m = E / c^2 = 13 ~\mathrm{TeV} / (3 \times 10^8 \frac{\mathrm{m}}{\mathrm{s}})^2 \approx 2.3 \times 10 ^{-23} ~\mathrm{kg}$.&#xA;Using the definition of the Schwarzschildradius $r_s = \frac{2GM}{c^2}$, our black hole would have the size of $r_s \approx 3.4 \times 10^{-50} m$&#xA;It would be pretty small (and irrelevant)&lt;/p&gt;&#xA;" OwnerUserId="10611" LastActivityDate="2016-02-07T20:18:37.730" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13596" PostTypeId="1" CreationDate="2016-02-07T21:27:29.083" Score="6" ViewCount="184" Body="&lt;p&gt;Is there a relation between the direction of rotation of black holes or neutron stars and the magnetic dipole moment of BH or neutron stars? BTW, are there different directions of this two parameters for suns?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to be sure there are or there are not the direction of rotation (which one can expressed by an arrow on the rotation axis) and the direction of the magnetic field (expressed by an arrow from the south to the north pole (both arrows are simple conventions)) parallel only. I'm not sure that the direction of the magnetic field is observable at all. And the comment from Zibadawa Timmy about pulsars perhaps a good extension to my question about BH, neutron stars and Suns.&lt;/p&gt;&#xA;" OwnerUserId="4242" LastEditorUserId="4242" LastEditDate="2016-02-08T04:47:42.790" LastActivityDate="2017-06-14T18:01:47.553" Title="Are there collected data about the direction of rotation of black holes and the direction of the magnetic field?" Tags="&lt;black-hole&gt;&lt;rotation&gt;&lt;magnetic-field&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13600" PostTypeId="1" AcceptedAnswerId="13608" CreationDate="2016-02-08T03:41:33.587" Score="1" ViewCount="232" Body="&lt;p&gt;The Sun is a fiery mass of colliding particles.  We still know very little about what it is.  Could it possibly be the output of the sun fluctuates, like our weather, from one extreme to another.  From hot durations to cold durations and back again. I think global warming is the result of higher amounts of solar energy being released from the Sun; that's why it's getting hotter here.  I believe the question is important to all people on planet Earth, par excellence; is humanity, namely our consumption of fossil fuels, the cause of our predicament? &lt;/p&gt;&#xA;" OwnerUserId="5306" LastEditorUserId="7472" LastEditDate="2018-02-02T00:07:29.653" LastActivityDate="2018-02-02T00:07:29.653" Title="Are we so sure global warming is a result of humans burning fossil fuels?" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;climate&gt;" AnswerCount="1" CommentCount="11" ClosedDate="2016-02-08T17:58:30.757" ContentLicense="CC BY-SA 3.0" />
  <row Id="13602" PostTypeId="1" CreationDate="2016-02-08T12:06:02.697" Score="5" ViewCount="418" Body="&lt;p&gt;The mankind had to work some centuries to infer the real configuration of solar system, starting from greeks, Ptolemeus, until Copernicus, Galilei, Kepler, Newton etc. Is there any planet where we could better/faster determine the configuration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For example, having a moon is an advantage. Maybe having two moons(or none) or having a thinner atmosphere or being closer/farther to the sun may helped.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of the question starts from inferring the form of the planet(round) until the discovery and orbits of all eight planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: The question ignores the fact that life is not possible on another planet. It is from pure astronomical viewpoint. On Neptune you have a different sky, so other questions/answers.&lt;/p&gt;&#xA;" OwnerUserId="7991" LastActivityDate="2016-04-06T04:42:07.563" Title="What planet is better than earth to infer solar system configuration?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;history&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13603" PostTypeId="1" AcceptedAnswerId="13633" CreationDate="2016-02-08T13:45:51.847" Score="2" ViewCount="307" Body="&lt;p&gt;If predictions are correct, the Milky Way and Andromeda are set to collide in around 4 billion years. If, when this occurs, a Quasar is formed by matter being accreted to the common galactic center (perhaps with the two supermassive black holes forming a binary), what would be the outcome for life in the new galaxy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's known that galactic collisions were much more common in the early universe; most Quasars are observed at high redshifts. It isn't clear if there are other conditions necessary to produce a quasar other than just the galactic collision itself, so obviously this is just speculation, but it is clear that Quasars are some of the most energetic systems in the Universe; if one were ignited so close to home, would every trace of life in the galaxy go out in a blaze of glory?&lt;/p&gt;&#xA;" OwnerUserId="10716" LastActivityDate="2016-02-09T16:05:15.377" Title="What would be the outcome for life in our galaxy if the merger of the Milky Way and Andromeda creates a Quasar?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;life&gt;&lt;quasars&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13604" PostTypeId="2" ParentId="11257" CreationDate="2016-02-08T13:51:06.623" Score="1" Body="&lt;p&gt;(DISCLAIMER) I am not an expert on this subject, and I only hope that my answer will attract better, and more complete ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Meeus book that @Gerard Ashton mentioned is a great resource for amateurs, but I am worried whether it is accurate enough for, let's say, eclipse or transit predictions. Also I glanced at the site linked in the question and saw something like &quot;A live view of Plutonian system&quot;. To be honest I don't think that the Meeus book can enable you to do that kind of thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, serious calculations of planetary positions are done today with the help of dynamical ephemerides developed and maintained by specialized agencies, such as &lt;a href=&quot;https://en.wikipedia.org/wiki/VSOP_(planets)&quot; rel=&quot;nofollow&quot;&gt;VSOP87&lt;/a&gt; by the Bureau des Longitudes in France, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Jet_Propulsion_Laboratory_Development_Ephemeris&quot; rel=&quot;nofollow&quot;&gt;DE431&lt;/a&gt; by Jet propulsion laboratory of NASA. The NASA &lt;a href=&quot;http://ssd.jpl.nasa.gov/?horizons&quot; rel=&quot;nofollow&quot;&gt;Horizons&lt;/a&gt; system is an online interface from which you can get the planets' positions. (It's powered by some version of the DE ephemeris, I think.) It is well documented, so feel free to check out their &lt;a href=&quot;http://ssd.jpl.nasa.gov/?horizons_doc&quot; rel=&quot;nofollow&quot;&gt;documentation&lt;/a&gt; if you want to have a go.&lt;/p&gt;&#xA;" OwnerUserId="10691" LastActivityDate="2016-02-08T13:51:06.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13606" PostTypeId="1" AcceptedAnswerId="13619" CreationDate="2016-02-08T15:41:01.440" Score="20" ViewCount="1611" Body="&lt;p&gt;For a lot of my uninformed life, I have doubted the existence of gravitons or even that gravity is an actual &quot;force&quot; (like electromagnetism). This is because my vision of general relativity was that mass curves space such that objects are still traveling in a &quot;straight line&quot; when being acted upon by &quot;gravity&quot;, so that no &quot;force&quot; is needed. I know now that this is a naive view, but I'm not 100% sure why. I was thinking the other day that just the fact that gravity follows an inverse square law implies that it is a force carried by particles (falling off in flux intensity due to the geometry of 3D space). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question would be: Does the fact that gravity follows an inverse square law naturally fall out of general relativity equations or is it an assumption used when developing the equations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, just now, I had the thought that other forces may curve space also (just in higher dimensions).&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="37191" LastEditDate="2021-02-16T08:54:01.897" LastActivityDate="2021-02-16T08:54:01.897" Title="Existence of gravitons?" Tags="&lt;gravity&gt;&lt;cosmology&gt;&lt;graviton&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13607" PostTypeId="2" ParentId="13606" CreationDate="2016-02-08T16:33:52.787" Score="13" Body="&lt;p&gt;Gravity is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Fictitious_force&quot; rel=&quot;nofollow noreferrer&quot;&gt;fictitious force&lt;/a&gt;, actually, much like centrifugal force.  In a free falling frame of reference it disappears.  In general relativity (GR) gravity is just a result of (differential) geometry: space-time curvature.  The inverse square law is just the low energy approximation, but the actual equation for gravity derived from GR is more complex than that.  The massive success of Newtonian gravity tells us that any model of gravity must be approximated by the classical inverse square law at low energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether GR does that by (Einstein's) design or something else is a matter of personal opinion.  Einstein definitely knew he had to get approximately Newtonian gravity at low energies, so he would have discarded or modified any ideas that failed this criterion.  However, there are &lt;a href=&quot;https://en.wikipedia.org/wiki/Inverse-square_law#Justification&quot; rel=&quot;nofollow noreferrer&quot;&gt;standard arguments for why gravity must obey an inverse square law&lt;/a&gt;, at least in low energy situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in GR &lt;a href=&quot;https://en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor&quot; rel=&quot;nofollow noreferrer&quot;&gt;a lot more than just mass contributes to the gravitational field&lt;/a&gt;.  Spin and electric charge, for example, and quite importantly: energy (the famous $E=mc^2$ tells us how to express a mass as energy, so we can treat these things on a common footing).  So, yes, all forces and particles contribute to gravity.  &lt;a href=&quot;https://physics.stackexchange.com/a/22878/55483&quot;&gt;Even photons&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GR itself makes no predictions (or requirements) for the existence of any new particles outside of the standard model, such as gravitons.  GR and quantum mechanics (QM) are famously incompatible: in extreme situations where both GR and QM are relevant (neutron stars and black hole formation, for example), they stop making sense rather quickly.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Graviton#Gravitons_and_renormalization&quot; rel=&quot;nofollow noreferrer&quot;&gt;Especially GR&lt;/a&gt;.  &quot;Gravitons&quot; and assorted variations are hypothetical particles that are proposed to resolve this issue by creating a quantum theory of gravity.  The only &quot;evidence&quot; we have for them at this stage is that our two most massively successful theories about the workings of the universe, GR and QM, are so painfully incompatible.  So we know these theories are flawed (aka wrong) and that some other theory is necessary that can handle these situations, while also incorporating all of the successes of QM and GR&amp;mdash;they're amazingly accurate when only one of them is particularly relevant, after all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exactly what that theory &lt;em&gt;is&lt;/em&gt; is an ongoing and substantial research area.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-02-08T16:33:52.787" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13608" PostTypeId="2" ParentId="13600" CreationDate="2016-02-08T17:33:37.837" Score="4" Body="&lt;p&gt;We know a lot about the sun. It isn't the main cause of global warming, as we can monitor exactly how much energy it is producing. The sun's output can fluctuate though, and we have yet to measure how badly this affects the climate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So what is the main cause of global warming?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main cause of global warming is emissions of C02 (carbon dioxide) into the atmosphere. C02 is odorless, colourless and non-toxic, however, it is the main cause of global warming. C02 is a greenhouse gas and  can stay in our atmosphere for about 20 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C02 causes something called the greenhouse effect. The greenhouse effect is the idea that long wave energy which is meant to go back to space, doesn't go back to space, and is instead captured by greenhouse gases such as C02. This causes lots of heat that should go off into space to be trapped inside the earth's atmosphere causing your temperature to rise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are there any other greenhouse gases?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other greenhouse gases do exist, carbon dioxide is just the main one that causes the global warming. Other greenhouse gases include water vapor, methane, nitrous oxide and ozone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Does global warming happen only on Earth?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Global warming happens all over the  solar system, planets like venus have many greenhouses gases in their atmosphere too, infact, on venus greenhouse gases alone make the planet 465 degrees celsius hot, which is way over the boiling point of water - 100 degrees celsius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Do I disagree?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, as there is too little evidence to prove that the sun fluctuates enough energy to gradually increase our climate dramatically, and we have also proven the effects of greenhouse gases in a planets atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="9645" LastActivityDate="2016-02-08T17:33:37.837" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13609" PostTypeId="2" ParentId="13602" CreationDate="2016-02-08T17:55:00.457" Score="4" Body="&lt;p&gt;&lt;strong&gt;I will argue Mars.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A smaller diameter makes it easier to determine the round shape of the planet, and determine the diameter more exactly.&lt;/li&gt;&#xA;&lt;li&gt;A thinner atmosphere and almost no clouds are better conditions for sky observation.&lt;/li&gt;&#xA;&lt;li&gt;The small orbital radius of the Mars moons makes it easy to figure out the distance to them. (For example, Phobos is only in the sky about a third of the time seen from equator). Also, the fact that there are two of them makes two data points to discover &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler&amp;#39;s_laws_of_planetary_motion&quot; rel=&quot;nofollow&quot;&gt;Kepler's third law&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Closer vicinity of the asteroid belt calls for an earlier discovery of objects with an irregular orbit.&lt;/li&gt;&#xA;&lt;li&gt;The Moon is a much closer example of something orbiting another object than Galileo’s discovery of the Jovian moons. &quot;Mars is the centre of the universe&quot; disproved.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10172" LastActivityDate="2016-02-08T17:55:00.457" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13610" PostTypeId="2" ParentId="13582" CreationDate="2016-02-08T17:57:02.380" Score="1" Body="&lt;p&gt;I think this is suggesting that the star is 3 pc from us (the observer), and you need to find the distance of the planet from it's star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the 3 pc was the distance of the planet from the star, then the question is just asking you to convert parsecs to AU, which is very easy and would make the information about the angular radius useless. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is the information you have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Distance form observer to star: 3pc&lt;/li&gt;&#xA;&lt;li&gt;Observed radius of orbit: 0.1 arcsec&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You want to use this to find the distance from the planet to the star, using trigonometry and converting to the correct units&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2016-02-08T17:57:02.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13611" PostTypeId="2" ParentId="13567" CreationDate="2016-02-08T18:20:26.897" Score="0" Body="&lt;p&gt;Caveat: I'm not a cosmologist, so this answer may not reflect the forefront of scientific knowledge/accuracy, but I have some knowledge so thought I'd share it and hopefully someone can correct me if I'm wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Big Bang theory states that everything is moving away from everything else, so it must have started closer together. We know from the Cosmic Microwave Background that everything was very dense and hot and small. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may have heard that the universe started in a singularity. This means one point at which all matter was in the same place. In this state, all information in matter is lost. No particle can be labelled as having a position or size or spin, because every particle is in the same state as every other particle. This means we lose all possibility of having any information about what happened before this. There are many theories of what &lt;em&gt;could&lt;/em&gt; have been before this, but no matter how simple/beautiful/mathematically rigorous the theory, it is physically impossible to know anything about it so they are impossible to test. In some definitions this makes these theories not actually science, as the scientific method involves making testable predictions.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2016-02-08T18:20:26.897" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13612" PostTypeId="1" CreationDate="2016-02-08T18:48:18.933" Score="2" ViewCount="147" Body="&lt;p&gt;Do we have a few decades worth of surface temperature data to compare?&lt;/p&gt;&#xA;" OwnerUserId="10720" LastActivityDate="2016-02-10T12:50:08.247" Title="Are any other planetary bodies warming up as well in the Solar System?" Tags="&lt;fundamental-astronomy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13614" PostTypeId="1" CreationDate="2016-02-08T20:28:36.057" Score="2" ViewCount="88" Body="&lt;p&gt;The question is related to this about &lt;a href=&quot;https://astronomy.stackexchange.com/questions/13596/are-there-collected-data-about-the-direction-of-rotation-of-black-holes-and-the?noredirect=1#comment20105_13596&quot;&gt;Are there collected data about the direction of rotation of black holes and the direction of the magnetic field?&lt;/a&gt;. Perhaps it is observable for pulsars?&lt;/p&gt;&#xA;" OwnerUserId="4242" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2017-01-13T22:00:10.897" Title="Can we determine where the north and south magnetic poles are on stars and black holes?" Tags="&lt;black-hole&gt;&lt;magnetic-field&gt;&lt;pulsar&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13615" PostTypeId="1" CreationDate="2016-02-08T23:47:11.540" Score="3" ViewCount="13240" Body="&lt;p&gt;Is there any evidence that supports or disproves the Oscillating Universe theory? What kind of observations could provide this evidence?&lt;/p&gt;&#xA;" OwnerUserId="10690" LastEditorUserId="2365" LastEditDate="2016-03-11T18:49:53.367" LastActivityDate="2019-08-21T18:28:16.857" Title="Are there any studies that prove/disprove the Oscillating Universe Theory?" Tags="&lt;fate-of-universe&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13616" PostTypeId="1" CreationDate="2016-02-08T23:56:02.130" Score="1" ViewCount="75" Body="&lt;ol&gt;&#xA;&lt;li&gt;Is there proof in either the CMB or other observed astronomical phenomena for Big Crunch?&lt;/li&gt;&#xA;&lt;li&gt;Could there have been a partial Big Crunch maybe like a cluster of Black Holes colluding to create a disappearing zone in the Universe?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10690" LastActivityDate="2016-02-08T23:56:02.130" Title="Is there any support for the Big Crunch Theory? Could there have been a partial Big Crunch in the past?" Tags="&lt;fate-of-universe&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ClosedDate="2016-10-04T15:05:39.263" ContentLicense="CC BY-SA 3.0" />
  <row Id="13617" PostTypeId="1" AcceptedAnswerId="13630" CreationDate="2016-02-09T00:17:29.350" Score="0" ViewCount="82" Body="&lt;p&gt;At the time of the big bang if there was an original entity, what were it's attributes(if applicable) - mass, electric charge, color charge or spin? Can the current matter and dark energy together account for the original entity or was some of the original stuff lost?&lt;/p&gt;&#xA;" OwnerUserId="10690" LastActivityDate="2016-02-09T15:03:19.890" Title="What was the entity/particle/state at the time of the Big bang?" Tags="&lt;early-universe&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13618" PostTypeId="2" ParentId="13606" CreationDate="2016-02-09T00:54:33.367" Score="6" Body="&lt;h1&gt;First, the fact that gravity falls off $1/r^2$ is visible in the metric.&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The metric describes the curvature of the space. For space around a massive object this is the Schwarzchild metric&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathrm{d}s^2 = -\left(1-\frac{r_s}{r}\right)\mathrm{d}t^2 + \left(1-\frac{r_s}{r}\right)^{-1}\mathrm{d}r^2 + r^2(\mathrm{d}\theta^2 + \sin^2\theta  \ \mathrm{d}\phi^2)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, if $r \gg r_s$ this looks like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathrm{d}s^2=-\mathrm{d}t^2+\mathrm{d}r^2+r^2(\mathrm{d}\theta^2 + \sin^2\theta \ \mathrm{d}\phi^2)&#xA;$$&#xA;which is the metric for flat space. So effectively the space gets flatter and flatter at a rate of $1/r^2$, which is the inverse square that you're looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But where does the Schwarzchild metric come from? Without getting into the gritty maths, it can be proven that it is the unique metric that possesses spherical symmetry, without which nothing would make much sense. This is called Birkhoff's theorem.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The little afterthought on your question takes some more thought&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I want to talk about where gravitons come from, but first lets talk about curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to measure the curvature of a space one way of doing it is to move in some closed loop, ending up back where you started. If the space is curved though, you won't be facing the same direction (this idea is called parallel transport)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/3/3e/Connection-on-sphere.png&quot; alt=&quot;Parallel Transport&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we're parallel-transporting a tangent vector, as in the picture. We get a tangent vector from the derivative at a point (a slighly special derivative called a covariant derivative, because the space is curved). Let's take the tangent vector and move forwards then left. And we try again this time moving left then forwards. We end up the same point both ways, but like in the picture, the derivatives will be different in some way. We summarise this up with a commutator (where $D$ is the covariant derivative) like so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;[D_\mu,D_\nu] = D_\mu D_\nu - D_\nu D_\mu\neq 0&#xA;$$&#xA;It basically means &quot;doing it one way is not the same as doing it the other&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's back up a tiny bit and talk about how electromagnetism and other forces are typically discussed, using quantum field theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We describe the theory in terms of a Lagrangian, for a fermion (like an electron) it looks like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathcal{L}=\bar\psi(i\gamma^\mu D_\mu-m)\psi&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I take the field $\psi$ and give it a transformation&#xA;$$&#xA;\psi\to\psi'=e^{i\xi(x)}\psi&#xA;$$&#xA;then the Lagrangian will remain unchanged. This type of transformation belongs to a group called $U(1)$. We say that the Lagrangian possesses $U(1)$ symmetry. Notice that this $D_\mu$ is in there again? Its the same thing, a covariant derivative, here in QED as well. We can try taking a commutator again&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;[D_\mu,D_\nu] = -iF_{\mu\nu}\psi&#xA;$$&#xA;where&#xA;$$&#xA;F_{\mu\nu} = \partial_\mu A_\nu -  \partial_\nu A_\mu&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this we form the complete QED (the quantum theory of electrodynamics) Lagrangian&#xA;$$&#xA;\mathcal{L}=\bar\psi(i\gamma^\mu D_\mu-m)\psi-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't get bogged down in the maths. The point is very simple. See the $A_\mu$? Its a new field, we had to introduce it to make things work. In QED this field corresponds to a photon (particles are quanta of a field, like a small bump in the field). We had to introduce it &lt;strong&gt;becuase we had curvature&lt;/strong&gt;. How do I know we have curvature? &lt;strong&gt;Becuase the covariant derivatives do not commute&lt;/strong&gt;, just like in GR, above. This time though, the curvature is not of phyical space, its of an abstract object called the $U(1)$ gauge bundle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you're totally on the right track when you say that other forces may curve space. Its nice that gravity curves space-time, its very physical and easy to imagine, for the other forces its not so simple to picture, even though its fundamentally the same.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Anyway, back to GR&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;If you want the full picture of Einstein's gravity you do some maths and arrive at something called the Einstein-Hilbert action (an action is just an integral over a Lagrangian), one tidy object that sums up the whole theory&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;S=\int R \sqrt{g} \ \mathrm{d}^4x&#xA;$$&#xA;where $R$ comes (more or less) from the commutator of covariant derivatives we saw at the top. When talking about QED I brushed over the fact that its a quantum theory (it is). This EH action, however, does not describe a quantum theory. So, you might say, lets make it one! Hold on a second though, because it doesn't actually work. The problem is something called renormalisability - QED is renormalisable, GR is not. This is the root of the incompatibility between GR and quantum field theory. If we could carry out the resultant qunatum particle would be a graviton. You're right to doubt their existance as they haven't yet been observed, however...&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Two versions of the same thing&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;We saw QED, which desribes particles of light, photons. They are quantised. Then we saw how in many ways GR and QED are very similar. We can't properly quantise GR but if we could we would have gravitons, exactly like photons popped out in QED. The duality between QED (and other gauge theories, QCD, etc) is clear, which leads a lot of people to believe that probably should have gravitons, even if they have not yet been observed, nor consistently formulated.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;A note on other theories&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There are many theories where gravitons are present from first principles without the problems of renormalisability, string theory or supergravity for instance.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;A note on errors in the above&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Sorry, I'm tired and rambling. Please point them out if you find them!&lt;/p&gt;&#xA;" OwnerUserId="10724" LastEditorUserId="9301" LastEditDate="2016-10-20T18:18:19.433" LastActivityDate="2016-10-20T18:18:19.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13619" PostTypeId="2" ParentId="13606" CreationDate="2016-02-09T02:19:09.790" Score="10" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;For a lot of my uninformed life, I have doubted the existence of gravitons or even that gravity is an actual &quot;force&quot; (like electromagnetism).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gravity is a force like electromagnetism, but it does have a special property in that all test particles fall the same way in a gravitational field, no matter their composition. This means that inertial masses and gravitational masses are the same (or at least universally proportional, so we can use units in which they are equal), and we are free to interpret gravitational freefall as inertial motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of quantum field theory, it is actually a theorem that at low energies, massless spin-2 particles must couple to all energy-momentum equally, regardless of particle species. In other words, the equivalence principle of general relativity is a provable theorem for gravitons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, we can also interpret general relativity as a massless spin-2 field on a flat background spacetime, but because of this universality, the background will be unobservable by any experiment. That's why relativists don't tend to do this, as it makes the geometric interpretation is more convenient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, quantized general relativity is very badly behaved if one tries to take them to arbitrarily energy scales. Physically, this means that some new physics must come in before then to fix it. However, this kind of situation is hardly unique to gravity, quantizing which still makes sense as an effective field theory at lower energies; cf. &lt;a href=&quot;http://relativity.livingreviews.org/Articles/lrr-2004-5/&quot;&gt;living review by Cliff P. Burgess&lt;/a&gt;. The tension between general relativity and quantum mechanics is often overstated in popular descriptions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My question would be: Does the fact that gravity follows an inverse square law naturally fall out of general relativity equations or is it an assumption used when developing the equations?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The inverse-square part falls out by itself, but the specific constant of proportionality needs an additional assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one considers a general field equation $G_{\mu\nu} = \kappa T_{\mu\nu}$, where $T_{\mu\nu}$ is the stress-energy tensor that is assumed to be symmetric and covariantly conserved, then the Einstein tensor $G_{\mu\nu} \equiv R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R$ is the unique scale-invariant solution that can built from the metric. This requirement means that only terms that are second-order in derivatives of the metric are allowed, and it is broken by e.g. cosmological constant term $\Lambda g_{\mu\nu}$, as this introduces a length $\Lambda^{-1/2}\sim 10^{10}\,\mathrm{ly}$ into the theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other ways of developing the Einstein field equation, e.g. via the Einstein-Hilbert action, which don't need specific assumptions about the stress-energy tensor. Regardless, the role of the Newtonian limit is in fixing the value of the otherwise undetermined constant $\kappa = 8\pi G/c^4$. If you're only interested in a Newton-like inverse-square relationship, then &lt;em&gt;that alone&lt;/em&gt; doesn't need any additional assumptions about trying to match Newtonian gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a timelike vector field $u$, which can be interpreted as the four-velocities of some family of observers, we can write the time-time projection of an equivalent form of the Einstein field equation, $R_{\mu\nu} = \kappa(T_{\mu\nu}-\frac{1}{2}g_{\mu\nu}T)$, as&#xA;$$R_{00} \equiv R_{\mu\nu}u^\mu u^\nu = \frac{1}{2}\kappa(\rho+3p)\text{,}$$&#xA;where $\rho$ is the energy density and $p$ is the average of the principal stresses as measured by an observer with four-velocity $u$. For non-relativistic matter, the stress terms are negligible compared to the energy density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way the Newtonian limit is typically discusses is to use the weak-field approximation, $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$ with $|h_{\mu\nu}|\ll 1$, to show that&#xA;$$\frac{1}{2}\kappa\rho\approx R_{00} = R^\alpha{}_{0\alpha 0}\approx\partial_\alpha\Gamma^{\alpha}_{00}\approx-\frac{1}{2}\nabla^2h_{00}\text{,}$$&#xA;which then has the form of Poisson's equation for Newtonian gravitational potential in terms of matter density $\rho_\text{m}$, i.e. $\nabla^2\Phi = 4\pi G\rho_\text{m}$. For slow-moving test particles, the geodesic equation reduces to the Newtoni an equation of motion:&#xA;$$\frac{\mathrm{d}^2\mathbf{x}}{\mathrm{d}t^2} = \frac{1}{2}\nabla h_{00} = -\nabla\Phi\text{.}$$&#xA;Another way to think of this is to write down the proper time of freefalling particle and show that extremizing it is equivalent to extremizing  $\int\left(\frac{1}{2}v^2+\frac{1}{2}h_{00}\right)\mathrm{d}t$, which is the action action (per mass) of a particle subject to Newtonian gravity whenever $h_{00} \approx -2\Phi/c^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may be interested in &lt;a href=&quot;http://math.ucr.edu/home/baez/einstein/node6a.html&quot;&gt;this simpler derivation&lt;/a&gt; of Newton's law of gravitational around a spherically symmetric body, based on the geometric interpretation of the Ricci curvature as the acceleration of the volume of a small ball of initially comoving test particles.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And, just now, I had the thought that other forces may curve space also (just in higher dimensions).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This was done for electromagnetism by Kaluza and Klein shortly after GTR, but it turns out that it's not a directly useful way to think about other forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, we can think of Riemann curvature in general relativity as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Curvature_form&quot;&gt;curvature form&lt;/a&gt; of the Levi-Civita connection on the tangent bundle of a given manifold, with structure group $\mathrm{O}(1,n)$. But in this language, the electromagnetic field strength is the curvature of the connection $ieA_\mu$ over a line bundle with structure group $\mathrm{U}(1)$. The other non-gravitational forces are similarly described by &lt;a href=&quot;https://en.wikipedia.org/wiki/Yang%E2%80%93Mills_theory&quot;&gt;Yang-Mills theory&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the other forces already have a description in which they're caused by a curvature, just not of spacetime. So while gravity is different from them, it's not different enough to consider it in some sense 'less real' than the others.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2016-02-09T02:19:09.790" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13620" PostTypeId="1" AcceptedAnswerId="13623" CreationDate="2016-02-09T03:42:17.713" Score="-1" ViewCount="187" Body="&lt;p&gt;Does  parallel universe really exists or just science fiction which ain't possible? Can we ever know?&lt;/p&gt;&#xA;" OwnerUserId="10726" LastEditorUserId="37191" LastEditDate="2021-02-01T21:51:12.267" LastActivityDate="2021-02-01T21:51:12.267" Title="Is parallel universe possible" Tags="&lt;cosmology&gt;&lt;universe&gt;&lt;space&gt;&lt;multiverse&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ClosedDate="2016-02-10T21:39:57.247" ContentLicense="CC BY-SA 4.0" />
  <row Id="13621" PostTypeId="2" ParentId="13620" CreationDate="2016-02-09T05:11:11.713" Score="0" Body="&lt;p&gt;As much as parallel universes go, there's a theory for this known as the Multiverse theory, part of the String theory.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In String theory, the Multiverse is a theory in which our universe is&#xA;not the only one; many universes exist parallel to each other. These&#xA;distinct universes within the multiverse theory are called parallel&#xA;universes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;There are cosmic inflation models that support this as said &lt;a href=&quot;https://en.wikipedia.org/wiki/Eternal_inflation#Inflation_and_the_multiverse&quot; rel=&quot;nofollow noreferrer&quot;&gt;here in Wikipedia&lt;/a&gt;. A quote from there:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;It's hard to build models of inflation that don't lead to a&#xA;multiverse. It's not impossible, so I think there's still certainly&#xA;research that needs to be done. But most models of inflation do lead&#xA;to a multiverse, and evidence for inflation will be pushing us in the&#xA;direction of taking [the idea of a] multiverse seriously. Alan Guth[7]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now to address your point if we can ever know if a parallel universe exists exists.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Opinion&lt;/strong&gt;: Possible but unlikely. I don't think that we are close to achieving this.&lt;/p&gt;&#xA;" OwnerUserId="7926" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-02-09T05:11:11.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13622" PostTypeId="2" ParentId="1443" CreationDate="2016-02-09T06:47:17.390" Score="4" Body="&lt;p&gt;You always look into the past. If you look into mirror, you see yourself as you looked moment before. Specifically this is how long before, in seconds where $d$ is the distance to mirror in meters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d*2}{300000}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If such mirror faced earth and was far enough, we would be able to see past indeed. Actually &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment&quot; rel=&quot;nofollow&quot;&gt;there's a tiny mirror facing Earth on Moon&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I can't believe nobody posted this one yet:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://imgs.xkcd.com/comics/geese.png&quot; title=&quot;Anyway, that's a common misconception. Geese live for a long time; all the ones we can see will probably keep flying around for billions of years before they explode.&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1135" LastActivityDate="2016-02-09T06:47:17.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13623" PostTypeId="2" ParentId="13620" CreationDate="2016-02-09T09:34:51.107" Score="2" Body="&lt;p&gt;Multiple and parallel universes are a feature of a number of &quot;theories&quot;, including most famously (and probably before string theory), the &lt;a href=&quot;https://en.wikipedia.org/wiki/Many-worlds_interpretation&quot; rel=&quot;nofollow&quot;&gt;many worlds interpretation&lt;/a&gt; of quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The attraction of the many worlds or multiverse ideas is that they offer a ready explanation of why we appear to live in a uiverse that has been &quot;fined tuned&quot; for our existence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The disadvantage of the &quot;theory&quot; is that there does not appear to be any readily available means to falsify it. For that reason many scientists view it as unscientific.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-02-09T09:34:51.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13624" PostTypeId="1" AcceptedAnswerId="13639" CreationDate="2016-02-09T09:50:45.937" Score="8" ViewCount="350" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9pJC6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9pJC6.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a picture of a jet of particles exiting the supermassive black hole in the center of Pictor A galaxy at nearly the speed of light. &lt;a href=&quot;http://www.astronomy.com/news/2016/02/blast-from-black-hole-in-a-galaxy-far-far-away?utm_source=SilverpopMailing&amp;amp;utm_medium=email&amp;amp;utm_campaign=ASY_News_Sub_160205_Final%20remainder&amp;amp;utm_content=&amp;amp;spMailingID=24659256&amp;amp;spUserID=MTE2MjkyOTM4OTcxS0&amp;amp;spJobID=741050037&amp;amp;spReportId=NzQxMDUwMDM3S0&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This event is supposedly created from a black hole of 40 million solar masses. The &lt;a href=&quot;http://futurism.com/distant-supermassive-black-hole-emits-giant-death-rays-larger-galaxy/&quot; rel=&quot;nofollow noreferrer&quot;&gt;full quote&lt;/a&gt; here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The newly released picture shows the terrifying power of an actual&#xA;  supermassive black hole located in the galaxy Pictor A, some 500&#xA;  million light years distant.  &lt;strong&gt;This black hole is about 40 million&#xA;  times the mass of the Sun (that’s about 10 times larger than the black&#xA;  hole at the center of our own galaxy)&lt;/strong&gt;, and if it were to replace the&#xA;  Sun within our Solar System, its event horizon would swallow the&#xA;  orbits of both Mercury and Venus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now a quick search to &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_most_massive_black_holes&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; would show a abundance of supermassive black holes many, many times larger than that supermassive black hole as well as a few that are around the same mass as that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my question: What set of special conditions that surround this particular supermassive black hole such that it is capable of creating this spectacular event?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit for clarity: By this I meant the jet and the hotspot.&lt;/p&gt;&#xA;" OwnerUserId="7926" LastEditorDisplayName="user1569" LastEditDate="2016-02-10T16:31:53.023" LastActivityDate="2016-02-10T16:31:53.023" Title="What conditions would lead to this event around the black hole in the Pictor A galaxy?" Tags="&lt;supermassive-black-hole&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13625" PostTypeId="1" CreationDate="2016-02-09T10:47:50.420" Score="4" ViewCount="162" Body="&lt;p&gt;If general relativity is understood to mean that mass curves space such that objects are still traveling in a straight line, why does mass naturally move closer toward's the center of other masses?&lt;/p&gt;&#xA;" OwnerUserId="10729" LastActivityDate="2016-02-27T03:01:57.270" Title="Why does mass naturally move closer toward's the center of other masses?" Tags="&lt;gravity&gt;&lt;general-relativity&gt;&lt;mass&gt;&lt;movement&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13626" PostTypeId="2" ParentId="13602" CreationDate="2016-02-09T12:39:21.563" Score="4" Body="&lt;p&gt;I think Mercury has some of the points that Mars has. Plus:&lt;br&gt;&#xA;- a highly eccentric orbit, making it more obvious that planets have ellptical orbits,&lt;br&gt;&#xA;- no moon, which is great to avoid falling into the pitfall of thinking about being at the center of the universe,&lt;br&gt;&#xA;- a very long solar day (about 6 earth months or two mercurian years), leaving plenty of time to study planetary motions&lt;/p&gt;&#xA;" OwnerUserId="10730" LastActivityDate="2016-02-09T12:39:21.563" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13627" PostTypeId="2" ParentId="13602" CreationDate="2016-02-09T13:28:07.073" Score="4" Body="&lt;p&gt;Venus, I propose, because it (almost) doesn't rotate. The winning (aristotelian) argument for geocentrism before Galileo was the idea that everybody's' hats would fly off by the (ethereal) air drag if the Earth rotated any faster than a horse in gallop (let alone at Earth's crazy 465 m/s compared to Venus' 1.8 m/s). Long nights too, as previously said. And with Mercury still as an inner planet, the faces of which to give further Galilean clues.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2016-02-09T13:36:02.713" LastActivityDate="2016-02-09T13:36:02.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13628" PostTypeId="2" ParentId="13625" CreationDate="2016-02-09T13:29:34.883" Score="3" Body="&lt;p&gt;Think at the surface of a ball. A straight line within this curved surface is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Great_circle&quot; rel=&quot;nofollow&quot;&gt;great circle&lt;/a&gt;.&#xA;Take some point on the ball. Draw two great circles (apparent straight lines in the surface) starting in this point, with some non-zero initial angle. The two apparent straight lines intersect at the opposite side of the ball.&#xA;Wrongly assuming the surface of the ball being flat would result in the conclusion, that the lines must be curved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same applies to our three-dimensional space. We intutively assume it as flat, such that two lines starting at an observer with some angle should never intersect elsewhere. But in some cases they do; so we intuitively think the lines must be curved. But according to general relativity, it's actually our space-time which is curved by masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, a ballistic curve looking like a parabola is actually a straight line in the curved space-time. So, two objects moving in different directions in e.g. Earth's orbit can eventually hit each other despite lack of drag or propulsion. They both move on a straight line (&lt;a href=&quot;https://en.wikipedia.org/wiki/Geodesic&quot; rel=&quot;nofollow&quot;&gt;geodesic&lt;/a&gt;) in a curved space-time. The curved space-time allows these to lines to intersect later. Our three-dimensional Euclidean intuition interpretes this as curved trajectories.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2016-02-09T13:29:34.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13630" PostTypeId="2" ParentId="13617" CreationDate="2016-02-09T15:03:19.890" Score="2" Body="&lt;p&gt;The notions of today's universe don't make much sense for the (hypothetical) pre-Planck epoch of the universe. All four known fundamental forces should have been united. Matter didn't exist. It can form only in cooler conditions. Even cause and effect don't work well below Planck scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure you eventually get can be treated in terms of a groupoid, as described e.g. in &lt;a href=&quot;http://arxiv.org/pdf/gr-qc/9906072.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;. Groupoids are related to &lt;a href=&quot;https://en.wikipedia.org/wiki/Group_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;groups (algebra)&lt;/a&gt;. Groups actually are - in some sense - very simple &lt;a href=&quot;https://en.wikipedia.org/wiki/Groupoid&quot; rel=&quot;nofollow&quot;&gt;groupoids&lt;/a&gt;. You may construct some groupoids from disjoint groups by connecting them via &lt;a href=&quot;https://en.wikipedia.org/wiki/Isomorphism&quot; rel=&quot;nofollow&quot;&gt;isomorphisms&lt;/a&gt;. So you get kind of a network of abstract entities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now try to adjust quantum mechanics and General Relativity, such that it fits with these structures, meaning without the smooth space and time, as usually assumed for standard physics. Let known or extrapolated physics of the Planck- and GUT (grand unified theory) epoch emerge from the pre-Planck physics to start the actual Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's possible to construct physical theories for a pre-Planck era. But it's very hard to find observational or experimental evidence to test these theories with respect to actual physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those pre-Planck groupoid theories can be defined in a way that they don't need a predecessor epoch, e.g. by eliminating the macroscopic cause and effect paradigm, which is based on a notion of an &lt;a href=&quot;https://en.wikipedia.org/wiki/Arrow_of_time&quot; rel=&quot;nofollow&quot;&gt;arrow of time&lt;/a&gt;.&#xA;Causality and a concept of before/after on isomorphic entities doesn't make much sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although, there should be some mechanism which eventually disrupted a strictly non-local pre-Planck configuration.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2016-02-09T15:03:19.890" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13631" PostTypeId="2" ParentId="13624" CreationDate="2016-02-09T15:43:56.353" Score="1" Body="&lt;p&gt;It's infalling matter, as described in &lt;a href=&quot;http://futurism.com/distant-supermassive-black-hole-emits-giant-death-rays-larger-galaxy/&quot; rel=&quot;nofollow&quot;&gt;the article&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The new image shows how the material being consumed by the black hole—stars, planets, interstellar gas, unwary astronauts (kidding)—releases a tremendous surplus of energy as it swirls around the event horizon in a massive accretion disk.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or in &lt;a href=&quot;http://www.astronomy.com/news/2016/02/blast-from-black-hole-in-a-galaxy-far-far-away?utm_source=SilverpopMailing&amp;amp;utm_medium=email&amp;amp;utm_campaign=ASY_News_Sub_160205_Final%20remainder&amp;amp;utm_content=&amp;amp;spMailingID=24659256&amp;amp;spUserID=MTE2MjkyOTM4OTcxS0&amp;amp;spJobID=741050037&amp;amp;spReportId=NzQxMDUwMDM3S0&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A huge amount of gravitational energy is released as material swirls towards the event horizon, the point of no return for infalling material. This energy produces an enormous beam, or jet, of particles traveling at nearly the speed of light into intergalactic space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The infalling matter, mostly dust and gas, but also stars, is accelerated to nearly the speed of light. That's a giant amount of energy. Part of it is emitted as electromagnetic radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other factor is surrounding dust. If all this happens in a dense cloud of dust, you don't see it well from out outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next: Are you sure, other supermassive black holes don't show similar features? &lt;a href=&quot;https://en.wikipedia.org/wiki/Quasar&quot; rel=&quot;nofollow&quot;&gt;Quasars&lt;/a&gt; near the edge of the visible universe are visible, since their respective jet points towards us.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2016-02-09T15:43:56.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13632" PostTypeId="2" ParentId="13612" CreationDate="2016-02-09T15:52:43.223" Score="1" Body="&lt;p&gt;Pluto is currently cooling down, since its distance from the Sun is increasing now, a few years after &lt;a href=&quot;https://en.wikipedia.org/wiki/Perihelion_and_aphelion&quot; rel=&quot;nofollow&quot;&gt;perihelion&lt;/a&gt;. That's seasonal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nasa.gov/centers/ames/research/2007/marswarming.html&quot; rel=&quot;nofollow&quot;&gt;Mars has been suspected to warm up&lt;/a&gt;. But as far as I know, that's questioned in the meanwhile.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2016-02-09T15:52:43.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13633" PostTypeId="2" ParentId="13603" CreationDate="2016-02-09T16:05:15.377" Score="2" Body="&lt;p&gt;Quasars have very pointed jets perpendicular to the rotational axis of the accretion disc around the central black hole. As long as this jet doesn't happen to point into the galactic plane, the cosmic radiation of these jets won't be a too serious problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a collision of two galaxies can become an eventual problem in a different way. A merger of galaxies triggers star formation and sweeps most of the galaxy free of interstellar dust and gas. So, after a short (hundreds of millions of years) period of time, we'll get lots of new stars. But they'll eventually exceed their main sequence lifetime. Due to lack of interstellar matter only very few new stars will form afterwards. That's a difficult settings for formation of planets and life.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2016-02-09T16:05:15.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13634" PostTypeId="2" ParentId="13590" CreationDate="2016-02-09T16:35:45.913" Score="2" Body="&lt;p&gt;In some sense, the answer is trivially 'yes':&#xA;Due to spacetime curvature, there exist no &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_geodesics#Circular_orbits_and_their_stability&quot; rel=&quot;nofollow&quot;&gt;circular orbits&lt;/a&gt; below $\frac{3}{2}r_s$, with $r_s$ the Schwarzschild radius.&#xA;So whenever you get an object sufficiently dense, that its radius is below its $\frac{3}{2}r_s$, the mass equivalent to any nonzero rotational energy overcomes the centrifugal force of an orbit at $\frac{3}{2}r_s$.&#xA;Below $\frac{3}{2}r_s$ even zero rotational energy overcomes the centrifugal force, since the latter is then pointing inward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't require a Black Hole.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2016-02-09T16:35:45.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13635" PostTypeId="1" CreationDate="2016-02-09T21:03:14.897" Score="2" ViewCount="1102" Body="&lt;p&gt;I am trying to obtain the distance covered by an object in orbit around the earth within a specified amount of time past its perigee passage. The object is in an elliptical orbit with eccentricity 0.2 and has a semi-major axis of 9600km. Where would I look to find the objects position 90 mins after it passes its perigee? Thanks for any advice&lt;/p&gt;&#xA;" OwnerUserId="10735" LastActivityDate="2016-03-14T03:02:34.090" Title="Determining distance from semi-major axis and eccentricity" Tags="&lt;orbit&gt;&lt;astrophysics&gt;&lt;fundamental-astronomy&gt;&lt;orbital-elements&gt;&lt;stellar-astrophysics&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13636" PostTypeId="1" AcceptedAnswerId="13648" CreationDate="2016-02-09T22:43:57.233" Score="3" ViewCount="392" Body="&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1501.07270&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; and &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/0004-637X/816/1/21/pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; state without discussion that the presence of lithium absorption in the stars they observe as strong evidence of these stars being pre-main sequence stars. Because they state this without any discussion and justification, I assume that it's a commonly known and accepted fact.  The second of the two papers also state without discussion or justification that some observed strong H$\alpha$ emission was also evidence of the stars being pre-main sequence stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Googling around I was able to get a partial answer for why this is so.  Pre-main sequence stars retain their primordial lithium as they collapse, but at some point around when they become main sequence stars, the lithium in their photospheres is destroyed by fusing with protons. I haven't been able to find anything describing why H$\alpha$ emission is associated with pre-main sequence stars, other than this has been known for a very long time (since before at least 1945).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is it about becoming a main sequence star that causes the lithium to fuse with protons near the surface?&lt;/li&gt;&#xA;&lt;li&gt;How fast is the lithium destroyed, and does this destruction occur right when the star enters the main sequence?  I'm wondering how definite the presence of lithium absorption is in saying whether a star has entered the main sequence or not.&lt;/li&gt;&#xA;&lt;li&gt;Why is H$\alpha$ emission such a prominent feature of pre-main sequence stars?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="128" LastEditorUserId="128" LastEditDate="2016-02-10T15:06:11.390" LastActivityDate="2016-02-10T15:06:11.390" Title="Why do pre-main sequence stars show lithium in absorption?" Tags="&lt;star&gt;&lt;star-formation&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13637" PostTypeId="5" CreationDate="2016-02-09T22:57:26.290" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2016-02-09T22:57:26.290" LastActivityDate="2016-02-09T22:57:26.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13638" PostTypeId="4" CreationDate="2016-02-09T22:57:26.290" Score="0" Body="Behaviour and properties of light, and its interaction with matter, particularly in detection of light." OwnerUserId="2270" LastEditorUserId="746" LastEditDate="2016-02-10T19:09:10.560" LastActivityDate="2016-02-10T19:09:10.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13639" PostTypeId="2" ParentId="13624" CreationDate="2016-02-10T00:29:45.820" Score="3" Body="&lt;p&gt;This is a clear example of an &lt;a href=&quot;https://en.wikipedia.org/wiki/Astrophysical_jet&quot; rel=&quot;nofollow&quot;&gt;astrophysical jet&lt;/a&gt;, in this case, most likely a relativistic jet. In short, an accretion disk forms around a black hole (supermassive or otherwise). Matter is pulled towards the black hole and further energized, before being accelerated into a jet emanating from the black hole's poles. Two different mechanisms have been proposed for the formation of jets:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Blandford%E2%80%93Znajek_process&quot; rel=&quot;nofollow&quot;&gt;Blandford-Znajek process&lt;/a&gt;&lt;/strong&gt; requires that a magnetic field forms (from the accretion disk) that is centered around the black hole. Charged particles then move along the field lines, into jets. I recently wrote an answer about the details (see &lt;a href=&quot;https://astronomy.stackexchange.com/questions/13570/how-does-an-accreting-black-hole-acquire-magnetic-fields/13580#13580&quot;&gt;How does an accreting black hole acquire magnetic fields?&lt;/a&gt;). For this process to work, you need an accretion disk. It is generally considered the most likely explanation for black hole jets.&lt;/li&gt;&#xA;&lt;li&gt;The &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Penrose_process&quot; rel=&quot;nofollow&quot;&gt;Penrose process&lt;/a&gt;&lt;/strong&gt; takes rotational kinetic energy from the ergosphere outside the event horizon and gives it to particles moving in jets. Note that this does not rely as heavily on the accretion disk as the Blandford-Znajek process does. For this process to work, you need a rotating black hole surrounded by some matter, likely in a disk.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The hotspot is, to me, much more interesting. It reminds me of structures seen around young stars: &lt;a href=&quot;https://en.wikipedia.org/wiki/Bipolar_outflow&quot; rel=&quot;nofollow&quot;&gt;bipolar outflows&lt;/a&gt; (streams of gas that can form shock waves) and &lt;a href=&quot;https://en.wikipedia.org/wiki/Herbig%E2%80%93Haro_object&quot; rel=&quot;nofollow&quot;&gt;Herbig-Haro objects&lt;/a&gt; (the results of shock waves from relativistic jets. Obviously, the mechanisms are different, so no clear analogy can be drawn. But what is interesting about bipolar outflows and Herbig-Haro objects is that the shock waves produced therein result from collisions with the interstellar medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a similar mechanism were to cause the shock waves by the hotspot, then we could conclude that the jets have hit the intergalactic medium. But I don't think this is necessarily the case, in part because of just how long these jets are prior to the formation of the hotspot. One would think that if the hotspot and shock waves are because of collisions with the intergalactic medium, the jets would be much shorter, because they would likely have reached higher density regions of it sooner. So that's why I find it interesting, and why I can't give you a good reason as to why the hotspot formed where it did, or the precise reason for it being there at all.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2016-02-10T00:29:45.820" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13644" PostTypeId="1" AcceptedAnswerId="13645" CreationDate="2016-02-10T07:22:28.823" Score="5" ViewCount="6751" Body="&lt;p&gt;If we take a look at stars more massive than the Sun, their densities vary a lot. UY Scuti is an extremely low-density star that's only &lt;strong&gt;8.5x&lt;/strong&gt; more massive than the Sun, but is &lt;strong&gt;1000-2000x&lt;/strong&gt; its size. r136a1, on the other hand, is &lt;strong&gt;256x&lt;/strong&gt; more massive than the Sun, but is only &lt;strong&gt;30x&lt;/strong&gt; its size. Neither of them are main-sequence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun has a lower mass than both of them, yet it's also smaller than both of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't seem like mass makes a big difference. Both reach hydrostatic equilibrium at completely different sizes. r136a1 has a much stronger gravitational pull than UY Scuti, but it should also have much more radiation pressure, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how can we determine the density of a star? Is there a formula?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-02-10T08:08:17.117" LastActivityDate="2016-02-10T08:33:17.393" Title="How do star densities work?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;mass&gt;&lt;stellar-dynamics&gt;&lt;density&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13645" PostTypeId="2" ParentId="13644" CreationDate="2016-02-10T08:33:17.393" Score="5" Body="&lt;p&gt;The mean density of the star is really only defined by the formula $\bar\rho=M/V=3M/4\pi R^3$.  The radius of a star is a generally a very complicated function of a star's other properties.  When we determine the radius in stellar models, it's only because we've solved equations that describe the structure of the whole star, and read off the value at what we define as the surface. So no simple formula in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, one can derive the approximate functional dependence for stars of various evolutionary states through the principle of &lt;em&gt;homology&lt;/em&gt;. i.e. assuming that stars of a certain type are just rescaled versions of each other. Glancing at my old course notes, on the upper main sequence, where stars burn hydrogen principally through the CNO cycle and have radiative envelopes dominated by electron-scattering opacity, we derived $R\propto M^{15/19}$. The same principle (but with different assumptions about the star) is used to determine the location of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hayashi_track&quot;&gt;Hayashi track&lt;/a&gt; for pre-main-sequence stars, along which $R\propto M^{-7}T^{49}$.  Particular formulae can be found for different types of star but the relationships between $M$ and $R$ vary wildly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither the two stars you mentioned are typical main-sequence stars. R136a1 is a Wolf-Rayet star, which is basically a star that has blasted away most of its hydrogen envelope.  Mass-radius relations are usually strongly dependent on mean molecular weight, which is higher without hydrogen, so the relations break down (or, rather, would have to be derived separately). But usually higher mean molecular weight gives a more compact star. UY Scuti has probably finished burning hydrogen in its core and has moved off the main sequence. So again, it'll follow a different relation.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2016-02-10T08:33:17.393" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13646" PostTypeId="1" AcceptedAnswerId="13651" CreationDate="2016-02-10T10:59:29.963" Score="5" ViewCount="340" Body="&lt;p&gt;Before reading this question, you should read &lt;a href=&quot;http://www.ligo.org/news/media-advisory.php&quot; rel=&quot;noreferrer&quot;&gt;this paper.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;From what I understood, we managed to detect gravitational waves for the first time.&lt;/p&gt;&#xA;&lt;p&gt;Is this discovery going to help us understand something we haven't understood yet? Dark matter? Galaxy movements? FTL travel? Gravity manipulation, Interstellar style?&lt;/p&gt;&#xA;&lt;p&gt;Though I understand the discovery, I'm a little confused about what it will actually improve for science.&lt;/p&gt;&#xA;&lt;p&gt;So, what are the doors this discovery will open ?&lt;/p&gt;&#xA;" OwnerUserId="9679" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-02-11T22:59:00.453" Title="What do gravitational waves allow us to understand?" Tags="&lt;gravity&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13648" PostTypeId="2" ParentId="13636" CreationDate="2016-02-10T11:27:03.003" Score="4" Body="&lt;p&gt;As pre-main sequence (PMS) stars contract towards the main sequence, their cores become hotter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lithium is turned into helium by proton capture reactions. These are initiated &lt;em&gt;in the core&lt;/em&gt; at temperatures of around 3 million K, so much lower than for hydrogen burning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same time, PMS stars, or at least the type of low-mass pre-main sequence stars that are being referred to here, are fully convective and this mixes material inside the star rapidly and effectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the sequence of events is that the PMS star contracts; the core reaches $3\times 10^{7}$ K at some point before the main sequence; Li is destroyed by proton capture; the Li-depleted is completely mixed with the rest of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way Li can be rapidly and totally destroyed on a timescale that depends on the mass of the PMS star. It is as short as 10 Myr for a star like the Sun, increasing to about 100 Myr for a star of $0.1 M_{\odot}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the basic picture, but there is a wrinkle for stars with $M&amp;gt;0.4M_{\odot}$, where the core becomes radiative rather than convective &lt;em&gt;before&lt;/em&gt; they reach the main sequence, but &lt;em&gt;after&lt;/em&gt; Li burning has commenced. The radiative core pushes outwards and its  base then drops below $3\times 10^{6}$ K. This means that although Li destruction runs to completion in the core, at least some Li can be preserved in the outer layers of such stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Li is therefore an extremely complicated function of age and mass. But for the purposes of identifying &lt;em&gt;low-mass&lt;/em&gt; PMS stars we can clearly say that anything that shows strong lithium signatures is probably younger than 100 million years if it is below about $0.5 M_{\odot}$ (or equivalently is a K- or M-type star).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want much more detail, see my review article on &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0411111&quot;&gt;Pre-main sequence Lithium depletion&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The H$\alpha$ emission of PMS stars is connected with their youth. It can arise in two ways. The first is associated with accretion from a circumstellar disc. This is caused by gas falling from the disk onto the star and heating up. The H alpha emission can be very strong and very doppler broadened ($&amp;gt;300$ km/s). As pre-main sequence discs only survive for 10 million years or less, then this is a very clear signature of a pre main sequence star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, young stars that have lost their discs also show strong (though not as strong as accretion-related emission) H$\alpha$ emission that is caused by non-radiative, magnetic heating of their chromospheres. This magnetic activity is associated with convective, rapidly rotating stars. Pre main sequence stars possess both these properties. Thus PMS stars must exhibit H$\alpha$ emission of some sort.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-02-10T11:27:03.003" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13649" PostTypeId="1" AcceptedAnswerId="13654" CreationDate="2016-02-10T12:09:29.870" Score="4" ViewCount="1679" Body="&lt;p&gt;When stars run out of hydrogen, they explode (though they also use heavier elements for some time) and form nebulae. In the nebulae, new stars are born which use hydrogen as their fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is from where do these new stars get hydrogen for their fuel?&lt;/p&gt;&#xA;" OwnerUserId="10741" LastEditorUserId="2270" LastEditDate="2016-02-10T14:20:40.407" LastActivityDate="2016-02-11T09:39:54.693" Title="Where do new stars get their hydrogen from?" Tags="&lt;star-formation&gt;&lt;star&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ClosedDate="2016-02-10T16:22:58.637" ContentLicense="CC BY-SA 3.0" />
  <row Id="13650" PostTypeId="2" ParentId="13612" CreationDate="2016-02-10T12:38:19.660" Score="2" Body="&lt;p&gt;The short answer to your question is &quot;decades&quot;, only for Mars, and only a two Mars decades (about 4 Earth decades)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if your question was related to climate change, as there were a number of extremely unscientific arguments made that Pluto and Mars are getting warmer, &quot;&lt;em&gt;and there are no SUV's on Mars&lt;/em&gt;&quot; . . . so, maybe it's not the CO2 but those claims weren't science.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silly articles and bad arguments aside, global temperature on other planets is a legit scientific question and there were scientific articles that said Pluto is getting warmer (2002, story below) and Mars is getting warmer (2007, also below), so there is some truth to this.   I'll give an answer as best I can.  Each planet deserves it's own bit of research, so this will be longish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pluto&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an &lt;a href=&quot;http://news.mit.edu/2002/pluto&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; on Pluto, the title is iffy but the article is good.   It's from 2002, so it's long before New Horizon's fly-by, it's even before New Horizon was launched, but &quot;pluto is getting warmer&quot; was a real scientific story in 2002, and Mars, in 2007 was real science too, but some of the conclusions drawn, loosely based on those two stories were ridiculously bad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gist of the article is, we can't see Pluto's atmosphere well, even with Hubble, but they were able to measure some aspects of Pluto's atmosphere by watching what happens when a star passes behind Pluto and that was measured twice, in 1988 and again in 1997.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pluto's Perihelion was in 1989 and most everyone figured that the little ball of ice/dwarf planet would get colder after 1989 when it passed it's closest point to the sun and stated moving further away but the readings said otherwise, and that was one of those really cool &quot;oh, wow, things are more interesting than we thought they were&quot; moments.  Pluto's atmosphere had apparently gotten thicker and it was about 2 degrees warmer.  Not what anyone expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, the sun couldn't make Pluto 2 degrees warmer while it moved further away without making the Earth 20 degrees warmer at the same time, so, clearly, something else was at play and it wasn't the sun warming Pluto.  Generally speaking, the warmest periods aren't necessarily precisely timed to closest to the sun, and there's a lag time caused (on Pluto) by sublimation of the frozen gases on it's surface, which thickens the atmosphere adds trace amounts of Methane and can warms the planet past Pluto's perihelion, just as on the Earth, June 21 isn't the warmest day on the Northern Hemisphere, nor December 21st the coldest, and 12:00 noon isn't the hottest time.  Could that seasonal &quot;lag time&quot;, apply over a number of Earth years on Pluto?   It's plausible.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we should remember, that's just two data points, which isn't nearly enough.  Or, 3 data points now with New Horizons, but it's still just 3 data points spanning 27 years on a planet where one year is 248 earth-years, so, there's a lot we don't know about seasons on Pluto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;New Horizons got better data on Pluto's atmosphere and frozen surface than we've ever had before, and what I remember reading about it, though I'm very far from well read on the subject, is that Pluto has a much more dynamic atmosphere and surface than anyone expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Gas Giants&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you asked: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do we have a few decades worth of surface temperature data to compare?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Surface temperature isn't well defined on gas giant planets because we can't see the surface.   Also (I think all 4, I know for sure Jupiter and Neptune), they give off more heat internally than they receive from the sun.  Temperature on the visible surfaces / upper atmospheres of the gas giants has more to do with circulation than anything else.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mars&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a number of articles on Mars warming, but there's still a lot we don't know about Mars' climate.   It's not too difficult using modern technology and with Mars' thin atmosphere, to read surface temperature on Mars by measuring the infra red light that comes from it's surface.  The surface temperature on Mars was first measured in 1920 (see &lt;a href=&quot;https://en.wikipedia.org/wiki/History_of_Mars_observation&quot; rel=&quot;nofollow&quot;&gt;4th paragraph from the top&lt;/a&gt;), and NASA says here that they've been taking accurate surface temperature readings on Mars for 30 years (article from May 2007), so it's pretty safe to say we have about 38-39 Earth years (20 or so Mars years) of reasonably good global temperature readings on Mars and a few decades more than that with less accuracy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's many problems with reaching any kind of conclusion on Mars based on that data though.  Mars has significantly greater eccentricity than Earth, so it may be prone to larger temperature swings than Earth and just 20 Mars years of data is too few.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Winter to Summer on Mars' enough CO2 can freeze to reduce it's atmosphere by as much as 25% and that can drive temperature changes and mars is also prone to planet wide dust storms that can last months.  These can have the effect of blocking sunlight and after settling, darkening the ice caps, which lowers albedo.  Like Pluto, Mars' atmosphere, climate, temperature and orbit are dynamic and there's still a lot we don't know about how they worth together to form a predictable climate on Mars.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an FYI, dust storms are nothing like in the movie the Martian, but they can cause a visible haze over the planet that can obscure viewing it's surface.   A nice short article on Mars' dust storms &lt;a href=&quot;http://www.universetoday.com/14892/mars-dust-storms/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.nasa.gov/feature/goddard/the-fact-and-fiction-of-martian-dust-storms&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's kind of cool that Mars forms an annual &quot;cloud&quot;, not unlike how parts of Earth form annual snow and ice covers in winter that thaw in Summer, but we don't know nearly enough about Mars, even with 20 Mars years of observation, to make good long term predictions on Mars' climate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Venus&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus, as far as I know, pretty much never changes.  It's the same temperature, Winter, Summer, Pole, Equator.  There might be some slight variation of Venus temperature depending on elevation, like the gas giants, but I'm not sure.   Venus traps heat so effectively that is changes very little, in fact, I've never seen one article mentioning Venus' surface temperature changing as much as 1 degree.   Venus reflects most of the sunlight that hits it away anyway, so it would be less prone to solar variation anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I won't do Mercury cause I don't know enough about it, but that's 7 out of 8 anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;. . . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your question is about having a good understanding of climates on other planets, then decades of surface temperature data wouldn't be enough even if we did have it.  Centuries of accurate surface temperature data, you could begin to recognize patterns, for the inner 4 planets anyway, but decades isn't enough to differentiate between a baseline, a natural fluctuation or a transition period.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone tells you there are no SUV's on mars, so how could Mars be warming, get away from them as fast as you can.  They're not interested in science.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-02-10T12:50:08.247" LastActivityDate="2016-02-10T12:50:08.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13651" PostTypeId="2" ParentId="13646" CreationDate="2016-02-10T12:48:45.123" Score="8" Body="&lt;p&gt;I think the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational-wave_astronomy#Science_potential&quot; rel=&quot;noreferrer&quot;&gt;wikipedia&lt;/a&gt; page gives a reasonable overview of why gravitational waves give us a window on the universe that is either not observable or is complementary to the view we get from electromagnetic waves (or neutrinos).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let's have a specific example. The potential of LIGO, as we have seen today, is to detect the gravitational wave signatures of merging black hole binaries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power radiated in gravitational waves strongly depends on the orbital separation of the black holes, it also depends on their masses. The frequency of the radiation will change as the black holes spiral together. The &quot;noise&quot; that this makes in the final few seconds as the black holes merge is the &quot;chirp&quot; that is detected by LIGO. Have a look (listen!) to what a chirp might &quot;sound&quot; like at this web page from &lt;a href=&quot;http://www.sr.bham.ac.uk/gwastro/what-are-we-looking-for&quot; rel=&quot;noreferrer&quot;&gt;University of Birmingham&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact characteristics of the chirp (duration, frequency etc.) can for instance yield the masses of the black holes and of their final configuration and may tell us about spin rates of the black holes. Yet these same mergers may produce zero electromagnetic wave signatures that could be seen by conventional telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, because the amplitude and the frequency of the waves depend differently on the masses and &lt;em&gt;distance&lt;/em&gt; to the GW source, these binary chirps acts as &quot;standard candles&quot; in a similar way to type Ia supernovae. In other words, measurements of the chirp independently gives the masses of the merging black holes &lt;em&gt;and&lt;/em&gt; the distance to them.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-02-11T22:59:00.453" LastActivityDate="2016-02-11T22:59:00.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13653" PostTypeId="1" AcceptedAnswerId="13655" CreationDate="2016-02-10T13:06:17.663" Score="5" ViewCount="9291" Body="&lt;p&gt;I am a student of class 9 and recently came to know about kepler's law of planetary motion and got confused about the foci and elliptical orbits of planets and some stars.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2016-06-04T17:33:03.857" Title="Why do planets move in an elliptical orbit?" Tags="&lt;eccentric-orbit&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ClosedDate="2016-04-16T18:43:22.883" ContentLicense="CC BY-SA 3.0" />
  <row Id="13654" PostTypeId="2" ParentId="13649" CreationDate="2016-02-10T13:24:53.950" Score="6" Body="&lt;p&gt;Stars only burn hydrogen in their core, where the temperature gets high enough for nuclear reactions to occur. They end their lives when they run out of fuel in the core, but lots of hydrogen still exists in their envelopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun will burn 5-10% of its mass before exhausting its core. When this happens, the core will contract due to the radiation pressure disappearing. It then starts to burn hydrogen in a shell around the core. Eventually, when the Sun dies, it will have burned less than half of its hydrogen. Larger stars burn an even smaller fraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that, when stars die they still leave hydrogen behind for the next generation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxies can still run of of gas, though. Since after all, each $M_\odot$ of star formed burns &lt;em&gt;of the order of&lt;/em&gt; $1\,M_\odot$, if a galaxy isn't fueled with new gas it will become depleted on a timescale of order $1$ over its &lt;em&gt;specific star formation rate&lt;/em&gt; sSFR, which is its star formation rate SFR measured in Solar masses per year, divided by its stellar mass $M_*$ in Solar masses:&#xA;$$&#xA;t_\mathrm{depl} \sim \frac{1}{\mathrm{sSFR}}=\frac{M_*/M_\odot}{\mathrm{SFR}/M_\odot\,\mathrm{yr}^{-1}}.&#xA;$$&#xA;For instance, a $10^9\,M_\odot$ galaxy with a star formation rate of $10\,M_\odot\,\mathrm{yr}^{-1}$ will become depleted of gas in roughly $10^8$ years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gas-depleted galaxies do exist, and the more depleted they are, the lower the star formation rate is (e.g. &lt;a href=&quot;http://arxiv.org/abs/0912.5243&quot; rel=&quot;nofollow&quot;&gt;Rose et al. 2010&lt;/a&gt;), but in general the timescale is longer than the above, since galaxies also accrete gas from the surrounding circumgalactic medium.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-02-11T09:39:54.693" LastActivityDate="2016-02-11T09:39:54.693" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13655" PostTypeId="2" ParentId="13653" CreationDate="2016-02-10T14:40:18.743" Score="5" Body="&lt;p&gt;Not sure if you're looking for a more mathematical answer or just the &quot;why&quot;, but to answer the why, I'll start with some history on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everyone who worked out a model for the Solar System, from Aristotle to Copernicus, liked circles.  Even though Copernicus correctly reasoned that the Earth moved around the Sun and not the Sun around the Earth, he continued to use circles in his models of the motion of the planets.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After Copernicus, Tycho Brahe, funded by the King of Denmark, had the best equipment at the time for observing the motion of the stars and planets and he was able to make star charts that were ten times as accurate as anyone before him. Brahe used equipment like &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/b/be/Tycho-Brahe-quadrant_01.png&quot; rel=&quot;nofollow&quot;&gt;this mural quadrant&lt;/a&gt;, and a large private observatory to take extremely accurate records.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kepler, who was a better mathematician than Brahe, desperately wanted to get his hands on Brahe's star charts and the use of his observatory and equipment (so much so that when Brahe died, there were rumors that Kepler had poisoned him, though that probably didn't happen). When Kepler finally had everything at his disposal, he was able to work things out and study the Solar System more accurately. However, he still didn't know why the planets moved in ellipses; he'd only worked out that the ellipses fit the movement so well that it almost certainly had to be true, but he had no idea why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kepler, in fact, didn't care for ellipses.  He liked circles better, but he couldn't deny that ellipses worked.  &lt;a href=&quot;http://www.polaris.iastate.edu/EveningStar/Unit2/unit2_sub4.htm&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nobody knew why planets moved in ellipses until Isaac Newton was asked that question and had to invent calculus to answer it.  Calculus explains why planets orbit in ellipses, and that's the real answer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &quot;calculus&quot; isn't a satisfying answer, a way to kind of explain it would be to throw a penny out of a space shuttle (which isn't a good idea, but let's say you do). As the penny falls towards the Earth, it falls faster and faster (if we ignore air resistance) until it hits the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you fling the penny from the space shuttle at a much faster speed and at a different angle, so that it only gets &lt;strong&gt;near Earth&lt;/strong&gt; and misses the planet, it would actually start to orbit Earth. It would fall faster and faster until it passes the Earth, and then, like shooting a bullet into the air, the penny will slow down as it flies past the Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Kepler's second law, the penny's fastest speed is at the point closest to the Earth (&lt;a href=&quot;https://en.wikipedia.org/wiki/Apsis&quot; rel=&quot;nofollow&quot;&gt;the perigee&lt;/a&gt;). That's essentially how objects in orbits work: as they move closer to the body they orbit, they accelerate faster and faster. Our penny will get so fast that, once it comes around the planet, it will be flung very far away, which will then slow it down. This is what creates an elliptical orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its motion is like a spring, falling toward the planet then flying away, but at the same time, orbiting in a circular motion with the spring motion, with 1 period per orbit. That motion of moving closer and then further in each orbit forms an ellipse.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It makes the most sense if you think of the velocity being greatest at the closest point and lowest at the furthest point. The low velocity moves it closer while the high velocity moves it back out further.  The total energy of the object in orbit (kinetic energy plus potential energy) remains constant. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="9301" LastEditDate="2016-06-04T17:33:03.857" LastActivityDate="2016-06-04T17:33:03.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13657" PostTypeId="1" AcceptedAnswerId="13659" CreationDate="2016-02-10T15:00:56.617" Score="6" ViewCount="4178" Body="&lt;p&gt;This &lt;a href=&quot;https://www.youtube.com/watch?v=kJkVegBsNyE&quot; rel=&quot;noreferrer&quot;&gt;video&lt;/a&gt; states that NASA has determined that the far side of the moon's crust is thicker than the near side's crust. How did they determine this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note I'm not asking for theories or hypotheses on how this came to be, but rather if the difference in thickness indeed is measurable scientifically today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;[So off-topic: If it is in fact true, I favour the theory that the centrifugal force of the earth's gravity along with the fact that the far side is always pointing outwards, is responsible for this and not some other moon smashing into our moon etc. as is shown in that video.]&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="10747" LastEditorUserId="10747" LastEditDate="2016-02-11T09:08:05.313" LastActivityDate="2017-05-31T05:37:17.707" Title="Why is it said that the far side of the moon's crust is thicker?" Tags="&lt;the-moon&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13659" PostTypeId="2" ParentId="13657" CreationDate="2016-02-10T17:30:47.440" Score="8" Body="&lt;p&gt;The near side and far side have very different topography. Compared to the near side, the far side is far more rugged and has far fewer maria. The near side in places has very thin crust (the maria).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than these surface features, the biggest indicator of a disparity in crustal thickness is the two kilometer offset between the Moon's center of figure and it's center of mass. This is typically attributed to a thicker crust on the far side of the Moon, consistent with the observations of the different surfaces of the near side and far side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NASA discovered that the center of figure (geometrical center) and the center of mass are offset from one another in the late 1960s. The simple act of observing how a probe orbits an object gives clues regarding the interior of that object. Scientists know, for example, that Mars has a partially molten core from observing the orbits of the many probes that have been sent to Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the Moon, NASA sent 16 unmanned probes to the Moon in the 1960s prior to the Apollo 8 mission. The first four of these, Ranger 6, 7, 8, and 9, were impactors. Their impacts the Moon indicated the Moon's center of mass was displaced from its center of figure by 2-3 kilometers. Later missions confirmed this, and also confirmed that the Moon's near side has five large mass concentrations (&quot;mascons&quot;).  The Apollo 15, 16, and 17 missions carried laser altimeters that gave very precise measurements of the distance to the Moon's surface. These further improved the estimate of the Moon's center of figure / center of mass offset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Aside: Read &lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2006/06nov_loworbit/&quot; rel=&quot;nofollow&quot;&gt;Bizarre Lunar Orbits&lt;/a&gt; for a description of how those lunar mascons change how things orbit the Moon.)&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="10760" LastEditDate="2016-02-11T20:27:53.637" LastActivityDate="2016-02-11T20:27:53.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13660" PostTypeId="2" ParentId="13657" CreationDate="2016-02-10T19:03:56.373" Score="5" Body="&lt;p&gt;You may find &lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2014/07/01/the_moon_s_two_faces_why_are_they_so_different.html&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; by Phil Plait or &lt;a href=&quot;http://earthsky.org/space/dark-side-of-the-moon-mystery-solved&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt; on EarthSky.org interesting.&lt;br&gt;&lt;br&gt;Basically, a new hypothesis proposes that, after the impact that created the moon, it got tidally locked to the Earth in only a few (~100) days. Because the Earth was still hot (2,500°C) from the residual heat of impact, it cooked the near side but let the far side cool more rapidly, thus forming a thicker crust than the near side.&lt;br&gt;&lt;br&gt;I like this hypothesis better than the secondary splat hypothesis, because it's neater and seems to follow what would have been a more natural sequence of events.&lt;br&gt;&lt;br&gt;I know this specific question was not asked but thought this hypothesis was germane to the discussion.&lt;/p&gt;&#xA;" OwnerUserId="7108" LastEditorUserId="2618" LastEditDate="2016-02-10T19:29:02.703" LastActivityDate="2016-02-10T19:29:02.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13661" PostTypeId="1" AcceptedAnswerId="13667" CreationDate="2016-02-10T19:51:02.127" Score="2" ViewCount="44" Body="&lt;p&gt;Because of the expansion of the universe, light from a far galaxy is redshifted. The expansion of the universe will make the wavelength of light longer. But is the colour of such light the same for us, as for an galaxy which is between us and the far galaxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I first thought that a further galaxy's speed was higher than ours, but I don't think that's right it is relative. But is the speed of the far galaxy from a view of the further galaxy higher than from our view? I don't think so because the universe is expanding faster the further you look. So for the further galaxy the speed of the far galaxy is less high than from our view because the far galaxy is further away from us. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is the conclusion right that the colour of the same wave is more red for us on earth than from the further galaxy?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="2365" LastEditDate="2016-02-10T22:08:41.757" LastActivityDate="2016-02-11T08:09:39.457" Title="Is the colour of a wave from a far galaxy the same for us as for a galaxy which lies between?" Tags="&lt;light&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13662" PostTypeId="1" AcceptedAnswerId="13665" CreationDate="2016-02-10T20:07:15.167" Score="5" ViewCount="359" Body="&lt;ol&gt;&#xA;&lt;li&gt;Is it just the momentum from the Big Bang? If yes, how can it account for the acceleration of the expansion?&lt;/li&gt;&#xA;&lt;li&gt;Why isn't the collective gravitational force slowing it down?&lt;/li&gt;&#xA;&lt;li&gt;What part does dark energy play in the expansion?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10690" LastActivityDate="2016-02-10T21:26:14.537" Title="What is driving the expansion of the Universe?" Tags="&lt;universe&gt;&lt;expansion&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13663" PostTypeId="2" ParentId="13585" CreationDate="2016-02-10T20:18:50.527" Score="4" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Lagrangian_point&quot; rel=&quot;nofollow&quot;&gt;Lagrangian point&lt;/a&gt; $L_2$ is very close to the most distant point from Earth with an umbra.&#xA;$L_2$ is like the radius of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;nofollow&quot;&gt;Hill sphere&lt;/a&gt; at $r=a\sqrt[3]{\frac{m}{3M}}$ for circular orbits, with $m$ the mass of Earth, $M$ the mass of the Sun, and $a$ the distance Earth-Sun. The ratio $\frac{m}{3M}$ of the Earth and the triple mass of the Sun is almost exactly $10^{-6}$, the cubic root hence $0.01$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diameter ratio of Earth and Sun is about $1/109$. Therefore the umbra of Earth ends near $92\%$ the distance to $L_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to another bonus question would then be: If Earth would be $9\%$ larger in diameter, but with the same mass, its umbra would end almost exactly at $L_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth's orbit isn't perfectly circular, but the aphel/perihel ratio of about $1.04$ is insufficient to question the result qualitatively.&#xA;The error of the implicite assumptions $\tan x=x=\sin x$ is negligible at the considered level of accuracy.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2016-02-10T20:44:48.460" LastActivityDate="2016-02-10T20:44:48.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13664" PostTypeId="1" AcceptedAnswerId="13757" CreationDate="2016-02-10T21:23:53.480" Score="4" ViewCount="318" Body="&lt;p&gt;I've been interested in astronomy for a while but only recently purchased my first set of binoculars (7x50). I've been learning various constellations as well as spotting some other objects with the help of my planisphere. I live in the suburbs of a big Australian city, but far enough out that the light pollution isn't too bad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've already been looking at objects like the Orion Nebula and Pleiades, but hoped some experts might be able to give me a list of objects that would be worth hunting down with my binoculars. Ideally some that would be visible now (it's Summer here), but some for other times of the year would also be fine.&lt;/p&gt;&#xA;" OwnerDisplayName="user10636" LastActivityDate="2016-02-14T21:16:15.710" Title="What Are Some Good Southern Hemisphere 7x50 Binocular Targets?" Tags="&lt;amateur-observing&gt;&lt;binoculars&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13665" PostTypeId="2" ParentId="13662" CreationDate="2016-02-10T21:26:14.537" Score="3" Body="&lt;p&gt;As far as I understood it, no one really knows. It is observable that the universe is expanding with increasing speed, therefore there has to be a form of energy driving it. We can meassure (or at least try) the amount of energy and find $\Omega_\lambda = 69.11\%$ (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Lambda-CDM_model#Parameters&quot; rel=&quot;nofollow&quot;&gt;$\lambda$CDM-Model&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To your questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: I wouldn't imagine it as Big Bang momentum, for it is also changing. After the Big Bang, there was (or at least had to be, to make our comological model work) a period of inflation, meaning rapid expansion, which then slowed down again. I can't think of an explanation which includes some kind of finite momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: On small scales (speaking of galaxies), where densities are high, expansion doesn't act. Gravity is working against it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3: It is responsible for the expansion. How exactly is not understood. (Little personal comment form my side. If a physicist calls something 'dark', it mostly likely means it is not understood)&lt;/p&gt;&#xA;" OwnerUserId="10611" LastActivityDate="2016-02-10T21:26:14.537" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13666" PostTypeId="2" ParentId="13661" CreationDate="2016-02-10T22:04:41.563" Score="1" Body="&lt;p&gt;The light from distant galaxies is redshifted because they are moving away from us and the more distant the galaxy is the faster it moves, and so the more the light is redshifted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A galaxy between us and a very distant galaxy is also moving away from us, but less fast. From the point of view of that galaxy the light of the distant galaxy is redshifted less than from our position, because the distant galaxy is not moving away from it as fast. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The light would appear less red in &quot;a galaxy which lies between&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-02-10T22:04:41.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13667" PostTypeId="2" ParentId="13661" CreationDate="2016-02-10T22:05:02.867" Score="2" Body="&lt;p&gt;If I understand you right, you're asking whether or not the redshift of the photons emitted from a far-away galaxy happens the instant it leaves the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Redshift is gradual…&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If so, the answer is no. The redshifting of photons happen gradually as they travel through the expanding Universe. You can find the derivation &lt;a href=&quot;https://en.wikipedia.org/wiki/Redshift#Mathematical_derivation&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; where you'll see that every infinitesimally small increase $da$ of the scale factor $a$ of the Universe (its &quot;size&quot;) increases the photon's redshift by an amount $dz$, or, in terms of wavelength, by an amount $d\lambda$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If galaxy $B$ lies at redshift $z_\mathrm{B}$, then an observer in galaxy $A$ at redshift $z_\mathrm{A}$ lying between us and $B$ (so that $z_\mathrm{A}&amp;lt;z_\mathrm{B}$) would measure $B$'s redshift to be&#xA;$$&#xA;z_\mathrm{B\,seen\,from\,A} = \frac{1+z_\mathrm{B}}{1+z_\mathrm{A}} - 1,&#xA;$$&#xA;which is less than $z_\mathrm{B}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/c2JfG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c2JfG.png&quot; alt=&quot;zAB&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;…at least in our Universe&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The redshifting is not due to the source moving away from us. If the expansion hadn't been gradual, but we instead lived in a crazy universe that were static when the distant galaxy emitted the light, and static when we observe, but somehow expanded suddenly by some factor in the meantime, then we would still observe a redshift, even though the galaxy were static both when it emitted the light and when we observed it.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-02-11T08:09:39.457" LastActivityDate="2016-02-11T08:09:39.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13668" PostTypeId="2" ParentId="13653" CreationDate="2016-02-10T22:24:34.460" Score="1" Body="&lt;p&gt;You probably know how things fall when they are thrown: A ball follows a curved path known as a parabola, the path of a ball is not circular, and it is certainly not a circle centred on the middle of the Earth. So you don't expect things moving under gravity always to move in circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets move under the action of gravity, just as if they were giant balls thrown around the sun, and they follow curved paths that are not circles. The exact shape can be calculated (&lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;though the maths is complex&lt;/a&gt;) and the shape is one of a family of curves: either an ellipse or a hyperbola.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stable orbits of the planets are all ellipses, and usually nearly circular. Comets can have orbits that are long thin ellipses, or are hyperbolic, in which case the comet doesn't orbit, but passes the sun and has enough energy to escape from the sun forever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may wonder why the planets orbits are nearly circular, but that is &lt;a href=&quot;https://physics.stackexchange.com/questions/2501/why-are-the-orbits-of-planets-in-the-solar-system-nearly-circular&quot;&gt;another question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-02-10T22:24:34.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13671" PostTypeId="2" ParentId="13664" CreationDate="2016-02-11T10:13:16.407" Score="4" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Plaiedes -  a jewel box&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Coal Sack - in Crucis, there's a globular cluster I recall there somewhere. Hours of fascination.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Messier objects&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Jupiter moons &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Saturn's rings and moons,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;? Lycrae&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Large and Small Magellanic Clouds - extra-galactic remnants on the southern pole of the Milky Way&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Just for starters. By the time you work through all these, you'll start to gain an appreciation of your own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Happy Hunting!!!&lt;/p&gt;&#xA;" OwnerUserId="10752" LastActivityDate="2016-02-11T10:13:16.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13672" PostTypeId="1" CreationDate="2016-02-11T14:19:23.720" Score="-1" ViewCount="181" Body="&lt;p&gt;Dark lines in a continuum spectrum is common. But are there bright lines too? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the stimulated emission process, the light can also be amplified, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, we should get bright lines instead of dark lines.&lt;/p&gt;&#xA;" OwnerUserId="10519" LastEditorUserId="10519" LastEditDate="2016-02-11T14:56:15.627" LastActivityDate="2017-01-02T14:06:25.113" Title="Bright lines in a continuum spectrum" Tags="&lt;spectroscopy&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13673" PostTypeId="1" AcceptedAnswerId="13674" CreationDate="2016-02-11T14:49:47.123" Score="1" ViewCount="68" Body="&lt;p&gt;Today the LIGO consortium are having a press conference to update us on their search for gravitational waves. How should I watch the announcement?&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2016-02-11T14:51:48.707" Title="Where can I watch the announcement from LIGO about gravitational waves?" Tags="&lt;gravitational-waves&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13674" PostTypeId="2" ParentId="13673" CreationDate="2016-02-11T14:51:38.720" Score="2" Body="&lt;p&gt;The event will be live streamed on youtube here: &lt;a href=&quot;https://www.youtube.com/user/VideosatNSF/live&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/user/VideosatNSF/live&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The will also be tweeting: &lt;a href=&quot;https://twitter.com/ligo&quot; rel=&quot;nofollow&quot;&gt;https://twitter.com/ligo&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A video of the conference will also be available afterwards if you miss it. It starts at 10:30 EST, which is 3:30 GMT.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2016-02-11T14:51:38.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13675" PostTypeId="2" ParentId="13673" CreationDate="2016-02-11T14:51:48.707" Score="0" Body="&lt;p&gt;This is the easiest place: &lt;a href=&quot;https://www.youtube.com/user/VideosatNSF/live&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/user/VideosatNSF/live&lt;/a&gt;, or alternatively it will probably be on most news websites.&lt;/p&gt;&#xA;" OwnerUserId="10437" LastActivityDate="2016-02-11T14:51:48.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13677" PostTypeId="2" ParentId="13635" CreationDate="2016-02-11T15:32:42.540" Score="1" Body="&lt;p&gt;Supposing that the mass of the object is negligible compared with the mass of the Earth, you can derive the orbital period $T$ from the 3rd Keplero's law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{T^2}{a^3} = \frac{4\pi^2}{G(m_E + m_b)} \approx \frac{4\pi^2}{Gm_E},$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a$ is the semi-major. With $T$, for each time istant you also know the mean anomaly $M$, given by (suppose $t = 0$ at perigee):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M(t) = \frac{2\pi}{T}t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving numerically the Keplero's equation for the eccentric anomaly $E$ (where $e$ is the eccentricity)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M = E - e\sin E$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then use the following equation to derive the true anonaly $\nu$, which is the angle between the direction of periapsis and the current position of the body, as seen from the Earth:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \cos \nu = \frac{\cos E - e}{1 - e\cos E}$&#xA;and&#xA;$ \sin \nu = \frac{\sqrt{1-e^2}\sin E}{1 - e\cos E}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance from the Earth is just given by the orbit equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ r = \frac{a(1-e^2)}{1 + e\cos \nu}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm not wrong with calculation, it should be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ T = 9364 s = 2.6 h.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ M(90min) = 207.60°$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ E(90min) = 203.11°$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \nu(90min) = 198.95°$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ r = 11'366 km$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the covered distance you should calculate the line integral of the orbit equation.&lt;/p&gt;&#xA;" OwnerUserId="10759" LastActivityDate="2016-02-11T15:32:42.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13678" PostTypeId="5" CreationDate="2016-02-11T15:56:53.943" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2016-02-11T15:56:53.943" LastActivityDate="2016-02-11T15:56:53.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13679" PostTypeId="4" CreationDate="2016-02-11T15:56:53.943" Score="0" Body="Questions about the measurement of light waves whereby the wavelength is classified by its position in the electromagnetic spectrum." OwnerUserId="876" LastEditorUserId="876" LastEditDate="2016-05-05T13:59:07.793" LastActivityDate="2016-05-05T13:59:07.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13681" PostTypeId="1" CreationDate="2016-02-11T18:59:13.753" Score="1" ViewCount="126" Body="&lt;p&gt;With today's announcement of the historic detection of gravitational waves from the merger of 36 solar mass and 29 solar mass black holes 1.3 billion light years away, one can not help but wonder how such a black hole binary system was formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has there ever been any observation of a hypergiant (class 0 or Ia star) binary system in the optical or other EM radiation? I would expect that these observations would be rare give the short lifespan of these stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a black hole binary formed from a binary of population 3 stars with masses over 600 solar mass, how long would one expect that they would remain in orbit around one another before they merge?&lt;/p&gt;&#xA;" OwnerUserId="7786" LastEditorUserId="7786" LastEditDate="2016-02-11T19:54:31.103" LastActivityDate="2016-02-11T19:54:31.103" Title="The black hole binary that was detected by advanced LIGO - how do such hypergiant binaries form?" Tags="&lt;black-hole&gt;&lt;binary-star&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13682" PostTypeId="1" AcceptedAnswerId="13683" CreationDate="2016-02-11T20:08:22.473" Score="4" ViewCount="188" Body="&lt;ol&gt;&#xA;&lt;li&gt;I read in an article announcing the detection of gravitational waves by LIGO that it will be possible to detect them from the Big Bang. Is this true?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10690" LastActivityDate="2016-02-15T23:39:27.910" Title="Is it possible to get a glimpse of the Big Bang through gravitation waves?" Tags="&lt;big-bang-theory&gt;&lt;early-universe&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13683" PostTypeId="2" ParentId="13682" CreationDate="2016-02-11T23:21:44.633" Score="7" Body="&lt;p&gt;Gravitational waves from the big bang may be &quot;heard&quot; but not by LIGO. The waves emitted at or around the inflationary epoch of the big bang are expected to be at much lower frequencies (milli-Hz or lower) than those announced today by LIGO. There are various sources of noise that make LIGO insensitive to GWs at frequencies below about 10 Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will take space-based interferometers like the proposed LISA, with longer interferometer arms and well away from terrestrial sources of noise to stand &lt;em&gt;a chance&lt;/em&gt; of detecting such GWs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they are detected - they might &quot;sound&quot; something like this (if upshifted into the audible range) - &lt;a href=&quot;http://www.ligo.org/science/GW-Overview/sounds/stochastic.wav&quot; rel=&quot;noreferrer&quot;&gt;from the LIGO website&lt;/a&gt;. It sounds like white(ish) noise because of the broad continuum of frequencies expected.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-02-15T23:39:27.910" LastActivityDate="2016-02-15T23:39:27.910" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="13684" PostTypeId="1" AcceptedAnswerId="13707" CreationDate="2016-02-12T02:16:05.643" Score="4" ViewCount="265" Body="&lt;p&gt;I'm a non-scientist. Today's announcement of LIGO's detection of GW waves has been heralded as &quot;huge&quot; –on the same level as the Higgs Boson detection. I realize this is the last remaining part of Einstein's theory of general relativity to be directly observed. Is that why it's such a big deal? Does this mean the theory is now on pace to become the  law of general relativity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side note: I had the privilege to visit and take a general tour LIGO in WA in 2000 or so and found it quite fascinating from a layman's pov. I've been waiting for this day ever since. &lt;/p&gt;&#xA;" OwnerUserId="10117" LastEditorUserId="10117" LastEditDate="2016-02-12T02:34:48.177" LastActivityDate="2016-02-12T14:58:04.583" Title="Why is the detection of gravitational waves such a &quot;Big&quot; deal?" Tags="&lt;gravitational-waves&gt;" AnswerCount="3" CommentCount="1" ClosedDate="2016-02-14T16:17:05.800" ContentLicense="CC BY-SA 3.0" />
  <row Id="13685" PostTypeId="1" AcceptedAnswerId="13801" CreationDate="2016-02-12T04:01:55.090" Score="4" ViewCount="111" Body="&lt;p&gt;In late 2015, the LIGO project team announced the detection of gravitational waves. The detector is (very basically) a laser measuring the distance between two mirrors in vacuum, over a large-enough distance to detect very small changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two mirrors are installed on Earth, and a couple kilometers apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe, perhaps mistakenly, that tectonic activities (if only vibrations) may change the mirror distance over time, so the LIGO project has probably used some model and technology to nullify their effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can that be done? If there is any mistake there, I wonder about how to validate the experiment results.&lt;/p&gt;&#xA;" OwnerUserId="10762" LastActivityDate="2016-02-23T00:25:27.737" Title="How to compensate the effect of tectonic activity in devices like LIGO?" Tags="&lt;gravitational-waves&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13687" PostTypeId="1" CreationDate="2016-02-12T07:08:22.450" Score="2" ViewCount="48" Body="&lt;p&gt;When reading through articles about discovering gravitational waves, I got basic understanding of events that happened. But one thing wasn't clear for me. They say that two black holes collided &lt;strong&gt;1.3 billion years&lt;/strong&gt; ago. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how can we know, that this collision happened 1.3 billion years ago?&lt;/p&gt;&#xA;" OwnerUserId="10764" LastActivityDate="2016-02-12T09:03:47.363" Title="Measuring time and distance in space" Tags="&lt;black-hole&gt;&lt;space-time&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13688" PostTypeId="1" AcceptedAnswerId="13692" CreationDate="2016-02-12T07:19:23.057" Score="30" ViewCount="5368" Body="&lt;p&gt;Of course everyone knows by now of the &lt;a href=&quot;http://www.iflscience.com/space/gravitational-waves-observed-first-time&quot;&gt;detection of Gravitational waves&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, since General Relativity and Quantum Mechanics &lt;a href=&quot;http://io9.gizmodo.com/why-cant-einstein-and-quantum-mechanics-get-along-799561829&quot;&gt;don't get along&lt;/a&gt;, can we say now that this detection proves that Quantum Mechanics doesn't actually apply and that General Relativity did prevail?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question: how can we identify the ripple's origin (let's say whether it's a result of the big bang or another big event)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 16-2-2016&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was reading an &lt;a href=&quot;http://www.iflscience.com/space/seeking-source-those-gravitational-waves&quot;&gt;article&lt;/a&gt; today and I thought I'd share it here; It's basically saying that without a third detector we can't triangulate the signal. &#xA;Some scientists tried ways to observe the light of the event directly after the observations of the wave but they couldn't detect the merger simply because it's too far away or too faint to be observed with our current technology.&lt;/p&gt;&#xA;" OwnerUserId="10304" LastEditorUserId="941" LastEditDate="2016-02-17T00:11:14.013" LastActivityDate="2016-03-04T22:29:34.340" Title="Quantum Mechanics after the detection of Gravitational Waves" Tags="&lt;general-relativity&gt;&lt;astrophysics&gt;&lt;gravitational-waves&gt;&lt;quantum-field-theory&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13689" PostTypeId="1" AcceptedAnswerId="13691" CreationDate="2016-02-12T07:56:02.263" Score="4" ViewCount="1421" Body="&lt;p&gt;Can a black hole which has an enormous mass and spins so fast cause spacetime to to be stretched to is absolute limits that it  start so rip?  &lt;/p&gt;&#xA;" OwnerUserId="10765" LastActivityDate="2017-08-11T08:27:55.807" Title="Can a black hole rip spacetime" Tags="&lt;black-hole&gt;&lt;space-time&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13690" PostTypeId="2" ParentId="13685" CreationDate="2016-02-12T07:57:22.377" Score="6" Body="&lt;p&gt;From what I remember from a visit/tour I took of the LIGO facility in WA back in 2000, they said something to the effect that non-GW events (earthquakes, car driving down the road, etc) have a completely different wave signatures or whatever than what was theorized with GW. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, detections in WA are compared against what the LIGO in LA may have detected at the same time (or within ~10 milliseconds–time for light/GW to travel between the two observatories?). Any tectonic activity would take much longer to propagate between the two sites, thus eliminating it as a GW.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure there's more to it than that, but that's the basic gist as my non-scientist mind recalls from 15 years ago.&lt;/p&gt;&#xA;" OwnerUserId="10117" LastEditorUserId="10762" LastEditDate="2016-02-22T07:43:25.410" LastActivityDate="2016-02-22T07:43:25.410" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13691" PostTypeId="2" ParentId="13689" CreationDate="2016-02-12T08:07:16.210" Score="8" Body="&lt;p&gt;There is a useful model of spacetime as a rubber sheet that is bent by masses laying on it. But it should be remembered that this is an analogy (&lt;a href=&quot;https://xkcd.com/895/&quot; rel=&quot;noreferrer&quot;&gt;Obligatory xkcd)&lt;/a&gt; and most analogies fail if pushed too far. Spacetime isn't made of something that can rip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rotating black hole, a &quot;Kerr black hole&quot; is stranger than a static one, as it pulls spacetime around it and can accelerate objects passing close to it. Even so there are no rips, no singularities outside the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in another way, at the centre of every black hole there is a singularity, and a singularity is a single point &quot;rip&quot; in spacetime. General relativity can be used to predict what happens to spacetime around the black hole, and even inside the event horizon. But at the singularity the gravity becomes infinite, and at that one point, spacetime doesn't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is suggested that a &lt;a href=&quot;http://www.scientificamerican.com/article/naked-singularities-extreme-physics-special/&quot; rel=&quot;noreferrer&quot;&gt;&quot;naked&quot; singularity&lt;/a&gt;, not surrounded by an event horizon, cannot exist. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-02-12T08:36:59.007" LastActivityDate="2016-02-12T08:36:59.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13692" PostTypeId="2" ParentId="13688" CreationDate="2016-02-12T08:33:30.207" Score="27" Body="&lt;p&gt;No more than the observation of light waves disproves quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light has properties of both a particle and a wave. At low energies, the particle nature of light is hard to detect: radio waves are made of photons, but individual radio wave photons are pretty hard to detect. I'm not sure that we have directly detected individual photons with energies below the infrared band. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational waves (probably) also have both a wave and a particle nature. The gravitational field is probably quantised. But at the frequencies and sensitivity at which LIGO operates, individual quanta cannot be measured. So this detection does not prove the ascendency of GR over QM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anything, understanding extreme events like black hole mergers might lead to a theoretical understanding of the quantum nature of gravity.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="9301" LastEditDate="2016-02-12T20:33:28.477" LastActivityDate="2016-02-12T20:33:28.477" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13693" PostTypeId="1" AcceptedAnswerId="13763" CreationDate="2016-02-12T08:44:22.700" Score="8" ViewCount="485" Body="&lt;p&gt;Does a spinning black hole have a stronger gravity than a non spinning one.After all, a spinning black hole is stretching space-time more, so does that make a difference is gravity?&lt;/p&gt;&#xA;" OwnerUserId="10765" LastActivityDate="2016-02-15T10:13:55.687" Title="Spinning black hole vs non spinning black hole" Tags="&lt;gravity&gt;&lt;black-hole&gt;&lt;space-time&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13695" PostTypeId="2" ParentId="13684" CreationDate="2016-02-12T09:00:45.713" Score="2" Body="&lt;p&gt;I can't think of any direct applications of this detection, but it's a huge achievement for fundamental research.&#xA;As you said, it was the last bit of Einstein's GR to experimentally confirm&lt;/p&gt;&#xA;" OwnerUserId="10611" LastActivityDate="2016-02-12T09:00:45.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13696" PostTypeId="2" ParentId="13687" CreationDate="2016-02-12T09:03:47.363" Score="1" Body="&lt;p&gt;A model is fitted to the signal that depends on various input parameters. The input of the model that fits the signal best is then considered the &quot;most favored&quot; model, while the input of models that fit better than a given threshold are used to infer &lt;em&gt;confidence intervals&lt;/em&gt; on the various parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model has to simultaneously reproduce as well as possible the frequency and the amplitude of the signal. Among the input parameters are the black hole masses ($36_{-4}^{+5}\,M_\odot$ and $29_{-4}^{+4}\,M_\odot$ ) and (luminosity) distance ($410_{-180}^{+160}\,\mathrm{Mpc}$), corresponding to a light travel time of roughly 1.3 Gyr, depending on the assumed cosmological density parameters (they used those of &lt;a href=&quot;http://arxiv.org/abs/1502.01589&quot; rel=&quot;nofollow&quot;&gt;Planck 2015&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2016-02-12T09:03:47.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13697" PostTypeId="2" ParentId="13684" CreationDate="2016-02-12T09:23:59.680" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I realize this is the last remaining part of Einstein's theory of general relativity to be directly observed. Is that why it's such a big deal?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Correct: though gravity waves were predicted for a long time, their final direct measurement adds a bit more reinforcement to relativity theory, the behaviour of spacetime, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important thing is, we know there must be more &quot;out there&quot; than just relativity and quantum mechanics - there has to be something happening at the crossover point between the two (Quantum gravity and stuff like that).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next stage may be to look in more detail at gravity wave events using the growing LIGO network and its more sensitive successors, and see if gravitational waves behave exactly as predicted, or if there are any gaps, which may point to new physics...&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2016-02-12T09:23:59.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13698" PostTypeId="1" AcceptedAnswerId="13703" CreationDate="2016-02-12T10:32:35.073" Score="1" ViewCount="347" Body="&lt;p&gt;This might be a silly question but if light travels at the speed of light and time is frozen in that speed then why the stars we see could be dead already and not just instantly disappearing from the sky the moment it explodes/dies? &lt;/p&gt;&#xA;" OwnerUserId="10766" LastActivityDate="2016-08-30T00:35:54.360" Title="When stars explode why can we still see them?" Tags="&lt;star&gt;&lt;light&gt;&lt;speed&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13699" PostTypeId="1" CreationDate="2016-02-12T11:19:48.620" Score="0" ViewCount="392" Body="&lt;p&gt;Can a black hole collide with the Sun? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What will happen in that case? Will the Sun survive?&lt;/p&gt;&#xA;" OwnerUserId="10767" LastEditorUserId="9301" LastEditDate="2016-02-12T13:42:29.817" LastActivityDate="2016-02-12T15:24:04.133" Title="Can black holes collide with the Sun? In that case what will happen?" Tags="&lt;the-sun&gt;&lt;black-hole&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2016-02-14T17:17:53.267" ContentLicense="CC BY-SA 3.0" />
  <row Id="13700" PostTypeId="2" ParentId="2470" CreationDate="2016-02-12T11:29:33.867" Score="1" Body="&lt;p&gt;One alternative we can think of about the shape of the universe, is a multi-dimensional sphere. In which we can start a journey from a point, and continue in a straight path (which will be bent in an extra dimension), and at some point of time, may be billions of light years later, you will end up in the same spot as you started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives the feeling the universe is infinite, but infact, it's a multi-dimensional sphere of a finite radius. Just like how the Earth which was supposed to be an infinite plane, turned out to be a finite sphere, the universe which was supposed to be an infinite 3d space, could be a finite n-sphere.&lt;/p&gt;&#xA;" OwnerUserId="10768" LastActivityDate="2016-02-12T11:29:33.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13701" PostTypeId="2" ParentId="13699" CreationDate="2016-02-12T12:06:07.107" Score="3" Body="&lt;p&gt;Yes, in theory. But;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, this is very unlikely to happen during the life of the sun. If it did happen it would be Bad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many black holes are much more massive than the sun, even though they are much smaller. If a black hole were to enter the sun, some the matter that forms the sun would form into a accretion disc around the black hole and eventually fall into the black hole. The rest of the sun would be utterly disrupted by the massive release of energy as matter falls towards the black hole.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solar system would be wrecked. The Earth, even if it survived intact, would not be able to support life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately black holes are rare, and space is BIG. The chances of a black hole coming anywhere near the sun in the next billion years or so is very very small. There are more pressing things to worry about.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-02-12T12:06:07.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13702" PostTypeId="2" ParentId="13688" CreationDate="2016-02-12T12:08:14.070" Score="18" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Another question, how can we identify the ripple's origin (let's say that if it's the result from the big bang or another big event)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(I'm just answering this part of the question, as James has already answered the main part about GR vs QM.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LIGO have produced &lt;a href=&quot;https://www.ligo.caltech.edu/image/ligo20160211b&quot; rel=&quot;nofollow noreferrer&quot;&gt;an image&lt;/a&gt; which shows their best estimate of where these two black holes were:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/PA21I.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PA21I.jpg&quot; alt=&quot;&amp;quot;Where the Gravitational Waves Came From&amp;quot; by LIGO&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All they can say is, somewhere in the southern sky. In the future a network of more detectors will allow such events to be pinpointed much more precisely.&lt;/p&gt;&#xA;" OwnerUserId="7491" LastEditorUserId="7491" LastEditDate="2016-02-12T14:26:23.530" LastActivityDate="2016-02-12T14:26:23.530" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13703" PostTypeId="2" ParentId="13698" CreationDate="2016-02-12T12:15:22.203" Score="5" Body="&lt;p&gt;Because we are not moving at the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From our point of view, light travels at the speed of light, and so events on a distant star are not visible until the light reaches us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the point of view of the photon, the universe is 2 dimensional (it is flattened by infinite Lorentz contraction) and there is no passing of time (infinite time dialation). The star and you are at the same point. Photons have a weird point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately we don't have to worry about that, because we have mass and so we are not moving at the speeed of light. We have a 3d universe and we have a dimension of time. Thank Higgs we have mass!&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-02-12T12:15:22.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13704" PostTypeId="1" AcceptedAnswerId="13705" CreationDate="2016-02-12T13:14:42.497" Score="2" ViewCount="1311" Body="&lt;p&gt;Quote from &lt;a href=&quot;https://www.ligo.caltech.edu/news/ligo20160211&quot; rel=&quot;nofollow noreferrer&quot;&gt;LIGO's news release&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;By looking at the time of arrival of the signals—the detector in&#xA;Livingston recorded the event &lt;strong&gt;7 milliseconds&lt;/strong&gt; before the detector in&#xA;Hanford—scientists can say that the source was located in the Southern&#xA;Hemisphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;GPS coordinates of LIGO Livingston: &lt;a href=&quot;https://www.google.at/maps/place/LIGO+Livingston/@30.5630018,-90.7763949,17z/data=!3m1!4b1!4m2!3m1!1s0x8626e835b7d97141:0x7f7b0db13ffa9864&quot; rel=&quot;nofollow noreferrer&quot;&gt;30.5630018,-90.7763949&lt;/a&gt;&lt;br /&gt;&#xA;GPS coordinates of LIGO Hanford: &lt;a href=&quot;https://www.google.at/maps/place/LIGO+Hanford+Observatory/@46.4554032,-119.4092701,831m/data=!3m1!1e3!4m2!3m1!1s0x54985dc1a03257c3:0x70ac2e7c3e7fc470!6m1!1e1&quot; rel=&quot;nofollow noreferrer&quot;&gt;46.4554032,-119.4092701&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Distance: &lt;a href=&quot;http://boulter.com/gps/distance/?from=46.4554032%20-119.4092701&amp;amp;to=30.5630018%20-90.7763949&amp;amp;units=k&amp;amp;sort=distance&quot; rel=&quot;nofollow noreferrer&quot;&gt;3030.13 km&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now if the gravitational waves propagate outwards at the speed of light a difference of 7ms implies the following maximum possible detector distance:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;299792458 * 0.007 / 1000 = 2099 km&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Where am I wrong ?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;As pointed out by &lt;a href=&quot;https://astronomy.stackexchange.com/users/2365/&quot;&gt;james-k&lt;/a&gt; in the comments below, the calculated 2099 km is the minimum distance and not the maximum ... these waves seem to have caused some mental derangement ;)&lt;/p&gt;&#xA;" OwnerUserId="10676" LastEditorUserId="37191" LastEditDate="2021-02-12T21:28:15.147" LastActivityDate="2021-02-12T21:28:15.147" Title="Gravitational wave detection time difference between LIGO Livingston and LIGO Hanford" Tags="&lt;gravitational-waves&gt;&lt;ligo&gt;" AnswerCount="5" CommentCount="6" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="13705" PostTypeId="2" ParentId="13704" CreationDate="2016-02-12T13:39:38.797" Score="5" Body="&lt;p&gt;Here's how my non-scientist mind envisions it. I draw a straight line between the two LIGO sites on a map. Then I take another straight line (like a straight edge/ruler) that represents the GW coming in.  If the GW line is coming from the south exactly parallel with the line I drew on the map, then both sites would detect the &quot;chirp&quot; at exactly the same time (0 milliseconds difference). If the GW line comes from the south exactly perpendicular to my drawn line, hitting the LA site first, then chirp would be detected at each site at approx 10 milliseconds (?) apart (assuming that's how long it takes for light to travel between the two sites in a straight line).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I envision the GW &quot;line&quot; came in from the south somewhere between parallel and perpendicular (closer to perpendicular) in relation to my drawn line. Someone else smarter than I can figure out the maths and correct angles, numbers, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that make any sense?&lt;/p&gt;&#xA;" OwnerUserId="10117" LastActivityDate="2016-02-12T13:39:38.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13707" PostTypeId="2" ParentId="13684" CreationDate="2016-02-12T14:58:04.583" Score="10" Body="&lt;p&gt;The way I see it, there are three reasons it's such a big deal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first, as you say, is the (further) confirmation of Einstein's theory of gravity. Newtonian gravity doesn't have gravitational waves. Their existence was already quite established by Hulse and Taylor's discovery and analysis of the binary pulsar &lt;a href=&quot;https://en.wikipedia.org/wiki/PSR_B1913%2B16&quot;&gt;PSR B1913+16&lt;/a&gt;.  The system's orbit is decaying almost &lt;em&gt;exactly&lt;/em&gt; as predicted by General Relativity.  (Note that in the classic figure, the curve is &lt;em&gt;not&lt;/em&gt; a fit: that is the prediction!) The direct detection of gravitational waves, however, better confirms that the waves behave as we expect they would, in the sense that the observations match the so-called &quot;chirp&quot; that we expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second reason is because this is now a new way of measuring things in the Universe. Note that the announcement told us (roughly) the masses of the two black holes.  It's actually very difficult to weigh black holes!  For example, &lt;a href=&quot;http://adsabs.harvard.edu/abs/2010ApJ...725.1918O&quot;&gt;this 2010 paper&lt;/a&gt; discusses constraints on 23 stellar mass black holes, all from the fact that they're in binary systems, and the motion of the partner can be observed.  Even then those are mostly mass functions that still depend on the inclination of the orbits, which is often unknown.  So the gravitational waves give us a nice measurement, not just confirmation that the phenomenon exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the method proven, there are some fascinating potential applications.  One of my favourites is that LIGO (or similar) could potentially detect the formation of a black hole in a core-collapse supernova.  Just like the detection of neutrinos from SN1987A confirmed some of our basic ideas about these events, detecting the gravitational waves from the birth of a black hole would inform our understanding further still.  And it might even tell us, for example, how massive the black hole is at birth.  There's no end to the novel ways that our theories could be tested!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final reason is simply the astonishing technical achievement.  They measured motions of at most a few thousandths of a &lt;em&gt;femto&lt;/em&gt;meter over the 4km arms.  That's mind-bogglingly precise.  Hopefully this will give more impetus to other gravitational wave detectors.  Here's hoping that &lt;a href=&quot;https://en.wikipedia.org/wiki/Evolved_Laser_Interferometer_Space_Antenna&quot;&gt;eLISA&lt;/a&gt; will fly!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parallel with the Higgs boson is quite apt.  In both cases, we built a big machine to find something that we were pretty sure should be there. (Though the LHC was a tad more expensive...)  Now it's a case of using the new tool to observe the Universe, and seeing if the waves behave as we think they should. And who knows, maybe some completely unexpected signal will appear to confound us completely.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2016-02-12T14:58:04.583" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13708" PostTypeId="2" ParentId="13699" CreationDate="2016-02-12T15:24:04.133" Score="1" Body="&lt;p&gt;The Sun would drown in the endless gravity well of the black hole, if they collide head on. The black hole would radiate nothing, unless from a disc surrounding it if more matter happens to fall into it. Sun grazing comets would make it flash a bit now and then. The sunlight would be gone, trapped beyond the horizon of spacetime. The planets would freeze air to ice as they continue in their orbits, disturbed and quite a bit faster since the new Sun+BH center is much more massive, without any sunlight hitting them. But perfect constant night sky conditions for amateur astronomers btw :-)&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-02-12T15:24:04.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13709" PostTypeId="1" AcceptedAnswerId="18580" CreationDate="2016-02-12T15:50:24.517" Score="6" ViewCount="12574" Body="&lt;ul&gt;&#xA;&lt;li&gt;Why does this relationship involves the absolute magnitude of the sun and its luminosity?&lt;/li&gt;&#xA;&lt;li&gt;How to derive an expression relating the absolute magnitude of a star to its luminosity?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So according to the magnitude equation, $$m - M = 2.5\log\left(\frac{d^2}{d_0^2}\right)$$ &#xA;$$\Rightarrow M = m - 2.5\log\left(\frac{d^2}{d_0^2}\right)\ \ (eq1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and Luminosity is $$L = 4\pi(d^2) \times f$$&lt;br&gt;&#xA; $$\Rightarrow d^2 = \frac{L}{4\pi \times f}\ \ (eq2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging eq2 to eq1 would have seem reasonable, but how is absolute magnitude of the sun and its luminosity used?&lt;/p&gt;&#xA;" OwnerUserId="10696" LastEditorUserId="2365" LastEditDate="2016-02-13T21:47:10.453" LastActivityDate="2019-04-28T21:29:27.850" Title="Relationship between absolute magnitude of a star and its luminosity?" Tags="&lt;astrophysics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13710" PostTypeId="2" ParentId="13688" CreationDate="2016-02-12T16:25:26.033" Score="20" Body="&lt;p&gt;The impact of this measurement on the status of quantum gravitation is exactly zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proper statement of the incompatibility of general relativity and quantum mechanics is that &lt;em&gt;the quantum field theory of general relativity is not renormalizable&lt;/em&gt;. Renormalizability essentially means that the theory is well-defined at all energy scales, which seems like a reasonable demand on a proposed fundamental theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what we know is that taking classical general relativity and quantizing it, we do not get a fundamental theory of quantum gravitation. This does nothing to rule out other proposed quantum theories of gravitation, for example, LQG or string theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the way physics works is that new theories must reduce to old ones in the domains of applicability of the old theories. Whatever the correct quantum theory of gravitation, its low-energy limit should be quantized general relativity, and the classical limit of that is classical general relativity. It's just not true that you have to choose between general relativity or quantum mechanics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this measurement of a prediction of classical general relativity does absolutely nothing to show that no quantum mechanical model of gravitation exists. It couldn't, because we already have a quantum mechanical model of gravitation: quantized general relativity. It's not as &quot;nice&quot; as we would like, but that really only rules it out as the &lt;em&gt;fundamental&lt;/em&gt; theory.&lt;/p&gt;&#xA;" OwnerUserId="10772" LastEditorUserId="10772" LastEditDate="2016-02-12T16:32:29.153" LastActivityDate="2016-02-12T16:32:29.153" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13711" PostTypeId="1" AcceptedAnswerId="13718" CreationDate="2016-02-12T16:39:38.687" Score="3" ViewCount="2494" Body="&lt;p&gt;Most people consider it common knowledge that the Sun's &quot;movement&quot; in the sky is only &lt;em&gt;perceived&lt;/em&gt; due to the &lt;em&gt;Earth's&lt;/em&gt; &quot;spin and movement&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on this, you'd think that the stars in the night sky (when viewed with time lapse photography) would take a path in the sky similar to the sun. After all, both the sun and stars are practically stationary over the course of one day, given their accepted distances from the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet, time lapse photography shows this is not the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare these photos:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;a href=&quot;https://www.google.com/search?q=time+lapse+sun&amp;amp;tbm=isch&quot; rel=&quot;nofollow&quot;&gt;Time Lapse of Sun&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;a href=&quot;https://www.google.com/search?q=time+lapse+stars&amp;amp;tbm=isch&quot; rel=&quot;nofollow&quot;&gt;Time Lapse of Stars&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) &lt;a href=&quot;https://www.google.com/search?q=time+lapse+moon&amp;amp;tbm=isch&quot; rel=&quot;nofollow&quot;&gt;Time Lapse of Moon&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Earth's spin is the primary reason for &quot;movement&quot; of things (farther than clouds in the sky), then &lt;strong&gt;why do the sun and moon move through the sky in a similar fashion that is totally different from the sky-path of the stars?&lt;/strong&gt; The Earth is said to spin around 360 degrees every 24 hours. It seems to me that everything in the sky should move in one direction where I live (Texas).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me (in Texas), it seems the stars should sweep from horizon to horizon over the course of the Earth's night-spin (like the sun and moon do). Yet, the time lapse photos show circular paths in the sky for the stars. That would make sense at the poles, but not in Texas!?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Lonnie Lee Best&lt;/p&gt;&#xA;" OwnerUserId="10773" LastEditorUserId="10773" LastEditDate="2016-02-13T03:13:21.690" LastActivityDate="2021-04-30T07:39:53.197" Title="Is the Earth Really Spinning? (honest question)" Tags="&lt;the-moon&gt;&lt;the-sun&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13712" PostTypeId="1" CreationDate="2016-02-12T16:51:58.293" Score="-2" ViewCount="73" Body="&lt;p&gt;Coming from this question &lt;a href=&quot;https://astronomy.stackexchange.com/questions/13596/are-there-collected-data-about-the-direction-of-rotation-of-black-holes-and-the&quot;&gt;Are there collected data about the direction of rotation of black holes and the direction of the magnetic field?&lt;/a&gt; I want to know: How the magnetic field and the rotation of the planets and the sun inside our planetary system are related? Any tabellary overview is welcome.&lt;/p&gt;&#xA;" OwnerUserId="4242" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-08-03T19:05:03.050" Title="How the magnetic field and the rotation of the planets and the sun inside our planetry system are related" Tags="&lt;rotation&gt;&lt;magnetic-field&gt;&lt;planetary-formation&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13714" PostTypeId="2" ParentId="13567" CreationDate="2016-02-12T19:20:11.150" Score="0" Body="&lt;p&gt;We don't really know what happened before the big bang, but it is logical to think that just like the the mass was released when the big bang happened, it also got crushed to that single point, and since energy can't be created or destroyed, there was also mass before the big bang. And this is also a lot similar to a black hole, leading us to think that the big bang might have been a black hole&lt;/p&gt;&#xA;" OwnerUserId="10765" LastActivityDate="2016-02-12T19:20:11.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13716" PostTypeId="1" AcceptedAnswerId="13737" CreationDate="2016-02-12T21:50:08.070" Score="16" ViewCount="874" Body="&lt;p&gt;According to the accepted answer on &lt;a href=&quot;https://astronomy.stackexchange.com/questions/131/is-the-moon-moving-further-away-from-earth-and-closer-to-the-sun-why?newreg=c97ffe82214b40f5bc69c703df67d44b&quot;&gt;Is the moon moving further away from Earth and closer to the Sun? Why?&lt;/a&gt;, the moon is receding from Earth because tidal forces and friction cause energy to be lost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, according to &lt;a href=&quot;http://ligo.org/science/GW-Inspiral.php&quot; rel=&quot;noreferrer&quot;&gt;LIGO's website&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As the two masses rotate around each other, their orbital distances&#xA;  decrease&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;as they lose energy radiating &lt;em&gt;inspiral gravitational waves&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do the bodies move apart in the Earth-moon case but closer together in the black hole case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If both opposing phenomena are present but a different one is stronger in the two cases, what determines the fate of a system?&lt;/p&gt;&#xA;" OwnerUserId="10776" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2019-03-22T19:13:50.313" Title="Why do Earth and moon move apart but binary black holes move closer?" Tags="&lt;the-moon&gt;&lt;gravity&gt;&lt;black-hole&gt;&lt;tidal-forces&gt;&lt;gravitational-waves&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13717" PostTypeId="2" ParentId="13424" CreationDate="2016-02-12T22:09:53.417" Score="2" Body="&lt;p&gt;The Earth atmosphere protects us from small impacts from both asteroids and man made objects. This is well known from meteoroids, where meteoroids as large as a few tens of meters in diameter usually fail to penetrate into the lower atmosphere because they get fragmented and dispersed at high altitude. Fragmentation height depends mainly on the meteoroid strength, only strongest irons reach the surface in one piece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could extrapolate from meteoroids to man made objects that even large do not have the same physical strength than a massive meteoroid body. This means than nearly all man machine objects will disintegrate before reaching earth. Heavy metallic ones (iron) will disintegrate at lower altitudes than lighter ones such glass and plastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two comments:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This is not true for objects not having reached orbit, e.g. rocket 1st or 2nd stage engines will fall without disintegrating.&lt;/li&gt;&#xA;&lt;li&gt;Even if they disintegrate, nuclear powered satellites will cause some level of radioactive pollution, however widely scattered in the atmosphere.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Although the risk to human beings walking on earth is next to nil, there is a real risk of impact from man made objects littering Earth orbits to other spatial vehicles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: &quot;The Impact Hazard&quot;, David Morrison, Clark R. Chapman and Paul Slovic, in &quot;Hazards due to Comets and Asteroids&quot;, T. Gehrels, Ed., 1994, The University of Arizona Press&lt;/p&gt;&#xA;" OwnerUserId="10374" LastActivityDate="2016-02-12T22:09:53.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13718" PostTypeId="2" ParentId="13711" CreationDate="2016-02-12T23:08:16.770" Score="15" Body="&lt;p&gt;You are already starting to get it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;That would make sense at the poles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What about one meter from the poles?&lt;br&gt;&#xA;Or a kilometre?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as you can see the celestial pole in the sky, you can see the stars revolve around it at night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us see if you are able to see the celestial pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Texas was about 30 degrees north last time I checked:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1TmStm.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1TmStm.png&quot; alt=&quot;you can!&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That explains the circular movement of the stars, the Sun and the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is true for all locations on the Earth, except for the equator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uHQ5wm.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uHQ5wm.png&quot; alt=&quot;cannot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the Earth spinning? That depends, you can always choose a frame of reference that suits you. However, only one of them are non-rotating, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Inertial_frame_of_reference&quot; rel=&quot;noreferrer&quot;&gt;Inertial frame.&lt;/a&gt; In all the others we have fictitious forces acting, like centrifugal or Coriolis forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can test if the Earth rotates by &lt;a href=&quot;https://en.wikipedia.org/wiki/Foucault_pendulum&quot; rel=&quot;noreferrer&quot;&gt;watching a pendulum throughout a day&lt;/a&gt;. The pendulum would then seem to slowly rotate during this period of time, meaning some fictitious &quot;force&quot; is acting on it. That means that we are located in a rotating frame of reference, and thus the Earth rotates.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="10172" LastEditDate="2016-02-13T10:13:15.347" LastActivityDate="2016-02-13T10:13:15.347" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13719" PostTypeId="1" CreationDate="2016-02-12T23:30:43.080" Score="1" ViewCount="44" Body="&lt;p&gt;This is almost identical to an unanswered question already asked, but I want to put it a different way. The announced LIGO event was from a localized source. I assume that gravitational waves from the Big Bang would come from &quot;everywhere&quot; like the CMB, so I don't see how LIGO or the next generation observatory (ie. eLISA) would detect this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing else I would like to know, if anyone is aware: I watched an hour long webcast on You Tube and the question I had during the whole time was asked just as the webcast went &quot;off the air&quot;. My question is that, since this announcement is about five months after the detection and since it happened &quot;right away&quot;, has anything else been detected over the last five months? (Has LIGO been continuously observing? I know that the upgrade is still in progress and sensitivity should get about three times better.)&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2016-02-12T23:30:43.080" Title="Will we see gravitational waves from the big bang?" Tags="&lt;gravitational-waves&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2016-02-15T09:31:18.800" ContentLicense="CC BY-SA 3.0" />
  <row Id="13720" PostTypeId="2" ParentId="13635" CreationDate="2016-02-13T02:56:33.583" Score="1" Body="&lt;p&gt;@Dario_Panarello I think what you're saying is correct for a &#xA;non-rotating geocentric observer, but the earth's radius is fairly &#xA;large compared to the orbit, so I don't think the geocentric &#xA;approximation works well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have an answer, but I think the solution looks something like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sA4oo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sA4oo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the light blue circle is the Earth, the small blue dot is the &#xA;geocenter, the black dot in the blue circle is the center of the &#xA;ellipse, the black ellipse is the orbit of the satellite, and the two &#xA;black dots on the ellipse are the perigee and final positions of the &#xA;satellite respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even allowing for the Earth's rotation in the 90 minute timefram, I'm not sure anyone on Earth could see the satellite both at perigee and at its final location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm working on a more complete answer at &lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/STACK/bc-solve-astronomy-13635.m&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/STACK/bc-solve-astronomy-13635.m&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2016-02-13T02:56:33.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13721" PostTypeId="1" CreationDate="2016-02-13T07:30:56.267" Score="1" ViewCount="103" Body="&lt;p&gt;I understand that in recent discovery gravitational waves have been detected. But knowing space time is not static it probably changes constantly in many places in the universe due to changes in distribution of massive bodies. For example, could the detected wave just be a slight oscillation in Gravity due to Jupiter fly by, or moon, or something similar?&lt;/p&gt;&#xA;" OwnerUserId="692" LastActivityDate="2016-02-13T09:37:03.127" Title="What caused gravitational waves detected by LIGO?" Tags="&lt;space-time&gt;&lt;gravitational-waves&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13722" PostTypeId="2" ParentId="13721" CreationDate="2016-02-13T09:03:29.810" Score="4" Body="&lt;p&gt;It couldn't be Jupiter or the moon because &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The frequency of the detected waves is too fast. The detector found something that rotated several hundred times a second, and then stopped.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The amplitude is too big. Even though Jupiter is close, the gravitational waves it produces in its orbit are extraordinarily weak. Far too weak for us to detect them (good thing too, because if it was producing powerful gravitational radiation, it would start to spiral into the sun)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We know one thing that can produce a chirp of this frequency and amplitude, and that is a black hole merger. In the last few moments before merging, these two black holes converted about three solar masses into gravitational radiation: That is a colossal amount of energy. About $$500,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000$$Joules.  That is why for one moment it was the &quot;brightest&quot;  point in the gravitational sky&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note. standard form is for wimps)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-02-13T09:03:29.810" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13723" PostTypeId="2" ParentId="13721" CreationDate="2016-02-13T09:37:03.127" Score="2" Body="&lt;p&gt;The signal found was quasi-periodic, with a rapidly increasing frequency - characteristic of a merging binary system (a so-called &quot;chirp&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The characteristic frequency of the GWs of a &quot;binary&quot; system are at twice the orbital period. So the frequency &quot;emitted&quot; by the Jupiter-Sun system would be wayyyy to low to be detected by LIGO, which is sensitive down to around 10 Hz (ie orbital periods of 0.2 s or shorter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These sorts of periods are only found in binaries involving stellar-sized black holes and neutron stars just before they merge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interesting thing about the power produced by a binary is that it goes as mass to the power of 5. Another point is that we detect the wave amplitude, which only falls as the inverse distance to the source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus LIGO is most likely to see the merger of the rarest, most massive black hole systems at very large distances.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-02-13T09:37:03.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13725" PostTypeId="1" CreationDate="2016-02-13T10:22:59.117" Score="11" ViewCount="345" Body="&lt;p&gt;Now that LIGO has finally measured gravitational waves using a huge laser interferometer, to me, the question remains, why was it possible? As it is explained in many news articles, gravitational waves are similar to water waves or electromagnetic waves, they just do not exist in a medium like water or space, but space-time itself is the transport medium. If space-time itself gets contracted and expanded by the gravitational waves, so does any means of measurement. The ruler you use for measurement (the laser beam) gets deformed while the wave travels through the measuring device. Otherwise the &quot;ruler&quot; had to live outside of space-time, but there is no outside. If space-time was a cup filled with pudding, on which we had painted a straight line with 10 marks, pushing into the pudding slightly with our thumb does bend the line, but for us, there remain 10 marks on the line, because to measure the extension, we had to use a ruler, outside of our space-time (pudding) to measure, let's say, 11 marks. But, well, there is no outside. I assume the same happens not only to the 3 spacial dimensions but also to the time dimension. Because they &quot;did it&quot;, what am I missing?&lt;/p&gt;&#xA;" OwnerUserId="10788" LastActivityDate="2019-04-25T16:14:49.407" Title="Why can we detect gravitational waves?" Tags="&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13726" PostTypeId="1" AcceptedAnswerId="13731" CreationDate="2016-02-13T12:15:05.617" Score="2" ViewCount="293" Body="&lt;p&gt;We all know LIGO has detected gravitational wave(GW) directly. &lt;a href=&quot;https://journals.aps.org/prl/pdf/10.1103/PhysRevLett.116.061102&quot; rel=&quot;nofollow noreferrer&quot;&gt;prl&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/aVhCW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aVhCW.png&quot; alt=&quot;fig.1 of prl&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The LIGO team argue they detected the merger of two solar mass black holes($36M\odot+29M\odot-3M\odot$) at z=0.09. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems GW has Doppler shift too. There is a GW spectrum with emission or absorption lines like optical spectrum?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure.2 of the PRL paper seems to be very simple. Generally, people take their masses derived seriously? There may be other models which can give different masses. It is different from Taylor binary which was observed many years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does a 100HZ GW mean?  That is the frequency of the two arms shrink and extend?&lt;a href=&quot;https://i.stack.imgur.com/rtWIH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rtWIH.png&quot; alt=&quot;below&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does the frequency stand for in the binary system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discovery can not be confirmed. DO you think the team win a Nobel prize only after other facilities like LISA find GW again?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditDate="2016-02-15T15:15:56.453" LastActivityDate="2016-02-15T15:15:56.453" Title="How to derive the redshift of GW150914?" Tags="&lt;gravitational-waves&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13727" PostTypeId="2" ParentId="13716" CreationDate="2016-02-13T13:30:48.590" Score="7" Body="&lt;p&gt;The recession rate of the Moon from the Earth is given as 38.04 mm per year, due to tidal forces, according to Wikipedia.  A good description (with diagrams) about how this occurs can be found &lt;a href=&quot;http://www.themcdonalds.net/richard/astro/papers/602-tides-web.pdf&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbital decay rate due to gravitational radiation can be determined by $$\frac{\mathrm{d}r}{\mathrm{d}t} = - \frac{64}{5}\, \frac{G^3}{c^5}\, \frac{(m_1m_2)(m_1+m_2)}{r^3}$$&lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_wave#Orbital_decay_from_gravitational_radiation&quot; rel=&quot;noreferrer&quot;&gt;Wikpedia:Gravitational wave&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works out to approximately 2nm per year, or 7 orders of magnitude smaller.  So, yes, both tidal recession and orbital decay via gravitational waves are expected to be happening, but tidal recession is a far larger effect.&lt;/p&gt;&#xA;" OwnerUserId="10789" LastEditorUserId="2365" LastEditDate="2016-02-13T22:03:22.043" LastActivityDate="2016-02-13T22:03:22.043" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13728" PostTypeId="1" CreationDate="2016-02-13T13:48:41.433" Score="2" ViewCount="181" Body="&lt;p&gt;What if those gravitational waves detected would be actually so strong that the distance of those mirrors moved would be larger? Not 1/1000th of proton but rather one millimeter, or one meter or even more? What would happen to us? How big would these black holes had to be and how much energy would be released in such a process?&lt;/p&gt;&#xA;" OwnerUserId="9677" LastActivityDate="2016-02-15T23:37:06.340" Title="What would happen if those gravitational waves were much stronger?" Tags="&lt;gravitational-waves&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13729" PostTypeId="2" ParentId="13728" CreationDate="2016-02-13T14:23:46.703" Score="0" Body="&lt;p&gt;We would have been able to detect them sooner, for sure, but they could eventually be strong enough to disrupt massive things like the fusion core in our sun. &lt;/p&gt;&#xA;" OwnerUserId="10790" LastActivityDate="2016-02-13T14:23:46.703" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13730" PostTypeId="2" ParentId="13726" CreationDate="2016-02-13T17:40:52.973" Score="2" Body="&lt;p&gt;Too many questions in one question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The title question: The frequency of the gravitational waves as a function of time give an estimate of the black hole masses; the peak strain amplitude of the waves then tells you how far away the source is based on the expected strain amplitude from a merging binary with the previously derived masses (the amplitude decreases inversely with distance); Hubble's law then gives you the redshift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second question - a coalescing binary gives a signal at twice the orbital frequency of the binary, but that frequency rapidly changes (increases) as the binary components get closer together and merge. The redshift of the source cannot be derived directly from the observed frequencies, it is derived as I explained above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third question, yes the 100 Hz would mean the arms expand and contract 100 times per second. The binary would be orbiting 50 times per second to produce this signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model that is fitted comes straight out of general relativity. The free parameters are the masses (and spins) of the black holes and the distance to the source. What alternative model were you thinking of?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this result will end up in a Nobel prize. The detection &lt;em&gt;was&lt;/em&gt; confirmed by two detectors and I would think they will find more in thecoming months - this discovery was found in the first 16 days of data. Further, there will soon be other ground based detectors that are equally capable (e.g. VIRGO).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-02-13T17:40:52.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13731" PostTypeId="2" ParentId="13726" CreationDate="2016-02-13T17:48:06.023" Score="3" Body="&lt;p&gt;That is a lot of questions, but I can take them in order:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It seems GW has Doppler shift too. There is a GW spectrum with emission or absorption lines like optical spectrum? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, GW has Doppler shift too, as they are travelling at a finite speed, but no, there are no emission or absorption lines we can detect it from. In EM radiation those are caused by the fact that light is quantized, such that certain wavelengths can be absorbed or emitted by atoms. As far as we know, gravity is not quantized, or the effect is too small to be measured. Any irregularity in the signal is then going to be continuous, and not wavelength specific. Detecting a spectra in gravitational waves is unlikely, but would strengthen the theories of &lt;a href=&quot;https://en.wikipedia.org/wiki/Quantum_gravity&quot; rel=&quot;nofollow&quot;&gt;quantum gravity&lt;/a&gt; and the graviton.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;people take their masses derived seriously?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The method for the mass estimate is described well by @Rob Jeffries, but an even simpler point can be made: There are no other ways to measure their mass more accurately.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What does a 100HZ GW mean?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hertz is the frequency, 1/s. That is the number of waves per second. 100Hz is comparable to normal sound waves.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What does the frequency stand for in the binary system?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As two objects revolve around each other, they would make two waves per orbit,resulting in a double frequency of the system's rotation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;DO you think the team win a Nobel prize only after other facilities like LISA find GW again?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is almost certainty ending in a Nobel prize, and they do not necessarily have to wait till after the results are confirmed by other experiments. That said, verification by peer researchers is one of the core principles of science.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="10172" LastEditDate="2016-02-13T17:53:53.743" LastActivityDate="2016-02-13T17:53:53.743" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13732" PostTypeId="2" ParentId="13725" CreationDate="2016-02-13T18:01:40.460" Score="13" Body="&lt;p&gt;The short answer is that waves that are &quot;in the apparatus&quot; are indeed stretched. However the &quot;fresh waves&quot; being produced by the laser are not. So long as the &quot;new&quot; waves spend much less time in the interferometer than it takes to expand them (which takes roughly 1/gravitational wave frequency), then the effect you are talking about can be neglected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Details:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an &lt;em&gt;apparent&lt;/em&gt; paradox: you can think about the detection in two ways. On the one hand you can imagine that the lengths of the detector arms change and that the round-trip travel time of a light beam is subsequently changed and so the difference in the time-of-arrival of wavecrests translates into a phase difference that is detected in the interferometer. On the other hand you have the analogy to the expansion of the universe - if the arm length is changed, then isn't the wavelength of the light changed by &lt;em&gt;exactly&lt;/em&gt; the same factor and so there can be no change in the phase &lt;em&gt;difference&lt;/em&gt;? I guess this latter is your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well clearly, the detector works so there must be a problem with the second interpretation. There is an excellent discussion of this by &lt;a href=&quot;http://scitation.aip.org/content/aapt/journal/ajp/65/6/10.1119/1.18578&quot; rel=&quot;noreferrer&quot;&gt;Saulson 1997&lt;/a&gt;, from which I give a summary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interpretation 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the two arms are in the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; directions and the incoming wave the &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; direction, then the metric due to the wave can be written&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ds^2 = -c^2 dt^2 + (1+ h(t))dx^2 + (1-h(t))dy^2,$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$h(t)$&lt;/span&gt; is the strain of the gravitational wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For light travelling on geodesic paths the metric interval &lt;span class=&quot;math-container&quot;&gt;$ds^2=0$&lt;/span&gt;, this means that (considering only the arm aligned along the x-axis for a moment)&#xA;&lt;span class=&quot;math-container&quot;&gt;$$c dt = \sqrt{(1 + h(t))}dx \simeq (1 + \frac{1}{2}h(t))dx$$&lt;/span&gt;&#xA;The time taken to travel the path is therefore increased to&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\tau_+ = \int dt = \frac{1}{c}\int (1 + \frac{1}{2}h(t))dx$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the original arm is of length &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; and the perturbed arm length is &lt;span class=&quot;math-container&quot;&gt;$L(1+h/2)$&lt;/span&gt;, then the time difference for a photon to make the &lt;em&gt;round&lt;/em&gt; trip along each arm is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \Delta \tau = \tau_+ - \tau_- \simeq  \frac{2L}{c}h$$&lt;/span&gt;&#xA;leading to a phase difference in the signals of&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\Delta \phi = \frac{4\pi L}{\lambda} h$$&lt;/span&gt;&#xA;&lt;strong&gt;This assumes that &lt;span class=&quot;math-container&quot;&gt;$h(t)$&lt;/span&gt; is treated as a constant for the time that the laser light is in the apparatus.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interpretation 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In analogy with the expansion of the universe, the gravitational wave &lt;em&gt;does&lt;/em&gt; change the wavelength of light in each arm of the experiment. However, only the waves that are &lt;em&gt;in the apparatus&lt;/em&gt; as the gravitational wave passes through can be affected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that &lt;span class=&quot;math-container&quot;&gt;$h(t)$&lt;/span&gt; is a step function so that the arm changes length from &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$L+h(0)/2$&lt;/span&gt; instantaneously. The waves that are just arriving back at the detector will be unaffected by this change, but subsequent wavecrests will have had successively further to travel and so there is a phase lag that builds up gradually to the value defined above in interpretation 1. The time taken for the phase lag to build up will be &lt;span class=&quot;math-container&quot;&gt;$2L/c$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then what about the waves that enter the apparatus later? For those, the laser &lt;em&gt;frequency&lt;/em&gt; is unchanged and as the speed of light is constant, then the wavelength is unchanged. These waves travel in a lengthened arm and therefore experience a phase lag exactly equivalent to interpretation 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, the &quot;buildup time&quot; for the phase lag is short compared with the reciprocal of the frequency of the gravitational waves. For example the LIGO path length is about 1,000 km, so the &quot;build up time&quot; would be 0.003 s compared with the reciprocal of the &lt;span class=&quot;math-container&quot;&gt;$\sim 100$&lt;/span&gt; Hz signal of 0.01 s and so is relatively unimportant when interpreting the signal (the detection sensitivity of the interferometer is indeed compromised at higher frequencies because of this effect). &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-04-25T16:14:49.407" LastActivityDate="2019-04-25T16:14:49.407" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="13733" PostTypeId="1" AcceptedAnswerId="13734" CreationDate="2016-02-13T18:16:04.910" Score="6" ViewCount="318" Body="&lt;p&gt;Have we actually seen a black hole out there? By &quot;seen&quot; I mean &quot;recorded a em-spectrum image of gravitational lensing caused by object believed to be black hole&quot;. &lt;/p&gt;&#xA;" OwnerUserId="9496" LastEditorUserId="9496" LastEditDate="2016-02-13T19:00:23.550" LastActivityDate="2019-04-10T18:09:58.783" Title="Have we seen a black hole?" Tags="&lt;black-hole&gt;&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13734" PostTypeId="2" ParentId="13733" CreationDate="2016-02-13T19:14:25.657" Score="9" Body="&lt;p&gt;The answer &lt;strong&gt;was&lt;/strong&gt; no prior to 10 April 2019. There were no resolved images at any wavelength of black holes or black hole candidates that demonstrated their lensing effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There were of course lensing images due to massive objects that probably have black holes at their centres (e.g. &lt;a href=&quot;http://arxiv.org/abs/1002.4991&quot; rel=&quot;nofollow noreferrer&quot;&gt;Courbin et al. 2010&lt;/a&gt; and see below), but that is not the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Fv9bS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fv9bS.jpg&quot; alt=&quot;A quasar acting as a gravitational lens&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A quasar acting as a gravitational lens - Courbin et al. (2010)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this has changed with the release of a reconstructed millimetre-wave image of the supermassive black hole at the centre of M87.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image below, obtained by the &quot;Event Horizon Telescope&quot; (a global interferometer) is formed by intense light from accreting matter being lensed by a 6.5 billion solar mass black hole. The dark circle is known as the &quot;photon shadow&quot; and is about two times the size of the black hole event horizon. The bright ring is caused by the lensing of light from around and behind the black hole. It is thought to be asymmetric and off-centre because of the black hole rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/obYbF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/obYbF.jpg&quot; alt=&quot;Event Horizon Telescope Image of M87 black hole&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-04-10T18:00:55.097" LastActivityDate="2019-04-10T18:00:55.097" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="13735" PostTypeId="2" ParentId="13704" CreationDate="2016-02-13T19:40:17.320" Score="3" Body="&lt;p&gt;The addition of more detectors will immensly help event localization.  To get an understanding of the calculation look at the old LORAN navigation system.  This system used synchronized transmitters, and the receiver measured the difference in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The receiving device measured the time between signals from multiple sources, that chirp at the same time. If you hear the chirps at the same time, from two of the synchronized transmitters then you are on the plane equidistant between the sources.  The plane is perpendicular to the line between the sources, and at the halfway point.  If you plot the intersection of the plane, and the surface of the earth it creates a great circle.  You are on the great circle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Signals that have a difference in time of reception form a curved surface.  As it cuts through the earth it defines a curved line on the surface (hyperbolic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the signals come at different times then you are closer to one than the other.  The charts used for determining position had lines of constant time difference plotted.  You listened to one pair of transmitters, got your time difference, found the line on the chart that represents that delay.  Do this for another pair of transmitters, get that difference, find the line for that pair with that delay, and where the lines cross is about where you are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is repeated for multiple pairs, and finding all the intersections (the best two lines are those crossing at right angles to reduce the error possible in your timers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the same thing in reverse.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Get more listener pairs and we can start to isolate the region of space of the signal source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="10795" LastActivityDate="2016-02-13T19:40:17.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13737" PostTypeId="2" ParentId="13716" CreationDate="2016-02-13T22:20:12.003" Score="14" Body="&lt;p&gt;Here is how the tides move the moon away from the Earth:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon orbits the earth, and there is a difference in gravitational force between the the side of the Earth nearest the moon, and the side far from the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This difference in force tends to pull the Earth into a oval shape with its long axis pointing towards the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the Earth is also spinning, and this spinning moves the axis of the oval forward, so the oval doesn't point towards the moon, but a little ahead of it. So there is a bump on the Earth, and it is permanently a little in front of the moon. This bump has mass and it pulls the moon towards it, So the moon is being pulled forwards. The moon pulls the Earth back. So the Earth's spin is slowed, but the moon gains energy and moves a little further from the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon has also been slowed down to the extent that the same side always faces Earth, and it will remain locked like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational waves, on the other hand, imply an emmision of energy from the binary system, and as energy is lost, the black holes spiral in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no tides on black holes because there is nothing there to be pulled into an oval shape. The event horizon is not a solid surface. The mass of a black hole is concentrated entirely at the singularity, there is no structure to be deformed into a bump.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-02-13T22:20:12.003" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13738" PostTypeId="1" CreationDate="2016-02-14T03:43:25.447" Score="3" ViewCount="157" Body="&lt;p&gt;Does the big (0.012 Earth masses) Moon of Earth clear away NEAs, Earth orbit crossing asteroids, in a significant way? Venus and Mars don't have large moons, do they therefore have larger or smaller population of near asteroids than Earth would have if it were in the same orbit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outer planets have large moons, but also lots of captured asteroids, trojans and centaurs. Does a large moon even help gathering such objects, rather than ejecting them? Moonless Mercury and Venus seem to be pretty clean.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2016-02-14T03:50:32.753" LastActivityDate="2016-02-15T23:52:28.260" Title="What effect does the Moon have on the near Earth asteroid population?" Tags="&lt;the-moon&gt;&lt;asteroids&gt;&lt;near-earth-object&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13742" PostTypeId="1" AcceptedAnswerId="13745" CreationDate="2016-02-14T09:44:27.523" Score="5" ViewCount="694" Body="&lt;p&gt;I know that we are able to &quot;see back into time&quot; by looking very far away through telescopes such as the Hubble telescope, but my question is, wouldn't you run into at least a couple of stars that were blocking your vision when you were trying to look even farther? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a big Boulder in front of another smaller Boulder  and if we are looking from the side of the larger Boulder, we wouldn't be able to see the smaller one because the bigger one was in the way. Hopefully that simplifies my question. If it is as simple as that, wouldn't we miss stars and not be able to view them?&lt;/p&gt;&#xA;" OwnerUserId="10807" LastActivityDate="2016-02-14T17:27:14.057" Title="Why when we look through a telescope in space, do the billions of stars not block our view from seeing further?" Tags="&lt;star&gt;&lt;universe&gt;&lt;telescope&gt;&lt;observational-astronomy&gt;&lt;space&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13744" PostTypeId="1" AcceptedAnswerId="13752" CreationDate="2016-02-14T10:08:30.597" Score="3" ViewCount="3976" Body="&lt;p&gt;According to wiki the cosmic event horizon is 5 Gpc but the age of the universe is about 13,7 billion years. Usually light from far galaxys needed at most 13,7 billion years to reach the earth so the event horizon should be the same. But in wiki the event horizon is 5*3,2billionly=16 billion ly. So is this difference just a matter of rounding or old data or is there something else what causes the differences?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmological_horizon#Event_horizon&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Cosmological_horizon#Event_horizon&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10445" LastActivityDate="2019-09-20T22:54:36.897" Title="Why is there a difference between the cosmic event horizon and the age of the universe?" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13745" PostTypeId="2" ParentId="13742" CreationDate="2016-02-14T10:10:43.457" Score="9" Body="&lt;p&gt;OK, Imagine the stars at distance x block an area of the sky. At a distance of 2x there should be four times as many stars, but they would seem four times smaller in terms of area covered. Thus, the blocking grows linearly until a significant part of the sky is blocked. The stars within 20 light years (excluding the Sun), blocks approximately $4.3 \cdot 10^{-16}$ of the sky. Even if we scale that up to the radius of the universe, 45 billion light years, only about $2 \cdot 10^{-6}$ of the sky is covered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even that tiny number is an overestimate, as we live in an extremely star-dense part of the universe, our galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space is simply too big.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="10172" LastEditDate="2016-02-14T17:27:14.057" LastActivityDate="2016-02-14T17:27:14.057" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13747" PostTypeId="2" ParentId="11136" CreationDate="2016-02-14T11:02:40.953" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://arxiv.org/pdf/1503.04451.pdf&quot; rel=&quot;nofollow&quot;&gt;This recent paper&lt;/a&gt; by Napier et al. indeed concludes that centaur comets break up into many pieces large enough to cause mass extinction events on Earth. Since objects orbit in the same way regardless of mass, I suppose that dangerously big comet debris are more common in meteor streams, and that major impacts are more common during meteor showers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small meteoroidal fragments do have their trajectories changed by Solar heating and the Solar wind, but since Earth crossing centaur fragments seem to be cleared out within only thousands of years, there should be no major difference between the large and tiny objects' orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also seems as if mass extinctions might be caused by meteor showers depositing dust particles in the atmosphere. Suggested by &lt;a href=&quot;http://meteor.uwo.ca/~pbrown/nature03881.pdf&quot; rel=&quot;nofollow&quot;&gt;Klekociuk et al.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2016-02-14T11:32:32.373" LastActivityDate="2016-02-14T11:32:32.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13748" PostTypeId="2" ParentId="13688" CreationDate="2016-02-14T11:33:34.143" Score="5" Body="&lt;p&gt;At the announcement press conference (2/11/2016), Kip Thorne said that the detection puts an upper limit on the rest mass of the graviton. They determined this limit by looking at distortions of the detected signal waveform compared to the idealized signal produced by computer simulations. The upper limit from the publication is $m_{graviton} &amp;lt; 1.2  × 10^{−22} \frac{eV}{c^2}$ or $1.9 × 10^{−41} kg$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refs:&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=vy5vDtviIz0&amp;amp;feature=youtu.be&amp;amp;t=1h5m23s&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=vy5vDtviIz0&amp;amp;feature=youtu.be&amp;amp;t=1h5m23s&lt;/a&gt;&#xA;&lt;a href=&quot;https://journals.aps.org/prl/pdf/10.1103/PhysRevLett.116.061102&quot; rel=&quot;nofollow&quot;&gt;https://journals.aps.org/prl/pdf/10.1103/PhysRevLett.116.061102&lt;/a&gt; (page 8)&lt;/p&gt;&#xA;" OwnerUserId="10809" LastEditorUserId="-1" LastEditDate="2016-03-04T22:29:34.340" LastActivityDate="2016-03-04T22:29:34.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13749" PostTypeId="1" AcceptedAnswerId="13750" CreationDate="2016-02-14T13:57:26.483" Score="0" ViewCount="2518" Body="&lt;p&gt;So a black hole is a something that has enormous gravity, therefore mass accelerates faster towards it. But if the black hole double  it's mass, will it's event horizon increase in length , or will it decrease?&lt;/p&gt;&#xA;" OwnerUserId="10765" LastActivityDate="2017-03-02T04:51:58.940" Title="Does the event horizon of a black hole increase or decrease by adding mass?" Tags="&lt;black-hole&gt;&lt;mass&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13750" PostTypeId="2" ParentId="13749" CreationDate="2016-02-14T14:33:53.647" Score="4" Body="&lt;p&gt;The radius of the event horizon ($r_\mathrm{s}$) is directly proportional to the mass of the black hole (M). More exactly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_\mathrm{s} = \frac{2 G M}{c^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black holes whose merger was detected by LIGO would each have been about 90 km in radius, and after merger, a little less than 180km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;nofollow&quot;&gt;Read about the Schwarzchild radius&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-02-14T14:33:53.647" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13751" PostTypeId="2" ParentId="12884" CreationDate="2016-02-14T14:36:57.913" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/C1nNL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/C1nNL.jpg&quot; alt=&quot;Image showing the revolution of earth&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eastward movement mentioned in the book that you're reading means that if you view the earth revolving around the sun from a vantage point in the solar system and if you kept the north pole upwards in your frame of reference, you would observe the earth going towards the right, around the sun. And if you moved to a point in the solar system where you can see the north pole of the earth as well as of the sun, you would see the earth moving anti clockwise in its orbit. &lt;/p&gt;&#xA;" OwnerUserId="10812" LastActivityDate="2016-02-14T14:36:57.913" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13752" PostTypeId="2" ParentId="13744" CreationDate="2016-02-14T15:03:24.043" Score="6" Body="&lt;p&gt;From the link you provide:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The particle horizon differs from the cosmic event horizon, in that the particle horizon represents the largest comoving distance from which light could have reached the observer by a specific time, while the event horizon is the largest comoving distance from which light emitted now can ever reach the observer in the future. The current distance to our cosmic event horizon is about 5 Gpc, well within our observable range given by the particle horizon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your confusion somes from the fact that you are mixing to different horizons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;particle horizon&lt;/strong&gt; is the sphere centered on us that has a radius equal to the distance that light can travel in 13.8 Gyr (the age of the Universe). That is, light that was &lt;em&gt;emitted when the Universe was born&lt;/em&gt; at a point on that horizon, reaches us today. Note that, because the Universe is expanding, the distance is not 13.8 Glyr, or 4.2 Gpc, as one might naively expect, but in fact 46.3 Glyr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;cosmic event horizon&lt;/strong&gt; is also a sphere centered on us, which is the boundary inside which light, if it is &lt;em&gt;emitted today&lt;/em&gt;, may still reach us sometime in the future. If it is emitted outside this horizon, the expansion of the Universe ensures that the light will never reach us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As time goes, we will see light that came from farther and farther away, and thus the distance to the particle horizon is &lt;em&gt;in&lt;/em&gt;creasing. Meanwhile, the accelerated expansion of the Universe ensures that light emitted from distant galaxies in the future are able to reach us only if they are increasingly closer. Thus fewer and fewer galaxies will be inside the event horizon. But since space is at the same time expanding, the boundary will asymptotically reach a finite size (~17 Glyr).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see the particle and the event horizons as blue and red lines in the plot in &lt;a href=&quot;https://physics.stackexchange.com/users/24142/pulsar&quot;&gt;@Pulsar&lt;/a&gt;'s great answer &lt;a href=&quot;https://physics.stackexchange.com/questions/60519/can-space-expand-with-unlimited-speed/63780#63780&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 16.5 Glyr that the distance to the event horizon is today is sort of a coincidence. It has nothing to do with the age of the Universe. It only depends on the future expansion of the Universe, which in turn depends on the densities of the components of the Universe (&lt;span class=&quot;math-container&quot;&gt;$\Omega_\mathrm{b},\Omega_\mathrm{DM},\Omega_\Lambda$&lt;/span&gt;, etc.). If the Universe had been dominated by matter (or radiation), then there would be no event horizon: No galaxy, ever-so far away would &lt;em&gt;not&lt;/em&gt; be visible to us, if we just had the patience to wait. A galaxy is 10,000 billion lightyears away? Just wait long enough (exactly how long depends on the actual density).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, our Universe happens to be dominated by dark energy, which accelerates the expansion without boundaries. This unfortunately means that the light leaving today from a galaxy 17 Glyr away will be carried away by the expansion faster than it can travel toward us. In contrast, the light emitted today from a galaxy 15 Glyr away will travel in our direction, but will nonetheless initially move away from us due to the expansion. However, its journey toward us makes this expansion rate smaller and smaller (since the expansion rate increases with distance from us), and after a period of time it will have traveled so far that it has overcome expansion and starts decreasing its distance from us and eventually reach us after 100 Gyr or so.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2019-09-20T22:54:36.897" LastActivityDate="2019-09-20T22:54:36.897" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="13753" PostTypeId="1" CreationDate="2016-02-14T19:01:11.147" Score="1" ViewCount="60" Body="&lt;p&gt;The SDSS (Sloan Digital Sky Survey) ImgCutout webservice should return jpeg images for a given field that is defined by the parameters. The service can be tested here: &lt;a href=&quot;http://skyservice.pha.jhu.edu/DR12/ImgCutout/ImgCutout.asmx?op=GetJpeg&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://skyservice.pha.jhu.edu/DR12/ImgCutout/ImgCutout.asmx?op=GetJpeg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple (and working) get query is given here:&#xA;&lt;a href=&quot;http://skyservice.pha.jhu.edu/DR12/ImgCutout/getjpeg.aspx?ra=132.749&amp;amp;dec=11.656&amp;amp;scale=0.79224%20%20%20&amp;amp;width=400&amp;amp;height=400&amp;amp;opt=&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://skyservice.pha.jhu.edu/DR12/ImgCutout/getjpeg.aspx?ra=132.749&amp;amp;dec=11.656&amp;amp;scale=0.79224%20%20%20&amp;amp;width=400&amp;amp;height=400&amp;amp;opt=&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you enter the same information (the parameters) in the form of the first link and click on invoke you will get a black image only. You can test the base64 encoded result here: &lt;a href=&quot;http://www.askapache.com/online-tools/base64-image-converter/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.askapache.com/online-tools/base64-image-converter/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Description of how to reproduce my findings:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Go to &lt;a href=&quot;http://skyservice.pha.jhu.edu/DR12/ImgCutout/ImgCutout.asmx?op=GetJpeg&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://skyservice.pha.jhu.edu/DR12/ImgCutout/ImgCutout.asmx?op=GetJpeg&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Fill parameter 'ra_' with '132.749'&lt;/li&gt;&#xA;&lt;li&gt;Fill parameter 'dec_' with '11.656'&lt;/li&gt;&#xA;&lt;li&gt;Fill parameter 'scale_' with '0.79224'&lt;/li&gt;&#xA;&lt;li&gt;Fill parameter 'width_' with '400'&lt;/li&gt;&#xA;&lt;li&gt;Fill parameter 'height_' with '400'&lt;/li&gt;&#xA;&lt;li&gt;Press 'Invoke' button&lt;/li&gt;&#xA;&lt;li&gt;Convert the content in the &lt;code&gt;&amp;lt;base64Binary&amp;gt;&lt;/code&gt; tag from base64 to a jpeg&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Following this steps I always get a black image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now the question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I require an additional parameter (non empty string) for either &lt;code&gt;imgtype_&lt;/code&gt; or &lt;code&gt;imageField_&lt;/code&gt; in order to not get the correct image? What are this parameters good for anyway?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Is there a better place this question? It would be nice if there was a tag for SDSS)&lt;/p&gt;&#xA;" OwnerUserId="6712" LastEditorUserId="6712" LastEditDate="2017-01-05T12:53:33.650" LastActivityDate="2017-01-05T12:53:33.650" Title="SDSS ImgCutout webservice returns black image for dr12" Tags="&lt;observational-astronomy&gt;&lt;sky-survey&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13754" PostTypeId="1" CreationDate="2016-02-14T19:43:23.663" Score="1" ViewCount="135" Body="&lt;p&gt;The horizon problem states that because of the speed of information can't be faster than light and the distances between far regions are too big there is a problem because they look quite the same. Therefor, the inflation is introduced. &#xA;But what I don't understand is why those regions can't be the same without inflation or something like that? They have the same origin and the universe was in the beginning too empty to make a difference for the direction you went. So in my simple understanding, it is rather expected that they should be the same but why are cosmologists think different about that?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="10445" LastEditDate="2016-02-14T20:37:16.330" LastActivityDate="2016-02-15T01:37:40.037" Title="Why can't different regions of the universe be the same as they have the same origin?" Tags="&lt;universe&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13755" PostTypeId="1" CreationDate="2016-02-14T20:55:00.493" Score="4" ViewCount="955" Body="&lt;p&gt;What is the best approach to determine the ellipticity of galaxies in the SDSS DR12. I have read this &lt;a href=&quot;http://www.sdss.org/dr12/algorithms/classify/&quot; rel=&quot;nofollow noreferrer&quot;&gt;page&lt;/a&gt;. Are those really good methods? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do flux-weighted second moments (as given in the stokes parameters) really give the correct value for ellipticity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I am not sure what table field they actually refer too (in the &lt;a href=&quot;http://skyserver.sdss.org/dr12/en/help/browser/browser.aspx#&amp;amp;&amp;amp;history=description+Galaxy+V&quot; rel=&quot;nofollow noreferrer&quot;&gt;Galaxy&lt;/a&gt; view) when talking about the method using the Adaptive Moments. For example where can I find the &lt;code&gt;m_rr_cc&lt;/code&gt; value for the method described &lt;a href=&quot;http://www.sdss.org/dr12/algorithms/classify/#photo_adaptive&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have calculated the ellipticity of a spiral galaxiy with two different methods using data from SDSS DR7. First I used the method using the stokes parameters and second I used the values for isoA and isoB. Both calculations where done in the &lt;em&gt;r-band&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data can be obtained with the following CAS Query:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;SELECT&#xA;    g.objid as objId,&#xA;    /*  stoke parameters*/&#xA;    q_r,u_r,&#xA;     /*  iso_a, iso_b */&#xA;    isoA_r,isoB_r&#xA;FROM Galaxy AS g&#xA;WHERE&#xA;    g.objid = 587722982832013381&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;yielding the result&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;objId: 587722982832013381   &#xA;q_r : -0.1206308    &#xA;u_r: 0.002988584    &#xA;isoA_r: 169.3949    &#xA;isoB_r: 84.78848&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The ellipticity by stoke parameters is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;e  = 1 - \frac{b}{a} = 1 - \frac{1 - \sqrt{Q^2 + U^2}}{1 + \sqrt{Q^2 + U^2}}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ellipticity from isoA and isoB is very given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;e  = 1 - \frac{isoB}{isoA}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I get completely different results when I use isoA and isoB or the stoke parameters. I get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ellipticity from stoke parameters: 0.2153498356&#xA;ellipticity from 1-(isoA/isoB): 0.4994626166&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The results can be reproduced with the following python code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import math&#xA;&#xA;def ellipticityStokes(q, u):&#xA;    e = 1 - ((1-math.sqrt((q**2) + (u**2)))/(1+math.sqrt((q**2) + (u**2))))&#xA;    return e&#xA;&#xA;def ellipticityNormal(a, b):&#xA;    e = 1 - (b/a)&#xA;    return e&#xA;&#xA;q = -0.1206308&#xA;u = 0.002988584 &#xA;&#xA;isoA = 169.3949  &#xA;isoB = 84.78848&#xA;&#xA;print(str(ellipticityStokes(q,u))) #0.2153498356239003&#xA;print(str(ellipticityNormal(isoA,isoB))) #0.4994626166431221&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Looking at the image&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gD8HT.jpg&quot; width=&quot;200&quot; height=&quot;200&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(and compare with this &lt;a href=&quot;https://ned.ipac.caltech.edu/level5/Sept11/Buta/Buta5.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;website from Buta&lt;/a&gt;) I would say the value 0.4994626166 (from isoA and isoB) makes more sense. The value calculated from the stokes parameters is just wrong I think. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the mistake here? I would expect both calculations to yield similar results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However for newer data releases isoA and isoB are not available any more because they seem not to trust those value? What should I do with newer data releases as DR12?&lt;/p&gt;&#xA;" OwnerUserId="6712" LastEditorUserId="9301" LastEditDate="2016-06-29T04:39:39.043" LastActivityDate="2016-06-29T04:39:39.043" Title="How to determine the ellipticity of galaxies in SDSS" Tags="&lt;galaxy&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13757" PostTypeId="2" ParentId="13664" CreationDate="2016-02-14T21:16:15.710" Score="5" Body="&lt;p&gt;For anyone else who might be asking a similiar question, I found a great resource for this. The PDFs found at the URL below contain a good list of binocular, naked-eye and telescope targets for each month of the year in both hemispheres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.skymaps.com/downloads.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.skymaps.com/downloads.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user10636" LastActivityDate="2016-02-14T21:16:15.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13758" PostTypeId="2" ParentId="13754" CreationDate="2016-02-14T21:57:18.850" Score="0" Body="&lt;p&gt;There is a fundamental (but disputed) idea in Cosmology called &quot;the Cosmological Principle&quot; (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmological_principle&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Cosmological_principle&lt;/a&gt;) which states, at a sufficiently large scale, the universe looks the same everywhere at the same epoch. And there is a lot of evidence to suggest that it does. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the inflation theory states that very small, quantum level, variations in the universe a very small time after the Big Bang should be reflected in observations of the universe at a large scale today because inflation means that these parts of the universe never had time to exchange information (or photons) to equalise their energy in the very earliest moments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an analogy (perhaps a bit strained) - if you had a room in which there was a radiator giving out heat and then the radiator was switched off, after a few hours you would expect the temperature in the room to be the same everywhere, because the warm air would mix with the colder air and the temperature would equalise. But imagine if there was some sort of barrier of distance between the warm parts and the cold parts - then the cold parts would stay cold and the warm parts would stay warm for much longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The task cosmologists have set themselves is to find the imprint of these patterns in the density of the observed universe (as energy and mass are equivalent the hotter/more energetic parts of the early universe should have a bit more mass in them today).&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2016-02-14T21:57:18.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13759" PostTypeId="1" AcceptedAnswerId="13765" CreationDate="2016-02-14T22:58:09.323" Score="0" ViewCount="131" Body="&lt;p&gt;I'm attempting to build a MMSN, I've been trying for about... a week to get an intuition on the scaling factors for calculating the mass of the disks, I'm not entirely sure how to think about it. I feel as if the concept is extremely simple but that I'm thinking of it incorrectly. Can someone explain what exactly these scaling factors account for and how one would calculate them? I'm not looking for a solution, I'm looking for intuition on how these scaling factors are found. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understand is that you pick an element, (Fe), and compare that to the solar abundance (does this mean the abundance of Fe in the sun? From there I would look at the abundance of Fe in the cores of the planet and compare them somehow to get some sort of scaling factor. From what I understand Fe makes .14% of the suns mass, so if I know lets say that neptune has a mass of 17m(earth) how would I start finding the scaling factor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! &lt;/p&gt;&#xA;" OwnerUserId="10817" LastActivityDate="2016-02-15T12:42:49.137" Title="Minimum Mass Solar Nebulae Scaling Factors" Tags="&lt;mass&gt;&lt;accretion-discs&gt;&lt;the-sun&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13760" PostTypeId="2" ParentId="13754" CreationDate="2016-02-15T01:37:40.037" Score="2" Body="&lt;p&gt;We don't really know what happened at the time $t=0$. It doesn't really make sense to say that everything &quot;touched each other&quot; at $t=0$, particularly if the Universe is infinite. But we can calculate the distance that a photon — and hence the maximum distance that any information — can travel in the expanding Universe in a given time. This calculations depends only on the Hubble parameter, and the densities of the various components of the Universe, and whether you start your calculation at $t=0$ or $t=10^{-12}\,\mathrm{s}$ makes little difference after a second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is related to your previous &lt;a href=&quot;https://astronomy.stackexchange.com/questions/13744/why-is-there-a-difference-between-the-cosmic-event-horizon-and-the-age-of-the-un&quot;&gt;question about the cosmic event horizon&lt;/a&gt; which was also about the &lt;em&gt;particle horizon&lt;/em&gt;, i.e. the distance that light has been able to travel since the Big Bang. If you were present when the CMB was emitted, you could also calculate your particle horizon. It would of course be much smaller than today, since light at this time had only traveled for 380,000 years. If the Universe hadn't been expanding, this horizon would of course be 380,000 lightyears, but due to the expansion, it is quite a lot larger, roughly 850,000 lightyears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that, at the time of the CMB, regions farther apart than 850 kly had not had the chance to be in causal contact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also calculate how large an angle 850 kly at the CMB would suspend if observed today. It turns out to be 1.7º. That is, if the Universe had just been expanding like we observe its expansion today — i.e. depending only on the densities of the known components — patches on the CMB map separated by an angle $\theta &amp;gt; 1.7^\circ$ shouldn't look the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But they do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inflation solves this problem by saying that the Universe initially went through a much, much faster expansion, such that regions much, much farther apart than the 850 kly have been in causal contact. Still, even with inflation, the particle horizon is not infinite. So on scales much, much larger than the observable Universe, it may be inhomogeneous.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-02-15T01:37:40.037" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13761" PostTypeId="1" AcceptedAnswerId="13762" CreationDate="2016-02-15T03:48:32.753" Score="0" ViewCount="95" Body="&lt;p&gt;I'm looking for some reference listing spectral lines outside the visible spectrum.  (In particular, I'm looking for an element or compound that emits strongly in the 700-900nm range.)&lt;/p&gt;&#xA;" OwnerUserId="1987" LastEditorUserId="10172" LastEditDate="2016-02-15T17:34:19.083" LastActivityDate="2016-02-16T08:33:08.210" Title="Reference for broader spectral lines?" Tags="&lt;spectra&gt;&lt;infrared&gt;&lt;resource-request&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13762" PostTypeId="2" ParentId="13761" CreationDate="2016-02-15T09:01:39.070" Score="3" Body="&lt;p&gt;&lt;strong&gt;Assuming your question is a reference request:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I now about one resource that has a very broad range of spectral data, the&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.nist.gov/PhysRefData/ASD/lines_form.html&quot; rel=&quot;nofollow&quot;&gt;NIST Atomic Spectra Database Lines Form&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has worked great for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.lib.utexas.edu/chem/info/spectra.html&quot; rel=&quot;nofollow&quot;&gt;This page&lt;/a&gt; lists a couple of different alternatives. For the near IR case you are interested in, several collections are listed &lt;a href=&quot;https://sites.google.com/site/anthonycrastospectroscopy/ir-spectral-data&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="10172" LastEditDate="2016-02-15T12:32:54.270" LastActivityDate="2016-02-15T12:32:54.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13763" PostTypeId="2" ParentId="13693" CreationDate="2016-02-15T10:13:55.687" Score="4" Body="&lt;p&gt;It makes a difference, but not to the strength of the gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Around a spinning black hole (or any other spinning mass) space time is dragged. This effect has been measured around the (spinning) Earth by &lt;a href=&quot;https://einstein.stanford.edu/&quot; rel=&quot;nofollow&quot;&gt;Gravity Probe B&lt;/a&gt;. It would cause a plumb line not to point directly at the centre of the Earth, but slightly forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a rotating black hole, there is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Ergosphere&quot; rel=&quot;nofollow&quot;&gt;region&lt;/a&gt; close to the event horizon, but outside it, in which the gravity (a combination of downwards and sideways accelerations) is so strong that it is not possible to remain stationary, relative to a distant observer, for to do so one would have to travel faster than the speed of light (relative to the local spacetime)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar situation exists in a non-rotating black hole, but only behind the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a rotating black hole does not have stronger gravity, but it is different: It is sideways. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-02-15T10:13:55.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13764" PostTypeId="1" AcceptedAnswerId="13768" CreationDate="2016-02-15T10:36:19.167" Score="4" ViewCount="135" Body="&lt;p&gt;If I understand it well enough, gravitational waves cause ripples in the space-time fabric. To our eyes, that would look like those rippling-effects on amateur videos. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does a source have to be, in order for the effects of gravitational waves to be visible, here, on Earth. &lt;/p&gt;&#xA;" OwnerUserId="10824" LastActivityDate="2020-09-04T18:28:56.197" Title="How far should the source be, for the gravitation waves to be visible on Earth?" Tags="&lt;general-relativity&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13765" PostTypeId="2" ParentId="13759" CreationDate="2016-02-15T12:07:06.160" Score="1" Body="&lt;p&gt;&lt;strong&gt;Attempt to do an intuitive explanation:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun was formed from a cloud of matter. A small amount of that matter remained around the Sun, and had a similar composition. This was the solar nebulae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the composition of the planets are now not the same as the Sun. This is explained by that some of the matter was collected by the Sun, and some of it was blown out of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;That was only the light stuff!&lt;/strong&gt; like hydrogen and helium. Most iron and other metals are conjectured to have remained in orbit, forming the planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the Sun is representative for the original composition (partly at least, stellar fusion has skewed the hydrogen/helium ratio). So if iron is 0.14% of the Sun, that is also the original abundance in the nebulae. &lt;strong&gt;So all the iron in the planetary bodies is 0.14% of the original nebulae mass&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From that you can calculate the original mass. It is a 'minimum' because some of the iron may have escaped as well.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="10172" LastEditDate="2016-02-15T12:42:49.137" LastActivityDate="2016-02-15T12:42:49.137" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13766" PostTypeId="1" CreationDate="2016-02-15T15:17:51.260" Score="2" ViewCount="161" Body="&lt;p&gt;What would happen if a strong gravitational wave passed through a human body or any other kind of material? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape of the human body would change?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the human body be distorted apart? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, would the person survive?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="2270" LastEditDate="2016-02-16T10:39:47.507" LastActivityDate="2016-02-16T10:39:47.507" Title="The effect of gravitational wave is like tidal forces?" Tags="&lt;tidal-forces&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13767" PostTypeId="2" ParentId="13766" CreationDate="2016-02-15T15:50:21.530" Score="6" Body="&lt;p&gt;It depends what you mean by &quot;very strong&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The announced first detection (14th September 2015) was considered a very strong signal and that still only stretched the 4 km detector arms by less than the width of a proton (a tiny particle inside the nucleus of an atom). This wave would have no effect on a person.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could hypothesis a stronger wave that would stretch and squish a person, I'm not sure what effect this would have but there are no objects in nature capable of producing waves this strong, even something as close as our sun becoming a supernova wouldn't be that strong, so it will never happen in real life.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2016-02-15T15:50:21.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13768" PostTypeId="2" ParentId="13764" CreationDate="2016-02-15T16:52:13.333" Score="11" Body="&lt;p&gt;The waves pass by at the speed of light. So you you would'nt see ripples, they would pass too fast, and remember the waves would be passing through you too. The wavelength was (relativly) long about 3000km. The wave doesn't pass you, you are inside the wave.&lt;/p&gt;&#xA;&lt;p&gt;The amplitude of the waves detected by LIGO was small, one part in &lt;span class=&quot;math-container&quot;&gt;$10^{21}$&lt;/span&gt;, Now while the intensity of the waves follows an inverse square law, because intensity is proportional to amplitude squared, the amplitude is inversely proportional to distance from source.&lt;/p&gt;&#xA;&lt;p&gt;We are familiar with strains that are one part in 1000, that is a variation of one millimetre for every metre. To get that we would have to be a lot closer to the source, &lt;span class=&quot;math-container&quot;&gt;$10^{18}$&lt;/span&gt; times closer. Since the black holes were about 1 billion light years distant, to get a strain of 1 part in 1000, the black hole merger would have to be &lt;strong&gt;about 10000km away&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now if you are 10000km from a merging pair of 30 solar mass black holes, the strain of gravitational radiation is the least of your problems. If it is any consolation, your problems won't lost long.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="35024" LastEditDate="2020-09-04T18:28:56.197" LastActivityDate="2020-09-04T18:28:56.197" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="13769" PostTypeId="1" AcceptedAnswerId="13770" CreationDate="2016-02-15T19:56:05.217" Score="0" ViewCount="291" Body="&lt;p&gt;The title perhaps gives the question away, but would anyone know/be able to roughly approximate the angular distance/apparent separation between AE Aurigae and Orion's belt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the value of $42^\circ$, from visual estimation but would like a way to calculate this.&lt;/p&gt;&#xA;" OwnerUserId="2091" LastEditorUserId="2365" LastEditDate="2016-02-15T20:39:54.850" LastActivityDate="2016-02-15T20:39:54.850" Title="Angular distance/apparent separation between AE Aurigae and Orion's belt" Tags="&lt;distances&gt;&lt;positional-astronomy&gt;&lt;astrometry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13770" PostTypeId="2" ParentId="13769" CreationDate="2016-02-15T20:22:50.373" Score="1" Body="&lt;p&gt;Given the cords of AE aurigae:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;RA 05h 16m 18.1s DEC +34° 18' 49&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And Alnitak in Orion's belt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;RA 05h 40m 45.52666s DEC  −01° 56′ 34.2649″&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get an angular separation of 34.83°&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By this formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cos(A) = \sin(\mathrm{DEC}_1)\cdot \sin(\mathrm{DEC}_2) + \cos(\mathrm{DEC}_1)\cdot \cos(\mathrm{DEC}_2)\cdot \cos(\mathrm{RA}_1 - \mathrm{RA}_2)$&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="2365" LastEditDate="2016-02-15T20:37:28.580" LastActivityDate="2016-02-15T20:37:28.580" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13771" PostTypeId="2" ParentId="13728" CreationDate="2016-02-15T23:37:06.340" Score="2" Body="&lt;p&gt;Based on a post by an Astrophysicist my understanding is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Gravitational waves would still be faint to detect irrespective of the size of the Black Holes. BTW aren't black holes meant to be more concentrated rather than bloated?&lt;/li&gt;&#xA;&lt;li&gt;The distance from ground zero is what matters rather than the size of the colliding black holes and even if they were a billion times closer they would still not measure up to 1 mm.&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;There wouldn't be significant impact from the waves themselves rather the Black Holes are the ones that would do the damage.&lt;/li&gt;&#xA;&lt;li&gt;For the shift to be 1 mm we would require a force enormously stronger than that of a gravitational wave.  &lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Here's the link to the article: &lt;a href=&quot;http://www.forbes.com/sites/briankoberlein/2016/02/13/could-gravitational-waves-ever-be-strong-enough-to-feel/#1605e7fa4aac&quot; rel=&quot;nofollow&quot;&gt;http://www.forbes.com/sites/briankoberlein/2016/02/13/could-gravitational-waves-ever-be-strong-enough-to-feel/#1605e7fa4aac&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Disclaimer&lt;/strong&gt;: All my points are a result of my learning from the above link rather than any native knowledge or original research that I profess to have done.&lt;/p&gt;&#xA;" OwnerUserId="10690" LastActivityDate="2016-02-15T23:37:06.340" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13772" PostTypeId="2" ParentId="13738" CreationDate="2016-02-15T23:52:28.260" Score="2" Body="&lt;p&gt;I think you answered your own question with your second paragraph. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Newton's law of gravity &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the Gravitational force is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.&lt;/li&gt;&#xA;&lt;li&gt;So the biggest body in the neighborhood has the most pulling power provided it agrees with the equation.&lt;/li&gt;&#xA;&lt;li&gt;As a consequence Jupiter and the outer planets should win out over Earth and keep them from hitting Earth or at least divert the collision course. Especially since the NEA come from their part of the cosmos.&lt;/li&gt;&#xA;&lt;li&gt;I suppose the same logic would also work with the comets. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Please see the following link for more:&#xA;&lt;a href=&quot;http://earthsky.org/space/is-it-true-that-jupiter-protects-earth&quot; rel=&quot;nofollow&quot;&gt;http://earthsky.org/space/is-it-true-that-jupiter-protects-earth&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="10690" LastActivityDate="2016-02-15T23:52:28.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13773" PostTypeId="1" CreationDate="2016-02-16T04:25:29.360" Score="4" ViewCount="58" Body="&lt;p&gt;This question pertains to &lt;a href=&quot;http://www.bbc.com/news/science-environment-31058529&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; article which talks about why the BICEP2 measurements of B-mode polarization in Cosmic microwave background radiation turned out to be noise from galactic stardust. They go on to add that the Plank data of the noise does not lend to getting a good B-mode estimate from the BICEP2 data.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, the results of the joint assessment would suggest that&#xA;  whatever signal BICEP2 detected, it cannot be separated at any&#xA;  significant level from the spoiling effects. In other words, the&#xA;  original observations are equally compatible with there being no&#xA;  primordial gravitational waves. &quot;This joint work has shown that the&#xA;  detection of primordial B-modes is no longer robust once the emission&#xA;  from galactic dust is removed,&quot; Jean-Loup Puget, principal&#xA;  investigator of Panck's HFI instrument, said in the Esa statement.&#xA;  &quot;So, unfortunately, we have not been able to confirm that the signal&#xA;  is an imprint of cosmic inflation.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can someone explain why(from a signal processing perspective) exactly the detection of primordial B-modes is not robust even after the emission from galactic dust is removed? And what are the proposed workarounds to tackle this problem?&lt;/p&gt;&#xA;" OwnerUserId="5395" LastEditorUserId="7982" LastEditDate="2019-07-25T13:24:35.637" LastActivityDate="2019-07-25T13:24:35.637" Title="Problem with denoising the BICEP2 data?" Tags="&lt;gravitational-waves&gt;&lt;cosmological-inflation&gt;&lt;cmb&gt;&lt;post-processing&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13775" PostTypeId="2" ParentId="13761" CreationDate="2016-02-16T08:33:08.210" Score="1" Body="&lt;p&gt;Both potassium (766.5+769.9 nm) and sodium (818.3+819.4 nm) have strong resonance lines (doublets) in this part of the spectrum. There are also the calcium triplet lines at 849.8+854.2+866.2 nm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the strongest atomic absorption lines seen in the spectra of solar-type stars or cooler at 700-900 nm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more specific answer would require the spectral type of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do something more general and/or extensive you would need to get hold of a good line list (a line list alone cannot easily tell you which lines will appear strongest in a stellar spectrum) and a spectral synthesis programme like SME or MOOG.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-02-16T08:33:08.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13776" PostTypeId="1" CreationDate="2016-02-16T10:39:22.327" Score="4" ViewCount="434" Body="&lt;p&gt;What would a sundial look like in a binary star system? What is the path drawn by the shadows? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It probably depends on the system, the sizes of the stars, their distance... If you know a binary star system for which a sundial would look different from what we have on Earth, we can take it as an example!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: thank you to barycarter for bringing to attention that the Kepler binary stars would look like one star from the surface of their planets. However, there must be, somewhere, a binary system where both stars are visible, such that the shadows are affected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT2 : A Google search shows that this question has been answered before, but all links to the answer are dead. - I searched &quot;binary solar system clock&quot;, found a link by a &quot;Charmaine Pauls&quot; - &#xA;But after using &quot;&lt;em&gt;sundial&lt;/em&gt;&quot; instead, I found &lt;a href=&quot;https://fr.pinterest.com/pin/421016265145961006/&quot; rel=&quot;nofollow&quot;&gt;this pintrest artwork&lt;/a&gt;, but it is not much help. It doesn't seem to be scientifically motivated.&lt;/p&gt;&#xA;" OwnerUserId="10836" LastEditorUserId="10836" LastEditDate="2016-02-25T18:21:42.347" LastActivityDate="2016-03-13T16:29:33.243" Title="Sundial (solar clock) in a binary star system" Tags="&lt;binary-star&gt;&lt;clock&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13777" PostTypeId="1" CreationDate="2016-02-16T11:56:12.107" Score="2" ViewCount="123" Body="&lt;p&gt;If I had a long stick that was millions of light years long, and with a width of about the Sun, and I then swiped it through the center of the milky way galaxy, what would the probability be of me hitting a star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ignore the speed of light. Assume it is possible to just swipe it through instantaneiously.&lt;/li&gt;&#xA;&lt;li&gt;Assume that the stars in the galaxy are equally distributed. Obviously if you were to swipe right through the middle you would hit the galactic center.&lt;/li&gt;&#xA;&lt;li&gt;The purpose of this thought experiment is just to help visualise the density vs size of the galaxy.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4238" LastEditorUserId="4238" LastEditDate="2016-02-17T08:52:38.747" LastActivityDate="2016-02-17T14:22:57.473" Title="visualisation of galaxy density" Tags="&lt;galaxy&gt;&lt;milky-way&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13778" PostTypeId="2" ParentId="13755" CreationDate="2016-02-16T12:29:10.560" Score="2" Body="&lt;p&gt;Regarding your original question about the &quot;correct value for ellipticity&quot;, keep in mind that a galaxy has no objective ellipticity. The inserted image nicely shows that the light distribution of a real galaxy is way more complicated than a smooth profile with isophotes of identical ellipticity. In other words, the ellipticity will in fact be defined through the measurement process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flux-weighted second moments (which use the &lt;em&gt;pixel brightness&lt;/em&gt; as weighting function) or adaptive moments (which use an elliptical &lt;em&gt;Gaussian&lt;/em&gt; weighting function) are two ways of doing this. And there might be nothing wrong with the observed difference in ellipticity measurements, as the galaxy light profile is very different from a Gaussian!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flux-weighted measurement is more sensitive to the ellipticity at inner radii, while the adaptive moments give more weight to the disk (assuming that the algorithms converge well).&lt;/p&gt;&#xA;" OwnerUserId="9150" LastEditorUserId="9150" LastEditDate="2016-02-16T12:54:53.047" LastActivityDate="2016-02-16T12:54:53.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13779" PostTypeId="1" CreationDate="2016-02-16T13:28:50.367" Score="2" ViewCount="166" Body="&lt;p&gt;Shortly after the news of the discovery of gravity waves, my physics professor explained the significance of the discovery by comparing it to a medical X-ray machine. He said that gravity waves can be used to detect objects that cannot be seen using telescopes. Therefore, could gravity wave detectors be used to confirm the existence of the theorized ninth planet? Even though it doesn't have a high enough albedo to be seen, it would still be detectable gravitationally, wouldn't it?&lt;/p&gt;&#xA;" OwnerUserId="10711" LastEditorUserId="661" LastEditDate="2016-02-17T17:39:08.193" LastActivityDate="2016-02-17T22:47:09.917" Title="Could gravity waves be used to confirm the existence of a ninth planet?" Tags="&lt;gravity&gt;&lt;planet&gt;&lt;gravitational-waves&gt;&lt;9th-planet&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13780" PostTypeId="2" ParentId="13779" CreationDate="2016-02-16T13:50:42.387" Score="13" Body="&lt;p&gt;No, they can't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity waves from a small, simple object moving slowly are very, very faint, to the point of being undetectable with current (or foreseeable) technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The waves that have been detected come from the merger (a very fast movement in the last orbits) of two black holes (two very big masses). And they were just detected over the noise level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do not confuse that with the possibility to detect the gravitational influence of such a planet over other bodies. That is how Neptune and Pluto were detected, indeed.&lt;/p&gt;&#xA;" OwnerUserId="473" LastEditorUserId="473" LastEditDate="2016-02-17T06:43:24.850" LastActivityDate="2016-02-17T06:43:24.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13781" PostTypeId="2" ParentId="13777" CreationDate="2016-02-16T15:00:56.737" Score="-1" Body="&lt;p&gt;Well Urmmmmm... its a weird question but to put it one way, you wouldn't hit any as our lovely black hole would consume this stick before you could hit any. Apart from that the chance is actually high, there are so many stars in our galaxy(and even more near the centre) that you could hit on. But lets say if you to count the black hole, but it didn't get consumed, it would take so long (because of the black holes gravity) that you would die of old age before you managed to swing down through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit anyone who can improve on my theory!&lt;/p&gt;&#xA;" OwnerDisplayName="user10841" LastEditorUserId="9645" LastEditDate="2016-02-17T14:22:57.473" LastActivityDate="2016-02-17T14:22:57.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13782" PostTypeId="1" AcceptedAnswerId="13783" CreationDate="2016-02-16T15:19:15.370" Score="34" ViewCount="6590" Body="&lt;p&gt;I am trying to find an original video/image of what LIGO actually saw, but all I can find is artist renditions of gravitational waves.&lt;/p&gt;&#xA;" OwnerUserId="10844" OwnerDisplayName="Scott Taylor" LastEditorUserId="37191" LastEditDate="2021-02-12T21:30:09.687" LastActivityDate="2021-02-14T20:23:37.097" Title="What did LIGO Actually See? (Gravitational waves discovery)" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;gravitational-waves&gt;&lt;ligo&gt;" AnswerCount="7" CommentCount="8" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13783" PostTypeId="2" ParentId="13782" CreationDate="2016-02-16T16:26:40.897" Score="36" Body="&lt;p&gt;The actual image isn't much. I was able to find it from &lt;a href=&quot;http://www.sciencemag.org/news/2016/02/gravitational-waves-einstein-s-ripples-spacetime-spotted-first-time&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Science&lt;/em&gt;&lt;/a&gt;, and this is all it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CahjT.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CahjT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's just a ripple, seen at slightly different times from two different observatories. The shift fits perfectly by shifting it by the speed of light difference in their locations. Thus is the proof of gravity waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted that the reason there are two instruments is to provide a cross check against other vibration sources. Each observatory works by detecting vibrations on a 4 km scale, down to a very small order of magnitude (1/10,000 the width of a proton). When the two are compared, then one can assume the signal must have come from a non-local source, which only Gravity Waves fit that definition.&lt;/p&gt;&#xA;" OwnerUserId="182" LastActivityDate="2016-02-16T16:26:40.897" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13785" PostTypeId="2" ParentId="13782" CreationDate="2016-02-16T19:58:23.180" Score="1" Body="&lt;p&gt;I don't know whether it's interesting for you, but here is the link of the paper that was published about those observations: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://journals.aps.org/prl/pdf/10.1103/PhysRevLett.116.061102&quot; rel=&quot;nofollow&quot;&gt;http://journals.aps.org/prl/pdf/10.1103/PhysRevLett.116.061102&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the answer above is quite straightforward! What the paper says (in short) is that LIGO has observed a transient gravitational-wave signal, and these observations match with the predictions of waveform derived by General Relativity for the system involving two black holes.&lt;/p&gt;&#xA;" OwnerUserId="10829" LastEditorUserId="10829" LastEditDate="2016-02-16T23:18:23.750" LastActivityDate="2016-02-16T23:18:23.750" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13786" PostTypeId="1" AcceptedAnswerId="13792" CreationDate="2016-02-16T20:23:03.227" Score="3" ViewCount="425" Body="&lt;p&gt;What methods exist to calculate the ellipticity of galaxies and what are their drawbacks? I have asked this &lt;a href=&quot;https://astronomy.stackexchange.com/q/13755/6712&quot;&gt;question about ellipticity in the SDSS&lt;/a&gt; but I want to know about general methods for cases where I have just an image of a galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do so far know these methods:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;ellipticity from stoke parameters or the flux-weighted second moment as described &lt;a href=&quot;http://www.sdss.org/dr12/algorithms/classify/#photo_stokes&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/li&gt;&#xA;&lt;li&gt;from adaptive moments as described &lt;a href=&quot;http://www.sdss.org/dr12/algorithms/classify/#photo_adaptive&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; (which would allow me to do shear calibration).&lt;/li&gt;&#xA;&lt;li&gt;fit an ellipse on a particular isophote as decribed &lt;a href=&quot;http://www.sdss.org/dr12/algorithms/classify/#photo_iso&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My question is whether there are further methods I am not aware of. Does there also exist a method when I fit an ellipse to all isophotes? And how would I calculate the 'average' from it (such that the spherical center does not contribute to much)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to learn about the pro and cons of each method. I am especially interessted in methods that get close to what people would estimate by eye. I would also be glad to hear about references (textbooks or papers) for the different methods. &lt;/p&gt;&#xA;" OwnerUserId="6712" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-02-17T11:54:22.007" Title="What methods exist to calculate the ellipticity of galaxies" Tags="&lt;galaxy&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13787" PostTypeId="1" AcceptedAnswerId="13789" CreationDate="2016-02-16T20:39:32.923" Score="2" ViewCount="3571" Body="&lt;p&gt;I found this page: &lt;a href=&quot;http://phys.org/space-news/&quot; rel=&quot;nofollow&quot;&gt;http://phys.org/space-news/&lt;/a&gt; . It seems to me pretty good page with interesting news and so on and I like that it is free. But I noticed in comments people arguing about the validity of the information and so on. Are they just &quot;trolls&quot; or is this page really bad source of information, if that is so, how much can I trust it?&#xA;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="9677" LastActivityDate="2016-02-18T00:55:16.387" Title="Is phys.org/space-news reliable source?" Tags="&lt;astrophysics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13788" PostTypeId="2" ParentId="13782" CreationDate="2016-02-17T00:46:01.940" Score="14" Body="&lt;p&gt;LIGO didn't &quot;see&quot; anything. It monitors the relative lengths of the paths taken by two laser beams in vacuum pipes about 4km long (although the laser path consists of about 75 trips up and down the arms) and at right angles to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A gravitational wave, travelling at the speed of light, changes the ratio of these lengths (one gets shorter, one gets larger, then they swap) by about plus or minus  1 part in $10^{21}$ (a billion trillion) about 30-200 times per second as it passes through the instrument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole event lasted about 0.3 seconds and the trace (which has been all over the news) simply records the fraction by which the length of the arms changes as a function of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The event was (nearly) simultaneously recorded by two almost identical setups in different parts of the USA. The detection of the same signal in both detectors rules out a local cause of the disturbance, and the small time delay between the detections allows a rough location of the gravitational wave source on the sky.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-02-17T01:13:48.680" LastActivityDate="2016-02-17T01:13:48.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13789" PostTypeId="2" ParentId="13787" CreationDate="2016-02-17T02:20:56.487" Score="6" Body="&lt;p&gt;I've been reading &lt;a href=&quot;http://phys.org/&quot;&gt;phys.org&lt;/a&gt; for years.&#xA;It's right up there with &lt;a href=&quot;http://physics.aps.org/&quot;&gt;physics.aps&lt;/a&gt;&#xA;and &lt;a href=&quot;http://physicsworld.com/&quot;&gt;physicsworld&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All good sources that provide decent refs.&#xA;Comments are often where people who like to argue, argue. Sometimes they are useful, often not.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2016-02-17T02:20:56.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13790" PostTypeId="2" ParentId="13777" CreationDate="2016-02-17T11:11:01.517" Score="4" Body="&lt;p&gt;For your experiment, you can take it 2D, so let's ingnore the thickness of Milky Way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we have 400 billions stars(the real number is between 100B-400B) in a circle with 100k ly(the real number is between 100k-160k). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of the circle is pi*50k*50k ~= 7800M sqly = 7.8 billions sqly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have then ~51 randomly stars per square ly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you do not sting the Milky Way with your stick, but cut through it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you swipe through the middle, you have 51*100k stars in a width of 1 ly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you have a chance to hit a star per 1/5.100.000 ly which means 1.855.000km. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the diameter of sun(of your stick) is 1.395.000km. Taking diameter of sun as medium diameter of a star(this is not true), you have a chance of 1.4/1.8 = 77% to hit a star. If you go with the &quot;low density&quot; numbers, 100B stars and 160k ly diameter, you'll get ~31%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My conclusion is you often hit a star.&lt;/p&gt;&#xA;" OwnerUserId="7991" LastActivityDate="2016-02-17T11:11:01.517" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13791" PostTypeId="1" CreationDate="2016-02-17T11:19:31.003" Score="4" ViewCount="249" Body="&lt;p&gt;After the discovery of gravitational waves by LIGO last week the team behind the Fermi gamma-ray telescope released a &lt;a href=&quot;http://arxiv.org/abs/1602.03920&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; that showed a soft gamma ray burst was detected only 0.4s after the gravitational wave was detected at LIGO, and it was also located in the right area of the sky. This could be evidence of a gravitational wave source with a detected EM counterpart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately though the Fermi group resist claiming a full discovery as no electromagnetic signature is expected form a binary black hole merger, but whats interesting is that I saw another &lt;a href=&quot;http://arxiv.org/pdf/1602.04735v1.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; today that claimed a GRB could be caused by a binary black hole merger inside a massive rapidly rotating star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is such a system realistic or even possible?&lt;/p&gt;&#xA;" OwnerUserId="10437" LastEditorUserId="10437" LastEditDate="2016-02-18T11:33:49.427" LastActivityDate="2016-04-26T19:43:12.017" Title="Could a pair of binary black holes form within a star?" Tags="&lt;black-hole&gt;&lt;astrophysics&gt;&lt;gravitational-waves&gt;&lt;gamma-ray-bursts&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13792" PostTypeId="2" ParentId="13786" CreationDate="2016-02-17T11:54:22.007" Score="2" Body="&lt;p&gt;The first two methods mentioned in the question are really similar. They both use moments of the light distribution, only the weighting is different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another family of methods is 2D profile fitting to the image, with more or less sophisticated luminosity profile models. A simple case would be to fit a &lt;a href=&quot;https://en.wikipedia.org/wiki/Sersic_profile&quot; rel=&quot;nofollow&quot;&gt;Sérsic profile&lt;/a&gt;. But instead, one could make a composite model with two or more elliptical profiles (say one exponential disk and a &lt;a href=&quot;https://en.wikipedia.org/wiki/De_Vaucouleurs%27_law&quot; rel=&quot;nofollow&quot;&gt;de Vaucouleur&lt;/a&gt; bulge). This would allow to separate the ellipticity of the disk from the ellipticity of the bulge. A widely used software for this is &lt;a href=&quot;https://users.obs.carnegiescience.edu/peng/work/galfit/galfit.html&quot; rel=&quot;nofollow&quot;&gt;GALFIT&lt;/a&gt; by Chien Peng.&lt;/p&gt;&#xA;" OwnerUserId="9150" LastActivityDate="2016-02-17T11:54:22.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13793" PostTypeId="2" ParentId="13782" CreationDate="2016-02-17T13:37:26.607" Score="29" Body="&lt;p&gt;First of all, I think your question belies a misunderstanding of the nature of the LIGO observatories.  The nature of the detectors is that they act like a microphone, as opposed to a camera.  What that means is that they are sensitive to gravitational waves coming in from most directions, but don't have an ability to distinguish where the waves came from.  By using multiple detectors (which is also necessary for a confident detection) the difference in time between the detectors can be used to give some idea of the location of the source. It also means that the output of the detectors is a single stream of data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image from &lt;a href=&quot;https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.116.061102&quot; rel=&quot;noreferrer&quot;&gt;the paper in Physical Review Letters&lt;/a&gt; (not behind a paywall) is a better summary of what LIGO heard than the current accepted answer.  I'll explain the panes from top to bottom.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The top panes show the 'raw' signals measured in the two detectors with the H1 data overlayed on the L1 data on the right.&lt;/li&gt;&#xA;&lt;li&gt;The second row of panes shows a number of different simulations of what general relativity (Einstein's theory) predicts for the gravitational waves.  These simulations are how LIGO is able to claim that they know that the wave was caused by two merging black holes.&lt;/li&gt;&#xA;&lt;li&gt;The third row of panes is the 'raw' data minus the simulations.&lt;/li&gt;&#xA;&lt;li&gt;The bottom panes are simply another way of plotting the 'raw' data called a time-frequency plot.  Time is on the x axis and frequency is on the y axis.  To a person from the field this signal is the most recognizable characteristic of a merger, known as a chirp.  As time moves forward, the frequency shifts higher.  You can actually listen to the 'raw' chirp &lt;a href=&quot;https://www.youtube.com/watch?v=TWqhUANNFXw&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VFyQx.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VFyQx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="821" LastEditorUserId="821" LastEditDate="2017-06-05T11:53:14.020" LastActivityDate="2017-06-05T11:53:14.020" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13794" PostTypeId="1" CreationDate="2016-02-17T13:40:23.350" Score="1" ViewCount="171" Body="&lt;p&gt;So they finally found gravitational waves.  That's cool and I trust is the case, but how did they confirm it originated 1.3 billion years ago and that two blackholes merging caused it and what distance it came from? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure if the extent of variation in the laser wavelength can tell us all this above information.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any clues ?&lt;/p&gt;&#xA;" OwnerUserId="10864" OwnerDisplayName="Gautam" LastEditorUserId="-1" LastEditDate="2016-02-18T19:16:00.423" LastActivityDate="2016-02-18T19:16:00.423" Title="How did LIGO verify that the gravitational waves originated 1.3 billion years ago from two specific black hole collision?" Tags="&lt;gravity&gt;&lt;black-hole&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2016-02-17T15:15:29.450" ContentLicense="CC BY-SA 3.0" />
  <row Id="13795" PostTypeId="2" ParentId="13794" CreationDate="2016-02-17T14:41:50.557" Score="3" Body="&lt;p&gt;The &lt;a href=&quot;http://journals.aps.org/prl/pdf/10.1103/PhysRevLett.116.061102?page=2#temp%3Aintralink-f1&quot; rel=&quot;nofollow&quot;&gt;original paper&lt;/a&gt; states the methodology used to determine the location of the gravity waves. The bottom line is that they were detected in two different locations. Using the difference in timing, one can estimate the location in the sky to an area of about 600 degrees squared where the origin of the wave could have been. The two signals had an offset of 7 ms. Pure speed of light differences would have been 10 ms. This can be used to effectively identify a circle in the sky where the merger might have happened. Of course, this doesn't reveal the distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance was figured by a number of things. First of all, the frequency pattern was used to estimate the mass of each black hole. Using these masses, one can find the estimated total power of the event. Using this estimate, and seeing the estimate of the amplitude from the signal, one can use the inverse squared law to estimate the distance to said event.&lt;/p&gt;&#xA;" OwnerUserId="182" LastActivityDate="2016-02-17T14:41:50.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13796" PostTypeId="1" CreationDate="2016-02-17T16:13:26.827" Score="1" ViewCount="179" Body="&lt;p&gt;In the history of the universe, there has been a time, where intelligent life had no chance to develop (e.g. in the early universe, when no habitable planets have formed) and there is a time today where we know, that at least one intelligent lifeform has evolved in our universe/galaxy. Between these two times many events took place in order to provide the conditions, that (intelligent) life can evolve. So I wonder myself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the time, when the conditions in the universe became life-friendly, has the human race as intelligent life evolved early, late or something in between?&lt;/p&gt;&#xA;" OwnerUserId="9147" LastActivityDate="2016-02-17T17:47:20.793" Title="How early/late is the human race as intelligent life in the universe/milkyway?" Tags="&lt;life&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2016-02-18T05:46:53.977" ContentLicense="CC BY-SA 3.0" />
  <row Id="13798" PostTypeId="2" ParentId="13782" CreationDate="2016-02-17T17:40:25.790" Score="7" Body="&lt;p&gt;According to the &lt;a href=&quot;https://losc.ligo.org/s/events/GW150914/GW150914_tutorial.html&quot; rel=&quot;noreferrer&quot;&gt;GW150914 tutorial&lt;/a&gt;, this is what Advanced LIGO L1 and H1 detectors originally saw:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nbSkr.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nbSkr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can download the raw data from this tutorial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other answers show already processed (whitened, filtered, shifted by 7 ms, inverted) waveforms.&lt;/p&gt;&#xA;" OwnerUserId="10867" LastActivityDate="2016-02-17T17:40:25.790" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13799" PostTypeId="2" ParentId="13796" CreationDate="2016-02-17T17:47:20.793" Score="3" Body="&lt;p&gt;I think this is a very subjective question depending on your point of view. If you assume that the history of the universe is 13.8 billion years old, then humans have only been around for 200,000 years which makes us very late at evolving. But of course the entire history of the universe hasn't been written yet!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I always find these sort of calendar/24hr clock representations very insightful too. If you look at this image below, modern &quot;intelligent&quot; humans have been around for the last 6 minutes of the year, if the age of the universe was 365 days long, so not very long at all!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H3dOe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H3dOe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, I think there is another answer to your question, but becuase we don't know how long the universe/milky way will last for, its impossible to say or indeed ever know for how long intelligent life could actually evolve for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm afraid i'm going to have to say somewhere in between too, as we simply cant see into the future.&lt;/p&gt;&#xA;" OwnerUserId="10437" LastActivityDate="2016-02-17T17:47:20.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13800" PostTypeId="2" ParentId="13688" CreationDate="2016-02-17T18:37:03.437" Score="1" Body="&lt;p&gt;Though the twin discovery of Gravitational Waves and Black Hole merger might not affect directly the status of QM it might indirectly bring new &quot;surprises&quot; For example, in this link:&#xA;&lt;a href=&quot;http://news.discovery.com/space/weve-detected-gravitational-waves-so-what-160213.htm&quot; rel=&quot;nofollow&quot;&gt;http://news.discovery.com/space/weve-detected-gravitational-waves-so-what-160213.htm&lt;/a&gt;&#xA;They comment that:&#xA;&quot;For some reason, the final spin of the black hole is slower than expected, indicating that the two black holes collided at a low speed, or they were in a collision configuration that caused their combined angular momentum to counteract each other. “That is very curious; why would nature do that?” said Lehner.&quot;&#xA;And the final comment is:&#xA;&quot;This early puzzle could be down to some basic physics that hasn’t been considered, but more excitingly it could reveal some “new” or exotic physics that is interfering with the predictions of general relativity&quot;.&#xA;Wow! &quot;Interfering with general relativity&quot; is a polite way of suggesting that it might be wrong. So maybe QM might come to the rescue of Gen.Relativity rather than the other way around.&lt;/p&gt;&#xA;" OwnerUserId="10869" LastActivityDate="2016-02-17T18:37:03.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13801" PostTypeId="2" ParentId="13685" CreationDate="2016-02-17T20:31:09.157" Score="8" Body="&lt;p&gt;There are four aspects to this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You suspend the mirrors and other components using the best suspension mechanism that has ever been built. This is very effective at isolating the experiment.&lt;/li&gt;&#xA;&lt;li&gt;There are two detectors thousands of km apart. To claim a genuine signal it should be seen identically in both in less than the light travel time between them.&lt;/li&gt;&#xA;&lt;li&gt;Many of the astrophysical GW sources (e.g. merging black holes) produce characteristic signals that look nothing like random seismic activity - for instance a correlated increase in frequency and amplitude terminating in a &quot;chirp&quot;.&lt;/li&gt;&#xA;&lt;li&gt;Lastly, LIGO is not immune to seismic noise - it is this that partly limits its sensitivity below frequencies of around 30 cycles per second. The seismic noise component at higher frequencies is much smaller.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The way that LIGO isolates the astrophysical signal from the various noise components that are present is described in &lt;a href=&quot;http://arxiv.org/abs/1602.03839&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; by the LIGO and VIRGO collaborations. A crude characterisation of the process is that you cross-correlate your signal against a library of expected signatures from astrophysical GW sources and look for a high degree of correlation (which you wouldn't expect in random noise).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The characterisation of the noise (including seismic noise), a description of how it is assessed and monitored (including how data is rejected when deemed too noisy) is found in a further &lt;a href=&quot;http://arxiv.org/abs/1602.03844&quot; rel=&quot;nofollow&quot;&gt;LIGO publication&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="10762" LastEditDate="2016-02-23T00:25:27.737" LastActivityDate="2016-02-23T00:25:27.737" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13802" PostTypeId="1" AcceptedAnswerId="13808" CreationDate="2016-02-17T21:04:22.753" Score="-1" ViewCount="252" Body="&lt;p&gt;It is hard to imagine that the universe could have a flat curvature while I've read that it is at this moment the most likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you look around the universe is expanding in all directions, so it looks like a spherical ball shaped universe. Nevertheless the curvature is until now measured flat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how to understand that something that has in all directions galaxies look flat. So I thought perhaps they mean that is in the 5th dimension, in analogy with an ant who can't see that the line he is walking on is actually 3 dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I read that If the true value of the cosmological curvature parameter is larger than $10^{−3}$ we will be able to distinguish between the three models (spherical, hyperbolic and flat) even now. So I think how is something in the 5th dimension measured as it is beyond our capability of experiencing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So somewhere I miss something but what?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="2365" LastEditDate="2016-02-17T21:56:36.407" LastActivityDate="2016-02-18T12:23:08.273" Title="Is the flat curvature of the universe in the 5th dimension?" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13803" PostTypeId="1" AcceptedAnswerId="13804" CreationDate="2016-02-17T21:16:36.213" Score="16" ViewCount="300" Body="&lt;p&gt;You may have heard in the news that the LIGO experiment recently detected a gravitational wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though I'm not an astronomer, the paper is a good read and mostly accessible. The detection of the gravitational wave is one thing, but the black hole merger is quite new to me. From the data I could collect in the paper and this site, the source is estimated at 1.3 billion years, and the chirp lasted only a few milliseconds.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: &lt;strong&gt;what is the frequency of events of this size order?&lt;/strong&gt;&#xA;Is there any estimate of the density of such events in the universe?&lt;/p&gt;&#xA;" OwnerUserId="10877" LastEditorUserId="2270" LastEditDate="2016-03-04T14:44:28.417" LastActivityDate="2016-03-04T14:44:28.417" Title="Frequency of gravitational wave detection" Tags="&lt;gravitational-waves&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13804" PostTypeId="2" ParentId="13803" CreationDate="2016-02-17T21:37:11.107" Score="10" Body="&lt;p&gt;&lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0264-9381/27/8/084006/pdf;jsessionid=00AFCD541625E3C533C93F5B30DC6B17.c4.iopscience.cld.iop.org&quot;&gt;Harry (2009)&lt;/a&gt; cited several different sources stated that, as far we know, the rate of &lt;em&gt;detectable&lt;/em&gt; events will be&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;40 neutron star mergers per year&lt;/li&gt;&#xA;&lt;li&gt;30 10 M$_\odot$ black hole mergers per year&lt;/li&gt;&#xA;&lt;li&gt;10 neutron star/black hole mergers per year&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is within a radius of about 200 Mpc. This cannot, however, be used to extrapolate the total rate of such events, because of detection bias - the more massive the objects, the more easy they are to detect. The same thing happens with exoplanets, but for different reasons (e.g. planets that are more massive or closer to their stars are easier to detect via transit or by radial velocity methods).&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-02-17T22:56:49.213" LastActivityDate="2016-02-17T22:56:49.213" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13805" PostTypeId="2" ParentId="13779" CreationDate="2016-02-17T22:47:09.917" Score="1" Body="&lt;p&gt;The maths on &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_wave#Basic_mathematics&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; give a way of calculating the the amplitude of graviational waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One detail is that for the waves to be detected by a ligo apparatus, one needs to be far enough from the source of the waves. The distance required depends on the frequency. For planet 9, if it has orbit of 10000 years, one would need to be more than $2500\pi$ light years distant. That is nearly 10000 light years away. This value is &quot;R&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we can use the formula:&#xA;$$h_{+} =-\frac{1}{R}\, \frac{G^2}{c^4}\, \frac{4m_1 m_2}{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the above value of R, the known values of the gravitation constant and speed of light, and the masses of the sun and a hypothesised value for the mass and semi-major axis of planet 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That gives a strain of $10^{-32}$.  That is far far far below what is detectable. It also would have a frequency of 10000 years or so. So you would be trying to detect an oscillation comparable to the plank length, you would need at least 10000 years of observation, and your detector has to be built on the other side of the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not possible.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-02-17T22:47:09.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13806" PostTypeId="2" ParentId="13782" CreationDate="2016-02-17T22:48:25.640" Score="5" Body="&lt;p&gt;The actual mechanism of measurement that LIGO used is laser interferometry, so one reasonable interpretation of what LIGO &quot;saw&quot; would be the interference pattern caused by the gravity waves, which would &quot;look&quot; something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/u6pXH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/u6pXH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I could not find an image of the actual laser interference that LIGO mentioned; it's probably too small for photography anyway. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the other graphs people are linking do are just graphs of the interference pattern data. Showing a graph of the LIGO data as an answer to this question is like showing an image histogram as an answer to the question, &quot;What does the Hubble space telescope see?&quot;&lt;/p&gt;&#xA;" OwnerUserId="10878" LastEditorUserId="10878" LastEditDate="2016-03-03T16:27:40.680" LastActivityDate="2016-03-03T16:27:40.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13807" PostTypeId="2" ParentId="13803" CreationDate="2016-02-17T22:49:57.323" Score="6" Body="&lt;p&gt;The rate of detected gravitational waves of this amplitude or detections of gravitational waves due to the merger of black hole binaries are both unknown quantities for the moment. Measuring these is partly the purpose of the experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rates of detection can be converted into a rate of mergers per unit volume in space and these can be compared with models and predictions. The aLIGO collaboration have released their first post-detection paper on that very topic - &lt;a href=&quot;https://dcc.ligo.org/P1500262/public&quot; rel=&quot;noreferrer&quot;&gt;Abbott et al. (2016)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The high mass of the discovered black holes implies that they either formed in a metal-poor environment from the core-collapse of massive stars or that they formed from the merger of smaller black holes in dense clusters. The range of rates previously predicted for the mergers of such objects covered a huge range because of the massive uncertainties in the production rate and mechanisms for binary formation of these objects, and lay in the range from 0 to about 1000 per year per cubic Gigaparsec.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate actually implied by the &lt;em&gt;single&lt;/em&gt; detection of a BH binary merger (from only 16 days of data) at $z=0.09$ is somewhere between 2 and 400 per year per cubic Gigaparsec. So at present, this doesn't really rule out a lot, but it is expected that with a few further detections in the coming months the uncertainty on this number will come down rapidly.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-02-17T22:49:57.323" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13808" PostTypeId="2" ParentId="13802" CreationDate="2016-02-17T22:59:08.600" Score="3" Body="&lt;p&gt;Several topics here:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;No 5th dimension. Not even a 4th (spatial)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You're on the right track: The Universe being &quot;flat&quot; does &lt;em&gt;not&lt;/em&gt; mean &quot;like a table top&quot;, just like it being &quot;spherical&quot; and &quot;hyperbolic&quot; don't mean like a ball or a saddle. Those terms are just 2D analogies, whereas the Universe has three spatial dimensions. Like any analogy, they are great for understanding some things, but shouldn't be taken too far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, the sum of the angles of a triangle, the fate of parallel lines, and the volume of some region in space can be more easily visualized in 2D than in 3D. You can understand that a triangle going North Pole→Congo→Indonesia→North Pole has 270º rather than 180º. But whereas the 2D surface of Earth curves in 3D space, a 4th spatial dimension is not needed for our 3D Universe to curve (not to speak of a 5th). And imaging why a triangle consisting of what seems to be completely straight lines from Earth→&lt;a href=&quot;https://en.wikipedia.org/wiki/GRB_090423&quot; rel=&quot;nofollow&quot;&gt;GRB090423&lt;/a&gt;→&lt;a href=&quot;https://en.wikipedia.org/wiki/EGSY8p7&quot; rel=&quot;nofollow&quot;&gt;EGSY8p7&lt;/a&gt;→Earth should &lt;em&gt;not&lt;/em&gt; have 180º is difficult, but nonetheless perfectly possible.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The Universe is not a ball with you in the center&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;When you look around in the Universe, it looks like a spherical ball with you in the center, but that's only because what you see is everything from where light has had the time to reach you in 13.8 Gyr, i.e. since Big Bang. Everything inside that sphere is called &lt;em&gt;the observable Universe&lt;/em&gt;, and it's bound by the so-called &lt;em&gt;particle horizon&lt;/em&gt;. In the 2D analogy, you can say that Earth looks like a flat disk, but only because that's how far you can see. Again, don't take this analogy too seriously, since the reason here is just the curvature of Earth, and not that Earth has only existed for $10^{-4}\,\mathrm{s}$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Curvature&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The dynamical and geometrical properties of the Universe depend on its expansion rate ($H_0$), the densities of its components (the &quot;Omegas&quot; of baryons, dark matter, dark energy, radiation, etc.), and its &quot;intrinsic&quot; curvature, which can also be written like a density parameter $\Omega_k$. The latter is found to be $0.000\pm0.005$ (&lt;a href=&quot;http://arxiv.org/abs/1502.01589&quot; rel=&quot;nofollow&quot;&gt;Ade et al. 2015&lt;/a&gt;), but as you say, if it's less than $\sim10^{-4}$, we may never know (&lt;a href=&quot;http://arxiv.org/abs/0901.3354&quot; rel=&quot;nofollow&quot;&gt;Vardanyan et al. 2009&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-02-18T12:23:08.273" LastActivityDate="2016-02-18T12:23:08.273" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13809" PostTypeId="1" CreationDate="2016-02-17T22:59:43.540" Score="2" ViewCount="762" Body="&lt;p&gt;new questioner here so please be gentle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do the singularities of merging black holes actually merge together?  I mean there are 2 infinitely small objects (singularities) that are trying to find each other but for them to merge they have to join at the EXACT same point in space.  I know they can get close, but to expect them to align perfectly with zero tolerance seems impossible.&lt;/p&gt;&#xA;" OwnerUserId="10879" LastActivityDate="2016-02-24T19:07:16.447" Title="Do black hole singularities actually merge?" Tags="&lt;black-hole&gt;&lt;singularity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13810" PostTypeId="1" AcceptedAnswerId="13814" CreationDate="2016-02-17T23:06:13.207" Score="1" ViewCount="1518" Body="&lt;p&gt;Can someone explain to me how I can generate a Position-Velocity (PV) diagram from a galaxy's radial velocity field? I understand that usually PV diagrams are generated across the major axis of the galaxy, but I am not quite sure why they appear as a distribution. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xUolL.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xUolL.gif&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1986" LastEditorUserId="7472" LastEditDate="2018-12-25T12:34:52.920" LastActivityDate="2018-12-25T12:34:52.920" Title="Create Position-Velocity diagram from a velocity field" Tags="&lt;galaxy&gt;&lt;kinematics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="13811" PostTypeId="2" ParentId="13809" CreationDate="2016-02-17T23:17:05.520" Score="5" Body="&lt;p&gt;Infinities are hard to bend the mind around, but in this case, the merge is not impossible. Yes, the distance between them must reach zero in order to make the black holes merge, but the rate of which energy is lost to gravitational waves also increases when they get close to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are therefore dealing with a $\frac{\mathrm{infinity}}{\mathrm{infinity}}$ problem, where a finite &lt;a href=&quot;https://en.wikipedia.org/wiki/Limit_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;limit&lt;/a&gt; may, and in this case does exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do not forget your $\lim$'s!&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="2365" LastEditDate="2016-02-24T19:07:16.447" LastActivityDate="2016-02-24T19:07:16.447" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13812" PostTypeId="1" CreationDate="2016-02-17T23:45:29.157" Score="2" ViewCount="50" Body="&lt;p&gt;I sometimes encounter telescopes/binoculars with a decent objective diameter but are very difficult to use because they only work when the eye is looking down the central axis (i.e. in the centre of the field) or close to it; if you try to focus on an object towards the edge of the field of view it goes black.  Is there a name for this tolerance?  It doesn't sound related to &lt;a href=&quot;https://en.wikipedia.org/wiki/Exit_pupil&quot; rel=&quot;nofollow&quot;&gt;exit pupil&lt;/a&gt; but probably is something to do with &lt;a href=&quot;https://en.wikipedia.org/wiki/Eye_relief&quot; rel=&quot;nofollow&quot;&gt;eye relief&lt;/a&gt; - i.e. the distance of eye to the eyepiece lens.&lt;/p&gt;&#xA;" OwnerUserId="10880" LastEditorUserId="616" LastEditDate="2016-04-23T00:13:30.690" LastActivityDate="2016-04-23T00:13:30.690" Title="Telescope optical tolerance from central axis" Tags="&lt;telescope&gt;&lt;binoculars&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13813" PostTypeId="1" CreationDate="2016-02-18T00:29:19.080" Score="3" ViewCount="332" Body="&lt;p&gt;I was thrilled by the recent announcement of the detection of gravitational waves. The media has done a great job explaining them and the theory behind them and I get the concept fairly well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if it's a warping of spacetime, not a particle phenomenon,  (or is it somehow?) How fast do these waves travel through space? Is it  constant or variable? Slower than the speed of light or even faster? If spacetime has no resistance per se, the what would keep them from traveling infinitely fast?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd love to understand it better and hope my question makes sense.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="2531" LastEditDate="2016-02-20T18:56:09.287" LastActivityDate="2017-10-17T15:34:54.527" Title="How fast can and do gravitational waves travel?" Tags="&lt;gravitational-waves&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13814" PostTypeId="2" ParentId="13810" CreationDate="2016-02-18T00:50:20.013" Score="4" Body="&lt;p&gt;Such diagrams are usually called &lt;a href=&quot;https://en.wikipedia.org/wiki/Galaxy_rotation_curve&quot; rel=&quot;nofollow noreferrer&quot;&gt;rotation curves&lt;/a&gt;, and show the velocity of stars/gas &lt;em&gt;in a disk galaxy viewed edge-on&lt;/em&gt; along the line of sight (LOS) as a function of distance from the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the $x$ axis gives the distance of the stars from the center ($x=0$) of the galaxy, and the $y$ axis gives the measured velocities at a given distance. Along a given LOS you'll find not a single velocity, but a (Gaussian-like) distribution of velocities, both because there is a certain scatter in velocities, but also because you only measure the component of the velocities projected onto your LOS, while both speed and direction change along the LOS, as seen here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8usas.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8usas.png&quot; alt=&quot;rotvel&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If&lt;/em&gt; they all had the same velocity along the LOS, the plot you show would be a thin line; instead it is smeared out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to mass being concentrated in the center, there's a steep increase in velocity at low $x$. If there were only the mass you can see, the velocity would decrease at larger $x$ due to the larger distance from the central mass. The reason it stay more or less constant is thought to be the extra, invisible mass known as &lt;em&gt;dark matter&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason the left side of the plot is similar to the right, but mirrored not only across the $y$ axis but also the $x$ axis, is that on this side, the stars move &lt;em&gt;toward&lt;/em&gt; you, while on the right side they move &lt;em&gt;away&lt;/em&gt; from you. The whole galaxy evidently moves away from you at $500\,\mathrm{km}\,\mathrm{s}^{-1}$, since this is the velocity at $x=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to recap: How do you generate such a plot? You measure the distribution of velocities along many lines of sight in the galactic plane, from one side ($-R_\mathrm{gal}$) to the other ($+R_\mathrm{gal}$). For instance, at $x=14\,\mathrm{kpc}$, you'd measure $V=675\pm12\,\mathrm{km}\,\mathrm{s}^{-1}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U5JtJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U5JtJ.png&quot; alt=&quot;los&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, images like this is made by using a grism which disperses the light according to its wavelength, together with a slit placed along the galactic plane, such that you block the light from the rest of the field. This is called &lt;em&gt;slit spectroscopy&lt;/em&gt;, in contrast to &quot;normal&quot; spectroscopy where you block all light except for a little hole placed on some object (e.g. a star), or integral field spectroscopy, where you get the whole spectrum behind every pixel in an image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sw7C3.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sw7C3.gif&quot; alt=&quot;Slit&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-02-18T08:19:57.650" LastActivityDate="2016-02-18T08:19:57.650" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13815" PostTypeId="2" ParentId="13787" CreationDate="2016-02-18T00:55:16.387" Score="0" Body="&lt;p&gt;Because of the .org, you probably can trust it, because it's an organization. Remember, anyone can write a comment.&lt;/p&gt;&#xA;" OwnerUserId="10881" LastActivityDate="2016-02-18T00:55:16.387" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13816" PostTypeId="1" CreationDate="2016-02-18T03:06:00.690" Score="3" ViewCount="1399" Body="&lt;p&gt;I'm having hard times understanding the equation of Drake, i can´t find the number of the variable that refers to the stars with planetary systems (fp)&lt;/p&gt;&#xA;" OwnerUserId="10882" LastEditorUserId="7982" LastEditDate="2021-02-19T01:29:21.170" LastActivityDate="2021-02-19T01:29:21.170" Title="what is the percentage of stars with planetary systems?" Tags="&lt;star-systems&gt;&lt;drake-equation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13817" PostTypeId="1" AcceptedAnswerId="14255" CreationDate="2016-02-18T05:23:54.103" Score="1" ViewCount="205" Body="&lt;p&gt;I'm once again confused by the twin paradox. Let's say I am on an interstellar starship flying at 0.6c from a star 30 light years away from Earth to a star 50 light years away from Earth and 40 light years from each other. How would you estimate the difference between the time on the ship when you reach the destination versus the time a person on Earth thinks it should be? Now let's say I travel back to the original star (In both circumstances, I accelerate at 1g until reaching 0.6c.) I would assume that I could use the twin paradox formula to determine the difference in elapsed time on the ship from that measured by a clock left at the first star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's say I'm traveling in the same manner (1g acceleration until reaching 0.6c and 1g deceleration for the same duration until arrival) beginning at Earth, traveling to 16 stars and returning to Earth. I am thinking that the time difference between the ship clock (date) and Earth time would be based on the distance and acceleration/ deceleration segments of each leg of the journey independent of the components of the travel vectors in the direction (toward or away from the Earth). Am I right??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would Earth send messages to the ship based on conventional measurements (ie Physics 101)? I would want to determine the difference between the time on Earth when the message is sent and the ship time when the message is received, considering of course the time of transit of the message (at light speed).&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="2365" LastEditDate="2016-03-11T18:44:43.793" LastActivityDate="2018-08-18T13:33:21.847" Title="Special Relativity as Applied to an Interstellar Starship?" Tags="&lt;special-relativity&gt;&lt;interstellar-travel&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13818" PostTypeId="1" AcceptedAnswerId="13822" CreationDate="2016-02-18T06:32:17.120" Score="6" ViewCount="501" Body="&lt;p&gt;If I understand the principle of operation of operation of LIGO, it detects relative distortions of the two perpendicular arms. So if both arms are distorted the same way, nothing would be spotted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, any wave coming in the direction 45 degrees between the arms (both ways), or from above or from below - or generally any direction in the symmetry plane 45 degrees to the arms - would be completely invisible, and anything close to it would be very muted as the relative distortion would be minimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that correct?&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="37191" LastEditDate="2021-02-12T21:29:34.850" LastActivityDate="2021-02-12T21:29:34.850" Title="Does LIGO have a blind spot?" Tags="&lt;gravitational-waves&gt;&lt;ligo&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13819" PostTypeId="2" ParentId="13813" CreationDate="2016-02-18T06:39:06.247" Score="4" Body="&lt;p&gt;Gravitational waves travel at the same speed as the speed of light in a vacuum, in general relativity. In special relativity, this is the maximum speed for any interaction in nature.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;If gravity was faster than the speed of light, then predictions we make according to the bending of light would be incorrect.&lt;/strong&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;If gravity was faster than the speed of light, then the orbits and positions of the planets would be completely different.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Those are my two main points, to help me prove to you that gravity travels at the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, &lt;em&gt;light is not the only thing in the universe which travels at the speed of light.&lt;/em&gt; It is about all massless particles! Photons and gluons are two good examples I can think of of massless particles, I'm not sure about gravitons though - but I'll still include them.&lt;/p&gt;&#xA;" OwnerUserId="9645" LastActivityDate="2016-02-18T06:39:06.247" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13821" PostTypeId="2" ParentId="13816" CreationDate="2016-02-18T07:49:46.783" Score="4" Body="&lt;p&gt;For the purposes of the Drake equation you may as well assume that every star has a solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At present, the exact fraction is unknown since the search techniques are limited to finding planets with certain characteristics. For example, transit searches tend to find close-in, giant planets; doppler shift surveys are also most sensitive to massive planets with short orbital periods, and so-on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can try and account for this incompleteness, but although some examples have been found, we are in the dark about the fraction of stars that have an Earth-sized planet at distances further than the Earth is from the Sun, or the fraction of stars that have any kind of planet orbiting beyond where Saturn is in our solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Drake equation, you need to know the fraction of stars with planetary systems times the number of &quot;habitable&quot; planets per system is. There have been attempts to estimate this &lt;a href=&quot;http://www.nasa.gov/content/nasa-kepler-results-usher-in-a-new-era-of-astronomy&quot; rel=&quot;nofollow&quot;&gt;from Kepler data&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possibly the best estimate at present is from &lt;a href=&quot;http://m.pnas.org/content/110/48/19273.abstract&quot; rel=&quot;nofollow&quot;&gt;Petigura et al. (2013)&lt;/a&gt;, who estimate that &lt;strong&gt;22%&lt;/strong&gt; of &lt;em&gt;sun-like&lt;/em&gt; stars have an earth-sized (1-2 Earth radii) planet that receives between 0.25-4 times the radiative flux of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously this is a lower limit because it doesn't include planets smaller than the Earth. It also doesn't include the moons of giant planets, which may also be habitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We still know very little about this fraction when it comes to M-dwarf stars, which are the most common type of star in the Galaxy...&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-02-18T12:32:08.603" LastActivityDate="2016-02-18T12:32:08.603" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13822" PostTypeId="2" ParentId="13818" CreationDate="2016-02-18T11:53:14.477" Score="5" Body="&lt;p&gt;&lt;a href=&quot;http://arxiv.org/pdf/1304.0670.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This paper&lt;/a&gt; should tell you all you need to know about how LIGO observes and localizes gravitational waves, and in particular its blind spots. The map below is taken from the paper and shows the observability of the system when in the  HLV configuration (Hanford-LIGO, Livingston-LIGO &amp;amp; Virgo). To triangulate a source by definition you need at least 3 measurements so that is why Virgo is included, although the two LIGO detectors can make very broad estimates for source localisation on their own (GW150914 was assigned a 600 square degree error box). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The (x) denotes a blind spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UzMbR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UzMbR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10437" LastActivityDate="2016-02-18T11:53:14.477" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13823" PostTypeId="1" AcceptedAnswerId="13825" CreationDate="2016-02-18T15:07:09.270" Score="3" ViewCount="706" Body="&lt;p&gt;Please excuse me for a lay question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we know, to calculate the orbital velocity, we take the mass of the orbiting body, the mass of the body being orbited and the distance between the two bodies into account. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we don't really care about the natural satellite(s) (of a planet) in the calculations. Why are they insignificant? Why isn't the whole system involved in the calculations?&lt;/p&gt;&#xA;" OwnerUserId="10298" LastEditorUserId="10298" LastEditDate="2016-02-18T19:22:37.803" LastActivityDate="2016-02-18T19:22:37.803" Title="Does natural satellite(s) of a planet affects its orbital velocity around a star?" Tags="&lt;star&gt;&lt;planet&gt;&lt;orbital-elements&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13824" PostTypeId="1" AcceptedAnswerId="13829" CreationDate="2016-02-18T15:58:38.347" Score="3" ViewCount="2104" Body="&lt;p&gt;I would like to find a table where I can, say, look for K5 dwarfs and read off that stars of that type have on average an absolute magnitude of X.XX in the infrared L band and Y.YY in the G band. I've looked everywhere and I can't find a compilation of this kind of information. Does any such thing exist?&lt;/p&gt;&#xA;" OwnerUserId="10892" LastActivityDate="2018-09-25T22:48:13.450" Title="Table of absolute magnitudes of stars by spectral type and photometric band" Tags="&lt;absolute-magnitude&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13825" PostTypeId="2" ParentId="13823" CreationDate="2016-02-18T16:15:48.080" Score="3" Body="&lt;p&gt;Yes, it does affect the planet's orbital velocity. Both planet and satellite move around the planet-satellite center of mass. That center of mass is orbiting the star with some constant velocity (let's assume circular orbits to keep things simple) but if you're only looking at the planet, then the orbital speed around the star will wobble up and down around the because of its additional motion around the planet-satellite center of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For most planet-satellite systems this effect is very small since usually the satellite is very, very light compared to the planet. Only when you have a genuine double system like Earth-Moon or Pluto-Charon does this effect start to matter. For transiting exoplanet systems they're trying to infer the existence of exomoons by exploiting the effect this will have on the transit timing of the planet. So far no exomoons have turned up though.&lt;/p&gt;&#xA;" OwnerUserId="10892" LastActivityDate="2016-02-18T16:15:48.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13826" PostTypeId="2" ParentId="13823" CreationDate="2016-02-18T17:43:29.903" Score="2" Body="&lt;p&gt;One way to look at this question is to not think of the Moon orbiting the Earth but look at it as two bodies orbiting each other, then it becomes obvious that orbital velocities of two objects around each other does affect the orbital velocity around the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon and Earth both orbit the barycenter between them, which is inside the earth, but in the direction of the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon orbits the barycenter about 3,640 km/hr (slightly slower than it orbits the Earth I think), so the corresponding earth's movement for 81 times the mass is an ellipse 1/81st the size, giving the Earth an elliptical orbital velocity of about 45 km/hr, which, being in an elliptical motion is a vector addition of velocity and so it's only about plus or minus 45 km/hr at full or no moon, when the motion is relatively parallel with, or 180 degrees opposite to the Earth's motion around the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;see picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ElllG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ElllG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This velocity variation has a period of about 27.3 days (&lt;a href=&quot;http://community.dur.ac.uk/john.lucey/users/lunar_sid_syn.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;sidereal&lt;/a&gt; not synodic) with an average diameter of 1/81st the Moon's orbital diameter, or about 9,400 KM, which is about the same distance the Earth orbits around the sun in just over 5 minutes, so, the effect is tiny, and is probably tiny for every planet-satellite system, but feel free to calculate Jupiter-Ganymede or Pluto-Charon if you like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the 30,000 km/s approximation, we get 30,045 km/hr at no moon and 29,955 km/hr at full moon or about 3/10ths of 1%, peak to troth, every 13.65 days or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparison, The Earth's &lt;a href=&quot;http://www.space.com/12161-fourth-july-earth-sun-distance-aphelion.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;perihelion&lt;/a&gt; is about 3.28% closer to the sun than it's aphelion, which using Kepler's equal areas law and an approximation of area = 1/2 base x height, the Earth's orbital velocity is about 3.28% greater at perihelion, or about 11 times as much variation in every 182.62 days of elliptical orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the most part, the Earth's orbital velocity isn't that relevant.  The Earth's position, not it's velocity, determines which stars or planets you see at night and if the Earth's spans slightly more or slightly less of it's orbit than usual, over a 24 hour period because of the Moon, 5 minutes of fluctuation over 13 days is only relevant to the most rigorous of astronomers.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-02-18T19:16:33.163" LastActivityDate="2016-02-18T19:16:33.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13827" PostTypeId="2" ParentId="13824" CreationDate="2016-02-18T17:54:19.917" Score="1" Body="&lt;p&gt;There are a lot of stars [citation needed] and they are organised into catalogs by various organisations, well known catalogs include the &lt;a href=&quot;http://www.usno.navy.mil/USNO/astrometry/optical-IR-prod/icas&quot; rel=&quot;nofollow&quot;&gt;UNSO catalog&lt;/a&gt; (produced by the united states Naval office) and the Guide Star Catalog (used by Hubble to accurately posistion itself)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the best for you is the Hipparcos catalog of stars, as it can be browsed quite easily by spectral type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://heasarc.gsfc.nasa.gov/db-perl/W3Browse/w3table.pl?tablehead=name%3Dhipparcos&amp;amp;Action=More%20Options&quot; rel=&quot;nofollow&quot;&gt;HEASARC Browse (Catalog search)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-02-18T17:54:19.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13828" PostTypeId="1" AcceptedAnswerId="13832" CreationDate="2016-02-18T19:26:01.203" Score="2" ViewCount="351" Body="&lt;p&gt;&lt;a href=&quot;http://canburytech.net/DriftAlign/DriftAlign_3.html&quot; rel=&quot;nofollow&quot;&gt;Canbury Tech&lt;/a&gt; has a good background on King tracking rate, which approximates the average sidereal motion of a refracted star across the sky. Most amateur telescope mounts have some sort of notion of this &quot;average&quot; rate, which is generally about 15.037 arc seconds per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, King actually produced a formula that would allow the rate to be precisely calculated for a given set of sky coordinates. Modern microprocessor controlled mounts ought to be able to make use of this technique, but many don't. As the author of telescope drive software, I'd like to incorporate this feature into my work, but I don't seem to be able to locate the forumula. &lt;a href=&quot;http://www.bbastrodesigns.com/equatTrackingRatesCalc.html&quot; rel=&quot;nofollow&quot;&gt;BBAstroDesigns&lt;/a&gt; has a calculator for this on their web site, but the don't show the formula anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone enlighten me as to how to compute this value?&lt;/p&gt;&#xA;" OwnerUserId="10894" LastActivityDate="2016-02-19T07:55:04.713" Title="What is the formulat to compute King tracking rate for a given set of topocentric coordinates?" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13829" PostTypeId="2" ParentId="13824" CreationDate="2016-02-18T19:40:50.493" Score="4" Body="&lt;p&gt;There are a couple of standard papers containing the table you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kenyon &amp;amp; Hartmann (1995) &lt;a href=&quot;http://adsabs.harvard.edu/abs/1995ApJS..101..117K&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/1995ApJS..101..117K&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Table A5 contains many colours for stars as a function of spectral type. You need to combine this with something that gives absolute V magnitude along the main sequence, like that of &lt;a href=&quot;https://materials.springer.com/lb/docs/sm_lbs_978-3-540-31378-6_1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schmidt-Kaler (1982)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative, that has colours for both main sequence and pre main sequence stars (though the absolute magnitude of a PMS star is age-dependent) is found in Pecaut &amp;amp; Mamajek (2013).&#xA; &lt;a href=&quot;http://arxiv.org/abs/1307.2657&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://arxiv.org/abs/1307.2657&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Online table at&#xA; &lt;a href=&quot;http://vizier.cfa.harvard.edu/viz-bin/VizieR-3?-source=J/ApJS/208/9/table5&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://vizier.cfa.harvard.edu/viz-bin/VizieR-3?-source=J/ApJS/208/9/table5&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-09-25T22:48:13.450" LastActivityDate="2018-09-25T22:48:13.450" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="13830" PostTypeId="2" ParentId="7996" CreationDate="2016-02-19T06:03:16.343" Score="1" Body="&lt;p&gt;As someone commented, you are actually short a term for fully defining the orbit.  The &quot;azimuth&quot; as you describe is commonly defined as a &quot;longitude of perihelion&quot;, and you're forgetting the orientation of the inclination, commonly defined as &quot;longitude of the ascending node.&quot; (Both of these values are included in the first link in the previous posted answer) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more detailed look at the math which uses the defining values, JPL made a whitepaper which goes into the math, and also allows you to account for changes in the orbital terms.  (Most of the math you're probably looking for is at the end of the document, beginning about section 8.10, on page 25)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;ftp://ssd.jpl.nasa.gov/pub/eph/planets/ioms/ExplSupplChap8.pdf&quot; rel=&quot;nofollow&quot;&gt;ftp://ssd.jpl.nasa.gov/pub/eph/planets/ioms/ExplSupplChap8.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10820" LastActivityDate="2016-02-19T06:03:16.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13831" PostTypeId="2" ParentId="13242" CreationDate="2016-02-19T06:55:30.527" Score="1" Body="&lt;p&gt;For the observer, they would see the person falling into the black hole almost in slow motion as their body seemed to stretch itself because of light distortions from the immense gravitational pull of the black hole. When you reach the horizon, the observer would see your body &quot;freeze&quot; and then slowly disappear, almost fading away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For you, the person falling in, you would either see your body stretching like silly puddy or falling normally depending on the size of the black hole. You would then fall normally until you reached the singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all theoretical of course &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.bbc.com/earth/story/20150525-a-black-hole-would-clone-you&quot; rel=&quot;nofollow&quot;&gt;BBC The strange fate of a person falling into a black hole&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10807" LastActivityDate="2016-02-19T06:55:30.527" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13832" PostTypeId="2" ParentId="13828" CreationDate="2016-02-19T07:55:04.713" Score="1" Body="&lt;p&gt;Too long for a comment. Possibly helpful resources:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.bbastrodesigns.com/lib/&quot; rel=&quot;nofollow&quot;&gt;http://www.bbastrodesigns.com/lib/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.bbastrodesigns.com/lib/coordLib.js&quot; rel=&quot;nofollow&quot;&gt;http://www.bbastrodesigns.com/lib/coordLib.js&lt;/a&gt; (file with actual formulas)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://canburytech.net/DriftAlign/Equations.html&quot; rel=&quot;nofollow&quot;&gt;http://canburytech.net/DriftAlign/Equations.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://canburytech.net/GoQat/index.html&quot; rel=&quot;nofollow&quot;&gt;http://canburytech.net/GoQat/index.html&lt;/a&gt; (free software that implements this function)&lt;/li&gt;&#xA;&lt;li&gt;&quot;A manual of celestial photography: Principles and practice for those interested in photographing the heavens Hardcover – 1931&quot; and it's 1988 reprint (ISBN 0933346468 for the latter, both available on Amazon), the original source.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2016-02-19T07:55:04.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13833" PostTypeId="2" ParentId="12843" CreationDate="2016-02-19T08:36:41.167" Score="4" Body="&lt;p&gt;This is an area of active research. The current Cold Dark Matter (CDM) paradigm predicts bottom-up formation of halos, whereby smaller halos coalesce in to the larger halos we observe indirectly today through X-ray measurements of clusters. This suggests that small dark matter halos wouldn't survive to the present day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There has &lt;a href=&quot;http://astrobites.org/2011/06/06/dark-matter-micro-haloes-in-the-milky-way/&quot; rel=&quot;nofollow&quot;&gt;been research&lt;/a&gt; in to the possibility of mini haloes surviving, but interactions with dense baryonic structures, such as stars, as well as streaming by an irregular galactic potential tend to wipe out any small overdensities in the distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if such mini haloes did exist in the early universe they could have contributed to the collapse of primordial gas to form Population III stars. &lt;a href=&quot;http://arxiv.org/abs/0801.4924&quot; rel=&quot;nofollow&quot;&gt;Simulations have been carried out&lt;/a&gt; to investigate such a model, with Population III stellar masses comparable to those expected $(\sim100 M\odot)$.&lt;/p&gt;&#xA;" OwnerUserId="941" LastEditorUserId="941" LastEditDate="2016-04-26T10:05:26.200" LastActivityDate="2016-04-26T10:05:26.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13834" PostTypeId="1" AcceptedAnswerId="13840" CreationDate="2016-02-19T09:24:07.563" Score="6" ViewCount="241" Body="&lt;p&gt;How evasive is the Earth to Catastrophic Meteors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://neartalk.com/ss/2016-02-19_001_1041x905.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google says&lt;/a&gt; the Earth is approximately 92.96 million miles from the Sun. It also says the suns radius is 432,474 miles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, it is 93,392,474 miles from the Earth to the center of the Sun:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;92,960,000 + 432,474 = 93,392,474 miles&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I've heard the Earth's orbit is elliptical, but to keep things simple, let's just say the orbit is a perfect circle. Therefore the radius of this circle is 93,392,474 miles. This would mean the diameter of our orbit would be 186,784,948 miles:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;93,392,474 x 2 = 186,784,948 miles&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The circumference of a circle is π x the diameter of a circle. Therefore, the number of miles in Earth's orbit would be about 586,504,736.72 miles:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3.14 x 186,784,948 = 586,504,736.72 miles&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So, the Earth travels about 586,504,736.72 miles in one year. Since there are 365 days in year, that would mean that Earth travels about 1,606,862 miles in one day.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;586,504,736.72 ÷ 365 = 1,606,862.292383562&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;That would mean the Earth travels about 66,953 miles per hour:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1,606,862.292383562 ÷ 24 = 66952.595515982&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And this would mean the Earth travels about 1,116 miles per minute:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;66952.595515982 ÷ 60 = 1115.876591933&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Lastly, this would mean the Earth travel about 18.6 miles per second:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1115.876591933 ÷ 60 = 18.597943199&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://neartalk.com/ss/2016-02-19_003_1041x870.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google says&lt;/a&gt; the diameter of the Earth is 7,917.5 miles. So, as fast as the Earth is traveling, a catastrophic meteor has about a 7 minute window to hit target Earth.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;7917.5 ÷ 18.597943199 = 425.719119329 seconds&#xA;425.719119329 ÷ 60 = 7.095318655 minutes&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What are the odds of a catastrophic meteor hitting the Earth? Or, maybe a better question: What are the odds of the Catastrophic Earth hitting a poor meteor!?&lt;/p&gt;&#xA;" OwnerUserId="10773" LastEditorUserId="10773" LastEditDate="2020-12-02T09:01:41.210" LastActivityDate="2020-12-03T09:11:35.200" Title="Earth Versus Catastrophic Meteor" Tags="&lt;earth&gt;&lt;comets&gt;&lt;rotation&gt;&lt;speed&gt;&lt;meteor&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="13835" PostTypeId="1" AcceptedAnswerId="13838" CreationDate="2016-02-19T13:37:22.613" Score="9" ViewCount="2086" Body="&lt;p&gt;Can't black holes just be super dense objects? They could still be black (having the color of black never really required special physics, after all) and have a really strong gravitational field. If we suspect that it actually absorbs light due to its gravity then it is possible that there is gravity strong enough to capture light, not letting any radiate out when matter falls into this object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just having a hard time accepting that anything can exist with an infinitely large property as it would lead to infinitely large mass leading to infinitely large forces... that would just destroy the universe infinitely fast, wouldn't it? So I think a supermassive object can be a badass just as well without having to be a singularity.&lt;/p&gt;&#xA;" OwnerUserId="10910" LastEditorUserId="2531" LastEditDate="2016-02-19T20:53:33.673" LastActivityDate="2018-03-03T20:43:35.350" Title="Are black holes really singularities?" Tags="&lt;gravity&gt;&lt;black-hole&gt;&lt;singularity&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13836" PostTypeId="1" CreationDate="2016-02-19T14:02:22.710" Score="0" ViewCount="98" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&#xA;Approximately once in 5 years people on Earth can observe bright comets whose nuclei have radius about 1km. The orbits of such comets are close to parabolic. Assuming that these nuclei are uniformly filling the volume of spherical Oort cloud with radius 10000 AU, estimate the number of large comet nuclei and mass of Oort cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to even start this question because I don't get how the comets uniformly fill Oort cloud yet on Earth we can only observe comets once every 5 years (unless there are some things blocking our views of these comets?). Would appreciate any help.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Relativity should not be needed in this question... &lt;/p&gt;&#xA;" OwnerUserId="10911" LastActivityDate="2016-04-19T16:07:53.953" Title="Astronomy Olympiad Gravitation Qn" Tags="&lt;astrophysics&gt;&lt;newtonian-gravity&gt;&lt;oort-cloud&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13837" PostTypeId="2" ParentId="13836" CreationDate="2016-02-19T14:56:48.173" Score="1" Body="&lt;p&gt;Presumably you're supposed to start by calculating the minimum orbital period a comet can have for a 10,000 AU aphelion distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then divide that period (number of years) by five, assuming we see on average one of these every five years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That will give you an estimate for how many there are out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I would also tell the person setting the question that there are some pretty wild assumptions implied in the question!)&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2016-02-19T14:56:48.173" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13838" PostTypeId="2" ParentId="13835" CreationDate="2016-02-19T15:23:28.567" Score="4" Body="&lt;p&gt;This answer is to some degree opinion-based. I share your scepticism about the existence of strict mathematical singularities as General Relativity would predict. This is mainly, because the assumption of a strict singuarity ignores quantum theory. One approach to overcome the singularity is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravastar&quot; rel=&quot;nofollow&quot;&gt;Gravastar&lt;/a&gt;. Related is a &lt;a href=&quot;http://arxiv.org/abs/1401.6562&quot; rel=&quot;nofollow&quot;&gt;Planck star&lt;/a&gt;. Both approaches try to overcome the paradoxes near the singuarity. A full answer might eventually be provided by a still to be defined &lt;a href=&quot;https://en.wikipedia.org/wiki/Quantum_gravity&quot; rel=&quot;nofollow&quot;&gt;quantum gravity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another difficulty for &quot;real&quot; black holes is rotation. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_metric&quot; rel=&quot;nofollow&quot;&gt;Schwarzschild solution&lt;/a&gt; is not to be expected to occur in real-world black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead a &lt;a href=&quot;https://en.wikipedia.org/wiki/Kerr_metric&quot; rel=&quot;nofollow&quot;&gt;Kerr solution&lt;/a&gt; (or might be a &lt;a href=&quot;https://en.wikipedia.org/wiki/Kerr%E2%80%93Newman_metric&quot; rel=&quot;nofollow&quot;&gt;Kerr-Newman&lt;/a&gt; metric) will get closer to real astronomical objects, hence including at least rotation, and might be some residual electric charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A massless particle travelling with the speed of light results in an undefined energy according to Special Relativity. Nature opens a physics dedicated to massless particles by allowing them to take any energy to resolve this undefined range. In a similar way nature may open a new kind of physics inside black holes to resolve the paradox between General Relativity and quantum theory.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2016-02-19T15:23:28.567" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13839" PostTypeId="2" ParentId="13835" CreationDate="2016-02-19T15:25:05.857" Score="6" Body="&lt;p&gt;&lt;strong&gt;What is a singularity?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A singularity is a point in the universe where matter is infinitely dense. The singularity is at the centre of the black hole, and is often hiding behind an event horizon. So basically, a singularity is a point in space where a lot of matter is squeezed together in a very small space. When you have a singularity &lt;strong&gt;matter&lt;/strong&gt; does not go to infinity. &lt;strong&gt;Density&lt;/strong&gt; is what goes to infinity. Density and matter are different things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A singularity is often created in a supernova explosion. A supernova is where a star is caused when there is so much matter in the centre of a star (at the end of its life) that the star cannot support its own gravitational force, and it collapses in on itself and explodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second way a supernova can happen is in a binary star system - where if there is a &lt;em&gt;white dwarf&lt;/em&gt;, it steals matter from the other star and eventually it accumulates so much matter that it explodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A singularity is caused by the big stars, not small ones. Small stars that explode in a supernova create something called neutron stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So what is a black hole?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A black hole is a area in space with a gravitational field so strong not even light can escape. Black holes don't have the colour black, they don't have a colour because we can't see any light coming back from them. So no, light is not absorbed. Black holes can also eventually dissipate, due to a mechanism called &lt;em&gt;Hawking Radiation&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are multiple types of black holes including:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Supermassive black holes (often found at the centre of galaxies)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Stellar black holes&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Primordial black holes&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Rotating (Kerr) black hole&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Finally, there are 3 main areas inside a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZvhG8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZvhG8.jpg&quot; alt=&quot;Diagram of a black hole&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This diagram can show us many things, for example, how black holes don't have a gravitational pull which goes across the whole universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a final note, we can work out the area where even light cannot escape by using the &lt;em&gt;Schwarzschild radius&lt;/em&gt;. Seen as you said this answer shouldn't include relativity, I won't include any maths either. The Schwarzschild radius, in summary, is where the event horizon is in a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for providing a awkwardly long answer, I was having trouble understanding what you didn't understand, and I thought that if I explained a singularity - then I'd surely help you, sorry if I've just said things you already know. I've provided some Wikipedia links if you want to have a look at some of the terms I left in italics. You might also want to look at &lt;em&gt;black hole thermodynamics&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schwarzschild radius&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Ergosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ergosphere&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hawking radiation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/White_dwarf&quot; rel=&quot;nofollow noreferrer&quot;&gt;White dwarf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Black_hole_thermodynamics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Black hole thermodynamics&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9645" LastEditorUserId="9645" LastEditDate="2016-02-20T10:07:52.507" LastActivityDate="2016-02-20T10:07:52.507" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13840" PostTypeId="2" ParentId="13834" CreationDate="2016-02-19T15:36:42.163" Score="4" Body="&lt;p&gt;There is nothing &lt;em&gt;wrong&lt;/em&gt; with your maths, but it isn't relevant to the question of the frequency of impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you ask about &lt;em&gt;comets&lt;/em&gt;, but do you also want to include &lt;em&gt;asteroids&lt;/em&gt;, since these are rather more common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also ask about &lt;em&gt;catastrophic&lt;/em&gt;, and this is a poorly defined word. A Tunguska scale event is a local disaster, but has no significant global effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Torino scale classifies objects according to their threat. A &quot;10&quot; on the Torino scale signifies a object carrying an energy of more than 10000 Megatons of TNT and capable of causing a global catastrophe. Such events occur, on average of less than once in 100,000 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Within the next 100 or so years the probability is considerably lower than $10^{-5}$ in each year, as many of the potentially hazardous objects have been observed, and their orbits shown not to intersect with Earth. So while a naive estimation would suggest that the chance of an impact in the next 100 years is about 1 in 1000, the actual probability is rather less than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you restrict to only comets, the probability is reduced further.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-02-19T15:36:42.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13841" PostTypeId="1" CreationDate="2016-02-19T16:11:28.103" Score="4" ViewCount="213" Body="&lt;p&gt;imagine two black holes approaching each other in a way that they will perform just very close flyby. You have a spaceship just below the event horizon of one black hole in line between centers of both black holes all the time.&#xA;When the black holes are closest to each other you should find yourself outside of the event horizon for a moment as the influence of one black hole should cancel out some part of the influence of the another black hole ( in similar fashion as the influences cancel out in the exact middle of a planet ); therefore, enabling you to escape. However, current physics says that nothing can leave the event horizon. Where is the flaw in this scenario?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UjtTV.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UjtTV.png&quot; alt=&quot;drawing_of_the_described_scenario&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10912" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-03-12T21:34:07.683" Title="Where is the flaw in this hypothetical concept of escaping event horizon using another's blackhole flyby?" Tags="&lt;black-hole&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13843" PostTypeId="2" ParentId="13841" CreationDate="2016-02-19T16:42:22.580" Score="6" Body="&lt;p&gt;In short, it doesn't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the event horizon is defined as the region of space which you can't escape from, If you are between two black holes, and you can escape then by definition you are not inside the event horizon. In the case when there are two black holes, the black hole is not a Schwazchild black hole (which is a solution of Einstein's equations for a single point, uncharged non-spinning mass) and the event horizon is not spherical. To determine the shape of the event horizon for two masses you need to use numerical solutions of the equations, you can do this yourself &lt;a href=&quot;http://www.linuxjournal.com/content/numeric-relativity-einstein-toolkit&quot;&gt;with the Einstein Toolkit&lt;/a&gt;, but the computing requirements are very high. A supercomputer is required to get very far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the exact situation you draw, it is likely that the release of graviational radiation would lead to a merger rather than a pass but you would need to run the simulation to confirm this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can't just remain stationary inside the event horizon. If you started inside it you can't stay in the same place and wait for a the pass of the other black hole to pull you out. Inside the event horizon spacetime is flowing towards the singularity and even if the shape of the event horizon is changed by the passing of another black hole, nothing passes from inside to out. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-02-19T16:42:22.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13844" PostTypeId="2" ParentId="13773" CreationDate="2016-02-19T16:44:11.030" Score="3" Body="&lt;p&gt;B-modes in the raw CMB data can be caused by primordial gravitational waves, as well as by cosmic dust. After subtracting the cosmic dust polarization from the BCEP2 data, the residual signal is too weak with respect to noise to be statistically sufficiently significant to claim a discovery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or stated in a different way: The observed B-modes can be explained away by cosmic dust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't prove the absence of primordial gravitational waves. But it doesn't provide sufficient evidence for their presence either.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2016-02-19T16:44:11.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13845" PostTypeId="2" ParentId="13835" CreationDate="2016-02-19T18:12:17.620" Score="4" Body="&lt;p&gt;I think you are starting off from the premise that there is some way of stabilising an extremely dense object inside the event horizon of a black hole. In the &lt;em&gt;classical&lt;/em&gt; theory of General Relativity the singularity is inevitable. Once inside an event horizon an object is compelled to move towards, and reach, the singularity in a finite time, in the same way that you are compelled to move forward in time whether you like it or not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore it is impossible to have something in &lt;em&gt;classical&lt;/em&gt; GR that looks like a black hole and has an event horizon, but which does not form some sort of singularity. However, we know that classical GR must break down on extremely small (quantum) scales so it is entirely possible that &lt;em&gt;something&lt;/em&gt; happens to prevent singularities in a quantum theory of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one is perfectly free to invent a version of General Relativity that allows for some sort of avoidance of a singularity. But there are (at least) two requirements. (1) It should explain all the other things that classical GR explains perfectly well. (2) It should have observable consequences outside the event horizon, because otherwise it's pointless navel gazing (IMO).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB: Wormholes and such-like may be possible in (classical GR) rotating black holes, but my limited understanding is that these form despite the singularity (which is no longer a point). In other words, the singularity is still there, but matter does not inevitably end up in it (but goes somewhere else!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stuff you've written about infinite mass, infinite force etc. is just wrong. The mass of a black hole is well-defined as are its gravitational effects. A finite mass still arises from taking the integral of an infinite density over zero volume.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-02-20T10:30:42.550" LastActivityDate="2016-02-20T10:30:42.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13846" PostTypeId="2" ParentId="11915" CreationDate="2016-02-19T19:32:10.200" Score="1" Body="&lt;p&gt;Are you asking about the PSF (point-spread function)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some simulations at &lt;a href=&quot;http://www.stsci.edu/jwst/software/webbpsf&quot; rel=&quot;nofollow&quot;&gt;http://www.stsci.edu/jwst/software/webbpsf&lt;/a&gt; ; there are some basic images available at that site as well as a downloadable package you can use to compute the PSF for a particular instrument and wavelength.  Since the telescope hasn't been fully assembled yet these are based on simulations.&lt;/p&gt;&#xA;" OwnerUserId="10914" LastActivityDate="2016-02-19T19:32:10.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13847" PostTypeId="2" ParentId="13835" CreationDate="2016-02-19T22:39:57.777" Score="1" Body="&lt;p&gt;There's many good answers to this on a physical level, but I don't see this one addressed, so I'll give an answer briefly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm just having a hard time accepting that anything can exist with an&#xA;  infinitely large property as it would lead to infinitely large mass&#xA;  leading to infinitely large forces...that would just destroy the&#xA;  universe infinitely fast, wouldn't it? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If black holes have infinities, which is only possible if they have a singularity, then those infinities are, by definition, infinity small and not infinite from any distance away, even a fraction of an atom's distance away.   So &quot;destroy the universe by infinite forces&quot;, sure, if the universe was smaller than an atom, or maybe, inside the event horizon where it couldn't do anything but fall towards the singularity, but at any safe and reasonable distance, black holes aren't dangerous and not a threat to the universe.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So I think a supermassive object can be a badass just as well without&#xA;  having to be a singularity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Gravastars to the rescue! Just what I needed. Finally, some theorists&#xA;  who are not comfortable with infinitely nasty objects populating our&#xA;  universe!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;OK for starters, the laws of physics are what they are, and they don't care what we think.  As to things being &quot;nasty&quot;, that's in the eye of the beholder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2000 years ago, &quot;hellfire&quot; was magma seen coming out of the occasional volcano and hell was inside earth.  Today, the inside of the Earth generates a magnetic field that protects us and gives us plate tectonics, which is really useful for life bearing planets.   The inside of the Earth hasn't changed, but our perception of it has changed enormously.  Now we love the inside of the Earth, but 2,000 years ago people feared it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;100 years ago, everyone thought the Universe was the Milky way and 80 years ago, they thought the Universe was eternal, until that lousy Hubble suggested the Universe had a beginning and, well, so much for things being eternal.   If Hubble had said that 300 years earlier he'd have been burned at the stake, no question about it, with a plaque that stated &quot;blasphemer&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the problem with point of view.  It's not a complete picture.  Black holes make a great &quot;boogie man&quot;, as something that eats everything and cannot be escaped, but that's only part of the big picture.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Einstein himself found black holes a distasteful idea (He wasn't too keen on quantum mechanics either), so Einstein imagined the Universe had some physical law that prevented black holes from ever forming and that might even be the case, but honestly, how different is it to be slowly squashed around the event horizon over an eternity vs quickly squashed in the singularity or whatever goes on in the center.  From a certain point of view, it's pretty much the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some theories have gone so far as to suggest black holes are magical places with entire baby &lt;a href=&quot;https://www.insidescience.org/content/every-black-hole-contains-new-universe/566&quot; rel=&quot;nofollow&quot;&gt;universes inside of them&lt;/a&gt;.  I find that more creative thinking than science, but the truth is, what happens inside the event horizon stays inside the event horizon and nobody knows.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;On black holes in general&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They're very useful.   The formation of black holes, and the gravitational collapse and rebound off the collapse, creates and distributes heavy elements across the galaxy and in the collapse that creates a black hole or Neutron star, something like 90% of the matter of the star gets blown off, recycled, if you will, back into the galaxy and only 10% or so, forms the collapsed core.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can think of the black hole as a nasty little remnant of a star's death, but I find the fact that large stars recycle and distribute so much of their matter across the galaxy to be highly cool.  Supermassive black holes also help in the formation of galaxies, so the simple truth is, black holes are very useful, even if you wouldn't want to cross paths with one.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now on what happens inside a black hole, there's some good answers to that already and I don't want to make this too long, especially since I'm a layman, but I find the speculation of that exotic region inside the event horizon to be great fun to think about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally don't believe in singularities.  I (THINK), that the wave and field nature of quantum mechanics and the fact that empty space has properties which, for example, particle anti particle pairs can form essentially out of nothing, (which make hawking radiation possible), I think there's probably some kind of exotic space of never a full singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think there's what might be called a physical material inside a black hole.  I think things would behave differently than that, more like the exotic nature of a proton or electron than the physical nature of a surface, but that's just my thoughts on the subject.   Without a quantum theory of gravity, it's a bit like the blind looking at a map though.  Nobody knows.  (too long?)&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-02-19T22:39:57.777" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13849" PostTypeId="1" CreationDate="2016-02-20T16:00:37.127" Score="-3" ViewCount="151" Body="&lt;p&gt;Is Gravity in the vaccum of space decelerating or exspanding? please explain why,when and how in detail, Along with a credible reference. If Science has not concluded either please give a reference as well.&lt;/p&gt;&#xA;" OwnerUserId="9326" LastEditorUserId="9326" LastEditDate="2016-02-21T07:57:10.497" LastActivityDate="2016-02-21T07:57:10.497" Title="What is the current state of gravity in the vacuum of space?" Tags="&lt;gravity&gt;" AnswerCount="2" CommentCount="7" ClosedDate="2016-02-21T05:26:18.427" ContentLicense="CC BY-SA 3.0" />
  <row Id="13850" PostTypeId="1" CreationDate="2016-02-20T16:34:26.113" Score="5" ViewCount="339" Body="&lt;p&gt;If I want to write a date in the Gregorian calendar, I can write it in the ISO 8601 format; today would be 2016-02-20. The advantage is that there are many parsers available to let computers read in this kind of date. But what if I want to write a Julian calendar date, as astronomers often do? Is there a similar widely-accepted format, and selection of parsers, to use with the Julian calendar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I'm at it, what if it's an ancient date where we don't know how many solar days ago it was; instead it is an astronomical phenomenon calculated in Terrestrial Time (more or less atomic time) or by dating decay of radioactive elements? Is there a standard way of writing such a date?&lt;/p&gt;&#xA;" OwnerUserId="7385" LastActivityDate="2016-02-26T19:56:52.987" Title="Is there a variation of ISO 8601 for julian calendar dates?" Tags="&lt;time&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13851" PostTypeId="1" CreationDate="2016-02-20T17:09:49.490" Score="-1" ViewCount="126" Body="&lt;p&gt;I would like to know how to figure out the moon's phase for July 4, 1937 at 9:34 pm. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="10920" LastActivityDate="2016-02-20T17:58:26.747" Title="How to figure out a moon phase for July 1937" Tags="&lt;moon-phases&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13852" PostTypeId="2" ParentId="13851" CreationDate="2016-02-20T17:48:55.963" Score="3" Body="&lt;p&gt;There exist websites that do exactly that for you, like &lt;a href=&quot;http://aa.usno.navy.mil/data/docs/MoonPhase.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Phases of the Moon &lt;/a&gt;. For your parameters, the phase was between the last quarter and new Moon, so just a tiny edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Visualized:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/huZu4m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/huZu4m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="10172" LastEditDate="2016-02-20T17:58:26.747" LastActivityDate="2016-02-20T17:58:26.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13853" PostTypeId="2" ParentId="12942" CreationDate="2016-02-20T19:38:33.473" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi?CGISESSID=f5439cd5581088af3a5e7bbe99a0549d&amp;amp;s_type=1#top&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; you can choose your target body and center of observation. If you want cartesian coordiantes use &quot;Vector Table&quot;.&lt;/p&gt;&#xA;" OwnerUserId="5431" LastActivityDate="2016-02-20T19:38:33.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13854" PostTypeId="2" ParentId="13813" CreationDate="2016-02-20T20:02:26.733" Score="1" Body="&lt;p&gt;The most basic reason that gravitational waves travel at the speed of light can be reduced to that in the far-field (i.e. far away from the source), their speed should be independent of their source. If you like the only speed at which waves can travel through a vacuum is at the speed of light because it is the only invariant speed in a vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are right though in that gravitational waves in general relativity are not particles or any other kind of matter but part of the spacetime metric. The way this is tackled is to take the metric $g$ and decompose it in to a background metric $g'$ and a perturbation $h$ such that $g = g'+h$. It is usual for the background metric to be the flat Minkowski metric and the speed of gravitational waves is the propagation speed of $h$ in $g'$.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastActivityDate="2016-02-20T20:02:26.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13855" PostTypeId="2" ParentId="13835" CreationDate="2016-02-20T20:22:25.440" Score="2" Body="&lt;p&gt;It is common to describe a singularity as a point of infinite density, but really they are more general than that: they are some sort of pathological behaviour in the spacetime metric. Whether singularities are actual physical phenomena or they just demonstrate the point at which a theory (in this case general relativity) can no longer describe nature is a matter of debate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Penrose's singularity theorem demonstrates that under a certain set of very reasonable assumptions that the gravitational collapse of an object like a star inevitably leads to a singularity. Note though the proof is of geodesic incompleteness (i.e. the time-lines of some free-falling object end abruptly). This is not (necessarily) the same as the apparent 'infinite density' curvature singularities that occur in idealized black hole solutions like the Schwarzschild metric.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastEditorUserId="2092" LastEditDate="2016-02-20T20:44:45.893" LastActivityDate="2016-02-20T20:44:45.893" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13856" PostTypeId="1" CreationDate="2016-02-20T22:52:21.303" Score="-6" ViewCount="146" Body="&lt;p&gt;I heard on a you tube video that a new theory came out in 2015 . The theory is that there was no &quot;Big Bang&quot; and that the Universe has no beginning .&lt;/p&gt;&#xA;" OwnerUserId="2562" LastActivityDate="2016-02-20T23:39:20.803" Title="Origin of the Universe" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2016-02-21T10:30:17.190" ContentLicense="CC BY-SA 3.0" />
  <row Id="13857" PostTypeId="2" ParentId="13856" CreationDate="2016-02-20T23:39:20.803" Score="1" Body="&lt;p&gt;Not entirely sure as to what question you're asking exactly but many theories of the origin of the universe have risen for many decades now. But as you would know, the Big Bang theory is the most accepted one. Since I'm typing this from my phone, I won't go into detail but I did once write an essay on this. &#xA;There are obviously parties that are against the theory and parties that are for it. But some of the other theories of origination sometimes fail to cover certain aspects or even suggest of multiple dimensions beyond what we already believe exists. &#xA;But none are as well researched or factual as the Big Bang theory. &#xA;Hope that helps :) &lt;/p&gt;&#xA;" OwnerUserId="10908" LastActivityDate="2016-02-20T23:39:20.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13859" PostTypeId="2" ParentId="13849" CreationDate="2016-02-21T01:58:44.920" Score="2" Body="&lt;p&gt;The only known theoretical way to stop gravity is &quot;&lt;a href=&quot;https://physics.aps.org/story/v2/st7&quot; rel=&quot;nofollow&quot;&gt;antigravity&lt;/a&gt;&quot;, as needed to stabilize a wormhole. For this you need &lt;a href=&quot;https://en.wikipedia.org/wiki/Negative_mass&quot; rel=&quot;nofollow&quot;&gt;negative energy&lt;/a&gt;, which has been observed only on small scales and in small amounts.&#xA;For any practically relevant scenarios gravity sums up. Observational evidence for &quot;antigravity&quot; is negligible compared to usual gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in some sense the accelerating universe due to &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_energy&quot; rel=&quot;nofollow&quot;&gt;dark energy&lt;/a&gt; limits the range of the gravity of a mass, since gravity propagates with the speed of light. For particles outside the &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_cosmological_horizons&quot; rel=&quot;nofollow&quot;&gt;cosmological event horizon&lt;/a&gt; of the considered mass, the field of gravity of the considered mass will never reach those particles.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2016-02-21T01:58:44.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13860" PostTypeId="2" ParentId="13849" CreationDate="2016-02-21T02:07:18.533" Score="2" Body="&lt;p&gt;Newton's third law is that if a body A exerts a force on body B, then body B exerts an equal (in magnitude) and opposite (in direction) force on body A. In Newtonian gravity when body A exerts a gravitational force on body B, then body B exerts an equal gravitational force on body A. As both forces are attractive they have opposite directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General relativity doesn't treat gravity as a force, but Newton's third law can be demonstrated in the Newtonian limit of GR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In big bang cosmology the initial expansion is not due to gravity or any other &quot;force&quot;, rather it comes from the initial conditions of the Universe. Matter and radiation have positive energy density and non-negative pressure which both contribute to their gravity which acts to slow down the rate of expansion (the Hubble parameter). However it has been demonstrated that the Universe's rate of expansion is in fact increasing which means that there must exist some form of dark energy which has a negative pressure which is large enough in proportion to its (positive) energy density to accelerate the expansion of the Universe.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastActivityDate="2016-02-21T02:07:18.533" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13861" PostTypeId="1" CreationDate="2016-02-21T11:57:13.820" Score="2" ViewCount="101" Body="&lt;p&gt;An observation could validate the predictions made by a previous theory. If something unpredicted is observed, then a new theory which is compatible with the observation should make predictions which later are confirmed by new observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if no such new predictions are possible to make? What is the scientific status of a theory which after the fact is only compatible with all data, but which cannot make any new predictions? If it would be considered weak, does this mean that discoveries through observation can actually destroy science?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a valid argument for delaying the construction of better observatories until theories have matured given all existing data? Kind of comparable with the forward planetary protection argument which says that human spaceflight to Mars should wait until it is certain that there's no life there which could go extinct when it meets Earthly life.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="6" LastEditDate="2019-01-29T14:47:49.490" LastActivityDate="2019-01-29T14:47:49.490" Title="Could science be lost if a phenomena is observed before predicted?" Tags="&lt;observational-astronomy&gt;&lt;prediction&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2016-02-22T07:50:19.690" ContentLicense="CC BY-SA 3.0" />
  <row Id="13864" PostTypeId="1" CreationDate="2016-02-21T20:14:05.877" Score="2" ViewCount="1154" Body="&lt;p&gt;&quot;A few minutes into the expansion, when the temperature was about a billion (one thousand million) kelvin and the density was about that of air, neutrons combined with protons to form the universe's deuterium and helium nuclei in a process called Big Bang nucleosynthesis. Most protons remained uncombined as hydrogen nuclei. As the universe cooled, the rest mass energy density of matter came to gravitationally dominate that of the photon radiation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After about 379,000 years the electrons and nuclei combined into atoms (mostly hydrogen); hence the radiation decoupled from matter and continued through space largely unimpeded. This relic radiation is known as the cosmic microwave background radiation.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question now is, when you count back starting with the cosmic background radiation which is now in far red (2,7K) at what time was there for the first time visible light, or was is right after the creation of atoms about 379,000 after the big bang?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastActivityDate="2016-02-23T14:15:33.080" Title="When was visible light for the first time created?" Tags="&lt;big-bang-theory&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13866" PostTypeId="2" ParentId="13864" CreationDate="2016-02-21T22:13:51.567" Score="4" Body="&lt;p&gt;The best model we have is that of electro-weak unification. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The photon is an quantum of the electromagnetic field, and as long as there has been an electromagnetic field, excitation of the field have existed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;weak&quot; nuclear interaction is another field, (it has the effect of changing up quarks to down quarks, and so is responsible for beta decay of radioactive material). Quanta of the weak field are rather different from photons. They are charged and have mass, but at very high energies (a million billion degrees or so) the electromagnetic and weak fields are unified into a single &quot;electroweak&quot; interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But after the universe had cooled (it took sometime between $10^{-36}$ and $10^{-32}$ seconds) the weak and electromagnetic fields split, and the first true photons formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These photons didn't get very far before colliding with other particles that were being formed. As you note, it took nearly 400000 years until the universe cooled enough to become transparent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you mention visible light. The photons formed after the split would have had very high energies, we would call most of them gamma rays if they existed today. But as photon of all energies could be formed, a small fraction would have been in the visible range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At recombination, when the universe became transparent, the temperature was about 3000K, at which temperature most photons are in the infra red range, but enough are visible to have given the universe a very bright orange-red colour, similar to the colour of a lightbulb.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-02-21T22:13:51.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13867" PostTypeId="2" ParentId="13861" CreationDate="2016-02-21T22:16:56.647" Score="2" Body="&lt;p&gt;The question leaves me confused on how to interpret it.  As some of the comments have noted, observations have routinely predated a theory that (accurately) predicts them.  In astronomy and cosmology this has tended to be the main driving force behind seeking new theories.  Yet I sense a more subtle question concerning how scientists would view a theory that predicts everything so far, but nothing new.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately I think there is a confusion concerning the typical mantra for adopting a new theory as the main theory for physics/science/whatever: if a new theory doesn't make any new testable predictions, and so can only reproduce what existing theories already predict, then preference is given to simpler and existing theories.  This has nothing to say about a single theory.  It's strictly about multiple theories being compared to each other.  It does not imply that a theory, viewed entirely on its own, that makes no new predictions is undesirable.  It simply means if we're going to replace theory X with theory Y, then theory Y better predict more stuff than X does.  It doesn't care if X does or doesn't predict testable things that have yet to be proven/disproven.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2016-02-21T22:16:56.647" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13869" PostTypeId="2" ParentId="6609" CreationDate="2016-02-22T12:29:24.230" Score="1" Body="&lt;p&gt;I think you're asking, &quot;If we know an object's shape, can we determine if it is in in hydrostatic equilibrium?&quot; If so, one might wonder if astronomers classify basketballs or ball bearings as being in hydrostatic equilibrium, since they're so spherical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below about 100 km in radius, the answer is in general no. Given some population of randomly lumpy objects (like asteroids), some of them will be close to the shape of a sphere purely on accident. The composition matters too -- an object of this size made of hydrogen gas would assume a spherical shape from hydrostatic equilibrium, but an object made of rock might not (like Mathilde below). We could make better predictions given detailed knowledge of the materials and environment of the object, but this is not always possible, as you mentioned. For small objects, intermolecular and atomic forces dominate gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lbSaN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lbSaN.jpg&quot; alt=&quot;Mathilde, from the wikipedia page on asteroids.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you get to a certain size of object, it becomes much easier to make a prediction about hydrostatic equilibrium. This is still very much context-dependent, and you still get complications from material composition, temperature, etc. However, atomic binding forces have some fixed strength, but gravity scales like mass. Given ordinary astrophysical materials, we can be very sure that a body like Jupiter is under hydrostatic equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can make some order of magnitude estimates by assuming that the atomic interaction energy must be at least as large as the thermal energy (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1995MNRAS.277...99H&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hughes and Cole 1995&lt;/a&gt;). If you check out equation 5 of that paper, you'll see an explicit expression for a radius that divides spherical objects and nonspherical ones. At some mass, the binding atomic energy becomes dwarfed by the gravitational potential, and you always get a spherical object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;tl;dr -- small objects no, big objects yes, medium objects may require detailed modeling.&lt;/p&gt;&#xA;" OwnerUserId="608" LastActivityDate="2016-02-22T12:29:24.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13870" PostTypeId="1" CreationDate="2016-02-22T16:25:57.360" Score="3" ViewCount="570" Body="&lt;p&gt;I just learned about precession of the equinox, i.e. the Earth's axis tilts one degree every 72 years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it takes 72 years/degree x 360 degrees =  25,920 (years in one cycle)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many theories to this, i.e.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Sun's gravity is affecting the tilt of Earth's axis&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.bibliotecapleyades.net/esp_precession.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sun itself is revolving around another star, and it takes 25,920 years&lt;/a&gt;, hence the slow tilt of Earth's axis&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I googled whether the Sun revolves around something else, and according to &lt;a href=&quot;http://starchild.gsfc.nasa.gov/docs/StarChild/questions/question18.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA.gov&lt;/a&gt;, Sun revolves around center of Milky Way it it takes 230 million years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I am confused between precession of the equinox and Sun revolving around another star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does precession of the equinox have anything to do with the Sun revolving around another star? Or is it simply the Sun's gravity acting on the Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the explanation according to modern science? I understand there were astronomers in the past who deduced that the Sun revolved around another star and it took 25,920 years.&lt;/p&gt;&#xA;" OwnerUserId="10935" LastEditorUserId="18643" LastEditDate="2019-07-30T11:37:25.663" LastActivityDate="2019-07-30T11:37:25.663" Title="Precession of the Equinox" Tags="&lt;the-sun&gt;&lt;gravity&gt;&lt;earth&gt;&lt;galaxy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="13872" PostTypeId="2" ParentId="13870" CreationDate="2016-02-22T19:46:00.747" Score="8" Body="&lt;p&gt;The sun does not orbit another star. A star, even a very small brown dwarf, would have been seen, if not in visible light then in the infra-red by the &lt;a href=&quot;https://www.nasa.gov/mission_pages/WISE/main/index.html&quot; rel=&quot;noreferrer&quot;&gt;WISE&lt;/a&gt; survey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there is no mystery about precession. The Earth is non-spherical, it is wider around the equator, and because the Earth is tilted with respect to the ecliptic, there is an asymmetry in the force of the sun, and the moon, on the Earth. This force pulls the axis of the Earth's rotation in a circle. The basic maths of this is not so hard, and is in the Wikipedia page on &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_precession&quot; rel=&quot;noreferrer&quot;&gt;Axial precession&lt;/a&gt;. This explanation of axial precession was known to Newton. There is nothing controversial about it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun does orbit the galaxy it takes about 230000 years to complete one orbit. This has nothing to do with axial precession.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The page you link to is worthless. It offers an &quot;alternate explanation&quot; of a phenomenon that is very well understood, and pretends that there is some mystery, where there is none. It the uses this to justify a lot of astrological nonsense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To repeat. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are not &quot;many theories&quot; explaining Axial precession. &lt;/li&gt;&#xA;&lt;li&gt;Axial precession is a well understood phenomenon, due to the Earth not being a sphere and the axis of the Earth not being perpendicular to its orbit &lt;/li&gt;&#xA;&lt;li&gt;The sun does not orbit another star. We know this because we could have seen it.&lt;/li&gt;&#xA;&lt;li&gt;No astrology based on the presence of another star is valid.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-02-22T20:51:51.927" LastActivityDate="2016-02-22T20:51:51.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13874" PostTypeId="1" CreationDate="2016-02-22T20:39:58.420" Score="-1" ViewCount="110" Body="&lt;p&gt;The number of atoms in the whole Universe seems to be finite, yet the Universe is infinite in dimensions? How it is possible in the Universe?&lt;/p&gt;&#xA;" OwnerUserId="10850" LastActivityDate="2016-02-22T21:53:49.673" Title="Size of the Universe" Tags="&lt;astrophysics&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2016-02-24T14:49:41.337" ContentLicense="CC BY-SA 3.0" />
  <row Id="13875" PostTypeId="2" ParentId="13874" CreationDate="2016-02-22T21:33:49.297" Score="3" Body="&lt;p&gt;The number of atoms in the observable universe is finite. But the observable universe is also finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ultimate topology of the whole Universe is unknown. It is not known if the Universe is finite or not. Both options seem hard to stomach, and since it concerns the unobservable, perhaps it lies outside of science...perhaps not.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-02-22T21:33:49.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13876" PostTypeId="1" AcceptedAnswerId="13879" CreationDate="2016-02-22T21:43:20.387" Score="11" ViewCount="482" Body="&lt;p&gt;Gravity is the curvature of spacetime, and its effects travel at lightspeed. However, space is expanding; eventually, light from distant galaxies will become more and more redshifted, and we will no longer be able to see them (&lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_cosmological_horizons#Future_horizon&quot;&gt;source&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, there is a limit to how far we could ever possibly see, since light too far will never reach us due to the rapid expansion of space...or at least, if I'm understanding this correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, gravitational waves travel at lightspeed. So, after enough time, when an object's light no longer reaches us, will its gravity no longer affect us either?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better rephrasing is: at a certain point in time, will the gravity of any extremely distant object — even the most massive stars, black holes, or &lt;em&gt;galaxies&lt;/em&gt; — simply not affect us whatsoever, in the slightest?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2016-02-23T07:35:30.377" Title="Will gravitational waves too far away ever reach us?" Tags="&lt;gravity&gt;&lt;expansion&gt;&lt;gravitational-waves&gt;&lt;dark-energy&gt;&lt;fate-of-universe&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13877" PostTypeId="2" ParentId="13874" CreationDate="2016-02-22T21:53:49.673" Score="2" Body="&lt;p&gt;I think you misunderstanding an important concept. There are a finite number of atoms in the observable universe — that is, the part we can see (notice the lower case &quot;u&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When people say &quot;the universe&quot;, they can often refer to the observable universe. Sloppy, I know. The whole Universe (notice the capital &quot;u&quot; here) may be infinite. Also, don't know where you got the idea that the Universe has an infinite number of dimensions.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2016-02-22T21:53:49.673" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13879" PostTypeId="2" ParentId="13876" CreationDate="2016-02-23T07:35:30.377" Score="8" Body="&lt;p&gt;The answer here is very similar to if you were asking about light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle gravitational waves might allow us to fractions of a second after the big bang. Electromagnetic waves can see back to where the cosmic background radiation formed, about 400,000 years after the big bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are right, the universe has expanded. At the present epoch it is estimated that the observable universe, containing objects that have emitted light or GWs that may reach us now, is around 46 billion light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it seems quite likely that the universe continues well beyond this horizon, and sources beyond this horizon can never have emitted light or GWs that will reach us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the relevant section of wikipedia (&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Observable_universe&quot;&gt;https://en.m.wikipedia.org/wiki/Observable_universe&lt;/a&gt;) points out, the detection of GWs extends our view very slightly. We cannot &quot;see&quot; with electromagnetic waves beyond 45.7 billion light years because of the &quot;fog&quot; of the cosmic microwave background,  but GWs can penetrate this fog allowing us (in principle) to see signals from objects currently 46.6 billion light years away.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-02-23T07:35:30.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13881" PostTypeId="2" ParentId="13864" CreationDate="2016-02-23T09:36:15.297" Score="2" Body="&lt;p&gt;To expand a bit on James Kilfiger's great answer:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The peak wavelength of the cosmic microwave background&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The wavelength of the cosmic microwave background (as well as any other photon) is stretched proportionally to the scale factor $a(t) = 1/(1+z)$ (&quot;the size&quot;) of the Universe, where $t$ is the age of the Universe and $z$ is the redshift of light emitted at $t$ and observed today.&#xA;Thus, at redshift $z$ the peak of the CMB spectrum was lying at a wavelength&#xA;$$&#xA;\lambda_\mathrm{max}(z) = \frac{\lambda_\mathrm{max,0}}{1+z}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today, the CMB spectrum peaks at $\lambda_\mathrm{max,0} \simeq 0.2\,\mathrm{cm}$. If we take visible light to lie in the range 400-700 nm, the maximum of the CMB spectrum must have been in the visible range at redshifts&#xA;$$&#xA;z \,\,=\,\, \frac{\lambda_\mathrm{max,0}}{\lambda_\mathrm{max}(z)} - 1 \,\,=\,\, \frac{0.2\,\mathrm{nm}}{\{4\,\verb+-+\,7\}\times10^{-5}\,\mathrm{nm}} - 1 \,\,\sim\,\, \{2700\,\verb+-+\,4700\}.&#xA;$$&#xA;This can be converted to an age of the Universe by integrating (numerically) the &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Friedmann equation&lt;/a&gt;. Assuming a &lt;a href=&quot;http://arxiv.org/abs/1502.01589&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck 2015&lt;/a&gt; cosmology, I get that &lt;strong&gt;the peak of the CMB spectrum was in the visible range when the Universe was between roughly 50,000 and 125,000 years old&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The color of the Universe&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;However, as James Kilfiger also mentions, even though the peak lies outside the visible range, a fraction of the spectrum is still inside. Its color depends on the response of the cones of the human eye and follows approximately the so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Planckian_locus&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planckian locus&lt;/a&gt; in the &lt;a href=&quot;https://en.wikipedia.org/wiki/CIE_1931_color_space&quot; rel=&quot;nofollow noreferrer&quot;&gt;CIE 1931 color space&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o4yQ2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o4yQ2.jpg&quot; alt=&quot;ColorTemp&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the plot above, the locus traces the color of the Universe as a function of temperature (&lt;em&gt;thin numbers&lt;/em&gt;), and age of the Universe in kiloyears (kyr; &lt;em&gt;bold numbers&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the Universe started out as bluish (when $T\gtrsim10^4\,\mathrm{K}$), became whitish at an age of around 200,000 years (when $T\sim5\verb+-+6000\,\mathrm{K}$), and then gradually went over orange and red at $t\sim1\,\mathrm{Myr}$ before fading into the infrared. This evolution is completely analogous to the colors of stars of a given temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ages have been calculated with Python's &lt;code&gt;CosmoloPy.distances.age()&lt;/code&gt;, which I think is not very accurate at such small ages, but it does give the approximate age. Also, whether or not the CMB was visible &lt;em&gt;to the naked eye&lt;/em&gt; — i.e. without a telescope to enlarge the light-collecting area — depends on the intensity at a given point.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-02-23T14:15:33.080" LastActivityDate="2016-02-23T14:15:33.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13883" PostTypeId="1" AcceptedAnswerId="13905" CreationDate="2016-02-23T10:55:52.587" Score="4" ViewCount="248" Body="&lt;p&gt;When an object in the space comes close to a massive body (like a black hole or a neutron star), we see from Earth that its speed significantly increases while it is close to the body. As we know, this is due to space distortion around massive objects known as gravity. But also, gravity causes time dilation, so an observer near a black hole will experience less time spans between same events than a far-away observer. So, the question is: if we travel on an asteroid and come close to a black hole, will we experience an increase in our speed, or the time dilation will compensate space distortion and we'll feel the same movement speed as before (i. e. no acceleration)?&lt;/p&gt;&#xA;" OwnerUserId="10948" LastEditorUserId="10948" LastEditDate="2016-03-01T12:15:28.417" LastActivityDate="2016-03-01T12:51:09.330" Title="When you travel fast around a black hole, do you experience high speed yourself?" Tags="&lt;black-hole&gt;&lt;space-time&gt;&lt;time-dilation&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13884" PostTypeId="1" CreationDate="2016-02-23T12:17:13.453" Score="6" ViewCount="421" Body="&lt;p&gt;I find hints around the web, e.g. &lt;a href=&quot;http://www.cosmodome.net/mirrordome.php&quot;&gt;http://www.cosmodome.net/mirrordome.php&lt;/a&gt; that Celestia can be projected on a dome using a spherical mirror projection configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that I can't find any documentation on how to configure Celestia to warp the generated image, so it can be used in such a projection setup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Stellarium\Nightshade I found some references, e.g. &lt;a href=&quot;http://paulbourke.net/dome/stellarium/&quot;&gt;from spherical projection guru, Paul Bourke,&lt;/a&gt; and I expected to find similar resources for Celestia.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Can Celestia be configured for a spherical mirror projection system?&lt;/li&gt;&#xA;&lt;li&gt;Can you point me to documentation on how to set up Celestia for a spherical mirror projection system?&lt;/li&gt;&#xA;&lt;li&gt;Is there a way to warp Celestia with 3rd party tools and then project the images?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10949" LastEditorUserId="2153" LastEditDate="2016-02-25T23:15:51.400" LastActivityDate="2016-03-03T19:30:20.557" Title="How to configure Celestia to work with a spherical mirror projection on a dome" Tags="&lt;software&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13885" PostTypeId="1" AcceptedAnswerId="13888" CreationDate="2016-02-23T15:21:27.660" Score="2" ViewCount="204" Body="&lt;p&gt;Usually they say the spectrum of a black body is a continuum and emits radiation at every frequency. Is it really a continuum like a line or are there (small) intervals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, can a body emit at 700nm and also at 400nm and everything in between, or at very small intervals like at 400.000001nm and 400.000002nm etc.? So is it really like a wave or line which implies a kind of infinity of wavelengths or is it more or less like discreet lengths although very close to each other? Or is it not really known because measuring is in a certain way always with discreet numbers (like 400.0001)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, it should always be at certain discreet numbers, firstly because the electrons from which the radiation comes only have a discreet values, and secondly because collisions between atoms/protons/electrons create energy at certain set wavelengths. But that is just my interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the reality of black body radiation?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="2270" LastEditDate="2016-02-23T17:22:08.727" LastActivityDate="2016-02-23T19:48:21.323" Title="In what manner is the spectrum of a black body really a continuum?" Tags="&lt;light&gt;&lt;spectra&gt;&lt;radiation&gt;&lt;black-body&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13886" PostTypeId="1" AcceptedAnswerId="13887" CreationDate="2016-02-23T16:02:39.823" Score="0" ViewCount="127" Body="&lt;p&gt;Where could I find detailed data from stars? Like spectrum, x ray and so on ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because I would like to do some data analysis on it. &lt;/p&gt;&#xA;" OwnerUserId="10950" LastEditorUserId="2365" LastEditDate="2016-02-23T20:05:06.903" LastActivityDate="2016-02-23T20:05:06.903" Title="Stellar data for data mining" Tags="&lt;star&gt;&lt;data-analysis&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2016-02-24T11:05:48.350" ContentLicense="CC BY-SA 3.0" />
  <row Id="13887" PostTypeId="2" ParentId="13886" CreationDate="2016-02-23T16:14:36.477" Score="3" Body="&lt;p&gt;The Sloan Digital Sky Survey website has a good page with lots of links on it to retrieve spectra in different ways, you can access it &lt;a href=&quot;https://www.sdss3.org/dr9/spectro/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. The most useful of these is probably &lt;a href=&quot;http://dr9.sdss3.org/&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;. This is just one of many places you can search but its a good starting point if your just curious about analysing spectra and not too picky about the source.&lt;/p&gt;&#xA;" OwnerUserId="10437" LastActivityDate="2016-02-23T16:14:36.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13888" PostTypeId="2" ParentId="13885" CreationDate="2016-02-23T19:36:01.163" Score="2" Body="&lt;p&gt;A blackbody spectrum is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Dean comments, there are no true blackbodies. But this is not really the issue here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's true that most emitted photons from, say, a star, which is quite close to a blackbody, is the result of some discrete process. E.g. an electron is brought to an excited state by a collision, the atom de-excites, and a photon with the wavelength corresponding to the given transition is emitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But two mechanisms make the total spectrum of the star truly continuous:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Firstly, ionized atoms recombine with free electrons. The electron will go to any state, ground or excited, and with various probabilities. The energy of the photon emitted is equal to the energy it would take to remove the electron from the atom (i.e. ionize it again), &lt;em&gt;plus&lt;/em&gt; the kinetic energy it had before recombination. And this kinetic energy can have any value (although if it's too large, the probability of recombination decrease).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Secondly, even without recombinations, you will not observe only the discrete lines. The reason is that the atoms of the blackbody are not stationary, but move with a (&quot;thermal&quot;) velocity $v=\sqrt{k_\mathrm{B}T/m}$, where $k_\mathrm{B}$ is Boltzmann's constant, $T$ is the temperature, and $m$ is the mass of the atom. This motion causes a Doppler broadening of the emitted lines, resulting in a continuous, and Gaussian$^\dagger$, probability distribution.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;Actually the probability distribution is a convolution of the Gaussian random motion, and the Lorentzian natural line profile, resulting in a so-called Voigt profile. This profile is dominated by the Gaussian in the line center, and only becomes Lorentzian in the wings.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-02-23T19:48:21.323" LastActivityDate="2016-02-23T19:48:21.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13889" PostTypeId="1" AcceptedAnswerId="13897" CreationDate="2016-02-24T00:50:47.203" Score="5" ViewCount="21364" Body="&lt;p&gt;I'm considering picking up some new eyepieces for my telescope and am interested in comparing 2-inch and 1.25-inch eyepieces. Are there any disadvantages to 2-inch eyepieces? Are 2-inch eyepieces only advantageous with longer eyepieces? &lt;/p&gt;&#xA;" OwnerUserId="10953" LastActivityDate="2017-01-10T22:06:51.060" Title="What are the advantages and disadvantages of a 2-inch eyepiece versus a 1.25-inch eyepiece?" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13891" PostTypeId="2" ParentId="1032" CreationDate="2016-02-24T02:09:56.827" Score="0" Body="&lt;p&gt;The reason is there are too many variables!&#xA;Any signal that passes by a sun or pulsar can be totally wiped out by the strong electromagnetic noise they produce.  Another problem is that decay of radio signals even without interference is not a simple thing to calculate, it really requires testing in the environment the signal will encounter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway it's really all for naught because radio waves are completely useless for interstellar communication. They are moving way too slow for any meaningful use. Even the nearest solar system would take 4 years to send a signal and 4 more years to get a reply.&lt;/p&gt;&#xA;" OwnerUserId="10955" LastActivityDate="2016-02-24T02:09:56.827" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13892" PostTypeId="2" ParentId="13889" CreationDate="2016-02-24T05:41:19.250" Score="4" Body="&lt;p&gt;I found a good explanation &lt;a href=&quot;https://stargazerslounge.com/topic/138540-2-inch-vs-125-inch/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. Pretty much sums it up:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In simple layman terms there are a few differences other than the obvious 3/4&amp;quot; larger diameter in the barrel size. Typically 2&amp;quot; Eyepieces are used for lower magnification and or wider field of view. A 2&amp;quot; barrel will allow for a wider Apparent Field Of View through the Eyepiece that a 1.25&amp;quot; wouldn't normally allow. So if you had a 2&amp;quot; and 1.25&amp;quot; 32mm EP, the AFOV in the 2&amp;quot; would be greater, probably much greater.&lt;/p&gt;&#xA;&lt;p&gt;There is another benefit to the 2&amp;quot; barrel that it also allows for larger lens elements to be used. In other words a larger eye lens so that even at higher magnifications / short focal lengths your not looking through a peep hole as you would in your standard 1.25&amp;quot; plossl. Even in a short focal length EP's you can be looking through an eye lens that is 1.25&amp;quot; in diameter. Another advantage to many of them is a more comfortable eye relief so you don't have to paste your eye to the EP.&lt;/p&gt;&#xA;&lt;p&gt;And probably the most obviouse difference is the cost.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-02-24T05:41:19.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13893" PostTypeId="1" AcceptedAnswerId="13896" CreationDate="2016-02-24T13:00:37.317" Score="0" ViewCount="388" Body="&lt;p&gt;So, if you search for &lt;a href=&quot;https://en.wikipedia.org/wiki/Henyey_track&quot; rel=&quot;nofollow&quot;&gt;Henyey tracks on wikipedia&lt;/a&gt; (I know, the shame of it!) you will come across this statement: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Henyey track is characterized by a slow collapse in near hydrostatic equilibrium. They are approaching the main sequence almost horizontally in the Hertzsprung–Russell diagram (i.e. the luminosity remains almost constant).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ignoring the bad grammar, I take umbrage with the following: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the luminosity remains almost constant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would say that the apparent horizontal track associated with the Henyey track is more due to the fact that in a typical HR diagram the temperature axis ranges over about one order of magnitude, while the luminosity axis may vary over five or more orders of magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to frame this as a question, would you agree or disagree with the statement on the site or with what I have stated (i.e., that the Henyey track is an apparent horizontal track caused by the nuance in the magnitude scales of luminosity in comparison with the temperature)?&lt;/p&gt;&#xA;" OwnerUserId="2091" LastEditorUserId="2365" LastEditDate="2016-02-24T19:00:45.883" LastActivityDate="2016-02-24T19:00:45.883" Title="Description of Henyey tracks on wikipedia incorrect?" Tags="&lt;stellar-evolution&gt;&lt;stellar-astrophysics&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13895" PostTypeId="1" CreationDate="2016-02-24T13:44:34.373" Score="5" ViewCount="826" Body="&lt;p&gt;What happens to the hairiness/information on the horizons of two black holes if they collide?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After Hawkin the information of the matter which has fallen into the blackhole is encoded on the surface of the event horizon. What happens to the information of both blackholes when the event horizons merge?&lt;/p&gt;&#xA;" OwnerUserId="102" OwnerDisplayName="Quonux" LastEditorUserId="102" LastEditDate="2016-02-24T21:12:21.257" LastActivityDate="2016-09-12T05:46:32.090" Title="What happens to the information on the event horizons of two merging black holes?" Tags="&lt;black-hole&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13896" PostTypeId="2" ParentId="13893" CreationDate="2016-02-24T17:31:50.947" Score="5" Body="&lt;p&gt;The wording provides a contrast with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hayashi_track&quot; rel=&quot;noreferrer&quot;&gt;Hayashi track&lt;/a&gt; phase that immediately precedes it, where the luminosity decreases by orders of magnitude with little change in effective temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are correct though, the Henyey track for low-mass stars is not horizontal in the HR diagram, however it is a critical point that the luminosity does not change &lt;em&gt;very much&lt;/em&gt; (less than a factor of 3) as the effective temperature increases, since it implies that the star slowly shrinks, and by the virial theorem, the core temperature gradually rises and is ultimately able to sustain nuclear fusion.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-02-24T17:31:50.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13897" PostTypeId="2" ParentId="13889" CreationDate="2016-02-24T19:22:26.273" Score="10" Body="&lt;p&gt;When you look into the eyepiece, you've noticed that the image is round, framed by a black ring. That ring is called the &lt;strong&gt;field stop&lt;/strong&gt;, and it's an actual round piece of metal or plastic inside the eyepiece. Its function is to limit the image to the size where the image quality in that eyepiece is acceptable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to remove the field stop, the image outside of it would start to look bad, and its edges would be fuzzy and barely usable. Also, most people prefer a neat, sharp edge to the field of view, so there's the esthetic element to consider also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of the field stop is given by two parameters:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the &lt;strong&gt;focal length&lt;/strong&gt; of the eyepiece, measured in mm&lt;/li&gt;&#xA;&lt;li&gt;the &lt;strong&gt;apparent field of view&lt;/strong&gt; of the eyepiece, or how wide the image appears when you look in the eyepiece, measured in degrees of angle&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Longer focal length eyepieces will naturally have bigger field stops, keeping everything else in proportion. Eyepieces with a wider field of view will also require a bigger field stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;IMPORTANT:&lt;/strong&gt; the field stop, obviously, must be smaller than the diameter of the barrel of the eyepiece, otherwise it would not fit in. This is what dictates the lower limit for the diameter of the barrel. The barrel must be bigger than the field stop. You need to increase the barrel diameter when the field stop grows too big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when you stick to short focal length eyepieces with a narrow field of view, the required field stop is pretty small, and it can easily fit into a 1.25&quot; barrel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But with longer focal length eyepieces, especially with modern eyepieces with a wide field of view, the field stop can grow pretty large, and eventually becomes bigger than the 1.25&quot; barrel. That's when you need to move up to a 2&quot; barrel. That's what drives the need for larger size barrels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long time ago, there used to be eyepieces with 0.965&quot; barrel diameter, because back then opticians didn't know how to make wide field eyepieces. Then the field of view kept increasing, so opticians moved up to 1.25&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually the newer 2&quot; standard was adopted, as the field of view kept growing. At 32mm focal length, a Plossl eyepiece with a 52 degree field of view can fit in a 1.25&quot; barrel, but a more recent 68 degree design requires 2&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays there's a new standard emerging, with a 3&quot; barrel diameter, imposed by the very wide field eyepiece designs invented recently. At 30mm focal length, an eyepiece with an 82 degree field of view could use a 2&quot; barrel, but if you demand 100 degrees FoV at 30mm, you need to move up to 3&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, larger barrel diameters make heavier and more expensive eyepieces. So you need to mix and match. At smaller focal lengths, the 1.25&quot; barrel works well. For longer focal length eyepieces, as you move up in terms of the size of the field of view, eventually you must switch to 2&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only use Plossl eyepieces, you can have all of them in the 1.25&quot; format, because even at the longest focal lengths that make sense in practice, the Plossl field of view is narrow enough to not require a big field stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a practical example, my collection of eyepieces currently is about evenly split between 1.25&quot; and 2&quot;, with the 2&quot; barrels being reserved for eyepieces over 18mm focal length. All my eyepieces have an 82 degree field of view.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Of course, with a bigger barrel the eyepiece is larger, there's more glass in it, it's heavier and it can be more expensive. This is why manufacturers try to use the smaller barrels when it makes sense, and switch to bigger barrels only when physics and math force them to take this step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need a certain focal length combined with a certain field of view, and that requires a bigger barrel, then by all means use that size. It's kind of a mix-and-match situation.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2017-01-10T22:06:51.060" LastActivityDate="2017-01-10T22:06:51.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13898" PostTypeId="1" AcceptedAnswerId="13899" CreationDate="2016-02-24T19:23:42.183" Score="0" ViewCount="107" Body="&lt;p&gt;(1) Please go through following image&#xA;&lt;a href=&quot;http://skyserver.sdss.org/dr12/en/tools/chart/navi.aspx?ra=126.61547&amp;amp;dec=27.89242&quot; rel=&quot;nofollow&quot;&gt;Possible Super Red Giant?&lt;/a&gt;&#xA;However sdss shows it as galaxy! Am I missing something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) &lt;a href=&quot;http://www.google.co.in/sky/#latitude=69.82067508546967&amp;amp;longitude=-36.44439697265625&amp;amp;zoom=10&amp;amp;Spitzer=0.00&amp;amp;ChandraXO=0.00&amp;amp;Galex=0.00&amp;amp;IRAS=3.00&amp;amp;WMAP=0.00&amp;amp;Cassini=0.00&amp;amp;slide=1&amp;amp;mI=-1&amp;amp;oI=-1&quot; rel=&quot;nofollow&quot;&gt;Google Sky shows object in latitude and logitude format like this.&lt;/a&gt;&#xA;Is there any way to convert those numbers to Right Ascension (ra) and Declination (dec) numbers or opposite of that  so that I can spot google sky data on SDSS and SDSS object on Google Sky?&lt;/p&gt;&#xA;" OwnerUserId="10961" LastEditorUserId="10961" LastEditDate="2016-02-24T19:37:09.870" LastActivityDate="2016-02-24T22:13:03.080" Title="SDSS. Google Sky and DeCaLs" Tags="&lt;amateur-observing&gt;&lt;positional-astronomy&gt;&lt;observational-astronomy&gt;&lt;deep-sky-observing&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13899" PostTypeId="2" ParentId="13898" CreationDate="2016-02-24T22:13:03.080" Score="2" Body="&lt;p&gt;The Sloan Digital Sky Survey has saturation limits at around 13th or 14th magnitude. You are looking at $\phi$ 1 Cancri, which is a 6th magnitude star. It is hardly surprising that SDSS fails to morphologically classify it correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No idea what your second question is about. You have given a link which has a latitude and longitude in the url. Longitude and latitude are not valid ways of expressing the coordinates of a celestial object. When I open this link, I see an image of your star and in the bottom left of the screen there is a display which gives the RA and Dec of where the cursor is.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-02-24T22:13:03.080" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13900" PostTypeId="1" CreationDate="2016-02-25T03:07:27.910" Score="2" ViewCount="363" Body="&lt;p&gt;What exactly is meant when we say the universe is expanding? I understand stars and galaxies are getting farther and farther away from us, but are individual particles (atoms, molecules, anything found on Earth, etc.) drifting away from each other as well? If no, then why not? If yes, then shouldn't it, at some point, get to a point when particles can no longer bond with each other?&lt;/p&gt;&#xA;" OwnerUserId="10964" LastActivityDate="2016-02-28T09:49:57.343" Title="What exactly is meant by &quot;expanding universe&quot;?" Tags="&lt;expansion&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2016-02-29T15:54:21.920" ContentLicense="CC BY-SA 3.0" />
  <row Id="13901" PostTypeId="2" ParentId="13900" CreationDate="2016-02-25T04:08:08.483" Score="4" Body="&lt;p&gt;When we say the Universe is expanding, we really mean that space is expanding. The Universe could be infinite in size, but there continues to be more and more space between objects. Essentially, objects are moving away from each other because more and more space is being created between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the early Universe, matter was much closer together than it is now; this density caused extreme temperatures and no hadronic matter could form. The first expansion is thought to have occurred because of how hot and energetic everything was, and spacetime itself was expanded in an event we call the Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As objects spread out from each other during the inflationary epoch, the Universe began to cool down. Now, dark energy is the main culprit for its accelerating expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question, yes, dark energy is making all particles move away from each other, although farther objects move away from each other at a faster rate. Why is that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, the only thing holding all atoms, stars, planetary systems and galaxies together are the four fundamental interactions — the strong force, weak force, electromagnetic force, and gravity. In space, the main attracting interaction is gravity. As such, dark energy will cause farther objects (which are less affected by our gravity) to move away at a faster and faster rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of it like you are holding a dog with a leash, and the ground begins to spread out from below you. You and your dog are being moved away in opposite directions, but you both stay close together because of the strong leash.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, your mention of atoms being torn apart is actually the basis of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Rip&quot; rel=&quot;nofollow&quot;&gt;Big Rip hypothesis&lt;/a&gt;. It states that dark energy will become more and more abundant until even closeby objects separate from one another. First, galaxies will be disbanded as the stars move away from each other in all directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much later, the gravity of planetary systems won't be enough to hold them together. Later still, planets and stars will be disbanded into their molecules. This will eventually continue as the amount of dark energy increases, until not even atoms can be held together by the electromagnetic force, and the subatomic particles will be moved apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the time the expansion of the Universe exceeds lightspeed at the subatomic scale, no particle will ever be able to interact with one another. The Universe would dissolve into countless lonely particles that won't be able to do anything. Now, the Big Rip hypothesis is only one of the three most known fates of the Universe. Currently, the most accepted theory is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Heat_death_of_the_universe&quot; rel=&quot;nofollow&quot;&gt;heat death of the Universe&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-02-28T09:49:57.343" LastActivityDate="2016-02-28T09:49:57.343" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13902" PostTypeId="1" AcceptedAnswerId="13904" CreationDate="2016-02-25T10:55:20.063" Score="5" ViewCount="662" Body="&lt;p&gt;I heard of one that occurred in 2008 in the Nubian Desert in Sudan, how often does it happen? Is there a catalogue of all hits or are they very rare?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related link (This is what inspired my interest not defining what I'm looking for)&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/2008_TC3&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/2008_TC3&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the time of asking I had not thought about the distinction between hitting the atmosphere and hitting the Earth.&lt;/p&gt;&#xA;" OwnerUserId="969" LastEditorDisplayName="user5434" LastEditDate="2017-05-06T22:15:52.690" LastActivityDate="2017-05-06T22:15:52.690" Title="How often do meteorites hit the Earth?" Tags="&lt;asteroids&gt;&lt;impact&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13903" PostTypeId="1" AcceptedAnswerId="13906" CreationDate="2016-02-25T14:49:17.623" Score="4" ViewCount="145" Body="&lt;p&gt;In astronomy, solar abundances are often calculated or tabulated per $10^{12}$ H atoms. I understand that in case of the Sun this can be done because H atoms are in majority or are present for a scale comparision. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can the same element abundance be calculated for an example meteorite which does not have any volatile gas such as H? In particular, I am referring to the table 1 of &lt;a href=&quot;http://astro.uni-tuebingen.de/~rauch/TMAP/grevessesauval2001.pdf&quot; rel=&quot;nofollow&quot;&gt;http://astro.uni-tuebingen.de/~rauch/TMAP/grevessesauval2001.pdf&lt;/a&gt;. It lists Solar abundance and lists meteorites abundance as a comparision. How have they calculated the element abundance for a meteorite in the $log_{10} \epsilon_{H} = 12$ scale?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I have a sample rock which has its's composition given in Weight $\%$ of various oxides such as $SiO_{2}, MgO, FeO$ etc. I have to convert these into elemental abundances in $log_{10} \epsilon_{H} = 12$  scale. How should I do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="10971" LastActivityDate="2016-02-25T23:46:28.337" Title="How are the element abundances calculated for a meteorite in the Hydrogen log10 scale?" Tags="&lt;the-sun&gt;&lt;cosmology&gt;&lt;meteorite&gt;&lt;elemental-abundances&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13904" PostTypeId="2" ParentId="13902" CreationDate="2016-02-25T14:53:49.513" Score="11" Body="&lt;p&gt;There is definitely not a catalogue of all meteorites hitting Earth. For instance, the ones falling in desert areas and in the ocean aren't found, and even the ones falling in more populous regions are easily mistaken for normal rocks.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Meteorite rate&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Meteorites come in all sizes, from sand grains to dinosaur-annihilating rocks.&#xA;The size distribution of these rocks (as well as most other things in space) follow an approximate power law, meaning that the smaller the meteorites you consider, the more often they hit Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Estimating the total number of meteorites hitting Earth is difficult, but one study (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1996MNRAS.283..551B&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bland et al. 1996&lt;/a&gt;), looking at the composite of the earth on three arid site and considering the weathering over time, finds that between 36 and 116 meteorites over 10 g hit Earth every year per $10^6\,\mathrm{km}^2$. Multiplying by the surface area of Earth $A_\oplus$, that's &lt;strong&gt;roughly 20,000 to 60,000 meteorites hitting Earth per year&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Meteor rate&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;UPDATE: You want to know the rate of incoming &lt;em&gt;meteors&lt;/em&gt;, i.e. all rocks that hit Earth's atmosphere. By definition, a meteor&lt;em&gt;ite&lt;/em&gt; is a &lt;em&gt;meteor&lt;/em&gt; that lands on the surface of Earth. Before it enters the atmosphere and becomes a meteor, you can call it asteroid, meteoroid, rock, pebble, or dust grain, depending on its size and your mood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1996Natur.380..323T&quot; rel=&quot;nofollow noreferrer&quot;&gt;Taylor et al. (1996)&lt;/a&gt; have monitored incoming meteors with the AMOR radar in New Zealand and detected in one year 350,000 trails from incoming meteors of size 10–100 $\mu$m vaporizing in ~100 km's height. I don't know exactly how large an area AMOR monitors, but from simple trigonometry, and as an order-of-magnitude estimate, if the events happen in $h \sim 100\,\mathrm{km}$'s height, the radar can &quot;see&quot; a distance $d=\sqrt{h^2+2R_\oplus h} \sim 1100\,\mathrm{km}$, and thus covers an area $A\sim4\times10^6\,\mathrm{km}^2$. Thus, again multiplying by $A_\oplus$, this amounts to &lt;strong&gt;the order of 40 to 50 million meteors per year&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Killer meteor rate and the Sudan asteroid&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The meteors discussed above is probably not what you had in mind. As mentioned, the bulk of these &quot;rocks&quot; are actually more like sand grains. So what is the rate of devastating killer-rocks from space? &lt;a href=&quot;http://adsabs.harvard.edu/abs/2002Natur.420..294B&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brown et al. (2002)&lt;/a&gt; fit a power law to several different data sets (&lt;a href=&quot;https://en.wikipedia.org/wiki/Lincoln_Near-Earth_Asteroid_Research&quot; rel=&quot;nofollow noreferrer&quot;&gt;LINEAR&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Spacewatch&quot; rel=&quot;nofollow noreferrer&quot;&gt;Spacewatch&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Near-Earth_Asteroid_Tracking&quot; rel=&quot;nofollow noreferrer&quot;&gt;NEAT&lt;/a&gt; and others), and find that the number $N$ of meteors above a diameter $D$ in meters is roughly&#xA;$$&#xA;N(&amp;gt;D) \simeq 37 D^{-2.7}.&#xA;$$&#xA;The following plot is from their paper, with the long black line showing their fit, and the short black lines indicating the uncertainty:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Cbdsz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cbdsz.png&quot; alt=&quot;meteors&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mention the asteroid &lt;a href=&quot;https://en.wikipedia.org/wiki/2008_TC3&quot; rel=&quot;nofollow noreferrer&quot;&gt;2008 TC&lt;sub&gt;3&lt;/sub&gt;&lt;/a&gt;, which had a diameter of 4.1 m.&#xA;According to the fit, &lt;strong&gt;such an event happens roughly $1/(37[4.1\,\mathrm{m}]^{-2.7})\sim$ once per year&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also see that the largest meteor hitting Earth in a century would be $(37\times100\,\mathrm{years})^{1/2.7}\sim20\,\mathrm{m}$, equivalent to $\sim500\,\mathrm{kiloton\,TNT}$.&#xA;Similarly, hundred-meter-sized objects ($\sim1\,\mathrm{megaton\,TNT}$) should hit us roughly once per 10,000 years. By extrapolation of their fit beyond observational data, &lt;strong&gt;objects of $D&amp;gt;1\,\mathrm{km}$ — or 100 gigaton TNT — should hit Earth on average once every ~3.4 million years&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-02-29T23:12:54.187" LastActivityDate="2016-02-29T23:12:54.187" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13905" PostTypeId="2" ParentId="13883" CreationDate="2016-02-25T18:13:01.617" Score="2" Body="&lt;p&gt;If you jump out an aeroplane, do you experience high speed? Many people would say you do, since your velocity relative to the ground increases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the importance of the phrase &quot;relative to the ground&quot; since to describe speed you always need to say what you are measuring the velocity relative to. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple choice is to measure your velocity relative to a distant observer. In this case then, yes as you pass close to the black hole your velocity relative to the distant observer increases. As you approach the event horizon, you velocity relative to the observer will approach the speed of light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on how close you pass the the event horizon, you will experience significant time dialation, so that more time will pass in the observer's frame than in yours. This means that if you manage to escape the gravitational pull of the black hole, and return to the observer, you will have found that the observer has aged compared to you.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-02-25T18:13:01.617" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13906" PostTypeId="2" ParentId="13903" CreationDate="2016-02-25T19:16:53.660" Score="4" Body="&lt;p&gt;Meteorite abundances are referenced to their silicon abundance and then that is bootstrapped onto the hydrogen scale by assuming that the silicon abundance of meteorites is the same as that in the Sun. See for example &lt;a href=&quot;http://arxiv.org/abs/0909.0948&quot; rel=&quot;nofollow&quot;&gt;Asplund et al. (2009)&lt;/a&gt;. &#xA;This means there is an additional uncertainty of a couple of hundredths of a dex when comparing meteoritic and solar photospheric abundances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to your second question is work out the abundances relative to Si and then use the solar Si/H ratio to put them on the hydrogen scale.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-02-25T23:46:28.337" LastActivityDate="2016-02-25T23:46:28.337" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13907" PostTypeId="1" AcceptedAnswerId="13909" CreationDate="2016-02-26T04:11:12.547" Score="5" ViewCount="1496" Body="&lt;p&gt;This is problem 1-4 from Principles of Stellar Evolution and Nucleosynthesis by Clayton:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming at the Earth a characteristic velocity of 400km/s and density of 10amu/cm$^{3}$ for the solar wind, calculate the rate of mass loss for the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There weren't any formulas about this in the section so I made a stab at it with dimensional analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dM}{dt} = \frac{\rho V}{\Delta t} = \rho v A$$&#xA;$$\frac{dM}{dt} = \left( \frac{10 amu}{cm^{3}} \right) \left( \frac{400km}{s} \right) \left( \frac{4 \pi (6.96e10 cm)^{2}}{1} \right) \left( \frac{10^{5}cm}{km} \right) \left(\frac{10^{-24} g}{1 amu} \right) \left( \frac{M_{\odot}}{2 \times 10^{33} g} \right) \left( \frac{3600s}{hr} \right) \left( \frac{24 hr}{day}\right)\left(\frac{365day}{yr}\right)$$&#xA;$$\frac{dM}{dt} = 3.84 \times 10^{-19} M_{\odot} / yr $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the answer given in the book is $0.4 \times 10^{-13} M_{\odot} / yr$. So, I'm off by about five magnitudes. Can anyone point out where I went wrong and/or point me in the correct direction?&lt;/p&gt;&#xA;" OwnerUserId="10977" LastActivityDate="2016-02-26T11:55:32.867" Title="Rate of Mass Loss from the Solar Wind" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;solar-wind&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13908" PostTypeId="2" ParentId="13907" CreationDate="2016-02-26T06:40:33.813" Score="5" Body="&lt;p&gt;The velocity and the density are at the earth position, so the area term must include the earth-sun distance instead of the solar radius.&lt;/p&gt;&#xA;" OwnerUserId="608" LastActivityDate="2016-02-26T06:40:33.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13909" PostTypeId="2" ParentId="13907" CreationDate="2016-02-26T11:25:21.343" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\left( \frac{4 \pi (6.96e10\,\text{cm})^{2}}{1} \right)$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the primary source of your error. Your value of 6.96×10&lt;sup&gt;10&lt;/sup&gt; cm is the radius of the Sun. The problem specifically said &quot;&lt;em&gt;Assuming at the Earth ...&lt;/em&gt;&quot;. You need to calculate the flux through the surface of a sphere whose radius is about one astronomical unit rather than one solar radius. The astronomical unit is 149597870700 meters (exactly), or about 1.5×10&lt;sup&gt;13&lt;/sup&gt; cm. This error alone makes your value low by a factor of about 50000. The remaining factor of two results mostly from using 10&lt;sup&gt;-24&lt;/sup&gt; grams per amu.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dimensional analysis can only take you so far. While your result is dimensionally correct, you didn't think enough about the nature of the problem.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2016-02-26T11:55:32.867" LastActivityDate="2016-02-26T11:55:32.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13910" PostTypeId="1" AcceptedAnswerId="13913" CreationDate="2016-02-26T15:27:34.287" Score="9" ViewCount="6296" Body="&lt;p&gt;I recently purchased a &lt;a href=&quot;http://www.celestron.com/browse-shop/astronomy/telescopes/powerseeker-114eq-telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Celestron PowerSeeker 114EQ&lt;/a&gt; and with it I got three lenses (which I assume are low quality), a 4mm, a 20mm and a 3 x Barlow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hsq71.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hsq71.jpg&quot; alt=&quot;4mm, Barlow and 20mm&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can get pretty decent views of the moon but a very blurry, small view of Jupiter with the 4mm and Barlow.  I can see the color of the storm rings for example, but barely.  I've not had a chance to have a go at Saturn yet.  A view similar to this...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iURCN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iURCN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been reading up on this and was looking at purchasing a 4-8mm Plössl lens but I'm not sure if the lens is my weak spot or some other aspect of the telescope.&#xA;Another thing to note is that I've been using it in a pretty bright street, looking up very close to street lights.  I've not had a chance to take out somewhere with less light pollution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what I'm asking is, how much difference would a lens upgrade make, specifically to my image quality?&lt;/p&gt;&#xA;" OwnerUserId="10980" LastEditorUserId="10980" LastEditDate="2016-02-29T18:24:24.480" LastActivityDate="2016-02-29T18:24:24.480" Title="How much of a difference do good lenses make?" Tags="&lt;jupiter&gt;&lt;newtonian-telescope&gt;&lt;telescope-lens&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13911" PostTypeId="1" CreationDate="2016-02-26T16:35:23.283" Score="0" ViewCount="899" Body="&lt;p&gt;We have harnessed and learned to generate and control sound waves, light waves, water waves, and electromagnetic waves in general.  What would be needed for science to be able to generate and use gravitational waves in the same way? What would be the outcome?&lt;/p&gt;&#xA;" OwnerUserId="4146" LastActivityDate="2016-02-27T14:13:46.430" Title="What would harnessing of gravitational waves look like?" Tags="&lt;gravitational-waves&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13912" PostTypeId="1" AcceptedAnswerId="13914" CreationDate="2016-02-26T17:16:58.047" Score="4" ViewCount="355" Body="&lt;p&gt;I saw this video at &lt;a href=&quot;http://www.space.com/32055-mystery-fast-radio-burst-space-explosions-location.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Space.com&lt;/a&gt; and I noticed that at 00:05 the animation illustrates waves propagating from space into &lt;a href=&quot;https://en.wikipedia.org/wiki/Parkes_Observatory&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Dish&lt;/a&gt; at Parks Observatory, but at 00:50 the direction of propagation is reversed! Would there be some reason for transmitting - for example to create an &lt;a href=&quot;https://en.wikipedia.org/wiki/Laser_guide_star&quot; rel=&quot;nofollow noreferrer&quot;&gt;artificial&lt;/a&gt; &quot;star&quot; for wavefront correction analogous to &lt;a href=&quot;https://en.wikipedia.org/wiki/Adaptive_optics&quot; rel=&quot;nofollow noreferrer&quot;&gt;adaptive optics&lt;/a&gt; at visible wavelengths, or could it be simply a mistake in the animation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some GIFs from screenshots. You can check the &lt;a href=&quot;http://www.space.com/32055-mystery-fast-radio-burst-space-explosions-location.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;video&lt;/a&gt; at the indicated times. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;circa 00:05&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NdD8W.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NdD8W.gif&quot; alt=&quot;at 00:05&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;circa 00:50&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/n8Oon.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/n8Oon.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2016-02-27T01:36:11.853" LastActivityDate="2016-02-27T01:36:11.853" Title="Why is this video showing radio waves transmitted from a radio telescope?" Tags="&lt;radio-astronomy&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13913" PostTypeId="2" ParentId="13910" CreationDate="2016-02-26T17:24:53.533" Score="10" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;...a very blurry, small view of Jupiter with the 4mm and Barlow...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Be aware that a 4mm eyepiece &lt;strong&gt;and&lt;/strong&gt; a 3x barlow at the same time will give you a very high magnification - too high!  For regular Jupiter viewing I would suggest you stick to 100x or 200x at most, unless the air is exceptionally still. (After a few sessions you'll find out what &quot;Still&quot; air means.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of beginner scopes like this are sold with barlows that tend to give too much power, my advice is to keep it packed away for most sessions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'd stick to using that 4mm on its own, jupiter will look quite small but with practice you can usually tease a bit more detail out of the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have one of these scopes, but my experience of comparing cheap modern eyepieces with expensive ones is that the lower cost ones are generally OK these days. 4mm is quite a high power so the extra magnification will cause a lot of blurriness and it will take patience to get the focusing at its sharpest, so don't panic if everything is looking fuzzy at the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a first stage I would suggest a few things that are nothing to do with the eyepieces at all...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Collimation - look this up, it just means adjusting the two mirrors&#xA;so your eye is looking right down the tube in a straight line. For a&#xA;long focal length scope like yours, it's not likely to be a problem&#xA;unless one of them is wildly out of line.&lt;/li&gt;&#xA;&lt;li&gt;Tube currents/thermal behaviour - on almost any cold night, when&#xA;the tube and main mirror are still warm from being indoors, rising&#xA;air currents in the tube will mess up your image and make it&#xA;shimmery, at high power. Low power will look OK. It might take an hour or so for the image to improve (a guess)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Anyway, in summary my guess is changing the eyepieces straight away won't make a dramatic difference. Eyepiece makers will say otherwise of course :)&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2016-02-26T17:24:53.533" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13914" PostTypeId="2" ParentId="13912" CreationDate="2016-02-26T18:02:45.580" Score="2" Body="&lt;p&gt;I guess it's a mistake in the animation. At radio wavelengths turbulence doesn't affect the measurments anymore and above 200 MHz the ionosphere doesn't play a role yet, so I guess the notion of a guide star is unnecessary here.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2016-02-26T18:02:45.580" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13915" PostTypeId="1" CreationDate="2016-02-26T18:49:46.237" Score="4" ViewCount="452" Body="&lt;p&gt;I see explanations likes this:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Galaxies and other sources, then, are not strictly expanding away from each other but rather are attached to the fixed grid on the expanding fabric of spacetime. Thus, the galaxies give us the impression of moving away from each other. (&lt;em&gt;&lt;a href=&quot;http://www.scientificamerican.com/article/where-is-the-universe-exp/&quot; rel=&quot;noreferrer&quot;&gt;Scientific American&lt;/a&gt;&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;p&gt;[...] space is expanding [...] the most correct way to think about it would be that the grid points are getting farther apart. (&lt;em&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/95538/what-is-the-meaning-of-the-expansion-of-space&quot;&gt;Physics.se&lt;/a&gt;&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;How have we determined that it is some underlying grid or space itself that is expanding rather than the objects just moving apart on a static grid of spacetime?&lt;/p&gt;&#xA;" OwnerUserId="10982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-03-04T17:52:44.840" Title="Distinction between metric expansion and objects just moving apart from each other?" Tags="&lt;cosmology&gt;&lt;space-time&gt;&lt;expansion&gt;&lt;cosmological-inflation&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13916" PostTypeId="2" ParentId="13911" CreationDate="2016-02-26T19:10:42.750" Score="4" Body="&lt;p&gt;The mass of fundamental particles, like the proton or electron is very very small in comparison to the charge. This means that the motion of electrons in atoms can produce significant amounts of high frequency electromagnetic radiation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum effects stop electrons from falling into the protons they orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To produce and control gravitational radiation in significant amounts you would need something very small, but very massive: a microscopic pair of black holes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such things would be rather hard to manage. In short it is beyond any conceivable current or future technology.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-02-26T19:10:42.750" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13917" PostTypeId="2" ParentId="13850" CreationDate="2016-02-26T19:51:29.317" Score="2" Body="&lt;p&gt;Astronomers use &quot;Julian day numbers&quot; to indicate time. The Julian day number is the number of days since Noon GMT on January 1st 4713BC (in the Julian calendar). A time is then a single real number. For example, 00:30:00.0 UT January 1, 2013 (Gregorian), is 2456293.520833 (wikipedia)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So except to specify the epoch, the Julian calendar is not in use by astronomers to specify dates. There is no international standard equivalent to iso8601 for representing dates in the Julian calendar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ancient dates are often expressed as years before present, where &quot;present&quot; is 1950. 1950 is chosen as it preceeds most of the atmospheric nuclear testing that dumped massive amounts of C14 into the atmosphere and made carbon dating only valid for years before 1950.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-02-26T19:56:52.987" LastActivityDate="2016-02-26T19:56:52.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13918" PostTypeId="1" AcceptedAnswerId="13925" CreationDate="2016-02-27T00:17:09.560" Score="3" ViewCount="106" Body="&lt;p&gt;Are exceptionally large, bright, meteors (such as Tunguska or Chelyabinsk) more common during common meteor showers like the Perseids, or the Leonids? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, why is that? Do the meteor showers simply not carry any more massive pieces?&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="10953" LastEditDate="2016-02-27T09:29:58.743" LastActivityDate="2016-02-27T17:01:14.263" Title="Do massive bolide entries coincide with meteor showers?" Tags="&lt;meteor&gt;&lt;meteorite&gt;&lt;meteor-shower&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13919" PostTypeId="2" ParentId="13625" CreationDate="2016-02-27T03:01:57.270" Score="0" Body="&lt;p&gt;The short answer is that the Einstein field equation relates the energy, momentum, and stress of matter to curvature in such a way so that it would be true. We could certainly imagine other relationships between matter and curvature in which the opposite would be the case, or some other stranger things; they just don't resemble how our universe works. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will provide an overview of some facts regarding differential geometry and how they connect to this question. Depending on your background, this may be insufficient as a full explanation, but at least it will provide you with pointers about what to follow up on (although most of it would be more relevant either in math.SE or physics.SE, depending).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;On a (pseudo-)Riemannian manifold, a curve $\gamma$ is an affine geodesic whenever it can be parametrized in such a way as to keep the same velocity vector $u = \dot{\gamma}$, or slightly more formally, whenever its velocity vector parallel-transported along the curve remains the same. Equivalently, one can think of affine geodesics are curves of zero acceleration along themselves, which is expressed by the &lt;em&gt;geodesic equation&lt;/em&gt;:&#xA;$$\nabla_uu = 0\text{.}$$&#xA;There is an obvious intuitive connection between this and &lt;em&gt;inertial motion&lt;/em&gt;, so we can think of these geodesics as describing the motion of free test particles. Unless the length along the geodesic is zero (which can happen on a pseudo-Riemannian manifold), we can take the length along it to be the affine parametrization. Later in physics, the length along the worldline of a massive particle is the time measured by a clock going along it--its proper time $\tau$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have you have two nearby geodesics with the same velocity $u$ but separated by a vector $n$ (i.e., this vector connects points of equal affine parameter), we can talk about the relative acceleration between those geodesics. This is known as the &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Geodesic_deviation&quot; rel=&quot;nofollow&quot;&gt;geodesic deviation&lt;/a&gt;&lt;/em&gt; equation, and in terms of the Riemann curvature $R$, one of its forms would be:&#xA;$$\nabla^2_{u,u}n + R(n,u)u = 0\text{,}$$&#xA;or equivalently in coordinates:&#xA;$$\frac{\mathrm{D}^2n^\alpha}{\mathrm{d}\tau^2} + \underbrace{R^\alpha{}_{\mu\beta\nu}u^\mu u^\nu}_{\mathbb{E}^\alpha{}_\beta}\,x^\beta = 0\text{,}$$&#xA;where the &lt;em&gt;tidal tensor&lt;/em&gt; $\mathbb{E}^\alpha{}_\beta$ is sometimes called the 'electric' part of the Riemann curvature tensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're not familiar with how tensors work, for now just imagine the Riemann tensor as a mathematical machine that takes three vectors and spits out another one in a certain way. In four dimensions, the tensor has $256$ components, although only $20$ of them are algebraically independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, once you get used to the idea that nearby geodesics can deviate from one another due to the presence of curvature, we can ask a similar question: suppose you have a small ball (size $\epsilon$) of test particles, all initially comoving with the same velocity $u$. Since their geodesics will have some relative acceleration, the ball generally won't stay of the same volume (or stay a ball, for that matter). In an inertial frame comoving with the test particles, we can consider the initial fraction acceleration of its volume in the limit of zero size of the ball:&#xA;$$\lim_{\epsilon\to 0}\left.\frac{\ddot{V}}{V}\right|_{t=0}\!= -\underbrace{R^\alpha{}_{\mu\alpha\nu}}_{R_{\mu\nu}} u^\mu u^\nu = -\mathbb{E}^\alpha{}_{\alpha}\text{,}$$&#xA;where the Ricci curvature $R_{\mu\nu}$ is the trace of the Riemann curvature, and the whole expression is equivalent to the trace of the tidal tensor. A more general situation is described by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Raychaudhuri_equation&quot; rel=&quot;nofollow&quot;&gt;Raychaudhuri equation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The takeaway is this: the Ricci curvature tells us whether small balls of initially comoving test particles will expand or contract, i.e. move towards or away from each other. (NB: There's almost no physics in this section--other than borrowing some vocabulary to help frame these issues in a way that will more readily connect to physics, these are facts of differential geometry.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To have physical content, we need to not only interpret spacetime as a four-dimensional manifold, but also say something about what the curvature is in terms of physical quantities. Given the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor&quot; rel=&quot;nofollow&quot;&gt;stress-energy-momentum tensor&lt;/a&gt; $T_{\mu\nu}$ and its trace $T$, the Einstein field equation provides the Ricci curvature:&#xA;$$R_{\mu\nu} = \kappa\left(T_{\mu\nu} - \frac{1}{2}g_{\mu\nu}T\right)\tag{EFE}\text{,}$$&#xA;where $\kappa = 8\pi G/c^4$ is a constant. So does that mean that matter attracts or not? Well, it still depends: locally, we can say that under gravitational freefall, a small ball of initially comoving test particles with initial four-velocity $u$ will contract whenever $R_{\mu\nu}u^\mu u^\nu\geq 0$. Substituting in the above Einstein field equation, we get the &lt;a href=&quot;https://en.wikipedia.org/wiki/Energy_condition#Strong_energy_condition&quot; rel=&quot;nofollow&quot;&gt;strong energy condition&lt;/a&gt;, which basically boils down to 'gravity is always locally attractive'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, even though the strong energy condition might not necessary hold (and gravity can be repulsive, e.g. on the cosmological scale due to dark energy), it holds under ordinary circumstances. That's because in the local inertial frame comoving with $u$, the trace of the stress-energy tensor is $T = -\rho + 3p$, where $p$ is the pressure (pretending for simplicity that we're dealing with a perfect fluid; otherwise, we can take average of the principal stresses), while $\rho = T_{\mu\nu}u^\mu u^\nu$ is the energy density. Therefore, the time-time projection of the Einstein field equation is (in units of $G=c=1$):&#xA;$$\mathbb{E}^\alpha{}_\alpha \equiv R_{\mu\nu} u^\mu u^\nu = 4\pi(\rho+3p)\text{.}$$&#xA;Since energy density and pressure have equivalent units, to compare, ordinary water has a density of $(1\;\mathrm{kg}/\mathrm{L})c^2 = 9.0\times 10^{19}\,\mathrm{Pa} = 8.9\times 10^{14}\,\mathrm{atm}$. Ordinary matter has &lt;em&gt;overwhelmingly&lt;/em&gt; higher energy density than stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above time-time projection of the Einstein field equation is the direct analogue of the Gauss' law for Newtonian gravity (matter density) or electromagnetism (charge density). In the weak-field limit (small deviation from flatness), we can use this to recover Poisson's equation for Newtonian gravity; see also part of &lt;a href=&quot;https://astronomy.stackexchange.com/questions/13606/existence-of-gravitons/13619#13619&quot;&gt;this&lt;/a&gt; answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the Einstein field equations guarantees that not only is ordinary matter locally attractive, but that in the weak-field limit, it is essentially Newtonian as well.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2016-02-27T03:01:57.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13920" PostTypeId="1" AcceptedAnswerId="16153" CreationDate="2016-02-27T07:11:25.620" Score="2" ViewCount="301" Body="&lt;p&gt;Recently I've been trying to spot a couple of nebulae, IC59 and IC1318. The skies at the site are modestly dark (4.5 on the Bortle scale), I'm using an 8&quot; scope, and I've slowly scanned around the areas endlessly with a low magnification/wide angle eyepiece but I haven't been able to find either of them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is a filter likely to significantly help in this situation? Would a nebulae filter be more helpful than a light pollution filter?&lt;/p&gt;&#xA;" OwnerUserId="10953" LastActivityDate="2016-09-15T08:45:21.607" Title="How useful are filters for spotting nebulae?" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;nebula&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13921" PostTypeId="2" ParentId="13911" CreationDate="2016-02-27T14:13:46.430" Score="1" Body="&lt;p&gt;To put this in perspective, consider a fifty-plus year old human whose main source of exercise for the last twenty-plus years has been walking back and forth between work and the parking spot where the person parks his or her car. Let's strap that person to a bicycle connected to a generator. The energy output of that feeble source of energy easily exceeds the feeble 200 watt gravitational waves produced by the Earth's orbit about the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitation is extremely feeble compared to electromagnetism. The electrostatic repulsion between a pair of electrons is 10&lt;sup&gt;45&lt;/sup&gt; times stronger than is the gravitation attraction between a pair of electrons. Gravitational waves are in turn extremely feeble compared to gravitation itself. I mentioned the Earth's orbit about the Sun. The total mechanical energy of that orbit is about 10&lt;sup&gt;34&lt;/sup&gt; times greater than the paltry amount of energy lost due to gravitational waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a Kardashev level III civilization, a civilization that has learned to harness the equivalent of the energy output of an entire galaxy. Aside: Humanity isn't even at Kardashev level I. That's another few hundred years in the future. Also note that the Kardashev scale is logarithmic. Karadshev level I is near future science fiction. Kardashev level II is also in the realm of science fiction. Kardashev level III? That's beyond science fiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even that Kardashev level III civilization would not harness gravitational waves. It would instead go out of their way to avoid them. A Kardashev level III civilization might, for example, intentionally feed matter to a supermassive black hole at the heart of a galaxy to make that black hole become an active galactic nucleus. The energy output of an AGN is immense, exceeding that of a normal galaxy. The gravitational waves produced by an AGN are minuscule compared to the electromagnetic output, by a factor of about 10&lt;sup&gt;-80&lt;/sup&gt; or so.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2016-02-27T14:13:46.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13922" PostTypeId="2" ParentId="13918" CreationDate="2016-02-27T14:22:54.823" Score="3" Body="&lt;p&gt;In historical times we do not have enough data to do serious statistics about meteors of the size of the Chelyabinsk meteor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_meteor_air_bursts&quot; rel=&quot;nofollow&quot;&gt;list of meteor air bursts&lt;/a&gt; there are only six events within two orders of magnitude of the Chelyabinsk meteor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{lll}&#xA;\hline&#xA;\text{Date} &amp;amp; \text{Location} &amp;amp; \text{Energy [TJ]}\\&#xA;\hline&#xA;\text{1490-04-04} &amp;amp; \text{Qingyang district, China} &amp;amp; unknown\\&#xA;\text{1908-06-30} &amp;amp; \text{Tunguska event} &amp;amp; 42000–63000\\&#xA;\text{1930-08-13} &amp;amp; \text{Curuca River Area, Brazil} &amp;amp; 38–20920\\&#xA;\text{1963-08-03} &amp;amp; \text{Prince Edward Islands} &amp;amp; 740–1490\\&#xA;\text{2009-10-08} &amp;amp; \text{South Sulawesi, Indonesia} &amp;amp; 130–210\\&#xA;\text{2013-02-15} &amp;amp; \text{Chelyabinsk meteor} &amp;amp; 2100\\&#xA;\hline&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that for the Tunguska event it is not proven to date that it had been caused by a meteor. &lt;/p&gt;&#xA;" OwnerUserId="10568" LastEditorUserId="10568" LastEditDate="2016-02-27T17:01:14.263" LastActivityDate="2016-02-27T17:01:14.263" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13923" PostTypeId="1" CreationDate="2016-02-27T09:44:48.853" Score="6" ViewCount="422" Body="&lt;p&gt;Advanced LIGO seems to operate only intermittently. Is there a schedule for at what times it will be able to register new gravity waves? Even if no formal schedule is available, what main factors determine the scheduling?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose VIRGO in Italy, to get started in 2016, will operate mostly simultaneously with aLIGO. Or would its design require another type of schedule?&lt;/p&gt;&#xA;" OwnerUserId="1030" OwnerDisplayName="LocalFluff" LastEditorUserId="1056" LastEditDate="2016-02-29T18:18:04.380" LastActivityDate="2016-02-29T18:18:04.380" Title="What is the schedule for science runs of aLIGO (and VIRGO)?" Tags="&lt;observational-astronomy&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13924" PostTypeId="2" ParentId="13923" CreationDate="2016-02-27T16:34:30.393" Score="4" Body="&lt;p&gt;There is no set date due to the complexity of the machines and their upgrades but &lt;a href=&quot;http://www.ligo.org/science/Publication-ObservingScenario/index.php&quot;&gt;this&lt;/a&gt; is the closest you will get to an answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;O1 is planned to last for four months, closing mid-January 2016. Then work will start on upgrading the instruments for our second observing run, which is called O2; those upgrades will be informed by what we have learned about the instruments during O1. O2 will start in 2016 and last around six months. Hopefully, around this time Advanced LIGO will be joined by Advanced Virgo.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They have no fixed date for the second observing run (O2) but it should be up an running later this year (my guess would be June-July time).&lt;/p&gt;&#xA;" OwnerUserId="10437" LastActivityDate="2016-02-27T16:34:30.393" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13925" PostTypeId="2" ParentId="13918" CreationDate="2016-02-27T16:59:14.143" Score="2" Body="&lt;p&gt;Meteor showers are associated with dust trails left by comets, There are larger clumps, but most cometary dust is pretty small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most large impactors are small asteroids: This was certainly the case in Chelyabinsk, and is the likely the case with Tunguska. Asteroids are not associated with cometary orbits and the impacts are sporadic, not concentrated in showers. Each year there are &lt;a href=&quot;http://neo.jpl.nasa.gov/fireballs/&quot; rel=&quot;nofollow&quot;&gt;10s of significant bolides&lt;/a&gt; They don't correlate with meteor showers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Tunguska did occur during a regular shower, the Beta Taurids, and some have hypothesised that it may have be an fragment of Comet Encke, but analysis of the impact site suggests that this was a coincidence.)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-02-27T16:59:14.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13926" PostTypeId="2" ParentId="13915" CreationDate="2016-02-27T23:51:09.643" Score="1" Body="&lt;h1&gt;Time will show&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;...although that is a very long time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the grid of the universe was static, objects moving away from us should continue to move away from us with a continues velocity. That is not the case from observed data: Things farther away from us seems to move away faster. This is known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble&amp;#39;s_law&quot; rel=&quot;nofollow&quot;&gt;Hubble's law&lt;/a&gt;, and is one of the basic concepts of modern cosmology. The model of an expanding metrics fit that data, in that when there is more space fabric between us and the far object, there are more space that stretches, and therefore the velocity is higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That things move away from us, and have an accelerating velocity, should suggest a universe that expands faster and faster. There are however other forces limiting this growth, most notably gravity. The fate of the growth of the universe is determined by that unknown balance, that must also consider &lt;a href=&quot;http://science.nasa.gov/astrophysics/focus-areas/what-is-dark-energy/&quot; rel=&quot;nofollow&quot;&gt;dark matter and energy&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnitude, or even the validity of those models is the current frontier of astronomy, and cosmology in general. Nothing is settled yet.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastActivityDate="2016-02-27T23:51:09.643" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13927" PostTypeId="1" CreationDate="2016-02-28T04:25:01.473" Score="7" ViewCount="123" Body="&lt;p&gt;Space expands everywhere, also here. And time is inseparable from space. Does this mean that time also &quot;expands&quot; as in changing its pace? Is the changing rate of time also astronomically observable? How did time behave during the radical inflation shortly after Big Bang?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-02-29T18:03:56.763" Title="Does the accelerating expansion of spacetime mean that the pace of time is changing?" Tags="&lt;space-time&gt;&lt;cosmological-inflation&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13928" PostTypeId="2" ParentId="13915" CreationDate="2016-02-28T09:42:23.807" Score="0" Body="&lt;p&gt;For one, the expansion of space allows objects to move away from us faster than light. Einstein said nothing can move faster than light through space; however, space itself can expand between objects faster than light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were actually a force pushing these objects away, they would not recede from us faster than light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, according to QFT, there does not seem to be a universally (always) repulsive force. Whether forces can be repulsive or not depends on the spin of their mediating field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A scalar (spin-0) force is universally attractive, as is a spin-2 force, while a spin-1 is attractive for different charges and repulsive for like charges. So the electromagnetic, the weak and the strong force can be repulsive, while gravity cannot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of this pattern we can assume that there are no universally repulsive forces that can be currently described by fields in QFT.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2016-02-28T09:42:23.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13929" PostTypeId="2" ParentId="13927" CreationDate="2016-02-28T11:45:51.110" Score="4" Body="&lt;p&gt;To talk about 'the rate of time', we essentially need at least two different time coordinates. For example, this happens in special-relativistic time dilation, which is equivalent to $\mathrm{d}t'/\mathrm{d}t$ across two different inertial frames. Fortunately, we can do something similar here.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Space expands everywhere, also here. And time is inseparable from space. Does this mean that time also &quot;expands&quot; as in changing its pace? ... In a similar way that the expansion of space is compared relative to, well, to itself I suppose.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A spatially isotropic and homogeneous universe has the metric in the form&#xA;$$\mathrm{d}s^2 = -\mathrm{d}t^2 + a^2(t)\mathrm{d}\Sigma^2\text{,}$$&#xA;where $a(t)$ is the scale factor and $\mathrm{d}\Sigma^2$ is the metric of an isotropic and homogeneous Riemannian manifold: the 'open' hyperbolic $3$-plane, the flat Euclidean $3$-space, or the 'closed' $3$-sphere (or real projective $3$-space, but that's usually not considered because it's non-orientable). If the scale factor is ever zero in the past, the cosmological time for this is conventionally chosen to be $t = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cosmological time measures the proper time of an observer at rest relative to the bulk of the matter in the universe, so in some sense it's the most intuitive choice of a time coordinate, but like all coordinates, it's not sacred. We can, for example, define a &lt;em&gt;conformal time&lt;/em&gt; coordinate $\eta$ such that $\mathrm{d}\eta = \mathrm{d}t/a$, in which the metric takes the form&#xA;$$\mathrm{d}s^2 = a^2(\eta)\left[-\mathrm{d}\eta^2 + \mathrm{d}\Sigma^2\right]\text{,}$$&#xA;and so all of the dimensions of spacetime are affected by cosmic expansion in the same way. Therefore, I think conformal time satisfies the requirements in your question, although it is not measured by any local clock.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the changing rate of time also astronomically observable?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The scale factor is astronomically observable, and $\mathrm{d}\eta/\mathrm{d}t = 1/a$, so yes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How did time behave during the radical inflation shortly after Big Bang?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The conformal time essentially uses the particle horizon as a measure of time, i.e. the furthest distance from which an ideal lightlike signal could have travelled since $t = 0$ in order to reach the observer by the present time. During inflation, the particle horizon rapidly expanded.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditDate="2016-02-29T18:03:56.763" LastActivityDate="2016-02-29T18:03:56.763" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13930" PostTypeId="1" AcceptedAnswerId="13967" CreationDate="2016-02-28T13:14:52.027" Score="10" ViewCount="621" Body="&lt;p&gt;This &lt;a href=&quot;https://astronomy.stackexchange.com/a/13914/7982&quot;&gt;Question and Answer&lt;/a&gt; got me thinking. If atmospheric &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_seeing&quot; rel=&quot;nofollow noreferrer&quot;&gt;seeing&lt;/a&gt; at visible wavelengths is the result of refractive index inhomogeneity, would it also be a similar problem for mm to cm wavelengths? From a quick search, the index of refraction of air at STP is about 1.0003 (visible) and 1.0002 (radio).&lt;/p&gt;&#xA;&lt;p&gt;If it is not, is there a way to understand quantitatively why it is not a problem?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RfvZr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RfvZr.png&quot; alt=&quot;File:Atmos struct imaging.svg&quot; /&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/s4UAs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/s4UAs.png&quot; alt=&quot;File:Atmospheric seeing r0 t0.svg&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Sources: &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Atmos_struct_imaging.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Atmospheric_seeing_r0_t0.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2020-07-10T02:14:12.420" LastActivityDate="2020-07-10T02:14:12.420" Title="Would Adaptive Optics be Useful in Radio Astronomy?" Tags="&lt;radio-astronomy&gt;&lt;angular-resolution&gt;&lt;photography&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13932" PostTypeId="1" AcceptedAnswerId="13956" CreationDate="2016-02-28T17:45:08.177" Score="2" ViewCount="495" Body="&lt;p&gt;In the theory of multi-verses, we can imagine countless other universes where the constants of nature, including the amount of dark energy in a volume of space and possibly even shapes of hidden dimensions vary from one universe to another. Is there sound reasoning to believe that the mathematics of nature (or laws of nature) would be the same in other universes?&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2016-03-01T19:21:38.353" Title="Physical laws in other universes" Tags="&lt;cosmology&gt;&lt;multiverse&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13933" PostTypeId="1" CreationDate="2016-02-28T18:48:37.393" Score="2" ViewCount="83" Body="&lt;p&gt;The frequency of the gravity waves detected by aLIGO recently were in the audible sound spectra. Actually spanning over half of the keys of a piano, I've been told, during 2/10 or so seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would a human ear sense a gravity wave if powerful enough? The waves going through empty space from the event horizons, through the walls of a spacecraft and through the ears of astronauts. Would one actually hear a &quot;zip&quot; sound because space makes waves inside our ears?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-04-28T17:38:08.590" Title="Would humans hear gravity waves from a binary BH fusion nearby?" Tags="&lt;gravitational-waves&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13934" PostTypeId="1" AcceptedAnswerId="13941" CreationDate="2016-02-29T00:43:16.003" Score="4" ViewCount="81" Body="&lt;p&gt;I know spectroscopy of light in visible wavelengths is very effective for studying and determining the chemical composition of bodies within the solar system and bright objects outside of it. However, can this same method be used with exoplanets? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In looking about some related articles, I've learned that it is possible, if a planet passes between us and its star, to filter out the star's light leaving only what passes through the planet's atmosphere.&lt;br /&gt;&#xA;&lt;strong&gt;But what if there is no such occultation?&lt;/strong&gt; Can we filter out the star's light in other wavelengths like UV or infrared and look at just the planet via its own emitted radiation (assuming the planet is large and hot as these are what we are most able to detect at this time)?&lt;/p&gt;&#xA;" OwnerUserId="10995" LastEditorDisplayName="user1569" LastEditDate="2016-03-01T00:49:49.803" LastActivityDate="2016-03-01T00:49:49.803" Title="Can visible wavelength spectroscopy study an exoplanet's chemical composition directly?" Tags="&lt;exoplanet&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13935" PostTypeId="1" CreationDate="2016-02-29T09:09:16.803" Score="3" ViewCount="151" Body="&lt;p&gt;I read about the LIGo experiment recently, but i didn't understand their significance. Although, it's clear to me that the gravitational waves do exist and are caused by large accelerating bodies in space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but how are they significant to Humans as we do not have any mechanisms for harnessing these waves and Don't know what use they may be of.&lt;/p&gt;&#xA;" OwnerUserId="10996" LastActivityDate="2016-02-29T16:01:30.353" Title="What are gravitational waves actually?" Tags="&lt;solar-system&gt;&lt;gravity&gt;&lt;black-hole&gt;&lt;gravitational-waves&gt;" AnswerCount="2" CommentCount="0" ClosedDate="2016-03-01T01:46:04.007" ContentLicense="CC BY-SA 3.0" />
  <row Id="13936" PostTypeId="2" ParentId="13935" CreationDate="2016-02-29T10:49:25.187" Score="1" Body="&lt;p&gt;We &lt;em&gt;do&lt;/em&gt; know what they may be use of. We were able to detect the merger of two black holes, their masses prior to merger, and the mass after merger by observing the gravitational waves produced in the process.&lt;/p&gt;&#xA;" OwnerUserId="10997" LastActivityDate="2016-02-29T10:49:25.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13937" PostTypeId="2" ParentId="13930" CreationDate="2016-02-29T15:34:00.027" Score="0" Body="&lt;p&gt;The purpose of adaptive optics is to reach or approach the &lt;a href=&quot;https://en.wikipedia.org/wiki/Diffraction-limited_system&quot; rel=&quot;nofollow&quot;&gt;diffraction limit&lt;/a&gt; of the system, which is the maximum resolution achievable due to the wave nature of electromagnetic radiation.  The formula for the diffraction limit (in radians) is approximately $\lambda / D$.  For a 30-meter radio telescope observing the 21-centimeter line, this works out to 0.007 radians, or about 24 arcminutes.  This is much larger than the sub-arcsecond diffraction limit of an optical telescope; no matter what you do with your telescope you can't do better than this, so seeing is simply not a factor for single-dish radio astronomy.&lt;/p&gt;&#xA;" OwnerUserId="10914" LastActivityDate="2016-02-29T15:34:00.027" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13938" PostTypeId="2" ParentId="13935" CreationDate="2016-02-29T16:01:30.353" Score="3" Body="&lt;p&gt;It's a &quot;ripple&quot; in spacetime. Imagine the traditional picture of planets as marbles sitting on a sheet--heavier bodies push the sheet down farther and deeper. If you place a bowling-ball on one end of the sheet, the entire sheet is affected, but not instantly. It takes an amount of time for the sheet to be moved--this movement can be seen as a wave, with a wavefront, a speed, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, massive gravitational bodies don't magically appear in space, (that we know of... maybe God is planning to drop a giant marble in our Solar System) but gravitational waves are still generated by any change in placement of mass or any acceleration (Einstein's theories that show that gravity and acceleration are indistinguishable). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you get gravitational waves caused by movements of planets around the sun, etc. That's going to be relatively smooth though, so hard to detect. What we really need is something that creates sharp waves--two giant black holes which are orbiting each other at an incredibly fast RPM is perfect since they're massive and will create a lot of high/low waves as they rotate that we can then detect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we improve the detection technology, we may be able to use gravitational waves to detect all sorts of things besides black holes, it just depends on how much we can measure. The nice thing is that you don't need &quot;line of sight&quot; to measure with a gravity wave. So if God drops a giant marble on the other side of the Sun, we could detect it even though we couldn't see it.&lt;/p&gt;&#xA;" OwnerUserId="158" LastActivityDate="2016-02-29T16:01:30.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13939" PostTypeId="1" CreationDate="2016-02-29T16:13:07.483" Score="-5" ViewCount="421" Body="&lt;p&gt;I am a beginner / novice and I have got very little knowledge about astronomy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please tell me whether 'Planets' revolves around all the 80+ constellations or not&lt;/p&gt;&#xA;" OwnerUserId="11000" LastActivityDate="2016-02-29T16:45:14.170" Title="Does planets revolves around all the 88 constellations?" Tags="&lt;solar-system&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2016-03-02T15:16:29.770" ContentLicense="CC BY-SA 3.0" />
  <row Id="13940" PostTypeId="2" ParentId="13939" CreationDate="2016-02-29T16:45:14.170" Score="2" Body="&lt;p&gt;Planets does not go through all the constellations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The path the Sun and the planets (approximately) follows is called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ecliptic&quot; rel=&quot;nofollow noreferrer&quot;&gt;ecliptic&lt;/a&gt;. The constellation this line goes through is known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Zodiac&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zodiac&lt;/a&gt;, but the current number is actually 13.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A star map, with the ecliptic drawn:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fj69s.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fj69s.png&quot; alt=&quot;star map&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10172" LastActivityDate="2016-02-29T16:45:14.170" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13941" PostTypeId="2" ParentId="13934" CreationDate="2016-02-29T20:43:37.830" Score="2" Body="&lt;p&gt;This can be done by high-contrast imaging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently it is only feasible for a very limited class of systems (e.g. &lt;a href=&quot;https://en.wikipedia.org/wiki/HR_8799&quot; rel=&quot;nofollow&quot;&gt;HR 8799&lt;/a&gt;) since it is difficult to achieve the required angular resolution (0.1 arcsecs for exoplanets at a distance 10 parsec) with today's adaptive optics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Future large telescopes such as the &lt;a href=&quot;https://en.wikipedia.org/wiki/European_Extremely_Large_Telescope&quot; rel=&quot;nofollow&quot;&gt;European Extremely Large Telescope&lt;/a&gt; are designed for direct imaging as well as visible and near-infrared spectroscopy of exoplanets. &lt;/p&gt;&#xA;" OwnerUserId="10568" LastActivityDate="2016-02-29T20:43:37.830" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13943" PostTypeId="1" AcceptedAnswerId="24382" CreationDate="2016-03-01T03:28:27.177" Score="4" ViewCount="889" Body="&lt;p&gt;If we assume a planet the size of Earth with an equally dense atmosphere of nitrogen and methane, rather than oxygen, and it's in the goldilocks zone. Would a hypothetical alien species be able to use oxygen for lanterns and stoves the same way we use methane (or natural gas, etc)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, what primary differences in ratios and/or densities of gasses would have to exist? If not, why not?&lt;/p&gt;&#xA;" OwnerUserId="1768" LastActivityDate="2018-02-05T19:42:38.677" Title="Oxygen in a methane atmosphere akin to methane in oxygen atmosphere?" Tags="&lt;planetary-atmosphere&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13945" PostTypeId="1" CreationDate="2016-03-01T06:45:48.170" Score="-13" ViewCount="510" Body="&lt;p&gt;Can there be life inside a black hole is it possible that we are living inside a massive black hole right now?&#xA;This is because i see no reason why we cant be inside a super massive black hole the only thing is that we cannot escape it?&#xA;I HAVE ALSO HEARD THAT THERE CAN BE WHITE HOLES OF WHICH MATTER OF BLACK HOLE CAN BE ESCAPED?&lt;/p&gt;&#xA;" OwnerUserId="11007" LastActivityDate="2016-03-01T13:30:14.283" Title="Can there be life in black hole?" Tags="&lt;supermassive-black-hole&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2016-03-01T17:03:39.203" ContentLicense="CC BY-SA 3.0" />
  <row Id="13946" PostTypeId="2" ParentId="13945" CreationDate="2016-03-01T06:56:28.387" Score="1" Body="&lt;p&gt;Atoms don't exist inside a black hole. Even a neutron star has enough gravity to squish all the quarks into neutrons. A black hole has even more gravity. We would not see all the normal atoms and molecules and empty space we see around us if we were inside a black hole.&lt;/p&gt;&#xA;" OwnerDisplayName="user10250" LastActivityDate="2016-03-01T06:56:28.387" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13948" PostTypeId="1" AcceptedAnswerId="13950" CreationDate="2016-03-01T08:37:28.297" Score="12" ViewCount="60399" Body="&lt;p&gt;There will be a solar eclipse soon at my area... naturally I want to watch this with my son.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it safe for us to watch the solar eclipse's reflection in the water? There is a swimming pool near my house and I plan to watch it there.. is it safe to do so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember vividly doing this with my dad when I was little, but some website says we are not supposed to watch it in a bucket of water..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is it ok to watch it in the water?&lt;/p&gt;&#xA;" OwnerUserId="11008" LastEditorUserId="11008" LastEditDate="2016-03-02T04:57:03.940" LastActivityDate="2017-08-15T01:03:17.927" Title="Is it safe to watch solar eclipse's reflection in water?" Tags="&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13949" PostTypeId="2" ParentId="13943" CreationDate="2016-03-01T10:32:27.760" Score="2" Body="&lt;p&gt;I agree with Cipher's answer, but just to add, from a certain point of view, Oxygen could be stored and used like a fuel in a situation like that.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as far as aliens doing so, I have some doubt that you can have life (intelligent life if they're using fuel) on a planet with a Nitrogen/CH4 atmosphere cause I don't see how that would actually work and how the resperatory cycles would balance out.  I looked into that a little bit and it gets pretty complicated, so, I don't want to say it's impossible, but I suspect it's unlikely, but, if we take your scenario and work with the assumption that there's intelligent life on a Goldilocks planet that breathes N2/CH4 (maybe throw some CO2 in there), on a rocky planet with oceans, I see no reason why they couldn't use Oxygen as fuel, at least, running &quot;oxygen lines&quot; inside their alien homes to make fire, perhaps to cook with - sure.  It seems like it would be a useful convenience.  You don't gain much energy pulling Oxygen from Silicate rock or Iron Oxide or from water and then using it to burn in a methane atmosphere, so I don't think it would be an energy source, unless they use bacteria to produce oxygen and then, maybe.  Oxygen wouldn't concentrate underground the way hydrocarbons do, so in that sense, it wouldn't be drill-able, but as a transportable &quot;fuel like&quot; substance, it would work.  I see no reason why that couldn't be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we ever set up an astronaut colony on Titan, which, while much farther away from the earth than Mars, it has several advantages, one being, the surface pressure is manageable, so structures wouldn't need to be able to contain a significant fraction of 14.4 PSI.  They'd need to be mostly air tight, but compared to the structural integrity needed to not lose atmosphere to a vacuum, that's easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronauts/colonists on Titan would be able to collect rocky CO2 and H2O from the Moon's surface and use that to grow plants inside and in time, they would probably have an abundance of oxygen and under certain circumstances, that Oxygen could be used in a similar way to fuel.   I think the scenario the way you wrote it is very unlikely, but oxygen being used like fuel on a methane or hydrogen abundant atmosphere, more likely some kind of astronaut colony, there's no reason why that couldn't be done that I can think of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS - That's on the speculative side for an answer here, but that's my thoughts on the question.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-03-01T10:32:27.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13950" PostTypeId="2" ParentId="13948" CreationDate="2016-03-01T12:08:39.620" Score="8" Body="&lt;p&gt;If it's not safe to look at the sun's reflection in the water normally then it's no safer during an eclipse because even if 1% of the sun is showing this is more than enough light to seriously damage your eyes. The best way to view the sun, if you haven't ordered your eclipse glasses yet, is to poke a hole in a piece of paper and project an image of the sun onto another, like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CjvjP.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CjvjP.jpg&quot; alt=&quot;Enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are instructions &lt;a href=&quot;http://www.jpl.nasa.gov/edu/learn/project/how-to-make-a-pinhole-camera/&quot; rel=&quot;noreferrer&quot;&gt;on how to do this here on the JPL website&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10437" LastEditorUserId="332" LastEditDate="2016-12-17T23:13:47.503" LastActivityDate="2016-12-17T23:13:47.503" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13951" PostTypeId="2" ParentId="13883" CreationDate="2016-03-01T12:51:09.330" Score="1" Body="&lt;p&gt;If we move around the black hole speed of the end will become more than initial point because time remains stretched near it.relatively we don't see much difference but in other frame of reference it seems quite stationary .&lt;/p&gt;&#xA;" OwnerUserId="9458" LastActivityDate="2016-03-01T12:51:09.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13953" PostTypeId="1" CreationDate="2016-03-01T16:11:28.310" Score="4" ViewCount="1375" Body="&lt;p&gt;I know it's practical to hand grind a convex spherical mirror and that it's practical to make a &lt;em&gt;concave&lt;/em&gt; parabolic mirror from a spherical one. But as I understand it, the procedure for doing so depend on tracking progress by using interference patters generated by focusing light off the mirror, and that procedure clearly can't be used (unmodified) with a convex mirror.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short of floating the blank in a bath of index of refraction matched optical fluid and apply the concave procedure to the back side, does anyone know of a way to do that sort of grinding?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I've never actually run across a procedure for finishing a hyperbolic mirror for either convex or concave, though I haven't spent much time searching on that one. &lt;/p&gt;&#xA;" OwnerUserId="11011" LastActivityDate="2017-03-08T13:23:37.570" Title="Is it practical to hand grind a convex parabolic or hyperbolic mirror?" Tags="&lt;optics&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13954" PostTypeId="1" CreationDate="2016-03-01T17:51:20.237" Score="5" ViewCount="2542" Body="&lt;p&gt;Can I use a chrome painted dish TV antenna as a parabolic reflector for my &lt;strong&gt;optical&lt;/strong&gt; telescope? Given that glass mirrors are quite expensive, I am thinking of using a Dish TV antenna and coating its reflecting surface with chrome paint to give it a mirror finish. But I couldn't find any instance of it online. I &lt;strong&gt;did&lt;/strong&gt; find several examples of people using a dish antenna for making an amateur radio telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if there is a reason why we can't use it in optical telescope.&lt;/p&gt;&#xA;" OwnerUserId="11012" LastEditorDisplayName="user5434" LastEditDate="2016-10-30T02:38:34.350" LastActivityDate="2018-01-31T07:59:21.650" Title="Dish antenna as parabolic mirror for OPTICAL telescope?" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;photography&gt;&lt;optics&gt;&lt;newtonian-telescope&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13955" PostTypeId="2" ParentId="13954" CreationDate="2016-03-01T18:05:28.363" Score="6" Body="&lt;p&gt;It's not possible I'm afraid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Optical wavelengths (light) are typically of a wavelength under a micron, and an optical surface needs to be accurate to this level or better to be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radio wavelengths are typically 10-20cm or longer, and an adequate reflector can be made with surfaces accurate to a few centimetres (at a guess as I don't know the exact tolerance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also - I once experimented using a frequently mentioned substitute, a convex shaving mirror(!) as a reflecting telescope.  Even that was so poor it was unusable at the lowest possible magnification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(But it is possible to make your own reflector out of reasonably stable glass - it just won't be as big as a satellite dish! Search around for &quot;Amateur telescope making&quot;.)&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2016-03-01T18:05:28.363" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13956" PostTypeId="2" ParentId="13932" CreationDate="2016-03-01T19:21:38.353" Score="3" Body="&lt;p&gt;First of all it has to be stated that it is difficult to have a scientific discussion about multiple universes. The problem with multiple universes is that it can be logically deducted from assumptions and hypotheses, but not tested nor empirically verified. This makes discussions around this topic, like string theory, highly &quot;religious&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are different multiverse theories coming from cosmology, physics and philosophy. One of the theories comes from the cosmological theory of cosmic inflation. Cosmic inflation states that there are always fluctuations on the microscopic level and this results in energy waves that can form into matter and anti-matter pairs. The cosmic inflation now states that at the beginning of our universe there was a cosmological constant or a big vacuum that is by nature unstable and caused the microscopic fluctuations to happen on a cosmic size and thus resulted in a rapid expansion of the early universe. This cosmic inflation is a variation of the (hot) big bang theory, that explains a number of observations in the universe that are not explained by the (hot) big bang theory. What it doesn't explain is that if I have an energy fluctuation at microscopic level, it can happen that this energy wave splits into two matter pairs or an anti-matter/matter pair. Later these pairs annihilate each other and what has happened at microscopic level evens out on macroscopic level. However, when particles appear at microscopic level, what needs to happen to let these particles exist for a longer time to have stars evolve from this matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now back to the multiple universes. The going theory now is that if such a thing can happen to our universe, such a thing can happen in the endless vastness that contains our universe and multiple universes can start and exists next to each other. These multiple universes might also explain that when I have a cosmic fluctuation, I have two universes that come into existence. One with the matter and one with the nati-matter. Bear in mind!!!! These are all philosophical hypotheses and none of it can be verified, falsified or empirically tested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I now have made it logically possible to have multiple universes coming from different cosmic fluctuations, I can now have different universes. For instance, I can have for each matter universe a corresponding anti-matter universe to get the energy balance evened out. Now if I have several universes with each different compositions, I can also have universes where, at least, the laws of nature apply differently. In a universe without dark matter, laws of physics apply differently. Whether the &quot;laws of nature&quot; are totally different in other universes, is difficult to scientifically proof. Especially with theories that are build on our laws of nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you think we're already on thin ice here, philosophers can make it even more unreal. If I have cosmic fluctuations that can happen under different  circumstances, which result in universes with different compositions that then have a different application of the laws of nature, I can equally well argue that there will be universes where certain laws of nature don't exists, while there are some particles that don't exist in that universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to answer your question briefly, there are theories that say there are other universes that have different laws of nature. However, multiverse theories are highly speculative and are difficult to proof. So, whether these theories are right, we'll probably will never know.&lt;/p&gt;&#xA;" OwnerUserId="6599" LastActivityDate="2016-03-01T19:21:38.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13957" PostTypeId="1" AcceptedAnswerId="13966" CreationDate="2016-03-02T03:11:24.017" Score="1" ViewCount="1719" Body="&lt;p&gt;I was just wondering what units I could scale down to (or or shouldn't) for working out values like Local Solar Times, or Local Time etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the papers I'm looking at, I can't see any mention of SI's or ISO's so I'm a little lost as to what to use as a reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ade&lt;/p&gt;&#xA;" OwnerUserId="11016" LastActivityDate="2016-03-02T14:21:53.433" Title="Local Standard Time Meridian - LSTM (Units)" Tags="&lt;space-time&gt;&lt;the-sun&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13958" PostTypeId="2" ParentId="13953" CreationDate="2016-03-02T05:15:27.307" Score="3" Body="&lt;p&gt;If you're asking in regards to testing methods, as indicated in the comments, the simplest setup for interferometrically testing convex conic mirrors is with a Hindle Test, shown below in a figure from the University of Arizona College of Optical Sciences.  This setup can achieve a perfect null after adjusting the reference sphere to be focused at the focus of the test optic - the catch is that the sphere needs to be larger than your test optic, with a hole through it as shown.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U5jbv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U5jbv.png&quot; alt=&quot;Hindle Test&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In industry, it is much more common to use an aperture-stitching interferometer for small quantities of non-research-level optics.  Larger, more precise, and higher quantity aspheres may use a set of nulling optics, or a diffractive/holographic element to create a null wavefront, as covered in better detail in the link below, which is a slideset from U of Arizona's optical fabrication and testing course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://fp.optics.arizona.edu/jcwyant/Short_Courses/SIRA/7-TestingAsphericSurfaces.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://fp.optics.arizona.edu/jcwyant/Short_Courses/SIRA/7-TestingAsphericSurfaces.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're feeling especially ambitious, there is a concept for measuring surface form of a mirror by displaying points on a monitor and using an HD camera to see where the reflections come from, thus telling you the angle of the optical surface at that location.  The data is then integrated to form a full surface map.  In theory, this system could be developed at low cost, with relatively high performance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.osapublishing.org/ao/abstract.cfm?uri=ao-13-11-2693&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.osapublishing.org/ao/abstract.cfm?uri=ao-13-11-2693&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this (or at least some of it) helps!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-J&lt;/p&gt;&#xA;" OwnerUserId="10820" LastActivityDate="2016-03-02T05:15:27.307" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13959" PostTypeId="2" ParentId="13954" CreationDate="2016-03-02T05:27:59.853" Score="6" Body="&lt;p&gt;I would have added this as a comment (not enough rep yet, I'm afraid)...&#xA;To elaborate on Andy's answer, the first reason is that the surface of the satellite dishes are too coarse to form any kind of image.  Polished optical surfaces are smoothed to a polish (generally much smoother than one would achieve in polishing a car, though).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other problems come from the precision of the form - the paraboloid of the satellite dish may have errors on the order of millimeters, because radio waves (which satellites use) don't care about that much error.  In fact, satellite dishes are EXACTLY the same as radio telescopes, except the receiver cone is designed for frequencies used by communication satellites instead of frequencies of interest to astronomy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for your interest, old satellite dishes ARE well-suited to function as rough solar collectors (for a solar oven) or as an acoustic telescope (position a microphone or speaker where the receiver cone normally lies)!&lt;/p&gt;&#xA;" OwnerUserId="10820" LastActivityDate="2016-03-02T05:27:59.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13962" PostTypeId="1" AcceptedAnswerId="13964" CreationDate="2016-03-01T14:33:24.650" Score="3" ViewCount="861" Body="&lt;p&gt;With the recent results of LIGO, I read a lot how limited the detection capabilities where and how one could only expect to measure the most extreme events like black hole mergers and supernovae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get the black hole mergers, since here we see an ever higher frequency gravitational interference of two large bodies. But how can a supernova cause a gravitational wave? All the mass is still there, it is just thrown out  isn't it? So from far away shouldn't the mass still be concentrated on a relatively small place and thus the gravity from the recently exploded star should remain unchanged, right?&lt;/p&gt;&#xA;" OwnerUserId="10632" OwnerDisplayName="choeger" LastActivityDate="2020-12-16T19:19:51.323" Title="How can a supernova cause a gravitational wave?" Tags="&lt;gravity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13963" PostTypeId="2" ParentId="13962" CreationDate="2016-03-02T13:54:20.887" Score="2" Body="&lt;p&gt;When I first read about this (supernovae causing gravitational waves) that was the same thought I had. But I read up about it and found that no supernovae explosion is perfectly spherically symmetric and so therefore the center of mass/energy will move with the explosion, causing a short burst of gravitational waves as the retarded gravitational potential catches up with the new center of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is an example of what these signals would look like if detected by LIGO. They don't produce that increasing frequency signal you mentioned, instead its simply a short sharp change in the strain momentarily (note the timescale along the bottom too, the recent black hole merger event lasted about 0.2s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/axrK5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/axrK5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10437" LastActivityDate="2016-03-02T13:54:20.887" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13964" PostTypeId="2" ParentId="13962" CreationDate="2016-03-02T14:01:08.640" Score="9" Body="&lt;p&gt;The key here is that to produce gravitational waves, the supernova explosion must not have perfect spherical symmetry. Technically, what you need is an &quot;accelerating&quot; gravitational quadrupole moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is quite unlikely that supernovae explosions will be symmetric. Asymmetries can be produced by binary companions, magnetic fields, rotation...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The characteristics of a &quot;burst&quot; source of gravitational waves will be quite unlike the signature of coalescing binary black holes. The magnitude of the signal is expected to follow the collapsing core of the star and reach a peak at something called the &quot;core bounce&quot;, where the equation of state of the gas &quot;stiffens&quot; and a shock wave propagates outwards. The signal is thus likely to gradually increase in amplitude for some tens of milli-seconds, followed by an abrupt spike at the bounce, followed by a &quot;ring-down&quot; as the core (a proto-neutron star) settles down towards a spherically symmetric configuration over the course of a further 10 milli-seconds or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The frequencies of the waves will have a broad range governed by the characteristic frequency implied by $(G\rho)^{1/2}$, which is the inverse of the freefall timescale, where $\rho$ is the density at the time when the GWs are emitted. For a typical core collapse, the densities could range from $10^{14}$ to a few $10^{17}$ kg/m$^3$ over the course of the event, producing GW frequencies of tens to thousands of Hz - exactly in the range at which aLIGO is sensitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because GWs pass through the envelope of the star unimpeded then the GW signature of a supernova probes the very heart of a supernova explosion and should be seen some hours before the visible signature is apparent. On the other hand any neutrino emission from a core collapse supernova should be detected at a very similar time to the GW waves. Any delay here could fix the currently poorly known neutrino mass.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-03-02T14:01:08.640" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13965" PostTypeId="1" CreationDate="2016-03-02T14:05:45.977" Score="7" ViewCount="459" Body="&lt;p&gt;I am writing a work of fiction about a world where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length of a year is 335 days.&#xA;The world has two Moons.&#xA;Moon A is Larger and further away with a 78 day lunar orbit.&#xA;Moon B is smaller and closer with a 31 day orbit.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How often would both moons be full at the same time, overlapping each other like a bullseye?&lt;/li&gt;&#xA;&lt;li&gt;How often would lunar eclipses occur for Moon A?&lt;/li&gt;&#xA;&lt;li&gt;How often would lunar eclipses occur for Moon b?&lt;/li&gt;&#xA;&lt;li&gt;How often would both eclipse at the same time?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is a work of fiction, so liberties can be taken.  I just want it to be close.&lt;/p&gt;&#xA;" OwnerUserId="11021" LastActivityDate="2016-03-07T08:18:49.200" Title="Eclipse Cycle Calculations Needed" Tags="&lt;solar-eclipse&gt;&lt;lunar-eclipse&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13966" PostTypeId="2" ParentId="13957" CreationDate="2016-03-02T14:21:53.433" Score="2" Body="&lt;p&gt;In a time, the word &quot;standard&quot; normally means the civil time, recognized by the government in the place of interest. I'm not sure about other countries, but in general conversation in the USA, &quot;standard&quot; also means that daylight saving time is not in effect. &quot;Daylight&quot; time (as in &quot;Eastern Daylight Time&quot;) means the civil time, recognized by the government, and daylight saving time is in effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These can be expressed in the ISO 8601 format; the standard gives the example &quot;15:27:46+01:00&quot; meaning the time zone is one hour ahead of (east of) Greenwich.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In astronomy, time can also be the local time of the observer, ignoring the existence of time zones. This is local mean time, and is obtained from Universal Time by adjusting for the longitude of the observer. Usually the exact form of Universal Time that would be used would be UT1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Local apparent solar time is based on the actual position of the sun; it is the time given by a sundial. It can be about 15 minutes ahead of or behind local mean time.&lt;/p&gt;&#xA;" OwnerUserId="7385" LastActivityDate="2016-03-02T14:21:53.433" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13967" PostTypeId="2" ParentId="13930" CreationDate="2016-03-02T17:14:27.297" Score="8" Body="&lt;p&gt;In fact, the techniques of adaptive optics are already being used in radio astronomy.  They are implicit in the basic imaging algorithms (e.g., CLEAN) used to produce maps from radio interferometers.  In those cases, they are usually being used to correct for the artificial structure introduced by the way the interferometer samples the sky, rather than for structure imposed by the intervening material.  But at low frequencies (1 GHz and below, certainly) they are also used to correct for the artificial structure imposed on the incoming radio wavefronts as they pass through the ionosphere.  Current large low-frequency instruments (such as the LWA and LOFAR) rely heavily on these methods.  &lt;/p&gt;&#xA;" OwnerUserId="11023" LastActivityDate="2016-03-02T17:14:27.297" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13968" PostTypeId="1" AcceptedAnswerId="13973" CreationDate="2016-03-02T17:36:12.670" Score="2" ViewCount="238" Body="&lt;p&gt;Which planets and moons have we accidentally (or purposely) seeded with life?  I would expect this to be any planet or moon we have sent an unsterilized (or not fully sterilized) lander to, or crashed anything into.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am speaking largely of bacteria and other microorganisms that might have hitched a ride on our spacecraft, or any life that was deliberately taken to and left on another world (which I am not aware).  I know we went to the moon, we at the very least &quot;seeded the moon&quot; temporarily with life (humans) and probably bacteria on the space suits / lander / equipment we sent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wondering where we have &quot;seeded&quot; / brought life (even if it doesn't / didn't survive there).&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="2618" LastEditDate="2017-03-07T00:46:59.903" LastActivityDate="2017-03-07T00:46:59.903" Title="What planets and moons have we seeded with life (if any)?" Tags="&lt;planet&gt;&lt;earth&gt;&lt;life&gt;&lt;natural-satellites&gt;&lt;origin-of-life&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13969" PostTypeId="2" ParentId="13968" CreationDate="2016-03-02T17:44:06.290" Score="1" Body="&lt;p&gt;My best guess (though I don't know about every mission) is that we have at least seeded our moon with life when we visited it.  We have also likely seeded Venus, Mars, and Jupiter with the various probes we have sent to each.  As far as I know, we have only done flybys or put probes in orbit for the other planets and moons, so they have likely not been seeded yet.  I would like clarification / confirmation, and other possible locations.  I know we now generally sterilize our probes, but the procedure could be incomplete, and I don't believe we have always sterilized them in the past.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2016-03-02T17:44:06.290" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13970" PostTypeId="2" ParentId="13965" CreationDate="2016-03-02T18:05:54.343" Score="0" Body="&lt;p&gt;Here is my attempt, &lt;strong&gt;assuming the orbits are circular, and along the same plane&lt;/strong&gt;.  Otherwise, calculations become more complex, and those variables would have to be specified.  Size of the planet and size of the moons would also impact the calculation (e.g. they might never fully eclipse if the planet is fairly small / dense).  Also note that Moon A may eclipse Moon B and Moon B may eclipse Moon A depending on their size.  I do not consider this in my calculations (I only consider the planet eclipsing them).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1.How often would both moons be full at the same time, overlapping each other like a bullseye?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not including the planets orbit, 31 is a prime number, and 78 breaks down to the factors of 2, 7, 7.  Since there is no shared common denominator, it does not appear these will synch up more frequently than 78 * 31 =&#xA;&lt;strong&gt;2418 days&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2.How often would lunar eclipses occur for Moon A?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;About 78 days&lt;/strong&gt; (+- about 78/335 days due to the planet orbiting the sun)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3.How often would lunar eclipses occur for Moon b?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;About 31 days&lt;/strong&gt; (+- about 31/335 days due to the planet orbiting the sun)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;4.How often would both eclipse at the same time?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Same answer as #1, which I calculate at &lt;strong&gt;2418 days&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I welcome refinement / correction of my calculations, this is a &quot;rough attempt&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="1386" LastEditDate="2016-03-02T19:54:43.417" LastActivityDate="2016-03-02T19:54:43.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13972" PostTypeId="2" ParentId="11096" CreationDate="2016-03-03T04:54:38.850" Score="1" Body="&lt;p&gt;In this case, &lt;a href=&quot;http://www.wolframalpha.com/&quot; rel=&quot;nofollow&quot;&gt;Wolfram|Alpha&lt;/a&gt; has you covered.  For example, look at the result for &lt;a href=&quot;http://www.wolframalpha.com/input/?i=which%20constellation%20was%20the%20sun%20in%20Jan%201%201400&quot; rel=&quot;nofollow&quot;&gt;&quot;which constellation was the Sun in Jan 1 1400&quot;&lt;/a&gt;.  That said, I'm not sure this calculate includes the precession of the Earth's rotation axis, since I would've expected it to get my astrological birth sign right in 500 BC.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2016-03-03T04:54:38.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13973" PostTypeId="2" ParentId="13968" CreationDate="2016-03-03T05:31:02.560" Score="5" Body="&lt;p&gt;This would seem to be constrained to planets and moons earth has landed on or impacted at all, as we have no way of telling after the fact whether a lander was perfectly sterilized. That list is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Planets:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Mercury&lt;/em&gt;: &lt;a href=&quot;https://en.wikipedia.org/wiki/MESSENGER&quot; rel=&quot;nofollow&quot;&gt;USA's MESSENGER&lt;/a&gt; in 2015 de-orbited and crashed upon completion of mission. Contamination is unlikely because of high impact and temperatures.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Venus&lt;/em&gt;: &lt;a href=&quot;https://en.wikipedia.org/wiki/Venera&quot; rel=&quot;nofollow&quot;&gt;13 missions by the USSR&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Pioneer_Venus_Multiprobe&quot; rel=&quot;nofollow&quot;&gt;1 by the USA&lt;/a&gt; (1966-1985). Contamination unlikely due to harsh environment.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Mars&lt;/em&gt;: 12 missions, multi-national origins. Contamination plausible.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Jupiter&lt;/em&gt;: &lt;a href=&quot;https://en.wikipedia.org/wiki/Galileo_(spacecraft)#Spacecraft&quot; rel=&quot;nofollow&quot;&gt;The Galileo missions&lt;/a&gt; (2), by the USA Contamination unlikely, because of pressure.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Moons:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Earth's moon&lt;/em&gt;: Far too many to count, myriad landings. Contamination plausible.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Saturn's moon Titan&lt;/em&gt;: &lt;a href=&quot;https://en.wikipedia.org/wiki/Huygens_(spacecraft)&quot; rel=&quot;nofollow&quot;&gt;Huygens probe&lt;/a&gt;, landed by the ESA (2005) Contamination Plausible.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Asteroids/Comets:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Eros&lt;/em&gt;: &lt;a href=&quot;https://en.wikipedia.org/wiki/NEAR_Shoemaker&quot; rel=&quot;nofollow&quot;&gt;2001 landing by USA&lt;/a&gt;. Contamination possible.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Itokawa&lt;/em&gt;: &lt;a href=&quot;https://en.wikipedia.org/wiki/Hayabusa&quot; rel=&quot;nofollow&quot;&gt;2005 landing by Japan&lt;/a&gt;. Contamination possible.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Comet 9P/Tempel 1&lt;/em&gt;: &lt;a href=&quot;https://en.wikipedia.org/wiki/Deep_Impact_(spacecraft)&quot; rel=&quot;nofollow&quot;&gt;2005 landing by USA&lt;/a&gt;. Contamination is unlikely because of high impact speeds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Comet 67P/Churyumov–Gerasimenko&lt;/em&gt;: &lt;a href=&quot;https://en.wikipedia.org/wiki/Rosetta_(spacecraft)&quot; rel=&quot;nofollow&quot;&gt;2014 landing by ESA&lt;/a&gt;. Contamination possible.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11031" LastEditorUserId="11031" LastEditDate="2016-03-03T14:01:31.767" LastActivityDate="2016-03-03T14:01:31.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13974" PostTypeId="1" AcceptedAnswerId="13976" CreationDate="2016-03-03T05:34:11.123" Score="1" ViewCount="258" Body="&lt;p&gt;How far outside the visible spectrum could an optical telescope feasibly operate?&lt;/p&gt;&#xA;" OwnerUserId="11031" LastActivityDate="2016-03-03T18:06:22.223" Title="What is the longest observable wavelength of light using an optical telescope?" Tags="&lt;telescope&gt;&lt;light&gt;&lt;optics&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13975" PostTypeId="1" AcceptedAnswerId="13979" CreationDate="2016-03-03T07:10:31.190" Score="17" ViewCount="1414" Body="&lt;p&gt;As we know, some stars can fuse heavy elements to produce carbon and oxygen after using up hydrogen if they are massive, and the remnant of stars are usually hot, my question is, if a star fuse to produce carbon and oxygen, does it also burn to produce carbon dioxide?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="2365" LastEditDate="2016-03-03T18:04:02.093" LastActivityDate="2016-03-03T18:04:02.093" Title="Do star remnants actually burn?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13976" PostTypeId="2" ParentId="13974" CreationDate="2016-03-03T07:17:49.643" Score="5" Body="&lt;p&gt;I guess it depends what you mean by an optical telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you just mean any telescope that can record images at visible light wavelengths, then it is possible to use these (with the appropriately cooled instruments) to make observation at so-called mid-infrared wavelengths of $\sim 20$ $\mu$m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See for example on the Gemini telescopes&#xA;&lt;a href=&quot;https://www.gemini.edu/sciops/instruments/midir-resources&quot;&gt;https://www.gemini.edu/sciops/instruments/midir-resources&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The background noise at such wavelengths is extremely high - the telescope and sky look hot at these wavelengths - so these instruments have largely been superseded by space observatories like Spitzer and WISE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost all large &quot;optical&quot; telescopes are routinely used to make observations out to $2.4\ \mu$m (the K band). &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-03-03T07:17:49.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13977" PostTypeId="2" ParentId="13965" CreationDate="2016-03-03T07:49:20.323" Score="3" Body="&lt;p&gt;Frankly, four bodies do not periodically line up, even if the orbits are planar, unless there exist a common multiplum. Because all your periods are &lt;strong&gt;integers&lt;/strong&gt;, that holds, but be aware that your double alignments, like the double solar eclipse or double lunar eclipse is not periodic if one of the periods does not &lt;strong&gt;exactly&lt;/strong&gt; follow that integer system. (so it does not hold for 31.01 days, for example. How you handle quazi-periods is talked about &lt;a href=&quot;https://physics.stackexchange.com/a/232918/102747&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;How often would both moons be full at the same time, overlapping each other like a bullseye?  &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Generally non-periodic, but can only happen at the &lt;a href=&quot;https://en.wikipedia.org/w/index.php?title=Orbital_period&amp;amp;redirect=no&quot; rel=&quot;nofollow noreferrer&quot;&gt;synodic period&lt;/a&gt; of the two moons, that is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Iu5t9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Iu5t9.png&quot; alt=&quot;synodic&quot;&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{2418}{47}$ for the two moons. However, that must also happen when both of them is on the same line as the Sun, in other words when The relative synodic period between one of the moons and the sun, and the moons' synodic period both divides the requested period. For the inner moon and the Sun, the synodic period is $\frac{10385}{304}$. Since 47 does not have any factors in common with 304, the first common multiplum appears at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{10385}{304} \cdot \frac{2418}{47} = \frac{12555465}{7144}$$, or 1757.484 days. (@Jonathan gets 2418, because his model consider the Sun to stay static in the sky, but it does in fact revolve around one time a year.) The exact same period is also valid for your &lt;strong&gt;question 4&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;How often would lunar eclipses occur for Moon A?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, the answer is the synodic period of the Sun and the moon, already stated to be $\frac{10385}{304}$, or 34.161 days. That is a little bit longer than the Moon's orbital period, because the planet has moved a little in its orbit, slightly changing the Sun's position in the sky.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;How often would lunar eclipses occur for Moon B?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Synodic period here too, $\frac{26130}{257}$ = 101.673 days.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-03-07T08:18:49.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13979" PostTypeId="2" ParentId="13975" CreationDate="2016-03-03T10:27:53.203" Score="16" Body="&lt;p&gt;Stars don't &quot;burn&quot;, they undergo &lt;em&gt;nuclear&lt;/em&gt; reactions that do not involve atoms and chemistry at all. The temperatures in the interiors of stars, and certainly in the interiors of stellar remnants like C/O white dwarfs, are &lt;em&gt;far&lt;/em&gt; too hot (millions of degrees) for electrons to bind to nuclei and far too hot for molecules to survive. The energy required to break up carbon dioxide is 5.5 eV, which is easily provided unless a gas is cooled well below 10,000K. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only parts of a star where chemistry can occur is in the outer atmospheres where temperatures can drop to thousands of degrees and where atoms and partially ionised atoms can exist. Here, yes, then it is possible for carbon atoms and oxygen atoms to interact, but that would mostly produce carbon &lt;em&gt;mon&lt;/em&gt;oxide. This molecule is produced and can survive between temperatures of about 1100K and 3500K in the atmospheres of cool M-dwarfs and brown dwarfs and the atmospheres of red giant stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These chemical reactions are utterly negligible in terms of their energetics compared with the fusion reactions that power a star.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-03-03T10:36:13.093" LastActivityDate="2016-03-03T10:36:13.093" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13981" PostTypeId="2" ParentId="13884" CreationDate="2016-03-03T13:32:52.260" Score="1" Body="&lt;p&gt;According to Paul Bourke's page on &lt;a href=&quot;http://paulbourke.net/dome/screencapture/&quot; rel=&quot;nofollow&quot;&gt;screen capture&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;While only the commercial version of Celestia supports fisheye, it also does not support warping of fisheye (although some versions have been created that do). The solution outlined here has been tested successfully with a fisheye supported version of Celestia.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems that Celestia in its free version is not providing any support for dome projection and the commercial version requires tweaking before you can use its support for fisheye projection and convert it to Spherical Mirror Projection.&lt;/p&gt;&#xA;" OwnerUserId="10949" LastActivityDate="2016-03-03T13:32:52.260" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13982" PostTypeId="1" AcceptedAnswerId="13984" CreationDate="2016-03-03T14:25:39.737" Score="4" ViewCount="97" Body="&lt;p&gt;I was reading the paper about LIGO's detection of gravitational waves, and I wanted to know what the Schwarzschild Radius did. For example, I feel like the &lt;a href=&quot;https://www.ligo.caltech.edu/video/ligo20160211v4&quot;&gt;simulation here&lt;/a&gt; is oversimplified, and that reality wouldn't actually have that simple bump together of the two radii.&lt;/p&gt;&#xA;" OwnerUserId="11031" LastActivityDate="2016-03-03T16:13:22.940" Title="When black holes such as those detected by LIGO merge, in what ways does the Schwarzschild Radius warp?" Tags="&lt;gravity&gt;&lt;black-hole&gt;&lt;space-time&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13983" PostTypeId="1" CreationDate="2016-03-03T15:07:27.657" Score="2" ViewCount="45" Body="&lt;p&gt;We all have heard about the problem with space junk in low earth orbit. NASA has a plan to put an asteroid in orbit around the moon. Would it be possible to equip propulsion systems onto an asteroid in order to maneuver it into an elliptical orbit around the earth, with the goal of collecting this junk?&lt;/p&gt;&#xA;" OwnerUserId="11036" LastEditorUserId="11031" LastEditDate="2016-03-03T15:45:49.033" LastActivityDate="2016-03-03T18:00:18.543" Title="Would it be possible to equip an asteroid to collect space junk in Earth's orbit?" Tags="&lt;orbit&gt;&lt;earth&gt;&lt;asteroids&gt;&lt;orbital-migration&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2016-03-03T18:16:08.150" ContentLicense="CC BY-SA 3.0" />
  <row Id="13984" PostTypeId="2" ParentId="13982" CreationDate="2016-03-03T16:13:22.940" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=Zt8Z_uzG71o&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; is a better simulation SXS did for the LIGO detection (still frame below), the two event horizons (Schwarzschild Radii as you say) just simply come together and merge quite effortlessly like two bubbles in a bath coming together and merging into one. I don't know if you were expecting something more dramatic than this but there is no reason to suggest its a violent event.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/17WLL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/17WLL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10437" LastActivityDate="2016-03-03T16:13:22.940" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13985" PostTypeId="2" ParentId="13983" CreationDate="2016-03-03T16:41:49.473" Score="5" Body="&lt;p&gt;This is plausible, and might even be a good idea if used right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, NASA has been working on plans for an asteroid redirect mission, called &lt;a href=&quot;https://en.wikipedia.org/wiki/Asteroid_Redirect_Mission&quot;&gt;ARM&lt;/a&gt;. While it remains to be seen whether this will be approved, and they plan to put it into lunar orbit, this is a hypothetical scenario anyways so I'll ignore that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, putting an asteroid into earth orbit is a little bit of a difficult subject. Sure, you could do it, but messing up would net you a whole lot of problems. I presume it would have to go into high earth orbit if it goes into orbit at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This rules out using the asteroid directly like a satellite, but that would have been impractical even if it was sanctioned by everyone with enough influence to have an opinion on earth. From there the satellite would be best used as a base of operations, from which you could send drones to collect junk and keep refueling stations. I'd also include a telescope or two, but that's personal preference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a list of targets though, &lt;a href=&quot;http://stuffin.space/?search=&quot;&gt;here&lt;/a&gt; would be a good place to look.&lt;/p&gt;&#xA;" OwnerUserId="11031" LastEditorUserId="11031" LastEditDate="2016-03-03T16:46:54.967" LastActivityDate="2016-03-03T16:46:54.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13986" PostTypeId="1" AcceptedAnswerId="13988" CreationDate="2016-03-03T16:52:40.193" Score="2" ViewCount="244" Body="&lt;p&gt;The LIGO complex is very large, but I assume that there is a very small chamber that actually does the science since they are measuring very small sheers. How big around is the actual evacuated chamber?&lt;/p&gt;&#xA;" OwnerUserId="11039" LastEditorUserId="11031" LastEditDate="2016-03-03T17:42:39.920" LastActivityDate="2016-03-03T17:42:39.920" Title="What are the dimensions of LIGO Detector evacuated chambers?" Tags="&lt;black-hole&gt;&lt;space-time&gt;&lt;gravitational-waves&gt;&lt;instruments&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13987" PostTypeId="2" ParentId="13986" CreationDate="2016-03-03T17:03:52.387" Score="2" Body="&lt;p&gt;LIGO detectors have two 4 km arms, and everything in those arms in suspended in a vacuum. The actual dimensions of those arms I was unable to find, but &lt;a href=&quot;https://physics.aps.org/featured-article-pdf/10.1103/PhysRevLett.116.061102&quot; rel=&quot;nofollow&quot;&gt;the published paper here&lt;/a&gt; has more information for you to peruse.&lt;/p&gt;&#xA;" OwnerUserId="11031" LastActivityDate="2016-03-03T17:03:52.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13988" PostTypeId="2" ParentId="13986" CreationDate="2016-03-03T17:11:22.703" Score="12" Body="&lt;p&gt;The arms are $4\,\mathrm{km}\,\times\, 1.2\,\mathrm{m}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the &lt;a href=&quot;https://www.ligo.caltech.edu/page/vacuum&quot;&gt;LIGO webpage&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The 1.2 m diameter beam tubes were created in 19-20 m-long segments, rolled into a tube with a continuous spiral weld. While a mathematically perfect cylinder will not collapse under pressure, any small imperfection in a real tube would allow it to buckle (a crushed vacuum tube would be catastrophic). To prevent collapse, LIGO's tubes are supported with stiffener rings that provide a significant layer of resistance to buckling under the extreme pressure of the atmosphere. The tubes must withstand these stresses for at least 20 years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2016-03-03T17:11:22.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13989" PostTypeId="2" ParentId="13983" CreationDate="2016-03-03T18:00:18.543" Score="0" Body="&lt;p&gt;Low Earth Orbital space is a very big volume. And even though there are lots of pieces of junk, they are distributed throughout this large volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pieces of junk are all move at orbital velocity. That means they are moving 8 times faster than a bullet. To catch a piece of orbital junk you would need to match your orbit to the junk, to get your relative velocity down to close to zero. That would require lots and lots of fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm not sure how having an asteroid helps here. Perhaps you imagine the junk hitting the asteroid, and being destroyed.  The result of junk hitting the asteroid would be a shower of small fragments - not great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would still have the problem steering the asteroid into a meeting with a piece of junk. Asteroids are big and heavy and hard to manoeuver, you would need fuel, and lots of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short this does not seem to be a feasible solution to the very real problem of space junk.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-03-03T18:00:18.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13990" PostTypeId="2" ParentId="13974" CreationDate="2016-03-03T18:06:22.223" Score="3" Body="&lt;p&gt;It depends on the telescope.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the telescope is fully reflective (no transparent glass elements), it should be able to operate well-beyond visual wavelengths, with its transmission limited primarily on your mirror costing material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.thorlabs.com/images/TabImages/Concave_metallic_mirror_8_degree_AOI.gif&quot; rel=&quot;nofollow&quot;&gt;http://www.thorlabs.com/images/TabImages/Concave_metallic_mirror_8_degree_AOI.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any telescope with refractive (transparent glass) elements however, will quickly lose performance outside of the visual range due to transmission, but also from how the glass index changes with wavelength, which causes non-visual light to be defocused (an effect which increases the further you get from visual wavelengths).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally speaking, the defocused issue of refractive telescopes can be compensated if you have enough focus travel. You will see &quot;chromatic aberration&quot; becoming much more significant very quickly, as telescopes are only designed to correct this over visual wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, the limiting factor of refractive telescopes is the transmission of the glass in the lenses.  Many glasses will transmit reasonably well from 400-2000nm, though some have worse performance and some have better.&lt;/p&gt;&#xA;" OwnerUserId="10820" LastActivityDate="2016-03-03T18:06:22.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13991" PostTypeId="1" AcceptedAnswerId="14020" CreationDate="2016-03-03T18:38:13.473" Score="1" ViewCount="353" Body="&lt;p&gt;I recently inherited some Fortran 77 code that calls the &lt;a href=&quot;http://stsdas.stsci.edu/documents/SUG/UG_49.html&quot; rel=&quot;nofollow&quot;&gt;IMFORT&lt;/a&gt; interface to write data out to an imh and pix files. I could not get this to compile on a modern compiler and it is not supported. So I decided to set off on writing a Fits writer in Fortran. I was able to make one but is ugly and does not properly handle the end card. I tried googling for a Fortran Fits writer but came up empty. Is there a library for astronomical Frotran codes that contains Fits writer exist out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The code may be better suited for stack overflow, but this is specialized to astronomy since no other science seems to use fits, so I figured I would be better off starting here.&lt;/p&gt;&#xA;" OwnerUserId="11039" LastActivityDate="2020-06-13T12:43:44.037" Title="Fortran Fits writer/reader" Tags="&lt;data-analysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13994" PostTypeId="1" CreationDate="2016-03-03T21:25:46.263" Score="6" ViewCount="764" Body="&lt;p&gt;According to Wikipedia, &quot;On 20 January 2016, researchers Konstantin Batygin and Michael E. Brown at Caltech announced calculation-based evidence of a massive ninth planet in the Solar System.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If visual evidence or another researcher &lt;strong&gt;confirms&lt;/strong&gt; the existence of Planet Nine, does the &lt;em&gt;actual&lt;/em&gt; discovery warrant naming rights, or would this go to our Caltech friends?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know from &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10213/naming-of-the-planets-of-the-solar-system&quot;&gt;this question's answer&lt;/a&gt;, that: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Since the beginning of the 20th century the IAU is setting the standards for naming celestial bodies including planets. However, the &lt;strong&gt;IAU doesn't set the names&lt;/strong&gt;, it only sets the standards.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Does the 'discoverer,' or the 'confirmer' get to name (theoretical) Planet Nine?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7370" LastActivityDate="2016-11-05T11:03:49.860" Title="Who gets to name Planet Nine if it is confirmed?" Tags="&lt;planet&gt;&lt;naming&gt;" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13995" PostTypeId="2" ParentId="13994" CreationDate="2016-03-03T21:45:29.783" Score="4" Body="&lt;p&gt;The cited answer is correct insofar as the IAU is not the body that makes names of celestial bodies official. However, it's authority is implicit because it is recognized by the astronomical community and the public as the foremost authority on astronomical naming, and thus the names the IAU adopts are the ones most commonly used in the scientific community (excluding objects referred to by their catalog numbers).&lt;/p&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.iau.org/public/themes/naming/&quot; rel=&quot;nofollow noreferrer&quot;&gt;one of their information pdfs&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The IAU has been the official arbiter of planetary and satellite naming since its inception in 1919. The IAU’s decisions are officially adopted by the nearly 11 000 professional astronomers who are its members, coming from more than 90 countries.&lt;/p&gt;&#xA;&lt;p&gt;. . .&lt;/p&gt;&#xA;&lt;p&gt;The IAU does not consider itself as having a monopoly on the naming of celestial objects — anyone can in theory adopt names the way they choose. However, given the publicity and emotional investment associated with these discoveries, worldwide recognition is important and the IAU offers its unique experience for the benefit of a successful public naming process (which must remain distinct, as in the past, from the scientific designation issues).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The IAU may reach out to the public for naming suggestions, or it may taken case-by-case suggestions by individuals. So while Batygin and Brown can call the ninth planet whatever they want - &lt;a href=&quot;http://www.nature.com/news/evidence-grows-for-giant-planet-on-fringes-of-solar-system-1.19182&quot; rel=&quot;nofollow noreferrer&quot;&gt;they've informally referred to it as &amp;quot;Phattie&amp;quot;&lt;/a&gt; - public consensus will largely rest on the results of the IAU's decision, which could, of course, end up taking a suggestion of the pair.&lt;/p&gt;&#xA;&lt;p&gt;Anyone can come up with a suggestion, although &lt;a href=&quot;http://xkcd.com/1253/&quot; rel=&quot;nofollow noreferrer&quot;&gt;sometimes that's a terrible idea&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-03-04T01:06:29.890" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13996" PostTypeId="1" CreationDate="2016-03-03T21:54:16.283" Score="0" ViewCount="237" Body="&lt;p&gt;I know of two methods for finding exoplanets: the transit method and the radial velocity method. These two methods work as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Transit&lt;/strong&gt;: we observe stars and watch for when a planet obstructs the light from the star.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Radial Velocity&lt;/strong&gt;: detecting the planet by observing the motion of the star and using Kepler's law.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Are these methods also used for discovering black-holes? What are the pros and cons of these methods, I think that the con of transit is that it's difficult to see the black hole. A pro is that it doesn't matter how many stars there are around the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the Radial Velocity method, a pro is that it's easy to detect but requires exactly one companion star (I used this &lt;a href=&quot;http://www.stronggravity.eu/how-to-measure-black-hole-mass/&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="11044" LastEditorUserId="2365" LastEditDate="2016-03-11T18:40:26.960" LastActivityDate="2016-07-22T22:43:24.210" Title="Blackhole Finding Techniques" Tags="&lt;black-hole&gt;&lt;astrophysics&gt;&lt;planetary-transits&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13997" PostTypeId="2" ParentId="13994" CreationDate="2016-03-03T21:54:24.447" Score="3" Body="&lt;p&gt;This would probably be a unique situation. Never in the era of the IAU has the existence of a space body been predicted prior to discovery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard rule is that priority in naming goes to the team that discovers an object. In the scenario you describe, discovery would mean locating the object and determining its orbit. The statistical evidence for a planet nine is interesting, but far from overwhelming, and the team cannot be said to have discovered a planet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However as this would be a unique situation, the IAU may choose to ignore its own protocols, and to consult more widely than normal, to seek a consensus among astronomers on the name. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-03-03T21:54:24.447" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13998" PostTypeId="1" CreationDate="2016-03-04T02:06:28.147" Score="2" ViewCount="131" Body="&lt;p&gt;I know some measures must exist, but I have no idea what the programs are, what organizations are involved, or how effective they are.&lt;/p&gt;&#xA;" OwnerUserId="11031" LastActivityDate="2016-03-16T05:23:42.157" Title="What measures are currently in place to detect asteroids on possible collision courses with earth?" Tags="&lt;solar-system&gt;&lt;earth&gt;&lt;asteroids&gt;&lt;near-earth-object&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13999" PostTypeId="1" AcceptedAnswerId="14002" CreationDate="2016-03-04T02:17:26.683" Score="8" ViewCount="525" Body="&lt;p&gt;I understand basically that the universe is homogenous (looks the same from every point) and I was told in my astro class that it's supposed to have no center, but how that works boggles my mind. I can't get over the feeling that from some point in the universe a civilization could look in a certain direction and see a sign indicative of the direction towards the edge or the center, like younger galaxies or a lack of galaxies moving a certain direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Thanks to Sir Cumference for highlighting some things that are unclear about my question. Sure, to a certain extent you might be able to think of the universe as being without a center, but if you think of a ball growing larger from nothing, and you pick a random point in the growing sphere, you can always say that the point is further from some points on the sphere than others, or in other words, that there is a direction towards a &quot;center&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other details you can give that are related at all would be extremely welcome.&lt;/p&gt;&#xA;" OwnerUserId="11031" LastEditorUserId="9301" LastEditDate="2016-03-05T14:04:08.547" LastActivityDate="2016-07-14T21:43:05.897" Title="How does the concept of a universe with no center work?" Tags="&lt;galaxy&gt;&lt;universe&gt;&lt;space&gt;&lt;space-time&gt;&lt;expansion&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14000" PostTypeId="2" ParentId="13998" CreationDate="2016-03-04T02:39:45.417" Score="4" Body="&lt;p&gt;&lt;a href=&quot;http://neo.jpl.nasa.gov/&quot; rel=&quot;nofollow&quot;&gt;This site&lt;/a&gt; is an asteroid tracking program started up by NASA. It seems to be fairly effective at detecting asteroids. Read up on &lt;a href=&quot;http://www.jpl.nasa.gov/news/news.php?feature=4816&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; for further information about NASA's plan.&lt;/p&gt;&#xA;" OwnerUserId="7926" LastEditorUserId="11031" LastEditDate="2016-03-04T03:20:21.713" LastActivityDate="2016-03-04T03:20:21.713" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14001" PostTypeId="2" ParentId="13803" CreationDate="2016-03-04T03:16:39.870" Score="2" Body="&lt;p&gt;As you may have guessed, this question is of great interest to the LIGO team. Simultaneously with the publication of the paper you mentioned announcing the discovery, the LIGO team submitted a number of companion papers with further details about the discovery, and predictions.  One of these addresses your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1602.03842&quot; rel=&quot;nofollow&quot;&gt;The Rate of Binary Black Hole Mergers Inferred from Advanced LIGO Observations Surrounding GW150914&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their event rate estimation method considers both GW150914, and another significantly weaker (and less statistically significant) event.  They consider a number of models for how the event rate might depend on system properties, and ask what the observations of GW150914 and the other candidate event imply for the overall rate.  The results vary from model to model, but they chose models they felt could roughly bracket astrophysically plausible behavior.  As summarized in their abstract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Considering only GW150914, assuming that all BBHs in the universe have the same masses and spins as this event, imposing a false alarm threshold of 1 per 100 years, and assuming that the BBH merger rate is constant in the comoving frame, we infer a 90% credible range of $2-53 \, \mathrm{Gpc}^{-3} \,&#xA;\mathrm{yr}^{-1}$ (comoving frame). Incorporating all triggers that pass the search threshold while accounting for the uncertainty in the astrophysical origin of each trigger, we estimate a higher rate, ranging from $6-400 \,&#xA;\mathrm{Gpc}^{-3} \, \mathrm{yr}^{-1}$ depending on assumptions about the BBH mass distribution. All together, our various rate estimates fall in the conservative range $2-400 \, \mathrm{Gpc}^{-3} \, \mathrm{yr}^{-1}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that the paper is submitted, not published, i.e., still under peer review.  Speaking as someone with expertise in such calculations, some aspects of the method look fishy to me, so I think it's worth checking back on the article in a few weeks for revisions.  It doesn't take fancy methodology to see that the order of magnitude here (a few to ~100 per cubic gigaparsec per year) is in the right ballpark.  But the paper presents a methodology that could make more detailed and more precise estimates and predictions as data accumulate, so it's important to make sure the methodology is sound.&lt;/p&gt;&#xA;" OwnerUserId="11047" LastActivityDate="2016-03-04T03:16:39.870" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14002" PostTypeId="2" ParentId="13999" CreationDate="2016-03-04T03:20:58.330" Score="12" Body="&lt;p&gt;When we talk about the universe, we are really talking about one of two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The observable universe, which is everything we can possibly see.&lt;/li&gt;&#xA;&lt;li&gt;The Universe, which is everything that has ever existed, currently exists, and will exist.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The observable universe has its own center, usually the Earth. It is a spherical region of everything that we can see, essentially anything whose light has reached us. We usually refer to this when we say things like &quot;there are $10^{86}$ atoms in the universe.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, everyone has their own observable universe, and it can change depending on where you are. An exoplanet far away has its own observable universe, and can receive light from different places. Essentially, you are the center of your own observable universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume you're talking about the latter, though. The Universe (notice the capital &quot;U&quot;) is all of space and time and its contents. Anything that has existed, will exist, and currently exists is part of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Universe is thought to be infinitely large, so it can't have a center. The center of something is the point equidistance from the edges, but if something spans infinitely long, it would just keep going. It wouldn't have an edge, and thus it wouldn't have a center. You couldn't find the point equidistant from the edges if it just spans infinitely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might ask, &quot;then where did the Big Bang start? Surely it must've been the center of the Universe, right?&quot; Well, you can say the Big Bang happened everywhere. Before the Big Bang, matter filled most of the Universe's emptiness. It was essentially dense, and so got extremely hot to the point where no hadrons could form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is thought that this temperature caused space itself to expand. Essentially, more space was created in between all the matter, until everything was able to cool down. This is what we call the Big Bang. It didn't happen at a certain point, but rather it happened everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: I understand your confusion. Let me just clear some common misconceptions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Universe is not like a ball. Rather, you can think of it like a flat grid, and its &quot;expansion&quot; just means that the distances between objects on the grid are getting larger. In essence, more space is being created between the objects. That's what we mean by expansion — that objects are moving away from each other, since more space is being created between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an easy analogy: imagine you are walking your dog. Suddenly, the ground begins expanding between you. You and your dog will separated and continue receding away from each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's essentially happening everywhere: space is expanding between everything, so we are drifting away from other galaxies. The Universe is infinite, and we can constantly drift apart from other objects because space is being created in between us. Here's a GIF I made that might help you get it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/amQeZ.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/amQeZ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see how the galaxies drift apart as the space between them increases. And this happens everywhere in the Universe. So let's keep that in mind and clear up the Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Big Bang did not happen at a single point, nor did the Universe begin at a single point. The Universe is, and has always been, infinite. The Big Bang was just when the Universe's expansion really began — that is, when objects started drifting away from each other. The Universe was still infinite, but there was less space between the matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This density caused the Universe to get extremely hot and expand. So here, space itself was distorted and began to expand. More and more space was created between the matter, and still is now (although now it is mainly due to dark energy instead of heat).&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-03-04T08:08:22.080" LastActivityDate="2016-03-04T08:08:22.080" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14003" PostTypeId="1" CreationDate="2016-03-04T04:33:59.173" Score="3" ViewCount="241" Body="&lt;p&gt;I've heard from a number of people that the Universe is infinite, and from my perspective, I can't see how this can actually be known, especially given that the Universe started out a finite size and grows at a finite rate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Regarding the possible similarity of my question: I want to know how&#xA;  it could be and how we would know, rather than if it is. This is&#xA;  significantly different.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11031" LastEditorUserId="9301" LastEditDate="2016-03-05T14:03:15.903" LastActivityDate="2016-05-02T14:38:22.783" Title="How could we tell if the Universe is infinite?" Tags="&lt;universe&gt;&lt;cosmology&gt;&lt;space-time&gt;&lt;expansion&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2016-03-05T15:00:00.780" ContentLicense="CC BY-SA 3.0" />
  <row Id="14006" PostTypeId="2" ParentId="14003" CreationDate="2016-03-04T08:17:21.983" Score="4" Body="&lt;p&gt;We don't know for sure, but it certainly fits into our theories. There is, of course, no way to actually test if the Universe is infinite, but right now we think it is likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you read my updated answer on your other post, the Universe has always been infinite in size. I explain over there how it actually works: space is created in between everything, and thus one could say the Universe is expanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These objects can actually drift away from each other faster than the speed of light. That is, light from them eventually won't make it to us, since they'll be drifting away too quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this doesn't actually go against Einstein's theory that the speed of light is the fastest thing in the Universe. Einstein said that nothing can travel through space faster than light — but here, space itself is actually being created between the objects. Distances are increasing because space itself is dilating, and thus we can drift apart from other objects faster than light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really, there is no limit (as far as I know) to how fast we can drift away. Farther objects will keep drifting faster and faster away, since our gravity has a much weaker effect on them.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-05-02T14:38:22.783" LastActivityDate="2016-05-02T14:38:22.783" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14007" PostTypeId="1" CreationDate="2016-03-04T11:04:43.433" Score="-3" ViewCount="969" Body="&lt;p&gt;Since the universe is always expanding, that means there's an edge right? So what are some theories for outside of that growing edge.&lt;/p&gt;&#xA;" OwnerUserId="9827" LastActivityDate="2016-03-07T20:15:14.870" Title="What happens outside of the universe?" Tags="&lt;universe&gt;&lt;space&gt;&lt;expansion&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2016-03-06T22:03:56.073" ContentLicense="CC BY-SA 3.0" />
  <row Id="14008" PostTypeId="1" CreationDate="2016-03-04T12:25:54.030" Score="4" ViewCount="119" Body="&lt;p&gt;What will the benefits and advantages be of the upcoming generation of telescopes such as LSST, E-ELT and JWST? And more broadly where will this leave astronomy in 10-20 years time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't a direct duplicate of (&lt;a href=&quot;https://astronomy.stackexchange.com/questions/884/what-are-the-next-planned-space-telescopes&quot;&gt;What are the next planned space telescopes?&lt;/a&gt;) because I want to know what the aims and benefits of such telescopes are, and what new ones might have been planned since.&lt;/p&gt;&#xA;" OwnerUserId="10437" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-03-04T13:17:44.507" Title="3000th Question: What does the future of astronomy look like with the next generation of telescopes?" Tags="&lt;telescope&gt;&lt;observational-astronomy&gt;&lt;future&gt;" AnswerCount="0" CommentCount="6" ClosedDate="2016-03-04T14:42:13.137" ContentLicense="CC BY-SA 3.0" />
  <row Id="14009" PostTypeId="1" CreationDate="2016-03-04T15:09:38.620" Score="1" ViewCount="3813" Body="&lt;p&gt;I know the sun doesn't go down but we are moving and is it true that the sun goes down to the west? I've read on many webpages that this is not entirely true.&lt;/p&gt;&#xA;" OwnerDisplayName="user11055" LastActivityDate="2016-04-05T03:43:18.503" Title="Is it true that the sun goes down to the west?" Tags="&lt;the-sun&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14010" PostTypeId="1" CreationDate="2016-03-04T16:14:47.910" Score="7" ViewCount="6720" Body="&lt;p&gt;I was wondering if and how astronomers have/had detected the molecule Tetrahydrocannabinol (THC), commonly known for it's effects in marijuana, in space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had previously seen a hoax regarding the drug being detected on an exoplanet, but I wanted to know if there was any possibility of it.&lt;/p&gt;&#xA;" OwnerUserId="11031" LastEditorUserId="11031" LastEditDate="2016-03-04T17:03:15.880" LastActivityDate="2021-03-05T17:59:23.873" Title="Has THC been detected in space?" Tags="&lt;telescope&gt;&lt;space&gt;&lt;astrochemistry&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14012" PostTypeId="2" ParentId="14007" CreationDate="2016-03-04T17:37:47.220" Score="1" Body="&lt;p&gt;There are several theorie of what is outside of the universe. The problem with all these theories is that there is no empirical way to falsify or proof these theories and therefore have a high level of speculation and what theory you think best suites your believes. The theories can be classified into single universe theories and multiverse theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The single universe theories state that our universe lies in a vast / infinite universe that is much larger than what we can see now. We are limited in seeing beyond the hubble volume, since the speed of light prevents us from seeing things beyond the distance that light has travelled since our known universe started to exist. There are several theories on what lies beyond the hubble volume, which vary from nothing to dark matter structures that have a gravitational pull on the observable universe. In these theories the universe is not expanding, but the hubble volume is expanding with the age of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The multiverse universe is actually a variation of the single universe theory. It also states that there is a large vastness, however, filled with multiple universes. Either in a single universe there are bubbles of universes where each bubble is created by a big bang. The other is spun out of the string theory, which states that there are always energy ripples in a perfect vacuum, which can lead to the existence of multiple universes. The theory is &lt;a href=&quot;https://astronomy.stackexchange.com/questions/13932/physical-laws-in-other-universes/13956#13956&quot;&gt;here&lt;/a&gt; further explained. Another theory is the fecund theory, which states that each black hole in the universe spawns a new universe with slightly different physical laws. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to answer your question concisely. &quot;What happens outside of the universe?&quot; Whatever you believe.&lt;/p&gt;&#xA;" OwnerUserId="6599" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-03-07T20:15:14.870" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14014" PostTypeId="2" ParentId="13915" CreationDate="2016-03-04T17:52:44.840" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How have we determined that it is some underlying grid or space itself that is expanding rather than the objects just moving apart on a static grid of spacetime?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;General relativity tells us that the geometry of spacetime is dynamic, being affected by the matter content and its motion. So a fixed, static background spacetime is by its nature some sort of idealized edge case at best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, let's try it. Imagine that in flat, static Minwkoski spacetime with spherical coordinates $(T,R,\theta,\phi)$, there is an expanding spherically symmetric cloud of galaxies expanding from a center at $T = 0$, each with some velocity $v$, such that the galaxies have a negligible effect on the background geometry or each others' velocity. Thus the radial coordinate of each galaxy is $R = vT$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Parametrizing by the rapidity $\eta$ at the given radial coordinate ($v = \tanh \eta$) and the time measured by the galaxy $t$ as given by special-relativistic time dilation ($t = T/\gamma = T/\cosh\eta$), in those coordinates, the Minwkoski metric becomes&#xA;$$\begin{eqnarray*}&#xA;\mathrm{d}s^2 &amp;amp;=&amp;amp; -\mathrm{d}T^2 + \mathrm{d}R^2 + R^2\mathrm{d}\Omega^2\\&#xA;&amp;amp;=&amp;amp; -\mathrm{d}t^2 + t^2\left(\mathrm{d}\eta^2 + \sinh^2 \eta\,\mathrm{d}\Omega^2\right)\text{,}&#xA;\end{eqnarray*}$$&#xA;which is a linearly expanding spatially hyperbolic universe where $\eta$ plays the role of the radial coordinate (up to some dimensionful factor, i.e., $r = r_0\eta$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We sure didn't get far! A spherically symmetric explosion of galaxies in a fixed, static Minkowski spacetime is equivalent to a spatially hyperbolic universe that in which space itself is expanding, with a scale factor $a\propto t$ in terms of cosmological time $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a homogeneous and isotropic universe, the time-projection of the Einstein field equation is the first Friedmann equation,&#xA;$$\frac{\dot{a}^2+k}{a^2} = \frac{8\pi G\rho + \Lambda}{3}\text{,}$$&#xA;so in the case of vanishing cosmological constant ($\Lambda = 0$) and negligible energy density ($\rho = 0$), its solution $a(t) = a_0\pm it\sqrt{k}$ is real only if $k&amp;lt;0$, implying an open, spatially hyperbolic universe; conventionally, $k\in\{-1,0,+1\}$, as only the sign of $k$ is important.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why propose an expanding underlying grid instead of some other mystery force that accelerates objects away faster the farther away they are from us? How would we distinguish between those two scenarios?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If one introduces an extra force and fine-tunes things well enough, you probably can't. However, this is an incredibly silly thing to do: you would need to postulate that (1) gravity &lt;em&gt;simply does not work&lt;/em&gt; on the cosmological scale, and that (2) that its effects are exactly mimicked by an extra force that we have absolutely no evidence for. This doesn't achieve anything, not even philosophical pandering, because we would still know that general relativity is correct on smaller scales, so spacetime geometry would still be dynamic on those scales.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2016-03-04T17:52:44.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14016" PostTypeId="1" CreationDate="2016-03-04T19:27:26.073" Score="2" ViewCount="487" Body="&lt;p&gt;I have heard that the JWST will be capable of looking at the early stages of the universe. However I have also heard that it won't be capable of detecting the Oort Cloud that supposedly provides comets for the solar system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is the JWST able to see millions of light years away but cannot see a giant spherical cloud of rocks around the solar system?&lt;/p&gt;&#xA;" OwnerUserId="10297" LastEditorUserId="11031" LastEditDate="2016-03-04T20:02:13.530" LastActivityDate="2016-03-04T21:57:35.213" Title="Power of the James Webb Space Telescope" Tags="&lt;telescope&gt;&lt;fundamental-astronomy&gt;&lt;deep-sky-observing&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2016-03-05T14:59:50.967" ContentLicense="CC BY-SA 3.0" />
  <row Id="14018" PostTypeId="1" CreationDate="2016-03-04T20:13:03.220" Score="16" ViewCount="1013" Body="&lt;p&gt;Without quantum tunneling our Sun wouldn't be hot or massive enough to produce the energy it does at the moment. So what would have been the temperature or mass of our Sun without quantum tunneling of protons to maintain the same energy we receive from our Sun? &lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="2531" LastEditDate="2016-03-06T13:37:05.967" LastActivityDate="2016-03-07T16:08:39.617" Title="What would the Sun be like if nuclear reactions could not proceed via quantum tunneling?" Tags="&lt;the-sun&gt;&lt;stellar-astrophysics&gt;&lt;nucleosynthesis&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14019" PostTypeId="1" CreationDate="2016-03-04T20:26:51.720" Score="6" ViewCount="140" Body="&lt;p&gt;Discovery in Astronomy/Astrophysics (of astronomical objects) vs discovery of physical phenomena in experimental Physics (as such) - do they differ in required burden of evidence (with the notion, that astronomy/astrophysics deals with real physically existing astronomical objects, while the aim of experimental physics per se is mostly to verify rather somewhat abstract concepts of theoretical physics)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps answering my question in some definite way would require some analysis with historical perspective and consideration of precedences ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This generalized question of mine was motivated by more specific case featured in &lt;a href=&quot;https://physics.stackexchange.com/q/236107/25575&quot;&gt;https://physics.stackexchange.com/q/236107/25575&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="10866" LastEditorUserId="1344" LastEditDate="2019-07-15T18:00:52.077" LastActivityDate="2019-07-15T18:00:52.077" Title="Discovery in Astronomy vs one in Physics - do they differ in required burden of evidence?" Tags="&lt;astrophysics&gt;&lt;history&gt;&lt;resource-request&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14020" PostTypeId="2" ParentId="13991" CreationDate="2016-03-04T20:44:35.727" Score="2" Body="&lt;p&gt;It looks like &lt;a href=&quot;http://heasarc.gsfc.nasa.gov/fitsio/fitsio.html&quot; rel=&quot;nofollow&quot;&gt;CFITSIO&lt;/a&gt; has an &lt;a href=&quot;http://heasarc.gsfc.nasa.gov/docs/software/fitsio/c/f_user/fitsio.html&quot; rel=&quot;nofollow&quot;&gt;interface for F77&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm always thinking about programmer time vs. computation time in research settings. If your data is reasonably small and simple in structure, I'd be very tempted to have Fortran write to a text file, and then use something like astropy to turn the text file into a FITS file. It's slow and icky, but it takes fifteen minutes of thinking if you've used astropy before. &lt;/p&gt;&#xA;" OwnerUserId="608" LastActivityDate="2016-03-04T20:44:35.727" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14021" PostTypeId="1" CreationDate="2016-03-04T21:15:35.250" Score="0" ViewCount="141" Body="&lt;p&gt;I believe the current estimated age of the Universe is around 14 billion years.  I just read in another post on here that the diameter of the Universe is around 90 billion light years.  This doesn't make sense to me, shouldn't the radius of the Universe be equal to it's age?  Unless the explosion caused particles to spread out faster than the speed of light? Is that possible? I thought the speed of light was a speed limit that nothing can exceed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if there is something I'm not understanding here or if I'm trying to compare apples to oranges.  I'm only an astronomy enthusiast.&lt;/p&gt;&#xA;" OwnerUserId="11059" LastActivityDate="2016-03-05T09:34:57.493" Title="Regarding the age / size of the Universe" Tags="&lt;universe&gt;&lt;light&gt;&lt;speed&gt;&lt;size&gt;" AnswerCount="2" CommentCount="6" ClosedDate="2016-03-06T10:42:54.993" ContentLicense="CC BY-SA 3.0" />
  <row Id="14022" PostTypeId="2" ParentId="14021" CreationDate="2016-03-04T21:36:04.930" Score="1" Body="&lt;p&gt;The Universe is estimated to be around 14 billion years old, yes. That said, you seem to be talking about the observable universe. The idea is that space is actually capable of expanding faster than the speed of light, as space itself is not matter. This expansion of space is described by something called &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/astro/hubble.html&quot; rel=&quot;nofollow&quot;&gt;Hubble's law&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11031" LastEditorUserId="10437" LastEditDate="2016-03-05T06:10:03.813" LastActivityDate="2016-03-05T06:10:03.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14023" PostTypeId="2" ParentId="14016" CreationDate="2016-03-04T21:57:35.213" Score="7" Body="&lt;p&gt;It all comes down to the brightness of objects (not their size). For all intents and purposes we can assume that the most distant galaxies and the small, but much closer, objects in the Oort clouds are unresolved point sources.The Oort cloud objects are too faint to see, with JWST, but it should be able to see bright galaxies and quasars even at 13 billion light years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;JWST will have a magnitude limit of something like &lt;a href=&quot;http://www.stsci.edu/jwst/science/sensitivity&quot;&gt;30th AB magnitude&lt;/a&gt; at red and near-infrared wavelengths. It turns out that quasars at Z~10 can have brightnesses that enable them to be detected at these magnitudes and brighter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, but now work out how bright an Oort cloud object is seen in reflected light from the Sun. Taking a solar luminosity and calculating the flux at say the nearest hypothesised Oort cloud objects at about 2000 au from the Sun, we get  $1.4\times 10^3/(2000^2) = 3.5\times10^{-4}$ W/m$^2$. Now let's be generous and say that as much as 10 per cent of this is reflected (a very high albedo) and that it is a big Oort object with a radius of 20 km. If this acts as a Lambertian reflector, then it will receive $3.5\times 10^{-4} \times \pi \times 20000^2 = 4.4\times 10^{5}$ W and it will radiate $4.4\times 10^{4}$ W isotropically back into a hemisphere towards the inner solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now reversing the calculation. After this light has travelled back 2000 au to the JWST, the flux received will be $4.4 \times 10^4 /(2\pi [2000 \times 1.5\times 10^{11}]^2) = 4.9 \times 10^{-25}$ W/m$^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can compare this with the original flux of the Sun at the Earth. This is $1.4\times 10^{3}$ W/m$^2$. Now assuming that the reflection doesn't change the solar spectrum, we can take the ratio of the solar flux at earth to the reflected light from the Oort cloud object to calculate that the Oort cloud object is 68.6 magnitudes fainter than the Sun as seen from the Earth (or close to the Earth - JWST will not be in orbit around the Earth) in any magnitude band you care to choose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the near infrared, the Sun has an apparent magnitude of around -27.5. This means that &lt;strong&gt;the Oort cloud object would have a near infrared apparent magnitude of 41 and would thus be 11 magnitudes too faint to be seen by JWST&lt;/strong&gt;. (And remember we chose a big object that was as close as an Oort cloud object is expected it to be &lt;em&gt;and&lt;/em&gt; gave it a high albedo), &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider that 30th mag detection threshold. For a solar-type spectrum this equates to a flux threshold of $\sim 10^{-20}$ W/m$^2$. Ignoring the cosmological nuances of proper distance etc., then something at 13 billion light years would have to have a luminosity of a mere $10^7$ solar luminosities to produce this kind of flux - the brightness of a modest galaxy. Of course there is also the extreme redshift to consider, so this calculation is way too optimistic, but even allowing for an order of magnitude or two, one should easily be able to see the brightest galaxies and quasars at 30th magnitude.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-03-04T21:57:35.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14024" PostTypeId="2" ParentId="14021" CreationDate="2016-03-04T23:09:41.833" Score="8" Body="&lt;p&gt;The expansion rate of space is not itself the reason that the radius $R_\mathrm{Uni}$ of the observable Universe is larger than 14 billion lightyears (Gly). Just the fact that space expands is the reason. If space did &lt;em&gt;not&lt;/em&gt; expand, then $R_\mathrm{Uni}$ would be the expected 14 Gly, as this is the distance that light can travel in the 14 billion years (Gyr) since the Big Bang. But since space has expanded in the meantime, then the distance to some particle that emitted a photon that we observe today, has been continuously increasing during the 14 Gyr, and hence $R_\mathrm{Uni}&amp;gt;14\,\mathrm{Gly}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An analogy to this is a worm on a rubber band that is stretched while it crawls along. The distance between its starting point and it's end point is not just a matter of how fast it crawls, but also of how much you stretch the rubber band.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term &quot;space expands faster than light&quot; is a bit deceptive. Space expands *homogolously*$^1$, meaning that a given point in space recedes from you at a speed depending on its distance from you. If a galaxy is 1 Mpc (= 3.26 Mly) from you, it recedes at $70\,\mathrm{km}\,\mathrm{s}^{-1}$. If it's 2 Mpc from you, it recedes at $140\,\mathrm{km}\,\mathrm{s}^{-1}$. And so on. At some distance the recession velocity becomes larger than $c$, and in fact galaxies at a distance of $R_\mathrm{Uni}$ recede at more than $3c$. This doesn't violate special relativity which says that nothing can travel through space faster than $c$, because the galaxies do not travel &lt;em&gt;through&lt;/em&gt; space. They lie approximately still in space, but space itself simply expands, i.e. the distance between everything increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like three worms on a rubber band, with 1 cm between them. If you stretch the band to double length, the distance from worm #1 to #2 is 2 cm, while from #1 to #3 it's 4 cm.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^1$&lt;sub&gt;Homogolous is not the same as homogeneous. Whereas the latter means &quot;the same everywhere&quot;, referring to some physical quantity like density, homogolous means &quot;proportional to distance&quot;.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-03-05T09:34:57.493" LastActivityDate="2016-03-05T09:34:57.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14026" PostTypeId="2" ParentId="14019" CreationDate="2016-03-05T08:03:29.143" Score="6" Body="&lt;p&gt;There is no central authority in science. There is no council that sets the standards. The criteria for a discovery are the same: You publish your findings, and your peers accept your results.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is the 5 sigma rule in particle physics. Perhaps you were thinking of this. But that is not an official rule, instead it's a convention among particle physics. And it's not applicable to every field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things are never cut and dry. What does &quot;Peers&quot; mean. What if only most are convinced. Does it count? Discoveries are usually judged in retrospect. After the dust has settled and the debate s are over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;//edit&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the bar set so high in physics?  When &quot;discovering&quot; a particle there is no way to &quot;see&quot; it, instead you observe a mass of data and see a statistical discrepancy. Cern produces masses and masses of data, and it is sifted for anything unusual. And with so much data, the chance of seeing &lt;em&gt;something&lt;/em&gt; unusual is actually quite high. (imagine searching for repetitions of '9' in the digits of pi - if you search far enough you can be sure to find a string of 6 9s even though the chance of a random string having 6 9s is very low)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a mass of data, and the opportunity to repeat experiments it makes sense to set the bar very high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare with an observation, such as the &quot;chirp&quot; at Ligo. We can't repeat it at will, the data is there: something happened that is consistent with a black hole merger. No other theory has been proposed that can explain the observations. The observation is less dependent on a statistical finding after running multiple experiments, but a single direct observation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fields in which a result could be explained by chance, then a statistical analysis is done, and published. This contributes to the quality of a finding, and so the number of people you will convince. And convincing your peers is the only criteria that counts in the end. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-03-05T16:58:38.977" LastActivityDate="2016-03-05T16:58:38.977" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="14027" PostTypeId="2" ParentId="14009" CreationDate="2016-03-05T08:55:29.207" Score="1" Body="&lt;p&gt;Earth rotates. That gives us the feeling of Sunrise and Sunset. Actually there are no such things. See Below animated image. Earth completes one rotation in a day (24hrs). (Actually 1 rotation of earth is called a &quot;Day&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image also depicts the cause of &lt;a href=&quot;https://en.wikipedia.org/wiki/Solstice&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solstice&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Equinox&quot; rel=&quot;nofollow noreferrer&quot;&gt;Equinox&lt;/a&gt;, earth's tilted axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zWhQT.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zWhQT.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1669" LastEditorUserId="1669" LastEditDate="2016-03-05T14:27:24.000" LastActivityDate="2016-03-05T14:27:24.000" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14028" PostTypeId="2" ParentId="14018" CreationDate="2016-03-05T12:38:37.100" Score="16" Body="&lt;p&gt;Short answer: Without tunnelling, stars like the Sun would never reach nuclear fusion temperatures; stars less massive than around $5M_{\odot}$ would become &quot;hydrogen white dwarfs&quot; supported by electron degeneracy pressure. More massive objects would contract to around a tenth of a solar radius and commence nuclear fusion. They would be hotter than &quot;normal&quot; stars of a similar mass, but my best estimate is that they have similar luminosities. Thus it would not be possible to get a stable nuclear burning star with 1 solar luminosity. Stars of 1 solar luminosity could exist, but they would be on cooling tracks, much like brown dwarfs are in the real universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very interesting hypothetical question. What would happen to a star if you &quot;turned off&quot; tunnelling. I think the answer to this is that the pre-main-sequence stage would become significantly longer. The star would continue to contract, releasing gravitational potential energy in the form of radiation and by heating the core of the star. The virial theorem tells us that the central temperature is roughly proportional to $M/R$ (mass/radius). So for a fixed mass, as the star contracts, its core gets hotter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are then (at least) two possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The core becomes hot enough for protons to overcome the Coulomb barrier and begin nuclear fusion. For this to happen, the protons need to get within about a nuclear radius of each other, let's say  $10^{-15}$ m. The potential energy is&#xA;$e^2/(4\pi \epsilon_0 r) = 1.44$ MeV or $2.3\times 10^{-13}$ J.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The protons in the core will have a mean kinetic energy of $3kT/2$, but some small fraction will have energies much higher than this according to a Maxwell-Boltzmann distribution. Let's say (and this is a weak point in my calculation that I may need to revisit when I have more time) that fusion will take place when protons with energies of $10 kT$ exceed the Coulomb potential energy barrier. There will be a small numerical uncertainty on this, but because the reaction rate would be highly temperature sensitive it will not be an order of magnitude out. This means that fusion would not begin until the core temperature reached about $1.5 \times 10^{9}$ K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Sun, fusion happens at around $1.5\times 10^7$ K, so the virial theorem result tells us that stars would need to contract by about a factor of 100 for this to happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the gravity and density of such a star would be much higher than the Sun, hydrostatic equlibrium would demand a very high pressure gradient, but the temperature gradient would be limited by convection, so there would need to be an extremely centrally concentrated core with a fluffy envelope. Working through some simple proportionalities I think that the luminosity would be almost unchanged (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Mass%E2%80%93luminosity_relation&quot; rel=&quot;noreferrer&quot;&gt;luminosity-mass relation&lt;/a&gt; but consider how luminosity depends on radius at a fixed mass), but that means the temperature would have to be hotter by a factor of the square root of the radius contraction factor. However, this could be academic, since we need to consider the second possibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) As the star shrinks, the electrons become degenerate and contribute degeneracy pressure. This becomes important when the phase space occupied by each electron approaches $h^3$. There is a standard bit of bookwork, which I am not going to repeat here - you can find it something like &quot;The Physics of Stars&quot; by Phillips - which shows that degeneracy sets in when&#xA;$$\frac{ 4\pi \mu_e}{3h^3}\left(\frac{6G R\mu m_e}{5}\right)^{3/2} m_u^{5/2} M^{1/2} = 1,$$&#xA;where $\mu_e$ is the number of mass units per electron, $\mu$ is the number of mass units per particle, $m_e$ is the electron mass and $m_u$ is an atomic mass unit. If I've done my sums right this means for a hydrogen gas (let's assume) with $\mu_e=1$ and $\mu = 0.5$ that degeneracy sets in when&#xA;$$ \left(\frac{R}{R_{\odot}}\right) \simeq 0.18 \left(\frac{M}{M_{\odot}}\right)^{-1/3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, when the star shrinks to the size of $\sim$ Jupiter, its interior will be governed by electron degeneracy pressure, not by perfect gas pressure. The significance of this is that electron degeneracy pressure is only weakly dependent (or independent for a completely degenerate gas) on temperature. This means that the star can cool whilst only decreasing its radius very slightly. The central temperature would &lt;em&gt;never&lt;/em&gt; reach the high temperatures required for nuclear burning and the &quot;star&quot; would become a hydrogen white dwarf with a final radius of a few hundredths of a solar radius (or a bit smaller for more massive stars).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second possibility must be the fate of something the mass of the Sun. However, there is a cross-over point in mass where the first possibility becomes viable. To see this, we note that the radius at which degeneracy sets in depends on $M^{-1/3}$, but the radius the star needs to shrink to in order to begin nuclear burning is proportional to $M$. The cross-over takes place somewhere in the range 5-10 $M_{\odot}$. So stars &lt;em&gt;more&lt;/em&gt; massive than this could commence nuclear burning at radii of about a tenth of a solar radius, without their cores being degenerate. An interesting possibility is that at a few solar masses there should be a class of object that contracts sufficiently that nuclear ignition is reached when the core is substantially degenerate. This might lead to a runaway &quot;hydrogen flash&quot;, depending on whether the temperature dependence of the reaction rate is extreme enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best question of the year so far. I do hope that someone has run some simulations to test these ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: As a postscript it is of course anomalous to neglect a quantum effect like tunnelling, whilst at the same time relying on degeneracy pressure to support the star! If one were to neglect quantum effects entirely and allow a star like the Sun to collapse, then the end result would surely be a classical black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A further point that would need further consideration is to what extent radiation pressure would offer support in stars that were smaller, but &lt;em&gt;much&lt;/em&gt; hotter.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-03-07T16:08:39.617" LastActivityDate="2016-03-07T16:08:39.617" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14030" PostTypeId="1" AcceptedAnswerId="14034" CreationDate="2016-03-05T16:36:56.900" Score="3" ViewCount="1571" Body="&lt;p&gt;Here is a bit of a newbie question, if you please. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun is constantly ejecting small particles via solar winds that come into contact with planets and other objects in our solar system. Do these particles contribute to the planets's atmospheres? Or do they do more harm than good? (define: good = contribute).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do solar winds contain any larger atoms than helium. For instance, Oxygen? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intuition on this issue is mixed since we have Mars with a very thin atmosphere, but Venus with a very thick atmosphere, neither having a very strong magnetosphere to protect them from solar winds.&lt;/p&gt;&#xA;" OwnerUserId="10566" LastActivityDate="2016-03-06T05:22:00.087" Title="How do solar winds affect the atmospheric composition and density of planets?" Tags="&lt;planetary-atmosphere&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14032" PostTypeId="1" AcceptedAnswerId="14040" CreationDate="2016-03-05T21:59:39.580" Score="5" ViewCount="655" Body="&lt;p&gt;I am developing an astronomy software and need to know the color of each planet in the Solar System, when observed with the naked eye. I cannot find that information after googling for a while. Is there any good source? a RBG color associated to each planet would be ideal but it can be any other color system.&lt;/p&gt;&#xA;" OwnerUserId="11050" LastActivityDate="2021-05-21T18:18:43.890" Title="Color of planets" Tags="&lt;planet&gt;&lt;observational-astronomy&gt;&lt;luminosity&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14033" PostTypeId="2" ParentId="14032" CreationDate="2016-03-06T00:22:11.480" Score="-3" Body="&lt;p&gt;First look up what elements the planets are composed of. Then figure out how the elements should react due to the planets atmosphere,light scattering,spin,sun reflection from natural satellites if one or many is present,gravity wave overlaps calculations.&#xA;  Finally how the human eye perceives them. But may I add the human eye will never be able to view the colors in there true form, The technology used to view the planets will not correlate to the human eye.&#xA;  So a composite of what the human eye should observe must be composed. &quot;The amount of details reflects the amount of workmanship, that then could produce a Masterpiece. Its up too you&quot;&lt;/p&gt;&#xA;" OwnerUserId="9326" LastActivityDate="2016-03-06T00:22:11.480" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14034" PostTypeId="2" ParentId="14030" CreationDate="2016-03-06T05:04:08.780" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Do these particles contribute to the planets' atmospheres? Or do they&#xA;  do more harm than good? (define: good = contribute).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That depends on the size (mass/escape velocity) of the planet.   See &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_escape&quot; rel=&quot;noreferrer&quot;&gt;atmospheric escape&lt;/a&gt; and pretty picture&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2RADR.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2RADR.png&quot; alt=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/4/4a/Solar_system_escape_velocity_vs_surface_temperature.svg/512px-Solar_system_escape_velocity_vs_surface_temperature.svg.png&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the 4 inner planets, the solar wind strips gas from the planets.  This process if fairly straight forward.   It has more to do with heat than traditional &quot;blowing&quot;, and when the when the outer atmosphere gas molecules reach a certain temperature they can achieve escape velocity.   Earth mostly just loses hydrogen and helium and very little of it's heavier gases.   Venus over time has lost most of it's Methane and Water Vapor, the key difference between Earth and Venus is that Earth has a magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars has lost most of it's atmosphere and what atmosphere Mars retains may be related to the thawing and re-freezing of it's icy poles, and Mercury's atmosphere is so thin that it's atmosphere actually is little more than a loosely retained collection of solar wind particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 4 outer planets don't lose much to the solar wind and the Solar wind (I would think) adds to rather than takes away from their total mass, but this addition is very small compared to the mass of a planet. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do solar winds contain any larger atoms than helium. For instance,&#xA;  Oxygen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, but only trace amounts.  The Solar wind is 92% hydrogen and 8% helium.  &lt;a href=&quot;http://solar-center.stanford.edu/FAQ/Qsolwindcomp.html&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;.  Very close to the universal average ratio.   Solar particles are so hot that when they're ejected from the sun the electrons and atomic nuclei are largely separated so you get a stream of charged particles, electrons and protons being the most common and Alpha Particles (Helium Nuclei) 3rd.  This means that any planets with a magnetic field, those particles get caught in planet's magnetic field.   That's why a magnetic field is considered very important in preserving a planet's atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the captured solar wind particles would probably get deposited into the Planet's atmosphere over time but I'm not clear on that process and I don't know the percentages.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting back to the Sun, it has about 2% &quot;heavy&quot; elements in it, see &lt;a href=&quot;http://www.space.com/17170-what-is-the-sun-made-of.html&quot; rel=&quot;noreferrer&quot;&gt;article&lt;/a&gt;, and by &quot;heavy&quot; I mean, Oxygen, Carbon, Neon, others, but gravity draws the heavier elements towards the center of the sun.   Also, the helium the sun creates is largely formed near the center of the sun, so the outer parts where the solar wind gets ejected has a more standard/universal balance of about 92% hydrogen by element.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My intuition on this issue is mixed since we have Mars with a very&#xA;  thin atmosphere, but Venus with a very thick atmosphere, neither&#xA;  having a very strong magnetosphere to protect them from solar winds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Venus mass is sufficient to make a difference.  Atmospheric escape has to do with escape velocity from the planets and Venus' escape velocity is over twice Mars'.   &lt;a href=&quot;http://nssdc.gsfc.nasa.gov/planetary/factsheet/&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;.  Venus' atmosphere is also sufficiently thick that it generates it's own magnetic protection of sorts, (I've read that, didn't see a good article to reference just now).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter has an enormous magnetic field, which casts a pretty big net, much larger than the planet itself and that captures solar wind particles, but compared to the mass of Jupiter, the captured solar particles, even with the large magnetic net, is teeny-tiny and makes no real difference in the mass of the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, when the sun approaches it's red giant stage, it's ejections will significantly increase, so it's possible that, 5 billion or so years from now the sun will feed Jupiter a stream of particles in a measurable way and Jupiter will grow larger and perhaps, visibly glow due to the particles it receives.  Jupiter might be a fun planet to watch when our sun goes red giant, perhaps from the safe distance of a colony on Saturn's moon titan.   But that's just speculation.   :-)   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will never turn into a star, it's far far far too light for that, but Jupiter could still be fun to watch (from a safe distance) 5 billion years from now.   (My answers always tend to be too long). . . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corrections  welcome.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-03-06T05:22:00.087" LastActivityDate="2016-03-06T05:22:00.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14035" PostTypeId="1" AcceptedAnswerId="14036" CreationDate="2016-03-06T05:30:43.590" Score="5" ViewCount="50" Body="&lt;p&gt;Why do we need to build telescopes that can view in the infrared? What exactly do we need to see in the infrared region? I know this seems like a really broad question but I'm just curious about it. &lt;/p&gt;&#xA;" OwnerUserId="11069" LastActivityDate="2016-03-06T08:23:21.070" Title="Infrared Region Telescopes" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14036" PostTypeId="2" ParentId="14035" CreationDate="2016-03-06T08:23:21.070" Score="5" Body="&lt;p&gt;Three main reasons.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Lots of things in the universe are cool, much cooler than the Sun. Therefore they emit most of their radiation in the infrared spectrum. Examples might include brown dwarfs, giant exoplanets, circumstellar discs.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Lots of things in the universe are distant and highly redshifted. Thus even if they emit their radiation at visible wavelengths, we receive it at infrared wavelengths.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Lots of things in the universe are hidden by dust, which effectively absorbs visible light. Dust is much more transparent to infrared radiation, so we can see beyond it. An example would be the centre of our Galaxy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-03-06T08:23:21.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14037" PostTypeId="2" ParentId="13999" CreationDate="2016-03-06T09:53:14.917" Score="3" Body="&lt;p&gt;I suspect that what the class told you was that the universe might be infinite. We currently have no way of knowing if that was true or not. The issue is that there is no way of knowing when inflation stopped outside of our Hubble sphere. I believe that there is some evidence in the CMBR that it has a minimum size which is several times the Hubble radius, as there would be resonances that aren't apparent in the spectrum.  However there is no upper limit. Note, however, that this theory does include the possibility that the universe wasn't always infinite but that it was a finite size early. &lt;/p&gt;&#xA;" OwnerUserId="10800" LastActivityDate="2016-03-06T09:53:14.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14038" PostTypeId="2" ParentId="11295" CreationDate="2016-03-06T11:17:51.313" Score="1" Body="&lt;p&gt;We know that there is no known object that will collide with the Earth in the next hundred years or so, and is large enough to do substantial damage. &lt;strong&gt;We don't know if there are undiscovered objects that can collide with the Earth&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having made an observation of an asteroid, over a period of time, it is a relatively simple process to predict its future path. Each asteroid, as it is discovered, has its orbit calculated, and no discovered object has an orbit that intersects with that of Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there are known to be many more objects that we haven't discovered yet. It is possible that there is a undiscovered object that will collide with the Earth. It is unlikely because large objects are rare, but not impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The survey can never be complete, as comets enter the inner solar system from very great distances, and cannot be seen until they do.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-03-06T11:17:51.313" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14040" PostTypeId="2" ParentId="14032" CreationDate="2016-03-06T12:05:33.167" Score="8" Body="&lt;p&gt;Here are some values found by taking the hue from images, and adjusting the brightness to fit the albedo: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury #&lt;a href=&quot;http://www.htmlcsscolor.com/hex/1A1A1A&quot; rel=&quot;noreferrer&quot;&gt;1a1a1a&lt;/a&gt; Yes it is really that dark&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus #&lt;a href=&quot;http://www.htmlcsscolor.com/hex/e6e6e6&quot; rel=&quot;noreferrer&quot;&gt;e6e6e6&lt;/a&gt; or perhaps a bit darker&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth tricky as it is a mix of colors, and changes over the year&#xA;seems to average out as about #&lt;a href=&quot;http://www.htmlcsscolor.com/hex/2F6A69&quot; rel=&quot;noreferrer&quot;&gt;2f6a69&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars #&lt;a href=&quot;http://www.htmlcsscolor.com/hex/993d00&quot; rel=&quot;noreferrer&quot;&gt;993d00&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter #&lt;a href=&quot;http://www.htmlcsscolor.com/hex/b07f35&quot; rel=&quot;noreferrer&quot;&gt;b07f35&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn #&lt;a href=&quot;http://www.htmlcsscolor.com/hex/b08f36&quot; rel=&quot;noreferrer&quot;&gt;b08f36&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uranus #&lt;a href=&quot;http://www.htmlcsscolor.com/hex/5580aa&quot; rel=&quot;noreferrer&quot;&gt;5580aa&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neptune #&lt;a href=&quot;http://www.htmlcsscolor.com/hex/366896&quot; rel=&quot;noreferrer&quot;&gt;366896&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might find these surprisingly dark. Planets look like bright dots against the dark sky&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-07-04T18:30:13.973" LastActivityDate="2016-07-04T18:30:13.973" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14041" PostTypeId="1" CreationDate="2016-03-06T12:05:46.857" Score="8" ViewCount="305" Body="&lt;p&gt;Is it possible to determine what daytime the Chicxulub impactor striked the Earth 66 mya? Was it day or night, what was the current phase of the Moon etc.? Can sediments, rhytmites or something else hold this information?&lt;/p&gt;&#xA;" OwnerUserId="11072" LastEditorUserId="24184" LastEditDate="2021-03-21T17:32:41.393" LastActivityDate="2021-03-21T18:02:24.773" Title="Time of day of the K-Pg asteroid impact?" Tags="&lt;astrophysics&gt;&lt;chicxulub&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14043" PostTypeId="1" AcceptedAnswerId="14050" CreationDate="2016-03-06T15:30:26.943" Score="2" ViewCount="645" Body="&lt;p&gt;The news media recently reported that galaxy GN-z11, formed just 400 million years after the big bang, is much further away than originally thought at perhaps 13.2 billion light years.  They say, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The key to the discovery was precisely measuring the shift of the&#xA;  galaxy's light into longer, redder wavelengths, which correspond to&#xA;  how far the photons had traveled before reaching Hubble's eye.&#xA;  (Reuters).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, suppose I'm standing next to, but safely away from, some train tracks.  As the train approaches, sounding its whistle constantly, the pitch will appear to increase until it reaches me, upon which I will briefly hear the whistle's 'rest pitch' (the pitch of the whistle as if I were aboard the train), then it would decrease as the train moves away.  Thus, the lower the pitch, the further away the train is.  I hope that's right.  So, if we know the rest frequency of galaxy GN-z11, its measured frequency indicates how far away it is from earth. Trouble is, how do we know what the rest frequency of light coming from this distant galaxy is?   &lt;/p&gt;&#xA;" OwnerUserId="5306" LastEditorUserId="5306" LastEditDate="2016-03-06T20:04:09.320" LastActivityDate="2016-03-09T16:33:53.623" Title="How do we know galaxy GN-z11 is as far away as it is?" Tags="&lt;galaxy&gt;&lt;distances&gt;&lt;doppler-effect&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14045" PostTypeId="1" CreationDate="2016-03-06T19:19:39.427" Score="3" ViewCount="192" Body="&lt;p&gt;As far as I know galaxies and clusters form along so-called &quot;galactic filaments&quot; which are influenced by a dark matter web that organises them in to this position in the universe. So it looks like dark matter attracts baryonic matter by a kind of gravity. But is it also true the other way around, and that 'our' mass has gravitational forces on dark matter?&#xA;And if not why does it only go one way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a study scientists have observed the four colliding galaxies of cluster Abell 3827 and discovered that one of the envelopes of dark matter in the galaxy behind it encloses. The backlog in the system stands at 5,000 light years (50,000 million million kilometers - the NASA space probe Voyager would do 90 million years to cover that distance.&#xA;Expected to create some lag between a galaxy and its associated dark matter when dark matter affects itself, however small, by other forces than the gravity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;https://www.youtube.com/watch?v=qtfTUPhXutQ&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=qtfTUPhXutQ&lt;/a&gt; and&#xA;&lt;a href=&quot;http://spacetelescope.org/news/heic1506/&quot; rel=&quot;nofollow&quot;&gt;http://spacetelescope.org/news/heic1506/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="10445" LastEditDate="2016-03-06T20:01:02.157" LastActivityDate="2016-03-06T20:01:02.157" Title="Is dark matter influenced by gravity of baryonic matter?" Tags="&lt;dark-matter&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14046" PostTypeId="1" CreationDate="2016-03-06T19:19:45.780" Score="0" ViewCount="229" Body="&lt;p&gt;I have seen a lot of information about moon phase calculation, but need an accurate moon shape calculator/algorithm given the observer's position and time. This implies that the illuminated portion may be rotated, and not only have a right-left orientation as traditionally depicted.&lt;/p&gt;&#xA;" OwnerUserId="11050" LastActivityDate="2016-03-06T22:15:48.097" Title="Moon SHAPE calculator" Tags="&lt;observational-astronomy&gt;&lt;moon-phases&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14047" PostTypeId="2" ParentId="14045" CreationDate="2016-03-06T19:31:01.170" Score="6" Body="&lt;p&gt;Dark matter attracts baryons not only by a &quot;kind of gravity&quot;, but simply by gravity. Everything that has mass (even if it's only relativistic mass like light) attracts everything else by gravity, and thus baryons also attract dark matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason dark matter &quot;organizes&quot; galaxies is just that there is much more of it, roughly 7 times more, to be specific. So, on large scales, the dynamics of galaxies are dominated by dark matter. But because baryons can dissipate energy, it may collapse in clumps — such as molecular clouds — which are hence dominated by baryons.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2016-03-06T19:31:01.170" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14048" PostTypeId="2" ParentId="14007" CreationDate="2016-03-06T21:17:06.360" Score="2" Body="&lt;p&gt;There are a number of ways of answering this question. Here are a few.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly we can simply state that, by definition, the universe is everything and therefore there is simply nothing beyond it. Not nothing as in empty space, but nothing at all, not even space (or it could be that the universe is infinite.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, we could look at what we mean when we say that the universe is expanding. What this really means is that the space-time metric is growing. It does not mean that the universe is expanding into some void, but that distances inside the universe are increasing with time. My favourite parallel is with an balloon being blown up. Points on the surface get further away without the addition of any more material to the balloon. (It is, of course, an imperfect parallel as the balloon is expanding in a medium).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thirdly, and now we are touching on metaphysics (in the sense that it may be impossible to test some or any of these theories), we can say that &quot;our&quot; universe is one of many and that, actually it may well have an edge which even now is crashing into some other universe, or perhaps we are inside a black hole inside another universe. These multiverse theories are ten-a-penny but they are not without scientific merit, possibly.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2016-03-06T21:17:06.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14049" PostTypeId="2" ParentId="14046" CreationDate="2016-03-06T22:15:48.097" Score="3" Body="&lt;p&gt;There is an accurate depiction of the moon phase and shape in &lt;a href=&quot;http://www.stellarium.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellarium&lt;/a&gt; (see below) that you might find useful. You can select your observing location and time anywhere on Earth (or another planet if you so wish). Then you can just search for the moon in the search window (if you cant spot it in the sky), and you have a real representation of how the moon looks at that specific time and location. However if you need to repeat this calculation many times then this isn't the best solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WCd64.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WCd64.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10437" LastActivityDate="2016-03-06T22:15:48.097" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14050" PostTypeId="2" ParentId="14043" CreationDate="2016-03-06T23:06:32.257" Score="5" Body="&lt;p&gt;So GN z-11 is the latest &quot;furthest away&quot; galaxy. It has a claimed redshift measurement of $z=11.1$, meaning that we are seeing the light it emitted about 400 Myr after the big bang (dependent on an assumed set of cosmological parameters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the redshift measurement, the discoverers used grism (relatively low resolution) spectroscopy in the near infrared. What they were looking for is the rest-frame Lyman alpha continuum break, which would be redshifted into this wavelength range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lyman alpha continuum break is caused by the absorption of nearly all ultraviolet photons with wavelengths smaller than 121 nm. These high energy photons are capable of photoionising the neutral hydrogen present in the intergalactic medium at a range of lower redshifts. This neutral hydrogen is present in abundance at redshifts greater than 6, as it had not yet been &lt;em&gt;re-ionised&lt;/em&gt; by quasars and starlight. The consequence is that no light is expected to reach us from rest-frame wavelengths shortwards of 121 nm, but a galaxy's light &lt;em&gt;can&lt;/em&gt; reach us from rest-frame wavelengths that are longer than this. When one &lt;em&gt;observes&lt;/em&gt; the spectrum, we see flux at long wavelengths which suddenly cuts off at shorter wavelengths. The wavelength of the break is $\lambda = 121 \times (1+z)$ nm, where $z$ is the redshift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is this Lyman alpha continuum break that has been identified at an observed wavelength of 1470 nm. This leads to the redshift estimate of $z= (1470/121)-1 = 11.1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The details are presented in &lt;a href=&quot;http://arxiv.org/abs/1603.00461&quot;&gt;Oesch et al. (2016)&lt;/a&gt;. The continuum break is the only thing visible in the spectrum. The authors are confident that this is what it is because their previous broadband photometry had given them an estimated redshift of $&amp;gt;10$ (which is why they observed this candidate in the first place).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-03-06T23:06:32.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14052" PostTypeId="2" ParentId="13965" CreationDate="2016-03-07T03:14:15.187" Score="0" Body="&lt;p&gt;NOTE: I am using a &quot;geocentric&quot; frame of reference, where both the&#xA;moons and the sun orbit the planet, and am creating an arbitrary xy&#xA;coordinate system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We note from @Hohmannfan's answer that (answering your questions out&#xA;of order for simplicity):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Moon B will eclipse the sun every $\frac{10385}{304}$ (~ 34.16)&#xA;days. In this time period, the sun completes $\frac{31}{304}$th of&#xA;an orbit and Moon B completes $1\frac{31}{304}$ orbits, lapping the&#xA;sun once.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Moon A will eclipse the sun every $\frac{26130}{257}$ (~ 101.67)&#xA;days. The sun will complete $\frac{78}{257}$ of an orbit, and Moon A&#xA;will lap it by completing $1\frac{78}{257}$ orbits.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Moon B will overlap Moon A once every $\frac{2418}{47}$ (~ 51.44)&#xA;days, in which Moon A will complete $\frac{31}{47}$ of an orbit and&#xA;Moon B will lap it by completing $1\frac{31}{47}$ orbits.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, as @Hohmannfan notes, there's no guarantee that the moons&#xA;will be &lt;em&gt;full&lt;/em&gt; when they overlap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also no guarantee that the two moons will &lt;em&gt;ever&lt;/em&gt; both eclipse&#xA;the sun at the exact same time, although they will get arbitrarily&#xA;close to doing so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the $\frac{2418}{47}$ days between two successive lunar overlaps, the sun&#xA;moves $\frac{2418}{47} \times \frac{1}{335}$ of an orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As above, the moons have advanced $\frac{31}{47}$ of an orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, compared to the sun, the moons have advanced $\frac{31}{47} -&#xA;\frac{2418}{47} \times \frac{1}{335}$ or $\frac{7967}{15745}$ of an&#xA;orbit (this number is surprisingly close to $\frac{1}{2}$ but that's&#xA;just a coincidence).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This happens between every pair of overlaps, so the sun's angular&#xA;distance (in orbits) from the overlapping moons is $\frac{7967&#xA;n}{15745}+r$ where $r$ is the angular distance at a specific overlap&#xA;and $n$ is any integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the overlapping moons to eclipse the sun $\frac{7967 n}{15745}+r$&#xA;must be an integer. If $r$ is irrational, this can never happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the angular distance can get arbitrarily small, even to the&#xA;point where an observer wouldn't realize the double moon eclipse isn't&#xA;100% perfect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By a similar argument, you can show the two full moons will get&#xA;arbitrarily close to overlapping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOW&lt;/strong&gt;, if we make the simplifying assumption that both moons are&#xA;eclipsing the sun at year 0 (perhaps your astronomer-priests have&#xA;decided this unusual occurence is a good time to start numbering the&#xA;years, and believe zero (not one) is a good first year), we can make&#xA;some other calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the moons line up every $\frac{2418}{47}$ days and the sun and&#xA;Moon B line up every $\frac{10385}{304}$ days, all three will line up&#xA;(to form a double moon eclipse of the sun) on the least common&#xA;multiple of these numbers, or 810,030 days (which would be exactly&#xA;2418 of your years, and note that 2418 is the product of the two lunar&#xA;orbits in days). In this time:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Moon A will have completed exactly 10,385 orbits.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Moon B will have completed exactly 26,130 orbits.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As above, the sun will have completed exactly 2,418 orbits.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As it turns out, there can never be a perfect double full moon bullseye:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Moon B will be full on day $\frac{10385}{608}$ (~ 17.08), at which&#xA;point it will have completed $\frac{335}{608}$ of an orbit and the&#xA;sun will have completed $\frac{31}{608}$ of an orbit, so Moon B will&#xA;have gained half an orbit on the sun, which is required for a full&#xA;moon. After that, the moon will be full every $\frac{10385}{304}$&#xA;days, the period of time it takes the sun to complete&#xA;$\frac{31}{304}$ orbits, and Moon B to complete $1\frac{31}{304}$&#xA;orbits.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By similar calculation, Moon A will be full on day&#xA;$\frac{13065}{257}$ (~ 50.84) and every $\frac{26130}{257}$ days&#xA;thereafter.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To find when they're both full at the same time, we solve this&#xA;linear Diophantine equation:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$\frac{10385 n}{304}+\frac{10385}{608}=\frac{26130 m}{257}+\frac{13065}{257}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where n and m are integers. This reduces to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n\to \frac{47424 m+15745}{15934}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, $47424 m$ is always even, so $47424 m+15745$ is always&#xA;odd. Since the denominator ($15934$) is even, you are dividing an odd&#xA;number by an even number, and the result can never be an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this doesn't tell the full story. For example, if we compute&#xA;the positions on day $\frac{34987576465283}{92766720}$ (~ 377156.55),&#xA;we find:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Moon B is at 122.5656 degrees.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Moon A is at 122.5581 degrees, only ~ 27 arcseconds away.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The sun is at 302.5658 degrees, 179.9998 degrees from moon B, and&#xA;179.9924 degrees from moon A (~ 28 arcseconds from opposition).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In other words, this is pretty close to a double full moon, even&#xA;though it's not exact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a similar vein, even though double solar eclipses only occur once&#xA;every 810,030 days, there are several close calls:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;\begin{array}{cc}&#xA;                   \text{Day} &amp;amp; \text{Sep (')} \\&#xA;                   -810030.00000 &amp;amp; 0.00 \\&#xA;                   -754313.10860 &amp;amp; 0.91 \\&#xA;                   -698596.21710 &amp;amp; 1.82 \\&#xA;                   -642879.32570 &amp;amp; 2.73 \\&#xA;                   -587162.43420 &amp;amp; 3.64 \\&#xA;                   -531445.54280 &amp;amp; 4.55 \\&#xA;                   -475728.65130 &amp;amp; 5.47 \\&#xA;                   -445735.13160 &amp;amp; 7.29 \\&#xA;                   -420011.75990 &amp;amp; 6.38 \\&#xA;                   -390018.24010 &amp;amp; 6.38 \\&#xA;                   -364294.86840 &amp;amp; 7.29 \\&#xA;                   -334301.34870 &amp;amp; 5.47 \\&#xA;                   -278584.45720 &amp;amp; 4.55 \\&#xA;                   -222867.56580 &amp;amp; 3.64 \\&#xA;                   -167150.67430 &amp;amp; 2.73 \\&#xA;                   -111433.78290 &amp;amp; 1.82 \\&#xA;                   -55716.89145 &amp;amp; 0.91 \\&#xA;                   0.00000 &amp;amp; 0.00 \\&#xA;                   55716.89145 &amp;amp; 0.91 \\&#xA;                   111433.78290 &amp;amp; 1.82 \\&#xA;                   167150.67430 &amp;amp; 2.73 \\&#xA;                   222867.56580 &amp;amp; 3.64 \\&#xA;                   278584.45720 &amp;amp; 4.55 \\&#xA;                   334301.34870 &amp;amp; 5.47 \\&#xA;                   364294.86840 &amp;amp; 7.29 \\&#xA;                   390018.24010 &amp;amp; 6.38 \\&#xA;                   420011.75990 &amp;amp; 6.38 \\&#xA;                   445735.13160 &amp;amp; 7.29 \\&#xA;                   475728.65130 &amp;amp; 5.47 \\&#xA;                   531445.54280 &amp;amp; 4.55 \\&#xA;                   587162.43420 &amp;amp; 3.64 \\&#xA;                   642879.32570 &amp;amp; 2.73 \\&#xA;                   698596.21710 &amp;amp; 1.82 \\&#xA;                   754313.10860 &amp;amp; 0.91 \\&#xA;                   810030.00000 &amp;amp; 0.00 \\&#xA;                  \end{array}&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The table above lists all near-eclipses within 7.5 minutes of arc,&#xA;where day is the number of days from year 0 (including days before&#xA;year 0), and sep is the maximum separation (in minutes of arc) of any&#xA;two of Moon A, Moon B, and the sun. Note that days $0$ and $\pm&#xA;810030$ are perfect eclipses, as expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the closest we get to double full moons is below. In this&#xA;case, sep is (in minutes of arc) the maximum of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the angular distance of Moon A from opposition&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the angular distance of Moon B from opposition&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the angular distance between Moon A and Moon B&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$&#xA;\begin{array}{cc}&#xA;                   \text{Day} &amp;amp; \text{Sep (')} \\&#xA;                   -797168.29790 &amp;amp; 10.29 \\&#xA;                   -767174.80850 &amp;amp; 8.92 \\&#xA;                   -711457.91490 &amp;amp; 7.55 \\&#xA;                   -655741.02130 &amp;amp; 6.17 \\&#xA;                   -600024.12770 &amp;amp; 4.80 \\&#xA;                   -544307.23400 &amp;amp; 3.43 \\&#xA;                   -488590.34040 &amp;amp; 2.06 \\&#xA;                   -432873.44680 &amp;amp; 0.69 \\&#xA;                   -377156.55320 &amp;amp; 0.69 \\&#xA;                   -321439.65960 &amp;amp; 2.06 \\&#xA;                   -265722.76600 &amp;amp; 3.43 \\&#xA;                   -210005.87230 &amp;amp; 4.80 \\&#xA;                   -154288.97870 &amp;amp; 6.17 \\&#xA;                   -98572.08511 &amp;amp; 7.55 \\&#xA;                   -42855.19149 &amp;amp; 8.92 \\&#xA;                   -12861.70213 &amp;amp; 10.29 \\&#xA;                   12861.70213 &amp;amp; 10.29 \\&#xA;                   42855.19149 &amp;amp; 8.92 \\&#xA;                   98572.08511 &amp;amp; 7.55 \\&#xA;                   154288.97870 &amp;amp; 6.17 \\&#xA;                   210005.87230 &amp;amp; 4.80 \\&#xA;                   265722.76600 &amp;amp; 3.43 \\&#xA;                   321439.65960 &amp;amp; 2.06 \\&#xA;                   377156.55320 &amp;amp; 0.69 \\&#xA;                   432873.44680 &amp;amp; 0.69 \\&#xA;                   488590.34040 &amp;amp; 2.06 \\&#xA;                   544307.23400 &amp;amp; 3.43 \\&#xA;                   600024.12770 &amp;amp; 4.80 \\&#xA;                   655741.02130 &amp;amp; 6.17 \\&#xA;                   711457.91490 &amp;amp; 7.55 \\&#xA;                   767174.80850 &amp;amp; 8.92 \\&#xA;                   797168.29790 &amp;amp; 10.29 \\&#xA;                  \end{array}&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Even though you said this was fiction, note that it's highly&#xA;unlikely that the moons' orbital period will be an exact multiple of&#xA;the planets day. The only exception to this is if the moon(s) are&#xA;tidally locked, in which case the orbital period will equal exactly&#xA;one day.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Similarly, it's unlikely the planet's orbital period would be an&#xA;exact multiple of its rotation period (ours certainly isn't).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is an interesting problem in general, and I am writing&#xA;&lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/MATHEMATICA/bc-orrery.m&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/MATHEMATICA/bc-orrery.m&lt;/a&gt;&#xA;to solve a similar problem:&#xA;&lt;a href=&quot;https://physics.stackexchange.com/questions/197481/&quot;&gt;https://physics.stackexchange.com/questions/197481/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-03-07T03:14:15.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14054" PostTypeId="1" AcceptedAnswerId="14055" CreationDate="2016-03-07T10:35:07.620" Score="14" ViewCount="1886" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Comet&quot;&gt;Comets&lt;/a&gt; orbiting the sun have a &lt;a href=&quot;https://en.wikipedia.org/wiki/Comet#Tails&quot;&gt;&lt;em&gt;tail&lt;/em&gt;&lt;/a&gt; that points away from the Sun, due to the solar wind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Strictly speaking, a comet can have both a type I tail - ion tail, and a type II tail - a dust tail. This question is about the type II tail - the dust tail). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I understand that some of the source of the dust is volatile materials in the comet oxidising due to the comet's proximity to the sun. But surely some of these comets have been orbiting for hundreds of thousands of years (if not millions). Why do they still have unspent fuel? Why are they still dusty? Hasn't all the dust blown off by now?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;strong&gt;Why do comets appear to have a continuous supply of dust?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2894" LastActivityDate="2016-03-07T20:46:22.193" Title="Why do comets appear to have a continuous supply of dust?" Tags="&lt;solar-system&gt;&lt;the-sun&gt;&lt;comets&gt;&lt;solar-wind&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14055" PostTypeId="2" ParentId="14054" CreationDate="2016-03-07T13:55:47.323" Score="15" Body="&lt;p&gt;The rate of loss of mass from a comet is perhaps surprisingly low. A paper &lt;a href=&quot;http://www.lpl.arizona.edu/sites/default/files/alllpl/AfrPaperPubl.pdf&quot;&gt;The calculation of $Af\rho$ and mass loss rate for comets&lt;/a&gt; gives a rate for a &quot;typical&quot; comet at 1.29AU as 153kg/s. A typical comet has a mass of about $10^{13}kg$. Even if the comet is constantly active (and they are not) it would have enough material for over 2000 years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact comets are only active when the are near perihelion, and they are only near perihelion for a small fraction of their orbit. Put it together and you have the potential for an active life of hundreds of thousands of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, however, comets do run out of volatile substances, at which point they become inactive, and appear as small dark asteroids. An example of which is the &lt;a href=&quot;http://www.nasa.gov/feature/halloween-skies-to-include-dead-comet-flyby&quot;&gt;object 2015 TB145&lt;/a&gt;, which passed the Earth on Halloween 2015. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: not &quot;oxidising&quot;, but &quot;vaporising&quot;, the ices in the comet are changing into gas, but there is no burning.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-03-07T20:46:22.193" LastActivityDate="2016-03-07T20:46:22.193" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14056" PostTypeId="1" AcceptedAnswerId="14058" CreationDate="2016-03-07T16:07:42.123" Score="1" ViewCount="101" Body="&lt;p&gt;I was introduced to mean free path in my astro physics class, &#xA;and the professor posts a question for us to ponder: &quot;Why do we need the &quot;mean&quot; in mean free path?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This further makes me to wonder what is the mean free path for, and how is it used?&lt;/p&gt;&#xA;" OwnerUserId="10696" LastActivityDate="2016-03-07T16:58:49.983" Title="Why do we need the &quot;mean&quot; in mean free path?" Tags="&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14058" PostTypeId="2" ParentId="14056" CreationDate="2016-03-07T16:58:49.983" Score="2" Body="&lt;p&gt;In every context we talk about &quot;mean free path&quot;, we talk about particles. Particles usually bounce around a lot, from collision to collision. In between two bounces a particle moves in a straight line. That is the expected behaviour from Newton's laws of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Between one collision and the next, a particle therefore moves a distance. That is the &quot;free path&quot;. So why do we need to include &quot;mean&quot;? The thing is, you can never be sure about how far the particle moves after a collision. It might as well immediately bounce into another particle, as well as move a considerable distance. However, if you measure really many particles, their average is going to be pretty stable and predictable. That gives a nice, usable number. The &quot;mean&quot; here is for average. (In case you wonder, yes, this &lt;a href=&quot;https://en.wikipedia.org/wiki/Mean_free_path#Mean_free_path_in_optics&quot; rel=&quot;nofollow&quot;&gt;applies to optics too&lt;/a&gt;, but the concept requires a little more abstraction).&lt;/p&gt;&#xA;" OwnerUserId="10172" LastActivityDate="2016-03-07T16:58:49.983" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14059" PostTypeId="1" AcceptedAnswerId="14062" CreationDate="2016-03-07T17:58:25.653" Score="2" ViewCount="1006" Body="&lt;p&gt;I need to calculate a percentage difference of a B-V colour index between its estimated and actual value. So I tried doing this by &lt;em&gt;difference between values/actual value x 100&lt;/em&gt;. However as B-V values can be on either side of zero this did not give representative answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if anyone knows a way of calculating percentages of values close to zero or on either side of zero their help would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks     &lt;/p&gt;&#xA;" OwnerUserId="11090" LastActivityDate="2016-03-07T21:47:04.960" Title="How to calculate B-V colour index value percentage difference" Tags="&lt;star&gt;&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14061" PostTypeId="1" AcceptedAnswerId="14064" CreationDate="2016-03-07T21:37:47.580" Score="4" ViewCount="838" Body="&lt;p&gt;I was gazing through my binoculars at Jupiter last night and noticed 3 (maybe 4) tiny dots in line with it. A first I thought they might have been background stars but quickly realised their colour was similar to that of Jupiter and their brightness was larger than expected considering their size. As a result, I'm pretty certain I was looking at Jupiter's moons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably, I was seeing the largest of its moons (Io, Europa, Ganymede, Callisto). They were far too tiny to see any details, obviously, but would it have been only those moons I could see? I assume the rest are too small. And is seeing those 4 moons as tiny bright dots through my 7x50s feasible?&lt;/p&gt;&#xA;" OwnerDisplayName="user10636" LastActivityDate="2016-03-07T22:28:14.873" Title="Which of Jupiter's moons did I most likely see?" Tags="&lt;amateur-observing&gt;&lt;jupiter&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14062" PostTypeId="2" ParentId="14059" CreationDate="2016-03-07T21:47:04.960" Score="3" Body="&lt;p&gt;$B-V$ corresponds to the base 10 logarithm of a flux ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B-V = -2.5 \log \left(\frac{f_B}{f_V}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So trying to guess what you are trying to calculate, it is the percentage change in the blue to visible flux ratio?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In which case the percentage change is&#xA;$$ p = \frac{ 10^{-(B-V)_2/2.5} - 10^{-(B-V)_1/2.5}}{10^{-(B-V)_1/2.5}}\times 100$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A percentage change can of course be negative.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-03-07T21:47:04.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14063" PostTypeId="2" ParentId="14061" CreationDate="2016-03-07T22:11:59.297" Score="2" Body="&lt;p&gt;Yes, you have definitely seen the Galilean moons, as they are named after their discoverer — Io, Europa, Ganymede, and Callisto. Those four moons are so bright that you would in fact be able to see them with the naked eye if they hadn't been just next to the much brighter Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, the fifth brightest Moon, Amalthea, requires a telescope of &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10212/what-kind-of-telescope-do-i-need-to-see-most-of-the-jupiters-moons/10215#10215&quot;&gt;at least 25 cm diameter&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-03-07T22:11:59.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14064" PostTypeId="2" ParentId="14061" CreationDate="2016-03-07T22:28:14.873" Score="5" Body="&lt;p&gt;The galilean moons are visible in binoculars, indeed. Yesterday evening (assuming you're in the western hemisphere), 3 of them were visible as separate dots. See the chart here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=jupiter+galilean+moons+on+march+6+2016+at+9pm+pst&quot;&gt;http://www.wolframalpha.com/input/?i=jupiter+galilean+moons+on+march+6+2016+at+9pm+pst&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can adjust the query to reflect your time zone, and the actual date/time of the observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also smartphone apps that allow you to determine the position of the moons at any given time. On Android I use the Moons of Jupiter app.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2016-03-07T22:28:14.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14065" PostTypeId="2" ParentId="13105" CreationDate="2016-03-08T05:37:34.380" Score="-2" Body="&lt;p&gt;The Big Bang did not happen at a single point, nor did the Universe begin at a single point. The Universe is, and has always been, infinite. The Big Bang was just when the Universe's expansion really began — that is, when objects started drifting away from each other. The Universe was still infinite, but there was less space between the matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This density caused the Universe to get extremely hot and expand. So here, space itself was distorted and began to expand. More and more space was created between the matter, and still is now (although now it is mainly due to dark energy instead of heat).&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2016-03-08T05:37:34.380" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14066" PostTypeId="1" CreationDate="2016-03-08T05:58:26.207" Score="4" ViewCount="138" Body="&lt;p&gt;Are there any coordinates for the Black Hole system, or galaxy for which the LIGO observations, observed the gravity wave? Any coordinates of the black hole system for which the gravity wave signal has been observed? &lt;/p&gt;&#xA;" OwnerUserId="11102" LastActivityDate="2016-03-08T08:59:59.303" Title="What star system/Galaxy did LIGO observe the gravity wave?" Tags="&lt;gravity&gt;&lt;black-hole&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14067" PostTypeId="2" ParentId="14066" CreationDate="2016-03-08T07:06:06.950" Score="5" Body="&lt;p&gt;The location of GW150914 was rather poorly constrained by the 6.9 ms time difference between the event when it was observed by the two separated LIGO detectors at Livingston and Hanford.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further details are here &#xA;&lt;a href=&quot;https://www.ligo.caltech.edu/image/ligo20160211b&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.ligo.caltech.edu/image/ligo20160211b&lt;/a&gt; and this site provides a visualisation of where the waves came from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AyEAe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AyEAe.jpg&quot; alt=&quot;Location of GW150914&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The location OF GW150914 can only be constrained to be somewhere within 600 square degrees with 90% confidence. There are millions of possible host galaxies within this region.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-03-08T08:59:59.303" LastActivityDate="2016-03-08T08:59:59.303" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14068" PostTypeId="1" CreationDate="2016-03-08T07:41:29.910" Score="3" ViewCount="324" Body="&lt;p&gt;Which one is better?&#xA;Features: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)Magnification: 18x - 90x &#xA;Eyepiece Diameter: 5mm (18x), 10mm (60x) &#xA;Objective Diameter: 50mm &#xA;Focal Length: 36cm &#xA;Tripod Height: 38cm &#xA;Main tube Color: Silver &#xA;Net Weight: 1.78kg &#xA;Case Size: 43cm(L) x 10cm(W) x 26cm(H)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)Optical Design: Refractor Mount Type: Altazimuth Ideal Usage: Astronomical and land observation Focal Length: 500 mm Aperture: 40 mm Focal Ratio: 13 Highest Useful Magnification: 94x Finderscope: 2 x 20 Lens Coating: Fully coated Light Gathering Power: 33x Limiting Stellar Magnitude: 10.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) 2x Barlow Lens and Enhanced Stability The Celestron Powerseeker 40AZ black comes with two extra eyepieces, one which is 20 mm and magnifies up to 25x and the other one is 8 mm which magnifies up to 63x. It also includes a 2x Barlow lens that almost doubles the magnifying capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4)Meade NG60-SM Altazimuth Refractor Telescope Meade's value priced NG60-SM Altazimuth Refractor is an affordable entry level telescope that features an easy to use Altazimuth mount with slow motion controls for precise tracking. The complete package includes a sturdy metal tripod, a red dot viewfinder, two 1.25 inch eyepieces and a star diagonal, and a software DVD with instructional video. The NG60-SM Refractor Telescope comes disassembled in a compact box, but the instructional DVD video guides you through all the steps required for assembly. Go ahead and try it out in the daytime, that's the best time to align the red-dot finder scope while looking at a distant tree or telephone pole. The optics of Meade's NG60-SM produce an image that is right side up but the diagonal mirror reverses the image left-to-right. That's no problem most of the time, but an optional correct image diagonal is available. The low power 25mm eyepiece produces a magnification 28X which is just right for spotting the Moon or the planets, while the 9mm eyepiece (78X magnification) can be used to zoom in for more detail. The MH25 eyepiece at 28X shows a lovely view of the Lunar disk in a dark sky, while the MH9 eyepiece at 78X shows literally hundreds of craters on the Moon and begins to show the rings of Saturn and the cloud bands of Jupiter. For even better views it's easy to add better eyepieces. My best view of Saturn, for example, came with an optional 6.4mm Super Plossl eyepiece, about 110X magnification. The Altazimuth mount included with Meade's NG60-SM is lighter and easier to use than an Equatorial mount, yet the slow motion knobs make it easy to keep objects in view. The rotation of the Earth causes the Moon or planets to appear to drift out of the eyepiece, but the slow motion knobs make it easy to keep the telescope on target, even with a high power eyepiece. The telescope is sensitive to vibration, however, so a high power eyepiece can be difficult to focus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So which 1 is best?????&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanna see Saturn its rings,Jupiter and its moons,mars,venus, basically planets and their moons&lt;/p&gt;&#xA;" OwnerUserId="11051" LastEditorUserId="11051" LastEditDate="2016-03-08T08:30:13.063" LastActivityDate="2016-04-07T09:53:31.487" Title="Telescope choice" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14069" PostTypeId="2" ParentId="14068" CreationDate="2016-03-08T09:39:52.277" Score="1" Body="&lt;p&gt;If your choice is between a 40mm refractor and a 60mm refractor, I would say the 60mm is likely to be better. I had (still have) a 40mm and it was OK for a beginner but dim at higher powers, and those maximum powers were probably around 70 times (very rough estimate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A note about the star diagonal: these reverse left to right as mentioned in the question, but this is not a problem for astronomy. After a little while you learn to swap left for right when aiming the scope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 60mm refractor will show Jupiter as a fairly small bright circle at 80x, in still conditions two or more horizontal bands will often be visible. The moons will be visible in a low power eyepiece as small points, and their positions move from night to night. (Note we wouldn't expect to see any detail in the moons, just their positions.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either of the small refractors you mentioned will show the moon quite well, there is a lot to see especially when the air is still.&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2016-03-08T09:39:52.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14070" PostTypeId="1" CreationDate="2016-03-08T12:59:12.103" Score="2" ViewCount="89" Body="&lt;p&gt;I am trying to find application that would be capable of notifying me about &#xA;&lt;a href=&quot;http://spotthestation.nasa.gov/&quot; rel=&quot;nofollow&quot;&gt;visible ISS pases&lt;/a&gt; for specified location and/or &lt;a href=&quot;http://heavens-above.com/IridiumFlares.aspx&quot; rel=&quot;nofollow&quot;&gt;visible Iridium flares&lt;/a&gt; for specified location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This software must be capable of running on Ubuntu.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I strongly prefer active open-source projects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that there are multiple mobile apps but I want something that may run on Linux PC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To avoid &quot;this is subjective&quot; - I will be happy about any software fulfilling this requirements and criteria are not subjective.&lt;/p&gt;&#xA;" OwnerUserId="10434" LastActivityDate="2016-03-08T15:54:34.617" Title="Notifier about visible ISS passes and/or Iridium flares for Ubuntu PC" Tags="&lt;software&gt;&lt;artificial-satellite&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14071" PostTypeId="2" ParentId="14070" CreationDate="2016-03-08T15:54:34.617" Score="1" Body="&lt;p&gt;Here is something that I wrote for my personal purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fetches iridium flare details based on your heavens-above credentials and displays a notification in Ubuntu. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/astronomersiva/satellite-passes-notifier&quot; rel=&quot;nofollow&quot;&gt;Github&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had one for ISS passes as well. Will find it and push it to the repo soon.&lt;/p&gt;&#xA;" OwnerUserId="1853" LastActivityDate="2016-03-08T15:54:34.617" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14072" PostTypeId="1" AcceptedAnswerId="14078" CreationDate="2016-03-08T15:56:06.147" Score="0" ViewCount="138" Body="&lt;p&gt;In a nutshell, the theory of relativity says that we cannot know that the ether is moving because if we were on the sun, the earth would seem to be moving and visa-versa. Apparently, we can't find &lt;code&gt;absolute&lt;/code&gt; motion; we can only find &lt;code&gt;relative&lt;/code&gt; motion.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How do we know that indeed the earth moves? &lt;/li&gt;&#xA;&lt;li&gt;What is relative motion?&lt;/li&gt;&#xA;&lt;li&gt;What is absolute motion? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11105" LastEditorUserId="2365" LastEditDate="2016-03-08T20:54:21.583" LastActivityDate="2016-03-09T01:54:17.063" Title="How can we tell that the sun is moving with theories such as the theory of relativity?" Tags="&lt;general-relativity&gt;&lt;geocentrism&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14073" PostTypeId="2" ParentId="14072" CreationDate="2016-03-08T16:54:25.413" Score="2" Body="&lt;p&gt;We can measure absolute motion, as this is motion with respect to the CMB, and what do you know, they both move (the Sun and the Earth).&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2016-03-08T16:54:25.413" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14074" PostTypeId="2" ParentId="14072" CreationDate="2016-03-08T16:54:52.967" Score="3" Body="&lt;p&gt;We don't know. All that we know is that the Earth moves relative to the Sun and relative to other things in the wider universe. That is what we mean by relative motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A measurement of absolute motion requires you to define an &lt;em&gt;absolute reference frame&lt;/em&gt;. We don't have one of those - you can choose your reference frame to suit your purposes. The beauty of relativity is that one of its axioms is that the laws of physics remain unchanged whichever reference frame you pick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conrad suggests using the cosmic microwave background as an absolute frame of reference. Whilst I agree that this is a useful frame of reference - it demonstrates with what velocity we (or any other object) are moving with respect to the local co-moving volume; there is no sense in which this is absolute or has any special properties beyond its convenience. The laws of physics remain unchanged whatever your velocity with respect to the cosmic microwave background.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-03-08T17:04:26.263" LastActivityDate="2016-03-08T17:04:26.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14075" PostTypeId="1" CreationDate="2016-03-08T19:15:34.670" Score="2" ViewCount="1206" Body="&lt;p&gt;Please forgive the simplicity of my question.  I have only had basic science courses.  Common sense seems to indicate that planetary orbits should be very close to circular, any perturbations having been eliminated over the ages.  Could it be that the composition of the planet be influencing the pull of gravity to produce a different gravitational attraction due to its mass?  I mean that if a planet were composed all of one substance, would the planetary orbit be perfectly circular?  I think about earth having constantly shifting water and gasses and even solids that affect its own rotation.  Might these shifts also affect the orbit of the planet around the sun, or does the total mass of the earth and moon together determine the orbit?&lt;/p&gt;&#xA;" OwnerUserId="11106" LastActivityDate="2016-03-09T03:31:17.793" Title="Why aren't planet orbits circular?" Tags="&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2016-03-09T05:13:51.490" ContentLicense="CC BY-SA 3.0" />
  <row Id="14077" PostTypeId="2" ParentId="12974" CreationDate="2016-03-08T21:20:03.427" Score="3" Body="&lt;p&gt;Yes, provided the star is big enough, the &quot;Jupiter&quot; is big enough and the planet is small enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The details are discussed herein: &quot;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/3595/is-there-a-ceiling-for-stable-l4-or-l5-masses&quot;&gt;Is there a ceiling for stable L4 or L5 masses?&lt;/a&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarise: The sun needs to be 25 times more massive than the giant planet, and the small planet needs to have negligible mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worth noting that the minor body could not be said to be dominating its orbit, and so would be considered a dwarf planet by the IAU.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-03-08T21:20:03.427" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14078" PostTypeId="2" ParentId="14072" CreationDate="2016-03-09T01:54:17.063" Score="1" Body="&lt;p&gt;There are some simple answers to your questions, and some (historic) controversy.  I'll answer the simple elements first (but backwards by your ordering of the questions), and then describe the nature of the controversy.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;What is absolute motion?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Absolute motion is motion that is referenced (compared to) a feature that you can say is standing still.  For example, how do you know the speed of a car?  You compare it to something that is standing still.  &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;What is relative motion?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Relative motion is motion that is referenced (compared to) something that may be moving.  For example, when driving, you may find that the car next to you is edging ahead of you.  You can say that it is moving 5 miles per hour faster than you - that is its relative speed.  (Note, it also has an &quot;absolute&quot; speed, say 65 miles per hour, while you have an &quot;absolute&quot; speed of 60 miles per hour.  Here I use &quot;absolute&quot; in the sense of comparing speed to the surface of the Earth, not a universal absolute.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This brings us to the definition of a reference frame, which is critical for understanding the later arguments.  A reference frame is a perspective from which one can begin to measuring things.  It is a defined zero point from which you can measure distance, speed etc.  So, in the example above, you have a reference point, yourself in the car, and you see the other car moving ahead of you at 5 miles an hour, and you also see the ground moving past you at 60 miles an hour.  The driver of the car also has a reference point, himself, and sees you as moving backwards at 5 miles an hour, and the ground moving past at 65 miles an hour.  Someone sitting on the ground also has a reference point, seeing two cars moving away at 60 and 65 miles an hour respectively. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How do we know that indeed the earth moves?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We compare the Earth to other things.  We can compare it to the sun, and therefore know that it is both moving around itself (rotating) and around the sun (revolving).  We compare the sun to the milky way, noting that the entire solar system is orbiting around the center of our galaxy.  We compare the galaxy to other galaxies, and note that our galaxy is drifting towards our local great attractor.  And (this is where the controversy of the other answers comes into play) we can compare our speed to the equivalent motion of an object at rest with the Cosmic Microwave Background Radiation (called CMBR for short), and can even deduce how fast the local great attractor is moving compared to that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the controversy?  The controversy is around the question, &quot;Is there a universal reference frame?&quot; In other words, is there such thing as a point in the universe that we can say is the zero point that we measure all positions and velocities from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Historically, this was fought out between Einstein and Lorentz.  Before the Michelson-Morley experiments, it was thought that light traveled through a ghostly material called an &quot;aether&quot; that wasn't moving, which could be considered the universal reference frame.  The Michelson-Morley experiments were set up to determine the Earth's relative motion through the aether, but it didn't find any.  Lorentz created a mathematical method that could account for this, but it was missing a physical basis.  Then along came Einstein, who replaced the assumption of an aether with the assumption that light travels at a constant speed.  From this he developed the now famous Special Theory of Relativity (although, that is the name it has subsequently come to be called). This was a little too far for Lorentz, and he developed a counter theory, which slowly transformed into a very similar theory to Einstein's, with one exception - Lorentz believed that there was an, as yet undetectable, zero reference frame for the universe.  Here is what he said in &lt;a href=&quot;https://en.wikipedia.org/wiki/Hendrik_Lorentz&quot; rel=&quot;nofollow&quot;&gt;1910&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Provided that there is an aether, then under all systems x, y, z, t, one is preferred by the fact, that the coordinate axes as well as the clocks are resting in the aether. If one connects with this the idea (which I would abandon only reluctantly) that space and time are completely different things, and that there is a &quot;true time&quot; (simultaneity thus would be independent of the location, in agreement with the circumstance that we can have the idea of infinitely great velocities), then it can be easily seen that this true time should be indicated by clocks at rest in the aether. However, if the relativity principle had general validity in nature, one wouldn't be in the position to determine, whether the reference system just used is the preferred one. Then one comes to the same results, as if one (following Einstein and Minkowski) deny the existence of the aether and of true time, and to see all reference systems as equally valid. Which of these two ways of thinking one is following, can surely be left to the individual.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Einstein maintained that not only was a universal reference frame undetected, it was unnecessary, and Occam's razor should rule (i.e. that if something is unnecessary for the theory to work, then it shouldn't be included in the theory).  It is interesting that both theories actually predict the same things, but in the end, Einstein won. Most physicists these days would state that a universal reference frame does not exist.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since that time, two key things have happened that impact this.  First is the discovery of the expansion of the universe. This means that, even if we were to make an arbitrary universal zero position and time, other reference frames that start out at stationary compared to this reference frame will start to move away, &lt;strong&gt;even though any object there has undergone no acceleration&lt;/strong&gt;. This greatly complicates any attempt to set a universal reference frame, an makes it much easier to consider that none exist.  Tick one for Einstein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second thing was the discovery of the CMBR. This is the relic of a change in the entire universe at around 380,000 years after the big bang.  At that point, the universe went from opaque (where light couldn't travel very far without encountering some matter to interact) to transparent (where some light could travel consistently without interruption until someone like you measures it).  Two interesting (and relevant) features were found about this CMBR - it comes from every direction (consequently it is universal, i.e. can be seen from anywhere), and it is very consistent in intensity from every direction (termed isotropic).  This means that anyone, anywhere, can measure their speed relative to the local knowledge of the universe.   Furthermore, this effective reference frame can be used to measure velocity irrespective of the expansion of the universe, if one considers certain interpretations about comoving reference frames. Although this is very different to a universal reference frame of an aether, it is certainly, in one sense, a universal reference frame.  Tick one to Lorentz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what does this mean?  Does it mean that Einstein was wrong?  No.  Einstein was concerned with the physics of light and electrodynamics, of the relationship between time and position and velocity.  The basis of Einstein's argument was that, no matter how fast you go relative to any other object, the physics of what you measure locally will be the same. The original concept of a universal frame implied otherwise, and Einstein has been proven experimentally correct over that original concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the concept of using the CMBR to define a universal reference frame is a slightly different idea than the original concept (and more in line with Lorentz's final theory).  It does not imply any change in the local physics due to relative speed.  But it is, as suggested by Rob Jeffries above, a useful reference frame.  I would argue it is a special reference frame, in a manner that science has created special zero points throughout history (e.g. sea level, 0 degrees latitude, binding energy etc).  None of these zero points change the physics of what is measured locally, but they make it much easier to compare measurements at different locations etc.  In the same way, the CMBR reference frame makes it easier to answer a range of questions about non-local physics.  For example, what is the age of the universe?  (The age is maximum at the CMBR reference frame.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in conclusion, when we see some arguments between physicists about whether the CMBR is an absolute or universal reference frame, it is usually because one side or the other is interpreting the term in the same manner as the original concept, with all its incorrect implications, rather than the lesser concept of a natural definer of the local speed to a universal feature of the universe.&lt;/p&gt;&#xA;" OwnerUserId="10800" LastActivityDate="2016-03-09T01:54:17.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14079" PostTypeId="2" ParentId="14075" CreationDate="2016-03-09T02:22:04.523" Score="2" Body="&lt;p&gt;They &lt;strong&gt;are&lt;/strong&gt; close to circular, although we might quibble about the meaning of &quot;close&quot;. Except for Mercury, all planetary orbits have eccentricities below 0.1. That's close enough to a circle to actually pass for a circle at a superficial examination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;&#xA;Mercury 0.2056&#xA;Venus   0.0068&#xA;Earth   0.0167&#xA;Mars    0.0934&#xA;Jupiter 0.0484&#xA;Saturn  0.0542&#xA;Uranus  0.0472&#xA;Neptune 0.0086&#xA;&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out these ellipses. The first one has an ecc. of 0.1. Looks like a circle, doesn't it? Even the second one (ecc = 0.2) kind of looks like a circle, too, if you don't examine it too long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/r0pPr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/r0pPr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for why they are not exactly circular, down to the last digit? There will always be perturbations that will squish perfect circles a little. Planets interact with each other, and pull each other off perfect circular orbits. It's very unlikely that in a complex system like ours you'll find highly circular planetary orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, again, the differences between the current planetary orbits and perfect circles are for the most part quite tiny, and you should think of them as &quot;basically circles&quot;, unless you're an astronomer or a NASA engineer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comets, OTOH, those do tend to move on narrow elliptic orbits.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2016-03-09T02:37:32.610" LastActivityDate="2016-03-09T02:37:32.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14080" PostTypeId="2" ParentId="14075" CreationDate="2016-03-09T03:31:17.793" Score="2" Body="&lt;p&gt;While this has been answered before, I'll touch on a few of your specific questions. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Common sense seems to indicate that planetary orbits should be very&#xA;  close to circular, any perturbations having been eliminated over the&#xA;  ages.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This seemed logical to a lot of people including some big brains throughout history like Aristotle, Ptolomy, even Copernicus and Kepler himself thought circles were neater than ellipses, but he couldn't deny that the carefully detailed planet charts and the mathematical calculations worked so much better with ellipses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It wasn't until Newton that they knew why.  Newton worked it out that ellipses are stable orbits, and in fact a circle is a type of ellipse with zero eccentricity.  Somebody said once, and I'm repeating, setting a dial exactly at zero is very hard because you're always going to be off by a tiny fraction.  Setting the dial between 0 and 1 is easy.  That's why all orbits are ellipses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now when you say perturbations, that usually refers to a 3rd body in the system.   For example, Earth-Sun, the Earth orbits the sun in an ellipse.  If you have Earth and Venus orbiting the Sun, Venus and earth perturb each other's ellipse.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Could it be that the composition of the planet be influencing the pull&#xA;  of gravity to produce a different gravitational attraction due to its&#xA;  mass? I mean that if a planet were composed all of one substance,&#xA;  would the planetary orbit be perfectly circular?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gravity does change a bit at different points in an orbit for the reasons you say, planets aren't uniform.  The Moon is especially unbalanced for example so this is more apparent with orbits around the moon, but all planets have some degree of non uniform mass and that has some tiny effects but that's not the cause for ellipses though, in fact, non-uniform mass will slightly warp an elliptical orbit.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I think about earth having constantly shifting water and gasses and&#xA;  even solids that affect its own rotation. Might these shifts also&#xA;  affect the orbit of the planet around the sun, or does the total mass&#xA;  of the earth and moon together determine the orbit?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We think of water as sloshing around and causing drag cause that's our experience when we spin or move something with water in it, but that's not what happens in orbit.  The Earth in orbit around the sun, for example, everything on the earth including the oceans are all falling around the sun together, so there's no &quot;shifting&quot; going on, in the sense that I think you mean.  There are tidal bulges, but the effect of the tidal bulges on the shape of an orbit is pretty small.   In fact, there's a curious side effect to tidal bulges, they tend to circularize orbits, over time.  They actually have the opposite effect of what you suggest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to cover this at a basic science level.   Corrections / clarifications welcome.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-03-09T03:31:17.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14081" PostTypeId="2" ParentId="12992" CreationDate="2016-03-09T06:10:24.317" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How many known comets are known to have had their orbits substantially&#xA;  changed by a single flyby?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Probably nearly all if not all of the (traditional) &lt;a href=&quot;https://en.wikipedia.org/wiki/Comet#Short_period&quot; rel=&quot;nofollow noreferrer&quot;&gt;short period comets&lt;/a&gt; were deflected in closer to the sun by one of the gas giant planets.  The reason for that is simple.  Comets can't form in orbits with a near-sun fly-by so nearly all of them had to be deflected inwards by a gas giant planet.   From the Wikipedia article quoted above.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Based on their orbital characteristics, short-period comets are&#xA;  thought to originate from the centaurs and the Kuiper belt/scattered&#xA;  disc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Short Period Comets also have a limited lifespan.  Halley's for example, with a perihelion of 0.6 AU, which is closer to the sun than Venus, it loses a significant amount of ice every time it flies past the sun and Halley's may lose all of it's ice within a hundred or so more fly-bys.  &lt;a href=&quot;http://www.thenakedscientists.com/HTML/questions/question/1000112/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most Short Period comets are also planet crossers because they have highly elliptical orbits and planet crossing orbits aren't long term stable either.   So of the 586 known short period comets, it's safe to say they have an lifespan of (probably) less than a million years when averaged out.  Also, if you consider that Jupiter can work both ways, it can pull comets out of the inner solar system as well as toss them in, then you can work out that they would need to be replaced at least fairly often, perhaps on the scale of every 100 years or so, perhaps on the scale of decades.  (there's probably more accurate estimates out there, my estimate is on the rough side)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a great deal of variation with individual comets life spans though.   For example, a Halley-type orbit (see picture) doesn't cross Jupiter's orbit due to the 18 degree inclination, but Halley type comets are more rare (75 known per Wikipedia article above).   Jupiter family short period comets are more common (511 known), and those are also more likely to get jostled by Jupiter since they cross Jupiter's orbit.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wv4uz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wv4uz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source of picture &lt;a href=&quot;http://lifeng.lamost.org/courses/astrotoday/CHAISSON/AT314/HTML/AT31402.HTM&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia also mentions &lt;a href=&quot;https://en.wikipedia.org/wiki/Main-belt_comets&quot; rel=&quot;nofollow noreferrer&quot;&gt;Main-belt comets&lt;/a&gt;, but those aren't what I think of as traditional comets as they orbit within the asteroid belt and don't have highly elliptical orbits.   Because the Asteroid belt is mostly stable in a kind of gravitational zone between Jupiter and Mars, those &quot;comets&quot; may have been there for billions of years, and are only beginning to show comet like features as the sun has increased in luminosity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most common type of comet, &lt;a href=&quot;https://en.wikipedia.org/wiki/Comet#Long_period&quot; rel=&quot;nofollow noreferrer&quot;&gt;long period comets&lt;/a&gt;, per the article, some 4,600 of them and about 10 new ones discovered every year and many many more exist but have not been discovered.  The Oort cloud has perhaps more than a trillion &quot;potential&quot; comets, or cosmic icy objects 1 KM or greater in diameter.   &lt;a href=&quot;https://en.wikipedia.org/wiki/Oort_cloud#Structure_and_composition&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know enough to say how Oort cloud objects are directed toward the inner solar system on average.   With so many objects, it's possible that some Oort cloud objects naturally formed with high enough eccentricity to pass through the inner solar-system once an orbit.  Some might be deflected by objects like Planet Nine (assuming it exists) or by repeated passes of dwarf planets or by the occasional pass of a rogue planet or the very rare close pass of another star, but it's way above my paygrade to try to estimate by which method Oort cloud comets are usually directed into the inner solar-system.  A percentage of those probably pass close enough to Jupiter or one of the other gas giants for a measurable deflection but I couldn't even guess what percentage.   Since these are the most common comets, it's very hard to give an estimate to your question in relation to these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Loosely related, but &lt;a href=&quot;https://en.wikipedia.org/wiki/Comet_Shoemaker%E2%80%93Levy_9#Jupiter-orbiting_comet&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shoemaker-Levy was captured&lt;/a&gt; into Jupiter's orbit around 1960 or 1970.  It's hard to know what it's orbit was before then because the calculations are difficult, but it had to be a Jupiter crosser of some kind.  It's apogee (.33 AU) extended outside of Jupiter's &lt;a href=&quot;https://en.wikipedia.org/wiki/Sphere_of_influence_%28astrodynamics%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sphere of influence&lt;/a&gt;, which works out to about .22 AU, but it was a stable enough orbit to make a few cycles before crashing into Jupiter in 1994.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter's sphere of influence (.44 AU in diameter) is barely over 1% of it's orbital journey around the sun, but over hundreds or thousands of orbits, 1% is enough to clean out or deflect most of the objects that cross it's path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Comet_ISON&quot; rel=&quot;nofollow noreferrer&quot;&gt;Comet Ison&lt;/a&gt;, a long period comet was also (virtually) destroyed when it passed around the sun close enough to break apart in November 2013.   Comets seem to die fairly frequently.   Fortunately (or perhaps unfortunately), there's trillions more where they came from.   :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that wasn't too vague an answer, but I think the short answer, based on deduction is that events like the one you mention with Jupiter deflecting Comet 67P are reasonably common, maybe every few decades, maybe once a century.   They're not rare events.   Earth will measurably deflect a small asteroid in 2029.   Asteroids are much more common than comets, but Earth is also much smaller than Jupiter.  Icy bodes are likely to pass close enough to Jupiter to have a significant deflection every so often and there's about a 50/50 chance on whether that deflection will push them closer to the sun or further away with a small chance that they will be caught in Jupiter's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that wasn't too general.   Since this had gone unanswered, I thought I'd give it a shot.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-03-09T06:10:24.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14082" PostTypeId="1" AcceptedAnswerId="14083" CreationDate="2016-03-09T09:27:18.143" Score="35" ViewCount="7805" Body="&lt;p&gt;Stars twinkle because their light has to squeeze through several different layers of the Earth's atmosphere. So why doesn't the moon twinkle as well?&lt;/p&gt;&#xA;" OwnerUserId="9510" LastEditorUserId="7982" LastEditDate="2021-02-28T11:50:04.503" LastActivityDate="2021-02-28T11:50:04.503" Title="Why doesn't the moon twinkle?" Tags="&lt;the-moon&gt;&lt;light&gt;&lt;atmospheric-effects&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14083" PostTypeId="2" ParentId="14082" CreationDate="2016-03-09T12:01:24.280" Score="64" Body="&lt;p&gt;The first handful of hits on Google actually return incomplete and even wrong answers (e.g. &quot;Because the Moon is much brighter&quot; which is plain wrong, and &quot;Because the Moon is closer&quot; which is incomplete [see below]). So here's the answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you mention, when light enters our atmosphere, it goes through several parcels of gas with varying density, temperature, pressure, and humidity. These differences make the refractive index of the parcels different, and since they move around (the scientific term for air moving around is &quot;wind&quot;), the light rays take slightly different paths through the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;strong&gt;Stars are point sources&lt;/strong&gt;&#xA;&#xA;&lt;p&gt;Stars are immensely far away, effectively making them point sources. When you look at a point source through the atmosphere, the different paths taken from one moment to another makes it &quot;jump around&quot; — i.e. it twinkles (or &lt;em&gt;scintillates&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The region in which the point source jumps around spans an angle of the order of an arcsecond. If you take a picture of a star, then during the exposure time, the star has jumped around everywhere inside this region, and thus it's no longer a point, but a &quot;disk&quot;.&lt;/p&gt;&#xA;&#xA;&lt;strong&gt;…the Moon is not&lt;/strong&gt;&#xA;&#xA;&lt;p&gt;The same is true for the Moon, but since the Moon (as seen from Earth) is much larger (roughly 2000 times larger, to be specific) than this &quot;seeing disk&quot; as it's called, you simply don't notice it. However, if you are observing details on the Moon through a telescope, then the seeing puts a limit on how fine details you can see. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is even true for planets. The planets you can see with the naked eye span from several arcsec up to almost an arcmin. Although they &lt;em&gt;look&lt;/em&gt; like point sources (because the resolution of the human eye is roughly 1 arcmin), they aren't, and you will notice that they don't twinkle (unless they're near the horizon where their light goes through a thicker layer of atmosphere).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image below may help understanding why you see the twinkling of a star, but not of the Moon (greatly exaggerated):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wp9Ct.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wp9Ct.png&quot; alt=&quot;twinkle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: Due to the comments below, I added the following paragraph:&lt;/p&gt;&#xA;&#xA;&lt;strong&gt;Neither absolute size, nor distance is important in itself. Only the ratio is.&lt;/strong&gt;&#xA;&#xA;&lt;p&gt;As described above, what makes a light source twinkle depends on its &lt;em&gt;apparent&lt;/em&gt; size compared to the seeing $s$, i.e. its &lt;em&gt;angular diameter&lt;/em&gt; $\delta$ defined by the ratio between its absolute diameter $d$ and its distance $D$ from Earth:&#xA;$$&#xA;\delta = 2 \arctan \left( \frac{d}{2D} \right)&#xA;\simeq \frac{d}{D}\,\,\,\mathrm{for\,small\,angles}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\delta \lesssim s$, the object twinkles. If it's larger, it doesn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, saying that the Moon doesn't twinkle because it's close is an incomplete answer, since for instance a powerful laser 400 km from Earth — i.e. 1000 times &lt;em&gt;closer&lt;/em&gt; than the Moon — would still twinkle because it's small. Or vice versa, the Moon &lt;em&gt;would&lt;/em&gt; twinkle even at the distance it is, if it were just 2000 times smaller.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Finally, to achieve good images with a telescope you not only want to put it at a remote site (to avoid light pollution), but also — to minimize the seeing — at high altitudes (to have less air) and at particularly dry regions (to have less humidity). Alternatively you can just put it in space.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="9533" LastEditDate="2016-03-10T21:59:57.417" LastActivityDate="2016-03-10T21:59:57.417" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="14084" PostTypeId="1" CreationDate="2016-03-09T12:13:48.853" Score="2" ViewCount="118" Body="&lt;p&gt;I've obtained the coordinates of brightest pixels from different images (.fits files) using Python. I want to get the flux of each pixel. How do I go about finding out the flux?&lt;/p&gt;&#xA;" OwnerUserId="11112" LastActivityDate="2016-04-08T21:17:48.377" Title="Given a collection of pixels, how to deduce the flux from them?" Tags="&lt;star&gt;&lt;telescope&gt;&lt;observational-astronomy&gt;&lt;exoplanet&gt;&lt;light&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14086" PostTypeId="2" ParentId="14082" CreationDate="2016-03-09T15:49:31.283" Score="6" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Twinkling&quot; rel=&quot;nofollow&quot;&gt;wikipedia page on twinkling&lt;/a&gt;, aka scintillation, covers it quite succinctly; it boils down to the fact that distant stars are sufficiently distant to be a point source of coherent light. Solar planets and Luna are close enough to have a resolvable diameter while being visible, which means that their light is not coherent like a point source's might be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, the threshold at which a distant source of light becomes an effective point source is going to be a function of its size and distance, relative to the aperture size of the viewing device (in this case, the human eye). You could effectively think of it as a cylinder between the aperture and the perimeter of the light source: when that cylinder is sufficiently narrow when passing through the atmosphere, you get visible twinkling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important to note that scintillation is not the mirage effect, which is caused by temperature gradients in the atmosphere and causes the 'swimming' effect. Scintillation does not displace the apparent position of the light source, instead resulting in variations of brightness and colour. The actual mechanism of scintillation results from plane-wave light and atmospheric turbulence causing interference in that light's wavefront. This is clearly demonstrated by &lt;a href=&quot;http://apod.nasa.gov/apod/ap110428.html&quot; rel=&quot;nofollow&quot;&gt;this image from NASA&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11114" LastEditorUserId="11114" LastEditDate="2016-03-21T10:35:40.403" LastActivityDate="2016-03-21T10:35:40.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14088" PostTypeId="1" AcceptedAnswerId="14090" CreationDate="2016-03-09T16:17:46.677" Score="2" ViewCount="109" Body="&lt;p&gt;This is related to the question: &quot;Can dark matter decrease the Jeans length?&quot; If dark matter assists in any way with star formation, shouldn't it be detectable by precisely measuring velocities in young open clusters? In particular, could the motions of stars in the Pleiades all be explained without the presence of dark matter?&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="7997" LastEditDate="2016-03-11T05:31:58.327" LastActivityDate="2016-03-11T07:40:08.037" Title="Could dark matter assist with star formation?" Tags="&lt;dark-matter&gt;&lt;star-formation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14090" PostTypeId="2" ParentId="14088" CreationDate="2016-03-09T17:09:04.073" Score="6" Body="&lt;p&gt;Strange you should ask - I am currently working on a paper on NGC 2516 - &quot;the southern Pleiades&quot; and looking at the dynamical status of the stars compared with the distribution of visible mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our conclusion is that the radial velocities of the stars are in virial equilibrium, with velocity dispersions that are entirely consistent with the mass that is present in the stars that are visible. So, no need for dark matter there - you could probably rule out any contribution of more than about 50% of the visible stars. (NB the point of the work is not to look for dark matter, is is not expected - see below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one of the first very detailed measurement for an open cluster (like the Pleiades), because the velocity precisions required have to be considerably better than the velocity dispersion. And for a sparse cluster like the Pleiades, this is $&amp;lt;1$ km/s, which is challenging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one old paper by &lt;a href=&quot;http://adsabs.harvard.edu/abs/1970AJ.....75..563J&quot; rel=&quot;nofollow&quot;&gt;Jones (1970)&lt;/a&gt; that uses proper motions in the Pleiades to estimate a dynamical mass based on virial equilibrium of $690\ M_{\odot}$, which compares with $470\ M_{\odot}$ from directly counting stars. But Jones points out that the latter number is a lower limit (they certainly couldn't see very low mass stars and brown dwarfs in their study) and the former number has large uncertainties. So there is no real evidence for any &quot;dark matter&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dynamics of globular clusters are easier to measure - they are more massive and have much higher velocity dispersions. The general picture is reasonable agreement between velocity dispersions and inferred mass distributions from counting the visible stars. There are of course uncertainties, but there can be no large dark matter component. Limits as small as $&amp;lt;6$% dark matter have been set in some clusters (e.g. &lt;a href=&quot;http://arxiv.org/abs/1210.7787&quot; rel=&quot;nofollow&quot;&gt;Ibata et al. 2012&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first Gaia satellite results, due in about 18 months time, will blow this field apart. We will have exquisitely precise tangential velocities for $\sim 1000$ stars in the Pleiades.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark matter is not expected to play a role in small scale (star cluster) formation. When clusters form, the dissipation of energy in gas interactions is what allows them to end up as bound systems. Dark matter is dissipationless. As the escape velocity of these clusters is 1-10 km/s, compared to velocities of hundreds of km/s for datk matter, there is not even expected to be any gravitational concentration of dark matter. This is quite different from galaxy formation, which is associated with pre-existing dark matter structures.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-03-11T07:40:08.037" LastActivityDate="2016-03-11T07:40:08.037" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14093" PostTypeId="1" CreationDate="2016-03-09T19:25:26.123" Score="8" ViewCount="680" Body="&lt;p&gt;The mass of a star is limited to about a couple of hundred solar masses. The fusion rate scales strongly with density (which is why the most massive stars have extremely short lifetimes) so if a star was massive enough (above the Eddington limit) then the radiation pressure would blow it apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a black hole also has radiation, but by definition it is not capable of escaping due to gravity. But is there in some way a limit of the size. Imagine that all the matter in the universe formed a black hole. Should that be possible or is there a law which forbids creating it?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="10437" LastEditDate="2016-03-09T22:56:34.033" LastActivityDate="2020-06-23T00:14:49.553" Title="Is there a limit to the size of a black hole?" Tags="&lt;black-hole&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14094" PostTypeId="2" ParentId="14093" CreationDate="2016-03-09T19:35:59.437" Score="6" Body="&lt;p&gt;General Relativity does not limit the size of a black hole, and we have good (indirect) observations of black holes with a mass of more than a billion suns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes are not object that are held together by gravity, and which could be ripped apart if forces opposing that gravity were increased. They are (in GR) singularities: a complete collapse of the matter to a point. There is no &quot;object&quot; inside a black hole that could be emitting radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if all the matter in the universe were to be a black hole, then that singularity would be in all our futures. We wouldn't &quot;see&quot; the singularity because it is in the future, never in the past. In fact a universe in which is one black hole could look rather like the one in which we are living. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-03-09T19:35:59.437" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14095" PostTypeId="2" ParentId="11244" CreationDate="2016-03-09T22:10:58.840" Score="3" Body="&lt;p&gt;There is a symmetry in the rotating gravitational field, which means that capture of an asteroid to the L4 is just as likely as to the L5 Lagrange point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of Mars the split is 1:6, and a simple binomial model suggests a probability of 0.0625 (the probability of a 1:6 or 0:7 split given the hypothesis that they are distributed randomly is 0.0625) This doesn't give a reason to suppose there is anything unusual happening, and as noted in the comments, there is no favouring of L5 when other planets are considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted by UserLTK: &quot;Earth has an L4 asteroid, none in L5. Uranus also has an L4, no L5 and Neptune has more L4 than L5 objects.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conclusion is that &lt;em&gt;No, the greater number of L5 martian trojans is just a random effect&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-03-11T21:04:38.533" LastActivityDate="2016-03-11T21:04:38.533" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14098" PostTypeId="1" CreationDate="2016-03-10T01:11:58.920" Score="0" ViewCount="108" Body="&lt;p&gt;Have we ever observed, or better yet, recorded, an impact from a comet/meteorite/fragment/etc on another celestial body? For example, in all its time orbiting Mars, has the MRO observed anything impacting the surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the other objects in our solar system that we've observed (aside from gas giants) are either speckled or virtually covered with craters, but I've never heard mention of actually seeing one of these impacts happen. At a basic level I'm curious if we've ever actually directly seen one, but what would be even better is if there's a publicly-available video of it happening.&lt;/p&gt;&#xA;" OwnerUserId="7519" LastActivityDate="2016-03-16T10:31:29.667" Title="Celestial impacts" Tags="&lt;solar-system&gt;&lt;meteorite&gt;&lt;impact&gt;&lt;crater&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14099" PostTypeId="2" ParentId="14098" CreationDate="2016-03-10T02:44:09.053" Score="4" Body="&lt;p&gt;Yes, there is an observed impact of a comet into a celestial body. A fairly recent one in 1994 is the comet which impacted into Jupiter, &lt;a href=&quot;http://www.space.com/19855-shoemaker-levy-9.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shoemaker-Levy-9&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/L2YWn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L2YWn.png&quot; alt=&quot;Comet impacting into Jupiter&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a &lt;a href=&quot;https://www.youtube.com/watch?v=CiLNxZbpP20&quot; rel=&quot;nofollow noreferrer&quot;&gt;video&lt;/a&gt; of this event happening.&lt;/p&gt;&#xA;" OwnerUserId="11015" LastEditorUserId="10172" LastEditDate="2016-03-10T04:15:01.987" LastActivityDate="2016-03-10T04:15:01.987" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14100" PostTypeId="2" ParentId="14093" CreationDate="2016-03-10T05:37:44.937" Score="5" Body="&lt;p&gt;There's no theoretical limit.  If you had enough energy to move stars or galaxies, you could in theory keep feeding a black hole until it became enormously large, larger even than the Milky way for example.  But there are practical limits past which black holes are unlikely to grow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two reasons for this are that 1), black holes aren't efficient at taking in matter.  They can spit out as much as 90% of the energy from the matter that falls into them, and 2) once they reach a certain size, black holes are too large to form accretion disks, so matter tends to orbit around them rather than funnel into them.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://scitechdaily.com/black-holes-could-grow-as-large-as-50-billion-suns-before-their-food-crumbles-into-stars-research-shows/&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt; and &lt;a href=&quot;http://www.sciencedaily.com/releases/2008/09/080909095133.htm&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your 2nd question&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Imagine that all the matter in the universe formed a black hole.&#xA;  Should that be possible or is there a law which forbids creating it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've pondered this one myself and I have no idea the answer.  Is there a size past which Dark Energy would overcome gravitation?   Dark energy operating inside the black hole might overcome the gravitation past a certain size, but that's just my novice speculation and I think the black hole would need to be billions of light years across for that to happen.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know the answer to that one.  I'd be curious if anyone does though.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-03-10T05:37:44.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14101" PostTypeId="1" AcceptedAnswerId="14116" CreationDate="2016-03-10T07:35:24.550" Score="4" ViewCount="330" Body="&lt;p&gt;As I know, some stars produces carbon, nitrogen and oxygen when they are old and become white dwarfs or neutron stars. And even their surface gravity is strong, it may rotate very fast so that the surface gravity is diminished, hence materials such as rocks and water can present at the surface normally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, is it possible for life (eg:bacteria) to survive on the equator of white dwarf or neutron star if it is cooled and rotate very fast? Or is it even possible for the equator of white dwarfs or neutron stars become the origin of life?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastActivityDate="2020-11-27T11:38:03.050" Title="Can life survive on the equator of cooled and fast rotating white dwarf or neutron star?" Tags="&lt;life&gt;&lt;neutron-star&gt;&lt;white-dwarf&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14102" PostTypeId="1" AcceptedAnswerId="14103" CreationDate="2016-03-10T08:01:49.427" Score="22" ViewCount="6585" Body="&lt;p&gt;Does turning the colour of the sky blue need more luminous light? Does it depend on luminosity or some other factors are also responsible for this phenomenon? Why can't the moon light turn the sky blue even a little bit (at least the area near the disc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks  &lt;/p&gt;&#xA;" OwnerUserId="7964" LastEditorUserId="7982" LastEditDate="2021-02-28T11:49:29.777" LastActivityDate="2021-02-28T11:49:29.777" Title="Why can't moon light (reflected sun light) turn the sky blue?" Tags="&lt;light&gt;&lt;atmospheric-effects&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14103" PostTypeId="2" ParentId="14102" CreationDate="2016-03-10T09:42:09.633" Score="28" Body="&lt;p&gt;The simple answer is that it does, but it's not bright enough to be visible to the naked eye. Earth's atmosphere scatters the moon light just like sunlight.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The full moon (like the sun) fills about 1/2 of 1 degree of the sky, the entire sky being 180 degrees, give or take, so the full moon fills less than 1 part in 100,000 of the night sky, so there simply isn't enough blue light to be visible over the brighter stars even with the brightest full moon. Our eyes are very good at seeing variations in brightness, but not that good. . . . and, for what it's worth, the night sky has always appeared to have a dark bluish tint to me, but that might just be my brain playing tricks on me because logically I know it's there. I'm not sure whether it's actually visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a good sized telescope, moonlight scattering acts as a form of light pollution. Telescope users know that you get better visuals when there's no moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Light_pollution#Measurement_and_global_effects&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="209" LastEditDate="2016-08-03T12:28:11.180" LastActivityDate="2016-08-03T12:28:11.180" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14105" PostTypeId="2" ParentId="14102" CreationDate="2016-03-10T10:13:09.100" Score="5" Body="&lt;p&gt;userLTK nailed it, but I'll add the answer to the last part of your question. Close to the &#xA;Moon we often do see scattered light. This is a phenomenon called &lt;a href=&quot;https://en.wikipedia.org/wiki/Mie_scattering&quot; rel=&quot;nofollow&quot;&gt;Mie scattering&lt;/a&gt; where it cannot be assumed that the scatterers are much smaller than the wavelength of light (water droplets etc). Mie scattering is roughly wavelength independent and much stronger in the forward scattering direction. Therefore any &quot;glare&quot; around the Moon is usually the same colour as the Moon itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sometimes-seen &lt;a href=&quot;https://en.wikipedia.org/wiki/22%C2%B0_halo&quot; rel=&quot;nofollow&quot;&gt;halo&lt;/a&gt; (at about 22 degrees) is a different phenomenon caused by refraction through ice crystals in the upper atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-04-26T06:47:35.537" LastActivityDate="2016-04-26T06:47:35.537" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14106" PostTypeId="2" ParentId="14101" CreationDate="2016-03-10T15:24:18.603" Score="1" Body="&lt;p&gt;It is quite difficult to say about the presence of life, however it cannot be neglected as we know by researches that even on earth there are also many place where life was supposed to be impossible to exist but life exist . Now see average temperature on the surface of the dwarf star is about 2 times of our Sun's temperature and have more gravity and also the gases and liquids will not present at this much temperature hence it is very very difficult for life to exist there. &lt;/p&gt;&#xA;" OwnerUserId="9231" LastEditorUserId="10657" LastEditDate="2016-05-09T14:45:02.807" LastActivityDate="2016-05-09T14:45:02.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14107" PostTypeId="1" CreationDate="2016-03-10T15:28:11.420" Score="1" ViewCount="1308" Body="&lt;p&gt;If I am to apply at a telescope for viewing objects, how to determine (easily) what is the minimum and maximum possible RA and DEC visible from that telescope during that time of the year. For eg, take the telescope at paranal&#xA;(24.6258°S, 70.4310°W) and if I observe anytime from October to March. &lt;/p&gt;&#xA;" OwnerUserId="11140" LastActivityDate="2016-05-11T16:55:09.337" Title="How to find maximum and minimum right acsension and declination based on the telescope's location?" Tags="&lt;declination&gt;&lt;ascension&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14108" PostTypeId="1" AcceptedAnswerId="14113" CreationDate="2016-03-10T15:28:16.083" Score="1" ViewCount="24" Body="&lt;p&gt;I understand how infrared spectroscopy works for determining molecules in the atmosphere of a transiting object (like Venus from Earth). But how does this work for objects that don't pass between us and the sun? If the IR spectra is an absorption spectra, is the light emitted by a planet, like Uranus, enough for us to determine its atmoshpere's composition via IR spectroscopy?&lt;/p&gt;&#xA;" OwnerUserId="10995" LastActivityDate="2016-03-10T19:04:53.563" Title="Infrared Astronomy for non transiting objects?" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14109" PostTypeId="2" ParentId="11318" CreationDate="2016-03-10T15:45:26.683" Score="1" Body="&lt;h1&gt;Your First Question - Can it be done from the ground?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Unfortunately making maps that have this level of detail from the ground is not possible with current telescopes. There are 2 problems; first, it takes a lot of magnification to resolve Pluto and even more to resolve surface features. Second, from the ground you need more than just magnification, you need EXTREMELY stable skies or that will wash out the surface features. Getting just one super sharp image is not enough, you need several to ensure that you have high enough signal to noise to believe the surface features are real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why &lt;a href=&quot;http://www.boulder.swri.edu/~buie/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Marc Buie&lt;/a&gt;, who used the High Resolution Channel on the &lt;a href=&quot;http://www.stsci.edu/hst/acs/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Advanced Camera for Surveys&lt;/a&gt; on HST to create the image you put in your question. Unfortunately, even that level of resolution is not possible anymore as HRC is no longer functional. These kind of maps can be &lt;a href=&quot;http://www.boulder.swri.edu/~buie/pluto/mapstory.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;very hard to make&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been other maps of pluto, both from the ground and using HST. Here is a compilation of several using a compilation of ground based photometry, the Fine Optical Camera on HST and the ACS HRC. As you can see from the ground based photometry you can make a map, but it is very poor resolution compared to what HST can do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8Va0M.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8Va0M.png&quot; alt=&quot;Other Pluto Maps&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been other &lt;a href=&quot;http://hubblesite.org/newscenter/archive/releases/1996/09/&quot; rel=&quot;nofollow noreferrer&quot;&gt;HST press releases about those other images&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Second Question - How well do the maps match?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;There is a good &lt;a href=&quot;https://maas.museum/observations/2015/07/07/pluto-new-horizons-images-vs-the-2010-buie-maps/&quot; rel=&quot;nofollow noreferrer&quot;&gt;blog post&lt;/a&gt; about how they match up from the museum of applied arts and sciences at the Sydney Observatory. But the bottom line is that they match up pretty well, but not perfectly. This is expected as we expected volitle transport on the surface so the surface features will change. As you can see in the image below, that there were a lot of changes from the map made in 1994 to the one made in 2003. So one would expect changes going out to 2015.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://maas.museum/app/uploads/sites/6/2015/07/HSTmap.vs_.NHImage.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://maas.museum/app/uploads/sites/6/2015/07/HSTmap.vs_.NHImage.jpg&quot; alt=&quot;New Horizons and Hubble Images&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, yes your intuition is valid, but Pluto's surface is dynamic so it will be changing on you!&lt;/p&gt;&#xA;" OwnerUserId="11039" LastActivityDate="2016-03-10T15:45:26.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14110" PostTypeId="1" AcceptedAnswerId="14112" CreationDate="2016-03-10T18:01:13.017" Score="2" ViewCount="188" Body="&lt;p&gt;I am an amateur astronomer currently working on Spitzer telescope dataset with python (astropy and aplpy). From the examples in astropy's website, I understood that the FITS data is a way to represent image data with human-readable headers.  However, I now have a fits image ( SPITZER_I2_46469632_0000_0000_2_bcd.fits ) which is a 3D array of dimensions (64, 32, 32). I believe the second and third coordinates to be the length and height of the image in pixels. What about the first dimension? How do I interpret such images and &lt;code&gt;NaN&lt;/code&gt;s in the data?&lt;/p&gt;&#xA;" OwnerUserId="11142" LastActivityDate="2016-03-10T18:53:51.653" Title="3D array in FITS data" Tags="&lt;exoplanet&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14111" PostTypeId="2" ParentId="14102" CreationDate="2016-03-10T18:53:41.123" Score="13" Body="&lt;p&gt;It does actually, but the human eye can't see it. But long exposure photography can see it easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lfMNz.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lfMNz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or &lt;a href=&quot;https://i.imgur.com/PsOWCua.jpg&quot; rel=&quot;noreferrer&quot;&gt;this photo&lt;/a&gt;, taken about three hours after sunset and lit by a nearly-full moon.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1900" LastEditDate="2016-03-11T07:59:13.803" LastActivityDate="2016-03-11T07:59:13.803" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14112" PostTypeId="2" ParentId="14110" CreationDate="2016-03-10T18:53:51.653" Score="1" Body="&lt;p&gt;It appears to me that the full Spizter frame should normally be 256x256 but, if observing a very bright source or when requiring high temporal resolution, a different observing mode is performed wherein only 32x32 pixels are read from the CCD. However, what happens is that the exposure times are short and a total of 64 frames are observed in total, creating a data cube of 64x32x32. If you want, you can treat each individial 32x32 frame as its own image, or else stack them all together for the full image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://irsasupport.ipac.caltech.edu/index.php?/Knowledgebase/Article/View/118/0/what-is-different-about-taking-irac-data-with-the-subarray-mode-as-opposed-to-full-array-or-hdr-mode-and-what-are-the-main-differences-in-data-processing-and-data-products-for-irac-subarray&quot; rel=&quot;nofollow&quot;&gt;Full details can be found here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-03-10T18:53:51.653" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14113" PostTypeId="2" ParentId="14108" CreationDate="2016-03-10T19:04:53.563" Score="1" Body="&lt;p&gt;Yes it is. As long as light is produced by a source and then passes through a medium, we can perform spectroscopy on it. When considering a planet like Uranus, most of the light that it emits is actually reflected light or re-emitted light from the Sun. Uranus produces very little light by its own merit. For light that is directly reflected (by that I mean the light from the Sun goes into the atmosphere, hits molecules, and gets scattered out of the atmosphere again), this is directly equivalent to the standard transiting planets you mention in your question. For light that is re-emitted, you still see absorption spectra, but the black body background has changed. In this scenario, the light is absorbed lower in the atmosphere and re-emitted at a longer wavelength, but that emitted light must still try to leave the atmosphere and will experience some absorption along the way, creating absorption lines.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-03-10T19:04:53.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14115" PostTypeId="1" CreationDate="2016-03-10T20:30:05.870" Score="1" ViewCount="141" Body="&lt;p&gt;In the GR Einstein's interpretation was in terms of a deformation of the geometry of the space ending in a curvature of spacetime. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've thought that mass could do this by interacting with other mass and causes gravitational waves which makes the space curved. But a prof told me that this isn't the case. For example the gravitational waves caused between the earth and the sun is just a very tiny effect which causes spiraling earth to the sun which is of little effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So mass is establishing the geometry of the space and the time. But the question remains how is mass doing that (curving spacetime)?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="10445" LastEditDate="2016-03-10T20:56:55.130" LastActivityDate="2016-03-13T09:13:27.600" Title="How is mass curving spacetime?" Tags="&lt;gravity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14116" PostTypeId="2" ParentId="14101" CreationDate="2016-03-10T20:57:11.133" Score="3" Body="&lt;p&gt;I am going to attempt a weak answer, mods feel free to delete it, but I'm fairly certain I'm right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shortly - no. It's not possible. Even if you balance gravity and centrifugal force perfectly at ground level at the equator, they will very, very quickly become imbalanced as soon as you move north, south, or up from there. So quickly in fact that the gradients may be too big even for a human being not moving at all. Maybe if you're laying down, with your body oriented along the equator, but even then I think the gradients would be too big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe bacteria would survive, briefly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure the math could be done quite easily to estimate the gradients. This is based entirely on intuition.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2016-03-10T21:02:35.887" LastActivityDate="2016-03-10T21:02:35.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14118" PostTypeId="1" AcceptedAnswerId="14182" CreationDate="2016-03-10T21:46:29.307" Score="4" ViewCount="271" Body="&lt;p&gt;Do we know of any hot jupiters that could be orbiting a red dwarf (or, more probably, orbiting a barycenter between the two)? Is this scenario even physically possible?&lt;/p&gt;&#xA;" OwnerUserId="7461" LastActivityDate="2016-03-18T07:23:04.590" Title="Are there any hot jupiters orbiting red dwarfs?" Tags="&lt;orbit&gt;&lt;exoplanet&gt;&lt;star-formation&gt;&lt;red-dwarf&gt;&lt;hot-jupiter&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14119" PostTypeId="1" AcceptedAnswerId="14120" CreationDate="2016-03-11T07:25:47.047" Score="1" ViewCount="648" Body="&lt;p&gt;I'm looking for a link to a table of visible stars that is open and available to everyone. It should have magnitude, RA, dec, and possibly some indication of color. This will be used to produce somewhat realistic night skies as a backdrop for showing the motion of the planets in the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not so interested in names, constellations, etc. These are of course useful to know and possibly to plot, but what I'm primarily after is the information necessary to illustrate star position, brightness, and some color information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit:&lt;/strong&gt; to reiterate, &quot;&lt;em&gt;...table of visible stars that is open and available to eveyone.&lt;/em&gt;&quot; I'm assuming that in 2016 the visible stars are not behind a paywall, am I wrong? Approx RA, dec, mag - are these available for open access and usage?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2016-03-12T01:25:39.260" LastActivityDate="2016-03-12T01:38:45.920" Title="Open access table of visible stars with magnitude, coordinates, and possibly color?" Tags="&lt;astrometry&gt;&lt;star-catalogues&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14120" PostTypeId="2" ParentId="14119" CreationDate="2016-03-11T07:49:45.850" Score="4" Body="&lt;p&gt;The Hipparcos catalogue by van Leeuwen (2007) contains all the information you require, plus estimates of distance from parallax. It is open and free to use for scientific purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR?-source=I%2F311&quot; rel=&quot;nofollow&quot;&gt;http://vizier.u-strasbg.fr/viz-bin/VizieR?-source=I%2F311&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The direct page that describes the catalogue contents and ftp site is &lt;a href=&quot;http://cdsarc.u-strasbg.fr/viz-bin/Cat?I/311&quot; rel=&quot;nofollow&quot;&gt;http://cdsarc.u-strasbg.fr/viz-bin/Cat?I/311&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tables themselves are at &lt;a href=&quot;ftp://cdsarc.u-strasbg.fr/pub/cats/I/311&quot; rel=&quot;nofollow&quot;&gt;ftp://cdsarc.u-strasbg.fr/pub/cats/I/311&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-03-12T01:38:45.920" LastActivityDate="2016-03-12T01:38:45.920" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14121" PostTypeId="1" CreationDate="2016-03-11T08:28:18.960" Score="2" ViewCount="99" Body="&lt;p&gt;The Ligo website says &quot;Detecting the relic gravitational waves from the Big Bang will allow us to see farther back into the history of the Universe than ever before.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find this puzzling. How can a Gravitational Wave originating at about the same time and place as the particles that make up their detector (ie in the Big bang) be detected?&#xA;To use a simple analogy - an explosion in an electronics warehouse causes a flash of light and pieces of electronics gear to be scattered. These pieces are then assembled to make a camera which is able to photograph the explosion. I am confused! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answers suitable for a high school physics class greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="11153" LastEditorUserId="2270" LastEditDate="2016-03-11T16:58:45.503" LastActivityDate="2016-03-11T16:58:45.503" Title="Gravitational Waves and the Big bang" Tags="&lt;light&gt;&lt;general-relativity&gt;&lt;big-bang-theory&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14122" PostTypeId="2" ParentId="14121" CreationDate="2016-03-11T09:11:22.567" Score="4" Body="&lt;p&gt;Firstly, Big Bang didn't happen at a point in space, away from which we are traveling. Big Bang was the creation of space. This space has been expanding ever since, so that the distances between everything increases, but Big Bang happened right where you are, where the Andromeda galaxy is, where &lt;a href=&quot;https://en.wikipedia.org/wiki/GN-z11&quot;&gt;GN-z11&lt;/a&gt; is, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Universe has evolved ever since. If we want to know how galaxies look 13.8 billion years (Gyr) after Big Bang, we can just look around in our neighborhood. If we look too far away, we don't see 13.8 Gyr old galaxies, because the light has taken some time to reach us. Thus, if we want to see galaxies that are 12.8 Gyr old, we simply look 1 billion lightyears away; if we want to se 10 Gyr old galaxies, we look (roughly) 3.8 billion lightyears away, and so on.$^\dagger$ In this way, we look back in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look away far enough, we would in principle be able to look 13.8 Gyr back in time. However, we face a problem in that until the Universe was 380,000 years old, it was opaque to light (why this is so is another story). It wasn't opaque to gravitational waves, however. And since lots of GWs are thought to have originated during the epoch called &lt;a href=&quot;https://en.wikipedia.org/wiki/Inflation_%28cosmology%29&quot;&gt;inflation&lt;/a&gt;, which were responsible for the expansion of the Universe, and which took place a fraction of a second after the creation of space, we say that GWs offer the possibility of looking all the way back to Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;This is somewhat imprecise, since galaxies weren't created instantly after Big Bang, and since all galaxies weren't created at the same time. But for the sake of the arument, let's pretend they were.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2016-03-11T09:11:22.567" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14123" PostTypeId="2" ParentId="14115" CreationDate="2016-03-11T12:47:55.387" Score="1" Body="&lt;p&gt;In general relativity, the curvature of spacetime is an interpretation of  the model for gravity: Starting with the assumptions of gravity being equivalent to acceleration, and the speed of light being constant, a mathematical framework that describes gravity is developed. The mathematical conclusion of this model is that the distances between points is dependent on the distribution of mass and energy. A simple way of describing this is to say that spacetime is curved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In GR, the apparent force of gravity is explained by the curvature of spacetime. It is reasonable to ask if there is a more fundamental theory that can explain the curvature. &lt;a href=&quot;http://www.superstringtheory.com/blackh/blackh4.html&quot; rel=&quot;nofollow&quot;&gt;String theory&lt;/a&gt; provides one possibility. In string theory it is possible to deduce the existence of a particle that would be a graviton, and the from that the field equations that describe gravity. In other words, string theory can predict exactly the same mathematical formulae that arise from curved spacetime in GR. However, on very small scales, string theory makes a series of corrections to the field theory, suggesting that the curvature of spacetime is actually an approximation, which is valid on the large scale, to a string theory of gravity.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-03-13T09:13:27.600" LastActivityDate="2016-03-13T09:13:27.600" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14125" PostTypeId="1" CreationDate="2016-03-11T19:35:44.060" Score="13" ViewCount="1294" Body="&lt;p&gt;NASA released this &lt;a href=&quot;http://www.nasa.gov/image-feature/goddard/2016/hubble-sees-a-legion-of-galaxies/&quot; rel=&quot;noreferrer&quot;&gt;image&lt;/a&gt; of deep space. The way light shifting vis-a-vis universe expansion is &lt;strong&gt;supposed&lt;/strong&gt; to work is by making the furthest galaxies appear mostly red-shifted. But NASA's image shows somewhat of an even balance between blue and red shift. Why is that?&lt;/p&gt;&#xA;" OwnerUserId="11160" LastEditorUserId="1056" LastEditDate="2016-03-12T17:19:30.060" LastActivityDate="2018-08-16T19:25:05.973" Title="Why are there so many seemingly blue-shifted galaxies in deep space" Tags="&lt;redshift&gt;&lt;deep-sky-observing&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14126" PostTypeId="2" ParentId="14125" CreationDate="2016-03-11T20:51:17.210" Score="6" Body="&lt;p&gt;There are two factors here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, to construct an image such as this, several optical filters are used, and then these are then merged to give an image. The colours that you see are therefore not the &quot;true&quot; colours (they couldn't be, the light you see is far too faint for the human eye) but are chosen to bring out details in the image. A galaxy that is blue in the image might not actually be blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, red shift means that all the light is shifted, If a galaxy is emitting a broad spectrum of light, the ultraviolet is shifted to the visible, as the red is shifted to the infrared, the apparent colour doesn't change much. So the apparent colour, red or blue isn't telling you about the red-shift of each galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To measure the red shift you need a detailed spectrum. A spectrum will include spectral lines, which have known frequencies. By noting the shift of these lines the redshift is measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The different colours you see here are probably a consequence of the type of stars in each galaxy. Galaxies with more young stars appear bluer, and this is probably emphasised in the image (but remember it isn't &quot;true colour&quot;)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-03-11T20:51:17.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14127" PostTypeId="2" ParentId="14125" CreationDate="2016-03-11T21:00:41.603" Score="16" Body="&lt;p&gt;You cannot gauge the redshift of a galaxy by looking at a false colour image. The images taken through different filters are stacked and colourised to suit. You can say that the blue galaxies are indeed bluer than the red galaxies, but there is no absolute scale with which to judge redshift by eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, there is no detail in the NASA web page, but the ACS and WFC3 cameras have near infrared capabilities. So I would think that this image is a visual false-colour image of information that extends well redward of what the eye can perceive. So even the things that look blue might have a spectrum that peaks at redder wavelengths, whilst anything that looks red might actually be infrared!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, beyond this, in order to judge what the redshifted appearance of a galaxy would be, you would need to know what the galaxy looked like with zero redshift. That is, light that is emitted in the ultraviolet could be redshifted into the visible part of the spectrum. It is entirely possible that visual colour of a redshifted galaxy would not change very much at all if the galaxy emitted lots of UV light in &lt;em&gt;its&lt;/em&gt; frame of reference or equally, if it emitted lots of very red light that was then redshifted out of the telescope's sensitivity range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, there could be some genuine astrophysics going on. Many distant galaxies are bluer than nearby galaxies because they are undergoing intense star formation. Massive star forming regions emit copious UV light that is redshifted into the optical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, finally! Many of the galaxies in the picture will be quite close and will not be very redshifted.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-03-11T21:00:41.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14128" PostTypeId="2" ParentId="13462" CreationDate="2016-03-11T21:27:15.923" Score="2" Body="&lt;p&gt;This news story is based on an &lt;a href=&quot;http://science.sciencemag.org/content/351/6272/482.full?ijkey=qgiw36Ak9gSes&amp;amp;keytype=ref&amp;amp;siteid=sci&quot; rel=&quot;nofollow&quot;&gt;article from &lt;em&gt;Science&lt;/em&gt;&lt;/a&gt; The image is also taken from the same &lt;em&gt;Science&lt;/em&gt; magazine article. The text here is a procedural text, it describes the process of using a trapezium to make distance-speed-time calculations, for approximation of the motion of Jupiter along the ecliptic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should look at the original article in science and the &lt;a href=&quot;http://amor.cms.hu-berlin.de/~ossendrm/cuneiform.html&quot; rel=&quot;nofollow&quot;&gt;author's website&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-03-11T21:34:54.900" LastActivityDate="2016-03-11T21:34:54.900" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14129" PostTypeId="1" AcceptedAnswerId="14130" CreationDate="2016-03-12T12:32:08.890" Score="2" ViewCount="356" Body="&lt;p&gt;Today known most massive star, &lt;a href=&quot;http://space.about.com/od/stars/tp/The-Top-10-Most-Massive-Stars.htm&quot; rel=&quot;nofollow&quot;&gt;R136a1&lt;/a&gt;  weighs approximately 256 times the mass of our sun, how did supermassive black holes, with a mass of 10 billion times the sun form?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand, that when a star becomes black hole, its density increases, but does its mass also grow?&lt;/p&gt;&#xA;" OwnerUserId="11169" LastEditorUserId="2365" LastEditDate="2016-03-12T16:35:59.767" LastActivityDate="2018-01-04T16:29:36.707" Title="Why are black holes that massive?" Tags="&lt;star&gt;&lt;black-hole&gt;&lt;mass&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14130" PostTypeId="2" ParentId="14129" CreationDate="2016-03-12T14:24:17.847" Score="3" Body="&lt;p&gt;Here comes positive feedback - if something is heavy it tends to attract more things so it is even heavier and heavier. At one point it will clean its surroundings. This is the process for those supermassive black holes to create. In middle of galaxies there is lot of stuff (stars) that can be &quot;eaten&quot; by these huge black holes. Plus they had a lot of time to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT (thanks for questioning the answer I was very sure with it but there are some problems):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Rob Jeffries said, the mass could not be acquired this way (probably not all of it) because of radiation pressure: when stuff falls towards black hole in accretion disk there is lot of heat created which pushes the rest of gas away. See Eddington limit. &lt;a href=&quot;https://en.wikipedia.org/wiki/Eddington_luminosity&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Eddington_luminosity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So after checking some pages those are the theories I found:&#xA;&lt;a href=&quot;http://science.nasa.gov/astrophysics/focus-areas/black-holes/&quot; rel=&quot;nofollow&quot;&gt;http://science.nasa.gov/astrophysics/focus-areas/black-holes/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;One possible mechanism for the formation of supermassive black holes involves a chain reaction of collisions of stars in compact star clusters that results in the buildup of extremely massive stars, which then collapse to form intermediate-mass black holes. The star clusters then sink to the center of the galaxy, where the intermediate-mass black holes merge to form a supermassive black hole.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another source:&#xA;&lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/S/Supermassive+Black+Hole&quot; rel=&quot;nofollow&quot;&gt;http://astronomy.swin.edu.au/cosmos/S/Supermassive+Black+Hole&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Stellar black holes result from the collapse of massive stars, and some have suggested that supermassive black holes form out of the collapse of massive clouds of gas during the early stages of the formation of the galaxy. Another idea is that a stellar black hole consumes enormous amounts of material over millions of years, growing to supermassive black hole proportions. Yet another, is that a cluster of stellar black holes form and eventually merge into a supermassive black hole.&quot;&lt;/p&gt;&#xA;" OwnerUserId="9677" LastEditorUserId="9677" LastEditDate="2016-03-12T21:47:11.243" LastActivityDate="2016-03-12T21:47:11.243" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14131" PostTypeId="2" ParentId="14107" CreationDate="2016-03-12T16:20:01.003" Score="1" Body="&lt;p&gt;In the Southern hemisphere, the maximum declination you can see is&#xA;90-L where L is your latitude. The minimum is -90, since you can see&#xA;the south celestial pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the earth rotates and revolves, you can see any right ascension &#xA;within those declinations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only exception is that you ordinarily can't see stars that are too &#xA;close to the Sun. However, the Sun moves enough between October and &#xA;March that all right ascensions will be visible at night sometime &#xA;between October and March. &lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2016-03-12T16:20:01.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14132" PostTypeId="2" ParentId="14129" CreationDate="2016-03-12T16:31:44.887" Score="7" Body="&lt;p&gt;The mass of a object does not increase when it collapses into a black hole. So a supermassive black hole must have started off quite small, and then grown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formation and growth of supermassive black hole is not settled science. Supermassive black holes probably started as large stellar mass black holes (The very earliest stars could have been very large, and were almost entirely free of any heavier elements, and could have collapsed to form black holes with a mass of about 100 suns) The black holes then grew as more matter fell into them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a strong correlation between the size of a SMBH and its galaxy. Its not clear why this is, but it suggests that the environment that drove galaxy growth, also contributed to the growth of the black hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This article on the &lt;a href=&quot;http://arxiv.org/pdf/0903.0547.pdf&quot;&gt;Growth of supermassive black holes across cosmic time&lt;/a&gt; may interest you.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-03-12T16:38:28.920" LastActivityDate="2016-03-12T16:38:28.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14133" PostTypeId="1" AcceptedAnswerId="14653" CreationDate="2016-03-12T17:38:16.107" Score="0" ViewCount="971" Body="&lt;p&gt;If we sent someone on a path that passed as near as possible to a black hole without getting pulled in, how far into the future would they go due to time dilation?  Let's assume the black hole is 5 solar masses (I am assuming the mass will affect the calculation).  I would like to know from two considerations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;An observer that can survive anything (e.g. ideal case)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The forces that a human can tolerate (e.g. a human would not be able to get as close to a black hole as would be ideal) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="1386" LastEditDate="2016-04-23T14:46:05.120" LastActivityDate="2016-04-23T20:14:20.870" Title="How far into the future can we go by traveling close to a black hole?" Tags="&lt;black-hole&gt;&lt;space-time&gt;&lt;general-relativity&gt;&lt;special-relativity&gt;&lt;time-dilation&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14136" PostTypeId="2" ParentId="13704" CreationDate="2016-03-12T23:29:02.033" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Updated to use reported timing confidence intervals instead of trying&#xA;  to infer those from reported sky location resolution uncertainties.&#xA;  The latter approach was misleading because the sky location has been further&#xA;  resolved using more information than just the arrival time difference.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Beware that I am neither an astronomer nor involved with aLIGO.&#xA;I just use classic euclidean geometry here,&#xA;combined with occasional transformations between cartesian&#xA;and spherical coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have two observatories, $L$ (Livingston) and $H$ (Hanford)&#xA;and a far-away point $E$ emitting a (gravitational wave) signal.&#xA;Let us assume that the signal propagates at constant speed $c$&#xA;through low-density matter, including planet Earth.&#xA;The straight-line distance between $L$ and $H$ (passing through Earth)&#xA;is $|HL| \approx c\cdot 10.0\cdot10^{-3}\,\mathrm{s}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L$ receives the signal from $E$ about&#xA;$t_{ELH}\approx 7\cdot10^{-3}\,\mathrm{s}$ before $H$.&#xA;This implies that $E$ is a distance&#xA;$d = c\,t_{ELH}$ closer to $L$ than to $H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assuming large-scale euclidean geometry and looking at the&#xA;plane containing $E,L,H$, the set of all candidate points $E$&#xA;where $|EH|-|EL| = d$ is one branch of a hyperbola, namely&#xA;the branch closer to $L$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GDk39.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GDk39.png&quot; alt=&quot;Hyperbola&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, setting&#xA;$$\begin{align}&#xA;    r &amp;amp;= \frac{d}{2}    &amp;amp; r_1 &amp;amp;= \frac{|HL|}{2} &amp;amp; R &amp;amp;= |LE|&#xA;\\  L &amp;amp;= (r_1,0)        &amp;amp; H &amp;amp;= (-r_1,0)         &amp;amp; E &amp;amp;= (x,y)&#xA;\end{align}$$&#xA;we get the cartesian hyperbola equation&#xA;$$\frac{x^2}{r^2} - \frac{y^2}{r_1^2-r^2} = 1$$&#xA;and in polar coordinates, viewed from $L$:&#xA;$$R = \frac{r_1^2 - r^2}{r - r_1\cos\alpha}$$&#xA;For very large $R/r_1$, $|\alpha|$ must be very close to its asympotic limit:&#xA;$$\begin{align}&#xA;    \cos\alpha &amp;amp;\approx \frac{r}{r_1} = \frac{d}{|HL|} = \frac{t_{ELH}}{t_{HL}}&#xA;\\  \text{where}\quad&#xA;    t_{HL} &amp;amp;= \frac{|HL|}{c}&#xA;    \approx 10.0\cdot 10^{-3}\,\mathrm{s}&#xA;    \qquad\text{(time directly from $H$ to $L$)}&#xA;\end{align}$$&#xA;In 3D this means that $E$ lies on one of the two sheets of a hyperboloid&#xA;of revolution whose axis of revolution is the line $HL$.&#xA;In particular, in $L$'s view of the sky, partially hidden under the horizon,&#xA;there should be a fairly thin circular band that covers all projected&#xA;candidate points for $E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solid angle of view enclosed by the circular band is&#xA;$$  A_{\text{ste}} = 2\pi(1 - \cos\alpha)&#xA;\approx 2\pi\left(1 - \frac{t_{ELH}}{t_{HL}}\right)$$&#xA;Uncertainties in $t_{ELH}$ yield uncertainties in $\cos\alpha$&#xA;which widen the circular band.&#xA;Thus the circle becomes an annulus.&#xA;The solid angle of view covered by that annulus measures&#xA;$$\begin{align}&#xA;    A_{\text{ste.max}} - A_{\text{ste.min}}&#xA;    = 2\pi(\cos\alpha_{\text{min}} &amp;amp;{}- \cos\alpha_{\text{max}})&#xA;    \approx 2\pi\frac{t_{ELH\text{.max}} - t_{ELH\text{.min}}}{t_{HL}}&#xA;\\  \text{briefly:}\quad&#xA;    \Delta A_{\text{ste}} &amp;amp;\approx 2\pi\frac{\Delta t_{ELH}}{t_{HL}}&#xA;\end{align}$$&#xA;Substituting reported data for $t_{ELH}$, with $\text{min}$ and $\text{max}$&#xA;referring to 90% probability intervals, we have:&#xA;$$\begin{array}{rcrrr}&#xA;    \text{entity} &amp;amp; \text{unit} &amp;amp; \text{nom} &amp;amp; \text{min} &amp;amp; \text{max} &amp;amp; \Delta&#xA;\\\hline&#xA;    t_{ELH} &amp;amp; 10^{-3}\,\mathrm{s}   &amp;amp; 6.9   &amp;amp; 6.5   &amp;amp; 7.4   &amp;amp; 0.9&#xA;\\  \cos\alpha &amp;amp; 1                  &amp;amp; 0.69  &amp;amp; 0.65  &amp;amp; 0.74  &amp;amp; 0.09&#xA;\\  90^\circ - \alpha &amp;amp; {}^\circ    &amp;amp; 44    &amp;amp; 41    &amp;amp; 48    &amp;amp; 7&#xA;\end{array}$$&#xA;And thus the annulus, when projected to a unit (celestial) sphere, covers&#xA;$$\Delta A_{\text{ste}} \approx 0.18\pi \approx (1.9\cdot10^3)^{\circ\circ}$$&#xA;(The ${}^{\circ\circ}$ means&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Square_degree&quot; rel=&quot;nofollow noreferrer&quot;&gt;square degrees&lt;/a&gt;.)&#xA;In contrast, the reported value of the &lt;em&gt;credible region&lt;/em&gt; varies between&#xA;$140^{\circ\circ}$ (50% probability) and $590^{\circ\circ}$ (90% probablity).&#xA;This is because large parts of the annulus could be ruled out based on&#xA;additional information such as signal strength ratios.&#xA;But those refinements are not the topic of this thread.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For easier visualization, let us move to a point $B$ on Earth's surface&#xA;where the center of the annulus is in the zenith,&#xA;and the annulus itself is at constant altitude $90^\circ - \alpha$,&#xA;as given in the above table.&#xA;To that end, let $O$ be the location of Earth's center and require&#xA;that the 3D ray $OB$ points in the same direction as the ray $HL$.&#xA;This determines the latitude and longitude of $B$.&#xA;Note that the parallel shift from $HL$ to $OB$ incurs a parallax,&#xA;but that is negligible because the distance to $E$ is so large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idealizing our planet to a sphere and doing the calculations&#xA;(transforming the locations of $H$ and $L$ from spherical coordinates to&#xA;cartesian coordinates, computing the difference vector, normalizing it,&#xA;and transforming back to spherical coordinates), I find $B$ to be at&#xA;$$B\colon\ 27^\circ 21'54.22''\mathrm{S}\ 38^\circ 36'33.45''\mathrm{W}&#xA;= 27.365061^\circ\mathrm{S}\ 38.609293^\circ\mathrm{W}$$&#xA;which is &lt;a href=&quot;https://www.google.de/maps/place/27%C2%B021%2754.2%22S+38%C2%B036%2733.5%22W/@-27.0915311,-38.9169065,4z/data=!4m2!3m1!1s0x0:0x0&quot; rel=&quot;nofollow noreferrer&quot;&gt;some 700km southeast of Rio de Janeiro in the Atlantic&lt;/a&gt;.&#xA;If you had been there at &lt;code&gt;2015-09-14 09:50:45 UTC&lt;/code&gt;&#xA;and had spun around yourself, drawing a band at constant altitude&#xA;$90^\circ - \alpha \approx 44^\circ$ and width $7^\circ$ in the sky,&#xA;that would have marked all the directions&#xA;where the signal source $E$ could have been.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the sky view simulated, I have written a tiny script &lt;code&gt;gw150914.sts&lt;/code&gt; for&#xA;&lt;a href=&quot;http://nightshadesoftware.org/projects/nightshade-legacy/wiki&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nightshade&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# Display the sky at the time of the aLIGO GW150914 observation&#xA;# from an Earth-based position chosen such that the set of possible sky&#xA;# positions of the source form a circular annulus with azimuth-independent&#xA;# altitude = asin(delta_t/10ms), about 44 deg.&#xA;clear&#xA;timerate rate 0&#xA;date utc 2015-09-14T09:50:45&#xA;flag show_tui_datetime on&#xA;set home_planet &quot;Earth&quot;&#xA;moveto lat -27.365061 lon -38.609293 alt 0 heading 0 duration 2&#xA;wait duration 2&#xA;flag azimuthal_grid on&#xA;zoom fov 120 duration 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This gives the view below. I have added dashed green circles to delimit&#xA;the annulus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CF9KI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CF9KI.png&quot; alt=&quot;Nightshade view with added indication of the annulus&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But do not expect to find likely candidate spots within that annulus.&#xA;The annulus hides the radial depth of the projected space volume.&#xA;A range for that radial depth is given by the estimated bounds&#xA;for the distance $R=|LE|$. The corresponding flat space volume is&#xA;$$\Delta V = \frac{1}{3}\Delta A_{\text{ste}}&#xA;\left(R_{\text{max}}^3 - R_{\text{min}}^3\right)$$&#xA;Plugging in the reported values, again from 90% probability intervals,&#xA;$$\begin{align}&#xA;\Delta A_{\text{ste}} &amp;amp;\approx 590^{\circ\circ} \approx 0.0572\pi&#xA;&amp;amp;   R_{\text{min}} &amp;amp;\approx 0.23\cdot10^{9}\,\mathrm{pc}&#xA;&amp;amp;   R_{\text{max}} &amp;amp;\approx 0.57\cdot10^{9}\,\mathrm{pc}&#xA;\end{align}$$&#xA;yields&#xA;$$\Delta V \approx 10^{25}\,\mathrm{pc}^3$$&#xA;To get a very rough estimate of the number of galaxies in that volume,&#xA;let us multiply $\Delta V$ with a rough galaxy density estimate of $10^{-20}\,\mathrm{pc}^{-3}$,&#xA;which results in about $10^5$ galaxies in that volume.&#xA;My conclusion from this back-of-the-envelope calculation is&#xA;that we cannot hope to localize $E$ further, unless given more information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things would have been a bit different if we had had a third aLIGO observatory&#xA;distant from the line $HL$ operating and capable of detecting the signal&#xA;from $E$.&#xA;Then we could draw three annuli, take intersections and thus&#xA;determine the direction of the signal within one reasonably small spot&#xA;on the celestial sphere.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Further reading&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;As &lt;a href=&quot;https://www.ligo.caltech.edu/image/ligo20160211b&quot; rel=&quot;nofollow noreferrer&quot;&gt;this LIGO post&lt;/a&gt; tells&#xA;us, parts of the annulus could be ruled out by considering additional features&#xA;of the data such as varying signal strength.&lt;/li&gt;&#xA;&lt;li&gt;A more detailed description of the probability distributions for the results&#xA;is given in &lt;a href=&quot;https://dcc.ligo.org/public/0122/P1500218/012/GW150914_parameter_estimation_v13.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this LIGO paper&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;More information about the methods used to further narrow the credible&#xA;sky regions of the source location is given&#xA;in &lt;a href=&quot;http://dx.doi.org/10.1007/lrr-2016-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Addendum&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;An equivalent script &lt;code&gt;gw150914.ssc&lt;/code&gt;&#xA;for &lt;a href=&quot;http://stellarium.sourceforge.net&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellarium&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;// Display the sky at the time of the aLIGO GW150914 observation&#xA;// from an Earth-based position chosen such that the set of possible sky&#xA;// positions of the source form a circular annulus with azimuth-independent&#xA;// altitude = asin(delta_t/10ms), about 44 deg.&#xA;&#xA;var obs_lat = -27.365061;&#xA;var obs_long = -38.609293;&#xA;var obs_alt = 0;&#xA;var obs_locname = &quot;Atlantic, 700km SE from Rio de Janeiro&quot;;&#xA;var obs_planet = &quot;Earth&quot;;&#xA;var obs_date = &quot;2015-09-14T09:50:45&quot;;&#xA;var duration = 2;&#xA;&#xA;core.clear(&quot;deepspace&quot;);&#xA;core.setTimeRate(0);&#xA;core.setDate(obs_date, &quot;utc&quot;);&#xA;core.setObserverLocation(obs_long, obs_lat, obs_alt, duration,&#xA;                         obs_locname, obs_planet);&#xA;core.wait(duration);&#xA;core.moveToAltAzi(&quot;90&quot;, &quot;0&quot;, duration);&#xA;core.wait(duration);&#xA;GridLinesMgr.setFlagAzimuthalGrid(true);&#xA;StelMovementMgr.zoomTo(120, duration);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And yes, &lt;code&gt;core.setObserverLocation&lt;/code&gt; expects longitude before latitude.&lt;/p&gt;&#xA;" OwnerUserId="11168" LastEditorUserId="11168" LastEditDate="2016-03-20T16:16:18.883" LastActivityDate="2016-03-20T16:16:18.883" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14137" PostTypeId="2" ParentId="14125" CreationDate="2016-03-13T04:33:43.600" Score="0" Body="&lt;p&gt;I used image search to find a version with more &lt;a href=&quot;https://www.spacetelescope.org/images/potw1610a/&quot; rel=&quot;nofollow&quot;&gt;detail&lt;/a&gt; (Google Image Search is great for this sort of thing) about how it was taken, the RGB channels in the mage were taken with IR, Visual, and Blue filters respectively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main galaxy cluster in it, &lt;a href=&quot;https://en.wikipedia.org/wiki/Abell_2744&quot; rel=&quot;nofollow&quot;&gt;Abel 2744&lt;/a&gt;, is only at a red-shift of 0.3; so it's colors aren't that far removed from what we could see visually (with much better eyes).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, just looking at the picture doesn't give you any way to tell the difference between a galaxy that's relatively near but dim or much farther but bright.  However I suspect a majority of the tiny blue galaxies are smaller members of Abel 2744 not extremely distant objects that would need a major blue shift to cancel the red shift from ex.&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2016-03-13T04:33:43.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14138" PostTypeId="1" CreationDate="2016-03-13T06:41:07.800" Score="1" ViewCount="941" Body="&lt;p&gt;Assuming I only have a simple telescope, how can I determine the size of the moon, its angular velocity and distance from earth?&lt;/p&gt;&#xA;" OwnerUserId="6610" LastActivityDate="2016-03-13T08:13:06.767" Title="The moon: how to determine its size, speed and distance from earth using a simple telescope?" Tags="&lt;the-moon&gt;&lt;telescope&gt;&lt;distances&gt;&lt;size&gt;&lt;velocity&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14139" PostTypeId="2" ParentId="14138" CreationDate="2016-03-13T08:04:18.650" Score="1" Body="&lt;p&gt;Finding the distance is tricky. To find the distance you need to carefully note the position of the moon (relative to the stars ) from two different places, *at the same time *.  The moon will appear to be in a slightly different position from the two viewpoints (an effect called paralax). That done, finding the distance, and size is a simple exercise in trigonometry, and Khan academy has a page which shows &lt;a href=&quot;https://www.khanacademy.org/partner-content/nasa/measuringuniverse/measure_the_solarsystem/a/parallax-distance&quot; rel=&quot;nofollow&quot;&gt;hoebto find the distance to the moon&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Greater accuracy could be obtained by measuring the time that a star is hidden by the moon, from two locations, and using that to determine the position of the moon to greater accuracy than possible by direct measurements.  Hipparchus apparently used a solar eclipse to obtain two positions atbthe same time needed for calculation of paralax. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing the distance makes finding the diameter and speed simple to do, by observing the angular size, and using a little trig.  Working with photographs takenby the telescope is convenient but not strictly necessary.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-03-13T08:13:06.767" LastActivityDate="2016-03-13T08:13:06.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14140" PostTypeId="2" ParentId="10114" CreationDate="2016-03-13T11:23:23.393" Score="3" Body="&lt;p&gt;Watched the video, and I thought it was really cool seeing the modeling of different young solar-systems.  The orbital resonances they were talking about were for smaller objects that form debris fields in resonance with a giant planet.   Since they only looked at models for the inner planets, the closest example to the bit in the video with our current solar-system would be Jupiter's &lt;a href=&quot;https://en.wikipedia.org/wiki/Hilda_family&quot; rel=&quot;nofollow&quot;&gt;Hildas&lt;/a&gt;.  For that specific part of the video, they only showed one inner most giant planet.  For more than one, Resonances like that are unlikely to form between giant planets, like in our solar system where most of the resonance objects fall inside Jupiter or outside Neptune.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mathematics gets too complicated for me to know for sure, but as I understand it, planets in resonance is neither stable nor unstable, unless the aphelion of one planet is in resonance with the perihelion of another, in which case the planets could drawn together I would think.  That's my amateur answer to your question.  I wouldn't think orbital resonance affects stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When one planet is dominant and the other quite small, then resonances appear largely stable (Neptune/Pluto for example), provided they never get close to each other, which Neptune-Pluto don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter and Saturn were thought to have entered orbital resonance at one point in the early solar-system and that, from what I've read, affected other objects in the solar system but didn't affect their relation to each other much.  I think the only way to get any kind of answer is to run long term simulations on various model.   In short, planets can be stable both in resonance with each other, and not in resonance and planets can be unstable, again, either in resonance or not in resonance, though if they have unstable orbits, they won't stay in resonance for long.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-03-13T11:23:23.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14141" PostTypeId="2" ParentId="13776" CreationDate="2016-03-13T16:29:33.243" Score="3" Body="&lt;p&gt;The sun forms a useful basis for measuring time because to a reasonable approximation it stays put, and the rotation of the Earth causes it to traverse the sky once every 24 hours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Provided the planet is spinning, a normal sundial can be used, but may not be quite as regular as on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A binary stellar system can have one of three possible configurations: Two stars in a close binary, with a planet orbiting at a much greater distance. This is the &quot;Tattooine&quot; configuration. Everything has a double shadow. A regular sundial can be used but it is not as regular, as the positions of the two stars changes as they orbit each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second configuration is a planet orbiting a star with a second star at a much greater distance. If Jupiter had been a star this would be the second configuartion.  The secondary star would be much weaker, and a regular sundial could be used. The shadow from the secondary would just be ignored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last configuration is like the second, except the planet is in orbit around the smaller star. The more distant star is stronger, and compariable in brightness. In this situation it is quite likely that the planet is tidally locked to the star, so a sundial isn't useful for telling the time. The slow motion of the distant star might give a long day (which could last many years) to the side of the planet that faces away from the nearby star, followed by a long cold night. This would be a challenging planet on which to live.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-03-13T16:29:33.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14144" PostTypeId="1" AcceptedAnswerId="14149" CreationDate="2016-03-14T00:22:52.827" Score="1" ViewCount="220" Body="&lt;p&gt;When Phobos meets its predicted doom in a few 10s of millions of years, whether it breaks up or collides with the surface of Mars, will this change have any noticeable effects (permanent or temporary) on the Earth?&lt;/p&gt;&#xA;" OwnerUserId="2929" LastActivityDate="2016-03-14T05:08:25.153" Title="When Phobos collides with mars (or breaks up), will it have any noticeable effects on Earth?" Tags="&lt;earth&gt;&lt;mars&gt;&lt;phobos&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14145" PostTypeId="1" AcceptedAnswerId="14151" CreationDate="2016-03-14T00:46:35.007" Score="1" ViewCount="239" Body="&lt;p&gt;In my astronomy class I learned that a only low-mass stars (&amp;lt; 0.5 Msun) will contract, and then become degenerate, until it is a white dwarf. However, we also learned about the Chandrasekhar limit, which states white dwarfs have a maximum mass of 1.4 Msun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the starting star has less than 0.5 Msun, how does the resulting white-dwarf reach up to 1.4 Msun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I perhaps wrong in my assumption that the mass of the resulting white-dwarf depends on the mass of the original star?&lt;/p&gt;&#xA;" OwnerUserId="11183" LastActivityDate="2016-03-14T14:38:22.600" Title="How can a low-mass star increase its mass to 1.4 Msun?" Tags="&lt;star&gt;&lt;white-dwarf&gt;" AnswerCount="2" CommentCount="8" ClosedDate="2016-03-14T13:23:13.717" ContentLicense="CC BY-SA 3.0" />
  <row Id="14146" PostTypeId="2" ParentId="14145" CreationDate="2016-03-14T02:37:09.993" Score="0" Body="&lt;p&gt;From these 2 sites: &lt;a href=&quot;http://www.whitedwarf.org/education/ask/&quot; rel=&quot;nofollow&quot;&gt;whitedwarf.org&lt;/a&gt; and &lt;a href=&quot;http://imagine.gsfc.nasa.gov/science/objects/dwarfs1.html&quot; rel=&quot;nofollow&quot;&gt;imagine.gsfc.nasa.gov&lt;/a&gt; both states that stars less than 8 times the solar masses would eventually become a white a white dwarf. So, depending on the mass of the star, a white dwarf having 1.4 solar masses is possible.&lt;/p&gt;&#xA;" OwnerUserId="7926" LastActivityDate="2016-03-14T02:37:09.993" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14147" PostTypeId="1" CreationDate="2016-03-14T03:20:42.343" Score="3" ViewCount="309" Body="&lt;p&gt;This questions concerns the longitudinal aspect of the Equation of Time, also called the Equation of the Center. In some sources the equation looks like the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nu - M = 2\varepsilon \sin M$       (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\nu$ is the True Anomaly of the Sun's position from the Earth, $M$ is the Mean Anomaly, and $\varepsilon$ is the eccentricity of the Earth's orbit (0.0167). $\nu - M$ is the difference between the Sun's actual angle and the and the angle that would exist if the Earth's orbit were circular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other sources have a first-order approximation that looks like the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time deviation (minutes) = $-7.655 \sin d$       (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where d is the day of the year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My difficulty is reconciling these two equations. None of the sources actually make this connection explicit. I assume that the difference has to do with converting angles and times, and I have tried various approaches to making the numbers work out, without avail. I would appreciate it if someone could tell me how the value of -7.655 minutes is derivable from $2 \varepsilon$ and or point me at a resource that demonstrates the connection between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note I realize that both (1) and (2) are approximations. At this stage, I am trying to understand the situation in its simplest form before adding refinements.&lt;/p&gt;&#xA;" OwnerUserId="8092" LastEditorUserId="2365" LastEditDate="2016-03-14T17:35:57.750" LastActivityDate="2018-06-29T13:48:05.550" Title="Equation of the Center constant factor" Tags="&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14148" PostTypeId="1" CreationDate="2016-03-14T04:29:21.757" Score="26" ViewCount="7079" Body="&lt;p&gt;I am only 12 years old and I'm constantly wondering and trying understand how gravity really works. On YouTube everyone always talks about objects warping space time around themselves and uses the analogy of a trampoline. I still don't understand gravity because if space were like a trampoline, then earth would be spiraling in towards the sun along with all the other planets, right? So could someone explain to me how gravity really works without the trampoline analogy?&lt;/p&gt;&#xA;" OwnerDisplayName="user11186" LastEditorUserId="6593" LastEditDate="2016-03-14T14:05:00.233" LastActivityDate="2016-03-15T18:08:48.027" Title="How does gravity really work" Tags="&lt;gravity&gt;" AnswerCount="5" CommentCount="19" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14149" PostTypeId="2" ParentId="14144" CreationDate="2016-03-14T05:08:25.153" Score="5" Body="&lt;p&gt;No. Phobos is small - just 11 Km across - the size of a small city. Mars (and Phobos) is so far away that a Phobos impact will not affect Earth much. (Mars ranges from 100 billion meters away to nearly 400 billion meters depending on its and Earth's position in their orbits.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Phobos hits the Roche limit as it will break up and become a thin light-grey planetary ring around Mars for a few million years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it breaks up, (almost) all the debris should stay near Mars because the escape velocity from its orbit around Mars is greater than the velocity imparted to boulders of Phobos as it self-destructs.&lt;/p&gt;&#xA;" OwnerUserId="11175" LastActivityDate="2016-03-14T05:08:25.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14150" PostTypeId="2" ParentId="14148" CreationDate="2016-03-14T05:59:14.927" Score="20" Body="&lt;p&gt;Why are objects not &lt;em&gt;escaping&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider first an object with velocity, and no gravity in action:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Nue9em.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nue9em.png&quot; alt=&quot;escaping?&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, that blue object will become more and more distant, if it continues in the same direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it does not continue in the same direction, after a while, the gravity of the big black object has changed its course:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nCjECm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nCjECm.png&quot; alt=&quot;new course&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That happens again, and again and again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dw5MBm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dw5MBm.png&quot; alt=&quot;repeat&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question is: Why does not the object spiral in? You are perhaps thinking that as it comes closer, the gravity becomes stronger, and therefore the object is forced to come even closer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when it falls closer, its velocity increases. As we have seen, the velocity of the objects tries to make it escape. So when it is closer, it has more velocity to counteract the increased gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Just in case of a more literal interpretation of your question, the trampoline in the original analogy causes friction, and therefore spiralling, but space is a vacuum.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="10172" LastEditDate="2016-03-14T22:45:20.420" LastActivityDate="2016-03-14T22:45:20.420" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14151" PostTypeId="2" ParentId="14145" CreationDate="2016-03-14T07:08:43.217" Score="2" Body="&lt;p&gt;You were taught wrong. Stars of up to about 8 solar masses will end up as white dwarfs. But it is only their &lt;em&gt;cores&lt;/em&gt; that become degenerate and end up as the white dwarf. The rest of the envelope is lost during the giant phase due to a dense wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There &lt;em&gt;is&lt;/em&gt; a non-linear, but probably monotonic, relation between the initial progenitor and final white dwarf masses (see below, from &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013MmSAI..84...58K&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kalirai 2013&lt;/a&gt;) - the Sun will likely end as a 0.5 solar mass white dwarf, but in normal stellar evolution, degenerate white dwarf stars can only be produced up to about 1.25 solar masses by the most massive progenitors. Any more massive than this and it is likely that the core does not become degenerate before igniting and burning through the heavier nuclear fuels. The most massive, probably single, white dwarf known is &quot;WD 33&quot; in the cluster NGC 2099 and has a mass of $1.28^{+0.05}_{-0.08}\ M_{\odot}$, is likely made of an O/Ne mixture, and had an estimated progenitor mass on the main sequence of $&amp;gt;3.5\ M_{\odot}$ (&lt;a href=&quot;http://arxiv.org/abs/1603.00471&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cummings et al. (2016)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to get a more massive white dwarf, up to the Chandrasekhar mass (about 1.38 solar masses for a C/O or O/Ne white dwarf in general relativity), it almost certainly needs to &lt;em&gt;accrete&lt;/em&gt; mass from a close binary companion or be the result of some sort of merger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the leading candidate to explain type Ia supernova. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gSvkx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gSvkx.png&quot; alt=&quot;Initial final mass relation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-03-14T14:38:22.600" LastActivityDate="2016-03-14T14:38:22.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14152" PostTypeId="2" ParentId="14148" CreationDate="2016-03-14T07:54:02.793" Score="3" Body="&lt;p&gt;The trampoline analogy is useful if you understand gravity within a General Relativity framework. The conceptual problem there is that actually the space-time is wrapped in 4, not in 3 dimensions, i.e. including time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, when the Earth rotates around the Sun, it loses a very tiny amount of energy in form of gravitational waves. So, the Earth is actually spiraling towards the Sun. The thing is that this gravitational wave emission is so small, that by the time we observe any considerable spiraling, the Earth and the Sun would have already ceased to exist. Much before that, the Solar System becomes unstable due to chaotic effects already contained in classical Newtonian mechanics. &lt;/p&gt;&#xA;" OwnerUserId="11188" LastActivityDate="2016-03-14T07:54:02.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14153" PostTypeId="2" ParentId="14148" CreationDate="2016-03-14T10:22:47.203" Score="30" Body="&lt;p&gt;First of all: &quot;How gravity really works&quot; is a deep question, and any serious scientist would quickly concede that all we have is an incomplete working model. You certainly have heard about  &lt;a href=&quot;https://en.wikipedia.org/wiki/General_relativity&quot;&gt;General Relativity&lt;/a&gt;; the first image on the page is your trampoline. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our working model, General Relativity, is &lt;strong&gt;working&lt;/strong&gt; because it explains a lot of observations very nicely. (Careful, here is another deep question lingering: &quot;Explains&quot; means that we can predict some observations from other observations with the model of gravity we have in our  mind. It does not necessarily mean that we understand the &quot;real nature&quot; of the underlying issues.) But we are very confident that the model is working over a wide range of observations. One of the last &quot;first-time&quot; observations which followed the predictions and thus gave us more confidence in the model was the two black holes colliding lately. Lately? Well, billions of years ago. We just learned about it lately.  Here is a link to a &lt;a href=&quot;http://www.nytimes.com/2016/02/12/science/ligo-gravitational-waves-black-holes-einstein.html&quot;&gt;New York Times article with an impressive video&lt;/a&gt;. (I think one can still read a limited number of Times articles for free, so try it out.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our model of gravity is &lt;strong&gt;incomplete&lt;/strong&gt; because it doesn't connect well to the model of nature we have for other things (elementary particles, quantum physics). For a while (like 70 years or so) it didn't connect at all; Einstein himself completely failed to connect the dots, which was probably not encouraging since he had received the Nobel Price for laying one of the foundations of quantum physics &lt;em&gt;and&lt;/em&gt; was the obvious authority about gravity. If he couldn't do it, who could?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm not mistaken, the physicists today are making progress, slowly. This connection between quantum physics and gravity is one of the main unresolved problems in modern physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last, let me address your concern about the planets spiraling into the sun. This idea probably comes from actual balls on an actual trampoline spiraling in, I suppose. You probably know that the balls lose speed due to friction, much the same way you slow down on your bike when you stop pedaling. Some of the kinetic energy is transformed into heat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you know what? You are right. Given enough time, &lt;em&gt;the planets would eventually fall into the sun.&lt;/em&gt; Low-flying satellites fall back to earth after a few years, because there are still traces of atmosphere slowing them down out there. The reason is that there is &quot;friction&quot; in the wider sense involved in all large-scale processes in the universe. That is actually one of the fundamental physical principles making up the world we know. It's just that the near-vacuum between the planets doesn't provide that much friction, and the planets are fairly massive bodies with an enormous mass and kinetic energy. It will take a long long time for them to lose enough energy that they'll be so close as to touch the sun. (Perhaps too long to happen at all.) In fact, over human life times the planets, moons and stuff are almost perfect examples for movement without friction. But in the &lt;em&gt;astronomical time scale&lt;/em&gt; -- billions of years --, there certainly &lt;em&gt;is&lt;/em&gt; friction. For example, the moon is showing us always the same side because friction slowed its rotation so that the rotation is now &quot;locked&quot; with its orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line: The idea that gravity bends space and time &quot;explains&quot; all large-scale observations so far; the &quot;trampoline&quot; is a good model for a 2-dimensional &quot;space&quot;, i.e. a surface, if you ignore friction. &lt;/p&gt;&#xA;" OwnerUserId="11190" LastEditorUserId="11190" LastEditDate="2016-03-15T15:31:33.207" LastActivityDate="2016-03-15T15:31:33.207" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14155" PostTypeId="1" CreationDate="2016-03-14T18:38:41.513" Score="3" ViewCount="1127" Body="&lt;p&gt;I was musing today about black holes, and came across what seems to be a contradictory fact about black holes. If any matter were to actually fall into a black hole, it seems like it would need to gain infinite energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is seen by considering an object close to the event horizon of a black hole. If we consider escape velocity for a particle at the event horizon, it has an escape velocity of the speed of light. The energy required to achieve this escape velocity would therefore be infinite. As we approach the event horizon, it seems like we should be able, then, to define an arbitrarily small distance across which the particle gains an arbitrary amount of energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the object approaches arbitrarily close to the event horizon, it seems like its relativistic mass would therefore grow arbitrarily high, resulting in a corresponding increase in the gravity of the black hole, which seems impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we extrapolate back to the formation of the black hole, it then seems like it should be impossible for any particles to fall into the black hole, even as it forms. Instead, mass should be distributed in a probability cloud whose density asymptotically approaches zero at the event horizon of the black hole. New particles could then enter this cloud of incredibly dense (but not black hole dense) matter, where they would be captured rather than falling past the event horizon, within which there would be a vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this were true, there would be no black holes, only hollow spheres of radiation and matter which externally look like black holes, since their gravity would be great enough that particles entering the sphere would exit at an incredibly slow rate. Interestingly, such spheres wouldn't suffer from the apparent information paradox that black holes do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That or I'm missing an obvious explanation for why that's all wrong, which seems more likely because I'm neither a physicist nor an astronomer. What's the critical flaw in this argument?&lt;/p&gt;&#xA;" OwnerUserId="11201" LastActivityDate="2016-03-14T23:02:29.423" Title="Does conservation of energy make black holes impossible?" Tags="&lt;black-hole&gt;&lt;astrophysics&gt;" AnswerCount="2" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="14156" PostTypeId="1" CreationDate="2016-03-14T20:19:11.720" Score="1" ViewCount="414" Body="&lt;p&gt;Recently, I obtained my heatmap from a sky survey. But now, I'm trying to get a signal to noise map from my heatmap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My heatmap looks like :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9o6g5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My colour bar indicates the number of stars per pixel (I assume, but I'm not totally sure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several steps in order to get my &quot;S/N&quot; map, but firstly, I need to get mean value from pixels around pixel i. To do that, I must draw a circle around the pixel i (circle with a radius about 10 arcmin, for exemple).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To know where I am, that's to say on which pixel I'm working. I scripted a red cross which indicates the pixel. I want to determine the mean value around this cross for each pixel in the heatmap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I searched on this website and I found this post: Get pixel value after a colormap. I think that it will be interesting in my case also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I really haven't an idea to script this. After that, I need to calculate lots of things like gaussian RMS etc .. but for that, it will be easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my script : &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    # -*- coding: utf-8 -*-&#xA;    #!/usr/bin/env python&#xA;&#xA;    from astropy.io import fits&#xA;    from astropy.table import Table&#xA;    from astropy.table import Column&#xA;    from astropy.convolution import convolve, Gaussian2DKernel&#xA;    import numpy as np&#xA;    import scipy.ndimage as sp&#xA;    import matplotlib.pyplot as plt&#xA;&#xA;&#xA;&#xA;            ###################################&#xA;            # Importation du fichier de champ #&#xA;            ###################################&#xA;&#xA;    filename = '/home/valentin/Desktop/Field169_combined_final_roughcal.fits_traite_traiteXY_traiteXY_final'&#xA;&#xA;    print 'Fichier en cours de traitement' + str(filename) + '\n'&#xA;&#xA;    # Ouverture du fichier à l'aide d'astropy&#xA;    field = fits.open(filename)         &#xA;&#xA;    # Lecture des données fits  &#xA;    tbdata = field[1].data             &#xA;&#xA;&#xA;            #######################################&#xA;            # Parametres pour la carte de densité #&#xA;            #######################################&#xA;&#xA;    # Boite des étoiles bleues :&#xA;    condition_1 = np.bitwise_and(tbdata['g0-r0'] &amp;gt; -0.5, tbdata['g0-r0'] &amp;lt; 0.8 )    # Ne garder que les -0.4 &amp;lt; (g-r)0 &amp;lt; 0.8&#xA;    condition_final = np.bitwise_and(tbdata['g0'] &amp;lt; 23.5, condition_1)      # Récupere les valeurs de 'g0' &amp;lt; 23.5 dans les valeurs de blue_stars_X&#xA;&#xA;    Blue_stars = tbdata[condition_final]&#xA;&#xA;    RA_Blue_stars = Blue_stars['RA']                        # Récupere les valeurs de 'RA' associées aux étoiles bleues&#xA;    DEC_Blue_stars = Blue_stars['DEC']                      # Récupere les valeurs de 'DEC' associées aux étoiles bleues&#xA;&#xA;&#xA;    # Boite des étoiles tres bleues :&#xA;    condition_2 = np.bitwise_and(tbdata['g0-r0'] &amp;gt; -0.5, tbdata['g0-r0'] &amp;lt; 0.2 )&#xA;    condition_final2 = np.bitwise_and(tbdata['g0'] &amp;lt; 23.5, condition_2)&#xA;&#xA;    Very_Blue_stars = tbdata[condition_final2]&#xA;&#xA;    RA_Very_Blue_stars = Very_Blue_stars['RA']                      # Récupere les valeurs de 'RA' associées aux étoiles bleues&#xA;    DEC_Very_Blue_stars = Very_Blue_stars['DEC']&#xA;&#xA;    # ==&amp;gt; La table finale avec le masque s'appelle Blue_stars &amp;amp; Very_Blue_stars&#xA;&#xA;            ##################################################################&#xA;            # Traçage des différents graphiques de la distribution d'étoiles #&#xA;            ##################################################################&#xA;&#xA;&#xA;    fig1 = plt.subplot(2,2,1)&#xA;    plt.plot(tbdata['g0-r0'], tbdata['g0'], 'r.', label=u'Etoiles du champ')&#xA;    plt.plot(Blue_stars['g0-r0'], Blue_stars['g0'], 'b.', label =u'Etoiles bleues')&#xA;    plt.plot(Very_Blue_stars['g0-r0'], Very_Blue_stars['g0'], 'k.', label =u'Etoiles tres bleues')&#xA;    plt.title('Diagramme Couleur-Magnitude')&#xA;    plt.xlabel('(g0-r0)')&#xA;    plt.ylabel('g0')&#xA;    plt.xlim(-1.5,2.5)&#xA;    plt.ylim(14,28)&#xA;    plt.legend(loc='upper left')&#xA;    plt.gca().invert_yaxis()&#xA;&#xA;    fig1 = plt.subplot(2,2,2)&#xA;    plt.plot(RA_Blue_stars, DEC_Blue_stars, 'b.', label =u'Etoiles bleues', alpha=0.15)&#xA;    plt.title('Carte de distribution des etoiles bleues')&#xA;    plt.xlabel('RA')&#xA;    plt.ylabel('DEC')&#xA;    plt.legend(loc='upper left')&#xA;&#xA;    fig1 = plt.subplot(2,2,3)&#xA;    plt.plot(RA_Very_Blue_stars, DEC_Very_Blue_stars, 'r.', label =u'Etoiles tres bleues',alpha=0.4)&#xA;    plt.title('Carte de distribution des etoiles tres bleues')&#xA;    plt.xlabel('RA')&#xA;    plt.ylabel('DEC')&#xA;    plt.legend(loc='upper left')&#xA;&#xA;    fig1 = plt.subplot(2,2,4)&#xA;    plt.plot(RA_Blue_stars, DEC_Blue_stars, 'b.', label =u'Etoiles bleues', alpha=0.15)&#xA;    plt.plot(RA_Very_Blue_stars, DEC_Very_Blue_stars, 'r.', label =u'Etoiles tres bleues',alpha=0.4)&#xA;    plt.title('Carte de distribution des etoiles bleues et tres bleues')&#xA;    plt.xlabel('RA')&#xA;    plt.ylabel('DEC')&#xA;    plt.legend(loc='upper left')&#xA;&#xA;            ######################################################################&#xA;            # Traçage des différents graphiques de la carte de densité d'étoiles #&#xA;            ######################################################################&#xA;&#xA;&#xA;    # Carte de densité des étoiles bleues pour 1 pixel de 1 arcmin^2 (bins = 180)&#xA;&#xA;    X_Blue_stars = Blue_stars['X']&#xA;    Y_Blue_stars = Blue_stars['Y']&#xA;&#xA;    heatmap, xedges, yedges = np.histogram2d(X_Blue_stars, Y_Blue_stars, bins=180) # bins de 180 car 3° de champ en RA = 180 arcmin de champ en RA&#xA;    extent = [xedges[0], xedges[-1], yedges[0], yedges[-1]]&#xA;&#xA;    RotatePlot = sp.rotate(heatmap,90)&#xA;&#xA;    plt.clf()&#xA;    #fig = plt.subplot(2,2,1)&#xA;    plt.imshow(RotatePlot, extent=extent, interpolation='none')&#xA;    plt.colorbar()&#xA;    plt.title('Carte de densite des etoiles bleues (non lisse)')&#xA;    plt.xlabel(&quot;X&quot;)&#xA;    plt.ylabel(&quot;Y&quot;)&#xA;&#xA;&#xA;    # Carte de densité lissée (par convolution avec une gaussienne 2 sigma) des étoiles bleues pour 1 pixel de 1 arcmin^2 (bins = 180)&#xA;    # ==&amp;gt; Avec Astropy&#xA;&#xA;    plt.clf()&#xA;    fig = plt.figure(1)&#xA;    smoothed_heatmap = plt.imshow(convolve(RotatePlot, Gaussian2DKernel(stddev=2)), interpolation='nearest')&#xA;    plt.colorbar()&#xA;    plt.plot(100,120,'rx', markeredgewidth=3, markersize=10)&#xA;    plt.title('Carte de densite des etoiles bleues lisse (astropy)')&#xA;    plt.xlabel(&quot;X (arcmin)&quot;)&#xA;    plt.ylabel(&quot;Y (arcmin)&quot;)&#xA;    plt.xlim(0,180)&#xA;    plt.ylim(0,180)&#xA;    print smoothed_heatmap.cmap(smoothed_heatmap.norm(X_Blue_stars[100],Y_Blue_stars[120]))&#xA;&#xA;    #plt.savefig('/home/valentin/Desktop/Final.png')&#xA;    plt.show()&#xA;&#xA;    print &quot;Création du Diagramme&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="11206" LastEditorUserId="2270" LastEditDate="2016-05-14T19:17:00.720" LastActivityDate="2016-06-13T20:15:42.020" Title="Create a signal to noise map from heatmap" Tags="&lt;star&gt;&lt;sky-survey&gt;&lt;python&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14157" PostTypeId="2" ParentId="14155" CreationDate="2016-03-14T20:35:32.913" Score="0" Body="&lt;p&gt;Let's reformulate your argument as follows (using idealized assumptions for simplicity):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;An object some distance away from the black hole has potential energy equal to the kinetic energy it would have at the event horizon.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This is the same amount of energy that would be required to pull the object from the event horizon back to its original position (and velocity).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We would need to accelerate the object to the escape velocity to return it to its position.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since the escape velocity is greater than &lt;em&gt;c&lt;/em&gt;, it would take an infinite amount of energy to return it to its position.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The incorrect point is &lt;strong&gt;3&lt;/strong&gt;. We can pull an object out of a gravitational field slowly by exerting a force on it the whole way. We would only need to accelerate it to the escape velocity if we wanted it to escape &quot;on its own&quot;, i.e. without us exerting a force to keep pushing it away from the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I'm less sure of my answer after doing some back-of-the-envelope calculations. It would be helpful for a real expert to weigh in. I'll leave my answer here anyway.&lt;/p&gt;&#xA;" OwnerUserId="11205" LastEditorUserId="11205" LastEditDate="2016-03-14T20:59:25.030" LastActivityDate="2016-03-14T20:59:25.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14158" PostTypeId="2" ParentId="14156" CreationDate="2016-03-14T22:06:27.637" Score="1" Body="&lt;p&gt;You are making life difficult for yourself. Make the image &lt;em&gt;be&lt;/em&gt; the number of stars per pixel and make sure your pixels are big enough to contain say at least 10 stars in each pixel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At that point you can assume you have &lt;a href=&quot;https://en.wikipedia.org/wiki/Poisson_distribution&quot; rel=&quot;nofollow&quot;&gt;Poissonian statistics&lt;/a&gt; where the variance is equal to the expectation value. To a first approximation, what you have observed is the most likely expectation value, so the &quot;noise&quot; is the square root of the variance, which is the square root of how many stars are in each pixel. Thus your signal-to-noise map is simply the square root of the number of stars per pixel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It maybe that your map has been normalised to give number of stars per unit angular area of the pixel. In which case, simply multiply by the pixel area and then take the square root.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-03-14T22:06:27.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14159" PostTypeId="2" ParentId="14155" CreationDate="2016-03-14T23:02:29.423" Score="5" Body="&lt;p&gt;For simplicity, let's consider a Schwarzschild black hole, so that the spacetime is spherically symmetric and static. In particular, the Schwarzschild time $t$ coordinate gives a direction in which the geometry 'stays the same' (a Killing vector field), and its inner product with the orbital four-velocity $u$ is conserved:&#xA;$$\epsilon = -g(u,\partial_t) = \left(1-\frac{2M}{r}\right)\frac{\mathrm{d}t}{\mathrm{d}\tau}\text{,}$$&#xA;where $\tau$ is the proper time of the orbiting particle. One can think of this as the specific (per-mass) energy, including mass-energy: a particle escaping to infinity that becomes asymptotically at rest with respect to stationary observers would have $\epsilon = 1$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If we consider escape velocity for a particle at the event horizon, it has an escape velocity of the speed of light. The energy required to achieve this escape velocity would therefore be infinite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The energy as measured by whom? Imagine a family of stationary observers everywhere surrounding the black hole, or at least along an infalling particle's trajectory. Those observers measure the speed and energy of the particle as it falls past them. As the particle nears the horizon, they will report speeds arbitrarily close to the speed of light and arbitrarily high energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to a far-away stationary observer, those observers near the horizon are experiencing increasingly divergent gravitational time dilation, and the orbital energy of the particle stays constant. Flinging a particle of mass $m$ into the black hole will increase the mass of the black hole by $m\epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What's the critical flaw in this argument?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All energy gravitates, so I would say the main flaw is forgetting to add the gravitational potential energy of the particle to the mass of the black hole as well. Or better put, you should be concerned with the orbital energy, not just its mass+kinetic parts ('relativistic mass').&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this situation, you have a well-defined conserved orbital energy. If you insist on measuring the Lorentz factor according to close-by stationary observers, then yes, it diverges to $+\infty$, but then you would have to admit a gravitational potential energy term that diverges to $-\infty$, because their sum must resolve to be the orbital energy.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2016-03-14T23:02:29.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14160" PostTypeId="1" CreationDate="2016-03-15T02:47:30.830" Score="3" ViewCount="190" Body="&lt;p&gt;I've never understood why it's spacetime and not space and time. In other words why aren't space and time spoken of as two different things. Can someone explain to me why this is&lt;/p&gt;&#xA;" OwnerDisplayName="user11186" LastActivityDate="2016-03-15T21:35:11.280" Title="Why is space and time spoken of as one thing" Tags="&lt;space-time&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14161" PostTypeId="2" ParentId="14160" CreationDate="2016-03-15T03:24:46.737" Score="7" Body="&lt;p&gt;There are different words for different aspects of space. For example, consider: &lt;em&gt;length&lt;/em&gt;, &lt;em&gt;width&lt;/em&gt;, and &lt;em&gt;height&lt;/em&gt;. Other words include &lt;em&gt;depth&lt;/em&gt; and &lt;em&gt;breadth&lt;/em&gt;. We can speak of them as different things if we choose to, but we generally consider them to be part of unified concept of &lt;em&gt;space&lt;/em&gt;. Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's because we understand that these words just pick out measurements along specific directions in space set by context or the speaker's viewpoint, and that there is no intrinsic difference between those directions. Depending on which way they are facing, what's &lt;em&gt;length&lt;/em&gt; to one person is &lt;em&gt;width&lt;/em&gt; to another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More formally, there is a symmetry between directions in space. We can rotate freely and the intrinsic properties of space look just the same, e.g. the Euclidean metric (infinitesimal distance formula)&#xA;$$\mathrm{d}s^2 = \mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2$$&#xA;is invariant under the &lt;a href=&quot;https://en.wikipedia.org/wiki/Orthogonal_group&quot;&gt;orthogonal group&lt;/a&gt; of rotations, $\mathrm{O}(3)$. Together with spatial translations (which are like rotations about a point at infinity), this forms the &lt;a href=&quot;https://en.wikipedia.org/wiki/Euclidean_group&quot;&gt;Euclidean group&lt;/a&gt; of isometries, $\mathrm{ISO}(3)$. Because of this symmetry that &quot;intermixes&quot; directions of space, we think of those different directions as being unified one into one thing: space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In special relativity, spacetime has a Minkowski metric (in units of $c=1$)&#xA;$$\mathrm{d}s^2 = -\mathrm{d}t^2 + \mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2\text{,}$$&#xA;and it is invariant under the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lorentz_group&quot;&gt;Lorentz group&lt;/a&gt; $\mathrm{O}(1,3)$ that act like rotations in spacetime; together with the translations (in space and in time), the full isometry group is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Poincar%C3%A9_group&quot;&gt;Poincaré group&lt;/a&gt;. In fact, the usual Lorentz transformation (say, boosting along the $x$-axis) is exactly a rotation with by a hyperbolic angle in the $tx$ plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why people began to think of &lt;em&gt;spacetime&lt;/em&gt; as one unified whole. The reasons are essentially the same as why people think of &lt;em&gt;space&lt;/em&gt; as one unified whole despite it having different directions: there is a symmetry that &quot;intermixes&quot; those directions, albeit slightly differently in spacetime because of the different sign of the temporal and spatial directions in the metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general relativity, this is generalized further, but the details aren't immediately relevant here.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2016-03-15T03:24:46.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14162" PostTypeId="2" ParentId="14148" CreationDate="2016-03-15T04:46:21.230" Score="1" Body="&lt;p&gt;Great question!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you heard of Newton's First Law? It says that an object in motion continues moving at the same speed and in the same direction &lt;em&gt;unless acted upon by a force&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we roll a ball along the ground, it will eventually stop. Before Newton, many people believed that everything slows down by itself. Newton's insight was that this is not true, and actually the only reason a rolling ball will slow down is because the ground and the air rub or push against the ball to slow it down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a trampoline, a ball will rub against the trampoline material and against the air, which slows it down. This is the only reason that the ball ends up spiralling towards the centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When there is nothing to slow the object down, it won't spiral towards the middle, it will just keep going round and round forever. In space there is (almost) nothing to slow an object down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you find this hard to believe, you can write a computer program to do all the calculations and see what happens! I have made &lt;a href=&quot;https://jsfiddle.net/0erknpk8/&quot; rel=&quot;nofollow&quot;&gt;an example simulation&lt;/a&gt; for you. You will see that without friction, the planet will end up where it started each time it goes around the sun. If you change the planet's initial yspeed from 20 to 40 and then click &quot;Run&quot; up the top you will see a more circular orbit. You can change other things and see what happens. I hope you find this useful!&lt;/p&gt;&#xA;" OwnerUserId="11212" LastActivityDate="2016-03-15T04:46:21.230" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14163" PostTypeId="1" AcceptedAnswerId="14167" CreationDate="2016-03-15T06:21:45.910" Score="7" ViewCount="2561" Body="&lt;p&gt;As we know, we can find rocks from Mars on Earth, how about Venus and Mercury? Is that not found yet? Or it is impossible to find them because they are closer to the Sun, and debris won't go far away from the Sun?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="14094" LastEditDate="2019-06-16T04:11:31.430" LastActivityDate="2019-06-16T04:11:31.430" Title="Can we find rocks from Venus or Mercury on Earth?" Tags="&lt;mars&gt;&lt;venus&gt;&lt;mercury&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14164" PostTypeId="2" ParentId="14163" CreationDate="2016-03-15T07:55:39.380" Score="1" Body="&lt;p&gt;Possible, but extremely unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rocks thrown off Mercury by an impact would not only have to clear Mercury's escape velocity to get off the surface of that planet, but would also have to go faster than the escape velocity from the Sun's gravity well at Mercury's distance from the Sun.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The escape velocity from Mercury's gravity well is 4.25 Km per&#xA;second.&lt;/li&gt;&#xA;&lt;li&gt;The escape velocity from the Sun's gravity well at Mercury's&#xA;distance is 67.7 Km per second.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That means a rock would have to go faster than 67.7. Kps to even get close to Earth's orbit. Very few rocks will be going that fast, so most rocks knocked off Mercury by an impact will end up orbiting the Sun at the same distance as Mercury. Eventually those rocks will fall back onto Mercury.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same thinking goes rocks knocked off Venus's surface, but the numbers are a little different than Mercury's.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The escape velocity from Venus's gravity well is 10.3 Km per second.&lt;/li&gt;&#xA;&lt;li&gt;The escape velocity from the Sun's gravity well at Venus's distance is 49.5 Km per second.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11175" LastActivityDate="2016-03-15T07:55:39.380" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14165" PostTypeId="2" ParentId="14148" CreationDate="2016-03-15T16:32:08.737" Score="-2" Body="&lt;p&gt;Neutrino String Induction-Refraction is the cause of gravity. Some will say that neutrinos are insignificant, but Dirac, Hawking, and Tyson think otherwise and most discount the effect of a charged particle traveling at the speed of light. Keep in mind that no one can, or has proved that mass is a property of matter, more of an effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go to www.themechanismofreality.com , this site explains exactly how gravity works. Every Physicist who examines this agrees that this is correct. From CERN to the University of Beijing's Physic Department agree that this is a 'Fantastic connection between graviton physics and string theory&quot;! This was also confirmed , indirectly, by LIGO and the gravity wave announcement. Enjoy!&lt;/p&gt;&#xA;" OwnerUserId="11227" LastEditorUserId="11227" LastEditDate="2016-03-15T17:19:49.260" LastActivityDate="2016-03-15T17:19:49.260" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14167" PostTypeId="2" ParentId="14163" CreationDate="2016-03-15T19:43:49.130" Score="9" Body="&lt;p&gt;You can think of it in terms of &lt;a href=&quot;https://en.wikipedia.org/wiki/Hohmann_transfer_orbit&quot; rel=&quot;noreferrer&quot;&gt;Hohmann transfer orbits&lt;/a&gt;, which define the minimum $\Delta v$ that needs to be applied to bring something from one orbital radius to another orbital radius when orbiting a massive body. This calculation takes into account that the two objects have Keplerian orbits where the objects &lt;em&gt;begins&lt;/em&gt; with at least the orbital speed of the initial orbital radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hohmann $\Delta v$ is given by&#xA;$$\Delta v = \sqrt{\frac{GM_{\odot}}{r_1}} \left(\sqrt{\frac{2r_2}{r_1 +r_2}} -1 \right),$$&#xA;where $r_1$ and $r_2$ are the initial and final radius from the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conceptually what you have to do is give a rock enough energy to escape from the planet and then give it an additional $(m/2)(\Delta v)^2$ of kinetic energy to get it to transfer into the other orbit. If the ejection speed is $v_{ej}$ then&#xA;$ v_{eg} &amp;gt; \sqrt{(\Delta v)^2 + v_{esc}^2},$&#xA;where $v_{esc}$ is the escape velocity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numbers for Mercury $\rightarrow$ Earth are $\Delta v = 9.2$ km/s and for Mars $\rightarrow$ Earth $\Delta v = -2.6$ km/s (you have to slow it down to allow it to fall inwards).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The escape velocities for these planets are 4.35 km/s and 5 km/s respectively (so almost the same).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means you need to give a rock more kinetic energy to get it to Earth from Mercury as from Mars. In the case of Mars, the transfer kinetic energy is almost negligible once the rock can escape Mars' gravity. In the case of Mercury, the rock needs to be given an initial ejection velocity of $&amp;gt; \sqrt{9.2^2 + 4.3^2}= 10.1 $ km/s. This compares with $&amp;gt; \sqrt{2.4^2 + 5^2}= 6.5$ km/s for Mars. At lower ejection speeds most of the ejected objects will be reaccreted by the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Against this, the leading theory to explain migration of rocks between planets is high velocity impacts. Objects falling from much further out will hit Mercury with greater speeds than Mars and impart greater energies to the ejecta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A paper addressing the possibility of Mecurean meteorites was presented by &lt;a href=&quot;http://arxiv.org/abs/0801.4038&quot; rel=&quot;noreferrer&quot;&gt;Gladman &amp;amp; Coffey (2008)&lt;/a&gt;. They concluded that once ejection speeds are large enough ($\sim 10$ km/s) to produce Earth-crossing ejecta, that significant accretion of meteorites should take place. Several per cent of high speed ejecta should impact the Earth (or its atmosphere at least) within 30 million years. This compares with an efficiency a factor of 2-3 higher for Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various reports and speculations that at least one meteorite in existing collections (NWA7325, pictured) &lt;em&gt;may&lt;/em&gt; have come from Mercury. See &lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/2013/03211549-lpsc-hermean-meteorite.html?referrer=https://www.google.co.uk/&quot; rel=&quot;noreferrer&quot;&gt;here for example&lt;/a&gt;. It appears that the main problem is getting agreement on what the chemical signatures of such meteorites are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gCmwP.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gCmwP.jpg&quot; alt=&quot;NWA7325&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accretion of material from Venus is a different matter. The required ejection velocities are higher because the escape velocity for Venus 10.4 km/s. But more importantly, drag in the dense Venusian atmosphere would prevent anything emerging from the planet with anything like these speeds.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-03-15T19:51:03.997" LastActivityDate="2016-03-15T19:51:03.997" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14168" PostTypeId="1" AcceptedAnswerId="14171" CreationDate="2016-03-15T21:15:09.403" Score="2" ViewCount="621" Body="&lt;p&gt;Say I have a FITS file with an observed field of stars, each one with x,y coordinates given in pixels. Assume the image is aligned so that the x axis is the right ascension, and the y axis the declination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a tutorial to transform the pixel coordinates of each star to equatorial, even if the transformation is approximate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'v been looking for a while but found no tutorial/book/site that enumerates the steps needed to produce this transformation. &lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2016-03-15T23:50:34.233" Title="Transform pixel coordinates (in FITS file) to equatorial" Tags="&lt;astrometry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14169" PostTypeId="2" ParentId="14160" CreationDate="2016-03-15T21:35:11.280" Score="-2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is space and time spoken of as one thing?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because it's how Minkowski presented it in 1908. See &lt;a href=&quot;https://en.wikisource.org/wiki/Translation:Space_and_Time&quot; rel=&quot;nofollow noreferrer&quot;&gt;Space and Time&lt;/a&gt; where you can read this: &lt;em&gt;&quot;Henceforth, space for itself, and time for itself shall completely reduce to a mere shadow, and only some sort of union of the two shall preserve independence&quot;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I've never understood why it's spacetime and not space and time. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're right to do so. Because spacetime is actually a mathematical abstraction that people mistake for reality. The Earth is surrounded by space, not spacetime. Light moves through space, not spacetime. See &lt;a href=&quot;https://physics.stackexchange.com/a/133821/76162&quot;&gt;this&lt;/a&gt; physics stack exchange answer where relativist Ben Crowell says &lt;em&gt;&quot;Objects don't move through spacetime. Objects move through space&quot;.&lt;/em&gt; When you move fast through space your measurements of space and time change, but it's &lt;em&gt;you&lt;/em&gt; that's changed, not space, and not the things in space. For example a star doesn't actually length-contract to a discoid shape just because &lt;em&gt;you&lt;/em&gt; step on the gas. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In other words why aren't space and time spoken of as two different things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are. The thing to appreciate is that whilst a lot of popscience articles blithely talk about spacetime, people like Einstein were very particular about saying space when they meant space. See for example Einstein's &lt;a href=&quot;http://einsteinpapers.press.princeton.edu/vol7-trans/192?highlightText=%22neither%20homogeneous%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;1920 Leyden Address&lt;/a&gt; where he referred to the continuum of space-time and yet described a gravitational field as space that's neither homogeneous nor isotropic. The distinction between the two tends to get lost. In a gravitational field space is inhomogeneous, and the inhomogeneity diminishes with distance in a non-linear fashion. When you plot it using say light-clocks at different elevations, you see a curve on your plot. Your plot depicts curved spacetime. That's the map. But inhomogeneous space is the territory. See &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0256-307X/25/5/014/meta;jsessionid=58E90A5C820387FA2AAD6535D3562F76.c1.iopscience.cld.iop.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;Inhomogeneous Vacuum: An Alternative Interpretation of Curved Spacetime&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can someone explain to me why this is?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By and large it's because people don't have a clear understanding of relativity. Space and time are not on an equal footing. I can hop forward a metre but you can't hop forward a second. There is no real rotation when you accelerate, it's just an abstract mathematical rotation, wherein your &quot;local motion&quot; is of necessity reduced by your macroscopic motion through space because the total rate of motion can't exceed c, because of the wave nature of matter. See my time dilation answer &lt;a href=&quot;https://physics.stackexchange.com/questions/241772/what-is-time-dilation-really/242014#242014&quot;&gt;here&lt;/a&gt; on stack exchange. Since when you look at what clocks do  you appreciate that the time is a cumulative measure of local motion, you're actually better off thinking in terms of space and motion rather than space and time.  &lt;/p&gt;&#xA;" OwnerUserId="9772" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-03-15T21:35:11.280" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14170" PostTypeId="1" CreationDate="2016-03-15T22:07:15.270" Score="5" ViewCount="132" Body="&lt;p&gt;I imagine the answer is unknown for the entire galaxy, but I would appreciate a notion of the scales and locations of regions where astronomers have ventured at least a guess to the distribution of organic matter in interstellar dust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Updated&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am specifically looking for plots showing the distribution of some selected organic compounds (doesn't really matter which ones right now) within a given section of the sky.  Or, if possible, within a given arc and range with respect to the galactic center (again, if possible).  &lt;/p&gt;&#xA;" OwnerUserId="11231" LastEditorUserId="11264" LastEditDate="2018-02-23T14:06:48.060" LastActivityDate="2018-02-23T14:06:48.060" Title="What is the distribution of organic compounds in the Milky Way Galaxy?" Tags="&lt;milky-way&gt;&lt;interstellar-medium&gt;&lt;dust&gt;&lt;astrochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14171" PostTypeId="2" ParentId="14168" CreationDate="2016-03-15T22:16:58.143" Score="2" Body="&lt;p&gt;Assuming that each (square) pixel has the same angular scale (not a given if the field of view is large) of $\theta$ degrees/pixel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the declination (in degrees): $\delta \simeq \delta_0 + \theta y$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right ascension (in degrees): $\alpha \simeq \alpha_0 - \theta x/\cos \delta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\alpha_0$ and $\delta_0$ are the RA and Dec at $x=0$, $y=0$. [The minus sign is there because right ascension increases towards the left of a sky image.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approximatimation becomes poor as the field of view gets larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A slightly more complex, but accurate, approach is described in &lt;a href=&quot;http://gtn.sonoma.edu/data_reduction/astrometry.php&quot; rel=&quot;nofollow&quot;&gt;http://gtn.sonoma.edu/data_reduction/astrometry.php&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-03-15T23:50:34.233" LastActivityDate="2016-03-15T23:50:34.233" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14172" PostTypeId="2" ParentId="14170" CreationDate="2016-03-15T23:42:05.957" Score="2" Body="&lt;p&gt;The answer depends on which you are talking about (you mentioned both organic compounds and organic matter, which are two completely different things). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Organic compounds we do not have a clear definition of, but is most agreed upon to contain carbon atoms, either on their own (C), bonded with at least one hydrogen atom (C-H), or bonded with at least one other carbon atom (C-C). Compounds such as Methane (C-H4), Ethanol (C2-H6-O), and  Glucose (C6-H12-O6) are textbook examples of organic compounds. Methane has been discovered on several objects &lt;a href=&quot;https://en.wikipedia.org/wiki/Methane#Extraterrestrial_methane&quot; rel=&quot;nofollow&quot;&gt;within our solar system&lt;/a&gt;, including Mars, Venus, and Saturn. Methane has also been discovered on the exoplanet &lt;a href=&quot;https://en.wikipedia.org/wiki/HD_189733_b&quot; rel=&quot;nofollow&quot;&gt;HD 189733b&lt;/a&gt; (though it is the only example we have of methane on an exoplanet), and often exists in interstellar clouds, showing that it is quite common throughout our galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Organic matter refers to matter composed of organic compounds, as part of an organism. While we are searching for organic matter as proof of extraterrestrial life, we have yet to discover any organic matter or clear evidence for such matter outside of our own planet.&lt;/p&gt;&#xA;" OwnerUserId="11233" LastActivityDate="2016-03-15T23:42:05.957" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14173" PostTypeId="2" ParentId="14098" CreationDate="2016-03-16T05:12:22.913" Score="1" Body="&lt;p&gt;At least some of the 'active asteroids' recently observed are most likely the result of impacts. One example is 2010 A2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AxSLy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AxSLy.jpg&quot; alt=&quot;active asteroid 2010 A2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details, see &lt;a href=&quot;https://en.wikipedia.org/wiki/P/2010_A2_(LINEAR)&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/P/2010_A2_(LINEAR)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7144" LastActivityDate="2016-03-16T05:12:22.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14174" PostTypeId="2" ParentId="13998" CreationDate="2016-03-16T05:23:42.157" Score="1" Body="&lt;p&gt;NASA doesn't have its own asteroid tracking program. They are rather coordinating activities and provide funding for asteroid research and discovery programs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most successful dedicated asteroid discovery programs at the moment are Pan-STARRS (&lt;a href=&quot;http://pan-starrs.ifa.hawaii.edu/public/&quot; rel=&quot;nofollow&quot;&gt;http://pan-starrs.ifa.hawaii.edu/public/&lt;/a&gt;) and the Catalina Sky Survey (&lt;a href=&quot;http://www.lpl.arizona.edu/css/&quot; rel=&quot;nofollow&quot;&gt;http://www.lpl.arizona.edu/css/&lt;/a&gt;); see &lt;a href=&quot;http://neo.jpl.nasa.gov/stats/&quot; rel=&quot;nofollow&quot;&gt;http://neo.jpl.nasa.gov/stats/&lt;/a&gt; for some discovery statistics. Both these programs, as well as many others, run telescope that survey the night sky for yet unknown asteroids. On average, both programs together find 4 new near-Earth asteroids (and many others) per night. &lt;/p&gt;&#xA;" OwnerUserId="7144" LastActivityDate="2016-03-16T05:23:42.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14175" PostTypeId="1" CreationDate="2016-03-16T09:23:00.120" Score="2" ViewCount="1394" Body="&lt;p&gt;I was wondering if it's possible to measure star distance using parallax at home with a small telescope (Something like a 3-inch reflector)? Or is this not really practical?&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2016-03-16T17:52:58.797" Title="Measuring star distance by parallax using a small telescope" Tags="&lt;distances&gt;&lt;parallax&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14176" PostTypeId="2" ParentId="14098" CreationDate="2016-03-16T10:31:29.667" Score="3" Body="&lt;p&gt;Asteroid impacts on the Moon are monitored by researchers and amateurs. &lt;a href=&quot;http://www.nasa.gov/centers/marshall/news/lunar/#.VukzIfkrJhE&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Meteoroid Environment Office&lt;/a&gt; has a list of candidate impacts I stumbled upon, Lunar Monitoring Program is another. &lt;a href=&quot;https://www.youtube.com/watch?v=IYloGuUZCFM&quot; rel=&quot;nofollow noreferrer&quot;&gt;Videos and photos&lt;/a&gt; of Lunar impacts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aJ2Cu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aJ2Cu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/9OBRv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9OBRv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;em&gt;One impact photo to the left. All recorded impacts 2005-2015 to the right above.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-03-16T10:31:29.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14177" PostTypeId="2" ParentId="14175" CreationDate="2016-03-16T10:36:18.623" Score="1" Body="&lt;p&gt;Short answer - not really, parallax for the closest stars is right on the limits of resolution for a good ground-based amateur equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nearest star would show a parallax angle of under an arcsecond.  (The &lt;a href=&quot;https://en.wikipedia.org/wiki/Parsec&quot; rel=&quot;nofollow&quot;&gt;Parsec&lt;/a&gt; even takes its definition based on one arcsecond of parallax.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ground based amateur observations are probably limited to an arcsecond at most, so it's probably not possible to measure them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A special (and totally fictional) case - if a nearby star like Alpha Centauri were exactly between us and a much more distant star, and it were possible to see them both without Centauri overpowering that distant star, then &lt;strong&gt;maybe&lt;/strong&gt; it would be possible to observe them as a close double.  That's not true in practice however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prompted by Rob's much better answer, the earliest parallax determination I've been able to find was using &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the 6.2-inch (157.5 mm) aperture Fraunhofer heliometer at Königsberg &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Heliometer&quot; rel=&quot;nofollow&quot;&gt;Heliometer&lt;/a&gt; article on Wikipedia.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In purely aperture terms, that's an amateur-sized instrument (with some admittedly very precise measuring gear built in.) So my suggestion now is to ignore my initially doubtful response and try measurements, perhaps using a camera at fairly high magnification as Rob suggests...&lt;/p&gt;&#xA;" OwnerUserId="7491" LastEditorUserId="7491" LastEditDate="2016-03-16T17:52:58.797" LastActivityDate="2016-03-16T17:52:58.797" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14178" PostTypeId="1" AcceptedAnswerId="14179" CreationDate="2016-03-16T13:07:40.550" Score="1" ViewCount="75" Body="&lt;p&gt;I'm writing a research paper over outer space (yeah, yeah, broad topic) and I'm wondering what the proper terminology for nebula clouds is.  This is an excerpt from my paper showing how I'm going to use the term:  &quot;Outer space is the host of all the types of stars and galaxies and a lot of other interesting phenomena, such as black holes, supernovas, and ________.&quot;  Long story short, would I use &quot;nebula clouds&quot; or &quot;nebulae&quot;?&lt;/p&gt;&#xA;" OwnerUserId="11240" LastActivityDate="2016-03-16T14:41:55.033" Title="What's the proper terminology for nebula clouds?" Tags="&lt;nebula&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14179" PostTypeId="2" ParentId="14178" CreationDate="2016-03-16T13:18:09.370" Score="0" Body="&lt;p&gt;&quot;nebula clouds&quot; sound a little bit redundant, because, is there any nebula that is &lt;em&gt;not&lt;/em&gt; a cloud?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would stick to &quot;nebulae&quot;. (And if you want to be consistent, you should be writing &quot;supernovae&quot; too).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not an expert on this topic, but I have never seen any use of &quot;nebula cloud&quot; before.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastActivityDate="2016-03-16T13:18:09.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14180" PostTypeId="1" CreationDate="2016-03-16T13:56:03.927" Score="1" ViewCount="275" Body="&lt;p&gt;In a previous question (&lt;a href=&quot;https://astronomy.stackexchange.com/questions/14168/transform-pixel-coordinates-in-fits-file-to-equatorial&quot;&gt;Transform pixel coordinates (in FITS file) to equatorial&lt;/a&gt;), &lt;a href=&quot;https://astronomy.stackexchange.com/users/2531/rob-jeffries&quot;&gt;Rob&lt;/a&gt; supplied a &lt;a href=&quot;http://gtn.sonoma.edu/data_reduction/astrometry.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;great source&lt;/a&gt; that describes the process of obtaining celestial coordinates for objects in an observed field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The source mentions the &lt;em&gt;tangent point&lt;/em&gt; in its description of the Standard Coordinates system:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;These are the projection of the RA and DEC of an object onto the tangent plane of the sky, i.e., the plane tangent to the sphere of the sky that intersects the sky at the tangent point (which is &lt;strong&gt;usually somewhat offset from the image center&lt;/strong&gt;). The X coordinate is aligned with RA and the Y coordinate is aligned with DEC. The &lt;strong&gt;origin of the standard system is at the tangent point in the image&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would've assumed that the tangent point (ie: the origin of the standard coordinates system) would be located at the center of the observed field. Apparently this is not correct, hence the questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;del&gt;Why is the tangent point usually offset from the image center?&lt;/del&gt; (See Edit below)&lt;/li&gt;&#xA;&lt;li&gt;How is the tangent point $(\alpha_0, \delta_0)$ obtained?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found the answer to the first question in &lt;a href=&quot;https://www.amazon.ca/The-Handbook-Astronomical-Image-Processing/dp/0943396824/191-9927439-8084045&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Handbook of Astronomical Image Processing, Berry, R. &amp;amp; Burnell, J. (Willmann-Bell, Second Edition 2005)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Chapter 9 of this book, Astrometry is discussed. Standard coordinates are introduced in page 253 and the effects that disturb the tangent point $(\alpha_0, \delta_0)$ are discussed in page 254:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In real life, however, $(\alpha_0, \delta_0)$ is inevitably somewhat off the center of the image, and the x and y axes of the detector will be rotated through some angle. (...) In addition to displacement and rotation, the detector might be slightly tilted relative to the incoming light; the manufacturer's figures for the pixel dimensions might be slightly inaccurate at the CCD's operating temperature; and in the case of scanned photographs, the axes may not be perfectly orthogonal and equally scaled.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QlZvj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QlZvj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Figure 9.2, copyright of the Authors)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in page 258 (after explaining how to obtain X,Y from known positions of reference stars in the frame) the authors convert the X,Y coordinates to equatorial coordinates &lt;strong&gt;assuming that $(\alpha_0, \delta_0)$ are &quot;now known&quot;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I might have missed something obvious, but I don't see where these coordinates are obtained in the preceding pages. Question 2 thus, still stands.&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-03-16T23:10:36.127" Title="Obtain tangent point (origin) of the Standard Coordinates system" Tags="&lt;astrometry&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14181" PostTypeId="2" ParentId="14178" CreationDate="2016-03-16T14:41:55.033" Score="0" Body="&lt;p&gt;The term &quot;nebula&quot; has been used historically for virtually any astronomical object that was fuzzier than stars. Specifically, galaxies which to the naked eye and in small telescopes looked like elongated clouds were called nebulae (e.g. the Andromeda Nebula), even though they definitely aren't clouds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays, the term is mostly used in the context of &lt;em&gt;planetary nebulae&lt;/em&gt;, which are the beautifully colored death breaths of low-mass stars, lighted up by a central white dwarf. These objects were originally thought to be planets, but has nothing to do with that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other types of nebulae are usually called by their more descriptive names, e.g. molecular clouds, HII regions, and supernova remnants. The only objects I can think of that are often called nebulae, are &lt;em&gt;reflection nebulae&lt;/em&gt;, which are the clouds of dust and gas surrounding a young star cluster.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2016-03-16T14:41:55.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14182" PostTypeId="2" ParentId="14118" CreationDate="2016-03-16T15:17:01.117" Score="4" Body="&lt;p&gt;I found one &quot;hot Jupiter&quot; in the Kepler data (Kepler 45b). The star is a M dwarf with an effective temperature of 3820K. The planet has an estimated mass of 160.5 M(Earth) and radius of 10.76 R(Earth). This gives a density of about 0.8 g/cm2 which is consistent. The planet is located at approximately 0.03 AU from the star with an orbital eccentricity of 0.11. Also, it may be interesting that the star has a very high metallicity of Fe/H = 0.28. (This data is from &lt;a href=&quot;http://kepler.nasa.gov/Mission/discoveries/&quot; rel=&quot;noreferrer&quot;&gt;http://kepler.nasa.gov/Mission/discoveries/&lt;/a&gt; ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It must be said, however, that M dwarfs only made up about 4.3% of the initial Kepler study which found 82 &quot;Jupiter size&quot; planets (as of my tally in April of 2015). Thus, there is an indication that there are less &quot;hot Jupiters&quot; around M dwarfs, but sample size is way too small. There does seem to be a correlation between stars hosting &quot;hot Jupiters&quot; and their metallicity (more common around higher Fe/H stars).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My gut feeling is that &quot;hot Jupiters&quot; in general are uncommon. They are just the easiest thing to find.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2016-03-16T15:17:01.117" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14183" PostTypeId="2" ParentId="14175" CreationDate="2016-03-16T17:22:02.467" Score="1" Body="&lt;p&gt;I think what you need to do is have a CCD imager on a telescope with a large f-ratio, such that each pixel on the detector covers a small angle on the sky - I would say at most 0.25 arcseconds. The field of view also needs to be wide enough that you can get many faint stars in the same CCD image. You must have a telescope able to track the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then what you do is you take pictures on the nights of best seeing, making sure that you do not saturate the neaby star you are interested in, but getting reasonable signal to noise on the many faint stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you need to astrometrically calibrate your images so that you can estimate the apparent position of the star of interest with respect to the (faint, and hence assumed) background stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have seeing of 2 arcseconds, then the best centroiding precision you might reasonably expect to achieve is a precision of around 1/10 of the seeing disc or 0.2 arcseconds. If you have that sort of data repeated a number of times over the course of say 2 years (because you need to fit a solution which includes the annual proper motion as well as the annual parallax), then I reckon you might be able to measure parallaxes for stars out to a few parsecs (maybe ten?) with no problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with a three inch reflector is that you will need to do very long exposures in order to get lots of faint background stars in your image and then you may have problems with tracking accuracy.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-03-16T17:22:02.467" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14185" PostTypeId="1" AcceptedAnswerId="14190" CreationDate="2016-03-16T19:45:01.270" Score="6" ViewCount="1612" Body="&lt;p&gt;White dwarfs usually do not collapse, as they have electron degeneracy pressure due to the Pauli exclusion principle. However, if one accretes mass beyond the Chandrasekhar limit, it is energetically favorable for the electrons to combine with protons and form neutrons. This gives us a neutron star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, neutron stars usually do not collapse into black holes due to neutron degeneracy pressure. How is it possible that beyond the LOV limit, the Pauli exclusion principle no longer prevents the collapse? Shouldn't it still prevent neutrons, which are fermions, from being compressed together any further?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen answers involving quark stars, but those are purely hypothetical. What is the most accepted explanation for this?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-05-19T20:21:25.373" LastActivityDate="2017-10-29T00:15:47.737" Title="How could a neutron star collapse into a black hole?" Tags="&lt;black-hole&gt;&lt;general-relativity&gt;&lt;stellar-evolution&gt;&lt;neutron-star&gt;&lt;degenerate-matter&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14186" PostTypeId="2" ParentId="14185" CreationDate="2016-03-16T20:03:48.857" Score="0" Body="&lt;p&gt;I can see why your drawing the comparison with white dwarfs, but in reality we don't yet understand the effects of gravity and pressure well enough at those densities to be able to know for sure if an accreting Neutron Star will collapse directly into a black hole or go into some intermediate phase like the quark stars you mentioned. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I've always been taught the accepted view is that the Pauli exclusion principle only works so far, and once that is overcome by gravity the star collapses into a singularity and forms a black hole. In my time studying on my undergraduate course I heard no mention of quark stars or any other intermediate step (not that I was looking though).&lt;/p&gt;&#xA;" OwnerUserId="10437" LastActivityDate="2016-03-16T20:03:48.857" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14187" PostTypeId="2" ParentId="14185" CreationDate="2016-03-16T21:43:41.970" Score="2" Body="&lt;p&gt;In Layman's terms, the Pauli exclusion principal wouldn't need to be overcome to form the black hole.  A Neutron star of a certain size will shrink below it's Schwarzschild radius naturally.   That's not hard to see.  In fact, like white dwarfs, Neutron stars grow smaller in radius as they gain mass.  The maximum mass wouldn't be much more than 2.5 or so solar masses past which the Neutron Star couldn't avoid becoming a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relativistic effects get complicated, such as what precisely happens at the 100% time dilation and beyond.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;__ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as to what happens inside the black hole, there's two general points I can make.  One is, as the neutrons (quark matter, whatever it is), grows more compact the weight and force to compact it further keeps increasing.  That's fairly obvious.  It almost becomes the unstoppable force (weight and gravitation) vs the immovable object (Pauli exclusion) question.  The problem with knowing exactly what happens is essentially the singularity problem.  The math breaks down.   I don't think anyone knows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way that I like to look at it, is Gluons, like photons, move at the speed of light.  Inside a black hole, Gluons, like Photons would be drawn towards the center, not able to fly outwards and that property might greatly shrink the size of a Proton or Neutron down to the size of . . . maybe an Electron?? but again, who knows?   Maybe some kind of quantum tunneling keeps the size of the Neutrons somewhat consistent but the gravitational escape velocity exceeding the speed of light could greatly reduce the more standard/observed size of the Neutrons.  (I think).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know you asked for the most accepted explanation and I've only touched on this from a layman's POV, so, hopefully someone with a bigger brain than me will answer this one more precisely to your specific question.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-03-16T21:43:41.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14188" PostTypeId="1" CreationDate="2016-03-16T22:12:05.477" Score="0" ViewCount="2539" Body="&lt;p&gt;I've seen websites that show the ratios of the 10 most common elements but they compare them by relative mass. I think it's more interesting to know the relative abundances of atomic nuclei because that is what you need to know to understand why the universe has the chemical makeup that it does. What are the 10 most abundant elements in the universe by number of atomic nuclei?&lt;/p&gt;&#xA;" OwnerUserId="11247" LastEditorUserId="10953" LastEditDate="2016-03-17T05:29:07.087" LastActivityDate="2016-03-18T10:28:38.797" Title="What are the 10 most abundant elements in the universe by number of atomic nuclei?" Tags="&lt;galaxy&gt;&lt;universe&gt;&lt;cosmology&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14189" PostTypeId="1" AcceptedAnswerId="14191" CreationDate="2016-03-16T22:20:51.567" Score="5" ViewCount="499" Body="&lt;p&gt;Do the colorful pictures of the universe show how it looks like that in reality or is it just a visual effect that is added later.&lt;/p&gt;&#xA;" OwnerUserId="11248" LastEditorUserId="2365" LastEditDate="2016-03-17T17:56:01.407" LastActivityDate="2016-03-17T17:56:01.407" Title="Are all the photos of the universe really so colorful or is it just visual effects that are added later?" Tags="&lt;photography&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14190" PostTypeId="2" ParentId="14185" CreationDate="2016-03-16T23:03:02.613" Score="14" Body="&lt;p&gt;The scenario you describe &lt;em&gt;may&lt;/em&gt; occur. On the other hand it may actually be that neutronisation in a white dwarf is the trigger for a thermonuclear type Ia supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may be misunderstanding the Pauli Exclusion Principle (PEP).The PEP states that no two fermions can occupy the same quantum state, &lt;strong&gt;not&lt;/strong&gt; that they cannot occupy the same space or be compressed to whatever density you like. The quantum states here consist of two spin states for every possible momentum state. In a degenerate gas, all these states are filled up to the Fermi energy. All that happens when the neutron star gets smaller (or collapses), is that the Fermi energy just keeps increasing as the neutron density climbs, and the neutron degeneracy pressure just keeps increasing as a consequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in General Relativity, pressure (like mass/energy) is a source of gravitational curvature and actually &lt;em&gt;increases&lt;/em&gt; the required pressure gradient needed to support the star. At a certain threshold radius - a small factor larger than the Schwarzschild radius, a point of instability is reached where increasing the pressure is actually counter-productive.  Beyond this, you can make the pressure as large as you like and it will not prevent the formation of a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even inside the BH there is not necessarily a problem with the PEP. You can compress fermions to infinite density so long as they can have infinite momentum.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-03-16T23:12:21.883" LastActivityDate="2016-03-16T23:12:21.883" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14191" PostTypeId="2" ParentId="14189" CreationDate="2016-03-16T23:18:41.020" Score="11" Body="&lt;p&gt;There is no straight yes or no answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some objects do have significant radiation in visible part of the spectrum which would make them colorful if we would be close enough while observing them (our eye needs large number of photons to distinguish color so that means, astronomy wise, that object needs to be either very bright or very close or both).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand lot of them are being adjusted for color later. Usually most beautiful images are mosaics of several images from different filters (even from parts of spectrum our eye cant see) where each image is given a specific color based on filter used and then later all of them get combined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to learn more check this very helpful paper, which has some tutorials as well:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.stsci.edu/stsci/resources/imaging/papers/rector2007/files/Rector2005.pdf&quot;&gt;http://www.stsci.edu/stsci/resources/imaging/papers/rector2007/files/Rector2005.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Btw, as a curiosity - images from rovers on Mars are mainly black and white and are colorized later (thats why the rovers usually have a color wheel on them for calibration purposes as &lt;a href=&quot;http://www.universetoday.com/wp-content/uploads/2012/02/621670main_pia15289-43_946-710.jpg&quot;&gt;seen here&lt;/a&gt;). This lead to &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomy_on_Mars#The_color_of_the_sky&quot;&gt;uncertainties of color of Mars sky&lt;/a&gt; which turned into a long lasting debate.&lt;/p&gt;&#xA;" OwnerUserId="7094" LastActivityDate="2016-03-16T23:18:41.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14192" PostTypeId="2" ParentId="14188" CreationDate="2016-03-17T01:36:42.527" Score="4" Body="&lt;p&gt;All right, so I took the first list on wikipedia listing the 10 most common elements by mass in parts per million, and did what Rob recommended and here's what I got.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements#Abundance_of_elements_in_the_Universe&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements#Abundance_of_elements_in_the_Universe&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen - 739000amu(H)/1amu(H)=739000 H atoms &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Helium - 240000amu(He)/4amu(He)=60000 He atoms&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxygen - 10400amu(O)/16amu(O)=650 O atoms &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon - 4600amu(C)/12amu(C)=383 C atoms&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neon - 1340amu(Ne)/20.1amu(Ne)=66 Ne atoms &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron - 1090amu(Fe)/55.845amu(Fe)=19.5 Fe atoms &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitrogen - 960amu(N)/14amu(N)=68.5 N atoms &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silicon - 650amu(Si)/28.1amu(Si)=23 Si atoms&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnesium, - 580amu(Mg)/24.3amu(Mg)=24 Mg atoms &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sulfur - 440amu(S)/32.1amu(S)=13.7 S atoms &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So therefore, the 10 most common elements in the universe by atom, with the relative ratios between them, is;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.Hydrogen (739000)&#xA;2.Helium (60000)&#xA;3.Oxygen (650)&#xA;4.Carbon (383)&#xA;5.Nitrogen (68.5)&#xA;6.Neon (66)&#xA;7.Magnesium (24)&#xA;8.Silicon (23)&#xA;9.Iron (19.5)&#xA;10.Sulfur (13.7)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone sees a mistake that I made with my reasoning or calculations then please point it out.&lt;/p&gt;&#xA;" OwnerUserId="11247" LastEditorUserId="1288" LastEditDate="2016-03-17T10:40:55.093" LastActivityDate="2016-03-17T10:40:55.093" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14196" PostTypeId="2" ParentId="12078" CreationDate="2016-03-17T16:02:33.037" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://wfirst.gsfc.nasa.gov/&quot; rel=&quot;nofollow&quot;&gt;WFRIST/AFTA&lt;/a&gt; is probably the closest thing to Hubble that is going up. It has a similar 2.4 meter mirror will perform wide-field imaging and surveys of the near infrared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will have a Wide Field Instrument (WFI) which operates in the 0.7 - 2.0 micron range as well as a Coronagraphic Instrument (CGI) which operates in the 0.4 - 1.0 micron range. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since most of the telescope hardware (called AFTA (Astrophysics-Focused Telescope Assets)) was donated by the National Reconnaissance Office (NRO) to NASA, a lot of the telescope hardware already exists and does not need to be built so it's likely to be a 'quickly built' mission.&lt;/p&gt;&#xA;" OwnerUserId="11039" LastActivityDate="2016-03-17T16:02:33.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14197" PostTypeId="2" ParentId="12078" CreationDate="2016-03-17T16:33:08.640" Score="1" Body="&lt;p&gt;Although the JWST is often touted as the successor to the HST, this isn't technically true, as Rob pointed out the JWST is designed to work in the infrared not optical wavelengths. One possible, more direct, successor could be the ATLAST (Advanced Technology Large-Aperture Space Telescope) telescope, it would not only be able to observe in the optical but also infrared and ultraviolet. The primary mirror is planned to be between 8-16.8m in diameter and would be able to achieve up to 2'000 times better angular resolution than the Hubble can at present. It would make use of the newly developed Space Launch System (SLS) to place it into an L2 orbit (also where JWST is heading), and there are also plans to allow servicing missions to the telescope every 5-7 years (similar to how the HST was serviced in orbit) to enable astronomers to upgrade the instruments on-board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vYJJv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vYJJv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this telescope still hasn't left the drawing board and is unlikely to meet its 2025 launch date, funding for such a large telescope is guaranteed to cause a delay in construction as it would probably require an international collaboration just like the HST was jointly commissioned by ESA and NASA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See:&lt;a href=&quot;http://www.stsci.edu/atlast&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.stsci.edu/atlast&lt;/a&gt; for references.&lt;/p&gt;&#xA;" OwnerUserId="10437" LastActivityDate="2016-03-17T16:33:08.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14198" PostTypeId="2" ParentId="14189" CreationDate="2016-03-17T17:54:53.653" Score="5" Body="&lt;p&gt;Many astronomical objects are too dim to be easily seen. There is no place in the universe where M42 (between Orion's legs) would look to the naked eye like it does in photos: it is too dim, and our eyes are not good at detecting colour of dim objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike a regular camera, which has pixel detectors for red green and blue light, most coloured astronomical images are made by using a sequence of filters in front of the camera lens. The filters only let a particular colour through. The result is several greyscale images, when several images have been made of the same object, they are mixed together, by assigning a colour to each. So &quot;red&quot; may represent infrared, for example, a colour that cannot be seen by humans. The assignment of image colours to filters is called a pallet, and can make a big difference to the appearance of an image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hubble images of &quot;the pillars of creation&quot; popularised a particular pallet, and Hubble site has some details on &lt;a href=&quot;http://hubblesite.org/gallery/behind_the_pictures/meaning_of_color/eagle.php&quot; rel=&quot;noreferrer&quot;&gt;how they create the colours in images&lt;/a&gt;.  The hubble pallet uses &quot;blue&quot; for a mid range filter, yellow green (this colour is produced by oxygen ions), &quot;green&quot; for and orangy red filte: it picks up the light from hydrogen ions, and &quot;red&quot; for a intense red, fading towards the infra-red (sulphur). By using &quot;false colour&quot; details of the distribution of ions can be picked out better, and it makes a more attractive image.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-03-17T17:54:53.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14199" PostTypeId="1" AcceptedAnswerId="14201" CreationDate="2016-03-17T21:00:21.890" Score="1" ViewCount="66" Body="&lt;p&gt;What happens with electromagentic waves that are emitted into &#xA;intergalactic space, but reach not any object because of inflation? Lets assume that photon A and photon B are emitted. Photon B hits some object, photon A enters a gigantic void or simply has nothing in its path, and heads for the horizon, so to say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relativity tells that A and B do not &quot;feel&quot; time passing. From our point of view B travels a finite time, while in the frame of B the object is reached instantly. However, since A will be eternally underway, what is going to happen? We have kind of a zero*infinity travelling time in the frame of A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do our models make statements about this? Or is there some explanation&lt;/p&gt;&#xA;" OwnerUserId="11258" LastEditorUserId="2365" LastEditDate="2016-03-18T17:04:47.327" LastActivityDate="2016-03-18T17:04:47.327" Title="Light that travels eternally because of inflation?" Tags="&lt;cosmology&gt;&lt;general-relativity&gt;&lt;cosmological-inflation&gt;&lt;special-relativity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14200" PostTypeId="2" ParentId="13326" CreationDate="2016-03-17T22:22:14.570" Score="4" Body="&lt;p&gt;Yes, gravitational waves can be redshifted. The answer of this question explains why: &lt;a href=&quot;https://physics.stackexchange.com/questions/137292/can-gravitational-waves-be-red-shifted&quot;&gt;https://physics.stackexchange.com/questions/137292/can-gravitational-waves-be-red-shifted&lt;/a&gt;&#xA;The answer to the question above basicly states that doppler effect applies to any wave that propagates at light speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actualy measuring the redshift is much more difficult, because neither redshift nor rest-frame mass can be extracted from the signal. Only the combination of the two can be obtained - M(1+z). This is called the redshifted mass. Deriving actual redshift from redshifted mass is quite complicated as this problem is relatively new and no efficient solution exists yet. This article here describes the derivation of the redshift in detail:&#xA;&lt;a href=&quot;http://arxiv.org/pdf/1312.1862v2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://arxiv.org/pdf/1312.1862v2.pdf&lt;/a&gt;&#xA;From it you can see that it is indeed possible to derive redshift from redshifted mass, but the method is really imprecise as it has a massive 10 - 20% uncertanity.&lt;/p&gt;&#xA;" OwnerUserId="10228" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-03-17T22:22:14.570" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14201" PostTypeId="2" ParentId="14199" CreationDate="2016-03-17T23:12:08.010" Score="1" Body="&lt;p&gt;Infinities are a type of singularity. As soon as you introduce an actual infinity into any kind of mathematical model you can't expect it to make useful predictions. Mathematical models can't deal with infinite time. Similarly, using a frame that moves at the speed of light, with infinite rapidity, is singular.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the particular case you describe, in a regular frame, the photon becomes progressively redshifted by the expansion of spacetime, its energy (to an observer in a inertial frame) becomes lower and lower.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-03-18T17:04:04.167" LastActivityDate="2016-03-18T17:04:04.167" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14202" PostTypeId="1" CreationDate="2016-03-18T01:56:22.460" Score="1" ViewCount="88" Body="&lt;p&gt;Scientists say that gravity warps space and time. They also say that the more gravity there is the slower time moves. Why and/or how does gravity do this.&lt;/p&gt;&#xA;" OwnerDisplayName="user11260" LastActivityDate="2016-03-18T10:16:30.003" Title="How does gravity affect spacetime" Tags="&lt;gravity&gt;" AnswerCount="2" CommentCount="0" ClosedDate="2016-03-20T08:38:55.413" ContentLicense="CC BY-SA 3.0" />
  <row Id="14203" PostTypeId="1" AcceptedAnswerId="14254" CreationDate="2016-03-18T04:19:55.433" Score="0" ViewCount="86" Body="&lt;p&gt;I would like to know how do astronomers estimate the mass of a star that has been ejected from its galaxy.  What kind of accuracy can be achieved?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="2531" LastEditDate="2016-03-24T12:14:52.427" LastActivityDate="2016-03-24T12:14:52.427" Title="How to determine the mass of a runaway star?" Tags="&lt;star&gt;&lt;mass&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14204" PostTypeId="2" ParentId="14202" CreationDate="2016-03-18T04:28:44.940" Score="0" Body="&lt;p&gt;I think you mean to say that gravity is the warping of space/time by matter/energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity is the result of mass/energy OCCUPYING or &quot;living in&quot; space time. Gravity is the curvature of space/time by mass (I use mass here, but energy also warps space/time's geometry).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time is relative -- dependent on your position/vel in space/time -- because of the constancy of the speed of light. Rather than explain it here, simply see &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Time_dilation&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Time_dilation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9648" LastActivityDate="2016-03-18T04:28:44.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14207" PostTypeId="2" ParentId="138" CreationDate="2016-03-18T07:29:58.060" Score="4" Body="&lt;p&gt;Yes the sun rotates and it takes about 26.24 days.They are many methods to determine the rotation periods but most common one is by observing sunspots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a link with a detailed explanation &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_rotation&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Solar_rotation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11261" LastActivityDate="2016-03-18T07:29:58.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14210" PostTypeId="1" AcceptedAnswerId="14220" CreationDate="2016-03-18T09:58:13.173" Score="0" ViewCount="264" Body="&lt;p&gt;Will Pi have the same value on the surface of say Jupiter, compared to the value of PI on Earth. Or does Pi vary with gravity?&lt;/p&gt;&#xA;" OwnerUserId="9136" LastActivityDate="2016-03-18T14:29:20.857" Title="Does Pi change when gravity changes" Tags="&lt;gravity&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2016-03-18T22:05:33.130" ContentLicense="CC BY-SA 3.0" />
  <row Id="14212" PostTypeId="1" CreationDate="2016-03-18T10:07:00.110" Score="1" ViewCount="152" Body="&lt;p&gt;In this &lt;a href=&quot;http://earthsky.org/space/incredible-solar-eclipse-footage-from-flight&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; about the &lt;a href=&quot;http://blog.alaskaair.com/alaska-airlines/news/eclipse-flight/&quot; rel=&quot;nofollow noreferrer&quot;&gt;AS870 flight&lt;/a&gt; that flew through the umbra of the March 8-9 2016 solar eclipse, there is a link to a &lt;a href=&quot;https://www.youtube.com/watch?v=hMvTZaqMDx8&quot; rel=&quot;nofollow noreferrer&quot;&gt;YouTube video&lt;/a&gt; showing the eclipse recored by a handheld camera in the cabin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this is personal video and not recorded for analysis, but the images - especially during the zoomed part - remind me of an annular eclipse, rather than a total eclipse. Below are some screenshots - the weak Fresnel reflections in the cabin window act a bit like attenuators (ND filters) and give less saturated, though distorted views. Basically they all look like donuts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first thinking was that if it was a total eclipse on the ground, it couldn't be annular at 35,000 feet closer to the moon. But thinking and looking again, the sun is near the horizon, and the plane at about 150 West longitude is not necessarily closer to the sun than observers on the ground were in Indonesia circa 120 East.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is - is that just corona, or is the plane farther away enough from the moon at this time and location to make the eclipse annular? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be helpful to read about &lt;a href=&quot;https://astronomy.stackexchange.com/a/849/7982&quot;&gt;hybrid eclipses&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;note:&lt;/strong&gt; I'm looking for a quantitative answer, not an opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;There is another version of the video on YouTube with more views, but the narrative doesn't hold a candle to double rainbow guy &lt;a href=&quot;https://youtu.be/OQSNhk5ICTI?t=24&quot; rel=&quot;nofollow noreferrer&quot;&gt;original&lt;/a&gt; and &lt;a href=&quot;https://www.youtube.com/watch?v=MX0D4oZwCsA&quot; rel=&quot;nofollow noreferrer&quot;&gt;musical version&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/etrea.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/etrea.png&quot; alt=&quot;Alaska Airlines 870 Eclipse 1&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/CwIdT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CwIdT.png&quot; alt=&quot;Alaska Airlines 870 Eclipse 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a screenshot of a typical AS870 flight from flightradar24.com, pretty much between 152 and 158 degrees west.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3gcbt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3gcbt.png&quot; alt=&quot;screen shot from flightradar24.com&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is from the second link in the top line - the Alaska Airlines &lt;a href=&quot;http://blog.alaskaair.com/alaska-airlines/news/eclipse-flight/&quot; rel=&quot;nofollow noreferrer&quot;&gt;blog entry&lt;/a&gt; for the flight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iWIc7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iWIc7.jpg&quot; alt=&quot;Alaska Airlines Blog image of AS870 eclipse path&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How an eclipse could possibly appear annular from a plane over one place on the planet while appearing total on the ground at another place on the planet 5,000 kilometers away (note, these occur at different times - the shadow is moving relative to earth and the earth is rotating):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/P6ERZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P6ERZ.png&quot; alt=&quot;eclipse geometry&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-03-18T17:48:33.627" Title="Why does the eclipse in this video look annular?" Tags="&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14213" PostTypeId="1" AcceptedAnswerId="19597" CreationDate="2016-03-18T10:15:21.317" Score="4" ViewCount="199" Body="&lt;p&gt;There are some pyramid shaped observatories around the world, like the &lt;a href=&quot;https://imgur.com/a/owHJD&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solar Lab Pyramid&lt;/a&gt; in the Canary Islands or &lt;a href=&quot;http://evk2cnr.org/cms/en/evk2cnr_committee/pyramid&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Pyramid International Laboratory&lt;/a&gt; in Nepal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering, why a pyramid? Because of its relative stability, or is there something else?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; &lt;a href=&quot;https://imgur.com/a/owHJD&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solar Lab Pyramid&lt;/a&gt; in the Canary Islands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/graBR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/graBR.jpg&quot; alt=&quot;Solar Lab Pyramid Canary Islands&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; &lt;a href=&quot;http://evk2cnr.org/cms/en/evk2cnr_committee/pyramid&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Pyramid International Laboratory&lt;/a&gt; in Nepal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Stv7q.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Stv7q.jpg&quot; alt=&quot;Pyramid International Laboratory Nepal&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7554" LastEditorUserId="7982" LastEditDate="2017-11-11T10:16:32.567" LastActivityDate="2017-11-11T10:16:32.567" Title="Pyramid shaped observatories" Tags="&lt;observatory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14214" PostTypeId="2" ParentId="14202" CreationDate="2016-03-18T10:16:30.003" Score="1" Body="&lt;p&gt;In fact time always &quot;moves&quot; at the same speed to any given observer. It appears to move at different speeds to an external observer, but you cannot experience slowed time as such - except in the sense that you might observe external clocks ran faster than the watch you had your wrist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An attempt to explain: The supposition - very strongly supported by the experimental evidence (the latest discovery of gravity waves being the jewel in the crown) is that we live in a four dimensional spacetime. An individual always experiences time as &quot;moving&quot; at a constant speed, but as has been remarked by others, externally gravity appears to add curvature to the &quot;straight line&quot; we are moving along on the time dimension and so we &quot;gain distance&quot; in this direction more slowly when seen externally.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2016-03-18T10:16:30.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14215" PostTypeId="2" ParentId="11447" CreationDate="2016-03-18T10:22:47.470" Score="0" Body="&lt;p&gt;Not really, the shape of the supernova remnant is mainly caused by the structure and biometry of the star. At the time of explosion the layers of the star expands&#xA;and high speeds which result in a supernova remnant. A good example is the crab nebula in constellation Taurus.&lt;/p&gt;&#xA;" OwnerUserId="11261" LastEditorUserId="10172" LastEditDate="2016-03-18T11:10:24.970" LastActivityDate="2016-03-18T11:10:24.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14217" PostTypeId="1" AcceptedAnswerId="14243" CreationDate="2016-03-18T10:29:24.830" Score="3" ViewCount="359" Body="&lt;p&gt;Almost all celestial objects are rotating in some way about their axes. Why is this and should we expect the rotation to slow as the universe ages?&lt;/p&gt;&#xA;" OwnerUserId="551" LastEditorUserId="823" LastEditDate="2016-03-18T14:00:22.547" LastActivityDate="2016-03-21T05:32:21.627" Title="Why is the universe full of spinning objects?" Tags="&lt;planet&gt;&lt;rotation&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14218" PostTypeId="1" AcceptedAnswerId="14629" CreationDate="2016-03-18T11:06:34.227" Score="1" ViewCount="105" Body="&lt;p&gt;I'd like to show a demonstration of Runge Kutta integration of real systems using examples of interesting multiple stars, where &quot;interesting&quot; means you can see through a small telescope that they are at least double. I've selected three:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Gamma Andromedae (Almach)&lt;/li&gt;&#xA;&lt;li&gt;Alpha Geminorum (Castor)&lt;/li&gt;&#xA;&lt;li&gt;Zeta Ursae Majoris (Mizar and Alcor)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;How can I find a state vector (set of relative positions and velocities) - or something similar - for these multiple star systems? An open access link (for everyone) would be great, but a journal reference is fine as well if that's what it takes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the periods are long and there will be uncertainty - I'd just like to use whatever is known and available, for this demonstration.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2016-03-21T15:45:18.570" LastActivityDate="2016-04-21T23:38:16.783" Title="State vectors of &quot;interesting&quot; multiple stars" Tags="&lt;orbit&gt;&lt;binary-star&gt;&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14220" PostTypeId="2" ParentId="14210" CreationDate="2016-03-18T14:29:20.857" Score="3" Body="&lt;p&gt;$\pi$ is just a number. It doesn't vary with &lt;em&gt;anything&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio of the circumference to the diameter of a circle varies with geometry, so in the general case, this definition doesn't hold. But there are other definitions, such as&#xA;$$&#xA;\pi = \int_{-1}^{1}\frac{dx}{\sqrt{1-x^2}},&#xA;$$&#xA;or&#xA;$$&#xA;\mathrm{The\,smallest\,positive\,number} \, x \, \mathrm{for\,which}\, \arccos(2x) = 0.&#xA;$$&#xA;The latter definition can also be expressed a power series, which is independent of geometry.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2016-03-18T14:29:20.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14221" PostTypeId="1" CreationDate="2016-03-18T14:49:31.833" Score="1" ViewCount="1456" Body="&lt;p&gt;I've been scouring the internet for a straightforward way to approximate the distance between Earth and Mars at a given time. I hope to achieve an accuracy of about 100km. &lt;/p&gt;&#xA;" OwnerUserId="11265" LastEditorUserId="2365" LastEditDate="2016-03-20T14:49:51.830" LastActivityDate="2016-03-20T14:49:51.830" Title="Approximate distance from Earth to Mars at a given moment with reasonable (+/-100km) accuracy?" Tags="&lt;distances&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14222" PostTypeId="2" ParentId="14221" CreationDate="2016-03-18T15:26:51.590" Score="3" Body="&lt;p&gt;Do not underestimate the astronomers! 100 km? That's absurd. Saturn('s center of mass) is located to within one mile's distance from the Sun. Mars is of course even much better located.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://theskylive.com/mars-tracker&quot; rel=&quot;nofollow&quot;&gt;Here's a site I googled&lt;/a&gt; that seems to give current distance to Mars to within single kilometers. I don't know how that site makes up its numbers, but it is certainly possible to measure it like that. With mutual radio communication, there are really good opportunities for precis distance measuring. 135,834,832 km just now! Pretty close and that's why the ExoMars space probe was launched towards Mars a few days ago. And &lt;a href=&quot;http://cseligman.com/text/planets/marsoppositions.htm&quot; rel=&quot;nofollow&quot;&gt;here is&lt;/a&gt; a table with the conunction distances just past and soon coming up.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2016-03-18T15:33:06.247" LastActivityDate="2016-03-18T15:33:06.247" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14223" PostTypeId="1" AcceptedAnswerId="14225" CreationDate="2016-03-18T16:07:55.083" Score="3" ViewCount="1735" Body="&lt;p&gt;I have been trying to understand Kepler's law of planetary motion for a while. So I was thinking, if it would even apply to the moon orbiting around the earth. Does the moon too orbit in an ellipse too? Does its orbit have the same properties as the earth's orbit around the sun?&lt;/p&gt;&#xA;" OwnerUserId="11255" LastActivityDate="2016-03-18T20:20:27.000" Title="Can we apply Kepler's law of planetary motion to the moons motion around the earth?" Tags="&lt;orbit&gt;&lt;the-sun&gt;&lt;the-moon&gt;&lt;planet&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14224" PostTypeId="2" ParentId="14212" CreationDate="2016-03-18T16:42:12.537" Score="2" Body="&lt;p&gt;It is quite clear from the film (and the awful soundtrack) that a &lt;em&gt;total&lt;/em&gt; eclipse was observed. The Sun's corona and prominences &lt;em&gt;cannot&lt;/em&gt; be seen during an annular eclipse. Nor is there a sudden fading or drastic brightening at the beginning and end of an annular eclipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The accompanying article talks about &quot;totality&quot; and removing safety glasses to look at the Sun during totality, and that the period of totality was 1 min 53 seconds; the (frankly, annoying) narrator shouts the word &quot;totality&quot; about 23 times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The eclipse was observed towards the end period of totality (what I mean by that is the shadow is tracking west to east across the Earth's surface. That means the Sun was above the western horizon as seen by the plane, and at 5:35pm, would be 10 degrees above the horizon. That means that in order to observe this event the plane must have been closer to the moon than a point on the surface of the Earth where totality would have been observed at the same time. That means the period of totality would have been slightly longer and the moon would have had a slightly larger angular size as seen from the plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I had the exact latitude and longitude I could tell you just how much longer the eclipse was. Messing about with Stellarium, I can get a reasonable match of the eclipse time at W157 58 35, N31 22 00. At 10,000m the eclipse is total and lasts about 70s. At sealevel at the same latitude it lasts maybe a few seconds longer (it's difficult to be sure). But the sea level region of totality observed at the same time would have been somewhat further east (but still along the path where totality was observed), where the period of totality would have been shorter. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-03-18T17:48:33.627" LastActivityDate="2016-03-18T17:48:33.627" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14225" PostTypeId="2" ParentId="14223" CreationDate="2016-03-18T17:23:19.773" Score="5" Body="&lt;p&gt;Yes. Kepler's laws apply to orbits of the Earth, and the &lt;strong&gt;moon's orbit is elliptical&lt;/strong&gt;, with an eccentricity of 0.055&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or at least nearly elliptical. In Newton's theory of gravity, if there are two particles in orbit around each other then they will have an elliptical orbit. If there are more than two bodies then the gravity of other bodies will perturb the orbits from exactly elliptical. A more accurate description of the moon's orbit is &quot;Elliptical + perturbations&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most significant perturbations come from the sun, and from the fact that the Earth is not a perfect sphere. Other planets and tides also have effects on the Moon's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two wiki pages you should read: &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbit_of_the_Moon&quot; rel=&quot;nofollow&quot;&gt;Orbit of the moon&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_Theory&quot; rel=&quot;nofollow&quot;&gt;Lunar Theory&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-03-18T20:20:27.000" LastActivityDate="2016-03-18T20:20:27.000" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14226" PostTypeId="2" ParentId="14221" CreationDate="2016-03-18T17:39:54.417" Score="3" Body="&lt;p&gt;The python library, &lt;a href=&quot;http://rhodesmill.org/pyephem/&quot; rel=&quot;nofollow&quot;&gt;pyephem&lt;/a&gt; has this function.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import ephem&#xA;km = 149597871   # 1AU in km&#xA;m = ephem.Mars()&#xA;m.compute('2016/3/18')&#xA;print(m.earth_distance*km)&#xA;&#xA;136729311.6250654&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-03-18T18:24:20.583" LastActivityDate="2016-03-18T18:24:20.583" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14228" PostTypeId="2" ParentId="10284" CreationDate="2016-03-18T18:34:13.147" Score="1" Body="&lt;p&gt;An asteroid that remains outside Earth's atmosphere, and therefore does not impact Earth, will have no effect on the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asteroids, even biggish ones like the dinosaur killer, are pretty small compared to the Earth. They do not have enough mass to graviationally affect earth's orbit noticeably. And if they miss earth, the pass quickly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concern would be that if an asteroid can pass close to Earth, that it may collide in a future orbit. The asteroid's path will be changed by the earth's gravity. However, a very close encounter will enable astronomers to get a very high quality position on the rock, and so be able to predict its future path with very great accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comet tails aren't a worry. We passed through Halley's tail in 1910, the effect on the Earth was precisely nothing. ☄&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-03-18T20:16:46.507" LastActivityDate="2016-03-18T20:16:46.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14229" PostTypeId="1" AcceptedAnswerId="14259" CreationDate="2016-03-18T23:38:27.477" Score="0" ViewCount="391" Body="&lt;p&gt;I have this problem: I want to calculate the declination of the star in the year 5000. And I know &lt;strong&gt;RA&lt;/strong&gt; (right ascension) and &lt;strong&gt;Dec&lt;/strong&gt; (declination) of the star right now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate it using spherical trigonometry.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I know that the declination changes over time due to the precession&#xA;  and somewhere I read that I should use spherical triangle, but I am not&#xA;  really sure how to use it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thank you for your ideas.&lt;/p&gt;&#xA;" OwnerUserId="10592" LastEditorUserId="10592" LastEditDate="2016-03-19T18:11:43.380" LastActivityDate="2016-03-21T15:51:07.617" Title="Declination change due to precession" Tags="&lt;coordinate&gt;&lt;declination&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14233" PostTypeId="1" CreationDate="2016-03-19T05:04:03.207" Score="2" ViewCount="358" Body="&lt;p&gt;Assuming no collision happens, do all elliptical galaxies eventually become spiral? Also, when a galaxy becomes spiral or barred spiral, can it switch between these shapes or is it set forever?&lt;/p&gt;&#xA;" OwnerUserId="11272" LastEditorUserId="31410" LastEditDate="2021-01-12T16:04:09.163" LastActivityDate="2021-01-12T16:04:09.163" Title="Is there a strict chronological order that determines the shape of galaxies?" Tags="&lt;galaxy&gt;&lt;classification&gt;&lt;galactic-evolution&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14234" PostTypeId="2" ParentId="14233" CreationDate="2016-03-19T07:35:32.463" Score="3" Body="&lt;p&gt;Galaxies evolve from spiral to elliptical.  The spirals are formed by patterns of new star formation in the disk surrounding the bulge, which contains mostly old stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As galaxies run out of hydrogen gas clouds, which is the raw material from which stars are formed, then no new stars form, and with no new star formation, the complex structures of the arms are lost. Elliptical galaxies are made of old stars and have little new star formations. They were formed when spiral galaxies merged, causing a burst of star formation, and leaving a bulge of old stars and little gas to make new ones. The ultimate fate of the milky way and the andromeda galaxy is to merge and form a giant elliptical galaxy.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-03-19T07:35:32.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14236" PostTypeId="1" CreationDate="2016-03-19T10:46:39.190" Score="-4" ViewCount="4188" Body="&lt;p&gt;We live in NW Tasmania in a country town and tonight out walking I pointed out a long cluster of stars to my sister as they made an unusual shape but then we realised they were moving really fast - we stood and watched them pass over us and decided to record it but sadly the moon was too bright to get  a good recording. Any ideas on what it was? very intrigued as haven't seen anything like it before.&lt;/p&gt;&#xA;" OwnerUserId="11274" LastActivityDate="2017-01-08T00:23:09.010" Title="Cluster of fast moving stars" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="8" ClosedDate="2017-01-09T14:51:13.273" ContentLicense="CC BY-SA 3.0" />
  <row Id="14238" PostTypeId="1" AcceptedAnswerId="14240" CreationDate="2016-03-19T11:40:19.180" Score="0" ViewCount="101" Body="&lt;p&gt;Is there an equation that can give me the possibile distance of the planet from his star knowing planet's temperature (or other datas)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found &lt;a href=&quot;http://www.planetarybiology.com/calculating_habitable_zone.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; method for calculating the Habitable Zone but it's based only on star's properties. &lt;/p&gt;&#xA;" OwnerUserId="11275" LastEditorUserId="2365" LastEditDate="2016-03-20T16:04:40.187" LastActivityDate="2016-03-20T16:04:40.187" Title="Is it possible to know if a planet is located in the Habitable Zone knowing it's properties?" Tags="&lt;planet&gt;&lt;exoplanet&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14239" PostTypeId="2" ParentId="2441" CreationDate="2016-03-19T15:31:54.050" Score="27" Body="&lt;p&gt;Yes, you are absolutely right, from OUR VIEWPOINT it does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Kip Thorne's book &quot;Black Holes and Time Warps: Einstein's Outrageous Legacy.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Like a rock dropped from a rooftop, the star’s surface falls downward (shrinks inward) slowly at first, then more and more rapidly. Had Newton’s laws of gravity been correct, this acceleration of the implosion would continue inexorably until the star, lacking any internal pressure, is crushed to a point at high speed. Not so according to Oppenheimer and Snyder’s relativistic formulas. Instead, as the star nears its critical circumference, its shrinkage slows to a crawl. The smaller the star gets, the more slowly it implodes, until it becomes frozen precisely at the critical circumference. No matter how long a time one waits, if one is at rest outside the star (that is, at rest in the static external reference frame) one will never be able to see the star implode through the critical circumference. That is the unequivocal message of Oppenheimer and Snyder’s formulas.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Is this freezing of the implosion caused by some unexpected, general relativistic force inside the star? No, not at all, Oppenheimer and Snyder realized. Rather, it is caused by gravitational time dilation (the slowing of the flow of time) near the critical circumference. Time on the imploding star’s surface, as seen by static external observers, must flow more and more slowly, when the star approaches the critical circumference, and correspondingly everything occurring on or inside the star including its implosion must appear to go into slow motion and then gradually freeze.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“As peculiar as this might seem, even more peculiar was another prediction made by Oppenheimer and Snyder’s formulas: Although, as seen by static external observers, the implosion freezes at the critical circumference, it does not freeze at all as viewed by observers riding inward on the star’s surface. If the star weighs a few solar masses and begins about the size of the sun, then as observed from its own surface, it implodes to the critical circumference in about an hour’s time, and then keeps right on imploding past criticality and on in to smaller circumferences.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“By looking at Oppenheimer and Snyder’s formulas from the viewpoint of an observer on the star’s surface, one can deduce the details of the implosion, even after the star sinks within its critical circumference; that is one can discover that the star gets crunched to infinite density and zero volume, and one can deduce the details of the spacetime curvature at the crunch.” P217-218&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, so from our perspective all the matter will be clustered around the critical circumference and no further. That's fine, this shell in theory can exert all the forces required on the external universe such as gravitational attraction, magnetic field etc. The point like singularity which is in the indefinite future of the black hole, (from our point of view) indeed in the indefinite future of the universe itself could not exert such forces on this universe. This singularity is only &quot;reached&quot; as an observer rides in past the critical circumference and, through the process of time dilation, reaches the end of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is obviously an area of active research and thinking.  Some of the greatest minds on the planet are approaching this issue in different ways but so far have not reached a consensus but intriguingly a consensus appears to be beginning to emerge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sciencealert.com/stephen-hawking-explains-how-our-existence-can-escape-a-black-hole&quot; rel=&quot;noreferrer&quot;&gt;http://www.sciencealert.com/stephen-hawking-explains-how-our-existence-can-escape-a-black-hole&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stephen  Hawking said at a conference in August 2015 that he believes that &quot;information is stored not in the interior of the black hole as one might expect, but on its boundary, the event horizon.&quot; His comment refers to the resolution of the &quot;information paradox,&quot; a long-running physics debate in which Hawking eventually concedes that the material that falls into a black hole isn't destroyed, but rather becomes part of the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read more at: &lt;a href=&quot;http://phys.org/news/2015-06-surface-black-hole-firewalland-nature.html#jCp&quot; rel=&quot;noreferrer&quot;&gt;http://phys.org/news/2015-06-surface-black-hole-firewalland-nature.html#jCp&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the mid-90s, American and Dutch physicists Leonard Susskind and Gerard 't Hooft also addressed the information paradox by proposing that when something gets sucked into a black hole, its information leaves behind a kind of two-dimensional holographic imprint on the event horizon, which is a sort of ‘bubble’ that contains a black hole through which everything must pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What occurs at the event horizon of a black hole is very hard to understand. What is clear, and what proceeds from General Relativity, is that from the viewpoint of an external observer in this universe, any infalling matter cannot proceed past the critical circumference.  Most scientists then change the viewpoint to explain how, from the viewpoint of an infalling observer, they will proceed in a very short period of time to meet the singularity at the centre of the black hole. &#xA;This has given rise to the notion that there is a singularity at the centre of every black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this is is an illusion, as the time it will take to reach the singularity is essentially infinite to us in the external universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the matter cannot proceed past the critical circumference is perhaps not an “illusion” but very real. The matter must from OUR VIEWPOINT become a “shell” surrounding the critical circumference. It will never fall through the circumference while we remain in this universe. So to talk of a singularity inside a black hole is incorrect.  It has not happened yet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The path through the event horizon does lead to a singularity in each case, but it is indefinitely far in the future in all cases. If we are in this universe, no singularity has yet been formed. If it has not been formed yet, where is the mass?  The mass is exerting pull on this universe, correct?  Then it must be IN this universe.  From our point of view it must be just this side of the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ASTONISHINGLY IT MAY BE POSSIBLE TO PROVE THIS.  The recent announcement of gravitational waves detected on the merger of 2 black holes was accompanied by an unverified but potentially matching gamma ray burst from the same area of the sky. This is inexplicable from the conventional viewpoint which holds that all the matter would be compressed into a singularity and would be incapable of coming out again.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 2 black holes merge and emit gamma rays… the above is certainly an explanation which is also consistent with General Relativity. The mass never quite made it through the event horizon (from our viewpoint) and was perturbed by the huge violence of the merger, some escaping. It may be a deep gravitational well, but a very powerful gamma ray should just be able to escape given the right kick (attraction by an even larger black hole approaching).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further more refined observations of similar events, which are likely to be reasonably frequent, may provide more evidence.  There is not likely to be any other credible explanation.&lt;/p&gt;&#xA;" OwnerUserId="11278" LastEditorUserId="11278" LastEditDate="2016-03-24T22:26:32.457" LastActivityDate="2016-03-24T22:26:32.457" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14240" PostTypeId="2" ParentId="14238" CreationDate="2016-03-19T16:09:51.363" Score="1" Body="&lt;p&gt;The habitable zone is, roughly, defined as the region around a star in which a planet could have liquid surface water. So &lt;strong&gt;if the surface temperature of the planet is between 0 and 100*C then the planet is in the habitable zone&lt;/strong&gt;. The converse, however, is not true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this definition is that the sort of chemical reactions that are needed for life work best when there are complex molecules dissolved in some kind of solvent, and the best polar solvent is H2O.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know the distance a planet is from its star, the temperature can be estimated from the properties of the star and the atmosphere of the planet. Not every planet in the habitable zone actually has liquid water. Venus, Earth and Mars are all in the Habitable zone of the sun, since they could all have liquid water if they had the right atmosphere. But of course only Earth actually does have liquid water in any substantial amounts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So just being in the habitable zone, doesn't make a planet certainly suitable for life, and even if water does exist on the surface, it does not mean that humans could inhabit the planet.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-03-19T21:59:11.543" LastActivityDate="2016-03-19T21:59:11.543" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14241" PostTypeId="2" ParentId="2524" CreationDate="2016-03-19T16:24:32.740" Score="-1" Body="&lt;p&gt;You are correct, the universe you were leaving would seem to speed up and ultimately the whole history of the universe would occur as you cross the critical circumference and onwards to the point like singularity. Galaxy mergers would take place, other black holes would coalesce with yours and so on. The singularity you finally reach would in fact be one and the same with the singularity at the end of the universe. There is only one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding Hawking radiation, from the viewpoint of an outside observer, who would see you frozen in time (or at least the constituent matter and energy in your body) at or around the critical circumference, this may not actually take place.  If matter is frozen by time dilation, there will be no quantum fluctuations taking place and nothing falling through the critical circumference into the black hole, therefore no Hawking radiation taking place. Looked at from the viewpoint of an infalling observer it will take a very short time to cross the critical circumference, therefore the significance of quantum fluctuations in this period of time would seem to be small.&lt;/p&gt;&#xA;" OwnerUserId="11278" LastActivityDate="2016-03-19T16:24:32.740" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="14242" PostTypeId="1" CreationDate="2016-03-19T19:32:37.183" Score="4" ViewCount="3222" Body="&lt;p&gt;I see that there are many maps of a single galaxy, the Milky Way, showing the positions of the stars within our galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could not find a map of Galaxies, however. Is there such a map? In lieu of that I suppose I could make my own map if I knew celestial coordinates of all the nearby galaxies and the distances to those galaxies. Is there such a data table readily available?&lt;/p&gt;&#xA;" OwnerUserId="2800" LastActivityDate="2016-03-19T20:58:23.807" Title="Is there a map of the galaxies?" Tags="&lt;galaxy&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14243" PostTypeId="2" ParentId="14217" CreationDate="2016-03-19T19:45:43.307" Score="4" Body="&lt;p&gt;It comes from angular momentum. Angular momentum is a conserved quantity of physics. That means that the sum of angular momentum of the universe is constant, even though some parts of the universe may transfer angular momentum to other parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do not know the total amount of angular momentum of the universe, but from observations we know that it is not uniformly distributed. This explains why celestial objects rotate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the total amount of angular momentum of the universe would be zero it would be theoretically possible that rotation could cease in the future. However, nobody knows the ultimate fate of the universe, so this is purely speculative.&lt;/p&gt;&#xA;" OwnerUserId="10568" LastActivityDate="2016-03-19T19:45:43.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14245" PostTypeId="2" ParentId="14242" CreationDate="2016-03-19T20:58:23.807" Score="10" Body="&lt;p&gt;A map of all galaxies gets kind of unwieldy, like a map of all stars in the milky way or a map of every house in the country, or every grain of sand on a beach . . . you get the idea. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start here - Local Group&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UsM0M.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UsM0M.jpg&quot; alt=&quot;https://upload.wikimedia.org/wikipedia/commons/5/57/5_Local_Galactic_Group_%28ELitU%29.png&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Local_Group&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then Virgo Supercluster&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GM1A6.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GM1A6.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Virgo_Supercluster&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then local superclusters&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rp1FO.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rp1FO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pics-about-space.com/virgo-supercluster-of-galaxies?p=2&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And an &lt;a href=&quot;http://www.huffingtonpost.com/2014/09/05/laniakea-supercluster-galaxy-tully_n_5763274.html&quot; rel=&quot;noreferrer&quot;&gt;article&lt;/a&gt;, even if it's a summary it's very much worth reading, with a more recent map with a &quot;ginormous&quot; supercluster that includes the milky way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xqhNV.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xqhNV.jpg&quot; alt=&quot;http://i.huffpost.com/gen/2023458/thumbs/o-LANIAKEA-900.jpg?6&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and here's &lt;a href=&quot;http://www.atlasoftheuniverse.com/nearsc.html&quot; rel=&quot;noreferrer&quot;&gt;another&lt;/a&gt;, more info &lt;a href=&quot;http://imagine.gsfc.nasa.gov/features/cosmic/nearest_superclusters_info.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest of these maps is some 520 million light years across, so this is just a tiny part of the entire known universe, which (depending on how you measure) is either 27.5 billion light years across or 84 billion light years across.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth pointing out, I haven't actually answered your question, and that was deliberate, cause I think the nearest 520 million light years is enough and it's all we have a really good picture of anyway (as far as I know).  The farther out you go the more holes and inaccuracies there will be in the map.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also some closer stuff that's blocked by the Milky way so we can't see it, like the &lt;a href=&quot;https://en.wikipedia.org/wiki/Great_Attractor&quot; rel=&quot;noreferrer&quot;&gt;great attractor&lt;/a&gt;.  No worries about superclusters eating our galaxy one day because dark energy expansion keeps these super-clusters from merging most of the time.  Andromeda will merge with the Milky way in about 4 billion years though but we're not expected to ever merge with the great attractor, even as it pulls our galaxy towards it, the distance between us is growing.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-03-19T20:58:23.807" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14246" PostTypeId="5" CreationDate="2016-03-19T22:16:09.060" Score="0" Body="&lt;p&gt;Naturally, astronomical discoveries may have an impact on space exploration. For example, spectroscopic analysis of asteroids could have a bearing on a target for robotic exploration of the asteroid belt. However, in general most questions with this tag will be either closed or transferred to our sister site &lt;a href=&quot;https://space.stackexchange.com&quot;&gt;Space Exploration Stack Exchange&lt;/a&gt;, where they may find better answers.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="7472" LastEditDate="2018-10-01T12:06:55.543" LastActivityDate="2018-10-01T12:06:55.543" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14247" PostTypeId="4" CreationDate="2016-03-19T22:16:09.060" Score="0" Body="If you are using this tag please consider whether your question would be more appropriate on the space exploration stack exchange." OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-03-19T22:16:09.060" LastActivityDate="2016-03-19T22:16:09.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14249" PostTypeId="1" CreationDate="2016-03-20T00:36:36.273" Score="-1" ViewCount="138" Body="&lt;p&gt;If a human was naked a few hundred feet from the sun, floating, with no atmosphere, not affected by the vacuum of space or UV radiation. Would the sun burn him/her? Or not have any effect?&lt;/p&gt;&#xA;" OwnerUserId="11282" LastEditorUserId="2365" LastEditDate="2016-03-20T12:48:12.080" LastActivityDate="2016-03-20T12:48:12.080" Title="Could the sun burn a human floating next to it?" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2016-03-20T13:37:28.170" ContentLicense="CC BY-SA 3.0" />
  <row Id="14250" PostTypeId="2" ParentId="14249" CreationDate="2016-03-20T10:50:34.950" Score="3" Body="&lt;p&gt;From the conditions you give, yes. Absolutely. The photon flux would simply transfer too much energy for the molecules in your body to stay fixed (the radiation would be immense). Another potential way to re-phrase this question would be: &quot;How close could a human, without protection from any form of radiation, get to the Sun before their skin burns?&quot;&lt;/p&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2016-03-20T10:50:34.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14251" PostTypeId="2" ParentId="13515" CreationDate="2016-03-20T15:29:56.020" Score="2" Body="&lt;p&gt;Nobody likes the idea of naked singularities, as they would have a toxic effect on causality. If a singularity existed that was not separated from us by an event horizon, then not only would the future not be predictable, but the past would not be fixed. Like the grandfather paradox, it wouldn't make sense, so it cant exist. The trouble is that GR doesn't implicitly rule them out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a black hole is spinning fast enough, or has enough charge then it seems a naked singularity could form. Naively you may think that the centrifugal force in the first case, and the electromagnetic force in the second are sufficient to overcome gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I particular if $G^2M^4/c^2 &amp;lt; J^2$, where J is angular momentum, M is the mass, G is the gravitation constant, and c is the speed of light. The there will be a naked singularity. For a small, stellar mass black hole would need an angular momentum of $10^{42} \mathrm{kg m^2 s^{-1}}$ to lose its event horizon&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-03-20T15:29:56.020" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14252" PostTypeId="1" CreationDate="2016-03-20T15:49:31.797" Score="1" ViewCount="224" Body="&lt;p&gt;Is dark matter trapped near / in black holes, like regular matter is, or does something different happen with dark matter?&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2016-12-09T19:58:23.863" Title="How does dark matter interact with black holes?" Tags="&lt;gravity&gt;&lt;black-hole&gt;&lt;dark-matter&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14253" PostTypeId="2" ParentId="14252" CreationDate="2016-03-20T15:57:06.720" Score="4" Body="&lt;p&gt;It interacts gravitationally.That's all there is to it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a big difference though to the way that normal and dark matter interact with black holes - dark matter is far less likely to be captured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given two lumps of matter, one normal one dark, &lt;em&gt;with the same mass and angular momentum with respect to the black hole&lt;/em&gt;; only the normal matter is capable of shedding its angular momentum (normally an accretion disc is involved), which allows its orbit to shrink enough to be captured (within three Schwarzschild radii). Dark matter is dissipationless; if it has too much angular momentum it won't get captured.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-03-20T15:57:06.720" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14254" PostTypeId="2" ParentId="14203" CreationDate="2016-03-20T22:26:41.403" Score="1" Body="&lt;p&gt;OK, I'll assume you are talking about &lt;a href=&quot;https://www.nasa.gov/mission_pages/hubble/science/runaway-star.html&quot; rel=&quot;nofollow&quot;&gt;this star&lt;/a&gt; or some other high-mass &quot;runaway star&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most massive stars are born in rich clusters and are often part of multiple system. Is not unusual for such stars to get ejected from these regions are high velocities, either as a result of dynamical interactions in multiple systems or when they are companions to stars that explode as supernovae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steps to estimate the mass are: (i) Obtain very detailed optical and UV spectroscopy. From this you estimate the spectral type of the star, it's effective temperature and its (current) mass-loss rate. (ii) You combine this with a known distance (the star in the link was in the Large Magellanic Cloud, so the distance was known fairly accurately) to estimate a luminosity. (iii) You compare the luminosity, temperature and mass-loss rate with the predictions of a specialised high-mass stellar evolutionary model, which includes mass-loss and rotation. The comparison yields an estimate of the mass and age of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The details are found in &lt;a href=&quot;http://adsabs.harvard.edu/abs/2010ApJ...715L..74E&quot; rel=&quot;nofollow&quot;&gt;Evans et al. (2010)&lt;/a&gt;. It is probably fair to say that the estimate of $90\ M_{\odot}$, is uncertain by a few tens of solar masses.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-03-20T22:26:41.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14255" PostTypeId="2" ParentId="13817" CreationDate="2016-03-21T00:42:04.333" Score="1" Body="&lt;p&gt;In the first example, you certainly could use the calculation of special relativity to measure the difference in the time shown on a clock which was making the high speed journey and one which was at rest. The details may make the calculation trickier, but we can use the flat minowsky space of the clock at rest to do the calculation, so it is relatively straightforward. [see what I did there?]. It would be a factor of about 1.6 so the journey that would take a little over 67 years would, on board ship, seem to take 67*1.6 = 41 years. (I have ignored the time spent accelerating and decelerating, in order to keep the maths simple. Earth would observe the ship leaving 30years after it did (light distance) and arriving after 67 years, however, since the light must travel further, the ship would appear to arrive after 67+50-30 years = 87 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the multiple journey the calculations are just longer. The direction of travel doesn't affect time dialation, only the speed. So it would again be straightforward to calculate the total time dialation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be quite possible for Earth to communicate with the ship. But those the messages would not arrive for up to 50 years. Messages would be certainly out of date.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-03-21T00:42:04.333" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14256" PostTypeId="1" AcceptedAnswerId="14267" CreationDate="2016-03-21T02:24:38.657" Score="8" ViewCount="681" Body="&lt;p&gt;If my understanding is correct, Chury, the best explored comet so far is rather on the smallish side as comets go. Most comets observable with naked eye were so because they flew close to Earth, not because they were so big. I wonder, though, how big are the biggest ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's get it in two variants:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How big is the biggest known comet? (mass, diameter)&lt;/li&gt;&#xA;&lt;li&gt;Is there an estimate, or a theoretical limit on how big a comet can be? (for whatever reason, e.g. becoming a planet, breaking up due to tidal forces or whatever)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="83" LastEditorUserId="38" LastEditDate="2016-03-21T12:25:53.103" LastActivityDate="2021-02-09T00:15:13.083" Title="How big do comets get?" Tags="&lt;comets&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14257" PostTypeId="2" ParentId="14217" CreationDate="2016-03-21T05:32:21.627" Score="7" Body="&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/user/minutephysics&quot;&gt;The MinutePhysics channel&lt;/a&gt; on YouTube made &lt;a href=&quot;https://www.youtube.com/watch?v=tmNXKqeUtJM&quot;&gt;a wonderful short animation&lt;/a&gt; that answers your very question. It explains how a cloud of gas and dust with zero net angular momentum collapses down to form a flat spinning object such as a galaxy or our Solar System.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer involves a few things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The universe conserves angular momentum in every interaction between&#xA;physical objects.&lt;/li&gt;&#xA;&lt;li&gt;Mathematics says a big cloud of gas and dust must have exactly one plane in which the overall angular momentum is zero.&lt;/li&gt;&#xA;&lt;li&gt;When particles of gas and dust collide, any angular momentum they have outside that plane is cancelled, so over millions of years and (countless numbers of collisions) the cloud ends up flattening.&lt;/li&gt;&#xA;&lt;li&gt;But as they collide, their angular momenta within that one plane are&#xA;conserved, so the flattened system ends up spinning.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11175" LastActivityDate="2016-03-21T05:32:21.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14258" PostTypeId="1" CreationDate="2016-03-21T15:13:41.390" Score="4" ViewCount="1208" Body="&lt;p&gt;Imagine a photon reaching the Hubble Space Telescope today had originally been emitted from a star in the early universe 13 billion years ago. Einstein’s Special Relativity tells us, traveling at the speed of light, that from the photon’s frame of reference, no time will have passed for the photon between the moment it was emitted from the star and the instant it was recorded by the Hubble telescope. In addition, at the speed of light, the length contraction (or Lorentz contraction) is said to shrink the distance between these two objects (the ancient star and the Hubble telescope) to zero. But herein lies the apparent paradox in two parts: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In the early universe 13 billion years ago when the star first emitted the photon, there was no Hubble Telescope for the photon to instantaneously collide with from its frame of reference. In other words, how can the photon instantly collide with something which won't be invented for 13 billion years? &lt;/li&gt;&#xA;&lt;li&gt;Given the complete length contraction of all points in space to zero at the speed of light, how can a single photon ever be absorbed at &lt;em&gt;any one&lt;/em&gt; particular point when, if all distance between points in the universe is zero from the photon’s frame of reference, the photon will hit &lt;em&gt;each and every point&lt;/em&gt; throughout the universe simultaneously, not just one? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;How to explain this apparent paradox in laymen’s terms.&lt;/p&gt;&#xA;" OwnerUserId="11288" LastEditorUserId="15770" LastEditDate="2020-05-03T18:40:20.060" LastActivityDate="2020-05-03T18:40:20.060" Title="Photon Paradox?" Tags="&lt;time&gt;&lt;observable-universe&gt;&lt;hubble-telescope&gt;&lt;special-relativity&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="14259" PostTypeId="2" ParentId="14229" CreationDate="2016-03-21T15:51:07.617" Score="0" Body="&lt;p&gt;I'm not absolutely sure if it is correct, but I use information from this site: &lt;a href=&quot;http://star-www.st-and.ac.uk/~fv/webnotes/chapt9a.htm&quot; rel=&quot;nofollow&quot;&gt;http://star-www.st-and.ac.uk/~fv/webnotes/chapt9a.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I convert equatorial coordinates (RA = α, Dec = δ) to ecliptic coordinates (λ, β)  using this formula.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;sin(β) = sin(δ) cos(ε) - cos(δ) sin(ε) sin(α)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;cos(λ) cos(β) = cos(α) cos(δ)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;ε = Axial tilt (23,5°)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that due to precession only λ changes (β remains same for ecliptic coordinates). So I add the angle of precession to the λ. Then I convert the new λ and β back to the equatorial coordinates.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;sin(δ) = sin(β) cos(ε) + cos(β) sin(ε) sin(λ) &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;cos(λ) cos(β) = cos(α) cos(δ)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10592" LastActivityDate="2016-03-21T15:51:07.617" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14260" PostTypeId="1" CreationDate="2016-03-21T16:23:55.743" Score="0" ViewCount="315" Body="&lt;p&gt;According to Newton's laws, the trajectories for the two-body problem are conics: either ellipses, or parabolas or hyperbolas. Of course periodic motions require Ellipses and in the Solar system these elliptic curves are a pretty good approximation of the trajectory of each planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is possible that the trajectory of a very fast comet could be an hyperbola or a parabola with the Sun at a focus; in that case the motion would not be periodic and the comet would be visible near the Sun only once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: was there ever any observation of a  comet on a hyperbolic (or parabolic) trajectory (with the Sun at a focus)? Same question for different stellar objects. It would be certainly very difficult to observe, since the comet would be visible only one time at its perihelion and would go at infinity after this with no return.&lt;/p&gt;&#xA;" OwnerUserId="11289" LastActivityDate="2016-03-21T21:46:06.340" Title="Parabolic or hyperbolic trajectories" Tags="&lt;observational-astronomy&gt;&lt;newtonian-gravity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14261" PostTypeId="2" ParentId="14258" CreationDate="2016-03-21T16:57:36.990" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;1) In the early universe 13 billion years ago when the star first emitted the photon, there was no Hubble Telescope for the photon to instantaneously collide with from its frame of reference. In other words, how can the photon instantly collide with something which won't be invented for 13 billion years?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This isn't relevant. Regardless of any &quot;clock&quot; running locally on the photon, it is still travelling for a long time (in our frame of reference) to reach its target (our telescope).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact if this paradox were so, it would be impossible to see anything. If say a distant probe transmits a radio signal to us, travelling for an hour or more at the speed of light, our radio receiver would have to be switched on ready to receive long before the message reached it.  Again, not true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there simply isn't a paradox.&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2016-03-21T16:57:36.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14263" PostTypeId="2" ParentId="14260" CreationDate="2016-03-21T21:46:06.340" Score="1" Body="&lt;p&gt;Yes, and it is not uncommon for an orbit have an eccentricity close to one. The wikipedia site, linked in a comment above, notes C/1980 E1, which entered the inner solar system with an eccentricity close to one, but had a close encounter with jupiter and was accelerated. It left the inner solar system with a eccentricity of 1.05, and so is on a hyperbolic trajectory, and will escape from the sun's gravity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbits that are highly hyperbolic are very unlikely. Comets formed as part of the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are not really harder to spot than any other comet. A comet takes many months to make its passage through the inner solar system. There is plenty of time for them to be spotted, especially if you have probes like SOHO or NEAT&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-03-21T21:46:06.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14264" PostTypeId="2" ParentId="14258" CreationDate="2016-03-21T22:12:34.060" Score="10" Body="&lt;p&gt;And that is why you don't do the calculations in a frame that is moving at lightspeed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have two observers that are moving relative to each other you can use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lorentz_transformation&quot; rel=&quot;noreferrer&quot;&gt;Lorentz transformation&lt;/a&gt; to change between their frames of reference. But if one of the observers is a photon the lorentz transformation becomes singular, because $\gamma$ is infinite. Simply, &lt;strong&gt;you can't take a photon's point of view&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would be much better off not thinking of a photon as a classical massless particle, but as a quantum mechanical particle. In the QM setting, you can take two points in spacetime and say that there is a certain amplitude (a value that relates to probability) for the photon to be at those two points. There is no need to think of it as ever &lt;em&gt;being&lt;/em&gt; at intermediate points. I suggest you look up Feymann's book on &quot;QED&quot; for a readable account of this. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-03-21T22:12:34.060" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14265" PostTypeId="1" AcceptedAnswerId="14266" CreationDate="2016-03-21T23:37:13.427" Score="4" ViewCount="1785" Body="&lt;h1&gt;Question:&lt;/h1&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Apart from a Solar Eclipse, How much time is needed until a Waxing Crescent Moon be seen following a New Moon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Would the time of year be significant? The Vernal Equinox + 1 Month.&lt;/li&gt;&#xA;&lt;li&gt;Could it be seen the same night that it occurs?&lt;/li&gt;&#xA;&lt;li&gt;Could it be seen in the day when it occurs?&lt;/li&gt;&#xA;&lt;li&gt;Would this be affected if the viewer was in a rural location, without lights? Or would city lights help?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;h1&gt;Context:&lt;/h1&gt;&#xA;&lt;p&gt;Ancient Lunar calendars depended on observation of the New Moon, (like political / religious obligations).&lt;/p&gt;&#xA;&lt;p&gt;What was the potential margin of error? Hours? a Day? (Not considering bad-weather.)&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/New_moon&quot; rel=&quot;nofollow noreferrer&quot;&gt;From Wikipedia&lt;/a&gt; - A New Moon Cannot be Seen:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In astronomy, new moon is the first phase of the Moon, when it orbits as seen from the Earth, the moment when the Moon and the Sun have the same ecliptical longitude. The Moon is not visible at this time except when it is seen in silhouette during a solar eclipse when it is illuminated by earthshine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Related:&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/6072/is-it-possible-for-a-person-to-not-see-the-new-moon-at-different-places-on-earth&quot;&gt;Is it possible for a person to not see the new moon at different places on earth?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/630/from-which-country-or-area-is-the-new-moon-visible-first?rq=1&quot;&gt;From which country or area is the new moon visible first?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11293" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-03-22T23:55:38.170" Title="How Soon Could a Waxing Crescent Moon Be Seen?" Tags="&lt;observational-astronomy&gt;&lt;time&gt;&lt;moon-phases&gt;&lt;the-moon&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14266" PostTypeId="2" ParentId="14265" CreationDate="2016-03-22T02:43:56.630" Score="3" Body="&lt;p&gt;From the US Naval Observatory: &lt;a href=&quot;http://aa.usno.navy.mil/faq/docs/crescent.php&quot;&gt;Crescent Moon Visibility&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Naked-eye sightings as early as 15.5 hours after New Moon have been reliably reported while observers with telescopes have made reliable reports as early as 12.1 hours after New Moon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;USNO is usually a very reliable site for info like this.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2016-03-22T02:43:56.630" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14267" PostTypeId="2" ParentId="14256" CreationDate="2016-03-22T05:30:54.360" Score="6" Body="&lt;p&gt;I suppose size limit of the comet is still under process but astronomers have found a few large ones though!&lt;br /&gt;&#xA;The comet Holmes which appeared in 2007 became brighter and also swelled up in size. Its nuclear size is only 3.6km (2.2 miles), but it released so much dust and gas so that it expanded to about 869,900 miles (1.4 million kilometers) i.e larger than the sun.&lt;br /&gt;&#xA;The largest comet in terms of nuclear size was discovered in the year 1729, comet Sarabat which had a nucleus of about &lt;span class=&quot;math-container&quot;&gt;$100 {\rm km}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11283" LastEditorUserId="37191" LastEditDate="2021-02-08T21:37:21.083" LastActivityDate="2021-02-08T21:37:21.083" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14268" PostTypeId="2" ParentId="13289" CreationDate="2016-03-22T06:50:02.137" Score="-1" Body="&lt;p&gt;The number of stars and planets in the milky way can only be given in approximate values.There are about 100-400 billion stars and as of 2016 there are about 2097 planets discovered from 1337 planetary systems. &lt;/p&gt;&#xA;" OwnerUserId="11283" LastEditorUserId="11283" LastEditDate="2016-03-23T06:16:17.510" LastActivityDate="2016-03-23T06:16:17.510" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14269" PostTypeId="1" AcceptedAnswerId="14270" CreationDate="2016-03-22T06:55:53.727" Score="0" ViewCount="111" Body="&lt;p&gt;Is there any popular star or one which we can see with our naked eyes in our milky way that is atleast 90% sun-like(mass,radius,spectra,luminosity).The important thing is it must not be a binary.&lt;br&gt;&#xA;Its okay if its not visible to our naked eyes because our sun's absolute magnitude is +4.5!.&lt;/p&gt;&#xA;" OwnerUserId="11283" LastEditorUserId="11283" LastEditDate="2016-03-22T07:42:29.700" LastActivityDate="2016-03-22T15:46:06.470" Title="Sun like star in our milkyway?" Tags="&lt;star&gt;&lt;milky-way&gt;&lt;star-systems&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14270" PostTypeId="2" ParentId="14269" CreationDate="2016-03-22T07:15:17.243" Score="5" Body="&lt;p&gt;There's a whole Wikipedia page about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_analog&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Solar_analog&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't want Alpha Cen A, then 18 Sco might be the one.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2270" LastEditDate="2016-03-22T15:46:06.470" LastActivityDate="2016-03-22T15:46:06.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14272" PostTypeId="1" CreationDate="2016-03-22T13:10:23.733" Score="1" ViewCount="266" Body="&lt;p&gt;I often wonder why the Drake equation gets taken so seriously and why popular scientists in general say &quot;it's probable' life exists elsewhere in the cosmos.&#xA;My problem is that the mechanism for abiogenesis is as yet unknown. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't know the likelihood that RNA like molecules can assemble from chance or for that matter that this is the source of life on this planet (as opposed to coming via a comet). We do know that we have no record of it recurring and as yet Miller-uri style experiments have not yielded results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument I keep hearing from Dawkins, Sagan and many others is that the universe is vast and life occurred relatively quickly in our planet's history which suggests it is common.&#xA;I detect a logical fallacy based on the inductive reasoning of using one occurrence in order to establish a pattern. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of it this way. If I've never played cards and get dealt my first hand which is a royal flush I might think its quite likely based on my past hand that I will get another in the next deal. This doesn't change if the deck contains a hundred suites and the odds of a royal flush is astronomical - if my first hand is a royal flush.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly the vastness of the cosmos tells you nothing about the likelihood of other life if you don't the probability of its occurrence.It could be very likely or so rare as to only occur once. &#xA;Does anybody have an opinion on this?&lt;/p&gt;&#xA;" OwnerUserId="11301" LastEditorUserId="7982" LastEditDate="2021-02-19T01:28:25.853" LastActivityDate="2021-02-19T01:28:25.853" Title="Is the Drake equation based on a logical fallacy?" Tags="&lt;cosmology&gt;&lt;drake-equation&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2016-03-23T17:25:38.290" ContentLicense="CC BY-SA 3.0" />
  <row Id="14273" PostTypeId="2" ParentId="14272" CreationDate="2016-03-22T13:26:09.690" Score="8" Body="&lt;p&gt;To directly answer the question, the Drake Equation is not based on a &quot;logical fallacy&quot; (which means a different thing, philosophically). The logic is fine, the real problem is we don't have any meaningful data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding abiogenesis (the first formation of some simple life): as you correctly state, we know only that it happened once (here, presumably but that doesn't matter). However the argument that it happened quite quickly on our planet is meaningless statistically, because that's just a sample of one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If anyone thinks this answer is too waffly or off topic just say so and I'll delete it by the way; I think the question is a fair one but we just can't present any data, just reason.)&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2016-03-22T13:26:09.690" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14275" PostTypeId="1" AcceptedAnswerId="14278" CreationDate="2016-03-22T15:01:20.353" Score="1" ViewCount="120" Body="&lt;p&gt;Simple question -- why is declination positive in the northern hemisphere?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seems like it should've been called &quot;inclination&quot; since the word originated north of the equator.&lt;/p&gt;&#xA;" OwnerUserId="11302" LastActivityDate="2016-03-23T10:09:52.727" Title="Why is declination positive in the northern hemisphere?" Tags="&lt;coordinate&gt;&lt;declination&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14276" PostTypeId="1" CreationDate="2016-03-22T15:15:18.663" Score="4" ViewCount="1111" Body="&lt;p&gt;I'd like to calculate the positions of the sun and moon in the ECEF coordinate system using the SOFA (&lt;a href=&quot;http://www.iausofa.org/current_C.html&quot; rel=&quot;nofollow&quot;&gt;http://www.iausofa.org/current_C.html&lt;/a&gt;) libraries given a Julian day input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, I've got this process for the sun:  (1) convert Jd (UTC) to TAI, to TT, then to TDB, then (2) calculate earth position/velocity in heliocentric coordinates (iauEpv00), then (3) negate the terms to get ECI.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) I'm not sure if the process above is correct, especially since calculating TDB time requires a &quot;dtr&quot; parameter, which apparently you have to have TDB times to estimate using iauDtdb().&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) I'm not sure how to convert the resulting vector from ECI to ECF, which is probably just a time-based rotation of the longitude, but how much time is elapsing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) Not sure where to even start with the moon.&lt;/p&gt;&#xA;" OwnerUserId="11302" LastActivityDate="2020-06-10T06:26:21.680" Title="Solar / lunar positions in ECEF using SOFA C libraries" Tags="&lt;the-sun&gt;&lt;the-moon&gt;&lt;positional-astronomy&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14278" PostTypeId="2" ParentId="14275" CreationDate="2016-03-23T10:09:52.727" Score="2" Body="&lt;p&gt;After quite an extensive search of dictionaries and books on astronomical nomenclature I came across &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1942PASP...54...77W&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; that has a comprehensive review of celestial coordinate systems. And upon reading page 84 (8/14) I came across the answer to your question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would appear that the reason declination is used, as opposed to inclination, is that declination was first interpreted as a &quot;deviation from standard&quot; not an inclination of angular distance from a given point as we know it today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &quot;deviation from standard&quot; was first used to describe the motion of the sun  along the ecliptic that &quot;deviated&quot; from the then widely accepted notion of all heavenly bodies following the line of the equator. It goes into more detail in the article but it would appear that when translated from Greek to Latin and then Latin to Arabic, the precise meaning of the word changed and evolved so that now the term declination refers to the angular distance of any celestial body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source:&lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1942PASP...54...77W&amp;amp;db_key=AST&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1942PASP...54...77W&amp;amp;db_key=AST&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10437" LastActivityDate="2016-03-23T10:09:52.727" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14279" PostTypeId="1" CreationDate="2016-03-23T14:08:57.710" Score="0" ViewCount="599" Body="&lt;p&gt;Can a human in a normal astronaut suit designed for the moon, land on Mercury and walk on it's surface. If not what would happen to him/her?&lt;/p&gt;&#xA;" OwnerUserId="11313" LastActivityDate="2016-03-23T16:03:27.667" Title="Can a human land on Mercury?" Tags="&lt;mercury&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2016-03-23T23:26:50.813" ContentLicense="CC BY-SA 3.0" />
  <row Id="14280" PostTypeId="1" AcceptedAnswerId="14285" CreationDate="2016-03-23T14:37:37.200" Score="3" ViewCount="363" Body="&lt;p&gt;I was studying our solar system and all celestial objects are depicted as round objects, including black holes.. worm holes as hollow cylinders as a pathway across two universes.. even the galaxy is show circular.. Why does this happen? &lt;/p&gt;&#xA;" OwnerUserId="11315" LastActivityDate="2016-03-23T21:15:56.477" Title="Why are planets round?" Tags="&lt;solar-system&gt;&lt;planet&gt;" AnswerCount="1" CommentCount="7" ClosedDate="2018-01-26T15:35:04.087" ContentLicense="CC BY-SA 3.0" />
  <row Id="14283" PostTypeId="2" ParentId="14279" CreationDate="2016-03-23T16:03:27.667" Score="2" Body="&lt;p&gt;No sir/madam. An astronaut or a cosmonaut cannot walk on Mercury's surface. The reason is plainly simple. During daytime, its temperature surges up to 430 degree C making it 2nd hottest planet after Venus. Whereas, night time temperatures are as low as -180 degree C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite of having ice caps at its polar regions which might seem attractive, its totally not feasible to land there and then mine the frozen ice to make up a base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a much more descriptive read, kindly refer this link : &lt;a href=&quot;http://www.space.com/28356-how-to-live-on-mercury.html&quot; rel=&quot;nofollow&quot;&gt;What it would be to live on Mercury&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11317" LastActivityDate="2016-03-23T16:03:27.667" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14284" PostTypeId="1" CreationDate="2016-03-23T19:13:19.607" Score="5" ViewCount="491" Body="&lt;p&gt;I was reading about the impending collision of our galaxy with the Andromeda galaxy, and learned that there is a small possibility that our solar system could be ejected during the collision. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that orbits within our solar system remain unchanged (it is ejected as a whole), would floating in intergalactic space cause any significant changes in the living conditions on earth? &lt;/p&gt;&#xA;" OwnerUserId="11197" LastActivityDate="2016-04-09T23:08:21.780" Title="Would being ejected from the Milky Way Galaxy have any major impact on life on Earth?" Tags="&lt;life&gt;&lt;intergalactic-space&gt;" AnswerCount="2" CommentCount="10" FavoriteCount="1" ClosedDate="2016-04-11T12:33:03.307" ContentLicense="CC BY-SA 3.0" />
  <row Id="14285" PostTypeId="2" ParentId="14280" CreationDate="2016-03-23T19:16:37.497" Score="6" Body="&lt;p&gt;Gravity makes objects compress to their centers of mass, since gravity extends in all directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For planets, gravity continues compressing the object until the rocks cannot be compressed anymore, since the pressure will fight against gravity. At this point (called hydrostatic equilibrium), the object has become a spheroid. One of the requirements for an object to be called a &quot;planet&quot; is that it has reached hydrostatic equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For black holes, the shape of their event horizons really depend on whether the black hole is spinning or not. For non-spinning black holes, gravity extends in all directions, so the event horizon will become spherical in shape. For spinning black holes, it will likely be an oblate spheroid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most galaxies are only spherical when they are forming. Their angular momentum usually causes them to flatten out over time. However, one part of galaxies, called the &quot;halo&quot;, is spherical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wormholes are purely hypothetical, so I won't really entertain that.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-03-23T21:15:56.477" LastActivityDate="2016-03-23T21:15:56.477" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14287" PostTypeId="2" ParentId="4725" CreationDate="2016-03-23T21:56:02.930" Score="9" Body="&lt;p&gt;Let us assume that $N$ stars have ever been born in the Milky Way galaxy, and given them masses between 0.1 and 100$M_{\odot}$. Next, assume that stars have been born with a mass distribution that approximates to the Salpeter mass function - $n(m) \propto m^{-2.3}$. Then assume that all stars with mass $m&amp;gt;25M_{\odot}$ end their lives as black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if $n(m) = Am^{-2.3}$, then&#xA;$$N = \int^{100}_{0.1} A m^{-2.3}\ dm$$&#xA;and thus $A=0.065N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of black holes created will be &#xA;$$N_{BH} = \int^{100}_{25} Am^{-2.3}\ dm = 6.4\times10^{-4} N$$&#xA;i.e 0.06% of stars in the Galaxy become black holes. NB: The finite lifetime of the galaxy is irrelevant here because it is much longer than the lifetime of black hole progenitors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I follow the other answers by scaling to the number of stars in the solar neighbourhood, which is approximately 1000 in a sphere of 15 pc radius $\simeq 0.07$ pc$^{-3}$. I assume that as stellar lifetime scales as $M^{-2.5}$ and the Sun's lifetimes is about the age of the Galaxy, that almost all the stars ever born are still alive. Thus, the black hole density is $\simeq 4.5 \times 10^{-5}$ pc$^{-3}$ and so there is one black hole within 18 pc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, so why might this number be wrong? Although the number is very &lt;em&gt;insensitive&lt;/em&gt; to the assumed &lt;em&gt;upper&lt;/em&gt; mass limit of stars, it is &lt;em&gt;very sensitive&lt;/em&gt; to the assumed lower mass limit. This could be higher or lower depending on the very uncertain details of the late stellar evolution and mass-loss from massive stars. This could drive our answer up or down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some fraction $f$ of these black holes will merge with other black holes or will escape the Galaxy due to &quot;kicks&quot; from a supernova explosion or interactions with other stars in their dense, clustered birth environments (though not all black holes require a supernova explosion for their creation). We don't know what this fraction is, but it &lt;em&gt;increases&lt;/em&gt; our answer by a factor $(1-f)^{-1/3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if they don't escape, it is highly likely that black holes will have a much higher velocity dispersion and hence spatial dispersion above and below the Galactic plane compared with &quot;normal&quot; stars. This is especially true considering most black holes will be very old, since most star formation (including massive star formation) occurred early in the life of the Galaxy, and black hole progenitors die very quickly. Old stars (and black holes) have their kinematics &quot;heated&quot; so that their velocity and spatial dispersions increase. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I conclude that black holes will therefore be under-represented in the solar neighbourhood compared with the crude calculations above and so you should treat the 18pc as a lower limit to the &lt;em&gt;expectation value&lt;/em&gt;, although of course it is possible (though not probable) that a closer one could exist.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-02-28T19:26:18.507" LastActivityDate="2018-02-28T19:26:18.507" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14288" PostTypeId="2" ParentId="14236" CreationDate="2016-03-24T14:21:42.587" Score="1" Body="&lt;p&gt;It is highly likely that you might have observed an artificial satellite, most commonly of Iridium series which can flare as bright as -6 magnitude. You can have a check on it anytime. &lt;a href=&quot;http://www.heavens-above.com/&quot; rel=&quot;nofollow&quot;&gt;Heavens Above&lt;/a&gt; is one of the most trusted sources that can notify you depending on your location. Even NASA is having such service that can notify you about ISS passes. A dozen of mobile apps are there for assistance too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think it was visible for a short time like meteors are, then you are free to report it here : &lt;a href=&quot;http://www.amsmeteors.org/fireballs/fireball-report/&quot; rel=&quot;nofollow&quot;&gt;AMS Fireball Report&lt;/a&gt; or here : &lt;a href=&quot;http://fireballs.imo.net/members/imo/report_intro&quot; rel=&quot;nofollow&quot;&gt;IMO Fireball Report&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't feel embarrassed if you might have mistaken a jet to that of a moving star. We all have faced that in our initial days of sky observation. Hope this helps  &lt;/p&gt;&#xA;" OwnerUserId="11317" LastActivityDate="2016-03-24T14:21:42.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14289" PostTypeId="1" AcceptedAnswerId="14661" CreationDate="2016-03-24T15:32:48.803" Score="3" ViewCount="272" Body="&lt;p&gt;I live in north India. About 24° latitude. I can see a very Bright star in sky. Its towards east and high up. Is it Venus , Jupiter, star or anything else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its not Sirius. I used Orion belt for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Date: 24 March , 2016&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time : 8 :00 p.m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 11: 00 pm.&#xA;Its directly upwards. Angle b/w moon and sun ~60°.&lt;/p&gt;&#xA;" OwnerUserId="11322" LastEditorUserId="11322" LastEditDate="2016-03-24T17:32:54.103" LastActivityDate="2016-04-25T03:40:01.553" Title="Which Planet/Star is it?" Tags="&lt;planet&gt;&lt;star-gazing&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14290" PostTypeId="1" CreationDate="2016-03-24T17:12:10.937" Score="1" ViewCount="158" Body="&lt;p&gt;Does the orbit of the sun (and other stars within the Milky way) around the galaxy result directly from radial gravitational force toward the objects in the center of the galaxy, or is there additional forces at play? &lt;/p&gt;&#xA;" OwnerUserId="11197" LastActivityDate="2016-03-24T19:48:44.603" Title="Is the motion of the Sun around the Galaxy a result of gravitational pull?" Tags="&lt;galaxy&gt;&lt;galactic-dynamics&gt;&lt;celestial-mechanics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14292" PostTypeId="2" ParentId="14289" CreationDate="2016-03-24T17:29:21.090" Score="1" Body="&lt;p&gt;Thank you all!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its not Sirius, &#xA;I found Sirius a bit far in south west using orion belt in west. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its not Venus, just because of timing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its not flickering and is brightest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, it ought to be Jupiter.&lt;/p&gt;&#xA;" OwnerUserId="11322" LastActivityDate="2016-03-24T17:29:21.090" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14293" PostTypeId="2" ParentId="14290" CreationDate="2016-03-24T17:30:07.147" Score="1" Body="&lt;p&gt;Stars in the Milky Way orbit the Milky Way because of all of the mass that comprises the Milky Way. That other stuff includes stars, other ordinary matter such as gas clouds, and perhaps dark matter. I used &quot;perhaps&quot; because dark matter has not yet been confirmed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General relativity does not explain the orbits of stars about a galaxy with the assumption that the only form of mass is ordinary matter. This leads to two possibilities: Either general relativity (and by extension, Newtonian gravity) is somehow wrong at galactic scales, or mass comes in flavors other than ordinary matter. The former leads to alternative formulations of gravitation such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Modified_Newtonian_dynamics&quot; rel=&quot;nofollow&quot;&gt;MOND&lt;/a&gt;. The latter leads to alternative forms of mass such as dark matter. Even though the existence of dark matter is yet to be confirmed, the vast majority of astrophysicists lean toward the latter assumption, that ordinary matter is not the only form of matter.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2016-03-24T17:30:07.147" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="14294" PostTypeId="2" ParentId="14290" CreationDate="2016-03-24T19:48:44.603" Score="2" Body="&lt;p&gt;I would say the answer to your question is yes and no. Yes that the orbit of our Sun in the Milky Way is due solely to gravitational forces as you suggest, but no in that they are not all radial towards the center of the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motion of the Sun around the center of our galaxy can be broken into two motions. The first is an &quot;azimuthal&quot; orbital motion around the center of the galaxy due to radial gravitational forces from mass interior to our orbit (be it stars, gas/dust, dark matter, dragons, etc.). The other type of motion is an oscillation into and out of the plane of the milky way. This is also caused by gravitational forces, albeit not radial forces from the center of the milky way. As the sun moves &quot;above&quot; the plane of the Milky Way, there is more mass below than above and it gets pulled &quot;down&quot;. When it is in the plane it has residual energy from &quot;falling down&quot; and continues to fall &quot;below&quot; the plane where it then gets pulled back &quot;up&quot;. This type of oscillatory motion is constant and analogous to a mass on a spring.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-03-24T19:48:44.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14295" PostTypeId="2" ParentId="14289" CreationDate="2016-03-24T20:06:30.300" Score="0" Body="&lt;p&gt;My guess is that it would be the star &lt;a href=&quot;https://en.wikipedia.org/wiki/Altair&quot; rel=&quot;nofollow&quot;&gt;Altair&lt;/a&gt;. That is the brightest star in the region of the sky you indicate at your time and location. Altair is one of the brightest stars in the sky and highly visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to try and confirm yourself, try checking out an interactive planetarium &lt;a href=&quot;http://neave.com/planetarium/&quot; rel=&quot;nofollow&quot;&gt;like this one&lt;/a&gt;. Put in your location, set to the correct time, and check out what the sky looks like.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-03-24T20:06:30.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14296" PostTypeId="1" AcceptedAnswerId="14300" CreationDate="2016-03-24T21:10:40.127" Score="1" ViewCount="133" Body="&lt;p&gt;After the Big Bang, dark matter clusters and build the seeds for visible matter, that then forms stars, planet, you, me, everybody...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But do we really need dark matter to build cluster first. What if spacetime isn't flat per se on cosmlogical sclaes but rather crumbled or heterogenous, like a 4-dimensional version of folded cloth. We don't need dark matter, if we accept that space time didn't came along at the big bang in a clean way, as mathematical text books on general relavity suggest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to support this idea with astronomical data of galaxy mergers?&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2017-10-04T13:50:37.877" Title="Initially non-flat space-time makes dark matter obsolete$\dots$" Tags="&lt;space-time&gt;&lt;dark-matter&gt;&lt;galaxy-cluster&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14297" PostTypeId="1" AcceptedAnswerId="14299" CreationDate="2016-03-24T22:17:05.733" Score="2" ViewCount="71" Body="&lt;p&gt;If not, would it be possible to approximate the maximum heat energy a star could lose before the change became irreversible?&lt;/p&gt;&#xA;" OwnerUserId="11325" LastEditorUserId="11325" LastEditDate="2016-03-24T22:35:12.223" LastActivityDate="2016-03-24T23:59:19.610" Title="If a star were to suddenly lose nearly all of its stored heat, would it be able to return to its normal state?" Tags="&lt;star&gt;&lt;stellar-evolution&gt;&lt;star-formation&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2016-03-26T03:35:03.560" ContentLicense="CC BY-SA 3.0" />
  <row Id="14298" PostTypeId="2" ParentId="14289" CreationDate="2016-03-24T23:43:21.033" Score="0" Body="&lt;p&gt;Try this link &lt;a href=&quot;http://www.ucolick.org/~bolte/AY4_00/weblab/project/chart.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ucolick.org/~bolte/AY4_00/weblab/project/chart.html&lt;/a&gt; . It's a good one and it's free. Do you see a triangle of bright stars looking up? If so they are probably Altair, Vega and Arcturus. You need to use longitude also. I guessed 80 deg. East. Here are the apparent brightnesses in magnitudes of the 3 stars above east to west: Altair +0.76, Vega +0.026 and Arcturus -0.05. Vega is slightly north of the other two. Therefore, I think your star is the brightest star in the northern hemisphere, Arcturus.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="7997" LastEditDate="2016-03-25T00:01:34.610" LastActivityDate="2016-03-25T00:01:34.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14299" PostTypeId="2" ParentId="14297" CreationDate="2016-03-24T23:59:19.610" Score="3" Body="&lt;p&gt;If you suddenly sucked all the heat out of a normal star like the Sun, nuclear reactions would cease, the gas pressure would fall to zero. The star would collapse, initially on a freefall timescale (about 1 hour for the Sun).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The released gravitational potential energy would half be radiated away and half would go into re-heating the interior. The temperature would rapidly rise, the collapse would be halted, nuclear reactions would restart, the star would slowly expand again and a thermal and hydrostatic equilibrium should be restored on a Kelvin-Helmholtz timescale (about 10 million years for the Sun).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to see this, is to compare the current thermal content of a star with its capacity for generating new heat by nuclear reactions. e.g. approximate the Sun as a uniform ball of ideal gas at an average temperature of a few million degrees. The thermal content is roughly $10^{41}$ J. The Sun generates this much energy in nuclear reactions in only 8 million years - coinciding with the timescale to reestablish an equilibrium. Or you could compare it with the gravitational potential energy of around $2\times 10^{41}$ J. i.e. There is enough gravitational potential energy to reheat the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, not everything would be a continuation. The collapse and reheating would be similar to the star repeating its pre main sequence phase. During this phase the star becomes fully convective and so the core material would get completely mixed with the rest of the star again. This would have the effect of rejuvenating the star as a new zero age main sequence star, but with a slightly higher helium abundance. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-03-24T23:59:19.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14300" PostTypeId="2" ParentId="14296" CreationDate="2016-03-25T02:59:51.157" Score="2" Body="&lt;p&gt;I think there a few problems with this idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, as Stan Liou indicates, the early universe expanded extremely rapidly, potentially through hyper-inflation during the inflationary epoch. While we're still looking for observational evidence that this period occurred (e.g., the failed BICEP2 results a few years ago), the existence of such an epoch early in the universe neatly explains away several problems with the standard Big Bang theory (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Inflation_%28cosmology%29#Motivations&quot; rel=&quot;nofollow noreferrer&quot;&gt;the horizon problem, monopole problem, and the flatness problem&lt;/a&gt;). Assuming such a period existed, then any region of space-time that was &quot;crumpled&quot; would have been quickly and cleanly smoothed out during the hyper-inflation. Such crumpled-ness would not have been around long enough to affect cluster growth and structure. What's more, we very clearly measure that the Universe is flat over large, cosmic scales so it seems unlikely that there are constant and pervasive wrinkles as you suggest there could be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second issue I potentially see with your proposal is that it would imply a top-down approach to galaxy formation. That is, first you get huge clusters of mass (conglomerated by your &quot;crumpled&quot; space-time), then you get fragmentation and individual galaxies can form, then galaxies fragment into individual stars, etc. In other words, big structures form first, then smaller structures derive from the bigger structures. The opposite formation model would be &quot;bottom-up&quot; formation where first small things clump, then those clump into larger bodies and soon you have galaxies clumping into clusters. I don't think there is definitive evidence that galaxy formation is either top-down or bottom-up, but simulations seem to me to suggest bottom-up is more likely (of course those simulations use dark matter). I'll leave you to think about that one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last point of contention I have is that dark matter plays more of a role than just providing extra gravity. It contributes to the total mass-energy of the universe. Current theoretical models perform extremely well when we include dark matter. I believe a strong piece of evidence for this fact is our ability to theoretically match the &lt;a href=&quot;http://www.nature.com/nature/journal/v440/n7088/fig_tab/nature04803_F2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;CMB power spectrum&lt;/a&gt; using a Lambda-CDM model. I don't see that crumpled space-time could reproduce this result nearly as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you'd also have a very hard time describing a mechanism that could allow &quot;crumpling&quot; to exist in the space-time of our universe which exactly matches the observed gravitational effects in galaxies. We can easily see that they appear to have a halo of matter which follows something akin to an &lt;a href=&quot;https://en.wikipedia.org/wiki/Navarro%E2%80%93Frenk%E2%80%93White_profile&quot; rel=&quot;nofollow noreferrer&quot;&gt;NFW halo&lt;/a&gt; or else an &lt;a href=&quot;https://en.wikipedia.org/wiki/Einasto_profile&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einasto profile&lt;/a&gt;. If you can find away that crumpled space-time would naturally form to match these profiles and somehow not be smoothed out by hyper-inflation, I'm all ears.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2017-10-04T13:50:37.877" LastActivityDate="2017-10-04T13:50:37.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14301" PostTypeId="2" ParentId="12385" CreationDate="2016-03-25T04:13:58.417" Score="1" Body="&lt;p&gt;See this Physics Stack Exchange Question/ Answer: &lt;a href=&quot;https://physics.stackexchange.com/questions/141839/why-heavy-elements-dont-sink-to-the-core&quot;&gt;https://physics.stackexchange.com/questions/141839/why-heavy-elements-dont-sink-to-the-core&lt;/a&gt; . You don't need volcanics nor meteorites to keep many heavy elements in the crust (and mineable). Gold is one of the elements that bonds to iron, however, and sinks, which is why it is rare and valuable.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-03-25T04:13:58.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14302" PostTypeId="1" CreationDate="2016-03-25T08:06:24.990" Score="1" ViewCount="67" Body="&lt;p&gt;It was predicted from Newtonian physics &lt;a href=&quot;http://cosmictimes.gsfc.nasa.gov/teachers/guide/1919/guide/gravity_bends_starlight.html&quot; rel=&quot;nofollow&quot;&gt;already in the 18th century&lt;/a&gt; that gravity should bend light, but not by as much as general relativity predicts. This was first confirmed during a solar eclipse in 1919. Surely, solar eclipses were imaged long before that. Didn't anyone notice that stars near the Sun were displaced? Hadn't anyone thought of testing the newtonian prediction of bent light?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-03-25T11:57:03.877" Title="Could the Sun's bending of light be measured on photographic plates before Einstein's prediction?" Tags="&lt;space-time&gt;&lt;general-relativity&gt;&lt;history&gt;&lt;photography&gt;&lt;newtonian-gravity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14304" PostTypeId="1" AcceptedAnswerId="18817" CreationDate="2016-03-25T08:21:53.457" Score="6" ViewCount="428" Body="&lt;p&gt;24 minutes into &lt;a href=&quot;http://www.ustream.tv/recorded/84853974&quot; rel=&quot;noreferrer&quot;&gt;this van Karman lecture&lt;/a&gt; by Dr. Tamppari, the slide below is shown with symbols for transient events on Mars. Red dots represent meteoroid impacts which have occurred while MRO has been in Mars' orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;New impacts are almost only found at lower latitudes. And longitudinally there are two equally sized gaps at the equator. Is this because of some observational bias or is there a natural explanation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5Rsfm.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5Rsfm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-10-21T18:20:18.417" Title="What explains the distribution of new impact craters on Mars?" Tags="&lt;mars&gt;&lt;impact&gt;&lt;crater&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14306" PostTypeId="2" ParentId="755" CreationDate="2016-03-25T11:38:02.350" Score="1" Body="&lt;p&gt;Our sun takes about 225-250 million years to complete one galactic year (also called a cosmic year),so from the age of the sun it has completed approximately &lt;strong&gt;19 galactic years&lt;/strong&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="11283" LastActivityDate="2016-03-25T11:38:02.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14307" PostTypeId="2" ParentId="14302" CreationDate="2016-03-25T11:57:03.877" Score="2" Body="&lt;p&gt;I suggest reading the &lt;a href=&quot;http://rsta.royalsocietypublishing.org/content/roypta/220/571-581/291.full.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; on the 1919 expedition to get a clearer picture of why they did it at that time and why it hadn't been done before. From reading chapter 2 I think the main reason was the astrophotography equipment required for the experiment and the alignment of bright enough stars close to the sun to observe the effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course before Einsteins prediction in 1911 no one had any specific reason to observe the stars close to a solar eclipse before and during totality, it might seem obvious now but that's hindsight for you.&lt;/p&gt;&#xA;" OwnerUserId="10437" LastActivityDate="2016-03-25T11:57:03.877" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14309" PostTypeId="1" CreationDate="2016-03-25T19:09:09.990" Score="2" ViewCount="126" Body="&lt;p&gt;I'm creating a C++ program that calculates the ecliptic coordinates based on the formulae from &lt;a href=&quot;https://en.wikipedia.org/wiki/Position_of_the_Sun&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; But, my calculations appear off. The mean anomaly for today, for example, should be 80.4-something; both my program and Google calculate approx. 79.2 - the result of (357.528 + 0.9856003*5927) % 360 (where 5927 is the number of days since January 1, 2000, GMT) &lt;/p&gt;&#xA;" OwnerUserId="11332" LastEditorUserId="11332" LastEditDate="2016-03-25T19:17:56.963" LastActivityDate="2016-03-30T15:42:57.140" Title="Are Wikipedia's sun ecliptic-coordinate formulae accurate?" Tags="&lt;the-sun&gt;&lt;positional-astronomy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14310" PostTypeId="1" AcceptedAnswerId="14657" CreationDate="2016-03-25T21:08:09.563" Score="0" ViewCount="1564" Body="&lt;p&gt;I have a latitude / longitude / time of a location on Earth, I need to calculate the Right Ascension / Declination that's overhead at the time provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I've found out that Right Ascension is the same as the sidereal time for the location but I haven't found any information about declination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to set my overhead viewing location in the web portal of World Wide Telescope.  It defaults to &quot;now&quot; but I want to show 10PM and there's no way (that I've found) to set time through the javascript interface.  I can however set Right Ascension and Declination, hence my question.&lt;/p&gt;&#xA;" OwnerUserId="11333" LastActivityDate="2016-04-24T13:20:10.567" Title="Calculating Right Ascension and Declination from Latitude / Longitude / Time" Tags="&lt;declination&gt;&lt;ascension&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14311" PostTypeId="1" CreationDate="2016-03-26T01:14:33.267" Score="0" ViewCount="485" Body="&lt;p&gt;Scientists used to use following reasoning:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;most galaxies are red-shifted&lt;/em&gt; $\implies$ &lt;em&gt;there was a 'Big Bang'&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why this is being considered valid since not every galaxy has such property?&lt;/p&gt;&#xA;" OwnerUserId="11085" LastEditorUserId="11085" LastEditDate="2016-03-26T01:39:05.363" LastActivityDate="2016-03-27T02:18:02.423" Title="Why Milky Way and Andromeda are being drawn together if there was 'Big Bang'?" Tags="&lt;big-bang-theory&gt;&lt;redshift&gt;&lt;galactic-dynamics&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14312" PostTypeId="2" ParentId="14311" CreationDate="2016-03-26T07:27:20.133" Score="6" Body="&lt;p&gt;The evidence for expansion is that the redshift is proportional to distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The redshift of a galaxy can be divided into two components: that due to the cosmological expansion, which stretches the wavelength of light whilst it travels towards us; and a peculiar motion with respect to the cosmological expansion, which causes a straightforward doppler shift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The former term increases with distance - this is known as Hubble's law. The redshift here is always positive (ie always a stretching of wavelength). It is this that tells us the universe is expanding. The latter term is caused by the gravitational effects of other nearby galaxies and clusters of galaxies on the galaxy in question. It is typically of order a few hundred km/s and can be positive or negative (ie it can cause a redshift &lt;em&gt;or&lt;/em&gt; blueshift).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Redshift due to cosmological expansion completely dominates at distances greater than a few hundred million light years. Up till then (and Andromeda is only 2 million light years distant), peculiar motions can result in blueshifts for some galaxies.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-03-26T07:27:20.133" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14313" PostTypeId="1" CreationDate="2016-03-26T08:50:11.360" Score="0" ViewCount="135" Body="&lt;p&gt;By total eclipse I mean the Moon and Sun are seen equal in size, neither of them is bigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the Moon's average distance is 384000 km and the Sun's average distance is 149.5 million km. But when this distances occurs, are the Moon and Sun seen equal in size? (from Earth of course)&lt;/p&gt;&#xA;" OwnerUserId="11336" LastEditorUserId="2365" LastEditDate="2016-03-26T22:51:00.350" LastActivityDate="2016-03-28T15:14:56.280" Title="At the moment of total solar eclipse what are exact distances of Moon &amp; Sun to the Earth" Tags="&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14314" PostTypeId="1" CreationDate="2016-03-26T21:14:02.837" Score="2" ViewCount="124" Body="&lt;p&gt;We know the existence of dark matter because we can test its gravitational mass (e.g. in gravitational lens) but, since we cannot see this matter, how we can be sure that it has an inertial mass, and it is the same as the gravitational mass? In other words, does the principle of equivalence apply also to dark matter?&lt;/p&gt;&#xA;" OwnerUserId="7741" LastEditorUserId="2365" LastEditDate="2016-03-26T22:33:14.540" LastActivityDate="2016-03-26T22:47:33.923" Title="Dark matter inertial mass" Tags="&lt;dark-matter&gt;&lt;mass&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14315" PostTypeId="2" ParentId="14314" CreationDate="2016-03-26T22:47:33.923" Score="3" Body="&lt;p&gt;If the inertial mass is not equal to the gravitational mass, it would be equivalent to the gravitational constant, G, being different, or perhaps non-constant, for dark matter. If that were so, the manner in which dark matter would orbit the milky way would be different, leading to a different distribution of dark matter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://web.mit.edu/redingtn/www/netadv/specr/012/node7.html&quot; rel=&quot;nofollow&quot;&gt;We know from the rotation curve of the milky way roughly how dark matter is distributed in the milky way&lt;/a&gt;. Our models are consistent with the inertial mass of dark matter being equal to the gravitational mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is, of course, far from &quot;proof&quot;. As we don't even know what dark matter is, it is impossible to be certain of any of its properties. And the equivalence of inertial and graviational mass is not &quot;proved&quot; even for normal matter (but no experiment has ever detected a difference) But it would be exceedingly surprising if gravity affected dark matter differently, and there would need to be strong evidence for that. As it stands, there is none.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-03-26T22:47:33.923" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14316" PostTypeId="2" ParentId="14313" CreationDate="2016-03-26T23:15:09.613" Score="4" Body="&lt;p&gt;First off, your definition of &quot;total solar eclipse&quot; is different from that generally used. A total solar eclipse occurs when the moon is directly in front of the sun and &lt;em&gt;is bigger&lt;/em&gt; so it blocks the whole of the photosphere. Also note that the sun and moon are both somewhat oblate (not perfect spheres, but flattened) and not exactly the same shape, but for the sake of estimation I'll use the diameter information from wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon has an average diameter of 3,474 km, and the sun has an average diameter of 1.392684 million km. So the Sun is 400.9 times larger than the moon. When the sun is 149600000 km distant, then if the moon is $149600000/400.9 = 373200$km, the sun and the moon would appear equal in size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the moon is 384000km from the centre of the Earth, the distance to the Earth's surface is smallest at the sub-lunar point, about 378000km, but still greater than 373200, so when the Sun and Moon are both at their average distance, an annular eclipse occurs. Annular eclipses are slightly more common than total eclipses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if the distance from an observer to the moon is closer than 1/400.9 of the distance to the sun, then a total eclipse is possible, and this allows you to generate infinitely many solutions to the question &quot;How far is the moon during a total solar eclipse&quot;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-03-28T15:14:56.280" LastActivityDate="2016-03-28T15:14:56.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14317" PostTypeId="2" ParentId="14309" CreationDate="2016-03-26T23:33:44.590" Score="0" Body="&lt;p&gt;The formulae on wikipedia are correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you haven't explained where you have &quot;80.4&quot; from, this answer is a &quot;best effort&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value you have calculated is close, so there may be a small error in your code. My guess is that you haven't correctly calculated the time as a Julian Day Number. The Julian Day changes at noon UT (probably as this means that the Julian Day doesn't change during the European night) Whereas most other systems of dates start and end at midnight. This can easily introduce a 12 hour error into your calculations. When you say &quot;number of days since Jan 1 2000, are you starting your counting at noon or midnight?&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-03-26T23:33:44.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14318" PostTypeId="2" ParentId="14311" CreationDate="2016-03-27T02:18:02.423" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Scientists used to use following reasoning:&lt;br&gt;&#xA;  &lt;em&gt;most galaxies are red-shifted&lt;/em&gt; $\implies$ &lt;em&gt;there was a 'Big Bang'&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is an oversimplification, and a rather large one at that. There are multiple lines of evidence that all lead to the concept of the Big Bang, one of which is that galaxies tend to have a redshift that is strongly proportional to distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the correlation between distance and redshift is very strong, it is not perfect. Nearby galaxies do not exhibit this behavior. The reason is simple. That redshift is proportional to distance means that nearby galaxies would have a very small redshift due to recession. (Ordinary) gravitation can easily overwhelm this tiny recessional velocity, and that is exactly what happens.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2016-03-27T02:18:02.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14319" PostTypeId="1" AcceptedAnswerId="14326" CreationDate="2016-03-27T13:27:32.133" Score="34" ViewCount="7413" Body="&lt;p&gt;A couple days ago, I zoomed in with my 30x optical zoom camera, and after some exposure adjustments, a bright star in the night sky turned into this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LDHQl.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LDHQl.jpg&quot; alt=&quot;Bright star with (planets)?&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are those other planets or other stars?  Or is that a lens effect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The bright object in question was ~60 degrees above the horizon, and ESE of me (East-south-east). I took the picture on 3-25-16 from Madison, Wisconsin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Question answered, more clear picture added FYI.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gt7bj.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gt7bj.jpg&quot; alt=&quot;More clear pic - Enjoy!&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11343" LastEditorUserId="5278" LastEditDate="2016-12-13T19:40:16.243" LastActivityDate="2016-12-13T19:40:16.243" Title="Did I see another planet?" Tags="&lt;planet&gt;&lt;amateur-observing&gt;" AnswerCount="3" CommentCount="11" FavoriteCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="14320" PostTypeId="2" ParentId="14319" CreationDate="2016-03-27T15:59:35.783" Score="35" Body="&lt;p&gt;That should be Jupiter and his 4 Galilean moons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are usually very well visible even with very cheap equipment and a nice experience for amateur astronomy.&#xA;On your picture 2 of them seem missing, maybe they were cut-off by the field-of-view, or possibly as a commenter pointed out, they might be behind the planet.&lt;br&gt;&#xA;You can test that notion actually, as Io runs around Jupiter pretty fast and can show significant movement relative to the planet on a timespan of ~2-3 hours (a typical observing session). So if after 2 hours another 'diamond' appears behind Jupiter, you can be pretty sure it to be the volcanic moon Io.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditDate="2016-03-27T17:32:45.113" LastActivityDate="2016-03-27T17:32:45.113" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14321" PostTypeId="2" ParentId="14319" CreationDate="2016-03-27T16:31:37.370" Score="21" Body="&lt;p&gt;This appears to be Jupiter and two of its four &quot;Galilean&quot; moons, being the four discovered by Galileo with his telescope in 1610. I searched with Wolfram Alpha (&lt;a href=&quot;http://www.wolframalpha.com/input/?i=jupiter+moon+configuration+march+25th+2016+9pm+US+central+time&quot; rel=&quot;noreferrer&quot;&gt;http://www.wolframalpha.com/input/?i=jupiter+moon+configuration+march+25th+2016+9pm+US+central+time&lt;/a&gt;) to try and determine which moons you were looking at, and the answer was quite interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Callisto was too close to Jupiter (as viewed from Earth) to be seen in your image. The outermost object is definitely Ganymede, and the other object appears to be a combination of Europa and Io, as these were very close (again, as viewed from Earth), so much so that they appear as one object in the image. Hope this helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qYdRo.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qYdRo.jpg&quot; alt=&quot;Screenshot of WolframAlpha results&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11347" LastEditorUserId="11347" LastEditDate="2016-12-02T14:31:27.483" LastActivityDate="2016-12-02T14:31:27.483" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14322" PostTypeId="2" ParentId="13895" CreationDate="2016-03-27T18:47:10.280" Score="0" Body="&lt;p&gt;Not every scientist agrees that information is &quot;encoded&quot; on the surface of a black hole.  Many scientists believe black holes actually destroy information. In fact, Stephen Hawking and Kip Thorne made &lt;a href=&quot;https://en.wikipedia.org/wiki/Thorne%E2%80%93Hawking%E2%80%93Preskill_bet&quot; rel=&quot;nofollow&quot;&gt;a famous wager&lt;/a&gt; against John Preskill about whether information is destroyed by black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest (and in my opinion, most likely) answer to your question is that black hole event horizons don't encode any information at all. The black destroys it. Which is why we say &quot;black holes have no hair&quot;. Once you make a black hole out of any material, you can no longer tell what went into it. If you make one of photons or neutrinos or neutrons or whatever, all you know after the black hole forms is the amount of mass/energy it contains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when two black holes collide, their event horizons still contain zero information. Zero from the first black hole plus zero from the second.&lt;/p&gt;&#xA;" OwnerUserId="11175" LastActivityDate="2016-03-27T18:47:10.280" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14323" PostTypeId="2" ParentId="12178" CreationDate="2016-03-27T18:56:59.490" Score="10" Body="&lt;p&gt;Mars definitely has far less Iron than Earth. &lt;a href=&quot;https://en.wikipedia.org/wiki/Mars&quot; rel=&quot;noreferrer&quot;&gt;Mars has 10.7% of Earth's mass&lt;/a&gt;. On the other hand, &lt;a href=&quot;https://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements#Abundance_of_elements_in_the_Earth&quot; rel=&quot;noreferrer&quot;&gt;Iron comprises 32% of Earth&lt;/a&gt; since there is so much Iron in its inner core, outer core, and the mantle. That means if Mars was made entirely of Iron (which it is not), Earth would still have more than 3 times as much Iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are right that Mars has more Iron at its surface than Earth so it looks very red. Earth has many other metals on its surface (not to mention water to make it look blue and plants to make it look green).&lt;/p&gt;&#xA;" OwnerUserId="11175" LastEditorUserId="11175" LastEditDate="2017-05-06T19:01:25.153" LastActivityDate="2017-05-06T19:01:25.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14324" PostTypeId="2" ParentId="13791" CreationDate="2016-03-27T19:01:59.487" Score="1" Body="&lt;p&gt;I doubt two black holes could form within a single star, but they might form from a &lt;a href=&quot;https://en.wikipedia.org/wiki/Contact_binary&quot; rel=&quot;nofollow&quot;&gt;contact binary&lt;/a&gt;. That's a pair of stars so close that they touch and share a single atmosphere. For the stars that made the black holes for the LIGO gravitational wave discovery, just consider two very massive stars that are touching or nearly touching. They would share a single gaseous envelope for an atmosphere. Each one separately formed a black hole when it died, and then their two black holes merged.&lt;/p&gt;&#xA;" OwnerUserId="11175" LastActivityDate="2016-03-27T19:01:59.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14325" PostTypeId="1" AcceptedAnswerId="14337" CreationDate="2016-03-27T19:33:30.190" Score="2" ViewCount="83" Body="&lt;p&gt;I'm currently trying to collect the data to run an N-body simulation for 11 of the galaxies in the Local Group where proper motions are known, however I don't understand how to get the required line of sight velocities for the simulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have redshift data from NED for the required galaxies, however by judging what sources tell me, this won't be the required information for the simulation due to our motion around the Milky way's galactic centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, one source has given the NED redshift as 229 km/s for Carina Dwarf, but the corrected line of sight velocity is 14.3 km/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know how to convert these redshifts into the desired corrected versions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!!&lt;/p&gt;&#xA;" OwnerUserId="11127" LastEditorUserId="11127" LastEditDate="2016-03-27T21:48:30.857" LastActivityDate="2016-03-28T13:26:41.837" Title="How do I modify redshifts to gain corrected line of sight velocities?" Tags="&lt;galaxy&gt;&lt;redshift&gt;&lt;galaxy-cluster&gt;&lt;velocity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14326" PostTypeId="2" ParentId="14319" CreationDate="2016-03-27T19:43:05.467" Score="58" Body="&lt;p&gt;You don't say what time you were looking. Here is a screenshot from Stellarium at 10pm Wisconsin time on 25th March 2016. Jupiter is in the ESE, but the altitude is a bit lower than 60 degrees. Seems fairly conclusive. You were seeing Ganymede and a Europa/Io combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J9DNs.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J9DNs.jpg&quot; alt=&quot;Jupiter&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-03-27T19:43:05.467" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14327" PostTypeId="2" ParentId="12178" CreationDate="2016-03-27T22:57:15.960" Score="5" Body="&lt;p&gt;Unless 100+ years of studying solar system formation didn't get us anywhere then you'd actually predict that mars would have less iron than earth (in relative terms). This is because the further out in the solar system you go, the lighter the materials that make up the celestial bodies become. You start off with the inner planets for example, which are made of rock and metal. At its most extreme is mercury, which has a metallic core that makes up the vast majority of its volume. As you start to move to the outer parts, there's much less metal. Even further out rocky material starts to peter until you're just left with frozen ices and covalent compounds. This pattern exists because when the solar system formed the heavier materials condensed near the centre while the lighter gases were blown away to the edges. Mars, being the last terrestrial planet, should have formed in an orbit that had less iron and heavy materials than earth and all the other rocky bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, its good to bet that on average there is more relative iron on mars surface than there is on earth. This would be because despite it's lower abundance of iron overall, it's much more spread out among the lithosphere rather than being entirely concentrated deep in its core like on earth. Mars being smaller planet might not have ever built up enough heat during its formation to completely liquefy its surface, which a planet needs in order for the heavier materials to sink to the centre, making differentiated layers such as the ones we see on earth like the crust, core and mantle etc. Don't get me wrong, mars does have these differentiated layers, but they probably aren't as pronounced as they are here on earth. This is what I think we can safely say is thus the reason for mar's red surface, despite it's location in the solar system.&lt;/p&gt;&#xA;" OwnerUserId="11247" LastActivityDate="2016-03-27T22:57:15.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14328" PostTypeId="1" CreationDate="2016-03-28T02:29:27.153" Score="3" ViewCount="130" Body="&lt;p&gt;During the collapse of a star at the commencement of supernovae, the intense pressures force fusion and fission to create the heavier elements and release vast amounts of energy. As this energy/mass is released - the gravitational field must also decrease in an instant, and as this &quot;Warping of space time&quot; is instantly released and returns to flat space (Or flatter space) - does this create or contribute to the explosion of the supernovae itself? I have always been puzzled by the concept of the collapsed outer layers &quot;Bouncing off of an in-compressible core&quot; as the source of the explosion. Isn't this instantaneous &quot;Relaxation of the curvature of the space&quot; not what we see as the explosion?&lt;/p&gt;&#xA;" OwnerUserId="11310" LastActivityDate="2016-03-28T13:10:06.490" Title="Gravitational field released during supernovae" Tags="&lt;general-relativity&gt;&lt;supernova&gt;&lt;gravitational-waves&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14331" PostTypeId="2" ParentId="8435" CreationDate="2016-03-28T08:36:59.210" Score="-3" Body="&lt;p&gt;If the Earth's axis is always pointing toward Polaris then the Earth's axis has to continually change throughout year over a 186,000,000 mile orbit distance change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the tilt in December is 23.4 degrees then, March would be less tilt and an added tilt toward the sun, June would be the smallest Earth tilt, October would be less tilt and and added tilt toward the sun. Without this Earth tilt change Polaris would only be aligned one day a year in December.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can easily prove this yourself.&lt;/p&gt;&#xA;" OwnerUserId="11357" LastEditorUserId="2365" LastEditDate="2016-03-28T15:07:43.543" LastActivityDate="2016-03-28T15:07:43.543" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14332" PostTypeId="1" AcceptedAnswerId="14334" CreationDate="2016-03-28T09:12:29.117" Score="1" ViewCount="129" Body="&lt;p&gt;NASA &lt;a href=&quot;http://www.nasa.gov/feature/jpl/comet-flying-by-earth-observed-with-radar-and-infrared&quot; rel=&quot;nofollow noreferrer&quot;&gt;published a picture&lt;/a&gt; of a comet the other day. The image shows the comet being lit from above. See NASA's picture below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3LoGD.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3LoGD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since it is a radar image, I would have assumed to get a shading on all sides and grazing angles, like velvet or scanning electron microscopy. Or alternatively to have the sides facing the radar dish being shaded, and the edges being dim. For the velvety look, compare to this picture from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LK9xg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LK9xg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why is the comet's image shaded only from above?&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2016-03-28T11:06:19.790" Title="Why are radar images of comets shaded only on one side?" Tags="&lt;comets&gt;&lt;photography&gt;&lt;radar&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14333" PostTypeId="2" ParentId="14328" CreationDate="2016-03-28T09:29:15.060" Score="1" Body="&lt;p&gt;As a star collapses, there is a massive release of energy. A supernova releases $10^{44}$ Joules, equivalent to a mass of $10^{27}kg$ which is a lot. But compared to the mass of the sun ($2\times10^{30}$kg) it is not so much: 0.0005 solar masses. So there is a change in the gravitational field, but this is not the cause or a major contribution to the supernova. The field doesn't change &quot;in an instant&quot; Most of the energy is carried away by neutrinos at just under the speed of light, and changes to the gravitational fields propagate at the speed of light (there is some gravitational radiation emitted too)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supernovae are powered principally by the release of gravitational potential energy as the core falls in on itself. The explosion we see is only a small fraction of the energy released, most is in the form of neutrinos. Supernovae are so ridiculously energetic that you can't trust your intuition, but computer models are becoming better at &lt;a href=&quot;https://www.youtube.com/watch?v=2RxIwtxdEnQ&quot; rel=&quot;nofollow&quot;&gt;predicting the dynamics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-03-28T09:29:15.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14334" PostTypeId="2" ParentId="14332" CreationDate="2016-03-28T11:06:19.790" Score="4" Body="&lt;p&gt;There is a very good blog article &lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/2011/3248.html?referrer=https://en.wikipedia.org/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; that explains this effect but basically the Doppler shifting of the radar return means that certain parts of the asteroid light up more than others in the observing wavelengths from Earth. Its better explained in the article&gt;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taken from the article &quot;How radio telescopes get &quot;images&quot; of asteroids&quot;&#xA;Posted by Emily Lakdawalla:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As an asteroid rotates, some parts of it are moving toward us, while other parts are moving away. As the broadcast radio wavefronts hit the part of the asteroid that is moving toward us, the asteroid smacks into each wavefront faster than it would if it were not rotating. The speed of the wavefronts does not change, because the speed of light is constant, so the wavefronts end up being packed closer together. This is a Doppler shift. The asteroid has taken the broadcast wavelength and reflected it at a shorter wavelength from the parts of the asteroid that are rotating toward us. On the other side of the asteroid, which is rotating away, the opposite thing happens; each arriving wavefront smacks into the asteroid a little later than it would if the asteroid were not rotating, so the reflected waves are spread farther apart.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/2011/3248.html?referrer=https://en.wikipedia.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.planetary.org/blogs/emily-lakdawalla/2011/3248.html?referrer=https://en.wikipedia.org/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10437" LastActivityDate="2016-03-28T11:06:19.790" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14335" PostTypeId="2" ParentId="14328" CreationDate="2016-03-28T13:10:06.490" Score="3" Body="&lt;p&gt;What you are missing is that the &lt;a href=&quot;https://en.wikipedia.org/wiki/Shell_theorem&quot; rel=&quot;nofollow&quot;&gt;shell theorem&lt;/a&gt;, which says that for a spherically symmetric mass distribution (we can count the pre- and post-supernova states as this for the sake of argument), that the gravitational field at some distance $d$ from the mass is the same as if all the mass were concentrated at the centre of the distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the gravitational field due to the pre-supernova star is unchanged after the supernova at a distance $d$ until some of the mass (or equivalently in General Relativity, energy) as travelled &lt;em&gt;beyond&lt;/em&gt; a distance $d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a core-collapse supernova, if the pre-supernova star was say $15 M_{\odot}$, then what might happen is that $1.4 M_{\odot}$ is left as a neutron star remnant, whilst $10^{46}$ J of energy is released - mostly in the form of neutrinos travelling at almost the speed of light and the envelope of the star expanding outwards at tens of thousands of km/s and a kinetic energy of about $10^{44}$ J.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The neutrinos carry an equivalent rest mass of a mere $0.05M_{\odot}$, so if you were on a (indestructable) planet in orbit around the supernova, then yes, after the main neutrino pulse has passed there would be a small decrease in the gravitational force felt towards the centre of the planet's orbit (not instantaneously, the neutrino pulse lasts some tens of seconds) that would result in an outward acceleration and  the orbit widening slightly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometime later (the orbital radius divided by the supernova ejecta speed) the main mass loss from the supernova would pass by and this would result in a drastic decrease in the gravitational force and a drastic widening of the orbit.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-03-28T13:10:06.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14336" PostTypeId="1" CreationDate="2016-03-28T13:23:57.063" Score="17" ViewCount="1274" Body="&lt;p&gt;If farthest galaxies run away from us with acceleration making them exceed speed of light, we should expect them to disappear from sky among time with increasing quantity. Did we observe this? Can we indicate next galaxies to eliminate and their time of decline?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question concerns galaxies moving with all speed ranges, not only ones greater than speed of light.&lt;/p&gt;&#xA;" OwnerUserId="11085" LastEditorUserId="9301" LastEditDate="2016-06-29T04:39:08.633" LastActivityDate="2017-09-29T12:10:10.090" Title="Are there any galaxies which fell out of sight horizon due to cosmic expansion?" Tags="&lt;galaxy&gt;&lt;observational-astronomy&gt;&lt;expansion&gt;&lt;observable-universe&gt;&lt;horizon&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14337" PostTypeId="2" ParentId="14325" CreationDate="2016-03-28T13:26:41.837" Score="2" Body="&lt;p&gt;You need a model for the motion of the Sun with respect to the Milky Way centre. You then have to subtract the component of this that is resolved towards the galaxy in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solar motion around the Galaxy is somewhat uncertain, but is roughly (11, 240, 7) km/s when expressed as a vector aligned with Galactic coordinates (i.e. towards the Galactic centre, tangential to the Galactic centre and perpendicular to the Galactic disk). &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-03-28T13:26:41.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14338" PostTypeId="2" ParentId="14336" CreationDate="2016-03-28T15:06:49.253" Score="21" Body="&lt;p&gt;No. In fact the opposite is the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(See the last paragraph for an intuitive explanation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a common misbelief that galaxies receding faster than the speed of light are not visible to us. This is not the case; we easily see galaxies moving at superluminal velocities. This does not — as I think most people would think — contradict the theory of relativity, which says that nothing can travel &lt;em&gt;through&lt;/em&gt; space faster than $c$. Galaxies do not travel &lt;em&gt;through&lt;/em&gt; space (except with &lt;a href=&quot;https://astronomy.stackexchange.com/a/22653/5264&quot;&gt;small velocities of 100-1000 km/s&lt;/a&gt;); rather, space itself is expanding, causing distances between the galaxies to increase.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;We see &quot;super-luminal&quot; galaxies&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The recession velocity $v_\mathrm{rec}$ of a galaxy is given by Hubble's Law:&#xA;$$&#xA;v_\mathrm{rec} = H_0 \, d,&#xA;$$&#xA;where $H_0 \simeq 67.8\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$ is the Hubble constant (&lt;a href=&quot;https://arxiv.org/abs/1502.01589&quot; rel=&quot;noreferrer&quot;&gt;Planck Collaboration et al. 2016&lt;/a&gt;). This law implies that galaxies farther away than&#xA;$$&#xA;r_\mathrm{HS} \equiv \frac{c}{H_0} \simeq 4400\,\mathrm{Mpc} \simeq 14.4 \, \mathrm{Gly}\,\,\mathrm{(&quot;\!\!Giga\mbox{-}lightyears\!\!&quot;)}&#xA;$$&#xA;recede faster than $c$. Here, the subscript &quot;HS&quot; is chosen because the ragion within which galaxies recede slower than $c$ is called the &quot;Hubble sphere&quot;. Objects at a distance of $r_\mathrm{HS}$ have a redshift of $z\simeq1.6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a photon emitted from a distant galaxy (say, &lt;a href=&quot;https://en.wikipedia.org/wiki/GN-z11&quot; rel=&quot;noreferrer&quot;&gt;GN-z11&lt;/a&gt; at redshift $z=11.1$) in the past, in the direction of the Milky Way (MW). What special relativity tells us is that &lt;em&gt;locally&lt;/em&gt;, the photon always travels through space at $v=c$. Initially, the photon thus increases it distance from GN-z11 at velocity $c$. However, &lt;em&gt;even though the photon travels toward us, its distance to MW increases&lt;/em&gt;, due to the expansion of the Universe. As the photon increases its distance to GN-z11, the same expansion causes it to recede from GN-z11 at an ever-increasing velocity. Moreover, as it travels toward MW, it will slowly &quot;overcome&quot; the expansion until it reaches the point where $v_\mathrm{rec} = c$. For an infinitesimally small period, it will stand will wrt. MW, after which it will begin to travel faster and faster as measured from MW. Eventually, its velocity — still in MW's reference frame — will reach $c$, at which point it will have reached MW.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, even though GN-z11 and MW recede from each other at $v_\mathrm{rec} = 2.2c$, we are still able to see it. What is perhaps even more counterintuitive is that when GN-z11 emitted the light we see today, it receded &lt;em&gt;even&lt;/em&gt; faster, at $v_\mathrm{rec} \sim 4c$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;We see more and more distant galaxies&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There is, however, a limit to how fast a galaxy visible to us can recede, given by the distance $r_\mathrm{PH}$ that light has had the time to travel since the Universe was created. Light comes to us from all directions, so we're situated in the center of a sphere of radius $r_\mathrm{PH}$. This sphere is called &quot;the observable Universe&quot;, and its surface (which is not a physical thing) is called the &lt;em&gt;particle horizon&lt;/em&gt; (hence the subscript &quot;PH&quot;). Galaxies at the particle horizon are receding at $v_\mathrm{rec}\simeq3.3c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As time goes by, light from ever-more-distant galaxies$^\dagger$ will reach us; that is $r_\mathrm{PH}$ increases. In other words, &lt;em&gt;the observable Universe always increases in size, and no galaxy visible today will ever leave the observable Universe, no matter its speed&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since future observable galaxies will be more and more redshifted, their light will eventually shift out of the visible range and into longer and longer radiowaves. Furthermore, the time between each detected photon will increase, so they will be dimmer and dimmer, and thus in practice, they &lt;em&gt;will&lt;/em&gt; disappear.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Intuitive explanation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A good analogy for better understanding why light can reach us from a galaxy that recedes faster than light, is the &quot;worm on a rubber band&quot;: Attach an (infinite stretchable) rubber band (of length, say, 10 cm) to a wall and walk away at any constant speed you choose, e.g. 1 m/s. Before you start, put your pet worm at the end near the wall. It wants to get back to you, and starts crawling at 1 cm/s, i.e. 100× slower than you. Will it ever reach you? If you look at it from the perspective of the wall, both you and the worm move away, but whereas you recede at a constant speed, the worm, although slower in the beginning, accelerates because it moves on the rubber band, but the part of the rubber band between the worm and the wall increases in size. The rest of the rubber band of course also increases in size, but that doesn't matter — as long as you have a constant speed, and the worm accelerates, it &lt;em&gt;will&lt;/em&gt; reach you (although in this example, it will take the worm $10^{26}$ billion years, at which point it may have lost its patience. But if you walk at only 10 cm/s, it will take just 6 hours).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this analogy, you're the MW, the wall is GN-z11, and the worm is a photon. Now if you don't walk at a constant speed, but &lt;em&gt;also&lt;/em&gt; accelerate (this is an analogy of the effect of dark energy), the worm may or may not reach you, depending on your speeds. Just like there is a limit to how distant galaxies we will ever be able to see.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;Note that since large distances also means looking back in time (since the light has spent a long time traveling), we actually don't see galaxies this far away, as they hadn't formed this early in history. We do however see the gas from which the galaxies were born, as far back as 380,000 years after Big Bang.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2017-09-29T12:10:10.090" LastActivityDate="2017-09-29T12:10:10.090" CommentCount="24" ContentLicense="CC BY-SA 3.0" />
  <row Id="14339" PostTypeId="2" ParentId="14336" CreationDate="2016-03-28T16:20:10.453" Score="7" Body="&lt;p&gt;As time passes, there are galaxies that are currently not in the observable universe which will become observable But this is not a sudden winking on. Instead, over hundreds of millions of years we will see a proto galaxy evolve into a mature galaxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example there is a  &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0512396&quot;&gt;&quot;blob&quot; of hydrogen&lt;/a&gt; that some interpret as being the accretion of hydrogen onto a dark matter halo.  If this interpretation is correct, then the galaxy that eventually forms from it is outside the observable universe. But it won't remain so. Over billions of years the hydrogen will have formed stars, and the galaxy will be in our observable universe. We don't see the sudden appearance of a new galaxy, rather we see the evolution over billions of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an effect of greater red-shifting. Ultimately galaxies will begin to retreat fast enough that they are red-shifted below the level of detectability. It is suggested that in about 2 trillion years only local galaxies will be visible. This again is not a rapid process(!) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we do not observe galaxies disappearing over a cosmic horizon, and do not expect to do so.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-03-28T16:54:37.307" LastActivityDate="2016-03-28T16:54:37.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14340" PostTypeId="1" CreationDate="2016-03-28T20:03:05.403" Score="1" ViewCount="387" Body="&lt;p&gt;FYI my application is not astronomy related, but I know astronomy folks have a lot of knowledge of photodetection and spectrometry instrumentation, so I am seeking this community's help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in doing both spectral and RGB imaging. I'm wondering if I can do both with just an RGB CMOS camera... The way I would approach the spectral imaging is to illuminate my sample with monochromatic light one wavelength at a time and then convert the images to grayscale. Each grayscale image would represent the intensity of light reflecting from my sample at a particular wavelength. I would love for this to work, but I am doubtful that the converted image will actually represent what a grayscale camera would record. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bit of an odd scenario... Wasn't sure how to search the web for this. Any help is appreciated! Thanks!&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2016-03-29T02:03:47.920" LastActivityDate="2016-04-26T15:43:54.427" Title="Photodetector Question: Does converting an RGB image to grayscale produce the same result as using a grayscale image detector?" Tags="&lt;photography&gt;&lt;spectrometry&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14341" PostTypeId="1" AcceptedAnswerId="14343" CreationDate="2016-03-28T22:45:02.903" Score="2" ViewCount="314" Body="&lt;p&gt;On the Heavens above app for Android and maybe iOS, it shows negative magnitude values for Iridium satellites, but for other ones, it shows positive.  Here is a screenshot&lt;a href=&quot;https://i.stack.imgur.com/wYmPf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wYmPf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this?&lt;/p&gt;&#xA;" OwnerUserId="11343" LastActivityDate="2016-03-29T18:51:19.873" Title="Magnitude, satellite flare and the Heavens Above app" Tags="&lt;satellite&gt;&lt;flare&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14343" PostTypeId="2" ParentId="14341" CreationDate="2016-03-29T01:44:30.233" Score="5" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnitude_(astronomy)&quot; rel=&quot;nofollow&quot;&gt;astronomy &quot;magnitude&quot; scale&lt;/a&gt; works backwards: smaller numbers indicate brighter objects.  Back in the days before precision measurements of brightness, stars were categorized by eye, with the brightest being &quot;stars of the first magnitude&quot;.  When more precise measurement became possible, this scale was retained, and extended into the negative numbers for very bright objects like Venus, the Sun, and a few of the brightest stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Iridium satellites have &lt;a href=&quot;https://en.wikipedia.org/wiki/Satellite_flare#Iridium_flares&quot; rel=&quot;nofollow&quot;&gt;enormous, mirror-like antenna arrays&lt;/a&gt;.  When one of them is angled correctly, it will reflect sunlight straight at you, producing an incredibly bright flare visible even in broad daylight.  Very few other satellites have large reflective surfaces other than their solar panels, and solar panels are kept pointed straight at the Sun, so they never generate flares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give some points of comparison, the Iridum flare listed in your screenshot, at magnitude -5.2, is comparable to Venus at its brightest, visible during daylight if you know where to look.  The other satellites listed, with magnitudes in the 2-3.5 range, have brightnesses between that of Polaris, and that of the dimmer stars making up Ursa Minor.&lt;/p&gt;&#xA;" OwnerUserId="1559" LastEditorUserId="1559" LastEditDate="2016-03-29T18:51:19.873" LastActivityDate="2016-03-29T18:51:19.873" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14344" PostTypeId="1" CreationDate="2016-03-29T02:05:16.053" Score="1" ViewCount="90" Body="&lt;p&gt;Each year while observing voids, astronomers discover more and more unexpected galaxies within them.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How do we know the cosmic microwave background radiation is not the weak light from distant galaxies? &lt;/li&gt;&#xA;&lt;li&gt;How do we know cosmic microwave background radiation is not stars' light reflected by dark objects?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11085" LastEditorUserId="2365" LastEditDate="2016-03-30T19:13:15.990" LastActivityDate="2016-03-30T19:13:15.990" Title="How do we know background radiation does not come from undiscovered galaxies?" Tags="&lt;cmb&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14346" PostTypeId="1" CreationDate="2016-03-29T14:49:40.993" Score="0" ViewCount="123" Body="&lt;p&gt;I IT developer, and I would code a small program that simulates the orbit of two bodies, the problem of two bodies eventually. &#xA;I have no problem on the part programming but I have a little trouble with the calculations to be done to retrieve the position of the second body that is in orbit around the first. &#xA;Could someone expliqer calculations me or give me a site with a concrete example, such calculations to do to have the orbit of the moon around the earth or the earth around the sun?&lt;/p&gt;&#xA;" OwnerUserId="10958" LastEditorUserId="37191" LastEditDate="2021-02-12T21:31:56.800" LastActivityDate="2021-02-12T21:31:56.800" Title="Simulates Orbit" Tags="&lt;orbit&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2016-03-30T13:26:12.210" ContentLicense="CC BY-SA 3.0" />
  <row Id="14347" PostTypeId="1" CreationDate="2016-03-29T15:20:54.240" Score="0" ViewCount="46" Body="&lt;p&gt;Neutrino's are very small though they have mass. Because of that it is probably that the don't have a speed faster than photons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is it in all events, like supernova's or else events, where it is possible that we first receive their neutrino's instead of their light? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps gravity or other interactions could change something in their way to earth?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastActivityDate="2016-03-29T16:14:42.143" Title="Are there events in universe which we receive first from their neutrino's instead of their photons?" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14348" PostTypeId="2" ParentId="14347" CreationDate="2016-03-29T16:14:42.143" Score="2" Body="&lt;p&gt;Yes.  When a core collapse supernova happens, it takes a few hours for the shock wave to reach the surface and &quot;break out&quot; (when we first &quot;see&quot; it). Meanwhile, the neutrinos can escape nearly directly once they reach the top of the core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, as you wrote, photons travel slightly faster, so if the SN is far enough away, the photons might overtake these neutrinos.  However, I have not done the calculation to see if the universe is big enough.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2016-03-29T16:14:42.143" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14349" PostTypeId="2" ParentId="14340" CreationDate="2016-03-29T16:39:36.693" Score="2" Body="&lt;p&gt;Really, all cameras/detectors are greyscale detectors.  Every pixel in a CCD detector puts out a single number for the number of photons it detected. To create an RGB image you need three filters.  If you have 3 detectors, you can put a different filter in front of each and get 3 numbers at once.  If you have 1 detector, then you take 3 measurements with a different filter each time.  To create the color image, one combines these values but you have freedom to weight them as you like.  Typically, one tries to imitate the sensitivity of the eye, but that is a complicated issue since the sensitivity of the eye to different colors changes with total intensity and the filters can not exactly reproduce the eye's bandpasses.  So, there is no right way to do this.  Some people like red more than blue, so they weight the R-band more. Once, you get aR+bG+cB, you loose the information you need to derive R, G, and B separately.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to do your experiment properly, you need to first get the response function (response vs wavelength) of the filters + CCD of your camera at each wavelength that you are using.  You do this by imaging without the sample.  Then repeat with the sample and divide &quot;with&quot; by &quot;without&quot;, where &quot;with&quot; is the output in the R, G, or B channel that has the most response in the &quot;without&quot; case.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2016-04-26T15:43:54.427" LastActivityDate="2016-04-26T15:43:54.427" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14350" PostTypeId="2" ParentId="14346" CreationDate="2016-03-29T17:25:14.413" Score="0" Body="&lt;p&gt;The equations of motion are just second order ordinary differential equations. They can be solved numerically by any of the usual methods, However, for two bodies an exact solution can be found, that solution was known to Kepler. To model the trajectory you need to know the orbital period and the eccentricty (e) of the orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know the  period of orbit of a body, then the  &quot;Mean anomaly (M)&quot; is the angle time/period *2*pi radians (it increases uniformly from zero to 2pi in one orbital period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The find the Eccentric anomaly (&quot;E&quot;, the angle made by the body, the center of the elliptical orbit and the point of periapsis when the body is closest to the sun)  you solve $M=E-e\sin(E)$ (it can be solved by newton's method quickly, though convergence is fastest for roughly circular orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then know the body is on a ellipse, with the sun at one focus and you have calculated the ray on which the body is found at a given time. This gives you the position of the body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a rough implementation of this in a &lt;a href=&quot;https://gist.github.com/zeimusu/7a2e71de2301df893731775398df7fe5&quot; rel=&quot;nofollow&quot;&gt;python gist&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-03-29T17:25:14.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14351" PostTypeId="1" CreationDate="2016-03-29T17:47:01.583" Score="8" ViewCount="1157" Body="&lt;p&gt;Why do telescopes use mirrors that simply &lt;em&gt;reflect&lt;/em&gt; photons, when they instead could be covered with large sensors to &lt;em&gt;register&lt;/em&gt; them? Reflection is all good and well, all thanks to silver and beryllium for that. But wouldn't it be better to electronically register the photons directly instead of after having them bounce around between stupid mirrors? Would any data be lost in a pure CCD-telescope without any mirrors or lenses?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Couldn't a large wired CCD light sensor send the detected signals further in a smarter way than a stupid physically reflecting surface can do? It's the same photons and the telescope itself doesn't generate any new information about the distant galaxies it reflects. Why physically bend mirrors for adaptive optics, instead of bending the raw binary data with an algorithm for the same effect?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2016-03-29T17:54:16.357" LastActivityDate="2021-03-08T06:26:23.473" Title="Could mirrors be replaced with CCDs?" Tags="&lt;telescope&gt;&lt;photometry&gt;" AnswerCount="7" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14352" PostTypeId="2" ParentId="13386" CreationDate="2016-03-29T17:56:11.610" Score="6" Body="&lt;p&gt;Robert Hooke's &lt;a href=&quot;https://books.google.com/books?id=ISP_gRwuz94C&amp;amp;pg=PA255&amp;amp;dq=twinkling%20stars%20air&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwiBjJG1t-bLAhVH8CYKHakfD5cQ6AEILTAD#v=onepage&amp;amp;q=twinkling%20stars%20air&amp;amp;f=false&quot;&gt;Micrographia&lt;/a&gt; from 1664 has a detailed discussion of &quot;seeing&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The table [of contents], which is at the end of the book, explains that pages 230-232 discuss:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;that the Air near the Earth is composed of parts of differing density...this property produces the effects of waving and dancing of Bodies; and of the twinkling of the Stars&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11382" LastActivityDate="2016-03-29T17:56:11.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14353" PostTypeId="2" ParentId="14351" CreationDate="2016-03-29T19:55:24.517" Score="9" Body="&lt;p&gt;To answer your question, we need to first show the job each mirror is doing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First up, the Newtonian (lovingly called the &quot;Newt&quot;, and invented by Sir Ike Newton):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Reflecting_telescope#/media/File%3ANewtonian_telescope2.svg&quot;&gt;https://en.m.wikipedia.org/wiki/Reflecting_telescope#/media/File%3ANewtonian_telescope2.svg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two mirrors in this design, not surprisingly labeled as primary and secondary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The job of the primary mirror is NOT merely to reflect light, but to concentrate the diffuse photons onto a much smaller point.  This makes really dim objects brighter, and is the first step in magnification.  (Further magnification is done by the eyepiece, which is similar to a small refracting telescope.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the Newt, the secondary mirror reflects the now concentrated photons to a more convenient point for viewing.  Without the secondary mirror your head would get in the way of the view.  A secondary mirror is not necessary, and in fact many telescopes will place instruments, such as CCD's, at this &quot;prime focus&quot; point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the Hubble Space Telescope, the secondary mirror reflects the concentrated photons to the scope's instruments, where they can work their magic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all reflecting telescope designs the primary mirror uses the laws of physics to give the end user, whether it's the human eye, or research gear, as many concentrated photons as possible, maximizing what we can see/detect.  The bigger the primary mirror, the more concentrated the photons, and the more we have to work with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to seeing what we call &quot;the dim fuzzies&quot;, bigger IS better!&lt;/p&gt;&#xA;" OwnerUserId="11373" LastEditorUserId="38" LastEditDate="2016-03-29T20:07:58.577" LastActivityDate="2016-03-29T20:07:58.577" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14354" PostTypeId="2" ParentId="14147" CreationDate="2016-03-29T19:58:45.667" Score="2" Body="&lt;p&gt;I don't have a complete answer, but I do know where the $7.6$ factor comes from: It equals&#xA;$$(2×0.0167)×(24×60)/(2 \pi).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0.0167$ comes from the eccentricity of the Earth, and the factor of $2$ comes from the above equation. $24×60$ converts from days to minutes, and $2 \pi$ is the number of radians in a complete revolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I still don't completely understand is why these coordinate conversions need to be done, specifically the division by $2 \pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks everyone for your help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I now have a somewhat better explanation for the computation in this answer. &lt;code&gt;2ε sin(M)&lt;/code&gt; is the number of radians that the mean Sun is behind the apparent Sun on Noon on a day with mean anomaly &lt;code&gt;M&lt;/code&gt;. Dividing the &lt;code&gt;2ε&lt;/code&gt; factor by &lt;code&gt;2π&lt;/code&gt;gives the fraction of the rotation &lt;em&gt;of the Earth&lt;/em&gt; (i.e. one day). When the result is multiplied by the number of minutes in a &lt;em&gt;sidereal&lt;/em&gt; day &lt;code&gt;(23×60+54)&lt;/code&gt; yields the desired coefficient &lt;code&gt;7.63345&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8092" LastEditorUserId="8092" LastEditDate="2018-06-29T13:48:05.550" LastActivityDate="2018-06-29T13:48:05.550" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14355" PostTypeId="2" ParentId="14351" CreationDate="2016-03-29T22:21:34.567" Score="3" Body="&lt;p&gt;I actually found a concept of 2D easy-to-scale telescope some time ago (&lt;a href=&quot;http://www.lockheedmartin.com/us/innovations/011916-webt-spider.html&quot; rel=&quot;nofollow&quot; title=&quot;here&amp;#39;s the link&quot;&gt;here's the link&lt;/a&gt;).&#xA;I guess we will slowly abandon refractor telescopes, for, as I understand, we are pushing them to their limits right now and it is getting really hard to make a bigger one (because of how hard it is to make a sufficiently sized mirror of the quality needed).&#xA;BUT it worth noting, that I am not an expert in any way, so I suggest someone, who knows more about the topic, edit this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: There's a really good point in the DJohnM's answer, so I thought I'll add that the thing I linked here (SPIDER) isn't just a big 2D array of CCDs; it actually does have a tiny lense over each of the detectors and each of those measures light in a bunch of different wavelength, so it can preserve information about direction and wavelength of light. So the answer to initial question is no, we can't just build a CCD array instead of full-sized telescope, but the idea to make telescopes scalable in two dimensions instead of three seems to be a good one and threre are people working on it.&lt;/p&gt;&#xA;" OwnerUserId="11387" LastEditorUserId="11387" LastEditDate="2016-03-30T05:53:55.300" LastActivityDate="2016-03-30T05:53:55.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14356" PostTypeId="1" AcceptedAnswerId="14361" CreationDate="2016-03-29T23:28:34.463" Score="4" ViewCount="236" Body="&lt;p&gt;As it is explained in &lt;a href=&quot;https://youtu.be/blSTTFS8Uco&quot; rel=&quot;nofollow&quot;&gt;this video&lt;/a&gt;, one of cosmic inflation's observable effects is the homogeneity of our universe. Inflation allows two points on the different sides of the observable universe to be causally connected at some point in time, so they may exchange their mass densities and temperatures, which are then going to be about the same for both of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I do understand, that this effect doesn't have to end on the edge of the observable universe and might go on for whatever distance the rate of inflation will make it to, I do not understand how this could be valid for the entire Universe if it is infinite. From my guessing, if the Universe is infinite, there will allways be two points, that was never causally connected, no matter how fast it expanded at the inflation period. That will mean, that, while the Universe should be homogeneous on the sufficiently large scales, it doesn't have to on the scales &lt;strong&gt;even larger&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess I either don't fully understand the idea of cosmic inflation or don't get the particular way the Universe is infinite in.&lt;/p&gt;&#xA;" OwnerUserId="11387" LastEditorUserId="11387" LastEditDate="2016-03-30T05:16:23.837" LastActivityDate="2016-03-30T11:39:00.627" Title="How can cosmic inflation make an infinite universe homogeneous?" Tags="&lt;cosmology&gt;&lt;cosmological-inflation&gt;&lt;cosmological-principle&gt;&lt;homogeneity&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14358" PostTypeId="2" ParentId="14351" CreationDate="2016-03-30T03:42:41.497" Score="17" Body="&lt;p&gt;The CCD has no way of recording the &lt;strong&gt;&lt;em&gt;direction&lt;/em&gt;&lt;/strong&gt;, the point in the sky, from which a photon is coming. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you point your mirror-less telescope at the Moon.  &lt;strong&gt;Every&lt;/strong&gt; point on the moon's surface would be reflecting photons onto &lt;strong&gt;every&lt;/strong&gt; part of the CCD &lt;em&gt;at the same time.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You've just created an expensive, sensitive, ambient light meter. There would be no image information whatsoever.&lt;/p&gt;&#xA;" OwnerUserId="420" LastActivityDate="2016-03-30T03:42:41.497" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14360" PostTypeId="2" ParentId="12230" CreationDate="2016-03-30T04:37:46.583" Score="0" Body="&lt;p&gt;I've only read one, but it was my upper level undergraduate text and I thought it was excellent. It was also called &quot;Introduction to Stellar Astrophysics&quot; and consists of Two Volumes:&#xA; 1) Basic Stellar Observations and Data&#xA; 2) Stellar Atmospheres&#xA;The author is Erika Bohm-Vitense.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2016-03-30T04:37:46.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14361" PostTypeId="2" ParentId="14356" CreationDate="2016-03-30T11:39:00.627" Score="6" Body="&lt;p&gt;Inflation is used to explain why the &lt;em&gt;observable&lt;/em&gt; universe is extremely homogeneous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without inflation, we can do the following &lt;em&gt;crude&lt;/em&gt; calculation. The cosmic microwave background was formed about 300,000 years after the big bang, at a redshift of about 1100. Thus causally connected regions at the epoch of CMB formation would have a radius of $\sim 300,000$ light years, which has now expanded by a factor of 1100 to be $3.3\times 10^{8}$ light years in radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be compared with the radius of the observable universe, which is currently around 46 billion light years. This means that causally connected regions should only be $\sim 4 \times 10^{-7}$ of the observable universe, or equivalently, patches of CMB of $\sim 2$ degrees radius on the sky are causally connected. This is clearly not the case as the variations in the CMB are no more than about 1 part in $10^{5}$ across the whole sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inflation solves this by allowing previously causally connected regions to inflate to become larger than the entire observable universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You appear to understand this quite well, so I am not entirely clear what your question is. We cannot know whether the &lt;em&gt;entire universe&lt;/em&gt; is homogeneous, since we cannot measure it. The cosmological principle is an &lt;em&gt;assumption&lt;/em&gt; that appears to hold &lt;em&gt;approximately&lt;/em&gt; true in the observable universe, but need not apply to the universe as a whole. Indeed it is not absolutely true in the observable universe otherwise it would be quite uninteresting, containing no galaxies, clusters, or other structure. I think the only requirement on inflation is that it blows up a patch of causally connected universe so that it becomes &lt;em&gt;much&lt;/em&gt; bigger than the observable universe at the current epoch.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-03-30T11:39:00.627" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14362" PostTypeId="2" ParentId="12230" CreationDate="2016-03-30T12:44:38.727" Score="0" Body="&lt;p&gt;There's another good and relatively short book (so it's maybe fit as introductory reading) from R. Kippenhahn - Stellar Structure and Evolution.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2016-03-30T12:44:38.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14363" PostTypeId="2" ParentId="14309" CreationDate="2016-03-30T15:42:57.140" Score="1" Body="&lt;p&gt;Assuming that you used the time of 170900 EDT(-4 UT), which is 230900 UT, the following is my math:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the Julian Date for 25Mar2016 at 230900, which is 2457473.464583, be set to JD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the formula (on Wikipedia:n = JD - 2451545.0 ), your n is 5928.464583.&#xA;Then, taking the second formula ( g = 357.528° + 0.9856003° * n ) you get g=357.528°+5843.0964715441749 = 6200.3544715441749 mod 360 = +80.35447°.  Were you to round up this would give +80.4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without how you handled your date/time in c++ and your rounding, I would wager that is were the error occurred.&lt;/p&gt;&#xA;" OwnerUserId="11399" LastActivityDate="2016-03-30T15:42:57.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14364" PostTypeId="2" ParentId="14344" CreationDate="2016-03-30T15:54:15.080" Score="4" Body="&lt;p&gt;Assuming you mean the Cosmic Microwave Background ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CMB is &lt;em&gt;entirely&lt;/em&gt; in a very narrow range of the microwave part of the spectrum. Galaxy emissions are all over the spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CMB comes uniformly from all over the sky. It's like looking at painting that is all one color. Galaxy light looks like pinpoints on a dark background. (Nearby galaxies are not pinpoints, but those at cosmological distances look that way.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CMB matches the exact wavelength we expect for light that is heavily redshifted from the moment of last scattering. That moment was hundreds of millions of years before galaxies ever formed. Light from oldest galaxies are at a redshifted wavelength from 500 million years after the last scattering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the CMB was due to early starlight reflecting off dark objects, you would see a very different spectrum. It would contain many absorption lines as the dark objects absorbed some wavelengths better than others. Also the reflected starlight would be across a wide range of frequencies, unlike the CMB which has a very narrow range.&lt;/p&gt;&#xA;" OwnerUserId="11175" LastActivityDate="2016-03-30T15:54:15.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14365" PostTypeId="2" ParentId="14284" CreationDate="2016-03-30T15:57:51.530" Score="0" Body="&lt;p&gt;I can think of two ways this benefits life.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Less chance of a passing star disrupting our Oort Cloud and sending&#xA;deadly comets crashing into the Earth.&lt;/li&gt;&#xA;&lt;li&gt;Less chance of nearby supernova destroying our ozone layer.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if our Solar System passes near &lt;a href=&quot;https://en.wikipedia.org/wiki/Astrophysical_jet#Relativistic_jet&quot; rel=&quot;nofollow&quot;&gt;a relativistic jet&lt;/a&gt; emanating from the central black hole of a galaxy, Earth will get hit with high speed particles that will disrupt the magnetosphere and ozone layer. Definitely not good for any life on Earth at that time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting ejected from the galaxy also means there is less chance of future humans exploring the galaxy by hopping to nearby stars. We'd only have our solar system to explore and colonize. There would not be another star system within a hundred thousand light years.&lt;/p&gt;&#xA;" OwnerUserId="11175" LastEditorUserId="11175" LastEditDate="2016-04-09T23:08:21.780" LastActivityDate="2016-04-09T23:08:21.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14366" PostTypeId="2" ParentId="10623" CreationDate="2016-03-30T16:50:19.197" Score="1" Body="&lt;p&gt;As far as I know, the system with the most stellar members is the &lt;a href=&quot;http://galacticcenter.astro.ucla.edu/science.html&quot; rel=&quot;nofollow&quot;&gt;black hole at the center of the Milky Way&lt;/a&gt; with 8 stars in close orbit. Of course, David Hammen is correct, and if there are other orbiting stars with much longer periods we'd have trouble determining their orbits until they started their close approaches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the other extreme, is a galaxy a star system? I assume no, but just thought I'd ask, since you have not defined the term. We know the orbits of quite a few stars in the Milky Way, but that's probably not what you mean.&lt;/p&gt;&#xA;" OwnerUserId="11401" LastEditorUserId="11401" LastEditDate="2016-03-30T17:17:26.917" LastActivityDate="2016-03-30T17:17:26.917" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14367" PostTypeId="1" AcceptedAnswerId="14368" CreationDate="2016-03-30T16:54:04.557" Score="2" ViewCount="173" Body="&lt;p&gt;I wondered how time would be dilated for an object in the middle of 2 black holes, however &#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_time_dilation&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Gravitational_time_dilation&lt;/a&gt; only provides a formula for one source of gravity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i tried approximating it by multiplying the time dilation factors, however when i used 1 single source with twice the mass the results differed from 2 objects with half the mass quite drastically&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the examples:&lt;br&gt;&#xA;mass $1,00E+036$kg distance $3000000000$m&lt;br&gt;&#xA;time:$0,7105905766$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;mass $2,00E+036$kg distance $3000000000$m&lt;br&gt;&#xA;time:$0,099387802$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;approximation: $0,7105905766*0,7105905766=0,5049389676&#xA;$&#xA;which is off by a factor of 5 and would be even more off the closer the distance gets to the schwarzschild radius&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so how can the time dilation of multiple objects be approximated ?&lt;/p&gt;&#xA;" OwnerUserId="10326" LastActivityDate="2016-03-30T19:11:22.087" Title="gravitational time dilation multiple sources" Tags="&lt;gravity&gt;&lt;black-hole&gt;&lt;time-dilation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14368" PostTypeId="2" ParentId="14367" CreationDate="2016-03-30T19:11:22.087" Score="2" Body="&lt;p&gt;In the static weak-field approximation, the metric is approximated in terms of the the Newtonian gravitational potential $\Phi$ as:&#xA;$$\mathrm{d}s^2 = -\left(1+2\Phi\right)\mathrm{d}t^2 + (1-2\Phi)\mathrm{d}S^2\text{,}$$&#xA;where $\mathrm{d}S^2$ is the metric for Euclidean $3$-space. A more general situation requires one to solve a multi-body problem in general relativity, although the &lt;a href=&quot;https://en.wikipedia.org/wiki/Parameterized_post-Newtonian_formalism&quot; rel=&quot;nofollow&quot;&gt;parametrized post-Newtonian formalism&lt;/a&gt; provides a canonical approximation scheme. So the completely general answer to your question is &quot;it's very difficult problem of numerical general relativity that doesn't have a neat analytic solution.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, at large distances from slowly-moving objects (i.e., large compared to their Schwarzschild radii, we can approximate a multi-body time dilation relative to a stationary observer at infinity as just given by the sum of their gravitational potentials:&#xA;$$\frac{\mathrm{d}\tau}{\mathrm{d}t} = 1 - \frac{1}{c^2}\sum_k\frac{GM_k}{r_k} \text{.}$$&#xA;Note that this is consistent with the Schwarzschild formula for large radial coordinates, since by the Taylor-MacLaurin expansion,&#xA;$$\frac{\mathrm{d}\tau}{\mathrm{d}t} = \sqrt{1-\frac{2GM}{rc^2}} = 1 - \frac{GM}{rc^2} + \mathcal{O}\left(\frac{R^2}{r^2}\right)\text{,}$$&#xA;where $R = 2GM/c^2$ is the Schwarzschild radius. This also means that simply multiplying the factors is a pretty good approximation as long as those factors are close to $1$.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2016-03-30T19:11:22.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14369" PostTypeId="1" AcceptedAnswerId="14373" CreationDate="2016-03-30T22:48:11.363" Score="4" ViewCount="513" Body="&lt;p&gt;I'm currently going through Binney &amp;amp; Tremaine (2008) on my own to learn about stellar dynamics. I also have been perusing additional online resources such as &lt;a href=&quot;http://www.scholarpedia.org/article/N-body_simulations_(gravitational)&quot;&gt;this scholarpedia wiki&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often when distinguishing between collision-less vs collisional stellar systems, the virial theorem is invoked along with the equations for &quot;crossing time&quot; (also known as &quot;dynamical time&quot;) and &quot;relaxation time.&quot; A large galaxy is said to be collision-less because its relaxation time is many orders of magnitude higher than its age, whereas a dense stellar system (e.g., a globular cluster) is collisional because its relaxation time is less than its age. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what is the relationship between this so-called &quot;relaxed&quot; state and virial, dynamical, and thermodynamic equilibrium? What do the three different kinds of equilibria intuitively mean? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I have heard that large galaxies are assumed to be in virial equilibrium and then people derive &quot;dynamical masses&quot; (why not &quot;virial masses&quot;?). What would it take and/or mean for a large elliptical galaxy to be not just in virial equilibrium, but also in dynamical or thermodynamic equilibrium?&lt;/p&gt;&#xA;" OwnerUserId="607" LastActivityDate="2016-03-31T18:00:46.687" Title="Stellar systems: what is the difference between virial, dynamical and thermodynamic equilibrium?" Tags="&lt;galaxy&gt;&lt;galactic-dynamics&gt;&lt;star-systems&gt;&lt;stellar-dynamics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14370" PostTypeId="1" CreationDate="2016-03-31T06:06:21.313" Score="2" ViewCount="113" Body="&lt;p&gt;If one day we have enough technology to push the Earth away a bit further from the sun to reduce global warming, is it true that it will start to distort the orbits of Mars and Venus, giving them a larger eccentricity and finally colliding with sun or each other?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="823" LastEditDate="2016-03-31T08:53:48.787" LastActivityDate="2016-03-31T08:53:48.787" Title="What would happen if we changed Earth's orbit?" Tags="&lt;orbit&gt;" AnswerCount="0" CommentCount="5" ClosedDate="2016-03-31T12:35:36.560" ContentLicense="CC BY-SA 3.0" />
  <row Id="14371" PostTypeId="2" ParentId="6589" CreationDate="2016-03-31T07:39:55.883" Score="1" Body="&lt;p&gt;The principle of conservation of energy tells us that the amount of energy in the universe is constant. Therefore we can conclude the energy that once made up a star that turned into a black hole cannot be destroyed. So yes according to the principle of conservation of energy a black hole contains energy. &lt;/p&gt;&#xA;" OwnerUserId="11403" LastActivityDate="2016-03-31T07:39:55.883" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14372" PostTypeId="2" ParentId="13225" CreationDate="2016-03-31T17:44:05.560" Score="1" Body="&lt;p&gt;Many of the moons of Jupiter and Saturn are tidally locked and probably move ice from their equatorial regions to their polar regions. Ganymede is the best example where you see bright polar ice caps and a dark equatorial zone.  A migration process occurs on all of these moons in which light from the sun gets absorbed by water ice molecules and sends them on random jumps.  After a long random walk, they eventually end up near the pole where they stay for a long time even if they are not in a perpetual shadow, and they stay there for nearly forever if they are.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, if the moon has a very thick coating of ice everywhere, this process may not have dug down deep enough, so it is still covered with ice everywhere, in which case it is hard to see the effect happening.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2016-04-26T15:38:37.757" LastActivityDate="2016-04-26T15:38:37.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14373" PostTypeId="2" ParentId="14369" CreationDate="2016-03-31T18:00:46.687" Score="2" Body="&lt;p&gt;&lt;strong&gt;Thermal equilibrium&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermal equilibrium relies strongly on the idea of &lt;a href=&quot;https://en.wikipedia.org/wiki/Equipartition_theorem&quot; rel=&quot;nofollow&quot;&gt;equipartition of (kinetic) energy&lt;/a&gt;. In a stellar system, this means that the total kinetic energy is divided evenly amongst all the stars. This doesn't imply that the velocities are all the same; they can't, because not all the masses are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dynamical equilibrium&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dynamical equilibrium means that over dynamical timescales, the system is stable - basically, it will not succumb to core collapse due to a gravothermal instability. Note that it may not be possible for a system to reach thermal equilibrium, even if it is in dynamical equilibrium. In a system with two main types of stars, it must satisfy the Spitzer stability condition (see &lt;a href=&quot;http://arxiv.org/pdf/astro-ph/0111057v1.pdf&quot; rel=&quot;nofollow&quot;&gt;Fregeau et al. (2001)&lt;/a&gt; and &lt;a href=&quot;http://modest15s.net/wp-content/uploads/2015/04/Spera.pdf&quot; rel=&quot;nofollow&quot;&gt;these results&lt;/a&gt;):&#xA;$$\left(\frac{M_2}{M_1}\right)\left(\frac{m_2}{m_1}\right)^{3/2}&amp;lt;0.16$$&#xA;where $M_1$ and $M_2$ are the total masses of types 1 and 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Virial equilibrium&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Virial equilibrium comes about when the system satisfies the &lt;a href=&quot;https://en.wikipedia.org/wiki/Virial_theorem&quot; rel=&quot;nofollow&quot;&gt;virial theorem&lt;/a&gt; (see &lt;a href=&quot;https://cds.cern.ch/record/432240/files/0003390.pdf&quot; rel=&quot;nofollow&quot;&gt;Meylan (2000)&lt;/a&gt;), i.e.&#xA;$$2\langle T\rangle+\langle V\rangle=0$$&#xA;where $T$ and $V$ are kinetic and potential energies.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2016-03-31T18:00:46.687" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="14374" PostTypeId="1" AcceptedAnswerId="14377" CreationDate="2016-03-31T20:27:31.443" Score="2" ViewCount="7517" Body="&lt;p&gt;The atmosphere of a planetary body (assuming it has an atmosphere) is described as being made up of distinct layers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, Earth, Saturn and Jupiter all have a stratosphere and a troposphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is it that defines where one layer stops and another starts?&lt;/p&gt;&#xA;" OwnerUserId="11412" LastEditorUserId="11412" LastEditDate="2016-04-01T15:00:59.597" LastActivityDate="2016-04-01T17:13:53.090" Title="What is it that distinguishes one atmospheric layer from another?" Tags="&lt;planetary-atmosphere&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14375" PostTypeId="2" ParentId="14374" CreationDate="2016-04-01T00:36:22.747" Score="1" Body="&lt;p&gt;The problematic notion here is the &quot;distinct&quot; layers, those effectively don't exist.&#xA;Earth's atmosphere is a continuum, as any atmosphere is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This stems from the hydrostatic pressure law $\nabla P = - g \rho$, that gives together with the ideal gas law $P = \rho k_B T / \mu$ an exponentially decaying solution with height, for pressure as well as for density. Here P is the pressure, $\rho$ is the local mass-density, g is the local gravitational acceleration, $k_B$ the Boltzmann constant and $T$ the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there are some caveats to &quot;layers&quot; as a manner of characterizing the atmosphere:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;An &lt;em&gt;inversion layer&lt;/em&gt; can be a distinct layer where the temperature decrease with height is inversed. Thus the temperature increases for a bit, then decreases again. Therefore there will be exactly two points where the temperature increase is 0. Between those two points the definition of a distinct layer makes sense.&lt;/li&gt;&#xA;&lt;li&gt;In a similar way the whole atmospheric temperature on a global scale has a funny height-dependency (see e.e. &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/9/9d/Comparison_US_standard_atmosphere_1962.svg&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;). Again one can define layers between the points where something interesting in the T-profile happens. This is where the characterization into troposphere, stratosphere, etc. stems from.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now we can choose variables other than temperature to characterize the atmosphere. Like ionization state, mixture state, dynamical movements, chemical makeup and then define layers that make sense only when talking about this particular variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summarizing&lt;/strong&gt;: Layers can be defined, even physically meaningful, but only in the context of a particular variable.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2016-04-01T00:36:22.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14376" PostTypeId="1" AcceptedAnswerId="14400" CreationDate="2016-04-01T01:13:34.520" Score="0" ViewCount="262" Body="&lt;p&gt;Is it just a coincidence that the two major expansionary periods occur close to periods of nucleosynthesis?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Big Bang nucleosynthesis occurred very close to the inflationary period.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Supernova and stellar nucleosynthesis created heavier elements as the stars died, and the universe underwent increased acceleration.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Further thoughts:&#xA;&lt;br&gt;&#xA;The metaphor for the expansion of the universe is often described as an inflating balloon, but could it be equally valid to think of expansion as matter falling into its own gravity well, away from infinity minus other matter within its gravity field. The effect being that matter radiates space itself as gravity pulls itself away from infinity. Just as heavier elements create a stronger gravity field, heavier elements would accelerate away from infinity minus other matter within its gravity field, at a faster rate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, on a cosmic scale, atomic decay would release the confined energy and the energy would return to infinity, creating the appearance of a contracting universe that restarts the Big Bang.  If this idea is reality it would create a geodesically complete cosmology.&lt;/p&gt;&#xA;" OwnerUserId="9124" LastActivityDate="2016-04-02T18:33:01.353" Title="Is nucleosynthesis responsible for the expansion of the universe?" Tags="&lt;big-bang-theory&gt;&lt;cosmological-inflation&gt;&lt;dark-energy&gt;&lt;nucleosynthesis&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14377" PostTypeId="2" ParentId="14374" CreationDate="2016-04-01T01:31:24.077" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is it that defines where one layer stops and another starts?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Temperature. More specifically, it's whether temperature rises or falls with increasing altitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the troposphere, temperature generally decreases with increasing altitude, at an average rate of 6.4 °C/km (the environmental lapse rate). This decrease stops at the tropopause, the boundary between the troposphere and the stratosphere. The ozone layer is in the stratosphere, making temperatures in the stratosphere increase with increasing altitude. While this boundary is a bit fuzzy, it is still very real. It takes an incredibly strong thunderstorm (think hurricanes, storms powerful enough to spawn tornados, and very tall and strong storms in the Inter-Tropical Convergence Zone) to penetrate that boundary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Temperatures in the stratosphere stop rising at the stratopause, the very fuzzy boundary between the stratosphere and the mesosphere. Like the troposphere, temperatures in the mesosphere fall (and fall very sharply) with increasing altitude. Thermodynamics makes the boundary between the stratosphere and mesosphere very different (and not nearly as clear) as the boundary between the troposphere and the mesosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boundary between the mesosphere and the thermosphere is similar to the boundary between the troposphere and stratosphere. Temperatures rise with increasing altitude in the thermosphere. This change from falling temperatures in the mesosphere to rising temperatures in the thermosphere makes for a rather stable boundary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something else happens near that boundary between the mesosphere and the thermosphere. Long-lived gases are fairly well-mixed in the troposphere, stratosphere, and mesosphere thanks to turbulence. Gases in the thermosphere and exosphere act more like a bunch of individual particles rather than a gas. Unlike the dense layers below, gases in the thermosphere and exosphere are differentiated. The makeup tends toward lighter and lighter particles (e.g., helium and hydrogen) with increasing altitude. Eventually, all one finds are hydrogen and helium. These are the gases that escape from the atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2016-04-01T01:31:24.077" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14378" PostTypeId="1" AcceptedAnswerId="36230" CreationDate="2016-04-01T03:39:02.277" Score="7" ViewCount="81" Body="&lt;p&gt;I found this article &lt;a href=&quot;http://www.epj-conferences.org/articles/epjconf/pdf/2011/01/epjconf_ohp2010_03005.pdf&quot;&gt;http://www.epj-conferences.org/articles/epjconf/pdf/2011/01/epjconf_ohp2010_03005.pdf&lt;/a&gt; from 2011 announcing that two dozen secondary eclipses of &quot;hot Jupters&quot; have been seen in Kepler data. I even think I have seen a light curve showing phase changes of a transiting planet. This makes me think that there should be evidence in the Kepler data for non-transiting planets since the amplitude of the light curve due to planetary phase changes shouldn't be much different from that of a secondary eclipse as long as there is a fairly high inclination (say over 60 degrees).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is anybody aware of such an examination of the data?&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2020-05-18T17:09:11.617" Title="Has the Kepler data been examined for light curves due to phase changes of non-transitting hot Jupiters and other?" Tags="&lt;exoplanet&gt;&lt;kepler&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14379" PostTypeId="1" CreationDate="2016-04-01T07:31:26.490" Score="0" ViewCount="68" Body="&lt;p&gt;I understand the collapse of a star during a supernova, as the outward pressure cannot compete with the gravitational field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it conceivable, that a relaxation of the space-time curvature is what expands the space and looks like an explosion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have often wondered, to extrapolate, if a &quot;release&quot; of the curvature of space itself, or in an extreme example a massive black hole releasing all of its potential energy in an instant (In a way not yet understood) could be in fact the (or a) big bang.&lt;/p&gt;&#xA;" OwnerUserId="11310" LastEditorUserId="2365" LastEditDate="2016-04-01T08:47:44.000" LastActivityDate="2016-04-01T08:57:44.120" Title="My supernovae question" Tags="&lt;supernova&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2016-04-06T15:40:23.347" ContentLicense="CC BY-SA 3.0" />
  <row Id="14380" PostTypeId="2" ParentId="14379" CreationDate="2016-04-01T08:57:44.120" Score="0" Body="&lt;p&gt;No. The precise details of collapse of a supernova are not perfectly understood, in the way we understand (for example) the orbits of the planets. However the energy comes from the release of gravitational potential energy. The &quot;relaxation of spacetime&quot; is not significant. We don't understand supernovae perfectly. But we have a good model of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should re-read Rob's answer to your previous question, not least because he got the order of magnitude of the energy release in a supernova correct. He also points out your error in thinking of the gravitational field changing &quot;in an instant&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning the big bang. No, as has been often pointed out, the big bang was not a explosion. There is no analogy between a supernova and the big bang.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-04-01T08:57:44.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14381" PostTypeId="1" AcceptedAnswerId="14382" CreationDate="2016-04-01T09:50:05.120" Score="10" ViewCount="2040" Body="&lt;p&gt;Mercury looks like the Moon, and so it makes me think about a question: is it possible that Venus and Mercury were the a same planet originally, and a giant impact with that planet made it split into Mercury and Venus (like with the Moon and Earth)?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="2153" LastEditDate="2016-04-01T23:21:03.357" LastActivityDate="2016-04-02T21:04:10.983" Title="Is it possible that Mercury was originally the moon of Venus after a giant impact?" Tags="&lt;venus&gt;&lt;natural-satellites&gt;&lt;mercury&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14382" PostTypeId="2" ParentId="14381" CreationDate="2016-04-01T13:34:16.040" Score="7" Body="&lt;p&gt;This was originally going to be a comment, but it ran too long, so I'm making it an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some models argue that the scenario of a satellite of Venus escaping like this is unlikely. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2006DPS....38.0703A&quot;&gt;Alemi &amp;amp; Stevenson (2006)&lt;/a&gt; have explored the possibility of a prior Venusian moon, starting from the assumption that Venus would not have been able to avoid a &lt;a href=&quot;https://en.wikipedia.org/wiki/Giant_impact_hypothesis&quot;&gt;giant impact&lt;/a&gt;. Here's their sequence of events:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A large body collides with Venus in a similar manner to the proposed Earth-Theia collision.&lt;/li&gt;&#xA;&lt;li&gt;Debris from the impact moves outwards into a disk surrounding Venus,&lt;/li&gt;&#xA;&lt;li&gt;A moon coalesces from the disk, and begins to slowly recede because of &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_acceleration&quot;&gt;tidal acceleration&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Another large body hits Venus. It reduces Venus's angular momentum, reversing its rotation.&lt;/li&gt;&#xA;&lt;li&gt;The moon spirals into Venus as it undergoes tidal deceleration, finally colliding with it again.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;One of the tricky things about testing this model is that the authors say that there would not necessarily have been drastic composition changes, meaning that it would be hard to analyze the planet's surface and see if there is evidence supporting the double impact hypothesis. So far, there have not been tests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is certainly true that Venus could have suffered other impacts - the model does not preclude that. There are a couple problems with Mercury arising from such a collision:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Other impacts could have ended up with the same result as the original moon.&lt;/li&gt;&#xA;&lt;li&gt;The chances of many more impacts aren't too high.&lt;/li&gt;&#xA;&lt;li&gt;Solar tides would likely have destabilized the orbit of any moon larger than a few kilometers in diameter (see &lt;a href=&quot;http://arxiv.org/abs/0906.2781&quot;&gt;Sheppard &amp;amp; Trujillo (2009)&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/MESSENGER&quot;&gt;&lt;em&gt;MESSENGER&lt;/em&gt;&lt;/a&gt; determined that &lt;a href=&quot;http://messenger.jhuapl.edu/soc/highlights122011.php&quot;&gt;Mercury has a high potassium/thorium ratio&lt;/a&gt; on its surface, which would seem to disprove any events involving extremely high temperatures, including any giant impact variant.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of course, if we accept that Venus could have captured a moon, only the third objection remains - still a strong point against the survival of a satellite, even by itself.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-04-02T21:04:10.983" LastActivityDate="2016-04-02T21:04:10.983" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14383" PostTypeId="2" ParentId="14374" CreationDate="2016-04-01T17:13:53.090" Score="0" Body="&lt;p&gt;The stratosphere and the troposphere are defined by the variation in temperature with height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Within a troposphere, the temperature drops as altitude increases, and drops fast enough for convection to occur. Whereas within a stratosphere, the temperature rises with increasing altitude. Since convection requires that the rate of temperature cooling with height (the &quot;lapse rate&quot;) exceeds the rate at which a gas cools due to a reduction in pressure, in the stratosphere the atmosphere will be stable. &lt;em&gt;Tropos&lt;/em&gt; means &quot;turning&quot; and &lt;em&gt;stratos&lt;/em&gt; means layered. The stratosphere is warmed by the absorbtion of solar radiation, particularly UV radiation (in the ozone layer), whereas the troposphere is warmed mostly from the ground, or in the case of Jupiter, from Jupiter's internal heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is that in troposphere the atmosphere will be well mixed, and clouds will form, whereas in a stratosphere there will be less mixing and far fewer clouds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Jupiter the tropopause (the point of lowest temperature) occurs at a pressure of about 0.1 bars. (&lt;a href=&quot;http://www.windows2universe.org/jupiter/lower_atmosphere.html&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-04-01T17:13:53.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14384" PostTypeId="2" ParentId="8485" CreationDate="2016-04-01T18:50:26.777" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does that suggest that about a half of the stars in the observable universe could not belong to any galaxy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not really.  A key sentence in the article is &quot;The best interpretation is that we are seeing light from stars outside of galaxies but in the same dark matter halo&quot;.  So the stars are still within the dark matter halo of a galaxy, but are outside the boundary of the galaxy if the dark matter halo is not considered.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore the &quot;intrahalo light&quot; explanation is just one of two possible explanations according to &lt;a href=&quot;http://arxiv.org/pdf/1601.02514v1.pdf&quot; rel=&quot;nofollow&quot;&gt;Updated analysis of near-infrared background fluctuations&lt;/a&gt; which explains:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two scenarios have been proposed to interpret the clustering excess. The first advocates the contribution from intrahalo light (IHL), i.e. relatively old stars stripped from&#xA;  their parent galaxies following merging events. These stars&#xA;  therefore reside in between dark matter halos and constitute a low-surface brightness haze around galaxies. The IHL&#xA;  is expected to come mostly from low redshifts (1 + z &amp;lt;&#xA;  ∼ 1.5)&#xA;  systems (Cooray et al. 2012b; Zemcov et al. 2014).  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The second scenario is instead based on the presence&#xA;  of a class of early, highly obscured accreting black holes&#xA;  of intermediate mass (∼ 10^4−6M⊙) at z &gt;&#xA;  ∼ 13 (Yue et al.&#xA;  2013b, 2014). As a suitable mechanism to produce such&#xA;  objects does exist – the so called Direct Collapse Black&#xA;  Holes (DCBH, for a concise overview of the problem see&#xA;  Ferrara et al. 2014), and the interpretation of the super-&#xA;  massive black holes observed at z = 6 seemingly requires&#xA;  massive seeds (Volonteri &amp;amp; Bellovary 2011), such hypothesis seems particularly worth exploring.   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Both scenarios successfully explain the observed clustering excess, albeit with apparently demanding requirements. In fact, if the excess is to be explained by intra-&#xA;  halo light, then a large fraction of the stars at low-z must&#xA;  reside outside systems that we would normally classify as&#xA;  “galaxies” (Zemcov et al. 2014). On the other hand, in the&#xA;  DCBH scenario the abundance of seed black holes produced&#xA;  until z ∼ 13 must represent a sizeable fraction of the estimated present-day black hole abundance, as deduced from&#xA;  local scaling relations (Kormendy &amp;amp; Ho 2013) and recently&#xA;  revised by Comastri et al. (2015). However, it is important&#xA;  to outline that both scenarios are not in conflict with any&#xA;  known observational evidence&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11382" LastEditorUserId="11382" LastEditDate="2016-04-01T18:57:16.047" LastActivityDate="2016-04-01T18:57:16.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14385" PostTypeId="2" ParentId="10859" CreationDate="2016-04-01T21:14:50.400" Score="2" Body="&lt;p&gt;I know this is an old question, but recent developments in astronomy have caused this to become relevant again. Early in 2016 there was a paper published by Konstantin Batygin and Michael E. Brown (&lt;a href=&quot;http://arxiv.org/abs/1601.05438&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) which indicates the possible existence of a new planet, dubbed &quot;Planet 9&quot;. Subsequent work by other authors have lent more evidence to its possible existence, e.g., a paper by Fienga et al. (&lt;a href=&quot;http://arxiv.org/abs/1602.06116&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) which concludes that including Planet 9 in orbital dynamic models decreases residuals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This being said, the original paper specifically states&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...our calculations suggest that a perturber on a $a'\sim700\:AU$, $e'\sim0.6$ orbit would have to be somewhat more massive (e.g. a factor of a few) than $m'=10\:M_\oplus$ to produce the desired effect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Such a planet is certainly massive enough to have cleared out its orbit under normal circumstances. For reference, Neptune has a mass of $17\:M_\oplus$ while Uranus has a mass of $14.5\:M_\oplus$. The special feature of this planet is that it is located $700\:AU$ distant from the Sun which is in a very ambiguous and unknown region of our solar system. It is not clear at all how much, if any, debris in the form of comets resides out there. There are models suggesting a disk (referred to as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hills_cloud&quot; rel=&quot;nofollow&quot;&gt;Hills Cloud&lt;/a&gt;) exists at those distances, implying that this new planet could reside inside an extended cometary disk. Of course this is only speculation, but it is a distinct possibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the above-mentioned disk exists, we have to consider a few scenarios.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Planet 9 was &quot;inserted&quot; into this disk by being expelled from the inner solar system early on in its formation. This is a possible cause for its current position and suggests one of two outcomes.&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It has existed here long enough that it has cleared out its orbit, thus making its planetary discriminant well within the &quot;planetary&quot; side of the definition.&lt;/li&gt;&#xA;&lt;li&gt;It has either not existed in the disk long enough to clear the orbit, or has achieved some sort of equilibrium such that it can't now clear the orbit (considering most of the clearing occurs during formation). Here is where you may run into an ambiguous case. &lt;em&gt;For all intents and purposes we may consider this a planet, but there remains the possibility that, despite it being more massive than the Earth, it could formally be considered a dwarf planet or else exist in this fuzzy region as you suggest.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;Planet 9 formed at its current orbit (plus or minus some migration).&#xA;In this scenario I think it is clear that it would have cleared its&#xA;orbit and resoundingly be considered a true planet.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now these are all wild suppositions. I don't have any of the math to back it up, and I'm not sure the numbers to plug into the various equations are well known, or even exist. I'm merely remarking on the possibility that the situation you asked about now seems like it could exist.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2016-04-01T21:21:40.207" LastActivityDate="2016-04-01T21:21:40.207" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14386" PostTypeId="1" AcceptedAnswerId="14398" CreationDate="2016-04-01T21:37:28.190" Score="3" ViewCount="1084" Body="&lt;p&gt;After finding out that there is a solar system with three stars, I did extensive research. I couldn't get a definitive answer. What would happen if they collided with each other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there would be total destruction everywhere, but I would like to know what, including more scientific short- and long- term effects, would happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can't explain three stars, two stars colliding would be okay. I'm still an amateur at astronomy, so please don't make the answer too complicated.&lt;/p&gt;&#xA;" OwnerUserId="11369" LastEditorUserId="11369" LastEditDate="2016-04-02T14:30:49.587" LastActivityDate="2016-04-02T17:03:09.517" Title="What happens if two or three stars collide with each other?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14387" PostTypeId="1" CreationDate="2016-04-01T21:43:36.533" Score="7" ViewCount="5612" Body="&lt;p&gt;My understanding of a white dwarf star is that, although it hasn't happened yet due to the age of the universe, eventually as it loses energy due to radiating heat and tidal forces, it will cool to become a black dwarf - essentially a black ball of carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens when a neutron star eventually cools? What is left, and do we know what it would look like?&lt;/p&gt;&#xA;" OwnerUserId="11419" LastActivityDate="2018-05-22T08:25:34.083" Title="What happens over time as a neutron star cools?" Tags="&lt;neutron-star&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14388" PostTypeId="1" CreationDate="2016-04-01T22:18:48.503" Score="4" ViewCount="73" Body="&lt;p&gt;When we first observe a new heavenly body (it could be a new star, asteroid, etc.,or a minor planet in our own solar system), are there any procedures set in place for establishing a system of longitudinal meridians? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being that a prime meridian is an arbitrary concept that you can pick and establish anywhere, is a location decided based on a physical feature, or perhaps from the first points of data we gathered when making detailed observations of the heavenly bodies for the first time? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about heavenly bodies with no easily discernible or non-stationary features (Gas Giants)? Also does the IAU regulate this process? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have always wondered if the USSR and the US shared common reference points for locations such as on the moon. &lt;/p&gt;&#xA;" OwnerUserId="11397" LastEditorUserId="10172" LastEditDate="2016-06-22T20:26:53.337" LastActivityDate="2017-01-08T19:45:11.163" Title="Protocol for establishing longitudinal meridians on other heavenly bodies" Tags="&lt;coordinate&gt;&lt;iau&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14389" PostTypeId="2" ParentId="14387" CreationDate="2016-04-01T23:09:52.343" Score="4" Body="&lt;p&gt;To begin with, it's important to discuss just how a neutron star cools. There are several related mechanisms involved in the main stage of a neutron star's thermal evolution, and they mainly rely on neutrino emission (primary reference &lt;a href=&quot;http://arxiv.org/abs/1501.04397&quot; rel=&quot;noreferrer&quot;&gt;Lim et al. (2015)&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First and foremost is the family of &lt;a href=&quot;https://en.wikipedia.org/wiki/Urca_process&quot; rel=&quot;noreferrer&quot;&gt;Urca processes&lt;/a&gt;. The direct Urca process (Durca) is&#xA;$$\text{baryon }1\to\text{baryon }2+\text{lepton}+\text{antineutrino}$$&#xA;$$\text{baryon }2+\text{lepton}\to\text{baryon }1+\text{neutrino}$$&#xA;The baryons here are typically neutrons and protons, and the lepton is an electron, with its corresponding antineutrino. However, it can also happen with &lt;a href=&quot;http://www.jinaweb.org/docs/msu.pdf&quot; rel=&quot;noreferrer&quot;&gt;pions, kaons, and quark matter&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next up is &lt;a href=&quot;https://en.wikipedia.org/wiki/Bremsstrahlung&quot; rel=&quot;noreferrer&quot;&gt;bremsstrahlung&lt;/a&gt;. This version is unlike the more commonly known variant where radiation is emitted from electrons. Instead, it is nucleon-nucleon bremsstrahlung, when two neutrons, two protons, or a proton and a neutron interact to produce a neutrino/antineutrino pair. In fact, one of the main differences between this bremsstrahlung and the Urca process is the lack of electrons in the former reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Cooper_pair&quot; rel=&quot;noreferrer&quot;&gt;Cooper pair&lt;/a&gt; decay takes place in neutron or proton pairs. The formation of the pairs is made possible by the extreme conditions inside the neutron star, which cause superfluidity and influence many of the other processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These three mechanisms take place in the core of the neutron star and throughout it. However, some other pathways take place only in the crust. They include&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$nn$-bremsstrahlung, which is characterized in part by free neutrons&lt;/li&gt;&#xA;&lt;li&gt;Bremsstrahlung through the interaction of an electron with a nucleus ($eZ$ bremsstrahlung)&lt;/li&gt;&#xA;&lt;li&gt;The decay of &lt;a href=&quot;https://en.wikipedia.org/wiki/Plasmon&quot; rel=&quot;noreferrer&quot;&gt;plasmons&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Electron-positron annihilation&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now that we know the processes, what are the results?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above processes are actually only dominant in the second of three stages in a neutron star's life (see &lt;a href=&quot;http://cds.cern.ch/record/479566/files/0012122.pdf&quot; rel=&quot;noreferrer&quot;&gt;Yakovlev et al.&lt;/a&gt;). The first is the internal relaxation stage, lasting from ~10-10&lt;sup&gt;3&lt;/sup&gt; years, ending when the neutron star reaches thermal relaxation. The second is this neutrino cooling stage, lasting to ~10&lt;sup&gt;5&lt;/sup&gt; years. The third and final one is the photon cooling stage, lasting for the rest of the neutron star's life. At this point, the star is simply getting cooler and cooler. It is not known how long this stage can last, but it doesn't appear that any boundary has been placed on it.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2016-04-01T23:09:52.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14390" PostTypeId="2" ParentId="13895" CreationDate="2016-04-02T00:43:36.053" Score="-5" Body="&lt;p&gt;I'm not a scientist, just someone who likes to read and think about this stuff. I think mass would be transferred. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reasoning is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Black holes aren't holes, they are masses.&lt;/strong&gt; Gravity comes from mass. Each black hole has immense gravitational pull as they likely have incredibly dense masses in them. &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;However, they do not have equal gravity.&lt;/strong&gt; So what happens when two masses with enormous gravity pull on each other? The stronger pull wins.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;They aren't static either&lt;/strong&gt;. Black holes do grow as they pull in more mass, so that implies that they can grow and so can probably shrink. &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;I don't think black holes are inescapable.&lt;/strong&gt; Light cannot escape but light has a constant speed while the force of gravity is theoretically infinite. Think about it, they grow instead of becoming an infinitely dense singular point. That means that mass resists the pull to further contract into a denser ball. So, if a black hole's gravity can be resisted, why not from a bigger gravitational pull pulled matter away from a smaller gravitational pull?&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;There's no such thing as empty space.&lt;/strong&gt; We are slowly realizing that empty space is likely filled with dark matter and other hard to detect substances. So the space that appears black isn't empty. So your scenario is still mass acting upon mass.&lt;/li&gt;&#xA;&lt;li&gt;I believe it has been shown that black holes have merged, so that process likely started when two event horizons collided.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Again, I'm no scientist, but it seems quite likely that mass is exchanged. &lt;/p&gt;&#xA;" OwnerUserId="10247" LastActivityDate="2016-04-02T00:43:36.053" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14391" PostTypeId="1" CreationDate="2016-04-02T02:34:23.220" Score="0" ViewCount="1238" Body="&lt;p&gt;I need to know the definition of deep space.  I am an amateur astronomer, and am experiencing some difficulty understanding this.&lt;/p&gt;&#xA;" OwnerUserId="11421" LastActivityDate="2016-04-03T04:06:22.543" Title="What is the definition of &quot;deep space&quot;?" Tags="&lt;distances&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14392" PostTypeId="2" ParentId="14391" CreationDate="2016-04-02T04:49:49.697" Score="1" Body="&lt;p&gt;I wonder if perhaps you're overthinking things a little bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Places in the ocean that are very far below the Earth's surface are referred to as &quot;deep waters&quot;. Deep space could be defined in the exact same way, but in the opposite direction. &lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2016-04-02T04:49:49.697" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14393" PostTypeId="2" ParentId="14391" CreationDate="2016-04-02T06:24:45.417" Score="3" Body="&lt;p&gt;&lt;strong&gt;Deep space&lt;/strong&gt; could be simply defined, as a part of the universe, which is lying beyond the Earth-Moon system. However, when speaking about &lt;strong&gt;deep space objects&lt;/strong&gt; we generally mean stars, star clusters and nebulae, which are outside the Solar System.&lt;/p&gt;&#xA;" OwnerUserId="11048" LastActivityDate="2016-04-02T06:24:45.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14394" PostTypeId="2" ParentId="14387" CreationDate="2016-04-02T07:39:33.240" Score="10" Body="&lt;p&gt;The cooling history of a neutron star can be divided into an extremely rapid neutrino cooling phase, followed by an indefinitely long cooling phase due to the emission of photons from its surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key to understanding neutron star cooling is to realise that degenerate neutrons possess almost no thermal energy, even when extremely hot - because the Pauli Exclusion principle forbids the neutrons from losing kinetic energy and falling into an already-occupied quantum state. Any cooling processes therefore act to reduce the temperature of the neutron star very quickly. Secondly, the interior of a neutron star is almost isothermal, due to the high thermal conductivity of degenerate gases, but the temperature at the surface is smaller by about a factor of 100 or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a neutron star first forms, its internal temperature exceeds 10 billion kelvin, its surface temperature would be 100 million degrees and emit hard X-rays. Nevertheless, the small size of the neutron star limits radiative losses and instead it is neutrinos escaping from the interior that can cool it by a factor of around 100 in a few thousand years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The neutrino losses proceed by something called the &lt;em&gt;modified URCA&lt;/em&gt; process (somewhat faster direct URCA or processes involving pions/kaons maybe important in the first seconds/minutes of more massive neutron stars, but are blocked by complete neutron degeneracy and the comparative paucity of protons in the interior): cycles of neutron beta decay and inverse beta decay on protons produce anti-neutrinos and neutrinos. Bystander particles are required to conserve momentum. The temperature dependence of this process is $\sim T^{8}$, so if the temperature falls by a factor 100, the neutrino cooling rate falls by a factor of $10^{16}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, photon cooling from the surface, whilst initially negligible &lt;em&gt;compared to neutrino cooling&lt;/em&gt;, falls only as $T^{4}$. After around 10,000-100,000 years, when the neutron star surface has cooled to about a million degrees, photon cooling, through soft X-ray emission dominates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If nothing else happened to an isolated neutron star, it would continue to cool, such that $T \propto t^{-1/2}$. The &lt;em&gt;vast majority of the estimated 1 billion neutron stars in our Galaxy, are in this state&lt;/em&gt;, whereas most of the pulsars we can observe are young and in the neutrino-cooling phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How cold can they get? The formula above suggests their temperature halves when their age quadruples. Following this, they could cool to the temperature of the Sun in around a billion years. However, there are reheating processes that can occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isolated neutron stars can accrete material from the interstellar medium. This may be very little, since most neutron stars move fast. The gravitational potential energy is partly released as heat when it impacts the neutron star surface. Then, the low heat capacity of neutrons works in the opposite direction and comparatively little accretion is required to balance the photon losses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, neutron stars are born as fast rotators and spin down. Angular momentum must be transferred outwards from the fluid interior to the outer crust. This is not a frictionless process. Thus some of the rotational kinetic energy can keep the star warm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thirdly, the neutron star has a huge magnetic field that gradually dissipates. Some of that energy may end up as heat in the neutron star through simple Ohmic dissipation of currents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line is that the thermal emission from neutron stars older than about 100,000 years has not been observed, so we simply don't know how cool they might get. Some more information and references can be found &lt;a href=&quot;https://physics.stackexchange.com/questions/156050/position-of-neutron-stars-in-h-r-diagrams/156072#156072&quot;&gt;in the Physics SE answer&lt;/a&gt; where I discuss where old neutron stars might appear on the HR diagram.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-05-22T08:25:34.083" LastActivityDate="2018-05-22T08:25:34.083" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14397" PostTypeId="1" AcceptedAnswerId="14419" CreationDate="2016-04-02T09:58:00.800" Score="1" ViewCount="218" Body="&lt;p&gt;I originally went to math.stackexchange.com and they suggested I try here as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to figure out how to calculate times of moonrise and moonset for a given position on the earth on a particular date. I'm using Vallado's Fundamentals of Astrodynamics and Applications because there is an example in the book that lays out a simplified process. I'm stuck at the point where I'm trying to calculate Longitude of the ecliptic for the Moon. The formula is below where T = -0.013634497.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;λEcliptic = 218.32° + 481,267.8813T + 6.29Sin(134.9 + 477,198.85T) - 1.27Sin(259.2 - 413,335.38T) + 0.66Sin(235.7 + 890,534.23T) + 0.21Sin(269.9 + 954,397.70T) - 0.19Sin(357.5 + 35,999.05T) - 0.11Sin(186.6 + 966,404.05T)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The expected answer is -0.8412457°. However, I'm unable to figure out how to get this answer. My calculations are below: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) 218.32° + 481,267.8813T = -6343.525&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2) 6.29Sin(134.9 + 477,198.85T) = 5.963779&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;3) -1.27Sin(259.2 - 413,335.38T) = -0.900843&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;4) 0.66Sin(235.7 + 890,534.23T) = -0.292285&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;5) 0.21Sin(269.9 + 954,397.70T) = -0.126871&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;6) -0.19Sin(357.5 + 35,999.05T) = 0.138211&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;7) -0.11Sin(186.6 + 966,404.05T) = 0.054722&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Simply adding or subtracting gets me -6338.688731. How can I get the value -0.8412457° from this. My trigonometry is rusty and I'm not sure how to get the right answer. Any help would be greatly appreciated. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="11424" LastActivityDate="2016-04-04T02:25:43.263" Title="Longitude of the Ecliptic" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;rotation&gt;&lt;ecliptic&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14398" PostTypeId="2" ParentId="14386" CreationDate="2016-04-02T17:03:09.517" Score="2" Body="&lt;p&gt;The collision of another star with our sun would be a very effective way of ending all life on Earth. Fortunately stars are so far apart that the the chance of it happening in the life of the sun is very close to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in the densely packed cores of globular clusters the rate of collision is much higher, and there may be a collision somewhere in the milkyway once every 10000 years or so. Three way collisions don't happen (or are exceptionally unlikely) but would be similar to two two-star collisions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result varies depending on the size of the stars, their density, the relative velocity, and whether a collision is head on or not. Generally a denser star is likely to come out of a collision in better shape. So a red giant, could have most of its very rarified outer layers pulled off and the core escaping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if two main sequence stars, like our sun, collide, they first compress each other. Some of the stellar envelope escapes perpendicular to the collision (think of jam being squeezed out of a sandwich). But within a few hours the two stars have mixed and merged, and   a new, larger star formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mixing adds lots more hydrogen to the core of the star, which causes it to rejuventate. The first evidence of stellar collisions is in globular clusters, in which there are some large blue stars that seem younger than all the other stars in the cluster (Blue stragglers)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A white dwarf and main sequence star collision is bad news for the star. The collision causes the core of the star to heat up and a massive release of thermonuclear energy. Plenty more than is needed to completely disrupt the star. The star is completely destroyed in the process. However the white dwarf is relatively unscathed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;White dwarf-white dwarf collisions can, depending on the size result in gravitational collapse to a neutron star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper &lt;a href=&quot;http://adsabs.harvard.edu/full/2002ASPC..263....1S&quot; rel=&quot;nofollow&quot;&gt;Stellar collisions, mergers and their consequences&lt;/a&gt; is a very readable summary.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-04-02T17:03:09.517" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14399" PostTypeId="1" AcceptedAnswerId="14401" CreationDate="2016-04-02T17:24:39.897" Score="3" ViewCount="334" Body="&lt;p&gt;If I understand it correctly, the apparent horizon is the boundary between where light directed outwards will move outwards, and light directed outwards will move inwards. Even if light is moving away from the black hole outside of the apparent horizon, it will not escape to infinity if it is within the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The event horizon is essentially the apparent horizon in the future, if the apparent horizon continues growing. Theoretically, the two horizons should eventually meet up...again, if I understand it correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My confusion is with how the two form. According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_horizon#Differences_from_the_.28absolute.29_event_horizon&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the simple picture of stellar collapse leading to formation of a black hole, an event horizon forms before an apparent horizon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My main question is, how can the event horizon form &quot;before&quot; the apparent horizon? I understand that one can slice the Schwarzschild geometry so that there is no apparent horizon at all, but if an apparent horizon exists, I still don't understand how or why it would form &lt;strong&gt;after&lt;/strong&gt; the event horizon.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2016-04-02T21:36:29.367" Title="How does an absolute horizon form before the apparent horizon?" Tags="&lt;gravity&gt;&lt;black-hole&gt;&lt;space-time&gt;&lt;general-relativity&gt;&lt;supermassive-black-hole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14400" PostTypeId="2" ParentId="14376" CreationDate="2016-04-02T18:33:01.353" Score="2" Body="&lt;p&gt;Big bang nucleosynthesis did not happen &quot;close to the inflationary period&quot;.&#xA;inflation happened at about $10^{-36}$s. Nucleosynthisis happened from about 10seconds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In relative terms, the ratio of the age of the universe now, to 10 seconds is much much smaller than the ratio of 10 seconds to $10^{-36}$s. And, significantly, nucleosynthesis occurred after inflation, so can't be the cause of inflation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for dark energy, nobody has any real model for it. The test for any theory is how well it does empirically. As it stands you have advanced a pet theory, without any mathematical model or analysis.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-04-02T18:33:01.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14401" PostTypeId="2" ParentId="14399" CreationDate="2016-04-02T20:38:36.967" Score="3" Body="&lt;p&gt;I guess you are talking about black hole formation here. If we take the &lt;a href=&quot;https://physics.aps.org/story/v13/st23&quot; rel=&quot;nofollow&quot;&gt;Oppenheimer-Snyder&lt;/a&gt; model for the spherically symmetric collapse of a star, then an event horizon forms first followed later by an apparent horizon that is at, or interior to, the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The event horizon is the surface behind which light rays will never reach an infinite distance from the black hole. The apparent horizon is the surface behind which outwardly directed light rays will move inward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two will coincide for a static black hole, but for a forming black hole with a growing mass, it is possible for an outwardly directed photon to be outside the apparent horizon, but inside the event horizon, because the black hole mass is inevitably larger in the future if some of its future mass is still beyond its final Schwarzschild radius. As the event horizon forms at the centre of the star and moves outwards, then it is clear that the apparent horizon, which is always interior to the event horizon, must form later. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the very simplified Oppenheimer-Snyder model, featuring an initially uniform star comprising of pressureless &quot;dust&quot;, the apparent horizon first forms just as the surface of the collapsing star coincides with the event horizon and the apparent horizon is therefore always coincident with the event horizon. In more realistic models the apparent horizon forms a little earlier and then moves outwards to coincide with the event horizon as the surface of the collapsing star moves inside the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB: Note that whilst the apparent horizon can be defined at any point in time, the &quot;event horizon&quot; referred to above is a theoretical ideal, since we cannot preclude that the black hole might accrete some more mass in the future!&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-04-02T21:36:29.367" LastActivityDate="2016-04-02T21:36:29.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14402" PostTypeId="2" ParentId="14391" CreationDate="2016-04-03T04:06:22.543" Score="1" Body="&lt;p&gt;It's not in any sort of academic context, but in relation to space exploration &lt;a href=&quot;http://www.nasa.gov/exploration/systems&quot; rel=&quot;nofollow&quot;&gt;NASA&lt;/a&gt; defines deep space as such:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Deep space is the vast region of space that extends beyond our Moon, to Mars and across our solar system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, not exactly an academic context, but &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1579129781&quot; rel=&quot;nofollow&quot;&gt;Govert Schiling&lt;/a&gt; (the author of &lt;em&gt;Deep Space: Beyond the Solar System to the End of the Universe and the Beginning of Time&lt;/em&gt;) in relation to astronomy defines deep space differently:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In a sense, everything beyond Earth’s atmosphere is deep space. But more typically it’s a phrase that has been used by observational and amateur astronomers alike for extended objects beyond the Solar System. I decided to provide a brief introductory section on the Sun and the planets, but the emphasis is really on the objects in our Galaxy and the wider Universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So this more or less echoes what some others have contributed, but provides additional context.&lt;/p&gt;&#xA;" OwnerUserId="6" LastActivityDate="2016-04-03T04:06:22.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14404" PostTypeId="1" CreationDate="2016-04-03T10:16:21.530" Score="1" ViewCount="100" Body="&lt;p&gt;The actual time for one day is 23hrs, 56mins, 4.1secs right? Then how can the clocks and watches can show perfect time? &#xA;I mean, if I observe the sun rise at 6:00am this day, tomorrow I should observe it before 6:00 or I can say at 5:56 approximately as our measuring devices follow 12+12 i.e., 24hr day. But in practice again it will be 6:00am when the sun rises. How it is possible?&lt;/p&gt;&#xA;" OwnerUserId="11429" LastActivityDate="2016-04-08T23:01:02.120" Title="How the watches are showing perfect time?" Tags="&lt;space-time&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14405" PostTypeId="2" ParentId="14404" CreationDate="2016-04-03T11:07:11.033" Score="5" Body="&lt;p&gt;Those 23hrs, 56mins, 4.1secs are actually the &lt;a href=&quot;https://en.wikipedia.org/wiki/Sidereal_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;sidereal&lt;/a&gt; rotation period of Earth. It is how much time the Earth needs to perform one rotation relative to the stars. But that is not exactly the same as solar time. By the time the Earth has completed one rotation, it has also moved a little forward in its orbit around the Sun. Thus, it needs a few more minutes of rotation to catch up so the Sun is at the same location of the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uI2MHm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uI2MHm.png&quot; alt=&quot;solar time&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="616" LastEditDate="2016-04-08T23:01:02.120" LastActivityDate="2016-04-08T23:01:02.120" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14406" PostTypeId="1" AcceptedAnswerId="19675" CreationDate="2016-04-03T12:25:53.483" Score="3" ViewCount="283" Body="&lt;p&gt;Last night, I used my new Sony WX500 camera to get long exposure pictures of the sky from my backyard.  I used the &quot;Heavens Above&quot; app to find out when satellite flares would happen and I even got a great picture of the ISS passing overhead at roughly -2mag.  I was able to identify many flares that I had gotten pictures of when reviewing the pictures today.  However, this is where it gets odd.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some moving &quot;things&quot; didn't match what was supposed to show according to the app.  Some were accidentally caught in the pictures I took, and I'm sure there aren't any flares that the app showed me.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could they be?  Shooting stars?  Forgotten or broken satellites?&lt;/p&gt;&#xA;" OwnerUserId="11343" LastEditorUserId="10172" LastEditDate="2016-06-22T20:28:24.417" LastActivityDate="2017-05-08T21:10:39.967" Title="Unknown moving objects" Tags="&lt;photography&gt;&lt;satellite&gt;&lt;flare&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ClosedDate="2017-10-30T13:18:55.623" ContentLicense="CC BY-SA 3.0" />
  <row Id="14407" PostTypeId="1" CreationDate="2016-04-03T13:29:05.953" Score="1" ViewCount="258" Body="&lt;p&gt;Can we determine the physical structure of a black hole by observing its gravitational effect on objects in orbit?  There are three possibilities that I see, and would like to test:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Singularity at the center - The traditional view that there is a singularity at the center (e.g. all mass is concentrated in the center).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A structure where the black hole is filled throughout with matter inside the event horizon - My proposal, due to gravitational time dilation slowing / stopping time (e.g. a singularity will not form until an infinite amount of time has passed).  If the answer to my question on this subject here is correct, we should observe either possibility 2 or 3:  &lt;a href=&quot;https://astronomy.stackexchange.com/questions/2441/does-matter-accumulate-just-outside-the-event-horizon-of-a-black-hole&quot;&gt;Does matter accumulate just outside the event horizon of a black hole?&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A hollow shell, with matter only on / near the outside edge of the event horizon - My proposal, combined with the &quot;vapor bubble&quot; behavior, as hinted at in some of the answers to the linked question.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Would these three possibilities differ gravitationally in ways that could be observed (e.g. by observing orbiting stars), or would they be identical from the outside?  If they would produce different results, I would love to see what result it is that we actually observe!  We have observed stars orbiting the super massive black hole at the center of the galaxy.  &lt;a href=&quot;http://scitechdaily.com/astronomers-discover-star-orbiting-the-black-hole-at-center-of-the-milky-way/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://scitechdaily.com/astronomers-discover-star-orbiting-the-black-hole-at-center-of-the-milky-way/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-05-06T13:31:39.997" Title="Could we verify the structure of a black hole by observing an orbiting object?" Tags="&lt;gravity&gt;&lt;black-hole&gt;&lt;supermassive-black-hole&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14408" PostTypeId="1" AcceptedAnswerId="14413" CreationDate="2016-04-03T15:03:01.317" Score="1" ViewCount="76" Body="&lt;p&gt;Did they just put a ruler on the photographic plate? Quite some precision must have been required to discover Neptune from its influence on Uranus for example.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-04-03T19:43:28.427" Title="How was precision astrometry done before digital imaging?" Tags="&lt;history&gt;&lt;photography&gt;&lt;astrometry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14409" PostTypeId="2" ParentId="14408" CreationDate="2016-04-03T15:38:42.287" Score="0" Body="&lt;p&gt;Micrometers and &lt;a href=&quot;https://en.wikipedia.org/wiki/Blink_comparator&quot; rel=&quot;nofollow&quot;&gt;blink comparators&lt;/a&gt; allowed considerable precision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pluto was discovered with the aid of the latter.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2016-04-03T15:38:42.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14410" PostTypeId="2" ParentId="2362" CreationDate="2016-04-03T16:43:18.667" Score="2" Body="&lt;p&gt;Yes, Frank Drake looked into this - see page 91 here&#xA;&lt;a href=&quot;http://www.lpi.usra.edu/lunar/strategies/objectives/ast_observations_moon.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.lpi.usra.edu/lunar/strategies/objectives/ast_observations_moon.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He calculated that you could make a telescope with diameter 30 km using steel, and up to 60 to 90 km with stronger materials. And no problem with wind loading on the Moon.&lt;/p&gt;&#xA;" OwnerUserId="11432" LastActivityDate="2016-04-03T16:43:18.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14411" PostTypeId="1" AcceptedAnswerId="14422" CreationDate="2016-04-03T17:14:01.930" Score="1" ViewCount="75" Body="&lt;p&gt;It's clear the universe is expanding because light from distant bodies, like galaxies, are shifted towards the red side of the spectrum.  In other words, the light measured is more red than it should be when compared to laboratory results.  This is great for measuring the distance to other bodies, but can't we also measure their velocities of departure by measuring how fast the light from these bodies is becoming more and more red over time?  That is, v is directly proportional to the change of frequency divided by the time elapsed?&lt;/p&gt;&#xA;" OwnerUserId="5306" LastActivityDate="2016-04-04T08:14:29.603" Title="Can we tell how fast bodies are moving away by measuring their frequency?" Tags="&lt;distances&gt;&lt;velocity&gt;&lt;doppler-effect&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14412" PostTypeId="1" CreationDate="2016-04-03T17:36:15.763" Score="0" ViewCount="181" Body="&lt;p&gt;Is there any python/IDL code available to create synthetic test images of stars with and without background noise?&lt;/p&gt;&#xA;" OwnerUserId="11433" LastActivityDate="2018-07-20T22:21:09.090" Title="How to create synthetic test images of stars?" Tags="&lt;astrophysics&gt;&lt;radio-astronomy&gt;&lt;photometry&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14413" PostTypeId="2" ParentId="14408" CreationDate="2016-04-03T19:43:28.427" Score="5" Body="&lt;p&gt;Prior to digital imaging then photographic plate negatives were analysed with scanning microdensitometers to produce astrometric catalogues. Many of these catalogues are still in use today, they are valuable sources of early epoch positions that enable proper motion measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details you could look at the descriptions of the SuperCosmos project &lt;a href=&quot;http://ssa.roe.ac.uk//&quot; rel=&quot;noreferrer&quot;&gt;http://ssa.roe.ac.uk//&lt;/a&gt; (which is based on Schmidt plates) or the UCAC4 catalogue, &lt;a href=&quot;http://ad.usno.navy.mil/ucac/readme_u4v5&quot; rel=&quot;noreferrer&quot;&gt;http://ad.usno.navy.mil/ucac/readme_u4v5&lt;/a&gt;, which uses plates to get positions and proper motions for faint stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the good old days, before even these catalogues existed, astronomy groups would have copies of the whole Schmidt and Palomar sky surveys. You would put the relevant plate on a massive, concrete-based, hydraulic X,Y measuring machine, with a binocular microscope. You would measure sets of Standard stars, get a 6-coefficient fit to convert x and y into RA and Dec,then find your objects on the plate, measure x,y, calculate RA and Dec. Then snap a polaroid to use as a finder chart at the telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was doing this as late as 1995 before the advent of the Digitised Sky Survey. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-04-03T19:43:28.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14417" PostTypeId="2" ParentId="14411" CreationDate="2016-04-03T22:31:43.033" Score="1" Body="&lt;p&gt;When light is emmitted by an object that is moving away from us, the light is shifted to the red. This is an example of the doppler effect. So the &lt;strong&gt;red shift measures radial velocity, not distance&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By estimating the distance to galaxies, Edwin Hubble found that the velocity of galaxies correlates with their distance. So by measuring red shift you can estimate the distance of a galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance is only an estimate, and there is cosiderable uncertainty in our knowledge of the distance of galaxies. Hundreds of millions of light years of uncertainty. The distance to even the fastest moving galaxies doesn't change fast enough to be measured. Even if a galaxy is moving close to the speed of light, in one year it would only have travelled 1 light year, much less than the uncertainty in the estimate of the distance. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-04-04T07:35:58.230" LastActivityDate="2016-04-04T07:35:58.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14418" PostTypeId="1" AcceptedAnswerId="14427" CreationDate="2016-04-04T00:10:11.003" Score="1" ViewCount="650" Body="&lt;p&gt;Given two bodies of equal mass in an elliptical orbit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know they will be orbiting about a common center of mass, i.e. the barycenter. But, do the velocities have to be equal in magnitude and opposite in direction (normal to R when R is their distance from each other) for the orbit to be stable? I think, if the velocity of one mass were to vary with respect to the other it would create a moving barycenter in which the two masses would collide or throw on another out of the orbit, but I can't find any mathematical verification of this. &lt;/p&gt;&#xA;" OwnerUserId="10995" LastActivityDate="2016-04-04T20:26:00.383" Title="Two body orbit of equal masses" Tags="&lt;orbit&gt;&lt;celestial-mechanics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14419" PostTypeId="2" ParentId="14397" CreationDate="2016-04-04T02:25:43.263" Score="1" Body="&lt;p&gt;There's a mistake in the book (see page 280 on &lt;a href=&quot;https://books.google.be/books/about/Fundamentals_of_Astrodynamics_and_Applic.html?id=PJLlWzMBKjkC&amp;amp;redir_esc=y&quot; rel=&quot;nofollow&quot;&gt;google books&lt;/a&gt;): the values for $\lambda_{ecl}$ and $\phi_{ecl}$ should be swapped. Your result is correct:&#xA;$$&#xA;-6338.688^\circ\text{ mod } (360^\circ) = -218.688^\circ,&#xA;$$&#xA;which is the value that is erroneously listed under $\phi_{ecl}$.&lt;/p&gt;&#xA;" OwnerUserId="11439" LastActivityDate="2016-04-04T02:25:43.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14420" PostTypeId="1" CreationDate="2016-04-04T04:11:15.573" Score="0" ViewCount="218" Body="&lt;p&gt;Could you help me sorting this list in terms of total mass carried by these classes of objects?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The star (Sun) - it's very big but only one.&lt;/li&gt;&#xA;&lt;li&gt;The planets (incl. dwarf planets). Certainly beaten by the Sun, but...&lt;/li&gt;&#xA;&lt;li&gt;the asteroids, meteorites, comets, all that rubble. They aren't heavy individually but there's a lot of them.&lt;/li&gt;&#xA;&lt;li&gt;the interplanetary medium. It's very low density, but it overshadows the rest by far in terms of volume occupied.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm not sure where to attach the moons - some of them are larger than some dwarf planets, some are definitely more asteroid-sized. I guess - whoever decides to answer this, is free to pick where they'd like to include the moons (or maybe to split them between the two classes, say, attaching the ones with hydrostatic equilibrium to dwarf planets, and bundling the rest with the rubble)?&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2016-04-04T07:40:04.190" Title="Which class of objects in Solar System carries more mass?" Tags="&lt;solar-system&gt;&lt;mass&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14421" PostTypeId="2" ParentId="14420" CreationDate="2016-04-04T07:01:18.890" Score="1" Body="&lt;p&gt;The sun by far occupies the largest part of the solar system's mass. The mass of the sun happens to be approximately $(1.98855±0.00025)×10^{30}$ kg. Which is immensely huge. By comparison the inter-planetary medium though may occupy a large volume is nowhere dense enough to even compare to the mass of the sun which is approximately 99% of the solar system's mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of all planets combined by comparison happens to be approximately $2.6634×10^{27}$kg.&#xA;As far as interplanetary matter is concerned, its density is to low and its volume is pretty hard to define so mass remains a highly variable quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moons and asteroids account only for a fraction of the mass of the solar system though they happen to be present in large numbers. &lt;/p&gt;&#xA;" OwnerUserId="11441" LastEditorUserId="2365" LastEditDate="2016-04-04T07:40:04.190" LastActivityDate="2016-04-04T07:40:04.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14422" PostTypeId="2" ParentId="14411" CreationDate="2016-04-04T08:02:25.593" Score="2" Body="&lt;p&gt;Your understanding is correct. The doppler shift observed from a galaxy is the sum of its peculiar velocity with respect to the &quot;Hubble flow&quot; and the redshift due to the Hubble flow, which is caused by the expansion of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no direct way from a spectrum to separate these two components - they have the same qualitative result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, the expansion of the universe (or a change in the peculiar velocity) could be directly measured by looking for a change in redshift with time, which would depend on the cosmological parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an extremely small effect and is confused by the peculiar motions of individual galaxies. Nevertheless, measuring this redshift drift is one of the prime goals of the Codex Instrument on the E-ELT (see Pasquini et al. 2010, &lt;a href=&quot;http://esoads.eso.org/abs/2010Msngr.140...20P&quot; rel=&quot;nofollow&quot;&gt;http://esoads.eso.org/abs/2010Msngr.140...20P&lt;/a&gt; )&#xA;using Lyman alpha absorption systems towards distant quasars. This experiment is also planned for the Square Kilometre Array, using the 21cm line (Kloeckner et al. 2015 &lt;a href=&quot;http://arxiv.org/abs/1501.03822&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1501.03822&lt;/a&gt; ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases, to overcome the experimental uncertainties (eg at 21 cm, it amounts to line drifts of 0.1 Hz over a decade), then observations of millions of galaxies must be combined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no prospect of measuring this effect in an individual galaxy, furthermore I fear your understanding of cosmological redshift is flawed. The dependence on distance is a statistical average, not an absolute dependence. Individual galaxies are moving in individual gravitational potentials from objects around them. This gives them their peculiar velocities with respect to the flow. This velocity could increase or decrease as a galaxy got further away, but is never expected to be large enough to be detectable on human timescales for any individual galaxy. In addition, any change in &lt;em&gt;peculiar velocity&lt;/em&gt; should average to zero when looking at millions of galaxies, leaving the redshift drift due to the expansion.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-04-04T08:14:29.603" LastActivityDate="2016-04-04T08:14:29.603" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14423" PostTypeId="1" AcceptedAnswerId="14425" CreationDate="2016-04-04T14:28:11.303" Score="1" ViewCount="149" Body="&lt;p&gt;I am writing a book with the following scene in one chapter: It is trickling and my protagonists are outside looking into the night sky, looking to the stars. Reviewing the scene I am wondering: Is this possible? &lt;strong&gt;Can you see the stars when it is trickling, or even when it rains?&lt;/strong&gt; Are there images showing this?&lt;/p&gt;&#xA;" OwnerUserId="11443" LastActivityDate="2016-04-04T14:38:13.013" Title="Can you see the stars when it is trickling?" Tags="&lt;star&gt;&lt;amateur-observing&gt;&lt;night-sky&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14425" PostTypeId="2" ParentId="14423" CreationDate="2016-04-04T14:38:13.013" Score="2" Body="&lt;p&gt;Yes you can. You just need a partly clear sky (to see the stars) and part cloud (to produce the rain).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example of suitable conditions, look for pictures of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Moonbow&quot; rel=&quot;nofollow&quot;&gt;moonbow&lt;/a&gt;.  (I've only seen one once, personally.)&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2016-04-04T14:38:13.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14426" PostTypeId="2" ParentId="13615" CreationDate="2016-04-04T20:22:30.593" Score="0" Body="&lt;p&gt;The evidence suggests, strongly, that the expansion of the universe is accelerating. If an oscillating universe theory ruled that out then this would, in turn, dismiss this theory. As it is, it does strongly indicate that the idea that gravitational attraction would eventually slow the expansion of the universe and then send the expansion into reverse, is a false one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, you really need to be more specific in your question: &lt;strong&gt;which&lt;/strong&gt; theory of an oscillating universe is it that you are asking about?&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2016-04-04T20:22:30.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14427" PostTypeId="2" ParentId="14418" CreationDate="2016-04-04T20:26:00.383" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I know they will be orbiting about a common center of mass, i.e. the&#xA;  barycenter. But, do the velocities have to be equal in magnitude and&#xA;  opposite in direction (normal to R when R is their distance from each&#xA;  other) for the orbit to be stable?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The orbital velocities do not have to be and in general are not equal in magnitude. What is equal is the &lt;em&gt;angular velocity&lt;/em&gt;, that is the angular rate (e.g., in $rad/sec$) that two bodies will be orbiting their common barycenter. The orbital radius, $r$, orbital velocity, $v$, and angular velocity, $\omega$, are related by the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega = v/r$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that these are all scalar quantities and $v$ can be thought of as the component of the velocity vector which is perpendicular to $r$. Since conservation of angular momentum implies that $\omega$ remain constant for each body, individually, then we know that $v/r$ must also be constant which implies that bodies which orbit farther away from the barycenter must necessarily be orbiting faster and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now of course that simplifying explanation works for two bodies. Once you throw in more than two bodies, things become more complicated and the barycenter can move around, causing more complex motions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I think, if the velocity of one mass were to vary with respect to the&#xA;  other it would create a moving barycenter in which the two masses&#xA;  would collide or throw on another out of the orbit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a subtly different question. If one of the two masses had a varying orbital velocity, that would imply it was gaining or losing energy by some mechanism. This can occur through things like tidal interactions as it does for our Earth and Moon. As stated above, the angular velocity must remain constant which implies that the body whose velocity is changing must also migrate towards or away from the barycenter, potentially resulting in a collision or escape. Since the Moon's orbital velocity is being sped up through tidal interactions, it is moving farther away as a result. Another example might be two black holes emitting gravitational waves as they orbit which propagates away energy and thus they orbital velocities, causing them to get closer together until they collide.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-04-04T20:26:00.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14428" PostTypeId="2" ParentId="14009" CreationDate="2016-04-05T03:43:18.503" Score="1" Body="&lt;p&gt;I live in an 'L' shaped building on the 14th floor, so I have a pretty clear view of the horizon. Up until March 20th of this year, the sun was setting behind the other wing of the building. My living room window faces roughly NW, and on March 21st, I was able to see the sun at the horizon for the first time this year. Each day, the sun will set a little further northwest until it's setting almost directly opposite my window. Then it will begin to retreat until one day in September, when it will begin to set behind the other wing of the building again.&lt;/p&gt;&lt;p&gt;This is the first year I took note of the day that sunset was first visible, and I'm planning to note which day is the last day I can see the sun set.&lt;/p&gt;&#xA;" OwnerUserId="9516" LastActivityDate="2016-04-05T03:43:18.503" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14431" PostTypeId="2" ParentId="1724" CreationDate="2016-04-05T16:06:36.800" Score="0" Body="&lt;p&gt;We can assume that the stars are equal in mass, and their orbits are circular&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbital speed is 80000 m/s and at an orbital period of 10 months (or $2.628\times 10^7$s) the length of the orbit is $2.1024\times 10^{12}$ m or 14.05 AU The radius of the orbit therefore is $14.05/\tau$ = 2.237AU. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The version of Keplers law given is $$T^2 =\frac{a^3}{m_1+m_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;substituting $T^2 =(10/12)^2 = 0.6944$ (divide by 12 to convert to years)  and $a^3= 11.19$ gives $$m_1+m_2 = \frac{11.19}{0.6944} = 16.12\  \mathrm{solar\ mass}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $m_1=m_2$, the mass of each star is 8.06 solar masses, or $1.6\times 10^{31}$kg&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-04-05T16:06:36.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14432" PostTypeId="1" AcceptedAnswerId="14433" CreationDate="2016-04-05T19:30:41.543" Score="12" ViewCount="1213" Body="&lt;p&gt;If dark matter has gravity just like normal matter, does that mean it can also form planets, solar systems and so on?&#xA; Any answer will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="10765" LastActivityDate="2016-04-06T08:22:04.270" Title="Can dark matter be found in the shape of planets, galaxies etc.?" Tags="&lt;gravity&gt;&lt;dark-matter&gt;&lt;mass&gt;&lt;planet&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14433" PostTypeId="2" ParentId="14432" CreationDate="2016-04-05T20:22:24.547" Score="18" Body="&lt;p&gt;Planets and stars, no. Globular clusters and galaxies, yes.&lt;/p&gt;&#xA;&lt;h2&gt;Small scales&lt;/h2&gt;&#xA;&lt;p&gt;To condense into such relatively compact objects as planets, stars, and even the more diffuse star-forming clouds, particles need to be able to dissipate their energy. If they don't do this, their velocities prohibit them from forming anything.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Normal&amp;quot; particles, i.e. atoms, do this by colliding. When atoms collide, they're excited, and when they de-excite, they emit radiation which leaves the system, carrying away energy. In this way, an ensemble of particles can relax into a less energetic system, eventually condensing into e.g. a star. Additionally, the collisions cause more energetic particles to donate energy to the less energetic ones, making the ensemble reach &lt;em&gt;thermodynamic equilibrium&lt;/em&gt;, i.e. all particles have the same energy on average.&lt;/p&gt;&#xA;&lt;p&gt;Dark matter is, by definition, unable to collide and radiate, and hence, on such small scales as stars and planets, particles that enters a potential well with a given energy will maintain that energy. They will thus accelerate toward the center, then decelerate after its closest approach to the center, and finally leave the system with the same energy as before (if it was unbound to begin with). This makes it impossible for collisionless matter to form such small objects.&lt;/p&gt;&#xA;&lt;h2&gt;Large scales&lt;/h2&gt;&#xA;&lt;p&gt;On the scale of galaxies, however, various relaxation mechanisms allows dark matter to form structure. This is the reason that in pure N-body simulations of the Universe, such as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Millennium_Run&quot; rel=&quot;noreferrer&quot;&gt;Millennium Simulation&lt;/a&gt;, you will see galaxies. The sizes of these structures depend on the resolution, but are measured in millions of Solar masses.&lt;/p&gt;&#xA;&lt;p&gt;The relaxation mechanisms include:&lt;/p&gt;&#xA;&lt;strong&gt;Phase mixing&lt;/strong&gt;&#xA;&lt;p&gt;This is sort of like galaxy arms winding up, but in phase space rather than real space.&lt;/p&gt;&#xA;&lt;strong&gt;Chaotic mixing&lt;/strong&gt;&#xA;&lt;p&gt;This happens when particles come so close that their trajectories diverge exponentially.&lt;/p&gt;&#xA;&lt;strong&gt;Violent relaxation&lt;/strong&gt;&#xA;&lt;p&gt;The two mechanisms listed above assume a constant gravitational potential &lt;span class=&quot;math-container&quot;&gt;$\Phi$&lt;/span&gt;, but as the systems relaxes, &lt;span class=&quot;math-container&quot;&gt;$\Phi$&lt;/span&gt; changes, giving rise to an additional relaxation process. For instance, more massive particles tend to transfer more energy to their lighter neighbors and so become more tightly bound, sinking towards the center of the gravitational potential. This effect is known as mass segregation and is particularly important in the evolution of globular star clusters.&lt;/p&gt;&#xA;&lt;strong&gt;Landau damping&lt;/strong&gt;&#xA;&lt;p&gt;For a perturbation/wave with velocity &lt;span class=&quot;math-container&quot;&gt;$v_p$&lt;/span&gt;, if a particle comes with &lt;span class=&quot;math-container&quot;&gt;$v\gg v_p$&lt;/span&gt;, it will overtake the wave, first gaining energy as it falls into the potential, but later losing the same amount of energy as it climb up again. The same holds for particles with &lt;span class=&quot;math-container&quot;&gt;$v\ll v_p$&lt;/span&gt; which are overtaken by the wave.&#xA;However, particles with &lt;span class=&quot;math-container&quot;&gt;$v\sim v_p$&lt;/span&gt; (i.e. that are near resonance with the wave) may experience a net gain or loss in energy.&#xA;Consider a particle with &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; slightly larger than &lt;span class=&quot;math-container&quot;&gt;$v_p$&lt;/span&gt;. Depending on its phase when interacting with the wave, it will be either accelerated and move away from resonance, or decelerated and move closer to resonance. The latter interact more effectively with the wave (i.e. be decelerated for a longer time), and on average there will thus be a net transfer of energy from particles with &lt;span class=&quot;math-container&quot;&gt;$v \gtrsim v_p$&lt;/span&gt; to the wave. The opposite is true for particles with &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; slightly smaller than &lt;span class=&quot;math-container&quot;&gt;$v_p$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;You can read more about these mechanisms in Mo, Bosch, &amp;amp; White's &lt;a href=&quot;https://www.astro.umd.edu/%7Erichard/ASTRO620/MBW_Book_Galaxy.pdf&quot; rel=&quot;noreferrer&quot;&gt;Galaxy Formation and Evolution&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-04-06T07:56:13.920" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14435" PostTypeId="2" ParentId="13602" CreationDate="2016-04-06T04:42:07.563" Score="4" Body="&lt;p&gt;I would propose Enceladus.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Enceladus has diameter smaller than even Mercury.&lt;/li&gt;&#xA;&lt;li&gt;From the nearside of a tidally-locked moon, the parent planet, Saturn, would appear nearly stationary. Being stationary Saturn might be viewed as the “center of the universe” instead.&lt;/li&gt;&#xA;&lt;li&gt;Being on a &lt;em&gt;moon&lt;/em&gt; as oppose to the &lt;em&gt;parent planet&lt;/em&gt;, with other moons orbiting along with your own, it would seem more apparent that the other celestial bodies did not orbit around your planet/moon.&lt;/li&gt;&#xA;&lt;li&gt;If their were theories proposed that the Saturn was the center of the universe they may disproved by watching the other planets go around &lt;em&gt;the Sun&lt;/em&gt;. Which would be easier to notice as a telescope wouldn't be required. Quite unlike the Galilean moons which require a telescope to see.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11460" LastActivityDate="2016-04-06T04:42:07.563" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14436" PostTypeId="1" AcceptedAnswerId="14515" CreationDate="2016-04-06T06:13:34.733" Score="14" ViewCount="268" Body="&lt;p&gt;If the &quot;9 th planet&quot; is ejected by Jupiter or Saturn, I think the perihelion should be near Jupiter and Saturn, but now the predicted orbit is very far away from Netpune, why would that happen?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastActivityDate="2016-04-12T16:11:43.367" Title="If &quot;9 th planet&quot; is ejected by Jupiter or Saturn, why does it have a very far perihelion?" Tags="&lt;9th-planet&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14437" PostTypeId="2" ParentId="13467" CreationDate="2016-04-06T07:32:30.203" Score="0" Body="&lt;p&gt;To say the &quot;speed of light&quot; is (mostly) about light is a common misunderstanding of that speed. It's actually the speed of causality, and it has far more implications that merely how fast photons propagate. It is also the conversion factor from mass to energy. (E = mc^2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When space expands the photons within it don't suddenly go faster to cover the expansion. Think of it this way. Let's say you have a beetle crawling across a rubber sheet and you stretch that rubber sheet to double its original size. Does the beetle suddenly start crawling twice as fast? No, the beetle still plods along at its same speed. Just like light crossing the universe. The universe expands and light has to take more time.&lt;/p&gt;&#xA;" OwnerUserId="11175" LastActivityDate="2016-04-06T07:32:30.203" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14439" PostTypeId="1" AcceptedAnswerId="14440" CreationDate="2016-04-06T08:46:26.937" Score="1" ViewCount="273" Body="&lt;p&gt;Does lack of a very big crater support the idea that Venus never had big moons before? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, why can't I find a very big crater from the moon impact?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="2270" LastEditDate="2016-04-06T16:03:10.013" LastActivityDate="2016-04-06T16:03:10.013" Title="If Venus had moon before and it fell back to Venus, why don't I see a unique, very big crater?" Tags="&lt;venus&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14440" PostTypeId="2" ParentId="14439" CreationDate="2016-04-06T10:12:22.557" Score="5" Body="&lt;p&gt;Venus has volcanism and weather, and those can cover or erode craters.  There are relatively &lt;a href=&quot;https://en.wikipedia.org/wiki/Venus&quot;&gt;few impact craters on Venus today&lt;/a&gt;, indicating that the surface of Venus is at most 600 million years old -- meaning that any older surface features have been erased.  The Earth's surface is even younger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the &lt;a href=&quot;https://en.wikipedia.org/wiki/Late_Heavy_Bombardment&quot;&gt;Late Heavy Bombardment&lt;/a&gt;, in the very early history of the Solar System, there was a period of time when many asteroids hit the Moon, creating many large impact craters.  It seems likely that the Earth and Venus got hit by many asteroids then, too.  Yet, no evidence of these craters is visible on Earth or Venus today. So, if Venus had a moon and that moon crashed into Venus thousands of millions of years ago, then the impact crater would likely have been erased by now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if Venus got hit by a sufficiently large object, then the energy of the impact might melt the entire surface of Venus, so that no lasting crater can form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, absence of very big craters from Venus is not evidence that Venus never had big moons.&lt;/p&gt;&#xA;" OwnerUserId="9401" LastActivityDate="2016-04-06T10:12:22.557" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14441" PostTypeId="1" CreationDate="2016-04-06T14:09:57.960" Score="0" ViewCount="218" Body="&lt;p&gt;A new class of Neutron stars has been found in the Andromeda galaxy (M31), This is the first time a Spinning Neutron star has ever been detected by astronomers in M31. The article states-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;“pulsars” can be found in stellar couples, with the neutron star cannibalizing its neighbor. This can lead to the neutron star spinning faster and to pulses of high-energy X-rays from hot gas being funneled down magnetic fields onto the neutron star.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;“It could be what we call a ‘peculiar low-mass X-ray binary pulsar’ — in which the companion star is less massive than our Sun — or alternatively an intermediate-mass binary system with a companion of about two solar masses,” said Paolo Esposito of INAF-Istituto di Astrofisica Spaziale e Fisica Cosmica, Milan, Italy.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;While the precise nature of the system remains unclear, the data imply that it is unusual and exotic.&lt;/strong&gt;&#xA;&lt;a href=&quot;http://www.astronomy.com/news/2016/04/andromedas-first-spinning-neutron-star-has-been-found&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.astronomy.com/news/2016/04/andromedas-first-spinning-neutron-star-has-been-found&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question-&lt;/strong&gt;&#xA;  Could the companion of this Spinning Neutron Star not be a star at all but a Black Hole? This might explain the unusual spin and that all neutron stars do not consume other binary companions stars but rather a Black Hole is in the process of consuming the neutron stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Black Hole can devour all of the outer energy of any active star leaving only the dense core of heavy metals that will eventually cool off entirely creating the rouge brown dwarf. &#xA;  Supernovae never derived from dying neutron stars but rather from Binary star Collisions.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/0FABb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0FABb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="9326" LastEditorUserId="10437" LastEditDate="2016-04-06T15:44:02.367" LastActivityDate="2016-04-06T18:30:15.097" Title="Explanation for the first spinning neutron star detected in the Andromeda galaxy?" Tags="&lt;black-hole&gt;&lt;space&gt;&lt;neutron-star&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14442" PostTypeId="1" CreationDate="2016-04-06T15:05:28.340" Score="10" ViewCount="1534" Body="&lt;p&gt;Perhaps it is hard to known for sure because neutrinos are very difficult to detect, although they do go through the earth in very large numbers. But are they passing through the earth equally from all directions or is it really depending on particular events in the universe. It's probable that the sun generates a lot of them, so perhaps in the daytime we receive more of them than at night. But besides the sun's radiation, are the neutrinos coming from everywhere? Or is it possible that the majority of them comes from one direction, and perhaps even has an influence on the earth's movement?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="-1" LastEditDate="2016-04-06T20:31:48.787" LastActivityDate="2016-04-07T19:02:07.313" Title="Is the earth bombarded equally in all directions by neutrinos?" Tags="&lt;universe&gt;&lt;neutrinos&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14444" PostTypeId="1" CreationDate="2016-04-06T16:08:11.123" Score="0" ViewCount="76" Body="&lt;p&gt;I appreciate that a Einstein–Rosen bridge has never been observed but since there are some valid theories in General Relativity I thought I'd ask this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the start and end point of a Einstein–Rosen bridge change in an expanding Universe or would the whole thing stretch but retain it's start and end position?&lt;/p&gt;&#xA;" OwnerUserId="11466" LastActivityDate="2016-04-06T19:26:28.637" Title="Would a Einstein–Rosen bridge change size and/or position in an expanding Universe?" Tags="&lt;general-relativity&gt;&lt;astrophysics&gt;&lt;wormhole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14445" PostTypeId="2" ParentId="14442" CreationDate="2016-04-06T16:35:44.737" Score="15" Body="&lt;p&gt;There are only two types of neutrino source that are &quot;bright&quot; enough to be reliably detected. The sun and nearby supernovae. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The source of solar neutrinos is nuclear fusion, which is also the source of most of the star's energy. Neutrinos also spread out in all directions, so their intensity follows an inverse square law. So the amount of neutrinos is proportional to the brightness of the star. With current detectors no star is bright enough to be observed except for the sun. Other stars do produce neutrinos, and stellar neutrinos come from everywhere (probably more from the milky way) but there are not enough of them to be detected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supernovae in the milky way and neighbouring galaxies produce ridiculous amounts of neutrinos, and a spike in neutrinos was observed from SN1987A the closest recent supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the sun is the brightest source of neutrinos you might think that the Earth would block out neutrinos during night. However, neutrinos pass right through the Earth almost without noticing. The Earth is transparent to neutrinos. So we detect as many neutrinos at night as during the day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing is certain, &lt;strong&gt;there is absolutely no effect on the Earth's rotation or anything else from neutrinos&lt;/strong&gt;, they just pass right through.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-04-06T16:35:44.737" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="14446" PostTypeId="2" ParentId="14441" CreationDate="2016-04-06T18:30:15.097" Score="3" Body="&lt;p&gt;In the article it notes that the partner has less mass than the sun, or possibly up to about twice the mass of the sun (quite a range of uncertainty there!) It's not clear, from the article how they made these calculations, measurement of doppler shifts to get orbital velocity I suppose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass suggested there is not especially small, there are pulsar binary systems in which the secondary is one hundredth the mass of the sun, and could be considered to be planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it does rule out a black hole companion, which would have a mass of 3 solar masses at the least, (and more likely more than 5). There is no need to hypothesis a black hole to explain the spin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last few sentences in your question don't make any sense to me.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-04-06T18:30:15.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14447" PostTypeId="2" ParentId="14444" CreationDate="2016-04-06T19:26:28.637" Score="1" Body="&lt;p&gt;The latter is closest to the truth, although I wouldn't use the phrasing &quot;stretch&quot;. The &quot;mouths&quot; of the wormhole are (more or less) fixed in comoving coordinates (i.e. the coordinate system that expands with the Universe, and in which galaxies lie approximately still). But the bridge is sort of outside our three-dimensional space, and doesn't necessarily follow the same expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the mouths retained a fixed physical distance, they would accelerate in comoving coordinates beyond bound, eventually moving through space at superluminal velocities, which is forbidden. For example, consider the Milky Way and the galaxy &lt;a href=&quot;https://en.wikipedia.org/wiki/GN-z11&quot; rel=&quot;nofollow&quot;&gt;GN-z11&lt;/a&gt; which today lies at a distance of 32 billion lightyears from us. Roughly 13.3 billion years ago, the size of the Universe was 1/10 of today's value; that is $a=0.1$, and the distance to GN-z11 was only 3.2 billion lightyears. If at that time you created a wormhole$^\dagger$ bridging MW and GN-z11, &lt;em&gt;and if the distance between the mouths were fixed in physical coordinates&lt;/em&gt;, then today, at least one of the mouths would have moved so far outside its host galaxy that it must have traveled superluminally.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;Creating a wormhole 13.3 billion years ago is left as an exercise for the reader.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2016-04-06T19:26:28.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14448" PostTypeId="2" ParentId="14340" CreationDate="2016-04-06T20:02:49.810" Score="1" Body="&lt;p&gt;OP here. For those who care, I think I figured out a solution; for my particular situation at least. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure that as long as I use reference images for each wavelength I will be able to measure the reflectivity of my samples using RGB images converted to grayscale. Much like in spectrophotometry, I will be calculating the following ratio:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;sample(wav)/ref(wav) = reflect(wav)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;sample(wav)&lt;/strong&gt; is a grayscale pixel value as a function of wavelength for images of the sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ref(wav)&lt;/strong&gt; is a grayscale pixel value as a function of wavelength for images of the reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;reflect(wav)&lt;/strong&gt; is the reflectivity of the sample as a function of wavelength for an arbitrary pixel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FYI my samples are Silicon wafers and the reference I will be using will be an Aluminum mirror. Both are specularly reflective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I will attempt to prove that this makes sense mathematically...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say at a particular wavelength there is some intensity reflected from the sample, &lt;strong&gt;I_sample(wav)&lt;/strong&gt;, and some intensity reflected from the reference, &lt;strong&gt;I_ref(wav)&lt;/strong&gt;. We want to measure something proportional to each of these intensities at each wavelength. Let's see if the RGB camera is going to screw us over or not...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red, green, and blue filters each have there own transmission as a function of wavelength. Let's call these R(wav), G(wav), and B(wav). Thus the R,G,B values (0-255) that get recorded for the sample are proportional to R(wav)*I_sample(wav), G(wav)*I_sample(wav), B(wav)*I_sample(wav). And similarly for the reference we have R(wav)*I_ref(wav), G(wav)*I_ref(wav), B(wav)*I_ref(wav). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting these images to grayscale involves weighting the R, G, B values in the following manner: gray = aR+bG+cB. Which will give us the following grayscale pixel values:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;sample(wav):&lt;/strong&gt; a x R(wav) x I_sample(wav) + b x G(wav) x I_sample(wav) + c x B(wav) x I_sample(wav)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ref(wav):&lt;/strong&gt; a x R(wav) x I_ref(wav) + b x G(wav) x I_ref(wav) + c x B(wav) x I_ref(wav)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A factor of &lt;strong&gt;a&lt;em&gt;R(wav) + b&lt;/em&gt;G(wav) + c*B(wav)&lt;/strong&gt; can be pulled out of each, and then if the ratio of &lt;strong&gt;sample(wav)/ref(wav)&lt;/strong&gt; is calculated the common terms cancel leaving &lt;strong&gt;I_sample(wav)/I_ref(wav)&lt;/strong&gt;. This is reflectivity! Voila! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this is correct... I have to go, but will continue to think about this, and will edit if I realize anything is wrong. &lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2016-04-07T01:07:18.240" LastActivityDate="2016-04-07T01:07:18.240" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14449" PostTypeId="2" ParentId="14442" CreationDate="2016-04-06T20:09:36.613" Score="8" Body="&lt;p&gt;In addition to neutrinos from the Sun and other discrete sources in the Universe (see James's answer), there is also expected to be a &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_neutrino_background&quot; rel=&quot;nofollow&quot;&gt;cosmic neutrino background&lt;/a&gt;. Although this is yet to be detected (efforts are underway), its expected properties are reasonably well understood. The neutrinos &quot;decoupled&quot; from the universe seconds after the big bang at temperatures $&amp;gt;10^{10}$K. As the universe expands, the de Broglie wavelength of these neutrinos (which are not massless) lengthens with it, such that the neutrinos are expected to have a temperature of $&amp;lt;2$K today. There are 112 of these cosmic neutrinos per cubic centimetre per neutrino flavour (probably 3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The C$\nu$B is analogous to the cosmic microwave background in a number of ways, but (a) it hasn't been detected; (b) it is cooler; (c) because neutrinos have a small but non-zero mass, the C$\nu$B neutrinos are likely &lt;em&gt;non-relativistic&lt;/em&gt; today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This latter point is important to your question. On large scales we expect the neutrino background to have an asymmetry due to the motion of the Earth through the universe with respect to the co-moving standard of rest. This is exactly the same global dipole asymmetry seen in the cosmic microwave background. However, non-relativistic neutrinos are also anisotropic because they are &lt;em&gt;much&lt;/em&gt; more affected by gravitational fields. In particular they should be &lt;em&gt;gravitationally focused&lt;/em&gt; by the Sun, such that the Earth receives more neutrino flux when the Earth is &quot;leeward&quot; of the Sun with respect to its motion with respect to the co-moving rest frame. This will produce an annual modulation in any non-directional neutrino flux amplitude of a few tenths of a per cent (&lt;a href=&quot;http://arxiv.org/abs/1404.0680#&quot; rel=&quot;nofollow&quot;&gt;Safdi et al. 2014&lt;/a&gt;)  and might allow a dectection of the C$\nu$B to be confirmed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On top of this there may be other anisotropies caused by the acceleration of C$\nu$B neutrinos by massive galaxies and clusters of galaxies, that should lead it to be much more inhomogeneous and anisotropic than the cosmic microwave background. Overdensities with respect to the average of factors of 10 or more are possible (see section 2.2 of &lt;a href=&quot;http://journal.frontiersin.org/article/10.3389/fphy.2014.00030/full&quot; rel=&quot;nofollow&quot;&gt;Yanagisawa 2014&lt;/a&gt;), but it depends on exactly what the neutrino mass is.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2365" LastEditDate="2016-04-07T19:02:07.313" LastActivityDate="2016-04-07T19:02:07.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14450" PostTypeId="1" CreationDate="2016-04-06T20:46:30.787" Score="0" ViewCount="198" Body="&lt;p&gt;I presume that the photons from the CMB approach the earth from all directions, otherwise we couldn't detect them with a picture where it is present everywhere in the universe with a tiny anisotropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the CMB is a result of the big bang. Another result of the big bang is that in all directions we see galaxies moving away from us, like we are in the center of the universe (which is probably not the case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why are all the galaxies (except a few) moving away from us while the CMB is approching us. What is the reason for that difference? Or aren't those photons coming from everywhere?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="2270" LastEditDate="2016-04-07T15:01:09.390" LastActivityDate="2016-04-09T20:15:12.853" Title="How is it possible that the CMB approaches the earth from all directions?" Tags="&lt;cosmology&gt;&lt;big-bang-theory&gt;&lt;cmb&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14451" PostTypeId="2" ParentId="14450" CreationDate="2016-04-06T20:54:42.853" Score="2" Body="&lt;p&gt;You are comparing apples and oranges. The CMB is like the light from a distant galaxy, which while the galaxy is receding the light approaches us. CMB is the light emitted by universe at recombination. The matter that emitted the CMB is/was receding from us with a very high red-shift.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2016-04-06T20:54:42.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14452" PostTypeId="1" CreationDate="2016-04-06T21:01:30.533" Score="0" ViewCount="175" Body="&lt;p&gt;Do all / most / any other moons orbiting a planet synchronize their rotation so as to become tidally locked with it, or is the earth a special case?&lt;/p&gt;&#xA;" OwnerUserId="11470" LastEditorUserId="10437" LastEditDate="2016-04-06T23:31:23.980" LastActivityDate="2016-04-07T22:18:24.253" Title="Far side of the moon question" Tags="&lt;the-moon&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14453" PostTypeId="1" CreationDate="2016-04-06T01:13:40.870" Score="-2" ViewCount="148" Body="&lt;p&gt;It has been suggested that space and beyond is infinite, but admittedly the same hypothesis was made by virtue of the fact that it is physically impossible to prove otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we know that there is no border at the end of &quot;infinite&quot; space?&lt;/p&gt;&#xA;" OwnerDisplayName="gerentstein banzon" LastEditorUserId="2153" LastEditDate="2016-04-06T21:51:20.370" LastActivityDate="2016-04-07T09:29:55.073" Title="How do we know that there is no border at the end of &quot;infinite&quot; space?" Tags="&lt;cosmology&gt;&lt;space&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2016-04-11T03:31:54.540" ContentLicense="CC BY-SA 3.0" />
  <row Id="14454" PostTypeId="1" CreationDate="2016-04-06T22:13:05.657" Score="0" ViewCount="3655" Body="&lt;p&gt;The sun travels in an arc across the sky. Where I live, during the summer the arc rises higher, but in the winter the arc is lower. Once might speak of highest altitude or elevation the sun reaches during any given day. In the old days this was called &quot;high noon&quot; when the sun had reached the highest point in the sky for that day. The town clock would be set to 12 at this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a term for the elevation of the sun in the sky?&lt;/p&gt;&#xA;" OwnerUserId="2800" LastActivityDate="2016-07-06T15:06:32.103" Title="Term for altitude of the sun?" Tags="&lt;the-sun&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14455" PostTypeId="1" AcceptedAnswerId="14466" CreationDate="2016-04-07T02:31:39.817" Score="6" ViewCount="588" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thorne%E2%80%93%C5%BBytkow_object&quot;&gt;Thorne–Żytkow objects&lt;/a&gt; (TZOs) form from collisions between neutron stars and main sequence or giant stars. Ultimately, the neutron star becomes the &quot;core&quot; of the giant star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, could this also happen with white dwarfs, if the combined mass is below the TOV limit? White dwarfs are degenerate stars, just like neutron stars, so would they be able to replace the star core in the same way? Would anything happen to the white dwarf if the combined mass is beyond the Chandrasekhar limit?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="37191" LastEditDate="2021-03-18T10:25:36.587" LastActivityDate="2021-03-18T10:25:36.587" Title="If a white dwarf collides with a giant star, could it create a TZO?" Tags="&lt;star&gt;&lt;neutron-star&gt;&lt;white-dwarf&gt;&lt;stellar-dynamics&gt;&lt;thorne-zytkow-object&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14456" PostTypeId="2" ParentId="14454" CreationDate="2016-04-07T06:10:28.040" Score="1" Body="&lt;p&gt;The term for when the sun is at its highest is &quot;solar noon&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Noon#Solar_noon&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Noon#Solar_noon&lt;/a&gt; It is the moment the sun crosses the meridian. The meridian is an imaginary half-circle that goes from the northernmost point on the horizon to the highest point directly above you and then to the southernmost point on the horizon. The sun is always at its highest point in its daily path when it crosses the merdian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're right that in earlier centuries, each town would set its clocks at noon for when the sun was at its highest point. Those of us who are familiar with astronomy and history would call that type of timekeeping &quot;apparent solar time&quot;. You can't guarantee an exact 24 hour day with &quot;apparent solar time&quot;, so some people used &quot;mean solar time&quot; instead. (Historical note: Having each town choose its local noon became a nightmare for railway companies that wanted to set train schedules. So the railway companies urged the creation of 4 time zones in America and encouraged each small town to use &quot;standard time&quot; instead of &quot;apparent solar time&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as rgettman already mentioned, the term for the sun's height above the horizon is &quot;altitude&quot;. Astronomers use the term &quot;elevation&quot; for a person's distance above sea level.&lt;/p&gt;&#xA;" OwnerUserId="11175" LastActivityDate="2016-04-07T06:10:28.040" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14457" PostTypeId="2" ParentId="14452" CreationDate="2016-04-07T06:17:27.360" Score="1" Body="&lt;p&gt;Many, many moons throughout our solar system are tidally locked. &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_locking&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Tidal_locking&lt;/a&gt; This wikipedia page lists 34 moons throughout our solar system that are known to be tidally locked, and another 26 that are suspected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very common for moons within a certain distance of their planets to be tidally locked. If they are too far away, they won't be tidally locked. Astronomers have even deduced that some exoplanets are tidally locked, but I have no idea how they know that. This leads to some interesting questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is the distance at which moons are likely to be tidally locked?&lt;/li&gt;&#xA;&lt;li&gt;At what distance are moons less likely to be tidally locked?&lt;/li&gt;&#xA;&lt;li&gt;How can we know an exoplanet is tidally locked to its star?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11175" LastActivityDate="2016-04-07T06:17:27.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14458" PostTypeId="2" ParentId="14455" CreationDate="2016-04-07T06:36:43.927" Score="1" Body="&lt;p&gt;The outer layers of the star will fall onto the white dwarf, forming an accretion disk as the star stuff spirals in. Like this image shows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ulE45.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ulE45.jpg&quot; alt=&quot;Main sequence star dumping matter onto a white dwarf&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the white dwarf steals enough material to get bigger than 1.4 solar masses, the dwarf will become a &lt;a href=&quot;https://en.wikipedia.org/wiki/Supernova#Normal_Type_Ia&quot; rel=&quot;nofollow noreferrer&quot;&gt;Type Ia supernova&lt;/a&gt;. This is probably why we never (rarely?) see Thorne–Żytkow objects made from a white dwarf instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the white dwarf actually collides with its companion star before it explodes as a supernova, I could imagine it would fall into the center as the stellar material drags down the orbit of the white dwarf. Maybe then it will form something like a Thorne–Żytkow object. Or maybe it will accumulate enough material as it falls in to create a neutron star, at which point it blows itself apart in Type II supernova.&lt;/p&gt;&#xA;" OwnerUserId="11175" LastActivityDate="2016-04-07T06:36:43.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14459" PostTypeId="2" ParentId="14455" CreationDate="2016-04-07T06:40:01.837" Score="2" Body="&lt;p&gt;If you slam a white dwarf into a main-sequence star or red giant such that the white dwarf becomes the core you'd get... a red giant (or supergiant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps that sounds odd but basically the cores of low-mass red giants are electron degenerate.  They roughly consist of deep convective envelopes sitting on top of (helium) white dwarfs separated by a thin but intense shell of material undergoing nuclear reactions.  Of course, most real white dwarfs are of the carbon-oxygen variety, so I imagine what you'd really get is something more like an asymptotic giant branch (AGB) star.  But these are also basically red giants too.  Off-hand, I can't say whether the second burning shell (hydrogen into helium) will be present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, composition wise things might be a bit different to ordinary red giants, red supergiants or AGB stars but the overall structure will be similar: a dense, degenerate core shrouded in a big sparse envelope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the Chandrasekhar limit, that only defines what happens to a degenerate object.  If the white dwarf is massive enough, then once in equilibrium it would presumably start burning carbon and oxygen into heavier metals, just like a red supergiant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a final note, a quick search didn't turn up any literature on this. I know of material about mergers between two white dwarfs or two non-degenerate stars but I couldn't find anything on WD+MS (or similar) mergers.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2016-04-07T06:40:01.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14460" PostTypeId="2" ParentId="14453" CreationDate="2016-04-07T09:29:55.073" Score="0" Body="&lt;p&gt;We do not know what lies beyond the limits of our instruments, by definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all we know, there could be a big sign saying &quot;This area under construction.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, that doesn't seem very probable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most probable answer is &quot;more of the same&quot;.  After all, there is no reason the bit of the universe we can see should be different from other places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you keep adding &quot;more of the same&quot; to the universe, you get the infinite universe you have seen described.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is just a theory, but it seems like a pretty probable one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also easier to do calculations on. Having a border introduces all kinds of problems like galaxies near the border only getting gravitational forces from one side only.  With an infinite universe these forces come from all directions and cancel each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is still just a theory, but it is one that is (relatively) easy to work with.&lt;/p&gt;&#xA;" OwnerUserId="9866" LastActivityDate="2016-04-07T09:29:55.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14461" PostTypeId="2" ParentId="14452" CreationDate="2016-04-07T09:58:44.423" Score="2" Body="&lt;p&gt;Moons are not always tidally locked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are formed with a certain spin angular momentum that will be dragged by asymmetries on the hosting planetary body. Those asymmetries then backreact onto the moon and drag it's spin to become synchronized with the orbit angular momentum.&lt;br&gt;&#xA;How rapidly this process happens is roughly a function of the tidal forces &#xA;$$F_{Tid} = G \frac{m_{Moon} \cdot m_{Planet}}{r^3} d$$&#xA;exerted by those asymmetries. Here $d$ is the moon size, $r$ the distance between planet and moon and the other quantities are masses and the gravitational constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time $t_{lock}$ this locking takes can be estimated by comparing the excess angular momentum $L_{ex} = r^2m_{Moon}(\omega_{actual} - \omega_{orbit})$ with those acting forces:&#xA;$$t_{lock} \approx L_{ex} / \left(r F_{Tid} \right) \approx \frac{r^5}{ d \cdot m_{Planet}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter equation makes it fairly obvious that this is not a process that happens at a certain magical orbit or at a certain time. It is an asymptotic process, and the only coincidal thing is that we observe any given moon at a time where it may not have lost enough spin angular momentum yet.&lt;br&gt;&#xA;This should hopefully also clarify some confusion from the comments:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Yes, the tidal locking time is also dependent on the mass of the host-planet, but much more sensitive to the planet-moon-distance. Note that this derivation is independent of the moon's mass, and only dependent on it's size. The literature knows much more sophisticated variants of this, that also incorporate a (realistic) weak dependency on the moon's mass.&lt;/li&gt;&#xA;&lt;li&gt;For Hot Jupiter exoplanets it is thought they'd be in tidal lock when the asteroseismologic age of the system is bigger than the derived tidal lock time.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Concluding&lt;/strong&gt;&lt;br&gt;&#xA;1.) Any moon would be tidally locked if we could wait long enough.&lt;br&gt;&#xA;2.) There are much more effects / torques to consider for a detailed understanding than I did here.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditDate="2016-04-07T22:18:24.253" LastActivityDate="2016-04-07T22:18:24.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14463" PostTypeId="1" AcceptedAnswerId="14465" CreationDate="2016-04-07T16:17:31.140" Score="1" ViewCount="577" Body="&lt;p&gt;How large and bright would Betelgeuse appear if it were as close to Earth as Sirius, before and after it goes supernova?&lt;/p&gt;&#xA;" OwnerUserId="9124" LastActivityDate="2016-04-07T21:01:39.073" Title="What would be the apparent magnitude of Betelgeuse if it were as close to Earth as Sirius?" Tags="&lt;distances&gt;&lt;astrophysics&gt;&lt;size&gt;&lt;luminosity&gt;&lt;red-giant&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14464" PostTypeId="2" ParentId="14463" CreationDate="2016-04-07T19:19:14.180" Score="2" Body="&lt;p&gt;Given a distance of $d_1 = 640~\mathrm{ly}$ for Betelgeuze and $d_2 = 8.6~\mathrm{ly}$ for Sirius the difference in the apparent magnitude for Betelgeuze at distances $d_1$ and $d_2$ that is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_1 - m_2 = 5~\mathrm{mag} \cdot log_{10}(\frac{d_1}{d_2})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is $9.4~\mathrm{mag}$. With magnitude $m_1 = 0.45~\mathrm{mag}$ we get $m_2 = -8.9~\mathrm{mag}$ for Betelgeuze at the distance of Sirius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is about $860$ times brighter than Sirius ($-1.46~\mathrm{mag}$). It would be visible during daytime&lt;/p&gt;&#xA;" OwnerUserId="10568" LastEditorUserId="10568" LastEditDate="2016-04-07T19:28:36.657" LastActivityDate="2016-04-07T19:28:36.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14465" PostTypeId="2" ParentId="14463" CreationDate="2016-04-07T21:01:39.073" Score="2" Body="&lt;p&gt;The distance to betelgeuse is poorly known, so we don't actually know how bright it is with much accuracy measurements of its parallax by satellite give a distance of 197 parsecs &lt;em&gt;+/- 45 parsecs&lt;/em&gt; (1 parsec is 3.26 light years). The absolute magnitude (the brightness if it were 10 parsecs distant) is estimated to be -5.85. This is based on both the distance and models of how bright red supergiants &quot;should&quot; be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To convert from absolute (M) to apparent (m) magnitudes at distance D one can use the formula:&#xA;$$m-M = 5 ((\log_{10}{D}) - 1)$$&#xA;And at a distance of Sirius (2.54 parsecs) that gives a brightness of -8.7, bright enough to cast shadows, as bright as a crescent moon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A supernova would be a lot brighter. A typical supernova has an absolute magnitude of between -15 and -20, with the low end more likely. Assuming an absolute magnitude of about -16, gives an apparent magnitude of about -19. This would be a bright as interior lighting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the supernova, there might be a neutron star which, if it was like the crab pulsar, might have an absolute magnitude of about 4, and at the distance of sirius appear to have a magnitude of about 1, just a little less bright than Betelgeuse is now.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-04-07T21:01:39.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14466" PostTypeId="2" ParentId="14455" CreationDate="2016-04-07T22:16:15.870" Score="3" Body="&lt;p&gt;Thorne and Żytkow's &lt;a href=&quot;http://adsabs.harvard.edu/abs/1977ApJ...212..832T&quot; rel=&quot;nofollow noreferrer&quot;&gt;original paper&lt;/a&gt; on TŻOs actually opens with a comparison of TŻOs and the type of object you mention, with a  white dwarf degenerate core instead of a neutron star degenerate core. They note that the equilibrium states - essentially, stable configurations - of such combinations lie near the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hayashi_track&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hayashi track&lt;/a&gt; (actually acting a bit like AGB stars, in some cases), indicating high metallicity, as is the case with TŻOs.&lt;/p&gt;&#xA;&lt;p&gt;These objects generate energy the same way TŻOs do: matter is accreted by the core, releasing gravitational potential energy, and the red giant envelope continues some fusion, although, of course, core fusion has been substantially disrupted by the arrival of the new degenerate core. The main difference in energy production are the ratios between nuclear contributions to luminosity and gravitational contributions to luminosity:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$L_{\text{nuc}}/L\approx0.99,\quad L_{\text{grav}}/L\approx0.01\quad\text{for white dwarf core}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$L_{\text{nuc}}/L\approx0.04,\quad L_{\text{grav}}/L\approx0.96\quad\text{for neutron core}$$&lt;/span&gt;&#xA;Why the difference? &lt;span class=&quot;math-container&quot;&gt;$L_{\text{grav}}$&lt;/span&gt; is proportional to&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{GM_c}{R_cc^2}$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$_c$&lt;/span&gt; refers to values for the core. The masses and radii of neutron stars differ drastically from those of white dwarfs. This becomes less important in the case of supergiant TŻOs (i.e. &lt;span class=&quot;math-container&quot;&gt;$M&amp;gt;10 M_{\odot}$&lt;/span&gt;), because convection cycles &amp;quot;burned&amp;quot; nuclear fuel back outwards into the envelope, and so energy ratios become more like those found in the case of a white dwarf core.&lt;/p&gt;&#xA;&lt;p&gt;This difference in energy production ratios also means that the objects will remain in roughly stable states for different amounts of time; red giants with white dwarf cores can survive in equilibrium for at least an order of magnitude or more as long as TŻOs.&lt;/p&gt;&#xA;&lt;p&gt;One interesting thing to note is that TŻOs and red giants with white dwarf cores may share some of the same problems when it comes to stability. The envelopes are expected to be composed similarly and act similarly, with the potential difference in nuclear fusion rates, and so the same dynamical instabilities are possible in both cases. However, Thorne and Żytkow state that they find this possibility unlikely.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="31410" LastEditDate="2020-11-29T00:28:30.023" LastActivityDate="2020-11-29T00:28:30.023" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14467" PostTypeId="1" CreationDate="2016-04-07T23:22:27.690" Score="4" ViewCount="311" Body="&lt;p&gt;If the VLT (very large telescope) can use interferomerty to act as a 200 meter mirror, then why are they trying to build the 39 meter E-ELT? What advantage does the E-ELT have, if the VLT can act as a 200 meter mirror?&lt;/p&gt;&#xA;" OwnerUserId="11479" LastActivityDate="2019-01-12T22:26:55.623" Title="VLT interferometry vs E-ELT?" Tags="&lt;telescope&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14468" PostTypeId="1" CreationDate="2016-04-08T06:00:27.953" Score="1" ViewCount="172" Body="&lt;p&gt;Is there some substance on Venus or some kind of material, because neither bodies have any source of light.&lt;/p&gt;&#xA;" OwnerUserId="11481" LastEditorUserId="10437" LastEditDate="2016-04-08T12:57:00.450" LastActivityDate="2020-03-25T02:48:25.660" Title="What makes the Moon and Venus shine?" Tags="&lt;the-moon&gt;&lt;venus&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14470" PostTypeId="2" ParentId="14467" CreationDate="2016-04-08T07:43:51.237" Score="7" Body="&lt;p&gt;Ok, not sure where you get 200m from, the maximum baseline for the VLT interferometer is 130m. Nevertheless, your question still stands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of reasons. Firstly, it is important to understand that with relatively limited number of baselines (4 VLTs = 6 baselines) one does not get actual images out of the interferometer, but rather &quot;visibilities&quot; that enable one to work out whether sources are double, extended or whatever. Second, the interferometry only works at infrared wavelengths. This is a technical issue that &lt;em&gt;might&lt;/em&gt; be solved in the next decade or so, but a general rule of thumb is that the longer the baseline, the longer the wavelengths you will be limited to use. Thirdly, the light gathering power is limited to the light that can be gathered by the constituent telescopes. I think this also limits the kinds of instruments you can use since the required delay lines and optical components of the system make everything inefficient. Fourth, there are limitations of the abilities of optical/IR interferometers caused by turbulence in the atmosphere. This limits them to very small fields of view and the performance does not reach the ideal case. It also means that you are limited to bright objects because individual exposure times need to be very short.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The E-ELT is designed to work at optical &lt;em&gt;and&lt;/em&gt; infrared wavelengths and will gather enormous amounts of light. Not everything in astronony needs spatial resolution; sometimes what you need is just a bigger light bucket. Examples include: deep imaging of the universe, gathering spectra from faint objects; accumulating enough photons to do transmission spectroscopy using exoplanet transits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's horses for courses really. At optical and IR wavelengths the future will likely be big telescopes in space and even bigger interferometers in space.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-04-08T12:45:16.610" LastActivityDate="2016-04-08T12:45:16.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14471" PostTypeId="2" ParentId="14467" CreationDate="2016-04-08T07:45:00.150" Score="2" Body="&lt;p&gt;The quantity of light gathered is the important difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The VLT and its kind can use interferometric methods to achieve the resolution of a much higher telescope - that allows detection of fine detail that previously couldn't be seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Large single aperture scopes like the E-ELT don't offer such a high resolution, but they collect a larger quantity of light so can produce acceptable images of much fainter objects. Part of the E-ELT's core goals is imaging of very distant galaxies, and for this it doesn't need exceptional resolution, but it needs very sensitive light gathering.  (&quot;Light&quot; including part of the infra-red region too, though that's separate question.)&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2016-04-08T07:45:00.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14472" PostTypeId="1" CreationDate="2016-04-08T07:51:36.480" Score="0" ViewCount="152" Body="&lt;p&gt;If we want the full moon to be 2 times more bright,how we should change the distance between the sun and the earth? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help. &lt;/p&gt;&#xA;" OwnerUserId="11483" LastActivityDate="2016-04-08T08:46:58.787" Title="The relation between the light of full moon and the distance between the sun and the earth" Tags="&lt;the-sun&gt;&lt;the-moon&gt;&lt;light&gt;&lt;luminosity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14473" PostTypeId="2" ParentId="14472" CreationDate="2016-04-08T08:46:58.787" Score="2" Body="&lt;p&gt;The strength of the light from the Sun scales with the inverse square of distance [note 1]. That means that we would need to have the Earth-Moon system at $\frac{1}{\sqrt{2}}$ (approx 0.7) AU for the full Moon to be twice as bright [note 2].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 1: The fact that the Sun is not a point source of light has only a very minor effect on the scaling.&#xA;Note 2: Technically, it will be slightly brighter because more of the absorbed radiation will be radiated away in the visible spectrum. This is however also a negligible effect.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastActivityDate="2016-04-08T08:46:58.787" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14474" PostTypeId="1" CreationDate="2016-04-08T10:49:52.563" Score="2" ViewCount="68" Body="&lt;p&gt;LIGO has a mind boggling sensitivity. What would it take to directly measure the expansion of spacetime, along the laser beam? I suppose it could be done at interplanetary scale.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-04-08T10:49:52.563" Title="Can the expansion of spacetime be directly measured with laser interferometry (like GW can)" Tags="&lt;expansion&gt;&lt;gravitational-waves&gt;&lt;interferometry&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14475" PostTypeId="1" CreationDate="2016-04-08T11:11:52.243" Score="9" ViewCount="683" Body="&lt;p&gt;The rotations of comets and asteroids are affected by outgassing volatilities. Io is very volcanically active. Does this give Io a varying orbit and a slow rotation? Tidal forces should take longer than the effect of a large eruption. Tidal locking means that any volcano alway pushes Io in the same direction (radially, tangential or inclined) accumulating the same kind of orbital change during all of its its active lifetime.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="15545" LastEditDate="2018-05-08T13:27:56.567" LastActivityDate="2018-06-10T18:05:40.340" Title="Is Io's orbit or rotation affected by its volcanism?" Tags="&lt;celestial-mechanics&gt;&lt;volcanism&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="14476" PostTypeId="1" CreationDate="2016-04-08T14:52:49.063" Score="3" ViewCount="789" Body="&lt;p&gt;As beginner, I would like to understand how planets exhibit retrograde motion.&lt;/p&gt;&#xA;&lt;p&gt;When does the retrograde motion start for a planet like Mars and how long (time limit) planets will appear to be in such motion?&lt;/p&gt;&#xA;&lt;p&gt;Any good URL or simple explanation will be helpful.&lt;/p&gt;&#xA;" OwnerUserId="11000" LastEditorUserId="37191" LastEditDate="2021-04-16T08:04:24.850" LastActivityDate="2021-04-16T08:04:24.850" Title="Retrograde motion of planets: reason and time?" Tags="&lt;planet&gt;&lt;solar-system&gt;&lt;retrograde-orbit&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14477" PostTypeId="2" ParentId="14468" CreationDate="2016-04-08T15:19:51.313" Score="6" Body="&lt;p&gt;The sun is an excellent source of light for all the planets and moons, although it gets a little dodgy way out near Pluto. The moon reflects about 10% of the sunlight that hits it, that's why we see it. Venus' albedo is about 0.75, what with its clouds and all. That means 75% of the sunlight that hits it is reflected back into space for us to see.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2016-04-08T15:19:51.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14478" PostTypeId="1" CreationDate="2016-04-08T15:28:29.033" Score="2" ViewCount="501" Body="&lt;p&gt;Many times we see in the sky that most of the meteor shower trails vanish when they are about to strike the earth several feet above the ground. Why does this happen? They could strike the earth's surface (still emitting light until they strike).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6tHy7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6tHy7.jpg&quot; alt=&quot;A meteor&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pic source: &lt;a href=&quot;https://imgur.com/afiZGCm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://imgur.com/afiZGCm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7964" LastEditorUserId="10437" LastEditDate="2016-04-08T16:30:15.643" LastActivityDate="2020-09-15T07:43:45.323" Title="Why do most meteors vanish just before hitting the ground?" Tags="&lt;planetary-atmosphere&gt;&lt;meteor&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14479" PostTypeId="2" ParentId="14276" CreationDate="2016-04-08T16:42:32.573" Score="2" Body="&lt;p&gt;A few responses to my own question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) TDB time isn't critical because it only varies from TT by a few milliseconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) The resulting vector was not ECI but the vector from the earth to the sun in the celestial frame.  I needed to build the celestial to terrestrial rotation matrix using iauC2t00a() (assuming zero polar motion since it's generally less than 20 meters), and then rotate the resulting pv (position, velocity) vector-pair into the terrestrial frame using iauRxpv().&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) (Still no clue on this one.)&lt;/p&gt;&#xA;" OwnerUserId="11302" LastActivityDate="2016-04-08T16:42:32.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14480" PostTypeId="2" ParentId="14476" CreationDate="2016-04-08T17:01:17.020" Score="3" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_retrograde_motion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; on apparent retrograde motion seems to have a table containing exactly what you are looking for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FVfG1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FVfG1.png&quot; alt=&quot;retrograde motion table&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The apparent motion is observed when the projection of a planet is compared to the star background. There are no real change in the planetary orbits, but when for example the Earth catches up with Mars, it seems to move retrograde relative to the background because of our own change of view. (Like how it appears that you are scrolling further down on this page when the background moves upwards.)&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="10172" LastEditDate="2016-04-08T17:07:50.557" LastActivityDate="2016-04-08T17:07:50.557" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14481" PostTypeId="1" AcceptedAnswerId="14485" CreationDate="2016-04-08T19:45:22.420" Score="1" ViewCount="565" Body="&lt;p&gt;I'm currently working with proper motions and have needed to convert my galactic longitude and latitude into right ascension and declination (ie from galactic to equatorial coordinates). This is fine since NED have a tool to do this, however once I convert into Cartesian coordinates I do not get the same results as I had for the galactic-cartesian conversion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the conversion from galactic to cartesian is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x = Dcos(l)cos(b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y = Dcos(b)sin(l)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z = Dsin(b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where D is the distance to the galaxy in question in my case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also for the equatorial to cartesian transformation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x = Dcos(RA)cos(DEC)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y = Dcos(DEC)sin(RA)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z = Dsin(DEC)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have done this for both my galactic coordinates and converted equatorial coordinates, but I do not get the same x,y and z. Should this be happening?&lt;/p&gt;&#xA;" OwnerUserId="11127" LastEditorUserId="11127" LastEditDate="2016-04-08T20:28:03.867" LastActivityDate="2016-04-08T21:39:53.373" Title="Question on Equatorial and Galactic coordinate systems" Tags="&lt;coordinate&gt;&lt;positional-astronomy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14482" PostTypeId="2" ParentId="13704" CreationDate="2016-04-08T20:14:27.487" Score="0" Body="&lt;p&gt;On Feb 12 James wrote:&quot;A difference of 7ms would imply a minimum separation of the sites of about 2000 km, not a maximum.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculated minimum separation of the sites is around 3001.65 km compared with the estimated maximum of 3030.12 km indicated by gollum that same day. The angular separation between Livingston and Hanford detectors is around 27.25 degrees from the center of our planet using  an average Earth's radius of 6371.0 Km and gives this result. If the detected GW150914 gravitational waves propagated outwards at more speed that the speed of light a difference of 7ms is in accord with this MINIMUM possible detector distance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The GW's velocity on 2015/09/14/09:50:45 UT based on this argument would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3001.65 km/2099.0 km)*c ~ 1.43 times the speed of light &lt;/p&gt;&#xA;" OwnerUserId="11491" LastActivityDate="2016-04-08T20:14:27.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14483" PostTypeId="2" ParentId="14478" CreationDate="2016-04-08T20:33:30.530" Score="7" Body="&lt;p&gt;Only very few meteors actually make it anywhere near the surface of Earth; most burn up 75–100 km above the surface. From your point of view, however, the curvature of Earth's surface may make it look as if they get much closer, and even fall below the horizon. But depending on where you live, the horizon often has quite a lot more background light (e.g. from cities far away). That means that when they get near the horizon, &lt;em&gt;which you may interpret as &amp;quot;several feet above the ground&amp;quot;&lt;/em&gt;, the seemingly disappear.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="9527" LastEditDate="2020-09-15T07:43:45.323" LastActivityDate="2020-09-15T07:43:45.323" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14484" PostTypeId="2" ParentId="14481" CreationDate="2016-04-08T21:11:43.510" Score="0" Body="&lt;p&gt;That is expected as the x,y,z axes of the equatorial coordinate system don't point in the same direction as those of the galactic coordinate system. To compare you would have to change basis from equitorial to galactic  coordinates, which is effectively what NED does.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-04-08T21:11:43.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14485" PostTypeId="2" ParentId="14481" CreationDate="2016-04-08T21:39:53.373" Score="1" Body="&lt;p&gt;Yes, because the Milky Way galaxy and the Earth's equator are in different planes, currently about 63 degrees apart. To avoid confusion, rename the galactic Cartesian coordinates to (u,v,w). The transformation from (u,v,w) to (x,y,z) can be expressed as multiplication by a rotation matrix.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-04-08T21:39:53.373" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14486" PostTypeId="1" CreationDate="2016-04-09T03:41:07.137" Score="5" ViewCount="908" Body="&lt;p&gt;I want to know how high does light pollution typically reach into the sky &lt;strong&gt;straight up&lt;/strong&gt;? Has anyone ever researched this because I wonder if being at 10000ft elevation would reduce the amount of light pollution from nearby cities because of the height assuming that light pollution weakens the higher you go. How high would it go &lt;strong&gt;straight up&lt;/strong&gt; the sky (&lt;em&gt;not&lt;/em&gt; to the sides about how far I would have to travel to have no light pollution disturbing the night sky)?&lt;/p&gt;&#xA;" OwnerUserId="11494" LastEditorUserId="37191" LastEditDate="2021-05-16T22:01:15.353" LastActivityDate="2021-05-16T22:01:51.257" Title="How high does light pollution reach into the sky?" Tags="&lt;light&gt;&lt;light-pollution&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14487" PostTypeId="2" ParentId="14486" CreationDate="2016-04-09T06:56:03.470" Score="5" Body="&lt;p&gt;Light pollution occurs because light from the ground refects off atoms in the atmosphere. So you can reduce light pollution either by getting away from  light, or getting above the atmosphere. 50% of the atmosphere lies below 5500m,  if you can get 5500 m high, you half light pollution.  Getting high also improves clarity, and reduces the disturbance caused by atmospheric turbulence, which is why most major telescopes are situated on mountain tops.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="37191" LastEditDate="2021-05-16T22:01:51.257" LastActivityDate="2021-05-16T22:01:51.257" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14489" PostTypeId="2" ParentId="14450" CreationDate="2016-04-09T20:15:12.853" Score="2" Body="&lt;p&gt;The Big Bang happened everywhere and the recombination of electrons with nuclei - the thing that causes the CMB - also happened everywhere. So in every direction you look you can see the radiation from that recombination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that recombination also happened a long time ago and so, given light has a constant velocity, the places where we can see the recombination take place are also a long way away. And, as a result of the expansion of space-time, that radiation is very red-shifted, so instead of the recombination looking like it is happening at roughly 4000 degrees Kelvin, it looks like something happening at 3 degrees Kelvin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That expansion is also what makes it look like the distant galaxies are all moving away from us and the further away the faster they seem to be moving.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2016-04-09T20:15:12.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14490" PostTypeId="2" ParentId="11491" CreationDate="2016-04-09T20:27:23.897" Score="2" Body="&lt;p&gt;I suggest you use the JPL Horizons system. You can investigate how Charon's orbital elements change with time. The short answer is, not surprisingly, not by much. The values for the eccentricity I see hover around 0.00015 and those for the orbital inclination relative to Pluto's equatorial plane are about 0.09 degrees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow&quot;&gt;http://ssd.jpl.nasa.gov/horizons.cgi&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11498" LastActivityDate="2016-04-09T20:27:23.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14491" PostTypeId="2" ParentId="1438" CreationDate="2016-04-09T20:30:41.057" Score="2" Body="&lt;p&gt;One thing that seems to be clear is that HgMn stars have only an extremely weak net longitudinal magnetic field component, if any. &lt;a href=&quot;http://www.ast.obs-mip.fr/users/donati/pub/weak.pdf&quot; rel=&quot;nofollow&quot;&gt;Shorlin et al. (2002)&lt;/a&gt; did an early survey of HgMn, Am, and Ap stars, and detected no longitudinal magnetic fields in the former, with a median 1$\sigma$ uncertainty of 39 Gauss. &lt;a href=&quot;http://journals.cambridge.org/download.php?file=%2FIAU%2FIAU6_S272%2FS1743921311010349a.pdf&amp;amp;code=d09eb28ac837ffb701f7ea49a2a202a6&quot; rel=&quot;nofollow&quot;&gt;Makaganiuk et al. (2010)&lt;/a&gt; also found $B_z$ values of 0 in the stars they surveyed, with a higher precision - a 1$\sigma$ uncertainty of 0.81-10 Gauss, varying between stars. Other studies also yielded precisions of less than a few Gauss for some stars (see mentions by &lt;a href=&quot;https://uu.diva-portal.org/smash/get/diva2:450464/FULLTEXT01.pdf&quot; rel=&quot;nofollow&quot;&gt;Makaguniak (2011)&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some reports have found longitudinal values in the 10s to 100s of Gauss, but as &lt;a href=&quot;http://www.astro.uu.se/~oleg/papers/moscow13.pdf&quot; rel=&quot;nofollow&quot;&gt;Kochukhov&lt;/a&gt; notes, subsequent inveistigations have failed to confirm these findings, which have had extremely high uncertainties. One example is &lt;a href=&quot;http://arxiv.org/abs/1208.2910&quot; rel=&quot;nofollow&quot;&gt;Hubrig et al. (2012)&lt;/a&gt;, the paper you cite, which claimed to have found weak longitudinal and quadratic fields in several stars, including HD 65949. &lt;a href=&quot;http://www.diva-portal.org/smash/get/diva2:644466/FULLTEXT01.pdf&quot; rel=&quot;nofollow&quot;&gt;Kochukhov et al. (2013)&lt;/a&gt; then found no longitudinal fields on the star, to within a few Gauss, and &lt;a href=&quot;http://arxiv.org/abs/1309.2158?context=astro-ph.SR&quot; rel=&quot;nofollow&quot;&gt;Bagnulo et al. (2013)&lt;/a&gt; attributed to 2012 findings to instrument error, leading to flawed data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-longitudinal magnetic fields have not been observed in much detail (small-scale longitudinal fields have not yet been ruled out, either, by large-scale global ones appear to be nonexistent), and complicated ones could still exist. Kochukhov et al. (2013) do say that they have ruled out large so-called tangled magnetic fields, but small-scaled ones are still possible, according to &lt;a href=&quot;http://www.ta3.sk/caosp/Eedition/FullTexts/vol27no3/pp296-303.pdf&quot; rel=&quot;nofollow&quot;&gt;Hubrig&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing worth noting is that the vast majority of these studies, including the one you referenced, which has been disputed, are focused on B-type HgMn stars, in part because fewer A-type HgMn stars have been discovered.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2016-04-09T20:30:41.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14492" PostTypeId="1" CreationDate="2016-04-09T21:28:43.903" Score="6" ViewCount="2450" Body="&lt;p&gt;I've been looking, unsuccessfully, for hours for a simple set of equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Input&lt;/strong&gt;: &lt;br&gt;&#xA;RA and Dec of an object&lt;br&gt;&#xA;Observers Lat and Lng&lt;br&gt;&#xA;Current Time&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Output&lt;/strong&gt;: &lt;br&gt;&#xA;Rise Time &lt;br&gt;&#xA;Transit Time&lt;br&gt;&#xA;Set Time&lt;br&gt;&#xA;(and, hopefully) max elevation at transit &lt;br&gt;&#xA;(and, hopefully) current alt and az of object&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like this should be a simple relationship, but I haven't been able to find it.&lt;br&gt;&#xA;If an equation simply gives the current alt/az for an object, I guess I could solve for az and then find the maximum for a given time range. &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions&lt;/strong&gt;: &lt;br&gt;&#xA;1. Anybody know of such equations? Or, even better, some sample code. &lt;br&gt;&#xA;2. If not, can anybody point me in the right direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="11500" LastActivityDate="2020-08-31T00:17:11.147" Title="Need Simple equation for Rise, Transit, and Set time" Tags="&lt;coordinate&gt;&lt;celestial-mechanics&gt;&lt;declination&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14493" PostTypeId="2" ParentId="14284" CreationDate="2016-04-09T22:18:11.800" Score="0" Body="&lt;p&gt;The Solar System moving through intergalactic space would have no impact on life. However, in order for the sun to be ejected out of the Galaxy, a very close encounter with a stellar mass object would be necessary. That would likely have a huge negative impact on life on earth, at a minimum perturbing many objects in the Oort Cloud into the inner solar syste. At the other end of the spectrum, the orbit of earth could be significantly altered or the earth could even be ejected from the solar system, depending on the geometry of the encounter. &lt;/p&gt;&#xA;" OwnerUserId="11498" LastActivityDate="2016-04-09T22:18:11.800" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14494" PostTypeId="2" ParentId="14492" CreationDate="2016-04-09T23:20:04.880" Score="3" Body="&lt;p&gt;The calculations are not trivial, but they are encapsulated in software libraries such as pyephem, which has &lt;a href=&quot;http://rhodesmill.org/pyephem/rise-set.html&quot; rel=&quot;nofollow&quot;&gt;examples of finding rise, set and transit&lt;/a&gt;&#xA;If you want to understand how these are calculated, you can readnthe source, which is based on the xephem application. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One detail which complicates the calculation is the refraction caused by the atmosphere, which can change the moment of setting by several minutes.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-04-09T23:20:04.880" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14495" PostTypeId="2" ParentId="14492" CreationDate="2016-04-09T23:48:39.253" Score="2" Body="&lt;p&gt;If you want a website rather than a software package like James suggests, I recommend use the INGT &lt;a href=&quot;http://catserver.ing.iac.es/staralt/index.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;object visibility tool&lt;/a&gt; which has all 3 inputs you request and produces an elevation plot with rise and set times, maximum altitude etc. An example is shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wLQKd.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wLQKd.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10437" LastActivityDate="2016-04-09T23:48:39.253" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14496" PostTypeId="2" ParentId="13084" CreationDate="2016-04-10T00:14:49.670" Score="3" Body="&lt;p&gt;Navigating the solar system, you'd probably track your probe using Earth-based radio telescopes in the equatorial ICRF, though you might describe the overall trajectory in terms of the ecliptic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Navigating at sea, you might measure the Sun's position in horizontal coordinates at a time in GMT to determine your own geographic coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Practical uses in backyard astronomy:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;horizontal: telling someone where to look relative to an object on the horizon; pointing a telescope on an altitude/azimuth mount&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;equatorial: finding or plotting an object in a star atlas; choosing atlas pages to use on a given night; pointing a telescope on an equatorial mount&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;ecliptic: importing comet or asteroid orbital elements into planetarium software; planets are easiest to observe when their longitude differs most from the Sun's&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;galactic: low latitudes have more Milky Way star clusters and nebulae; high latitudes have more external galaxies&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;supergalactic: pondering the distribution of galaxy clusters?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-04-10T00:14:49.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14497" PostTypeId="1" AcceptedAnswerId="14498" CreationDate="2016-04-10T03:09:25.493" Score="0" ViewCount="110" Body="&lt;p&gt;If I were to go out on a clear night, in an area with low light pollution, and take a photo of the stars, could an expert analyze the photo and figure out what date the photo is from? We can assume a powerful camera, but not reaching up to the strength of a telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question stems from a larger project. It would be very convenient for me if it were possible to take a photo of the stars to tell the time. Even more convenient if you could narrow it down to within 10 minutes, but if that's not possible I'm wondering how close we can get. Can you tell the year? The millennium? Do you need to know which location on Earth the photo was taken from? Does it help if you already know the year in which the photo was taken?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can assume access to modern computing, existing databases, state-of-the-art software, etc. (no supercomputers)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="11501" LastActivityDate="2016-04-10T07:42:22.810" Title="How accurately can you tell the time using a photo of the stars?" Tags="&lt;photography&gt;&lt;time&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14498" PostTypeId="2" ParentId="14497" CreationDate="2016-04-10T07:42:22.810" Score="5" Body="&lt;p&gt;To make a clock you need something that changes predictably with time, and stars are nearly constant, making them poor clocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stars do move across the sky each night, but since the morning stars of winter become the evening stars of  spring, and the sky looks different in different locations you need to know the date of the image, the location of the photo, and the exact direction of the photo. That could give sub 10 min accuracy. Unfortunately while date and location may often be recorded with a photo,  exact direction is not, unless the horizon is visible in the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without that information, and it is possible to 'get lucky'. For example, if there are two or more planets in the shot, their locations could be measured to give at least the date of the photo. Similarly if a rare event like a nova is visible that can offer dating information. If the shot happened to catch a bright meteor, and the date was known well enough, you could get almost 1 second accuracy. Variable stars might also give the chance of at least dating the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With no planets you can use the very slow proper motion of the stars. With a good enough telescopic shot of the night sky you can be quite accurate, but with a wide view of a camera you wont be able to get enough accuracy to give any useful date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exposure time of a typical camera is to short to capture any stars. If you just point and shoot at tye sky, you just get black. To capture images of stars  you would need a timed exposure.  A telescope would help too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you photographer wanted to create an image which could be timed accurately, and they had time to prepare for it, or it was a lucky shot this might be possible, otherwise not.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-04-10T07:42:22.810" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14499" PostTypeId="2" ParentId="13582" CreationDate="2016-04-10T14:03:29.377" Score="0" Body="&lt;p&gt;3 parsecs is the distance from us (Earth, Sun etc) to the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a distance of 1 parsec, the apparent movement of a star wrt the background, as the Earth orbits on its 1 AU orbit is 1 arc second.&#xA;So, from 1 parsec distant, the earth would move by 1 arcsec each 6 months. It woukd have an apparevt angular radiusof 0.5 arc sec.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this star is 3 parsecs distant, so an orbit with radius 1au would appear to have a angular radius  3 times smaller (here I am approximating, it is valid because the angle is small) or 1/6  arc sec. The planet in the question has a smaller orbit, 6/10 of the Earth, or &lt;strong&gt;0.6&lt;/strong&gt; au&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The convenient units, and tge the small angles mean that we can avoid trionometry in the solution.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-04-10T14:09:40.817" LastActivityDate="2016-04-10T14:09:40.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14501" PostTypeId="2" ParentId="12629" CreationDate="2016-04-10T16:29:10.847" Score="3" Body="&lt;p&gt;Earth gains mass through the impact dust from space, we see the meteors that form as this material impacts the atmosphere. We gain 40000 to 80000 tonnes a year. Even if ut &quot;burns up&quot; the mass is still added to the atmosphere.  However, hydrogen is lost from the atmosphere, and this amounts for rather  more, about 100000 tonnes a year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resjult is a net loss of about 20000 to 60000 tonnes per year. This is miniscule compared to the mass of the Earth. It is many order of magnitude less than the uncertainty in the mass. There are no direct consequences of this loss if mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other thing that could change the mass od the earth (eg space probes or cosmic rays) have even less effect.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-04-10T16:29:10.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14502" PostTypeId="2" ParentId="11879" CreationDate="2016-04-11T02:51:02.217" Score="3" Body="&lt;p&gt;Consider a nearly circular orbit. On average, $ v = 2 \pi r / T $, and the gravitational force balances the centrifugal force:&#xA;$$ {G M m \over r^2} = {m v^2 \over r} = {4 \pi^2 m r \over T^2} $$&#xA;Solve for $G$ and substitute values for Earth's orbit around the Sun:&#xA;$$ G = {4 \pi^2 r^3 \over M T^2 } = 4 \pi^2 {\mathrm{AU}^3 \over M_{\odot} \ \mathrm{y}^2} $$&#xA;If you want day units instead, use a conversion factor to cancel the years: &#xA;$$ G = 39.5 {\mathrm{AU}^3 \over M_{\odot} \ \mathrm{y}^2}&#xA;\left( {1 \ \mathrm{y} \over 365.25 \ \mathrm{d}} \right)^2&#xA;= 2.96 \times 10^{-4} {\mathrm{AU}^3 \over M_{\odot} \ \mathrm{d}^2} $$&#xA;The factor-label method also works if you start with the MKS value:&#xA;$$ G = 6.67 \times 10^{-11} \mathrm{m^3 \over kg \ s^2}&#xA;\left( {1 \ \mathrm{AU} \over 1.496 \times 10^{11} \ \mathrm{m}} \right)^3&#xA;\left( {1.99 \times 10^{30} \ \mathrm{kg} \over 1 \ M_{\odot}} \right)&#xA;\left( {86400 \ \mathrm{s} \over 1 \ \mathrm{d}} \right)^2&#xA;= {2.96 \times 10^{-4} {\mathrm{AU}^3 \over M_{\odot} \ \mathrm{d}^2}} $$&#xA;Validation tests will tell whether your simulation works correctly. Set up a variety of simple systems where you have a clear idea of what should happen, and then check what actually does happen.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-04-11T02:51:02.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14503" PostTypeId="1" AcceptedAnswerId="14504" CreationDate="2016-04-11T03:55:04.877" Score="-5" ViewCount="152" Body="&lt;p&gt;Estimate the mass of the star given this formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{L}{L_\mathrm{\odot}} = \left(\frac{M}{M_\mathrm{\odot}}\right)^{3.5}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $L= 2.752\times10^{28}\,\mathrm{W}$, how do I find the mass of the star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help.&lt;/p&gt;&#xA;" OwnerUserId="11509" LastEditorUserId="9301" LastEditDate="2017-01-04T16:03:52.833" LastActivityDate="2017-01-04T16:03:52.833" Title="Finding Mass of Star with only Luminosity" Tags="&lt;fundamental-astronomy&gt;&lt;luminosity&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2017-01-05T23:30:34.720" ContentLicense="CC BY-SA 3.0" />
  <row Id="14504" PostTypeId="2" ParentId="14503" CreationDate="2016-04-11T04:48:49.373" Score="2" Body="&lt;p&gt;Sounds like homework (correct me if that assumption is wrong, as you may want a somewhat different answer then.), so here is a &lt;strong&gt;hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clean up the equation first, to something that looks more like an equation for mass:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{L}{L_{sun}} = \left(\frac{M}{M_{sun}}\right)^{3.5}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{L}{L_{sun}}\right)^{\frac{1}{3.5}} = \frac{M}{M_{sun}}$$&lt;/p&gt;&#xA;" OwnerUserId="10172" LastActivityDate="2016-04-11T04:48:49.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14505" PostTypeId="2" ParentId="11879" CreationDate="2016-04-11T11:20:01.050" Score="0" Body="&lt;p&gt;I programmed one by myself some time ago. I used full SI units in combination with doubles (64-bit floating numbers). They work great for the scale of our solar system and are still extremly precise.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        var sun = new Star();&#xA;        sun.Position = new Vector3D(0,0,0);&#xA;        sun.Velocity = new Vector3D(0,0,0);&#xA;        sun.Mass = 1.998855e30;&#xA;&#xA;        var earth = new Planet();&#xA;        earth.Mass = 5.9722e24;&#xA;        earth.Position = new Vector3D(0, 149.6e9, 0);&#xA;        earth.Velocity = new Vector3D(29780,0,0);&#xA;&#xA;        system.World.Objects.Add(sun);&#xA;        system.World.Objects.Add(earth);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you are intersted, my code for this simulation is on github: &lt;a href=&quot;https://github.com/RononDex/Simulation&quot; rel=&quot;nofollow&quot;&gt;https://github.com/RononDex/Simulation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simulation itself is in the subfolder &lt;code&gt;Simulation.Testing&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="10271" LastActivityDate="2016-04-11T11:20:01.050" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14506" PostTypeId="1" AcceptedAnswerId="14509" CreationDate="2016-04-11T18:38:50.290" Score="3" ViewCount="158" Body="&lt;p&gt;The chemical evolution of galaxies is an important way to learn about their formation and stellar/gaseous constituents. Many galaxies show narrow emission lines at optical wavelengths (3500-9000 Angstroms) from ionized elements (e.g., [O II] at 3727 Angstroms) and recombination Balmer lines (e.g., H$\alpha$ at 6563 Angstroms). The origin of the ionized emission line flux can depend on the amount of the element present in the galaxy, and also on the nature of the ionizing source (i.e., the oft-mentioned &quot;shape of the ionizing spectrum&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a few questions that I cannot seem to find clear answers to in textbooks, publications and review papers. My main reference for the following facts/questions is Kewley &amp;amp; Dopita 2002, ApJS, 142. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the difference between metallicity and abundance, and what impact would these two parameters have on the ionized emission line flux from a galaxy (assuming that the ionizing spectrum is kept constant)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are claims that some ionized lines (e.g., the auroral line [O III] at 4363 Angstroms) would become much weaker as the metallicity increases. Why? I would think that higher metallicity means more Oxygen and, assuming we keep the ionization spectrum fixed, more likely for more [O III] to exist. Should this not increase the [O III] 4363AA flux? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Similarly, some excitation line ratios (e.g., ([O II] 3727 + [O III] 5007)/H$\beta$, commonly known as $R_{23}$) are known to decrease with increasing abundance (measured via log(O/H)+12). Again, why? I would think if the metallicity or oxygen abundance is higher, then there is an increased chance of getting more [O II] and [O III] emission. Also, assuming that metallicity is higher, the amount of hydrogen should decrease, and this would decrease the $H\beta$ flux in the $R_{23}$ ratio above, thus increasing the ratio at high abundance. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="607" LastActivityDate="2016-04-12T19:02:40.313" Title="How can ionized emission line flux decrease as a function of increasing metallicity or abundance?" Tags="&lt;galaxy&gt;&lt;star-formation&gt;&lt;spectroscopy&gt;&lt;active-galaxy&gt;&lt;metallicity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14507" PostTypeId="1" CreationDate="2016-04-11T23:30:21.733" Score="3" ViewCount="61" Body="&lt;p&gt;Since Mars is only weakly magnetized does Mars not suffer from  ground induced currents or are they far worse than on Earth?&lt;/p&gt;&#xA;" OwnerUserId="11520" LastEditorUserId="11520" LastEditDate="2016-04-11T23:35:53.363" LastActivityDate="2017-01-17T16:05:14.933" Title="Ground Induced Currents on Mars" Tags="&lt;mars&gt;&lt;magnetic-field&gt;&lt;solar-storm&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14508" PostTypeId="2" ParentId="14492" CreationDate="2016-04-12T03:17:04.837" Score="5" Body="&lt;p&gt;I'm not sure it qualifies as &quot;simple&quot;, but, using &#xA;&lt;a href=&quot;http://idlastro.gsfc.nasa.gov/ftp/pro/astro/hadec2altaz.pro&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://idlastro.gsfc.nasa.gov/ftp/pro/astro/hadec2altaz.pro&lt;/a&gt; (and some &#xA;additional calculations/simplifications): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ &#xA;   \begin{array}{|c|c|c|c|} &#xA;\hline &#xA;    \text{Event} &amp;amp; \text{Time} &amp;amp; \phi  &amp;amp; Z \\ &#xA;\hline &#xA;    \text{Any} &amp;amp; \text{t} &amp;amp; \tan ^{-1}(\cos (\lambda ) \sin (\delta )-\cos &#xA;      (\delta ) \cos (\alpha -t) \sin (\lambda ),\cos (\delta ) \sin (\alpha &#xA;      -t)) &amp;amp; \tan ^{-1}\left(\sqrt{(\cos (\lambda ) \sin (\delta )-\cos (\delta &#xA;      ) \cos (\alpha -t) \sin (\lambda ))^2+\cos ^2(\delta ) \sin ^2(\alpha &#xA;      -t)},\cos (\delta ) \cos (\lambda ) \cos (\alpha -t)+\sin (\delta ) \sin &#xA;      (\lambda )\right) \\ &#xA;\hline &#xA;    \text{Rise} &amp;amp; \alpha -\cos ^{-1}(-\tan (\delta ) \tan (\lambda )) &amp;amp; \tan &#xA;      ^{-1}\left(\sec (\lambda ) \sin (\delta ),\cos (\delta ) \sqrt{1-\tan &#xA;      ^2(\delta ) \tan ^2(\lambda )}\right) &amp;amp; 0 \\ &#xA;\hline &#xA;    \text{Transit} &amp;amp; \alpha  &amp;amp;  &#xA;\begin{cases} &#xA;                   \delta &amp;gt;\lambda  &amp;amp; 0 \\ &#xA;                   \delta =\lambda  &amp;amp; \text{Zenith} \\ &#xA;                   \delta &amp;lt;\lambda  &amp;amp; \pi  &#xA;                  \end{cases} &#xA;    &amp;amp; \frac{\pi }{2}-\left| \delta -\lambda  \right| \\ &#xA;\hline &#xA;    \text{Set} &amp;amp; \alpha +\cos ^{-1}(-\tan (\delta ) \tan (\lambda )) &amp;amp; \tan &#xA;      ^{-1}\left(\sec (\lambda ) \sin (\delta ),-\cos (\delta ) \sqrt{1-\tan &#xA;      ^2(\delta ) \tan ^2(\lambda )}\right) &amp;amp; 0 \\ &#xA;\hline &#xA;    \text{Lowest Point} &amp;amp; \alpha +\pi  &amp;amp;  &#xA;\begin{cases} &#xA;                   \delta &amp;gt;-\lambda  &amp;amp; 0 \\ &#xA;                   \delta =-\lambda  &amp;amp; \text{Nadir} \\ &#xA;                   \delta &amp;lt;-\lambda  &amp;amp; \pi  &#xA;                  \end{cases} &#xA;    &amp;amp; \left| \delta +\lambda  \right|-\frac{\pi }{2} \\ &#xA;\hline &#xA;   \end{array} &#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\phi$ is the azimuth of the object &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$Z$ is the altitude of the object above the horizon &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\alpha$ is the right ascension of the object &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\delta$ is the declination of the object &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\lambda$ is the latitude of the observer &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$t$ is the current local sidereal time &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note the two-argument form of arctangent is required so that the &#xA;results are in the correct quadrant: &#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Two-argument_variant_of_arctangent&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Two-argument_variant_of_arctangent&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional caveats: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If $\left| \delta -\lambda \right|&amp;gt;\frac{\pi }{2}$, the object is &#xA;always below the horizon, and the equations for rising time and &#xA;setting time will not work. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $\left| \delta +\lambda \right|&amp;gt;\frac{\pi }{2}$, the object is &#xA;always above the horizon (circumpolar), and the equations for rising &#xA;and setting time will also not work. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The measurements above are in radians. You can convert $\pi \to &#xA;  180 {}^{\circ}$ for degrees.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Because we use the local sidereal time, the longitude doesn't &#xA;appear in any of the formulas above. However, we do need it to find &#xA;the local sidereal time, as below. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To find the local sideral time $t$ in radians, we use &#xA;&lt;a href=&quot;http://aa.usno.navy.mil/faq/docs/GAST.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://aa.usno.navy.mil/faq/docs/GAST.php&lt;/a&gt; and make some substitions &#xA;to get: &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$t = 4.894961212735792 + 6.30038809898489 d + \psi$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\psi$ is your longitude in radians, and $d$ is the number of &#xA;days (including fractional days) since &quot;2000-01-01 12:00:00 UTC&quot;. Traditionally, we use $\phi$ for longitude, but I'm already using it in the formulas above for azimuth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you combine the formula for local sidereal time and &#xA;azimuth/altitude and assume excessive precision, you get my answer to &#xA;&lt;a href=&quot;https://astronomy.stackexchange.com/a/8415/21&quot;&gt;https://astronomy.stackexchange.com/a/8415/21&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional computations for these results at: &#xA;&lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/STACK/bc-rst.m&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/STACK/bc-rst.m&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was going to add some graphs to show how the altitude is NOT a sine wave and how the azimuth is NOT a straight line (although you might expect them to be), but they turned out not to be terribly instructive/helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also be able to get simpler formulas if you set $t$ to be the &quot;hour angle&quot; (which is $\alpha-t$ in the current setup).&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-04-12T03:17:04.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14509" PostTypeId="2" ParentId="14506" CreationDate="2016-04-12T07:38:01.150" Score="1" Body="&lt;h2&gt;Metallicity and abundance&lt;/h2&gt;&#xA;&#xA;&lt;strong&gt;Metallicity&lt;/strong&gt;&#xA;&#xA;&lt;p&gt;Without specifying a given metal, the term &quot;metallicity&quot; — abbreviated $Z$ — usually refers to the total metallicity of all elements, i.e. the &lt;em&gt;mass&lt;/em&gt; fraction of all metals to the total mass of some ensemble of elements, e.g. a star, a cloud of gas, a galaxy, etc. (as usual, the term &quot;metal&quot; refers to all elements that are not hydrogen or helium). For instance, the mass of all metals in the Sun, divided by the Sun's mass, is 0.02:&#xA;$$&#xA;Z_\odot \equiv \frac{M_\mathrm{C} + M_\mathrm{N} + M_\mathrm{O} + \ldots}{M_\odot} = 0.02.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes we talks about the metallicity of a given element, e.g. oxygen. The mass fraction of oxygen in the Sun is 0.005 (i.e. oxygen comprises 1/4 of all metals by mass), so we say $Z_\mathrm{O} = 0.005$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately it is not uncommon to implicitly talk about the metallicity of an object, &lt;em&gt;divided by Solar metallicity&lt;/em&gt;, such that a galaxy which has one-tenth of Sun's metallicity is said to have $Z=0.1$, rather than $Z=0.002$.&lt;/p&gt;&#xA;&#xA;&lt;strong&gt;Abundance&lt;/strong&gt;&#xA;&#xA;&lt;p&gt;The term &quot;abundance&quot; is only used for a single element. It basically expresses the same thing as metallicity, and is often used interchangeably, but is expressed in terms of the number $N$ of element nuclei, and as the ratio not to &lt;em&gt;all&lt;/em&gt; nuclei but to &lt;em&gt;hydrogen&lt;/em&gt; nuclei. For &lt;a href=&quot;https://astronomy.stackexchange.com/questions/11836/why-you-use-log-to-measure-metallicity-in-galaxies&quot;&gt;wacky historical reasons&lt;/a&gt;, we also take the logarithm and add a factor of 12. Taking again oxygen as an example, the mass fraction of 0.005 corresponds to a nuclei fraction of roughly $5\times10^{-4}$, so we say that the abundance of oxygen is (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2009CoAst.158..151G&quot; rel=&quot;nofollow noreferrer&quot;&gt;Grevesse (2009)&lt;/a&gt;)&#xA;$$&#xA;A(\mathrm{O}) \equiv \log \left( \frac{N_\mathrm{O}}{N_\mathrm{H}} \right) + 12 = 8.7.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Metallicity of a given species vs. total metallicity&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In general, the ratio of a given element to all metals is &lt;em&gt;roughly&lt;/em&gt; constant. That is, various elements are produced by stars approximately by the same amount. But various processes may cause elements to exist in various forms. For instance, metals deplete to dust, but some elements tend not to form dust, e.g. Zn. For this reason, Zn is a better proxy of the total metallicity than, e.g. Mg, since half of the Mg may be locked up in dust. &lt;/p&gt;&#xA;&#xA;&lt;strong&gt;Metals increases cooling&lt;/strong&gt;&#xA;&#xA;&lt;p&gt;Elements also appear in various excitation states, which depend on various processes. The lines you mention, [O II] and [O III], arise from collisionally ionized oxygen, which subsequently recombines (in my first answer I wrote, wrongly, that it was &lt;em&gt;excited&lt;/em&gt;), and thus depend on the temperature of the gas. As the metallicity of the gas in a galaxy increases, the ratio of the intensity of these lines to that of hydrogen lines (e.g. H$\beta$) first increases, as expected. However, the increased metallicity also allows the gas to cool more efficiently. The reason is that metals have many levels through which electron can&#xA;&quot;cascade&quot; down.&#xA;If the electron recombines to the level where it was before, a photon of the same energy will be emitted, which itself may radiatively ionize another atom. But the many levels in metals makes de-excitation to intermediate level more probable, such that the electron cascades down, emitting several low-energy (infrared) photons, which are incapable of ionizing atoms and thus escape. The result is that energy leaves the system, i.e. the system is cooled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This in turn means that, above a certain metallicity threshold — which is specific to a given species — the abundance of the collisionally excited lines begin to decrease. The following figure is taken from &lt;a href=&quot;http://adsabs.harvard.edu/abs/2002RMxAC..12...62S&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stasińska (2002)&lt;/a&gt;, and shows the turnover for the two oxygen lines:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JOnpQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JOnpQ.png&quot; alt=&quot;oxygen&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that measuring the metallicity of a single species in general gives two solutions for the total metallicity. Luckily, as the turnover is different for different elements, measuring the metallicity for several species can constrain the total metallicity.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-04-12T19:02:40.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14510" PostTypeId="2" ParentId="11441" CreationDate="2016-04-12T08:21:33.753" Score="1" Body="&lt;p&gt;They should only be almost perfectly flat. If you tried to add material out of the main axis, due to orbit mechanics it would fall in towards the plane of the rings, and then once closer, gravitational dampening would reduce its oscillation over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has this made sense?&lt;/p&gt;&#xA;" OwnerUserId="11525" LastEditorUserId="823" LastEditDate="2016-04-12T09:36:37.763" LastActivityDate="2016-04-12T09:36:37.763" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14511" PostTypeId="1" AcceptedAnswerId="14514" CreationDate="2016-04-12T11:18:57.770" Score="9" ViewCount="1318" Body="&lt;p&gt;Planets lie on the same plane because of the accretion disk formed during the Protostar stage, as I read in this &lt;a href=&quot;https://astronomy.stackexchange.com/questions/130/why-do-the-planets-in-our-solar-system-orbit-in-the-same-plane/11428#comment22008_137&quot;&gt;question&lt;/a&gt;.&#xA;&lt;br&gt;&#xA;I also read about the collision of particles in the gas cloud causing the overall spin to be in just one direction. But what decides the direction in which the accretion disk spins relative to the direction of the core?&#xA;&lt;br&gt;&#xA;(I'm thinking that it might be because of some primary conditions - maybe the direction in which the core is spinning.)&lt;/p&gt;&#xA;" OwnerUserId="11524" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-04-12T13:06:17.570" Title="What decides the direction in which the accretion disk spins?" Tags="&lt;star&gt;&lt;protostar&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14512" PostTypeId="2" ParentId="14511" CreationDate="2016-04-12T11:47:21.427" Score="2" Body="&lt;p&gt;Well the spin of the accretion disc is relative. We consider the top of anything as North. What if the the top was to be the South. It depends on perspective. Something rotating clockwise when seen from north would rotate counter clockwise when seen from south.&lt;/p&gt;&#xA;" OwnerUserId="11527" LastActivityDate="2016-04-12T11:47:21.427" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14513" PostTypeId="2" ParentId="14511" CreationDate="2016-04-12T12:41:51.137" Score="0" Body="&lt;p&gt;There's some angular momentum to the initial gas cloud, which is a bunch of gas particles with varying velocities. The angular momentum of this will not add up to zero usually, so as the cloud coalesces the resultant accretion disk will have a spin.&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2016-04-12T12:41:51.137" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14514" PostTypeId="2" ParentId="14511" CreationDate="2016-04-12T13:06:17.570" Score="8" Body="&lt;p&gt;Stellar systems are born from clouds of turbulent gas. Although &quot;turbulence&quot; means that different parcels of gas move in different directions, the cloud have some overall, net angular momentum. Usually a cloud gives birth to multiple stellar systems, but even the subregion forming a given system has a net, and non-vanishing (i.e. $\ne0$), angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Parcels moving in opposite directions will collide, and friction will cause the gas to lose energy, such that the cloud contracts. Eventually subclouds moving in one direction will &quot;win over&quot; subclouds moving in other directions such that everything moves in the same direction, keeping the original angular momentum (minus what is ejected e.g. through jets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the central star will rotate in the same direction as the circumstellar disk and that, in general, the planets that form subsequently, will also not only orbit the star in the same direction, but also spin in the same direction around their own axes. This is called &lt;em&gt;prograde rotation&lt;/em&gt;. Sometimes, however, collisions between bodies may cause a planet or asteroid to spin in the opposite direction. This is called &lt;em&gt;retrograde rotation&lt;/em&gt; and is the case for Venus and Uranus.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2016-04-12T13:06:17.570" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14515" PostTypeId="2" ParentId="14436" CreationDate="2016-04-12T13:43:11.797" Score="8" Body="&lt;p&gt;This is certainly a problem for any hypothesis that form planet 9 by ejecting it from the inner solar system. The issue (for those wondering) is that if you simply scatter an object out of the inner solar system, but it is retained in a bound elliptical orbit, then it should come back! The proposed planet 9 needs to have a perhelion distance beyond 200 au and at least a moderate eccentricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution might be found in a paper by &lt;a href=&quot;http://arxiv.org/abs/1603.08010&quot;&gt;Bromley &amp;amp; Kenyon (2016)&lt;/a&gt;, who address this very point. In their model they scatter a &quot;super-Earth&quot; from 5-15 au in the inner solar system, at an early stage of the solar system formation when the Sun is still surrounded by a gaseous disc. They run through a suite of simulation with different mass planets, different gas surface densities and different rates of evolution and dissipation of the gas disc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outcome is that there are sets of parameters that lead to eccentric orbits with large perihelion distances that are caused by damping by &lt;a href=&quot;https://en.wikipedia.org/wiki/Dynamical_friction&quot;&gt;dynamical friction&lt;/a&gt; from the gas disc. What is required are long-lived, low-density gas discs, or more short-lived, massive gas discs that clear from the inside-out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same paper also summarises a number of other possibilities that could explain the large perihelion distance and eccentricity of putative planet 9. These are: (i) in situ formation from a ring of solids - though the orbit would tend to be more circular; (ii) the nearby passage of another star could perturb the orbit, provide a boost to the perihelion distance and increase the semi-major axis, perhaps not by enough, but this mechanism could be more effective if the planet was in an eccentric orbit to begin with; (iii) tides, from the Galaxy, or more likely, from the birth cluster of the Sun could have influenced the orbit of a scattered planet and produced its high eccentricity and perihelion distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gist of the discussion is that whilst the authors think these other things are possible, they favour their own gas-drag model (naturally!).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-04-12T16:11:43.367" LastActivityDate="2016-04-12T16:11:43.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14516" PostTypeId="1" CreationDate="2016-04-12T17:19:41.030" Score="0" ViewCount="4426" Body="&lt;p&gt;What is the reason for that scale? Is it because otherwise defining an maximum would be too hard(?). Why do objects that are (apparently) brighter get assigned a smaller number (and down to negative)?&lt;/p&gt;&#xA;" OwnerUserId="11531" LastActivityDate="2016-04-12T23:56:26.800" Title="How can apparent magnitude be negative?" Tags="&lt;observational-astronomy&gt;&lt;amateur-observing&gt;&lt;fundamental-astronomy&gt;&lt;apparent-magnitude&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14518" PostTypeId="1" AcceptedAnswerId="14524" CreationDate="2016-04-12T18:31:12.077" Score="4" ViewCount="757" Body="&lt;p&gt;In Tonry &amp;amp; Davis (1979), they describe spectroscopic redshift measurement via correlating with templates at known redshift. In Section IIIa, they say &quot;Because the spectra are binned linearly with ln$\lambda$, a velocity redshift is a uniform linear shift.  Why is this the case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the relevant paragraph from the paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bvY80.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bvY80.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11533" LastActivityDate="2021-02-24T22:15:33.663" Title="Why do linear velocity redshifts correspond to linear pixel shifts when the spectra are binned in constant log wavelength?" Tags="&lt;redshift&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14519" PostTypeId="2" ParentId="14516" CreationDate="2016-04-12T18:45:50.823" Score="2" Body="&lt;p&gt;Apparent magnitude is measure of how bright an object appears to an observer on Earth, meaning it's a function of both the object's intrinsic luminosity and its distance from us.  The concept of magnitudes dates back to the Ancient Greeks, when stars in the sky were categorized into six magnitudes (the brightest being 1 and the faintest being 6).  Each successively lower magnitude was twice as bright as the one before, meaning the scale was logarithmic.  We still use magnitudes for historical reasons, though the scale was later standardized to use the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_x - m_{x,0} = -2.5\log_{10}(\frac{F_x}{F_{x,0}})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m_x$ and $F_x$ are the magnitude and flux of the object of interest and $m_{x,0}$ and $F_{x,0}$ are the magnitude of flux of a reference object (where usually Vega is used to define the 0 point in magnitude).  This means any object that appears brighter than Vega has a negative magnitude.  There is no limit to how bright an object can appear, so there is no lower limit to magnitudes.  The sun, for example, being the brightest object in our sky, has a magnitude of roughly -27.&lt;/p&gt;&#xA;" OwnerUserId="11533" LastEditorUserId="11533" LastEditDate="2016-04-12T23:56:26.800" LastActivityDate="2016-04-12T23:56:26.800" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14520" PostTypeId="2" ParentId="14516" CreationDate="2016-04-12T18:57:35.840" Score="2" Body="&lt;p&gt;The magnitude scale is a logarithmic scale.  An increase of 1 magnitude corresponds to a decrease in brightness of about 2,5 times dimmer. Vega, a bright star has a magnitude of 0, so any star that is brighter than Vega would have a magnitude that is less than 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an odd system; the reason for it is historical. The ancient Greeks ordered stars by their brightness into categories : the first magnitude  stars were the brightest the sixth magnitude were the dimmest. When it became possible to exactly measure the light intensity from stars, the scale was chosen so as to approximate the traditional magnitudes, but with this formal measurements of brightness, the logarithmic nature of the scale made it inevitable that the brightest objects would have a magnitude that was below 1, or even negative.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-04-12T18:57:35.840" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14521" PostTypeId="2" ParentId="13013" CreationDate="2016-04-12T21:37:19.020" Score="5" Body="&lt;p&gt;The short answer is &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Dynamical_friction&quot; rel=&quot;noreferrer&quot;&gt;dynamical friction&lt;/a&gt;&quot;: massive objects moving through a field of less massive objects create a &quot;wake&quot; which pulls back on them, leading to loss of energy. Because the SMBHs are much more massive than the stars, the molecules and atoms of the gas, and the dark matter particles (whatever they might be), they are especially prone to this. The net effect is for the SMBHs to lose energy and settle into the center of the (combined) system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once they form a binary, they can also lose energy via 3-body encounters with stars near the (combined) galaxy center: a star interacts with the SMBH binary and gains energy (usually ejected from the galaxy core), while the binary loses energy by shrinking. Massive elliptical galaxies often have low-density stellar &quot;cores&quot;, which are usually assumed to be the relics of one or more rounds of SMBH-binary mergers. If there is lots of gas in the galaxy center, they binary can also shrink via gravitational interactions with the gas.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2016-04-12T21:37:19.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14522" PostTypeId="2" ParentId="11441" CreationDate="2016-04-13T00:06:18.717" Score="4" Body="&lt;p&gt;There is an explanation for why rings flatten out &lt;a href=&quot;https://physics.stackexchange.com/a/6552/102747&quot;&gt;here&lt;/a&gt;. The general mechanism is that particles collide, and gets a very uniform momentum. Thus, any set-up giving unusually thick rings is in essence &quot;cheating&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some ways:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moons can cause &lt;a href=&quot;https://en.wikipedia.org/wiki/Rings_of_Saturn#D_Ring&quot; rel=&quot;nofollow noreferrer&quot;&gt;spiral waves&lt;/a&gt; in the rings, giving them more of a structure in the z direction. The ones known in Saturn's rings has a modes amplitude of just 10-100 m, but larger Moons can easily increase that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way is simply having massive rings. Then they can not get more flattened, as there are no more empty space to remove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A tilted ring relative to the Planets orbit around the star is going to experience tidal forces, as long as the radius of the rings is some notable fraction of the planet's orbital radius. From the context that sparked the question, that is not a suitable mechanism though, along with the possibility of having a so low density that particle collisions are rare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However, more promising:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Rings_of_Jupiter&quot; rel=&quot;nofollow noreferrer&quot;&gt;halo ring&lt;/a&gt; of Jupiter is estimated to be around 12500 km thick (about the same as the diameter of the Earth), and are very fine dust kept from condensing into a disc by both the magnetic fields of Jupiter, and by iterations with the Galilean Moons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have four planets with rings in the solar system, so the sample size is quite small. Applying some small-sample-size statistical methodology, in this case an unusual application of the &lt;a href=&quot;https://en.wikipedia.org/wiki/German_tank_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;German Tank Problem&lt;/a&gt;, we can give a rough but realistic maximum thickness of a ring:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$N \approx m+\frac{m}{k}-1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $m$ is the highest observed value, and $k$ the sample size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modified slightly to get a non-integer version that makes some sense, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{max}_{\text{thickness}} \approx 12500 \; \mathrm{km}+\frac{12500 \; \mathrm{km}}{4} \approx 16000 \; \mathrm{km}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By no means a very certain limit, but at least about what can obtain from what we know.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-10-17T16:44:41.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14523" PostTypeId="1" AcceptedAnswerId="14527" CreationDate="2016-04-13T05:04:20.697" Score="0" ViewCount="124" Body="&lt;p&gt;I've &lt;a href=&quot;http://www.ted.com/talks/jeremy_kasdin_the_flower_shaped_starshade_that_might_help_us_detect_earth_like_planets&quot; rel=&quot;nofollow&quot;&gt;heard&lt;/a&gt; that some astronomers are building an artifical solar-eclipse ship, that should help us to dim the diffraction of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why don't we just wait till night? What is the function of the eclipse-ship really?&lt;/p&gt;&#xA;" OwnerUserId="11538" LastEditorUserId="2270" LastEditDate="2016-04-13T12:17:20.420" LastActivityDate="2016-04-13T12:17:20.420" Title="Why would astronomers want to eclipse the Sun?" Tags="&lt;the-sun&gt;&lt;telescope&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14524" PostTypeId="2" ParentId="14518" CreationDate="2016-04-13T05:43:53.477" Score="6" Body="&lt;p&gt;All this means is that you need to bin your spectra in equal intervals of log wavelength for each pixel to be a constant interval in velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First consider the case whereeachpixel is worth a constant interval in linear wavelength. Here we have&#xA;$$ \frac{\Delta \lambda}{\lambda} = \frac{\Delta v}{c},$$&#xA;and the $\Delta v$ represented by each pixel &lt;em&gt;depends on $\lambda$, which changes across the spectrum.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now if you bin in equal increments of &#xA;$$\Delta \log \lambda = \frac{\Delta \lambda}{\lambda} = \frac{\Delta v}{c}$$ &#xA;and each pixel has the same fixed velocity interval, $\Delta v = c \Delta \log\lambda$, independent of $\lambda$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This log wavelength binning is a prerequisite for cross-correlation procedures that yield velocities.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-04-13T05:43:53.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14525" PostTypeId="1" CreationDate="2016-04-13T09:43:05.867" Score="6" ViewCount="812" Body="&lt;p&gt;How do you target a radio telescope on the precise object you wish to observe? You can point it in the general direction but how do you get the information from the exact point in the sky that you are investigating?&#xA;This seems self evident with an optical telescope but not with a radio telescope.&lt;/p&gt;&#xA;" OwnerUserId="11542" LastEditorUserId="10172" LastEditDate="2016-06-15T05:08:11.087" LastActivityDate="2018-06-19T16:24:20.337" Title="Radio telescope targeting" Tags="&lt;observational-astronomy&gt;&lt;radio-astronomy&gt;&lt;radio-telescope&gt;" AnswerCount="7" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14526" PostTypeId="2" ParentId="12410" CreationDate="2016-04-13T09:57:28.130" Score="1" Body="&lt;p&gt;The main factor is the velocity of the encounter. The higher the relative velocity between the two galaxies, the easier it is for them to pass through each other without being slowed down enough for a proper merger to take place, and without being strongly distorted by the encounter. (In a very high speed encounter, the two galaxies will spend almost no time close enough for tidal forces to be effective.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In clusters of galaxies, where the typical velocities are around 1000 km/s, mergers are rare; in small groups, where the velocities are around 100 km/s, mergers are more common. So the Milky Way and Andromeda are pretty much destined to merge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During a merger it's possible for stars in the outer parts of the galaxies to be ejected by tidal forces, but the central regions of both galaxies should definitely merge. Thus, the supermassive black holes (one in each central region of the original galaxies) will end up in the center, too. (Unless you have something exotic like a merger where one of the galaxies already has an unmarked binary SMBH; then you &lt;em&gt;might&lt;/em&gt; get one of the three SMBHs ejected via a 3-body interaction. But I suspect that's pretty unlikely.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2016-04-13T09:57:28.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14527" PostTypeId="2" ParentId="14523" CreationDate="2016-04-13T10:04:52.307" Score="3" Body="&lt;p&gt;As Conrad Turner notes, that talk is about a device for blocking the light from &lt;em&gt;another&lt;/em&gt; star (one you think might have planets orbiting around it). It's not for blocking the light of the Sun!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you try to look at a planet orbiting around another star, the glare from the star makes it very hard to see the planet. By placing a specially shaped device in front of the telescope (in this case, thousands of kilometers in front), between the telescope and the star you're interested in, you block most of the light from the star, but not the light from any planets near that star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/New_Worlds_Mission&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/New_Worlds_Mission&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2016-04-13T10:04:52.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14529" PostTypeId="2" ParentId="14525" CreationDate="2016-04-13T16:35:35.550" Score="1" Body="&lt;p&gt;I'm assuming your question is about fixed dish radio telescopes, like Arecibo.  (For what it's worth, there are non-fixed radio telescopes, such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Green_Bank_Telescope&quot; rel=&quot;nofollow&quot;&gt;Green Bank Telescope&lt;/a&gt;, that target like any other telescope.)  Arecibo uses a spherical reflector instead of the traditional parabolic reflector, and a mobile suspended structure above the primary reflector containing a secondary and tertiary reflector.  Moving this structure allows the telescope to aim at different parts of the sky.  It's still limited, however, due to the fixed dish, to roughly 30 degrees from the zenith.&lt;/p&gt;&#xA;" OwnerUserId="11533" LastActivityDate="2016-04-13T16:35:35.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14530" PostTypeId="2" ParentId="13467" CreationDate="2016-04-13T16:39:38.220" Score="2" Body="&lt;p&gt;In physics the &quot;speed&quot; of anything depends on the coordinate system you choose, since speed is measured as change in coordinate position in some interval of coordinate time. Even in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Special_relativity&quot; rel=&quot;nofollow&quot;&gt;special theory of relativity&lt;/a&gt;, which doesn't take into account gravity and hence involves no spacetime curvature, the notion that the speed of light is always equal to the same constant (labeled $c$ in physics and astronomy) would only true in a special class of coordinate systems known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Inertial_frame_of_reference&quot; rel=&quot;nofollow&quot;&gt;inertial frames&lt;/a&gt;, it is quite possible to define a &quot;non-inertial&quot; coordinate system in special relativity such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Rindler_coordinates&quot; rel=&quot;nofollow&quot;&gt;Rindler coordinates&lt;/a&gt; in which the speed of light does &lt;em&gt;not&lt;/em&gt; have the same value $c$. In the &lt;a href=&quot;http://en.wikipedia.org/wiki/General_relativity&quot; rel=&quot;nofollow&quot;&gt;general theory of relativity&lt;/a&gt;, which models gravity in terms of mass/energy curving spacetime, you can only have &quot;local inertial frames&quot; defined on very small patches of spacetime (specifically, the limit as the size approaches zero)--see &lt;a href=&quot;http://www.einstein-online.info/spotlights/equivalence_principle&quot; rel=&quot;nofollow&quot;&gt;this article on the &quot;equivalence principle&quot;&lt;/a&gt; for the conceptual details of how local inertial frames can be defined by observers in freefall measuring events in their immediate neighborhood (like an observer looking at events within an elevator in freefall). Such observers will always measure the local speed of light within a vacuum in their local region to be equal to $c$, regardless of the larger-scale properties of the spacetime they're embedded in, like the &quot;expansion of space&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you try to define a global coordinate system on a large region of curved spacetime, this coordinate system is always a non-inertial one, so there is no guarantee that the coordinate speed of light in this coordinate system will be equal to $c$, and indeed the coordinate speed of light may vary from one region of spacetime to another depending on what coordinate system you choose (the equations of general relativity work in &lt;em&gt;all&lt;/em&gt; smooth coordinate systems, as long as you define the &lt;a href=&quot;http://en.wikipedia.org/wiki/Metric_tensor_(general_relativity)&quot; rel=&quot;nofollow&quot;&gt;metric&lt;/a&gt; correctly relative to your chosen coordinate system). In the basic model of curved spacetime in cosmology (the &lt;a href=&quot;http://en.wikipedia.org/wiki/Friedmann%E2%80%93Lema%C3%AEtre%E2%80%93Robertson%E2%80%93Walker_metric&quot; rel=&quot;nofollow&quot;&gt;FLRW model&lt;/a&gt;), the simplifying assumption is made that matter is a sort of uniform fluid filling all of space, so that if you pick the right definition of simultaneity (multiple definitions are always possible in relativity due to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Relativity_of_simultaneity&quot; rel=&quot;nofollow&quot;&gt;relativity of simultaneity&lt;/a&gt;), you will find that the density of this fluid is identical at every point in space at any given moment of cosmic time. This obviously isn't completely true to life, but it's expected that on large scales the density of matter is &lt;em&gt;close&lt;/em&gt; to uniform at any given cosmic time, so it's seen as a reasonable approximation. The expansion of space basically means that the density of the fluid gets lower as time passes, and that if two objects are at rest relative to the local fluid in their immediate neighborhood, then the proper distance between them will grow with time (&lt;a href=&quot;http://en.wikipedia.org/wiki/Comoving_distance#Uses_of_the_proper_distance&quot; rel=&quot;nofollow&quot;&gt;proper distance&lt;/a&gt; corresponds to what you would measure if you laid a bunch of short rulers end-to-end between the two objects at a particular moment in time, and then added up the distances).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it happens, this cosmological model has a further nice feature (as discussed in the 'proper distance' link above which is based on &lt;a href=&quot;http://adsabs.harvard.edu/abs/2004PASA...21...97D&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;, see p. 99 of the 'Full Refereed Journal' link). The most &quot;natural&quot; coordinate system to use in this model is one in which the time coordinate corresponds to the proper time measured by a set of observers who have been at rest relative to the cosmic fluid since the big bang, and the spatial coordinate is such that the coordinate distance between any such observers at a given time corresponds to their proper distance at that time. If you use such a system, it works out that the overall coordinate velocity of any object can be broken down into a sum of two velocities:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The &quot;recession velocity&quot; at any given space, which is the velocity that an observer &lt;em&gt;at rest relative to the cosmic fluid&lt;/em&gt; would be moving (the rate that their proper distance from the origin of the coordinate system is growing as a function of time, where we can assume the origin corresponds to our own location in space).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &quot;peculiar velocity&quot; of any object which is &lt;em&gt;not&lt;/em&gt; at rest relative to the cosmic fluid, which is just the same as the velocity of that object as measured in the local inertial frame of an observer at the same location who &lt;em&gt;is&lt;/em&gt; at rest relative to the cosmic fluid. So, &lt;em&gt;the peculiar velocity of a light ray must always be $c$.&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, if we know the recession velocity $v_{rec}$ at some distant location in space, then a light ray emitted directly towards us from that location will have an overall velocity $v_{rec} - c$ in this coordinate system, and a light ray emitted directly away from us will have an overall velocity $v_{rec} + c$. So from the perspective of this coordinate system, it makes sense to say as a shorthand that the light itself always travels at $c$, but space is also expanding away from us and this accounts for why the light is redshifted, and also why light originally emitted at distance $d$ won't necessarily take a time of $d/c$ to reach our own location. But this neat way of describing things is specific to both the cosmological model being assumed and the coordinate system used, things may not work out so neatly in other choices of spacetime or other coordinate systems. The only really general statement you can make about the speed of light is the one I mentioned earlier, that regardless of what global coordinate system you use and what the speed of a light ray works out to be in that system, it's always true that in a local inertial frame defined on a small patch of spacetime, light traveling through that patch always has a speed of $c$ as measured in that local frame.&lt;/p&gt;&#xA;" OwnerUserId="2437" LastActivityDate="2016-04-13T16:39:38.220" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14531" PostTypeId="1" AcceptedAnswerId="14532" CreationDate="2016-04-13T16:48:20.887" Score="2" ViewCount="502" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The Solar constant is the intensity of the solar radiation in the&#xA;  upper atmosphere. It's value is about $1400$ $Wm^{-2}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now we begin by stating that the Power radiated by the sun is about $3.9*10^{26}$W. Then we using the Formula $I=\frac{P}{4\pi d^2}$, we compute the Intensity. (Here d is $1.5*10^{11}$m). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the mathematics but have trouble in interpreting the meaning of the solar constant. The Intensity formula used above is generally depicted by authors as follows (as per my understanding): There is a sphere around the Sun of radius $d$ and if the power is radiated equally in all directions then each point on the Sphere receives an Intensity $I=\frac{P}{4\pi d^2}$. But the Earth is not a point object, so is this assumption valid? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, what do you mean by the upper atmosphere? I've come across another version: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Solar constant is the intensity of the solar radiation in when the sun is directly overhead. It's value is about $1400$ $Wm^{-2}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guess my confusion is due to a lack of an intuitive understanding of this concept and I would be grateful if anyone could provide me with one. &lt;/p&gt;&#xA;" OwnerUserId="11547" LastActivityDate="2016-04-13T18:39:00.343" Title="Conceptual Doubt regarding the calculation of the Solar Constant" Tags="&lt;radiation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14532" PostTypeId="2" ParentId="14531" CreationDate="2016-04-13T17:32:12.430" Score="5" Body="&lt;p&gt;Imagine you have a solar collector measuring one square meter, which can directly measure the power (energy per unit time) striking it. If there are layers of atmosphere above the collector, some of the incoming light is absorbed or reflected by particles in the atmosphere, so you won't measure as much power as if you place the collector in space at the distance of Earth's orbit, with no atmosphere between you and the Sun. But if you place the collector in the upper atmosphere of the Earth, the density of atmospheric particles above you is so small that they absorb/reflect only a negligible amount of power. Orientation is also an issue--you'll get the maximum power if you orient the plane of the collector so that it's perpendicular to the direction of the incoming rays. If it's not perpendicular, then if you approximate the incoming radiation as a set of parallel rays which each carry an equal amount of power (a &lt;a href=&quot;http://en.wikipedia.org/wiki/Ray_(optics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;standard approximation in optics&lt;/a&gt;) then fewer rays strike the collector when it's at some angle that's not perpendicular to the rays. All this is illustrated in the following diagram (from &lt;a href=&quot;http://www.itacanet.org/the-sun-as-a-source-of-energy/part-2-solar-energy-reaching-the-earths-surface/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;), with the two possible orientations of the solar collector illustrated in gray, and the light rays illustrated as black lines:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OBetI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OBetI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solar constant is defined to be the power you get on such a surface of unit area if it's not being obstructed by atmospheric particles, and if it's oriented perpendicular to incoming rays so that it receives the maximum possible power at that location. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One aspect of the orientation issue that's important to understand is that if you orient your solar collector so that it's parallel to the ground (so that it would be lying flat if it was actually at ground level), then it will only receive the maximum power if the surface of the Earth is itself perpendicular to incoming rays from the Sun at that point, which is &lt;em&gt;not&lt;/em&gt; true in the above diagram. It would only be true at the point on Earth that's the dead &quot;center&quot; of the Earth as seen by an observer looking at it from the Sun's position, or the point where the gray line connecting the centers of Earth and Sun intersects with surface of the Earth on this diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Osyh6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Osyh6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is also the only point on Earth where an observer on the surface would see the Sun &quot;directly overhead&quot;, at the very top of the apparent dome of the sky. So even if you stripped away the Earth's atmosphere, only at this point on the Earth would the power reaching a square meter of the surface be equal to the solar constant, the &lt;em&gt;average&lt;/em&gt; power per square meter over the entire illuminated half of the Earth's surface would be less than the solar constant since most of the surface is at an angle relative to incoming rays. Incidentally, the angle of the surface relative to incoming rays is also the main reason why at Northern latitudes, it's warmer in the summer (when the surface at a Northern latitude is closer to being perpendicular to oncoming rays) than in the winter (when the surface is at a more inclined angle relative to incoming rays), as illustrated in a diagram from &lt;a href=&quot;http://www.popsci.com/five-questions-about-winter-solstice-answered&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NdGAP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NdGAP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's true that since the Earth is not a point, an observer in the upper atmosphere directly above that point on the surface would not be at &lt;em&gt;exactly&lt;/em&gt; the same radius from the Sun as the radius of Earth's orbit (defined in terms of the position of the center of the Earth), but if you actually do the calculation of the solar constant at the radius of Earth's center vs. the radius of the point on the surface that's closest to the Sun, the difference is very small so it can be ignored if you're only stating the solar constant to nearest Watt/meter^2 or even the nearest tenth of a Watt/meter^2. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_luminosity#Nominal_Solar_Luminosity&quot; rel=&quot;nofollow noreferrer&quot;&gt;nominal solar luminosity&lt;/a&gt; is defined to be $3.828 \times 10^{26}$ and 1 AU is defined as 149597870700 meters, and if you use the formula you mentioned to calculate the solar flux at 1 AU it works out to 1361.1664654 watts per square meter. The radius of the Earth at the equator is about 6378 meters, so at an equinox when the Sun is directly overhead at the equator, an observer seeing the Sun directly overhead should be at a distance of 149597870700 - 6378 = 149597864322 meters, if you use the same formula to calculate the solar flux at this radius you get 1361.1665815, so it only makes a different if you want precision beyond the second decimal place.&lt;/p&gt;&#xA;" OwnerUserId="2437" LastEditorUserId="2437" LastEditDate="2016-04-13T18:39:00.343" LastActivityDate="2016-04-13T18:39:00.343" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14533" PostTypeId="1" CreationDate="2016-04-13T17:36:35.797" Score="4" ViewCount="260" Body="&lt;p&gt;I have a set of &lt;strong&gt;medium-band filters&lt;/strong&gt; and I would like to compute $A_{\lambda}/E(B-V)$ for each filter which are not reported in the literatures. The magnitudes of the objects in the photometry catalogue should be corrected for the galactic dust extinction using &lt;strong&gt;&lt;em&gt;Cardelli et al. 1989&lt;/em&gt;&lt;/strong&gt; law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In appendix B of &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/305772/pdf&quot; rel=&quot;nofollow&quot;&gt;D.J. Schlegel, D.P. Finkbeiner, &amp;amp; M. Davis (1998, ApJ, 500, 525)&lt;/a&gt; estimating the $A_{\lambda}/E(B-V)$ for filters is obtained by using this equation &#xA;$$&#xA;\Delta m_f=-2.5\frac{\int d\lambda W_f(\lambda)F(\lambda)10^{\frac{-A(\lambda)\Delta m_V}{2.5}}}{\int d\lambda W_f(\lambda)F(\lambda)}&#xA;$$&#xA;$\Delta m_V$ is the reddening magnitude of an elliptical galaxies and $F(\lambda)$ is the flux of source and $W_f(\lambda)$ is the response function of a filter. I am wondering how I can use  &lt;strong&gt;&lt;em&gt;Cardelli et al. 1989&lt;/em&gt;&lt;/strong&gt; law to estimate these quantities for different filters or there is an alternative approach?&lt;/p&gt;&#xA;" OwnerUserId="2231" LastEditorUserId="9301" LastEditDate="2016-06-29T04:39:52.230" LastActivityDate="2016-06-29T04:39:52.230" Title="Estimating galactic dust extinction for medium band filters" Tags="&lt;galaxy&gt;&lt;redshift&gt;&lt;dust&gt;&lt;photometry&gt;&lt;extinction&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14534" PostTypeId="2" ParentId="14533" CreationDate="2016-04-13T21:12:58.207" Score="4" Body="&lt;p&gt;The quantity you want is basically the extinction law, and is usually called $k(\lambda)$. An extinction law is a fit to several measurements of the extinction $A_\lambda$ in some direction (or an average of several directions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cardelli et al. (1989) provides different functional forms for the mean extinction law, parametrized in their Eq. 1 as&#xA;$$&#xA;\frac{A_\lambda}{A_V} = a(x) + \frac{b(x)}{R_V},&#xA;$$&#xA;where $x$ is the inverse wavelength in $\mu\mathrm{m}^{-1}$, and the coefficients are given separately for IR, optical, UV, and FUV in Eqs. 2, 3, 4, and 5, respectively. The total-to-selective extinction $R_V\equiv A_V/E(B-V)$ takes different values for different lines of sight, but usually lies in the range 2.5 to 6, with 3.1 being a typical value in the Milky Way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the quantity you're interested in, simply convert your favorite wavelength to $x$, stick into Eq. 1, and multiply by $R_V$:&#xA;$$&#xA;k(\lambda) \equiv \frac{A_\lambda}{E(B-V)} = \frac{A_\lambda}{A_V} R_V.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2016-04-13T21:12:58.207" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14535" PostTypeId="1" AcceptedAnswerId="14539" CreationDate="2016-04-14T02:09:13.043" Score="5" ViewCount="2285" Body="&lt;p&gt;On the Sun's article on Wikipedia, there is an image showing how the Sun's brightness, radius and temperature have changed over time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iedcp.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iedcp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the past (and next) few billion years, I see the luminosity continuously rising, the temperature remaining about the same, and the radius also rising.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems bizarre to me. I would imagine that, for luminosity to increase, the temperature would have to increase. And for temperature to increase, the radius would have to &lt;em&gt;decrease&lt;/em&gt; (as the core becomes denser, temperatures rise).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why does it work this way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I would also appreciate a concrete explanation for why all three decreased at the beginning of the Sun's life, but suddenly began increasing.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2016-04-14T08:32:19.080" Title="Why is the Sun's brightness and radius increasing, but not its temperature?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;stellar-evolution&gt;&lt;stellar-astrophysics&gt;&lt;stellar-dynamics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14536" PostTypeId="1" AcceptedAnswerId="14543" CreationDate="2016-04-14T04:00:18.623" Score="0" ViewCount="87" Body="&lt;p&gt;As far as I know, Venus and Mercury have 0 flatting, but Mars and Earth have detectable flatting, and Venus and Mercury are both rotating slowly. I'm confused as to the relation between rotational speed and flatting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does rotation cause a rocky planet to be flat? Or zero flatting implies Venus and Mercury rotate slowly? Which is correct?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="2270" LastEditDate="2016-04-14T15:35:55.883" LastActivityDate="2016-04-14T15:35:55.883" Title="In rocky planets,does fast rotation cause flatting or low flatting imply slow rotation?" Tags="&lt;rotation&gt;&lt;terrestrial-planets&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14537" PostTypeId="1" AcceptedAnswerId="26974" CreationDate="2016-04-14T04:45:19.073" Score="2" ViewCount="176" Body="&lt;p&gt;Does a star, when collapsing into a neutron star, or even a black hole, pick up any significant relativistic mass due to its rapid collapse rate? &lt;/p&gt;&#xA;" OwnerUserId="11552" LastEditorUserId="2153" LastEditDate="2018-07-14T20:52:19.347" LastActivityDate="2018-07-17T13:01:54.560" Title="Does the mass of a collapsing star change?" Tags="&lt;star&gt;&lt;general-relativity&gt;&lt;mass&gt;&lt;stellar-remnants&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14539" PostTypeId="2" ParentId="14535" CreationDate="2016-04-14T08:32:19.080" Score="5" Body="&lt;p&gt;The effective temperature $T_\mathrm{eff}$ of a star, which is presumably what's been plotted, is defined through its relationship with the star's radius $R$ and luminosity $L$ by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L=4\pi R^2\sigma T_\mathrm{eff}^4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This comes from the assumption that the star radiates like a black body at the photosphere. While this isn't strictly true, it's quite accurate, and regardless, that's how we define the effective temperature. The actual surface temperature will be slightly different but also behave roughly as plotted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, even if $T_\mathrm{eff}$ is constant, the star expands if it grows brighter.  Also, you can see that the sensitivity to temperature is steeper than radius, so a moderate change in luminosity can be absorbed by a relatively small change in effective temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the luminosity is basically determined by the simple behaviour of the nuclear reactions in the core (in terms of temperature and density), the surface properties depend on how energy is being transported near the surface. For radiation, you have to consider what the opacity of the material is, which itself depends on ionization states and whatnot. It's easy enough to see why the luminosity grows (the core gets denser and also hotter, producing energy faster) but the determination of the surface properties is more complicated. For the Sun, it turns out the way shown in the plot after you solve all the equations with the relevant opacities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as an extreme counterexample to &quot;brighter means hotter and smaller&quot;, remember that red giants are much brighter but also much cooler!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I'm not sure the source of the data but I would guess the wiggle at the start is because of the star finishing its contraction onto the main sequence.  That is, before the first minimum, energy is being released by gravitational contraction.  After it, the energy from nuclear reactions starts to dominate.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2016-04-14T08:32:19.080" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14540" PostTypeId="1" AcceptedAnswerId="19443" CreationDate="2016-04-14T08:37:38.717" Score="7" ViewCount="117" Body="&lt;p&gt;What are the conditions for an ionosphere to form on a given planet? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Factors I could think of are &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Class and age of the star the planet orbits&lt;/li&gt;&#xA;&lt;li&gt;Distance to said star&lt;/li&gt;&#xA;&lt;li&gt;Density and composition of the planet's atmosphere&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11554" LastActivityDate="2018-10-20T20:18:43.693" Title="What are the conditions for an ionosphere to form?" Tags="&lt;planetary-atmosphere&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14542" PostTypeId="2" ParentId="14537" CreationDate="2016-04-14T12:21:56.780" Score="1" Body="&lt;p&gt;The bulk motion of the gas would contribute to the relativistic mass-energy of the compact object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;White dwarfs are not produced by collapse, so there is nothing to comment on here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutron stars are produced by collapse. Most of the kinetic energy is lost in the form of neutrinos, the rest would be thermalised within the neutron star, increasing the kinetic energy of its constituent particles (which become partially relativistic) and that energy does contribute to the gravitational mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a black hole forms then all the mass-energy that goes into its formation contributes to its gravitational mass.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-04-14T12:21:56.780" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14543" PostTypeId="2" ParentId="14536" CreationDate="2016-04-14T12:47:19.093" Score="3" Body="&lt;p&gt;The flattening of a planet is a function of both its spin rate and its structure. But for a series of planets of homologous structure the flattening depends on the spin rate. the faster it spins the greater the flattening (for spin rates typical of planets any way). So if a planet is spinning slowly it will display little polar flattening, similarly if it does not show flattening it should be spinning slowly.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2016-04-14T12:47:19.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14544" PostTypeId="1" AcceptedAnswerId="14545" CreationDate="2016-04-14T17:29:06.743" Score="0" ViewCount="103" Body="&lt;p&gt;Some years ago I looked up to the blue sky, around 2 PM, and I saw, for 2 or 3 seconds, something like a star fading out.&#xA;Could it be a Supernova or another phenomena? Is it highly possible?&lt;/p&gt;&#xA;" OwnerUserId="11558" LastActivityDate="2016-05-29T03:16:14.057" Title="What could be a star-like object that shines at daytime for few seconds? What's the probability to see it?" Tags="&lt;supernova&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14545" PostTypeId="2" ParentId="14544" CreationDate="2016-04-14T20:22:39.610" Score="3" Body="&lt;p&gt;Several satellites (but most notably Iridium) have large reflective panels.  If aligned with the sun properly, it can shine with sufficient brightness to be visible in the daytime.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An &lt;a href=&quot;https://en.wikipedia.org/wiki/Satellite_flare#Iridium_flares&quot; rel=&quot;nofollow&quot;&gt;Iridium Flare&lt;/a&gt; usually lasts only a few seconds.  It sounds quite consistent with your description.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be hard to tell in a blue sky with no nearby references, but an Iridium satellite would also be moving while fading out.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are sites that can show you upcoming flares for your area.  If you knew the exact date in the past, they could probably even &quot;post-dict&quot; what flares were available then and see if it matched.  &lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2016-04-14T20:22:39.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14546" PostTypeId="1" CreationDate="2016-04-14T21:54:50.560" Score="3" ViewCount="75" Body="&lt;p&gt;Suppose you had an optical spectrum of a galaxy, and you unambiguously detected emission lines from cooling ionized gas -- both forbidden lines like [O II] 3727AA and [O III] 4959,5007AA, as well as recombination lines like H$\beta$. So you have measured things like the integrated flux, equivalent width ($\equiv$ integrated flux / local continuum flux density), FWHM, etc. for your emission lines. Now, you want to know how much warm ionized gas is in the galaxy (or at least in the region in which your spectrum was taken) -- i.e., you want to go from your aforementioned Gaussian parameters like integrated flux to a gas mass in solar masses. How do people typically do this, and what assumptions do you have to make? Any references would be greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="607" LastEditorUserId="9301" LastEditDate="2016-06-29T04:37:59.080" LastActivityDate="2016-06-29T04:37:59.080" Title="How do people derive ionized gas mass from optical emission lines in a galaxy spectrum?" Tags="&lt;galaxy&gt;&lt;spectroscopy&gt;&lt;interstellar-medium&gt;&lt;gas&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14547" PostTypeId="1" AcceptedAnswerId="14602" CreationDate="2016-04-15T05:47:52.683" Score="1" ViewCount="427" Body="&lt;p&gt;Tonight while observing Jupiter I was able to see a five (what looked like) moons. I have observed Jupiter often before and only been able to make out the four galilean moons. It's my understanding that through my 8-inch scope it would be impossible for me to see any others. Switching to a widefield lens I was able to see Jupiter along with some nearby stars however the 5 moon-like points around it all still appeared... well, moon-like in comparison to the bright stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I observed Jupiter from about 2016-04-15 05:20 to 05:40 UTC and all five of the moon-like objects were present the whole time. What was I probably looking at?&lt;/p&gt;&#xA;" OwnerUserId="10953" LastActivityDate="2016-04-19T17:16:10.840" Title="What is the fifth moon-like object I saw around jupiter through my telescope?" Tags="&lt;planet&gt;&lt;amateur-observing&gt;&lt;jupiter&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14548" PostTypeId="2" ParentId="14547" CreationDate="2016-04-15T07:35:09.870" Score="4" Body="&lt;p&gt;A seventh magnitude star.  See this link:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cloudynights.com/topic/533968-jovian-events-414-1516/&quot; rel=&quot;nofollow&quot;&gt;http://www.cloudynights.com/topic/533968-jovian-events-414-1516/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further info: the star in question was probably HIP 54057 (also known as SAO 118636 or HD 95848.) On April 12th it was occulted by Jupiter (that is, Jupiter passed in front of it) for about three hours. So the star would have been in the area of Jupiter for a few days.&lt;/p&gt;&#xA;" OwnerUserId="7491" LastEditorUserId="7491" LastEditDate="2016-04-19T17:16:10.840" LastActivityDate="2016-04-19T17:16:10.840" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14549" PostTypeId="1" AcceptedAnswerId="14550" CreationDate="2016-04-15T08:00:40.737" Score="0" ViewCount="345" Body="&lt;p&gt;I have a Celestron 5se + canon 650d. They are mated with the celestron t-ring and the 93625 t-adaptor. &lt;a href=&quot;https://i.stack.imgur.com/SJdPq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SJdPq.jpg&quot; alt=&quot;picture below&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;I want to take magnified photos of planets with the camera. But if I add an eyepiece to the t-adaptor, everything gets blur.&lt;/p&gt;&#xA;&lt;p&gt;Here are some scenarios that highlight the problem:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;look at Jupiter with 25mm eyepiece in telescope: clear and focused image&lt;/li&gt;&#xA;&lt;li&gt;look at Jupiter with 6mm eyepiece in telescope: clear and focused zoomed image&lt;/li&gt;&#xA;&lt;li&gt;look at Jupiter with camera and no eyepiece inside the t-adaptor, in telescope: clear and focused image but small.&lt;/li&gt;&#xA;&lt;li&gt;look at Jupiter with camera and 6mm eyepiece inside the t-adaptor, in telescope: large unfocused ball&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;I dont know how to solve this.&lt;/p&gt;&#xA;&lt;p&gt;Clearly, the 93625 t-adaptor was made to accept an eyepiece and lock it with the thumbscrew, but why does it not result in focused images?&lt;/p&gt;&#xA;" OwnerUserId="11560" LastEditorUserId="1891" LastEditDate="2020-12-23T22:55:43.663" LastActivityDate="2020-12-23T22:55:43.663" Title="How do I get focussed images with my T-adaptor and 6mm eyepeice?" Tags="&lt;photography&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14550" PostTypeId="2" ParentId="14549" CreationDate="2016-04-15T10:48:59.247" Score="1" Body="&lt;p&gt;Using the eyepiece, and no camera lens, is a very different configuration to the camera+tube+scope (case 3).  The configuration you're after is &quot;Eyepiece projection&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know exactly what you have available, but I believe you will need to fit a T-threaded extension tube to the external thread on your T-adapter. Then fit the camera onto that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note - before you go out and buy a new extension you can probably do a hand-held test of this by the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Attach the adapter to the scope, with a low-power eyepiece in the adapter (low power makes the test easier to see.)&lt;/li&gt;&#xA;&lt;li&gt;Point the scope at anything distant (you can do this in daylight, but NOT at the sun of course!)&lt;/li&gt;&#xA;&lt;li&gt;Now hold the camera, by hand, a few inches behind the&#xA;eyepiece+adapter. Perhaps you can use a cardboard tube to keep the stray light out.&lt;/li&gt;&#xA;&lt;li&gt;See if you can adjust the scope's focus, and the camera position, to get a sharp image.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If this worked, then you can get a T-threaded extension tube to go on the back of your current adapter, to hold the camera in position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See for example this link:&#xA;&lt;a href=&quot;http://www.astronomysource.com/2011/10/13/eyepiece-projection/&quot; rel=&quot;nofollow&quot;&gt;http://www.astronomysource.com/2011/10/13/eyepiece-projection/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2016-04-15T10:48:59.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14551" PostTypeId="1" AcceptedAnswerId="14552" CreationDate="2016-04-15T13:36:53.243" Score="1" ViewCount="845" Body="&lt;p&gt;I'm looking for a free, open-source and Linux compatible alternative to the &lt;a href=&quot;https://www.aurigaimaging.com/index.php&quot; rel=&quot;nofollow&quot;&gt;RegiStar&lt;/a&gt; software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need it to automatically combine several images into a single frame, where each individual image shows a portion of a large observed region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Individual images also overlap by some amount, and need to be aligned and possibly rotated before superimposing them onto the final frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe RegiStar does all this but: a- it is not free, b- it is not open-source, and c- it seems to run on Windows only.&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="354" LastEditDate="2016-04-15T16:11:58.573" LastActivityDate="2016-04-15T16:33:23.793" Title="Open source alternative for RegiStar (align, rotate, and combine images into single frame)" Tags="&lt;photography&gt;&lt;software&gt;&lt;alignment&gt;" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="14552" PostTypeId="2" ParentId="14551" CreationDate="2016-04-15T15:12:12.673" Score="3" Body="&lt;p&gt;I've had success with &lt;a href=&quot;http://hugin.sourceforge.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hugin&lt;/a&gt;. It's a panorama creator, but it can be used to rotate, resize, and relocate images into a coherent whole. It has a few different automated profiles, but sometimes they are a bit off with star stacks. You can manually select the guide stars though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A really good program for after that is &lt;a href=&quot;https://www.gimp.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;GIMP&lt;/a&gt; and the plugin &lt;a href=&quot;http://gmic.eu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;G'MIC&lt;/a&gt; to stack and remove noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately it's not a single package like RegiStar, but it's free and can do quite a lot!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before&#xA;&lt;a href=&quot;https://i.stack.imgur.com/h6W8W.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h6W8W.jpg&quot; alt=&quot;Before&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After&#xA;&lt;a href=&quot;https://i.stack.imgur.com/NxMew.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NxMew.jpg&quot; alt=&quot;After&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11525" LastEditorUserId="11525" LastEditDate="2016-04-15T16:33:23.793" LastActivityDate="2016-04-15T16:33:23.793" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14553" PostTypeId="2" ParentId="12368" CreationDate="2016-04-15T16:00:03.197" Score="1" Body="&lt;p&gt;Most likely candidate is a meteor with the brightest part cut off by the shutter. You can see that it is increasing in brightness. Forgery can be eliminated due to various sightings, and it shouldn't a satellite due to the distance covered in half a second, not to mention that it would be illuminated by the sun in that portion of the sky.&lt;/p&gt;&#xA;" OwnerUserId="11525" LastActivityDate="2016-04-15T16:00:03.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14554" PostTypeId="2" ParentId="529" CreationDate="2016-04-15T16:05:08.490" Score="4" Body="&lt;p&gt;&lt;a href=&quot;http://math.ucr.edu/home/baez/end.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This page&lt;/a&gt; by physicist &lt;a href=&quot;http://en.wikipedia.org/wiki/John_C._Baez&quot; rel=&quot;nofollow noreferrer&quot;&gt;John Baez&lt;/a&gt; explains what will happen in the long term to bodies that aren't massive enough to collapse into black holes, like rogue planets and white dwarfs, assuming they don't cross paths with preexisting black holes and get absorbed. Short answer: they'll evaporate, for reasons unrelated to Hawking radiation. It's apparently just a thermodynamic matter, presumably due to the internal thermal energy of the body periodically causing particles on the surface to randomly get enough kinetic energy to achieve escape velocity and escape the body (the wiki article &lt;a href=&quot;http://en.wikipedia.org/wiki/Atmospheric_escape#Thermal_escape_mechanisms&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; mentions this is known as 'Jeans escape'). Here's the full discussion:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Okay, so now we have a bunch of isolated black dwarfs, neutron stars, and black holes together with atoms and molecules of gas, dust particles, and of course planets and other crud, all very close to absolute zero.&lt;/p&gt;&#xA;&lt;p&gt;As the universe expands these things eventually spread out to the point where each one is completely alone in the vastness of space.&lt;/p&gt;&#xA;&lt;p&gt;So what happens next?&lt;/p&gt;&#xA;&lt;p&gt;Well, everybody loves to talk about how all matter eventually turns to iron thanks to quantum tunnelling, since iron is the nucleus with the least binding energy, but unlike the processes I've described so far, this one actually takes quite a while. About &lt;span class=&quot;math-container&quot;&gt;$10^{1500}$&lt;/span&gt; years, to be precise. (Well, not too precise!) So it's quite likely that proton decay or something else will happen long before this gets a chance to occur.&lt;/p&gt;&#xA;&lt;p&gt;For example, everything except the black holes will have a tendency to &amp;quot;sublimate&amp;quot; or &amp;quot;ionize&amp;quot;, gradually losing atoms or even electrons and protons, despite the low temperature. Just to be specific, let's consider the ionization of hydrogen gas — although the argument is much more general. If you take a box of hydrogen and keep making the box bigger while keeping its temperature fixed, it will eventually ionize. This happens no matter how low the temperature is, as long as it's not exactly absolute zero — which is forbidden by the 3rd law of thermodynamics, anyway.&lt;/p&gt;&#xA;&lt;p&gt;This may seem odd, but the reason is simple: in thermal equilibrium any sort of stuff minimizes its free energy, E - TS: the energy minus the temperature times the entropy. This means there is a competition between wanting to minimize its energy and wanting to maximize its entropy. Maximizing entropy becomes more important at higher temperatures; minimizing energy becomes more important at lower temperatures — but both effects matter as long as the temperature isn't zero or infinite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;[I'll interrupt this explanation to note that any completely &lt;em&gt;isolated&lt;/em&gt; system just maximizes its entropy in the long term, this isn't true for a system that's in contact with some surrounding system. Suppose your system is connected to a much bigger collection of surroundings (like being immersed in a fluid or even a sea of cosmic background radiation), and the system can trade energy in the form of heat with the surroundings (which won't appreciably change the temperature of the surroundings given the assumption the surroundings are much larger than the system, the surroundings being what's known as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_reservoir&quot; rel=&quot;nofollow noreferrer&quot;&gt;thermal reservoir&lt;/a&gt;), but they can't trade other quantities like volume. Then the statement that the total entropy of system + surroundings must be maximized is equivalent to the statement that the system alone must minimize a quantity called its &amp;quot;Helmholtz free energy&amp;quot;, which is what Baez is talking about in that last paragraph--see &lt;a href=&quot;https://physics.stackexchange.com/a/66361/59406&quot;&gt;this answer&lt;/a&gt; or &lt;a href=&quot;http://scienceworld.wolfram.com/physics/HelmholtzFreeEnergy.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;. And incidentally, if they can trade both energy &lt;em&gt;and&lt;/em&gt; volume, maximizing the total entropy of system + surroundings is equivalent to saying the system on its own must minimize a slightly different quantity called its &amp;quot;Gibbs free energy&amp;quot; (which is equal to Helmholtz free energy plus pressure times change in volume), see &amp;quot;Entropy and Gibbs free energy&amp;quot; &lt;a href=&quot;http://2ndlaw.oxy.edu&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.]&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Think about what this means for our box of hydrogen. On the one hand, ionized hydrogen has more energy than hydrogen atoms or molecules. This makes hydrogen want to stick together in atoms and molecules, especially at low temperatures. But on the other hand, ionized hydrogen has more entropy, since the electrons and protons are more free to roam. And this entropy difference gets bigger and bigger as we make the box bigger. So no matter how low the temperature is, as long as it's above zero, the hydrogen will eventually ionize as we keep expanding the box.&lt;/p&gt;&#xA;&lt;p&gt;(In fact, this is related to the &amp;quot;boiling off&amp;quot; process that I mentioned already: we can use thermodynamics to see that the stars will boil off the galaxies as they approach thermal equilibrium, as long as the density of galaxies is low enough.)&lt;/p&gt;&#xA;&lt;p&gt;However, there's a complication: in the expanding universe, the temperature is not constant — it decreases!&lt;/p&gt;&#xA;&lt;p&gt;So the question is, which effect wins as the universe expands: the decreasing density (which makes matter want to ionize) or the decreasing temperature (which makes it want to stick together)?&lt;/p&gt;&#xA;&lt;p&gt;In the short run this is a fairly complicated question, but in the long run, things may simplify: if the universe is expanding exponentially thanks to a nonzero cosmological constant, the density of matter obviously goes to zero. But the temperature does not go to zero. It approaches a particular nonzero value! So all forms of matter made from protons, neutrons and electrons will eventually ionize!&lt;/p&gt;&#xA;&lt;p&gt;Why does the temperature approach a particular nonzero value, and what is this value? Well, in a universe whose expansion keeps accelerating, each pair of freely falling observers will eventually no longer be able to see each other, because they get redshifted out of sight. This effect is very much like the horizon of a black hole - it's called a &amp;quot;cosmological horizon&amp;quot;. And, like the horizon of a black hole, a cosmological horizon emits thermal radiation at a specific temperature. This radiation is called Hawking radiation. Its temperature depends on the value of the cosmological constant. If we make a rough guess at the cosmological constant, the temperature we get is about &lt;span class=&quot;math-container&quot;&gt;$10^{-30}$&lt;/span&gt; Kelvin.&lt;/p&gt;&#xA;&lt;p&gt;This is very cold, but given a low enough density of matter, this temperature is enough to eventually ionize all forms of matter made of protons, neutrons and electrons! Even something big like a neutron star should slowly, slowly dissipate. (The crust of a neutron star is not made of neutronium: it's mainly made of iron.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2437" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-04-18T20:20:10.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14555" PostTypeId="1" CreationDate="2016-04-15T18:03:01.207" Score="2" ViewCount="456" Body="&lt;p&gt;How can I calculate position (alt, az) of star for observer at some station (longitude, latitude) on specific date and time? I need full coordinates reduction with all elements included in calculation (proper motion of star, atmospheric pressure and temperature... )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried with pyephem but I am not sure can I finish.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import ephem&#xA;&#xA;polaris = ephem.readdb(&quot;Polaris,f|M|F7,2:31:48.704,89:15:50.72,2.02,1&quot;)&#xA;&#xA;polaris.compute('2016/3/1 16:22:56')&#xA;&#xA;print polaris.a_ra &#xA;print polaris.a_dec&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I also tried with astroplan and I think that I am closer to solution, but still don't know how to get coordinates after reduction and add proper motion. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import astropy.units as u&#xA;from astropy.coordinates import EarthLocation&#xA;from astropy.coordinates import SkyCoord&#xA;from pytz import timezone&#xA;from astroplan import Observer&#xA;from astropy.time import Time&#xA;from astroplan import FixedTarget&#xA;import numpy as np&#xA;import astropy.units as u&#xA;from astroplan.plots import plot_sky&#xA;from astroplan.plots import plot_parallactic&#xA;from astroplan.plots import plot_airmass&#xA;import matplotlib.pyplot as plt&#xA;from astroplan import FixedTarget&#xA;&#xA;&#xA;longitude = '21d33m20.4s'&#xA;latitude = '+43d08m24.6s'&#xA;elevation = 1150 * u.m&#xA;time = Time('2015-06-16 12:00:00')&#xA;&#xA;location = EarthLocation.from_geodetic(longitude, latitude, elevation)&#xA;&#xA;observer = Observer(name='Name',&#xA;               location=location,&#xA;               pressure=0.615 * u.bar,&#xA;               relative_humidity=0.11,&#xA;               temperature=0 * u.deg_C,&#xA;               timezone=timezone('Europe/Belgrade'),&#xA;               description=&quot;..&quot;)&#xA;&#xA;&#xA;&#xA;coordinates = SkyCoord('2h31m48.704s', '89d15m50.72s', frame = 'icrs')&#xA;polaris = FixedTarget(name='Polaris', coord=coordinates)&#xA;&#xA;plot_airmass(polaris, observer, time)&#xA;&#xA;ax = plt.gca()&#xA;box = ax.get_position()&#xA;ax.set_position([box.x0, box.y0, box.width * 0.8, box.height * 0.8])&#xA;&#xA;plt.legend(loc=1, bbox_to_anchor=(1.35, 1))&#xA;plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="11563" LastEditorUserId="2365" LastEditDate="2016-04-15T19:32:22.483" LastActivityDate="2016-04-15T19:32:22.483" Title="Coordinates reduction python" Tags="&lt;coordinate&gt;&lt;python&gt;" AnswerCount="0" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14556" PostTypeId="2" ParentId="13177" CreationDate="2016-04-15T20:32:59.990" Score="2" Body="&lt;p&gt;The Earth moves in an elliptic orbit around the sun (or around the barycenter). If, in helocentric coordinates the Earth is at position (x,y), then in Geocentric coordinates the position of the sun is in position (-x,-y) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the locus of the Sun in Geocentric coordinates exactly matches the locus of the Earth in Heliocentric. The path of the sun in geocentric coordinates is an Ellipse.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-04-15T20:32:59.990" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14557" PostTypeId="1" AcceptedAnswerId="14563" CreationDate="2016-04-16T09:17:39.227" Score="2" ViewCount="225" Body="&lt;p&gt;I just bought a new telescope with these specifications:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Aperture: 203 mm&lt;/li&gt;&#xA;&lt;li&gt;Focal Length: 1200 mm&lt;/li&gt;&#xA;&lt;li&gt;F/ratio: f/5.9&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I usually use a 28 mm eyepiece, giving me about 42.5x magnification, but I also have a 20 mm and a 10 mm eyepiece. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was trying to find Messier objects, but I only found the Orion Nebula, and even then it wasn't very impressive. I was wondering, what types of objects are see-able in my telescope? I read that Charles Messier found the Messier objects using a four-inch refractor, and I have an 8 inch reflecting telescope. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I tried to find things like the pinwheel galaxy, or the crab nebula, I only saw points of light. So basically, how do I see things like nebula and galaxies?&lt;br /&gt;&#xA;For reference, I live in the midwestern United States.&lt;/p&gt;&#xA;" OwnerUserId="11568" LastEditorDisplayName="user1569" LastEditDate="2016-04-16T13:58:19.727" LastActivityDate="2016-05-01T15:03:12.977" Title="How do I see more than just points in the sky?" Tags="&lt;galaxy&gt;&lt;telescope&gt;&lt;nebula&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14558" PostTypeId="1" CreationDate="2016-04-16T10:25:50.203" Score="3" ViewCount="72" Body="&lt;p&gt;Where can I find the observed frequency of triple star systems where the outer star is in a retrograde orbit?&lt;/p&gt;&#xA;" OwnerUserId="11569" LastActivityDate="2016-04-19T18:53:23.227" Title="Triple stars with retrograde orbits" Tags="&lt;star&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14559" PostTypeId="1" AcceptedAnswerId="14562" CreationDate="2016-04-16T14:34:04.717" Score="3" ViewCount="4052" Body="&lt;p&gt;How long would it take to reach the current edge of the reachable portion of the universe, with the following bounds in mind:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We figure out a way build a space ship to provide a constant 1 G acceleration for an extended period of time.&lt;/li&gt;&#xA;&lt;li&gt;We accelerate until we are halfway there, then decelerate past this point (so we stop when we get there.&lt;/li&gt;&#xA;&lt;li&gt;Our target is about 15 billion lightyears away.  As per &lt;a href=&quot;http://curious.astro.cornell.edu/about-us/104-the-universe/cosmology-and-the-big-bang/expansion-of-the-universe/616-is-the-universe-expanding-faster-than-the-speed-of-light-intermediate&quot; rel=&quot;nofollow&quot;&gt;http://curious.astro.cornell.edu/about-us/104-the-universe/cosmology-and-the-big-bang/expansion-of-the-universe/616-is-the-universe-expanding-faster-than-the-speed-of-light-intermediate&lt;/a&gt; galaxies past 4,740 megaparsecs are no longer reachable. Using this site &lt;a href=&quot;http://www.convertunits.com/from/megaparsecs/to/light+years&quot; rel=&quot;nofollow&quot;&gt;http://www.convertunits.com/from/megaparsecs/to/light+years&lt;/a&gt; to convert megaparsecs to lightyears means this &quot;unreachable&quot; point is currently 15,459,812,396.5002 lightyears away.  Let's simplify to 15 billion lightyears, which is still supposed to be reachable.&lt;/li&gt;&#xA;&lt;li&gt;We somehow work out all the details of life support, collision with particles at relativistic speeds, etc...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would like to know the time the traveler experiences, and the time that everyone else (e.g. those on earth) experience.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="1386" LastEditDate="2016-04-16T17:53:15.060" LastActivityDate="2020-06-17T10:40:54.947" Title="How long would it take to reach the edge of the reachable universe?" Tags="&lt;space-time&gt;&lt;special-relativity&gt;&lt;observable-universe&gt;&lt;time-dilation&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14560" PostTypeId="2" ParentId="14559" CreationDate="2016-04-16T15:14:33.963" Score="2" Body="&lt;p&gt;I welcome everyone else's input on both this answer and possible different solutions! This is what I came up with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, providing a 1G thrust the whole way (switching to decelerating half way there) would be a great way to provide &quot;artificial gravity&quot;.  I found a site that actually has scripts to help with such calculations!  Assuming the site is correct, here is what we come up with:&#xA;&lt;a href=&quot;http://www.cthreepo.com/lab/math1/&quot; rel=&quot;nofollow&quot;&gt;http://www.cthreepo.com/lab/math1/&lt;/a&gt;&#xA;Using the Long Relativistic Journeys calculation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Enter acceleration: 1G&lt;/li&gt;&#xA;&lt;li&gt;Enter light years to travel: 15000000000&lt;/li&gt;&#xA;&lt;li&gt;Time experienced by observer: Infinity (incorrect, and the site points out that due to a limitation in the java script, such large numbers do not calculate properly) &lt;/li&gt;&#xA;&lt;li&gt;Time experienced on Earth: 15010384285.855611&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, I noticed something interesting that might still get us there:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;For a trip of 15 light years, it shows the observer experiences about 5.54 years, and the earth time experienced is about 16.84 years.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For a trip of 150 light years, it shows 9.80 years for the observer, and 152.03 years for earth&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For a trip of 1500 light years, it shows 14.24 years for the observer, and 1502.97 years for earth&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For a trip of 15000 light years, it shows 18.70 years for the observer, and 15012.32 years for earth&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am noticing a pattern here... The time for the observer increases by about 4.5 years every time the distance is multiplied by 10.  The time experienced by earth grows closer to the number of light years as distance is increased too.  Now, let's see where the calculation breaks down, and see if this pattern is still happening before that point.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;It breaks down at 1.5 billion light years, so lets take a few steps back and see if we still see the pattern.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For a trip of 1,500,000 light years, it shows 27.63 years for the observer, and 1501040.36 years for earth.  Note that commas were added to easier see what number we are talking about, when entering numbers into the calculation, don't use commas.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For a trip of 15,000,000 light years, it shows 32.10 years for the observer, and 15010386.22 years for earth.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For a trip of 150,000,000 light years, it shows 35.61 years for the observer, and 150103844.77 years for earth.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Yes, the pattern still largely holds.  It seems be about 3.5 - 4.5 years for each additional factor of 10 for distance (and seems to be getting less time &quot;near the breaking point&quot;).  Let's assume based on this that each factor of 10 increase produces an approximate 4 year increase on the observer's time, and that the time experienced on earth is very close to the number of light years travelled.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;For a trip of 1.5 billion light years, this would mean 39.61 years experienced for the observer, and slightly over 1.5 billion years experienced on earth.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now for the big one and our answer, &lt;strong&gt;a 15 billion light year trip should take about 43.61 years for the observer, while slightly over 15 billion years would be experienced on earth!&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This has some astonishing implications:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;With such a ship, we could theoretically travel to the end of the reachable universe within a human lifetime (though 15 billion years would have passed on earth)!  &lt;/li&gt;&#xA;&lt;li&gt;We could also theoretically reach any point within this area in less time (so we could travel practically anywhere, to a neighboring galaxy, to a galaxy off in the distance, to the center of our galaxy, etc...) within a human lifetime (in less than 44 years)!&lt;/li&gt;&#xA;&lt;li&gt;We could time travel into the distant future by going out somewhere very distant at relativistic speeds, then coming back to earth!  Note: going back in time still hasn't been figured out yet as far as we know.  For large distances, the time travelled would be slightly more than double the distance in light years travelled (accounting for the return trip).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2016-04-16T15:14:33.963" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14561" PostTypeId="1" CreationDate="2016-04-16T15:31:34.640" Score="0" ViewCount="202" Body="&lt;p&gt;I just bought a new telescope. I tried it out, and I even woke up at 3am when the moon set so there was light pollution. I have an 8 inch dobsonian telescope with a focal length of 1200 mm. For some reason, I only see points of light in the sky, except for the moon and Jupiter and the Orion Nebula. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal is to complete a messier marathon, but first I need to be able to see the Messier objects. So my question is, what is the best way to look for Deep Sky Objects with my telescope? And how much does light pollution get in the way of looking for them? For reference, I live in the northern United States.&lt;/p&gt;&#xA;" OwnerUserId="11568" LastEditorUserId="25994" LastEditDate="2019-07-30T13:50:14.267" LastActivityDate="2019-07-30T13:50:17.103" Title="How do I see Deep Sky Objects in my telescope?" Tags="&lt;telescope&gt;&lt;messier-catalog&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2016-04-18T21:10:19.507" ContentLicense="CC BY-SA 4.0" />
  <row Id="14562" PostTypeId="2" ParentId="14559" CreationDate="2016-04-16T16:38:15.633" Score="8" Body="&lt;p&gt;Jonathan's answer is essentially correct, but as Rob Jeffries comments, he doesn't take into account that the Universe is expanding during the journey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The edge of the observable Universe is 47 billion lightyears (Gly) away. Even if you are a lightbeam, you cannot reach that point. The farthest you can go if departing today is roughly 5 Gpc, or 17 Gly, but this journey would of course take infinitly long (or else it wouldn't be &quot;the farthest you can go&quot;). This distance is probably what the linked article is referring to (I didn't read the article; it's very, very long).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in order for the answer to be any fun, you have to freeze the Universe, using magic, which is what Jonathan's calculator is doing. Here I'll just provide the analytical solution: In that case, the &lt;em&gt;proper&lt;/em&gt; time $\tau$ (i.e. the time experienced by the traveler) to reach a distance $x$ when traveling at a constant acceleration $a$ is&#xA;$$&#xA;\tau = \frac{c}{a} \cosh^{-1} \left( \frac{ax}{c^2} +1 \right),&#xA;$$&#xA;where $c$ is the speed of light. If you wish to decelerate after having reached halfway, you just divide $x$ by $2$ and multiply the result by $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you plug in the $x=15\,\mathrm{Gly}$ you request, you get roughly &lt;strong&gt;45 years&lt;/strong&gt;. To get to the edge of the Universe at 47 Gly actually only takes a few years more. The reason for this is simply that traveling at 1G gets you to (almost) the speed of light in only a couple of years, and hence you experience (almost) no time, no matter how far you go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time experienced for the Earthlings for the traveler at constant acceleration is given by&#xA;$$&#xA;t(\tau) = \frac{c}{a} \sinh \left( \frac{a \tau}{c} \right),&#xA;$$&#xA;which works out to 15 Gyr for the 15 Gly, and 47 Gyr for the observable Universe. The reason is simply that the traveler, from the point of view of the Earthlings, extremely fast reaches a speed which is almost the speed of light.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-05-02T07:15:12.193" LastActivityDate="2016-05-02T07:15:12.193" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14563" PostTypeId="2" ParentId="14557" CreationDate="2016-04-16T17:25:28.473" Score="4" Body="&lt;p&gt;Dark Skies are your friend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Messier was searching for comets, it was before electric lighting and the nights were dark in a way that is rarely seen today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at the &lt;a href=&quot;http://darksitefinder.com/maps/world.html&quot; rel=&quot;nofollow&quot;&gt;Dark sky map&lt;/a&gt;, and try to get somewhere in the blue or black region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then allow your eyes to become dark adjusted, and to look with the edge or your vision, or learn to do astrophotography.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And limit your expectations. Messier was looking for comets (little fuzzy blobs) and he made a list of the fixed little fuzzy blobs that might confuse him. Most Messier objects do look like fuzzy patches. You might take an account with &lt;a href=&quot;http://www.telescope.org/&quot; rel=&quot;nofollow&quot;&gt;telescope.org&lt;/a&gt; to see what can be done with a telecope 16 times larger than yours.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-04-16T17:35:39.020" LastActivityDate="2016-04-16T17:35:39.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14564" PostTypeId="1" AcceptedAnswerId="14566" CreationDate="2016-04-15T12:01:52.607" Score="0" ViewCount="1171" Body="&lt;p&gt;Specifically in this case, how large would Jupiter be seen in the Io's sky in comparison to the moon in the Earth's sky? How to calculate this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if somebody knows any online software which lets you input moons and planets to see how their sky looks like to also fill in the answer, it's appreciated.&lt;/p&gt;&#xA;" OwnerUserId="11562" OwnerDisplayName="Pablo" LastEditorUserId="11048" LastEditDate="2016-04-16T21:06:29.060" LastActivityDate="2016-04-19T15:11:34.797" Title="How to calculate a planet's apparent size when the planet is viewed from a moon in orbit around it?" Tags="&lt;planet&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14566" PostTypeId="2" ParentId="14564" CreationDate="2016-04-16T19:49:51.060" Score="3" Body="&lt;p&gt;By a straightforward bit of trigonometry, if the distance from the observer to the (centre of) the planet is x km, then the radius, r km, of the planet subtends $\arcsin(r/x)$, and so the angular size of the planet is twice this: $2\arcsin(r/x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Io x = 420,000 km, and the radius of Jupiter is r = 70,000 km, so the angular size is&#xA;$2\arcsin(1/6)=19^\circ$. For comparison, the moon from the Earth has an angular size of about half a degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.stellarium.org&quot; rel=&quot;nofollow&quot;&gt;Stellarium&lt;/a&gt; can be set up to view from other worlds of the solar system&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="616" LastEditDate="2016-04-19T15:11:34.797" LastActivityDate="2016-04-19T15:11:34.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14567" PostTypeId="1" CreationDate="2016-04-16T20:08:45.390" Score="1" ViewCount="277" Body="&lt;p&gt;What are the reasonable chain of reactions to form copper (63, 29) from ca(40, 20) during supernova explosion? And how it will happen? I do not understand the theory behind it. I thought that the reaction will end at Iron and now it does not make sense to me. &lt;/p&gt;&#xA;" OwnerUserId="11570" LastActivityDate="2016-04-16T21:12:27.733" Title="How to form Copper from Calcium in a supernova explosion?" Tags="&lt;supernova&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14568" PostTypeId="1" AcceptedAnswerId="14682" CreationDate="2016-04-16T20:29:57.643" Score="2" ViewCount="165" Body="&lt;p&gt;I have been calculating galaxy space velocities (where proper motions are known) in order to measure their orbits of the Milky Way using the method proposed in the appendix of &lt;a href=&quot;http://www.aanda.org/articles/aa/pdf/2011/01/aa13415-09.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.aanda.org/articles/aa/pdf/2011/01/aa13415-09.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question I have is: Do we put the Milky Way through the same calculation process to get an initial velocity, or do we have that it's (0,0,0) as it is essentially our origin point (ie the U velocity component would mean that the Milky Way is moving in the direction away from its own galactic centre which doesn't really make sense to me as this could be any direction)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have put it through the process and found an initial space velocity of the Milky Way of (-11.1,232.24,7.25) (my method differs slightly from the link since I have used an updated velocity vector for the motion of the sun with respect to the Local standard of Rest) which essentially just comes from correction solar and galaxy rotation motions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is this calculated vector correct for the Milky Way or should it be (0,0,0)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks &lt;/p&gt;&#xA;" OwnerUserId="11127" LastEditorUserId="9301" LastEditDate="2016-06-29T04:37:14.827" LastActivityDate="2016-08-28T05:58:14.773" Title="Question regarding the Milky Way when calculating galactic space velocities for galaxies" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;velocity&gt;&lt;galaxy-center&gt;" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="14569" PostTypeId="1" CreationDate="2016-04-16T20:59:26.733" Score="2" ViewCount="445" Body="&lt;p&gt;Stars have in the past passed within 1 light year from the Sun, and will do so in the future too. If they have Oort clouds, would there not be interactions between their Oort cloud objects and the planets of the Solar system? For example, the outer edge of the Sun's Oort cloud is approximately 100,000 AU away. So, if another star comes at a distance of 1 light year from the Sun, its Oort cloud should similarly go through the inner solar system. Does this actually happen?&lt;/p&gt;&#xA;" OwnerUserId="11572" LastActivityDate="2016-04-18T08:42:29.097" Title="Do other stars have Oort clouds?" Tags="&lt;oort-cloud&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2016-04-18T13:49:08.510" ContentLicense="CC BY-SA 3.0" />
  <row Id="14570" PostTypeId="2" ParentId="14567" CreationDate="2016-04-16T21:12:27.733" Score="4" Body="&lt;p&gt;Copper is not thought to be primarily &lt;em&gt;made&lt;/em&gt; in a supernova. It is thought to be mainly produced by the &lt;a href=&quot;https://en.wikipedia.org/wiki/S-process&quot; rel=&quot;nofollow noreferrer&quot;&gt;s-process&lt;/a&gt; of slow neutron capture onto iron-peak nuclei that already exist inside a star. These reactions are &lt;em&gt;endothermic&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The source of the neutrons is still somewhat debated, it could either be from the decay of $^{13}$C in relatively low-mass asymptotic giant branch stars or is more likely from the decay of $^{22}$Ne in more massive evolved stars (e.g. &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0004-637X/710/2/1557/meta;jsessionid=F521A33E8AC70B164FD9E8D19FDF5ED8.c3&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pignatari et al. 2010&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below you can see a typical route for producing copper from $^{56}$Fe. The axes of the plot are neutron number on the x-axis and proton (atomic) number on the y-axis. Three neutron captures are followed by a beta decay, a neutron capture, a beta decay, then 3 more neutron captures followed by another beta decay to form $^{63}$Cu.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net process is&#xA;$$ ^{56}{\rm Fe} + 7n \rightarrow \  ^{63}{\rm Cu} + 3e + 3\bar{\nu}_e$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The copper is then &lt;em&gt;distributed&lt;/em&gt; into the interstellar medium by a later supernova explosion in the same star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are no &quot;seed&quot; iron-peak nuclei (e.g. in the first/second generation of stars), then copper can be inefficiently produced by explosive &lt;a href=&quot;https://en.wikipedia.org/wiki/R-process&quot; rel=&quot;nofollow noreferrer&quot;&gt;r-process&lt;/a&gt; neutron capture during supernovae explosions. However, this would not contribute very much to the copper we see on the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2YjW2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2YjW2.jpg&quot; alt=&quot;s-process network&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-04-16T21:12:27.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14571" PostTypeId="1" AcceptedAnswerId="14573" CreationDate="2016-04-16T22:55:00.100" Score="0" ViewCount="1746" Body="&lt;p&gt;If you were to divide the present-day&#xA;universe into cubes with sides 10 million light-years long, each&#xA;cube would contain, on average, about one galaxy similar in size&#xA;to the Milky Way. Now suppose you travel back in time, to an era&#xA;when the average distance between galaxies is one quarter of its&#xA;current value, corresponding to a cosmological redshift of z = 3.&#xA;How many galaxies similar in size to the Milky Way would you&#xA;expect to find, on average, in cubes of that same size? In order to&#xA;simplify the problem, assume that the total number of galaxies of&#xA;each type has not changed between then and now. Based on your&#xA;answer, would you expect collisions to be much more frequent at&#xA;that time or only moderately more frequent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am very confused with this question. From my intuition, I take that volume would equal to distance^3. They have given us that the past distance would be 1/4 of current. Therefore, do I cube 1/4distance? Also how does the redshift of 3 play a part in the solution. Thanks for any help! I do not need an answer, just how to start.&lt;/p&gt;&#xA;" OwnerUserId="11509" LastActivityDate="2016-04-16T23:23:34.560" Title="Cosmological redshift and comparing past galaxy sizes" Tags="&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14572" PostTypeId="1" CreationDate="2016-04-16T23:21:48.400" Score="1" ViewCount="1245" Body="&lt;p&gt;I want to ask a question about the Three-torus model of the universe, as described in wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Three-torus_model_of_the_universe&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Three-torus_model_of_the_universe&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;is a proposed model describing the shape of the universe as a three-dimensional torus.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure i understand the maths of three-dimensional tori, but I want to ask:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this model posit that the shape of the universe is that of a ring doughnut? If so, is the universe contained inside the ring doughnut, or is it supposed to exist only on the surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or does the model say that the universe has the shape of a higher dimensional version of a ring doughnut? If so, does it exist inside the higher dimensional ring doughnut, or on its surface?&lt;/p&gt;&#xA;" OwnerUserId="6920" LastEditorUserId="6920" LastEditDate="2016-04-16T23:56:09.717" LastActivityDate="2016-09-27T08:02:03.267" Title="Three-torus model of the universe" Tags="&lt;cosmology&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14573" PostTypeId="2" ParentId="14571" CreationDate="2016-04-16T23:23:34.560" Score="0" Body="&lt;p&gt;Comoving separations go with the scale factor. Comoving volumes scale as $(1+z)^{-3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At $z=3$, that volume was 64 times smaller.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-04-16T23:23:34.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14574" PostTypeId="2" ParentId="14572" CreationDate="2016-04-17T01:03:52.740" Score="1" Body="&lt;p&gt;A three-torus is the boundary of the solid three torus, just like the two-torus is the surface of a solid donut. You can imagine it as a cubical room where each wall/ceiling/floor is a portal to the opposite-facing wall (i.e. the wall to your right is a portal that sends you to your left), but preserves orientation (when you walk out of the portal, your heart is still on your left). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also think of it as a world where your position is described by three coordinates (like x, y, and z in Euclidean space), but each coordinate corresponds to an angle on the unit circle. If you go far enough in one coordinate, you loop back to where you started (360 degrees is the same as 0 on a circle).&lt;/p&gt;&#xA;" OwnerUserId="608" LastActivityDate="2016-04-17T01:03:52.740" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14575" PostTypeId="2" ParentId="13462" CreationDate="2016-04-17T11:59:12.420" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://www.sciencemag.org/content/351/6272/482/suppl/DC1&quot; rel=&quot;nofollow&quot;&gt;This link&lt;/a&gt; has Dr. Ossendrijver interpretation of the five clay tablets but if you are interested in a second peer review I would suggest emailing Dr. Irving Finkel at the British Museum to give his opinion on the same tablets once he is also one of the handful of global economy experts on deciphering the cuneiform language but was not listed as a reference in the research paper. This second opinion could either add credibility to Dr. Ossendrijver paper, or possibly give a counter argument or theory on what the tablets reveal. There is a lot of subjective interpretation on these five tablets since they are fragments and none show the actual trapezoid figures.  &lt;/p&gt;&#xA;" OwnerUserId="11575" LastEditorUserId="10172" LastEditDate="2016-04-17T12:22:57.287" LastActivityDate="2016-04-17T12:22:57.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14576" PostTypeId="2" ParentId="14561" CreationDate="2016-04-17T13:28:03.090" Score="0" Body="&lt;p&gt;To see Deep Sky Objects you'll definitely have to head out of whatever town you're in by at least 5km, but the further the better. Next, you'll have to have a good knowledge of where to look for them or a guide. You should be able to see a fair few once you've done that, especially if you've already seen the Orion Nebula.&lt;/p&gt;&#xA;" OwnerUserId="11525" LastEditorUserId="25994" LastEditDate="2019-07-30T13:50:17.103" LastActivityDate="2019-07-30T13:50:17.103" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14577" PostTypeId="2" ParentId="12539" CreationDate="2016-04-17T13:42:33.647" Score="1" Body="&lt;p&gt;I've seen this question for the past few days and have shied away from it because it seemed too difficult, but here goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;550AU is the focal point of the sun, so theoretically anything on the opposite side would be focused into a perfectly coherent image right? Unfortunately due the density of the sun (Mass to radius ratio specifically) only radio waves that just miss the sun would be focused to that point, all other shorter wavelengths don't diffract enough and simply 'collide' with the sun or form a lense ring at a greater distance. So radio-observation only. Explained at this article &lt;a href=&quot;https://physics.stackexchange.com/a/25501&quot;&gt;https://physics.stackexchange.com/a/25501&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A star such as a neutron star however would make an excellent candidate for stellar lensing because it has a high density and low radius, meaning much shorter wavelength radiation is able to go around the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This wouldn't be particularly effective at discovering new exoplanets, only observing them in more detail.&lt;/p&gt;&#xA;" OwnerUserId="11525" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-04-17T15:19:20.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14578" PostTypeId="1" CreationDate="2016-04-17T21:58:59.473" Score="4" ViewCount="3279" Body="&lt;p&gt;Simple question, but can't seem to find the answer anywhere. It can't be nuclear fusion because nuclear fusion occurs as a result of the heat. And it can't be because of gravity because it's believed that the core of black holes are nearly absolute zero. So what makes stars hot? &lt;/p&gt;&#xA;" OwnerUserId="11579" LastActivityDate="2016-04-18T09:02:19.857" Title="What Makes Stars Hot?" Tags="&lt;star&gt;&lt;heat&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14579" PostTypeId="2" ParentId="14578" CreationDate="2016-04-17T22:25:27.797" Score="2" Body="&lt;p&gt;Before the nuclear fusion in the core starts, the heat of the star comes from the contraction of the original nebula. When the matter comes closer together, the potential energy of it decreases, just like when you drop a rock. Energy is however constant, so it has to go somewhere. That &quot;somewhere&quot; is the heat in the newborn star.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastActivityDate="2016-04-17T22:25:27.797" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14581" PostTypeId="2" ParentId="14569" CreationDate="2016-04-18T08:42:29.097" Score="0" Body="&lt;p&gt;The other question has some very good detailed answers, worth reading, but briefly and more specific to your question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Would there not be interactions between their Oort cloud objects and&#xA;  the planets of the Solar system? For example, the outer edge of the&#xA;  Sun's Oort cloud is approximately 100,000 AU away. So, if another star&#xA;  comes at a distance of 1 light year from the Sun, its Oort cloud&#xA;  should similarly go through the inner solar system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While very little is actually known, it's not crazy to think that smaller stars might have smaller oort clouds and larger stars, larger ones, though there's a great deal of uncertainty on what an average oort cloud is, how far an average one extends, and how densely packed they are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First part first, how often does a star pass within one light year of our sun?&lt;/strong&gt;    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not very often.   Ballpark, maybe every million to couple million years or so.  The last time was 70,000 years ago, a small star, 15% the mass of the sun.  See &lt;a href=&quot;https://en.wikipedia.org/wiki/Scholz&amp;#39;s_star&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the fly-by section of that same wiki-article, it says a star passes that close (0.82 light years or less) about once every 9 million years.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scholz's star is quite small, only about 15% the mass of our sun, and it's unclear if it had an Oort cloud that extended .82 light years at all, so while it passed through our Sun's Oort cloud, it's unclear if the inner solar-system passed through it's Oort cloud, assuming it has one.  It's also possible that the sun stole much of Scholz's star's oort cloud during the fly-by, but that's just speculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did Scholz's star send Oort cloud objects heading towards Earth?  That's a good question and the answer is, probably, but we don't know, because if it did, it would take longer than 70,000 years for any re-directed Oort cloud comets to reach the inner solar-system.   By the same article, an estimated 2 million years for re-directed Oort cloud objects to reach the inner solar system, so, in about 1.93 million years (give or take), we'll start to see what Scholz's sent our way.    :-)   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1.36 million years, &lt;a href=&quot;https://en.wikipedia.org/wiki/Gliese_710#Computing_and_details_of_the_closest_approach&quot; rel=&quot;nofollow&quot;&gt;Gliese 710&lt;/a&gt; could pass within 1 light year of Earth.  The estimate is 1.1 light years plus/minus .577 light years.  Gliese 710 is larger than Scholz's, about 60% the mass of the sun but we know nothing about how large or how dense it's Oort cloud might be.   But, very generally, it seems likely that the Sun has passed through another stars Oort cloud at some point, perhaps every 1-10 million years, perhaps every hundred million years or so, but it seems reasonable to me that has happened and will continue to happen, on rare occasion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2nd question, what would happen if the inner planets passed through another star's Oort cloud.&lt;/strong&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we use our &lt;a href=&quot;https://en.wikipedia.org/wiki/Oort_cloud#Structure_and_composition&quot; rel=&quot;nofollow&quot;&gt;Oort cloud&lt;/a&gt; as a model, and, warning, these estimates are enormously ballpark, but, a trillion Oort cloud objects 1 KM or larger and the Oort cloud extends between 100,000 and 200,000 AU (lets say 150,000 as a middle estimate), that works out to one 1 KM or larger object per cubic region of space about 15 AU across.   That's 3/4ths the distance between the Sun and Neptune, so if you figure just one, 1 KM or larger comet or object in a 3D space over such a vast distance, most would pass by the inner planets unaffected.   The occasional one would get it's orbit measurably altered with a fly-by of one of the gas giant planets and rarely, there would be an impact.  Now presumably there's more smaller stuff in Oort clouds too, so there should be smaller impacts more frequently, though there's almost nothing that's actually known about how dense Oort clouds are, so anything beyond &quot;infrequent&quot;, is impossible to say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Passing through another Stars Oort cloud should take tens of thousands, perhaps 100,000 years years or so, so there probably would be the occasional impact on a planet here and there, but on a human time scale, significant impacts should be very rare.    &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-04-18T08:42:29.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14582" PostTypeId="2" ParentId="14578" CreationDate="2016-04-18T09:02:19.857" Score="6" Body="&lt;p&gt;Stars do not &lt;em&gt;get&lt;/em&gt; hot because of nuclear fusion, they &lt;em&gt;become&lt;/em&gt; hot enough to sustain nuclear fusion and this process &lt;em&gt;maintains&lt;/em&gt; their temperatures. Nuclear fusion actually &lt;em&gt;stops&lt;/em&gt; a star getting hotter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Protostars (before nuclear fusion) &lt;em&gt;get&lt;/em&gt; hot because of a well known statistical relationship between the gravitational potential energy of a gas and the internal kinetic energy of the particles that make up the gas. [In an ideal gas, the kinetic energy of the particles is directly proportional to the temperature of the gas.] This is known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Virial_theorem#In_astrophysics&quot;&gt;virial theorem&lt;/a&gt;, which says that twice the summed kinetic energy of particles ($K$) plus the gravitational potential energy ($\Omega$, which is a negative quantity for a bound object) equals zero.&#xA;$$ 2K + \Omega = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can write down the &lt;em&gt;total&lt;/em&gt; energy of the system as&#xA;$$ E_{tot} = K + \Omega$$&#xA;and hence from the virial theorem that&#xA;$$E_{tot} = \frac{\Omega}{2},$$&#xA;which is also negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we now &lt;em&gt;remove&lt;/em&gt; energy from the system, for instance by allowing the gas to radiate away energy, such that $\Delta E_{tot}$ is &lt;em&gt;negative&lt;/em&gt;, then we see that &#xA;$$\Delta E_{tot} = \frac{1}{2} \Delta \Omega$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\Omega$ becomes &lt;em&gt;more negative&lt;/em&gt; - which is another way of saying that the protostar attains a &lt;em&gt;more collapsed&lt;/em&gt; configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oddly, at the same time, we can use the virial theorem to see that&#xA;$$ \Delta K = -\frac{1}{2} \Delta \Omega = -\Delta E_{tot}$$&#xA;is &lt;em&gt;positive&lt;/em&gt;. i.e. the kinetic energies of particles in the gas (&lt;strong&gt;and hence their temperatures&lt;/strong&gt;) actually become hotter. In other words, the gas has a negative heat capacity. But a hotter temperature usually means more radiation is produced and if the energy losses continue, then so does the collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process is ultimately arrested in a star by the onset of nuclear fusion. This replaces the radiative losses with nuclear energy and the star attains a quasi-equilibrium that lasts as long as it has nuclear fuel to burn.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-04-18T09:02:19.857" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14583" PostTypeId="2" ParentId="6562" CreationDate="2016-04-18T11:13:45.143" Score="0" Body="&lt;p&gt;I'm not sure what you mean by radius of the universe. Do you mean radius of the observable universe or total radius?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, the expansion of the universe is determined by the outwards pressure of dark energy and the residual expansion from the Big Bang itself, which is resisted by the gravitational attraction of all mass and energy in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There would probably be an equation that describes a possible relation between expansion of the universe and the the total entropy.&lt;/p&gt;&#xA;" OwnerUserId="11525" LastActivityDate="2016-04-18T11:13:45.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14584" PostTypeId="2" ParentId="11447" CreationDate="2016-04-18T11:20:41.783" Score="1" Body="&lt;p&gt;The factors that affect the shape of supernova remnants include:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Interstellar medium (the gas and particles between stars)&lt;/li&gt;&#xA;&lt;li&gt;Planets that surrounded the star&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Magnetic field&lt;/strong&gt; (in the same way magnetic fields cause auroras)&lt;/li&gt;&#xA;&lt;li&gt;Mass distribution of the star before supernova&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are probably more, but that's all I can think of at the moment&lt;/p&gt;&#xA;" OwnerUserId="11525" LastActivityDate="2016-04-18T11:20:41.783" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14587" PostTypeId="1" AcceptedAnswerId="14605" CreationDate="2016-04-18T14:38:02.943" Score="1" ViewCount="146" Body="&lt;p&gt;I'm quite certain that there isn't any actual images of a transitory event of Jupiter or Saturn across the solar disk from say e.g.,s the Voyager probe, Pioneer 10/11 or even recent New Horizons mission? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another related question I have, what would the approximate transit of Jupiter look like as seen from Saturn. The distance between Saturn and Jupiter is approximately 4-5AU, and the diameter of Jupiter is one tenth that of the Sun's. Are we looking at perhaps a transit which covers 10-15% of the solar disk?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;N.B.&lt;/strong&gt; Please note I understand that to observe a transit of Jupiter we have to be superior to it in the Solar System, hence I am asking for images taken by deep Solar System probes only. &lt;/p&gt;&#xA;" OwnerUserId="2091" LastEditorUserId="2091" LastEditDate="2016-04-19T20:15:47.693" LastActivityDate="2016-04-19T20:15:47.693" Title="Were there any images of a transitary event of Jupiter or Saturn as imaged by a deep probe mission" Tags="&lt;solar-system&gt;&lt;jupiter&gt;&lt;saturn&gt;&lt;planetary-transits&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14588" PostTypeId="2" ParentId="6562" CreationDate="2016-04-18T19:12:57.187" Score="1" Body="&lt;p&gt;Holographic entropy bounds are upper bounds for the &lt;em&gt;maximum&lt;/em&gt; amount of entropy a given region can have, rather than its current entropy--for example, the spherical region of space just large enough to enclose the Earth and its atmosphere would currently contain a much lower entropy than a spherical black hole of the same radius, whose entropy is given by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bekenstein_bound&quot; rel=&quot;nofollow&quot;&gt;Bekenstein bound&lt;/a&gt;. The &lt;a href=&quot;http://www.scholarpedia.org/article/Bekenstein_bound&quot; rel=&quot;nofollow&quot;&gt;scholarpedia article on the Bekenstein bound&lt;/a&gt; mentions the issue of cosmological entropy bounds:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Problems with bound (1) or the holographic bound are known to occur in extreme situations. Both fail when the gravitational potential is large (strong self-gravity), e.g. system already collapsed inside a black hole. In an infinite universe the holographic bound fails when applied to a sufficiently large region. In a closed (finite) universe the specification of R or bounding area becomes ambiguous (Bousso 1999, 2002). The covariant entropy bound corrects these and other shortcomings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So apparently the &quot;covariant entropy bound&quot; (also known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Bousso%27s_holographic_bound&quot; rel=&quot;nofollow&quot;&gt;Bousso's holographic bound&lt;/a&gt;) is the thing to look into when considering the maximum entropy of the universe (and I assume you just mean the observable universe here rather than whatever may lie beyond it). &lt;a href=&quot;http://physics.princeton.edu/strings2014/slides/Bousso.pdf&quot; rel=&quot;nofollow&quot;&gt;This set&lt;/a&gt; of lecture slides by &lt;a href=&quot;http://en.wikipedia.org/wiki/Raphael_Bousso&quot; rel=&quot;nofollow&quot;&gt;Raphael Bousso&lt;/a&gt;, the physicist who originally proposed the covariant entropy bound, says that this bound is &quot;conjectured to hold in arbitrary spacetimes, including cosmology.&quot; But it also says &quot;If correct, origin must lie in quantum gravity&quot;, probably meaning one can't derive this fact in a non-conjectural way from basic laws of physics without having a complete theory of &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_gravity&quot; rel=&quot;nofollow&quot;&gt;quantum gravity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an introduction to this stuff, I recommend &lt;a href=&quot;http://www.phys.huji.ac.il/~bekenste/Holographic_Univ.pdf&quot; rel=&quot;nofollow&quot;&gt;this Scientific American article&lt;/a&gt; that discusses various holographic bonds, including the Bekenstein bound along with Bousso's conjectured more general bound (the article was written by Bekenstein himself). Though it is from 2003 so I'm not sure what kind of theoretical progress has been made since then. Anyway, here's what it says about Buosso's entropy bound:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 1999 Raphael Bousso, then at Stanford, proposed a modified holographic bound, which has since been found to work even in situations where the bounds we discussed earlier cannot be applied. Bousso’s formulation starts with any suitable 2-D surface; it may be closed like a sphere or open like a sheet of paper. One then imagines a brief burst of light issuing simultaneously and perpendicularly from all over one side of the surface. The only demand is that the imaginary light rays are converging to start with. Light emitted from the inner surface of a spherical shell, for instance, satisfies that requirement. One then considers the entropy of the matter and radiation that these imaginary rays traverse, up to the points where they start crossing. Bousso conjectured that this entropy cannot exceed the entropy represented by the initial surface—one quarter of its area, measured in Planck areas. This is a different way of tallying up the entropy than that used in the original holographic bound.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Bousso’s bound refers not to the entropy of a region at one time but rather to the sum of entropies of locales at a variety of times: those that are “illuminated” by the light burst from the surface. Bousso’s bound subsumes other entropy bounds while avoiding their limitations. Both the universal entropy bound and the ’t Hooft-Susskind form of the holographic bound can be deduced from Bousso’s for any isolated system that is not evolving rapidly and whose gravitational field is not strong. When these conditions are overstepped—as for a collapsing sphere of matter already inside a black hole—these bounds eventually fail, whereas Bousso’s bound continues to hold. Bousso has also shown that his strategy can be used to locate the 2-D surfaces on which holograms of the world can be set up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But again, this bound should be the upper limit on the entropy of a given region, not necessarily the actual entropy in that region, so this type of holographic bound shouldn't imply that when we increase the entropy of the universe, the universe needs to grow. &lt;a href=&quot;http://arxiv.org/abs/1003.3012&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; specifically discusses how most matter-containing regions in cosmology apart from black holes would fail to &quot;saturate&quot; the covariant entropy bound, meaning that the actual entropy in the region is smaller than the maximum possible according to the bound.&lt;/p&gt;&#xA;" OwnerUserId="2437" LastEditorUserId="2437" LastEditDate="2016-04-18T19:34:31.860" LastActivityDate="2016-04-18T19:34:31.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14589" PostTypeId="1" AcceptedAnswerId="14590" CreationDate="2016-04-19T01:50:23.910" Score="1" ViewCount="176" Body="&lt;p&gt;Recently I have been trying to get a good look at Jupiter through my 6&quot; Orion reflector, but the image seems to always come out as an oblong white streak. I've tried changing the lenses I'm using, but that only changes the size of the oblong streak. I looked at the moon with the same lenses to see if it was a general problem with the reflector, but the image of the moon came out clear as usual. What sort of problems could be causing this? &lt;/p&gt;&#xA;" OwnerUserId="11585" LastActivityDate="2016-04-19T05:01:23.277" Title="Jupiter appearing odd through a 6&quot; reflector" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;jupiter&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14590" PostTypeId="2" ParentId="14589" CreationDate="2016-04-19T05:01:23.277" Score="3" Body="&lt;p&gt;Most likely would be the the main mirror needs collimating. On big things like the moon you won't notice bad collimation as much as something small like a planet or star.&lt;/p&gt;&#xA;" OwnerUserId="11525" LastActivityDate="2016-04-19T05:01:23.277" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14591" PostTypeId="1" AcceptedAnswerId="14606" CreationDate="2016-04-19T08:23:35.983" Score="2" ViewCount="705" Body="&lt;p&gt;I think Titan may be just captured by Saturn, not formed with Saturn together, because:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It has so much nitrogen and methane, which is similar to Pluto and Triton&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Except Titan, other moons of Saturn are very small, which other moons may be ejected by Titan during capturing process, similar to Triton and Neptune&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;According to Nice model, Neptune swapped position with Uranus, while Saturn pulled Jupiter away from the Sun, so I think the orbit of Neptune and Saturn is quite close at past, so if Neptune can capture kuiper object, Saturn also can&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is it possible that Titan was a Kuiper object?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastActivityDate="2016-04-19T21:48:49.167" Title="Is it possible that Titan is a kuiper object captured by Saturn?" Tags="&lt;saturn&gt;&lt;titan&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14592" PostTypeId="1" AcceptedAnswerId="14595" CreationDate="2016-04-19T08:33:53.523" Score="5" ViewCount="772" Body="&lt;p&gt;Ever since I observed the depictions of the Solar System, I was obsessed with the question of why the gas giants (outer planets) have very large orbits, compared to the planets that are closer to the Sun. Is it because of their mass or something I have ignored?&lt;/p&gt;&#xA;" OwnerUserId="11587" LastEditorUserId="2153" LastEditDate="2016-04-20T00:53:00.460" LastActivityDate="2016-07-17T14:30:37.490" Title="Why do the gas giants in the Solar System have comparatively large orbits compared to the inner planets?" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;planet&gt;&lt;gas-giants&gt;&lt;orbital-migration&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14593" PostTypeId="2" ParentId="14592" CreationDate="2016-04-19T09:45:57.063" Score="2" Body="&lt;p&gt;When the solar system formed, there was an accretion disc, spinning around the newborn Sun.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JrxXT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JrxXT.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The Sun was emitting radiations, which pushed the lighter materials of the accretion disk (gases) further away, and kept the heavier materials (rocks) much closer. This is why gas giants are almost always further away than rocky planets. Almost.&lt;/p&gt;&#xA;&lt;p&gt;In the case of &amp;quot;hot Jupiters&amp;quot;, astronomers think that they simply spiraled their way to their star, &amp;quot;eating&amp;quot; smaller planets on their way. It doesn't happen in our solar system because Saturn holds Jupiter back.&lt;/p&gt;&#xA;&lt;p&gt;Exact source nowhere to be found on the internet, but everything I wrote here comes from a very interesting movie I watched in &amp;quot;La Cité de L'Espace&amp;quot; (Space City) in Toulouse, France. It is a famous exposition / museum about astronomy and space exploration.&lt;/p&gt;&#xA;" OwnerUserId="9679" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-04-19T09:45:57.063" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14594" PostTypeId="1" CreationDate="2016-04-19T11:05:14.493" Score="0" ViewCount="170" Body="&lt;p&gt;I understand that binary star systems may cause type Ia supernovae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a question regarding supernova explosion in a binary system. What is the last nuclear reaction before massive explosion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am thinking of two 2 close stars of 3M and 6M. &lt;/p&gt;&#xA;" OwnerUserId="11577" LastEditorUserId="823" LastEditDate="2016-04-22T09:22:29.463" LastActivityDate="2016-04-22T10:12:40.590" Title="What is the last nuclear reaction in a binary system before supernova?" Tags="&lt;supernova&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14595" PostTypeId="2" ParentId="14592" CreationDate="2016-04-19T12:34:57.557" Score="2" Body="&lt;p&gt;As one gets farther from the sun, the gravity from the planets themselves becomes relatively stronger. So if a body were in an orbit between, for example, Jupiter and Saturn, those two planets would soon make the intermediate body's orbit change. This is conceptually related to the latest definition of &quot;a planet&quot;; A planet must clear its general vicinity of other stable orbiting bodies. &lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2016-04-19T12:34:57.557" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14596" PostTypeId="1" CreationDate="2016-04-19T13:01:50.987" Score="0" ViewCount="57" Body="&lt;p&gt;Regarding matter that has the highest redshifts, do we see such matter in every general direction we look (relative to Earth: up down left right front back)?&lt;/p&gt;&#xA;" OwnerUserId="11589" LastActivityDate="2016-04-23T11:00:09.423" Title="Locations of farthest matter" Tags="&lt;universe&gt;&lt;redshift&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14597" PostTypeId="2" ParentId="14558" CreationDate="2016-04-19T15:37:39.990" Score="-1" Body="&lt;p&gt;I would say that this sort of system would be near impossible. Both the mechanics of triple star systems being very complicated and the fact that have the outermost star would have to be captured, not from the original local stellar material, no easy feat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the system would be virtually non existent, especially in the long term due to large gravitational wave energy decay (thanks LIGO) and general instability.&lt;/p&gt;&#xA;" OwnerUserId="11525" LastActivityDate="2016-04-19T15:37:39.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14598" PostTypeId="1" AcceptedAnswerId="14601" CreationDate="2016-04-19T15:56:20.957" Score="2" ViewCount="1367" Body="&lt;p&gt;So I just learned that CMB redshift is 1100 regardless where we look (up down left right). According to Hubble's Law that makes it around 46 billion light years away, making it the farthest matter away from us. If CMB redshift is the same regardless which direction we look, then how could that be unless we're in the center of the universe?&lt;/p&gt;&#xA;" OwnerUserId="11589" LastActivityDate="2016-04-21T14:30:54.233" Title="Are we center of universe?" Tags="&lt;universe&gt;&lt;redshift&gt;&lt;cmb&gt;" AnswerCount="3" CommentCount="2" ClosedDate="2016-04-19T18:49:06.443" ContentLicense="CC BY-SA 3.0" />
  <row Id="14599" PostTypeId="2" ParentId="14598" CreationDate="2016-04-19T16:35:43.470" Score="1" Body="&lt;p&gt;We're at the centre of the &lt;strong&gt;observable&lt;/strong&gt; universe - but this is because everything (when observed over very long distances) is expanding away from everything else, so we are at the centre of what we observe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spherical shape of the observable universe is really an optical illusion, in conjunction with the large-scale expansion of the universe.  We simply can't see anything that is apparently receding at (close to) light speed, so everything beyond a certain range is but off from our view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual universe - well that's probably much bigger than the observable universe. What shape it is, and what conditions are like in other regions, we can only guess at right now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I have to lie down now, my head is beginning to spin.)&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2016-04-19T16:35:43.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14600" PostTypeId="2" ParentId="14598" CreationDate="2016-04-19T16:37:58.940" Score="1" Body="&lt;p&gt;In a way you could argue that we are the centre of the universe. From our stand point that is. As a matter of fact every other standpoint has the right to argue that they are the centre of the universe because the redshift is always the same regardless where in the universe you are. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So imagine observing the redshift of a different galaxy. Imagine there is a galaxy behind the first one at exactly twice the distance from earth. Now from earth this galaxy would appear to be more redshifted (I know the galaxy in the foreground obstructs the view but just in theory). &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;          (      far galaxy   (redder)&#xA;           )&#xA;&#xA;&#xA;&#xA;           (     close galaxy (red)&#xA;            )&#xA;&#xA;&#xA;&#xA;&#xA;            O      Earth (observer)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you lived in the closer galaxy, you would observe earth redshifted by the same amount as an observer on earth would see you redshifted. This seems logical. The interesting thing however is that the far away galaxy will &lt;strong&gt;not&lt;/strong&gt; be redshifted the same amount as viewed from earth. &#xA;In fact, the redshift of earth and the second galaxy will be exactly identical when viewed from the galaxy between them.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;          (      far galaxy   (red)&#xA;           )&#xA;&#xA;&#xA;&#xA;           (     close galaxy (observer)&#xA;            )&#xA;&#xA;&#xA;&#xA;&#xA;            O      Earth (red)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is because all points in the universe are moving away from each other and not just all points moving away from one point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may have heard this analogy before but if you draw spots on a balloon and blow it up, they behave exactly like galaxies moving away from each other. No matter which point you observe all points seem to be moving away from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while this may not be entirely accurate in cosmological terms you can imagine the galaxies to be the spots on the balloon and the balloon to be the fabric of space.&lt;/p&gt;&#xA;" OwnerUserId="10321" LastActivityDate="2016-04-19T16:37:58.940" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14601" PostTypeId="2" ParentId="14598" CreationDate="2016-04-19T16:42:51.857" Score="3" Body="&lt;p&gt;At the time when the universe was ~380,000 years old it was completely filled with a plasma and light. Light could only travel very short distances (I forget the specifics, but think a couple meters at the most) before being absorbed. So the entire universe was a fog, if you were there and there was an object 10 meters in front of you, you couldn't see it because all the light it emitted or reflected would be reabsorbed before it reaches your eyes. There are two things to keep in mind: &lt;strong&gt;1) This fog disappeared across the entire universe simultaneously (plus or minus some time, but effectively simultaneously); at the same point in time after the Big Bang, 380,000 years after the Big Bang, the fog dissipated due to recombination. And 2) Light takes time to travel.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's say you are sitting in the fog-filled universe a few seconds before recombination happens. Light travels about 1 foot per nanosecond. So if the fog disappeared and there was a street lamp 10 feet in front of you, you wouldn't receive the first light from that lamp for 10 nano seconds. All of the light the lamp was emitting before recombination was just reabsorbed, but after recombination the first photons to leave the lamp no longer have anything to run into, so they continue on until hitting your eyes. A street lamp 20 feet away wouldn't be visible for 20 nanoseconds after recombination. The light of the plasma itself is effectively a street lamp at every location in space, so due to the 2 facts above, you would see successively farther street lamps in a shell centered on you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After recombination happened and the entire universe became transparent. The fog is gone, but the farther you look the farther back in time you look, so you can only see back as far as when the fog existed, and the farthest you can see is the fog itself. After 13.8 billion years you can see all the way out to a shell of radius 13.8 billion light years around you (ignore expansion of the universe for now), and what you see is the light of the plasma (the fog) that was filling the universe 13.8 billion years ago, the CMB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This scenario is the same if the universe is infinite in extent. We are not the center of the universe, we are simply the center of the &lt;em&gt;observable&lt;/em&gt; universe, as is anyone else. Aliens looking towards Earth but are at the current location of edge of the CMB in some direction will not see Earth, they will see the last light of the CMB itself at our location when it was emitted 13.8 billion years ago; we are the edge of &lt;em&gt;their&lt;/em&gt; observable universe.&lt;/p&gt;&#xA;" OwnerUserId="11591" LastEditorUserId="11591" LastEditDate="2016-04-21T14:30:54.233" LastActivityDate="2016-04-21T14:30:54.233" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14602" PostTypeId="2" ParentId="14547" CreationDate="2016-04-19T17:14:35.797" Score="4" Body="&lt;p&gt;&lt;a href=&quot;http://www.stellarium.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellarium&lt;/a&gt; identifies the mystery &quot;moon&quot; as a star, &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=HIP+54057&quot; rel=&quot;nofollow noreferrer&quot;&gt;HIP 54057 a.k.a. HD 95848&lt;/a&gt;. Your telescope probably showed it on the right instead of the left. The location on Earth is insignificant since you provided the time in UTC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter passed directly in front of the same star on 2016-04-12, but the &lt;a href=&quot;http://www.hellas-astro.gr/sites/default/files/jupocc2016_AChristou.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;occultation&lt;/a&gt; was only visible from the Eastern Hemisphere.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/G6IYd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G6IYd.png&quot; alt=&quot;Screenshot of Jupiter and HD 95848&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-04-19T17:14:35.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14603" PostTypeId="2" ParentId="14558" CreationDate="2016-04-19T18:53:23.227" Score="2" Body="&lt;p&gt;The usual recipe used in the population synthesis literature is that triple stars comprise ~10% of all stellar systems and the mutual inclination is uniform in the cosine.  Most of these systems have a large enough semi-major axis ratio that the inner binary is dynamically decoupled from the outer binary, so retrograde orbits will be just as likely as prograde orbits.  So, to an order of magnitude, ~5% of stellar systems should be triples with retrograde orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things get more complicated if you are interested in more details.  More massive stars have higher multiplicities, so the fraction is probably somewhat higher for them.  For more compact systems, dynamical effects like the Kozai-Lidov mechanism will become important and can lead to prograde orbits becoming retrograde and vice versa.  Moreover, in compact systems, retrograde orbits are slightly more stable than prograde orbits, so they are likely to be a little more overrepresented, though I'm not aware of any observational evidence that there is any great difference between the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want more details on the observations of multiple stellar systems you should read &lt;a href=&quot;http://adsabs.harvard.edu/abs/2010ApJS..190....1R&quot; rel=&quot;nofollow&quot;&gt;Raghavan et al. (2010)&lt;/a&gt;.  However, I don't think that they measured mutual inclinations.  In general those measurements are very difficult to do.  Andrei Tokovinin has done a lot of great observational work on triple stars as well, so you might want to look at Tokovinin (2014) and Tokovinin (2008) among other papers.  There's also a good review by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013ARA%26A..51..269D&quot; rel=&quot;nofollow&quot;&gt;Duchene &amp;amp; Kraus (2013)&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="10371" LastActivityDate="2016-04-19T18:53:23.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14604" PostTypeId="2" ParentId="107" CreationDate="2016-04-19T19:17:10.587" Score="1" Body="&lt;p&gt;Just for fun I did a limited analysis of Jupiter size planets found using radial velocity or astrometry (mostly the former) techniques with the following criteria: 1) Data from the on-line sources &lt;a href=&quot;http://exoplanet.eu/catalog/bd%2B15_2375_b/&quot; rel=&quot;nofollow&quot;&gt;http://exoplanet.eu/catalog/bd%2B15_2375_b/&lt;/a&gt; and &lt;a href=&quot;http://www.exoplanets.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.exoplanets.org/&lt;/a&gt; giving planets discovered through Apr. 2016. 2) Msini: 0.5-10 M(J) 3) Distance to star less than 58 parsecs 4) Planets out to 5200 day periods (or 6AU) 5) Only data from K,G or F dwarfs on main sequence (V or IV/V). This was an attempt to minimize observational bias.                                                I still found a pronounced peak in Jupiters with periods less than 50 days, but less pronounced if plotting against distance (sma). However, I also found peaks in numbers of Jupiters at slightly farther out than Earth and at a distance about the same as the beginning of the asteroid belt in the solar system.The Earth peak is much wider and goes away in the period histogram. This data is dominated by planets orbiting G stars 89 of 139 (107 of 171 planets). Interestingly (or not) there is a peak in Jupiters that are in &quot;multiple Jupiter systems&quot; around the orbit of Venus that &quot;washes out&quot; the Earth peak. The percenatage of &quot;hot Jupiters&quot; depends on definition, but 22 of 171 have period less than 50 days and 13 have orbits less than 0.1AU.    If anyone is interested in more detail of my results please e-mail me.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2016-04-19T19:17:10.587" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14605" PostTypeId="2" ParentId="14587" CreationDate="2016-04-19T20:12:29.850" Score="2" Body="&lt;p&gt;Cassini is supposed to have used its &lt;a href=&quot;http://www.jpl.nasa.gov/news/news.php?release=2012-406&quot; rel=&quot;nofollow&quot;&gt;VIMS&lt;/a&gt; spectrometer to observe a Venus transit from Saturn in &lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2012/20dec_transitofvenus/&quot; rel=&quot;nofollow&quot;&gt;December 2012&lt;/a&gt;. NASA said it was a first. They were more interested in an absorption spectrum than an image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.shatters.net/celestia/&quot; rel=&quot;nofollow&quot;&gt;Celestia&lt;/a&gt; can model such events.&#xA;Here is a video of a simulated &lt;a href=&quot;https://www.youtube.com/watch?v=z26KecBXcIM&quot; rel=&quot;nofollow&quot;&gt;grazing Jupiter transit from Saturn&lt;/a&gt;.&#xA;There are similar videos of Saturn transits &lt;a href=&quot;https://www.youtube.com/watch?v=V6w_6aehJO0&quot; rel=&quot;nofollow&quot;&gt;from Uranus&lt;/a&gt; and &lt;a href=&quot;https://www.youtube.com/watch?v=VMQD-bu-W3U&quot; rel=&quot;nofollow&quot;&gt;from Neptune&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transits are rare events because two planets must be near the intersection of their orbital planes at the same time.&#xA;A &lt;a href=&quot;https://www.fourmilab.ch/documents/canon_transits/&quot; rel=&quot;nofollow&quot;&gt;numerical integration study&lt;/a&gt; by John Walker found only 1413 transits of Jupiter and 201 of Saturn from outer planets in 250000 years.&#xA;A similar, smaller study could determine whether a given spacecraft could have caught a transit on its way through the interplanetary void, but I would not expect a positive result.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-04-19T20:12:29.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14606" PostTypeId="2" ParentId="14591" CreationDate="2016-04-19T21:48:49.167" Score="4" Body="&lt;p&gt;Titan is roughly ten times more massive than Pluto or Eris, the most massive known Kuiper Belt objects (KBOs). (Titan is in fact more massive than any other moon in the Solar System except Ganymede.) It would be a rather strange coincidence if the object that was by far the most massive of the KBOs was in orbit around &lt;em&gt;Saturn&lt;/em&gt;, well interior to the Kuiper Belt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Titan's orbit is prograde (it orbits in the same direction that Saturn rotates) and barely tilted with respect to Saturn's equator (less than half a degree), the rings, and most of the other moons. That strongly suggests it formed out of an accretion disk around Saturn, as the Galilean moons formed around Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Titan also has an extremely dense atmosphere, unlike any of the KBOs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's rather unlikely that Titan is a captured KBO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Neptun's moon Triton, by the way, &lt;em&gt;is&lt;/em&gt; thought to be a captured KBO. But it is only slightly more massive than Eris or Pluto, and it has a peculiar retrograde orbit that is very difficult to explain if it formed around Neptune, but easier to explain if it was captured. And since Neptune is the furthest out of the (known) major planets, it's not &lt;em&gt;that&lt;/em&gt; surprising that it could have captured a KBO.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2016-04-19T21:48:49.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14607" PostTypeId="2" ParentId="13112" CreationDate="2016-04-19T22:49:54.597" Score="1" Body="&lt;p&gt;As awesome as space rain would be, the sun does not deposit any water molecules on earth. Solar wind consists of ionized particles propelled into space from the sun, and while some of these particles are hydrogen, and some are oxygen (&lt;a href=&quot;http://www.spaceref.com/news/viewsr.html?pid=35448&quot; rel=&quot;nofollow&quot;&gt;http://www.spaceref.com/news/viewsr.html?pid=35448&lt;/a&gt;), these are ionized particles therefore incapable of bonding with each other in the manner necessary to create water. Furthermore, even if these particles did lose enough kinetic energy to reach the energy level low enough to bond to one another, it is unlikely that they would be able to do so because their high velocity, coupled with the incredibly high chance of them missing the earth and the deflective prowess of the magnetosphere would preclude such processes. If somehow enough hydrogen and oxygen atoms were low energy enough and did arrive at the poles, I doubt that they would bond, because the reaction between hydrogen and oxygen to form water requires activation energy. &lt;/p&gt;&#xA;" OwnerUserId="11585" LastActivityDate="2016-04-19T22:49:54.597" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14608" PostTypeId="2" ParentId="14592" CreationDate="2016-04-20T00:50:43.223" Score="7" Body="&lt;p&gt;The current explanation for this is something called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Frost_line_(astrophysics)&quot; rel=&quot;nofollow&quot;&gt;frost line&lt;/a&gt; (which changes over time). At greater distances from the Sun, a body will receive less and less radiation, and so it will be colder than if it were closer to the Sun. Eventually, conditions become cold enough for &lt;a href=&quot;https://en.wikipedia.org/wiki/Volatiles&quot; rel=&quot;nofollow&quot;&gt;volatiles&lt;/a&gt; to condense into grains. These volatiles make it possible for large-scale accretion to happen, and thus giant planets are formed. They may become gas giants (like Jupiter and Saturn) or ice giants (like Uranus and Neptune), depending on the composition of the protoplanetary disk at a given distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A graphic (from &lt;a href=&quot;http://lasp.colorado.edu/education/outerplanets/solsys_planets.php#jovian&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) gives a better summary, although it's copyrighted.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-07-17T14:30:37.490" LastActivityDate="2016-07-17T14:30:37.490" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14609" PostTypeId="1" AcceptedAnswerId="14610" CreationDate="2016-04-20T01:16:41.723" Score="3" ViewCount="344" Body="&lt;p&gt;I came across a website saying that classical T Tauris have an &quot;EW(Hα) &gt; 5 angstroms&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know what &quot;EW&quot; means?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7746" LastActivityDate="2016-04-20T01:55:53.610" Title="What does &quot;EW(Hα)&quot; mean?" Tags="&lt;astrophysics&gt;&lt;star-formation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14610" PostTypeId="2" ParentId="14609" CreationDate="2016-04-20T01:55:53.610" Score="7" Body="&lt;p&gt;&quot;EW&quot; stands for Equivalent Width. The equivalent width of a spectral line is essentially the range of continuum one would integrate over to get the same flux as the spectral line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, this is the Hα line.&lt;/p&gt;&#xA;" OwnerUserId="891" LastActivityDate="2016-04-20T01:55:53.610" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14613" PostTypeId="1" CreationDate="2016-04-20T08:15:40.773" Score="1" ViewCount="4987" Body="&lt;p&gt;Of course it is not exactly known how big the universe is, but I thougt that the universe is about 100 billion lightyears in diameter. But if that is true can you also make an estimation of the volume of the universe?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastActivityDate="2017-10-14T01:52:53.523" Title="What is the volume of the universe?" Tags="&lt;universe&gt;" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14615" PostTypeId="2" ParentId="14594" CreationDate="2016-04-20T12:55:38.873" Score="1" Body="&lt;p&gt;I guess it depends on the type of Supernovae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Type 1a supernovae are produced by a white dwarf star orbiting a companion close enough that matter transfers from the companion to the surface of the white dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming only your star system of a 3M star and 6M star, the 6M star would become a white dwarf star first.  At the end of it's life, it would have been furiously converting hydrogen to helium, and helium to carbon via the C-N-O cycle and the triple alpha process (C-N-O being dominant in stars above 1.3M sol).  The highest reaction in this stars main sequence lifetime would have been the creation of oxygen, but oxygen is a catalyst in the C-N-O cycle, and therefore the last probable reaction in the star would be that of Helium in some form of shell burning in the outer core (more on that next).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The white dwarf, while orbting the companion, may siphon off hydrogen from the surface of the donor star.  This hydrogen will accumulate evenly over the surface of the dwarf star, slowly growing the mass of the dwarf.  Also, the hydrogen will be shell-burned into helium during this time increasing the density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If sufficient mass can be accumulated, then the carbon within the white dwarf can ignite in a runaway reaction - literally blowing the entire star apart.  These supernovae produce every element up to Nickel (Ni-56).&lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2016-04-20T12:55:38.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14616" PostTypeId="2" ParentId="1571" CreationDate="2016-04-20T13:21:47.587" Score="1" Body="&lt;p&gt;In retrospect, we now know that the peak brightness was indeed 10.5 during the first week of February, 2014:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.skyandtelescope.com/astronomy-news/observing-news/supernova-in-m82-passes-its-peak/&quot; rel=&quot;nofollow&quot;&gt;http://www.skyandtelescope.com/astronomy-news/observing-news/supernova-in-m82-passes-its-peak/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2016-04-20T13:21:47.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14619" PostTypeId="2" ParentId="14613" CreationDate="2016-04-20T17:19:01.643" Score="4" Body="&lt;p&gt;Cosmologists estimate the &lt;a href=&quot;https://en.wikipedia.org/wiki/Age_of_the_universe&quot; rel=&quot;nofollow&quot;&gt;age of the universe&lt;/a&gt; at 13.8 billion years.&#xA;The most distant objects we could theoretically observe emitted light that long ago; light from more distant objects hasn't reached us yet.&#xA;Due to expansion of the universe, that horizon is now about 46 billion light-years away.&#xA;If we assume uniform expansion in all directions, we can approximate the &lt;a href=&quot;https://en.wikipedia.org/wiki/Observable_universe&quot; rel=&quot;nofollow&quot;&gt;observable universe&lt;/a&gt; as a sphere of that radius, consistent with the diameter you'd heard.&#xA;I figure its volume as roughly $4 \times 10^{32}$ cubic light-years.&#xA;The shape and size of the &lt;em&gt;un&lt;/em&gt;observable universe are anyone's guess.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2016-04-20T18:58:00.460" LastActivityDate="2016-04-20T18:58:00.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14620" PostTypeId="2" ParentId="14613" CreationDate="2016-04-20T18:51:33.943" Score="6" Body="&lt;p&gt;Based on your comments, I think your confusion comes from having seen the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background#/media/File:Ilc_9yr_moll4096.png&quot; rel=&quot;noreferrer&quot;&gt;classic rugby ball-shaped image of the CMB&lt;/a&gt;.&#xA;The CMB &lt;em&gt;we&lt;/em&gt; observe is not from the whole cosmos, but only from a thin &lt;em&gt;and completely spherical&lt;/em&gt; shell centered on us, with a radius $R_\mathrm{CMB}\simeq45.6\,\mathrm{Gly}$ (using a &lt;a href=&quot;http://arxiv.org/abs/1502.01589&quot; rel=&quot;noreferrer&quot;&gt;Planck 2015&lt;/a&gt; cosmology). Just as the shell of Earth can be projected onto a rugby-shaped figure using a &lt;a href=&quot;https://en.wikipedia.org/wiki/Mollweide_projection&quot; rel=&quot;noreferrer&quot;&gt;Mollweide projection&lt;/a&gt;, so can the shell of the CMB. Here's a figure from &lt;a href=&quot;http://www.universeadventure.org/big_bang/cmb-origins.htm&quot; rel=&quot;noreferrer&quot;&gt;Universe Adventure&lt;/a&gt; that can help visualize:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tSbxw.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tSbxw.jpg&quot; alt=&quot;projection&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although we (still) can't see beyond the CMB, the observable Universe is all the way out to redshift $z=\infty$, while the CMB comes from $z\simeq1100$, but the difference is not big; $R_\mathrm{obs.Uni.} \simeq 47\,\mathrm{Gly}$. Thus, the volume of the &lt;em&gt;observable Universe&lt;/em&gt; is&#xA;$$&#xA;V = \frac{4\pi}{3}R_\mathrm{obs.Uni.}^3 = 435,\!000\,\mathrm{Gly}^3.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total Universe is probably much, much larger, and may in fact easily be infinite.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-04-21T10:56:55.017" LastActivityDate="2016-04-21T10:56:55.017" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14622" PostTypeId="1" CreationDate="2016-04-21T08:16:46.897" Score="0" ViewCount="317" Body="&lt;p&gt;I have a Celestron astromaster130 eq. While viewing through its eyepiece, or using a smartphone camera for photography purpose there is always a definitive no light region( it is circular in curvature). I make sure that I look directly at the eyepiece, but this tunneling makes it really difficult to view things at times. I have googled this, but am not sure of what this effect is called. Is it a general problem or is there a need to rectify some fault in the telescope itself. If not, what are the ways in which this can be effectively reduced?&#xA;Thanks. &#xA;Edits are most welcome. &lt;/p&gt;&#xA;" OwnerUserId="551" LastActivityDate="2016-04-23T09:28:36.580" Title="Tunnel vision while viewing through a reflector" Tags="&lt;telescope&gt;&lt;naked-eye&gt;&lt;telescope-lens&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14623" PostTypeId="1" CreationDate="2016-04-21T11:57:58.507" Score="0" ViewCount="166" Body="&lt;p&gt;Assuming the Big Bang was 13.8 Billion Years ago: Is it possible to observe a Galaxy's redshift showing distances slightly greater than that, like 48 Billion Light Years away? I understand that in a static universe, light traveling in a straight line would only be visible at less than 46.6 Bly. But, can Hubble's constant distort this number? Also, do we assume some error due to the light taking a non-linear path because of lensing? If so, is it possible to have an error of several hundred million years?&lt;/p&gt;&#xA;" OwnerUserId="10995" LastEditorUserId="11525" LastEditDate="2016-04-22T08:35:40.170" LastActivityDate="2016-04-22T08:35:40.170" Title="Seeing a galaxy (quasar) greater than 46.6 billion light years away" Tags="&lt;cosmology&gt;&lt;distances&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14624" PostTypeId="2" ParentId="14623" CreationDate="2016-04-21T12:23:30.060" Score="1" Body="&lt;p&gt;The edge of the observable universe is actually 46.6 billion light years away, despite the Big Bang being only 13.8 Billion years ago. This is because the light which we are now receiving as the furthest visible &lt;em&gt;stuff&lt;/em&gt; had to travel through ever expanding space in between, being redshifted down into what we call the Cosmic Microwave Background Radiation (CMBR). There is a little bit further than that which we are technically receiving, but it has been redshifted infinitely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see anything further away than 46.6 Bly, it would have had to existed literally before time itself, or travelled faster than the speed of light. Two highly improbable things&lt;/p&gt;&#xA;" OwnerUserId="11525" LastEditorUserId="11525" LastEditDate="2016-04-21T13:04:31.837" LastActivityDate="2016-04-21T13:04:31.837" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14625" PostTypeId="2" ParentId="14622" CreationDate="2016-04-21T12:40:21.587" Score="2" Body="&lt;p&gt;This is completely normal behaviour. What you are seeing is the edge of the projection angle for the lense. This is probably sort of what you're seeing? :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lXIjq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lXIjq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically this is more noticable with cheaper eyepieces as the individual elements are smaller, and have a smaller field of view.&lt;/p&gt;&#xA;" OwnerUserId="11525" LastEditorDisplayName="user1569" LastEditDate="2016-04-21T13:44:10.037" LastActivityDate="2016-04-21T13:44:10.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14626" PostTypeId="2" ParentId="6523" CreationDate="2016-04-21T14:37:26.540" Score="2" Body="&lt;p&gt;If you mean the escape velocity of the Earth, then no. Any meteor that enters the earth's atmosphere and burns up or collides with the earth must have had higher than the escape velocity, purely because they were not from that one body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless of course as you say they have ricocheted of another object in orbit sufficiently reducing the kinetic energy to below escape levels&lt;/p&gt;&#xA;" OwnerUserId="11525" LastActivityDate="2016-04-21T14:37:26.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14627" PostTypeId="2" ParentId="10422" CreationDate="2016-04-21T15:32:14.103" Score="2" Body="&lt;p&gt;Upon some google-ing and wiki-ing I found this image of a gravity map of the moon:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uYRn3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uYRn3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That scale up the top is measured in milli-Gal which is thousandths of a cm/s^2. For scale gravity is ~9.81m/s^2 which equals ~981000mGal. The difference between gravity at sea level and the top of mount everest is 2Gal, or 2000mGal, which is 0.2% of average gravity. On the moon, the variation seems to be about 1000mGal, with the average being 162000mGal, which is about 0.6% of average gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, on the surface, there is no noticeable difference. Not to a human anyway.&lt;/p&gt;&#xA;" OwnerUserId="11525" LastActivityDate="2016-04-21T15:32:14.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14628" PostTypeId="2" ParentId="14133" CreationDate="2016-04-21T16:53:47.370" Score="0" Body="&lt;p&gt;As far as we know, once beyond the event horizon, someone might experience all of time in the blink of an eye until the hawking radiation makes the black hole disappear and the person comes out again. More likely though you'd just get spaghettified..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all seriousness though, there is no real limitation, although they aren't really &lt;em&gt;transported&lt;/em&gt; to the future, the outside world just moves faster.&lt;/p&gt;&#xA;" OwnerUserId="11525" LastActivityDate="2016-04-21T16:53:47.370" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="14629" PostTypeId="2" ParentId="14218" CreationDate="2016-04-21T23:38:16.783" Score="1" Body="&lt;p&gt;The USNO Sixth Catalog of Orbits of Visual Binary Stars (&lt;a href=&quot;http://www.usno.navy.mil/USNO/astrometry/optical-IR-prod/wds/orb6&quot; rel=&quot;nofollow&quot;&gt;ORB6&lt;/a&gt;)&#xA;has the orbital elements you need.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\gamma$ And is WDS 02039+4220&lt;/li&gt;&#xA;&lt;li&gt;$\alpha$ Gem is WDS 07346+3153&lt;/li&gt;&#xA;&lt;li&gt;$\zeta$ UMa is WDS 13239+5456&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Double star expert Bruce MacEvoy &lt;a href=&quot;http://www.handprint.com/ASTRO/bineye3.html&quot; rel=&quot;nofollow&quot;&gt;explains&lt;/a&gt; orbital elements in maybe enough detail to help you work out some state vectors.&#xA;My search also turned up a &lt;a href=&quot;http://link.springer.com/chapter/10.1007/978-1-4614-3945-5_7&quot; rel=&quot;nofollow&quot;&gt;book chapter&lt;/a&gt; by Andreas Alzner.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-04-21T23:38:16.783" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14630" PostTypeId="2" ParentId="11878" CreationDate="2016-04-22T01:55:52.793" Score="1" Body="&lt;p&gt;With the axis of rotation of the earth matching the axis of rotation of the stars, and the equator of the earth matching the equator of the stars, any place along the equator would match what you have described. However due to closeness to the horizon, the polar-most stars wouldn't be visible. If you were able to get a hundred metres or so above the ground, with nothing in the surrounding landscape, you could see all of the stars in the sky over a year's time.&lt;/p&gt;&#xA;" OwnerUserId="11525" LastActivityDate="2016-04-22T01:55:52.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14631" PostTypeId="1" AcceptedAnswerId="14636" CreationDate="2016-04-22T06:53:02.317" Score="4" ViewCount="2220" Body="&lt;p&gt;This is the opposite of &lt;a href=&quot;https://astronomy.stackexchange.com/questions/8105/can-you-see-man-made-lights-on-the-dark-side-of-the-earth-from-the-surface-of-th&quot;&gt;another question&lt;/a&gt;. That question is about whether you could see cities on Earth if you were standing on the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's there are cities on the Moon and you're standing on the Earth on a clear night. Could you see the city lights?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're looking at Earth cities from the Moon, your line of sight is not affected by atmospheric turbulence because the Moon is airless. But if you are on Earth, then you have to look through the turbulent atmosphere to see faint lights. &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_seeing&quot; rel=&quot;nofollow noreferrer&quot;&gt;So you won't have to put up with the turbulence that makes stars look blurry or twinkley from the Earth&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you're on the Earth, any city lights on the Moon might look blurry or twinkley from our atmospheric turbulence. If it helps, imagine yourself at the top of a mountain so there is less atmosphere to look through and hence less turbulence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this picture of the Moon (and Venus in the background), imagine cities on dark part of the Moon's surface facing Earth. Those city lights should be easier to see than cities within the lit crescent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9pNNX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9pNNX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11175" LastEditorUserId="7982" LastEditDate="2021-02-28T11:48:16.257" LastActivityDate="2021-02-28T11:48:16.257" Title="Can you see city lights on the Moon from Earth?" Tags="&lt;the-moon&gt;&lt;amateur-observing&gt;&lt;atmospheric-effects&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14633" PostTypeId="1" AcceptedAnswerId="14634" CreationDate="2016-04-22T12:32:47.207" Score="1" ViewCount="387" Body="&lt;p&gt;Is it possible that a brown dwarf object could be orbiting our star and undetected so far by wise surveys?&lt;/p&gt;&#xA;" OwnerUserId="11615" LastActivityDate="2016-04-22T13:17:35.890" Title="Could there be a brown dwarf in our solar system?" Tags="&lt;solar-system&gt;&lt;brown-dwarf&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2016-04-22T14:53:36.683" ContentLicense="CC BY-SA 3.0" />
  <row Id="14634" PostTypeId="2" ParentId="14633" CreationDate="2016-04-22T13:17:35.890" Score="2" Body="&lt;p&gt;The nemesis theory proposes that a low mass star or brown dwarf in highly elliptical orbit is a companion to our sun as a solution to the cyclical mass extinction problem (&lt;a href=&quot;http://www.theatlantic.com/science/archive/2015/11/the-next-mass-extinction/413884/&quot; rel=&quot;nofollow&quot;&gt;http://www.theatlantic.com/science/archive/2015/11/the-next-mass-extinction/413884/&lt;/a&gt;) and (&lt;a href=&quot;http://www.space.com/22538-nemesis-star.html&quot; rel=&quot;nofollow&quot;&gt;http://www.space.com/22538-nemesis-star.html&lt;/a&gt;). Scientists noticed that some mass extinctions follow a seemingly cyclical pattern, so researchers proposed that the orbit of a small companion star (or a brown dwarf) was disturbing comets or asteroids as it approached aphelion and flinging them towards the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As interesting as this would be, WISE's infrared scans of the sky have revealed nothing and there is currently no evidence supporting the theory. Furthermore the orbit of any companion would have to be very large and would therefore be very unstable and would likely be detectable. The Astrobiologist David Morrison stated of the theory: &quot;&lt;strong&gt;(T)he Sun is not part of a binary star system. There has never been any evidence to suggest a companion. The idea has been disproved by several infrared sky surveys, most recently the WISE mission. If there were a brown dwarf companion, these sensitive infrared telescopes would have detected it.&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;" OwnerUserId="11585" LastActivityDate="2016-04-22T13:17:35.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14635" PostTypeId="1" AcceptedAnswerId="14640" CreationDate="2016-04-22T13:57:31.483" Score="6" ViewCount="336" Body="&lt;p&gt;I have learned that it is not possible to watch the upcoming Mercury transit with the plain eye (using special filter glasses).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to watch the transit with improvised devices (like a camera obscura)?&lt;/p&gt;&#xA;" OwnerUserId="11617" LastActivityDate="2019-11-15T03:02:34.983" Title="Watching the Mercury transit with improvised devices" Tags="&lt;amateur-observing&gt;&lt;mercury&gt;&lt;planetary-transits&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14636" PostTypeId="2" ParentId="14631" CreationDate="2016-04-22T15:53:45.703" Score="7" Body="&lt;p&gt;It is possible but unlikely. Here is a really good '&lt;em&gt;Science 2.0&lt;/em&gt;' article about the possibility (&lt;a href=&quot;http://www.science20.com/robert_inventor/could_you_see_moon_city_lights_or_a_greenhouse_from_earth_just_for_fun-157480&quot;&gt;http://www.science20.com/robert_inventor/could_you_see_moon_city_lights_or_a_greenhouse_from_earth_just_for_fun-157480&lt;/a&gt;). Essentially, you likely wouldn't see the light on Moon settlements because there would need to have many thousands of bright lights and windows- both of these would be impractical on a moon base (windows let in radiation and are expensive, and bright lights waste valuable power). Furthermore, in order to reach a magnitude of brightness to constitute as visible, there would have to be a very large amount settlements concentrated together.&lt;/p&gt;&#xA;" OwnerUserId="11585" LastActivityDate="2016-04-22T15:53:45.703" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14637" PostTypeId="1" AcceptedAnswerId="14639" CreationDate="2016-04-22T16:54:06.903" Score="3" ViewCount="252" Body="&lt;p&gt;Are there any known &lt;strong&gt;spectral lines&lt;/strong&gt; shifted by ~1100? If not, then how certain is mainstream that the CMB has a redshift of ~1100? All I see is a blackbody radiation curve void of spectral lines.&lt;/p&gt;&#xA;" OwnerUserId="11589" LastActivityDate="2016-04-22T19:34:37.613" Title="Evidence of CMB redshift" Tags="&lt;universe&gt;&lt;redshift&gt;&lt;cmb&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14639" PostTypeId="2" ParentId="14637" CreationDate="2016-04-22T19:34:37.613" Score="3" Body="&lt;p&gt;The redshift of the CMB is not measured, it is calculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The brief explanation is that as the universe expands and cools, it becomes energetically favourable to form bound atoms. The reduction in free electrons allows the universe to become transparent and photons escape as a blackbody radiation field and eventually form the CMB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature at recombination, $T$, can be calculated using well understood physics (see &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Recombination_(cosmology)&quot; rel=&quot;nofollow&quot;&gt;https://en.m.wikipedia.org/wiki/Recombination_(cosmology)&lt;/a&gt;) and if the current temperature $T_0$ is known, then the redshift is simply given by $ z = T/T_0 -1$.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-04-22T19:34:37.613" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14640" PostTypeId="2" ParentId="14635" CreationDate="2016-04-22T19:41:46.137" Score="5" Body="&lt;p&gt;Mercury's angular diameter on transit day will be 12 arcseconds.&#xA;A camera obscura using a 12 mm aperture could &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_resolution&quot; rel=&quot;nofollow noreferrer&quot;&gt;resolve&lt;/a&gt; it; one lens from +0.75 diopter reading glasses, if you can get them, will project a bright 12 mm image of the Sun at a distance of 1.33 m.&#xA;Note that a larger aperture or a shorter focal length will make the Sun image hotter than direct sunlight unless you add a filter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Test with sunspots before relying on it for Mercury.&#xA;If it counts as improvised, projection with 7x35 binoculars easily showed me the 2012 transit of Venus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: I tried both of the above with the Nov 2019 Mercury transit.&#xA;With handheld wobbling, neither showed it.&#xA;Someone else's 50mm binoculars did, probably because they were mounted on a tripod.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-11-15T03:02:34.983" LastActivityDate="2019-11-15T03:02:34.983" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14642" PostTypeId="1" CreationDate="2016-04-22T22:22:15.710" Score="1" ViewCount="74" Body="&lt;p&gt;I am a high school student (going into 11th grade next year), who will be taking an introduction to astronomy and astrophysics as part of my core curriculum. As such, I would like to get a head start now, but I am not sure where to start.  Are there any specific learning resources that are recommended? Books, videos, documentaries? &lt;/p&gt;&#xA;" OwnerUserId="11619" LastEditorUserId="11619" LastEditDate="2016-04-23T05:56:53.697" LastActivityDate="2016-04-23T10:12:52.073" Title="A good place to start?" Tags="&lt;amateur-observing&gt;&lt;cosmology&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2016-04-25T16:48:12.207" ContentLicense="CC BY-SA 3.0" />
  <row Id="14643" PostTypeId="2" ParentId="6523" CreationDate="2016-04-23T00:31:45.510" Score="0" Body="&lt;p&gt;Edited.  I thought about this wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A slowly spiraling in orbit would hit the earth at not much more than orbital speed, which is escape velocity over the square root of 2, but not a meteor approaching from a distance, so I thought about this the wrong way.   Answer is no.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-04-25T01:20:07.747" LastActivityDate="2016-04-25T01:20:07.747" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14645" PostTypeId="2" ParentId="14622" CreationDate="2016-04-23T09:28:36.580" Score="2" Body="&lt;p&gt;In a telescope (actually, in any optical system), the last optical element creates a circular area called the &quot;exit pupil&quot;, this area must line up with the pupil in your eye (the &quot;entrance pupil&quot; for the optic system called your eye). If the two pupils don't line up, you won't see any of the light the telescope gathers.&lt;/p&gt;&#xA;" OwnerUserId="11625" LastActivityDate="2016-04-23T09:28:36.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14646" PostTypeId="2" ParentId="14642" CreationDate="2016-04-23T10:12:52.073" Score="1" Body="&lt;p&gt;Find a local astronomy club; they will help you get started, learn the sky, and show you different types of equipment. Also, read &quot;Turn Left At Orion&quot;, H.A. Rey's &quot;A New Way to See Them&quot; and &quot;Find The Constellations&quot; to learn the sky. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And get a pair of 10x50 binoculars. These are the best &quot;telescope&quot; (actually, two telescopes!) to start with: inexpensive, wide-field of view to find objects (anti-soda straw effect), and 100x the light gathering power of the naked eye.&lt;/p&gt;&#xA;" OwnerUserId="11625" LastActivityDate="2016-04-23T10:12:52.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14647" PostTypeId="2" ParentId="14596" CreationDate="2016-04-23T11:00:09.423" Score="2" Body="&lt;p&gt;It's not possible to see photons from &quot;beyond&quot; the cosmic microwave background (CMB), because (assuming our cosmology is generally correct) the CMB is the result of recombination of electrons with protons in the early universe. Photons from before that era were absorbed in the plasma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially then, the earliest matter we can see (in photonic terms) is this glowing hot hydrogen, but massively red-shifted so it looks like hydrogen at 3K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see the CMB in every direction.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2016-04-23T11:00:09.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14648" PostTypeId="1" AcceptedAnswerId="14651" CreationDate="2016-04-23T15:05:15.370" Score="1" ViewCount="1739" Body="&lt;p&gt;Like the question says, are there electrons inside an active star, or has all the matter been so thoroughly ionized / plasmarized that the electrons are long gone? I ask because it would make sense when thinking about the magnetic field of a star; naked protons (neutrons are of no magnetic concern) would create a very one-way magnetic effect when circling inside the star, while protons + electrons would seem to counteract one another, magnetically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Come to think of it, I guess the same question could be asked for the Earth's liquid core, considering its temperatures...?&lt;/p&gt;&#xA;" OwnerUserId="11627" LastActivityDate="2016-04-23T17:28:23.403" Title="Are there any electrons inside the sun / a star?" Tags="&lt;astrophysics&gt;&lt;stellar-astrophysics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14649" PostTypeId="1" AcceptedAnswerId="14652" CreationDate="2016-04-23T15:18:22.010" Score="4" ViewCount="236" Body="&lt;p&gt;I understand why hydrogen and helium are the most common elements in the universe. I also understand iron, because that's where stellar fusion stops. But why are (specifically) oxygen, carbon and neon more common than iron? I would expect iron to either outrank them, or some steady flow of atomic nnumbers to lead up to iron. These three, and the scattershot of seemingly random elements trailing behind iron, seem a very odd mix!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements&quot;&gt;https://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11627" LastActivityDate="2016-04-23T19:45:24.903" Title="Why are certain elements so common?" Tags="&lt;elemental-abundances&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14650" PostTypeId="2" ParentId="14648" CreationDate="2016-04-23T15:30:29.937" Score="4" Body="&lt;p&gt;Yes, there are electrons inside stars, they have not gone anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must say your new theory of electromagnetic force does not make any sense to me, but there are actually a few cases where electrons &quot;disappears&quot; from a star:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, the nuclear fusion of 4 hydrogen atoms into one helium atom actually consumes electrons. Let us count the particles before and after:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;4 electrons -4&lt;/li&gt;&#xA;&lt;li&gt;4 protons +4&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;After:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2 electrons -2&lt;/li&gt;&#xA;&lt;li&gt;2 protons +2&lt;/li&gt;&#xA;&lt;li&gt;2 neutrons 0&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Please note that the overall charge has not changed, and &lt;strong&gt;can not&lt;/strong&gt; possibly do that by any means. The charge is always preserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other way electrons can disappear from a star is through the formation of a neutron star. There, the particles has are pressed together so strongly that the protons and electrons have merged into neutrons. Note how the charge is preserved here too.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastActivityDate="2016-04-23T15:30:29.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14651" PostTypeId="2" ParentId="14648" CreationDate="2016-04-23T17:28:23.403" Score="6" Body="&lt;p&gt;Plasma consists of ionised particles. The electrons are not bound to their nuclei and are free to move around within the star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, they are not free to leave the star. Without the electrons to maintain a neutral charge, there would be massive electrical repulsion from all the positive protons and helium nuclei, the result would probably be something like the formation of a black hole larger than the observable universe... The &quot;What if&quot;, blog considered something like this in &lt;a href=&quot;https://what-if.xkcd.com/140/&quot;&gt;Proton Earth, electron moon&lt;/a&gt; and concluded that it would be one of very few things that could actually destroy the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, the sun contains about $10^{57}$ electrons. One for every proton.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-04-23T17:28:23.403" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14652" PostTypeId="2" ParentId="14649" CreationDate="2016-04-23T19:45:24.903" Score="5" Body="&lt;p&gt;This is a very broad question: its answer involves the full details of stellar evolution, Galactic chemical evolution and nuclear physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll limit myself to the following observations:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The elements you mention (actually  are all &quot;islands of nuclear stability&quot;. Whilst their binding energy per nucleon is not as high as that of iron, it is a little higher than elements immediately around them in the periodic table. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Nuclear fusion reactions are exothermic (up to iron), but they require energy to initiate them (in a similar way that you need to get coal hot before it will burn). This ignition becomes more difficult the higher the atomic number (the number of protons) in a nucleus. This means that it is not a given that nuclear reactions will simply proceed to the element with the highest binding energy. If there is a way to prevent the material in a star reaching this ignition temperature, then the reactions will not proceed through to iron.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This means that even in a very massive star, it is only the core where burning runs to completion in the iron peak elements. The majority of the star is still hydrogen and helium, with the layers outside the core containing the remnants of burning of lighter elements. It is these products that get blown into space in a supernova and mostly consists of products of helium burning (C/O) and alpha capture onto these elements (Ne, Mg, Si, S, Ar, Ca),  with each being slightly harder to produce than the last and accounting basically for their relative abundances. Most iron would end up trapped in the neutron star or black hole remnant.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Iron is mostly produced following thermonuclear detonation of C/O white dwarfs - a completely different production path. Hence the abundance of Fe is decoupled from things like Oxygen, and more dependent on the rate of type Ia supernovae, governed by binary interactions and statistics than the details of nuclear physics.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Carbon and Nitrogen are mainly produced in lower mass stars that never become supernovae and where conditions never reach ignition temperatures to produce heavier elements by fusion. There are many more of these than stars which end as supernovae.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, the relative abundances of the elements is a very complicated issue that has no simple answer, but where the broad details are well understood in terms of the processes I mention above.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-04-23T19:45:24.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14653" PostTypeId="2" ParentId="14133" CreationDate="2016-04-23T20:14:20.870" Score="4" Body="&lt;p&gt;The way that you have specified the question, the answer is as far as you like. You simply put your spaceship into any orbit around the black hole and wait.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more sensible question is what is the largest time dilation factor that can be accomplished - i.e. that maximises your travel time into the future for a given amount of proper time experienced on the spaceship. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This in turn is governed by how close to the black hole you can come and still tolerate the tidal forces. If you don't put a limit on this (your first case), then the answer is again infinite; you can hover as close to the event horizon as you like, using an enormous amount of rocket fuel, and the time dilation (see below) can be arbitrarily large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your second case is more realistic. Roughly we can say that the tidal acceleration across a body of length $l$ is given by $GMl/4r^3$, where $M$ is the black hole mass and $r$ is the distance from the black hole. If we make this acceleration equal to say $1 g$, and your body length $l \sim 1$m, then for a $5M_{\odot}$ black hole $r \simeq 2500$ km (well outside the Schwarzschild radius of 15 km).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could &quot;hover&quot; at this radius, then the time dilation factor would be&#xA;$$\frac{\tau}{\tau_0} = \left( 1 - \frac{2GM}{rc^2}\right)^{1/2},$$&#xA;where $\tau$ is the time interval on a clock on the spaceship and $\tau_0$ is the time interval well away from the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $M=5M_{\odot}$ and $r = 2500$ km, this factor is 0.9970.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the spaceship is in a &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_time_dilation#Circular_orbits&quot; rel=&quot;nofollow&quot;&gt;circular orbit&lt;/a&gt;, the factor is $(1 - 3GM/rc^2)^{1/2} = 0.9955$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These factors are perhaps not as big as you might have imagined!&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-04-23T20:14:20.870" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14654" PostTypeId="1" AcceptedAnswerId="14656" CreationDate="2016-04-24T01:42:06.930" Score="1" ViewCount="738" Body="&lt;p&gt;Stars go from the main sequence phase to the red giant branch due to the depletion of hydrogen in the core. As a result, the star contracts and shell hydrogen fusion begins, which apparently produces much more energy. &lt;strong&gt;Why does this produce more energy than core hydrogen fusion?&lt;/strong&gt; The Wikipedia article on &lt;a href=&quot;https://en.wikipedia.org/wiki/Red_giant&quot; rel=&quot;nofollow&quot;&gt;red giants&lt;/a&gt; says that this is because of higher temperatures, but why does shell hydrogen fusion occur at higher temperatures than core hydrogen fusion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;After the helium flash in low-medium mass stars, does shell hydrogen fusion continue to occur?&lt;/strong&gt; If so, then why does there a drop in luminosity from the helium flash to the horizontal branch? And if not, then why not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly to the first question, &lt;strong&gt;in the AGB phase, why does shell helium fusion produce so much more energy than core fusion?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why does the early AGB phase derive most of its energy from shell helium fusion and not shell hydrogen fusion?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7746" LastEditorUserId="2365" LastEditDate="2016-04-24T06:23:46.633" LastActivityDate="2016-04-24T08:55:37.487" Title="Why does shell fusion produce more energy than core fusion?" Tags="&lt;stellar-evolution&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14655" PostTypeId="2" ParentId="13083" CreationDate="2016-04-24T07:37:55.607" Score="2" Body="&lt;p&gt;Being struck by a piece of rock that is pebble sized or bigger could do damage to any base. Fortunately such objects are rare. Their danger could not be entirely avoided, but would be just part of the overall risk of such a mission. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Micrometeorites would hit any exposed base, just as they have hit the &lt;a href=&quot;http://news.nationalgeographic.com/news/2011/10/111006-meteor-shower-meteors-draconids-nasa-storm-moon-space-science/&quot; rel=&quot;nofollow&quot;&gt;ISS&lt;/a&gt;  and the &lt;a href=&quot;http://rebrn.com/re/how-is-the-iss-not-taking-damage-from-the-perseid-meteor-shower-514972/&quot; rel=&quot;nofollow&quot;&gt;space shuttle&lt;/a&gt; in the past. The base would be make strong enough to survive very small impacts. The base could be built partially underground, which would also reduce cosmic radiation. And in would be possible to isolate parts of the base, so a breach doesn't cause destruction of the entire base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, astronauts have to accept dangers that most people would find intolerable. The ISS was estimated to have a 5% chance of catastrophic failure over its lifespan. 4% of astronauts have died in a spacecraft. Breach due to micrometeorite impact would be one of the many risks.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-04-24T07:37:55.607" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14656" PostTypeId="2" ParentId="14654" CreationDate="2016-04-24T08:55:37.487" Score="4" Body="&lt;p&gt;Ultimately this is more of an overly long comment, as I think a more satisfying and complete answer would properly explain things in a more concrete fashion&amp;mdash;more of a &quot;it has to do this because...&quot; answer than my &quot;it can do this because...&quot; one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short of the answer to the first question is that helium fusion needs ~25 times the temperature that hydrogen fusion does.  The proton-proton chain initates around $4\times 10^6$ Kelvin, whereas helium fusion doesn't begin until around $10^8$ Kelvin.  So when the main stage ends and the helium core contracts and the temperature rises, the &quot;edge&quot; of the core can have temperatures well in excess of the minimum hydrogen fusion temperature, and so a shell around it can have temperatures well beyond it.  Fusion rates are (approximately) polynomial in temperature, with the degree depending on the reaction in question, so small increases in temperature can produce substantially more fusion.  The gravitational force is strong enough to overwhelm the pressure from the induced fusion, and so will contract the surrounding shell to temperatures in excess of the minimum needed.  This is basically what happens in the cores of main sequence massive stars (relative to, say, the Sun).  Actually, our own Sun's energy is mostly from the proton-proton chain and has a core temperature of around $1.57\times 10^7$ Kelvin, nearly four times the minimum necessary.  And still, the core needs to be nearly 10 times hotter than that to initiate Helium fusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second question, the short of the answer is that the core has undergone thermal expansion after the Helium flash, and so occupies the temperature ranges most conducive to a strong hydrogen fusion rate.  The material outside the core is now at lower temperatures and pressure, so the fusion rate is reduced substantially.  So the energy output comes principally from the core, and the helium fusion at near minimum temperatures releases less energy than the hydrogen shell at (well) beyond minimum temperatures did.  Thus the star overall produces less energy and contracts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining questions are explained in similar fashion: one has to pay attention to the sensitivity of reaction rates to temperature, and what the temperatures in those shells actually are.  The sensitivities are different for each reaction chain, and the temperatures can go well beyond the minimum necessary.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2016-04-24T08:55:37.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14657" PostTypeId="2" ParentId="14310" CreationDate="2016-04-24T13:20:10.567" Score="2" Body="&lt;p&gt;The declination of the point overhead (zenith) is the same as the observer's latitude. The RA of the point transiting zenith at any given time is the equivalent of the local sidereal time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Alternatively, the local sidereal time is RA of the observer's meridian.)&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2016-04-24T13:20:10.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14658" PostTypeId="1" AcceptedAnswerId="14659" CreationDate="2016-04-24T14:35:52.100" Score="0" ViewCount="771" Body="&lt;p&gt;I've been searching on Internet about how to calculate which RA is in zenith at a given location in a given time but I haven't find anything (or maybe I don't know how to search).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using my star finder I have not draw any conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate which RA is on zenith on a given location in a given time?&lt;/p&gt;&#xA;" OwnerUserId="7566" LastActivityDate="2016-04-24T23:19:55.337" Title="Calculate right ascension in Zenith at this moment in my location" Tags="&lt;ascension&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14659" PostTypeId="2" ParentId="14658" CreationDate="2016-04-24T23:19:55.337" Score="0" Body="&lt;p&gt;Right ascension at Zenith should always be your local sidereal time.&lt;/p&gt;&#xA;" OwnerUserId="11525" LastActivityDate="2016-04-24T23:19:55.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14660" PostTypeId="1" AcceptedAnswerId="14665" CreationDate="2016-04-25T02:45:07.510" Score="2" ViewCount="157" Body="&lt;p&gt;This &lt;a href=&quot;http://www.nytimes.com/video/science/100000003087764/echoes-of-light.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NYTimes video&lt;/a&gt; shows what I think is a series of real Hubble images which have been &quot;morphed&quot; artistically to appear to be continuously expanding. (also see &lt;a href=&quot;http://www.nytimes.com/2014/09/03/science/space/the-v838-monocerotis-star-still-has-astronomers-heads-exploding.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The V838 Monocerotis Star Still Has Astronomers’ Heads Exploding&lt;/a&gt;) If I understand correctly it is the light-echo, not actual material that is doing the expanding at this rate. They also give four &lt;a href=&quot;http://www.nytimes.com/2006/06/06/science/space/06find.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;discrete images&lt;/a&gt; available in 2006 (shown below) and you can find a link to a video there which also seems to be made of just four images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is (very roughly speaking) tomography of the cloud - the parabolic &quot;shell of reflection&quot; expands and each image highlights a view of the material in that shell. That seems to be what the diagram illustrates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;my question is:&lt;/strong&gt; why weren't there more images taken of this - technically speaking - &lt;em&gt;awesome&lt;/em&gt; event, enough to make an animation without the artistic &quot;morphing&quot;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;note:&lt;/strong&gt; I'm not asking &quot;why do you think there weren't...&quot;. I'm looking for the actual reason. Are these &quot;found images&quot; - nobody thought of doing it and later just pulled these from archives, or where these thought to be sufficient? Are there more images available?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some links to video I have found&lt;/p&gt;&#xA;&#xA;&lt;p&gt;YouTube:&#xA;&lt;strong&gt;&quot;Hubble: Timelapse of V838 Monocerotis (2002-2006)&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=U1fvMSs9cps&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=U1fvMSs9cps&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...which can also be seen at &lt;a href=&quot;http://www.spacetelescope.org/videos/heic0617a/&quot; rel=&quot;nofollow noreferrer&quot;&gt;spacetelescope.org&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.spacetelescope.org/videos/heic0617a/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.spacetelescope.org/videos/heic0617a/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... and this &lt;a href=&quot;https://en.wikipedia.org/wiki/V838_Monocerotis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; links to this video&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/1/1b/Evolution_of_the_light_echo_around_V838_Monocerotis_%28Heic0617a%29.ogv&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://upload.wikimedia.org/wikipedia/commons/1/1b/Evolution_of_the_light_echo_around_V838_Monocerotis_%28Heic0617a%29.ogv&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and here are some &quot;stills&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Images from &lt;a href=&quot;http://www.nytimes.com/2006/06/06/science/space/06find.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NYTimes&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wj71s.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wj71s.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2016-05-05T07:39:40.143" LastActivityDate="2016-05-05T07:39:40.143" Title="V838 Monocerotis &quot;light-echo&quot; images morphed into nice video, but why so few original images?" Tags="&lt;light&gt;&lt;dust&gt;&lt;hubble-telescope&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14661" PostTypeId="2" ParentId="14289" CreationDate="2016-04-25T03:40:01.553" Score="3" Body="&lt;p&gt;The site &lt;a href=&quot;https://in-the-sky.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;in-the-sky.org&lt;/a&gt; has a wide variety of functions and options. In Planetarium mode, I chose a random city at about 24N and in the middle, which helps to get the correct UTC + 05:30 India Time Zone, and then just put in the time and date and turned on alt/az grid.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is likely to have been Jupiter, as you suspect. Below are two screen shots - 20:00 and 23:00 local time, with Jupiter climbing from 41 to about 73 degrees above the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PRiJo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PRiJo.png&quot; alt=&quot;8PM Indian Standard Time&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9Bwh3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9Bwh3.png&quot; alt=&quot;11PM Indian Standard Time&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2016-04-25T03:40:01.553" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14663" PostTypeId="1" AcceptedAnswerId="14664" CreationDate="2016-04-25T06:46:38.183" Score="2" ViewCount="603" Body="&lt;p&gt;Very quick and simple one today. What would be the best/optimal pinhole size for a pinhole projector to observe the transit of Mercury on May 9th?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to get the optimum between resolution and brightness...has anyone any experience in this area?&lt;/p&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2016-04-25T21:20:35.577" Title="Pinhole projector for the Transit of Mercury" Tags="&lt;observational-astronomy&gt;&lt;amateur-observing&gt;&lt;mercury&gt;&lt;planetary-transits&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14664" PostTypeId="2" ParentId="14663" CreationDate="2016-04-25T07:18:53.433" Score="5" Body="&lt;p&gt;I've just rewritten this answer - @MikeG &lt;a href=&quot;https://astronomy.stackexchange.com/questions/14663/pinhole-projector-for-the-transit-of-mercury/14664#comment22309_14664&quot;&gt;caught&lt;/a&gt; a glaring error by pointing out a really basic handy relationship called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_resolution&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rayleigh criterion&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align} {\theta}_R \approx1.22 \frac{\lambda}{D}. \end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's better to read the (or any) article, but very briefly, the angular resolution is roughly the ratio of the wavelength to the diameter of a circular aperture. You can apply this equally well to pinhole-only imaging, or to a system which images by focusing with curved mirrors or lenses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury's diameter is about 4900 km and since it will be on the line between the sun and the earth, the distance will be about 150,000,000 minus 58,000,000 or 92,000,000km. In that case the angular width of Mercury will be about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align} {\theta}_{merc} \approx \frac{4.9 \times 10^3\ \mathrm{km}}{9.2 \times 10^7\ \mathrm{km}} \approx 0.000052 \ \mathrm{rad} \approx 0.0031°  \approx 11 \ \mathrm{arcsec}.  \end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to even poorly resolve Mercury as a dark fuzzy dot, you'd like the diffraction width to be equal or less than the angular width. If you set the two angles equal and let $\lambda$ = 580 nm, you get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align} D \approx 1.22\ \frac{5.8 \times 10^{-7}\ \mathrm{m}}{\theta_R}  \approx \frac{7.1 \times 10^{-7}}{5.2 \times 10^{-5}}\ \mathrm{m}   \approx 14\ \mathrm{mm}\ (minimum)   \end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the light is essentially parallel, your geometrical resolution on the screen is about the diameter of the pinhole. To make a 5.2E-05 radian object 14mm on a screen the screen would have to be VERY far away:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align} L_{to screen} \approx \frac{D}{\theta_{merc}}   \approx  \frac{0.014}{0.000052} \approx 270 \ meters! \end{align} &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can try that, use a 14 or 20mm &quot;pinhole&quot; at 270 meters away, but I think the light will be far to faint to see. I once did something similar to see a solar eclipse. It may have been a 10mm &quot;pinhole&quot; but I'm sure I wasn't &lt;em&gt;that&lt;/em&gt; far away. I used household mirrors to bring the light indoors to a &lt;em&gt;very&lt;/em&gt; dark area, and it worked great.  But that was at most only abour 30 meters!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you would really like to try it, here are some tests you can do ahead of time:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Search to see if someone has done it and given details&lt;/li&gt;&#xA;&lt;li&gt;Try some simple experiments using the full sun. If the fuzziness at the edge of the sun's image really seems to be 300 times smaller than the sun's image, then that's a &lt;em&gt;suggestion&lt;/em&gt; that it &lt;em&gt;might&lt;/em&gt; work.&lt;/li&gt;&#xA;&lt;li&gt;Check the internet for daily sun images and see if you can find a sunspot. Then test your setup and see if the sunspot suggests your resolution looks as good as 1/300.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-04-25T21:20:35.577" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14665" PostTypeId="2" ParentId="14660" CreationDate="2016-04-25T08:36:38.737" Score="5" Body="&lt;p&gt;To get time on the Hubble Space Telescope is a difficult, peer-reviewed process. The time is usually about ten times oversubscribed. The scientific justification to get more images must be more than &quot;it would make a nice movie&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The submission of proposals for HST time is a confidential process; so I don't know whether the principal investigators for these images asked for more at different epochs and were turned down, or whether they thought this many images were sufficient for the science they wanted to accomplish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no technical reason why they couldn't have obtained more images. In fact more images &lt;em&gt;were&lt;/em&gt; obtained (see &lt;a href=&quot;http://hla.stsci.edu/hlaview.html#Inventory%7CfilterText%3D%24filterTypes%3D%7Cquery_string=V838%20Mon&amp;amp;posfilename=&amp;amp;poslocalname=&amp;amp;posfilecount=&amp;amp;listdelimiter=whitespace&amp;amp;listformat=degrees&amp;amp;RA=106.021120&amp;amp;Dec=-3.846190&amp;amp;Radius=0.200000&amp;amp;inst-control=all&amp;amp;inst=ACS&amp;amp;inst=ACSGrism&amp;amp;inst=WFC3&amp;amp;inst=WFPC2&amp;amp;inst=NICMOS&amp;amp;inst=NICGRISM&amp;amp;inst=COS&amp;amp;inst=WFPC2-PC&amp;amp;inst=STIS&amp;amp;inst=FOS&amp;amp;inst=GHRS&amp;amp;imagetype=best&amp;amp;prop_id=&amp;amp;spectral_elt=&amp;amp;proprietary=both&amp;amp;preview=1&amp;amp;output_size=256&amp;amp;cutout_size=12.8%7Cra=&amp;amp;dec=&amp;amp;sr=&amp;amp;level=&amp;amp;image=&amp;amp;inst=ACS%2CACSGrism%2CWFC3%2CWFPC2%2CNICMOS%2CNICGRISM%2CCOS%2CWFPC2-PC%2CSTIS%2CFOS%2CGHRS&amp;amp;ds=&quot; rel=&quot;noreferrer&quot;&gt;this link&lt;/a&gt; to the Hubble archive) at later epochs between 2005 and 2012. A problem I guess with incorporating these images into the movie sequence were that the light echo gets fainter, more diffuse and starts to exceed the size of the detector, which would necessitate more than one pointing. They were also taken with a different camera, which might raise aesthetic compatibility problems. I show below a fairly crude colour composite I extracted from the archive for an observation in 2007.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RcPL6.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RcPL6.jpg&quot; alt=&quot;V838 Mon in February 2007&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-04-25T08:36:38.737" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14666" PostTypeId="1" CreationDate="2016-04-25T18:16:01.193" Score="4" ViewCount="990" Body="&lt;p&gt;It's my understanding that the Sun (uppercase) is used to refer to our sun/star, because it is The Sun. Latin name is Sol, hence the Solar System. Even the tag (lowercase sun) on this post has the caption &quot;Questions regarding the closest star to Earth, at the centre of the Solar System.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other planetary systems have their own stars, which I understand are called a sun, in the context of that planetary system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried my best to find a credible source that exoplanets indeed have their own suns, and that the term &quot;sun&quot; doesn't just refer to our Sun. I have only found this one website that backs this up: &lt;a href=&quot;http://curious.astro.cornell.edu/physics/78-the-universe/stars-and-star-clusters/general-questions/346-what-is-the-difference-between-a-star-and-a-sun-beginner&quot; rel=&quot;noreferrer&quot;&gt;http://curious.astro.cornell.edu/physics/78-the-universe/stars-and-star-clusters/general-questions/346-what-is-the-difference-between-a-star-and-a-sun-beginner&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to be common knowledge in the scientific community that this is correct, but I can't find anything else online to back this up. Help?&lt;/p&gt;&#xA;" OwnerUserId="11637" LastEditorUserId="11637" LastEditDate="2016-04-29T01:11:51.053" LastActivityDate="2020-07-05T07:45:17.557" Title="Is a planetary system star's referred to as their sun?" Tags="&lt;the-sun&gt;&lt;terminology&gt;" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14667" PostTypeId="2" ParentId="14666" CreationDate="2016-04-25T20:44:01.303" Score="4" Body="&lt;p&gt;When language developed there was a need to refer to the &quot;the bright shiny thing that makes the days light&quot; and to the &quot;sparkly things in the night sky&quot;. These things were very obviously different and so received different names.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://solar-center.stanford.edu/FAQ/Qsunasstar.html&quot; rel=&quot;nofollow&quot;&gt;Several ancient greeks, and some renaissance Italians were executed for suggesting that the sun might be a star&lt;/a&gt;. And it wasn't until the middle of the 17th century that the idea that the sun was a star became &quot;settled science&quot;. And not until the late 20th century was it apparent that other stars have their own planetary systems from which they could be viewed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our language hasn't really caught up. But there is nothing wrong with referring to &quot;stars from the perspective of an orbiting planet&quot; as suns. &#xA;For example a nasa blog describes Kepler16-b as a planet &quot;&lt;a href=&quot;http://www.nasa.gov/mission_pages/kepler/multimedia/images/Kepler-16_transit-art.html&quot; rel=&quot;nofollow&quot;&gt;Where the sun sets twice&lt;/a&gt;&quot; The important thing is to be clear, ask &quot;would my readers know if I am referring to The Sun or a sun in this context&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-04-25T20:44:01.303" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14668" PostTypeId="1" AcceptedAnswerId="14670" CreationDate="2016-04-25T22:11:26.510" Score="3" ViewCount="1898" Body="&lt;p&gt;Reading on Wikipedia I saw that Titan is 80% more massive than the earth's moon but has only 85% the surface gravity. Why is this?&lt;/p&gt;&#xA;" OwnerUserId="10953" LastActivityDate="2020-10-18T20:56:13.947" Title="Why does Titan have lower surface gravity than the Moon when Titan is more massive?" Tags="&lt;gravity&gt;&lt;titan&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14670" PostTypeId="2" ParentId="14668" CreationDate="2016-04-25T22:34:51.360" Score="15" Body="&lt;p&gt;Surface gravitational acceleration on an object with mass &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; and radius &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; is given by&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;g = \frac{GM}{R^2} \propto G\rho R&#xA;$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\rho \propto M/R^3$&lt;/span&gt; is the density of the object.&#xA;If one body has smaller surface &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; than another, it must have smaller density &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt;, smaller radius &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt;, or both.&#xA;Titan is larger than Earth's Moon, so your observation about its surface gravity means Titan must be less dense than the Moon.  Wikipedia confirms:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$R_\text{Titan} = 1.5 R_\text{Moon}$&lt;/span&gt;, but&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\rho_\text{Moon} = 3.34\rm\,g/cm^3$&lt;/span&gt; while Titan has only &lt;span class=&quot;math-container&quot;&gt;$\rho_\text{Titan} = 1.88\rm\,g/cm^3$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1498" LastEditorUserId="1498" LastEditDate="2020-10-18T20:56:13.947" LastActivityDate="2020-10-18T20:56:13.947" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="14671" PostTypeId="1" AcceptedAnswerId="14676" CreationDate="2016-04-26T01:44:48.067" Score="5" ViewCount="280" Body="&lt;p&gt;How big will Apophis appear in the sky at its closest point of approach in 2029? Will it compare  to the moon? How long will it linger as it passes?&lt;/p&gt;&#xA;" OwnerUserId="11641" LastActivityDate="2019-06-22T16:42:41.907" Title="How big will Apophis appear?" Tags="&lt;observational-astronomy&gt;&lt;amateur-observing&gt;&lt;asteroids&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14672" PostTypeId="2" ParentId="14666" CreationDate="2016-04-26T01:47:00.203" Score="2" Body="&lt;p&gt;When I was very young, I didn't know that the sun in the sky was just another example of the stars I saw at night. I just wasn't surrounded by people who talked about such things. Later, when I learned that the sun was also a star, I was glad to know there was a star close enough to study. The term 'sun' was used to describe &lt;i&gt;our&lt;/i&gt; star, even before people realized it was a star. So the term 'sun' is a generic word that might follow us to the stars and be used in everyday conversation to refer to the local day stars. 'Sol' is what the Romans would say whenever they talked about the sun. Currently 'sol' is being used to refer to a rotation of a planet other than Earth in our solar system. For instance on Mars a 'sol' is about 24 hours and 38 minutes, and on Jupiter a 'sol' would be about 9 hours and 50 something minutes. There's no reason why we can't refer to a rotation of the Earth as a 'sol', but we already call it a 'day'.&lt;/p&gt;&#xA;" OwnerUserId="9516" LastActivityDate="2016-04-26T01:47:00.203" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14673" PostTypeId="2" ParentId="14666" CreationDate="2016-04-26T02:44:53.023" Score="6" Body="&lt;p&gt;The technical name (in English) for the Sun is not Sol, which is just Latin for sun. The technical name for the Sun is the Sun. Another body in the sky has a similarly boring name, the Moon. There's one more boringly named object: in the Solar System: The Earth. Note the use of &amp;quot;the&amp;quot; (a definite article) and the use of capitalization to indicate a specific object.&lt;/p&gt;&#xA;&lt;p&gt;The problem with Sun, Moon, and Earth is that we have been using these names (or their predecessors) for thousands of years. For example, sun, sol, ἥλιος (helios), and a bunch of other names for that very bright object in the sky whose presence distinguishes day from night all derive from the same proto-Indo-European word &lt;a href=&quot;https://en.wiktionary.org/wiki/Reconstruction:Proto-Indo-European/s%C3%B3h%E2%82%82wl%CC%A5&quot; rel=&quot;nofollow noreferrer&quot;&gt;sóh₂wl̥&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;As an end note, from &lt;a href=&quot;http://curious.astro.cornell.edu/about-us/159-our-solar-system/the-sun/the-solar-system/4-what-are-the-names-of-the-earth-moon-sun-and-solar-system-beginner&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://curious.astro.cornell.edu/about-us/159-our-solar-system/the-sun/the-solar-system/4-what-are-the-names-of-the-earth-moon-sun-and-solar-system-beginner&lt;/a&gt;,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You may read or hear people using Luna for the Moon, or Terra or Gaia for the Earth, or Sol for the Sun, but in English-speaking countries, these are poetic terms, often seen in science fiction stories, but not used by astronomers in scientific writing. In some countries where Romance languages are spoken, these terms are the official names.&lt;/p&gt;&#xA;&lt;p&gt;It's also interesting to note that most astronomers do not call our galaxy the Milky Way in technical writing--they call it the Galaxy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-04-26T02:44:53.023" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14675" PostTypeId="1" AcceptedAnswerId="14680" CreationDate="2016-04-26T04:03:23.330" Score="1" ViewCount="119" Body="&lt;p&gt;I did a time-lapse animation looking towards the sunrise from Sydney, Australia on 25th of April.  You can see it here around 1m 25s in (that link should jump to just before it appears very near where the Sun is about to rise):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=lqr_KuLd2-c&amp;amp;t=1m25s&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=lqr_KuLd2-c&amp;amp;t=1m25s&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It then proceeds towards the right of screen over around the next 21 seconds (around 10&amp;#189; minutes) before disappearing off the right. You might need to send the video to full screen to see the small white 'bullet of haze' that is moving against a very wide background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of my first thoughts was that it might be a meteor skimming the upper atmosphere, but I was less confident of that given it seemed to be coming from the general direction of the Sun, and I'd thought that was a direction unlikely to produce meteors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That was until I read &lt;a href=&quot;http://www.timeanddate.com/astronomy/meteor-shower/eta-aquarids.html&quot; rel=&quot;nofollow&quot;&gt;Eta Aquarid Meteor Shower in 2016&lt;/a&gt; which states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The best time to view the Eta Aquarids is in the early mornings, right before dawn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, at least that pins down the same time of day, if not the direction from which it came.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the likelyhood that this feature is a piece of comet debris that skimmed Earth's atmosphere?&lt;/p&gt;&#xA;" OwnerUserId="1840" LastEditorUserId="1840" LastEditDate="2016-04-26T04:10:12.077" LastActivityDate="2016-04-27T13:34:47.957" Title="Does this video catch an Eta Aquarid - Earth skimming meteor?" Tags="&lt;comets&gt;&lt;meteor&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14676" PostTypeId="2" ParentId="14671" CreationDate="2016-04-26T06:09:06.547" Score="5" Body="&lt;p&gt;Current estimates put Apophis's diameter around &lt;a href=&quot;http://www.esa.int/Our_Activities/Space_Science/Herschel_intercepts_asteroid_Apophis&quot; rel=&quot;nofollow noreferrer&quot;&gt;325 m&lt;/a&gt; and its 2029-04-13 approach about &lt;a href=&quot;https://newton.spacedys.com/neodys/index.php?pc=1.1.8&amp;amp;n=99942&quot; rel=&quot;nofollow noreferrer&quot;&gt;38000 km&lt;/a&gt; from the center of the Earth. I figure an angular size &amp;lt;= 2 arcsec, almost starlike even if you manage to track it in a telescope.&#xA;&lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi?find_body=1&amp;amp;body_group=sb&amp;amp;sstr=99942&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ephemerides&lt;/a&gt; show it at apparent magnitude 3.5 or so just before closest approach - visible but not outstanding to the unaided eye at night - and spending about 36 hours within 1 lunar distance of the Earth.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-06-22T16:42:41.907" LastActivityDate="2019-06-22T16:42:41.907" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14677" PostTypeId="2" ParentId="13551" CreationDate="2016-04-26T07:12:51.450" Score="6" Body="&lt;p&gt;I think it's a fun question, and I can answer some of it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If their event horizons don't &quot;touch&quot; I'd assume their courses will&#xA;  change considerably but it would be a normal flyby - right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While the relativistic speeds and time dilation might change things a bit, a flyby is still a flyby. They would either fly past each other, the dense objects would change direction as a result, but the combined kinetic energy and momentum would remain constant.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second option is they enter into orbit around each other.   Pretty much the same with any two massive objects passing near each other.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But what happens if their event horizons &quot;touch&quot;, if their event&#xA;  horizons &quot;overlap&quot; just a few meters?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With classic (non Kerr) black holes, if the event horizons touch the black holes have no option but to merge.  The event horizon is directly proportional to the radius, so when the two event horizons touch it, as that happens, no longer two black holes but one black hole.   The mass of the two objects combined has an event horizon that circles both of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Kerr black holes this question is more complicated and I'm not sure but I think it's the same. Once the event horizons touch it's no longer two separate black holes but one bigger one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On the one hand both objects have a massive kinetic energy with&#xA;  opposite directional vectors. And on the other hand nothing can escape&#xA;  below the gravitational radius.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you're talking about event horizons, gravity always wins.  Kinetic energy cannot exceed the speed of light, so it loses.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where does all the kinetic energy go ? Would the two black holes&#xA;  extremely deform, like creating a long string before they collapse&#xA;  into spherical form?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A long string doesn't make any sense.  A temporarily non-round event horizon as the holes merge seems possible, but think of a merger of two stretched balls, not a string.   Each black hole is centered around it's &quot;singularity&quot;, those shapes might warp some as the holes collide or spiral into each other.  As an FYI, direct collisions would be rare.  Spiraling into each other is much more common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinetic energy gets absorbed inside the black hole.  It can't, by definition of the black hole, escape.   Total momentum is likely preserved.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And what if in a similar scenario one of the two objects is a neutron&#xA;  star. Would the neutron star be ripped apart when it scrapes the black&#xA;  hole on high-speed flyby ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A few points to consider regarding a black hole. Anything that falls within a black hole's photon-sphere is unlikely to ever escape, so one of the magical distances to consider is 1.5 times the radius of the black hole where all orbits are destined to fall into the black hole.   If the black hole and neutron star are of similar size and the neutron star passes close to the event horizon and relativistic speed, then there's an interesting scenario where some of the neutron star (at close to the speed of light) would be outside the photon-sphere and have the momentum to escape and some of it would not.   Neutron stars are bound together very tightly, but the tidal forces would be very strong and it's possible that some of the neutron star material could get torn off in this hypothetical scenario.  Would the entire neutron star be torn apart?  Maybe.  It's hard to say.  The math in that hypothetical is way above my pay-grade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the neutron star is much smaller than the black hole, then it's likely inside the photon-sphere and it will spiral into the black hole, perhaps breaking apart in the process into an accretion disk, perhaps not, but it will get absorbed into the black hole if it's entirely within the photon-sphere, even at nearly the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with this hypothetical is that it's pretty much impossible.  Objects the size of stars don't accelerate at nearly the speed of light relative to other near by stellar mass objects.  What actually happens is described in the first answer to your question, the two dense objects would either have sufficient relative velocity to fly past each other or they would get caught into a mutual orbit that would likely lead to a merger.  In the actual universe, a close to the event horizon fly-by between a neutron star and a black hole would result in a merger of the two.  In theory, with enough kinetic energy the neutron star could fly close to and still fly past the black hole, but that much relative velocity probably never happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People smarter than me should feel free to correct anything I got wrong, but I thought this one was a fun one to answer. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="18643" LastEditDate="2019-05-23T06:39:11.763" LastActivityDate="2019-05-23T06:39:11.763" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14678" PostTypeId="1" AcceptedAnswerId="14699" CreationDate="2016-04-26T10:14:05.247" Score="2" ViewCount="1294" Body="&lt;p&gt;Say we have a relativistic fluid/gas, as we have in some astrophyical systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let us write:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$e$ - energy density in the fluid's rest frame.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$P$ - pressure in the fluid's rest frame.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$n$ - number density in the fluid's rest frame.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$m$ - mass of the particles.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I know that for the non-relativistic case we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e=nmc^2+\frac{1}{\hat{\gamma}-1}P$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\hat{\gamma}$ is the adiabatic index. $\hat{\gamma}=1+\frac{2}{f}$ for a gas with $f$ degrees of freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the ultra-relativstic case we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e=3P$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is what is $e(P,n)$ for a relativstic case (which is the general case of the 2 limits shown above)? I would also like to know how to derive it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Is the following way the correct way to do it ? :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number density of particles is:&#xA;$$n=\int_{0}^{\infty} n_p(p) dp $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure is:&#xA;$$P=\int_{0}^{\infty} \frac{1}{3} p v(p) n_p(p) dp $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy density is:&#xA;$$e=\int_{0}^{\infty} \epsilon(p) n_p(p) dp $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n_p(p)= (2s+1)\frac{1}{ e^{({\epsilon(p)-\mu})/{k_B T}}+(-1)^{2s+1} } \frac{4\pi p^2}{h^3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $s$ is the spin of the particles, for electrons $s=\frac{1}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \epsilon(p)=(m^2c^4+p^2c^2)^{\frac{1}{2}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v(p)= \frac{d\epsilon}{dp}=\frac{p}{m}\left(1+\left(\frac{p}{mc}\right)^2\right)^{-\frac{1}{2}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From calculating the three integrals above we can finally obtain $e(P,n)$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Can anyone confirm this is the proper way to do it, or am I missing something here?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It seems as if those integrals cannot be solved analytically - is this&#xA;true?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Perhaps in this case there is no explicit formula for $e(P,n)$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11644" LastEditorUserId="9533" LastEditDate="2016-04-29T12:23:17.933" LastActivityDate="2016-04-29T12:23:17.933" Title="What is the equation of state for a relativistic fluid/gas?" Tags="&lt;special-relativity&gt;&lt;gas&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14679" PostTypeId="1" AcceptedAnswerId="14681" CreationDate="2016-04-26T12:02:36.133" Score="0" ViewCount="713" Body="&lt;p&gt;The link below describes how an array of inexpensive cameras can create very detailed images. Would this work for astronomy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://graphics.stanford.edu/papers/CameraArray/&quot; rel=&quot;nofollow&quot;&gt;https://graphics.stanford.edu/papers/CameraArray/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5432" LastActivityDate="2016-04-26T14:57:03.273" Title="Can I use an array of inexpensive cameras as an alternative to a telescope?" Tags="&lt;photography&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14680" PostTypeId="2" ParentId="14675" CreationDate="2016-04-26T14:27:50.990" Score="1" Body="&lt;p&gt;Things entering the atmosphere (whether a meteor or an atmosphere-skimming object) travel too fast to be visible for 10 minutes.  Normally a pass of a satellite (at above atmosphere height) takes 2-3 minutes and most meteors are over in seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even the famous &lt;a href=&quot;https://en.wikipedia.org/wiki/1972_Great_Daylight_Fireball&quot; rel=&quot;nofollow&quot;&gt;daylight fireball&lt;/a&gt; from the 1970s was over in a very short time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would guess what you have there is a plane at high altitude, though it's moving a bit slower than I would expect. (I'm open to suggestions.)&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2016-04-26T14:27:50.990" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14681" PostTypeId="2" ParentId="14679" CreationDate="2016-04-26T14:57:03.273" Score="2" Body="&lt;p&gt;Yes it would.  However, it also depends on your definition of &quot;Detail&quot;.  For some, it's a tight shot where the whole frame is only a few dozen arcseconds wide, and others think of detail as being wide field, but also &quot;deep&quot; - or having a very high signal to noise ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want wide field, and a high signal to noise ratio, then yes, a multi-camera array will work well, and you'd be able to accumulate a large number of high integration time images in a relatively short amount of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's a breakdown of what you and others who may not know need to know about deep sky astrophotography:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main goal for all photos is to produce as visually pleasing image as possible.  A single image taken of any target will accumulate light from many sources.  The first, and obvious source is the light from whatever target you're trying to photograph.  The second, less obvious source is light pollution from man-made sources, the moon, and natural sky glow.  The third source is the electronic and thermal noise of the camera itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically, all of these sources are &quot;signal&quot; - however two of the three are unwanted, so we will call the light pollution signal and the camera signal 'noise'.  Longer exposures produce more signal from what you want, but also more noise from what you don't want.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luckilly, noise can be reduced by using several techniques in conjunction with each other.  First, you need to take long photos.  The signal will increase linearly with time.  A 60s photo with a signal of &quot;x&quot; will have a tenth of the signal as a 600s photo.  Noise, however, increases with the square root of the exposure.  Using the same example, the 600s photo will have a signal to noise ratio that is about 3.2 (sqrt of 10) times better than the single 60s photo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Increasing the Signal to noise ratio by taking longer exposures is the first step to producing 'detailed' wide field images.  Camera noise can be further reduced by taking many images.  An array of cameras makes this trivial.  Two cameras will produce twice the images as one camera, and 10 cameras will produce 10 times the images as a single camera.  That's obvious.  Why does it matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we stack images, the more we have, the better the stacking algorithms can preserve the signal we want to keep, and reduce the camera noise we don't.  Once again, the signal to noise ratio mathematics work out the same as above.  1 image will have a certain signal to noise ratio, but 10 images will have a signal to noise ratio that can be about 3.2 times higher - because the total integration time has increased ten fold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you're stacking photos, you should be calibrating them as well (dark frames, bias frames, etc) - which really reduce the amount of noise per sub-frame.  Because of this, your actual SNR boost by stacking should in theory be a lot higher than 3.2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If photographed a target across 60 degrees of the sky (30 before meridian and 30 after) then you'd be able to acquire photos for 4 hours.  One camera taking 10 minute photos would get 24 photos in the 4 hour period of time.  If you had 5 cameras, you'd get 96 photos.  72 additional photos in the same amount of time is a lot of extra data to help reduce the noise, and really let the faint details stand out above the background - especially if all of your photos were well calibrated.&lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2016-04-26T14:57:03.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14682" PostTypeId="2" ParentId="14568" CreationDate="2016-04-26T16:22:25.877" Score="1" Body="&lt;p&gt;v$_r$ and &lt;strong&gt;μ&lt;/strong&gt; are radial velocities and proper motion as seen from the Sun (ie heliocentric)  and the equations in the appendix convert to v in the $S_o$ frame which is the center of the Galaxy.  So, the motion of the MW is (0,0,0).  If you want to, you can solve for v$_r$ and &lt;strong&gt;μ&lt;/strong&gt; for the MW by plugging v=(0,0,0) into the equation, and find the heliocentric proper motion and radial velocity of the Galactic center. But it is not needed.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2016-04-26T16:22:25.877" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14683" PostTypeId="1" CreationDate="2016-04-26T21:32:45.507" Score="5" ViewCount="222" Body="&lt;p&gt;The title pretty much explains this one. Can I use a specific solar optical eye piece, or other optical device, in conjunction with the pinhole projector method to focus the Transit of Mercury? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would anyone have any specific recommendations for specifications, etc if it is possible to achieve this... &lt;/p&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2016-09-26T18:58:52.470" Title="Pinhole Projector: Can I use a specific solar optical eye piece to focus the Transit of Mercury?" Tags="&lt;amateur-observing&gt;&lt;the-sun&gt;&lt;planetary-transits&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14684" PostTypeId="1" AcceptedAnswerId="14692" CreationDate="2016-04-26T22:17:58.630" Score="13" ViewCount="1697" Body="&lt;p&gt;Everything in the Solar System revolves around the &quot;barycenter&quot;: the overall center of mass. This barycenter is not in the center of the Sun. Some articles and essays I've read go so far as to suggest that the position of the barycenter does not have a set of fixed coordinates within the System: it fluctuates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well. Since everything, including the Sun, revolves around this barycenter, the Sun must have its own orbit around it. &lt;strong&gt;What does it look like? How large is it? How elliptic?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In my research, I have tried and failed to establish whether the barycenter is within the Sun or outside it. Either way, an orbit is an orbit).&lt;/p&gt;&#xA;" OwnerUserId="9510" LastActivityDate="2018-11-12T20:21:05.880" Title="What does the Sun's orbit within the Solar System look like?" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14685" PostTypeId="1" AcceptedAnswerId="14689" CreationDate="2016-04-27T06:42:40.590" Score="1" ViewCount="164" Body="&lt;p&gt;I know gas giants have moons, just like star have planets, how about &lt;a href=&quot;https://en.wikipedia.org/wiki/Kuiper_belt&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kuiper belts&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/Oort_cloud&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oort Cloud&lt;/a&gt;? Is there any corresponding structure in gas giants? If not, why don't they have?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="37191" LastEditDate="2020-12-30T13:34:59.917" LastActivityDate="2020-12-30T13:34:59.917" Title="Is there Kuiper belt/Oort cloud like structure in gas giants?" Tags="&lt;gas-giants&gt;&lt;oort-cloud&gt;&lt;kuiper-belt&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14686" PostTypeId="1" CreationDate="2016-04-27T07:01:32.577" Score="2" ViewCount="372" Body="&lt;p&gt;I used to go to my roof to see the ISS, but it is not always visible. Then it appears again but does it appear with the exact same orbit as I saw before?&#xA;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="11652" LastEditorUserId="2270" LastEditDate="2016-04-27T15:26:12.727" LastActivityDate="2016-04-27T15:26:12.727" Title="How much time does the ISS take to repeat its orbit?" Tags="&lt;orbit&gt;&lt;artificial-satellite&gt;&lt;iss&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14687" PostTypeId="2" ParentId="11983" CreationDate="2016-04-27T07:13:33.507" Score="1" Body="&lt;p&gt;This is a more recent review &lt;a href=&quot;http://arxiv.org/abs/1209.4688&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1209.4688&lt;/a&gt; from someone hoping to understand population III by observing their supernovae&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since CR7 was discovered, there's been a bunch of papers on whether it could be pop III - &lt;a href=&quot;http://arxiv.org/abs/1506.07173&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1506.07173&lt;/a&gt; for example. Try searching for &quot;CR7 population III&quot;.&lt;/p&gt;&#xA;" OwnerUserId="11654" LastActivityDate="2016-04-27T07:13:33.507" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14688" PostTypeId="2" ParentId="11348" CreationDate="2016-04-27T07:15:39.397" Score="5" Body="&lt;p&gt;Edit: I found you a very relevant paper!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/pdf/astro-ph/0007341v1.pdf&quot;&gt;http://arxiv.org/pdf/astro-ph/0007341v1.pdf&lt;/a&gt; seems to be solving exactly your problems for sources with arbitrary angular separation $\alpha$. Equations (12) and (14) will give you the comoving distance $\chi_2'$ and redshift $z_2'$ between them at the epoch when $Q_1$ received light from $Q_2$. You will have to solve these equations numerically to find the solutions, which will be more complicated depending on the cosmological model you use.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Have a read of these notes &lt;a href=&quot;http://arxiv.org/abs/astro-ph/9905116&quot;&gt;http://arxiv.org/abs/astro-ph/9905116&lt;/a&gt;. They are my go-to when trying to work out cosmological distances.&lt;/p&gt;&#xA;" OwnerUserId="11654" LastEditorUserId="11654" LastEditDate="2016-04-27T18:03:47.627" LastActivityDate="2016-04-27T18:03:47.627" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14689" PostTypeId="2" ParentId="14685" CreationDate="2016-04-27T07:40:15.080" Score="3" Body="&lt;p&gt;The gravitational region around the planets isn't hard to calculate, sometimes called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Sphere_of_influence_%28astrodynamics%29&quot; rel=&quot;nofollow&quot;&gt;Sphere of Influence&lt;/a&gt;, sometimes called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;nofollow&quot;&gt;Hill Sphere&lt;/a&gt;.  They're calculated differently but they define pretty much the same idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual long term stable region is somewhere around 50% of the Hill Sphere.   A gas giant like Jupiter simply doesn't have a large enough stable region that it has gravitational control of to have it's own Kuiper Belt or Oort Cloud equivalent.  Also, proximity to the sun and the gravitational stability of the trojan points is a factor.  What actually happens with a Sun-Jupiter system is that the stable regions are the &lt;a href=&quot;https://en.wikipedia.org/wiki/Jupiter_trojan&quot; rel=&quot;nofollow&quot;&gt;trojan points&lt;/a&gt; and (because of Saturn) the Cis-Resonance orbits but not trans-resonance (Saturn is too big and too close to Jupiter, so it disrupts trans-resonance).    Jupiter's crowded Trojan points and Hildas (mostly Cis-resonance) are a little bit like it's kuiper-belt equivalent and that's a result of the sun-jupiter system, the sun being dominant and jupiter, a thousandth the mass of the sun, but large enough to be dominant among the planets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth pointing out that at Jupiter distance from the sun, the solar system is quite a bit more crowded and has largely flattened out into a plane, unlike the Kuiper belt, so it's not a real comparison to the kuiper belt, but that's where orbiting objects tend to collect, cause those are the most stable regions.   The Kuiper belt and Oort cloud aren't really regions as they encompass the entire sphere around the sun at certain distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be possible, I would think, for a gas giant in the right situation to have something like it's own Kuiper belt and Oort Cloud, but it would need to be hugely distant from it's star and far from other large planets and preferably quite massive.   The limitations would still be no more than 1/2 of that planet's hill sphere, but it's theoretically possible, I would think, under the right circumstances.   Even so, such a situation, the Trojan points and trans and cis orbitals might still be more common than any possible kuiper belt or oort cloud equivalent, so I'm just speculating that it might be possible.   I'm not 100% sure.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-04-27T07:40:15.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14691" PostTypeId="2" ParentId="14686" CreationDate="2016-04-27T08:30:42.020" Score="3" Body="&lt;p&gt;Objects in orbit pass over the surface above what's commonly called their &lt;a href=&quot;https://en.wikipedia.org/wiki/Ground_track&quot; rel=&quot;nofollow noreferrer&quot;&gt;ground track&lt;/a&gt;. For objects in low earth orbit, though they go around the earth every 1.5 or more hours, the Earth rotates beneath them so they don't trace a simple &quot;circle&quot; over the same points on land.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/GPmQU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GPmQU.jpg&quot; alt=&quot;Ground track - NASA image from Wikipedia&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbit is around the Earth in a fixed plane, so we are likely (but not guaranteed) to see the satellite pass over 12 or 24 hours later.&lt;/p&gt;&#xA;" OwnerUserId="7491" LastEditorUserId="7491" LastEditDate="2016-04-27T09:33:29.430" LastActivityDate="2016-04-27T09:33:29.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14692" PostTypeId="2" ParentId="14684" CreationDate="2016-04-27T10:05:17.620" Score="17" Body="&lt;p&gt;In case anyone can't follow the links in my comment (above), here are the two pictures I mentioned. From: &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Solar_system_barycenter.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://web.archive.org/web/20150524071045/https://qph.is.quoracdn.net/main-qimg-118a1175c48f6ff2afd0183dd0040dd4?convert_to_webp=true&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/0/0c/Solar_system_barycenter.svg/463px-Solar_system_barycenter.svg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/0/0c/Solar_system_barycenter.svg/463px-Solar_system_barycenter.svg.png&quot; alt=&quot;Motion of the barycenter of the Solar System relative to the Sun&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/re77M.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/re77M.gif&quot; alt=&quot;track of the solar centre&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first claims to show the track of the solar system barycentre in the heliocentric reference frame. The outer yellow circle marks the photosphere of the Sun. The second plot claims to show the track of the centre of the Sun in the barycentric reference frame. The yellow circle shows the photosphere of the Sun to scale. As you can see, the plots are actually (almost) the same! Given that to go from one frame to the other is just a translation, I suppose they can both be right providing the x and y axes are defined appropriately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the questions posed: &quot;What does it look like&quot; - it looks like these two pictures. &quot;How large is it?&quot; As you can see, the maximum separation between the barycentre and the solar centre appears to be about 2 solar radii over the timescale covered by these plots, but is as small as a tenth of a solar radius (e.g. in 1950). &quot;How elliptic?&quot; Not at all really, it is a complicated superposition caused by the orbits mainly of Jupiter and Saturn, but all the planets contribute to a greater or lesser extent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The barycenter is calculated from the instantaneous positions of all the discrete masses in the solar system. I do not know for sure, but I assume that it includes all of the planets, and that everything else is negligible at the scale of the thickness of the line.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="24837" LastEditDate="2018-11-12T20:21:05.880" LastActivityDate="2018-11-12T20:21:05.880" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="14693" PostTypeId="2" ParentId="11900" CreationDate="2016-04-27T11:42:07.847" Score="2" Body="&lt;p&gt;Long ago I was part of a group that flew high altitude balloons to study x-rays from pulsars etc. How much you detect and at what energy depends upon how much atmosphere is above your detector. Lower energy x-rays are absorbed more than higher energy x-rays. Realistically you won't detect anything except maybe the Sun at energies below 20 keV and at altitude below 25km. We aimed for altitudes near 40km. To detect high energy x-rays the cheapest method is to use a proportional counter. Not something you can buy off-the-shelf. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the Sun only emits a lot of high energy x-rays during solar flares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting experiment would be to fly a geiger counter see &lt;a href=&quot;http://earthtosky.net/&quot; rel=&quot;nofollow&quot;&gt;http://earthtosky.net/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10761" LastActivityDate="2016-04-27T11:42:07.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14694" PostTypeId="1" AcceptedAnswerId="14698" CreationDate="2016-04-27T13:04:06.750" Score="4" ViewCount="743" Body="&lt;p&gt;Don't hate on me if I am asking a very basic and straightforward thing. I have a few questions about black holes and neutron stars.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;What is the mass range (in terms of solar masses) for a main sequence star to end its life as . . .&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A) a neutron star?&lt;/li&gt;&#xA;&lt;li&gt;B) a black hole?&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Is it there a (practically observed or proven) possible method for a main sequence star to form a neutron star (or black hole) at the end of its life cycle &lt;em&gt;without undergoing through the process of supernova&lt;/em&gt;?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, please explain or guide me some article on this matter.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;What is the mass range of a main sequence star to end up as a pair instability supernova?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the range of pair instability supernova overlaps with that of neutron star or black hole forming supernovae, how do we determine what type of end a star would have?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11657" LastEditorUserId="2153" LastEditDate="2016-04-27T22:28:52.113" LastActivityDate="2017-04-20T00:15:00.160" Title="Stellar mass limits for Neutron Star and Black Holes" Tags="&lt;black-hole&gt;&lt;supernova&gt;&lt;stellar-evolution&gt;&lt;mass&gt;&lt;neutron-star&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14695" PostTypeId="2" ParentId="14675" CreationDate="2016-04-27T13:34:47.957" Score="1" Body="&lt;p&gt;As Andy mentioned, it seems far too slow to be any space debris, and also a bit slow to be a plane. I feel like it might have been a lense flare from the sun. Try setting the camera up in the same spot for another video and see what happens :)&lt;/p&gt;&#xA;" OwnerUserId="11525" LastActivityDate="2016-04-27T13:34:47.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14696" PostTypeId="2" ParentId="11999" CreationDate="2016-04-27T14:04:15.187" Score="2" Body="&lt;p&gt;A Celestron Nexstar 130 SLT is not suitable for spectroscopy and the only photography that it will be mush use for would be afocal or DSLR imaging of the Moon (or Sun with a suitable filter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more serious work you will need a equatorial mounted scope with a better focuser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exactly which one to choose will depend of your budget and what you want to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you say the computer does not work can you be more specific.&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-04-27T14:04:15.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14698" PostTypeId="2" ParentId="14694" CreationDate="2016-04-27T15:26:52.760" Score="9" Body="&lt;p&gt;A succinct summary of supernova types is given in the following image based on &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/375341/pdf;jsessionid=58FD0BF8C78226B96F7F4220EA0F8935.c3.iopscience.cld.iop.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heger et al. (2003)&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/1/14/Supernovae_as_initial_mass-metallicity.svg&quot; alt=&quot;&quot;&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;Image courtesy of Wikipedia user Fulvio 314 under &lt;a href=&quot;https://creativecommons.org/licenses/by-sa/3.0/deed.en&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Creative Commons Attribution-Share Alike 3.0 Unported license&lt;/a&gt;. The graph is based on the graph in Fig. 1 of the linked paper.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pair instability realm is upwards of ~100 solar masses, though it is metallicity-dependent (Question 3). As Figure 1 (below) shows, neutron stars form in the mass range of &gt;9 solar masses - again, this is metallicity-dependent (Question 1a). Starting at around 25 solar masses, black holes will form (Question 1b).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/hwwkWf6.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is thought to be possible for a black hole to form without a supernova (see the section of the graph marked &quot;direct black hole&quot;). This has not been confirmed observationally, although there are some possibilities. I've written about this in more detail &lt;a href=&quot;https://astronomy.stackexchange.com/a/20787/2153&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2017-04-20T00:15:00.160" LastActivityDate="2017-04-20T00:15:00.160" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14699" PostTypeId="2" ParentId="14678" CreationDate="2016-04-27T19:08:02.710" Score="2" Body="&lt;ul&gt;&#xA;&lt;li&gt;Your equations seem correct. Note, you can also get $v$ without having to differentiate, from $E=\gamma mc^2$ and $p=\gamma mv$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here are some notes on relativistic fluids as related to stellar interiors, including derivation of these equations and their solutions in the non- and ultra-relavistic limits &lt;a href=&quot;http://www.jb.man.ac.uk/~smao/starHtml/equationState.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.jb.man.ac.uk/~smao/starHtml/equationState.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Correct, the integrals cannot be solved analytically (only in the two limits) so there is no general analytic formula&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11654" LastEditorUserId="11654" LastEditDate="2016-04-27T23:28:24.263" LastActivityDate="2016-04-27T23:28:24.263" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14700" PostTypeId="1" CreationDate="2016-04-27T22:30:16.090" Score="1" ViewCount="464" Body="&lt;p&gt;first post here.  As I'm new, StackExchange won't let create or use the wcsaxes tag.  wcsaxes looks like the most appropriate tool for the job, but astropy is closely related.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the title says it all, but I'll give a little more detail.  I have a bunch of sources in (RA, Dec) and want to plot them in the simplest possible projection (i.e. square, but if this is not possible we can make allowances).  I want to see the geometry of the Earth over my region of interest, mostly to identify the Earth's magnetic field lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following code gets me close, but I get this error:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;AttributeError: 'NoneType' object has no attribute 'to_geodetic'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If I change &quot;altaz&quot; to &quot;galactic&quot;, I get a Galactic coordinates grid over the points, which is what I want, but obviously in the wrong coordinate frame.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#!/usr/bin/env python2&#xA;&#xA;import numpy as np&#xA;from astropy.wcs import WCS&#xA;from astropy.time import Time&#xA;import matplotlib.pyplot as plt&#xA;&#xA;&#xA;# time = Time(2606629, format=&quot;jd&quot;, location=(&quot;116.670810d&quot;, &quot;-26.756528d&quot;)).iso&#xA;w = WCS(naxis=2)&#xA;w.wcs.ctype = [&quot;RA---MER&quot;, &quot;DEC--MER&quot;]&#xA;# w.wcs.dateavg = time&#xA;&#xA;ra_min = 0&#xA;ra_max = 15&#xA;dec_min = -45&#xA;dec_max = -15&#xA;N = 1000&#xA;&#xA;sim_ra  = np.random.uniform(ra_min, ra_max, size=N)&#xA;sim_dec = np.random.uniform(dec_min, dec_max, size=N)&#xA;&#xA;fig = plt.figure()&#xA;ax = fig.add_axes([0.1, 0.1, 0.9, 0.9], projection=w)&#xA;overlay = ax.get_coords_overlay('altaz')&#xA;&#xA;overlay[0].set_ticks(color='white')&#xA;overlay[1].set_ticks(color='white')&#xA;overlay[0].set_axislabel('Longitude')&#xA;overlay[1].set_axislabel('Latitude')&#xA;overlay.grid(color='black', linestyle='solid', alpha=0.5)&#xA;&#xA;plt.scatter(sim_ra, sim_dec)&#xA;plt.xlabel('RA')&#xA;plt.ylabel('Dec')&#xA;plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I played a little with trying to get the observation time into the WCS header (note that the actual time is artificial, but should work regardless), without success.  Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="11660" LastActivityDate="2016-04-27T22:30:16.090" Title="Plot an AltAz grid over a square grid of RADec points" Tags="&lt;python&gt;&lt;astropy&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14701" PostTypeId="2" ParentId="14683" CreationDate="2016-04-28T08:11:47.377" Score="4" Body="&lt;p&gt;The whole point of a pin hole projector is that it doesn't need any optics just a pin hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are after magnification then that is related to the distance you project the solar image behind the pin hole, with a projection distance of 1000mm the solar disc will have a diameter of about 9mm. Increasing the distance will increase the size of the image but also reduce its brightness. Increasing the size of the pin hole will increase the brightness but reduce the sharpness of the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose you could use a handheld magnifying glass to look at a small image but would not personally recommend it as unless you are extremely careful there is always a risk of fire or even blinding yourself or someone else when using a lens near an unfiltered solar observation. There is a famous story of an experienced astronomer setting his beard on fire because he left the cap off his telescopes finder. &lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-04-28T08:11:47.377" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14702" PostTypeId="2" ParentId="13933" CreationDate="2016-04-28T10:07:09.910" Score="2" Body="&lt;p&gt;As your ear is extremely small compared to the wave length of the gravity waves, remember they travel at the speed of light not the speed of sound. The structures in the ear would not respond, also the ear would change in size (by a  ridiculously small amount) itself, thus nullifying the effect anyway. LIGO works by comparing the light travel time of two laser beams at precise right angles, your ear is not directional like that.&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-04-28T10:07:09.910" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14703" PostTypeId="2" ParentId="13112" CreationDate="2016-04-28T14:18:23.767" Score="2" Body="&lt;p&gt;Perhaps I have always been misunderstanding the concept of space weather, rain from space now that's a cool idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seriously though, the solar wind particles are at such a a high temperature that molecules cannot form, instead it is an ionised gas containing mainly Hydrogen and Helium and only very tiny traces of other elements.&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-04-28T14:18:23.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14704" PostTypeId="2" ParentId="6379" CreationDate="2016-04-28T14:33:34.827" Score="1" Body="&lt;p&gt;If you can provide examples of numerical methods in GR you've seen/heard of that would help focus the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the article you linked to: &quot;The technique keeps track of a vast number of quarks and gluons by describing the space and time inside a proton with a set of points that make up a 4D lattice&quot;. This almost gets to the main issue with Numerical Relativity. There is no natural computational grid on which to simulate space-time. The whole game with GR is that gravity is space-time so first you have to simulate the space-time and then you have to simulate the objects (neutron stars, black holes, gravitational waves) on top. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the links below go into, its very difficult to create a consistent computational grid since the physical space-time your trying to simulate for a black hole has &quot;funny&quot; things in it like singularities, or an event horizon pas which we can't really know what's going on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this article: &lt;a href=&quot;http://astronomy.com/magazine/2016/02/putting-einstein-to-the-test?page=1&quot; rel=&quot;nofollow&quot;&gt;http://astronomy.com/magazine/2016/02/putting-einstein-to-the-test?page=1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;does a good job of summing up the field, and its quite accessible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For something more rigorous please see: &lt;a href=&quot;http://arxiv.org/pdf/1010.5260v2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/1010.5260v2.pdf&lt;/a&gt;&#xA;That paper gets into some of the math behind the article linked to above.&lt;/p&gt;&#xA;" OwnerUserId="11665" LastActivityDate="2016-04-28T14:33:34.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14705" PostTypeId="2" ParentId="13933" CreationDate="2016-04-28T17:38:08.590" Score="1" Body="&lt;p&gt;Gravity would warp your entire body, or at least tug on it, because &lt;em&gt;bodies&lt;/em&gt; that aren't bound together by gravity largely resist it's deformation, at least that's the major point I'm working on here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, it should pull on some structure in your ear, be it the bones or the tiny Philae (I believe that's the right word, edit if not), either way it results in a deviation from their original position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on proximity you might also be able to feel it, in likeness to standing in front of a large speaker. However the ears would be far more sensitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope I've answered your question :)&lt;/p&gt;&#xA;" OwnerUserId="11525" LastActivityDate="2016-04-28T17:38:08.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14706" PostTypeId="2" ParentId="14407" CreationDate="2016-04-28T19:43:02.687" Score="2" Body="&lt;p&gt;The comments from @userLTK, and @Lacklub are correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets assume there is an object of radius $R$ and mass $M$, from a Newtonian point of view, if you are at another radius $r$, such that $r &amp;gt; R$, then there is no difference in the gravitational field  experience by an object at $r$ if the mass spread across a shell of radius $R$ or if its concentrated anywhere between $r=0$ and $r=R$. GR doesn't do much to change this, and in fact if $r &amp;gt;&amp;gt;R$, then the result is of course exactly the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to black holes, all taken from from Wiki &lt;a href=&quot;https://en.wikipedia.org/wiki/Black_hole&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Black_hole&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Physical Properties&lt;/strong&gt;&#xA;&quot;The simplest static black holes have mass but neither electric charge nor angular momentum...This means that there is no observable difference between the gravitational field of such a black hole and that of any other spherical object of the same mass.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically if you're outside the event horizon (not that'd you know where it was) you're experience with the black hole is the same as with a planet or star of that mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again from same wiki article:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Singularity&lt;/strong&gt;&#xA;&quot;At the center of a black hole, as described by general relativity, lies a gravitational singularity, a region where the spacetime curvature becomes infinite... It can also be shown that the singular region contains all the mass of the black hole solution&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;it can be shown&quot; references page 204 of Carroll, Sean M. (2004). Spacetime and Geometry. I don't have my copy with me right now so I can't look it up, but I would say I remember reading at in Carroll back in the day. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, let me just add that, once something passes into the event horizon of a black hole, there's no getting back. So we really have to ask ourselves how any information about the &quot;stuff&quot; inside the event horizon would come to us? I like your idea of getting some indirect evidence and perhaps gravitational waves will shed some light on this topic but I don't suppose there is any known way to getting direct access to anything beyond the event horizon.&lt;/p&gt;&#xA;" OwnerUserId="11665" LastEditorUserId="11665" LastEditDate="2016-05-06T13:31:39.997" LastActivityDate="2016-05-06T13:31:39.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14707" PostTypeId="1" AcceptedAnswerId="14719" CreationDate="2016-04-29T04:06:17.057" Score="0" ViewCount="85" Body="&lt;p&gt;Why do we need 7 polar satellites for a navigation system like the &lt;a href=&quot;http://www.isro.gov.in/irnss-programme&quot; rel=&quot;nofollow&quot;&gt;Indian Regional Navigation Satellite System (IRNSS)&lt;/a&gt;?  &lt;/p&gt;&#xA;" OwnerUserId="2948" LastEditorDisplayName="user1569" LastEditDate="2016-04-30T21:53:49.870" LastActivityDate="2016-04-30T21:53:49.870" Title="Polar satellites for Global Navigation Systems" Tags="&lt;satellite&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14708" PostTypeId="2" ParentId="14707" CreationDate="2016-04-29T09:55:56.370" Score="1" Body="&lt;p&gt;The satellites used in Indian Navigational System or so called Indian Regional Navigation Satellite System are not polar satellites.Three of them are Geostationary satellites and other four are Geosynchronous satellites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polar satellites revolve around the earth in North-South direction, where as these satellites revolve around with the same speed as that of the earth's rotation so that they appear to be fixed on sky for observers on earth. This is exactly what is needed for navigation systems.&lt;/p&gt;&#xA;" OwnerUserId="5422" LastActivityDate="2016-04-29T09:55:56.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14711" PostTypeId="1" AcceptedAnswerId="14713" CreationDate="2016-04-29T16:22:12.973" Score="7" ViewCount="278" Body="&lt;p&gt;I will soon finish a master's degree in computer science. I have been interested in astronomy for some time, which has included playing around with a Newtonian reflector telescope and reading about CCD-cameras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I checked some local universities, and was amazed that they have only 10 or so graduates each year (1-2 get a PhD and the rest a master's degree). How many jobs are there for these graduates? Compared to other fields, how tough is the competition in astronomy? Is it different for some relatively new areas, e.g. space neuroscience or astrobiology?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am considering to apply for an astronomy PhD, so any input would be helpful.&lt;/p&gt;&#xA;" OwnerUserId="6991" LastEditorUserId="6991" LastEditDate="2016-04-29T16:30:03.017" LastActivityDate="2016-04-29T17:30:06.920" Title="How much competition there is for jobs in astronomy compared to other fields of science?" Tags="&lt;telescope&gt;&lt;astrobiology&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14712" PostTypeId="1" CreationDate="2016-04-29T17:27:40.090" Score="1" ViewCount="636" Body="&lt;p&gt;Some meteorites just blaze away, streaking through the sky. This is clearly explained, as they either slow down enough for the flame to cease, or ablate in their entirety, turning to vapor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But some, like the Chelyabinsk meteorite, explode violently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What effect is responsible for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can it be that with their irregular surface acting as a turbine they enter a spin of increasing speed, and once the centrifugal force reaches a certain level, it causes the meteorite to break apart rapidly, hurling parts of it in all directions at speeds far exceeding its speed so far?&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2016-05-02T07:31:15.427" Title="Why do meteorites explode?" Tags="&lt;meteorite&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14713" PostTypeId="2" ParentId="14711" CreationDate="2016-04-29T17:30:06.920" Score="7" Body="&lt;p&gt;The competition for permanent positions in astronomy is very tough.  The field as a whole produces roughly ~200 Ph.Ds per year, but there are usually only a handful (say ~10) tenure-track positions that open up every year.  So perhaps ~5% of Ph.Ds end up in tenure-track positions in astronomy.  There are more permanent positions in astronomy that aren't tenure track positions, but not too many more, probably enough to support about ~20% of Ph.Ds.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for non-permanent positions in astronomy, there are plenty of those.  The funding situation for postdoctoral positions is such that, while there is still a great deal of competition for open positions (stemming partially from the fact that almost everyone applies to almost every open job --- I exaggerate, but only slightly), there are enough open positions, that almost all astronomy Ph.Ds who want a postdoc will be able to get one.  But those positions only last for an average of three years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the job prospects for astronomy Ph.Ds who decide to leave the field of astronomy (like myself) are quite good.  Those holding an astronomy degree have one of the lowest unemployment rates of anyone &lt;a href=&quot;http://money.cnn.com/2013/01/04/news/economy/jobs-lowest-unemployment/&quot;&gt;(around 0.3%)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a few links to some papers on the state of the job market in astronomy:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/0712.2820&quot;&gt;http://arxiv.org/abs/0712.2820&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/0805.2624&quot;&gt;http://arxiv.org/abs/0805.2624&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/1512.02223&quot;&gt;http://arxiv.org/abs/1512.02223&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10371" LastActivityDate="2016-04-29T17:30:06.920" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14714" PostTypeId="2" ParentId="14712" CreationDate="2016-04-29T18:07:13.483" Score="2" Body="&lt;p&gt;The Chelyabinsk meteor didn't explode.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact meteors in general usually don't explode, this is a common misconception.&lt;br&gt;&#xA;When an object at orbital speeds ($\sim11 km/s$) enters the atmosphere, this speed corresponds to extremely high mach-numbers (M$\sim 30$). This generates a strong shock-front in front of the meteor which causes the meteor to heat up enourmously.&lt;br&gt;&#xA;To give a feeling of how important this heat-up is, you can look up on &lt;a href=&quot;https://en.wikipedia.org/wiki/Hypersonic_speed&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; that already at Mach-numbers of 2-3 there are severe engineering problems in not letting aircraft overheat. At Mach-numbers of 30, the air in the shock is so hot that it locally ionizes the air and easily evaporates the solids away from the falling meteor. Upon reaching the ground one hears this shock as loud thunder.&lt;br&gt;&#xA;The other thing that happened with the Chelyabinsk meteor is that it broke up in mid-flight. This caused the total solid surface of the meteor that is exposed to the shock to momentarily increase strongly. More surface exposure to the heat causes more surface to glow. This lasts until the smaller fragments that resulted from the breakup evaporate themselves.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taken together, the mach-shock and the increase in glowing surface area upon break-up give the impression of an &quot;explosion&quot;.&lt;br&gt;&#xA;Also for an explosion a reservoir of energy would need to be suddenly tapped, which is just not there.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2016-04-29T18:07:13.483" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="14715" PostTypeId="1" AcceptedAnswerId="14717" CreationDate="2016-04-29T19:01:38.150" Score="4" ViewCount="205" Body="&lt;p&gt;The Earth is a very smooth sphere, and the Sun even more so, with only minor fluctuations. I am wondering: are larger stars even rounder? Intuitively, that seems self evident, but I am not so sure. For instance, the hydrostatic equilibrium causes larger stars to be much less dense than red dwarfs. So what is the most important factor for how round a star is, a higher mass, or less activity? The most prominent cause of irregularities is of course the rotation rate of the star, which is pretty much independent of size. Ignoring that, do larger stars have smaller a deviations from the ellipsoid relative to their size?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it seems like the &quot;other than rotation rate&quot; criterion is not really meaningfull, I now terminate it.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="10172" LastEditDate="2016-04-30T06:44:20.063" LastActivityDate="2020-12-15T00:35:58.733" Title="Are larger stars rounder?" Tags="&lt;star&gt;&lt;hydrostatic-equilibrium&gt;&lt;stellar-structure&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14717" PostTypeId="2" ParentId="14715" CreationDate="2016-04-29T22:45:10.767" Score="4" Body="&lt;p&gt;In terms of mean angular velocity, the distribution of rotation rates among main sequence stars is well known. Allen (1963) compiled data on mass, radius, and equatorial velocity, which was then expanded upon by &lt;a href=&quot;http://adsabs.harvard.edu/abs/1965Obs....85..166M&quot; rel=&quot;nofollow noreferrer&quot;&gt;McNally (1965)&lt;/a&gt;, who focused on angular velocity and angular momentum. It became clear that angular velocity increases from low rates for spectral types of G and below before rising to a peak around type A stars and then slowly decreasing.&lt;/p&gt;&#xA;&lt;p&gt;Equatorial velocity continues increasing to mid-B type stars, before slowly decreasing, but because of the increased radii of O and B type main sequence stars, the peak in angular velocity occurs before this. As part of &lt;a href=&quot;http://catdir.loc.gov/catdir/samples/cam032/99016740.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jean-Louis Tassoul's &lt;em&gt;Stellar Rotation&lt;/em&gt;&lt;/a&gt; notes, many O type stars have rotational periods similar to that of the G-type stars like the Sun!&lt;/p&gt;&#xA;&lt;p&gt;The distribution is not smooth and uniform (McNally noticed a strange discontinuity in angular momentum per unit mass right for A0 and A5 stars; see his Figure 2); &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0303631&quot; rel=&quot;nofollow noreferrer&quot;&gt;Barnes (2003)&lt;/a&gt; observed two distinct populations in open clusters, consisting of slower rotators (the I sequence) and faster rotators (the C sequence). Stars may migrate from one sequence to another as they evolve. Interestingly enough, stars on the I sequence lose angular momentum &lt;span class=&quot;math-container&quot;&gt;$J$&lt;/span&gt; faster than stars on the C sequence:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\mathrm{d}J}{\mathrm{d}t}\propto-\omega^n,\quad\text{where}\begin{cases}&#xA;    n=3\text{ on the I sequence}\\&#xA;    n=1\text{ on the C sequence}\\&#xA;  \end{cases}$$&lt;/span&gt;&#xA;Here, of course, &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; is angular velocity. These results obey &lt;a href=&quot;http://adsabs.harvard.edu/abs/1972ApJ...171..565S&quot; rel=&quot;nofollow noreferrer&quot;&gt;Skumanich's law&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Oblateness can be determined from mass, radius, and angular velocity &lt;a href=&quot;https://en.wikipedia.org/wiki/Equatorial_bulge&quot; rel=&quot;nofollow noreferrer&quot;&gt;as&lt;/a&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$f=\frac{5\omega^2R^3}{4GM}$$&lt;/span&gt;&#xA;Using this and McNally's data, some quick calculations get me the following table:&#xA;|Spectral type|&lt;span class=&quot;math-container&quot;&gt;$f/f(O5)$&lt;/span&gt;|&#xA;|--|-------|&#xA;|O5 | 1    |&#xA;|B0 | 1.28 |&#xA;|B5 | 1.84 |&#xA;|A0 | 1.67 |&#xA;|A5 | 1.35 |&#xA;|F0 | 0.482|&#xA;|F5 | 0.0387|&#xA;|G0 | 0.000314|&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2365" LastEditDate="2020-12-15T00:35:58.733" LastActivityDate="2020-12-15T00:35:58.733" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="14718" PostTypeId="1" CreationDate="2016-04-29T23:59:31.857" Score="2" ViewCount="485" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BfYBZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BfYBZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took an image of Jupiter through my 8&quot; Dobsonian Telescope, attaching a DSLR and a 1.25&quot; Barlow Lens where the eyepiece goes, as shown in this video: &lt;a href=&quot;https://www.youtube.com/watch?v=reFxoF3XoaU&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=reFxoF3XoaU&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Through numerous online sources, I learnt that to find the angle of view of my image, the magnification of my image needed to be calculated, and that this value along with the given field view for my eyepiece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.rocketmime.com/astronomy/Telescope/Magnification.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.rocketmime.com/astronomy/Telescope/Magnification.html&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above website answers the following question, the calculations of which I attempted to mimic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My first telescope was a Meade 6600 -- they don't make it any more -- it's a 6-inch f/5 Newtonian scope. It came with a 25mm eyepiece. So... what was the magnification I was getting with this scope?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here are the calculations I did in hopes of getting the above result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\textrm{Diameter} = 8'' = 203.2\ \textrm{mm}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{\textrm{ratio}} = \frac{\textrm{focal length of objective}}{203.2\ \textrm{mm}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\therefore \textrm{focal length of objective} = 203.2 \cdot 5.9 = 1200\ \textrm{mm}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\textrm{Magnification} = \frac{\textrm{focal length of objective}}{\textrm{focal length of eyepiece}} = \frac{1200}{x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As shown in the video (the first link above), I didn't use an eyepiece to take my picture - I used a barlow lens, a couple of adapters, and a DSLR. So at this point, I am not sure what value to use for the &quot;focal length of the eyepiece.&quot; How can I proceed to calculate the magnification?&lt;/p&gt;&#xA;" OwnerUserId="9650" LastEditorUserId="2365" LastEditDate="2016-04-30T08:03:33.517" LastActivityDate="2021-03-21T11:11:30.170" Title="Magnification of a telescope" Tags="&lt;optics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14719" PostTypeId="2" ParentId="14707" CreationDate="2016-04-30T00:28:01.117" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://www.isro.gov.in/irnss-programme&quot; rel=&quot;nofollow&quot;&gt;IRNSS&lt;/a&gt;&#xA;is intended to serve an area spanning 100 degrees in longitude by 80 degrees in latitude.&#xA;Anywhere in that service area, at any time, a receiver needs to compare distances to four of the satellites.&#xA;To determine latitude, the satellites can't all be above the equator, so some of the orbits are inclined 29$^\circ$. Those satellites oscillate north-south in pairs. Polar orbits aren't needed; global systems serve the poles with orbits inclined only 55$^\circ$.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-04-30T00:28:01.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14720" PostTypeId="2" ParentId="10875" CreationDate="2016-04-30T02:36:29.730" Score="1" Body="&lt;p&gt;In principle a &lt;a href=&quot;https://en.wikipedia.org/wiki/Fiberscope&quot; rel=&quot;nofollow&quot;&gt;fiberscope&lt;/a&gt; eyepiece could work but not very well.&#xA;By putting one end of the cable flush with the field stop of an eyepiece (few allow this) and the other end exactly in the telescope's focal plane, you'd get a fiber-pixelated image.&#xA;A binoviewer would require two cables, difficult to align.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-04-30T02:36:29.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14722" PostTypeId="2" ParentId="14718" CreationDate="2016-04-30T03:59:41.633" Score="6" Body="&lt;p&gt;The idea of magnification is not relevant to astro-photography, what is relevant is the image scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image scale depends on the focal length of the objective and the size of the sensor. The Barlow effectively modifies the focal length, so if you have a $B\times$ Barlow and an objective of focal length $f_O$ the effective focal length is $f_e=B\times f_O$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your sensor has side of length $l$ then the field of view that corresponds to that side is (in radian) $\theta_{rad}=l/f_e$ or in degrees $\theta_{deg}=(l/f_e)\times(180/\pi)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if your sensor has a short side length of $\approx 0.015 \mbox{m}$ and a the telescope has a focal length of $1.2 \mbox{m}$ and you are using a $\times2$ Barlow you have the short side of the image corresponds to a field of view of $\approx 21$ minutes of arc.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2016-05-02T16:54:19.143" LastActivityDate="2016-05-02T16:54:19.143" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="14723" PostTypeId="1" CreationDate="2016-04-30T07:09:01.273" Score="1" ViewCount="11484" Body="&lt;p&gt;I am a beginner.  Please let me know whether earth has retrograde motion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it does not, let me know the reason&lt;/p&gt;&#xA;" OwnerUserId="11000" LastActivityDate="2016-04-30T08:14:20.517" Title="Does earth have retrograde motion?" Tags="&lt;earth&gt;&lt;planet&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14724" PostTypeId="2" ParentId="14723" CreationDate="2016-04-30T07:56:06.547" Score="3" Body="&lt;p&gt;It does not have a retrograde motion. It has a prograde rotation, and a prograde orbit of the Sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to look down on the solar system from far above the north pole you would see the planets orbiting the sun counter-clockwise. And the planets, including the earth also spinning counter clockwise. The Moon also orbits the Earth counter-clockwise.  All this motion is in the same direction, and called prograde.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a couple of exceptions: Venus spins clockwise. Venus spins in a retrograde motion. There is a moon of Neptune, Triton, that orbits retrograde. Uranus spins on its side. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is that most objects rotate the same way is that the planets and the sun formed out of the same cloud of gas and so have the same direction of spin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another sense of &quot;retrograde&quot;. As the Earth overtakes the other planets they seem to move backwards in the sky for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the inner planets, Earth would appear to move rather like mars, with periods of retrograde motion. From the outer planets Earth would move like Venus, moving from being a &quot;morning star&quot; to an &quot;Evening star&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-04-30T08:14:20.517" LastActivityDate="2016-04-30T08:14:20.517" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14725" PostTypeId="1" CreationDate="2016-04-30T08:03:58.657" Score="-7" ViewCount="1510" Body="&lt;p&gt;According to Michio Kaku, in his book Hyperspace, he said that in this Hyperspace, there exist a total of 10 dimensions that govern the laws of physics. Is it possible that we too live in a Hyperspace, and if so, would gravity be the 4th dimension that would govern the laws of physics we are yet to discover?&lt;/p&gt;&#xA;" OwnerUserId="6601" LastActivityDate="2016-04-30T16:19:13.030" Title="Could gravity be the 4th dimension?" Tags="&lt;gravity&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2016-04-30T21:53:13.087" ContentLicense="CC BY-SA 3.0" />
  <row Id="14726" PostTypeId="2" ParentId="14725" CreationDate="2016-04-30T16:19:13.030" Score="3" Body="&lt;p&gt;In &lt;a href=&quot;https://en.wikipedia.org/wiki/String_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;some models of quantum gravity&lt;/a&gt; there are &lt;a href=&quot;https://en.wikipedia.org/wiki/String_theory#Extra_dimensions&quot; rel=&quot;nofollow noreferrer&quot;&gt;10 dimensions&lt;/a&gt;. 7 of these are curled up and very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/14148/how-does-gravity-really-work?rq=1&quot;&gt;Gravity&lt;/a&gt; is a force it is not a dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If these models of quantum gravity is correct then we do live in these 10 dimension. If these models are correct then we do live in &quot;hyperspace&quot;, but only 3 dimensions are macroscopic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dimensions don't govern the laws of physics.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-04-30T16:19:13.030" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14728" PostTypeId="1" AcceptedAnswerId="14755" CreationDate="2016-05-01T00:39:12.400" Score="0" ViewCount="650" Body="&lt;p&gt;With every improvement of measurement we are able to determine the age of the Universe more exactly. I just wondered how much and what means it would need to narrow the range of age so that it would be possible to determine the exact date of the Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that this would take a lot of assumptions for the sake of simplicity - I am just curious; this being said: let us assume that all current theories hold and no new side effects or new theories would evolve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read &lt;a href=&quot;https://astronomy.stackexchange.com/questions/676/age-of-the-universe&quot;&gt;this post&lt;/a&gt; and some physics articles about the matter, but my question focuses on the accuracy gap between our current best telescopes and the accuracy that would be needed to narrow down the age to &quot;24 earth hours&quot;.&lt;/p&gt;&#xA;" OwnerUserId="11683" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-05-03T04:38:17.610" Title="Birthday of the Universe" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14729" PostTypeId="1" AcceptedAnswerId="14731" CreationDate="2016-05-01T01:39:58.270" Score="0" ViewCount="118" Body="&lt;p&gt;I was reading the abstract of a paper regarding attempts at detection of a molecular precursor to the amino acid glycine, and I came across concepts (and notation) that I found confusing. If someone could describe in layman's terms (I've taken an astrophysics class so I'm not totally unaware) that would be great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the question I refer to the following section of a paper's abstract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;we derive &lt;strong&gt;upper limit column densities&lt;/strong&gt; of NH$_2$OH of $\leq 1.4 \times 10^{13} \mathrm{cm}^{-2}$&#xA;  and $\leq 1.5 \times 10^{13} \mathrm{cm}^{-2}$ toward the B1 and B2 shocks, respectively, and&#xA;  upper limit relative abundances of $N_\mathrm{NH_2OH}/N_\mathrm{H_2}$&#xA;  $\leq 1.4 \times 10^{-8}$ and $\leq 1.5 \times 10^{-8}$, respectively.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The paper in question is called &lt;em&gt;CSO and CARMA Observations of &lt;a href=&quot;https://en.wikipedia.org/wiki/L1157&quot; rel=&quot;nofollow&quot;&gt;L1157&lt;/a&gt;. I. A Deep Search for Hydroxylamine (NH$_2$OH)&lt;/em&gt;, and located &lt;a href=&quot;https://ui.adsabs.harvard.edu/#abs/2015ApJ...812...76M/abstract&quot; rel=&quot;nofollow&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11031" LastEditorUserId="616" LastEditDate="2016-05-01T05:00:00.050" LastActivityDate="2016-05-01T07:09:13.560" Title="What does this information about &quot;upper limit column densities&quot; and &quot;upper limit relative abundances&quot; mean?" Tags="&lt;astrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14731" PostTypeId="2" ParentId="14729" CreationDate="2016-05-01T07:09:13.560" Score="2" Body="&lt;p&gt;A column density is just the integral of a (number) density along the line of sight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often, you would get the same observational absorption or emission signature from a number density of $n$ molecules cm$^{-3}$, spread along a line of sight through a cloud of diameter $x$, as you would with a density of $0.1n$ spread over $10x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, all we can infer is the product or, more accurately, the integral of $n\ dx$, and this is known as the column density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the paper you refer to, I doubt they know the thickness of the shock they are investigating, so a column density is all that can be determined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relative abundance is more straighforward; it is just the ratio of the number densities of the two labelled species.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-05-01T07:09:13.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14733" PostTypeId="1" AcceptedAnswerId="14734" CreationDate="2016-05-01T11:56:00.690" Score="19" ViewCount="2888" Body="&lt;p&gt;I'm curious to know what the sky would look like without any other galaxies out there. How much do other galaxies factor into the stars we see? Does the Milky Way account for most of them? Would the night sky look normal? Or would it be very empty?&lt;/p&gt;&#xA;" OwnerUserId="11693" LastActivityDate="2016-05-01T18:19:46.953" Title="What would the night sky look like if the Milky Way were the only galaxy in the universe?" Tags="&lt;fundamental-astronomy&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14734" PostTypeId="2" ParentId="14733" CreationDate="2016-05-01T13:06:45.003" Score="23" Body="&lt;p&gt;I feel sure this is a repeat, but couldn't immediately find it. The only things in the night sky we can see (with the naked eye) that are &lt;em&gt;not&lt;/em&gt; part of our own Galaxy are (on a good night) the Andromeda galaxy and the Large and small Magellanic clouds. Every individual star brighter than $V=6$ and visible to the naked eye is in the Milky Way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it would hardly look any different.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-05-01T13:06:45.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14735" PostTypeId="2" ParentId="14557" CreationDate="2016-05-01T15:03:12.977" Score="0" Body="&lt;p&gt;Do you have a star chart to allow you to star hop from a known star to smaller stars to your target? Most messier objects are not naked eye visible and with the small field of view of a telescope, unless you know where to look, you're not likely to find something by blindly hunting around. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need dark skies, as most DSOs are small and dim. &lt;/p&gt;&#xA;" OwnerUserId="912" LastActivityDate="2016-05-01T15:03:12.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14737" PostTypeId="2" ParentId="14733" CreationDate="2016-05-01T18:19:46.953" Score="18" Body="&lt;p&gt;It wouldn't be too apparent, but there are a few objects that you can see in good viewing conditions with the naked eye that would disappear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here they are in order of brightness. I marked the objects in a reddish color.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Large_Magellanic_Cloud&quot; rel=&quot;noreferrer&quot;&gt;Large Magellanic Cloud&lt;/a&gt;, apparent magnitude 0.9, located in the constellation Dorado. Only visible from the southern hemisphere.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/EZqpI.png&quot; alt=&quot;Large Magellanic Cloud&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Small_Magellanic_Cloud&quot; rel=&quot;noreferrer&quot;&gt;Small Magellanic Cloud&lt;/a&gt;, apparent magnitude 2.7, located in the constellation Tucana. Only visible from the southern hemisphere.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Orabf.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Orabf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Andromeda_Galaxy&quot; rel=&quot;noreferrer&quot;&gt;Andromeda Galaxy&lt;/a&gt;, apparent magnitude 3.4, located in the constellation Andromeda.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/MQw0Q.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MQw0Q.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And that's it. Seriously. Sure, there are a few others, but they're exceedingly difficult to see. &lt;/p&gt;&#xA;" OwnerUserId="11031" LastActivityDate="2016-05-01T18:19:46.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14740" PostTypeId="1" AcceptedAnswerId="14747" CreationDate="2016-05-01T23:02:24.057" Score="6" ViewCount="845" Body="&lt;p&gt;The Chandrasekhar limit is the maximum mass of a stable white dwarf star. Beyond this, a carbon-oxygen white dwarf will typically explode in a type 1a supernova, due to the nuclear reactions at those temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard that oxygen-neon-magnesium white dwarfs, on the other hand, will not ignite. Rather, electron capture becomes energetically favorable and they become neutron stars. If I understand correctly, this also happens for carbon-oxygen white dwarfs in binaries, if the white dwarf has most of the mass in the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this? Can anyone walk me through, step by step, what happens if a white dwarf (of both compositions) exceeds the Chandrasekhar limit, and how each differs?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://arxiv.org/abs/astro-ph/9701225&quot;&gt;https://arxiv.org/abs/astro-ph/9701225&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2016-05-10T20:09:58.997" Title="Why does the Chandrasekhar limit affect white dwarfs differently?" Tags="&lt;supernova&gt;&lt;mass&gt;&lt;neutron-star&gt;&lt;white-dwarf&gt;&lt;degenerate-matter&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14742" PostTypeId="1" CreationDate="2016-05-01T23:25:19.363" Score="0" ViewCount="151" Body="&lt;p&gt;Is it POSSIBLE that we are a binary system? I have been doing research into this topic and it seems we may infact be a binary system. If we are, what effects would a massive body comming into our system be? And are these effects happening now?&lt;/p&gt;&#xA;" OwnerUserId="10954" LastActivityDate="2016-05-02T09:02:36.880" Title="Binary Star question" Tags="&lt;binary-star&gt;&lt;star-systems&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2016-05-02T14:39:11.300" ContentLicense="CC BY-SA 3.0" />
  <row Id="14744" PostTypeId="2" ParentId="14712" CreationDate="2016-05-02T07:31:15.427" Score="1" Body="&lt;p&gt;Meteorites don't explode. To understand them properly we need to know it's stages.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Meteoroid&lt;/strong&gt; &lt;em&gt;(small piece of rock travelling in outer space towards a planet or satellite)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Meteor&lt;/strong&gt; &lt;em&gt;(This is when meteoroid reaches the planets atmosphere and is travelling)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Meteorite&lt;/strong&gt; &lt;em&gt;(This is after the meteor has fallen on the surface)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As I mentioned above meteorites don't explode. But, when a meteor is big enough and is still travelling in atmosphere, it produces bright flame around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a common misconception that the flame is produces is due to friction of meteor with atmosphere. But, the actually when the meteor travels in atmosphere, it compresses the air below it. When gas is compressed heavily it produces incandescence and glows. This is what exactly happening here also.&lt;/p&gt;&#xA;" OwnerUserId="5422" LastActivityDate="2016-05-02T07:31:15.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14745" PostTypeId="2" ParentId="14740" CreationDate="2016-05-02T08:58:59.433" Score="3" Body="&lt;p&gt;There are a variety of white dwarfs with various compositions, and analysing how they detonate in a supernova (or not) is an topic under investigation. A simple model, described in &quot;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/12845/how-is-the-first-detonation-in-supernove-type-ia-triggered/12848&quot;&gt;How is the first detonation in Supernove type Ia triggered?&lt;/a&gt;&quot; is of a helium shell initially igniting and that setting off the carbon in the core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this type of Type 1a supernova there is no neutron star formed, as the star is completely destroyed. The mass at which a white dwarf will undergo a type 1a supernova is just below the chandrasekhar limit, so neutronisation won't occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in some white dwarfs with an atypical composition (as you note a Mg-Ne-O white dwarf) it is possible for the star to avoid detonation, and reach the chandrasekhar limit, and so for electron capture to occur, and a neutron star to form. It is worth noting that there is not definite observation of a white dwarf collapsing to a neutron star (where as there a lots of observations of type 1a supernovae) however these &quot;accretion induced collapse&quot; scenarios may explain some magnetars and short gamma ray burst &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/pdf/1512.00678v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Progenitors of the Accretion-Induced&#xA;Collapse&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the two scenarios are, &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Accretion occurs onto a Carbon-oxygen WD. The pressure and temperature in the core of a white dwarf increases until thermonuclear reactions begin (at about 1.38 solar masses). Since the white dwarf is degenerate it can't expand to reduce the rate of thermonuclear reactions, and the entire star detonates and is destroyed.&lt;/li&gt;&#xA;&lt;li&gt;Accretion occurs onto a ONeMg WD. The star reaches 1.44 solar masses, electron degeneracy is no longer sufficient to prevent collapse. Electron capture occurs and the star collapses to neutron star.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-05-02T08:58:59.433" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14746" PostTypeId="2" ParentId="14742" CreationDate="2016-05-02T09:02:36.880" Score="1" Body="&lt;p&gt;Since every object in the known universe which has mass has an infinite gravitational field every object in the universe affects one another. &#xA;Every star in our surrounding affects Our solar system but the relative proximity of the sun masks these effects to us. &#xA;However if the sun would be a part of a binary system of stars&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Our planet should have been affected by it because a star is a much larger mass when compared to the planets.&lt;/li&gt;&#xA;&lt;li&gt;Since the star and our sun interact with each other they would produce a noticeable wobble. &lt;/li&gt;&#xA;&lt;li&gt;Any such star would definitely be very close to the sun so as to be able to influence it and hence easily visible yet no evidence has been found for this. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So for now one can say that we and the sun are not part of any binary star system which has been (if it exists) undiscovered till recently&lt;/p&gt;&#xA;" OwnerUserId="11587" LastActivityDate="2016-05-02T09:02:36.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14747" PostTypeId="2" ParentId="14740" CreationDate="2016-05-02T09:36:00.440" Score="3" Body="&lt;p&gt;Whether a white dwarf responds to the accretion of material by exploding or collapsing depends on the competition between energy being released in fusion reactions and energy being locked away by endothermic electron capture (neutronisation) reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is thought that most white dwarfs of moderate mass have a C/O composition. They will need to accrete a lot of mass to get to a density (at about $4\times 10^{13}$ kg/m$^3$, reached at $1.38M_{\odot}$ in a non-rotating WD) where neutronisation becomes energetically feasible. It is possible, that before this happens, that fusion reactions are ignited (due to high density, rather than temperature). The threshold density for ignition is lower for nuclei with lower atomic number (He &amp;lt; C &amp;lt; O) and the ignition threshold densities for He and C are probably lower than the neutronisation threshold for C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a C/O WD that has accreted a lot of matter, ignition could take place in C at the core, or it could be triggered in He (at even lower densities) at the base of a deep accreted shell of material. The electron degenerate matter has a pressure that is independent of temperature, leading to runaway fusion and the complete destruction of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;O/Ne/Mg WDs are made as the final stages of more massive stars ($8-10M_{\odot}$) and are born as remnants with much higher mass $&amp;gt;1.2M_{\odot}$ than typical C/O WDs. More massive WDs are smaller, with higher density. The neutronisation thresholds for O, Ne, Mg are only $1.9\times10^{13}$, $6\times 10^{12}$ and $3\times 10^{12}$ kg/m$^3$ respectively (all lower than for C). This means that a O/Ne/Mg WD may have to accrete very little mass to reach this central density, begin neutronisation, which leads to collapse. In addition if such densities are insufficient to trigger C burning in a C/O WD, then they certainly won't be high enough to trigger burning in O/Ne/Mg because of stronger coulomb repulsion. Further, if little mass is accreted, then there won't be a deep envelope of accreted material in which to ignite burning off-centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all these reasons, O/Ne/Mg WDs may be more likely to collapse than explode (the collapse would cause a type of core-collapse supernova though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Actually looking at the paper you reference (which is a bit dated), although some of the numbers have changed slightly, the semi-quantitative argument I give above is exactly how it is explained there. So I'm not sure whether my answer helps you.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-05-10T20:09:58.997" LastActivityDate="2016-05-10T20:09:58.997" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14748" PostTypeId="1" AcceptedAnswerId="14749" CreationDate="2016-05-02T12:50:17.900" Score="1" ViewCount="121" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/q/11096/7982&quot;&gt;This question&lt;/a&gt; got me thinking of an interview of, or a talk by &lt;a href=&quot;https://en.wikipedia.org/wiki/Neil_deGrasse_Tyson&quot; rel=&quot;nofollow noreferrer&quot;&gt;Neil deGrasse Tyson&lt;/a&gt; that I saw somewhere in the YouTube universe, giving a question one might ask to possibly challenge literal interpretation of one's astrological sign. For some reason I can't find it now, but I believe he said that currently the sun does not actually pass through all twelve zodiac constellations any more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm &lt;em&gt;not looking for&lt;/em&gt; opinions or any particulars about Astrology, please, &lt;em&gt;just the science!&lt;/em&gt; (if there is some) If someone is familiar with the statement, I'd like to hear if I've got it right, and if it's due to proper motion, precession, redrawing/redefinition of constellation boundaries, or something else.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-05-02T13:37:12.797" Title="Does the sun now 'bypass&quot; some of the original zodiac constellations?" Tags="&lt;constellations&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14749" PostTypeId="2" ParentId="14748" CreationDate="2016-05-02T13:10:47.880" Score="2" Body="&lt;p&gt;The ecliptic still passes through the original 12 constellations, although Earth's precession has shifted the Sun's apparent position by roughly one constellation. That is, if your birthday is today, May 2, an astrologer would claim you were Taurus, whereas in reality the Sun is currently positioned in Aries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, the ecliptic now passes through the constellation Ophiuchus (The Serpent Bearer). You can see the &quot;historical&quot; and the &quot;true&quot; constellations e.g. &lt;a href=&quot;http://www.ast.cam.ac.uk/~frosales/html/zodiac/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2016-05-02T13:10:47.880" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14750" PostTypeId="1" CreationDate="2016-05-02T16:21:49.800" Score="4" ViewCount="285" Body="&lt;p&gt;What is the actual meaning of the statement 'Dust is optically thin in the Far Infrared (FIR) over most of the Galaxy'? Kindly Help&lt;/p&gt;&#xA;" OwnerUserId="11434" LastActivityDate="2016-05-03T07:55:01.000" Title="Why dust is optically thin in Far Infrared wavelengths?" Tags="&lt;astrophysics&gt;&lt;star-formation&gt;&lt;dust&gt;&lt;interstellar-medium&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14751" PostTypeId="1" AcceptedAnswerId="14772" CreationDate="2016-05-02T16:37:02.967" Score="1" ViewCount="75" Body="&lt;p&gt;I am interested in a sun-like star 138 light years away that is in the EUVE catalog. A table shows that a photon count of 2.6+/-0.5 photons per kilosecond at 100 angstroms has been detected for this star. I can not determine if this is high or not. I am even having a hard time finding what solar emission is at this wavelength so i can attempt the math to compare with the sun. Can somebody help me here?&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2016-05-04T11:30:28.410" Title="How to interpret extreme ultraviolet/ x-ray data in the literature?" Tags="&lt;star&gt;&lt;data-analysis&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14752" PostTypeId="2" ParentId="14750" CreationDate="2016-05-02T18:18:28.697" Score="6" Body="&lt;p&gt;The term &quot;optically thin&quot; means that the &lt;em&gt;optical depth&lt;/em&gt; is small. The optical depth is a measure of the opacity of a medium, in this case dust, experienced by light traveling through that medium, and is defined as&#xA;$$&#xA;\tau \equiv n \, r \, \sigma,&#xA;$$&#xA;where $n$ is the density of the particles in question, $r$ is the distance traveled through the medium, and $\sigma = \sigma(\lambda)$ is the cross section of the particles, which is dependent on the wavelength $\lambda$ of the photons. If $\tau\ll1$, the medium is said to be optically thin, while if $\tau\gg1$, it is said to be optically thick. The fraction of the light that is extinguished by the journey through the medium is $e^{-\tau}$, so the two cases indicate when the light is mostly transferred and mostly extinguished, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cosmic dust is composed of particles spanning a large range of sizes. Photons interacting with the dust are either scattered in another direction, or absorbed, depending on the albedo of the dust. But in both cases, a photon has a larger probability of interacting with a dust grain which is comparable to its wavelength. Because of the size distribution, an ensemble of dust grains hence has a characteristic &lt;em&gt;extinction curve&lt;/em&gt;. In the figure below (modified from &lt;a href=&quot;http://arxiv.org/abs/0907.2698&quot; rel=&quot;nofollow noreferrer&quot;&gt;Laursen et al. 2009&lt;/a&gt;), I plotted the (functional fits to the observed) extinction curves of dust in the Small (&lt;em&gt;dashed&lt;/em&gt; line) and Large (&lt;em&gt;solid line&lt;/em&gt;) Magellanic Clouds$^\dagger$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Yvb8E.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Yvb8E.png&quot; alt=&quot;ext&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extinction is here given in terms of &quot;cross section per hydrogen atom&quot;, but you can just think of it as an average dust grain cross section. The colors show the ultraviolet (&lt;em&gt;purple&lt;/em&gt;) and infrared (&lt;em&gt;red&lt;/em&gt;) regions of the spectrum.&#xA;You can see that the extinction, or the cross section, is largest for UV photons, and as you go to longer wavelengths, the cross section decreases sharply and is very small in the far infrared ($\lambda \gtrsim 30 \, \mu\mathrm{m}$; the exact definitions depend on your field of interest).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what your statement refers to. But in fact it's badly phrased, because the optical depth is also a function of density and distance, so for instance for a distance $r = 1 \, \mathrm{cm}$ through a dust cloud, the optical depth is $\ll1$ for all photons.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;Extinction curves in other galaxies look similar. In particular, the extinction along most sightlines in the Milky Way features the same bump around $\lambda \simeq 2175$ Å. The origin of this bump still isn't well understood, but may be partly due to graphites and/or PAHs.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-05-03T07:55:01.000" LastActivityDate="2016-05-03T07:55:01.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14753" PostTypeId="1" AcceptedAnswerId="14754" CreationDate="2016-05-02T18:34:45.343" Score="0" ViewCount="492" Body="&lt;p&gt;I am currently self teaching myself some basic astronomy. My book claims, if a constellation is observed for exactly 6 months during a year, then it lies on the celestial equator. I have failed to understand why this must be true. Shouldn't the visible duration depends on the position of the observer? For example, for an observer at equator, wouldn't all constellations (accept those at the polaris) be visible for exactly half a year? Also, is the constellation at the NCP always visible for people in the northern hemisphere? Thank you in advance!&lt;/p&gt;&#xA;" OwnerUserId="11710" LastActivityDate="2016-05-02T19:59:54.397" Title="Simple Question about Constellation" Tags="&lt;constellations&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14754" PostTypeId="2" ParentId="14753" CreationDate="2016-05-02T19:50:59.757" Score="2" Body="&lt;p&gt;Taken at face value it's not correct that points on the equator are visible for exactly six months.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, Orion is on the celestial equator. It is true that Orion will be above the horizon for half the day (12 hours), but during the summer when it is above the horizon, so is the sun, so it can't be seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But Orion is visible from  Northern Europe between the middle of August (when it rises shortly before the sun) and the middle of April (when it sets shortly after dusk) That is rather more than six months. Exact dates can't be given as visibility during twilight depends on factors like the clarity of the atmosphere, and stars don't wink off as the sun rises, they fade out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you always observe at the same time then points on the celestial equator will be above the horizon for six months of the year, but just being above the horizon doesn't make it visible, when close to the horizon the thicker atmosphere (and more prosaically, trees and buildings) will make a constellation effectively obscured. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that the north celestial pole is always above the horizon, as are many of the constellations around it. From my location that includes Cassiopeia, Ursa minor, most of Ursa major. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, for Northern Hemisphere observers, the further north a point is, the longer it will remain visible.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-05-02T19:59:54.397" LastActivityDate="2016-05-02T19:59:54.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14755" PostTypeId="2" ParentId="14728" CreationDate="2016-05-03T04:38:17.610" Score="1" Body="&lt;p&gt;The age of the universe as we know t is very hard to determine. Current estimates put down the age f universe to 13.5 billion years +/- 0.8 billion. This is a huge a margin of error when compared to the margin or error you speak of. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also it will never ever be possible to measure the age of universe because:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The universe started according to the big bang theory and if we take it as true it gives us a near infinite radiation density. We could measure the expansion and try to estimate the age but it is still hazy as we have no way knowing the initial density of the universe.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is always an uncertainty in any measurement. and measuring something so far away(The edge of the universe) will have a lot of uncertainty. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These measures of today may still have a huge margin of error but they are the best we have till date. And to narrow the error down to about even a few years would take us a lot of time and tremendous development in technologies. So basically we have a very slim chance of determining the precise age of the universe in the foreseeable future&lt;/p&gt;&#xA;" OwnerUserId="11587" LastActivityDate="2016-05-03T04:38:17.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14756" PostTypeId="2" ParentId="213" CreationDate="2016-05-03T04:41:50.523" Score="-1" Body="&lt;p&gt;Asteroids and Meteoroids are very different in size. Asteroids are huge usually about few kilometers big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas Meteoroids can be so small from centimeters to about few meters wide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both are very different in size. Sometimes, when two asteroids collide they might throw away few percentage of their rock which falls into planets as meteorites.&lt;/p&gt;&#xA;" OwnerUserId="5422" LastActivityDate="2016-05-03T04:41:50.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14758" PostTypeId="2" ParentId="4761" CreationDate="2016-05-03T15:04:33.107" Score="1" Body="&lt;p&gt;In my opinion yes, mass affects volume.  Instead of a cube, imagine a pyramid with a 3 sided base.  All edges have the same length defined by the time to reflect light back and forth between the apexs.  The number of smaller such pyramids that could be fit inside would increase with mass because the path of light would be bent.  Bent light means bent space, as well as bent spacetime.  They are not independent.  Per Einstein,  the speed of light is not constant when mass is present and depends on the coordinates.  This amounts to more space. I believe that it was Eddington who first tried to quantify how much space is added by a given amount of mass. &lt;/p&gt;&#xA;" OwnerUserId="11717" LastActivityDate="2016-05-03T15:04:33.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14759" PostTypeId="1" AcceptedAnswerId="14760" CreationDate="2016-05-03T19:48:00.083" Score="3" ViewCount="1412" Body="&lt;p&gt;Beyond the Chandrasekhar limit, white dwarfs become extremely hot. As a result, previously unfusable carbon can become fusable, causing nuclear reactions. This leads to a thermal runaway and ultimately a type Ia supernova.&lt;/p&gt;&#xA;&lt;p&gt;However, this doesn't seem to happen in neutron stars. They can keep gaining mass until they become black holes. After a certain point, shouldn't the immense temperatures cause a thermal runaway like in white dwarfs, leading to their explosion?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="31410" LastEditDate="2020-12-15T22:10:34.857" LastActivityDate="2020-12-15T22:10:34.857" Title="Why can't neutron stars ignite and explode?" Tags="&lt;supernova&gt;&lt;stellar-evolution&gt;&lt;neutron-star&gt;&lt;temperature&gt;&lt;degenerate-matter&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14760" PostTypeId="2" ParentId="14759" CreationDate="2016-05-03T20:14:36.073" Score="9" Body="&lt;p&gt;In a white dwarf, the dense matter is not in its lowest energy configuration. Energy can still be extracted from the white dwarf material by fusion, provided it can be ignited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What &lt;em&gt;exothermic&lt;/em&gt; nuclear reactions would there be that could take place in a neutron star? The bulk of the material is in the form of neutrons with a small number of protons and electrons. At these densities, &lt;em&gt;that is&lt;/em&gt; the most stable equilibrium composition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a neutron star gains mass in a gradual way, then the most likely course of events will be that its radius will decrease (that is what happens in objects supported by degenerate matter) until it reaches a General Relativistic instability where its collapse to a black hole is inevitable (when $R$ is somewhere between 1.25 and 2 times the Schwarzschild radius). It is possible that neutrons may transform before that into additional hadronic degrees of freedom or into quark matter, but these are endothermic processes that suck kinetic energy out of the neutron gas and only hasten the collapse.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-05-03T20:14:36.073" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="14761" PostTypeId="1" CreationDate="2016-05-03T20:36:05.553" Score="4" ViewCount="430" Body="&lt;p&gt;I am an early graduate student in astronomy and have hard time understanding why do post-MS stars move up the RGB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what I understand about post main sequence evolution of stars. As their hydrogen core is exhausted, the core shrinks under its own gravity. This lets a region of previously too cold hydrogen enter hotter regions, thus starting a hydrogen shell burning process. While this process can keep the luminosity constant, it expands the envelope, hence lowering the temperature and moving the star to the right in the HR diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then suddenly, the star starts moving up on the RGB: slight or no temperature change, but sudden increase in luminosity. What makes the star move up the HR diagram all of the sudden? This is before the helium flash when the core starts burning helium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="11673" LastActivityDate="2016-05-04T06:41:49.513" Title="Why do post main sequence stars enter the red giants branch?" Tags="&lt;star&gt;&lt;main-sequence&gt;&lt;red-giant&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14762" PostTypeId="1" AcceptedAnswerId="15022" CreationDate="2016-05-04T00:01:17.753" Score="1" ViewCount="3173" Body="&lt;p&gt;I saw a list recently of the moons of the various planets and the discoverer(s) and date of discovery. Earth's Moon was listed as Unknown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know you don't 'discover' a huge disc in the sky. But still, what is the first recorded reference to the Moon being a satellite of the Earth? That someone knew it is smaller and orbiting? Celestial Sphere does not count.&lt;/p&gt;&#xA;" OwnerDisplayName="user11722" LastEditorUserId="9301" LastEditDate="2016-05-04T12:55:03.363" LastActivityDate="2019-07-19T10:34:33.443" Title="What is the first recorded reference to the Moon being a satellite of the Earth?" Tags="&lt;the-moon&gt;&lt;amateur-observing&gt;&lt;history&gt;" AnswerCount="3" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="14763" PostTypeId="2" ParentId="14751" CreationDate="2016-05-04T00:28:46.773" Score="1" Body="&lt;p&gt;If you're looking at Table 2 &quot;EUVE Deep Survey&quot; from the &lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/bib_query?1996ApJS..102..129B&quot; rel=&quot;nofollow&quot;&gt;Second EUVE Source Catalog&lt;/a&gt;, the cheap way out would be to compare the star of interest to other stars of similar spectral type in that table, correcting for apparent magnitude.&#xA;To compare to the Sun, part 4 &quot;Flux Calibration&quot; and Figure 3 &quot;Deep-survey effective area vs. wavelength models&quot; may help.&#xA;Then you could apply a similar response function to the 5-20 nm band of a reference solar spectrum which includes extreme UV. I found a usable one in Table 2 of this &lt;a href=&quot;http://lasp.colorado.edu/home/wp-content/uploads/2011/07/Woods-Solar-UV-Variability.pdf&quot; rel=&quot;nofollow&quot;&gt;book chapter&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-05-04T00:28:46.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14766" PostTypeId="2" ParentId="14761" CreationDate="2016-05-04T02:47:10.363" Score="1" Body="&lt;p&gt;As the hydrogen supply in the core is exhausted, the pressure supporting against gravitational collapse lessens, and the core begins to collapse, which causes the inner temperature to rise.  As a result, hydrogen in the less-processed regions outside the core starts to burn in a shell.  Stellar models predict that, at this stage, there is a huge expansion of the outer layers of the star (the understanding of which is difficult to grasp intuitively, but it's robustly predicted by the equations of stellar structure).  The increase in luminosity, due to the gravitational contraction and the hydrogen shell burning, moves the star up on the H-R diagram, while the increase in radius lowers the effective temperature, moving the star to the right on the diagram.  This is the so-called &quot;red giant phase&quot;.&lt;/p&gt;&#xA;" OwnerUserId="11533" LastActivityDate="2016-05-04T02:47:10.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14768" PostTypeId="2" ParentId="14761" CreationDate="2016-05-04T06:41:49.513" Score="2" Body="&lt;p&gt;If your question is why the star starts moving up the red giant branch, it's in essence because of the behaviour of the surface opacity and the development of a substantial convective envelope in order to meet the surface boundary condition. It's basically the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hayashi_track&quot; rel=&quot;nofollow&quot;&gt;Hayashi track&lt;/a&gt; in reverse. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can say this because if you create a model of a star and artificially suppress convection in the equations, then the model just keeps getting colder and colder at roughly constant luminosity, instead of ascending the giant branch.  See e.g. Fig. 2 of &lt;a href=&quot;http://arxiv.org/abs/0902.0406&quot; rel=&quot;nofollow&quot;&gt;Stancliffe et al. (2009)&lt;/a&gt;. Note that the star still becomes large in terms of radius, just not bright.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2016-05-04T06:41:49.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14769" PostTypeId="1" CreationDate="2016-05-04T10:07:25.720" Score="2" ViewCount="644" Body="&lt;p&gt;I have an image with some sources in it. I need to get the pixel size and beam size of that image. How can I calculate it? &lt;/p&gt;&#xA;" OwnerUserId="11434" LastActivityDate="2020-02-02T13:01:55.347" Title="How to calculate the pixel size and beam size of an image?" Tags="&lt;data-analysis&gt;&lt;photometry&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14770" PostTypeId="1" CreationDate="2016-05-04T10:52:14.117" Score="4" ViewCount="431" Body="&lt;p&gt;I read that thermal emission/radiation are the ones whose spectra is similar to that of a Black body radiation. Also thermal radiation depends solely on the temperature of the object. How can we consider Free-Free emission and spectral line emission as thermal emission processes? kindly explain&lt;/p&gt;&#xA;" OwnerUserId="11434" LastActivityDate="2016-05-04T16:10:23.947" Title="Why Free-Free emission is regarded as Thermal emission?" Tags="&lt;radiation&gt;&lt;dust&gt;&lt;radiative-transfer&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14771" PostTypeId="2" ParentId="14770" CreationDate="2016-05-04T11:21:02.797" Score="6" Body="&lt;p&gt;Thermal radiation $\neq$ blackbody radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermal radiation is radiation that comes from a system where an equilibrium has been reached, where the various energy states are occupied according to the Boltzmann distribution and the particle velocity distributions are Maxwellian at some given temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That does not necessarily imply that the radiation field is in equilibrium with the matter at the same temperature. It is this latter equilibrium that is required in order for the radiation field to approximate to a blackbody.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the following statements are true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blackbody radiation is thermal radiation. Thermal radiation is not necessarily blackbody radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of thermal, but non-blackbody, radiation would be  &lt;a href=&quot;https://en.wikipedia.org/wiki/Bremsstrahlung#Thermal_bremsstrahlung&quot;&gt;thermal bremsstrahlung&lt;/a&gt;. The particle velocities in the gas follow a Maxwellian distribution at a characteristic temperature and this determines the spectrum of the emitted free-free radiation. However, the bremsstrahlung radiation is able to escape from the gas without further interaction - in other words the gas is &quot;optically thin&quot;. The result is a thermal bremsstrahlung spectrum that is not at all like a blackbody spectrum, especially at frequencies $&amp;lt;kT/h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the bremsstrahlung process is purely caused by the interaction of free electrons and ions. If you allow there to be atoms and partially ionised atoms in the gas, then it is feasible for those to be in equilibrium at the same temperature. The population of the various energy states in the atoms and ions is determined by this temperature, and transitions between these levels produces spectral emission lines in a completely predictable, temperature-dependent, way.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-05-04T16:10:23.947" LastActivityDate="2016-05-04T16:10:23.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14772" PostTypeId="2" ParentId="14751" CreationDate="2016-05-04T11:30:28.410" Score="2" Body="&lt;p&gt;The EUVE catalog gives you a count-rate in a certain EUV band. To convert this into a flux, you need a flux conversion factor to go from counts per second to energy flux at the Earth per second. This conversion factor, also often known as an &quot;ECF&quot;, depends on the intrinsic spectrum of the source and also how much absorption (known as the hydrogen column density) there is between Earth and the source. So there is no one answer that can be given.&#xA;I&#xA;Your best port of call is the &lt;a href=&quot;https://heasarc.gsfc.nasa.gov/docs/software/tools/pimms_ug4.html#euve&quot; rel=&quot;nofollow&quot;&gt;PIMMS flux conversion tool&lt;/a&gt;, which has an EUVE module and can do these flux conversions for you if you know something about the spectrum and the absorption. I'd try it for you but I couldn't immediately see how to do it for EUVE, though it claims the information is there...&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-05-04T11:30:28.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14773" PostTypeId="1" AcceptedAnswerId="14774" CreationDate="2016-05-04T14:10:47.063" Score="0" ViewCount="135" Body="&lt;p&gt;When historically did we realize that the Sun is a star, like all the others?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An answer was posted, then the question was put on hold. I would like to answer:&lt;br&gt;&#xA;If indeed someone asserted that the Sun was a star 2500 years ago, that is rather humbling. That it took until about 100 years ago for it to be proven is even more humbling!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People level a lot of claims against science and scientists. But if a field of knowledge &lt;strong&gt;takes over 100 generations to prove a true statement&lt;/strong&gt;, perhaps they have a right to criticize. So, I think that science needs to do a better job of proving its theories. Either that, or people need to have more common sense, not to recognize something that has been staring billions of people in the face for all time. Perhaps, we should all be a bit more humble about the state of knowledge, even in obvious areas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe if we had taken the people who proposed this fact, and the concept of atoms, seriously at the time, we would be on our way to &lt;em&gt;other&lt;/em&gt; stars by now instead of still having only visited that big white thing nearby (forgot the name just now) and that not for the past 40 years.&lt;/p&gt;&#xA;" OwnerDisplayName="user11722" LastEditorDisplayName="user11722" LastEditDate="2016-05-04T18:23:18.877" LastActivityDate="2016-05-04T18:23:18.877" Title="When was the nearest star discovered?" Tags="&lt;the-sun&gt;&lt;amateur-observing&gt;&lt;history&gt;" AnswerCount="1" CommentCount="12" ClosedDate="2016-05-04T15:18:00.650" ContentLicense="CC BY-SA 3.0" />
  <row Id="14774" PostTypeId="2" ParentId="14773" CreationDate="2016-05-04T14:24:47.730" Score="1" Body="&lt;p&gt;After a quick google, &lt;a href=&quot;http://solar-center.stanford.edu/FAQ/Qsunasstar.html&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; page pops up showing that Anaxagoras was the first documented person to make the link between the sun and stars being the same thing.&lt;/p&gt;&#xA;" OwnerUserId="11525" LastActivityDate="2016-05-04T14:24:47.730" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14776" PostTypeId="1" CreationDate="2016-05-04T15:16:06.250" Score="0" ViewCount="381" Body="&lt;p&gt;&quot;The solar elevation angle is the altitude of the sun, the angle between the horizon and the centre of the sun's disc.&quot; - &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_zenith_angle&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The meridian is when the sun is directly facing earth and is at exactly 90° overhead at some point along the longitudinal line at a given location. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question is, if the sun is 93 million miles away and travels overhead at 90° between the Tropic of Capricorn and the Tropic of Cancer, how is it that there are Meridian solar elevation angles reported from relatively nearby cities on Earth of less than 89.99°?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take for instance Calgary, Canada as of today May, 04 2016 -- estimated ground distance from Calgary to where the Sun is at 90° on Calgary's longitudinal line is ~3000 miles. The solar elevation angle (aka &quot;altitude&quot;) is reported by the timeanddate website to be 55°! - &#xA;&lt;a href=&quot;http://www.timeanddate.com/astronomy/canada/calgary&quot; rel=&quot;nofollow&quot;&gt;TimeandDate&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reversing the angle, using right triangle math: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The base angle, our solar elevation angle, of a right angle triangle is: &#xA;&lt;br&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;solar elevation angle&quot;= arctangent (h/a) &#xA;&lt;br&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;89.998151749049 = arctangent (93000000/3000) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So using right triangle math alone with the presumption of the Sun being 93 million miles away and base distance of 3000 miles &lt;strong&gt;we get an expected solar elevation angle of 89.998151749049°&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That makes sense, since if something is 93 million miles away from us and you think of it as a triangle, then the distance between any place on earth and the location on earth where the sun is at exactly 90° overhead -- you should expect the other angle of the triangle to be at near 90°, connected by a relatively tiny sliver of distance in a much elongated triangle to the center point of the distant sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we go the other direction and leave out the presumption that the sun is 93 million miles away. We take h as unknown, we know the reported solar elevation angle of 55°, we can again use right triangle math: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Let a=3000, solar elevation angle=55°&lt;br&gt;&#xA; &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;h = a * tangent (&quot;solar elevation angle&quot;)&lt;br&gt; &#xA; &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;4,284.4440202263 = 3000 * tangent(55°)&lt;br&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So according to my calculations, based on the advertised solar elevation angle of 55° at Calgary, Canada on May, 04 2016 at Meridian, at a ground distance of 3000 miles from where the Sun is at exactly 90° overhead --&amp;nbsp;&amp;nbsp;&lt;strong&gt;the Sun is actually only 4,284.4440202263 miles high!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;** &lt;strong&gt;Update&lt;/strong&gt; **: The math would work out for a sphere Earth with a radius of 4000 miles, using the exact solar position and adding  corrective angles due to a presumption of a spherical curvature: &lt;br&gt;&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Calgary, Canada: 51°03′N 114°04′W&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Location of Sun at Calgary Meridian: 16°23′N 114°04′W&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Distance = 2395.386 miles&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Radius of Earth=4000 miles&lt;br&gt; &#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Arc distance = 4790.772 miles&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Width of arc = 4509.52059 miles&lt;br&gt; &#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Angle to center of the Earth=68.622754110849°&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;68.622754110849°/2=34.31137705542450°&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This math utilized calculations that may have had a bit less acurracy but to make it work you need to add a bandaid padding of approximately 34.31137705542450° to Calgary,Canada's advertised angles. If you do you arrive at 55°+34.31137705542450° = 89.31137705542450° which is close to what I would expect to see. If, we are truly living on a Flat Earth, then the calculations would be telling us that the Sun height is only 4,284.4440202263 miles high and we can use simple trignometry.&lt;br&gt;&#xA;&lt;a href=&quot;http://www.handymath.com/cgi-bin/arc18.cgi&quot; rel=&quot;nofollow&quot;&gt;Arc Calculator&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.movable-type.co.uk/scripts/latlong.html&quot; rel=&quot;nofollow&quot;&gt;Coordinate Distance Calculator&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the math is difficult for you, you can check with an online triangle calculator such as: &lt;a href=&quot;http://keisan.casio.com/exec/system/1273849674&quot; rel=&quot;nofollow&quot;&gt;Ke!san&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="11727" LastEditorUserId="11727" LastEditDate="2016-05-05T13:32:47.670" LastActivityDate="2016-05-05T13:32:47.670" Title="Solar Elevation Angles -- Anomaly?" Tags="&lt;the-sun&gt;&lt;angular-resolution&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14777" PostTypeId="2" ParentId="14776" CreationDate="2016-05-04T15:53:06.413" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So using right triangle math alone with the presumption of the Sun being 93 million miles away and base distance of 3000 miles we get an expected solar elevation angle of 89.998151749049°. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the wrong way to do it. A point 3000 miles south will be pointing in a very different direction, because the earth is round and only about 8000 miles in diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth is around 25,000 miles in circumference. A point 3000 miles away is 0.12 times the circumference away. (3000/25000.) The circumference is equivalent to 360 degrees (assuming the Earth is a perfect circle, which is almost true for its polar section.) So 0.12 times 360 degrees is a difference of forty-something degrees. This probably accounts for your difference (very roughly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional comment about roughness of these calculations: in fact the Earth is not a perfect sphere. Drawing a great circle along the meridian, the Earth will not appear perfectly round because the equator bulges slightly (and the poles are flattened).  It's only something like 30-40 miles difference, but bear in mind this has a slight effect when calculating angles by dividing the circumference, as I did above.  (Still better than assuming a straight line though, which is what I think the question is doing.)&lt;/p&gt;&#xA;" OwnerUserId="7491" LastEditorUserId="7491" LastEditDate="2016-05-04T16:09:23.880" LastActivityDate="2016-05-04T16:09:23.880" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14779" PostTypeId="2" ParentId="14776" CreationDate="2016-05-04T23:53:39.230" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WtwVE.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WtwVE.gif&quot; alt=&quot;[[image31]]&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your confusion may stem from the way we sometimes draw the Sun (not to &#xA;scale) as though it's close to the Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, as you note, the lines of vision to the Sun are nearly parallel: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mrAVr.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mrAVr.gif&quot; alt=&quot;[[image32]]&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the angles (measured from the red line, the horizon, to the &#xA;arrow) are still different, but for entirely different reasons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've actually stolen @Andy's answer (and @Tanenthor's comment), but &#xA;I'm hoping the diagram makes it easier to understand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Re the parallel lines, the 11810x60 image below may help explain: even &#xA;though the lines eventually converge (on an object about 10 times the &#xA;distance to the moon, still nowhere near the distance to the Sun), &#xA;they are nearly parallel when leaving Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/e6UHe.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/e6UHe.gif&quot; alt=&quot;[[image33]]&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2016-05-05T01:30:19.223" LastActivityDate="2016-05-05T01:30:19.223" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14780" PostTypeId="2" ParentId="14660" CreationDate="2016-05-05T06:49:51.277" Score="1" Body="&lt;p&gt;The V838 Monocerotis &lt;em&gt;expansion&lt;/em&gt; (not a supernova) and the observation of the subsequent &quot;spectacular&quot; light echo was quite a notable event! From &lt;a href=&quot;http://www.nature.com/nature/journal/v422/n6930/abs/nature01508.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nature &lt;strong&gt;422&lt;/strong&gt;, 405-408 (27 March 2003)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xU4aE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xU4aE.png&quot; alt=&quot;Nature Cover&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/TSrZZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TSrZZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0004-6256/135/2/605&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronom. J. &lt;strong&gt;135&lt;/strong&gt;, 2, 2008&lt;/a&gt; or &lt;a href=&quot;http://arxiv.org/abs/0711.1495&quot; rel=&quot;nofollow noreferrer&quot;&gt;ArXiv&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;em&gt;.&lt;strong&gt;Galactic light echoes are extremely rare. The only other known example of extent similar to that of V838 Mon was the echo produced by Nova GK Persei 1901&lt;/strong&gt; (Kapteyn 1902; Perrine 1902; Ritchey 1902). Following early misunderstandings, light-echo geometry was properly described by Couderc (1939), and more recent discussions are given by many authors, including Chevalier (1986), Felten (1991), Sparks (1994), Sugerman (2003), and references therein&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It was the sole topic of an international conference &lt;a href=&quot;http://www.ing.iac.es/conferences/v838mon/&quot; rel=&quot;nofollow noreferrer&quot;&gt;photo from here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BgHSZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BgHSZ.jpg&quot; alt=&quot;Conference&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0608222&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; it's noted that the Hubble observations were inserted into the observing schedule using the Director's Discretionary time, since the peer review process is too slow to accommodate observations of transient events. While there is no mention of any reason why observations were not more frequent, nor continued in 2003, one can speculate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;em&gt;Based on the highly structured appearance of the initial ground-based images, &lt;strong&gt;our team proposed for Director’s Discretionary (DD) time on HST for a program of direct imaging and imaging polarimetry&lt;/strong&gt;. The team members are as follows: S. Starrfield (Arizona State University); Z. Levay, N. Panagia, W. Sparks, B. Sugerman, R. White, and myself (STScI); A. Henden (AAVSO); M. Wagner (University of Arizona); R. Corradi (Issac Newton Group); U. Munari (Padova University); L. Crause (SAAO); and M. Dopita (ANU)&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;&lt;em&gt;We &lt;strong&gt;received HST observing time at five epochs in 2002 through DD allocations: April, May, September, October, and December&lt;/strong&gt;. All of the observations were made with the Advanced Camera for Surveys (ACS), which had been installed in HST during SM3b in March 2002. I need not emphasize to this audience &lt;strong&gt;how extraordinarily unfortunate it is that no HST observations were obtained during 2003—the loss of this opportunity is truly incalculable&lt;/strong&gt;. However, the echoes were imaged twice in 2004 through the Hubble Heritage program, in February and October. More happily, the HST Cycle 14 allocation committee did award our team observing time for an intensive HST imaging campaign from October 2005 to January 2006, and we also have two more epochs of observations scheduled in Cycle 15 for late 2006 and early 2007&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nature.com/nature/journal/v422/n6930/fig_tab/nature01508_F2.html#figure-title&quot; rel=&quot;nofollow noreferrer&quot;&gt;Figure 2&lt;/a&gt; of the Nature paper describes the preservation of the actual light curve (history) within the structure of the light-echo shell:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YztnU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YztnU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;strong&gt;FIGURE 2. HST images of the light echoes&lt;/strong&gt;&#xA;  The apparently superluminal expansion of the echoes as light from the outburst propagates outward into surrounding dust is shown dramatically. Images were taken in 2002 on 30 April (a), 20 May (b), 2 September (c) and 28 October (d). Each frame is 83&quot; times 83&quot;; north is up and east to the left. Imaging on 30 April was obtained only in the B filter, but B, V and I were used on the other three dates, allowing us to make full-colour renditions. The time evolution of the stellar outburst (Fig. 1) is reflected by structures visible in these colour images. In b, for example, note the series of rings and filamentary structures, especially in the upper right quadrant. &lt;strong&gt;Close examination shows that each set of rings has a sharp, blue outer edge, a dip in intensity nearer the star, and then a rebrightening to a redder plateau. Similar replicas of the outburst light curve are seen propagating outwards throughout all of the colour images&lt;/strong&gt;.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again from &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0004-6256/135/2/605&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronom. J. &lt;strong&gt;135&lt;/strong&gt;, 2, 2008&lt;/a&gt; or &lt;a href=&quot;http://arxiv.org/abs/0711.1495&quot; rel=&quot;nofollow noreferrer&quot;&gt;ArXiv&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yT35t.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yT35t.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Figure 2.&lt;/strong&gt; Images representing the degree of linear polarization, p, for each of the four epochs of data shown in Figure 1. Image scales and orientations are the same as in Figure 1. The image stretch is linear, ranging from black representing zero linear polarization to full white representing ~50% linear polarization. These images illustrate the apparent outward motion of a ring of highly polarized light in the light echo.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Abstract&lt;/strong&gt;&#xA;  Following the outburst of the unusual variable star V838 Monocerotis in 2002, a spectacular light echo appeared. A light echo provides the possibility of direct geometric distance determination, because it should contain a ring of highly linearly polarized light at a linear radius of ct, where t is the time since the outburst. We present imaging polarimetry of the V838 Mon light echo, obtained in 2002 and 2005 with the Advanced Camera for Surveys on board the Hubble Space Telescope, which confirms the presence of the highly polarized ring. Based on detailed modeling that takes into account the outburst light curve, the paraboloidal echo geometry, and the physics of dust scattering and polarization, we find a distance of 6.1 ± 0.6 kpc. The error is dominated by the systematic uncertainty in the scattering angle of maximum linear polarization, taken to be θmax = 90° ± 5°. &lt;strong&gt;The polarimetric distance agrees remarkably well with a distance of 6.2 ± 1.2 kpc obtained from the entirely independent method of main-sequence fitting to a sparse star cluster associated with V838 Mon&lt;/strong&gt;. At this distance, V838 Mon at maximum light had MV  sime −9.8, making it temporarily one of the most luminous stars in the Local Group. &lt;strong&gt;Our validation of the polarimetric method offers promise for measurement of extragalactic distances using supernova light echoes.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2016-05-05T07:36:30.720" LastActivityDate="2016-05-05T07:36:30.720" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14781" PostTypeId="1" AcceptedAnswerId="14783" CreationDate="2016-05-05T07:37:01.597" Score="1" ViewCount="191" Body="&lt;p&gt;Drawing from &lt;a href=&quot;https://astronomy.stackexchange.com/a/14780/7982&quot;&gt;this answer&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The V838 Monocerotis &lt;em&gt;expansion&lt;/em&gt; (not a supernova) and the observation of the subsequent &quot;spectacular&quot; light echo was quite a notable event! From &lt;a href=&quot;http://www.nature.com/nature/journal/v422/n6930/abs/nature01508.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nature &lt;strong&gt;422&lt;/strong&gt;, 405-408 (27 March 2003)&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xU4aE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xU4aE.png&quot; alt=&quot;Nature Cover&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/TSrZZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TSrZZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hubble images are quite spectacular (discussed further in &lt;a href=&quot;https://astronomy.stackexchange.com/a/14780/7982&quot;&gt;that answer&lt;/a&gt;). I've included them again below. The color images are 83x83 arcsecconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A later &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0608222&quot; rel=&quot;nofollow noreferrer&quot;&gt;conference paper&lt;/a&gt; laments the fact that no HST images could be obtained the following year (2003):  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;em&gt;We &lt;strong&gt;received HST observing time at five epochs in 2002 through DD allocations: April, May, September, October, and December&lt;/strong&gt;. All of the observations were made with the Advanced Camera for Surveys (ACS), which had been installed in HST during SM3b in March 2002. I need not emphasize to this audience &lt;strong&gt;how extraordinarily unfortunate it is that no HST observations were obtained during 2003—the loss of this opportunity is truly incalculable&lt;/strong&gt;. However, the echoes were imaged twice in 2004 through the Hubble Heritage program, in February and October. More happily, the HST Cycle 14 allocation committee did award our team observing time for an intensive HST imaging campaign from October 2005 to January 2006, and we also have two more epochs of observations scheduled in Cycle 15 for late 2006 and early 2007&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question is:&lt;/strong&gt; Why could't observations from one or more of the ground telescopes that are using &lt;a href=&quot;https://en.wikipedia.org/wiki/Adaptive_optics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Adaptive Optics&lt;/a&gt; been used to at least &lt;em&gt;supplement the data&lt;/em&gt; in 2003? I understand they may not provide the same quality, but not even close? Not even for the polarization mapping? Is it related to the extended area of the target? Is a field of 80 arcsec too large to correct all at once? Is the pattern too complex, or is it the lack of an isolated bright &quot;guide star&quot; the problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;note:&lt;/strong&gt; I need a reasoned and hopefully quantitative answer, not your &quot;best guess&quot; or speculation. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nature.com/nature/journal/v422/n6930/fig_tab/nature01508_F2.html#figure-title&quot; rel=&quot;nofollow noreferrer&quot;&gt;Figure 2&lt;/a&gt; of the Nature paper describes the preservation of the actual light curve (history) within the structure of the light-echo shell:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YztnU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YztnU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;strong&gt;FIGURE 2. HST images of the light echoes&lt;/strong&gt;&#xA;  The apparently superluminal expansion of the echoes as light from the outburst propagates outward into surrounding dust is shown dramatically. Images were taken in 2002 on 30 April (a), 20 May (b), 2 September (c) and 28 October (d). Each frame is 83&quot; times 83&quot;; north is up and east to the left. Imaging on 30 April was obtained only in the B filter, but B, V and I were used on the other three dates, allowing us to make full-colour renditions. The time evolution of the stellar outburst (Fig. 1) is reflected by structures visible in these colour images. In b, for example, note the series of rings and filamentary structures, especially in the upper right quadrant. &lt;strong&gt;Close examination shows that each set of rings has a sharp, blue outer edge, a dip in intensity nearer the star, and then a rebrightening to a redder plateau. Similar replicas of the outburst light curve are seen propagating outwards throughout all of the colour images&lt;/strong&gt;.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0004-6256/135/2/605&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronom. J. &lt;strong&gt;135&lt;/strong&gt;, 2, 2008&lt;/a&gt; or &lt;a href=&quot;http://arxiv.org/abs/0711.1495&quot; rel=&quot;nofollow noreferrer&quot;&gt;ArXiv&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yT35t.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yT35t.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Figure 2.&lt;/strong&gt; Images representing the degree of linear polarization, p, for each of the four epochs of data shown in Figure 1. Image scales and orientations are the same as in Figure 1. The image stretch is linear, ranging from black representing zero linear polarization to full white representing ~50% linear polarization. These images illustrate the apparent outward motion of a ring of highly polarized light in the light echo.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2018-10-26T06:37:30.127" Title="Why weren't the Hubble light-echo images of V838 Monocerotis supplemented by ground telescopes?" Tags="&lt;light&gt;&lt;dust&gt;&lt;hubble-telescope&gt;&lt;adaptive-optics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14782" PostTypeId="1" CreationDate="2016-05-05T08:44:10.907" Score="-7" ViewCount="464" Body="&lt;p&gt;I really want to know whether NASA's Moon mission really was a hoax, or if it is a fact. Since there are many points and scenarios which proves it as being filmed in a film studio with controlled conditions and not exactly on from moon's surface.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I also personally believe that we can not send human beyond &quot;Van Allen Radiation Belt&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Also mission was announced in 1961 and in just 8-9 years we were able to send humans to the Moon.(really, I think is not possible with today's technology as well. It will at least take 10-15 years to actually do it today).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The ISS is orbiting at a few hundred km, and still it has lot of risk factors involved. Crew members are really had near death &lt;a href=&quot;https://en.wikipedia.org/wiki/International_Space_Station_maintenance&quot; rel=&quot;nofollow&quot;&gt;experiences&lt;/a&gt;&#xA;while carrying out maintenance.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;And we are able to go to 370,300 km away from earth and safely return back (what ! ! ! really not possible considering today's technology as well-at least for humans).&#xA;So are these Moon missions a hoax, or a forced lie ? &#xA;I don't want to consider any political aspect here (like between the US and Russia etc.) I just want to know whether humans were on the Moon or not ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11734" LastEditorUserId="10172" LastEditDate="2016-05-05T09:53:13.020" LastActivityDate="2016-05-05T10:48:58.310" Title="Was NASA's moon mission completely fake?" Tags="&lt;earth&gt;&lt;space-travel&gt;" AnswerCount="2" CommentCount="11" ClosedDate="2016-05-05T13:19:53.213" ContentLicense="CC BY-SA 3.0" />
  <row Id="14783" PostTypeId="2" ParentId="14781" CreationDate="2016-05-05T09:33:41.657" Score="3" Body="&lt;p&gt;Adaptive optics does not work (or at least did not 10+ years ago) on ground based astronomical telescopes at visible wavelengths. No images remotely comparable to HST could be obtained for faint&lt;span class=&quot;math-container&quot;&gt;$^*$&lt;/span&gt; objects from the ground in the BVI wavebands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I checked for papers discussing adaptive optics at other wavelengths and couldn't find any. The field of view is a factor. Back in 2003 there wouldn't have been instruments that could have covered V838 Mon at IR wavelengths with NIR AO observations. It can just about be done now at &lt;a href=&quot;https://www.gemini.edu/?%20q=node/10235&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gemini South&lt;/a&gt;, but field of view is still an issue for AO visible wavelength observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^*$&lt;/span&gt; NB &quot;Lucky imaging&quot; or speckle imaging can work from the ground for resolving binaries etc, but is not so good for imaging faint, extended objects. Having said that, I have seen some images of things like planetary nebulae taken with Lucky Imaging, but that is now and that was then. Nobody was doing it in 2003.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-10-26T06:37:30.127" LastActivityDate="2018-10-26T06:37:30.127" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="14784" PostTypeId="2" ParentId="14782" CreationDate="2016-05-05T09:48:16.003" Score="3" Body="&lt;p&gt;While &quot;was the Moon landing a hoax?&quot; is starting to be a tiresome genre, the points you make is somewhat legitimate. I am not going to address the issue in general, (Best to go &lt;a href=&quot;https://skeptics.stackexchange.com/questions/1128/has-man-walked-on-the-moon&quot;&gt;here&lt;/a&gt; if you want that) as others have done that better before. But here you go for your issues specifically:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&quot;Van Allen Radiation Belt&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Yes, it exists, and the radiation is about 10000 times more intense than on Earth. But in a transfer orbit towards the Moon, they where only inside it for a few hours. Sure, that still corresponds to a few years of radiation on Earth, but as you may be aware of, that is not enough to kill you.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Not enough time?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The decision in 1961 to send humans to the Moon did not come all of a sudden. It is not like all the technology required had to be invented after that. For instance, the Soviet &lt;a href=&quot;https://en.wikipedia.org/wiki/Luna_3&quot; rel=&quot;nofollow noreferrer&quot;&gt;Luna 3&lt;/a&gt; probe had already performed a flyby of the Moon, taking the first images of the far side. From there, it is pretty much just scaling things up, and make a suitable landing craft. (That is still expensive, though.)&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Risk&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Who said there was no risk involved? Out of the seven missions intended to land on the Moon, one of them, the Apollo 13, barely managed to return after an explosion on board. That is a (although the sample size is small) 14% risk of failure, much higher than any of the missions to the ISS. Also, the space station has operated for years, in contrast to the just over a week Apollo missions.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Distance&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In space, there is nothing to stop you if you have velocity in one direction. Therefore, you can literary travel for ever, with only a modest amount of speed. Therefore, distance is not a good way to measure difficulty. What really matters is how much &lt;a href=&quot;https://en.wikipedia.org/wiki/Delta-v&quot; rel=&quot;nofollow noreferrer&quot;&gt;change in velocity&lt;/a&gt; you need. The ISS travels at about 7800 m/s, and a transfer orbit towards the Moon is only 40% more than that.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="-1" LastEditDate="2017-04-13T12:46:44.910" LastActivityDate="2016-05-05T10:04:56.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14786" PostTypeId="2" ParentId="14782" CreationDate="2016-05-05T10:48:58.310" Score="0" Body="&lt;p&gt;As @Hohmannfan pointed out, but in addition, there is solid proof that humans have landed on the moon, with several major incidents with the various Apollo missions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the missions put a reasonably large retro-reflector array on the surface of the moon to reflect back lasers with the purpose of measuring the moon's distance to the earth, accurate to within millimetres. That alone I consider to be the biggest most undeniable piece of evidence.&lt;/p&gt;&#xA;" OwnerUserId="11525" LastActivityDate="2016-05-05T10:48:58.310" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14787" PostTypeId="5" CreationDate="2016-05-05T13:34:02.600" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2016-05-05T13:34:02.600" LastActivityDate="2016-05-05T13:34:02.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14788" PostTypeId="4" CreationDate="2016-05-05T13:34:02.600" Score="0" Body="Questions regarding the subject of taking photographs of the night sky using digital or film imaging systems as well as the development and processing of the images." OwnerUserId="876" LastEditorUserId="9301" LastEditDate="2016-10-04T20:45:10.723" LastActivityDate="2016-10-04T20:45:10.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14791" PostTypeId="1" AcceptedAnswerId="14792" CreationDate="2016-05-05T18:17:54.557" Score="-5" ViewCount="1267" Body="&lt;p&gt;Is it possible for the Sun and the Moon to crash into each other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, under what conditions? If no, why?&lt;/p&gt;&#xA;" OwnerUserId="11739" LastEditorUserId="6" LastEditDate="2016-05-05T18:21:20.443" LastActivityDate="2016-05-05T22:37:44.330" Title="Is it possible for the Sun and the Moon to crash into each other?" Tags="&lt;the-sun&gt;&lt;the-moon&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2016-05-11T23:29:47.867" ContentLicense="CC BY-SA 3.0" />
  <row Id="14792" PostTypeId="2" ParentId="14791" CreationDate="2016-05-05T18:44:12.407" Score="4" Body="&lt;p&gt;Technically speaking, only the Moon would crash into the Sun.  That being said, it's certainty one of many possibilities in the far distant future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About 7.6 billion years from now, the Sun will have ballooned to a large enough size to engulf much of the inner solar system.  One of the consequences of growing into a small red giant will be mass lost due to a rapid sheding of material through a stronger solar wind.  This mass loss will allow all of the planets to be pulled on by the Sun less, and they can achieve a higher orbit.  In Earth's case, it could be as much as 1.2 AU from the center of the Sun.  Also, there will eventually be a large increase in the solar wind density where Earth orbits the Sun.  This solar wind and chromosphere will create drag on Earth while in motion around the Sun, and will rob the planet of its angular momentum (orbital velocity).  The Earth will slowly begin to transition into a smaller orbit, and some sources think that this may lead to the Earth falling into the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this scenario, Earth-Moon system is gravitationally stable, and the Earth has sufficient gravity to hold onto the Moon during this time - Earth will pull the Moon down with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is not known to me is if the drag from the solar wind and chromospher will cause the Moon to lose enough angular momentum to spiral inwards, ultimately merging with the Earth before the Earth is engulfed by the Sun.  This would be a spectacular sight, as the moon would first be ripped apart by the tidal interaction with Earth (inside the Roche limit, or about 11,500 miles from the center of the Earth).  Our Moon would resemble a string of moonlets, each eventually falling to Earth.  Either way, the result is the same - the Moon will only exist within the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.smithsonianmag.com/smart-news/earth-will-die-a-hot-horrible-death-when-the-sun-expands-and-swallows-us-and-now-we-know-what-that-looks-like-28965223/?no-ist&quot; rel=&quot;nofollow&quot;&gt;http://www.smithsonianmag.com/smart-news/earth-will-die-a-hot-horrible-death-when-the-sun-expands-and-swallows-us-and-now-we-know-what-that-looks-like-28965223/?no-ist&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.space.com/3373-earth-moon-destined-disintegrate.html&quot; rel=&quot;nofollow&quot;&gt;http://www.space.com/3373-earth-moon-destined-disintegrate.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="876" LastEditDate="2016-05-05T22:37:44.330" LastActivityDate="2016-05-05T22:37:44.330" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14793" PostTypeId="2" ParentId="14791" CreationDate="2016-05-05T18:47:48.830" Score="2" Body="&lt;p&gt;No. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon orbits the Earth at less than 405,000 km. The Earth orbits the sun at more than 147,100,000km. The orbits are stable over billions of years.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon never comes less than 146,700,000km from the sun&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible for the moon to pass in front of the sun, that is called a solar eclipse.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-05-05T18:47:48.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14794" PostTypeId="1" AcceptedAnswerId="14802" CreationDate="2016-05-05T23:19:12.573" Score="1" ViewCount="269" Body="&lt;p&gt;a fresh astronomy enthusiast here.&#xA;Recently I've been reading upon how immensely far other stars are from our solar system and how is it possible that we may be seeing stars that are already dead. My question is - if there is a star that is a millions of light years away from us and we are seeing it shine tonight, but in reality that star died long long time ago (perhaps also millions of years ago), is it possible that matter from that star somehow travelled through space to our solar system and at some point in history it collided with the Earth and now we're &quot;made&quot; of this star (made as in that star stuff is present here on Earth)?&#xA;To me, I don't see reason why not, because even though matter cannot travel faster than light, my intuition tells me that the answer to my question is yes, it is possible that this could happen, (if only in theory) given enough distance. But maybe I'm not considering some important points which are obvious to more experienced folk.&lt;/p&gt;&#xA;" OwnerUserId="11744" LastEditorUserId="11744" LastEditDate="2016-05-06T06:47:36.530" LastActivityDate="2016-05-06T22:04:33.297" Title="Are we made of stars we're seeing?" Tags="&lt;star&gt;&lt;light&gt;&lt;distances&gt;&lt;matter&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2016-05-13T01:53:11.330" ContentLicense="CC BY-SA 3.0" />
  <row Id="14795" PostTypeId="1" AcceptedAnswerId="14825" CreationDate="2016-05-06T03:06:19.230" Score="9" ViewCount="475" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Adaptive_optics#In_astronomy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Adaptive Optics&lt;/a&gt; (AO) techniques allow ground based observatories to dramatically improve resolution by actively compensating for the effects of &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_seeing&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomical Seeing&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atmospheric effects are quite variable in both time and location. A parameter called &lt;em&gt;Isoplanatic Angle&lt;/em&gt; (IPA) is used to express the angular extent over-which a given wavefront correction optimized for one point (usually a guide star, artificial or natural) will be effective. As an example, Table 9.1 in this &lt;a href=&quot;http://www.gmto.org/Resources/GMT-ID-01470-Chapter_9_AO.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Giant Magellan Telescope&lt;/a&gt; resource shows values for IPA scaling almost linearly (actually: $\sim\lambda^{6/5}$)  from 176 arcseconds at a wavelength of 20 microns to only 4.2 arcseconds at 0.9 microns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This suggests an IPA of 2 to 3 arcseconds for visible wavelengths, which taken by itself is not a killer limitation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it seems almost all currently active AO work is done exclusively in various infrared wavelengths, &lt;a href=&quot;https://astronomy.stackexchange.com/a/14783/7982&quot;&gt;apparently down to 0.9 microns but no further&lt;/a&gt;. (AO is also implemented computationally to &lt;a href=&quot;https://astronomy.stackexchange.com/a/13967/7982&quot;&gt;array data in radioastronomy&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this because the observed wavelength needs to be longer than the guide star monitoring wavelength? Because it is simply much harder and there is always Hubble above the atmosphere for visible work so it's not worth the extra effort, or is there another more fundamental reason?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not looking for speculation or opinion, I'd like a quantitative explanation (if that applies) - hopefully with a link for further reading - thanks!&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-10-25T23:23:51.787" LastActivityDate="2018-10-25T23:23:51.787" Title="Why aren't ground-based observatories using adaptive optics for visible wavelengths?" Tags="&lt;radio-astronomy&gt;&lt;angular-resolution&gt;&lt;infrared&gt;&lt;adaptive-optics&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14796" PostTypeId="1" AcceptedAnswerId="14801" CreationDate="2016-05-06T03:37:21.557" Score="-4" ViewCount="299" Body="&lt;p&gt;I was using this formula to calculate the orbital period of a satellite in days: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$T = \sqrt{(4\times \pi^2)\times \frac{R^3}{GM_{center}}}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where R^3 is the radius of the orbit, or distance of the semi-major axis, G is the gravitational universal constant, and M-center is the mass of the object being orbited. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm attempting to calculate the orbital period for a planet that has a diameter of $5,124$ kilometers, or $\frac{804,500,000}{157}$ meters, and a mass of $5.526\times 10^{13}$ Kilograms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The star, which substitutes M-center, has a mass of 3.978*10^30 Kilograms. The length of the semi-major axis is $2.3AU$, or $3.44\times 10^{11}$ meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When substituted, I get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$T = \sqrt{\frac{(4\times \pi^2)\times(3.44\times 10^{11})^3}{(6.674\times 10^{-11})\times(3.978\times 10^{30})}}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I simplified this too: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$T = \sqrt{\frac{(1.607094708736\times 10^{33})}{(6.674\times 10^{-11})\times (5.5616^{13})}}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which further simplifies too:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;T = sqrt(1.607094708736*10^33)/(2.656092*10^20)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Something definitely seems wrong at this point. But, this comes out to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$T = \sqrt{(6.0505988073304689747192491826337340724643574093066053\times 10^{12})}$ &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$T = 2.459796\times 10^{6}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I find it hard to believe that it takes a planet that long to orbit around a star that is only 2.3AU away from it. Obviously, I am doing something very wrong, but I simply do not know where. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somewhere, I am missing something. &lt;/p&gt;&#xA;" OwnerUserId="11745" LastEditorUserId="15347" LastEditDate="2017-01-04T15:18:43.027" LastActivityDate="2017-01-04T15:18:43.027" Title="Calculating Orbital Period" Tags="&lt;astrophysics&gt;&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14797" PostTypeId="1" AcceptedAnswerId="14798" CreationDate="2016-05-06T08:34:53.787" Score="5" ViewCount="308" Body="&lt;p&gt;White dwarfs consists mostly of carbon and oxygen. In my opinion, they are too hot to contain these elements in molecular form and hence chemical reactions does not happen (I think resulting CO2 will also decompose under such high temperature and pressure). I have two questions related to that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) After a few billion years, can the WD cool down enough to sustain chemical reactions which will result in production of CO2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Given, there is very high pressure and temperature, after a few billion years what will be the dominant allotrope of carbon (Do we have, in future, diamonds in the sky ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I strongly doubt given the extreme density of white dwarfs( I think density of white dwarfs wont change over time), if any of the scenario is possible. &lt;/p&gt;&#xA;" OwnerUserId="10657" LastActivityDate="2016-05-06T15:04:47.190" Title="Chemical reactions in white dwarf and carbon allotropes" Tags="&lt;white-dwarf&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14798" PostTypeId="2" ParentId="14797" CreationDate="2016-05-06T09:38:55.853" Score="6" Body="&lt;p&gt;White dwarfs are objects the size of the Earth, but with a mass more similar to the Sun. Typical internal densities are $10^{9}$ to $10^{11}$ kg/m$^{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;White dwarfs are born as the contracting core of asymptotic giant branch stars that do not quite get hot enough to initiate carbon fusion. They have initial central temperatures of $\sim 10^{8}$K, that swiftly (millions of years) drops to a few $10^{7}$K due to neutrino emission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important point to make is that after that, the interior of a white dwarf is almost &lt;em&gt;isothermal&lt;/em&gt;. This is because the degenerate electrons that provide the pressure support also have extremely long mean free paths for scattering interactions and thus the thermal conductivity is extremely high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exterior of the white dwarf is a factor of 100 cooler than the interior. The temperature drop happens over a very thin shell (perhaps 1% of the outer part of the white dwarf), where the degenerate gas transitions to becoming non-degenerate at the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This outer layer acts like an insulating blanket and makes the cooling timescales of white dwarfs very long. From interior temperatures of say $3\times 10^{7}$ K it takes a billion years or so to cool to $5\times 10^{6}$ K and then another 10 billion years to cool to around $10^{6}$K, and such white dwarfs, which must have arisen from the first stars that were born with progenitor masses of 5 to 8 solar masses, &lt;em&gt;will be the coolest white dwarfs in the Galaxy&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At these temperatures there is no possibility of the carbon undergoing chemical reactions, it is completely ionised; the carbon and oxygen nuclei are in a crystalline lattice at these densities, surrounded by a degenerate electron gas. There &lt;em&gt;is&lt;/em&gt; evidence that crystallisation does take place, via asteroseismology of some pulsating massive white dwarfs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The details of the crystalline structure in these objects is unknown, and the subject of theoretical investigation. However, diamond is pure carbon and white dwarfs are expected to be a carbon/oxygen mixture. A further complication is that the process of crystallisation may be accompanied by gravitational separation of the carbon and oxygen, so that the inner core is more oxygen-rich than the outer core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Original ideas were that the crystalline form would be body-centred cubic (bcc), but other more complex possibilites are opened up by the mixture of carbon and oxygen. bcc Carbon would be a new allotrope of carbon and not like diamond - it is a denser way of arranging the nuclei.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: To answer a point in the comments. Even if you were to wait trillions of years and allow white dwarfs to cool to the thousands or even hundreds of degrees that you might think would allow electrons to recombine and for chemistry to occur, that is not how it works. In the degenerate electron gas, the typical Fermi energy of the electrons is an MeV or so, compared with the eV-keV of bound electron states, and this is completely independent of the temperature. So the high electron number densities ensure that they will never recombine with the carbon nuclei (a theory first developed by &lt;a href=&quot;http://www.jstor.org/stable/97037?seq=1#fndtn-page_scan_tab_contents&quot; rel=&quot;nofollow&quot;&gt;Kothari 1938&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-05-06T15:04:47.190" LastActivityDate="2016-05-06T15:04:47.190" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14799" PostTypeId="2" ParentId="14794" CreationDate="2016-05-06T13:53:17.143" Score="1" Body="&lt;p&gt;Earth has &lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_year&quot; rel=&quot;nofollow&quot;&gt;orbited the galaxy 16 or more times&lt;/a&gt; since the planet formed. Given that even stars in a single open cluster usually have somewhat different velocities relative to galactic center, it's highly doubtful that we can see stars made from the exact same starting material as earth. As for the supernova that delivered the compression wave to our stellar nursery, it's likely too dim by now to be seen for more a few dozen parsecs, and might be located nearly anywhere.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2016-05-06T13:53:17.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14800" PostTypeId="2" ParentId="14795" CreationDate="2016-05-06T15:35:05.513" Score="3" Body="&lt;p&gt;The simple answer for the wavelength part is that performance of AO systems degrades the shorter in wavelength you look. The basics of what happens is as you go to shorter the wavelengths of light, you need a finer plate scale to detect variations in seeing which requires very expensive (and in some cases non-existant) hardware. You also need a higher AO frequency (ability to measure the light and deform/refocus the telescope) to account for the higher frequency of light, this again takes very expensive hardware if it exists at all at the frequency required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because some of the basic calculations (not taking &lt;a href=&quot;ftp://ftp.bioeng.auckland.ac.nz/pub/pub/jtur044/references/fitting/NOLL1976.pdf&quot; rel=&quot;nofollow&quot;&gt;Zernike polynomials&lt;/a&gt; into account) are based on the &lt;a href=&quot;http://www.telescope-optics.net/Strehl.htm&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Strehl ratio&lt;/strong&gt;&lt;/a&gt; and &lt;a href=&quot;http://flatwavefronts.com/uploads/papers/IsThatReallyYourStrehlRatio.pdf&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; (ratio of peak intensity of an aberrated image compared to perfect image) to figure out what the intensity of the source should be and the &lt;a href=&quot;https://www.noao.edu/image_gallery/text/fwhm.html&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;FWHM&lt;/strong&gt;&lt;/a&gt; (Full-Width Half Max - width of the profile of light at half intensity) to essentially measure where the light should be. Both of these measurements are wavelength dependent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basic further reading can be found at &lt;a href=&quot;http://www.ing.iac.es/Astronomy/science/neon08/neon08_AO_Rutten.pdf&quot; rel=&quot;nofollow&quot;&gt;The Isac Newton Group of Telescopes&lt;/a&gt;. Much more in-depth reading can be found at the &lt;a href=&quot;http://fp.optics.arizona.edu/jcwyant/Zernikes/Zernikes.pdf&quot; rel=&quot;nofollow&quot;&gt;university of Arizona Optics department&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11039" LastEditorUserId="11039" LastEditDate="2016-05-06T18:20:42.813" LastActivityDate="2016-05-06T18:20:42.813" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14801" PostTypeId="2" ParentId="14796" CreationDate="2016-05-06T17:16:16.273" Score="1" Body="&lt;p&gt;Keep the units with the numbers, even if you convert units. $G$ has units too; since you used the $\mathrm{m}^3 \mathrm{kg}^{-1} \mathrm{s}^{-2}$ value and put $R$ in meters and $M_{\mathrm{center}}$ in kilograms, your $T$ should be in seconds. Sanity check: by Kepler's third law $T^2 \propto R^3$, a 2.3 AU orbit around the Sun (half the mass of your central star) should take 3.5 years.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-05-06T17:16:16.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14802" PostTypeId="2" ParentId="14794" CreationDate="2016-05-06T22:04:33.297" Score="4" Body="&lt;p&gt;Stars really are immensely far, however it's a &lt;a href=&quot;https://xkcd.com/1342/&quot; rel=&quot;nofollow&quot;&gt;common misconception&lt;/a&gt; that the stars that you can see are millions of light years away. Most of the visible stars are a few tens to a few hundreds of light years away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it is possible that there are stars that have exploded in a supernova: Eta Carinae looks to be approaching the end, and is 7500 light years away (so we see it as it was 7500 years ago). It's one of the most powerful stars in the milky way, but it is so distant that it looks like a faint star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is impossible for matter from an Eta Carinae supernova to reach the solar system before light from the supernova, because matter must be slower than light. No matter how far the star is, whether near or far, light beats matter in any race. In fact, Eta Carinae is too remote for any matter from its eventual supernova to ever reach the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;starstuff&quot; of which we are made was present in the molecular cloud that collapsed to form the solar system about 4.7 billion years ago, that molecular cloud was enriched with material from ancient stars that died during the 8 billion years that elapsed before the formation of the sun.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-05-06T22:04:33.297" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14803" PostTypeId="1" AcceptedAnswerId="14820" CreationDate="2016-05-06T23:43:41.637" Score="0" ViewCount="104" Body="&lt;p&gt;From what I understand of current models, the bulge of the galaxy formed first, and thus, would contain older population II like stars. Currently, however, the halo has a higher population of stars with lower metalicity than the bulge. Is this because the density  of stars in the bulge is much higher? My thinking is that this is a densely populated region of massive hot stars that die quickly and create a newer region of next generation population I stars that are metal rich. The inverse would be true for the halo where stars are less densely packed. Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="10995" LastActivityDate="2016-05-08T12:57:26.470" Title="Metalicity and age of bulge stars vs halo" Tags="&lt;star&gt;&lt;stellar-evolution&gt;&lt;metallicity&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2016-05-11T20:23:02.197" ContentLicense="CC BY-SA 3.0" />
  <row Id="14804" PostTypeId="1" CreationDate="2016-05-07T02:40:02.623" Score="1" ViewCount="589" Body="&lt;p&gt;If we were to project the ecliptic out as an infinite plane, what visible star (besides the sun) would lie closest to it?&lt;/p&gt;&#xA;" OwnerUserId="2800" LastActivityDate="2016-05-07T23:05:50.480" Title="What visible star is closest to the ecliptic?" Tags="&lt;ecliptic&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14805" PostTypeId="1" AcceptedAnswerId="14811" CreationDate="2016-05-07T03:05:50.140" Score="0" ViewCount="300" Body="&lt;p&gt;Let's assume 'solar eclipse' occurred&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, at what degree 'moon' will be in 'with reference to' earth (and)   at what degree 'moon' will be in 'with reference to' sun&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, by 'how many degrees' (out of 360 deg) the moon would have orbited 'earth' and 'sun' in the event of 'solar eclipse'?&lt;/p&gt;&#xA;" OwnerUserId="11000" LastEditorUserId="2365" LastEditDate="2017-05-25T18:23:00.360" LastActivityDate="2017-05-25T18:23:00.360" Title="How far does the moon move relative to the sun during a solar eclipse" Tags="&lt;solar-eclipse&gt;&lt;the-sun&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14806" PostTypeId="2" ParentId="1715" CreationDate="2016-05-07T06:10:26.047" Score="5" Body="&lt;p&gt;Using microlensing the MOA (Microlensing Objects in Astrophysics), OGLE (Optical Gravitational Lensing Experiment) groups have found many free-floating planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stars, free floating planets etc are all moving around the center of our galaxy. They are moving at different speeds, so very occasionally a foreground object passes through the direct line of sight to a background star. When this happens the gravity of the foreground objects acts like a lens magnifying the image of the background star. As the alignment gets better the brightness of the background object appears to brighten. It fades again as the alignment worsens. &#xA;With very close alignments the apparent brightness of the background star can increase a 1000 fold. The duration of the the rise and fall depends mostly on the mass of the foreground objects. For Jupiter mass planets it's about 4 days, for earth mass planets its on the order of several hours. For a single forground object the rise and fall in brightness of the background star is a very smooth and well known shape. If the foreground object is part of a binary this curve gets distorted with extra bumps, dips and other anomalies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that microlensing does not need to detect any light from the foreground object, so it could be in order of mass, unlit planet free from any star, a very dim star, normal star, white dwarf, neutron star or even a black hole&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The MOA and Ogle groups monitor millions of stars per night. They find 1000+ microlensing events per year. A small fraction of these are less than a day in length, and have no signs of extra bumps and wiggles. So they are from free-floating planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However measuring the mass of a single lens requires a lot of observations and second order effects. If the background star is large in angular diameter the microlensing light curve it distorted. Modeling these distortions with a estimate of the type of background star, yields and estimate of the mass of the lensing object. If observing the same event from 2 places, it is possible to measure a time delay between when the light arrive at each place. This yields an estimate of the distance to the foreground object. This with knowing the type of the background star yields a mass estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A free-floating planet with a moon has possibly been found.&#xA;see the MOA web site &lt;a href=&quot;http://www.phys.canterbury.ac.nz/moa/&quot; rel=&quot;noreferrer&quot;&gt;http://www.phys.canterbury.ac.nz/moa/&lt;/a&gt; for more details about finding rogue planets using microlensing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Kepler spacecraft and microlensing teams are in a joint campaign, the main aim of which is to detect and then measure the mass of free-floating planet. Because Kepler is far from earth there is a substantial delay between it's light-curves and those measured from earth. See &#xA;&lt;a href=&quot;http://www.nasa.gov/feature/ames/kepler/searching-for-far-out-and-wandering-worlds&quot; rel=&quot;noreferrer&quot;&gt;http://www.nasa.gov/feature/ames/kepler/searching-for-far-out-and-wandering-worlds&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10761" LastActivityDate="2016-05-07T06:10:26.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14807" PostTypeId="2" ParentId="14804" CreationDate="2016-05-07T15:00:03.043" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Regulus&quot; rel=&quot;nofollow&quot;&gt;Regulus ($\alpha$ Leo)&lt;/a&gt; is the one to beat, with ecliptic latitude $\beta$ = +0.46$^\circ$ and apparent V magnitude +1.4.&#xA;Some dimmer stars are closer to the ecliptic, e.g.&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Alpha_Librae&quot; rel=&quot;nofollow&quot;&gt;$\alpha$ Lib&lt;/a&gt; ($\beta$=+0.33$^\circ$, V=+2.7),&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Delta_Geminorum&quot; rel=&quot;nofollow&quot;&gt;$\delta$ Gem&lt;/a&gt; ($\beta$=-0.18$^\circ$, V=+3.5), and&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Delta_Cancri&quot; rel=&quot;nofollow&quot;&gt;$\delta$ Cnc&lt;/a&gt; ($\beta$=+0.08$^\circ$, V=+3.9).&#xA;For more suggestions, poke around in Stellarium with the Ecliptic turned on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want linear distance from the ecliptic plane, multiply the distance from us by $\sin \beta$.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2016-05-07T21:04:03.513" LastActivityDate="2016-05-07T21:04:03.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14808" PostTypeId="1" AcceptedAnswerId="28676" CreationDate="2016-05-07T15:46:39.433" Score="5" ViewCount="3735" Body="&lt;p&gt;If we were to make telescopes with Fresnel lenses instead of regular lenses, would it be more practical?  Would this also allow for much bigger telescopes to be made?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example of the type of lens I am talking about from &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Fresnel_lens-1.svg&quot; rel=&quot;noreferrer&quot;&gt;https://commons.wikimedia.org/wiki/File:Fresnel_lens-1.svg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bRwH9.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bRwH9.png&quot; alt=&quot;Here is an example of a Fresnel Lens&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="1386" LastEditDate="2016-05-07T18:06:09.913" LastActivityDate="2018-12-08T21:17:42.147" Title="Would a telescope that uses Fresnel lenses be more practical than using regular lenses?" Tags="&lt;telescope&gt;&lt;telescope-lens&gt;" AnswerCount="6" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14809" PostTypeId="2" ParentId="14805" CreationDate="2016-05-07T16:09:30.000" Score="0" Body="&lt;p&gt;Total and annular solar eclipses occur where the Sun-Earth-Moon angle is 0$^\circ$.&#xA;Where that angle is smaller than 0.5$^\circ$ but not 0$^\circ$, there is a partial solar eclipse.&#xA;The Sun-Moon-Earth angle at those times is 180$^\circ$.&#xA;Due to the Moon's orbital inclination, most new moons miss eclipse alignment, some by as much as 5$^\circ$.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-05-07T16:09:30.000" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14810" PostTypeId="2" ParentId="14808" CreationDate="2016-05-07T17:50:23.277" Score="9" Body="&lt;p&gt;The main advantage of a Fresnel lens is its reduced mass compared to a normal lens. Its disadvantages include difficulty of manufacture and diffraction from the discontinuities in the aperture, these I suspect are killers given the practicality of the alternatives, reflecting telescopes, to a Fresnel lens refactor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diffraction will give reduced contrast in the image, already noticeable in a reflector with partially obstructed aperture, but may well be worse with a Fresnel lens. But the difficulty in manufacture may well be the real limit given the relative ease of manufacturing large mirrors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I'm at it; it seems as thought making a achromatic Fresnel multiplet might be quite fiddly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We might also note where there is no alternative, systems analogous to Fresnel lenses are used. The example I have in mind are &lt;a href=&quot;http://imagine.gsfc.nasa.gov/science/toolbox/xray_telescopes1.html&quot; rel=&quot;noreferrer&quot;&gt;grazing incidence x-ray telescopes&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2016-05-07T18:20:32.937" LastActivityDate="2016-05-07T18:20:32.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14811" PostTypeId="2" ParentId="14805" CreationDate="2016-05-07T18:14:32.550" Score="0" Body="&lt;p&gt;If I understand your question correctly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon and sun have a visual angle of ~32 Arcminutes. So during a total solar eclipse, from when it begins to when it ends, the moon traverses ~64 Arcminutes (one edge touching to edge of the sun to the opposite side at the end which is two diameters) relative to the sun. But in actuality it will have travelled further with respect to your viewing location.&lt;/p&gt;&#xA;" OwnerUserId="11525" LastEditorUserId="11525" LastEditDate="2016-05-08T07:10:47.023" LastActivityDate="2016-05-08T07:10:47.023" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14812" PostTypeId="2" ParentId="14804" CreationDate="2016-05-07T20:04:10.127" Score="2" Body="&lt;p&gt;HIP 76880 = $\kappa$ Librae (V=4.72) has an Ecliptic latitude of -0.019 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the winner amongst all stars in the Hipparco/Tycho catalogue with a Hipparcos magnitude &amp;lt;6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mean physical distance, rather than angular distance, then the sine of the ecliptic latitude must be multiplied by a distance estimate for the star. This cannot be conclusive, the distances to many naked eye stars are very uncertain, a small fraction of Hipparcos parallaxes are too uncertain to be useful in this regard.&#xA;However, from those with decent parallaxes then &lt;strong&gt;HIP 3765 = HD 4628&lt;/strong&gt; (V=5.72) has an absolute distance of 0.036 pc above the ecliptic plane.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-05-07T23:05:50.480" LastActivityDate="2016-05-07T23:05:50.480" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14813" PostTypeId="2" ParentId="14808" CreationDate="2016-05-07T20:31:14.477" Score="3" Body="&lt;p&gt;A group at LLNL have worked on it with DARPA funding. They had a folding glass system called &lt;a href=&quot;https://str.llnl.gov/str/March03/Hyde.html&quot; rel=&quot;nofollow&quot;&gt;Eyeglass&lt;/a&gt; in 2003 and a flexible membrane system called &lt;a href=&quot;https://str.llnl.gov/january-2013/britten&quot; rel=&quot;nofollow&quot;&gt;MOIRE&lt;/a&gt; in 2013. They embraced the diffraction and put it to work, totally unlike refractive optics as we know them.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-05-07T20:31:14.477" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14815" PostTypeId="1" CreationDate="2016-05-08T03:43:03.297" Score="1" ViewCount="93" Body="&lt;p&gt;I went through Prof. Hitoshi Murayama's online lecture where he describes why the universe is flat. One of the evidences for a flat universe is the evenness of CMB radiation on all directions. My understanding is, that a uniform CMB pattern only indicates that inflation or early expansion of the universe was more or less equal on all directions. Why is this cited for the claim that the universe is flat, or in what sense is the universe 'flat'?    &lt;/p&gt;&#xA;" OwnerUserId="5395" LastActivityDate="2016-05-08T07:08:35.080" Title="Intuitive explanation for why the universe is flat" Tags="&lt;cosmological-inflation&gt;&lt;cmb&gt;&lt;observable-universe&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14816" PostTypeId="2" ParentId="14815" CreationDate="2016-05-08T07:08:35.080" Score="4" Body="&lt;p&gt;The CMB lets us measure how close to flat the universe is right now.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, inflation tries to explain how we got from whatever the early universe was to right now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motivation for the latter being that even extremely small deviations from perfect flatness in the early universe should have resulted in very obvious deviations from flatness today.  That we don't see obvious deviations then suggests the universe was amazingly close to perfectly flat early on, and such perfect tuning seems bizarre. Inflation basically lets a (in some sense &quot;very&quot;) non-flat early universe get flattened to this precision, thereby resolving the &quot;problem&quot; of having an almost perfectly flat early universe.  It takes the situation from an extremely narrow range of initial conditions for the flatness of the universe and gives us a broad range of equally valid possibilities, instead.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inflation is also commonly used to explain the remarkable homogeneity of the CMB, which suggests that parts of the universe that cannot ever have received light speed signals from each nevertheless achieved thermal equilibrium (an at-most light speed process).  Inflation throws in a super-luminal expansion to allow these regions to be within light speed communications of each other in the very early universe, with just enough time to achieve thermal equilibrium, and then inflates them beyond light speed communication.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2016-05-08T07:08:35.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14817" PostTypeId="1" CreationDate="2016-05-08T07:52:21.333" Score="4" ViewCount="1718" Body="&lt;p&gt;It is surprising that even late in 1920, people do not know there are galaxies besides the Milky Way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that all the stars visible to the naked eye are near to us? &lt;/p&gt;&#xA;" OwnerUserId="11754" LastActivityDate="2016-05-28T22:38:55.213" Title="Do all the stars visible to the naked eye belong to Milky Way?" Tags="&lt;galaxy&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14818" PostTypeId="2" ParentId="14817" CreationDate="2016-05-08T09:00:14.750" Score="12" Body="&lt;p&gt;All stars visible &lt;em&gt;individually&lt;/em&gt; to the naked eye belong to the Milky Way. &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_galaxies#Naked-eye_galaxies&quot;&gt;A handful&lt;/a&gt; of galaxies are visible to the naked eye, but here you only see the combined light of millions to billions of stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only exception is when a massive star ends its life, exploding as a supernova. This can be visible to the naked eye for a brief perdiod of time. Most notably, the gamma-ray burst &lt;a href=&quot;https://en.wikipedia.org/wiki/GRB_080319B&quot;&gt;GRB 080319B&lt;/a&gt; was visible to the naked eye for roughly 30 seconds in 2008, despite its distance from us being $\sim10^5$ times the size of the Milky Way.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="7982" LastEditDate="2016-05-08T09:34:35.490" LastActivityDate="2016-05-08T09:34:35.490" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14819" PostTypeId="2" ParentId="14808" CreationDate="2016-05-08T12:21:21.450" Score="4" Body="&lt;p&gt;In addition to the diffraction issues pointed out by Conrad Turner, a Fresnel lens would suffer from all the other problems that large lenses in telescopes suffer from:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Chromatic aberration&lt;/li&gt;&#xA;&lt;li&gt;Differential sagging under its own weight (&quot;differential&quot; because how it sags will vary depending on how the telescope is pointed)&lt;/li&gt;&#xA;&lt;li&gt;Thermal expansion and contraction&lt;/li&gt;&#xA;&lt;li&gt;The inability to correct for the previous two variations by programmatically adjusting the shape of the lens (which can be done for &lt;em&gt;mirrors&lt;/em&gt; by attaching mechanical actuators to the back side of the mirror)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2016-05-08T12:21:21.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14820" PostTypeId="2" ParentId="14803" CreationDate="2016-05-08T12:57:26.470" Score="2" Body="&lt;p&gt;Your intuition is largely correct: the key is that the proto-bulge region had a deep enough potential well so that the supernovas couldn't expel the remaining gas, and so new stars could form out of the gas (enriched by the supernova ejecta) in a continuing cycle. In the low-mass, isolated protogalactic clouds which probably contributed to the halo, the initial round of supernovas ejected most of the gas (including the original gas that hadn't yet formed stars) -- thus, little opportunity to form more stars (out of higher-metallicity gas) on a continuing basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's really the total mass in a given region that matters. For example, the density of stars in the central regions of a globular cluster is pretty high, but the mass of the cluster as a whole isn't enough to keep most of the gas when its massive stars go supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Side note: the traditional terms &quot;Population I&quot; and &quot;Population II&quot; aren't used all that much any more, since age and metallicity can vary continuously and aren't always strongly associated.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2016-05-08T12:57:26.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14821" PostTypeId="1" AcceptedAnswerId="14822" CreationDate="2016-05-08T14:27:47.887" Score="3" ViewCount="3314" Body="&lt;p&gt;Obviously all planets are not in same perfect plane. Because if at all they were in the exact same plane, it would mean that Mercury transits and Venus transits would not be that rare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any explanation would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="11255" LastEditorUserId="823" LastEditDate="2016-05-13T19:44:04.857" LastActivityDate="2021-02-10T17:03:54.117" Title="Why aren't all planets in the same plane?" Tags="&lt;orbit&gt;&lt;mercury&gt;&lt;horizon&gt;&lt;planetary-transits&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14822" PostTypeId="2" ParentId="14821" CreationDate="2016-05-08T17:38:47.900" Score="4" Body="&lt;p&gt;Your reasoning is correct: if Mercury orbited in the same plane as Earth, we'd see it transit the Sun every 4 months or so. In fact these orbital planes are inclined 7 degrees to each other, and the other major planets' orbits are inclined 1 to 3 degrees relative to Earth's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The planets perturb each other's orbits slightly, so no planet's orbit is perfectly planar.&#xA;However, the Solar system average plane of all orbits and rotations is&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Invariable_plane&quot; rel=&quot;nofollow&quot;&gt;invariable&lt;/a&gt;,&#xA;and most individual planets' orbits will remain near it for millions of years.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-05-08T17:38:47.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14823" PostTypeId="1" CreationDate="2016-05-08T20:11:03.743" Score="0" ViewCount="75" Body="&lt;p&gt;As I understand it, light that is emitted from a source is not imparted with the motion of the source and so always follows a &quot;straight line&quot;. If this is correct, I am having a difficult time conceiving how the Lunar Laser Ranger experiments can detect photons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this experiment, a pulse of light is aimed towards a retro-reflector and it is reflected back. It takes about 2.5 seconds round trip and due to diffraction, the returning pulse covers a circle of approximately 20kms in diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth's orbital velocity is approximately 30km/s and in the time the pulse takes to make the round trip, the detector would be 75kms further along the Earth's orbital path. If my first paragraph is accurate, and the light pulse is not imparted with the Earth's orbital velocity, then how is detection of the returning light pulse achieved?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this makes sense and thank you.&lt;/p&gt;&#xA;" OwnerUserId="11759" LastActivityDate="2016-06-08T03:11:22.457" Title="LLR &amp; Orbital Motion" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;earth&gt;&lt;light&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14824" PostTypeId="2" ParentId="14823" CreationDate="2016-05-08T20:24:55.243" Score="1" Body="&lt;p&gt;Well it kind of makes sense, but what is important here is the Earth's velocity relative to the Moon, not the Sun!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly: the Moon-Earth separation is 400,000 km and a complete orbit takes 27 days. Thus the relative orbital speed is more like 1 km/s and so a 20km diameter beam easily covers the expected relative motion of about 2.5 km during the 2.5s round trip. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-05-08T20:24:55.243" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14825" PostTypeId="2" ParentId="14795" CreationDate="2016-05-08T22:09:34.753" Score="10" Body="&lt;p&gt;There's a pretty good discussion at &lt;a href=&quot;http://www.eso.org/sci/facilities/develop/ao/what_ao.html&quot; rel=&quot;noreferrer&quot;&gt;this page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several factors at work:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The smaller isoplanatic angle, as you note. This limits how much of the sky you can observe with AO, since your target needs to be within the isoplanatic angle of a bright enough references star. (Even with laser guide stars, there is still a need for a reference star for &quot;tip/tilt&quot; correction.) The difference in angular area on the sky means that the area of the sky that can theoretically be observed with AO will be about 20 times larger in the near-IR than in the optical, just from the difference in isoplanatic angle.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The effects of turbulence are stronger and have shorter timescales in the optical. This has three effects:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. The corrective optics (e.g., deformable mirror) need to have more movable parts (&quot;a near-perfect correction for an observation done in visible light (0.6 microns) with an 8-m telescope would require ~ 6400 actuators, whereas a similar performance at 2 microns needs only 250 actuators.&quot;) and need to operate on a faster timescale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. In addition to the electromechanical complexity, you'll have to do far more in the way of calculations to drive all those actuators, and on a shorter timescale. So the computing power required goes up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. In order to provide the inputs for the corrective computations, you need to observe the reference star on a much finer angular scale (&quot;A large number of actuators requires a similarly large number of subapertures in the wavefront sensor, which means that for correction in the visible, the reference star should be ~ 25 times brighter than to correct in the infrared.&quot;). This limits how much of the sky you can do AO for even more: a star that might be bright enough in the near-IR to correct a region 20-30 arcsec wide isoplanatic patch won't be bright enough to correct the corresponding 5-arcsec-wide isoplanatic patch in the visible.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In order to make corrections, you need to observe the reference object in the optical. This is easy to do with a near-IR setup using an optical/IR beamsplitter: send the optical light to the AO equipment and send the near-IR light to the near-IR instrument. In the optical, you use an optical beamsplitter to send half the light to the instrument and the the other half to the AO equipment. This means that the AO equipment only gets half the light it would if it were used with a near-IR instrument, which makes it (even) harder to do the corrections.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Finally, there is an issue unrelated to the AO itself, which is that you need different science instruments depending on whether you're working in the optical or the near-IR. Optical instruments use silicon CCDs for detection; these are only sensitive out to about 0.9-1 microns. Near-IR instruments use different detectors (usually HgCdTe-based), which are good from about 1-3 microns. (Near-IR instrument also need a different design to reduce contamination from thermal emission from the telescope and optics for observations at wavelengths longer than 2 microns.) So in practice the choice has been: combine AO with a near-IR instrument and get good performance with affordable/feasible technology, or combine AO with an optical instrument and get very limited performance with more expensive (or even, until recently, unattainable) technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, there &lt;i&gt;are&lt;/i&gt; some optical AO systems starting to appear, such as &lt;a href=&quot;http://arxiv.org/abs/1407.5098&quot; rel=&quot;noreferrer&quot;&gt;MagAO on the Magellan telescope&lt;/a&gt; (which has both an optical instrument &lt;em&gt;and&lt;/em&gt; a near-IR instrument, and can correct for both simultaneously).&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2016-05-09T09:23:20.730" LastActivityDate="2016-05-09T09:23:20.730" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="14826" PostTypeId="1" AcceptedAnswerId="14836" CreationDate="2016-05-08T23:42:00.553" Score="0" ViewCount="1478" Body="&lt;p&gt;In multiple articles I have seen the age of a star, within the milky way, referred to as its redshift (typically denoted by $z$). I know that $z$ can be calculated as $z=\frac{\lambda_{obsv} - \lambda_{emit}}{\lambda_{emit}}$ but what I can't find is how this is then used to calculate the age of a star within our galaxy? What characteristic does the redshift show us (spin, velocity, temperature, etc.) and how is this used mathematically to calculate age? Or is it just a method to determine where it falls on the HR diagram?&lt;/p&gt;&#xA;" OwnerUserId="10995" LastEditorUserId="11031" LastEditDate="2016-05-09T13:22:34.883" LastActivityDate="2016-05-12T07:46:07.483" Title="Redshift to calculate age of stars" Tags="&lt;stellar-evolution&gt;&lt;redshift&gt;" AnswerCount="1" CommentCount="12" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14827" PostTypeId="1" CreationDate="2016-05-08T23:45:30.720" Score="5" ViewCount="454" Body="&lt;p&gt;I was curious about how much brighter it would be and how much further we could see.&lt;/p&gt;&#xA;" OwnerUserId="11031" LastActivityDate="2016-12-26T10:33:28.097" Title="What would the night sky look like if the interstellar medium didn't exist to absorb or block light?" Tags="&lt;light&gt;&lt;interstellar-medium&gt;&lt;night-sky&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14828" PostTypeId="1" CreationDate="2016-05-09T03:21:21.607" Score="5" ViewCount="164" Body="&lt;p&gt;The first measurements of the (absolute) size of the solar system was made using the Transit of Venus, an event that arguably will only happen twice in a lifetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transits of Mercury occur far more often. Why have astronomers had to wait for Venus Transits to make historical measurements of the size of the solar system?&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2016-05-10T17:29:47.777" Title="Why was the size of the solar system not defined by Mercury Transits?" Tags="&lt;solar-system&gt;&lt;history&gt;&lt;planetary-transits&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14829" PostTypeId="2" ParentId="14827" CreationDate="2016-05-09T06:29:41.347" Score="7" Body="&lt;p&gt;The main effect would be that the Milky Way would become much more prominent and asymmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment, our view into the Galactic plane is limited to around 1000-3000 parsecs by dust. If you look at the Galactic latitude distribution of naked eye (&lt;a href=&quot;https://astronomy.stackexchange.com/questions/6005/arent-there-more-naked-eye-visible-stars-in-the-milky-way-plane/10260#10260&quot;&gt;Aren&amp;#39;t there more naked-eye-visible stars in the Milky Way plane?&lt;/a&gt; ) you see that most naked eye stars are much closer than this and there is only a modest concentration towards the Galactic plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I think this means is that there would not be a big increase in the numbers of resolved naked eye stars. It is the sensitivity of our eyes, rather than dust that limits how far we can see them. However, the numbers of unresolved stars in the Milky Way would be greatly increased. The number of stars in the resolution element in your eye would increase as distance squared, but the light received decreases as distance squared. This means each &quot;shell&quot; you add contributes equally to the observed brightness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun is in a 30000pc diameter Galactic disc, about 8000pc from the centre. Thus I estimate that going from being able to see stars to 2000pc, seeing them to anywhere from around 7000pc (away from the Galactic centre) to 23000pc (through the Galactic centre) will increase the brightness of the Milky Way by factors of 3-10, strongly concentrated towards the Galactic plane (more so than now), depending on which way you look. This asymmetry will be increased by the increasing stellar density towards the Galactic centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, the &quot;bulge&quot; would become more prominent. This is a pseudo-spherical region of diameter 4000pc, with a greater density of stars, centred on the Galactic centre. The bulge is mainly hidden by dust now, but without dust we would see a brighter, roughly circular blob towards the Galactic centre (Saggitarius), with an angular diameter of 30 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some idea of the change in the visual appearance to the Milky Way can be gained by comparing an all-sky image at visible wavelengths (affected by dust) with a 2MASS infrared version (much less affected by dust). See how the &quot;width&quot; of the Milky Way becomes more concentrated and the distribution of stars becomes smoother, with a prominent bulge at the Galactic centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/d8IZN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/d8IZN.jpg&quot; alt=&quot;Optical image of the MW&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ERoEE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ERoEE.jpg&quot; alt=&quot;IR image of the MW&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-12-26T10:33:28.097" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14832" PostTypeId="1" AcceptedAnswerId="14859" CreationDate="2016-05-09T09:27:58.183" Score="3" ViewCount="153" Body="&lt;p&gt;Kepler has found some systems with multiple planets and shown that it's likely that 7% of sun-like stars have Earth-like planets (&lt;a href=&quot;https://www.youtube.com/watch?lv=FSHYpN4Zej8&quot; rel=&quot;nofollow noreferrer&quot;&gt;video&lt;/a&gt;).  I don't think we know enough about extra solar planets to know how planetary systems are organized on average, but the multiple-planet systems are few and would be ones where the orbits of the planets are close to the star and/or flat.  In our own solar system for instance in the small strip of the sky where an observer could see the Earth transit, there are only two small sections where they could also see Venus transit, and if they did happen to see both the Earth and Venus, they would not be able to see any of our other planets transit (up to Saturn at least).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote a &lt;a href=&quot;http://plnkr.co/edit/O5zsZmUKkktATt2maJZX?p=preview&quot; rel=&quot;nofollow noreferrer&quot;&gt;small program&lt;/a&gt; to show roughly the visible area of sky where each planet could be seen to transit.  Shown is 4 degrees above or below the ecliptic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/I4gKi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I4gKi.png&quot; alt=&quot;graph of visibility&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there just not enough known about the variability of inclinations in orbits of planets in general to speculate about what percent of confirmed candidate planets might have siblings with different inclinations that we cannot see?  If other planetary systems had varying inclinations in their orbits like ours, it looks like for every 'Earth' they found, any other planets would only have about a 7% chance of also being visible.  Is there any reason to think that other systems with multiple planets might have less or more variability in the inclinations of the orbits of their planets?&lt;/p&gt;&#xA;" OwnerUserId="343" LastEditorUserId="343" LastEditDate="2016-05-09T09:41:02.867" LastActivityDate="2016-05-13T01:40:12.647" Title="Likelihood of extra planets in systems identified by Kepler?" Tags="&lt;kepler&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14833" PostTypeId="1" CreationDate="2016-05-09T15:40:29.300" Score="1" ViewCount="198" Body="&lt;p&gt;I just finished an introductory astrophysics course$^1$ and I have a lingering question that I can't seem to resolve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We learned that for the first few hundred million years, the universe was pretty boring and not much interesting$^2$ happened. We also learned, correctly or not, that the first stars started forming somewhere between 500 million and 600  million years after the Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also learned how to use redshift values to calculate age, and we talked about how the highest $z$-number we've discovered is some galaxy (GN-z11) at roughly $z=11$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &lt;a href=&quot;http://www.astro.ucla.edu/~wright/CosmoCalc.html&quot; rel=&quot;nofollow&quot;&gt;this calculator&lt;/a&gt;, we calcultaed that this galaxy apparently formed about 410 million years after the Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, this galaxy seems to be older than when astrophysicists think star formation happened. How can this possibly be? Clearly one of my assumptions is wrong, so is it:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Stars actually started forming &lt;em&gt;before&lt;/em&gt; 500 million years post-Big Bang.&lt;/li&gt;&#xA;&lt;li&gt;Using the UCLA calculator to calculate age is technically incorrect.&lt;/li&gt;&#xA;&lt;li&gt;A galaxy doesn't need stars to be considered a galaxy.&lt;/li&gt;&#xA;&lt;li&gt;Some other assumption I made is wrong which makes this post invalid.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As a follow-up question, what happens if we keep finding galaxies at higher $z$-numbers? At what point do we need to reconsider our theory about what happened in the &quot;early&quot; universe?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^1$I'm not an astrophysics major, so forgive any blatant falsehoods in this post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^2$On a macro-scale, at least.&lt;/p&gt;&#xA;" OwnerUserId="11769" LastActivityDate="2017-09-05T20:04:24.737" Title="How can we explain high redshift numbers?" Tags="&lt;redshift&gt;&lt;star-formation&gt;&lt;age&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14834" PostTypeId="1" AcceptedAnswerId="14835" CreationDate="2016-05-09T16:30:41.987" Score="2" ViewCount="192" Body="&lt;p&gt;I was wondering whether there is an easy way to approximate the number of stars in a Globular Cluster (GC) with 10^5 solar masses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can one, for instance just assume the GC is made of sun-like stars and therefore has 10^5 stars? &#xA;Or is this too simple? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your answers! &lt;/p&gt;&#xA;" OwnerUserId="11715" LastActivityDate="2016-05-09T17:38:11.147" Title="How many stars are there in a Globular Cluster of 10^5 solar masses?" Tags="&lt;observational-astronomy&gt;&lt;globular-clusters&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14835" PostTypeId="2" ParentId="14834" CreationDate="2016-05-09T17:38:11.147" Score="6" Body="&lt;p&gt;The mean mass of the initial mass function is ~0.3 solar masses (&lt;a href=&quot;http://arxiv.org/abs/1212.0939&quot;&gt;Maschberger 2013&lt;/a&gt;).  A GC with 10^5 solar masses will therefore have ~3 x 10^5 stars in it.  Since the most massive stars will already have died in a GC, the mean mass of the present-day mass distribution of the GC will actually be slightly lower, implying a slightly larger number of stars, but this will be a relatively small correction (certainly not more than a factor of 3). &lt;/p&gt;&#xA;" OwnerUserId="10371" LastActivityDate="2016-05-09T17:38:11.147" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14836" PostTypeId="2" ParentId="14826" CreationDate="2016-05-09T20:45:01.153" Score="4" Body="&lt;p&gt;The redshift that is referred to is not a Doppler redshift, but a cosmological redshift. The difference is that the former is caused by the source moving through space, while the latter is caused by the &quot;stretching&quot; of the wavelength of the light as it travels through space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cosmological redshift happens gradually along its journey, and hence it is a measure of the distance to the source. But since traveling through space takes time, it is also a measure of the &lt;em&gt;lookback time&lt;/em&gt; to the source — that is, the time that has passed since its emission. This notion of the term can thus be used as a timeline for phenomena in the Universe, and perhaps somewhat confusingly, it is sometimes used this way even when referring to local phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, Earth was formed 4.54 billion years (&quot;Gyr&quot;) ago. If some unrelated galaxy emitted light at the same time, and if that light reaches us today, then that galaxy must be at a particular distance$^\dagger$. During its journey it has been redshifted to $z \simeq 0.42$. Thus we may say that Earth was formed at redshift $0.42$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, stars (in the Milky Way or elsewhere) that are said to have formed at redshift $z=10$ and $z=15$, can also be said to have formed 13.3 Gyr and 13.5 Gyr ago, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the figure below, I plotted the relation between cosmological redshift and lookback time (assuming a &lt;a href=&quot;http://arxiv.org/abs/1502.01589&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck 2015&lt;/a&gt; cosmology). Now hug each other, Rob and NotSoSN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DnKjt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DnKjt.png&quot; alt=&quot;t_z&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;5.4 billion lightyears, actually. Not just 4.54 billion lightyears, because space has expanded in the meantime.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-05-12T07:46:07.483" LastActivityDate="2016-05-12T07:46:07.483" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14837" PostTypeId="2" ParentId="14833" CreationDate="2016-05-10T07:07:59.873" Score="2" Body="&lt;p&gt;There is evidence that this assumption is incorrect.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Stars actually started forming before 500 million years post-Big Bang.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is some evidence that &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_population#Population_III_stars&quot; rel=&quot;nofollow&quot;&gt;Population III stars&lt;/a&gt; formed as early as 200 million years after the Big Bang. The oldest known star, &lt;a href=&quot;https://en.wikipedia.org/wiki/HD_140283&quot; rel=&quot;nofollow&quot;&gt;the Methuselah Star&lt;/a&gt;, dates back to 200 million years after the Big Bang. It is a second generation star (judging from its metal content), so that means a generation of very large stars must have lived and died before this one was born.&lt;/p&gt;&#xA;" OwnerUserId="11175" LastActivityDate="2016-05-10T07:07:59.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14838" PostTypeId="2" ParentId="14817" CreationDate="2016-05-10T07:23:50.163" Score="0" Body="&lt;p&gt;There was only one time in recorded history when a star was visible to the naked eye and not within our galaxy. That was &lt;a href=&quot;https://en.wikipedia.org/wiki/SN_1987A&quot; rel=&quot;nofollow&quot;&gt;Supernova 1987A&lt;/a&gt;. When the astronomer, Ian Shelton, first saw it through a telescope, he was amazed at how bright it looked. Thinking it might be naked-eye visible, he stepped out of the observatory and saw it within his own eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since then, all the stars you see are nearby stars within our galaxy. All naked-eye visible stars are within a 17 thousand light years. Most are actually within a few hundred light years, but there are some bright stars within the &lt;a href=&quot;https://en.wikipedia.org/wiki/47_Tucanae&quot; rel=&quot;nofollow&quot;&gt;47 Tucanae globular cluster&lt;/a&gt; that are 17 thousand light years away.&lt;/p&gt;&#xA;" OwnerUserId="11175" LastActivityDate="2016-05-10T07:23:50.163" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14839" PostTypeId="1" CreationDate="2016-05-10T11:06:15.403" Score="1" ViewCount="240" Body="&lt;p&gt;For a design project I'm looking for elevation data for Mars' surface - the idea is to replicate a surface area of Mars as a physical model, using the elevation data to create a very rough 3d model of that particular area's topography.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where can I find such data?&lt;/p&gt;&#xA;" OwnerUserId="11773" LastActivityDate="2016-05-10T11:48:28.233" Title="Topography / elevation data of Mars" Tags="&lt;mars&gt;&lt;data-analysis&gt;&lt;topographic-map&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14840" PostTypeId="2" ParentId="14839" CreationDate="2016-05-10T11:48:28.233" Score="1" Body="&lt;p&gt;Try this link &lt;a href=&quot;http://geo.pds.nasa.gov/missions/mgs/megdr.html&quot; rel=&quot;nofollow&quot;&gt;http://geo.pds.nasa.gov/missions/mgs/megdr.html&lt;/a&gt;&#xA;This should have the data you're looking for, though you may need to process it a bit to get it into a format hat will be usable.&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-05-10T11:48:28.233" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14842" PostTypeId="1" AcceptedAnswerId="18759" CreationDate="2016-05-10T12:22:00.373" Score="2" ViewCount="235" Body="&lt;p&gt;We have discovered about 2500 radio pulsars. Part of them show pulsations in other bands like X-ray and gamma ray. The radiation mechanism remains unclear now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder whether there are radio pulsations caused by the spins of white dwarfs. It is the best to give reference papers.&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="2153" LastEditDate="2016-10-17T15:20:36.890" LastActivityDate="2017-02-08T15:00:04.787" Title="Are there radio white dwarfs?" Tags="&lt;radio-astronomy&gt;&lt;pulsar&gt;&lt;white-dwarf&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14843" PostTypeId="2" ParentId="14828" CreationDate="2016-05-10T17:29:47.777" Score="10" Body="&lt;p&gt;It is indeed possible to measure the AU using transits of Mercury, and Edmund Halley tried to do just that in 1677.  However, there are two advantages to a transit of Venus.  The first is that during the transit, Venus is only 0.28 AU from Earth, whereas Mercury is about 0.7 AU away.  This makes the parallactic effect twice as big.  The second difficulty is that you need to measure the time between second contact and third contact (i.e., the first and last time that the planet is entirely in front of the Sun).  Mercury is so small that it is hard to determine when second and third contact are.  (In fact the limiting effect for the measurement of the AU during the transit of Venus was the so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Black_drop_effect&quot;&gt;&quot;black drop effect&quot;&lt;/a&gt;, which made it impossible to measure the times of second and third contact to a precision less than about ten seconds.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A former professor of mine from when I was in graduate school has a really excellent writeup of the history of these measurements:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit4/venussun.html&quot;&gt;http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit4/venussun.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10371" LastActivityDate="2016-05-10T17:29:47.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14844" PostTypeId="1" AcceptedAnswerId="14845" CreationDate="2016-05-10T22:09:18.637" Score="1" ViewCount="377" Body="&lt;p&gt;Why did Mercury not appear to transit through the middle of the sun (the equator)? I initially thought that this was because of the ecliptic, but, then I thought that actually we'd see it above the solar equator, wouldn't we? So I'm confused. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;https://www.youtube.com/watch?v=AhWMOkrzKzs&quot; rel=&quot;nofollow&quot;&gt;video of the transit&lt;/a&gt; from NASA shows exactly what I'm talking about. Mercury is shown relatively low on the solar disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking specifically for why it was visible in the bottom half of the sun, not the top half?&lt;/p&gt;&#xA;" OwnerUserId="11780" LastEditorUserId="11780" LastEditDate="2016-05-11T12:20:25.093" LastActivityDate="2016-05-11T19:42:08.270" Title="Why did Mercury not appear to transit through the middle of the sun?" Tags="&lt;solar-system&gt;&lt;planetary-transits&gt;&lt;ecliptic&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14845" PostTypeId="2" ParentId="14844" CreationDate="2016-05-10T22:37:48.100" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_inclination&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mercury's orbit is inclined 7 degrees&lt;/a&gt; relative to Earth's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This drawing is exaggerated. When Mercury is &quot;between&quot; us and the Sun it might be below, above, or passing through the Sun's disc from our point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/r1i2U.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/r1i2U.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.universetoday.com/14009/orbit-of-mercury/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Another post from Universe Today&lt;/a&gt; explains this also.&lt;/p&gt;&#xA;" OwnerDisplayName="user10250" LastEditorDisplayName="user10250" LastEditDate="2016-05-11T14:58:34.530" LastActivityDate="2016-05-11T14:58:34.530" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14847" PostTypeId="1" AcceptedAnswerId="14848" CreationDate="2016-05-11T10:05:24.547" Score="0" ViewCount="602" Body="&lt;p&gt;I am currently following a class of observational astronomy lab. I will present a brief description of the method in first and then proceed to the question, so anyone is welcomed to read the entire post or move directly to the problem. The current question is about the Moving-Cluster method, for which I quote from Wikipedia&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In astrometry, the moving cluster method and the closely related convergent point method are means, primarily of historical interest, for determining the distance to star clusters. They were used on several nearby clusters in the first half of the 1900s to determine distance. The method is now largely superseded by other, usually more accurate distance measures.*&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZppoO.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZppoO.gif&quot; alt=&quot;An image for the Convergence point of a Cluster&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can find more here &lt;a href=&quot;http://pages.uoregon.edu/soper/Stars/movingcluster.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://pages.uoregon.edu/soper/Stars/movingcluster.html&lt;/a&gt; . A brief description of the method is like that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From photographs taken, say, 10 years apart, we can see that a star of the cluster has moved. (It has proper motion.) Let us suppose that the proper motion is an angle μ. Then the angle θ is the angle between the convergent point and the star or equally the angle between the radial and tangential speed of the moving star&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IgIXf.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IgIXf.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one knows the angle θ, he/she will be able to find the movement of the star and in the end, calculate it' s distance, through deduced relations beginning with $ v_t =v_r \tan θ.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Question&lt;/em&gt;&lt;/strong&gt;&#xA;I know the &lt;em&gt;declination&lt;/em&gt; and the &lt;em&gt;right ascension&lt;/em&gt; of both the star and the convergent point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem as is to find the angle θ. In the exercise I have it is requested to find the angle between the star and the convergent point, that is θ , using spherical trigonometry. My mind goes on to using the cosine relationship:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \cos a = \cos b \cos c + \sin b \sin c \cos A ,$$ for the triangle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Vz9RO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vz9RO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, although it may be simple, I cannot understand where this triangle is to be implemented. What are it' s angles in my problem. Wherever I have looked, the angle θ is taken as granted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, should I use something like the triangle? If yes, what the triangle should be? Should the Sun be on one angle as the image above with the convergent point? Should I define another triangle? In the end, for whatever triangle I have thought, how will I calculate the needed sides? Note that I know the &lt;em&gt;right ascension&lt;/em&gt; and the &lt;em&gt;declination&lt;/em&gt; of both the convergent point and the star's.&lt;/p&gt;&#xA;" OwnerUserId="11783" LastActivityDate="2016-05-11T16:49:17.400" Title="Moving-Cluster method for determination of the distance of Hyades. A starter problem" Tags="&lt;distances&gt;&lt;positional-astronomy&gt;&lt;observational-astronomy&gt;&lt;astrometry&gt;&lt;star-cluster&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14848" PostTypeId="2" ParentId="14847" CreationDate="2016-05-11T16:49:17.400" Score="1" Body="&lt;p&gt;The angle you are looking for&#xA;$$\cos \theta = \sin \delta_1 \sin \delta_2 + \cos \delta_1 \cos \delta_2 \cos (a_1-a_2),$$&#xA;where $(a_1, \delta_1)$ are the RA and Dec of the star and $(a_2,\delta_2)$ the RA and Dec of the convergent point.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-05-11T16:49:17.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14850" PostTypeId="2" ParentId="14844" CreationDate="2016-05-11T19:42:08.270" Score="1" Body="&lt;p&gt;As Mercury is inclined by 7 degrees relative to the orbit of the Earth (the ecliptic) at any given time it may be above the ecliptic or below. However there are two times during its orbit when it crosses the ecliptic. These are known as &quot;nodes&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A conjunction occurs when the planet is vertically above or below the sun. A superior conjunction happens when the planet is on the far side of the sun, an inferior conjunction happens when the planet is closer to the Earth than the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a node occurs at about the same time as an inferior conjunction, then the planet will appear to travel in front of the sun. Whether it travels through the middle of the solar disk, or through the upper or lower part, depends on whether the exact time of the node occurs just before the conjunction or just after. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In May 2016, Mercury crossed the ecliptic at about 08:00 (utc), and had it inferior conjuction at about 15:00. It was a little below the ecliptic by the time of conjunction, but still close enough to cross the solar disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact position of Mercury also depends on the location you are viewing from on Earth, a fact that Edmond Halley noticed could be exploited to calculate the distance to the planet, and by extension, the distance to the sun.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-05-11T19:42:08.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14855" PostTypeId="2" ParentId="14476" CreationDate="2016-05-12T04:10:15.237" Score="3" Body="&lt;p&gt;Here's a nice animation from the Washington Post:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yJSGd.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yJSGd.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1232" LastActivityDate="2016-05-12T04:10:15.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14856" PostTypeId="1" AcceptedAnswerId="14857" CreationDate="2016-05-12T07:48:34.450" Score="2" ViewCount="205" Body="&lt;p&gt;Suppose a brown dwarf requires only one more hydrogen atom to become massive enough to start fusion, what will happen if we add a hydrogen atom to it?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It fuses until the mass is dropped and then fusion stops suddenly&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It starts to fuse until all hydrogen use up&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Which one is correct?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="10437" LastEditDate="2016-05-12T13:48:28.700" LastActivityDate="2016-05-12T13:56:39.847" Title="Do stars tend to fuse all hydrogen even when the mass is small?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14857" PostTypeId="2" ParentId="14856" CreationDate="2016-05-12T13:20:20.310" Score="6" Body="&lt;p&gt;The difference between a brown dwarf and a star is not a sharp boundary. A brown dwarf is simply a ball of gas where the (small) fusion rate is incapable of providing a significant fraction of the luminosity (which is mainly provided by gravitational contraction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A star will contract and reach a minimum luminosity, whilst a brown dwarf's luminosity will monotonically decrease throughout its life. The core becomes degenerate and it is this that provides the pressure that supports a brown dwarf, even though it becomes colder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add some mass to a brown dwarf, then the result depends when you add it. If you add it early during the strong contraction phase then you might get a low-mass star. If it is added after the brown dwarf is degenerate, then it could contract further, but do so without increasing the temperature enough to ignite hydrogen strongly. This is an interesting problem that deserves a model simulation!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If fusion can start to any degree, then convective mixing in these very low mass objects will very gradually (&lt;em&gt;trillions&lt;/em&gt; of years), turn nearly all their H to He.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-05-12T13:56:39.847" LastActivityDate="2016-05-12T13:56:39.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14858" PostTypeId="1" AcceptedAnswerId="14866" CreationDate="2016-05-12T14:11:41.213" Score="1" ViewCount="145" Body="&lt;p&gt;I have received a new Celestron Powerseeker 127 (a 5' short Newtonian), however it came without a tripod mount. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideas or suggestions on how to procure the right mount our a suitable alternative would be most welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for sourcing tips rather than advice on how to choose a tripod.&lt;/p&gt;&#xA;" OwnerUserId="4114" LastEditorUserId="4114" LastEditDate="2016-05-13T04:22:16.937" LastActivityDate="2016-05-21T21:20:44.860" Title="Tripod mount missing!" Tags="&lt;telescope&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14859" PostTypeId="2" ParentId="14832" CreationDate="2016-05-12T18:01:13.827" Score="1" Body="&lt;p&gt;There are many planets that Kepler couldn't detect because the ecliptic is inclined so no transit occurs from the perspective of Earth. We are much more likely to see a transit if the transiting planet is close to the star, and so in Kepler's sample of stars, a disproportionate number of the planets discovered orbit very close to the star. It is probable that most stars have planets, and many of those planets orbit much further out than the average in the Kepler sample. And, moreover, the stars which have been found to have planets are likely to have others, orbiting further from the sun, and inclined, so they are not detectable by transits. Kepler has not detected Earth size planets orbiting in the habitable zone of sun-like stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far we do not have the technology to detect a planetary system that mirrors the Solar System. Our alien friends, if they are very close to the ecliptic may detect transits of Earth but would likely miss the other planets. And there is no reason to suspect that the Solar system is not typical in the range of inclination of its planets.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2031" LastEditDate="2016-05-13T01:40:12.647" LastActivityDate="2016-05-13T01:40:12.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14860" PostTypeId="1" CreationDate="2016-05-12T18:21:19.473" Score="2" ViewCount="236" Body="&lt;p&gt;Astronomic observations in geodesy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well understood that sun and star positions may be expressed in Greenwich hour angle and declination. Or Right Ascension Declination. And may be published reckoned to Universal Time (UT), or TT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An approximation of UT can be obtained by apply a correction (DUT) to UTC. That results in an expression of time described as UT1. DUT time offsets are published and broadcast (WWV shortwave radio) in 0.1&quot; increments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for daily precise (better than 0.1&quot;) DUT corrections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current broadcast DUT correction is -0.2&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have retrieved IERS Bulletin B, which has daily values of UT1-UTC in ms. Current value is -21.996 ms. Is it just me? Is -21.996 ms = -0.022&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is there something in the IERS Bulletin B that I'm missing?&lt;/p&gt;&#xA;" OwnerUserId="11798" LastActivityDate="2016-08-10T21:51:20.257" Title="UTC to UT1 time corrections" Tags="&lt;telescope&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14862" PostTypeId="2" ParentId="14860" CreationDate="2016-05-12T20:23:31.800" Score="1" Body="&lt;p&gt;I think you may be looking at the wrong date? -21.9964 is the first value in the current bulletin B but that is for 3/2/2016. The current bulletin B only seems to go up to 5/1/2016, with -134.3179 as the value. I think you may instead want bulletin A, which shows -0.154825s for today (5/12/2016)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://maia.usno.navy.mil/ser7/ser7.dat&quot; rel=&quot;nofollow&quot;&gt;http://maia.usno.navy.mil/ser7/ser7.dat&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10005" LastActivityDate="2016-05-12T20:23:31.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14864" PostTypeId="1" CreationDate="2016-05-12T17:09:52.200" Score="0" ViewCount="21" Body="&lt;p&gt;Astronomic observations in geodesy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well understood that sun and star positions may be expressed in Greenwich hour angle and declination. Or Right Ascension Declination. And may be published reckoned to Universal Time (UT), or TT. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An approximation of UT can be obtained by apply a correction (DUT) to UTC. That results in an expression of time described as UT1. DUT time offsets are published and broadcast (WWV shortwave radio) in 0.1&quot; increments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for daily precise (better than 0.1&quot;) DUT corrections. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current broadcast DUT correction is -0.2&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have retrieved IERS Bulletin B, which has daily values of UT1-UTC in ms. Current value is -21.996 ms. Is it just me? Is -21.996 ms = -0.022&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is there something in the IERS Bulletin B that I'm missing?&lt;/p&gt;&#xA;" OwnerUserId="11798" OwnerDisplayName="user18317" LastActivityDate="2016-05-13T00:25:52.210" Title="DUT time offsets to UTC" Tags="&lt;planet&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" ClosedDate="2016-05-13T00:32:15.803" ContentLicense="CC BY-SA 3.0" />
  <row Id="14865" PostTypeId="1" AcceptedAnswerId="14868" CreationDate="2016-05-13T06:16:32.390" Score="3" ViewCount="245" Body="&lt;p&gt;Not an expert, hence the question. To my knowledge the magnetic field of the earth travels south to north. Wouldn't there be a bias on whether the charged particles causing the auroras are negatively or positively charged? If each pole attracts the opposite, how come they're the same?&lt;/p&gt;&#xA;" OwnerDisplayName="user11801" LastEditorUserId="2270" LastEditDate="2016-05-13T11:02:27.543" LastActivityDate="2016-05-13T11:02:27.543" Title="How come the northern and southern lights are the same?" Tags="&lt;night-sky&gt;&lt;aurora&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14866" PostTypeId="2" ParentId="14858" CreationDate="2016-05-13T07:39:44.793" Score="3" Body="&lt;p&gt;A high end photographic tripod would probably work but it will be difficult to make small tracking adjustments. So a proper astronomical mount would be much better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An EQ2 mount would just about manage a scope of that size however an EQ3-2 would be a lot more stable and have the advantage that you can either get a goto version initially or upgrade at a later date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are on a tighter budget and/or don't want an equatorial mount then something like a Celestron Heavy Duty Alt/Az Tripod or Skywatcher AZ3 Alt Az Mount would be suitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the mounts mentioned are available from numerous online retailers but shop around as prices can vary.  &lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-05-13T07:39:44.793" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14868" PostTypeId="2" ParentId="14865" CreationDate="2016-05-13T10:21:37.513" Score="6" Body="&lt;p&gt;As most people are aware when they play with bar magnets, magnetic poles repel magnets of the same polarity and attract opposites. Magnets have two polarities: north and south. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electric charges are classified as positive and negative, and also attract opposites and repel charges of the same sign. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, magnetic poles and electric charges do not match up in the way your question suggests - it's not like North is positive and South is negative and they attract electric charges. Electric and magnetic fields are connected and interact (in fact, both are produced by moving electric charges) but the signs/polarity do not interact in the way you are imagining..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aurorae (northern and southern lights) are produced because charged particles are streaming toward the Earth all the time in the solar wind. Charged particles (positive or negative) are blocked by magnetic fields - the easiest way to imagine this is to draw the Earth's magnetic field like a bar magnet and the charged particles cannot cross the field lines. At the poles the field is not parallel to the Earth's surface (blocking charged particles) but perpendicular to it, so the charged particles can move into the atmosphere without crossing magnetic field lines. In the atmosphere they interact with particles in the air and create the beautiful lights that we see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in short, the lights are the same at the north and south poles because it is the same kind of particles interacting with out atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2016-05-13T10:21:37.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14869" PostTypeId="1" AcceptedAnswerId="14876" CreationDate="2016-05-13T12:19:03.923" Score="4" ViewCount="22754" Body="&lt;p&gt;I wanted to make a telescope with DIY things lying around in home. I read up that the aperture was a very important aspect and thus bought a &lt;strong&gt;convex lens&lt;/strong&gt; with &lt;strong&gt;100mm aperture&lt;/strong&gt; and &lt;strong&gt;200mm focal length&lt;/strong&gt;, to use it as an &lt;strong&gt;objective lens&lt;/strong&gt;. Apart from that, I have the following lenses lying around:-&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;1 convex lens    -   focal length 20 cm             aperture - 10 cm &lt;/li&gt;&#xA;&lt;li&gt;2 convex lenses  -   focal length 17 cm             aperture - 5 cm&lt;/li&gt;&#xA;&lt;li&gt;1 convex lens    -   focal length 2.2 cm            aperture - 2.5 cm&lt;/li&gt;&#xA;&lt;li&gt;1 concave lens   -   focal length 8.33 cm           aperture - 5.5 cm&lt;/li&gt;&#xA;&lt;li&gt;1 convex lens    -   focal length 200 cm            aperture - 5.5 cm&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I can also buy lenses of range of Power(In Dioptres) from (-14D) - (+14D) of aperture range of 5.5 cm to 7 cm, if suggested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made a telescope using lens 1(as objective) and lens 3(as eyepiece) which on calculation, gave a magnification of 9x, but according to stellafane.org and &lt;a href=&quot;http://skyandtelescope.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://skyandtelescope.com&lt;/a&gt; the minimum magnification is 14x and that I was 'wasting' the light gathered by the objective lens. This telescope showed me a bright Mars, Jupiter and Saturn, but as dots. Upon seeing the Moon, I could just see it a little bigger than what would be seen with the naked eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know the way to make the telescope, which I would like to use to observe planets and DSOs if possible. I am ready to make separate telescopes also, if suggested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know which lenses should I use, or buy and the way to use it. I live in a suburb and it is not very lighted up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way I could use the lens 1 in a telescope which I could use to view the planets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have accepted @JamesScreech 's answer and am on my way to making it. Apart from that, I would also like to use the big aperture of lens 1, for viewing DSOs, But, it has a very short focal length(just 200mm), and therefore, I am finding it hard to get any eyepiece lens which can &quot;completely accommodate the exit pupil of my scope&quot;(as said on skyandtelescope )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6WRyM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6WRyM.png&quot; alt=&quot;the screenshot of skyandtelescope page&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any way to get such a lens, or make an eyepiece barrell sort of, using a combination of lenses mentioned above or buying. And how to use it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="11807" LastEditorUserId="11807" LastEditDate="2016-05-14T06:07:03.610" LastActivityDate="2016-05-14T15:35:46.717" Title="How to make a telescope for viewing planets, moon and DSOs using a convex lens of aperture 100 mm and focal length 200 mm and other lenses at home?" Tags="&lt;telescope&gt;&lt;optics&gt;&lt;refractor-telescope&gt;&lt;telescope-making&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14870" PostTypeId="2" ParentId="14869" CreationDate="2016-05-13T16:12:24.030" Score="3" Body="&lt;p&gt;You would be better off using lens 5 and 3. Though don't expect much, a simple single lens for the objective and for the eyepiece will give a lot of chromatic aberration (false colour). Refracting telescopes usually use at least two lens of different glass types (crown &amp;amp; flint often in a &lt;a href=&quot;https://en.wikipedia.org/wiki/Achromatic_lens&quot; rel=&quot;nofollow&quot;&gt;doublet&lt;/a&gt;) and &lt;a href=&quot;https://en.wikipedia.org/wiki/Eyepiece&quot; rel=&quot;nofollow&quot;&gt;eyepieces&lt;/a&gt; usually have 4+ lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to see craters on the Moon, the 4 Galilean Satellites of Jupiter, a couple of belts on Jupiter and Saturn's rings. You wont be able to reduce chromatic aberration as simply as that, the type of glass (it's refractive index) and curvature of the lens all need to be matched. You would be much better off buying a ready built objective and eyepieces if you want to make a refractor yourself, though it would probably be cheaper to buy a ready made telescope.&lt;/p&gt;&#xA;" OwnerUserId="11658" LastEditorUserId="616" LastEditDate="2016-05-13T18:09:38.730" LastActivityDate="2016-05-13T18:09:38.730" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14871" PostTypeId="1" AcceptedAnswerId="14872" CreationDate="2016-05-14T01:09:29.410" Score="8" ViewCount="3019" Body="&lt;p&gt;I know the atmospheric pressure on Mars is less than that of Earth, and so is its gravity.  However, I know that the deeper you go (e.g. in a cave or a hole that is dug), the more atmospheric pressure you would experience.  How deep would you have to go on Mars (e.g. below the surface) for the atmospheric pressure to be equal to the pressure we experience on the surface of the earth?&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2021-02-19T23:27:50.700" Title="At what depth on Mars would the atmosphere have equal pressure of that on Earth?" Tags="&lt;planetary-atmosphere&gt;&lt;mars&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14872" PostTypeId="2" ParentId="14871" CreationDate="2016-05-14T06:56:39.760" Score="9" Body="&lt;p&gt;Nasa has a &lt;a href=&quot;https://www.grc.nasa.gov/www/k-12/airplane/atmosmrm.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;atmospheric model&lt;/a&gt; of mars: &#xA;$$0.699 *e^{-0.00009 h} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A naive application of this model, solving for a pressure of 101 kPa, gives a depth of -55 km. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Armstrong limit depth (at which water boils at body temperature) is -24km&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model assumes constant temperature, and gravity (it doesn't correct for the fact that at 55 km below the surface you would be well into the martian mantle and the temperature would be very much higher, and deep enough for a measurable difference in gravity). There's no &quot;goldilocks depth&quot; at which you would only need an oxygen supply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These depths are not achievable with current technology. The deepest mines on Earth are about 4km deep, and even the Kola superdeep borehole only managed 12km&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-08-09T10:18:18.620" LastActivityDate="2018-08-09T10:18:18.620" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="14874" PostTypeId="1" AcceptedAnswerId="14877" CreationDate="2016-05-14T11:07:07.737" Score="-9" ViewCount="398" Body="&lt;p&gt;In 1969 America landed on the moon, and at that time our technology was not as good as it is now. So my question is, we have much more advanced technology than in 1969, so why are we not utilising this technology and why are the USA and/or Russia not going to the Moon any more? With our current technology we should able go to The Moon daily...&lt;/p&gt;&#xA;" OwnerUserId="11792" LastEditorUserId="10172" LastEditDate="2016-05-18T09:07:43.067" LastActivityDate="2016-05-18T09:07:43.067" Title="Why aren't people going to The Moon any more?" Tags="&lt;the-moon&gt;&lt;moonlanding&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ClosedDate="2016-05-15T11:27:27.097" ContentLicense="CC BY-SA 3.0" />
  <row Id="14875" PostTypeId="1" AcceptedAnswerId="14880" CreationDate="2016-05-14T13:39:59.957" Score="5" ViewCount="5747" Body="&lt;p&gt;What is the temperature 55 km (34.18 miles) beneath the surface of Mars?  The reason I ask is that I want to know if it might be habitable for a possible future colony if they could dig that deep (which might be easier in the lower gravity environment).  If the answer to my previous question is correct, this is the depth that the atmospheric pressure would be equivalent to that on Earth. &lt;a href=&quot;https://astronomy.stackexchange.com/questions/14871/at-what-depth-on-mars-would-the-atmosphere-have-equal-pressure-of-that-on-earth&quot;&gt;At what depth on Mars would the atmosphere have equal pressure of that on Earth?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2020-09-30T12:20:09.947" Title="What is the temperature 55 km beneath the surface of Mars?" Tags="&lt;mars&gt;&lt;temperature&gt;&lt;geology&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14876" PostTypeId="2" ParentId="14869" CreationDate="2016-05-14T15:35:46.717" Score="2" Body="&lt;p&gt;I generally agree with the answer above, but have a couple more insights which might help you if you decide to proceed with trying to make your own scope...&#xA;The lens pairs that James mentioned (crown and flint) are known as a doublet.  Glass has two key properties in play here - its index of refraction (how much it bends light) and its dispersion (how much that bending changes over color).  The lens pair balances a strongly convex crown (low index and low dispersion) with a weakly concave flint (high index and high dispersion).  The dispersions are designed to cancel out, while you want the curvature of the convex crown to overpower the concave flint in terms of index, so it still has some ability to focus.  The design also inherently lends itself toward long focal lengths which are desirable in telescope objectives.&lt;br&gt;&#xA;Eyepieces, due to their short-desirable focal lengths necessitate more lenses which allow you to balance the chromatic aberration, and also address other optical aberrations which come into play with such a short focal length (distortion, astigmatism, coma, and spherical aberration being the main concerns).  There are well-established design forms which are often used for making well-corrected eyepieces, some of which can be found right on wikipedia: &lt;a href=&quot;https://en.wikipedia.org/wiki/Eyepiece#Eyepiece_designs&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Eyepiece#Eyepiece_designs&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing to consider, which I don't think has been mentioned is selection of focal lengths and apertures.  A 5cm aperture is plenty sufficient to view the Galilean moons, and probably some bright DSOs if you're well-corrected, and if your focal lengths are well-chosen.  The system magnification is the ratio of focal lengths between the objective lens and eyelens/eyepiece. (200cm/2.2cm = 90.9x).  This means that something like the Galilean moons, which have a max extent of about 1/8 degree, would be magnified to have an apparent extent of 11 degrees (much easier to resolve).&#xA;Your aperture selection (particularly of your objective) will determine the light-gathering ability.  But the magnification applies here too, so if you have a 5cm objective at 91x, your &quot;exit pupil&quot; will only be 0.55mm diameter, which is tiny compared to your eye's aperture.  You'd still be able to see the object, but your eyes will easily accommodate up to a 30cm objective aperture (3mm exit pupil).  Keep in mind, there is a tradeoff between aperture and aberrations, so unless you're designing a very well-aligned 2- or 3-element objective, you may want to stick with a max objective aperture of 50-75mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of alignment, don't just set the lenses a certain distance apart and expect to see an image.  You will need to allow for some adjustment, which is probably easier looking at a distant object during the day.  After you form an image, you may need to adjust the centration and tilt of the eyelens to form the sharpest image to optimize your alignment.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that said, a small aperture, high-end refractive (glass) telescope can perform better than a reflective telescope of the same aperture.  But as aperture increases, the cost of the materials and impact of aberrations makes refractive telescopes vastly inferior to reflective (mirror) telescopes.  Since the design for these only necessitates 1 powered mirror and an off-the-shelf eyepiece for $50 or less, the best bang for your buck will definitely be a reflective telescope.  Sorry if that's not what you're hoping to hear, but it's why most telescopes on the market today are reflective.&lt;/p&gt;&#xA;" OwnerUserId="10820" LastActivityDate="2016-05-14T15:35:46.717" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14877" PostTypeId="2" ParentId="14874" CreationDate="2016-05-14T15:50:02.117" Score="7" Body="&lt;p&gt;I see from your comment on another post that you are a so-called skeptic, so no answer that is given will satisfy your irrational beliefs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simple reason is that it is very expensive; nothing to do with technological capabilities. And as there is little to be gained scientifically, then astronomers/geologists etc. would rather spend the money elsewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Politicians are unwilling to ask taxpayers/voters to finance further manned (or womanned) trips to the Moon. For instance each space shuttle mission cost around a billion dollars. The cost of the Apollo programme in today's money would easily exceed 100 billion dollars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Priorities may change when a Chinese person walks on the Moon...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also worth noting that neither the USA or Russia have the launch capabilities of a Saturn V or similar at the moment. &lt;a href=&quot;https://space.stackexchange.com/questions/5459/russian-manned-moon-landing-capability-today&quot;&gt;https://space.stackexchange.com/questions/5459/russian-manned-moon-landing-capability-today&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2016-05-15T08:16:34.323" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14879" PostTypeId="2" ParentId="14875" CreationDate="2016-05-14T21:20:44.267" Score="1" Body="&lt;p&gt;Nobody knows it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;55 km depth is probably between crust and mantle, with &lt;a href=&quot;http://web.mit.edu/hufaul/www/UPubs/NimmoMars13.pdf&quot; rel=&quot;nofollow&quot;&gt;estimated&lt;/a&gt; temperatures well above 500 K.&lt;/p&gt;&#xA;" OwnerUserId="10568" LastActivityDate="2016-05-14T21:20:44.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14880" PostTypeId="2" ParentId="14875" CreationDate="2016-05-14T21:34:29.667" Score="7" Body="&lt;p&gt;If we look at Mars' possible geothermal gradient (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Geothermal_gradient&quot; rel=&quot;noreferrer&quot;&gt;Earth&lt;/a&gt;'s) which is about 25 °C per km.   Using the low estimate of Mars's gradient to be 1/4 that of Earth's &lt;a href=&quot;http://www.lpi.usra.edu/meetings/geomars2001/pdf/7044.pdf&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;, that's a bit over 6° C per km. so 55 km, 330° C.  Added that to Mars' average surface temperature of -55 C, you're talking 275° C or 527° F at 55 km underground, and that's a low estimate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Estimates of the thickness of Mars' crust are often less than 55 km and if you're talking about digging into mantle or even, half way to the mantle, that's both an unsustainable dig and unlivable temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Mars#Internal_structure&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The average thickness of the planet's crust is about 50 km (31 mi),&#xA;  with a maximum thickness of 125 km (78 mi)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/83wKn.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/83wKn.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.psrd.hawaii.edu/Oct12/Mantle-of-Mars.html&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, small point, but the 55km estimate assumes uniform temperature.  The simpler pressure calculations don't take into account temperature fluctuation.   With warmer temps and warmer lighter air at the bottom of the dig, which would be inevitable, you'd probably need 60 km, maybe 65 km of depth to achieve 1 atm.  There's also circulation issues when you have cold heavier atmosphere above warm lighter atmosphere and there's drainage/flooding problems for the lowest part of a valley, so to avoid flooding, you might want to keep the surface temperature below freezing, to avoid runoff and silt build-up, or dig even deeper so there's a lake of sorts that catches the melting ground-ice around the dig and the living area is above the lake.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't mean that deep valley living on mars will never be possible, but it's likely impossible with it's current atmosphere.  With extensive terraforming, tapping gases from underground and/or from it's ice-caps or by crashing comets into Mars (that would take a lot of comets), over time, enough of an atmosphere could be created where deep valley living might work.    A simulated atmospheric pressure can also be created with a bubble-dome of sorts, or one of my favorites, a stretch-wrap around the entire planet, but for now that's way beyond our means.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the near term, living underground on Mars is probably the way to go.   Dig 50 or 100 feet underground and you'd have the means to create a near vacuum seal, which on Mars, you'd want to do, you could create earth like air pressure and you'd be protected from solar and cosmic rays and access to plenty of underground water.   underground on Mars is likely the way to go, at least, in the near future.   There's no shortages of articles on this.  Here's one &lt;a href=&quot;http://www.huffingtonpost.com/2011/12/13/life-on-mars-we-could-be-martians_n_1144790.html&quot; rel=&quot;noreferrer&quot;&gt;6&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Humans can also live at less than 1 atm.  &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_highest_cities_in_the_world&quot; rel=&quot;noreferrer&quot;&gt;People have settled as high as 16,700 feet&lt;/a&gt;, which is somewhere in the neighborhood of 55% of 1 atm.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-05-15T01:58:06.500" LastActivityDate="2016-05-15T01:58:06.500" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14881" PostTypeId="2" ParentId="14525" CreationDate="2016-05-15T05:32:37.617" Score="2" Body="&lt;p&gt;All telescope mounts are calibrated with several known target objects, but to initially line up with said objects it is possible to use eyeballing and then moving the radio-telescope around until reaching a peak value from instruments, which would indicate you are pointing directly at the object. I'm not sure how the professional observatories manage it, but it would be a similar process.&lt;/p&gt;&#xA;" OwnerUserId="11525" LastActivityDate="2016-05-15T05:32:37.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14882" PostTypeId="1" CreationDate="2016-05-15T10:02:06.350" Score="4" ViewCount="78" Body="&lt;p&gt;In chapter 2.2 of &lt;em&gt;Astrophysics of Gaseous Nebulae and AGN&lt;/em&gt; Ostriker and Ferland claim that, as far as ionization is concerned, one can assume all atoms to be in the ground state in a Strömgren Sphere setting (static, homogeneous $n\approx 10/\mathrm{cm}^3$, isothermal $T\approx 10^4\,\mathrm{K}$ pure hydrogen cloud around a single star). While I do not doubt the validity of this assumption, I have some trouble with how they justify it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They estimate the inverse ionization rate for ground state hydrogen at a typical distance from the star, which is of course the lifetime against ionization from the ground state $\tau_\mathrm{ion}^{1^2S}$. Then they say that the estimate also holds for ionizations from excited levels and they find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau_\mathrm{ion}\gg\tau$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\tau_\mathrm{ion}$ and $\tau$ are the typical lifetime against excitation and the typical lifetime of excited states respectively.&#xA;So far so good. In the next step, however, they end their chain of argumentation by saying that consequently any atom in an excited state will have enough time to decay to the ground state before it can be ionized, so all ionizations are from ground state. That is my understanding at least.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem with this is the following: suppose I have some mechanism that keeps a good fraction of my atoms in an excited state. In that case, even if $\tau_\mathrm{ion}\gg\tau$, I will have ionizations from this excited level simply because it is always populated. So what one should really look at in my opinion is the lifetime of an atom against excitation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first question is: Am I missing something or did I get something wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And secondly: If my argumentation makes sense, one has to compare the lifetime against excitation with $\tau$, right? Where can I find the corresponding cross sections for radiative and collisional excitation? I was only able to find ionization cross sections.&lt;/p&gt;&#xA;" OwnerUserId="11806" LastEditorUserId="9301" LastEditDate="2016-06-12T22:04:44.870" LastActivityDate="2016-09-10T23:50:58.557" Title="Population of excited H levels in a Strömgren Sphere" Tags="&lt;interstellar-medium&gt;&lt;hydrogen&gt;&lt;plasma-physics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14884" PostTypeId="1" CreationDate="2016-05-15T22:38:02.570" Score="2" ViewCount="462" Body="&lt;p&gt;Is there a limit to how massive a planet's ring can be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is, how does it depend on the mass of the planet, presence/absence of moons, distance to the star or other parameters?&lt;/p&gt;&#xA;" OwnerUserId="11826" LastActivityDate="2016-05-17T22:46:51.483" Title="Mass limit of planetary ring" Tags="&lt;astrophysics&gt;&lt;planetary-ring&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14885" PostTypeId="1" CreationDate="2016-05-16T02:02:27.240" Score="0" ViewCount="248" Body="&lt;p&gt;If a universe is finite and is not expanding at a speed equal to or greater than c, what happens when light or another form of electromagnetic radiation traveling reaches the boundary?&lt;/p&gt;&#xA;" OwnerUserId="11825" LastEditorUserId="11525" LastEditDate="2016-05-16T14:36:37.063" LastActivityDate="2016-12-08T13:50:04.530" Title="In a finite universe, what happens when light reaches the boundary?" Tags="&lt;universe&gt;&lt;light&gt;&lt;radiation&gt;" AnswerCount="2" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="14887" PostTypeId="1" CreationDate="2016-05-16T11:41:46.427" Score="4" ViewCount="267" Body="&lt;p&gt;Is there any proof for space being created?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because I guess &lt;strong&gt;an explanation for the dark energy&lt;/strong&gt; (dark = our understanding of it) could be that &lt;em&gt;there is a gigantic amount of matter/dark matter that we cannot see (the gray clumps in the picture) that are attracting our matter (the known universe in the middle) =&gt; of course the universe is &quot;expanding&quot; with an acceleration.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I talking non-sense?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LUuBH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LUuBH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11828" LastActivityDate="2020-10-18T11:05:42.193" Title="Is there any proof of space being created?" Tags="&lt;universe&gt;&lt;expansion&gt;&lt;dark-energy&gt;&lt;matter&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14888" PostTypeId="2" ParentId="14887" CreationDate="2016-05-16T12:31:41.753" Score="4" Body="&lt;p&gt;As is always the case in physics, there is no proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if your scenario were true, it would have to be rather fine-tuned in order to create the observed expansion of the Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, the expansion is observed to be highly &lt;em&gt;isotropic&lt;/em&gt;, i.e. the expansion rate is the same in all directions. Hence, your lumps couldn't really look like your drawing, but would have to lie in a shell around our Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the matter farthest from us would achieve a larger acceleration since it were closer your surrounding matter. In fact the opposite is observed; in the &quot;local&quot; Universe, the expansion accelerates, whereas in the distant Universe (which due to the finite speed of light also means the &lt;em&gt;early&lt;/em&gt; Universe), the expansion actually &lt;em&gt;de&lt;/em&gt;celerated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, we would have to be located near the center of the Universe which, while not impossible, would be regarded as a highly improbable coincidence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fourth, preferably you would have to come up with some mechanism that could result in such a configuration of matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark energy, at least in the form of the cosmological constant which is usually assumed, is so far (!) the simplest explanation for the observed fact that the expansion of the Universe is accelerating, and that the geometry of space seems to be flat. But several other models that fit the observations exist, and although I personally think dark energy makes the most sense — partly because this is so far from my field that I'll have to rely on what I read — I think most cosmologists are rather open to the possibility that we may one day have to severely reshape or even reject this model.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2016-05-16T12:31:41.753" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14889" PostTypeId="1" CreationDate="2016-05-16T15:02:43.743" Score="31" ViewCount="10312" Body="&lt;p&gt;The density of the Sun is $1410~\frac{\text{kg}}{\text{m}^{3}}$ and Mercury's is $5430~\frac{\text{kg}}{\text{m}^{3}}$, but shouldn't the Sun be denser? Because when the Solar System was forming, there was a big disk of debris, and depending on the density of the debris it went closer or further from the centre, which then formed the planets, but the Sun is in the centre, and it's less dense than Mercury, why?&lt;/p&gt;&#xA;" OwnerUserId="11830" LastEditorUserId="1929" LastEditDate="2016-05-17T13:54:04.587" LastActivityDate="2017-04-24T14:56:58.383" Title="Why is the Sun's density less than the inner planets?" Tags="&lt;solar-system&gt;&lt;the-sun&gt;&lt;mercury&gt;&lt;density&gt;" AnswerCount="6" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14890" PostTypeId="2" ParentId="14889" CreationDate="2016-05-16T15:10:24.827" Score="29" Body="&lt;p&gt;Fusion inside of a star affects the sun's density (which does not happen with a planet). It produces an outward pressure that balances against the attraction of gravity, thereby reducing the density as long as the star is burning. Once a star the mass of the sun is no longer able to sustain fusion, what is left is a white dwarf which is in fact much denser than Mercury.&lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorDisplayName="user1569" LastEditDate="2016-05-19T14:40:01.203" LastActivityDate="2016-05-19T14:40:01.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14891" PostTypeId="2" ParentId="14889" CreationDate="2016-05-16T16:25:17.797" Score="45" Body="&lt;p&gt;The sun isn't the same density all the way through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to MSFC's &lt;a href=&quot;http://solarscience.msfc.nasa.gov/interior.shtml&quot; rel=&quot;noreferrer&quot;&gt;solar interior page&lt;/a&gt;, the core density at the centre of the sun is a whopping 150,000 kg/m$^3$. Surrounding it the radiative zone is around 20,000 - 200 kg/m$^3$ (already less dense than water). Eventually at the edge is the convective zone - the density at the part that we see is much less dense than our own air...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So although the Sun's average density isn't very remarkable, the core is the densest place in the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wcfNB.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wcfNB.png&quot; alt=&quot;Sun - cross section view&quot;&gt;&lt;/a&gt;(Sun cross section from &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Sun_poster.svg&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia.org&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="7491" LastEditorUserId="7491" LastEditDate="2016-05-17T15:00:45.883" LastActivityDate="2016-05-17T15:00:45.883" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14893" PostTypeId="2" ParentId="14887" CreationDate="2016-05-16T17:34:44.690" Score="9" Body="&lt;p&gt;This is an intriguing proposition, but I would ask how your hypothesis explains that the universe appears to be &lt;strong&gt;flat&lt;/strong&gt;? That is with &lt;span class=&quot;math-container&quot;&gt;$\Omega_M + \Omega_\Lambda = 1$&lt;/span&gt;. The evidence for this comes from measurements of the cosmic microwave background, yet if we sum up all the matter (including dark matter), we only arrive at &lt;span class=&quot;math-container&quot;&gt;$\Omega_M \sim 0.3$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;I do not think that your proposed arrangement both allows an accelerating expansion of the visible part of the universe (which note, can only be gravitationally influenced by &amp;quot;outside regions&amp;quot; that are within causal contact with it) and for it to be geometrically flat.&lt;/p&gt;&#xA;&lt;p&gt;In addition, I note you are attempting to explain the expansion in some sort of Newtonian sense - I suppose your clumps of far-away dark matter are meant to be &amp;quot;stretching&amp;quot; the observable universe? But that isn't what happens in GR and any Newtonian interpretation would have to counter the problem that if the matter outside the observable universe was uniformly distributed (i.e. in a spherical shell), then it would exert exactly zero force on everything inside it (the shell theorem); but if it were not symmetric then you have to explain why the expansion of the universe appears to be (very close to) isotropic observationally.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2020-10-17T12:51:23.403" LastActivityDate="2020-10-17T12:51:23.403" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14895" PostTypeId="2" ParentId="14889" CreationDate="2016-05-16T18:43:46.443" Score="21" Body="&lt;p&gt;The density of matter depends not only on its composition, but also on temperature and pressure.  It's not meaningful to say that substance &lt;em&gt;A&lt;/em&gt; is denser than substance &lt;em&gt;B&lt;/em&gt; without specifying the conditions under which the comparison is being made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a simple everyday example, at room temperature (and pressure) water is significantly denser than air.  But heat them both above 100 &amp;deg;C, and the water evaporates and actually becomes considerably &lt;em&gt;less&lt;/em&gt; dense than air, even at the same temperature and pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ideal_gas_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;ideal gas law&lt;/a&gt;, the density of different gases at a given temperature and pressure is approximately proportional to their average molecular mass.  The molecular mass of water is only about half that of diatomic oxygen and nitrogen, which are the main components of air on Earth, and thus water vapor is &lt;a href=&quot;https://chemistry.stackexchange.com/questions/16055/why-do-we-use-helium-in-balloons/16059#16059&quot;&gt;only about half as dense&lt;/a&gt; as air at the same temperature and pressure.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface temperature of Mercury is less than 1000 &amp;deg;C (and the interior temperature should not be much greater), and it mostly consists of metals and silicate minerals (i.e. rock) that are solid or liquid at those temperatures.  The Sun's temperature, meanwhile, is over 5000 &amp;deg;C at the surface (photosphere), and a &lt;em&gt;lot&lt;/em&gt; hotter deeper inside.  If you could heat Mercury up to the same temperature as the Sun, most of the rocks and metals it consists of would evaporate, and would become a lot less dense.  So a lot of the density difference simply comes down to the fact that Mercury is a lot cooler than the Sun, and thus able to stay solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reason why the Sun is less dense than Mercury is that the Sun contains a lot of lightweight hydrogen gas (which has both a very low molecular weight and a very low evaporation point), while Mercury has almost no hydrogen at all.  The main reason for this is that the Sun's heat and the solar wind have effectively blown away any hydrogen and other volatile low-density substances that Mercury might once have had (or that might've existed in its general area while the solar system was forming).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun itself can retain hydrogen due to its enormous gravity (but even so, it loses about one billion kilograms of it per second; that's basically what the &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_wind&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar wind&lt;/a&gt; I mentioned above mostly is).  Mercury, however, is much smaller, and thus its gravity isn't strong enough to hold onto its own hydrogen so close to the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Basically the same thing happened to Venus, Earth and Mars, which is why these inner planets didn't turn into huge balls of hydrogen gas like Jupiter and Saturn did.  However, Earth and Venus were both big enough, and located far enough from the Sun, that they could hang on to other slightly less volatile substances &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Earth#Evolution_of_Earth.27s_atmosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;like water and air&lt;/a&gt;.  Mars is located even further from the Sun, but is also a lot smaller than Earth, which is the main reason why it today has &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Mars&quot; rel=&quot;nofollow noreferrer&quot;&gt;only a very thin atmosphere&lt;/a&gt; of carbon dioxide, and very little if any water.)&lt;/p&gt;&#xA;" OwnerUserId="464" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:06.147" LastActivityDate="2016-05-16T18:43:46.443" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14896" PostTypeId="2" ParentId="14882" CreationDate="2016-05-16T19:43:10.163" Score="1" Body="&lt;p&gt;There just aren't any other mechanisms to keep the population excited in a Stromgren sphere.  In principle collisions can also lead to excitations, but if you do an order of magnitude calculation you'll find that the time between collisions is much longer than the time to receive an ionizing photon.  In a laboratory setting you might be able to coerce some particles into remaining in an excited state using the Quantum Zeno effect, but that won't happen in interstellar space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So without something to keep the atoms in an excited state, even if you start with everything in the excited state, in time $\tau$ you will find that everything has decayed to the ground state, with the exception of some very small fraction $\tau_{\textrm{ion}} / \tau$, which may have moved up to a higher energy state.  Assuming the lifetimes of those very excited states are comparable, after a few more times $\tau$, you will find that even those will have decayed to the ground state.&lt;/p&gt;&#xA;" OwnerUserId="10371" LastActivityDate="2016-05-16T19:43:10.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14897" PostTypeId="1" AcceptedAnswerId="14899" CreationDate="2016-05-16T19:50:00.943" Score="0" ViewCount="118" Body="&lt;p&gt;If put enough weight on a particular point on Earth's surface disturbing the balance between hemispheres, is it possible that the Earth's spin could change like an unbalanced spinning top?&lt;/p&gt;&#xA;" OwnerUserId="11835" LastEditorUserId="1203" LastEditDate="2016-05-17T02:25:08.260" LastActivityDate="2019-02-19T20:48:09.470" Title="Does weight influence Earth's spin?" Tags="&lt;earth&gt;&lt;amateur-observing&gt;&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2016-05-16T21:27:06.713" ContentLicense="CC BY-SA 3.0" />
  <row Id="14898" PostTypeId="1" CreationDate="2016-05-16T19:55:33.607" Score="3" ViewCount="323" Body="&lt;p&gt;Per definition a dwarf planet is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a celestial body that (a) is in orbit around the Sun, (b) has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape1, (c) has not cleared the neighbourhood around its orbit, and (d) is not a satellite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia page for &lt;a href=&quot;https://en.wikipedia.org/wiki/%28225088%29_2007_OR10&quot; rel=&quot;nofollow noreferrer&quot;&gt;225088 Gonggong, former 2007 OR10&lt;/a&gt; states three possibilities for its dimensions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;1535+75−225 km&lt;/li&gt;&#xA;  &lt;li&gt;1280+210−210 km&lt;/li&gt;&#xA;  &lt;li&gt;1200+300−200 km&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which is all minimum 1000km in diameter. So how can you still doubt the hydrostatic equilibrium for such a large body?&lt;/p&gt;&#xA;" OwnerUserId="5501" LastEditorUserId="5501" LastEditDate="2020-05-28T14:52:36.420" LastActivityDate="2020-12-28T17:43:11.710" Title="Why 225088 Gonggong is not a dwarf Planet yet?" Tags="&lt;dwarf-planets&gt;&lt;hydrostatic-equilibrium&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="14899" PostTypeId="2" ParentId="14897" CreationDate="2016-05-16T20:34:21.320" Score="3" Body="&lt;p&gt;The Earth &lt;strong&gt;does&lt;/strong&gt; spin like an unbalanced top. The Earth's rotation axis is not fixed. It instead moves in a complex manner due to a combination of external torques exerted by the Moon and Sun, a torque-free nutation due to the oblate shape of the Earth, and also due to changes on and in the Earth.&lt;/p&gt;&#xA;&lt;p&gt;The torque-induced motions are called precession and nutation, distinguished by period. The largest and slowest of these motions is the axial precession. This causes the Earth's rotation axis to trace out a cone over the course of 26000 years.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GYtyX.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GYtyX.gif&quot; alt=&quot;&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;https://solarsystem.nasa.gov/basics/bsf16-9.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;nasa.gov&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;The torque-induced nutations are also cyclical motions induced by the Moon and the Sun. These are much smaller in magnitude and have a much shorter period. The largest of these has a magnitude of about 20 arc seconds and a period of 18.6 years. All other nutation terms have much smaller magnitude and have shorter period.&lt;/p&gt;&#xA;&lt;p&gt;The torque-free nutation would have a period of about 305 days if the Earth was solid. The oceans, the atmosphere, and the outer core alter this. The Chandler wobble has a period of about 433 days and a magnitude of less than an arc second. Because the Chandler wobble isn't as predictable as are precession and nutation, it's lumped into a catch-all category called &amp;quot;polar motion.&amp;quot; The redistribution of water over the course of a year (e.g., snow on Siberia in the winter but not in the summer) results in a yearly component of the polar motion.&lt;/p&gt;&#xA;&lt;p&gt;There are lots and lots of other factors, all small. Polar motion is observed after the fact.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-02-19T20:48:09.470" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="14901" PostTypeId="1" CreationDate="2016-05-16T20:40:08.470" Score="3" ViewCount="657" Body="&lt;p&gt;I am working on an online application to visualize the positions of the planets and to calculate their ephemerides (you find it &lt;a href=&quot;http://planeten.decocode.de/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, but it is not complete yet). I do not have much knowledge in astronomy and I am learning new stuff every day while developing the code. A few questions came to my mind that I hope to get answers for here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing I do not quite understand is the effect of the earth's axial precession. I understand the way it causes the vernal point to move along the ecliptic within the period of the precession of about 25770 years. But since the longitude of the ascending nodes of each planet's orbit depend on the vernal point and remain almost constant over time it seems like all the orbits change their position relative to the stars at the same rate as the vernal point does. But I cannot believe that the orientation of the orbits is related to the precession of the earth's axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see the effect in the application: When you move the topmost time slider for years you can see the zodiac rotate and the orbits staying in place because the vernal point is fixed with the scale. Shouldn't the orbits move together with the stars? But then the longitude of the ascending nodes wouldn't be constant anymore... I am confused. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="11838" LastEditorUserId="11838" LastEditDate="2016-05-17T04:52:50.257" LastActivityDate="2018-04-09T13:56:42.563" Title="What is the effect of the axial precession on the orientation of the planet's orbits?" Tags="&lt;solar-system&gt;&lt;planet&gt;&lt;fundamental-astronomy&gt;&lt;orbital-elements&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14902" PostTypeId="2" ParentId="14901" CreationDate="2016-05-16T21:51:54.073" Score="1" Body="&lt;p&gt;When talking about the longitude of ascending node you must be very careful to define the reference plane that you are using.  As you state, due to the Earth's axial precession, the First Point of Aries moves along the ecliptic over ~26,000 years.  This is because the celestial equator is slowly precessing about the ecliptic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in the case of the orbital elements of other planets, the longitude of ascending node is again with respect to the First Point of Aries.  As your intuition suggests, the orbits of the outer planets won't be influenced by the precession of the Earth's equinox.  This means that the longitudes of ascending node will not be constant over long time periods.  If we suppose that the physical orbits are fixed (i.e., the planets are not perturbing each other), the change in the longitudes of ascending node will be entirely due to the Earth's precession.  When you take both these changes into account, you will find that they cancel each other out and the position of these orbits remains fixed with respect to the distant stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see in the orbital elements provided by NASA that they are given with respect to the J2000.0 epoch.  In other words, the orbital elements are provided for a &lt;em&gt;single&lt;/em&gt; point in time, and to get the orbital elements today you must take into account the Earth's precession.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ssd.jpl.nasa.gov/txt/p_elem_t1.txt&quot; rel=&quot;nofollow&quot;&gt;http://ssd.jpl.nasa.gov/txt/p_elem_t1.txt&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10371" LastActivityDate="2016-05-16T21:51:54.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14904" PostTypeId="2" ParentId="14884" CreationDate="2016-05-17T01:59:07.090" Score="3" Body="&lt;p&gt;A couple of points on this.  While faint rings can exist outside the Roche Limit (like Saturn's G and E rings) - photo is too big to link but click &lt;a href=&quot;https://en.wikipedia.org/wiki/Rings_of_Saturn#/media/File:Saturn%27s_Rings_PIA03550.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.  rings of any significant mass are only going to reside inside the Roche Limit, so the practical limit of the mass of a planet's ring is about the mass of the largest moon a planet is likely to have, roughly speaking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ring around a planet can form in a couple of ways, by giant impact, by a moon passing inside the Roche limit and breaking apart or by accretion, like how Enceladus feeds Saturn's faint e-ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, rings don't have to be particularly massive to be impressive.  The mass of Saturn's rings estimated &lt;a href=&quot;https://en.wikipedia.org/wiki/Rings_of_Saturn#Physical_characteristics&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; (and there's some uncertainty to that estimate, but close enough for our purposes)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Based on Voyager observations, the total mass of the rings was&#xA;  estimated to be about 3 x 10^19 kg. This is a small fraction of the&#xA;  total mass of Saturn (about 50 ppb) and is just a little less than the&#xA;  moon &lt;a href=&quot;https://en.wikipedia.org/wiki/Mimas_%28moon%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mimas&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the mass of Saturn's rings is the mass of a pretty small moon, a little under 200 km in radius, but that wasn't your question, so, moving on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A planet with rings &lt;a href=&quot;http://www.cnn.com/2015/01/28/tech/giant-planetary-ring-system/&quot; rel=&quot;nofollow noreferrer&quot;&gt;200 times the size&lt;/a&gt; of Saturn (40,000 times the surface area) was observed and if we assume the same density as Saturn's rings (which is a very bad assumption to make, but this is just an approximation), 1.2 x 10^24, or about twice the mass of Mars.   Now we're getting somewhere, but there's a few points to make on our friend J1407B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, it may not be a planet at all.  It's mass estimate is 10-40 jupiters which suggests it's in the brown dwarf category.   2nd, it's a young solar system, just 16 million years old, so the ring system may still be in the process of forming a moon system and not be a permanent ring at all.  &lt;a href=&quot;http://www.sci-news.com/astronomy/science-super-saturn-j1407b-massive-ring-system-02436.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.   Also, same article, it mentions that parts of the ring system block 95% of the sun's light, so it's a far denser system than Saturn and there's a moon in there thought to be &lt;a href=&quot;https://en.wikipedia.org/wiki/1SWASP_J140747.93-394542.6#1SWASP_J140747.93-394542.6_b&quot; rel=&quot;nofollow noreferrer&quot;&gt;0.8 times the mass of earth&lt;/a&gt;, but I still don't think that should counts since it may be an accretion disk remaining from the formation of that solar-system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we set the planetary mass limit to be about 13 Jupiter masses, which is roughly the cut-off between heavy Jupiter and brown dwarf, and we take some estimates of the size of moons to planets, during formation, it's unlikely that moons can be more than maybe a ratio of one to several hundred, due to limitations in angular momentum and likely size of formation.   In our solar system, the largest moon to planet ratio, not counting giant impacts or captures is about 1 to 4,200 (Saturn to Titan).  Jupiter to Ganymede is 1 to about 12,800 and Tritan to Neptune (and Tritan may be a captured moon), about 1 to 4,800.   If we assume a heavy Jupiter of about 13 Jupiter masses and a moon in a decaying orbit that breaks up with a mass ratio of 1 to 400 - and needless to say, that's very ballpark, then this theoretical estimate works out to a practical limit around a heavy jupiter to have a moon system of roughly 10 earth masses.  (1/400th the mass of a 13 jupiter mass planet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there are creative ways to increase it, like, lets say there are two moons, perhaps in Trojan points to each other, and they both spiral in gradually.   Or, lets say a passing object of greater mass flies past, but the problem with a passing object is that, due to escape velocity speed, such an object, if it passes within the Roche Limit, it would have a highly elliptical orbit, which wouldn't lend itself to a ring system, since the momentum of the elliptical orbit would only pass through and not stay inside the Roche sphere.  What you need for a good sized ring system to form is a slowly decaying circular orbit, not a recent capture of a large passing object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Shoemaker Levy 9 which passed through but quickly went back outside Jupiter's Roche limit.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dw4Ke.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dw4Ke.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we imagine a scenario of a double planet, which is probably a rare scenario but not impossible, and over time due to proximity and tidal forces, the two planets form a decaying circular orbit around each other, in this theoretical, lets say we have two heavy Jupiter orbiting each other, one of 12 Jupiter masses and the other of 6 Jupiter masses (purely theoretical) and the 6 Jupiter mass planet begins to break up as it enters the Roche limit of the 12 Jupiter mass one.   There's 2 problems with assuming you'd get a ring with half the mass of the planet.   One is, hydrogen and helium gas are likely at least 80% of the mass of the planet that's breaking apart, perhaps 90% and hydrogen and helium don't easily bind into ice particles, so over time, much of that gas is likely to get blown away by the solar wind.  You're left with a rocky and icy mass of quite a bit less than 6 Jupiter masses, probably less than 1.  The other problem, with an orbiting object that large, it doesn't break apart all at once, but it's denser core stays together and spirals closer to the planet while the lighter ices around the surface which do begin to form the ring, can get spun off and flung either into the planet or flung outside the Roche limit, as the denser core of the smaller of the two planet moved slowly closer, there would be a vacuum cleaner effect wrecking havoc on the very same ring that planet was forming.  For smaller moons, this effect can be seen around saturn as tiny moons create tiny breaks in Saturns' rings.  With a much more massive core of the planet that was forming a moon, you'd get orbital chaos and it wouldn't be a good scenario to form a ring, even with all that available ring material.   You'd still probably end up with a ring, but I think only a small percentage of the original smaller planet's mass would survive the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar problem arises with giant impacts.  If the impact is too large, the planet breaks apart.   If it's not quite that large, like the giant impact on earth was thought to be, 4.4 billion years ago, then debris is blown outside the Roche Limit and forms a moon, in our case 1/81st the mass of the earth, though at the time it might have been a bit more, maybe 1/72-74 or so, and there were probably two moons initially not one, but I digress.   The point is, to have debris blown off the planet but staying inside the Roche limit so it forms a ring, not a moon, you need a smaller impact than the one that formed our moon, and a smaller impact implies less debris, so with Rocky worlds, what does that mean, 1/100th?   1/200th?  In there somewhere.  With gas giants the impact method is even worse due to the outer layer of gas giants being so much hydrogen that you don't get good material for a dense permanent ring structure from giant impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there are practical limits to the size of a ring system you're likely to see (if we ignore young solar-systems still information like J1407b.   It's a very ballpark just barely educated guess, but I think the practical limit for a ring system is probably pretty tiny compared to the mass of the planet, like a ratio of 1 to 200 or so, which if you have a planet the mass of 13 Jupiters, that's still quite a bit of mass.   If we use the 1 to 200 ratio and 13 jupiters, that's the mass of 20 earths.  I have a very hard time seeing how a ring system around a planet could get much more massive than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in theory, like, lets say we want to build a ring system, and we design this enormous snow blower and we blow ice crystals around a planet, just for fun.   A ring system would probably remain stable well beyond that mass.   I don't have the means to calculate where instability might come into play, but if you build your own ring system around a planet, you might be able to get the ring system as massive as 1/10th the mass of the planet, maybe a bit more before some kind of gravitational instability took over.   I don't think you're ever likely to see that in a natural situation, but theoretically I think it could be done.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to answer your questions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;does it depend on the mass of the planet?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;absolutely.  The mass of the planet (and it's density) determines the size of it's Roche limit, which defines how far the ring system can extend.   As pointed out above, planets can have rings beyond the Roche limit but only faint ones.   Dense/Thick rings only form inside the Roche Limit and the more massive the planet, the larger the Roche Limit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;presence/absence of moons?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Small moons don't matter much.  Saturn has small moons inside it's rings and they create small breaks in it's ring system.  Large moons, particularly if they are close to the planet, can create gravitational disruptions and wouldn't be good for a permanent ring system.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;distance to the star or other parameters?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Beyond the frost line is best, cause that's where ices don't melt.  Ices make up more mass of our solar system and probably most solar systems than rocky material and they break apart more easily, so statistically speaking, a ring system should do better far away from the sun, in our case, at least Jupiter Distance.   Jupiter's enormous magnetic field might also not be long term ring friendly, so in addition to far from the sun you'd want a planet without too strong a magnetic field that's shooting high velocity charged particles through the ring.  Here's a fun article about how Saturn's magnetic field erodes it's &lt;a href=&quot;http://earthsky.org/space/its-raining-on-saturn-and-rings-are-the-cause&quot; rel=&quot;nofollow noreferrer&quot;&gt;ring&lt;/a&gt;.  Jupiter would melt any ice ring it had much faster than Saturn does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that wasn't too long and it's more my trying to work it out than a definitive answer, but until we get much better telescopes, there might not be a definitive answer to this one. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-05-17T22:46:51.483" LastActivityDate="2016-05-17T22:46:51.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14905" PostTypeId="2" ParentId="14889" CreationDate="2016-05-17T07:43:29.297" Score="12" Body="&lt;p&gt;I'd say the most important answer is because &lt;strong&gt;the volume of stars is counted differently than for (inner) planets&lt;/strong&gt;.&lt;br&gt;For the former, most of the gas surrounding the dense core is counted. The latter don't have significant enough amounts of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is even more pronounced with larger stars.&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/VY_Canis_Majoris#Controversy&quot; rel=&quot;nofollow noreferrer&quot;&gt;VY Canis Majoris&lt;/a&gt;: &quot;With an average density of 0.000005 to 0.000010 kg/m3, the star is &lt;em&gt;a hundred thousand times less dense than the atmosphere of Earth&lt;/em&gt; (air) at sea level. It is also undergoing strong mass loss with the outer layers of the star &lt;em&gt;no longer gravitationally bound&lt;/em&gt;&quot;&lt;br&gt;&#xA;Yeah, &lt;strong&gt;less density than the air outside the ISS&lt;/strong&gt;, and still a part of the star's volume.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/0Nevk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0Nevk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; The star is farting gas like nobody's business, and a huge part of that still counts into its diameter. The Sun is no different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously we're &lt;strong&gt;not using the same metric&lt;/strong&gt;, so there is &lt;strong&gt;no point comparing the values&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11844" LastEditorUserId="11844" LastEditDate="2017-04-24T14:56:58.383" LastActivityDate="2017-04-24T14:56:58.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14906" PostTypeId="2" ParentId="14889" CreationDate="2016-05-17T11:29:40.013" Score="3" Body="&lt;p&gt;All the other answers address the density of the sun, but I feel that none of them actually addresses the OP's misconception. OP seems to think denser material should sink, but this is not the case. Thus Pluto is denser than Uranus, but orbits further out. There is nothing strange about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is that orbital energy is conserved indefinitely unless there is some kind of interaction. A planet feels &quot;weightless&quot; just like an astronaut in a space station, because it is in freefall towards the centre of mass of the solar system. Unless it interacts with another body, &lt;strong&gt;matter, regardless of its density, will continue to orbit at the same distance from the centre of mass of the solar system&lt;/strong&gt;, as a consequence of conservation of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Density only becomes an issue when objects come into physical contact, and a body receives a push from another body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus &lt;strong&gt;in an orbiting spacecraft, dense objects just float around &quot;weightlessly&quot; and do not &quot;fall&quot; to the &quot;bottom.&quot;&lt;/strong&gt; Both the air and the objects in the spaceship are experiencing gravity, but they are falling at the same rate, so they do not push each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;When the spacecraft is on the ground&lt;/strong&gt;, the Earth's surface pushes up on the spacecraft, and prevents it from accelerating towards the centre of the earth. Under these circumstances, &lt;strong&gt;the denser objects, if unconstrained, will fall towards the floor of the spacecraft, displacing the less dense air&lt;/strong&gt;. When they hit the floor, they receive a push from it, preventing their continued fall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In space objects do not push each other by physical contact, so density makes no difference.&lt;/strong&gt; A trillion tons of iron and a trillion tons of silica may have different volumes, but they have the same mass, therefore so long as their interactions with the rest of the solar system are purely gravitational, both will behave identically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;On the other hand, matter that has coalesced into a planet, sun, or moon will become stratified by density. In the case of a moon or rocky planet this is almost entirely due to the denser materials sinking and forcing the more voluminous ones to rise.&lt;/strong&gt; In the case of the sun or a gas giant the core will also be denser due to compression. In addition to contact forces, friction is also present. Note also that &lt;strong&gt;friction is necessary for orbital decay&lt;/strong&gt;: without it satellites will orbit at the same height indefinitely.&lt;/p&gt;&#xA;" OwnerUserId="9819" LastEditorUserId="9819" LastEditDate="2016-05-17T22:07:53.123" LastActivityDate="2016-05-17T22:07:53.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14907" PostTypeId="2" ParentId="14898" CreationDate="2016-05-17T11:46:57.050" Score="3" Body="&lt;p&gt;As the comments already say, an object being a dwarf planet is a matter of convention. If the IAU says it's a dwarf planet, it's a dwarf planet. Otherwise, it's not. The requirements you are listing from Wikipedia are the IAU criteria for pronouncing objects as dwarf planets, but that does not mean that all objects fulfilling these criteria are dwarf planets. (don't ask me for the reasonings of the IAU, i'm not an expert on astronomer's brain twists)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a lot more examples of objects that could be dwarf planets, and are considered as being that by many researchers, despite these objects not having the status of dwarf planet. Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_possible_dwarf_planets&quot; rel=&quot;nofollow&quot;&gt;List of possible dwarf planets&lt;/a&gt; has about 200 objects facing the same issue as 2007 OR10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: as someone in the comments mentioned, not all objects on this list are actual candidates for dwarf planets (which is not what I meant, but I might indeed have implied this)&lt;/p&gt;&#xA;" OwnerUserId="11824" LastEditorUserId="11824" LastEditDate="2016-05-17T20:43:38.433" LastActivityDate="2016-05-17T20:43:38.433" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14908" PostTypeId="1" AcceptedAnswerId="14909" CreationDate="2016-05-17T14:05:21.313" Score="15" ViewCount="3267" Body="&lt;p&gt;Inspired by this answer to the question, &lt;a href=&quot;https://astronomy.stackexchange.com/a/14891/11851&quot;&gt;Why is the Sun's density less than the inner planets?&lt;/a&gt;, what is the most dense object in the universe?&lt;/p&gt;&#xA;" OwnerUserId="11851" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2019-06-04T17:57:07.790" Title="What is the most dense object in the universe?" Tags="&lt;universe&gt;&lt;density&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14909" PostTypeId="2" ParentId="14908" CreationDate="2016-05-17T14:18:52.993" Score="19" Body="&lt;p&gt;Let us define this as the largest &lt;em&gt;observable&lt;/em&gt; density of a stable object, in order to exclude black holes which may have  a very large (infinite) density at their centers or objects collapsing towards a black hole status.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we restrict the definition in this way, then the answer should be the core of the most massive neutron star that we know about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At present there are a couple of neutron stars with mass of about $2M_{\odot}$ (&lt;a href=&quot;http://arxiv.org/abs/1010.5788&quot;&gt;Demorest et al. 2010&lt;/a&gt;; &lt;a href=&quot;https://arxiv.org/abs/1304.6875&quot;&gt;Antoniadis et al. 2013&lt;/a&gt;. Depending on the exact composition and equation of state at their centres these should have densities of around $2 \times 10^{18}$ kg/m$^{3}$ at their centers and average densities of $\sim 10^{18}$ kg/m$^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that these densities are around 3 times the density of a proton or neutron or 5-10 times the density of nuclei at zero pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, the density of a single electron is much higher.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-05-17T16:41:07.223" LastActivityDate="2016-05-17T16:41:07.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14910" PostTypeId="1" AcceptedAnswerId="14911" CreationDate="2016-05-17T15:28:17.897" Score="0" ViewCount="79" Body="&lt;p&gt;Suppose an amateur astronomer finds himself watching Solar Eclipse. He can see the stars around totality. Is Sun's GR effect strong enough for the amateur astronomer to be able to tell that a particular star is appearing a few arc-degrees (or arcminutes, or arcseconds) away from where it should?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, what is the minimum requirement to be able to witness GR for yourself during a Solar Eclipse?&lt;/p&gt;&#xA;" OwnerUserId="4099" LastActivityDate="2017-05-07T20:47:32.393" Title="Can we measure/detect GR light bending during solar eclipse using ordinary equipment?" Tags="&lt;general-relativity&gt;&lt;solar-eclipse&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14911" PostTypeId="2" ParentId="14910" CreationDate="2016-05-17T16:11:59.413" Score="3" Body="&lt;p&gt;The deflection you are looking for is $\sim 1.7$ arc seconds, which should be within the seeing limits at a good site. It should also be within the resolution limits of a telescope with $\ge200$mm diameter primary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that you will not be able to see it with the naked eye, you will need to be able to photography the star field during the eclipse and again when it is visible at night and then do some arduous plate measuring and data reduction to detect the shifts of stars in the field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So amateur equipment is up to detecting the bending of light around the Sun in a total eclipse, but if by witness you mean see the stars move with the naked eye the answer is almost certainly no.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2016-05-17T16:11:59.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14912" PostTypeId="1" CreationDate="2016-05-17T16:33:29.493" Score="1" ViewCount="48" Body="&lt;p&gt;Is it possible to observation changes in the barycenter of the solar system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are graphics showing changes barycenter but they are made on the basis of computer simulation (E.g.: &lt;a href=&quot;http://img.bhs4.com/c1/c/c1ce801bf6423a3dbb86bf7289bc954d8c4be0b1_large.jpg&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://qph.is.quoracdn.net/main-qimg-118a1175c48f6ff2afd0183dd0040dd4?convert_to_webp=true&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible? If so, how?&lt;/p&gt;&#xA;" OwnerUserId="11856" LastActivityDate="2016-05-17T16:33:29.493" Title="The observation of changes barycenter of the solar system" Tags="&lt;solar-system&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14913" PostTypeId="1" AcceptedAnswerId="14914" CreationDate="2016-05-17T17:52:26.517" Score="-5" ViewCount="211" Body="&lt;p&gt;Has Science observed the entire process of a star being created when the area of creation was just a void in space? Or is there still missing data that has not been entirely observed and recorded?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asking for a simple yes or no if science has fully observed pre birth/Void in space and the entire birth of a star filling the void and what star was created.&#xA;Please no theory's only scientific fact with a reference thank you.&lt;/p&gt;&#xA;" OwnerUserId="9326" LastActivityDate="2016-05-17T18:38:11.900" Title="Has Science observed and recorded a stars birth?" Tags="&lt;star-systems&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14914" PostTypeId="2" ParentId="14913" CreationDate="2016-05-17T18:38:11.900" Score="6" Body="&lt;p&gt;The star formation process from giant molecular cloud to unobscured protostar is thought to take about a million years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, there are very few large scale physical processes that occur in the universe on a human timescale. Nevertheless we are sophisticated enough to understand that you do not necessarily have to see something happening to know that it has occurred and work out how it happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or are you asking whether all the separate phases of the star formation process have been observed in different places? The answer to that is broadly yes. The rarest (shortest) phases is the initial collapse to a &quot;core&quot; that is embedded within an obscuring molecular cloud. Nevertheless, such objects can be seen at sub-mm and radio wavelengths.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-05-17T18:38:11.900" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14917" PostTypeId="1" CreationDate="2016-05-18T01:42:16.880" Score="0" ViewCount="182" Body="&lt;p&gt;If Jupiter's gravity is what kept the asteroid belt from forming a planet, then why didn't the sun's gravity keep the inner planets from forming?&lt;/p&gt;&#xA;" OwnerUserId="11867" LastActivityDate="2016-05-20T22:45:13.340" Title="How does large body gravity affect planet formation?" Tags="&lt;planet&gt;&lt;asteroids&gt;&lt;jupiter&gt;&lt;formation&gt;&lt;the-sun&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14918" PostTypeId="2" ParentId="14917" CreationDate="2016-05-18T04:56:04.193" Score="0" Body="&lt;p&gt;During the early days of the solar system there was a high abundance of matter in the early solar system. Easily enough to form the planets and a lot of matter to spare. Also during the initial stages of formation of the solar system matter was distributed very densely when compared to what the density is today. At that point there were lots of asteroids and other small bodies very close together therefore at that stage local gravitational fields played a major role in formation of the planets. rather than the sun&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the asteroid belt. It cannot form into a planet because the matter density is very low and its proximity to the giant Jupiter makes it very vulnerable to its gravitational field. The gravitational fields between the asteroids are simply not enough to coalesce them together into a planet due to very large distances (between the asteroids) involved&lt;/p&gt;&#xA;" OwnerUserId="11587" LastActivityDate="2016-05-18T04:56:04.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14920" PostTypeId="1" CreationDate="2016-05-18T09:45:25.607" Score="2" ViewCount="1716" Body="&lt;p&gt;I am going through the theory of saha equation but I am not able to understand the concept of ionisation state? What are ionization states actually?&lt;/p&gt;&#xA;" OwnerUserId="11434" LastActivityDate="2016-05-18T09:45:25.607" Title="What is meant by ionization states?" Tags="&lt;astrophysics&gt;&lt;fundamental-astronomy&gt;&lt;spectroscopy&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14921" PostTypeId="1" AcceptedAnswerId="14923" CreationDate="2016-05-18T09:51:32.370" Score="16" ViewCount="3223" Body="&lt;p&gt;Wolfram Alpha's query: &lt;a href=&quot;http://www.wolframalpha.com/input/?i=Sun+Earth+distance+in+AU&quot; rel=&quot;noreferrer&quot;&gt;Sun Earth distance in AU&lt;/a&gt; gives:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1.012 au  (astronomical units)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They definitely use this value in other places too - e.g. Earth-Sun $L_1$ distances from Earth and from Sun don't add up to 1&amp;nbsp;AU - they are 0.01009&amp;nbsp;AU and 1.001&amp;nbsp;AU respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contrarily, Google gives &lt;a href=&quot;https://www.google.pl/search?q=earth+sun+distance+in+AU&amp;amp;ie=utf-8&amp;amp;oe=utf-8&amp;amp;gws_rd=cr&amp;amp;ei=2DY8V92SGerN6QTFjYDQBA&quot; rel=&quot;noreferrer&quot;&gt;Earth Sun distance in AU&lt;/a&gt; = 1.000 AU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top search results for &lt;a href=&quot;https://www.google.pl/search?q=Sun+Earth+distance+1.012+AU&amp;amp;ie=utf-8&amp;amp;oe=utf-8&amp;amp;gws_rd=cr&amp;amp;ei=rzc8V-mxKOrP6ASe6YWYBQ&quot; rel=&quot;noreferrer&quot;&gt;Sun Earth distance 1.012 AU&lt;/a&gt; yield a page with a big heading of &quot;Paranormal UFO Aliens Wow!&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the 1.012&amp;nbsp;AU value an error in Wolfram Alpha, or is it some little-known adjustment, redefinition or other quirk of the unit or the Solar System?&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="332" LastEditDate="2016-05-19T01:49:43.877" LastActivityDate="2016-06-05T23:41:10.687" Title="Is the Earth-Sun distance 1.012 AU?" Tags="&lt;units&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14923" PostTypeId="2" ParentId="14921" CreationDate="2016-05-18T13:26:41.627" Score="28" Body="&lt;p&gt;As of this moment (2016 May 18, 13:15 UTC), the Earth is 1.0116 astronomical units from the Sun. WA is smart enough to know that &quot;Sun Earth distance in AU&quot; is a time-dependent question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: The stricken text that follows from my original answer is incorrect. I am leaving it present (but stricken) for the sake of humility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;It interpreted your query to mean &lt;a href=&quot;http://www.wolframalpha.com/input/?i=Sun+Earth+distance+today+in+AU&quot; rel=&quot;nofollow&quot;&gt;Sun Earth distance today in AU&lt;/a&gt;, and because today is 2016 May 18 and because you didn't specify a time of day, it picked noon (UTC), it in turn interpreted your query to mean &lt;a href=&quot;http://www.wolframalpha.com/input/?i=Sun+Earth+distance+on+2016+May+18+at+noon+UTC+in+AU&quot; rel=&quot;nofollow&quot;&gt;Sun Earth distance on 2016 May 18 at noon UTC in AU&lt;/a&gt;.&lt;/s&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The correct answer&lt;/strong&gt;: I happened to ask WA the very question posed in the OP a couple of minutes apart and got two different answers (1.014 and 1.015 AU), and this did not occur across a day boundary. WA apparently interpreted your query to mean the current distance between the Sun and the Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;You have to be very careful of what you ask WA.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better query is to ask WA &lt;a href=&quot;http://www.wolframalpha.com/input/?i=What+is+the+mean+distance+between+the+earth+and+the+sun+in+AU%3F&quot; rel=&quot;nofollow&quot;&gt;What is the mean distance between the earth and the sun in AU?&lt;/a&gt; That query will give you an answer of 1.0000010178 au. An even better question is &lt;a href=&quot;http://www.wolframalpha.com/input/?i=What+is+the+semimajor+axis+of+Earth&amp;#39;s+orbit%3F&quot; rel=&quot;nofollow&quot;&gt;What is the semimajor axis of Earth's orbit?&lt;/a&gt; That will give you an answer of 1.00000011 au. Note the extra zero, but also note that for some reason, the precision is reduced.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2016-06-04T00:09:06.667" LastActivityDate="2016-06-04T00:09:06.667" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14924" PostTypeId="1" AcceptedAnswerId="14928" CreationDate="2016-05-18T14:34:57.550" Score="2" ViewCount="309" Body="&lt;p&gt;Hypothetically, let's say we had a gas giant that continued to accrete mass. I've heard that the cores of gas giants are electron degenerate. So if the planet continued to accrete mass and the core mass went beyond the Chandrasekhar limit, what would happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In white dwarfs, the result is dependent on the composition. Carbon-oxygen white dwarfs will undergo carbon fusion, leading to a type 1a supernova. Oxygen-magnesium-neon white dwarfs will undergo rapid oxygen fusion, leading to a rapid ignition and supernova but leaving behind a neutron degenerate core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So would the composition of a gas giant's core play a similar role, if it went beyond the Chandrasekhar limit? What would happen if, say, Jupiter somehow accreted a core mass beyond the limit?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="2365" LastEditDate="2018-07-27T22:34:10.830" LastActivityDate="2018-07-27T22:34:10.830" Title="What would happen to a gas planet if its core mass goes beyond the Chandrasekhar limit?" Tags="&lt;jupiter&gt;&lt;gas-giants&gt;&lt;degenerate-matter&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14925" PostTypeId="2" ParentId="14924" CreationDate="2016-05-18T14:52:00.597" Score="6" Body="&lt;p&gt;Gas giants like Jupiter consists mainly of Hydrogen and some Helium. If you gradually add mass to these planets then core temperatures will rise gradually and a stage will come where they will ignite like normal stars. Unlike white dwarfs where accretion of mass leads to type 1A supernova.  &lt;/p&gt;&#xA;" OwnerUserId="10657" LastActivityDate="2016-05-18T14:52:00.597" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14926" PostTypeId="5" CreationDate="2016-05-18T15:28:09.440" Score="0" Body="&lt;p&gt;At high densities, matter will be compressed so much that the Pauli exclusion principle will act to counter any further compression. This causes the particles to form a degenerate gas (aka a Fermi gas). Degenerate gases play a vital role in stellar evolution (e.g. the He flash), brown dwarfs, very low-mass stars, white dwarfs, neutron stars, and gas giants. Degenerate gases differ from normal gases in numerous ways:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Particles in a degenerate gas cannot all occupy the lowest energy levels and therefore they exert a pressure.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eDn2v.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eDn2v.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a normal (perfect) gas, particles occupy energy levels according to the Maxwell-Boltzmann distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a degenerate gas, all the lowest energy levels are filled up. The Pauli Exclusion Principle forbids two fermions from occupying the same quantum state. Therefore for each energy/momentum state there can only be two fermions (spin-up and spin-down). Every level between the lowest possible energy (the rest mass energy of a fermion) and the Fermi energy is occupied. The Fermi energy is a function of the number density of fermions in the gas. A gas can be considered &quot;completely degenerate&quot; if the kinetic energy at $E_F$ is $\gg kT$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the particles have non-zero momentum, even if they are cooled towards zero temperature, then degenerate gases exert a degeneracy pressure.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Degenerate gases will almost never expand from temperature increases&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In a normal (perfect) gas, an increase in temperature increases the pressure. However, in a degenerate gas, temperature increases will lead to very little change in the degeneracy pressure until $kT$ becomes comparable with the kinetic Fermi energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This phenomenon allows white dwarfs and neutron stars to cool at almost constant radius; it slows the contraction of gas giant planets; it also allows explosive, runaway nuclear burning in the cores of low-mass stars during the &quot;helium flash&quot; and in type Ia thermonuclear supernovae.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Degenerate gases have high thermal and electrical conductivity and low viscosity.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It is difficult for a fermion in a degenerate gas to lose energy in dissipative scattering interactions, since there are no free lower energy states for it to fall into. This means that the mean free paths of fermions in a degenerate gas are long. This leads to high thermal conductivity, low viscosity and, where charged fermions are involved, high electrical conductivity. In certain circumstances, degenerate gases can become superfluids or superconductors. These phenomena are found in neutron stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that degenerate gases tend to be isothermal.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Degenerate gases have low specific heat capacity.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As it is difficult to extract the energy from a fermion in a degenerate gas, their specific heat capacity is very low. In gases where the electrons are degenerate but the ions are not (e.g. in a white dwarf, or the core of a star) this lowers the heat capacity of the gas with respect to a perfect gas, because the electrons do not contribute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a neutron star where all the particles form degenerate gases, the heat capacity becomes &lt;em&gt;much&lt;/em&gt; smaller than the perfect gas case. This is why neutron stars cool extremely quickly.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="2531" LastEditDate="2016-05-20T08:51:30.037" LastActivityDate="2016-05-20T08:51:30.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14927" PostTypeId="4" CreationDate="2016-05-18T15:28:09.440" Score="0" Body="Questions regarding matter at densities so high that gravitational contraction is countered by the Pauli exclusion principle." OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-05-18T15:53:31.933" LastActivityDate="2016-05-18T15:53:31.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14928" PostTypeId="2" ParentId="14924" CreationDate="2016-05-18T15:48:12.010" Score="4" Body="&lt;p&gt;@Knu8 was right that adding mass to a gas giant will turn it into a star long before the gas giant could become a white dwarf or neutron star. But that works if you add fusionable material such as hydrogen. If you add something that can't fuse into a heavier element, such as Iron or Tin, the matter just keeps accreting onto the gas giant until the planet collapses under its own weight until electron-degeneracy pressure holds it up. That's when it becomes a white dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks @Knu8 for the right answer. I gave you an upvote for that. I posted a separate answer to describe what happens if you add heavier elements. &lt;/p&gt;&#xA;" OwnerUserId="11175" LastActivityDate="2016-05-18T15:48:12.010" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14929" PostTypeId="1" CreationDate="2016-05-18T16:28:08.783" Score="1" ViewCount="240" Body="&lt;p&gt;I have a story I wanna write but I want to be sure it's not completely scientifically irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there is  &lt;strong&gt;black hole&lt;/strong&gt; modelisation called the Kerr black holes, in which there is a limit around black holes called &lt;strong&gt;the event horizon&lt;/strong&gt;. It's a spatial limit ; no light ray crossing this limit will ever be able to escape the gravity of the black hole. That's the definition of event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By extension, I guess any physical object (such as a spaceship) crossing the event horizon will be doomed to orbiting around the black hole until it gets destroyed. My question is about &lt;strong&gt;the time you can spend between the moment you cross the event horizon&lt;/strong&gt; (= you are doomed) &lt;strong&gt;and the moment you actually die&lt;/strong&gt; (for example by the tide effect, which basically destroys bodies because of the difference of attraction force between the feet and the head of a human body)? Could this moment last... 100 years ? 1000 years ? &#xA;(for my story the more the better)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know black holes studies are very theoretical, I'd just like to avoid any huge scientific plot holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am quite unskilled in that domain of astronomy, about time dilatation, etc. So if anyone has any idea of the order of magnitude of the time a object can spend beyond a black hole event horizon before it gets destroyed ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your answers :)&lt;/p&gt;&#xA;" OwnerUserId="11881" LastActivityDate="2016-05-18T18:00:33.840" Title="How long can you be trapped orbiting around a black hole for?" Tags="&lt;black-hole&gt;&lt;light&gt;&lt;horizon&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14930" PostTypeId="2" ParentId="14889" CreationDate="2016-05-18T16:40:47.270" Score="2" Body="&lt;p&gt;Simple answer. The sun is mostly hydrogen with an atomic weight of 1. Mercury is mostly (70%) metal such as iron (with an atomic weight of 55). Iron has a head start on density. For hydrogen to equal iron in density, 55 hydrogen atoms would have to be compressed in the space of a single iron atom. This happens in the core of the sun, but not in the entire sun.&lt;/p&gt;&#xA;" OwnerUserId="11882" LastActivityDate="2016-05-18T16:40:47.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14931" PostTypeId="1" CreationDate="2016-05-18T16:42:25.190" Score="2" ViewCount="193" Body="&lt;p&gt;I recently read some articles and pages on Solar Sails, and the possibility of using lasers to propel spacecraft, and I had a relatively outlandish idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be possible for a star to be outputting enough light that planet-sized (or smaller) objects caught in it's gravitational pull would not cross a certain threshold? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sort of 'reverse blackhole' if you will, where the event horizon is an area that the matter(not light) can not proceed further into the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would assume this to look like a star with a large ring around it, equidistant from any point of the star's center, where matter would coalesce. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, would it be possible that an object could remain 'stationary' (ie. not in orbit and no acceleration in any direction beyond the pull of gravity) around this star without fear of gravity pulling it into the star? &lt;/p&gt;&#xA;" OwnerDisplayName="user11883" LastActivityDate="2016-05-18T19:30:53.437" Title="Solar Pressure and Gravity" Tags="&lt;star&gt;&lt;gravity&gt;&lt;light&gt;&lt;astrophysics&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14932" PostTypeId="2" ParentId="14931" CreationDate="2016-05-18T17:39:21.180" Score="2" Body="&lt;p&gt;While I'm lacking the formulae, the applicable surface area of a body would have to be great enough, and it's density low enough that it's gravity remains below that threshold. So a smaller body is much more likely. If you think about it, most dust from near the sun gets pushed out by the pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There wouldn't be a specific point where you would get the matter buildup, it'd be more like the Roche limit where the gravity and density are the factors of the distance from the larger body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without running calculations I'm also unsure as to whether it would in fact be a stable arrangement but more like an L1 style Lagrange point, where it requires some energy every now and then otherwise it drifts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edits are welcome, I feel that I haven't used the best language available here and may have rambled.&lt;/p&gt;&#xA;" OwnerUserId="11525" LastActivityDate="2016-05-18T17:39:21.180" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14933" PostTypeId="2" ParentId="14929" CreationDate="2016-05-18T18:00:33.840" Score="1" Body="&lt;p&gt;You're right to mention time dilation, because the point of view is important. The standard metaphor is that of an astronaut (A) falling towards the event horizon, wearing a watch, while another astronaut (B) watches from a distance. Astronaut A will simply fall through the horizon and hit the singularity, but B will never actually see A cross the event horizon, instead they will see A's watch run slower and slower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of survival time, bigger is better. Getting too close to a really small black hole could rip you apart with tidal forces without even having to cross the event horizon (just as moons and planets exert tidal forces on each other) - but with a supermassive one you'd not even notice you'd crossed it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a relatively &quot;standard&quot; black hole of about 30 solar masses the infalling astronaut will have around 0.0001s between crossing the EH and hitting the singularity (from their point of view....see above, B will never see them cross the EH).&lt;/p&gt;&#xA;" OwnerUserId="11884" LastActivityDate="2016-05-18T18:00:33.840" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14934" PostTypeId="2" ParentId="14931" CreationDate="2016-05-18T19:30:53.437" Score="2" Body="&lt;p&gt;Both the radiation and gravity follow an inverse square law, so there is no &quot;horizon&quot; at which gravity would overcome light radiation, they both get weaker at the same rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radiation pressure affects smaller objects disproportionately. This is one factor in the formation of comet tails: dust and gas is pushed away from the heavier nucleus, which is relatively unaffected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very small objects can already be pushed from our sun, I did a &lt;a href=&quot;https://www.wolframalpha.com/input/?i=G*(mass%20of%20sun)*(x%20kg)%2F(1AU)%5E2%20%20%3D%20((x%2F5000)%5E(2%2F3)%20*%200.000009)N&quot; rel=&quot;nofollow&quot;&gt;quick estimate&lt;/a&gt; and found that dust with a mass of less than $10^{-16}\,kg$ the mass of a bacteria, would be more affected by the light of the sun than the gravity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a planet to be affected more by the radiation than the gravity of a star would be beyond realistic (though might make a good &quot;xkcd 'what if'&quot;. Not only would be be enough to vaporise a planet, it would be enough to unbind the star. It would be a supernova.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-05-18T19:30:53.437" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14935" PostTypeId="1" AcceptedAnswerId="14940" CreationDate="2016-05-18T21:06:09.840" Score="2" ViewCount="542" Body="&lt;p&gt;According to numerous articles, solar flares have done damage to the power grid on Earth in the past: &lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2008/06may_carringtonflare/&quot; rel=&quot;nofollow&quot;&gt;http://science.nasa.gov/science-news/science-at-nasa/2008/06may_carringtonflare/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;How bad can we expect a severe solar flare to impact the power grid / electronics?  Would a bad flare also break electronics (e.g. computers), or &quot;just&quot; the power grid?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there anything that can be done to reduce the effects?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2017-04-29T11:36:33.570" Title="How bad could we reasonably expect a solar flare to impact earth, and what can be done to mitigate the impact?" Tags="&lt;earth&gt;&lt;solar-flare&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14937" PostTypeId="1" AcceptedAnswerId="14939" CreationDate="2016-05-19T03:29:13.980" Score="0" ViewCount="187" Body="&lt;p&gt;There is evidence of a global resurface event on Venus: the craters are young and evenly distributed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible that there is another reason: Venus had a very deep ocean in the past, and that hindered formation of craters on the young planet&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus is believed to have had a large amount of water when it formed, but the water molecule has been broken down due to lacking magnetic field for protection, and the hydrogen released has escaped from the planet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My hypothesis is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In ancient times, Venus had a very deep ocean&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The very deep ocean prevents formation of impact craters on the ancient planet, so no old impact craters are found on the surface of Venus&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Later, the water in the ocean was broken down into hydrogen and oxygen  by solar activity, so the water level decreased.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When most water had escaped from planet, asteroids can strike the crust directly and form impact craters.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As most impact craters can only be formed after the water escaped, so we can only find young craters at the surface of Venus.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is my hypothesis possible?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="2365" LastEditDate="2016-05-21T06:37:18.867" LastActivityDate="2016-05-23T14:41:02.830" Title="Can young surface of Venus be caused by early oceans instead of global resurface event?" Tags="&lt;venus&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14938" PostTypeId="2" ParentId="14917" CreationDate="2016-05-19T06:17:10.420" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;why didn't the sun's gravity keep the inner planets from forming?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The short answer is, the asteroid belt orbits the sun, it doesn't orbit Jupiter, and you shouldn't expect the same effect with 2 different relations.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun is the gravitational object that the Asteroid belt orbits.  Jupiter is not and it has a stabilizing effect on the material that orbits around it.  Jupiter has a (mostly) destabilizing effect on nearby material that orbits around the sun, though there is some stability with orbital resonance and trojan points.   The reason why that is gets a little bit mathy, but I'll cover the basics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a solar system forms, probably out of a combination of strong solar wind from a supernova and a dense enough gas cloud, most of the solar-system material forms the young sun and maybe 2%-4% remains in orbit and over time, spirals and flattens out into a disk.  (see cool &lt;a href=&quot;https://www.youtube.com/watch?v=tmNXKqeUtJM&quot; rel=&quot;nofollow&quot;&gt;video&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun's gravity stabilizes that disk.  There's no reason for it to disburse it, though the young and usually strong solar wind can blast it and clear out the closer regions of smaller particles and any ices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter is entirely different.  Jupiter and other planets which form inteh disk, tends to eat everything in their path.  That's one of the definitions of a planet.  It clears out it's orbital area.  These planets on their own would remain in orbit around the sun, but they can interact over time with each other.   Pretty much anything in Jupiter's path, once Jupiter gets large enough, either gets eaten or gets thrown into a very different orbit, either further our or closer in.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The asteroid belt is (as Suhrid Mulay points out) a relatively low density region of space that doesn't have enough mass to coalesce into it's own planet, but it might not have always been that low density and that's probably only part of the reason.  The other part of the reason is the proximity of Jupiter which tends to disrupt anything that orbits too close to it and Jupiter may have at one point, moved in quite a bit closer to Mars before moving back out.  Tossing much of the material in that region far away.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets don't do well forming too close to other planets because they will gravitationally disturb each other.  A planet the size of Jupiter had pretty significant reach when it comes to disturbing other planets and planet formation.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-05-20T22:45:13.340" LastActivityDate="2016-05-20T22:45:13.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14939" PostTypeId="2" ParentId="14937" CreationDate="2016-05-19T08:14:42.270" Score="2" Body="&lt;p&gt;Our best guess is that if Venus had an ocean, it did 4 billion years ago. Then it had the runaway greenhouse effect due to its thick atmosphere. The runaway greenhouse effect would have boiled away the oceans if the planet ever had oceans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several guesses about why its surface is so young (300 million to 600 million years old). One is that the planet get so hot the entire surface melts. It has no plate tectonics and it has a super thick atmosphere to trap the heat. All that heat melts the surface every half billion years or so. Another reason is that it has many very large volcanoes. (167 volcanoes over 100 km across.) That's far more very large volcanoes than Earth has. Those volcanoes resurface the planet and keep it young.&lt;/p&gt;&#xA;" OwnerUserId="11175" LastActivityDate="2016-05-19T08:14:42.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14940" PostTypeId="2" ParentId="14935" CreationDate="2016-05-19T12:11:13.747" Score="4" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Space_weather&quot; rel=&quot;nofollow&quot;&gt;Space Weather&lt;/a&gt; is the field within heliophysics that tries to understand this Sun-Earth relationship. When a Solar Flare occurs multiple things may happen and how it interacts with the Earth and our technology depend on different factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first effect, the X-ray radiation produced by an Solar Flare affects the ionosphere and therefore the radio communications. This effect is almost immediate and there's no way to attenuate or prevent it. We can forecast the likelihood of a flare to happen on the day but still we don't have the knowledge enough to exactly predict when it will happen. The radiation from the flare it takes around 8 minutes to get to Earth, and at the time we see it the radio communications have already been affected. This effects mostly to VLF bands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a solar flare high energetic particles and coronal mass ejection could be produced, these two events have different effects on Earth and whether interacts with Earth or not depends on where it happens on the Sun and the properties of the solar wind at that time. High energetic particles (also known as Solar Energetic Particle events - SEPs) could arrive around 30 minutes after the flare happened, and Coronal Mass Ejections (CMEs) between 2 or 3 days. These times depends on the properties of the events, not all are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CMEs could be somehow predicted, and are these the ones that could affect, between others, the power grid. It won't affect your computer. The reason of why it affects the power grids is due to a induced electric current happened by the interaction of the CME's magnetic field produces in the Earth's magnetic field. The electric current finds it easier to travel through the power grid than through the Earth itself (the cables are better conductors). In simple words, when that happens the electric grid &quot;thinks&quot; that there is an increase of demand and tries to compensate it generating more electricity and that can burn them - as &lt;a href=&quot;http://www.nasa.gov/topics/earth/features/sun_darkness.html&quot; rel=&quot;nofollow&quot;&gt;it happened in Canada in 1989&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To reduce it effects there are a lot of engineering work around these systems and I'm not an expert in such topic to be able to explain them properly. However, a warning to the power grids of a couple of days in advance helps. The difficult bit is how to be exact in the forecasting. CMEs sometime are deflected by the solar wind and just a flank of it &quot;touches&quot; the Earth, other times small ones get accelerated and have a larger impact than expected. So, to be able to reduce the effects we need to have a better understanding of their origin and surroundings, understand their physics and be able to model what they are doing. There are few models out there (e.g., &lt;a href=&quot;http://ccmc.gsfc.nasa.gov/&quot; rel=&quot;nofollow&quot;&gt;CCMC&lt;/a&gt;) but at the moment we need to keep researching to improve these further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answering to &quot;how bad&quot; a &lt;em&gt;solar storm&lt;/em&gt; (not just a flare) can affect Earth is hard to quantified, in UK it has been classified as one of the highest priority natural hazards in the National Risk Register, and similarly in other countries. An event like Carrington's (the one you linked) could have a lot of impact in our current technological civilization and it's been estimated in many years and trillions of US dollars to restore the damage (see &lt;a href=&quot;http://www.raeng.org.uk/publications/reports/space-weather-full-report&quot; rel=&quot;nofollow&quot;&gt;Royal Academy of Engineer report on Space weather&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="137" LastActivityDate="2016-05-19T12:11:13.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14941" PostTypeId="1" AcceptedAnswerId="14948" CreationDate="2016-05-19T18:48:38.167" Score="1" ViewCount="190" Body="&lt;p&gt;Are there commonly used names for objective directions within a spiral galaxy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Central and peripheral describe objective directions, because the observer's frame of reference doesn't affect whether something is closer or farther from the middle.  &quot;Clockwise&quot; or &quot;counterclockwise&quot; don't work, because they depend on which side you're viewing from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there names that can describe one object's location relative to another object, either in the galaxy's direction of spin or against it?  Or words to describe whether something is farther or closer out along a particular arm of a spiral galaxy? &lt;/p&gt;&#xA;" OwnerUserId="11896" LastActivityDate="2016-05-20T14:41:17.517" Title="Direction Names Within a Galaxy" Tags="&lt;galaxy&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14942" PostTypeId="2" ParentId="14941" CreationDate="2016-05-19T19:09:15.367" Score="2" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Supergalactic_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;galactic coordinate system&lt;/a&gt; is usually used for galaxies. It is based on the galactic plane — approximately where the galactic disc is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It uses galactic latitude (for the y-value of the objects) and galactic longitude (x and z values, based on the disc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a &lt;a href=&quot;https://en.wikipedia.org/wiki/Supergalactic_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;supergalactic coordinate system&lt;/a&gt; used for superclusters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its equator is an invisible line called the supergalactic plane, which is the preferred distribution of galaxies in a supercluster.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-05-20T14:41:17.517" LastActivityDate="2016-05-20T14:41:17.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14944" PostTypeId="2" ParentId="2416" CreationDate="2016-05-19T22:45:13.417" Score="2" Body="&lt;p&gt;I was able to get the Cartesian orbital vectors for all the major bodies from HORIZON at the J2000 epoch only.  I could extend the coverage forward thru time. It’s easy to get data overload doing this. My simulation is modeled using the Laws of Gravitation and Motion alone.  This gives results that are surprisingly close to those published.  Running the solar system backwards (by reversing the velocity vectors) has given me the initial vectors back to 1900.  This is all I needed and the results were close enough for my purposes.  I still have the CSV  files.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also have had all sorts of problems with the horizons interface.  For instance changing the date had no effect on the value of the vectors. i.e.: all specified start dates have the same values.  Lately, I have not been able to duplicate this feat.  There are obviously some serious problems with this interface, especially lately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the data I got was correct because it correlates, perfectly, with published events, e.g.: the recent transit of Mercury.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I too am still looking for this type of data.&lt;/p&gt;&#xA;" OwnerUserId="11899" LastActivityDate="2016-05-19T22:45:13.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14945" PostTypeId="2" ParentId="14935" CreationDate="2016-05-19T22:52:25.323" Score="2" Body="&lt;p&gt;Satellites have been damaged before. GPS has been effected before. Power grids surge during magnetic disturbances. Pipelines energize too. March 13 1989 was a whopper. There's a lot of reporting online of this event. Some papers are still being written on the event. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nasa.gov/topics/earth/features/sun_darkness.html&quot; rel=&quot;nofollow&quot;&gt;http://www.nasa.gov/topics/earth/features/sun_darkness.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if a bigger, say a magnitude more severe, take the above and extrapolate from there. With our greater dependence on satellites for military, commerce, and daily living, a major loss of connectivity would occur and may be prolonged if the satellites are damaged, and orbits disturbed by the solar wind. Power grids, electric rail service, all could be down for prolonged periods if transformers are overloaded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ionosphere could be reduced and that would impact life. &lt;/p&gt;&#xA;" OwnerUserId="11798" LastActivityDate="2016-05-19T22:52:25.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14946" PostTypeId="2" ParentId="14921" CreationDate="2016-05-19T23:07:52.413" Score="0" Body="&lt;p&gt;The earth sun distance is complex. So, as was said, is date specific. An AU is 'average' distance of a mean tropical year. And the earth's elliptical orbit is complicated in that the earth/moon barycenter orbits the solar system's barycenter. So the distance between barycenters may also be a matter of precision depending on the application. &lt;/p&gt;&#xA;" OwnerUserId="11798" LastActivityDate="2016-05-19T23:07:52.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14948" PostTypeId="2" ParentId="14941" CreationDate="2016-05-20T06:32:43.527" Score="2" Body="&lt;p&gt;There are no &quot;commonly&quot; used words to describe these things, but there are apt words developed in SciFi books.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The adverbs, &lt;strong&gt;spinward&lt;/strong&gt; and &lt;strong&gt;antispinward&lt;/strong&gt; would seem to do the trick. Inwards and ouwards will do for radial motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Movement towards towards the end of a spiral arm would be outwards and antispinward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trailing and leading can also be used to state relative position with respect to Galactic rotation (for objects not too far separated).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-05-20T06:32:43.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14949" PostTypeId="1" AcceptedAnswerId="14953" CreationDate="2016-05-20T07:50:07.260" Score="3" ViewCount="775" Body="&lt;p&gt;Type 1a supernovae are used as standard candles. Are there other high energy cosmic events which can be used as standard candles. For example, collapse of neutron star into a black hole is also similar to type 1a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a galaxy, there are far more white dwarfs(candidates of type 1a) compared to neutron stars(I read somewhere that our galaxy contains around 200 million of them). Also, the energy released in neutron star collapse is not in visible spectrum(I don't think so that is the criteria for &quot;standard candles&quot;). Are these the reasons that neutron star collapse is not qualified as another type of standard candle?&lt;/p&gt;&#xA;" OwnerUserId="10657" LastActivityDate="2016-05-23T13:07:50.600" Title="Cosmic events as standard candles" Tags="&lt;neutron-star&gt;&lt;white-dwarf&gt;" AnswerCount="2" CommentCount="10" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14950" PostTypeId="1" CreationDate="2016-05-20T08:00:46.183" Score="0" ViewCount="569" Body="&lt;p&gt;As I know, spring equinox of 2016 is on 2016/3/20 4:30 GMT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I execute following statements in python:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;```&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from astropy.time import Time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from astropy.coordinates import get_sun&lt;/p&gt;&#xA;&#xA;&lt;p&gt;t=Time('2016-03-20 4:30:00')&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s=get_sun(t)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;```&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect to get s.ra.degree and s.dec.degree very close to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I got s.ra.degree=0.7052336 and s.dec.degree=0.30536325.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope some one can help me. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="11904" LastActivityDate="2016-05-20T12:31:22.447" Title="How to interprete function value of get_sun in astropy" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14951" PostTypeId="2" ParentId="14949" CreationDate="2016-05-20T09:00:09.033" Score="4" Body="&lt;p&gt;Gravitational wave sources, such as close, compact binaries can be treated as standard candles. Actually a better metaphor would be &quot;standard sirens&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As was recently demonstrated by the recent detection of a gravitational wave signal from a merging black hole binary, the amplitude of the signal, combined with its frequency behaviour give you both the characteristics of the merging black hole &lt;em&gt;and&lt;/em&gt; the distance to the source (independently of its redshift).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such events &lt;em&gt;are&lt;/em&gt; the most powerful events in the universe (I think more powerful than Gamma Ray bursts) - in this case $3M_{\odot}$ of mass was converted into gravitational waves in a fraction of a second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my knowledge, the collapse of a formed neutron star into a black hole has never been witnessed (at any wavelength).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Well, what do I know - it turns out that there are certain classes of Gamma Ray Burst (GRB) events that may be explained by the accretion onto a neutron star in a close binary system, causing it to collapese (see &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014A%26A...569A..39R&quot; rel=&quot;nofollow&quot;&gt;Ruffini et al. 2014&lt;/a&gt;). They are often called &quot;binary driven hypernovae&quot;. Although there is a possibility that these sources could become standard candles, the understanding of the relevant mechanisms is not sufficient for them to be considered so yet.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-05-20T10:49:03.253" LastActivityDate="2016-05-20T10:49:03.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14952" PostTypeId="1" AcceptedAnswerId="16186" CreationDate="2016-05-20T10:36:38.623" Score="4" ViewCount="600" Body="&lt;p&gt;I've slightly modified the title to try to attract some attention. If we call sunlight &quot;white&quot; and limb darkening is a result of seeing deeper at normal incidence and shallower at oblique incidence, then &lt;em&gt;the center end edge of the solar disk viewed from earth should appear to have slightly different effective color temperatures, since the temperature is varying rapidly with depth&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to have an approximate expression for the wavelength-dependent limb darkening of the sun in the visible spectrum, either a relatively simple analytical expression that I can understand (with the appropriate coefficients for the sun) or just some linear images of the sun in various bands in the visible wavelengths so I can try to make one of my own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;https://astronomy.stackexchange.com/q/12331/7982&quot;&gt;question&lt;/a&gt; received this helpful &lt;a href=&quot;https://astronomy.stackexchange.com/questions/12331/linear-limb-darkening-coefficient-u#comment17382_12331&quot;&gt;comment&lt;/a&gt; which links &lt;a href=&quot;http://www.astro.keele.ac.uk/jkt/codes/jktld.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;to here&lt;/a&gt; but honestly I can't make my way through that to a practical expression I can use. The &lt;a href=&quot;https://i.stack.imgur.com/MMyje.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; article is not helping me much either, except for the &lt;a href=&quot;https://en.wikipedia.org/wiki/File:2012_Transit_of_Venus_from_SF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;image there&lt;/a&gt;. I plotted scans of RGB but by the time an image gets into the internet, things like sRGB and gamma mean it may not be linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;update:&lt;/strong&gt; At the &lt;a href=&quot;http://sdo.gsfc.nasa.gov/data/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solar Dynamics Observatory (SDO)&lt;/a&gt; website, I just found the image &lt;a href=&quot;http://sdo.gsfc.nasa.gov/assets/img/latest/latest_1024_HMIIC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;sdo.gsfc.nasa.gov/assets/img/latest/latest_1024_HMIIC.jpg&lt;/a&gt;. The color gradient of the limb darkening seems very similar to the Wikimedia image below. I've discovered that it is called a &quot;colorized intensitygram&quot; and the color gradient is purely artificial - the data is single channel intensity. The limb darkening is certainly real (compare to the artificially &quot;flattented&quot; display!)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.solarham.net/latest_imagery/hmi1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.solarham.net/latest_imagery/hmi1.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/a8KdC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a8KdC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I appreciate (the existence of) the complexities and intricacies of photon transport, instrumental effects, and color perception, but I am &lt;em&gt;just starting&lt;/em&gt; to do astronomically correct animations, so please for right now, something imperfect, or not absolutely correct is good enough for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;20 pixel wide averages of horizontal (-) and vertical (--) &quot;scans&quot; across the center of this 600x600 pixel &lt;a href=&quot;https://en.wikipedia.org/wiki/File:2012_Transit_of_Venus_from_SF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;image from here&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;note:&lt;/strong&gt; It seems this is a somewhat futile example, as the image is likely to be a monochrome continuum image with false color!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/28nHI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/28nHI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-06-10T04:53:51.240" Title="Is the Sun slightly blue in the center? - Wavelength-dependent limb darkening of the Sun" Tags="&lt;stellar-astrophysics&gt;&lt;the-sun&gt;&lt;photography&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14953" PostTypeId="2" ParentId="14949" CreationDate="2016-05-20T12:04:27.070" Score="5" Body="&lt;p&gt;&lt;sub&gt;Intro for the uninformed: A &lt;em&gt;standard candle&lt;/em&gt; is an important concept in astronomy, helping to map out distances in the Universe. Since the observed flux &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; of a light source decreases with distance &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; by a known factor (&lt;span class=&quot;math-container&quot;&gt;$r^2$&lt;/span&gt;), if we know its intrinsic luminosity &lt;em&gt;L&lt;/em&gt;, we can calculate the distance. For large distances, where bright sources are needed, we usually use &lt;em&gt;supernovae&lt;/em&gt; (SNe). But the luminosity of a SN depends on the mass of its progenitor star which is not known in general. However, for a specific type of SNe — &amp;quot;type Ia&amp;quot; — it &lt;em&gt;is&lt;/em&gt; known: This type are SNe that explode when a white dwarf accreting mass from a companion star exceed the mass threshold for explosion of &lt;span class=&quot;math-container&quot;&gt;$1.4\,M_\odot$&lt;/span&gt;.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;In addition to the gravitational waves discussed by Rob Jeffries, I can mention the following candidates for standard candles:&lt;/p&gt;&#xA;&lt;h2&gt;Type II supernovae&lt;/h2&gt;&#xA;&lt;p&gt;Type Ia SNe are so similar in luminosity because they all have (almost) the same mass when they go off. But there is also evidence that type II SNe can act as standard candles. As is also to some extent the case with Ia, their lightcurves (how the luminosity changes with time) are not identical, but can be standardized using the so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Phillips_relationship&quot; rel=&quot;nofollow noreferrer&quot;&gt;Philips relation&lt;/a&gt; (see e.g. &lt;a href=&quot;http://arxiv.org/abs/0910.1590&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kasen &amp;amp; Woosley 2009&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;h2&gt;GRB supernovae&lt;/h2&gt;&#xA;&lt;p&gt;Gamma-ray bursts as a whole are too diverse to be used as standard candles, but when an associated SN is detected, it becomes a standard(izable) candle (&lt;a href=&quot;http://arxiv.org/abs/1407.3506&quot; rel=&quot;nofollow noreferrer&quot;&gt;Li &amp;amp; Hjorth 2014&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;h2&gt;Quasars&lt;/h2&gt;&#xA;&lt;p&gt;My favorite candidate are &lt;em&gt;quasars&lt;/em&gt;. This technique doesn't rely on the Philips relation. Quasars are caused by gas accreting onto a supermassive black hole in the center of galaxies, resulting in an &amp;quot;active galactic nuclus&amp;quot; (AGN) with extreme energy outputs (easily over &lt;span class=&quot;math-container&quot;&gt;$10^{12}\,L_\odot$&lt;/span&gt;, and even up to &lt;span class=&quot;math-container&quot;&gt;$10^{14}$&lt;/span&gt;–&lt;span class=&quot;math-container&quot;&gt;$10^{15}\,L_\odot$&lt;/span&gt;; &lt;a href=&quot;https://arxiv.org/abs/astro-ph/9908052&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ibata et al. 1999&lt;/a&gt;). It turns out that there is a correlation between the absolute luminosity of the AGN and the size of its &lt;em&gt;broad-line region&lt;/em&gt; (BLR), i.e. the region around the quasar where fast-moving gas clouds absorb the continuum&lt;span class=&quot;math-container&quot;&gt;$^1$&lt;/span&gt; of the quasar and emit lines&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt;, e.g. H&lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; (&lt;a href=&quot;http://arxiv.org/abs/1109.4632&quot; rel=&quot;nofollow noreferrer&quot;&gt;Watson et al. 2011&lt;/a&gt;). The reason is that the size of the BLR is determined by the depth that the ionizing radiation from the quasar can penetrate into the BLR, which is proportional to the square root of the luminosity. The figure below (from Watson et al. 2011) shows the relation between distances (&lt;span class=&quot;math-container&quot;&gt;$D_L$&lt;/span&gt;) determined by this technique and distances obtained from type Ia SNe.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hW3hQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hW3hQ.png&quot; alt=&quot;DL&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;strong&gt;Advantages over supernovae&lt;/strong&gt;&#xA;&lt;p&gt;A huge advantage of quasars over SNe is that they don't disappear after a few weeks, meaning that if e.g. we want to refine some measurement, we can go back and observe it again at any time. Another advantage is that quasars, being so luminonous, can be detected out to much larger distances (roughly to &lt;span class=&quot;math-container&quot;&gt;$z\simeq4$&lt;/span&gt;) than SNe (which are only observed ut to &lt;span class=&quot;math-container&quot;&gt;$z\simeq2$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;strong&gt;Reverberation mapping&lt;/strong&gt;&#xA;&lt;p&gt;Since quasars are so far away, the BLR, being less than a parsec in size, cannot be resolved. But luckily, a technique call &amp;quot;reverberation mapping&amp;quot; allows us to determine the size:&lt;/p&gt;&#xA;&lt;p&gt;The spectrum of the quasar consists of a continuum with spectral lines. Quasars vary in luminosity on rather short timescales. If we measure a quasar's luminosity regularly over some period of time, we get a so-called &amp;quot;lightcurve&amp;quot;. But since the lines are created at a distance from the source, a given &amp;quot;bump&amp;quot; in the lightcurve (i.e. a temporary increase in luminosity) does't show up in the continuum and the lines at the same time. Instead, there is a delay, corresponding to the extra distance that the light had to travel from the quasar to the cloud reflecting&lt;span class=&quot;math-container&quot;&gt;$^3$&lt;/span&gt; it.&lt;/p&gt;&#xA;&lt;p&gt;In the figure below, &lt;em&gt;blue&lt;/em&gt; shows the continuum, while &lt;em&gt;red&lt;/em&gt; shows the lines. The lines are seen to lag behind the continuum, since they first had to travel from the quasar (&lt;em&gt;black&lt;/em&gt;) to the clouds (&lt;em&gt;magenta&lt;/em&gt;).&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wFL0M.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wFL0M.png&quot; alt=&quot;reflect&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Since the clouds lie at a range of distances, they exhibit different time lags, effectively broadening the line:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PGYwc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PGYwc.png&quot; alt=&quot;lightcurve&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;So, in the figure above, showing observed flux as a function of time, the light in the &lt;em&gt;line&lt;/em&gt; increased in luminosity roughly 1.5 days later than the light in the &lt;em&gt;continuum&lt;/em&gt;. This means that the BLR is roughly 1.5 lightdays (or ~250 AU) in radius.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^1$&lt;/span&gt;&lt;sub&gt;I.e. the continuous and relatively featureless spectrum of many different atomic transitions and physical processes.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt;&lt;sub&gt;I.e. features in the spectrum resulting from strong atomic transitions. For instance, an eletron falling from the second to the first excited states of a hydrogen atom emits a photon with the wavelength 6563 Å, called &amp;quot;H&lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;&amp;quot;.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^3$&lt;/span&gt;&lt;sub&gt;The light isn't really &amp;quot;reflected&amp;quot;. Instead, it is the high-energy photons (UV and X-rays) of the continuum that ionize the atoms in the clouds. When the ions recombine, they emit spectral lines.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-05-23T13:07:50.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14954" PostTypeId="2" ParentId="14950" CreationDate="2016-05-20T12:31:22.447" Score="1" Body="&lt;p&gt;Looking at the astropy documentation, the Sun's position is calculated in GCRS co-ordinates these are intern expressed in J2000.0 and not JNow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference is therefore due in part to precession between J2000.0 and JNow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will also find that the time of the equinox is not exactly 4:30 but has been rounded. The algorithm used to get to 4:30 itself may also have precision limitations due to all sorts of reasons. The same will be true of the calculated position of the Sun at a given time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These errors will all add up to give a slightly incorrect figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cartes du Ciel (a well known planatarium application) calculates the Sun's location at 4:30 as being:&#xA;JNow&#xA;RA: 00h00m01.71s &#xA;DE:+00°00'01.4&quot;&#xA;J2000&#xA;RA: 23h59m11.85s&#xA;DE:-00°05'23.6&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it looks like most off the error is due to limits of the Astropy library.&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-05-20T12:31:22.447" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14956" PostTypeId="1" CreationDate="2016-05-20T18:24:04.580" Score="2" ViewCount="40" Body="&lt;p&gt;I am looking to get some information on the adverse effects of urban growth/light pollution on the oldest/traditional observatories of the world (europe).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone recommend me books or articles to read?&lt;/p&gt;&#xA;" OwnerUserId="11910" LastEditorUserId="2365" LastEditDate="2016-05-21T06:21:39.977" LastActivityDate="2016-05-21T06:21:39.977" Title="Adverse effects of urban growth on old astronomy observatories? Need article/book" Tags="&lt;light-pollution&gt;&lt;observatory&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2016-10-17T15:32:26.660" ContentLicense="CC BY-SA 3.0" />
  <row Id="14957" PostTypeId="1" AcceptedAnswerId="14963" CreationDate="2016-05-20T19:58:57.713" Score="0" ViewCount="70" Body="&lt;p&gt;Does the latitude of your position on earth change the appearance of the cresent like it changes the position of other objects visible in the sky? Does it rotate it?&lt;/p&gt;&#xA;" OwnerUserId="11911" LastActivityDate="2016-05-21T11:36:18.320" Title="Appearance of the Cresent" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2016-05-23T14:40:00.250" ContentLicense="CC BY-SA 3.0" />
  <row Id="14959" PostTypeId="1" CreationDate="2016-05-20T23:58:22.383" Score="2" ViewCount="85" Body="&lt;p&gt;I took many images like the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K4UVL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K4UVL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As many of you may recognise, this is a picture of Jupiter and its four Galilean moons. Using these images, I am aiming to calculate the mass of Jupiter by finding the orbital distance and orbital period of each of these moons, and then plugging those values into Kepler's Third Law. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this requires a seemingly tedious process of the following sort:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Locate the centre of mass of Jupiter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Measure the distance to each of the planets&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a lot of data (many, many pictures). I am hoping that there is a relatively easier (or even more accurate) method of doing all of the above than using software like Gimp to do everything for each image manually. Can anyone recommend anything?&lt;/p&gt;&#xA;" OwnerUserId="9650" LastActivityDate="2016-07-20T12:53:41.370" Title="Astronomy Calculations Tool" Tags="&lt;jupiter&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14960" PostTypeId="1" AcceptedAnswerId="14961" CreationDate="2016-05-21T08:32:09.290" Score="4" ViewCount="503" Body="&lt;p&gt;I don't know much about planets outside our solar system. Do we even know the limit of the size of how large a planet or satellite can be? The largest planet I know of is Jupiter, but could there be planets as large as our Sun or even larger? Is there a theoretical limit to the size of a planet?&lt;/p&gt;&#xA;" OwnerUserId="5265" LastEditorUserId="2531" LastEditDate="2016-05-21T10:42:26.790" LastActivityDate="2016-05-21T10:42:26.790" Title="Which is the largest planet ever observed?" Tags="&lt;planet&gt;&lt;exoplanet&gt;&lt;size&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14961" PostTypeId="2" ParentId="14960" CreationDate="2016-05-21T10:10:01.070" Score="14" Body="&lt;p&gt;Yes, there is a limit. Anything with a &lt;em&gt;mass&lt;/em&gt; larger than about 13 times that of Jupiter would be called a brown dwarf (a failed star), though whether such an object would consist entirely of gas, or had a rocky/icy core as is probable for most giant planets, is not presently observable. Any larger than about 75 Jupiter masses and we would just call it a star. The &lt;em&gt;exact&lt;/em&gt; definition of what a planet is (especially, the 13 Jupiter mass boundary) is still disputed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the bonafide planets that have been detected and confirmed, the catalogue at exoplanets.org lists &lt;a href=&quot;http://exoplanets.org/detail/Kepler-435_b&quot; rel=&quot;noreferrer&quot;&gt;Kepler-435b&lt;/a&gt; as the one with the largest measured radius (although its radius error bar overlaps with that of other planets). The quoted radius is $1.99 \pm 0.18$ times that of Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most giant planets have very similar radii for masses between about 0.5 and 10 times that of Jupiter. The reason for this is that they are largely supported by electron degeneracy pressure. The diversity in the radii (between about 0.7 and 2 times that of Jupiter) of such planets is not yet fully understood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plot below shows mass vs radius for &quot;planets&quot;. The smaller (probably rocky/icy) planets do show a trend of increasing radius with mass (the solid line is where a theoretical relationship for rocky/icy planets has been used to estimate the mass from the radius). The gas giants above about 0.5 Jupiter masses show no trend and a small scatter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bZPne.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bZPne.png&quot; alt=&quot;Mass versus radius&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-05-21T10:20:51.793" LastActivityDate="2016-05-21T10:20:51.793" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14962" PostTypeId="2" ParentId="14959" CreationDate="2016-05-21T11:32:53.537" Score="1" Body="&lt;p&gt;Try &lt;a href=&quot;http://jupos.org/gh/download.htm&quot; rel=&quot;nofollow&quot;&gt;WinJupos&lt;/a&gt;. It can be used to measure Jupiter (as well as other planets)&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-05-21T11:32:53.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14963" PostTypeId="2" ParentId="14957" CreationDate="2016-05-21T11:36:18.320" Score="1" Body="&lt;p&gt;Yes it does. If you are at a high latitude the crescent Moon will set looking like a reversed C orientation but in the tropics it will set more like a U orientation. Though the time of the year will also have some effect due to the angle of the ecliptic to the horizon.&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-05-21T11:36:18.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14967" PostTypeId="2" ParentId="14858" CreationDate="2016-05-21T21:20:44.860" Score="1" Body="&lt;p&gt;If you search for &quot;vixen/synta mounts&quot; you should find lots of compatible mounts and tripods.  There is a huge range of prices and level of features. Check out what the specifications say about their load capacity, relative to the weight of your scope.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2016-05-21T21:20:44.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14968" PostTypeId="1" AcceptedAnswerId="14970" CreationDate="2016-05-21T22:37:46.010" Score="0" ViewCount="161" Body="&lt;p&gt;A blue moon is so called for when we have 13 full moons in a year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Blue_moon&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Blue_moon&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to have 11 fun moons in a year? (Maybe if we included calendar changes, for example the start of the Gregorian calendar, this would be more likely)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is possible, what is this phenomenon called, and how rare are these years?&lt;/p&gt;&#xA;" OwnerUserId="11918" LastActivityDate="2016-05-22T09:27:06.627" Title="Is it possible for the year to have 11 full moons (i.e. the opposite of a blue moon)" Tags="&lt;the-moon&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14969" PostTypeId="2" ParentId="14968" CreationDate="2016-05-22T01:06:50.690" Score="7" Body="&lt;p&gt;Since a full moon happens once every 29 days (and that's not exact, but close enough), the longest period of time where you can have 11 full moons but not 12 is 11 x 29 + 28, or 347 days, so in a normal year, it's not possible.  There are slight variations in the time between full moons but not nearly enough to make up the extra 18 days needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you'd need a special circumstance.   Great Brittan's &lt;a href=&quot;http://www.adsb.co.uk/date_and_time/calendar_reform_1752/&quot;&gt;short year&lt;/a&gt; in 1752 wasn't sufficiently short, it had 11 days removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a phenomenon where not everywhere on Earth gets a full moon every 29 days.   In the North and South poles for example, where during summer they get 24 hours of sun and winter, no sun, a &lt;a href=&quot;https://planetarium.madison.k12.wi.us/mooncal/north-pole-moon.htm&quot;&gt;somewhat similar effect happens with the moon&lt;/a&gt;.   The Moon orbits the Earth at &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbit_of_the_Moon&quot;&gt;5 degrees off the ecliptic&lt;/a&gt; or basically, the plane between the Earth and the Sun and that makes the Moon's apparent movement across the sky a little more complicated than the Sun's, but because full moons only happen when the moon is roughly opposite the sun, it's not hard to see that when the sun is in the sky for 24 hours during the Polar summer, they don't get a full moon for several months because the full moon is below the horizon when the sun is in the sky 24 hours.   You might only see 6-9 full moons a year at the poles (if I was to guess), and there's probably a range of latitudes close to the poles where 11 full moons might be about the yearly average.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to split hairs even further, the Moon only takes an instant to pass the 180 degree point in relation to the Earth and Sun, so in a sense, a perfect 180 degree full moon is only visible to half the earth and by the time the other half of the earth turns to make that moon visible, the moon has already gone past 180 degrees and from a certain perspective, it's no longer full.   But, a full moon is generally regarded as a roughly 24 hour period, not an instantaneous moment, so I don't like that argument personally.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-05-22T06:49:29.790" LastActivityDate="2016-05-22T06:49:29.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14970" PostTypeId="2" ParentId="14968" CreationDate="2016-05-22T07:39:42.597" Score="1" Body="&lt;p&gt;It is nit possible for a year to have 11 full moons. Even if there is a full moon on dec 31, there will still be 12 full moons in the subsequent year. It is possible for there to be a calendar month (February ) without a full moon. The next time this will happen,  depending on your location, is &lt;a href=&quot;http://www.timeanddate.com/moon/phases/?year=2018&quot; rel=&quot;nofollow&quot;&gt;2018,&lt;/a&gt; as there will be no full moon during the month of February that year. Location is significant, as the start of February varies with timezone &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the Two full moons in a month definution of blue moon, 2018 will have a blue moon in both January and March., the march moon will also be the fourth full moon that season,&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-05-22T09:27:06.627" LastActivityDate="2016-05-22T09:27:06.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14971" PostTypeId="1" CreationDate="2016-05-22T10:36:20.567" Score="2" ViewCount="74" Body="&lt;p&gt;The 2nd answer to this &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1179/hypothetical-beyond-neptune-far-away-planets-orbiting-the-sun?rq=1&quot;&gt;question&lt;/a&gt; by MBR got me thinking about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some general guideline for residual heat a planet should emit, say 4.5 billion years after formation.   I'm defining heat by the radiant energy that leaves the planet, not internal heat or surface temperature for rocky bodies and assuming little or no solar input.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia &lt;a href=&quot;https://en.wikipedia.org/wiki/Planet_Nine#Radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;lists an estimate&lt;/a&gt; of planet 9's temperature to be 47 degrees K based on remaining heat of formation, but I think that would depend pretty significantly on it's size as well as whether it had any recent impacts.   I can't help but wonder if that's an accurate estimate.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter, as one example, emits about twice as much energy as it receives from the sun, and it's temperature is about 128 degrees Kelvin.  Is it safe to assume that Jupiter's upper cloud temperature would still be ~ 100 Kelvin even if it was in deep space, but otherwise having the same formation and history.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neptune is also a few degrees warmer than Uranus, so has got to be putting out a fair bit of internal heat as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously there's a few factors at play - size, whether there was a recent large impact, abundance of radioactive elements, efficiency of circulation (solid layers are likely more efficient at trapping heat than circulating fluid layers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm tempted to guess that there's too many unknowns on gas giant circulation, inner layers, composition and mass for there to be any good simple charts on this stuff, but I was curious if there was anything.   I tried google without success.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-05-22T16:50:08.900" Title="How much heat emanates from the planets after formation?" Tags="&lt;planet&gt;&lt;temperature&gt;&lt;9th-planet&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14972" PostTypeId="1" AcceptedAnswerId="14973" CreationDate="2016-05-22T19:18:01.547" Score="0" ViewCount="433" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Omicron_Persei&quot; rel=&quot;nofollow&quot;&gt;Omicron Persei&lt;/a&gt; is a binary star. I did not find information on how many planets it has and if we can actually determine it currently. However, I do know that most stars have planets orbiting around them and so I assume there are such planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those who are interested, Futurama featured the planet Omicron Persei 8 - the 8th planet in that system. I was curious as to if there is an 8th planet.&lt;/p&gt;&#xA;" OwnerUserId="11921" LastActivityDate="2016-05-23T08:50:52.700" Title="How many planets does Omicron Persei have?" Tags="&lt;star-systems&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14973" PostTypeId="2" ParentId="14972" CreationDate="2016-05-23T08:50:52.700" Score="2" Body="&lt;p&gt;Currently &lt;a href=&quot;http://exoplanet.eu/&quot; rel=&quot;nofollow&quot;&gt;http://exoplanet.eu/&lt;/a&gt; (which lists more exoplanets than the official NASA archive) lists no known planets of Omicron Persei. So currently we know of none, it is outside of the Kepler field so has no Kepler candidates either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though just because none are known that does not mean it has none, it may be that their mass and or orbital plane is such that it is very difficult to identify planets or no one has looked with a suitable instrument.&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-05-23T08:50:52.700" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14975" PostTypeId="1" AcceptedAnswerId="14994" CreationDate="2016-05-23T15:46:48.993" Score="5" ViewCount="315" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Kirkwood_gap&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kirkwood gaps&lt;/a&gt; in the asteroid belt are associated with orbital resonances with Jupiter yet planets seem to prefer resonant locations. Why don't asteroids accumulate near the &amp;quot;Kirkwood locations&amp;quot; rather than avoid them??&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="37191" LastEditDate="2021-01-25T22:26:37.973" LastActivityDate="2021-01-25T22:26:37.973" Title="Planetary orbital resonances" Tags="&lt;asteroid-belt&gt;&lt;orbital-resonance&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="14976" PostTypeId="1" CreationDate="2016-05-23T17:16:36.943" Score="1" ViewCount="155" Body="&lt;p&gt;I keep hearing astronomers saying there are observatories scanning the skies all day and night looking for near-Earth objects, extrapolating their trajectories. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If we were to get the news of an on-coming object, how early we will find out or how much time we will have?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Currently, people don't even think about it; Governments are cutting spending in their space exploration programs, let alone deflecting an impact-er.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under certain circumstances, would the notice from the watchdogs be enough for us to build something that would prevent the loss?&lt;/p&gt;&#xA;" OwnerUserId="4099" LastActivityDate="2016-05-25T14:14:29.743" Title="How ready are we to deal with a asteroid/comet/giant-meteor on a collision course with Earth?" Tags="&lt;asteroids&gt;&lt;impact&gt;&lt;near-earth-object&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14977" PostTypeId="5" CreationDate="2016-05-23T18:03:59.847" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2016-05-23T18:03:59.847" LastActivityDate="2016-05-23T18:03:59.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14978" PostTypeId="4" CreationDate="2016-05-23T18:03:59.847" Score="0" Body="Questions regarding the non-stellar objects gravitationally bound to a star or star system." OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-05-24T07:52:39.840" LastActivityDate="2016-05-24T07:52:39.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14979" PostTypeId="1" AcceptedAnswerId="14980" CreationDate="2016-05-23T19:34:52.257" Score="4" ViewCount="133" Body="&lt;p&gt;&lt;em&gt;Greetings!&lt;/em&gt; Based on &lt;a href=&quot;https://en.wikipedia.org/wiki/Epsilon_Eridani#Kinematics&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; and more precisely &lt;a href=&quot;https://arxiv.org/pdf/1004.1557v1.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;, it is said that, in about 31,500 years, the stars &lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR-5?-ref=VIZ574369b052ac&amp;amp;-out.add=.&amp;amp;-source=V/137D/XHIP&amp;amp;recno=16495&quot; rel=&quot;nofollow&quot;&gt;Epsilon Eridani&lt;/a&gt; and Luyten &lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR-5?-ref=VIZ57436a7d67ce&amp;amp;-out.add=.&amp;amp;-source=V/137D/XHIP&amp;amp;recno=36100&quot; rel=&quot;nofollow&quot;&gt;726-8AB&lt;/a&gt; will &quot;meet&quot; at a very close distance (less than 1 light year); I found that anecdote very interesting in the context of some SF writings I'm trying to do (not in English, quite obviously!) to bring some &quot;subtle consistency&quot; in it, but I was unable to either answer nor calculate with my very little knowledge of astronomy this simple question: to what distance will those stars be from the Solar System in 31,500 years?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much for your help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: oh, a friend of mine told me that my question could lack of useful information, such as the current distance of each of those stars from the Solar System, so here there are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Current distance from Epsilon Eridani to Solar System:&lt;/strong&gt; 10.5ly&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Current distance from Luyten 726-8 to Solar System:&lt;/strong&gt; 8.7ly&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Other info about Epsilon Eridani:&lt;/strong&gt; declination 9.46° south of the celestial equator (no idea if this can help) | Right Ascension 03h 32m 55.84496s | Radial velocity +15.5 ± 0.9 km/s&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Other info about Luyten 726-8:&lt;/strong&gt; declination –17° 57′ 01.8″ | Right Ascension      01h 39m 01.54s | Radial velocity +29.0 km/s&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[EDIT]&lt;/strong&gt; &#xA;As suggested in the comments, I added the Right Ascension and Radial Velocity for those two stars and I'm looking for the &quot;Velocity vectors&quot; (even if I have no idea what this means)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[EDIT 2]&lt;/strong&gt;&#xA;Added links to the Extended Hipparcos Compilation (XHIP) &lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR-5?-ref=VIZ574369b052ac&amp;amp;-out.add=.&amp;amp;-source=V/137D/XHIP&amp;amp;recno=16495&quot; rel=&quot;nofollow&quot;&gt;Epsilon Eridani&lt;/a&gt; and Luyten &lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR-5?-ref=VIZ57436a7d67ce&amp;amp;-out.add=.&amp;amp;-source=V/137D/XHIP&amp;amp;recno=36100&quot; rel=&quot;nofollow&quot;&gt;726-8AB&lt;/a&gt; pages with all sort of info I don't understand, but there are things such as &quot;Heliocentric Velocity&quot; on 3 axis, maybe it can help. &lt;em&gt;(Got the XHIP thing thanks to the following thread: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/2148/position-and-velocity-vectors-of-nearby-stars&quot;&gt;Position and velocity vectors of nearby stars?&lt;/a&gt;)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="11929" LastEditorUserId="11929" LastEditDate="2016-05-23T20:46:42.207" LastActivityDate="2016-05-23T21:30:36.353" Title="In 31.5kyr, Epsilon Eridani and Luyten 726-8 will be &lt; 1ly close; but how distant from the Solar system?" Tags="&lt;star&gt;&lt;movement&gt;&lt;kinematics&gt;&lt;future&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14980" PostTypeId="2" ParentId="14979" CreationDate="2016-05-23T21:30:36.353" Score="4" Body="&lt;p&gt;These stars are currently moving away from us, in the case of Epsilon Eridani, at 15.5 km/s. In 31500 years they will be further away than they are now, the distance can be calculated by applying Pythagoras' theorem to the distances and velocities you mention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that Eps Eri will be about 12.2 light years from Earth, and Lutyen 726-8 will be 12.6 light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They won't be particularly close together in the sky, I estimate about 3 degrees, and they will be even dimmer than they are now, though Eps Eri will still be a naked eye object.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-05-23T21:30:36.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14982" PostTypeId="1" CreationDate="2016-05-24T01:56:24.847" Score="2" ViewCount="457" Body="&lt;p&gt;I was trying to calculate the diameter of Jupiter from a picture I took of it. Here's the information I was able to get that I needed to calculate the diameter:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Focal Length of Telescope: 1.2 m&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Multiplier for Barlow Lens: 2x&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Camera CCD (Sensor) dimensions: 14.8 x 22.2 (mm)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I found the diameter to be 135 pixels in my image. Here's what I did:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;f_e &amp;amp;= \text{Focal Length $\cdot$ Barlow Multiplier}\\&#xA; &amp;amp;= 1.2 \cdot 2\\&#xA; &amp;amp;= 2.4&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, to calculate the field of view:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Field of View} = \frac{\text{Sensor Size}}{f_e} \cdot \frac{180}{\pi}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the sensor size is the length in millimetres of the size of the camera CCD (I used the horizontal axis, or the longer side, to get the 135 pixels, meaning that I'd use 22.2 mm), and the second fraction is there solely to convert radians into degrees. Substituting my values, I got:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\text{Field of View} &amp;amp;= \frac{0.0222}{2.4} \cdot \frac{180}{\pi}\\&#xA;&amp;amp;= 0.53^\circ &#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As my entire image is 4271.8 pixels wide, and I know the angular field of view, I used ratios to get the field of view of the 135-pixelled Jupiter to be $\theta = 0.017^\circ$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, using basic trigonometry, I was able to calculate the radius of jupiter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan \left(\frac{\theta}{2}\right) = \frac{\text{Radius of Jupiter}}{\text{Distance to Jupiter}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The photo was taken on March 22, 2016 - Wolfram Alpha gives 667.6 million as the distance to Jupiter on that day:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan \left(\frac{0.017^\circ}{2}\right) = \frac{\text{Radius of Jupiter}}{667.6 \cdot 10^6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\therefore \text{Radius of Jupiter} = 97\ 577\ \mathrm{km}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Googling the radius of Jupiter gives $69\ 911\ \mathrm{km}$ as the result. Is there any reason why my value is so significantly off? Anything wrong with my procedure, anything I'm missing? I was not able to to upload this image I took as the file size was too large, however, the image does not lack clarity and I am completely stumped as to why my result is off by such a significant amount. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help will be greatly appreciated, thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="9650" LastActivityDate="2016-05-25T02:05:43.557" Title="Calculating the Diameter of Jupiter through Image" Tags="&lt;photography&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14984" PostTypeId="2" ParentId="14982" CreationDate="2016-05-24T07:37:53.247" Score="1" Body="&lt;p&gt;A simpler approach would be to use a planetarium program to calculate the angular size of Jupiter at the date/time you want then use a bit of trig. to work out the angular size of each pixel of the camera&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tan(theta) = pixel size (in mm) over effective focal length in mm (2400mm)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry I don't know how to use the equation editor)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then divide one by the other and you have the diameter of Jupiter in pixels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: This assumes that the barlow is operating at 2x, as the barlow magnification factor is dependent on the distance of the camera sensor from the barlow's lens (and the focal length of the lens) this will probably not be accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if Jupiter is 38&quot; in diameter and your pixel size is 5.6 microns&#xA;1 pixel = about 0.48&quot; so&#xA;Jupiter will be about 79 pixels in diameter.&lt;/p&gt;&#xA;" OwnerUserId="11658" LastEditorUserId="11658" LastEditDate="2016-05-24T07:43:12.810" LastActivityDate="2016-05-24T07:43:12.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14986" PostTypeId="2" ParentId="14982" CreationDate="2016-05-24T07:59:12.920" Score="2" Body="&lt;p&gt;Your barlow is marked &quot;2x&quot; but I'll bet it isn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is your camera mounted on the barlow? More importantly, what distance is between the camera's focal plane and the barlow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An experiment to try (or at least to think about).  Get an eyepiece extension tube (or improvise something) and put it between the camera and barlow, increasing the spacing. You will have to refocus a bit and you will get a different (I guess bigger) image size, even though the barlow is still &quot;2x&quot;. To get an exact &quot;2x&quot; barlow you might need a particular spacing, and &lt;strong&gt;eyepieces are normally closer to the eyepiece than cameras&lt;/strong&gt; because of the extra bulk of the camera.  (I think this is the most likely cause of your discrepancy; your camera is currently mounted further back than an eyepiece would be so projection is making the image larger.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second experiment to try: without the barlow, photograph the moon and see how close you get to its expected size.  If it's close, then our assumptions about the barlow are in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third thing to try: are you absolutely sure about the distance from Earth to Jupiter? (It's &lt;em&gt;probably&lt;/em&gt; OK but I'm always suspicious about online calculators.)&lt;/p&gt;&#xA;" OwnerUserId="7491" LastEditorUserId="7491" LastEditDate="2016-05-24T09:37:10.677" LastActivityDate="2016-05-24T09:37:10.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14987" PostTypeId="2" ParentId="8167" CreationDate="2016-05-24T09:36:06.273" Score="-1" Body="&lt;p&gt;Earth diameter = 6,371 km&#xA;Moon diameter = 1,737 km&#xA;ratio 3.6 : 1&#xA;Moon seen from Earth has size X.&#xA;so&#xA;Earth seen from Moon has size 3.6X x 3.6X making it look roughly 13 times bigger on the sky than Moon looks on Earths sky.&#xA;I wonder how come they forgot about this...&lt;/p&gt;&#xA;" OwnerUserId="11936" LastActivityDate="2016-05-24T09:36:06.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14988" PostTypeId="1" AcceptedAnswerId="14989" CreationDate="2016-05-24T12:32:11.730" Score="13" ViewCount="1559" Body="&lt;p&gt;Neutron stars can have small atmospheres. However, they also have extremely strong gravitationally pulls. Shouldn't the all the gas molecules be drawn to the star's surface, and become solids under the immense pressure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe I'm thinking about this the wrong way, but I don't see how it could be possible.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="7982" LastEditDate="2021-02-28T11:27:52.453" LastActivityDate="2021-02-28T11:27:52.453" Title="How can neutron stars have gaseous atmospheres?" Tags="&lt;star&gt;&lt;gravity&gt;&lt;neutron-star&gt;&lt;stellar-atmospheres&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14989" PostTypeId="2" ParentId="14988" CreationDate="2016-05-24T13:29:53.467" Score="12" Body="&lt;p&gt;Gravity is only important insofar that it is capable of compressing the material to high densities. Whether that material is capable of solidifying depends on the competition between Coulombic potential energy and the thermal energy of the particles. The former increases with density, the latter increases with temperature. A dense plasma can still be a gas if it is hot enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rough formula for the exponential scale height of the atmosphere is&#xA;$$ h = \frac{kT}{\mu m_u g},$$&#xA;where $T$ is the temperature of the gas, $m_u$ is an atomic mass unit, $\mu$ is the number of atomic mass units per particle and $g$ is the surface gravity, with $g = GM/R^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a typical neutron star with $R=10$ km, $M= 1.4M_{\odot}$, we have $g=1.86\times 10^{12}$ m/s$^2$. The atmosphere could be a mixture of ionised helium ($\mu=4/3$) or perhaps iron ($\mu = 56/27$), so let's say $\mu=2$ for simplicity. The temperature at the surface of the neutron star will change with time; typically for a young pulsar, the surface temperature might be $10^{6}$ K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives $h = 2$ mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this not a &quot;solid&quot;? Because the thermal energy of the particles is larger than the coulombic binding energy in any solid lattice that the ions could make. That is not the case in the solid surface below the atmosphere because the density grows very rapidly (from $10^{6}$ kg/m$^{3}$ to more than $10^{10}$ kg/m$^{3}$ (where solidification takes place) only a few cm in, because the scale height is so small. Of course the temperature increases too, but not by more than a factor of about 100. After that, the density is high enough for electron degeneracy, and the material becomes approximately isothermal and at a small depth the &quot;freezing temperature&quot; falls below the isothermal temperature.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-05-24T19:30:45.247" LastActivityDate="2016-05-24T19:30:45.247" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="14991" PostTypeId="2" ParentId="8167" CreationDate="2016-05-24T15:50:30.787" Score="1" Body="&lt;p&gt;The Earth is clearly visible from the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nasa.gov/sites/default/files/styles/full_width_feature/public/images/296636main_1241_full_full.jpg?itok=ixcY9zFS&quot; rel=&quot;nofollow&quot;&gt;http://www.nasa.gov/sites/default/files/styles/full_width_feature/public/images/296636main_1241_full_full.jpg?itok=ixcY9zFS&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the linked image, Harrison Schmitt, the mission specialist on Apollo 17, stands next to the American flag, with the Earth in the background, the image was taken by mission commander, Gene Cernan. Schmitt claims to have taken the &quot;blue marble image&quot; during the journey to the moon.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-05-24T15:50:30.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14993" PostTypeId="1" AcceptedAnswerId="14995" CreationDate="2016-05-24T17:34:56.840" Score="2" ViewCount="118" Body="&lt;p&gt;I’ve got a question about white dwarfs and oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read in a book that a temperature of 100 million degrees is required to fuse Helium in the core of a red giant. The Helium fuses into Carbon by the triple-alpha process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It’s also written that, after a temperature of 350 million degrees is reached, the core of a red giant ceases to be degenerate. So, the core is able to expand and its temperature is controlled. The core won't reach a temperature of more than 350 million.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my teacher has said that a temperature of 600 million degrees is needed to form Oxygen from Carbon: Carbon fuses into Neon and by photodisintegration gives Oxygen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how can there be Oxygen in white dwarfs, if we don't reach a temperature higher than 350 million degrees? By what process can oxygen be formed in a red giant? &lt;/p&gt;&#xA;" OwnerUserId="11940" LastEditorUserId="2365" LastEditDate="2016-05-24T18:49:00.180" LastActivityDate="2016-05-24T22:10:39.700" Title="How can white dwarf form Oxygen ? (Temperature problem)" Tags="&lt;white-dwarf&gt;&lt;red-giant&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14994" PostTypeId="2" ParentId="14975" CreationDate="2016-05-24T19:43:49.413" Score="8" Body="&lt;p&gt;This is actually a very subtle question, much more so than the answers to the similar questions provided in the comments give it credit for.  When I was in graduate school at Ohio State I routinely asked this question to visiting dynamicists and invariably got different answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The very basic answer is that if you have two sufficiently strong resonances sufficiently close together, then the resonance will be unstable.  Otherwise, the resonance will be stable.  But what determines &quot;sufficiently strong&quot; and &quot;sufficiently close&quot; is where things get very complicated quickly.  A basic criterion is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Chirikov_criterion&quot; rel=&quot;noreferrer&quot;&gt;Chirikov criterion&lt;/a&gt;.  (The &lt;a href=&quot;http://www.scholarpedia.org/article/Chirikov_criterion&quot; rel=&quot;noreferrer&quot;&gt;Scholarpedia article&lt;/a&gt; is somewhat more detailed.)  However, the Chirikov criterion is not universally valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have overlapping resonances, then an object gets bounced back and forth between these two resonances chaotically.  These different resonances perturb the orbit in different ways, and eventually they will perturb the orbit into an unstable orbit, thus leading to depletion of the resonance.  If a resonance is &quot;distant&quot; from other resonances, then the resonance tends to keep objects locked in place, leading to an excess of objects in the resonance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the resonances in the asteroid belt are fairly close together, which leads to them being unstable.  The Kirkwood Gaps are the most prominent manifestation of these instabilities.  For example, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Alinda_family&quot; rel=&quot;noreferrer&quot;&gt;Alinda family&lt;/a&gt; of asteroids are in a 1:3 resonance with Jupiter, and are very close to a 4:1 resonance with the Earth.  This leads to instability, and hence very few asteroids in this family.  However, in the outer Solar System, the resonances are generally far apart, and so are mostly stable.  The &lt;a href=&quot;https://en.wikipedia.org/wiki/Plutino&quot; rel=&quot;noreferrer&quot;&gt;plutinos&lt;/a&gt; are one example of such a stable resonance, being in a 3:2 resonance with Neptune.&lt;/p&gt;&#xA;" OwnerUserId="10371" LastActivityDate="2016-05-24T19:43:49.413" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14995" PostTypeId="2" ParentId="14993" CreationDate="2016-05-24T19:43:56.247" Score="2" Body="&lt;p&gt;Some oxygen is produced during CNO cycle processing of hydrogen, starting with carbon nuclei. Oxygen is also produced by alpha capture onto carbon nuclei at temperatures well below 350 million K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of these occur in and around the cores of low mass stars before they become white dwarfs. Neon production is not required.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-05-24T22:10:39.700" LastActivityDate="2016-05-24T22:10:39.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14996" PostTypeId="1" CreationDate="2016-05-24T20:52:59.290" Score="7" ViewCount="488" Body="&lt;p&gt;&lt;a href=&quot;http://what-if.xkcd.com/150/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This XKCD what-if&lt;/a&gt; talks about rainbows on planets in a binary star system.   It points out that there are two types:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;&lt;strong&gt;circumbinary&lt;/strong&gt;&lt;/em&gt; planets, where the planet orbits far from and around both stars&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;&lt;strong&gt;[the other kind of planet]&lt;/strong&gt;&lt;/em&gt;, where the planet orbits one of the stars and the other star is distant.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gSX15.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gSX15.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h1&gt;What is the proper name for [the other kind of planet] ?&lt;/h1&gt;&#xA;" OwnerUserId="1106" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-05-25T17:44:24.660" Title="What's the name for [the other kind of planet] in a binary star system?" Tags="&lt;planet&gt;&lt;exoplanet&gt;&lt;binary-star&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14997" PostTypeId="2" ParentId="14996" CreationDate="2016-05-24T22:11:17.653" Score="8" Body="&lt;p&gt;I don't know about names for the planets, specifically, but the orbits are called &lt;a href=&quot;https://en.wikipedia.org/wiki/Binary_star#Planets&quot;&gt;S-type and P-type&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;S-type&lt;/strong&gt;: The planet orbits around one star, and the host star has a binary companion (i.e., &quot;the other kind&quot; in the XKCD comic)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;P-type&lt;/strong&gt;: The planet orbits around both stars of the binary&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10371" LastActivityDate="2016-05-24T22:11:17.653" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14998" PostTypeId="1" CreationDate="2016-05-25T06:59:49.123" Score="-1" ViewCount="120" Body="&lt;p&gt;So, telescopes are like time machines. If we get an image and data for analysis from a planet say 1000 light years away, we are essentially looking 1000 years back in time. Now, if we notice presence of liquid water on it and somehow notice there is life on it, how do we know it still exists. There are so many possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It existed 1000 years back, then asteroid came and destroyed it. Or, the evolution on that planet was 10K times faster than what we have and life on that planet already existed to a better planet somewhere else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if we send signals, we will have to wait for 2000 years to communicate assuming they send response instantaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, have the smarter beings on our planet already figured out a solution to this? I am sure they have thought of this problem already.&lt;/p&gt;&#xA;" OwnerUserId="11947" LastActivityDate="2016-05-25T19:17:26.970" Title="Communicating with distant planet with life possibility" Tags="&lt;telescope&gt;&lt;exoplanet&gt;" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="15001" PostTypeId="1" CreationDate="2016-05-25T11:27:32.400" Score="6" ViewCount="362" Body="&lt;p&gt;Super massive black holes are usually millions of times more massive than stellar mass black holes. I have read on SE and over internet that 3M☉ worth of energy is released in black hole mergers (not sure which type of merger). I have tow related questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the difference in energy released from a stellar mass black hole merger vs. a supermassive black hole merger?  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How much energy will be released in the super massive black hole merger when Milky way and Andromeda will merge in few billion years?  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10657" LastEditorUserId="10657" LastEditDate="2016-05-27T14:36:38.950" LastActivityDate="2016-05-28T13:56:09.840" Title="Difference in energy released in stellar mass black hole merger and supermassive black hole merger" Tags="&lt;black-hole&gt;&lt;milky-way&gt;&lt;supermassive-black-hole&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15002" PostTypeId="2" ParentId="14976" CreationDate="2016-05-25T14:14:29.743" Score="2" Body="&lt;p&gt;The short answer is &quot;not very, but we're getting better&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of particularly large Earth orbit intersecting bodies, kilometers across and bigger, they're fairly well known and tracked. We'd probably get months to years of warning. Technologically, we've probably got the capability of diverting such an object if we have enough warning, but it'd likely be the most complex human endeavor ever. Blowing it up is unlikely to help, we'd need a nice long shove from as far away as possible to divert it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bigger problem is smaller bodies in the tens to hundreds of metres. Some of these are known about and tracked, but plenty more aren't. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Chelyabinsk_meteor&quot; rel=&quot;nofollow&quot;&gt;Chelyabinsk meteor&lt;/a&gt; was about 20 metres across and, in terms of size/density/speed/impact angle wasn't that far off being a city killer. We never saw it coming. (Ironically, a very similar but unrelated 30m object was being tracked through a near-miss that same day.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd suggest that a city-killer impact with little or no warning is still quite possible. Something capable of taking out, say, a small country we'd probably have some warning of, but probably not enough. A mass-extinction event we'd probably have lots of warning for, but doing anything about it would be a very big challenge.&lt;/p&gt;&#xA;" OwnerUserId="11884" LastActivityDate="2016-05-25T14:14:29.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15003" PostTypeId="2" ParentId="14996" CreationDate="2016-05-25T17:44:24.660" Score="2" Body="&lt;p&gt;Astronomy currently only gives us the names circumbinary (left picture) and Trojan planet (unpictured, planet is around a Lagrange point), but from circumbinary we can extrapolate other names:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;circumunary - revolves around one star&lt;/li&gt;&#xA;&lt;li&gt;circumtrinary/circumternary - revolves around a trinary/ternary star system&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Technically, there is another term that actually has some use: circumsystem. This is used for objects that revolve around a multiple star system regardless of the number of stars within the system.&lt;/p&gt;&#xA;" OwnerUserId="6" LastActivityDate="2016-05-25T17:44:24.660" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15004" PostTypeId="2" ParentId="14998" CreationDate="2016-05-25T19:17:26.970" Score="2" Body="&lt;p&gt;If a planet is 1000 light years away, there is no way that we could be sure that there is life on it &quot;now&quot;. However there is a probabilistic argument: Life takes many millions of years to develop, and the chance of a disaster being so great as to sterilise a planet is very small: case in point, the Earth hasn't been sterilised since the last common ancestor of all living things existed, nearly 4 billion years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no way of knowing that life would continue to exist on a planet 1000 light years away during the many millennia that it would take for a probe to reach the distant planet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interstellar travel would take so long that we have to accept that no matter when we begin, we will not finish in our lifetime, and so we would do so for the benefit of our descendents. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its not clear what you mean by &quot;evolution is 10x faster&quot; since evolution doesn't have a clearly defined speed.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-05-25T19:17:26.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15005" PostTypeId="1" AcceptedAnswerId="15007" CreationDate="2016-05-25T23:54:55.340" Score="9" ViewCount="293" Body="&lt;p&gt;In this recent &lt;a href=&quot;http://www.bbc.com/news/science-environment-36381581&quot; rel=&quot;noreferrer&quot;&gt;BBC article&lt;/a&gt; I read the one-sentence paragraph:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Development on the mirrors - in particular, a very complex quaternary mirror - continues apace.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and became interested in the following questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;What is&lt;/em&gt; a quaternary mirror&lt;/li&gt;&#xA;&lt;li&gt;Why does the E-ELT need one&lt;/li&gt;&#xA;&lt;li&gt;What is it about this mirror that is &quot;very complex&quot;?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;note:&lt;/em&gt; the E-ELT is the European Extremely Large Telescope nearing final design, and to be installed in Chile's Atacama Desert (host to many other telescopes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A drawing of the E-ELT from from &lt;a href=&quot;https://en.wikipedia.org/wiki/European_Extremely_Large_Telescope&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; (note scale of humans):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KjoAL.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KjoAL.jpg&quot; alt=&quot;E-ELT&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An image from the BBC article:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GWAl2.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GWAl2.jpg&quot; alt=&quot;BBC image of E-ELT vs Big Ben&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2016-05-26T00:25:23.247" LastActivityDate="2016-05-26T03:42:22.317" Title="What is a quaternary mirror and why does the E-ELT need one?" Tags="&lt;telescope&gt;&lt;observatory&gt;&lt;e-elt&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15006" PostTypeId="1" AcceptedAnswerId="15012" CreationDate="2016-05-26T00:42:08.327" Score="3" ViewCount="304" Body="&lt;p&gt;In this recent &lt;a href=&quot;http://www.bbc.com/news/science-environment-36381581&quot; rel=&quot;nofollow noreferrer&quot;&gt;BBC article&lt;/a&gt; I read that the European Extremely Large Telescope or E-ELT is in final design and is planned to be on line by 2024, and that (of course) it will rely heavily on &lt;a href=&quot;https://en.wikipedia.org/wiki/Adaptive_optics#In_astronomy&quot; rel=&quot;nofollow noreferrer&quot;&gt;adaptive optics&lt;/a&gt; (AO) technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a &lt;a href=&quot;https://astronomy.stackexchange.com/a/14825/7982&quot;&gt;great answer&lt;/a&gt; and discussion related to my previous question, it seems that while most AO is actually implemented for infrared imaging, work is being done to push the technology to visible wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the E-ELT have AO available for visible light imaging in the beginning, will it be phased in later, or is there currently no plan for visible light AO?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;note: &lt;a href=&quot;https://en.wikipedia.org/wiki/Adaptive_optics#In_astronomy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Adaptive Optics&lt;/a&gt; (AO) techniques allow ground based observatories to dramatically improve resolution by actively compensating for the effects of &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_seeing&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomical Seeing&lt;/a&gt; through dynamically deformable optics. AO is also implemented in radio astronomy computationally applied when data from arrays is combined, as described in &lt;a href=&quot;https://astronomy.stackexchange.com/a/13967/7982&quot;&gt;this answer&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A drawing of the E-ELT from from &lt;a href=&quot;https://en.wikipedia.org/wiki/European_Extremely_Large_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; (note scale of humans):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KjoAL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KjoAL.jpg&quot; alt=&quot;E-ELT&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-05-27T13:53:12.677" Title="Will the E-ELT use Adaptive Optics at visible wavelengths?" Tags="&lt;angular-resolution&gt;&lt;adaptive-optics&gt;&lt;e-elt&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15007" PostTypeId="2" ParentId="15005" CreationDate="2016-05-26T03:42:22.317" Score="11" Body="&lt;p&gt;You may know that a standard Newtonian telescope has two mirrors, they are called the primary and secondary mirror.&lt;/p&gt;&#xA;&lt;p&gt;The E-ELT has five mirrors: The quaternary mirror is simply &amp;quot;mirror number four&amp;quot;, counting in the direction the light enters the scope.&lt;/p&gt;&#xA;&lt;p&gt;It's complex because that's where the adaptive optics sits:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The quaternary mirror has an approximate diameter of 2.4-m (2380x2340mm). It is a flat adaptive mirror, with up to 8000 actuators, thereby allowing the surface to be readjusted at very high time frequencies. [...]&lt;/p&gt;&#xA;&lt;p&gt;This mirror will correct in real time for high order wavefront errors (e.g. atmosphere, wind shake, low spatial frequency telescope errors) and small amplitude residual tip-tilt corrections.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Source: The European Southern Observatory's &lt;a href=&quot;http://www.eso.org/sci/facilities/eelt/telescope/mirrors/&quot; rel=&quot;noreferrer&quot;&gt;page on the E-ELT optics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9682" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-05-26T03:42:22.317" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15008" PostTypeId="1" CreationDate="2016-05-26T10:10:33.660" Score="1" ViewCount="1135" Body="&lt;p&gt;When looking for a formula to convert polar ecliptic geocentric coordinates of an object to equatorial coordinates I find various sources that give these formulae (like &lt;a href=&quot;https://en.wikipedia.org/wiki/Celestial_coordinate_system#Equatorial_.E2.86.90.E2.86.92_ecliptic&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Declination $δ = \arcsin(\cos ε \times \sin β + \sin ε \times \cos β \times \sin λ)$&lt;br&gt;&#xA;Right ascension $α = \arctan((\cos ε \times \sin λ - \sin ε \times \tan β) / \cos λ)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where&lt;br&gt;&#xA;β = ecliptic geocentric latitude&lt;br&gt;&#xA;λ = ecliptic geocentric longitude&lt;br&gt;&#xA;ε = obliquity of the ecliptic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when applying these formulae I get results like in the following list where β = 0° and ε = 23.4°:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; λ       δ        α&#xA;  0   0.0000   0.0000&#xA; 45  16.3095  42.5443&#xA; 90  23.4000  90.0000&#xA;135  16.3095 -42.5443&#xA;180   0.0000  -0.0000&#xA;225 -16.3095  42.5443&#xA;270 -23.4000  90.0000&#xA;315 -16.3095 -42.5443&#xA;360  -0.0000  -0.0000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The values for declination seem good, but right ascension values seem to lack some sort of adjustment to the quadrant of the full circle (just a guess). But nowhere did I find any mentioning of this. Can you help? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="11838" LastEditorUserId="9301" LastEditDate="2016-10-19T14:15:38.693" LastActivityDate="2017-03-12T20:06:25.277" Title="Converting ecliptical to equatorial coordinates" Tags="&lt;fundamental-astronomy&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15010" PostTypeId="2" ParentId="660" CreationDate="2016-05-27T06:49:48.560" Score="4" Body="&lt;p&gt;My career path:3-year Bachelor's degree in &quot;Physics with Astrophysics&quot;; PhD in X-ray astronomy; 5-years as a postdoctoral research assistant (two separate posts); got a lectureship at a UK university doing teaching and research in Physics and Astrophysics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is reasonably typical. These days, the content of the first degree is not so important - Physics, Astrophysics, Applied Maths all would be ok. &quot;Astronomy&quot; would put you at a disadvantage, since the implication is a non-physical, observational approach; but you would have to look at the course content.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A masters degree or 4-year first degree is usually necessary to get onto PhD programmes in the best places (this has changed since my day). Doing your PhD quickly and writing several publications is usually necessary to proceed any further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The normal next step is to get a postdoctoral position; preferably somewhere other than your PhD institute. Then after 2-3 years of producing more research papers (2-3 per year), you could try for personal research fellowships. If you can get one of these, or perhaps a second/third postdoc position, and your research is going well and is productive, then there is a few year window in which to get into a tenured or tenure-track position. Getting some teaching experience at this stage is probably important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For someone on a &quot;normal&quot; career path, it would be unusual to get a University position before the age of 30 (i.e. 8-9 years after your first degree). The large majority of people with a PhD in Astrophysics do not end up doing that for a living.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-05-27T06:49:48.560" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15012" PostTypeId="2" ParentId="15006" CreationDate="2016-05-27T13:53:12.677" Score="2" Body="&lt;p&gt;From looking at the &lt;a href=&quot;http://www.eso.org/public/usa/teles-instr/e-elt/&quot; rel=&quot;nofollow noreferrer&quot;&gt;E-ELT website&lt;/a&gt;, it appears that at first light, the AO will only work for near-IR. Specifically, the instrument that can use AO is the &lt;a href=&quot;http://www.eso.org/public/usa/teles-instr/e-elt/e-elt-instr/micado/&quot; rel=&quot;nofollow noreferrer&quot;&gt;MICADO instrument&lt;/a&gt;. The description page for this instrument states&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;MICADO, or the Multi-Adaptive Optics Imaging Camera for Deep&#xA;Observations, is one of the first-light instruments for the European&#xA;Extremely Large Telescope (E-ELT) instrument and takes the Adaptive&#xA;Optics technique to the next level. It will be the first dedicated&#xA;imaging camera for the E-ELT and works with the multi-conjugate&#xA;adaptive optics module, MAORY.&lt;/p&gt;&#xA;&lt;p&gt;MICADO will equip the E-ELT with a first light capability for&#xA;diffraction limited imaging at near-infrared wavelengths.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.eso.org/public/usa/teles-instr/e-elt/e-elt-instr/maory/&quot; rel=&quot;nofollow noreferrer&quot;&gt;MAORY instrument&lt;/a&gt; will be the specific AO instrument which, at first light will operate at&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;wavelengths from 0.8–2.4µm&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So in short, the answer appears to be that the E-ELT will only have AO capabilities for near-IR at first light.&lt;/p&gt;&#xA;&lt;p&gt;As to whether or not there are plans for visible AO capabilities, I couldn't find anything specific to that question. Given that first light for this project is 8 years away at best, I think that plans for future upgrades are tentative at best, and likely not easily accessible to those not on the project. I will say that while the MAORY instrument appears to be the intended AO system at first light, there is listed a second AO instrument called ATLAS that will eventually make its way onto the telescope. While I can find no specific details of this particular instrument, its possible that this will provide for visible AO.&lt;/p&gt;&#xA;&lt;p&gt;On a tangentially related note, this telescope will also have active optics that will help to improve their imaging capabilities.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-05-27T13:53:12.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15013" PostTypeId="1" AcceptedAnswerId="15014" CreationDate="2016-05-27T14:55:38.240" Score="0" ViewCount="334" Body="&lt;p&gt;I try to get the azimuth of an object from its equatorial coordinates using this formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a = \arctan2(\sin(θ - α), \sin φ * \cos(θ - α) - \cos φ * \tan δ)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where&lt;br&gt;&#xA;φ = geographic latitude of the observer (here: 0°)&lt;br&gt;&#xA;θ = sidereal time (here: 0°)&lt;br&gt;&#xA;δ = declination&lt;br&gt;&#xA;α = right ascension&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made two JavaScript functions to implement this calculation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  obliq = deg2rad(23.44);  // obliquity of ecliptic&#xA;  lat2  = deg2rad(0);  // observer's latitude&#xA;  lmst  = deg2rad(0);  // siderial time&#xA;&#xA;  function equatorial(lat, lon) {&#xA;    // returns equatorial from ecliptic coordinates&#xA;    dec = Math.asin( Math.cos(obliq) * Math.sin(lat) + Math.sin(obliq) *&#xA;          Math.cos(lat) * Math.sin(lon));&#xA;    ra  = Math.atan2(Math.cos(obliq) * Math.sin(lon) - Math.sin(obliq) * Math.tan(lat),&#xA;          Math.cos(lon));&#xA;    ra += 2 * Math.PI * (ra &amp;lt; 0);&#xA;    return [dec, ra];&#xA;  }&#xA;&#xA;  function horizontal(lat, lon) {&#xA;    // returns horizontal from ecliptic coordinates&#xA;    coords = equatorial(lat, lon);&#xA;    dec = coords[0];  // δ&#xA;    ra  = coords[1];  // α&#xA;    alt = Math.asin(Math.sin(lat2) * Math.sin(dec) + Math.cos(lat2) *&#xA;          Math.cos(dec) * Math.cos(lmst - ra));&#xA;    azm = Math.atan2(Math.sin(lmst - ra), Math.sin(lat2) * Math.cos(lmst - ra) -&#xA;          Math.cos(lat2) * Math.tan(dec));&#xA;    azm += 2 * Math.PI * (azm &amp;lt; 0);&#xA;    return [alt, azm];&#xA;  }&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I cannot see any error, but I get strange results for azimuth (a) as can be seen in this table (the other values seem correct):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   λ       δ         α         h         a&#xA;    0    0.0000    0.0000   90.0000    0.0000&#xA;   15    5.9094   13.8115   75.0000  246.5600&#xA;   30   11.4723   27.9104   60.0000  246.5600&#xA;   45   16.3366   42.5357   45.0000  246.5600&#xA;   60   20.1510   57.8186   30.0000  246.5600&#xA;   75   22.5962   73.7196   15.0000  246.5600&#xA;   90   23.4400   90.0000    0.0000  246.5600&#xA;  105   22.5962  106.2804  -15.0000  246.5600&#xA;  120   20.1510  122.1814  -30.0000  246.5600&#xA;  135   16.3366  137.4643  -45.0000  246.5600&#xA;  150   11.4723  152.0896  -60.0000  246.5600&#xA;  165    5.9094  166.1885  -75.0000  246.5600&#xA;  180    0.0000  180.0000  -90.0000  248.3079&#xA;  195   -5.9094  193.8115  -75.0000   66.5600&#xA;  210  -11.4723  207.9104  -60.0000   66.5600&#xA;  225  -16.3366  222.5357  -45.0000   66.5600&#xA;  240  -20.1510  237.8186  -30.0000   66.5600&#xA;  255  -22.5962  253.7196  -15.0000   66.5600&#xA;  270  -23.4400  270.0000   -0.0000   66.5600&#xA;  285  -22.5962  286.2804   15.0000   66.5600&#xA;  300  -20.1510  302.1814   30.0000   66.5600&#xA;  315  -16.3366  317.4643   45.0000   66.5600&#xA;  330  -11.4723  332.0896   60.0000   66.5600&#xA;  345   -5.9094  346.1885   75.0000   66.5600&#xA;  360   -0.0000  360.0000   90.0000   68.3079&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Does anyone see the error? Thank you.&lt;/p&gt;&#xA;" OwnerUserId="11838" LastActivityDate="2016-05-27T17:48:11.073" Title="Calculating azimuth from equatorial coordinates" Tags="&lt;fundamental-astronomy&gt;&lt;coordinate&gt;&lt;azimuth&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15014" PostTypeId="2" ParentId="15013" CreationDate="2016-05-27T17:48:11.073" Score="1" Body="&lt;p&gt;&lt;code&gt;lat2&lt;/code&gt; = 0 and &lt;code&gt;lmst&lt;/code&gt; = 0 place the vernal equinox at the zenith, so most points on the ecliptic &lt;em&gt;should&lt;/em&gt; have azimuth either &lt;code&gt;obliq&lt;/code&gt; degrees north of east or &lt;code&gt;obliq&lt;/code&gt; degrees south of west. Try other values for &lt;code&gt;lat2&lt;/code&gt; or &lt;code&gt;lmst&lt;/code&gt; and see if the resulting azimuths make sense.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-05-27T17:48:11.073" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15015" PostTypeId="1" CreationDate="2016-05-27T20:47:06.587" Score="2" ViewCount="298" Body="&lt;p&gt;Sorry for such a newbie question. I searched and read multiples sources about supernova but I'm not completely sure if what is called &quot;supernova&quot; is the explosion, the resulting celestial body or the event as whole (explosion plus resulting celestial body).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be considered as incorrect if for some reason the explosion of a star is called as &quot;supernova&quot;?&lt;/p&gt;&#xA;" OwnerUserId="11967" LastActivityDate="2016-05-28T17:47:16.460" Title="&quot;Supernova&quot; is the explosion or the resulting celestial body? Is it incorrect to call the explosion &quot;supernova&quot;?" Tags="&lt;supernova&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15016" PostTypeId="2" ParentId="15015" CreationDate="2016-05-27T21:03:28.393" Score="5" Body="&lt;p&gt;&lt;em&gt;A supernova is the explosion&lt;/em&gt;, visible on Earth as a apparently new star that appears to brighten over a few days, and then fade. There have been no visible supernovae in our galaxy for several hundred years, and we are well overdue another one, though they are quite often visible in other galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the supernova there will be a &lt;em&gt;remnant&lt;/em&gt;, a small nebula of hot gas that expands, eventually taking the form of a ring. In the centre of the remnant there may be &lt;em&gt;neutron star&lt;/em&gt; or a &lt;em&gt;black hole&lt;/em&gt;. A spinning neutron star may be a &lt;em&gt;pulsar&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supernovae that occur when a old, giant star explodes at the end of its life can produce a neutron star or black hole. But some supernovae form when a white dwarf accumulates matter from another star onto its surface, and then explodes, in which case the star is completely destroyed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term &quot;nova&quot; was previously used for all stellar explosions. However the term &quot;nova&quot; is nowadays limited to smaller explosions, caused by the fusion of hydrogen on white dwarf (but no sufficient build up of hydrogen to cause the white dwarf to be destroyed)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-05-27T21:03:28.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15017" PostTypeId="2" ParentId="15001" CreationDate="2016-05-28T10:55:09.100" Score="-2" Body="&lt;p&gt;Super massive and stellar mass black holes are different in mass and size.. Since we cannot describe nor we can be certain in either of their mass or size .. It's completely impossible (for the time being) to answer whether or not how much difference there will be regarding their energy release and the amount it releases..Yes,the energy is high  3 solar mass .. but the energy cannot be determined regarding there are various factors which leads to the determination of Mass and energy.&lt;/p&gt;&#xA;" OwnerUserId="11970" LastActivityDate="2016-05-28T10:55:09.100" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15018" PostTypeId="2" ParentId="15001" CreationDate="2016-05-28T13:22:23.937" Score="2" Body="&lt;p&gt;Too long for a comment, but it's an incomplete answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravity waves detected were from a stellar mass &lt;a href=&quot;https://en.wikipedia.org/wiki/Binary_black_hole&quot; rel=&quot;nofollow&quot;&gt;binary black hole&lt;/a&gt; merger, sometimes abbreviated bbh for binary black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two black holes are thought to be about 36 and 29 solar masses with a final combined mass of about 62.  So, roughly 5% of the black hole mass turned into into gravitational wave energy.   &lt;a href=&quot;http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.116.061102&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and from Wikipedia article (which can be taken with a grain of salt)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As the orbiting black holes give off these waves, the orbit decays,&#xA;  and the orbital period decreases. This stage is called binary black&#xA;  hole inspiral. The black holes will merge once they are close enough.&#xA;  Once merged, the single hole settles down to a stable form, via a&#xA;  stage called ringdown, where any distortion in the shape is dissipated&#xA;  as more gravitational waves.&lt;a href=&quot;http://www.iflscience.com/space/two-supermassive-black-holes-merge&quot; rel=&quot;nofollow&quot;&gt;3&lt;/a&gt; In the final fraction of a second the&#xA;  black holes can reach extremely high velocity, and the gravitational&#xA;  wave amplitude reaches its peak.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The existence of stellar-mass binary black holes (and gravitational&#xA;  waves themselves) were finally confirmed when LIGO detected GW150914&#xA;  (detected September 2015, announced February 2016), a distinctive&#xA;  gravitational wave signature of two merging stellar-mass black holes&#xA;  of around 30 solar masses each, occurring about 1.3 billion light&#xA;  years away. In its final moments of spiraling inward and merging,&#xA;  GW150914 released around 3 solar masses as gravitational energy,&#xA;  peaking at a rate of 3.6×1049 watts — more than the combined power of&#xA;  all light radiated by all the stars in the observable universe put&#xA;  together.&lt;a href=&quot;http://www.iflscience.com/space/two-supermassive-black-holes-merge&quot; rel=&quot;nofollow&quot;&gt;3&lt;/a&gt;[4][5] Supermassive binary black hole candidates have been&#xA;  found but as yet, not categorically proven&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Supermassive bbh are thought to have been observed, see &lt;a href=&quot;http://www.iflscience.com/space/two-supermassive-black-holes-merge&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and we may see one in our lifetime, not sure how often they happen.  But beyond that, I have no way of knowing how much more energetic a supermassive BBH merger would be other than to say, I would think, a whole lot, as it would have to scale upwards by some formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the case of Andromeda and the Milky Way, we're talking Andromeda's 100 million solar mass black hole and the Milky way's 4-6 million solar masses.  Maybe someone here can estimate, but if the energy output is even 5%-10% of the smaller of the two, that's still a few hundred thousand solar masses of energy, an absolutely crazy energy output.  Even if it's far smaller than a few hundred thousand solar masses, it's still enormous energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also not sure if it's a dangerous form of energy output as it basically squashes things back and forth maybe a couple times before returning them to normal.  I have no idea what the safe distance is from that kind of merger and gravity wave creation.    Interesting question though.  Fun to think about.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-05-28T13:56:09.840" LastActivityDate="2016-05-28T13:56:09.840" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15019" PostTypeId="2" ParentId="15015" CreationDate="2016-05-28T17:47:16.460" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it incorrect to call the explosion “supernova”?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes and no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Better said, the explosion is the very first part of a supernova. While the explosion lasts for but a few seconds to a few hundreds of seconds, a supernova can last for hundreds of days. What we see visibly as a supernova are the after effects of that explosion. The explosion can create lots and lots of stuff moving at very high velocities, and lots and lots of highly radioactive nuclei.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the star had outgassed material prior to the explosion, the highly kinetic material produced by the explosion runs into that previously outgassed material and makes it glow. This takes some time to cool down. The radioactive material produced during the short course of the explosion proper takes time to decay to stable elements. This radioactive decay eventually produces gamma rays, some of which is absorbed by the nearby material, heating it, thereby eventually producing thermal radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a type Ia supernova produces a large amount of nickel-56. This decays to cobalt-56 with a half-life of about 6 days, which in decays to iron-56 with a half-life of about 77 days. The heating that results from these decays is what we see, and because it is so predictable, this is makes type Ia supernovae a fantastic standard candle.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2016-05-28T17:47:16.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15020" PostTypeId="2" ParentId="12890" CreationDate="2016-05-28T19:27:25.870" Score="2" Body="&lt;p&gt;For some mysterious symmetry reasons, we can describe particles with a theory that involves the representations of Lie groups (which we call the standard model). Early universe cosmology involves this mathematics to some extent, since it's thought that there should be a unification of forces from the high energies after the Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also some cosmological implications to various quantum gravity theories. I've heard a lot about something called the AdS/CFT correspondence, which involves connections between algebra and geometry. &lt;/p&gt;&#xA;" OwnerUserId="608" LastActivityDate="2016-05-28T19:27:25.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15021" PostTypeId="2" ParentId="14817" CreationDate="2016-05-28T22:38:55.213" Score="1" Body="&lt;p&gt;For decades the variable star S Doradus, in the Large Magellanic Cloud, was considered to be the most intrinsically luminous star known.  But even though the Large Magellanic Cloud is the second or third closest external galaxy, S Doradus is still too far away from Earth to be seen with the naked eye from Earth.  O fall the stars that always or sometimes are intrinsically brighter than S Doradus, only the variable Eta Carinae, in our own Milky Way Galaxy, is close enough to Earth that it is sometimes visible with the naked eye. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Omega Centauri, or NGC 5139, was cataloged as a star by Ptolemy in his &lt;em&gt;Almagest&lt;/em&gt; about 150 AD and by Bayer in the &lt;em&gt;Uranometria&lt;/em&gt; of 1603 - Bayer named it Omega Centauri.  It is at a distance of about 15,800 light years from Earth.  There is a theory that Omega Centauri is a remnant of a dwarf galaxy that as captured by our galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus Omega Centauri is an object that was listed and named as a star and may be the remnant of a former galaxy that has been captured by our own galaxy.  Thus so far it seems to &lt;strong&gt;ALMOST&lt;/strong&gt; fit the definition of an extra-galactic star visible from Earth with the naked (Human) eye. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course Edmund Halley noticed that Omega Centauri  was not a star as early as 1677. Today it is classified as a globular Star cluster in our own galaxy (and possibly the remnant of the core of a dwarf galaxy).  The light that makes Omega Centauri visible to the naked eye on Earth comes from the light emitted by hundreds of thousands or millions of stars, not one single star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though a globular star cluster has the light of tens of thousands to millions of stars, and our Milky Way Galaxy has over a hundred globular clusters, only a few of them are visible from Earth with the naked eye (47 Tucanae mentioned by RichS is another, and it was also mistaken for a single star at first).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a single star that was as far away as Omega Centauri 15,800 light years away, or 47 Tucanae 17,000 light year away, would have to shine as bright as tens or hundreds of thousands, and maybe even millions, of ordinary stars to appear just barely visible to the naked eye like those two clusters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Canis Major Dwarf Galaxy is believed to be 25,000 light years from earth and the nearest external galaxy, if it really is a galaxy.  The Saggitarius Dwarf Elliptical Galaxy is about 70,000 light years from Earth and the Large Magellanic Cloud is about 163,000 light years from Earth.  A star in the Large Magellanic Cloud, about ten times as far away as Omega Centauri or 47 Tucanae, would have to be as bright as millions to hundreds of millions of ordinary stars to be seen by the naked eye from Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose it is theoretically possible that one of the only about 6,000 stars visible with the naked eye from Earth might actually be in one of the 2 or 3 closest galaxies or floating alone outside of the disc of our galaxy.  But it would have to be a supergiant or hypergiant star, and astronomers would have to somehow not notice oddities in its spectrum that would point to it being so rarely luminous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would estimate that the odds against that would be &quot;astronomical&quot;.&lt;/p&gt;&#xA;" OwnerUserId="11975" LastActivityDate="2016-05-28T22:38:55.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15022" PostTypeId="2" ParentId="14762" CreationDate="2016-05-29T02:37:49.267" Score="4" Body="&lt;p&gt;From before the dawn of history people naturally assumed that the sky was a solid dome above the flat earth.  The dome was assumed to rotate once a day, so the stars were assumed to be lights attached to the dome.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone who assumed that the sky dome was opaque had to assume that the sun and the wandering planets were nearer than the sky dome.  I think that it was usually assumed the sky dome was opaque and colored blue, thus making the sky seem blue when it reflected the light of the sun during the day, thus making the sun closer than the sky dome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those who kept records of astronomical observations soon noticed that the moon occulted or passed in front of stars, planets, and the sun (solar eclipses), and thus was closer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everyone assumed that the sun and the moon were tiny, until traveler's reports showed that they had the same apparent diameters every place that as visited, and so were about equally far away from every place that was visited.  Thus as the size of the known world grew larger and larger, the minimum possible size and distance of the moon and the sun grew larger and larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea of a spherical earth was proposed and gradually accepted by Greek philosophers between the sixth and third centuries BC.  Thus the earth became known as Earth, a sphere instead of a flat disc, and the dome of the sky became a hollow spherical shell around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Hellenistic times the diameter of the Earth was calculated with reasonable accuracy, as well as the distance to the moon and thus its diameter.  So it became known that the moon was about a quarter the diameter of the Earth and over sixty Earth radii distant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of solar eclipses, it had been known since prehistoric times that the moon was closer than the sun, and keepers of astronomical records of events when the moon occulted (passed in front of) stars and planets knew that the moon was closer than planets and stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone who assumed geocentrism, that the Earth was the enter of the universe, naturally assumed that every object revolved around the Earth, including the moon, and that the moon was thus the closest satellite of the Earth.  A few ancient Greek philosphers supported the Heliocentric theory, that the Earth and the planets revolved around the sun.  Some of them could have believed that the moon also revolved around the sun, but as far as I know all heliocentric believers also had the moon revolve around the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the last few centuries BC there were many educated persons who believed that the Earth, the moon, and the sun were giant balls of rock (and thus somewhat similar objects), that the sun was a giant burning rock, and that the stars and planets might also be giant burning rocks far, far away from the Earth and appearing as dots as seen from Earth.  Since everybody already since prehistoric times believed that the moon revolved around the Earth, by the last few centuries BC educated people in Western civilization believed that the moon was what we now call a natural satellite of the Earth, though some of them believed and some did not that the sun and planets were also natural satellites of the Earth.&lt;/p&gt;&#xA;" OwnerUserId="11977" LastActivityDate="2016-05-29T02:37:49.267" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15023" PostTypeId="1" CreationDate="2016-05-29T02:38:33.053" Score="3" ViewCount="643" Body="&lt;p&gt;This question is about the LIGO gravitational wave signals' frequency. The signals start from about 35 Hz to peak at about 250 Hz, giving evidence of gravitational waves. The question is about the exact frequencies the waves' detection started to the exact peak of the wave. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems the LIGO graphs show the peak is a little &#xA;above  &lt;a href=&quot;http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.116.061102#fulltext&quot; rel=&quot;nofollow&quot;&gt;250 Hz&lt;/a&gt;. Are there tables published by LIGO that put the start and peak frequencies as 35 to 250 Hz? The graph in the paper by &lt;a href=&quot;http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.116.061102#fulltext&quot; rel=&quot;nofollow&quot;&gt;Abbott, et al&lt;/a&gt; shows the start and peak (the start is around 36-37  Hz, the peak seems to be at around 256 Hz) - are there exact numbers (tables) for the start and peak, that one can reference? &lt;/p&gt;&#xA;" OwnerUserId="11102" LastEditorUserId="37191" LastEditDate="2021-02-12T21:28:38.873" LastActivityDate="2021-02-12T21:28:38.873" Title="LIGO gravitational wave chirp signal frequency" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;gravitational-waves&gt;&lt;ligo&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15024" PostTypeId="2" ParentId="14544" CreationDate="2016-05-29T03:16:14.057" Score="0" Body="&lt;p&gt;I saw a similar &quot;ufo&quot; several decades ago, before there were Iridium satellites.  Possibly it was another type of satellite with large reflective panels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking it over, I later decided that if an airplane reflected the sunlight directly toward me at one moment of a turn, it might make a similar flash.&lt;/p&gt;&#xA;" OwnerUserId="11978" LastActivityDate="2016-05-29T03:16:14.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15026" PostTypeId="1" CreationDate="2016-05-29T11:10:07.167" Score="3" ViewCount="317" Body="&lt;p&gt;The waves detected by LIGO were basically subatomic distortions of space-time. Could there be much greater distortions closer to a source. Would they become perceptible to a human in any way and what would it do to us?&lt;/p&gt;&#xA;" OwnerUserId="11982" LastActivityDate="2016-05-29T19:28:00.533" Title="How large could gravitational waves get and what effect could they have on us?" Tags="&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15027" PostTypeId="1" AcceptedAnswerId="15062" CreationDate="2016-05-29T15:46:20.003" Score="1" ViewCount="331" Body="&lt;p&gt;I'm developing a planetarium software and I need the R.A. and DEC that represents every constellation area in the sky. I think I only need R.A. and DEC for each vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where can I find it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm searching on Internet and I can't find it. Maybe because I'm not doing the right search.&lt;/p&gt;&#xA;" OwnerUserId="7566" LastActivityDate="2016-06-01T06:19:31.953" Title="R.A. and DEC. for Constellation area vertices" Tags="&lt;constellations&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15028" PostTypeId="1" CreationDate="2016-05-29T16:01:55.050" Score="1" ViewCount="478" Body="&lt;p&gt;I continue working on a planetarium software and I want to draw the lines that the draw the constellation's figure:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6RtSE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6RtSE.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Is there a catalogue or something similar with the stars which are the vertex of the figure?&lt;/p&gt;&#xA;" OwnerUserId="7566" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-06-27T04:41:31.237" Title="Which stars do I have to use to join with a line to draw constellation?" Tags="&lt;constellations&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="0" ClosedDate="2018-06-27T13:30:37.577" ContentLicense="CC BY-SA 3.0" />
  <row Id="15029" PostTypeId="1" AcceptedAnswerId="15040" CreationDate="2016-05-29T19:20:53.427" Score="7" ViewCount="503" Body="&lt;p&gt;I am new to astronomy and I found that the telescopic dataset can be stored in the &lt;a href=&quot;https://en.wikipedia.org/wiki/FITS&quot; rel=&quot;nofollow&quot;&gt;FITS&lt;/a&gt; format instead of JPEG. Why do we have a special image format just for scientific images?&lt;/p&gt;&#xA;" OwnerUserId="11988" LastEditorUserId="1056" LastEditDate="2016-05-30T20:54:26.243" LastActivityDate="2016-05-30T20:54:26.243" Title="Why do we use FITS format for scientific images especially in astronomy? How is it different from formats such as JPEG, PNG etc?" Tags="&lt;telescope&gt;&lt;astrophysics&gt;&lt;fundamental-astronomy&gt;&lt;data-analysis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15030" PostTypeId="2" ParentId="15026" CreationDate="2016-05-29T19:28:00.533" Score="6" Body="&lt;p&gt;Wave strain scales as $1/r$; the waves that were detected came from a source about a Gpc away and had a strain of $10^{-20}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a strain that would affect things at the 1% level, you would need to be $10^{18}$ times closer, or at about 30,000 km from the merging black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tidal force due to the black holes is roughly $GMmh/4r^3$, where $h$ is your height, $m$ your mass and $M\sim 60M_{\odot}$ the combined mass of the black holes. Assuming 1.8m height and 80kg mass, the tidal force stretching you is around 50N. So I think you would notice this more than the gravitational wave, and even more so if you were to get any closer...&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-05-29T19:28:00.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15031" PostTypeId="1" AcceptedAnswerId="15050" CreationDate="2016-05-29T19:39:46.450" Score="0" ViewCount="104" Body="&lt;p&gt;For cold dark matter, density profiles are well known and easy to find information about - eg. NFW, Burkert, Einasto, and others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for some reason I couldn't find explicit expressions for the density profiles for Hot and Warm dark matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to know what are $\rho_{_{HDM}}(r)$,$\rho_{_{WDM}}(r)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in the differences between cold/hot/warm dark matter and why do we use Cold dark matter ($\Lambda$-CMD  model) and not Hot or Cold dark matter. I need to back my claims with formal quantitative analysis and not just a qualitative explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can recommend about other aspects which differentiate the models, or a good review paper on the subject, I would welcome it.&lt;/p&gt;&#xA;" OwnerUserId="11644" LastEditorUserId="2270" LastEditDate="2016-05-31T19:01:41.257" LastActivityDate="2016-06-22T12:06:38.377" Title="Warm and Hot dark matter density profiles" Tags="&lt;dark-matter&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15032" PostTypeId="2" ParentId="15029" CreationDate="2016-05-29T22:40:03.383" Score="3" Body="&lt;p&gt;FITS has a lot of nice features which formats like PNG don't have, like astronomical coordinates (where in the sky does the image correspond to?) and storing multiple exposures. Most traditional image formats also aren't as flexible. For example, FITS supports floating point data storage, where pixels on an image can have precise values like 1.2324241, whereas PNG has 256 possible values in each of the four channels (rgba). &lt;/p&gt;&#xA;" OwnerUserId="608" LastActivityDate="2016-05-29T22:40:03.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15034" PostTypeId="1" AcceptedAnswerId="15035" CreationDate="2016-05-30T10:04:10.740" Score="19" ViewCount="2185" Body="&lt;p&gt;Astronomers recently claim that there is evidence for a 9th planet. As far as I understood it, this is mainly based on coinciding orbital parameters of a few Kuiper belt objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could the 9th planet be a virtual planet? I mean this in the sense that we see the gravitational influence of a multi-body system. Could the 9th planet simply be the barycenter of the Kuiper belt and its apparent orbit some form of precession?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from the question, if this was a possible or likely scenario, how could we distinguish a &quot;real&quot; 9th planet from a virtual one, except by the obvious answer if we can observe it directly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Forgive me, if this is a stupid or trivial question or if this was answered before. I am not a professional astronomer, yet I have a physics/engineering background. I did a quick internet search to answer this question but I simply may have used the wrong terms or sources.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4073" LastActivityDate="2016-09-01T00:12:26.080" Title="Could the 9th planet be virtual?" Tags="&lt;planet&gt;&lt;solar-system&gt;&lt;9th-planet&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15035" PostTypeId="2" ParentId="15034" CreationDate="2016-05-30T10:56:25.987" Score="17" Body="&lt;p&gt;The ninth planet can absolutely be a &quot;virtual&quot;, in the way you describe it, meaning observed data indicating the gravitational influence by an object is not actually caused by such an object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple visualisation of this is the case of a two-body system, where we observe two objects orbiting a common barycentre. From the observed data, one can get the impression that there is an object in the barycentre, pulling both of your observed objects against it.&lt;sup&gt;1&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a barycentric way to explain the observations gets less likely as you add more bodies. Be aware that the current indications of &quot;planet 9&quot; is based on a mere correlation of some not-so-accurate data, from a small sample size, heavily biased by observation constraints.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;how could we distinguish a &quot;real&quot; 9th planet from a virtual one,&#xA;  except by the obvious answer if we can observe it directly?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is only by its gravitational influence alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a rule of thumb, incomplete and inaccurate data &lt;strong&gt;always&lt;/strong&gt; imply the risk of causing virtual objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should also be noted too that observations in the Kuiper belt generally can not yield how a orbit changes, due to the extremely long revolution times. observing &quot;Interactions&quot; are actually mostly about tracing an orbit back from its current state vectors, to find probable interaction events in the past. That is literary to &lt;em&gt;ask&lt;/em&gt; for data artefacts and virtual objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;Before anyone objects about how easy it is to falsify that: &lt;em&gt;Yes&lt;/em&gt; it is easy to determine that there can not actually be an object there, based purely on observing the two bodies. &lt;strong&gt;But&lt;/strong&gt;, that requires 1. accurate data about the motion of the two objects, and 2. their mass. Our current data about the Kuiper belt fails to give good enough data on both points.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastActivityDate="2016-05-30T10:56:25.987" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15036" PostTypeId="1" AcceptedAnswerId="19557" CreationDate="2016-05-30T11:27:45.693" Score="0" ViewCount="54" Body="&lt;p&gt;The supermassive black hole in the Milky Way is covered by dust as seen from here. Is it common that SMBH's are covered in dust in their galactic disks? Would that dust absorb and disperse a jet from an active galactic nucleus, so that the atmospheres of Earth like planets would be protected?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AGN jets build bubbles perpendicular to the galactic plane, but the jet could be oriented in any direction, right? The bubbles are shaped by general stellar winds from the disk, redirecting the jet streamed material towards the galactic poles, if I get that rigt.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-12-30T09:43:17.867" Title="Does the dust around SMBH's protect habitable planets from jets?" Tags="&lt;supermassive-black-hole&gt;&lt;habitable-zone&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15037" PostTypeId="1" AcceptedAnswerId="15039" CreationDate="2016-05-30T14:47:38.357" Score="3" ViewCount="221" Body="&lt;p&gt;Exoplanets are planets that are located outside our solar system - whether that be orbiting a star or drifting past one. Now, the closest star to us is Alpha Centauri which is just over four light-years away. So, how can astronomers detect gases in the atmospheres of planets over this distance with any degree of accuracy (or at all)?&lt;/p&gt;&#xA;" OwnerUserId="4160" LastEditorUserId="2270" LastEditDate="2016-05-31T19:00:45.043" LastActivityDate="2016-05-31T19:00:45.043" Title="How do astronomers detect gases that are in the atmosphere of exoplanets?" Tags="&lt;exoplanet&gt;&lt;planetary-atmosphere&gt;&lt;gas&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2020-12-04T18:39:58.273" ContentLicense="CC BY-SA 3.0" />
  <row Id="15039" PostTypeId="2" ParentId="15037" CreationDate="2016-05-30T16:00:25.173" Score="6" Body="&lt;p&gt;It is only possible to detect gases in transiting Exoplanets. The spectrum of the star is taken when the Exoplanet is not in transit and again when in transit. The differences between the spectrums is due to a small amount of light from the parent star passing through the upper atmosphere of the Exoplanet and being absorbed by gases in the atmosphere.&#xA;The effect is VERY small and difficult to detect.&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-05-30T16:00:25.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15040" PostTypeId="2" ParentId="15029" CreationDate="2016-05-30T17:21:10.757" Score="11" Body="&lt;p&gt;File formats tend to be industry/field-specific, with the format, tools, and expectations of the field coevolving to become more dependent on each other over time. JPEG co-evolved with amateur digital photography, PNG co-evolved with the web. Likewise, FITS co-evolved with astronomical data processing, and so is naturally more suited for that purpose than formats that had entirely different goals and communities involved in their development.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Pedantic nit: JPEG is the name of the compression method and the group that designed it. The file format is technically &quot;JFIF&quot;, but everybody colloquially calls it JPEG.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an image format, FITS has a number of desirable qualities that are lacking in JFIF and PNG, that are crucial for scientific data:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Storage of more bits per pixel (CCDs can records 12, 14, or more bits), and also floating point values.&lt;/li&gt;&#xA;&lt;li&gt;Storage of arbitrary number of data channels (scientific data may have many, or other, frequency bands than the RGB that JPEG and PNG are limited to).&lt;/li&gt;&#xA;&lt;li&gt;No lossy compression as is typical (though not strictly required) for JPEG.&lt;/li&gt;&#xA;&lt;li&gt;Higher resolution (JPEG/JFIF, for example, is limited to 65,535 pixels in each direction), and FITS is also capable of storing 3D data volumes.&lt;/li&gt;&#xA;&lt;li&gt;Support for unlimited metadata in the header, for example the sky coordinates, information about the telescope, etc. JPEG and PNG don't have the fields and aren't set up to record arbitrary metadata.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Also, the use of FITS for astronomy predates the existence of both JPEG/JFIF and PNG -- FITS was standardized in 1981, JPEG in 1992, PNG in 1996. So even if those formats were suitable (which they are not), by the time they were invented there was already widespread use and existence of astronomical image processing tools that were geared to FITS files (and growing archives of astronomical data in FITS format), so it would have been a major effort to switch formats that would never be undertaken unless a new format not only met but exceeded all the prior requirements of the field.&lt;/p&gt;&#xA;" OwnerUserId="366" LastActivityDate="2016-05-30T17:21:10.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15042" PostTypeId="1" AcceptedAnswerId="15045" CreationDate="2016-05-30T20:28:52.910" Score="9" ViewCount="1308" Body="&lt;p&gt;&lt;em&gt;I know that if you exceed orbital velocity, you will never fall-back to the planet.  My question is not about orbits.  It's about brute-force propulsion to achieve altitude.  I'm using an intentionally slow velocity to help illustrate my point.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine I have a rocket with very efficient fuel storage.  My rocket can store enough energy to accelerate to 100kph shortly after leaving the ground, and continue to maintain that speed (100kph) for a very long period of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My rocket just goes straight up.  It doesn't try to enter an orbit.  As it leaves the atmosphere, it can throttle-back because there's no air resistance.  As it continues to gain altitude in inter-planetary space, it can throttle-back even more because Earth's gravitational influence diminishes with distance.  It just maintains enough throttle to continue moving away from Earth at 100kph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some point, Earth's gravitational influence would be moot, as other bodies (Jupiter, Sun), would gain relative influence.  Eventually, far outside the solar system, even the Sun's influence would be insignificant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My rocket never achieved escape velocity, but it sure did escape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that my fuel supply could last long enough, and I wasn't concerned about travel time, could this method allow my rocket to &quot;leave&quot; without achieving escape velocity?&lt;/p&gt;&#xA;" OwnerUserId="11993" LastActivityDate="2019-08-02T00:45:15.600" Title="Can I leave a planet without achieving escape velocity?" Tags="&lt;gravity&gt;&lt;escape-velocity&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15043" PostTypeId="1" AcceptedAnswerId="15049" CreationDate="2016-05-30T21:06:58.563" Score="5" ViewCount="1280" Body="&lt;p&gt;After a Sun-sized protostar forms, its core will become denser over time due to radiation. The core eventually gets dense and hot enough for hydrogen fusion to take place. In the late phases of the star's life, the core will continue fusing hydrogen until it runs out and outer layers begin fusion hydrogen, while dumping helium into the core. The core will get massive enough that it will begin to contract, and it will eventually become electron degenerate due to the high densities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as the core became denser and denser, it reached temperatures at which fusion could take place, but it did &lt;em&gt;not&lt;/em&gt; become dense enough for the matter to become electron degenerate, right? Only later in its life, when the star's hydrogen ran out, would it become dense enough for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So then why would gas giant cores be electron degenerate, but not become hot enough for nuclear fusion? Shouldn't they start fusing before they can become electron degenerate, as stars do? Am I misunderstanding this concept entirely, or is there more to it than what I described?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2016-05-31T16:38:13.103" Title="Why is the core of a gas giant supported by electron degeneracy pressure instead of nuclear fusion?" Tags="&lt;star&gt;&lt;gas-giants&gt;&lt;density&gt;&lt;protostar&gt;&lt;degenerate-matter&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15044" PostTypeId="2" ParentId="15042" CreationDate="2016-05-30T21:10:44.257" Score="12" Body="&lt;p&gt;The basic concept here, that you instead of relying on having a large enough velocity that the Earth can not pull you back in time just have a low and constant velocity, and keeps thrusting to counteract gravity instead, is not flawed by default, given your assumptions&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Assuming that my fuel supply could last long enough, and I wasn't&#xA;  concerned about travel time, could this method allow my rocket to&#xA;  &quot;leave&quot; without achieving escape velocity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt; escape velocity decreases by distance, so even though you are travelling at only 100kph, that is eventually going to be above escape velocity, as the escape velocity goes towards zero when the distance goes towards infinity. You can not really escape a gravitational field, so the only meaningful metric is if you have enough kinetic energy to not be pulled back. That is not possible by your constraint of not having escape velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a sense though, you can escape the Earth's gravitational field, using the common approximation of &lt;a href=&quot;https://en.wikipedia.org/wiki/Sphere_of_influence_%28astrodynamics%29&quot;&gt;sphere of influence&lt;/a&gt;. It defines a region for Earth's gravitational influence, and ignores it outside of it. The border of the SOI can be reached without ever going at escape velocity.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastActivityDate="2016-05-30T21:10:44.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15045" PostTypeId="2" ParentId="15042" CreationDate="2016-05-30T21:50:36.170" Score="10" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Escape_velocity&quot; rel=&quot;noreferrer&quot;&gt;Escape velocity&lt;/a&gt; is how fast you must go in order to keep moving away indefinitely &lt;em&gt;without&lt;/em&gt; additional thrust. If the Earth were the only major body in the system, and you moved away from it at 100 km/h for 1180 years, you'd be 6.9 AU away. Since Earth escape velocity at that distance is only 100 km/h, you could then stop the engine and coast to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we add the Sun to this oversimplified system, you could escape it by maintaining 100 km/h until you were 36 light-years away, which would take 390 million years. In the real universe, of course, the Earth's and Sun's neighbors make their spheres of influence much smaller.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2016-05-31T19:57:29.103" LastActivityDate="2016-05-31T19:57:29.103" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15046" PostTypeId="2" ParentId="15027" CreationDate="2016-05-30T22:27:18.330" Score="0" Body="&lt;p&gt;They are defined on the coordinate grid of 1875; precession has shifted and rotated them since then, and a map for the present epoch must account for that. In addition to the vertices, you may need to interpolate a few points along the edges.&#xA;&lt;a href=&quot;http://cdsarc.u-strasbg.fr/viz-bin/Cat?VI/49&quot; rel=&quot;nofollow&quot;&gt;Davenhall and Leggett 1989&lt;/a&gt; provide boundary coordinates for epochs 1875 and 2000.&#xA;For determining which constellation contains a given point, &lt;a href=&quot;http://cdsarc.u-strasbg.fr/viz-bin/Cat?VI/42&quot; rel=&quot;nofollow&quot;&gt;Roman 1987&lt;/a&gt; provides regions and lookup code.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-05-30T22:27:18.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15047" PostTypeId="2" ParentId="15043" CreationDate="2016-05-30T22:59:16.513" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So then why would gas giant cores be electron degenerate, but not become hot enough for nuclear fusion?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Degeneracy isn't an on/off switch. It's a quantum mechanical aspect of pressure that is always present, just as is thermal pressure. A substance is highly degenerate if degeneracy pressure completely dominates over thermal pressure, non-degenerate if degeneracy pressure is negligible compared to thermal pressure.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Shouldn't they start fusing before they can become electron degenerate, as stars do?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While gas giants are too small for gravitational collapse to heat the core to a temperature sufficient to result in fusion, they're not too small for the gravitational collapse to make degeneracy pressure a very significant aspect of pressure. The density at the center of Jupiter is about half that at the center of the Sun, but the temperature is only about 1/600 of that at the center of the Sun. This means that degeneracy pressure dominates at the center of Jupiter but that thermal pressure dominates at the center of the Sun.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2016-05-30T22:59:16.513" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15049" PostTypeId="2" ParentId="15043" CreationDate="2016-05-31T08:06:58.980" Score="6" Body="&lt;p&gt;The test to see whether degeneracy pressure is going to be significant is to compare $kT$ with the Fermi energy $E_F$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Fermi energy is the energy level up to which all energy states would be occupied in a completely degenerate fermion gas. It is given by (for non-relativistic conditions)&#xA;$$ E_F = \frac{h^2}{2m}\left(\frac{3}{8\pi}\right)^{2/3} n^{2/3},$$&#xA;where $m$ and $n$ are the mass and number density of the fermions (in this case, electrons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $E_F \gg kT$, then the gas can be considered completely degenerate and the electrons exert a non-relativistic degeneracy pressure. If $E_F \sim 10 kT$ the electrons are partially degenerate.&#xA;In &lt;em&gt;both&lt;/em&gt; these circumstances, the pressure exerted by the electrons is much higher than they would have in a perfect gas, because a large fraction of electrons occupy high energy (and momentum) states. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Degeneracy can be achieved either by having a low temperature or a high number density of fermions. In a white dwarf, electrons are completely degenerate because the number density of electrons is extremely high. In a gas giant, the electron density is nowhere near as high, but then the temperature is a lot lower and so they reach a state of partial degeneracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure of a partially degenerate gas is higher than that of a perfect gas at the same density and temperature. More importantly, the pressure is only very weakly dependent on temperature. As a gas giant radiates away its potential energy, it is able to do so whilst only contracting very slightly and the core does not become hot enough to initiate nuclear fusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that degeneracy pressure and thermal pressure are not two different things. Both arise merely as a consequence of particles having a distribution of momenta and obeying (in this case) Fermi-Dirac statistics to determine how they occupy the possible energy states. The expression $P = nkT$ is simply a convenient approximation for fermions that holds only when $E_F \ll kT$ and the quantum nature of the particles is not apparent.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-05-31T16:38:13.103" LastActivityDate="2016-05-31T16:38:13.103" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15050" PostTypeId="2" ParentId="15031" CreationDate="2016-05-31T15:03:41.133" Score="-1" Body="&lt;p&gt;I think you're going to have a difficult time finding this. The reason being, nobody has really bothered studying non-CDM density profiles because it was pretty clear from the onset that CDM was the (most promising) answer. Why spend time further analyzing a model that looks wrong? I will say though, that there are issues with CDM models (e.g., where are all the &quot;satellite&quot; halos that CDM predicts), and because of that, some people have turned to WDM or a Cold-Warm dark matter model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do we know these non-CDM models are wrong, you may ask. The answer for that comes from N-body simulations. There have been a number of continually impressive simulations over the decades that model the formation of large structure in our universe (see &lt;a href=&quot;https://arxiv.org/ftp/arxiv/papers/1210/1210.0544.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; for example). These simulations have included various forms of dark matter in an attempt to discover how different types of dark matter affect the formation of structure and content in our universe. The resounding answer seems to always be that hot dark matter just produces an incompatible universe from what we actually observe in ours. There's no point in thoroughly modeling a hot dark matter density profile when it is clear that the HDM universe is drastically different from our own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest that if you want to mathematically compare the various dark matter models, that you look at the power spectra from them. A good place to start might be &lt;a href=&quot;http://arxiv.org/pdf/astro-ph/9605198v1.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-05-31T15:03:41.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15051" PostTypeId="1" CreationDate="2016-05-31T15:06:36.750" Score="5" ViewCount="3388" Body="&lt;p&gt;Given that moons commonly orbit planets, why do we never encounter a comet orbiting a planet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would happen (exactly, in detail) if a large one did settle into a stable orbit around the Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry... just making the question a bit clearer!  The accepted answer will describe where all the water ends up.&lt;/p&gt;&#xA;" OwnerUserId="12000" LastEditorUserId="12000" LastEditDate="2016-05-31T17:04:16.693" LastActivityDate="2016-06-01T15:40:48.103" Title="Can a comet orbit a planet?" Tags="&lt;comets&gt;&lt;water&gt;" AnswerCount="4" CommentCount="10" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15052" PostTypeId="2" ParentId="15051" CreationDate="2016-05-31T15:30:49.690" Score="10" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Given that moons commonly orbit planets, why do we never encounter a comet orbiting a planet?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By definition, it would no longer be a comet, but rather a moon (or more properly a satellite). Comets are icy bodies that orbit the Sun, satellites are any body which orbits a larger body than itself, other than the Sun. There are many examples of asteroids and comets that have been captured and become satellites, although it is not easy to prove with 100% certainty that a satellite is indeed captured and did not form in place. Some familiar (potential) examples might be Phobos and Deimos, around Mars, or the various moons around Pluto such as Nix, Hydra, Kerberos, and Stix.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What would happen (exactly, in detail) if a large one did settle into a stable orbit around the Earth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have to take exception to this question by pointing out that it is impossible for a comet to &quot;settle into a stable orbit&quot; around the Earth. Even basic physics will tell you conservation of energy and momentum disallow capturing comets by them &quot;settling&quot;. In order for one body to capture another, you need a third body in the system to conserve angular momentum. What are the chances that two comets come by Earth at the same time and with exactly the correct orbital properties such that they can interact and cause one of them to be captured? I'm not going to do the math but I can guarantee the chances are diminishingly small. The moon could play the role of the third body, but again, you need precise orbital characteristics which seems unlikely to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we disregard that point and just say that somehow a comet has been put into orbit around the Earth, be it a nearly miraculous capture or else put there by human intervention or by some other effect. What would happen to it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to that depends on where its orbit is. It seems unlikely to me that a comet could survive for long in some orbits. If it was very close to the Earth, it would eventually fall onto the planet through friction with our atmosphere. If it was farther out, you'd have to consider how the Earth-Moon system affected its dynamics. Possibly there could be some stable resonant orbit between the Earth, Moon, and comet. It is hard to say what the long term results would be as an analytical solution to the three body problem is impossible. The only way to answer this is through a simulation.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2016-06-01T15:40:48.103" LastActivityDate="2016-06-01T15:40:48.103" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="15053" PostTypeId="2" ParentId="15051" CreationDate="2016-05-31T16:17:56.897" Score="15" Body="&lt;p&gt;Its very unlikely for a comet to become a satellite of an inner solar system planet. Much less likely than it is for an asteroid. Most asteroids are on fairly circular orbits, and so the relative velocity between asteroids and planets is quite low. In comparison comets have very elliptical orbits, and their relative velocities to the planets are much larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an asteroid to be captured it must lose momentum. This is possible, though rare. For example, a binary asteroid can be captured if it is separated by tidal forces. For an comet with much more momentum, the chance of being captured is much much lower. Asteroids are captured by the Earth moon system, but not into stable orbits, they don't stay long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it did occur, the comet would still be active, with a coma of gas, which would be visible just like a very nearby comet. It wouldn't be particularly bright, since the surface brightness of a comet doesn't depend on distance from the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over time the comet would run out of volatiles and become more or less indistinguishable from a captured asteroid. If it were in the Earth's orbit it probably wouldn't last that long, as there are not many orbits that are stable in the long term around the Earth, due to perturbations from the moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dust and gas, including water vapour, will initally remain in orbit, forming a faint ring. It will, over time, be disrupted, and either end up in the atmosphere, on the moon, or ejected from the system. A comet doesn't contain enough water to make a difference to the Earth's ecosystem.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-05-31T17:12:00.510" LastActivityDate="2016-05-31T17:12:00.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15054" PostTypeId="2" ParentId="8841" CreationDate="2016-05-31T18:12:57.190" Score="1" Body="&lt;p&gt;In a published scientific paper, the significance of any error bars should be explained. If this has not been clearly stated then peer review is not doing it's job. However, in some fields there may be unstated conventions which slip through without explanation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my own field (star formation) plots which shows averages over a large data set would show means with error bars representing one standard deviation, while plots which show single measured values might have error bars representing calculated uncertainties on the measurement, for example due to detection limits or instrumental uncertainties.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2016-05-31T18:12:57.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15055" PostTypeId="2" ParentId="15023" CreationDate="2016-05-31T18:20:01.617" Score="3" Body="&lt;p&gt;A fact sheet published by the LIGO collaboration at the time of the announcement of GW150914 (that's the official name of the first detection) gives the peak frequency as &quot;~250 Hz&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Abbott et al. paper also quotes a value of 35 to 250 Hz in the abstract (I haven't checked for more precise measurements later in the paper, I'm sure the OP or other interested parties can do that independently from the link.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would say that a value of 250 can be quoted as accurate and referenced to Abbott et al. 2016.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2016-05-31T18:20:01.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15056" PostTypeId="1" AcceptedAnswerId="15068" CreationDate="2016-05-31T19:52:03.183" Score="1" ViewCount="132" Body="&lt;p&gt;On the TRAPPIST telescope, the filter wheels are described to have a filter with the following description:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;NIR Luminance I+z' (&gt;700nm)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After a bit of googling, I wasn't able to find what the specifics of this filter are (its FWHM and mean wavelength).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go about finding these qualities of this filter?&lt;/p&gt;&#xA;&#xA;&#xA;" OwnerUserId="12004" LastActivityDate="2016-06-01T09:53:08.523" Title="What does the filter name I+z' mean" Tags="&lt;light&gt;&lt;observational-astronomy&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15057" PostTypeId="2" ParentId="15051" CreationDate="2016-05-31T20:01:55.627" Score="4" Body="&lt;p&gt;Yes.  See &lt;a href=&quot;http://adsabs.harvard.edu/abs/1993Rise...74..224V&quot; rel=&quot;nofollow&quot;&gt;Split comet Shoemaker-Levy 9 became a satellite of Jupiter&lt;/a&gt; &lt;em&gt;Ríse hvezd&lt;/em&gt;, Vol. 74, p. 224-225&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;a href=&quot;https://books.google.com/books?id=6aQRBwAAQBAJ&amp;amp;pg=PA1456&amp;amp;lpg=PA1456&amp;amp;dq=Split%20comet%20Shoemaker-Levy%209%20became%20a%20satellite%20of%20Jupiter&amp;amp;source=bl&amp;amp;ots=9Peu_X29Cj&amp;amp;sig=EXfEBUzDU6gp1Ji4orDjjbF2BTY&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwjY29O1jIXNAhVGVyYKHTRdANoQ6AEIKzAC#v=onepage&amp;amp;q=Split%20comet%20Shoemaker-Levy%209%20became%20a%20satellite%20of%20Jupiter&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;The Century of Space Science&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;comet Shoemaker-Levy 9 which was probably captured into a Jupiter-centric orbit and thus, temporarily, became a satellite of Jupiter before splitting into at least 21 piece and crashing into Jupiter's atmosphere&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And according to &lt;a href=&quot;http://adsabs.harvard.edu/full/1996AstL...22..771P&quot; rel=&quot;nofollow&quot;&gt;Disintegration of fragments Q1 and Q2 of comet Shoemaker-Levy 9 during the fall to Jupiter&lt;/a&gt; &lt;em&gt;Astronomy Letters&lt;/em&gt;, Volume 22, November 1996, pp.771-779:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The comet was captured by Jupiter and became its satellite in 1929 ± 9 according to Chodas and Yeomans (1995) and in 1959 according to the calculations Chemetenko and Medvedev (1994). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11382" LastEditorUserId="11382" LastEditDate="2016-06-01T11:43:55.573" LastActivityDate="2016-06-01T11:43:55.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15059" PostTypeId="2" ParentId="15051" CreationDate="2016-06-01T00:35:26.980" Score="2" Body="&lt;p&gt;&lt;strong&gt;It is absolutely possible for a planet to 'capture' a comet&lt;/strong&gt;, Mars has already captured two asteroids and the only difference between the asteroids and comets are what they are made of. Plus, there are theories that believe that &lt;strong&gt;Saturn's rings were formed by a captured comet&lt;/strong&gt; that strayed too close to the gas giant and thus, was ripped apart into millions of pieces of rock and ice (Assuming theory is accurate). Note that this is much more likely to happen to the outer planets due to their &lt;strong&gt;immense gravitational influence&lt;/strong&gt; and &lt;strong&gt;their distance from the sun&lt;/strong&gt;. Distance matters because comets generally go faster when they get nearer to the sun which makes it &lt;strong&gt;harder for a planet to capture comets&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although, technically it wouldn't be a comet anymore but a satellite.&lt;/p&gt;&#xA;" OwnerUserId="4160" LastActivityDate="2016-06-01T00:35:26.980" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15060" PostTypeId="1" AcceptedAnswerId="15083" CreationDate="2016-06-01T00:45:22.017" Score="1" ViewCount="112" Body="&lt;p&gt;I know that astronauts are constantly being protected from radiation from the sun and the cosmos. They are normally very safe from the harmful effects of radiation but what if their protection wasn't effective at all? What would be the effects on the human body if it is exposed to radiation? &lt;/p&gt;&#xA;" OwnerUserId="4160" LastActivityDate="2016-06-01T18:31:35.003" Title="What would happen if an astronaut orbiting Earth was exposed to radiation?" Tags="&lt;orbit&gt;&lt;earth&gt;&lt;radiation&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2016-06-21T21:47:26.090" ContentLicense="CC BY-SA 3.0" />
  <row Id="15061" PostTypeId="1" CreationDate="2016-06-01T04:21:02.890" Score="0" ViewCount="158" Body="&lt;p&gt;I know a single star is almost surely more massive than a single brown dwarf, but consider a case: a very small star is orbiting a very massive brown dwarf which has a very massive moon, such that the total mass of that brown dwarf and its moon is larger than that of the star. Not counting capturing, is this type of system possible (formed naturally)?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="7491" LastEditDate="2016-06-01T14:14:07.463" LastActivityDate="2016-06-01T23:37:38.737" Title="Is it possible to have a star orbiting around a brown dwarf?" Tags="&lt;star&gt;&lt;orbit&gt;&lt;brown-dwarf&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15062" PostTypeId="2" ParentId="15027" CreationDate="2016-06-01T06:19:31.953" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/users/2365/james-kilfiger&quot;&gt;James Kilfiger&lt;/a&gt; has suggested me to expand my comment as answer, and here it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Searching I have found the data in the international astronomical union website. In &lt;a href=&quot;http://iau.org/public/themes/constellations&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://iau.org/public/themes/constellations&lt;/a&gt; you can find a TXT for each constellation with vertex's R.A. and DEC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of that page there is an explanation about the fields on that text files. Their coordinates are J2000.&lt;/p&gt;&#xA;" OwnerUserId="7566" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-06-01T06:19:31.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15063" PostTypeId="1" AcceptedAnswerId="15065" CreationDate="2016-06-01T07:56:12.593" Score="8" ViewCount="8648" Body="&lt;p&gt;I have some 16 x 50 binoculars that my parents bought me years ago. Recently I have tried to do some basic lunar and planetary observations with them but I am seriously struggling with shaking. This was most noticeable tonight while attempting to observe the ISS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I accept that keeping the image of a moving target stable is going to be next to impossible, I would like to know how I could reduce my shaking while observing the moon and other objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is anyone able to provide some simple techniques to reduce shaking?&lt;/p&gt;&#xA;" OwnerUserId="11994" LastActivityDate="2019-01-25T08:46:49.617" Title="How do I prevent or reduce shake when observing with binoculars?" Tags="&lt;amateur-observing&gt;&lt;binoculars&gt;" AnswerCount="7" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15064" PostTypeId="2" ParentId="15061" CreationDate="2016-06-01T07:59:51.690" Score="1" Body="&lt;p&gt;&lt;strong&gt;Yes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a brown dwarf count as a star in this case, the solution is as easy as a smaller brown dwarf orbiting a larger one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, it is still possible if you have two brown dwarfs orbiting close to each other, both just too small to be red dwarfs, together out-massing an orbiting red dwarf, just large enough to be counted as a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I have no example of such a system, the Alpha Centauri system has to closely orbiting stars, with a more distant star orbiting, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Luhman_16&quot; rel=&quot;nofollow&quot;&gt;Luhman 16&lt;/a&gt; is a system with two brown dwarfs orbiting each other.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastActivityDate="2016-06-01T07:59:51.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15065" PostTypeId="2" ParentId="15063" CreationDate="2016-06-01T08:07:28.643" Score="18" Body="&lt;p&gt;Don't hold the binoculars in your hands. Humans are made of meat. wobbly wobbly meat. There's apparently devices that let you mount binoculars on tripods - (this &lt;a href=&quot;https://www.google.com.sg/?client=firefox-b-ab#q=binoculars%20tripod%20mount&quot;&gt;google search would be a start&lt;/a&gt;). Those and a tripod would probably be helpful in decreasing shake. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose it would affect mobility a &lt;em&gt;little&lt;/em&gt; but that's a tradeoff. &lt;/p&gt;&#xA;" OwnerUserId="12011" LastActivityDate="2016-06-01T08:07:28.643" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15066" PostTypeId="1" AcceptedAnswerId="15084" CreationDate="2016-06-01T08:43:00.757" Score="5" ViewCount="759" Body="&lt;p&gt;I know that normal planets don't get any larger than Jupiter (or 2 Jupiter radii if hot), as adding more gas just increases density, not radius, until you reach the point of being a brown dwarf star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That compressability, I suppose, is a property of the overall average material that forms star systems, as beyond a critical size it accretes the ambiant gas as well as dust in the cloud.  That is, it's mostly hydrogen and a quarter helium with a bit of other stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about a body made of other stuff, such as rock or metal? Without worrying about planetary formation processes, just that atoms are piled together and feel self gravity. Perhaps it needs to be grown slowly so it has time to cool before adding more. Otherwise, no special techniques: just what would happen if material of suitable composition was heaped together?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that normal ideas of minerals would not exist under that kind of pressure, even if carefully allowed to cool. But can elements other than H and He still compress their volume in the same way, or would a rocky world be able to reach sizes of &lt;em&gt;millions&lt;/em&gt; of miles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about more exotic cases, like the &quot;puffy&quot; planet I heard of that has the density of styrofoam?&lt;/p&gt;&#xA;" OwnerUserId="7142" LastEditorUserId="7142" LastEditDate="2016-06-02T00:34:46.970" LastActivityDate="2016-06-02T17:48:27.617" Title="How large (that is, radius) could a planet be?" Tags="&lt;planetary-formation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15067" PostTypeId="2" ParentId="15034" CreationDate="2016-06-01T08:51:26.750" Score="4" Body="&lt;p&gt;That's an interesting way to put it.  Indeed, since bodies spend most of their time at the slow end of an eccentric orbit, the &lt;strong&gt;extended lobes will attract each other&lt;/strong&gt;, torquing the orbits into a birdsnest shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.seti.org/weeky-lecture/bizarre-orbits-minor-planets-beyond-neptune&quot; rel=&quot;nofollow&quot;&gt;this presentation&lt;/a&gt; by Dr. Madigan.  I'm really disappointed that the more scientifically literate press (like SciAm) doesn't cover this idea, but hypes the large distant bodies idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(See slide at 15:58 in the video. But, it's the explaination up to that point that describes the mechanism.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, the particular clustering of orbit shapes and orientations can be explained through self-interaction of the long-term averages of the bodies' orbits: the apahelion acts as a &quot;virtual object&quot;, as you put it, and attract each other.&lt;/p&gt;&#xA;" OwnerUserId="7142" LastEditorUserId="7142" LastEditDate="2016-09-01T00:12:26.080" LastActivityDate="2016-09-01T00:12:26.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15068" PostTypeId="2" ParentId="15056" CreationDate="2016-06-01T09:53:08.523" Score="3" Body="&lt;p&gt;They must be referring to two different filters:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the (Bessel) I filter which has a central wavelength of $\lambda \simeq 800\,\mathrm{nm}$  and a width of $\Delta\lambda\simeq 150\,\mathrm{nm}$, and&lt;/li&gt;&#xA;&lt;li&gt;the (Sloan) z' filter, which has $\lambda,\Delta\lambda \simeq 970,255\,\mathrm{nm}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hence their transmission is in the near infrared (&quot;NIR&quot;), and they're both above 700 Å.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact transmission curves vary a little depending on the producer, so if you want to calculate exact magnitudes, for instance, you need to find the transmission curves for the telescope in question. From the &lt;a href=&quot;http://www.ati.ulg.ac.be/TRAPPIST/Trappist_main/Equipment.html&quot; rel=&quot;nofollow&quot;&gt;TRAPPIST equipment website&lt;/a&gt;, it seems they use filters from &lt;a href=&quot;http://www.astrodon.com&quot; rel=&quot;nofollow&quot;&gt;Astrodon&lt;/a&gt;. Astrodon don't seem to provide filter curves for these exact filter, but if you're okay with a good approximation, you can find the curves here:&#xA;&lt;a href=&quot;http://svo2.cab.inta-csic.es/theory/fps/index.php?id=NOT/ALFOSC.Bes_I&amp;amp;&amp;amp;mode=browse&amp;amp;gname=NOT&amp;amp;gname2=ALFOSC#filter&quot; rel=&quot;nofollow&quot;&gt;Bessel I&lt;/a&gt; and&#xA;&lt;a href=&quot;http://www.aip.de/en/research/facilities/stella/instruments/data/sloanugriz-filter-curves&quot; rel=&quot;nofollow&quot;&gt;Sloan z'&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2016-06-01T09:53:08.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15069" PostTypeId="2" ParentId="15063" CreationDate="2016-06-01T09:57:26.123" Score="6" Body="&lt;p&gt;Ideally a tripod as Journeyman suggests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another idea is to use a different grip. Use &lt;strong&gt;both hands&lt;/strong&gt; to hold the right hand objective housing. Allow the left hand objective to rest on the back of your left hand/wrist.  Also if possible hold your elbows in closer to your body (this is really only possible if you're not looking very high up.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means your two hands are in a &quot;relatively&quot; rigid triangular arrangement. It's still far inferior to using a tripod, but it's a bit better than just holding the binoculars in the traditional way...&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2016-06-01T09:57:26.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15070" PostTypeId="2" ParentId="15063" CreationDate="2016-06-01T10:19:52.817" Score="9" Body="&lt;p&gt;Tripods are good, as are monopods. These aren't as stable as tripods but are easier to lug around as they have 1/3 of the number of legs. When using a monopod you form the other two legs of the &quot;tripod&quot; to create the stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other options:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Lean your elbows on a wall or something else at the right height. This is more useful when looking at terrestrial objects, but you should be able to adapt the stance for looking the heavens.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Sit down. Take a deck chair out with you and set it to recline as far as is comfortable. Then you can rest your elbows on your chest to give some more support.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Wrap the binocular's strap around one arm so that the binoculars are now rigidly held in your hand. This will reduce (but not eliminate) the shake from your wrist.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="111" LastEditorUserId="111" LastEditDate="2019-01-25T08:46:49.617" LastActivityDate="2019-01-25T08:46:49.617" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="15071" PostTypeId="1" AcceptedAnswerId="15072" CreationDate="2016-06-01T11:37:35.610" Score="1" ViewCount="87" Body="&lt;p&gt;Using spectroscopy the chemical composition of exoplanets atmosphere is determined. As a scientist, what spectral lines would you look for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which elements are relevant in pursue of a second earth?&lt;/p&gt;&#xA;" OwnerUserId="12014" LastEditorUserId="12014" LastEditDate="2016-06-01T11:50:22.673" LastActivityDate="2016-06-01T19:51:04.747" Title="Which elements are an indication of habitable exoplanets?" Tags="&lt;exoplanet&gt;&lt;spectroscopy&gt;&lt;planetary-transits&gt;&lt;astrobiology&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15072" PostTypeId="2" ParentId="15071" CreationDate="2016-06-01T12:25:25.303" Score="2" Body="&lt;p&gt;If you are asking about second earth for us Carbon based life forms then atmospheric composition of &quot;second earth&quot; should be similar to Earth. Primarily Nitrogen and Oxygen. Some green house gases like CO2, CH4, water vapors for temperature regulation. Water Vapors can also be sign of active water cycle. Some O3 will be nice for UV protection. &lt;/p&gt;&#xA;" OwnerUserId="10657" LastActivityDate="2016-06-01T12:25:25.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15073" PostTypeId="2" ParentId="15063" CreationDate="2016-06-01T12:30:03.433" Score="4" Body="&lt;p&gt;I always recommend &lt;a href=&quot;https://astronomy.stackexchange.com/questions/8325/what-to-look-for-in-a-tripod-for-binoculars/8329#8329&quot;&gt;a good tripod&lt;/a&gt;. That said, there are a few tricks to stabilizing binoculars in human hands.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Hold your elbows together, under the binocs and pressed up against your belly. This reduces the leverage that the binoculars have on your stable body by half.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Hold something heavy on/under the binoculars. In the army I would take to full M-16 magazines and stack them &lt;em&gt;under&lt;/em&gt; the binoculars, holding them all together. The increased mass reduced vibration. Note however that it does become more difficult to hold the binocs for long periods of time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Stand in a pistol stance. Legs spread apart, slightly wider than the shoulders and knees very slightly bent. Shoulders back.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The way that I would test my stability is to find lettering far from where I stand. I could very measurably increase the distance that I could read from by doing those three things. You would likely not have such a test available at night for astronomy observations, though.&lt;/p&gt;&#xA;" OwnerUserId="209" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-06-01T12:30:03.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15074" PostTypeId="1" AcceptedAnswerId="15079" CreationDate="2016-06-01T12:46:35.430" Score="5" ViewCount="1339" Body="&lt;p&gt;Recently, &lt;strong&gt;over a thousand exoplanets were confirmed in one huge sweep&lt;/strong&gt; which is a huge achievement and in the mix &lt;strong&gt;a dozen or so possibly habitable planets were confirmed&lt;/strong&gt; as well. This made me wonder, Are there any common characteristics of (possibly) habitable planets? If so, what are they? Could scientists use these characteristics to locate them? Or are they using these characteristics already?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; Answers can be theoretical as well!&lt;/p&gt;&#xA;" OwnerUserId="4160" LastActivityDate="2016-06-01T14:47:03.947" Title="Are there any common characteristics of habitable planets?" Tags="&lt;exoplanet&gt;&lt;earth-like-planet&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15075" PostTypeId="2" ParentId="15063" CreationDate="2016-06-01T13:17:06.670" Score="3" Body="&lt;p&gt;For binoculars, your best option isn't just a simple tripod, but a &lt;a href=&quot;http://www.telescope.com/catalog/product.jsp?productId=103106&quot; rel=&quot;nofollow&quot;&gt;parallelogram mount&lt;/a&gt; (link for representation only, not a product recommendation) on a tripod.  While many standard to slightly larger than standard binoculars will have an adapter to attach them to a standard tripod; except for looking at things near the horizon it won't be comfortable to use.  The problem is that because they're short the binocular eye pieces will often be too close to the tripod legs themselves for comfortable head positioning.  If you're showing things to other people a parallelogram mount has the added bonus of being adjustable in height without changing what its pointed at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're hand holding the most important thing you can do is to use both hands and brace your arms in some way.  The top rail of a tall fence or a table (if you're seated) often work well.  If you don't have anything else you can use your body itself by resting your upper arms against your chest to form a cantilever brace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A last option, although not one I particularly recommend for other reasons (RSI injuries in particular) would be to get a job filling jars in an artisan scale canning/bottling plant.  A friend of mine does that and after several years has bulked up his arms to the point he was able to use my &lt;a href=&quot;http://www.telescope.com/Binoculars/Astronomy-Binoculars/Orion-15x70-Astronomy-Binoculars-with-Tripod-Adapter/c/5/sc/72/p/99637.uts&quot; rel=&quot;nofollow&quot;&gt;15x70&lt;/a&gt; binoculars with 1 hand the way an average person could do with a light weight 5x35 pair.&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2016-06-01T13:17:06.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15076" PostTypeId="2" ParentId="15074" CreationDate="2016-06-01T14:04:58.533" Score="4" Body="&lt;p&gt;Some of the possible characteristics are listed below.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Distance of planet from its host star. Is the planet in Goldie-lock zone?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Atmospheric composition. Presence of Oxygen, Nitrogen, CO2, water vapors. &#xA;Thick atmosphere can shield from radiation from host star.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mass of the planet. surface gravity.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is the planet tidally locked with its host star. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Orbit of planet around its host star. Highly elliptical orbit can be a problem(just a guess).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Characteristics mentioned above are relatively easier to identify with current technologies than mentioned below:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Presence of liquid water on surface.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Wind speed. (Very high wind speeds like on Uranus can make life impossible).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Tectonic activities, active volcanoes, solid crust for us land dwellers. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Magnetosphere. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10657" LastActivityDate="2016-06-01T14:04:58.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15077" PostTypeId="1" CreationDate="2016-06-01T14:27:16.550" Score="1" ViewCount="127" Body="&lt;p&gt;In time different during the first second after Big Bang than it is now? Inflation had a dramatic effect on space, and spacetime is one thing. Was the properties of time affected too? Curved spacetime around black holes cause funny effects for outside observers, infalling objects appear to never reachthe event horizon. The very early universe was even denser. When we talk about the first second, are such effects accounted for?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-06-01T18:21:21.617" Title="Was time different before the great inflation?" Tags="&lt;space-time&gt;&lt;big-bang-theory&gt;&lt;cosmological-inflation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15078" PostTypeId="1" AcceptedAnswerId="18598" CreationDate="2016-06-01T14:33:39.380" Score="3" ViewCount="401" Body="&lt;p&gt;Wasn't spacetime as much curved as a black hole directly after Big Bang, because mass was so densly packed? Wasn't everything like an event horizon and how could things expand across it? Could inflation happen in a black hole too so that its content does become accessible to the outside universe?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-10-06T07:22:58.063" Title="How could universe inflate itself out of the very dense and curved early spacetime? Could it happen in a black hole too?" Tags="&lt;black-hole&gt;&lt;space-time&gt;&lt;big-bang-theory&gt;&lt;cosmological-inflation&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15079" PostTypeId="2" ParentId="15074" CreationDate="2016-06-01T14:47:03.947" Score="9" Body="&lt;p&gt;I think I will answer your question from two vantage points. Firstly, what do we have to measure about a new exoplanet to consider it potentially habitable? Keep in mind that in detecting these 1000 or so new exoplanets, we've only gone so far as to show that they (statistically) exist and to measure some very very basic properties of them. It is on these very basic properties that we make our judgement of habitability and further research and observation is necessary to confirm these suppositions. When a new exoplanet is announced, we generally know the following information, which we use to determine habitability:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;Distance from the star.&lt;/em&gt; This is a crucial piece of the puzzle because it tells us generally how hot or cold the planet is. We presume that life cannot exist on planets which are boiling hot or freezing cold. This information can be ascertained from either the radial velocity method or the transit method.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Size of the planet.&lt;/em&gt; Note here, size generally refers to diameter or radius. This is important because we want to generally know what type of planet we've detected. Gas giants are, well ... giant and inhospitable to life as we know it. We want to make sure the planet we've found is small enough that it is likely terrestrial. Although size is not a guarantee that it is terrestrial, its a piece of the puzzle. We can measure size by the transit method, however there is some degree of uncertainty to this process because we need to know the size of the host star to a good degree of accuracy, which is not always possible.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Mass of the planet.&lt;/em&gt; This is a crucial component in determining habitability. Size alone may tell us that a planet is what we call a &quot;Super Earth&quot; or &quot;Sub Neptune&quot; in that it is slightly larger than our own planet. The problem with that is, if the planet sits on the size border between Earth and Neptune, how are we to know whether it is rocky or gaseous? This can be resolved by looking at the mass, which is ascertained by the radial velocity method. The mass combined with size, gives us density and density is a surefire way to know if the planet is made of dense material or light gases.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;That's really about it for basic information gleaned from simple planet detection methods. If you have distance from star, size, and mass (all of which you can get from the RV and transit methods) then you can presume that the planet is Earth-like and at a good temperature. That's enough to claim potential habitability, but not enough to actually show it can support human life. For that, you need to investigate the planet further and show it meets a variety of other conditions. As of know, some of those conditions are measurable with enough work, some aren't. These conditions may include, but are not limited to:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;An atmosphere with water in it.&lt;/em&gt; We have the technology to measure this fact, but I'll say that it is very very difficult to do so and I don't know if anyone has really successfully shown water exists in an exoplanetary atmosphere (some have convincing evidence though). Generally, the process to do this is to measure the radius of the planet as a function of wavelength. Different wavelengths of light will refract through the atmosphere differently (just like in a prism) and so you measure a different radius of the planet (radius being the physical size of the planet plus the extent of the atmosphere) at different wavelengths. From this information, you can model what the atmosphere must be made of to cause the radius to change depending on wavelength. The issue with this process is that is requires removing all possible false detections and having a great signal to noise which is very hard.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;A magnetosphere.&lt;/em&gt; We believe this is a necessity for life. Every second of every day our planet is being bombarded with high energy particles from our Sun. Most of these particles are harmlessly deflected by our magnetosphere and so we're protected. Planets without this protection would have a hard time hosting life as that life would be destroyed by radiation (or at least life like us). There is currently no way to detect whether a planet has a magnetosphere, although I believe there are potential ideas on the chalkboard.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;The presence of a moon.&lt;/em&gt; This one is a bit iffy. Some people will argue our Moon has been instrumental to our development and for life in generally. The argument generally goes that the Moon has helped stabilize our tilt and thus kept our seasons and global weather well regulated. While we might be able to exist on a planet without such conditions, some believe life cannot begin on such a harsh planet without that regulating moon. There is also the benefit that the Moon has shielded us from a large number of impacts by taking the blow itself. Again, we could live on the planet, but the lack of a moon might not foster growth of life naturally. Moons, at this time, are out of our reach to detect.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are numerous other factors for what helps define whether a planet is habitable, be it by us or in hosting new life. The list could go on and on, but these are the major points I think.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-06-01T14:47:03.947" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15080" PostTypeId="2" ParentId="15077" CreationDate="2016-06-01T14:51:34.880" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Curved spacetime around black holes cause funny effects for outside observers, infalling objects appear to never reachthe event horizon. The very early universe was even denser. When we talk about the first second, are such effects accounted for?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These time dilation effects are (always) accounted for. An observer at any &quot;place&quot; in the inflating universe would experience time at his, her or its own rate. Several observers might find they had experienced different rates of time if they gathered together afterwards and compared notes, however.&lt;/p&gt;&#xA;" OwnerUserId="7491" LastEditorUserId="7491" LastEditDate="2016-06-01T15:22:41.117" LastActivityDate="2016-06-01T15:22:41.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15082" PostTypeId="2" ParentId="15077" CreationDate="2016-06-01T18:21:21.617" Score="3" Body="&lt;p&gt;Food for thought on density and time dilation.  Density isn't gravity.  We tend to think it is, but it's not.  Variation in density gives us gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have an infinite universe, very young, hot and dense, but it's uniform, then you have no time dilation.  One way to think about this is to ask if light passing through the universe is red shifted or blue shifted.  If the universe is uniform and infinite, it's the same in all directions around the photon, so the photon is neither red nor blue shifted, so, contrary to what our intuition about relativity tells us, there might not have been a great deal of time dilation in the very young universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of-course the ray of light passing through expanding space is enormously red-shifted, but that doesn't affect time dilation.  The gravitational red or blue shift in the very young universe might still be close to zero meaning very little time dilation anywhere.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the early universe grew less uniform then you had time dilation variations in denser parts of the early Universe but with FTL expansion and gravity limited by the speed of light, it's unclear if there ever was massive time dilation in the young universe (at least, unclear to me, if somebody here knows 100%, feel free to post and/or correct me). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we assume the universe is finite and flat and the big bang happened at a point, which, generally isn't the model most people like these days, then you could run into enormous time dilation inside the tiny young and dense universe, but it still expanded FTL, so the calculations get weird and too hard for me.   If the expansion of space happens faster than gravity can keep up, gravitational time dilation might not apply, at least not until after the rate of expansion slowed down enough for gravity to permeate the local regions.  I couldn't even begin to run the math on this and you'd have different answers for different models too, I'd think.   I agree with Andy though, it would need to be accounted for or the models wouldn't be much good.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-06-01T18:21:21.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15083" PostTypeId="2" ParentId="15060" CreationDate="2016-06-01T18:31:35.003" Score="1" Body="&lt;p&gt;I guess that depends on how much radiation they're getting. I should also point out that astronauts are not 100% protected from radiation, as discussed in &lt;a href=&quot;https://space.stackexchange.com/questions/1034/how-are-the-astronauts-in-the-iss-protected-from-solar-flares&quot;&gt;this question&lt;/a&gt;. One of the most common affects on actual astronauts is the &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/19580503&quot; rel=&quot;nofollow noreferrer&quot;&gt;development of cataracts&lt;/a&gt;. They can also receive &lt;a href=&quot;http://www.nasa.gov/mission_pages/station/research/experiments/137.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;damage to their nervous system&lt;/a&gt; over time, among other effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it appears that you're asking what would happen to a person who was exposed to a huge amount of radiation, such as if they went into space with a suit that did little to protect from radiation. I think to answer that, we just have to look at cases where humans have actually been exposed to huge amounts of radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One interesting example that comes to mind is &lt;a href=&quot;https://en.wikipedia.org/wiki/Anatoli_Bugorski&quot; rel=&quot;nofollow noreferrer&quot;&gt;Anatoli Bugorski&lt;/a&gt; who was a Russian particle physicist who accidentally had his head inside a particle accelerator when it was on. His head was subjected to huge amounts of radiation. The results of this are described as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The left half of Bugorski's face swelled up beyond recognition and,&#xA;  over the next several days, started peeling off, revealing the path&#xA;  that the proton beam (moving near the speed of light) had burned&#xA;  through parts of his face, his bone and the brain tissue underneath.&#xA;  As it was believed that he had received far in excess of a fatal dose&#xA;  of radiation, Bugorski was taken to a clinic in Moscow where the&#xA;  doctors could observe his expected demise. However, Bugorski survived&#xA;  and even completed his Ph.D. There was virtually no damage to his&#xA;  intellectual capacity, but the fatigue of mental work increased&#xA;  markedly. Bugorski completely lost hearing in the left ear and only&#xA;  a constant, unpleasant internal noise remained. The left half of his&#xA;  face was paralyzed due to the destruction of nerves.&lt;a href=&quot;https://space.stackexchange.com/questions/1034/how-are-the-astronauts-in-the-iss-protected-from-solar-flares&quot;&gt;1&lt;/a&gt; He was able to&#xA;  function well, except for the fact that he had occasional complex&#xA;  partial seizures and rare tonic-clonic seizures.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now imagine what could happen if a person were to be exposed in a similar fashion to highly energetic particles in space, but throughout their body and not just their face. I expect that a person could not survive such an exposure and that their body would be destroyed. Even if they survived, it is highly likely they would live with crippling medical conditions and very likely develop cancer later in life.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2016-06-01T18:31:35.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15084" PostTypeId="2" ParentId="15066" CreationDate="2016-06-01T18:59:27.767" Score="8" Body="&lt;p&gt;Puffy planets tend to be Jupiter or Saturn like, probably lower mass than Jupiter, perhaps lower metalicity but the most important factor is heat.  Either close to the sun or recently formed.   Heat expands gas planets.  You're correct that as you add more mass the planet of Jupiter mass tends not to grow larger, but if there's enough internal heat, gas giant planets can get a bit larger than Jupiter.   Planets as much as 2 Jupiter Radii have been observed (though there's some inaccuracy in those estimates), but growing larger than Jupiter is largely a factor of high temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rocky worlds will never grow as large as Jupiter.  The greater mass will prevent it, and past a certain mass, a rocky world is unlikely to remain what we consider rocky.  Above a certain mass it holds onto hydrogen which is the most abundant gas in the universe, and that would give the massive rocky world an appearance more like a gas giant.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to answer your question in theory, if you had a rocky world with Jupiter mass and a negligible atmosphere, it could never get close to Jupiter diameter.  The mass crushes the inside of the planet.  Mercury for example is made of denser material than the earth.  Higher Iron content, but it's less dense than the Earth because the Earth's mass crushes its rocky mantle and metallic core.  When you start to get a Jupiter-mass rocky world, the crushing becomes significant and you could never build a rocky world close to as big as Jupiter.  As with hydrogen, it would reach a certain maximum size, then it would begin to get smaller under the crush of gravity.   Even so called &quot;non compactible&quot; material, does compact at the pressure inside large planets.   Iron has a density of 7.874 g/cm&lt;sup&gt;3&lt;/sup&gt; and slightly less than that at high temperature, but the Earth's metallic core has a density of 12.6&amp;ndash;13.0, and that's primarily due to crushing.  When you've reached Jupiter mass, the crushing and density is significantly greater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At about the mass of the sun the planet would be smaller than the Earth, and it would basically resemble a white dwarf star, and no longer be distinguishable as a rocky world.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can add a few links to back this up if needed and if someone wants to run the math or give an answer with more detail, feel free.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="7142" LastEditDate="2016-06-02T00:32:58.773" LastActivityDate="2016-06-02T00:32:58.773" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15085" PostTypeId="2" ParentId="15071" CreationDate="2016-06-01T19:42:39.997" Score="4" Body="&lt;p&gt;While not strictly relevant to your question, I'm very much looking forward to what the James Webb Telescope might tell us about exo-planet atmospheres.   That's probably the thing I'm most looking forward to in astronomy discoveries over the next 5-10 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one thing I've heard that they are looking for specifically is a combination of Oxygen (which implies Photosynthesis) and CH4 which, in combination with oxygen implies bacterial digestion of plant matter (aka cow farts).   Those 2 things together are of particular interest because CH4 and O2 tend to interact with each other, so, if not regularly produced they aren't likely to both be present, at least, in measurable amounts.   Other factors like plate tectonics, Methane escaping from the ground or under the oceans, comet impacts and perhaps even lightning might make such readings a little bit unreliable, but as I understand it, that combo, CH4 and O2 is the big one to look for.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other interesting things might be a CO2/O2 ratio, also implying photosynthesis and perhaps telling us something about plate tectonics or O2 respiration.   Variation in H20 content which could tell us something about weather, if the planet has freezing/thawing periods like Mars or vast deserts and few oceans and little water or occasional hurricanes.   Dust particles could tell us something about the soil or volcanic activity.  I would think NH3 content in abundance might be an indicator of little or no life, but I'm not positive there.  Water with ammonia oceans might be possible and still have life, though we might find that not a good planet to live on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally habitable is a term that needs a bit of exploration.  Lets use the Earth as an example some 2 billion years ago, there was Oxygen in the atmosphere, there was CO2 in the atmosphere, there were oceans, there was a magnetic field, there was even an ozone layer, so it had everything Knu8 listed, but, there were no trees, there wasn't even any soil.  There might have been a thin microbial layer living on the rocky surface or, if too hot, perhaps in caves.  There were no fish, there wasn't even any good plants or krill like things growing in the oceans for fish to eat.  It was mostly one celled life mostly just in the oceans and perhaps lakes.  Earth 2 billion years ago might not be our first choice for habitation.  It was also prone to photosynthesis driven snowball earth periods as well as having no soil and no trees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also the question as to whether we'd want to inhabit an &quot;earth-like&quot; planet.  Bacteria can be pretty hostile and if there's life on the planet, it might not be safe to settle, and there would also be ethical questions.   If there's no life, then the Ethics isn't an issue, but there's likely a lot more terraforming.   It's all relatively moot given the difficulty of that kind of space travel, but &quot;habitable&quot; is a more difficult question than it first appears.   Our technology might be able to make a dead planet habitable over time, where as a planet with life on it might appear more habitable but could pose unforeseen problems.  I'm not sure which would be more ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not an expert in this, but what to look for in exoplanet atmospheres for habitability is a little complicated and I've only touched on it briefly.   Fun thing to think about though.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-06-01T19:51:04.747" LastActivityDate="2016-06-01T19:51:04.747" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15086" PostTypeId="1" AcceptedAnswerId="15087" CreationDate="2016-06-01T20:31:31.520" Score="14" ViewCount="4463" Body="&lt;p&gt;The vastness of space brings me a sense of chilliness even though I have never experienced it, although I wish to. &lt;strong&gt;Just how cold is interstellar space (on average)?&lt;/strong&gt; How is this even measured? I mean you can't just stick a thermometer in space, right? &lt;/p&gt;&#xA;" OwnerUserId="4160" LastEditorUserId="4160" LastEditDate="2016-06-02T19:02:35.097" LastActivityDate="2016-06-09T09:18:02.027" Title="How cold is interstellar space?" Tags="&lt;space&gt;&lt;temperature&gt;&lt;interstellar-medium&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15087" PostTypeId="2" ParentId="15086" CreationDate="2016-06-01T21:39:46.067" Score="23" Body="&lt;p&gt;You &lt;em&gt;can&lt;/em&gt; stick a thermometer in space, and if it is a super-high-tech one, it might show you the temperature of the gas. But since the interstellar medium (ISM) is so dilute, a normal thermometer will radiate energy away faster than it can absorb it, and thus it won't reach thermal equilibrium with the gas.&#xA;It won't cool all the way to 0 K, though, since the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;noreferrer&quot;&gt;cosmic microwave background radiation&lt;/a&gt; won't allow it to cool further than 2.7 K, as described by David Hammen.&lt;/p&gt;&#xA;&lt;p&gt;The term &amp;quot;temperature&amp;quot; is a measure of the average energy of the particles of a gas (other definitions exist e.g. for a radiation field). If the gas is very thin, but particles move at the same average speed as, say, at the surface of Earth, the gas is still said to have a temperature of, say, 27º C, or &lt;span class=&quot;math-container&quot;&gt;$ 300\,\mathrm{K}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The ISM consists of several different phases, each with their own physical characteristics and origins. Arguably, the three most important phases are (see e.g. &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0106359&quot; rel=&quot;noreferrer&quot;&gt;Ferrière 2001&lt;/a&gt;):&lt;/p&gt;&#xA;&lt;strong&gt;Molecular clouds&lt;/strong&gt;&#xA;&lt;p&gt;Stars are born in dense molecular clouds with temperatures of just 10-20 K. In order for a star to form, the gas must be able to collapse gravitationally, which is impossible if the atoms move too fast.&lt;/p&gt;&#xA;&lt;strong&gt;The warm neutral medium&lt;/strong&gt;&#xA;&lt;p&gt;The molecular clouds themselves form from gas that is neutral, i.e. not ionized. Since most of the gas is hydrogen, this means that it has a temperature of roughly &lt;span class=&quot;math-container&quot;&gt;$10^4\,\mathrm{K}$&lt;/span&gt;, above which hydrogen tends to get ionized.&lt;/p&gt;&#xA;&lt;strong&gt;The hot ionized medium&lt;/strong&gt;&#xA;&lt;p&gt;Gas that accretes onto the galaxy in its early phases tend to have much larger temperature, of roughly &lt;span class=&quot;math-container&quot;&gt;$10^6\,\mathrm{K}$&lt;/span&gt;. Additionally, the radiative feedback from the hot stars (O and B), and the kinetic and radiative energy injected by supernova explosions ionize and heat gas bubbles that expand. This gas comprises the hot ionized medium.&lt;/p&gt;&#xA;&lt;strong&gt;Cooling&lt;/strong&gt;&#xA;&lt;p&gt;The reason that the ISM is so sharply divided into phases, as opposed to just being a smooth mixture of particles of all sorts of energies, is that gas cools by various physical processes that have a rather temperature-specific efficiency.&#xA;&amp;quot;Cooling&amp;quot; means converting the kinetic energy of particles into radiation that is able to leave the system.&lt;/p&gt;&#xA;&lt;em&gt;Hot gas&lt;/em&gt;&#xA;&lt;p&gt;Very hot gas is fully collisionally ionized and thus cools mainly through free electron emitting Bremsstrahlung. This mechanism becomes inefficent below &lt;span class=&quot;math-container&quot;&gt;$\sim10^6\,\mathrm{K}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;em&gt;Warm gas&lt;/em&gt;&#xA;&lt;p&gt;Between &lt;span class=&quot;math-container&quot;&gt;$10^4\,\mathrm{K}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$10^6\,\mathrm{K}$&lt;/span&gt;, recombinations (i.e. electrons being caught by ions) and collisonal excitation and subsequent de-excitation lead to emission, removing energy from the system.&#xA;Here the metallicity&lt;span class=&quot;math-container&quot;&gt;$^\dagger$&lt;/span&gt; of the gas is important, since various elements have different energy levels.&lt;/p&gt;&#xA;&lt;em&gt;Cool gas&lt;/em&gt;&#xA;&lt;p&gt;At lower temperatures, the gas is almost fully neutral, so recombinations cease to have any influence. Collisions between hydrogen atom become too weak to excite the atoms, but if molecules or metals are present, it is possible through fine/hyperfine lines, and rotational/vibrational lines, respectively.&lt;/p&gt;&#xA;&lt;p&gt;The total cooling is the sum of all these processes, but will be dominated by one or a few processes at a given temperature. The figures below from &lt;a href=&quot;http://adsabs.harvard.edu/abs/1993ApJS...88..253S&quot; rel=&quot;noreferrer&quot;&gt;Sutherland &amp;amp; Dopita (1993)&lt;/a&gt; shows the main cooling processes (left) and the main cooling elements (&lt;em&gt;right&lt;/em&gt;), as a function of temperature:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zlt2p.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zlt2p.png&quot; alt=&quot;processes/elements&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The thick line show the total cooling rate. The figure below, from the same paper, shows the total cooling rate for different metallicities. The metallicity is a logarithmic scale, so [Fe/H] = 0 means Solar metallicity, and [Fe/H] = –1 means 0.1 times Solar metallicity, while &amp;quot;nil&amp;quot; is zero metallicity.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/s2Zc1.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/s2Zc1.png&quot; alt=&quot;total&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Since these processes don't cover equally the full temperature range, the gas will tend to reach certain &amp;quot;plateaus&amp;quot; in temperatures, i.e. it will tend to occupy certain specific temperatures. When gas cools, it contracts. From the ideal gas law, we know that the pressure &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; is proportional to the product of the density &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; and the temperature &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;. If there's pressure equilibrium in the ISM (which there isn't always, but in many cases is a good assumption), then &lt;span class=&quot;math-container&quot;&gt;$nT$&lt;/span&gt; is constant, and thus if a parcel of hot ionized gas cools from &lt;span class=&quot;math-container&quot;&gt;$10^7\,\mathrm{K}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$10^4\,\mathrm{K}$&lt;/span&gt;, it must contract to increases its density by a factor &lt;span class=&quot;math-container&quot;&gt;$10^3$&lt;/span&gt;. Thus, cooler clouds are smaller and denser, and in this way the ISM is divided up in its various phases.&lt;/p&gt;&#xA;&lt;p&gt;So, to conclude, interstellar space is not as cold as you may think. However, being extremely dilute, it is difficult to &lt;em&gt;transfer&lt;/em&gt; heat, so if you leave your spaceship, you will radiate away energy faster than you can absorb it from the gas.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^\dagger$&lt;/span&gt;&lt;sub&gt;In astronomy, the term &amp;quot;metal&amp;quot;, refers to all elements that are not hydrogen or helium, and &amp;quot;metallicity&amp;quot; is the fraction of gas that consists of metals.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-06-03T19:46:44.033" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15088" PostTypeId="2" ParentId="15086" CreationDate="2016-06-01T21:43:07.133" Score="9" Body="&lt;p&gt;The title of the question asks about interstellar space, but the body asks about the interstellar medium. These are two very different questions. The temperature of the interstellar medium varies widely, from a few kelvins to over ten million kelvins. By all accounts, the vast majority of the interstellar medium is at least &quot;warm&quot;, where &quot;warm&quot; means several thousand kelvins.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I mean you can't just stick a thermometer in space, right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can if you have &lt;em&gt;Star Trek&lt;/em&gt; or &lt;em&gt;Star Wars&lt;/em&gt; technology. Assuming an old-style bulb thermometer released in a place far removed from a star, the temperature of that thermometer would drop rather quickly, eventually stabilizing at about 2.7 kelvin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regard to a macroscopic object such as an old-style thermometer or a human in a spacesuit, there's a big difference between the temperature of interstellar space and the temperature of interstellar medium. Even if the local interstellar medium is in the millions of kelvins, that macroscopic object will still cool to about 2.7 kelvin because there's no substance to that hot interstellar medium. The density of the interstellar medium is so very, very low that radiation losses completely dominate over conduction from the medium. The interstellar medium can be very hot precisely because it is a gas (gases are a bit weird), and because it is extremely tenuous (extremely tenuous gases are beyond weird).&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2016-06-01T21:43:07.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15089" PostTypeId="2" ParentId="15066" CreationDate="2016-06-01T22:26:03.720" Score="2" Body="&lt;p&gt;The maximum possible radius of a planet is probably unknown to date because current observation techniques not only will give us a biased view of the distribution of planet radii but it is also difficult to tell planets and brown dwarfs apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However &lt;a href=&quot;http://www.exoplanet.eu/diagrams/&quot; rel=&quot;nofollow noreferrer&quot;&gt;current data&lt;/a&gt; suggest a limit of roughly 2 Jupiter radii.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Xsudy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xsudy.png&quot; alt=&quot;The Extrasolar Planets Encyclopaedia&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10568" LastActivityDate="2016-06-01T22:26:03.720" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15090" PostTypeId="2" ParentId="15061" CreationDate="2016-06-01T23:37:38.737" Score="1" Body="&lt;p&gt;NO&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The differences between a brown dwarf and a low mass star are based upon their mass. A brown dwarf does not have central temperatures and pressures to create energy using hydrogen fusion. The mass is too low to create these conditions. A star by definition is an object where energy is created using hydrogen fusion. This happens when the core temperature/pressure are high enough. This requires a higher mass. The exact mass boundaries between planet, brown dwarf and star are not fully determined. They also depend somewhat on composition. Even so there is no real chance of having a brown dwarf more massive than the lowest mass star.  The masses could be similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of orbits: any pair of bodies orbit around their common center of mass (barycenter). They orbit about each other. When the two objects are the same mass the barycenter is half way between them. When say one star is ten times the mass the barycenter is nearer (1/10th the distance) the more massive star.&#xA;Think of a see-saw with unequal weight people on it. Now the less massive object (say planet) orbits about a much more massive object (star) the barycenter is inside the star. So we can say the planet orbits around the star.  In realality they are both orbiting around the barycenter.&lt;/p&gt;&#xA;" OwnerUserId="10761" LastActivityDate="2016-06-01T23:37:38.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15092" PostTypeId="1" CreationDate="2016-06-02T08:43:06.093" Score="1" ViewCount="546" Body="&lt;p&gt;I am new to using SDSS image files and have noticed that some of the values in the FITS file are negative. Can somebody explain to me what these values actually mean?&lt;/p&gt;&#xA;" OwnerUserId="12038" LastActivityDate="2016-06-02T20:56:01.500" Title="SDSS Image FITS files have negative values. What do these negative values mean?" Tags="&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15093" PostTypeId="2" ParentId="15063" CreationDate="2016-06-02T09:59:15.607" Score="1" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Grip the binoculars as far out as you can. Shake is actually angular movement of the binoculars. Abstractly speaking your binoculars are a single-sided lever with the swivel-point where the eye-pieces are pressed against your face, while the hands cause angular movement. The longer the lever is, the smaller the angular deflection for a specific movement of the hands becomes. And with smaller angular deflection, the shake becomes smaller. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Press your elbows down against something solid, or — lacking that — push them against your rib cage. If you can fixate your elbows, then you restrict the movement of the hands greatly. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11474" LastEditorUserId="11474" LastEditDate="2016-06-02T10:05:02.663" LastActivityDate="2016-06-02T10:05:02.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15097" PostTypeId="2" ParentId="15092" CreationDate="2016-06-02T19:56:40.247" Score="1" Body="&lt;p&gt;Without more information about the specific nature of the fits file, my immediate guess is that you're using a file that has already been corrected for things like bias and flat fielding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that for a simple exposure (what we might call the &quot;science&quot; image), the fits image should have only positive values as each pixel is a (proxy) count of how many photons hit it and how can you have a negative number of photons hit your detector? But often astronomers need to clean up their science image from various sources of noise. One such example is cleaning it with a bias frame. A bias frame is an exposure with the shutter closed and the exposure time set to zero. Essentially you just pull the signal from the CCD without taking a picture. In a perfect world, the CCD will be blank and empty and there will be nothing to read out if you never exposed, but in the real world things are messy and you have lingering electrons all over the place. Reading a bias frame allows you to get some estimate of this noise that can inherently exist in a CCD. An astronomer will then take this bias frame and subtract it from the science image to remove this noise, but of course this can result in your science image having negative pixel values. This usually isn't a problem though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astrophoto.net/calibration.php&quot; rel=&quot;nofollow&quot;&gt;This resource&lt;/a&gt; looks pretty comprehensive about the basics of cleaning science frames.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-06-02T19:56:40.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15098" PostTypeId="2" ParentId="15092" CreationDate="2016-06-02T20:56:01.500" Score="1" Body="&lt;p&gt;Without further details I would guess your fits file is using 16 bit signed integers between -32768 and +32767.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In which case you should find that the fits header item called bzero has been set to 32768.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The true integer values are given by your pixel value*bscale + bzero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very common format for raw data from CCD cameras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand if you are talking about processed &lt;em&gt;sky subtracted&lt;/em&gt; SDSS images, then of course some pixels are negative - the &lt;em&gt;average&lt;/em&gt; pixel in a &quot;sky&quot; region will be zero, so a substantial fraction &lt;em&gt;must&lt;/em&gt; be negative.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-06-02T20:56:01.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15099" PostTypeId="1" AcceptedAnswerId="15101" CreationDate="2016-06-03T02:15:21.773" Score="0" ViewCount="152" Body="&lt;p&gt;I know &quot;super-earths&quot; are quite common in other planetary systems, but there are none in the Solar System. Is it due to the material that could have created a single large super-earth being used to create four smaller terrestrial planets in the Solar System?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="2365" LastEditDate="2016-06-03T06:37:09.547" LastActivityDate="2016-06-03T13:40:11.530" Title="Is the material that could have formed a &quot;Super-earth&quot; divided among the four terrestrial planets" Tags="&lt;solar-system&gt;&lt;exoplanet&gt;&lt;terrestrial-planets&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15101" PostTypeId="2" ParentId="15099" CreationDate="2016-06-03T13:40:11.530" Score="3" Body="&lt;p&gt;I think the answer to your question lies with Jupiter, be it directly or indirectly, the gas giant is now believed to have had a large influence on the way the inner planets formed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Kte1g.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Kte1g.jpg&quot; alt=&quot;Image Credit: NASA, ESA, AND A. SIMON/GSFC&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many planetary scientists believe that the reason we only have 4 small inner planets today and not any &quot;super-Earth&quot; sized ones is because of Jupiter, and how it interacted with the circumstellar disk and protoplanets at the time of the solar system formation. There may well have been one or two super-Earths that formed in the early solar system, but models suggest that they would have been sent crashing into the Sun by the inward advances of Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this blog &lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2015/03/25/jupiter_menace_gravity_may_have_wiped_out_early_inner_solar_system.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; which explains it in more detail. And there is also another &lt;a href=&quot;http://mnras.oxfordjournals.org/content/early/2016/03/04/mnras.stw431.short&quot; rel=&quot;nofollow noreferrer&quot;&gt;MNRAS article&lt;/a&gt; that goes into more detail about the physics and orbital mechanics behind what might have happened in the early solar system.&lt;/p&gt;&#xA;" OwnerUserId="10437" LastActivityDate="2016-06-03T13:40:11.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15102" PostTypeId="1" AcceptedAnswerId="15103" CreationDate="2016-06-03T18:31:11.237" Score="0" ViewCount="349" Body="&lt;p&gt;In the original book, the classic 1962 movie and the modern 2001 re-make for TV. The store &quot;The day of the Triffits&quot; depicts a meteor shower that happen all around the Earth at the same time, and leaves everyone who sees it blind the following days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the nature of how the Earth rotates, moves around the Sun and a meteor field (required to create a shower of many meteorites). Is it really possible for such an event to happen everywhere at the same time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is kind of an important plot in the story, because if one side of the planet goes blind first they can warn the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1G92h.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1G92h.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2016-06-03T19:38:06.127" Title="Can a meteor shower happen simultaneously all around the planet?" Tags="&lt;meteor&gt;&lt;meteor-shower&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15103" PostTypeId="2" ParentId="15102" CreationDate="2016-06-03T19:38:06.127" Score="2" Body="&lt;p&gt;Realistically, no, that cannot happen. Most meteor showers that we experience on Earth - the most famous being the &lt;a href=&quot;https://en.wikipedia.org/wiki/Perseids&quot; rel=&quot;nofollow&quot;&gt;Perseids&lt;/a&gt; and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Leonids&quot; rel=&quot;nofollow&quot;&gt;Leonids&lt;/a&gt; - are a result of comets passing roughly through our orbit and leaving behind debris that was burned off as the comet passed by the Sun. We then come along, sweeping through this debris and from our perspective on Earth, we see a shower of meteors raining down on us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That implies two important things. First, we primarily see meteors as we sweep through the debris from comets. They are not objects which have made a trajectory for Earth and hit us head on. Second, we only see meteors when we're on the side of the planet facing the orbital direction of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, for the sake of the story, one could envision an alien race of sufficiently advanced technology surrounding our planet with asteroid and comet debris and subsequently &quot;dropping&quot; these meteroids on us from everywhere all at once. I'm not sure if aliens are involved in the story you've linked, but aside from third party intervention, there's now way you're going to see meteors from every place on Earth all at the same time.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-06-03T19:38:06.127" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15105" PostTypeId="1" AcceptedAnswerId="15108" CreationDate="2016-06-03T22:21:42.520" Score="1" ViewCount="787" Body="&lt;p&gt;As you probably know, &lt;strong&gt;the planets have pretty predictable orbits&lt;/strong&gt; around the sun but &lt;strong&gt;what exactly causes the complex movement of celestial bodies in the sky from Earth's surface?&lt;/strong&gt; Mars is an excellent example of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/kbynKfNfHk4&quot; rel=&quot;nofollow&quot;&gt;Mars example&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4160" LastEditorUserId="4160" LastEditDate="2016-06-04T11:18:31.987" LastActivityDate="2016-06-04T11:18:31.987" Title="What causes celestial bodies to move like they do from Earth's surface?" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15106" PostTypeId="2" ParentId="13532" CreationDate="2016-06-04T02:09:13.480" Score="-1" Body="&lt;p&gt;The Great Attractor at the heart of the Laniakea Supercluster meets the criteria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light does not make for a good sign post. If I continually accelerate I am always changing reference frames but my perspective on the speed of light will always be the same. How can you tell what ABSOLUTE frame of reference an observers is in when the absolute signpost of lightspeed looks the same from every relative perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if there was Dark Flow motion in common with the whole visible universe how would we detect it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would detect it by looking for the DARK MASS bloated by time dilation due to the dark flow motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore finding extra DARK MASS in galactic spin stability implies additional time dilation due to dark flow.&lt;/p&gt;&#xA;" OwnerUserId="12051" LastActivityDate="2016-06-04T02:09:13.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15108" PostTypeId="2" ParentId="15105" CreationDate="2016-06-04T07:43:10.263" Score="2" Body="&lt;p&gt;The apparent motion of planets is complex but predictable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The apparent motion is due to the combination of three different motions:      &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The rotation of the Earth,&lt;/li&gt;&#xA;&lt;li&gt;the motion of the Earth around the Sun,&lt;/li&gt;&#xA;&lt;li&gt;and the motion of the planet around the Sun. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The Planets move in ellipses, slightly perturbed by other planets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These motions are predictable, and their combination is straightforward trigonometry. The resultant motion is complex, the planets appear to loop relative to the stars, but entirely predictable. We know exactly where the planets will be in the sky for any time for many thousands of years in the future or the past. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-06-04T09:09:14.360" LastActivityDate="2016-06-04T09:09:14.360" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15109" PostTypeId="1" AcceptedAnswerId="15111" CreationDate="2016-06-04T11:49:42.123" Score="1" ViewCount="585" Body="&lt;p&gt;If the universe keeps expanding, is it possible to travel to the edge of the universe?&lt;/p&gt;&#xA;" OwnerUserId="12050" OwnerDisplayName="SaravInfern" LastEditorUserId="2153" LastEditDate="2016-06-04T17:05:22.823" LastActivityDate="2016-06-04T23:01:13.777" Title="If the universe keeps expanding, can we travel to the edge of the universe?" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2016-06-06T14:14:49.207" ContentLicense="CC BY-SA 3.0" />
  <row Id="15110" PostTypeId="1" AcceptedAnswerId="15115" CreationDate="2016-06-04T16:19:35.063" Score="7" ViewCount="4275" Body="&lt;p&gt;I found this on the Wikipedia page on &lt;a href=&quot;https://en.wikipedia.org/wiki/Supernova&quot; rel=&quot;nofollow noreferrer&quot;&gt;supernovae&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The total supernova rate in our galaxy is estimated to be about 4.6 per century, or one every 22 years, although &lt;strong&gt;we haven't actually observed one&lt;/strong&gt; for several centuries.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That sounds like a freak occurrence to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It made me curious about the numbers so I refreshed my knowledge on the Poisson distribution. It is not specified how many centuries we have not seen a supernova, but let's assume it's only two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If on average 9.2 supernovae happen per 2 centuries then the chance of having exactly 9 in any two century period can be calculated with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$p(k) = {\lambda^k \times e^{-\lambda} \over k!}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Filling in &lt;span class=&quot;math-container&quot;&gt;$k=9$&lt;/span&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$p(9) = {9.2^9 \times e^{-9.2} \over 9!}=0.1315$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chance that there is not a single supernova in a two century period is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$p(0) = {9.2^0 \times e^{-9.2} \over 0!}=0.00010104$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's about a 1 in 10,000 chance. If &quot;several&quot; means 3 centuries, then the chance of not having any supernova is as little as 1 in 100,000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So from this I can only come to the conclusion that supernovae did in fact happen in the Milky Way in the last two centuries, but that we didn't &lt;em&gt;see&lt;/em&gt; any of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now I'm confused. I read about supernovae that have happened in galaxies billions of light-years from here that lit up as the brightest source of light in the sky for days. Surely we would be able to see &lt;em&gt;all&lt;/em&gt; of the supernovae that happen in our Milky Way (no more than 100,000 light years away) with the naked eye, let alone with telescopes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are some supernovae so much more &lt;em&gt;super&lt;/em&gt; than others? Are all of the Milky Way's supernovae visible to the naked eye?&lt;/p&gt;&#xA;" OwnerUserId="12058" LastEditorUserId="12058" LastEditDate="2019-04-12T09:29:57.573" LastActivityDate="2019-04-12T09:29:57.573" Title="Why don't we see more supernovae in our galaxy?" Tags="&lt;light&gt;&lt;distances&gt;&lt;supernova&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="15111" PostTypeId="2" ParentId="15109" CreationDate="2016-06-04T17:03:29.763" Score="5" Body="&lt;p&gt;When we talk about the universe, we are really talking about one of two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The observable universe, which is everything we can possibly see.&lt;/li&gt;&#xA;&lt;li&gt;The Universe, which is everything that has ever existed, currently exists, and will exist.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The observable universe has its own center, usually the Earth. It is a spherical region of everything that we can see, essentially anything whose light has reached us. We usually refer to this when we say things like &quot;there are $10^{86}$ atoms in the universe.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, everyone has their own observable universe, and it can change depending on where you are. An exoplanet far away has its own observable universe, and can receive light from different places. Essentially, you are the center of your own observable universe. As more light reaches us, our observable universe will grow (or expand, if you will). If this is what you are referring to, then your answer is &lt;a href=&quot;https://astronomy.stackexchange.com/questions/14559/how-long-would-it-take-to-reach-the-edge-of-the-reachable-universe&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are referring to the latter, then there is a totally different answer and &quot;expansion&quot; refers to something completely different. The Universe (notice the capital &quot;U&quot;) is all of space and time and its contents. Anything that has existed, will exist, and currently exists is part of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Universe is thought to be infinitely large, so it can't have a center. The center of something is the point equidistance from the edges, but if something spans infinitely long, it would just keep going. It wouldn't have an edge, and thus it wouldn't have a center. You couldn't find the point equidistant from the edges if it just spans infinitely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Universe is not like a ball. Rather, you can think of it like a flat grid, and its &quot;expansion&quot; just means that the distances between objects on the grid are getting larger. In essence, more space is being created between the objects. That's what we mean by expansion — that objects are moving away from each other, since more space is being created between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an easy analogy: imagine you are walking your dog. Suddenly, the ground begins expanding between you. You and your dog will separated and continue receding away from each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's essentially happening everywhere: space is expanding between everything, so we are drifting away from other galaxies. The Universe is infinite, and we can constantly drift apart from other objects because space is being created in between us. Here's a GIF I made that might help you get it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/amQeZ.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/amQeZ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see how the galaxies drift apart as the space between them increases. And this happens everywhere in the Universe. The Universe is infinite, but more and more space is being created between matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fun fact: These objects can actually drift away from each other faster than the speed of light. That is, light from them eventually won't make it to us, since they'll be drifting away too quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this doesn't actually go against Einstein's theory that the speed of light is the fastest thing in the Universe. Einstein said that nothing can travel through space faster than light — but here, space itself is actually being created between the objects. Distances are increasing because space itself is dilating, and thus we can drift apart from other objects faster than light.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-06-04T23:01:13.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15112" PostTypeId="1" AcceptedAnswerId="15113" CreationDate="2016-06-04T18:02:29.797" Score="30" ViewCount="1503" Body="&lt;p&gt;I came across &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Sn_discoveries.gif&quot; rel=&quot;noreferrer&quot;&gt;this gif&lt;/a&gt; showing supernova discoveries from the late 19th century to 2010. Here's the data in 2010:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/ox3akzG.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that there's a prominent region shaped like an inverted U in which there are few detection points, which I have crudely enclosed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/TvrDXi8.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this? Is it an artifact of detectors simple not observing that part of the sky, or is there some underlying astronomical reason?&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2016-06-05T18:16:48.200" Title="Why is there a gap in this image of supernova discoveries?" Tags="&lt;supernova&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15113" PostTypeId="2" ParentId="15112" CreationDate="2016-06-04T18:32:42.270" Score="46" Body="&lt;p&gt;The coordinate system in this image is RA and Dec. It is a coordinate system which uses the Earth's equator (projected onto the sky) as its midline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inverted U is the Milky Way. The Milky Way is full of dust and gas, and blocks our view of galaxies (and supernovae) behind it. There is enough dust in the plane of the galaxy to block our view in that direction. For example the galaxy &lt;a href=&quot;http://apod.nasa.gov/apod/ap160129.html&quot;&gt;IC 342&lt;/a&gt; is one of the nearest galaxies, and would be brilliant if it were not close to the galactic plane. There may be other galaxies that are completely hidden.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our galaxy's bulk not only hides supernovæ that are in other galaxies, it also hides most of the supernovæ that occur in the Milky Way&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="823" LastEditDate="2016-06-05T18:16:48.200" LastActivityDate="2016-06-05T18:16:48.200" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15114" PostTypeId="2" ParentId="15110" CreationDate="2016-06-04T19:34:03.827" Score="2" Body="&lt;p&gt;Some supernovæ are more super than others, but that is not the reason that we have not observed one in 300 years. There have been 5 or 6 naked eye supernovae in the last 1000 years, There have certainly been others (such as Cassiopeia A and &lt;a href=&quot;https://en.wikipedia.org/wiki/G1.9%2B0.3&quot; rel=&quot;nofollow&quot;&gt;Supernova remnant G1.9+0.3&lt;/a&gt; But they have have been hidden by the dense dust and gas in the plane of the galaxy. Some regions of the centre of the galaxy have 30 magnitudes of extinction in visible light. That means that a SN with a magnitude of 0, would appear to have magnitude 30, and be essentially invisible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while we have been unlucky: we should probably expect to have seen a couple of supernovae since Kepler's star, the probability is not so very small.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-06-04T19:34:03.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15115" PostTypeId="2" ParentId="15110" CreationDate="2016-06-04T19:34:27.553" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So from this I can only come to the conclusion that supernovae did in&#xA;  fact happen in the Milky Way in the last two centuries, but that we&#xA;  didn't see any of them.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But now I'm confused. I read about supernovae that have happened in&#xA;  galaxies billions of light-years from here that lit up as the&#xA;  brightest source of light in the sky for days. Surely we would be able&#xA;  to see all of the supernovae that happen in our Milky Way&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One problem with estimating how many supernovae have happened in the Milky Way in the last two centuries, is that the remnant from a supernova is far dimmer than the nova itself and far harder to find. Several supernovae could have happened in the Milky Way in the last 2 centuries and remain undiscovered — and just for clarity, we define the date of a supernova by the date the light from the explosion reaches the Earth, so saying saying a nova &quot;happened&quot; in the last 200 years refers to the date the light from the event reached Earth, not the actual date of the event, which you probably already know, but just to clarify. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for argument's sake, let's say that a supernova's light reached the Earth about 50 years ago, but it took place on the far side of the galaxy. To find that, we'd have to look for a nebula on the far side of the galaxy and that's a hard thing to see. Similar to looking for the theoretical Planet Nine, finding old supernova remnants on the other side of the Milky Way takes a lot of looking. Even with modern telescopes, it's still a needle in a haystack, and especially if the view is blocked by dust like much of the Milky Way is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Milky Way supernova remnant was discovered in 1985, &lt;a href=&quot;https://en.wikipedia.org/wiki/G1.9%2B0.3&quot; rel=&quot;noreferrer&quot;&gt;supernova remnant G1.9+0.3&lt;/a&gt;. It was thought to have &quot;happened&quot; around 1868, though it went unobserved and probably wasn't visible to the naked eye at the time.   There's probably been several others more recent than that one.  G1.9+0.3 would have been visible if not for interstellar dust. From article above:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It was a type Ia supernova believed to have exploded about 25,000&#xA;  years ago, and the signal began reaching Earth around 1868. The light&#xA;  from the supernova would have been visible to 19th century&#xA;  astronomers, had it not been obscured by the dense gas and dust of the&#xA;  Galactic Center.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure the 4.6 supernovae per century from your article is accurate. It might be, but the number I'm used to hearing is about one per century. But regardless of which number is actually correct, it still doesn't imply high mathematical improbability because many Milky Way novae would have gone unnoticed if they were far enough away. In short, I agree with what you said here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I can only come to the conclusion that supernovae did in fact happen in the Milky Way in the last two centuries, but that we didn't see any of them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's a related &lt;a href=&quot;http://www.universetoday.com/120006/why-cant-we-see-the-center-of-the-milky-way/&quot; rel=&quot;noreferrer&quot;&gt;article on Milky Way visibility&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observation of a distant galaxy nova is possible if we have telescopes looking in that direction. A nova is much more detectable at the time it goes nova.  Much less so, years later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today, however, with &lt;a href=&quot;https://en.wikipedia.org/wiki/Supernova_Early_Warning_System&quot; rel=&quot;noreferrer&quot;&gt;neutrino detection in 7 locations around the globe&lt;/a&gt;, I think it's virtually impossible that we'd miss a supernova in the Milky Way, so we have a good chance of seeing one in our lifetime and it's likely that none have occurred since somewhere around 1980. A &lt;a href=&quot;https://en.wikipedia.org/wiki/SN_1987A#Neutrino_emissions&quot; rel=&quot;noreferrer&quot;&gt;supernova was detected in the Andromeda galaxy in 1987&lt;/a&gt; by that method, and our neutrino detection has improved since then to give us early warning and pinpoint location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as visibility, size matters, but what matters more is how close and how much dust is in the way. Most of the recorded supernovae were quite bright and fairly noticeable to someone who was familiar with the stars in the sky (&lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_supernovae#List&quot; rel=&quot;noreferrer&quot;&gt;list of known supernovae&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eight Milky Way novae have been recorded by history and observed by the naked eye in the last 2,000 years, well below the number that should have happened in that time. Five of those eight had brightness greater than $-3$, which is brighter than Jupiter and would have been immediately noticed by anyone familiar with star charting. Two others had magnitudes around zero, which is still brighter than most stars. &lt;a href=&quot;https://en.wikipedia.org/wiki/SN_386&quot; rel=&quot;noreferrer&quot;&gt;SN386&lt;/a&gt; was less bright, but is still easily visible and recorded by Chinese astronomers. And finally, &lt;a href=&quot;https://en.wikipedia.org/wiki/Cassiopeia_A&quot; rel=&quot;noreferrer&quot;&gt;Cassiopeia A&lt;/a&gt; was quite dim, but it was still observable. Only about 10% (ballpark estimate) of the Milky Way is close enough and unobstructed enough to provide supernovae that would visibly get noticed.  Most would have gone unnoticed until recently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that's not too wordy, I can try to clean up if needed.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="9301" LastEditDate="2016-06-04T22:48:26.247" LastActivityDate="2016-06-04T22:48:26.247" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15116" PostTypeId="2" ParentId="15110" CreationDate="2016-06-04T20:28:01.973" Score="7" Body="&lt;p&gt;Firstly, the figure you quote is disputed and must have very substantial uncertainties. The same wikipedia article also cites 3 per century.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, and more importantly, you have not accounted for dust obscuration. Most type II supernovae will occur in massive stars close to the Galactic plane and will be afflicted by huge amounts of extinction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume an peak absolute magnitude of about -18 and visual extinction of 2 mag per kpc in the galactic plane, then a supernova would have to be within a few kpc to be a notable naked eye object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider the Milky Way disc to be of radius 15kpc, then a horizon at say 3kpc only encompasses 4% of the possible volume. Thus if there are say 4 per century, only 0.16 per century would be notable naked eye objects.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-06-04T20:28:01.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15117" PostTypeId="2" ParentId="10252" CreationDate="2016-06-04T20:39:44.567" Score="3" Body="&lt;p&gt;At any one time, an average observer can see about 2,500 stars in a clear dark sky. Note that eyesight varies and sharp-eyed individuals may be able to see a half-magnitude dimmer stars than the average eye (&lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_magnitude&quot; rel=&quot;nofollow&quot;&gt;apparent magnitude&lt;/a&gt; is a scale in which each integer is $2.51$ ($100^{0.2}$) times brighter or dimmer than the next consecutive integer.) A very dark sky may enable magnitude $+7.5$ or even $+8$ stars to be seen, but in a typical &quot;dark&quot; non-urban sky the limit is often $+5.5$ to $+6.5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supernovae can potentially be seen as far as 13 billion light-years (ly) away, essentially from the 'edge' of our observable universe. So, it's not saying much to say that a supernova might be seen across a distance of 100,000 ly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some types of supernovae can be over $-22$ in &lt;a href=&quot;https://en.wikipedia.org/wiki/Absolute_magnitude&quot; rel=&quot;nofollow&quot;&gt;absolute magnitude&lt;/a&gt; — where absolute magnitude is defined as their apparent magnitude if they were observed from a 32.6 ly distance (to be honest, I'm not sure if supernovae absolute brightness is defined in exactly this same way). By way of comparison, the Sun's abs. mag. is $+4.8$ (lower numbers indicate brighter stars).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the dust and interstellar medium, the possible distance we can see is just a small fraction of the size of the Milky Way. In reality, few stars are bright enough to be seen over 400 ly away. Deneb which has an estimated distance of 2,600 ly (but may be as close as 1,550 ly, the large uncertainty is due to its variation in brightness). Only 6 visible stars are thought to be farther from us than 1,000 ly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to sum up, looking inwards into the Milky Way, our visibility is very restricted to the nearest 1,000–2,000 ly (while the MW disk's radius is 50,000 to 90,000 ly and we're about 27,000 ly on this side of center, but only the brightest stars are visible from more than about 400 ly away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While looking away, we can see the Andromeda Galaxy (but not its individual stars) which is 2,500,000 ly from us. In other words, most people blame dust for the poor visibility, but that's mostly only relevant for telescopes. For us, our limited eye-sight is the real barrier (not to mention the dearth of really dark skies).&lt;/p&gt;&#xA;" OwnerUserId="12059" LastEditorUserId="9301" LastEditDate="2016-06-04T22:08:04.913" LastActivityDate="2016-06-04T22:08:04.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15118" PostTypeId="1" AcceptedAnswerId="16160" CreationDate="2016-06-04T22:01:46.003" Score="3" ViewCount="249" Body="&lt;p&gt;The ancient Chinese story of the Weaver Girl (Altair) and the Cowherd (Vega) tell of starcrossed lovers who are only allowed to meet once a year on the 7th day of the 7th (Chinese) lunar month, roughly around early August, when magpies form a bridge across the silver river (the Milky Way.) This is celebrated in the Qixi festival in China (August 9th in 2016.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the timing, I'm suspicious that the magpies represent the Perseid meteor shower (August 12th in 2016.)  But I just got my first planisphere, and Altair and Vega are pretty far away from Perseus (although Perseus is still on the Milky Way.)  The Wiki article and others that I've looked into on Weaver Girl and Cowherd does not connect the magpies to any meteor shower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the Weaver Girl and Cowherd story is pretty old - up to 2600 years according to Wikipedia.  I don't think the Perseids have always been in Perseus due to the precession of the equinox, but 2600 years ago maybe the &quot;Perseid&quot; shower was closer to the southern triangle, or is that too far away to be explained only by precession of the equinox?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm very new to astronomy and I'm not Chinese, but I love this story and I'm looking for more information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="12060" LastEditorUserId="12060" LastEditDate="2016-06-05T13:38:15.437" LastActivityDate="2016-06-07T19:44:35.860" Title="Was the radiant of the Perseid meteor shower ever near Vega thousands of years ago?" Tags="&lt;amateur-observing&gt;&lt;history&gt;&lt;meteor-shower&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15119" PostTypeId="1" AcceptedAnswerId="16121" CreationDate="2016-06-05T00:52:21.817" Score="7" ViewCount="1148" Body="&lt;p&gt;This is a basic question, but it's been bugging me. In the Wikipedia article for &lt;a href=&quot;https://en.wikipedia.org/wiki/Lithium_burning&quot; rel=&quot;nofollow&quot;&gt;lithium burning&lt;/a&gt;, it states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Stars, which by definition must achieve the high temperature (2.5 × 10^6 K) necessary for fusing hydrogen, rapidly deplete their lithium. This occurs by a collision of lithium-7 and a proton producing two helium-4 nuclei. The temperature necessary for this reaction is just below the temperature necessary for hydrogen fusion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would imagine that with lithium having more protons, it would have a stronger Coulomb repulsion and require higher temperatures to fuse with hydrogen. Now, this article is pretty sketchy since it cites no sources, and I would normally dismiss this. But according to &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/304151/fulltext/35775.text.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, lithium burning occurs in protostars, before hydrogen fusion even takes place. How could lithium fuse with hydrogen at temperatures that low?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-06-05T13:08:00.630" LastActivityDate="2016-06-05T13:08:00.630" Title="Why does lithium fuse at lower temperatures than hydrogen?" Tags="&lt;star&gt;&lt;temperature&gt;&lt;stellar-astrophysics&gt;&lt;hydrogen&gt;&lt;nucleosynthesis&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16121" PostTypeId="2" ParentId="15119" CreationDate="2016-06-05T12:26:16.223" Score="7" Body="&lt;p&gt;The slowest reaction rate in the pp chain determines how quickly hydrogen can &quot;burn&quot; in the core of a sun-like star. That rate-determining step is actually the fusion of two protons to form deuterium via the diproton and a weak interaction decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fusion of lithium, whereby it fuses with a proton and then splits into two Helium nuclei is actually part of the PPII series of reactions that convert helium 3 into helium 4. It does not involve weak interactions and therefore the cross-section at a given temperature is much higher than the first step in the pp chain, regardless of the stronger coulomb repulsion between the reactants. It is therefore initiated at lower temperatures - as you say, Li &quot;burning&quot; occurs in low-mass pre main sequence stars well before hydrogen &quot;burning&quot; commences (but after deuterium burning).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-06-05T12:26:16.223" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16122" PostTypeId="1" AcceptedAnswerId="16145" CreationDate="2016-06-05T13:51:54.683" Score="3" ViewCount="1368" Body="&lt;p&gt;This VLA image is beautiful, and instructive, but I don't understand it fully. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CNqFY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CNqFY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This version is from an article in &lt;a href=&quot;http://gizmodo.com/jupiters-mysterious-interior-is-coming-into-focus-1779754843&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gizmodo&lt;/a&gt; where the credit reads: &lt;em&gt;A radio image of Jupiter captured by the Very Large Array. Image: Imke de Pater, Michael H. Wong, Robert J. Sault&lt;/em&gt; and which cites the recent article in Science: &lt;a href=&quot;http://science.sciencemag.org/content/352/6290/1198&quot; rel=&quot;nofollow noreferrer&quot;&gt;Peering through Jupiter’s clouds with radio spectral imaging&lt;/a&gt;. However, that paper does not actually show this image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another version of the image can be found in Berkeley News: &lt;a href=&quot;https://news.berkeley.edu/2016/06/02/new-radio-map-of-jupiter-reveals-whats-beneath-colorful-clouds/&quot; rel=&quot;nofollow noreferrer&quot;&gt;New radio map of Jupiter reveals what’s beneath colorful clouds&lt;/a&gt; and gives more information how the image was produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kR6Uf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kR6Uf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A radio image of Jupiter from the VLA at three wavelengths: 2 cm in blue, 3 cm in gold, and 6 cm in red. A uniform disk has been subtracted to better show the fine banded structure on the planet. The pink glow surrounding the planet is synchrotron radiation produced by spiraling electrons trapped in Jupiter’s magnetic field. Banded details on the planet’s disk probe depths of 30-90 km below the clouds. This image is averaged from 10 hours of VLA data, so the fine details seen in the other maps are smeared here by the planet’s rotation. Image by Imke de Pater, Michael H. Wong (UC Berkeley), Robert J. Sault (Univ. Melbourne).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would like to know more about how the image was processed. What does it look like &lt;em&gt;before&lt;/em&gt; the uniform disk is subtracted - is the radio emission from the region around the planet actually as strong as the planet disk itself, as this image suggests? If there is a reference for this image I'd also like to read it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the mean time, it &lt;em&gt;really is&lt;/em&gt; beautiful!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image is sometimes paired with the following image, taken here from &lt;a href=&quot;https://www.sciencenews.org/article/jupiter%E2%80%99s-stormy-weather-no-tempest-teapot&quot; rel=&quot;nofollow noreferrer&quot;&gt;Science News&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gOB6D.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gOB6D.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this image resolves lateral detail, I'm guessing it might be the same 10 hours of data, but processed with a moving frame and corrected for distortion as it wraps around during the rotation of the planet, while the global image above is smeared by rotation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was it? Where can I read more about this?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2016-06-05T14:07:52.143" LastActivityDate="2016-06-06T12:12:47.087" Title="What exactly is this VLA radio image of Jupiter showing?" Tags="&lt;jupiter&gt;&lt;radio-astronomy&gt;&lt;data-analysis&gt;&lt;magnetic-field&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16123" PostTypeId="2" ParentId="15118" CreationDate="2016-06-05T14:11:27.103" Score="2" Body="&lt;p&gt;Okay, this turned out to be quite a long answer describing a few arguments I've com up with. The short answer is: maybe, maybe not, but I think it's possible. There are arguments to be made for either side of this discussion, and I don't think you you can ever be absolutely sure unless you find an expert on both astronomy and Chinese folk tales (which I don't claim to be), or invent a time machine to go back 2600 years and check for yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most obvious argument to be made in favor of the hypothesis that these magpies are the Perseid meteor shower, is that folk tales referencing stars are very likely to be based on astronomical occurrences during that period (such as a meteor shower, perhaps).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The (in my mind) most obvious argument against your hypothesis is what you suggest yourself: Perseus is quite far away from Vega and Altair. However, Perseus is the radiant of the shower, which means that all perseids will appear to come out of Perseus. Because of this, it is perfectly possible for a lot of perseids to appear between Altair and Vega, parallel to the Milky Way. It is, however, impossible for them to go from Vega to Altair, or vice versa, since they wouldn't come out of Perseus anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You suggest that this could be explained by the precession of the equinox, but a quick lookup on Stellarium (which is a great (and free) tool if you're into astronomy) tells me that even 2600 years ago Perseus was quite far away from Vega and Altair (actually, that's quite obvious, since the precession of the equinox causes stars to move relative to the extension of the north pole, but not relative to each other, as far as I know).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another argument to be made against this hypothesis is that the perseids are by far not spectacular enough to be seen as a bridge over the Milky Way, even when they all appear in the required spot. During the yearly peak of the perseids, you are expected to see about 1-2 meteors per minute, which can be noticeable at best. This argument is then countered by the fact that the ZHR (how much meteor you see per hour) is not the same every year, but can peak on years when Earth passes through a fresh dust cloud left behind by the comet that causes the shower (which, in this case, is the comet Swift-Tuttle, if you're interested). Therefore, it is perfectly possible that the perseid meteor shower was extremely active 2600 years ago, causing the creation of this story. Such peaks are somewhat rare, but not too rare for this hypothesis to actually make sense (in fact, we will be seeing such a peak in 2028).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the dates of the festival and the peak of the perseids are a few days apart and not exactly the same is not quite an argument against your hypothesis, since it is perfectly possible for the date of the festival to have moved slightly due to differences between the (old) Chinese and Western calendars. I'm not an expert on calendars, however, so I can't say that for sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given all these uncertainties, your hypothesis (along with what you mentioned yourself in your own comments) seems plausible, apart from one thing: we don't know whether Swift-Tuttle was already active 2600 years ago. If it wasn't, the perseid meteor shower obviously did not exist (note that comets don't exist forever: some gravitational variation causes them to exit the Oort cloud and enter the solar system). We don't know when exactly Swift-Tuttle started being a comet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is only a weak argument, however, and when it comes to astromical hypotheses, the most convincing argument is usually a principle which is referred to as Ockham's razor: if given multiple explanations, choose the one that requires the least amount of assumptions (or the most plausible ones).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, since the date of creation (2600 years ago) is very vague (the uncertainty on this must be at least a hundred years), I think it is perfectly possible to assume that the perseids peaked somewhere during that period, causing the creation of this story. &lt;/p&gt;&#xA;" OwnerUserId="11824" LastActivityDate="2016-06-05T14:11:27.103" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16124" PostTypeId="1" CreationDate="2016-06-05T14:17:29.027" Score="4" ViewCount="195" Body="&lt;p&gt;Some X-ray sources in nebulae are caused by hot materials in the millions of degrees that formed in supernovae. What's the cooling rate of this stuff? How long does it stay this hot?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we would take our solar system for example, the time from the formation of the nebula (by a supernova?) until it was cool enough to form solid planet was only a few million years. That's very quick, for instance relative to the age of the Earth. How ephemeral are these X-ray sources? I would expect them to not last as long as IR sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an order of magnitude for the lifetime of the emission? Does it take millions of years? Tens of millions? Tens of thousands of years?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A comment suggests it doesn't really cool down, but diffuse to the interstellar medium. How long does it take for it to diffuse until there isn't a clear X-ray source?&lt;/p&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="13068" LastEditDate="2018-10-13T00:03:27.477" LastActivityDate="2021-01-22T00:18:44.793" Title="How long do hot X-ray sources last?" Tags="&lt;observational-astronomy&gt;&lt;spectra&gt;&lt;nebula&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="16126" PostTypeId="1" AcceptedAnswerId="16134" CreationDate="2016-06-05T16:51:09.883" Score="3" ViewCount="770" Body="&lt;p&gt;I am not sure whether this question is right or not so bear with me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand 'Sun rise' and 'Sun set' is caused by the Earth's rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read that 'Sun light' takes around 8 minutes to reach Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every city in the world has its own 'Sun rise and Sun set' timings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets assume the 'Sun rises' at 6 AM in 'City A'.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that 'City A' sees/feels the light of the Sun which was in the sky @ 5:52 AM ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, does 'Sun rise' and 'Sun set' timings change according to 'light years'?&lt;/p&gt;&#xA;" OwnerUserId="11000" LastEditorUserId="10437" LastEditDate="2016-06-06T11:44:35.097" LastActivityDate="2016-06-06T11:44:35.097" Title="Question related to Sun rise and Sun set times" Tags="&lt;solar-system&gt;&lt;the-sun&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16129" PostTypeId="1" CreationDate="2016-06-05T19:04:25.967" Score="13" ViewCount="3116" Body="&lt;p&gt;I just learned about Brown Dwarfs, they are &quot;failed&quot; stars, they narrowly missed the stellar mass mark. I learned that &lt;a href=&quot;http://www.space.com/12714-coldest-failed-stars-brown-dwarfs-wise.html&quot;&gt;Y Dwarfs have temperature as low as 80 Fahrenheit (The first one found by WISE observatory)&lt;/a&gt;, why aren't those not called a Jupiter-like planets without a host star? Why do we call them a star and designate a separate category for them (T or Y)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok suppose T and Y dwarfs are okay, let's leave them alone. But why can't Jupiter be a Y Dwarf who is in the binary relationship with the Sun? Sun-Jupiter's Barycenter is just outside the Sun, we could call them in orbit with each other correct?&lt;/p&gt;&#xA;" OwnerUserId="4099" LastActivityDate="2016-06-05T20:36:44.200" Title="Why can't our Sun be a binary with Jupiter as a T or Y dwarf?" Tags="&lt;jupiter&gt;&lt;binary-star&gt;&lt;brown-dwarf&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16130" PostTypeId="2" ParentId="16126" CreationDate="2016-06-05T19:21:02.967" Score="4" Body="&lt;p&gt;I love Einstein's explanation of &quot;events&quot;. Not the exact words but he stipulated that an event has not happened until the light of that event has reached you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the Sunrise time is when you actually see the disc of the Sun, not when it has actually been in your line of sight for a person who might be standing on the surface of Mercury observing you and Sun both.&lt;/p&gt;&#xA;" OwnerUserId="4099" LastEditorUserId="4099" LastEditDate="2016-06-05T20:07:32.727" LastActivityDate="2016-06-05T20:07:32.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16131" PostTypeId="1" AcceptedAnswerId="16138" CreationDate="2016-06-05T19:52:38.803" Score="3" ViewCount="80" Body="&lt;p&gt;I have access to my university's telescope, &lt;a href=&quot;https://en.wikipedia.org/wiki/Dearborn_Observatory&quot; rel=&quot;nofollow&quot;&gt;Dearborn Observatory&lt;/a&gt;, an &lt;strong&gt;18.5 inch refractor&lt;/strong&gt; on the shore of Lake Michigan, just north of Chicago (yes, it's an atrocious location, but the telescope still works fantastically given this), and I was wondering if it would be possible to detect exoplanet transits.  I have access to a CCD.  Obviously, normally this would take quite a while, but I was looking at &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_exoplanets&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's list of transiting exoplanets&lt;/a&gt; and several of them have periods of less than 10 hours or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any good resources about the math for this?  Or is the approximation that size of the change in light observed by the transit is just using the ratio of the area that the planet occludes to the area of the host star a reasonably good approximation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is observing the transits of any known exoplanets completely unfeasible with this size of telescope?  From previous nights of photographing the sky, it seems like the telescope's limit is roughly magnitude 13-15, beyond which the signal-to-noise ratio is just too low to get meaningful data.&lt;/p&gt;&#xA;" OwnerUserId="9504" LastEditorUserId="9504" LastEditDate="2016-06-06T05:57:51.353" LastActivityDate="2016-06-06T05:57:51.353" Title="Capability of observing transits with terrestrial telescopes of various sizes?" Tags="&lt;exoplanet&gt;&lt;planetary-transits&gt;&lt;refractor-telescope&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16132" PostTypeId="1" AcceptedAnswerId="16146" CreationDate="2016-05-03T15:44:27.260" Score="16" ViewCount="716" Body="&lt;p&gt;As a general science (hence Wikipedia!) reader, the latest information I have is that &lt;strong&gt;60&lt;/strong&gt; galaxies &quot;have been found and counted&quot; inside the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bo%C3%B6tes_void&quot; rel=&quot;noreferrer&quot;&gt;Boötes void&lt;/a&gt;, this of 1997.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) What is the latest count of galaxies &quot;known&quot; inside the Boötes void?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Boötes void is a ball 250 million light-years across. (Say, .002 of the width of the visible universe?)  It seems - unbelievable - it has only &lt;em&gt;60&lt;/em&gt; galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;60&quot; value (or some new number since 1997), maybe I am misunderstanding it, or it is misreported, or has a different meaning?  Galaxies are counted in the thousands or tens of thousands, it seems hard to believe in a cosmological space, the number &quot;60&quot; would come up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what is (2) the thinking on how &lt;strong&gt;many galaxies there likely are&lt;/strong&gt; in the Boötes Void? An order of magnitude.  Is it &quot;about 100,&quot; or &quot;about a trillion,&quot; or what?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you (3) normally quote galaxy density in?  Galaxies per Gpc&lt;sup&gt;3&lt;/sup&gt;, or?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, I am trying to get a (4) &lt;strong&gt;comparative sense of galactic density in the Boötes void versus elsewhere&lt;/strong&gt;. What is the galaxy density in the Boötes void, versus the universe in general, or if it's more meaningful versus perhaps superclusters, walls, or?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it &quot;one tenth as dense&quot;?  One zillionth?  One half?  Just how void is that sucker?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Further...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BruceS below has usefully pointed out that, indeed, the &quot;60&quot; galaxies are anyway &lt;em&gt;only in a small tube which apparently cuts through&lt;/em&gt; the Boötes Void.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that, incredibly, the actual main body of the Boötes Void - putting aside that tube - &lt;strong&gt;&lt;em&gt;is literally empty, no galaxies at all&lt;/em&gt;&lt;/strong&gt;??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That seems totally incredible.  I feel we really need a galaxy-counting astronomer to clarify the nature of the Boötes Void.&lt;/p&gt;&#xA;" OwnerUserId="13071" OwnerDisplayName="Joe Blow" LastEditorUserId="9301" LastEditDate="2016-06-29T04:38:57.290" LastActivityDate="2016-06-29T04:38:57.290" Title="Boötes Void numbers, 2016" Tags="&lt;galaxy&gt;&lt;cosmology&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16133" PostTypeId="2" ParentId="16132" CreationDate="2016-06-05T19:47:41.467" Score="2" Body="&lt;p&gt;First, notice that the distribution of galaxies in the universe is not uniform. Not only they are located mostly along filaments or walls, but also the density among them varies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far 60 (up to 2013 &lt;a href=&quot;http://asd.gsfc.nasa.gov/blueshift/index.php/2013/07/30/jasons-blog-next-stop-voids/&quot; rel=&quot;nofollow&quot;&gt;http://asd.gsfc.nasa.gov/blueshift/index.php/2013/07/30/jasons-blog-next-stop-voids/&lt;/a&gt;) galaxies have been discovered in the Boötes Void and all of those are found in a tube shape running through the void. If we are to use a rough estimate of about 1 galaxy every 10 million light years (4 times farther than Andromeda) there should be approximately 2,000 galaxies in the Boötes Void. So a good order of magnitude estimate would be that the galaxy density is about 3% that of a &quot;typical&quot; region.&lt;/p&gt;&#xA;" OwnerDisplayName="bruce smitherson" LastActivityDate="2016-06-05T19:47:41.467" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16134" PostTypeId="2" ParentId="16126" CreationDate="2016-06-05T20:35:03.623" Score="2" Body="&lt;p&gt;Sunrise and sunset times are based on when an observer at the location in question would see the first or last limb of the sun appear or disappear; they do not take into consideration the light-speed delay between the Earth and the sun. They do, however, take into consideration the average refraction of the atmosphere, so sunrise occurs a bit earlier than plain line-of-sight would predict as would sunset occur a bit later.&lt;/p&gt;&#xA;" OwnerUserId="7108" LastActivityDate="2016-06-05T20:35:03.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16135" PostTypeId="2" ParentId="16129" CreationDate="2016-06-05T20:36:44.200" Score="30" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But why can't Jupiter be a Y Dwarf who is in the binary relationship with the Sun?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are two reasons: One is that Jupiter is too small to have ever undergone fusion of any sort. To qualify as a brown dwarf, an object needs to be large enough to have undergone deuterium fusion in its core. This requires a mass of at least 13 Jupiter masses. The other is that Jupiter formed by planetary formation mechanisms rather than by gravitational collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is debate on what to call objects of less than 13 Jupiter masses that formed by gravitational collapse. Are these sub brown dwarfs or free-floating gas giants, and does it even matter? There is also debate on what to call objects above 13 Jupiter masses that formed by planetary formation mechanisms as opposed to by gravitational collapse. Are these brown dwarfs or hyper planets, and once again, does it even matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, Jupiter is not a brown dwarf.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2016-06-05T20:36:44.200" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16136" PostTypeId="2" ParentId="16126" CreationDate="2016-06-05T20:42:48.893" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I understand 'Sun rise' and 'Sun set' is caused by 'earth rotation'.&#xA;  I read 'Sun light' takes around 8 minutes to reach earth&#xA;  Every city in this world has its own 'Sun rise and sun set' timings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is all correct. The apparent motion of the sun in the sky is mostly due to the rotation of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Lets assume 'sun arises' at 6 AM in 'City A'.&#xA;  Does it mean that 'City A' sees / feels the light of Sun which was in the sky @ 5:52 AM ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It means that at 6AM, light from the sun shines on city A. That light had left the sun at about 5:52. At 5:52, city A was in the shadow of the Earth. The time that the light left the sun is fairly irrelevant to the timing of sunrise. A 5:52 the light left the sun, in the 8 minutes that followed the Earth rotated by about 2 degrees and by 6 AM city A was just in the right position to catch a little of the light. We don't change the timings. We would not say that the sun rose at 5:52, because at 5:52 the sun appears to be below the horizon. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-06-05T20:42:48.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16138" PostTypeId="2" ParentId="16131" CreationDate="2016-06-05T22:42:59.693" Score="4" Body="&lt;p&gt;A bright, nearby star with a large exoplanet would be best. For example, the first star to have a transiting planet observed was &lt;a href=&quot;https://en.wikipedia.org/wiki/HD_209458&quot;&gt;HD 209458&lt;/a&gt;, in Pegasus. It has a magnitude of 7.65.  When the large &quot;hot jupiter&quot; transits the star it dims by a &lt;a href=&quot;https://ned.ipac.caltech.edu/level5/March03/Livio/Livio3_2.html&quot;&gt;relative flux of 0.984&lt;/a&gt;. That corresponds to a change in magnitude of about 0.016. I.e. it changes from a magnitude 7.65 star to a 7.67 star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know the radii you can calculate the relative flux during a transit by assuming the planet blocks all the light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radius of the planet HD209458b is about 100000km and the star 800000km, so the planet has an area 1/64 of the area of the star's disc, and as expected 1-1/64 = 0.984: which is in good agreement with the light curve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That won't be noticeable with the naked eye, and is challenging for an amateur. A couple of links: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.britastro.org/vss/ccd_photometry.htm&quot;&gt;http://www.britastro.org/vss/ccd_photometry.htm&lt;/a&gt; general information on photometry. (summary: you need some kit)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://brucegary.net/book_EOA/EOA.pdf&quot;&gt;http://brucegary.net/book_EOA/EOA.pdf&lt;/a&gt; (summary: it can be done, with a 14 inch telescope and Arizonan dark skies.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-06-05T22:42:59.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16139" PostTypeId="2" ParentId="14921" CreationDate="2016-06-05T23:41:10.687" Score="1" Body="&lt;p&gt;I searched the same thing and Wolfram returned the following note:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Assuming &quot;earth&quot; is a word | Use as a planet instead&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And so I did, and when searching for the distance from earth to the sun using earth as a planet, Wolfram returned a list of earth's orbital parameters in the bottom of the page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this list of parameters, I clicked in the link &quot;sources&quot; to see where this data was coming from. Among the sources was &lt;a href=&quot;http://ssd.jpl.nasa.gov/&quot; rel=&quot;nofollow&quot;&gt;this JPL website&lt;/a&gt; with all sort of solar system data you can imagine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think that Wolfram either connects to JPL's website to get the sun-earth distance in real time or uses their tables to calculate it.&lt;/p&gt;&#xA;" OwnerUserId="13074" LastActivityDate="2016-06-05T23:41:10.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16140" PostTypeId="1" CreationDate="2016-06-05T20:18:30.107" Score="13" ViewCount="1729" Body="&lt;p&gt;I'm not sure if this post should be in the physics subject forum, but this seems to fit here too. I have been reading a book about chemistry and how the universe came to be with the theory of stars creating most of the elements we know of. I began wondering how the Earth came to be in orbit around the Sun. I understand that we are in a constant free fall towards the Sun, but our motion causes us to &quot;miss&quot; the Sun stopping us from burning up. However, I'm confused how we were put into motion in the first place. My best guess is that we were captured by the Sun's gravity, and were put into orbit, which would mean the Earth was once a meteor, right? If someone could please explain this, it'd be very helpful because I can't seem to find the answer anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I'm just a 9th grader, so I don't have much knowledge in astrophysics.&lt;/p&gt;&#xA;" OwnerDisplayName="bobthebuilder" LastEditorUserId="10437" LastEditDate="2016-06-06T11:37:49.533" LastActivityDate="2018-08-07T19:50:37.217" Title="How did the Earth come to be in orbit around the sun?" Tags="&lt;solar-system&gt;&lt;earth&gt;&lt;planetary-formation&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16141" PostTypeId="2" ParentId="16140" CreationDate="2016-06-05T22:15:32.190" Score="2" Body="&lt;p&gt;Earth used to be many, many asteroids and meteorites. The larger asteroids would pull meteorites and smaller asteroids, crash with them, and bundle into a bigger and bigger clump, that eventually became so heavy, that rock began behaving like liquid under the pressures, forming the neat sphere Earth is now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That all rubble was already orbiting the Sun when that happened; some of it arrived from outside the Solar System, but most was just a nebula of gases and other matter that coalesced into the current solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for &quot;falling and missing&quot;... that's a semi-accurate description of orbital mechanics. Taking the distances into account, &quot;missing&quot; the central body is actually quite easy; look at the sky at night - some of the brighter stars are actually other planets in our system. That's it - these tiny dots. The Sun is bright, but it's also tiny in the sky. There's a lot of room to &quot;miss&quot;, and instead of crashing, to fly in an elliptic trajectory around the central body - passing it by, faster when you're close, then getting ejected &quot;around the other side&quot; to slow down, and fly away, only to return down the same (missing) path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now... how comes Earth's, and most planets' orbits are not elongated ellipses, as is most common with most randomly moving objects in gravitational field - but pretty near circles - that's a subject for a different question and a longer debate.&lt;/p&gt;&#xA;" OwnerUserId="83" OwnerDisplayName="SF." LastActivityDate="2016-06-05T22:15:32.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16142" PostTypeId="2" ParentId="16140" CreationDate="2016-06-05T22:21:49.303" Score="2" Body="&lt;p&gt;Planetary systems are &lt;a href=&quot;https://en.wikipedia.org/wiki/Formation_and_evolution_of_the_Solar_System&quot; rel=&quot;nofollow&quot;&gt;formed from clouds of gas and dust&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravity of the cloud's mass holds it together; the densest part at the center of the cloud collapses until it's dense enough to begin nuclear fusion, turning into a star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things further out randomly collide and stick until a few of them become big enough to have significant gravity of their own. When those &quot;planetesimals&quot; collide with each other, on average their orbits become more circular. After billions of years of this, you wind up with relatively few bodies in nearly circular orbits: the planets of a solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Capture of bodies arriving from outside the solar system does happen, but formation from a proto-stellar gas cloud is the more common way planets are formed.&lt;/p&gt;&#xA;" OwnerUserId="2573" OwnerDisplayName="Russell Borogove" LastActivityDate="2016-06-06T00:36:06.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16143" PostTypeId="2" ParentId="16131" CreationDate="2016-06-06T04:53:49.460" Score="4" Body="&lt;p&gt;An 18.5 inch telescope is easily large enough to collect the photons you require. So long as you have a good CCD camera, with a pixel size on the sky that is considerably smaller than the &quot;seeing&quot; and you take a great deal of care in how you make and analyse the observations. Exposure times can be minutes if necessary, because a typical transit lasts a couple of hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I think is necessary though, is to pick targets where there is another star of similar (or maybe you can get away with 1-2 mag fainter) brightness in the same CCD field of view as the target. This can be used (so long as it isn't a variable on hour timescales) to compensate for changes in atmospheric conditions and extinction.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-06-06T04:53:49.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16144" PostTypeId="2" ParentId="16140" CreationDate="2016-06-06T09:58:48.140" Score="22" Body="&lt;p&gt;The picture was so much cleaner 20 to 25 years ago. I'll present that nice clean picture first. Stars form from the gravitational collapse of huge clouds of interstellar gas. Those gas clouds inevitable have some net non-zero angular momentum. This forces the gas cloud to change shape from being more or less spherical to being disk-shaped. (Why? That's a different question. Ask it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this protoplanetary disk continued to feed mass to the growing protostar, it also set the stage for the formation of planets. The gas cloud was mostly primordial hydrogen and helium, but it also contained heavier elements thanks to stellar fusion and supernovas in the billions of years that preceded the formation of our solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those heavier elements behave quite differently than do hydrogen and helium. They have chemistry. The planets started as microscopic clumps of mass of these heavier elements, bound together chemically. These microscopic clumps occasionally collided, eventually forming larger clumps of mass. These larger clumps in turn collided with one another, forming even larger clumps of mass. Eventually the clumps became large enough that they interacted gravitationally, making them grow even larger. This process continued, eventually forming protoplanets, and then planetary embryos, and finally planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Temperature in the protoplanetary disk were high near the forming protostar but dropped precipitously with increasing distance from the protostar. At some point, volatiles such as water, ammonia, methane, and carbon dioxide become as solid as rock. This is the ice line, aka the snow line or frost line. Asteroids inside of Ceres' orbit tend to be rocky. Asteroids outside of Ceres' orbit tend to be icy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets that form outside the ice line can grow very quickly and then they can grow very, very large. The stuff that comprises the protoplanetary disk orbits the growing protostar at something other than the rate suggested by Kepler's laws thanks to the pressure of all that stuff in the disk. Thanks to the square-cube law, larger objects aren't as subject to that pressure. Those larger objects orbit at a Keplerian rate. Planets that form outside the ice line grow very quickly and then sweep up gas and ice because they are orbiting at a different speed than the immediate surroundings. The result is gas giants such as Jupiter and Saturn and further out, ice giants such as Uranus and Neptune. Planetary growth is a much more difficult process and much slower process inside the ice line. That's why Mercury, Venus, the Earth, and Mars are rocky and so much smaller than Jupiter, Saturn, Uranus, and Neptune.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;That's the pretty picture. The not so pretty picture:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Why are Mercury and Mars so much smaller than Venus and the Earth?&lt;br&gt;&#xA;Simulations suggest that the rocky planets should all be more or less the same size. That is not the case in our own solar system, let alone elsewhere.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How could Uranus and Neptune have formed?&lt;br&gt;&#xA;Simulations cannot recreate Uranus and Neptune at their current distances from the Sun. The material in the protoplanetary disk should have been too sparse at those distances to form large planets.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Much, much worse, what's the deal with all the weird exoplanets scientists have found?&lt;br&gt;&#xA;Scientists have found Jupiter-sized objects orbiting very close to their sun, Neptune-sized objects orbiting where the simple model would have only rocky planets forming, and planets in highly inclined (and sometimes retrograde) orbits that don't make sense.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These simulations (which have become to be very good) and the plethora of exoplanets have pushed the theory of how planets form back into the “that’s funny” stage. (&quot;The most exciting phrase to hear in science, the one that heralds new discoveries, is not “Eureka!” but “That’s funny...”&quot;, a quote widely attributed to Isaac Asimov.)&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2016-06-06T10:04:17.603" LastActivityDate="2016-06-06T10:04:17.603" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16145" PostTypeId="2" ParentId="16122" CreationDate="2016-06-06T12:12:47.087" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A uniform disk has been subtracted to better show the fine banded structure on the planet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This means the surface of the planet is far brighter (in the raw data) than the surrounding magnetosphere. The planet itself has been darkened considerably in the image, otherwise it would be a pale circle and show little or no detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://public.nrao.edu/news/pressreleases/vla-jupiter-atmosphere&quot; rel=&quot;nofollow&quot;&gt;press release&lt;/a&gt; from the VLA's site itself doesn't even show the wide field image however; it concentrates on the image showing detail of the spot itself. It mentions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;an innovative data-reduction technique to &quot;unsmear&quot; the data from many hours of observing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is an &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0612769v1&quot; rel=&quot;nofollow&quot;&gt;earlier 2006 paper&lt;/a&gt; by some of the same researchers showing earlier attempts, with a much more coarse radio map of Jupiter. This goes into more detail about the earlier image processing techniques though (requires good knowledge of Fourier analysis etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: I can't read the current &lt;em&gt;Science&lt;/em&gt; paper in full, it's behind a paywall.)&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2016-06-06T12:12:47.087" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16146" PostTypeId="2" ParentId="16132" CreationDate="2016-06-06T12:39:17.750" Score="9" Body="&lt;p&gt;An &lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/nph-abs_connect?db_key=AST&amp;amp;db_key=PRE&amp;amp;qform=AST&amp;amp;arxiv_sel=astro-ph&amp;amp;arxiv_sel=cond-mat&amp;amp;arxiv_sel=cs&amp;amp;arxiv_sel=gr-qc&amp;amp;arxiv_sel=hep-ex&amp;amp;arxiv_sel=hep-lat&amp;amp;arxiv_sel=hep-ph&amp;amp;arxiv_sel=hep-th&amp;amp;arxiv_sel=math&amp;amp;arxiv_sel=math-ph&amp;amp;arxiv_sel=nlin&amp;amp;arxiv_sel=nucl-ex&amp;amp;arxiv_sel=nucl-th&amp;amp;arxiv_sel=physics&amp;amp;arxiv_sel=quant-ph&amp;amp;arxiv_sel=q-bio&amp;amp;sim_query=YES&amp;amp;ned_query=YES&amp;amp;adsobj_query=YES&amp;amp;aut_logic=OR&amp;amp;obj_logic=OR&amp;amp;author=&amp;amp;object=&amp;amp;start_mon=&amp;amp;start_year=&amp;amp;end_mon=&amp;amp;end_year=&amp;amp;ttl_logic=AND&amp;amp;title=bootes+void&amp;amp;txt_logic=OR&amp;amp;text=&amp;amp;nr_to_return=200&amp;amp;start_nr=1&amp;amp;jou_pick=NO&amp;amp;ref_stems=&amp;amp;data_and=ALL&amp;amp;group_and=ALL&amp;amp;start_entry_day=&amp;amp;start_entry_mon=&amp;amp;start_entry_year=&amp;amp;end_entry_day=&amp;amp;end_entry_mon=&amp;amp;end_entry_year=&amp;amp;min_score=&amp;amp;sort=SCORE&amp;amp;data_type=SHORT&amp;amp;aut_syn=YES&amp;amp;ttl_syn=YES&amp;amp;txt_syn=YES&amp;amp;aut_wt=1.0&amp;amp;obj_wt=1.0&amp;amp;ttl_wt=0.3&amp;amp;txt_wt=3.0&amp;amp;aut_wgt=YES&amp;amp;obj_wgt=YES&amp;amp;ttl_wgt=YES&amp;amp;txt_wgt=YES&amp;amp;ttl_sco=YES&amp;amp;txt_sco=YES&amp;amp;version=1&quot; rel=&quot;noreferrer&quot;&gt;ADS search&lt;/a&gt; on refereed astronomical papers including the words &quot;Boötes Void&quot; in the title returns only two papers in this millennium, in 2001 and 2002, and even these do not report any new data, but use data from the beginning of the 1990's. I can't find any newer references for the number density of galaxies in the Boötes Void in particular, but a typical value is roughly one-tenth of the average number density in the Universe.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Theoretical approach&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I previously &lt;a href=&quot;https://astronomy.stackexchange.com/questions/8903/how-many-galaxies-are-there-in-the-hercules-corona-borealis-great-wall&quot;&gt;answered the similar question&lt;/a&gt; of the number of galaxies in an &lt;em&gt;over&lt;/em&gt;density. This answer is based on an anlytical fit to (observationally calibrated) simulations of galaxy halo formation, giving the so-called &lt;em&gt;halo mass function&lt;/em&gt;, i.e. the number of galaxy halos per halo mass. The total number $N_\mathrm{gal}$ of galaxies in a volume $V$ can be expressed as&#xA;$$&#xA;N_\mathrm{gal} = n_\mathrm{gal} \times V \times \delta,&#xA;$$&#xA;where $n_\mathrm{gal}$ is the average number density of galaxies in the Universe, and $\delta$ is the relative density in the volume. As I wrote in the other answer, $n_\mathrm{gal}$ is a number that impels you to define a lower threshold of galaxy size. The reason is that the lower you go, the more there are, and there is no formal threshold for how larger a clump of stars you need before you call it a galaxy. But as in the other answer, for the sake of this discussion we may use Small Magellanic Cloud-sized galaxies as our lower minimum. In that case, with $\delta\sim0.1$ and $V\sim236,\!000\,\mathrm{Mpc}^3$, the total number becomes&#xA;$$&#xA;N_\mathrm{gal,Böotes} = 0.17\,\mathrm{Mpc}^{-3} \, \times \, 236,\!000\,\mathrm{Mpc}^3 \, \times \, 0.1 \, \simeq \, 4000\,\mathrm{galaxies}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;strong&gt;Units of galaxy number density&lt;/strong&gt;&#xA;&#xA;&lt;p&gt;The way I write this result also answers your question 3: Galaxy number densities are almost always written in $\mathrm{Mpc}^{-3}$. In theoretical/numerical work, you'll often see the factor $h^3$ in front of the unit. This is simply the Hubble constant divided by 100 (i.e. $h=0.7$ for $H_0 = 70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$), allowing people to compare results more easily without knowing the exact value of $H_0$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Observational approach&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The observations of the Boötes Void are old, and seem to have been carried out on 1m-class telescopes. Hence, they won't be able to observe the smallest galaxies. In addition to telescope specifications, weather, etc., the exact detection limit (in terms of a limiting magnitude $m_\mathrm{lim}$) depends on how long they integrate (i.e. expose). Without reading the old papers in detail, I can't say what this is, but a typical value for such surveys would be, very approximately, $m_\mathrm{lim} \sim 20$ (if someone has a more realistic value than this, please edit). That is, objects fainter (i.e. &lt;em&gt;larger&lt;/em&gt; values due to the backward astronomical system) than $m=20$ would be missed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance to the Boötes Void implies a &lt;a href=&quot;https://en.wikipedia.org/wiki/Distance_modulus&quot; rel=&quot;noreferrer&quot;&gt;distance modulus&lt;/a&gt; of $\mu\sim37$, so the minimum absolute magnitude is&#xA;$$&#xA;M_\mathrm{lim} = m_\mathrm{lim} - \mu \simeq -17,&#xA;$$&#xA;which is somewhere between the Small and the Large Magellanic Cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure below (from &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0411364&quot; rel=&quot;noreferrer&quot;&gt;Wyder et al. 2005&lt;/a&gt;) shows the local-Universe &lt;a href=&quot;https://en.wikipedia.org/wiki/Luminosity_function&quot; rel=&quot;noreferrer&quot;&gt;luminosity function&lt;/a&gt; for UV-selected galaxies.&#xA;That is, it shows the number density of galaxies at a given magnitude. For instance, it shows (with the green dashed line) that the number density (here called $\Phi$) of galaxies with magnitudes around $M=-17$ is roughly $10^{-2.5}\sim0.003\,\mathrm{Mpc}^{-3}\,\mathrm{mag}^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tIoX5.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tIoX5.png&quot; alt=&quot;UVLF&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating over magnitudes from the $M=-17$ doesn't change the 0.003 much, since the density of brighter galaxies quckly declines. I get 0.004, i.e. the number density of galaxies at least as bright as $M=-17$ is $0.004\,\mathrm{Mpc}^{-3}$, smaller than the theoretical result above by 1.5 orders of magnitude. Multiplying this by the volume $V$ and the relative density $\delta$ yields $N_\mathrm{gal,Böotes} \simeq 100$ galaxies, not far from the 60 you quote.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To conclude, the number 60 seems in rough accordance with what is expected observationally, but theoretically, we'd expect there to be many more galaxies (although they're very small).&lt;/p&gt;&#xA;&#xA;&lt;strong&gt;Location of the galaxies&lt;/strong&gt;&#xA;&#xA;&lt;p&gt;They detected galaxies seem to be lying in a &quot;tube&quot; extending across the void. In general, galaxies and the underlying dark matter mass field tend not to be evenly distributed, but to form knots, sheets and filaments, separated by voids. My guess is that this &quot;tube&quot; is such a filament. Outside this filament, the void is voidier, but not completely void. There will still be galaxies, although few and small.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-06-07T10:26:02.763" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="16147" PostTypeId="1" CreationDate="2016-06-06T13:42:40.827" Score="11" ViewCount="352" Body="&lt;p&gt;As I was curious about the exact identical comment to my title here: &lt;a href=&quot;https://astronomy.stackexchange.com/a/16135/10102&quot;&gt;https://astronomy.stackexchange.com/a/16135/10102&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as it seems no one asked it yet, I'm simply interested in the difference of the both, so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the distinction between &quot;planetary formation mechanisms&quot; and &quot;gravitational collapse&quot;? &lt;/p&gt;&#xA;" OwnerUserId="10102" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-06-09T16:17:09.557" Title="What's the distinction between &quot;planetary formation mechanisms&quot; and &quot;gravitational collapse&quot;?" Tags="&lt;planetary-formation&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16148" PostTypeId="1" AcceptedAnswerId="16149" CreationDate="2016-06-06T15:46:19.367" Score="6" ViewCount="548" Body="&lt;p&gt;I have recently come across this so called &quot;Red Supergiant problem&quot; in the literature, a phrase that was coined by &lt;a href=&quot;https://arxiv.org/abs/0809.0403&quot;&gt;Stephen Smartt in 2009&lt;/a&gt; in reference to why red supergiants with masses ∼16-30M⊙ have not been identified as progenitors of Type IIP supernovae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it astronomers have observed red supergiants with masses between 8-25M⊙ but only those below ∼16M⊙ have been proven to have undergone a Type IIP supernovae. With the minimum mass required to produce a Wolf-Rayet star at least 25-30M⊙ these red supergiants arent massive enough to be evolving into WR stars either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There could be systematic errors with the models or calculations and it is well known that calculating the mass of a supernova progenitor post-eruption can be quite a difficult task, but assuming these calculations are correct it leaves a big hole in our understanding of stellar evolution. So my question is, what can explain this gap?&lt;/p&gt;&#xA;" OwnerUserId="10437" LastActivityDate="2016-06-06T20:38:48.580" Title="What are the possible solutions to the Red Supergiant problem?" Tags="&lt;supernova&gt;&lt;stellar-evolution&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16149" PostTypeId="2" ParentId="16148" CreationDate="2016-06-06T20:38:48.580" Score="6" Body="&lt;p&gt;Solutions to the Red Supergiant Problem can be either observational or physical, and to date, both types have been proposed. Recent data and improved computer modeling have both helped make the later stages of red supergiant evolution clearer, leading to possible solutions to the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2012IAUS..279..419W&quot;&gt;Walmswell &amp;amp; Eldridge (2012)&lt;/a&gt; suggested an observational solution, namely, that circumstellar dust resulting from mass loss via the strong stellar winds of a red supergiant could lead to strong extinction in some cases, making it hard or impossible to detect the resulting supernovae. Based on data collected on 18 supernovae, they found an upper limit for the mass of a Type IIP  supernova progenitor (at 90% confidence) to be around $27 M_{\odot}$, which would explain the dearth of detections of progenitors in the $16$-$30 M_{\odot}$ range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The observational difficulties are moot if there is a physical reason behind the problem, i.e. if there is some mechanism preventing these red supergiants from undergoing Type IIP supernovae. &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/2041-8205/717/1/L62/meta&quot;&gt;Yoon &amp;amp; Cantiello (2010)&lt;/a&gt; suggested that red supergiant stellar winds could be increased to &quot;superwinds&quot; by envelope pulsations. This would lead to periods of larger mass loss rates than expected, followed by periods of smaller mass loss rates. This would correspond to oscillating around states of instability due to the pulsations, and would be prominent only in red supergiants of masses greater than or equal to about $19M_{\odot}$. The result could be drastic, with stars losing well over half their masses. The star could then evolve toward a Type Ib or Type IIb supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smartt et al. did consider the possibility of envelope loss in the paper you cited, but dismissed it as being unfeasible for a star in the mass range. However, it appears that they did not consider the superwinds induced by pulsations, making this solution feasible once more. It's also worth mentioning that the authors suggested direct collapse to black holes with at the most a very faint supernova; this idea had been explored in the past.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2016-06-06T20:38:48.580" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16150" PostTypeId="2" ParentId="15078" CreationDate="2016-06-06T21:21:50.707" Score="6" Body="&lt;p&gt;Great question! Sorry for this huge response, and it might not be a satisfying answer, but it'll address your questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sadly, as with most of astronomy, the Big Bang is surrounded in mystery. It is one of the biggest uncertainties out there, and you'll come to this realization if you've researched it enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of what we can get out of the Big Bang is through extrapolation. If we observe the expansion of the Universe, we can presume that at one point, there was far less space between all the matter in the Universe. This is where things begin to break down, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Through General Relativity — which explains our modern understanding of gravity — and our extrapolation, we would reach the phenomena you're describing; the entire Universe would have originated as an infinitely small point (called a singularity), from which matter was incredibly hot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, General Relativity begins to break down at those levels, since its equations cannot explain the conditions of such a universe accurately. For one, as you hint at, an infinite density and potentially infinite mass would likely have an infinitely strong gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no place for such infinities in a proper mathematical description of the universe, and at that point, Einstein's equations — the basis for predicting the evolution of the cosmos — collapse. These equations provide us with unreliable results and we run into some of the problems you are describing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum mechanics is more fit for such small scales, but it fails to explain how gravity acts and what its role was at the time. Thus, this tells us that QM and GR are necessarily incomplete. Their equations cannot describe the Big Bang singularity in its entirety, especially since GR (which deals with large, high-mass phenomena) and QM (which deals with small, low-mass phenomena) are incompatible and cannot describe small, high-mass phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we're left with a lot of unanswered questions: did the Big Bang create the Universe, or was it simply an event in the Universe's history? Why would the Universe be compressed to such a small scale? Would the idea of an infinitely dense universe go against the presumption of an infinite universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There has long been hope that these questions can be explained within a theory of quantum gravity. Quantum gravity attempts to unify general relativity and the concepts of quantum theory (QM and QFT). These hopes have recently taken on a more concrete shape, with the help of what is called loop quantum gravity, and its applications to cosmology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum gravity is still unproven and hypothetical. String theory also tries to unify GM and QFT, but is unfalsifiable. Thus, the Big Bang is a point in time that astronomers really know little about. It's unfortunate and surprising, but it's something physicist are trying to overcome.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-10-06T01:11:32.610" LastActivityDate="2016-10-06T01:11:32.610" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16152" PostTypeId="2" ParentId="16147" CreationDate="2016-06-07T08:06:08.373" Score="4" Body="&lt;p&gt;&quot;Gravitational collapse&quot; is not the generally used name of any planet formation model it simply refers to the collapse of a cloud of gas or collection of objects under the influence of their mutual gravity. &quot;Planetary formation mechanisms&quot; simply refers to mechanisms which form planets, all of which involve the mutual gravitational attraction of the planetary components at some phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of the answer you refer to, there &quot;gravitational collapse&quot; refers to where a small object (like a brown dwarf) forms through the same direct collapse and fragmentation process that forms stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets on the other hand are thought to form &lt;em&gt;in the disc&lt;/em&gt; that is inevitably created around a protostar during the star formation process. This might be through something called gas disc instability, or the more likely model at present is something known as the &lt;a href=&quot;http://www.scholarpedia.org/article/Planetary_formation_and_migration#Giant_planet_formation&quot; rel=&quot;nofollow&quot;&gt;core accretion model&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-06-09T16:17:09.557" LastActivityDate="2016-06-09T16:17:09.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16153" PostTypeId="2" ParentId="13920" CreationDate="2016-06-07T08:15:16.467" Score="7" Body="&lt;p&gt;With an 8&quot; scope, a filter will very likely give you better results than observing without a filter. Although a filter does block light, the crucial aspect is that a filter increases &lt;strong&gt;contrast&lt;/strong&gt; (by blocking light pollution and extraneous wavelengths of light more than the nebula), thereby allowing you to spot low contrast diffuse nebulae (like IC59 and IC1318) much more easily. This is in fact more critical for visual observation than for photography because it is possible to increase contract in post-production with photography. You will find that visual astronomers go to great lengths to increase contrast--baffling, flocking, premium mirrors, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light pollution filters are broadband, meaning they &lt;strong&gt;allow&lt;/strong&gt; all light except light emitted by streetlights. From darker skies (like bortle 4.5), this will not give dramatic results. I would recommend using an OIII filter to start with. The OIII filter is a narrowband filter, meaning it &lt;strong&gt;cuts&lt;/strong&gt; all light except at a very narrow wavelength range from ionized oxygen. Very often OIII filters make the difference between being able to see an object and not being able to see it, even from very dark skies. As an excellent test of the OIII filter, check out the Veil Nebula too; from your skies, this object would be rather difficult without an OIII but rather stunning with an OIII filter.&lt;/p&gt;&#xA;" OwnerUserId="13087" LastActivityDate="2016-06-07T08:15:16.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16154" PostTypeId="1" AcceptedAnswerId="16178" CreationDate="2016-06-07T13:54:35.437" Score="1" ViewCount="88" Body="&lt;p&gt;first doubt here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a fixed camera, whose coordinates I know, pointing to the sky. After the first image, I performed a astrometry calculation from nova.astrometry.net and I got, among other things, the Ra and Dec of the center of view. I would like to use the calibration information for other pictures without needing to calibrate again, e. g. copy the calibration info and just update Ra and Dec of the next pictures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought I could do so transforming the Ra and Dec to Az and Alt using the Julian Date of the first event(the calibrated one), and then go back to Ra and Dec using the Julian Date of the second event. But I get wrong results. I tried between different pictures and sometimes the error is bigger some time smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using the functions eq2hor and hor2eq of Astrolib (IDL), which takes in account precession and nutation, so I assume than there is a gap in my knowledge and procedure, rather than a programming mistake. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="13088" LastEditorUserId="13088" LastEditDate="2016-06-07T14:06:59.960" LastActivityDate="2016-06-09T12:10:31.410" Title="Pointing direction movement of a fixed camera" Tags="&lt;coordinate&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16155" PostTypeId="1" AcceptedAnswerId="16159" CreationDate="2016-06-07T07:02:42.177" Score="5" ViewCount="346" Body="&lt;p&gt;If we can tell a planet exists by the dimming of the brightness of a star when it passes in front of our view of the star does that mean we cannot see a planet if it does not cross in front of our view? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If planet of a distance star goes around the star rather than in front of it than we cannot see that planet. Correct? &lt;/p&gt;&#xA;" OwnerUserId="13092" OwnerDisplayName="1.21 gigawatts" LastActivityDate="2016-06-07T20:21:56.907" Title="Confirm: We can only notice planets if their orbit passes in front of their star and our view" Tags="&lt;planet&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16156" PostTypeId="2" ParentId="16155" CreationDate="2016-06-07T07:25:01.943" Score="7" Body="&lt;p&gt;This is not true in general. There are other &lt;a href=&quot;https://en.wikipedia.org/wiki/Methods_of_detecting_exoplanets&quot;&gt;methods to detect a planet&lt;/a&gt; that do not rely on a direct line of sight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are not completely wrong: Direct observation (e.g. of a planet's atmosphere) requires a line of sight. We can however notice the presence of a planet by indirect means: If the planet causes an effect on an observable body (e.g. tidal forces on its star) we can deduce its existence from this (regular) effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, we could also use the gravitational waves a planet should emit, although no practical method exists to do so (yet).&lt;/p&gt;&#xA;" OwnerUserId="10632" OwnerDisplayName="choeger" LastActivityDate="2016-06-07T07:25:01.943" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16157" PostTypeId="1" AcceptedAnswerId="16158" CreationDate="2016-06-07T15:22:16.717" Score="3" ViewCount="88" Body="&lt;p&gt;It came up in conversation today that we were nearly at 21st June, and the days would shortly be getting, um, shorter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We were trying to think of the the term that means &quot;When the sun is at the highest point it can attain in the sky &lt;em&gt;during the year&lt;/em&gt;&quot;, as opposed to during the day, (which is the solar noon). We have terms like Perigee and Apogee, it seems (to me) that there should be a name for this as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there?&lt;/p&gt;&#xA;" OwnerUserId="13090" LastActivityDate="2016-06-07T17:08:30.633" Title="Is there a name for the maximum in local solar elevation angle?" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16158" PostTypeId="2" ParentId="16157" CreationDate="2016-06-07T17:08:30.633" Score="3" Body="&lt;p&gt;I think this is an indirect answer, but the name for this event is &lt;strong&gt;solstice&lt;/strong&gt;. In the strictest definition, the solstice is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;either of the two times in the year, the summer solstice and the winter solstice, when the sun reaches its highest or lowest point in the sky at noon, marked by the longest and shortest days.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By definition, the solstice is the exact point in time when the Earth's axial tilt is pointed as close to (or as far from) the Sun as possible and by extension, the Sun is at its highest (or lowest) point in the sky during the year. If you're in the northern hemisphere, this point is the summer solstice, whereas it is the winter solstice for the southern hemisphere.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-06-07T17:08:30.633" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16159" PostTypeId="2" ParentId="16155" CreationDate="2016-06-07T17:25:29.250" Score="8" Body="&lt;p&gt;The answer by choeger provides a correct outlook, but I wanted to provide a bit more detail on some exoplanet detection methods that have actually been used to find planets we couldn't see. Now I interpret &quot;see&quot; in the manner posed by the question in that &quot;see&quot; references observing a transit of the star by the planet. The questioner is correct that we cannot &quot;see&quot; the planet in this manner if the alignment is not perfectly correct. So how else might we detect the planet?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;Radial Velocity Method&lt;/em&gt; - We can try to detect its gravitational effects on the star through a method known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Radial_velocity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Radial Velocity method&lt;/a&gt;. Both the star and planet(s) will be orbiting their common center of mass. That means if we watch the star, we can see it orbit that center of mass as well. If we see the star &quot;wobbling&quot;, we can infer what mass must exist in that system to cause the wobble. I should point out that we are only able to see radial wobbling (that is, wobbling of the star towards and away from us) which means that while we can still detect planets which are not directly aligned with the star as they would be for the transit method, they are much harder to detect. I also want to point out that the answer by choeger mentioned tidal forces, but that is actually &lt;em&gt;not&lt;/em&gt; something we can use to detect planets or measure their effects on their host stars.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Direct Observation&lt;/em&gt; - This is exactly what it sounds. With some clever techniques and the right conditions, we can take a picture of the planet itself. One pretty famous example was the direct imaging of Fomalhaut b. Shown below is an image over a few years as it travels through its orbit.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GdOIY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GdOIY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;em&gt;Microlensing&lt;/em&gt; - This uses the concept of gravitational mass acting as a lens to background, luminous objects by bending spacetime, and consequently, the path of light through space. If a planet passes in front of a background object (not its host star), it will cause a slight magnification of that background object. I should say that this requires a chance alignment and so it is not a repeatable measurement, meaning there is often little chance that we can actually confirm that we saw a planet. Below is a light curve of a background object and you can see it grow brighter as the main star passes in front of the background object. Pay attention to the little blip on the right which indicates the planet also passes in front of the background object.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jOT4T.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jOT4T.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2016-06-07T20:21:56.907" LastActivityDate="2016-06-07T20:21:56.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16160" PostTypeId="2" ParentId="15118" CreationDate="2016-06-07T19:14:55.243" Score="2" Body="&lt;p&gt;There is a discussion of this at &lt;a href=&quot;http://cosmoquest.org/forum/archive/index.php/t-107733.html&quot; rel=&quot;nofollow&quot;&gt;http://cosmoquest.org/forum/archive/index.php/t-107733.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, the perseids occur on the same date of the sidereal year, which differs from the tropical year due to precession. This causes a change of about one day in 70 years to the date of a meteor shower. 2600 years causes a change of about 37 days compared to the current calendar: The perseids would be at the end of June or the start of July, in the Gregorian calendar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the Chinese Calendar isn't Gregorian: it is Lunisolar, based on astronomical observation of the solstice and the months, with intercalendary months added as required. The date of the the Qixi festival varies from year to year. Meteor showers were noted in China (see my link below), and may be incorporated into myth, but that would be speculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position of the radiant would move with the stars, so the radiant would still be in Perseus 2600 years ago, and not near Vega. But Perseus (and Vega) would be in a slightly different position in the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may be interested in article about &lt;a href=&quot;http://arxiv.org/pdf/astro-ph/0501216.pdf%3Forigin%3Dpublication_detail&quot; rel=&quot;nofollow&quot;&gt;Early meteor shower observations in Japan, Korea and China&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-06-07T19:44:35.860" LastActivityDate="2016-06-07T19:44:35.860" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16161" PostTypeId="1" AcceptedAnswerId="16162" CreationDate="2016-06-07T19:16:54.640" Score="1" ViewCount="66" Body="&lt;p&gt;Are there good references of astronomical texts where we get a first-hand experience of astronomers uncovering astronomical gems using maths alone and not using observation. Observation may have been used later to validate the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;https://books.google.com/books/about/In_Search_of_Planet_Vulcan.html?id=PO7yC6BtQ54C&amp;amp;printsec=frontcover&amp;amp;source=kp_read_button#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; book, there are several examples of astronomical jewels that were uncovered by the professionals. However the book shies away from describing the actual maths that aided the astrophysicists to make their conclusions and predictions that were later confirmed by experimenters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, how does a professional astronomer come to conclusions that there must be one more planet at so and so location? Are there books that uncover the exact line of thinking and maths that is involved  behind the works of the professionals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not looking for a book which solves a toy problem rigorously using Newton's laws and then concludes that the 21st century astronomers use similar techniques at grander scales in their works. Though this is what is happening, i am looking for a more transparent description of how the pioneer discoverers make astronomical predictions with surety. The ideal place to look would be the research papers of these pioneers but is there a book that does that for many such achievements?&lt;/p&gt;&#xA;" OwnerUserId="13091" LastActivityDate="2016-06-07T19:33:09.117" Title="Books describing first-hand experience of astrophysicists finding planets without skipping actual maths" Tags="&lt;astrophysics&gt;&lt;fundamental-astronomy&gt;&lt;resource-request&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16162" PostTypeId="2" ParentId="16161" CreationDate="2016-06-07T19:33:09.117" Score="1" Body="&lt;p&gt;The discovery and recapture of Ceres, and the brilliant discovery and use by Gauss of the Normal distribution, the method of least squares, and a method that would now be thought of as a type of fast fourier transform to predict the position of the planet to within 0.5 degrees, based on just 6 observations is one of the great achievements of &quot;pen and paper astronomy&quot;. Gauss, though we now think of him as a pure mathematician, held the post of Professor of Astronomy in Göttingen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Discovery of the First Asteroid, Ceres: Historical Studies in Asteroid Research&lt;/em&gt; by Clifford Cunningham, ISBN 978-3-319-21777-2 tells this story in some detail, but &lt;em&gt;ouch&lt;/em&gt; its not cheap! nearly £100 on dead trees.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-06-07T19:33:09.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16163" PostTypeId="2" ParentId="15086" CreationDate="2016-06-07T22:58:00.210" Score="1" Body="&lt;p&gt;This is a historically important issue and I think it is worth adding a bit about this history to the excellent responses given above.  The story illustrates the physical meaning of &quot;&lt;strong&gt;temperature of space&lt;/strong&gt;&quot;.  In 1940, McKellar (PASP, vol 52. p187) identified some strange interstellar lines, previously seen by Adams in 1939 in the spectrum of a star, as lines due to the rotation of CN and CH molecules.  These lines were at the time unique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their relative intensities could only be understood if the rotation (ie. spin) was due to the collisions of the molecules with photons at a temperature of 2.7K.  A year later he revised this to 2.3K. For obvious reasons he referred to this as the &quot;&lt;strong&gt;spin temperature&lt;/strong&gt;&quot;: the temperature derived from spinning molecules.  No other source suggested itself, and it was not until 1966, after the discovery of the cosmic background radiation, that McKellar's interpretation was linked with the cosmic background radiation at 2.725K.  McKellar had found a &quot;&lt;strong&gt;thermometer in space&lt;/strong&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ironically, Hoyle in 1950 criticized Gamow's 1949 view of a hot big bang by saying that the Gamow theory would provide a higher temperature to space than allowed by McKellar's analysis.&lt;/p&gt;&#xA;" OwnerUserId="5301" LastActivityDate="2016-06-07T22:58:00.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16164" PostTypeId="1" CreationDate="2016-06-08T03:11:19.010" Score="0" ViewCount="216" Body="&lt;p&gt;Can a star eject a liquid or a gas that has not been ionized? I am implying that these gases could be a source of reusable energy.&lt;/p&gt;&#xA;" OwnerUserId="9326" LastEditorUserId="823" LastEditDate="2016-06-09T17:40:56.510" LastActivityDate="2016-06-09T18:37:52.977" Title="Can a star eject a liquid or a gas that has not been ionized?" Tags="&lt;star&gt;&lt;star-systems&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16165" PostTypeId="2" ParentId="16164" CreationDate="2016-06-08T06:52:49.527" Score="2" Body="&lt;p&gt;There are a variety of scenarios in which a star can emit gases, although that's probably not the best way to think about it. A better way to visualize it is in terms of mass loss rather than gas emission. Inevitably, most or all of a stars mass must eventually go &quot;back to the galaxy in which they are in.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_mass_loss&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Stellar_mass_loss&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a star the size of our Sun or thereabouts, towards the end of its life it will form a red giant. After this stage, its outer layers will &quot;puff out&quot; creating a planetary nebula. This is essentially the star emitting most of its material out to its surroundings, contributing its mass to the formation of new stars later on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other route, the supernova, happens when a star much more massive than the sun starts fusing iron and suddenly collapses, rebounding and creating a large shockwave which expels most of its material out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, certain kinds of stars like Wolf-Rayet stars or (possibly) R Coronae Borealis variables can expel matter due to the fact that they are unstable and can suddenly eject part of their outer layers.&lt;/p&gt;&#xA;" OwnerUserId="13085" LastActivityDate="2016-06-08T06:52:49.527" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16166" PostTypeId="1" CreationDate="2016-06-08T07:17:50.103" Score="0" ViewCount="123" Body="&lt;p&gt;My curiosity was pricked by the an edit made to this post: &lt;a href=&quot;https://astronomy.stackexchange.com/q/16147/10102&quot;&gt;What&amp;#39;s the distinction between &amp;quot;planetary formation mechanisms&amp;quot; and &amp;quot;gravitational collapse&amp;quot;?&lt;/a&gt; sadly the edit was changing the original questions intend, so I had to roll it back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the question of the edit itself felt interesting to me aswell, so I'm reposting it here in another question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What &quot;planetary formation mechanisms&quot; other than &quot;gravitational collapse&quot; exist?&lt;/p&gt;&#xA;" OwnerUserId="10102" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-06-08T10:04:38.003" Title="What other “planetary formation mechanisms” exist beyond “gravitational collapse”?" Tags="&lt;planetary-formation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16167" PostTypeId="2" ParentId="16166" CreationDate="2016-06-08T10:04:38.003" Score="4" Body="&lt;p&gt;&quot;Gravitational collapse&quot; is not the generally used name of any planet formation model. Gravitational disk instability may be what is meant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of the answer you refer to, the other, and probably dominant planet formation is known as the &lt;a href=&quot;http://www.scholarpedia.org/article/Planetary_formation_and_migration#Giant_planet_formation&quot;&gt;core accretion model&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course gravity plays a crucial role here too, but the distinction between this and direct collapse of an instability in a gas disc (what is perhaps being referred to as gravitational collapse) is that the major phase of gas accretion takes place &lt;em&gt;after&lt;/em&gt; the initial formation of a rocky/icy core. This may account for the correlation between planet frequency and host star metallicity.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-06-08T10:04:38.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16169" PostTypeId="2" ParentId="15086" CreationDate="2016-06-08T17:18:35.607" Score="2" Body="&lt;p&gt;Just one further complication.  It is possible to set up &quot;refrigerators&quot; in interstellar space.  These are situations which are effectively the opposite of masers - the energy levels of the material involved (in this case, formaldehyde) can end up behaving as if they are cooler than the surroundings.  As a result, you can see formaldehyde in absorption against the cosmic microwave background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just another example of the fact that, at the low densities of interstellar space, you have to look at the details of how individual atoms and molecules are behaving, because they are only poorly linked by collisions to the surroundings.  And that makes for some neat effects.&lt;/p&gt;&#xA;" OwnerUserId="11023" LastActivityDate="2016-06-08T17:18:35.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16171" PostTypeId="2" ParentId="16140" CreationDate="2016-06-09T00:56:13.107" Score="1" Body="&lt;p&gt;The Nice theory of planetary formation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Created in Nice, France&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also answers the not so pretty picture (except for Planet 9). In the beginning, the sun, as a protostar, sucked up mass from a nebula and became a rapidly spinning star. The matter flattened out into a disk (Why? See &lt;a href=&quot;https://physics.stackexchange.com/questions/25950/why-are-some-galaxies-flat&quot;&gt;Why are some galaxies flat&lt;/a&gt;.) The Nice theory explains the Late Heavy Bombardment, after most matter had already coalesced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Grand Tack&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Around 5 billion years ago, the Late Heavy Bombardment happened as Jupiter and Saturn locked into a strong resonace, which brought them closer to the Sun. As this happened, icy debris that was still present in the outer solar system was dragged inward to the inner solar system. This prompted a bona fide &quot;Grand Attack&quot; on the inner solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Grand Attack&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Grand Attack happened after Jupiter got out of the resonance, flinging it (and Saturn) out back into the outer solar system. The debris it dragged into the solar system locked into swarms of debris that would have siphoned enough energy off of any &lt;a href=&quot;http://kepler.nasa.gov/Mission/discoveries/&quot; rel=&quot;nofollow noreferrer&quot;&gt;super-Earths&lt;/a&gt; to drag them into the sun. The remaining debris coalesced into the four planets of the inner solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest of the theory is consistent with what everyone else already states&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why are Mercury and Mars so much smaller than Venus and the Earth?&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What's the deal with all the weird exoplanets scientists have found?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have made any mistakes in conventions, please tell me. Us sixth-graders are very inexperienced on StackExchange.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10161" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-06-09T14:43:20.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16172" PostTypeId="1" CreationDate="2016-06-09T01:42:46.940" Score="5" ViewCount="164" Body="&lt;p&gt;I was wondering if theoretically the universe expanding affects speed or acceleration due to gravity, compared to if space weren't accelerating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I have it in my head, something is falling towards a planet with gravity G acting upon it. It's been falling for T amount of time and has gone X distance with Y left to go. Currently the universe is expanding, so the distance between its starting and ending point is a very tiny bit longer when it arrives than when it started, either affecting its distance traveled at time T, or its speed at time T.  Compare this to a non-expanding universe where the distance between start and end remains the same, no matter what point in time you choose. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that local gravity forces and atomic forces greatly overpower expansion forces so galaxies, solar systems, planets, and watermelons don't just fly apart. I'm just wondering if the value of either gravitational acceleration, or an object's speed could theoretically be changed slightly due to cosmic expansion. &lt;/p&gt;&#xA;" OwnerUserId="13098" LastActivityDate="2021-03-15T16:02:20.257" Title="Theoretically could the expansion of the universe affect acceleration due to gravity?" Tags="&lt;gravity&gt;&lt;space-time&gt;&lt;expansion&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16173" PostTypeId="2" ParentId="16147" CreationDate="2016-06-09T02:43:14.760" Score="2" Body="&lt;p&gt;Gravitational collapse is usually meant in the context of lots of small particles falling into a large gravity well:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. Galaxies (Clusters, Superclusters, cosmic filaments, too, I think) start off as the gravitational collapse of the hydrogen in a large volume of space (only if the density was perfectly uniform (not possible to be perfect) would there be no collapse).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. Portions of molecular clouds (hydrogen (and helium) gas and dust) gravitationally collapse into the protostellar disk which collapses into the protostar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. All stars are in the process of gravitational collapse. This collapse is slowed down (stopped) by their fusion ignition and heating, but eventually that will end and collapse will continue for whatever is left from the fusion and the supernova. White dwarfs form, and they'll be stable about a trillion of years (we think) and then become cool black dwarfs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D. Gravitational collapse is also used to describe the start of a black hole. In this case, no forces, not heat, not electron degeneracy, can overcome the gravitational attraction. Just today I saw a report of a dark rain observation - the inflow of matter directly onto a black hole, when normally we expect an accretion disk to form and for matter to spiral in, heating up enormously in the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E. It's possible to form a planet via gravitational collapse, but generally larger objects (stars) are what result. This also means that while gravitational collapse might be a contributor to planetary formation, we THINK the major factors are collisions and accretion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how it goes. Hydrogen is all over the Universe. Some of it collapses into stars and galaxies. Stars form and blow up, spewing dust all over the place. Dust and gas (H and He) form into clouds because of the explosions pushing them around (a supernova is to a molecular cloud as a leaf blower is to leaves).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational collapse of very large regions of the molecular cloud ensues. Protostellar disk forms, star in center forms, and stuff in disk collides and accretes into planets (as star forms, things begin to heat up, meaning planets close to center will not have much hydrogen, while planets far away will have lots of volatile gases (Gas giants).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This picture is being questioned because so many Gas Giant sized exo-planets are being found close to their stars...stay tuned, the jury is out (we think these Hot Gas Giants got there by orbital perturbation, and didn't form &quot;close&quot;, but its a very active question.)&lt;/p&gt;&#xA;" OwnerUserId="12059" LastEditorUserId="6" LastEditDate="2016-06-09T14:13:05.790" LastActivityDate="2016-06-09T14:13:05.790" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16174" PostTypeId="1" AcceptedAnswerId="16177" CreationDate="2016-06-09T09:09:01.463" Score="1" ViewCount="835" Body="&lt;p&gt;I'm just a guy who is interested in all topics around space, so forgive me if it is a dumb question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read some things about asteroid mining in the past and I think it's a really interesting topic, as it opens up a lot of possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one thing came to my mind. Humans tend to pushing things a little too far sometimes when we want to reach a goal. So, imagine, we are getting really good at off-earth mining at a larger scale and in a larger timespan, we continue to mine thousands and thousands of asteroids. Wouldn't that affect the gravitational balance of our star system at some point, or is the gravitational influence of an asteroid belt negligible? I am talking about extreme examples, like removing nearly the whole asteroid belt beyond Mars. Also, what would happen to all the stuff we don't need from an asteroid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure I'm not the first one who thinks of something like this, but I found nothing about this particular topic.&lt;/p&gt;&#xA;" OwnerUserId="13105" LastActivityDate="2016-06-09T15:52:18.840" Title="Could asteroid mining affect gravitational balance?" Tags="&lt;gravity&gt;&lt;asteroids&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16175" PostTypeId="2" ParentId="15086" CreationDate="2016-06-09T09:18:02.027" Score="0" Body="&lt;p&gt;The cosmic background of neutrinos is at a temperature of ~1.95K, below that of the cosmic background photons at 2.7K.  There is no inconsistency here because those neutrinos were once in equilibrium with the photons just before the photons got heated by the annihilating electrons (~1 second after big bang).  The loss of electrons caused the neutrinos to decouple from the photons at that point and are no longer in equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the &quot;temperature of space&quot; depends on whether you cite the photon or neutrino temperature, and what you measure depends on what kind of thermometer you use.  The curvature of space time can also be associated with a temperature, but that's another story.&lt;/p&gt;&#xA;" OwnerUserId="5301" LastActivityDate="2016-06-09T09:18:02.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16176" PostTypeId="2" ParentId="16174" CreationDate="2016-06-09T10:39:14.317" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So, imagine, we are getting really good at off-earth mining at a larger scale and in a larger timespan, we continue to mine thousands and thousands of asteroids.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First off, that is a problem for our children's children's children to solve. It is not our problem. The quantity of materials mined from space so far is a bit over 380 kilograms, the 382 kilograms returned from the Moon by the Apollo missions, the 300 grams of materials returned from the Moon by automated Soviet spacecraft, 7 particles of interstellar dust returned from comet Wild 2 by the Stardust spacecraft, and 1500 particles of dust returned from asteroid Itokawa by the Japanese Hayabusa spacecraft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the foreseeable future, there is very little stuff in asteroids that is worth anything on Earth. There is no point in mining commodity metals in space and bringing that material down to Earth. Doing so would be a money-losing proposition. Mining precious metals in vast quantities would turn those expensive precious metals into worthless commodity metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that might be of value on Earth is helium 3, but that is so very, very rare that bringing it down to Earth will have zero effect on the Earth's mass. This of course depends on making fusion viable; that's been cited as being twenty years in the future since the 1960s. Current estimates are well beyond 20 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The foreseeable future (at most 50 years from now; anything beyond that is science fiction) of asteroid mining is using those materials mined in space in space. That opens the door to extracting substances that are extremely common on Earth. The same economics that dictates that mining precious metals in space doesn't make sense economically means that sending consumable up into space doesn't make much sense economically. For now, the most valuable substances to be mined in space are volatiles such as water and methane. This assumes a sizable space infrastructure that needs those substances.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Wouldn't that affect the gravitational balance of our star system at some point, or is the gravitational influence of an asteroid belt negligible?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The asteroid belt accounts for about a billionth (one part in 10&lt;sup&gt;9&lt;/sup&gt;) of the mass of the solar system. The gravitational influence of the asteroid belt is negligible.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2016-06-09T10:44:27.190" LastActivityDate="2016-06-09T10:44:27.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16177" PostTypeId="2" ParentId="16174" CreationDate="2016-06-09T10:43:25.980" Score="2" Body="&lt;p&gt;Something to remember is that the total mass of asteroids in the asteroid belt is tiny compared to the masses of the planets. Ceres, the largest body in the asteroid belt by a wide margin, is only 0.00015 times the mass of Earth, Ceres itself accounts for about a third the belt's mass:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.sciencedaily.com/terms/asteroid_belt.htm&quot; rel=&quot;nofollow&quot;&gt;Asteroid belt facts from ScienceDaily&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't been able to find any papers or articles that suggest the residents of the asteroid belt have any measurable perturbation on the orbits of Jupiter and Mars (the other way around is a different story!). If the asteroids that spend most of their time near Mars and Jupiter were to disappear, I doubt there would be any observable effect on the orbits of the planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider that comets and asteroids commonly pass close to Earth and the Moon and their influence on Earth's orbit is negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for what we would do with what we don't need from an asteroid, there are a few options I could think of based on the size of the target.&#xA;With smaller asteroids, if there is a risk that any changes to its orbit (such as an impact or accidental explosion) that would send it close to Earth, a complete controlled demolition of it might be warranted, or simply attach large chemical rockets to it and push it away. In the near-future, it is likely that even if we do start mining asteroids, provided we haven't dangerously altered its orbit, they will simply be abandoned afterwards.&lt;/p&gt;&#xA;" OwnerDisplayName="user10106" LastEditorDisplayName="user10106" LastEditDate="2016-06-09T15:52:18.840" LastActivityDate="2016-06-09T15:52:18.840" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16178" PostTypeId="2" ParentId="16154" CreationDate="2016-06-09T12:10:31.410" Score="1" Body="&lt;p&gt;As has been noted, the camera, if truly fixed, will point at spot with fixed declination. The RA it points to will be a simple function (namely so many hours and minutes ahead/behind) of the &lt;em&gt;local&lt;/em&gt; sidereal time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either your camera is not truly fixed (a bit of shake) or (possibly more likely), you have not correctly determined the local sidereal time.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2016-06-09T12:10:31.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16179" PostTypeId="1" AcceptedAnswerId="16180" CreationDate="2016-06-09T12:35:55.770" Score="3" ViewCount="101" Body="&lt;p&gt;I have a small amateur telescope, and would like to defeat the atmospheric seeing conditions as much as is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortuitously, I happen to live fairly close to this lovely patch of bare elevated rock:&lt;a href=&quot;https://i.stack.imgur.com/ZO48R.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZO48R.jpg&quot; alt=&quot;Mt Cardigan&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a &lt;a href=&quot;https://en.wikipedia.org/wiki/Mount_Cardigan&quot; rel=&quot;nofollow noreferrer&quot;&gt;granite peak&lt;/a&gt; about 3,155ft (960m) above sea level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this peak be expected to have improved seeing, as compared to the seeing from an adjacent valley on the same night?&lt;/p&gt;&#xA;" OwnerUserId="1367" LastEditorUserId="7982" LastEditDate="2021-02-28T11:47:30.227" LastActivityDate="2021-02-28T11:47:30.227" Title="Does the summit of a small mountain offer better seeing?" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;planetary-atmosphere&gt;&lt;atmospheric-effects&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16180" PostTypeId="2" ParentId="16179" CreationDate="2016-06-09T13:01:14.957" Score="4" Body="&lt;p&gt;In terms of steadiness of the air and hence stability of the image (ie ability to see fine detail on planets or split close double stars), it could go either way, depending on the atmospheric conditions, including wind speed and direction, and the local topography.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, valleys can sometimes provide better seeing because they provide some protection from atmospheric conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, and all other things being equal, you will probably have a higher limiting magnitude (ie be able to see fainter objects) from the top of the mountain. &lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2016-06-09T13:01:14.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16181" PostTypeId="1" CreationDate="2016-06-09T13:50:58.483" Score="1" ViewCount="416" Body="&lt;p&gt;Is this term official jargon/wording for 'Dark Matter'?&#xA;If so, why include the word &quot;Fluid&quot; if there is supposedly no viscosity in actual space. It would be like saying 'Hey, want to go swim in my 'gravel' pool?&quot;&#xA;&lt;a href=&quot;http://www.dailymail.co.uk/sciencetech/article-3013284/Is-dark-matter-FLUID-Mystery-particles-mapped-galaxy-clusters-results-suggest-not-particles-all.html&quot; rel=&quot;nofollow&quot;&gt;http://www.dailymail.co.uk/sciencetech/article-3013284/Is-dark-matter-FLUID-Mystery-particles-mapped-galaxy-clusters-results-suggest-not-particles-all.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13107" OwnerDisplayName="NormLDude" LastEditorUserId="13107" LastEditDate="2016-06-09T14:16:52.900" LastActivityDate="2017-06-05T14:40:17.517" Title="What is 'Dark Fluid'?" Tags="&lt;dark-matter&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16183" PostTypeId="2" ParentId="16181" CreationDate="2016-06-09T17:04:26.907" Score="2" Body="&lt;p&gt;No, &amp;quot;dark fluid&amp;quot; is not a synonym for dark matter.&lt;/p&gt;&#xA;&lt;p&gt;The Daily Mail article derives from a &lt;a href=&quot;http://actu.epfl.ch/news/galaxy-clusters-collide-dark-matter-still-a-myst-2/&quot; rel=&quot;nofollow noreferrer&quot;&gt;press release by the EPFL&lt;/a&gt; and an article in &lt;a href=&quot;http://science.sciencemag.org/content/347/6229/1462.full&quot; rel=&quot;nofollow noreferrer&quot;&gt;Science magazine&lt;/a&gt;, neither of which mention the word 'fluid'.&lt;/p&gt;&#xA;&lt;p&gt;We don't know what dark matter is yet. There are several theories. &lt;a href=&quot;http://www.digitaljournal.com/science/is-dark-matter-a-fluid-its-particles-aren-t-particles-at-all/article/429310&quot; rel=&quot;nofollow noreferrer&quot;&gt;One of them&lt;/a&gt; compares dark matter to a fluid:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Normally speaking, subatomic particles exchange momentum when they interact. Consequently, galactic collisions should also lead to interaction by dark matter particles and by studying it, we would have a better idea of what dark matter really is.&lt;/p&gt;&#xA;&lt;p&gt;To verify whether dark matter is made up of particles, the researchers explored two possible outcomes. The first, was that particles of dark matter often interact, but don't exchange much momentum. The second was that they seldom interact, but exchange a lot of momentum.&lt;/p&gt;&#xA;&lt;p&gt;If the first case was true, dark matter would slow down after the collision, since the greater frequency of the interactions would have drag-like effect. If the second case was proven to be true, then dark matter would likely be dispersed into space.&lt;/p&gt;&#xA;&lt;p&gt;Surprisingly, the study discovered that the dark matter particles from two colliding galaxies, simply pass through each other, suggesting that dark matter particles do not interact with themselves, &lt;em&gt;but have a more fluid-like character instead&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;(emphasis mine). So they're not saying dark matter is exactly like a fluid, they're saying dark matter has some properties that are comparable to the behavior of fluids.&lt;/p&gt;&#xA;" OwnerUserId="361" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-06-09T17:04:26.907" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16184" PostTypeId="1" CreationDate="2016-06-09T19:56:13.527" Score="1" ViewCount="147" Body="&lt;p&gt;Usually an orbit is considered as an acceleration. So if the Earth has a speed of about 100.000km/h around the sun. &#xA;Now I think there is some relativity involved, so from which point of view shall we observe those speeds? But lets imagine that we stand on mars which stays still compared to the Earth and the sun, only the Earth is moving. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shall we measure a different speed when the Earth is thrown into space? Would it be more of less than 100.000km/h? Or is there something missing?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="14021" LastEditDate="2016-09-14T14:34:59.180" LastActivityDate="2016-09-14T14:34:59.180" Title="Would the (orbitting) speed of the Earth change if suddenly the sun disappear and the Earth is thrown into space?" Tags="&lt;gravity&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16185" PostTypeId="2" ParentId="16184" CreationDate="2016-06-09T20:59:01.047" Score="3" Body="&lt;p&gt;Velocity is a vector quantity with both a magnitude and direction. The magnitude of velocity is speed. During the Earth's nearly circular orbit, the speed changes little, but the direction of motion changes, and this change of direction is the acceleration. The Earth is always accelerating towards the Sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we speak of velocity one should mention the frame of reference: we measure the velocity of the Earth wrt the the Sun. The sun is therefore by definition &quot;at rest&quot; (we should actually use the centre of mass of the Earth-Sun system, an also include the gravitational effect of the other planets, I pass over these details) This frame approximates an inertial frame of reference &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars doesn't stand still relative to either the sun or the Earth. We can use a frame of reference in which Mars is at rest, but this is then a non-inertial frame, it rotates, so in this frame both the Earth and Sun are moving, and accelerating due to the non-inertal motion of the frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we (magically) remove the sun (but not the frame of reference in which the sun is at rest) then there will be no force on the Earth, and it will move in a straight line at constant velocity, relative to the frame in which the sun was at rest. It would move at the same speed as its orbit, about 100,000 km/h (in that frame of reference)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-06-09T20:59:01.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16186" PostTypeId="2" ParentId="14952" CreationDate="2016-06-09T21:18:22.260" Score="1" Body="&lt;p&gt;You might find this paper helpful:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.physics.hmc.edu/faculty/esin/a101/limbdarkening.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.physics.hmc.edu/faculty/esin/a101/limbdarkening.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look specifically at Eq. 5 for the wavelength dependence.  To get the intensity, use Equation 1.  The value of $u$ can be found in the caption to Figure 2.&lt;/p&gt;&#xA;" OwnerUserId="10371" LastActivityDate="2016-06-09T21:18:22.260" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16187" PostTypeId="1" AcceptedAnswerId="16188" CreationDate="2016-06-09T21:19:20.937" Score="3" ViewCount="237" Body="&lt;p&gt;Please excuse if this has been asked and I just didn't find it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found several references that indicate that the inclination of Earth's orbit varies over a period of ~70,000 years, but I can't seem to find anything that tells what the range of variation is. I know that it's currently at 1.57° to the invariable plane, but I'm looking for the value of the maximum tilt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="13109" LastEditorUserId="6" LastEditDate="2019-01-29T14:47:23.940" LastActivityDate="2019-01-29T14:47:23.940" Title="What is the magnitude of variation of Earth's orbital inclination?" Tags="&lt;orbital-migration&gt;&lt;inclination&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16188" PostTypeId="2" ParentId="16187" CreationDate="2016-06-09T22:39:40.147" Score="2" Body="&lt;p&gt;In a &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/375560/fulltext/55714.fg10.html&quot; rel=&quot;nofollow&quot;&gt;solar system model&lt;/a&gt; by &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/375560/fulltext/55714.text.html&quot; rel=&quot;nofollow&quot;&gt;Varadi Runnegar and Ghil&lt;/a&gt; Earth's inclination varies between 0 and about 0.05 radians (or about 3 degrees)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variation is rather chaotic, but you may note that there seems to have been a qualitative change in the pattern about 60-70 million years ago&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-06-09T22:39:40.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16189" PostTypeId="1" AcceptedAnswerId="16190" CreationDate="2016-06-10T06:47:04.433" Score="1" ViewCount="116" Body="&lt;p&gt;I was wondering what kinds of stars would be visible in M31 and other nearby galaxies using a moderate (~8&quot;) telescope and a CCD camera. This would be a limiting magnitude of around 17.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have heard of LBVs (Luminous Blue Variables). However, I would like to know what spectral class of stars would most likely be visible (would it be mainly OBA type stars)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The context of this question is in the search for extragalactic transiting exoplanets. Current amateur detectors have the capability to detect planets with host star apparent magnitudes of 14. I was wondering if it might be possible to find exoplanets in M31, M33 or even the LMC/SMC with these limits. An advantage to searching for planets in M31 would be the very limited search region--an amateur astronomer would only have to perform photometry on the M31 region over several months or years to find a planet, rather than having to search a much larger region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other issues with detecting transiting planets around very luminous stars? Would the percent dip in brightness be far less for a Jupiter sized planet? How would this affect our ability to detect extragalactic transiting exoplanets?&lt;/p&gt;&#xA;" OwnerUserId="13085" LastActivityDate="2016-06-10T12:07:44.533" Title="What kinds of stars would it be possible to observe in M31 with a small amateur telescope?" Tags="&lt;amateur-observing&gt;&lt;exoplanet&gt;&lt;photometry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16190" PostTypeId="2" ParentId="16189" CreationDate="2016-06-10T12:07:44.533" Score="1" Body="&lt;p&gt;LBV stars are very powerful, very large and bright, and even though they may be near the end of their life, still young - since they don't survive for long. These are problematic.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their powerful solar winds will clear the region around them, meaning that planets are less likely to form close to the star. It is much less likely for a LBV star to be aligned correctly to allow a transit to be visible from Earth. A hot Jupiter close to such a star would rapidly lose mass due to the extreme temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a transit does occur, because the star is so large, it obscures a small proportion of the light, so a the change in brightness is small. And of course they are intrinsically variable themselves, so detecting the very small signal is harder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since they are still young so planets may not have finished forming by the time they go supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We haven't discovered any planets orbiting LBVs in the Milkyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most stars that have known planets are F,K and G spectral type, this is partly sample bias (we look for planets around sun-like stars) But such stars would not be visible in M31 using amateur equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An 8in telescope has a limiting magnitude of about 14, (E. Hubble was observing Cepheids with a mag of 16+) but you are already  close to the limit of the telescope, and distinguishing the very small changes is consequently is harder. Given the difficulties, it is perhaps unsuprising that we haven't found many planets more that 10000 ly away.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-06-10T12:07:44.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16191" PostTypeId="1" AcceptedAnswerId="16194" CreationDate="2016-06-10T12:28:21.250" Score="3" ViewCount="2428" Body="&lt;p&gt;This is a basic question, but I may as well ask it. I had always thought that the core of a main-sequence star is defined as the part hot enough for nuclear fusion. Some dictionaries seem to agree with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, many books say that protostars and pre-main-sequence stars also have cores, though they are not undergoing hydrogen fusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what exactly is the definition of a stellar core? If it doesn't necessarily have to be hot enough for fusion, what separates it from the rest of the star?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2016-06-10T13:01:07.740" Title="Definition of stellar core?" Tags="&lt;star&gt;&lt;temperature&gt;&lt;stellar-astrophysics&gt;&lt;core&gt;&lt;stellar-structure&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16192" PostTypeId="1" AcceptedAnswerId="16363" CreationDate="2016-06-10T12:56:56.750" Score="4" ViewCount="1275" Body="&lt;p&gt;I'd like to know a &lt;strong&gt;breakdown of the mass of the Milky Way, by the major categories&lt;/strong&gt;. So, something like&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Stars       ... 3%&#xA;Cool nebula ... 7%&#xA;Gray matter ... 12%&#xA;Etc&#xA;Etc&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This &lt;strong&gt;order-of-magnitude knowledge&lt;/strong&gt; that everyone should know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, when talking of the galaxy, we can, obviously, think of either just &lt;strong&gt;the visible disk-bulge&lt;/strong&gt;, or perhaps, &lt;strong&gt;a greater unit&lt;/strong&gt; of the galaxy sitting in a &quot;bulb&quot; (of - perhaps - dark matter (?), gas (?), dust (?), halo stars (?) or something else): a brief clear answer in a few sentences would expose these concepts.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Surprisingly, this information is nowhere to be found in one place as an overview.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I appreciate that the largest chunk may be dark matter (is that right?); but I don't even have a &lt;em&gt;ballpark idea&lt;/em&gt; for the rest. It's the sort of knowledge that is obvious to professionals, but surprisingly not outlined in any of the usual popular science sources.&lt;/p&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="13071" LastEditDate="2016-06-25T12:56:39.077" LastActivityDate="2016-06-25T12:56:39.077" Title="Milky Way mass fraction breakdown review" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;mass&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16193" PostTypeId="2" ParentId="16191" CreationDate="2016-06-10T12:57:38.957" Score="3" Body="&lt;p&gt;You are right; when referring to stars such as the Sun, the core is usually defined as the volume in the centre where fusion is taking place. In the case of larger stars with fusion occurring in multiple layers, the core is still taken to be the area in the middle where nuclear fusion is producing the heaviest elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With protostars or pre-main-sequence stars, I haven't found any articles recently that specifically refer to cores as part of the star's structure. It is possible they are using the word 'core' to just mean 'the centre'. If you're able to provide links to where you have seen these terms used, we might be able to offer some context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that bodies not hot enough to begin hydrogen fusion, such as brown dwarves, may have layered and differentiated structures akin to Jupiter, in which case, the 'core' is the solid or near-solid region in the centre where accreted heavier elements eventually accumulate. The core's boundary would then be taken as the region where one would transition from heavy elements to the compressed metallic hydrogen that surrounds it.&lt;/p&gt;&#xA;" OwnerDisplayName="user10106" LastActivityDate="2016-06-10T12:57:38.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16194" PostTypeId="2" ParentId="16191" CreationDate="2016-06-10T13:01:07.740" Score="6" Body="&lt;p&gt;I think that there isn't a strict answer to this question. However, I believe the answer is that there's a difference between the core of a hydrogen-burning star and the core of a protostar or star-forming, gas cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a hydrogen-burning star, the core, as you say, is the region of the star where fusion is taking place. This is surrounded by the radiative and convective zones (which one comes first depends on the mass of the star). For true stars, the concept of the core is well defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For things like protostars and star-forming gas clouds, the concept of the core is less well defined and more a nomenclature for a region of the object rather than a strict definition. I don't think it is correct to assume that because one is talking about the core of a protostar, that they are referring to the core as it is defined for a hydrogen-burning star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what does it mean to be a core in a protostar or star-forming gas cloud? I will first point out that protostars can still be fusing. In the pre-main sequence stage, most stars will be &lt;a href=&quot;https://en.wikipedia.org/wiki/Deuterium_fusion&quot;&gt;fusing deuterium&lt;/a&gt; and we can consider the core to be the region where this is occurring. A protostar does not become a fully fledged star until it begins hydrogen burning, but deuterium fusion requires lower temperatures and pressures than the full proton-proton chain of hydrogen fusion (of which deuterium fusion is a part) and so can occur in protostars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We might also distinguish a core from outer regions based on a density or compositional change. If we plot up something like the density (or really any property that changes with radius), we are likely to see a drastic jump signifying some sort of boundary. Cores for star-forming regions may be defined by highly dense regions, high temperature regions, or regions &quot;where the physics changes&quot; because we've passed some critical threshold at some radius. These cores are less well defined and can't be likened to stellar cores very well.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-06-10T13:01:07.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16195" PostTypeId="1" AcceptedAnswerId="16196" CreationDate="2016-06-10T13:29:12.723" Score="2" ViewCount="79" Body="&lt;p&gt;All over the Universe, there are (gravitationally bound) binary star systems. Considering the huge number of brown dwarfs, are there currently known double systems of gravitationally bound binary brown dwarfs? Given their comparatively low gravitation, how long would they need to melt together to a larger object?&#xA;What happens if binary brown dwarfs collide? Is this anyway a matter of any interest?&lt;/p&gt;&#xA;" OwnerUserId="11010" LastActivityDate="2016-08-18T02:32:40.873" Title="Are there currently observed systems of close (narrow) binary brown dwarfs?" Tags="&lt;binary-star&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16196" PostTypeId="2" ParentId="16195" CreationDate="2016-06-10T15:59:10.560" Score="3" Body="&lt;p&gt;You've got a lot of questions here, so I'll try to answer them one by one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there currently known double systems of gravitationally bound binary brown dwarfs?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Certainly there are. I don't study low-mass stars so I'm not up on the literature enough to know any good examples off the top of my head. Kozaky suggested the &lt;a href=&quot;https://en.wikipedia.org/wiki/Luhman_16&quot; rel=&quot;nofollow&quot;&gt;Luhman 16 system&lt;/a&gt; in his comment and a quick google search also brought up the triple star system &lt;a href=&quot;https://en.wikipedia.org/wiki/Epsilon_Indi&quot; rel=&quot;nofollow&quot;&gt;Epsilon Indi&lt;/a&gt; with a central K type star being orbited (at ~1460 AU) by a binary brown dwarf system. Perhaps some other more knowledgeable person can suggest more systems of interest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given their comparatively low gravitation, how long would they need to melt together to a larger object?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That depends on a variety of factors, including their respective masses and initial distance apart. As two objects orbit one another, they continually release gravitational waves due to their interacting gravitational fields. These waves carry away energy and to conserve energy, the stars must lose angular momentum, thus getting closer. The time necessary for two stars of mass $m_1$ and $m_2$, with an initial radius $r$ to fall into each other is &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_wave#Binaries&quot; rel=&quot;nofollow&quot;&gt;given by&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t = \frac{5}{256} \frac{c^5}{G^3} \frac{r^4}{(m_1m_2)(m_1+m_2)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c$ is the &lt;a href=&quot;https://www.google.com/search?q=speed%20of%20light&amp;amp;ie=utf-8&amp;amp;oe=utf-8&quot; rel=&quot;nofollow&quot;&gt;speed of light&lt;/a&gt; and $G$ is the &lt;a href=&quot;https://www.google.com/search?q=gravitational%20constant&amp;amp;ie=utf-8&amp;amp;oe=utf-8&quot; rel=&quot;nofollow&quot;&gt;gravitational constant&lt;/a&gt;. We can apply some basic numbers and calculate an order of magnitude time. If I assume $r = 1\:AU=1.5\times10^{11}\:m$ and $m_1 = m_2 = 2\times10^{28}\:kg$ (10 times the mass of Jupiter) then I find the timescale for collision into one object is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t = 5\times10^{30}\:s \approx 100\:billion\:trillion\:years$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is obviously way longer than the age of the Universe and effectively means that two Brown Dwarfs will never collapse. This fact could have been seen from the equation. Notice that the mass of the objects is in the denominator. That means that with lower mass means longer timescales and the fact that the distance is to the fourth power means the distance is a much stronger controlling factor here. You need huge masses to offset that $r^4$ for you to get a collision on reasonable timescales. Even doing a best case scenario of $r = 0.1\:AU$ and $m_1=m_2=10^9\:kg$ still gives me $\sim100\:thousand\:trillion\:years$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What happens if binary brown dwarfs collide?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not much? I don't know if anyone has ever really tried to work out the theoretical work for such a problem. Given the answer above, the only chance two brown dwarfs are going to collide is by a freak chance that they're on a collision course in space and not in a slowly degrading orbit, but the chances of that are pretty minuscule as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One situation that may be of related interest is to consider regions of star formation. While it is somewhat contested, some people believe stars can form &quot;bottom up&quot;. That means smaller objects form first, then those coalesce into larger objects and on upwards until you end up with big, bright stars. In this scheme, you would first form many many brown dwarfs in a tight area and these would dynamically interact and collide with each other to form larger stars. This isn't the same scenario of two lone binary brown dwarfs colliding, but its the best you're going to get if you want to see colliding brown dwarfs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this scenario, I think that since brown dwarfs aren't fusing and their main energy output comes from core collapse (by some &lt;a href=&quot;https://en.wikipedia.org/wiki/Kelvin%E2%80%93Helmholtz_mechanism&quot; rel=&quot;nofollow&quot;&gt;Kelvin-Helmhotz mechanism&lt;/a&gt;), it isn't going to be a hugely explosive result (comparatively) to smash two together. They will of course burst out with energy in the process, but it won't be as dramatic as two fully-fledged stars merging. And that's probably a good thing. If you had a star forming region with many brown dwarfs that were slowly coalescing into larger stars, you wouldn't want each merger to blow out huge energies as it would prevent further stars from coalescing.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this anyway a matter of any interest?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From an academic standpoint, I'm sure it is. As I said, I don't study this topic, but understanding the affects of brown dwarf collisions would be very useful for fleshing out star forming theories. I'm sure one could think of numerous other academic benefits for studying such a topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a practical standpoint, I can't say understanding brown dwarf collisions immediately benefits humanity. But as with any topic in science, you never know how knowledge gained today may help in the future.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="9301" LastEditDate="2016-08-18T02:32:40.873" LastActivityDate="2016-08-18T02:32:40.873" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16197" PostTypeId="1" CreationDate="2016-06-10T16:44:40.560" Score="2" ViewCount="881" Body="&lt;p&gt;With the discovery of the Higgs Boson as the mass carrier, what is the possibility at the centre of a black hole being so dense and gravity so infinite that the structure of a proton or neutron are ripped apart into their fundamental elements by these extreme energies that Higgs Bosons are massed together at the centre, while the remaining massless subatomic particles are just outside this layer, similar to an onion  (Similar to a star as more elements fuse together they become more dense, and gravity forces them to the centre again like an onion)&lt;/p&gt;&#xA;" OwnerUserId="13122" LastEditorUserId="37191" LastEditDate="2021-02-12T21:31:41.217" LastActivityDate="2021-02-12T21:31:41.217" Title="Higgs Bosons in the centre of Black holes or extremely large stars" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;astrophysics&gt;&lt;quantum-field-theory&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2016-06-13T09:12:32.270" ContentLicense="CC BY-SA 3.0" />
  <row Id="16198" PostTypeId="1" AcceptedAnswerId="16200" CreationDate="2016-06-10T19:36:35.650" Score="4" ViewCount="155" Body="&lt;p&gt;I know that the solar system is tilted 62.6° to the plane of the galaxy.  I'm curious whether this angle changes over time, and what are the extents and time frame of any such oscillations if they exist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon reflection, I'm assuming that since such an oscillation would require a change in the overall angular momentum of the entire solar system, and that would necessitate a collective re-orientation of the orbit of every planet, dwarf planet, SSSB, etc. related to the Sun that it's unlikely that any such mechanism exists?&lt;/p&gt;&#xA;" OwnerUserId="13109" LastEditorUserId="13109" LastEditDate="2016-06-10T20:07:53.000" LastActivityDate="2020-02-13T03:46:04.570" Title="Oscillation in tilt of the solar system" Tags="&lt;solar-system&gt;&lt;galaxy&gt;&lt;ecliptic&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16199" PostTypeId="2" ParentId="16197" CreationDate="2016-06-10T22:12:21.100" Score="3" Body="&lt;p&gt;No. That's not what a black hole is. A black hole is a vacuum solution to general relativity. In other words a black hole is just mass, without a &quot;thing&quot; left to be massive. The mass is collapsed to a singularity, and is surrounded by an event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you pass over the event horizon, every path you can take in will lead you to the singularity. In this sense the singularity is more like a point in time than a point in space. You can't &quot;see&quot; the singularity, since it is always in your future (you can only see things that are in the past)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no &quot;thing&quot; at the singularity, There are no protons neutrons or higgs bosons, and any that are formed will find themselves at the singularity, and ceasing to exist in a short time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So black holes are very very weird. They can't be easily understood my analogy. They are understood in terms of Einstein's mathematical theory of general relativity. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-06-10T22:12:21.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16200" PostTypeId="2" ParentId="16198" CreationDate="2016-06-10T22:15:28.907" Score="3" Body="&lt;p&gt;The tidal field of the Galaxy does lead to the oscillation of the plane of very wide binaries.  The mechanism of this oscillation is identical to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kozai_mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kozai-Lidov mechanism&lt;/a&gt; (the only difference is that in the case of KL oscillations the tidal field is generated by the averaged orbit of a tertiary stellar companion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you run the numbers, the timescale for these oscillations is extremely long for any of the planets (I forget how long exactly, but much longer than the age of the Universe).  It's somewhat shorter for comets in the Oort Cloud, only a few billion years, because their orbits are much larger and so would tend to experience larger a larger tidal effect.&lt;/p&gt;&#xA;" OwnerUserId="10371" LastEditorUserId="7982" LastEditDate="2020-02-13T03:46:04.570" LastActivityDate="2020-02-13T03:46:04.570" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="16201" PostTypeId="1" AcceptedAnswerId="16227" CreationDate="2016-06-11T00:51:07.483" Score="18" ViewCount="603" Body="&lt;p&gt;Consider say this outstanding diagram from another question,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mZkT8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mZkT8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To begin with galaxies have a typical density D. Intergalactic space has a much lower density d.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we know: in galaxies (with density D) most of the stuff is dark matter.  Let's say for convenience it is 10:1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the ratio of dark matter &lt;em&gt;in&lt;/em&gt; galaxies is (we'll say) 10:1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here's what I can not find explained anywhere:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) In intergalactic space. First of all simply &lt;em&gt;is there&lt;/em&gt; any dark matter?  Or, do we believe that dark matter exists, simply, only in galaxies? Not between?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) If there is dark matter between galaxies. Do we know or believe: what is the ratio of dark matter out there?  In fact, is it just the same ratio (say, 10:1) as inside galaxies? Or, is it perhaps even more dominant, or perhaps much less dominant?  What is the ratio out there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) If there is dark matter everywhere. Galaxies exist in filaments and walls.  Does dark matter do this? or is it evenly-spread?&lt;/p&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="16506" LastEditDate="2020-08-25T12:27:21.107" LastActivityDate="2020-08-25T12:27:21.107" Title="Conventional matter to dark matter ratio, outside of galaxies" Tags="&lt;galaxy&gt;&lt;dark-matter&gt;&lt;intergalactic-space&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16202" PostTypeId="2" ParentId="925" CreationDate="2016-06-11T03:59:35.347" Score="-2" Body="&lt;p&gt;Well, this is my relatively inaccurate size comparison of galaxies. IC 1101 is 6 million LY at this scale, NGC 263 is 2.3 million LY, and there are some other galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Vf8JK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vf8JK.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13124" LastEditorUserId="2153" LastEditDate="2016-06-11T13:43:03.400" LastActivityDate="2016-06-11T13:43:03.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16203" PostTypeId="2" ParentId="16197" CreationDate="2016-06-11T06:29:07.643" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;With the discovery of the Higgs Boson as the mass carrier, what is the&#xA;  possibility at the centre of a black hole being so dense and gravity&#xA;  so infinite that the structure of a proton or neutron are ripped apart&#xA;  into their fundamental elements by these extreme energies that Higgs&#xA;  Bosons are massed together at the centre,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The center of a black hole is full of uncertainties.  There's likely enough energy at the center of a black hole to create Higgs Boson particles by high energy collisions, or, other energy exchanges or whatever goes on inside there, but nobody knows for sure.  There's a lot of uncertainty as to exactly what happens inside a black hole.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;while the remaining massless subatomic particles are just outside this&#xA;  layer, similar to an onion (Similar to a star as more elements fuse&#xA;  together they become more dense, and gravity forces them to the centre&#xA;  again like an onion)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's an error in this question.  No subatomic particles are massless.  subatomic particles can be without restmass, but they must have energy and energy has mass.    Inside a black hole, both rest mass particles and particles with no rest mass are still drawn towards the center.  It's bad physics to think of the inside of a black hole as &quot;layers&quot; going from more dense to less dense.   That model breaks down once gravitational acceleration exceeds the speed of light.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every other large/massive non-black-hole object in the universe that's gravitationally bound tends to form some degrees of density layering, even Neutron Stars, but black holes don't necessarily follow that model and thinking of the center of a black hole as different layers of density is probably a bad way to think about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to add that the Higgs (which gives rest mass by interaction between some particles and the Higgs field) and Gravity, might not be as strong as you think.  More on that &lt;a href=&quot;https://profmattstrassler.com/2012/10/15/why-the-higgs-and-gravity-are-unrelated/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, and some background reading on the Higgs &lt;a href=&quot;https://profmattstrassler.com/articles-and-posts/the-higgs-particle/the-higgs-faq-2-0/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and some loosely related questions on whether the Higgs particle happens in nature &lt;a href=&quot;https://physics.stackexchange.com/questions/32105/does-the-higgs-boson-or-the-125-gev-boson-if-its-not-exactly-the-higgs-as-pre&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://physics.stackexchange.com/questions/37975/do-higgs-bosons-happen-in-nature-all-the-time-rarely-or-do-they-only-happen-wh&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-06-11T06:29:07.643" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16204" PostTypeId="1" AcceptedAnswerId="16233" CreationDate="2016-06-11T17:11:44.637" Score="6" ViewCount="462" Body="&lt;p&gt;It seems fairly certain there is a large black hole, about 25,000 ly from Earth roughly in the center of our galaxy, in the direction of the Sag. A radio source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;• In fact, &lt;strong&gt;is it thought to be literally at the center of gravity of our galaxy?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does we know, or is this uncertain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;• Indeed, do we actually know where the CG of the galaxy is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I observe that the term &quot;Galactic Center&quot; seems to be an astronomical term. (&lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_Center#Distance_to_the_Galactic_Center&quot; rel=&quot;noreferrer&quot;&gt;Example&lt;/a&gt;.) Can an astronomer tell me: does that term mean what we believe to be the CG, or rather, is it a nominal mapping point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;• If the black hole IS exactly at the CG of the galaxy ... why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would seem to me a big BH could form any old place, and perhaps still be orbiting or oscillating the CG - right?  If the SBH is literally the CG of there galaxy, is there a reason for that?  (Like, &quot;SBH can only form at the CG&quot; .. or &quot;stellar collisions can only happen at the CG&quot; .. or whatever, those are only wild guesses.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or indeed, is the SBH the &quot;main player&quot; in the galaxy - just as the solar system formed around the sun (so it would be silly to ask &quot;why is the sun at the center&quot;), did the SBH sort of co-form with the galaxy (or something?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any event, is the SBH at the actual CG of our galaxy? Do we know?&lt;/p&gt;&#xA;" OwnerUserId="13071" LastActivityDate="2016-06-15T11:56:10.050" Title="Is our central black hole actually at the CG of the galaxy?" Tags="&lt;supermassive-black-hole&gt;&lt;galactic-dynamics&gt;" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="16206" PostTypeId="1" AcceptedAnswerId="16208" CreationDate="2016-06-12T13:40:26.677" Score="3" ViewCount="435" Body="&lt;p&gt;Temperature is a measure of the average kinetic energy of particles in a given place (correct me if I am wrong) so there must be a definite limit to heating like there is an absolute zero to cooling. And what process could achieve this temperature?&lt;/p&gt;&#xA;" OwnerUserId="4160" LastActivityDate="2016-11-21T03:12:29.220" Title="Is there a limit to how hot an object an get?" Tags="&lt;temperature&gt;&lt;heat&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ClosedDate="2016-11-21T14:56:08.040" ContentLicense="CC BY-SA 3.0" />
  <row Id="16208" PostTypeId="2" ParentId="16206" CreationDate="2016-06-12T14:04:43.250" Score="5" Body="&lt;p&gt;First, let's get a bit of background. &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Absolute_hot&quot; rel=&quot;nofollow noreferrer&quot;&gt;Absolute hot&lt;/a&gt;&quot; is the concept that a maximum temperature &lt;strong&gt;exists&lt;/strong&gt;. It describes your question exactly. For a long time, no one was able to figure out whether an absolute hot exists, but in the 20th century major revolutions in theoretical physics gave us answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Absolute hot&quot; isn't the name of the hottest temperature, but it's the concept that one exists. So what is our answer — is there a highest possible temperature that fits the concept of an &quot;absolute hot&quot;? Well, &lt;strong&gt;yes&lt;/strong&gt;. But first we must talk about heat and temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photons act both like particles and waves. As we add more energy, the photon's wavelength decreases, and its frequency increases. Thus, the wavelength of the light emitted by hot objects generally depends on the temperature of the object. Hotter objects will emit more energetic photons with shorter wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to modern physics, a Planck length is the smallest meaningful length in the Universe (the reason for that is an entirely different question). If we keep raising the temperature of our object, nonstop, the wavelength of the photons it emits will decrease until it reaches a Planck length. Our object would reach a temperature called &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Planck_temperature&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck temperature&lt;/a&gt;&lt;/strong&gt;. That's about $1.417×10^{32} \; \mathrm{K}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planck temperature is theoretically the hottest possible temperature. At that point, the object could not theoretically get any hotter, as the photon's wavelength cannot decrease beyond a Planck length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what happens if we were to add &lt;em&gt;more&lt;/em&gt; energy to the object? Would it go beyond Planck temperature? Well...God knows. At that point, we wouldn't even call it temperature anymore. Nobody knows what exactly would happen, but it certainly wouldn't match the definition of temperature.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-11-21T03:12:29.220" LastActivityDate="2016-11-21T03:12:29.220" CommentCount="26" ContentLicense="CC BY-SA 3.0" />
  <row Id="16209" PostTypeId="2" ParentId="16206" CreationDate="2016-06-12T22:52:49.057" Score="2" Body="&lt;p&gt;As I stated in my comment in Sir Cumference's answer, I might just be naive on the subject, but I believe that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the density of states (the number of states per possible energy) for a given system can be taken to be effectively continuous, one can define the inverse temperature:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{k_BT} = \beta = \frac{dS}{dE}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where E is the system's energy, and S is the system's entropy ($ S= \ln(\Omega(E) $), where $\Omega(E)$ is the density of states (a function of the energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This means that for a given density of states of a system: if the derivative of its natural log has any extrema (local maxima &lt;em&gt;or&lt;/em&gt; minima), then the system will have &lt;em&gt;infinite&lt;/em&gt; temperature at those energies (at the locations of the extrema--remember that $\Omega$ is a function of the energy).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, by this definition: at energies where the slope of a system's entropy, S, is negative, the system is defined to have a negative energy (which ironically is more energetic than states with positive temperature!)... although to be honest: I am not totally convinced that systems at negative temperatures are actually at equilibrium, which (I would think) indicates that temperature is ill-defined... but definitely a curiosity worth some consideration.&lt;/p&gt;&#xA;" OwnerUserId="10809" LastActivityDate="2016-06-12T22:52:49.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16210" PostTypeId="1" CreationDate="2016-06-12T23:15:07.833" Score="1" ViewCount="99" Body="&lt;p&gt;I'm dealing with a homologously contracting star with Mass M, Radius R and a gravitational binding energy of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E = -a GM^2 / R $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a is a constant). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was looking for a way to find an expression for L(t), the luminosity as a function of time. Unfortunately, my attempts using time derivatives have not taken me far. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your help! &lt;/p&gt;&#xA;" OwnerUserId="11715" LastActivityDate="2016-06-13T11:31:35.723" Title="Luminosity L(t) for a homologously contracting star" Tags="&lt;star-formation&gt;&lt;stellar-astrophysics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16211" PostTypeId="1" AcceptedAnswerId="16228" CreationDate="2016-06-12T23:56:22.007" Score="1" ViewCount="1092" Body="&lt;p&gt;The gravitational collapse of a gas cloud can be described by the Jeans Criterion for mass, radius and density of the gas cloud, which is (c stands for cloud): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M_J = (\frac{5kT}{G \mu m_H})^{3/2} (\frac{3}{4 \pi \rho_c}) ^{1/2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_J = (\frac{15kT}{4 \pi G \mu m_H \rho_c})^{1/2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho_J = (\frac{3}{4 \pi M_c^2}) (\frac{5kT}{G \mu m_H}) ^{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to show that they are equivalent and simply used &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho = \frac{m}{V} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;on the criteria above an rearranged the equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However this doesn't work. What am I doing wrong?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your help! &lt;/p&gt;&#xA;" OwnerUserId="11715" LastEditorUserId="9301" LastEditDate="2016-06-13T04:15:14.780" LastActivityDate="2016-06-13T18:14:47.910" Title="How to show that the Jeans Criterion for Mass, Radius and Density are equivalent?" Tags="&lt;star&gt;&lt;star-formation&gt;&lt;stellar-astrophysics&gt;&lt;gas&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16212" PostTypeId="2" ParentId="16210" CreationDate="2016-06-13T00:16:40.203" Score="3" Body="&lt;p&gt;Fairly sure this is standard bookwork if you are talking about a PMS star on the Hayashi track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiate your GPE wrt time, assuming that the mass is constant. Take half of this as the luminosity (via the virial theorem). This gives $L$ in terms of $dR/dt$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then if you are on the Hayashi track you can assume surface temperature $T$ is constant and Stefan's law gives you $dR/dt$ in terms of $L$, $T$ and $dL/dt$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute that in and integrate the resulting differential equation to get $L(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this approach, I managed to get&#xA;$$ L = \left(\frac{\alpha GM^2}{6}\right)^{2/3} \left(4\pi \sigma T^4\right)^{1/3} t^{-2/3},$$&#xA;where $T$ is the temperature of the Hayashi track in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't want to make the Hayashi track assumption, you can say that $L = AM^B T^C$ defines tracks in the HR diagram and this can be used in addition to Stefan's law to eliminate $dT/dt$ and give your result in terms of constants $A$, $B$, $C$. These can come from polytropic theory or by fitting the $L,M,T$ function to numerical calculations. (E.g. see Jackson &amp;amp; Jeffries 2014 &lt;a href=&quot;https://arxiv.org/abs/1404.0683&quot; rel=&quot;nofollow&quot;&gt;https://arxiv.org/abs/1404.0683&lt;/a&gt; ).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-06-13T11:31:35.723" LastActivityDate="2016-06-13T11:31:35.723" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16215" PostTypeId="1" CreationDate="2016-06-13T05:44:02.030" Score="2" ViewCount="121" Body="&lt;p&gt;As we know the atmosphere of Neptune looks more active than Uranus', is this due to the tidal forces caused by the retrograde orbit of Triton around the planet?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="10437" LastEditDate="2016-06-13T09:27:28.287" LastActivityDate="2016-12-30T09:04:19.093" Title="Does Triton trigger the active atmosphere of Neptune?" Tags="&lt;neptune&gt;&lt;triton&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16216" PostTypeId="1" AcceptedAnswerId="16229" CreationDate="2016-06-13T09:57:30.463" Score="6" ViewCount="638" Body="&lt;p&gt;What is the age difference between Jupiter &amp;amp; Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which planet was formed first, or were they formed at the same time?&lt;/p&gt;&#xA;" OwnerUserId="11792" LastEditorUserId="16685" LastEditDate="2019-11-05T17:33:55.193" LastActivityDate="2019-11-05T17:34:21.520" Title="What is the Age Difference Between Earth &amp; Jupiter?" Tags="&lt;solar-system&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="16217" PostTypeId="1" AcceptedAnswerId="16222" CreationDate="2016-06-13T10:25:42.177" Score="1" ViewCount="1858" Body="&lt;p&gt;From the theory of relativity, if any object is moving with the speed of light or more than the speed of light, then the time stops or seemed to be stopped for that object. That means, that the speed of time for an object moving with the speed of light is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what will be the speed of time for an object in the space, travelling at a negligible speed(i.e.,zero) or approximately stationary.&lt;/p&gt;&#xA;" OwnerUserId="13128" LastActivityDate="2016-06-14T15:10:26.463" Title="What is the speed of time?" Tags="&lt;space-time&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16221" PostTypeId="2" ParentId="16217" CreationDate="2016-06-13T11:43:33.143" Score="2" Body="&lt;p&gt;The speed of time is constant for everyone regardless of the reference frame. If (A) moves with the speed of light he won't notice that time is going slower. But for another person (B) not in the same reference frame, he sees his time is going slower. &lt;/p&gt;&#xA;" OwnerUserId="13132" LastEditorUserId="10437" LastEditDate="2016-06-14T15:10:26.463" LastActivityDate="2016-06-14T15:10:26.463" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16222" PostTypeId="2" ParentId="16217" CreationDate="2016-06-13T11:54:37.880" Score="5" Body="&lt;p&gt;The expression &quot;speed of time&quot; doesn't make much sense, since speed is the rate of change of position with time, and &quot;time&quot; doesn't have a position. Hence it doesn't have a speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to know the amount of time dilation at a given speed, it is answered by special relativity. $t' = t\gamma $ where $t$ and $t'$ are the amount of time that passes in the two frames, and&#xA;$$\gamma = \frac{1}{\sqrt{1-v^2/c^2}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At low speeds, $\gamma$ is approximately 1, and at high speeds it approaches $\infty$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To fit with your notion of the speed of time as being the relative rate of passing of time in the two frames, an expression for the speed of time would be $t/t' = 1/\gamma$, which is close to zero at high speeds, and close to 1 for low speeds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the speed of time at low speed is equal to 1. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-06-13T11:54:37.880" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16225" PostTypeId="1" CreationDate="2016-06-13T14:15:45.900" Score="-7" ViewCount="368" Body="&lt;p&gt;Let's think of Saturn, because of its density. Can there be a place near the center of the planet where the gravity upwards is more than towards the center, and with an atmosphere that is more suitable for life. &#xA;Yes, I am calling the hollow Earth extreme, but is it that for a giant gas planet too. Can it be possible?&lt;/p&gt;&#xA;" OwnerUserId="13119" LastEditorUserId="10172" LastEditDate="2016-06-13T18:52:11.773" LastActivityDate="2016-06-13T21:19:07.877" Title="Can a gas giant be hollow?" Tags="&lt;planet&gt;&lt;gravity&gt;&lt;gas-giants&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16227" PostTypeId="2" ParentId="16201" CreationDate="2016-06-13T17:17:37.407" Score="5" Body="&lt;p&gt;The short answer to your question is no, there is not dark matter between galaxies, at least at any appreciable level. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is ordinary matter between galaxies, called the intergalactic medium (IGM).  The IGM is an extremely dilute, hot gas that pervades the space between galaxies.  The typical density of the IGM is about one hydrogen atom per cubic meter and its temperature is somewhere around 10^6 K.  These high temperatures keep the IGM in pressure equilibrium with any galaxies and prevent it from collapsing onto galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark matter, however, is cold, and lacks the pressure support of the gas in the IGM, so it condenses onto galaxies. [1] One interesting difference between dark matter and ordinary matter is that dark matter lacks any sort of cooling mechanism.  As a result, when ordinary matter collapses onto a galaxy, it can cool and condense into a relatively compact space and later form stars and planets.  The dark matter, however, remains extended because it has no way to shed its gravitational potential energy.  Galaxies therefore exhibit a substantial dark matter halo, where the ratio of dark matter to ordinary matter increases substantially a few galactic radii away from the center.  Within the galaxy itself, ordinary matter is dominant by at least an order of magnitude, but the density of ordinary matter falls off exponentially while the dark matter falls off only with the cube of distance, leaving dark matter dominant a few galactic radii away from the galaxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One interesting question is whether or not dark matter halos exist in intergalactic space without a corresponding host galaxy at the center.  As far as we can tell, the answer to that seems to be no, or at least, such structures must be quite small and exceedingly rare.  There are dwarf galaxies (only 10^6 solar masses or so) which have enormous dark matter halos (much out of proportion to their size, relative to the Milky Way), but nearly every halo seems to have at least some ordinary matter present, no matter how feeble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cosmic voids are an even more extreme example.  Voids are dominated by dark energy and therefore expand more rapidly than the rest of the universe.  Voids tend to push any matter (dark or otherwise) into walls that separate them from surrounding voids.  In a real sense, the universe is made of voids, and everything we know lies on the walls and filaments that separate them.  &lt;a href=&quot;http://adsabs.harvard.edu/abs/2008ApJ...686...53T&quot; rel=&quot;noreferrer&quot;&gt;This paper&lt;/a&gt; examined dark matter halos in cosmic voids and essentially found that there weren't any.  In other words, they found that dark matter halos in cosmic voids have host galaxies just as often as dark matter halos outside of cosmic voids (which is to say, almost always). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1]: To be absolutely pedantic, I wouldn't doubt that there are a few dark matter particles whizzing about in intergalactic space.  But I would be surprised if its density exceeded the (already puny) density of ordinary matter in the IGM.&lt;/p&gt;&#xA;" OwnerUserId="10371" LastActivityDate="2016-06-13T17:17:37.407" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16228" PostTypeId="2" ParentId="16211" CreationDate="2016-06-13T18:14:47.910" Score="0" Body="&lt;p&gt;What you want to do is use the relation between Mass, Radius, and Density. The proper expression should be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho_c = \frac{M_c}{(4/3) \pi R_c^3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your three Jeans conditions are $M_c &amp;gt; M_J$, $R_c &amp;gt; R_J$, and $\rho_c &amp;gt; \rho_j$. By using the relation above, you can transform any of the three conditions into one of the others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, going from $M_c &amp;gt; M_J \rightarrow R_c &amp;gt; R_J$ is done as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M_c = \frac{4}{3}\pi R_c^3 \rho_c &amp;gt; (\frac{5kT}{G \mu m_H})^{3/2} (\frac{3}{4 \pi \rho_c}) ^{1/2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, solving for $R_c$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_c^3 &amp;gt; (\frac{5kT}{G \mu m_H})^{3/2} (\frac{3}{4 \pi \rho_c}) ^{3/2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_c &amp;gt; (\frac{15kT}{4 \pi G \mu m_H \rho_c})^{1/2} \equiv R_J$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other transformations follow in a similar manner.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-06-13T18:14:47.910" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16229" PostTypeId="2" ParentId="16216" CreationDate="2016-06-13T18:26:02.097" Score="12" Body="&lt;p&gt;It is generally thought that Jupiter &quot;formed&quot; first via the core accretion process. An icy core forms first and then sweeps up the gas around it. This process is generally thought to take from a few million to ten million years. After that, the solar system gas was dissipated and Jupiter was basically as it is today, but larger and hotter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth formed on a longer timescale in the maelstrom of rocky planetesimals that made up the inner solar system. An essential part of the process was the collision that formed the moon, which probably occurred between 30 and 60 million years after the protosun began to form. Further heavy bombardment of the Earth that may have significantly changed its surface and atmosphere continued for some tens of millions of years after that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wikipedia page on planet formation in the solar system looks ok to me&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Formation_and_evolution_of_the_Solar_System&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Formation_and_evolution_of_the_Solar_System&lt;/a&gt;&#xA;I may add some more academic references subsequently.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="16685" LastEditDate="2019-11-05T17:34:21.520" LastActivityDate="2019-11-05T17:34:21.520" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16230" PostTypeId="2" ParentId="16201" CreationDate="2016-06-13T18:26:37.900" Score="12" Body="&lt;p&gt;Let me see if I can answer at least some of this.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes, there is dark matter between galaxies. This is demonstrated by the fact that in galaxy groups and clusters, you need &lt;i&gt;more&lt;/i&gt; dark matter than is found in the galaxies themselves to explain what's going on: in terms of why the groups or clusters are gravitationally bound in spite of the extreme velocities of the galaxies within the groups/clusters; why the X-ray-emitting gas in clusters is so hot and high-pressure and yet still confined to the cluster; and in order to explain the gravitational lensing of background galaxies by clusters. (Also supported by observations of things like the Sunyaev-Zeldovich effect in clusters.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In fact, the oldest evidence for dark matter was the extremely high velocities of galaxies within clusters as measured by Fritz Zwicky back in the 1930s. This is also one of the failure modes of Modified Newtonian Dynamics as an alternative to dark matter: even if you think you can get rid of dark matter &lt;i&gt;within&lt;/i&gt; galaxies, you still end up needing dark matter within galaxy clusters, in addition to the galaxies themselves.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, of course, cosmological models indicate that dark matter accounts for about 25% of the critical density of the universe. In order to have that much dark matter, you &lt;i&gt;have&lt;/i&gt; to have more dark matter than what's inside individual galaxies.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;If there is dark matter between galaxies. Do we know or believe: what is the ratio of dark matter out there?&quot; I think the simplest way to answer this would be to take the dark-matter/baryonic-matter ratio for the universe as a whole, as determined by current cosmological observations and models. This gives a ratio of about 5:1.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Galaxies exist in filaments and walls. Does dark matter do this? or is it evenly-spread?&quot; Yes, dark matter does this. The reason we see galaxies in filaments, clusters, and walls is &lt;i&gt;because&lt;/i&gt; the dark matter has gravitationally clumped into those structures; on this scale, the galaxies are basically along for the ride, following the gravitational pull of what the dark matter is doing. The reason for galaxies existing in &lt;em&gt;filaments&lt;/em&gt; is because of the uneven distribution of the density of both regular and dark matter throughout the universe. An overdensity leads to the matter collapsing to form (for instance) a galaxy, but the overdensity can be asymmetric. This leads to the collapse happening faster in one particular direction, causing galaxies to form in filaments (see e.g. &lt;a href=&quot;https://arxiv.org/pdf/1404.3956.pdf&quot; rel=&quot;noreferrer&quot;&gt;this summary paper&lt;/a&gt; for an overview).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://wwwmpa.mpa-garching.mpg.de/galform/virgo/millennium/&quot; rel=&quot;noreferrer&quot;&gt;This page&lt;/a&gt; has images from a large-scale cosmological simulation (the Millennium Simulation). Most of the images showing filamentary structure on different scales are showing the dark matter &lt;i&gt;only&lt;/i&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="13193" LastEditDate="2016-06-20T06:46:33.727" LastActivityDate="2016-06-20T06:46:33.727" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16231" PostTypeId="2" ParentId="16225" CreationDate="2016-06-13T21:19:07.877" Score="9" Body="&lt;p&gt;&lt;a href=&quot;http://www.sparknotes.com/physics/gravitation/potential/section3.rhtml&quot;&gt;Newton's shell theorem&lt;/a&gt; proves that inside a gas giant, any layers that are further than you from the centre have a zero gravitational effect on you. So if you are inside a gas giant (and by some magic not dead) the only gravity comes from the layers that are closer to the centre. So the gravitational pull is always down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If fact, the pressure increases as you go down, as the weight of the layers above you causes pressure (just as the the Earth's air is more dense the lower you go). Since the gas is under extreme pressure, it would fill up any gap that formed, (just like the air fills a vacuum on Earth) In a gas giant the pressures are extreme, there is certainly no way that a hole or cavity could ever form.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-06-13T21:19:07.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16232" PostTypeId="1" CreationDate="2016-06-14T04:21:02.837" Score="3" ViewCount="150" Body="&lt;p&gt;I am a beginner, please pardon my non-use of proper astronomical terms. I would appreciate proper edits to my questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have learned that when the $\frac{\mathrm{d}F}{\mathrm{d}r}$ is a very big negative number, that would mean that the far side will have much less gravitational tug than the near side of a satellite. The differential gravity or $\frac{\mathrm{d}F}{\mathrm{d}r}$ will be as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}F}{\mathrm{d}r} = \frac{-2 GMm}{r^3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.wolframalpha.com/input/?i=-2%20*%20(Gravitational%20Constant)%20*%20(mass%20of%20Earth)%20*%20(Mass%20of%20Moon)%20%2F%20(Semimajor%20axis%20of%20moon)%5E3&quot; rel=&quot;nofollow&quot;&gt;If we compute $\frac{\mathrm{d}F}{\mathrm{d}r}$ of Earth-Moon system&lt;/a&gt; we get $-1.308 \times 10^{-5} \ \mathrm{M_o^2G/AU^3} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was surprised to know that Ganymede, being almost 3 times as far away from Jupiter as our Moon is from Earth, still is tidally locked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.wolframalpha.com/input/?i=-2*(Gravitational%20Constant)%20*%20(mass%20of%20Jupiter)%20*%20(Mass%20of%20Ganymede)%20%2F(Ganymede%20orbital%20distance)%5E3&quot; rel=&quot;nofollow&quot;&gt;If we compute $\frac{\mathrm{d}F}{\mathrm{d}r}$ of Jupiter-Ganymede system&lt;/a&gt; we get $-1.565 \times 10^{-6} \ \mathrm{M_o^2G/AU^3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then wondered what is the $\frac{\mathrm{d}F}{\mathrm{d}r}$ of Sun-Earth that our Earth is not tidally locked to it and &lt;a href=&quot;https://www.wolframalpha.com/input/?i=-2*(Gravitational%20Constant)%20*%20(mass%20of%20Sun)%20*%20(Mass%20of%20Earth)%20%2F(Earth%20orbital%20distance)%5E3&quot; rel=&quot;nofollow&quot;&gt;I found $-2.422 \times 10^{-8} \ \mathrm{M_o^2G/AU^3}$&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is a very small negative number so the difference in tug is not significant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the minimum $\frac{\mathrm{d}F}{\mathrm{d}r}$ (or I guess &quot;maximum&quot;) value or a boundary value at which the object can be tidally locked and beyond that, the tidal locking won't occur?&lt;/p&gt;&#xA;" OwnerUserId="4099" LastEditorUserId="9301" LastEditDate="2016-10-17T15:19:00.320" LastActivityDate="2017-01-04T22:15:24.907" Title="What is the cut-off point of $\frac{\mathrm{d}F}{\mathrm{d}r}$ beyond which no tidal locking will occur?" Tags="&lt;tidal-forces&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16233" PostTypeId="2" ParentId="16204" CreationDate="2016-06-14T10:00:19.907" Score="4" Body="&lt;p&gt;In simple terms: yes, the Milky Way's supermassive black hole (SMBH) is at the center of the galaxy, we know approximately where the center is (but not terribly precisely), and we should expect the SMBH to be there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can define the Galactic Center using the orbits of stars and gas: what is the (average) center of their orbits? You can also use the spatial distribution of the same (which is probably a bit closer to what you're thinking of as a &quot;center of gravity&quot; or &quot;center of mass&quot;). This is never going to give you an exact answer, and since the galaxy is not perfectly symmetric or unchanging, there will always be some uncertainty and variation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a 2D &lt;i&gt;coordinate&lt;/i&gt; definition, which is the point on the sky where Galactic latitude and longitude both = 0. This is intended to approximate the direction toward the GC, but since it was defined back in the 1950s, no one is worried if modern attempts to measure the (3D) center of the galaxy might be slightly different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as we can tell, yes, the SMBH is at the Galactic Center. If a SMBH &lt;i&gt;did&lt;/i&gt; form outside the central region of a galaxy, dynamical friction would make it spiral into the center of the galaxy (and damp any radial orbital oscillations as well). This would also cause a SMBH from another (smaller) galaxy that merged with the Milky Way to spiral into the GC and merge with the SMBH there; see my answer to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/13013/why-do-we-believe-that-the-super-massive-black-holes-at-the-centers-of-two-mergi?rq=1&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The SMBH is not the &quot;main player&quot; in our galaxy: it has a mass of about 4 million solar masses, which you can compare with the total stellar mass of about 50 billion solar masses. (And a dark-matter mass of roughly one trillion solar masses.) Nonetheless, there is evidence that SMBHs &quot;co-form&quot;, in some sense, with their host galaxies. In particular, the mass of SMBHs is fairly tightly correlated with several properties of the host galaxy (e.g, the mass of the &quot;bulge&quot;, or the velocity dispersion of stars in the inner part of the galaxy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edited to fix a typo.]&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-06-15T11:56:10.050" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16234" PostTypeId="1" CreationDate="2016-06-14T12:19:03.330" Score="2" ViewCount="155" Body="&lt;p&gt;There's really two questions I'm asking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, when maps are produced of stars and galaxies, are they just produced as we see them now from our viewpoint? Or are they based on the motion of those bodies? Eg Galaxy X is 1 million light years away and we know it's moving at Y speed, so it's real position now is actually 1 million years worth of movement along that trajectory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, are those things taken into account when doing other studies of things like dark matter? Ie things are actually further apart than what we see because we're seeing them as they were Z million years ago. I would expect that would affect calculations for things like how much 'missing' matter there was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To try to further clarify this second question, simplifying things to 3 points on a triangle... if we're at A and we observe points B and C which are X billion light years apart, then the effect of gravity on the speed of B and C moving apart would have been greater in the past when they were closer together and furthermore the effects of the current gravitational force will take more than X billion years to reach the other object. As far as I understand, dark matter is a way to account for observed gravitational effects that can't be accounted for in other ways. When astronomers do these calculations, can I assume that they have taken into account the time delay in the effects of gravity between galaxies? Or have I in my naivety, stumbled on some profound insight! (I doubt it.)&lt;/p&gt;&#xA;" OwnerUserId="13140" LastEditorUserId="13140" LastEditDate="2016-06-15T23:37:40.367" LastActivityDate="2016-06-20T06:52:32.073" Title="Is time taken into account by astronomers when creating and using maps of the universe?" Tags="&lt;space-time&gt;&lt;expansion&gt;&lt;dark-matter&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16235" PostTypeId="1" CreationDate="2016-06-14T13:42:47.100" Score="1" ViewCount="192" Body="&lt;p&gt;I just wanted to double check I had something correct. If an object, say a nebula, is measured to roughly have a diameter or extent of, say, 250 arcseconds, would its diameter in parsec simply therefore be 250 parsecs?&lt;/p&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2016-06-22T13:46:01.560" Title="A quick (elementary) check on arsecond conversion to sky distance in parsecs" Tags="&lt;distances&gt;&lt;parallax&gt;&lt;parsec&gt;" AnswerCount="2" CommentCount="26" ContentLicense="CC BY-SA 3.0" />
  <row Id="16236" PostTypeId="2" ParentId="16235" CreationDate="2016-06-14T14:31:59.440" Score="2" Body="&lt;p&gt;The pedagogical answer is this.  You're confusing the&lt;/p&gt;&#xA;&lt;h1&gt;width&lt;/h1&gt;&#xA;&lt;p&gt;of something, with it's&lt;/p&gt;&#xA;&lt;h1&gt;parallax!&lt;/h1&gt;&#xA;&lt;p&gt;Here's the fixed background&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sP8tT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sP8tT.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here's something that happens to have width ...&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rjaZc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rjaZc.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here's something that has parallax ...&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BXxqu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BXxqu.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;That's the deal!&lt;/p&gt;&#xA;&lt;p&gt;Note that width can just be measured, using one printout of the photo and a wooden ruler.&lt;/p&gt;&#xA;&lt;p&gt;But parallax means &lt;em&gt;two&lt;/em&gt; photos taken 180 days apart, and you measure how much the thing has &lt;em&gt;moved&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;(The mind-boggling &lt;a href=&quot;http://sci.esa.int/gaia/&quot; rel=&quot;nofollow noreferrer&quot;&gt;GAIA&lt;/a&gt; does &lt;em&gt;not actually use&lt;/em&gt; wooden rulers!)&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;One huge confusion with &amp;quot;parsecs&amp;quot; is this:&lt;/p&gt;&#xA;&lt;p&gt;Parallax means the distance &lt;strong&gt;an object moves in two photos&lt;/strong&gt; take six months apart. That technique can be used with objects &lt;strong&gt;from 1 to about 100 parsecs away&lt;/strong&gt; from us.&lt;/p&gt;&#xA;&lt;p&gt;However, astronomers (for convenience) still use the measure &amp;quot;parsecs&amp;quot; even when the distance &lt;strong&gt;is far more than 100 parsecs&lt;/strong&gt;.  And, they even use the measure &amp;quot;parsecs&amp;quot; for things like &amp;quot;size&amp;quot;, which just have &lt;strong&gt;no connection at all&lt;/strong&gt; to parallax photo measurements.&lt;/p&gt;&#xA;&lt;p&gt;If you hear that something is &amp;quot;&lt;strong&gt;5000&lt;/strong&gt; parsecs away&amp;quot;, it is inconceivable the distance was determined using parallax photo measurements!!  And, if you hear that something is &amp;quot;1000 parsecs &lt;strong&gt;wide&lt;/strong&gt;&amp;quot;, that just has no connection at all to parallax photo measurements (it's a width - nothing to do with the distance from Earth!!)&lt;/p&gt;&#xA;&lt;p&gt;In the question at hand here, the &amp;quot;width&amp;quot; of an object was &lt;strong&gt;quoted in parsecs&lt;/strong&gt;. The OP was originally assuming that, since parsecs were mentioned in relation to the width, you could use something related to the &lt;strong&gt;technique of parallax photo measurement&lt;/strong&gt;, in relation to the width.  In fact, parallax photo measurement just has &lt;strong&gt;no connection, whatsoever&lt;/strong&gt; to width. You literally might as well use smell or taste - heh!  The confusion is that parsecs happen to be also used (very confusingly) to measure things like width: but remembering that the technique of parsecs (parallax photo measurement) has absolutely no connection at all to width, and no connection to anything more than 100 ps from us.&lt;/p&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-06-21T22:26:00.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16237" PostTypeId="1" AcceptedAnswerId="24045" CreationDate="2016-06-15T02:11:31.700" Score="3" ViewCount="322" Body="&lt;p&gt;There is a short Wikipedia article &lt;a href=&quot;https://en.wikipedia.org/wiki/Focal_Plane_Arrays&quot; rel=&quot;nofollow noreferrer&quot;&gt;Focal Plane Arrays&lt;/a&gt; that enumerates some projects, but my question is more along the lines of what is (at least) nearly complete or in &quot;first light&quot; phase, even if not commissioned yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to differentiate between focal plane arrays (an array of multiple feeds and amplifiers) used in &lt;em&gt;single dish telescopes&lt;/em&gt;, and those integrated into &lt;em&gt;multiple dish arrays&lt;/em&gt;, because I'm particularly curious about single dishes being used for spatial information, or even true imaging. Ideally, the answer will give some information for each case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of &lt;em&gt;single dish instruments&lt;/em&gt;, are the elements - roughly speaking - used as pixels? Despite the longer wavelength, it's still optics and &lt;em&gt;it is&lt;/em&gt; a telescope. If there are N individual, uncoupled feeds, does one build up an image roughly N times faster? Is the relative phase between the feeds ever used (for &lt;em&gt;single dish instruments&lt;/em&gt;)?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For those like me who aren't already familiar with focal plane arrays, here is a random picture from one of the links I found in a quick internet search. It's from the &lt;a href=&quot;http://www.atnf.csiro.au/research/multibeam/.overview.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Parkes 21cm Multibeam Receiver&lt;/a&gt;, has (had?) 13 receivers and sat at the focus of the 64m dish. The photo is dated 1997 - I have a hunch there's been some development in this technique in the intervening 20 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is this actually the only one?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lE5ke.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lE5ke.png&quot; alt=&quot;Parkes Radio Telescope 21cm Focal Plane Array in 1997&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/T15Ao.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/T15Ao.png&quot; alt=&quot;Parkes Radio Telescope 21cm Focal Plane Array in 1997&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; The Parkes array is still in use as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/URU8o.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/URU8o.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Above:&lt;/strong&gt; Superposition of the half-power beam widths of the 13-element array of the Parkes 21-cm Mutlibeam Receiver, as used in &lt;a href=&quot;http://arxiv.org/abs/1602.07477&quot; rel=&quot;nofollow noreferrer&quot;&gt;a study of a Fast Radio Bursters&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's likely the image is a screen shot from from &lt;a href=&quot;https://www.nature.com/articles/nature17140&quot; rel=&quot;nofollow noreferrer&quot;&gt;The host galaxy of a fast radio burst&lt;/a&gt; Nature volume 530, pages 453–456 (25 February 2016), Keane et al. I can't find my archived copy now, but instead see Phys.org's &lt;a href=&quot;https://phys.org/news/2016-02-fast-radio-discovery-universe.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;New fast radio burst discovery finds 'missing matter' in the universe&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the caption there reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This image shows the field of view of the Parkes radio telescope on the left. On the right are successive zoom-ins in on the area where the signal came from (cyan circular region). The image at the bottom right shows the Subaru image of the FRB galaxy, with the superimposed elliptical regions showing the location of the fading 6-day afterglow seen with ATCA. Image Credit: D. Kaplan (UWM), E. F. Keane (SKAO).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-10-22T13:34:42.893" LastActivityDate="2019-04-27T06:39:40.933" Title="What is the highest granularity focal-plane array on a dish radio telescope? Or is this the ONLY ONE?" Tags="&lt;radio-astronomy&gt;&lt;photography&gt;&lt;radio-telescope&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="16238" PostTypeId="1" AcceptedAnswerId="16242" CreationDate="2016-06-15T02:30:42.470" Score="2" ViewCount="321" Body="&lt;p&gt;That actually sums up my question nicely: &lt;em&gt;How did single dish (or single receiver) radio telescopes originally generate images&lt;/em&gt;? - or at least 2D intensity maps or contour plots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Early radio telescopes were actually not much more than a &lt;a href=&quot;https://en.wikipedia.org/wiki/Directional_antenna&quot; rel=&quot;nofollow&quot;&gt;Directional Antenna&lt;/a&gt; pointing up, connected to a sensitive receiver with various filters, maybe an &lt;a href=&quot;https://en.wikipedia.org/wiki/Intermediate_frequency&quot; rel=&quot;nofollow&quot;&gt;IF stage&lt;/a&gt; or two, but no demodulation. You measured the &quot;noise&quot; signal strength as a function of time on a chart recorder - pen and ink. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there were radio images generated and published &lt;em&gt;well before&lt;/em&gt; there were high granularity interferometers and computational correlators. The early ones were contour plots, on 2D plotters - again pen and ink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;How was this done? How did, say, a single dish antenna with a single feed generate radio maps&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit:&lt;/strong&gt; It doesn't necessarily have to be a dish antenna - it's the &lt;em&gt;single receiver&lt;/em&gt; part I'm interested in.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="37191" LastEditDate="2021-04-13T13:58:50.477" LastActivityDate="2021-04-13T13:58:50.477" Title="How did single dish (or single receiver) radio telescopes originally generate images?" Tags="&lt;observational-astronomy&gt;&lt;radio-astronomy&gt;&lt;photography&gt;&lt;radio-telescope&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16239" PostTypeId="1" AcceptedAnswerId="18751" CreationDate="2016-06-15T03:00:24.307" Score="3" ViewCount="176" Body="&lt;p&gt;For this question, I think a good working definition of &lt;a href=&quot;https://en.wikipedia.org/wiki/Co-orbital_configuration&quot; rel=&quot;nofollow&quot;&gt;co-orbital configuration&lt;/a&gt; would be two bodies orbiting around a third much larger body in a 1:1 resonance and where neither mass is negligible. In other words, you can't really say that either one is &lt;em&gt;at&lt;/em&gt; the other's $L_3$ because they are interacting with each other more as equals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this article in the &lt;a href=&quot;https://www.newscientist.com/article/dn20160-two-planets-found-sharing-one-orbit/&quot; rel=&quot;nofollow&quot;&gt;New Scientist&lt;/a&gt; in my first on-line search, but it has a retraction at the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for example, here is a theoretical analysis of stability in an ArXiv preprint on the subject: &lt;a href=&quot;http://arxiv.org/abs/1004.0726&quot; rel=&quot;nofollow&quot;&gt;Dynamics of two planets in co-orbital motion&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the caveats above (which I think are fairly standard for the term &quot;co-orbital planet&quot;) have any been discovered? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit:&lt;/strong&gt; Have any new candidates been at least identified or proposed?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2016-06-27T01:23:23.443" LastActivityDate="2016-10-17T14:52:11.203" Title="Have any co-orbital exoplanet pairs been discovered (and not subsequently retracted)?" Tags="&lt;planet&gt;&lt;exoplanet&gt;&lt;orbital-mechanics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16240" PostTypeId="1" CreationDate="2016-06-15T04:33:01.483" Score="6" ViewCount="267" Body="&lt;p&gt;This question is about design aspects of large radio telescope dishes which allow them to flex under the influence of gravity as they change elevation angle, and still maintain good optical performance at short wavelengths - on the order of a centimeter or less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What is a radio &quot;homology telescope&quot; and is the 500m dish in China one?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was reading about radio astronomy for background on &lt;a href=&quot;https://astronomy.stackexchange.com/q/16238/7982&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;https://astronomy.stackexchange.com/q/16237/7982&quot;&gt;this&lt;/a&gt; question. In this &lt;a href=&quot;http://www.cv.nrao.edu/course/astr534/RadioTelescopes.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;informative and entertaining&lt;/a&gt; NRAO page I saw the term &lt;em&gt;homology telescope&lt;/em&gt; for the first time, so I immediately select-right-click-serched it. At first I was confused by these Mathematics Stackexchange and Stanford Bio-X links:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://math.stackexchange.com/q/1323553/284619&quot;&gt;homology of mapping telescope of a monoid&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://math.stackexchange.com/q/1341729/284619&quot;&gt;action of a monoid on a mapping telescope&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://biox.stanford.edu/research/seed-grants/genome-space-telescope-bringing-dna-sequence-homology-21st-century&quot; rel=&quot;nofollow noreferrer&quot;&gt;Genome Space Telescope: Bringing DNA Sequence Homology into the 21st Century&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;until I realizes that these quit different &lt;em&gt;usages&lt;/em&gt; of homology and/or telescope :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the &lt;a href=&quot;https://books.google.com.tw/books?id=DoDBBAAAQBAJ&amp;amp;lpg=PA191&amp;amp;ots=vRxGPaU7AH&amp;amp;dq=homology%20telescope&amp;amp;pg=PA188#v=onepage&amp;amp;q=homology&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;following explanation&lt;/a&gt; in &lt;em&gt;Tools of Radio Astronomy&lt;/em&gt; by  Thomas Wilson, Kristen Rohlfs and Susanne Huettemeister (Hüttemeister) particularly concise and helpful:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;7.5 The Practical Design of Parabolic Reflectors&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;7.5.1 General Considerations&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Measurements of the mechanical properties of an antenna are of importance for its performance. This is especially true if the antenna deforms homologously. By &lt;em&gt;Homology&lt;/em&gt; it is meant that, at various elevations, the main reflector deforms from one paraboloid into another. Today, homology is an intrinsic part of the design of all symmetric reflectors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later in the same subsection:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The design of an offset paraboloid has, however, complications. Since the design has less symmetry homology is more difficult to achieve and therefore active, real time adjustments of the surface are needed if the design limit of 7 mm or perhaps 3mm wavelength is to be reached. This will be accomplished by an actuator system controlled in real time by a laser measuring system. But for longer wavelengths, the GBT will not require active surface adjustment (Fig. 7.10).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is a screen capture of Figure 7.10 from &lt;a href=&quot;https://books.google.com.tw/books?id=DoDBBAAAQBAJ&amp;amp;lpg=PA191&amp;amp;ots=vRxGPaU7AH&amp;amp;dq=homology%20telescope&amp;amp;pg=PA188#v=onepage&amp;amp;q=homology&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;google&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eJSJz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eJSJz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;note:&lt;/strong&gt; There are a number of nice photos and some discussion of the Green Bank Telescope and discussion of it's homology (homologusness?) on the &lt;a href=&quot;http://www.cv.nrao.edu/course/astr534/RadioTelescopes.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NRAO page&lt;/a&gt; where I started, as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So here is the question-cluster&lt;/strong&gt; that I'm currently stuck on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If off-axis telescopes require active mechanical adjustments to maintain a homologous paraboloid figure, do the on-axis telescopes mentioned do it passively - &lt;em&gt;by mechanical design&lt;/em&gt; alone? They naturally droop - &lt;em&gt;by design&lt;/em&gt; - differently at different elevations (different angles above the horizon) in order to retain a paraboloid figure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) For the passive design, is it a) &lt;em&gt;the same&lt;/em&gt; paraboloid but just a different section, b) or another paraboloid with the same focal point, or c) has a different focal point and therefore requires axial movement of the secondary mirror or feeds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) is the nearly-completed 500m dish in China (&lt;a href=&quot;https://en.wikipedia.org/wiki/Five_hundred_meter_Aperture_Spherical_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;FAST&lt;/a&gt;) a &lt;em&gt;homology telescope&lt;/em&gt;? I know it has a lot of articulation and works off axis, but the word &lt;em&gt;spherical&lt;/em&gt; is in the name and the receiver/feed translates laterally. &lt;em&gt;Is it still using the principle of homology&lt;/em&gt;, but in a different way?   &lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2020-11-24T01:06:56.047" LastActivityDate="2020-11-24T01:06:56.047" Title="What is a radio &quot;homology telescope&quot; and is the 500m dish in China one?" Tags="&lt;radio-astronomy&gt;&lt;terminology&gt;&lt;radio-telescope&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16241" PostTypeId="1" CreationDate="2016-06-15T07:01:04.317" Score="1" ViewCount="68" Body="&lt;p&gt;I had this image, a picture of Jupiter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GdFGw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GdFGw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to measure the pixel diameter of Jupiter. As evident, the image is quite blurry, and thus, many value of equally justified pixel diameters can be obtained. To eliminate this I posterised the image (which is technically, quote from Gimp, &quot;designed to intelligently weigh the pixel colors of the selection or active layer and reduce the number of colors while maintaining a semblance of the original image characteristics&quot;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SgIxY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SgIxY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the above result, I obtained three pixel diameters of Jupiter (one from each of the coloured circles), and with a little bit of calculation, realised that the yellow circle gave rise to the most accurate value of Jupiter's diameter (which was surprisingly off by less than one percent!). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My questions are:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1)&lt;/strong&gt; Why is my original image like that? Is it due to some form of light scattering?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2)&lt;/strong&gt; Is there a better method of obtaining an accurate pixel diameter than posterising the image&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3)&lt;/strong&gt; Is the posterised result expected? Are there names that can be given to the three distinct spherical objects (which seem to resemble shadows, like the umbra, prenunbra, antumbra, etc.)? Is there a reason why the middle posterised object gives the most accurate value?&lt;/p&gt;&#xA;" OwnerUserId="9650" LastActivityDate="2016-06-15T14:06:31.207" Title="Data Analysis: Image Sharpening through Posterisation" Tags="&lt;data-analysis&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16242" PostTypeId="2" ParentId="16238" CreationDate="2016-06-15T07:19:12.157" Score="6" Body="&lt;p&gt;They scan the object, if you point the dish a a point in the sky as the Earth rotates the dish scans across astronomical objects, then move the dish to point at a slightly different position and let it scan across the object again, and again. After a while you can re-construct an image from the scan lines in a similar way to analogue TV.&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-06-15T07:19:12.157" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16243" PostTypeId="2" ParentId="16234" CreationDate="2016-06-15T07:48:36.977" Score="2" Body="&lt;p&gt;For any local map of galaxies, the relative velocity is slow enough as to be irrelevant.    Andromeda and the Milky way are roughly 2.5 million light years apart and they are moving towards each other currently at about 1 light year every 2,270 years (&lt;a href=&quot;http://phenomena.nationalgeographic.com/2014/03/24/scientists-predict-our-galaxys-death/&quot; rel=&quot;nofollow&quot;&gt;based on 402,000 km/h&lt;/a&gt;).   In the 2.5 million years that it takes for Andromeda's light to reach us, it's moved less than 0.05% closer to us, so any adjustment on the maps to account for Andromeda's distance relative to our galaxy would be very small and not worth adjusting unless you're actually plotting it's movement in great detail.   For most maps, what we see, even 2.5 million years old, is close enough and that's true for all local galaxies.  That said, your intuition is correct, the light we see and the pictures made of Andromeda are where it was 2.5 million years ago.  The more interesting movement is that it's rotated a handful of degrees in that time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For maps over a billion light years across or of the known universe, then it's a little different because the relative movement is sufficient to measurably have changed the map in the time it took the light to reach us.   That said, calculating where galaxies would be now vs where they were when the light left them is rather a lot of work.   What's commonly done, rather than show distance for maps of that size is that they're usually coded with a redshift percentage instead of the more familiar distance ratio key.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Redshift(V&lt;sub&gt;H&lt;/sub&gt;/c)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://www.cfa.harvard.edu/image_archive/2011/52/hires.jpg&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, lower right corner.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize I've not answered your 2nd question on missing matter.  I'm not sure what you're asking. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-06-15T08:21:59.290" LastActivityDate="2016-06-15T08:21:59.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16244" PostTypeId="2" ParentId="16241" CreationDate="2016-06-15T09:18:16.997" Score="3" Body="&lt;p&gt;1) your original image is blurred and possibly overexposed. I can't tell if the blurring is due to poor focus or inaccurate tracking. I think you really need much better, sharper images that show the true boundary of the planet before you can start to measure diameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) with a sharp image, there should be no need for posterising the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) &quot;Is there a reason why the middle posterised object gives the most accurate value&quot; - well we could claim that it marks the &quot;average&quot; estimated diameter, but in my opinion the answer is: coincidence.  If you posterised it to give, say, four bands instead of three, which would you take as your &quot;true&quot; diameter? And without knowing the real diameter of Jupiter in advance, how would you ever know?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: by the way, I noted your previous question on this subject. Did you read and use those answers? How sure your magnified image is really the size you think? You could try to verify this by photographing a double star of known separation...&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2016-06-15T09:18:16.997" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16245" PostTypeId="1" AcceptedAnswerId="16249" CreationDate="2016-06-15T09:36:31.000" Score="6" ViewCount="178" Body="&lt;ul&gt;&#xA;&lt;li&gt;The Sun is losing some of its mass during explosions.&lt;/li&gt;&#xA;&lt;li&gt;The Gravitational force between Sun and Earth will change (probably decrease).&lt;/li&gt;&#xA;&lt;li&gt;If Earth maintains its speed, it will change its orbit.&lt;/li&gt;&#xA;&lt;li&gt;Hence, Earth will decelerate&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is this true?&lt;/p&gt;&#xA;" OwnerUserId="13149" LastEditorUserId="2365" LastEditDate="2016-06-15T18:32:52.290" LastActivityDate="2016-06-15T18:55:50.153" Title="Is Earth slowing down in order to stay on its orbital track?" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16246" PostTypeId="2" ParentId="16245" CreationDate="2016-06-15T10:06:56.063" Score="6" Body="&lt;p&gt;As the Sun's mass is reduced and it's gravitational attraction is thus reduced the Earth will &lt;strong&gt;slowly&lt;/strong&gt; spiral out from the Sun and yes slow down (to preserve angular momentum). But the effect is ridiculously small (do the math yourself) it will not be measurable directly.&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-06-15T10:06:56.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16247" PostTypeId="2" ParentId="16241" CreationDate="2016-06-15T13:29:44.230" Score="2" Body="&lt;p&gt;I would do this bare hands so to speak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read the image into a package that allows you to do arithmetic on the data. Sum the colour channels. Then threshold at something like half maximum and count the number $N$ of pixels above threshold. You might want to look at the result at this point to make sure the processing has not captured a feature that is not wanted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is something like posterisation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can define the half maximum equivalent diameter as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D=2(N/π)^{1/2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing this, in GNU Octave, with your top image I get $D\approx 394.6$ pixels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would also help if the pixel values were not saturated, but then I suppose we have to work with the image we are given, at least it means we don't have to worry about limb darkening or contrast stretching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GNU Octave session:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt;&amp;gt; i=double(imread(&quot;H:/Desktop/Jupiter.png&quot;));&#xA;&amp;gt;&amp;gt; ii=i(:,:,1)+i(:,:,2)+i(:,:,3);&#xA;&amp;gt;&amp;gt; idd=(ii&amp;gt;max(max(ii))/2);&#xA;&amp;gt;&amp;gt; N=sum(sum(idd))&#xA;N =  122287&#xA;&amp;gt;&amp;gt; D=2*sqrt(N/pi)&#xA;D =  394.59&#xA;&amp;gt;&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2016-06-15T14:06:31.207" LastActivityDate="2016-06-15T14:06:31.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16248" PostTypeId="2" ParentId="14525" CreationDate="2016-06-15T18:17:52.297" Score="4" Body="&lt;p&gt;Large radio telescope have pretty good pointing accuracy:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the individual dishes of the &lt;a href=&quot;https://science.nrao.edu/facilities/vla/docs/manuals/oss/performance/sensitivity&quot; rel=&quot;nofollow noreferrer&quot;&gt;VLA are accurate&lt;/a&gt; to about 10 arcsec.&lt;/li&gt;&#xA;&lt;li&gt;the giant &lt;a href=&quot;http://www.jodrellbank.net/visit/whats-here/lovell-telescope/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lovell telescope&lt;/a&gt; at Jodrell Bank has a similar accuracy.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The second critical parameter is the &lt;a href=&quot;http://www.cv.nrao.edu/course/astr534/Interferometers1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;beam width&lt;/a&gt;. Beam width depends highly on frequency:&lt;/p&gt;&#xA;&lt;p&gt;beamwidth = wave length/dish diameter&lt;/p&gt;&#xA;&lt;p&gt;When you use several dishes in an interferometer, you can increase their effective accuracy and &lt;a href=&quot;https://science.nrao.edu/facilities/vla/docs/manuals/oss/performance/resolution&quot; rel=&quot;nofollow noreferrer&quot;&gt;decrease their collective beam width&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;A bit more info on the mechanics of the Lovell telescope:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A control computer calculates the required drive rates to follow each radio source. The drive motors are servo-controlled, so there is a continuous check that the correct rate has been achieved. The position of the telescope is constantly monitored and fed back to control computer to ensure that the telescope is pointing correctly.&lt;/p&gt;&#xA;&lt;p&gt;For good tracking the pointing accuracy should be about a twentieth of the resolution. Since the resolution is proportional to the wavelength being received (see below), it follows that the pointing accuracy is more critical at shorter wavelengths. The control computer is able to correct for pointing errors caused by the telescope bowl sagging under its own weight as it moves up and down. In this way the pointing errors can be kept to about 10 arcsec.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So, &lt;a href=&quot;https://en.wikipedia.org/wiki/Ward_Leonard_control&quot; rel=&quot;nofollow noreferrer&quot;&gt;servo motors&lt;/a&gt; and presumably, calibration are what enables this accuracy.&lt;/p&gt;&#xA;&lt;p&gt;The Lovell telescope has 2 elevation drive motors with gearboxes on each side of the dish. These can be driven so that one motor pulls the dish, and the dish drags along the second motor. This eliminates gear backlash in the system: the two gear trains are &amp;quot;wound up&amp;quot; in opposite directions. This means they can make adjustments without gear backlash interfering. (source: a video presentation that runs in the visitor center at Jodrell Bank)&lt;/p&gt;&#xA;&lt;p&gt;Further reading: &lt;a href=&quot;https://archive.org/stream/TheStoryOfJodrellBank/Piper-theStoryOfJodrellBank_djvu.txt&quot; rel=&quot;nofollow noreferrer&quot;&gt;the story of Jodrell Bank&lt;/a&gt;, and a &lt;a href=&quot;http://www.rfcafe.com/references/radio-electronics/jodrell-bank-radio-telescope-february-1958-radio-electronics.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Radio Electronics article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="361" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-07-27T20:53:11.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16249" PostTypeId="2" ParentId="16245" CreationDate="2016-06-15T18:55:50.153" Score="6" Body="&lt;p&gt;The sun is slowly losing mass, partly from the conversion of mass to energy (which then escapes as neutrinos and light) and partly from the solar wind (particularly in coronal mass ejections). Solar wind accounts for a loss of about 1.5 million tons per second, fusion accounts for 4 million tons per second. However compared to the mass of the sun (about 2 million billion trillion tonnes) it is tiny&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth isn't on rail tracks, so as the sun loses mass, it will change its orbit. Each year it moves about 1.6cm further from the sun, and since more distant orbits are slower, it will also slow down. After a billion years the Earth will have slowed to 99.999% of its current speed.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-06-15T18:55:50.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16250" PostTypeId="1" CreationDate="2016-06-15T19:13:52.107" Score="1" ViewCount="810" Body="&lt;p&gt;Gravity is related to mass, and gravity is an attracting force on every body. Then what happens when bodies only attract each other..without repulsion they may strike the sun. I think there must be a repulsive force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question arises, if only attraction is there, then why does the moon rotate elliptically around the earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And Mars comes closer to earth after 3 years?&lt;/p&gt;&#xA;" OwnerUserId="13154" LastEditorUserId="6" LastEditDate="2016-06-15T19:22:18.353" LastActivityDate="2016-10-26T18:33:20.027" Title="If the planets only attract, then how can they stay in their orbit?" Tags="&lt;planet&gt;&lt;solar-system&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16251" PostTypeId="2" ParentId="16250" CreationDate="2016-06-15T19:32:36.390" Score="8" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/8254/do-planets-repel&quot;&gt;Gravity does not repel&lt;/a&gt;, but there is another &quot;force&quot; at work here. The angular momentum of the original protoplanetary disk is conserved. In other words, just because the cloud condensed into planets doesn't mean it is going to stop spinning. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we orbit artificial objects, we speed them up until they reach a velocity that sustains orbit--so again there is another force at work, just not gravity. Gravity keeps objects in orbit (rather than letting them fly off into space). Imagine if you spin a &lt;a href=&quot;https://en.wikipedia.org/wiki/Yo-yo&quot; rel=&quot;nofollow noreferrer&quot;&gt;yo-yo&lt;/a&gt; around in the air. The tension of the string is like gravity, keeping the yo-yo in orbit, but if you let go of the string then the other &quot;force&quot; (its momentum) will cause it to fly off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Technically, momentum is not a force. It is the product of the mass and velocity of the system--a conserved quantity, but the object required a force to bring it up to that velocity. In the case of a planet, it is the astronomical forces that acted on the protoplanetary disk. In the case of the yo-yo, it was your applied force to spin it.)&lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-06-15T19:32:36.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16252" PostTypeId="1" AcceptedAnswerId="16258" CreationDate="2016-06-15T21:06:35.313" Score="8" ViewCount="309" Body="&lt;p&gt;For example; is there any possibility for the moon to always have a side facing the Sun whilst orbiting Earth? And if so then what would the day cycle be like?&lt;/p&gt;&#xA;" OwnerUserId="13156" LastEditorUserId="2618" LastEditDate="2017-03-07T00:46:40.733" LastActivityDate="2018-07-22T12:48:21.827" Title="Is it possible for a moon to continuously have a side facing its star whilst orbiting a planet?" Tags="&lt;the-sun&gt;&lt;orbit&gt;&lt;gravity&gt;&lt;natural-satellites&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16253" PostTypeId="1" CreationDate="2016-06-15T21:25:35.300" Score="3" ViewCount="104" Body="&lt;p&gt;I am looking for images of astronomical objects taken at the ground level of the actual object, by something on the actual object, i.e. the Mars Rover.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the complete list and accurate?.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Earth&lt;/li&gt;&#xA;&lt;li&gt;Mars&lt;/li&gt;&#xA;&lt;li&gt;Titan&lt;/li&gt;&#xA;&lt;li&gt;Venus&lt;/li&gt;&#xA;&lt;li&gt;67P/Churyumov-Gerasimenko&lt;/li&gt;&#xA;&lt;li&gt;Moon&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(I am not interested in images of objects take from the ground level of Earth).&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="13157" LastEditorUserId="13157" LastEditDate="2016-06-16T08:17:52.260" LastActivityDate="2016-06-16T08:17:52.260" Title="Which astronomical objects do we have images of, taken at ground level?" Tags="&lt;photography&gt;" AnswerCount="0" CommentCount="7" ClosedDate="2016-06-16T14:16:51.120" ContentLicense="CC BY-SA 3.0" />
  <row Id="16254" PostTypeId="2" ParentId="16252" CreationDate="2016-06-15T22:57:12.490" Score="6" Body="&lt;p&gt;The moon is gravitationally locked to the Earth. Because gravitational field depends on distance, the Earth's gravity does not affect the moon uniformly: the farthest side is not attracted as strongly as the closest side, these are tidal effects. This causes the heaviest side of the moon to face the earth, i.e. the moon is tidally locked to the earth. The sun is much farther away (~360 times). That means that the difference in the gravitational field vector from the sun across the moon (i.e. tidal effects) is much lower. Therefore, if an object is gravitationally bound, it will be so to the object it orbits. If we were close enough to the sun that the moon would be affected more by the tidal effects of the sun than those of the Earth, we would be so close that the moon would no longer orbit the Earth because it would be pulled away by the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So no, a moon will not be tidally locked to a star. However, if a moon were perfectly spherically symmetric (and solid), it would not be able to be tidally locked. Then, by &lt;em&gt;extreme&lt;/em&gt; coincidence it &lt;em&gt;could&lt;/em&gt; have a sidereal rotation period similar to the orbital period of its planet around the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are objects that are tidally locked to stars, &lt;s&gt;for example Mercury. Mercury orbits the sun in 87 days, and it rotates about its axis in 87 days. That means that on one side of Mercury it is always day, and on the other side it is always night, there is no day-night cycle on Mercury.&lt;/s&gt; Mercury is actually in orbital resonance, see below.&lt;/p&gt;&#xA;" OwnerUserId="13159" LastEditorUserId="13159" LastEditDate="2018-07-22T12:48:21.827" LastActivityDate="2018-07-22T12:48:21.827" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="16255" PostTypeId="1" CreationDate="2016-06-16T08:00:32.160" Score="4" ViewCount="601" Body="&lt;p&gt;The gravity wave detections (GW150914 and GW151226) have both been ambiguous in their exact position, the LIGO papers (&lt;a href=&quot;https://www.ligo.caltech.edu/image/ligo20160211b&quot; rel=&quot;nofollow&quot;&gt;GW150914&lt;/a&gt; and &lt;a href=&quot;http://journals.aps.org/prl/pdf/10.1103/PhysRevLett.116.241103&quot; rel=&quot;nofollow&quot;&gt;GW151226&lt;/a&gt;) give no precise information about the location of the origin of the waves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of GW150914, the approximate location is in the area known as the Magellanic Clouds. In case of GW151226, the parsecs are known, but no precise location. There are two detectors (in Hanford and Livingstone,USA) that made gravitational wave detection possible, does one need a third detector(or more) to pinpoint the position of the origin of these waves? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words can triangulation from three detectors pinpoint the location  or are there other techniques for pinpointing the origin of gravitational waves? &lt;/p&gt;&#xA;" OwnerUserId="11102" LastEditorUserId="2270" LastEditDate="2016-06-16T14:05:50.697" LastActivityDate="2019-01-24T10:53:53.473" Title="Techniques for locating origin of gravitational waves" Tags="&lt;gravity&gt;&lt;black-hole&gt;&lt;gravitational-waves&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16256" PostTypeId="1" CreationDate="2016-06-16T11:21:39.860" Score="1" ViewCount="168" Body="&lt;p&gt;A recent &lt;a href=&quot;http://www.nytimes.com/2016/06/12/opinion/sunday/yes-there-have-been-aliens.html?_r=0&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; about the existence of extraterrestrial life claimed that &quot;unless the probability for evolving a civilization on a habitable-zone planet is less than one in 10 billion trillion, then we are not the first.&quot;&#xA;It continues to say that &quot;a probability for civilizations to form of one in 10 billion per planet was considered highly pessimistic.&quot;&#xA;Here's a &lt;a href=&quot;http://online.liebertpub.com/doi/full/10.1089/AST.2015.1418&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; to the actual study. The author of the article is a coauthor of the study.&#xA;Could someone go through the study and explain the source of the probabilities and their assessments (optimistic vs pessimistic.) &lt;/p&gt;&#xA;" OwnerUserId="13136" LastEditorUserId="13136" LastEditDate="2016-06-19T23:42:41.230" LastActivityDate="2016-06-19T23:42:41.230" Title="why was the probability of one in 10 billion per planet considered highly pessimistic by a recent study on aliens?" Tags="&lt;extra-terrestrial&gt;" AnswerCount="2" CommentCount="19" ClosedDate="2016-06-19T09:01:50.757" ContentLicense="CC BY-SA 3.0" />
  <row Id="16257" PostTypeId="2" ParentId="16256" CreationDate="2016-06-16T12:38:05.087" Score="0" Body="&lt;p&gt;The main reason for this is the new data we aquired with the Kepler telescope and our findings about the goldilocks zone (where temperature is right for water to be liquid on the planet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the last decade a lot of discoveries, like the highly likely liquid water on the moon europe (which is way outside of the goldilock zone), the findings of amino acids in stelar clouds, etc, lead to the understanding that the chemical ingrediants and the chemical processes for life are possible in way more places than we thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding to this, the data from the Kepler telescope suggests that around 25% of all solar systems have a potentially habitable planet in some way (which makes that around 60-100 billion potential habitable planets in the milky way alone)&lt;/p&gt;&#xA;" OwnerUserId="10271" LastEditorUserId="9301" LastEditDate="2016-06-18T12:55:44.257" LastActivityDate="2016-06-18T12:55:44.257" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="16258" PostTypeId="2" ParentId="16252" CreationDate="2016-06-16T12:42:31.547" Score="8" Body="&lt;p&gt;I worked out the math, it's pretty straight forward.   If we take the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;noreferrer&quot;&gt;Hill sphere&lt;/a&gt;, which is an estimate for the furthest possible orbit, and I'm going to just run the math on circular orbits.  Eliptical orbits are harder to put in resonance anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the simple Hill Sphere formula.  $3\frac{r^3}{a^3} = \frac{m}{M}$, where $a$ is the semi major axis planet to sun and $r$ is the distance moon to the planet, little $m$, mass of the planet, big $M$, mass of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate tidal forces, the force changes with the 3rd power of the radius, so for equal tidal force on a moon from both the star it's planet orbits and the planet it orbits, $\frac{r^3}{R^3} = \frac{m}{M}$, and that rather neatly puts the equal tidal forces outside the outer edge of the Hill Sphere by a factor of the cube root of 3, or about 1.44&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outer half of the Hill Sphere isn't long term stable anyway in real systems, so for the Moon to experience a stronger tidal force from the star it's planet orbits than it experiences from it's planet, it would need to be well outside the stable orbital region around the planet, by at least a multiple of 2.88.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no way for any moon in any system to be tidally locked to it's star and not it's planet.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-06-20T23:53:39.613" LastActivityDate="2016-06-20T23:53:39.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16262" PostTypeId="1" CreationDate="2016-06-16T16:02:01.750" Score="4" ViewCount="96" Body="&lt;p&gt;This recent paper in Nature &lt;a href=&quot;http://www.nature.com/ncomms/2016/160614/ncomms11851/full/ncomms11851.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;A new type of solar-system material recovered from Ordovician marine limestone&lt;/a&gt; is available to read on line, and &lt;a href=&quot;http://www.nature.com/ncomms/2016/160614/ncomms11851/fig_tab/ncomms11851_F3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Figure 3&lt;/a&gt; is already widely publicized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xDuvp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xDuvp.jpg&quot; alt=&quot;Figure 3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Comparison of Δ17O versus ε54Cr of achondrites, and ordinary chondrites fallen on Earth in recent time with Österplana 065 (Öst 65) and fossil L-chondritic meteorites Österplana 018, 029 and 032 (Öst 18, Öst 29 and Öst 32 in figure). The fields for ordinary chondrites (OC), Mars (SNCs), Earth and earth-likes, Vesta (HEDs), brachinites, ureilites, winonaites (win), acapulcoites (acp)/lodranites/ungrouped achondrites (ung) and a newly identified unique metal-rich chondrite are marked with select representative samples with available data. Symbol colours indicate meteorite type or grouping. Note that the carbonaceous chondrites and affiliated achondrites plot outside the field, with highly positive ε54Cr. The ε54Cr literature values are from refs 12, 18, 19, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, and Δ17O literature values are from refs 10, 12, 25, 30, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The paper is long and careful, and in any scientific field of classification, there is a huge amount of work and thought and discussion that is sometimes difficult for those outside to understand without the necessary background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last two sentences of the abstract read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;...This may be the first documented example of an ‘extinct’ meteorite, that is, a meteorite type that does not fall on Earth today because its parent body has been consumed by collisions. The meteorites found on Earth today apparently do not give a full representation of the kind of bodies in the asteroid belt ~500 Myr ago.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this part of the image, without knowing a lot about meteorite taxonomy but a little bit about error bars, the image alone doesn't convince me of much at all by itself. There are good mathematical techniques to try to find clustering in data, and probably some methodologies to apply to calculate the probability that the clustering is likely to arise by random chance or is likely to be &quot;real&quot; (which is just another way to say - albeit loosely - not likely to arise by random chance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JyCwJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JyCwJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing those colorized islands probably come from statistical analysis of a very large number of meteorites, but I don't know that for sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these color islands generally accepted and their shapes published independently somewhere? Is there a plot where I can see &lt;em&gt;all of the points&lt;/em&gt; that were used to identify these clustering? How exactly were these contours drawn?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, is there a statistical analysis using these error bars and contours that express the probability that &lt;em&gt;the extinct meteorite specimen described in the paper&lt;/em&gt; is part of the distribution or not? Of course a number by itself is weak because of the softness of the assumptions, but is there one and if so how was it calculated?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2016-06-17T02:24:37.280" LastActivityDate="2016-06-17T02:24:37.280" Title="&quot;Extinct Species&quot; of meteorite - how quantitative is the statistical analysis?" Tags="&lt;data-analysis&gt;&lt;meteorite&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16263" PostTypeId="1" CreationDate="2016-06-16T17:01:33.730" Score="1" ViewCount="31" Body="&lt;p&gt;I'm trying to make light curves for Pluto using old data. Most papers give magnitudes &quot;reduced to standard viewing geometry&quot; of &lt;em&gt;r&lt;/em&gt; = 39.5 AU, &lt;em&gt;&amp;Delta;&lt;/em&gt; = 38.5 AU, and &lt;em&gt;&amp;alpha;&lt;/em&gt; = 1 degree, where &lt;em&gt;r&lt;/em&gt; is heliocentric distance, &lt;em&gt;&amp;Delta;&lt;/em&gt; is geocentric distance, and &lt;em&gt;&amp;alpha;&lt;/em&gt; is phase angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When papers give apparent magnitudes, I'm left to perform the reductions myself. I'm fine with the adjustment for &lt;em&gt;r&lt;/em&gt; and &lt;em&gt;&amp;Delta;&lt;/em&gt;, but how do I reduce to phase angle of 1 degree?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also really appreciate a good source if you have it on hand.&lt;/p&gt;&#xA;" OwnerUserId="13168" LastActivityDate="2016-07-02T15:33:17.497" Title="How do I reduce planetary photometry to standard viewing geometry using phase angle?" Tags="&lt;pluto&gt;&lt;photometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16264" PostTypeId="2" ParentId="16255" CreationDate="2016-06-16T17:05:50.600" Score="3" Body="&lt;h3&gt;Short answer&lt;/h3&gt;&#xA;&lt;p&gt;Three detectors would improve the ability to pinpoint the direction of the gravitational waves drastically.&lt;/p&gt;&#xA;&lt;h3&gt;Long answer&lt;/h3&gt;&#xA;&lt;p&gt;Finding the direction of the source of a gravitational wave, as with most things in science, is harder than it might initially seem. I'd like to point out that the process used in determining direction is not precisely a form of triangulation (even if it had 3+ detectors) because triangulation involves determining precise position from knowing distances to the object from 3 or more locations. That's not really what's happening here.&lt;/p&gt;&#xA;&lt;p&gt;I'll outline the basics for determining the direction and how that method might be improved with more detectors, but you can also read this &lt;a href=&quot;http://www.thephysicsmill.com/2016/03/06/direction-ligos-gravitational-waves/&quot; rel=&quot;nofollow noreferrer&quot;&gt;excellent source&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;Time of Arrival&lt;/h2&gt;&#xA;&lt;p&gt;The main way we can determine direction is time of arrival at the various detectors. The currently active detectors are located some 10 milli-lightseconds apart. That means one of them is going to detect the wave first, then sometime later, the other one detects it. Given that the wave has to be traveling at the speed of light, you can determine the direction it was traveling from such that it traversed the distance between the two detectors in the time difference measured.&lt;/p&gt;&#xA;&lt;p&gt;But, this presents a problem because this method has degeneracies with only 2 detectors. Really, all you can calculate is the angle of the direction of the wave with respect to the line connecting your two detectors. That only gives you a cone on the sky of where the wave could've come from and you can't pinpoint the azimuthal angle without more information. This concept can be seen in the image below showing the probable direction. You can see they've limited it further, but their shape still takes part of a circle (the projection of that cone on the sky).&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5c1tO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5c1tO.jpg&quot; alt=&quot;location of wave&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;What if you added another detector?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;As stated above, you have a degeneracy with only 2 detectors, but adding a third would break that degeneracy. You'd turn your cone into a point and you'd know much more precisely where the source came from. More detectors would allow you to pinpoint that location even more precisely, but only 3 are necessary to remove all degeneracies.&lt;/p&gt;&#xA;&lt;p&gt;Now, of course I've been describing how it would work &amp;quot;in theory&amp;quot;. In practice, observations are usually much more messy. Any measurement comes with error, so even with 3 detectors you're not going to be able to pinpoint the location exactly. You'll more likely be able to pinpoint it to a small patch in the sky. That's where other direction detection techniques can be used to refine that location.&lt;/p&gt;&#xA;&lt;h1&gt;Detector Sensitivity&lt;/h1&gt;&#xA;&lt;p&gt;The way in which the experiment works is that a mass is set up which can length contract under the influence of gravitational waves. This tiny contraction of space-time (and thus the mass) by the passing is measurable with precise lasers and interferometry. However, the lasers can only measure the contraction of the mass in one direction. The degree to which the mass contracts in that direction depends on the direction in which the wave is hitting it. This is an oversimplification, but if the modulation of the wave (that is the direction in which it goes &amp;quot;up&amp;quot; and &amp;quot;down&amp;quot;) is aligned perfectly with the direction in which the contractions are being measured, you measure the full effects of the wave. If the wave comes in at some angle to that, you're still measuring the effects, but you'll only be measuring some part of the effect, reduced by the sine of the angle.&lt;/p&gt;&#xA;&lt;p&gt;This process of wave direction detection requires at least 2 detectors, but more is better for beating down the noise.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-06-16T17:05:50.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16265" PostTypeId="2" ParentId="16255" CreationDate="2016-06-16T17:21:01.677" Score="1" Body="&lt;p&gt;Firstly the skymap for the GW150914 event covers some 600 square degrees, of which the Large Magellanic cloud is only a small fraction of that, and furthermore the estimated distance to this merger is around 400Mpc, orders of magnitude larger than the distance to the LMC (~50kpc) &lt;a href=&quot;https://losc.ligo.org/s/events/GW150914/GW150914-FactSheet-BW.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;[source]&lt;/a&gt;. So dont assume this is where it came from because the error boxes are so large we cannot pinpoint it to a single galaxy yet.&lt;/p&gt;&#xA;&lt;p&gt;In regards to your question, there was a very good illustration at yesterdays press conference that outlined the improved localization that adding a 3rd detector (Virgo) would to triangulating the position of the gravitational wave source. Unfortunately I cannot find a copy of David's presentation from yesterday but this is one of the images he showed, with a comparison of the error boxes if Virgo was operational too (not shown).&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KMFcK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KMFcK.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Further to this they also confirmed the plans to add a 4th (KAGRA) and 5th (LIGO India) gravitational wave observatories by 2025, thus further improving the localisation of the sources. Improvements and upgrades of the current detectors is also excpected to improve the localisation slightly too.&lt;/p&gt;&#xA;" OwnerUserId="10437" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-06-16T17:21:01.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16266" PostTypeId="1" AcceptedAnswerId="16273" CreationDate="2016-06-16T20:24:56.030" Score="1" ViewCount="1055" Body="&lt;p&gt;Suppose I have a wavelength array for a spectrum in units of Angstroms. Suppose further that the wavelength has &quot;uniform logarithmic spacing&quot; such that if I just take the difference in Angstroms between neighboring array elements i and i+1, this won't be the same as the difference in Angstroms between any other two neighboring elements j and j+1. However, if I first took the log10 of the wavelength array, then the difference between any two neighboring logarithmic array elements i and i+1 would be equal to the difference between any other logarithmic array elements j and j+1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I know a priori that this wavelength array has a velocity offset of X km/s, which I basically want to remove. How do I apply a velocity shift of -X km/s to this wavelength array which has uniform logarithmic spacing?&lt;/p&gt;&#xA;" OwnerUserId="607" LastEditorUserId="607" LastEditDate="2016-06-16T20:30:38.867" LastActivityDate="2016-06-17T12:57:07.937" Title="How do I apply a velocity shift to a wavelength array with uniform logarithmic spacing?" Tags="&lt;redshift&gt;&lt;spectroscopy&gt;&lt;observational-astronomy&gt;&lt;velocity&gt;&lt;wavelength&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16267" PostTypeId="1" AcceptedAnswerId="16268" CreationDate="2016-06-17T06:19:50.210" Score="1" ViewCount="150" Body="&lt;p&gt;As we know, rogue planets don't orbit around a star, how about galaxy? Can a rouge planet not orbit around a galaxy?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastActivityDate="2016-06-17T07:59:20.700" Title="Can a rogue planet neither orbit around a star nor a galaxy?" Tags="&lt;rogue-planet&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16268" PostTypeId="2" ParentId="16267" CreationDate="2016-06-17T07:21:23.603" Score="3" Body="&lt;p&gt;They would be enormously difficult to observe, but with a big enough &lt;a href=&quot;http://www.universetoday.com/121961/how-do-stars-go-rogue/&quot; rel=&quot;nofollow&quot;&gt;gravitational assist&lt;/a&gt; a rogue planet certainly could be ejected from a galaxy and become a &quot;double rogue&quot; so to speak, or intergalactic rogue planet is probably a better term.  Rogue (or intergalactic) stars have been observed.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Intergalactic_star&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; and &lt;a href=&quot;http://www.sciencemag.org/news/2014/11/rogue-stars-outside-galaxies-may-be-everywhere&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;.  The same gravitational forces that create intergalactic stars absolutely should create inter-galactic rogue planets and in fairly significant numbers.  The merging of 2 galaxies should create millions of them.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-06-17T07:59:20.700" LastActivityDate="2016-06-17T07:59:20.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16269" PostTypeId="2" ParentId="16267" CreationDate="2016-06-17T07:25:03.627" Score="3" Body="&lt;p&gt;There is no reason why it couldn't, though the chances of it happening are small. There are hyper velocity stars that are not gravitationally bound to the Milkyway, if a planet is ripped away from its star due to a close encounter with another star or black hole it could be given enough velocity to reach a galaxies escape velocity (about 537 km/s).&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-06-17T07:25:03.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16271" PostTypeId="1" AcceptedAnswerId="16275" CreationDate="2016-06-17T09:24:32.277" Score="0" ViewCount="115" Body="&lt;p&gt;Are there any planets not orbiting a celestial body which can support life despite the temperature? Also what is the absolute minimum temperature that life can survive in?&lt;/p&gt;&#xA;" OwnerUserId="13158" LastActivityDate="2016-06-17T15:27:49.200" Title="Survival on a rogue planet" Tags="&lt;rogue-planet&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2016-06-17T18:30:51.133" ContentLicense="CC BY-SA 3.0" />
  <row Id="16272" PostTypeId="2" ParentId="16256" CreationDate="2016-06-17T11:50:58.603" Score="0" Body="&lt;p&gt;The original intent of the Drake equation was to estimate the number of extant radio-emitting intelligent species in the galaxy. If the nearest extant radio-emitting intelligent species is in some other galaxy cluster (i.e., hundreds of millions of light years away), it doesn't really matter as far as the Drake equation is concerned. The intent was to find local intelligences, ones we could hear, and possibly talk to. The number of intelligent species in the observable universe, ever, is pushing the Drake equation a bit too far.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Here's my question: why was the probability of one in 10 billion per planet considered highly pessimistic?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ward and Brownlee, the authors of the rare Earth hypothesis, would beg to differ. They argued that the Drake equation is fundamentally flawed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on how one spins the numbers, we might be completely surrounded by alien intelligent species that for some reason have left us alone. Or we might truly be alone in the galaxy. Or we might even alone in our galaxy cluster. This wide variation across many orders of magnitude makes the Drake equation non-scientific. It has zero predictive power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, going so far as to say that we are the first intelligent species to have occurred across the observable universe? Even Ward and Brownlee would say &quot;no&quot; to that. Their hypothesis is that intelligent life is rare rather than unique.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2016-06-17T11:50:58.603" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16273" PostTypeId="2" ParentId="16266" CreationDate="2016-06-17T12:57:07.937" Score="2" Body="&lt;p&gt;If you have bins of equal log wavelength increment then a velocity shift corresponds directly to a pixel shift. That is the reason for binning in equal log wavelength increments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See  &lt;a href=&quot;https://astronomy.stackexchange.com/questions/14518/why-do-linear-velocity-redshifts-correspond-to-linear-pixel-shifts-when-the-spec?rq=1&quot;&gt;this question&lt;/a&gt; for a short proof. All you need to do is work out what each pixel corresponds to in terms of a velocity.&#xA;$$\Delta V  = c\, \Delta \log \lambda$$&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-06-17T12:57:07.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16274" PostTypeId="2" ParentId="16271" CreationDate="2016-06-17T13:48:46.527" Score="1" Body="&lt;p&gt;We have not yet discovered any planets that are certainly capable of supporting life, (except Earth) and the few rogue planets that we do believe that we have found have been gas giants, no good for life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is unlikely that life could exist on a rogue planet. The surface of a terrestrial planet that was not in orbit around a star would be exceedingly cold, as cold as Pluto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only place for life to start would be in a layer of liquid water, similar to that on Enceladus. If the core were sufficiently warm, it could maintain a liquid water layer, under a thick layer of ices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No such rogue planet is known to exist.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-06-17T13:48:46.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16275" PostTypeId="2" ParentId="16271" CreationDate="2016-06-17T15:27:49.200" Score="1" Body="&lt;p&gt;IHMO, I think it is possible into any planet with a hot core, that means if there is life on some of this planets they should be on the subterran part of the planet on underneath the deep freezed oceans, underground seas and caves, that means too that the life we will find on such planets will be quite shiny and probably looking like much of the animals we find at deep see here on earth.&lt;/p&gt;&#xA;" OwnerUserId="10950" LastActivityDate="2016-06-17T15:27:49.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16276" PostTypeId="1" AcceptedAnswerId="16277" CreationDate="2016-06-17T17:13:52.133" Score="2" ViewCount="141" Body="&lt;p&gt;We are translating some illustrated book for kids about astronomy, and then have a question as in the title.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Referring to Wikipedia etc., It appears that &quot;binary stars&quot; represents the star systems of two stars, whereas &quot;multiple stars&quot; does those of three or more stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, our question is whether it is an abuse to use &quot;multiple stars&quot; for binary stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="13173" LastActivityDate="2016-06-17T17:51:24.907" Title="Could the word &quot;multiple stars&quot; include binary stars?" Tags="&lt;binary-star&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16277" PostTypeId="2" ParentId="16276" CreationDate="2016-06-17T17:51:24.907" Score="8" Body="&lt;p&gt;I'm not aware of any official (i.e., IAU) definition of a multiple star system.  However, as someone who used to do research in that field, I would interpret multiple star system as encompassing systems with two or more stars.  So a binary system, a triple system, and a quadruple system are all multiple star systems.&lt;/p&gt;&#xA;" OwnerUserId="10371" LastActivityDate="2016-06-17T17:51:24.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16278" PostTypeId="1" AcceptedAnswerId="16279" CreationDate="2016-06-17T19:12:39.573" Score="2" ViewCount="460" Body="&lt;p&gt;I was taking some photo stacks yesterday, and I also took a plenty of random ones (pointed not at a specific object) too at 200mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I've ended with a shot but I have no idea where I pointed my camera at, and because of the narrow angle of 200mm, I'm lost without any guiding star/object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This happened to me before too, so I was wondering whether there is a software or a website that I can upload a photo, and it will analyze the objects, cross match them with known objects and detect which part of the sky it is? Or is there a method that I can use to figure out by myself (remember that there are no well-known bright stars in the image, at least not that I can identify by myself)?&lt;/p&gt;&#xA;" OwnerUserId="13176" LastActivityDate="2016-06-20T00:31:00.977" Title="How to find which part of the sky a photo contains?" Tags="&lt;photography&gt;&lt;software&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16279" PostTypeId="2" ParentId="16278" CreationDate="2016-06-17T20:31:13.253" Score="7" Body="&lt;p&gt;Have you tried &lt;a href=&quot;http://nova.astrometry.net&quot;&gt;nova.astrometry.net&lt;/a&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They set up a web service for doing more or less what you're talking about.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2016-06-17T20:31:13.253" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16280" PostTypeId="2" ParentId="14901" CreationDate="2016-06-17T22:51:29.440" Score="1" Body="&lt;p&gt;Relative to the time-dependent &lt;em&gt;equinox of date&lt;/em&gt;,&#xA;the longitudes of a planet's ascending node and perihelion increase by 1.4 degrees per century due to precession,&#xA;plus or minus smaller amounts due to perturbation by other planets.&#xA;Relative to the fixed &lt;em&gt;equinox of a standard epoch&lt;/em&gt;,&#xA;those longitudes only change due to perturbation.&#xA;Another explanation of this distinction is in the Wikipedia article on &lt;a href=&quot;https://en.wikipedia.org/wiki/Ecliptic_coordinate_system#Primary_direction&quot; rel=&quot;nofollow&quot;&gt;ecliptic coordinates&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your application, the longitude scale refers to the equinox of date for the time sliders.&#xA;The vernal equinox of standard epoch J2000.0 is in astronomical Pisces regardless of time.&#xA;The aphelion of Pluto's orbit should also stay in the same constellation over time, unaffected by precession of the Earth's axis.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2016-06-18T02:47:22.603" LastActivityDate="2016-06-18T02:47:22.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16281" PostTypeId="1" AcceptedAnswerId="16284" CreationDate="2016-06-18T12:36:18.820" Score="2" ViewCount="190" Body="&lt;p&gt;I recently read a bit about &quot;grand tack&quot; hypothesis. It was extremely interesting and it seems to reasonably well explain many features of our solar system. At the same time it supposedly is consistent with our knowledge of other planet systems, which I don't understand. What exactly happened in solar system and at which point , that is so rare out in the other systems? Combination of Jupiter/Saturn showing up at the right places and the right time? Isn't it common that the biggest gas giants appear somewhere over the edge of frost line? &lt;/p&gt;&#xA;" OwnerUserId="13178" LastEditorUserId="2365" LastEditDate="2016-06-18T20:06:16.917" LastActivityDate="2016-06-19T16:16:57.263" Title="What starting circumstances have to be for &quot;grand tack&quot; hypothesis to work, and why they are lacking in other planet systems(as far as we know)?" Tags="&lt;exoplanet&gt;&lt;jupiter&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16282" PostTypeId="2" ParentId="16250" CreationDate="2016-06-18T12:39:27.197" Score="6" Body="&lt;p&gt;It's simply that&lt;/p&gt;&#xA;&lt;h1&gt;we are moving fast.&lt;/h1&gt;&#xA;&lt;p&gt;It's that simple.  The Earth is moving VERY fast around the sun, so it is not &amp;quot;dragged in&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;If for some bizarre reason the Earth &lt;strong&gt;slowed down&lt;/strong&gt; ... it would move in closer to the Sun.  If it stopped moving - it would fall directly in to the sun.  It's that simple.&lt;/p&gt;&#xA;&lt;p&gt;Note that this is &lt;em&gt;exactly&lt;/em&gt; how satellites work.  (I mean the satellites we launch to use for TV communications and the like.)  If they need the satellite to move down closer, they simply slow it down a bit. To go up higher, you just speed it up a bit.&lt;/p&gt;&#xA;&lt;p&gt;So it's that simple.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;I think there must be a repulsive force.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;There's no repulsive force ... we are quite simply moving very fast!&lt;/p&gt;&#xA;&lt;p&gt;Note that, quite simply, say you want a rocket to &amp;quot;get away from&amp;quot; the gravitational power of the Earth ... what do you do? Simply make it go very fast!  It's that simple.&lt;/p&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-10-26T18:33:20.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16284" PostTypeId="2" ParentId="16281" CreationDate="2016-06-18T14:39:32.720" Score="5" Body="&lt;p&gt;We do not yet know whether our solar system is a common or rare outcome of the planetary process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statistical properties of exoplanetary systems are &lt;em&gt;severely&lt;/em&gt; affected by observational biases. It is &lt;em&gt;far&lt;/em&gt; easier to detect close-in giant planets; objects like Saturn, at Saturn-like distances from their host star, have not been found due to the limitations of the observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is therefore far too early to say whether the Grand Tack model may or may not be applicable to a wide variety of exoplanetary systems. What we do know, is that something drastically different happened in the case of those systems that have a Jupiter-sized planet within an astronomical unit or so of their parent star; but these are found in only a few per cent of stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an addendum, it is also the case, that despite taking 10+ years of data to establish them, there are now many cases of Jupiter-like planets in Jupiter-like orbits, though they are probably not that common. The overall conclusion at present is that the solar system does not look that unusual, although the lack of close-in planets larger than the Earth is notable (see &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0004-637X/810/2/105/meta;jsessionid=F4481348562BEAA3C8011C029B07B129.c1.iopscience.cld.iop.org&quot; rel=&quot;nofollow&quot;&gt;Martin &amp;amp; Livio 2015&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-06-19T16:16:57.263" LastActivityDate="2016-06-19T16:16:57.263" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16286" PostTypeId="1" AcceptedAnswerId="16287" CreationDate="2016-06-18T21:40:21.723" Score="6" ViewCount="3931" Body="&lt;p&gt;Suppose I have a template stellar population spectrum (say, from Bruzual &amp;amp; Charlot 2003) which runs from like 1000 Angstroms to 160,000 Angstroms and which has x-axis wavelength units of Angstroms and y-axis flux density units of $L_{\odot}$ per Angstrom. Now, let's say I want to get the integrated luminosity within a bandpass that runs from $\lambda_1$ to $\lambda_2$ Angstroms, and I have another &quot;spectrum&quot; which gives me the filter response function for this photometric bandpass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that I'm supposed to &quot;convolve the stellar template spectrum with the bandpass response function&quot; in order to get the integrated stellar luminosity within my bandpass, but how do I do this exactly (preferably in python)? The stellar template and filter response spectra don't have the same number of data points and are not defined on the same wavelength grid (the stellar template spectrum obviously extends much further than the filter response function). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like I could interpolate the filter response spectrum onto the wavelength array of the stellar template spectrum, and assign a value of 0 to the interpolated response spectrum for wavelengths outside $\lambda_1$ to $\lambda_2$. Then I would be able to do a simple element-wise multiplication, but this does not seem like the correct approach using traditional &quot;convolution.&quot; &lt;/p&gt;&#xA;" OwnerUserId="607" LastEditorUserId="941" LastEditDate="2018-04-27T11:08:01.617" LastActivityDate="2018-04-27T11:08:01.617" Title="How can I convolve a template spectrum with a photometric filter response spectrum?" Tags="&lt;spectroscopy&gt;&lt;stellar-astrophysics&gt;&lt;luminosity&gt;&lt;photometry&gt;&lt;python&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16287" PostTypeId="2" ParentId="16286" CreationDate="2016-06-18T22:38:47.970" Score="5" Body="&lt;p&gt;Your feeling is right: You shouldn't convolve the spectrum and the filter, you should only multiply so that flux outside the bandpass is suppressed. Subsequently you integrate the resulting function over wavelength, so that flux density (in energy/time/area/wavelength) becomes flux (in energy/time/area).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply setting the flux to 0 outside $\lambda_1$ and $\lambda_2$ (or, equivalently, just integrating from $\lambda_1$ to $\lambda_2$) corresponds to a &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Top-hat_filter&quot; rel=&quot;nofollow&quot;&gt;top-hat filter&lt;/a&gt;&quot;. Most realistic filters are more smooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, something like (untested):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;from scipy.integrate import simps&#xA;&#xA;lamS,spec = np.loadtxt('spectrum.dat',unpack=True) #Two columns with wavelength and flux density&#xA;lamF,filt = np.loadtxt('filter.dat'  ,unpack=True) #Two columns with wavelength and response in the range [0,1]&#xA;filt_int  = np.interp(lamS,lamF,filt)              #Interpolate to common wavelength axis&#xA;filtSpec  = filt_int * spec                        #Calculate throughput&#xA;flux      = simps(filtSpec,lamS)                   #Integrate over wavelength&#xA;print 'Total flux is {0:8.3e}'.format(flux)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;strong&gt;Magnitude&lt;/strong&gt;&#xA;&#xA;&lt;p&gt;I'm wondering, however, if &quot;the total luminosity within the bandpass&quot; is really what you're interested in. This quantity doesn't really bear any physical significance for the source, as it depends on the particular filter. Usually, one would normalize to the filter, thus getting the &lt;em&gt;magnitude&lt;/em&gt;, which is independent of the filter shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &lt;a href=&quot;https://en.wikipedia.org/wiki/AB_magnitude&quot; rel=&quot;nofollow&quot;&gt;AB magnitude&lt;/a&gt; system, the magnitude $m_\mathrm{AB}$ of a source is (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1983ApJ...266..713O&quot; rel=&quot;nofollow&quot;&gt;Oke &amp;amp; Gunn 1983&lt;/a&gt;; note that they make a sign error in their own definition):&#xA;$$&#xA;m_\mathrm{AB} = -2.5 \log f_\nu -48.6,&#xA;$$&#xA;where the average $f_\nu$ in $\mathrm{erg} \,\mathrm{s}^{-1} \,\mathrm{cm}^{-2} \,\mathrm{Hz}^{-1}$ is given by&#xA;$$&#xA;f_\nu = \frac{1}{c}\frac{\int ST\lambda\,d\lambda}{\int T / \lambda\,d\lambda},&#xA;$$&#xA;where $S$ is the spectrum with units $\mathrm{erg} \,\mathrm{s}^{-1} \,\mathrm{cm}^{-2}$ Å$^{-1}$, $T$ is the filter curve (un-normalized as above), $\lambda$ is the wavelength in Ångström, and $c$ is the speed of light in Å $\mathrm{s}^{-1}$. That is, a flat spectrum (in $\nu$) of height $f_\nu$ would have the same integrated flux over $T$ as $S$ would.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;from scipy.integrate import simps&#xA;c_AAs     = 2.99792458e18                          # Speed of light in Angstrom/s&#xA;lamS,spec = np.loadtxt('spectrum.dat',unpack=True) #Two columns with wavelength (in Angstrom) and flux density (in erg/s/cm2/AA)&#xA;lamF,filt = np.loadtxt('filter.dat'  ,unpack=True) #Two columns with wavelength and response in the range [0,1]&#xA;filt_int  = np.interp(lamS,lamF,filt)              #Interpolate to common wavelength axis&#xA;I1        = simps(S*T*lam,lam)                     #Denominator&#xA;I2        = simps(  T/lam,lam)                     #Numerator&#xA;fnu       = I1/I2 / c_AAs                          #Average flux density&#xA;mAB       = -2.5*np.log10(fnu) - 48.6              #AB magnitude&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-06-21T06:56:05.733" LastActivityDate="2016-06-21T06:56:05.733" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="16290" PostTypeId="1" CreationDate="2016-06-19T18:30:02.833" Score="9" ViewCount="1228" Body="&lt;p&gt;I am getting conflicting accounts everywhere. I heard astronomers saying &lt;code&gt;Polaris&lt;/code&gt; is a 3-star system. Some say they are binary but are optical-doubled with a 3rd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.amazon.com/s/ref=dp_byline_sr_book_1?ie=UTF8&amp;amp;text=Philip%20Plait%20Ph.D.&amp;amp;search-alias=books&amp;amp;field-author=Philip%20Plait%20Ph.D.&amp;amp;sort=relevancerank&quot; rel=&quot;noreferrer&quot;&gt;Phil Plait&lt;/a&gt; a renowned astronomer wrote in his book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471409766&quot; rel=&quot;noreferrer&quot;&gt;&lt;code&gt;Bad Astronomy&lt;/code&gt;&lt;/a&gt; that &lt;code&gt;Polaris&lt;/code&gt; is sextuple star system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/O00VT.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O00VT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above text reads &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... but due to its proximity to the NCP it has taken on the popular name of Polaris. The star itself is actually rather interesting; it's really a multiple star consisting of at least six stars in orbit around each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While in his &lt;code&gt;Crash Course Astronomy Series&lt;/code&gt; &lt;a href=&quot;https://youtu.be/pIFiCLhJmig?t=4m16s&quot; rel=&quot;noreferrer&quot;&gt;He said &lt;code&gt;Polaris&lt;/code&gt; was a Pentuple star system&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I am not saying &lt;code&gt;Phil&lt;/code&gt; is doing anything wrong. I am actually a fan of him, enjoyed reading the books he authored. He may have said one or the other by mistake. I tried asking him this question, it does not look like there is an effective way to submit him an Errata or ask him questions like these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is, how many stars are &lt;code&gt;Polaris&lt;/code&gt; and are they all gravitationally bound to each other or some of them are just &lt;code&gt;Optically&lt;/code&gt; grouped?&lt;/p&gt;&#xA;" OwnerUserId="4099" LastActivityDate="2016-06-20T11:09:26.157" Title="How many actual stars are Polaris?" Tags="&lt;binary-star&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16291" PostTypeId="2" ParentId="16290" CreationDate="2016-06-19T20:49:03.017" Score="9" Body="&lt;p&gt;Polaris consists of multiple stars, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;α UMi Aa is the main star. It is a supergiant, and a cepheid variable. α UMi Ab and α UMi B are smaller (but still larger than the sun) and both are in orbit with α UMi Aa, (the former is close in, the latter is further out).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two further stars: α UMi C and D. These stars are thought to have formed before α UMi Aa, and so are not considered to be physically associated with α UMi A. There are other stars that appear nearby, one might be physically associated: J021006.2+891626 J021454.3+890852 are their catalogue numbuers. They are faint, but appear to be the same age as Polaris.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, Polaris is known to have 3 stars: Aa, Ab and B. The stars C and D are not physically related to Polaris. There may be one or two more stars that are distant dwarfs and may share Polaris's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0004-6256/139/5/1968&quot;&gt;http://iopscience.iop.org/article/10.1088/0004-6256/139/5/1968&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-06-20T11:09:26.157" LastActivityDate="2016-06-20T11:09:26.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16292" PostTypeId="2" ParentId="16290" CreationDate="2016-06-19T22:31:41.823" Score="8" Body="&lt;p&gt;It's worth noting that in many cases, if not most, we &lt;em&gt;simply don't know the exact answer&lt;/em&gt; to such excellent questions as the one you ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the book you mention (ISBN-13: 978-0471409762) was written around 2000 and appeared in 2002: that's a really &lt;strong&gt;long time ago&lt;/strong&gt; in terms of all the amazing new instruments which have become available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, while the Hubble telescope was operational when that book was written, the amazing GAIA telescope (&lt;a href=&quot;https://en.wikipedia.org/wiki/Gaia_(spacecraft)&quot;&gt;link&lt;/a&gt;), which would perhaps be relevant to your question, only started operating around 2014.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The youtube video you mention seems to be more up to date - a couple years ago. But then, the author (even if he's an expert) may not especially be keeping up with the latest research and thinking on the particular star system you mention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's worth bearing in mind that many of these questions, &lt;strong&gt;we just don't know&lt;/strong&gt; - it's something that is being actively studied now!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting point to consider: we're not even really sure of the &lt;strong&gt;distance&lt;/strong&gt; to the Polaris system - for goodness sake!  That may be a surprise.  It looks like estimates of the distance - and they are just estimates - range from roughly 350 to 450 light years.  We could be out by &lt;em&gt;a hundred light years&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your specific question: how many stars are in the system. Don't forget when you look at a specific star, X, let's say it is 500 ly away.  In the same &quot;shot&quot; you see a vast number of stars &lt;strong&gt;right next to&lt;/strong&gt; star X. Some of those may be &quot;next door to us&quot; and some may be tremendously further away. It is by no means a &quot;sure thing&quot; to determine if two stars which appear right next to each other, are actually anywhere near each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A good example of that is ... you know the Andromeda Galaxy, which you can even see with your eye. At first, astronomers had no clue if it was a thing (sort of like a star, but cloudy), and as close to us as the other stars, or, if it was an enormous object an incredible distance away. Obviously we now know it is  truly huge thing an unbelievably long way away ... but, we only learned that in about 1920 - !!!  Not even 100 years ago.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a short point on your specific question: note that the book by Plait is some 15+ years old, and the video talk is more recent; in the first instance you can go with the info in the video talk. It's a good example of how astronomy knowledge is dramatically changing all the time.&lt;/p&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="13071" LastEditDate="2016-06-19T22:43:29.637" LastActivityDate="2016-06-19T22:43:29.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16293" PostTypeId="2" ParentId="16278" CreationDate="2016-06-20T00:31:00.977" Score="2" Body="&lt;p&gt;This type of software is called plate-solving software. There are several other software other than astrometry.net that can achieve this as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pinpoint.dc3.com/&quot; rel=&quot;nofollow&quot;&gt;http://pinpoint.dc3.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astrosurf.com/pulgar/elbrus/elbrusin.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.astrosurf.com/pulgar/elbrus/elbrusin.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To name a few&lt;/p&gt;&#xA;" OwnerUserId="13085" LastActivityDate="2016-06-20T00:31:00.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16295" PostTypeId="2" ParentId="4738" CreationDate="2016-06-20T00:47:36.543" Score="3" Body="&lt;p&gt;You need a diffraction grating in order to split the sunlight into a spectrum. It turns out even a normal CD or DVD because of the pits that they have to encode data work very well as high resolution diffraction gratings. You could make a simple enclosure with a CD and a slit like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://coolcosmos.ipac.caltech.edu/cosmic_games/spectra/makeGrating.htm&quot; rel=&quot;nofollow&quot;&gt;http://coolcosmos.ipac.caltech.edu/cosmic_games/spectra/makeGrating.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should allow you to observe the sun's spectrum, although I would also recommend that you be very careful when observing the sun directly. Preferably use a very narrow slit--this will both make it safer and make the absorption lines much clearer. &lt;/p&gt;&#xA;" OwnerUserId="13085" LastActivityDate="2016-06-20T00:47:36.543" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16298" PostTypeId="2" ParentId="16234" CreationDate="2016-06-20T06:52:32.073" Score="3" Body="&lt;p&gt;For the first part of your question, &lt;em&gt;@userLTK&lt;/em&gt; has already provided a very good explanation, and I believe that by introducing redshift, the second part of your question is also partially answered. I will try to expand a little on it, hoping that I have understood your question correctly. To start off, I'm going to go ahead and quote my own thesis (not yet published);&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[W]hen moving to celestial distances we quickly realise that time measurement ambiguity also arises. To combat this we tend to not describe distances to objects directly, but relate these properties to an object’s relative line-of-sight velocity with respect to us.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is why, when &lt;a href=&quot;https://www.nasa.gov/feature/goddard/2016/hubble-team-breaks-cosmic-distance-record&quot; rel=&quot;nofollow&quot;&gt;NASA earlier this year announced the Hubble Space Telescope having observed the most distant galaxy yet&lt;/a&gt;, the &lt;em&gt;distance&lt;/em&gt; to the galaxy was never mentioned. The distance is not a direct observable, and has to be calculated based on certain assumptions. One direct observable we &lt;em&gt;do&lt;/em&gt; have, is the line-of-sight velocity, which we express in terms of &lt;em&gt;redshift&lt;/em&gt;. By describing objects in terms of redshift, we simultaneously have a &quot;distance&quot; measurement to the objects, as well as knowing how fast they move away from us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we make observations and try to learn more about things like the energy content of the universe (radiation, regular matter, dark matter, dark energy), the redshifts of the objects we observe are crucial. We consider various models of the universe, and see how they compare with what we observe on different redshifts. I very quickly found back to a &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/519988/pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; written by my supervisor some years back; figure 7 in there shows you that kind of comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I am working on something which I hope will lead to us being able to see whether dark energy evolves over time. In doing that, knowing the redshifts of the objects are crucial to my work, and I constantly need to keep in mind the implications of the distant objects' movement away from us. It is simply a factor I need to multiply or divide by in my calculations, to make sure I am comparing the correct things with each other. &lt;/p&gt;&#xA;" OwnerUserId="13193" LastActivityDate="2016-06-20T06:52:32.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16303" PostTypeId="1" AcceptedAnswerId="16305" CreationDate="2016-06-19T02:11:20.213" Score="11" ViewCount="691" Body="&lt;p&gt;Under a question I asked a few days ago &lt;a href=&quot;https://space.stackexchange.com/q/16738/12102&quot;&gt;Have there been any documented mini-moons since 2006 RH120?&lt;/a&gt; @Hobbes &lt;a href=&quot;https://space.stackexchange.com/questions/16738/have-there-been-any-documented-mini-moons-since-2006-rh120#comment42638_16738&quot;&gt;mentioned&lt;/a&gt; the &lt;a href=&quot;http://www.jpl.nasa.gov/news/news.php?feature=6537&quot; rel=&quot;noreferrer&quot;&gt;recent news&lt;/a&gt; about 2016 HO3 - a near-Earth asteroid that &lt;em&gt;stays&lt;/em&gt; near Earth because its orbit around the sun is coupled to Earth. Thus the term &lt;a href=&quot;https://en.wikipedia.org/wiki/Quasi-satellite&quot; rel=&quot;noreferrer&quot;&gt;quasi-satellite&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=SbbAnVU4rmY&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=SbbAnVU4rmY&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;In this &lt;a href=&quot;https://www.youtube.com/watch?v=SbbAnVU4rmY&quot; rel=&quot;noreferrer&quot;&gt;NASA JPL video&lt;/a&gt; (above) the view is rotating around the sun following Earth. You can see the earth move slightly closer and farther from the sun since the Earth's orbit is not quite circular.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=2mVfE_qmQAg&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=2mVfE_qmQAg&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=2mVfE_qmQAg&quot; rel=&quot;noreferrer&quot;&gt;This video&lt;/a&gt; (above) from &lt;a href=&quot;http://arksky.org/calendar/alerts/714-what-is-it-the-strange-new-object-2016-ho3&quot; rel=&quot;noreferrer&quot;&gt;http://arksky.org/calendar/alerts/714-what-is-it-the-strange-new-object-2016-ho3&lt;/a&gt; shows a projection of 2016 HO3's motion against the stars as seen from Earth's location, but in a fixed direction. You can see the the sun and planets tend to follow the ecliptic, while 2016 HO3 does a figure-eight every year.&lt;/p&gt;&#xA;&lt;p&gt;The NASA JPL &lt;a href=&quot;http://www.jpl.nasa.gov/news/news.php?feature=6537&quot; rel=&quot;noreferrer&quot;&gt;news brief&lt;/a&gt; states:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The asteroid's orbit also undergoes a slow, back-and-forth twist over multiple decades. &amp;quot;The asteroid's loops around Earth drift a little ahead or behind from year to year, but when they drift too far forward or backward, Earth's gravity is just strong enough to reverse the drift and hold onto the asteroid so that it never wanders farther away than about 100 times the distance of the moon,&amp;quot; said Chodas. &amp;quot;The same effect also prevents the asteroid from approaching much closer than about 38 times the distance of the moon. In effect, this small asteroid is caught in a little dance with Earth.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;note: &lt;em&gt;Paul Chodas is the manager of NASA's Center for Near-Earth Object (NEO) Studies&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The item goes on to say:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Asteroid 2016 HO3 was first spotted on April 27, 2016, by the Pan-STARRS 1 asteroid survey telescope on Haleakala, Hawaii, operated by the University of Hawaii's Institute for Astronomy and funded by NASA's &lt;a href=&quot;https://www.nasa.gov/planetarydefense/overview&quot; rel=&quot;noreferrer&quot;&gt;Planetary Defense Coordination Office&lt;/a&gt;. The size of this object has not yet been firmly established, but it is likely larger than 120 feet (40 meters) and smaller than 300 feet (100 meters).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;First, I think it's quite cool that there even exists a Planetary Defense Office. It must be fun to go to work every day knowing that you are &lt;em&gt;defending the planet&lt;/em&gt;!&lt;/p&gt;&#xA;&lt;p&gt;I am interested in the story behind &amp;quot;...first spotted...&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Pan-STARRS&quot; rel=&quot;noreferrer&quot;&gt;Pan STARRS&lt;/a&gt; telescope PS1 (&lt;a href=&quot;http://pan-starrs.ifa.hawaii.edu/public/home.html&quot; rel=&quot;noreferrer&quot;&gt;http://pan-starrs.ifa.hawaii.edu/public/home.html&lt;/a&gt;) is managed by the &lt;a href=&quot;http://ps1sc.org/&quot; rel=&quot;noreferrer&quot;&gt;PS1 Consortium&lt;/a&gt;. The original plan was for four telescopes, the status of PS2 is unclear. PS1 is a 1.8 meter telescope with a &lt;a href=&quot;http://pan-starrs.ifa.hawaii.edu/public/design-features/camera-small.htm&quot; rel=&quot;noreferrer&quot;&gt;Gigapixel camera&lt;/a&gt; at the relatively fast f/4.4 focus, corrected for a very large 3° FOV, which means it's of the &amp;quot;&lt;a href=&quot;http://pan-starrs.ifa.hawaii.edu/public/project-status/gallery-mirrors.htm&quot; rel=&quot;noreferrer&quot;&gt;primary with giant hole&lt;/a&gt;&amp;quot; variety (e.g. &lt;a href=&quot;https://en.wikipedia.org/wiki/Large_Synoptic_Survey_Telescope&quot; rel=&quot;noreferrer&quot;&gt;LSST&lt;/a&gt;). Typically it records an image every 30-60 seconds, and can record the entire night sky every three days.&lt;/p&gt;&#xA;&lt;p&gt;I looked at the entry for 2016 HO3 (SPK-ID: 3752445) in the NASA JPL &lt;a href=&quot;http://ssd.jpl.nasa.gov/sbdb.cgi?sstr=2016+HO3&amp;amp;orb=1&quot; rel=&quot;noreferrer&quot;&gt;Small Body Data Browser&lt;/a&gt; and saw this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LElDP.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LElDP.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The solution (shown on June 18) was calculated on June 11 2016 by an &amp;quot;O. Matic&amp;quot; :) using data going back to 2004.&lt;/p&gt;&#xA;&lt;p&gt;As I said above, I am interested in the story behind &amp;quot;...first spotted...&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;What were the sequence of events? Does &amp;quot;spotted&amp;quot; mean identified/flagged as an unknown object? Did it trigger a search of a database of earlier survey images from other telescopes? Is there a list available of the 80 total observations used? I'm curious if a twelve year old &amp;quot;first known image of 2016 HO3&amp;quot; exists.&lt;/p&gt;&#xA;" OwnerUserId="7982" OwnerDisplayName="uhoh" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-02-16T05:58:51.053" Title="How was Earth's &quot;quasi-satellite&quot; 2016 HO3 &quot;first spotted&quot; and it's orbit determined?" Tags="&lt;observational-astronomy&gt;&lt;asteroids&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16304" PostTypeId="2" ParentId="16303" CreationDate="2016-06-20T18:55:59.130" Score="5" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Phil_Plait&quot; rel=&quot;noreferrer&quot;&gt;Phil Plait&lt;/a&gt;'s blog, Bad Astronomy, &lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2016/06/17/asteroid_2016_ho3_has_an_orbit_very_similar_to_earth_s.html&quot; rel=&quot;noreferrer&quot;&gt;states&lt;/a&gt; (emphasis mine)&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;One of them was just discovered recently by astronomers: the asteroid 2016 HO3.&lt;/p&gt;&#xA;&lt;p&gt;. . .&lt;/p&gt;&#xA;&lt;p&gt;Here’s how that works. The asteroid was first seen in April 2016 in observations of the sky taken by the Pan-STARRS observatory, designed to look for asteroids and comets that get close to Earth. That’s recent enough that a really good orbit for it is hard to determine, but &lt;strong&gt;it turns out it was seen in older observations&lt;/strong&gt; (those can be found by tracing the orbit backward and checking if any observations of it were archived), providing a much longer baseline and therefore a better orbit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In other words, it seems like a basic approximation of the orbit was created, and the astronomers looked at images taken of areas near where the asteroid should have been, according to the calculations. The calculations could then be refined, and further previous sightings could be investigated.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-06-20T18:55:59.130" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16305" PostTypeId="2" ParentId="16303" CreationDate="2016-06-20T21:01:56.970" Score="5" Body="&lt;p&gt;Here is the IAU Minor Planet Center's list of known &lt;a href=&quot;http://www.minorplanetcenter.net/db_search/show_object?utf8=%E2%9C%93&amp;amp;object_id=K16H03O&quot; rel=&quot;noreferrer&quot;&gt;observations of 2016 HO3&lt;/a&gt;.&#xA;The Pan-STARRS people located observations from 2011-2015 in their archive.&#xA;The 2004 observation occurred in an &lt;a href=&quot;http://www.sdss.org/&quot; rel=&quot;noreferrer&quot;&gt;SDSS&lt;/a&gt; survey image (DR3?).&#xA;It may be possible to retrieve an old SDSS image showing the asteroid.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-06-20T21:01:56.970" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16307" PostTypeId="1" AcceptedAnswerId="16310" CreationDate="2016-06-21T09:45:08.537" Score="2" ViewCount="470" Body="&lt;p&gt;If somehow our moon was destroyed (for example by an asteroid) what would happen to Earth? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the Earth be affected by the absence of the moon? Would humans be able to live normal lives without the moon?&lt;/p&gt;&#xA;" OwnerUserId="11792" LastEditorUserId="2365" LastEditDate="2016-06-21T13:04:56.980" LastActivityDate="2016-10-26T01:05:53.590" Title="What would Earth be like without the Moon?" Tags="&lt;the-moon&gt;&lt;earth&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16308" PostTypeId="2" ParentId="16307" CreationDate="2016-06-21T10:43:34.237" Score="5" Body="&lt;p&gt;&lt;strong&gt;Short term:&lt;/strong&gt; The loss of oceanic tides. Very dark nights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Long term:&lt;/strong&gt; Erratic wobble of the earth's axis relative to the sun as it no longer has the moon to keep things stable, massive chaotic weather changes as a result possibly culminating in worldwide mass extinction that may never be recovered from.&lt;/p&gt;&#xA;" OwnerUserId="11009" LastActivityDate="2016-06-21T10:43:34.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16309" PostTypeId="1" AcceptedAnswerId="16317" CreationDate="2016-06-21T12:23:23.933" Score="3" ViewCount="545" Body="&lt;p&gt;In my fictional world I have selected all parameters to maximize the Hills Sphere of my planet. The planet has three times Earth's mass and it is located in 2 AU from its star that's 1.4 times more massive than our Sun. &lt;br/&gt;I assume that there are more planets on the system but in such way to have minimum interaction with my planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to have three  moons on a 1:2:4 resonance. I also want:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the first one to be as large as our moon to the sky&lt;/li&gt;&#xA;&lt;li&gt;the second 1.5 times larger&lt;/li&gt;&#xA;&lt;li&gt;and the third, half our moon's size.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To achieve this I have assumed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the first moon having 1.1 lunar radius,&lt;/li&gt;&#xA;&lt;li&gt;the second 2.3&lt;/li&gt;&#xA;&lt;li&gt;and the third 1.1.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Their orbits are 18, 36, 72 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really care about their density and I assumed that their masses are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 lunar mass (probably rocky)&lt;/li&gt;&#xA;&lt;li&gt;3 lunar mass (water or ice world)&lt;/li&gt;&#xA;&lt;li&gt;0.3 lunar mass (probably ice world)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt; I guess that to be more stable i should have eccentricities smaller than 0.05 and inclination equal to zero.&#xA;&lt;br/&gt; I am not sure about the spin of my moons, is there a chance being tidal locked?&#xA;&lt;br/&gt;&lt;br/&gt;This is my effort to achieve maximum stability of the system, what parameters should I reconsider to make it more stable? I am willing to change almost everything even the resonance, but I want them to be visible from my planet in similar way moon is to Earth.&#xA;&lt;br/&gt;&lt;br/&gt;Have in mind this question is about a fictional world for a fantasy story so even the minimum possibilities for that senario to happen is enough for me.&#xA;So any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;P.S I have asked a similar question on worldbuild.stackexchange but they directed me also here for a better answer. &#xA;&lt;br/&gt;&lt;/p&gt;&#xA;" OwnerUserId="13202" LastActivityDate="2016-06-21T20:53:08.180" Title="Could this three moons system be stable?" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;gravity&gt;&lt;natural-satellites&gt;&lt;earth-like-planet&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16310" PostTypeId="2" ParentId="16307" CreationDate="2016-06-21T13:23:29.373" Score="7" Body="&lt;p&gt;Destroying the moon is not easy. It's pretty big. Has quite a lot of gravity and is sitting in the Earth's gravity well. It gets hit by a lot of asteroids and doesn't get destroyed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon has a binding energy of about $10^{29}$ joules. That is a lot of energy. Even if you got the largest asteroid (ceres) to collide with the moon you would not get close to destroying it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would need something like the full power of the sun for about 15 minutes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If something massive enough to disrupt the moon were to hit it, the energy of that explosion, and the massive amount of moon rocks that would subsequently hit the Earth would probably be enough to melt the surface of the Earth. It would be the end of life on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another way to get rid of the moon. Get a wizard to magic it away. The effect of not having a moon would then be pretty minor in the short term (no tides) and fairly significant in the long term. Human life could continue as normal for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon stabilizes the rotational axis of the Earth, without the moon the angle would vary more (over thousands of years) this might mean more extreme ice ages and perhaps ultimately a snowball Earth scenario.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-06-21T13:23:29.373" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16311" PostTypeId="1" AcceptedAnswerId="16313" CreationDate="2016-06-21T16:38:10.857" Score="85" ViewCount="14459" Body="&lt;p&gt;I've heard from a few sources&lt;sup&gt;*&lt;/sup&gt; recently that the Sun is a 1,000th generation star, meaning it had a thousand stars that came before it based on its heavy-element content.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that earlier stars' supernovae created the heavier elements and those were incorporated into the Sun when it formed and we can calculate the generation based on that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is based on the age of the universe and the age of the Sun and the average age of stars in the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the universe is ~13.7 billion years old, the sun is ~4.6 billion years old (and has ~5 billion years of life left), which means it was formed when the universe was ~9.1 billion years old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 1,000 stars to live in that span of time, those stars would have to have an average lifespan of ~9 million years (not including time between stars). This equates to each of those thousand stars having an average mass of ~100-150 solar masses so their lives would be sufficiently short to not go over that ~9 million-year lifetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically, there were 1,000 stars that lived and died before our Sun, all of which were huge monsters. This just all seems so improbable. How can this be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;* TV shows, I don't remember which ones&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="2950" LastEditorDisplayName="user1569" LastEditDate="2018-11-06T10:04:07.503" LastActivityDate="2018-11-06T10:04:07.503" Title="How can there be 1,000 stellar ancestors before our Sun?" Tags="&lt;the-sun&gt;&lt;stellar-evolution&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="20" ContentLicense="CC BY-SA 4.0" />
  <row Id="16312" PostTypeId="2" ParentId="16309" CreationDate="2016-06-21T17:01:07.280" Score="2" Body="&lt;p&gt;I see no reason why this wouldn't work.  The innermost Galilean moons are in a 1:2:4 resonance so it's clearly a stable orbital configuration.  They could all be tidally locked if you want, but the planet itself can't be tidally locked with all of them.  If the planet were tidally locked it would likely be locked with the innermost moon.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also of note, the Galilean moons all have eccentricity less than 0.01 and inclinations less than 0.5 degrees.  &lt;/p&gt;&#xA;" OwnerUserId="10371" LastActivityDate="2016-06-21T17:01:07.280" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16313" PostTypeId="2" ParentId="16311" CreationDate="2016-06-21T17:23:39.693" Score="143" Body="&lt;p&gt;The Sun is actually a &lt;strong&gt;THIRD&lt;/strong&gt; generation star. What I mean by this is that there are chemical elements in the Sun that were made inside another star, but that star itself can only have made those elements because it had material in it that must also have been made inside previous, second generation stars. Eventually we get back to the first generation stars, born out of primordial gas from the big bang that contained almost no heavy elements (those beyond helium) at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is quite a mouthful, so let me explain using an example - barium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is barium in the Sun. We can tell that by looking at the spectrum and seeing absorption lines due to barium. But barium cannot be made in the Sun. The barium is made via the &lt;a href=&quot;https://en.wikipedia.org/wiki/S-process&quot; rel=&quot;noreferrer&quot;&gt;s-process&lt;/a&gt;, which involves the slow capture of neutrons onto the nuclei of iron-peak elements. This happens during the &lt;a href=&quot;https://en.wikipedia.org/wiki/Asymptotic_giant_branch&quot; rel=&quot;noreferrer&quot;&gt;asymptotic red giant branch phase&lt;/a&gt; of stellar evolution, and the Sun has 6 billion years or so before it gets to that point. [NB: not even half the abundance of chemical elements beyond iron are produced by supernovae explosions$^1$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, before the Sun, there must have existed a star - probably an intermediate mass star - which evolved to become a giant, made barium in its interior, then lost its envelope through a massive wind into the interstellar medium, and that material was incorporated into the protosun. Such stars (between, say, 2 and 10 solar masses) would have much shorter lifetimes than the Sun$^2$, so plenty of time for them to live and die before the Sun was born.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But wait a minute! That previous star must have &lt;em&gt;already&lt;/em&gt; had iron-peak elements in its interior to act as a &quot;seed&quot; for the s-process production of barium. These were not and could not be made inside that star. They must have been made in a previous star, probably a massive star, that burned through all the nuclear fusion stages before exploding as a supernova, casting heavy elements, including iron-peak elements, into the interstellar medium. This previous star could also have had its own (metal-rich) ancestors, but ultimately as we go back in time we reach a point where the previous star was a &lt;em&gt;first&lt;/em&gt; generation star, made from primordial H/He gas, with almost no heavy elements. These first generation (a.k.a. population III stars, just to be confusing) were probably very massive and short lived - a few million years. They would be born when the universe was a few hundred million years old and we can see no examples of them in our Galaxy today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To try and define more precisely what I mean by &quot;generation&quot;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First generation - made from primordial big bang material.&lt;/li&gt;&#xA;&lt;li&gt;Second generation - a star made only from the detritus of dying first    generation stars, enriched in heavy elements but lacking in primary s-process elements.&lt;/li&gt;&#xA;&lt;li&gt;Third generation - a star made from material already enriched in heavy elements and including elements that are produced in the s-process inside previous second (or third) generation stars.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So that is why I claim the Sun can be classed as a &quot;third generation star&quot; - it contains atoms/nuclei that must have been inside &lt;em&gt;at least&lt;/em&gt; two previous stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you should not take this too literally. There are grains of material trapped inside meteorites that consist of solids that were already present in the pre-solar material. These are important because these grains were thought to have formed in &lt;em&gt;individual&lt;/em&gt; stellar events and their isotopic compositions can be studied. These tell us that the Sun formed from material that has been inside &lt;em&gt;many&lt;/em&gt; different stars of different types.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stellar evolution and nucleosynthesis calculations tell us the same story. For example, whilst most of our oxygen was made in massive stars that underwent a core collapse supernova, such events do not produce that much carbon. The C/O ratio tells us that most of our carbon comes via the winds from intermediate mass AGB stars. Heavy elements like uranium may be dominantly produced in neutron star collisions, but others like barium and strontium are not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The details of how many ancestors have contributed to the Sun has no simple answer. Much of the solar hydrogen and helium could be pristine; some will have been through more than one star. Heavier elements (bar some lithium) will have been through at least one star. The fact that we have s-process elements like Ba, Sr, La and Ce, which are formed by neutron capture onto iron-peak elements, tells us those have been through &lt;em&gt;at least&lt;/em&gt; two stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, these are &lt;em&gt;vast&lt;/em&gt; underestimates. Mixing in the interstellar medium is reasonably effective. The material spewed out from supernovae and stellar winds 5-12 billion years ago has had plenty of time to mix throughout the Galaxy before the Sun's birth. Turbulence and shear instabilities, driven by the winds and supernovae from massive stars, should distribute material on galactic-length scales in a billion years or less (&lt;a href=&quot;http://adsabs.harvard.edu/full/1995A%26A...294..432R&quot; rel=&quot;noreferrer&quot;&gt;Roy &amp;amp; Kunth 1995&lt;/a&gt;; &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0208441&quot; rel=&quot;noreferrer&quot;&gt;de Avillez &amp;amp; Mac Low 2003&lt;/a&gt;), though local inhomogeneities associated with nearby recent events can persist over $10^{8}$ years. If this is the case, then the Sun is the product of the $\sim$ billion stars that died before it was born.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason you are confused with your lifetime argument is that you have ignored the possibility of the Sun being made from stars that lived &lt;em&gt;at the same time&lt;/em&gt; in different parts of the Galaxy. The material that they ejected near the end of their lives has just been thoroughly mixed up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^1$ The rest are produced by the s-process in intermediate mass AGB stars; through nova events on white dwarfs; or perhaps, in the case of the heavier elements, through the collision of neutron stars (see &lt;a href=&quot;https://physics.stackexchange.com/questions/7131/origin-of-elements-heavier-than-iron-fe?lq=1&quot;&gt;this Physics SE question&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^2$ A rough expression for the lifetime of a star is $10 (M/M_{\odot})^{-5/2}$ billion years.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="23547" LastEditDate="2018-07-05T18:28:12.877" LastActivityDate="2018-07-05T18:28:12.877" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="16315" PostTypeId="1" CreationDate="2016-06-21T18:20:09.023" Score="1" ViewCount="103" Body="&lt;p&gt;Suppose I have a large thin disk of rotating gas in a galaxy -- the disk has a maximum inclination-corrected $V_{max}$ and a maximum radial extent of $R_{max}$ corresponding to that $V_{max}$ measurement. The virial theorem tells me that the total (aka dynamical) mass enclosed by that rotating disk (including stars, dark matter, etc.) is $V_{max}^2 R_{max} / G$, where $G$ is Newton's gravitational constant. Here I have assumed that the virial coefficient is 1, but the virial coefficient can vary widely based on the geometry of the problem. For example, it can be almost 10 for dwarf galaxies (an order of magnitude!), or a factor of $\sim3-5$ for dispersion-dominated rather than rotation-supported systems like elliptical galaxies where you're using the stars to probe the dynamical mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is simple: for a simple rotating disk of gas as above (assume it is thin and obviously not dispersion-supported, which I think is one of the simplest possible geometries for a question like this), what is the virial coefficient? I have seen coefficients of both 1 and 2, but I have no idea which derivation is correct. Adding to the problem is that some derivations seem to be using the virial theorem to give an order of magnitude estimate which means they leave out discussion of the virial coefficient altogether. &lt;/p&gt;&#xA;" OwnerUserId="607" LastEditorUserId="9301" LastEditDate="2016-06-29T04:35:59.523" LastActivityDate="2016-06-29T04:35:59.523" Title="Virial coefficient when computing dynamical mass enclosed by a rotating galactic disk of gas" Tags="&lt;galaxy&gt;&lt;galactic-dynamics&gt;&lt;interstellar-medium&gt;&lt;stellar-dynamics&gt;&lt;kinematics&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16316" PostTypeId="1" CreationDate="2016-06-21T20:20:26.293" Score="1" ViewCount="53" Body="&lt;p&gt;A planet is taught to us (at primary school) as &lt;strong&gt;a ball of rock that orbits around the Sun&lt;/strong&gt; and is &lt;strong&gt;similar to our own planet, Earth&lt;/strong&gt; (in as a ball of rock). But of course, humans have to be more specific than that so there must be some internationally agreed on requirements to define a planet by an official committee, right? I know of an example of this is when in 2006, &lt;strong&gt;these requirements were enforced and Pluto got demoted from planet status to dwarf planet status.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4160" LastActivityDate="2016-06-21T20:20:26.293" Title="What are the requirements to become a planet?" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;pluto&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2016-06-21T21:10:32.047" ContentLicense="CC BY-SA 3.0" />
  <row Id="16317" PostTypeId="2" ParentId="16309" CreationDate="2016-06-21T20:53:08.180" Score="2" Body="&lt;p&gt;Nothing in your description sounds wildly implausible. I'll just go through and extend some of your properties to make sure they make sense though.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The planet has three times Earth's mass&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll assume that your new planet is Earth-like in composition and density. That implies that it's radius should be about $R_p \approx 1.4 \: R_\oplus$. Let's put that aside to use for later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's look at your moons.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;the first one to be as large as our moon to the sky&lt;/li&gt;&#xA;  &lt;li&gt;the second 1.5 times larger&lt;/li&gt;&#xA;  &lt;li&gt;and the third, half our moon's size.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;To achieve this I have assumed:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;the first moon having 1.1 lunar radius,&lt;/li&gt;&#xA;  &lt;li&gt;the second 2.3&lt;/li&gt;&#xA;  &lt;li&gt;and the third 1.1.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first moon, you want to be as large as our current Moon in the sky and you say it has a physical radius of $R_{M1} = 1.1\:R_{M}$. Our current moon subtends about $30\:arcsec = 8.73\times10^{-3}\:rad$ on average. You can calculate the distance, $d_{M1}$, the moon must be from the planet to have the same apparent size as our moon with the following equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d = \frac{R}{tan(\delta/2)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $R_{M1}$ for $R$ and $8.73\times10^{-3}\:rad$ for $\delta$, your moon must be $d \approx 4\times10^8\:m = 44.5\:R_p$. Your first moon must be about 44.5 planetary radii out. Compare this to the Moon's distance of about 60 planetary radii out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we want to know how long it would take for such a moon to orbit, given the planet's mass, the moon's mass, and orbital distance. You can get that from Kepler's third law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = \sqrt{\frac{4\pi^2}{G(M_p + M_{M1})}d^3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can say $M_p = 3\:M_\oplus$ (as you specified) and we calculated $d$ just now. We only need to specify the mass of the moon. You provided the desired masses for your moons, so we're all set. Note $G = 6.67\times10^{-11} m^3kg^{-1}s^{-2}$ is the gravitational constant. I find that $P_{M1} = 1.44\times10^6\:s=16\:days$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's stop right here for a second now. We've reached a point where your numbers are in conflict. Given your planet and your first moon's apparent size, mass, and radius, we found that it would have to orbit the planet in 16 days. If you want to keep the 1:2:4 resonance, then you need your other moons to orbit in 32 and 64 days, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I won't go through the rest of the math, I'll leave that up to you. But what you can do, if you want to ensure everything is consistent with real physics, is say I know how long my remaining two moons need to orbit for (i.e., you know $P$ for them), then at what distance must they orbit (i.e., what is the value of $d$ for them)? Work Kepler's third law backwards to get that. Then, given your desired apparent sizes in the sky, determine how large must they be physically by working the angular size equation backwards. You'll get new radii for the moon. You might then want to verify that the new mass and radii for your moons correspond to densities that match the desired moon types. E.g., if your third moon is going to be an ice moon, it should have a density of $\sim1\:g/cm^3$. You can play with your numbers until you get a system that matches your requires and fits the equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;An important note: All these equations use MKS units. That is, masses should be in kilograms ($kg$), distances and sizes in meters ($m$), time in seconds ($s$), and angles in radians ($rad$).&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-06-21T20:53:08.180" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16318" PostTypeId="1" CreationDate="2016-06-21T22:16:51.903" Score="1" ViewCount="86" Body="&lt;p&gt;I was reading how certain strong magnetic fields can block or alter radio waves so bearing this in mind would it be possible for the Sun's magnetosphere to block attempts to establish radio contact from other solar systems if the signal is weak enough?&lt;/p&gt;&#xA;" OwnerUserId="13215" LastActivityDate="2016-06-22T07:19:15.303" Title="Could the Sun's magnetosphere block extraterrestrial contact?" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16319" PostTypeId="2" ParentId="16235" CreationDate="2016-06-22T01:08:57.757" Score="0" Body="&lt;p&gt;Only if it happens to be 206 kpc away, &lt;em&gt;i.e.&lt;/em&gt; 1/4 of the distance from here to the Andromeda galaxy. A parsec is the distance at which 1 AU makes an angle of 1 arcsecond. If the nebula is 100 parsecs away and has an angular diameter of 250 arcseconds, then its diameter is&#xA;$$100 \ \mathrm{pc} \times 250 \ \mathrm{as} = 25000 \ \mathrm{AU} = 0.12\,\mathrm{pc}$$&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2016-06-22T13:46:01.560" LastActivityDate="2016-06-22T13:46:01.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16320" PostTypeId="2" ParentId="16318" CreationDate="2016-06-22T07:19:15.303" Score="1" Body="&lt;p&gt;The Sun's magnetosphere is too weak to have a significant effect on extra solar radio waves. If it did, we wouldn't have any radio telescopes as it would block the signals they detect.&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-06-22T07:19:15.303" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16324" PostTypeId="2" ParentId="15031" CreationDate="2016-06-22T12:06:38.377" Score="1" Body="&lt;p&gt;There are indeed models for halos made of warm dark matter. The profiles will depend on exactly what the dark matter is and what its temperature is. I am no expert, but a starting point could be a paper by Vinas et al. (2012), who claim that the profiles are much flatter in the centre than those of CDM halos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1202.2860&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1202.2860&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hot dark matter is defined as such during the early universe and will be nearly uniformly distributed at the epochs of early structure formation. However at later epochs, hot dark matter with rest mass can cool down. For example, the cosmic neutrino background would have been highly uniform when it decoupled and when  most galaxies formed. But because of the non-zero neutrino rest mass and the expansion of the universe, these neutrinos are now non-relativistic and will be affected by cosmic structures and become &quot;clumpy&quot;. The same will be true for any dark matter particle with a rest mass greater than about 0.1 eV.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-06-22T12:06:38.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16326" PostTypeId="1" CreationDate="2016-06-22T14:59:33.047" Score="1" ViewCount="59" Body="&lt;p&gt;Like the Metonic cycle, except for position in the sky, not phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Original question: If I look up at the moon tonight, how long before it is in the same apparent position, and same phase, again?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realized I can calculate this myself, if only I know two piece of information. Unfortunately I do not know the name of the second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question should be answered by the product of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the Metonic cycle, and&lt;/li&gt;&#xA;&lt;li&gt;the cycle where the moon's azimuth and altitude recur on a given day&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What is the name of this second cycle?&lt;/p&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="13237" LastEditDate="2016-06-22T19:29:03.177" LastActivityDate="2016-06-22T19:44:35.723" Title="What is the name of the cycle whereby the lunar azimuth and altitude recur on a given day?" Tags="&lt;the-moon&gt;&lt;moon-phases&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16327" PostTypeId="1" AcceptedAnswerId="16332" CreationDate="2016-06-22T15:56:41.080" Score="1" ViewCount="82" Body="&lt;p&gt;I've been reading about satellite observations (such as &lt;a href=&quot;http://sattrackcam.blogspot.co.uk/2016/06/updated-orbit-for-mentor-7-nrol-37.html&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;).&#xA;The writer uses terms such as these to define the orbit:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 41584U 16036A   16167.96105997 0.00000000  00000-0  00000+0 0    07 &#xA;2 41584   7.5055 353.7008 0046333  41.2140 319.1375  1.00195548    05 &#xA;&#xA;rms 0.004 deg      from 9 obs June 14.70 - June 16.79  (2.09 day arc) &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What does each variable refer to, and what's the easiest way to view these on a map (either online or in an app of some kind?)&lt;/p&gt;&#xA;" OwnerUserId="4146" LastActivityDate="2016-06-23T02:55:44.690" Title="How to understand orbit definitions" Tags="&lt;orbital-elements&gt;&lt;satellite&gt;&lt;artificial-satellite&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16328" PostTypeId="2" ParentId="16326" CreationDate="2016-06-22T16:24:44.160" Score="1" Body="&lt;p&gt;You are looking for the Metonic cycle.  This is almost exactly 19 years, and represents 235 synodic months.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have edited the detailed question so that it is out of sync with the original Title.  The other period you need is the Draconic month, which is the period between the moon crossing the ecliptic (in the same direction).&lt;/p&gt;&#xA;" OwnerUserId="11822" LastEditorUserId="11822" LastEditDate="2016-06-22T19:44:35.723" LastActivityDate="2016-06-22T19:44:35.723" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16329" PostTypeId="2" ParentId="13250" CreationDate="2016-06-22T21:44:32.327" Score="1" Body="&lt;p&gt;You and your phone will be destroyed before you get to the black hole, but &#xA;If we pretend that you and your phone can survive being in the vicinity of a black hole ... And let's suppose this is a regular black hole of about 5 to 10 solar masses, and you are falling in, you are not trying to orbit it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you fall towards the black hole, at first there is no problem. As you get closer to the event horizon you will accelerate, but until the last second or so the relativistic effects are quite small. Your phone signal will be redshifted, so the receiver will have to adjust the frequency that they are listening in on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last second before you fall in will last forever. Your signal will become redshifted further and further, your apparent rate of speaking will slow down and the last second before you pass the Event horizon will, to the outside observer, last forever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You however are doomed. You pass the Event horizon and reach the singularity in a fraction of a second.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-06-22T21:44:32.327" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16330" PostTypeId="1" AcceptedAnswerId="16334" CreationDate="2016-06-22T21:52:27.500" Score="1" ViewCount="79" Body="&lt;p&gt;I was wondering about what happens when a black hole forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean, a black hole forms when a star collapses, so when an enormous mass of protons, electrons, neutrons etc collapses together.&lt;br&gt;&#xA;But when it does, all those particles should come &quot;in contact&quot;, so what happens when electrons and protons come in contact?&lt;br&gt;&#xA;Shouldn't they annihilate each other releasing energy?&lt;br&gt;&#xA;In the end, shouldn't the singularity be composed only by the mass of particles without an electrical charge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not a professional, I just find astronomy fascinating and I like to read about it, so forgive me if I said something obviously wrong :)&lt;/p&gt;&#xA;" OwnerUserId="13248" LastEditorUserId="2365" LastEditDate="2016-06-23T05:40:15.760" LastActivityDate="2016-06-23T05:40:15.760" Title="Black holes and material interaction" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16331" PostTypeId="2" ParentId="13996" CreationDate="2016-06-22T21:57:59.887" Score="1" Body="&lt;p&gt;We detect black holes mostly by their effects on nearby matter. The &quot;radial velocity&quot; method is important after x-ray observations suggest a likely black-hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matter in orbit around a black hole can form an accretion disc. This disc will become very hot, and glow brightly in X-ray radiation. If we see X-ray radiation coming from a star we look at the radial velocity of the star, and use this to infer the mass of the orbiting object. Since a neutron star cannot exist at more than 3 solar masses, if the object is more than that it must be a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compared to planets, black holes are rare. Only about 20 candidates are known in the Milky way.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-06-22T21:57:59.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16332" PostTypeId="2" ParentId="16327" CreationDate="2016-06-22T23:53:37.837" Score="3" Body="&lt;p&gt;These are known as&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Two-line_element_set&quot; rel=&quot;nofollow&quot;&gt;two-line elements&lt;/a&gt;&#xA;(see also &lt;a href=&quot;http://spaceflight.nasa.gov/realdata/sightings/SSapplications/Post/JavaSSOP/SSOP_Help/tle_def.html&quot; rel=&quot;nofollow&quot;&gt;nasa.gov&lt;/a&gt;,&#xA;&lt;a href=&quot;http://www.satobs.org/element.html&quot; rel=&quot;nofollow&quot;&gt;satobs.org&lt;/a&gt;).&#xA;On both lines, the first field after the line number is a NORAD ID,&#xA;which you can use to query a database such as&#xA;&lt;a href=&quot;http://www.n2yo.com/satellite/?s=41584&quot; rel=&quot;nofollow&quot;&gt;N2YO&lt;/a&gt;.&#xA;The second line gives geocentric&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_elements&quot; rel=&quot;nofollow&quot;&gt;orbital elements&lt;/a&gt;&#xA;for the epoch given in the first line.&#xA;The last line looks like an error estimate by observation analysis software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first convenient visualization service I found was&#xA;&lt;a href=&quot;http://www.orbitalpredictor.com/home/41584/&quot; rel=&quot;nofollow&quot;&gt;Orbital Predictor&lt;/a&gt;,&#xA;where I naively pasted the above TLE data with a name less appropriate than &quot;Mentor 7&quot; or &quot;2016-036A.&quot;&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2016-06-23T02:55:44.690" LastActivityDate="2016-06-23T02:55:44.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16333" PostTypeId="1" AcceptedAnswerId="16337" CreationDate="2016-06-23T01:59:50.107" Score="1" ViewCount="431" Body="&lt;p&gt;Please correct me if I am wrong as I may have made some incorrect assumptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay so we know that at some stage of &quot;nearness to a black hole&quot;, light is no longer reflected back at us from the black hole itself. Some scientists have proposed the speed of light and it's relationship to relativity is natures way of adjusting the rate at which time passes seemingly to disallow &quot;weirdness that could result&quot; .. for example the fast moving object would become more massive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also know that the mass of the black hole causes a nearby observer to behave as if he/she were moving at incredible speeds and subsequently slows time for the observer &quot;relatively speaking of course&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that from the perspective of the observer light is traveling to and away from him/her at it's normal rate even though effectively the observer is traveling very fast. Imagine however what would happen to the frequency of the light. The red/blue shift would undoubtedly be obscene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we here on earth are &quot;relatively speaking&quot; experiencing time passing at our normal &quot;earth rate&quot;. I can't help but imagine that immense gravity near a black hole is distorting time and its darkness is a result of either extreme red/blue shift or the light is somehow phased or altered by the temporal distortion. Yes gravity is the origin of the darkness but is the direct cause or an indirect relationship where the rate of time passing is the culprit?&lt;/p&gt;&#xA;" OwnerUserId="13251" LastActivityDate="2018-08-26T21:57:27.763" Title="Black hole darkness a result of gravity or temporal distortion?" Tags="&lt;black-hole&gt;&lt;space-time&gt;&lt;redshift&gt;&lt;special-relativity&gt;&lt;time-dilation&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16334" PostTypeId="2" ParentId="16330" CreationDate="2016-06-23T02:31:02.130" Score="1" Body="&lt;p&gt;Electrons and protons do not annihilate each other, since protons have a much greater mass than electrons. Only matter and antimatter does, since a particle and its antimatter counterpart have the exact same mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, under the right conditions, an electron and proton can combine. This forms a neutron, since the charges will cancel each other out. This process is called electron capture, and it's the reason why neutron stars form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not an expert on black holes (I haven't studied GR), but I know that a lot of black holes form from neutron star collapses.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2016-06-23T02:31:02.130" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16336" PostTypeId="1" CreationDate="2016-06-21T14:56:58.043" Score="4" ViewCount="269" Body="&lt;p&gt;I've been researching this but so far have not found an answer.  If you were standing on the top of the rim of Shackleton Crater at the lunar south pole, would you be able to see any of Earth? I can't find anything about the height of the rim, although nearby Malapert Mountain is apparently visible from Earth so you could see Earth from Malapert.  Shackleton, though, I'm not so sure.  Any insights appreciated. KPRedmond&lt;/p&gt;&#xA;" OwnerDisplayName="K. P. Redmond" LastActivityDate="2016-06-25T19:39:18.163" Title="Would a viewer see Earth from the rim of Shackleton Crater?" Tags="&lt;planet&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16337" PostTypeId="2" ParentId="16333" CreationDate="2016-06-23T13:19:22.710" Score="-1" Body="&lt;p&gt;Gravity is in fact a side-effect or result of spacetime distortion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider this simple diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SetZh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SetZh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blue grid representing timespace, and the orange sphere being a high-mass object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving &lt;strong&gt;through&lt;/strong&gt; space means going from one tile to another, so when light passes into the distorted dip in spacetime it not only has to simply pass from square to square, but do so whilst those squares are being pulled in towards the massive object.&#xA;For a planet or average star, this distortion is negligible as far as light is concerned, but the dip a Black Hole would make would be many, many times deeper than the above diagram - space would be flowing in to the hole as fast, or faster than light can travel along said space to pass through it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like trying to climb up a downwards escalator that always moves faster than you can climb it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But spacetime flows, it &lt;em&gt;should&lt;/em&gt; eventually be dragged out, like pulling a table cloth out from under a load of dishes, but the mass of a black hole is so great, timespace essentially becomes pinned, like jamming your fork into the tablecloth as your friend tries to pull it away; time comes to a standstill at the epicentre of the hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important to note that at no point does light slow down travelling through the vacuum of space here, it's just space is moving in the opposing direction preventing the light from escaping in any meaningful amount beyond the expected level of Hawking Radiation. &lt;/p&gt;&#xA;" OwnerUserId="11009" LastActivityDate="2016-06-23T13:19:22.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16338" PostTypeId="1" AcceptedAnswerId="16341" CreationDate="2016-06-23T13:58:36.100" Score="1" ViewCount="462" Body="&lt;p&gt;I need your help! I am having some weird tracking issues with my tripod (Advanced VX). When I setup and calibrated my scope yesterday I took this picture of NGC6888 (90s exposure): &lt;a href=&quot;https://i.stack.imgur.com/Ft7cC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ft7cC.jpg&quot; alt=&quot;http://imgur.com/3siYrNX&quot;&gt;&lt;/a&gt;&#xA;As you can see this image looks quite good tracking wise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However due to unknown reasons it seems that 1 in 3 images is always bad and has tracking issues, like this one: &lt;a href=&quot;https://i.stack.imgur.com/ZrMuf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZrMuf.jpg&quot; alt=&quot;http://imgur.com/NNduJZS&quot;&gt;&lt;/a&gt;&#xA;This movement always occours in the same direction, it seems that for some reason one axis of the mount is moving faster periodically, causing these tracking issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Funny is that is seems to be occouring periodically. I always get 4 images good, 2 images bad, 4 images good, 2 images bad, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried fiddeling around with the anti-backlash settings (I never touched them before) but that didnt fix it. My current anti-backlash settings are: RA +- 55, Dec +- 60&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know where those tracking issues could come from and how to fix them? Note I also have a 5s mirror-lock so it should not come from vibrations of the camera.&lt;/p&gt;&#xA;" OwnerUserId="10271" LastActivityDate="2016-06-23T16:09:05.997" Title="Having some weird issues with my tracking (Advanced VX)" Tags="&lt;photography&gt;&lt;alignment&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16339" PostTypeId="1" AcceptedAnswerId="16340" CreationDate="2016-06-23T14:00:23.063" Score="8" ViewCount="2762" Body="&lt;p&gt;It has to something with the positioning of sun and galactic center of the milky way but I wasn't able to understand how for a given star, its co-ordinates are calculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some explanation would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="11988" LastActivityDate="2016-06-23T15:09:06.727" Title="What is the Galactic Latitude and Galactic Longitude? How is different from the latitude and longitude we use on Earth?" Tags="&lt;milky-way&gt;&lt;positional-astronomy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16340" PostTypeId="2" ParentId="16339" CreationDate="2016-06-23T14:18:46.240" Score="16" Body="&lt;p&gt;The Galactic Coordinate system is a longitude-latitude coordinate system that is used to define the positions of objects in space, most commonly objects within our own galaxy. It uses the center of our galaxy as the focal point (i.e., where we consider $(0^{\circ},\:0^{\circ})$), much like we use the position just off Africa as the focal point of the Earth's lon-lat system. You can see a visual representation of this system below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IG5EE.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IG5EE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reference point is of course from our solar system. Longitude is measured in the plane of the galaxy, going positive as you sweep to the left (thus following a right-hand rule). The symbol $\ell$ is generally used for galactic longitude. Latitude is measured above or below the plane, going positive &quot;above&quot; the plane and negative &quot;below&quot;. We define above and below based on the right hand rule as applied to our galaxy's rotation. The symbol $b$ is generally used for galactic latitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone says an object is at galactic coordinate position $(\ell,b) = (180^{\circ},\:+45^{\circ})$, then the object will be located directly opposite the galactic center and above the plane of the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from being a lon-lat system, just as the Earth's is, it does not share any features with Earth's lon-lat system. That is, there's generally no relation between the two and the definition of one is not affected by the other.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2016-06-23T15:09:06.727" LastActivityDate="2016-06-23T15:09:06.727" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16341" PostTypeId="2" ParentId="16338" CreationDate="2016-06-23T15:03:18.683" Score="3" Body="&lt;p&gt;Looks like you have a periodic error issue. The worm period on the AVX is about 9mins I think. The issue has a period of 90sec * 6 = 540sec (9min).&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-06-23T15:03:18.683" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16342" PostTypeId="2" ParentId="16338" CreationDate="2016-06-23T16:09:05.997" Score="1" Body="&lt;p&gt;This is not a solution to the problem, just an idea which may help diagnose it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try setting up with the mount deliberately aligned &lt;strong&gt;slightly&lt;/strong&gt; off the pole. Then a time exposure with &quot;perfect&quot; tracking should record stars as a straight line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once set up like this take a single long exposure for, say, half an hour and see if the trails on the image are straight with regular bumps (that would probably indicate periodic error as James suggests), or if something more random is going on.&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2016-06-23T16:09:05.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16343" PostTypeId="2" ParentId="13587" CreationDate="2016-06-23T16:45:26.913" Score="2" Body="&lt;p&gt;When a body spins around another body due to gravity and maintains a consistent orbit, we can know clearly two things about the body: &#xA;1. The speed of the orbit&#xA;2. The average radius of the orbit&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about it. The Moon spins around the Earth due to gravity of the Earth. Right? And it's orbit always stays the same(when averaged). For it to stay in this balance where it isn't flying away from the Earth, requires that it be traveling at a speed and distance that equal out the effect of gravity on the object. So by knowing this speed(orbital speed) and distance(orbital radius), we can determine something very useful about the gravitational effect of the Earth on the Moon. From the gravitational effect, we can then determine the mass of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding the mass of the Earth otherwise is quite difficult because we don't know exactly what is inside, and there are too many factors involved. We can't just put the Earth on a big scale. Also, determining it from gravity is flawed because there are too many factors that cause high uncertainty(Earth is not a perfect sphere, the moon exhibits gravity too, atmospheric pressure, centrifugal force, etc.). BUT we DO know the distance of the Moon from the Earth very accurately using Trigonometry. We also know the time the Moon takes to go around the Earth as we can easily measure this(lunar cycle). So if you take the orbital speed squared and multiply it by the orbital radius of the Moon's spin around the Earth, you obtain the Geocentric Gravitational Constant of the Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using that constant, we can determine the mass of the Earth. But we also need to know the Gravitational Constant(G) to make this determination. However, G has been difficult to determine with a high level of certainty. This is because the experiments to determine G have so far given inconsistent results. Either G is changing OR the factors in the experiments have not all been ruled out, likely the latter. You can read more on Geocentric Gravitational Constant here: &lt;a href=&quot;https://en.wikipedia.org/wiki/Standard_gravitational_parameter&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Standard_gravitational_parameter&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13264" LastActivityDate="2016-06-23T16:45:26.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16344" PostTypeId="1" AcceptedAnswerId="16347" CreationDate="2016-06-23T18:00:33.053" Score="1" ViewCount="146" Body="&lt;p&gt;This article, &lt;a href=&quot;http://phys.org/news/2014-12-binary-terrestrial-planets.html&quot; rel=&quot;nofollow&quot;&gt;http://phys.org/news/2014-12-binary-terrestrial-planets.html&lt;/a&gt;, suggests binary planets could orbit each other at a distance of only three planet radii. For two earth-like planets, that is approximately 12,000 miles. Does this distance seem accurate? How long would it take for these planets to orbit their barycenter?&lt;/p&gt;&#xA;" OwnerUserId="13267" LastActivityDate="2016-06-23T21:43:20.573" Title="Distance and orbital period for terrestrial binary planets" Tags="&lt;planet&gt;&lt;orbital-elements&gt;&lt;orbital-mechanics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16346" PostTypeId="2" ParentId="16333" CreationDate="2016-06-23T21:29:39.447" Score="1" Body="&lt;p&gt;As an object falls towards a black hole, light emitted by the object will become red-shifted by the gravity. As photons rise out of the gravity well of the black hole they lose energy, they can't slow down but their wavelength is stretched. The object will go dark as it approaches the event horizon due to gravitational redshifting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photons starting from the event horizon would be redshifted to zero energy, and no photons can travel from inside the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also will observe gravitational time dilation, the last second before the object reaches the event horizon will be stretched out indefinitely. The two effects are all part of the same relativistic theory of gravity. You can't unpick the two effects.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-06-23T21:29:39.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16347" PostTypeId="2" ParentId="16344" CreationDate="2016-06-23T21:43:20.573" Score="2" Body="&lt;p&gt;The formula for orbital period is $$T= 2\pi\sqrt{\frac{a^3}{G \left(M_1 + M_2\right)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your example, $a=19000000$ metres, and $M_1 = M_2 = 6e24$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which gives an orbital period of just over 5 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether such an arrangement is likely depends on the nature of the the formation of the binary planet, and how the orbital period and rotation rates are affected by tides.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-06-23T21:43:20.573" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16350" PostTypeId="1" AcceptedAnswerId="16351" CreationDate="2016-06-24T05:24:08.577" Score="2" ViewCount="607" Body="&lt;p&gt;I feel like this is going to be a duplicate of this question, &lt;a href=&quot;https://astronomy.stackexchange.com/questions/11709/astrophotography-unable-to-achieve-focus-with-current-setup&quot;&gt;Astrophotography - Unable to achieve focus with current setup&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the suggestions to go to a straight adapter doesn't apply in my case. I have a Canon Rebel XT, with a 2&quot; 48mm adapter &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B00T3P5ABC&quot; rel=&quot;nofollow&quot;&gt;https://www.amazon.com/gp/product/B00T3P5ABC/ref=oh_aui_detailpage_o01_s00?ie=UTF8&amp;amp;psc=1&lt;/a&gt;, on my Orion XT10 dob, &lt;a href=&quot;http://www.telescope.com/Telescopes/Dobsonian-Telescopes/IntelliScope-Dobsonians/Orion-SkyQuest-XT10i-IntelliScope-Dobsonian-Telescope/pc/1/c/12/sc/27/p/102013.uts&quot; rel=&quot;nofollow&quot;&gt;http://www.telescope.com/Telescopes/Dobsonian-Telescopes/IntelliScope-Dobsonians/Orion-SkyQuest-XT10i-IntelliScope-Dobsonian-Telescope/pc/1/c/12/sc/27/p/102013.uts&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I attach the adapter to the camera and focus on a planet or star, I can't achieve focus. My focus adjuster on the scope can't get close enough. It's as if the camera is just out of range of the focuser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what the issue is and would love to hear some suggestions. I'm just a noob getting started and have a lot to learn so I apologize at all the beginner questions here.&lt;/p&gt;&#xA;" OwnerUserId="13272" LastActivityDate="2016-06-24T07:32:45.713" Title="Unable to achieve focus" Tags="&lt;photography&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16351" PostTypeId="2" ParentId="16350" CreationDate="2016-06-24T07:32:45.713" Score="3" Body="&lt;p&gt;This is a common problem with Newtonian type telescopes and DSLRs, the sensor in the camera is too far back to reach focus. As mentioned in the comments a low profile focuser may help or move the main mirror up the tube, I've seen this done on some Skywatcher Newtonians to solve similar problems.&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-06-24T07:32:45.713" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16352" PostTypeId="1" CreationDate="2016-06-24T10:38:59.940" Score="0" ViewCount="216" Body="&lt;p&gt;I want to solve this exercise from Spherical Astronomy by Smart:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that, in latitude 45°, the interval between the moment at which a star's azimuth is 90° east and the moment of setting is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't understand it, it says prove that it is constant for a star, or it is constant for every star with different declinations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="11484" LastEditorUserId="11143" LastEditDate="2016-06-24T13:10:21.760" LastActivityDate="2016-06-24T14:37:55.880" Title="an exercise from Spherical Astronomy by Smart" Tags="&lt;declination&gt;&lt;azimuth&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16353" PostTypeId="1" AcceptedAnswerId="16356" CreationDate="2016-06-24T13:38:01.977" Score="2" ViewCount="722" Body="&lt;p&gt;Recently LIGO discovered gravitational waves caused by two black holes that were orbiting each other, and then collapsed into one black hole. A few months later, we find out that this actually happens quite often. However, these collisions are far enough away from Earth that no one can feel the gravitational waves, and it actually takes very precise measurement to even determine that they happen. So my question is, &lt;strong&gt;how close would a black hole collision have to be for us to feel the gravitational waves pass through Earth?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, is this even possible? As in, would the distance have to be so short that we'd get swallowed up before the black holes collided? And, if we were to feel such gravitational waves, what would they feel like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Let's say that the black holes have the same mass as Earth, giving their event horizons a radius of about 9 mm (the size of a dime). So two Earth-sized black holes that are stuck in an orbit finally collapse is the scenario here.&lt;/p&gt;&#xA;" OwnerUserId="13275" LastEditorUserId="13275" LastEditDate="2016-06-24T14:12:01.130" LastActivityDate="2016-06-24T15:48:49.690" Title="How close would merging black holes have to be to feel gravitational waves?" Tags="&lt;gravity&gt;&lt;black-hole&gt;&lt;earth&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ClosedDate="2016-06-27T13:20:13.113" ContentLicense="CC BY-SA 3.0" />
  <row Id="16355" PostTypeId="2" ParentId="16352" CreationDate="2016-06-24T14:37:55.880" Score="3" Body="&lt;p&gt;Constant for every star with different declination's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though you would have to assume the the declination is such that it can have an azimuth of 90 each, not all stars will. Depending on your latitude some stars will never get to an azimuth of 90, they will either be always be further north or south. &lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-06-24T14:37:55.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16356" PostTypeId="2" ParentId="16353" CreationDate="2016-06-24T15:48:49.690" Score="5" Body="&lt;p&gt;It's a bit hard to answer this question as there are some ambiguities and certain parameters of the setup aren't defined, but I can try to give you a general idea.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If we were to feel such gravitational waves, what would they feel like?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll start here as we need to have some well established meaning for how we can feel a wave before calculating what conditions produce such a wave. However, this is the ambiguous part. A gravitational wave passing over us would effectively just be felt as a differential gravitational field, which can induce spaghettificiation as Andy points out. One side of your body might feel a stronger pull than the other being that one side is experiencing slightly different gravity than the other. But I have to say that you'd need extreme waves to have this effect. The frequency of the waves would have to be extremely high for the wavelength to be on the order of the human body's size. If the wavelengths are much longer than your body, the change, even from a powerful gravitational wave, would probably be too slow for you to detect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll note that the frequency of the waves detected by LIGO were about $250\:Hz$ for the first detection (at the peak) and $450\:Hz$ for the second detection (again at the peak). That implies a wavelength of $\sim600\:km$ at the shortest. At these extreme wavelengths, the gravity differential is negligible. You'd need wavelengths on the order of a meter which implies frequencies of $10^9-10^{10}\:Hz$ which is right near the peak of astrophysical possibilities. So already your proposal of Earth-sized black holes is out. There's no way they can produce these extreme waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now we run into another problem. In order to make the wavelengths on the order of the human body such that you can actually feel the differential, you've made the frequency extremely high. Could your body even notice gravitational changes happening in $10^{-10}\:sec$? I doubt it. That'd be like me blasting you with hot air, then cold, but flipping back and forth between the two at a rate of 10 billion times a second. I have a feeling everything would just cancel out and you wouldn't really feel anything, but it's hard to say.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How close would a black hole collision have to be for us to feel the gravitational waves pass through Earth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's ignore the issue of whether you'd ever feel a wave and look at the main question. I'll just consider raw energy in the wave and hope that somehow equates to feeling it. The power output of two orbiting bodies of mass $M$ (assuming the same mass for simplicity), and orbiting at a radius of $r$ &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_wave#Power_radiated_by_orbiting_bodies&quot;&gt;is given by&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{grav} = \frac{32}{5}\frac{G^4}{c^5}\frac{2M^5}{r^5}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $G=6.67\times10^{-11}\:m^3kg^{-1}s^{-2}$ is the gravitational constant and $c=3\times10^8\:m/s$ is the speed of light. Now $P_{grav}$ is the total power radiated out in energy per second in the form of gravitational waves. You of course would not be experiencing this entire power though because this is radiated in all directions. The power is not radiated spherically symmetrically, but for simplicity's sake, let's just assume it is. That means we can say the power you experience given your area $A$ and distance $d$ from the system is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{exp} = P_{grav}\frac{A}{4\pi d^2} = \frac{32}{5}\frac{G^4}{c^5}\frac{2M^5}{r^5}\frac{A}{4\pi d^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Play with some numbers for this and see what is necessary to have an appreciable power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will point out though that at the distance you'd have to be where this power became appreciable, it would be overwhelmed by the tidal forces involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in short, no I don't think you'll ever be able to feel gravitational waves, no matter the circumstance.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-06-24T15:48:49.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16357" PostTypeId="2" ParentId="16336" CreationDate="2016-06-24T17:49:50.323" Score="4" Body="&lt;p&gt;The moon has a monthly motion called &lt;em&gt;&lt;a href=&quot;http://earthsky.org/space/how-much-of-the-moon-can-we-see-from-earth-lunar-libration&quot; rel=&quot;nofollow&quot;&gt;libration&lt;/a&gt;&lt;/em&gt; Of interest to you is libration in latitude. The orbit isn't exactly in the plane of the ecliptic, and the rotation axis is slightly different from each. This means that about a week after passing its ascending node, you can see up to about 6.5 degrees past the moon's South pole. Two weeks later, the south pole is invisible, but then north pole can be seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the point of view of an observer on Shackelton it would mean that over the course of a month the Earth would slowly rise and then set again.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-06-25T19:39:18.163" LastActivityDate="2016-06-25T19:39:18.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16358" PostTypeId="1" CreationDate="2016-06-24T21:13:15.583" Score="0" ViewCount="767" Body="&lt;p&gt;If Earth's axis of rotation were tilted 90 degrees, what type of seasons would be observed at the equator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My prediction is that there would be no seasons, as the orientation of the equator would be the same as if the Earth &lt;/p&gt;&#xA;" OwnerUserId="13277" LastActivityDate="2016-06-25T06:03:21.893" Title="How would seasons at the equator be if the Earth were tilted 90 degrees?" Tags="&lt;rotation&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2016-06-27T15:56:38.830" ContentLicense="CC BY-SA 3.0" />
  <row Id="16360" PostTypeId="2" ParentId="16358" CreationDate="2016-06-25T03:13:10.433" Score="3" Body="&lt;p&gt;I assume you mean &quot;what if the Earth's equator was perpendicular to it's orbit?&quot; in which case the seasons would be greatly exaggerated at the poles and day and night would only really mean something for about half the year if you're nearer the equator. Temperature differential would be much more significant than it is now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless you mean that the axis of Earth's rotation is exactly perpendicular with respect to it's orbit, in which case there would be no seasonal variation whatsoever. (Not including Malkovitch cycles and solar cycles)&lt;/p&gt;&#xA;" OwnerUserId="11525" LastEditorUserId="2365" LastEditDate="2016-06-25T05:44:44.240" LastActivityDate="2016-06-25T05:44:44.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16363" PostTypeId="2" ParentId="16192" CreationDate="2016-06-25T06:06:20.560" Score="3" Body="&lt;p&gt;I should clear this up: the Milky Way isn't a completely well-defined body. Galaxies shouldn't be thought of as a single body — they are more like a collection of various material and stars. That said, I can only give you a wonky estimate of what &lt;em&gt;might&lt;/em&gt; be the composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;Milky Way&quot;, I'm assuming you mean only the Milky Way's disc and halo (and not &lt;em&gt;everything&lt;/em&gt; that orbits the Galactic Center, like the satellite galaxies). That alone is hard to define, but let's get a rough estimate. The entire mass of the Milky Way is a bit uncertain, but it falls between $0.8–1.5×10^{12}M_☉$. Let's average that at about $1.2×10^{12}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sagittarius A*, the black hole at the center of our galaxy, has a mass of roughly $4,000,000M_☉$. That would make it only ~0.0003% of the galaxy! A few sources state that dark matter composes about 90% of the Milky Way's mass.&#xA;So it has a mass of approximately $1.08×10^{12}M_☉$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want to dissect the rest, we'll have to make generalizations. From some sources like &lt;a href=&quot;http://members.nova.org/~sol/chview/chv5.htm&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; one, about 75% of MW stars are M type stars, 15% are F-K stars and the remaining 10% are mostly degenerate stars. Knowing that there are 100 billion stars in our galaxy, we can take their average masses and &lt;em&gt;kind of&lt;/em&gt; conclude that 3% of the Milky Way's mass is from stellar objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the mass of gases in the Milky Way, &lt;a href=&quot;http://www.astro.cornell.edu/academics/courses/astro1101/lectures/21MilkyWay.pdf&quot; rel=&quot;nofollow&quot;&gt;this source&lt;/a&gt; states that both atomic gas and molecular gas would individually make up 0.25% of the Milky Way's mass. Combined, we get 0.5%. The rest is buried in a sea of contradictions and mystery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, here's the best breakdown I could make:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Sagittarius A*: 0.0003%&#xA;Dark matter: 90%&#xA;Stellar objects: 3%&#xA;Gases and nebulae: 0.5%&#xA;Everything else: 6.5%&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Again, it is extremely difficult to break down the mass of the components of a galaxy. This is the best you can really do without combining tremendous generalizations with shoddy math and sources.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-06-25T12:54:50.700" LastActivityDate="2016-06-25T12:54:50.700" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="16364" PostTypeId="1" CreationDate="2016-06-25T08:57:22.590" Score="0" ViewCount="225" Body="&lt;p&gt;As a black body a star emits a continuous spectrum of light and other radiation. I've learned they are capable of knowing of what elements the stars are made of because of the spectral lines. But how do they distinguish those lines from other radiations?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="2270" LastEditDate="2016-06-25T16:46:06.057" LastActivityDate="2016-06-25T16:46:06.057" Title="How do astronomers distinguish the spectral lines?" Tags="&lt;star&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16365" PostTypeId="2" ParentId="16364" CreationDate="2016-06-25T10:42:43.057" Score="5" Body="&lt;p&gt;The lines show as either peaks above the black body emission (for emission lines) or troughs below the black body emission (for absorption lines). Though stars more often have absorption lines and nebulae emission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JdLke.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JdLke.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &#xA;&lt;a href=&quot;https://i.stack.imgur.com/DTJKx.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DTJKx.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-06-25T10:42:43.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16366" PostTypeId="1" CreationDate="2016-06-25T11:30:23.027" Score="0" ViewCount="91" Body="&lt;p&gt;One of the most massive Supermassive Black Hole observed is the one at the centre of the galaxy NGC1600 with a mass of 17 billion suns. It would have a density of ∼0.01kg/m^3, or one part in 100,000 times the density of water, or 1% the density of earth atmosphere at sea level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if there is so much gas why isn't the bh shrinking and becoming more dense as gravity would pull everything together?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastActivityDate="2016-06-26T05:42:04.287" Title="How can black holes be sometimes so gaseous?" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16367" PostTypeId="1" CreationDate="2016-06-25T15:33:33.847" Score="-1" ViewCount="190" Body="&lt;p&gt;When we look to the distant farthest reaches of the universe we see light that was emitted at the big bang 14 billion years ago.  But the universe was tiny back then so that light, which is only reaching us now, was emitted pretty much right next to us (in galactic terms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's been traveling for 14 billion years at the speed of light and is only reaching us right now, despite only having set off from right next door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard explanation of how it's covered such a tiny distance begins with &quot;expansion&quot;, which means the light has been swimming against the tide effectively and since that doesn't fit properly, it's augmented by inventing inflation which is basically a huge rush of Hubbly expansion at the outset, to increase the distance away from us before it began swimming towards us in accordance with Hubble's law which we now observe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we instead explain how it's covered such a tiny distance by saying that we are all departing the big bang at or near the speed of light because we are actually for example something like Hawking Radiation, then we don't need &quot;Expansion&quot; do we?  Because the CMB is just the light that was emitted near parallel to us and has been traveling through space next to us all this time.  So it's quite natural that it is only now reaching us.  And then we don't need to invent inflation, do we, to make Hubble's law and the Big Bang fit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And since it's been traveling so close to parallel all this time it seems natural that it should have such a low level of energy - unlike light emitted right next to us recently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then we have a natural reason for mass-energy equivalence don't we, because everything's moving at the speed of light anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then we have a natural reason for quantum theory don't we because if everything around us is moving nearly parallel at the speed of light then only its direction can change, not its speed so we would expect to find that if we impose a speed vector on objects which only have direction, their degrees of freedom will be over-specified and we won't be able to define their parameters all at once because they don't really exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then it becomes obvious that &quot;perceived speed&quot; is simply an aggregation of how parallel the component parts of something are, so simple objects containing minimal component parts must travel at the speed light while compound things can move more slowly because they have component parts moving in different, changing directions around each other and their overall motion is the combination of the parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then that would predict that a photon can't age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then everything seems to make a lot more sense because when we look out at the &quot;Big Bang&quot; in distant space and see it receding at the speed of light, well that's what we know it to be doing anyway so whoopdedoo who cares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And because the universe is now just a set of paths in 4-dimensional space with no &quot;speed&quot;, it should be described by a rotation group SO(3,1) and oh hang on haven't I seen that somewhere before, like the set of transformations in relativity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then we would all have to be made of light.  So how do we design an experiment in which we bounce beams of appropriately formed light off each other itself at exceptionally low angles of incidence in order to prove that they do, to the surprise of almost the entire physics community, bounce off each other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And how did we get in such a pickle inventing all this inflation and dark matter?&lt;/p&gt;&#xA;" OwnerUserId="12000" LastActivityDate="2016-06-25T22:29:54.903" Title="Why is the CMB not simply travelling parallel to us?" Tags="&lt;general-relativity&gt;&lt;big-bang-theory&gt;&lt;expansion&gt;&lt;cosmological-inflation&gt;&lt;cmb&gt;" AnswerCount="1" CommentCount="8" ClosedDate="2016-07-02T17:32:38.570" ContentLicense="CC BY-SA 3.0" />
  <row Id="16368" PostTypeId="1" AcceptedAnswerId="16369" CreationDate="2016-06-25T16:25:33.190" Score="2" ViewCount="136" Body="&lt;p&gt;How to make random positions in Equatorial coordinate system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I divide Dec between -90 and 90 evenly, the space distribution is not random.&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2016-06-26T12:47:24.503" Title="a random distribution of RA&amp;Dec" Tags="&lt;coordinate&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16369" PostTypeId="2" ParentId="16368" CreationDate="2016-06-25T17:39:22.713" Score="6" Body="&lt;p&gt;Well, your problem is that you have a sphere. To compensate for the polar declination skew, you just calculate $$\sin^{-1}(\mathtt{rand})$$&#xA;Where $\mathtt{rand}$ is a number in an evenly divided -1 to 1 range. Alternatively, you can use another trigonometric function based on what you have available, but you get the idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RA can obviously be divided evenly.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="2365" LastEditDate="2016-06-25T19:17:41.807" LastActivityDate="2016-06-25T19:17:41.807" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16370" PostTypeId="1" CreationDate="2016-06-25T18:46:48.900" Score="1" ViewCount="50" Body="&lt;p&gt;Nothing is showing up in the viewfinder of camera when I try to do eyepiece projection photography.could I have something set up wrong?&lt;/p&gt;&#xA;" OwnerUserId="13282" LastActivityDate="2016-06-25T19:59:26.153" Title="Eyepiece projection photography" Tags="&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2016-06-27T14:10:30.940" ContentLicense="CC BY-SA 3.0" />
  <row Id="16371" PostTypeId="2" ParentId="16366" CreationDate="2016-06-25T19:37:53.940" Score="2" Body="&lt;p&gt;A black hole is not a solid or gaseous body. It is a shape that space-time can take. Now the event horizon is sometimes treated as being the &quot;edge&quot; of the black hole, but there is nothing at this edge. The black hole can't shrink its event horizon, since the event horizon is not a physical object&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no &lt;em&gt;matter&lt;/em&gt; inside the black hole (for this reason the shape of space-time around a black hole is sometimes called a &quot;vacuum solution&quot; of Einstein's equations). The &lt;em&gt;mass&lt;/em&gt; of the black hole is concentrated at the singularity (which, due to the curving of space time you can imagine as more like a point in time that you can't avoid if you cross the event horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the radius of the black hole's event horizon is proportional to the mass, the volume of the black hole is proportional to the cube of the mass. If you double the mass, you increase the volume by a factor of 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Density is mass/volume so if you double the mass you change the density by 2/8=0.25, so the more massive a black hole is, the lower is the average density of the region inside the event horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does mean that if you did manage to find a gas cloud with 17 billion times the mass of the sun (much bigger than any known nebula - Orion Nebula M43 is only a few thousand solar masses) and you were to somehow compress it to a sphere with 0.1 atm pressure, it would undergo gravitational collapse and form a black hole.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-06-26T05:42:04.287" LastActivityDate="2016-06-26T05:42:04.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16372" PostTypeId="2" ParentId="16370" CreationDate="2016-06-25T19:59:26.153" Score="1" Body="&lt;p&gt;The problem may be focus and or a faint image in the viewfinder, try looking at a distant object in daytime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course unless you are using a dslr type camera the viewfinder will not be looking through the main optics anyway so wont show anything.&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-06-25T19:59:26.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16373" PostTypeId="2" ParentId="16367" CreationDate="2016-06-25T22:29:54.903" Score="3" Body="&lt;p&gt;Some of your questions seem to be genuine enough, so I'll address those:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the CMB was emitted, it was emitted everywhere in the Universe, in all directions. That means that we see it at all times, but the part that we see at a given time comes from an ever-increasing distance. Today that distance is roughly 47 billion lightyears (Gly). It is also redshifted, today by a factor of $z \simeq 1100$. That means that when it was emitted, the Universe was a factor 1100 smaller (in all three directions). Hence, that particular part of the Universe that emitted the light that we see today was $47\,\mathrm{Gly} \,/\, 1100 \simeq 43 \,\mathrm{Mly}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the CMB is observed from all directions, only light from small patches on the sky can be regarded as being parallel. Obviously light from the left of you and above you is not parallel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CMB hasn't really been &quot;right on our heels&quot; since it was emitted, any more than, say, the cars overtaking you when you riding your bike are right on your heels. You're just constantly being overtaken by new CMB photons. The part we see today was actually carried away from us by expansion for the first 4 billion years or so, before it overtook expansion and started reducing its distance from us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inflation, as James Kilfiger comments, is something completely unrelated, taking place when the Universe was some $10^{-33}$ s old, as opposed to the CMB which was emitted when the Universe was 380,000 years old.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2016-06-25T22:29:54.903" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16375" PostTypeId="1" AcceptedAnswerId="16376" CreationDate="2016-06-26T09:53:38.200" Score="0" ViewCount="204" Body="&lt;p&gt;Can any one explain to me What will happen to Earth if its has two moons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same Distance &amp;amp; two sides of earth (North &amp;amp; south)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will it be good for humans or bad?&lt;/p&gt;&#xA;" OwnerUserId="11792" LastEditorUserId="13071" LastEditDate="2016-06-26T13:58:27.193" LastActivityDate="2016-06-26T17:47:22.297" Title="Two moons and one Earth?" Tags="&lt;the-moon&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ClosedDate="2016-07-01T13:43:16.970" ContentLicense="CC BY-SA 3.0" />
  <row Id="16376" PostTypeId="2" ParentId="16375" CreationDate="2016-06-26T10:05:45.213" Score="1" Body="&lt;p&gt;If we had two moons in opposition of each other the only short term effect would be amplified tides. Long term it will accelerate the decay of Earth's rotational period.&lt;/p&gt;&#xA;" OwnerUserId="11525" LastActivityDate="2016-06-26T10:05:45.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16377" PostTypeId="1" AcceptedAnswerId="16385" CreationDate="2016-06-26T13:24:52.900" Score="3" ViewCount="249" Body="&lt;p&gt;Globular clusters lie in the galactic halo, outside of the disc. However, galaxies are more or less a collection of material and objects — why is it, then, that most stars form a plane due to the angular momentum, but some patches of stars do not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, globular clusters often contain some of the earliest stars formed in a galaxy, right? How come they do not flatten, yet a lot of the older stars do?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2016-06-26T21:30:52.730" Title="Why don't globular clusters flatten with a galactic disc?" Tags="&lt;star&gt;&lt;gravity&gt;&lt;galaxy&gt;&lt;galactic-dynamics&gt;&lt;globular-clusters&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16378" PostTypeId="1" AcceptedAnswerId="16397" CreationDate="2016-06-26T14:02:29.067" Score="4" ViewCount="125" Body="&lt;p&gt;There is a video explaing the transiting exoplanet light curve — &lt;a href=&quot;https://www.youtube.com/watch?v=RrusIZaWDW8&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=RrusIZaWDW8&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear to my why the curve goes down when the planet is between the observer and the star, but I don't understand why the curve goes a little bit down when the the planet is behind the star. I was expected that this should not change the light curve at all.&lt;/p&gt;&#xA;" OwnerUserId="13286" LastEditorUserId="9301" LastEditDate="2016-06-28T18:44:22.797" LastActivityDate="2016-06-28T18:44:22.797" Title="Why the light curve goes down when the planet is behind the star?" Tags="&lt;star&gt;&lt;planet&gt;&lt;observational-astronomy&gt;&lt;exoplanet&gt;&lt;light&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16379" PostTypeId="1" AcceptedAnswerId="16381" CreationDate="2016-06-26T15:19:37.860" Score="26" ViewCount="20583" Body="&lt;p&gt;Theoretically, planets would have an approximately equal chance of going one way in their orbit or another but in reality, this is not the case (at least in our solar system). &lt;strong&gt;Why is this?&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="4160" LastActivityDate="2020-10-20T07:14:38.730" Title="Why do the planets orbit in the same direction?" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;solar-system&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16380" PostTypeId="1" CreationDate="2016-06-26T15:26:21.003" Score="3" ViewCount="633" Body="&lt;p&gt;In the beginning stars only consisted of the hydrogen element and due to nuclear fusion of those elements in stars and supernova's more heavy elements were created. Because of that, like our Sun, the stars were made not only of hydrogen but also oxygen and carbon etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now is it expected that in future more stars will be made of more heavy elements or are there causes/laws which prohibit stars forming of e.g. stars made of elements without hydrogen.&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="2531" LastEditDate="2016-06-26T21:05:05.387" LastActivityDate="2016-06-26T21:49:12.963" Title="Is the composition of stars in future made of more and more heavy elements?" Tags="&lt;star&gt;&lt;elemental-abundances&gt;&lt;astrochemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16381" PostTypeId="2" ParentId="16379" CreationDate="2016-06-26T15:52:56.143" Score="44" Body="&lt;p&gt;The same reason (almost) all of them rotate in the same direction: because of the &lt;strong&gt;conservation of angular momentum.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before a star and its planets exist, there’s just a cloud of disorganized gas and small molecules. The Solar System formed from such a cloud around 4.6 billion years ago. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On that scale, there is some small amount of rotation within the cloud. It could be caused by the gravity of nearby stellar objects, local differences in mass as the cloud churns, or even the impact of a distant supernova. The point is, all molecular clouds have at least a little rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a large system like a molecular cloud, each particle has some angular momentum, and it all adds together across a very wide area. That’s a lot of momentum, and it is conserved as the cloud continues to collapse under its own gravity. That angular momentum also flattens the cloud, which is the reason why the Solar System is near-planar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/d3kRf.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/d3kRf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the cloud finally collapses, it forms a star and shortly after planets. However, angular momentum is always conserved. That's why planets all follow the same orbit, and why almost all of them rotate in the same direction. There's nothing to turn them the other direction, so they will continue spinning in the same direction as the original gas cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few exceptions, though. Whenever objects formed in such a way that sent them orbiting the opposite direction, they usually collided with objects going in the same direction as the original cloud. This destroyed any outlying objects or sent them in the same direction as the original cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, two huge exceptions are planets Venus and Uranus. Uranus spins on an axis of almost 90-degrees (on its side). Venus meanwhile spins the opposite direction as Earth and the other planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases there is strong evidence that these planets were struck by large objects at some point in the distant past. The impacts were large enough to overcome the angular momentum of the bodies, and give them a different spin. There are also a range of other theories; for example, some astronomers think that Venus may have been flipped upside-down. Point is, there were irregular events that happened to both of these planets.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-06-26T21:07:13.107" LastActivityDate="2016-06-26T21:07:13.107" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16382" PostTypeId="2" ParentId="16375" CreationDate="2016-06-26T17:47:22.297" Score="3" Body="&lt;p&gt;An orbit as you describe is not stable. Even if the two moons are intially exactly the same distance from the Earth, small perturbations from the sun and other planets will cause one of them to speed up and the other to slow down. Eventually either one of them will hit the other, or one of them will hit the Earth. Whichever happens, it's not good.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-06-26T17:47:22.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16384" PostTypeId="2" ParentId="16380" CreationDate="2016-06-26T21:26:19.317" Score="4" Body="&lt;p&gt;The initial stars were made of hydrogen &lt;em&gt;and&lt;/em&gt; helium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These enriched the interstellar medium (ISM) with some chemical elements right across the periodic table, when massive primordial stars ended their lives as supernovae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subsequent generations of stars continue to enrich the ISM, if their lives are short enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the general gist of what you suggest is true, but the rate of enrichment is &lt;em&gt;very&lt;/em&gt; slow. The Sun was born 4.5 billion years ago, but sun-like stars born within the last few million years also have a metallicity consistent with the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this? Well, star formation is quite inefficient; only about 10 percent of a collapsing gas cloud ends up in stars. Then a star only processes about 10-20 percent of its mass as nuclear fuel; and then only a fraction of this is returned to the ISM in the form of winds and supernova ejecta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a more enriched ISM requires bursts of massive star formation in dense environments. Such conditions occurred early in the life of the Milky Way and perhaps will again when we collide with the gas-rich Andromeda galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, star formation will continue into the future; the stars will &lt;em&gt;very&lt;/em&gt; slowly get more metal-rich; there is nothing to prevent more metal-rich stars forming; perhaps the main consequence would be a small reduction in the maximum possible stellar mass (due to the increased gas opacity and consequent radiation pressure).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-06-26T21:26:19.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16385" PostTypeId="2" ParentId="16377" CreationDate="2016-06-26T21:30:52.730" Score="6" Body="&lt;p&gt;Globular clusters &lt;em&gt;formed&lt;/em&gt; whilst the gas of the proto Milky Way was still approximately spherically distributed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gas forms a dissipative system that loses energy and collapses (within the first billion years) to a disk whilst conserving angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formed stars and clusters are essentially collisionless so the halo stars continue to have a spherical distribution, whilst more recently formed stars formed from gas already in a disk and so continue to be in a disk.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-06-26T21:30:52.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16386" PostTypeId="2" ParentId="16380" CreationDate="2016-06-26T21:49:12.963" Score="2" Body="&lt;p&gt;Just to add, while I think Rob Jeffries answer covers it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now is it expected that in future more stars will be made of more&#xA;  heavy elements or are there causes/laws which prohibit stars forming&#xA;  of e.g. stars made of elements without hydrogen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While this is unlikely to happen because Hydrogen will stick around as the most abundant element for a very very very long time.   It will exist in gas clouds, and expelled by stars in coronal mass ejections and stellar nebulas from stars that are too small to go supernova and gas giant planets and very slow burning red dwarf stars, so a star and a galaxy without hydrogen is unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, in theory, without hydrogen stars wouldn't burn neatly at all.    Hydrogen is unique in the way it can fuse slowly in the core of a star and at comparatively low temperature.   Helium fusion in smaller stars is a associated with a &lt;a href=&quot;https://en.wikipedia.org/wiki/Helium_flash&quot; rel=&quot;nofollow&quot;&gt;helium flash&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Carbon_detonation&quot; rel=&quot;nofollow&quot;&gt;carbon fusion&lt;/a&gt; in smaller stars is basically a &lt;a href=&quot;https://en.wikipedia.org/wiki/Type_Ia_supernova&quot; rel=&quot;nofollow&quot;&gt;type 1a supernova&lt;/a&gt;.   Very large stars could fuse helium and carbon for a relatively short stellar life-span but stars of that size are rare.  Smaller stars are much more common, and without hydrogen, they wouldn't behave like stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The increase in heavy elements, called &lt;a href=&quot;https://en.wikipedia.org/wiki/Metallicity&quot; rel=&quot;nofollow&quot;&gt;metallicity&lt;/a&gt; of stars, does slowly increase over time.   It increases faster in the center of galaxies than the spiral arms.   Our sun for example contains about &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/tables/suncomp.html&quot; rel=&quot;nofollow&quot;&gt;2% by mass&lt;/a&gt; heavy elements.   Metallicity in this case means carbon or heavier elements, not metals.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-06-26T21:49:12.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16387" PostTypeId="1" AcceptedAnswerId="16389" CreationDate="2016-06-26T21:55:15.100" Score="8" ViewCount="237" Body="&lt;p&gt;Is so-called Planet Nine (given it exists) observable in principle? By &quot;observable in principle&quot;, I mean &quot;if we knew &lt;em&gt;exactly&lt;/em&gt; where to look, would we be able (from a technological standpoint) to get an image of the purported celestial body&quot;?&lt;/p&gt;&#xA;" OwnerUserId="13292" LastEditorUserId="13292" LastEditDate="2016-06-26T22:23:22.453" LastActivityDate="2020-07-16T04:55:33.813" Title="Is Planet Nine observable in principle?" Tags="&lt;observational-astronomy&gt;&lt;9th-planet&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16389" PostTypeId="2" ParentId="16387" CreationDate="2016-06-26T22:40:30.907" Score="8" Body="&lt;p&gt;If they knew exactly where to point Mr. Hubble, then yes, it should be easily visible, though at that distance, it would still be blurry, not a clean image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The estimated &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_magnitude&quot;&gt;apparent magnitude&lt;/a&gt; of Planet Nine varies based on which website you believe, Google provides ranges from 20 to 25 with &lt;a href=&quot;https://en.wikipedia.org/wiki/Planet_Nine&quot;&gt;Wikipedia&lt;/a&gt; saying &gt;22.  Higher numbers mean less visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hubble can see things up to apparent magnitudes of about &lt;a href=&quot;https://lcogt.net/spacebook/what-apparent-magnitude/&quot;&gt;30&lt;/a&gt;, so, yes, it should be easily visible by Hubble or (perhaps) even more visible by an infra-red telescope or radio telescope that knows where to look.  But in any case, with current technology all images would still be blurry.   Hubble's pictures of Pluto are blurry and Pluto's much closer.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-06-26T22:40:30.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16390" PostTypeId="1" CreationDate="2016-06-26T23:51:17.710" Score="5" ViewCount="374" Body="&lt;p&gt;Wikipedia says the definition of &lt;a href=&quot;https://en.wikipedia.org/wiki/Outer_space#Interplanetary_space&quot; rel=&quot;nofollow&quot;&gt;interplanetary space&lt;/a&gt; is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the region dominated by the interplanetary medium, which extends out&#xA;  to the heliopause where the influence of the galactic environment&#xA;  starts to dominate over the magnetic field and particle flux from the&#xA;  Sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While it defines the &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_System#Boundaries&quot; rel=&quot;nofollow&quot;&gt;solar system&lt;/a&gt; as the region in which the sun is the primary gravitational influence&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Sun's gravitational field is estimated to dominate the&#xA;  gravitational forces of surrounding stars out to about two light years&#xA;  (125,000 AU).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, if the region outside the heliopause but inside the solar system isn't interplanetary space, then what is it?&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2674" LastEditDate="2016-06-27T00:27:47.307" LastActivityDate="2017-01-03T15:45:14.310" Title="What is the word for space that is in the solar system, but outside the heliopause?" Tags="&lt;solar-system&gt;&lt;interstellar-medium&gt;&lt;oort-cloud&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16391" PostTypeId="1" AcceptedAnswerId="18768" CreationDate="2016-06-27T01:57:07.747" Score="8" ViewCount="440" Body="&lt;p&gt;What I mean is, with a central mass orbits are relatively simple, but orbits around the galaxy are different, in essence as the star orbits through the dark matter halo, the further it moves away from the center of the galaxy the greater the gravitational field, so while orbits tend to slow down as kinetic energy is turned into potential energy, stars are effectively orbiting around more mass the further they are from the center of the galaxy, which might speed them up. The equal areas over equal time rule would seem to no longer apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, ignoring up and down motion relative to the plane of the galaxy, is there a common shape for galactic orbits? Is there a formula? I think eccentricity would still apply, but I don't think the orbits would be ellipses, so it would be a different kind of eccentricity. Would these orbits tend to circularize (in an ideal dark matter cloud, ignoring perturbations. Obviously perturbations would lead to, in a practical sense, unpredictable orbits, not neat and tidy formulas. I gather from a show I caught on the science channel that the Milky Way's magnetic field also plays a role in forming its spiral arms, but I'm asking in an ideal, mathematical sense, what shape would orbits be through a galaxy, from a 2D point of view, ignoring up and down motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It occurred to me that the question might be more simply asked by a black hole in effect &quot;orbiting&quot; through a planet or even a low-velocity neutrino (if there are any low-velocity neutrinos). The inside a planet orbit is quite different than the Kepler orbit, because gravity increases as the object moves further away from the center. The overall orbit might still be somewhat like an ellipse but the laws for such an orbit would be different, you would see the highest acceleration at the furthest points from the center though you'd probably still see the highest velocity at the closest point to the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I'm mostly just curious if those shape has been worked out and what it looks like.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="9301" LastEditDate="2016-10-17T17:04:56.090" LastActivityDate="2016-10-19T05:07:10.587" Title="What is the shape (along the plane, not up-down) of stellar orbits in flat spiral galaxies" Tags="&lt;orbit&gt;&lt;milky-way&gt;&lt;orbital-mechanics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16392" PostTypeId="2" ParentId="16379" CreationDate="2016-06-27T12:26:44.690" Score="6" Body="&lt;p&gt;Sir Cumference's answer is great. Molecular clouds are generally thousands of times more massive than the Solar System, and since they're less dense they're much much larger in volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't know where our Solar System originated from, and we don't know how many other stars were born in the same cloud, probably hundreds or even thousands (just recently 1 or 2 stars were suggested to be sisters of Sol, but the jury is, as far as I know, still out on that).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, either due to interstellar winds, magnetic fields, supernovae explosions, or some other difference in average density, a volume of our mother molecular cloud began to collapse due to gravity being just a bit more in some areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more the cloud became concentrated, the more the gravitational attraction increased, so the faster it collapsed. While dust and gas collide, the whole system conserves energy and momentum (as it is an isolated system) , and thus is naïve to assume that planet orbits should be random — which means &lt;em&gt;any&lt;/em&gt; which way, you seem to have assumed that space is two dimensional, and the most random arrangement would be a flat disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Nope.&lt;/strong&gt; It would be a sphere... like a swarm of flies around something stinky. When we program a computer to model a swarm of random dust and gas collapsing, it turns out that due to chance it will select a preferred direction. A random dust cloud will collapse into a disk with most of the particles orbiting in the same direction (this ignores possible effects from the Milky Way influencing the process, so even without the molecular cloud orbiting the center of the Milky Way, disk formation will occur).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that these answers are tentative: most of the gravity in the Milky Way is of dark matter, and we're still working on understanding how that influences star formation and until we know a lot more about dark matter, we can't be sure our computer models are correct. Generally, we prefer models that give results similar to the actual way our Solar System is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But guess what? The thousands of exoplanets we've discovered have far more &quot;hot Jupiters&quot; (gas giants very close to their stars) than we expected. So we are adjusting our models. One popular idea is that planets had a lot more collisions than we used to think. This means more planets in very close to the Star, and more planets actually ejected from the star system. Who knows, perhaps that's where Theia came from.&lt;/p&gt;&#xA;" OwnerUserId="13300" LastEditorUserId="4160" LastEditDate="2016-06-27T20:59:15.390" LastActivityDate="2016-06-27T20:59:15.390" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16394" PostTypeId="1" CreationDate="2016-06-27T18:03:05.113" Score="2" ViewCount="335" Body="&lt;p&gt;A nebula is made of dust (elements heavier than He and H) and gas (H and He).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars are made when nebulae's collapse and hydrogen begins to fuse. 99% of nebula material goes into making the star. The left-over 0.01 of debris creates planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does 99% of nebula material that makes the star only consist of gas (He/H) and the 0.01% is the dust which makes the planets?&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Question Clarification:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are the constituents of the 99% of mass that makes the star and the 0.01% mass that makes the planets the same?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e.   &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Does both the 99% and 0.01% compose of mainly He/H and a small portion of metals (meaning the star contains heavier elements like Si and Fe etc before fusion?)&lt;br&gt;&#xA;&lt;strong&gt;OR&lt;/strong&gt;  &lt;/li&gt;&#xA;&lt;li&gt;is 99% of the nebula composed of He/H, which collapses to form the star, and 0.01% consists of heavier elements that make the planets?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13308" LastEditorUserId="13308" LastEditDate="2016-06-28T11:12:47.480" LastActivityDate="2016-06-28T16:30:34.067" Title="Star formation: When a nebula collapses, is it only the gasses that form the star?" Tags="&lt;planet&gt;&lt;star-formation&gt;&lt;nebula&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16395" PostTypeId="1" AcceptedAnswerId="16396" CreationDate="2016-06-27T18:54:12.283" Score="14" ViewCount="9113" Body="&lt;p&gt;Universe Sandbox 2 is the second installment of Universe Sandbox that is available on Steam. It is an educational simulation game based on astronomy and anything space, it ranges from supernova animations to climate and atmospheric simulations (and of course collisions) but do any of these features hold water? Do they truly resemble the dynamics of space?&lt;/p&gt;&#xA;" OwnerUserId="4160" LastEditorUserId="7982" LastEditDate="2021-03-08T00:02:19.647" LastActivityDate="2021-03-08T00:02:19.647" Title="Is Universe Sandbox 2 realistic?" Tags="&lt;space&gt;&lt;n-body-simulations&gt;&lt;universe-sandbox&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16396" PostTypeId="2" ParentId="16395" CreationDate="2016-06-27T19:40:42.510" Score="13" Body="&lt;p&gt;Not sure if these kinds of questions are allowed here, but I may as well answer it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Universe Sandbox 2 is mostly realistic. Its accuracy degrades as you increase the flow of time. The slower the time, the more accurate the calculations. This&#xA;means that if you want to simulate the Solar System over a long period of time, it won't be very accurate. If you want to simulate day-long events, like planets crashing, it might be better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, Universe Sandbox 2 uses 300-year old Newtonian physics, rather than general relativity. Newtonian physics just requires n-body mechanics, so it is much easier to implement. General relativity requires simulating the spacetime itself. That is, taking your simulation space, discretizing it to a hi-res 3-D grid and checking the effect that each and every point in that grid has on all neighboring points at every timestep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of simulating N number of bodies, they would be simulating a huge number of points. They would start with some initial data of the shape of spacetime and then see how it evolves according to the Einstein equations, which are 10 highly non-linear partial differential equations. Accurate general relativity simulations require supercomputers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, Newtonian physics can get the job done for the most part. The developers seem interested in incorporating some GR concepts, like gravity propagating at the speed of light, spinning black holes and ways to better approximate GR.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-07-02T17:38:37.370" LastActivityDate="2016-07-02T17:38:37.370" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16397" PostTypeId="2" ParentId="16378" CreationDate="2016-06-27T20:19:09.470" Score="5" Body="&lt;p&gt;Think of when the planet is at the &quot;side&quot; - there's a little bit of light from the planet (ie, reflecting off the planet) shining towards us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it be due to that little bit of light - when the planet is behind the star, it no longer reflects towards us?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe that's the effect you have in mind?&lt;/p&gt;&#xA;" OwnerUserId="13071" LastActivityDate="2016-06-27T20:19:09.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16399" PostTypeId="1" CreationDate="2016-06-27T22:04:42.500" Score="7" ViewCount="2367" Body="&lt;p&gt;As &lt;a href=&quot;http://adsabs.harvard.edu/full/1947PASP...59..232W&quot;&gt;pointed out here&lt;/a&gt;, the limit of naked eye visibility for people with extremely good night vision is about magnitude 8.5. Neptune at opposition is much brighter at magnitude 7.7, so it should be visible to people with good eyesight. Astronomer &lt;a href=&quot;https://en.wikipedia.org/wiki/Brian_A._Skiff&quot;&gt;Brian Skiff&lt;/a&gt;, made an attempt about a decade ago, &lt;a href=&quot;http://www.freelists.org/post/az-observing/First-attempt-at-Neptune-nakedeye&quot;&gt;he failed to spot it&lt;/a&gt;. He said that it should be relatively easy to spot from the Southern Hemisphere as it is high in the sky viewed from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But surely many other people with good eyesight who are capable of seeing faint stars of magnitude 8 or even dimmer who work at the Paranal observatory will have given this a try? &lt;/p&gt;&#xA;" OwnerUserId="2185" LastEditorUserId="9301" LastEditDate="2016-08-26T17:59:33.407" LastActivityDate="2016-08-26T17:59:33.407" Title="Have there been successful attempts at spotting Neptune with the naked eye?" Tags="&lt;naked-eye&gt;&lt;apparent-magnitude&gt;&lt;neptune&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16400" PostTypeId="2" ParentId="16399" CreationDate="2016-06-28T00:10:35.953" Score="13" Body="&lt;p&gt;Well, having seen Neptune and identifying Neptune are two totally different things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's tackle this one step at a time. Your link says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From these observations it would appear that, if seen upon a &lt;strong&gt;perfectly black background&lt;/strong&gt;, a star of magnitude approximately $8.5$ would be at the limit of unusually good vision.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The problem with this is that there are stars, diffused sunlight and &lt;a href=&quot;https://en.wikipedia.org/wiki/Airglow&quot; rel=&quot;nofollow&quot;&gt;airglow&lt;/a&gt; that prevent the night sky from being near-perfectly black. This has always been a problem, and it's only gotten worse with air pollution — meaning that we would be less likely to see Neptune unaided nowadays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that in the past, there &lt;em&gt;have&lt;/em&gt; been a few people who have seen Neptune, but none that would take note of it. Do you know how many objects of magnitude less than $7.8$ are there in the sky? The answer is, approximately 30 thousand! Good luck finding Neptune. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even with telescopes, Neptune can be hard to point out. Galileo actually discovered Neptune 200 years before it was recognized as a planet, but he mistook it for a star. Point is, even aided eyes often cannot identify Neptune.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the reason Curtis could see such dim objects is because he had exceptionally good vision and was aided by tools:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Without using the diaphragms and apertures, Curtis was able to see stars of magnitude $6.5$ directly against the background of the sky. With their aid he easily observed stars in the magnitude range $7.2$ to $7.4$. Four stars between magnitudes $7.9$ and $8.1$ were usually seen with little difficulty...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without these tools, he probably wouldn't have seen it at all. Not to mention, Neptune's brightness fluctuates — at its brightest, it is $7.8$; at its dimmest, it is $8$. That might not seem like much, but when finding Neptune, it could make a big difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless, even in ideal conditions, it is generally agreed that Neptune is the only planet that cannot be seen with the naked eye. Most sources state that you need to at least use binoculars to see Neptune.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-07-02T22:28:58.437" LastActivityDate="2016-07-02T22:28:58.437" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16402" PostTypeId="1" AcceptedAnswerId="16410" CreationDate="2016-06-28T05:25:53.017" Score="8" ViewCount="1253" Body="&lt;p&gt;In general, fermions form a degenerate gas under high density or extremely low temperature. It's clear that white dwarf stars are supported by electron degeneracy pressure. However, there are still a significant number of protons in a white dwarf. Under those high densities, do the protons form a degenerate gas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, are white dwarfs supported by proton degeneracy pressure as much as they are by electron degeneracy pressure?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="2531" LastEditDate="2016-06-28T17:00:37.150" LastActivityDate="2017-02-13T07:36:35.043" Title="Are white dwarf stars supported by proton degeneracy as well?" Tags="&lt;gravity&gt;&lt;white-dwarf&gt;&lt;density&gt;&lt;degenerate-matter&gt;&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16403" PostTypeId="1" CreationDate="2016-06-28T11:18:49.197" Score="4" ViewCount="2554" Body="&lt;p&gt;I've heard that the only difference between a star and a planet is mass, meaning that if planets accredited enough mass they would too become stars.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Does this mean that the star started off as a planet?&lt;/li&gt;&#xA;&lt;li&gt;Stars require hydrogen fusion and earth has little H. Could the earth become a star if more mass was added, but its relatively low abundance of hydrogen remained the same?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13308" LastActivityDate="2019-11-18T15:48:53.283" Title="Could the earth become a star if more mass was added?" Tags="&lt;star&gt;&lt;planet&gt;&lt;the-sun&gt;&lt;star-formation&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16404" PostTypeId="2" ParentId="16403" CreationDate="2016-06-28T12:15:09.010" Score="6" Body="&lt;p&gt;A star does not start off as a planet; you have a large cloud of gas that is collapsing in on itself due to gravity. The majority of the gas goes towards creating the star (more than 99% in the case of our Solar System). However, gravitational collapses can occur several places in the gas cloud, and some of the gas will contribute towards the collapse of far smaller over-density seedlings. If the mass of the smaller collapsing object is large enough, gravity will pull the object together into a sphere, satisfying one of the three criteria we have to classify an object as a planet (see e.g. &lt;a href=&quot;https://astronomy.stackexchange.com/questions/212/what-physical-features-determine-if-a-planet-is-a-major-minor-or-dwarf-planet?rq=1&quot;&gt;this question&lt;/a&gt; for more information about such criteria). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The very large mass of the gas which is going to become a star allows it to collapse in a way where the pressure becomes great enough that the central parts of the cloud start fusing together hydrogen -- or burning, if you wish. You do need simple gases for this type of ignition to occur, as even higher temperatures are needed for more complex atoms to fuse together. This is because fusion becomes less and less energy effective as the atoms become more complex. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I have no direct sources to refer you to, you can be decently sure that the Earth wouldn't become a star if it increased in mass without increasing it's amount of hydrogen substantially. &lt;em&gt;However&lt;/em&gt;, with the great amount of hydrogen present in the Universe compared to the other elements, a scenario like that would not occur, and Earth's hypothetical accumulation of matter would realistically be of gas which would be used for the effective fusion we find in stars.&lt;/p&gt;&#xA;" OwnerUserId="13193" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-06-28T17:01:44.843" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16405" PostTypeId="2" ParentId="16402" CreationDate="2016-06-28T12:27:53.500" Score="7" Body="&lt;p&gt;Proton degeneracy is not important, because its effect is much smaller -- much like nuclear particles in theory also are dictated by gravity, but the electromagnetic and nuclear forces are dominating, since they are much stronger. Proton degeneracy is weaker than electron degeneracy due to the far greater mass of the proton compared to the electron. &lt;a href=&quot;https://en.wikipedia.org/wiki/Degenerate_matter&quot;&gt;The Wikipedia article on degenerate matter&lt;/a&gt; explains this very well;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because protons are much more massive than electrons, the same momentum represents a much smaller velocity for protons than for electrons. As a result, in matter with approximately equal numbers of protons and electrons, proton degeneracy pressure is much smaller than electron degeneracy pressure, and proton degeneracy is usually modeled as a correction to the equations of state of electron-degenerate matter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13193" LastActivityDate="2016-06-28T12:27:53.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16406" PostTypeId="2" ParentId="16394" CreationDate="2016-06-28T12:35:39.817" Score="6" Body="&lt;p&gt;The very first stars to form indeed consisted essentially only of hydrogen and helium. When stars die, they leave behind them more massive atoms, as you say. These heavier elements are too incorporated in newer stars when they form. This results in stars which start out with a lower portion of hydrogen and helium, thus making them somewhat less effective; see for example &lt;a href=&quot;http://www.physlink.com/education/askexperts/ae465.cfm&quot;&gt;this question and answer&lt;/a&gt;. The Sun, for instance, is a so-called &lt;em&gt;third generation star&lt;/em&gt;, which means that it is made up of materials which have already been used at least twice since the Big Bang, and thus naturally contains heavier elements it has not produced itself. &lt;/p&gt;&#xA;" OwnerUserId="13193" LastEditorUserId="13193" LastEditDate="2016-06-28T16:30:34.067" LastActivityDate="2016-06-28T16:30:34.067" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16407" PostTypeId="2" ParentId="16403" CreationDate="2016-06-28T15:42:58.563" Score="8" Body="&lt;p&gt;Regarding the title: &lt;strong&gt;Yes.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does this mean that the star started off as a planet?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, a star could technically start out as a planet, if it accreted enough mass. However, this is extremely unlikely, since the planet would need to be 80x the mass of Jupiter for it to undergo nucleosynthesis.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Stars require hydrogen fusion and earth has little H. Could the earth become a star if more mass was added, but its relatively low abundance of hydrogen remained the same?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The most important components for a star is the composition and mass. As it stands, Earth doesn't have nearly the concentration of hydrogen for a star to form, even if we kept adding mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most stars work by fusing the lightest element, hydrogen. To fuse heavier elements requires a much, much higher temperature, since the more protons the element has, the more it will repel other ones instead of fusing. Yes, massive enough stars can fuse heavier elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, Earth would inevitably grow in hydrogen, so it would eventually have a significant amount. Above a certain size, given ambient conditions (like radiation pressure, light intensity from the star, etc), Earth would become massive enough that it starts capturing hydrogen gas from the interplanetary medium, and becomes a gas giant (as called2voyage pointed out). Where this happens isn't precisely known, but it's roughly 10x Earth's mass. This means that, theoretically, a terrestrial planet &lt;em&gt;could&lt;/em&gt; form a star if you increased its mass ever further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I can hear you thinking, &quot;What if Earth were surrounded by a vacuum instead? If we kept adding mass, would it become a star?&quot; Well theoretically, the planet would eventually reach temperatures where most of its elements could fuse. For example, the &quot;star&quot; could sustain itself on oxygen and silicon fusion, but not on iron fusion.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-06-28T16:35:25.040" LastActivityDate="2016-06-28T16:35:25.040" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16408" PostTypeId="1" CreationDate="2016-06-28T16:43:45.737" Score="2" ViewCount="113" Body="&lt;p&gt;On the Earth it is difficult to detect radiation below 30MHz from space due to our atmosphere. But why do they want to measure this radiation behind the moon? It has something to do with the big bang, but isn't the frequency of CMB not around 160GHz?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Chinese are planning to launch a satellite with a 30 MHz antenna on it in 2018.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aren't there already enough decent measurements of the CMB? So what would they like to find concerning the Big Bang?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="13426" LastEditDate="2016-07-13T17:32:02.300" LastActivityDate="2016-08-12T21:02:46.363" Title="What is the reason the Chinese want to measure the &lt;30Mhz radiation behind the moon?" Tags="&lt;big-bang-theory&gt;&lt;radio-astronomy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16409" PostTypeId="1" CreationDate="2016-06-28T16:58:41.810" Score="0" ViewCount="192" Body="&lt;p&gt;I need to calculate the relative angular path between two planets in the geocentric view or in other words the number of degrees of a circle one planet travels away from another planet around the Earth. In the heliocentric system I simply subtract the longitude of one planet from the longitude of the other and add n*360 if they made full cycles in between. Would the same work for the geocentric coordinates? Does the loop a retrograding planet makes in the sky actually count as its path? Should I consider the retrograde movement as an optical illusion or include it in the formula and how?&lt;/p&gt;&#xA;" OwnerUserId="13319" LastActivityDate="2016-06-30T13:16:18.370" Title="Does the retrograde movement affect the true angular distance between 2 planets?" Tags="&lt;planet&gt;&lt;solar-system&gt;&lt;coordinate&gt;&lt;ascension&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16410" PostTypeId="2" ParentId="16402" CreationDate="2016-06-28T17:22:24.277" Score="10" Body="&lt;p&gt;Straightforwardly no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a start there are almost no free protons inside a white dwarf. They are all safely locked away in the nuclei of carbon and oxygen nuclei (which are bosonic).&#xA;There are a few protons near the surface, but not in sufficient numbers to be degenerate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us assume though that you were able to build a hydrogen white dwarf that had equal numbers of free protons and electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The density at which the electrons become degenerate is set by the requirement that their Fermi (kinetic) energy exceeds $kT$. The Fermi energy is given by&#xA;$$E_F = \frac{p_F^{2}}{2m} = \left(\frac{3}{8\pi}\right)^{2/3} n^{2/3} \left(\frac{h^2}{2m}\right),$$&#xA;where $n$ is the number density (which would be the same for protons and electrons), but $m$ is the mass of a proton or electron, which is different by a factor 1800.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus for a given white dwarf temperature, the electrons become degenerate at number densities a factor of $(m_p/m_e)^{3/2} = 78,600$ times lower than do protons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if we were to compress a hypothetical hydrogen white dwarf to the point where the protons were also degenerate (which for a typical white dwarf interior temperature of $10^{7}$ K, would need mass densities considerably in excess of $10^{12}$ kg/m$^{3}$), the (ideal) degeneracy pressures would then be given by&#xA;$$ P = \frac{h^2}{20m}\left(\frac{3}{\pi}\right)^{2/3} n^{5/3}$$&#xA;and so we immediately see that the degeneracy pressure due to protons would be $\sim 1800$ times less than that due to the same number density of electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we push up to higher densities then both the protons and electrons will become &lt;em&gt;relativistically&lt;/em&gt; degenerate. In this case the pressure becomes &lt;em&gt;independent of the particle mass&lt;/em&gt;. However, the Fermi energies of the particles would now be high enough ($&amp;gt;1$ GeV !) to block beta decay and instigate neutronisation. The protons and electrons start combining to form neutrons and an n,p,e fluid is formed where the neutrons vastly outnumber the protons and electrons. This in fact prevents the protons ever becoming relativistic, even at neutron star densities, and the contributions of electrons to the pressure is &lt;em&gt;always&lt;/em&gt; orders of magnitude higher than for the protons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;handwaving&lt;/strong&gt; way of understanding this is that the degeneracy pressure depends on the &lt;em&gt;product&lt;/em&gt; of the momentum and velocity of the particles. In turn, the momentum of the fermions depends on how much they are compressed through the uncertainty principle. For a given particle number density, the separation $\Delta x$ is the same for protons and electrons and therefore the uncertainty principle says that the $\Delta p \sim \hbar/\Delta x$ of the momentum is also the same. This is another way of saying that the Fermi momentum does &lt;em&gt;not&lt;/em&gt; depend on the mass of the fermion; however, for a given fermion momentum, the velocity clearly does! Therefore the degeneracy pressure must be lower by approximately the mass ratio of the fermions&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-02-13T07:36:35.043" LastActivityDate="2017-02-13T07:36:35.043" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16411" PostTypeId="5" CreationDate="2016-06-28T18:31:01.587" Score="0" Body="&lt;p&gt;A white dwarf, also known as a degenerate dwarf, is a type of stellar remnant. It forms from the pressured core during the death of medium stars not capable of exerting enough gravity to overcome electron resistance. White dwarfs have masses on average of about 0.5 to 0.6 Solar Masses, although there are some exceptions. A white dwarf exerts faint luminosity, that will fade over time as the light it emits are from thermal temperature which will run out over time. A white dwarf mainly contains helium with possible heavy elements such as carbon and oxygen, and a debatable coating of hydrogen. If a white draws enough mass (i.e. 1.4 solar masses) to pass the Chandrasekhar limit, fusing begins in its interior while electron resistance is overcome by gravity and the white dwarf explodes into a type Ia supernova. Once a white dwarf cools, it becomes a black dwarf. No black dwarfs exist due to the cooling time being longer than the current age of the universe.&lt;/p&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="31410" LastEditDate="2020-12-11T12:33:58.517" LastActivityDate="2020-12-11T12:33:58.517" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16412" PostTypeId="4" CreationDate="2016-06-28T18:31:01.587" Score="0" Body="Questions about electron degenerate stellar remnants." OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-06-28T18:39:35.413" LastActivityDate="2016-06-28T18:39:35.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16413" PostTypeId="1" AcceptedAnswerId="16414" CreationDate="2016-06-28T19:16:08.660" Score="6" ViewCount="31410" Body="&lt;p&gt;I just bought a Celestron Travel Scope 70 and it comes with a 20mm and a 10mm eyepiece. I get a good view of the Moon but I am not able to see Jupiter, Saturn or Mars as expected. I read about the Barlow lens but I am not sure how much magnification is required to view these planets. So based on my telescope, which lens should I go for? And should I also get Red/Blue filters to view them?&lt;/p&gt;&#xA;" OwnerUserId="13324" LastEditorUserId="9301" LastEditDate="2016-06-28T20:23:24.797" LastActivityDate="2018-08-16T12:31:10.807" Title="Which lens and magnification is best to see planets?" Tags="&lt;planet&gt;&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16414" PostTypeId="2" ParentId="16413" CreationDate="2016-06-28T20:20:44.497" Score="11" Body="&lt;h2&gt;Eyepieces&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://stargazerslounge.com/topic/113230-highest-useful-magnification/&quot; rel=&quot;noreferrer&quot;&gt;highest useful magnification&lt;/a&gt; of the Celestron Travel Scope 70 is about 165x. However, you are unlikely to find that much magnification to be usable except under very ideal conditions (&quot;good seeing&quot; i.e. clear atmospheric conditions, dark skies with no light pollution, viewing on a night with a new moon, etc). Even then the image at the eyepiece is going to be dim at that magnification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any telescope can in theory achieve an unlimited magnification. But in practice, its magnifying power is limited by the amount of light the telescope is taking in, which is to say, its aperture. The telescope is only collecting so much light. Every increase in magnification is zooming in on a smaller and smaller area, which means it is looking at a smaller and smaller fraction of the original amount of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On your telescope, the aperture is 70mm. This is the main thing that dictates how bright of an object you can hope to see at the eyepiece. There is another stat that is important to figuring out what eyepieces are likely to show you, though, and that is the telescope's focal length. Your scope has a 400mm focal length. &lt;a href=&quot;http://astronomyonline.org/Science/TelescopeMagnification.asp&quot; rel=&quot;noreferrer&quot;&gt;Total magnification&lt;/a&gt; is the focal length of the telescope divided by the focal length of the eyepiece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your telescope came with two eyepieces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;400mm (telescope focal length) / 20mm (eyepiece focal length) = 20x magnification&lt;br&gt;&#xA;400mm (telescope focal length) / 10mm (eyepiece focal length) = 40x magnification&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the smaller the eyepiece focal length, the greater the final magnification. If you tried an 8mm eyepiece, you'd get a 50x magnification. To get a 65x magnification, you'd need somewhere around a 6.2mm eyepiece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Interesting aside: Since the magnification also depends on the telescope's focal length, note that each eyepiece will have a different magnification depending on the telescope it is in.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before you go buying any extra eyepieces, though, I'd strongly recommend you go see what you can see with the equipment you already have! You should certainly be able to see Jupiter, Saturn, and Mars. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Jupiter will appear as a disc. You can see its four Galilean moons, Io, Europa, Ganymede, and Callisto, so-named after the first person to see them, Galileo. They appear as points of light around the disc of Jupiter. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For Saturn, you should be able to make out the rings poking out either side of Saturn's disc. They look like tiny little teacup handles. You can also see Saturn's largest moon, Titan. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mars is a little bit less exciting, but it should be visible as a reddish disc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you do opt to buy another eyepiece, one thing to pay attention to is eye relief. Eye relief is how far away from the eyepiece you can be to still see a decent image. Eyepieces with shorter eye relief are cheaper, but they are also harder to use. An eye relief around 6mm to 8mm is much more comfortable for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for me, I can never help myself: for every telescope I've owned, I always buy an eyepiece that pushes the limits of my telescope, just to see what it looks like. But from experience, I can tell you it is not a great view. Still, it can be fun to see what your equipment can do. But before you jump all the way to an eyepiece that gives you 130x magnification (or whatever), I'd say to shoot solidly for the middle range of your scope (in your case, maybe 60x - 80x). If you can afford it, buy a mid range eyepiece with decent eye relief and field of view (FOV). You'll be much happier with the view through this eyepiece than you will an eyepiece that maxes out the magnification.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Filters&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Do yourself a favor and at least pick up a lunar filter. They filter a certain percentage of the light and allow the rest through without modifying the color. The moon is so bright, it will make your eyes water trying to look at it through even a modest telescope. Lunar filters make looking at the moon a much more enjoyable experience.&lt;/p&gt;&#xA;" OwnerUserId="12030" LastEditorUserId="12030" LastEditDate="2016-06-29T15:00:31.510" LastActivityDate="2016-06-29T15:00:31.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16418" PostTypeId="1" CreationDate="2016-06-29T21:58:23.863" Score="4" ViewCount="683" Body="&lt;p&gt;if the Cosmic microwave background radiation is coming towards us in the speed of light and all matter and energy was singular at the big bang, doesn't that mean we somehow moved faster than the speed of light (otherwise the CMBR would have already passed us not be coming toward us?)&lt;/p&gt;&#xA;" OwnerUserId="13330" LastActivityDate="2016-07-07T11:38:48.080" Title="Are we moving faster than the Cosmic microwave background radiation?" Tags="&lt;cmb&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16419" PostTypeId="2" ParentId="16418" CreationDate="2016-06-29T22:11:47.957" Score="1" Body="&lt;p&gt;The CMBR is coming at us at the speed of light -- but not from a distinct point.  It comes from everywhere.  It is distributed evenly throughout the universe because the universe used to be a lot closer together before the period of inflation.  Inflation was the epoch from 10^-36 seconds after the big bang to 10^-33 after where the universe increased in volume by a factor of about 10^78.&lt;/p&gt;&#xA;" OwnerUserId="13331" LastActivityDate="2016-06-29T22:11:47.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16422" PostTypeId="2" ParentId="16409" CreationDate="2016-06-30T00:39:29.953" Score="2" Body="&lt;p&gt;In our moving Earth-based point of view,&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_retrograde_motion&quot; rel=&quot;nofollow&quot;&gt;retrograde apparent motion&lt;/a&gt;&#xA;means a planet's geocentric ecliptic longitude is temporarily decreasing&#xA;instead of increasing as usual.&#xA;The word &quot;apparent&quot; avoids confusion with a&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Retrograde_and_prograde_motion&quot; rel=&quot;nofollow&quot;&gt;retrograde orbit&lt;/a&gt;.&#xA;All major planets' orbits are prograde,&#xA;with heliocentric longitude increasing at all times&lt;sup&gt;1&lt;/sup&gt;.&#xA;Most retrograde-orbiting Solar system objects are comets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As seen from Earth, the apparent angular separation&#xA;between two planets near the ecliptic&#xA;is approximately&lt;sup&gt;2&lt;/sup&gt; the difference&#xA;between their geocentric longitudes at that point in time,&#xA;regardless of their rates of change in longitude.&#xA;Even if you plot this value versus time,&#xA;your calculation need not break apparent motion down&#xA;into positive and negative cases.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Except where longitude wraps around from 360$^\circ$ to 0$^\circ$.&lt;/li&gt;&#xA;&lt;li&gt;When longitude difference is small (near conjunction), latitude difference is also significant.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2016-06-30T13:16:18.370" LastActivityDate="2016-06-30T13:16:18.370" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16423" PostTypeId="2" ParentId="16418" CreationDate="2016-06-30T08:21:22.980" Score="3" Body="&lt;p&gt;You seem to struggle with a very common thing; the understanding that The Big Bang is not something that happened &lt;em&gt;in&lt;/em&gt; space, but rather &lt;em&gt;to&lt;/em&gt; space. It could be good for you to read some answers to misconceptions about The Big Bang &lt;a href=&quot;https://www.cfa.harvard.edu/seuforum/faq.htm#e1&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, or the answer to a question about it on &lt;a href=&quot;https://physics.stackexchange.com/questions/136860/did-the-big-bang-happen-at-a-point&quot;&gt;Physics Stack Exchange&lt;/a&gt;. The Big Bang caused everything in the universe to move away &lt;em&gt;from each other&lt;/em&gt;; there is &lt;em&gt;no central point&lt;/em&gt; from which everything is receding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first few hundred thousand years of our Universe, photons and matter particles were coupled in a photon-particle soup. They were all so close together, that they constantly bumped into each other, not being able to travel far at all before changing directions. As the universe expanded and the temperature of the soup became colder, the photons and matter particles were able to travel further before interacting with each other. When the temperature became low enough for protons and electrons to form hydrogen, the number of matter particles the photons had to interact with became even lower, allowing the photons to travel even further before having to interact with matter particles. It was approximately at this point in time that the CMB originated; the CMB is simply the light of the photons which interacted a very last time around the time photons and matter particles stopped being a photon-particle soup (we call this time the &lt;em&gt;recombination epoch&lt;/em&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe was expanding everywhere both at the very start of its existence and during the recombination epoch, as well as now. The release of the CMB photons thus happened everywhere, and the photons of the CMB travelled in all directions. As the universe expands, the photons travel new distances, and we are constantly observing the CMB photons, which simply are reaching us from different places of origin in the universe -- which, for all we know, may be infinitely large, so we may just be seeing these kind of photons until the end of time.&lt;/p&gt;&#xA;" OwnerUserId="13193" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-06-30T08:21:22.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16424" PostTypeId="2" ParentId="16409" CreationDate="2016-06-30T10:35:43.160" Score="2" Body="&lt;p&gt;If you are calculating the angular distance between planets, as viewed from Earth, then of course you need to include the retrograde motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The natural way of doing this would be to first use heliocentric coordinates to calculate the positions of both planets and the Earth relative to the sun (You could do this using Kepler's laws). Then change your coordinate system to the calculated position of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you are using a calculated position of the Earth, the apparent retrograde motion (which is caused by the movement of the Earth) is taken care of, and doesn't need any special handling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have found the position of the planets, you can find the angle between them by straightforward spherical geometry. (Since the planet's don't all orbit in exactly the same plane, just subtracting longitudes is less accurate, but that is the same whether you are working geo- or helio-centrically.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's possible, but much harder, to work in geocentric coordinates throughout. A good approximation can be found using Ptolemy's work. However Keplerian ellipses are more accurate and simpler. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-06-30T10:35:43.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16425" PostTypeId="1" AcceptedAnswerId="16426" CreationDate="2016-07-01T03:15:04.723" Score="3" ViewCount="275" Body="&lt;p&gt;So I am thinking about orbiting telescopes tonight and I was thinking that it would probably be beneficial to keep them in the shadow of the Earth or some other space body to avoid light interference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, I am thinking about looking off in all directions around Earth. I would imagine staying in a shadow would help the most. But that at the very least there would be a range of degrees around the direction of the sun that you could not observe until the Earth changed orbital position so that the Sun's light was no longer interfering when looking in a specific astronomical direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct in any of these assumptions? I believe there are telescopes that don't sit in the shadow of a body, but I do think that would greatly limit the range of area it could look at during a given timeframe.&lt;/p&gt;&#xA;" OwnerUserId="13339" LastActivityDate="2016-07-04T14:06:24.563" Title="Would space-based telescopes need to remain in the shadow of Earth to avoid light interference from the sun?" Tags="&lt;orbit&gt;&lt;telescope&gt;&lt;photography&gt;&lt;space-telescope&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16426" PostTypeId="2" ParentId="16425" CreationDate="2016-07-01T07:25:35.423" Score="4" Body="&lt;p&gt;Its a nice idea, but it wouldn't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To stay in the Earth's shadow a telescope would need to orbit the Earth in the same time the earth orbited the Sun so that it always stayed on the opposite side. For its orbit to be this long it would need to be a long way out, remember geosynchronous orbit (one orbit in 23h 56min 4sec) is at a distance of 42,164 km. For a 1 year orbit the telescope would be so far away that its would not be in orbit around the Earth but the Sun, as it would have a much larger gravitational influence. It would also be too far away to be in the Earth's shadow cone.&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-07-01T07:25:35.423" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16427" PostTypeId="1" AcceptedAnswerId="16433" CreationDate="2016-07-01T09:35:42.347" Score="5" ViewCount="22166" Body="&lt;p&gt;I haven't been to south pole but can the Polaris still be viewed if the viewer is in the south pole? Or this question makes no sense at all?&lt;/p&gt;&#xA;" OwnerUserId="11924" LastActivityDate="2018-09-02T20:43:58.407" Title="Can you still see Polaris even if you are in the south pole?" Tags="&lt;amateur-observing&gt;&lt;star-gazing&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16428" PostTypeId="2" ParentId="16427" CreationDate="2016-07-01T09:59:04.450" Score="9" Body="&lt;p&gt;While the majority of the celestial sky is visible on both hemispheres, you are not able to see Polaris on the south pole, since Polaris is pointing directly towards the north pole. I know that during winter time, you can definitely just see the plough/big dipper (part of the Ursa Major constellation) as far south as Uluru/Ayers Rock in Australia, but that is not enough to see the northern star. The northern star will generally speaking disappear below the horizon when you are at around the equator. There is no &quot;southern star&quot; to orient yourself on the southern hemisphere, but there are a number of rules of thumb to find the south celestial pole on the southern hemisphere as well, several of them involving the &quot;southern cross&quot;, or the Crux constellation. &lt;/p&gt;&#xA;" OwnerUserId="13193" LastActivityDate="2016-07-01T09:59:04.450" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16429" PostTypeId="2" ParentId="16427" CreationDate="2016-07-01T10:52:54.613" Score="4" Body="&lt;p&gt;At the South pole no stars north of the celestial equator will be visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general equations of the declination limit at a given latitude:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;declination limit = $latitude - 90$    (for the northern hemisphere, $latitude &amp;gt; 0$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;declination limit = $90 + latitude$    (for the southern hemisphere, $latitude &amp;lt; 0$)&lt;/p&gt;&#xA;" OwnerUserId="11658" LastEditorUserId="10172" LastEditDate="2016-07-01T12:58:04.693" LastActivityDate="2016-07-01T12:58:04.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16431" PostTypeId="2" ParentId="16425" CreationDate="2016-07-01T13:37:27.340" Score="4" Body="&lt;p&gt;The mindboggling GAIA spacecraft-telscope &lt;em&gt;sort of&lt;/em&gt; does what you suggest! That's kind of the closest concept to what you describe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as explained in JamesS answer, if you were that far away you wouldn't be in orbit around the Earth.  But, GAIA's L2-like orbit is kind of the closest thing to what you say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/e5zUn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/e5zUn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explanation: &lt;a href=&quot;https://en.wikipedia.org/wiki/Gaia_(spacecraft)#Launch_and_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Gaia_(spacecraft)#Launch_and_orbit&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that &lt;strong&gt;&lt;em&gt;just as you say&lt;/em&gt;&lt;/strong&gt;, light interference is a huge problem for instruments this insanely sensitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GAIA has a problem where, apparently some tiny little bits of the sun-shade (like small hairs) were poking around, and that is enough to cause stray light; also some ice particles seem to be causing stray light.  &lt;a href=&quot;http://blogs.esa.int/gaia/2014/04/24/commissioning-update/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://blogs.esa.int/gaia/2014/04/24/commissioning-update/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="13071" LastEditDate="2016-07-03T14:39:04.383" LastActivityDate="2016-07-03T14:39:04.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16432" PostTypeId="2" ParentId="11191" CreationDate="2016-07-01T15:58:34.117" Score="0" Body="&lt;p&gt;Here I found description, formulas and an example of calculating equatorial coordinates change &lt;a href=&quot;http://www.astronexus.com/a-a/motions-long-term&quot; rel=&quot;nofollow&quot;&gt;http://www.astronexus.com/a-a/motions-long-term&lt;/a&gt;. You'll need proper motion (in Dec and RA), distance to star and star's radial velocity, all this data and Dec and RA for J2000 are available on the same page (&lt;a href=&quot;http://www.astronexus.com/hyg&quot; rel=&quot;nofollow&quot;&gt;http://www.astronexus.com/hyg&lt;/a&gt;), HYG database contains data from several star catalogues. Unfortunately I haven't tried this calculations yet so I don't know if you get same results as stated in Stellarium.&lt;/p&gt;&#xA;" OwnerUserId="13346" LastActivityDate="2016-07-01T15:58:34.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16433" PostTypeId="2" ParentId="16427" CreationDate="2016-07-01T19:21:01.597" Score="12" Body="&lt;p&gt;Currently Polaris is at a declination of a bit over 89 degrees, which means that no one south of 1 degree south latitude can see Polaris. That's almost all of the Southern hemisphere, let alone the South Pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polaris won't be the North Star forever, thanks to axial precession. In about 13000 years or so, Polaris will have a declination of about 46 degrees or so (twice the 23 degree axial tilt). Polaris will thus be visible in 13000 years or so as a wintertime star to all of Africa, all of Australia, and most of South America, but none of Antarctica.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After millions of years, proper motion may make Polaris visible over Antarctica. But then again, being a yellow supergiant, its unlikely that Polaris will be visible anywhere (without a telescope). It will instead be dead.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2016-07-01T19:21:01.597" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16435" PostTypeId="2" ParentId="16263" CreationDate="2016-07-02T15:33:17.497" Score="1" Body="&lt;p&gt;Phase functions are nonlinear in general, but Pluto presents such a small range of phases to Earth that a linear approximation will do:&#xA;about 0.03 mag/deg for Pluto and 0.08 mag/deg for Charon, or 0.04 mag/deg for the two together.&#xA;The &lt;a href=&quot;https://books.google.com/books?id=VcY7iYJwJZoC&amp;amp;pg=PA211&amp;amp;lpg=PA211&amp;amp;dq=Pluto+phase+coefficient&amp;amp;source=bl&amp;amp;ots=YDfEZAxWP3&amp;amp;sig=5nEneBnP92ozpOp2Ia-I1E3EFoo&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwiD4ZvMi9XNAhXCJB4KHURdCQIQ6AEITDAG#v=onepage&amp;amp;q=Pluto%20phase%20coefficient&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Stern and Tholen book&lt;/a&gt; summarizes the estimates up to 1997. &#xA;A search for &lt;a href=&quot;https://www.google.com/search?q=Pluto+phase+coefficient&quot; rel=&quot;nofollow&quot;&gt;Pluto phase coefficient&lt;/a&gt;&#xA;yields some of the papers behind those and more recent estimates.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-07-02T15:33:17.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16436" PostTypeId="1" CreationDate="2016-07-02T16:17:24.773" Score="0" ViewCount="679" Body="&lt;p&gt;Is this picture of a digital camera being oversaturated by bright light source i.e. the Sun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://farm4.staticflickr.com/3110/3147227521_396e16bcfc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://farm4.staticflickr.com/3110/3147227521_396e16bcfc.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Y1nYN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y1nYN.jpg&quot; alt=&quot;oversaturated?&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13352" LastEditorUserId="10172" LastEditDate="2016-07-02T16:25:40.730" LastActivityDate="2016-07-02T17:04:28.103" Title="Is this picture of a digital camera being oversaturated by bright light source i.e. the Sun?" Tags="&lt;the-sun&gt;" AnswerCount="2" CommentCount="5" ClosedDate="2016-07-03T21:45:39.520" ContentLicense="CC BY-SA 3.0" />
  <row Id="16437" PostTypeId="2" ParentId="16436" CreationDate="2016-07-02T16:38:20.677" Score="1" Body="&lt;p&gt;That does seems like the most probable explanation. Pointing a camera directly towards the Sun is &lt;a href=&quot;https://photo.stackexchange.com/q/4016&quot;&gt;not exactly healthy for it&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may want to try to take some other pictures and study them to see it some permanent damage is present. (That is, if this is your camera anyway.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a close up look the artefact can look somewhat like some common painting or spray tool in common editing software, but it is hard to tell if it is fake. I could not find any meta-data like &quot;created with GIMP&quot;. It says though that the image is almost 8 years old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It could be an optical artefact, as it appears like the image is taken through a window. (Please give some more information on this if you have it.) Any astronomical phenomenon seems highly unlikely.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="-1" LastEditDate="2017-04-13T12:44:06.413" LastActivityDate="2016-07-02T16:38:20.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16438" PostTypeId="2" ParentId="16436" CreationDate="2016-07-02T16:54:41.960" Score="2" Body="&lt;p&gt;It's an artifact from the digital sensor and/or the image processing in some cameras. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been a number of digital cameras with this issue. E.g. &lt;a href=&quot;https://www.cinema5d.com/sony-fixing-sony-a7s-ii-black-sun-bug/&quot; rel=&quot;nofollow&quot;&gt;here is a Sony camera&lt;/a&gt; where the sun turned black with a specific setting for the picture profile. (In the example video, it's the S-log picture profile at 0:49.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sony called it a bug, and &lt;a href=&quot;https://www.cinema5d.com/sony-issues-a7s-ii-firmware-update-fixing-black-hole-sun-problem/&quot; rel=&quot;nofollow&quot;&gt;released a fix&lt;/a&gt; for it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This artifact only affects a relatively small number of cameras, and it makes sense to think of it as a bug in the camera image processing: Yes, the sensor is oversaturated, but oversaturation is better represented as all white rather than all black. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have an affected camera, it's not just the sun that turns black, it's all oversaturated highlights, like &lt;a href=&quot;http://www.eoshd.com/2016/03/canon-c300-mark-ii-suffers-sun-spot-issue/&quot; rel=&quot;nofollow&quot;&gt;this example with car headlights&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9682" LastEditorUserId="9682" LastEditDate="2016-07-02T17:04:28.103" LastActivityDate="2016-07-02T17:04:28.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16440" PostTypeId="2" ParentId="11284" CreationDate="2016-07-03T12:35:18.837" Score="5" Body="&lt;p&gt;In any system of orbiting bodies, the length of time two bodies spend in proximity determines the mutual gravitational perturbation of each body on the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If both bodies are orbiting in the same direction, the inner body (closer to the central mass [primary star, for instance]) will overtake the outer one, because it is orbiting at a faster angular velocity. Because both bodies are orbiting in the same direction, they will have a great deal of time close to one another, and more time mutually to disturb one another. This is especially pronounced when the two orbits are adjacent and close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if two bodies are orbiting in opposite directions, they spend very little time in close proximity, so gravitational perturbation is minimized. Passing one another happens very quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By analogy, compare the amount of time two vehicles spend close to one another on a highway when one is passing the other and going in the same direction. A passenger in one could have time to study the other vehicle in great detail. Then, compare two vehicles moving in opposite directions; any passenger could only gain only a brief glimpse of the other vehicle up close. The amount of time for a passenger to study the other vehicle is analogous to the amount of gravitational disturbance one orbiting body will have on another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar problem occurs when the orbital periods of two bodies are simple ratios of one another (orbital resonance). The repeated tugging of both bodies on one another at the same place in their orbits creates an additive distortion at that location in their orbits. This problem would occur in both situations -- orbiting in the same direction and opposing directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though the reasoning is simple, the exact amount of perturbation is quite complex. A &quot;simple&quot; 3-body problem requires the use of advanced mathematics (celestial mechanics).&lt;/p&gt;&#xA;" OwnerUserId="13358" LastActivityDate="2016-07-03T12:35:18.837" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16441" PostTypeId="1" CreationDate="2016-07-03T16:13:22.200" Score="1" ViewCount="871" Body="&lt;p&gt;The changes of the sunrise and sunset times not expire regularly in a straight line but according to a sinusoid. Around the solstice (summer solstice on June 21st and winter solstice on December 21), the day length changes the least. The difference in sunrise and sunset in the days around the solstice is only a few tens of seconds per day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At equinox (March 20 and September 23), the length of day and night on earth is everywhere the same. The difference in day length from day to day around the equinox is changing rapidly. The difference in the length of day can rise around the time of equinox to 3 minutes per day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This asymmetrical lengthening and shortening of the days is because the &quot;middle&quot; of the day, every day a little shifts. This has to do with the fact that the earth does not describe a exact circle orbit around the sun, but an elliptical orbit. Because the Earth's axis at an angle to the orbit around the sun takes the shorter and longer are the days asymmetrical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the elliptical orbit, I understand that, for example,the sun is the farthest from  the earth in the summer (on northern hemispere). Because of the (second) law of areas of Kepler is the speed of the earth there  the slowest. In itself, I understand then that at that time the difference in day length also is shorter. That is during June 21 (summer solstice). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But so is apparently also on the winter solstice on 21 December. Precisely at that time the earth is closest to the Sun (perihelion) so the earth has a higher speed. But because of the higher rate would you expect the day lengthening/shortening would therefore be greater. However, it appears not to be so. How is that possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question in short terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sunrise in june on the 21 at 5:30 and 22 at 5:31 etc&#xA;sunrise in march on the 21 at 7:30 and 22 at 7:33&#xA;So the difference is in june one minute and in march it is 3 minutes. And the question is what causes that the difference is bigger in march then in june. See &lt;a href=&quot;http://www.timeanddate.com/sun/france/paris&quot; rel=&quot;nofollow&quot;&gt;http://www.timeanddate.com/sun/france/paris&lt;/a&gt; for a bigger disquisition&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="10445" LastEditDate="2016-07-03T20:16:36.300" LastActivityDate="2016-07-04T22:34:12.100" Title="What are the exact causes that differences in sunrises are shorter around june and december but long in march and september?" Tags="&lt;solar-system&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16443" PostTypeId="2" ParentId="16441" CreationDate="2016-07-03T18:36:26.647" Score="1" Body="&lt;p&gt;The next aphelion is July 4 at UT 16:24. Next perihelion Jan 4, 2017. Depending on latitude, the length of daylight varies over a year. Since the Earth's spin is NOT locked to its rotation, where the Sun is in the sky on the same date for consecutive years will not be the same.&#xA;I am unable to find a table of day lengths to the second, so I can neither confirm nor deny the accuracy of your vague claim - that is since you didn't provide specific evidence, nor cite any source, I am left with little to work with.&#xA;You noted that the date of maximum axial tilt is NOT the date of apses? Given two non-linear factors, I'm surprised you expect the result to be exactly symmetrical. You also need to be careful writing about differences of differences. There is a better language for that than English: it's called mathematics.&lt;/p&gt;&#xA;" OwnerUserId="13361" LastActivityDate="2016-07-03T18:36:26.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16444" PostTypeId="2" ParentId="12214" CreationDate="2016-07-03T19:16:23.483" Score="1" Body="&lt;p&gt;A distribution is a function mapping from one (or more) variables onto other  (often a single, usually non-negative) variable(s). Generally, matter distribution is the relationship between location and mass density (sometimes averages over macroscopic distances, sometimes objects which are assumed to be point masses). Typically in astronomy the radial distance from some center point is the (most significant) independent variable. You should have some familiarity with euclidean, polar, conical, and spherical coordinate systems. Most 'well behaved' distributions → 0 as distance → ∞.&lt;/p&gt;&#xA;" OwnerUserId="13362" LastActivityDate="2016-07-03T19:16:23.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16445" PostTypeId="2" ParentId="16441" CreationDate="2016-07-03T19:42:15.297" Score="2" Body="&lt;p&gt;If I understand you correctly, you ask why does the length of the day not change much around the winter solstice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no mystery here: The rate of change of day length has (almost) nothing to do with the elliptical orbit of the Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In autumn the days get shorter; the rate of change of day length is negative. In spring the days get longer; the rate of change is positive. &#xA;At midwinter solstice the rate of change of day length is zero: The length of the day doesn't change.  Exactly the same is true in midsummer. Mathematicians would say that the &quot;day length function&quot; is stationary on those dates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has nothing to do with the elliptical orbit. Every smooth function is stationary at its maxima and minima. The reason is not astronomical but mathematical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the Earth is furthest from the sun (and therefore moving slowest) on July 4th.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-07-03T19:42:15.297" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16446" PostTypeId="2" ParentId="16441" CreationDate="2016-07-03T21:43:48.337" Score="0" Body="&lt;p&gt;The phenomenon you are asking about has nothing to do with the shape of Earth's orbit. Although it does affect sunrise/sunset times, it isn't the major influence on rate of change of sunrise/sunset time. It has everything to do with the Earth's axial tilt and the Sun's aspect to the Earth at different latitudes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Day length is determined by how much the Earth's axis is tilted toward or away from the Sun. For some mid-latitude location, say New York, London, Tokyo, etc., we will have a normal day and night of some length, varying with the season.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the perspective of a Sun-Earth line, the Earth's axis appears to wobble, with each pole uniformly describing a circle, and completing one rotation each year (bear in mind that the Earth's axis is fixed in space and only appears to wobble because our frame of reference is rotating in the opposite direction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the &quot;wobble&quot; is projected onto the Sun-Earth line, it will appear as a tilt that smoothly varies toward and away from the Sun. Imagine watching someone on a carousel from a distance. They travel around in a circle at a steady speed, but from your bystander perspective, they seem to be coming toward you, turning, and then receding. As they turn, they seem to slow, until, facing sideways for an instant, they are neither advancing toward you nor receding. The change in tilt between Earth and Sun is analogous to that person on the carousel.&lt;/p&gt;&#xA;" OwnerUserId="1983" LastEditorUserId="1983" LastEditDate="2016-07-04T22:34:12.100" LastActivityDate="2016-07-04T22:34:12.100" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16448" PostTypeId="2" ParentId="16441" CreationDate="2016-07-04T06:12:17.310" Score="0" Body="&lt;p&gt;Because of an apparent language barrier, I'm not sure if the question is about the cause of the seasons or about the equation of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the questioner, do not make the mistake of assuming that conditions specific to Paris apply worldwide. It is now the time to go skiing in Chile, New Zealand, and Australia. While early July is the peak of summer in the northern hemisphere, it is the peak of winter in the southern hemisphere. Skiing fanatics with lots of disposable income take advantage of this fact and spend November to April in the northern hemisphere and then spend May to October in the southern hemisphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape of the Earth's orbit has very little to do with the seasons. It is the Earth's axial tilt that drives the seasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;The other possible interpretation of the question is the time from one local noon to the next. This is the subject of the equation of time. Unlike the seasons, the time from one local noon to the next is the more or less the same from 66 degrees south latitude to 66 degrees north latitude. (Regions outside of those latitude limits can see days or nights that are several 24 hour days long.) Due to the Earth's axial tilt and it's slightly eccentric orbit, the time as measured by a clock between successive local noons varies somewhat over the course of a year.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2016-07-04T06:12:17.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16449" PostTypeId="2" ParentId="16425" CreationDate="2016-07-04T11:51:00.407" Score="1" Body="&lt;p&gt;As Joe Blow explains, there is an orbit, at the L2 point, 1.5 million km directly further away from the Sun/Earth, where a satellite could remain permanently in the Earth's shadow, though it would not be completely shadowed - the Sun would still appear as a narrow ring around the blocking Earth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be a great advantage in terms of cooling the spacecraft and in terms of avoiding scattered light entering the instrument. It would allow a greater fraction of the sky to be observed at any one time, though there would still be a zone of avoidance close to the Earth/Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, all the satellites (WMAP, Planck, Herschel, Gaia) that have exploited L2 have been specifically placed in &quot;Lissajous orbits&quot; that orbit around the L2 point and are &lt;em&gt;never&lt;/em&gt; in the Earth's shadow. The reason is that you need the sunlight for power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neverthless, L2 still offers a very stable thermal environment and it is still the case that a much larger proportion of the sky is viewable than from a lower Earth orbit.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-07-04T13:53:37.237" LastActivityDate="2016-07-04T13:53:37.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16450" PostTypeId="1" AcceptedAnswerId="16453" CreationDate="2016-07-04T11:56:48.737" Score="6" ViewCount="447" Body="&lt;p&gt;What color is earthshine (e.g the average color of earth that shines on the moon)?  I am making a solar system exploration game, and would like to know this for the lighting of my moon level at night, with a &quot;full earth&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: With your answers, this is what I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lqhRh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lqhRh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;Here is a video including the moon level, bathed in earthshine as the specified color in the answer at &quot;night time&quot; (clip is 12 seconds into the video):&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=HKGCtkps8gk&amp;amp;list=PL9Bk1v7iJfdCpRaQ6vTGYOM-lXAO7RAQ1&amp;amp;index=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=HKGCtkps8gk&amp;amp;list=PL9Bk1v7iJfdCpRaQ6vTGYOM-lXAO7RAQ1&amp;amp;index=1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If interested in my game I am making, here is a link to it:&#xA;&lt;a href=&quot;https://jonathanlcopeland.wixsite.com/planetary&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://jonathanlcopeland.wixsite.com/planetary&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="1386" LastEditDate="2018-08-24T01:27:24.873" LastActivityDate="2018-08-24T01:27:24.873" Title="What color is earthshine?" Tags="&lt;the-moon&gt;&lt;earth&gt;&lt;light&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="16451" PostTypeId="2" ParentId="16450" CreationDate="2016-07-04T12:31:16.773" Score="2" Body="&lt;p&gt;It will depend on the face of the Earth that is reflecting the Sun light. It will be a mixture of blue from oceans, white from clouds / poles, brown from deserts and green from forests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the Earth has more ocean than land it will usually be dominated by blue (ocean) and white (cloud) making a pale blue.&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-07-04T12:31:16.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16452" PostTypeId="2" ParentId="16425" CreationDate="2016-07-04T14:06:24.563" Score="1" Body="&lt;p&gt;Light pollution isn't as much of a problem in space as it is on Earth because you're in a vacuum.  Imagine making a small black disk and holding it up at arms-length to cover the Sun....it's not going to simulate the effects of an eclipse, you're not going to get to see the corona and background stars because the rest of the sky is still flooded with light - or, more accurately, the Earth's &lt;em&gt;atmosphere&lt;/em&gt; is flooded with light as it scatters off water vapour, dust and the like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, you &lt;em&gt;can&lt;/em&gt; cover the Sun up in space and see the corona etc in exactly that way, the STEREO solar observatories do it. As long as you're careful about blocking &lt;strong&gt;direct&lt;/strong&gt; sunlight from hitting your lens, and make sure there's no reflective bits of spaceship in the frame, you're not going to get light pollution in the same way you would on Earth.&lt;/p&gt;&#xA;" OwnerUserId="11884" LastActivityDate="2016-07-04T14:06:24.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16453" PostTypeId="2" ParentId="16450" CreationDate="2016-07-04T15:07:50.620" Score="9" Body="&lt;p&gt;In my answer to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/14032/color-of-planets&quot;&gt;Color of planets&lt;/a&gt; I estimated the colour of the Earth as #&lt;a href=&quot;http://www.htmlcsscolor.com/hex/2F6A69&quot; rel=&quot;noreferrer&quot;&gt;2f6a69&lt;/a&gt; (based on averaging the colour in various images of Earth, and adjusting for albedo), but noted that it is the hardest of the planets to give a single colour to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were on the near side of the moon with no sun the cones in your eyes would not be fully active: you would be have &lt;a href=&quot;https://en.wikipedia.org/wiki/Mesopic_vision&quot; rel=&quot;noreferrer&quot;&gt;Mesopic vision&lt;/a&gt;. Also your eyes would adapt to the blue light conditions. As a result you probably wouldn't notice blueness of the light much.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-07-04T18:31:03.603" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16454" PostTypeId="1" AcceptedAnswerId="16475" CreationDate="2016-07-05T00:28:58.027" Score="0" ViewCount="1019" Body="&lt;p&gt;According to the Serbian geophysicist Milutin Milankovitch, there are three elements that make an ice age possible:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Eccentricity (orbital shape):  Varying between 0.000055 and 0.0679 over the course of 100,000 years (1.0 being a perfect circle).&lt;/li&gt;&#xA;&lt;li&gt;Obliquity (axial tilt):  Varying between 22.1 and 24.5 degrees over the course of 41,000 years.&lt;/li&gt;&#xA;&lt;li&gt;Axial precession (change in the orientation of the rotational axis on a rotating body):  Polaris being the North Star for a total of 26,000 degrees.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The question is how connected those three elements are, in the event someone wants to change the numbers in a worldbuilding process to make either a longer or shorter ice age:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the duration and extent of eccentricity affect those of obliquity and precession?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or does the duration and extent of obliquity affect those of eccentricity and precession?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or does the duration and extent of precession affect those of eccentricity and obliquity?&lt;/p&gt;&#xA;" OwnerUserId="10085" LastActivityDate="2016-07-06T00:11:46.610" Title="The Three Elements in the Milankovitch Cycles" Tags="&lt;orbital-elements&gt;&lt;orbital-migration&gt;&lt;orbital-mechanics&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16455" PostTypeId="1" AcceptedAnswerId="16456" CreationDate="2016-07-05T01:16:35.960" Score="2" ViewCount="998" Body="&lt;p&gt;I have a spectral cube (in FITS format) whose spectral dimension is in ångströms. The sampling along the spectral dimension is 0.28A (CDELT=0.28). The observation in the cube is Ha emission of a galaxy at redshift 0.1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I convert the sampling to km/s?&lt;/p&gt;&#xA;" OwnerUserId="1986" LastEditorDisplayName="user1569" LastEditDate="2016-08-04T13:25:04.170" LastActivityDate="2016-08-04T13:25:04.170" Title="Converting ångström spectral dimension to galaxy speed (km/s)" Tags="&lt;spectroscopy&gt;&lt;spectra&gt;&lt;units&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16456" PostTypeId="2" ParentId="16455" CreationDate="2016-07-05T03:08:07.283" Score="3" Body="&lt;p&gt;If you want bins that have an equal step in km/s then you will have to rebin in steps of equal log wavelength. On this scale the rest wavelength of H alpha is at zero km/s and then each bin will represent a increment in velocity with respect to that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://astronomy.stackexchange.com/questions/16266/how-do-i-apply-a-velocity-shift-to-a-wavelength-array-with-uniform-logarithmic-s?rq=1&quot;&gt;How do I apply a velocity shift to a wavelength array with uniform logarithmic spacing?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-07-05T03:08:07.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16457" PostTypeId="1" CreationDate="2016-07-05T05:49:32.343" Score="2" ViewCount="474" Body="&lt;p&gt;Even though our Earth moves on average 67,000 miles per hour around the Sun, we cannot feel its movement since its speed is fairly constant. More precisely, its change in speed is very low and our bodies cannot detect that low of acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if instead a hypothetical human observer were placed on a planet with a very eccentric orbit, say 0.99? Would that human be able to feel the planet's acceleration as it approached its host star, and as it receded from its host star? And if a human observer would not be able to feel any acceleration even at e=0.99, what is the minimum eccentricity, if any, that a planet must have in order for a hypothetical human observer to feel the effects of its acceleration and deceleration?&lt;/p&gt;&#xA;" OwnerUserId="13374" LastActivityDate="2016-08-04T17:21:07.997" Title="Would we be able to feel the acceleration of a planet with a highly eccentric orbit?" Tags="&lt;planet&gt;&lt;eccentric-orbit&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16458" PostTypeId="1" AcceptedAnswerId="16460" CreationDate="2016-07-05T07:24:30.980" Score="2" ViewCount="599" Body="&lt;p&gt;Several sites claim NASA has confirmed a second moon orbiting earth. This is the &lt;strong&gt;Asteroid 2016 HO3&lt;/strong&gt;. One source states The newfound quasi-satellite/moon-2016 HO3, is likely larger than 120 feet (40 m) and smaller than 300 feet (100 m).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.physics-astronomy.com/2016/06/nasa-just-confirmed-that-earth-has-new.html#.V3tdaNIrLIU&quot; rel=&quot;nofollow&quot;&gt;http://www.physics-astronomy.com/2016/06/nasa-just-confirmed-that-earth-has-new.html#.V3tdaNIrLIU&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&#xA;Is this claim true circling the internet and can anyone post a link from NASAs offical web site to conclude this is true?&lt;/p&gt;&#xA;" OwnerUserId="9326" LastEditorUserId="9326" LastEditDate="2016-07-05T18:03:03.410" LastActivityDate="2019-04-07T19:43:50.150" Title="Has NASA confirmed earth now has a second moon orbiting the planet?" Tags="&lt;orbit&gt;&lt;near-earth-object&gt;&lt;the-moon&gt;&lt;asteroids&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16459" PostTypeId="2" ParentId="16457" CreationDate="2016-07-05T07:32:56.693" Score="-1" Body="&lt;p&gt;First I think you have miss understood some basic principals. Although the Earth's &lt;strong&gt;speed&lt;/strong&gt; (a scalar quantity) is nearly constant its velocity is not, acceleration is a change in &lt;strong&gt;velocity&lt;/strong&gt; (a vector quantity). Due to its orbital motion the acceleration of the Earth is constantly changing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the actual question, the acceleration would be so small compared to that of gravity (between the person and the planet) that the effect will not be noticeable.&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-07-05T07:32:56.693" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16460" PostTypeId="2" ParentId="16458" CreationDate="2016-07-05T07:37:58.683" Score="5" Body="&lt;p&gt;It's not true; this object is not a moon of Earth's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jpl.nasa.gov/news/news.php?feature=6537&quot; rel=&quot;noreferrer&quot;&gt;Here's a NASA/JPL announcement&lt;/a&gt; of it. &lt;a href=&quot;https://en.wikipedia.org/wiki/(469219)_2016_HO3&quot; rel=&quot;noreferrer&quot;&gt;2016 HO3&lt;/a&gt; is the &quot;name&quot; being used for it at the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They're calling it a &quot;quasi-moon&quot;. It's in an orbit that is in resonance with Earth's but is not permanently bound. (It's not the first object to be found with such properties too.)&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2016-07-05T07:37:58.683" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16461" PostTypeId="1" CreationDate="2016-07-05T07:58:59.800" Score="0" ViewCount="144" Body="&lt;p&gt;Where we are, and in the conventional galaxies etc. that we see, there is a whole lot of relatively uniform space but in isolated places within that space, there are pockets of concentrated &quot;mass&quot; which is highly organised into all the objects we observe and study, and out of which we are made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that there were regions of space where that is all turned on its head, and there were huge vast swathes of relatively uniform matter, and within that matter there are pockets of &quot;space&quot; which is highly organised into some other form of object, out of which other objects and beings are made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously on the first, and simplest test - the question of &quot;how much&quot; dark matter there is, this theory passes the empirical test of reality with flying colours, i.e. astronomers seem to observe much more mass in the regions of hypothesised dark matter in accordance with what this conjecture implies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What other testable predictions does this conjecture make, and how well do they fit our observations?&lt;/p&gt;&#xA;" OwnerUserId="12000" LastActivityDate="2016-08-04T18:40:56.787" Title="Could dark matter be &quot;inside-out&quot; reality?" Tags="&lt;dark-matter&gt;&lt;dark-energy&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16463" PostTypeId="1" AcceptedAnswerId="16465" CreationDate="2016-07-05T09:19:37.397" Score="2" ViewCount="192" Body="&lt;p&gt;The Sun moves at the speed of 220km/s around the galactic center in a circular orbit. The sun is located at about 30,000 light years from the galactic center. The Sun orbits around the center of the Milky Way once every 225 million years. The period of time is called a cosmic year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curious about the speed at which the star closest to galactic center revolves around galactic center. Has this been calculated or approximated. Also what is the time it takes to complete one revolution for the star around galactic center or what is the cosmic year length for the star that is closest to galactic Center .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have taken reference from &lt;a href=&quot;http://www.bibliotecapleyades.net/ciencia/ciencia_astrosciences07.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.bibliotecapleyades.net/ciencia/ciencia_astrosciences07.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10430" LastEditorUserId="10430" LastEditDate="2016-07-08T14:26:33.137" LastActivityDate="2016-07-08T14:26:33.137" Title="What may be the period and speed of revolution of the star that is closes to galactic centre?" Tags="&lt;star&gt;&lt;cosmology&gt;&lt;galaxy-center&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16465" PostTypeId="2" ParentId="16463" CreationDate="2016-07-05T11:34:22.823" Score="5" Body="&lt;p&gt;Star S2 reaches maximum velocities of 5,000 km/s according to &lt;a href=&quot;http://www.eso.org/public/news/eso0226/&quot;&gt;this ESO page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbital period is given as a little over 15 years and &lt;a href=&quot;https://arxiv.org/pdf/astro-ph/0306220v1.pdf&quot;&gt;this paper&lt;/a&gt; gives a peri-center (closest approach to the black hole) of 17 light hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's another star in the area called &lt;a href=&quot;https://en.wikipedia.org/wiki/S0%E2%80%93102&quot;&gt;S0-102&lt;/a&gt; which has a shorter period, but its orbit is less eccentric - so its closest approach to the black hole is further out at 36 light hours. Its maximum velocity will &lt;em&gt;probably&lt;/em&gt; be less than S2 but I haven't been able to find a figure.&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2016-07-05T11:34:22.823" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16466" PostTypeId="1" CreationDate="2016-07-05T13:54:31.407" Score="6" ViewCount="163" Body="&lt;p&gt;On Jupiter and Saturn, aurora is seen around the planet's poles, so that implies that there must be a kind of magnetic field around those planets. Perhaps even the moon Europa could have a magnetic field created by salty water (Does it?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Earth we have two radiation belts, an inner and outer belt. One contains electrons and the other mostly protons. Do other planets also have these two radiation belts?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="2365" LastEditDate="2016-07-07T14:34:06.900" LastActivityDate="2016-07-13T15:42:11.397" Title="Do other planets with a magnetic field also have two Van Allen belts?" Tags="&lt;planet&gt;&lt;radiation&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16467" PostTypeId="1" CreationDate="2016-07-05T15:20:40.550" Score="2" ViewCount="231" Body="&lt;p&gt;I attended a talk last week about Type 1a supernovae with the focus being on double-degenerate systems as potential progenitors for them. We were shown a graph of White Dwarf (WD) mass distribution like the one below (&lt;a href=&quot;http://inspirehep.net/record/1184318/plots&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;), and it got me thinking, because even though they are very faint objects they are also very numerous in our galaxy. &#xA;So this led me to wonder what is the furthest known WD from Earth given that they are so faint but so numerous. I have looked for an answer online but the closest I got was a Type 1a progenitor at redshift 2 called &quot;SN UDS10Wil&quot; but this is not what I am looking for because this is not observable from Earth, it is just inferred from the supernova classification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6LnnU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6LnnU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10437" LastActivityDate="2016-07-11T07:11:13.440" Title="What is the most distant observable White Dwarf known?" Tags="&lt;white-dwarf&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16469" PostTypeId="1" CreationDate="2016-07-05T16:36:41.997" Score="1" ViewCount="364" Body="&lt;p&gt;I have a pair of binoculars with 10x magnification, 50mm objective diameter, and 6.5 degree (real) angular field of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking into buying solar filters for it (something like &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B00KVVFBV6&quot; rel=&quot;nofollow&quot;&gt;these&lt;/a&gt;) so I can view the sun. If I do so, will I be able to see most/many sunspots and planetary transits? Or is this simply not a powerful enough instrument to be worth investing in solar filters for it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like the idea of being able to check out the sun sometimes, but I don't know if I'm just getting way out of my league here and would need more complicated and expensive equipment to make it at all worthwhile. Thought I'd ask some experts before buying anything.&lt;/p&gt;&#xA;" OwnerUserId="13389" LastEditorUserId="1378" LastEditDate="2016-07-05T21:53:11.047" LastActivityDate="2016-07-05T21:55:02.353" Title="What solar features would I see at 10x magnification?" Tags="&lt;amateur-observing&gt;&lt;the-sun&gt;&lt;binoculars&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16470" PostTypeId="1" CreationDate="2016-07-05T17:01:53.943" Score="4" ViewCount="3134" Body="&lt;p&gt;I wonder, how accurate the lunar calendar is?&#xA;Can we predict the new moons precisely for tens of years in future?&#xA;If we know all factors that affects the movement of the moon, then why it is not possible to predict it?&lt;/p&gt;&#xA;" OwnerUserId="13388" LastActivityDate="2017-02-22T13:46:18.903" Title="lunar calendar accuracy for predicting new moons" Tags="&lt;moon-phases&gt;&lt;the-moon&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16471" PostTypeId="2" ParentId="16461" CreationDate="2016-07-05T17:46:40.640" Score="1" Body="&lt;p&gt;&lt;em&gt;As a disclaimer: This answer does not include references, and is mostly based on my first intuition when reading the question. I can see that it can potentially be an interesting question to discuss.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we live in a solar system, in a galaxy, in a galaxy cluster. That in itself means that there is a lot of matter around. We are also (at least very likely) living in an &lt;em&gt;overdensity&lt;/em&gt; in the universe, meaning that there is extremely much matter where we are compared to the average in the universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume that we do live in an underdensity of the universe, that would mean that there is more matter present in the universe on average compared to where we are. However, if &lt;em&gt;pockets&lt;/em&gt; of space in a distribution of regular matter was dark matter, then it sounds like there would be a larger amount of regular matter than dark matter in the universe, something which does not agree with our estimates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume for a minute that it is the other way around (that we live in an area with a lot of &lt;em&gt;matter&lt;/em&gt;, not a lot of &lt;em&gt;space&lt;/em&gt;), then this theory will struggle with explaining some of the things that point towards the existence of dark matter. One such thing is the rotation curves of galaxies, which indicate that there is more mass presence than there is seen. I expect that there would not be enough space within a galaxy to create such &quot;pockets&quot; of space. Say, however, that we assume this was possible, then the voids we observe between galaxies and galaxy clusters would be even more gigantic pockets of space, which would mean that there would be an enormous amount of dark matter, which I would think would be inconsistent with what we see from the CMB. &lt;/p&gt;&#xA;" OwnerUserId="13193" LastActivityDate="2016-07-05T17:46:40.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16473" PostTypeId="2" ParentId="16469" CreationDate="2016-07-05T20:52:19.477" Score="2" Body="&lt;p&gt;You will definitely see some sunspots at 10x; in exceptional cases sunspots are big enough to be visible with the naked eye through the sunset haze near horizon, but many sunspots are visible in a small instrument like a 10x binoc with proper filtering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun is at low activity right now, but still a few sunspots are showing up once in a while. And the activity should increase gradually in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planetary transits - yes. However, the next transit of Venus will take place in a century or so. Mercury transits more often, and the filters will allow you to see it - a tiny dot on the face of the Sun. A bit higher magnification might be beneficial, but anyway even at 10x you should see the planet, even if it's rather small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, I recommend solar filters based on the Baader solar filter film (just google it). They are neutral-density (do not change the color of the image), and are recommended and have been tested with instruments from small binocs to dobsonian telescopes. Make sure the filters are firmly attached to the instrument, and are free of defects.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2016-07-05T21:55:02.353" LastActivityDate="2016-07-05T21:55:02.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16474" PostTypeId="2" ParentId="16458" CreationDate="2016-07-05T23:16:16.397" Score="5" Body="&lt;p&gt;2016 HO3 does not really orbit Earth.&#xA;It orbits the Sun in such a way that it happens to loop around Earth at distances of&#xA;&lt;a href=&quot;http://www.skyandtelescope.com/astronomy-news/2016-ho3-a-new-minimoon-for-the-earth/&quot; rel=&quot;noreferrer&quot;&gt;0.1 to 0.25 AU&lt;/a&gt;.&#xA;Earth's gravitational sphere of influence is much smaller,&#xA;about 0.01 AU at most; outside that radius, the Sun is the dominant attractor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Minor Planet Center has an&#xA;&lt;a href=&quot;http://www.minorplanetcenter.net/db_search/show_orbit?utf8=%E2%9C%93&amp;amp;object_id=2016+HO3&amp;amp;epoch=2016-07-31.0&amp;amp;number=469219&amp;amp;designation=2016+HO3&amp;amp;name=&amp;amp;peri=307.2276436&amp;amp;m=297.53221&amp;amp;node=66.5132056&amp;amp;incl=7.7714415&amp;amp;e=0.104143476&amp;amp;a=1.00123&amp;amp;commit=Interactive+Orbit+Sketch&quot; rel=&quot;noreferrer&quot;&gt;orbital diagram&lt;/a&gt;&#xA;which you can examine in 3D.&#xA;Earth's orbit is light blue.&#xA;2016 HO3's orbit is orange to the north of Earth's orbital plane and medium blue to the south.&#xA;A dark gray line passes through the Sun and the asteroid's perihelion and aphelion.&#xA;It's a moderately eccentric, moderately inclined orbit around the Sun with a 1.00 year period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the course of its solar orbit, the asteroid alternately moves slower, outside, and to the north of Earth's orbit; and faster, inside, and to the south.&#xA;In a rotating frame of reference fixed on Earth,&#xA;as in the &lt;a href=&quot;https://www.youtube.com/watch?v=SbbAnVU4rmY&quot; rel=&quot;noreferrer&quot;&gt;video&lt;/a&gt; you've seen,&#xA;this looks like an oddly shaped, highly inclined, retrograde loop around Earth, but it is not due to Earth's gravity.&#xA;Earth's slight influence helps to maintain this coincidental pattern&#xA;but is not nearly enough to capture the asteroid.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-07-05T23:16:16.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16475" PostTypeId="2" ParentId="16454" CreationDate="2016-07-06T00:01:18.247" Score="4" Body="&lt;p&gt;The primary driver of ice ages over the history of the Earth is much more Earth science than astronomy.  The modern ice age cycle (last 2.6 million years or so) is only possible with relatively low levels of CO2 in the atmosphere and land masses near the North Pole where glaciers can form.  Other factors in the modern ice age period are thought to be the formation of the Isthmus of Panama, perhaps the relatively permanent antarctic ice sheet, and just maybe, the formation of the Himalayas as well.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of circumstances required on Earth that make milankovich driven Ice ages possible are pretty complicated and have to be set just about right.  When they are just right, then the Milankovich cycles, which are quite weak, can take over and be the drivers of ice ages.   Prior to 2.6 million years ago, the last time the Earth went through a period of rising and receding ice ages was 440 million years ago when most of the land was on the South Pole.  CO2 was much more abundant, but the single continent sitting over the south pole made the formation of ice sheets possible.   Over most of Earth's history, the Milankovich cycles haven't governed ice ages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, onto the astronomy bits of this question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The question is how connected those three elements are, in the event&#xA;  someone wants to change the numbers in a worldbuilding process to make&#xA;  either a longer or shorter ice age:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's a strong correlation between the Milankovich cycles and the length of recent ice ages, so the answer here is yes, if you were able to adjust the &quot;numbers&quot; then the length of ice ages would change accordingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 3 Milankovich cycles combine into two primary patterns, the 40,000 year cycle and the 100,000 year cycle (more on that shortly).  These were observed first by geologists and later tied to Milankovich cycles.  What starts and stops ice ages is complicated and there's a few moving parts, so there's not 100% consensus on Milankovich cycles driving ice ages, but the correlation is strong and as far as I've read, this idea is largely agreed upon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ice age Cycles&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From about 2.3 million years ago to about 800,000 years ago, the 40,000 year ice age cycle was dominant, and for the last 800,000 years, the 100,000 year cycle has been dominant.  The 40,000 year cycle is thought to be the stronger of the two, and one theory as to why the 100,000 year cycle became dominant 800,000 years ago is that the Earth grew slightly cooler, perhaps with the glaciation of Greenland.   See &lt;a href=&quot;https://en.wikipedia.org/wiki/100,000-year_problem&quot;&gt;100,000 year problem&lt;/a&gt;.  The 40,000 year cycle still plays a role even within the 100,000 year cycle currently the dominant one.  Article on that &lt;a href=&quot;http://www.livescience.com/6937-ice-ages-blamed-tilted-earth.html&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axial tilt.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axial tilt is on a pretty consistent 41,040 year cycle and it's strongly tied to the arguably somewhat misnamed, 40,000 year cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axial tilt is the cause of seasons on Earth, so when the Axial tilt is high (over 24 degrees), the seasonal variation increases, meaning, warmer summers and colder winters on both hemispheres.  When it's low (below 23), it means more mild seasons.  While this doesn't change how much sunlight hits the earth, it does change climate on Earth for other reasons,  The overall result is that the more mild seasons lead to a cooler earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://en.wikipedia.org/wiki/Milankovitch_cycles#Axial_tilt_.28obliquity.29&quot;&gt;Wikipedia&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This trend in forcing, by itself, tends to make winters warmer and&#xA;  summers colder (i.e. milder seasons), as well as cause an overall&#xA;  cooling trend.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason for this, briefly, is that it still snows in high latitudes in the Northern Hemisphere during mild winters and snow and Ice takes longer to melt during mild summers.  If this persists over enough years, Snow cover can build up, which causes &lt;a href=&quot;https://en.wikipedia.org/wiki/Albedo&quot;&gt;Albedo&lt;/a&gt; to rise and reflect more sunlight back into space, and less summer melting of ice, over many years, can lead to an ice age.  Once an ice age gets going it's self perpetuating, sometimes referred to as feedback mechanisms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axial Tilt reached a maximum about 8,700 BCE and it will reach a minimum in about 11,800 CE.  The warming of the Earth from the Axial tilt maximum isn't sudden though, it begins several thousand years before the maximum and continues for several thousand years after.   This fits nicely with the ending of the last ice age around 10-11,000 BCE.  We're currently in a mid-point in the Axial tilt cycle and it's gradually approaching a cooling effect, though, at least for now, the effect of man made climate change is far bigger than the gradual Axial Tilt cooling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axial Precession&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precession is a bit more complicated.  The Earth is on a 26,000 year &quot;wobble&quot;,  that moves the North Star and other stars around in a circle in the night sky, but what matters with this Milankovich cycle is what month the Earth is closest to the sun, or, at &lt;a href=&quot;https://en.wikipedia.org/wiki/Aphelion&quot;&gt;Perihelion&lt;/a&gt;.  The change in Perihelion over time works on roughly a 21,000 year cycle, because it combines Axial precession with &lt;a href=&quot;https://en.wikipedia.org/wiki/Apsidal_precession&quot;&gt;Apsidal precession&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently the Earth is closest to the Sun in January, but that will gradually change and in about 10,500 years, the Earth will be closest to the Sun in July.  When the Sun is closer, the Earth gets slightly warmer, so the current position of this cycle leads to milder seasons in the Northern Hemisphere (which can lead to cooling) and more extreme seasons in the Southern Hemisphere.   Unlike Axial Tilt which affects both hemispheres in the same way, precession affects each hemisphere in the opposite way to the other.   This, in effect either adds to Axial Tilt's effect or subtracts from it.   As pointed out above, the simplest factor is that milder seasons in the Northern Hemisphere that are ice-age friendly and more extreme seasons are not.   So when perihelion is in January, that somewhat cools the Earth and when it's in July it somewhat warms it.  We're currently in a &quot;cold&quot; period of this cycle, we were in a warm one 10,500 years ago, which also times well with the end of the last ice age. This cycle is moving towards a gradual warming and more extreme Northern Hemisphere seasons currently working in opposition to the Axial tilt effect.  It's cycle is roughly twice as fast and so this also ties in nicely with the 40,000 year cycle too, as these 2 periods (21,000 years and 41,000 years) are close to in sync.  These two cycles are thought to be key factors in driving the greening and droughting of the Sahara desert.  Article &lt;a href=&quot;http://www.astrobio.net/news-exclusive/how-earths-orbital-shift-shaped-the-sahara/&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/North_African_climate_cycles&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Eccentricity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth's orbital period and semi-major axis of it's orbit remain roughly unchanged.   What changes with eccentricity is how far the focal point (the Sun is at one of the focal points of the ellipse), moves from the center of the elipse.   The effect of the Earth's orbital eccentricity changing is two-fold.  First, Axial precession only matters if the Earth's orbit is an ellipse.   If the Earth's orbit is circular, the precession Milankovich cycle has no effect.  Precession has maximum effect when the Earth's orbit is at it's most elliptical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 2nd effect is that a more elliptical orbit means that, because planets move more slowly in their orbit when they are further from the sun (Equal areas in equal time, per Kepler's law), a more elliptical orbit means that over the calendar year, the Earth gets less sunlight because it's average distance form the sun over time is slightly higher.  So, low eccentricity means slightly warmer, high eccentricity, slightly colder, even though the planet passes through an entire orbit every year, the total energy the Earth receives does drop slightly when the Earth's orbit is most eliptical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, greater eccentricity amplifies the effect of the Axial precession, which when it's at it's highest, tends towards ending ice ages, so this effect operates in both directions, which is part of the reason why the 100,000 year cycle is smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In effect, eccentricity multiplies the effect of precession, but it has an additional effect that works in opposition to the precession ice age driver.  (Does that make sense? - I can try to clean up if not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precession and Axial Tilt can add together or cancel each other out.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and there's other effects.  How much sunlight is reflected off the high albedo south pole, how much sunlight warms the antarctic ocean, how strong the Atlantic current is.   The entire picture is hugely complicated, but from an astronomical point of view, eccentricity is the only one that can reduce or increase total solar energy hitting the Earth.   Axial tilt and precession have no effect on how much sunlight hits the earth, they only drive the variation between the seasons.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the duration and extent of eccentricity affect those of obliquity&#xA;  and precession?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The extent of eccentricity directly increases the seasonal variation effect of precession.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Or does the duration and extent of obliquity affect those of&#xA;  eccentricity and precession?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, the obliquity doesn't affect either of those, but the effect of obliquity can add to, or cancel out the effect of precession.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Or does the duration and extent of precession affect those of&#xA;  eccentricity and obliquity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Precession doesn't effect eccentricity (which way the Earth is pointed obviously doesn't effect eccentricity), but precession can add to or subtract from Obliquity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the gist of it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can run some numbers if you like, but that gets even longer.  The sunlight energy that reaches the earth and how it varies by axial tilt is longitude dependent, so any calculations on changes in energy by changes in axial tilt will vary based on longitude.  At many latitudes, the summer solar energy is more than double what it is for winter.  Even a one or two degree in the angle of axial tilt variation can make a measurable difference in the seasonal variation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precession energy variation isn't longitude dependent, but it varies every day with the earth's orbit.   Some bad math on it, with the current eccentricity of about 0.0167, what that means is that at perihelion, the Earth is about .9833 AU from the sun and at aphelion, about 1.0167 AU. Because solar energy diminishes with the square of the distance, the roughly 3.4% variation in distance from the sun works out to about 7% more solar energy hitting the Earth perihelion to anhelion.   When the Elliptical orbit peaks, that ratio exceeds 20% and when it's minimized, less than 2%, but those are ratios by season and location, the total energy the Earth receives from the sun doesn't change, only seasonal variability changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's as far as I'll go with the numbers, as far as the timing, that's pretty straight forward, some of that covered above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axial Tilt, when at a maximum it tends to melt glaciers and leads to warming, this last happened 10,700 years ago and fits with the end of the last ice age.   Minimal tilt can lead to cooling and the earth will enter an Axial Tilt Minimum in about 9,800 years.   The entire cycle, high to low back to high lasts 41,040 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precession is significantly amplified with high eccentricity and significantly reduced with low eccentricity.   Earth is currently in a low eccentricity so the precession effect for the last 10,000 years or so has been relatively small.  10,500 years from now precession will be at peak seasonal variation in the Northern Hemisphere, which will work against the Axial Tilt's cooling.   In about 21,000 years, Precession will again be working towards more mild seasons in the Northern Hemisphere, eccentricity will be high and the Axial Tilt will be past it's minimum, but still in a low point.   So, taken together, we might see an ice age begin to form somewhere around 18-20,000 years in the future, but we have little to worry about.  It's not difficult to generate enough atmospheric CO2 to prevent that ice age from happening.   We're already well over the necessary amount of CO2 to prevent a coming ice age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, Eccentricity was at a minimum about 9,000 years ago and will reach a maximum in about 27,000 years in the future.  Eccentricity is the least consistent of the cycles, it varies between about 90,000 years and 125,000 years and there's a larger, 400,000 year cycle on top of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If parts of that aren't clear, I can try to tidy up, or just ask.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-07-06T00:11:46.610" LastActivityDate="2016-07-06T00:11:46.610" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16478" PostTypeId="1" AcceptedAnswerId="16482" CreationDate="2016-07-06T02:09:45.827" Score="7" ViewCount="1619" Body="&lt;p&gt;If Earth started spinning fast enough, would the centrifugal force eventually overcome Earth's gravity enough enough to fling someone (let's assume 75 kg) into orbit or into space? If so, how fast would it need to spin?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being that I'm not a scientist, I'm hoping for a simple answer like, &quot;one rotation per xx hours/mins/seconds needed to throw someone off Earth&quot; or something to that effect. Complicated equations and/or jargon is fine for the sciencey types, of course. I just probably won't understand it too much. &lt;/p&gt;&#xA;" OwnerUserId="10117" LastActivityDate="2020-06-13T14:15:53.087" Title="How fast would Earth need to rotate to fling someone off due to centrifugal force?" Tags="&lt;gravity&gt;&lt;earth&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16479" PostTypeId="2" ParentId="16478" CreationDate="2016-07-06T02:43:58.410" Score="13" Body="&lt;p&gt;If it would rotate fast enough to throw a person off, it would throw everything off - dirt, houses, cars, etc., as the mass of an object is irrelevant - the centrifugal force (pulling out) and the gravitational force (pulling in) are both proportional to an object's mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically the whole thing would explode out at the same time. To calculate we equate the force of gravity $mg$ with the centrifugal force $mr\omega^2$, where $\omega$ is the angular velocity in radians per second &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = mg = mr\omega^2 $$ $$\Rightarrow g = r\omega^2 $$ $$\Rightarrow \omega = \sqrt{g/r} \approx 0.0012/s.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Insert the radius of the earth $r$ and the gravitational constant $g$, take the square root, and you have the necessary angular speed $\omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\omega = 2 \pi/T$, you get $T = 2 \pi/\omega \approx 1.5 \mathrm{hours}$ [I did that in my head, so it might be wrong].&lt;/p&gt;&#xA;" OwnerUserId="10084" LastEditorUserId="2365" LastEditDate="2016-07-06T05:38:49.870" LastActivityDate="2016-07-06T05:38:49.870" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16482" PostTypeId="2" ParentId="16478" CreationDate="2016-07-06T12:13:44.970" Score="13" Body="&lt;p&gt;Aganju's answer is excellent, but I'd like to add one thing: The radius used is Earth's radius, which is applicable if you're at the equator. At higher latitudes, the effective radius is smaller by a factor $\cos\theta$, where $\theta$ is the latitude in radians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For iMerchant, who seem to be located in Vancouver at&#xA;$$\theta = 49.3^{\circ}\times\pi/180 = 0.86\,\mathrm{rad},$$&#xA;the relevant radius is&#xA;$$&#xA;R_\mathrm{eff} = R_\oplus\cos\theta = 4150\,\mathrm{km},&#xA;$$&#xA;so at the $T=1.4\,\mathrm{h}$ of Aganju's answer (1.5 h is just a little too slow), the centrifugal (or &lt;a href=&quot;https://xkcd.com/123/&quot; rel=&quot;noreferrer&quot;&gt;-petal&lt;/a&gt;, if you prefer) acceleration would be only $g_\mathrm{spin} = 6.5\,\mathrm{m}\,\mathrm{s}^{-2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, whereas Earth's acceleration is directed toward the center of Earth, in Vancouver the centrifugal acceleration would be directed at an angle $\theta$ toward South. Thus, only a component $g_\mathrm{spin} \cos\theta = 4.2\,\mathrm{m}\,\mathrm{s}^{-2}$ would point upward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not enough to send iMerchant into space. However, a component $g_\mathrm{spin} \sin\theta = 4.9\,\mathrm{m}\,\mathrm{s}^{-2}$, or roughly half a G, would point horizontally toward South, enough to send him tumbling down the street, or at least walk awkwardly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/T5LKT.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/T5LKT.png&quot; alt=&quot;Rotating Earth&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-07-06T18:44:33.337" LastActivityDate="2016-07-06T18:44:33.337" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16483" PostTypeId="1" CreationDate="2016-07-06T13:17:00.353" Score="1" ViewCount="192" Body="&lt;p&gt;Has the velocity of photons, which are coming from black holes, already been measured? I am not aware of such measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for the question is the following: If space-time near black holes has some extra dimensions, than photons created in this higher dimensional space-time might have non-zero components of the wave-vector in these extra dimensions. On the earth however we see just our 3 or 4 components of the wave-vector, so we might see just the projection of the original wave-vector onto our (3,1)-dimensional space-time. In this case the velocity of the photons might be less than the standard velocity of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have in mind a special model of our universe. I believe, that our universe is the horizon of a (4,2)-dimensional space-time. On this horizon (which is our universe) we do not see the extra dimensions, however, they should be visible near black holes. So, in this model, black holes are the windows into the extra dimensions. &lt;/p&gt;&#xA;" OwnerUserId="13397" LastEditorUserId="13397" LastEditDate="2016-07-06T13:41:46.623" LastActivityDate="2016-07-07T12:33:14.107" Title="Has the velocity of photons coming from black holes already been measured?" Tags="&lt;black-hole&gt;&lt;light&gt;&lt;general-relativity&gt;&lt;horizon&gt;&lt;photons&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16485" PostTypeId="2" ParentId="14454" CreationDate="2016-07-06T15:06:32.103" Score="1" Body="&lt;p&gt;A word for the highest point (not elevation) is &lt;em&gt;zenith&lt;/em&gt;. See: &lt;a href=&quot;http://www.merriam-webster.com/dictionary/zenith&quot; rel=&quot;nofollow&quot;&gt;http://www.merriam-webster.com/dictionary/zenith&lt;/a&gt; , second definition: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the highest point reached in the heavens by a celestial body&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So 'solar zenith' would describe the highest point the sun reaches.'&lt;/p&gt;&#xA;" OwnerUserId="11879" LastActivityDate="2016-07-06T15:06:32.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16488" PostTypeId="1" AcceptedAnswerId="18029" CreationDate="2016-07-07T00:03:00.027" Score="3" ViewCount="1227" Body="&lt;p&gt;I'm working on a project for fun where I represent some sleep data geographically. For a given day, I have a date, a time for falling asleep that night, and a time for waking up the next day. The idea is something like &lt;a href=&quot;https://xkcd.com/448/&quot; rel=&quot;nofollow&quot;&gt;this xkcd comic&lt;/a&gt;. My attempts at taking the inverse of the sunrise/sunset equations I've found online have not been fruitful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, I would like to find an equation or algorithm to give a rough location on Earth with a specific sunset time on some specific day and a specific sunrise time the next day. I am not too concerned with accuracy since this is just for fun, so I'm hoping elevation information isn't necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a programmer, so the goal is really to be able to type something like &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt;&amp;gt;&amp;gt;&amp;gt; locate(date = '2014-08-27', sunset = '10:00 PM', sunrise = '7:30 AM')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and get back&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Latitude: 49.887, Longitude: 96.141&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I could program it myself if I just understood enough astronomy. Any help is greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="13404" LastEditorUserId="13404" LastEditDate="2016-08-25T19:15:12.320" LastActivityDate="2018-10-10T12:34:30.390" Title="Inverse of the sunrise equation - finding locations with a given sunrise time on a given day" Tags="&lt;earth&gt;&lt;rotation&gt;&lt;the-sun&gt;&lt;movement&gt;&lt;orbital-mechanics&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16489" PostTypeId="2" ParentId="16483" CreationDate="2016-07-07T04:54:45.697" Score="2" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Photons can't come from the singularity of a black hole, or from beyond the event horizon.&lt;/li&gt;&#xA;&lt;li&gt;While some very good candidates for black holes exist, none have been certainly observed.&lt;/li&gt;&#xA;&lt;li&gt;We have never observed Hawking radiation: it remains theoretical.&lt;/li&gt;&#xA;&lt;li&gt;We have observed radiation from the accretion disk of likely black holes, however the matter in this disc is orbiting in 'normal' spacetime.&lt;/li&gt;&#xA;&lt;li&gt;Direct measurement of the speed of individual photons is pretty much out of the question. We can measure the speed of light by creating photons at one point and then timing when those photons arrive at another. You can't measure a photon's passing, as to measure it you would have to interact with it. If you interact with it, it is no longer the same photon. This leads to...&lt;/li&gt;&#xA;&lt;li&gt;When we speak of a photon travelling from one point to another, remember that it is a quantum process: the photon interacts with matter and with itself. Photons are not classical particles. Everything about them needs to be considered through quantum electrodynamics.&lt;/li&gt;&#xA;&lt;li&gt;The flashes from pulsars are coming from a region of intense gravity. They do propagate at the speed of light: pulsars measured from different locations flash at slightly different times due to light travel time. There is nothing special about the space-time outside of a black hole as compared to that around a neutron-star, only a difference of degree. Gravity is still pretty intense around a neutron star, $g\approx10^{12}\mathrm{ms^-2}$&lt;/li&gt;&#xA;&lt;li&gt;Gravitational redshift from white dwarfs is well known. Gravitational redshift is predicted by relativity, which assumes a constant speed of light. &lt;/li&gt;&#xA;&lt;li&gt;Constant speed of light is pretty fundamental to relativity. If a theory breaks this, a lot of other physics is broken (exceptional claims require exceptional evidence)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-07-07T12:33:14.107" LastActivityDate="2016-07-07T12:33:14.107" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16490" PostTypeId="2" ParentId="16488" CreationDate="2016-07-07T07:27:55.657" Score="2" Body="&lt;p&gt;Due to the complexity of the problem (see my comment above) and the fact that you only want a rough answer why not try calculating the times of Sun rise / set at say one degree intervals for the whole Earth and then compare these with the times you have. If one degree is not near enough you could use a smaller interval and also use a range in the comparison (say +/- 5 mins).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exactly how fine your location spacing and time range is will depend on how accurate you want your answers and both could easily be tweaked until you are satisfied with the results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approach may not be the most efficient, but with the speed of modern computer shouldn't take very long. &lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-07-07T07:27:55.657" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16491" PostTypeId="2" ParentId="646" CreationDate="2016-07-07T08:52:52.963" Score="0" Body="&lt;p&gt;Traveling at light speed has implications for both space and time. In addition to no experience of time, photons perceive no space in the direction of their travel. Thus their &quot;instantaneous&quot; space travel covers zero distance. In other words, each photon perceives your eyeball to be attached to the photosphere of Alpha Centauri, thus allowing for a very short travel time...&lt;/p&gt;&#xA;" OwnerUserId="13412" LastActivityDate="2016-07-07T08:52:52.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16492" PostTypeId="1" AcceptedAnswerId="16587" CreationDate="2016-07-07T09:15:24.250" Score="0" ViewCount="409" Body="&lt;p&gt;If not, when can a white dwarf be more massive than a neutron star?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="10172" LastEditDate="2016-07-07T09:58:55.803" LastActivityDate="2016-07-15T16:32:25.263" Title="Is the lower mass limit of a neutron star the same as the upper mass limit of a white dwarf?" Tags="&lt;neutron-star&gt;&lt;white-dwarf&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16493" PostTypeId="2" ParentId="16478" CreationDate="2016-07-07T09:25:10.863" Score="2" Body="&lt;p&gt;Joe Blow commented about the orbital speed of satellites, correctly. As shown by Aganju, the earth would have to rotate 16 times faster in order to levitate equatorial residents. It's current rotational speed of 0.5 km per sec would then increase to 8 km per s (5 miles/s). We should all recognize this speed as that of orbiting satellites.&lt;/p&gt;&#xA;" OwnerUserId="13413" LastActivityDate="2016-07-07T09:25:10.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16497" PostTypeId="2" ParentId="16418" CreationDate="2016-07-07T11:38:48.080" Score="0" Body="&lt;p&gt;Going back to the beginning, we used to have CMBR right here in our vicinity. It, indeed, has been moving away from us for the last 13.7 billion years. The CMBR we do see now is reaching us from space that emitted it 13.7 billion years ago. Observers over THERE can now see our own radiation that departed our vicinity 13.7 billion years ago as their CMBR, ie, this is a symmetric situation.&lt;/p&gt;&#xA;" OwnerUserId="13419" LastActivityDate="2016-07-07T11:38:48.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16498" PostTypeId="2" ParentId="12749" CreationDate="2016-07-07T11:56:28.013" Score="1" Body="&lt;p&gt;If all that exists originated in the big bang, as per the begining of your question, why then would you be searching for the oldest part of something that was globally formed at the same instant? I wouldn't ;-) Thankfully, the finite light speed allows us to view objects as they were during younger times at increasing distances.&lt;/p&gt;&#xA;" OwnerUserId="13421" LastActivityDate="2016-07-07T11:56:28.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16499" PostTypeId="1" CreationDate="2016-07-07T14:34:16.310" Score="2" ViewCount="1102" Body="&lt;p&gt;If a planet is tidally-locked, can it have an obliquity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. the Earth's rotation is being slowed by tidal forces.  A long way in the future it will be tidally locked.  As it slows, will the obliquity also be eroded or will other forces (such as the orbit of the moon) allow it to maintain an obliquity?&lt;/p&gt;&#xA;" OwnerUserId="13424" LastEditorUserId="13424" LastEditDate="2016-07-07T14:41:16.170" LastActivityDate="2016-07-09T15:17:33.720" Title="Can a tidally-locked planet maintain an obliquity" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;exoplanet&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16500" PostTypeId="7" CreationDate="2016-07-07T14:42:01.073" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="6" LastEditDate="2016-07-07T14:42:01.073" LastActivityDate="2016-07-07T14:42:01.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16501" PostTypeId="1" AcceptedAnswerId="16504" CreationDate="2016-07-07T15:44:45.173" Score="2" ViewCount="115" Body="&lt;p&gt;We know that stellar radiation pressure balances the gravitational compressive forced of a star. Are there other factors which resist such a collapse? Also, if radiation pressure balances gravitational compression, shouldn't the process of shrinking be a quasistatic process making stars Eternally Collapsing Objects (ECOs)?&lt;/p&gt;&#xA;" OwnerUserId="13414" LastEditorUserId="13426" LastEditDate="2016-07-07T20:46:25.610" LastActivityDate="2016-07-07T20:46:25.610" Title="Why aren't stars like Eternally Collapsing Objects?" Tags="&lt;radiation&gt;&lt;stellar-dynamics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16504" PostTypeId="2" ParentId="16501" CreationDate="2016-07-07T16:30:47.197" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;We know that stellar radiation pressure balances the gravitational compressive forced of a star.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We do not know that. Degeneracy pressure, thermal pressure, and radiation pressure are what collectively balance gravitation in a star. Degeneracy pressure dominates over the other two in white dwarfs and neutron stars. Thermal pressure dominates in low mass stars such as the Sun. It's only in very massive stars where radiation pressure dominates.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Shouldn't the process of shrinking be a quasistatic process making stars (Eternally Collapsing Objects)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are forgetting about fusion. A star is more or less in equilibrium thanks to a number of negative feedbacks. (Negative feedback in a control system is good. Positive feedback is very bad. The &quot;screech!&quot; sound you occasionally hear from microphones is positive feedback.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the fusion rate increases slightly in a region near the center of a star. This increases the temperature and thereby the pressure, which in turn causes the core to expand, which in turn causes the density and temperature in the core to fall, which in turn decreases the fusion rate. The reverse applies for a slight decrease in the fusion rate. Similar negative feedbacks apply for changes in temperature, pressure, and density. A main sequence star is self regulating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things get a bit trickier when a star leaves the main sequence, but that is a different question. &lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2016-07-07T19:11:40.673" LastActivityDate="2016-07-07T19:11:40.673" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16505" PostTypeId="2" ParentId="16499" CreationDate="2016-07-07T16:39:12.153" Score="1" Body="&lt;p&gt;The Obliquity of earth has been oscillating(approximately) with a mean period of 41,040 years. Theoretically, it has been justified that in absence of the moon, earth obliquity might become unstable. It is also a known fact that the moon is slowly moving away. By the time earth is tidally locked, the moon's effect might be significantly less, resulting in obliquity shifts of about 20-25 degrees.&lt;/p&gt;&#xA;" OwnerUserId="13414" LastActivityDate="2016-07-07T16:39:12.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16506" PostTypeId="2" ParentId="16467" CreationDate="2016-07-07T16:54:09.093" Score="2" Body="&lt;p&gt;Based on a Simbad search, it looks like the most distant white dwarf is WD J0106-0014, which was measured in &lt;a href=&quot;http://adsabs.harvard.edu/abs/2011ApJS..197...38D&quot; rel=&quot;nofollow&quot;&gt;Debes et al. 2011&lt;/a&gt; to be at 6442 pc.  Here's the Simbad page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?mescat.distance=on&amp;amp;Ident=%401372027&amp;amp;Name=WD+J0106-0014&amp;amp;submit=display+selected+measurements#lab_meas&quot; rel=&quot;nofollow&quot;&gt;http://simbad.u-strasbg.fr/simbad/sim-id?mescat.distance=on&amp;amp;Ident=%401372027&amp;amp;Name=WD+J0106-0014&amp;amp;submit=display+selected+measurements#lab_meas&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10371" LastActivityDate="2016-07-07T16:54:09.093" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16508" PostTypeId="1" AcceptedAnswerId="16513" CreationDate="2016-07-07T20:45:25.523" Score="2" ViewCount="417" Body="&lt;p&gt;After discovering &lt;a href=&quot;https://physics.stackexchange.com/questions/216276/what-does-a-blackbody-sound-like&quot;&gt;this question&lt;/a&gt; exploring the sound of a blackbody, I started wondering about the sound of the Cosmic Microwave Background radiation from the Big Bang, specifically what the current &lt;em&gt;pitch&lt;/em&gt; might be at its peak frequency and how that might evolve over time due to redshifting. Here's what I've been able to deduce so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;pitch_{CMB} = 160.4\,GHz \approx D_{33} = 157.7\,GHz&#xA;\\&#xA;pitch_{CMB}\,lowered\,by\,30\,octaves = {160.4\, GHz \over 2^{30}} = 149.38\, Hz \approx D_3 = 146.83\,Hz&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Building on this result:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&lt;li&gt;How much time will need to elapse before the CMB redshift will cause this pitch to drop by an octave (or even by a semitone)?&lt;/li&gt;&#xA;&lt;li&gt;How much variability would this pitch have depending on the dipole anisotropy of the CMB from our perspective as well as due to more generalized anisotropies in the CMB?&lt;/li&gt;&#xA;&lt;li&gt;What pitch would the peak value of this blackbody have had at earlier points in the history of the Universe?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I've tried to apply some of the information in &lt;a href=&quot;https://astronomy.stackexchange.com/questions/8763/does-the-cosmic-microwave-background-change-over-time&quot;&gt;this question&lt;/a&gt; and elsewhere, but would greatly appreciate hearing from someone with more expertise in the subject area if this unorthodox comparison has piqued your interest like it has mine.&lt;/p&gt;&#xA;" OwnerUserId="13402" LastEditorUserId="13402" LastEditDate="2016-07-08T14:45:29.320" LastActivityDate="2019-01-26T06:05:16.890" Title="Future redshifting and effect on the 'pitch' of CMB radiation" Tags="&lt;cosmology&gt;&lt;big-bang-theory&gt;&lt;redshift&gt;&lt;cmb&gt;&lt;black-body&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16509" PostTypeId="1" CreationDate="2016-07-07T20:55:19.353" Score="1" ViewCount="161" Body="&lt;p&gt;I understand that a neutron star is dense enough that adding more matter will increase the amount of degenerate matter directly, and the limit to its size is about 1.4 solar masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if an object were formed &lt;em&gt;not&lt;/em&gt; by an explosion that crushes it down to neutron matter, is it necessarily so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given normal atoms of any desired species, and it is piled up carefully so it does not get hot. When it gets around one solar mass, will it necessarily collapse into a neutron star, or can it get more massive?  Will such an object eventually collapse catastrophically or can it have a degenerate matter core and a substantial thickness of less-compressed matter on top and normal matter on top of that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the largest possible mass of a cold solid object?  Can it be several solar masses?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &lt;em&gt;cold&lt;/em&gt; I mean that it’s not a star, puffed up by ongoing consumption of energy. Whatever it's made out of is not being “used up”.&lt;/p&gt;&#xA;" OwnerUserId="7142" LastEditorUserId="7142" LastEditDate="2016-07-08T20:08:34.070" LastActivityDate="2016-07-08T21:30:50.653" Title="How massive can a cold solid object be?" Tags="&lt;astrophysics&gt;&lt;temperature&gt;&lt;degenerate-matter&gt;" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="16510" PostTypeId="1" AcceptedAnswerId="16517" CreationDate="2016-07-07T22:12:16.700" Score="1" ViewCount="1426" Body="&lt;p&gt;The &lt;a href=&quot;http://iraf.noao.edu/iraf/web/&quot; rel=&quot;nofollow&quot;&gt;IRAF&lt;/a&gt; package is &lt;em&gt;old&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been looking around for a more modern software to replace it in the processes of CCD data reduction and photometry, but haven't been able to find any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest I've found is the &lt;a href=&quot;http://www.stsci.edu/institute/software_hardware/pyraf&quot; rel=&quot;nofollow&quot;&gt;PyRAF&lt;/a&gt; tool, but this seems more like a Python wrapper around IRAF rather than a replacement for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some new software I might've miss or is IRAF really the only option even today?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Add&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I forgot to mention this, but I'm looking for tools that work under Linux and are free (open source + no charge), if possible. I will not pay (neither for a Windows license nor for a software package) to get rid of IRAF.&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="354" LastEditDate="2016-07-08T15:03:10.270" LastActivityDate="2019-04-21T22:56:25.177" Title="Data reduction and photometry without IRAF?" Tags="&lt;software&gt;&lt;photometry&gt;" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16511" PostTypeId="2" ParentId="16510" CreationDate="2016-07-08T07:22:24.593" Score="0" Body="&lt;p&gt;AIP4Win has a comprehensive photometry package, also I believe MaximDL (though more expensive) has some as well.&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-07-08T07:22:24.593" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16513" PostTypeId="2" ParentId="16508" CreationDate="2016-07-08T14:31:54.110" Score="3" Body="&lt;p&gt;The wavelength of the CMB increases linearly with the scale factor $a$, which is defined as 1 today, so the &quot;sound&quot; of the CMB has dropped an octave when the Universe has doubled in size (in all three directions), i.e. when $a = 2$, which will happen when it is roughly 25 billion years old (see e.g. Fig 1 of &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0310808&quot; rel=&quot;nofollow noreferrer&quot;&gt;Davis &amp;amp; Lineweaver 2004&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnitude of the dipole of the CMB is $3.4\,\mathrm{mK}$ (&lt;a href=&quot;http://arxiv.org/abs/astro-ph/9312056&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kogut et al. 1993&lt;/a&gt;), i.e. $\sim10^{-3}$ of the peak value. The primordial fluctuations are even lower, roughly a factor $10^{-5}$ of the peak value. The dipole is partly due to our motion round the Milky Way, and so will fluctuate somewhat with the period of a Galactic year, which is 225 million years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure shows how the peak frequency evolved in the past. Its current value is 160.4 Ghz. To calculate how a human would perceive the sound at a given time, I suppose you'll have to multiply the spectrum with the frequency-dependent transmission function of the human ear, but that's physiology which I know nothing about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EX0pT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EX0pT.png&quot; alt=&quot;CMBfreq&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure was produced by calculating the frequency at a given time as&#xA;$$&#xA;\nu(t) = \frac{160.4\,\mathrm{GHz}}{a},&#xA;$$&#xA;where $a$ runs from $1/1100$ to $4$,&#xA;and calculating the age by integrating the &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Friedmann equation&lt;/a&gt;, assuming a &lt;a href=&quot;http://arxiv.org/abs/1502.01589&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck 2015&lt;/a&gt; cosmology.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-07-08T14:52:28.700" LastActivityDate="2016-07-08T14:52:28.700" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16514" PostTypeId="2" ParentId="16470" CreationDate="2016-07-08T16:37:08.353" Score="3" Body="&lt;p&gt;The answer depends on what you mean with &quot;precisely&quot;.  The &lt;a href=&quot;http://adsabs.harvard.edu/abs/1983A&amp;amp;A...124...50C&quot; rel=&quot;nofollow&quot;&gt;ELP theory&lt;/a&gt; seems to have an accuracy of ~10&lt;sup&gt;-6&lt;/sup&gt;&amp;deg; in longitude over a century, which comes down to about a millisecond for phases.&lt;/p&gt;&#xA;" OwnerUserId="286" LastActivityDate="2016-07-08T16:37:08.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16515" PostTypeId="2" ParentId="16509" CreationDate="2016-07-08T17:03:07.057" Score="4" Body="&lt;p&gt;The limit to the neutron star mass is &lt;em&gt;at least&lt;/em&gt; $2M_{\odot}$, since at least two have measured masses as large as this (e.g. &lt;a href=&quot;http://www.nature.com/nature/journal/v467/n7319/full/nature09466.html&quot; rel=&quot;nofollow&quot;&gt;Demorest et al. 2010&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to your question is the same as the answer to what is the maximum mass of a neutron/quark star, since if you compress matter of any sort, this is ultimately what it will become.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to this question is also &lt;em&gt;unknown&lt;/em&gt; and depends on the uncertain equation of state (the relationship between pressure, density and composition) at ultra-high densities, but must be somewhere between the $2M_{\odot}$ I mentioned above and an upper limit of around $3.5M_{\odot}$, which is imposed by General Relativity and an equation of state where the speed of sound equals the speed of light (the hardest possible equation of state). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether these highly compressed objects are &quot;solid&quot; is also a matter of debate and research. The conventional picture of an old-ish white dwarf is that it is indeed a crystalline solid. The bulk of the interior of neutron stars of moderate mass is almost certainly a fluid of neutrons along with some protons and electrons. The neutrons &lt;em&gt;may&lt;/em&gt; solidify at very high densities and this is thought to be one of the &quot;harder&quot; options for the equation of state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer a comment by @zephyr - cold in this context means that the Fermi energies of degenerate fermion species are very much higher than $kT$. White dwarfs and neutron stars (older than a few seconds) are &quot;cold&quot;, despite having interior temperatures, in the case of the latter, of a billion degrees. Making them even colder does not change the size/density of the object but does enable the interior of a white dwarf to crystallise.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-07-08T21:30:50.653" LastActivityDate="2016-07-08T21:30:50.653" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16516" PostTypeId="2" ParentId="20906" CreationDate="2016-07-08T17:18:29.943" Score="3" Body="&lt;p&gt;I did this manually some years ago by listing the BSC numbers of stars and drawing lines between them.  The resulting data file can be found at the bottom of this page: &lt;a href=&quot;http://hemel.waarnemen.com/download/#cetera&quot; rel=&quot;nofollow&quot;&gt;BSC constellation lines&lt;/a&gt;.  As @james-screech mentioned, there is no standard way to draw the constellations, so this is just one (my) interpretation.  To give you some idea, maps (in Dutch) using these data can be found here: &lt;a href=&quot;http://hemel.waarnemen.com/kaarten/vastekaart.php?f=noordpool.jpg&quot; rel=&quot;nofollow&quot;&gt;NP&lt;/a&gt;, &lt;a href=&quot;http://hemel.waarnemen.com/kaarten/vastekaart.php?f=equator_08-00-16.jpg&quot; rel=&quot;nofollow&quot;&gt;Eq 0h&lt;/a&gt;, &lt;a href=&quot;http://hemel.waarnemen.com/kaarten/vastekaart.php?f=equator_20-12-04.jpg&quot; rel=&quot;nofollow&quot;&gt;Eq 12h&lt;/a&gt;, &lt;a href=&quot;http://hemel.waarnemen.com/kaarten/vastekaart.php?f=zuidpool.jpg&quot; rel=&quot;nofollow&quot;&gt;SP&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="286" LastActivityDate="2016-07-08T17:18:29.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16517" PostTypeId="2" ParentId="16510" CreationDate="2016-07-08T21:47:29.530" Score="1" Body="&lt;p&gt;I suspect that everything you want and more is available and written in python or has python wrappers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astropy.org/&quot; rel=&quot;nofollow&quot;&gt;Astropy&lt;/a&gt;&#xA;&lt;a href=&quot;https://github.com/astropy/ccdproc&quot; rel=&quot;nofollow&quot;&gt;ccdproc&lt;/a&gt;&#xA;&lt;a href=&quot;https://pypi.python.org/pypi/photutils&quot; rel=&quot;nofollow&quot;&gt;photutils&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-07-08T22:01:30.437" LastActivityDate="2016-07-08T22:01:30.437" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16518" PostTypeId="2" ParentId="16470" CreationDate="2016-07-08T23:08:45.590" Score="3" Body="&lt;p&gt;The motion of the moon is known with great accuracy, and so the exact moment of a new moon (the moment that the sun and moon have the same longitude in the sky) can be calculated with great accuracy for tens of thousands of years into the future. This is the astronomical meaning of New moon, and the only one which is relevant here. The uncertainty is due to effects such as the non-spherical nature of the Earth's gravity and the exact magnitude of perturbations of the orbit by other bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless there is an eclipse, the moon is not visible at the moment of the new moon, and doesn't become visible until some time later (&lt;a href=&quot;https://astronomy.stackexchange.com/questions/14265/how-soon-could-a-waxing-crescent-moon-be-seen?rq=1&quot;&gt;How Soon Could a Waxing Crescent Moon Be Seen?&lt;/a&gt;). The time when the moon can be seen for the first time depends on lots of factors, such as atmospheric clarity, weather and latitude, these can't be predicted from day to day.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2017-02-22T13:46:18.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16519" PostTypeId="2" ParentId="20906" CreationDate="2016-07-08T23:23:55.233" Score="1" Body="&lt;p&gt;You can find the constellations here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.iau.org/public/themes/constellations/&quot; rel=&quot;nofollow&quot;&gt;http://www.iau.org/public/themes/constellations/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at the gif for each constellation and you will find out which stars are connected (there is also a text file for the boundaries of each constellation in RA and Dec, though this is not what you want).&lt;/p&gt;&#xA;" OwnerUserId="2339" LastActivityDate="2016-07-08T23:23:55.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16521" PostTypeId="2" ParentId="16499" CreationDate="2016-07-09T13:53:15.970" Score="3" Body="&lt;p&gt;Well, I can't do the math, but if an intuitive answer is OK:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If a planet is tidally-locked, can it have an obliquity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm going to cheat and give you an answer using moons, because many moons are are tidally locked to their planets.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Io is a funny one, it's slightly eccentric orbit and proximity to Jupiter causes it to resurface pretty regularly.   It's axial tilt isn't listed and might not be known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Europa_(moon)&quot; rel=&quot;nofollow&quot;&gt;Europa&lt;/a&gt;.  Axial tilt  0.1°&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Ganymede_(moon)&quot; rel=&quot;nofollow&quot;&gt;Ganymede&lt;/a&gt;.   Axial tilt 0–0.33° &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Callisto_(moon)&quot; rel=&quot;nofollow&quot;&gt;Callisto&lt;/a&gt;.  Axial tilt zero&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Titan_(moon)&quot; rel=&quot;nofollow&quot;&gt;Titan&lt;/a&gt;, to Saturn,  Axial tilt zero&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Triton_(moon)&quot; rel=&quot;nofollow&quot;&gt;Triton&lt;/a&gt;, to Neptune,  Axial tilt zero&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our &lt;a href=&quot;https://en.wikipedia.org/wiki/Moon&quot; rel=&quot;nofollow&quot;&gt;Moon&lt;/a&gt; is a bit of an oddball, it has an Axial Tilt that's 1.5424° to it's orbit.   click here for &lt;a href=&quot;https://en.wikipedia.org/wiki/Moon#/media/File:Earth-Moon.PNG&quot; rel=&quot;nofollow&quot;&gt;picture&lt;/a&gt;.   Why our Moon varies a bit more than the others, I'm not sure, but it might be the proximity to the sun which keeps the Moon from orbiting more directly over Earth's equator and causes some imperfection in it's tidal lock.  (Kind of a guess, I'm not sure on this part)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, simple answer, Tidal locking with solid bodies by definition, the same side of the planet or moon faces the object it orbits all of the time.   Obliquity doesn't make much sense if the planet's primary rotation is in sync with it's orbit around an object.   (Tidal Locked planets do in fact rotate).  So, it seems counter intuitive that a planet could have a high obliquity and at the same time, be tidally locked to the object it orbits.  Mercury and Venus, work as examples for this common sense argument too.  Mercury is, in a sense tidally locked but due to it's eccentric orbit it's in a 3:2 ratio.  Mercury has 0 obliquity and Venus appears to be on it's way to being tidally locked to the sun and it has a 177.36 degree obliquity, but that's essentially 2.64 degrees, if Venus wasn't rotating in the opposite direction of the other planets.   As Venus moves into being tidally locked to the sun, it's rotation should approach 0 (or 180) obliquity, the slowing of it's rotation in comparison to it's rotation matching it's orbit suggests that this is the only likely outcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to check the axial tilt of other moons, more listed &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_locking#List_of_known_tidally_locked_bodies&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there may be some factors that could, just maybe, lead to exceptions, perhaps A planet with very high inclination might be a wild card.  Or a rotation that's in a 2:1 resonance with the tidal locking.  A binary star system could have some odd orbits too (See Pluto's and Charon's, satellite Nix &lt;a href=&quot;http://www.space.com/29559-pluto-moons-weird-orbit-chaos.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  Now, Nix isn't spherical, so it's prone to much more pronounced oddities, but a planet around a binary star could exhibit some strange orbital resonances, but I suspect using a binary star system to find an exception isn't what you had in mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm 90%-95% confident in saying that obliquity should tend towards zero as the planet becomes tidally locked to a single star that it orbits.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A long way in the future it will be tidally locked. As it slows, will&#xA;  the obliquity also be eroded or will other forces (such as the orbit&#xA;  of the moon) allow it to maintain an obliquity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a tricky one because if the sun wasn't going to go Red Giant, the Earth would eventually become tidally locked to the Moon (in 10-50 billion years), and not the sun.   If the Earth didn't have a moon, then it could get tidally locked to the sun, in, oh, I don't know, 100-300 billion years as a guess.   The Moon is the dominant tidal force on the Earth, so in it's current set-up, the Earth can't get tidally locked to the sun.   If the Moon was smaller, then maybe somebody could run the numbers on this question.   If the Moon was 1/3rd it's current mass then the Lunar and Solar tidal forces would be similar, and . . . then it becomes an interesting question.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Curious sidebar, while the Moon currently stabilizes Earth's obliquity, that's not always going to be the case.  As the Moon moves further from the Earth, the Earth's obliquity is expected to show much more pronounced fluctuations, but that doesn't have anything to do with the Earth being tidally locked to the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.  Corrections welcome.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-07-09T15:17:33.720" LastActivityDate="2016-07-09T15:17:33.720" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16523" PostTypeId="2" ParentId="699" CreationDate="2016-07-09T16:51:49.317" Score="5" Body="&lt;p&gt;This question is little different from asking why, if the Sun is a ball of Hydrogen, it hasn't already exploded (or burned out). It is difficult for me to take it seriously. But I'll try. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methane and Oxygen are gases and will react at room temperature and pressure in certain concentration ranges and when exposed to a spark (or other high energy density discharge; a flame, friction, etc.). We call these limits the LEL and UEL (abbreviations for Upper and Lower Explosion Limits) for hydrogen which change as the atmospheric concentration of Oxygen changes. A tragic example of this was the enormously stupid decision by NASA to use pure oxygen in the Apollo space capsules. During testing of Apollo 1, three astronauts burned to death in a flash fire in which the extreme oxygen concentration allowed almost all plastic materials (wire jackets, clothing, etc.) to burn intensely. An urban myth I have heard, but can neither confirm nor contradict, is that if you held a cigarette lighter under your finger in a pure O2 atmosphere, it would start to burn, and continue to burn even after the lighter was put out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature of a gas is an indication of how fast the atoms/molecules are going. As you lower the temperature, the average speed declines. For the $\mathrm{CH_4+O_2}$ reaction to proceed, you need the methane and the oxygen molecules' collision to impact hard enough to break chemical bonds. As temperature drops, this becomes increasingly unlikely. OTOH, the reaction also releases energy, so if there is enough energy both from the speed of the molecules and the energy released from the reaction products to induce more that one additional reaction, then you will have a chain reaction and the result will depend on whether the supply of both molecule types is large enough to sustain the reaction. It can self-quench, stay in a steady state (form a flame) or explode. Note that while on Earth, the most common reaction in the air is using oxygen as the oxidizer (one of the reactants), methane can react with many other gasses, for instance $\mathrm{F_2}$, fluorine or $\mathrm{Cl_2}$, Chlorine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there is virtually no oxidizer present in Titan's atmosphere, methane can't burn, even if the temperature, pressure, and concentration were high enough.&lt;/p&gt;&#xA;" OwnerUserId="13442" LastEditorUserId="2365" LastEditDate="2016-07-09T19:15:07.763" LastActivityDate="2016-07-09T19:15:07.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16524" PostTypeId="2" ParentId="646" CreationDate="2016-07-09T17:51:06.183" Score="3" Body="&lt;p&gt;There's a little bit too much sophistry in some of the answers to this question. While it is true a photon doesn't experience time, the OP was asking about light emitted from Proxima Centuri as observed from Earth. Since PC is 4 light-years away, the light took 4 years to reach us - since neither we nor the Centuri system are traveling, relative to each other, at relativistic speeds (near light-speed; somewhere, depending on context, between ~5% of c and 20% of c is where we start talking about the speeds being relativistic). Some of the light was absorbed (by dust, or ions) between there and here and was probably re-emitted as infrared light, but most of it (the visible part) traveled the whole way uninterrupted, so yes it did begin its journey 4 years ago. But note that Proxima isn't visible to the naked eye, so its not a very robust example.&lt;/p&gt;&#xA;" OwnerUserId="13443" LastActivityDate="2016-07-09T17:51:06.183" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16525" PostTypeId="1" AcceptedAnswerId="16528" CreationDate="2016-07-09T21:43:03.870" Score="0" ViewCount="2536" Body="&lt;p&gt;Planets sweeps out equal areas in equal times. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the area being calculated here the area of a triangle? You draw a straight line from the center of the sun to the earth at point A. The Earth moves to Point B and you have another straight line from point A to point B and the finally another straight line from point B to the Sun? Even though the Earth makes an arc, I guess we make it a straight line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess my question is whether I have understood the second law correctly? That the area of a triangle was being calculated in an elliptical orbit, no matter how elliptical the orbit is. &lt;/p&gt;&#xA;" OwnerUserId="13444" LastActivityDate="2020-11-12T11:07:54.487" Title="Kepler's second law of motion" Tags="&lt;planet&gt;&lt;kepler&gt;&lt;johannes-kepler&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16526" PostTypeId="2" ParentId="16525" CreationDate="2016-07-09T22:02:38.973" Score="4" Body="&lt;p&gt;That is not correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area is the total area between the two radius lines, so there is a curved side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you have two points almost 180 degrees from each other. Using just a triangle, the area is close to zero. Now, two points placed closely together can have the same area in between them. Then you have an equal are, but &lt;em&gt;not&lt;/em&gt; equal time, so your hypothesis can not be correct.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastActivityDate="2016-07-09T22:02:38.973" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16527" PostTypeId="1" CreationDate="2016-07-09T23:29:36.383" Score="0" ViewCount="6066" Body="&lt;p&gt;I have always been observing something that looks like a star but not as bright as a star moving from south to north at the same time every night. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is it? It cannot be a star because I know stars do not move.&lt;/p&gt;&#xA;" OwnerUserId="13445" LastEditorUserId="10172" LastEditDate="2016-07-10T15:35:50.503" LastActivityDate="2016-07-10T15:35:50.503" Title="What are the small things, look like stars, moving in the sky?" Tags="&lt;artificial-satellite&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2016-07-11T12:12:18.340" ContentLicense="CC BY-SA 3.0" />
  <row Id="16528" PostTypeId="2" ParentId="16525" CreationDate="2016-07-10T00:41:27.767" Score="2" Body="&lt;p&gt;Hohmannfan's answer is correct, but as I understand your question, you do understand the general idea correctly and the answer to your question is yes, no matter how eccentric the orbit, the planet spans equal areas over equal time, at least nearly perfectly. (more on that later)&lt;/p&gt;&#xA;&lt;p&gt;You're mistake is in calling the sectors &amp;quot;triangles&amp;quot;.    They're sectors of an ellipse from a focal point, not from the center.  I'm not sure what the mathematical term is for those shapes, but calling them triangles, while convenient, is incorrect, calling them sectors of an ellipse is problematic too because a sector without clarification implies it's initiated from the center point, not a focal point.   So, near as I could find, &amp;quot;Sector from a focal point of an ellipse&amp;quot; is correct, but long winded.  &amp;quot;Kepler area&amp;quot; would probably do the trick.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; just puts up a picture and calls them &amp;quot;shaded areas&amp;quot; or &amp;quot;shaded sectors&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://raw.githubusercontent.com/gist/PM2Ring/8c07b38fdf153f3f6e2a1d3e43890c23/raw/b1238c7da17b3ee26005ddb1b5381ebac26a7dad/KeplerEllipse2.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JTTWU.png&quot; alt=&quot;Kepler Ellipse&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;As for deviations from Kepler's law.  Mercury is close enough to the sun that relativity caused an observed deviation from it's predicted elliptical orbit.  Late 19th and early 20th century astronomers tried to explain that by adding a planet (&lt;a href=&quot;https://en.wikipedia.org/wiki/Vulcan_(hypothetical_planet)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vulcan&lt;/a&gt;), but they couldn't find that theoretical planet.   Einstein's laws later explained Mercury's observed deviation.&lt;/p&gt;&#xA;&lt;p&gt;Also, planets pull on each other and that's been known at least since Newton's time, as he did a fair bit of work on it but never solved it to his satisfaction.  Earth, for example, speeds up if Venus and Jupiter are ahead of it, and slows down when they are behind it.  These are called orbital perturbations and are simple in principal, quite complicated mathematically.   An unexplained deviation in Uranus' elliptical orbit lead to the discovery of the planet Neptune.  That's not a flaw in Kepler's law.  His law is so exact that when a deviation to an elliptical orbit is observed, we know there's another gravitational body out there and we even know about where to look.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="16685" LastEditDate="2020-11-12T11:07:54.487" LastActivityDate="2020-11-12T11:07:54.487" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="16529" PostTypeId="1" AcceptedAnswerId="16533" CreationDate="2016-07-10T09:25:15.993" Score="3" ViewCount="875" Body="&lt;p&gt;I heard that Saturn has 62 moons with confirmed orbits, 53 of which have names and only 13 of which have diameters larger than 50 kilometers. Some come within 1km of each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do these moons keep in Saturn's orbit without hitting each other, and someday will these moons collide with each other and be destroyed?&lt;/p&gt;&#xA;" OwnerUserId="11792" LastEditorUserId="2365" LastEditDate="2016-07-10T10:14:11.573" LastActivityDate="2016-07-25T14:13:48.417" Title="Will Saturn's Moons be destroyed?" Tags="&lt;solar-system&gt;&lt;saturn&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16530" PostTypeId="2" ParentId="16527" CreationDate="2016-07-10T10:23:00.577" Score="4" Body="&lt;p&gt;A starlike point visibly moving across the sky is a satellite. The biggest and brightest is the international space station, but there are many others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They can be seen any night that is clear, usually shortly after dusk, or before dawn. &lt;a href=&quot;http://www.heavens-above.com/&quot;&gt;http://www.heavens-above.com/&lt;/a&gt; allows you to see predictions of when satellites will be overhead.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-07-10T10:23:00.577" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16531" PostTypeId="2" ParentId="16529" CreationDate="2016-07-10T12:06:11.730" Score="8" Body="&lt;p&gt;The various moons are on orbits that are mostly a lot more than 1km apart from other moons. The orbits are stable, so the moons just keep going round Saturn on the same orbit. The moons are far enough apart that they have negligible gravitational influence on other moons. So the moons never get close to each other and don't collide. There are moons, like Daphnis and Atlas that can come with 1000km of each other, but they are very small. No moons come within 1km of each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some exceptions to this: Janus and Epimetheus orbit at nearly the same distance from Saturn. If they stayed like this they would collide within each other within about 4 years. However as one moon starts to catch up with the other, it is pulled forwards, causing it to move out, and slow down. Letting the other moon move inwards and faster. The two moons essentially swap orbits. Although there is only about 50km between the orbits, the two moons never approach closer than 10000km, so are not at any risk of collision. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moon collisions can take place, one theory on the origin of the rings is they they were formed from a moon that broke up, either from a collision or by Saturn's gravity (It's not the most popular or recent theory). Collisions have never been observed, nor are any forecast to occur for as long as we can predict.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the 62 moons, remember that some of them are extremely small: a few tens of metres across, and there is no clear division between &quot;small moon&quot; and &quot;large clump of ring material&quot;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-07-10T12:06:11.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16533" PostTypeId="2" ParentId="16529" CreationDate="2016-07-10T14:27:53.283" Score="4" Body="&lt;p&gt;Indeed, the situation is temporary on a scale of hundreds of millions of years, with reoccuring instabilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moons don't run into each other in the short term, but the orbits affect each other and are affected by The sun and other bodies in the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.seti.org/weeky-lecture/orbital-dynamics-suggests-recent-formation-saturns-moons&quot; rel=&quot;nofollow&quot;&gt;this presentation by Matija Cuk&lt;/a&gt; for details. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Recently the age of Saturn's moons interior to Titan, previously thought to be as old as Saturn, also became actively debated. I will show how computer simulations of the past orbital dynamics of Saturn's moons Tethys, Dione and Rhea can tell us how long they have been around. &lt;strong&gt;It appears that the inner moons and rings of Saturn are only about 100 million years&lt;/strong&gt;, equivalent to the Cretaceous period on Earth. I will also discuss how the present moons likely originated from debris resulting from a major orbital instability in which the previous generation of icy moons was destroyed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You ask “someday will these moons collide with each other and be destroyed?” so the answer is &lt;strong&gt;yes&lt;/strong&gt;, at least the inner group. So having 62 (a lot of them) is not the issue; the outer part is stable over billions of years.&lt;/p&gt;&#xA;" OwnerUserId="7142" LastActivityDate="2016-07-10T14:27:53.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16534" PostTypeId="1" CreationDate="2016-07-10T16:28:27.677" Score="7" ViewCount="2812" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Sphere_of_influence_(black_hole)&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Sphere_of_influence_(black_hole)&lt;/a&gt; the gravitational sphere of influence of a supermassive black hole is really limited compared to the size of its hosting galaxy, why then is there one at the center of every galaxy?&lt;/p&gt;&#xA;" OwnerUserId="2244" LastEditorUserId="2244" LastEditDate="2016-07-11T21:52:31.413" LastActivityDate="2016-09-03T22:48:27.873" Title="Why is there a supermassive black hole at the center of every galaxy?" Tags="&lt;black-hole&gt;&lt;galaxy&gt;&lt;cosmology&gt;&lt;supermassive-black-hole&gt;&lt;galaxy-center&gt;" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="16535" PostTypeId="2" ParentId="10463" CreationDate="2016-07-10T21:50:21.150" Score="4" Body="&lt;p&gt;An infalling observer cannot remain stationary inside the event horizon of a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to them, there is a short period of proper time before they meet whatever lies at the centre of the black hole (a singularity in GR).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the observer looks back, they do not see an infinite amount of time pass by in the outside universe. In fact for the type of black hole where they might (a) survive and (b) have time to make any observations (i.e. a supermassive black hole) then they would see the outside universe only for a relatively short time longer than their experienced proper time and certainly for nowhere near as long as the Hawking radiation evaporation timescale. So the possibility that the black hole is (slowly) evaporating is of no consequence to their final moments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The necessary material and diagrams to understand the previous paragraph have been covered before on both Physics and Astronomy SE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/82678/does-someone-falling-into-a-black-hole-see-the-end-of-the-universe&quot;&gt;https://physics.stackexchange.com/questions/82678/does-someone-falling-into-a-black-hole-see-the-end-of-the-universe&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/2524/would-time-go-by-infinitely-fast-when-crossing-the-event-horizon-of-a-black-hole&quot;&gt;Would time go by infinitely fast when crossing the event horizon of a black hole?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-07-10T21:50:21.150" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16536" PostTypeId="1" AcceptedAnswerId="18789" CreationDate="2016-07-11T04:57:42.200" Score="7" ViewCount="3862" Body="&lt;p&gt;My title doesn't do my question justice in explaining, but the kind of satellite system I'm thinking of is two rings on two separate axis, much like (if you've seen it) the film 'Treasure Planet' shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wnc1D.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wnc1D.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is no, as the ring of a planet matches it's axis, but I was wondering whether or not some other phenomenon or influence could cause this formation.&lt;/p&gt;&#xA;" OwnerUserId="13458" LastEditorUserId="-1" LastEditDate="2016-07-11T18:59:56.387" LastActivityDate="2016-10-20T11:40:46.127" Title="Could a cross-ringed planet exist?" Tags="&lt;planetary-ring&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16537" PostTypeId="2" ParentId="16536" CreationDate="2016-07-11T06:42:16.133" Score="10" Body="&lt;p&gt;While such a scenario would be unlikely, all you'd need is an existing ring and an icy moon to enter inside the Roche limit at a measurably different inclination. I wouldn't want to guess how long a set-up like that would be stable, but it's possible to have one.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="38" LastEditDate="2016-07-11T21:45:15.393" LastActivityDate="2016-07-11T21:45:15.393" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16538" PostTypeId="2" ParentId="16467" CreationDate="2016-07-11T07:11:13.440" Score="3" Body="&lt;p&gt;There is the unusual white dwarf &lt;a href=&quot;http://www.aanda.org/articles/aa/abs/2015/12/aa27261-15/aa27261-15.html&quot; rel=&quot;nofollow&quot;&gt;RX J0439.8−6809&lt;/a&gt;.  It seems to be a carbon-oxygen white dwarf with a temperature of around 250,000 K!  It is 9200 pc away.  You might want to look for other PG 1159 stars to see how far they can be detected.&lt;/p&gt;&#xA;" OwnerUserId="7937" LastActivityDate="2016-07-11T07:11:13.440" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16539" PostTypeId="2" ParentId="16534" CreationDate="2016-07-11T07:57:40.357" Score="0" Body="&lt;p&gt;The gravitational &quot;pull&quot; of a black hole is not limited. In fact the gravitational force of attraction of every particle in the Universe effects every other particle in the Universe, irrespective of their distance of separation or mass. Or to be more exact every particle distorts the fabric of time/space throughout the Universe. The distortion may me small the further away from the particle you get but it is still there according to General Relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum theory will put a of limit on things, in that the when the distortion is reduced to the Plank length it will be lost in the general background noise.&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-07-11T07:57:40.357" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16540" PostTypeId="1" CreationDate="2016-07-11T12:12:55.077" Score="0" ViewCount="62" Body="&lt;p&gt;Why are planets moving in the first place? Gravity causes them to orbit, but why move at all? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;note&lt;/em&gt;: this is not a question on why planets orbit each other, I know the reason for that is gravity. What I'm asking is why do planets move at all? For example, if you turned off gravity, why would the planets carry on moving apart rather than be stationary? &lt;/p&gt;&#xA;" OwnerUserId="13308" LastActivityDate="2016-07-11T14:27:12.853" Title="Why aren't planetary bodies static?" Tags="&lt;movement&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2016-07-11T17:06:06.087" ContentLicense="CC BY-SA 3.0" />
  <row Id="16541" PostTypeId="1" AcceptedAnswerId="16544" CreationDate="2016-07-11T13:07:46.583" Score="2" ViewCount="207" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FCUb8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FCUb8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In relation to the image above, although the topic relates to math, how can I visualise the situation described in the final equation presented, i.e., the declinations, ascensions, solar and lunar radii. A diagram would be of great help.&lt;/p&gt;&#xA;" OwnerUserId="13460" LastActivityDate="2016-07-13T01:56:12.813" Title="Solar Eclipse: How it occurs" Tags="&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16542" PostTypeId="2" ParentId="16540" CreationDate="2016-07-11T14:27:12.853" Score="1" Body="&lt;p&gt;It's simply because the sun and planets were formed out of a big pile of dust.  &lt;strong&gt;Originally the dust was spinning&lt;/strong&gt;.  So once the dust became planets, &lt;strong&gt;it kept spinning around&lt;/strong&gt;.  That's all there is to it - that's why the solar system is spinning around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's was once a pile of dust that was spinning a bit; it's still spinning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then ask,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;For example, if you turned off gravity, why would the planets carry on moving?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;THat's just the same as asking &quot;if I spin something around on a string, and break the string, why does it keep moving?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For that matter, it's the same as asking &lt;strong&gt;&quot;What is momentum?&quot;&lt;/strong&gt;  So, if you push something ...... why does it keep moving?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this stage in history, we have &lt;strong&gt;utterly no clue, at all, what the heck time, space, matter, and momentum are&lt;/strong&gt;.  &quot;Why does momentum do what it does?&quot; is for now just one of those super-deep questions like &quot;What is time&quot; or &quot;So what caused the big bang&quot; or &quot;What is gravity&quot; or &quot;What's the explanation for this quantum stuff?&quot; or indeed ... &quot;What is momentum&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &lt;strong&gt;why do the planets keep moving?: answer &quot;momentum&quot;&lt;/strong&gt;.  If you want to know &quot;what causes momentum?&quot;, that is one of the basic total mysteries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For now, nobody has a clue. You may as well ask ... what is space, what is time, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding momentum, you might like to read up about the so-called &quot;Mach's conjecture&quot;. The famous Einstein was, like you, fascinated by the question &quot;WTF is momentum anyway?&quot; One sort of general thinking-point originating with this smart guy called Mach is that momentum could have something to do with &quot;all the other mass in the universe&quot;. Nobody has a clue about this, and it's just a vague general idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isaac Newton was a pretty smart guy (if a bit whacky), and in the end he found gravity so mysterious, he just put it down to God. He probably found momentum as mysterious, and he was one of the first guys to think about it clearly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting point: actually everything astronomers look at (all galaxies, all structures of galaxies) in fact does not (!) behave the way small things (our solar system, as in your question) behaves in terms of gravity and momentum.  This is usually explained by invisible unknown matter (&quot;dark matter&quot;) or for a few scientists, that gravity works differently than we think presently.  So the fact is with issues like gravity/momentum you ask about ... not only do we have no clue &quot;why momentum works&quot; but when you look through a telescope, issues like &quot;orbits&quot; work totally differently anyway!&lt;/p&gt;&#xA;" OwnerUserId="13071" LastActivityDate="2016-07-11T14:27:12.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16544" PostTypeId="2" ParentId="16541" CreationDate="2016-07-12T01:28:30.497" Score="0" Body="&lt;p&gt;For a small region not far from the equator,&#xA;an &lt;a href=&quot;https://en.wikipedia.org/wiki/Equirectangular_projection&quot; rel=&quot;nofollow noreferrer&quot;&gt;equirectangular projection&lt;/a&gt; will do.&#xA;If you want celestial north to be up,&#xA;declination should increase from bottom to top,&#xA;and right ascension should increase from right to left.&#xA;To correct the aspect ratio, scale the RA axis by cos(dec).&#xA;Then the sun and moon are circles of the respective angular radii, without the RA scale factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the equation for $z$, $\Delta$ is the declination of the sun, $\delta$ is the declination of the moon, and $a$ is the difference in their right ascensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/j2Lh0.png&quot; alt=&quot;visualization of partial solar eclipse&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2016-07-13T01:56:12.813" LastActivityDate="2016-07-13T01:56:12.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16546" PostTypeId="1" AcceptedAnswerId="16611" CreationDate="2016-07-12T15:38:36.930" Score="9" ViewCount="3966" Body="&lt;p&gt;The atmospheric composition of both planets are very similar. Then why are Jupiter's bands more visible (North and South equatorial belt, etc.):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Fcdvb.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fcdvb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whilst Saturn's are not: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xE13v.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xE13v.jpg&quot; alt=&quot;except for the storm of course&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(except for the storm of course)&lt;/p&gt;&#xA;" OwnerUserId="9949" LastEditorUserId="1056" LastEditDate="2016-07-19T06:21:25.640" LastActivityDate="2016-07-19T06:21:25.640" Title="Why are Saturn bands much fainter than Jupiter's?" Tags="&lt;planet&gt;&lt;jupiter&gt;&lt;planetary-atmosphere&gt;&lt;saturn&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16547" PostTypeId="1" AcceptedAnswerId="16548" CreationDate="2016-07-12T15:44:29.973" Score="5" ViewCount="469" Body="&lt;p&gt;In the past, weren't many astronomical telescope mirrors periodically removed and 'resilvered', and wasn't this actually done using silver?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing that this is much less common now - new telescopes are so large and complex that routine invasive procedures like that needed for maintenance of a silver coating would not be acceptable, and mirrors in many if not most older telescopes would have been 'updated' to aluminum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was silver used in the past because aluminum coating was too technically difficult, or because silver has slightly better reflectivity in the red, or some other reason?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;note:&lt;/em&gt; I'm only asking about 'optical' telescopes meaning NIR &amp;amp; Vis, not the longer wavelength IR telescopes &lt;a href=&quot;https://physics.stackexchange.com/a/197079/83380&quot;&gt;where IR emissivity is a concern&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-07-12T17:10:03.447" Title="Was the use of silver on 'optical' telescope mirrors more common in the past? Why?" Tags="&lt;telescope&gt;&lt;observatory&gt;&lt;mirror&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16548" PostTypeId="2" ParentId="16547" CreationDate="2016-07-12T17:10:03.447" Score="7" Body="&lt;p&gt;Aluminium coating is a relatively recent process - it became available around the 1920s or 1930s. The Hale telescope arrived just in time to take advantage of this new technology.  (It requires a reasonably good vacuum to work, which probably explains why it took a while to come along.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before that - around the mid-1800s - various chemical &quot;silvering&quot; processes, usually involving real silver, were used. These weren't ideal as silver tarnished over time - the silver literally oxidised on exposure to air.  (Aluminium also reacts with air, but forms a thin oxide layer which doesn't harm the image and prevents further oxidation.)  Also some of these silvering processes were either toxic or explosive in application...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before that - the first telescope mirrors were made of metal that was cast and ground to the required curve then highly polished. The material was &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Speculum_metal&quot;&gt;Speculum&lt;/a&gt;&quot; a copper-tin alloy. They were poor in quality; far less reflective than later mirrors.  &lt;a href=&quot;http://parsonstown.info/engineering/telescopes&quot;&gt;This account&lt;/a&gt; gives a nice description of casting mirrors for the Parsonstown telescope using over a ton of molten metal. Two mirrors were necessary for the larger telescopes because the mirror in use had to be taken out and re-polished so often...&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2016-07-12T17:10:03.447" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16550" PostTypeId="1" AcceptedAnswerId="16553" CreationDate="2016-07-12T17:40:41.760" Score="3" ViewCount="225" Body="&lt;p&gt;I have a question. I studied &lt;a href=&quot;https://en.wikipedia.org/wiki/Bondi_accretion&quot; rel=&quot;nofollow&quot;&gt;Bondi's model for spherical accretion&lt;/a&gt;. I know the assumptions made in the model are big and that they are rarely representative of what happens in astrophysics. What I wanted to ask is: is there a situation that suits this model even approximately, or it is a total idealization? Is there always an accretion disk, or may there be a situation where the accretion is spherical?&lt;/p&gt;&#xA;" OwnerUserId="13436" LastEditorUserId="2153" LastEditDate="2016-07-12T21:52:44.533" LastActivityDate="2016-07-13T13:36:27.240" Title="Spherical Accretion (Bondi Model)" Tags="&lt;accretion-discs&gt;&lt;accretion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16551" PostTypeId="1" AcceptedAnswerId="16554" CreationDate="2016-07-12T18:26:05.983" Score="12" ViewCount="190" Body="&lt;p&gt;2015 RR245 was recently announced, in a highly eccentric orbit in the KBO. From what I understand, the so called &quot;Planet 9&quot; was predicted because these objects tended to be pointed in a particular direction. How is this object pointed relative to the other KBOs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7uJTS.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7uJTS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/B9i2Z.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B9i2Z.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="182" LastActivityDate="2016-07-13T07:37:22.753" Title="How does 2015 RR245 compare to the other KBOs predicting &quot;Planet 9&quot;?" Tags="&lt;9th-planet&gt;&lt;dwarf-planets&gt;&lt;kuiper-belt&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16552" PostTypeId="1" AcceptedAnswerId="16555" CreationDate="2016-07-12T18:53:18.090" Score="4" ViewCount="195" Body="&lt;p&gt;Since about 1990 astronomers been able to detect planets around other stars, using a couple different techniques, which is amazing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By this point, do we know whether planets form &lt;strong&gt;only around single systems, only around multiple systems, or both&lt;/strong&gt;??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Surprisingly you can't find this general issue explained in popular writing. I think it's &quot;obvious background knowledge&quot; to experts, hence just doesn't get outlined explicitly in popular writing.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as of mid 2016:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) About how many exoplanets do we know of (say, with good certainty),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) In fact, how many (if any) of these are around single-star systems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(iii) How many (if any) are around multiple-star systems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If indeed it is possible to have planets in multiple systems: is it just a planet going around one of the stars, or is &quot;all the way around both&quot; known?&lt;/p&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="13071" LastEditDate="2016-07-13T17:36:13.237" LastActivityDate="2016-09-25T01:58:56.657" Title="Do planets form around lone stars, multiple stars, or both? Do we know this yet?" Tags="&lt;exoplanet&gt;&lt;binary-star&gt;" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="16553" PostTypeId="2" ParentId="16550" CreationDate="2016-07-12T21:51:43.733" Score="3" Body="&lt;p&gt;One example is that of the core of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Thorne%E2%80%93%C5%BBytkow_object&quot; rel=&quot;nofollow&quot;&gt;Thorne–Żytkow object&lt;/a&gt;, or TŻO. It consists of a high-metallicity M class red giant or supergiant with a neutron star at its core. Generally, TŻOs form when a neutron star collides with a suitable red giant/supergiant. The neutron star settles into the center of the other star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, the main source of energy for the TŻO comes from gravitational contraction of the envelope as gas from inside the star falls toward the surface of the neutron star. In this case, the change in mass $\dot{M}$ of the neutron star is (see &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1977ApJ...212..832T&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow&quot;&gt;Thorne &amp;amp; Żytkow (1977)&lt;/a&gt;):&#xA;$$\dot{M}=4\pi r^2\rho v_{\text{in}}\mathscr{R}\tag{1}$$&#xA;The difference between $(1)$ and the underlying equation behind Bondi's formula is the factor of $\mathscr{R}$, a correction factor for relativistic redshift, which is given by&#xA;$$\mathscr{R}=\sqrt{1-\frac{2GM_{tc}}{c^2r}}\tag{2}$$&#xA;where $M_{tc}$ is the total core mass. This formula may look familiar; it comes from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_metric&quot; rel=&quot;nofollow&quot;&gt;Schwarzschild metric&lt;/a&gt;. You need $\mathscr{R}$ to properly describe accretion onto any compact object. This is something needed not just in the case of TŻOs. However, Bondi's formula can be used nonetheless, if you multiply by $\mathscr{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, Bondi accretion can be used without relativistic compensations &lt;a href=&quot;https://ned.ipac.caltech.edu/level5/March09/Edgar/frames.html&quot; rel=&quot;nofollow&quot;&gt;in a number of ways&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mass transfer in certain binary systems, where an envelope moves matter from one star to another.&lt;/li&gt;&#xA;&lt;li&gt;Protostellar clusters, where many young protostars move about, accreting gas.&lt;/li&gt;&#xA;&lt;li&gt;Galaxy clusters, where galaxies move through the intergalactic medium (IGM).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-07-13T13:36:27.240" LastActivityDate="2016-07-13T13:36:27.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16554" PostTypeId="2" ParentId="16551" CreationDate="2016-07-13T07:37:22.753" Score="8" Body="&lt;p&gt;It was the first thing I thought too, but I would say it won't count in the statistics.  The earlier &lt;a href=&quot;http://arxiv.org/pdf/1601.05438v1.pdf&quot;&gt;paper by Batygin &amp;amp; Brown&lt;/a&gt; was talking about bodies orbiting further out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.minorplanetcenter.net/db_search/show_object?object_id=2015%20RR245&quot;&gt;2015 RR245 orbits&lt;/a&gt; between 33.7 and 129 AU.  The &quot;cluster&quot; of bodies considered in the paper have aphelion distances from 150 AU upwards.  So I think it would be a mistake to try and fit this new body into the existing figures.&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2016-07-13T07:37:22.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16555" PostTypeId="2" ParentId="16552" CreationDate="2016-07-13T15:29:07.537" Score="3" Body="&lt;p&gt;There are currently 3424 planets identified (including those in the Solar System and unconfirmed exoplanets). Of these, 3310 are confirmed exoplanets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out of the 3424, 180 are in binary (or multiple) star systems. Of those, 28 are circumbinary. I am not certain if any circumternary planets have been identified, though I know some circumternary disks have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is more difficult to identify circum-&lt;em&gt;n&lt;/em&gt;-ary planets than those that only revolve around one star, so these numbers might not precisely represent the actual proportions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.openexoplanetcatalogue.com/&quot; rel=&quot;nofollow&quot;&gt;Open Exoplanet Catalogue&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6" LastActivityDate="2016-07-13T15:29:07.537" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16556" PostTypeId="2" ParentId="16466" CreationDate="2016-07-13T15:42:11.397" Score="1" Body="&lt;p&gt;Yes, &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetosphere_of_Jupiter&quot; rel=&quot;nofollow&quot;&gt;Jupiter&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetosphere_of_Saturn&quot; rel=&quot;nofollow&quot;&gt;Saturn&lt;/a&gt; have their own radiation belts.  Jupiter's is extremely powerful, which is why all of &lt;a href=&quot;http://www.nasa.gov/image-feature/goddard/2016/infographic-juno-built-to-withstand-intense-radiation-environments&quot; rel=&quot;nofollow&quot;&gt;Juno's&lt;/a&gt; sensitive electronics had to be radiation hardened and shielded inside a &lt;a href=&quot;http://www.nasa.gov/mission_pages/juno/news/juno20100712.html&quot; rel=&quot;nofollow&quot;&gt;vault&lt;/a&gt;.  It's also why Juno is going to be on a polar orbit doing flybys of Jupiter, instead of orbiting around it closely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These magnetic bands are indeed separated into inner and outer portions.  From &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetosphere_of_Jupiter#Size_and_shape&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Jupiter's magnetosphere is traditionally divided into three parts: the inner, middle and outer magnetosphere. The inner magnetosphere is located at distances closer than 10 RJ from the planet. The magnetic field within it remains approximately dipole, because contributions from the currents flowing in the magnetospheric equatorial plasma sheet are small. In the middle (between 10 and 40 RJ) and outer (further than 40 RJ) magnetospheres, the magnetic field is not a dipole, and is seriously disturbed by its interaction with the plasma sheet (see magnetodisk below).&#xA;  Jupiter's inner magnetic belt was &lt;a href=&quot;http://www.nasa.gov/mission_pages/juno/multimedia/pia03478.html&quot; rel=&quot;nofollow&quot;&gt;mapped by Galileo&lt;/a&gt;, showing it contains high-energy electrons.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetosphere_of_Jupiter#Role_of_Io&quot; rel=&quot;nofollow&quot;&gt;Eruptions from Io&lt;/a&gt; result in the middle portion of the magnetosphere containing huge amounts of ions of various elements.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetosphere_of_Saturn#Size_and_shape&quot; rel=&quot;nofollow&quot;&gt;Saturn's&lt;/a&gt; magnetosphere is likewise divided into separate parts.  However, there are 4 regions to Saturn's magnetic field, and they are separated by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetosphere_of_Saturn#Magnetospheric_regions&quot; rel=&quot;nofollow&quot;&gt;kinds of particles they contain&lt;/a&gt;.  These particles result from interactions with Saturn's various moons, as well as its ring system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Galileo probe discovered that Europa does produce &lt;a href=&quot;https://en.wikipedia.org/wiki/Europa_(moon)#Composition&quot; rel=&quot;nofollow&quot;&gt;a weak magnetic field&lt;/a&gt;, as a result of its subsurface ocean.  This ocean has an induced current created from Europa moving through Jupiter's magnetosphere, which in turn creates a magnetic field.  This magnetic field is actually one of the things which caused many scientists to theorize a subsurface ocean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, these belts are not &quot;Van Allen&quot; belts.  That name is reserved strictly for Earth's radiation belts.&lt;/p&gt;&#xA;" OwnerUserId="13426" LastActivityDate="2016-07-13T15:42:11.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16557" PostTypeId="2" ParentId="16408" CreationDate="2016-07-13T15:58:32.603" Score="2" Body="&lt;p&gt;So the instrument in question is actually an antenna being built and managed by &lt;a href=&quot;http://www.asianscientist.com/2016/07/tech/astron-isis-moon-mission-chinese-satellite-chang-e-4/&quot; rel=&quot;nofollow&quot;&gt;Dutch scientists&lt;/a&gt;.  The Chinese satellite will carry it behind the moon.  It will make measurements at 30 MHz.  As you have said in your question, this portion of the EM spectrum is &lt;a href=&quot;http://www.nrl.navy.mil/rsd/7210/7213/lfra-from-space&quot; rel=&quot;nofollow&quot;&gt;largely unexplored&lt;/a&gt;, thanks to us not being able to view at this wavelength from Earth.  There are a &lt;a href=&quot;http://sci.esa.int/science-e/www/object/doc.cfm?fobjectid=53829&quot; rel=&quot;nofollow&quot;&gt;number of possible things&lt;/a&gt; (pdf of a PowerPoint presentation) to do with such an antenna.  However, with regards to the origin of the universe, at such a frequency, they can make&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;pathfinder measurements of the red-shifted HI line that originates from&#xA;   before the formation of the first stars (dark ages, reionization)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is one of the ways to examine the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrogen_line#In_cosmology&quot; rel=&quot;nofollow&quot;&gt;state of the early universe&lt;/a&gt;, before stars had begun to form.  The afterglow of heat from the Big Bang had faded, and most of the universe was composed of Hydrogen.  However, this hydrogen had yet to form stars, so everything was dark - hence the name &quot;the dark ages&quot;. However, examining this HI line will allow scientists to map the early universe just after recombination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is being done from behind the moon, because the astronomers want to use the Moon to block radio transmissions from Earth which might interfere with the signal they receive.&lt;/p&gt;&#xA;" OwnerUserId="13426" LastEditorUserId="13426" LastEditDate="2016-07-13T17:17:22.717" LastActivityDate="2016-07-13T17:17:22.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16558" PostTypeId="2" ParentId="11539" CreationDate="2016-07-13T18:47:31.243" Score="1" Body="&lt;p&gt;I saw this &lt;a href=&quot;https://www.youtube.com/watch?v=QuzndVmiuv8&amp;amp;feature=youtu.be&amp;amp;t=704&quot; rel=&quot;nofollow&quot;&gt;presentation&lt;/a&gt; of the National Academy of Science 2015 where Dr. Robin Canup explains, illustrated by simulations, that many large moons formed sequentially around giant planets. But their growing mass thru accretion caused perturbations in the gaseous planetary disk (&quot;type I interaction&quot;) that drew them inwards to collide with the planet within a million years. The total mass of all the satellites to the planet's mass remains fairly constant given (partly) the density of the Solar Nebula. The larger moons crash into the planet while new moons are growing larger. The process stops when the Solar Nebula has been exhausted. So I suppose speculatively that different exoplanetary systems might have different moon size ceilings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Impact formation size of Earth's Moon, in the middle of the Jovian moons, is a coincidence according to this model.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2016-07-13T18:57:14.523" LastActivityDate="2016-07-13T18:57:14.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16559" PostTypeId="1" CreationDate="2016-07-13T19:46:44.773" Score="2" ViewCount="102" Body="&lt;p&gt;I know that there is plenty of theory to predict the size of a neutron star:  my question is whether or not there are any reliable size estimates based on observation. Please note that, to be reliable, such an observation would need to have also a good estimate for the neutron star's distance from us, which must necessarily be reckoned from some independent source, i.e., it must &lt;em&gt;not&lt;/em&gt; be reckoned from the apparent size of the object, or from theory, otherwise one is guilty of circular reasoning!&lt;/p&gt;&#xA;" OwnerUserId="13290" LastEditorUserId="2153" LastEditDate="2016-07-13T21:22:44.677" LastActivityDate="2016-07-15T14:28:50.627" Title="Are there any reliable optical measurements of the radius of a neutron star?" Tags="&lt;observational-astronomy&gt;&lt;neutron-star&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16561" PostTypeId="1" CreationDate="2016-07-13T20:29:28.110" Score="0" ViewCount="125" Body="&lt;p&gt;If one can estimate the mass of our universe, might one also calculate its ultimate size, (after it fully expands), based on the Schwarzschild formula for the radius of a black hole? Please note that this simple formula &#xA;$$R=\frac{2GM}{c^2}$$ &#xA;informs us that larger black holes are less dense. So, if it turns out that our universe is in fact a gigantic black hole, then one would expect it to be extremely low-density!&lt;/p&gt;&#xA;" OwnerUserId="13290" LastEditorUserId="2153" LastEditDate="2016-07-13T21:18:30.357" LastActivityDate="2016-07-14T16:39:53.873" Title="Can one calculate the size (i.e. the radius) of our universe?" Tags="&lt;black-hole&gt;&lt;universe&gt;&lt;supermassive-black-hole&gt;&lt;observable-universe&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2016-07-15T13:16:15.167" ContentLicense="CC BY-SA 3.0" />
  <row Id="16562" PostTypeId="2" ParentId="16559" CreationDate="2016-07-13T21:15:05.480" Score="3" Body="&lt;p&gt;Yes, the neutron star &lt;a href=&quot;https://en.wikipedia.org/wiki/RX_J1856.5-3754&quot; rel=&quot;nofollow&quot;&gt;RX J1856.5-3754&lt;/a&gt; has an observed* radius of 17 km. After accounting for general relativity, its actual radius is calculated as 14 km. It is not the only &lt;a href=&quot;https://en.wikipedia.org/wiki/Category:Neutron_stars&quot; rel=&quot;nofollow&quot;&gt;neutron star&lt;/a&gt; whose radius is known.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;* It's worth mentioning how astronomers arrived at this radius. Technically, fitting the X-ray data to a blackbody arrives at a calculation of a 5 km radius, which seems unusually small, so for a time it was believed that this star may be something more exotic, like a quark star. However, after atmospheric modeling based on temperature, a more accurate estimate of the emission radius was derived to be 17 km. Once gravitational redshift was accounted for, the value of 14 km was yielded. See &lt;a href=&quot;https://arxiv.org/pdf/astro-ph/0612145v1.pdf&quot; rel=&quot;nofollow&quot;&gt;Magnetic Hydrogen Atmosphere Models and the Neutron Star RX J1856.5−3754 - Wynn C. G. Ho, et al.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorUserId="6" LastEditDate="2016-07-15T14:28:50.627" LastActivityDate="2016-07-15T14:28:50.627" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16563" PostTypeId="2" ParentId="16561" CreationDate="2016-07-13T21:42:11.897" Score="1" Body="&lt;p&gt;If I understand the question the answer is, no, you can't estimate the final size of the Universe by calculating the schwarzschild radius of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the Universe is not known, and may be infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not know if the observable universe will ever stop expanding, however the evidence suggests that it won't, and so won't have an &quot;ultimate&quot; size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the observable universe is of the order $10^{53}kg$, mostly comprised of dark energy and dark matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A black hole with that mass would have a radius of $2MG/c^2 = 10^{26}m$, which as has been observed is on the same order of magnitude as the (comoving) radius of the observable universe. Whether this has any cosmological implications is not clear. It doesn't imply that the Universe is a black hole. It doesn't say anything about the Universe at all, as this calculation was for the observable universe. It doesn't say anything about the &quot;ultimate size&quot; of the universe or the observable universe (which is expected to continue expanding without limit) &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-07-13T21:42:11.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16564" PostTypeId="1" CreationDate="2016-07-13T23:57:46.380" Score="1" ViewCount="84" Body="&lt;p&gt;With all of the footage we have of total lunar eclipses, has NASA, ESA, or any other space program ever taken a photo or video of it from the Moon itself? &lt;/p&gt;&#xA;" OwnerUserId="13480" LastActivityDate="2016-07-14T00:05:04.550" Title="Has anyone filmed a total lunar eclipse from the Moon's perspective?" Tags="&lt;the-moon&gt;&lt;lunar-eclipse&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16565" PostTypeId="2" ParentId="16564" CreationDate="2016-07-14T00:05:04.550" Score="5" Body="&lt;p&gt;What can you say but &quot;Wow...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.space.com/32245-total-solar-eclipse-video-dscovr.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.space.com/32245-total-solar-eclipse-video-dscovr.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;The moon's dark shadow travels across Earth's face in spectacular new views of Tuesday's (March 8) [2016] total solar eclipse captured by a satellite in deep space.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aUKFX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aUKFX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry my answer is off by about 700,000 miles compared to what you asked :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm fairly sure there did not happen to be, a time when NASA had cameras on the moon, when, there was a lunar eclipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that cameras today (such as this one on &lt;code&gt;DSCOVR&lt;/code&gt;) are unbelievably better than cameras from &lt;code&gt;Apollo&lt;/code&gt; era, so this would blow away any merely 700,000 miles closer imagery captured during the Apollo era.&lt;/p&gt;&#xA;" OwnerUserId="13071" LastActivityDate="2016-07-14T00:05:04.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16566" PostTypeId="1" AcceptedAnswerId="16575" CreationDate="2016-07-14T06:18:30.463" Score="5" ViewCount="255" Body="&lt;p&gt;Suppose a hot Jupiter is collided into parent star due to tidal force, does the life of the star becomes longer due to extra hydrogen, or becomes shorter due to extra mass?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastActivityDate="2016-07-14T21:51:18.850" Title="If a hot Jupiter collides with the parent star, does it enhance or reduce the life of the star?" Tags="&lt;star&gt;&lt;hot-jupiter&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16569" PostTypeId="1" AcceptedAnswerId="16574" CreationDate="2016-07-14T13:34:08.483" Score="3" ViewCount="191" Body="&lt;p&gt;The thought &lt;strike&gt;crossed&lt;/strike&gt; kidnapped my mind one night almost a year ago. I was laying down, but I was very interested in thinking about the secrets of a black-hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://news.nationalgeographic.com/news/2014/02/140218-black-hole-blast-explains-big-bang/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Nat. Geo. - News article&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does the current state of research tell us about the idea that black holes might form universes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't forget to backup your answer!&lt;/p&gt;&#xA;" OwnerUserId="13316" LastEditorUserId="6" LastEditDate="2016-07-14T13:39:48.787" LastActivityDate="2016-07-14T19:13:29.193" Title="How well supported are the theories of black-holes containing their own universes?" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16570" PostTypeId="1" AcceptedAnswerId="16576" CreationDate="2016-07-14T13:51:33.250" Score="4" ViewCount="169" Body="&lt;p&gt;What is the history behind the name that we gave to &lt;a href=&quot;https://en.wikipedia.org/wiki/Grand_design_spiral_galaxy&quot;&gt;that class&lt;/a&gt; of galaxies?&lt;/p&gt;&#xA;" OwnerUserId="2969" LastEditorUserId="6" LastEditDate="2016-07-14T13:54:22.747" LastActivityDate="2016-07-14T22:31:19.207" Title="Where did the grand design spiral galaxy designation come from?" Tags="&lt;galaxy&gt;&lt;classification&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16574" PostTypeId="2" ParentId="16569" CreationDate="2016-07-14T18:50:07.947" Score="3" Body="&lt;p&gt;More and more of modern theoretical physics deals with unphysical models and ideas. The only basis we have of predicting what is happening beyond an event horizon, ANY event horizon, is what we understand about what happens around us, on this side of that horizon. Feel perfectly safe in making up anything you wish about what the Physics is like beyond the event horizon, we'll never be able to contradict you. We'll never be able to falsify any claim made for events on the other side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Purple people eaters? Why not? Ghosts and Gods? Sure. 1+1 = 3? Yup, that too. The event horizons in nature are the natural limits to our Universe. The edge, if you will. Unless and until we are able to observe consequences of the claims you make (called verifiable predictions) you're not dealing with Science, at least not what I regard as Science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now, we know of 4 types of edges to our Universe:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The edge at ~1E-44 seconds after the hypothetical Creation Event&lt;/li&gt;&#xA;&lt;li&gt;The event horizon of any black hole&lt;/li&gt;&#xA;&lt;li&gt;the event horizon at about 15-20 billion light-years distance (at which space is expanding at greater than the speed of light and&lt;/li&gt;&#xA;&lt;li&gt;the edge of our Observable Universe (from which no signals will ever reach us).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Physics is about what goes on on this side of these &quot;edges&quot;. Metaphysics, philosophy, religion, opinion are what you may use to deal with those things beyond, but don't call it physics or science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I am sympathetic to the idea that there may be an (as yet unknown) physical law which prevents singularities from actually forming. In fact, we have only weak indirect evidence that event horizons actually exist, and absolutely none that singularities form. Generally, when a physical theory and its mathematical expression leads to infinities, we change the theory - see renormalization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to know about multiverses or what is beyond the event horizons of black holes? OK. I want to know about the color of the nail polish that the tortoise carrying the Universe on its back uses. Both are equally valid areas of inquiry. See for example &lt;a href=&quot;https://en.wikipedia.org/wiki/Multiverse#Proponents_and_skeptics&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Multiverse#Proponents_and_skeptics&lt;/a&gt; ... and links found there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should also mention that you know an article which talks about events prior to time zero (which I call the Creation Event) are metaphysics and/or religious. None of them, I repeat: none of them are based on known Physics. (Why? because our known physics breaks down at about 10E-44 seconds, extrapolating back to t=0 isn't possible.) Sure they make an interesting story, but they should be regarded in the same way as any fiction. You get to choose which you like and which you don't without any regard for ever having to justify your choices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much of human knowledge is about narratives which allow us to understand our place in the Universe. It is part of our human nature to be curious and seek out narratives that make sense of our world. The set of scientific narratives differ from this set because the goal is different. They doesn't require our understanding (for instance, Feynman said: &quot;I think I can safely say that nobody understands quantum mechanics.&quot; (1965) Little has changed since then.), they only require our ability to apply them to make/verify observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Creating a satisfying narrative does not require the scientific method(s) be enforced. But ignore them at your peril.&lt;/p&gt;&#xA;" OwnerUserId="13487" LastEditorUserId="6" LastEditDate="2016-07-14T19:13:29.193" LastActivityDate="2016-07-14T19:13:29.193" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16575" PostTypeId="2" ParentId="16566" CreationDate="2016-07-14T21:51:18.850" Score="4" Body="&lt;p&gt;The short answer is pela's comment: more mass equals shorter life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a bit of a longer explanation...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way the life of the star could be prolonged is if the new material can be transported to the core.  This requires convection.  Most stars are not fully convective, having distinct layers of either convective or radiative zones. The net effect of the collision in this case is to simply add more mass to the star as a whole while adding no new fuel to the core (or a burning shell); ergo, shorter lifespan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But some stars do (or may) have fully convective stages, such as the main sequences of very small (aka red dwarf) stars and very large stars with enough metals for a dominant CNO cycle.  So in such a situation the material of the colliding body would be transported to (and from) the core, giving more fuel to burn.  However, the star still has the same, increased mass, regardless of where the &quot;new mass&quot; currently is inside the star, resulting in higher core temperature. The net effect of which is that the increase to the fusion rate exceeds the proportional mass increase.  So, shorter lifespan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic rule here being that core temperatures are dictated by the gravity well of the star, which is dictated by the mass (and a few other GR contributions, but usually predominantly mass).  The more mass, the deeper the well. So adding more mass just means the core (and areas near it) are in a deeper gravity well, meaning the temperatures go up.  Stellar fusion reaction rates are approximately non-linear polynomials in the temperature; some go as far as being proportional to $T^{40}$, the 40-th power of the temperature!  It's this non-linearity which results in the fundamental relation of the short answer: more mass equals shorter life.  Increasing the mass by a given proportion increases the fusion rates by a (much) larger proportion.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2016-07-14T21:51:18.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16576" PostTypeId="2" ParentId="16570" CreationDate="2016-07-14T22:23:00.127" Score="4" Body="&lt;p&gt;The term can be traced back to the theoretical work of C.C.Lin in explaining the spirals as due to density waves in galactic discs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://adsabs.harvard.edu/full/1970IAUS...38..377L&quot;&gt;paper from 1970&lt;/a&gt; discusses the &quot;Existence of 'Grand Design'&quot;. Note the use of &quot;scare quotes&quot;, suggesting that the term was new and would not be widely recognised by his readers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://Chia-Chiao%20Lin&quot;&gt;Chia-Chiao Lin&lt;/a&gt; was an important Chinese/American Mathematician. It is reasonable to believe that the term was coined by him or one of the mathematicians that he worked with at MIT. And was in use since about 1964, when the first of a series of papers &quot;On the structure of Spiral Galaxies&quot; was published.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-07-14T22:31:19.207" LastActivityDate="2016-07-14T22:31:19.207" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16577" PostTypeId="1" AcceptedAnswerId="41680" CreationDate="2016-07-15T03:17:47.490" Score="3" ViewCount="172" Body="&lt;p&gt;Of course the &lt;a href=&quot;http://phys.org/news/2016-07-biggest-galactic-dark-energy.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;newly announced SDSS-III data and maps&lt;/a&gt; from the &lt;a href=&quot;https://i.stack.imgur.com/ZR16m.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sloan Digital Sky Survey&lt;/a&gt; with 1.2 million objects, along with all previous data sets are openly available, and I'm sure there are many tools to access, work with, and view the data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the simplest way to extract a list of coordinates of galaxies in order to just try to visualize it on my own? There may be visualization tools also, if you want to add a link that would be great, but this question is about getting a list of coordinates so I can look at density or even try to plot one dot per galaxy in some slice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing the coordinates might be available in RA, dec, and redshift, and possibly also some calculated/inferred x, y, z. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use python, but I'm not familiar with &lt;a href=&quot;http://www.astropy.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;AstroPy&lt;/a&gt; yet, so if it can be done, albeit inefficiently and/or inaccurately by my writing a straightforward python script on my laptop, that would be the most helpful answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit:&lt;/strong&gt; if doing this in AstroPy is failry easy, I do have an Anaconda installation, and therefore already have at least a basic AstroPy installation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a graphic (below), &lt;a href=&quot;http://phys.org/news/2016-07-biggest-galactic-dark-energy.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;from here in Phys.org&lt;/a&gt; attributed there to Daniel Eisenstein and SDSS-III.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZR16m.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZR16m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another graphic (below), &lt;a href=&quot;http://phys.org/news/2016-07-biggest-galactic-dark-energy.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;from here in Phys.org&lt;/a&gt; attributed there to Jeremy Tinker and SDSS-III.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9E8Pl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9E8Pl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2016-09-14T14:13:56.573" LastActivityDate="2021-02-26T01:14:31.373" Title="Are 3D coordinate data from Sloan DSS-III available &amp; easily accessible to non-pros?" Tags="&lt;galaxy-cluster&gt;&lt;python&gt;&lt;sky-survey&gt;&lt;astropy&gt;&lt;sloan-dss&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16578" PostTypeId="1" CreationDate="2016-07-15T03:47:02.997" Score="4" ViewCount="246" Body="&lt;p&gt;Will the astrometric precision of the Gaia space telescope be able to detect the gravitational influence of cold old &lt;em&gt;solitary&lt;/em&gt; neutron stars on the movements of stars? At least in a statistical sense to estimate how common they are around here, which could say something about how large the population III stars were.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if an inactive neutron star could be precisely located thanks to its gravitational influence, would it be possible to observe it directly?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2016-07-15T08:55:15.293" LastActivityDate="2016-07-25T15:32:53.380" Title="Will Gaia detect inactive neutron stars?" Tags="&lt;observational-astronomy&gt;&lt;neutron-star&gt;&lt;orbital-mechanics&gt;&lt;gaia&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16580" PostTypeId="1" AcceptedAnswerId="16585" CreationDate="2016-07-15T07:05:31.487" Score="2" ViewCount="276" Body="&lt;p&gt;In solar system,most ring systems seem almost lies exactly on the equator of planet, is it possible that the plane of a ring system is not lie on the equator of the planet?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="13359" LastEditDate="2016-07-15T09:47:18.320" LastActivityDate="2016-07-15T21:52:30.527" Title="Do planetary rings always follow rotational axis?" Tags="&lt;planetary-ring&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16581" PostTypeId="1" CreationDate="2016-07-15T07:58:00.620" Score="-5" ViewCount="152" Body="&lt;p&gt;What would happen when a planet consumed all its fuel and its core extincts.&#xA;Would anything change on the planet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this ever happen before its star did so?&#xA;If not, why?&lt;/p&gt;&#xA;" OwnerUserId="10102" LastEditorUserId="2365" LastEditDate="2018-08-01T10:19:13.800" LastActivityDate="2018-08-01T10:19:13.800" Title="What would happen when the earth consumed all its fuel?" Tags="&lt;planet&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ClosedDate="2016-07-15T13:12:52.573" ContentLicense="CC BY-SA 3.0" />
  <row Id="16582" PostTypeId="2" ParentId="16581" CreationDate="2016-07-15T10:00:25.207" Score="1" Body="&lt;p&gt;A planet does not have any fuel. Only stars &quot;burn&quot; fuel either in their cores or concentric shells around their cores.&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-07-15T10:00:25.207" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16583" PostTypeId="1" AcceptedAnswerId="28811" CreationDate="2016-07-15T10:31:10.263" Score="8" ViewCount="684" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/16548/7982&quot;&gt;This answer&lt;/a&gt; nicely summarizes the general picture of historical use of silver for mid to large size telescope mirrors, and the shift to aluminum once good quality vacuum pumps and chambers suitable for this process were available. A freshly evaporated, atomically clean aluminum surface will can sometimes violently react with oxygen, so this has to be done under high vacuum, and vented carefully with inert gas first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When did the term &quot;resilvering&quot; generally start to actually refer to aluminum coating, and not silver? Was that actually the case by the 1960's or 1970's? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I remember, evaporated aluminum will form a protective oxide layer (&quot;native oxide&quot;) of the order of 10A thick within seconds or minutes that is self limiting - no further oxygen can penetrate this layer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is so, why was re-silvering actually necessary? The cost, risk, and downtime associated with removing the primary and other mirrors every few years sounds like something you wouldn't do unless absolutely necessary.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2021-03-10T00:05:48.567" Title="When did &quot;resilvering&quot; large telescope mirrors actually refer to aluminization, and why was it necessary?" Tags="&lt;telescope&gt;&lt;observatory&gt;&lt;mirror&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16584" PostTypeId="2" ParentId="16581" CreationDate="2016-07-15T11:54:10.127" Score="2" Body="&lt;p&gt;The core of the Earth is hot because of its own density and radioactive materials. Basically, the more you compress an object, the hotter it gets. It does not undergo fusion, since only massive (stellar) objects do so. A significant amount of radioactive decay happens in the core, which is mainly the source of its heat, but no fusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, stars are hot because of their own densities, but at a much larger scale, and sans the nuclear decay. This is especially evident when stars are born as non-fusing protostars; they are prevented from getting smaller because as they do so, their core will get hotter. Hotter objects actually expand, so protostars are in a state of equilibrium; if they get any smaller, the temperatures will increase and they will become larger again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, protostars slowly get smaller and smaller over time. This is because they lose energy as they emit light. Light emission is, in fact, a source of energy loss. So the star will continue to collapse slowly, and once the core reaches a certain density, it will begin fusing hydrogen. This produces radiation pressure, which counters the force of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, let me clear up something: nucleosynthesis in stars does not cause heat, but rather it prevents the star's gravitational collapse. It's the star's density that causes such high temperatures. If a star runs out of &quot;fuel&quot;, that spells out the last phases of its life — as far as I know, the end result reaches extremely high temperatures (supernovae, white dwarfs, etc.)&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-07-15T14:46:21.310" LastActivityDate="2016-07-15T14:46:21.310" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16585" PostTypeId="2" ParentId="16580" CreationDate="2016-07-15T12:24:13.953" Score="7" Body="&lt;p&gt;Nearly always.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The asymmetric gravity of a spinning (and hence oblate) planet will induce tides that can pull small moons and ring particles into orbit around its equator in a few million years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had a moon that was for some reason not orbiting around the equator, and it broke up (for example by passing the Roche limit) then the resulting ring would not be around the equator at first. However it is unlikely that a moon would be in such an orbit in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very large moons (such as the the Moon) and those that are captured minor planets (Triton) can be in an inclined orbit. Triton is large, and was captured from the Kuiper belt. There has not been enough time for Triton to correct its orbit. Triton orbits backwards too. Such moons are comparatively far from their host planets, and so not at risk of being broken up to form a ring.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-07-15T21:52:30.527" LastActivityDate="2016-07-15T21:52:30.527" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16586" PostTypeId="2" ParentId="16492" CreationDate="2016-07-15T15:23:47.023" Score="5" Body="&lt;p&gt;No, the two limits are not the same - there is some range of masses that both white dwarfs (WDs) and neutron stars (NSs) can have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Chandrasekhar_limit&quot; rel=&quot;noreferrer&quot;&gt;Chandrasekhar mass limit&lt;/a&gt; suggests that WDs cannot be more massive than about $1.4\,M_\odot$.  However, this is true for &lt;em&gt;non-rotating&lt;/em&gt; WDs.  Rapidly rotating WDs might be &lt;a href=&quot;http://esoads.eso.org/abs/2004A%26A...419..623Y&quot; rel=&quot;noreferrer&quot;&gt;as massive as $2\,M_\odot$&lt;/a&gt;.  Accretion in a binary system could both spin up a WD and increase its mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NSs that have not undergone accretion are regularly observed with &lt;a href=&quot;http://adsabs.harvard.edu/abs/2012ApJ...757...55O&quot; rel=&quot;noreferrer&quot;&gt;masses around $1.3\,M_\odot$&lt;/a&gt;, while NSs who have 'recycled pulsars' have an observed mean mass of $\sim 1.5\,M_\odot$ according to the same paper.&lt;/p&gt;&#xA;" OwnerUserId="286" LastActivityDate="2016-07-15T15:23:47.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16587" PostTypeId="2" ParentId="16492" CreationDate="2016-07-15T16:24:55.243" Score="5" Body="&lt;p&gt;The smallest, precisely measured mass for a neutron star is now $1.174 \pm 0.004 M_{\odot}$ - &lt;a href=&quot;http://arxiv.org/abs/1509.08805&quot; rel=&quot;nofollow noreferrer&quot;&gt;Martinez et al. (2015)&lt;/a&gt;. The theoretical lower limit is more like $0.1M_{\odot}$, but there are no obvious formation channels to produce such an object. See  &lt;a href=&quot;https://physics.stackexchange.com/questions/143166/what-is-the-theoretical-lower-mass-limit-for-a-gravitationally-stable-neutron-st&quot;&gt;https://physics.stackexchange.com/questions/143166/what-is-the-theoretical-lower-mass-limit-for-a-gravitationally-stable-neutron-st&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The highest mass for a stable white dwarf (commonly called the Chandrasekhar mass) is theoretically about $1.39 M_{\odot}$ for a helium or carbon white dwarf (and a little bit lower for oxygen or neon white dwarfs), but can be increased somewhat by rotation. The observation of type Ia supernovae is strong circumstantial evidence that this limit is reached and then exceeded, probably by mass transfer onto a smaller white dwarf. The most massive, probably single, white dwarf known/measured is &quot;WD 33&quot; in the cluster NGC 2099 and has a mass of $1.28^{+0.05}_{-0.08}\ M_{\odot}$ (&lt;a href=&quot;http://arxiv.org/abs/1603.00471&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cummings et al. 2016)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, both observationally and theoretically, the maximum mass of a white dwarf is higher than the minimum mass for a neutron star.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-07-15T16:32:25.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16588" PostTypeId="1" CreationDate="2016-07-15T16:33:28.313" Score="2" ViewCount="554" Body="&lt;p&gt;Describing a galaxy 70 million light years away, I consider it incorrect to state for example &quot;it has an active black-hole at its center&quot;. It did 70 million years ago, who knows what is out there now. I could understand using the present tense for objects within the solar system and even nearby (less than say 100 ly) but not for such distant objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some convention to use present-tense for all observations, or am I being a pedant?&lt;/p&gt;&#xA;" OwnerUserId="7979" LastActivityDate="2016-07-15T19:36:11.320" Title="Why do astronomers not use past tense when discussing observations of the universe" Tags="&lt;terminology&gt;" AnswerCount="3" CommentCount="11" ClosedDate="2016-07-16T07:23:52.003" ContentLicense="CC BY-SA 3.0" />
  <row Id="16590" PostTypeId="2" ParentId="16588" CreationDate="2016-07-15T18:07:21.920" Score="4" Body="&lt;p&gt;You're being a pedant and are not necessarily correct!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The present tense could be clearly defined as what is experienced at the time of observation. To write that it had a black hole 70 million years ago would (pedantically) only be correct at one instant of time, not correct when the reader read the sentence and even ignoring that pedantry, not necessarily true for gravitationally lensed light where the light has not travelled along a straight path!&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-07-15T18:07:21.920" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16591" PostTypeId="1" CreationDate="2016-07-15T18:09:55.960" Score="4" ViewCount="107" Body="&lt;p&gt;The flyby anomaly is an occasional, unpredictable, and as yet unexplained discrepancy in acceleration of satellites when they pass close by the Earth.  It causes an increase of up to 13 mm/s during flybys but we have no understanding of it and how to predict it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The laws of centripetal force predict that due to rotation, the more dense molten material within the Earth might accumulate at the equator, since there it can move further from the axis of rotation.  This will, of course be counteracted by gravity.  Nevertherless, it will be an effect present at the equator more so than at the poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This predicts that we should expect to see a flyby anomaly of some magnitude and that it will be correlated with proximity of a flyby to the equator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody provide anomaly data and / or correlate it with angle of incidence in order to test this hypothesis?&lt;/p&gt;&#xA;" OwnerUserId="12000" LastActivityDate="2016-07-15T18:09:55.960" Title="Do differentials in the Earth's density cause the flyby anomaly?" Tags="&lt;gravity&gt;&lt;earth&gt;&lt;satellite&gt;&lt;near-earth-object&gt;" AnswerCount="0" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16592" PostTypeId="2" ParentId="16588" CreationDate="2016-07-15T18:38:21.223" Score="0" Body="&lt;p&gt;This is a great question and observation, and in fact it is not just distant events this applies to.  When I look in the road and see my car parked outside, I should be thinking &quot;it &lt;em&gt;was&lt;/em&gt; outside&quot;, half a picosecond ago.&quot;  I think it's just out of convenience that we use the present to describe all states as we perceive them now.  But in actual fact we know nothing of &quot;now&quot; other than our own thoughts and feelings.  All external stimulus is transmitted to us via various different media and through our senses, all of which take time to reach us, so if we are to be accurate, we never know anything about the present.&lt;/p&gt;&#xA;" OwnerUserId="12000" LastActivityDate="2016-07-15T18:38:21.223" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16594" PostTypeId="2" ParentId="16588" CreationDate="2016-07-15T19:36:11.320" Score="0" Body="&lt;p&gt;I wasn't surprised, after reading your title, to find that you used the word &quot;now&quot; in the body of your post. It's difficult to ask your question unless you believe that there exists something external to and independent of you which &quot;now&quot; captures. There isn't. Well, let me qualify that: as long as we're all living within a light second of each other (Mars varies between 3 and 22 light-minutes from Earth, the Moon is ~1 light-seconds away) and we're not (disparately) subject to relativistic forces, we can all share an approximate &quot;now&quot; among ourselves. Not only does our language and much of our thinking involve this now concept, but we're probably hard-wired to treat it as a fact. Seeing someone use some other norm is disconcerting and unintuitive. Lectures and books on relativity usually discuss space-time diagrams. These diagrams include light-cones and separate the Universe into 3 parts: the time-like part, the space-like part and the light-like (or null-interval) surface. For Modern Physics, which prefers frame invariant language, the logical meaning of &quot;now&quot; is the point of intersection of the forward and past light cones. This gives us an observer dependent &quot;now&quot;, and allows us to describe an observation from a black hole 70 million light years away as being BOTH 70 million years away and 70 million light-years distant. Any claim about what that black hole is doing &quot;now&quot; in the popular sense, is unfalsifiable (for the next 70 million years or so) and isn't very interesting (and very much useless) to empirical science. Its (popular) narrative, however, suffers in translation from the Astronomical/Cosmological &quot;now&quot; to the popular one.&lt;/p&gt;&#xA;" OwnerUserId="13487" LastActivityDate="2016-07-15T19:36:11.320" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16595" PostTypeId="2" ParentId="16583" CreationDate="2016-07-15T19:58:53.420" Score="8" Body="&lt;p&gt;Amateur telescope and mirror maker here. Not sure if I qualify as a &quot;citable source&quot; but anyway, here it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All metals will eventually tarnish. It may take a long time, but it will happen. The process is not entirely chemical always. Sometimes it's purely mechanical (abrasion). Other times it's in between. Surface phenomena are complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even gold-coated mirrors will tarnish. You could use gold for infrared observations. But no matter how much care you take of it, eventually you will have to re-coat it. Sure, gold lasts a pretty long time, but not even gold is forever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aluminum is covered with a natural layer of colorless sapphire (aluminum oxide), which protects the metal from further destruction - sapphire is very hard and very inert chemically. But oxygen still diffuses through sapphire (albeit extremely slowly), and so do other corrosion agents like sulphur oxides, marine salt, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, yes, replacing silver with aluminum was an improvement. But not even aluminum is forever. In practice, you will maybe double the lifetime of the reflective layer when you migrate from silver to aluminum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted that silver is still used. Aluminum is most popular nowadays, but you'll still find silver-coated mirrors (its reflectivity is higher than aluminum), gold-coated (for infrared), etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are further coating techniques, like layering silicon oxide on top of metal, which further improve the lifetime of any kind of metal.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2019-04-18T08:33:02.813" LastActivityDate="2019-04-18T08:33:02.813" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="16596" PostTypeId="2" ParentId="16577" CreationDate="2016-07-15T22:07:20.757" Score="3" Body="&lt;p&gt;SDSS DR12 &lt;a href=&quot;http://www.sdss.org/dr12/data_access/#CatalogData&quot; rel=&quot;nofollow&quot;&gt;Catalog Data&lt;/a&gt;&#xA;looks like a good starting point,&#xA;apparently pretty open to those willing and able to figure it out.&#xA;Their&#xA;&lt;a href=&quot;http://compute.sciserver.org/dashboard/Home/Index&quot; rel=&quot;nofollow&quot;&gt;SciServer Compute&lt;/a&gt;&#xA;site hosts Jupyter notebooks to query CasJobs in SQL.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Large Scale Structure galaxy catalog under&#xA;&lt;a href=&quot;http://www.sdss.org/dr12/data_access/vac/#BOSSValue-AddedCatalogs&quot; rel=&quot;nofollow&quot;&gt;BOSS value added catalogs&lt;/a&gt;&#xA;may also be relevant.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-07-15T22:07:20.757" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16597" PostTypeId="1" CreationDate="2016-07-16T08:50:27.130" Score="5" ViewCount="268" Body="&lt;p&gt;Many stars are part of constellations. Due to dark matter the stars have velocities which are nearly the same unregardless their radius of orbit around the centre of the Milky Way. Also, it takes about 200 million years to orbit this centre. But nevertheless the angle from which we look at those stars could change a bit during this orbit.&#xA;Regarding all these effects, could you say that constellations don't change at all or does it just take a long while?  And of course stars could run out of energy and explode and vanish, so that we don't see them anymore. But is this the only way the constellations could change?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="10445" LastEditDate="2016-07-16T09:31:05.613" LastActivityDate="2016-07-17T09:53:31.260" Title="Do constellations change?" Tags="&lt;star&gt;&lt;galaxy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16598" PostTypeId="2" ParentId="16597" CreationDate="2016-07-16T09:21:49.853" Score="9" Body="&lt;p&gt;The constellations change slowly, over thousands of years, due to the &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Proper_motion&quot; rel=&quot;noreferrer&quot;&gt;proper motion&lt;/a&gt;&lt;/em&gt; of the stars with respect to the Sun.  The main cause of these motions is the orbit of the stars around the centre of our Galaxy (with or without dark matter).  &lt;a href=&quot;https://youtu.be/WpOyYjhJR_U&quot; rel=&quot;noreferrer&quot;&gt;This video&lt;/a&gt; shows these changes for the constellation of Ursa Major over 400,000 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way the constellations can change is indeed through &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_evolution&quot; rel=&quot;noreferrer&quot;&gt;stellar evolution&lt;/a&gt;&lt;/em&gt; $-$ either the birth or death (or brightening/dimming) of stars.  However, the timescale for stellar evolution for naked-eye stars is typically hundreds of millions to billions of years (the Sun's life expectancy is around 12 billion years), so that this effect is much slower (and less well predictable).&lt;/p&gt;&#xA;" OwnerUserId="286" LastEditorUserId="286" LastEditDate="2016-07-17T09:53:31.260" LastActivityDate="2016-07-17T09:53:31.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16599" PostTypeId="2" ParentId="16597" CreationDate="2016-07-16T16:11:29.253" Score="6" Body="&lt;p&gt;Let me make some more general points to clear up some misconceptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun orbits around the Milky Way at a speed of around 230 km/s with respect to the Galactic centre (GC), taking around 200 million years to complete a circuit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars at a similar radius to the Sun from the GC orbit at &lt;em&gt;similar&lt;/em&gt;, but not identical speeds and may have peculiar motions inwards and outwards or upwards and downwards. These &quot;peculiar&quot; velocities with respect to the Sun have an order of magnitude of tens of km/s and increase as stars get older. In other words, there is a &lt;em&gt;dispersion&lt;/em&gt; of velocities, and each star in a constellation has its own peculiar velocity with respect to the Sun and other stars in the constellation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stars that make up the visible constellations are generally within around a thousand light years (a few hundred parsecs) of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take a star that is 100 light years away, moving at 10 km/s tangentially to our line of sight, then that star will move tangentially about 3 light years in 100,000 years. This corresponds to a change in angle on the sky of nearly 2 degrees (4 times the diameter of the full moon). Nearer (and usually brighter) stars will be more badly affected. This gives you an idea of the sort of timescales over which you might expect typical constellations to &lt;em&gt;visibly&lt;/em&gt; change, and it is much shorter than both the orbital timescale of the Sun around the Galaxy and stellar evolutionary lifetimes, which are longer than 10 million years for all but the most massive stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomers have to routinely take these motions into account by applying corrections for the &lt;a href=&quot;https://en.wikipedia.org/wiki/Proper_motion&quot;&gt;proper motion&lt;/a&gt; of nearby stars, when locating them automatically with telescopes.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-07-16T16:11:29.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16600" PostTypeId="1" AcceptedAnswerId="16601" CreationDate="2016-07-16T18:10:10.293" Score="0" ViewCount="185" Body="&lt;p&gt;I need to know does SDSS spectrum contains error spectrum as well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sdss.org/dr12/spectro/spectro_basics/#AbouttheSpectra&quot; rel=&quot;nofollow&quot;&gt;http://www.sdss.org/dr12/spectro/spectro_basics/#AbouttheSpectra&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above link says, further HDU's contain error spectrum and than tells about the inverse variance. Does inverse variance contains error spectrum? If yes, how can I extract that, if not where can I find error spectrum?&lt;/p&gt;&#xA;" OwnerUserId="11679" LastActivityDate="2016-07-16T21:41:53.193" Title="Extraction of error spectrum in Sloan Digital Sky Survey" Tags="&lt;spectroscopy&gt;&lt;spectra&gt;&lt;quasars&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16601" PostTypeId="2" ParentId="16600" CreationDate="2016-07-16T21:41:53.193" Score="1" Body="&lt;p&gt;The variance is the square of the standard deviation, so you can calculate 1$\sigma$ errors of the spectrum by taking the square root of one over the inverse variance of HDU1 (&lt;a href=&quot;https://data.sdss.org/datamodel/files/SPECTRO_REDUX/RUN2D/PLATE4/spPlate.html#hdu1head&quot; rel=&quot;nofollow&quot;&gt;https://data.sdss.org/datamodel/files/SPECTRO_REDUX/RUN2D/PLATE4/spPlate.html#hdu1head&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You will get infinity for bad pixels since their inverse variance is set to zero.)&lt;/p&gt;&#xA;" OwnerUserId="457" LastActivityDate="2016-07-16T21:41:53.193" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16603" PostTypeId="1" AcceptedAnswerId="16712" CreationDate="2016-07-16T23:21:30.257" Score="0" ViewCount="89" Body="&lt;p&gt;The &lt;a href=&quot;http://minorplanetcenter.net/mpec/RecentMPECs.html&quot; rel=&quot;nofollow&quot;&gt;Minor Planet Electronic Circulars&lt;/a&gt; (MPECs)&#xA;for July 15 through July 18 reported an unusually high number of re-observations of known trans-Neptunian objects (TNOs).&#xA;These were found in the&#xA;&lt;a href=&quot;http://pan-starrs.ifa.hawaii.edu/public/&quot; rel=&quot;nofollow&quot;&gt;Pan-STARRS&lt;/a&gt; archive;&#xA;most had not otherwise been observed since discovery.&#xA;Is this activity part of the &lt;a href=&quot;http://www.findplanetnine.com/&quot; rel=&quot;nofollow&quot;&gt;search for Planet 9&lt;/a&gt;,&#xA;another observing program such as &lt;a href=&quot;http://www.ossos-survey.org/&quot; rel=&quot;nofollow&quot;&gt;OSSOS&lt;/a&gt;,&#xA;a result of software improvements,&#xA;or just someone clearing a backlog?&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2016-07-27T15:52:23.213" LastActivityDate="2016-07-27T17:46:51.557" Title="Observed data: Why so many MPECs about TNOs lately?" Tags="&lt;asteroids&gt;&lt;kuiper-belt&gt;&lt;tno&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16605" PostTypeId="1" AcceptedAnswerId="16610" CreationDate="2016-07-17T00:12:35.000" Score="3" ViewCount="117" Body="&lt;p&gt;Recently, a fascinating triple-star system with an exoplanet has been &lt;a href=&quot;http://science.sciencemag.org/content/early/2016/07/06/science.aaf9671&quot; rel=&quot;nofollow&quot;&gt;discovered&lt;/a&gt; (&lt;a href=&quot;https://arxiv.org/abs/1607.02525&quot; rel=&quot;nofollow&quot;&gt;arxiv&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm very surprised about the temperature of the exoplanet: it is roughly 850 kelvin, even though its orbit is ~82 AU away from the main star (~1.5 $M_☉$), and &gt;250 AU from the other two stars. The planet itself has ~4 jupiter-masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why can such a planet be so hot, for instance compared to Jupiter? I would have expected it to be very cold because it loses a lot of energy via radiation (that's actually the reason why the group discovered it via direct imaging), and it does not get much energy from its stars.&lt;/p&gt;&#xA;" OwnerUserId="2662" LastEditorUserId="9301" LastEditDate="2016-07-17T02:11:34.040" LastActivityDate="2016-07-17T08:22:49.750" Title="Why is the exoplanet HD 131399Ab so hot?" Tags="&lt;exoplanet&gt;&lt;temperature&gt;&lt;star-systems&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16607" PostTypeId="1" AcceptedAnswerId="16657" CreationDate="2016-07-17T02:50:49.570" Score="5" ViewCount="578" Body="&lt;p&gt;Performance of astronomical mirrors that are exposed to the environment for long periods of time can experience a slow degradation of performance. They usually need to receive some kind of regular maintenance, which may involve cleaning and/or re-coating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While &lt;a href=&quot;https://astronomy.stackexchange.com/q/16583/7982&quot;&gt;I'm told that&lt;/a&gt; the aluminum degrades and must be replaced from time to time, this is not consistent with my understanding of the behavior of aluminum in air. I'm pretty sure that a freshly evaporated layer of pure aluminum will quickly develop a native oxide layer of perhaps 20Å of aluminum oxide as amorphous $\text{Al}_2 \text{O}_3$ (alumina) which is then pretty much impermeable to diffusion of anything else the atmosphere might expose it to, including various pollutants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can understand that it might in some cases ultimately be simpler to strip and replace the aluminum after a number of cleanings &lt;em&gt;due to accumulated damage from the cleaning&lt;/em&gt;, but is there any evidence that the aluminum must be replaced due to chemical degradation of the aluminum itself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this article, &lt;a href=&quot;http://articles.adsabs.harvard.edu//full/1997PASP..109..303M/0000303.000.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Reflectivity Degradation Rates of Aluminum Coatings at the CFHT&lt;/em&gt;, Magrath, B. 1997, Pub. Ast. Soc. Pac. &lt;strong&gt;109&lt;/strong&gt;, 303&lt;/a&gt; that seems to show that - at least at one observatory location - a representative test mirror exposed to the observing environment (in the dome) that was washed once a month showed no significant degradation in reflectivity over 32 months, similarly to a mirror that remained protected but not cleaned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the test mirror that was exposed and not washed showed a continuous loss in reflectivity, dropping by about 10% over the same period. A fourth test mirror undergoing regular $\text{CO}_2$ cleaning lost about 5% reflectivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a small, limited test of course, less than three years, only one location, one wavelength, and only one parameter (reflectivity). But it gives a bit of support to the idea that &lt;em&gt;while mirrors get dirty and require regular cleaning, the aluminum itself may not actually deteriorate.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So I'm asking:&lt;/strong&gt; &lt;em&gt;Do primary mirrors in large observatories undergo regular removal and re-coating of the aluminum? Why?&lt;/em&gt;, where &quot;why&quot; addresses the specific need for the removal and replacement. If these primaries actually do receive regular re-coatings of aluminum, is it because the aluminum actually degrades?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for an answer that carries some backup references. If just for example, your answer is of the form &lt;em&gt;You silly goose, everybody knows aluminum tarnishes like the Dickens!&lt;/em&gt;, please back it up with a reference to measurements of aluminum tarnishing. Thanks! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://articles.adsabs.harvard.edu//full/1997PASP..109..303M/0000303.000.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Magrath 1997&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;GIF (&quot;blink comparator&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XIvTX.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XIvTX.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://global.kyocera.com/fcworld/charact/chemistry/chemiresist.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://global.kyocera.com/fcworld/charact/chemistry/chemiresist.html&lt;/a&gt; - alumina can potentially be remarkably resistant to chemical attack compared to other materials - note logarithmic scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RftTf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RftTf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The individual figures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/abvnq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/abvnq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Abn42.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Abn42.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9yDgk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9yDgk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-07-21T10:57:29.583" Title="Do primary mirrors in large observatories undergo regular removal and re-coating of the aluminum? Why?" Tags="&lt;telescope&gt;&lt;observatory&gt;&lt;mirror&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16608" PostTypeId="1" AcceptedAnswerId="17899" CreationDate="2016-07-17T06:01:44.217" Score="3" ViewCount="307" Body="&lt;p&gt;What are the velocity, mass, and charge distribution of the solar wind.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Near the earth within the magnetosphere in the ecliptic&lt;/li&gt;&#xA;&lt;li&gt;Near the earth but outside the magnetosphere in the ecliptic&lt;/li&gt;&#xA;&lt;li&gt;Outside the ecliptic at 1 AU&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I would like to understand the energy content of the solar wind, and how it compares with the solar radiation(solar constant is $1360\, Wm^{-2}$)?&lt;/p&gt;&#xA;" OwnerUserId="13500" LastActivityDate="2016-08-12T14:51:57.257" Title="Energy carried by solar wind" Tags="&lt;the-sun&gt;&lt;solar-wind&gt;&lt;photons&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16610" PostTypeId="2" ParentId="16605" CreationDate="2016-07-17T08:14:16.157" Score="4" Body="&lt;p&gt;It's a giant planet and it's young. p.3 of the paper you reference gives a likely age of 16 million years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat arises from gravitational contraction and it would be a similar temperature even if it were further away from the stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a ball of gas contracts, the virial theorem establishes that half the gravitational potential lost is radiated away and half heats up the gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter would also have been much hotter in the past and even now emits more radiation than it receives from the Sun. The rate of contraction (and hence the planet's luminosity and temperature) decreases as the planet ages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Direct imaging surveys for giant planets deliberately target young stars for this reason.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-07-17T08:22:49.750" LastActivityDate="2016-07-17T08:22:49.750" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16611" PostTypeId="2" ParentId="16546" CreationDate="2016-07-17T08:25:43.517" Score="9" Body="&lt;p&gt;I'll give this one a shot. Correction is welcome.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Upper atmosphere temperature.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not just elements that give a planet color, but the temperature of elements. When we examine what a planet looks like, we're basically talking about reflected sunlight from the planet's surface or atmosphere. With Earth, its atmosphere is transparent enough that its surface can be seen from space. However, it is the only planet in the Solar System where that is the case. The other planets have thick atmospheres, so all we see is basically just high altitude clouds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as I understand it, the pictures of planets are often enhanced to make any distinctions more apparent. So if you actually flew above Jupiter, its lines might not be quite so distinct as the pretty pictures you'll usually find on the web. Regardless, that doesn't change your question; its lines are still far, far more distinct than Saturn's are and Galileo was able to see its Red Spot, so I think it's safe to say its lines are pronounced, even if the pictures we get are doctored a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So anyway, Jupiter is at the right distance from the Sun to undergo atmospheric transition from ice to gas. As &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Jupiter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; puts it,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The upper ammonia clouds visible at Jupiter's surface are organized in&#xA;  a dozen zonal bands parallel to the equator and are bounded by&#xA;  powerful zonal atmospheric flows (winds) known as jets. The bands&#xA;  alternate in color: the dark bands are called belts, while light ones&#xA;  are called zones. Zones, which are colder than belts, correspond to&#xA;  upwellings, while belts mark descending air. The zones' lighter&#xA;  color is believed to result from ammonia ice; what gives the belts&#xA;  their darker colors is not known with certainty&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ammonia ice, like all ice, is very reflective, so the colder bands with ice are lighter. Ammonia gas is &lt;a href=&quot;https://en.wikipedia.org/wiki/Ammonia&quot; rel=&quot;nofollow noreferrer&quot;&gt;transparent&lt;/a&gt;, but anyone who has ever flown over the ocean knows that if you have enough of a transparent thing (water), it has a distinct color. While the quote from Wikipedia above says the reason for the darker colors is uncertain, less ice in its uppermost atmosphere means less reflected light and darker color.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn, Neptune and Uranus are far enough away from the Sun where they always have ice in their upper most atmospheres, so they have less variation in color.  Earth's clouds are also primarily ice (not water vapor), so that is basically the answer. Jupiter is the right distance from the Sun for its upper atmosphere to transition and have distinct bands, some with ice, some without.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;But why are the bands in straight lines?&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is due to the &lt;a href=&quot;http://csep10.phys.utk.edu/astr161/lect/earth/coriolis.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Coriolis effect&lt;/a&gt;.  Viewed from above, the Coriolis effect creates bands that line up with the equator.  Jupiter and Saturn both rotate quite quickly (9.5 and 10.8 hours respectively), so both have strong Coriolis effects.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uLRbw.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uLRbw.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter's atmosphere isn't just moving along those visible lines; it's circulating up from the warmer lower parts of the atmosphere to higher parts by convection. This is because there's a lot of heat to be transferred. Jupiter (along with Saturn, Uranus and Neptune) are all radiating more heat into space than they are receiving from the Sun, so while the heat from the Sun plays a role in not freezing Jupiter's upper atmosphere over warmer bands, it is the Coriolis effect that creates the bands. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn has bands too (as you know). They are not as visible because both the warm bands and cold bands of Saturn are icy. See articles &lt;a href=&quot;http://www.nature.com/nature/journal/v453/n7192/fig_tab/nature06897_F1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://saturn.jpl.nasa.gov/resources/2743/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and article with false color picture &lt;a href=&quot;http://www.nasa.gov/multimedia/imagegallery/image_feature_778.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, crediting Saturn's colors to ammonia ice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for bands to form, all you need is relatively fast rotation. But for highly visible bands, the gas giant planet needs to be the right distance from the Sun and/or have the right amount of internal heat. It's all about the temperature.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel that I have to punctuate this with a &quot;probably&quot;, because I can't swear that Saturn has ammonia ice all around its upper atmosphere, but I'm pretty sure that is the primary difference between Saturn's more uniform look (Uranus and Neptune too). It is primarily the ammonia that gives these planets their color, even though both Saturn and Jupiter are 99% hydrogen and Helium. Straight gas molecules (O2, N2, H2) and noble gases tend to have very little interaction with visible light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a &lt;a href=&quot;http://www.exoclimes.com/topics/the-colour-of-giant-planets-in-the-solar-system/&quot; rel=&quot;nofollow noreferrer&quot;&gt;fun article&lt;/a&gt; on gas giant planet's color. The science of what gives an atmosphere a specific color, however, is quite complex and above my pay-grade. Also, I also found this &lt;a href=&quot;http://www.teachastronomy.com/astropedia/article/Thermal-Radiation-from-Gas-Giant-Planets&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; interesting, though I can't swear to the accuracy of it.  Saturn radiates a surprising amount of heat given that it's 1/3rd the mass of Jupiter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, &lt;a href=&quot;http://pds-rings.seti.org/saturn/earthbased/displaysaturn.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt; may provide a more accurate but less sexy picture of Saturn than we're used to seeing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nyFD1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nyFD1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final note, on Jupiter's dark spots and dark bands. Wikipedia said the reason for the darkness is unknown (as quoted above). While I think the lack of atmospheric ammonia ice is very much a part of the answer, &lt;a href=&quot;https://astronomy.stackexchange.com/questions/7980/why-jupiters-great-red-spot-is-reddish?rq=1&quot;&gt;here's&lt;/a&gt; an alternate explanation for Jupiter's dark red spot and (perhaps) to some extent, its dark bands. Jupiter gets, on average, about 3.4 times the solar radiation per square meter than Saturn gets. To add to that, it might have more extensive atmospheric mixing going on than Saturn and more organic compounds in its upper atmosphere, which, combined with UV rays, could well play a role in Jupiter's darker lines too.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-07-19T00:13:51.630" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16612" PostTypeId="2" ParentId="8279" CreationDate="2016-07-17T11:41:53.613" Score="1" Body="&lt;p&gt;A well-known analytical planetary theory is &lt;a href=&quot;http://esoads.eso.org/abs/1988A%26A...202..309B&quot; rel=&quot;nofollow&quot;&gt;VSOP&lt;/a&gt;.  The paper describes six versions, most of which are &lt;em&gt;heliocentric&lt;/em&gt;.  One of them (E) is &lt;em&gt;barycentric&lt;/em&gt; and has an extra (ninth) body: the Sun.  Hence, for VSOP you have the choice of using the Solar System's barycentre as the origin and computing the position of an extra body, or of computing heliocentric positions of the planets directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use a different theory, I would expect something similar: a heliocentric version and/or a barycentric version with equations for the Sun.&lt;/p&gt;&#xA;" OwnerUserId="286" LastActivityDate="2016-07-17T11:41:53.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16613" PostTypeId="1" CreationDate="2016-07-17T14:08:11.987" Score="1" ViewCount="259" Body="&lt;p&gt;Is it coïncidence that the mass of a planet has about its maximum in the middle planets (Jupiter), while the mass of the closest and furthest planets has a minimum (Mercury and Pluto)?&lt;/p&gt;&#xA;" OwnerUserId="10659" LastEditorUserId="2153" LastEditDate="2016-07-17T21:39:38.370" LastActivityDate="2016-07-17T21:39:38.370" Title="A question about the distribution of mass among the planets in the Solar System" Tags="&lt;planet&gt;&lt;solar-system&gt;&lt;mass&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16614" PostTypeId="2" ParentId="16613" CreationDate="2016-07-17T14:34:18.227" Score="4" Body="&lt;p&gt;First, Pluto isn't a planet, and hasn't been for about a decade. It's a &lt;a href=&quot;https://en.wikipedia.org/wiki/Dwarf_planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;dwarf planet&lt;/a&gt;, and is better grouped with other small, rocky objects beyond Neptune's orbit, such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Haumea&quot; rel=&quot;nofollow noreferrer&quot;&gt;Haumea&lt;/a&gt;. These objects, and other minor planets, have more in common with objects in the Kuiper Belt and Oort Cloud than the eight planets of the Solar System.&lt;sup&gt;1&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we see a different pattern: Four relatively low-mass planets (Mercury, Venus, Earth and Mars) in the inner Solar System (the four &lt;a href=&quot;https://en.wikipedia.org/wiki/Terrestrial_planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;terrestrial planets&lt;/a&gt;) and four relatively high-mass planets (Jupiter, Saturn, Uranus, and Neptune) farther out (the four &lt;a href=&quot;https://en.wikipedia.org/wiki/Giant_planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;giant planets&lt;/a&gt;). The question now is why the two groups are arranged as they are, why the terrestrial planets are close to the Sun while the giant planets are further away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote about this in &lt;a href=&quot;https://astronomy.stackexchange.com/a/14608/2153&quot;&gt;my answer to Why do the gas giants in the Solar System have comparatively large orbits compared to the inner planets?&lt;/a&gt;. It boils down to something called &lt;a href=&quot;https://en.wikipedia.org/wiki/Frost_line_(astrophysics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;the frost line&lt;/a&gt;, a rough boundary before which ices of water, ammonia, methane, and other compounds cannot form. Beyond it, they can form ice crystals (not just water ice, of course), which made it easier to accrete large amounts of material from the protoplanetary disk. This leads to the more massive gas giants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, this explains the mass distribution of the planets, although there are exceptions (e.g. &lt;a href=&quot;https://en.wikipedia.org/wiki/Hot_Jupiter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hot Jupiters&lt;/a&gt;), which often result form planet-planet interactions, rather than forming at distances inconsistent with the frost line.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1 I exclude &lt;a href=&quot;https://en.wikipedia.org/wiki/Planet_Nine&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planet Nine&lt;/a&gt; from this list, as its existence has yet to be verified.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-07-17T14:34:18.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16615" PostTypeId="2" ParentId="10246" CreationDate="2016-07-17T15:10:17.270" Score="2" Body="&lt;p&gt;The magnitude of the Moon without the &lt;em&gt;opposition surge&lt;/em&gt; is given by a simple equation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$m = -12.73 + 1.49 \cdot |\psi| + 0.043 \cdot \psi^4,$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\psi$&lt;/span&gt; is the &lt;em&gt;phase angle&lt;/em&gt; in radians &lt;a href=&quot;http://esoads.eso.org/abs/1976asqu.book.....A&quot; rel=&quot;nofollow noreferrer&quot;&gt;(Allen 1976)&lt;/a&gt;.  This can be converted to flux using &lt;span class=&quot;math-container&quot;&gt;$m \propto -2.5 \cdot ^{10}\!\log F$&lt;/span&gt;,&#xA;where &lt;span class=&quot;math-container&quot;&gt;$^{10}\!\log$&lt;/span&gt; denotes the 10-base logarithm (as opposed to e.g. the natural log).&lt;/p&gt;&#xA;&lt;p&gt;For the increase in flux due to the variable distance and opposition surge, I multiply F with the factor&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\left(\frac{\Delta_0}{\Delta}\right)^2  \cdot  \max\left(1, 1.35 - 2.865 \cdot |\psi| \right),$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\Delta_0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\Delta$&lt;/span&gt; are the mean and current distance to the Moon, respectively.  However, I can't find a reference for the last term at the moment, and I can't remember where I obtained it.  I'll post that information if I find it.&lt;/p&gt;&#xA;" OwnerUserId="286" LastEditorUserId="286" LastEditDate="2020-11-06T16:49:13.107" LastActivityDate="2020-11-06T16:49:13.107" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16616" PostTypeId="1" CreationDate="2016-07-17T17:27:47.157" Score="1" ViewCount="48" Body="&lt;p&gt;I came across this cool tool which encodes geographic coordinates into 3 dictionary words - table.lamp.chair - for example. You can see their map here &lt;a href=&quot;https://map.what3words.com/&quot; rel=&quot;nofollow&quot;&gt;https://map.what3words.com/&lt;/a&gt;. It divides the world into a grid of 57 trillion 3mx3m squares and each one is assigned with one of these three-word addresses.&#xA;It made me curious as to whether this sort of thing could work for communicating the locations of stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be pretty simple to assign unique three-word addresses to stars in stellar databases. What would be even more powerful is if we could encode celestial coordinates into three-word addresses, and create an interactive &quot;what3words&quot; space map. But would this be at all useful?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The finite surface area of the earth allows us to create the grid of 57 trillion squares. Can we work with the infinite nature of the celestial sphere to create something similar?&lt;/p&gt;&#xA;" OwnerUserId="13505" LastActivityDate="2016-07-17T19:27:28.607" Title="Could celestial coordinates be encoded into three-word combinations?" Tags="&lt;coordinate&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16617" PostTypeId="2" ParentId="16616" CreationDate="2016-07-17T19:27:28.607" Score="1" Body="&lt;p&gt;Possible, yes. In fact since you can just use their algorithm. Sky locations expressed a RA and Dec map directly to coordinates expressed as Longitude and Latitude. The celestial sphere is no more infinite than the terrestrial one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each square seems to take about 0.1 arcseconds, which is very accurate, but it does raise a problem: The stars move. Barnards star has proper motion of over 10 arcseconds per year. This is an extreme example, but all the nearby stars move, so their addresses would change. The algorithm is designed not to give nearby locations similar names. So each year Barnards star would have 10 completely different addresses. Other stars would change address due to parallax. If you don't account for precession, all stars would be being renamed all the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the purpose of the three word addresses is to give addresses to the address-less, in a way that is memorable. Is there a need for addresses of sky locations? We have a cute way of naming larger regions of the sky (the 88 constellations) and stars have names (or catalog numbers) that don't change due to their real or apparent motion. If you need to refer to a location you can use the numeric coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: this would not be useful, either for science (since the stars would keep moving address) or for outreach (the Constellations do very well for the whole sky)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-07-17T19:27:28.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16619" PostTypeId="1" AcceptedAnswerId="16620" CreationDate="2016-07-17T22:32:20.543" Score="0" ViewCount="152" Body="&lt;p&gt;I don't know much about the astronomy or the universe but enjoy reading articles from time to time as they pop up. Just finished reading an article on CNN that discussed how a South African telescope has discovered hundreds of galaxies, some 200 million light years away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This got me thinking: Are these galaxies and or some of their stars we see in telescopes still there? Let's say STAR X is confirmed to be 200 million light years away. What if it completely exploded 1 million years ago? Wouldn't we need to wait 199 million years before we could detect that? &lt;/p&gt;&#xA;" OwnerUserId="13509" LastActivityDate="2016-07-17T23:58:28.080" Title="Do far away stars that we observe today still exist?" Tags="&lt;galaxy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16620" PostTypeId="2" ParentId="16619" CreationDate="2016-07-17T23:44:11.190" Score="0" Body="&lt;p&gt;Simple answer is yes. It will take light from an explosion that happened 1 mya at a distance of 200 Mly, about 199 million years to reach &quot;us&quot; - not that any of &quot;us&quot; will be around then. The speed of light is the fastest speed we can get information from anywhere in the Universe. There's nothing special in asking about what happens to that star after that light starts out on its journey here. The fact is, we can't know, until we are able to gather more light from that same source. So, that star would appear to explode if you were looking up at it 199 million years from now (keeping in mind that both the Solar System and the Milky Way are also moving through space, so the exact time it takes the light to get here will be different than the time it took the light we are now seeing from that star to have gotten here.)&lt;/p&gt;&#xA;" OwnerUserId="13487" LastEditorUserId="13487" LastEditDate="2016-07-17T23:58:28.080" LastActivityDate="2016-07-17T23:58:28.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16621" PostTypeId="1" AcceptedAnswerId="29073" CreationDate="2016-07-18T03:32:23.567" Score="1" ViewCount="126" Body="&lt;p&gt;The final configuration of the &lt;a href=&quot;https://en.wikipedia.org/wiki/MeerKAT&quot; rel=&quot;nofollow noreferrer&quot;&gt;MeerKAT radio array&lt;/a&gt; will be 64 dishes. In the &lt;a href=&quot;http://www.ska.ac.za/releases/20160716.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;press release of July 16, 2016&lt;/a&gt; a radio image is shown using a completed sub-set of 16 dishes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which 16 are these? I can find documents on-line with coordinates for 64 location, (sometimes numbered 1 through 64, &lt;a href=&quot;https://arxiv.org/pdf/0910.2935.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;but not here&lt;/a&gt;), but I can't figure out which 16 are used to produce this image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found images of KAT-7 and MeerKAT in google maps - maybe someday we can get daily ground truth updates from &lt;a href=&quot;https://www.planet.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planet Labs&lt;/a&gt;, but for right now I'll stick to stackexchange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image &lt;a href=&quot;http://www.ska.ac.za/releases/20160716.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;from the press release&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UVYyk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UVYyk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/l9OXm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l9OXm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Above:&lt;/strong&gt; MeerKAT &lt;em&gt;array core&lt;/em&gt; from Google Maps at (30.7136109S, 21.4399576E). The remaining sites are farther away and can be seen if one zooms out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xs0Rd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xs0Rd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Above:&lt;/strong&gt; KAT-7 array from Google Maps at (30.7217011S, 21.4070518E)&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-04-09T06:38:22.700" LastActivityDate="2019-04-09T06:38:22.700" Title="Which 16 antenna locations were used this MeerKAT radio image?" Tags="&lt;radio-astronomy&gt;&lt;radio-telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16623" PostTypeId="1" AcceptedAnswerId="16629" CreationDate="2016-07-18T06:55:35.857" Score="1" ViewCount="181" Body="&lt;p&gt;There were so many Apollo Missions, and they landed on the moon (so they say). When astronauts came back to Earth from the moon, they brought Moon rock samples. Where are these moon rocks/samples now, and where can we see them? &lt;/p&gt;&#xA;" OwnerUserId="11792" LastEditorUserId="6" LastEditDate="2016-07-18T13:04:09.903" LastActivityDate="2016-07-18T13:07:21.753" Title="Where are the Moon rocks from Apollo?" Tags="&lt;moonlanding&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ClosedDate="2016-07-18T13:08:23.357" ContentLicense="CC BY-SA 3.0" />
  <row Id="16624" PostTypeId="1" AcceptedAnswerId="16625" CreationDate="2016-07-18T08:58:12.207" Score="3" ViewCount="396" Body="&lt;p&gt;Is it true that stars start with a rocky core, then collect hydrogen by the gravity of that rocky core, just like planets collecting materials?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastActivityDate="2016-07-18T13:57:30.323" Title="Do stars start with rocky core?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16625" PostTypeId="2" ParentId="16624" CreationDate="2016-07-18T09:00:39.797" Score="7" Body="&lt;p&gt;No, stars do not have a rocky core. They start as clouds of gas, become gravitationally unstable and collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some planets are thought to start this way too - though probably not the ones in our solar system, because the outer three three giant planets have solid cores and Jupiter is likely to also. The two modes of giant planet formation are reviewed in detail by &lt;a href=&quot;https://arxiv.org/abs/1006.5486&quot;&gt;D'Angelo et al. (2010)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-07-18T13:57:30.323" LastActivityDate="2016-07-18T13:57:30.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16626" PostTypeId="2" ParentId="11743" CreationDate="2016-07-18T09:35:25.667" Score="2" Body="&lt;p&gt;I'd like to add more inputs on the physics behind telescope resolution. The telescope resolution is given by its primary mirror size which produces a &quot;quasi&quot; punctual image for each light source it sees, that is called a airy disk. The airy disk size is inversely proportional to the telescope primary mirror diameter and like pixels on a LCD screen the lower the pixel size the better the resolution. However, the limiting factor on the ground is the atmospheric turbulence, this is why huge telescope of few meters wide cannot see better than a 500mm amateur telescope resolution wise (see Fried parameter). So in order to reach the telescope full resolution potential we need to build adaptive optics or better we could send the telescope to space. &lt;/p&gt;&#xA;" OwnerUserId="13514" LastEditorUserId="13514" LastEditDate="2016-07-18T11:43:41.853" LastActivityDate="2016-07-18T11:43:41.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16627" PostTypeId="1" AcceptedAnswerId="16628" CreationDate="2016-07-16T16:37:55.950" Score="0" ViewCount="748" Body="&lt;p&gt;Imagine viewing the solar system from the outside:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SoM4h.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SoM4h.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the sun is quite bright, would you still be able to see stars in the background, or does the sunlight disable us to do so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider this photo:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TPbrz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TPbrz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the light from the sun disable us from seeing stars in the background or have they simply been removed in an editor after the fact of being taken (the photo)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously there needs be some way of seeing stars from space, since looking at photos from space stations (or even from Earth) those are full of stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, viewing the solar system from above (from the outside), can you see stars in the background or does the sunlight simply block them out? &lt;/p&gt;&#xA;" OwnerUserId="13520" OwnerDisplayName="D. Ataro" LastActivityDate="2020-03-31T13:49:27.677" Title="Can stars be seen in the background when looking directly at our solar system?" Tags="&lt;solar-system&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16628" PostTypeId="2" ParentId="16627" CreationDate="2016-07-16T17:10:48.637" Score="4" Body="&lt;p&gt;Stars are much less bright than the earth in full sunlight as per your picture. So with the exposure required to give that picture of earth, no, you won't see stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But move the camera a bit to the side, so earth is not in shot and you can see stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same goes for the sun - make sure sunlight is not entering the lens, and you can easily see stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sunlight doesn't block any light out (unless you want to see a star that is actually behind the sun) it's just so bright that you need to make sure it isn't in shot.&lt;/p&gt;&#xA;" OwnerUserId="43" OwnerDisplayName="Rory Alsop" LastActivityDate="2016-07-16T17:10:48.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16629" PostTypeId="2" ParentId="16623" CreationDate="2016-07-18T13:07:21.753" Score="7" Body="&lt;p&gt;It's well-explained on NASA'a web site ... &lt;a href=&quot;http://curator.jsc.nasa.gov/Lunar/index.cfm&quot;&gt;http://curator.jsc.nasa.gov/Lunar/index.cfm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About &lt;strong&gt;380 kg&lt;/strong&gt; of rocks and soil were brought back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The &lt;strong&gt;lunar sample building&lt;/strong&gt; at Johnson Space Center is the chief repository for the Apollo samples.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://curator.jsc.nasa.gov/Lunar/laboratory_tour.cfm&quot;&gt;http://curator.jsc.nasa.gov/Lunar/laboratory_tour.cfm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About &lt;strong&gt;three hundred&lt;/strong&gt; moon rocks were given as gifts by the US to cities and countries around the world, and you can easily find that list, and hence go see them all. (More fun than PokemonGo!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_Apollo_lunar_sample_displays&quot;&gt;List showing where to see moon rocks&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can usually very easily see one at the local &quot;national museum&quot; in your country.  I've seen maybe five all over the world at the places listed there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that at least 100 have &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_sample_displays#Current_fate&quot;&gt;gone missing&lt;/a&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few moon rocks in the wild and there were &lt;strong&gt;cases of theft&lt;/strong&gt; - they would be worth millions of bucks ... &lt;a href=&quot;http://www.natgeotv.com/ca/million-dollar-moon-rock-heist&quot;&gt;TV show about that&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13071" LastActivityDate="2016-07-18T13:07:21.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16630" PostTypeId="1" AcceptedAnswerId="16632" CreationDate="2016-07-18T13:40:02.400" Score="20" ViewCount="3061" Body="&lt;p&gt;To cite the German newpaper article &lt;a href=&quot;http://www.heise.de/newsticker/meldung/Astronomen-beobachten-erwachendes-Schwarzes-Loch-2743638.html&quot; rel=&quot;noreferrer&quot;&gt;Astronomen beobachten erwachendes Schwarzes Loch&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Das Materie-Monster sitzt den Angaben zufolge im Herzen der 42 Millionen Lichtjahre entfernten Polarring-Galaxie NGC 660, deren Aktivität innerhalb weniger Monate Hunderte Male zugenommen hatte.&lt;/p&gt;&#xA;&lt;p&gt;Erst wenn die Massemonster große Mengen Materie verschlucken, werden sie aktiv. Bei diesem Prozess wird so viel Energie frei, dass die Materie hell aufleuchtet, bevor sie im Schwarzen Loch verschwindet und ein Teil von ihr in Form von Jets weit ins Weltall hinaus geschleudert wird.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This translates roughly to:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;According to the data the matter-monster is in the middle of the 42 million light years far away polar ring galaxy NGC 660, whose activity has increased a lot in only a few months.&lt;/p&gt;&#xA;&lt;p&gt;Only when these Matter-monsters swallow big amounts of matter, they become active. &lt;strong&gt;This process releases so much energy, that it brightly enlightens the matter, before it disappears inside the black hole. A part of the matter is flung out in the universe in the form of jets.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;My physics teacher told me once, that a black hole is just a very small-sized and heavy object which has so much gravity that nothing at all, not even light can't escape of its gravity. This explanation is also supported by this &lt;a href=&quot;https://astronomy.stackexchange.com/questions/70/if-nothing-travels-at-the-speed-of-light-except-light-how-can-a-black-hole-als&quot;&gt;SE.astronomy - If nothing travels at the speed of light, except light, how can a black hole also pull light into itself?&lt;/a&gt; question.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If a &amp;quot;normal&amp;quot; (not supermassive) black hole can already prevent light to escape, how can matter which is pulled into the black hole can produce energy/light which can't escape the gravity of the black hole?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How can a supermassive black hole be able to pull matter, but not the light photons of the energy?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Additionally: Why is part of the matter which is pulled into the black hole flung (i.e. &lt;strong&gt;accelerated&lt;/strong&gt;) into the universe? I understand why this matter is maybe divided - the acceleration increases as far as i know quadratic, i.e. the differences in terms of the actual acceleration, in dependence of the location may be so huge that the matter can't be held together. But i don't understand why a part of the matter is accelerated in the exact &lt;strong&gt;opposite&lt;/strong&gt; direction, as a force &lt;strong&gt;bigger&lt;/strong&gt; as the gravity of the supermassive black hole would be needed. Therefore: Why is a part of the matter accelerated in the opposite direction (i.e. out of the black hole's gravity) than the other part?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;em&gt;Note: My physics education is rather limited. I know a bit about the Newtonian gravity and a bit about the theory of Conservation of energy. But that's all I know about physics.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="13515" LastEditorUserId="-1" LastEditorDisplayName="user1569" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2020-02-29T13:51:49.743" Title="How can a supermassive black hole cause so much energy to enlighten its matter when its massive gravity prevents light to escape?" Tags="&lt;gravity&gt;&lt;black-hole&gt;&lt;light&gt;&lt;supermassive-black-hole&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16631" PostTypeId="2" ParentId="16630" CreationDate="2016-07-18T14:07:00.573" Score="14" Body="&lt;p&gt;The light comes from well &lt;em&gt;outside&lt;/em&gt; the black hole event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matter cannot fall into a black hole without first losing most of its angular momentum (otherwise it would just continue to orbit the black hole). This is accomplished by the outward transfer of angular momentum by viscosity (and other means) in an accretion disc surrounding the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the matter gets closer to the black hole it also loses gravitational potential energy and this goes into (i) heating the gas and (ii) radiation from the gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At about 3 times the &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schwarzschild radius&lt;/a&gt; of the black hole, the matter encounters the &lt;a href=&quot;https://en.wikipedia.org/wiki/Innermost_stable_circular_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;innermost stable circular orbit&lt;/a&gt;, which is the closest that anything with mass can make a stable orbit around a black hole. It is usually assumed that from there the material plummets into the black hole and is &quot;lost&quot; from our universe - increasing the mass of the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So all the radiation comes from orbiting material that is at least 3 times the Schwarzschild radius from the black hole. There is no problem for light to &quot;escape&quot; from this position, although it is strongly redshifted by both gravity and the transverse relativistic doppler effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue about &quot;jets&quot; has been covered by another question: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/809/why-do-black-holes-have-jets-and-accretion-disks?rq=1&quot;&gt;Why do black holes have jets and accretion disks?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-07-18T14:07:00.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16632" PostTypeId="2" ParentId="16630" CreationDate="2016-07-18T14:09:56.867" Score="23" Body="&lt;p&gt;It is quite correct that a black hole has so much mass that light cannot escape from a region around the black hole. The edge of this region is called the event horizon. If you cross an event horizon you are never coming back. That applies equally to light, and matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Around the black hole there may be matter in orbit. Since the Black hole has such strong gravity, the speed of the orbiting matter will be very fast. In fact it will be close to the speed of light. This high speed gives it lots of energy. The matter will form a disc, called an accretion disc, around the black hole, and collisions in this disc will cause the matter to heat up, to millions of degrees. At these temperatures, the disc will glow with X-rays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the part of the disc closest to the black hole, matter will be falling in from the disc, but before it reaches the black hole it can get enough energy to be ejected out, at very high speed, close to the speed of light. It gets ejected at right angles to the disc, at the poles of the black hole. These are the &quot;jets&quot;. Intense radiation is produced along these jets. Blazars are distant supermassive black holes with jets that are pointed right at us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the black hole itself is &quot;black&quot;, but the matter orbiting around it may be very bright.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-07-19T19:46:47.363" LastActivityDate="2016-07-19T19:46:47.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16633" PostTypeId="1" AcceptedAnswerId="16634" CreationDate="2016-07-18T14:11:18.117" Score="7" ViewCount="653" Body="&lt;p&gt;I came across &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1938MNRAS..99....4E&amp;amp;defaultprint=YES&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;a 1938 paper by Arthur Eddington&lt;/a&gt;&lt;sup&gt;1&lt;/sup&gt; in which he discusses stellar polytropic models where the polytropic index &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; varies inside the star, decreasing from a value &lt;span class=&quot;math-container&quot;&gt;$n_2$&lt;/span&gt; at the surface to a value &lt;span class=&quot;math-container&quot;&gt;$n_1$&lt;/span&gt; at the center. He specifically mentions that the case of &lt;span class=&quot;math-container&quot;&gt;$n_1=1.5$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$n_2=3.5$&lt;/span&gt; should represent &quot;ordinary gaseous stars&quot; quite well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In his conclusion, he states&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We have seen that actual stars are likely to be represented by a model with &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; continually increasing outwards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;78 years later, we have better data about the structure of the Sun, thanks in part to helioseismology. So I have two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Do researchers use Eddington's variable index treatment in theoretical models?&lt;/li&gt;&#xA;&lt;li&gt;Does Eddington's model better fit current data from the Sun than the standard polytropic model with &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; constant throughout the star?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I ask partly because I know that the standard polytropic model (with &lt;span class=&quot;math-container&quot;&gt;$n=3$&lt;/span&gt;) begins to fail in the outer regions of a star, thanks in part to convection, as shown below, and I wondered if even a simple two-phase model could suitably describe a convective envelope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JOWT0.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JOWT0.gif&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;Graphs from &lt;a href=&quot;http://www.astro.utu.fi/~cflynn/Stars/l9.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1 Eddington mentions earlier work by &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1932MNRAS..92..610M&amp;amp;defaultprint=YES&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Milne (1932)&lt;/a&gt;, who uses a slightly more complex analysis. I haven't yet read Milne's paper, so I'm not familiar with its conclusions and use.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="7472" LastEditDate="2018-12-25T12:34:56.927" LastActivityDate="2018-12-25T12:45:18.297" Title="Does Eddington's variable polytropic index better fit data from the Standard Solar Model?" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16634" PostTypeId="2" ParentId="16633" CreationDate="2016-07-18T16:56:15.127" Score="4" Body="&lt;p&gt;The definition of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Polytrope&quot; rel=&quot;nofollow noreferrer&quot;&gt;polytrope&lt;/a&gt; is one for which the pressure and density are related by&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ P = K \rho^{(n+1)/n}$$&lt;/span&gt;&#xA;The pressure is thus assumed to be independent of temperature and this simplifies the equations of stellar structure so that certain analytic solutions and approximations are possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is however a  big assumption and is generally only accurate for stars/planets supported by degeneracy pressure, where &lt;span class=&quot;math-container&quot;&gt;$1.5&amp;lt;n&amp;lt;3$&lt;/span&gt; depending on whether the degeneracy is non-relativistic or relativistic. The other instance is where the energy transport is by adiabatic convection, which gives an independent relationship between &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; and means that for a completely ionized gas that &lt;span class=&quot;math-container&quot;&gt;$P \propto \rho^{5/3}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$n=1.5$&lt;/span&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A further polytropic approximation can also be made where the energy transport is predominantly radiative and where it is assumed that the radiation pressure is a fixed proportion of the total pressure. In these circumstances &lt;span class=&quot;math-container&quot;&gt;$P \propto \rho^{4/3}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$n \simeq 3$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a bit of a primer &lt;a href=&quot;http://www2.astro.psu.edu/users/rbc/a534/lec15.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, but most stellar astrophysics textbooks will go through this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Eddington standard model assumes a radiative (&lt;span class=&quot;math-container&quot;&gt;$n=3$&lt;/span&gt;) polytrope. The vast majority of the Sun, &lt;em&gt;by mass&lt;/em&gt;, is indeed radiative and the polytropic approximation works reasonably well. However the outer parts of the Sun (beyond about 70% of the radius) are convective and so a single polytropic model does not fare so well. So what you can do is have a &quot;mixed polytropic model&quot;, where you divide the star up into polytropic shells with different polytropic indices, each forming the boundary conditions for the ones above and below.&#xA;A double polytropic model for the Sun, with two zones that have &lt;span class=&quot;math-container&quot;&gt;$n_1\simeq 3$&lt;/span&gt; in the interior and &lt;span class=&quot;math-container&quot;&gt;$n_2\simeq 1.5$&lt;/span&gt; in the outer convection zone &lt;em&gt;should&lt;/em&gt; be much more accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But modern astrophysical calculations do not rely on the analytic approximations of polytropes and the Lane Emden equation. They &lt;em&gt;numerically&lt;/em&gt; solve the coupled differential equations of stellar structure, subject to the boundary conditions imposed by an atmosphere on the top of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Unashamedly tempted by the bounty, I did a little bit more research. I came across some sort of high-level set of lecture notes by &#xA;&lt;a href=&quot;http://www.rfrench.org/astro/papers/P110-HET611-RobertFrench.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Robert French&lt;/a&gt; of Swinburne University. He discusses modelling the Sun with composite polytropes at some length. On p.8 he describes how the matching criteria at the boundaries between polytropic zones is actually quite tricky and you have to let the matching radius and central density float in order to get a smooth join. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conclusion he arrives at is that a combination of a &lt;span class=&quot;math-container&quot;&gt;$n_1=3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$n_2=1.5$&lt;/span&gt; actually appears to do a &lt;em&gt;worse&lt;/em&gt; job of fitting the entire Sun than modelling with a single average polytrope (although with &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; varied depending on which physical quantity you want on the y-axis - e.g. &lt;span class=&quot;math-container&quot;&gt;$\bar{n}=3.11$&lt;/span&gt; for density vs radius). If on the other hand you allow the polytropic indices and the interface radius be free parameters in order to get a good match to the standard solar model, you can get a &lt;em&gt;really&lt;/em&gt; good fit with &lt;span class=&quot;math-container&quot;&gt;$n_1=3.94$&lt;/span&gt; in the radiative zone, &lt;span class=&quot;math-container&quot;&gt;$n_2=1.58$&lt;/span&gt; in the convection zone and an interface at 68% of the solar radius. (Plot reproduced below - compare these with the plots in your question!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZeY4n.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZeY4n.png&quot; alt=&quot;Composite polytrope model of the Sun&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-12-25T12:45:18.297" LastActivityDate="2018-12-25T12:45:18.297" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="16635" PostTypeId="2" ParentId="11743" CreationDate="2016-07-18T19:22:25.863" Score="8" Body="&lt;p&gt;You've pretty much answered it yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very large telescopes are expensive to build, and there are diminishing returns for large instruments operating under Earth's atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air turbulence (what is known as &quot;seeing&quot;) limits the resolving power of the telescope - its ability to distinguish small details and make high resolution images. It's an essentially random phenomenon, so sometimes it gets better, but generally speaking large instruments are more strongly affected by turbulence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light pollution is the glow caused by all artificial light sources (city lights, industrial lights), which are making it difficult to see very faint, very distant objects. It is drowning the faint objects in artificial glow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bigger the telescope is, the better its resolving power, and the better its light gathering ability (the ability to see faint objects). But seeing and light pollution are affecting both. You could fight seeing with adaptive optics, with decent results. You could fight light pollution by installing the instrument far from cities. Or you could fight both by launching the instrument into space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these solutions cost money. Large instruments are expensive. Adaptive optics are expensive. Running an observatory in the middle of nowhere also adds costs. Finally, launching things into space is not cheap either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, each one of these methods is used in one case or another. For each kind of application, astronomers need to make a decision for what methods they could use to improve performance, and then match that up with the money available to the project. It's a complex overall decision that takes a long time, lots of discussions, often politics gets involved, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what gets built is a compromise between what astronomers want, and what is actually doable from a financial, political, and technological standpoint.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;BTW, &quot;magnification&quot; is not a parameter that astronomers use to define an instrument. Any instrument can give you any magnification you want - just use a different eyepiece. The relevant parameter here is the resolving power - the ability of the instrument to distinguish fine details. This depends on the instrument size - the diameter (or &quot;aperture&quot;) of the telescope. Bigger scope = better resolving power (all else being equal). As I said above, seeing (air turbulence) is a major limiting factor here.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2016-07-18T19:22:25.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16636" PostTypeId="2" ParentId="10948" CreationDate="2016-07-18T19:24:35.880" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I assume you have an algorithm to compute position of the Sun, since you can compute rise and set times.  You will need to compute the position of the Moon in a similar manner.  Then, convert both positions to geocentric ecliptical coordinates (ecliptic latitude and longitude) and use some numerical bisection method (root finder) to find where the difference of the longitudes of the two objects has a root (i.e., becomes zero).  The corresponding JD provides the moment of the geocentric conjunction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Calculation of the altitude is a standard procedure - you'll need to convert your ecliptic coordinates to equatorial (right ascension and declination) and then to horizontal (azimuth and altitude).  Finally, correct for the horizontal parallax to convert your geocentric coordinates to topocentric coordinates for the location of the observer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="286" LastActivityDate="2016-07-18T19:24:35.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16637" PostTypeId="2" ParentId="10904" CreationDate="2016-07-18T20:23:11.520" Score="1" Body="&lt;p&gt;Yes, this is the correct approach.  The $h$ in the equation is the altitude above the horizon of the object at which you consider it to be rising or setting.  This is typically non-zero, because of atmospheric refraction, and, in the case of the Sun or the Moon, because of their finite diameters.  In your case, the object 'rises' when it climbes above $h=30^\circ$ and 'sets' when it drops below that altitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\left|\cos(LHA)\right|&amp;gt;1$, there is no solution, because your object never crosses the $h=30^\circ$ line.  $LHA$ is the &lt;em&gt;local hour angle&lt;/em&gt;, and you can find the &lt;em&gt;local sidereal time&lt;/em&gt; $\theta$ using&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$LHA = \theta - \alpha,$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\alpha$ is the right ascension of your object.&lt;/p&gt;&#xA;" OwnerUserId="286" LastActivityDate="2016-07-18T20:23:11.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16638" PostTypeId="1" AcceptedAnswerId="16639" CreationDate="2016-07-19T00:32:14.637" Score="3" ViewCount="262" Body="&lt;p&gt;I was talking with a friend about how slowly the star field changes (based on the speed that we are moving through the galaxy) and I started to wonder about a star's visible size.  They are basically the pixels that make up our sky.  This made me wonder, how many stars (let's use the north star as a reference.) would it take (lined up side by side) to draw line that appeared solid across the middle of the sky, perpendicular to the horizon which appeared solid?&lt;/p&gt;&#xA;" OwnerUserId="7895" LastActivityDate="2016-07-19T01:18:41.190" Title="How many stars would it take to draw a line across the middle of the sky that appeared solid?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16639" PostTypeId="2" ParentId="16638" CreationDate="2016-07-19T01:18:41.190" Score="8" Body="&lt;p&gt;This is more of a question of human perception, than astronomy. I was going to answer your question with this: &quot;One, it just needs to be close to Earth.&quot; But, I decided it wasn't THAT funny. Anyway, stars are essentially point sources as far as our eyes are concerned. Typical visual resolution is about 0.02° or 0.0003 radians. Assuming from horizon to horizon is 180° (or π radians) that calculates out to roughly 10,000 stars. I'd probably increase that by 50% or 100% to be sure. You do understand that you can't line stars up side by side, I hope. There's no need, its about the angular distance, not absolute distance between them, that matters. They can be light years apart as long as they appear to be within about 1 arcminute of one another, our eyes will see them as a single object - subject to the psychological aspects of keeping color and apparent magnitude roughly uniform as well.&lt;/p&gt;&#xA;" OwnerUserId="13487" LastActivityDate="2016-07-19T01:18:41.190" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16640" PostTypeId="2" ParentId="16608" CreationDate="2016-07-19T08:59:26.343" Score="1" Body="&lt;p&gt;I'm not sure such a detailed answer to your question is available.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://esoads.eso.org/abs/2000asqu.book.....C&quot; rel=&quot;nofollow noreferrer&quot;&gt;This book&lt;/a&gt; cites &lt;a href=&quot;http://esoads.eso.org/abs/1988ASSL..142...11N&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt; as a source for the mass loss due to the solar wind:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\dot{M} \sim 2.5 \times 10^{-14}\,M_\odot/yr$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://esoads.eso.org/abs/1992adps.book.....L&quot; rel=&quot;nofollow noreferrer&quot;&gt;This book&lt;/a&gt; cites &lt;a href=&quot;http://esoads.eso.org/abs/1977soiv.conf..351F&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt; as the source for the following data at 1 AU:&lt;/p&gt;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;Kinetic energy density of the solar wind:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{1}{2}N_\mathrm{p}m_\mathrm{p}v^2 = 1.44 \pm 0.09 \times 10^{-8}\,$&lt;/span&gt;erg cm&lt;span class=&quot;math-container&quot;&gt;$^{-3}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Thermal energy density for protons, electrons and helium atoms:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{3}{2} N k T \approx 4.8 \pm 3.2 \times 10^{-10}\,$&lt;/span&gt;erg cm&lt;span class=&quot;math-container&quot;&gt;$^{-3}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Wind velocity:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$v_\mathrm{w} = 468 \pm 116\,$&lt;/span&gt;km s&lt;span class=&quot;math-container&quot;&gt;$^{-1}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;The numbers in 2. and 3. indicate that the thermal energy contributes for about 3%, and can be ignored given the uncertainties involved.  Using 1. and 4. we get a kinetic 'wind luminosity' of&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$~~~ L_\mathrm{w} \approx \frac{1}{2} \dot{M} v_\mathrm{w}^2 \approx 1.7 \times 10^{27}\,$&lt;/span&gt;erg s&lt;span class=&quot;math-container&quot;&gt;$^{-1} \approx 4.5 \times 10^{-7} L_\odot.$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="286" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-07-19T08:59:26.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16642" PostTypeId="1" CreationDate="2016-07-19T09:18:43.973" Score="3" ViewCount="435" Body="&lt;p&gt;A blue moon is two full moons in the same month.  How many times will a blue moon occur in the average lifetime of say 70 years?&lt;/p&gt;&#xA;" OwnerUserId="12000" LastEditorUserId="616" LastEditDate="2016-07-19T11:31:05.337" LastActivityDate="2016-07-21T03:29:57.203" Title="How often is a blue moon?" Tags="&lt;the-moon&gt;&lt;moon-phases&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16643" PostTypeId="2" ParentId="16642" CreationDate="2016-07-19T10:52:11.617" Score="3" Body="&lt;p&gt;As months are not of fixed length it is only possible to calculate an average.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average frequency of a blue moon can be calculated as follows. It is the period of time it would take for an extra synodic orbit of the moon to occur in a year. Given that a year is approximately 365.2425 days and a synodic orbit is 29.5309 days, then there are about 12.368 synodic months in a year. For this to add up to another full month would take 1/0.368 years. Thus it would take about 2.716 years, or 2 years, 8 months and 18 days for another blue moon to occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for a 70 year life span the average number of blue Moons would be 70/2.716 or 25.77.&lt;/p&gt;&#xA;" OwnerUserId="11658" LastEditorUserId="11658" LastEditDate="2016-07-19T11:34:44.407" LastActivityDate="2016-07-19T11:34:44.407" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16644" PostTypeId="1" CreationDate="2016-07-19T14:26:40.520" Score="1" ViewCount="1697" Body="&lt;p&gt;Here’s a scenario:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you’re up in space and you’re far enough way that you can see the planet earth. Whilst looking down, you can see  blue sky, blue water and green land and white clouds and darker shades of different colors. This is because our sun is reflecting light off our planet that we can see these colors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you’re up in space and you’re far enough way that you can see the planet Jupiter. If you looked, would you be able to see colors in the same way we’ve seen from Juno’s pictures. I see red, white, orange… The sun is so much further away, how can the colors we see be just as bright/visible to us? The same question holds true for planets outside our solar system. They’re so far way from there light source, yet I can see colors on them. How come the distance of the light source doesn’t affect the brightness of the colors I see on these planets? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like if I was to turn on a light bulb in a huge dark room and shine that light in the middle, the surrounding areas would be much brighter than the furthest point of that room/and at some point there would be no light reaching an even further point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bear in mind I’m a basic idiot when it comes to this stuff. &lt;/p&gt;&#xA;" OwnerUserId="13533" LastActivityDate="2016-07-19T15:19:53.273" Title="Do far away planets reflect light in the same way our moon/earth reflects light?" Tags="&lt;the-sun&gt;&lt;light&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16645" PostTypeId="2" ParentId="16644" CreationDate="2016-07-19T15:19:53.273" Score="3" Body="&lt;p&gt;The short answer is yes, the outer planets of the Solar System reflect light in much the same way that the Earth does, but exoplanets are too far away for us to see clearly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you looked at Jupiter, you would be able to see the same colors that show up in Juno's photographs.  Even though the sun is less bright out there, it is still bright enough to illuminate the surface.  At Earth, the sun's irradiance (its intensity) is ~1361 W/m$^2$.  At &lt;a href=&quot;http://nssdc.gsfc.nasa.gov/planetary/factsheet/jupiterfact.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jupiter&lt;/a&gt;, the solar irradiance is ~50.26 W/m$^2$.  This is about the intensity of a common lightbulb.  Going further out, the solar irradiance at Pluto is ~0.873 W/m$^2$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this is not a very high intensity.  When you move away from the sun, things get darker and darker.  However, we have special instruments to counteract that.  We use telescopes to gather up large amounts of light and focus them, so that we are able to get a clear picture.  That is how telescopes work.  We have pictures of Jupiter from the Earth, most notably from the &lt;a href=&quot;http://hubblesite.org/gallery/album/solar_system/jupiter/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble Space Telescope&lt;/a&gt;, and you can see that they are, indeed in color. (The black dot in the middle of the Great Red Spot is the shadow of Ganymede, one of Jupiter's moons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MiFTU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MiFTU.jpg&quot; alt=&quot;A true-color photograph of Jupiter taken by the Hubble Space Telescope&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are plenty of &lt;a href=&quot;http://science.howstuffworks.com/telescope.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;resources online&lt;/a&gt; to learn how telescopes work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for planets outside our Solar System, we can't actually get pictures of them.  Any images you've seen of exoplanets are actually drawings made by artists.  They're so far away that we just don't have the resolving power to see them clearly.&lt;/p&gt;&#xA;" OwnerUserId="13426" LastActivityDate="2016-07-19T15:19:53.273" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16646" PostTypeId="1" AcceptedAnswerId="16647" CreationDate="2016-07-19T16:27:40.517" Score="10" ViewCount="1726" Body="&lt;p&gt;If in the distant future a ship were somehow able to move beyond the galaxy and into intergalactic space, how far would it have to go before we could look out a window and see the Milky Way in its entirety, rather than just edge-on as we see it from Earth?&lt;/p&gt;&#xA;" OwnerUserId="13535" OwnerDisplayName="Aerovistae" LastActivityDate="2019-06-11T07:53:48.787" Title="How far into intergalactic space would you have to go to see the Milky Way?" Tags="&lt;photography&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16647" PostTypeId="2" ParentId="16646" CreationDate="2016-07-19T16:49:58.730" Score="17" Body="&lt;p&gt;We can calculate the distance we need to be from an object in order for it to be any arbitrary &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_diameter&quot; rel=&quot;nofollow noreferrer&quot;&gt;angular size&lt;/a&gt; using the formula &lt;span class=&quot;math-container&quot;&gt;$D=\frac{d}{2\tan(\frac12\delta)}$&lt;/span&gt;.  In this case, &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; is the distance from the object, &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; is its diameter, and &lt;span class=&quot;math-container&quot;&gt;$\delta$&lt;/span&gt; is the angular size.  The Milky Way's diameter is 100,000 light years.  If we want the Milky Way to appear as large as the Sun appears from the Earth (0.5&lt;span class=&quot;math-container&quot;&gt;$^\circ$&lt;/span&gt;), we would need to be 5,700,000 light years away.  If we want it to appear much larger, we could be closer.  For the Milky Way to be the size of a 1 meter circle held at arm's length (90&lt;span class=&quot;math-container&quot;&gt;$^\circ$&lt;/span&gt;), we would need to be 50,000 light years away.&lt;/p&gt;&#xA;" OwnerUserId="13426" LastEditorUserId="13426" LastEditDate="2019-06-11T07:53:48.787" LastActivityDate="2019-06-11T07:53:48.787" CommentCount="15" ContentLicense="CC BY-SA 4.0" />
  <row Id="16648" PostTypeId="1" CreationDate="2016-07-19T17:21:05.390" Score="3" ViewCount="64" Body="&lt;p&gt;As reported in this science article &lt;a href=&quot;http://phys.org/news/2016-07-center-milky-galaxy.html&quot; rel=&quot;nofollow&quot;&gt;X Marks the spot Centre of Milky Way&lt;/a&gt;, a X shaped feature is confirmed centered on our galaxy's centre, when looking on from within the plane of the Milky Way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of the presented explanation of this feature in this article, I would think the X shape could also be explained by historical influence on star formation, and to a degree star type, of polar outflowing material from the central galactic black hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least assuming, if it were true, that its rotational axis over the eons of growth has varied about the axis perpendicular to the plain of the Milky Way. I am suggesting a precession type movement over its lifetime, rather like that of earths rotational axis. The orientation of its rotational axis impacted by large mass inflows at various ties, in particular early in its life when in falling mass to black hole total mass ratio was lesser than present times. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would think there would also be likely some gravitational interaction, like wobble, of the inner swirling mass of the interstellar material within the galactic plane as the mass of the central black hole increases non-uniformally with episodes of in-falling material, such as during earlier galaxy mergers as part of our galaxies formation process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Outflow of material from the black hole, high energy particles of mass and radiation, would from my understanding have influence on stellar formation from changing the characteristics of the interstellar material from which the stars are later formed, as well as the effects of compression of material triggering gravitational collapse for star formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of a precession type movement of a rotational body within a medium, the X shape seems so distinctly characteristic.  It seems to be a self evident logical possible explanation to consider. At least from a non-professional and intuitive point of view!  Yet such is not mentioned as explanation in the article. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why should such a model be excluded as explanation for this X feature?&lt;/p&gt;&#xA;" OwnerUserId="13359" LastActivityDate="2016-07-19T17:21:05.390" Title="Galaxy centre X structure - could this reflect impact of black hole at galactic centre" Tags="&lt;milky-way&gt;&lt;supermassive-black-hole&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16649" PostTypeId="1" CreationDate="2016-07-19T19:48:37.223" Score="2" ViewCount="120" Body="&lt;p&gt;My mom had a star named for my dad who died a few years back.   I told her that I'd try to find an image of that star and send it to her.  My dad's Birthday was 3/31/1931 if that helps.  Would love some help with this.  Thanks so much, in advance.  :-)&lt;/p&gt;&#xA;" OwnerUserId="13536" LastActivityDate="2016-07-20T15:48:12.813" Title="Can I get help getting image of Celestial Body at RA 2.220 and Declination +15.59?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16650" PostTypeId="1" CreationDate="2016-07-19T19:55:22.553" Score="8" ViewCount="113" Body="&lt;p&gt;Yesterday the K2 mission detected transit events of objects passing in front of a number of M dwarfs which could turn out to be rocky planets. If some of these planets host atmospheres, could the JWST detect IR absorption lines from molecules in these atmospheres?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the expected limiting distance for these stars to allow for such measurements by the JWST?&lt;/p&gt;&#xA;" OwnerUserId="7786" LastEditorUserId="7491" LastEditDate="2016-07-20T07:08:12.913" LastActivityDate="2017-07-01T22:09:24.823" Title="What is the projected range of the JWST to be able to detect exoplanet atmospheres?" Tags="&lt;exoplanet&gt;&lt;planetary-atmosphere&gt;&lt;infrared&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16652" PostTypeId="2" ParentId="12283" CreationDate="2016-07-20T05:36:46.047" Score="2" Body="&lt;p&gt;That is called a corona. It is caused by diffraction of light by thin clouds with small but evenly sized particles. &lt;a href=&quot;http://apod.nasa.gov/apod/ap150615.html&quot; rel=&quot;nofollow&quot;&gt;The best coronas&lt;/a&gt; show multiple coloured bands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a similar, but unrelated, phenomenon, &lt;a href=&quot;http://apod.nasa.gov/apod/ap150811.html&quot; rel=&quot;nofollow&quot;&gt;called a halo&lt;/a&gt;, which is caused by high altitude ice crystals with a regular hexagonal shape. It is much larger, 22 degrees wide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both these can be seen around the sun as well as the moon, and are fairly common, more common than rainbows (for example). &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-07-20T05:36:46.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16655" PostTypeId="2" ParentId="16649" CreationDate="2016-07-20T14:52:07.340" Score="2" Body="&lt;p&gt;I think your star is an 8th magnitude star in Aries: properly catalogued as HD 14610. It is an F5 white dwarf, a little larger and brighter than the sun, but about 300ly away, and too dim to see without a telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is located at RA 02h 22m 01s, dec +15* 59' 30''&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An image from the sky surveys: &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=HD++14610&quot; rel=&quot;nofollow&quot;&gt;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=HD++14610&lt;/a&gt; (click on &quot;Interactive Aladin-lite to get a wider view)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Others have touched on the unofficial nature of these &quot;star naming&quot; services, which some feel are fraudulent.  However if it makes your mum happy.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-07-20T15:48:12.813" LastActivityDate="2016-07-20T15:48:12.813" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16656" PostTypeId="1" CreationDate="2016-07-20T16:52:40.627" Score="1" ViewCount="97" Body="&lt;p&gt;I'm trying to use model atmospheres such as Kurucz or MARCS for the first time personally and I'm wondering if it is better to wrap it with IDL or Python in terms of writing the code, and getting the values I wanted from the model atmospheres into arrays for printing into tables. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially is it better to approach this issue using IDL or with Python, or is it just more of a preferential thing if anything?&lt;/p&gt;&#xA;" OwnerUserId="13543" LastEditorUserId="13543" LastEditDate="2016-07-20T18:09:32.623" LastActivityDate="2016-07-21T13:22:05.830" Title="Interfacing FORTRAN programs for model atmospheres with IDL or Python" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16657" PostTypeId="2" ParentId="16607" CreationDate="2016-07-20T17:44:58.957" Score="4" Body="&lt;p&gt;This might not be the type of telescopes your are thinking about, but as IACTs (Imaging Atmospheric Cherencov Telescopes) in the end also measure light in (or near) the optical range, their mirrors are of similar build. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important arrays (H.E.S.S., VERITAS and MAGIC) have, at least partially, mirrors made of glass or with a glass surfaces, coated on&#xA;the front surface with aluminium (Al) and an additional single protective layer (e.g. SiO$_2$ , Al$_2$O$_3$ ). But as they are exposed to the environment the whole time without any luxury such as protecting domes or similar, the coating degenerates pretty fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of H.E.S.S., which is located in the Khomas Highland of Namibia, the mirrors have to endure all kinds of environmental factors, beginning with the sand from the dessert, extreme temperature changes between day and night to curious birds picking at their reflected images. So after five years the optical efficiency of the mirrors was so bad, that one could actually see ones hand through from the mirror. To prolong the lifespan of the instrument, the mirrors were refurbished with a slightly different aluminium coating. In the following image the time of the mirror recoating is marked with the blue shaded area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ueGMu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ueGMu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image source: &lt;a href=&quot;https://www.mpi-hd.mpg.de/hd2012/pages/presentations/Stegmann.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.mpi-hd.mpg.de/hd2012/pages/presentations/Stegmann.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The degeneration for the mirrors of the other observatories mentioned above where of similar timescales and they also did some recoating campaigns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further information see for example the following papers, which contain some overview about the different possible techniques: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.brera.inaf.it/~canestrari/pubblicazioni/pubbl44.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.brera.inaf.it/~canestrari/pubblicazioni/pubbl44.pdf&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://arxiv.org/pdf/1406.0622&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/pdf/1406.0622&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you ever happen to be in Namibia, go visit the World's largest optical telescope:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pZeH9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pZeH9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image source: &lt;a href=&quot;https://www.lsw.uni-heidelberg.de/projects/hess/HESS/hessmultnu.phtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.lsw.uni-heidelberg.de/projects/hess/HESS/hessmultnu.phtml&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="950" LastEditorUserId="950" LastEditDate="2016-07-21T10:57:29.583" LastActivityDate="2016-07-21T10:57:29.583" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16658" PostTypeId="1" AcceptedAnswerId="16666" CreationDate="2016-07-21T02:24:19.823" Score="4" ViewCount="197" Body="&lt;p&gt;I know it is possible that a star system ejects a planet out of the star system and the planets becomes rogue planets and orbit around the galaxy. Is it also possible that a planet gains enough speed because of ejecting and escapes out of the galaxy?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="2153" LastEditDate="2016-07-21T13:54:30.643" LastActivityDate="2016-07-21T20:42:31.750" Title="Is it possible that a star system ejects an object out of the galaxy?" Tags="&lt;rogue-planet&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16659" PostTypeId="2" ParentId="16658" CreationDate="2016-07-21T03:20:26.050" Score="0" Body="&lt;p&gt;To my understanding, yes. This is not to say that eventually it may return to its original host galaxy over billions of years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would think the frequency of such events would be higher during galaxy mergers/collisions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recall a report of an ejected supermassive black hole identified that had been ejected from its host galaxy. &lt;/p&gt;&#xA;" OwnerUserId="13359" LastActivityDate="2016-07-21T03:20:26.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16660" PostTypeId="1" AcceptedAnswerId="16662" CreationDate="2016-07-21T03:26:00.533" Score="1" ViewCount="109" Body="&lt;p&gt;The moon, for example, appears very bright in the night sky, but quite dark from photographs on/near its surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, Mars and Venus appear as bright dots in the sky but in closer photographs they're quite dark.&lt;/p&gt;&#xA;" OwnerUserId="6610" LastActivityDate="2016-07-22T06:48:40.123" Title="Why celestial objects appear brighter in the night sky than in photographs from closer proximity?" Tags="&lt;light&gt;&lt;night-sky&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16661" PostTypeId="2" ParentId="16642" CreationDate="2016-07-21T03:29:57.203" Score="0" Body="&lt;p&gt;Be aware of the other kind of event also referred to as &quot;a blue moon&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some large volcanic events, with volcanic ash sent high into the stratosphere and circulating widely around the planet, the moon has appeared blueish in hue rather than the usual reddish, such as seen with total eclipses of the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These &quot;blue moon&quot; events are very rare, and can not be predicted. &lt;/p&gt;&#xA;" OwnerUserId="13359" LastActivityDate="2016-07-21T03:29:57.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16662" PostTypeId="2" ParentId="16660" CreationDate="2016-07-21T06:59:54.157" Score="6" Body="&lt;p&gt;The Moon, Mars, Venus are seen against a dark sky using an eye whose aperture is adjusted to the average brightness of the visual field. So bright small sources are over-exposed and so appear bright.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photographs of planetary surfaces are taken with aperture and exposure set for proper exposure. So appear closer to how they would look if they filled your eye's field of view and so the aperture would be set to properly expose the surface.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2016-07-21T06:59:54.157" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16665" PostTypeId="2" ParentId="16656" CreationDate="2016-07-21T13:22:05.830" Score="2" Body="&lt;p&gt;After looking into the various wrappers and ways Python and IDL work with FORTRAN I came to the conclusion that it is more of a preference which language you use. However as people continuously add astronomy support to Python this may change, but as of now, it seems to be a question of preference. &lt;/p&gt;&#xA;" OwnerUserId="13543" LastActivityDate="2016-07-21T13:22:05.830" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16666" PostTypeId="2" ParentId="16658" CreationDate="2016-07-21T13:45:52.277" Score="6" Body="&lt;p&gt;Intergalactic planets have never been observed.  They're too small and it's too great a distance, but &lt;a href=&quot;https://en.wikipedia.org/wiki/Intergalactic_star&quot; rel=&quot;nofollow noreferrer&quot;&gt;intergalactic stars&lt;/a&gt; have, and if stars can be ejected from a galaxy, planets can to.  The orbital mechanics is the same for both.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The escape velocity of the galaxy is impressive.  For solar-systems, where most of the mass is in the star in the center, the escape velocity is just about the orbital velocity times the square root of two.   &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/vesc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Milky way excape velocity calculation is larger and more complicated because most of the mass is in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_matter_halo&quot; rel=&quot;nofollow noreferrer&quot;&gt;dark matter halo&lt;/a&gt; and very much spread out, not localized in the center.   The center is still more dense, the mass is far more distributed so the square root of two rule doesn't apply.  Using the Sun as an example, the orbital velocity of our sun is about 220 km/s and it's escape velocity is 537 km/s.  &lt;a href=&quot;https://space.stackexchange.com/questions/3948/escape-velocity-for-our-galaxy&quot;&gt;Source&lt;/a&gt;.   As objects moves further towards the edge of the Milky way, the escape velocity decreases and the orbital velocity generally increases, so at close to the edge of the Milky way, you'd get something close to the square root of two ratio again.  That means an object close to the edge of the Milky way only needs to add 41.4% of it's velocity to escape.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity assists depend on the relative velocity of the larger object to the &quot;stationary Milkyway&quot;, so, if our solar-system was to fly close enough to very massive star and get a big gravity assist, assuming the massive star has a similar orbital velocity around the Milky way, 220 km/s, that could add as much as 440 km/s to our solar-system's velocity which would be enough to escape the Milky way.   Only large stars have enough gravity for that kind of kick, but there's enough large stars in our galaxy that they shouldn't have a hard time ejecting objects that fly reasonably close to them.   You probably need at least a half dozen or dozen or so solar masses and a close fly-by, but, statistically, it should happen from time to time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia article points out that galactic collisions could be responsible for these stellar ejections, and galactic collisions probably do eject a great deal more material compared to a galaxy in a peaceful spiral arm rotation, but it's still possible for &lt;a href=&quot;http://solarsystem.nasa.gov/basics/grav/primer.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravity assists&lt;/a&gt; to cause objects to be ejected even from a galaxy not currently colliding with anohter, because twice the orbital velocity of the large star plus the orbital velocity of the approaching star or planet, when combined is greater than escape velocity, at least for the outer parts of the spiral galaxy.   Needless to say, most gravity assists don't line up perfectly, and half of them push an object further in, not further out, but statistically, and with multiple assists, it's very likely that stars and rogue planets get ejected from the Milky-way, it's probably happening as I write this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By some estimates, there might be a &lt;a href=&quot;https://cosmosmagazine.com/space/lone-stars-half-stars-universe-may-lie-outside-galaxies&quot; rel=&quot;nofollow noreferrer&quot;&gt;HUGE number of objects&lt;/a&gt; that doesn't belong to any galaxy.    &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2016-07-21T13:45:52.277" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16668" PostTypeId="2" ParentId="16658" CreationDate="2016-07-21T20:42:31.750" Score="0" Body="&lt;p&gt;I'd like to point out another interesting side of this question. You specifically ask&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it also possible that a planet gains enough speed because of ejecting and escapes out of the galaxy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This question doesn't specify the means of ejecting which opens the possibilities for interesting results. The normal process of ejection, causing rogue planets, is a gravitational collision (note the difference from physical collision) between two planets in a given stellar system, usually during the formation period. The ejection speed of the ejected planet relies on the mass of the two &quot;colliding&quot; planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it varies widely depending on where you are in the galaxy, the escape velocity from our rough position in the galaxy &lt;a href=&quot;https://space.stackexchange.com/questions/3948/escape-velocity-for-our-galaxy&quot;&gt;can be calculated&lt;/a&gt; to be $\sim500\:km/s$. This is really fast and more than double our orbital speed around the galaxy. Planetary collisions are highly unlikely to cause such an ejection speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if two colliding planets in a stellar system can't cause one to be ejected outside our galaxy, how might it happen? The answer is stellar collisions. If the ejected speed relies on mass, throw some more mass in there. We have observed numerous examples of what we call &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_kinematics#Hypervelocity_stars&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hypervelocity Stars&lt;/a&gt;. These are stars moving throughout our galaxy at velocities on the order of $\sim1000\:km/s$ or more and are thus on an escape trajectory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not completely known exactly how these stars achieve such a monstrous velocity, but they seem to originate from dense stellar regions such as the core of the galaxy or stellar clusters. This implies the cause is from chance stellar collisions where one star just so happens to get a huge velocity boost. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this can happen to stars, why not planets? It would likely be a very rare occurrence, but I can imagine a scenario where a planet forms around a star in one of these dense regions, that star gravitationally interacts with another star and with the proper setup, the planet is ejected at high speeds in the process.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2016-07-21T20:42:31.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16669" PostTypeId="1" CreationDate="2016-07-22T04:48:36.160" Score="4" ViewCount="103" Body="&lt;p&gt;Is it possible to measure to measure isotopic abundance of remote astronomical objects - ie measurement without having a sample to feed into a mass spectrometer? Do different isotopes show any differences in absorption or emission spectra which can be used for this purpose?&lt;/p&gt;&#xA;" OwnerUserId="13125" LastActivityDate="2016-07-22T08:43:30.163" Title="Is remote measurement of isotope ratios possible, or must a sample be aquired" Tags="&lt;telescope&gt;&lt;observational-astronomy&gt;&lt;light&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16670" PostTypeId="2" ParentId="16660" CreationDate="2016-07-22T06:48:40.123" Score="1" Body="&lt;p&gt;To add to Conrad Turner, if you look at the moon during the day against a bright sky, then i looks quite &quot;washed out&quot; and much more like the photographs taken from near its surface.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2016-07-22T06:48:40.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16671" PostTypeId="2" ParentId="16669" CreationDate="2016-07-22T08:43:30.163" Score="4" Body="&lt;p&gt;Yes, isotopic ratios can be readily established by spectroscopy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecules where the constituent atoms differ by one mass unit have different vibrational and rotational energy levels and the gaps between them depend on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Reduced_mass&quot;&gt;reduced mass&lt;/a&gt; of the molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in a diatomic molecule&#xA;$$ \mu = \frac{m_1 m_2}{m_1 + m_2}$$&#xA;and the vibrational frequencies depend on $\mu^{-1/2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This technique is used for example to look at carbon monoxide feature in the infrared and establish the ratio of $^{13}$C/$^{12}$C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Isotopic_shift&quot;&gt;isotopic shift&lt;/a&gt; for atoms is much smaller in general, but is still detectable. The different mass of the nucleus changes the reduced mass of the electron (a little) and this changes the wavelength of the characteristic transition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting example of this has been the search for $^6$Li in the spectra of population II stars (e.g. &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0510636&quot;&gt;Asplund et al. 2006&lt;/a&gt;). This isotope of Li should be produced in the big bang along with the more common $^7$Li but is more readily burned in stellar interiors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main optical resonance line of lithium occurs at 670.8 nm. The difference between the  wavelength of this line (actually, it is an unresolved doublet) is 0.015nm. This separation cannot be resolved in stellar spectra and instead the shape of the absorption line has to be modelled to estimate the $^6$Li/$^7$Li ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The isotopic shifts become &lt;em&gt;smaller&lt;/em&gt; for heavier elements.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-07-22T08:43:30.163" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16672" PostTypeId="5" CreationDate="2016-07-22T12:45:42.637" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2016-07-22T12:45:42.637" LastActivityDate="2016-07-22T12:45:42.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16673" PostTypeId="4" CreationDate="2016-07-22T12:45:42.637" Score="0" Body="Questions regarding the use of a reflective surface, for example in optics." OwnerUserId="10172" LastEditorUserId="10172" LastEditDate="2016-07-22T13:16:07.260" LastActivityDate="2016-07-22T13:16:07.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16674" PostTypeId="1" AcceptedAnswerId="16675" CreationDate="2016-07-22T13:20:31.517" Score="6" ViewCount="783" Body="&lt;p&gt;The temperature of the Sun's atmosphere, also referred to as the solar corona, is known to be hot even hotter than the Sun's surface. What is its temperature and how is it measured? How does it compare to other stars and how does it vary across the Sun?&lt;/p&gt;&#xA;" OwnerUserId="117" LastEditorUserId="13663" LastEditDate="2016-08-18T14:56:11.890" LastActivityDate="2016-08-18T14:56:11.890" Title="What is the temperature of the solar atmosphere (the corona) and how is it measured?" Tags="&lt;the-sun&gt;&lt;spectroscopy&gt;&lt;plasma-physics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16675" PostTypeId="2" ParentId="16674" CreationDate="2016-07-22T14:24:58.090" Score="10" Body="&lt;p&gt;This is a rather broad question and this will not be a fully comprehensive answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no single temperature to the solar corona. The coronal temperature varies by an order of magnitude from place-to-place. It is hottest ($\sim 10^{7}$ K) in magneticloops undergoing flares, which tend to be anchored in low latitude regions. It is coolest (a bit less than $10^{6}$ k in coronal holes which are usually at high latitudes. It is hard even to come up with some average number, because it is not altogether clear how the various pieces should be weighted in any average. It is also cyclic - in that magnetic activity, and average coronal temperatures vary with all the other magnetic phenomena on the Sun with a roughly 11-year period. If you weight it by &lt;a href=&quot;https://astronomy.stackexchange.com/questions/13429/what-is-emission-measure&quot;&gt;emission measure&lt;/a&gt;, which is density squared times volume, then the solar corona is between $10^{6}$K at solar minimum to about $3\times 10^{6}$K at solar maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature of coronal plasma can be measured using X-ray spectroscopy. If you assume that the plasma can be represented by some equilibrium temperature, then you can make a prediction of what the X-ray/EUV spectrum will look like. To some extent the spectrum will also depend on the composition of the plasma and also on its density, but these produce comparatively minor differences compared to changes in temperature and can be simultaneously estimated in most cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For stars you essentially do the same thing but with worse data. Here you are getting some sort of average temperature for the whole corona - there is no spatial resolution. X-ray satellite observatories like &lt;a href=&quot;http://chandra.si.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chandra&lt;/a&gt; and &lt;a href=&quot;http://www.cosmos.esa.int/web/xmm-newton&quot; rel=&quot;nofollow noreferrer&quot;&gt;XMM-Newton&lt;/a&gt; have grating spectrographs and detectors with energy sensitivity that can record X-ray spectra from stars. You then use software like &lt;a href=&quot;https://heasarc.gsfc.nasa.gov/xanadu/xspec/&quot; rel=&quot;nofollow noreferrer&quot;&gt;XSPEC&lt;/a&gt; along with an optically thin plasma model, such as that of &lt;a href=&quot;https://heasarc.gsfc.nasa.gov/xanadu/xspec/manual/XSmodelMekal.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mewe-Kaastra-Liedahl&lt;/a&gt;, (and not forgetting to include absorption by the interstellar medium) to get the best fitting temperature for the measured spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as how the Sun compares with other stars, an excellent starting point is the review by &lt;a href=&quot;http://arxiv.org/abs/0712.1763&quot; rel=&quot;nofollow noreferrer&quot;&gt;Guedel (2007)&lt;/a&gt;. A short summary would be that the temperature of a stellar corona depends on it's rotation rate - the faster it rotates, the hotter it is. Because (single) stars lose angular momentum and hence rotate more slowly as they age, then both X-ray luminosity and (average) coronal temperature decrease with age. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/astro-ph/0503546&quot; rel=&quot;nofollow noreferrer&quot;&gt;Telleschi et al. (2005)&lt;/a&gt; studied solar analogues at a variety of ages. They found a relationship that the average temperature&#xA;$$\langle T \rangle = 1.16\times10^{7} P_{\rm rot}^{-0.48}\ {\mathrm K},$$&#xA;where the rotation period $P_{\mathrm rot}$ is in days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that rotation period and age are linked by a &quot;gyrochronology&quot; relation roughly of the form $P_{\rm rot} \simeq 27 (t/4.5\ {\rm Gyr})^{1/2}$, for $t \geq 100$ Myr, this means that&#xA;$$\langle T \rangle = 2.4\times10^{6} (t/4.5\ {\rm Gyr})^{-0.24}\ {\mathrm K}.$$&#xA;At ages below 100 Myr, the coronae &quot;saturate&quot; at the luminosity and temperature of a 100 Myr star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, stars with photospheric temperatures cooler than the Sun tend to have hotter coronae, whilst those with warmer photospheres have cooler coronae. Stars with spectral types A basically don't have coronae, and hotter O and B stars have coronal winds that are heated by completely different mechanisms to the solar corona.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-07-22T21:17:36.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16677" PostTypeId="1" AcceptedAnswerId="16679" CreationDate="2016-07-23T04:29:45.273" Score="3" ViewCount="217" Body="&lt;p&gt;Wikipedia says that there are now 67 &lt;em&gt;known&lt;/em&gt; moons of Jupiter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It got me wondering: &lt;strong&gt;is it really possible in our day and age that there are more, but we haven't observed them yet?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, there is a Jovian moon discovered as recently as in 2011, so I wouldn't be amazed if the answer was 'yes'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really though, choosing Jupiter for this question was a bit arbitrary on my part, as the same question can be asked for other more distant and bigger planets of our solar system, like Saturn.&lt;/p&gt;&#xA;" OwnerUserId="13292" LastEditorUserId="2618" LastEditDate="2017-03-07T00:45:55.460" LastActivityDate="2017-03-07T00:45:55.460" Title="More than 67 Jovian moons?" Tags="&lt;observational-astronomy&gt;&lt;jupiter&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16678" PostTypeId="1" AcceptedAnswerId="16680" CreationDate="2016-07-23T04:36:18.520" Score="11" ViewCount="569" Body="&lt;p&gt;Neutron stars and black holes are hard to detect when they are solitary, and there seems to be big uncertainties about how common they are. White dwarfs are much easier to detect and the nearest one is Sirius B only 2.6 parsec from here. Should we expect to have exotic company even closer by? How likely is it that we have a yet undetected compact star remnant nearby, as in as close as our nearest active star? What are the odds of spotting one of them nearby?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could a yet undetected one be discovered? Could one of the upcoming sky surveying telescopes catch it or would one have to rely on a rare microlensing event? How would it then be observed? Would such an exotic object, say, just one parsec away, give important insights in physics given its relativistic effects and strange composition?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2019-10-16T14:28:33.620" Title="How far away is the nearest compact star remnant likely to be?" Tags="&lt;black-hole&gt;&lt;observational-astronomy&gt;&lt;neutron-star&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16679" PostTypeId="2" ParentId="16677" CreationDate="2016-07-23T05:22:40.320" Score="7" Body="&lt;p&gt;Yes, there probably are more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a minor problem of definition. Unlike the case for &quot;planet&quot;, the IAU has not produced a definition of &quot;moon&quot;. There is no requirement to be round, or have cleared an orbit. So any chunk of rock or ice that we can see orbiting a planet can reasonably be called a moon, and Jupiter has, over the years, picked up a lot of stuff: Small asteroids that it has captured from solar orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is likely that there are other, sub-1km sized objects in orbit around Jupiter, that we haven't noticed yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Juno is, as of 2016, in orbit around Jupiter. However her main focus of study is the planet itself, not hunting for moons. The recent moon discoveries have been made by terrestrial observations. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-07-23T05:22:40.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16680" PostTypeId="2" ParentId="16678" CreationDate="2016-07-23T09:42:16.333" Score="16" Body="&lt;p&gt;There can be no closer white dwarf. The &lt;a href=&quot;https://public.nrao.edu/news/pressreleases/cold-white-dwarf&quot; rel=&quot;noreferrer&quot;&gt;coolest, oldest&lt;/a&gt; white dwarfs (3000K), would be rare, but are still luminous enough &lt;span class=&quot;math-container&quot;&gt;$6\times10^{-6} L_{\odot}$&lt;/span&gt; to have been easily detected at distances closer than Sirius. At the distance of Sirius, such an object would have a visual magnitude of around 12-13 and would be brighter at near infrared wavelengths where all sky surveys such as 2MASS would definitely have spotted it from its parallax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutron stars and black holes could be almost undetectable but are expected to be roughly 10 and 100 times rarer respectively. Calculated as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us assume that &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; stars have ever been born in the Milky Way galaxy, and given them masses between 0.1 and 100&lt;span class=&quot;math-container&quot;&gt;$M_{\odot}$&lt;/span&gt;. Next, assume that stars have been born with a mass distribution that approximates to the Salpeter mass function - &lt;span class=&quot;math-container&quot;&gt;$n(m) \propto m^{-2.3}$&lt;/span&gt;. Then assume that all stars with mass &lt;span class=&quot;math-container&quot;&gt;$m&amp;gt;25M_{\odot}$&lt;/span&gt; end their lives as black holes, all stars with &lt;span class=&quot;math-container&quot;&gt;$8&amp;lt;m/M_{\odot}&amp;lt;25$&lt;/span&gt; end their lives as neutron stars and about &lt;em&gt;half&lt;/em&gt; the stars with &lt;span class=&quot;math-container&quot;&gt;$0.9&amp;lt;m/M_{\odot}&amp;lt;8$&lt;/span&gt; end their lives as white dwarfs (the other half are still alive as main sequence stars, as are all stars born with lower masses).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if &lt;span class=&quot;math-container&quot;&gt;$n(m) = Am^{-2.3}$&lt;/span&gt;, then&#xA;&lt;span class=&quot;math-container&quot;&gt;$$N = \int^{100}_{0.1} A m^{-2.3}\ dm$$&lt;/span&gt;&#xA;and thus &lt;span class=&quot;math-container&quot;&gt;$A=0.065N$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of black holes created will be &#xA;&lt;span class=&quot;math-container&quot;&gt;$$N_{BH} = \int^{100}_{25} Am^{-2.3}\ dm = 6.4\times10^{-4} N$$&lt;/span&gt;&#xA;i.e 0.064% of stars in the Galaxy become black holes. NB: The finite lifetime of the galaxy is irrelevant here because it is much longer than the lifetime of black hole progenitors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a similar way, the number of neutron stars &#xA;&lt;span class=&quot;math-container&quot;&gt;$$N_{NS} = \int^{25}_{8} Am^{-2.3}\ dm = 2.6\times10^{-3}N$$&lt;/span&gt;&#xA;and the number of white dwarfs&#xA;&lt;span class=&quot;math-container&quot;&gt;$$N_{WD} = 0.5\times \int^{8}_{0.9} Am^{-2.3}\ dm = 0.027 N$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we use these results as scaling factors to apply to the local stellar population. There are about 1000 &quot;normal&quot; stars in a sphere of 15 pc radius, thus a density of 0.07 pc&lt;span class=&quot;math-container&quot;&gt;$^{-3}$&lt;/span&gt;. Thus one uses the results above to calculate the density of compact remnants and then take &lt;span class=&quot;math-container&quot;&gt;$(3/4\pi n)^{1/3}$&lt;/span&gt; as an estimate of the average distance to one of them. This gives an expectation value of 18 pc to the nearest black hole, 11 pc to the nearest neutron star and 5 pc to the nearest white dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the distance to the nearest white dwarf is roughly as expected. For reasons discussed in &lt;a href=&quot;http:///astronomy.stackexchange.com/questions/4725/statistically-what-would-the-average-distance-of-the-closest-black-hole-be&quot; rel=&quot;noreferrer&quot;&gt;my answer to this related question&lt;/a&gt; the distance calcuated to the nearest black hole and neutron star remnants is likely to be an &lt;em&gt;underestimate&lt;/em&gt; because many escape from the Galaxy or have very high velocity dispersions and much larger Galactic scale heights than normal stars. So whilst it is &lt;em&gt;possible&lt;/em&gt; that an unseen one exists closer than Sirius, it is highly unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could such an object be detected? An old, cold neutron star or black hole could be completely undetectable at all wavelengths of electromagnetic radiation - though it could be fruitful to examine carefully any candidate detections [see below] for signs of X-ray emission due to accretion from the interstellar medium). But your question has I think the correct suggestion. The objects would likely have a substantial proper motion and so there is a decent chance that you would see a &quot;moving&quot; gravitational lensing signature. This would still be very small unless the object just happened to pass directly in front of a background star - but such a microlensing event would be transient and may not be observed. More likely is that Gaia would pick up the subtle shifts in the positions of background stars changing over the 5 years of its mission. As per your other question: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/16578/will-gaia-detect-inactive-neutron-stars?rq=1&quot;&gt;Will Gaia detect inactive neutron stars?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-10-16T14:28:33.620" LastActivityDate="2019-10-16T14:28:33.620" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="16681" PostTypeId="1" AcceptedAnswerId="16682" CreationDate="2016-07-23T16:08:52.903" Score="1" ViewCount="105" Body="&lt;p&gt;I'm modelling the solar system. I'm using data from &lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow&quot;&gt;Nasa&lt;/a&gt;.&#xA;To do this I'm using the barycenter of the planets Jupiter, Saturn, Uranus, Neptune. However the website fails to give the Gravitational Parameter $\mu$ for these barycenters, even though it does do so for Pluto's barycenter.&#xA;To make matters more confusing, the $\mu$'s given at &lt;a href=&quot;https://en.wikipedia.org/wiki/Standard_gravitational_parameter&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; are slightly different, though I assume these are just out of date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'd like to know if there's an as up to date resource as possible available for these parameters. Paywalls should be OK as long as they are of major publications.&#xA;$$\frac{\mathrm{AU}^3}{\mathrm{M}_\odot\cdot\mathrm{day}^2}$$&#xA;-form also good.&lt;/p&gt;&#xA;" OwnerUserId="12043" LastActivityDate="2016-07-23T22:24:25.063" Title="Jupiter, Saturn, Uranus, Neptune system GM" Tags="&lt;mass&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16682" PostTypeId="2" ParentId="16681" CreationDate="2016-07-23T22:07:18.173" Score="3" Body="&lt;p&gt;In the &lt;a href=&quot;http://ilrs.gsfc.nasa.gov/docs/2014/196C.pdf&quot; rel=&quot;nofollow&quot;&gt;JPL report on ephemerides DE430 and DE431&lt;/a&gt;,&#xA;on page 49, table 8 gives the mass parameters they used for the major planets.&#xA;They cite the&#xA;&lt;a href=&quot;http://dx.doi.org/10.1007/s10569-011-9352-4&quot; rel=&quot;nofollow&quot;&gt;IAU 2009 system of astronomical constants&lt;/a&gt;.&#xA;There may be newer estimates, but to verify your own simulation against HORIZONS, you might as well use the same constants they did.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-07-23T22:07:18.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16684" PostTypeId="1" AcceptedAnswerId="16685" CreationDate="2016-07-24T07:10:47.913" Score="2" ViewCount="203" Body="&lt;p&gt;so I recently wrote a basic code to analyze the distance between two points which are stored in two separate excel files. Now that I have that code working its time to apply the legitimate use and use this code to analyze not just 2 dimensions (x,y) but three (ra,dec,z). In the data sheets now I have stored the coordinates of two separate galaxies.  Heres the amended code below:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import math&#xA;import pandas&#xA;file1 = pandas.read_excel('Book1.xlsx')&#xA;file2 = pandas.read_excel('Book2.xlsx')&#xA;file1['RA_diff'] = file2['RA'] - file1['RA']&#xA;file1['DEC_diff'] = file2['DEC'] - file1['DEC']&#xA;file1['Z_diff'] = file2['Z'] - file1['Z']&#xA;dist = file1.apply(lambda row: math.hypot(row['RA_diff'], row['DEC_diff', row['Z_diff']]),   axis=1)&#xA;if dist.values &amp;gt;= .5:&#xA;    print 'no match'&#xA;elif dist.values &amp;lt;= .5:&#xA;    print True, dist&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This code works great with just the RA and DEC but not no much with the Z added in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My hope is that somebody please give me some pointers on how to shorten this whole process by using the method:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from astropy.coordinates import SkyCoord&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Basically: How do I use this tool to my advantage in reading in (from two separate excel documents) two separate coordinates of (ra, dec, z) to determine if the objects have any proximity to one another?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My research has turned up rather fruitless to analyze the points in this way specifically as I mostly simply don't understand the help pages. This is my first ever coding project so please be gentle in your advice, recommendations, help, feedback or ANYTHING else you might be able to offer. Thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="13554" LastActivityDate="2016-07-24T08:10:08.333" Title="Using SkyChoord to read in multiple coordinates." Tags="&lt;galaxy&gt;&lt;distances&gt;&lt;data-analysis&gt;&lt;python&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16685" PostTypeId="2" ParentId="16684" CreationDate="2016-07-24T08:10:08.333" Score="2" Body="&lt;p&gt;From the &lt;a href=&quot;http://docs.astropy.org/en/stable/coordinates/index.html&quot; rel=&quot;nofollow&quot;&gt;documentation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt;&amp;gt;&amp;gt; c1 = SkyCoord(ra=10*u.degree, dec=9*u.degree, distance=10*u.pc, frame='icrs')&#xA;&amp;gt;&amp;gt;&amp;gt; c2 = SkyCoord(ra=11*u.degree, dec=10*u.degree, distance=11.5*u.pc, frame='icrs')&#xA;&amp;gt;&amp;gt;&amp;gt; c1.separation_3d(c2)  &#xA;&amp;lt;Distance 1.5228602415117989 pc&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The rest of the code is just reading the excel files and printing the results. What you have now is incorrect, even without the &quot;z&quot; coordinate. You can't calculate the separation of two points in spherical coordinates by Pythagoras' theorem: eg (30*, 0*) and (30*, 359*) should be very close. You need to convert to cartesian coordinates first, which is what the astropy module does.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-07-24T08:10:08.333" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16687" PostTypeId="1" AcceptedAnswerId="16688" CreationDate="2016-07-24T18:02:13.947" Score="2" ViewCount="945" Body="&lt;p&gt;I found this picture and I am really wondering if it is possible for the moon to get as close to the earth as this picture shows?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;( Super moon above Rio De Janeiro) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VPapg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VPapg.jpg&quot; alt=&quot;Super moon above Rio De Janeiro&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13571" LastEditorUserId="2270" LastEditDate="2016-07-27T15:00:08.223" LastActivityDate="2019-03-22T06:21:39.287" Title="Is this a real picture of the moon?" Tags="&lt;positional-astronomy&gt;&lt;the-moon&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16688" PostTypeId="2" ParentId="16687" CreationDate="2016-07-24T18:34:55.370" Score="7" Body="&lt;p&gt;I don't know whether this  picture is a fake, but you could get this sort of image without cheating, if it were taken a long way away from Rio, and the magnification cranked up.  In other words, get far enough away from Rio that its angular diameter is about 1 minute, and take the picture from there.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2016-07-24T18:34:55.370" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16689" PostTypeId="1" CreationDate="2016-07-24T22:45:44.997" Score="0" ViewCount="85" Body="&lt;p&gt;I'm becoming interested in the currently identified stars in the universe. There is a wide swath where much fewer named stars are shown. Why is this?&#xA;The image below is a mollweide projection of the universe, with the earth at its center. It shows about 230,000 named stars located at their right ascension/declination points.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/6rkMR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6rkMR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13507" LastActivityDate="2016-07-25T14:43:11.940" Title="Parts of the Universe Shows Fewer Named Stars" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="8" ClosedDate="2016-07-25T18:15:37.453" ContentLicense="CC BY-SA 3.0" />
  <row Id="16694" PostTypeId="1" CreationDate="2016-07-25T05:19:02.513" Score="1" ViewCount="74" Body="&lt;p&gt;I'm trying to locate three stars inside Cassiopeia, gam cas, &lt;a href=&quot;https://en.wikipedia.org/wiki/Alpha_Cassiopeiae&quot; rel=&quot;nofollow&quot;&gt;Schedar&lt;/a&gt;, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Beta_Cassiopeiae&quot; rel=&quot;nofollow&quot;&gt;Caph&lt;/a&gt;. I have found &lt;a href=&quot;https://en.wikipedia.org/wiki/Gamma_Cassiopeiae&quot; rel=&quot;nofollow&quot;&gt;gamma Cas&lt;/a&gt; inside the Tycho-2 catalog supp1.dat file, however when I look for Schedar or Caph I have no luck. I converted their RA and dec to decimal degree, Schedar = (10.12, 56.53) and Caph = (2.29, 59) and used the &lt;a href=&quot;http://en.wikipedia.org/wiki/Vim_%28text_editor%29&quot; rel=&quot;nofollow&quot;&gt;Vim&lt;/a&gt; command &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;:/|002\.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;to find their location in the star catalog, but I haven't found any stars that match the RA, dec, or even visual brightness, which are 2.24 for Schedar and 2.28 for Caph. I'm kind of stuck at this point and not sure where to head from here.&lt;/p&gt;&#xA;" OwnerUserId="13574" LastEditorUserId="332" LastEditDate="2016-07-28T07:36:15.097" LastActivityDate="2016-08-27T07:52:44.497" Title="Can't find Cassiopias Schedar or Caph in Tycho-2 catalog supplement 1" Tags="&lt;star-catalogues&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16695" PostTypeId="2" ParentId="16694" CreationDate="2016-07-25T08:29:32.870" Score="0" Body="&lt;p&gt;The supplemental catalog lists stars (and other point sources) that are found in Tycho1 or Hippacos, but not in Tycho2, and is much smaller than the main catalog with about 17000 record compared with over 2.5 million in the main catalog.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I can't but think that you are doing something very wrong in searching a data file like Tycho with vi and a regular expression. You should be using the dedicated tools designed to work with the files, such as the wcstools program &lt;code&gt;scat&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Schedar is in the Tycho catalog with reference TYC 3663-2668-1&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-07-25T08:29:32.870" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16696" PostTypeId="2" ParentId="13165" CreationDate="2016-07-25T09:47:41.467" Score="5" Body="&lt;p&gt;Here is how you do it &quot;properly&quot; for the Hipparcos data. As Warrick correctly points out, what you have done in your question is massively biased towards giant and supergiant stars, which actually form the very small minority of stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must form a &lt;em&gt;volume limited sample&lt;/em&gt;. To do this, sort the stars by distance (1/parallax) and choose a cut off point. Your sample will always be incomplete, but the greater your distance cut off, the more incomplete it will be and it will become incomplete for more intrinsically luminous stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a philosophical problem to solve here in terms of what you are trying to achieve. The vast majority of stars in the Galaxy are faint M dwarfs with an absolute magnitude of $&amp;gt;10$. As Hipparcos is only complete to around magnitude 10-11, you will only get these M dwarfs in your subsample if you limit yourself to 10pc. But you will find that this sample contains no evolved stars (too rare) and no white dwarfs (too faint).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &#xA;This has piqued my interest again, so I have a practical (approximate) solution, based on a two-part process. The first part involves a &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0143-0807/24/2/303&quot; rel=&quot;noreferrer&quot;&gt;paper I wrote&lt;/a&gt; (actully an undergraduate experiment) based on the nearest 1000 stars to the Sun (from the Gliese &amp;amp; Jahreiss catalogue CNS3). This sample is roughly complete down to mid M-dwarfs, so everything else I say, and the results I give, only applies to a sample of stars more massive than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at this volume-limited sample of 1000 nearby stars you can straight away say something about the relative numbers of different types of stars in the Galactic disk (saying something about stars anywhere else in the Galaxy is filled with much more uncertainty). A colour-magnitude diagram is shown below, and from this we see that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun is among the brightest of stars - brighter than 95% of other stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About 6% of the population are white dwarfs (though a few faint, old white dwarfs may still be missing from the sample). This makes sense. If you integrate a typical mass function assuming only stars more massive than about 1$M_{\odot}$ have even had time to become white dwarfs, then this is what you get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only 0.9% of the population are giants. The reason for this is that only a small proportion of stars are massive enough to have evolved into giants on the lifetime of the Galaxy. But once they are there, their lifetimes are short compared to the main sequence phase and most have become white dwarfs (see above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a handful of objects, maybe 0.5%, that could be classified as subdwarfs, between the main sequence and the white dwarfs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in broad terms: 92.5% of stars (above $\sim 0.2M_{\odot}$) are main sequence (class V), 6% are white dwarfs, 1% are giants (class III) and 0.5% are subdwarfs (class VI).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NeRqe.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NeRqe.png&quot; alt=&quot;Nearest 1000 stars&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no very massive stars or supergiants in the immediate solar neighbourhood. That is because they are very rare. To get a better estimate we need to look at a larger volume limited sample. To do this I took all the stars (about 7000) from the Hipparcos catalogue that are closer than 50pc and assumed this was &lt;em&gt;complete&lt;/em&gt; to just below a solar luminosity and assumed that those stars with an absolute magnitude brighter than the Sun (1949 stars with $M_V&amp;lt;4.5$ represent 5% of the &lt;em&gt;total&lt;/em&gt; population in this volume. The absolute magnitude vs colour diagram for this sample is shown below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of these 1949 luminous stars, I find roughly 190 are giants - giving a giant fraction of $5\times 190/1940=0.5$%, in reasonable agreement with the nearby star sample based on smaller numbers. There are still NO supergiants, even in this larger sample. Thus supergiants have a frequency of $\leq 5 \times 1/1949 = 0.0025$%. i.e. less than 1 star per 40,000 is a supergiant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/It5A3.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/It5A3.png&quot; alt=&quot;Hipparcos CMD of 7000 stars closer than 50pc&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-10-02T18:18:51.603" LastActivityDate="2016-10-02T18:18:51.603" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16698" PostTypeId="2" ParentId="16689" CreationDate="2016-07-25T14:43:11.940" Score="0" Body="&lt;p&gt;The database used for the above was for Deep Sky Objects, not stars.&#xA;When I changed the database the correct one, then the following image is distribution of 120,000 named stars. Thanks for flagging my error.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ZTUFS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZTUFS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13507" LastActivityDate="2016-07-25T14:43:11.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16699" PostTypeId="2" ParentId="16578" CreationDate="2016-07-25T15:32:53.380" Score="4" Body="&lt;p&gt;I will throw out a suggestion that could hardly be considered an &quot;answer&quot;, but it would be interesting if someone could put flesh on the bones of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I first read your question I thought you were asking whether gravitational perturbations from an unseen neutron star would cause a measurable effect on the positions of other stars. I think the answer to this is certainly no unless the neutron star was so close to another star that they were (or could be practically considered as) a binary system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Binary systems featuring neutron stars could definitely exist (they won't be common though) and Gaia should see them if they are close enough and the companion star is bright enough. A variant of this would be a neutron star that just happens to drift (probably at high speed) close to another star, producing a notable change in the photocentre of the visible star over the 5-year Gaia mission. This is unlikely because close encounters between unrelated stars are very unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then I started thinking about lensing. Even the presence of planets in our solar system causes a gravitational lensing signature - the positions of stars are distorted by $\sim 1-10$ microarcseconds close to the positions of planets and this is detectable by Gaia. The distortion can be spotted because of course the planets &lt;em&gt;move&lt;/em&gt; with respect to the background stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In GR, the lensing angular deflection is&#xA;$$\alpha = \frac{4GM}{c^2r},$$&#xA;where $r$ is the closest approach of the light to the lensing object of mass $M$. For a lensing object at distance $d$ and a background star at an angular separation of $\theta$ from it, then&#xA;$$\alpha = 0.04 \left(\frac{M}{M_{\odot}}\right)\left(\frac{d}{\rm 1\ pc}\right)^{-1} \left(\frac{\theta}{\rm 1\ arcsec}\right)^{-1}\ \ \mu{\rm \ arcsec}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutron stars are of order 1-2$M_{\odot}$; the closest they might come is a few pc. The &lt;a href=&quot;http://www.cosmos.esa.int/web/gaia/science-performance#astrometric%20performance&quot; rel=&quot;nofollow&quot;&gt;astrometric precision&lt;/a&gt; of Gaia will be about 20 microarcsec for a star with $V =15$, falling to about a milli-arcsec at $V\sim 20$.&#xA;The signature of a neutron star moving with high proper motion might be that it passes very close in the foreground to a bright star, causing its position to wobble backwards and then forwards by an amplitude given by the above formulae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a 20 microarcsecond signature that might be detectable, a neutron star at one pc would have to pass within 0.003 arcseconds of a $V=15$ star.  The surface density of such stars varies according to Galactic latitude but is around 1000 per square degree for stars of $V\leq 15$ (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1980ApJS...44...73B&quot; rel=&quot;nofollow&quot;&gt;Bahcall &amp;amp; Soneira 1980&lt;/a&gt;) at intermediate Galactic latitudes. If the neutron star has a proper motion like Barnard's star of around 10 arcseconds/year, then it traverses 50 arcseconds during the Gaia mission. If background stars are evenly spaced, the chance of coming within 0.003 arcseconds of one is about $10^{-5}$ (if I've done my sums right).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, almost no chance of seeing this.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-07-25T15:32:53.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16700" PostTypeId="2" ParentId="16694" CreationDate="2016-07-25T15:57:12.770" Score="1" Body="&lt;p&gt;Alpha Cas (Schedar) is  TYC 3663-2668-1 or HIP 3179 in the Hipparcos catalogue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=Schedar&amp;amp;NbIdent=1&amp;amp;Radius=2&amp;amp;Radius.unit=arcmin&amp;amp;submit=submit+id&quot; rel=&quot;nofollow&quot;&gt;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=Schedar&amp;amp;NbIdent=1&amp;amp;Radius=2&amp;amp;Radius.unit=arcmin&amp;amp;submit=submit+id&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beta Cas (Caph) is not in the Tycho-2 catalogue but is in the Hipparcos catalog as HIP 746&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=Caph&amp;amp;NbIdent=1&amp;amp;Radius=2&amp;amp;Radius.unit=arcmin&amp;amp;submit=submit+id&quot; rel=&quot;nofollow&quot;&gt;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=Caph&amp;amp;NbIdent=1&amp;amp;Radius=2&amp;amp;Radius.unit=arcmin&amp;amp;submit=submit+id&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gamma Cas is also does not have a Tyc number but is listed as HIP 4427 in the Hipparcos catalogue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=Gamma+Cas&amp;amp;NbIdent=1&amp;amp;Radius=2&amp;amp;Radius.unit=arcmin&amp;amp;submit=submit+id&quot; rel=&quot;nofollow&quot;&gt;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=Gamma+Cas&amp;amp;NbIdent=1&amp;amp;Radius=2&amp;amp;Radius.unit=arcmin&amp;amp;submit=submit+id&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All three of these stars are right at the bright limit of the Tycho-2 catalogue.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-07-25T15:57:12.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16703" PostTypeId="1" CreationDate="2016-07-26T07:40:53.720" Score="4" ViewCount="130" Body="&lt;p&gt;&lt;a href=&quot;https://twitter.com/CassiniSaturn/status/757599817316966400&quot; rel=&quot;noreferrer&quot;&gt;This tweet&lt;/a&gt; from CassiniSaturn shows an image of Saturn's rings being optically distorted by the dense atmosphere. It is explained more fully &lt;a href=&quot;https://saturn.jpl.nasa.gov/resources/7394/&quot; rel=&quot;noreferrer&quot;&gt;on this page&lt;/a&gt; and is available &lt;a href=&quot;http://photojournal.jpl.nasa.gov/catalog/PIA20491&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The refraction is visible because at at least some depth, the density and path length is sufficient to bend the light significantly while the density and composition is still sufficiently transparent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly speaking, what is the density where this is taking place? Is it rarefied compared to earth but simply a very very long path length, or is this happening at a density far larger than Earth's surface atmosphere? How large is this refraction -  like 0.1 degrees or 10 degrees? I can't picture the actual geometry very well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one may be a bit of a judgement call, unless this image happens to be further documented somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1Q4rY.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1Q4rY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; cropped portion of the NASA's Cassini image &lt;a href=&quot;http://photojournal.jpl.nasa.gov/catalog/PIA20491&quot; rel=&quot;noreferrer&quot;&gt;from here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2021-02-19T23:11:22.927" LastActivityDate="2021-02-19T23:11:22.927" Title="Refraction by Saturn's atmosphere - how dense is it here?" Tags="&lt;saturn&gt;&lt;planetary-atmosphere&gt;&lt;cassini&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16704" PostTypeId="1" AcceptedAnswerId="16705" CreationDate="2016-07-26T12:38:43.577" Score="35" ViewCount="3989" Body="&lt;p&gt;Shouldn't the Moon appear as bright as a full Moon seen at midnight from Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FNXzZ.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FNXzZ.jpg&quot; alt=&quot;Moon photobombs Earth&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The photo was taken by DSCOVR at Lagrange point 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the picture, The Moon appears dark gray. Of course the Earth appears bright, reflecting sunlight from clouds and water. The Moon's surface is gray and should reflect less light than the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be irrelevant that we see the far side, since the reflectivity of the Moon's surface should be the same on the far side as the side that faces the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The midnight full Moon appears much, much brighter as seen from Earth than it does in this picture, despite the fact that the amount of sunlight reflecting from the surface of the Moon is the same in both instances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the photo was taken with 3 separate exposures of red, blue and green, but this should not affect the brightness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why does it appear so dull?&lt;/p&gt;&#xA;" OwnerUserId="1700" LastEditorUserId="1700" LastEditDate="2016-07-26T15:32:11.537" LastActivityDate="2016-08-27T04:34:30.557" Title="Why does the Moon appear gray when passing between the Sun and the Earth?" Tags="&lt;the-moon&gt;&lt;earth&gt;&lt;light&gt;&lt;satellite&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16705" PostTypeId="2" ParentId="16704" CreationDate="2016-07-26T13:18:26.467" Score="41" Body="&lt;p&gt;That's what it really would look like if you were there with DSCOVR. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Albedo&quot; rel=&quot;noreferrer&quot;&gt;albedo&lt;/a&gt; of the Moon is only about &lt;a href=&quot;https://en.wikipedia.org/wiki/Moon&quot; rel=&quot;noreferrer&quot;&gt;0.136&lt;/a&gt;, about half of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth&quot; rel=&quot;noreferrer&quot;&gt;Earth's average albedo&lt;/a&gt;. Of course the part with clouds is higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was shocked too, but it was explained in written copy that accompanied  the release of the original image. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Shouldn't the Moon appear as bright as a full Moon seen at midnight from Earth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It does. If the moon were a diffuse, white ball, a full moon would be about seven times brighter!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you watch the image or GIF, the Moon is roughly the same brightness as central Australia or the Sahara region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phil Plait explains well in &lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2016/07/12/dscovr_sees_moon_passing_in_front_of_earth_from_space.html&quot; rel=&quot;noreferrer&quot;&gt;Bad Astronomy&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a lot &lt;a href=&quot;http://www.asterism.org/tutorials/tut26-1.htm&quot; rel=&quot;noreferrer&quot;&gt;to read here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I just ran across these images of astronauts on the surface while reading &lt;a href=&quot;https://space.stackexchange.com/a/17950/12102&quot;&gt;this answer&lt;/a&gt;. Their suits are not 100% white to begin with, but the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_soil&quot; rel=&quot;noreferrer&quot;&gt;Lunar soil&lt;/a&gt; - at least in these locations - is significantly darker. It is close to the same color as the (presumably) nearly-black radiator fins for the heat sink of the RTG unit (2nd photo) at the astronaut's foot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hGZSD.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hGZSD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; &quot;Buzz Aldrin carries the EASEP.&quot; from &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Buzz_Aldrin_carries_the_EASEP.jpg&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OVZLJ.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OVZLJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; &quot;Astronaut Alan L. Bean from Apollo 12, put the Plutonium 238Pu Fuel from the Lunar Module into the SNAP 27 RTG&quot; from &lt;a href=&quot;https://en.wikipedia.org/wiki/Apollo_Lunar_Surface_Experiments_Package#/media/File:Putting_the_Plutonium_238_fuel_into_the_SNAP_27.jpg&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.imgur.com/YBwGQBd.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/YBwGQBd.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Image from NASA/NOAA from &lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2016/07/12/dscovr_sees_moon_passing_in_front_of_earth_from_space.html&quot; rel=&quot;noreferrer&quot;&gt;Bad Astronomy&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2016-08-27T04:34:30.557" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16706" PostTypeId="1" AcceptedAnswerId="16708" CreationDate="2016-07-26T13:21:59.637" Score="2" ViewCount="202" Body="&lt;p&gt;I learnt that the escape velocity is given by $$v_e = \sqrt{\frac{2GM}{r}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I want to launch a rocket from the earth into space and want to calculate the escape velocity $v_e$ (I guess without air resistance). Which value of $r$ do I have to use:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The mean radius of the earth...&lt;/li&gt;&#xA;&lt;li&gt;... or the distance between the center of mass of the earth and the launch location?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(I have basic classical (and quantum) mechanics knowledge, but I am very new to orbital mechanics.)&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="13570" LastActivityDate="2016-07-26T15:42:13.073" Title="Is the radius used in the formula for the escape velocity the average radius of the celestial object or the radius at the starting location?" Tags="&lt;velocity&gt;&lt;orbital-mechanics&gt;&lt;escape-velocity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16707" PostTypeId="1" AcceptedAnswerId="16803" CreationDate="2016-07-26T13:42:55.510" Score="8" ViewCount="625" Body="&lt;p&gt;&lt;strong&gt;note:&lt;/strong&gt; Solar Probe+ is now officially &lt;a href=&quot;https://en.wikipedia.org/wiki/Parker_Solar_Probe&quot; rel=&quot;nofollow noreferrer&quot;&gt;Parker Solar Probe&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;In &lt;a href=&quot;https://space.stackexchange.com/q/17498/12102&quot;&gt;https://space.stackexchange.com/q/17498/12102&lt;/a&gt; I asked about the &lt;a href=&quot;http://solarprobe.jhuapl.edu/common/content/SolarProbePlusFactSheet.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solar Probe Plus&lt;/a&gt; mission. Reading a &lt;a href=&quot;http://solarprobe.jhuapl.edu/mission/docs/SolarProbePlus2008.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;2008 report&lt;/a&gt; I saw that there is an elevated exposure to high velocity dust near the sun. I understand the high velocity, anything in orbit that close will be moving fast, but I don't understand why there is dust there.&lt;/p&gt;&#xA;&lt;p&gt;Simply put - wouldn't the gravity pull it in or the solar wind blow it away?&lt;/p&gt;&#xA;&lt;p&gt;What is the origin of the refractory carbon and silicates? Is this orbiting leftover material from the formation of the solar system, or is it falling in from farther out, or is it left-over carbon and silicon that fell into the early forming sun and is now being blown back out?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;4.3.5 Micrometeoroid and Dust. Solar Probe+ will encounter dust particles ranging in diameter from submicron up to several hundred microns and consisting of highly refractory carbon and silicate species with a typical bulk density of ~2.5 g/cm2. The particles will be traveling at relative speeds as high as 350 km/s. To define the shielding requirements for Solar Probe+, a dust model was developed based primarily on the work of Mann et al. (2004). The model employs the following assumptions...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BtQ13.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BtQ13.png&quot; alt=&quot;Figure 4-6. Predicted dust environment at 0.1 AU (20 Rs) (from Mann et al. 2004 and Ishimoto, 2000)&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; screen shot from &lt;a href=&quot;http://solarprobe.jhuapl.edu/mission/docs/SolarProbePlus2008.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;SolarProbePlus2008&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ymUaM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ymUaM.png&quot; alt=&quot;Screen shot from NASA's Solar Probe Plus Fact Sheet&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; screen shot from NASA's &lt;a href=&quot;http://solarprobe.jhuapl.edu/common/content/SolarProbePlusFactSheet.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solar Probe Plus Fact Sheet&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aBM7A.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aBM7A.jpg&quot; alt=&quot;illustration of NASA's Solar Probe Plus from http://solarprobe.jhuapl.edu/spacecraft/&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; illustration of NASA's Solar Probe Plus from &lt;a href=&quot;http://solarprobe.jhuapl.edu/spacecraft/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://solarprobe.jhuapl.edu/spacecraft/&lt;/a&gt; Sun is up - to state the obvious!&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2020-12-16T22:33:17.953" LastActivityDate="2020-12-16T22:33:17.953" Title="What is the origin of the dust near the sun?" Tags="&lt;the-sun&gt;&lt;star-formation&gt;&lt;planetary-formation&gt;&lt;dust&gt;" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="16708" PostTypeId="2" ParentId="16706" CreationDate="2016-07-26T15:42:13.073" Score="1" Body="&lt;p&gt;Just to provide an official answer, the radius $r$ in that equation is formally the distance between the center of masses between the two objects. That does come with some assumptions of course, but for most standard physics problems that concept is perfectly fine. As you stated, the equation does not assume anything about air resistance, and to extend that, it does not assume anything about continued propulsion. If you're at some radius $r$ from the center of the Earth and you have some velocity $v&amp;gt;v_e$, then just coasting along in your trajectory will allow you to eventually escape the Earth's gravitational pull (ignoring the obvious case where you intersect the Earth with your trajectory).&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-07-26T15:42:13.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16709" PostTypeId="2" ParentId="16704" CreationDate="2016-07-26T15:46:16.183" Score="11" Body="&lt;p&gt;I think the confusion comes down to what &quot;brightness&quot; means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general we don't really measure (either with our eyes or our cameras) the absoloute brightness of objects. Instead we measure the brightness relative to other objects in the scene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the photo of the earth and the moon taken from between the earth and the sun the moon looks dark grey because the sunlit earth is brighter than the sunlit moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the night sky the moon looks white because it's bright compared to the night sky and the nighttime earth (even with some artificial lighting the nighttime earth is not lit up to anything like daytimes levels..&lt;/p&gt;&#xA;" OwnerUserId="11128" LastEditorUserId="11128" LastEditDate="2016-07-26T18:45:47.573" LastActivityDate="2016-07-26T18:45:47.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16710" PostTypeId="2" ParentId="16703" CreationDate="2016-07-26T16:37:31.003" Score="4" Body="&lt;p&gt;This is a hard question to answer without hard numbers. When you look at this image, you can clearly see that the refractive power of the atmosphere changes as you look closer to the limb of Saturn. The rings start off being refracted slightly, then get more refracted as you get closer to the limb. That essentially implies the portion of the ring closer to the limb is traveling through lower and denser regions of Saturn's atmosphere and thus being refracted more. In that sense, there isn't a single answer here since there is a range of refraction going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first approach in considering this problem is to consider &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Optical_depth&quot; rel=&quot;nofollow&quot;&gt;optical depth&lt;/a&gt;&lt;/em&gt;. In case you don't know what this is, optical depth, usually denoted by $\tau$, is a dimensionless number which is a measure of how much material light must pass through and interact with when traveling through a medium. The way optical depth is defined is such that a medium with $\tau\lt1$ means light can get through, while $\tau\gt1$ means light cannot get through. A more conceptual way of thinking about optical depth is to say that it represents, on average, how many atoms a particle of light will hit when trying to travel through a medium. That is, if $\tau=0.5$, then about half the light particles will hit something and be lost, while the other half will get through unscathed. If $\tau = 2$, then on average the light particles will hit two atoms and most light will be scattered/absorbed out of your line of sight, making you unable to see through that medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simplified definition of optical depth is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau=\sigma N \ell$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\sigma$ is the collision cross section of the atoms in the medium, in $m^2$, $N$ is the number density of atoms in the medium, in $m^{-3}$, and $\ell$ is the path length of light through that medium, in $m$. So let's finally take a look at your questions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Roughly speaking, what is the density where this is taking place? Is it rarefied compared to earth but simply a very very long path length, or is this happening at a density far larger than Earth's surface atmosphere?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ultimately, you're asking what $N$ and $\ell$ are such that $\tau=1$, since this would be the point where the refraction stops and you don't see any more of the rings. Unfortunately, this is difficult to answer based on that image alone. At best you can look up a radial density profile for Saturn such as in &lt;a href=&quot;http://sirius.bu.edu/aeronomy/Nagy%20et%20al%202009%20-%20Saturn%20After%20Cassini-Huygens%20book%20chapter.pdf&quot; rel=&quot;nofollow&quot;&gt;Nagy et al. 2009&lt;/a&gt; (ref. fig. 8.1). Note in that paper, they refer to height as the height above the 1 bar pressure level. This paper also gives you an idea for common atoms within the upper atmosphere and you can look up $\sigma$ for these atoms. Predominantly though, I imagine hydrogen will dominate the interactions so using the cross section of hydrogen will be a good approximation. Then assume some sort of approximate path length $\ell$ based on the geometry (maybe use the formula for &lt;a href=&quot;http://mathworld.wolfram.com/CircularSegment.html&quot; rel=&quot;nofollow&quot;&gt;chord length&lt;/a&gt;). Putting actual numbers into this problem could be a research project all to its own, but you can play around with what you consider to be good values for $N$ and $\ell$ such that you're constrained by the fact that $\tau=\sigma N\ell=1$. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How large is this refraction - like 0.1 degrees or 10 degrees?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is even harder to answer. The only way I know to determine something like this is to run a full ray tracing model of light through Saturn's atmosphere. I can hazard a guess that the most refraction you're seeing there is on the order of 10 degrees. This is supported by &lt;a href=&quot;http://arxiv.org/pdf/1510.03430v1.pdf&quot; rel=&quot;nofollow&quot;&gt;Dalba et al. 2015&lt;/a&gt; who actually did a ray tracing analysis of light through Saturn (in regards to planetary transits) and found refraction angles on that order (ref. fig. 2).&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-07-26T16:37:31.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16711" PostTypeId="1" CreationDate="2016-07-26T20:00:27.577" Score="4" ViewCount="318" Body="&lt;p&gt;The age of the Universe can be estimated from taking the inverse of the Hubble constant: $t_\text{universe} = 1/H_0 =d/v.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Yiugw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Yiugw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me this method assumes that any given galaxy has been receding at a constant apparent velocity for the lifetime of the Universe. However, by including data from larger &amp;amp; larger distances, it can be seen that Hubble's law doesn't hold. The data suggest that the rate of expansion was actually slower in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7bl6l.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7bl6l.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it appropriate to use only nearby galaxies to estimate the age of the Universe using the Hubble time when the rate of expansion isn't constant?&lt;/p&gt;&#xA;" OwnerUserId="13595" LastActivityDate="2020-05-13T11:26:24.733" Title="Why can we trust Hubble Time if the rate of expansion is not constant?" Tags="&lt;universe&gt;&lt;expansion&gt;&lt;dark-energy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16712" PostTypeId="2" ParentId="16603" CreationDate="2016-07-26T20:38:41.080" Score="2" Body="&lt;p&gt;This is the result of a new way to investigate the images taken by the PanSTARRS survey, see&#xA;&lt;a href=&quot;http://arxiv.org/abs/1607.04895&quot; rel=&quot;nofollow&quot;&gt;Weryk et al. 2016&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second batch of new discoveries was issued on July 26.&#xA;This has been planned for several years, and now there's enough data to get useful results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that Gareth Williams from the MPC has currently some technical problems with the most recent MPECs (O360 and higher). If these links don't work for you, it's probably because of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further information, see&#xA;the &lt;a href=&quot;http://www.ps1sc.org/blog/2011/06/22/searching-for-clues-in-the-outer-solar-system/&quot; rel=&quot;nofollow&quot;&gt;PS1 Science Consortium blog&lt;/a&gt;&#xA;and &lt;a href=&quot;http://adsabs.harvard.edu/abs/2015DPS....4721112H&quot; rel=&quot;nofollow&quot;&gt;Holman et al. 2015&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13596" LastEditorUserId="616" LastEditDate="2016-07-27T17:46:51.557" LastActivityDate="2016-07-27T17:46:51.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16716" PostTypeId="2" ParentId="16711" CreationDate="2016-07-26T22:40:53.230" Score="6" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$H_0^{-1}$&lt;/span&gt; is only a rough estimate for the age of the universe and you have correctly identified the reasons why not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A correct age estimation relies on knowing &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt; &lt;em&gt;and&lt;/em&gt; the densities of matter and dark energy, so that the past expansion history of the universe can be correctly modelled. Even this relies on an assumption about how dark energy behaves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more interesting question is why a value of &lt;span class=&quot;math-container&quot;&gt;$H_0 \simeq 70$&lt;/span&gt; km s&lt;span class=&quot;math-container&quot;&gt;$^{-1}$&lt;/span&gt;/ Mpc gives an &lt;span class=&quot;math-container&quot;&gt;$H_{0}^{-1}$&lt;/span&gt; of 14 billion years, which is within a few per cent of the current best estimate for the age of the universe of 13.8 billion years. In the &lt;span class=&quot;math-container&quot;&gt;$\Lambda$&lt;/span&gt;CDM cosmology model, the reason for this cosmic coincidence is that the universal expansion underwent a period of decelaraton up until about 4 billion years ago, when it started to accelerate again (the red curve on the plot below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, a tangent to the red curve, showing the size of the universe versus time, at the present epoch almost goes to &lt;span class=&quot;math-container&quot;&gt;${\rm size}=0$&lt;/span&gt; about 14 billion years ago. If we were to go backwards or forwards in time by 5 billion years, the agreement between &lt;span class=&quot;math-container&quot;&gt;$H_0^{-1}$&lt;/span&gt; and the age of the universe would not be so good. At earlier times &lt;span class=&quot;math-container&quot;&gt;$H_0^{-1}$&lt;/span&gt; would have overestimated the age of the universe, whereas at later times &lt;span class=&quot;math-container&quot;&gt;$H_0^{-1}$&lt;/span&gt; will underestimate the age of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lI70D.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lI70D.jpg&quot; alt=&quot;Expansion history of the universe&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2020-05-13T11:26:24.733" LastActivityDate="2020-05-13T11:26:24.733" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="16717" PostTypeId="1" AcceptedAnswerId="16788" CreationDate="2016-07-26T23:31:25.333" Score="5" ViewCount="303" Body="&lt;p&gt;In a &lt;a href=&quot;https://astronomy.stackexchange.com/q/16707/7982&quot;&gt;previous question&lt;/a&gt; I asked about the source of carbon and silicate dust that Solar Probe Plus will encounter in its close flyby of the Sun. It seems likely that most sources would include infall from outside the solar system or fragments resulting from collisions inside the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the Sun is believed to be about 0.4% carbon and 0.1% silicon by mass (see  &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/tables/suncomp.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://chemistry.about.com/od/geochemistry/a/sunelements.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;). I believe these represent the relative abundances during the formation of the Sun, but do not account for nucleosynthesis that early in the Sun's life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the distribution of carbon and silicon in the Sun is believed to be uniform, or does it have a radially dependent distribution? Is there be any kind of gravitational settling, i.e. heavier atoms tending to move toward the center as happens in planets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we can only directly measure elemental abundances directly near the surface using spectroscopic techniques, would the radial distribution come from some model or calculation?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-08-03T00:38:21.977" Title="How are heavier elements such as carbon and silicon distributed within the Sun?" Tags="&lt;the-sun&gt;&lt;elemental-abundances&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16718" PostTypeId="2" ParentId="16704" CreationDate="2016-07-27T04:41:58.193" Score="9" Body="&lt;p&gt;The other answers have pretty well covered the fact that the moon is made of dark gray stone.  But I wanted to mention a way that you can verify this on your own with nothing more sophisticated than a camera with a manual exposure mode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photographers have a rule of thumb called &lt;a href=&quot;https://www.slrlounge.com/photography-essentials-the-sunny-16-rule/&quot;&gt;&quot;sunny 16&quot;&lt;/a&gt;.  This was much more important back in the day when cameras didn't have sophisticated light meters and on-board computers.  Basically, the rule says that in direct sunlight, if you set your aperture to f/16, the correct shutter speed will be the reciprocal of your film speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon is a sunlit object, and the sunny 16 rule exposes sunlit objects so they look the way we normally perceive them in sunlight.  But if you apply the sunny 16 rule to the moon, it looks dark gray -- its &quot;true&quot; color and brightness as we would perceive it against a sunlit background of average terrestrial brightness.&lt;/p&gt;&#xA;" OwnerUserId="13599" LastActivityDate="2016-07-27T04:41:58.193" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16719" PostTypeId="1" AcceptedAnswerId="16735" CreationDate="2016-07-27T05:50:03.027" Score="8" ViewCount="834" Body="&lt;p&gt;The rings of Saturn and Jupiter are circular. Orbits can be elliptical, so can rings be elliptical?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="13591" LastEditDate="2016-07-27T23:33:24.103" LastActivityDate="2016-07-28T01:23:04.640" Title="Can there be elliptical rings around a planet?" Tags="&lt;planetary-ring&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16720" PostTypeId="1" AcceptedAnswerId="16721" CreationDate="2016-07-27T06:13:31.747" Score="-4" ViewCount="81" Body="&lt;p&gt;As we know, lunar eclipse appear red, if we can see Earth transit from Mars, does it look like a black dot surrounded by red light on the sun?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastActivityDate="2016-07-27T07:07:20.397" Title="Does Earth transit look like a black dot surrounded by red light?" Tags="&lt;lunar-eclipse&gt;&lt;planetary-transits&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16721" PostTypeId="2" ParentId="16720" CreationDate="2016-07-27T07:07:20.397" Score="2" Body="&lt;p&gt;Earth would look very much like Venus does during a transit: a black dot moving across the face of the sun. The Earth-Mars distance is greater than Venus Earth, so the dot would be slightly smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atmosphere, which would be brightly visible from the moon during an eclipse by the Earth, would be lost in the glare of the sun. It may be detectable with a sufficiently sensitive spectrometer, but not visually.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-07-27T07:07:20.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16722" PostTypeId="2" ParentId="16719" CreationDate="2016-07-27T08:42:15.490" Score="2" Body="&lt;p&gt;Rings are made up of a large number of individual particles, each in elliptical orbits. We see rings as circular because we don't see and follow the individual particles just the overall swarm of particles. Their individual orbits average out and just appear are a circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The individual orbits are constantly being changed due to collisions between particles, and gravitational influences for moons. Saturn's F ring is a good example of &lt;a href=&quot;http://www.nasa.gov/mission_pages/voyager/PIA_02292_07712.html&quot; rel=&quot;nofollow&quot;&gt;this.&lt;/a&gt; As can be seen from the images the ring is not a circle.&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-07-27T08:42:15.490" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16727" PostTypeId="2" ParentId="14683" CreationDate="2016-07-27T17:09:02.970" Score="2" Body="&lt;p&gt;A 10x25 or similar monocular should be able to project a usable image for a modest cost. Its theoretical&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_resolution&quot; rel=&quot;nofollow&quot;&gt;angular resolution&lt;/a&gt;&#xA;in yellow light would be 6 arcseconds, significantly smaller than Mercury at inferior conjunction.&#xA;Focused on a screen 50cm away, the Sun image would be 45mm wide and 30% as bright as direct sunlight.&#xA;I recommend surrounding the monocular with a sheet of cardboard to shade the screen.&#xA;Until the next transit in November 2019, it would probably show sunspots too.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-07-27T17:09:02.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16728" PostTypeId="2" ParentId="16704" CreationDate="2016-07-27T17:39:10.600" Score="3" Body="&lt;p&gt;Because if the photograph were taken with an exposure setting that made the moon appear shiny and bright white, the Earth is so much &lt;em&gt;more&lt;/em&gt; bright that it would completely wash out the scene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no way to objectively measure &quot;brightness&quot; of objects in a photograph without knowing how the image was captured.&lt;/p&gt;&#xA;" OwnerUserId="4155" LastActivityDate="2016-07-27T17:39:10.600" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16729" PostTypeId="1" AcceptedAnswerId="16730" CreationDate="2016-07-27T18:45:42.813" Score="9" ViewCount="4415" Body="&lt;p&gt;Galaxies come in different shapes and sizes. What factors determine the shape and size of a galaxy, and how can these change over time?&lt;/p&gt;&#xA;" OwnerUserId="11951" LastEditorUserId="6" LastEditDate="2016-07-27T21:11:33.820" LastActivityDate="2016-07-27T21:21:44.373" Title="Do galaxies change shape and size over time?" Tags="&lt;galaxy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16730" PostTypeId="2" ParentId="16729" CreationDate="2016-07-27T20:07:05.210" Score="15" Body="&lt;p&gt;The size of a galaxy is dependent on how the matter was distributed in the early universe, and how it collapsed under its own gravity to form clumps of (dark) matter that the galaxies would form around. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The very earliest galaxies are irregular, but as matter fell towards them, and they developed a consistent orbital direction they formed into a disc. Gravity waves in the disc then show up as spiral arms. Large galaxies that are actively forming stars tend to have this structure. Small galaxies may never reach the point of forming a disc and remain irregular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As galaxies collide and interact, the orbits of the stars are disrupted. The result of the merger of two large spiral galaxies is often a galaxy in which the stars orbit in all directions, which looks like an elliptical galaxy. Star formation in elliptical galaxies is often very low. Elliptical galaxies are the most evolved form of galaxies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time scale of this evolution is slow. It takes billions of years for galaxies to evolve from one form to another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The details of the process of galaxy formation and evolution are still uncertain. The exact nature of dark matter will be important in understanding how galaxies form. The role (if any) that black holes play in galaxy formation is uncertain.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-07-27T20:07:05.210" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16731" PostTypeId="2" ParentId="16729" CreationDate="2016-07-27T20:08:22.470" Score="11" Body="&lt;p&gt;Yes, definitely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxies grow through smooth accretion of matter (both dark and baryonic), and through mergers with other galaxies. Minor mergers (i.e. where one galaxy is much larger than the other) simply disrupt the smaller one without changing the shape of the larger (but may increase its star formation rate for a while). Major mergers, where galaxies are of comparable size, can completely &quot;wreck&quot; the galaxies. If the merging galaxies are spirals, they can sometimes settle down into a larger spiral galaxy, possibly with an enhanced central bulge. But sometimes, depending among other things on their orientation prior to merging, they become ellipticals instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Universe is expanding, galaxies were closer together in the past, and thus most of the growth has already taken place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small galaxies which undergo intense star formation, may blow out their gas due to stellar feedback, in this way getting rid of most of their gas and thus &lt;em&gt;de&lt;/em&gt;crease a little in size and &quot;extinguish&quot; themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxies have their origin from the clumps in the almost smooth primordial soup of gas and dark matter (and radiation). These clumps have a size distribution — the &lt;em&gt;halo mass function&lt;/em&gt; — which is a power law at small masses and with a exponential cut-off at large masses. Thus, the size of galaxies follow a similar distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure on the &lt;em&gt;left&lt;/em&gt; below (created with &lt;a href=&quot;http://hmf.icrar.org&quot; rel=&quot;noreferrer&quot;&gt;this tool&lt;/a&gt;) shows the distribution of dark matter halos, i.e. number per volume at a given mass &lt;em&gt;today&lt;/em&gt; (redshift $z=0$, solid line) and when the Universe was only two billion years old ($z=3$, dashed line). You can see that in early times, the large halos hadn't formed yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure on the &lt;em&gt;right&lt;/em&gt; (from my own cosmological simulation) shows how many stars there are in a (simulated) dark matter halo of a given mass. You can see that it follows an approximate power law, but with quite some scatter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/erVFt.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/erVFt.png&quot; alt=&quot;HMF_Mstar&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-07-27T21:21:44.373" LastActivityDate="2016-07-27T21:21:44.373" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16732" PostTypeId="1" CreationDate="2016-07-27T20:59:11.867" Score="3" ViewCount="4185" Body="&lt;p&gt;I've read that during the equinoxes, tides are at their maximum. Shouldn't this only apply to latitudes near the equator? I mean, when the solar declination is near to 0º and we have a full or new moon, the combined effect of the Sun's and Moon's gravity should create the highest tides, but only &lt;strong&gt;near the equator&lt;/strong&gt; (see in wikipedia the article Tides, which says that equinoctial tides are maximum)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true for other latitudes?&lt;/p&gt;&#xA;" OwnerUserId="9949" LastEditorUserId="9949" LastEditDate="2016-07-29T12:41:40.410" LastActivityDate="2016-09-10T17:48:13.690" Title="Why are tides biggest during the equinoxes?" Tags="&lt;the-sun&gt;&lt;the-moon&gt;&lt;rotation&gt;&lt;declination&gt;&lt;tidal-forces&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16734" PostTypeId="1" AcceptedAnswerId="16736" CreationDate="2016-07-27T23:11:31.437" Score="2" ViewCount="360" Body="&lt;p&gt;The Milky Way is a barred spiral galaxy, with a supermassive black hole at its center. &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Galactic_Center&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it is a barred spiral galaxy, it has a central bar-shaped structure composed of stars.&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Barred_spiral_galaxy&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my questions are -:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Do the bar and supermassive black hole overlap or cross each other?&lt;/li&gt;&#xA;&lt;li&gt;Do the stars from the bar get sucked inside the supermassive black hole?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11951" LastEditorUserId="2365" LastEditDate="2016-07-28T01:26:04.420" LastActivityDate="2017-09-25T21:30:45.117" Title="Black hole at Galactic Center of the Milky way and Bar of stars?" Tags="&lt;milky-way&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16735" PostTypeId="2" ParentId="16719" CreationDate="2016-07-28T01:23:04.640" Score="7" Body="&lt;p&gt;Short answer: no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many collisions within ring systems, and collisions always work, over time, to push orbits to a circular shape (or destroy the rings). Any deviations tend to be quickly corrected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the same reason, rings tend to be extremely flat and extremely thin. E.g. Saturn's rings are only dozens of meters thick; given their diameter of 300 thousand kilometers, if you cut a model of the rings from ordinary size printer paper, the paper model would be too thick. (would not represent the thickness of the rings to the correct scale)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another characteristic of rings is that they are equatorial. If they are not in plane with the equator, precession will drive them towards that configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could apply a perturbation that makes a ring non-circular, or non-flat, or non-equatorial, but then the system will quickly evolve towards circular, flat, and equatorial (or will be destroyed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.uwgb.edu/dutchs/AstronPowerPoint/Rings.ppt&quot;&gt;https://www.uwgb.edu/dutchs/AstronPowerPoint/Rings.ppt&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2016-07-28T01:23:04.640" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16736" PostTypeId="2" ParentId="16734" CreationDate="2016-07-28T01:37:03.727" Score="5" Body="&lt;p&gt;Lets get a sense of scale. The bar is about 30000 light years long, and contains the central bulge of the galaxy, a roughly spherical mass of stars with a diameter of 10000 ly.  It's big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The supermassive black hole is pretty big for a black hole, but only about 80 light seconds across. Compared to the size of the bar, it is tiny.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The supermassive black hole is found at the centre of the bulge, at the centre of the bar, and entirely within it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes a star does fall into the black hole. The black hole is currently inactive, that means that there is not much matter falling in every year. Each year, roughly 1/10000 of the mass of the sun falls into the black hole. This is mostly matter from the solar winds of nearby stars.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-07-28T16:55:24.713" LastActivityDate="2016-07-28T16:55:24.713" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16737" PostTypeId="1" CreationDate="2016-07-28T08:35:21.430" Score="0" ViewCount="277" Body="&lt;p&gt;Why does the light of objects such as the Sun, stars etc. take time to reach us? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. Why does sunlight take 8 minutes to reach us, instead of reaching our eyes as soon as it rises.&lt;/p&gt;&#xA;" OwnerUserId="13592" LastEditorDisplayName="user1569" LastEditDate="2016-07-28T12:33:22.207" LastActivityDate="2016-07-28T14:41:57.787" Title="Why does light from objects take time to reach us?" Tags="&lt;light&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2016-07-29T13:13:21.277" ContentLicense="CC BY-SA 3.0" />
  <row Id="16738" PostTypeId="2" ParentId="16737" CreationDate="2016-07-28T08:48:54.577" Score="3" Body="&lt;p&gt;For the very simple reason that light travels very fast, but not infinitely fast.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It takes light about 1 nanosecond to travel 30cm. That is so fast that you don't normally notice it. However it does take some time for light to reach your eyes, even from just across the room. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objects in space are a lot further away. It takes light about 1.2 seconds to reach the moon, and as you note it takes light 8 minutes to come from the Sun. The reason is that the sun is 150 million km away, and it takes about 8 minutes for light to go that far. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If light did have infinite speed, it would be a weird universe. Effects would happen at the same time as causes, so the basic nature of cause preceding effect would be broken. The &quot;speed=frequency*wavelength&quot; formula for light would be broken. Refraction and diffraction wouldn't exist. A photon would be emitted and absorbed at the same time. Could something that is absorbed at the same time as it is emitted actually exist? &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-07-28T09:32:56.783" LastActivityDate="2016-07-28T09:32:56.783" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16739" PostTypeId="1" AcceptedAnswerId="16740" CreationDate="2016-07-28T09:36:58.990" Score="0" ViewCount="187" Body="&lt;p&gt;In our garden, we have a little pool to enjoy our 2 children. It is 3 meters in diameter and contains some 4 m³ of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For cleaning-convenience every evening I go walking some 5 or 10 rounds in the pool all in the same direction, to set up a relatively fast rotating current in the pool. As a result, all of the remaining dirt (leaves, insects and seeds) are gathered together in the middle of the pool, where they can be easily removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the analogy to galaxies is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The pool represents the galaxy.     &lt;/li&gt;&#xA;&lt;li&gt;The water represents the space-time? (or all we cannot see in a galaxy? Or the galaxy as a whole?)&lt;/li&gt;&#xA;&lt;li&gt;The dirt represents barionic matter which is forced to be dense in&#xA;the center of the galaxy, hence building the super massive black&#xA;hole.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is clear that it is perhaps no good idea trying to map a 3m standup pool to a 100000 lightyear galaxy, but the observation seems to be worth a thought (and I do not think that I am the first one searching for analogies in that direction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question now is, could this be a valid analogy to the process of building supermassive black holes as a result of rotation?&lt;/p&gt;&#xA;" OwnerUserId="11010" LastEditorUserId="2365" LastEditDate="2016-07-28T10:11:10.363" LastActivityDate="2016-07-28T10:11:24.800" Title="Is a swimming pool analogy to building of black holes in the centers of galaxies valid?" Tags="&lt;black-hole&gt;&lt;galactic-dynamics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16740" PostTypeId="2" ParentId="16739" CreationDate="2016-07-28T10:11:24.800" Score="2" Body="&lt;p&gt;The process of an eddy gathering matter to the centre of a pool is due to the motion of water in a vortex. Roughly, if water is rotating  counterclockwise then a secondary motion will be set up, rotating clockwise about an axis parallel with the tangential motion of the water. This tends to push debris towards the centre of the vortex. (The same effect causes deposition on the inside of meanders)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space time doesn't work like this. There is very little spinning of spacetime around a galaxy, the amount of frame dragging is very small. And spacetime doesn't follow the Navier–Stokes equations, so there is no true vortex. The debris flows with the water, where as matter travels through spacetime. Your pool has a clearly defined &quot;up&quot;, the galaxy doesn't. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while the appearance is somewhat similar, the mechanism is different. Though we don't really understand the mechanism of formation and growth of supermassive black holes.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-07-28T10:11:24.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16741" PostTypeId="1" CreationDate="2016-07-28T10:52:17.007" Score="1" ViewCount="75" Body="&lt;p&gt;The wikipedia says that Voyager 1 cross the heliosphere, but is really outside the Solar System, or need to cross the Oort cloud?&lt;/p&gt;&#xA;" OwnerUserId="7940" LastActivityDate="2016-07-28T10:52:17.007" Title="Are the Voyager spacecraft outside of our Solar System?" Tags="&lt;solar-system&gt;&lt;oort-cloud&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2016-07-29T09:59:46.330" ContentLicense="CC BY-SA 3.0" />
  <row Id="16742" PostTypeId="2" ParentId="16737" CreationDate="2016-07-28T14:41:57.787" Score="0" Body="&lt;p&gt;I sense in your question that you wonder why speed of light is not infinite as many persons think it is. If the speed of light was infinite (no delay) then there would not have any refraction phenomenon and the immediate consequence is that your eye wouldn't be able to focus so you would be blind somehow.&lt;/p&gt;&#xA;" OwnerUserId="13514" LastActivityDate="2016-07-28T14:41:57.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16743" PostTypeId="1" AcceptedAnswerId="16747" CreationDate="2016-07-28T16:08:26.830" Score="1" ViewCount="103" Body="&lt;p&gt;Consider the motion of a planet around an oblate star. Because of the oblateness the star has a non-zero quadrupole moment which causes the orbital angular momentum of the planet to precess around the symmetry axis of the oblate star. My naive question is: is this precession rate exactly the same as the precession rate of the longitude of the ascending node or are there subtle differences between the two? &lt;/p&gt;&#xA;" OwnerUserId="2358" LastActivityDate="2016-07-28T19:31:59.513" Title="orbital angular momentum and the longitude of ascending node" Tags="&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16744" PostTypeId="1" CreationDate="2016-07-28T17:42:06.070" Score="1" ViewCount="469" Body="&lt;p&gt;I am confused about the definition of the period of an eclipsing binary, a binary configuration where one component crosses the other across the line of sight. When people say that an eclipsing binary has a period of T years do they mean that the period of the individual component is T years?&lt;/p&gt;&#xA;" OwnerUserId="2358" LastEditorUserId="7982" LastEditDate="2021-04-16T23:28:49.930" LastActivityDate="2021-04-16T23:28:49.930" Title="period of an eclipsing binary" Tags="&lt;time-domain-astronomy&gt;&lt;eclipsing-star-systems&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16747" PostTypeId="2" ParentId="16743" CreationDate="2016-07-28T19:31:59.513" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My naive question is: is this precession rate exactly the same as the precession rate of the longitude of the ascending node or are there subtle differences between the two?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They're the same, assuming the star's equator is the principal plane of the coordinate system. One way to calculate the longitude of ascending node as $\Omega = \arctan(h_x, -h_y)$, where $\arctan(a,b)$ is the two-argument inverse tangent function and $h_x$ and $h_y$ are the x and y components of angular momentum vector.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2016-07-28T19:31:59.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16748" PostTypeId="1" CreationDate="2016-07-29T00:06:17.797" Score="1" ViewCount="83" Body="&lt;p&gt;Is there an established astronomical term for the period of time between alignments of pairs of planets?  For example, Mars, Earth, and the Sun roughly align ever 780 Earth days.  In my writing, I've variously call this period the 'alignment cycle' or 'conjunction cycle' or 'opposition cycle', but I'd like to know if there's a correct term for this.&lt;/p&gt;&#xA;" OwnerUserId="13618" LastActivityDate="2016-07-29T00:22:43.120" Title="Term for the period between planet alignments?" Tags="&lt;planet&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16749" PostTypeId="2" ParentId="16748" CreationDate="2016-07-29T00:22:43.120" Score="2" Body="&lt;p&gt;The time between oppositions of Mars (or another planet) is called its synodic period. That is the length of time it takes for Mars to be in the same position in Earth's sky relative to the sun, and is about 780 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It contrasts with the sidereal period, which is the length of time the planet takes to orbit the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There isn't a fixed number of days between conjunctions, so I don't think this has a name.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-07-29T00:22:43.120" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16751" PostTypeId="1" CreationDate="2016-07-29T22:40:20.500" Score="5" ViewCount="275" Body="&lt;p&gt;I was reading about the &lt;a href=&quot;https://en.wikipedia.org/wiki/Helium_flash&quot; rel=&quot;noreferrer&quot;&gt;helium flash&lt;/a&gt;, the short but sudden onset of helium fusion in certain red giant stars. As I understand, the upper (nondegenerate) layers of the star absorb the energy as they quickly expand, which leads to the stabilization of the core as thermal pressure takes over, leading to the period of helium burning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was curious as to whether this could be generalized to the onset of fusion of heavier elements later in a more massive star's life, and I came across a reference to a hypothetical silicon flash in &lt;a href=&quot;http://adsabs.harvard.edu/abs/2015ApJ...810...34W&quot; rel=&quot;noreferrer&quot;&gt;Woosley &amp;amp; Heger (2015)&lt;/a&gt;. More digging led me to a passage in &lt;a href=&quot;https://books.google.com/books?id=DjeVdb0sLEAC&amp;amp;source=gbs_navlinks_s&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Fundamental Astronomy&lt;/em&gt;&lt;/a&gt; (page 250), which states&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In stars with masses around 10 $M_\odot$ either carbon or oxygen may be ignited explosively just like helium in relatively low-mass stars; there is a &lt;em&gt;carbon&lt;/em&gt; or &lt;em&gt;oxygen flash&lt;/em&gt;. This is much more powerful than the helium flash, and may make the star explode as a supernova.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's say that a star undergoes a relatively low-power carbon, oxygen, or silicon flash, and enough energy is absorbed by the outer layers such that the star does &lt;em&gt;not&lt;/em&gt; undergo a supernova. Would enough energy escape the outer layers of the star so that we be able to detect the flash using optical or neutrino telescopes? Obviously, this is distance-dependent, so I'll say that the hypothetical star is within five parsecs of Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As another way to put the question, how luminous would the flash be?&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-07-29T22:51:18.990" LastActivityDate="2016-09-13T00:30:38.267" Title="Could non-supernova carbon, oxygen, or silicon flashes be observed?" Tags="&lt;star&gt;&lt;observational-astronomy&gt;&lt;stellar-evolution&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16753" PostTypeId="1" AcceptedAnswerId="18142" CreationDate="2016-07-30T00:12:33.410" Score="10" ViewCount="28815" Body="&lt;p&gt;While taking pictures of the bubble nebula I noticed a very strange artifact on only one of my pictures. It can't be a satellite since its not a straight line and this is a 90s exposure that should make a line much longer. Also I doubt this is just readout noise from the camera (image is unstretched, unedited) since hot pixels don't form lines. So could this be a cosmic ray hit? If not does anyone have an idea what this could be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image was taken with my Canon 7D Mark II @ISO1600, 90s exposure.&#xA;Celestron C8 SGT (XLT), Celestron Advanced VX&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BNcJH.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BNcJH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IQMfI.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IQMfI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10271" LastActivityDate="2018-12-15T19:36:02.537" Title="Could this be a cosmic ray hit on my cameras sensor (CMOS, DSLR)?" Tags="&lt;photography&gt;&lt;cosmic-ray&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16754" PostTypeId="1" CreationDate="2016-07-30T04:22:14.223" Score="4" ViewCount="366" Body="&lt;p&gt;At PSE there are a few &lt;a href=&quot;https://physics.stackexchange.com/questions/129089/how-does-angular-momentum-transfer-between-a-planet-and-its-moon&quot;&gt;answers&lt;/a&gt; that explain that since the moon is responsible for tides and tides slow down spinning, L must be transferred to the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that if earth's L were reduced by an asteroid, this would not influence the moon's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you  clarify:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;does the change of $\omega$ of the Earth alter the orbit of moon only when that change is attributable to the moon or in any case?&lt;/li&gt;&#xA;&lt;li&gt;would there be tides if there were no moon?&lt;/li&gt;&#xA;&lt;li&gt;what is the mechanism through which angular momenta can exchange magnitude? This is apparently impossible since there is no connection or, we we consider the obsolete Netwonian gravity, the connection is radial and never tangential?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user13628" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-08-02T14:13:18.093" Title="Can mechanical angular momentum be transferred by gravity?" Tags="&lt;orbit&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16755" PostTypeId="1" AcceptedAnswerId="16756" CreationDate="2016-07-30T06:04:22.773" Score="10" ViewCount="2616" Body="&lt;p&gt;When looking at visible light coming from, say a star, scientists can determine by the shifting of the light towards the red or the blue end of the spectrum to determine if it is moving towards or away from earth. But they can also look at the light as it falls on the visible spectrum to determine the composition of said star. How do they determine if the light landing on the red end of the spectrum, for example, means it's moving away vs the composition of the star? How do they tell the difference? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, how is all this reconciled with the fact that a star's color is also based on it's temperature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: Question was expanded to include temperature after comments and updated answer subsequent to my original question asking only about motion vs composition.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="10117" LastEditorUserId="10117" LastEditDate="2016-08-02T01:10:26.903" LastActivityDate="2018-09-19T07:26:36.810" Title="How do scientists determine if a star's color is due to Doppler shift vs the star's composition vs temperature?" Tags="&lt;doppler-effect&gt;&lt;spectrometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16756" PostTypeId="2" ParentId="16755" CreationDate="2016-07-30T07:45:33.750" Score="13" Body="&lt;p&gt;The colour of a star is due mostly to its temperature. Except at extreme velocities, the red-shift won't affect the light enough to change the visible colour. To detect a red or blue shift you need some marker in the spectrum that can be measured accurately. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spectrum of light from a star will contain dark lines, called &lt;a href=&quot;https://en.wikipedia.org/wiki/Fraunhofer_lines&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fraunhofer lines&lt;/a&gt;, which are due to the absorption of light at particular wavelengths by gases in the star's atmosphere. The wavelengths at which these lines appear are known very finely, for example sodium causes a line at 589.592nm . If the star is moving, the light, along with the Fraunhofer lines is &lt;a href=&quot;http://lcogt.net/spacebook/redshift/&quot; rel=&quot;nofollow noreferrer&quot;&gt;red- or blueshifted&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Q4ylT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q4ylT.png&quot; alt=&quot;redshift&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The presence of these line shows the composition of the star's atmosphere.&#xA;Measurement of the position of the lines allows for accurate measurement of the Doppler shift of a star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cool, red star which is stationary will have the Fraunhofer lines in the same position as a hot stationary blue star. The motion of the star is measured by the position of the Fraunhofer lines, not the colour of the star.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="18643" LastEditDate="2018-09-19T07:26:36.810" LastActivityDate="2018-09-19T07:26:36.810" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="16758" PostTypeId="2" ParentId="12723" CreationDate="2016-07-30T10:03:38.203" Score="-3" Body="&lt;p&gt;This answer is just misleading.  The rings are not formed by the moons.  They are clearly formed by a force.  And I don't think its gravity.  Gravity forms spheres or spherical shells if as you suggest its the roche limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FYI saturns rings are about 30 feet thick and extend for millions of miles.  Gravity does not confine things to thin discs.  There is an magnetic force at work here.  And it has been demonstrated by plasma physicists.&lt;/p&gt;&#xA;" OwnerUserId="13632" LastActivityDate="2016-07-30T10:03:38.203" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16759" PostTypeId="2" ParentId="16754" CreationDate="2016-07-30T10:15:28.870" Score="3" Body="&lt;blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;does the change of &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; alter orbit of moon only when that change is attributable to the moon or in any case?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Angular momentum is a conserved quantity (more on this later). What this means is that if some object is undergoing a change in angular momentum, that angular momentum must necessarily be transferred to some other object or objects. The Earth is losing rotational angular momentum. It's rotation rate is slowing down. That means some mechanism must exist for transferring that angular momentum.&lt;/p&gt;&#xA;&lt;p&gt;The Earth's angular momentum would change if the Earth was struck by an asteroid, but that would be a very, very small change. The loss of gases from the top of the atmosphere is also represents a negligibly small transfer of angular momentum, as does the interaction of the Earth's magnetic field with the solar wind. The Earth is far too large to be subject to the YORP effect, and The only mechanism left is gravitation. Almost all of the Earth's loss of rotational angular momentum due to a slowing rotation rate is via gravitation. A small amount is transferred to the Sun and to the Earth's orbit about the Sun. All that is left is the Moon. That vast majority of the Earth's lost angular momentum goes into making the Moon recede from the Sun.&lt;/p&gt;&#xA;&lt;p&gt;Angular momentum is a conserved quantity. The conservation laws are very important in higher level physics (Lagrangian and Hamiltonian physics); they are assumed rather than derived. Even higher level physics derives the conservation laws from Noether's theorem. However, angular momentum can also be derived from the strong form of Newton's third law. Proving this is standard fare in every undergraduate and graduate classical mechanics text. For example, it's the subject of chapter 2 in &lt;em&gt;Classical Dynamics Of Particles And Systems&lt;/em&gt; by Marion et al. and it's the subject of chapter 1 (starting on page 5) in &lt;em&gt;Classical Mechanics&lt;/em&gt; by Goldstein et al. There are some nice online derivations as well, such as the week 1 lecture notes for &lt;a href=&quot;http://physics.oregonstate.edu/%7Eostroveo/COURSES/ph426&quot; rel=&quot;nofollow noreferrer&quot;&gt;a class on Central Forces at Oregon State University&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Another name for a force that obeys the strong form of Newton's third law is a &lt;em&gt;central force&lt;/em&gt;. A central force is one in which the force&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Acts between pairs of particles that comprise a system of particles,&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Obeys the weak form of Newton's third law (equal but opposite forces),&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is directed along or against the line connecting the two particles, and&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is a function of the radial distance between the particles only.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Examples include Hookean spring forces, Coulomb forces, and Newtonian gravitation. The gravitational force between a pair of particles is given by &lt;span class=&quot;math-container&quot;&gt;$\vec F = G M_1 M_2 / r^2 \hat r$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\hat r$&lt;/span&gt; is a unit vector pointing along the line that connects the two particles.&lt;/p&gt;&#xA;&lt;br&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;would there be tides if there were no moon?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Yes. The Sun also raises tides, but smaller in magnitude than those raised by the Moon. The heights of the tides raised by the Sun are a bit less than half those raised by the Moon.&lt;/p&gt;&#xA;&lt;br&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;what is the mechanism through which angular momenta can exchange magnitude? This is apparently impossible since there is no connection or, we we consider the obsolete Netwonian gravity, the connection is radial and never tangential?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;There are two issues with this thesis. One is that Newtonian gravity is not an obsolete theory; it remains widely use in space exploration and in astronomy. A better way to look at it is that Newtonian gravity has a limited domain of applicability.&lt;/p&gt;&#xA;&lt;p&gt;The other issue is the claim that Newtonian is purely radial. This is only true in the case of point masses or objects with a spherically symmetric mass distribution. It is not the case for objects with an aspherical mass distribution. For example, the Earth has a significant equatorial bulge due to its rotation. This equatorial bulge exerts a torque on satellites orbiting the Earth that causes the satellites' orbits to precess. Scientists take advantage of this. There's a special class of satellites, the sun synchronous satellites, whose orbits by design precess by 360 degrees over the course of a year.&lt;/p&gt;&#xA;&lt;p&gt;With regard to the Moon, the rotation of the Earth makes the tides raised by effectively lead the Moon by a small amount. While this isn't Newton's tidal bulge, it has a similar effect on the Moon. Averaged out over the course of time, the waters on the near side of the Earth pile up slightly in front of the Moon while the waters on the far side of the Earth pile up slightly behind the Moon. Because those far side waters are further from the Moon, the near side waters dominate gravitationally, causing the Moon to accelerate.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-08-02T14:13:18.093" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16763" PostTypeId="1" AcceptedAnswerId="16765" CreationDate="2016-07-31T06:04:51.233" Score="0" ViewCount="607" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_distance_(astronomy)&quot; rel=&quot;nofollow&quot;&gt;wiki&lt;/a&gt; says that average distance from earth increases 3.8 cm per year, can you tell how much of that is due to the &lt;em&gt;expansion of the universe&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;wiki also says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Through the action of tidal forces, angular momentum is slowly being&#xA;  &lt;strong&gt;transferred from Earth's rotation to the Moon's orbit&lt;/strong&gt;. The result is&#xA;  that Earth's rate of spin is imperceptibly decreasing (at a rate of&#xA;  2.3 milliseconds/century),[20] and the &lt;em&gt;lunar orbit is gradually expanding&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you give some details about this process, is the orbit changing shape, eccentricity? is the average orbital speed decreasing, by what value? is the spin angular velocity increasing, by what value?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I did not misunderstand it, the current view on tides says that tidal bulges exert a torque on the moon, &lt;em&gt;and L is tranferred to it&lt;/em&gt;, if that is correct, that L being transferred to its spin is increasing or decreasing it? Can you clarify also how a torque on the spin can produce a change of the orbit, which only implies a change (&lt;em&gt;a decrease&lt;/em&gt;, in fact) of its &lt;em&gt;orbital speed&lt;/em&gt; and not of the spinning speed? Can the spin (L) influence the orbital L, and how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I asked specific questions, which are not asked or answered there. In particular I am asking: if as they say the tidal/equatorial bulge increases spin angular momentum, how is this transferresin its turn to the orbital angulat momentum? Or, alternatively &lt;/li&gt;&#xA;&lt;li&gt;can gravity directly increase the orbital momentum of a satellite?&lt;/li&gt;&#xA;&lt;li&gt;Actually the answer, besides being incomplete, may be wrong because momentum/KE is not added , but &lt;strong&gt;subtracted&lt;/strong&gt; to the moon, since its orbit gets larger. There can be &lt;strong&gt;no transfer, since both earth and moon are slowing down&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;torque is not correctly explained&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user13628" LastEditorDisplayName="user13628" LastEditDate="2016-07-31T13:17:30.507" LastActivityDate="2016-07-31T16:37:35.553" Title="The Moon drifting away from the Earth" Tags="&lt;orbit&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2016-07-31T23:21:57.343" ContentLicense="CC BY-SA 3.0" />
  <row Id="16764" PostTypeId="1" AcceptedAnswerId="16779" CreationDate="2016-07-31T13:26:17.557" Score="2" ViewCount="276" Body="&lt;p&gt;Planets are more or less spherical, can you tell which one is the most asymmetric? Do they figure out such a property from an eccentric behaviour, or what?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a simple technical question: an asymmetric body gets a torque in a gravitational field, is the rotational energy/acceleration to be subtracted from the regular energy/acceleration? In other words: does an asymmetric planet get less radial acceleration than a perfect spherical one?&lt;/p&gt;&#xA;" OwnerDisplayName="user13628" LastActivityDate="2016-08-02T14:24:18.943" Title="What is the most asymmetric known planet?" Tags="&lt;orbital-mechanics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16765" PostTypeId="2" ParentId="16763" CreationDate="2016-07-31T16:04:22.487" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can you give some details about this process, is the orbit changing&#xA;  shape, eccentricity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This isn't a simple answer.   Also, do you want to know this for a simple 2 body system or the more complex, essentially 3 body system of Sun-earth-Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Earth and Moon were alone in the cosmos, far away from everything else, then over time, the tidal forces would circularize the Moon's orbit as it pushes the Moon away.  The reason for this is that the same force that slowly pushes the Moon away from the Earth pushes hardest when the Moon is closest, and least hard when the Moon is furthest away, so over a long period of time that circularizes it's orbit.  All 2 body systems where the moon is being pushed away tends to circularize.  This can be observed in our own solar-system.   Two body systems that are very far away from orbital disturbances, like Pluto-Charon or Neptune-Triton or even, Sun-Neptune, are highly circular orbits.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when you bring the Sun into the picture, everything changes.  The Moon orbits the Earth and the Earth Moon system orbits the sun, and in a very real sense, the Moon also orbits the sun, but it's within the Earth's stable region of it's hill sphere, so the Moon's orbit around the Earth will remain stable for billions of years.   But while it's stable, it's measurably inconstant.  The moon's orbit wobbles a great deal, being stretched and contracted based on it's position in relation to the Earth and Sun.  3 body mechanics is quite complex, even with stable orbits.  Isaac Newton was aware of this but he wasn't able to work it out.  Mathematicians who came later were able to figure out orbital interactions with more than 2 bodies, called perturbation theory.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, your question depends on whether you mean a 2 body system or a 3 body system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth's eccentricity, for example, changes on roughly a 100,000 year cycle, mostly due to Jupiter, but that effect is pretty slow, on a roughly &lt;a href=&quot;https://en.wikipedia.org/wiki/Milankovitch_cycles#Orbital_shape_.28eccentricity.29&quot; rel=&quot;nofollow noreferrer&quot;&gt;100,000 year cycle&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jXDUc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jXDUc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon's apogee and perigee happen every 27.3 days (&lt;a href=&quot;https://community.dur.ac.uk/john.lucey/users/lunar_sid_syn.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sidereal orbit&lt;/a&gt;) and it's eccentricity changes measurably every orbit.  In a sense, it's not very accurate to say the Moon even orbits around the Earth in an ellipse, it's too squashed and stretched for the elliptical model to be accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the Earth takes about 100,000 orbits to go through one elliptical period, the Moon needs only 6 or 7 orbits to do that.   Note how measurably it's perigee and apogee change every orbit.   &lt;a href=&quot;https://www.fourmilab.ch/earthview/pacalc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;See table&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a fair bit of information available on the Moon's irregular orbit, but I suspect that has almost nothing to do with your primary question.   In general, in the 3 body system, as the Moon moves away from the Earth it's orbit grows more erratic, so in a sense, because of the sun, the Moon's orbit grows more eccentric, over time, where, if the sun wasn't there, it would grow more circular.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is the average orbital speed decreasing, by what value? is the spin&#xA;  angular velocity increasing, by what value?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On average, as the Moon moves away it's orbital speed decreases.  More distant orbits have less kinetic energy, but again, with a 3 body system it's more complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, I think I see what you're asking on Spin.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;from Wiki &quot;angular momentum is slowly being transferred from Earth's&#xA;  rotation to the Moon's orbit.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There can be no transfer, since both earth and moon are slowing down&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You should look at the Earth-Moon as a closed system.  For a closed system, angular momentum remains constant, but angular momentum in this sense, is around the barycenter.   The Moon's angular momentum, in the sense that Wikipedia means isn't the Moon's spin but the Moon's orbit.  As the Moon moves away from the Earth, it's velocity decreases but it's angular momentum around the barycenter increases because angular momentum is a product of both velocity and distance from the center.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter, for example, has about 71% of the total angular momentum of the solar system.   See &lt;a href=&quot;http://www.astronoo.com/en/articles/angular-momentum.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt;, which I think is related to your question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;wiki says that average distance from earth increases 3.8 cm per year,&#xA;  can you tell how much of that is due to the expansion of the universe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Expansion is best understood if objects aren't orbiting around each other.  If the Moon was stationary and the Earth was stationary, and the only way to do that in our thought experiment is to give one of them a mass of zero so there's no orbit and no gravitational attraction.   We're breaking almost every rule in the book with this thought experiment, but bear with me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take the measured expansion of space to be 74.3 km/s over 3 million light years.   &lt;a href=&quot;http://www.space.com/17884-universe-expansion-speed-hubble-constant.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Average lunar distance 384,400 km.  And, 9.461 trillion km in 1 light year, so about 2.8 x 10^19 km in 3 million lightyears.  Roughly 74 trillion lunar distances. which works out kind of neatly, 74.3 km/s divided by 74 trillion gives us one trillionth of a KM/s, 31.5 million seconds in a year gives us about .0000315 km/year expansion between a stationary, no gravitational attraction moon and Earth, which equals about 3.15 cm per year of expansion of space over 1 lunar distance, which is curiously not far off from 3.8 cm per year increase in distance due to tidal forces, but the similarity is purely coincidental as well as irrelevant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(too long?)&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-07-31T16:37:35.553" LastActivityDate="2016-07-31T16:37:35.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16766" PostTypeId="1" CreationDate="2016-07-31T19:57:33.837" Score="1" ViewCount="247" Body="&lt;p&gt;Here, we are looking for a statistically significant difference between the mean redshifts of galaxies ahead and those behind, with reference to a prior-specified direction. For example, suppose that we were aboard an advanced spaceship that had been accelerating in that direction at (say) 1 $g$ for some years. How soon into our journey would we notice that the galaxies ahead were bluer than the galaxies behind? Clearly this depends on our existing model, based on prior observations, for the distribution of galaxies' redshifts, and on some reasonable kind of weighting for each individual galaxy (perhaps according to the cosine of angle from our direction of travel, but not according to the galaxy's distance). However, in this analysis, we are ignoring individual data about any particular galaxy acquired before our present observation. For simplicity, let us also suppose that there is no statistically significant difference between the mean redshifts forward and back in the line of our journey before we began it. For the sake of definiteness, if it is needed, let us take a significance level of 1%. &lt;/p&gt;&#xA;" OwnerUserId="13641" LastActivityDate="2016-08-10T13:27:59.363" Title="How fast would we have to go for the red(/blue)shifts of galaxies ahead of us to differ from those behind us with statistical significance?" Tags="&lt;redshift&gt;&lt;special-relativity&gt;&lt;observable-universe&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16767" PostTypeId="1" AcceptedAnswerId="16768" CreationDate="2016-08-01T05:50:19.020" Score="1" ViewCount="77" Body="&lt;p&gt;Suppose a terrestrial planet with no atmosphere is tidally locked and very near the star like a hot Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the far side still very cold or would crust would conduct heat from the near side to far side?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="2365" LastEditDate="2016-08-01T07:37:38.363" LastActivityDate="2016-08-01T07:50:42.527" Title="Does terrestrial planets conduct heat from near to far side?" Tags="&lt;terrestrial-planets&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16768" PostTypeId="2" ParentId="16767" CreationDate="2016-08-01T07:50:42.527" Score="3" Body="&lt;p&gt;The transfer of heat would be insignificant. Consider the Earth, the core is as hot as the surface of the sun. Some of this heat does get transferred to the surface, but it is only 0.03% of the energy received from the sun. Rock is a good insulator, most of the heat is transferred by convection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The far side of a tidally locked planet would be very cold. It would receive very little heating from the planet's interior or the near side.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-08-01T07:50:42.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16769" PostTypeId="1" AcceptedAnswerId="16771" CreationDate="2016-08-01T08:09:46.700" Score="5" ViewCount="1829" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_acceleration&quot; rel=&quot;nofollow noreferrer&quot;&gt;The wikipage on &amp;quot;tidal accelaeration&amp;quot;&lt;/a&gt; has this picture of the tidal bulge:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0g9pT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0g9pT.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;but says that:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;The average tidal bulge is synchronized with the Moon's orbit,&lt;/strong&gt; and&#xA;Earth rotates under this tidal bulge in just over a day. However,&#xA;Earth's rotation drags the position of the tidal bulge ahead of the&#xA;position directly under the Moon. As a consequence, there exists a&#xA;substantial amount of mass in the bulge that is offset from the line&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;On the other hand, this academical &lt;a href=&quot;http://www.lhup.edu/%7Edsimanek/scenario/tides.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt;&#xA;states the opposite:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;as we move with respect to both tidal bulge and moon, &lt;strong&gt;the moon&#xA;crosses our meridian before we experience the highest tide.&lt;/strong&gt;] How&#xA;early? Some books show misleading diagrams with the symmetry axis of&#xA;the tidal bulges making an angle of 30° or more with the moon. In&#xA;fact, the angle is only 3°, so the tides are late by about 24(3/360)60&#xA;= 12 minutes&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;through the centers of Earth and the Moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Can you say what is the real position of the bulge? it is surely a hard fact not open to interpretations. Do you have access to sites that say where is now the moon and where is the tidal bulge?&lt;/p&gt;&#xA;&lt;p&gt;According to the basic laws of conservation of momentum and energy, a mass moving to a greater radius should be slowed downm and if we add friction the lag of 12 minutes or more makes sense. If the highest point of the bulge is forward wrt to the vertical of the moon, can you explain what stronger force pushes the water in the direction of the spin?&lt;/p&gt;&#xA;&lt;p&gt;I found this &lt;a href=&quot;http://www.timeanddate.com/worldclock/sunearth.html?iso=20160801T0824&amp;amp;earth=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;site&lt;/a&gt; that gives the position of sun and moon, but doesn't show tides&#xA;&lt;a href=&quot;https://i.stack.imgur.com/4S4UF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4S4UF.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;This animation, on the other hand, shows that there is no bulge&lt;/p&gt;&#xA;&lt;p&gt;The red areas experience a rise over 1 m, the blue one a depression of over 1 m:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/N3FBf.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/N3FBf.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user13628" LastEditorUserId="37191" LastEditorDisplayName="user13628" LastEditDate="2021-02-03T12:13:29.923" LastActivityDate="2021-02-03T12:13:29.923" Title="What is the real position of the tidal bulge?" Tags="&lt;tidal-forces&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="16771" PostTypeId="2" ParentId="16769" CreationDate="2016-08-01T09:42:33.490" Score="3" Body="&lt;p&gt;There is no great contradiction. The offset of the tidal bulge is about 3 degrees. It is exaggerated in diagrams for clarity. The diagram is correct, but not to scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This causes the tides to be slightly &lt;strong&gt;late&lt;/strong&gt;. Imagine a person standing on the Earth of the diagram, with the moon directly overhead. The tidal bulge is on their left. The rotation of the Earth will take them towards the left (the moon is also orbiting but its motion is much slower), so a little later (12 min later) they will reach the maximum of the tide. The maximum is delayed by about 12 min.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actual flows of water around the coast are driven by this tidal bulge, but are complex effects of local topography. The actual flows of water are highly non-linear, including multiple &lt;a href=&quot;https://en.wikipedia.org/wiki/Amphidromic_point&quot; rel=&quot;nofollow noreferrer&quot;&gt;locations at which there is no tide&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UCa4v.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UCa4v.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-08-01T09:42:33.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16772" PostTypeId="2" ParentId="16769" CreationDate="2016-08-01T09:46:04.277" Score="1" Body="&lt;p&gt;As the same &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_acceleration&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt; states&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An equilibrium tidal bulge does not really exist on Earth because the continents do not allow this mathematical solution to take place. Oceanic tides actually rotate around the ocean basins as vast gyres around several amphidromic points where no tide exists. The Moon pulls on each individual undulation as Earth rotates—some undulations are ahead of the Moon, others are behind it, whereas still others are on either side. The &quot;bulges&quot; that actually do exist for the Moon to pull on (and which pull on the Moon) are the net result of integrating the actual undulations over all the world's oceans. Earth's net (or equivalent) equilibrium tide has an amplitude of only 3.23 cm, which is totally swamped by oceanic tides that can exceed one metre.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the cartoon that is drawn is an idealised scenario that gets messed with due to the real physical geography of the Earth and its oceans. The tides are also affected by the Sun. However, any asymmetry in the same sense as is shown in the cartoon will result in tidal torques and tidal accelerations in the sense that the Earth's rotation is slowed whilst the Earth-Moon system is tidally accelerating to larger separations and longer orbital period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;bulge&quot; leads the meridian position of the Moon in your diagram because there is a lag in the response to the tidal force. This is akin to a driven oscillator where the damping leads to a lag in the response with respect to the driving force. And the Earth simply rotates by a certain amount during the lag time. The periodicity of the driving force and response is 12h25m. The actual lag and phase shift will depend on local geography but should be longer than 12 minutes. &lt;a href=&quot;http://www.jalc.edu/mikolajsawicki/tides_new2.pdf&quot; rel=&quot;nofollow&quot;&gt;This source&lt;/a&gt; (which I &lt;em&gt;strongly&lt;/em&gt; recommend you read) says that in the Southern ocean - which has the &quot;cleanest response&quot; to the tidal driving - the bulge rises about 2-3 hours after the Moon crosses the meridian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand the rest of your question; the tides are raised by tidal forces due to the Moon (and Sun).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-08-01T11:24:55.023" LastActivityDate="2016-08-01T11:24:55.023" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16774" PostTypeId="1" CreationDate="2016-08-01T15:40:43.420" Score="2" ViewCount="48" Body="&lt;p&gt;With regard to the wave/particle duality of light, does Big Bang red shift affect both the frequency of light waves, and, the frequency of individual photons? I assume that the e=hf formula relates to individual photons, and not to light waves too?&lt;/p&gt;&#xA;" OwnerUserId="13650" LastActivityDate="2016-08-01T16:03:53.157" Title="Does Big Bang red shift affect the frequency of both light waves and individual photons?" Tags="&lt;cosmology&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16775" PostTypeId="2" ParentId="16774" CreationDate="2016-08-01T16:03:53.157" Score="3" Body="&lt;p&gt;Yes it affects both. If you treat light as waves then the wavelength from a source at a redshift $z$  is &quot;stretched&quot; by a factor of $1+z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photons emitted from a source at redshift $z$ have their energies reduced by a factor of $1+z$.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-08-01T16:03:53.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16776" PostTypeId="1" AcceptedAnswerId="17890" CreationDate="2016-08-02T02:49:23.520" Score="6" ViewCount="754" Body="&lt;p&gt;From what I gather, the observable universe is expanding at the speed of light. In addition, information cannot travel faster than the speed of light. It seems as though these two facts would imply that information cannot leave the observable universe. However, if that is true, it would either mean that the universe started spread apart, or that the observable universe is the entire universe. So, which of these explanations is correct?&lt;/p&gt;&#xA;" OwnerUserId="13656" LastEditorUserId="1891" LastEditDate="2019-01-27T13:25:34.770" LastActivityDate="2019-01-27T13:25:34.770" Title="Can information leave the observable Universe?" Tags="&lt;universe&gt;&lt;light&gt;&lt;observable-universe&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="16777" PostTypeId="1" CreationDate="2016-08-02T02:54:53.253" Score="1" ViewCount="303" Body="&lt;p&gt;I am looking at data used in a project to get the velocity and acceleration of a CME (Coronal Mass Ejection). I understand how the velocity was calculated, and for the acceleration, I get the v2-v1 part of the formula for acceleration that was used (ex. 306.5-331.2), but what time intervals (t2-t1) were used to calculate the accelerations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cmifs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cmifs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13657" LastActivityDate="2016-08-25T23:28:04.727" Title="How was the acceleration calculated for this CME (Coronal Mass Ejection)?" Tags="&lt;velocity&gt;&lt;coronal-mass-ejection&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16778" PostTypeId="2" ParentId="16777" CreationDate="2016-08-02T08:06:30.780" Score="1" Body="&lt;p&gt;The average velocity in the interval 8:05 to 8:36 is 331.2. So, use $v_1=$ 331.2 at $t_1=$ 08:20:30 under the assumption that acceleration is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average velocity between 8:36 and 9:27 is $v_2=$ 306.5, so use $t_2=$ 9:01:30&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acceleration is $(306.5-331.3)/(41\times60)=-0.01\ $km/s$^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same calculation gives me $0.029\ $km/s$^2$ and $-0.038\  $km/s$^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this just arithmetic error on the first value (using 51m instead of 41m), a small rounding issue on the second and the third I agree with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than that, then some unstated alternative interpolation scheme has been used.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-08-25T23:28:04.727" LastActivityDate="2016-08-25T23:28:04.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16779" PostTypeId="2" ParentId="16764" CreationDate="2016-08-02T10:41:39.273" Score="5" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;What is the most asymmetric known planet?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;That depends on what you mean by asymmetric, on what counts as planet, and on what parts of a planet are to be considered / excluded. The dominant feature of many planets is that they aren't quite spherical due to their rotation. If that counts then Saturn is the winner. Saturn's poles are 5900 km closer to Saturn's center of mass than are points on Saturn's equator.&lt;/p&gt;&#xA;&lt;p&gt;That oblateness is however exactly the shape an object such as Saturn is expected to take on. It means that Saturn is in an equilibrium shape. Moreover, it has a very definite axis of symmetry, the rotation axis. So that's probably not what is meant in the question as being asymmetric. All of the giant planets are quite symmetric by this standard. We need to look to the terrestrial planets.&lt;/p&gt;&#xA;&lt;p&gt;One concept that might be of aid is the center of mass - center of figure offset of each of the four terrestrial planets. Do the oceans count? Do the icecaps count? Excluding the oceans but not the ice leads to COM-COF offset of about 2100 meters. Including both leads to a COM-COF offset of about 800 meters, about the same as Mercury and Venus. Mars however has the Earth beat by far. Various estimates place the Mars COM-COF offset between 2.5 km and 3.3 km.&lt;/p&gt;&#xA;&lt;p&gt;Another approach is to look at how lumpy the planet's gravitational field is. Mars once again is the winner in this regard.&lt;/p&gt;&#xA;&lt;br&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I have a simple technical question: an asymmetric body gets a torque in a gravitational field, is the rotational energy/acceleration to be subtracted from the regular energy/acceleration? In other words: does an asymmetric planet get less radial acceleration than a perfect spherical one?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;No. In a system of particles bound by central forces such as self-gravitation, Coulomb forces, and chemical bonds, the acceleration of the system's center of mass is the net sum of the external forces acting on the individual particles, divided by the total mass.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-08-02T14:24:18.943" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16781" PostTypeId="1" CreationDate="2016-08-02T11:35:25.503" Score="1" ViewCount="304" Body="&lt;p&gt;Please give preference to those most widely used in the field, if possible.&lt;/p&gt;&#xA;" OwnerUserId="6786" LastEditorUserId="6" LastEditDate="2016-08-02T14:41:35.707" LastActivityDate="2016-08-16T20:33:54.173" Title="What programming languages does SETI (Search for ExtraTerrestrial Intelligences) most commonly use for signal processing?" Tags="&lt;radio-astronomy&gt;&lt;life&gt;&lt;extra-terrestrial&gt;&lt;software&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16782" PostTypeId="1" AcceptedAnswerId="16784" CreationDate="2016-08-02T12:09:05.980" Score="0" ViewCount="383" Body="&lt;p&gt;I noticed that last June the Sun was on the tropic of Cancer and moon on Madagascar:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VZu4e.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VZu4e.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Can you explain how that works and if they can get even farther in latitude?&lt;/p&gt;&#xA;" OwnerDisplayName="user13628" LastEditorUserId="-1" LastEditorDisplayName="user13628" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-08-02T12:43:55.653" Title="How far can the Sun and the Moon get?" Tags="&lt;orbit&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16783" PostTypeId="2" ParentId="16782" CreationDate="2016-08-02T12:32:16.407" Score="0" Body="&lt;p&gt;You mean the point on the Earth where the sun and moon are directly overhead?   Just as eclipses can happen, being on opposite sides of the Earth can happen too.   Generally speaking this is a &lt;a href=&quot;http://spaceplace.nasa.gov/eclipses/en/&quot; rel=&quot;nofollow noreferrer&quot;&gt;lunar eclipse&lt;/a&gt;.  Directly opposite, just like solar eclipses are relatively uncommon because the moon orbits the Earth on a different plane than the Earth's orbit around the sun, so the timing has to be just right for them to line up.  (see somewhat exaggerated diagram).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pm6HJ.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pm6HJ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you were to carefully draw the equator across the picture you posted, you'd see the Moon and Sun are not at opposite latitudes, which they would need to be to be as &quot;far apart as possible&quot;. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-08-02T12:32:16.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16784" PostTypeId="2" ParentId="16782" CreationDate="2016-08-02T12:35:07.753" Score="2" Body="&lt;p&gt;The Earth's equator is inclined to the plane of the solar system (called the ecliptic) by 23 degrees. In other words, the Sun's position, as seen from the Earth, varies between 23 degrees North (at the northern hemisphere summer solstice) and 23 degrees South (at northern hemisphere winter solstice), appearing to cross the Earth's equator twice (at the equinoxes). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tropic lines mark the maximum apparent solar move north or south.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon's orbit is inclined by about 5 degrees to the ecliptic so moves over a somewhat wider range of the sky as seen from the Earth. The Moon orbits the Earth every 28 days or so, so moves over that range each month.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At full Moon the Moon is opposite the Sun in the sky, so if the Sun was close to the tropic, then the Moon would be around the opposite tropic when full, and so on.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2016-08-02T12:35:07.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16786" PostTypeId="2" ParentId="87" CreationDate="2016-08-02T18:18:31.203" Score="3" Body="&lt;p&gt;The following are largely excerpts from an answer I wrote at &lt;a href=&quot;https://physics.stackexchange.com/a/258093/59023&quot;&gt;https://physics.stackexchange.com/a/258093/59023&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What are the main differences between solar flares and coronal mass ejections?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are several differences between &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_flare&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar flares&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Coronal_mass_ejection&quot; rel=&quot;nofollow noreferrer&quot;&gt;coronal mass ejections&lt;/a&gt; (CMEs), the latter of which involve large amounts (i.e., upwards of billions of tons) of matter (in the form of &lt;a href=&quot;https://en.wikipedia.org/wiki/Plasma_(physics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;plasma&lt;/a&gt;) to leave the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A solar flare is really described by the sudden enhancement of &lt;a href=&quot;https://en.wikipedia.org/wiki/X-ray&quot; rel=&quot;nofollow noreferrer&quot;&gt;x-rays&lt;/a&gt; from a small region in the solar corona.  They can produce streams of high energy particles, called solar energetic particles or &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_energetic_particles&quot; rel=&quot;nofollow noreferrer&quot;&gt;SEPs&lt;/a&gt; (CMEs can produce SEPs as well), but these are streams of magnetic field-aligned particles propagating away from the sun.  They can have energies up to ~GeV for ions and several ~MeV for electrons, but rarely higher.  Earth's atmosphere does a great job of shielding much of this and the surrounding magnetic field helps to &quot;deflect&quot; some of the incident particles as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A CME, on the other hand, expands nearly as fast as it propagates which means their radius of curvature by the time they reach Earth is nearly 1 &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_unit&quot; rel=&quot;nofollow noreferrer&quot;&gt;AU&lt;/a&gt;.  They are, in effect, a large &quot;magnetic piston&quot; (sometimes called a magnetic cloud or flux rope etc.) that piles-up plasma on their leading edge and can produce rather strong &lt;a href=&quot;https://en.wikipedia.org/wiki/Shock_waves_in_astrophysics&quot; rel=&quot;nofollow noreferrer&quot;&gt;collisionless shock waves&lt;/a&gt;.  The magnetic field orientation of the magnetic piston is very important, as its orientation relative to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth%27s_magnetic_field&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth's magnetic&lt;/a&gt; field can result in either a small &lt;a href=&quot;https://en.wikipedia.org/wiki/Aurora&quot; rel=&quot;nofollow noreferrer&quot;&gt;auroral&lt;/a&gt; response or a major &lt;a href=&quot;https://en.wikipedia.org/wiki/Geomagnetic_storm&quot; rel=&quot;nofollow noreferrer&quot;&gt;geomagnetic storm&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...what causes them...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Both are thought to ultimately arise from a process called &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetic_reconnection&quot; rel=&quot;nofollow noreferrer&quot;&gt;magnetic reconnection&lt;/a&gt; -- a process that results in a reconfiguration of the magnetic field topology and transformation of magnetic energy into bulk particle kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of flares, reconnection results in jets of highly energetic electrons and protons that slam into the upper solar atmosphere resulting in x-rays produced by &lt;em&gt;thick target&lt;/em&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Bremsstrahlung&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bremsstrahlung&lt;/a&gt; radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of CMEs, reconnection results in a release of a massive amount of plasma often confined within a &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetic_cloud&quot; rel=&quot;nofollow noreferrer&quot;&gt;magnetic cloud&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...what sort of damage can they cause...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See my previous answers at &lt;a href=&quot;https://physics.stackexchange.com/a/258093/59023&quot;&gt;https://physics.stackexchange.com/a/258093/59023&lt;/a&gt;, &lt;a href=&quot;https://physics.stackexchange.com/a/214509/59023&quot;&gt;https://physics.stackexchange.com/a/214509/59023&lt;/a&gt;, and &lt;a href=&quot;https://physics.stackexchange.com/a/149199/59023&quot;&gt;https://physics.stackexchange.com/a/149199/59023&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...what are the differences in their composition?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As I explained above, flares are just enhancements in x-rays so there is no composition.  CMEs generally have different charge state and heavy ion composition than the nominal &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_wind&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar wind&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13663" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-08-02T18:18:31.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16787" PostTypeId="2" ParentId="623" CreationDate="2016-08-02T19:00:16.183" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the largest recorded solar flare?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since @Moriarty already dealt with historical records, I will focus on the modern observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;X-Ray Classification&lt;/strong&gt;&lt;br&gt;&#xA;On April 2, 2001 the &lt;a href=&quot;https://en.wikipedia.org/wiki/Geostationary_Operational_Environmental_Satellite&quot; rel=&quot;nofollow&quot;&gt;GOES&lt;/a&gt; spacecraft recorded an X20 flare (see the following for flare classification: &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_flare#Classification&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Solar_flare#Classification&lt;/a&gt;).  Most flares fall below C-class (i.e., at least 100 times weaker) and X-class flares are the strongest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On November 4, 2003 there was an X20+ class flare.  Initially it was estimated to be an &lt;a href=&quot;http://thewatchers.adorraeli.com/2013/11/04/largest-ever-recorded-solar-flare-in-history-the-x-whatever-mega-flare/&quot; rel=&quot;nofollow&quot;&gt;X40&lt;/a&gt;, but was later re-classified as an &lt;a href=&quot;http://www.esa.int/Our_Activities/Space_Science/Extreme_space/The_biggest_solar_X-ray_flare_ever_is_classified_as_X28&quot; rel=&quot;nofollow&quot;&gt;X28&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;CME Speeds&lt;/strong&gt;&lt;br&gt;&#xA;In events that also release &lt;a href=&quot;https://en.wikipedia.org/wiki/Coronal_mass_ejection&quot; rel=&quot;nofollow&quot;&gt;coronal mass ejections&lt;/a&gt; (CMEs), we can classify the strength of the solar storm by the CME speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 1859 &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_storm_of_1859&quot; rel=&quot;nofollow&quot;&gt;Carrington event&lt;/a&gt; reached Earth in just ~17.67 hr, which is an average speed of ~2400 km/s (or ~5.3 million mph).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was a CME on &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/swe.20097/abstract&quot; rel=&quot;nofollow&quot;&gt;July 12, 2012&lt;/a&gt; observed by the &lt;a href=&quot;https://en.wikipedia.org/wiki/STEREO&quot; rel=&quot;nofollow&quot;&gt;STEREO&lt;/a&gt; spacecraft that had an observed speed of ~2500 $\pm$ 500 km/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The record CME speed was set in August 1972, where the CME speed was estimated at ~2850 km/s (or ~6.4 million mph).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Probability of Occurrence&lt;/strong&gt;&lt;br&gt;&#xA;There was a recent &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1029/2011SW000734/abstract&quot; rel=&quot;nofollow&quot;&gt;study&lt;/a&gt; that found the probability of another Carrington-like event is as high as ~12% in the next decade.&lt;/p&gt;&#xA;" OwnerUserId="13663" LastEditorUserId="13663" LastEditDate="2016-08-03T12:22:49.617" LastActivityDate="2016-08-03T12:22:49.617" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16788" PostTypeId="2" ParentId="16717" CreationDate="2016-08-03T00:23:08.427" Score="7" Body="&lt;p&gt;&lt;a href=&quot;http://www.sns.ias.edu/~jnb/SNdata/solarmodels.html&quot; rel=&quot;noreferrer&quot;&gt;A page from the Institute for Advanced Study&lt;/a&gt; links to data from various modifications to the Standard Solar Model. The newest given there is from &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0412440&quot; rel=&quot;noreferrer&quot;&gt;Bahcall&lt;sup&gt;1&lt;/sup&gt; et al. (2005)&lt;/a&gt;, which I'll use as an example. The authors' calculations depend on data from the Opacity project. Data from two models are available through the IAS link: the &lt;a href=&quot;http://www.sns.ias.edu/~jnb/SNdata/Export/BS2005/bs05_agsop.dat&quot; rel=&quot;noreferrer&quot;&gt;BS05(AGS, OP) model&lt;/a&gt; and the &lt;a href=&quot;http://www.sns.ias.edu/~jnb/SNdata/Export/BS2005/bs05op.dat&quot; rel=&quot;noreferrer&quot;&gt;BS05(OP) model&lt;/a&gt;; the difference lies in the main thing you're interested in: heavy element abundances. BS05(AGS, OP) has, for instance, a central&lt;sup&gt;2&lt;/sup&gt; $^{12}\text{C}$ mass fraction of $7.79\times10^{-6}$, while BS05(OP) has a central $^{12}\text{C}$ mass fraction of $1.05\times10^{-5}$. All of the central data from the major thermodynamic variables (e.g. pressure and temperature), as well as the mass fractions, are well within an order of magnitude in each model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, the general heavy element trend is the same in both models. $^{12}\text{C}$ increase by two to three orders of magnitude from the center. $^3\text{He}$ (rather than the more common $^4\text{He}$, which has a mass fraction given by $Y$) increases by one to two orders of magnitude. $^{14}\text{N}$ actually &lt;em&gt;decreases&lt;/em&gt; by two orders of magnitude, while $^{16}\text{O}$ remains relatively constant. The model gives no data for silicon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other new models agree (though not all), to less than an order of magnitude or so, with the data from Bahcall et al. Graphical representations can be found in e.g. &lt;a href=&quot;http://cos.colorado.edu/~kevinf/PAPERS/solmodel.pdf&quot; rel=&quot;noreferrer&quot;&gt;this paper&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aQQm7.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aQQm7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm unsure of the reason behind the $^3\text{He}$ spike; that may warrant a follow-up questions. My one guess is that this could be due to its role as an intermediate nucleus in the second step of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction&quot; rel=&quot;noreferrer&quot;&gt;p-p chain&lt;/a&gt;. However, the same trends for each element are followed, although the changes in mass fraction are &lt;em&gt;not&lt;/em&gt; constant at all radii $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your question about how the central figures were derived, the answer is that helioseismological measurements and neutrino fluxes are some of the best indicators of composition. This was used as the basis for the BS05(AGS, OP) and BS05(OP) models, as well as, in fact, most variants of the Standard Solar Model.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1 This is Bahcall's page, so, naturally, these are some of his papers. However, the results appear to be consistent with the calculations of others.&lt;/sup&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;2 Technically, this is from about 0.0016 solar radii, but this distance step is comparatively small.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-08-03T00:38:21.977" LastActivityDate="2016-08-03T00:38:21.977" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16791" PostTypeId="1" CreationDate="2016-08-03T05:19:40.190" Score="2" ViewCount="83" Body="&lt;p&gt;When there are accretion discs jets of particles or radiations will be emitted in the direction parallel to the rotational axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My beginner level astronomy textbook just says that it is due to magnetic field but does not explain why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can anyone explain the mechanism in layman's terms? Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2016-08-03T19:07:20.430" LastActivityDate="2016-08-03T19:07:20.430" Title="Accretion discs emitting jets" Tags="&lt;black-hole&gt;&lt;supermassive-black-hole&gt;&lt;accretion-discs&gt;&lt;galaxy-center&gt;&lt;quasars&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2016-08-03T14:28:41.640" ContentLicense="CC BY-SA 3.0" />
  <row Id="16793" PostTypeId="2" ParentId="16791" CreationDate="2016-08-03T09:22:55.637" Score="2" Body="&lt;p&gt;If you don't mind a layman's explanation, the first thing to understand is why the accretion disk is flat and that's a product of conservation of angular momentum, explained very well in this &lt;a href=&quot;https://www.youtube.com/watch?v=tmNXKqeUtJM&quot; rel=&quot;nofollow noreferrer&quot;&gt;video&lt;/a&gt;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we begin with a flat, rapidly rotating (or orbiting) disk of very hot matter, too hot for electrons to stay bound to atoms, so it's both hot and electrically charged, you get gonzo magnetic fields, which (by a process I don't understand well) creates magnetic field files that extends in twists both above and below the disc, see pretty pictures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NzQjt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NzQjt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o7a9Z.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o7a9Z.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as the matter from the disk tries to fall inside the black hole, if there's too much matter, it can't all squeeze inside at once because mostly, black holes are quite tiny and the pressure and magnetism shoots a lot of the matter back out, primarily through the middle of the magnetic field lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a kind of simple explanation anyway.  Correction is welcome. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-08-03T09:22:55.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16794" PostTypeId="1" CreationDate="2016-08-03T11:54:13.303" Score="1" ViewCount="58" Body="&lt;p&gt;I read on on the website of the SKA (&lt;a href=&quot;https://www.skatelescope.org/&quot; rel=&quot;nofollow&quot;&gt;https://www.skatelescope.org/&lt;/a&gt;) that SETI will sort of have &quot;passive&quot; access to the data that they capture on the array, sort of like a &quot;pigyback&quot; concept and search the data for signals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much of the data will they have access to? Everything? That would be literally petabytes of data per day that they would have to process. If so how do they plan to process all that data? Like a more effecient SETI@Home?&lt;/p&gt;&#xA;" OwnerUserId="10271" LastEditorUserId="10271" LastEditDate="2016-08-03T12:19:06.973" LastActivityDate="2019-11-16T22:44:59.163" Title="What information will SETI have access to from the SKA (Square kilometer array)" Tags="&lt;radio-astronomy&gt;&lt;software&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16795" PostTypeId="1" CreationDate="2016-08-03T11:57:34.430" Score="3" ViewCount="309" Body="&lt;p&gt;I will expect that gravitational lensing should turn an object into a ring image. But I saw in some situations that instead multiple images are formed. For example, the Einstein-cross in the image below, consisting of four images:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/CGVQy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CGVQy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it so?&lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2016-08-05T18:49:41.647" Title="Images of gravitational lensing" Tags="&lt;gravitational-lensing&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16796" PostTypeId="2" ParentId="10655" CreationDate="2016-08-03T11:59:28.000" Score="1" Body="&lt;p&gt;This sounds like a satelite to me. Satelites can also &quot;dissapear&quot; right above you, since the sun no longer reaches them and they fade into darkness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also 2 minutes is very well possible for a satelites to be visible that is in earths orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should also know, that there are hunderds of other satelites that can be as bright when observed from earth as the ISS is.&lt;/p&gt;&#xA;" OwnerUserId="10271" LastActivityDate="2016-08-03T11:59:28.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16797" PostTypeId="1" AcceptedAnswerId="16799" CreationDate="2016-08-03T12:01:15.407" Score="-1" ViewCount="206" Body="&lt;p&gt;i heard &amp;amp; saw in website &amp;amp; you tube most of the people saying that i Jupiter planet made of Gas..i like to know if we some how went to Jupiter &amp;amp; land a Space Ship cant we find a Hard Land with soil.&lt;/p&gt;&#xA;" OwnerUserId="11792" LastActivityDate="2016-08-03T13:47:08.723" Title="Is it Jupiter Only a Gas Giant?" Tags="&lt;planet&gt;&lt;jupiter&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ClosedDate="2016-08-03T14:53:32.123" ContentLicense="CC BY-SA 3.0" />
  <row Id="16798" PostTypeId="2" ParentId="16744" CreationDate="2016-08-03T12:03:00.457" Score="2" Body="&lt;p&gt;The period (at least in physics) is defined the time an oscilating system needs to get back to its starting point (for a sinus curve its 2*pi).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now when they say an eclipsing binary has a period of T years, it means that its brightness varies as seen from earth when they are eclipsing (thats how we detect eclipsing stars and planets by the way), and that the curve of its brightness is like an oscilation that gets back to its starting point every T years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The meaning of this is, that this system consisting of 2 stars will be exactly as you see it now again in T years.&lt;/p&gt;&#xA;" OwnerUserId="10271" LastActivityDate="2016-08-03T12:03:00.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16799" PostTypeId="2" ParentId="16797" CreationDate="2016-08-03T12:17:00.370" Score="4" Body="&lt;p&gt;Yes, you can't land in the usual sense. You just sink into the gas and will be destroyed by the high pressure at some point. In fact deep inside the gas may be even denser than the ship and so the ship will tend to float. Anyway if you really want to &quot;land&quot;, then assuming that somehow your ship survives all the adverse conditions and you have an engine to oppose the upthrust and to pass through the metallic hydrogen mantle, you will finally &quot;land&quot; at the solid core of the planet.&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2016-08-03T13:47:08.723" LastActivityDate="2016-08-03T13:47:08.723" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16800" PostTypeId="2" ParentId="14304" CreationDate="2016-08-03T12:37:41.053" Score="5" Body="&lt;p&gt;The simplest explanation for the latitudinal distribution would be that most &lt;a href=&quot;http://ssd.jpl.nasa.gov/?bodies&quot; rel=&quot;nofollow&quot;&gt;small solar system bodies&lt;/a&gt; that could impact one of the planets happen to orbit near the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ecliptic&quot; rel=&quot;nofollow&quot;&gt;ecliptic plane&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Mars&quot; rel=&quot;nofollow&quot;&gt;Mars&lt;/a&gt;' &lt;a href=&quot;http://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html&quot; rel=&quot;nofollow&quot;&gt;orbital inclination&lt;/a&gt; is only ~$1.85^{\circ}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The longitudinal &quot;clumpiness&quot; may be due to erosion rate differences, since the new impacts almost exclusively appear in the lighter shaded regions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update/Edits&lt;/strong&gt;&lt;br&gt;&#xA;@LocalFluff brought up an excellent point that made me re-examine my original thoughts on this matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some digging, it turns out that (not surprisingly) there has been a lot of work on impact craters on just Mars alone.  As for the distribution, there was a relatively recent study by &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1029/2011JE003967/abstract&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Robbins and Hynek&lt;/em&gt;, [2012]&lt;/a&gt; (doi:10.1029/2011JE003966)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They found that an interesting ratio of the depth-to-diameter of the impact craters.  This ratio peaks near low latitudes (i.e., within roughly $\pm30^{\circ}$ of the Martian equator) and generally decreases towards higher latitudes.  They argue that the likely cause of the larger diameter craters near the poles results from a shallower &lt;a href=&quot;https://en.wikipedia.org/wiki/Cryosphere&quot; rel=&quot;nofollow&quot;&gt;cryosphere&lt;/a&gt;.  They also find that the craters at high latitudes are more likely to be &lt;em&gt;complex&lt;/em&gt;, which &lt;code&gt;...are large and have a variety of interior morphologies such as wall terraces, central peaks, and flat floors...&lt;/code&gt;  Part of the reason for the different crater types, they argue, may be related to the substance in which the impactor hits, i.e., shallower ice at high latitudes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted that the morphology of impact craters depends strongly on the properties of the target (e.g., Martian surface) and the impactor, namely, its mass, velocity, and angle of incidence [e.g., &lt;em&gt;Barnouin et al.&lt;/em&gt;, 2012; &lt;em&gt;Horedt and Neukum&lt;/em&gt;, 1984; &lt;em&gt;Le Feuvre and Wieczorek&lt;/em&gt;, 2008; &lt;em&gt;Neukum and Ivanov&lt;/em&gt;, 1994; &lt;em&gt;Pike&lt;/em&gt;, 1988; &lt;em&gt;Schultz&lt;/em&gt;, 1988].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I still think that my original answer is partially correct in that most high-latitude impacts would be more of a glancing blow than impact and submersion into the target, which partly explains the larger diameter in some cases.  Further, as &lt;em&gt;Robbins and Hynek&lt;/em&gt;, [2012] show, the distribution of ~1-3 km diameter craters are clearly peaked within $\pm30^{\circ}$ of the Martian equator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below I have listed several relevant studies that I could find with a quick search.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;&lt;strong&gt;References&lt;/strong&gt;&lt;/h1&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Barlow, N.G., and C.B. Perez &quot;Martian impact crater ejecta morphologies as indicators of the distribution of subsurface volatiles,&quot; &lt;em&gt;J. Geophys. Res.&lt;/em&gt; &lt;strong&gt;108&lt;/strong&gt;(E8), pp. 5085, doi:10.1029/2002JE002036, 2003.&lt;/li&gt;&#xA;&lt;li&gt;Barnouin, O.S., &lt;em&gt;et al.&lt;/em&gt; &quot;The morphology of craters on Mercury: Results from MESSENGER flybys,&quot; &lt;em&gt;Icarus&lt;/em&gt; &lt;strong&gt;219&lt;/strong&gt;, pp. 414-427, 2012.&lt;/li&gt;&#xA;&lt;li&gt;Hartmann, W.K. &quot;Discovery of multi-ring basins - Gestalt perception in planetary science,&quot; In: Merill, R.B., Schultz, P.H. (Eds.), &lt;em&gt;Multi-ring Basins: Formation and Evolution&lt;/em&gt;, Pergamon Press, New York, NY, pp. 79-90, 1981.&lt;/li&gt;&#xA;&lt;li&gt;Horedt, G.P.,  and G. Neukum &quot;Planetocentric versus heliocentric impacts in the jovian and saturnian satellite system,&quot; &lt;em&gt;J. Geophys. Res.&lt;/em&gt; &lt;strong&gt;89&lt;/strong&gt;, pp. 10405-10410, 1984.&lt;/li&gt;&#xA;&lt;li&gt;Le Feuvre, M., and M.A. Wieczorek &quot;Nonuniform cratering of the terrestrial planets,&quot; &lt;em&gt;Icarus&lt;/em&gt; &lt;strong&gt;197&lt;/strong&gt;, pp. 291-306, 2008.&lt;/li&gt;&#xA;&lt;li&gt;Mouginis-Mark, P.J. &quot;Martian fluidized crater morphology: Varia- tions with crater size, latitude, altitude, and target material,&quot; &lt;em&gt;J. Geophys. Res.&lt;/em&gt; &lt;strong&gt;84&lt;/strong&gt;(B14), pp. 8011-8022, doi:10.1029/JB084iB14p08011, 1979.&lt;/li&gt;&#xA;&lt;li&gt;Neukum, G., and B.A. Ivanov, B.A. &quot;Crater size distributions and impact probabilities on Earth from Lunar, terrestrial-planet, and Asteroid cratering data,&quot; In: Gehrels, T., Matthews, M.S., Schumann, A.M. (Eds.), &lt;em&gt;Hazards Due to Comets and Asteroids&lt;/em&gt;, University of Arizona Press, Tuscon, AZ, pp. 359-416, 1994.&lt;/li&gt;&#xA;&lt;li&gt;Pike, R.J. &quot;Geomorphology of impact craters on Mercury,&quot; In: Vilas, F., Chapman, C.R., Matthews, M.S. (Eds.), &lt;em&gt;Mercury&lt;/em&gt;, University of Arizona Press, Tuscon, AZ, pp. 165-273, 1988.&lt;/li&gt;&#xA;&lt;li&gt;Robbins, S.J., and B.M. Hynek &quot;A new global database of Mars impact craters $\geq$1 km:  2. Global crater properties and regional variations of the simple-to-complex transition diameter,&quot; &lt;em&gt;J. Geophys. Res.&lt;/em&gt; &lt;strong&gt;117&lt;/strong&gt;(E6), E6001, doi:10.1029/2011JE003966, 2012.&lt;/li&gt;&#xA;&lt;li&gt;Schultz, P.H. &quot;Cratering on Mercury - A relook,&quot; In: Vilas, F., Chapman, C.R., Matthews, M.S. (Eds.), &lt;em&gt;Mercury&lt;/em&gt;, University of Arizona Press, Tuscon, AZ, pp. 274-335, 1988.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13663" LastEditorUserId="13663" LastEditDate="2016-08-03T18:32:24.890" LastActivityDate="2016-08-03T18:32:24.890" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16802" PostTypeId="2" ParentId="13712" CreationDate="2016-08-03T19:05:03.050" Score="1" Body="&lt;p&gt;The magnetic field of the sun is generated by the Magnetohydrodynamic dynamo powered by the rotation of the conductive plasma that makes the sun. The details are not well understood. Similarly the magnetic field of the Earth is generated by the movement of the outer core. We have quite good models of the Earth's field now (including how it can flip), but it is not completely understood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A non-rotating planet cannot sustain a substantial magnetic field.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-08-03T19:05:03.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16803" PostTypeId="2" ParentId="16707" CreationDate="2016-08-03T19:33:15.867" Score="6" Body="&lt;p&gt;There are two primary dust populations near 1 &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_unit&quot; rel=&quot;nofollow noreferrer&quot;&gt;AU&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Interplanetary_medium&quot; rel=&quot;nofollow noreferrer&quot;&gt;interplanetary&lt;/a&gt; dust (IPD) and &lt;a href=&quot;https://en.wikipedia.org/wiki/Interstellar_medium&quot; rel=&quot;nofollow noreferrer&quot;&gt;interstellar&lt;/a&gt; dust (ISD) [&lt;em&gt;Mann&lt;/em&gt;, 2010].  I also discussed dust observations in detail at &lt;a href=&quot;https://physics.stackexchange.com/a/160627/59023&quot;&gt;https://physics.stackexchange.com/a/160627/59023&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;&lt;strong&gt;Interplanetary Dust&lt;/strong&gt;&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;IPD of ~1 $\mu$m size drift sunward due to &lt;a href=&quot;https://en.wikipedia.org/wiki/Poynting%E2%80%93Robertson_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Poynting-Robertson drag&lt;/a&gt; while following roughly Keplerian orbits [e.g., &lt;em&gt;Malaspina et al.&lt;/em&gt;, 2014].  Closer to the sun, these particles break up due to collisions, &lt;a href=&quot;https://en.wikipedia.org/wiki/Sublimation_(phase_transition)&quot; rel=&quot;nofollow noreferrer&quot;&gt;sublimation&lt;/a&gt;/&lt;a href=&quot;https://en.wikipedia.org/wiki/Ablation&quot; rel=&quot;nofollow noreferrer&quot;&gt;ablation&lt;/a&gt;, and/or &lt;a href=&quot;https://en.wikipedia.org/wiki/Sputtering&quot; rel=&quot;nofollow noreferrer&quot;&gt;sputtering&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dust grains of ~0.1 $\mu$m size are the so-called &quot;$\beta$ meteorites&quot;, which travel away from the sun due to the imbalance of radiation pressure over gravity [&lt;em&gt;Mann&lt;/em&gt;, 2010].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smallest dust grains with $\ll$ 0.1 $\mu$m size, the so-called &lt;em&gt;nanograins&lt;/em&gt; or &lt;em&gt;nanodust&lt;/em&gt;, act like large &lt;a href=&quot;https://en.wikipedia.org/wiki/Pickup_Ion&quot; rel=&quot;nofollow noreferrer&quot;&gt;pickup ions&lt;/a&gt;, which are carried anti-sunward by the frame-dependent convective electric field (i.e., just the Lorentz force) produced when the dust grain moves relative to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_wind&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar wind&lt;/a&gt; flow (i.e., $\mathbf{E}_{conv} = - \mathbf{V}_{sw} \times \mathbf{B}_{sw}$, where the subscript $conv$($sw$) stands for convective(solar wind), and $\mathbf{V}$ and $\mathbf{B}$ are the bulk flow velocity and quasi-static magnetic field).  These particles can reach speeds in excess of 100 km/s relative to the sun [&lt;em&gt;Meyer-Vernet et al.&lt;/em&gt;, 2009].&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;&lt;strong&gt;Interstellar Dust&lt;/strong&gt;&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;ISD was first discovered by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ulysses_(spacecraft)&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Ulysses&lt;/em&gt;&lt;/a&gt; spacecraft, which is ~1 $\mu$m size and moves at ~26 km/s relative to the solar system barycenter.  More recent work [&lt;em&gt;Malaspina et al.&lt;/em&gt;, 2014] has found a relationship between dust impact count rates and ecliptic longitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason can be seen from the following.  The Earth's transverse speed about the sun is ~29 km/s.  Thus, when the Earth moves anti-parallel(parallel) to the ISD flow direction the relative dust-spacecraft speed is ~55(~3) km/s, which produced an enhanced(depressed) dust count rate.  This occurs because there is a threshold impact speed necessary to produce a large enough &lt;a href=&quot;https://en.wikipedia.org/wiki/Plasma_(physics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;plasma&lt;/a&gt; cloud (i.e., $\gtrsim$5-10 km/s depending on dust size) [&lt;em&gt;Meyer-Vernet et al.&lt;/em&gt;, 2009; 2014].&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the origin of the dust near the sun?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The primary sources of ~1 $\mu$m size near 1 AU are cometary debris trails, asteroids, planets, moons, and ISD [&lt;em&gt;Mann&lt;/em&gt;, 2010; &lt;em&gt;Zaslavsky&lt;/em&gt;, 2015].  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Simply put - wouldn't the gravity pull it in or the solar wind blow it away?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some are attracted by a combination of gravity and Poynting-Robertson drag while the smaller grains are either &quot;pushed&quot; out by radiation pressure (i.e., $\beta$ meteorites) or &quot;picked-up&quot; by the solar wind Lorentz force (i.e., nanodust).&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;&lt;strong&gt;References&lt;/strong&gt;&lt;/h1&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;D.M. Malaspina et al., &quot;Interplanetary and interstellar dust observed by the Wind/WAVES electric field instrument,&quot; &lt;em&gt;Geophys. Res. Lett.&lt;/em&gt; &lt;strong&gt;41&lt;/strong&gt;, pp. 266-272, doi:10.1002/2013GL058786, 2014.&lt;/li&gt;&#xA;&lt;li&gt;Mann, I. &quot;Interstellar Dust in the Solar System,&quot; &lt;em&gt;Annual Review of Astronomy and Astrophysics&lt;/em&gt; &lt;strong&gt;48&lt;/strong&gt;, pp. 173-203, doi:10.1146/annurev-astro-081309-130846, 2010.&lt;/li&gt;&#xA;&lt;li&gt;Meyer-Vernet, N., et al. &quot;Dust detection by the Wave instrument on STEREO: Nanoparticles picked up by the solar wind?,&quot; &lt;em&gt;Sol. Phys.&lt;/em&gt; &lt;strong&gt;256&lt;/strong&gt;, pp. 463-474, doi:10.1007/s11207-009-9349-2, 2009.&lt;/li&gt;&#xA;&lt;li&gt;Meyer-Vernet, N., et al. &quot;The importance of monopole antennas for dust observations: Why Wind/WAVES does not detect nanodust,&quot; &lt;em&gt;Geophys. Res. Lett.&lt;/em&gt; &lt;strong&gt;41&lt;/strong&gt;, pp. 2716-2720, doi:10.1002/2014GL059988, 2014.&lt;/li&gt;&#xA;&lt;li&gt;Zaslavsky, A. &quot;Floating potential perturbations due to micrometeoroid impacts: Theory and application to S/WAVES data,&quot; &lt;em&gt;J. Geophys. Res.&lt;/em&gt; &lt;strong&gt;120&lt;/strong&gt;, pp. 855-867, doi:10.1002/2014JA020635, 2015.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13663" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-08-04T00:55:37.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16804" PostTypeId="1" AcceptedAnswerId="16806" CreationDate="2016-08-03T22:25:01.847" Score="1" ViewCount="79" Body="&lt;p&gt;As the title states, will viewers in Sydney, Australia be able to see the Perseid meteor shower in August, 2016? Assume the viewer is well outside of the city and light pollution is a minimal problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.australiangeographic.com.au/news/2016/08/perseid-meteor-shower-to-put-on-special-show-up-north&quot; rel=&quot;nofollow&quot;&gt;Australian Geographic&lt;/a&gt;, the event will be visible in the northern parts of the country &amp;ndash; specifically, &quot;head north for the best chance to witness the event&quot; &amp;ndash; but it is not clear whether it will be possible to see anything in the southern parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If any viewing &lt;em&gt;is&lt;/em&gt; possible, what would be the recommended dates and time of night to see the most?&lt;/p&gt;&#xA;" OwnerDisplayName="user10636" LastEditorDisplayName="user10636" LastEditDate="2016-08-03T23:07:30.380" LastActivityDate="2016-08-04T00:01:17.590" Title="Will the 2016 Perseid meteor shower be visible in Sydney, Australia?" Tags="&lt;meteor-shower&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16806" PostTypeId="2" ParentId="16804" CreationDate="2016-08-04T00:01:17.590" Score="2" Body="&lt;p&gt;The &lt;a href=&quot;http://imo.net/calendar&quot; rel=&quot;nofollow&quot;&gt;International Meteor Organization calendar&lt;/a&gt; for 2016, page 12, gives the Perseid radiant declination as +58$^\circ$ and says &quot;regrettably, it cannot be properly viewed from most of the southern hemisphere.&quot;&#xA;At latitude -34$^\circ$ the radiant never rises above the horizon, so there may be fewer Perseid meteors than sporadic non-Perseids.&#xA;This does not change year to year as the AG article seems to suggest.&#xA;What does vary is interference from the Moon; this year it's full on the 18th, so the article is correct to recommend the morning or two before the predicted peak on August 12-13.&#xA;Also check your local weather forecast.&#xA;The predawn hours, e.g. 2 hours before sunrise, are generally best for meteors because the morning side is forward as the Earth orbits the Sun.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-08-04T00:01:17.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17807" PostTypeId="1" AcceptedAnswerId="17809" CreationDate="2016-08-04T06:10:01.763" Score="5" ViewCount="1355" Body="&lt;p&gt;How big are the tidal influences of planets (say, Jupiter) on the sun? Do they lead to any measurable effects, for example, do they influence the sunspots, or are they observable in any other way?&lt;/p&gt;&#xA;" OwnerUserId="13607" LastActivityDate="2016-08-04T18:39:01.820" Title="Tidal influences of planets on the sun" Tags="&lt;the-sun&gt;&lt;tidal-forces&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17809" PostTypeId="2" ParentId="17807" CreationDate="2016-08-04T09:37:54.313" Score="9" Body="&lt;p&gt;The basic tidal acceleration felt by a test mass near the surface of the Sun, due to a body of mass $m$ at a distance $r$ is given by&#xA;$$a_{\rm tidal} = 2\frac{Gm R}{r^3},$$&#xA;where $R$ is the radius of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can work out that Jupiter has the largest tidal effect on the Sun, &lt;em&gt;very closely followed&lt;/em&gt; by Venus, and both produce tidal forces that are only a factor of $\sim 2.2$ times greater than that produced by the Earth on the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distortion of the Sun could be modelled as following the consequent equipotentials (well inside the Roche lobe) in a binary system, but this is not analytically solvable and I could not locate any straightforward approximations. One way to judge the size of the effect would be to compare the tidal acceleration with the centrifugal acceleration at the Sun's surface $R\omega_{\rm Sun}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $a_{\rm tidal}$ for Jupiter is $3.7\times10^{-10}$ m/s$^2$, whereas the centrifugal acceleration (at the solar equator) is 7 orders of magnitude larger at 0.006 m/s$^2$. Given that the Sun is &lt;a href=&quot;https://physics.stackexchange.com/questions/208344/why-is-the-sun-almost-perfectly-spherical/208348#208348&quot;&gt;almost spherical&lt;/a&gt; to 1 part in 100,000, because the centrifugal acceleration is only 20 millionths of the Sun's surface gravity, we can safely assume that tides due to planets that are 7 orders of magnitude weaker do not distort the shape of the Sun to any degree worth worrying about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was surprised to find that the potential influence of planetary tides on solar activity is still an area of recent research. Suggestions that planetary tides might be responsible for sunspots or the sunspot cycle on the Sun go back a long way. &lt;a href=&quot;http://adsabs.harvard.edu/abs/1996Ap%26SS.243..393G&quot; rel=&quot;nofollow noreferrer&quot;&gt;Grandpierre (1996)&lt;/a&gt; suggested that planetary tides influence flows in the Sun and could be responsible for the 11-year solar cycle. A NASA report by &lt;a href=&quot;http://ntrs.nasa.gov/search.jsp?R=20070025111&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hung (2007)&lt;/a&gt;, although not peer-reviewed, does suggest a link between solar activity and planetary positions. The claim is made that major solar flares happen &quot;underneath&quot; the locations of the major tide-inducing planets (Jupiter, Venus, Earth, Mercury [in that order]). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I very quickly found a peer-reviewed paper in &quot;Solar Physics&quot; by &lt;a href=&quot;http://link.springer.com/article/10.1007/s11207-013-0288-6&quot; rel=&quot;nofollow noreferrer&quot;&gt;Seker (2013)&lt;/a&gt;, which discusses all this and reviews the evidence on sunspots. They conclude that the appearance of new sunspots is &lt;em&gt;not&lt;/em&gt; correlated at all with the positions of the planets. However, they do raise the caveat that because the magnetic fields producing the spots are produced at the tachocline between the radiative core and convective envelope of the Sun and take several days to buoyantly rise. Thus the longitude at which the spots appear and where they are generated, with respect to the positions of the planets, can change significantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the jury is still out, but this is quite &quot;fringe&quot; material. The solar cycle is certainly thought to be a product of the dynamo mechanism that produces the magnetic field. However, the location of sunspots definitely cannot be predicted with such models at the moment. Interest in this topic has been revived recently by claims of correlations between magnetic activity and the orbital positions of hot Jupiter-type exoplanets and even the suggestion that tidal effects or interactions between stellar and planetary magnetic fields might be responsible for increased magnetic activity (see &lt;a href=&quot;http://arxiv.org/abs/1408.6049&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lanza 2014 for a recent review&lt;/a&gt;). For these close-in exoplanets of course the tides are many orders of magnitude larger than in the solar system (see equation above).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-08-04T18:39:01.820" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17810" PostTypeId="2" ParentId="17807" CreationDate="2016-08-04T13:02:30.303" Score="1" Body="&lt;p&gt;The height of the tidal bulge raised on the Sun by the Earth is about $10^{-13}$ times the radius of the Sun, which is about a 20th of a millimetre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This presumes that the Sun is made of a frictionless inertialess fluid: it is a measure of the &lt;em&gt;driving force&lt;/em&gt; behind the tides).&lt;/p&gt;&#xA;" OwnerUserId="12016" LastActivityDate="2016-08-04T13:02:30.303" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17811" PostTypeId="1" AcceptedAnswerId="17819" CreationDate="2016-08-04T13:31:52.800" Score="3" ViewCount="145" Body="&lt;p&gt;In this nice NYTimes 'survey article' &lt;a href=&quot;http://www.nytimes.com/interactive/2016/07/05/science/nasa-space-missions.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Our Vast Solar System&#xA;and Its Many Explorers&quot;&lt;/a&gt; there is a &lt;a href=&quot;http://photojournal.jpl.nasa.gov/catalog/PIA19102&quot; rel=&quot;nofollow noreferrer&quot;&gt;NEOWISE image&lt;/a&gt; of &lt;a href=&quot;https://en.wikipedia.org/wiki/C/2014_Q2_(Lovejoy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Comet C/2014 Q2 (Lovejoy)&lt;/a&gt;. See &lt;a href=&quot;http://www.jpl.nasa.gov/spaceimages/details.php?id=PIA19102&quot; rel=&quot;nofollow noreferrer&quot;&gt;here also&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The NASA description of the image says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The red color is caused by the strong signal in the NEOWISE 4.6-micron wavelength detector, owing to a combination of gas and dust in the comet's coma.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I understand that this is a false color image, and the image at 4.6-micron is overlaid using red. &lt;strong&gt;So I should really ask:&lt;/strong&gt; why is &lt;a href=&quot;https://en.wikipedia.org/wiki/C/2014_Q2_(Lovejoy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Comet C/2014 Q2 (Lovejoy)&lt;/a&gt; (and probably many) comet actually bright at 4.6 microns &quot;...owing to a combination of gas and dust in the comet's coma.&quot; Is the gas fluorescing and the dust scattering? Or is this thermal radiation from the dust? The explanation in quotes doesn't really explain why &quot;&lt;em&gt;the red color is caused by the strong signal (around) 4.6-micron wavelength (is) owing to a combination of gas and dust in the comet's coma.&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iuZQp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iuZQp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; The more familiar green glow of this (and many) comet &lt;a href=&quot;https://en.wikipedia.org/wiki/C/2014_Q2_(Lovejoy)#/media/File:C2014_Q2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;from here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RMnzv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RMnzv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2016-08-04T16:03:42.420" LastActivityDate="2016-08-05T17:50:32.783" Title="Why is the &quot;green&quot; comet C/2014 Q2 (Lovejoy) red in this picture?" Tags="&lt;comets&gt;&lt;dust&gt;&lt;gas&gt;&lt;infrared&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17813" PostTypeId="2" ParentId="16457" CreationDate="2016-08-04T17:21:07.997" Score="2" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While you're on that planet, you are also in the same orbit around the star as the planet is. As long as you're in the same orbit as your vehicle (planet, spacecraft), you don't experience any acceleration relative to the vehicle. Whether you're on the planet, or you're just tagging along on the same orbit, is the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar question: in a satellite in a very eccentric orbit around the Earth, is any break in microgravity experienced in various parts of the orbit? Same answer - no. The satellite and everything in it follow the same orbit, so microgravity conditions are maintained.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now things would be different if you had a very eccentric orbit around a source of much more intense gravity, like a neutron star or a black hole. At the lowest point in the orbit, you could experience tidal forces: the things closer to center experience stronger gravity compared to things further away from center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for this to be experienced within a small spacecraft, or within the limits of one human body, the gradient of the field would have to be very strong indeed; the orbit would have to reach very low, very close to the neutron star or black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what you would experience would not be orbital &quot;acceleration&quot;, but just tidal force. Even so, in extreme cases the tidal force could be strong enough to destroy the ship or kill the passengers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the short story &quot;Neutron Star&quot; by Larry Niven for a literary depiction (fairly accurate physically) of this phenomenon. Unfortunately, the story is spoiled by the explanation above, but hopefully learning about gravity compensates for spoiling the story. :)&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2016-08-04T17:21:07.997" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17814" PostTypeId="1" CreationDate="2016-08-04T19:56:25.623" Score="7" ViewCount="1434" Body="&lt;p&gt;The other night, while playing Elite: Dangerous, I came across a rather strange celestial body - one I never imagined possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was a Brown Dwarf star with a &lt;em&gt;very large&lt;/em&gt; ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/67Pqd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/67Pqd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is something like this even remotely possible? Have such objects actually been seen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was also quite interesting that this star wasn't even the primary star in the system. It was orbiting a larger star (blue-white, I think), along with several planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mnzf8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mnzf8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditDate="2016-08-11T05:05:14.593" LastActivityDate="2020-02-11T16:15:12.443" Title="Do/Can Ringed Stars Exist?" Tags="&lt;star&gt;&lt;planetary-ring&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17815" PostTypeId="2" ParentId="17814" CreationDate="2016-08-04T20:41:09.920" Score="13" Body="&lt;p&gt;In a sense, that is what a &lt;a href=&quot;https://en.wikipedia.org/wiki/Circumstellar_disc&quot; rel=&quot;nofollow noreferrer&quot;&gt;circumstellar disk&lt;/a&gt; is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gTiCB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gTiCB.jpg&quot; alt=&quot;HL Tauri disk&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Source: &lt;a href=&quot;http://www.eso.org/public/archives/images/large/eso1436a.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;European Southern Observatory (ESO)&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are usually most noticeable around young stars as protoplanetary disks, disks that form planets. In the picture above, the gaps in the &quot;rings&quot; likely represent forming planets, similar to how the gaps in Saturn's rings indicate the presence of moons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our own star has a few &quot;rings&quot;, though likely not as visually impressive: the asteroid belt, the Kupier belt, the scattered disk, and the Hills cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, there are differences between these structures and the ring systems we are familiar with from planets. For example, rings are made of solid particles whereas protoplanetary disks contain a lot of gases. Rings may exist around &quot;mature&quot; planets, whereas protoplanetary disks represent an immature phase of planetary systems. Protoplanetary disks are not visible to the naked eye, whereas ring systems may be. In technical terms, a circumstellar disk is not a ring system. &lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorUserId="6" LastEditDate="2020-02-11T16:15:12.443" LastActivityDate="2020-02-11T16:15:12.443" CommentCount="19" ContentLicense="CC BY-SA 4.0" />
  <row Id="17816" PostTypeId="2" ParentId="16732" CreationDate="2016-08-04T21:03:42.153" Score="1" Body="&lt;p&gt;The following is a plausibility argument based on the symmetry of an assumed circular orbit of the earth around the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The earth's rotational axis is inclined to the ecliptic plane by an angle of about 23.4°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this reason idealized tidal bulges induced by the sun would travel around the earth's equator at the vernal and autumnal equinox whereas they would travel around circles of 23.4° northern &lt;em&gt;and&lt;/em&gt; southern longitude at summer and winter solstice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For symmetry reasons the tidal effects induced by the sun must have either a maximum or a minimum at the equinoxes. From this we conclude that also the combined tidal effects of sun and moon have either a maximum or a minimum at the equinoxes (on average).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the higher proportion of land to water at 23° northern latitude and a similar proportion at 23° southern latitude compared to the equator, it seems reasonable to assume a maximum at the equinoxes.&lt;/p&gt;&#xA;" OwnerUserId="10568" LastActivityDate="2016-08-04T21:03:42.153" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17817" PostTypeId="1" AcceptedAnswerId="17836" CreationDate="2016-08-04T21:19:18.000" Score="9" ViewCount="4486" Body="&lt;p&gt;If a galaxy is &lt;strong&gt;defined&lt;/strong&gt; as a collection of planetary systems (and all matter in between), and a planetary system is defined as a collection of planets circling a star (and all matter in between), and a planet (and other smaller bodies, like asteroids, moonlets, moons, etc) is basically just a very very, very large collection of minerals, stardust and gases (which make up the entire universe), &lt;strong&gt;which came first: The planets, planetary systems or galaxies?&lt;/strong&gt; &lt;em&gt;I've been given to understand the universe came first...but after that?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought I understood, at least the layman's basics, of the universe formation. However, today, I’m having trouble &lt;strong&gt;conceptualising the order&lt;/strong&gt; in which they would have been formed, &lt;strong&gt;IF&lt;/strong&gt; they are &lt;em&gt;defined&lt;/em&gt; by the smaller unit.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is this just me getting the semantics of classification and the actual chronology of each feature forming, mixed up?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;VERY Basically,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The universe is made up of a complex mix of everything (matter, antimatter, darkmatter, to name just a few), and expanding 'space'.&lt;/li&gt;&#xA;&lt;li&gt;A very, very, large area of 'space' is covered in very, very, tiny particles: made up of at first gases and later 'stardust'.&lt;/li&gt;&#xA;&lt;li&gt;A galaxy is formed of gravity-attracted gases, and after the very first stars went supernova and exploded there was also 'stardust', which combines to create 'clumps' of interstellar clouds (cloud nebulae).&lt;/li&gt;&#xA;&lt;li&gt;The cloud nebulae are the 'nursery' areas for star formation (both initial stars and later generation stars). &lt;/li&gt;&#xA;&lt;li&gt;The newly created star uses remnants of 'stardust' in the cloud nebulae to form planetary systems and then, a 'short' time later the orbiting planets (and other sized objects). &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So then the chronological order would be: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Universe (Largest scale) =&gt; gases (very small scale) =&gt; galaxy =&gt; interstellar clouds =&gt; &lt;strong&gt;stars =&gt; planets (combined to form planetary systems)&lt;/strong&gt;. And the tiny particles of gases and 'stardust', are involved in &lt;em&gt;nearly&lt;/em&gt; all scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qzJV2.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qzJV2.jpg&quot; alt=&quot;Universe components&quot;&gt;&lt;/a&gt;&#xA;Image taken from &lt;a href=&quot;http://www.spaceexplained.com/astronomy-explained/the-solar-system-versus-a-galaxy-versus-the-milky-way-versus-the-universe/&quot; rel=&quot;noreferrer&quot;&gt;spaceexplained&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;hr&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Just writing out this question has helped! I hope you don't mind me posting it anyway, as it might help others and &lt;strong&gt;I still need confirmation that I have&lt;/strong&gt;, hopefully, worked myself &lt;em&gt;out of&lt;/em&gt; my conundrum (not deeper).&lt;/p&gt;&#xA;" OwnerUserId="13680" LastEditorUserId="616" LastEditDate="2020-12-17T12:43:43.970" LastActivityDate="2020-12-17T12:43:43.970" Title="Which came first: Galaxies &lt;=&gt; Stars &lt;=&gt; Planets?" Tags="&lt;galaxy&gt;&lt;cosmology&gt;&lt;star-formation&gt;&lt;planetary-formation&gt;" AnswerCount="2" CommentCount="14" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17818" PostTypeId="2" ParentId="16795" CreationDate="2016-08-04T21:27:32.673" Score="6" Body="&lt;p&gt;As mentioned in the comments, a ring will (ideally) only occur if you have perfect alignment of source, lens, and observer. This is true for point sources, for extended sources things get more complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Double and quadruple images occur when the source is small and not in perfect alignment - typically a quasar, as in your example. For the sake of completeness, this would actually be triple or quintuple, but the third / fifth image is very close to the centre, overpowered by the lens, and highly demagnified (faint).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are examples of rings and almost-rings, but these tend to be extended sources, i.e. galaxies. It is more likely that some part of an extended source happens to be aligned with lens and observer than it is for a point source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YIGJP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YIGJP.jpg&quot; alt=&quot;Cosmic Horseshoe&quot;&gt;&lt;/a&gt;&lt;br /&gt;(Image of the &lt;a href=&quot;http://apod.nasa.gov/apod/ap111221.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cosmic Horseshoe&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example, if the host galaxy of that quasar could be seen, it might be an Einstein ring (or an oval, as the lens seems to be somewhat asymmetric).&lt;/p&gt;&#xA;" OwnerUserId="13681" LastEditorDisplayName="user1569" LastEditDate="2016-08-05T18:49:41.647" LastActivityDate="2016-08-05T18:49:41.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17819" PostTypeId="2" ParentId="17811" CreationDate="2016-08-04T21:56:39.250" Score="5" Body="&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1088/0004-637X/792/1/30&quot; rel=&quot;nofollow&quot;&gt;Mainzer et al. 2014&lt;/a&gt; characterize the performance of the reactivated NEOWISE.&#xA;Having run out of cryogenic coolant for the original WISE's 12 and 22 &amp;mu;m bands, it only detects in the 3.4 and 4.6 &amp;mu;m bands.&#xA;The comet looks red in the false-color infrared image because, relative to the 4.6 &amp;mu;m &quot;W2&quot; band, it emits less in the 3.4 &amp;mu;m &quot;W1&quot; band than stars do.&#xA;Figure 3 of the paper shows a similar image in which an asteroid also looks red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Section 4.1 discusses comets, attributing the 4.6 &amp;mu;m band detection to a combination of reflected sunlight and thermal radiation from dust, 4.26 &amp;mu;m emission from CO&lt;sub&gt;2&lt;/sub&gt;, and a little 4.67 &amp;mu;m emission from CO.&#xA;They estimate how much is due to gas vs. dust by comparison with ground-based observations near the same times, as in Figure 17.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2016-08-05T17:50:32.783" LastActivityDate="2016-08-05T17:50:32.783" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17820" PostTypeId="2" ParentId="16534" CreationDate="2016-08-04T22:14:18.817" Score="2" Body="&lt;p&gt;As some of the comments pointed out, there isn't a definite answer since this is still ongoing research. However, there are some things that might explain it:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Galaxy formation: Current theory says that dark matter is non-uniformly distributed in the early universe. Once radiation and baryonic matter decouple, matter flows towards the dark matter overdensities. We know that stars (Pop III) must form very early during this, because their light reionizes the universe. These early stars are thought to have a mass of up to 1000 solar masses. It's possible that some of the infalling gas collapses straight into a black hole.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Density profile: A galaxies' density rises sharply towards the centre. So, during the formation of the galaxy, the gas rushing inward might have formed a black hole without going through the process of star and supernova. Also, this means there's enough to feed the black hole once it's formed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Dynamic friction: Imagine a self-gravitating system of heavy and light objects (like a star cluster or, indeed, a galaxy). A heavy object passing through an ensemble of lighter ones will lose some of it's kinetic energy to them, which means it will &quot;sink&quot; towards the centre of the gravitational well. This is well observed in globular clusters. Problem: The timescale for this to happen is far too long to matter for large galaxies.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There seems to be a correlation between the mass of the central black hole and the mass of the elliptical galaxy or the bulge of the spiral galaxy. So, some process makes sure that the central black hole gets to grow along with its galaxy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Coming back to your question: There are two parts to it, one: why a black hole at all, two, why at the centre. The second part is simpler, it would get there eventually, and probably (let's say in a galaxy merger) quite quickly. For the first part, there is no definite answer but the points above are some indications as to why it's not too surprising that there should be one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a last point, black holes at the centres of other galaxies are not at all easy to find. Unless the black hole is active (quasar), there are at best very indirect detection methods (velocity dispersion in the central part etc.).&lt;/p&gt;&#xA;" OwnerUserId="13681" LastActivityDate="2016-08-04T22:14:18.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17821" PostTypeId="1" AcceptedAnswerId="17822" CreationDate="2016-08-04T22:23:36.607" Score="2" ViewCount="611" Body="&lt;p&gt;As in the title, If the universe is considered to be 'flat', what shape is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a disc, elliptical, square, rectangle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is this still unknown as we can only study a small segment of the entire universe, known as the observable universe?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This previous &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1299/what-is-the-physical-geometric-shape-of-the-universe&quot;&gt;related question&lt;/a&gt; did ask the same thing. However the OP was also asking about if the universe had a centre and if everything (galaxies, dust, stuff etc...) was evenly spread out only along the outer edges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;As far as I understand it, the Big Bang that started everything off isn't a typical 'explosion' with a central point. Shapley noted that it appeared as if galaxies are moving away from the Milky Way. We now realise that it is the space between galaxies that is expanding (and that we are not the centre of the universe). in 1998, Hubble then found that the far distant galaxies appear to be moving faster than closer galaxies. This means that the rate of expansion is accelerating. This acceleration rate is known as the Hubble constant.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I feel as if the answers focussed more on his sub questions. I found the comments to be more helpful than the answers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can I assume that the answer is still as summarized in the related questions comments?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;'We don't know?'&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astromax commented on the question,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Well, it is indeed flat, meaning Euclidean geometry can be used. Since light travels at a finite rate and the universe is not infinitely old, our 'bubble' of observable stuff within the universe would be in the shape of a sphere centered on us. However, the center of the sphere would change if you decided to move to a different part of the universe. Is the universe infinite? This is not known, since we cannot see passed our observational horizon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Accepted Answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be honest I didn't understand the first half of the accepted answer, and having attempted to read through the &lt;a href=&quot;http://arxiv.org/pdf/1303.5086v2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck paper&lt;/a&gt; that was linked, I can tell you that I don't know much about astronomy! I kept getting lost in all the new (to me) astonomical terms being thrown about. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read the second half of the accepted answer to be explaining that the universe has no centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gerald commented on the OP's question about the shape 'shortly' after the big bang,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Something else: Roughly resembling the 3-dimensional surface of 4-dimensional sphere, but not quite symmetrical. The precise shape is not exactly known, but probably not too much distorted, like a torus, a cube or a dodecahedron. This is still under investigation; more precise results are expected within a few years, when polarization of the CMB will be analysed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and went further to clarify the linked Planck paper&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Neither the circles-in-the-sky search nor the likelihood method ﬁnd evidence for a multiply-connected topology&quot; of the Planck paper, section 6.1 means it's not a torus-like or more complex object with holes.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Second Answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more popular second answer, definitely focussed more on answering the 'does the universe have a centre?' question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Helpful comments to my focus, Astromax commented &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, that the universe is &quot;infinite&quot; is not something anyone can prove. The expansion of space-time seems to occur everywhere, but that does not mean that the universe is infinite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stan Lou commented &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Being infinite doesn't imply having no boundary (not by itself, at least). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and ended with &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Yes, a flat 3-torus is a possible geometry, but the geometry does not have to be &quot;repetitive&quot; in the sense of wrapping around on itself. It doesn't have to be a torus. Beyond the horizon could be a Euclidean space instead of a flat torus. Or there could be pink unicorns. Or anything else. The point of the horizon is that we don't know what's beyond it. (But if one assumes that the universe is globally isotropic, then it can't be a torus.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the OP summarised it by saying &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So it seems like it boils down to this: It's probably flat, probably 3D. It might be edge-less, or not, we don't know. If yes, might be hypertorus and we could eventually do a round trip and came to starting point. If not, might be a ball, or anything else. We don't know, and can't imagine because of our limited scope. Is this at least remotely aligned with today's theories or are there still some gapping holes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The comments then end, as they moved the discussion to chat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can I assume that the answer is still as summarized. 'We don't know?'&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13680" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-08-06T17:05:54.630" Title="If the universe is 'flat', what shape is it?" Tags="&lt;universe&gt;&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2016-08-05T12:10:23.940" ContentLicense="CC BY-SA 3.0" />
  <row Id="17822" PostTypeId="2" ParentId="17821" CreationDate="2016-08-05T01:18:58.237" Score="6" Body="&lt;p&gt;Let me clear this up first. Shape of the Universe means the curvature of the Universe. The meaning of &quot;shape&quot; is not the same as in &quot;what is the shape of this object I'm holding&quot;. It doesn't mean that the Universe necessarily has a boundary or is finite in shape — it just describes its space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea that the Universe is flat means that the Pythagorean theorem holds up for spatial coordinates. If you've studied non-Euclidean geometry, you'll know that a triangle can have more or less than 180°, depending on the surface its drawn on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Universe is flat, we could draw a triangle between spatial coordinates, and the angles would add up to 180°. If the Universe is negatively curved, the angles would add up to less than 180°; if it were positively curved, the angles would be more than 180°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general relativity, mass, energy, momentum and a bunch of other properties of matter curve spacetime. Thus, most astronomers assume the curvature of space to be very close to zero, although they are not certain of its sign, meaning it may be slightly positive or negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we still don't know the exact curvature of the Universe. However, most astronomers postulate that the Universe is infinite in size. This means there is no boundary or &quot;triangular, rectangular, etc.&quot; shape. It cannot be viewed from the outside because the Universe is infinite — there's no shape for something that extends infinitely in all directions.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2016-08-05T01:18:58.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17826" PostTypeId="1" CreationDate="2016-08-05T16:11:57.717" Score="2" ViewCount="397" Body="&lt;p&gt;How did Edwin Hubble find distances using Cepheid variables as 'standard candles'?&lt;/p&gt;&#xA;" OwnerUserId="13561" LastEditorDisplayName="user1569" LastEditDate="2016-08-18T15:04:24.620" LastActivityDate="2016-08-18T15:04:24.620" Title="How does the cepheid variables help in finding distance" Tags="&lt;distances&gt;&lt;cepheids&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17827" PostTypeId="2" ParentId="17826" CreationDate="2016-08-05T17:04:10.633" Score="7" Body="&lt;p&gt;Why not read his &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1925Obs....48..139H&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot;&gt;original paper&lt;/a&gt;? It's fairly short by today's standards, despite being such a monumental turning point in our understanding of the size and nature of the Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short though, Hubble measured the distance to Cepheid's the same way anyone does. There is a distinct relation between the luminosity oscillation period of the star and the star's luminosity. If he can measure the star's oscillation period with good accuracy, he can derive the luminosity. If he can also then measure the brightness (i.e., how bright it appears to be to him) then he can calculate the distance (&lt;a href=&quot;https://www.e-education.psu.edu/astro801/content/l4_p4.html&quot;&gt;since luminosity, distance, and brightness are all related&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2016-08-05T19:44:31.150" LastActivityDate="2016-08-05T19:44:31.150" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17828" PostTypeId="2" ParentId="17817" CreationDate="2016-08-05T22:15:51.597" Score="-3" Body="&lt;p&gt;The short answer is we don't know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A longer pedantic answer would point out that use of a term without understanding what you mean by that term is never appropriate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term I feel is problematical is &quot;stardust&quot;. In our standard picture of big bang nucleosynthesis, we are optimistic (because of the good agreement between calculations and observations of cosmological abundances) that we understand that as the Universe cooled hydrogen, helium, and traces of other elements formed (lithium is the one most often mentioned). It is almost NEVER pointed out that if the Universe is composed of 27% dark matter and only 5% regular matter (mostly hydrogen) then these calculations are an abysmal failure: they don't predict the existence of dark matter! All this is ignored and we congratulate ourselves (or rather the popular media lionize science's &quot;accomplishments&quot;). You can't ignore most of the matter in the Universe and expect that your models will be accurate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether the first star preceded or followed the first galaxy is to some extent a matter of definition. To some extent. It isn't at all clear whether or not any black holes existed prior to galaxy/star formation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If super black holes existed, then anything within their gravitational range would comprise a galaxy (if you chose to use a reasonable definition of galaxy). What would exist then would be gas (and plasma), dark matter, and the beginning traces of what you'd probably say is stardust: the lithium, boron, etc. which would agglomerate as it collided with similar atoms. The density of this dust would be far too low, I think, to ever form planets without other objects gravity intervening. (Although I probably should never say never). OTOH, if the only black holes existing were sufficiently small, they'd probably evaporate with only transient effects on the particle dynamics. So, we can ignore stardust, most of that was formed IN the first (Generation III) stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: which came first; galaxies or stars? And the answer is: what exactly defines a galaxy? We believe there were density fluctuations which would lead to gravitational collapse of the gas surrounding them (and composing them). At what point does this motion qualify as being distinguishable from other motions (other volumes of collapse)? Perhaps you can guess that it will depend on both the dark matter present AND the velocity (temperature) of the gas. A careful analysis would probably involve escape velocities (and as you probably know, that depends on the dark matter distribution) and the point at which &quot;most&quot; gas was orbiting rather than just randomly colliding which would in turn rest on what &quot;most&quot; means (51%, 90%, 99%,...). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So define the term galaxy. If you define it in terms of stars, then the answer is easy: (some) stars came first. If you define it in terms of zones of gravitation, then it is possible that the answer is both. That is, in some cases hydrogen had enough time to collapse into a star before &quot;most&quot; gas was bound to a near by &quot;galaxy&quot; (probably proto-galaxy would be a better term here) and in other places the protogalaxy had defined itself before the first star lit up. We don't know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a mistake to assume that the Universe as we see it today is a reasonable approximation of the Universe as it would appear only 10 or 100 million years after the big bang. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our models aren't good enough yet (and probably won't be until we understand a lot more about what dark matter is and how it behaves) to be confident that protogalaxies existed before the first stars or that the first stars lit up before the zones I'm calling protogalaxies were separate systems.&lt;/p&gt;&#xA;" OwnerUserId="13686" LastEditorUserId="13680" LastEditDate="2016-08-07T01:24:00.503" LastActivityDate="2016-08-07T01:24:00.503" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17831" PostTypeId="1" CreationDate="2016-08-07T09:12:18.463" Score="1" ViewCount="369" Body="&lt;p&gt;We know that an object falling to the black hole, as observed from an observer far away, will freeze as it gets closer and closer to the event horizon. So my question is how can a black hole form in the first place? I mean we usually say thing a star collapses into a blackhole. But considering the effect of &quot;time-freezing&quot;, how could that be possible in the first place?&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2016-08-08T09:38:38.360" LastActivityDate="2016-08-08T09:38:38.360" Title="How can a black hole ever form?" Tags="&lt;black-hole&gt;&lt;time-dilation&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2016-08-08T13:21:17.270" ContentLicense="CC BY-SA 3.0" />
  <row Id="17832" PostTypeId="2" ParentId="17831" CreationDate="2016-08-07T11:44:22.650" Score="-3" Body="&lt;p&gt;After a supernovae explosion from stars whose mass is more than the Chandrasekhar limit, the remaining neutron star is so dense that it collapses into a one single singularity( A point where the density in infinite and is compressed into a point).the only way we can detect a black hole is by the entropy given by the black hole. so at the point of formation the black hole is not taking any matter into it.After the black hole is formed we will see any object going into it as frozen in spacetime&lt;/p&gt;&#xA;" OwnerUserId="13561" LastActivityDate="2016-08-07T11:44:22.650" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17833" PostTypeId="1" AcceptedAnswerId="17834" CreationDate="2016-08-07T11:45:14.087" Score="3" ViewCount="135" Body="&lt;p&gt;Why it is said that in standard Big Bang model space curvature grows with time when the Universe expands? My intuition is that it should be the opposite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw this in a book talking about the problem of standard Big Bang model and why we need inflation theory. It says that the Universe is observed to the nearly flat now. And because the space curvature grows with time, it should be even flatter in the past, which cannot be explained. But the inflation theory solves the problem because even if space is initially far from flat, inflation will smear out the curvature and make it flat.&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2016-08-07T12:37:54.143" LastActivityDate="2016-08-07T14:52:25.097" Title="Standard Big Bang model and space curvature" Tags="&lt;big-bang-theory&gt;&lt;cosmological-inflation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17834" PostTypeId="2" ParentId="17833" CreationDate="2016-08-07T14:41:37.790" Score="3" Body="&lt;p&gt;What exactly did the book say? It all depends on what you mean/define as &quot;curvature&quot;. What you describe appears to be a description of the behaviour of $\Omega$. Inflation does indeed drive $\Omega$ towards unity and simultaneously flattens space because the radius of curvature grows exponentially bigger. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\Omega &amp;lt; 1$ at some early epoch then it should decrease quickly with time such that $\Omega &amp;lt;&amp;lt; 1$ in the present day - this means that the universe has negative curvature, but does not mean it is becoming more curved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Friedmann equation, the curvature parameter $k$ is a constant $(1,0,-1)$&#xA;$$H^2 = \frac{8\pi G\rho}{3} - \frac{kc^2}{a^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the spatial curvature is $k/a^2$ and the radius of curvature is $a$ if $k=+1$. Thus as the universe expands and $a$ gets larger, any curvature becomes smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a little more detail - one can write the above equation in terms of the density parameter $\Omega$, the ratio of density to the critical density $3H^2/8\pi G$:&#xA;$$(\Omega ^{-1}-1)\rho a^{2}={\frac {-3kc^{2}}{8\pi G}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During inflation, the energy density $\rho c^2$ remains constant as $a$ grows exponentially. In order to keep the left hand side equal to the right hand side (which is just a collection of constants), then $\Omega$ must be driven very close to unity, while $k/a^2$ will tend towards zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After inflation then $\rho$ will vary with $a$ depending on whether the expansion is dominated by matter ($a^{-3}$) or radiation $(a^{-4})$. In both these cases $\rho a^2$ will decrease as the universe expands, such that if $k \ne 0$, then $(\Omega^{-1} -1)$ must increase, which means that $\Omega$ must either grow or shrink away from unity. But $k/a^2$ continues to get smaller as the universe expands. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-08-07T14:52:25.097" LastActivityDate="2016-08-07T14:52:25.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17835" PostTypeId="1" AcceptedAnswerId="17981" CreationDate="2016-08-07T18:39:44.663" Score="11" ViewCount="679" Body="&lt;p&gt;Apparently, there are clouds of &quot;dust&quot; between the stars. Would a starship have to fly around those clouds, trying to find &quot;tunnels&quot; between clouds, or are the interstellar clouds harmless for a spaceship?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm mainly thinking in terms of abrasion or (micro)collisions, not radiation, but would welcome information on the latter also.&lt;/p&gt;&#xA;" OwnerDisplayName="user2233" LastEditorUserId="2270" LastEditDate="2016-08-19T11:03:18.100" LastActivityDate="2017-03-02T20:27:14.567" Title="Would a spaceship have to evade interstellar clouds?" Tags="&lt;interstellar-medium&gt;&lt;interstellar-travel&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17836" PostTypeId="2" ParentId="17817" CreationDate="2016-08-07T20:00:37.620" Score="15" Body="&lt;p&gt;The structure we see in the Universe has formed from the gravitational collapse of the matter that was once an almost smooth density field of gas (&amp;quot;baryons&amp;quot;) and dark matter&lt;span class=&quot;math-container&quot;&gt;$^1$&lt;/span&gt;. The word &amp;quot;almost&amp;quot; is important here, for if it had been completely — or even non-completely but much more — smooth, then the collapse would not have had the time to happen before the expansion of space has diluted the matter enough to prevent any collapse, and we would never had come into existence.&lt;/p&gt;&#xA;&lt;p&gt;That is, the density field was slightly clumpy, and these clumps — or &lt;em&gt;overdensities&lt;/em&gt; — existed on all scales.&#xA;But calculating which clump sizes collapse &lt;em&gt;first&lt;/em&gt; — stellar-sized, galaxy-sized, cluster-sized, etc. — is far from trivial. Analytical attempts has to make several approximations, but can still make quite meaningful predictions which have subsequently been backed up and refined by numerical simulations, and by now, although many loose ends still exist, we have a rather good picture of structure formation:&lt;/p&gt;&#xA;&lt;h3&gt;Gravitational collapse&lt;/h3&gt;&#xA;&lt;p&gt;An overdensity is denoted by&#xA;&lt;span class=&quot;math-container&quot;&gt;$\delta\equiv\rho/\langle\rho\rangle - 1$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; is the density of the overdensity and &lt;span class=&quot;math-container&quot;&gt;$\langle\rho\rangle$&lt;/span&gt; is the average density.&#xA;The evolution of overdensities under the force of gravity can be calculated exactly for &lt;span class=&quot;math-container&quot;&gt;$\delta \ll 1$&lt;/span&gt; using linear perturbation theory, but when &lt;span class=&quot;math-container&quot;&gt;$\delta$&lt;/span&gt; becomes of order unity, the non-linear regime is entered, and severe approximations must be made, so one turns instead to numerical simulations. It turns out that if &lt;span class=&quot;math-container&quot;&gt;$\delta\gtrsim1.68$&lt;/span&gt; (i.e. if a region in space has a density which is 2.68 times the ambient density), it will collapse. The answer to your question is then given by what size of clumps first reach &lt;span class=&quot;math-container&quot;&gt;$\delta\gtrsim1.68$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h3&gt;First clumps&lt;/h3&gt;&#xA;&lt;p&gt;Primordial quantum fluctuations&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt; grew in size during the (admittedly still pretty blurry) inflation, a fraction of a second after Big Bang. In the young Universe dark energy was negligible, and the dynamics of the Universe was dominated by matter. Because dark matter comprises &lt;span class=&quot;math-container&quot;&gt;$\sim5/6$&lt;/span&gt; of the total amount, we can initially neglect the presence of gas, but when the density becomes very high, gas pressure builds up and counteracts the collapse.&lt;/p&gt;&#xA;&lt;p&gt;The overdensities amplified as matter started to collapse. It turns out that the density fluctuations are larger for smaller scales, so the smaller the clump, the sooner it will collapse. This results in the so-called &amp;quot;bottom-up&amp;quot; formation of structure, which is in contrast to what was originally thought; namely that galaxies formed &amp;quot;top-down&amp;quot; in a monolithic collapse (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1962ApJ...136..748E&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eggen, Lynden-Bell, &amp;amp; Sandage 1962&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;p&gt;However, this approach neglects both gas and the motion of dark matter particles (treating it as so-called &lt;em&gt;cold dark matter&lt;/em&gt;). Taking into account the effects of this puts a lower threshold for the masses of the structures of &lt;span class=&quot;math-container&quot;&gt;$\sim10^5$&lt;/span&gt;-&lt;span class=&quot;math-container&quot;&gt;$10^6M_\odot$&lt;/span&gt; (e.g. &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0604050&quot; rel=&quot;nofollow noreferrer&quot;&gt;Naoz et al. 2006&lt;/a&gt;; &lt;a href=&quot;http://arxiv.org/abs/0906.4372v1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Yoshida 2009&lt;/a&gt;). Hence the first structures that formed are believed to be minihalos roughly the mass of globular clusters.&lt;/p&gt;&#xA;&lt;h3&gt;Stars, galaxies, and clusters&lt;/h3&gt;&#xA;&lt;p&gt;Stars consist of collapsed gas and almost no dark matter, and the formation of a star thus needs the theory of hydrodynamics rather than just gravity. In order for matter to collapse to such dense structures as stars, it must get rid of much of its energy. This is not possible for the collisionless dark matter (at least in the normal sense, but &lt;a href=&quot;https://astronomy.stackexchange.com/a/14433/5264&quot;&gt;this post&lt;/a&gt;), but gas, which can collide and cool by radiating, is able to do so&lt;span class=&quot;math-container&quot;&gt;$^3$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Through radiative cooling, the minihalos thus fragmented further into first gas clouds and then the first stars when the Universe was a few 100 million years old. Subsequently conglomerations of stars merged into galaxies, which eventually formed sheets, filaments and clusters of galaxies.&lt;/p&gt;&#xA;&lt;h3&gt;Dust&lt;/h3&gt;&#xA;&lt;p&gt;The first stars were formed out of pure hydrogen and helium (and small amounts of lithium), and the material for making planets did not exist. But when these stars — which were very massive — exploded as supernovae and polluted the interstellar medium with metals&lt;span class=&quot;math-container&quot;&gt;$^4$&lt;/span&gt; and stardust&lt;span class=&quot;math-container&quot;&gt;$^5$&lt;/span&gt;, the formation of less massive stars became possible&lt;span class=&quot;math-container&quot;&gt;$^6$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h3&gt;Planets&lt;/h3&gt;&#xA;&lt;p&gt;Dust particles stick together to form pebbles, rocks, and planets. Planet formation is probably not possible for too massive stars&lt;span class=&quot;math-container&quot;&gt;$^7$&lt;/span&gt;, but with the formation of small stars this became possible.&lt;/p&gt;&#xA;&lt;h3&gt;Timeline&lt;/h3&gt;&#xA;&lt;p&gt;We can make the following timeline:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Minihalos&lt;/li&gt;&#xA;&lt;li&gt;Gas clouds&lt;/li&gt;&#xA;&lt;li&gt;Stars&lt;/li&gt;&#xA;&lt;li&gt;Galaxies&lt;/li&gt;&#xA;&lt;li&gt;Stardust&lt;/li&gt;&#xA;&lt;li&gt;Planets&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;However, note that the merging of minihalos to larger &amp;quot;galaxies&amp;quot; (without a significant amount of stars) also may happen at earlier epochs, and in particular that star formation (and dust and planets) is a continuous process which still takes place today (although the bulk of the star formation took place when the Universe approximately 3–6 billion years old (&lt;a href=&quot;http://arxiv.org/abs/astro-ph/9708220&quot; rel=&quot;nofollow noreferrer&quot;&gt;Madau et al. 1998&lt;/a&gt;)).&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^1$&lt;/span&gt;&lt;sub&gt;And radiation, which actually dominated the energy density until the Universe was &lt;span class=&quot;math-container&quot;&gt;$\sim50\,000$&lt;/span&gt; years old.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt;&lt;sub&gt;The term &lt;em&gt;primordial quantum fluctuations&lt;/em&gt; might be the coolest term in physics.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^3$&lt;/span&gt;&lt;sub&gt;There &lt;em&gt;are&lt;/em&gt; hypotheses of so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_star_(dark_matter)&quot; rel=&quot;nofollow noreferrer&quot;&gt;dark stars&lt;/a&gt;, but that's beyond the scope of this text (primarily because I don't understand them).&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^4$&lt;/span&gt;&lt;sub&gt;For an astronomer, the term &lt;em&gt;metal&lt;/em&gt; means any element heavier than helium. It's easier this way.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^5$&lt;/span&gt;&lt;sub&gt;Astronomers just call it &amp;quot;dust&amp;quot;, and use only the term &lt;em&gt;star&lt;/em&gt;dust when talking to non-astronomers, in order for them not to fall asleep.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^6$&lt;/span&gt;&lt;sub&gt;The reason is that with the many possible electronic transitions of metals, the gas has more ways to cool, and hence may collapse at earlier times, before the proto-star clump reaches high masses.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^7$&lt;/span&gt;&lt;sub&gt;Because the high radiation pressure of very massive stars blows away and/or destroys the dust (I think, but this is not my expertise).&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2020-12-17T08:45:07.100" LastActivityDate="2020-12-17T08:45:07.100" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="17837" PostTypeId="2" ParentId="17835" CreationDate="2016-08-08T00:03:27.027" Score="1" Body="&lt;p&gt;Related: &lt;a href=&quot;https://physics.stackexchange.com/questions/26326/how-dense-are-nebulae&quot;&gt;https://physics.stackexchange.com/questions/26326/how-dense-are-nebulae&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's compare nebulae to the air density where the ISS orbits, at 400 000 meters. According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Density_of_air#Altitude&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, air pressure at a given altitude is given by the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p = p_0 \left(1 - \frac{Lh}{T_0} \right)^\frac{gM}{RL}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or $101.325\left(1 - \frac{0.0065×400000}{288.15}\right)^{(9.80665×0.0289644)/(8.31447×0.0065)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google Calculator doesn't like this ^ but putting it in piece-by-piece gives -5737666.10745. Then, let's find the density with the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ρ = \frac{pM}{RT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or $\frac{-5737666.10745×0.0289644}{8.31447×2473.15}$ which is -8.08192432875. Unfortunately Wikipedia doesn't tell me what unit this number is in (just that it's a &quot;molar form&quot; and that it's density) so unfortunately I'm completely stuck here and I can't finish answering the question. Hopefully this partial answer helped someone make a full answer.&lt;/p&gt;&#xA;" OwnerUserId="13696" LastEditorUserId="9301" LastEditDate="2016-08-15T21:39:04.107" LastActivityDate="2016-08-15T21:39:04.107" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17838" PostTypeId="1" CreationDate="2016-08-08T01:44:00.547" Score="3" ViewCount="141" Body="&lt;p&gt;Is our universe noisy or silent? Can people hear them? When astronauts are out of the Earth and near an explosion, can they hear its noise. &lt;/p&gt;&#xA;" OwnerUserId="13697" LastEditorUserId="2531" LastEditDate="2016-08-08T11:33:11.813" LastActivityDate="2016-08-08T21:03:45.840" Title="Are there any sounds in space?" Tags="&lt;universe&gt;&lt;space&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17839" PostTypeId="1" AcceptedAnswerId="17848" CreationDate="2016-08-08T06:21:24.303" Score="3" ViewCount="312" Body="&lt;p&gt;As Uranus is less massive than Neptune, would this be evidence that it was formed further away from the sun than Neptune, and later it swaps orbit with Neptune to move closer to sun?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="2270" LastEditDate="2016-08-08T20:35:45.103" LastActivityDate="2016-08-08T20:35:45.103" Title="Does the lower mass of Uranus suggest it swapped position with Neptune?" Tags="&lt;neptune&gt;&lt;uranus&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17840" PostTypeId="2" ParentId="17838" CreationDate="2016-08-08T06:28:14.263" Score="1" Body="&lt;p&gt;As there is no air in space for sound to travel through, you can't hear anything outside of the small bubble you're in, e.g. space suit, space ship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is however 'noise' in the electromagnetic spectrum which can be picked up by electronic equipment, with this being an example &lt;a href=&quot;https://www.youtube.com/watch?v=suSNye-3tVo&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=suSNye-3tVo&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11525" LastActivityDate="2016-08-08T06:28:14.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17842" PostTypeId="2" ParentId="17838" CreationDate="2016-08-08T10:18:02.440" Score="4" Body="&lt;p&gt;There is sound, but you cannot hear it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) There is no sound in space:&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=ce7AMJdq0Gw&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; shows how sound gets fainter in an evacuated glass jar. The same happens with an explosion (or a shout) in space - there is no air to transmit the pressure waves, so no sound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) There is sound in space:&#xA;The interstellar and even intergalactic medium is not completely empty, even though it is much more empty than the best vacuum we can create on earth. Nonetheless, there can be pressure waves in those media.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Dvpai.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dvpai.jpg&quot; alt=&quot;The bullet cluster&quot;&gt;&lt;/a&gt; &lt;br /&gt;(The bullet cluster, taken from NASA's &lt;a href=&quot;http://apod.nasa.gov/apod/ap060824.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;APOD&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image shows the mass distribution of two colliding galaxy clusters in blue, and the X-ray gas in red. The galaxies in the clusters have passed through each other without much disturbance, but the intra-cluster gas has collided. The red cone to the right of the centre looks a lot like the sonic boom familiar from aircraft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more regular sound waves (repeating pattern as opposed to the one-off of an explosion), you might find those in the interplanetary gas / solar wind of oscillating stars (RR Lyrae or Delta Cephei type stars). I am not sure if anyone has ever measured this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, while human ears can pick up sound in the range of ca. $10^2$ to $10^4$ Hz, these sound waves would have a much much lower frequency. Also, the force exerted by these pressure waves on a human ear would be much too small to register. &lt;/p&gt;&#xA;" OwnerUserId="13681" LastEditorUserId="13681" LastEditDate="2016-08-08T21:03:45.840" LastActivityDate="2016-08-08T21:03:45.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17843" PostTypeId="2" ParentId="17839" CreationDate="2016-08-08T14:24:05.950" Score="1" Body="&lt;p&gt;No. The difference in mass is less than 20%, and for astronomers that counts as &quot;basically the same&quot; (for the masses, check &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_gravitationally_rounded_objects_of_the_Solar_System#Planets&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question implies that more massive planets should form closer to their star. However, there is as yet no definitive theory on the formation of planetary systems - that might come with the next generation of telescopes, (&lt;a href=&quot;http://www.eso.org/public/teles-instr/e-elt/&quot; rel=&quot;nofollow&quot;&gt;E-ELT&lt;/a&gt;, &lt;a href=&quot;http://www.tmt.org/&quot; rel=&quot;nofollow&quot;&gt;TMT&lt;/a&gt;, &lt;a href=&quot;http://www.gmto.org/&quot; rel=&quot;nofollow&quot;&gt;GMT&lt;/a&gt;, &lt;a href=&quot;http://jwst.nasa.gov/science.html&quot; rel=&quot;nofollow&quot;&gt;JWST&lt;/a&gt;) as they will allow us both to study more than one planetary system and better observe the early stages of planet formation.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastActivityDate="2016-08-08T14:24:05.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17844" PostTypeId="1" CreationDate="2016-08-08T14:36:36.017" Score="2" ViewCount="264" Body="&lt;p&gt;Some of brown dwarfs generate the enought amount of heat to warm their atmosphere up to 300 K. This is a condition to liquid water and, possibly, organic life forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is where are a some studies about life forms in BD atmosphere?&lt;/p&gt;&#xA;" OwnerUserId="13699" LastEditorUserId="2365" LastEditDate="2017-10-01T14:37:50.803" LastActivityDate="2017-10-01T14:37:50.803" Title="Life forms in brown dwarf atmosphere" Tags="&lt;life&gt;&lt;brown-dwarf&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17845" PostTypeId="1" CreationDate="2016-08-08T16:52:07.270" Score="3" ViewCount="215" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Related: &lt;a href=&quot;https://astronomy.stackexchange.com/q/17814/550&quot;&gt;Do/Can Ringed Stars Exist?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is there any particular difference, in behavior or properties, between a planetary ring system and a circumstellar disk? Or is the only real difference a matter of what's in the middle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question came up as a matter in the related discussion, particularly because the (game-simulated) object that inspired the question is a brown dwarf - something between a star and a gas giant. What I observed in-game appeared to be more similar to the rings around Saturn, than to the description and image provided of a circumstellar disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, are there properties of the rings/disks themselves which can differentiate one type from another? Or is the type of central body the only thing that defines them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the central body is the only factor, what should we call a ring around a brown dwarf, assuming one could exist in nature?&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-08-11T15:20:57.857" Title="Difference between &quot;planetary ring&quot; and &quot;circumstellar disk&quot;?" Tags="&lt;planetary-ring&gt;&lt;terminology&gt;&lt;disk&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17846" PostTypeId="1" CreationDate="2016-08-08T18:35:47.513" Score="4" ViewCount="262" Body="&lt;p&gt;I'm hoping someone can point me in the right direction for a beginner-friendly telescope for my family. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My husband has mentioned wanting a telescope off and on for a few years, but he's never delved into the idea enough to give specifics. We have an autistic 7-year-old who has developed a Special Interest in astronomy, and since it's our practice to delve into his SI's with him, it seems like a good time to take the plunge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pertinent information... &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;We live in a major metropolitan area -- in the suburbs about 30-45 minutes from downtown Chicago. We will be able to drive to a less light-polluted area from time to time, but realistically, we'll be seeing what we can see from the yard or second-story balcony most of the time. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;7yo will likely be most interested in objects seen with some degree of detail -- planets, stars, comets, our moon, other brighter objects. We adults would like to glimpse some of the dimmer objects as well, though we know they'll be faint. 7yo will likely be impressed by glimpsing those objects, as well, since he's a bit obsessed with the vastness of the universe at the moment.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We are true beginners. Think of people who have never touched a scope. But we're patient and smart, so we can learn. Which is kind of the point, I suppose.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;7yo is not always patient, so an easy, quick set-up would be desirable.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No children will be using the scope unsupervised, but we will be teaching 7yo how to use it while closely supervised. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If giving specific recommendations, we'd like to keep the cost under $500US.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13705" LastEditorUserId="2365" LastEditDate="2016-08-11T07:39:15.017" LastActivityDate="2018-02-04T21:01:11.417" Title="Suggestions for a beginner-friendly telescope" Tags="&lt;telescope&gt;" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17847" PostTypeId="2" ParentId="17846" CreationDate="2016-08-08T19:51:07.153" Score="4" Body="&lt;strong&gt;Cheap telescopes are just fine&lt;/strong&gt;&#xA;&lt;p&gt;I won't recommend a specific telescope, but I can tell you the following: I live in the middle of Copenhagen, and with my random 15 year old 4&amp;quot; telescope, which at that time cost \$300-$400, I can see the Moon's craters and mountain ridges, Jupiter's belts and moons, Saturn's ring, Venus' phases, and some nebulae. My telescope is manual, i.e. has no motor to track the sky, but I'd guess that today for the same price, you could get a telescope that does have a motor and a small computer that can find celestial objects, which makes things much easier.&lt;/p&gt;&#xA;&lt;strong&gt;Stay home&lt;/strong&gt;&#xA;&lt;p&gt;In my experience, going outside the city doesn't really add much to the experience (for amateur telescopes). To really see dark skies, you need to be 50-100 km away from the nearest large city, I'd say. Maybe that's not far from you, but when buying a telescope, you should also take into account that larger telescopes are more difficult to transport and set up, so if it's too large, you might end up never pulling yourself together to take it out.&lt;/p&gt;&#xA;&lt;strong&gt;Stay inside&lt;/strong&gt;&#xA;&lt;p&gt;If you take your telescope outside, it takes some time before it has reached the same temperature as the air, and during this time, there can be quite a lot of turbulence, which makes the image flicker. If I quickly want to show my 7 yo something, I simply use it from inside my living room, looking through the window. Two minutes, and he's impressed (until something else catches his attention).&lt;/p&gt;&#xA;&lt;strong&gt;What you see is not spectacular&lt;/strong&gt;&#xA;&lt;p&gt;Also, you should be prepared that there's no way that you can see anything like the colorful pictures that you can find on the internet. You will be able to see that Jupiter and Mars have slightly red colors, but apart from that, most things will look black/white. The reason is that the eye needs much more light than can be collected through a telescope in order to see colors. The largest telescope that I've looked through with the eye was a 31&amp;quot; telescope, and even that didn't reveal much colors.&lt;/p&gt;&#xA;&lt;p&gt;To see colors, you need to take a picture with a long exposure. Many telescopes have the possibility of attaching a camera to it, but for this you definitely need a telescope with a motor such that it moves along with the sky, or else everything will just leave long white trail across the image.&lt;/p&gt;&#xA;&lt;p&gt;You should also be prepared that planets and the like will look quite small. You could see something like this (very approximately to scale):&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iKbJ9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iKbJ9.png&quot; alt=&quot;img&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;strong&gt;Know what you see&lt;/strong&gt;&#xA;&lt;p&gt;The beauty of what you see in an amateur telescope, even if it's small and dim, lies in &lt;em&gt;knowing&lt;/em&gt; what it is that you see. Knowing that some nebula is 1000 lightyears away, and that its ever so slightly reddish color is hydrogen being illuminated by young, massive stars makes it much more fantastic to see.&lt;/p&gt;&#xA;&lt;p&gt;My telescope is of medium quality. What you get for an expensive one is probably more clear images, but the most important factor is the light-collecting area, i.e. a 5&amp;quot; will show you more than a 4&amp;quot;. I'm sure users like Florin Andrei can tell you more here.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-02-04T21:01:11.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17848" PostTypeId="2" ParentId="17839" CreationDate="2016-08-08T20:15:13.590" Score="3" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Nice_model&quot; rel=&quot;nofollow&quot;&gt;Nice model&lt;/a&gt; of solar system evolution proposes that both Uranus and Neptune migrated outward from their original orbits due to numerous encounters with planetesimals,&#xA;stimulated by a temporary 2:1 resonance between Jupiter and Saturn.&#xA;Simulations going back billions of years can put either Uranus or Neptune initially closer to the Sun.&#xA;&lt;a href=&quot;http://dx.doi.org/10.1086/522825&quot; rel=&quot;nofollow&quot;&gt;Desch 2007&lt;/a&gt;, assuming that the Nice model is correct, argues for a particular density profile of the protoplanetary disk and notes that a closer proto-Neptune fits that profile better.&#xA;Some solar system scientists have problems with the Desch model, but Uranus's smaller mass is consistent with it.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-08-08T20:15:13.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17850" PostTypeId="1" AcceptedAnswerId="17867" CreationDate="2016-08-08T21:20:53.150" Score="4" ViewCount="795" Body="&lt;p&gt;I understand that the meteors originated from the the comet Swift–Tuttle. Presumably they left the comet with some velocity relative to the solar system. As comets follow elliptical orbits presumably the meteors are generally also following an elliptical orbit. And as the earth follows a circular orbit, why do we keep meeting them every year despite being on different orbits? &lt;/p&gt;&#xA;" OwnerUserId="13708" LastActivityDate="2018-08-23T11:06:39.700" Title="Why do we keep orbiting through the Perseid meteors?" Tags="&lt;orbit&gt;&lt;meteor&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17853" PostTypeId="1" AcceptedAnswerId="17869" CreationDate="2016-08-09T02:34:18.430" Score="4" ViewCount="253" Body="&lt;p&gt;After reading the Astronomy Magazine article &lt;a href=&quot;http://www.astronomy.com/news/observing/2016/08/perseid-meteor-shower-set-for-its-best-show-in-nearly-20-years&quot; rel=&quot;noreferrer&quot;&gt;Perseid meteor shower set for its best show in nearly 20 years&lt;/a&gt; and NASA's &lt;a href=&quot;http://www.nasa.gov/topics/solarsystem/features/watchtheskies/perseid-meteor-shower-aug11-12.html&quot; rel=&quot;noreferrer&quot;&gt;Look Up! Perseid Meteor Shower Peaks Aug. 11-12&lt;/a&gt;, I understand that Jupiter's gravity is affecting this year's possible peak rate of Perseid's meteors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is - in simple terms - deflecting some of them out of their general orbit, a bit more towards earth, so that we intercept more than usual. But I am having difficulty visualizing this. Is this deflection in the plane of the ecliptic, or out of plane? At our closest approach, how far away are we from the orbit of this dust?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good diagram would really help!  I found this at &lt;a href=&quot;http://www.space.com/33671-perseids-get-a-boost-from-jupiter.html&quot; rel=&quot;noreferrer&quot;&gt;space.com&lt;/a&gt; credited to Jeremie Vaubaillon, but it's not helping me with the 'big picture'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PevJe.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PevJe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also found this &lt;a href=&quot;http://www.space.com/32868-perseid-meteor-shower-guide.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; and credited to NASA/JPL:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gYTiB.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gYTiB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2016-08-09T02:51:37.727" LastActivityDate="2016-08-10T04:30:55.550" Title="What (actually) is Jupiter doing to this year's Perseids meteor shower?" Tags="&lt;orbit&gt;&lt;comets&gt;&lt;meteor-shower&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17854" PostTypeId="1" CreationDate="2016-08-09T08:13:18.447" Score="2" ViewCount="365" Body="&lt;p&gt;...reach Earth surface and then are reflected passing the atmosphere again to reach outer space? And once in outer space can they travel indefinetely in it or the interplabetary plasma would stop them? This occurs only when when the plasma is excited? When it is not, these radio waves would be able to make it through?&lt;/p&gt;&#xA;" OwnerUserId="13715" LastEditorUserId="13663" LastEditDate="2016-08-10T13:03:29.093" LastActivityDate="2016-08-10T13:33:02.577" Title="Do some ELF (Extremely low frequency) radio waves pass the atmosphere" Tags="&lt;universe&gt;&lt;radio-astronomy&gt;&lt;plasma-physics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17855" PostTypeId="1" CreationDate="2016-08-09T09:22:42.933" Score="1" ViewCount="136" Body="&lt;p&gt;I'm a highschool student who is writing a research paper about plotting Period-Luminosity graph of the Cepheid variables by deriving distance through independent method such as trigonometric parallax and Baade-Wesselink method:&#xA;$$∆R = R(t) - R(t_0) = -p \int_{t_0}^t v(t)dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While trying to understand and use the Baade-Wesselink method, I faced some difficulties in terms of calculating linear radius variation. According to some papers, I need to integrate radial velocity with respect to time to get the linear radius variation from $t_0$ to $t$, where $t = t_0$ $+$ its period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got the linear radial velocity from the SIMBAD database. For example, for the V* AQ Pup (one of classical Cepheids), its radial velocity is 56.9km/s. Then how do I integrate this value from $t_0$ to $t$? Does this mean [56.9t]^t_t0 and after all $56.9t_1 - 56.9t_0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also, may I ask how can I get the value of $t_0$ and $t$? Does this mean &quot;proper motion&quot; in &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-basic?Ident=Hip++038965+&amp;amp;submit=SIMBAD+search&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;? &lt;/p&gt;&#xA;" OwnerUserId="13714" LastEditorDisplayName="user1569" LastEditDate="2016-08-18T16:47:51.810" LastActivityDate="2016-08-18T16:47:51.810" Title="Integrating radial velocity with respect to time" Tags="&lt;star&gt;&lt;distances&gt;&lt;velocity&gt;&lt;cepheids&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17856" PostTypeId="1" CreationDate="2016-08-09T09:52:32.043" Score="2" ViewCount="49" Body="&lt;p&gt;I do not have much knowledge about radio astronomy. I know that astronomers know an objects' shape by measuring the reflected photons that reach radiotelescopes, but can they measure also the absorbed ones for the same purpose?&lt;/p&gt;&#xA;" OwnerUserId="13716" LastEditorDisplayName="user15217" LastEditDate="2016-12-21T13:28:25.277" LastActivityDate="2016-12-21T13:28:25.277" Title="Do astronomers measure radio waves absorption to know the shape of a celestial body?" Tags="&lt;planet&gt;&lt;radio-astronomy&gt;&lt;photons&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2016-12-22T14:24:08.583" ContentLicense="CC BY-SA 3.0" />
  <row Id="17857" PostTypeId="1" CreationDate="2016-08-09T10:03:34.907" Score="1" ViewCount="55" Body="&lt;p&gt;I'm a highschool student and in the process of writing a research paper of deriving period-lumniosity relation of Cepheid variable independently of the pre-existing PL relation. Thus, I need to derive distance to each Cepheid variable independently by using, for example, trigonometric parallax and Baade-Wesselink method. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the process of using Baade-Wesselink method, I need to calculate the change in angular diameter, which is represented by ∆θ. The question is, how can I derive that value given fluxes from different bands (V,B, etc.. I'm using the SIMBAD database currently).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which definite equation do I need to use to derive it?&lt;/p&gt;&#xA;" OwnerUserId="13714" LastEditorUserId="10172" LastEditDate="2016-08-18T15:05:10.887" LastActivityDate="2016-08-18T15:05:10.887" Title="Change in angular diameter of Cepheid variables" Tags="&lt;luminosity&gt;&lt;angular-diameter&gt;&lt;cepheids&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17858" PostTypeId="1" CreationDate="2016-08-09T11:16:04.220" Score="5" ViewCount="418" Body="&lt;p&gt;Mercury is &lt;strong&gt;a&lt;/strong&gt;synchronously tidally locked so that all of its surface regularly sees the Sun. But an exoplanet which is synchronously locked to its star, could we tell that it is? It would have a hot side and a cold side. This should affect transit and occultation on some wavelengths, but does the symmetry of the movement degenerate any possibility to disentangle tidal locking from that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury seems to cool down during night time, and gas giants maybe distribute their temperature atmospherically. So their tidal locking might be hard to discern. But there are terrestrial exoplanets with much shorter orbital period which should have a cool side and a hot side permanently. That's my speculation. My basic curiosity is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is Mercury a freak or a standard in terms of tidally locked planets being asynchronously so?&lt;/strong&gt; And by observational support, not just models.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2016-08-09T11:21:32.340" LastActivityDate="2017-10-14T20:14:17.197" Title="(How) Can one determine if an exoplanet is synchronously tidally locked or not?" Tags="&lt;exoplanet&gt;&lt;observational-astronomy&gt;&lt;planetary-transits&gt;&lt;tidal-locking&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ClosedDate="2017-10-17T16:33:28.270" ContentLicense="CC BY-SA 3.0" />
  <row Id="17859" PostTypeId="1" AcceptedAnswerId="17860" CreationDate="2016-08-09T11:41:24.960" Score="4" ViewCount="5149" Body="&lt;p&gt;Assuming a 747 could fly around Jupiter at its top Earth speed (I know this assumption is unrealistic), how long would it take to fly around Jupiter once? &lt;/p&gt;&#xA;" OwnerUserId="11792" LastEditorUserId="11792" LastEditDate="2016-08-10T12:11:27.250" LastActivityDate="2019-03-14T22:08:30.670" Title="How long would it take a Boeing 747 to go around Jupiter?" Tags="&lt;planet&gt;&lt;jupiter&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17860" PostTypeId="2" ParentId="17859" CreationDate="2016-08-09T12:14:04.680" Score="9" Body="&lt;p&gt;It will depend on the speed of the jet, however:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter has an equatorial circumference of 449,200 km, &lt;a href=&quot;https://en.wikipedia.org/wiki/Boeing_747#Specifications&quot; rel=&quot;noreferrer&quot;&gt;a current-generation 747 has a top speed of 988 km/h&lt;/a&gt; so it will take about 455 hours or nearly 19 days (18.9) (this doesn't allow for re-fueling).&lt;/p&gt;&#xA;" OwnerUserId="11658" LastEditorUserId="2934" LastEditDate="2016-08-09T19:16:55.237" LastActivityDate="2016-08-09T19:16:55.237" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17861" PostTypeId="2" ParentId="17859" CreationDate="2016-08-09T13:01:00.090" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://what-if.xkcd.com/30/&quot; rel=&quot;nofollow&quot;&gt;Someone else&lt;/a&gt; answered a somewhat related question (&lt;strong&gt;can&lt;/strong&gt; you fly through Jupiters atmosphere?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the website:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Jupiter: Our Cessna can’t fly on Jupiter; the gravity is just too strong. The power needed to maintain level flight is three times greater than that on Earth. Starting from a friendly sea-level pressure, we’d accelerate through the tumbling winds into a 275 m/s (600 mph) downward glide deeper and deeper through the layers of ammonia ice and water ice until we and the aircraft were crushed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not sure if it would work with a stronger airplane. So, the answer is probably &quot;you can't do that&quot;.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastActivityDate="2016-08-09T13:01:00.090" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17862" PostTypeId="1" CreationDate="2016-08-09T14:33:32.410" Score="2" ViewCount="35" Body="&lt;p&gt;When does oscillation of interplanetary plasma occur? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When and what are the reasons of the excitation of this plasma that makes it oscillate? Does it oscillate when is not excited? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can extremely low radio frequencies pass through it when it is not oscillating?&lt;/p&gt;&#xA;" OwnerUserId="13718" LastEditorUserId="2365" LastEditDate="2016-08-09T22:38:35.137" LastActivityDate="2016-08-09T22:38:35.137" Title="When does oscillation of interplanetary plasma occurs?" Tags="&lt;planet&gt;&lt;photons&gt;&lt;radio&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17863" PostTypeId="1" AcceptedAnswerId="17865" CreationDate="2016-08-09T16:08:49.273" Score="2" ViewCount="244" Body="&lt;p&gt;I want to convert the list of Julian date values of observing brightness of a Cepheid variable into its phase.&lt;br /&gt;&#xA;The equation I found in the Internet is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{Phase}= \frac{\mathrm{HJD}-T_0}{P} - \mathrm{INT}\left[\frac{\mathrm{HJD}-T_0}{P}\right]$$&#xA;What is the $\mathrm{INT}$ in this equation? I only know that $T_0$ is the one reference point among the list of Julian date values and $P$ is the period. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what kind of criteria exist in determining the reference point?&lt;/p&gt;&#xA;" OwnerUserId="13714" LastEditorDisplayName="user1569" LastEditDate="2016-08-18T15:04:33.793" LastActivityDate="2017-09-06T06:46:29.677" Title="Deriving phase of a pulsating star from list of Julian date values of observing brightness" Tags="&lt;star&gt;&lt;time&gt;&lt;luminosity&gt;&lt;cepheids&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17865" PostTypeId="2" ParentId="17863" CreationDate="2016-08-09T19:59:03.890" Score="3" Body="&lt;p&gt;INT refers to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Floor_and_ceiling_functions&quot; rel=&quot;nofollow&quot;&gt;floor function&lt;/a&gt;. So $\mathrm{INT}(x)$ is the largest integer not exceeding $x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the function is used to get the fractional part of $\frac{\mathrm{HJD}-T_0}{P}$, i.e. the part after the decimal point. So if for example $\frac{\mathrm{HJD}-T_0}{P}=3.4$, then $\frac{\mathrm{HJD}-T_0}{P}-\mathrm{INT}\left(\frac{\mathrm{HJD}-T_0}{P}\right)=0.4$.&lt;/p&gt;&#xA;" OwnerUserId="9533" LastActivityDate="2016-08-09T19:59:03.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17866" PostTypeId="1" AcceptedAnswerId="17873" CreationDate="2016-08-09T22:53:50.543" Score="4" ViewCount="291" Body="&lt;p&gt;I've heard gas giants are supported because there is an equilibrium between thermal pressure and gravity. That is, if Jupiter were to suddenly begin compressing, temperatures would increase to the point that it would expand to its original size. Since it is in equilibrium, both the compression and thermal pressure counter one another, so its size never fluctuates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, Jupiter's average temperature is extremely low, so much so that its gases are electron degenerate. Is this a contradiction, or has the information above been just hearsay?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-08-10T01:15:57.837" LastActivityDate="2016-08-10T14:28:11.507" Title="Are gas giants supported by thermal pressure?" Tags="&lt;gravity&gt;&lt;temperature&gt;&lt;gas-giants&gt;&lt;gas&gt;&lt;degenerate-matter&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17867" PostTypeId="2" ParentId="17850" CreationDate="2016-08-10T02:40:33.173" Score="3" Body="&lt;p&gt;This is probably best understood by picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comet Swift-Tuttle is both on a highly elliptical orbit, and a highly inclined orbit, though the inclination of the orbit doesn't matter to your specific question, but it helps with the overall picture.   What really matters for Earth getting a meteor shower is that the two orbital paths intersect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hzEpL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hzEpL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes the &lt;strong&gt;periodic&lt;/strong&gt; meteor shower is the &lt;a href=&quot;https://www.astro.umd.edu/~msk/science/trails/&quot; rel=&quot;nofollow noreferrer&quot;&gt;dust trail&lt;/a&gt;, that follows (and precedes) the comet (not to be confused with dust tail - that's something else).   Any solid debris that gets blown off the comet at low velocity enters an orbit with a very similar orbital path to the comet and over hundreds or thousands of orbits, as the comet effectively gets broken up each time it passes the sun, the comet generates a dust or debris trail that largely follows it's entire orbit, at least, that's where the greatest concentration of particles is.   Not all meteor showers are periodic, but all periodic meteor showers happen in this way with a semi permanent orbiting ring of dust that the Earth flies through once a year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dust trail (generally not visible, too spread out, unless heated by a close pass to the sun)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nFqVN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nFqVN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.uni.edu/morgans/astro/course/Notes/section4/tails.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dust tail&lt;/a&gt; (can be visible), only exists as the comet passes close to the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_periodic_comets&quot; rel=&quot;nofollow noreferrer&quot;&gt;short period comet&lt;/a&gt;, unless it's very young, has a debris or dust trail that roughly spans it's entire orbit though the trail is most dense near the comet.  In fact, the density of the dust trail can tell us how long the comet has been on it's current orbit.   Most short period comets were deflected into their current orbits within the last 100,000 years or so.  Short period comets have relatively short life-spans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long period comets or &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_near-parabolic_comets&quot; rel=&quot;nofollow noreferrer&quot;&gt;near-parabolic comets&lt;/a&gt; are different because their orbits are much longer, though they can still generate trails during a pass close to the sun, they tend to be much more spread out given the enormous size of their orbits.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the Earth's orbit intersects Swift-Tuttle's debris trail once a year and the debris trail stays mostly in the same place, we get the Perseid meteor shower about the same time every year, and the meteors always come from the same direction in the sky.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not every short period comet has an orbit that intersects with the Earth's orbit, which is why not all comets are associated with meteor showers, in fact, the majority aren't.   3 dimensions are hard to draw in 2-D pictures, but Pluto and Neptune can be used as an example.  It's often said that they cross orbits, but they actually don't.  Pluto gets closer to the sun than Neptune, but only when Pluto is near it's &lt;a href=&quot;https://en.wikipedia.org/wiki/Perihelion&quot; rel=&quot;nofollow noreferrer&quot;&gt;perihelion&lt;/a&gt;, at which point it's well below (or well above, depending on orientation), Neptune's orbital plane.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting sidebar to this is that comets that give us meteor showers are also comets that are potential earth impacters.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-08-10T03:09:55.013" LastActivityDate="2016-08-10T03:09:55.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17869" PostTypeId="2" ParentId="17853" CreationDate="2016-08-10T04:30:55.550" Score="6" Body="&lt;p&gt;To put it in simple terms, a comet's debris trail or dust trail is like a ring of debris that generally follows the entire orbit of the comet, more concentrated close to the comet, more spread out further away, but it can be thought of, for simplicity, as an entire ring of debris, kind of, a miniature asteroid belt with a fixed elliptical path.   The picture below isn't perfect, cause it gives the impression that orbits are all the same when there's some spreading out, but it's the gist of it.  (Note, the debris field and comet in the picture below isn't Swift-Tuttle/Perseid)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/N3kR1.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/N3kR1.png&quot; alt=&quot;https://www.astro.umd.edu/~msk/science/trails/encke-trail.png&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this is that any debris that's knocked off the comet at a low velocity has pretty much the same orbit as the original comet, and over hundreds or thousands of orbits, it tends to fill up the entire comet's orbit with mostly tiny particles, some larger ones.  The debris field or dust trail from the comet generally follows in the comets orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This picture is a bit more accurate as to what the Earth is passing through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://lintvwpri.files.wordpress.com/2016/08/perseid_base4.png?w=650&amp;amp;h=366&quot; rel=&quot;noreferrer&quot;&gt;https://lintvwpri.files.wordpress.com/2016/08/perseid_base4.png?w=650&amp;amp;h=366&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more compact the dust trail or debris field, the more meteors, but the smaller the target the Earth has to hit, so odds are greater that the Earth will miss the field entirely.  The more spread out, the greater chance the Earth hits the debris field, but more spread out means we get fewer meteors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth's orbit crosses the Perseid (eliptical ring of debris) or Swift-Tuttle's debris field about the same time every year because Earth's and Swift-Tuttle's orbits nearly intersect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qMOeI.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qMOeI.png&quot; alt=&quot;http://www.amsmeteors.org/wp-content/uploads/2015/08/comet-path.png&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This year, because of the placement of Jupiter, the debris field (which we can kind of think of as a ring), got bent by Jupiter, and Earth is hitting closer to the center of the debris field which means, more meteors.   I can't say on which plane the field got bent, cause it depends on it's orbital relation to Jupiter and I couldn't find that, but it was the ring of debris that got bent, so Earth will hit it more directly this year than usual.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the ring of debris is pretty much permanent, Jupiter bends the ring every 11 years as it passes closest to it, and occasionally that bend works in earths' favor and gives us more meteors, so its a rare event that might not happen again for 11 years, it might be over 100 years before Jupiter does this for us again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.theverge.com/2016/8/9/12411218/meteor-shower-jupiter-august-11-thursday-night?yptr=yahoo&quot; rel=&quot;noreferrer&quot;&gt;article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Typically Earth just grazes by Swift-Tuttle’s debris field, but our&#xA;  planet will be even closer to the particle stream this year thanks to&#xA;  some help from our neighbor Jupiter. The gas giant occasionally gets&#xA;  close to the comet stream, and its immense gravity pushes the debris&#xA;  nearer to Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-08-10T04:30:55.550" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17870" PostTypeId="2" ParentId="16766" CreationDate="2016-08-10T13:27:59.363" Score="2" Body="&lt;p&gt;First off, I'm not going to calculate an answer to the &quot;significance level of 1%&quot; part. This is more a rule of thumb kind of answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're moving in a spaceship and want to see&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a statistically significant difference between the mean redshifts of galaxies ahead and those behind&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, the red- and blueshift due to our motion must exceed the peculiar motion of the individual galaxies. Cosmologists have been dealing with a very similar problem trying to measure the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble%27s_law&quot; rel=&quot;nofollow&quot;&gt;Hubble constant&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember from a cosmology class (can't find a source at the moment) that the &lt;a href=&quot;https://en.wikipedia.org/wiki/Coma_Cluster&quot; rel=&quot;nofollow&quot;&gt;Coma cluster&lt;/a&gt; is the first cluster whose redshift is dominated by the Hubble flow as opposed to peculiar motion. This is mainly due to the fact that the local group is moving towards the nearer Virgo cluster. From the above wikipedia link: Distance ca. $100 \textrm{MPc}$, apparent recession velocity ca. $7000 \textrm{km s}^{-1}$. So, at that speed, you would definitely see a systematic difference in redshift of galaxies ahead of and behind you. How long to get to that speed at constant acceleration of $10 \frac{m}{s^2}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; a = \frac{v}{t} \Rightarrow t = \frac{v}{a} = \frac{7\cdot 10^6 m s^{-1}}{10 m s^{-2}} = 7 \cdot 10^5 s \simeq 8\, \textrm{days.}&#xA;\end{equation}&#xA;Note that this is a worst-case scenario. People have used much closer galaxies to determine the Hubble constant. &lt;a href=&quot;http://esoads.eso.org/abs/2008A%26ARv..15..289T&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt;'s abstract says &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Due to the linearity of the expansion field the Hubble constant $\textrm{H}_0$ can be found at any distance $&amp;gt;4.5$ Mpc. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is all very local, so we'll just assume a Hubble constant of $\textrm{H}_0 = 70\textrm{km}\,\textrm{s}^{-1}\,\textrm{MPc}^{-1}$ (&lt;a href=&quot;http://esoads.eso.org/abs/2015arXiv150201589P&quot; rel=&quot;nofollow&quot;&gt;Planck&lt;/a&gt; says 67.8 and &lt;a href=&quot;http://esoads.eso.org/abs/2011ApJ...730..119R&quot; rel=&quot;nofollow&quot;&gt;Riess et al.&lt;/a&gt; say 73.8) and calculate redshift and hence apparent recession velocity at that distance using the Hubble law&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; v = \textrm{H}_0 \cdot d = 70\,\textrm{km}\,\textrm{s}^{-1}\,\textrm{MPc}^{-1} \cdot 4.5\, \textrm{MPc} = 315\,\textrm{km}\,\textrm{s}^{-1}\,.&#xA;\end{equation}&#xA;At constant acceleration of 1g you'd get to that speed in about 9 hours, but you might need more time to do your data analysis on the galaxy spectra you'll have to take.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastActivityDate="2016-08-10T13:27:59.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17871" PostTypeId="2" ParentId="17854" CreationDate="2016-08-10T13:33:02.577" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Do some ELF (Extremely low frequency) radio waves pass the atmosphere reach Earth surface and then are reflected passing the atmosphere again to reach outer space?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In space &lt;a href=&quot;https://en.wikipedia.org/wiki/Plasma_(physics)&quot; rel=&quot;nofollow&quot;&gt;plasmas&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Extremely_low_frequency&quot; rel=&quot;nofollow&quot;&gt;ELF&lt;/a&gt; is used to describe waves in the frequency band ~3 Hz to ~3 kHz.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Very_low_frequency&quot; rel=&quot;nofollow&quot;&gt;Very low frequency (VLF)&lt;/a&gt; is used to describe waves in the frequency band ~3 kHz to ~30 kHz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cutoff_frequency&quot; rel=&quot;nofollow&quot;&gt;cutoff frequency&lt;/a&gt; of the Earth's &lt;a href=&quot;https://en.wikipedia.org/wiki/Ionosphere&quot; rel=&quot;nofollow&quot;&gt;ionosphere&lt;/a&gt; prevents waves at such low frequencies from freely passing through it.  In plasmas, this tends to be any wave at or below the local electron &lt;a href=&quot;https://en.wikipedia.org/wiki/Plasma_oscillation&quot; rel=&quot;nofollow&quot;&gt;plasma frequency&lt;/a&gt;, which can be approximated as $f_{pe} \approx 8.9787 \ \sqrt{n_{e} \left[ cm^{-3} \right]}$ kHz (where the $\left[ cm^{-3} \right]$ implies that the electron number density, $n_{e}$, should be in units of $cm^{-3}$).  The densities in the ionosphere are ~$10^{3} - 10^{6} \ cm^{-3}$, which corresponds to $f_{pe}$ ~ 280 kHz to ~9 MHz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Early &lt;a href=&quot;http://adsabs.harvard.edu/abs/1953RSPTA.246..113S&quot; rel=&quot;nofollow&quot;&gt;studies&lt;/a&gt; found a relationship between &lt;a href=&quot;https://en.wikipedia.org/wiki/Lightning&quot; rel=&quot;nofollow&quot;&gt;lightning&lt;/a&gt; and a VLF emissions called &lt;a href=&quot;https://en.wikipedia.org/wiki/Whistler_(radio)&quot; rel=&quot;nofollow&quot;&gt;whistlers&lt;/a&gt;.  It is now known that high energy electron beams, associated with the lightning, escape into the &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetosphere&quot; rel=&quot;nofollow&quot;&gt;magnetosphere&lt;/a&gt;.  These beams radiate the whistlers through a plasma &lt;a href=&quot;https://en.wikipedia.org/wiki/Instability&quot; rel=&quot;nofollow&quot;&gt;instability&lt;/a&gt; (i.e., free energy due to a fast beam streaming through an otherwise homogeneous plasma).  So the free energy sources can escape the lower atmosphere, but the electromagnetic emissions at such low frequencies cannot.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And once in outer space can they travel indefinetely in it or the interplabetary plasma would stop them?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Generally, waves with $f &amp;lt; f_{pe}$ experience &lt;a href=&quot;https://en.wikipedia.org/wiki/Damping&quot; rel=&quot;nofollow&quot;&gt;damping&lt;/a&gt; as they propagate through a plasma.  This is because the wave fields oscillate slow enough to interact with the particles, thus adding an effective inertia term to the system.  ELF and VLF waves generally damp through &lt;a href=&quot;https://en.wikipedia.org/wiki/Electron_cyclotron_resonance&quot; rel=&quot;nofollow&quot;&gt;cyclotron&lt;/a&gt; and Landau resonances.  Note, however, that both of these resonances are also the processes responsible for wave growth in the presence of sufficient free energy (e.g., free energy from beams is often related to a Landau-like or &lt;a href=&quot;https://en.wikipedia.org/wiki/Cherenkov_radiation&quot; rel=&quot;nofollow&quot;&gt;Cerenkov&lt;/a&gt;-like emission).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This occurs only when when the plasma is excited?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not necessarily.  We can artificially induce VLF emissions using &lt;a href=&quot;https://en.wikipedia.org/wiki/Lualualei_VLF_transmitter&quot; rel=&quot;nofollow&quot;&gt;ground-based transmitters&lt;/a&gt;.  But otherwise yes, there needs to be free energy for the radiation of electromagnetic waves.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When it is not, these radio waves would be able to make it through?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See my comments above about the cutoff frequency.&lt;/p&gt;&#xA;" OwnerUserId="13663" LastActivityDate="2016-08-10T13:33:02.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17872" PostTypeId="1" CreationDate="2016-08-10T14:24:01.077" Score="0" ViewCount="146" Body="&lt;p&gt;I'm a high school student and currently I'm trying to use Baade-Wesselink method to independently derive period-luminosity relation of Cepheid variables. I noticed that to get the linear radius variation of a Cepheid, I need to integrate its radial velocity over its period. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, I don't know which program/software I should use to plot the best-fit line going through the points of the raw data of radial velocity over the phase. Even more, the thing is I need to integrate that best-fit line and plot the integrated curve as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody recommend good software that I can use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I've tried the integration with a program called &lt;a href=&quot;http://www.vernier.com/products/software/lp/&quot; rel=&quot;nofollow&quot;&gt;Logger Pro&lt;/a&gt; but it was not successful. )&lt;/p&gt;&#xA;" OwnerUserId="13714" LastEditorUserId="10172" LastEditDate="2016-08-18T15:06:26.933" LastActivityDate="2016-08-18T19:27:20.687" Title="Which software is appropriate to integrate radial velocity with respect to time" Tags="&lt;star&gt;&lt;distances&gt;&lt;velocity&gt;&lt;cepheids&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2016-08-22T20:05:05.937" ContentLicense="CC BY-SA 3.0" />
  <row Id="17873" PostTypeId="2" ParentId="17866" CreationDate="2016-08-10T14:28:11.507" Score="6" Body="&lt;p&gt;I am not sure what you mean by &quot;thermal&quot; pressure. Jupiter is supported by pressure, just like all objects that are in (approximate) hydrostatic equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That pressure is provided by your everyday, temperature-dependent Maxwell-Boltzmann ideal gas pressure in the outer parts, but the free electrons in the interior become degenerate and so in these regions I suppose you would more accurately describe the pressure as being due to (partial) electron degeneracy pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the very centre there may be a liquid or even a solid core. In the metallic hydrogen model then this would still be degenerate electrons contributing the pressure. For a solid, rocky core, well solids are rather incompressible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can only describe Jupiter as being in approximate hydrostatic equlibrium. It is losing energy from its surface; this energy is supplied from gravitational potential energy and Jupiter is shrinking at a rate calculated to be about 2 cm per year. As it does so, the interior gas becomes denser and more and more degenerate and the pressure will become more independent of temperature. As a result, the rate of contraction will slow down and Jupiter will tend towards the radius of a &quot;cold&quot; body of that composition.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-08-10T14:28:11.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17874" PostTypeId="2" ParentId="17845" CreationDate="2016-08-10T18:38:04.837" Score="2" Body="&lt;p&gt;Besides all the obvious properties (size, distance, central body, etc.), there are a few other differences between the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, planetary rings usually have a clearly defined boundary. This is because rings usually have shepherd moons — small moons that orbit near the inner or outer edges of rings or within gaps in the rings. The gravity of shepherd moons maintains a sharply defined edge to the ring; material that drifts closer to the shepherd moon's orbit is either deflected back into the body of the ring, ejected from the system, or accreted onto the moon itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Circumstellar disc boundaries can be more complicated. Usually, planets play a similar role to shepherd moons in ring systems, so we can define an inner and outer edge for the asteroid belt. However, discs that lie beyond the planets are trickier. The Kuiper belt's outer edge is approximately 47.8 AU from the Sun, at the 1:2 resonance. However, some objects have been detected up to 55 AU from the Sun. Meanwhile, scattered disc objects have extreme orbits and the disc's outer edge is not clearly defined. Distant SDOs have been proposed as Hills cloud objects, causing a vague ambiguity between the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, rings are much denser than circumstellar discs. As a result, collisions are relatively common between ring debris, while they are very rare in circumstellar discs.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2016-08-10T18:38:04.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17875" PostTypeId="1" AcceptedAnswerId="17876" CreationDate="2016-08-10T19:26:56.097" Score="58" ViewCount="10277" Body="&lt;p&gt;If nothing else wipes out human existence prior to this, at what point will the Sun make Earth uninhabitable for humans?&lt;/p&gt;&#xA;" OwnerUserId="13733" LastEditorUserId="2365" LastEditDate="2018-08-01T10:18:39.610" LastActivityDate="2018-08-01T10:18:39.610" Title="When will the Sun end all human life on Earth?" Tags="&lt;the-sun&gt;" AnswerCount="5" CommentCount="8" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="17876" PostTypeId="2" ParentId="17875" CreationDate="2016-08-10T19:59:26.727" Score="66" Body="&lt;p&gt;The Sun is gradually getting larger and brighter. In fact, as called2voyage pointed out, its brightness is increasing by 1% every 100 million years. You can see how the Sun will change in the future from this graph: (&lt;a href=&quot;https://en.wikipedia.org/wiki/Sun#Main_sequence&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/loYwr.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/loYwr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://adsabs.harvard.edu/abs/2009PNAS..106.9576L&quot; rel=&quot;noreferrer&quot;&gt;this paper&lt;/a&gt;, within 1 billion (short scale) years from now, the ever-increasing luminosity will have made Earth nearly uninhabitable. The average temperature will have reached 47°C, compared to its current 15°C. Essentially no water will be left either, except at the poles. This may allow for simple life to survive for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By 3.5 billion years, Earth will no longer resemble its current self. Its oceans, magnetic field and ozone layer and plate tectonics will be no more. Its surface temperature will skyrocket to roughly 1,330°C, hot enough to melt surface rock. No longer will our planet resemble a pale blue dot, and it will be more like Venus. Our planet is officially dead, along with all life on it. (&lt;a href=&quot;https://www.newscientist.com/article/mg14219191.900-science-fiery-future-for-planet-earth-/&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;~4.5 billion years from now, the Sun will become a red giant and possibly consume Earth. However, according to &lt;a href=&quot;http://adsabs.harvard.edu/abs/2003AsBio...3..317S&quot; rel=&quot;noreferrer&quot;&gt;this paper&lt;/a&gt;, it may heat up potentially habitable bodies like Triton, to the point where they would support life. Unfortunately, the Sun won't remain in this stage for long enough — life usually takes billions of years to develop.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-08-10T21:05:24.140" LastActivityDate="2016-08-10T21:05:24.140" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="17877" PostTypeId="2" ParentId="17846" CreationDate="2016-08-10T20:19:03.060" Score="2" Body="&lt;p&gt;There are several options for you out there, and many good amateur telescopes. I'll just throw some things out that may help in your search.&lt;/p&gt;&#xA;&lt;h3&gt;Telescopes&lt;/h3&gt;&#xA;&lt;p&gt;&lt;strong&gt;Brands&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;I'd suggest going for something like a &lt;a href=&quot;http://www.celestron.com/browse-shop/astronomy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Celestron&lt;/a&gt;. They're well established and reliable from my experience. Aside from Celestron, &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/B004WO8UWK&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Vixen telescope&lt;/a&gt; has been rated as great for kids.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;What to look for&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Don't just go immediately for the telescope with the highest quoted magnification. Making something bigger isn't always the answer. If your image is blurry, magnifying it just gives you a big blurry picture. You'll want a telescope with a good combination of magnification (which can be changed based on your eyepiece) and aperture size (the size of the primary lens/mirror). In general, bigger = better when it comes to aperture size. Bigger means you collect more light and thus the object will be better defined. Usually, when a telescope lists itself as &amp;quot;XXX Telescope, 4 inch&amp;quot;, that size is referring to the aperture diameter.&lt;/p&gt;&#xA;&lt;p&gt;As pela mentioned, there's also the possibility of getting telescopes with automatic tracking software and equipment. Keep in mind that objects in the sky are constantly on the move and this motion is magnified in a telescope. If you get your telescope pointed at an object, it may only take minutes or seconds before that object has left your field of view. Tracking equipment would be built into the telescope and will almost imperceptibly move the telescope with the motion of your object so that it can stay in your field of view for long periods of time. This is especially necessary if you intend to move into the amazing field of astrophotography.&lt;/p&gt;&#xA;&lt;h3&gt;Binoculars&lt;/h3&gt;&#xA;&lt;p&gt;Just to throw another option out there, often a good pair of binoculars is just as good as a cheaper telescope. You can still view many features on the moon, or see Jupiter's Galilean moons. If you go this route, I'd suggest getting a monopod or something to help you hold the binoculars steady. Note also this may be trickier for young children to properly view through as they may have a hard time sighting on a particular object.&lt;/p&gt;&#xA;&lt;h3&gt;Star software&lt;/h3&gt;&#xA;&lt;p&gt;It can be very useful to bring along, or have available, a computer with software that can tell you what is currently in the sky and where to look. Online examples (if you're at home with wifi) would include &lt;a href=&quot;http://neave.com/planetarium/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this planetarium&lt;/a&gt; or else the more robust &lt;a href=&quot;http://www.stellarium.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellarium software&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;h3&gt;Viewing tips&lt;/h3&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Check weather and cloud coverage before viewing. A good example might be &lt;a href=&quot;https://weather.com/weather/radar/interactive/l/l&quot; rel=&quot;nofollow noreferrer&quot;&gt;this map&lt;/a&gt;. Even if the sky looks clear, there may be tenuous cloud cover that can obscure viewing for most objects.&lt;/li&gt;&#xA;&lt;li&gt;Get a red flashlight. You'll want to see what you're doing in the dark, but using a normal flashlight, or your phone, will ruin your eyes. To really see fine details, your eyes need to be adjusted to the dark which takes 10-30 minutes. Using a red flashlight to illuminate will prevent your eyes from being over-exposed.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-08-11T20:00:05.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17878" PostTypeId="2" ParentId="17875" CreationDate="2016-08-10T22:28:31.600" Score="47" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;At what point will the sun make earth uninhabitable for humans?&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It nearly happened on May 23, 1967.&lt;/strong&gt;&lt;br&gt;&#xA;According to &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/2016SW001423/abstract&quot; rel=&quot;noreferrer&quot;&gt;this history paper&lt;/a&gt; released yesterday, summarized nicely &lt;a href=&quot;http://www.space.com/33687-solar-storm-cold-war-false-alarm.html&quot; rel=&quot;noreferrer&quot;&gt;here on Space.com&lt;/a&gt;, at the &lt;a href=&quot;https://news.agu.org/press-release/1967-solar-storm-nearly-took-us-to-brink-of-war/&quot; rel=&quot;noreferrer&quot;&gt;American Geophysical Union&lt;/a&gt; (paper publisher), and in &lt;a href=&quot;https://news.google.com/news/story?ncl=dcciObZkOd2cRTMcLfqXlLVVy2fkM&quot; rel=&quot;noreferrer&quot;&gt;many other media outlets&lt;/a&gt;, a powerful Earth-directed solar flare on that date jammed radar of the US Air Force's Ballistic Missile Early Warning System. Radar jamming like that, especially across all three sites, was considered an act of war, with the Soviet Union as the obvious culprit.  Nuke-laden planes prepared to go destroy the enemy and set off the nuclear war that likely would have rendered Earth uninhabitable for humans.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately for all of us, the US had invested resources in monitoring the sun several years earlier, and somebody figured out that this was a solar flare with just enough time to stop the nuclear deliveries and keep the Cold War cold. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.popularmechanics.com/space/a22265/solar-flare-nuclear-war/&quot; rel=&quot;noreferrer&quot;&gt;Popular Mechanics points out&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If that bulletin had been delayed a few minutes, those nuclear aircraft could have launched, and the solar flares would have made it impossible to communicate in the air. If those aircraft had launched, there would have been no way to call them back.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It also nearly happened on September 26, 1983.&lt;/strong&gt;&lt;br&gt;&#xA;With the superpowers again at hair trigger tensions, a rare alignment of &lt;em&gt;sunlight glinting off high altitude clouds&lt;/em&gt; and the highly elliptical orbits of Soviet early-warning satellites caused the detection system to report that five intercontinental ballistic missiles, likely carrying nuclear warheads, had been launched by the United States against the Soviet Union.  With only a short time to act on such a report, Soviet leaders likely would have launched a large nuclear strike against the US in retaliation for the detected attack.  However, the human on duty to receive this detection report was civilian-trained Stanislav Petrov, who was skeptical of the detection and correctly classified this as a false alarm.  (Five nuclear missiles seemed small for the kind of attack he expected from the US.)  This &lt;a href=&quot;https://en.wikipedia.org/wiki/1983_Soviet_nuclear_false_alarm_incident&quot; rel=&quot;noreferrer&quot;&gt;decision&lt;/a&gt; has been credited with preventing a large-scale nuclear war, which otherwise would have been triggered by misinterpreted sunlight. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So a possible answer is, &quot;&lt;strong&gt;the next time some tense socio-technical system we've built misinterprets the Sun.&lt;/strong&gt;&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/17875/when-will-the-sun-end-all-human-life-on-earth/17878?noredirect=1#comment27542_17878&quot;&gt;Thanks&lt;/a&gt; to user &lt;a href=&quot;https://astronomy.stackexchange.com/users/11832/js&quot;&gt;JS&lt;/a&gt; for the example from September 1983, and to all those who happened to be in the right place at the right time and helped keep us safe.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="7849" LastEditorUserId="7849" LastEditDate="2018-07-31T22:12:19.233" LastActivityDate="2018-07-31T22:12:19.233" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="17879" PostTypeId="2" ParentId="17875" CreationDate="2016-08-11T07:30:08.950" Score="13" Body="&lt;p&gt;According to Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Timeline_of_the_far_future#Future_of_the_Earth.2C_the_Solar_System_and_the_Universe&quot;&gt;timeline of the far future&lt;/a&gt; this will happen at least in &lt;strong&gt;800 million years&lt;/strong&gt; from now:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Carbon dioxide levels fall to the point at which C4 photosynthesis is no longer possible. Free oxygen and ozone disappear from the atmosphere. Multicellular life dies out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But it might already end 200 million years earlier:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Sun's increasing luminosity begins to disrupt the carbonate–silicate cycle; higher luminosity increases weathering of surface rocks, which traps carbon dioxide in the ground as carbonate. As water evaporates from the Earth's surface, rocks harden, causing plate tectonics to slow and eventually stop. Without volcanoes to recycle carbon into the Earth's atmosphere, carbon dioxide levels begin to fall. By this time, carbon dioxide levels will fall to the point at which C3 photosynthesis is no longer possible. All plants that utilize C3 photosynthesis (~99 percent of present-day species) will die.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13736" LastEditorUserId="13736" LastEditDate="2016-08-11T13:36:54.037" LastActivityDate="2016-08-11T13:36:54.037" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17880" PostTypeId="1" AcceptedAnswerId="17884" CreationDate="2016-08-11T07:42:55.907" Score="0" ViewCount="81" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0jpx9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0jpx9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is the equation used in the Baade-Wesselink method to independently get distances to Cepheid variables (in the process of deriving Period-Luminosity relation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the paper &lt;a href=&quot;http://arxiv.org/abs/1003.3854&quot; rel=&quot;nofollow noreferrer&quot;&gt;New Baade-Wesselink distances and radii for four metal-rich Galactic Cepheids&lt;/a&gt;, δθ in the equation accounts for a phase shift between the radial velocity curve and the angular diameter changes measured via surface-brightness relation. My question is how can I calculate this δθ value given that I have lists of radial velocity values and V-R values with the observed HJD dates (which is possible to be converted in terms of phase). &lt;/p&gt;&#xA;" OwnerUserId="13714" LastEditorDisplayName="user1569" LastEditDate="2016-08-18T15:05:52.213" LastActivityDate="2016-08-18T15:05:52.213" Title="How to calculate a phase shift in the process of deriving Baade-Wesselinke distance?" Tags="&lt;star&gt;&lt;distances&gt;&lt;velocity&gt;&lt;cepheids&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17881" PostTypeId="2" ParentId="17875" CreationDate="2016-08-11T10:48:40.443" Score="5" Body="&lt;p&gt;I read a &lt;a href=&quot;https://en.wikipedia.org/wiki/Death_from_the_Skies!&quot; rel=&quot;nofollow&quot;&gt;good book&lt;/a&gt; that covered the sun expanding scenario. I don't remember the time line but at some point in its lifetime the sun will grow larger and swallow Mercury and the earth will become too hot to sustain life. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is the scenario you're asking about and others have answered, but in the book he points out that over the vast time scales involved, it's not a very big deal for a space faring civilization to move the Earth out to a safe distance then back in when the sun shrinks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would have expected some kind of Star Trek level technology, but he describes how to use a simple &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravity_tractor&quot; rel=&quot;nofollow&quot;&gt;Gravity tractor&lt;/a&gt; (not to be confused with a Tractor Beam) to adjust comets flight paths to speed up the earth (the reverse of the gravity slingshot used to speed up our current space probes) moving it further away from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer your original question assuming the human life on earth is still advanced and spacefaring, human life on earth could last until the sun burns out and maybe a bit beyond that.&lt;/p&gt;&#xA;" OwnerUserId="4181" LastEditorUserId="9301" LastEditDate="2016-08-12T20:30:15.410" LastActivityDate="2016-08-12T20:30:15.410" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17882" PostTypeId="1" AcceptedAnswerId="17883" CreationDate="2016-08-11T11:29:20.520" Score="0" ViewCount="137" Body="&lt;p&gt;Also, in which altitude did fireball went thermal explosion?&lt;/p&gt;&#xA;" OwnerUserId="13746" LastActivityDate="2016-08-11T11:51:21.037" Title="How to calculate a meteor's entrance altitude to Earth's atmosphere?" Tags="&lt;meteor&gt;&lt;meteor-shower&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2016-08-11T13:26:57.120" ContentLicense="CC BY-SA 3.0" />
  <row Id="17883" PostTypeId="2" ParentId="17882" CreationDate="2016-08-11T11:51:21.037" Score="1" Body="&lt;p&gt;There is a example &lt;a href=&quot;http://ogleearth.com/2013/02/reconstructing-the-chelyabinsk-meteors-path-with-google-earth-youtube-and-high-school-math/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; of the process required. But basically you need observations of the track from two locations and then just perform some relatively simple trigonometric triangulation. The &lt;a href=&quot;http://www.nemetode.org/&quot; rel=&quot;nofollow&quot;&gt;Nemetode&lt;/a&gt; network regularly does this using images from a network of cctv cameras. &lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-08-11T11:51:21.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17884" PostTypeId="2" ParentId="17880" CreationDate="2016-08-11T13:54:44.223" Score="2" Body="&lt;p&gt;It is admirable that you are tackling this type of problem as a high school student, but I really think that some of these topics are out of your grasp currently, partly because of the highly technical nature, but also because of your lack of access to the necessary data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one thing, you say you have V-R colors as a function of time, but all these equations were calibrated to use V-K colors. From your linked paper, you're going to need equation 2 for your process to calculate $p$ and equation 3 for getting $\theta_0$. Unfortunately, these equations assume V-K colors. You can redo the analysis done in &lt;a href=&quot;http://www.aanda.org/articles/aa/pdf/2009/30/aa12333-09.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; to calculate $p$ and &lt;a href=&quot;http://www.aanda.org/articles/aa/pdf/2007/42/aa8225-07.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; to get $\theta_0$ for V-R colors, but it's not going to be easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For another, you're going to need measurements of the angular size as a function of time (and your cited paper suggests that only exists for a handful of Cepheids and was measured using the VLTI) as well as the RV as a function of time. The $\delta \theta$ that you reference is described exactly in the paper as &quot;a  phase shift  between  the  radial  velocity  curve  and  the  angular  diameter changes measured either interferometrically or via the SB relation.&quot; This means if you have $RV(t)$ and $\Delta \theta(t)$, you can determine a phase shift between the two curves, presumably by fitting some equation of the form $f(t) = a\ \mathrm{sin}(bt+c)$ to both equations and saying that $\delta \theta = c_{RV} - c_{\Delta \theta}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analysis you're trying to do requires very specific data and it seems that the relevant papers on this subject have already done the analysis on all stars in which all necessary data existed. You cannot easily apply this process to any Cepheid without significant work.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-08-11T13:54:44.223" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17885" PostTypeId="2" ParentId="17845" CreationDate="2016-08-11T15:17:17.447" Score="3" Body="&lt;p&gt;From &lt;a href=&quot;https://physics.stackexchange.com/a/197897/28199&quot;&gt;Physics SE&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A circumstellar disk is a set of objects that orbit about a star-like object and that collectively look disk-like in some way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://en.wikipedia.org/wiki/Ring_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A ring system is a disc or ring orbiting an astronomical object that is composed of solid material&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, a circumstellar disk may have gas, dust, asteroids, etc., but it must orbit a star-like object. A ring system &lt;em&gt;must&lt;/em&gt; have something solid--dust or moonlets--but it may orbit any astronomical object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than that, you specifically asked about a &lt;em&gt;planetary&lt;/em&gt; ring. That would mean a ring system that specifically had a planemo as the central body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, a disk around a brown dwarf could be described as a circumstellar disk (a brown dwarf is a star-like object) &lt;em&gt;or&lt;/em&gt; a ring system (as long as the disk contained solid material). It could not be called a planetary ring, because a brown dwarf is not a planet.&lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-08-11T15:20:57.857" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17886" PostTypeId="1" CreationDate="2016-08-11T16:12:42.293" Score="4" ViewCount="134" Body="&lt;p&gt;I'm setting up an observatory and I'm wondering what user-facing software would be best to use in terms of hardware control. There is a German mount, dome, focuser, filter wheel, various switches etc. All devices have / will have ASCOM drivers. For simplicity I'm setting up the operator PC to run Windows, but that is not set in stone (yet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried MaxIm DL and found it a pleasant surprise with all features it has available (e.g. dome slaving or autofocus).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some other software packages that I should try before deciding on (i.e. buying) one?&lt;/p&gt;&#xA;" OwnerUserId="7393" LastActivityDate="2016-08-11T16:12:42.293" Title="Observatory control software" Tags="&lt;software&gt;&lt;observatory&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17887" PostTypeId="1" AcceptedAnswerId="17888" CreationDate="2016-08-11T17:41:31.530" Score="-1" ViewCount="318" Body="&lt;p&gt;So I have two parameters, I have the distance to a galaxy of 879 kpc and the sky distance in degrees (0.131501). How would I then use these two bits of information to calculate the Galactocentric distance in kpc? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;N.B. for the two objects within M33, i.e., the centre of M33 and an undefined region of M33 are separated by 0.131501 degrees. Both objects can assume to have a distance from us of ~879 kpc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I want to convert the value for the distance between the two objects within M33 (0.131501 degrees) to kpc...&lt;/p&gt;&#xA;" OwnerUserId="2091" LastEditorUserId="6" LastEditDate="2016-08-11T18:47:34.613" LastActivityDate="2016-08-11T18:47:39.443" Title="How to calculate a Galactocentric distance of another galaxy" Tags="&lt;distances&gt;&lt;parallax&gt;&lt;parsec&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17888" PostTypeId="2" ParentId="17887" CreationDate="2016-08-11T18:04:25.443" Score="2" Body="&lt;p&gt;Basic angle calculations: $\mathrm{arc\ length} = r\theta$&#xA;(angle measured in radians) and for small angles the arc length approximates the chord length. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle in radians is therefore $0.131501\pi/180=0.00223$, so the separation is $0.00223\times 879= 2.02 \mathrm{kpc}$ &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2091" LastEditDate="2016-08-11T18:47:39.443" LastActivityDate="2016-08-11T18:47:39.443" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17890" PostTypeId="2" ParentId="16776" CreationDate="2016-08-11T20:26:53.813" Score="6" Body="&lt;p&gt;Good question! I will need to go into some cosmology to answer it, my main source being &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0310808&quot; rel=&quot;nofollow noreferrer&quot;&gt;Davis &amp;amp; Lineweaver (2003)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing &lt;em&gt;traveling through space&lt;/em&gt; should be able to leave the observable universe. However, information can and will inevitably leave our observable universe. How is this possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Universe is expanding.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we lived in a static universe, no objects would leave our observable universe. However, in our accelerating universe, space is being created between objects, with farther objects drifting away faster than closer ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Einstein said nothing can travel through space faster than light. However, space is literally being created between objects, so they can be pushed away from each other faster than light. So while the observable universe is expanding at the speed of light, distant objects are moving away from us even faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the result? Take a look at this image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/txwtv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/txwtv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This describes how the observable universe will evolve. The &lt;em&gt;Hubble sphere&lt;/em&gt; defines the area in which objects are receding from us at or below the speed of light. Beyond this, objects are far enough to recede from us faster than light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The light cone is the region of the Universe capable of influencing us today. Essentially, it is the observable universe, and its boundary is clearly marked. It is, as you mentioned, growing at the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The light cone does not contain the entire Hubble sphere, since photons from many objects receding faster than light cannot reach us yet. Conversely, the Hubble sphere doesn't contain the entire light cone. This is because photons emitted by objects outside of the Hubble sphere can crawl into regions of space that are receding from us a bit slower than the one from which they were emitted. They can then crawl into regions that expand slower than the last one, and so on. Eventually, the photons will reach regions of space that aren't expanding faster than light, and they can reach us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there is a distance beyond which space expands faster than the light can traverse it. This distance is called the cosmic event horizon. Light from beyond the event horizon will never reach us. Our observable universe grows because more photons come into our view. However, the observable universe will infinitely approach the event horizon, since light from beyond there can never reach us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, after an incomprehensibly long time, the observable universe would stop growing. At this point, it may finally be possible to travel through space beyond the observable universe.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2017-06-19T17:41:22.900" LastActivityDate="2017-06-19T17:41:22.900" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17892" PostTypeId="2" ParentId="17846" CreationDate="2016-08-11T23:49:20.727" Score="1" Body="&lt;p&gt;Sky and Telescope has a &lt;a href=&quot;http://www.skyandtelescope.com/astronomy-news/types-of-telescopes/&quot; rel=&quot;nofollow&quot;&gt;brief overview&lt;/a&gt; and a more &lt;a href=&quot;http://www.skyandtelescope.com/astronomy-equipment/how-to-choose-a-telescope/&quot; rel=&quot;nofollow&quot;&gt;comprehensive article&lt;/a&gt; on this topic.&#xA;Beyond an ample aperture and a stable mount, some choices are matters of preference.&#xA;Astronomy club public events are good opportunities to try a variety of equipment, with the owners' consent of course.&#xA;In addition to brands mentioned in other answers, &lt;a href=&quot;http://www.meade.com/&quot; rel=&quot;nofollow&quot;&gt;Meade&lt;/a&gt; and &lt;a href=&quot;http://www.telescope.com/&quot; rel=&quot;nofollow&quot;&gt;Orion&lt;/a&gt; make decent telescopes.&#xA;A motivated beginner could get a lot of mileage out of a $200 instrument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for what to look at, &lt;a href=&quot;http://skymaps.com/&quot; rel=&quot;nofollow&quot;&gt;Skymaps.com&lt;/a&gt; makes a monthly all-sky map which you can download and print free for personal use.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-08-11T23:49:20.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17893" PostTypeId="2" ParentId="101" CreationDate="2016-08-12T00:48:08.420" Score="2" Body="&lt;p&gt;I think about 3 factors that impact the quality of observing and astrophotography.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Seeing - as mentioned before, the turbulence in the atmosphere and ultimately, determines the resolution you can image or resolve by eye.  I would kill for 1 arc-second seeing at my site :-(&lt;/li&gt;&#xA;&lt;li&gt;transparency - humidity in the air, dust, fog clouds - all these things soak up photons as they travel across the universe and before they hit your eyeball.  Note that at some sites, poor transparency due to humidity, etc. might produce very good seeing because the atmosphere isn't mixing very much and is stable.  Similarly, if you observe nearer to the horizon vs. overhead near the zenith, you look through more atmosphere which makes your transparency and seeing problems worse.&lt;/li&gt;&#xA;&lt;li&gt;light pollution - from nearby (and maybe not so near..) lights emitting light up into the air.  As it scatter of dust, water, vapor, etc, it makes it difficult to see dim objects.  Some interaction with transparency - if there's a lot of water vapor in the air, it also scatters more light back into your instrument.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;At my observatory site, I had so-so seeing, but being out in the sticks, not much of a light pollution problem.  All these things are trade-offs..&lt;/p&gt;&#xA;" OwnerUserId="13766" LastActivityDate="2016-08-12T00:48:08.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17894" PostTypeId="2" ParentId="17872" CreationDate="2016-08-12T01:09:48.423" Score="2" Body="&lt;p&gt;I'm not sure what you are actually trying to do. Radial velocity isn't generally measured, it is computed from the projected velocity. But if I take you at your word, that your raw data is radial velocity, then that simplifies things. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know the period, $P$, and you know the $v$ AND if you know the theoretical relationship between them $v = f(p)$ then you can use linear least squares to compute the adjustable parameters of $f(p)$ by plotting $v$ (measured) vs $v$(calculated). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any 'best fit' equation is able to interpolate values of $v$ so all you need to divide up the period into small intervals and approximate the area under the curve using its two end-points (it is rarely necessary to use parabolic fits that also use the mid-points of those small intervals). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the problem of integration becomes one of simple geometry (you have a tall thin rectangle with a triangle on top of it for each interval). Let's think about a star with a 30 day period. That's ~40,000 minutes, so I'd probably use 5 or 8 minute intervals ...starting at minute 0 what's the area under the curve from t=0 to t=8 minutes? then from 8 to 16, all the way up to 43,200th minute. Sum of those areas is total area. An Excel spreadsheet can easily do that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, if you also use 5 minute intervals and compare the results you'll see how much better (hopefully NOT much) a smaller interval is...analytical integration is this taken to the limit of the interval approaching zero, but in real life, numerical integration is necessary for almost all 'interesting' problems.  Here are 2 links, fwiw: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.physicsforums.com/threads/finding-periodic-best-fit-equation-for-data-set.715638/&quot; rel=&quot;nofollow&quot;&gt;https://www.physicsforums.com/threads/finding-periodic-best-fit-equation-for-data-set.715638/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.aanda.org/articles/aa/pdf/2004/08/aah4718.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.aanda.org/articles/aa/pdf/2004/08/aah4718.pdf&lt;/a&gt; ← see page 537 and references noted there.&lt;/li&gt;&#xA;&lt;li&gt;The Institute for Astronomy (IFA) at the University of Hawaii &lt;a href=&quot;https://www.google.com/search?q=Baade-Wesselink+site%3Ahawaii.edu&quot; rel=&quot;nofollow&quot;&gt;has several pdfs on BW&lt;/a&gt; that may be helpful&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13767" LastEditorDisplayName="user1569" LastEditDate="2016-08-18T19:27:20.687" LastActivityDate="2016-08-18T19:27:20.687" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17895" PostTypeId="1" AcceptedAnswerId="17896" CreationDate="2016-08-12T02:20:11.680" Score="2" ViewCount="99" Body="&lt;p&gt;The peak viewing dates of the 2016 &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Perseids&quot; rel=&quot;nofollow&quot;&gt;Perseid Meteor Shower&lt;/a&gt; is said to be between the dates of August 11 through the early morning hours of August 12.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being a resident of the United States, I keep getting my peak viewing dates and times for my area and for the whole of the &lt;a href=&quot;http://www.space.com/33713-perseid-meteor-shower-us-weather-forecast-2016.html&quot; rel=&quot;nofollow&quot;&gt;United States&lt;/a&gt;.  There is never any mention of any other parts of the world having the capability to view this meteor shower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is the Perseid Meteor Shower visible to all parts of the world?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13770" LastActivityDate="2016-08-12T08:27:36.303" Title="Can you view the Perseid meteor shower anywhere on Earth?" Tags="&lt;meteor-shower&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17896" PostTypeId="2" ParentId="17895" CreationDate="2016-08-12T08:27:36.303" Score="3" Body="&lt;p&gt;You can see perseid meteors from all the Northern Hemisphere, and the Northern part of the Southern Hemisphere. Provided the sky is clear, the moon has set, and Perseus is above the horizon you will see meteors. This shower has a wide peak, so you can see them over several nights.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-08-12T08:27:36.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17897" PostTypeId="1" AcceptedAnswerId="17898" CreationDate="2016-08-12T09:24:19.850" Score="2" ViewCount="127" Body="&lt;p&gt;KIC 8462852, the F-star that inexplicably dims by up to 20% for short periods, is still getting alot of attention. A slow continuous dimming by about 0.3% a year has been identified &lt;a href=&quot;https://arxiv.org/pdf/1608.01316v1.pdf&quot; rel=&quot;nofollow&quot;&gt;recently&lt;/a&gt;. But all anomalies I know of come from the Kepler space telescope. It is an F3-star at 1500 ly distance. Shouldn't it be easy to observe its light curve with even relatively modestly sized ground based telescopes? Have such observations been done and have they confirmed any anomalies? You see, I suspect something weird went wrong with the Kepler telescope here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: There have been studies of its brightness on archived historic photographs. But I'm more interested in observations simultaneous as Kepler or since then, to confirm either a gradual dimming or short deep dips.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2016-08-12T10:09:43.270" LastActivityDate="2016-08-12T12:32:13.023" Title="What telescopes have observed anomalies in the light curve of Tabby's star, KIC 8462852?" Tags="&lt;observational-astronomy&gt;&lt;kepler&gt;&lt;planetary-transits&gt;&lt;photometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17898" PostTypeId="2" ParentId="17897" CreationDate="2016-08-12T12:32:13.023" Score="3" Body="&lt;p&gt;I do not know whether complimentary observations of the light curve are being done, but I will try to answer the rest of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it easy to observe the light curve with ground-based telescopes? No, it's quite difficult. From the &lt;a href=&quot;http://kepler.nasa.gov/Science/about/targetFieldOfView/&quot; rel=&quot;nofollow&quot;&gt;Kepler homepage&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since transits only last a fraction of a day, Kepler must monitor all target stars continuously. Their brightnesses must be measured at least once every few hours.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It goes on to explain how the observed field was chosen - no downtime due to it being too close to the sun, no obstruction by asteroids, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ground-based telescope has the obvious problem that it can't observe during the day, so you need several telescopes in different parts of the world, and combine their measurements. Since the distortion patterns, viewing conditions, instrument characteristics etc. are all different, combining the data is anything but trivial if you want a precision measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, a change in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Point_spread_function&quot; rel=&quot;nofollow&quot;&gt;point spread function&lt;/a&gt; between telescopes might introduce both systematic and statistical errors in your brightness measurement. Kepler is only one instrument and optically highly stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is KIC 8462852 bright enough for a small ground-based telescope? Certainly. Its magnitude is 11.7, while Kepler wants to observe stars brighter than 14th magnitude (link above). &lt;a href=&quot;http://www.cruxis.com/scope/limitingmagnitude.htm&quot; rel=&quot;nofollow&quot;&gt;This magnitude calculator&lt;/a&gt; estimates a limiting magnitude of 17 for a 1m mirror.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It takes time to set up an observation. Large telescopes like the &lt;a href=&quot;http://www.eso.org/public/teles-instr/paranal/&quot; rel=&quot;nofollow&quot;&gt;VLT&lt;/a&gt; accept requests for observation time twice a year. After your observing proposal is accepted, it will still take weeks or months until your observation is actually made. For smaller telescopes, it might be easier and quicker, but it will take time to set up a monitoring programme, even if telescope time is available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are ground-based, long-term monitoring programmes. &lt;a href=&quot;http://cosmograil.epfl.ch/&quot; rel=&quot;nofollow&quot;&gt;COSMOGRAIL&lt;/a&gt; is one that comes to mind. From their homepage:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Obtaining regular observing time on telescopes in good sites was (and is still) not easy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and they've been going for a number of years by now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it would be technically challenging but possible. Politically, it might not be so easy: Every observatory wants to make the best use of the observing time they have, and most are oversubscribed (more time asked for than available). You need a fairly solid science case to block several telescopes around the world for several years, but the result you'd get if all goes well is merely confirmation of someone else's result.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastActivityDate="2016-08-12T12:32:13.023" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17899" PostTypeId="2" ParentId="16608" CreationDate="2016-08-12T14:51:57.257" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What are the velocity, mass, and charge distribution of the solar wind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Velocity&lt;/strong&gt;&lt;br&gt;&#xA;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_wind&quot; rel=&quot;nofollow&quot;&gt;solar wind&lt;/a&gt; speed has a large range of variation, between ~250–820 km/s [e.g., &lt;em&gt;Chen et al.&lt;/em&gt;, 2014; &lt;em&gt;Gopalswamy&lt;/em&gt;, 2006; &lt;em&gt;Jian et al.&lt;/em&gt;, 2011, 2014; &lt;em&gt;Kasper et al.&lt;/em&gt;, 2012; &lt;em&gt;Maksimovic et al.&lt;/em&gt;, 1998; &lt;em&gt;Marsch&lt;/em&gt;, 1983; &lt;em&gt;McComas et al.&lt;/em&gt;, 2013; &lt;em&gt;Schwenn&lt;/em&gt;, 1983; &lt;em&gt;Stverak et al.&lt;/em&gt;, 2008, 2009] near the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ecliptic&quot; rel=&quot;nofollow&quot;&gt;ecliptic plane&lt;/a&gt;.  These values are not including interplanetary shocks, which can have speeds exceeding 2000 km/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed is generally higher at higher latitudes out of the ecliptic plane, tending to be over 650 km/s [e.g., &lt;em&gt;McComas et al.&lt;/em&gt;, &lt;a href=&quot;http://adsabs.harvard.edu/abs/2008GeoRL..3518103M&quot; rel=&quot;nofollow&quot;&gt;2008&lt;/a&gt;; &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013ApJ...779....2M&quot; rel=&quot;nofollow&quot;&gt;2013&lt;/a&gt;].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Number Density&lt;/strong&gt;&lt;br&gt;&#xA;The number density also has a large range of values, from ~2–90 $cm^{-3}$ [e.g., &lt;em&gt;Chen et al.&lt;/em&gt;, 2014; &lt;em&gt;Gopalswamy&lt;/em&gt;, 2006; &lt;em&gt;Jian et al.&lt;/em&gt;, 2011, 2014; &lt;em&gt;Kasper et al.&lt;/em&gt;, 2012; &lt;em&gt;Maksimovic et al.&lt;/em&gt;, 1998; &lt;em&gt;Marsch&lt;/em&gt;, 1983; &lt;em&gt;McComas et al.&lt;/em&gt;, 2013; &lt;em&gt;Schwenn&lt;/em&gt;, 1983; &lt;em&gt;Stverak et al.&lt;/em&gt;, 2008, 2009].  Again, these do not include interplanetary shocks or &lt;a href=&quot;https://en.wikipedia.org/wiki/Coronal_mass_ejection&quot; rel=&quot;nofollow&quot;&gt;coronal mass ejections (CMEs)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Charge State&lt;/strong&gt;&lt;br&gt;&#xA;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Alpha_particle&quot; rel=&quot;nofollow&quot;&gt;alpha particle&lt;/a&gt; to proton number density ratio varies between ~1-5%, depending on &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_cycle&quot; rel=&quot;nofollow&quot;&gt;solar cycle&lt;/a&gt; and solar wind speed [e.g., &lt;a href=&quot;http://adsabs.harvard.edu/abs/2012ApJ...745..162K&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Kasper et al.&lt;/em&gt;, 2012&lt;/a&gt;; &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014JGRA..119.1486S&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Schwadron et al.&lt;/em&gt;, 2014&lt;/a&gt;].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have also measured the ratio of $O^{7+}/O^{6+}$ and $C^{6+}/C^{5+}$, finding ~1-30% and ~20-200%, respectively [e.g., &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014JGRA..119.1486S&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Schwadron et al.&lt;/em&gt;, 2014&lt;/a&gt;].&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Near the earth within the magnetosphere in the ecliptic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The properties of the terrestrial magnetosphere vary so widely, you would need to narrow down this question.  For instance, the charge states are completely different (e.g., we observe $O^{1+}$ but not $O^{7+}$) but the number densities range from ~$10^{-2}-10^{3} \ cm^{-3}$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Near the earth but outside the magnetosphere in the ecliptic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See responses above to first part.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Outside the ecliptic at 1 AU&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We don't have any measurements near 1 AU that are at high latitudes.  Some spacecraft have done &lt;em&gt;out of ecliptic&lt;/em&gt; polar orbits with high apogees, but the heliocentric latitudes were still within ~$10^{\circ}$ of the ecliptic plane.  The notes above discuss our only real measurements out of the ecliptic by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ulysses_(spacecraft)&quot; rel=&quot;nofollow&quot;&gt;Ulysses&lt;/a&gt; spacecraft.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I would like to understand the energy content of the solar wind, and how it compares with the solar radiation(solar constant is 1360 W $m^{-2}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solar wind ram(dynamic) pressure is typically only ~1 nPa or $10^{-9} \ J \ m^{-3}$.  This is highly variable and can change in milliseconds (e.g., interplanetary shocks), but that would still only be $10^{-6} \ W \ m^{-3}$.  If we make a hand-wavy argument that this drops to zero in ~3 &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth_radius&quot; rel=&quot;nofollow&quot;&gt;$R_{E}$&lt;/a&gt; (i.e., upper bound on thickness of magnetosheath), then the power per unit area can be up to ~20 W $m^{-2}$.  However, I would not read too much into that number as the actual power dissipated per unit area is different for numerous reasons.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;&lt;strong&gt;References&lt;/strong&gt;&lt;/h1&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;C.H.K. Chen &lt;em&gt;et al.&lt;/em&gt;, &lt;em&gt;Geophys. Res. Lett.&lt;/em&gt; &lt;strong&gt;41&lt;/strong&gt;, pp. 8081, 2014.&lt;/li&gt;&#xA;&lt;li&gt;N. Gopalswamy, &lt;em&gt;Space Sci. Rev.&lt;/em&gt; &lt;strong&gt;124&lt;/strong&gt;, pp. 145, 2006.&lt;/li&gt;&#xA;&lt;li&gt;L.K. Jian &lt;em&gt;et al.&lt;/em&gt;, &lt;em&gt;Solar Phys.&lt;/em&gt; &lt;strong&gt;274&lt;/strong&gt;, pp. 321, 2011.&lt;/li&gt;&#xA;&lt;li&gt;L.K. Jian &lt;em&gt;et al.&lt;/em&gt;, &lt;em&gt;Astrophys. J.&lt;/em&gt; &lt;strong&gt;786&lt;/strong&gt;, pp. 123, 2014.&lt;/li&gt;&#xA;&lt;li&gt;J. C. Kasper &lt;em&gt;et al.&lt;/em&gt;, &lt;em&gt;Astrophys. J.&lt;/em&gt; &lt;strong&gt;745&lt;/strong&gt;, pp. 162, 2012.&lt;/li&gt;&#xA;&lt;li&gt;M. Maksimovic &lt;em&gt;et al.&lt;/em&gt;, &lt;em&gt;Geophys. Res. Lett.&lt;/em&gt; &lt;strong&gt;25&lt;/strong&gt;, pp. 1265, 1998.&lt;/li&gt;&#xA;&lt;li&gt;E. Marsch, &lt;em&gt;Fifth International Solar Wind Conference&lt;/em&gt; &lt;strong&gt;228&lt;/strong&gt;, pp. 355, 1983.&lt;/li&gt;&#xA;&lt;li&gt;D.J. McComas &lt;em&gt;et al.&lt;/em&gt;, &lt;em&gt;Geophys. Res. Lett.&lt;/em&gt; &lt;strong&gt;35&lt;/strong&gt;, pp. L18103, 2008.&lt;/li&gt;&#xA;&lt;li&gt;D.J. McComas &lt;em&gt;et al.&lt;/em&gt;, &lt;em&gt;Astrophys. J.&lt;/em&gt; &lt;strong&gt;779&lt;/strong&gt;, pp. 2, 2013.&lt;/li&gt;&#xA;&lt;li&gt;N.A. Schwadron &lt;em&gt;et al.&lt;/em&gt;, &lt;em&gt;J. Geophys. Res.&lt;/em&gt; &lt;strong&gt;119&lt;/strong&gt;, pp. 1486-1492, 2014.&lt;/li&gt;&#xA;&lt;li&gt;R. Schwenn, &lt;em&gt;Fifth International Solar Wind Conference&lt;/em&gt; &lt;strong&gt;228&lt;/strong&gt;, pp. 489, 1983.&lt;/li&gt;&#xA;&lt;li&gt;v. Stverak &lt;em&gt;et al.&lt;/em&gt;, &lt;em&gt;J. Geophys. Res.&lt;/em&gt; &lt;strong&gt;113&lt;/strong&gt;, pp. 3103, 2008.&lt;/li&gt;&#xA;&lt;li&gt;v. Stverak &lt;em&gt;et al.&lt;/em&gt;, &lt;em&gt;J. Geophys. Res.&lt;/em&gt; &lt;strong&gt;114&lt;/strong&gt;, pp. 5104, 2009.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13663" LastActivityDate="2016-08-12T14:51:57.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17900" PostTypeId="5" CreationDate="2016-08-12T16:21:52.210" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2016-08-12T16:21:52.210" LastActivityDate="2016-08-12T16:21:52.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17901" PostTypeId="4" CreationDate="2016-08-12T16:21:52.210" Score="0" Body="Questions about infrared astronomy, both observations (in the broadest sense) and instruments. The infrared part of the electromagnetic spectrum has a longer wavelength than visible light, but shorter than sub-mm or radio waves." OwnerUserId="13681" LastEditorUserId="13681" LastEditDate="2016-08-12T17:07:31.107" LastActivityDate="2016-08-12T17:07:31.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17903" PostTypeId="2" ParentId="17875" CreationDate="2016-08-12T18:51:54.540" Score="4" Body="&lt;p&gt;I would be very surprised if the sun ever ends all life on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)  If we care about the Earth we will move it.  If we are still around by then it certainly will be within our technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)  I would say the most likely outcome (assuming we survive) is that the Earth gets taken apart for building materials.  It's an awful lot of mass simply being used as a gravity source--incredibly inefficient.&lt;/p&gt;&#xA;" OwnerUserId="2036" LastActivityDate="2016-08-12T18:51:54.540" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17904" PostTypeId="2" ParentId="16408" CreationDate="2016-08-12T21:02:46.363" Score="4" Body="&lt;p&gt;Frequencies below 30 MHz have a hard time penetrating the so-called ionosphere - the series of layers of ionized atmosphere that exist at high altitudes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is good news for folks like me, amateur radio operators, because it allows us to speak into a transmitter at home, and let the ionosphere bounce the radio waves back to Earth, then again up into the ionosphere, and so on, eventually reaching another radio operator on the opposite side of Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's bad news if you're a radio astronomer trying to receive such frequencies from distant objects - the ionosphere simply bounces those waves back into space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, apparently, you could raise a receiver above the ionosphere, in a satellite, and have it receive signals from outer space. But there's a catch. At 30 MHz and below, the wavelengths are huge. 27 MHz - the wavelength is 10 meters. 14 MHz - wavelength is 20 meters. 3.5 MHz - 80 meters. And so on, the lower the frequency, the bigger the wavelength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of the antenna needs to be comparable to the wavelength if you want to receive signals from one direction only, so as to be able to tell where the signal is coming from. This also helps with efficiency. A very good antenna needs to be much bigger than the wavelength for it to be directional and efficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, even with the ionosphere, there's a lot of signal &quot;leaking&quot; out of the Earth, through ionosphere, into space; some of it is bounced back down, but some of it penetrates the ionosphere and escapes into space. A satellite in low orbit would be swamped by transmitters from Earth, completely erasing any signals from deep space. You'd need a humongous antenna to ignore the signals from the nearby Earth. This is not feasible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be a lot better if you could just put the Moon in between Earth and your antenna. Even these low frequency transmissions cannot penetrate the Moon, and can't go around the Moon either. So the Moon acts like a screen, casting a shadow on the antenna and eliminating parasite transmissions from Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now your satellite can receive low frequency transmissions from deep space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TLDR: Too much noise near Earth at these frequencies, need to hide the antenna behind the Moon.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2016-08-12T21:02:46.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17908" PostTypeId="1" CreationDate="2016-08-12T22:20:13.587" Score="0" ViewCount="653" Body="&lt;p&gt;I am currently running into some problems with plotting some stars locations in a Cartesian coordinate system.&#xA;The stars aren't in the right order and I'm not sure where I am going wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I first use the equation (where S is a vector)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;S = [cos(ra)*cos(dec), sin(ra)*cos(dec), sin(dec)]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;to obtain a vector S of the star.&#xA;I now want to obtain the x and y positions of the stars In terms of a 2d coordinate system so I use&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;c * sx = S * px / S* pz&#xA;c * sy = S * py / S* pz&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where C is the scale ratio of pixels to tangent of the angle of the camera and P is the vector center of the camera.(Im using this because I'm trying to find the distance between each pair of stars when a photo of the stars is taken by a camera)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;px = unit vector X&#xA;py = unit vector Y&#xA;pz = unit vector Z&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and I solve for sx and sy. When I plot the stars location with sx and sy however I do not get the right locations of the stars. Any guidance?&lt;/p&gt;&#xA;" OwnerUserId="13574" LastEditorUserId="13574" LastEditDate="2016-08-14T17:48:49.650" LastActivityDate="2016-08-14T17:48:49.650" Title="plotting stars in cartessian coordinates using the stars ra and dec produces the wrong location of the stars" Tags="&lt;star&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17909" PostTypeId="1" CreationDate="2016-08-12T22:35:43.237" Score="0" ViewCount="112" Body="&lt;p&gt;On a casual star walk, (for comparison of identifying different objects) I witnessed some meteors, quite a few satellites gliding on their smooth and straight trajectories, and some airplanes with their much brighter (sometimes blinking) lights, lower altitude and the characteristic sense of moving towards and away from me. Suddenly one of the satellites seemed as if its motion had been a bit jittery, and its path started to curve inwards in an asymmetrical, gentle and large C-shape. The airplanes made noise, this one did not. Visually perfectly identical with all of the normal satellites. I had never seen anything like that before so I wondered if it could've been some debris colliding into things, but probably not. The best explanation I have for now is that it apparently might have been a satellite on a Tundra or Molniya orbit, as their ground tracks aren't the usual straight or sine wave shaped trajectories that we're used to. Most satellite tracking websites either didn't seem to have tracking options for those special types at all, or you could only see their locations for now, not a custom time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm in the capital city area of Finland, and when I saw the object, the time was 23:10 on the UTC+3 time zone, 12th of August 2016. Of what I found, the only strange looking ground track that comes even close would be Molniya 1-62, drawing a loop above Finland. What I can't find is a website that would allow me to rewind time to check where the satellite was at 23:10. But whether or not I should keep searching for that info.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I first have this question: If I look at the satellite tracks going across a map, how do I determine how far a satellite would be visible when viewed from the ground?&lt;/strong&gt; If I looked at similar tracks for airplanes, it's much easier to understand that the ground area from which an airplane can be spotted doesn't span vast distances, but how this compares to a satellite is difficult to imagine as we don't perceive them in a three dimensional space, but spots gliding across the flat surface of a dome. &lt;/p&gt;&#xA;" OwnerUserId="9848" LastEditorUserId="2365" LastEditDate="2016-08-12T23:00:45.457" LastActivityDate="2016-10-13T19:57:06.820" Title="Actual observable horizon for satellites on an orbit line map?" Tags="&lt;satellite&gt;&lt;eccentric-orbit&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17910" PostTypeId="2" ParentId="17909" CreationDate="2016-08-12T23:14:35.693" Score="1" Body="&lt;p&gt;It is not that satellite. As it passes over Finland, it has an altitude of 27000km and a magnitude of about 12. You won't see it. All Molniya orbits are similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The website &lt;a href=&quot;http://www.satflare.com/track.asp?q=15214#TOP&quot; rel=&quot;nofollow&quot;&gt;Satflare&lt;/a&gt; to shows the region in which satellites are visible (it uses a blue ring around the satellite) Your satellite was so high that it was visible from nearly the whole hemisphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I won't try to identify your object, except to say that an object as described is not in orbit.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-08-14T10:33:56.547" LastActivityDate="2016-08-14T10:33:56.547" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17911" PostTypeId="1" AcceptedAnswerId="18095" CreationDate="2016-08-12T23:50:03.270" Score="5" ViewCount="2040" Body="&lt;p&gt;Hi I am trying to understand the concept of dex and how to use it in calculations. The usual definition is that it is the order of magnitude, so $10^{0.1}$ is $0.1$ dex. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to do a simple exercise of calculating the value of the RHS of Eqn 4 in this paper &lt;a href=&quot;http://arxiv.org/abs/0908.2102&quot; rel=&quot;noreferrer&quot;&gt;arxiv paper&lt;/a&gt;, the gammas are incomplete gamma functions and the value of $\alpha$, $\phi_∗$ and $M_∗$ are given in table 2. The units of $\phi_∗$ is $h_{70}^3$ Mpc$^{−3}$ dex$^{−1}$. The value of the integral is given in the following paragraph as $5.06×10^5 h_{70}^3$ $M_⊙$ Mpc$^{−3}$ (let's take only the first one). If I substitute the values from table 2 to the RHS of Eq 4 and then calculate I get $7.58 × 10^5$ $h_{70}^3$ Mpc$^{−3}$ dex$^{−1}$ . What do I do with the dex$^{-1}$? Is it because of the dex$^{-1}$ I am a little off? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 1:&lt;/strong&gt; Does the $d(\log(M_{bh}))$ in the integral take care of the $\rm{dex}^{-1}$, and my little off answer is due to rounding errors?&lt;/p&gt;&#xA;" OwnerUserId="7191" LastEditorUserId="7191" LastEditDate="2016-08-13T18:36:09.230" LastActivityDate="2016-08-27T13:27:37.823" Title="Understanding &quot;dex&quot;" Tags="&lt;supermassive-black-hole&gt;&lt;magnitude&gt;&lt;initial-mass-function&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17912" PostTypeId="2" ParentId="17908" CreationDate="2016-08-13T01:47:06.250" Score="1" Body="&lt;p&gt;A photograph of stars is essentially a&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Gnomonic_projection&quot; rel=&quot;nofollow&quot;&gt;gnomonic projection&lt;/a&gt;&#xA;of part of the&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Celestial_sphere&quot; rel=&quot;nofollow&quot;&gt;celestial sphere&lt;/a&gt;.&#xA;If $\vec P$ is computed for the photograph center (RA, Dec) in the same way as $\vec S$,&#xA;and the north celestial pole is&#xA;$\hat{\mathrm k} = (0, 0, 1)^{\mathrm T}$,&#xA;then the photograph has basis vectors perpendicular to $\vec P$ and each other,&#xA;$$\hat{\mathrm u} = \frac{\vec P \times \hat{\mathrm k}}{\| \vec P \times \hat{\mathrm k} \|},&#xA;\hat{\mathrm v} = \frac{\hat{\mathrm u} \times \vec P}{\| \hat{\mathrm u} \times \vec P \|}$$&#xA;and you can project $\vec S$ onto the photograph as&#xA;$$\left( \frac{\vec S \cdot \hat{\mathrm u}}{\vec S \cdot \vec P},&#xA;\frac{\vec S \cdot \hat{\mathrm v}}{\vec S \cdot \vec P} \right)$$&#xA;scaling as needed.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2016-08-14T02:50:34.580" LastActivityDate="2016-08-14T02:50:34.580" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17914" PostTypeId="5" CreationDate="2016-08-13T17:05:00.373" Score="0" Body="&lt;p&gt;The European Extremely Large Telescope is a 40m-class optical / near infrared telescope currently under construction in Chile. It is expected to start regular observations in 2024.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Features include a five-mirror optical design, fully integrated adaptive optics, and two instrument platforms that can each accommodate multiple instruments.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastEditorUserId="7472" LastEditDate="2018-10-18T18:52:42.343" LastActivityDate="2018-10-18T18:52:42.343" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="17915" PostTypeId="4" CreationDate="2016-08-13T17:05:00.373" Score="0" Body="Questions about the upcoming European Extremely Large Telescope." OwnerUserId="13681" LastEditorUserId="13681" LastEditDate="2016-08-14T03:03:15.020" LastActivityDate="2016-08-14T03:03:15.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17916" PostTypeId="5" CreationDate="2016-08-13T17:07:06.913" Score="0" Body="&lt;p&gt;Use this tag for questions about the &lt;a href=&quot;http://www.iau.org/about/&quot; rel=&quot;nofollow&quot;&gt;International Astronomical Union&lt;/a&gt; (IAU). &lt;/p&gt;&#xA;" OwnerUserId="9533" LastEditorUserId="9533" LastEditDate="2016-08-24T20:56:16.230" LastActivityDate="2016-08-24T20:56:16.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17917" PostTypeId="4" CreationDate="2016-08-13T17:07:06.913" Score="0" Body="Questions about the International Astronomical Union" OwnerUserId="13681" LastEditorUserId="9533" LastEditDate="2016-08-24T20:56:25.817" LastActivityDate="2016-08-24T20:56:25.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17918" PostTypeId="1" CreationDate="2016-08-13T19:09:26.327" Score="10" ViewCount="1368" Body="&lt;p&gt;As I understand, neutron stars are born as extremely bright, extremely fast spinning cores of stars dying in a supernova. However, several websites tell me that within a course of a few years, the surface temperature of a neutron star falls from several &lt;em&gt;trillion&lt;/em&gt; kelvin to &lt;em&gt;only&lt;/em&gt; a few million kelvins. Furthermore, with the passage of time, the spinning speed of the neutron star also decreases considerably.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This raises the question: what is the final destiny of a neutron star? Does it always stay so horribly magnetic, hot and fast spinning or does it keep degrading into some form of cold, extremely dense star core with a much weaker magnetic field or do some of its features (specially the magnetic field strength and spin) stay at hightened levels forever (or at least several hundreds of billions of years)?&lt;/p&gt;&#xA;" OwnerUserId="11657" LastEditorUserId="31410" LastEditDate="2021-04-21T17:57:52.147" LastActivityDate="2021-04-21T17:57:52.147" Title="What is the final destiny of a neutron star?" Tags="&lt;stellar-evolution&gt;&lt;neutron-star&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="17920" PostTypeId="2" ParentId="17918" CreationDate="2016-08-13T22:34:31.407" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;This raises the question: what is the final destiny of a neutron star?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Neutron stars cannot stay hot forever. Neutron stars cool because they radiate. (This is called radiational cooling.) Except for their gravitational field which distorts spacetime in the vicinity of a neutron star, most lone neutron stars slowly fade away over time, eventually becoming essentially invisible. One way of detecting those cold, lone neutron stars is to observe the gravitational lensing of stars behind them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regard to magnetic field and rotation, those too drop over time. A neutron star's rotation is what creates the magnetic field, but this magnetic field drains the rotation rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternate fate for neutron stars is to undergo gravitational collapse and form a black hole. This can happen in a number of ways. A massive neutron star can undergo collapse as a result of its slowing rotation rate. The initial rapid rotation staves off gravitational collapse, but that no longer works when the neutron star's rotation rate drops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some neutron stars are not isolated. They are instead members of multiple star systems. Neutron stars can draw material from a partner star and eventually become massive enough to undergo collapse. Finally, a few neutron stars orbit one another closely. The discover of this, the Hulse-Taylor binary, led to the 1993 Nobel Prize in physics. Those closely orbiting neutron stars emit gravitational waves, thereby causing the orbit to decay. Those neutron stars eventually collide, once again resulting in a gravitational collapse.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2016-08-15T11:32:46.120" LastActivityDate="2016-08-15T11:32:46.120" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17921" PostTypeId="1" AcceptedAnswerId="17925" CreationDate="2016-08-14T03:08:29.400" Score="1" ViewCount="41" Body="&lt;p&gt;I was thinking about the proposed ESA mission &lt;a href=&quot;https://www.elisascience.org/articles/elisa-mission/elisa-mission-gravitational-universe&quot; rel=&quot;nofollow&quot;&gt;eLISA&lt;/a&gt;, which is essentially a space version of the LIGO experiment, and it occurred to me that trailing three satellites behind the Earth's orbit seemed excessively complicated in comparison to using a reflector on the ground and another on the moon (if I am not mistaken, there is even already a mirror on the moon, from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment&quot; rel=&quot;nofollow&quot;&gt;Lunar Laser Ranging experiment&lt;/a&gt;) and a single satellite. Is there not a geosynchronous orbit with a line of sight to the moon at all times? It seems to me the added range would provide better sensitivity. Provided you know the exact distance to the reflector, you can calculate individual frequencies for the Earth and Moon lasers, downconvert the faster frequency through a PLL of sorts, and still use the resulting data in an interferometry analysis. Is there something major I am missing here that would make this arrangement more difficult than the proposed eLISA mission? It seems to shift a lot of the physical risk and cost of the mission to computational costs, which by comparison are much, much cheaper.&lt;/p&gt;&#xA;" OwnerUserId="11640" LastActivityDate="2016-08-14T10:16:33.127" Title="Giant variation of the proposed eLISA mission using reflectors on Earth and the Moon possible?" Tags="&lt;gravitational-waves&gt;&lt;space-telescope&gt;&lt;interferometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17922" PostTypeId="1" AcceptedAnswerId="18020" CreationDate="2016-08-14T05:00:53.543" Score="6" ViewCount="187" Body="&lt;p&gt;Given that the gravitational wave detected by LIGO was a very weak echo of a very distant event, could it have been &quot;deviated&quot; and distorted on its way here by a sufficiently massive black hole in between?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would a black hole, since it traps even light, and since gravitational waves travel at the speed of light, absorb some portion of the gravitational wave and leave a &quot;shadow&quot; in its propagation such that if a black hole was between your instrument and an incoming gravitational wave you would have no way to detect it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so wouldn't the mass distribution in the Universe look less homogeneous at larger scales (as observed), since gravitational waves would not deform space uniformly and would instead need to &quot;equalize&quot; by forming voids like the &lt;a href=&quot;https://en.wikipedia.org/wiki/Bo%C3%B6tes_void&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boötes void&lt;/a&gt;, and filament structures like the &lt;a href=&quot;https://en.wikipedia.org/wiki/Pisces%E2%80%93Cetus_Supercluster_Complex&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pisces–Cetus Supercluster Complex&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/Sloan_Great_Wall&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sloan Great Wall&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, how could a gravitational wave from a distant event possibly affect the gravity well of a black hole closer to the observer? It doesn't seem possible to me that the attenuated ripple in space time from an event so far away could pass right through the gravity well of a closer local black hole like a galactic supermassive black hole.&lt;/p&gt;&#xA;" OwnerUserId="11640" LastEditorUserId="2531" LastEditDate="2019-12-27T08:12:14.430" LastActivityDate="2019-12-27T08:12:14.430" Title="Would gravitational waves be subject to external gravitational perturbations?" Tags="&lt;black-hole&gt;&lt;mass&gt;&lt;gravitational-waves&gt;&lt;galaxy-cluster&gt;&lt;gravitational-lensing&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="17924" PostTypeId="1" AcceptedAnswerId="17926" CreationDate="2016-08-14T06:15:01.260" Score="1" ViewCount="285" Body="&lt;p&gt;Do we know of a celestial body which doesn't spin at all on its own axis? If not, why is it so?&lt;/p&gt;&#xA;" OwnerDisplayName="user13628" LastActivityDate="2020-02-14T13:50:08.800" Title="Are there bodies in the Universe that are not spinning?" Tags="&lt;orbit&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17925" PostTypeId="2" ParentId="17921" CreationDate="2016-08-14T10:16:33.127" Score="2" Body="&lt;p&gt;This would not be a giant version of eLISA, but a small (and complicated) one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance between the LISA satellites will be 1 million kilometers, while the moon is on average 380.000km from Earth. So that arm would be less than half as long. Geosynchronous orbit is at 36.000km, not even 10% of the distance to the moon. The difference in length between your proposed detector arms is much shorter than the &lt;a href=&quot;https://en.wikipedia.org/wiki/Coherence_length#Lasers&quot; rel=&quot;nofollow&quot;&gt;coherence length&lt;/a&gt; of any laser, so your interferometer would not work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further problems:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Positioning of mirror on Earth: Where do you want to put the mirror such that the moon is always visible in the sky?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Positioning of satellite: I doubt (without checking) that you can see the moon at all times from geosynchronous orbit. Maybe you could work with one of the Earh-Moon Lagrange points.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Constant mirror adjustment: Earth is rotating, and the Moon revolves around the Earth. So your interferometer arms would change angles constantly. (And you'd need to dig lots of tunnels straight through Earth, see above.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Vibration: The mirrors have to be as free from vibration as possible. That is much easier to achieve with a free floating satellite than on earth or the moon. (Did you know there are &lt;a href=&quot;https://en.wikipedia.org/wiki/Quake_(natural_phenomenon)#Moonquake&quot; rel=&quot;nofollow&quot;&gt;moon quakes&lt;/a&gt;?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ellipticity: The distance to the moon varies between 360.000km and just over 400.000km during its orbit. So you'd be constantly adjusting the interferometer. A constant drift is not necessarily problematic, but the huge difference in distance might be.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Lastly, it is not clear to me how you want to replace precision measurement by calculation. What is PLL? How do you get meaningful interference patterns with lasers of different wavelengths? Also note that the &quot;exact distance&quot;, if you want to count waves, must be better than half the wavelength of your laser, so a few hundred nanometers. At the distance of the moon, that is better than $1 : 10^{15}$ (hope that calculation is correct).&lt;/p&gt;&#xA;" OwnerUserId="13681" LastActivityDate="2016-08-14T10:16:33.127" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17926" PostTypeId="2" ParentId="17924" CreationDate="2016-08-14T13:18:36.880" Score="-4" Body="&lt;p&gt;Every object in the universe--from a basic star to an exotic black hole--spins, and the origin of that spin can be traced back to the very beginning of time. Within instants after the Big Bang, the primordial fireball of energy expanded at an incredible speed, then later cooled and solidified into all the matter in the universe. Had this fireball been uniform in all directions, everything we see today would be completely homogeneous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.scientificamerican.com/article/what-causes-objects-such/&quot; rel=&quot;nofollow&quot;&gt;http://www.scientificamerican.com/article/what-causes-objects-such/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10445" LastActivityDate="2016-08-14T13:18:36.880" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17927" PostTypeId="1" CreationDate="2016-08-14T17:51:50.223" Score="2" ViewCount="98" Body="&lt;p&gt;My high school teacher and I recently built a 10 ft. mesh dish transit radio telescope. In order to compete this year at Nationals for our science fair, we came up with the idea of using a bluetooth transmitter and receiver to transmit the radiowave's received from the telescope to the lab inside where my computer is located. I will be using MATLAB to take in the voltage input, analyze the radio frequencies, and create a 3D plot/image of the object were receiving from. The difference in our scope vs's other radio telescopes is its ability to rotate 360 degrees repeatedly to image the sky, taking in radio waves at different angles of ascension. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where I'm stuck is the 3D plotting of the images. What links/articles could you think of to assist with the mathematical transformation of these radio signals into rough images. Considering the meridian circle as our basis of rotation, will the repeated persay imaging of the sky allow for an increased arcsecond of resolution due to the fact the arrays of data from each rotation will often overlap? If needed, I can draw and attach a rough image of what the radio telescope looks like. But the dish just rotates repeatedly about the horizontal axis, rather than a back and forth transit telescope, say 30 to 150 degrees and back from 150 to 30 degrees again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What facts about astronomy, laws/theories, or anything else do you think would assist me in the creation of these images. I want to write the program from scratch, but if there is any good libraries out there to possibly write it in another language, that'd be nice to know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if my question is too broad and kinda ambiguous. If needed I can narrow it down to a single question, but would rather not be closed due to its ambiguity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks again, and I truly appreciate any assistance, whether it's merely a URL or an in depth description, anything helps!&lt;/p&gt;&#xA;" OwnerUserId="13808" LastActivityDate="2016-08-14T17:51:50.223" Title="Help With Transit Radio Telescope Data Analysis" Tags="&lt;telescope&gt;&lt;astrophysics&gt;&lt;radio-astronomy&gt;&lt;radio-telescope&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17930" PostTypeId="1" AcceptedAnswerId="17931" CreationDate="2016-08-15T03:32:38.930" Score="1" ViewCount="231" Body="&lt;p&gt;I know we can find meteorites from space in Antarctica because snow is white, and we can find them easily on snow. Is it true that meteorites can also be found elsewhere such as beaches, streets or even embedded in a concrete building. It is just very hard to distinguish them from plain rock on earth?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="2365" LastEditDate="2016-08-15T07:27:14.250" LastActivityDate="2016-08-15T08:14:53.103" Title="Are meteorites found everywhere?" Tags="&lt;asteroids&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17931" PostTypeId="2" ParentId="17930" CreationDate="2016-08-15T06:32:48.403" Score="4" Body="&lt;p&gt;Yes you can find them anywhere on Earth. The thing is that by the time they reach the Earth's surface, these meteorites are very small, usually ranging from microscopic to dust sized particles. Searching &lt;a href=&quot;http://science.howstuffworks.com/question486.htm&quot; rel=&quot;nofollow&quot;&gt;How big does a meteor have to be to make it to the ground?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In this sense, most all meteoroids that enter the atmosphere make it&#xA;  to the ground, in the form of microscopic dust.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you said it is easier to find meteorites in snow because it is white and contrasts with the usual colour of meteorites. It would be very hard to try and distinguish those small particles with the normal soil on Earth. As for finding one randomly in a concrete building... the chances are very low and it possibly would have done some noticeable damage to the building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rather simple experiment to prove that yes you can find it anywhere on earth, &lt;a href=&quot;http://io9.gizmodo.com/5984951/how-to-collect-meteorites-in-your-backyard&quot; rel=&quot;nofollow&quot;&gt;follow this example.&lt;/a&gt; This is to search for micrometeorites. I have done this before and I found it to be particularly inefficient as it is rather wind in my area, but hey you might find it different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can attempt to search for meteorites somewhere else but just don't expect to see large chunks of it lying around.&lt;/p&gt;&#xA;" OwnerUserId="7926" LastActivityDate="2016-08-15T06:32:48.403" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17932" PostTypeId="2" ParentId="17930" CreationDate="2016-08-15T08:14:53.103" Score="1" Body="&lt;p&gt;Antarctica is a &lt;a href=&quot;http://mineralsciences.si.edu/research/meteorites/antarctica/&quot; rel=&quot;nofollow&quot;&gt;good place for finding meteorites&lt;/a&gt; because it is undisturbed, and meteorites are the only rocks there. A meteorite that falls on the central ice fields gets covered with snow, and is carried by the ice towards the sea. They get exposed as the ice is eroded. The flowing of the ice concentrates meteorites in particular areas, and finally, as you note they are highly visible against the snow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meteorites are also found in other undisturbed locations, such as deserts. In the sahara, meteorite hunting has been a &lt;a href=&quot;http://www.galactic-stone.com/pages/sahara&quot; rel=&quot;nofollow&quot;&gt;profitable activity&lt;/a&gt;. These are not normally recent falls, but people are finding rocks that have fallen over thousands of years. It takes some skill to identify a meteorite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In build up areas the ground is disturbed and meteorites get lost. Meteors fall everywhere, but if a meteorite had fallen 1000 years ago near my home, it would be covered in undergrowth, covered by soil, ploughed over and then buried by houses. It is unlikely to be found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A meteorite that hits a concrete building won't embed in it. Meteorites are travelling at terminal velocity when they reach the ground, which is about 100mph: not that fast. The meteorite may shatter, or bounce off. It won't leave a classic crater as you see on the moon. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-08-15T08:14:53.103" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17933" PostTypeId="2" ParentId="101" CreationDate="2016-08-15T08:44:57.217" Score="2" Body="&lt;p&gt;Site quality is typically measured and stated in the following way.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Fraction of photometric nights (really no clouds/cirrus) per year and/or the fraction of night where spectroscopic observations can be done (some clouds, thin cirrus). The best sites are up at the 60% and 80% level for these. You would also include time lost due to high winds&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The median astronomical seeing and its distribution from night to night - the full width half maximum of astronomical images. At the best sites this is around 0.5 arcsec on the best nights.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Median extinction. The amount the atmosphere absorbs for a star at zenith. Typically quoted in the V-band and just below 0.1 mag at the best sites.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Brightness of the dark, moonless sky in mag per square arcsecond. Often quoted in the V or B bands. A value of 21.5-22 mag/sqarcsec is a good site.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are some other numbers to do with adaptive optics correction or observing in the infrared that I might add later, but the above are the main ones.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-08-15T08:44:57.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17934" PostTypeId="2" ParentId="17918" CreationDate="2016-08-15T11:36:02.357" Score="8" Body="&lt;p&gt;Neutron stars have extremely small heat capacities. That is because they consist largely of degenerate fermions and the heat capacity is further suppressed if, as expected, those fermions are in a superfluid state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has (at least) two consequences: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) they cool down extremely rapidly - neutrino emission processes are highly effective, in the first $10^5$ years or so of a neutron star's life, at reducing its interior temperature to a few $10^7$ K and the surface temperature to $&amp;lt;10^6$ K. After that, the dominant cooling process is photons emitted from the surface ($\propto T^4$) and neutron stars rapidly fade from view thereafter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) However, the low heat capacity also means that it is easy to keep a neutron star hot if you have any way of adding energy to it - such as viscous dissipation of rotation by friction, accretion from the interstellar medium or ohmic heating by magnetic fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No isolated neutron star surfaces have been measured with temperatures much below $10^6$ K - i.e. all observed isolated neutron stars are at young ages. The situation is summarised in section 5.7 of &lt;a href=&quot;https://arxiv.org/pdf/astro-ph/0402143v1.pdf&quot; rel=&quot;noreferrer&quot;&gt;Yakovlev &amp;amp; Pethick (2004)&lt;/a&gt;. Without any reheating, a neutron star would reach 100K in only a billion years - this is already utterly invisible. The reheating mechanisms must play &lt;em&gt;some&lt;/em&gt; role for older neutron stars, but as Yakovlev &amp;amp; Pethick state:  &quot;Unfortunately, no reliable observational data on the thermal states of such stars are available&quot;. In conclusion, nobody really knows at the moment what the long-term ($&amp;gt;10^6$ years) fate of neutron stars is in terms of their temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation with regard to spin and magnetic field is more secure. There are not the same mechanisms available to spin-up an &lt;em&gt;isolated&lt;/em&gt; neutron star or regenerate their magnetic fields. Both are expected to decay with time and indeed the spin-down rate and magnetic field strength are intimately connected, because the spin-down mechanism is the emission of magnetic dipole radiation. The magnetic field decays through the generation of currents that then ohmically dissipate (providing a source of heat) or perhaps more rapidly via currents generated by the Hall effect or through ambipolar diffusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For pure magnetic dipole radiation, one predicts $\dot{\Omega} \propto \Omega^3$. For typical surface magnetic field strengths of $10^8$ T, pulsars spin down to periods of around a few seconds in less than a million years, at which point the &quot;pulsar activity&quot; switches off and we can't see them any more, unless they are in binary systems and accreting matter in order to spin them up again. Unfortunately, there is very little observational evidence to pin down how fast magnetic fields decay (because we don't see old, isolated neutron stars!). The decay of B-field cannot be &lt;em&gt;very&lt;/em&gt; fast, certainly timescales are longer than $10^5$ years. Theoretical estimates of B-field decay timescales are more like billions of years. If this theory is right then neutron stars would continue to spin down very rapidly even after the pulsar mechanism has ceased.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-08-15T16:59:30.680" LastActivityDate="2016-08-15T16:59:30.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17935" PostTypeId="1" CreationDate="2016-08-15T13:30:59.287" Score="2" ViewCount="468" Body="&lt;p&gt;I'm writing a script to generate a star chart for a given observation point (latitude and longitude) at a given time.  I've found a database of stars with their right ascension and declination from J2000, and with that written a program that displays a star chart with the center being any right ascension and declination that I feed into it.  I've also written a program that will convert local time into local sidereal time. Now I just need to figure out how to determine the zenith for an observer in J2000 right ascension and declination given their longitude, latitude, and local sidereal time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found a wikipedia page that gives formulas for converting between celestial coordinate systems, but the formula for converting equatorial and horizontal coordinates doesn't seem to depend on the sidereal time, which I imagine it should for my case.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried finding a vector that represents the observer's zenith at the given time, then rotating it about the earth's rotation axis by the sidereal time, but that's giving incorrect results.&lt;/p&gt;&#xA;" OwnerUserId="13816" LastActivityDate="2016-08-15T15:43:01.283" Title="How can I map the zenith for an observer at a given time to equatorial coordinates (as would be measured in J2000)?" Tags="&lt;star&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17936" PostTypeId="1" CreationDate="2016-08-15T14:16:59.927" Score="3" ViewCount="278" Body="&lt;p&gt;I would like to build a gravity model (with 3D graphics) that would build the orbit of the planets based on orbital elements input and the gravity calculation for each element. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://daed-dev.on.br/~csiqueira/on-physics/src/examples/framework-example-kepler.html&quot; rel=&quot;nofollow&quot;&gt;My initial code is live here (Click the 0.005 to turn it into 0.05 and then click on restart time to run the orbits).&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example above has all planets on Solar System (up to Neptune), the Sun and the asteroid Apophis. For clarity, the Sun and planets other than earth have been hidden, but are still considered for gravity calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the orbits do not match the correct estimate intersection date. &lt;a href=&quot;http://neo.jpl.nasa.gov/apophis/&quot; rel=&quot;nofollow&quot;&gt;In this article&lt;/a&gt;, it says that Earth-Apophis close-encounters will happen on 13 April (meaning that the earth should be in the orbit's intersection point every 13 April, or very close to it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data from the program is taken from Nasa JPL.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ssd.jpl.nasa.gov/sbdb.cgi?sstr=99942;orb=1&quot; rel=&quot;nofollow&quot;&gt;Apophis Data&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ssd.jpl.nasa.gov/txt/aprx_pos_planets.pdf&quot; rel=&quot;nofollow&quot;&gt;Planets Data&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;The sun is positioned at the origin&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Start date is 2454441.5, as given in JPL Apophis data. Planets orbital elements have been corrected to this date, with the calculations in the above PDF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why the orbits intersection date is wrong? Should I be doing something else? Am I doing something wrong?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The above Apophis link also cites a &quot;Standard Dynamical Model&quot;. If anyone knows how to correctly create one, I believe this would answer my question as well.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TLDR: How can I correctly create planetary orbits from kepler elements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance if anyone can help me!&lt;/p&gt;&#xA;" OwnerUserId="2790" LastActivityDate="2016-08-15T20:34:32.127" Title="How to correctly create planetary orbits from kepler elements" Tags="&lt;gravity&gt;&lt;orbital-elements&gt;&lt;software&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17937" PostTypeId="1" AcceptedAnswerId="17941" CreationDate="2016-08-15T14:39:02.830" Score="2" ViewCount="110" Body="&lt;p&gt;A book I recently purchased describes Dirac's proposed variation of Newton's G, and why that specific proposal wouldn't work because the earth would have been too hot for the known history of life. A footnote explains that the luminosity of the sun is proportional to the seventh power of G, which seems incredible to me. I know that this has to be a simplification, since stellar dynamics are annoyingly complex, but if we put that aside, is this essentially correct, or is it a typo? What would be the theoretical justification for this in broad terms? (eg, &quot;Gravitational pressure is proportional to this exponent of G, but since that also shrinks the size of the star, which increases the pressure, it's actually this exponent of G, and the rate of fusion is proportional to the gravitational pressure squared, so you end up with 7&quot;).&lt;/p&gt;&#xA;" OwnerUserId="1311" LastEditorUserId="9301" LastEditDate="2016-08-26T17:21:37.123" LastActivityDate="2016-08-26T17:21:37.123" Title="Solar luminosity proportional to $G^7$?" Tags="&lt;luminosity&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17938" PostTypeId="2" ParentId="17935" CreationDate="2016-08-15T15:43:01.283" Score="2" Body="&lt;p&gt;Zenith right ascension = sidereal time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zenith declination = geographic latitude.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-08-15T15:43:01.283" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17939" PostTypeId="1" AcceptedAnswerId="17940" CreationDate="2016-08-15T16:17:10.280" Score="20" ViewCount="2519" Body="&lt;p&gt;I was bopping around YouTube and observed this &lt;a href=&quot;https://www.youtube.com/watch?v=8nHBGFKLHZQ&quot;&gt;enjoyably produced video&lt;/a&gt;. In it, when describing the behavior of a black hole with the mass of a US nickel, the narrator says, &quot;Its 5 grams of mass will be converted to 450 terajoules of energy, which will lead to an explosion roughly three times bigger than the bombs dropped on Hiroshima and Nagasaki combined.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of all the fun things illustrated there, that was the one whose pretense I didn't understand. Do black holes explode after they've radiated away all of their mass? Or would the &quot;explosion&quot; just come from the rapid pace at which the black hole would consume nearby matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Googling I've done thus far has provided no firm answer. The closest I've come is from the Wikipedia page on &lt;a href=&quot;https://en.wikipedia.org/wiki/Hawking_radiation&quot;&gt;Hawking Radiation&lt;/a&gt;, which states, &quot;For a black hole of one solar mass, we get an evaporation time of 2.098 × 10^67 years—much longer than the current age of the universe at 13.799 ± 0.021 x 10^9 years. But for a black hole of 10^11 kg, the evaporation time is 2.667 billion years. This is why some astronomers are searching for signs of exploding primordial black holes.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some other websites refer to the last &quot;explosion&quot; of the Milky Way's supermassive black hole being some 2 million years ago, but is that the same mechanic mentioned on the Wikipedia page? Or the YouTube video, for that matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance. =)&lt;/p&gt;&#xA;" OwnerUserId="13817" LastEditorUserId="13817" LastEditDate="2016-08-15T18:27:15.687" LastActivityDate="2016-11-02T00:04:41.667" Title="Explosions of black holes" Tags="&lt;black-hole&gt;&lt;explosion&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17940" PostTypeId="2" ParentId="17939" CreationDate="2016-08-15T16:37:05.473" Score="24" Body="&lt;p&gt;What this video is talking about is Hawking Radiation, as you've linked. Hawking Radiation is a proposed hypothetical (by no means verified or proven) way for a black hole to radiate its energy into space. The basic idea is that a black hole is nothing but mass/energy compressed to an infinitesimal point, which is radiating its energy into space over time. For large black holes (such as solar mass or bigger), this radiation process is tiny and the time taken to leak all the black hole's energy into space (and thus for the black hole to &quot;evaporate&quot;) is exceedingly long. For tiny black holes however, the time to radiate all the black hole's energy is exceedingly short.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can calculate how long it will take for a black hole of mass $m$ to evaporate (and release all its mass/energy) with the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t_{\text{ev}} = \frac{5120\pi G^2 m^3}{\hbar c^4} = (8.41\times 10^{-17}\:\mathrm{s}\:\mathrm{kg}^{-3})\:m^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $m = 5\:\mathrm{g}=0.005\:\mathrm{kg}$, you get $t_{\text{ev}} \simeq 4\times10^{-19}\:\mathrm{s}$. Now that means in this tiny amount of time, the black hole will radiate all of its mass/energy away and completely evaporate. But the output of all the energy in 5 g of mass is a huge output. Putting out 450 Terajoules of energy in $10^{-19}\:\mathrm{s}$ is basically just an explosion. You can determine the total energy output from the famous equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E=mc^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just plug in $m = 0.005\:\mathrm{kg}$ and $c = 3\times 10^8\:\mathrm{m/s}$ and you'll get $E=4.5\times 10^{14}\:\mathrm{J} = 450\:\mathrm{Terajoules}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in short, hypothetical calculations (not even theory at this point) suggest that a tiny black hole with the mass of a nickel would immediately explode out in a huge amount of energy. Whether such a black hole can form, or if such an evaporation would/could occur is still highly debated and ultimately unknown at this point.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="9301" LastEditDate="2016-11-02T00:04:41.667" LastActivityDate="2016-11-02T00:04:41.667" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17941" PostTypeId="2" ParentId="17937" CreationDate="2016-08-15T19:34:30.027" Score="6" Body="&lt;p&gt;I think the argument is the following. The central temperature can be estimated from a form of the virial theorem. At least dimensionally speaking,  total thermal energy $MkT/\mu$ (where $\mu$ is the average mass per particle) is proportional to gravitational potential energy $GM^2/R$. So $T \propto GM/R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for a given mass, standard polytropic theory for a star, says that &#xA;$$M \propto \rho_c^{(3-n)/2n} G^{-3/2}$$&#xA;$$R  \propto \rho_c^{(1-n)/2n} G^{-1/2}$$&#xA; where $n=3$ is an approximate polytropic index for the Sun and $\rho_c$ is the central density. See for example &lt;a href=&quot;https://astronomy.stackexchange.com/questions/16633/does-eddingtons-variable-polytropic-index-better-fit-data-from-the-standard-sol/16634#16634&quot;&gt;Does Eddington&amp;#39;s variable polytropic index better fit data from the Standard Solar Model?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting this together we have &#xA;$$ T \propto GMR^{-1} \propto G^2 \rho_c^{-1/3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power generation in the solar core from the pp cycle goes approximately as $L \propto \rho_c^2 T^4$ and using the proportionality for $T$ derived above, we have $L \propto G^8 \rho_c^{2/3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we note that for a $n=3$ polytrope, that $\rho _c$ and $M$ are independent, but that $\rho_c^{2/3} \propto G^{-1} R^{-2}$. So for a fixed solar radius we have $L \propto G^7$.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-08-16T06:36:23.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17942" PostTypeId="1" CreationDate="2016-08-15T19:37:02.723" Score="3" ViewCount="1068" Body="&lt;p&gt;I'm looking for a function that given a time (for example &lt;a href=&quot;https://en.wikipedia.org/wiki/Unix_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;Unix time&lt;/a&gt;) gives back the current distance between Mars and Earth in light-seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be a mathematical function or programming function (preferably in Python or pseudocode), but must not rely on an Internet connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is similar to &lt;a href=&quot;https://space.stackexchange.com/questions/17479/current-communication-delay-to-mars&quot;&gt;a question I asked erlier&lt;/a&gt;, but again I need to get the distance without an Internet connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sidenote: I'm planning on learning orbital mechanics, but I need this function earlier. :)&lt;/p&gt;&#xA;" OwnerUserId="13570" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2016-08-24T02:26:26.223" Title="Function for current distance between in light-seconds between Mars and Earth" Tags="&lt;distances&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17943" PostTypeId="2" ParentId="17936" CreationDate="2016-08-15T19:50:35.903" Score="2" Body="&lt;p&gt;If you're running your own N-body dynamics,&#xA;you can get initial Cartesian positions and velocities for each body in your system (including the Sun) by asking&#xA;&lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow&quot;&gt;JPL HORIZONS&lt;/a&gt;&#xA;for Ephemeris Type: Vectors.&#xA;I would choose Coordinate Origin: Solar System Barycenter so that the Sun does not drift away from the origin,&#xA;and Target Body: [Planet] Barycenter (or Sol) unless you are modeling the Earth and Moon as separate bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you still want to compute initial Cartesian positions and velocities from Keplerian elements, first do a sanity check: Earth should cross the X axis at the equinoxes and the Y axis at the solstices.&#xA;Second, those approximate elements for the major planets may not be precise enough for your purpose.&#xA;Instead you might ask HORIZONS for Ephemeris Type: Elements, and Time Span beginning with the epoch of your elements for Apophis.&#xA;You could also get elements for Apophis in a more convenient epoch this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that Keplerian elements are subject to dynamic evolution and are only valid for a limited time near their epoch.&#xA;If you ask HORIZONS for Apophis's orbital elements one week before and after its close approach to Earth, you can see that they are quite different. Of course the latter are also highly uncertain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the gory post-Newtonian details of the dynamical model the Apophis prediction article discussed extending, see JPL's&#xA;&lt;a href=&quot;http://ilrs.gsfc.nasa.gov/docs/2014/196C.pdf&quot; rel=&quot;nofollow&quot;&gt;notes on DE430 and DE431&lt;/a&gt;, part III.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2016-08-15T20:34:32.127" LastActivityDate="2016-08-15T20:34:32.127" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17944" PostTypeId="2" ParentId="17939" CreationDate="2016-08-15T20:49:07.197" Score="-1" Body="&lt;p&gt;No black hole can explode, no matter the size.  It can radiate energy outside its event horizon, but a proper &quot;explosion&quot; of energy from the black hole itself would have to exceed the speed of light, which is not currently possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if an object has the mass of a nickel, it is not a black hole.  More specifically, its event horizon is small enough to fit inside its own surface area which is true for all massive objects that are not black holes.  In fact, the event horizon of something with the mass of a nickel is literally immeasurably small (although NOT non-existent).  Nickels (and other objects of similar mass) do not have enough gravity to form a black hole.  I suspect perhaps you are confusing &quot;mass&quot; with &quot;size&quot;...?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if a process existed that could turn the mass of a nickel into the energy it embodies (per E=mc^2), then yes, it would be a tremendous amount of energy.  The only such process known to exist would be to have an object with half the mass of a nickel interact (touch) an antimatter-object with half the mass of a nickel.  The combined mass of the 2 objects would total the mass of the nickel, and would release the appropriate amount of pure energy with no mass leftover. &lt;/p&gt;&#xA;" OwnerUserId="13823" LastActivityDate="2016-08-15T20:49:07.197" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17945" PostTypeId="2" ParentId="17942" CreationDate="2016-08-15T21:50:45.470" Score="4" Body="&lt;p&gt;There is no simple function that is accurate. Both Earth and Mars are orbiting in ellipses (Kepler's first law). They are travelling at varying speeds (Kepler's second). To find the current distance between them you have to calculate their positions in three dimensions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words you have to do orbital kinematics. &lt;a href=&quot;https://astronomy.stackexchange.com/questions/8012/how-do-i-calculate-the-positions-of-objects-in-orbit&quot;&gt;How do I calculate the positions of objects in orbit?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If accuracy is not important. You can approximate the orbits with circles in the same plane. The Earths position is then&#xA;$$(\cos(2\pi t + \alpha), \sin(2\pi t +\alpha))$$ &#xA;and Mars is &#xA;$$(1.5\cos(2\pi t/1.882 +\beta), 1.5\sin(2\pi t/1.882 + \beta))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance between them is calculated by pythagoras&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distances are in AU and the time is in years (converting to lightseconds is then easy). $\alpha$ and $\beta$ are constants chosen to fit the positions of the planets at time 0. The values 1.5 and 1.882 are chosen because Mars's orbit is about 1.5 times bigger than Earth, and a Mars year is 1.882 times longer than Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The circular approximation however is poor for Mars, and this ignores the fact that Mars isn't in exactly the same orbital plane as the earth, so the values will only be rather approximate. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-08-15T21:50:45.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17946" PostTypeId="2" ParentId="17942" CreationDate="2016-08-16T02:34:03.720" Score="4" Body="&lt;p&gt;&lt;strong&gt;note:&lt;/strong&gt; This is an &quot;extended comment&quot; type answer, addressing two potential solutions that may not be exactly what you need, but are close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A thing which gives you positions of things in the solar system given a specified time is called an &lt;a href=&quot;https://en.wikipedia.org/wiki/Ephemeris&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ephemeris&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Originally they were tables of predictions (past and future) based on calculations. By the time of Newton, they became quite good, since Newton developed a method to iteratively solve Kepler's equation (e.g. &lt;em&gt;Newton's Method&lt;/em&gt;). See the question &lt;a href=&quot;https://hsm.stackexchange.com/q/3575/3869&quot;&gt;How did Newton and Kepler (actually) do it?&lt;/a&gt; and the answers there for a little more background.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The Python package &lt;a href=&quot;http://rhodesmill.org/pyephem/&quot; rel=&quot;nofollow noreferrer&quot;&gt;PyEphem&lt;/a&gt; has been around and well supported, and is the pythonic reincarnation of &lt;a href=&quot;http://www.clearskyinstitute.com/xephem/&quot; rel=&quot;nofollow noreferrer&quot;&gt;XEphem&lt;/a&gt;. I haven't used it, but I believe it keeps enough information about orbital parameters at certain epochs to generate an ephemeris, including some gravitational perturbations. In other words, it's much more than planets moving on fixed elliptical orbits around a fixed sun. So I believe it runs without internet connection.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I never used it because I was recommended to look at &lt;a href=&quot;http://rhodesmill.org/skyfield/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Skyfield&lt;/a&gt; and it's exactly what I needed. It downloads a standard JPL ephemeris that you choose, and then just uses it from your hard drive after that. However, in order to deal with leap seconds and other &lt;em&gt;time related&lt;/em&gt; effects, it &lt;em&gt;occasionally&lt;/em&gt; needs to check the internet for leap second information updates, since these are arbitrary.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I don't know if Skyfield has a mode to avoid that. Actually that's a good question. If you work with a timescale that doesn't have leap seconds, I am not sure if it will run in its current version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both Skyfield and PyEphem Python packages have been written and are maintained by @BrandonRhodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can allow an occasional connection to the internet (say every month or few months) then Skyfield is &lt;em&gt;extremely easy and pythonic&lt;/em&gt; to use. For example here is a script I used &lt;a href=&quot;https://space.stackexchange.com/a/16986/12102&quot;&gt;in this answer&lt;/a&gt;. If you want to convert time formats from system time, you can search my questions in stackoverflow and space exploration. If you want it for your local computers system time now, you just use &lt;code&gt;t = ts.now()&lt;/code&gt; directly instead of utc.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;&#xA;from skyfield.api import load&#xA;&#xA;data   = load('de421.bsp')&#xA;ts     = load.timescale()&#xA;t      = ts.utc(2016, 7, 5, 9, 50, 0)&#xA;&#xA;jupiter, earth  = data['Jupiter barycenter'], data['Earth']&#xA;jpos, epos      = jupiter.at(t).position.km, earth.at(t).position.km&#xA;d_instantaneous = np.sqrt(((jpos - epos)**2).sum())&#xA;&#xA;d_light = earth.at(t).observe(jupiter).distance().km  # where WAS Jupiter 48 minutes ago?&#xA;&#xA;clight = 299792.458  # km/s&#xA;&#xA;print &quot;d_instantaneous / c = &quot;, d_instantaneous/clight&#xA;print &quot;d_light / c =         &quot;, d_light/clight&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2016-08-24T02:26:26.223" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17947" PostTypeId="1" CreationDate="2016-08-16T05:59:48.187" Score="3" ViewCount="630" Body="&lt;p&gt;As we know, asteroids can be found in the Kuiper belt and Oort cloud, but surely the Sun only captures them, not produces them. I know that supernovae may produce heavy elements such as oxygen, silicon and iron, which is the common composition of asteroids, so is it true that asteroids are come from star remnants?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="9301" LastEditDate="2016-08-16T18:21:50.010" LastActivityDate="2018-08-23T08:49:49.070" Title="Do asteroids originally come from debris of star remnants?" Tags="&lt;star&gt;&lt;asteroids&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17948" PostTypeId="2" ParentId="17947" CreationDate="2016-08-16T08:30:00.627" Score="4" Body="&lt;p&gt;The asteroids (which are mostly found in the asteroid belt between Mars and Jupiter) and more distant, icy bodies found in the Kuiper belt and the Oort cloud were formed at the same time as the rest of the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun and the planets formed from a molecular cloud, containing mostly hydrogen and helium, but also some other elements that came from the remains of stars, including supernovae. This gas cloud collapsed under gravitation to form a star, surrounded by a disk of dust and gas. The matter in the disc formed the planets, and the Kuiper belt. More distant matter in an extended halo around the star, formed the the Oort cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the sun neither captures nor produces asteroids and other solar system bodies. They were formed at the same time as the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note &quot;Kuiper&quot; not &quot;Kepler&quot; belt, and the Oort cloud is the source of comets, not asteroids. Asteroids are rocky; Oort cloud bodies are made of ices.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-08-16T08:30:00.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17949" PostTypeId="2" ParentId="17947" CreationDate="2016-08-16T11:59:11.353" Score="2" Body="&lt;p&gt;Astrophysicists classify stars into 3 &quot;populations&quot;. Population (Pop) III is the oldest (first &lt;em&gt;generation&lt;/em&gt;) and is the one which formed from the elements produced in the Big Bang (H &amp;amp; He (and traces of others notably Li)). Pop II stars formed from the molecular clouds produced by Pop III, and our Sun, a Pop I star, was produced in a molecular cloud from the Pop II and Pop III stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's still a lot we don't know but we believe many Pop III stars were much larger than what is common today.  Most of the heavier elements are &lt;em&gt;not&lt;/em&gt; formed during a star's main sequence life-time, but are formed during eruptions and explosions which occur after a star has left the main sequence (during its &quot;death&quot; or old age). The length of a star's main sequence life-time is very sensitive to its mass; these Pop III stars lived fast and died young, spewing a lot of the metallic elements away into space. We have evidence that at least two or three supernovae contributed to the molecular cloud that the Sun formed in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun probably formed in a globular cluster (also known as a &quot;stellar nursery&quot;) and the evolution of these is actively being studied. We believe a small part of our parent molecular cloud (maybe 1 thousandth or less) collapsed into our Solar System. We believe that our Solar System probably has hundreds or thousands of sibling stars, but only a few candidate &quot;relatives&quot; have been suggested/identified so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the cloud collapsed due to it having a bit of excess mass density (compared to the cloud's average density), possibly due to stellar winds and supernovae shock waves compressing parts of the cloud, our protostellar disk formed. The heavier elements, including dust, due to the condensation, began to collide and form larger particles, these had increased gravity as well as larger size and tended to grow in size. Some became protoplanets, most never grew so large. Some of the protoplanets collided and broke apart, some of them coalesced into planets. What's left over from this is asteroids and comets and of course a lot of dust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more volatile elements and the smaller particles tend to be driven away from the Sun, both due to radiation pressure as well as due to the Solar Wind. The Asteroid Belt is believed to be the result of the failure of protoplanetary material to coalesce, because of the way Jupiter disturbed that region of space, but our study of the thousands of exo-planets is challenging some of what we believed were our most likely models of stellar system development and the story is far from settled.&lt;/p&gt;&#xA;" OwnerUserId="13833" LastEditorUserId="11264" LastEditDate="2018-08-23T08:49:49.070" LastActivityDate="2018-08-23T08:49:49.070" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="17951" PostTypeId="1" AcceptedAnswerId="17952" CreationDate="2016-08-16T17:15:30.457" Score="2" ViewCount="775" Body="&lt;p&gt;Hard science fiction writing question.  Premise: what natural resources are present in the Jovian system that would be necessary and desireable to an autonomous, self-sustaining human community?  Water on Europa, for example.  What else? Would like to keep this plausible without a lot of speculation.  Thank you. &lt;/p&gt;&#xA;" OwnerUserId="13836" LastActivityDate="2018-01-09T01:14:29.540" Title="What are the natural life sustaining resources available in the Jovian moon system?" Tags="&lt;jupiter&gt;&lt;planetary-systems&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17952" PostTypeId="2" ParentId="17951" CreationDate="2016-08-16T20:24:25.713" Score="2" Body="&lt;p&gt;The moons Europa, Ganymede, and Callisto all have substantial amounts of water ice on their surfaces, and for the latter two, also silicate rocks. There is also some CO2 in the ice as well as ammonia and methane. These give potential sources of carbon and nitrogen, and with enough energy you could make glucose and amino acids - Food.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Io is a crazy place, with lava flows, and sulphur covering the surface. Its not an easy place to mine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One potential source of energy is ${}^3\mathrm{He}$ This is a potentially useful fuel for a fusion reactor. It is found (in very low concentrations) in shaded parts of the moon. It may also be found on Jupiter's moons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minor moon Carme may have originated in the Kuiper belt (long ago) and have an interesting combination of carbon based substances. Amalthea has an interesting red colour - maybe carbon compounds again. There are quite a few moons that began as captured asteroids, but none seem to have been metallic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of problems with life on Jupiter's moons: Cold - No air - radiation - etc&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-08-16T20:24:25.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17953" PostTypeId="2" ParentId="16781" CreationDate="2016-08-16T20:33:54.173" Score="2" Body="&lt;p&gt;A significant portion of SETI's work is done through a program known as SETI@home, which takes signals that have been received and processes them for responses, such as recurring patterns. &lt;a href=&quot;http://setiathome.berkeley.edu/sah_porting.php&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a page on how to download its source code. I grabbed the most recent nightly tarball and it turns out the program is coded in C++. I offer this as evidence that C++ is the most commonly used programming language used to process SETI's workload.&lt;/p&gt;&#xA;" OwnerUserId="69" LastActivityDate="2016-08-16T20:33:54.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17955" PostTypeId="1" AcceptedAnswerId="17961" CreationDate="2016-08-17T06:21:25.087" Score="3" ViewCount="108" Body="&lt;p&gt;As we know, Triton has retrograde orbit , which is an stong evidence it is captured by Neptune. Suppose if a moon is captured, but results in prograde orbit, and the orbit becomes circular and lies on equator of the planet after billions of years, can we still know(or find) it is captured?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="2618" LastEditDate="2017-03-07T00:45:29.277" LastActivityDate="2017-03-07T00:45:29.277" Title="If a captured moon has prograde, circular and equatorial orbit, can we know it is captured?" Tags="&lt;orbit&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17956" PostTypeId="1" AcceptedAnswerId="28993" CreationDate="2016-08-17T08:40:45.393" Score="8" ViewCount="553" Body="&lt;p&gt;The Moon is tidally locked to Earth - a long time ago it was spinning, but after a time, the spin stopped - two factors stabilizing it: unequal mass distribution creating an equilibrium position where there's one &quot;preferred&quot; axis by which it faces Earth - an attractor, and the dynamic tidal forces acting as stress upon its structure, dissipating the energy - a friction, reducing the rotation speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, at one point in time, it would no longer complete a full spin, but instead reverse its turn direction, start spinning in opposite direction, with maximum spin speed around the time where the &quot;preferred&quot; axis faces Earth, then slowing down again, until its spin stops, and reverses direction again - each time the angle smaller as the dynamic forces act against the direction of rotation - but also, with the rotation angle and angular velocity dropping, the value of the dynamic forces dropping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a significant apparent wobble caused by the orbit's eccentricity, and possibly some actual continuous wobble, as the result - with the attractor axis missing Earth as result. But is there any long-term oscillation left over from the times when the Moon was spinning - a harmonic motion around its original spin axis, that would likely not be in tune with the orbital period? (I imagine the period of these oscillations would be very long; after all it's about reversing Moon's spin back and forth, by a force that isn't all that strong.)&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2019-07-20T08:08:47.160" Title="Is there any residual oscillation left from the Moon rotation?" Tags="&lt;the-moon&gt;&lt;tidal-locking&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17957" PostTypeId="1" AcceptedAnswerId="17960" CreationDate="2016-08-17T09:08:35.770" Score="6" ViewCount="235" Body="&lt;p&gt;Imagine if Jupiter orbited the Sun at the Earth's orbital distance of 1AU. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would a planet of this size orbiting the Sun take 1 Earth year (365 days) to complete an orbit, or would the size of Jupiter affect its rotation and orbit? &lt;/p&gt;&#xA;" OwnerUserId="11792" LastEditorUserId="10437" LastEditDate="2016-08-17T13:07:53.677" LastActivityDate="2016-08-17T19:22:48.600" Title="Jupiter orbiting time in Earth's orbit?" Tags="&lt;orbit&gt;&lt;earth&gt;&lt;jupiter&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17958" PostTypeId="2" ParentId="17956" CreationDate="2016-08-17T09:41:43.113" Score="0" Body="&lt;p&gt;Yes, the Moon it is actually oscillating , and actually it is not really tidally locked, infact it is slowly going far away from Earth (by the time the Moon will leave the orbit the sun would already have expire its life however).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's see this photo from &lt;a href=&quot;https://en.wikipedia.org/wiki/Moon&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Q3Hj5.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q3Hj5.gif&quot; alt=&quot;moon oscillation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this image the moon phases where accelerated, the full complete oscillation takes about 1 month to complete.&lt;/p&gt;&#xA;" OwnerUserId="13777" LastActivityDate="2016-08-17T09:41:43.113" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17959" PostTypeId="2" ParentId="17956" CreationDate="2016-08-17T09:52:54.543" Score="2" Body="&lt;p&gt;You may have gotten your reference frame wrong. The moon never stopped revolving around its own axis, it's just that one revolution takes as long as one orbit around the earth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So, at one point in time, it would no longer complete a full spin, but instead reverse its turn direction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, it just slowed down contiually until one side faced earth permanently (discounting &lt;a href=&quot;https://en.wikipedia.org/wiki/Libration&quot; rel=&quot;nofollow&quot;&gt;libration&lt;/a&gt;, which, as you say, is due to the ellipticity of the orbit.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tidal forces get weaker as tidal locking progresses, so the change in lunar spin would be greater at earlier times and then asymptotically approach zero as the moon becomes tidally locked.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastActivityDate="2016-08-17T09:52:54.543" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17960" PostTypeId="2" ParentId="17957" CreationDate="2016-08-17T13:32:57.697" Score="9" Body="&lt;p&gt;If Jupiter was at 1 AU from the Sun, its orbit about the Sun would be about 4 hours and 10 minutes shorter than a sidereal year. That's an effect, but it's not very much of one, due to Jupiter's mass being about 1/1000th of the Sun's mass. For a more significant effect, if our Moon was much smaller than it is, it would orbit the Earth about 4 hours and 1 minute longer than a sidereal month. The Moon's mass is about 0.0123 Earth masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The period of the Keplerian orbit of an object of negligible mass about some massive object is $$P = 2\pi\sqrt{\frac{a^3}{GM}}$$ In the above, $a$ is the semimajor axis length, $G$ is the Newtonian gravitational constant, and $M$ is the mass of the central object. When the orbiting object has non-negligible mass $m$, the above expression needs to be modified to $$P = 2\pi\sqrt{\frac{a^3}{G(M+m)}}$$&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2016-08-17T19:22:48.600" LastActivityDate="2016-08-17T19:22:48.600" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17961" PostTypeId="2" ParentId="17955" CreationDate="2016-08-17T13:43:21.677" Score="3" Body="&lt;p&gt;If a satellite has a prograde, circular(ish) orbit, it will certainly be much more difficult to determine if said satellite has been captured or not. However, there are other indications we could look for in determining if that body was captured or formed in place.&lt;/p&gt;&#xA;&lt;h3&gt;Composition&lt;/h3&gt;&#xA;&lt;p&gt;As a general rule, the composition of any given body in a stellar system is a function of its distance from the star. The reason for this has to do with the temperature of the primordial disk as a function of distance from the Sun. Early on, as all the objects in our solar system were forming, the began coalescing out of the stellar disk surrounding our Sun. However, in order for objects to solidify out of this disk, they had to be cool enough to do so. Very close to the Sun, it was of course extremely hot and molecules like water or methane could not precipitate into solids. Only materials like iron and silicate could do so. Much father out, where it is colder, things like iron and silicate can still solidify, but now its cold enough that you can pack on ice and methane. This process is evident by the fact that Mercury, our closest planet to the Sun, has a core of almost complete Iron which makes up 42% of the planet's volume (compared to the Earth's 17% by volume Iron core). A graph of this concept is shown below.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wdht8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wdht8.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;What this means for your question is that you can look at the composition of a given satellite and determine if that composition matches what we'd expect for an object forming at that location. If for example, you see a satellite around Saturn which has a lot of Iron and very little ice and methane, you can likely presume it formed closer in to the Sun and was captured by Saturn (how this might happen, I don't know but its just a simple example).&lt;/p&gt;&#xA;&lt;p&gt;In fact, the Moon's nearly identical composition to the Earth is a strong indicator that our Moon formed in place (and from debris from Earth) rather than being captured.&lt;/p&gt;&#xA;&lt;p&gt;Now of course measuring composition is no easy feat. You can determine something like that by analyzing any emissions the planet produces (such as with the water plumes from Europa) or else by flying a satellite very close to the body and measuring gravitational effects.&lt;/p&gt;&#xA;&lt;h3&gt;Cratering&lt;/h3&gt;&#xA;&lt;p&gt;This is a less reliable method, but can be telling. The number and distribution of craters on a body tells us a lot about its history. If we assume that the surface of a given satellite is not constantly changing (through some geological process) then the craters on the surface represent a record of past interactions with that satellite. One approach I can imagine is to statistically predict the number of craters that should exist on the body given that it formed at its current position, then compare to what we see. If the surface hasn't been changed for billions of years, then a difference in the expected cratering indicates the body hasn't formed in the way we thought it might and possibly formed elsewhere or under different conditions. I admit though that this method is much less telling and more of a guess than using composition.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-08-17T13:43:21.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17965" PostTypeId="1" AcceptedAnswerId="17968" CreationDate="2016-08-17T18:54:20.527" Score="1" ViewCount="470" Body="&lt;p&gt;What I mean by this, is imagine a person blowing up a balloon. As the balloon expands, the person blowing it up is also walking. At a much grander scale, apply this concept to the universe. As the universe expands, is it also moving like the balloon through infinite space time?&lt;/p&gt;&#xA;" OwnerUserId="13849" LastActivityDate="2016-08-17T20:39:49.000" Title="Is the universe moving through infinite space time as it expands?" Tags="&lt;universe&gt;&lt;space-time&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17966" PostTypeId="2" ParentId="17965" CreationDate="2016-08-17T19:23:46.063" Score="0" Body="&lt;p&gt;When we talk about the universe, we are really talking about one of two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The observable universe, which is everything we can possibly see.&lt;/li&gt;&#xA;&lt;li&gt;The Universe, which is everything that has ever existed, currently exists, and will exist.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The observable universe has its own center, usually the Earth. It is a spherical region of everything that we can see, essentially anything whose light has reached us. We usually refer to this when we say things like &quot;there are $10^{86}$ atoms in the universe.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, everyone has their own observable universe, and it can change depending on where you are. An exoplanet far away has its own observable universe, and can receive light from different places. Essentially, you are the center of your own observable universe. As more light reaches us, our observable universe will grow (or expand, if you will).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are referring to the latter, then there is a totally different answer and &quot;expansion&quot; refers to something completely different. The Universe (notice the capital &quot;U&quot;) is all of space and time and its contents. Anything that has existed, will exist, and currently exists is part of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we say the Universe is expanding, we don't mean it's like a ball that is constantly growing in size. We simply mean that the space between objects is increasing. The Universe can be infinite in size, but more and more space is being created between matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of thinking about this is saying that space is dilating. Imagine that everything in space is on a coordinate system. The coordinate system itself is actually dilating, and thus the objects on it are drifting apart. Here's a GIF I made to help you understand it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/amQeZ.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/amQeZ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see how objects are drifting apart from each other, due to the expansion of space. Thus, your question doesn't make much sense; the Universe is all of space and it's contents, so how could it move through more space?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2016-08-17T19:23:46.063" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="17967" PostTypeId="1" CreationDate="2016-08-17T06:32:18.467" Score="4" ViewCount="109" Body="&lt;p&gt;I remember I hear a theory that if a star has a companion star, we can know the mass of the star more easily, is that true? if it is true, while I understand that by measuring the period of the companion star, we can know the mass ratio of 2 stars,but how can the companion star also help to measure the mass of parent star?&lt;/p&gt;&#xA;" OwnerUserId="8049" OwnerDisplayName="Gstestso" LastActivityDate="2016-08-18T00:55:24.780" Title="Is it true that we can estimate the mass of a star more accurately if it has a companion star?" Tags="&lt;mass&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17968" PostTypeId="2" ParentId="17965" CreationDate="2016-08-17T20:08:34.173" Score="1" Body="&lt;p&gt;I think what you're asking is a very fundamental question that doesn't really have a definitive answer. I interpreted your question slightly differently that &lt;a href=&quot;https://astronomy.stackexchange.com/a/17966/11143&quot;&gt;Sir Cumference&lt;/a&gt;, so I'll answer it differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For people just learning about the expansion of the Universe, they are often taught about the analogy of the expanding balloon. The concept here is that you have a deflated balloon that you draw two dots on. These dots may represent two galaxies in the Universe defined by the &lt;strong&gt;surface&lt;/strong&gt; of the balloon. The important concept here is that the surface of the balloon represents the entire universe in this scenario. The balloon is then blown up, and you notice that because of the &quot;expansion&quot; of the universe, the dots (i.e., galaxies) are moving farther apart. This is shown in the diagram below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gthfd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gthfd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This concept is usually used to teach an important fact about our universe, namely that there is no center. At no point on the balloon can you saw that you are the center of the expansion because the expansion is happening everywhere in equal magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you seem to be doing though, is trying to take this analogy one step further. Even though the balloon universe (ballooniverse?) is defined on the &lt;strong&gt;surface&lt;/strong&gt; of the balloon and is thus two dimensional, there still exists a three dimensional space that the 2D ballooniverse is expanding into. So you question, what does that say about our 3D universe? Are we expanding into some 4D space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That, in my opinion, is a distillation of the fundamental concepts behind your question. Assuming I'm interpreting your question correctly, then I would have to say the answer is &lt;em&gt;I don't know&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a somewhat (unsatisfying) tautological answer in that the Universe is, &lt;em&gt;by definition&lt;/em&gt;, everything that exists. Therefore, there can't be a 3D Universe and also a 4D space into which our 3D Universe is expanding into, because that 4D space would necessarily be a part of our Universe, by definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more reasoned approach to this answer is to look at it from a mathematical perspective. If you begin to learn higher math (and by higher I mean upper undergraduate/graduate level math) you'll learn about &lt;a href=&quot;https://en.wikipedia.org/wiki/Manifold&quot; rel=&quot;nofollow noreferrer&quot;&gt;manifolds&lt;/a&gt;. Manifolds are basically mathematical objects that can define things like the surface of a balloon. What you'll find is that the definition for the surface of a balloon, using the concept of manifolds, does not require that the surface exist in 3D space, even if that manifold is expanding. It is perfectly possible to define a 2D manifold that has the same topology (fancy word for structure) as an expanding balloon, without &quot;embedding&quot; it in 3D space. This concept makes your question kind of moot. It allows for an expanding 3D universe, without requiring the existence of some 4D space to expand into. So ultimately, I'm claiming that it is possible for a 3D Universe to exist which both has no center and is expanding, just not into a 4th dimension.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-08-17T20:39:49.000" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17969" PostTypeId="2" ParentId="13143" CreationDate="2016-08-17T20:08:50.987" Score="2" Body="&lt;p&gt;Stellarium 0.14 or later has a checkbox under Configuration, Tools, Planetarium options:&#xA;Include &lt;a href=&quot;https://en.wikipedia.org/wiki/Nutation&quot; rel=&quot;nofollow&quot;&gt;nutation&lt;/a&gt;.&#xA;If I uncheck it, the RA and Dec of date for those stars on 2016/1/1&#xA;match yours within 0.2 arcsecond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some libraries distinguish between astrometric and apparent positions,&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Aberration_of_light&quot; rel=&quot;nofollow&quot;&gt;aberration&lt;/a&gt;&#xA;making up to 20 arcseconds difference.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2016-08-18T05:09:56.290" LastActivityDate="2016-08-18T05:09:56.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17970" PostTypeId="2" ParentId="17967" CreationDate="2016-08-17T22:56:54.753" Score="3" Body="&lt;p&gt;In general, masses can only be precisely estimated for stars in binary systems (asteroseismology and some other indirect techniques can yield estimates too).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two broad types of binary where this is possible - astrometric binaries, where the motion of both stars can be seen on the sky and where the distance to the binary is known (e.g. Sirius AB); or eclipsing binaries, where the stars eclipse each other and their motions are inferred from their line of sight velocities, measured by spectroscopy (e.g. YY Gem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are correct, that the relative motions of the stars only yields their mass ratio. What is needed to get the individual masses is something that fixes the &lt;em&gt;absolute sizes&lt;/em&gt;  of the orbits and the orbital period, which can then be used with Kepler's third law to give the total mass of both stars and hence combined with the mass ratio to give the masses of the individual components. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In astrometric binaries you can measure the angular size of the orbits and calculate an absolute size using the known distance. In close (unresolved) binaries, that size is inferred from the velocities and orbital periods, but also requires knowledge of the inclination of the orbital plane. This is known quite accurately (is close to 90 degrees) only for &lt;em&gt;eclipsing&lt;/em&gt; binaries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could present the Maths if you really like, but these are quite standard results.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-08-17T22:56:54.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17971" PostTypeId="2" ParentId="17967" CreationDate="2016-08-18T00:55:24.780" Score="2" Body="&lt;p&gt;I'll just fill in the blanks with the math that Rob Jeffries alluded to. There's a variety of scenarios you can consider, but I'll just run through the basics of two binary stars, where both stars are distinctly visible. Such an example is Antares, a well known and rather bright star in the northern sky. Such a star system is usually referred to as a &lt;a href=&quot;http://www.midnightkite.com/index.aspx?URL=Visual&quot; rel=&quot;nofollow noreferrer&quot;&gt;Visual Binary&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If you want to solve for the masses of both stars, you'll need two equations. If algebra has taught you anything, it's that you need as many equations as you have unknowns if you want your system to be completely determined.&lt;/p&gt;&#xA;&lt;h3&gt;Mass Ratio&lt;/h3&gt;&#xA;&lt;p&gt;To start with, consider two binary stars of mass &lt;span class=&quot;math-container&quot;&gt;$M_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$M_2$&lt;/span&gt; which are located at positions &lt;span class=&quot;math-container&quot;&gt;$\vec{r}_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\vec{r}_2$&lt;/span&gt;, respectively. The center of mass for that system, &lt;span class=&quot;math-container&quot;&gt;$\vec{R}$&lt;/span&gt;, is defined to be&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\vec{R} = \frac{\vec{r}_1 M_1 + \vec{r}_2 M_2}{\vec{r}_1 + \vec{r}_2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now, you'll notice I'm defining the positions of the stars without setting up a coordinate system. Currently &lt;span class=&quot;math-container&quot;&gt;$\vec{r}_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\vec{r}_2$&lt;/span&gt; have no physical meaning. I'm going to choose a convenient coordinate system such that the origin is at the center of mass. This means that &lt;span class=&quot;math-container&quot;&gt;$\vec{R}=0$&lt;/span&gt; and we find that&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{M_1}{M_2} = \frac{\vec{r}_2}{\vec{r}_1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;This basic concept gives you the mass ratio, if only you can measure the ratio of the physical distance each star is from the center of mass. This is easily achieved by noting that a physical distance is related to the angular distance, &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;, (something quite measurable with a telescope) by the relation &lt;span class=&quot;math-container&quot;&gt;$r=\alpha d$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; is the distance to the binary system. This allows us to say that&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{M_1}{M_2} = \frac{\alpha_2}{\alpha_1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;As stated, &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; is an easily measurable parameter (more or less) by simply imaging the binary system with a telescope over a long enough time period. There's a bit more involved in this since I haven't talked at all about the inclinations of the orbits, but this is the basic premise.&lt;/p&gt;&#xA;&lt;h3&gt;Mass Sum&lt;/h3&gt;&#xA;&lt;p&gt;Now that we have a way of getting the mass ratio, the other equation we'll look for is the sum of the masses. Anytime you hear sum of masses, you should immediately think Kepler's third law.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P^2 = \frac{4\pi^2}{G(M_1+M_2)}a^3$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;In this equation &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; is the orbital period, &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; is the &lt;a href=&quot;https://www.google.com/search?q=gravitational%20constant&amp;amp;ie=utf-8&amp;amp;oe=utf-8&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravitational constant&lt;/a&gt;, and &lt;span class=&quot;math-container&quot;&gt;$a = r_1 + r_2$&lt;/span&gt; is the semi-major axis of the reduced mass system (there's some more math and assumptions behind this, but let's roll with it for now). This problem is as simple as solving for the sum of the masses. However, we'll also note that &lt;span class=&quot;math-container&quot;&gt;$a = \alpha d$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\alpha = \alpha_1 + \alpha_2$&lt;/span&gt;. Solving for the sum of masses gives us&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M_1 + M_2 = \frac{4\pi^2}{GP^2}\alpha^3d^3$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;As stated in the mass ratio section, it is a bit more involved than this because I haven't said anything about the inclination of the system. If the system has an inclination, it will tend to muck up the math and make it more difficult, but the principle concepts will be the same.&lt;/p&gt;&#xA;&lt;h3&gt;Putting it together&lt;/h3&gt;&#xA;&lt;p&gt;You now have two equations and two unknowns. Let's solve for the masses.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M_1 = \frac{4\pi^2}{GP^2}d^3\alpha^2\alpha_2$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M_2 = \frac{4\pi^2}{GP^2}d^3\alpha^2\alpha_1$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-08-18T00:55:24.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17972" PostTypeId="1" CreationDate="2016-08-18T04:17:21.960" Score="3" ViewCount="62" Body="&lt;p&gt;I'm a high school student who is doing research about Cepheid variables. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm in the process of deriving distance to Cepheid using the Baade-Wesselink method, which utilises the surface brightness relation. The surface brightness relation is used to derive angular diameter variation of the cepheid from its V-K band data in this method. I found the equation relating angular diameter and the V-K band of Cepheid variables from the research paper &lt;a href=&quot;http://www.aanda.org/articles/aa/abs/2007/42/aa8225-07/aa8225-07.html&quot; rel=&quot;nofollow&quot;&gt;The projection factor, period-radius relation, and surface-brightness colour relation in classical cepheids&lt;/a&gt;, which is apparently:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log{θ}=(0.2672+0.0016)(V-K)+(0.5354+0.0012)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, to derive the angular diameter variation what kind of raw data that I should use for V-K band? May I directly use the maximum V-K value and minimum V-K value and put them into the equation individually, and interpret their difference as an angular diameter variation? Or do I need to get lists of V-K band value varying over time and put it into the equation to get angular diameter variation over time and graphically measure the maximum and minimum angular diameter variation?&lt;/p&gt;&#xA;" OwnerUserId="13714" LastEditorUserId="9301" LastEditDate="2016-10-17T14:57:50.330" LastActivityDate="2016-10-17T14:57:50.330" Title="Using the surface-brightness relation for Cepheids" Tags="&lt;distances&gt;&lt;photometry&gt;&lt;angular-diameter&gt;&lt;cepheids&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17974" PostTypeId="2" ParentId="12118" CreationDate="2016-08-18T13:28:25.583" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can a solar flare destroy every electronic item on earth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Extreme Limit&lt;/strong&gt;&lt;br&gt;&#xA;Let's start with an extreme upper limit, a supernova.  So we know that a &lt;a href=&quot;https://en.wikipedia.org/wiki/Supernova#Energy_output&quot; rel=&quot;nofollow noreferrer&quot;&gt;typical supernova&lt;/a&gt; will not eradicate our atmosphere, as I discussed in detail at:  &lt;a href=&quot;https://physics.stackexchange.com/a/233126/59023&quot;&gt;https://physics.stackexchange.com/a/233126/59023&lt;/a&gt;.  It would also not affect most of our technology unless it was really close (i.e., the sun would need to collapse, but our current understanding is that the sun will not undergo a supernova explosion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Power Grids&lt;/strong&gt;&lt;br&gt;&#xA;We also know that due to the properties of most power grids, the likelihood of numerous transformers suffering irreparable damage is quite low as well, as I discussed in detail at: &lt;a href=&quot;https://physics.stackexchange.com/a/149199/59023&quot;&gt;https://physics.stackexchange.com/a/149199/59023&lt;/a&gt;.  The reason is that if one major &lt;a href=&quot;https://en.wikipedia.org/wiki/Transformer&quot; rel=&quot;nofollow noreferrer&quot;&gt;transformer&lt;/a&gt; were to fail, the rest of the grid would &quot;trip&quot; causing a temporary wide spread blackout.  The grid could, in principle, recover shortly afterwards work just fine.  Thus, so long as the duration of the large and rapid changes in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth's magnetic field&lt;/a&gt; from external influences (e.g., &lt;a href=&quot;https://en.wikipedia.org/wiki/Coronal_mass_ejection&quot; rel=&quot;nofollow noreferrer&quot;&gt;coronal mass ejections&lt;/a&gt; or CMEs) have subsided, the rest of the grid would be fine.  We have several groups actively studying the effects of the sun on Earth, as I discussed at:  &lt;a href=&quot;https://physics.stackexchange.com/a/214509/59023&quot;&gt;https://physics.stackexchange.com/a/214509/59023&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solar Flares&lt;/strong&gt;&lt;br&gt;&#xA;Finally, contrary to what Hollywood might want you to believe, solar storms are not going to kill us (well, at least not directly).  I wrote a detailed answer at &lt;a href=&quot;https://physics.stackexchange.com/a/258093/59023&quot;&gt;https://physics.stackexchange.com/a/258093/59023&lt;/a&gt; that basically explains why the combination of Earth's atmosphere and magnetic field shield us from most solar threats.  Spacecraft and astronauts, however, could be in harms way during very strong storms.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this really true and if so, how would it happen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course, it is possible to damage nearly all electrical devices with an &lt;a href=&quot;https://en.wikipedia.org/wiki/Electromagnetic_pulse&quot; rel=&quot;nofollow noreferrer&quot;&gt;EMP&lt;/a&gt;, but that does not mean the sun is capable of such a feat nor that it is likely to happen.  If the power grid were to catastrophically fail (i.e., dozens of major transformer failures), many electronic devices would still be fine and useable.  It's just that we would all have to buy small generators to use them because the electricity in your home wouldn't be on otherwise.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Could we do anything to to prevent it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, it's rather simple.  If we know that a large &lt;a href=&quot;https://en.wikipedia.org/wiki/Geomagnetic_storm&quot; rel=&quot;nofollow noreferrer&quot;&gt;geomagnetic storm&lt;/a&gt; were about to occur, we could simply power down the grid.  The damage to transformers arises because the &lt;a href=&quot;https://en.wikipedia.org/wiki/Geomagnetically_induced_current&quot; rel=&quot;nofollow noreferrer&quot;&gt;ground induced currents&lt;/a&gt; produce a &lt;a href=&quot;https://en.wikipedia.org/wiki/DC_bias&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;DC offset&lt;/em&gt;&lt;/a&gt; in the voltage/current arriving at the transformer.  They are designed to convert power for a given frequency and amplitude.  Thus, if you shift a sine wave away from zero, the peaks will appear to the transformer as larger amplitude signals.  These will be &lt;a href=&quot;https://en.wikipedia.org/wiki/Clipping_(signal_processing)&quot; rel=&quot;nofollow noreferrer&quot;&gt;clipped&lt;/a&gt; initially but they create excess heat and internal stresses on the transformer.  Over time, the heat and stress cause the transformer to effectively tear it self apart (Note if you ever get to hear one explode, it's one of the loudest things you will likely hear.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is no power in the system, then the DC offset will have no real effect for numerous reasons.  One could, in principle, also employ &lt;a href=&quot;https://en.wikipedia.org/wiki/High-pass_filter&quot; rel=&quot;nofollow noreferrer&quot;&gt;high pass filters&lt;/a&gt; but I am guessing cost and maintenance would be limiting factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main vulnerability is our spacecraft in orbit and it's not the induced electromagnetic fields that will cause them issues.  Their biggest threat is high energy charged particles, which often come with large solar storms.  The details can be found in my answer at &lt;a href=&quot;https://physics.stackexchange.com/a/258093/59023&quot;&gt;https://physics.stackexchange.com/a/258093/59023&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13663" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-08-18T13:28:25.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17975" PostTypeId="1" CreationDate="2016-08-18T16:46:51.313" Score="1" ViewCount="114" Body="&lt;p&gt;The Hubble Space Telescope offers a &lt;a href=&quot;http://www.stsci.edu/hst/acs/documents/handbooks/cycle19/c07_obstechniques10.html&quot; rel=&quot;nofollow&quot;&gt;parallel obervation mode&lt;/a&gt;. Observing programmes such as &lt;a href=&quot;http://candels.ucolick.org/index.html&quot; rel=&quot;nofollow&quot;&gt;CANDELS&lt;/a&gt; and &lt;a href=&quot;http://www.stsci.edu/~postman/CLASH/Home.html&quot; rel=&quot;nofollow&quot;&gt;CLASH&lt;/a&gt; use &lt;a href=&quot;http://www.stsci.edu/hst/wfc3&quot; rel=&quot;nofollow&quot;&gt;WFC3&lt;/a&gt; and &lt;a href=&quot;http://www.stsci.edu/hst/acs&quot; rel=&quot;nofollow&quot;&gt;ACS&lt;/a&gt; in parallel fields. These are also the only instruments of which I have heard that they are being used in parallel. Here are my questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to observe with more than two instruments in parallel? If so, what is the limit, and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why aren't all observations done with as many instruments as possible turned on?  Granted, a lot of the data would be useless / difficult to analyze (like spectra of a random patch of sky), but on the other hand it might lead to serendipitous discoveries. HST won't last forever, so why not get the most use possible out of it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this is about the science instruments - I assume the guidance sensors will be working alongside the others anyway. (Or is that wrong?)&lt;/p&gt;&#xA;" OwnerUserId="13681" LastActivityDate="2016-08-24T12:33:40.603" Title="How many science instruments can be used in parallel with the Hubble Space Telescope?" Tags="&lt;observational-astronomy&gt;&lt;hubble-telescope&gt;&lt;instruments&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17976" PostTypeId="5" CreationDate="2016-08-18T17:16:51.757" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2016-08-18T17:16:51.757" LastActivityDate="2016-08-18T17:16:51.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17977" PostTypeId="4" CreationDate="2016-08-18T17:16:51.757" Score="0" Body="Questions about the visualization tool for FITS images and data." OwnerUserId="13681" LastEditorUserId="13681" LastEditDate="2016-08-18T19:53:57.410" LastActivityDate="2016-08-18T19:53:57.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17978" PostTypeId="5" CreationDate="2016-08-18T18:58:11.880" Score="0" Body="&lt;p&gt;A Cepheid variable star pulsates radially, thereby varying diameter, temperature and brightness, with a well-defined and stable period and amplitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strong relation between pulsation period and luminosity makes them good distance indicators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Cepheid_variable&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article on Cepheids&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user1569" LastEditorDisplayName="user1569" LastEditDate="2016-08-18T19:54:22.673" LastActivityDate="2016-08-18T19:54:22.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17979" PostTypeId="4" CreationDate="2016-08-18T18:58:11.880" Score="0" Body="Use when your question is specifically about Cepheid variables." OwnerDisplayName="user1569" LastEditorDisplayName="user1569" LastEditDate="2016-08-18T19:54:07.267" LastActivityDate="2016-08-18T19:54:07.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17980" PostTypeId="1" AcceptedAnswerId="17982" CreationDate="2016-08-19T09:09:28.763" Score="5" ViewCount="236" Body="&lt;p&gt;Sorry for my dumb terms. I'm interrested in astronomy but I have next to zero background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that a black hole absorbs everything, including light, could we ever discern one with visible light (and live to pass the information on)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/q/1206/13863&quot;&gt;This question&lt;/a&gt; contains one generated picture of a black hole and Google image shows visual representations that all look similar (black circle in the centre, light that's distorted around it and the further we go away from the center the background is less and less distorted), but are those actually representative? Is there really a black spot in our vision field at the place where the black hole is? Both the answer to the question and &lt;a href=&quot;http://www.techinsider.io/black-hole-photograph-event-horizon-telescope-algorithm-2016-6&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt; discards such images as &quot;fantasy&quot;, but they don't really answer the questions I'm asking, which are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Will we ever be able to discern a black hole with visible light and live?&lt;/li&gt;&#xA;&lt;li&gt;If we can, what is a currently acceptable description of what we'll see with visible light? I don't especially expect a picture, so a textual description is also good, but keep in mind it's in regards to visible light. Other questions on Astronomy SE explain what we can &quot;detect&quot; using radio waves or other means of observation.&lt;/li&gt;&#xA;&lt;li&gt;In 2014-2015 CERN said they might be creating micropscopic short-lived (10&lt;sup&gt;-27&lt;/sup&gt; seconds) black holes with the LHC (and this lead to a global nonsensical fear that it'd destroy the planet). Did they create any? If yes, was there any opportunity to picture them despite their tiny size and short life?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13863" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-08-22T05:12:32.247" Title="Can we discern black holes with visible light?" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17981" PostTypeId="2" ParentId="17835" CreationDate="2016-08-19T09:13:33.657" Score="10" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As has been commented, the amount of damage taken by an interstellar spaceship depends on its velocity $v$, as well as the number of gas and dust particles that it encounters on its way. This number is usually measured per area, in which case it's called &lt;em&gt;column density&lt;/em&gt; $N$, and is equal to the total distance $d$ traveled times the particle density $n$, i.e. $N = nd$. For instance, if a spaceship travels 1 lightyear ($10^{18}\,\mathrm{cm}$) through a region with a density of $10\,\mathrm{cm}^{-3}\!$, each square centimeter of the spaceship will encounter $10^{19}$ particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the faster you go, the farther you go, and the more dense regions you go through, the more your spaceship is damaged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Breakthrough_Starshot&quot; rel=&quot;noreferrer&quot;&gt;Breakthrough Starshot&lt;/a&gt; project aims to reach our nearest-neighbor stellar system $\alpha$ Centauri in ~20 years, with a gram-sized satellite reaching $0.20c$ by means of a light sail. Today, there was a paper by &lt;a href=&quot;http://arxiv.org/abs/1608.05284&quot; rel=&quot;noreferrer&quot;&gt;Hoang et al.&lt;/a&gt; calculating the amount of damage taken by such a satellite. The total column density of gas from Earth to $\alpha$ Cen is $\sim10^{17.5\mathrm{-}18}\mathrm{cm}^{-2}$, and assuming (fairly) a dust-to-gas ratio of 1% and a carbonaceous/silicate dust grain population with a &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0008146&quot; rel=&quot;noreferrer&quot;&gt;Weingartner &amp;amp; Draine (2001)&lt;/a&gt; size distribution, they calculate that this &lt;strong&gt;journey to $\alpha$ Cen will erode the surface of the spacecraft to a thickness of the order of 1 mm&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the damage is caused by dust, not gas, but in principle gas may slowly heat up the spaceship. However, at $v=0.2c$, as long as the density is $\lesssim10\,\mathrm{cm}^{-3}$, the temperature is insufficient to cause any melting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecular clouds — the dense clouds where stars are born — have densities from $10^2\,\mathrm{cm}^{-3}$ and even up to $10^6\,\mathrm{cm}^{-3}$, i.e. many orders of magnitude higher than the roughly $1\,\mathrm{cm}^{-3}$ found in the more dilute interstellar medium. To reach even more distant stars in a tolerable time, you would have to go faster than $0.2c$, and thus it seems that it is in fact a good idea to evade these clouds.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2016-08-19T09:13:33.657" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17982" PostTypeId="2" ParentId="17980" CreationDate="2016-08-19T11:08:50.790" Score="6" Body="&lt;p&gt;There are two &quot;usual&quot; ways to detect a black hole:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is part of a binary system (two stars closely rotating around each other) and dragging matter from the other star into an accretion disk, that disk will heat up. Energy distribution will tell us whether the accreting object is indeed a black hole or a neutron star instead. This is primarily visible in X-ray, so not an answer to your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the black hole has a mass of 100.000 to millions of times that of the sun (supermassive black hole), it will sit in the centre of a galaxy, and sometimes accrete matter. This is mostly visible in Radio, and again in X-ray. Still not an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how can we detect a black hole in the optical? Best chance is an indirect detection, like in the images you linked above. The one with the distorted stars in the background is not &quot;just fantasy&quot;, but a simulation based on our understanding of gravity. The problem when we try to see this is scale, and hence resolution. The event horizon of a stellar mass black hole is on the order of a few kilometers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's replace our sun by a black hole. At a distance of $1.5\cdot10^8$km, with a black hole diameter of 6km, we'd see it occupying an angle of&#xA;\begin{equation}&#xA; \alpha = \tan{\left(\frac{6}{1.5\cdot10^8}\right)} = \frac{6}{1.5\cdot10^8} = 2.3^{\circ}\cdot10^{-6} = 8''.3 \cdot 10^{-3}\,.&#xA;\end{equation}&#xA;So a solar mass black hole at the position of the sun would be 8 milli-arcseconds (mas) in size. A single 8m Telescope at the  &lt;a href=&quot;http://www.eso.org/public/teles-instr/paranal/&quot; rel=&quot;nofollow&quot;&gt;VLT&lt;/a&gt; has a resolution of 50mas. If we consider that the gravitational lensing effect takes place on scales larger than the Schwarzschild radius, it could just manage to see this effect. (The theoretical resolution limit for an 8m telescope at 500nm wavelength is $1.22 \lambda / d = 15$mas.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For real black holes at distances of 10s or 100s of lightyears, you would need a truly gigantic telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next option, again with gravitational lensing: The &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_microlensing&quot; rel=&quot;nofollow&quot;&gt;microlensing&lt;/a&gt; effect causes a background star to be temporarily magnified - it shines brighter for short period of time. A black hole passing in the foreground can cause this, and it is an optical observation, but it is very difficult to know that the lens was indeed a black hole and not some star or other compact object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last option, Hawking radiation: Hawking postulated that black holes might emit radiation inversely proportional to their size. For small enough black holes, this is a runaway process - they emit more as they get smaller, and because the emission causes them to shrink, they emit yet more again. Since Hawking radiation has a black body spectrum, the maximum of this emission will lie in the visible range for a brief period of time before going into UV, X-ray, and gamma radiation. The problem here is that no one has observed it yet, and you'd need a fairly small (but not too small) black hole close by. Also, you might get killed as the radiation gets harder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To your questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Possible, but probably not. You would need to get very close, but not close enough to get killed by tidal forces. Or you'd need to be able to identify a microlens as a black hole by some means other than lensing.&lt;/li&gt;&#xA;&lt;li&gt;Either lensing effects (see above), or a brief flash going from red to blue (probably too short to observe).&lt;/li&gt;&#xA;&lt;li&gt;These black holes would be mostly visible by the absence of other particles. You'd get less elementary particles than you would expect for the reaction,some energy and some momentum would be missing, and there would be characteristic gamma radiation. However, for CERN to announce this, this would have to have happened often enough to not be a statistical fluke (&lt;a href=&quot;http://www.physics.cornell.edu/2016/05/11/may-2016-physicists-abuzz-about-possible-new-particle-as-cern-revs-up/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; turned out to not be a detection after all.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13681" LastEditorUserId="9301" LastEditDate="2016-08-22T05:12:32.247" LastActivityDate="2016-08-22T05:12:32.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17984" PostTypeId="1" CreationDate="2016-08-20T00:10:50.463" Score="-6" ViewCount="310" Body="&lt;p&gt;We have only 12 months in our calendar and only as many Zodiac signs as the number of months in a year. Now that we know there are 13 constellations should we not have 13 months in the calendar instead of 12 months and have 13 Zodiac Signs by adding a sign for Ophiuchus as well in the following manner? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have 13 months in a year, each month consisting of 28 days, we would have neat 4 weeks a month round the year which would obviously make life much easier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But 13 months multiplied with 28 makes it only 364 days a year. So we would have to have one extra day besides 13 months each year. We may call it by any name. It could be even Xmas day or just any other day for a non-leap year. During leap years we would have to have two extra days instead of one day. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At present the period of Capricorn starts on Dec 22 and ends on Jan 19. Likewise other Zodiac periods also start from such staggered dates of various months and end on odd dates only. If we may have 13 months and may allow a year to start from Dec 22 instead of Jan 1 we would be able to resolve this anomaly also because then each Zodiac sign would begin on the 1st of each month and would end on the last day of each month only. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way we switched over from the British system of miles, fathoms, yards,feet and inches to the metric system all over the world this reform would also obviously facilitate everybody's life enormously.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So should we not set the ball rolling to reform our calendar as conceived by me?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deborah Houlding in her post “Understanding the Zodiac: and why there really are 12 zodiac signs” at &lt;a href=&quot;https://www.skyscript.co.uk/zodiac.html&quot; rel=&quot;nofollow&quot;&gt;https://www.skyscript.co.uk/zodiac.html&lt;/a&gt; has mentioned:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Ophiuchus is not a new constellation, it was catalogued with many others in Ptolemy's 2nd-century treatise, Almagest, (an astronomical text of huge historical standing), where it is referred to as Septentarius, &quot;the serpent holder&quot;. It was always realised that the ecliptic cuts through a small part of this constellation, but it is a very large constellation, the majority of which the zodiac fails to embrace, so it was not considered a &quot;zodiac constellation&quot;.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They knew that there were as many as 17 constellations including Ophiuchus but had deliberately chosen only twelve of them because 360 degrees could have not been equally divided into `13 sections. It indirectly implies had there been 390 degrees in a circle they would have perhaps included Ophiuchus also. Why was it necessary to divide the circle into equal widths at all? Some Zodiacs were known to cover smaller sections than the others. They could have as well accommodated even all the 17 constellations by allotting them actual widths instead of equal widths as has been done now for 13 constellations. So we may say the fact that they thought of allotting equal widths to all Zodiacs was the main culprit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just have a look at the current distribution, shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 ARIES = APRIL 19 – MAY 13    … 25 days&#xA;2 TAURUS = MAY 14 – JUNE 19   …  37 days&#xA;3 GEMINI = JUNE 20 – JULY 20     … 31 days&#xA;4 CANCER = JULY 21 – AUG 9    …. 21 days&#xA;5 LEO = AUGUST 10 – SEPTEMBER 15   …  37 days&lt;br&gt;&#xA;6 VIRGO = SEPTEMBER 16 – OCTOBER 30    … 45 days&#xA;7 LIBRA = OCTOBER 31 – NOVEMBER 22   …  23 days &#xA;8 SCORPIO = NOVEMBER 23 – NOVEMBER 29  … 7 days &#xA;9 OPHIUCHUS = NOVEMBER 30 – DECEMBER 17  … 18 days &#xA;10 SAGITTARIUS = DECEMBER 18 – JANUARY 18  … 31 days&#xA;11 CAPRICORN = JANUARY 19 – FEBRUARY 15   … 28 days&#xA;12 AQUARIUS = FEBRUARY 16 – MARCH 11    …  24 days (in non-leap years) and 25  days (in the leap years) &#xA;13 PISCES = MARCH 12 – APRIL 18  ..  38 days&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Source: www.cool-story.com)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How nicely 365 days of the non-leap years and 366 days of the leap years have been assigned to the 13 Zodiacs? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as of now even if we had 13 sidereal months a year, each month of 28 days, each Zodiac would obviously not stretch over full month any way. So by having 28 days’ months and 13 months a year we would be benefited only to the extent that each month would have precisely 4 weeks while it is not so under the Gregorian calendar of 12 synodic months.   &lt;/p&gt;&#xA;" OwnerUserId="13871" LastEditorUserId="13871" LastEditDate="2016-08-26T02:28:47.337" LastActivityDate="2016-08-26T02:28:47.337" Title="Should we not accommodate Ophiuchus also in Our Calendar?" Tags="&lt;astrophysics&gt;" AnswerCount="3" CommentCount="3" ClosedDate="2016-08-20T17:53:04.483" ContentLicense="CC BY-SA 3.0" />
  <row Id="17985" PostTypeId="2" ParentId="17984" CreationDate="2016-08-20T00:13:59.243" Score="3" Body="&lt;p&gt;No; generally speaking a month is based upon the natural period of the moon in orbit around the earth. The zodiac signs have nothing to do with it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.calendar-origins.com/calendar-origins.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.calendar-origins.com/calendar-origins.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13872" LastActivityDate="2016-08-20T00:13:59.243" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17986" PostTypeId="2" ParentId="17984" CreationDate="2016-08-20T00:44:56.383" Score="-3" Body="&lt;p&gt;If we have 13 months a year and 13 Zodiac Signs instead of 12 months and 13 Zodiac signs instead of 12 signs as suggested by me and have &quot;28-day months&quot; we would be benefited yet in one more way. We shall have precisely 4 weeks per month because 7 x 4 = 28. However, it would make for only 364 days in a year. So we would have to cover this flaw by redefining the year as a period of &quot;13 months and a leap-day&quot; for each non-leap year and two leap-days for each leap-year. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;in addition, if we may commence the 13 months-calendar from 22nd December instead of 1st January all Zodiac spells would begin on the 1st of each month and would end at the end of each month. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my suggestion merits consideration according to me.  &lt;/p&gt;&#xA;" OwnerUserId="13871" LastEditorUserId="13871" LastEditDate="2016-08-21T15:23:58.530" LastActivityDate="2016-08-21T15:23:58.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17987" PostTypeId="2" ParentId="17984" CreationDate="2016-08-20T04:37:09.150" Score="9" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no astrophysical sense in which there are &quot;really&quot; 13 constellations along the ecliptic. The astronomical constellations are &lt;em&gt;arbitrary&lt;/em&gt; regions of the sky, defined for convenience so we can refer to the locations of celestial objects (M31 in Andromeda, the nebula in Orion, and so forth). They are not designed, and are not suitable, for timekeeping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at a map of the sky, you'll see that the constellations that happen to intersect the ecliptic are not evenly spaced. If we based our calendar on them, some months would be only a few days long and others would be several weeks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 12 evenly spaced astrological &quot;signs&quot; are just as arbitrary as the astronomical constellations. Neither have any real physical basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to argue for a 13-month year, you won't be the first to do so. -- but you can't reasonably use the constellations took support the idea.&lt;/p&gt;&#xA;" OwnerUserId="53" LastActivityDate="2016-08-20T04:37:09.150" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17988" PostTypeId="1" CreationDate="2016-08-20T04:45:06.640" Score="-1" ViewCount="990" Body="&lt;p&gt;I would like to know whether the constellations that appear smaller in size from Earth appear small because they are located far away from Earth as compared to the constellations which appear larger.  &lt;/p&gt;&#xA;" OwnerUserId="13871" LastEditorUserId="14021" LastEditDate="2016-09-14T07:41:29.693" LastActivityDate="2016-09-14T07:41:29.693" Title="Distance of various constellations from Earth" Tags="&lt;distances&gt;&lt;constellations&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17989" PostTypeId="2" ParentId="17988" CreationDate="2016-08-20T07:54:19.963" Score="6" Body="&lt;p&gt;The stars in any given constellation can be at vastly different distances, there is no link between the distance and the size of a constellation. The constellation boundaries are purely artificial boundaries.&lt;/p&gt;&#xA;" OwnerUserId="11658" LastActivityDate="2016-08-20T07:54:19.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17990" PostTypeId="1" CreationDate="2016-08-20T14:28:05.863" Score="33" ViewCount="9721" Body="&lt;p&gt;I assume that about 99.9% of the sun-rays that do not fall on any planet or any other celestial body keep on traveling farther and farther unto infinity. Apparently such rays get lost. Keeping in mind the colossal energy Sun has produced since 4.5 billion years I am somehow reluctant to  reconcile myself to the idea that Nature would have allowed wastage of so much energy produced by the Sun. Nonetheless I want to get enlightened whether it got really lost or got utilized. If it got utilized I want to know how it may have got utilized at all and whether any sustainable evidence is available in support of any such finding?  &lt;/p&gt;&#xA;" OwnerUserId="13871" LastEditorUserId="9301" LastEditDate="2016-08-22T20:43:15.673" LastActivityDate="2016-08-22T20:43:15.673" Title="What happens to the 99.9 % of the sun-rays that do not fall on any planets or any other celestial body?" Tags="&lt;the-sun&gt;&lt;light&gt;&lt;astrophysics&gt;" AnswerCount="3" CommentCount="11" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17991" PostTypeId="2" ParentId="17990" CreationDate="2016-08-20T15:40:09.627" Score="34" Body="&lt;p&gt;You want nature to be frugal and efficient. You want all the energy of the sun to have a purpose. However what you &lt;strong&gt;want&lt;/strong&gt; nature to be like has no bearing on what it &lt;strong&gt;is&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The light from the sun is a colossal amount of energy in human terms, but very minor in comparison to the rest of the universe. The light that didn't fall onto anything left the solar system and was never &quot;used&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The root of your misunderstanding is that you think that the sun has a purpose. &lt;a href=&quot;http://zeus.colorado.edu/astr1040-toomre/Lectures/Lecture06--30jan14.pdf&quot; rel=&quot;noreferrer&quot;&gt;It is a ball of plasma that emits energy&lt;/a&gt;. See for example Ernst Mayr's &lt;a href=&quot;http://www.informationphilosopher.com/solutions/scientists/mayr/Teleology.pdf&quot; rel=&quot;noreferrer&quot;&gt;essay on Teleology&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-08-22T18:04:07.847" LastActivityDate="2016-08-22T18:04:07.847" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17992" PostTypeId="2" ParentId="17990" CreationDate="2016-08-20T17:35:20.077" Score="57" Body="&lt;p&gt;The light from the Sun spreads, at least initially, in an isotropic fashion into the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it gets further from the Sun, &lt;em&gt;some&lt;/em&gt; of that light will interact with the interstellar medium (ISM) and therefore &lt;em&gt;some&lt;/em&gt; of the energy emitted by the Sun will be used to excite atoms and molecules or even ionise some atoms. This will be the fate of almost all the light which is emitted from the Sun towards the plane of our Galaxy, which contains sufficient molecular gas and dust to block starlight travelling through it for any distance. We know this happens because we can &quot;see&quot; dark clouds in the Milky Way, that can be penetrated by longer wavelength radiation to reveal all the billions of Sun-like stars that lie behind them. Roughly speaking, about half the visible light from the Sun will be absorbed every 1000 light years when travelling in the Galactic plane, so it is essentially all absorbed within a few thousand light years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But most of the Sun's light is not travelling in the direction of the Galactic plane, and interstellar and intergalactic space has a very low density of gas and dust. The equivalent extinction number for the intergalactic medium is that light travels many billions of light years with almost no chance of being absorbed (see &lt;a href=&quot;https://arxiv.org/abs/1005.4406&quot;&gt;Zu et al. 2010&lt;/a&gt;). This means that most of the light from the Sun will travel to cosmological distances (billions of light years) over the course of the next billions of years. Indeed light emitted from the Sun shortly after its birth has already travelled 4.5 billion light years. We know this has happened and will happen, because &lt;em&gt;we&lt;/em&gt; can observe galaxies (the light from which is nothing more than the summation of light from many stars like the Sun) that are 4.5 billion (and more) light years away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the Sunlight travels towards cosmological distances, its wavelength is &quot;stretched&quot; by the expansion of the universe, becoming redder and redder. We know this happens because distant galaxies have redshifted spectra. If the universe keeps expanding, then its density will continue to decrease and there is little to stop the radiation from the Sun travelling on forever, with a wavelength that scales as the scale factor, $a$, of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we follow a co-moving and co-expanding cube containing the Sun's radiation as the universe expands. The total radiative energy inside that cube diminishes as $a^{-1}$ - that is, the energy content of the universe in the form of radiation from stars (and other sources) becomes energetically less important as the universe expands and appears to be being superseded by the energy contained in the vacuum itself (a.k.a. dark energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, most of the energy emitted by the Sun is not &quot;used&quot; for anything; it propagates into space, becoming more and more dilute.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-08-20T20:15:38.757" LastActivityDate="2016-08-20T20:15:38.757" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17994" PostTypeId="2" ParentId="17988" CreationDate="2016-08-20T23:38:34.223" Score="5" Body="&lt;p&gt;As already mentioned by James Screech, &quot;[t]he constellation boundaries are purely artificial boundaries&quot;. The stars in any constellation are usually not &quot;related&quot; to each other in any way. The &lt;a href=&quot;https://www.youtube.com/watch?v=lD-5ZOipE48&quot; rel=&quot;noreferrer&quot; title=&quot;3D rendering of Orion&quot;&gt;3D rendering of Orion&lt;/a&gt; suggested by pela is a good example illustrating this point. There are some exceptions to this rule; the Pleiades, or Seven Sisters, are stars found in an open cluster, meaning they actually do exist in physical proximity of each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of it this way: Imagine you take a picture of, say, trees in a park. When you are standing in the park, taking the picture, the trees will be of various distances to you, and have varying distances between each other. When you look at the picture, though, some of the trees may look like they are closer together than they are in reality, because the picture you are looking at is 2D; it's a matter of perspective. In my illustration below, the blue line is longer than the red line, but the distance to the smaller tree is much bigger in reality, simply because it's standing very far behind the tree in the foreground. &lt;a href=&quot;https://i.stack.imgur.com/dPpro.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dPpro.png&quot; alt=&quot;Perspective&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were to start drawing lines to trees which look like they are close to the leftmost tree in the picture, we would be far more likely to draw the red line than the blue one, and make that our &quot;constellation&quot;. The same happens when people looked up at the starry sky back in the day; they looked at stars that &lt;em&gt;appeared&lt;/em&gt; to be close to each other in the sky, because that could make up images they could &lt;em&gt;imagine&lt;/em&gt; and &lt;em&gt;see&lt;/em&gt;, in a similar way to how we may lie down on the grass during a sunny day, and &lt;em&gt;imagine&lt;/em&gt; shapes in the clouds above us. &lt;/p&gt;&#xA;" OwnerUserId="13193" LastEditorUserId="2365" LastEditDate="2016-08-21T01:31:49.083" LastActivityDate="2016-08-21T01:31:49.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17995" PostTypeId="1" CreationDate="2016-08-15T22:09:20.210" Score="0" ViewCount="99" Body="&lt;p&gt;How can it be that all galaxies and stars are moving away from each other yet they still collide?  From what I have researched it as an open universe (or many open universes) yet the idea of stars and galaxies colliding boggles me.&lt;/p&gt;&#xA;" OwnerDisplayName="Dave Best" LastEditorUserId="6" LastEditDate="2019-01-29T14:46:28.060" LastActivityDate="2019-01-29T14:46:28.060" Title="The expansion of the universe and how stars and galaxies still collide" Tags="&lt;galaxy&gt;&lt;expansion&gt;&lt;collision&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17996" PostTypeId="2" ParentId="17995" CreationDate="2016-08-16T00:38:38.790" Score="0" Body="&lt;p&gt;Simple, and direct.  Objects moving away from each other can still cross paths.&#xA;&quot;Expansion is a &lt;em&gt;general&lt;/em&gt; direction, not a specific flight path.&lt;/p&gt;&#xA;" OwnerUserId="16027" OwnerDisplayName="Joe" LastActivityDate="2016-08-16T00:38:38.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17997" PostTypeId="2" ParentId="17995" CreationDate="2016-08-16T00:53:37.350" Score="1" Body="&lt;p&gt;The galaxies and stars &lt;em&gt;are not&lt;/em&gt; all moving away from each other. The space between them is expanding, however this does not affect their trajectories. For instance the Andromeda Galaxy is set to cross paths with the Milky Way, after which we will join together as the unimaginatively named &quot;Milkdromeda&quot;. As space expands it grows faster and faster, this will eventually lead to an &quot;island effect&quot;, where the space between galaxies will expand faster than light, at which point everything outside our galaxy will go dark forever.&lt;/p&gt;&#xA;" OwnerDisplayName="PtAltaria" LastActivityDate="2016-08-16T00:53:37.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17998" PostTypeId="2" ParentId="17995" CreationDate="2016-08-21T03:28:18.410" Score="4" Body="&lt;p&gt;The expansion is overridden on a small scale by gravity. Our galaxy is not expanding, and the stars are bound together.  In fact, a whole &lt;em&gt;group&lt;/em&gt; of galaxies don't notice the overall expansion. &lt;/p&gt;&#xA;" OwnerUserId="7142" LastEditorUserId="7142" LastEditDate="2016-08-24T07:41:42.793" LastActivityDate="2016-08-24T07:41:42.793" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17999" PostTypeId="1" AcceptedAnswerId="18000" CreationDate="2016-08-21T08:14:17.640" Score="0" ViewCount="127" Body="&lt;p&gt;The Apollo missions took moving rovers to the Moon. How many moving rovers are presently on the Moon?&lt;/p&gt;&#xA;" OwnerUserId="11792" LastEditorUserId="7996" LastEditDate="2016-08-21T17:53:49.837" LastActivityDate="2016-08-22T20:10:57.360" Title="How many rovers now on Moon?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ClosedDate="2016-08-21T15:47:31.133" ContentLicense="CC BY-SA 3.0" />
  <row Id="18000" PostTypeId="2" ParentId="17999" CreationDate="2016-08-21T08:18:56.393" Score="4" Body="&lt;p&gt;Three lunar rovers and the Chinese Yutu rover. None of which are operational. &lt;a href=&quot;https://en.m.wikipedia.org/wiki/List_of_artificial_objects_on_the_Moon&quot; rel=&quot;nofollow&quot;&gt;https://en.m.wikipedia.org/wiki/List_of_artificial_objects_on_the_Moon&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-08-22T20:10:57.360" LastActivityDate="2016-08-22T20:10:57.360" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18001" PostTypeId="1" CreationDate="2016-08-21T17:59:43.737" Score="1" ViewCount="3223" Body="&lt;p&gt;Last night we saw a large cluster of what looked like exceptionally bright stars that moved, as a group, across the night sky in Lancaster, PA. It was about 9:15 pm, and they moved roughly from SW to NE. There were 25 or 30 of them. They moved as quickly as a satellite I once saw, but were bigger and brighter. They were soundless and did not blink, and the relationships between the lights changed slightly as they moved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't believe in UFOs, but can't we find any explanation via google. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can that many satellites move as a group?&lt;/p&gt;&#xA;" OwnerUserId="13899" LastEditorUserId="2153" LastEditDate="2016-08-22T15:51:19.780" LastActivityDate="2019-11-17T21:13:25.567" Title="Can satellites travel in clusters?" Tags="&lt;orbit&gt;&lt;amateur-observing&gt;&lt;artificial-satellite&gt;" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18002" PostTypeId="2" ParentId="17990" CreationDate="2016-08-21T19:16:05.400" Score="8" Body="&lt;p&gt;Entropy is a fundamental condition of our universe, and has been recognized as such since as long ago as Newton's laws of thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Entropy:&lt;/strong&gt; Order does not increase over time, it decreases, except locally with the expenditure of energy. This expenditure trades off increased disorder elsewhere for increased order locally, and the tradeoff is always negative: The amount of order gained is always smaller than the amount of disorder created.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each emitted photon either keeps moving until the end of Time as we know it, slowly losing energy as the cosmos expands, or else, it interacts with other particles along the way. However, the notion that those interactions, or the lack of interactions, implies greater and lesser degrees of Utility or of Purpose or of Destiny is a metaphysical one, not an astronomy question or even a physics question.&lt;/p&gt;&#xA;" OwnerUserId="13900" LastEditorUserId="13900" LastEditDate="2016-08-21T19:38:22.987" LastActivityDate="2016-08-21T19:38:22.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18003" PostTypeId="1" AcceptedAnswerId="18030" CreationDate="2016-08-21T19:52:21.430" Score="-4" ViewCount="2466" Body="&lt;p&gt;That's the question. I know about gamma ray bursts, is there anything more energetic?&lt;/p&gt;&#xA;" OwnerUserId="9949" LastActivityDate="2016-08-24T04:37:24.637" Title="Which are the most energetic events in the Universe?" Tags="&lt;cosmology&gt;&lt;astrophysics&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18004" PostTypeId="2" ParentId="18001" CreationDate="2016-08-21T20:39:02.787" Score="1" Body="&lt;p&gt;I can't answer your second question. Do you have a video of them? Why do you think they were in orbit? Are balloons a possibility? You say 25 or 30, did you observe them simultaneously or did they blink in and out? How long did you observe them, how far did they travel, and various other questions would need answers to even speculate... Now as far as your first question, the answer is no and yes. A ballistic orbit (not powered) will circle the Earth, and by &quot;circle&quot; I mean the path will be a great circle (which you can look up). This means that no two orbits (at the same altitude) are parallel everywhere. If they seem to be traveling together in parallel now, then their orbits must cross twice for every orbit around the Earth. So IF they were at the same altitude, they would have to be in orbital paths that intersected, so they would have to be staggered so as not to collide. At different altitudes, you can stack any number of satellites above one another. But it's very unlikely that this is what you saw, there's no reason to set up two dozen or more satellites in this type of configuration. Just based on your comments, perhaps what you saw was the break-up of a satellite. 25-30 satellites is an enormous number and the cost would also be extravagant (unless they were mini-satellites).&lt;/p&gt;&#xA;" OwnerUserId="13901" LastActivityDate="2016-08-21T20:39:02.787" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18005" PostTypeId="2" ParentId="18003" CreationDate="2016-08-21T21:17:48.097" Score="-2" Body="&lt;p&gt;It depends on the gamma ray burst, but it is possible for some quasars and some high end super/hypernovae to produce an equal or greater amount of energy (this is from the perspective of those that involve single stars or stellar remnants-we are leaving galaxies, galaxy clusters, etc. out of it)&lt;/p&gt;&#xA;" OwnerUserId="13872" LastActivityDate="2016-08-21T21:17:48.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18006" PostTypeId="2" ParentId="18001" CreationDate="2016-08-21T21:20:27.380" Score="1" Body="&lt;p&gt;As a guess, I'd say you were probably looking at a group of aircraft flying together, possibly helicopters. Usually you can tell aircraft at night because the have red and green lights and anti collision lights that blink, but I have occasionally observed aircraft flying with landing lights on, and could not make out the other lights until they were closer, or they had changed direction, making them stand out.&lt;/p&gt;&lt;p&gt;I've never seen a UFO, although I once mistook Venus for one.&lt;/p&gt;&#xA;" OwnerUserId="9516" LastActivityDate="2016-08-21T21:20:27.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18009" PostTypeId="1" AcceptedAnswerId="18017" CreationDate="2016-08-22T03:38:54.550" Score="0" ViewCount="90" Body="&lt;p&gt;Given that I got the equation relating the angular diameter(θ) and $V$–$K$ band value, which is known as $\logθ_0=(0.262±0.004)(V−K)_0+(0.547±0.006)$ from this &lt;a href=&quot;https://www.researchgate.net/publication/46390282_The_projection_factor_period-radius_relation_and_surface-brightness_colour_relation_in_classical_Cepheids&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt;, how do I get the 'change' in angular diameter value from that equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Can I just get the max and mix value of the $V$–$K$ value and insert each into the equation and get max θ and min θ and get the difference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Or I need to the whole lists of angular diameter value over each period and insert the lists of value into the equation and plot the graph and then get the difference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Or is there any other method to do this?&lt;/p&gt;&#xA;" OwnerUserId="13714" LastEditorUserId="9301" LastEditDate="2016-08-22T05:04:17.680" LastActivityDate="2016-08-22T19:48:03.130" Title="How do I get the change in angular diameter value?" Tags="&lt;photometry&gt;&lt;angular-diameter&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18014" PostTypeId="5" CreationDate="2016-08-22T15:15:56.740" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2016-08-22T15:15:56.740" LastActivityDate="2016-08-22T15:15:56.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18015" PostTypeId="4" CreationDate="2016-08-22T15:15:56.740" Score="0" Body="The application of ballistics and celestial mechanics to the practical problems concerning the motion of rockets and other spacecraft." OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-08-22T16:05:38.743" LastActivityDate="2016-08-22T16:05:38.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18017" PostTypeId="2" ParentId="18009" CreationDate="2016-08-22T19:48:03.130" Score="2" Body="&lt;p&gt;You can see from the paper you linked that they followed the procedure you outlined in option (2). Figure 4 of that paper shows the $\theta (t)$ for a few stars. They specifically state towards the end of section 6&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Figure 4 shows in the top panel the relation between the angular diameter predicted by the SB-relation versus radial variation, and in the bottom panel the angular diameter as a function of phase.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The SB-relation they're talking about is the one you cite in your question. So your process should be to take your list of $V-K$ colors over time and convert them to angular size using the surface brightness relation, then fit a function to that (via cubic spline, least squares, etc.) and determine the total angular variation over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will note also that the $\theta$ vs $V-K$ relation you've cited here is equation 10 in the linked paper which is &lt;em&gt;not&lt;/em&gt; a result of that paper. Their result is listed in equation 9 with equations 10 and 11 being results from other, previous work for the reader to compare. You should sure of which equation you want to use and cite the appropriate source.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-08-22T19:48:03.130" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18019" PostTypeId="2" ParentId="10177" CreationDate="2016-08-23T06:50:07.270" Score="1" Body="&lt;p&gt;For a more analytical and much much less philosophical version of an answer than MacUserT's, here it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to MacUserT, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now, what has this to do with the ending of the universe in respect to&#xA;  time? The theory you mentioned as heat death is such a transcendental&#xA;  theory. The theory states that the universe expands until the universe&#xA;  is too big to be heated by the energy provided by the matter in the&#xA;  universe. This theory models that the universe follow the same rules&#xA;  as any thermo dynamic system and therefore at a certain point in time&#xA;  all energy, matter and temperature are evenly distributed in this huge&#xA;  universe. Because all is evenly distributed no stars will be created&#xA;  and all processes come to a grinding halt. Since the state before, now&#xA;  and after are the same, the scientists now determine this as the end&#xA;  of time. The formula theories also end up making some theory that time&#xA;  will end.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My issue with this statement, lies in the facts that the eventual heat death of the universe is not so much a transcendental theory so much as it is a thermodynamic &lt;strong&gt;law&lt;/strong&gt;. The heat death of the universe is more closely attached to the lifetime of stars in the universe than the cooling of an evenly distributed temperature throughout space. As galaxies age, they contain less and less star forming regions, thereby creating less and less stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Less stars means less heat, and when the final stars and white dwarfs die and cool off respectively the entirety of the universe will finally be able to reach 0K. This is no insignificant timescale, keep in mind that WDs take about $10^{13}$ years just to cool from their initial states down to a temperature where they will no longer emit visible light, which means they could take another $10^{20}$ years to reach anything close to zero kelvin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the even distribution of temperature. Space already has a fairly homogeneous distribution of heat ~2.7K it's called the CMB, and it has been cooling since the big bang, and there is no source of heat that will allow for an increase of this temperature on a whole, or in one word... entropy. Space is far too vast and heat sources are far too ... far ... apart to ever affect anything other than extremely local conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the end of time issue. The idea of time involves the transmission of information. For example, if you send a radio transmission to the moon and back, you will state that some time has passed from the sending to the receiving of the transmission. This process is possible because there is a temperature above 0K, for the electromagnetic waves to propagate. Meaning that when the temperature of all space reaches 0K, there will be no energy available for electromagnetic waves to propagate. Thus no information can be transmitted, and time ... is at a standstill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the universe would not end by time stopping, because time stopping is a by product of the &quot;universe ending&quot; because its heat death.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a fun version of this question being &lt;em&gt;answered&lt;/em&gt;, check out &#xA;Issac Asimov's short story &lt;em&gt;The Last Question&lt;/em&gt; &lt;a href=&quot;http://www.physics.princeton.edu/ph115/LQ.pdf&quot; rel=&quot;nofollow&quot;&gt;Click here to go to Princeton's pdf version of it&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2232" LastActivityDate="2016-08-23T06:50:07.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18020" PostTypeId="2" ParentId="17922" CreationDate="2016-08-23T06:55:43.700" Score="6" Body="&lt;p&gt;Firstly, gravitational waves (GWs) are not an echo - we measure the direct signal.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process you describe here is known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_lens&quot; rel=&quot;noreferrer&quot;&gt;gravitational lensing&lt;/a&gt;, the deviation of (usually) light rays due to massive objects between the source and the observer.  This also applies to GWs.  The result will be similar - the direction of the waves can be changed, resulting in a distorted image of the source and possibly a 'brighter' image (stronger signal) due to 'gravitational focussing'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, LIGO and Virgo do not image GWs, they detect point sources.  Hence, there is no distorted image, but the 'luminosity' of the signal may be affected, resulting in a (somewhat) misjudged distance (although they use a 'luminosity distance', which is not necessarily the same as a current physical separation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as in the case of light, a black hole could trap some of the GW signal that would have reached the observer (gravitons that fall directly into the black hole), if it is directly between the source and the observer, but the gravitational focussing makes it likely that the resulting signal will be stronger rather than weaker.&lt;/p&gt;&#xA;" OwnerUserId="286" LastActivityDate="2016-08-23T06:55:43.700" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18021" PostTypeId="2" ParentId="16488" CreationDate="2016-08-23T07:25:00.837" Score="2" Body="&lt;p&gt;As @Andy says, at any time there is a great circle around the Earth where the Sun is currently at the horizon, either rising or setting (or both at two points in the (ant)arctics).  This great circle can be described by the coordinates of the point where the Sun is directly overhead (which is the vector pointing from the Earth's centre to that location on the surface, or to the Sun).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a great circle exists for your moment for sunset, and another one exists for your time for sunrise.  You are interested in one of the two intersection points of these two great circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two great circles lie in two planes through the Earth's centre, described by two of the vectors mentioned before.  The intersection of the two planes is a line through the centre, whose direction is another vector (the normalised cross product of the two normal vectors), which can be converted back to two antipode positions on the Earth's surface.  This solution would be rather exact if the Earth were a sphere.&lt;/p&gt;&#xA;" OwnerUserId="286" LastEditorUserId="286" LastEditDate="2016-08-24T05:55:22.800" LastActivityDate="2016-08-24T05:55:22.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18023" PostTypeId="1" AcceptedAnswerId="18025" CreationDate="2016-08-23T10:53:57.583" Score="2" ViewCount="329" Body="&lt;p&gt;Since there's very little atmosphere on Mars, it doesn't matter much whether the Sun is in the zenith or hanging just above the horizon: without the atmospheric &quot;filter,&quot; the sunlight would have roughly the same effect on all sunlight areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would the temperatures be like on the sunlit portion of the planet?&lt;/p&gt;&#xA;" OwnerUserId="9510" LastEditorUserId="11143" LastEditDate="2016-08-23T12:37:21.303" LastActivityDate="2016-08-23T13:13:41.563" Title="If Mars were tidally locked to the Sun, what would the temperatures on its surface be like?" Tags="&lt;the-sun&gt;&lt;mars&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18024" PostTypeId="1" AcceptedAnswerId="18026" CreationDate="2016-08-23T12:45:24.267" Score="4" ViewCount="299" Body="&lt;p&gt;I don't have any formal education in physics and I have found it hard to find a succinct and somewhat simple to comprehend explanation of what we think is the actual reality of a black hole. I'm asking after seeing a statement saying that the mass of the central supermassive black hole in the Phoenix Cluster was 20 billion solar masses and the diameter we think to be approximately 118 billion kilometres across.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is someone able to point me to a resource or give me a simple explanation as to what we think a black hole actually is in shape? Is it a solid sphere with a central point and a radius of approximately 59 billion kilometres? A spinning disc with a face that is 118 billion kilometres across and if so what are the other measurements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I thinking about this all wrong with regards to trying to understand it as a shape at all? If I understand correctly, mass doesn't give any indication of size as it is dependent on density. Is a black hole just basically a somewhat spherical body of mass like any other that just has an absolutely massive density and spin?&lt;/p&gt;&#xA;" OwnerUserId="13936" LastActivityDate="2016-08-23T17:40:09.397" Title="Succinct explanation of black hole mass, diameter, shape?" Tags="&lt;black-hole&gt;&lt;supermassive-black-hole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18025" PostTypeId="2" ParentId="18023" CreationDate="2016-08-23T12:55:36.713" Score="4" Body="&lt;p&gt;Tidal lock would mean only one side of Mars would be exposed to solar irradiation, so its temperature would be mostly defined by the equilibrium between the energy received from the Sun and the black-body radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the orbit eccentricity, solar irradiation reaching Martian surface varies between 493 and 717 W/m² (&lt;a href=&quot;http://ccar.colorado.edu/asen5050/projects/projects_2001/benoit/solar_irradiance_on_mars.htm&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;). According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law&quot; rel=&quot;nofollow&quot;&gt;Stefan-Boltzmann law&lt;/a&gt;, the temperature required for the black-body radiation to match these levels will vary between 305 and 335 K (32°C an 62°C), respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such temperatures will be reached in the area directly facing the Sun (so the Sun is seen in zenith), and gradually decline towards the background radiation temperature of space (about 3K) when approaching the dark side (where the Sun is just barely above the horizon). Such temperature distribution is due to the angle of incidence, not to the atmospheric &quot;filter&quot; or lack thereof.&lt;/p&gt;&#xA;" OwnerUserId="13926" LastEditorUserId="13926" LastEditDate="2016-08-23T13:13:41.563" LastActivityDate="2016-08-23T13:13:41.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18026" PostTypeId="2" ParentId="18024" CreationDate="2016-08-23T13:11:43.913" Score="5" Body="&lt;p&gt;I should start off by saying that black holes are enigmas. In describing our universe, we have the General Theory of Relativity (GR) to help us describe gravity and objects which are very massive. We also have quantum mechanics (QM) and quantum field theory to help us describe small objects such as particles. For the most part, these two realms are completely separate and these two important theories don't mix. After all, you don't need QM to describe the physics of orbiting the Sun, and you don't need GR to describe electrons orbiting an atomic nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A problem arises though, when you begin talking about objects which require both QM and GR, and black holes fall very nicely into this regime. Often we find that when we mix the two theories, we wind up with nonsensical answers. It is highly probable that our understanding of the mechanics of a black hole is incomplete. What we really need is a quantum gravity theory to talk about black holes and as yet we don't have one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, let me lay out the basic structure of a black hole based on what current theory suggests to be true. Fortunately, black holes are relatively simple in structure, as dictated by the &lt;a href=&quot;https://en.wikipedia.org/wiki/No-hair_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;No-Hair Theorem&lt;/a&gt; (which basically says you only need a few parameters to completely describe the state of a black hole, as opposed to what you'd need to fully describe the physics of a person or even an atom). The (somewhat simplified) anatomy of a black hole is shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2SavL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2SavL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this diagram there are a few important parts. The first being the &lt;strong&gt;singularity&lt;/strong&gt;. When you put a bunch of mass together, gravity acts to squeeze that mass into a ball. That is what drives the fusion in our Sun and makes our planet round. Normally there is some resistive force that stops the matter from collapsing completely, such as the structural integrity of rocks in the Earth or fusion in the Sun. At some point though, if you pack on enough mass, that resistive force can't hold out and gravity wins. In this case, all the matter collapses down as far as it can go. It becomes a point of matter known as a singularity. This singularity has no size, it is literally (and mathematically) a point. All the mass of the black hole exists here. So when they say a black hole is 20 billion solar masses, they mean that the singularity contains the matter equivalent to 20 billion Suns, all confined to that single point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other important concept for a black hole is the &lt;strong&gt;event horizon&lt;/strong&gt;. I should stress that the event horizon is &lt;em&gt;not&lt;/em&gt; a physical object in space. It is a mathematical boundary. Without getting into the nitty-gritty and in basic terms, the event horizon is where our current physics &quot;stops working&quot;. We cannot describe inside the event horizon with our current physics because we are in that region where both QM and GR are important and they don't play well together. The event horizon is also the boundary where, if crossed, you're trapped forever in the black hole. There's a lot more physics involved with the event horizon, but that's the gist. Usually when someone refers to the size of a black hole, they're talking about the event horizon. Thus if they say the black hole is 118 billion kilometers across, they mean that the event horizon has a diameter of 118 billion kilometers. Note, from the image above, that the radius of the event horizon is easily calculated from only the mass of the singularity (and a few physical constants).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are further components to a black hole such as the ergosphere or photon sphere, and often black holes have accretion disks of material around them. I'll leave those concepts to another question though.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2016-08-23T17:40:09.397" LastActivityDate="2016-08-23T17:40:09.397" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18027" PostTypeId="1" CreationDate="2016-08-23T15:11:40.000" Score="2" ViewCount="914" Body="&lt;p&gt;&lt;a href=&quot;http://hubblesite.org/hubble_discoveries/10th/vault/allabout.shtml&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; website claims:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The space telescope can detect objects as faint as 31st magnitude...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's referring to Hubble, but does not cite any source or math. How do you calculate the faintest (apparent) magnitude that a space telescope can detect?&lt;/p&gt;&#xA;" OwnerUserId="7449" LastActivityDate="2016-08-23T19:21:12.403" Title="What is the faintest magnitude a telescope can see?" Tags="&lt;telescope&gt;&lt;light&gt;&lt;optics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18028" PostTypeId="2" ParentId="18027" CreationDate="2016-08-23T16:11:07.263" Score="5" Body="&lt;p&gt;This is a term known at the limiting magnitude. This term refers simply to the faintest apparent magnitude your instrument can detect. Wikipedia has an article describing &lt;a href=&quot;https://en.wikipedia.org/wiki/Limiting_magnitude&quot; rel=&quot;nofollow&quot;&gt;limiting magnitude&lt;/a&gt;.  I pulled the info from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wiki article begins off explaining the basic way of finding limiting magnitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$5 \cdot \log_{10}\left(\frac{D_1}{D_0}\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if we simply plug in some values, 254 mm telescope and assuming the eye is 6 mm, we get an answer of 8.13.  This can be rounded to nearest base and we can say that it increases the magnitude by 8.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article goes into more description about the contrast of the given medium. With Earth, we have the atmosphere, lights, and pollution to worry about. The more you are able to reduce these effects, the better. That's why for telescopes at observatories, we place them in high places, usually away from populated areas to help reduce these worrisome effects. This helps increase the contrast between the object you are looking at and its medium (in astronomy the night sky).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the article it explains this with the formula below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m \cdot v = m_{nakedeye} - 2 + 2.5 \cdot \log_{10}(D \cdot P \cdot t)$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$D$ = objective or main mirror diameter in mm&lt;/li&gt;&#xA;&lt;li&gt;$P$ = power or magnification&lt;/li&gt;&#xA;&lt;li&gt;$t$ = transmission factor, usually 0.85–0.9&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Simply putting in our values from above and estimating a 10&quot; telescope has a 250x magnification with a transmission factor of 0.85, we get how the wiki article states that the answer of a 10&quot; telescope easily getting a magnitude 15 is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for observational laboratories and something like the Hubble telescope, they use additional factors, including image processing mechanisms that help reduce noise in the image and integration time of the object you are viewing.  This means using a sensor that can capture the image at low light levels AND while it moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really want to know more, &lt;a href=&quot;http://www.stsci.edu/hst/acs/documents/handbooks/cycle19/c01_intro.html&quot; rel=&quot;nofollow&quot;&gt;go here&lt;/a&gt;. This link actually explains in very technical detail the technology behind the imaging discovery efficiency.  Basically, it helps boost the limiting magnitude to detect extremely faint objects.&lt;/p&gt;&#xA;" OwnerUserId="13849" LastEditorUserId="9301" LastEditDate="2016-08-23T19:21:12.403" LastActivityDate="2016-08-23T19:21:12.403" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18029" PostTypeId="2" ParentId="16488" CreationDate="2016-08-24T04:13:23.657" Score="4" Body="&lt;p&gt;To avoid the headaches associated with finding roughly two dozen similar solutions in the zig-zagging time zone boundaries , &lt;strong&gt;you might want to stick with UTC&lt;/strong&gt;. If you want to deal with local time - it's going to get a little crazy if you also try to include daylight savings toggling on and off.&lt;/p&gt;&#xA;&lt;p&gt;The points on the earth where the center of the sun coincides with the horizon (&lt;em&gt;ignoring topography, oblateness, atmospheric refraction, the finite speed of light and other small effects&lt;/em&gt;) is just the circle on the earth where a cone drawn from the Sun intersects a spherical Earth.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5fyLc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5fyLc.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; A sphere inside a cone, from &lt;a href=&quot;http://mathcentral.uregina.ca/QQ/database/QQ.09.07/s/juan1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://mathcentral.uregina.ca&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;But the next step is hard - getting from that circle to the lat/lon coordinates on the surface, because the axis is tilted, and because the motion around the sun speeds up and slows down as the Earth revolves closer and farther from the Sun. These can be approximated by periodic functions with some coeficients, and that's what you will find in the math behind the first answer. For the second answer, I'll list two Python solutions - PyEphem and Skyfield. Both are easy to use, but you are separated from the actual math (in one case it's an ephemeris/table. The third answer is really a collection of NASA/JPL routines that are highly regarded but may involve more time for you to get up to speed compared to the Python packages.&lt;/p&gt;&#xA;&lt;h2&gt;Answer 1: Astronomical Algorithms&lt;/h2&gt;&#xA;&lt;p&gt;This is something you'll have to dig into a bit, but if you like to program, it may be exactly what you're looking for. The &lt;a href=&quot;http://www.gaisma.com/en/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gaisma website&lt;/a&gt; is one of my favorites on the internet - easy to use and presents a bunch of information in easy-to-understand graphics. Click around!&lt;/p&gt;&#xA;&lt;p&gt;I believe that this site uses algorithms from the collection found at &lt;a href=&quot;http://www.esrl.noaa.gov/gmd/grad/solcalc/calcdetails.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this NOAA site&lt;/a&gt;. Click around there as well. They provide Excel spreadsheets which contain the algorithms and other resources. The &amp;quot;main&amp;quot; resource is a collection of algorithms published in the book &lt;a href=&quot;http://www.willbell.com/math/mc1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomical Algorithms - Jean Meeus&lt;/a&gt;. From that Amazon page you can see that there are many similarly titled books. I'd recommend going to a library if possible, because it's (in my opinion) always good to go to libraries. However parts of these can be found on-line. For example, a few pages shown from the book &lt;a href=&quot;http://www.math.harvard.edu/%7Eknill/3dprinter/exhibits/bc/Meeus1988.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomical Formulae for Calculators (1988)&lt;/a&gt; include an interesting table of contents.&lt;/p&gt;&#xA;&lt;h2&gt;Answer 2: Python packages PyEphem and Skyfield&lt;/h2&gt;&#xA;&lt;p&gt;I'll copy a bit of the text from &lt;a href=&quot;https://astronomy.stackexchange.com/a/17946/7982&quot;&gt;this answer&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The Python package &lt;a href=&quot;http://rhodesmill.org/pyephem/&quot; rel=&quot;nofollow noreferrer&quot;&gt;PyEphem&lt;/a&gt; has been around and well supported, and is the pythonic reincarnation of &lt;a href=&quot;http://www.clearskyinstitute.com/xephem/&quot; rel=&quot;nofollow noreferrer&quot;&gt;XEphem&lt;/a&gt;. I haven't used it, but I believe it keeps enough information about orbital parameters at certain epochs to generate an ephemeris, including some gravitational perturbations. In other words, it's much more than planets moving on fixed elliptical orbits around a fixed sun. So I believe it runs without internet connection.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I never used it because I was recommended to look at &lt;a href=&quot;http://rhodesmill.org/skyfield/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Skyfield&lt;/a&gt; and it's exactly what I needed. It downloads a standard JPL ephemeris that you choose, and then just uses it from your hard drive after that. However, in order to deal with leap seconds and other &lt;em&gt;time related&lt;/em&gt; effects, it &lt;em&gt;occasionally&lt;/em&gt; needs to check the internet for leap second information updates, since these are arbitrary.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;I don't know if Skyfield has a mode to avoid that. Actually that's a good question. If you work with a timescale that doesn't have leap seconds, I am not sure if it will run in its current version.&lt;/p&gt;&#xA;&lt;p&gt;Both Skyfield and PyEphem Python packages have been written and are maintained by @BrandonRhodes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I've included a simple Python script just to illustrate how Skyfield can be used. If you are comfortable with coding without curly braces, I would strongly recommend you give this a try. It's incredibly powerful and Pythonic.&lt;/p&gt;&#xA;&lt;p&gt;This is just a starter - you need to add some better housekeeping to detect sunrises vs sunsets, and maybe a more global-type search in some cases. Actually there is some slightly tedious housekeeping necessary to make this work robustly.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;note:&lt;/strong&gt; you can turn on atmospheric refraction using the arguments in the &lt;code&gt;apparent()&lt;/code&gt; method. See the &lt;a href=&quot;http://rhodesmill.org/skyfield/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Skyfield API&lt;/a&gt; documentation for more info, and for a discussion about iterating using Skyfield methods - especially solving for times, see &lt;a href=&quot;https://stackoverflow.com/a/36380163/3904031&quot;&gt;this helpful answer&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1nFcb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1nFcb.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;def alt_lonlat(lon, lat, t):&#xA;    &#xA;    topo = earth.topos(lat, lon)&#xA;    &#xA;    alt, az, dist = topo.at(trise).observe(sun).apparent().altaz() ## apparent() args for atmospheric refraction&#xA;    &#xA;    return alt.degrees&#xA;&#xA;&#xA;from skyfield.api import load&#xA;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;import scipy.optimize as spo&#xA;&#xA;data  = load('de421.bsp')&#xA;ts    = load.timescale()&#xA;&#xA;# your example:  '2014-08-27', sunset = '10:00 PM', sunrise = '7:30 AM'&#xA;&#xA;trise  = ts.utc(2014, 8, 27,  7, 30, 0)&#xA;tset   = ts.utc(2014, 8, 27, 22,  0, 0)&#xA;&#xA;earth = data['earth']&#xA;sun   = data['sun']&#xA;&#xA;zerozero = earth.topos(0.0, 0.0)   # gotta start looking somewhere!&#xA;&#xA;alt, az, dist = zerozero.at(trise).observe(sun).apparent().altaz() ## apparent() args for atmospheric refraction&#xA;&#xA;print &amp;quot;at trise, JD = &amp;quot;, trise.tt&#xA;print &amp;quot;at (0N, 0E) Sun's altitude: &amp;quot;, alt.degrees, &amp;quot;azimuth: &amp;quot;, az.degrees&#xA;print &amp;quot;at (0N, 0E) Sun's distance (km): &amp;quot;, dist.km&#xA;&#xA;# Find points on equator where sun is on horizon (rise or set) at t=trise&#xA;&#xA;limits   = ((0, 180.), (180, 360.))&#xA;lonzeros = []&#xA;&#xA;for a, b in limits:&#xA;&#xA;    answer, info = spo.brentq(alt_lonlat, a, b,&#xA;                              args=(0.0, trise),&#xA;                              full_output = True )&#xA;&#xA;    if info.converged:&#xA;        lonzeros.append(answer)&#xA;        print &amp;quot;limits &amp;quot;, a, b, &amp;quot; converged! Found longitude (deg): &amp;quot;, answer&#xA;    else:&#xA;        print &amp;quot;limits &amp;quot;, a, b, &amp;quot;whaaaa?&amp;quot;&#xA;        lonzeros.append(None)&#xA;&#xA;# make some curves&#xA;&#xA;lats = np.linspace(-60, 60, 13)&#xA;&#xA;longis = []&#xA;for lon0 in lonzeros:&#xA;    lons = []&#xA;    for lat in lats:&#xA;        &#xA;        answer, info = spo.brentq(alt_lonlat, lon0-90, lon0+90,&#xA;                                  args=(lat, trise),&#xA;                                  full_output = True )&#xA;        if info.converged:&#xA;            lons.append(answer)&#xA;        else:&#xA;            lons.append(None)&#xA;&#xA;        lons = [(lon+180)%360.-180 for lon in lons]  # wraparound at +/- 180&#xA;&#xA;    longis.append(lons)&#xA;&#xA;plt.figure()&#xA;&#xA;for lons in longis:&#xA;    plt.plot(lons, lats)&#xA;&#xA;for lons in longis:&#xA;    plt.plot(lons, lats, 'ok')&#xA;&#xA;plt.xlim(-180, 180)&#xA;plt.ylim(-90, 90)&#xA;&#xA;plt.title(&amp;quot;at trise, JD = &amp;quot; + str(trise.tt))&#xA;&#xA;plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;h2&gt;Answer 3: SPICE Kernels&lt;/h2&gt;&#xA;&lt;p&gt;As pointed out by @barrycarter in &lt;a href=&quot;https://astronomy.stackexchange.com/questions/17942/function-for-current-distance-between-in-light-seconds-between-mars-and-earth/17946#comment27647_17946&quot;&gt;this comment&lt;/a&gt; below &lt;a href=&quot;https://astronomy.stackexchange.com/a/17946/7982&quot;&gt;this answer&lt;/a&gt;, the JPL SPICE Kernels are available. I'm not familliar with them but it's what NASA uses so it must be pretty good :)&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Appendix:&lt;/h2&gt;&#xA;&lt;p&gt;Here are some screenshots for London, UK (&lt;a href=&quot;http://www.gaisma.com/en/location/london.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gaisma from here&lt;/a&gt;):&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rJ0NO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rJ0NO.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DLyLA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DLyLA.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zIkQz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zIkQz.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H3teS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H3teS.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-10-10T12:34:30.390" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="18030" PostTypeId="2" ParentId="18003" CreationDate="2016-08-24T04:37:24.637" Score="1" Body="&lt;p&gt;Wikipedia has a page on this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Orders_of_magnitude_(energy)&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Orders_of_magnitude_(energy)&lt;/a&gt;    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would have guessed the recently observed merging of 2 black holes, which, while just hard to detect gravity waves, in terms of energy it was about the equivalent of the entire rest mass of our sun.  Two large (but not super-massive) black holes each about 30 solar masses, merged and released about 3 solar masses in energy in gravity waves.   That's a huge amount of energy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd have guessed wrong though, as an observed gamma ray burst was bigger.  (see link above for precise estimates).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory, since I assume you're talking about localized events, when 2 galaxies merge and their super-massive black holes collide, that's probably, hands down, the most energetic local event, far surpassing the largest gamma-ray bursts.  Gravity waves might not be what people think of as energetic, but they can be enormously so.    If you're talking more typical &quot;bright flash&quot;, photons and accelerated particles, then the winner would be the largest gamma ray bursts.    Hypernova's are up there, but not as large as the largest Gamma-ray Bursts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you want to get flexible on what &quot;local&quot; means.  The gravitational energy of two large galaxies falling towards each other might get you a larger number, but I'm a little fuzzy on whether that qualifies as a local event&quot;, which I've built this answer on.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-08-24T04:37:24.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18031" PostTypeId="1" CreationDate="2016-08-24T05:24:57.617" Score="2" ViewCount="447" Body="&lt;p&gt;I was using my telescope to look at a nebula, intell I seen a star like object that was moving at a moderate speed in a straight line and eventually dimmed out. My first thought that it was just a satellite but I could only see it through my telescope which makes me think that it was to small to be a satellite. I'm not sure.&lt;/p&gt;&#xA;" OwnerUserId="13942" LastEditorUserId="2365" LastEditDate="2016-08-25T18:36:55.853" LastActivityDate="2017-01-12T20:32:45.373" Title="Moving object seen through telescope, what is it?" Tags="&lt;artificial-satellite&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18032" PostTypeId="1" AcceptedAnswerId="18034" CreationDate="2016-08-24T11:06:31.557" Score="3" ViewCount="1897" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Where can I see - or even download - a fairly high resolution (~0.1 nm) solar spectrum at zero air mass - in other words from space without absorption and other features from the Earth's atmosphere? Of particular interest is the near infrared around 750-780nm. Is the &quot;A&quot; Fraunhofer line due to absorption by oxygen in the Sun's atmosphere, or in Earth's?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The background discussion below shows some examples, but these are not directly from scientific sources and I have a hunch they are from observations on Earth's surface below the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Studies of photosynthesis on Earth using spectrometers in satellites measure solar flux induced fluorescence from foliage during the day, taking advantage of certain bands in the solar spectrum that are relatively &quot;dark&quot;. &lt;a href=&quot;http://www.nasa.gov/content/goddard/seeing-photosynthesis-from-space-nasa-scientists-use-satellites-to-measure-plant-health&quot; rel=&quot;nofollow noreferrer&quot;&gt;GOME-2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Fraunhofer_lines#Discovery&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fraunhofer line&lt;/a&gt; labeled &quot;A&quot; below is produced by absorption by oxygen, but it is not clear to me if it is the oxygen in the Suns atmosphere or the Earth's atmosphere that's producing this line. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The images below are spectra of sunlight with various instruments of widely different resolutions, and because the absorption is narrow, the apparent depth is a function of resolution rather than reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/atWQl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/atWQl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; a low resolution spectrum &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Spectrum_of_blue_sky.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;from here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wqWwe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wqWwe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; An spectrogram of the Sun &lt;a href=&quot;http://www.noao.edu/image_gallery/html/im0600.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;from this NOAO website&lt;/a&gt; (N.A.Sharp, NOAO/NSO/Kitt Peak FTS/AURA/NSF) with the &quot;A&quot; line near 760nm appearing near the top, a little right of center. The original data is from the &quot;Fourier Transform Spectrometer at the McMath-Pierce Solar Facility at the National Solar Observatory on Kitt Peak, near Tucson, Arizona&quot; but seems to be formatted and colorized to look like an Echelle spectrogram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A video describing the use of the dark band for Earth science can be seen at &lt;a href=&quot;https://youtu.be/1XilneV3cJI&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/1XilneV3cJI&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2016-08-24T11:38:34.143" LastActivityDate="2016-08-24T14:26:11.710" Title="How deep is the &quot;A&quot; Fraunhofer line in the solar spectrum? Is it from the Sun's or Earth's atmosphere?" Tags="&lt;the-sun&gt;&lt;spectroscopy&gt;&lt;spectra&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18033" PostTypeId="2" ParentId="17975" CreationDate="2016-08-24T12:27:11.730" Score="2" Body="&lt;p&gt;There are two types of parallel observations: Coordinated Parallel and Pure Parallel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coordinated Parallel means part of the same proposal, so it has to have the same scientific justification: you have to come up with a good case for why the different observations would be useful for the same project. (This is case with CANDELS.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pure Parallel means something independent of the main observation, which I think is more what you're asking about. The proposal guidelines currently note that these are allowed &lt;em&gt;only&lt;/em&gt; when the primary observation is spectroscopy (with the COS or STIS instruments).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One factor is that you need &lt;em&gt;some&lt;/em&gt; specification for &lt;em&gt;how&lt;/em&gt; you use a given instrument: what filter, what exposure time, what spectroscopic grating, etc. This requires that someone consider a plausible science case for what are effectively random pointings. (And you need to make sure that you're not accidentally going to try taking a long exposure of Vega, which might actually damage an instrument.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, yes, I gather there &lt;em&gt;are&lt;/em&gt; data-storage and overhead limits, as well as the idea that running the data transmitters at maximum may shorten their lives. From the WFC3 Instrument Handbook (e.g., &lt;a href=&quot;http://www.stsci.edu/hst/wfc3/documents/handbooks/currentIHB/c06_uvis13.html#372067&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The primary restriction on parallel observations, both coordinated and&#xA;  pure, is that they must not interfere with the primary observations:&#xA;  they may not cause the primary observations to be shortened; and they&#xA;  must not cause the stored-command capacity and data-volume limits to&#xA;  be exceeded.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In order to prolong the life of the HST transmitters, the number of&#xA;  parallels acquired during each proposal cycle is limited. Proposers&#xA;  must provide clear and strong justification in order to be granted&#xA;  parallel observing time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(I think the practical answer to your &quot;how many instruments&quot; is two, though I haven't seen an explicit statement to that effect. &lt;a href=&quot;http://www.stsci.edu/hst/proposing/documents/cp/4_Observation_Types3.html#1874145&quot; rel=&quot;nofollow&quot;&gt;This page&lt;/a&gt; includes specific limitations for individual instruments -- e.g., which ones can be used in Pure Parallel mode and with what other instruments.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2016-08-24T12:33:40.603" LastActivityDate="2016-08-24T12:33:40.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18034" PostTypeId="2" ParentId="18032" CreationDate="2016-08-24T13:11:58.330" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Where can I see - or even download - a fairly high resolution (~0.1 nm) solar spectrum at zero air mass - in other words from space without absorption and other features from the Earth's atmosphere?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can download a normalized solar spectrum with 0.1 nm resolution from &lt;a href=&quot;http://bass2000.obspm.fr/solar_spect.php&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. However this is not at zero airmass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A solar spectrum with zero airmass can be found on &lt;a href=&quot;http://rredc.nrel.gov/solar/spectra/am0/&quot; rel=&quot;nofollow&quot;&gt;this website&lt;/a&gt; with the direct link to their data in an excel sheet &lt;a href=&quot;http://rredc.nrel.gov/solar/spectra/am0/E490_00a_AM0.xls&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. This spectrum appears to be at a 1 nm resolution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Of particular interest is the near infrared around 750-780nm. Is the &quot;A&quot; Fraunhofer line due to absorption by oxygen in the Sun's atmosphere, or in Earth's?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you look up the Fraunhofer A line (really it's a &quot;band&quot;), you'll see it is caused by molecular oxygen, namely $\mathrm{O_2}$. This should tell you right away that it is caused by our atmosphere as molecules do not easily exist in the atmosphere of the Sun. In our Sun's atmosphere, most species are ionized forms of single atoms such as $\mathrm{O\:IV}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Astronomy, spectral lines which are caused by the Earth's atmosphere and &lt;em&gt;not&lt;/em&gt; by the source object being observed, are referred to as Telluric lines (from the latin word tellus meaning Earth). I found &lt;a href=&quot;http://webusers.oact.inaf.it/gca/papers/telluric.pdf&quot; rel=&quot;nofollow&quot;&gt;this source&lt;/a&gt; which describes, in detail, the various telluric lines that you may encounter.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2016-08-24T14:26:11.710" LastActivityDate="2016-08-24T14:26:11.710" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18035" PostTypeId="1" CreationDate="2016-08-24T14:49:04.983" Score="1" ViewCount="310" Body="&lt;p&gt;I am trying to find out how to find out if a certain galaxy is located inside a cluster. For the famous ones this is quite easy (sometimes even on wikipedia!), but I haven't been able to do it for some less-known galaxies. I have tried to look them up on Simbad, but I couldn't find it written down anywhere (even for the ones I know to be in clusters). Also generic &quot;googling&quot; has not helped me. Do you have any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="13945" LastActivityDate="2016-10-16T16:15:40.000" Title="How to find out if a galaxy is in a specific cluster" Tags="&lt;galaxy&gt;&lt;data-analysis&gt;&lt;galaxy-cluster&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18036" PostTypeId="1" AcceptedAnswerId="18037" CreationDate="2016-08-24T17:25:16.657" Score="1" ViewCount="43" Body="&lt;p&gt;Last evening I assisted in an interesting phenomenon in my region (Eastern Europe). It was a dark sky with heavy rain and blinding lightning (which struck every 5 seconds or so), but NO thunder at all.&#xA;I was thinking that the explanation might be:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The clouds are very high in the sky and and the sound does not reach the land&lt;/li&gt;&#xA;&lt;li&gt;The wind deviated the sound somewhere else.&lt;/li&gt;&#xA;&lt;li&gt;It was not a storm and aliens were conducting an experiment in the atmosphere&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm not a meteorologist, so if you have a better explanation, please let me know it.&lt;/p&gt;&#xA;" OwnerUserId="13948" LastEditorUserId="9301" LastEditDate="2016-08-26T18:04:30.707" LastActivityDate="2016-08-26T18:04:30.707" Title="Storm with lightnings and no thunder" Tags="&lt;sky-survey&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2016-08-25T06:45:20.180" ContentLicense="CC BY-SA 3.0" />
  <row Id="18037" PostTypeId="2" ParentId="18036" CreationDate="2016-08-24T17:33:39.287" Score="0" Body="&lt;p&gt;&quot;No, it is not possible to have thunder without lightning. Thunder is a direct result of lightning. However, it IS possible that you might see lightning and not hear the thunder because it was too far away. Sometimes this is called “heat lightning” because it occurs most often in the summer. &quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here is more information regarding lightning:&#xA;&lt;a href=&quot;http://www.nssl.noaa.gov/education/svrwx101/lightning/faq/&quot; rel=&quot;nofollow&quot;&gt;http://www.nssl.noaa.gov/education/svrwx101/lightning/faq/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13948" LastActivityDate="2016-08-24T17:33:39.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18038" PostTypeId="1" CreationDate="2016-08-24T19:26:48.347" Score="2" ViewCount="1034" Body="&lt;p&gt;In one stellar system like our milkyway, all the planets orbits in the same direction. But are exoplanets orbitting in the same direction as we (earth) do, or is there a higher percentage which orbit in one or the other way? If so what could be the reason for that?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastActivityDate="2016-08-24T21:50:28.303" Title="Does all the exoplanets orbit the same direction?" Tags="&lt;orbit&gt;&lt;exoplanet&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18039" PostTypeId="1" AcceptedAnswerId="18042" CreationDate="2016-08-24T19:57:02.193" Score="30" ViewCount="5954" Body="&lt;p&gt;I know that the light coming from Proxima Centauri is not bright enough to make it naked-eye visible from the Earth. Is the Sun naked-eye visible from Proxima Centauri?&lt;/p&gt;&#xA;" OwnerUserId="13964" OwnerDisplayName="Patrick" LastEditorUserId="13591" LastEditDate="2016-08-25T19:06:08.047" LastActivityDate="2016-08-27T03:55:42.360" Title="Is the Sun visible from Proxima Centauri to human eyes?" Tags="&lt;light&gt;&lt;distances&gt;&lt;naked-eye&gt;&lt;apparent-magnitude&gt;&lt;absolute-magnitude&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18040" PostTypeId="1" CreationDate="2016-08-24T20:17:06.680" Score="0" ViewCount="158" Body="&lt;p&gt;Y dwarf stars have sub-zero temperatures, so I became curious how a star like these dies out and if is could go supernova.&lt;/p&gt;&#xA;" OwnerDisplayName="Dante Alejandro" LastEditorUserId="2153" LastEditDate="2016-08-24T20:32:13.670" LastActivityDate="2016-08-24T20:32:13.670" Title="Can a Y dwarf star undergo a supernova?" Tags="&lt;supernova&gt;&lt;brown-dwarf&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18041" PostTypeId="2" ParentId="18039" CreationDate="2016-08-24T20:29:04.540" Score="4" Body="&lt;p&gt;The key to this is the so called Absolute Magnitude, which represents the visual magnitude from a distance of 10 parsecs (about 32 light years). The sun is much brighter than Proxima Centauri.  It has an absolute magnitude of 4.8, and at a distance of 4 light years (the distance of Proxima), it would be be somwhat brighter than 1st mag, and so very easily visible with the unaided eye.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastEditorUserId="11822" LastEditDate="2016-08-25T20:43:47.983" LastActivityDate="2016-08-25T20:43:47.983" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18042" PostTypeId="2" ParentId="18039" CreationDate="2016-08-24T20:29:19.300" Score="38" Body="&lt;p&gt;Well, there's two things we'll need for this: &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_magnitude&quot;&gt;apparent magnitude&lt;/a&gt; (the brightness that an object appears to have) and &lt;a href=&quot;https://en.wikipedia.org/wiki/Absolute_magnitude&quot;&gt;absolute magnitude&lt;/a&gt; (the actual brightness an object has). Both of these scales are logarithmic, with brighter objects being lower and dimmer objects being higher. Astronomers have determined that the Sun's absolute magnitude is 4.83. Knowing this, we can find the apparent magnitude of the Sun from Proxima Centauri's location. Apparent and absolute magnitudes are related by the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M = m - 5 (\log_{10}{d}-1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $M$ is the absolute magnitude, $m$ is the apparent magnitude, and $d$ is the distance, in parsecs. Astronomers have determined that Proxima Centauri is 1.3 parsecs from us. So the apparent magnitude can be determined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = 4.83 + 5(\log_{10}{1.3}-1) ≈ +0.4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As made clear in this &lt;a href=&quot;http://adsabs.harvard.edu/full/1947PASP...59..232W&quot;&gt;paper&lt;/a&gt;, most humans can see objects with apparent magnitudes as dim as $5$ without using tools. So yes, it is certainly visible, and would be quite bright. It is between Procyon and Achernar, the 9th and 10th brightest star on Earth's night sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a comparison, Proxima Centauri has an apparent magnitude of $+11.13$ from Earth's perspective. If we want to compare the two, we could use this formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_b = 10^{0.4 x} = 10^{0.4×(11.13-0.4)} \approx 19588$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the Sun would appear almost 20,000 times brighter from Proxima Centuari than PC appears from the Sun.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-08-25T14:02:44.593" LastActivityDate="2016-08-25T14:02:44.593" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18043" PostTypeId="2" ParentId="18040" CreationDate="2016-08-24T20:31:18.627" Score="5" Body="&lt;p&gt;Stars of spectral class Y are &lt;a href=&quot;https://en.wikipedia.org/wiki/Brown_dwarf#Spectral_class_Y&quot;&gt;dim brown dwarfs&lt;/a&gt;, with temperatures of about 300-400 K (For comparison, Earth's average temperature is around 288 K, give or take, the surface of the Sun is 5,800 K, and the &lt;em&gt;core&lt;/em&gt; of the Sun is 15 million K). Their temperature range means that they don't fuse hydrogen in the same way that &quot;true&quot; stars do. They may fuse deuterium (&quot;heavy hydrogen&quot;); Y dwarfs may not fuse lithium, as other brown dwarfs do, because they are relatively low-mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that Y dwarfs (and brown dwarfs, as a whole) don't undergo normal stellar nuclear fusion means that they'll never undergo core-collapse supernovae, which typically happen when an extremely massive star gets to the point of iron fusion and then can no longer generate energy from fusion. Brown dwarfs work completely different; this is not at all a possibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, they will slowly radiate away their heat until they become extremely dim, via the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kelvin%E2%80%93Helmholtz_mechanism&quot;&gt;Kelvin-Helmholtz mechanism&lt;/a&gt;. For more information, see the basic example evolutionary tracks on &lt;a href=&quot;http://www.astro.cornell.edu/academics/courses/astro201/bd_evol.htm&quot;&gt;this page&lt;/a&gt;; if you're really feeling ambitious, you can try going through &lt;a href=&quot;http://www.astro.sunysb.edu/metchev/PHY688/Burrows_Liebert93_BDscience.pdf&quot;&gt;these notes&lt;/a&gt; to get a better feel for brown dwarf evolution.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2016-08-24T20:31:18.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18044" PostTypeId="2" ParentId="18038" CreationDate="2016-08-24T21:10:54.933" Score="1" Body="&lt;p&gt;All planets in a planetary system should orbit in the same direction. However, this direction varies among different planetary systems. The direction that planets will orbit, for any planetary system, depends on the molecular cloud that formed them. The reason for this is &lt;strong&gt;conservation of angular momentum&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before a star and its planets exist, there’s just a cloud of disorganized gas and small molecules. The Solar System formed from such a cloud around 4.6 billion years ago. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On that scale, there is some small amount of rotation within the cloud. It could be caused by the gravity of nearby stellar objects, local differences in mass as the cloud churns, or even the impact of a distant supernova. The point is, all molecular clouds have at least a little rotation. The direction depends on a number of factors, so some clouds will turn clockwise, and others will turn counterclockwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a large system like a molecular cloud, each particle has some angular momentum, and it all adds together across a very wide area. That’s a lot of momentum, and it is conserved as the cloud continues to collapse under its own gravity. That angular momentum also flattens the cloud, which is the reason why the Solar System is near-planar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/d3kRf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/d3kRf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the cloud finally collapses, it forms a star and shortly after planets. However, angular momentum is always conserved. That's why all the Solar System planets follow the same orbit, and why almost all of them rotate in the same direction. There's nothing to turn them the other direction, so they will continue spinning in the same direction as the original gas cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few exceptions, though. Whenever objects formed in such a way that sent them orbiting the opposite direction, they usually collided with objects going in the same direction as the original cloud. This destroyed any outlying objects or sent them in the same direction as the original cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, two huge exceptions are planets Venus and Uranus. Uranus spins on an axis of almost 90-degrees (on its side). Venus meanwhile spins the opposite direction as Earth and the other planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases there is strong evidence that these planets were struck by large objects at some point in the distant past. The impacts were large enough to overcome the angular momentum of the bodies, and give them a different spin. There are also a range of other theories; for example, some astronomers think that Venus may have been flipped upside-down. Point is, there were irregular events that happened to both of these planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, what I described applies for all planetary systems. To answer your question, it depends on how their cloud rotated. The cloud that formed some planetary systems may have turned counterclockwise, just like for the Solar System (at least, it appears counterclockwise from above Earth's north pole). The cloud for others may have turned clockwise. That's why not all exoplanets systems will have the same orbital direction as Earth; however, typically all planets in a planetary system will share the same orbital direction.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-08-24T21:50:28.303" LastActivityDate="2016-08-24T21:50:28.303" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18045" PostTypeId="2" ParentId="484" CreationDate="2016-08-24T21:23:41.343" Score="15" Body="&lt;p&gt;Earlier this year (2016), scientists used the radial velocity method to discover a planet orbiting Proxima Centauri: &lt;a href=&quot;https://en.wikipedia.org/wiki/Proxima_Centauri_b&quot; rel=&quot;noreferrer&quot;&gt;Proxima Centauri b&lt;/a&gt;. It was announced in &lt;a href=&quot;http://www.nature.com/nature/journal/v536/n7617/full/nature19106.html&quot; rel=&quot;noreferrer&quot;&gt;Anglada-Escudé et al. (2016)&lt;/a&gt;. Here are some of its basic properties, as reported by the authors of the paper and known as of August 2016:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mass: At least 1.3 Earth masses&lt;/li&gt;&#xA;&lt;li&gt;Semi-major axis: 0.05 AU&lt;/li&gt;&#xA;&lt;li&gt;Maximum eccentricity: 0.35&lt;/li&gt;&#xA;&lt;li&gt;Orbital period: 11.2 days&lt;/li&gt;&#xA;&lt;li&gt;Surface temperature: 234 K (see &lt;a href=&quot;http://www.nature.com/nature/journal/v536/n7617/fig_tab/nature19106_T1.html&quot; rel=&quot;noreferrer&quot;&gt;Table 1&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, as the authors note, the planet may not be habitable:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The habitability of planets like Proxima [Centauri] b—in the sense of sustaining an atmosphere and liquid water on its surface—is a matter of intense debate. The most common arguments against habitability are tidal locking, strong stellar magnetic fields, strong flares and high ultraviolet and X-ray fluxes; but none of these have been proved definitive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, further observations might help us learn more about the planet and determine if it could be habitable for any type of life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted that the controversy about &lt;a href=&quot;https://en.wikipedia.org/wiki/Alpha_Centauri#Planets&quot; rel=&quot;noreferrer&quot;&gt;planets around Alpha Centauri&lt;/a&gt; continued after this question was initially answered by user8. &lt;a href=&quot;https://en.wikipedia.org/wiki/Alpha_Centauri_Bb&quot; rel=&quot;noreferrer&quot;&gt;Alpha Centauri Bb&lt;/a&gt; was shown to be nonexistent by &lt;a href=&quot;https://arxiv.org/abs/1510.05598&quot; rel=&quot;noreferrer&quot;&gt;Rajpaul et al. (2015)&lt;/a&gt;. However, in another twist, observations by &lt;a href=&quot;https://arxiv.org/abs/1503.07528v1&quot; rel=&quot;noreferrer&quot;&gt;Demory et al. (2015)&lt;/a&gt; seemed to show the existence of Alpha Centauri Bc (at the time, Alpha Centauri Bb was still a possibility, so now &quot;Alpha Centauri Bc&quot; would actually be &quot;Alpha Centauri Bb&quot; if it existed), an Earth-like planet. However, there has been no more evidence to confirm the existence of this planet - and at any rate, Proxima Centauri is closer to the Solar System than Alpha Centauri, so Proxima Centauri b wins no matter what.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-08-25T15:39:05.097" LastActivityDate="2016-08-25T15:39:05.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18046" PostTypeId="2" ParentId="18001" CreationDate="2016-08-24T21:59:35.093" Score="3" Body="&lt;p&gt;Yes, they can.  The ships that visit the ISS often orbit near the station for several hours before docking.  Further, there are several satellites that travel in pairs or trios with other satellites.  The nearer the satellites are to earth, or the farther apart they are, the more difficult the task of maintaining spatial relationships between the spacecraft. This is because the spacecraft must be in the same orbital plane to travel together.  The most common configuration is where they are in the exact same orbit, with one satellites a few seconds ahead of the other along the same path.&lt;/p&gt;&#xA;" OwnerUserId="13950" LastActivityDate="2016-08-24T21:59:35.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18047" PostTypeId="1" AcceptedAnswerId="18060" CreationDate="2016-08-24T23:14:19.217" Score="4" ViewCount="486" Body="&lt;p&gt;I was thinking about &lt;a href=&quot;https://worldbuilding.stackexchange.com/questions/52696/a-monochrome-planet&quot;&gt;this question on Worldbuilding&lt;/a&gt; and the answers involving &lt;a href=&quot;http://farm4.static.flickr.com/3034/2494152178_3ef018011c_b.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;monochromatic ambient light&lt;/a&gt;, which got me wondering if there were any stars that emitted relative monochromatic light (a &quot;very&quot; narrow band, although I don't have a definition, but something narrow enough to produce the effect in that photo, atmospheric effects notwithstanding).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: Is there any type of star that emits an &lt;em&gt;extremely&lt;/em&gt; narrow band of visible light (not necessarily exclusively visible, non-visible wavelengths can be included) that we have either observed or theorized to exist? &quot;Narrow&quot; as in, on par with a &lt;a href=&quot;https://en.wikipedia.org/wiki/Sodium-vapor_lamp&quot; rel=&quot;nofollow noreferrer&quot;&gt;low pressure sodium-vapor lamp&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My Google searches are hindered by a lot of noise, mostly things like &lt;a href=&quot;http://www.alamy.com/stock-photo-seamless-monochromatic-star-pattern-102321627.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;monochromatic star patterns&lt;/a&gt;, and a few hits regarding non-visible radiation.&lt;/p&gt;&#xA;" OwnerUserId="2929" LastEditorUserId="-1" LastEditDate="2017-04-13T12:52:24.020" LastActivityDate="2016-08-26T01:07:19.967" Title="Is there a type of star that emits relatively monochromatic visible light?" Tags="&lt;star&gt;&lt;light&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18049" PostTypeId="2" ParentId="18035" CreationDate="2016-08-25T01:45:31.710" Score="0" Body="&lt;p&gt;Your question is not specific, so I'm afraid the answer is general too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Simbad/NED check to see what observations/survey data exists for you galaxies. Almost all large galaxy surveys, and a lot of smaller ones, have corresponding &quot;environment catalogues&quot; where the galaxies are associated to their corresponding group. By cross-matching the coordinates (ra,dec) and redshift (or v) of your galaxies with the members of groups and clusters in one or more of these catalogues you should be able to find a match.&lt;/p&gt;&#xA;" OwnerUserId="891" LastActivityDate="2016-08-25T01:45:31.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18052" PostTypeId="1" CreationDate="2016-08-25T02:59:48.800" Score="8" ViewCount="608" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/whbly.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/whbly.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Table 1 from &lt;a href=&quot;https://safe.nrao.edu/wiki/pub/ALMA/Sep12Agenda/ProcSPIE_Baudry_8452-43_A4_v3.2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Performance Highlights of the ALMA Correlators&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ALMA receivers use 3-bit ADCs for what would seem to be to be a high dynamic range application needing much finer quantization to get anything useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I found these sentences within the abstract of &lt;a href=&quot;http://www.imeko.org/publications/iwadc-2008/IMEKO-IWADC-2008-167.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;ADC bit number and input power needed, in new radio-astronomical applications&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Abstract- For the most part, so far radio astronomy observations have been performed in protected frequency bands, reserved by ITU for scientific purposes. This means that, ideally, only the amplified equivalent system noise is present at the end of the receiver chain (i.e. the ADC input). So, typically, only a few bits are necessary to describe the signal (VLBI signals are digitised with only 2 bits), but today astronomers, in order to get more sensitivity and to boldly observe where no one has observed before, would like to study the radio sky even outside the protected bands... &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I even found a 1-bit ADC in &lt;a href=&quot;http://arxiv.org/pdf/0801.1404.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Performance Measurements of 8-Gsps 1-bit ADCs Developed for Wideband Radio Astronomical Observations&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I am just missing something obvious, but I can't understand how a measurement requiring high dynamic range gets by using few-bit ADCs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit:&lt;/strong&gt; Is it possible that the actual conversion of analog to digital is done to a far higher precision than suggested by the number of bits? &lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-10-10T02:52:48.843" LastActivityDate="2020-11-08T19:12:10.910" Title="Why are the ALMA receivers' ADCs only 3-bits?" Tags="&lt;radio-astronomy&gt;&lt;radio-telescope&gt;" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="18053" PostTypeId="1" AcceptedAnswerId="18057" CreationDate="2016-08-25T06:29:34.420" Score="3" ViewCount="2456" Body="&lt;p&gt;If Jupiter were as far from us as Mars, how big would it appear in Earth's sky? Would it appear huge, or would it be as small as a star?&lt;/p&gt;&#xA;" OwnerUserId="11792" LastEditorUserId="9301" LastEditDate="2016-08-26T17:40:11.323" LastActivityDate="2016-08-27T10:13:02.767" Title="If Jupiter was placed in Mars' orbit, how would Jupiter appear from the naked eye?" Tags="&lt;orbit&gt;&lt;jupiter&gt;&lt;mars&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ClosedDate="2016-08-28T14:02:50.000" ContentLicense="CC BY-SA 3.0" />
  <row Id="18054" PostTypeId="2" ParentId="18039" CreationDate="2016-08-25T07:00:58.467" Score="5" Body="&lt;p&gt;Alpha Centauri A and B happen to be rather similar to Sol, and their absolute magnitudes are 4.38 and 5.71 respectively (Wikipedia).  Add them together and you get absolute magnitude 4.10 (the scale is logarithmic, and backward).  Sol, with absolute magnitude 4.83, should look 0.73 magnitude dimmer than &amp;alpha;Cen at the same distance, so magnitude +0.46, quite bright.&lt;/p&gt;&#xA;" OwnerUserId="7959" LastEditorUserId="7959" LastEditDate="2016-08-27T03:55:42.360" LastActivityDate="2016-08-27T03:55:42.360" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18055" PostTypeId="1" CreationDate="2016-08-25T07:35:23.957" Score="2" ViewCount="635" Body="&lt;p&gt;For planning purposes it would be good to have a calendar of upcoming (at least some years) solar eclipses. &lt;a href=&quot;http://www.andrade.org/eclipse/upcoming&quot; rel=&quot;nofollow&quot;&gt;The closest thing I've been able to find&lt;/a&gt; has too many other events in it to be useful. A &lt;a href=&quot;http://icalshare.com/calendars?query=solar%20eclipses&quot; rel=&quot;nofollow&quot;&gt;similar calendar&lt;/a&gt; is no longer reachable.&lt;/p&gt;&#xA;" OwnerUserId="466" LastEditorUserId="466" LastEditDate="2016-09-01T19:42:58.007" LastActivityDate="2016-09-01T19:42:58.007" Title="Where can I find a solar eclipse calendar compatible with Google Calendar?" Tags="&lt;solar-eclipse&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18057" PostTypeId="2" ParentId="18053" CreationDate="2016-08-25T09:03:56.963" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If we were to view Jupiter from Mars's orbit with the the naked eye, how big would it appear to be, as compared with its appearance from the Earth? Would Jupiter appear as big as the moon does from Earth? Would it appear to be a disk to the naked eye, or would it be star-like?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here are some back-of-the-envelope calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum apparent magnitude of Jupiter is -2.94, which is observed when it's closest to Earth (4.2 AU). Mars can get as close to Jupiter as 3.68 AU. Since the flux is inversely proportional to distance squared, Jupiter will be 30% brighter on Mars than it is on Earth, giving it an apparent magnitude of -3.23. It will still be a bright star-like dot outshined by both Martian moons, Phobos and Deimos with magnitudes of -9 and -5 respectively. It will certainly not be as bright as Earth's Moon seen from Earth, which can be as bright as −12.9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomy_on_Mars&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is a detailed article which describes how various celestial bodies look from Mars. They have a picture of Earth, Moon and Jupiter taken by Mars Orbiter Camera (MOC) in 2003, when both Earth and Jupiter were close to Mars:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3700p.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3700p.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13926" LastEditorUserId="13926" LastEditDate="2016-08-27T10:13:02.767" LastActivityDate="2016-08-27T10:13:02.767" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18058" PostTypeId="1" CreationDate="2016-08-25T09:31:56.437" Score="0" ViewCount="71" Body="&lt;p&gt;Preferably in terms of percentage. As a layman, visual apparent magnitude means little to me.&lt;/p&gt;&#xA;" OwnerUserId="13961" LastEditorUserId="7491" LastEditDate="2016-08-25T14:03:40.983" LastActivityDate="2016-08-25T14:33:30.287" Title="How bright will Sirius be when it's closest to the solar system?" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2016-08-27T21:53:28.033" ContentLicense="CC BY-SA 3.0" />
  <row Id="18059" PostTypeId="1" AcceptedAnswerId="18064" CreationDate="2016-08-25T09:47:13.313" Score="4" ViewCount="332" Body="&lt;p&gt;Can &lt;a href=&quot;https://en.wikipedia.org/wiki/Search_for_extraterrestrial_intelligence&quot; rel=&quot;nofollow noreferrer&quot;&gt;SETI&lt;/a&gt; certify whether or not  &lt;a href=&quot;https://en.wikipedia.org/wiki/Proxima_Centauri_b&quot; rel=&quot;nofollow noreferrer&quot;&gt;Proxima b&lt;/a&gt; is inhabited by beings using electromagnetic communication?    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &lt;em&gt;certify&lt;/em&gt; I mean by past, present or future observations using technologies available.&lt;/p&gt;&#xA;" OwnerUserId="6668" LastEditorUserId="14006" LastEditDate="2017-10-02T20:09:10.220" LastActivityDate="2019-07-20T07:13:56.983" Title="Can SETI certify whether or not Proxima b is inhabited by beings using electromagnetic communication?" Tags="&lt;exoplanet&gt;&lt;radio-astronomy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18060" PostTypeId="2" ParentId="18047" CreationDate="2016-08-25T10:17:04.537" Score="3" Body="&lt;p&gt;Like the other answer stated, stars usually emit a good approximation to &lt;a href=&quot;https://en.wikipedia.org/wiki/Black_body&quot; rel=&quot;nofollow&quot;&gt;black body radiation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, under certain conditions, a star can emit monochromatic microwaves. This is known as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Astrophysical_maser&quot; rel=&quot;nofollow&quot;&gt;MASER&lt;/a&gt;. For this to happen, the star has to be either very cool (since the emission is from molecular gas) or embedded in a star forming region - in the latter case, it's not the actual star emitting the radiation but the molecular cloud it is embedded in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the &quot;normal&quot; starlight is still present even if you have a MASER, and the emission is highly directional. So you might not see the MASER if you're looking from the wrong direction at the star, and it will not be purely monochromatic, but the MASER emission might be quite dominant.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastActivityDate="2016-08-25T10:17:04.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18061" PostTypeId="1" CreationDate="2016-08-25T10:36:30.467" Score="3" ViewCount="128" Body="&lt;p&gt;Of course scientists assume there would also be exoplanets in other galaxies besides our own Milkyway. &#xA;Is it true that our telescopes are not good enough to see planets in other galaxies, nor by brightness changes of the stars nor by inclination changes of stars. If so could the total mass of those galaxies be an indication of the presence of exoplanets, or is dark matter a problem for that?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastActivityDate="2017-12-15T16:11:12.267" Title="Are there any measurements of other galaxies indicating the existence of exoplanets?" Tags="&lt;galaxy&gt;&lt;exoplanet&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18062" PostTypeId="2" ParentId="18052" CreationDate="2016-08-25T10:56:16.990" Score="2" Body="&lt;p&gt;The resolution of ADCs is inversely related to their conversion time. Getting more bits requires the signal to travel through more circuitry, which takes time. This is why you can have those high-quality audio ADCs with 18 or 20 bit resolution, which operate at frequencies in kHz range, meaning each conversion can take several milliseconds. At 4GS/s you only have 250 picoseconds at your disposal, so you can only get 3 bits (and only 1 bit at 8GS/s).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;how a measurement requiring high dynamic range gets by using few-bit ADCs?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This depends on the nature of the measurement, but the typical solution is to make successive measurements and calculate the average.&lt;/p&gt;&#xA;" OwnerUserId="13926" LastEditorUserId="13926" LastEditDate="2016-08-25T11:13:18.337" LastActivityDate="2016-08-25T11:13:18.337" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18063" PostTypeId="1" CreationDate="2016-08-25T11:03:41.203" Score="1" ViewCount="119" Body="&lt;p&gt;For example: I want to know the distance between Alpha Orionis (Betelgeuse) and Beta Orionis (Rigel). Is this possible? You only know the distance from Earth. May be this is a math problem and not an Astronomy question.&lt;/p&gt;&#xA;" OwnerUserId="13962" LastActivityDate="2016-08-25T15:02:08.670" Title="Is it possible to calculate distance between two objects?" Tags="&lt;star&gt;&lt;distances&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18064" PostTypeId="2" ParentId="18059" CreationDate="2016-08-25T11:33:10.270" Score="7" Body="&lt;p&gt;Using &lt;em&gt;current technology&lt;/em&gt; (and by that I mean experiments and telescopes that are available now) we would probably be unable to detect life on Earth even if observed from a distance of 4 light years, which is the distance to Proxima Centauri. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;blind&quot; search could look for radio signatures and of course this is what SETI has been doing for lots of different stars, including, I imagine, Proxima Centauri. If we are talking about detecting &quot;Earth-like&quot; signals, then we must assume that we are not talking about deliberate beamed attempts at communication, and so must rely on detecting random radio &quot;chatter&quot; and accidental signals generated by our civilisation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I have remarked in the linked question: The SETI Phoenix project was the most advanced search for radio signals from other intelligent life. Quoting from &lt;a href=&quot;http://adsabs.harvard.edu/full/2000ASPC..213..451C&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cullers et al. (2000)&lt;/a&gt;: &quot;&lt;em&gt;Typical signals, as opposed to our strongest signals fall below the detection threshold of most surveys, even if the signal were to originate from the nearest star&lt;/em&gt;&quot;. Quoting from &lt;a href=&quot;http://adsabs.harvard.edu/abs/2001ARA%26A..39..511T&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tarter (2001)&lt;/a&gt;: &quot;&lt;em&gt;At current levels of sensitivity, targeted microwave searches could detect the equivalent power of strong TV transmitters at a distance of 1 light year (within which there are no other stars)...&lt;/em&gt;&quot;. The equivocation in these statements is due to the fact that we &lt;em&gt;do&lt;/em&gt; emit stronger beamed signals in certain well-defined directions, for example to conduct metrology in the solar system using radar. Such signals have been calculated to be observable over a thousand light years or more. But these signals are brief, beamed into an extremely narrow angle and unlikely to be repeated. You would have to be very lucky to be observing in the right direction at the right time if you were performing targeted searches. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has been suggested that new radio telescope projects and technology like the &lt;a href=&quot;https://en.wikipedia.org/wiki/Square_Kilometre_Array#Search_for_extraterrestrial_life&quot; rel=&quot;nofollow noreferrer&quot;&gt;Square Kilometre Array&lt;/a&gt; may be capable of serendipitously detecting radio &quot;chatter&quot; out to distances of 50 pc (&lt;span class=&quot;math-container&quot;&gt;$\sim 150$&lt;/span&gt; light years) - see &lt;a href=&quot;http://adsabs.harvard.edu/abs/2007JCAP...01..020L&quot; rel=&quot;nofollow noreferrer&quot;&gt;Loeb &amp;amp; Zaldarriaga (2007)&lt;/a&gt; - and so Proxima Centauri would be well within this range. This array, due to begin full operation some time after 2025 could also monitor a multitude of directions at once for beamed signals.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-07-20T07:13:56.983" LastActivityDate="2019-07-20T07:13:56.983" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="18065" PostTypeId="2" ParentId="18061" CreationDate="2016-08-25T11:41:13.343" Score="4" Body="&lt;p&gt;All discovered planets are in our Galaxy. The signature of a planet is too weak to be detected around the faint and unresolvable stars in even the nearest other galaxies. New telescopes like the Large Synoptic Survey Telescope will be capable of detecting transiting planets, and also microlensing events by planets, in the Magellanic clouds (e.g. see &lt;a href=&quot;https://arxiv.org/abs/1712.03986v1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mroz &amp;amp; Poleski 2017)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fraction of galactic mass that is tied up in exoplanets should be a tiny fraction of the total mass. If you add up the mass of all the planets in the solar system you end up with around 1% of a solar mass (mostly Jupiter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no indication that other stars have &lt;em&gt;many&lt;/em&gt; more planets than the solar system, or that the mass of the solar system planets are at all unusual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fraction of planets that end up not attached to stars is also likely to be quite small. Essentially some fraction of planets might be stripped away from their parents during the earliest phases of stellar evolution in a dense cluster (where the &lt;em&gt;minority&lt;/em&gt; of stars are formed actually). So free-floating planets are unlikely to contribute much and indeed this has been confirmed by experiments a decade and more ago to detect &quot;massive compact halo&quot; objects in our own galaxy, which were unable to attribute much of the dark, &lt;em&gt;baryonic&lt;/em&gt; matter to planets or the like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the dark matter problem is not just that there is matter that cannot be seen - it is that the majority of the matter in the universe is non-baryonic - that is, it is not made of the normal stuff that makes up stars and planets. So lots and lots of planets could not explain away most of the dark matter problem in any case.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-12-15T16:11:12.267" LastActivityDate="2017-12-15T16:11:12.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18066" PostTypeId="2" ParentId="18061" CreationDate="2016-08-25T12:13:39.827" Score="6" Body="&lt;p&gt;Yes, it is true that our telescopes are &lt;strong&gt;not&lt;/strong&gt; big/strong enough to detect planets in other galaxies. All we know at the moment are some few planets (not even complete systems!) around some of the stars in our closest galactic neighbourhood.&lt;br&gt;&#xA;But let me explain why this is so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Far away objects are very faint and thus send out few light-particles our way. To compensate this, we must build bigger and bigger telescopes to catch more light-particles from those faint objects. And at the moment we are limited until the 30-40metre class telescopes become active in ~2025.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there is a class of 'cosmic accidents' that allow telescopes to catch more light-particles. We call those accidents &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_microlensing&quot; rel=&quot;noreferrer&quot;&gt;micro-lensing&lt;/a&gt; events, as they briefly concentrate the light of a background star around a planet's gravitational sphere of influence.&#xA;The result is a spike of light in our telescopes. The shape of this spike informs us about the characteristics of the lensing star and planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put this into context with your question about other galaxies, I've plotted for you how far away we have discovered planets so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6c6U6.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6c6U6.png&quot; alt=&quot;distance vs. declination to planet-hosting stars&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we have on the x-axis the distance to the planet-hosting star, and on the y-axis the declination (one of the celestial coordinates or directions).&lt;br&gt;&#xA;In red I've painted the planets that we have discovered with microlensing, in blue are all other methods.&lt;br&gt;&#xA;The microlensing discoveries lie all on one line around dec = -30°, because this is the direction of the galactic center, where the polish OGLE project is looking.&lt;br&gt;&#xA;The famous US-satellite Kepler that uses the transit method has it's field of view around dec = +45°, and this is also the biggest blue bunch of discoveries.&#xA;The small bump in between is the fov of the french CoRoT-satellite that also did some transit-discoveries and pioneered the transit searches from space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So what do we learn from this graph?&lt;/strong&gt;&lt;br&gt;&#xA;We see that we can discover planets on a massive scale up to ~2000pc away.&#xA;Further away it's only possible with luck (while OGLE also looks at quite an impressive number of stars at any given time) and lensing. Now to put this into context, our milky way galaxy has a diameter of ~40.000parsec, and the next galaxy, the Andromeda galaxy is 1 million parsec away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that's gonna be hard, even with microlensing, and even harder with the other techniques you've mentioned.&lt;br&gt;&#xA;Although there is one &lt;a href=&quot;http://adsabs.harvard.edu/abs/2009MNRAS.399..219I&quot; rel=&quot;noreferrer&quot;&gt;event from 2009&lt;/a&gt; claiming to have found something that looks like a planet in their lensing-curve, I don't think this technique is mature enough to announce that extra-galactic planet searches.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2016-08-25T12:13:39.827" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18067" PostTypeId="2" ParentId="328" CreationDate="2016-08-25T12:31:08.960" Score="12" Body="&lt;p&gt;&lt;sup&gt;A decent part of this answer is based on the introduction to &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?2005IAUS..227..423K&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;noreferrer&quot;&gt;Kroupa &amp;amp; Weidner (2005)&lt;/a&gt;, though I’ve obviously gone into a lot more depth on all of the references.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our story starts, as do many concerning stellar astrophysics, with Sir Arthur Eddington. In his 1926 book, &lt;a href=&quot;https://ia800706.us.archive.org/28/items/TheInternalConstitutionOfTheStars/Eddington-TheInternalConstitutionOfTheStars.pdf&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;The Internal Constitution of the Stars&lt;/em&gt;&lt;/a&gt;, he derived the &lt;a href=&quot;https://en.wikipedia.org/wiki/Eddington_luminosity&quot; rel=&quot;noreferrer&quot;&gt;Eddington luminosity&lt;/a&gt;, the maximum luminosity $L$ a star of mass $M$ can reach (Chapter 6, pages 114-115). His derivation goes along the following lines:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I. Take the equation of hydrostatic equilibrium and the equation of radiative equilibrium:&#xA;$$\frac{\mathrm{d}P}{\mathrm{d}r}=-g\rho\tag{1a}$$&#xA;$$\frac{\mathrm{d}p_R}{\mathrm{d}r}=-\frac{k\rho H}{c}\tag{1b}$$&#xA;The relevant variables are pressure ($P$), radius ($r$), gravitational acceleration ($g$), density ($\rho$), radiation pressure ($p_R$), mass coefficient of absorption ($k$), radiative flux per time ($H$), and the speed of light ($c$). Combining $(1\text{a})$ and $(1\text{b})$ yields&#xA;$$\mathrm{d}p_R=\frac{kH}{cg}\mathrm{d}P\tag{1c}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;II. At some radius $r$, the luminosity $L_r$ and enclosed mass $M_r$ can be related by&#xA;$$\frac{L_r}{M_r}=\frac{\eta L}{M}\tag{2a}$$&#xA;where $L$ and $M$ are the luminosity and enclosed mass at the radius of the star, and $\eta$ is some function of $r$, increasing inward from $\eta(R)=1$ at the stellar radius $R$. Given that&#xA;$$H=\frac{L_r}{4\pi r^2}\tag{2b}$$&#xA;$$g=\frac{GM_r}{r^2}\tag{2c}$$&#xA;we have&#xA;$$\frac{H}{g}=\frac{L_r}{4\pi GM_r}\tag{2d}$$&#xA;Putting this back into $(1\text{c})$, we find&#xA;$$\mathrm{d}p_R=\frac{L\eta k}{4\pi cGM}\mathrm{d}P\tag{2e}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;III. As temperature and density increase towards the center of the star, so does the pressure due to the matter, $p_G$. Therefore, $\mathrm{d}p_G&amp;gt;0$. Furthermore, given that $P=p_G+p_R$, $\mathrm{d}p_R&amp;lt;\mathrm{d}P$. This means that $(2\text{e})$ yields&#xA;$$\frac{L\eta k}{4\pi cGM}&amp;lt;1\tag{3}$$&#xA;which is the criterion leading to the Eddington luminosity. There are, of course, other ways of getting this criterion, but I thought I’d give Eddington’s original one, in all its mathematical glory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a suitable mass-luminosity relation for massive stars, we can then establish the mass of a star at the Eddington limit. Eddington himself took it to be in the range of 60-70 solar masses ($M_{\odot}$), though today a value of somewhere around 120 solar masses is more appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us take a detour to a lesser-known figure, Paul Ledoux. &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1941ApJ....94..537L&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;noreferrer&quot;&gt;In 1941&lt;/a&gt;, Ledoux analyzed vibration modes in stars due to the usual perturbations in density, pressure, radius, temperature, etc. He came up with the stability condition of&#xA;\begin{align}&#xA;  A_k&amp;amp;=\!&#xA;  \begin{aligned}[t]&#xA; \int_0^M\frac{\delta \rho_k}{\rho}\biggl[(\Gamma_3-1)\delta_k\left\{\epsilon_1+\epsilon_2-\epsilon_3-\frac{\mathrm{d}}{\mathrm{d}m}[4\pi r^2(F_1+F_3)]\right\}&amp;amp;\\&#xA;  -\frac{2}{3}\delta_k\left[4\pi r^2\bar{C}\frac{\mathrm{d}P}{\mathrm{d}m}+\epsilon_2+\frac{\mathrm{d}}{\mathrm{d}m}[4\pi r^2F_2]\right]\biggr] \mathrm{d}m&amp;amp;&amp;lt;1\\&#xA;  \end{aligned}\\&#xA;\end{align}&#xA;for the $k$th mode of vibration. I’m not going to explain all the variables because that’s not quite important; the important takeaway is that Ledoux took turbulent pulsations into account. His conclusion is that an exact model “probably” would lead to a limit of about 100 solar masses; using certain inexact assumptions, he found a limit of 128 solar masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ledoux’s analysis laid the groundwork for the work of &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1959ApJ...129..637S&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;noreferrer&quot;&gt;Schwarzschild &amp;amp; Härm (1958)&lt;/a&gt;. Their stability criterion isn’t necessarily simpler, but it can be written more compactly. Specifically, the coefficient of stability, $K$, defined as&#xA;$$K=\frac{1}{2}\frac{L_P}{E_P}$$&#xA;must be negative to ensure stability against pulsations. A positive $K$ means that the pulsation amplitude is increasing; a negative $K$ means that the pulsation amplitude is decreasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_P$ is the pulsation’s energy, while $L_P$ is the rate of gain of pulsation energy and can be expanded as&#xA;$$L_P=\overbrace{L_{PN}}^{\text{nuclear}}-\overbrace{L_{PH}}^{\text{heat leakage}}-\overbrace{L_{PS}}^{\text{progressive waves}}$$&#xA;Here, $L_{PN}$ represents the rate of energy gained, while $L_{PH}$ and $L_{PS}$ represent the rate of energy lost. All of the above quantities can be calculated through some relatively simple expressions (see Equations 9-12 and 15-22). The upshot of all this is that $K$ becomes negative at birth for stars greater than 60 solar masses. This can be figured out by writing $L_P$ and $E_P$ as functions of mass, $M$, and age, $\tau$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, interestingly enough, the critical age ($\tau_{cr}$) can be written as a function of mass:&#xA;$$\tau_{cr}=0.05\left(\frac{M}{M_{\odot}}-60\right)$$&#xA;where $\tau_{cr}$ is in millions of years. This means that a star of, say, 62 solar masses (to take the authors’ example) will evolve to a stable state in a quarter of a million years. We can also determine whether or not, in this time, the star’s instability will become too great and destroy it. It turns out that this is the case for stars with masses greater than 65 solar masses - putting the upper limit for the mass of a star at 65 solar masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a graphical representation from their paper, Figure 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8Jpxn.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8Jpxn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even later work on the same topic was done by &lt;a href=&quot;http://adsabs.harvard.edu/abs/1970ApJ...162..947Z&quot; rel=&quot;noreferrer&quot;&gt;Ziebarth (1970)&lt;/a&gt;, among others, who extended the models to study different metallicities and compositions (Schwarzschild &amp;amp; Härm) focused largely on stars with compositions similar to that of the Sun). His computations found a wide range of upper mass limits - 10 solar masses for pure helium stars, and 200 solar masses for pure hydrogen stars. Most stars fall in the middle, and so will have different limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual formation of massive stars also puts constraints on the mass. Kroupa &amp;amp; Weidner mention &lt;a href=&quot;http://adsabs.harvard.edu/abs/1974A%26A....37..149K&quot; rel=&quot;noreferrer&quot;&gt;Kahn (1974)&lt;/a&gt;, who studied how radiation pressure from a protostar could drastically lower accretion rates, stopping the star from continuing to significantly grow. As applied to a young Population I star, his simplest model comes out to a limit of about 80 solar masses, although different models of the “cocoon” yield different results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’ll add one final note on theory. Population III stars, the hypothetical first stars in the universe, are expected to have been extremely massive; as such, they would be excellent candidates for testing the upper mass limits. According to simulations by &lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/bib_query?2011Sci...334.1250H&quot; rel=&quot;noreferrer&quot;&gt;Hosokawa et al. (2011)&lt;/a&gt;, mechanisms similar to those discussed by Kahn would have stopped accretion at stellar masses around 43 solar masses -  a surprisingly low figure, given the expectations of how massive Population III stars should be. Additionally, as argued by &lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/bib_query?2009Sci...325..601T&quot; rel=&quot;noreferrer&quot;&gt;Turk et al. (2009)&lt;/a&gt;, sufficiently massive stars could fragment; in the case studied, a 50 solar mass star broke apart into two smaller core fragments.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Something I realized just now, a couple months after writing this, is that all of this assumes that the star is spherically symmetric. Most stellar models involve spherically symmetric, non-rotating stars, which allows us to make some assumptions such that the equations of stellar structure depend purely on $r$, the radial coordinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have, however, seen stars - not stellar just simply stellar remnants like pulsars, but even main-sequence stars - that rotate rapidly and are thus non-spherical. &lt;a href=&quot;https://en.wikipedia.org/wiki/Vega&quot; rel=&quot;noreferrer&quot;&gt;Vega&lt;/a&gt;, for instance, has an equatorial radius 19% larger than its polar radius. If a star of mass $M$ is rotating, the equations of stellar structure should be different, and so some of the above results should also be different. I'm not sure how important this is for various theoretical limits.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="9301" LastEditDate="2016-10-25T17:46:39.207" LastActivityDate="2016-10-25T17:46:39.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18068" PostTypeId="2" ParentId="18058" CreationDate="2016-08-25T13:49:37.810" Score="5" Body="&lt;p&gt;From &lt;a href=&quot;https://books.google.co.uk/books?id=CEk6SZQK0C0C&amp;amp;pg=PA213&amp;amp;lpg=PA213&amp;amp;dq=relative%20motion%20of%20sirius&amp;amp;source=bl&amp;amp;ots=KMbqPxsA0q&amp;amp;sig=A9KHZhHCBcpwcrfTHf7TWQ7eBfU&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwjvqajR2NzOAhXIK8AKHVZ6D5wQ6AEIMjAD#v=onepage&amp;amp;q=relative%20motion%20of%20sirius&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Sirius: Brightest Diamond in the Night Sky&lt;/a&gt; by Jay B. Holberg we have current distance of Sirius ~8.5ly, closest approach will be ~6.7ly. The intensity of of light from Sirius assuming constant luminosity scales as distance squared, so the intensity (power per unit area at the eye) at closest approach will be: $\left(8.5/6.7 \right)^2 \approx 1.6$ times that today, which is a change of magnitude from about $-1.4$ to about $ -1.6$&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2016-08-25T14:33:30.287" LastActivityDate="2016-08-25T14:33:30.287" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18069" PostTypeId="2" ParentId="18063" CreationDate="2016-08-25T15:02:08.670" Score="1" Body="&lt;p&gt;If you know the distance of each star from earth, and the angle between the stars, then you can us the cosine rule to find their distance from each other.  See this for detail&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Law_of_cosines&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Law_of_cosines&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2016-08-25T15:02:08.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18070" PostTypeId="2" ParentId="18047" CreationDate="2016-08-25T15:30:19.513" Score="6" Body="&lt;p&gt;In some sense, yes.&lt;/p&gt;&#xA;&lt;strong&gt;UV continuum &lt;span class=&quot;math-container&quot;&gt;$\rightarrow$&lt;/span&gt; UV line&lt;/strong&gt;&#xA;&lt;p&gt;This is probably not the answer you're looking for, but massive stars (O and B stars) emit a very hard UV spectrum of light (a UV &amp;quot;continuum&amp;quot;, i.e. a broad range of wavelengths). Since these stars don't live for long (because they burn their fuel fast), they tend to be located in the gas clouds from which they were born. The UV light ionizes the enshrouding hydrogen, but the protons and electrons quickly recombine. If the electron goes directly to the ground state, another UV photon is emitted, capable of ionizing another hydrogen atom. In most cases, however, the electron &amp;quot;cascades&amp;quot; down multiple levels, emitting photons of different energies, which may excite other atoms.&lt;/p&gt;&#xA;&lt;p&gt;It turns out that for every 3 ionizing photons emitted by the star, 2 will eventually — after several interactions — result in the photon corresponding to the energy difference between the first excited state and the ground state; the so-called &amp;quot;Lyman &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;&amp;quot; photon, with a wavelength of 1216 Ångström (121.6 nm). Although there is &lt;em&gt;some&lt;/em&gt; broadening due to thermal motion of the atoms and, in particular, due to the &lt;em&gt;resonant scattering&lt;/em&gt; of Lyman &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; on neutral hydrogen, the result is that most of the light from these stars is converted into a single, very narrow (of the order of a few Ångström) emission line, i.e. very close to monochromatic light.&lt;/p&gt;&#xA;&lt;p&gt;These stars are probably rarely surrounded by planets (because the radiation pressure will tend to blow away the particles used to build planets), and even if they were, these processes happen farther away from the stars than the planets would be. But if you observe a young galaxy, whose spectrum is dominated by young stars, the Lyman &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; emission line is often the only light visible.&lt;/p&gt;&#xA;&lt;strong&gt;UV line &lt;span class=&quot;math-container&quot;&gt;$\rightarrow$&lt;/span&gt; visible line&lt;/strong&gt;&#xA;&lt;p&gt;The Lyman &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; line is still in the UV, and thus invisible to humans. However, since light is redshifted as it travels through the expanding Universe, Lyman &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;-emitting galaxies sufficiently far away will have their emission line carried into the visible range. Since the shortest wavelength we can see is 400 nm, it must be redshifted by a factor &lt;span class=&quot;math-container&quot;&gt;$400/121.6 = 3.3$&lt;/span&gt; (i.e. &lt;span class=&quot;math-container&quot;&gt;$z=2.3$&lt;/span&gt;), corresponding to a distance of 18.6 billion lightyears (but if it's farther away than 25.5 billion lightyears, it will shift into the infrared and be invisible again). Note, though, that this is only in principle; such galaxies are far too dim to be visible to the naked eye. Too see them, you must use a telescope and a camera.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-08-25T17:08:14.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18071" PostTypeId="2" ParentId="10301" CreationDate="2016-08-25T16:25:48.613" Score="3" Body="&lt;p&gt;You can, in fact, differentiate between the crossings of the instability strip in certain types of variable stars. There have been quite a few evolutionary models devoted to this, and many, as well as scores of observations, show that the pulsational period $P$ of a Cepheid variable changes during each crossing. Look at, for instance, the period changes of 60 galactic Cepheids, shown in Figure 2 of &lt;a href=&quot;http://cds.cern.ch/record/611292/files/0304053.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pietrukowicz&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NcIOI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NcIOI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first and third crossings occur when the star is evolving redwards (towards the right of the HR diagram), and show positive $\dot{P}$. The second crossing occurs when the stars are evolving bluewards (towards the left of the HR diagram), and show negative $\dot{P}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Models and observations by &lt;a href=&quot;http://arxiv.org/pdf/1407.8098.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Neilson et al. (2014)&lt;/a&gt; show the same thing (see Figure 1):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gD6Jt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gD6Jt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial large period change indicates the first crossing, the smaller negative period change indicates the second, and the third less drastic increase indicates the third.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar indicators should be present in further crossings (fourth and fifth crossings in the later stages of core and shell helium burnings), and while the amplitude of $\dot{P}/P^2$ would likely be different, the same general positive/negative trends should be present. For more information, see &lt;a href=&quot;http://adsabs.harvard.edu/abs/2004A&amp;amp;A...423..335T&quot; rel=&quot;nofollow noreferrer&quot;&gt;Turner &amp;amp; Berdnikov (2004)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There doesn't seem to be as much information on period changes in non-Cepheid variable stars, although &lt;a href=&quot;https://arxiv.org/abs/1310.0533&quot; rel=&quot;nofollow noreferrer&quot;&gt;Smith (2013)&lt;/a&gt; states that most Mira variables show only minute period changes, if any, and RR Lyrae variables may show period increases as they evolve redwards across the instability strip toward the end of their lives, as is the case with Cepheids. This, by the way, does not fully agree with evolutionary theory.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-08-27T17:02:43.513" LastActivityDate="2016-08-27T17:02:43.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18072" PostTypeId="1" CreationDate="2016-08-25T19:57:57.070" Score="3" ViewCount="514" Body="&lt;p&gt;If I do a quick Google search for &lt;code&gt;amsterdam sunrise&lt;/code&gt;, I immediately get the time back: &lt;code&gt;6:42 AM&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this time calculated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I mean how it is really calculated, not how Google get it, with is probably from a table. Of course I mean sunrises and sunsets in the future, not from the past. $:)$)&lt;/p&gt;&#xA;" OwnerUserId="13570" LastActivityDate="2016-08-25T19:57:57.070" Title="How is the time of sunrise and sunset determined?" Tags="&lt;the-sun&gt;&lt;time&gt;" AnswerCount="0" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18073" PostTypeId="1" CreationDate="2016-08-16T16:59:18.937" Score="11" ViewCount="754" Body="&lt;p&gt;I know about the equations to describe the orbit of a moon around a planet. I know the moon's semi-major axis and eccentricity, and the same for its host world with the star they orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any equation that tells me how much of the moon is illuminated at night, and possibly how bright, as seen from the planet?&lt;/p&gt;&#xA;" OwnerUserId="13837" OwnerDisplayName="Kalcipher23" LastActivityDate="2020-06-22T18:55:16.273" Title="How do I know, mathematically rather than from observation, if a moon is full?" Tags="&lt;orbital-mechanics&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18074" PostTypeId="1" AcceptedAnswerId="18078" CreationDate="2016-08-26T04:17:56.870" Score="10" ViewCount="1315" Body="&lt;p&gt;The recent news of the &lt;a href=&quot;http://arxiv.org/abs/1606.06291&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ultra Diffuse Galaxy (UDG) Dragonfly 44&lt;/a&gt; is an excellent example of what could be termed 'observe different' thinking. The dragonfly telescope is noted not for the size of its collective aperture, but for the absence of the diffracting effects of secondary mirrors and surface roughness that limit the contrast of dim objects in conventional telescopes when brighter sources are nearby. See &lt;a href=&quot;http://www.dunlap.utoronto.ca/instrumentation/dragonfly/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://petapixel.com/2015/05/27/this-telescope-uses-10-canon-lenses-worth-100000/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.astro.caltech.edu/~ycao/B&amp;amp;ETalks/vandokkum_bne.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6RZ1e.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6RZ1e.jpg&quot; alt=&quot;Dragonfly Telescope&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; image of a Dragonfly refractive array telescope &lt;a href=&quot;http://www.dunlap.utoronto.ca/instrumentation/dragonfly/&quot; rel=&quot;nofollow noreferrer&quot;&gt;from here&lt;/a&gt;. &lt;em&gt;Image: P. Van Dokkum; R. Abraham; J. Brodie&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eefC0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eefC0.jpg&quot; alt=&quot;Dragonfly 44 ultradiffuse galaxy&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; The Dragonfly 44 ultradiffuse galaxy &lt;a href=&quot;https://www.washingtonpost.com/news/speaking-of-science/wp/2016/08/25/a-new-class-of-galaxy-has-been-discovered-one-made-almost-entirely-of-dark-matter/&quot; rel=&quot;nofollow noreferrer&quot;&gt;from here&lt;/a&gt;. &quot;Dragonfly 44 is very faint for its mass and consists almost entirely of dark matter. (Pieter van Dokkum, Roberto Abraham, Gemini Observatory/AURA)&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once identified, the radial velocities of the stars in Dragonfly 44 were measured using DEIMOS on the Keck II telescope, in order to determine a value for the mass for the dim, ultra-diffuse galaxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started reading the &lt;a href=&quot;http://arxiv.org/abs/1606.06291&quot; rel=&quot;nofollow noreferrer&quot;&gt;ArXiv&lt;/a&gt; article but quicky got bogged down in the abstract. The very exciting result is that the luminosity and therefore total number of stars is much smaller than what one would expect from the mass obtained from the radial velocity measurements, suggesting that it is &lt;a href=&quot;https://www.washingtonpost.com/news/speaking-of-science/wp/2016/08/25/a-new-class-of-galaxy-has-been-discovered-one-made-almost-entirely-of-dark-matter/&quot; rel=&quot;nofollow noreferrer&quot;&gt;made almost entirely of dark matter&lt;/a&gt;. I wanted to see if I could understand how the mass was calculated, but I got stuck on the phrase &lt;em&gt;deprojected half-light radius&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone just outline how this calculation is done, and what that phrase actually means?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Recently a population of large, very low surface brightness, spheroidal galaxies was identified in the Coma cluster. The apparent survival of these Ultra Diffuse Galaxies (UDGs) in a rich cluster suggests that they have very high masses. Here we present the stellar kinematics of Dragonfly 44, one of the largest Coma UDGs, using a 33.5 hr integration with DEIMOS on the Keck II telescope. We find a velocity dispersion of 47 km/s, which implies a dynamical mass of M_dyn=0.7x10^10 M_sun &lt;strong&gt;within its deprojected half-light radius&lt;/strong&gt; (&lt;em&gt;my emphasis&lt;/em&gt;) of r_1/2=4.6 kpc. The mass-to-light ratio is M/L=48 M_sun/L_sun, and the dark matter fraction is 98 percent within the half-light radius. The high mass of Dragonfly 44 is accompanied by a large globular cluster population. From deep Gemini imaging taken in 0.4&quot; seeing we infer that Dragonfly 44 has 94 globular clusters, similar to the counts for other galaxies in this mass range. Our results add to other recent evidence that many UDGs are &quot;failed&quot; galaxies, with the sizes, dark matter content, and globular cluster systems of much more luminous objects. We estimate the total dark halo mass of Dragonfly 44 by comparing the amount of dark matter within r=4.6 kpc to enclosed mass profiles of NFW halos. The enclosed mass suggests a total mass of ~10^12 M_sun, similar to the mass of the Milky Way. The existence of nearly-dark objects with this mass is unexpected, as galaxy formation is thought to be maximally-efficient in this regime.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-12-09T01:13:52.317" LastActivityDate="2019-12-09T01:13:52.317" Title="What (actually) is the &quot; deprojected half-light radius&quot; of this almost-all-dark-matter Galaxy?" Tags="&lt;galaxy&gt;&lt;telescope&gt;&lt;galactic-dynamics&gt;&lt;velocity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="18075" PostTypeId="2" ParentId="1120" CreationDate="2016-08-26T05:26:29.100" Score="1" Body="&lt;p&gt;The shadow of a vertical stick would show the Sun's azimuth, but that's not really a sundial.&#xA;A proper sundial's gnomon is aligned with the celestial poles so that the shadow indicates the Sun's&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Hour_angle&quot; rel=&quot;nofollow&quot;&gt;hour angle&lt;/a&gt;.&#xA;Then the great circle of hour angle $\pm$90$^\circ$ is projected onto the sundial plane as an east-west line;&#xA;the shadow falls on that line at 6am or 6pm local apparent solar time regardless of the Sun's declination.&#xA;Unless the sundial is an equatorial type, the other&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sundial#Horizontal_sundials&quot; rel=&quot;nofollow&quot;&gt;hour marks&lt;/a&gt;&#xA;are spaced closer together around noon and farther apart around 6am and 6pm.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-08-26T05:26:29.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18076" PostTypeId="2" ParentId="18053" CreationDate="2016-08-26T05:50:24.560" Score="4" Body="&lt;p&gt;Since the question is actually asking for a size of Jupiter as seen from Mars, there is a very simple equation that will give you a good idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\alpha}{360^{\circ}}=\frac{\ell}{2\pi r} $$&#xA;$\alpha$ is the angle that Jupiter would subtend in the sky, or the angular size, and $\ell$ is the actual diameter, or linear size, of Jupiter. $2\pi r$ is the circumference of Jupiter's orbit about Mars. Now I know what you're thinking, that this doesn't make any sense because of the vast differences of Jupiter's orbit when it is close compared to when it is at opposition. It'll work out...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following are all approximations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars orbits the Sun at $\sim1.53AU$ and Jupiter at $\sim5.2AU$. Because $1AU \approx 150Mkm$, The difference in their orbits are roughly $550.5Mkm$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha=\frac{140000km}{2\pi(550.5\times10^6km)}\times360^{\circ}=0.015^{\circ} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is just the average angular size, at the average closest distance, but given that the only real tool to measure angular size with the naked eye (that isn't a sextant) is your hand, and your smallest digit (the pinky) at arms length marks out $1^{\circ}$ you'll never be able to tell the difference in the variation of size due to the change in distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now answering your other questions directly:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Jupiter would look larger, but not enough to actually tell a difference in size.&lt;/li&gt;&#xA;&lt;li&gt;As shown above, Jupiter would be much smaller than the moon appears to us from Earth, $0.5^{\circ}$.&lt;/li&gt;&#xA;&lt;li&gt;No disk, as this is very much a point-like size as far as human eyes are concerned. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2232" LastActivityDate="2016-08-26T05:50:24.560" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18077" PostTypeId="2" ParentId="18073" CreationDate="2016-08-26T06:07:19.260" Score="6" Body="&lt;p&gt;The Moon phases can be defined by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Phase_angle_(astronomy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;phase angle&lt;/a&gt; between the Sun, Moon and Earth; for example, at 0°, the Moon is defined as full, and at 180° it is defined as new. If you want to know how bright the Moon is at a given angle, we would use the phase angle to find the apparent and absolute magnitudes of the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Absolute magnitude, when referring to illuminated objects (objects that do not produce their own visible light), simply means their apparent magnitude if viewed from 1 AU away. This means it is almost entirely dependent on the object's phase angle. Right now, you're asking about how bright the Moon would seem to a person on Earth, so we'll find the apparent magnitude. The formula to find an illuminated object's apparent magnitude (in the Solar System), if we know its absolute magnitude $H$, is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = H + 2.5 \log_{10}{\left(\frac{d_{BS}^2 d_{BO}^2}{p(\chi) d_0^4}\right)}\!\,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $d_0$ is 1 AU, $\chi$ is the phase angle (in radians) and $p(\chi)$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Bond_albedo#Phase_integral&quot; rel=&quot;nofollow noreferrer&quot;&gt;phase integral&lt;/a&gt; (integration of reflected light). $d_{BO}$ is the distance between the observer and the body, $d_{BS}$ is the distance between the Sun and the body, and $d_{OS}$ is the distance between the observer and the Sun. This formula probably looks pretty scary, but it can be simplified with some approximations. First, we can approximate the phase integral as this:&#xA;$$p(\chi) = \frac{2}{3} \left( \left(1 - \frac{\chi}{\pi}\right) \cos{\chi} + \frac{1}{\pi} \sin{\chi} \right)$$&#xA;Where $\chi$ is the phase angle, in radians. In the Moon's case, we can set $H_{Moon} = +0.25$ (this is the absolute magnitude during a full moon), $d_{OS} = d_{BS} = 1$ AU and $d_{BO} = 0.00257$ AU. Now we get the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_{Moon} = 0.25 + 2.5 \log_{10}{\left(\frac{0.00257^2}{p(\chi)}\right)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now, we've got a formula that approximates the apparent magnitude of the Moon at any given phase angle. However, even though this gives a close approximation, it is not 100% accurate. Astronomers use empirically derived relationships to predict apparent magnitudes when accuracy is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a quick script I wrote to calculate the apparent magnitude, given any phase angle:&#xA;&lt;a href=&quot;https://jsfiddle.net/fNPvf/33429/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://jsfiddle.net/fNPvf/33429/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-11-28T15:10:10.930" LastActivityDate="2016-11-28T15:10:10.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18078" PostTypeId="2" ParentId="18074" CreationDate="2016-08-26T06:49:47.353" Score="12" Body="&lt;p&gt;The half light radius is the radius from within which half the luminosity emerges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Deprojected&quot; means that the authors must have fitted some model to the 2D distribution of light, which can then be mathematically deprojected to give them a 3D model for luminosity as a function of radius, that they can then integrate to give a number for the half light radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In section 3, the authors explain that they have done this by fitting a &quot;Sersic profile&quot; to the surface brightness distribution &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Sersic_profile&quot;&gt;https://en.m.wikipedia.org/wiki/Sersic_profile&lt;/a&gt;  The Sersic profile actually has the 2D half light radius as one of its parameters. But if you imagine looking through a ball of stars, this 2D measurement of the half light radius is an underestimate of the true 3D half light radius, because the surface brightness profile is more sharply peaked than the 3D stellar density distribution that produces it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors appear to approximately correct (deproject) this by multiplying the half light radius by 4/3. They also make a small correction for the non-sphericity of the galaxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deprojection factor depends (slightly) on the $n$ index of the Sersic profile and needs to be found by doing a numerical integral. The details can be found in the appendices of Wolf et al. (2010. &lt;a href=&quot;http://arxiv.org/abs/0908.2995&quot;&gt;http://arxiv.org/abs/0908.2995&lt;/a&gt; ), who also provide expressions to directly estimate mass from the projected half light radius and the line of sight velocity dispersion.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-08-26T06:59:14.677" LastActivityDate="2016-08-26T06:59:14.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18079" PostTypeId="1" AcceptedAnswerId="18132" CreationDate="2016-08-26T11:57:13.443" Score="5" ViewCount="203" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/llpUZ.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/llpUZ.jpg&quot; alt=&quot;IC5070 - The pelican nebula&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;This is my first try in false color imaging. The target was IC5070 - the pelican nebula. I took around 3 hours of data (around 1h for Ha, SII and OIII). Then after stacking did an LRGB combination in PixInsight using Ha for red, SII for green and OIII for blue. For luminance I used the stacked Ha image.&lt;/p&gt;&#xA;&lt;p&gt;Now my question is, why are most of these stars so green? It would suggest that they were really bright in the silicon spectrum right? If so wouldn't that mean that those stars are extremly silicon heavy?&lt;/p&gt;&#xA;&lt;p&gt;(Ignore the vertical blue-green line, its a stacking artifact resulting from a camera defect on those pixels)&lt;/p&gt;&#xA;" OwnerUserId="10271" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-08-30T10:14:11.290" Title="Why are the stars in my image so green (false color image, HSO palette)?" Tags="&lt;amateur-observing&gt;&lt;photography&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18081" PostTypeId="1" AcceptedAnswerId="18082" CreationDate="2016-08-26T12:53:18.750" Score="9" ViewCount="1853" Body="&lt;p&gt;Our clocks use 24 hours for a day, our planet takes 24 hours to rotate completely and our planet revolves around the sun in 365 days.&lt;/p&gt;&#xA;&lt;p&gt;Imagine at the start of the year, the side of Earth facing Sol is 'Side A' and the dark side is 'Side B'. The side of the sun that Earth sees is 'Side A' and the opposite side is 'Side B'.&lt;/p&gt;&#xA;&lt;p&gt;After half a year, wouldn't 'Side B' of Earth be facing 'Side B' of Sol, and thus the day/night cycle would be out of sync by 12 hours?&lt;/p&gt;&#xA;&lt;p&gt;I've been trying to get my head around this problem for ages and I can't seem to figure it out. It seems like the angle that the sun hits the Earth must change by about a degree per day until it's gone around 360 degrees. Would that not mean the clocks offset by 1/365th of 24 hours every day over a year?&lt;/p&gt;&#xA;" OwnerUserId="13979" LastEditorUserId="9582" LastEditDate="2021-03-28T13:30:20.523" LastActivityDate="2021-04-02T15:15:41.953" Title="Why do our clocks not go out of sync with the day over a year?" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;rotation&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="18082" PostTypeId="2" ParentId="18081" CreationDate="2016-08-26T13:24:04.020" Score="16" Body="&lt;p&gt;You are correct, and in fact the clocks &lt;em&gt;are&lt;/em&gt; offset. But most people never notice this because of the way days are defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you measure the &quot;day&quot; length by the stars (instead of the sun) you will see it's not 24 hours at all. It's around 23 hours 56 minutes. (Check this by going out at night, picking a bright star, and noting the exact time it passes due south. Then observe it the following night and note it passes 4 minutes earlier.)  By the way this shortened &quot;day&quot; is &lt;a href=&quot;http://www.bbc.co.uk/schools/gcsebitesize/science/triple_ocr_21c/further_physics/naked_eye_astronomy/revision/6/&quot;&gt;the sidereal day&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each day, the Earth shifts along in its orbit (by about a degree, as you correctly observed in the question). So if you go out and point at the Sun at Noon every day, you are pointing in a slightly different direction. &lt;em&gt;(I'll also neglect the Solar Analemma at this point too as it's just too confusing.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really, the &quot;day&quot; of 24 hours is not related to the true rotation period of Earth at all; loosely speaking it's just the average time between sunrises, or sunsets, or noons.&lt;/p&gt;&#xA;" OwnerUserId="7491" LastEditorUserId="7491" LastEditDate="2016-08-26T13:40:02.743" LastActivityDate="2016-08-26T13:40:02.743" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18083" PostTypeId="1" AcceptedAnswerId="18085" CreationDate="2016-08-26T13:50:00.803" Score="5" ViewCount="888" Body="&lt;p&gt;I was wondering how is it that we can see the Pillars of Creation (I know they're no longer there too) which are labeled as 6,500 light years away from earth and yet Proxima B is labeled at just 4.2 light years away and yet we cannot see it clearly and it was very difficult to find?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the Pillars of Creation are giant gas bubbles with different molecules that make it so visible and Proxima B is a relatively tiny planet that doesn't emit light but I cannot fathom how much greater the distance is from Proxima B and yet it is clearly much more visible. Shouldn't our current technology be able to view 'close up' images of Proxima B?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing here?&lt;/p&gt;&#xA;" OwnerUserId="13981" LastEditorUserId="9301" LastEditDate="2016-09-16T04:09:39.830" LastActivityDate="2016-09-17T19:07:56.677" Title="Why can we see the Pillars of Creation, but not Proxima B?" Tags="&lt;planet&gt;&lt;exoplanet&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18084" PostTypeId="2" ParentId="18083" CreationDate="2016-08-26T14:16:41.147" Score="5" Body="&lt;p&gt;First, let's talk about size. The Pillars of Creation aren't just &quot;giant gas bubbles&quot;; they're clouds of gas and dust &lt;a href=&quot;http://www.pbs.org/wgbh/nova/origins/hubb-16.html&quot;&gt;&lt;em&gt;four light-years&lt;/em&gt; long&lt;/a&gt;. That's enormous, about 36,000,000,000,000 kilometers. Proxima Centauri b, on the other hand, has a diameter of &lt;a href=&quot;http://phl.upr.edu/press-releases/proxb&quot;&gt;0.8-1.4 Earth diameters&lt;/a&gt;; one Earth diameter is approximately 12,700 kilometers. That's miniscule, in comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_diameter&quot;&gt;angular diameters&lt;/a&gt; of the objects. The general formula is&#xA;$$\delta=2\arctan\left(\frac{\text{diameter}}{2\cdot\text{distance}}\right)$$&#xA;where the angular diameter is in radians. The Pillars of Creation are 6,500 light-years away, conservatively, giving just them an angular length of $3.52\times10^{-2}$ degrees. Proxima Centauri b is 4 light-years away, with an angular diameter of about $2.02\times10^{-8}$ degrees. 6 orders of magnitude is nothing to sneeze at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, the Pillars of Creation are located in an &lt;a href=&quot;https://en.wikipedia.org/wiki/Emission_nebula&quot;&gt;emission nebula&lt;/a&gt;. Light from the young stars that have just been born ionizes gas in the Eagles Nebula, which in turn emits the light. The dark gas and dust in the Pillars of Creation stands out quite sharply in all of this, making it easy to see.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-08-26T14:25:19.697" LastActivityDate="2016-08-26T14:25:19.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18085" PostTypeId="2" ParentId="18083" CreationDate="2016-08-26T14:27:04.690" Score="5" Body="&lt;p&gt;Simply put, the Eagle Nebula (from which the Pillars of Creation originate) shines quite brightly, much more than Proxima Centauri (which is a red dwarf). First, we'll take a look at their absolute magnitudes (how bright an object actually is) and their apparent magnitudes (the brightness an object appears to have, based on its distance from us). Both are logarithmic, with lower numbers being brighter and higher numbers being dimmer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Eagle Nebula's absolute magnitude is a whopping -8.21. Meanwhile, Proxima has an absolute magnitude of 15.60. The formula relating absolute magnitudes and apparent magnitudes is:&#xA;$$M = m - 5 (\log_{10}{d}-1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $M$ is the absolute magnitude, $m$ is the apparent magnitude, and $d$ is the distance, in parsecs. Plug in those numbers, and we find that the Eagle Nebula has an apparent magnitude of 6. Meanwhile, Proxima Centauri has an apparent magnitude of 11.13.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we wanted to compare Proxima Centauri with the Eagle Nebula, we could use the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_b = 10^{0.4 x} = 10^{0.4×(11.13-6)} \approx 113$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the Eagle Nebula shines approximately 113 times brighter than Proxima Centauri. It's no wonder why Proxima Centauri b isn't visible to the naked eye — even its host star is too dim to see from the naked eye. Sure, the Alpha Centauri star system has other stars in it, but those are about 15,000 AU from Proxima, so they should have a minimal effect on the planet's brightness.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="53" LastEditDate="2016-09-17T19:07:56.677" LastActivityDate="2016-09-17T19:07:56.677" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18086" PostTypeId="2" ParentId="15110" CreationDate="2016-08-26T15:42:19.813" Score="-2" Body="&lt;p&gt;Also, you might be forgetting that the distances of the stars that exploded will cause us to see them at different times.  Say three stars exploded in a given year but one was 1,000 light years away another was 1,500 light years away and the third was 2,000 light years away.  We wouldn't see those novae here on Earth in the same year.  So, while it's quite possible that there are 2 to 3 novae a year, we may (i.e. likely will) see them many years apart.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the flip side, a star that exploded 100 years ago that was 100 light years away and another star that exploded 500 years ago and was 500 light years away would appear to us in the same year here on Earth.&lt;/p&gt;&#xA;" OwnerUserId="13983" LastEditorUserId="9301" LastEditDate="2016-08-26T17:20:35.923" LastActivityDate="2016-08-26T17:20:35.923" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18087" PostTypeId="2" ParentId="18073" CreationDate="2016-08-26T18:12:21.627" Score="4" Body="&lt;p&gt;Here's a practical approach - the algorithm and the equations are packaged as a software library.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Install PyEphem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rhodesmill.org/pyephem/&quot; rel=&quot;nofollow&quot;&gt;http://rhodesmill.org/pyephem/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Run it:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;$ python&#xA;Python 2.7.12 (default, Jun 29 2016, 14:05:02) &#xA;[GCC 4.2.1 Compatible Apple LLVM 7.3.0 (clang-703.0.31)] on darwin&#xA;Type &quot;help&quot;, &quot;copyright&quot;, &quot;credits&quot; or &quot;license&quot; for more information.&#xA;&amp;gt;&amp;gt;&amp;gt; import ephem&#xA;&amp;gt;&amp;gt;&amp;gt; moon = ephem.Moon(ephem.now())&#xA;&amp;gt;&amp;gt;&amp;gt; print moon.phase&#xA;32.316860199&#xA;&amp;gt;&amp;gt;&amp;gt; print(ephem.next_new_moon(ephem.now()))&#xA;2016/9/1 09:03:05&#xA;&amp;gt;&amp;gt;&amp;gt; print(ephem.next_full_moon(ephem.now()))&#xA;2016/9/16 19:05:05&#xA;&amp;gt;&amp;gt;&amp;gt; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;'phase' is between 0 (new moon) and 100 (full moon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More details:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rhodesmill.org/pyephem/tutorial.html&quot; rel=&quot;nofollow&quot;&gt;http://rhodesmill.org/pyephem/tutorial.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2016-08-26T18:12:21.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18088" PostTypeId="1" CreationDate="2016-08-26T18:33:55.380" Score="1" ViewCount="545" Body="&lt;p&gt;I've read a lot about the Lunar Nodes online (on Wikipedia, astro.com and a bunch of other websites), but there is one thing I cannot figure out and the sources don't seem to clearly explain: &lt;strong&gt;why are the nodes moving on a daily basis?&lt;/strong&gt; As the node is a point of intersection of Moon's orbit and the Ecliptic I understand that this point does not fall in the same place in space every time, but I would expect that once it's passed by the Moon it stays where the crossing occurred and its position gets updated on the next crossing of the Ecliptic, so in ephemeris I would expect to see the same values of the coords between the crossings, but instead I can see they're changing on the daily basis (ie. this ephemeris has 2 columns showing true and mean nodes' positions: &lt;a href=&quot;http://www.astro.com/swisseph/ae/2000/ae_2016.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.astro.com/swisseph/ae/2000/ae_2016.pdf&lt;/a&gt;).&#xA;I'm sure the ephemeris cannot be wrong, so I must be missing some details. I thought that finding the Moon's nodes would be as easy as finding the dates/times/longitudes when the Moon's latitude equals 0, but apparently this way I can only find the times and static points of the crossings of the Ecliptic, not the moving nodes as they're shown in ephemeris tables.&lt;/p&gt;&#xA;" OwnerUserId="13319" LastActivityDate="2016-08-26T23:21:14.447" Title="Why is the Lunar Node in a daily motion?" Tags="&lt;the-moon&gt;&lt;coordinate&gt;&lt;ecliptic&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18089" PostTypeId="2" ParentId="18081" CreationDate="2016-08-26T19:36:54.767" Score="7" Body="&lt;p&gt;I was looking at the answer provided by Andy and the comments attached to it, and together they add up to a reasonable answer.&lt;br&gt;&lt;br&gt;There are several ways to measure a day, the two most common are sidereal and solar. A sidereal day is the time it take earth to rotate 360°; a solar day is the amount of time it takes the earth to rotate such that the sun appears at the same meridian as the day before. (Sunrise and sunset aren't really good references, as they are influenced by atmospheric conditions that affect refraction.) The solar day, or more accurately the mean solar day, is the day our calendars measure. The mean solar day is exactly what it says, the average amount of time during the course of a year that it takes for the sun to appear over the same meridian from day to day.&lt;br&gt;&lt;br&gt;Unfortunately, the definition of a sidereal day is very often given as the definition of a day, which is totally inaccurate, and very confusing.&lt;br&gt;&lt;br&gt;The length of the solar day varies during the year because the of Earth's distance from the sun; it is closer in January and farther in July. Because it is closer in January; it is also moving faster, which means that the planet has to rotate more for the sun to appear over the same meridian.&lt;br&gt;&lt;br&gt;If you can get your head around this, the concept really isn't very complicated.&lt;/p&gt;&#xA;" OwnerUserId="7108" LastActivityDate="2016-08-26T19:36:54.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18090" PostTypeId="2" ParentId="18088" CreationDate="2016-08-26T23:21:14.447" Score="2" Body="&lt;p&gt;The nodes are the two points on the orbit of the moon at which the plane of the orbit crosses the elliptic. The position of the nodes change in a predictable way due to the perturbation by the sun (and the other planets). Since the orbit is constantly changing, the position of the longitude. The node is not the position where the crossing last occurred. Rather it is the intersection of two imaginary lines in the sky, the ecliptic and the orbit of the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position of the nodes are measured as an angle, clockwise from the first point of Aries, thus it is a longitude, or right ascension. On the first of September 2016 the longitude of the ascending node of the moon is 162.484 degrees, and decreasing slowly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The linked pdf indicates a value of 12♍41, apparently this means 12 degrees 41 minutes within Virgo. The notation is not used in astronomy. If you are interested in the scientific study of the sky you should be aware that astro.com is an astrology website. The presentation of ephemeris may not represent scientific practice. I consider modern astrology to be pseudoscience and fraudulent.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-08-26T23:21:14.447" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18091" PostTypeId="1" CreationDate="2016-08-27T03:55:51.887" Score="3" ViewCount="257" Body="&lt;p&gt;In an &lt;a href=&quot;https://worldbuilding.stackexchange.com/questions/52964/can-a-main-sequence-star-with-planets-exist-orbiting-a-magnetar/52970?noredirect=1#comment148757_52970&quot;&gt;answer on a different SE&lt;/a&gt; someone raised the point that “common” explainations published never discuss stars being &lt;em&gt;captured&lt;/em&gt;.  Presumably binaries are &lt;em&gt;formed&lt;/em&gt; as pairs, or team up while still in the birth cluster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems quite plausible that an aged star (or stellar reminant) can be paired with a main sequence star in its prime, having paired up well &lt;em&gt;after&lt;/em&gt; the old supernova was completed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is this mechanism of multi-body reassortment known to occur, presumed, or thought to not occur?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another mechanism &lt;em&gt;implied&lt;/em&gt; but not detailed in popular works is the possible situation with Proxima Centauri: a star going with the flow in the galaxy may find itself loosly attached to another star (or system), and this bond is so tenuous that it can easily be broken by &lt;em&gt;another&lt;/em&gt; passing star, so the distant star &lt;a href=&quot;https://en.wikipedia.org/wiki/Do-si-do&quot; rel=&quot;nofollow noreferrer&quot;&gt;do-si-do&lt;/a&gt;s to different partners over the course of billions of years.  &lt;strong&gt;Is this an accurate depiction of the real situation?&lt;/strong&gt;  This loose attachment out of the general flow can occur with stars of any mass, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, a star can be captured &lt;em&gt;with planets&lt;/em&gt;, right?  Naturally a close-enough encounter between stars (and reminants) could seriously disrupt a planetary system, but I want to check on the possibility that capture with planets is also a possibility.  I think that’s not a problem for the “loose attachment” case, and Proxima Cen may indeed have a planet.  How about the necessary mechanics of binary reassortment?  Is that necessarily so close and tight thatbplanets would be seriously disrupted, or is it plausible that planets would simply go along for the ride?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: &lt;a href=&quot;https://en.wikipedia.org/wiki/PSR_B1620-26_b#Evolutionary_history&quot; rel=&quot;nofollow noreferrer&quot;&gt;PSR B1620-26&lt;/a&gt;.  A neutron star captures a &lt;em&gt;star with planet&lt;/em&gt;, trading off its original companion.&lt;/p&gt;&#xA;" OwnerUserId="7142" LastEditorUserId="-1" LastEditDate="2017-04-13T12:52:24.020" LastActivityDate="2016-09-01T11:17:43.957" Title="Binary stars of different ages?" Tags="&lt;galaxy&gt;&lt;orbital-mechanics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18092" PostTypeId="1" AcceptedAnswerId="18108" CreationDate="2016-08-27T06:06:02.377" Score="4" ViewCount="533" Body="&lt;p&gt;Let's say we have already a base on an exoplanet (like Proxima B). Would it be possible to send information from that base to the earth with todays technology? Not to talk about communication with such a distance ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And how would it be done? Light or Radiowaves? Would it be possible to direct it so well, that some very small fraction of that wave could reach the earth? Or is it only a question of how much energy is needed? &lt;/p&gt;&#xA;" OwnerUserId="13986" LastActivityDate="2016-08-29T02:22:41.040" Title="Is communication with proxima b possible with todays technology?" Tags="&lt;exoplanet&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18094" PostTypeId="2" ParentId="18091" CreationDate="2016-08-27T08:46:22.503" Score="2" Body="&lt;p&gt;Stellar encounters are very rare outside dense stellar environments. Such dense stellar environments are only found in a star's birth cluster, where all the stars are coeval to a close approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the switching and capture you refer to is probably common in the first few million years of a star's life, but would occur with stars of a similar age. Thereafter, once a stellar system is &quot;in the field&quot;, encounters with other stars hardly occur at all, but are not impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not aware of any binary where the components are thought to have a very different age.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-08-27T08:46:22.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18095" PostTypeId="2" ParentId="17911" CreationDate="2016-08-27T13:27:37.823" Score="5" Body="&lt;p&gt;The &quot;Schechter-like&quot; Equation 3 in that paper is meant to give the value of black-hole mass density in units of &quot;number of black holes of a particular mass per volume per log black hole mass&quot;, with the volume in units of megaparsecs and the &quot;log black hole mass&quot; in units of 0.1 log solar-mass units -- i.e., 0.1 dex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the integral involves multiplying by $d \log M_{\rm BH}$, you are multiplying $(\log M_{\rm BH})^{-1}$ by $\log M_{\rm BH}$, and thus the &quot;dex&quot; units do indeed cancel out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the exact value, you're going to have trouble reproducing that. First of all, the value from the text that you quoted (&quot;5.06&quot;) is the result when they integrate over &lt;em&gt;all&lt;/em&gt; BH masses, not just the $10^6$ to $10^{10}$ limits in Equation 4. From their Table 3, the relevant comparison value would be 4.9 if you're just doing the integration in Equation 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the values they quote in Table 3 (and in the text for the full integration) are apparently not straightforward computations from their Equation 4 (or from the full integral). They go through a complicated exercise of computing this value 10001 times, each time tweaking the input coefficients of Equation &lt;em&gt;1&lt;/em&gt;, then applying some more using random Gaussian deviations, so they end up with a distribution of BH density values, not just a single one. They say&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The final SMBH density is defined as the median of this distribution&#xA;  with the 1σ error given by 68 percentile range&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2016-08-27T13:27:37.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18097" PostTypeId="1" AcceptedAnswerId="18098" CreationDate="2016-08-27T20:01:54.867" Score="1" ViewCount="546" Body="&lt;p&gt;I heard that the sky wouldn't be dark at night if the light of every star had already reached the earth.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Of which (currently invisible) star will the light reach the earth next and when will that be?&lt;/li&gt;&#xA;&lt;li&gt;What was the last star that we can now see at night to reach the earth and when was that?&lt;/li&gt;&#xA;&lt;li&gt;How can we know (1) if the light hasn't even reached us yet? Or can't we know and will a star suddenly appear without anyone being able to predict that?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13994" LastActivityDate="2016-09-08T08:28:06.517" Title="The light of which star is next to reach the earth?" Tags="&lt;star&gt;&lt;light&gt;&lt;speed&gt;&lt;night-sky&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18098" PostTypeId="2" ParentId="18097" CreationDate="2016-08-27T20:31:29.340" Score="5" Body="&lt;p&gt;I think you may be thinking of &lt;a href=&quot;https://en.wikipedia.org/wiki/Olbers%27_paradox&quot; rel=&quot;nofollow&quot;&gt;Olbers' paradox&#xA;&lt;/a&gt;. This supposes that if the universe were infinite in time and extent, and stars were more or less randomly placed, then every line of sight would end at a star, and so the sky would be as bright as the surface of a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paradox can be resolved by supposing that the universe is not infinite in time, but had a beginning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't see stars suddenly appearing as their light reaches us. What we see is stars in all stages of their development. When we look the nebulae where stars are born we see clumps of gas in the process of collapsing to form stars. These are called &lt;a href=&quot;https://en.wikipedia.org/wiki/Protostar&quot; rel=&quot;nofollow&quot;&gt;protostars&lt;/a&gt;. Now the process of stellar formation takes a long time: about a million years for a medium sized star. The start of a star is not a sudden &quot;turning on&quot;. Instead, as the core of the protostar shrinks and heats up, nuclear fusion starts; slowly at first, then increasing until it is able to prevent further collapse and heating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Protostars usually are dim in visible light, they can be found with infra-red telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the process is slow and there isn't a sudden turning on, your questions cannot be directly answered. However, there is an object &lt;a href=&quot;http://www.nasa.gov/content/goddard/nasa-satellites-catch-growth-spurt-from-newborn-protostar/&quot; rel=&quot;nofollow&quot;&gt;HOPS383&lt;/a&gt; which is 1400 light years distant, which between 2004 and 2008 had a &quot;growth spurt&quot; and became visible in infra-red for the first time. It can't be seen in visible light yet - so in a sense this protostar answers your question 1 and 2, in different wavelengths.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2031" LastEditDate="2016-08-28T14:22:47.260" LastActivityDate="2016-08-28T14:22:47.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18099" PostTypeId="2" ParentId="18092" CreationDate="2016-08-27T23:34:02.563" Score="1" Body="&lt;p&gt;Yes, it is absolutely possible. It would take a few years (more than 4) just to get our signal there though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would imagine that a very high powered x-ray laser pulsed at a high frequency implying 0s and 1s, in a binary code would be the only real way to ensure that the same information leaving here would arrive there. The signal received would be in a larger wavelength, so that would have to be taken into account when deciphering the message. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when the laser fired from JPL's observatory at the Galileo satellite when it was at Jupiter, the Galileo didn't have any trouble decoding the message. (see my profile pic) And that was only a (summed pulse) 40MW Nd:YAG laser (&gt;_o), I'm sure that a space based laser system firing a similar strength laser (in the x-ray band) wouldn't have too too much trouble getting a message out there to Proxima B, if there were an array large enough to collimate the beam. My first order approximation puts the primary required to be something on the order of 60+ meters... which isn't feasible with today's technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ipnpr.jpl.nasa.gov/progress_report/42-114/114Q.pdf&quot; rel=&quot;nofollow&quot;&gt;project paper&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2232" LastEditorUserId="2232" LastEditDate="2016-08-29T02:22:41.040" LastActivityDate="2016-08-29T02:22:41.040" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18100" PostTypeId="1" AcceptedAnswerId="18296" CreationDate="2016-08-28T08:40:08.913" Score="2" ViewCount="11267" Body="&lt;p&gt;What is the resolution in megapixels of the Hubble Telescope?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The HST cannot see the planet Pluto clearly. Why is that?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jmHDp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jmHDp.jpg&quot; alt=&quot;Pluto imaged by HST&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11792" LastEditorUserId="11792" LastEditDate="2016-09-10T08:22:41.583" LastActivityDate="2016-09-10T11:57:51.033" Title="What is the resolution in megapixels of the Hubble Telescope?" Tags="&lt;pluto&gt;&lt;hubble-telescope&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18101" PostTypeId="2" ParentId="18100" CreationDate="2016-08-28T09:00:44.473" Score="9" Body="&lt;p&gt;The size of the cameras that took the Pluto pictures is easy enough to find: the ACS Wide Field Camera has two 2K by 4K (hence 8 Mega pixel) CCDs, with a field of view of 202×202 arcsec. The high resolution HRC had a square 1 Megapixel camera with a 26×29 arcsec field of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that the images of Pluto are relatively poor, is that Pluto is small! The size of the camera is irrelevant; it is the size of the pixels and angular resolution that are vital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pixel size of the HRC (which I think is what the pictures are mainly due to) is 0.025 arcsec, which matches the angular resolution of 0.05 arcsec. The angular diameter of Pluto is about 0.1 arcsec. Hence you effectively have an image that is only made of 4×4 pixels!  I'm surprised it looks as good as it does - but that is because these are not the raw images (see below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: In response to comments I had a look at some details from the original paper describing these images, by Buie et al. (2010).  &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0004-6256/139/3/1128/meta&quot; rel=&quot;noreferrer&quot;&gt;http://iopscience.iop.org/article/10.1088/0004-6256/139/3/1128/meta&lt;/a&gt;  It seems that what is shown is not an actual image from the ACS, but is a best-fitting map of what Pluto looks like, obtained by modelling all the images available. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model image is described by a much finer pixel scale than the original images, but then blurred with a Gaussian filter. However, to quote from the paper, &quot;The width of the Gaussian filter was chosen to impose a limit on the spatial frequencies in the output image comparable to the diffraction limit of HST. &quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the angular resolution of the final &quot;reconstructed images&quot; is still similar to and limited by the angular resolution of the HST.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to hammer home the message. It really doesn't matter here how many pixels the HST cameras have, so long as they have enough to sample the angular resolution properly. At a 0.1 arcsec angular diameter as seen from Earth, Pluto has an angular size equivalent to trying to take a photo of a quarter from 50 km. It really isn't going to help you if you have the latest camera phone with a gazillion pixels unless you also connect it to a 2.4m telescope like HST with an angular resolution better than this.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-08-29T16:59:41.263" LastActivityDate="2016-08-29T16:59:41.263" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18102" PostTypeId="1" CreationDate="2016-08-28T09:43:16.870" Score="-1" ViewCount="71" Body="&lt;p&gt;If you traveled outward and outward in a straight line, indefinitely and pugnaciously, you would never arrive at an outer boundary. Instead, you would come back to where you began.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, in the same manner, &lt;strong&gt;can we assume to be able to observe the Milky Way in the long distance, anywhere in space&lt;/strong&gt;, given the right time and technology? It would be the past version of our self, of course, but is it possible theoretically or practically?&lt;/p&gt;&#xA;" OwnerUserId="14000" LastEditorUserId="2365" LastEditDate="2016-08-28T11:15:41.793" LastActivityDate="2016-08-28T11:15:41.793" Title="Can we observe the Milky Way on the edge of the universe?" Tags="&lt;cosmology&gt;&lt;space-time&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2016-08-30T06:23:00.263" ContentLicense="CC BY-SA 3.0" />
  <row Id="18103" PostTypeId="2" ParentId="18102" CreationDate="2016-08-28T11:15:09.497" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;you would come back to where you began&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is at least doubtful. Even if the Universe has the topology of a 3-sphere, there hasn't been enough time for light to completely travel around it, and since the Universe is expanding at an accelerating rate light would never have the time to return to its starting position. In fact it may well be that the Universe is infinite in extent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have looked for evidence that the universe might be smaller, at least in some directions, or have a topology that allows for light to return to its starting point. No such evidence has been found.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-08-28T11:15:09.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18104" PostTypeId="1" AcceptedAnswerId="18106" CreationDate="2016-08-28T12:33:06.693" Score="1" ViewCount="113" Body="&lt;p&gt;I was reading Gene Roddenberry's &lt;a href=&quot;http://leethomson.myzen.co.uk/Star_Trek/2_The_Next_Generation/Star_Trek_-_The_Next_Generation_Bible.pdf&quot; rel=&quot;nofollow&quot;&gt;guide for TNG writers&lt;/a&gt; and he mentions that that the time of ST:TOS, only 4% of the stars in our galaxy have been charted; this goes up to 19% by the time of TNG. This got me to wondering: what percentage of stars in the Milky Way galaxy have actually been catalogued?&lt;/p&gt;&#xA;" OwnerUserId="14002" LastActivityDate="2016-08-28T14:13:04.437" Title="Percentage of charted stars" Tags="&lt;star&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18105" PostTypeId="1" CreationDate="2016-08-28T14:03:41.367" Score="1" ViewCount="47" Body="&lt;p&gt;Recently the news announced the discovery of a planet with characteristics similar in size to earth with the possibility for liquid water, 4.2 light years away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if there was a civilisation of human beings, pretty much identical to our own civilisation, with the same population, radio output, distribution, etc then, could we detect them now?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, assuming the progression of radio/telescope technology here on earth continues at the same pace, how long would it be before we were aware something artificial was happening on that planet ( disregarding communications, language, etc etc )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If it helps, assume the planet is an identical clone of earth in every way. I'm more interested in if our civilisation was on our closest star would we know?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="14003" LastEditorUserId="14003" LastEditDate="2016-08-28T16:03:40.367" LastActivityDate="2016-08-28T16:03:40.367" Title="Detecting people on Proxima Cen b" Tags="&lt;radio&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2016-08-28T16:58:07.870" ContentLicense="CC BY-SA 3.0" />
  <row Id="18106" PostTypeId="2" ParentId="18104" CreationDate="2016-08-28T14:13:04.437" Score="3" Body="&lt;p&gt;The largest catalogues, such as Hubble's Guide Star Catalog or the US Navy USNO-B1.0 catalogue have entries for about 1 billion object (mostly stars) with magnitudes down to 21.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The galaxy itself has about &lt;a href=&quot;http://www.space.com/25959-how-many-stars-are-in-the-milky-way.html&quot; rel=&quot;nofollow&quot;&gt;100 billion stars&lt;/a&gt;, but there is considerable uncertainty about that value. The true value could be less or a lot more. Between 50 and 500 billion is the range of uncertainty&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have catalogued between 0.2% and 2% of the galaxy's stars with best estimate of &lt;strong&gt;about 1%&lt;/strong&gt; of stars have been catalogued.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-08-28T14:13:04.437" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18107" PostTypeId="1" AcceptedAnswerId="18111" CreationDate="2016-08-28T14:33:29.870" Score="4" ViewCount="700" Body="&lt;p&gt;I saw several chronologies of the trial online but most of them only mention the name of Urban 8 as involved in the actual court proceedings. How many judges were involved and who were they exactly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 1. Reading the small print at wiki I found the names of the three of the ten cardinals who did not vote to convict: Francesco Barberini, Laudivio Zacchia and Gaspare Borgia.  I still don't know the names of any cardinals that voted to convict.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 2. Material related to the trial can be found in &quot;&lt;a href=&quot;//hsm.stackexchange.com/q/5103&quot;&gt;Hidden agenda of the Galileo trial?&lt;/a&gt;&quot; on the &lt;em&gt;History of Science and Mathematics&lt;/em&gt; Stack Exchange site.&lt;/p&gt;&#xA;" OwnerUserId="14004" LastEditorUserId="14019" LastEditDate="2016-08-29T10:34:07.673" LastActivityDate="2016-09-02T18:54:32.883" Title="How many judges sat on Galileo's trial over heliocentrism and who were they exactly?" Tags="&lt;solar-system&gt;&lt;history&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18108" PostTypeId="2" ParentId="18092" CreationDate="2016-08-28T16:08:38.293" Score="6" Body="&lt;p&gt;This is a pretty complex question.  You might have better luck at space exploration than here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you asked about current technology, how NASA does it now is they have very large microwave receivers, 3 stations, 70 meters each and the transmission is in microwave frequencies.    Microwaves are likely the best wavelengths to send long range space communications because it's a comparatively low background noise and because it travels through Earth's atmosphere, so receivers can be built on the ground, which is much easier than building a 70 meter (or larger) dish in space.   I'm using this as a &lt;a href=&quot;http://www.ka9q.net/space.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;, it seems reliable, though it's apparently a hobbyist, not anything official.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This picture below is a page from NASA, showing the microwave window, which is used for communication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3aTK0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3aTK0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the &lt;a href=&quot;http://deepspace.jpl.nasa.gov/about/#&quot; rel=&quot;nofollow noreferrer&quot;&gt;microwave receivers&lt;/a&gt;, called the Deep Space Network (DSN), operated by NASA's Jet Propulsion Laboratory (JPL)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The precise band used is a little tricky (and I'm not an expert), but bands for satellite communications are divided up by countries.  Article on that &lt;a href=&quot;http://www.spaceacademy.net.au/spacelink/radiospace.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.   I would think they might try to use a band in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Water_hole_(radio)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Water Hole&lt;/a&gt;, 1.42 to 1.666 GHz, which is in the L band, but most deep spacecraft communication and photos sent from distant space crafts is done in the S band.    I'm not sure why that is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As distances double, the strength of the signal is reduced by four and the signal has to stand out against background radiation, which is constant, while the signal weakens by the square of the distance.   This isn't an easy problem over vast distances.   But when you say &quot;communicate&quot;, there's different kinds of communication, Morse Code or &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_binary_codes&quot; rel=&quot;nofollow noreferrer&quot;&gt;Binary Codes&lt;/a&gt; are very simple, only a few bytes per letter and messages can be transmitted that way at far greater distances.   NASA is generally interested in photographs, and our spacecraft relay pictures back to NASA by ones and zeros via microwave messages, this can be compared to internet connections for speed.  New Horizons, which recently flew past Pluto, takes about 1 hour to send 1 picture.   Very good &lt;a href=&quot;https://space.stackexchange.com/questions/9916/how-are-new-horizons-images-sent-back-to-earth&quot;&gt;write-up on that here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Mars Reconnaissance Orbiter (&lt;a href=&quot;https://www.quora.com/How-long-does-it-take-to-transmit-a-picture-from-the-Curiosity-Rover-back-to-Earth&quot; rel=&quot;nofollow noreferrer&quot;&gt;per Quora answer&lt;/a&gt;), can communicate with as fast as 2 million bits per second, but it's much closer.    How far and/or how fast (the 2 questions are related) a space craft can communicate with Earth depends on how big a transmitter they give it, how much power it has and how long it's power source is required to last.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'll try to answer your question badly, using Pluto and New Horizons as a baseline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;New Horizons is a relatively recent launch, (at least, compared to the Voyagers), it was launched in 2006, so it's using 10 year old technology.   It's currently a bit over &lt;a href=&quot;http://pluto.jhuapl.edu/Mission/Where-is-New-Horizons/index.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;35 astronomical units&lt;/a&gt; from earth and it's still sending pictures because it sends them very slowly.   This &lt;a href=&quot;http://www.airspacemag.com/daily-planet/new-horizons-still-only-halfway-through-its-download-pluto-180958617/?no-ist&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; is 5 months old, but it says that it took New Horizons about 8 months to send half of the pictures it took of Pluto and Charon (and Nix and Hydra and . . . the others), by the equivalent of a tediously slow internet connection, (from the article)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;New Horizons is transmitting data at a poky 2,000 bits per second (March 2016), and by summer that will drop to 1,500—a fraction of one percent of&#xA;  broadband data rates here on Earth. It takes about an hour for an&#xA;  average image file to arrive, from the first bit to the last.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's also relatively low power.   I couldn't find specifics on the wattage of it's transmitter/receiver unit, but the entire New Horizons power usage is currently about 200 watts &lt;a href=&quot;http://pluto.jhuapl.edu/Mission/Spacecraft/Systems-and-Components.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt; and there are several components on it that share that energy.   &lt;a href=&quot;http://pluto.jhuapl.edu/Mission/Spacecraft/Payload.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.  Obviously we have the means to build much more powerful transmitters than 30 or 50 watts, or whatever it uses, but that's all that was needed for this specific mission.   New Horizons is expected to operate for 20-25 years, till about 2030, at which point it will be over twice as far from Earth as it is now, and transmitting data to us at about 500 bits per second.   &lt;a href=&quot;https://space.stackexchange.com/questions/4827/how-long-can-probes-function-e-g-new-horizons-can-and-should-we-build-centur&quot;&gt;Source&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A final bit of information is the angle of the beam.   (from the first New Horizons link above):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The system includes two broad-beam, low-gain antennas on opposite&#xA;  sides of the spacecraft, used mostly for near-Earth communications; as&#xA;  well as a 30-centimeter (12-inch) diameter medium-gain dish antenna&#xA;  and a large, 2.1-meter (83-inch) diameter high-gain dish antenna. The&#xA;  antenna assembly on the spacecraft’s top deck consists of the high,&#xA;  medium, and forward low-gain antennas; this stacked design provides a&#xA;  clear field of view for the low-gain antenna and structural support&#xA;  for the high and medium-gain dishes. Operators aim the antennas by&#xA;  turning the spacecraft toward Earth. The high-gain beam is only 0.3&#xA;  degrees wide, so it must point directly at Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, onto (approximately) Proxima Centauri.   It's about 268,000 AU from Earth, which is about, 7,600 times further from Earth than New Horizons currently is.   That requires a beam, everything else being equal, some 57 million times stronger.   That's not going to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't find any good information on the technology.   If a beam could be tightened from 0.3 degrees wide to 0.1, or 0.01, the first would provide 9 times as strong a signal.  The 2nd, 900 times.   A .01 degree wide beam from Proxima Centauri would spread out to about 46 astronomical units by the time it reaches Earth, so I would think, the tighter they could make the beam the better, but at .01 degrees, the signal strength would still be about 1/60,000th the strength of New Horizons current transmission.   .001 degrees, and I have no idea if that's possible, but that tight a beam, you're getting into the &quot;this could work&quot; range, about 1/600th the strength of the New Horizon's signal, and with greater power behind the signal and, perhaps, a larger satalite dish built on Earth, then it's in the range of workable.  (but I have no idea if microwave beans can be sent over just .001 degrees)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably with a much more energetic transmitter, say a few thousand watts and a very tight beam and they might be able to get a good enough transmission for Morse code, but, I would think, no pictures, only a letter at a time, like the old telegraph communication.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we have the technology to do it today?   Well, to me, it looks pretty touch and go.   I think, maybe, if there was a human colony already there, with nuclear power plants and equipment to build a large transmitter/receiver then, if they can get the beam tight enough, we probably could exchange dots and dashes with today's technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we build a spacecraft that could communicate to us from 4.2 light years away - no, I don't believe we could do that with today's technology, not at that distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that said, communication is a far less difficult problem than actually getting there.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope my layman's answer isn't too far off.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2016-08-28T16:08:38.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18109" PostTypeId="2" ParentId="18107" CreationDate="2016-08-28T16:40:07.763" Score="6" Body="&lt;p&gt;&lt;a href=&quot;http://galileo.rice.edu/&quot; rel=&quot;nofollow&quot;&gt;The Galileo project&lt;/a&gt; has a pretty good and searchable account of his life. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://galileo.rice.edu/chron/galileo.html&quot; rel=&quot;nofollow&quot;&gt;Timeline&lt;/a&gt;  (scroll down to June 1633)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;With a formal threat of torture, Galileo is examined by the&#xA;  Inquisition. The next day he is sentenced to prison at the pleasure of&#xA;  the Inquisition and to religious penances. The sentence is signed by&#xA;  only seven of the ten cardinal-inquisitors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, there were ten cardinals.   Pope Urban was not one of the ten, though he was the driving force behind the trial.    &lt;a href=&quot;http://www.washingtonpost.com/wp-srv/national/horizon/sept98/galileo.htm&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ten cardinals sat in judgment of Galileo. Pope Urban VIII was not&#xA;  present in person, but he was there in spirit, for his personal&#xA;  feelings of anger and frustration were the driving force behind the&#xA;  extraordinary proceedings&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I couldn't, however, find any record of who the ten cardinals were, or which seven signed and which three didn't.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-08-29T00:00:47.453" LastActivityDate="2016-08-29T00:00:47.453" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18110" PostTypeId="1" CreationDate="2016-08-28T17:04:43.280" Score="3" ViewCount="49" Body="&lt;p&gt;So the Drake Equation, proposed by Frank Drake, is often cited as a means to estimate the number of intelligent civilizations that exist in the observable universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initial calculations by Drake came up with a number of about 10,000 civilizations. But refinements in the equation on how to interpret, and updated data, have recently led to this evaluation published in &lt;a href=&quot;http://www.astrobio.net/news-exclusive/at-last-how-many-alien-civilizations-are-there/&quot; rel=&quot;nofollow&quot;&gt;Astrobiology Magazine&lt;/a&gt;:    &quot;There is a 75% chance we could find ET between 1,361 and 3,979 light-years away&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what I'm more interested in has to do with our technology and efforts in detecting the first civilization in terms of estimating a mean time between observations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely given the result of the Drake equation, our current efforts, and possibly other factors to consider such as sampling theory there must be a way to calculate this. Has anyone pursued this? If so what are the results? hours?, days? years? centuries? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a calculation I believe would justify more concerted efforts and investment or otherwise abandonment of any hope. Or perhaps a different strategy to mitigate the assumptions on the calculation to reduce the mean time.&lt;/p&gt;&#xA;" OwnerUserId="14007" LastEditorUserId="7982" LastEditDate="2021-02-19T01:28:43.117" LastActivityDate="2021-02-19T01:28:43.117" Title="In SETI has anyone calculated an estimate of the mean time between observations?" Tags="&lt;observational-astronomy&gt;&lt;life&gt;&lt;estimate&gt;&lt;drake-equation&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18111" PostTypeId="2" ParentId="18107" CreationDate="2016-08-28T19:31:29.380" Score="9" Body="&lt;p&gt;The &lt;a href=&quot;https://books.google.co.uk/books?id=Z7xFT0aKT0YC&amp;amp;pg=PA142&amp;amp;lpg=PA142&amp;amp;source=bl&amp;amp;ots=1gJtQm9CpG&amp;amp;sig=3JwYc95Dt3NNh1AA9qzO7vn9sQE&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwidn9SY6uTOAhWD7xQKHUp7C8UQ6AEILDAC#v=onepage&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;seven who convicted&lt;/a&gt; were &#xA;Felice Centini, Desiderio Scaglia, Antonio Barberini, Berlinghiero Gessi, Fabrizio Verospi, Guido Bentivoglio and Marzio Ginetti. And as you note Francesco Barberini, Laudivio Zacchia and Gaspare Borgia did not. These three didn't vote to acquit, they were just absent from the signing of the condemnation - we don't know why&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All information from &lt;a href=&quot;https://books.google.co.uk/books?id=Z7xFT0aKT0YC&quot; rel=&quot;nofollow&quot;&gt;The Galileo Case: Trial, Science, Truth&lt;/a&gt; page 142&#xA;By Mario D'Addio. Gracewing Publishing, 2004 ISBN 9780852446652&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-08-29T08:26:09.580" LastActivityDate="2016-08-29T08:26:09.580" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18112" PostTypeId="1" AcceptedAnswerId="18116" CreationDate="2016-08-29T02:14:11.133" Score="10" ViewCount="2322" Body="&lt;p&gt;What is the exact measurement of a light year? I searched google for the answer in meters and came up with $9.461\cdot 10^{15}$ meters. When I calculated the answer considering $299\,792\,458 \;\text{m/s}$ as the speed of light, I came up with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;299\,792\,458 \times 365 \times 24 \times 3\,600 = 9\,454\,254\,955\,508\,926 \;\text{m}&#xA;$$ Why is there such a gap? Did I miss something to add in the equation or is it just wrong?&lt;/p&gt;&#xA;" OwnerUserId="7926" LastEditorUserId="-1" LastEditDate="2016-08-29T11:54:19.467" LastActivityDate="2016-09-02T09:02:34.937" Title="Exact measurement of a light year" Tags="&lt;light&gt;&lt;distances&gt;&lt;speed&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18113" PostTypeId="2" ParentId="18112" CreationDate="2016-08-29T02:37:59.423" Score="7" Body="&lt;p&gt;You're forgetting that one year is not 365 days, but 365.2422 or something close to that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That change will give you a number much much closer to the google provided number.&lt;/p&gt;&#xA;" OwnerUserId="2232" LastActivityDate="2016-08-29T02:37:59.423" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18116" PostTypeId="2" ParentId="18112" CreationDate="2016-08-29T08:38:39.753" Score="27" Body="&lt;p&gt;By convention, astronomy uses the &lt;strong&gt;Julian Year&lt;/strong&gt; for the computation of a &lt;a href=&quot;//en.wikipedia.org/wiki/Light-year&quot; rel=&quot;nofollow&quot;&gt;light year&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Although there are several different kinds of year, the IAU regards a year as a Julian year of 365.25 days (31.5576 million seconds) unless otherwise specified. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Light-year#Definitions&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; gives the length as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$31 557 600 s \times 299 792 458 m/s = 9 460 730 472 580 800 m$ (exactly)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason that it's exact (and not subject to experimental error) is that the metre itself is defined in terms of the speed of light, so the quantities are fixed by definition.&lt;/p&gt;&#xA;" OwnerUserId="14019" LastEditorUserId="14019" LastEditDate="2016-09-02T09:02:34.937" LastActivityDate="2016-09-02T09:02:34.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18117" PostTypeId="1" AcceptedAnswerId="18122" CreationDate="2016-08-29T10:45:36.940" Score="2" ViewCount="214" Body="&lt;p&gt;I just read in the EarthSky.org article &lt;a href=&quot;http://earthsky.org/space/asteroid-2016-qa2-august-27-28-2016&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hours after discovery, asteroid swept by&lt;/a&gt; that the near earth asteroid 2016 QA2 recently passed the earth at about 85,000 km. The article says that it is a member of the Aten group of asteroids. That article links to this very cool GIF &lt;a href=&quot;http://www.virtualtelescope.net/2016qa2_28aug2016.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.virtualtelescope.net/2016qa2_28aug2016.gif&lt;/a&gt;, however it's 40MB so it takes a while to load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Aten_asteroid&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia says&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Aten asteroids are defined by having a semi-major axis of less than one astronomical unit (AU), the distance from the Earth to the Sun. They also have an aphelion (furthest distance from the Sun) greater than 0.983 AU${}^2$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;${}^2$&lt;a href=&quot;http://neo.jpl.nasa.gov/neo/groups.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://neo.jpl.nasa.gov/neo/groups.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://neo.jpl.nasa.gov/neo/groups.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;That link&lt;/a&gt; shows the following table and drawings. The text says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;NEAs are divided into groups (Aten, Apollo, Amor) according to their perihelion distance (q), aphelion distance (Q) and their semi-major axes (a).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Are $q$ and $Q$ really perihelion and aphelion respectively? The table says Aten asteroids have $Q&amp;gt;0.983$ but in the drawing for Aten asteroids it says $Aphelion&amp;lt;1.0167$. I noticed these values could be reciprocals, so there may be something more going on here. But at least on the surface there seems to be some contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone confirm these definitions with an external source - and help me understand the basis for these limits that deviate from 1 AU by only  +/- 1.7%?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zkpJM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zkpJM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The image below is from this explanation: &lt;a href=&quot;http://www.permanent.com/apollo-amor-aten-near-earth-asteroids.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Near Earth Asteroid / NEO Classifications Based on Locations&lt;/a&gt;. It might be helpful for discussion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XhnY2.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XhnY2.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2016-08-29T11:16:11.187" LastActivityDate="2016-08-29T17:51:17.143" Title="What (actually) defines an Aten-class near-earth asteroid?" Tags="&lt;asteroids&gt;&lt;near-earth-object&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18118" PostTypeId="1" AcceptedAnswerId="18130" CreationDate="2016-08-29T11:45:48.243" Score="3" ViewCount="305" Body="&lt;p&gt;This question relates to the Crab nebula, the Crab Pulsar and the known concurrent supernovae event as recorded by Chinese astronomers in 1054.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering, I'm looking at the history of the discovery of the Crab pulsar...and I'm wondering what the reason is (or suspected to be) as to why European records show a discernible lack of observations?&lt;/p&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2016-08-30T08:57:02.977" Title="On the lack of observational records by European astronomers of the 1054 guest star" Tags="&lt;observational-astronomy&gt;&lt;supernova&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18119" PostTypeId="2" ParentId="18118" CreationDate="2016-08-29T12:31:18.487" Score="5" Body="&lt;p&gt;Of the three known medieval supernovae (in 1006, 1054, and 1181 AD), only SN 1006 was definitely recorded by Europeans, and even then, was only discussed in any detail in a single European source (the annals of the monastery of St. Gall in Switzerland). Although there are multiple accounts (some quite detailed) of the supernova of 1006 in Arabic/Islamic records, there is only one known Arabic record of SN 1054, and none for SN 1181. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since SN 1006 was by far the brightest of the three, the simplest explanation would be that SN 1054 was, like SN 1181, just not bright enough or long-lasting enough to merit sufficent widespread commentary in Europe. (Remember that China in the 11th Century was a much more advanced society than Europe of the same time, with a dedicated tradition of government-supported astronomical/astrological observations.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2016-08-29T12:31:18.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18120" PostTypeId="1" AcceptedAnswerId="18121" CreationDate="2016-08-29T15:05:49.993" Score="26" ViewCount="4807" Body="&lt;p&gt;How do we know laws of physics are the same throughout the universe? Intuitively I would say they would vary in two natural ways: the constants in the equations may vary or the math in the equations may vary. As a guess they could change over a long time. What is the furthest radius we can prove from earth, with absolute certainty, that the laws of physics do not vary? I am aware this may not be a radius but a more complex shape that cannot be simply described by a radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nearest answer I can think of for a radius is a guess. And that guess is based on the furthest physics experiment we have done from earth. Which I think is an experiment with mirrors on the moon. Therefore if we assume (I don't know if this assumption is totally 100% reasonable) all physics laws hold because this experiment works. Then the radius is to the moon. This doesn't give an concrete answer for the radius, merely an educated guess.&lt;/p&gt;&#xA;" OwnerUserId="14021" LastEditorUserId="7472" LastEditDate="2017-07-15T22:25:05.087" LastActivityDate="2018-02-18T02:54:25.073" Title="The laws of physics through out the universe" Tags="&lt;cosmology&gt;" AnswerCount="7" CommentCount="7" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18121" PostTypeId="2" ParentId="18120" CreationDate="2016-08-29T16:02:08.960" Score="23" Body="&lt;p&gt;Nothing can be proved &quot;with absolute certainty&quot;; that is not how science works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We adopt a working hypothesis that the constants of nature are exactly that; both constant in time and space. Then we conduct experiments that attempt to falsify that hypothesis or at least place limits on by how much things might vary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reasons that are explained in answers to &lt;a href=&quot;https://physics.stackexchange.com/questions/21721/what-is-the-proof-that-the-universal-constants-g-hbar-ldots-are-real&quot;&gt;this Physics SE question&lt;/a&gt; (see also &lt;a href=&quot;https://physics.stackexchange.com/questions/78684/is-it-possible-to-speak-about-changes-in-a-physical-constant-which-is-not-dimens&quot;&gt;this question&lt;/a&gt;), only the dimensionless parameters like the fine structure constant can be assessed for their variation - other constants like $G$, $c$ and $h$ are tied up in our system of (measuring) units so we are unable to say whether they are changing or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the example of the fine structure constant, observations of absorption lines towards distant quasars put strong limits on by how much this can have varied in space and time (the two are inseparable, since it takes finite time for information to travel to us). So you can find lots of different attempts to do this in the literature - I dug out a few. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2015MNRAS.452.4153A&quot; rel=&quot;nofollow noreferrer&quot;&gt;Albareti et al. (2015)&lt;/a&gt; say the variation is less than a couple of parts in a 100,000 out to a redshift of 1 (a lookback time of about 8 billion years or so. Similar constraints exist for experiments carried out in different parts of the solar system. On the other hand, some authors do claim variations of a few parts per million on similar lookback times or in different directions (&lt;a href=&quot;http://arxiv.org/abs/astro-ph/0612407&quot; rel=&quot;nofollow noreferrer&quot;&gt;Murphy et al 2008&lt;/a&gt;; &lt;a href=&quot;http://arxiv.org/abs/1202.4758&quot; rel=&quot;nofollow noreferrer&quot;&gt;King et al. 2012&lt;/a&gt;), but these claims are disputed by many, if not most workers in the field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a massive review of this topic by &lt;a href=&quot;http://arxiv.org/abs/1009.5514&quot; rel=&quot;nofollow noreferrer&quot;&gt;Uzan (2011)&lt;/a&gt;, which you could read - this really is a broad question. My summary would be - at the moment there is no convincing evidence for any variation in space and time.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-08-29T16:02:08.960" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18122" PostTypeId="2" ParentId="18117" CreationDate="2016-08-29T16:04:40.500" Score="2" Body="&lt;p&gt;An Earth-crossing asteroid has either&#xA;an aphelion larger than Earth's perihelion ($Q \gt q_\oplus$, 0.983 AU),&#xA;or a perihelion smaller than Earth's aphelion ($q &amp;lt; Q_\oplus$, 1.017 AU).&#xA;For those which have both, the semimajor axis $a$ determines whether it is an Aten or an Apollo - subject to change by a close encounter with Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This Minor Planet Center&#xA;&lt;a href=&quot;http://minorplanetcenter.net/blog/asteroid-classification-i-dynamics/&quot; rel=&quot;nofollow&quot;&gt;blog post&lt;/a&gt;&#xA;is consistent with the JPL table.&#xA;I think the diagram used the wrong inequality for Aten aphelia.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2016-08-29T17:51:17.143" LastActivityDate="2016-08-29T17:51:17.143" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18123" PostTypeId="2" ParentId="18120" CreationDate="2016-08-29T16:15:39.347" Score="38" Body="&lt;p&gt;Let's start in the middle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the furthest radius we can prove from earth, with absolute certaintity, that the laws of physics do not vary?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Zero. Proofs are found in mathematics and court rooms, and are impossible in natural science. The best we can do is have &lt;a href=&quot;https://en.wikipedia.org/wiki/Falsifiability&quot;&gt;falsifiable&lt;/a&gt; theories. This holds for every description of reality - there's no &quot;proof&quot; even for the Laws of Gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what could we observe that would tell us that physical constants or relationships between physical quantities are different in other parts of the universe, or at other times during its existence?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gravity: For galaxy clusters, we have independent mass measurements from several different sources that agree within their (admittedly large) error bars. Gravitational lensing, velocity dispersion of the member galaxies and X-ray temperatures are all in agreement. So the laws of gravity seem to work even at redshifts up to 0.5 or even higher.&lt;/li&gt;&#xA;&lt;li&gt;Atomic physics: We observe highly redshifted objects. The wavelength of the light emitted by these objects is made longer by the expansion of the universe. Observing redshifted spectral lines of different chemical elements (or molecules) tells us that atomic physics worked the same when and where this light was emitted. If the transition levels between electron orbits had changed over time, we would get different redshifts for the same objects depending on what element's spectral line we observe.&lt;/li&gt;&#xA;&lt;li&gt;Nucleosythesis: Shortly after the big bang, the temperature lowered such that protons and neutrons were no longer created and destroyed constantly. A free neutron has a half live of about 8.5 minutes before it decays into a proton and an electron. Our theories predict that we'd get a helium (2x proton, 2x neutron) content in the universe of about 25%. (The rest of the &quot;normal&quot; matter being essentially all hydrogen), and that is indeed what we observe. Now, the helium content is dependent both on the matter density at the time this took place and the half live of the neutron. From other observations (BAO come to mind) we are fairly certain that we got the matter density about right. Which leaves only a small wiggle room for the half live of the neutron, and hence for changes in the weak force.&lt;/li&gt;&#xA;&lt;li&gt;We've covered gravity, electromagnetism, and the weak force. I don't know any good test for the strong force.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For a change of natural laws over time, we can look at the isotope distribution in rocks here on earth. We should be able to tell whether the decay rate of various elements was different at earlier times by looking at how many of each of their decay products are around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, we cannot say with &quot;absolute certainty&quot;, but what we observe seems to indicate that natural laws are the same throughout the universe.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastEditorUserId="14032" LastEditDate="2016-08-29T22:08:05.263" LastActivityDate="2016-08-29T22:08:05.263" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="18124" PostTypeId="1" CreationDate="2016-08-29T22:50:57.633" Score="5" ViewCount="105" Body="&lt;p&gt;There seem to be irregularities in the motion of the earth, so we add &quot;leap seconds&quot; to adjust our time accordingly. Since 1972 there have been 26 leap seconds I think. If we extrapolate that backwards 0.5 leap seconds per year, it would be 1 leap minute in 120 years and 5 leap minutes in 600 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this be enough to throw off a prediction of an eclipse? In other words, could someone in 1416 predict an eclipse right now given the unpredictable errors caused by leap seconds?&lt;/p&gt;&#xA;" OwnerUserId="2800" LastActivityDate="2017-12-30T01:11:22.307" Title="Affect of leap seconds on the prediction of an eclipse" Tags="&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18125" PostTypeId="2" ParentId="13698" CreationDate="2016-08-30T00:35:54.360" Score="1" Body="&lt;p&gt;It is the same reason for why we see the Moon. If the moon exploded, we would see the entire thing happen. Objects in space don't 'instantly' disappear and even when far away, light always gets sent in the same order and we see what happened.&lt;/p&gt;&#xA;" OwnerUserId="14035" LastActivityDate="2016-08-30T00:35:54.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18128" PostTypeId="1" AcceptedAnswerId="18138" CreationDate="2016-08-30T08:18:56.727" Score="1" ViewCount="71" Body="&lt;p&gt;I don't know if this is the right place for asking this question, my apologies in advance. I would like to be updated on the latests news on space and I am looking for a website devoted to space news, or a blog. I was only able to find minor blogs which were not updated often. Could you suggest me a high quality blog/space related website that you know? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="13519" LastActivityDate="2016-08-30T14:51:09.417" Title="High quality Space-related blog/website?" Tags="&lt;space&gt;" AnswerCount="2" CommentCount="7" ClosedDate="2016-08-30T20:20:26.690" ContentLicense="CC BY-SA 3.0" />
  <row Id="18129" PostTypeId="2" ParentId="18128" CreationDate="2016-08-30T08:36:04.883" Score="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/&quot; rel=&quot;nofollow&quot;&gt;Emily Lakdawalla's blog at the Planetary Society&lt;/a&gt; is good for hard science news about our solar system, especially planetary exploration, of course. There are occasional guest authors as well.&lt;/p&gt;&#xA;" OwnerDisplayName="user1569" LastActivityDate="2016-08-30T08:36:04.883" CommentCount="0" CommunityOwnedDate="2016-08-30T13:58:21.943" ContentLicense="CC BY-SA 3.0" />
  <row Id="18130" PostTypeId="2" ParentId="18118" CreationDate="2016-08-30T08:48:56.060" Score="3" Body="&lt;p&gt;Apart from observational parameters (&lt;a href=&quot;https://astronomy.stackexchange.com/a/18119/1569&quot;&gt;Peter's answer&lt;/a&gt;), astronomy was not that 'hot' in that period. One main reason is indeed that the Christian church did not promote it.&lt;br /&gt;&#xA;Quoting from the section &lt;a href=&quot;https://en.wikipedia.org/wiki/History_of_astronomy#Medieval_Western_Europe&quot; rel=&quot;nofollow noreferrer&quot;&gt;Medieval Western Europe&lt;/a&gt; on the Wikipedia &lt;a href=&quot;https://en.wikipedia.org/wiki/History_of_astronomy&quot; rel=&quot;nofollow noreferrer&quot;&gt;History of astronomy&lt;/a&gt; page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;After the significant contributions of Greek scholars to the development of astronomy, it entered a relatively static era in Western Europe from the Roman era through the 12th century&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www2.astronomicalheritage.net/index.php/show-theme?idtheme=16&quot; rel=&quot;nofollow noreferrer&quot;&gt;an article on the Portal to the heritage of astronomy&lt;/a&gt; it wasn't until the 12th century that astronomy got a more modern scientific twist:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Chronologically, the Middle Ages are conventionally divided into an Early Medieval and a Later Medieval period. A chief marker of this division is the re-emergence of urban society in the 12th century, which was accompanied by several changes that transformed medieval astronomy. The first was the movement of astronomical study from monasteries and cathedrals to the emerging universities. Accompanying the rise of the universities was the change of the content of astronomical study, since both astronomy and geometry took on a renewed quantitative aspect founded on the study of ancient texts. The rise of urban life saw the development of learned and skilled professions, including architects who applied this astronomical knowledge in their work.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The introduction at the &lt;a href=&quot;https://sites.lib.byu.edu/worldhistory/science/collection-highlights/renaissance-astronomy/medieval-astronomy/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tom L. Perry Medieval Astronomy collection&lt;/a&gt; dates the revival of astronomy just around/after the 11th century:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Astronomy was revived in Europe during the eleventh century with the arrival of the astrolabe, a device for measuring the position of heavenly bodies which was introduced to Europe from the Islamic world. Planetary observations were useful for astrological predictions and for medical practice, since the planets and the zodiac were thought to correspond to the organs of the human body. Over the next century, the writings of ancient Greek astronomers like Ptolemy were rediscovered by the West, further enriching Europe’s knowledge of the universe. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sir Robert Wilsons's book &lt;em&gt;Astronomy through the ages&lt;/em&gt;, when discussing Islamic astronomy in the period 8th–15th century, has &lt;a href=&quot;https://books.google.nl/books?id=AoiJ3hA8bQ8C&amp;amp;pg=PA33&amp;amp;lpg=PA33&amp;amp;dq=european+astronomy+throughout+the+ages&amp;amp;source=bl&amp;amp;ots=EyZd_4EDiI&amp;amp;sig=TOvWrGJD0l9Pc9w9YiijHFq1tsU&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwj1uYOm3ujOAhVLCsAKHdeoA_wQ6AEIPjAF#v=onepage&amp;amp;q=european%20astronomy%20throughout%20the%20ages&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; to say about the influence of the church in western astronomy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;During the great period of Islamic astronomy, scientific endeavors and inquiry were virtually absent in Christian Europe, largely because of the lack of encouragement (some would say active discouragement) for such endeavors by the Christian church, which believed that thruth and spiritual guidance could come only from holy scripture and that natural knowledge and understanding was best revealed by ancient writings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user1569" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-08-30T08:57:02.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18131" PostTypeId="2" ParentId="18120" CreationDate="2016-08-30T09:50:03.800" Score="3" Body="&lt;p&gt;We can't know for certain. However, we can state with confidence what would be broken were it not to be true, provided a certain mathematical formulation is valid. This is &lt;em&gt;Noether's theorem&lt;/em&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Noether%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Noether%27s_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TL;DR what breaks is the conservation of linear momentum. If you consider that the laws of physics may vary with time rather than place, what breaks is the conservation of energy. Both subject to the constraint that a Lagrangian formulation is valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've encountered serious physicists discussing the possibility that that time-invariance might not hold for the early stages of the universe. The consequence would be non-conservation of energy on the largest cosmological scales, which is where the evidence for this conservation law is least strong. (We have to posit the existence of dark matter and dark energy, and also not all the universe is observable).&lt;/p&gt;&#xA;" OwnerUserId="14041" LastActivityDate="2016-08-30T09:50:03.800" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18132" PostTypeId="2" ParentId="18079" CreationDate="2016-08-30T10:14:11.290" Score="3" Body="&lt;p&gt;I do some narrow band imaging and believe the most likely cause is you have stretched the SII channel to far. Most likely it is the weakest of the three. Also, I never use a luminance channel on narrow band. But if you must, I would sum the three channels and not just use a single channel. &lt;/p&gt;&#xA;" OwnerUserId="14042" LastActivityDate="2016-08-30T10:14:11.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18133" PostTypeId="1" AcceptedAnswerId="18846" CreationDate="2016-08-30T13:02:33.183" Score="1" ViewCount="338" Body="&lt;p&gt;Would anyone be able to reference the galaxy in the background of this image of the Ring Nebula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ICFHt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ICFHt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's obviously not a Messier object, I'm more confident in it being in the New General Catalogue.&lt;/p&gt;&#xA;" OwnerUserId="2091" LastActivityDate="2016-10-24T21:14:09.007" Title="Galaxy behind the Ring Nebula" Tags="&lt;galaxy&gt;&lt;observational-astronomy&gt;&lt;messier-catalog&gt;&lt;ngc&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18134" PostTypeId="2" ParentId="18120" CreationDate="2016-08-30T13:04:18.357" Score="3" Body="&lt;p&gt;One problem with your question is that it is a bit of a paradox.  If a Law of Physics appears to vary depending on time/place being observed then, because of what it means to be a physical law, we've simply misunderstood the law itself or aren't observing all forces at work.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a super simple example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zA6Jf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zA6Jf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These people haven't found a place in the universe where gravity acts differently, they're simply being pushed harder by a fan than the gravity is pulling down on them.  Of course, if the only information you had about them was this picture you wouldn't know that and might think that gravity acts differently where they are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If scientists observe variances is how a law behaves and simply waves their hand saying &quot;oh the law works differently there&quot; then that isn't science anymore.  We'd want to know why the law appears to work differently in one place vs another.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One example that is maybe more to OP's point is dark energy.  We observe that the Universe is expanding at an increasing rate even though our Laws of Physics, specifically gravity predict that its expansion would cause it to decelerate.  Rather than shrugging their shoulders and saying &quot;well the Laws of Physics just work differently at the edge of the Universe&quot; scientists theorized something called dark matter to explain why the Universe's expansion is accelerating despite gravity.&lt;/p&gt;&#xA;" OwnerUserId="14044" LastEditorUserId="14044" LastEditDate="2016-08-30T22:12:39.647" LastActivityDate="2016-08-30T22:12:39.647" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18135" PostTypeId="1" CreationDate="2016-08-30T13:07:57.093" Score="2" ViewCount="61" Body="&lt;p&gt;I'm designing an amateur astronomy observational plan for GCSE Astronomy students here in the UK. I would like some advise to the suitability of what I have selected as the content.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For the beginner level I have decided to see if they can image the&#xA;various phases of the Moon across a 28 day period.&lt;/li&gt;&#xA;&lt;li&gt;For the intermediate level I would say they could image the Pleiades.&lt;/li&gt;&#xA;&lt;li&gt;For the advanced level I have decided they can create a mosaic of M31&#xA;from various narrow field-of-view images. (I would suggest imaging&#xA;the LMC or SMC but we are based here in the northern hemisphere).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Would this be suitable targets over the course of a year?&lt;/p&gt;&#xA;" OwnerUserId="2091" LastEditorUserId="2091" LastEditDate="2016-08-30T14:23:50.503" LastActivityDate="2016-09-01T09:33:15.720" Title="Amateur astronomy plan for GCSE students" Tags="&lt;amateur-observing&gt;&lt;fundamental-astronomy&gt;&lt;photography&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18137" PostTypeId="1" AcceptedAnswerId="18141" CreationDate="2016-08-30T14:10:44.877" Score="4" ViewCount="143" Body="&lt;p&gt;In thinking about exclusion options for where its not worth to look for habitable planets, the past came to my mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right at the beginning of the universe, there was no possibility for (carbon based) life supporting planets to form, simply because there was not enough material to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Population III stars burnt out within millions of years to leave some of these elements, but was it enough to form planets? I have some problem getting things into perspective, considering that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;there are so many processes that produce different kinds and amounts of isotopes&lt;/li&gt;&#xA;&lt;li&gt;these isotopes are spread out in various different ways (and thus also different concentrations from how they were created) throughout the interstellar medium&lt;/li&gt;&#xA;&lt;li&gt;it takes probably a lot of time to gravitationally condense enough metals into a planet once it was spewed out from e.g. a super nova&lt;/li&gt;&#xA;&lt;li&gt;the majority of stars are smaller so it would probably take a clustering of heavier stars to provide the conditions earlier in the universe.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would consider a similar distribution of metals like in earth as sufficient for life, and would concentrate only on those found in DNA (N,H,C,P,O), fats (C,H,O) and proteins (C,O,H,N) as well as most of the essential minerals (Ca,Cl,K,Na,Mg,P,S) and maybe at least a bunch of (possibly essential) trace elements (Co,Fe,I,Cu,Mn,Mo,Se,Zn,As,B,Cr,F,Rb,Te,V,Sn,Ni).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the best approximation that we can currently give on when these elements were abundant enough to condense into earth like rocks with earth like elemental distribution?&lt;/p&gt;&#xA;" OwnerUserId="9853" LastActivityDate="2016-08-30T16:13:12.370" Title="How early could life supporting planets been formed?" Tags="&lt;life&gt;&lt;elemental-abundances&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18138" PostTypeId="2" ParentId="18128" CreationDate="2016-08-30T14:51:09.417" Score="1" Body="&lt;p&gt;Your question for &quot;space&quot; news sites is somewhat vague as space can cover a wide variety of topics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found &lt;a href=&quot;http://innerspace.net/&quot; rel=&quot;nofollow&quot;&gt;Innerspace.net&lt;/a&gt;  to be a nice blog/news source primarily concerning itself with the space exploration and technology aspects of space. It often reports on recent launches (such as those by Space X), advancements in space technologies, and space ventures by other countries outside the US such as China and Russia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're looking for more sciencey topics of a higher caliber, try something like &lt;a href=&quot;https://www.sciencedaily.com/news/space_time/&quot; rel=&quot;nofollow&quot;&gt;Science Daily&lt;/a&gt;. While this covers a variety of scientific fields, it does also cover space news. In general, this site is a curation of articles by various sources. Usually each article is linked to some specific scientific research or published paper.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-08-30T14:51:09.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18139" PostTypeId="1" AcceptedAnswerId="18145" CreationDate="2016-08-30T15:55:11.143" Score="5" ViewCount="181" Body="&lt;p&gt;There's been a story in the news recently of a &lt;a href=&quot;http://www.centauri-dreams.org/?p=36248&quot;&gt;potential SETI candidate signal received from the vicinity of star HD164595&lt;/a&gt;.  What's the nature of this signal?  Is it a single pulse of energy, or does it change over time?  Can someone point me to a link or paper than shows it in more detail?&lt;/p&gt;&#xA;" OwnerUserId="4146" LastActivityDate="2016-10-25T23:55:51.787" Title="What was the nature of the signal received from the direction of HD164595?" Tags="&lt;seti&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18141" PostTypeId="2" ParentId="18137" CreationDate="2016-08-30T16:13:12.370" Score="4" Body="&lt;p&gt;You can get relatively high metallicity rather quickly in parts of the early universe -- especially some globular clusters and the centers of massive galaxies -- because star formation rates in those places were very high, which means lots of massive stars and several rounds of &quot;massive stars form, go supernova and seed surrounding gas with metals, new massive stars form out of enriched gas&quot; -- the timescale from formation to supernova can be as short as 2 million years for really massive stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how high a metallicity do you need to form planets, and how early did (parts of) the universe get there?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;One the one hand, we know of a star with a metallicity (iron&#xA;abundance) only a quarter of the Sun's ([Fe/H] = -0.63) with a&#xA;Neptune-mass planet, and Kepler has found roughly Earth-sized planets&#xA;&lt;a href=&quot;http://www.nature.com/nature/journal/v486/n7403/full/nature11121.html&quot; rel=&quot;nofollow&quot;&gt;around stars almost as metal-poor&lt;/a&gt;.&#xA;So you don't need very high metallicities to form planets.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On the other hand, there is evidence of gas around quasars enriched&#xA;to &lt;em&gt;above&lt;/em&gt; solar metallicity by redshifts of 6 to 7 (about 800&#xA;million years after the Big Bang). There are also rather old stars in&#xA;our galaxy with high metallicities: for example, &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014A%26A...565A...8C&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; estimates that&#xA;the stars in the globular cluster NGC 6258 are 10-12 billion years&#xA;old with near-solar metallicity.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the short answer is that you could have planets forming within the first 1-2 billion years after the Big Bang, and quite possibly within a few hundred million years after the first stars formed. Because the metals would mostly be from core-collapse supernova (as opposed to the iron-dominated products of Type Ia supernovae), there would be an excess of so-called &quot;alpha elements&quot; -- particularly oxygen, magnesium, neon, silicon, sulfer, argon, and calcium -- relative to iron; I don't think you'd have to worry about missing too many life-critical elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These planets would not all be very good places for &lt;em&gt;life&lt;/em&gt;, however, because they would be in dense regions with lots of stars and lots of star formation, meaning lots of supernovae going off nearby, and possibly an active quasar as well if we're talking about a galaxy center.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2016-08-30T16:13:12.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18142" PostTypeId="2" ParentId="16753" CreationDate="2016-08-30T16:31:19.357" Score="9" Body="&lt;p&gt;It's a track about 50 pixels long, at &lt;a href=&quot;https://photographylife.com/reviews/canon-7d-mark-ii&quot; rel=&quot;noreferrer&quot;&gt;4.1 $\mu m$ per pixel&lt;/a&gt; that's 200 $\mu m$ in the plane of the sensor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's about 1 pixel wide, quite lumpy, and seems to change direction a bit. It really fits the description of an ionizing particle. There are multiple possible sources, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure @JamesK has nailed it as an electron, but what else can we learn? There must be some way to get a &lt;em&gt;very rough&lt;/em&gt; idea of how many $e^- h^+$ pairs correspond to a fully exposed pixel at &quot;@ISO1600&quot;, then we could convert it to a dE/dx and see if it's minimum ionizing or if it's much higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going diagonally, with a density of 2.3 g/cm${}^3$ that's about 1.3 mg/cm${}^2$ areal density per pixel. A [minimum ionizing particle] at 1.5 MeV / g cm${}^2$ would deposit only about 2000 eV which would make less than 1000 $e^- h^+$ pairs. That might correspond to a bright pixel. It's not out of the question that it's a minimum ionizing particle.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2016-08-30T16:31:19.357" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18144" PostTypeId="1" AcceptedAnswerId="18148" CreationDate="2016-08-30T17:37:46.250" Score="2" ViewCount="172" Body="&lt;p&gt;I was reading this article &lt;a href=&quot;http://www.space.com/11380-big-bang-evidence-universe-trillion-years.html&quot; rel=&quot;nofollow&quot;&gt;finding evidence of Universe's origin&lt;/a&gt;, which describes that in 1 trillion years we may lose the ability to determine how the universe was created.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer seems to be using hypervelocity stars.  These are stars that have been ejected from their galaxy.  Using the expansion of the universe, we can use the velocity of the star to then determine evidence of the Big Bang. But how does this work?  Is there calculations we can do now that agree with our understanding of the universe and its age when working with hypervelocity stars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand finding a hypervelocity star is very hard to do. I guess my thought here is that astronomers have modeled this scenario and found that if we were to find a hypervelocity star, we could then use the scenario above to calculate the age.  I'm assuming the age calculated agrees with current estimates.&lt;/p&gt;&#xA;" OwnerUserId="13849" LastEditorUserId="9301" LastEditDate="2016-08-31T00:51:30.017" LastActivityDate="2016-09-15T01:30:26.460" Title="How can we use hypervelocity stars to determine the origins of the Universe?" Tags="&lt;universe&gt;&lt;cosmology&gt;&lt;big-bang-theory&gt;&lt;future&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18145" PostTypeId="2" ParentId="18139" CreationDate="2016-08-30T17:47:08.093" Score="6" Body="&lt;p&gt;The page you link to contains what is, to my knowledge, the most descriptive graph of the signal from the RATAN-600 data that is open to the public. Bursov et al. have not yet presented their results, which will be explained at the International Academy of Astronautics SETI Permanent Committee meeting late this September. Chances seem to be good that we won't see much more information until then.&lt;/p&gt;&#xA;&lt;p&gt;Here is the graph at a frequency of 11.1 GHz, as will be presented by Bursov et al. and as used in &lt;a href=&quot;https://seti.berkeley.edu/HD164595.pdf&quot; rel=&quot;noreferrer&quot;&gt;Croft et al. (2016)&lt;/a&gt;, who very quickly put together a paper with an analysis of some basic follow-up observations using the Green Bank Telescope:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lmYrE.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lmYrE.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The main blip lasted only for two to three seconds - two seconds seems to be the popularly cited figure. Since then, no other group looking at HD 164595 has detected a repeat of the signal, and Croft et al. were unable to find anything similar in records in the NVSS catalog. Continuum imaging revealed nothing, and the authors concluded that the signal may have been due to observational error - although a one-time event is not out of the question.&lt;/p&gt;&#xA;&lt;p&gt;Again, though, there is no more information forthcoming from the RATAN-600 group, and I expect this will be the case until the results are formally presented in full this September. Until then, this is all the data we have, but it appears to be the most important bit of data.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Update, 14 September 2016&lt;/h2&gt;&#xA;&lt;p&gt;The Russian Academy of Sciences recently released &lt;a href=&quot;https://www.sao.ru/Doc-en/SciNews/2016/Sotnikova/&quot; rel=&quot;noreferrer&quot;&gt;a statement&lt;/a&gt; saying&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Subsequent processing and analysis of the signal revealed its most probable terrestrial origin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I'm not aware if the &amp;quot;processing and analysis&amp;quot; was done by the RATAN team or an outside group, but it seems more and more like the signal is not in fact from an extraterrestrial civilization. We'll probably find out more details two weeks from now.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Update, 25 October 2016&lt;/h2&gt;&#xA;&lt;p&gt;I have yet to see any published results from the RATAN group after the IAA meeting about a month ago, and I haven't seen any articles on the findings. The IAA has &lt;a href=&quot;http://avsport.org/IAA/prog16.htm&quot; rel=&quot;noreferrer&quot;&gt;some information on the meeting&lt;/a&gt;, but it has yet to add details to its &lt;a href=&quot;http://avsport.org/IAA/minutes.htm&quot; rel=&quot;noreferrer&quot;&gt;Minutes&lt;/a&gt; page. Perhaps that will be released at a later date - and I assume it will be - but that could be a while. In the meantime, I'm taking the lack of information to mean that the RATAN group agrees with the analysis by Croft et al., and that the signal was a false positive.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-10-25T23:55:51.787" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18146" PostTypeId="1" AcceptedAnswerId="18159" CreationDate="2016-08-30T18:22:23.380" Score="1" ViewCount="1620" Body="&lt;p&gt;In watching this &lt;a href=&quot;https://www.youtube.com/watch?v=a2ta2poeJFI&quot; rel=&quot;nofollow&quot;&gt;Hubble video on Omega Centari&lt;/a&gt; it got me thinking about how dense a galaxy can get. Do we know how dense a galaxy can get?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any good resources on reading more about very dense galaxies? &lt;/p&gt;&#xA;" OwnerUserId="6870" LastActivityDate="2016-08-31T14:39:01.073" Title="What's the theoretical maximum density of a galaxy?" Tags="&lt;galaxy-center&gt;&lt;density&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18147" PostTypeId="2" ParentId="18124" CreationDate="2016-08-30T18:26:29.097" Score="2" Body="&lt;p&gt;It has been known that changes in earth rotation affect the location of eclipse paths for over a century.  The most active research in this has been Morrison and Stephenson.  See, for example &lt;a href=&quot;http://adsabs.harvard.edu/abs/2004JHA....35..327M&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/2004JHA....35..327M&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14050" LastActivityDate="2016-08-30T18:26:29.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18148" PostTypeId="2" ParentId="18144" CreationDate="2016-08-30T19:13:39.637" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How can we use hypervelocity stars to determine the origins of the Universe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First and foremost, I should clear something up: our knowledge of the Big Bang is incredibly limited, and so we don't even know the origins of the Universe today. General relativity and quantum mechanics both break down as you get closer and closer to $t_0=0$. All we really know is that &lt;em&gt;something&lt;/em&gt; happened 13.8 billion years ago. However, if we look at the expanding universe, we can extrapolate that, 13.8 billion years ago, space was much less dilated than it is now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General relativity suggests that the Universe would be packed into a singularity. This is not a good approach, since GR can't properly explain such conditions (QM is more fit to do so, but that alone can't explain how gravity acted), so we really haven't the slightest clue as to what happened 13.8 billion years ago. That said, we still know when the Big Bang (whatever it may have been) occurred, and that's what future astronomers may be able to find out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea the article brings up is this: by one trillion years from now, the expansion of space will have driven away any other galaxies from our line of view, and CMB radiation will have become too redshifted to be detectable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, hypervelocity stars will likely be detected outside of the Milky Way (or rather, Milkomeda). Once these stars get far enough from the Milkomeda's gravitational influence, we can observe their acceleration due to the expansion of space, just as we do for many galaxies today. The 4-page paper the article is based on, &lt;a href=&quot;https://arxiv.org/abs/1102.0007&quot; rel=&quot;nofollow noreferrer&quot;&gt;Loeb (2011)&lt;/a&gt;, actually makes this clear in a graph:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/C5lUS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/C5lUS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the stars get farther and farther from Milkomeda, the gravity's pull on the stars will weaken and they will be accelerated by the expansion of space. This will really become noticeable once they are about 2.3 megaparsecs from the galaxy. Of course, these stars will not be easily found, but Loeb mentions that future astronomers will likely have advanced their tools and instruments by that time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Loeb explains that, once we observe their redshifts as their distances increase, we can confirm Hubble's law. After this, we just need to measure the necessary parameters, and finally we can calculate the age of the Universe using the Friedmann equation:&#xA;$$t_0 = \frac{1}{H_0} F(\Omega_r,\Omega_m,\Omega_\Lambda,\dots)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the gist of it. The hypervelocity stars will be the only extragalactic objects detectable by then, and thus we will rely on them to confirm our measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, Loeb mentions that observing and analyzing the expansion of the Universe at that time would answer some of the uncertainties we have today:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The working assumption so far was that the vacuum energy density remains constant and does not decay to a lower energy state over hundreds of e-folding times. Measuring the dynamics of hypervelocity stars at late cosmic&#xA;  times provides the added benefit of testing whether the cosmological constant is truely [&lt;em&gt;sic&lt;/em&gt;] constant. Indications to the contrary might reveal new physics, such as that considered by evolving dark energy models.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, even though much of the Universe will not be observable by that time, we will ironically have more answers about cosmology than we do today, simply by observing the few extragalactic objects that &lt;em&gt;will&lt;/em&gt; be visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the general explanation of how we can find the time since the Big Bang. Remember that even now, we really have no idea about the origins of the Universe — it's a major unsolved problem in physics. All we know is something happened 13.8 billion years ago. Using hypervelocity stars, future humans would be able to determine the time since the Big Bang as well. Don't get your hopes up on humans surviving for 1 trillion years, though.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-09-15T01:30:26.460" LastActivityDate="2016-09-15T01:30:26.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18150" PostTypeId="1" CreationDate="2016-08-31T04:49:12.613" Score="19" ViewCount="2047" Body="&lt;p&gt;It takes an infinite amount of time for something to fall past the event horizon of a black hole from the perspective of someone outside the event horizon. Black holes also evaporate after a finite amount of time from an outsider's perspective  due to Hawking radiation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that you would never actually reach the event horizon if you fell into a black hole because the black hole would evaporate?&lt;/p&gt;&#xA;" OwnerUserId="14054" LastActivityDate="2020-05-27T15:11:05.460" Title="Would a black hole evaporate via Hawking radiation before you fall past the event horizon?" Tags="&lt;black-hole&gt;&lt;hawking-radiation&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18151" PostTypeId="2" ParentId="18150" CreationDate="2016-08-31T05:43:04.827" Score="6" Body="&lt;p&gt;I've asked this question to a couple of physicists a few days ago. Great minds think alike, huh?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, bear in mind that Hawking radiation is only hypothetical. It is &lt;em&gt;not&lt;/em&gt; theory. If we trust that hypothesis, this is what we can get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general relativity, black holes can be described through a number of approximations. For example, the Schwarzschild solution for a black hole describes it as an eternal object — not something that exists for some times, and doesn't exist for others. According to this solution, the event horizon must have always existed, and must remain existing eternally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Schwarzschild black holes approximate black holes very accurately, but as you can tell, they fail to explain how a black hole can form, and (assuming Hawking radiation is real) they don't explain how one could eventually evaporate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, that solution won't help us. I've kept looking for one that accurately describes an evaporating, creatable black hole, but I've found nothing. The conclusion I've come to, along with those I've asked, is that our question has a major problem: Hawking radiation is explained via quantum field theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, you can't simply use a GR solution for a black hole; you'd need some unholy mixture of quantum field theory and general relativity (keep in mind that both GR and QFT are incompatible in many situations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, it all comes down to how little we really know about black holes. It's not really possible to determine which solution is the best, and our inability to reconcile QFT with GR poses a big problem. The best answer I could give is &quot;nobody really knows what would happen if you kept approaching a black hole&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't know if we would reach the event horizon, we don't know if the black hole will evaporate. We simply don't understand them well enough to know what solution would work, or how we would put QFT into it. If we somehow managed to find an approximation that properly combines GR and QFT, I assume (but don't quote me on this) the situation you described would be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is possible, by the way, then we could confidently say that a black hole of any size could rip you apart through tidal forces. Tidal forces become weaker as the black hole's size increases, so one would assume a large enough black hole wouldn't rip you apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we take Hawking radiation into account, and if your proposed scenario were indeed correct, the black hole would shrink as it evaporates. Since it would get smaller at a faster rate as we approach the event horizon, it would soon be small enough to rip us apart.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-08-31T16:52:43.437" LastActivityDate="2016-08-31T16:52:43.437" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="18152" PostTypeId="2" ParentId="4793" CreationDate="2016-08-31T06:31:26.240" Score="3" Body="&lt;p&gt;From a purely mathematical point of view; &lt;a href=&quot;https://math.stackexchange.com/questions/36289/is-infinity-a-number&quot;&gt;infinity is not a number&lt;/a&gt;, use it as one at your peril.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Density is mass per volume, &lt;code&gt;mass / volume&lt;/code&gt;. Infinite hydrogen atoms gives you infinite mass. An infinite universe gives you infinite volume. &lt;a href=&quot;https://math.stackexchange.com/a/1406021/52721&quot;&gt;&lt;code&gt;Infinity / Infinity&lt;/code&gt; is indeterminate&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;code&gt;mass / volume&lt;/code&gt; is the formula for a &lt;em&gt;homogeneous&lt;/em&gt; substance. The universe is not homogeneous, &lt;em&gt;it is heterogeneous&lt;/em&gt;. The universe is different depending on where you look. Instead of a simple &lt;code&gt;mass / volume&lt;/code&gt; you'd divide the universe into infinitesimally small volumes of space, measure the mass (and thus density) in each, and take the average.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This side-steps the infinity problem and you can get a sensible answer... assuming you can make infinite measurements.&lt;/p&gt;&#xA;" OwnerUserId="2019" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:35.350" LastActivityDate="2016-08-31T06:31:26.240" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18154" PostTypeId="1" AcceptedAnswerId="18157" CreationDate="2016-08-31T09:01:43.590" Score="12" ViewCount="24464" Body="&lt;p&gt;When I was small, I read that Sun is fixed at the center of the solar system and that all the other planets rotate around it.&lt;/p&gt;&#xA;&lt;p&gt;But later I heard that even the Sun is not fixed; it moves. Is this true?&lt;/p&gt;&#xA;&lt;p&gt;Why had people previously thought that the Sun is fixed?&lt;/p&gt;&#xA;&lt;p&gt;Is this correct planets in helix path or spring like path?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sQEAG.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sQEAG.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14057" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-05-17T16:39:10.477" Title="Is the Sun in our solar system moving or stationary?" Tags="&lt;the-sun&gt;&lt;solar-system&gt;&lt;milky-way&gt;&lt;orbital-mechanics&gt;" AnswerCount="5" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18155" PostTypeId="2" ParentId="18154" CreationDate="2016-08-31T09:46:05.170" Score="8" Body="&lt;p&gt;The Sun orbits the Galactic center once approximately every 226 million years as the galaxy spins, which is also travelling through space towards an unknown source of gravity (that is currently blocked from view by the luminosity of the galaxy itself) known as &quot;The Great Attractor&quot;.&lt;/p&gt;&#xA;" OwnerUserId="11009" LastEditorUserId="11009" LastEditDate="2016-08-31T13:44:45.020" LastActivityDate="2016-08-31T13:44:45.020" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18156" PostTypeId="1" AcceptedAnswerId="18160" CreationDate="2016-08-31T10:22:35.507" Score="2" ViewCount="325" Body="&lt;p&gt;I note there are two terms about Galactic rest frame in literatures:&#xA;&lt;em&gt;Galactic Standard of Rest (GSR)&lt;/em&gt; and &lt;em&gt;Galactic Rest Frame (Grf)&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are they the same frame? If not, what's the difference?&lt;/p&gt;&#xA;" OwnerUserId="9967" LastActivityDate="2016-08-31T14:48:57.340" Title="What's the difference between GSR and Grf" Tags="&lt;coordinate&gt;&lt;astrometry&gt;&lt;frame-of-reference&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18157" PostTypeId="2" ParentId="18154" CreationDate="2016-08-31T10:39:44.643" Score="15" Body="&lt;p&gt;The Sun moves, even in the context of the solar system. Gravity of the planets (mostly Jupiter) pulls the Sun out of position with respect to the centre of gravity of the solar system. &lt;a href=&quot;https://en.wikipedia.org/wiki/Barycenter&quot;&gt;This wikipedia entry&lt;/a&gt; explains it in a lot more detail, and also explains that their common centre of gravity lies outside of the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This wobbling of a star due to planets orbiting them is also one method of &lt;a href=&quot;https://en.wikipedia.org/wiki/Methods_of_detecting_exoplanets#Radial_velocity&quot;&gt;how we detect planets around other stars&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the Sun is fixed within the context of the solar system is a fairly good approximation. You need accurate and long term observations to detect the wobble.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastEditorUserId="2531" LastEditDate="2016-08-31T11:04:58.050" LastActivityDate="2016-08-31T11:04:58.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18158" PostTypeId="1" CreationDate="2016-08-31T11:31:09.580" Score="3" ViewCount="612" Body="&lt;p&gt;Would a black hole from a star made of anti-matter be the same as a blackhole made of ordinary matter?.&lt;/p&gt;&#xA;" OwnerUserId="14013" LastActivityDate="2016-08-31T14:29:43.290" Title="Is a black hole made from a anti-matter star the same as a ordinary black hole?" Tags="&lt;astrophysics&gt;" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="18159" PostTypeId="2" ParentId="18146" CreationDate="2016-08-31T11:42:52.650" Score="1" Body="&lt;p&gt;As a general rule, galaxies aren't very dense at all, though precise numbers are impossible, rough estimates can be made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.universetoday.com/22790/mass-of-the-milky-way/&quot; rel=&quot;nofollow&quot;&gt;Mass of the Milky-way&lt;/a&gt;:  with it's dark matter halo, about 6 x 10^42 kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Volume of the Milky-way (&lt;a href=&quot;https://en.wikipedia.org/wiki/Orders_of_magnitude_(volume)#Astronomical&quot; rel=&quot;nofollow&quot;&gt;per Wikipedia&lt;/a&gt;) about 3.3 × 10^61 cubic meters, or, 3.3 x 10^52 cubic km.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That gives us a rough density of the milky way of about 1 kg for every 5 billion cubic km.   That's obviously not very dense.   There are probably denser galaxies out there and larger galaxies, but the overall density is going to be very low, even for the densest galaxies.   Space is, generally speaking, mostly empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in regions that are so called &quot;densely packed with stars&quot;, if you measure the density in the traditional way, mass to volume, you end up with very low density, 1 kg per thousands if not millions of cubic km.   On earth, that density of air would essentially be a vacuum.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-08-31T14:39:01.073" LastActivityDate="2016-08-31T14:39:01.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18160" PostTypeId="2" ParentId="18156" CreationDate="2016-08-31T13:30:42.273" Score="3" Body="&lt;h2&gt;Short Answer&lt;/h2&gt;&#xA;&lt;p&gt;They're the same.&lt;/p&gt;&#xA;&lt;h2&gt;Long Answer&lt;/h2&gt;&#xA;&lt;p&gt;&lt;strong&gt;Local Standard of Rest (LSR)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Let me start out by defining the dynamical Local Standard of Rest (dynamical LSR or also LSRD). This definition is taken straight from &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/0805304029&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;An Introduction to Modern Astrophysics, 2nd Ed. by Carrol &amp;amp; Ostlie&lt;/a&gt;, page 903 (emphasis theirs).&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;To investigate the motion of the Sun and other stars in the solar neighborhood, we first define the &lt;strong&gt;dynamical local standard of rest&lt;/strong&gt; (dynamical LSR) to be a point that is &lt;em&gt;instantaneously&lt;/em&gt; centered on the Sun and moving in a &lt;em&gt;perfectly circular orbit&lt;/em&gt; along the solar circle about the Galactic center.&lt;/p&gt;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&lt;p&gt;Note that once the LSR is chosen, the Sun immediately begins to drift away from it, implying that we would effectively need to redefine the reference point constantly. In reality this is not a significant problem because (fortunately) the 230-Myr orbital period of the LSR is very long compared to the time since modern telescopic observations began (and even longer compared to the lifetime of a typical research grant); consequently there has not been sufficient time for the effect to become noticeable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Galactic Standard of Rest (GSR)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now it is important to understand how the LSR coordinate frame is defined, because the Galactic Standard of Rest (GSR) is defined with respect to the LSR. You can read about it &lt;a href=&quot;http://www.atnf.csiro.au/people/Tobias.Westmeier/tools_hihelpers.php#restframes&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, but in effect, the LSR is a non-inertial coordinate frame co-rotating with the rotation of the galaxy. The GSR is defined in the same way, but removes that co-rotation by subtracting out the rotational velocity of &lt;span class=&quot;math-container&quot;&gt;$220\:km/s$&lt;/span&gt;. In the LSR frame, another star's velocity would essentially be with respect to our Sun but in the GSR, the star's velocity would be with respect to a fixed point in the galaxy and be a summation of that star's motion with respect to our Sun and it's orbital motion around the galaxy.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Galactic Rest Frame (GRF)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The GRF is identical to the GSR, although I've heard GSR more often. In &lt;a href=&quot;https://arxiv.org/pdf/1608.03003.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;, you can see they define the GRF to be&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;... the galactic-rest-frame (grf) proper motions ... which are those seen by an observer at the location of the Sun and at rest with respect to the Galactic center.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This is identical to the definition of the GSR above. &lt;a href=&quot;http://arxiv.org/pdf/astro-ph/0612705v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Another paper&lt;/a&gt; similarly defines the GRF.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Removing the contribution of the motion of the Sun and of the Local Standard of Rest to the measured proper motion produces a Galactic rest-frame proper motion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-08-31T14:48:57.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18161" PostTypeId="2" ParentId="18158" CreationDate="2016-08-31T14:29:43.290" Score="4" Body="&lt;p&gt;To put a formal answer to this question, a black hole made of anti-matter would be indistinguishable from a matter black hole. Black holes are dubious objects, but it seems likely that it makes no sense to talk about an &quot;anti-matter black hole&quot; as once the matter goes into the black hole, be it matter or anti-matter, it ceases to have properties that define it as matter or anti-matter. The &lt;a href=&quot;https://en.wikipedia.org/wiki/No-hair_theorem&quot; rel=&quot;nofollow&quot;&gt;No-Hair Theorem&lt;/a&gt; for black holes states that, among a few other properties, a black hole is defined only by it's mass and net charge. The matter state of individual particles which fell into the black hole play no part in that description.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, there is a vigorous debate as to &lt;a href=&quot;https://en.wikipedia.org/wiki/Black_hole_information_paradox&quot; rel=&quot;nofollow&quot;&gt;whether or not &quot;information&quot; is conserved for black holes&lt;/a&gt;. E.g., if an anti-matter particle fell into a black hole, could you, at a later time, pull out a particle of the same properties or would you get out something of completely randomly new properties, not tied to the original matter. This in part plays into your question because it indicates that there is still a lot of uncertainty regarding the loss of information (including if the particle is anti-matter or matter) as particles fall into a black hole and whether not feeding anti-matter into a black hole actually makes it an anti-matter black hole or just simply a black hole.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-08-31T14:29:43.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18162" PostTypeId="7" CreationDate="2016-08-31T14:34:48.723" Score="0" Body="&lt;ul&gt;&#xA;    &lt;li&gt;Specific issues with astronomy&lt;/li&gt;&#xA;    &lt;li&gt;Real problems or questions that you’ve encountered&lt;/li&gt;&#xA;&lt;li&gt;Experimental and theoretical problems in astronomy, astrophysics, cosmology, and planetary science&lt;/li&gt;&#xA;&lt;li&gt;Astronomical equipment, both professional and amateur&lt;/li&gt;&#xA;&lt;li&gt;The use of space probes, telescopes, and rovers for astronomical purposes&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://meta.astronomy.stackexchange.com/q/195/2153&quot;&gt;Certain resource requests&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="6" LastEditDate="2016-08-31T14:34:48.723" LastActivityDate="2016-08-31T14:34:48.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18163" PostTypeId="2" ParentId="18150" CreationDate="2016-08-31T14:37:15.190" Score="2" Body="&lt;p&gt;As it has been mentioned in a comment by @zephyr, the infinite time issue is actually a non-issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you move closer to a black hole, the &lt;em&gt;relative&lt;/em&gt; time to your point of view doesn't change in the same way that it does from a different reference frame. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at your own situation, everything would happen in &quot;real time&quot; however, everything observed about you and your situation would take &quot;and infinite amount of seconds for one second to pass.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also a contested &lt;em&gt;issue&lt;/em&gt; because there have been measurements and possibly observations (I'd have to fact check this) of stars &quot;falling&quot; into black holes. So the infinite time from the outside observers perspective is also not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, the best way to answer your question without going to much into theory, hypotheticals, or mad science, is simply to state: &lt;strong&gt;No, it does not mean that.&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="2232" LastActivityDate="2016-08-31T14:37:15.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18164" PostTypeId="7" CreationDate="2016-08-31T14:37:54.387" Score="0" Body="&lt;ul&gt;&#xA;    &lt;li&gt;Anything not directly related to astronomy&lt;/li&gt;&#xA;    &lt;li&gt;Questions that are primarily opinion-based&lt;/li&gt;&#xA;    &lt;li&gt;Questions with too many possible answers or that would require an extremely long answer&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://meta.astronomy.stackexchange.com/q/97/2153&quot;&gt;Questions about physics or space exploration that do not involve astronomy&lt;/a&gt; (e.g. how a certain rocket works)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://meta.astronomy.stackexchange.com/q/149/2153&quot;&gt;Astrology&lt;/a&gt; or astronomy-related pseudosciences&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="6" LastEditDate="2016-08-31T14:37:54.387" LastActivityDate="2016-08-31T14:37:54.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18165" PostTypeId="2" ParentId="18120" CreationDate="2016-08-31T15:49:34.813" Score="2" Body="&lt;p&gt;&quot;They (the laws of physics) would vary in two natural ways:&quot;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;the constants in the equations may vary or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possible. We are fairly certain about the values of constants up to smaller astronomical scales (sub-galaxy). On the galactic scale and beyond we have strange deviations from what we would expect. On the galactic scale we currently attribute the deviations to &quot;dark matter&quot; which to me seems little more than a placeholder for the unknown. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a universal scale the apparently accelerating expansion of the universe is usually attributed to a different placeholder for the unknown, &quot;dark energy&quot;; or it may be that general relativity as we understand it does not hold on large astronomical scales, so that for example the gravitational constant is not in fact a constant, or whatever. This is fairly strong evidence that what we think to know is wrong or incomplete, so the answer is &quot;at the universal scale we &lt;em&gt;know&lt;/em&gt; that we are wrong&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the math in the equations may vary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the one thing which we are fairly sure about: The math will not vary. It may be incomplete, or wrongly applied, or whatever; but the math is the one thing which does not vary.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let's also not forget that there is famously &quot;plenty of room at the bottom&quot;. We don't even know the number of dimensions at very small (sub-nuclear) scales, we don't know how the single threads of the fabric of space time are knit together, etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On a more speculative level this may not be the only universe but for example just one shard of a multiverse; Lee Smolin wrote about the idea of an evolution of universes. The other ones would have most likely different constants, or differ in some other funny way.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On an even more speculative level: If you ask Elon Musk and others, we live in the Matrix anyway, and all laws of nature are subject to change at the whim of a keystroke equivalent by the sys admin. Something like &lt;code&gt;/gamemode 1 qwerty10&lt;/code&gt;, and your credit card runs never empty.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11190" LastEditorUserId="11190" LastEditDate="2016-08-31T15:59:11.007" LastActivityDate="2016-08-31T15:59:11.007" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18166" PostTypeId="1" AcceptedAnswerId="18168" CreationDate="2016-08-31T16:28:47.823" Score="4" ViewCount="206" Body="&lt;p&gt;This was done with the moon: &lt;a href=&quot;http://www.space.com/25884-radar-telescopes-moon-subsurface-features.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Radio Dishes Peer Beneath Moon's Surface&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KPAtE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KPAtE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13718" LastEditorUserId="11143" LastEditDate="2016-08-31T20:27:29.150" LastActivityDate="2016-09-01T16:37:37.527" Title="Can radio telescopes such as arecibo image the subsurface of asteroids or planets?" Tags="&lt;the-moon&gt;&lt;radio-astronomy&gt;&lt;radio-telescope&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18167" PostTypeId="1" AcceptedAnswerId="18180" CreationDate="2016-08-31T20:43:19.603" Score="3" ViewCount="471" Body="&lt;p&gt;In a paper I came across the description of the signal-to-noise ratio (SNR) for observations with a spectograph. This was reported as $10\:1/\mathring{A}$. I am rather new to spectroscopy, so could someone clarify these units for me? Why is it used and how should I read it? &lt;/p&gt;&#xA;" OwnerUserId="6965" LastEditorUserId="6965" LastEditDate="2016-08-31T20:56:43.473" LastActivityDate="2016-09-01T20:15:22.383" Title="Signal-to-noise in inverse angstrom for spectroscopy?" Tags="&lt;data-analysis&gt;&lt;spectroscopy&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18168" PostTypeId="2" ParentId="18166" CreationDate="2016-08-31T20:43:57.917" Score="4" Body="&lt;p&gt;It is important to understand how this process works. The method described in your article is known as Bistatic Radar. In effect, a transmitter sends out a signal (generally a radio telescope in the microwave region) which hits the surface of some body and bounces off to be received back on Earth by a second, separate radio telescope. Now, because a microwave's wavelength is so long ($\sim 0.1-100\:cm$) the microwaves don't bounce off from the exact surface, but instead are able to penetrate slightly into the subsurface before being reflected. This means the receiver gets a reflection of the subsurface of the body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case you linked, they used the Arecibo Telescope to provide the transmitted signal to the Moon and the Green Bank Telescope to receive the signal. A similar process could be done to other bodies besides our Moon. However, you run into the problem that the farther away an object is, the harder it will be to detect the returning signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that most objects within our solar system (and certainly all objects outside it) are outside the distance where this method will work. I was able to find &lt;a href=&quot;https://public.nrao.edu/news/pressreleases/asteroid-second-time&quot; rel=&quot;nofollow&quot;&gt;an instance&lt;/a&gt; where someone used this method for an asteroid that passed by the Earth at a distance 11 times that of the Moon. I'm not sure what the limit is to how far away something must be before this method no longer works, but I imagine it wouldn't work even for Mars unless we seriously upgrade our technology and power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course your question seems to presuppose the transmitter and receiver are both on Earth. If your transmitter is much closer, say a satellite orbiting that planet, then it is certainly feasible. &lt;a href=&quot;http://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=5738316&quot; rel=&quot;nofollow&quot;&gt;Here is a source&lt;/a&gt; which discusses the concept of using Bistatic Radar probing via a satellite, which also goes into the history a bit on what has been done before.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-08-31T20:43:57.917" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18169" PostTypeId="1" AcceptedAnswerId="18170" CreationDate="2016-09-01T02:30:58.687" Score="7" ViewCount="7863" Body="&lt;p&gt;If we consider the two largest masses in our solar system - the Sun and Jupiter, by themselves, they will orbit a common barycenter which is somewhere offset from the Sun's center in the direction of Jupiter (just above the Sun's surface). If we add Saturn, things get more complicated, but there is still a barycenter, even if it follows a complex path with respect to the Sun's center (or the Sun has a complex wobble around the barycenter). Add Neptune and all the other masses in the solar system, and things get even more complicated. Neverthess, there is still a barycenter (often outside the limb of the Sun) about which the Sun does a complex dance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what is Earth actually orbiting? The Earth's orbit is described to be an ellipse; the &quot;center&quot; of an elliptical orbit occurs at one of the focal points of the ellipse. What is at the focal point of the elliptical path the Earth follows? Is it the solar system barycenter or is it the Sun's (wobbling) center of mass? In other words, is the Earth's orbit shifting as the Sun is tugged around, our distance from the Sun never varying by more than the eccentricity of the ellipse, or do we orbit the solar system barycenter, and our distance from the Sun varies by the sum of our orbital eccentricity plus the amount the Sun wobbles in its own orbit around the barycenter?&lt;/p&gt;&#xA;" OwnerUserId="1983" LastEditorUserId="1983" LastEditDate="2016-09-01T02:45:50.003" LastActivityDate="2020-10-27T01:04:34.010" Title="What point does Earth actually orbit?" Tags="&lt;orbit&gt;&lt;the-sun&gt;&lt;earth&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18170" PostTypeId="2" ParentId="18169" CreationDate="2016-09-01T09:25:59.307" Score="12" Body="&lt;p&gt;Short answer is &quot;The Sun&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Conrad notes, since when you include the effects of Jupiter, the Earth orbit is non-keplerian, notions of centre are not really defined. But you can ask, in Newtonian gravity, where the Earth's acceleration vector points. Now the acceleration of the Earth is due mostly to the Sun, partly to the moon and slightly to the other planets.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's ignore the moon (ie consider the motion of the Earth Moon barycentre)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The acceleration due to the sun is 4 orders of magnitude greater than that due to Jupiter. So if Jupiter is at right angles to the Earth, the acceleration vector of the Earth is slightly pulled away from the centre of the sun. But not by much, in fact it points to a point about 4000km from the centre. The sun has a radius of 700000km, so the point the Earth is orbiting is well inside the sun, and isn't the Sun Jupiter barycentre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see why the Earth doesn't orbit the barycentre, consider the motion of the Sun in a three body system (Sun-Jupiter-Earth) The Earth doesn't orbit the Sun-Jupiter barycentre just as the Sun doesn't orbit the Earth-Jupiter barycenter.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-09-01T09:25:59.307" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18171" PostTypeId="1" AcceptedAnswerId="18183" CreationDate="2016-09-01T09:29:17.097" Score="5" ViewCount="523" Body="&lt;p&gt;I've been wondering for a while now why some objects in the universe (such as stars, planets, etc.) are round, while others are flat (galaxies, planetary rings and protoplanetary disks).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know these shapes come to be because of gravity. In planets, the gravity from the core pulls everything to the centre, thus, rounding the planet (same with stars). But with Saturn, for example, why have its rings formed as rings and not as a spherical layer of ice surrounding the planet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same with the Kuiper belt and asteroid belt, I would presume the mechanics there are the same as with a planetary orbit, hence, why they are belts and not clouds. But then what about the oort cloud? Why is that a huge cloud in the outermost solar system and not a huge ring of ice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the solar system itself, I've read that the solar system is kind of egg-shaped, but shouldn't it be just like the galaxy (or the other way round)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried looking into this and wondering about what different laws of physics may apply in each case, but it's all really confusing.&lt;/p&gt;&#xA;" OwnerUserId="13730" LastActivityDate="2016-09-01T21:27:09.913" Title="Why are some universal entities round and others are flat?" Tags="&lt;solar-system&gt;&lt;galaxy&gt;&lt;universe&gt;&lt;planetary-ring&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18173" PostTypeId="2" ParentId="18171" CreationDate="2016-09-01T10:05:45.767" Score="3" Body="&lt;p&gt;Preliminary note: there is a continuity of more or less flattened shapes. No macroscopic objects are &lt;em&gt;really spherical&lt;/em&gt;, or &lt;em&gt;really flat&lt;/em&gt;. Take Jupiter: it has a rotation period around its axis of 10 hours, and you can clearly see that it is an ellipsoid. It's the &lt;a href=&quot;https://en.wikipedia.org/wiki/Centrifugal_force&quot; rel=&quot;nofollow&quot;&gt;centrifugal force&lt;/a&gt;, combined with the gravitational attraction, that shapes these objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to your question: Planets and stars are dense bodies, and rotate as single objects. On the other hand, the Oort cloud, Saturns rings, stellar clusters and galaxies are made of independent particles, free-floating in lots of empty space. Depending on the formation history and gravitational environment (e.g., Saturn and its moons) of these entities, the orbits of these individual particles can be more or less co-aligned, and form a more or less spherical clouds.&lt;/p&gt;&#xA;" OwnerUserId="9150" LastActivityDate="2016-09-01T10:05:45.767" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18174" PostTypeId="2" ParentId="1997" CreationDate="2016-09-01T10:18:26.080" Score="3" Body="&lt;p&gt;I posted an answer for this on Physics SE recently, but have also just had a query on this from another &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10017/does-the-sun-turn-around-a-big-star/10018?noredirect=1#comment28234_10018&quot;&gt;Astronomy SE answer&lt;/a&gt;, so I am adding this here for completness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can approximate the plane of the galaxy as a disk made up of stars and gas, with a density $\rho(|z|)$, that decreases with absolute distance $|z|$ from the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If then assume that the Sun is close enough to $z=0$ and that the &lt;em&gt;radial&lt;/em&gt; variation in $\rho$ was negligible enough to treat the disk as an infinite plane (this is not bad, the amplitude of the Sun's motion is only about 10% of the radial scale length of the disk density), then you could construct a little cylinder through the plane, with one face at   $z=0$, where $g=0$, and use &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Gauss%27s_law_for_gravity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gauss's law for gravity&lt;/a&gt; to estimate the gravitational acceleration at height $z$.&#xA;$$ g(z) \simeq -4\pi G \int_0^{z} \rho(z)\ dz$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is effectively decomposing the Sun's orbit into a radial/tangential orbit plus a vertical motion, which I treat here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\rho(z)$ approximates to an exponentially decaying function with a scale height of maybe 200-300 pc. If we are closer to $z=0$ than that, then the density is roughly a constant $\rho_0$. Putting this into the equation above, we see that &#xA;$$g(z) =-4\pi G\rho_0 z.$$&#xA;But this is just simple harmonic motion with an angular frequency $\sqrt{4\pi G\rho_0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The density of the disk near the Sun has been estimated to be 0.076 solar mass per cubic parsec (&lt;a href=&quot;http://arxiv.org/abs/astro-ph/9709022&quot; rel=&quot;nofollow noreferrer&quot;&gt;Creze et al. 1998&lt;/a&gt;). Using this value, we get an &lt;em&gt;approximate&lt;/em&gt; predicted oscillation period up and down through the disk plane of 95 million years. This is pretty close to the accepted value of 70 million years considering the approximations that have been made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of the question you ask I should add that the mass density I quote above is actually &lt;em&gt;derived&lt;/em&gt; from the positions and motions of stars in the solar neighbourhood. As the paper I referenced discusses, the value they obtain is close to that obtained by counting stars and adding the contributions of gas and dust. There is actually little evidence for dark matter &lt;em&gt;in the disk&lt;/em&gt; from these measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This result is &lt;em&gt;entirely consistent&lt;/em&gt; with the idea of a dark matter distribution that is 10x the visible mass but roughly spherically symmetric, and which accounts for the Milky Way rotation curve. Not much of this dark matter is in the disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the picture is not quite right. The Sun completes only about 3 vertical oscillations for every orbit around the Galactic centre.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-09-01T13:55:34.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18176" PostTypeId="1" AcceptedAnswerId="18186" CreationDate="2016-09-01T15:19:34.297" Score="4" ViewCount="1132" Body="&lt;p&gt;According to the orbital plane XYZ, I understand that the axial tilt is the angle between the planet axis rotation and any normal of the plane (actually Z in right handed system). BUT, this is not sufficient to define a vector direction, I need another angle into the XY plane. Where do I find this information?  Actually I believe it's relative to the direction of the periapsis. Am I right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following all the nice conversation on the answers, I maybe need to ask &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to get the axial tilt vector(x,y,z) relative to ecliptic or orbital plane, given ONLY ONE angle, i.e. 23.439281 deg for the Earth, 25.19 for Mars, etc.&lt;/p&gt;&#xA;" OwnerUserId="13989" LastEditorUserId="616" LastEditDate="2016-09-02T01:33:19.967" LastActivityDate="2019-12-08T23:48:39.107" Title="How to get the axial tilt vector(x,y,z) relative to ecliptic" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;rotation&gt;&lt;coordinate&gt;&lt;ecliptic&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18177" PostTypeId="2" ParentId="18166" CreationDate="2016-09-01T16:37:37.527" Score="3" Body="&lt;p&gt;Yes, but not Earth-based ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mars.nasa.gov/mro/mission/instruments/sharad/&quot; rel=&quot;nofollow noreferrer&quot;&gt;SHARAD&lt;/a&gt; (Shallow Radar) is an instrument on the Mars Reconnaissance Orbiter which performs subsurface imaging of Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an image from &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Polar_Cap_SHARAD_PIA13164.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; of deposit layers on Mars's north pole, taken by SHARAD:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Gvib1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gvib1.jpg&quot; alt=&quot;Mars north pole deposit layers&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6" LastActivityDate="2016-09-01T16:37:37.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18178" PostTypeId="2" ParentId="18167" CreationDate="2016-09-01T17:13:55.987" Score="1" Body="&lt;p&gt;Warning: I never worked with spectra, and I don't have a complete answer. Anyway, here's how I understand it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A spectrograph has a resolution $R = \Delta \lambda / \lambda$ that tells us how well it can distinguish light of different wavelengths. &lt;a href=&quot;http://www.eso.org/sci/facilities/paranal/instruments/vimos.html&quot; rel=&quot;nofollow&quot;&gt;VIMOS&lt;/a&gt; from the paper has a resolution from 200 to 2500. With the same flux from the source and for a low-resolution spectrograph, you get more photons per wavelength bin than for a high-resolution one. This means you'll also get better SNR. So, giving SNR / wavelength makes inherent sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where I'm very unsure: giving a SNR / $(0.1$nm$)$ should be equivalent to a magnitude limit, &lt;strong&gt;if&lt;/strong&gt; the spectral energy distribution is the same. Or it could be a minimum quality criterion on the spectrum - stop taking data as soon as the SNR is reached.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastActivityDate="2016-09-01T17:13:55.987" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18179" PostTypeId="1" AcceptedAnswerId="18181" CreationDate="2016-09-01T19:35:38.220" Score="3" ViewCount="194" Body="&lt;p&gt;Maybe a slightly weird question, and maybe more appropriate for some future geophysics SE, but I feel like this is what questions like &lt;a href=&quot;https://astronomy.stackexchange.com/questions/856/do-moons-have-moons&quot;&gt;Do moons have moons?&lt;/a&gt; are really driving at, and it just almost gets answered there.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How massive must a rocky body be to liquify adequately to eliminate internal mass concentrations that could disrupt natural satellite orbits?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand geothermal energy to be a function of both gravitation and fissionable material in the planetary body, so an ideal answer would at least state its assumptions in regards to the amount of energy from fissionable material present.&lt;/p&gt;&#xA;" OwnerUserId="14080" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-09-04T13:25:19.830" Title="How massive must a rocky body be to liquify adequately to eliminate internal mass concentrations that could disrupt natural satellite orbits?" Tags="&lt;orbit&gt;&lt;astrophysics&gt;&lt;natural-satellites&gt;&lt;geology&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18180" PostTypeId="2" ParentId="18167" CreationDate="2016-09-01T20:15:22.383" Score="2" Body="&lt;p&gt;A spectrum imaged onto a CCD or other detector consists of photons spread out along a wavelength axis. Signal-to-noise (S/N) depends on how many photons you have from the source, how many from the background, intrinsic noise from the electronics, etc. -- and what kind of bin you're adding up the photons in. If you summed up all the photons across the spectrum (over all wavelengths), you'd get one number with a very high S/N, but no spectral information. If you divided the spectrum into N different wavelength bins, you'd get a lower S/N per bin, but more spectral information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the S/N ratio you mention means is that &lt;em&gt;if&lt;/em&gt; you divided that spectrum into bins that were 1 Angstrom wide, you'd have enough photons in each bin for a S/N of 10 in each bin (at least for some region of the spectrum, probably near the central wavelength). You can then have some idea how the S/N would improve for bins larger than 1 Angstrom in size, or how it would get worse for smaller bins. S/N per Angstrom is conventional for optical spectroscopy, both because Angstroms are traditional and because moderate resolutions spectrographs often have scales of a few tenths to several Angstroms per pixel along the wavelength direction.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2016-09-01T20:15:22.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18181" PostTypeId="2" ParentId="18179" CreationDate="2016-09-01T20:49:50.110" Score="5" Body="&lt;p&gt;Planets and moons are never going to be completely homogenized such that their radial density is perfectly spherically symmetric. They're always going to have &lt;a href=&quot;https://en.wikipedia.org/wiki/Mass_concentration_(astronomy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;mass concentrations&lt;/a&gt; (mascons) with regions of more mass than average. These mascons, due to the extra mass, cause gravitational anomalies of higher gravity in those local regions. These mascons exist for all bodies, be it the Earth (e.g., Hawaii), the Moon (e.g., &lt;a href=&quot;https://en.wikipedia.org/wiki/Mare_Imbrium#Characteristics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mare Imbrium&lt;/a&gt;), or Mars (e.g., &lt;a href=&quot;https://en.wikipedia.org/wiki/Isidis_Planitia&quot; rel=&quot;nofollow noreferrer&quot;&gt;Isidis Planitia&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question is to ask, how massive must a body be such that it can liquify itself and eliminate these mass concentrations so as to not disturb satellite orbits. This is specifically an issue in regard to orbiting the Moon because its Hill sphere is so small that only very low orbits can exist, but the mascons in the Moon make most of these unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think though, you're asking the wrong question. As I said above, the Earth itself has mascons which are in fact caused by its liquid interior! The region around and below Hawaii has a slightly higher than average mass caused by a &lt;a href=&quot;https://en.wikipedia.org/wiki/Mantle_plume&quot; rel=&quot;nofollow noreferrer&quot;&gt;mantle plume&lt;/a&gt;, effectively an upwelling of less dense material from the core-mantle boundary. Aside from creating the mascon, it was this mantle plume which also created Hawaii, but that's beside the point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point here is that all bodies have mascons, whether they have liquid interiors like the Earth or primarily solid interiors like the Moon. The difference between orbiting the Earth and the Moon though, is the distance you can do it at. The reason mascons are such a problem for orbits around the Moon is that you have to orbit so close such that the effect of these mascons compared to the general field is a noticeable perturbation on the orbit. If you could orbit the Moon farther out, they wouldn't be a problem. The opposite applies for the Earth. Most satellites around the Earth are very far away such that the Earth's mascons don't have any noticeable effect. If you orbited close enough (and likely that would require orbiting inside the atmosphere which presents lots of other problems), the mascons would become a problem just as they are for lunar orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, shown below is what the Earth would look like if you magnified all the various mass/gravity anomolies (such a shape is known as the Earth's geoid). All those irregularities would perturb a satellite if it had to orbit the Earth as close as it would need to orbit the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9r5tR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9r5tR.jpg&quot; alt=&quot;Earth&amp;#39;s Geoid&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I kind of side-stepped your question, but I hope you'll agree that the question isn't really meaningful. All bodies have mascons to some degree or another, you can't get away from that.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="14094" LastEditDate="2016-09-04T13:25:19.830" LastActivityDate="2016-09-04T13:25:19.830" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18183" PostTypeId="2" ParentId="18171" CreationDate="2016-09-01T21:27:09.913" Score="3" Body="&lt;p&gt;I tend to see it as simply a consequence of symmetry, symmetry breaking and the balance between them.&#xA;The simplest is the spherical symmetry where every direction are the same, as good as another one (isotropy). It's the case for the Sun, planets, globular star clusters, ...&#xA;If there is some (sufficent) rotation involved, this isotropy will be broken, there will be a distinguished direction and this will give planar objects: spiral galaxies, planetary rings, ... or even sometimes linear objects like a relativistic jet.&#xA;Of course, when there is some rotation involved but not much strong you will have some intermediate objects: the Earth is an ellipsoid, elliptical galaxies, ...&lt;/p&gt;&#xA;" OwnerUserId="11521" LastActivityDate="2016-09-01T21:27:09.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18184" PostTypeId="1" AcceptedAnswerId="18187" CreationDate="2016-09-01T21:28:09.923" Score="5" ViewCount="1023" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xlWIh.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xlWIh.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is a posible periodic solution for a three-body problem, can anybody tell me where is the center of mass of the system? As we know, in a n-body problem the bodies orbit around their center of mass, but here I don't know where it is. Can anybody help me? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="9949" LastActivityDate="2016-09-02T13:26:55.177" Title="Question about the Figure-eight periodic planar three-body motion" Tags="&lt;n-body-simulations&gt;&lt;celestial-mechanics&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18185" PostTypeId="1" CreationDate="2016-09-01T23:04:12.410" Score="2" ViewCount="3652" Body="&lt;p&gt;If the planet and star in question are very similar to the Earth and the Sun, how can I calculate the planet's surface temperature by knowing the distance from the star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question was originally asked on the Worldbuilding stack exchange, and someone suggested I ask it here.&lt;/p&gt;&#xA;" OwnerDisplayName="user14083" LastEditorUserId="2365" LastEditDate="2016-09-02T15:56:02.903" LastActivityDate="2016-09-02T15:56:02.903" Title="Calculate planet's surface temperature by distance from star" Tags="&lt;temperature&gt;&lt;earth-like-planet&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2016-09-25T11:40:04.607" ContentLicense="CC BY-SA 3.0" />
  <row Id="18186" PostTypeId="2" ParentId="18176" CreationDate="2016-09-02T00:22:13.210" Score="4" Body="&lt;p&gt;Earth is a special case since the equatorial and ecliptic coordinate systems are defined in terms of its own rotation and orbit.&#xA;Earth's north pole vector in&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Equatorial_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;equatorial coordinates&lt;/a&gt; is&#xA;$$\vec N_{\oplus,eq} = (0, 0, 1)$$&#xA;To transform this to&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Ecliptic_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;ecliptic coordinates&lt;/a&gt;, we rotate about the $x$ axis by the obliquity $\varepsilon$ = 23.44$^\circ$ and get&#xA;$$\vec N_{\oplus,ecl} = (0, \sin \varepsilon, \cos \varepsilon) = (0, 0.3978, 0.9175)$$&#xA;In a spherical coordinate system, two angles define a unique direction.&#xA;In equatorial coordinates these are the right ascension $\alpha$ and declination $\delta$.&#xA;This &lt;a href=&quot;http://astropedia.astrogeology.usgs.gov/download/Docs/WGCCRE/WGCCRE2009reprint.pdf&quot; rel=&quot;nofollow&quot;&gt;IAU report&lt;/a&gt;, table 1,&#xA;gives $\alpha$ and $\delta$ values for each major planet's north pole as of 2000-01-01 and formulas to compute them for other years.&#xA;To convert these to rectangular form&#xA;$$(x_{eq}, y_{eq}, z_{eq}) =&#xA;(\cos \alpha \cos \delta,&#xA;\sin \alpha \cos \delta,&#xA;\sin \delta)$$&#xA;Then if you want J2000 ecliptic coordinates&#xA;$$(x_{ecl}, y_{ecl}, z_{ecl}) =&#xA;(x_{eq},&#xA;y_{eq} \cos \varepsilon + z_{eq} \sin \varepsilon,&#xA;z_{eq} \cos \varepsilon - y_{eq} \sin \varepsilon)$$&#xA;But if you want another planet's orbital plane to be the $xy$-plane,&#xA;then you also need the orbit's inclination and longitude of ascending node;&#xA;I leave that transformation as an exercise for the motivated reader.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-09-02T00:22:13.210" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18187" PostTypeId="2" ParentId="18184" CreationDate="2016-09-02T01:38:53.237" Score="5" Body="&lt;p&gt;There may be some three-body &lt;em&gt;periodic solutions&lt;/em&gt; that orbit around a common point, but in general they get a little crazy-looking and may not always contain an immediately obvious center of mass from casual observation. But of course it will always exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you watch closely, you'll see that whenever one body reaches the intersection point, all three are on a straight line with the other two equidistant (symmetric). If you look where this orbit was originally described mathematically, they probably defined the initial conditions with the green one in the middle - so that the &lt;code&gt;y&lt;/code&gt; coordinates of all there were zero and the line at that point is horizontal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At those moments (all six permutations are present) the center of mass is at the intersection point since the masses are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are no external forces, the center of mass doesn't change in between those moments, so &lt;em&gt;the center of mass is always at the intersection point.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uu0yA.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uu0yA.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2016-09-02T03:57:51.220" LastActivityDate="2016-09-02T03:57:51.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18188" PostTypeId="2" ParentId="18185" CreationDate="2016-09-02T03:24:21.717" Score="5" Body="&lt;p&gt;Neat and tidy calculations for this aren't possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The math behind effective temperature of a planet can be found &lt;a href=&quot;https://en.wikipedia.org/wiki/Effective_temperature#Surface_temperature_of_a_planet&quot;&gt;here&lt;/a&gt; and that's a straight forward calculation where you can input distance, solar energy, planet's albedo and get an average temperature.   Earth's effective temperature calculation comes to -21 °C, which obviously isn't accurate, but that's as far as the calculation can take you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Feedback mechanisms:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With essentially no change to distance from the sun, and only relatively small changes in tilt and eccentricity, Earth's temperature can fluctuate from a current 15 °C global average to about 5 °C as it goes in and out of ice ages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feedback mechanisms like albedo (ice cover is the key driver, but also deserts, surface area of the oceans), CO2 capture, which increases with colder oceans, can amplify small changes, leading to significant changes in surface temperature.   An orbital change, which should account for maybe 1 degree C, or an increase in CO2, which, by itself, captures less than 1 degree C, through feedback mechanisms, can lead to much larger temperature changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More on feedback mechamisms &lt;a href=&quot;https://en.wikipedia.org/wiki/Climate_change_feedback&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Land placement and Ocean circulation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 or 10 million years ago, Earth's average temperature was about 18-20 °C and that change was driven, primarily, by the closing of the Isthmus of Panama which affected ocean circulation.  Just the closing of the Isthmus made ice ages possible.  That's a pretty small change in the grand scheme of things, whether your planet has a near equatorial passage where oceans can flow or whether it doesn't, but that one thing can mean more than 10 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Antarctica drifted over the south pole (the last 30-35 million years or so) glaciers formed and the earth cooled, because as ice forms, more sunlight is reflected into space and oceans levels drop.  Oceans having the lowest albedo and ice the highest.  Also, as Antarctica driffs away from the south pole over the next 20-30 million years, Earth is expected to warm, unless another change occurs, like the continued absorption into land and the oceans of CO2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth can only enter an ice age when there's land near the poles.   The &lt;a href=&quot;https://en.wikipedia.org/wiki/Late_Ordovician_glaciation&quot;&gt;glacial period&lt;/a&gt; of 440-460 million years ago most of the land was over the south pole.  The sun was about 4% less luminous and the earth had much more CO2 (though CO2 levels did drop during that time, they didn't drop close to where they are now) and the orbit of the earth, 440 million years ago has some unknowns, but a large land mass over the south pole is thought to have played a key role in making that glaciation possible.  The tipping point of ice formation at the poles is particularly sensitive and a significant variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, even with very similar to earth like planets.  Just where the land is and how well the oceans circulate and whether the planet can form glaciers and sea ice, or not, can fluctuate perhaps 20 degrees C in global average temperature.  How much CO2, how much active volcanism, how much reflective material in the upper atmosphere - all variables.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even with identical earths, the feedback mechanisms make it difficult to predict what Earth's temperature would be if you pushed Earth 1 million or 1/2 million miles further from the sun.  The calculations for surface temperature wouldn't follow a simple mathematical formula.   Now, you'd still get colder as you moved the planet away, but one push might give you 1-2 degree and the next push might give you 5-10 and the one after that 0-1.  There would be no way to do a clean formula.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-09-02T04:05:30.180" LastActivityDate="2016-09-02T04:05:30.180" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18189" PostTypeId="2" ParentId="18184" CreationDate="2016-09-02T03:44:36.000" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;As we know, in a n-body problem the bodies orbit around their center&#xA;  of mass, but here I don't know where it is. Can anybody help me?&#xA;  Thanks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All orbits are n-body systems.  There's essentially no perfect 2 body systems anywhere in the universe.   Our solar-system is an N-body, not a 2 body, but nearly all the orbits in our solar system approximate 2 body orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long term stable N-body systems are very similar to 2-body systems.  Unstable or chaotic N-body systems like this one on Wikipedia, don't last very long.  They turn into stable systems and in the process, usually eject some objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BWs44.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BWs44.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/N-body_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I would be hesitant to try to prove this mathematically, the figure 8 orbit reminds me of balancing a pencil on it's tip.  Slight variations tend to grow over time and it destabilizes quickly.  (cool video on &lt;a href=&quot;https://www.youtube.com/watch?v=U3vAoJhIWms&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pencil-balancing&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-09-02T03:44:36.000" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18191" PostTypeId="1" CreationDate="2016-09-02T08:40:18.643" Score="4" ViewCount="5627" Body="&lt;p&gt;I am looking for a particular set of information to essentially make a star map for a game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I need:&lt;br&gt;&#xA;All stars within 11,462 light years of Earth.&lt;br&gt;&#xA;Their stellar classification.&lt;br&gt;&#xA;If they have any planets around them (hypothetical or otherwise)&lt;br&gt;&#xA;Their position using an x, y, z coordinate relative to earth or galaxy center (I don't really get declination, etc)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I need the extrapolated position/distance data, not the catalogs themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best one I can find is HYG, HabCat, and HYGHag is limited to to like 120,000 of 1,2 billion estimated stars... Are there any better?&lt;/p&gt;&#xA;" OwnerUserId="14084" LastEditorUserId="14084" LastEditDate="2016-09-03T04:31:34.763" LastActivityDate="2017-10-13T15:29:56.213" Title="Looking for star database" Tags="&lt;star-catalogues&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18192" PostTypeId="2" ParentId="18191" CreationDate="2016-09-02T09:24:23.340" Score="5" Body="&lt;p&gt;Best chance would be the &lt;a href=&quot;http://www.cosmos.esa.int/web/hipparcos/catalogue-summary&quot; rel=&quot;noreferrer&quot;&gt;Hipparcos catalogue&lt;/a&gt;. The first set of &lt;a href=&quot;http://www.cosmos.esa.int/web/gaia/release&quot; rel=&quot;noreferrer&quot;&gt;Gaia&lt;/a&gt; data will be released Mid-September 2016, but I don't know if it will be more accurate than Hipparcos already.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;All stars within 11,462 light years of Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That won't be easy. These catalogues have a magnitude (brightness) cutoff. Brighter, more massive stars can be seen further away than smaller, darker ones.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Their stellar classification.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Spectral type should be in there. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If they have any planets around them (hypothetical or otherwise)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Searching for Exoplanets is an ongoing project, and there is no complete catalogue in any sense of the word yet for any radius. Most Exoplanets were found with the Kepler mission, but that one looked in a fairly small part of the sky. What exactly do you mean by hypothetical planets?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Their position using an x, y, z coordinate relative to earth or galaxy center (I don't really get declination, etc)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll have to convert that yourself. If you know how to program, there are libraries for coordinate conversion. I would argue, though, that an Earth-centred spherical coordinate system is preferential for most applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distances are hard to measure. Hipparcos and Gaia use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Parallax&quot; rel=&quot;noreferrer&quot;&gt;parallax method&lt;/a&gt;. This gets less accurate with increasing distance. Also, distances are naturally measured in parsec, but the conversion to light years is trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gaia catalogue will have $10^9$ stars with positions and velocities. It will also be more accurate than Hipparcos. But the measurements are ongoing, and the final data release is planned for 2022.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastActivityDate="2016-09-02T09:24:23.340" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18193" PostTypeId="2" ParentId="18184" CreationDate="2016-09-02T13:26:55.177" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If this is a possible periodic solution for a three-body problem?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What you've posted here is the classic &quot;figure 8&quot; orbit for a 3 body system. Originally the 3-body system was an unsolvable mathematics problem, until people started using computers to do the math for us. Recently, a set of 13 &quot;stable&quot; 3-body orbits have been found which were &lt;a href=&quot;http://arxiv.org/pdf/1303.0181v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;published in this paper&lt;/a&gt;. A diagram illustrating these orbits (with the accompanying caption) is shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xbFIv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xbFIv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;FIG. 1: The (translucent) shape-space sphere, with its backside also visible here. Three two-body collision points (bold red circles) - punctures in the sphere - lie on the equator. (a) The solid black line encircling the shape sphere twice is the figure-8 orbit. (b) Class I.A butterfly I orbit (I.A.1). Note the two reflection symmetry axes. (c) Class I.B moth I orbit (I.B.1) on the shape-space sphere. Note the two reflection symmetry axes. (d) Class II.B yarn orbit (II.B.1) on the shape-space sphere. Note the single-point reflection symmetry. (e) Class II.C yin-yang I orbit (II.C.2) on the shape-space sphere. Note the single-point reflection symmetry. (f) An illustration of a real space orbit, the “yin-yang II” orbit (II.C.3a).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is hard to tell, but the gif you've posted is represented by (a) above. So I suppose the answer to your question is, &lt;strong&gt;yes, the figure 8 is a possible solution for the 3-body problem&lt;/strong&gt;. However, I should caution that there is no possible 3-body system which is stable for long periods, &lt;em&gt;especially&lt;/em&gt; if those three bodies have nearly equal masses. Our Sun/Earth/Moon system is only so stable because $M_{Sun}&amp;gt;&amp;gt;M_{Earth}&amp;gt;&amp;gt;M_{Moon}$ and they all effectively have binary orbits. For three, roughly equal mass stars, your system gets wildly unstable very quickly. Pretty much any perturbation will knock your system out.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can anybody tell me where is the center of mass of the system?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As &lt;a href=&quot;https://astronomy.stackexchange.com/a/18187/11143&quot;&gt;uhoh very nicely reasons out&lt;/a&gt;, the center of mass is in the center of the figure 8. Note however, that your system appears to show all bodies orbiting in a single plane. Obviously the real universe is 3 dimensions and very likely these objects would be orbiting in 3 dimensions, making the location of the center of mass more complicated. In the image above, you can see that the orbits oscillate up and down periodically, but they will do so in such a way that the center of mass is in the midplane of this oscillation, as well as the center of the figure 8.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-09-02T13:26:55.177" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18194" PostTypeId="1" AcceptedAnswerId="18195" CreationDate="2016-09-02T16:30:25.637" Score="8" ViewCount="1244" Body="&lt;p&gt;I understand from Wikipedia that, &quot;The length of the day, which has increased over the long term of Earth's history due to tidal effects, ...&quot;.  If the angular velocity of the earth is decreasing then the stored rotational energy is decreasing.  Where does the power/energy go?&lt;/p&gt;&#xA;" OwnerUserId="14085" LastEditorUserId="6" LastEditDate="2016-09-13T16:06:48.517" LastActivityDate="2016-09-13T16:06:48.517" Title="Decreasing rate of Earth's rotation: where does the power go?" Tags="&lt;earth&gt;&lt;rotation&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18195" PostTypeId="2" ParentId="18194" CreationDate="2016-09-02T17:34:31.387" Score="13" Body="&lt;p&gt;You're correct in assuming the net angular momentum of the system in question here will remain constant. The Moon's orbit around Earth is responsible for the slowing of Earth's rotation. This effect is &lt;em&gt;extremely&lt;/em&gt; small. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The decrease in Earth's angular momentum is transferred to the moon, which resultantly sees it's orbit accelerate. This acceleration also causes the moon to move further and further from the Earth. This trend will continue until they reach a common speed.&lt;/p&gt;&#xA;" OwnerUserId="9343" LastActivityDate="2016-09-02T17:34:31.387" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18196" PostTypeId="1" CreationDate="2016-09-02T18:52:32.257" Score="2" ViewCount="31" Body="&lt;p&gt;I am interested in figuring out which Messier globular clusters (plus 47 Tuc and Omega Cen) are in the Kepler/K2 fields, both fields that have already been observed and any that will be observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Poking around online I was able to figure out that &lt;a href=&quot;https://keplerscience.arc.nasa.gov/K2/GuestInvestigationsC02.shtml&quot; rel=&quot;nofollow&quot;&gt;Campaign 2 of K2 observed M4 and M80&lt;/a&gt;, but I would like to get a complete sample of globular clusters that have fallen or will fall into the Kepler/K2 fields, especially the Kepler field.  Are there online tools that could help me figure this out?&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2016-09-02T18:52:32.257" Title="How to find out which Messier objects are in the Kepler/K2 fields?" Tags="&lt;observational-astronomy&gt;&lt;kepler&gt;&lt;messier-catalog&gt;&lt;k2&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18197" PostTypeId="1" CreationDate="2016-08-28T11:39:27.373" Score="1" ViewCount="37" Body="&lt;p&gt;I saw several chronologies of the trial online but most of them only mention the name of Urban 8 as involved in the actual court proceedings. How many judges were involved and who were they exactly? Some useful information can be found &lt;a href=&quot;https://astronomy.stackexchange.com/questions/18107/how-many-judges-sat-on-galileos-trial-over-heliocentrism-and-who-were-they-exac&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="14004" OwnerDisplayName="Mikhail Katz" LastEditorUserId="6" LastEditDate="2020-06-11T13:22:25.503" LastActivityDate="2020-06-11T13:22:25.503" Title="How many judges sat on Galileo's trial and who were they exactly?" Tags="&lt;history&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" ClosedDate="2016-09-02T19:43:46.943" ContentLicense="CC BY-SA 3.0" />
  <row Id="18198" PostTypeId="1" AcceptedAnswerId="18206" CreationDate="2016-09-02T19:38:01.863" Score="3" ViewCount="450" Body="&lt;p&gt;In &quot;A Briefer History of Time&quot; by Stephen Hawking he writes that if we were to travel faster than the universe were expanding and for long enough (in a straight line) we would eventually come back to the point we started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What suggests that the universe is cyclic in nature as this suggests?&#xA;Or what led him to come to this conclusion?&lt;/p&gt;&#xA;" OwnerUserId="14062" LastActivityDate="2016-09-10T00:51:50.027" Title="What suggests the cyclic nature of the universe? Proposed by Stephen Hawking" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18199" PostTypeId="1" CreationDate="2016-09-02T20:40:40.883" Score="2" ViewCount="54" Body="&lt;p&gt;Even if our solar system is not in a stable orbit around the center of the galaxy (is it?), since the birth of our sun, how far has it traveled relative to the center of the galaxy?&lt;/p&gt;&#xA;" OwnerUserId="14087" LastActivityDate="2016-09-02T20:40:40.883" Title="How many times has the sun orbited the galaxy?" Tags="&lt;cosmology&gt;&lt;milky-way&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2016-09-03T21:35:45.663" ContentLicense="CC BY-SA 3.0" />
  <row Id="18201" PostTypeId="2" ParentId="18194" CreationDate="2016-09-03T07:14:28.817" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Where does the power/energy go?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It goes into heating the Earth and the Moon. That heat in turn spreads out into the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the Earth-Moon system comes very close to conserving angular momentum, it does not conserve mechanical energy. In fact, that angular momentum transfers from the Earth's rotation to the Moon's orbit means that the total mechanical energy of the Earth-Moon system is necessarily reduced. Mechanical energy is only conserved in isolated, non-dissipative systems. The tides in the Earth's oceans, in the Earth as a whole, and in the Moon as a whole means that the Earth-Moon system is dissipative.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2016-09-03T07:14:28.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18202" PostTypeId="1" CreationDate="2016-09-03T09:47:28.377" Score="3" ViewCount="161" Body="&lt;p&gt;If a Kepler telescope were built to focus on one point source star at a time, would it provide more information, or did Kepler capture virtually every photon from any given star anyway?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to study known transit events more closely one star at a time, to possibly detect moons or atmosphere or rings by resolving the transit light curve's shape better, is a larger aperture necessary or would a smaller field of view help? What field of view would be optimal for observing a single point source?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-09-13T17:24:06.530" Title="Can a telescope be built to focus on a point source star?" Tags="&lt;telescope&gt;&lt;optics&gt;&lt;planetary-transits&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18203" PostTypeId="1" CreationDate="2016-09-03T10:07:18.397" Score="3" ViewCount="346" Body="&lt;p&gt;Why are the Galactic Coordinates not aligned so that the Axes go through the Long Bar Y-axis aligned to it's longer side, positive with Sol, and the X-axis align to the center of the short length with Sol being in the negative? Z should obviously be on the galactic plane where it generally is I think...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious about whether there is a reason we shouldn't align it like this? It seems to make the most sense to me where as the current system, while I understand it, to me is just to Earth-centric, especially when we have something that stands out so much on the galactic scale you could use to figure out where you are with where as Where Sol is being 0,0,0 isn't really helpful to anyone unless you're already on Earth ^.^&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The other part of my question is less &quot;real&quot; and more of a &quot;I would like to know if there are any problems with doing this that I am unaware of&quot; thing...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 3 question areas...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1&lt;/strong&gt;&lt;br&gt;&#xA;I use a unit I call a Lida, It is 24 terameters long and roughly equivalent to a Light Day.&#xA;On this map distances will be measured in Lida, more specifically, KiloLida.&#xA;&lt;br&gt;1 Sector = &lt;span class=&quot;math-container&quot;&gt;$20 kL^3$&lt;/span&gt;&lt;br&gt;&#xA;1 Block = &lt;span class=&quot;math-container&quot;&gt;$200 kL^3$&lt;/span&gt;&lt;br&gt;&#xA;1 Grid = &lt;span class=&quot;math-container&quot;&gt;$2,000 kL^3$&lt;/span&gt;&lt;br&gt;&#xA;1 Octant = &lt;span class=&quot;math-container&quot;&gt;$20,000 kL^3$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of the bigger areas contains 1000 of the next smaller unit down and can easily be referenced by 3 digits. So if I said Grid 346 that would reference the Grid in that Octant where each of those digits are xyz starting from the Galactic Center. Lower numbers are always closer, higher are always further from the Galactic Center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why not use parsecs and light years? They're really not practical to use. Likewise, AU are weird, and other units you either end up with very long numbers or poor conversions. Also with this unit, I find it very easy to work out distance and time in my head. The answer is 1.07% off, but that's not really an issue when you're not trying to be 100% precise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why use 20kL rather than 10kL or 1kL not many stars in that volume from what I looked at before so it makes sense to use that as division and it also divides up nice to the higher divisions.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2&lt;/strong&gt;&lt;br&gt;&#xA;I use another unit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 brut = &lt;span class=&quot;math-container&quot;&gt;$1*10^{19} kg$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest reason I'm using this is that scientific notation doesn't work in emergencies or common parlance so it makes sense to me to have a larger base unit when dealing with these larger scales. &lt;span class=&quot;math-container&quot;&gt;$1*10^{19} kg$&lt;/span&gt;, if I remember right (cuz I came up with this a few years ago and haven't used it since), comes from when the reference point numbers on the wiki for scale started getting fishy looking to me. Not a hard science thing, but seemed like a good place to have something like this.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3&lt;/strong&gt;&lt;br&gt;&#xA;This is the way I label Stars, planets, moons, etc... because it makes sense. I keep saying that, but you know, it's true ^.^&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SO I - Star (or black hole)&lt;br&gt;&#xA;SO II - Spheroid&lt;br&gt;&#xA;   &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;  Brown Dwarf (Star?)&lt;br&gt;&#xA;   &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;  Gaseous Spheroid - 12,000+ km radius and above (Most of its size is from gases)&lt;br&gt;&#xA;   &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;  Terrestrial Spheroid - 1,500 to 12,000 km radius (Has a rocky mantle)&lt;br&gt;&#xA;   &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;  Icey Spheroid - 200m to 1,500 km radius (Has an ice-water mantle)&lt;br&gt;&#xA;SO III - Debris&lt;br&gt;&#xA;   &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;  Asteroid - Comprised mainly of rock&lt;br&gt;&#xA;   &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;  Comet - comprised mainly of ice-water  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Child Star (?) - SO I that orbits a SO I&lt;br&gt;&#xA;Planet - SO II that orbits a SO I&lt;br&gt;&#xA;Moon - SO II that orbits a SO II&lt;br&gt;&#xA;?? (currently also called a moon it seems) - SO III that orbits a SO II&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SO means Stellar Object.&#xA;both SO and Spheroid are kinda bad names considering nothing but stars are Stellar and stars are Spheroids, but whatever, I think it gets the point across. Though I wish I could think up something better.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Anyways, do any of these 3 things create problems that I don't know about or that could cause a problem that I possibly can't see due to being a pleb in these matters?&lt;/p&gt;&#xA;" OwnerUserId="14084" LastEditorUserId="25994" LastEditDate="2019-03-14T15:29:40.910" LastActivityDate="2019-03-14T15:29:40.910" Title="Stellar Mapping and Coordinates" Tags="&lt;star&gt;&lt;distances&gt;&lt;coordinate&gt;&lt;naming&gt;" AnswerCount="2" CommentCount="8" ClosedDate="2016-09-04T23:49:52.137" ContentLicense="CC BY-SA 4.0" />
  <row Id="18204" PostTypeId="2" ParentId="18203" CreationDate="2016-09-03T19:01:08.407" Score="1" Body="&lt;p&gt;The reason why we centre our galactic coordinates on the sun is that is where we are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What actually is the purpose of a coordinate system‽ I'd like to be able to point to a body and say that it is at coordinates X, Y, Z (with uncertainties in each direction)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we know the direction of the centre of the galaxy, but we don't know it's distance with accuracy. If we were to use the location of the centre of the galaxy for the galactic coordinates &lt;em&gt;we could not give accurate coordinates for any of the local stellar objects&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we had a warp drive and could travel halfway across the galaxy before lunch then it might make sense to use a different set of coordinates. However, as things stand there is nothing wrong with being Earth centric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have mentioned the remainder of the question in comments. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-09-03T19:01:08.407" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18205" PostTypeId="2" ParentId="18203" CreationDate="2016-09-03T21:45:50.827" Score="1" Body="&lt;p&gt;The Galactic coordinate system was established back in the 1950s, before most people had any idea there &lt;em&gt;was&lt;/em&gt; a bar in the Galaxy. And, as James K pointed out, it's a 2D coordinate system for describing the positions of things &lt;em&gt;as seen from the Earth&lt;/em&gt;. You can describe the positions &lt;em&gt;on the sky&lt;/em&gt; (2D coordinates) of just about everything, so you can give them reasonably accurate Galactic coordinates; we only know the distances to a few things, so trying to come up with with 3D positions is difficult at best, and impossible for lots of things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make things worse, the orientation of the bar is still a bit uncertain; different studies still disagree on its position angle at the level of 5 to 10 degrees (occasionally more). So while you could argue that a bar-relative 3D coordinate system might make sense for a galaxy-spanning civilization, it's way too early to try using something like that for contemporary astronomy.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2016-09-03T21:45:50.827" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18206" PostTypeId="2" ParentId="18198" CreationDate="2016-09-04T01:51:09.010" Score="5" Body="&lt;p&gt;There's nothing &quot;cyclic&quot; in the universe Hawking is describing. Rather, Hawking's thought experiment applies to a particular &lt;em&gt;shape&lt;/em&gt; of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Observable_universe&quot; rel=&quot;nofollow&quot;&gt;observable universe&lt;/a&gt; is, for practical purposes, a sphere 93 billion light years in diameter, with us at the centre. Observations suggest that this observable universe is pretty much isotropic and homogenous. If anything exists beyond this sphere, it's so far away that the light from it hasn't had time (i.e. the age of the universe) to reach us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's rational to assume that there &lt;strong&gt;&lt;em&gt;is&lt;/em&gt;&lt;/strong&gt; a universe beyond what we can see. This is called the global universe. Since we can't observe it, we can only theorise what its properties might be. There are three broad possibilities for the shape of an isotropic global universe:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;flat (no curvature)&lt;br&gt;&#xA;  closed (positively curved)&lt;br&gt;&#xA;  open (negativelycurved)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://map.gsfc.nasa.gov/universe/uni_shape.html&quot; rel=&quot;nofollow&quot;&gt;Current observations&lt;/a&gt; indicate that the universe is &lt;em&gt;flat&lt;/em&gt;, with only a 0.4% margin of error - in other words, if it's a simply-connected universe (e.g. not a torus) then it extends infinitely in all directions. Hawking's thought experiment, on the other hand, takes place in a &lt;strong&gt;positively curved&lt;/strong&gt; universe. This can be thought of as a three-dimensional hypersphere. One way to picture this is as though we're an atom in the surface fabric of a balloon: from where we are, everything looks flat and infinite, but in fact the flatness curves away to form a spherical shape. Importantly, this shape is &quot;closed&quot; - in other words, &lt;em&gt;finite&lt;/em&gt;, as is the surface area of a sphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What Hawking is suggesting is that if you set out in a straight line on the surface of this balloon, and so long as your movement away from the starting point is faster than the rate the balloon is expanding, then eventually you'll circumnavigate the balloon and end up where you began. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's purely meant as a demonstration of what a closed (positively curved) universe &quot;looks like&quot;. Given that our local, observable universe appears very close to flat, the curvature would have to be so small that the &quot;balloon&quot; would be truly immense; circumnavigating it would therefore take so long (say, a billion billion years) that our Milky Way would be unrecognisable, if it existed at all, by the time you returned to your starting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be fair to Hawking, in 1988 when his ground-breaking &lt;em&gt;A Brief History of Time&lt;/em&gt; was published, there was good reason to think that the universe was positively curved, since a truly flat universe (curvature exactly zero) seemed an improbable &quot;special case&quot;. However, the discovery in 1998 of the &lt;em&gt;accelerating&lt;/em&gt; expansion of the universe, which provided direct evidence of dark energy and firmly installed Lambda-CDM as the preferred model, rewrote the science and was probably one of the reasons Hawking decided to rewrite his book. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even so, when &lt;em&gt;A Briefer History&lt;/em&gt; was published in 2005, only the earliest data from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Wilkinson_Microwave_Anisotropy_Probe&quot; rel=&quot;nofollow&quot;&gt;Wilkinson Microwave Anisotropy Probe&lt;/a&gt; (WMAP) was available, putting the total density of the universe at 1.02 (allowing a realistic possibility of a slight positive curvature), whereas by 2012 the WMAP data had reduced this to 1.0027 (very close to flat).&lt;/p&gt;&#xA;" OwnerUserId="14094" LastEditorUserId="14094" LastEditDate="2016-09-10T00:51:50.027" LastActivityDate="2016-09-10T00:51:50.027" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18208" PostTypeId="1" CreationDate="2016-09-04T08:19:22.400" Score="2" ViewCount="473" Body="&lt;p&gt;The idea of a 'super massive white hole' is sometimes discussed regarding GR,SR and the destination of material 'lost' in a black hole. Is there any good theoretical or observational evidence for these SWH, or are they merely metaphysical speculations?&lt;/p&gt;&#xA;" OwnerUserId="14078" LastEditorUserId="1891" LastEditDate="2019-09-16T16:10:52.063" LastActivityDate="2019-09-16T16:10:52.063" Title="Is there anything like a supermassive white hole? How are they formed? And where are they present?" Tags="&lt;cosmology&gt;&lt;white-hole&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="18209" PostTypeId="1" AcceptedAnswerId="18212" CreationDate="2016-09-04T10:41:28.727" Score="2" ViewCount="58" Body="&lt;p&gt;Hopefully this will be a quick and easy question, with a quick and easy answer, but why is it that we observe a perigee Moon coincident with a full Moon approximately every 13 months? Is this related to the Earth-Moon dynamics, perhaps the eccentricity of the lunar orbit?&lt;/p&gt;&#xA;" OwnerUserId="2091" LastEditorUserId="2091" LastEditDate="2016-09-04T10:53:13.580" LastActivityDate="2016-09-05T05:45:51.670" Title="Observations of a perigee coincident with a full Moon" Tags="&lt;solar-system&gt;&lt;earth&gt;&lt;observational-astronomy&gt;&lt;the-moon&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18210" PostTypeId="2" ParentId="18208" CreationDate="2016-09-04T15:35:43.123" Score="11" Body="&lt;p&gt;White holes are a theoretical construction of General Relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite extensive searches, nothing has been found: not a supermassive white hole, not a stellar-sized white hole, nothing. Thus, the answer to your question is: on our current knowledge, they don't exist anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is unclear how could they be formed. The geometry of the Kerr-solution (i.e. rotating black holes) suggests a white hole could exist on the &quot;other side&quot; of a black hole, but currently it is pure speculation.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="14094" LastEditDate="2016-09-05T10:42:21.187" LastActivityDate="2016-09-05T10:42:21.187" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="18212" PostTypeId="2" ParentId="18209" CreationDate="2016-09-04T20:02:30.237" Score="2" Body="&lt;p&gt;The time between full moons is called the the synodic month, and the time between perigees is called the anomalistic month. The anomalistic month is shorter than the synodic month by about two days, (27.6 days and 29.5 days)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the anomalistic month is two days shorter, and there are nearly 28 days in a month, it takes about 14 months for the two cycles to come back into alignment. Thus there is a perigee at full moon every &lt;em&gt;14&lt;/em&gt; (not 13) months.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More accurately the beat period between the two cycles is 411.78443 days. See wikipedia: &lt;a href=&quot;https://en.wikipedia.org/wiki/Full_moon_cycle&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Full_moon_cycle&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="9301" LastEditDate="2016-09-05T05:45:51.670" LastActivityDate="2016-09-05T05:45:51.670" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18215" PostTypeId="1" CreationDate="2016-09-05T08:03:31.037" Score="2" ViewCount="255" Body="&lt;p&gt;All matter has a gravitational pull, a ball, a rock, a planet and a star - the bigger the matter the more it affects the space fabric around it thus having a greater gravitational pull.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see black holes in the same kind of concept - matter that warps the space fabric/time around it - however as we can physically see a ball, a rock, a planet and a star I fail to visualize what would be the matter that continues a black hole to operate because we cannot see any physical element to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EXAMPLE&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/SyfIr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SyfIr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A black hole is cause when a star implodes on itself and crushes its self to an extent that gravity has such an effect on space time that we know of nothing that produce such a force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I would like to know is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: Is there a known &quot;bottom&quot; of a black hole? If so, what happens to all of the matter it swallows? surely it would be impossible to crush something so small that it physically no longer exists - where would it go then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: How small does the core of a collapsed star crush itself into? i.e the size of an atom or smaller?&lt;/p&gt;&#xA;" OwnerUserId="7954" LastEditorUserId="7954" LastEditDate="2016-09-05T08:09:34.803" LastActivityDate="2016-09-05T08:09:34.803" Title="Is there mass i.e a physical bottom to a black hole?" Tags="&lt;gravity&gt;&lt;black-hole&gt;&lt;space-time&gt;&lt;space&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2016-09-05T14:07:13.307" ContentLicense="CC BY-SA 3.0" />
  <row Id="18216" PostTypeId="1" CreationDate="2016-09-05T09:40:51.467" Score="2" ViewCount="54" Body="&lt;p&gt;In the Gliese catalog there are a bunch of prefixes, such as GJ, GI, Wo, etc. My question is do the numbers overlap ever? I tried to look it up, but nothing I could find was conclusive to me.&lt;/p&gt;&#xA;" OwnerUserId="14084" LastActivityDate="2016-09-05T09:40:51.467" Title="Gliese Catalog Designation prefixes" Tags="&lt;star-catalogues&gt;" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18218" PostTypeId="1" CreationDate="2016-09-05T19:46:30.327" Score="1" ViewCount="1062" Body="&lt;p&gt;I need to understand how to calculate the angles of the sunrise and sunset viewed from a surface oriented against the azimuth and  the horizon plane, and for any day of the year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I have a surface at this position 40° 43' 30.2&quot;N 73° 59' 37.5&quot;W, oriented towards north with an azimuth angle of 174°(O° is exactly the geometric south), and with a tilt angle of 90°,so the surface is vertical, or perpendicular to the horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could I calculate when it get the sun and when it doesn't during the day? &#xA;What are these angles against the azimuth axis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my example, sun should hit the surface when it rises, then it disappears, it reappears at the end of the day before of its set, mainly in the summer season. So in this case, I need to know four angles: -ws,w1,w2,ws, where -ws, and ws are the regular angle of rise and set, but how to calculate when sun disappears (w1), and appears again to the surface (w2)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know the complete formulas to these calculations? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know any websites, or any web app to do exactly these type of calculations? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the regular case when the same surface in the same place and position but with its normal towards south? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not very interested in the theory but in the practical calculation. I tried the formulas on this website (&lt;a href=&quot;http://www.itacanet.org/the-sun-as-a-source-of-energy/part-3-calculating-solar-angles/&quot; rel=&quot;nofollow&quot;&gt;http://www.itacanet.org/the-sun-as-a-source-of-energy/part-3-calculating-solar-angles/&lt;/a&gt;) but it is not clear what happens if Azs=0°, that is when the surface is towards north. In the calculation of a and b for the formula 3.7 I get at the denominator sin(0)=0, times tan(beta=90°)=inf, if the surface is vertical!! How could I solve this problem?&lt;/p&gt;&#xA;" OwnerUserId="14114" LastEditorUserId="2270" LastEditDate="2016-09-13T16:07:05.087" LastActivityDate="2016-09-13T16:07:05.087" Title="Angles of sunrise and sunset viewed from a surface anyway oriented" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;observational-astronomy&gt;&lt;surface&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18219" PostTypeId="1" AcceptedAnswerId="18221" CreationDate="2016-09-05T21:27:39.353" Score="3" ViewCount="3041" Body="&lt;p&gt;I'm working on a star map program and I'm wondering what are the most common map projections used to print star maps...? I want to map of the entire visible sky from a certain point on the surface of Earth. I have all the visible stars and their Altitude-Azimuth, but I need to map this semi-sphere onto a circle in 2D. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to get something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JRBKp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JRBKp.png&quot; alt=&quot;Star map&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been playing around with equidistant and gnomonic (-1 RE) projections but they feel too distorted. What are the most common projections used for 2D, circular, entire-visible-sky star maps?&lt;/p&gt;&#xA;" OwnerUserId="14117" LastActivityDate="2016-09-05T22:35:20.917" Title="Map projections used for star maps" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18220" PostTypeId="1" AcceptedAnswerId="18225" CreationDate="2016-09-05T21:50:29.837" Score="7" ViewCount="3098" Body="&lt;p&gt;The solar system is thought to have a &lt;a href=&quot;https://en.wikipedia.org/wiki/Heliosphere#Heliopause&quot; rel=&quot;noreferrer&quot;&gt;heliopause&lt;/a&gt; which can be considered its edge for many purposes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;The heliopause is the theoretical boundary where the Sun's solar wind is stopped by the interstellar medium; where the solar wind's strength is no longer great enough to push back the stellar winds of the surrounding stars. This is the boundary where the interstellar medium and solar wind pressures balance.&lt;/blockquote&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Heliopause: the boundary between solar wind and interstellar wind where they are in equilibrium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is there a similar phenomenon at the interface between space dominated by the Milky Way and intergalactic space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is at the edge of the Milky Way galaxy?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user14116" LastEditorDisplayName="user14116" LastEditDate="2016-09-05T22:30:01.973" LastActivityDate="2019-07-03T00:30:36.167" Title="What is at the edge of the galaxy?" Tags="&lt;galaxy&gt;&lt;intergalactic-space&gt;" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18221" PostTypeId="2" ParentId="18219" CreationDate="2016-09-05T22:35:20.917" Score="6" Body="&lt;p&gt;The stereographic projection&#xA;(&lt;a href=&quot;https://en.wikipedia.org/wiki/Stereographic_projection&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;,&#xA;&lt;a href=&quot;http://mathworld.wolfram.com/StereographicProjection.html&quot; rel=&quot;noreferrer&quot;&gt;Wolfram MathWorld&lt;/a&gt;)&#xA;is commonly used for all-sky maps&#xA;(e.g. &lt;a href=&quot;http://www.skymaps.com/&quot; rel=&quot;noreferrer&quot;&gt;Skymaps.com&lt;/a&gt;,&#xA;&lt;a href=&quot;https://www.fourmilab.ch/yoursky/&quot; rel=&quot;noreferrer&quot;&gt;Your Sky&lt;/a&gt;).&#xA;As distance from the center of the map increases,&#xA;constellations grow larger but keep their shapes.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-09-05T22:35:20.917" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18222" PostTypeId="2" ParentId="18220" CreationDate="2016-09-06T00:15:09.617" Score="4" Body="&lt;p&gt;It is argued that the edge of the galaxy has the halo of dark matter at around 100,000 light years, but there are arguments that extend the galaxy from anywhere between it being 120,000 to 200,000. The major factor though for putting the edge at around 100,000 light years is that there is a major star density fall off for no known reason. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly the depth of the galaxy is said to be 10,000 ly at the bulge and 1,000 ly for the rest of the galactic plane. There is similarly no reason for the fall off in star density in that direction known. And there are massive xray and gamma ray structure coming out of the galactic center that is 50,000 ly on either side which you could argue is the real edge of the galaxy along that axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what lies at the edge of the galaxy depends on how you define the edge of the galaxy which could be: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;100k x 100k x 10k  &lt;/li&gt;&#xA;&lt;li&gt;100k x 100k x 100k&lt;/li&gt;&#xA;&lt;li&gt;200k x 200k x 10k &lt;/li&gt;&#xA;&lt;li&gt;200k x 200k x 100k&lt;/li&gt;&#xA;&lt;li&gt;120k x 120k x 10k&lt;/li&gt;&#xA;&lt;li&gt;120k x 120k x 100k&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Or you could use some other arbitrary number, but you'd need to make that argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I however think that a star density definition should be how we should define things &quot;officially&quot; because it is more or less how most people think of the galactic limits in the first place. Of course if you define it like that then you're dealing with issues like the New Outer Arm where the density falls off and then increases, the actual density isn't a perfect shape, and we might have to say that things we call 2 or 3 galaxies are actually 1 galaxy, such as the 2 dwarf galaxies orbiting the Milky Way or the where there exist 2 galaxies with a stream of stars between the two as they merge over millions/billions of years. &lt;/p&gt;&#xA;" OwnerUserId="14084" LastActivityDate="2016-09-06T00:15:09.617" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18223" PostTypeId="1" CreationDate="2016-09-06T03:05:49.507" Score="0" ViewCount="471" Body="&lt;p&gt;We know that the Earth, like a top, slowly precesses as it rotates. We know the Sun rotates, but does it precess as well? What about planetary orbits? Is the orientation of the ecliptic stable or will its inclination change over time? If so, would the orbital inclinations of the planets currently in the ecliptic change independently or in unison? We know the ecliptic is inclined with respect to the galactic plane. Would that have been a consequence of &quot;residual&quot; angular momentum when the solar system formed, or did it gradually take up its present orientation?&lt;/p&gt;&#xA;" OwnerUserId="1983" LastActivityDate="2016-09-06T05:58:43.390" Title="Does the ecliptic precess? Does the Sun's rotation precess?" Tags="&lt;ecliptic&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18224" PostTypeId="2" ParentId="18223" CreationDate="2016-09-06T05:58:43.390" Score="2" Body="&lt;p&gt;As you mentioned, the Earth's rotational axis undergoes a slow precession. This results from tidal torques caused largely by the Sun and the Moon. There would be no precession if the axial tilt with respect to the orbital angular momentum vectors was zero or if the Earth was spherical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun is almost spherical it's oblate to about $10^{-6}$, and is almost perfectly aligned with the ecliptic, it has a relatively slow rate of rotation, and all of the bodies that might cause a precession are far from the Sun, and because the gravitational gradient torque is an inverse cube relationship. All of these factors combined means the precession rate of the Sun is ridiculously small. So the short answer to your question is: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Yes, the Sun precesses.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Planetary orbits are a little different, I'm not exactly sure how to state the process for their precession without waving my hands about the room, so I won't attempt it. However, I will say that, as all things in the universe, I am sure the ecliptic is a dynamic system unto itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the ecliptic (dashed line) and the galactic plane (spotted strip) see the attached image.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/USBZE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/USBZE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;If the ecliptic and the galactic plane were aligned, the two would line up on top of one another, as the Sun's and other planet's paths would be roughly the same path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is still residual angular momentum within the solar system, which is how we arrived at our current orientation amongst the stars.&lt;/p&gt;&#xA;" OwnerUserId="2232" LastActivityDate="2016-09-06T05:58:43.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18225" PostTypeId="2" ParentId="18220" CreationDate="2016-09-06T09:55:14.330" Score="8" Body="&lt;p&gt;The short answer is: probably nothing much, because galaxies are very fuzzy objects without &quot;edges&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the stellar disk, it just fades out in density, with no evidence for a sharp cutoff (this is true of disk galaxies in general). There may or may not be a &quot;break&quot; at some radius, but this just marks the beginning of a steeper falloff in density, not an &quot;edge&quot; in any real sense. The &lt;em&gt;gas&lt;/em&gt; disk extends even further out (that is, further out than we can easily trace the stars in the disk), and doesn't show much evidence for a sharp cutoff either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also a round halo of stars, which is low density but extends to at least 400,000 light years; &lt;a href=&quot;http://www.skyandtelescope.com/astronomy-news/the-most-distant-milky-way-stars-070920142/&quot; rel=&quot;noreferrer&quot;&gt;two stars were recently discovered&lt;/a&gt;  which are plausibly part of the halo and located at distances of about 800,000 light years. (Note that there are &lt;em&gt;other galaxies&lt;/em&gt; -- such as the Large and Small Magellanic Clouds -- which are closer to us and to the center of the Milky Way, so you have to accept that individual galaxies can be found &quot;inside&quot; other galaxies.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then there's the halo of dark matter...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can think of two ways in which you &lt;em&gt;might&lt;/em&gt; talk about the outer boundary of an isolated galaxy (which the Milky Way sort of is) from a theoretical perspective. The first is the idea of a halo &quot;accretion shock&quot;, in which relatively cold gas falling onto the galaxy from intergalactic space runs into the galaxy's halo of hot gas (&quot;hot&quot; = millions of degrees), creating a shock where the infalling gas slows down and the local density of gas goes up. (This is a little bit analogous to the heliopause, though the direction of gas flow is the other way around: the heliopause is where the outflowing solar wind shocks into the more or less stationary interstellar gas.) The problem is that extensive, hot halos of gas are thought to form only around &lt;em&gt;massive&lt;/em&gt; galaxies, and it's not clear the Milky Way is massive enough to qualify. It also won't work for galaxies in massive groups and clusters, where the individual galaxies aren't accreting gas any more (the &lt;em&gt;cluster&lt;/em&gt; is what has the hot halo and the accretion shock).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other possibility is a region where the infalling &lt;em&gt;dark matter&lt;/em&gt; particles merge into the dark-matter halo (whose particles are orbiting within their common gravitational potential, and so are not &lt;em&gt;as a whole&lt;/em&gt; falling towards the center any more). Simulations suggest that as one moves inward the local density of dark matter starts to increase rather rapidly at this point (and the average velocity of the dark matter particles changes), but it's still a fairly smooth process (unlike gas, the dark matter particles aren't &quot;shocking&quot; into each other). There's no way of determining where this is for the Milky Way, other than something really vague like &quot;maybe 1 million light years&quot;?&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2016-09-06T09:55:14.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18226" PostTypeId="2" ParentId="18220" CreationDate="2016-09-06T10:15:34.420" Score="3" Body="&lt;p&gt;While the Solar wind is pretty much homogenous in all directions, the galaxy puffs out stellar winds from discrete and temporary star forming regions. Protostars and giant stars which go supernova young, punch &quot;chimneys&quot; through the interstellar medium in the galactic disk here and there. The Local Fluff is btw our region inside the &lt;a href=&quot;http://www.solstation.com/x-objects/chimney.htm&quot; rel=&quot;nofollow&quot;&gt;Local Chimney&lt;/a&gt;, a relatively low density region formed by pressure from the young and exploding stars in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Scorpius%E2%80%93Centaurus_Association&quot; rel=&quot;nofollow&quot;&gt;Scorpius-Centaurus Association&lt;/a&gt; 400 light years from here. And the supermassive black hole turns its jet on and off depending on when it accretes material. I don't think there's a pan-galactic Milky Way wind that looks anything like the Solar wind. But around an intensely starbursting quasar galaxy maybe?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-09-06T10:15:34.420" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18227" PostTypeId="1" AcceptedAnswerId="18233" CreationDate="2016-09-06T10:31:14.157" Score="1" ViewCount="177" Body="&lt;p&gt;Is it possible, and does it occur naturally, that the light from the same source is simultaneously polarized differently at different wavelengths? Light can be polarized by interstellar magnetic fields. Could it polarize light of different wavelengths differently? Does a pair of polarized sunglasses polarize radio waves and UV in the same way as visible light?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-09-06T16:33:31.393" Title="Is there a spectrum of polarization? Does polarization vary depending on wavelength, from one and the same source?" Tags="&lt;light&gt;&lt;optics&gt;&lt;polarization&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18228" PostTypeId="2" ParentId="18227" CreationDate="2016-09-06T10:45:02.377" Score="2" Body="&lt;p&gt;Yes, astronomers call it &lt;a href=&quot;https://en.wikipedia.org/wiki/Faraday_effect#Faraday_rotation_in_the_interstellar_medium&quot; rel=&quot;nofollow&quot;&gt;Faraday rotation&lt;/a&gt;. The polarization angle of radio waves changes dependent on the wavelength because of the galactic magnetic field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sunglasses do not polarize radio waves, the wavelength is too large. You can polarize radio waves with a wire mesh.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastActivityDate="2016-09-06T10:45:02.377" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18229" PostTypeId="1" AcceptedAnswerId="18280" CreationDate="2016-09-06T11:00:05.633" Score="6" ViewCount="4019" Body="&lt;p&gt;I'm looking for some sort of boundary data to be able to render the milky way on a star map, as visible from Earth. Something that looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qbhpu.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qbhpu.png&quot; alt=&quot;star map&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For that, I need something like a collection of RA hours and DEC degrees of the &quot;boundary points&quot; of what's visible from Earth (technically the Galactic Center), possibly with proper motion too. I'm not looking for precise luminosity data or anything like that, just the points of the blob on the sky that most resembles the Milky Way's shape and position from the Earth. It's important that I want to render the sky map for any given surface point on Earth, for any given time (within the last 100 years at least).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know of a database like that? I've been looking on VizieR but I couldn't find what I was looking for.&lt;/p&gt;&#xA;" OwnerUserId="14117" LastEditorUserId="14117" LastEditDate="2016-09-08T10:56:25.820" LastActivityDate="2016-09-09T18:50:33.743" Title="Milky Way position on the sky" Tags="&lt;milky-way&gt;&lt;data-analysis&gt;&lt;star-catalogues&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18231" PostTypeId="1" CreationDate="2016-09-06T14:31:01.053" Score="0" ViewCount="107" Body="&lt;p&gt;The meaning of the $t$ which appears in space-time intervals or metrics in general relativity. I concluded that $t$ was just a mathematical thing which allows us to label the &quot;space-time manifold&quot; and only proper time τ had a physical meaning too. On Wikipedia I also found that the coordinate time is not a time that could be measured by a clock located at the place that nominally defines the reference frame. What does it really mean?&lt;/p&gt;&#xA;" OwnerUserId="14078" LastEditorUserId="14078" LastEditDate="2016-11-11T17:06:16.003" LastActivityDate="2016-11-11T17:06:16.003" Title="What does the coordinate of time really mean?" Tags="&lt;space-time&gt;&lt;coordinate&gt;&lt;time-dilation&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18232" PostTypeId="1" CreationDate="2016-09-06T16:05:57.537" Score="1" ViewCount="72" Body="&lt;p&gt;While studying the techniques astronomers use to mitigate radio frequency interference (RFI) I found it's often assumed that astronomical signals are (band-limited) Gaussian noise. Is this a reasonable assumption for every type of astronomical observation? Even when searching EM emission from specific elements (e.g. hydrogen line)?&lt;/p&gt;&#xA;" OwnerUserId="14123" LastActivityDate="2016-09-06T16:05:57.537" Title="Can all astronomical signals be considered Gaussian noise?" Tags="&lt;radio-astronomy&gt;&lt;spectroscopy&gt;&lt;spectrometry&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18233" PostTypeId="2" ParentId="18227" CreationDate="2016-09-06T16:17:00.460" Score="3" Body="&lt;p&gt;Yes. Radiation may be polarized when it is scattered, and different media scatter different wavelengths differently.&lt;/p&gt;&#xA;&lt;p&gt;For instance, neutral hydrogen scatters Lyman &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; radiation (i.e. photons coming from the &lt;span class=&quot;math-container&quot;&gt;$2$$\rightarrow$$1$&lt;/span&gt; transition of hydrogen). The degree of polarization is coupled to the &amp;quot;phase function&amp;quot; which determines in which direction it is scattered. If a Ly&lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; photon is close to the line center (which is at 1215.67 Ångström), it is scattered isotropically (i.e. equal probability of being scattered in any given direction), and will not be polarized, but if it is more than a few Doppler widths from the line center, it is scattered &lt;em&gt;an&lt;/em&gt;isotropically, and will then be polarized. Maximum polarization occurs at 90º scattering, where &lt;span class=&quot;math-container&quot;&gt;$3/7 \simeq 43\%$&lt;/span&gt; of the light will be polarized.&lt;/p&gt;&#xA;&lt;p&gt;If the photons are even farther from the line center, they won't be scattered at all (because they're no longer in &lt;em&gt;resonance&lt;/em&gt; with the &lt;span class=&quot;math-container&quot;&gt;$2$$\rightarrow$$1$&lt;/span&gt; transition), and the radiation is hence unpolarized.&lt;/p&gt;&#xA;&lt;strong&gt;Example of polarization at a particular wavelength&lt;/strong&gt;&#xA;&lt;p&gt;The figure below, taken from &lt;a href=&quot;http://arxiv.org/abs/1108.3332&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hayes et al. (2011)&lt;/a&gt;, shows a so-called &amp;quot;Lyman &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; blob&amp;quot;, which is a source of Lyman &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; that is quite a lot larger than a galaxy. There was some debate whether the radiation is created outside a central galaxy from gas accreting onto the galaxy and cooling — a signature of galaxy formation — or whether it is centrally produced (from stars) radiation that escapes the galaxy and subsequently scatters on the surrounding gas. If the latter scenario is true, you would expect the Lyman &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; to be polarized to some degree in sort of a ring around the center, because this light would be scattered at roughly 90º toward you. In the figure, the blue cross marks the center, and the red lines' direction mark the direction of polarization and their length mark the degree of polarization (if it were fully polarized, they would have the length of the red bar in the upper right corner).&lt;/p&gt;&#xA;&lt;p&gt;Indeed it seems that the light is polarized around a central source.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5eB9G.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5eB9G.png&quot; alt=&quot;LAB1&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-09-06T16:33:31.393" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18234" PostTypeId="1" AcceptedAnswerId="18236" CreationDate="2016-09-06T17:23:17.570" Score="3" ViewCount="1024" Body="&lt;p&gt;When a meteor hit the surface of a planet, it creates an impact crater. When an meteor hit the surface of a planet at an angle, like it usually happens in the movies, what do you call the long &quot;ditch&quot; that is formed before the actual crater where the meteor came to rest?&lt;/p&gt;&#xA;" OwnerUserId="14124" LastActivityDate="2016-09-07T06:58:55.407" Title="What's the terminology of a meteor &quot;ditch&quot; when impact happens at an angle?" Tags="&lt;meteor&gt;&lt;terminology&gt;" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18235" PostTypeId="2" ParentId="18231" CreationDate="2016-09-06T17:37:44.033" Score="2" Body="&lt;p&gt;First, imagine that you're in an inertial reference frame. We can think of this as a sort of lattice throughout space, with a synchronized clock at each lattice point in space. Now imagine two events, $A$ and $B$, which occur at space-time coordinates $(t_A, x_A, y_A, z_A)$ and $(t_B, x_B, y_B, z_B)$. The reference frame has (again, synchronized) clocks at $(x_A, y_A, z_A)$ and $(x_B, y_B, z_B)$. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Coordinate_time&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;coordinate time&lt;/em&gt;&lt;/a&gt; $\Delta t_{AB}$ is the difference in time measured by the clock at $B$ and the clock at $A$, so that $\Delta t_{AB}=t_B-t_A$. If $\Delta t_{AB}=0$, the events are &lt;em&gt;simultaneous&lt;/em&gt; in that reference frame. However, an observer in a different inertial frame (say, in one with a boost $\vec{\beta}$ in the $x$-direction) might measure a different $\Delta t_{AB}$. This means that coordinate time is &lt;em&gt;frame-dependent&lt;/em&gt; - it depends on the reference frame you're in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Proper_time&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Proper time&lt;/em&gt;&lt;/a&gt; is sometimes called &quot;wristwatch time&quot;. Now, there's a single clock, and it measures the time between two events $A$ and $B$ on a single worldline. This proper time $\tau_{AB}$ does &lt;em&gt;not&lt;/em&gt; depend on your reference frame, but it does depend on the path you take from $A$ to $B$ - that is, it's &lt;em&gt;path-dependent&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to think of the difference is that coordinate time tells you something about one reference frame relative to another (because it's frame-dependent), while proper time tells you something about the worldline of the observer (because it's path dependent).&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-09-06T18:34:20.047" LastActivityDate="2016-09-06T18:34:20.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18236" PostTypeId="2" ParentId="18234" CreationDate="2016-09-06T18:06:23.953" Score="6" Body="&lt;p&gt;I think the answer to this is that it doesn't have a name because such &quot;ditches&quot; don't occur for real meteor impacts. In movies and tv shows, these ditches are often shown as being caused by the meteor first hitting the ground at some angle, then sliding across the ground, digging out a ditch along the way, until they come to their final resting point where the crater is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you stop to think about it, this scenario doesn't really make sense. These types of images (e.g., &lt;a href=&quot;https://nccasymposium.bournemouth.ac.uk/2005/mscca/DIGGINS_html/images/arm_paris_destroyed.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt; or &lt;a href=&quot;http://i1.wp.com/guardiansofgeek.com/wp-content/uploads/2014/12/Crater-City.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;) usually show a thin tail or ditch where the meteor first impacts which gets wider and deeper as the meteor scrapes on the ground, resulting in a final, large crater. But, the most energy a meteor imparts to the ground comes from the first instance it hits. If it were to scrape along the ground as portrayed in many movies, the crater would come from the point it hit the ground and the tail would be in the direction the meteor was traveling. The &quot;ditch&quot; or &quot;tail&quot; would get thinner and shallower as the meteor lost energy scraping across the ground. Generally the opposite is shown in movies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But all that is moot anyway because meteors don't scrape along the ground at impact. That would just violate conservation of momentum and energy. You really have one of two scenarios.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The meteor hits the ground with such force and energy that it mostly destroys itself and you wind up with just a crater. Such an example is the meteor crater in Arizona - in a stroke of brilliance called &lt;a href=&quot;https://en.wikipedia.org/wiki/Meteor_Crater&quot; rel=&quot;nofollow noreferrer&quot;&gt;Meteor Crater&lt;/a&gt;. I would hazard a guess that almost all meteor impacts are of this form, be it on Earth or elsewhere.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8cOza.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8cOza.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;The other option is that the meteor isn't destroyed when it hits the ground. In this case, rather than dragging or scraping on the ground, your meteor will skip like a rock on water and create another crater on its second impact. In theory, it will keep skipping until it is finally destroyed or runs out of energy. This scenario requires that the impacting meteor is hitting at a very shallow angle with very little force and generally means that it is hitting a smaller body (but larger than itself) such as an asteroid/comet/moon. An example would be the &lt;a href=&quot;https://en.wikipedia.org/wiki/Messier_(crater)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Messier Crater&lt;/a&gt; on the Moon, linked by Andy in his comment. These types of craters are unlikely though, as they require pretty specific conditions to occur.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/etgp2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/etgp2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2016-09-06T18:11:35.027" LastActivityDate="2016-09-06T18:11:35.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18237" PostTypeId="2" ParentId="18234" CreationDate="2016-09-06T18:06:52.077" Score="3" Body="&lt;p&gt;They considered a type of crater. If you dislike the word crater, you could describe it as an elongated impact structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are known on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Schiller_(crater)&quot; rel=&quot;nofollow&quot;&gt;Moon&lt;/a&gt; and &lt;a href=&quot;http://www.esa.int/Our_Activities/Space_Science/Mars_Express/Mars_s_mysterious_elongated_crater&quot; rel=&quot;nofollow&quot;&gt;Mars&lt;/a&gt;, but are rather rare. On the Earth, an impactor at a lower angle will have to travel through a lot of air before reaching the ground. Making it more likely that the impactor will be broken up, and destroyed in the air, and so not form a crater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elliptical craters are known on Earth, such as the &lt;a href=&quot;http://www.passc.net/EarthImpactDatabase/mattwilson.html&quot; rel=&quot;nofollow&quot;&gt;Matt Wilson Crater&lt;/a&gt; Or the debated &lt;a href=&quot;http://www.passc.net/EarthImpactDatabase/riocuarto.html&quot; rel=&quot;nofollow&quot;&gt;Rio Cuarto craters&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in very shallow impacts, you won't get &quot;ditch&quot; being dug. That is only for the movies.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-09-06T18:06:52.077" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18241" PostTypeId="2" ParentId="18234" CreationDate="2016-09-06T23:07:03.003" Score="2" Body="&lt;p&gt;This isn't scientific, but in Ringworld by Larry Niven, a spaceship crashes at an angle that produces your 'crater'. The wording used by Larry Niven is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;We are in a furrow plowed by our own landing.&quot;&lt;/p&gt;&#xA;" OwnerUserId="14129" LastEditorUserId="14124" LastEditDate="2016-09-06T23:47:58.393" LastActivityDate="2016-09-06T23:47:58.393" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18242" PostTypeId="1" CreationDate="2016-09-07T03:12:40.307" Score="5" ViewCount="482" Body="&lt;p&gt;How did the Milky Way and Andromeda Galaxy separate before they began their collision course?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that gravity is pulling together the two galaxies, but what separated them in the first place while gravity is pulling them together?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your consideration.&lt;/p&gt;&#xA;" OwnerUserId="14131" LastActivityDate="2016-09-08T08:53:05.060" Title="How did the Milky Way and Andromeda Galaxy separate before they began their collision course?" Tags="&lt;gravity&gt;&lt;galaxy&gt;&lt;cosmology&gt;&lt;milky-way&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18243" PostTypeId="1" CreationDate="2016-09-07T03:56:40.413" Score="4" ViewCount="720" Body="&lt;p&gt;If we imagine a sphere of radius one light-year centered on Earth, it contains only one star - our Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we imagine a sphere of radius ten light-years centered on Earth, it contains about 10 stars, according to &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_nearest_stars_and_brown_dwarfs&quot; rel=&quot;nofollow&quot;&gt;this wikipedia page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we imagine a sphere of radius one hundred light-years centered on Earth, about how many stars would it contain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we imagine a sphere of radius one thousand light-years centered on Earth, about how many stars would it contain?&lt;/p&gt;&#xA;" OwnerUserId="1983" LastActivityDate="2016-09-07T13:56:57.727" Title="What is the density of stars in our region of the galaxy, up to say, 1000 light years distant?" Tags="&lt;galaxy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18244" PostTypeId="2" ParentId="18242" CreationDate="2016-09-07T04:47:09.970" Score="0" Body="&lt;p&gt;The short answer is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Inflation&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The two galaxies formed a great distance apart from one another, almost a billion years after the big bang. They are as close today as they have ever been.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VwRBt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VwRBt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2232" LastActivityDate="2016-09-07T04:47:09.970" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18246" PostTypeId="2" ParentId="18243" CreationDate="2016-09-07T05:23:50.990" Score="-1" Body="&lt;p&gt;Sol is about in the middle betwen the center and the classically defined edge. We have a density of about 1/4ly stars. It gets much denser within 5kly of the center of the galaxy, not sure how the density falls towards the edge. We also have the spiral arms which are density waves which are less dense outside of them...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond this there isn't much more of an exact answer due to not be able to see beyond 100 or 1000pc of Sol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as density beyond that, just look up the equation for volume of a cylinder, you can get that from the density in the wiki page if you find the volume of a sphere or cylinder (which is what you need beyond 1000ly radius) and then just divide.&lt;/p&gt;&#xA;" OwnerUserId="14084" LastActivityDate="2016-09-07T05:23:50.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18247" PostTypeId="2" ParentId="70" CreationDate="2016-09-07T06:18:01.047" Score="-1" Body="&lt;p&gt;Well, I have no astronomical background, so in layman's terms, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If nothing travels at the speed of light, except light, how can a black hole also pull light into itself?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe OP is confused over how a gravity wave can capture light if both can travel at most at the speed C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not work quite that way. Say you move a massive BH from position X to position Y and an observer is at 1 LightYear away from Y. Then, the observer will not feel the gravitation effect of BH until 1 year because that's how much it will take a gravity waves to reach to observer. That is, its changes in gravitational field &quot;Delta&quot; propagate at the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As others have mentioned, gravity bends/shapes the spacetime fabric, so after one year, at observer, spacetime is bend according to the gravity of BH at Y. Thus, any light that passes near to the observer will bend as it follows the spacetime. Plus, the curvature of space BH create in their vicinity is so severe that space is wrapped completely around itself and hence light can not escape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refer to &lt;a href=&quot;http://clarkplanetarium.org/cosmic-quiz-why-cant-light-escape-a-black-hole/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; for simple explanation.&lt;/p&gt;&#xA;" OwnerUserId="14132" LastEditorUserId="69" LastEditDate="2016-11-09T16:45:51.750" LastActivityDate="2016-11-09T16:45:51.750" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18248" PostTypeId="2" ParentId="18234" CreationDate="2016-09-07T06:58:55.407" Score="1" Body="&lt;p&gt;Whilst it is admittedly not a scientific text, here's a quotation from the 1977 novel &lt;em&gt;Lucifers Hammer&lt;/em&gt; by Larry Niven and Jerry Pournelle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Dr. Sharps says that all craters are circular.&quot; Harvey posed above the&#xA;  cake pan with a lug nut in his fingers; he let it drop. Flour sprayed.&#xA;  &quot;Whatever the velocity or the mass or the angle of flight of a meteor,&#xA;  it leaves a circle. I think he's right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9235" LastActivityDate="2016-09-07T06:58:55.407" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18249" PostTypeId="1" AcceptedAnswerId="18251" CreationDate="2016-09-07T07:41:38.773" Score="17" ViewCount="2747" Body="&lt;p&gt;Is it possible (however unlikely) to have a meteorite miss Earth so narrowly and at such flat angle that it would enter atmosphere, travel few (dozens, hundreds, thousands?) kilometers getting as close as few kilometers to the ground without hitting anything and then leave atmosphere without any significant, leaving millions of humans staring agape at the sky?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or does gravity and air friction prevent this by either making such flat trajectory impossible or by slowing down meteorite enough to make it fall to the ground?&lt;/p&gt;&#xA;" OwnerUserId="14133" LastActivityDate="2016-09-07T08:15:59.383" Title="Meteor traveling through atmosphere without hitting Earth" Tags="&lt;meteorite&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18251" PostTypeId="2" ParentId="18249" CreationDate="2016-09-07T08:15:59.383" Score="19" Body="&lt;p&gt;yes it is possible,they are called Earth grazers or Earth-grazing fireball.&#xA;they are not rare but only few incidents are recorded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for more details read this Wikipedia page &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth-grazing_fireball&quot;&gt;https://en.wikipedia.org/wiki/Earth-grazing_fireball&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14102" LastActivityDate="2016-09-07T08:15:59.383" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18252" PostTypeId="1" CreationDate="2016-09-07T08:51:07.880" Score="1" ViewCount="94" Body="&lt;p&gt;I just want to know if I can be part of SETI project. I have gone through SETI website but still didn't get enough guidance. Apart from installing boinc software for letting my machine used as part of seti , is there any other way round through which I can explore something about Extra terrestrial life whether through institutions or as part of any community, I simply do not want to be part of community which believe green men are visiting earth but want to part of genuine one. I can give inputs in computing , programming , communications as engineer.&lt;/p&gt;&#xA;" OwnerUserId="14136" LastEditorUserId="2365" LastEditDate="2017-10-01T14:36:20.280" LastActivityDate="2017-10-01T14:36:20.280" Title="What are the steps I have to follow to join SETI programe from my location (India)?" Tags="&lt;seti&gt;" AnswerCount="0" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18253" PostTypeId="2" ParentId="18242" CreationDate="2016-09-07T10:08:23.593" Score="7" Body="&lt;p&gt;I thought it was a fun question, I gave it some thought.  Layman's version, so corrections are welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Chronology_of_the_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;very early universe&lt;/a&gt;, the first few seconds or so, it was quite dense.  This is fairly well publicized and the specifics and details don't really matter for this question, but the energy and stuff in the early universe that later formed the Milky-way and the energy and stuff that formed Andromeda were cosmic neighbors and in the first seconds of the universe, quite close to each other.  The basic Hubble rule is that things that are close (in cosmological terms) were always close, like the dots on a balloon that gets blown up.  Everything moves away from everything else, but things that were next to each other were always next to each other.  On smaller scales, like stars within galaxies or planets, this isn't true because of gravitational assists and different orbital paths, but its true for galaxies.   (there might be some small variation on this with &quot;orbiting&quot; dwarf galaxies, but it's still by and large, true).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another important point to make about the early universe is that it was much more uniform, so, because of the uniformity, gravity was pulling nearly equally in all directions.   The nearby somewhat more massive clouds of matter and energy didn't dominate gravitationally because there was similar mass in all directions.    You might think that a denser universe means stronger gravity in the direction of nearby matter, but that doesn't happen if the matter is spread evenly enough in all directions.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, generally speaking, in the very young universe, the stuff that became the Milky way and the stuff that became Andromeda started out much closer, but they didn't pull each other much early on because of uniformity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over time, local gravity overcame uniformity.  I couldn't tell you when that happened (maybe someone here can), but it took some time before the makings of galactic groups and clusters began to form.   We know from the background radiation map that the Universe was still quite uniform as late as 380,000 years after the big bang.   See picture: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3IQX4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3IQX4.jpg&quot; alt=&quot;http://abyss.uoregon.edu/~js/images/Full_m.jpg&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Edit due to correction).  And at 380,000 years, the known Universe is thought to be about 1/1100th it's current radius.   We know from the cosmic background radiation that at that time, the Universe was still largely uniform, though non-uniformity did lead to the formation of galaxies, clusters and super-clusters, so it was important, but it still took some time for local groups and galactic clusters to form.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://wmap.gsfc.nasa.gov/universe/bb_cosmo_fluct.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On this map, the hot regions, shown in red, are 0.0002 Kelvin hotter&#xA;  than the cold regions, shown in blue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;These cosmic microwave temperature fluctuations are believed to trace&#xA;  fluctuations in the density of matter in the early universe, as they&#xA;  were imprinted shortly after the Big Bang. This being the case, they&#xA;  reveal a great deal about the early universe and the origin of&#xA;  galaxies and large scale structure in the universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But once local gravity overtook uniformity, then galaxies can start drawing each other together more efficiently.   The spiderweb pattern of galaxies shows that there is a tendency for this to happen, as galaxies draw other galaxies towards them, even as space expands, so you get lines of more dense regions with more galaxies and pockets of empty space, mostly free of galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lFnDf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lFnDf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/198138/is-most-of-the-matter-in-the-observable-universe-within-galaxies&quot;&gt;Source of picture for further reading&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what basically happened with Andromeda and the Milky-way is that expansion combination with uniformity set them fairly far apart even as they were still forming, but they were still next door neighbors as far as galaxies go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both the Milky-way and Andromeda have absorbed smaller, closer dwarf galaxies, perhaps many times, so they weren't the closest 2, but they were the closest 2 non dwarf galaxies (at least, that we know of).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might think that galaxies could orbit other galaxies, but the nature of Hubble expansion doesn't give galaxies much tangential velocity towards other nearby galaxies, so basically, 2 galaxies are either moving towards each other due to mutual gravity or away from each other due to expanding space.   Now in a 3 or more galaxy system with similar masses, they can get some tangential velocity, but generally speaking, the direction between any 2 large galaxies is mostly either towards or away, not an orbit.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to keep in mind, once the uniformity effect is overtaken by local gravitation of local groups or clusters, is that massive objects at these distances don't exactly pull each other quickly.   It takes a long time for gradual acceleration to turn into relevant velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At our current distance from Andromeda (about 2.5 million light years), and it's current mass (about 1.5 trillion solar masses), it's G force on us, using the formula below (Math can be added if needed):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RUpD3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RUpD3.png&quot; alt=&quot;http://study.com/cimages/multimages/16/gravityequation.png&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;comes to about 1/2.8 trillionths of a meter per second squared.  That's less than 1/20,000th the gravitational pull Pluto has on Earth.   A pull that small is going to accelerate things towards each other extremely slowly, so (I think) for the first billion or few billion years, the expansion of space between the Milky way and Andromeda likely exceeded any gravitationally caused velocity between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The good news is, Andromeda and the Milky-way have had a long time to pull on each other.  About 13 billion years or so.   1/2.8 trillion m/s^2 over a billion years is works out to about 11 km/s.  Over 13 billion years, 143 km/s.   And, the actual velocity that the Milky-way is moving towards Andromeda isn't too far away from that, about 110 km/s.  That right there gives us a ballpark Netwonian estimate for how far apart they were when they started.   (Now, footnote, for objects of somewhat similar mass, you have to take the mass of both of them, not just the larger one, so it's closer to 15 or 16 km/s per billion years), but that doesn't change the overall numbers too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark Energy expansion (which currently, between the Milky-way and Andromeda is about 60 km/s), and that makes the calculations a bit more tricky, especially since dark energy expansion might not have been constant throughout the last 13 billion years.  Related article &lt;a href=&quot;http://www.universetoday.com/118367/why-is-andromeda-coming-towards-us/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the current rate they are moving together, about 110 km/s or, 250,000 mph or 1/2,680 c, In one billion years at current velocity Andromeda will be 375,000 light years closer (and, in reality, a bit more as it will continue to accelerate by mutual gravity and dark energy expansion will be reduced as they get closer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if we work backwards, a billion years ago it was a bit less than 375,000 light-years further away than it's current 2.5 million light year distance, with diminishing distances further away over previous billion year intervals.   The farther back you go, and difficulties with knowing how close they were when local gravity became a key factor over uniformity makes estimates pretty difficult, but they were probably never too much further apart than they are now.  Perhaps double, as a very rough guess.   Because gravitational acceleration drops off at the square of the distance, I find it hard to believe that Andromeda was ever much more than twice as far as it is now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would imagine, based on my assumptions above, that by the time the galaxies were recognizable as galaxies, say, when the universe was a billion or 2 billion years old (give or take), Andromeda and the Milky way were probably moving away from each other initially, but over time, the gravitational attraction was able to overcome overcome the expansion of space.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-09-07T12:23:14.263" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18254" PostTypeId="2" ParentId="18243" CreationDate="2016-09-07T13:56:57.727" Score="6" Body="&lt;p&gt;Here's a crude way of estimating the answers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The density of stars in the Milky Way disk decreases exponentially as you go out in radius and also exponentially as you go out in height above/below the disk midplane. The corresponding scale lengths are about 3 kpc (3 kiloparsecs = about 10,000 ly) radially and about 300 parsecs (about 1,000 ly) vertically (&quot;scale height&quot;); this means that if you go out in radius by about 10,000 ly, or up/down in height by about 1,000 ly, the density decreases by a factor of e (2.718...). Changing radius by about one tenth of the scale length or less, or changing height by about one tenth of the scale height or less, only changes the local density by about 10%, which isn't enough to worry about. That corresponds to about 300 parsecs (1000 ly) in radius, or about 100 ly in height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the for case of a 100-ly radius, you can assume the density everywhere is about the same as the local density suggested by the table you linked to. That density is 60 stars / ($(4/3) \, \pi \, 16.3^3$) = 0.0033 stars per cubic ly. For a sphere of radius 100 ly, you would then expect about 14,000 stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naively, a sphere with a radius of 1000 ly would then have about 14 million stars. But a sphere that big extends above and below the disk midplane enough so that it samples about 1 scale height. So you can't just re-use the local density. Without getting into the proper approach of integrating the density functions in 3D, we can make a crude estimate as follows: If we used the density at one scale height (1/e times the local density), then the sphere would have only about 4.5 million stars. If we split the difference, that's about 10 million stars, which is probably not too terribly wrong. (I've been ignoring things like stars belonging to the &quot;thick disk&quot; and stars belonging to the halo, but they won't contribute a whole lot; also, the Sun isn't exactly at the disk midplane, but about 60 or 70 ly above it...)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2016-09-07T13:56:57.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18257" PostTypeId="1" AcceptedAnswerId="18266" CreationDate="2016-09-07T18:13:27.530" Score="4" ViewCount="670" Body="&lt;p&gt;What is the biggest possible crater from an asteroid impact? Not asking for the biggest existing crater in the solar system. I should be able to find plenty of lists online. What I'm curious about is how big a crater can form before the impact results in both bodies reforming destroying near all evidence of the crater itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't mind erosion or normal geological processes  destroying the craters. That is just a matter of time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg when the Mars size object hit the Earth. Both were substantially altered and incorporated in each other. Earth experienced a near complete reformation (in the spherical sense). Also resulted in the moon's formation. Obviously that sort of impact doesn't leave an 'obvious' crater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I assume the makeup (mineral, size, speed, angle of impact etc) of both bodies will change the answer, so maybe a just a short list of sizes for the various &lt;em&gt;xommon&lt;/em&gt; combinations?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg what would be the biggest crater possible on Earth before the Earth experienced severe...Um, reshaping?&lt;/p&gt;&#xA;" OwnerUserId="13680" LastActivityDate="2016-09-08T11:42:57.760" Title="How big can a crater get?" Tags="&lt;asteroids&gt;&lt;crater&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18258" PostTypeId="1" AcceptedAnswerId="18265" CreationDate="2016-09-07T22:26:36.370" Score="0" ViewCount="275" Body="&lt;p&gt;During stellar nucleosynthesis a star will convert a portion of its hydrogen atoms into helium.  Is there a constant of how many tons of hydrogen is converted based on the mass of the star?  Is it based on a percentage of the star's mass? Or can it vary? A Nasa blurb stated that 600 million tons of hydrogen per second are consumed by our own sun. Is there a mathematical formula that can tell you how much any star would consume and convert to helium?&lt;/p&gt;&#xA;" OwnerUserId="13849" LastActivityDate="2016-09-10T05:10:27.107" Title="How much star can star fusion fuse if star fusion can fuse hydrogen?" Tags="&lt;star&gt;&lt;nucleosynthesis&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18259" PostTypeId="1" CreationDate="2016-09-07T23:20:51.440" Score="4" ViewCount="389" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Do any numbers in between &lt;strong&gt;80&lt;/strong&gt; and &lt;strong&gt;85&lt;/strong&gt; (or thereabouts) hold any &lt;em&gt;relatively important&lt;/em&gt; or &lt;em&gt;relatively obvious&lt;/em&gt; meaning in &lt;strong&gt;ancient&lt;/strong&gt; astronomy ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A few clarifications:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;By &lt;em&gt;ancient&lt;/em&gt; I mean &lt;em&gt;related to human antiquity&lt;/em&gt;. (Thus, for instance, although in modern times there are indeed 88 known constellations, almost half of them were unfortunately &lt;em&gt;unknown to the ancients&lt;/em&gt;). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;em&gt;meaning&lt;/em&gt; in question could refer, for instance, to the number of time periods in a specific time cycle, or to the number of stars in a certain constellation, etc. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By &lt;em&gt;relatively important&lt;/em&gt; or &lt;em&gt;relatively obvious&lt;/em&gt; I mean something as self-evident and meaningful  to an &lt;strong&gt;ancient&lt;/strong&gt; sage as the fact that there are roughly 360 days a year, or that the number of constellations in the northern and southern hemispheres are 21 and 15, respectively (as known to the ancient Greeks, for instance). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some background information:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About a century ago, an &lt;a href=&quot;http://en.wikipedia.org/wiki/Sarmizegetusa_Regia&quot; rel=&quot;nofollow noreferrer&quot;&gt;archaeological site&lt;/a&gt;, located in a clearing in the woods, and dating back to the first two centuries BC, was discovered at 45.62°N and 23.31°E, at an altitude of roughly 1,000 meters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.georggerster.com/sites/default/files/styles/default_xl/public/photos/159_RUMANIA_Sarmizegetusa.tif.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://www.georggerster.com/sites/default/files/styles/gallery_s/public/photos/159_RUMANIA_Sarmizegetusa.tif.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure, oriented on a NNW-SSE axis, consists of four concentric circles, the second and fourth of which (counted from the exterior towards the interior) have a clear calendrical and astronomical interpretation, pointing to a semester of 180 days, divided into exactly six months of thirty days each, and to a number of 21 + 13 constellations in the boreal and austral hemispheres, comparable to the figure given by the astronomer &lt;a href=&quot;http://en.wikipedia.org/wiki/Ptolemy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Claudius Ptolemy&lt;/a&gt; in his famous &lt;a href=&quot;http://en.wikipedia.org/wiki/Almagest&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Almagest&lt;/em&gt;&lt;/a&gt;. I was wondering whether the third circle, consisting of a number of no less than 82, and no more than 84 pillars, might not also have a similar meaning. Now, don't get me wrong, interpretations &lt;em&gt;abound&lt;/em&gt;, but most of them tend to be either slightly forced, or are outright pseudo-science. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an interesting aside, there were, initially, only 68 pillars still standing, which just so happened to be double the number of those which constitute the inner horseshoe (as it is called), making many archaeologists speculate that its builders were aware of the fact that after 68 semesters of 180 days, corresponding to a number of 21 + 13 = 34 years of 360 days, the calendrical error would amount to &lt;em&gt;almost&lt;/em&gt; an entire semester. All nice and tidy, but this proposed solution suffered a devastating blow when further excavations, performed by the son of the man who initially discovered the site (both were historians by profession) revealed that the number of pillars comprising the third circle was much higher than previously thought. Furthermore, a certain scientist has already started speculating that the number of elements should be about 84 &lt;em&gt;long before&lt;/em&gt; the excavation in question even happened in the first place! (He based his ideas on the fact that if the second circle represents the solar year, and the fourth one basically corrects it, then there should also be a refinement to the nice-but-inaccurate 30 day month as well). However, even this little gem of a theory has a &lt;em&gt;small&lt;/em&gt; and &lt;em&gt;insignificant&lt;/em&gt; scratch: One of the main reasons why 84 year lunar cycles were even considered in the first place was because its value is a multiple of 28, the latter representing the number of years in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Perpetual_calendar&quot; rel=&quot;nofollow noreferrer&quot;&gt;perpetual calendar&lt;/a&gt; synchronizing the &lt;em&gt;seven day week&lt;/em&gt; to the solar cycle... Just one &lt;em&gt;minor&lt;/em&gt; glitch, though: The civilization that built the structure had months of thirty days, divided into five &lt;em&gt;six-day weeks&lt;/em&gt;, as is evidenced by the clear and uniform distribution of the pillars forming the second circle. (Certainly, it is not &lt;em&gt;completely&lt;/em&gt; impossible for them to have borrowed such a time-correcting mechanism from &lt;em&gt;other&lt;/em&gt; neighboring cultures, but it's not &lt;em&gt;highly&lt;/em&gt; likely either). So, &lt;em&gt;herein&lt;/em&gt; lies the dilemma... &lt;/p&gt;&#xA;" OwnerUserId="14152" LastEditorUserId="2365" LastEditDate="2018-06-13T16:36:00.310" LastActivityDate="2018-07-16T09:48:56.963" Title="Numbers 80 to 85 in Ancient Astronomy" Tags="&lt;fundamental-astronomy&gt;&lt;history&gt;" AnswerCount="1" CommentCount="14" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="18260" PostTypeId="1" AcceptedAnswerId="18261" CreationDate="2016-09-08T06:40:28.703" Score="17" ViewCount="7226" Body="&lt;p&gt;As like Earth, does the moon have a day (24 hours). If it has, how many hours are in a moon day?.&lt;/p&gt;&#xA;" OwnerUserId="11792" LastEditorUserId="11792" LastEditDate="2016-09-25T14:24:53.340" LastActivityDate="2016-09-25T14:24:53.340" Title="Does the moon have days?" Tags="&lt;time&gt;&lt;the-moon&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18261" PostTypeId="2" ParentId="18260" CreationDate="2016-09-08T07:00:26.653" Score="21" Body="&lt;p&gt;Defining that a day is a rotation on the axis, 1 moon day is approximately equal to 27.3 Earth days. This link on &lt;a href=&quot;http://www.universetoday.com/48788/moon-rotation/&quot; rel=&quot;noreferrer&quot;&gt;moon rotation&lt;/a&gt; may interest you.&lt;/p&gt;&#xA;" OwnerUserId="7926" LastEditorUserId="10522" LastEditDate="2016-09-10T10:44:45.503" LastActivityDate="2016-09-10T10:44:45.503" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18262" PostTypeId="2" ParentId="18260" CreationDate="2016-09-08T08:12:03.503" Score="49" Body="&lt;p&gt;The answer depends on your definition of a day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you define a day as we usually define it in the Earth (time between the Sun is at noon or average time between sunrises, a 24 hours day in Earth), the length of a day in the Moon is the &lt;a href=&quot;https://global.britannica.com/topic/synodic-period&quot; rel=&quot;nofollow&quot;&gt;synodic period&lt;/a&gt; of the Moon and it takes &lt;a href=&quot;https://en.wikipedia.org/wiki/Moon&quot; rel=&quot;nofollow&quot;&gt;29.530589 days (29 d 12 h 44 min 2.9 s)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, if you define a day as rotation on own axis, that is, time between a far away star being at noon, you have what in Earth is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Sidereal_time&quot; rel=&quot;nofollow&quot;&gt;sidereal day&lt;/a&gt; (23 hours, 56 minutes in Earth). In the moon the day defined this way would be &lt;a href=&quot;https://en.wikipedia.org/wiki/Moon&quot; rel=&quot;nofollow&quot;&gt;27.321661 days (27 d 7 h 43.19 min 11.5 s)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please notice that since the Moon is tidally locked to the Earth, rotation periods equal orbital periods.&lt;/p&gt;&#xA;" OwnerUserId="14097" LastEditorUserId="14097" LastEditDate="2016-09-25T12:58:43.530" LastActivityDate="2016-09-25T12:58:43.530" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18263" PostTypeId="2" ParentId="18097" CreationDate="2016-09-08T08:28:06.517" Score="1" Body="&lt;p&gt;To fully answer your question, you have to take into account the fact that light travels at the speed of light ($\approx 300,000 \mathrm{km/s}$). The further you look, the 'earlier' you look. Imagine that you are watching a star that is $3\ \mathrm{Glyr}$ away (the distance traveled by light in 3 billion years), then the light you see has been emitted 3 billion years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question, you thus have to find the earliest star created in the Universe we can see. The part of the Universe we can see is restricted to the part located at $14\ \mathrm{Glyr}$ and less, because it is the longest distance light could have traveled since the beginning of the Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now the question is: when did the first star form? A &lt;a href=&quot;http://www.eso.org/public/archives/releases/sciencepapers/eso1524/eso1524a.pdf&quot; rel=&quot;nofollow&quot;&gt;recent paper&lt;/a&gt; made by astronomers reported that the first stars (which are also the farthest!) are to be found $800\ \mathrm{Myr}$ after the Big Bang, and they are the in &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmos_Redshift_7&quot; rel=&quot;nofollow&quot;&gt;CR7&lt;/a&gt; galaxy. With increasing time, very far and very dim stars will appear in the sky, but you won't be able to see them with your naked eyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be able to see a star appear in the sky with your naked eye, you probably have to think about supernovae. Supernovae are massive stars that explode at the end of their life, increasing dramatically their emitted light in a matter of hours. If the exploding star was too dim before exploding to be visible, it will look as if a bright star appeared in the sky once it has exploded. You can find a sunnary of the observation of these stars on wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="14154" LastActivityDate="2016-09-08T08:28:06.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18264" PostTypeId="2" ParentId="18242" CreationDate="2016-09-08T08:53:05.060" Score="0" Body="&lt;p&gt;Most people underlined the role played by the expansion of the universe, however, this expansion is not actually 'pushing' Andromeda away, it is only changing the frequency of the light emitted by Andromeda, so that we can associate a speed that would give the same Doppler effect (when an object is emitting a wave, like a fire truck, and going away, the received wave is at a lower frequency than the emitted one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more correct answer would be that the two galaxies are the result of a long accretion (of gas) and merging history (with another now disappeared smaller galaxy). Each of this event changed the speed of Andromeda and the Milky Way relative to a rest frame, so that the speed nowadays is the result of these numerous interactions, the gravitational pull of one on each other as well as the gravitational potential created by the local group (the group of galaxies in which belongs the Milky Way and Andromeda, containing for example the Great Magellanic Cloud). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want an illustration of what the formation of a galaxy look like, you can take a look at a simulation, like the Horizon-AGN simulation. There is a video (here: &lt;a href=&quot;http://www.horizon-simulation.org/movies/horizon-AGN_denseproj.avi&quot; rel=&quot;nofollow&quot;&gt;http://www.horizon-simulation.org/movies/horizon-AGN_denseproj.avi&lt;/a&gt;) in which you can see galaxies forming within a cosmological framework. In the movie, each bright dot is a forming galaxy. You can see many very complicated interactions, so that two galaxies close to each other today may have formed very far away and were brought close by.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of our two galaxies, because we are located in a group of galaxies you can expect that the initial locations of our two galaxies were actually very different compared to nowadays. The reason for that is that in the past, as very correctly explained by userLTK, distances were smaller so that the gravitational pull of objects was (in average on the local group) stronger than today. Therefore, the interaction of the matter that is now forming the group of galaxy in which lay the MW and Andromeda had  gravitational interactions somehow stronger so that the relative speed was higher. Because of that, our two galaxies have been traveling for 14 Gyr from some remote location to their current location, where they happen to be going one toward each other.&lt;/p&gt;&#xA;" OwnerUserId="14154" LastActivityDate="2016-09-08T08:53:05.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18265" PostTypeId="2" ParentId="18258" CreationDate="2016-09-08T10:49:45.810" Score="3" Body="&lt;p&gt;There's a rough equation based on a star's mass but the age and metalicity of the star are factors too.   As large stars grow older they grow hotter on the inside and as they grow hotter, they fuse hydrogen faster, but they also become less massive over their lifetimes and have less and less hydrogen over time, so there's two opposing factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rough equation is found here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astronomynotes.com/evolutn/s2.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.astronomynotes.com/evolutn/s2.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a star's luminosity increases with mass as $(\text{the star's mass})^p$. The&#xA;  value of the exponent p varies between 3 and 4. For the rare massive&#xA;  stars (M* &gt; 30 Msun), p = 3 and for the more common low-mass stars (M*&#xA;  &amp;lt; 10 Msun), p = 4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Loosely speaking, the luminosity is roughly equivalent to the rate of hydrogen fusion though the hydrogen fusion takes place inside the star and it takes some time for the energy to make it's way to the outside of the star and radiate away, the age of the star is far greater than the time it takes the newly created energy to escape, that it's still a pretty good ratio direct, luminosity to hydrogen fusion rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A star with 2 solar masses is about (4th power) 16 times as luminous, which means it fuses hydrogen about 16 times as fast.   When you get into significantly larger stars, the 4th power rule gradually becomes the 3rd power.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as far as age being a factor, our Sun during it's main sequence will be about twice (or a bit more than twice) as luminous when it's 8 or 9 billion years old as it was when it was 1 billion years old, which corresponds to a gradual doubling of hydrogen fusion rate.  This happens (basically) because the core retains heat and more heat makes fusion more likely to happen, even as the amount of hydrogen is significantly lower in the later stage of it's main sequence than the early stage.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="14097" LastEditDate="2016-09-09T21:41:35.550" LastActivityDate="2016-09-09T21:41:35.550" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18266" PostTypeId="2" ParentId="18257" CreationDate="2016-09-08T11:42:57.760" Score="3" Body="&lt;p&gt;Theoretically, if you're talking crater to planet size, I (think) a crater that covers half or a bit more than half of the planet is possible if you have a dense planet (Say, Mercury) and a lighter, more icy object (say, Titan).   Mercury would break Titan apart like a hammer hitting glass (er, maybe), but Titan, which is slightly larger than Mercury but weights only 41% as much, with a lower velocity impact, might not make the entire surface moulton and could leave an impact crater over 50% of the surface with debris scattered unevenly leaving something like an impact crater.   That's, granted, pretty speculative, but I think that type of impact, a rocky world and an icy one of similar size might provide a collision that could create an impact crater a bit larger than 50% of the planet's surface.   Such an impact would be interesting to observe, but . . . I probably won't see one in my life time, unless they get some very good telescopes out there and get lucky.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically if you count surface melt, it's arguably possible to have an impact that would melt 75% or more of the surface but leave 25% or less intact, but I don't consider surface melt the same thing as an impact crater, so I wouldn't count that, but that's optional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as largest known craters, if we measure crater diameter against planet or moon radius (I'm going to ignore depth cause depth is easier on lower gravity bodies), but just diameter to radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hershal on Mimas&lt;/strong&gt; (the death star crater)  (Props to Andy)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MDyIN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MDyIN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diameter of Hershal, 139 km.    Radius of Mimas, 198 km.   Ratio, 70%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there's an even bigger one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/South_Pole%E2%80%93Aitken_basin&quot; rel=&quot;nofollow noreferrer&quot;&gt;South Pole–Aitken basin&lt;/a&gt;&lt;/strong&gt; on the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pTm3t.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pTm3t.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Article &lt;a href=&quot;http://www.nasa.gov/centers/goddard/news/features/2010/biggest_crater.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It went undetected until relatively recently because it's large enough that it no longer looks like a crater, but it is, and it affected the Moon's density where one half of the moon has a measurably thicker crust than the other half. (The crater half, I presume, the crust is thinner).   This makes orbits around the Moon weird cause the moon is gravitationally quite lumpy.  (Proximity to the Earth also makes orbits around the Moon weird, but that's another story.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This ancient crater has a diameter of about 2,500 km compared to the radius of the moon of 1,737 km a ratio of about 144%, over twice the relative diameter and 4 times the relative size relative to the crater on Mimas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;strong&gt;Mars&lt;/strong&gt; might have an even bigger one.   See &lt;a href=&quot;http://www.space.com/5558-huge-impact-created-mars-split-personality.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.  It's size isn't listed, but Mars (like the Moon) has a &quot;split personality&quot; where one side has a measurably thinner crust than the other side and one possible, perhaps likely cause of that is a giant impact.   This &quot;Crater&quot; if we want to call it a crater covers about 1/2 the planet.   This is different than the theorized Earth-Theia impact, which liquefied the surface and left no visible crater, but it may have liquefied half the surface and not left a very visible impact.     One problem with measuring craters as they get this large is they look less like craters.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-09-08T11:42:57.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18267" PostTypeId="2" ParentId="18260" CreationDate="2016-09-08T11:48:15.073" Score="7" Body="&lt;p&gt;Another valid definition is the time between earthrises instead of sunrises, since the moon is in orbit around earth. In that case, there is no direct concept of a day, because the moon is tidally locked to earth. &lt;/p&gt;&#xA;" OwnerUserId="13958" LastActivityDate="2016-09-08T11:48:15.073" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18268" PostTypeId="1" AcceptedAnswerId="18271" CreationDate="2016-09-08T13:43:36.853" Score="6" ViewCount="143" Body="&lt;p&gt;When black holes are merging, and emitting gravitational waves, are we seeing those waves start when the event horizons are merging, or when the singularities are merging? &lt;/p&gt;&#xA;" OwnerUserId="6870" LastActivityDate="2016-09-09T09:16:47.823" Title="When are gravitational waves emitted during a merger of black holes?" Tags="&lt;black-hole&gt;&lt;gravitational-waves&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18269" PostTypeId="1" CreationDate="2016-09-08T13:57:59.093" Score="1" ViewCount="580" Body="&lt;p&gt;Could you use gravitational waves to communicate from inside the event horizon of a black hole to someone who was outside?&lt;/p&gt;&#xA;" OwnerUserId="6870" LastActivityDate="2016-09-09T17:06:58.967" Title="Could information be transmitted from inside a black hole using gravitational waves?" Tags="&lt;black-hole&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18270" PostTypeId="2" ParentId="18268" CreationDate="2016-09-08T14:36:07.340" Score="3" Body="&lt;p&gt;Gravitational waves are on peak when event horizon start merging. &#xA;The black holes will merge once they are close enough. Once merged, the single hole settles down to a stable form, via a stage called ringdown and other one  will be torn apart by the growing tidal forces in a process sometimes referred to as spaghettification or the &quot;noodle effect&quot;.&#xA;Any distortion in the shape is dissipated as more gravitational waves. In the final fraction of a second the black holes can reach extremely high velocity, and the gravitational wave amplitude reaches its peak.&#xA;to know more read this &lt;a href=&quot;https://en.wikipedia.org/wiki/Binary_black_hole#Merger&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Binary_black_hole#Merger&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14102" LastActivityDate="2016-09-08T14:36:07.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18271" PostTypeId="2" ParentId="18268" CreationDate="2016-09-08T14:51:20.127" Score="10" Body="&lt;p&gt;The gravitational waves are actually emitted as long as you have a quadrupole moment with a second time derivative that is non-zero. The quadrupole is defined as:&#xA;$$ Q_{ij}^{tt}(x) =\int\rho \left(x^i x^j - \frac{1}{3}\delta_{ij}r^2\right)\mathrm{d}^3x.$$&#xA;To simplify this, here is a rule of thumb: if you have a system that is not spherically symmetric and that is changing with time (like two black holes), you can expect gravitational waves to be emitted. For example, the system made of the Earth and the Sun is emitting ridiculously small gravitational waves (I think it is emitting a few Joules per year), but it &lt;em&gt;is&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of how hard the waves are to measure, you can only expect to see them on Earth for black holes in their last stage, just before they merge, when the emission of gravitational waves is at its peak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the LIGO detection, the beginning of the signal as observed by our detectors was in the inspiral phase. This is the phase when the black holes are getting closer one to each other and losing gravitational energy (through the emission of gravitational waves), before their horizons touch. The end of the phase is actually when the black holes get within their so-called ISCO (Innermost Stable Circular Orbit), which is the smallest radius a massless particle could orbit in a stable fashion. Anything inside this orbit will slowly spiral inward and fall into the black hole.&lt;/p&gt;&#xA;" OwnerUserId="14154" LastEditorUserId="7411" LastEditDate="2016-09-09T09:16:47.823" LastActivityDate="2016-09-09T09:16:47.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18272" PostTypeId="2" ParentId="18218" CreationDate="2016-09-08T16:51:05.857" Score="1" Body="&lt;p&gt;In other words, at what hour angle $h$ is the Sun's azimuth $\mathtt{Az}_s$ at a right angle to the wall's normal vector (84$^\circ$ or 264$^\circ$)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these formulas, which I've checked, from&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_azimuth_angle&quot; rel=&quot;nofollow&quot;&gt;Wikipedia: Solar azimuth angle&lt;/a&gt;,&#xA;$$\sin \phi_s = \frac{-\sin h \cos \delta}{\sin \theta_s} \\&#xA;\cos \phi_s = \frac{\sin \delta \cos \Phi - \cos h \cos \delta \sin \Phi}{\sin \theta_s}$$&#xA;where $\phi_s$ is the Sun's north-clockwise azimuth,&#xA;$\theta_s$ is the Sun's zenith angle (90$^\circ$ - $\mathtt{Alt}_s$),&#xA;$\delta$ is the Sun's declination,&#xA;and $\Phi$ is your geographic latitude.&#xA;Then $\tan \phi_s = \sin \phi_s / \cos \phi_s$, and&#xA;$$\phi_s = \mathtt{atan2}(-\sin h \cos \delta, \ \sin \delta \cos \Phi - \cos h \cos \delta \sin \Phi)$$&#xA;This formula yields results consistent with the&#xA;&lt;a href=&quot;http://www.esrl.noaa.gov/gmd/grad/solcalc/&quot; rel=&quot;nofollow&quot;&gt;NOAA Solar Calculator&lt;/a&gt; in my tests.&#xA;We need to solve for $h$, but how isn't yet obvious to me; maybe someone else can help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The relation between $\phi_s$ and $\mathtt{Az}_s$ depends on whether $\mathtt{Az}_s$ is south-clockwise, south-counterclockwise, or the more conventional north-clockwise.&lt;/li&gt;&#xA;&lt;li&gt;$h$ here is the same as $\omega$ in the ITACA article.&#xA;As you observed, formula 3.7 is undefined for a wall facing due north or south.&#xA;I find their azimuth notation confusing (is A&lt;sub&gt;AZ&lt;/sub&gt; an alias for A&lt;sub&gt;ZS&lt;/sub&gt; or something else?) and the $\alpha$ on the left side of equation 3.9 is a mistake.&lt;/li&gt;&#xA;&lt;li&gt;I believe ITACA formula 3.11 is correct. If we let $\cos \theta_i$ = 0, it reduces to something like the above where the surface is vertical, but it has the same difficulty of solving for $\omega$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2016-09-12T16:53:26.263" LastActivityDate="2016-09-12T16:53:26.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18275" PostTypeId="1" AcceptedAnswerId="18278" CreationDate="2016-09-08T23:38:41.670" Score="2" ViewCount="558" Body="&lt;p&gt;I was wondering if there could theoretically exist systems in which an Earth-like planet was in the habitable zone of one (or two) Sun-like star(s), while actually not orbiting around any star.&#xA;To be more precise, I started asking myself whether a planet like earth could be placed in the inner orbit of a black hole, while one or a pair of Suns would be revolving around the black hole too, on an outer orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is not whether such a system could actually exist somewhere, but only if it could &lt;em&gt;theoretically&lt;/em&gt; exist, even if it is probabilistically not likely to exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I found this sets of issues:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Planet must be far enough from the black hole ( &gt; Roche Limit )&lt;/li&gt;&#xA;&lt;li&gt;Harmful radiations from black hole's accretion disk&lt;/li&gt;&#xA;&lt;li&gt;Planet-Star relative distance must not vary too much...which I guess in turn means that the planet-blackhole distance must be small compared to the average blackhole-star distance&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Given these limitations, how would you choose the stars and blackhole masses and their relative distances to the planet, in order to have it suitable for complex, perhaps intelligent forms of life? Do you see other issues to this purpose?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below I add what I've been able to collect so far (more like a collection of unsolved questions..)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, about the issue #1, I thought about a rather lightweight blackhole, with M &amp;lt; 10 Solar masses. I once read in a thread that when considering a central body = 10 solar masses, and a satellite with a density of 5000 kg/m^3 (Earth's average density), then the Roche Limit would be as little as 0.8% of an AU. That would be conveniently small, though I guessed that using Earth's average density may not guarantee that the whole planet is kept together (it may lose its atmosphere...). So, just to be sure, would you rather increase this distance to 1/20 of an AU, perhaps less, perhaps more? I'm asking here because I'm not able to conduct these calculation myself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then again, I'm not an expert and I don't know if staying out of the Roche Limit is a sufficient condition for orbiting around a blackhole of this small size without things becoming nasty. I know of issue #2 though,and about that I read someone suggesting that If the planet were subjected to tidal lock, then at least the face looking away from the blackhole would be spared from the harming X rays. But how close a planet should get to a blackhole in order to experience a tidal lock within a reasonable amount of time? Would it likely be tidally locked if it stayed at 0.05 AU away from a 10 solar masses black hole? I guess the answer to be yes, but I wanted some validations from the experts here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then about issue #3 again, I thought that If there were two Suns (making another binary system of their own), then there would be also a larger insulating flux, perhaps allowing us to keep the Suns more far away (by a quick calculation I think I found that two Suns at 1.4 AU would give the same flux as one Sun at 1 AU). Another possibility is to have a brighter Star class, at expense of its life-time. So for simplicity let's say we can go up to 2 AU (maybe even more) without the Planet getting too cold, or the Stars too short-lived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summarizing, I now have a 10 Solar Masses black hole and a (likely) tidally locked earth-like planet orbiting at 0.05 AU from it. Tidal forces should still be small enough not to suck the planet's atmosphere. Perhaps having a planet slightly bigger than the Earth would also help in this account. Then we have a couple of Stars orbiting together around the black hole, at a distance of 2 AU. Relative distance from the twin Stars to the Planet is 2 +/- 0.05 AU. I guess that would be within the limits of the habitable zone, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One last curiosity: assuming the black hole has an &quot;average&quot; spin, would it be realistic if the two Stars had a tilted orbit with respect to the black hole spinning axis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry for the many questions, by any means please don't assume I'm expecting a single person to answer all of them.&lt;/p&gt;&#xA;" OwnerUserId="14168" LastEditorDisplayName="user5434" LastEditDate="2017-03-11T22:22:43.247" LastActivityDate="2017-03-11T22:22:43.247" Title="Complex life in binary black hole - Sun(s) system" Tags="&lt;black-hole&gt;&lt;binary-star&gt;&lt;habitable-zone&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18277" PostTypeId="2" ParentId="18275" CreationDate="2016-09-09T08:27:57.647" Score="0" Body="&lt;p&gt;To start with, a (non rotating) black hole and a star of the same mass creates the same gravitational potential when you are far away enough. Since we observe planets at $0.05\mathrm{AU}$ around stars of about the same mass as the sun, I think you can expect planets in this region around a similar mass black hole, according to stability criterion. Moreover, the smallest black holes you can physically create have masses of about $3-4 M_\odot$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are multiple issues here though. First is how do you physically bring the planet that close to the black hole? I see two possibilities. Either the planet formed directly in the accretion disk of the black hole, and staid there, or the planet formed further away (say around the binary star system) and was captured by the black hole. If you assume that at $0.05\mathrm{AU}$, the accretion disk of the black hole is not that different than the accretion disk around a protostar at early ages, then you can maybe create a planet there (I can think of a few scenario that would prevent it, but it would require a more important study). If the planet formed further away in the disk or in the binary system, it is possible to make it migrate inward. This is actually called ex-situ formation in planetary systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you imagine that the planet is tidally locked (you can try fiddling around with the formula $t_{\text{lock}} \approx \frac{\omega a^6 I Q}{3 G m_p^2 k_2 R^5}$ from &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_locking#Timescale&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; that gives the timescale for the planet to become tidally locked to get an estimate of how long it would take) and that there is no accretion disk anymore, I guess you could manage to make the system stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final question is then to get the whole system stable on a large. Even though it may extremely unlikely, I think there has to be some stable orbits in the system using 3 body resonance, but this is not my field of expertise. having one or two stars is not making a huge difference, between the flux received from them and the gravitational pull are both in $1/r^2$, so that two stars at 1.4AU give the same flux and the same mean gravitational pull than one star at 1AU. The only difference is in the stability of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To conclude, I guess that the system is very unlikely to exist. Yet if you manage to find a planet there, you'd require to have life:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the accretion disk of the black hole has to be cleared&lt;/li&gt;&#xA;&lt;li&gt;the planet should be tidally locked to the black hole to be protected from the radiations&lt;/li&gt;&#xA;&lt;li&gt;the planet should be trapped in a very unlikely stable orbit so that it doesn't travel around in the system AND it has time to get tidally locked to the black hole&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="14154" LastActivityDate="2016-09-09T08:27:57.647" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18278" PostTypeId="2" ParentId="18275" CreationDate="2016-09-09T13:16:24.807" Score="1" Body="&lt;p&gt;The Roche limit for a solar-mass (or 10-solar-mass) black hole is &lt;em&gt;much&lt;/em&gt; larger than the Schwarzschild radius, or even the last stable orbit (which is a few times the Schwarzschild radius). So the Roche radius is a good limit (and, yes, you're right about the limit for an Earth-like planet and a 10-solar-mass BH.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You needn't assume the BH has an accretion disk; I suspect most of the ones wandering around in our galaxy don't at the moment, just because accretion disks &lt;em&gt;accrete&lt;/em&gt; (go into the black hole) over time, and so you only see them when there's some relatively recent source of gas to create the disk. (Most of the Milky Way black holes we know about are in close binaries, where the companion star is close to the Roche limit and is a regular source of gas.) Now, in principle the stellar wind(s) from the star(s) in your system could contribute to a very weak, tenuous accretion disk, but in practice it's unlikely to produce enough radiation to be of any concern, especially as you're assuming a planet with an atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there's no need to worry about whether or not the planet is tidally locked to the black hole, &lt;em&gt;unless&lt;/em&gt; you want to specify the length of the day-night period. Of course, if the planet is only a few tenths of an AU away from the BH, it probably &lt;em&gt;will&lt;/em&gt; be tidally locked before very long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variation in distance from the star(s) you suggest should be fine in &quot;habitable zone&quot; terms. In fact, given that the planet orbits the BH quite rapidly (about 1.3 days for a 0.05 AU orbit around a 10 solar-mass BH), it won't spend much time at either extreme, so you could probably have it go inside and outside of the formal habitable zone and still be habitable, as long as the &lt;em&gt;average&lt;/em&gt; insolation is in the habitable range.  This would create some interesting day/night/seasonal effects, particularly in the tidally locked case: e.g., at a certain time of &quot;year&quot;, the &quot;eastern&quot; hemisphere would experience noon when the stars were only 1.95 AU away, while for the &quot;western&quot; hemisphere noon would happen when the stars were 2.05 AU away. Half a year later, the situation would be reversed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;One last curiosity: assuming the black hole has an &quot;average&quot; spin, would it be realistic if the two Stars had a tilted orbit with respect to the black hole spinning axis?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That really depends on how the system formed in the first place. &lt;em&gt;If&lt;/em&gt; the initial pre-BH star and the other star(s) formed from the same collapsing cloud, they would probably have similar spin and orbital angular momenta; if the spin of the BH is entirely inherited from its progenitor (which goes supernova to form the BH), then the orbital plane might be similar to the BH's spin. But if the system got put together later (maybe an encounter between a binary consisting of the BH and another star and the current binary pair, in which the BH's original companion star gets ejected), then the two stars could have a tilted orbit.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2016-09-09T13:16:24.807" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18279" PostTypeId="1" CreationDate="2016-09-09T14:07:36.887" Score="0" ViewCount="4318" Body="&lt;p&gt;Take the following planet:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Same distance from the main star as Earth&lt;/li&gt;&#xA;&lt;li&gt;Main star is the same size as the sun&lt;/li&gt;&#xA;&lt;li&gt;Planet is the size of Venus&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How would I calculate the length of the day on this planet? If you need more information, please ask.&lt;/p&gt;&#xA;" OwnerUserId="14187" LastEditorUserId="53" LastEditDate="2016-09-09T15:46:37.660" LastActivityDate="2019-04-27T15:21:18.360" Title="How would I calculate the length of the day of this planet?" Tags="&lt;planet&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18280" PostTypeId="2" ParentId="18229" CreationDate="2016-09-09T14:12:24.233" Score="5" Body="&lt;p&gt;There is a very nice project called &lt;a href=&quot;https://github.com/ofrohn/d3-celestial/&quot; rel=&quot;noreferrer&quot;&gt;d3-celestial&lt;/a&gt; by Olaf Frohn on github. In contains a data file describing the Milky Way as polygons, see &lt;a href=&quot;https://github.com/ofrohn/d3-celestial/#files&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. A demo showing this Milky Way can be found &lt;a href=&quot;http://armchairastronautics.blogspot.de/p/skymap.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. And even better, the source for this data is cited, pointing to the &lt;a href=&quot;http://www.skymap.com/milkyway_cat.htm&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Milky Way Outline Catalog&lt;/em&gt; by Jose R. Vieira&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on your project, the json format from d3-celestial might be easier to read than the one from Jose R. Vieira.&#xA;Note that you don't have to worry about these &quot;contours&quot; moving on a time scale of hundred years, but this is another question.&lt;/p&gt;&#xA;" OwnerUserId="9150" LastActivityDate="2016-09-09T14:12:24.233" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18281" PostTypeId="2" ParentId="18279" CreationDate="2016-09-09T14:22:02.090" Score="2" Body="&lt;p&gt;You can use Kepler's third law to get the length of a year&#xA;$$\frac{P^2}{a^3} = \frac{4 \pi^2}{G(M+m)},$$&#xA;where $P$ is the period of rotation, $a$ is the distance between the planet and the star, $M$ is the mass of the star, $m$ is the mass of the planet and $G$ is the universal gravity constant.&#xA;As the mass of Venus is roughly the same as the Earth's ($M_V = 0.815M_\oplus$), you will get something close but a bit larger than 365 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the length of a day on this planet, you need more information as the rotation rate of the planet which gives the length of a day cannot be constrained easily. For example on Earth, one day is 24 hours today, and the length of a day circa 620 million years ago has been estimated from rhythmites (alternating layers in sandstone) as having been about 21.9 hours. The origin of the evolution of the length of a day is the tidal interactions between Earth and the Moon (as well as to a smaller extent with the Sun) , an effect known as tidal acceleration (or in this case, deceleration).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also the reason why we only see one side of the moon: it is tidally locked to the Earth so that the Moon revolves around the Earth in exactly one moon day. In the very far future, the Earth will revolve around the Moon in exactly one 'future' day, so that one day will last 27.7 present days!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the initial rotation rate of the planet at its formation is largely unconstrained so that even though you give us a planet's mass and age and the properties of its orbit and its star, it will still be impossible to answer your question entirely.&lt;/p&gt;&#xA;" OwnerUserId="14154" LastEditorUserId="14154" LastEditDate="2016-09-09T14:29:22.937" LastActivityDate="2016-09-09T14:29:22.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18282" PostTypeId="2" ParentId="18279" CreationDate="2016-09-09T14:45:40.563" Score="4" Body="&lt;p&gt;It's almost completely unconstrained. There's a limit to how fast a planet can spin before it starts to break up, which is $p = 2 \pi / \sqrt{m G / r^{3}}$ (where $m$ is the planet's mass and $r$ is its radius). For something like the Earth (and Venus is pretty similar), this corresponds to about 1.4 hours. So the day has to be longer than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there's no limit to how &lt;em&gt;slowly&lt;/em&gt; the planet can rotate. If it's rotating slowly enough so that one rotation period = one year, then as seen from the planet's surface the star would never seem to move in its sky and the solar day (time from one noon to the next) would be infinite. This is what you get if the planet is 1:1 tidally locked to the star. &lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="14154" LastEditDate="2016-09-09T14:51:26.583" LastActivityDate="2016-09-09T14:51:26.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18283" PostTypeId="1" CreationDate="2016-09-09T16:02:58.087" Score="7" ViewCount="508" Body="&lt;p&gt;It’s well known that (lunar) tides on Earth result in a transfer of angular momentum from Earth proper to the Earth–Moon &lt;em&gt;orbital&lt;/em&gt; motion. That’s why the Moon resides now in a high Earth orbit, and Earth’s axial rotation is nowadays much slower than in the past (Hadean, Archean etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about effects of &lt;strong&gt;solar&lt;/strong&gt; tides, indeed? The angular momentum certainly must be transferred from Earth proper to the Earth–Sun orbital motion. Does it result in a tiny increase of (Earth+Moon)’s heliocentric orbital angular momentum over millenia? What about Mars and Ceres: does the solar tide brake their axial rotation and change slightly their orbits? And what about Venus: does &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0273117787902377&quot; rel=&quot;noreferrer&quot;&gt;its solar tidal&lt;/a&gt; braking explain the slowness of rotation completely?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum. The solar tide may not be confused with “dual” effect: planetary tide on the central star. It exists as well and has much significance for (extrasolar) planets on near orbits — but this question is not about tides on Sun. Ī̲’d accept an answer assuming gravity of the Sun to be spherically symmetric, neglecting Sun’s deviation from spherical symmetry, as well as rotation (even going to account for General Relativity). The solar tide is &lt;strong&gt;interaction between this gravity and the orbiting object.&lt;/strong&gt; By Newton’s laws, gravitation affects Sun’s momentum reciprocally (and, hence, velocity), and we may ignore here any effect on Sun beyond its translational motion (momentum/velocity/position).&lt;/p&gt;&#xA;" OwnerUserId="2287" LastEditorUserId="2287" LastEditDate="2016-09-11T16:22:06.417" LastActivityDate="2016-09-13T19:10:52.320" Title="Effects of the solar tide on planets" Tags="&lt;solar-system&gt;&lt;earth&gt;&lt;venus&gt;&lt;orbital-mechanics&gt;&lt;tidal-forces&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18284" PostTypeId="2" ParentId="18229" CreationDate="2016-09-09T16:06:41.970" Score="1" Body="&lt;p&gt;From memory (I don't have a copy here right now) the &lt;a href=&quot;http://www.hnsky.org/software.htm&quot; rel=&quot;nofollow&quot;&gt;HNSky&lt;/a&gt; package &lt;strong&gt;used to&lt;/strong&gt; have a hand-drawn &quot;supplement&quot; file for the milky way.  It contained a lot of data points arranged as RA &amp;amp; Dec points and was extracted from photographs, with a bit of correction and scaling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest you load and install one of the older versions still on HN Sky's page, such as 3.0.0, and look for the supplement file. (The newer version 3.2.3 does not seem to include this file.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is only position data - no proper motion, etc. as the question mentioned. So only a partial answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: You might need to ask the authors of that file for permission if you're going to release software based on this. No harm in looking at it for purely personal interest of course.)&lt;/p&gt;&#xA;" OwnerUserId="7491" LastEditorUserId="7491" LastEditDate="2016-09-09T16:13:02.927" LastActivityDate="2016-09-09T16:13:02.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18285" PostTypeId="2" ParentId="18269" CreationDate="2016-09-09T17:06:58.967" Score="1" Body="&lt;p&gt;A black hole is defined as a region of spacetime where the gravitational effects are so strong that the escape velocity within the event horizon is greater than the speed of light.  According to relativity, nothing can propagate faster than the speed of light, so at most gravitational waves would propagate at this speed.  According to particle physics, mediation of gravity would be controlled by massless particles called gravitons, which, by virtue of being massless, would travel at the same speed as photons, i.e. the speed of light.  Suffice to say gravitational waves (analogous to electromagnetic waves) would not be able to escape from inside the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question of information, in regards to black holes is open however. The No-Hair Theorem would seem to imply the destruction of information as the black hole consumes matter.  However, the holographic principle, if correct would imply that information regarding the matter that formed the black hole would be encoded on the 2-dimensional boundary of the event horizon.  This information could very well be exhausted by Hawking radiation or gravitational waves.  Since this information was outside of the event horizon this would be fully consistent with the formulation of black holes.&lt;/p&gt;&#xA;" OwnerUserId="13546" LastActivityDate="2016-09-09T17:06:58.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18286" PostTypeId="2" ParentId="18229" CreationDate="2016-09-09T17:32:23.683" Score="1" Body="&lt;p&gt;The planetarium software, Guide, to be found at ProjectPluto.com, has  a map of the Milky Way.  This is how the dataset is described &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;The Nebula Databank was compiled by Eric-Sven Vesting to evade&#xA;the problems that came with earlier bright nebula databases.  For&#xA;example, previous versions of Guide gathered nebula data from five&#xA;separate catalogs.  There were few cross-indexes from one catalog to&#xA;another, and no way to indicate that one designation applied to a&#xA;part of larger area with a different designation.  Also,  brightness&#xA;levels were applied in an inconsistent manner at best.&#xA;   The Nebula Databank contains explicit links between the various&#xA;nebula catalogs,  enabling Guide to show all designations for a&#xA;given object and to avoid drawing some objects twice (if they appeared&#xA;in separate catalogs under different names).  Also,  Eric-Sven Vesting&#xA;created the nebula isophotes used by Guide to indicate the shapes of&#xA;most prominent nebulae.  Better positional data was generated,  usually&#xA;by comparing catalog positions to actual RealSky images.&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6I9GQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6I9GQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;(I hope I have not violated any copyright on this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not clear where that database is, but perhaps you could contact either ProjectPluto or Eric-Sven Vesting directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no affiliation or connection, financial or otherwise, with this software.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2016-09-09T17:32:23.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18288" PostTypeId="2" ParentId="13895" CreationDate="2016-09-09T19:14:28.880" Score="1" Body="&lt;p&gt;How black holes encode or destroy information is an open question, which RichS touched on.  However, while he provided an answer consistent with the No Hair Theorem, I will provide an answer derived from the holographic principle.  I'd like to stress that both are equally valid, since we (as of yet) do not know enough about black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of the holographic principle, the information about the matter that formed the black hole is encoded in some manner (micro gravitational fluctuations maybe?) on the two-dimensional 'surface' of the event horizon.  It has been theorized that one could reconstruct this information by measuring the outgoing Hawking radiation, since this process decreases the radius of the black hole, hence the surface area of the event horizon, and subsequently the amount of information.  I would liken this to burning a book and then trying to reconstruct the book by measuring the properties of the ashes and radiated light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When two black holes merge, they form a black hole of smaller mass than the combined masses.  Once more the total surface area of the event horizons has decreased to that of the event horizon of the new black hole, so information must have been 'radiated'.  Black hole-black hole mergers are not believed to have any optical counterpart (burst of light for sake of simplicity) and the merging process is derived from relativity, so Hawking radiation is not a component of consideration.  Where could the information have gone?  Well, luckily enough these mergers do radiate gravitational waves, now proven by LIGO's recent discovery.  &lt;strong&gt;Thus the information, if it does in fact encode itself on the event horizon surfaces, could be radiated/lost by the gravitational waves created during the merger.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The above description sounds very 'hand-wavey', so I will expand on proposed theoretical method behind it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational radiation is generated by the changing quadrupole moment, caused by the two in-spiraling black holes.  However, just as for the case of electromagnetic radiation emitted by oscillating charges, the quadrupole moment contribution is only one part of the greater multipole expansion of the oscillating masses.  Oscillations of the the event horizon would then cause deviations of the system from the simple quadrupole approximation, and result in gravitational radiation from the high-order multipole terms.  This radiation falls off with distance faster as one moves to higher-orders, making measurement of these contributions much more difficult.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this is just one proposed solution to the black hole information paradox. &lt;/p&gt;&#xA;" OwnerUserId="13546" LastEditorUserId="13546" LastEditDate="2016-09-12T05:46:32.090" LastActivityDate="2016-09-12T05:46:32.090" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18289" PostTypeId="1" CreationDate="2016-09-09T19:18:57.067" Score="0" ViewCount="51" Body="&lt;p&gt;I need help with an exercise in the book that I can't tackle. I need to calculate how much time we have to intercept and eliminate a NEO. I got the following orbital parameters:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a=-2791.44 km (semi-major axis), &lt;/li&gt;&#xA;&lt;li&gt;i=22.3 deg /Hyperbolic), &lt;/li&gt;&#xA;&lt;li&gt;e=2.77 &lt;/li&gt;&#xA;&lt;li&gt;v=247.96 deg, &lt;/li&gt;&#xA;&lt;li&gt;r= 2,106,101.331 km (radius from Earth's center of mass)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My problem is that I don't know where to begin, but I did an attempt. I assume that the intercept with the target only can happen when the phase angle between Earth and the NEO are the same. &#xA;I know that one can calculate the velocity of the S/C with ($m_y$=gravitational param.):&#xA;$$v=\sqrt{(2m_y/r)-(m_y/a)}$$ &#xA;If we calculate the escape velocity of the Earth with $V_e=\sqrt{2GM_e/r}$, we will get the difference in velocity. With this information, we can calculate the time needed.  Anyone that can help me? &lt;/p&gt;&#xA;" OwnerUserId="14196" LastEditorUserId="2365" LastEditDate="2016-09-09T21:03:50.860" LastActivityDate="2016-09-10T08:42:53.603" Title="Intercept a NEO trajectory" Tags="&lt;near-earth-object&gt;&lt;orbital-mechanics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18290" PostTypeId="2" ParentId="18289" CreationDate="2016-09-09T21:38:37.333" Score="1" Body="&lt;p&gt;Assuming that you have a phaser array sufficiently powerful to pulverize the NEO into a harmless meteor storm,&#xA;I read this as a question of when an object with those geocentric orbital elements arrives at perigee.&#xA;However, I get a different value for $r$ if I treat that $\nu$ value as a true anomaly; it could be a mean anomaly or hyperbolic anomaly instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose your textbook covers this stuff in detail, but for other readers&#xA;these &lt;a href=&quot;http://control.asu.edu/Classes/MAE462/462Lecture05.pdf&quot; rel=&quot;nofollow&quot;&gt;lecture notes&lt;/a&gt;&#xA;by M. Peet at ASU, using $f$ instead of $\nu$, may help.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-09-09T21:38:37.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18293" PostTypeId="1" CreationDate="2016-09-10T08:20:03.627" Score="3" ViewCount="87" Body="&lt;p&gt;When discussing a rocky planet, is there any terminology to distinguish that you're talking about the whole thing, including the atmosphere, veruses that you're talking about the ocean + crust + mantle + core as a separate thing from the atmosphere?  Or do you have to depend upon context to make that distinction?&lt;/p&gt;&#xA;" OwnerUserId="14202" LastActivityDate="2016-09-10T13:05:26.920" Title="Terminology: planet plus atmosphere vs planet and atmosphere as separate things" Tags="&lt;planet&gt;&lt;planetary-atmosphere&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18295" PostTypeId="2" ParentId="18293" CreationDate="2016-09-10T11:04:58.557" Score="2" Body="&lt;p&gt;About the only time it could make any difference is when considering impacts on the planet. In that case, if a body enters the atmosphere, it is considered to have impacted the planet. For example the dust sized particles that form meteors don't reach the surface of the Earth, yet are still considered to have impacted the Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, quoted sizes of diameters normally don't include the atmosphere since the top of the atmosphere is fuzzy, and varies with temperature. For exoplanets, none of this matters since we don't have anything like enough resolution to determine atmospheric depth, or even the presence of an atmosphere in many cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that gas giant planets don't have a solid surface. The cloud tops are then used as a proxy for the surface.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-09-10T11:04:58.557" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18296" PostTypeId="2" ParentId="18100" CreationDate="2016-09-10T11:57:51.033" Score="3" Body="&lt;p&gt;Just to address some of the potential confusion in the original question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Talking about &quot;resolution&quot; in &quot;megapixels&quot; only makes sense if you're comparing different imaging chips in the &lt;em&gt;same optical system&lt;/em&gt;: the same angular field of view, with the same angular resolution limit, imaged onto the chip. So comparing megapixels between, say, different full-frame DSLRs (or full-frame mirrorless cameras) makes sense, because if you put the same size lens on a full-frame Nikon or Canon DSLR, they image same field of view onto their chips, with similar if not identical angular resolution limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This comparison breaks down if you try to compare even the same camera with different lenses, because then the field of view imaged onto the chip changes. For example, I can put a wide-angle lens (large field of view) on a full-frame DSLR and take a picture of a lion at a distance. I'll get a great landscape picture, but if I zoom in on the lion in the resulting image, the lion will be blurry, with only a few pixels covering the lion. I can put long telephoto lens (small field of view) on the &lt;em&gt;exact same camera&lt;/em&gt;, take a picture of the lion, and the lion will be very detailed and sharp, spanning lots of pixels (though I won't get the rest of the lanscape).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many claims about megapixels in consumer and phone cameras, while technically correct, are mostly marketing blather, because there's no attempt to compare the fields of view and thus the effective angular resolution. This is why you'll be able to take much higher resolution images with a 20-megapixel DSLR than you can take with your 41-megapixel phone camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to think about it is this: if we could put whatever chip the Lumia uses in Hubble, would we get sharper images? No. The Lumia chip has &lt;em&gt;physically&lt;/em&gt; smaller pixels -- probably something like 1 to 1.5 microns in size, compared to the 21 micron ACS-HRC pixels. But, as Rob Jeffries points out, the HRC pixels are already small enough to resolve at Hubble's angular resolution limit. The Lumia chip wouldn't show any more &lt;em&gt;detail&lt;/em&gt; than the HRC chip already does, it would just give you bigger blur per pixel. (There's also the fact that commercial imaging chips divide their pixels up between different filters to get R, G, and B images simultaneously, so in practice your &quot;41 megapixel&quot; camera only records about 13 or 14 megapixels per filter. Cameras on Hubble don't use per-pixel filters, so each single-filter image uses all the pixels.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2016-09-10T11:57:51.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18298" PostTypeId="2" ParentId="18293" CreationDate="2016-09-10T13:05:26.920" Score="2" Body="&lt;p&gt;Planet means usually the whole thing, as far as I'm aware.&#xA;Then there are specialized terms for what you've suggested:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cryosphere when talking about ice only&lt;/li&gt;&#xA;&lt;li&gt;Hydrosphere when it concerns water&lt;/li&gt;&#xA;&lt;li&gt;Lithosphere when talking about the crust that is still solid rock&lt;/li&gt;&#xA;&lt;li&gt;(there are more terms for soil, geology, atmospheric substructure...)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But all those terms have an obvious relation to the construction of the word 'atmosphere'. The only term I've heard for 'Earth minus atmosphere' is 'bulk Earth'.&lt;br&gt;&#xA;Those terms should be context free, so that you can directly, precisely refer to them when starting a conversation about them.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2016-09-10T13:05:26.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18303" PostTypeId="2" ParentId="16732" CreationDate="2016-09-10T17:12:04.893" Score="3" Body="&lt;p&gt;Only a &lt;strong&gt;syzygial&lt;/strong&gt; tide during/near an equinox is the strongest. This means, there must be either new or full moon. In general, syzygial tides are strong because three bodies (Earth, Moon, Sun) align near one line, and tidal effects of Moon and Sun on Earth become (nearly) collinear and sum to the maximal possible magnitude. The Moon’s orbit is inclined to the ecliptic by about 5° only (cos 5° ≈ 0.996), so two tides are aligned almost perfectly on a Moon’s syzygy during whichever season. The difference is that, during/near an equinox, &lt;strong&gt;this line&lt;/strong&gt; also &lt;strong&gt;lies in the equatorial plane&lt;/strong&gt; and rotational motion of Earth’s surface/hydrosphere/crust can direct the stuff to move along this tidal line to a maximal range possible. At least, on the equator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for variations in terrestrial latitudes, namely mid-latitudes vs equator, comparison between syzygial tides during solstice and equinox far from the equator is a mechanical, not astronomical problem. Tidal bulge should be stronger during solstice, but projection of velocity to the tidal line will be greater on equinox. A detailed analysis of particular body of liquid (and its shore) is required to compare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, a quadratural (where Earth–Moon and Earth–Sun lines are perpendicular) tide during an equinox is as weak as any quadratural tide. The solar tide basically cancels a part of the lunar tide. A quadratural tide during an equinox might even be the weakest, since lunar tide outmatches the solar one, but on an equinox and Moon’s quadrature simultaneously the Moon definitely lies away of the equatorial plane.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastEditorUserId="2287" LastEditDate="2016-09-10T17:48:13.690" LastActivityDate="2016-09-10T17:48:13.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18304" PostTypeId="2" ParentId="18169" CreationDate="2016-09-10T22:44:49.857" Score="3" Body="&lt;p&gt;The only correct answer: &lt;strong&gt;no particular point&lt;/strong&gt;, since Earth’s motion is complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the Earth orbits the Earth–Moon barycentre.&lt;br&gt;&#xA;Surprisingly. But this points lies inside the planet, somewhere in the mantle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, it would be a reasonable approximation to say that the Earth+Moon system orbits the system of Sun + inferior planets (Venus and Mercury). Of course, we shouldn’t exactly suppose their combined gravity emits from a single point. But, roughly and in average, it comes from the Sun+Mercury+Venus centre of mass (nearly the Sun’s geometric centre). At any time doesn’t any of these bodies pull Earth and Moon out of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also influences from outer planets, such as Jupiter. But we shouldn’t, at any level of simplification, think of it as coming from inside the Solar System. Direction of the Jupiter’s gravity makes full turn around the ecliptic each time it completes the orbit (≈ 12 years). It would be reasonable to say that Sun+Mercury+Venus+Earth+Moon+Mars and Jupiter orbit about their common centre of mass (somewhere not far from the Sun’s surface), approximately. But not all bodies are, evidently, included here. Generally, “what point does … orbit” depends heavily on definition of our “stationary” reference frame, and arguing about this makes no sense, in principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An “Earth orbits the Sun” approximation would ignore not only influence of Venus (that has little significance over a human lifetime), but, first and foremost, gravity of the Moon.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastActivityDate="2016-09-10T22:44:49.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18305" PostTypeId="1" AcceptedAnswerId="18313" CreationDate="2016-09-11T02:42:42.647" Score="1" ViewCount="470" Body="&lt;p&gt;If all the stars disappeared simultaneously (from the earth's reference frame), how long would it take for the night sky to go completely dark to our unaided eyes? Would we still be able to see anything after a thousand years?&lt;/p&gt;&#xA;" OwnerUserId="14211" LastEditorUserId="14211" LastEditDate="2016-09-11T13:53:48.570" LastActivityDate="2016-09-11T13:53:48.570" Title="What would we see if all stars disappeared simultaneously?" Tags="&lt;star&gt;&lt;amateur-observing&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18306" PostTypeId="2" ParentId="18305" CreationDate="2016-09-11T04:35:57.917" Score="1" Body="&lt;p&gt;&quot;The farthest star we can see with our naked eye is V762 Cas in Cassiopeia at 16,308 light-years away. Its brightness is magnitude 5.8 or just above the 6th magnitude limit.&quot; As per this information, if all the stars went out simultaneously, then it would take 16308 years for the sky to become completely dark to the naked eye. &#xA;Of course, stars which are farther away will still be visible to powerful telescopes.&lt;/p&gt;&#xA;" OwnerUserId="13414" LastActivityDate="2016-09-11T04:35:57.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18307" PostTypeId="1" AcceptedAnswerId="18317" CreationDate="2016-09-11T04:51:25.623" Score="0" ViewCount="85" Body="&lt;p&gt;I read one of the related questions here, but what i'm curious to know is how do we deflect imminent collision courses of comets, if we detect them at a later stage? I read somewhere else that there are systems which use powerful lasers (radiation pressure) to deflect them, but honestly, i don't understand why something like radiation pressure will work on something travelling at about 40km/s through space, and still manage to cause a large enough deflection. Are there other methods ?&lt;/p&gt;&#xA;" OwnerUserId="13414" LastActivityDate="2016-09-11T23:08:43.750" Title="safeguards against comets?" Tags="&lt;comets&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18308" PostTypeId="2" ParentId="18307" CreationDate="2016-09-11T05:39:46.127" Score="3" Body="&lt;p&gt;Fortunately it is not something that we have yet had to face.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diverting the impact of a comet would not be easy, especially if it was a newly discovered comet on a nearly parabolic orbit. We would probably have less than 2 years to prepare. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that the Earth is orbiting the sun at 30 km/s. So if we can delay the comet by three or four minutes, that turns a direct hit into a miss. But most of the plans to divert objects apply to asteroids whose orbit is well known and for which we could have tens of years in which to act. For a new comet there may not be time.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-09-11T05:39:46.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18309" PostTypeId="2" ParentId="18305" CreationDate="2016-09-11T06:09:35.947" Score="0" Body="&lt;p&gt;Depending upon how far a star is away from us, is exactly related to how long it would take for us to notice if it weren't there anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given your scenario, all stars instantaneously disappear, I will give a very short timeline. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At roughly 8 minuets and 25 seconds after the event, we notice the sun go dark.&lt;br&gt;&#xA;  After about 4 years and 146 days, $\alpha$ Centauri goes dark.&lt;br&gt;&#xA;  Sirius disappears in 8 years and 219 days.&lt;br&gt;&#xA;  $\varepsilon$ Eridani follows at 10 years 182 days.&lt;br&gt;&#xA;  ...&lt;br&gt;&#xA;  M31 (Andromeda Galaxy) disappears at 2.5 million years.&lt;br&gt;&#xA;  ...  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hope you see the pattern forming here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closer the star or other object, the sooner it will appear to go dark, or disappear. Therefore, I would assume that because every humans eyes are basically the same, roundish, with a roughly 5mm pupil. The most distant object is probably right around 2.9 million light years away, the galaxy M33.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that there will be no GRB within 30-60 seconds of all of the stars going dark, the sky would be completely dark to the unaided human eye in &lt;strong&gt;2.9 million years&lt;/strong&gt;. Not that humans will survive that long into the future, but that is a different problem all together.&lt;/p&gt;&#xA;" OwnerUserId="2232" LastActivityDate="2016-09-11T06:09:35.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18313" PostTypeId="2" ParentId="18305" CreationDate="2016-09-11T10:02:02.510" Score="2" Body="&lt;p&gt;The farthest object that we can see with our naked eyes as of now is the &lt;strong&gt;Andromeda Galaxy (2.6 million light years away)&lt;/strong&gt;, although the &lt;strong&gt;M83 (14.7 million light years away)&lt;/strong&gt; has also been reported to be seen with the naked eye. &lt;em&gt;(You may want to click here for more information on what the farthest objects we can see with our naked eyes, from &lt;a href=&quot;https://astronomy.stackexchange.com/a/2088/14160&quot;&gt;Universe Today&lt;/a&gt; or see this &lt;a href=&quot;http://www.universetoday.com/110858/how-far-can-you-see-in-the-universe/&quot; rel=&quot;nofollow noreferrer&quot;&gt;answer&lt;/a&gt; from this same site.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that just before the moment all the stars disappear, these two farthest objects are still emitting light, then we can know judging by their distance from us, that this light will only reach our eyes 2.6m years later (because 1 light year = light travelling for 1 year) from the Andromeda Galaxy, and 14.7m years from M83.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore to answer your question more directly: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would still see the stars (assuming this is what you are referring to by &quot;anything&quot;) &lt;strong&gt;for up to 2.6~14.7m years&lt;/strong&gt;, depending on what really is the farthest thing we can see in the sky.&lt;/p&gt;&#xA;" OwnerUserId="14160" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-09-11T10:02:02.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18314" PostTypeId="1" AcceptedAnswerId="18315" CreationDate="2016-09-11T12:03:47.437" Score="1" ViewCount="725" Body="&lt;p&gt;This seems like a pretty basic question, but I couldn't find a duplicate upon quick inspection. From what I understand about black holes they actually contain no matter. All of the mass that maintains the black hole (ie. keeps space warped) is found in the energy bound in the space itself. I may not be stating this quite right, but this is basically what Kip Thorne states in &quot;The Science of Interstellar&quot;. Therefore, it seems that to maintain angular momentum during black hole formation that space itself must &quot;spin&quot;. This would imply that space is actually &quot;spinning&quot; even in the vicinity of the Earth (to a much lesser degree of course). Is this true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, I don't take everything that Kip says as gospel truth. He states several times that Io is the innermost moon of Saturn in Chapter 7.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="2365" LastEditDate="2016-09-11T12:25:04.553" LastActivityDate="2016-09-12T08:19:49.380" Title="How does a black hole spin?" Tags="&lt;black-hole&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18315" PostTypeId="2" ParentId="18314" CreationDate="2016-09-11T12:46:16.053" Score="5" Body="&lt;p&gt;What we call a black hole is a solution of Einstein's equations of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know several exact solutions: Empty and flat space corresponds to special relativity. Other known solutions include that of a point mass, a charged point mass, and a solution, known as the Kerr solution in which spacetime around a point mass is being &quot;dragged&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This last solution can be interpreted as corresponding to a black hole that would form if a rotating mass were to collapse. As you note, a black hole contains no matter: it is a vacuum solution of Einstein's equations. However a black hole does have properties like mass and angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape of space around the Earth can be approximated the Kerr solution of a spinning black hole, with a radius of about 1cm. We would therefore expect some dragging of spacetime around the Earth. In 2008 the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravity_Probe_B&quot;&gt;Gravity B probe&lt;/a&gt; had measured the gravity around the Earth and confirmed that the Frame Dragging effect predicted was present and its results were consistent with  General relativity. Frame dragging around the Earth is fantastically small and the Gravity B probe was working right at its limit of sensitivity. It causes a gyroscope to deviate from its spin by 0.00001 degrees per year.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-09-11T12:46:16.053" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18316" PostTypeId="1" AcceptedAnswerId="18354" CreationDate="2016-09-11T13:14:31.370" Score="7" ViewCount="671" Body="&lt;p&gt;How did astronomers in the 18th and 19th centuries used to calculate a comet's or planet's orbit using observational data, given that this data is relative to a non static reference point (i.e. the Earth)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, &lt;a href=&quot;http://www.pafko.com/tycho/observe.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; one can find the measurements of Mars' declination by Tycho Brahe throughout two decades. Not only it seems very difficult to me to deduce the right curve from the original observations points, but also the peaks of the curve (which resembles a sinusoid) do not seem perfectly equally distant from each other.&lt;/p&gt;&#xA;" OwnerUserId="1447" LastActivityDate="2016-09-19T00:15:28.760" Title="Calculating orbits using observational data" Tags="&lt;orbit&gt;&lt;history&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18317" PostTypeId="2" ParentId="18307" CreationDate="2016-09-11T15:45:06.823" Score="2" Body="&lt;p&gt;Difficulty is hard to say, but the reason a laser might work is because the laser is supposed to melt one side of the comet, which, as it melts, gas particles fly off at relatively high velocity, which redirects the momentum of the comet somewhat.  It's not the pressure from the laser, it's the heat from the laser and the fact that comets are made largely from ice, which can be melted or &quot;zapped&quot; into fast moving jets of gas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The approach wouldn't be a laser on or orbiting Earth shooting comets from a great distance but spacecrafts flying up close to the comet and using lasers to target one side.  It would be difficult to fire a laser so pinpoint as to target one side of the comet from earth and create localized heat that would create the necessary directional out-gassing.  This &lt;a href=&quot;http://www.astrobio.net/topic/solar-system/meteoritescomets-and-asteroids/deflecting-asteroids-with-lasers/&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; proposes spacecraft, perhaps several, flying adjacent to the comet.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nuclear weapon would be way more effective, however, depending on how fast we'd needed to push it.  This &lt;a href=&quot;http://abcnews.go.com/blogs/politics/2013/05/nasas-actual-plan-to-deflect-an-approaching-asteroid/&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; mentions a pulse laser landing on the comet or asteroid as well as other methods that might work.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-09-11T21:17:04.857" LastActivityDate="2016-09-11T21:17:04.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18318" PostTypeId="2" ParentId="18307" CreationDate="2016-09-11T23:08:43.750" Score="1" Body="&lt;p&gt;Space is really very empty and large. The Earth moves one radius in orbit every 3½ minutes. There are half a million minutes per year. So a pretty tiny push is enough to make the difference between a hit and a miss, given that there are a few years' warning and prompt action.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-09-11T23:08:43.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18319" PostTypeId="1" AcceptedAnswerId="18321" CreationDate="2016-09-12T08:07:38.760" Score="1" ViewCount="146" Body="&lt;p&gt;I've just learned about the &lt;a href=&quot;https://en.wikipedia.org/wiki/Terzan_5&quot; rel=&quot;nofollow&quot;&gt;Terzan 5 globular cluster&lt;/a&gt;, which according to Wikipedia has one of the highest star densities in our Galaxy - $10^6\ M_\odot/\mathrm{pc}^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem is, I've no idea how to interpret this number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the question is, what are the distances between stars in that cluster on average?&lt;/p&gt;&#xA;" OwnerUserId="14218" LastEditorUserId="10172" LastEditDate="2016-09-12T09:24:09.337" LastActivityDate="2016-09-12T11:42:15.383" Title="How far apart are stars in Terzan 5 cluster?" Tags="&lt;distances&gt;&lt;globular-clusters&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18320" PostTypeId="2" ParentId="18314" CreationDate="2016-09-12T08:19:49.380" Score="2" Body="&lt;p&gt;Black holes are solutions of the Einstein equations, the fundamental equations of Einstein's relativity, that link together space-curvature to density of matter and energy. The black holes solutions are found in the void, when you let a single point of the space-time be singular (the 'center of the black hole'). If you further assume that the black hole is static, you find three solutions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the Schwarzschild solution, this is a non-rotating non-charged black hole, which is entirely described by its mass M ;&lt;/li&gt;&#xA;&lt;li&gt;the Kerr solution, this is a rotating non-charged black hole, entirely described by its mass and spin (how fast it rotates) ;&lt;/li&gt;&#xA;&lt;li&gt;the Kerr-Newmann solution, describing a rotating charged black hole, entirely described by its mass, spin and charge.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It is a reasonable interpretation to think about a black hole as being a void with an infinitely small singularity in which lays all the mass, momentum and charge. In this sense, a black hole is empty, except for the region not described by the space-time framework (and that is not subject to the Einstein equations). So it is correct to think that black holes are only distorted region of space-time. The space-time is distorted so that going from the infinity (where there is no mass and no spin) you can reach the singularity continuously. If you fall onto a rotating black-hole, you will end up with the same spin as the black hole once you hit the singularity, (almost) regardless of how much spin you had initially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Schwarzschild black holes force you to fall inward once you cross the 'horizon' of the black hole, regardless of how much acceleration you can produce. Kerr black holes have the same horizon, as well as a so-called 'ergosphere'. This 'sphere' is a region of space where everything including light is forced to rotate in the direction of the spin of the black-hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the Earth, you cannot directly use the Kerr solution to approximate the rotating Earth (as stated previously, this is a solution of the equations &lt;em&gt;in void&lt;/em&gt;) how much drag you will have, but this gives you the correct idea, as stated by @James K.&lt;/p&gt;&#xA;" OwnerUserId="14154" LastActivityDate="2016-09-12T08:19:49.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18321" PostTypeId="2" ParentId="18319" CreationDate="2016-09-12T08:34:14.027" Score="6" Body="&lt;p&gt;You have to assume a characteristic mass for the stars in the cluster. Usually, to get an approximate I go with stars of mass $1 M_\odot$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, given the density of $10^6\ M_\odot.\mathrm{pc}^{-3}$ you have a density of stars of $\rho_\star = 10^6\ \mathrm{pc}^{-3}$. In a box of $1\ \mathrm{pc}^{3}$, you have $10^6$ stars, so the characteristic distance between stars is simply given by the density of stars to the minus one third:&#xA;$$ d_\mathrm{carac} = \rho_\star^{-1/3} \approx 10^{-2}\ \mathrm{pc} \approx 2000\ \mathrm{AU}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum: if you want to do a more correct calculation, you should take into account the different masses of stars to estimate the correct number of stars, using for example a &lt;a href=&quot;https://en.wikipedia.org/wiki/Initial_mass_function#Chabrier_.282003.29&quot;&gt;Chabrier&lt;/a&gt; initial mass function and try to get the mass function at the age of the cluster. Because of the fact that stars of about $\geq 1 M_\odot$ reach their red giant branch and lose much of their material in less than the age of the cluster, it will decrease the mean star mass of the cluster to something between $0.1$ and $0.8 M_\odot$ ($0.3 M_\odot$ according to Rob Jeffries). However it won't change the order of magnitude of the mean distance between stars.&lt;/p&gt;&#xA;" OwnerUserId="14154" LastEditorUserId="14154" LastEditDate="2016-09-12T11:42:15.383" LastActivityDate="2016-09-12T11:42:15.383" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18322" PostTypeId="1" AcceptedAnswerId="18426" CreationDate="2016-09-12T09:23:30.203" Score="0" ViewCount="162" Body="&lt;p&gt;I've read many times that every galaxy contains a black hole at the center. So is the creation of a galaxy linked to black holes?&lt;/p&gt;&#xA;" OwnerUserId="14136" LastEditorUserId="10437" LastEditDate="2016-09-12T12:13:50.543" LastActivityDate="2016-09-21T18:55:10.300" Title="Does creation of galaxy depend on black holes?" Tags="&lt;black-hole&gt;&lt;galaxy&gt;&lt;galaxy-center&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18323" PostTypeId="2" ParentId="18322" CreationDate="2016-09-12T11:48:11.210" Score="0" Body="&lt;p&gt;Most of the galaxies are believed to host a super massive black hole. The black hole in return has a dramatic impact on the evolution of the galaxy, an effect usually called &quot;AGN feedback&quot;. Quoting &lt;a href=&quot;https://astrobites.org/2012/09/11/does-agn-feedback-trigger-star-formation/&quot; rel=&quot;nofollow&quot;&gt;W. Ishibashi, A. C. Fabian&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Active Galactic Nuclei (AGN), regions of extremely high luminosity at the centers of some galaxies, are thought to radiate due to accretion onto a supermassive black hole.  This emitted radiation, or feedback, sends energy into its environment and influences the evolution of the host galaxy.  The details of this feedback are not well understood.  We do, however, observe several relationships between properties of the host galaxy and the central black hole — the most prominent of which is the M-sigma relation.  This empirical correlation states that the black hole mass (M) scales as the fourth power of the velocity dispersion (sigma) of stars in the galactic bulge.  Clearly, there is some connection between the evolution of the black hole and that of its host galaxy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The topic of understanding how the supermassive black hole form in the first place, evolve and interact with their galaxy is still in debate among the scientific community, so it is impossible to fully answer your question yet.&lt;/p&gt;&#xA;" OwnerUserId="14154" LastActivityDate="2016-09-12T11:48:11.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18324" PostTypeId="1" AcceptedAnswerId="18325" CreationDate="2016-09-12T13:08:04.467" Score="6" ViewCount="1049" Body="&lt;p&gt;Since the direction of Vernal Equinox lies in the Pisces constellation, why is its symbol Aries? Shouldn't it be ♓ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a historical anachronism, only invalidated by axial precession of Earth?&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2016-09-12T15:31:02.730" Title="Why is the symbol of Vernal Equinox ♈?" Tags="&lt;coordinate&gt;&lt;history&gt;&lt;symbols&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18325" PostTypeId="2" ParentId="18324" CreationDate="2016-09-12T15:31:02.730" Score="8" Body="&lt;p&gt;As Wikipedia explains in the article on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Zodiac#Precession_of_the_equinoxes&quot; rel=&quot;noreferrer&quot;&gt;Zodiac&lt;/a&gt;, the Zodiac was originally developed with the assumption of fixed equinoxes. It is assumed this is because the Babylonians who developed the Zodiac were unaware of the precession of the equinoxes. At that time, about 2500 years ago, the vernal equinox did in fact lie in Aries. For this reason, in astrology, that time is referred to as &quot;the Age of Aries&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, in modern times these astrological concepts are scarcely important to astronomy, and in the few cases where the astronomical symbol for the vernal equinox is used, it is more practical to have a fixed symbol than a changing one.&lt;/p&gt;&#xA;" OwnerUserId="6" LastActivityDate="2016-09-12T15:31:02.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18326" PostTypeId="1" CreationDate="2016-09-12T15:38:21.380" Score="-7" ViewCount="604" Body="&lt;p&gt;I've heard on various programs (yes TV) with &quot;supposed&quot; experts in various fields that there is evidence which would suggest that the moon could be hollow. The experts ranged from scientists, sound/audio analysts experts, to most importantly the . . . &lt;em&gt;ancient astronaut theorists&lt;/em&gt;! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is to know if this has already been proven to no way be true, or how easily this could be debunked (or proven) and optionally a brief theory of what this could mean scientifically either way to prove or disprove, etc. for what this actually is . . . . . a story or could it be true.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The story&lt;/strong&gt; seemed to also suggest Earth, the size of its moon, and the&#xA;   position of the moon to the Sun from Earth is extraordinary rare and&#xA;   some unknown technology could have done this so precisely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SsKv8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SsKv8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9969" LastActivityDate="2016-09-12T19:59:15.927" Title="The Hollow Moon Story" Tags="&lt;earth&gt;&lt;the-moon&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2016-09-12T18:38:53.647" ContentLicense="CC BY-SA 3.0" />
  <row Id="18327" PostTypeId="2" ParentId="18326" CreationDate="2016-09-12T17:50:58.150" Score="4" Body="&lt;p&gt;The gap between your question and physical reality is probably too broad to bridge. You need to provide us with a peer-reviewed (meaning published in a peer reviewed journal) theory. Unless you can, we can't really take this any further. Some of the problems with your question are: I don't know what you mean by &quot;hollow&quot;; your use of &quot;evidence&quot;, &quot;proven&quot; and &quot;true&quot; shows you misunderstand the scientific method, and you invoke myths (ancient astronauts) as if such nonsense should be addressed scientifically!&#xA; The mass of the Moon is known (to within a reasonable degree of accuracy). There have been a number of sets of seismometers placed on the Moon, and any model you (or your &quot;experts&quot;) provide must be consistent with the data from those. The Moon is a large enough body so that over geological time frames, it has assumed an approximately spherical shape; any model which included a vacuum or gas filled interior would need to describe the materials of the shell which would &lt;em&gt;not&lt;/em&gt; flow under the force of gravity and which were strong enough to resist the collapse of such a structure. In addition to gravity, centripetal forces, thermal forces, and impacts from asteroids/comets would also have to be resisted. The currently accepted model which best fits the evidence is that the Moon's core is an iron-rich solid surrounded by liquid/molten material. In Science, it is not the responsibility of the &quot;status quo&quot; to disprove any (hare-brained nonsense or otherwise) idea, it is the responsibility of the author of the proposal to show how it either explains some facts better or explains more facts than what is now &quot;accepted&quot;. In general, we don't find it productive to hypothesize that the data is the result of some unnatural process. This can be a problem, especially when the data has been faked or manipulated; but that has been the rare exception. And since the physical world is self-consistent, erroneous data is, if important enough, usually identified as such. In other words, as LaPlace wrote:&quot;The weight of evidence for an extraordinary claim must be proportioned to its strangeness.&quot; or as Carl Sagan paraphrased: &quot;Extraordinary claims require extraordinary evidence&quot;. What is it you are claiming and what evidence are you using to support it?&lt;/p&gt;&#xA;" OwnerUserId="14222" LastEditorUserId="53" LastEditDate="2016-09-12T19:59:15.927" LastActivityDate="2016-09-12T19:59:15.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18328" PostTypeId="1" AcceptedAnswerId="18329" CreationDate="2016-09-12T19:36:55.213" Score="0" ViewCount="258" Body="&lt;p&gt;The following paragraph is a scientific fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we look distantly into the universe (with telescopes) we see galaxies &#xA;(or whatever else you see) as they were millions years &#xA;ago- in other words it is how they looked in the past (this is because of the speed of light).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is could we build a technology (using current science or more speculative/theoretical science) using this scientific fact too see the earth history or past to such a level of detail that we could see historical events (e.g. individual dinosaurs) and the famous people's faces (in history books) - as if we were standing face to face with them- of the long ago past? What's the best level of detail could we theoretically see?&lt;/p&gt;&#xA;" OwnerUserId="14021" LastEditorUserId="14021" LastEditDate="2016-09-12T21:43:26.817" LastActivityDate="2016-09-12T21:43:26.817" Title="Using a telescope to look at the Earth History in Some Detail" Tags="&lt;light&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2016-09-13T13:29:48.113" ContentLicense="CC BY-SA 3.0" />
  <row Id="18329" PostTypeId="2" ParentId="18328" CreationDate="2016-09-12T19:57:40.127" Score="4" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we want to see Earth as it was 1000 years ago.  Assume that someone has set up a perfect mirror 500 light-years away, so that we can actually see the light that left Earth 1000 years ago. (That's a &lt;strong&gt;really big&lt;/strong&gt; assumption.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best telescopes in the world can't see the Apollo landing sites from Earth. We didn't get decent images of the descent stages, which were left on the surface, until the Lunar Reconnaissance Orbiter sent back photos it took from Lunar orbit. See &lt;a href=&quot;http://www.nasa.gov/mission_pages/LRO/news/apollo-sites.html&quot; rel=&quot;nofollow&quot;&gt;http://www.nasa.gov/mission_pages/LRO/news/apollo-sites.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1000 light-years is about &lt;em&gt;20 billion&lt;/em&gt; times as far away as the Moon. There's no way we could see people's faces at that distance with current technology -- or with any reasonable future technology. (There are physical limits on the resolution of an optical telescope of a given size.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And all this assumes that we have that perfect mirror out there. As far as I know, nobody has set up such a mirror for us, and all the light that left Earth 1000 years ago is now 1000 light-years away, badly faded, and beyond our reach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's conceivable that we could develop faster-than-light travel (which may or may not be physically possible), go out there, build a telescope with a really big aperture, and point it back at Earth. But that's not likely to happen any time soon, and I wouldn't know how to determine how good an image we &lt;em&gt;might&lt;/em&gt; be able to get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And going back millions of years just makes the problem worse.&lt;/p&gt;&#xA;" OwnerUserId="53" LastActivityDate="2016-09-12T19:57:40.127" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18331" PostTypeId="2" ParentId="16751" CreationDate="2016-09-13T00:30:38.267" Score="-1" Body="&lt;p&gt;There is also something called a &quot;giant eruption&quot;, like eta Carinae, which is a type of &quot;failed supernova&quot; because it didn't destroy the star.  It is not known what causes giant eruptions, but perhaps one thing to think about is a flash of the type you mention.  Giant eruptions certainly are visible, eta Car got quite bright when it blew off much of its envelope.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-09-13T00:30:38.267" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18332" PostTypeId="2" ParentId="12482" CreationDate="2016-09-13T00:40:42.660" Score="2" Body="&lt;p&gt;There might be a problem with your calculation, because by the time a white dwarf made of helium gets up to 0.5 solar masses, not only are the surface layers hot enough to fuse hydrogen, they are actually hot enough to fuse helium!  That's what will happen to the Sun when it ceases to be a red giant.  You are talking about carbon white dwarfs, but they are similar to helium white dwarfs.  So hydrogen fusion at the surface of a white dwarf is a common result-- it is what causes a phenomenon called a &quot;classical nova.&quot;  So you are on the right track-- but what you are talking about is a well-known area of study.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-09-13T00:40:42.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18333" PostTypeId="2" ParentId="13260" CreationDate="2016-09-13T01:07:00.203" Score="1" Body="&lt;p&gt;The surface temperature of fully convective stars is set by their opacity mechanism, often emission when the H minus ion is created.  The rate the star emits light is not set by the contraction rate, it is the other way around-- the star loses heat at a rate set by its structure, and that's what sets the contraction rate.  (As said, the rate of gravitational energy release is always double the rate the star loses heat to starlight.)  The reason the surface temperature always has to be something like 3000 K is pretty complicated, but it seems to be due to the fact that this temperature is a kind of peak in the surface's ability to emit light.  Convection sets the temperature structure, but there are a range of possible solutions with different internal energy, so the history of heat loss picks out whichever solution has the right internal energy.  But just why the surface is always about the same T is still the trickiest part to understand, I am hoping to get a better handle on that eventually.  Ignore anything you see in introductory textbooks, they never get it right.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-09-13T01:07:00.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18337" PostTypeId="1" CreationDate="2016-09-13T15:53:24.557" Score="2" ViewCount="158" Body="&lt;p&gt;I am a complete novice in this world and I would like a telescope where I could have enough power to view the moon's surface, I don't expect to see the flag or the hover sending signs to me, but perhaps enough to view cliffs, and other details from the surface of it and perhaps from other places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I honestly don't even know what kind of telescope to purchase. I live in a capital city, so there is a lot of light around here. My apartment is high with a balcony, which is where I would place it and there is nothing in front of me from there.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What kind of telescope would fit my needs?&lt;/li&gt;&#xA;&lt;li&gt;Are there reputable online stores that I could consult?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I have a $500 budget for this and honestly, I have no clue about specs or anything, I am a completely novice in this field.&lt;/p&gt;&#xA;" OwnerUserId="14235" LastEditorUserId="14235" LastEditDate="2016-09-13T16:24:25.150" LastActivityDate="2016-09-23T15:44:04.717" Title="What type of telescope can show cliffs on the moon surface from a city location?" Tags="&lt;telescope&gt;&lt;surface&gt;&lt;telescope-lens&gt;" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="18338" PostTypeId="2" ParentId="18202" CreationDate="2016-09-13T17:24:06.530" Score="3" Body="&lt;p&gt;A smaller field of view would &lt;em&gt;not&lt;/em&gt; necessarily help. The data transmitted back to Earth by Kepler was already limited by telemetry constraints and the portions of the CCD that were read out were already focused on small windows around (many) selected stars. Having a field of view that is at least 10 arcsecond across would be sufficient. What is more important is to make sure your star images are well-sampled. This is a bit of a problem for Kepler. The CCD pixels are quite big (4 arcseconds) to give a big field of view, but this is similar to the angular resolution of the telescope. This means all the light can fall in basically one pixel, which is not ideal for precise photometry. If you were focusing on one star you would make the field of view smaller but have lots of CCD pixels across your star image. This improves the precision of the photometry considerably.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to construct a similar telescope designed to look at one star at a time, then in principle you could have a much higher telemetry rate for that star. Hence, if you had enough photons and a low CCD readout noise, you could take much shorter exposures and see finer detail in the light curve transits - maybe moons, rings, starspots etc. The standard Kepler cadence was 30 minutes for most stars and one minute for some of the brightest stars - see &lt;a href=&quot;http://keplergo.arc.nasa.gov/PyKEprimerResources.shtml&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt;. A more rapid cadence would allow a much better characterisation of the host star through &lt;a href=&quot;https://en.wikipedia.org/wiki/Asteroseismology&quot; rel=&quot;nofollow&quot;&gt;asteroseismology&lt;/a&gt; (e.g. mass and radius), which in turn leads to better characterisation of its planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To take advantage of the higher telemetry rates, it would be advantageous to have a larger aperture telescope, simply to collect enough photons. The angular resolution of the telescope is not that important if investigating obviously single , bright sources. You would need an absolutely enormous telescope (100m+) to resolve the disks or planetary systems of even nearby stars, so this is not really feasible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A larger telescope would also allow you to split the light by wavelength and investigate transits in a wavelength dependent way. This is really important for investigating atmospheric composition and chemistry. This is perhaps the only area where focusing on a single star might be useful - because you could design a slit spectrograph to monitor the transit spectrum. On the other hand, these days it is possible to design multi-object spectrographs using fibres or slit masks, so overall I can't see any real advantage to having a telescope that is capable of gathering data for only one star at a time.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-09-13T17:24:06.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18339" PostTypeId="1" CreationDate="2016-09-13T18:35:15.787" Score="4" ViewCount="1153" Body="&lt;p&gt;I don't need &quot;exact&quot; numbers, but I'm trying to create an auto-generator for a stellar map to crate stars that are &quot;somewhat&quot; accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for numbers on the average number and type of planets around each type of Star, and whether they could be a good source for mining or are they good places for life?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, like, I need an idea of how what is the average number of planets that are expected around a G class star, how many are expected to be Gas giants or terrestrial and what is the likelihood of them being in the habitable zone and containing life. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I base it on Sol I would suspect about 10 planets, 50% terrestrial and 50% gas giants, with 1 to 3 in the habitable zone and a 30% chance of life developing. I'm looking for something similar to this type of information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume this wouldn't be good for across the board or even for G class stars in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I guess I'm looking for estimated planetary statistics based on spectral type? I don't want a list of exoplanet and stars that I have to look through and then make a bad guess at, but a good estimation of these numbers based on known facts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/872/is-there-a-upper-limit-to-the-number-of-planets-orbiting-a-star&quot;&gt;This Question&lt;/a&gt; doesn't have a simple enough answer for me to grab onto to use effectively without just making too much stuff up anyways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/10249/average-number-of-exoplanets-in-a-system&quot;&gt;This Question&lt;/a&gt; seems to have pieces of information that I'd use to work something out, but it would likely by far more inaccurate than what other people would come up with and there might be better answer than that so I ask my question.&lt;/p&gt;&#xA;" OwnerUserId="14084" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-09-14T11:20:48.650" Title="Average number and type of planets orbiting given star types" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;exoplanet&gt;&lt;planetary-systems&gt;" AnswerCount="1" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="18340" PostTypeId="2" ParentId="18283" CreationDate="2016-09-13T19:10:52.320" Score="7" Body="&lt;p&gt;You're right; the effect of solar tides on the Earth would be to increase the Earth's (or Earth+Moon's) orbital angular momentum around the Sun. But it would be difficult to compute this properly, and probably impossible to &lt;em&gt;measure&lt;/em&gt; the effect, because of all the other small alterations to the Earth's orbit going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple way to see the effect (which I gather you've already worked out) is to imagine two equal-mass objects in circular orbit about their common center of mass (midway between the two), one effectively a point mass (e.g., black hole) and the other being the Earth. If all motion is prograde (Earth rotates in same direction as orbit) and the Earth rotates faster than the orbital period, then the Earth's tidal bulge will be slightly ahead of the Earth-Sun direction. Then the near-side bulge will experience a retarding force from the point mass (slowing the Earth's rotation) while simultaneously pulling on the point mass in approximately the same direction as the latter's orbital motion. So the point mass is accelerated, adding to the angular momentum and making the orbit larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that most attempts to estimate the solar tidal torque on the Earth end up with it begin about 5 to 10 times weaker than the lunar torque, so most of the slowdown in the Earth's rotation is still due to the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, in the real Solar System the Sun &lt;em&gt;also&lt;/em&gt; experiences a tidal bulge raised by the Earth, which (because the Sun rotates faster than the Earth's orbit) will accelerate the Earth forward in its orbit, &lt;em&gt;also&lt;/em&gt; adding to the orbital angular momentum. But then you have remember that Venus and Mercury will raise &lt;em&gt;larger&lt;/em&gt; tidal bulges on the Sun with different periods (relative to the Earth's orbit), and so the combined effect gets complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for Mars and Ceres: yes, in principle the same effect would apply. But it would be weaker for Mars (smaller planetary radius, larger distance from Sun) and &lt;em&gt;much&lt;/em&gt; weaker for Ceres (same reasons, except moreso). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, &lt;em&gt;all&lt;/em&gt; of the orbits are being affected by N-body interactions with the rest of the planets (especially Jupiter), which I strongly suspect means that any potential changes in orbital angular momentum due to solar tides would be washed out by the various N-body and resonant interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(My understanding is that solar tidal braking could explain the current rotations of both Mercury and Venus, though you really can't be sure because you don't know what their &lt;em&gt;initial&lt;/em&gt; rotations were.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2016-09-13T19:10:52.320" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18342" PostTypeId="1" AcceptedAnswerId="18345" CreationDate="2016-09-14T08:42:00.340" Score="0" ViewCount="113" Body="&lt;p&gt;I read that two huge black holes collided or merged and so it created gravitational waves billion years ago , So now we are able to detect it &amp;amp; detected it. So what is the speed of gravitational wave ? Is it faster than light ? and  are there many more to be detected ?&lt;/p&gt;&#xA;" OwnerUserId="14136" LastActivityDate="2016-09-14T10:55:34.340" Title="What is the speed of gravitational waves ?" Tags="&lt;black-hole&gt;&lt;supermassive-black-hole&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2016-09-14T18:35:22.110" ContentLicense="CC BY-SA 3.0" />
  <row Id="18345" PostTypeId="2" ParentId="18342" CreationDate="2016-09-14T10:55:34.340" Score="5" Body="&lt;p&gt;Gravitational waves travel at the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many possible sources of gravitational waves. The &lt;em&gt;two&lt;/em&gt; confirmed detections so far (14 Sep 2016) are merging black hole binary systems, but sources of (detectable) gravitational waves may include merging neutron stars, short period binaries containing neutron stars or white dwarfs, gamma ray bursters or supernovae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a brief introduction see &lt;a href=&quot;http://www.ligo.org/science/GW-Sources.php&quot; rel=&quot;noreferrer&quot;&gt;Gravitational Wave Sources&lt;/a&gt; at the ALIGO site.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-09-14T10:55:34.340" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18346" PostTypeId="2" ParentId="18339" CreationDate="2016-09-14T11:14:41.890" Score="2" Body="&lt;p&gt;I wrestled with this one a bit, but because there's no real scientific answer to it, your best bet might to look at the 6 or 8 best known other solar-systems and build off that.   There's no right answer, so a base off evidence approach is probably the way to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, if you want to base it off nonsense, but sound impressive, you could use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Titius%E2%80%93Bode_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Titius-Bode law&lt;/a&gt;.   But I think that's a bad idea, unless large gas giant planetary migration is kept to a minimum, because a migrating Jupiter would make mincemeat of that &quot;law&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest exoplanets to find are planets with short orbits and the easiest stars for us to find planets around are smaller stars.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With those limitations to your view, given that large planets are easier to see, it's hard to say how many smaller ones are out there or how many distant planets from their star.   There's basically zero information to build on regarding those 2 points.   &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A few generalities though, just for fun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rocky planets like the 4 inner planets in our solar system form inside the the &lt;a href=&quot;https://en.wikipedia.org/wiki/Frost_line_(astrophysics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Frost Line&lt;/strong&gt;&lt;/a&gt;.  These planets are very low water by percentage.    Now, you could look at the Earth and say &quot;there's tons of water on Earth&quot;, but there's really not, at least, not in terms of water to silicate ratio.   Much of Earth's water is on the surface, so we see lots of water, but Earth is basically Iron, Nickle and Silicates with less than 1/3 of 1% of water by volume (even less by mass).   By volume, (using the 860 mile diameter sphere pictured below and from this &lt;a href=&quot;http://water.usgs.gov/edu/earthhowmuch.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt;).  Earth is about 1 part in 800 water by volume.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EKPDK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EKPDK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be as much or more water trapped in Earth's mantle than in the oceans, so the ratio might be closer to 1 part in 400 or there abouts, but the point remains the same.   If Earth is 0.25% water by volume, it's mostly made of stuff that isn't water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ceres, by comparison, may be as much as 25% water.  See &lt;a href=&quot;http://dawn.jpl.nasa.gov/science/ceres.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.space.com/1526-largest-asteroid-fresh-water-earth.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.  That's 100 times as much water to dry material ratio.   Outside the Frost-line you're going to get very different types of planets.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over time, you have to figure atmospheric changes, &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_escape&quot; rel=&quot;nofollow noreferrer&quot;&gt;which planets will lose their atmosphere&lt;/a&gt; and which planets will heat up like a greenhouse.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets take 3 types of stars, one with 1 solar mass like our sun, one with 0.4 solar masses (red dwarf) and one with 2.5 solar masses (very bright star).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our sun has a lifespan of roughly 10 billion years and a frost-line, currently at about 2.7 astronomical units (roughly Ceres distance from the sun), but when the sun was young and the planets were forming, our sun was some 30% less luminous, so the frost-line would have been (.7)^.5 * 2.7, or about 2.2 AU.   Now different frozen gases have different frost-lines, but this is just an approximation.   The fact that Ceres is about 25% water even though it's losing water to empty space implies that it formed outside the frost-line where frozen water was abundant.   Ceres is just one example.  All the large moons of the outer planets are also ice and rocky composites.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, take our 0.4 solar mass red dwarf - a very common type of star.  It's luminosity (0.4)^4th power, is about 1/39th as bright, so it's frost-line would be a bit over six times closer in, less than 0.4 AU.   With less gravitation it might not have a much smaller disk of spiraling matter at the time of formation, so a much larger percentage of the material that forms our red dwarf's solar-system would probably form outside the frost-line, which would imply, fewer 99% dry rocky worlds and more wet (25% water) worlds.   We have many worlds like that in our solar system, but they're all moons or dwarf planets.   Ceres, Ganymede, Europa, Titan, Triton, many others are icy-rocky composites that formed beyond our solar-systems frost-line.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of the Kepler observed exoplanets have densities that suggest &quot;water-worlds&quot;, many of them larger than Earth.   This fits with the red-dwarf, closer in frost-line argument.   We have no way of knowing how many rocky worlds smaller stars have, but I think it's a safe bet that they have a lower ratio of 99% or greater dry material worlds like our 4 inner planets simply because of the smaller zone inside the frost-line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reverse is true for the 2.5 solar mass star.   It also has a short lifespan of 10 billion / (2.5)^3, or about 640 million years, so there's less time for the planets to form, get past any early bombardments and cool down.    And the frost-line for our 2.5 solar mass star would be about 14 AU when the star is in it's early life, some 6 times further than the frost-line in our solar-system, so that's likely a much greater &quot;dry&quot; region for planet formation.   You probably get a higher percentage of rocky worlds around larger stars, so a star 1.5 or twice as bright as our sun could (in theory) have 6 or 8 rocky worlds of varying sizes ranging from hot to cold.  At least, I think it could, though I'm pretty sure this has never been observed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gas giants are fairly simple in nature.   All a gas giant needs to form is sufficient gravity to retain it's hydrogen and helium and a source of hydrogen and helium to collect.   A gas giant is more likely to form outside the frost line, but there's no reason why a gas giant couldn't form inside the frost-line.  You'd need about 8 earth masses of a rocky world at Earth temperatures to have a sufficiently high escape velocity that it could retain hydrogen and Helium and in theory, grow into a gas giant, but further away from the star where there's less solar heating, the amount of mass needed goes down with cooler surface temperatures, and with the additinoal mass of ices outside the frost line, it's probably far easier for gas giants to form outside the frostline than inside.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next thing to consider is &lt;a href=&quot;https://en.wikipedia.org/wiki/Planetary_migration&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Planetary Migration&lt;/strong&gt;&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take our Solar-system for example.  There's a theory that Neptune and Uranus may have formed formed inside Jupiter and Saturn, but Jupiter and Saturn pulled them outwards as they drew inwards, perhaps because in the early solar-system there was a period when Jupiter and Saturn were in resonance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also possible that there were 3 large gas giant planets in our solar-system initially.   (not counting Neptune and Uranus as &quot;large&quot; gas giants).   The 3 gas giant model helps explain the formation and water content of the 4 inner planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it's also thought that Jupiter migrated inwards towards the sun, then back outwards.   If Jupiter had kept migrating inwards, which may be the case for several hot Jupiter planets that have been observed, then there's no telling what would have happened to the inner planets, perhaps scattered every which way, some out of the solar-system, pushed too close to the sun or swallowed by the migrating hot Jupiter.  A migrating hot Jupiter would wreck havoc with smaller inner planets as they crossed orbits.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it is, our own Jupiter may have tossed a large gas giant planet completely out of the solar-system and perhaps made Neptune and Uranus switch places.    (all this is just numerical model based hypothesis, but several articles have been written about these ideas).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a very cool &lt;a href=&quot;https://www.youtube.com/watch?v=7dRLvSzDHo8&amp;amp;list=PLZ4fxe261udW2ViV7L-MEbgLFWl0IL7qH&amp;amp;index=20&quot; rel=&quot;nofollow noreferrer&quot;&gt;video&lt;/a&gt;, if you have some time that runs models on planet formation on &quot;sol&quot; like systems, estimating planet size, location and including how much water the inner planets are likely to get from comets.   Earth, it's thought, got most of it's water from comets and asteroids, largely thanks to Jupiter's migrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you skip forward to about 20 minutes 40 seconds in, you'll see he's computer modeled several different scenarios on planet formation with different gas giant combinations.   The chart shows eccentricity on the y-axis and distance from the star on the x-axis.  As eccentricity goes to one (top of the y-axis), the object is either sent outside the solar-system or close enough to the sun to get vaporized, so basically the dots you see flying up and out of the chart are leaving the solar-system one way or another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are, ofcourse other factors, mostly unknowns, like how much material is likley to be available for planet formation when a cloud of matter forms into a solar-system.   Our sun, for example, is something like 99.8% of the mass of the solar system with just 0.2% making up all the other stuff, but I don't know and I don't think anybody knows if that 0.2% is standard, higher than average or lower, and how larger stars might have different ratios than smaller onces, or how much of the material that gets blown off a star in it's young life gets captured by it's orbiting, forming planets.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's entirely possible that more planetary material can lead to larger but fewer planets as larger planets tend to sweep out a bigger area where less planetary material over the same area might actually give you more smaller planets, but how many planets are likely to form in an average star system, based on size of the star, I don't think anyone knows.   But predictions can be modeled on what happens after the planets form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What Kepler can tell us:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most red dwarf stars have planets and at least 25% have planets in their habitable zone.   &lt;a href=&quot;http://www.space.com/24894-exoplanets-habitable-zone-red-dwarfs.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.   (Red Dwarfs are the easiest planets to look at and look for stars).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the article:   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The new finds imply that virtually all red dwarfs throughout the Milky&#xA;  Way have planets, and at least 25 percent of these stars in the sun's&#xA;  own neighborhood host habitable-zone &quot;super-Earths,&quot; researchers said.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A simple (2-5 inner, 3-5 outer) won't cut it if you're looking for accuracy, but you might be able to run some models, based on how much you (randomly) decide that the gas giants move around and how much material you randomly give the planets.   Heavier planets would need more space around them.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for mining.  Venus would be a nightmare - too acidic.   Water worlds I would think would be terrible.   No solid platform to build on.   Ice worlds also not great.  Planets with past volcanic activity would be ideal and smaller planets are probably best, as most of the energy with mining probably goes into lifting the material into space, not actually the mining part.   Smaller planets could more easily build a space elevator.   Mars (volcanoes), maybe Mercury would probably be the best mining planets in our solar-system, unless you're mining for 3-HE, in which case, I'd look at the gas giants.   The moon is (I think), a terrible place to mine 3-HE, it's got so little of it, but the Moon's advantage is that it's close. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that said, I'd just look up 6 or 8 fairly well studied solar systems and build based on those, add some common sense, like a hot Jupiter would remove half the smaller planets, and go with that, because there is no real knowledge on what the most likely scenarios and probabilities are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Too long? / too many guesses?.  I think about exoplanets a lot.  It's one of my favorite subjects.  I'm very much looking forward to the James Webb Telescope's launch.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-09-14T11:20:48.650" LastActivityDate="2016-09-14T11:20:48.650" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18347" PostTypeId="1" AcceptedAnswerId="18348" CreationDate="2016-09-14T13:03:00.063" Score="6" ViewCount="1462" Body="&lt;p&gt;I would think that the amount of energy produced by one supernova is much more than a pair of colliding black holes, no? Wouldn't that much material and energy output from the explosion generate a powerful gravitational wave?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="2270" LastEditDate="2016-09-14T14:34:51.330" LastActivityDate="2016-09-14T21:33:54.000" Title="Do supernovae produce powerful gravitational waves?" Tags="&lt;supernova&gt;&lt;gravitational-waves&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18348" PostTypeId="2" ParentId="18347" CreationDate="2016-09-14T14:09:26.627" Score="10" Body="&lt;p&gt;Actually the peak luminosity for the first binary black hole merger LIGO detected was &lt;span class=&quot;math-container&quot;&gt;$3.6 \times 10^{49} \,\text{Watts}$&lt;/span&gt; (&lt;a href=&quot;https://losc.ligo.org/s/events/GW150914/GW150914-FactSheet-BW.pdf&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;). Whereas the brightest supernova ever recorded was ASASSN-15lh which had a peak luminosity of &lt;span class=&quot;math-container&quot;&gt;$2 \times 10^{38} \,\text{Watts}$&lt;/span&gt; (&lt;a href=&quot;https://en.wikipedia.org/wiki/ASASSN-15lh&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;). In fact the energy radiated by the binary black hole merger during the &lt;span class=&quot;math-container&quot;&gt;$0.2 \,\text{s}$&lt;/span&gt; it took to merge was over 10 times the combined luminosity of the observable universe (&lt;a href=&quot;http://www.ligo.org/science/Publication-GW150914/index.php&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;p&gt;This translates to the total energy emitted to be &lt;span class=&quot;math-container&quot;&gt;$5.4 \times 10^{47} \,\text{J}$&lt;/span&gt; for the black hole merger and &lt;span class=&quot;math-container&quot;&gt;$(1.1 \pm 0.2) \times 10^{45} \,\text{J}$&lt;/span&gt; for the ASASSN-15lh supernova. So as you can see, the binary black hole merger released around 100 times more energy than the brightest known supernova.&lt;/p&gt;&#xA;&lt;p&gt;In relation to the gravitational waves emitted by supernovae, their wave-forms are not as well understood as the &amp;quot;chirps&amp;quot; you get from merging black hole binaries, however they are still expected to be strong enough to be detected in the local universe.&lt;/p&gt;&#xA;" OwnerUserId="10437" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-09-14T21:33:54.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18349" PostTypeId="2" ParentId="18347" CreationDate="2016-09-14T14:13:46.517" Score="2" Body="&lt;p&gt;Quick answer: &lt;strong&gt;no&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the problem the use of the word &quot;energy&quot; here. Energy comes in many forms, such as heat, electrical energy, kinetic energy...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy released by a supernova is kinetic (material being thrown out into space), electromagnetic (light created by atomic interactions), etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational waves are not like light waves. They are only created by the movement of mass. So the material being thrown off will produce gravitational waves, the spinning of a star or stellar remnant will produce gravitational waves, but the masses and speeds involved are so much lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes have 10 or 100 times the amount of mass of a single star, and binary black holes are orbiting each other incredibly fast (the peak speed of the GW150914 black holes was 60% the speed of light), so they will produce much stronger gravitational waves, but not as much light.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2016-09-14T14:13:46.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18350" PostTypeId="2" ParentId="18347" CreationDate="2016-09-14T15:11:34.880" Score="7" Body="&lt;p&gt;Gravitational waves are produced by aspherical accelerating systems. A common example of a system that would produce them is two stars orbiting each other; an example of a system that would &lt;em&gt;not&lt;/em&gt; produce them is a totally stationary, non-rotating spherical star. (More specifically, the production of radiation &lt;a href=&quot;https://en.wikipedia.org/wiki/Quadrupole_formula&quot; rel=&quot;noreferrer&quot;&gt;depends on the mass quadrupole moment and its time derivatives&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supernovae are, on the whole, not spherically symmetric. We can see a simple example of this in &lt;a href=&quot;https://en.wikipedia.org/wiki/Pulsar_kick&quot; rel=&quot;noreferrer&quot;&gt;pulsar kicks&lt;/a&gt;, where a pulsar resulting from a supernova is expelled at an enormous speed (generally several hundred km/s) that cannot be accounted for by the motion of the originals star alone. There must be some drastic asymmetry responsible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What precisely would cause the emission of gravitational waves in a supernova? There are quite a few potential culprits (see &lt;a href=&quot;http://arxiv.org/abs/1212.4250&quot; rel=&quot;noreferrer&quot;&gt;Ott et al. (2012)&lt;/a&gt; and &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0509456&quot; rel=&quot;noreferrer&quot;&gt;Kotake et al. (2005)&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Asymmetric core bounce, possibly linked to progenitor rotation&lt;/li&gt;&#xA;&lt;li&gt;Convection inside the resulting neutron star&lt;/li&gt;&#xA;&lt;li&gt;Interactions between the neutron star and the shock wave&lt;/li&gt;&#xA;&lt;li&gt;Asymmetric neutrino emission&lt;/li&gt;&#xA;&lt;li&gt;Aspherical gas outflows&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;On the whole, rotation will have a large impact, especially when it comes to core bounce and convection. This should be obvious; it creates an asymmetry before the collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will we be able to detect those waves? It seems likely. Figure 132 of Kotake et al. indicates that some of our present detectors should be sensitive enough to detect neutrino- and matter- based gravitational waves from close sources ($\sim10\text{ kpc}$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nAjUQ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nAjUQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, &lt;a href=&quot;http://adsabs.harvard.edu/abs/2015ApJ...813...39A&quot; rel=&quot;noreferrer&quot;&gt;LIGO hasn't been successful in this area&lt;/a&gt;, but it's quite possible for it to detect gravitational waves from supernovae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're interested, I know of at least &lt;a href=&quot;http://www.astro.princeton.edu/~burrows/gw.web/&quot; rel=&quot;noreferrer&quot;&gt;one site&lt;/a&gt; that has openly available data from theoretical simulations that you can compare to detection capabilities.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-09-14T15:38:08.990" LastActivityDate="2016-09-14T15:38:08.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18351" PostTypeId="1" AcceptedAnswerId="18352" CreationDate="2016-09-14T15:53:15.880" Score="3" ViewCount="120" Body="&lt;p&gt;I just noticed &lt;a href=&quot;https://keplerscience.arc.nasa.gov/change-to-campaign-16-field-position-and-delay-to-go-cycle-5-deadline.html&quot; rel=&quot;nofollow&quot;&gt;this announcement&lt;/a&gt; from the K2 mission that the field observed for Campaign 16 of the K2 mission would be changed so that the telescope would be &quot;forward facing&quot;, saying that &quot;observing in this mode allows for simultaneous observations from K2 and from Earth.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(an aside: Campaign 17 was originally planned to be a forward facing to facilitate supernova science, but it was discovered that the telescope may not have enough fuel to last through Campaign 17, hence the modification to Campaign 16 to still facilitate supernova science)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exactly does it mean to be forward facing?&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2016-09-14T16:09:51.153" Title="What does &quot;forward facing&quot; mean for space telescopes?" Tags="&lt;space-telescope&gt;&lt;kepler&gt;&lt;k2&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18352" PostTypeId="2" ParentId="18351" CreationDate="2016-09-14T16:09:51.153" Score="4" Body="&lt;p&gt;This can best be described by two slides from &lt;a href=&quot;http://www.iap.fr/microlensing2016/presentations/Geert-Barentsen.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this presentation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Forward-facing&quot; implies looking towards Earth in the spacecraft's orbit, in the direction of the spacecraft's velocity vector:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rY4SW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rY4SW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Backward-facing&quot; implies looking in the opposite direction, away from Earth and in the opposite direction to that of the spacecraft's velocity vector:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HnmSq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HnmSq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are from the C9 campaign, which also used a forward-facing orientation.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2016-09-14T16:09:51.153" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18353" PostTypeId="1" AcceptedAnswerId="18412" CreationDate="2016-09-14T18:51:42.843" Score="0" ViewCount="107" Body="&lt;p&gt;What does a distant observer see if two masses with a given velocity are close enough to each other that time is dilated ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scenario (using 2 point masses for simplicities sake):&#xA;Mass a: 1,00E+030 kg&#xA;Mass b: 2,00E+030&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kinetic energy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;mass a: 1,39037910021726E+046&lt;/p&gt;&#xA;&#xA;&lt;p&gt;mass b: 2,78075820043453E+046&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this would mean a velocity of&#xA;Velocity: 149896229 m/s [ = c/2]&#xA;for both masses&lt;/p&gt;&#xA;&#xA;&lt;p&gt;distance between masses: 3960 m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the formula for time dilation &lt;img src=&quot;https://wikimedia.org/api/rest_v1/media/math/render/svg/fc63dfccb6a0a7379475600a65b36b65fb7b88e5&quot; alt=&quot;formula&quot;&gt; the time would pass at a rate of ~0,5 at mass a and at a rate of ~0,8 at mass b&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if both masses are flying parralel in the same direction does the observer see mass a flying with a speed of &#xA;74948115 m/s [= c/4] and mass b flying with a speed of 119916983 m/s [= c/2 *0.8]&#xA;?&lt;/p&gt;&#xA;" OwnerUserId="10326" LastActivityDate="2016-09-20T09:15:21.340" Title="absolute speed of bodies as observed by a distant observer" Tags="&lt;time-dilation&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18354" PostTypeId="2" ParentId="18316" CreationDate="2016-09-14T20:24:15.280" Score="10" Body="&lt;p&gt;After researching a bit more, I think I can answer my own question (or at least share what I have found). When Kepler derived his laws using Tycho Brahe's observational data of Mars, he followed the procedure below:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;first, the orbital period of Mars must be known. This value was already known in the time of Kepler (aprox. 687 days) &lt;a href=&quot;http://faculty.uca.edu/saustin/3110/mars.pdf&quot; rel=&quot;noreferrer&quot;&gt;$^1$&lt;/a&gt;, and can be derived as follows: first, one takes notes of the synodic period of Mars (which &quot;[...] is the time that elapses between two successive conjunctions with the Sun–Earth line in the same linear order.&quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_period#Varieties_of_orbital_periods&quot; rel=&quot;noreferrer&quot;&gt;$^2$&lt;/a&gt;). This can be done by calculating the elapsed time between two consecutive observations in which Mars is in the same apparent position in the sky. Then to calculate the orbital period of Mars, one can use:$\dfrac{1}{T_{M}} = \dfrac{1}{T_{E}} - \dfrac{1}{T_{M\mathrm{syn}}}$,&lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_period#Synodic_period&quot; rel=&quot;noreferrer&quot;&gt;$^3$&lt;/a&gt; where $T_{M}$ is the orbital period of Mars, $T_{E}$ is the orbital period of Earth and $T_{M\mathrm{syn}}$ is the synodical period or Mars.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;knowing the orbital period, Kepler then selected some of Brahe's observations that are exactly 687 days apart from each other (that is, one Martian year apart). This means that regardless of Mars relative position in the sky, the planet will be in the same position in relation to the Sun:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6AmDr.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6AmDr.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;using triangulation, Kepler then calculated the distance of Mars to the Sun in terms of the distance of Earth to the Sun (1 AU). He did this for several positions of Mars.&lt;a href=&quot;http://faculty.uca.edu/saustin/3110/mars.pdf&quot; rel=&quot;noreferrer&quot;&gt;$^1$&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;while any three points define a circle, four or more do not necessarily do so. Kepler noticed that the collected data of distances between Mars and the Sun could be fit into an ellipse with the Sun in one of the focuses.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1447" LastEditorUserId="9301" LastEditDate="2016-09-19T00:15:28.760" LastActivityDate="2016-09-19T00:15:28.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18355" PostTypeId="2" ParentId="639" CreationDate="2016-09-15T01:31:09.703" Score="2" Body="&lt;p&gt;A major factor, is that Venus' volcanoes are still active. Mars's died millions of years ago. If they were still erupting, then Mars' atmosphere would be much thicker today.&lt;/p&gt;&#xA;&lt;p&gt;Edit: This may actually be true as referenced by a paper published July 20 2020 cited below.&lt;/p&gt;&#xA;&lt;p&gt;Gülcher, A.J.P., Gerya, T.V., Montési, L.G.J. et al. Corona structures driven by plume–lithosphere interactions and evidence for ongoing plume activity on Venus. Nat. Geosci. (2020). &lt;a href=&quot;https://doi.org/10.1038/s41561-020-0606-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://doi.org/10.1038/s41561-020-0606-1&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Also see in Phys.org:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://phys.org/news/2020-07-scientists-volcanoes-venus.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Scientists discover volcanoes on Venus are still active&lt;/a&gt; discussed the above Nat. Geosci.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://phys.org/news/2020-01-scientists-evidence-venus-volcanoes.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Scientists find evidence that Venus has active volcanoes&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14242" LastEditorUserId="7982" LastEditDate="2020-07-21T02:14:16.857" LastActivityDate="2020-07-21T02:14:16.857" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="18356" PostTypeId="2" ParentId="335" CreationDate="2016-09-15T02:07:02.337" Score="0" Body="&lt;p&gt;And note you'll see the same things with cars on a highway.  If your car, in the left lane, is moving faster than a car in the right lane, and both of you are approaching a long arching curve to the left, then look at the other car against the reference of a line of distant trees as you both enter that gradual curve.  When the other car is well ahead of you, the fact that it is rounding a corner will make it appear to advance against the distant trees.   But as you catch up and pass that car, still in the turn, you will see that car back up against the trees.  Then as you get well ahead, again the other car will appear to advance against the trees, if you are still in the curve by then.   This is called &quot;parallax.&quot;&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-09-15T02:07:02.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18357" PostTypeId="1" AcceptedAnswerId="18359" CreationDate="2016-09-15T07:47:48.810" Score="14" ViewCount="7720" Body="&lt;p&gt;Obviously a star would be a point source. A galaxy should be an irregular blob if close, but if it is far away then it would seem that a galaxy too would be just a point source. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the star and galaxy were both only detectable as point sources can astronomers tell them apart with redshift? By some other method?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A followup question...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What percentage of galaxies if our universe can we only detect as point sources? &lt;/p&gt;&#xA;" OwnerUserId="9570" LastEditorUserId="11464" LastEditDate="2016-09-15T14:11:25.783" LastActivityDate="2017-04-12T20:29:18.853" Title="Can the difference between a star and a galaxy which are point sources be detected?" Tags="&lt;star&gt;&lt;galaxy&gt;&lt;redshift&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18358" PostTypeId="2" ParentId="13920" CreationDate="2016-09-15T08:45:21.607" Score="1" Body="&lt;p&gt;I found this article, which suggested a UHC filter could cover more bases if you can only buy one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.prairieastronomyclub.org/resources/by-dave-knisely/filter-performance-comparisons-for-some-common-nebulae/&quot; rel=&quot;nofollow&quot;&gt;http://www.prairieastronomyclub.org/resources/by-dave-knisely/filter-performance-comparisons-for-some-common-nebulae/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14098" LastActivityDate="2016-09-15T08:45:21.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18359" PostTypeId="2" ParentId="18357" CreationDate="2016-09-15T09:48:20.370" Score="14" Body="&lt;p&gt;To distinguish galaxies from stars, you can use the spectrum. Roughly, stars have a black-body like spectrum with features depending on the absorption and emission on the line of sight and in the chromosphere of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxies on the other hand of a spectrum that is the composite of tons of stars. The spectrum will for example be much wider (ranging from smaller to larger wavelengths) because of the diversity in the spectra of stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at &lt;a href=&quot;http://www.atnf.csiro.au/outreach/education/senior/astrophysics/spectra_astro_types.html&quot;&gt;http://www.atnf.csiro.au/outreach/education/senior/astrophysics/spectra_astro_types.html&lt;/a&gt; if you want a quick overview of the differences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have a precise number about the number of galaxies we see as point source, but the answer varies greatly from one instrument to another one. If you try to observe a galaxy using radiotelescopes in interferometry, you can resolve much better scales than an Earth-based small visible telescope, etc…&lt;/p&gt;&#xA;" OwnerUserId="14154" LastActivityDate="2016-09-15T09:48:20.370" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18361" PostTypeId="2" ParentId="18357" CreationDate="2016-09-15T12:39:37.923" Score="13" Body="&lt;strong&gt;Even &amp;quot;round&amp;quot; galaxies look different from stars&lt;/strong&gt;&#xA;&lt;p&gt;cphyc's answers the question excellently: Spectroscopy is the answer, although since — as explained below — galaxies are &lt;em&gt;not&lt;/em&gt; point sources, the morphology of stars and galaxies is also different: even elliptical galaxies observed along one of their axes look different from stars. Although both are round, the way that their light falls off radially is different; stars' light decrease roughly as a normal distribution from the center and out (with some extra profile folded in which depends on the instrument), while the &lt;em&gt;surface brightness profile&lt;/em&gt; of galaxies decrease in a somewhat more complicated fashion (e.g. a &lt;a href=&quot;https://en.wikipedia.org/wiki/Sersic_profile&quot; rel=&quot;noreferrer&quot;&gt;Sérsic profile&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;strong&gt;Can galaxies be point sources?&lt;/strong&gt;&#xA;&lt;p&gt;Wrt. the fraction of galaxies that are point source, the answer is virtually none. Galaxies can almost always be resolved although, as cphyc also correctly says, not with any instrument. Radio and gamma-ray telescopes have very poor resolution, and at these wavelengths the sources usually cannot be resolved unless they're relatively nearby. But at optical wavelengths, as well as UV and IR, telescopes like the Hubble Space Telescope and even good ground-based telescopes can resolve ~all galaxies, unless they're so small that they're too dim to be seen anyway.&lt;/p&gt;&#xA;&lt;strong&gt;Angular diameter in an expanding Universe&lt;/strong&gt;&#xA;&lt;p&gt;The reason is a rather peculiar feature of the expanding Universe: A galaxy will look smaller and smaller, the farther away it is (as expected from everyday life), but only out to a certain distance, after which they will appear larger and larger. Why is this so? Because light moves with a finite speed, we observe galaxies as they were in the past — the more distant, the longer time ago. And since in an expanding Universe, &amp;quot;long time ago&amp;quot; also means closer, the angle that a galaxy spans on the sky is the angle that it spanned when it &lt;em&gt;emitted&lt;/em&gt; the light, not the angle it spans &lt;em&gt;today&lt;/em&gt;. That is, very distant galaxies emitted the light we see today when they were so close that they spanned a large angle.&lt;/p&gt;&#xA;&lt;p&gt;The exact relation between distance and the solid angle of a galaxy depends on the cosmology (i.e. the values of density parameters, Hubble constant, etc.). For the latest &lt;a href=&quot;http://arxiv.org/abs/1502.01589&quot; rel=&quot;noreferrer&quot;&gt;Planck measurements (2015)&lt;/a&gt;, a galaxy that is 1 kpc (~3000 lightyears) across — which would be considered a  small galaxy — spans an angle given by this figure:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5UPyK.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5UPyK.png&quot; alt=&quot;arcsec_kpc&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;You'll see that galaxies look smaller and smaller the farther there are away, until at a distance of roughly 15 billion lightyears, after which they look larger again. The most distant galaxy observed, &lt;a href=&quot;https://en.wikipedia.org/wiki/GN-z11&quot; rel=&quot;noreferrer&quot;&gt;GN-z11&lt;/a&gt;, is so far away that its light was emitted less than half a billion years after the Big Bang. With a radius of &lt;span class=&quot;math-container&quot;&gt;$0.6\pm0.3\,\mathrm{kpc}$&lt;/span&gt; (&lt;a href=&quot;http://arxiv.org/abs/1603.00461&quot; rel=&quot;noreferrer&quot;&gt;Oesch et al. 2016&lt;/a&gt;) it still spans 0.15 arcsec, which is resolvable by HST.&lt;/p&gt;&#xA;&lt;strong&gt;Decreasing surface brightness&lt;/strong&gt;&#xA;&lt;p&gt;Unfortunately, this effect also makes distant galaxies more difficult to detect. A galaxy only emits so much light, so distributing its light over, say, twice the angular diameter, makes it four times less bright.&lt;/p&gt;&#xA;&lt;p&gt;Thus, the problem of observing very distant galaxies is not that they're small, but that they're &lt;em&gt;dim&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-04-12T20:29:18.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18363" PostTypeId="2" ParentId="18357" CreationDate="2016-09-15T15:47:24.517" Score="12" Body="&lt;p&gt;Good answers have already been given, but I wanted to provide another way of looking at it. Take a look at the image below, which is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble_Ultra-Deep_Field#Hubble_eXtreme_Deep_Field&quot; rel=&quot;noreferrer&quot;&gt;Hubble Extreme Deep Field&lt;/a&gt; (XDF) $-$ for those who don't know, this is a small patch of the sky that Hubble has stared at for a total of 23 days over 10 years $-$ and you'll notice something interesting. It is clear to see that many of the larger objects are galaxies, but you'll see a great number of smaller pinpoints of light (nearly 5,500 of them) which are galaxies so far away that Hubble can barely resolve their extent and size. Now take a look at the bright object in the lower right quadrant. You should see that it has blue and red spikes around it, referred to as &lt;a href=&quot;https://en.wikipedia.org/wiki/Diffraction_spike&quot; rel=&quot;noreferrer&quot;&gt;diffraction spikes&lt;/a&gt;. This object is clearly a star and you can primarily tell because of the diffraction spikes. You don't see these diffraction spikes on the galaxies, even the galaxies that are tiny pinpoints. This is a relatively easy way to visually distinguish between a star and a galaxy when you're looking at it through a telescope where such diffraction spikes are expected to occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/icAUQ.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/icAUQ.jpg&quot; alt=&quot;Hubble Extreme Deep Field&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that visually, stars and galaxies look different, even if they're both tiny spots on the image. There will also be differences in the way they look in less perceptible ways. This concept is capitalized upon by a program widely used by astronomers, &lt;a href=&quot;http://www.astromatic.net/software/sextractor&quot; rel=&quot;noreferrer&quot;&gt;SExtractor&lt;/a&gt;, designed to be given an image of the sky and be able to differentiate between stars and galaxies. It uses these small differences between the ways galaxies and stars appear in images to figure out which is which. If you want more detailed information on how this program distinguishes between stars and galaxies, take a look at their &lt;a href=&quot;http://adsabs.harvard.edu/abs/1996A%26AS..117..393B&quot; rel=&quot;noreferrer&quot;&gt;published paper&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-09-15T15:47:24.517" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18365" PostTypeId="1" CreationDate="2016-09-15T20:52:21.433" Score="0" ViewCount="1816" Body="&lt;p&gt;Circumpolar star definition: the stars that remain perpetually above horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My explanation: Latitude affects the constellation we see and Polaris(the North star) has altitude of approximately 89.4 degree. For example, in Miami(25 degree latitude) the Polaris is 90-25=65 degree above horizon. This means that for any city above equator, people are able to see the Polaris.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My explanation is verbose and probably incorrect. Could someone provide a valid one?  &lt;/p&gt;&#xA;" OwnerUserId="14254" LastEditorUserId="14254" LastEditDate="2016-09-15T21:28:21.647" LastActivityDate="2016-09-16T21:51:38.297" Title="How to explain that Polaris is a circumpolar star in the sky of any North America city?" Tags="&lt;star&gt;&lt;constellations&gt;&lt;pole&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18366" PostTypeId="2" ParentId="18365" CreationDate="2016-09-15T21:27:50.343" Score="2" Body="&lt;p&gt;In Miami, Polaris will be approx 25 degrees above the horizon, not 65. Height above the horizon is equal to the latitude of the observer.  At the North Pole, Polaris is very nearly directly overhead, all the time, because the Earth's axis of rotation points very nearly towards it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to precession, Polaris will not always be so close to where the Earth's axis points.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastEditorUserId="11822" LastEditDate="2016-09-16T21:51:38.297" LastActivityDate="2016-09-16T21:51:38.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18367" PostTypeId="1" AcceptedAnswerId="18370" CreationDate="2016-09-15T22:49:13.597" Score="6" ViewCount="956" Body="&lt;p&gt;After watching this &lt;a href=&quot;https://www.youtube.com/watch?v=gqXcXNUu1lg&quot; rel=&quot;nofollow&quot;&gt;JPL video&lt;/a&gt; of a Cassini time lapse of Saturn, I wondered since being mostly gases, how do we know how long a day—one rotation period—is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a usable relation between how fast the upper most layer of gases is and the planet's solid core?&lt;/p&gt;&#xA;" OwnerDisplayName="user11801" LastEditorDisplayName="user15217" LastEditDate="2016-12-27T04:39:01.573" LastActivityDate="2016-12-27T04:39:01.573" Title="How is a day measured on a gas giant?" Tags="&lt;planet&gt;&lt;rotation&gt;&lt;saturn&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18368" PostTypeId="2" ParentId="18367" CreationDate="2016-09-15T23:08:36.880" Score="5" Body="&lt;p&gt;I thought for a long time that day's on gas giants were measured by waiting for a certain feature come around, like Jupiter's great red spot.  However, looking it up just now I learned that that it is done by measuring the magnetic field.  &lt;a href=&quot;http://www.universetoday.com/15084/how-long-is-a-day-on-jupiter/&quot; rel=&quot;noreferrer&quot;&gt;This article&lt;/a&gt; is specifically about Jupiter, but I imagine measuring Saturn is much the same, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Scientist were finally able to use radio emissions from Jupiter’s&#xA;  magnetic field to calculate the planet’s rotational period and speed.&#xA;  While other parts of the planet rotate at different speeds, the speed&#xA;  as measured by the magnetosphere is used as the official rotational&#xA;  speed and period.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm afraid I don't have an answer to your second question.&lt;/p&gt;&#xA;" OwnerUserId="14255" LastActivityDate="2016-09-15T23:08:36.880" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18369" PostTypeId="1" CreationDate="2016-09-16T00:52:32.143" Score="0" ViewCount="100" Body="&lt;p&gt;Is it an assumption that the speed of propagation of gravitational waves is exactly the same as the speed of light? If the speed of gravity is only slightly less than the speed of light, the effect might not show up over solar or intra galactic distances, but over extragalactic distances could it give the appearance that there is dark matter influencing motion?&lt;/p&gt;&#xA;" OwnerUserId="14256" LastActivityDate="2016-09-16T08:54:19.713" Title="Is there evidence for the speed of gravity?" Tags="&lt;dark-matter&gt;&lt;gravitational-waves&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2016-09-16T13:31:57.833" ContentLicense="CC BY-SA 3.0" />
  <row Id="18370" PostTypeId="2" ParentId="18367" CreationDate="2016-09-16T01:23:26.420" Score="10" Body="&lt;p&gt;That the rotation period of the bulk mass of a planet is estimated through something with the magnetic field is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let me maybe elaborate a bit in-depth on that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No planetary magnetic field is a dipole. Earth has strong magnetic anomalies over the Atlantic, there may be ionic disturbances by moons like Io, or your magnetic field is just really crazy, like Uranus and Neptunes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jyLjY.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jyLjY.jpg&quot; alt=&quot;Magnetic Field model, showing multipolarity&quot;&gt;&lt;/a&gt;&#xA;(A simple, low-order model of Earth's magnetic field, from GFZ Potsdam)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But no matter how it looks, it will usually be constant on lines of latitude and longitude. This in particular signifies that the field and its distortions are created somehow by the bulk mass of the planet, else the field would move relative to a surface map, in the case of Earth.&#xA;So all this talk just to establish the following: We can take the shapes of the magnetic field as if they were surface shapes, and by following them determine a rate of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the story doesn't end there. While this sounds nice in principle, to do this in practice, we would need to send a probe and do in-situ measurments of this changing magnetic landscape.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what can be done from afar, and from Earth is observe ionospheric patterns. Because the ionosphere of a planet is well, ionized gas, at super-low pressures (roughly on the order of microbar to nanobar) it is frozen-in with the magnetic field that makes the particles of the plasma gyrate. This means, the ionospheric plasma follows the magnetic landscape, now translating into a plasma landscape. This plasma landscape is then usually observable from Earth by the radio signals they emit. It is also possible to see this in the ultraviolet, if there is strong plasma activity through auroral bombardment of a planet's upper atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your &lt;strong&gt;second question&lt;/strong&gt;, that's a simple no. &#xA;First of all there is no 'uppermost' layer of atmosphere, as the atmosphere is continously decreasing in density with increasing distance from the planetary body in question.&lt;br&gt;&#xA;What you're possibly thinking about are the upper cloud layers, as they visibly end at some height.&lt;br&gt;&#xA;But wind speeds are much more determined by pressure gradients created by the central star heating the planet, and the atmospheres ability to cool.&lt;br&gt;&#xA;There is one class of planet, however where a relation between the wind speeds at roughly 0.1 bar pressure and magnetic dynamo might exist. Those are &lt;a href=&quot;https://en.wikipedia.org/wiki/Hot_Jupiter&quot; rel=&quot;noreferrer&quot;&gt;Hot Jupiters&lt;/a&gt; which host atmospheres that might be sufficiently ionized for its winds to be braked down by the planetary magnetic field.&lt;br&gt;&#xA;Then there would exist a limiting speed, given a certain planetary field. However this is still a field of active research, but it would give you 'a' relation between the two.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2016-09-16T01:23:26.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18371" PostTypeId="2" ParentId="18369" CreationDate="2016-09-16T07:27:35.647" Score="0" Body="&lt;p&gt;It has long been one possibility that gravity does not work as in general relativity, or as in the Newtonian limit that is relevant to the gravity in a galaxy.  Thus, many different modifications to gravity have been tried.  But the reason they are generally not very successful is that dark matter seems to alter gravity on many different scales, from within a galaxy, to over galaxy clusters, to the expansion of the universe as a whole.  Thus it generally requires multiple different modifications to fit them all, and that makes for a very clunky model of gravity-- one can fit anything if given enough free parameters.  That sounds like it would be the problem for trying something based on the speed of gravity-- the timescale for propagation of changes would be very different over the three scales I just mentioned, yet dark matter has a similar influence on all three scales.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-09-16T07:27:35.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18372" PostTypeId="2" ParentId="18367" CreationDate="2016-09-16T08:25:28.667" Score="3" Body="&lt;p&gt;There's already good answer provided, but it is worth noting that the rotation rate of a planet is ill defined, as many planets are unlikely to be uniformly rotating. For example, Hot Jupiters present zonal flows, so should the rotation rate of the planet be measured taking into account the flows or not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't even know if there's a solid core in gas giants, so the question of the rotation rate should be 'is there a way to measure the rotation rate profile of a planet' and the answer is yes, but it's really hard. This is for example one of the long-term objectives of the Juno Mission. By measuring the pressure, density and composition of the atmosphere at different radius, you can try to stick a &lt;a href=&quot;https://en.wikipedia.org/wiki/General_circulation_model&quot; rel=&quot;nofollow&quot;&gt;general circulation model&lt;/a&gt; that would give you the local speed hence the rotation rate. Of course, the picture is made harder when you consider magnetic fields.&lt;/p&gt;&#xA;" OwnerUserId="14154" LastActivityDate="2016-09-16T08:25:28.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18373" PostTypeId="2" ParentId="18369" CreationDate="2016-09-16T08:54:19.713" Score="0" Body="&lt;p&gt;The fact that gravitational waves propagate at the speed of light is not an assumption, but is &lt;em&gt;derived from&lt;/em&gt; the assumption that nothing travels faster than light. In General Relativity, you build a metric, which is a tool to be able to measure and locate events in space and time. Building this metric puts constrains on the speed of information, and a basic assumption is that this speed is at most $c$ and is equal to $c$ if the information carrier is mass-less (photons for example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity waves are small non-static perturbation of this metric in the void. So they are propagating information without any mass and travel at $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is a physical approach to your question, answering it correctly would involve much harder calculations to perturb the metric, obtain the propagation equation and study the speed of the waves. This is for example explained in the wikipedia article in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_wave#Linear_approximation&quot; rel=&quot;nofollow&quot;&gt;linear section&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can think about an example (correct me if I'm wrong). Think about a massive object collapsing into a black hole. This is an extremely energetic process, if light was able to get out of the forming black hole, you would see it shining like a extremely bright supernova. To my knowledge there is no evidence of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More theoretically, I think that the assumption that light travels faster than gravity is incompatible with the cosmic censorship hypothesis,&lt;/p&gt;&#xA;" OwnerUserId="14154" LastActivityDate="2016-09-16T08:54:19.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18374" PostTypeId="1" CreationDate="2016-09-16T16:10:07.540" Score="9" ViewCount="728" Body="&lt;p&gt;I was reading &lt;a href=&quot;http://spaceflight101.com/gaia/gaia-instrument-information/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this overview article&lt;/a&gt; about the Gaia spacecraft and I saw the following statement:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;These spectra provide radial velocity information that are used to study the kinematic and dynamic evolution of the Milky Way. Radial velocities are derived from three isolated Calcium lines at 849.8, 854.2 and 855.2nm. Other lines in the 847 to 874nm range can provide data on star composition, surface gravity, and metal abundance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;note:&lt;/strong&gt; as pointed out in the comments below, the third line is at &lt;em&gt;866.2nm, not 855.2nm&lt;/em&gt; - this is suggested to be a typo - the '855' number also shows up on &lt;a href=&quot;http://sci.esa.int/gaia/40129-payload-module/?fbodylongid=1916&quot; rel=&quot;nofollow noreferrer&quot;&gt;this ESA page&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The high resolution spectroscopy only seems to operate between 847 and 874nm, and &amp;quot;three isolated Calcium lines&amp;quot; are used to measure radial velocity.&lt;/p&gt;&#xA;&lt;p&gt;Do all stars have enough calcium in their atmosphere to produce strong enough features to measure the radial velocity so accurately? I had thought that there are some stellar populations have very little besides hydrogen and helium in their atmosphere.&lt;/p&gt;&#xA;&lt;p&gt;Are these always emission lines or absorption lines, or will there be some stars with one and some of the other? What fraction of stars simply won't have significant amounts of calcium?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GS4hZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GS4hZ.png&quot; alt=&quot;Gaia's Radial Velocity Spectrometer (ESA)&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Gaia's Radial Velocity Spectrometer from &lt;a href=&quot;http://spaceflight101.com/gaia/gaia-instrument-information/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, credit: ESA.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/A5eLx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/A5eLx.jpg&quot; alt=&quot;Gaia's Imaging system, including mirrors 4, 5 and 6, prisms, diffraction gratings, and CCD array&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Gaia's Imaging system, including mirrors 4, 5 and 6, prisms, diffraction gratings, and CCD array, from &lt;a href=&quot;http://spaceflight101.com/gaia/gaia-instrument-information/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, credit: EADS Astrium.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jVhdY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jVhdY.jpg&quot; alt=&quot;Gaia's Optical Module, including Radial Velocity spectrometer (gratings) and a focal field corrector; SAS Astrium&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Gaia's Optical Module, including Radial Velocity spectrometer (gratings) and a focal field corrector, from &lt;a href=&quot;http://spaceflight101.com/gaia/gaia-instrument-information/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, credit: SAS Astrium.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2021-04-03T02:00:58.037" LastActivityDate="2021-04-03T02:00:58.037" Title="Why does Gaia use only calcium NIR lines for stellar radial velocity measurements?" Tags="&lt;spectroscopy&gt;&lt;velocity&gt;&lt;stellar-atmospheres&gt;&lt;metallicity&gt;&lt;gaia&gt;" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="18375" PostTypeId="1" CreationDate="2016-09-16T18:39:52.693" Score="2" ViewCount="180" Body="&lt;p&gt;I know that planets have a &quot;wandering&quot; movement on the celestial sphere. I also know that their period around the Sun vary largely as compared with the period of the Earth (one sidereal year), according to Kepler's third Law. I'm also aware though that their translation planes are closely co-planar with the ecliptic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, there must exist a pattern of their position with respect to the fixed stars on the celestial sphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone give some links, rules or charts on how to find out the celestial path of the planets on the celestial sphere over the year(s), with respect to the constellations of fixed stars?&lt;/p&gt;&#xA;" OwnerUserId="14262" LastEditorUserId="14262" LastEditDate="2016-09-17T18:23:04.743" LastActivityDate="2018-09-20T12:30:46.137" Title="The path of the planets on the celestial sphere" Tags="&lt;planet&gt;&lt;celestial-mechanics&gt;" AnswerCount="0" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18376" PostTypeId="2" ParentId="18374" CreationDate="2016-09-16T18:48:00.660" Score="7" Body="&lt;p&gt;The ESA &lt;a href=&quot;http://sci.esa.int/gaia/40129-payload-module/?fbodylongid=1916&quot; rel=&quot;nofollow&quot;&gt;states it pretty clearly&lt;/a&gt; (although their figure of 855.2 nm is incorrect; it should be 866.2 nm):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The RVS wavelength range, 847-874 nm, has been selected to coincide with the energy-distribution peaks of G- and K-type stars which are the most abundant RVS targets. For these late-type stars, the RVS wavelength interval displays, besides numerous weak lines mainly due to Fe, Si, and Mg, three strong ionised calcium lines (at around 849.8, 854.2, and 855.2 nm).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using &lt;a href=&quot;https://en.wikipedia.org/wiki/Wien%27s_displacement_law&quot; rel=&quot;nofollow&quot;&gt;Wien's law&lt;/a&gt;, we can see that stars with these as peak wavelengths in this interval correspond to effective temperatures in the range of 3000-3500 K:&#xA;$$T=\frac{b}{\lambda_{\text{max}}}$$&#xA;$$\begin{array}{|c|c|}&#xA;\hline \text{Wavelength (nm)}&amp;amp;\text{Temperature (K)}\\&#xA;\hline 847 &amp;amp; 3431\\&#xA;\hline 849.8 &amp;amp; 3409\\&#xA;\hline 854.2 &amp;amp; 3392\\&#xA;\hline 866.2 &amp;amp; 3345\\&#xA;\hline 874 &amp;amp; 3315\\&#xA;\hline&#xA;\end{array}$$&#xA;In reality, the majority of the stars Gaia studies have the most intense emissions at effective temperatures higher than this; these peaks correspond to hot M-type stars, not K- or G- type stars. The Sun, for instance, has an effective temperature of about 5800 K, and many K-type stars have effective temperatures around 4000 K. However, the target stars still guarantee intense emissions in the relevant parts of the spectrum, and thus noticeable calcium lines.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-09-17T03:22:12.760" LastActivityDate="2016-09-17T03:22:12.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18377" PostTypeId="2" ParentId="18374" CreationDate="2016-09-16T19:08:55.350" Score="11" Body="&lt;p&gt;The Ca triplet in the near infrared are extremely strong resonance &lt;em&gt;absorption&lt;/em&gt; lines. They are by far the strongest features in the near infrared spectra of cool G,K,M type dwarfs and giants, which will be the majority of the stars observed by the Gaia RVS.&#xA;The Ca triplet lines are so strong that even in low metallicity halo stars, that have little Ca in their photospheres, these lines are &lt;em&gt;still&lt;/em&gt; strong enough to measure radial velocities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lines are much weaker and much broader for hotter O, B and A stars, and measuring radial velocities for these will be difficult and much less precise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can have a look at an atlas of the Gaia Ca triplet region for stars of  different spectral types in Figure 2 of Munari et al. (2001). &lt;a href=&quot;http://cds.cern.ch/record/531022/files/0109057.pdf&quot; rel=&quot;nofollow&quot;&gt;http://cds.cern.ch/record/531022/files/0109057.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should also add that these three lines are &lt;em&gt;not&lt;/em&gt; the only features used to determine the velocities, they are just the strongest features in the spectra of most stars.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-09-17T06:43:02.323" LastActivityDate="2016-09-17T06:43:02.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18378" PostTypeId="1" CreationDate="2016-09-16T19:11:02.603" Score="7" ViewCount="387" Body="&lt;p&gt;Is the composition of the Solar Wind known? I'm especially interested in the heavy metals present. Most accounts deny the existence of anything other than electrons, protons, and some alpha particles; but it seems to me that if we can identify most of the known elements in the Sun's photosphere, then they are likely to exist in the Solar Wind at some concentration. If not, what is the detection limit (ppm? ppb? ppt?) we have so far established? Have any limits been established for atom clusters (dust?)? As a secondary question, I guess that (neutral) hydrogen and (negative hydride) H(-) do not exist in the solar wind due to its temperature being too high. Is this correct, and if so, is it true throughout the heliosphere or does the solar wind cool enough for the formation of bound species?&lt;/p&gt;&#xA;" OwnerUserId="14263" LastActivityDate="2016-09-17T17:48:52.207" Title="What is the composition of the Solar Wind?" Tags="&lt;observational-astronomy&gt;&lt;solar-wind&gt;&lt;elemental-abundances&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18379" PostTypeId="2" ParentId="18374" CreationDate="2016-09-16T19:59:40.407" Score="6" Body="&lt;p&gt;According to &lt;a href=&quot;https://arxiv.org/abs/1010.0419&quot;&gt;Cropper and Katz 2011&lt;/a&gt; part 2.2,&#xA;the RVS working group considered other bands,&#xA;but the ~850 nm band is relatively unaffected by absorption in the Earth's atmosphere,&#xA;facilitating ground-based preparation and follow-up.&#xA;In addition to the strong Ca II triplet,&#xA;this band is rich in lines enabling study of astrophysical quantities other than radial velocity,&#xA;adding to the science return on the spectrometer investment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For type B and hotter stars, a small minority of the population,&#xA;they hope to get radial velocity from&#xA;the Paschen &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrogen_spectral_series&quot;&gt;hydrogen series&lt;/a&gt;,&#xA;which accounts for the wide troughs at 854.3, 859.6, and 866.3 nm in the top of&#xA;&lt;a href=&quot;http://cds.cern.ch/record/531022/files/0109057.pdf&quot;&gt;Munari 2001&lt;/a&gt; figure 2.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2016-09-16T21:17:02.320" LastActivityDate="2016-09-16T21:17:02.320" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18380" PostTypeId="1" CreationDate="2016-09-16T21:52:18.120" Score="1" ViewCount="325" Body="&lt;p&gt;The formula below is used to calculate the synodic period between two bodies orbiting a same third one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_period#Synodic_period&quot; rel=&quot;nofollow&quot;&gt; $$\dfrac{1}{T_\mathrm{syn}} = \dfrac{1}{T_{1}} - \dfrac{1}{T_{2}}$$&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula can be used, for instance, to deduce the period of a planet in the Solar System, given that $T_\mathrm{syn}$ can be measured (period between two consecutive observations of a planet in the same relative position in the sky) and taking $T_1$ as 365.26 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read that this formula is derived from:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w_{\mathrm{syn}} = w_{1} - w_{2} = \dfrac{2\pi}{T_\mathrm{syn}} = \dfrac{2\pi}{T_{1}} - \dfrac{2\pi}{T_{2}} \therefore \dfrac{1}{T_\mathrm{syn}} = \dfrac{1}{T_{1}} - \dfrac{1}{T_{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: which approximations are required for this result to hold true? Must the orbits be circular? Must these two bodies be in the same plane around the third? Also, considering the case of elliptical orbits but still assuming that the mass of the third body $M ≫ m_{1}$ and $M ≫ m_{2}$, would the synodic period of body 2 as observed from body 1 be indeed constant?&lt;/p&gt;&#xA;" OwnerUserId="1447" LastEditorUserId="9301" LastEditDate="2016-09-17T00:01:27.940" LastActivityDate="2018-09-04T08:20:35.997" Title="Which approximations are required for the synodic period formula?" Tags="&lt;orbit&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18382" PostTypeId="1" CreationDate="2016-09-17T10:55:14.020" Score="1" ViewCount="126" Body="&lt;p&gt;As opposed to spiral galaxies, which are rotationally supported, elliptical galaxies are said to be supported by random motions of their stars. I don't really have an intuitive understanding of this. How do random motions prevent a collapse?&lt;/p&gt;&#xA;" OwnerUserId="6965" LastActivityDate="2016-09-17T16:09:25.523" Title="Ellipticals supported by random motions" Tags="&lt;galaxy&gt;&lt;galactic-dynamics&gt;&lt;stellar-dynamics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18383" PostTypeId="2" ParentId="18378" CreationDate="2016-09-17T15:36:11.190" Score="8" Body="&lt;p&gt;We have very good data on the heavy metals in the solar wind &lt;a href=&quot;http://sohowww.nascom.nasa.gov/gallery/Particle/cel001.html&quot; rel=&quot;noreferrer&quot;&gt;from the Charge, Element, Isotope Analysis System (CELIAS) on SOHO&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4xJ8T.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4xJ8T.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of these elements were previously known; others were observed for the first time. The new ones are phosphorus, chlorine, potassium, titanium, chromium, and nickel. They were detected in smaller amounts than, say, carbon, oxygen or calcium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the seeming overabundance of heavier elements, with the peak at iron, might seem surprising. One would think that the lighter metals - carbon, oxygen, etc. - might appear in higher quantities, given their substantially lower masses, which make it easier for them to escape the Sun's atmosphere. However, this is due to experimental bias. The instruments on SOHO were designed to search for these particular heavier elements, which of course led to the detection of several new isotopes. We now have a much better picture of what heavy metals can be found in the solar wind, but the chart above should &lt;em&gt;not&lt;/em&gt; be taken as evidence of the relative quantities, as a whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, there actually &lt;em&gt;is&lt;/em&gt; some neutral hydrogen, both in the solar wind and the solar atmosphere (see &lt;a href=&quot;http://adsabs.harvard.edu/full/1994ApJ...420..913O&quot; rel=&quot;noreferrer&quot;&gt;Olsene et al. (2004)&lt;/a&gt; and &lt;a href=&quot;http://image.gsfc.nasa.gov/publication/document/2001_collier_etal_04.pdf&quot; rel=&quot;noreferrer&quot;&gt;Collier et al. (2001)&lt;/a&gt;). It should be present in the corona, and can be detected and modeled by Lyman $\alpha$ scattering. It was also detected and studied by the Low Energy Neutral Atom (LENA) imager on &lt;a href=&quot;https://en.wikipedia.org/wiki/IMAGE&quot; rel=&quot;noreferrer&quot;&gt;IMAGE&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-09-17T17:48:52.207" LastActivityDate="2016-09-17T17:48:52.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18384" PostTypeId="2" ParentId="18382" CreationDate="2016-09-17T16:09:25.523" Score="3" Body="&lt;p&gt;The same way rotation does.  If you look at each star, it is following some kind of orbit, so it has a motion.  The only difference between a spiral and an elliptical is that the motions in a spiral are all in the same plane and the same direction, but in the elliptical, they are all over the place.  So &quot;random velocity&quot; just means you look at a bunch of stars-- each one is still in an orbit, so it's no surprise it stays in orbit.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-09-17T16:09:25.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18386" PostTypeId="2" ParentId="6571" CreationDate="2016-09-17T21:43:09.633" Score="1" Body="&lt;p&gt;Wikipedia states that 99.9% of the mass originally in the asteroid belt was lost in the first 100 million years of the solar system - I am assuming after the protosun initiated fusion.  If so, that would put the potential mass at 4000X the moon, or almost twice Neptune's mass.  That would be one big planet!&lt;/p&gt;&#xA;" OwnerUserId="14276" LastActivityDate="2016-09-17T21:43:09.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18387" PostTypeId="1" CreationDate="2016-09-18T05:47:04.620" Score="4" ViewCount="185" Body="&lt;p&gt;A sentence from the abstract of &lt;a href=&quot;http://arxiv.org/abs/1112.3781&quot; rel=&quot;nofollow noreferrer&quot;&gt;The population of natural Earth satellites&lt;/a&gt; states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At any given time there should be at least one NES of 1-meter diameter orbiting the Earth. The average temporarily-captured orbiter (TCO; an object that makes at least one revolution around the Earth in a co-rotating coordinate system) completes (2.88 ± 0.82) rev around the Earth during a capture event that lasts (286± 18) d.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As far as I know, the only known, well documented temporarily captured Earth satellite is &lt;a href=&quot;https://en.wikipedia.org/wiki/2006_RH120&quot; rel=&quot;nofollow noreferrer&quot;&gt;2006 RH120&lt;/a&gt;, which was in within 1 million km of the Earth for about 1 year. It has an estimated size of 2 to 3 meters, a rotation period (about it's axis) of about 3 minutes (presumably &lt;a href=&quot;http://www.aanda.org/articles/aa/pdf/2009/09/aa10965-08.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;determined by photometry&lt;/a&gt;). According to Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2006 RH120 was discovered on 14 September 2006 by the 27-inch (690 mm) Schmidt camera of the Catalina Sky Survey in Arizona.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using data from the &lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL Horizons&lt;/a&gt; database, it seems 2006 RH120 was only about 850,000 km from Earth at the time it was discovered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deep sky surveys are extremely prolific at detecting small solar system bodies. There was also a &lt;a href=&quot;https://astronomy.stackexchange.com/a/6211/7982&quot;&gt;recent answer&lt;/a&gt; addressing the possibility that the Gaia spacecraft's could detect trojan asteroids associated with Earth's orbit, linking to the paper &lt;a href=&quot;http://arxiv.org/abs/1212.0268&quot; rel=&quot;nofollow noreferrer&quot;&gt;Detection of inner Solar System Trojan Asteroids by Gaia&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://astronomy.stackexchange.com/a/6211/7982&quot;&gt;question&lt;/a&gt; asked specifically about TCO's being detected by Gaia, but my question is more general: &lt;em&gt;How would a small TCO (temporarily captured orbiter) or other natural Earth satellite&lt;/em&gt; most likely &lt;em&gt;be detected?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are Earth-based sky surveys, space based visible and near-visible, and possibly even infrared/thermal detections possible, even Earth-based radar, and possibly others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any searches specifically targeting TCOs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RkyCJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RkyCJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; a simple plot of the orbit of 2006 RH120 from &lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL Horizions&lt;/a&gt; transformed to an Earth-centered Earth-Sun-synodic frame - red line with dots every 10 days, for the period when it was within a +/- 1.6 million km cube centered on Earth in this frame. The green thick band around Earth is the moons orbit during this 448 day period.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-02-16T05:59:56.003" LastActivityDate="2018-05-17T17:15:17.807" Title="How would a small TCO (temporarily captured orbiter) or other natural Earth satellite most likely be detected?" Tags="&lt;observational-astronomy&gt;&lt;natural-satellites&gt;&lt;space-telescope&gt;&lt;near-earth-object&gt;&lt;gaia&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18390" PostTypeId="1" AcceptedAnswerId="18396" CreationDate="2016-09-19T01:46:12.890" Score="6" ViewCount="572" Body="&lt;p&gt;The black hole &lt;a href=&quot;http://web.archive.org/web/20120507004507/http://www.cosmosmagazine.com/node/873&quot;&gt;GRS 1915+105 was observed to be spinning&lt;/a&gt; at near the theoretical &lt;a href=&quot;http://www.universetoday.com/109308/how-fast-do-black-holes-spin/&quot;&gt;maximum spin rate&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you had two such black holes spinning clockwise, and also orbiting each other clockwise, what would happen if they merged? How could the new black hole avoid breaking the rotational speed limit? Where would the extra angular momentum go?&lt;/p&gt;&#xA;" OwnerUserId="6870" LastActivityDate="2016-09-22T20:45:52.967" Title="Could the merger of two rotating black holes create another that breaks the rotational speed limit?" Tags="&lt;black-hole&gt;&lt;rotation&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18391" PostTypeId="1" AcceptedAnswerId="18394" CreationDate="2016-09-19T04:13:01.120" Score="4" ViewCount="3175" Body="&lt;p&gt;I know that JD0.0 is Jan. 1st 4713 BC at Noon, a year in the Julian calendar is 365.25 days and that the number after 'JD' is the number of days so...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to calculate the number of days between Jan1.2000AD and Jan1.4713BC and I do&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \text{# of years} = 4712+2000=6712 \text{ years (skipping the year 0)} $$&#xA;$$ \text{# of days} = 6712 \cdot 365.25 =2451558 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think it should be JD2451558. An extra 13 days. What's going on here? I thought maybe the length of a year in the Julian calendar is $2451545/6712=365.2480632\text{days}$ but I can't find anything to support that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine that people would have questioned whoever proposed J2000 as JD2451545.0 and got a satisfying answer but I can't find one! Help&lt;/p&gt;&#xA;" OwnerUserId="13724" LastActivityDate="2016-09-19T11:38:04.400" Title="Why is JD 2451545.0 = January 1, 2000, Noon instead of JD 2451558.0?" Tags="&lt;time&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18392" PostTypeId="2" ParentId="18390" CreationDate="2016-09-19T05:34:09.370" Score="5" Body="&lt;p&gt;I believe the orbital angular momentum would need to radiate away in gravitational waves as the merger proceeded.  I don't know if any of the spin angular momentum would also radiate away, or if you'd still end up with a maximally rotating black hole.  That's probably an oversimplification, one should be skeptical of anything but a full calculation.  Note, for example, that we would not want to overlook the kinetic energy of the orbit when we calculate the mass-energy of the system, so if some of that mass-energy ends up in the final black hole mass, it might not seem maximally rotating any more.  So although I'm not giving a calculation, I'm saying two things:&#xA;1) orbital decay will produce gravitational waves that might carry away angular momentum, and&#xA;2) orbital energy counts as mass-energy for the ultimate black hole, so it might not be maximally rotating even if you forget gravitational radiation.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-09-19T05:34:09.370" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18394" PostTypeId="2" ParentId="18391" CreationDate="2016-09-19T08:38:01.660" Score="10" Body="&lt;p&gt;This &lt;a href=&quot;https://en.wikipedia.org/wiki/Julian_day&quot;&gt;Wikipedia article&lt;/a&gt; states (correctly) that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The Julian Day Number (JDN) is the integer assigned to a whole solar&#xA;  day in the Julian day count starting from noon Greenwich Mean Time,&#xA;  with Julian day number 0 assigned to the day starting at noon on&#xA;  January 1, 4713 BC, proleptic &lt;a href=&quot;https://en.wikipedia.org/wiki/Julian_calendar&quot;&gt;Julian calendar&lt;/a&gt; (November 24, 4714 BC,&#xA;  in the proleptic &lt;a href=&quot;https://en.wikipedia.org/wiki/Gregorian_calendar&quot;&gt;Gregorian calendar&lt;/a&gt;)&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Julian year consists of 365.25 days while a Gregorian year consists of 365.2425 days.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the introduction of the Gregorian calendar, the difference between Gregorian and Julian calendar dates has increased by three days every four centuries (all date ranges are inclusive).  In addition, 10 days were added to the Julian calendar to obtain the Gregorian calendar at the time of the calendar reform. The current difference between the Gregorian and Julian calendar dates is 13 days: 14 September 2016 Gregorian date is 1 September 2016 (Julian date).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should explain the extra 13 days in your calculations.&lt;/p&gt;&#xA;" OwnerUserId="5537" LastActivityDate="2016-09-19T08:38:01.660" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18395" PostTypeId="2" ParentId="18391" CreationDate="2016-09-19T11:13:42.550" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So I think it should be JD2451558. An extra 13 days. What's going on here?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bl4ZG.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bl4ZG.gif&quot; alt=&quot;British calendar for September 1752, where the first three days were Tuesday September 1, Wednesday September 2, and Thursday September 14&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The switch from the Julian calendar to the Gregorian calendar involved two changes. One change was the frequency at which leap years occur. The Julian calendar had a leap year every four years, making a Julian year 365.25 days long on average. The Gregorian calendar has 97 leap years every 400 years, making a Gregorian year 365.2425 days long on average.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other change was a one-time shortening of some year to bring the calendar back in line with the seasons in the third century AD. Different countries did this differently, some as early as 1582, others as late as 1923. As depicted above, Great Britain made this change in 1752 by making 3 September 1752 to 13 September 1752 vanish (that's eleven days). Since that change, both 1800 and 1900 would have been leap years in the Julian calendar but were not in the Gregorian calendar. Eleven days lost in September 1752 plus two more days for no February 29th in 1800 or 1900 exactly accounts for your thirteen day discrepancy.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2016-09-19T11:38:04.400" LastActivityDate="2016-09-19T11:38:04.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18396" PostTypeId="2" ParentId="18390" CreationDate="2016-09-19T14:17:00.673" Score="10" Body="&lt;p&gt;The simple answer is that &lt;strong&gt;no, black holes cannot merge into a new black hole with a spin greater than the maximum allowed spin.&lt;/strong&gt; The reason being that such an act would cause the black hole to show us a &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Naked_singularity&quot; rel=&quot;noreferrer&quot;&gt;naked singularity&lt;/a&gt;&quot;. Black holes have an event horizon which shields their singularity from observation by the outside universe. As they spin faster, that event horizon shrinks. At some point, it completely disappears, causing the singularity to become &quot;naked&quot; and observable to the universe. Some believe that naked singularities cannot possibly form, mainly because if they could, causality and physics would break down. This concept is known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_censorship_hypothesis&quot; rel=&quot;noreferrer&quot;&gt;cosmic censorship hypothesis&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a statement concerning this matter in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/052151407X&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Numerical Relativity: Solving Einstein's Equations on the Computer&lt;/em&gt;&lt;/a&gt;, which I believe is relevant (emphasis mine).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Black hole spins that are aligned with the orbital angular momentum&#xA;  increase the binary's total angular momentum. If this total angular&#xA;  momentum exceeds the maximum angular momentum of a Kerr black hole,&#xA;  then &lt;strong&gt;the binary cannot merge until a sufficient amount of angular&#xA;  momentum has been radiated away&lt;/strong&gt;. Quite generally, we expect binaries&#xA;  with black hole spins aligned with the orbital angular momentum to&#xA;  merge more slowly than binaries with spins that are anti-aligned.&#xA;  This effect, sometimes &lt;strong&gt;referred to as &quot;orbital hang-up&quot;&lt;/strong&gt;, has been&#xA;  explored with numerical simulations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://arxiv.org/pdf/gr-qc/0604012v2.pdf&quot; rel=&quot;noreferrer&quot;&gt;One such numerical simulation&lt;/a&gt; looked at aligned and anti-aligned mergers and found that the aligned black holes took much longer to merge and radiated away more energy, in the form of gravitational waves, before they merged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another way to burn off some of that spin angular momentum too. When spinning black holes merge, they experience a &quot;kick&quot; in their linear momentum. In other words, they suddenly speed up in their motion through space. This kick is a result of converting some of the two distinct black holes' orbital and spin momentums into the merged black hole's linear momentum. Quoting the same book as above concerning these kicks:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Most of the initial calculations focused on black hole spins that are&#xA;  aligned or anti-aligned with the orbital angular momentum. The&#xA;  resulting kicks are several hundred km/s in magnitude, easily&#xA;  exceeding the maximum kick of approximately 175 km/s found for&#xA;  non-spinning black holes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In summation, two spinning black holes cannot merge into a black hole which is spinning faster than the maximal rotation. &lt;strong&gt;Gravitational waves and the linear momentum &quot;kick&quot; play an important part in helping binary black holes to lose spin energy, such that the merged black hole doesn't exceed a maximal rotation.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the really interesting question that should be asked is, how do two black holes know, before merging, that their merger cannot happen as it would violate the cosmic censorship hypothesis? How do they know they need to radiate away extra energy before they can merge? What is the mechanism that prevents them from merging? Only time can tell for now.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2016-09-19T16:20:49.730" LastActivityDate="2016-09-19T16:20:49.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18397" PostTypeId="1" AcceptedAnswerId="18398" CreationDate="2016-09-19T14:49:54.297" Score="2" ViewCount="1026" Body="&lt;p&gt;The Gaia DR1 was released middle of September with ~1.1 billion objects. The schema contains right ascension, declination and parallax, which could be used to calculate Cartesian co-ordinates (xyz).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overwhelming majority (&gt;99.998%) of objects do not contain an entry for parallax. I suspect this needs to be computed using other columns (e.g. phot_g_mean_mag?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you compute the parallax for an object when its not included?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data model is documented &lt;a href=&quot;http://gaia.esac.esa.int/documentation/GDR1/datamodel/Ch1/gaia_source.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; .&lt;/p&gt;&#xA;" OwnerUserId="14283" LastActivityDate="2018-01-08T23:00:21.407" Title="Calculating the distance of objects using the Gaia DR1" Tags="&lt;astrometry&gt;&lt;gaia&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18398" PostTypeId="2" ParentId="18397" CreationDate="2016-09-19T15:59:19.807" Score="1" Body="&lt;p&gt;You &lt;em&gt;could&lt;/em&gt; use phot_g_mean_mag (apparent magnitude), but you would have to assume an absolute magnitude for each star or galaxy you were interested in. From the data in that table, there's no way to do that. (Some of the &lt;em&gt;matched&lt;/em&gt; tables might give you a starting point: e.g., if you decided that Gaia source X was matched by SDSS source Y, then you could track down the SDSS measurements, which might let you decide what kind of star it was and thus guess at its absolute magnitude.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2016-09-19T15:59:19.807" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18399" PostTypeId="1" AcceptedAnswerId="18403" CreationDate="2016-09-19T17:41:18.413" Score="7" ViewCount="5315" Body="&lt;p&gt;Earth rotates around the sun and around its axis (A North, South axis) giving us days, nights and seasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any known planets that rotate with an axis pointing toward its sun so that one half of the planet is always facing it (the other half is always facing away from it)? Like Uranus, but not switching which side is facing the sun every 6 months.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Would this even be theoretically possible?)&lt;/p&gt;&#xA;" OwnerUserId="14287" LastEditorUserId="616" LastEditDate="2016-09-19T19:50:42.323" LastActivityDate="2016-09-20T08:02:43.817" Title="Are there any known planets whose axis is oriented such that one pole always faces its star?" Tags="&lt;planet&gt;&lt;rotation&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18400" PostTypeId="2" ParentId="18399" CreationDate="2016-09-19T18:03:54.203" Score="10" Body="&lt;p&gt;Uranus has an axial tilt of about &lt;a href=&quot;https://en.wikipedia.org/wiki/Uranus#Axial_tilt&quot;&gt;98 degrees&lt;/a&gt;, so I think this is about as close as you'll get for planets with known tilts.  However, you won't find one with a pole that always faces the sun, just because of the geometry of the situation.  With Uranus, for about 1/4 of its 84 (Earth year) year, it will have one pole more or less facing the sun, then for the next 1/4 it'll have both poles facing perpendicular to its orbit (away from the sun), then the next 1/4 it'll have the opposite pole facing the sun, then for the final 1/4 it'll have both poles facing away again.  This is a bit of an over simplification, but it captures the general idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the two poles point mostly the same direction in space**, and for part of its orbit one pole or the other is pointing more or less at the sun, but for the periods in between it will face away.  If you want a planet who always has the same side pointing at the sun, it has to rotate with a North/South pole similar to Earth, and it needs its &quot;day&quot; to exactly equal its &quot;year&quot;.  None of the planets in our solar system do this, however many moons do.  Earth's Moon, as well as all of Jupiter's and Saturn's major moons do this.  This is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_locking&quot;&gt;tidal locking&lt;/a&gt;.  Note that Mercury is on the list I linked, but it isn't perfectly tidally locked, it is in a 2/3 resonance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;**There is a little bit of wobble due to &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_precession&quot;&gt;precession&lt;/a&gt;, but this happens over long time scales and doesn't change by a huge amount.&lt;/p&gt;&#xA;" OwnerUserId="14255" LastActivityDate="2016-09-19T18:03:54.203" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18401" PostTypeId="2" ParentId="18399" CreationDate="2016-09-19T19:01:10.230" Score="4" Body="&lt;p&gt;&lt;em&gt;(Would this even be theoretically possible?)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I very much doubt it. What you're suggesting would involve incredible changes in angular momentum over the course of one orbital period. In the scenario you suggest, the angular momentum vector for the planet (the direction of its north pole) starts off pointing in one direction. Six months later, it's now pointing in the &lt;em&gt;opposite&lt;/em&gt; direction. Since angular momentum is a vector quantity, that would require an enormous amount of torque and energy to achieve. And then you have to do the same thing again for the second half of the orbit, and repeat the whole thing each orbit.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2016-09-19T19:01:10.230" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18402" PostTypeId="2" ParentId="18397" CreationDate="2016-09-19T19:08:11.510" Score="3" Body="&lt;p&gt;The early 1990s &lt;a href=&quot;http://www.cosmos.esa.int/web/hipparcos&quot; rel=&quot;nofollow&quot;&gt;Hipparcos&lt;/a&gt; mission&#xA;yielded parallaxes for 118000 stars (Hipparcos catalog) and positions without parallax for another 2.4 million (Tycho-2 catalog).&#xA;The Tycho-Gaia astrometric solution (TGAS) combines those data with preliminary Gaia observations to get 2 million parallaxes, &quot;only&quot; 17 times as many as Hipparcos, with better precision.&#xA;If this is a small fraction of the total number of star positions, it's not a deficiency; it's just easier to get a position than a parallax.&#xA;&lt;a href=&quot;http://arxiv.org/abs/1609.04303&quot; rel=&quot;nofollow&quot;&gt;Lindegren et al. 2016&lt;/a&gt; explain what they could or could not do with the first 14 months of Gaia observations.&#xA;Future Gaia data releases should have still more parallaxes.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-09-19T19:08:11.510" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18403" PostTypeId="2" ParentId="18399" CreationDate="2016-09-19T19:47:44.013" Score="13" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Would this even be theoretically possible?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. The law of conservation of angular momentum prevents this for a planet with reasonable inner structure in a gravitational field like that around the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angular momentum of the planet points in a certain direction. Changing this direction requires a torque perpendicular to the rotation axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a torque existed that could force the rotation axis of the planet to always point towards the sun, the corresponding force had to have a component perpendicular to the line from the sun to the planet. The force also had to change periodically with the orbit of the planet. Obviously such a force does not exist in the planet-sun-system.&lt;/p&gt;&#xA;" OwnerUserId="10568" LastEditorUserId="10568" LastEditDate="2016-09-19T19:56:35.720" LastActivityDate="2016-09-19T19:56:35.720" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18405" PostTypeId="1" AcceptedAnswerId="18407" CreationDate="2016-09-19T22:30:35.567" Score="7" ViewCount="2200" Body="&lt;p&gt;How big are the chunks of &lt;s&gt;rock&lt;/s&gt; ice that make up Saturn's rings? Are there many objects larger than pebble size?&lt;/p&gt;&#xA;" OwnerUserId="6870" LastEditorUserId="6870" LastEditDate="2016-10-25T02:03:53.410" LastActivityDate="2019-04-10T16:23:29.393" Title="Size of Saturn's ring material" Tags="&lt;saturn&gt;&lt;planetary-ring&gt;&lt;size&gt;&lt;density&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18406" PostTypeId="2" ParentId="18405" CreationDate="2016-09-19T22:56:28.277" Score="3" Body="&lt;p&gt;Saturn's rings are composed of chunks as &lt;a href=&quot;https://en.wikipedia.org/wiki/Rings_of_Saturn#Physical_characteristics&quot; rel=&quot;nofollow&quot;&gt;large as 1km in size&lt;/a&gt;, although the typical particle is tiny.  They are spread through an area on average 10 meters thick.  Also, Saturn's rings are nearly pure ice, not rocks.  I don't know if we have a count of &quot;how many&quot; objects are larger than pebble size, given the gigantic number of particles that make up the rings I think we only have counts of the smaller moons within the rings.&lt;/p&gt;&#xA;" OwnerUserId="14255" LastActivityDate="2016-09-19T22:56:28.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18407" PostTypeId="2" ParentId="18405" CreationDate="2016-09-19T23:40:07.923" Score="9" Body="&lt;p&gt;The vast majority of the particles in Saturn's rings are small, on the order of $\sim10^{-1}$ m or lower. The columnar number density, according to data from &lt;em&gt;Voyager 1&lt;/em&gt; and Earth-based observations, can be approximated as a function of particle radius by a power law for all particle radii $a$ in meters such that $0&amp;lt;a&amp;lt;1$, as can be seen on this log-log plot (Fig. 15.5, &lt;a href=&quot;http://www.people.virginia.edu/~rej/papers09/SaturnChap15Cuzzietal09.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cuzzi et al. (2009)&lt;/a&gt;; taken from Fig. 8 of &lt;a href=&quot;http://ac.els-cdn.com/0019103583901926/1-s2.0-0019103583901926-main.pdf?_tid=75c1c778-81b0-11e6-a807-00000aab0f02&amp;amp;acdnat=1474650737_4e8ef1651c14974c243c53d51b8c4f82&quot; rel=&quot;nofollow noreferrer&quot;&gt;Marouf et al. (1982)&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iIHv3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iIHv3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;Though it is only acknowledged in the original paper, the vertical axes for the three different ring regions have been shifted upward different amounts to fit all three on the same graph.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After $a=1$, there's a deviation from the law, and then a steep dropoff at about $a=3$. Obviously, particles larger than this exist, and they certainly play an important role in ring structure, but they're relatively rare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, the trends show that smaller particles are much more common, and thus while there are indeed particles larger than pebbles - some as big as boulders, perhaps, or bigger - they are certainly few in number. Most particles are extremely small, smaller than pebbles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This data covers observations from ring semi-major axes of $\sim$ 75,000 km to $\sim$ 135,000 km - a fairly big spread, covering most of the rings and ending near &lt;a href=&quot;https://en.wikipedia.org/wiki/Rings_of_Saturn#Roche_Division&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Roche Division&lt;/a&gt;. The paper doesn't have one single graph of particle number density of a given size at a given semi-major axis, but it does have several subdivided plots (Fig. 15.1 and 15.2) of &lt;a href=&quot;https://en.wikipedia.org/wiki/Optical_depth_(astrophysics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;optical depth&lt;/a&gt; as a function of distance from the center of Saturn, which should give you some helpful data on total number density, if you want to make some basic assumptions about mean particle radius. This data is a bit newer, from &lt;em&gt;Cassini&lt;/em&gt;, but the &lt;em&gt;Voyager 1&lt;/em&gt; data is just as helpful.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-09-23T17:17:57.370" LastActivityDate="2016-09-23T17:17:57.370" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18408" PostTypeId="1" CreationDate="2016-09-20T00:40:54.440" Score="3" ViewCount="96" Body="&lt;p&gt;The local weatherman declared that fall starts at exactly 10:21am (local time). Ignoring the fact that seasonal boundaries have multiple definitions (meteorological, astronomical, etc.), what he perhaps means to have said is that the fall equinox occurs at that moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably, that is the moment at which Earth's axis of rotation is exactly perpendicular to the Sun-Earth line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: to what degree of precision can this be measured? The weatherman's statement was to the minute. But is it actually measured to a finer precision? If so, plus or minus how many seconds? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the the solstices? Can they be measured to the same degree of accuracy? Are there factors which uniquely affect the accuracy of either measurement?&lt;/p&gt;&#xA;" OwnerUserId="1983" LastActivityDate="2016-09-20T17:15:58.197" Title="Measurement precision of celestial events" Tags="&lt;earth&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18409" PostTypeId="1" CreationDate="2016-09-20T06:02:20.283" Score="6" ViewCount="490" Body="&lt;p&gt;I'm trying to simulate a virtual/imaginary &quot;solar system&quot; in software--just a hobby project for now. Unfortunately this has made me realize exactly how much math I've forgotten since college.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complete accuracy isn't important, and I'm not looking to do anything sophisticated... no need for multi-body calculations, just a star and a planet will do. I just want something that a lightweight simulation or a game might do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll have a configuration file with some basic starting parameters for each planet. At the start of the simulation, time=0 seconds, all planets would be lined up in a row (planning on just placing them on the same axis at their perihelion distance from the star). I'd like the planets to move around their orbits at the correct speed given the parameters supplied. So if I decide to set time = 1e10, I need to be able to calculate the x and y positions of each planet in their orbit at that precise time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm flexible on WHICH parameters need to be provided. So if this becomes easier using some other method of defining the orbit, I'm open to it. Right now I'm thinking perihelion, mass, and time elapsed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Since this is a 3D simulation, eventually I'd also like to work inclination in, but I'm trying to start off simple for now, so inclinations will all be zero.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's some sample parameter data below in case it helps provide a better example.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Star mass = 2.00e30&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planet mass = 3.30e23&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planet perihelion = 5800000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Ex. Where would this planet be, in X/Y coordinates, at time=x?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone posts a formula I can just plug in I won't complain. But if I could just be pointed in the right direction for what to learn or refresh my memory on, I'd like to try and figure this out myself. So far I've been looking at Kepler's 3rd Law equations but I'm not sure if that's a good place to start. Feeling a little guilty that I seem to have forgotten virtually &lt;em&gt;all&lt;/em&gt; math I learned in college just over a decade ago, and I kind of want to reverse some of that decay.  Hopefully this makes sense!&lt;/p&gt;&#xA;" OwnerUserId="13805" LastEditorUserId="13805" LastEditDate="2016-09-26T20:54:24.700" LastActivityDate="2016-09-30T13:42:43.570" Title="Creating a basic &quot;fixed&quot; solar system to host a 3D space travel simulation" Tags="&lt;orbit&gt;&lt;celestial-mechanics&gt;" AnswerCount="3" CommentCount="18" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18410" PostTypeId="2" ParentId="18399" CreationDate="2016-09-20T07:22:49.117" Score="-1" Body="&lt;p&gt;In theory, there's no reason why a planet couldn't orbit close to the way you suggest.   It's probably extremely rare, but theoretically possible.    This gets a little tricky to explain, but an axis of rotation can be divided into perpendicular vectors.   There's no real benefit to doing it that way but it can be done, just like direction and velocity can be separated into 3 vectors, rotation can too, and the combination of the perpendicular vectors gives you the rotation of the object and it's axis of rotation.   That's discussed in a bit more detail &lt;a href=&quot;https://www.quora.com/On-how-many-unique-axes-can-a-sphere-rotate-simultaneously&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we use the Moon as an example.   The Moon is oriented so it's heavier side permanently points towards the Earth.  (the side of the Moon facing the Earth has a thinner crust and because crust is lighter than the mantle, the &quot;Earth side&quot; of the Moon has slightly greater density than the Far side.  That greater density is more attracted to the Earth than the less dense far side and over time, the dense side of the Moon was oriented to permanently face the earth, similar to how the dense side of a floating objects orients itself downwards when floating in water.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All planets, not being perfect spheres with perfectly layered density are unbalanced but the Moon is more lopsided than most.  Mars more than the rest of the planets.  Earth's uneven gravity has been measured in great detail, but in terms of effects on Earth's orbit and rotation, it's uneven gravity is pretty insignificant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now some objects experience significant &lt;a href=&quot;http://www.astronomynotes.com/solarsys/tidal-heating.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;tidal deformation&lt;/a&gt;, like Io and Enceladus which are measurably squeezed by their elliptical orbits as they move closer to and further from their planets, but that's something else and you probably wouldn't want that on a planet as the volcanic effects would be too big.   But what I'm talking about isn't a tidal bulge but a permanent imbalance of mass where one side of the Moon is denser than the other.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we figure the dense side of the Moon must always point to Earth due to tidal locking, that still leaves one way that the Moon can rotate without affecting the locking and that's along it's &lt;a href=&quot;https://en.wikipedia.org/wiki/Selenographic_coordinates&quot; rel=&quot;nofollow noreferrer&quot;&gt;90 degree longitude lines&lt;/a&gt;, and that, in theory could happen in addition to it's existing synchronous rotation that keeps the heavy side of the Moon pointed towards earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This wouldn't be two axis of rotation, it would be one axis of rotation where the East-West vector of the rotation would still be tidally locked but the rotation around the 90 degree longitude line wouldn't interfere with the tidal locking, so the effect, the familiar face would spin around in a circle but always face the earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could in theory do that artificially if we put a big train on the moon and ran it 24/7 in the same direction around the 90 degree longitude line.   Do that long enough and the moon would begin to spin.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1INmj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1INmj.png&quot; alt=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/2/29/Moon-map.png/400px-Moon-map.png&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for such a scenario to actually exist, you'd need a lot of luck, because the forces that tend to tidally lock a planet or moon tend to also reduce any other rotations other than the synchronous 1:1 tidally locked rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem is the Equatorial bulge which is a consequence of rotation, tends to have more mass and that would want to orient itself towards the planet, so for this to work you'd need a slow rotation and a small equatorial bulge where the added mass around the equatorial bulge was small enough to not alter the direction of the heavier side of the planet.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it's not hard to imagine a planet that's not tidally locked having just the right orbital velocity and position to do what you want, but in such a scenario with no tidal locking, it would be temporary.   Planets rotations tends to slow down over time, so perfectly matching without tidal locking would be coincidental and temporary.   More likely you'd have a very slow rotation, not a permanent side facing the planet, but a very gradual movement.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no planets (or moons) like you describe that we know of.  It's probably a hugely unlikely scenario that would probably only be approximated (a bit like Uranus). &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-09-20T08:02:43.817" LastActivityDate="2016-09-20T08:02:43.817" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18412" PostTypeId="2" ParentId="18353" CreationDate="2016-09-20T09:15:21.340" Score="1" Body="&lt;p&gt;In general relativity, then notion of space and time is &lt;em&gt;local&lt;/em&gt;. As underlined in the OP, it means that at a radius $r$ of a star of mass $M$, time will run at $$\sqrt{1-\frac{2GM}{rc^2}}$$ the rate at an infinite distance of the object. For example imagine that $1-2GM/rc^2 = 1/4$, so that at a distance $r$, time runs at half the rate compare to an observer at infinity. If you are initially with an observer at infinity, you travel to $r$ (we neglect the time distortion during the travel), stay here an hour and then go back. When you'll reach the observer once again, your local clock will be one hour late, because 1 hour at $r$ will have passed in two hours at $r=\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this distortion of space and time is &lt;em&gt;only local&lt;/em&gt;. A massive object, like a star is only impacting in its neighborhood space and time. A distant observer is not impacted by the distortion, so that time and distances are unaffected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question, an observer far away (for example on Earth) measuring by some meaning the speed of the two stars will actually see $c/2$ for both of them.&lt;/p&gt;&#xA;" OwnerUserId="14154" LastActivityDate="2016-09-20T09:15:21.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18413" PostTypeId="2" ParentId="18409" CreationDate="2016-09-20T09:41:26.417" Score="3" Body="&lt;p&gt;The easiest approach is to do a numerical integration of the equation of motion of your system. For that you need to remember that the gravitational force of $a$ on $b$ is $$\vec{F}_{a\rightarrow b} = -G\frac{M_a M_b}{r_{a\rightarrow b}^{3}} \vec{{r_{a\rightarrow b}}}$$&#xA;where $\vec{{r_{a\rightarrow b}}} = \vec{x_b} - \vec{x_a}$ and that&#xA;$$ M\vec{a} = \sum \vec{F}$$&#xA;where $M$ is the mass of your object, $\sum\vec{F}$ is the sum of the forces applied to the object and $\vec{a}$ is the acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do the simulation, you then have to follow this numerical scheme&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;set some initial conditions to your problem: initial position $\vec{x_p}$ of the planet and $\vec{x_\star}$ of the star and initial velocities $\vec{v_p}$ and $\vec{v_\star}$.&lt;/li&gt;&#xA;&lt;li&gt;compute the force $\vec{F}$ of the star on the planet and the accelerations of the planet and the star:&#xA;$$ -M_\star \vec{a_\star} = M_p \vec{a_p} = \vec{F_{\star\rightarrow p}}$$&lt;/li&gt;&#xA;&lt;li&gt;update the speed of the planet and the star using the respective acceleration:&#xA;$$ \vec{v}(t+\Delta t) = \vec{v}(t) + \Delta t\times \vec{a} $$&lt;/li&gt;&#xA;&lt;li&gt;update the positions of the planet and the star using their respective velocity:&#xA;$$ \vec{x}(t+\Delta t) = \vec{x}(t) + \Delta t\times \vec{v} $$&lt;/li&gt;&#xA;&lt;li&gt;recompute the distance between the planet and the star $\vec{r_{{\star\rightarrow p}}} = \vec{x_p} - \vec{x_\star}$ and go back to 2.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="14154" LastActivityDate="2016-09-20T09:41:26.417" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18414" PostTypeId="1" AcceptedAnswerId="18419" CreationDate="2016-09-20T15:24:50.500" Score="2" ViewCount="563" Body="&lt;p&gt;Forgive my 'amazing' MS-Paint skills, but this is along the lines of what i'm wondering:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gN5O6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gN5O6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any situation where Scenario A occurs, or is space &lt;em&gt;always&lt;/em&gt; dragged back out as in Scenario B?&lt;/p&gt;&#xA;" OwnerUserId="11009" LastEditorUserId="11009" LastEditDate="2016-09-21T10:07:25.893" LastActivityDate="2016-09-21T19:31:54.747" Title="Do black holes pull in and keep spacetime itself or merely warp it but ultimately let it pass through?" Tags="&lt;black-hole&gt;&lt;space-time&gt;&lt;mass&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18416" PostTypeId="2" ParentId="18408" CreationDate="2016-09-20T17:15:58.197" Score="3" Body="&lt;p&gt;This doesn't answer your question, but has two screenshots, so that's something :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you run &lt;a href=&quot;http://ssd.jpl.nasa.gov/?horizons&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ssd.jpl.nasa.gov/?horizons&lt;/a&gt; with these settings:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7mbQO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7mbQO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;part of the output looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zZEk7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zZEk7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows the Sun crosses the equinox point between the 8th and 9th second of 14:21 (which is 10:21 your local time). HORIZONS will even let you compute at half second intervals but not beyond that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use the &lt;a href=&quot;http://naif.jpl.nasa.gov/naif/tutorials.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://naif.jpl.nasa.gov/naif/tutorials.html&lt;/a&gt; library, you can compute the time down to the nanosecond or better, but this doesn't really help because we don't know the Sun's position, the Earth's position, or even the direction the North Pole is pointing to that level of precision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, not an answer, but some hopefully helpful comments.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2016-09-20T17:15:58.197" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18417" PostTypeId="1" AcceptedAnswerId="18418" CreationDate="2016-09-20T20:12:08.087" Score="3" ViewCount="363" Body="&lt;p&gt;In the final energy state of the universe, there exists a hypothesis that all matter will have fallen into black holes and neutron stars. If, through Hawking radiation, after all matter is consumed and black holes eventually do dissipate, what is left of our Universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may be describing Heat Death, but I'm not 100% sure. However, due to the Law of Conservation of Energy, I suspect that energy will be all that is left, only in equilibrium meaning no ability for energy to form matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the event I'm wrong and there exists no matter and no energy, would the term Universe still be applicable here?&lt;/p&gt;&#xA;" OwnerUserId="13849" LastActivityDate="2019-05-19T14:02:25.083" Title="What is left of a universe where no matter and energy exist and is there a term for this?" Tags="&lt;cosmology&gt;&lt;terminology&gt;&lt;fate-of-universe&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18418" PostTypeId="2" ParentId="18417" CreationDate="2016-09-20T20:51:36.917" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;However, due to the Law of Conservation of Energy, I suspect that energy will be all that is left, only in equilibrium meaning no ability for energy to form matter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The law of conservation of energy states that energy is conserved &lt;strong&gt;in a closed system&lt;/strong&gt;. This means the Universe, which is likely an open system, does not necessarily have to obey that conservation law. Not to mention, scientific laws are &lt;em&gt;not&lt;/em&gt; absolute, despite their names; they are entirely empirical, and only apply as far as we've observed. As Edwin Hubble put it, &quot;Beyond the limit of the observations, the form of [a] function, or relation, is speculative. For this reason, [a] law is empirical and it must remain empirical until it is explained by an accepted theory.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, let's get on with the answer. You've just described the Black Hole Era in the heat death scenario. If Hawking radiation occurs, then the black hole will continue to emit particles. During the last stages of its evaporation, a black hole will emit not only massless particles, but also heavier particles, such as electrons, positrons, protons, and antiprotons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once all black holes have evaporated, the Universe will be nearly empty, sans some particles that will almost never interract. Eventually, even the protons and antiprotons emitted by the black holes will decay into photons and leptons. Thus begins the Dark Era, when the Universe will be dominated by dark matter, electrons and positrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrons and positrons will rarely come together and form &lt;a href=&quot;https://en.wikipedia.org/wiki/Positronium&quot; rel=&quot;nofollow&quot;&gt;positronium atoms&lt;/a&gt; (in which the electron and positron are electromagnetically bound). Positronium decays very within nanoseconds, since the two particles annihilate. Thus, over time, the energy levels in the Universe will become lower and lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens next is a matter of speculation. According to &lt;a href=&quot;https://arxiv.org/abs/hep-th/0410270&quot; rel=&quot;nofollow&quot;&gt;Carroll &amp;amp; Chen (2004)&lt;/a&gt;, quantum tunnelling and quantum fluctuations could hypothetically create another Big Bang, approximately $10^{10^{10^{56}}}$ years from now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/astro-ph/9701131&quot; rel=&quot;nofollow&quot;&gt;Adams &amp;amp; Laughlin (1997)&lt;/a&gt; speculates that quantum effects will dominate once all macroscopic matter is gone. Essentially, the laws of &quot;macro-physics&quot; will break down, while quantum mechanical phenomena prevail. As a result, the smallest perturbations would make the biggest difference in this era, so there is no telling what may happen to space or time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, they speculate that if the Universe is a false vacuum, the vacuum may decay into a lower-energy state, creating a new &quot;child universe&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is also possible for the universe to spontaneously create “child&#xA;  universes” through a quantum tunneling process roughly analogous to&#xA;  that considered above (e.g., Sato et al., 1982; Hawking, 1987; Blau,&#xA;  Guendelman, &amp;amp; Guth, 1987). In this situation, a bubble of false vacuum&#xA;  energy nucleates in an otherwise empty space-time. If this bubble is&#xA;  sufficiently large, it will grow exponentially and will eventually&#xA;  become causally disconnected from the original space-time. In this&#xA;  sense, the newly created bubble becomes a separate “child universe”.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So yes, something as small as quantum tunnelling will potentially have an enormous effect on the future of the Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, we really have no idea what will happen after enough time, so there's no definitive answer to your question in the long run.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-09-27T12:56:58.427" LastActivityDate="2016-09-27T12:56:58.427" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18419" PostTypeId="2" ParentId="18414" CreationDate="2016-09-21T05:03:23.043" Score="1" Body="&lt;p&gt;Black holes orbit things and galaxies move within space, so it would be safe to assume that black holes move through space-time and do not hold it in any particular state or position. They simply drag their &quot;depression&quot; along with them as they go.&lt;/p&gt;&#xA;" OwnerUserId="2232" LastActivityDate="2016-09-21T05:03:23.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18420" PostTypeId="1" AcceptedAnswerId="18422" CreationDate="2016-09-21T09:53:19.370" Score="3" ViewCount="938" Body="&lt;p&gt;I was reading two sites before I posted the question here. &lt;a href=&quot;http://spaceplace.nasa.gov/review/dr-marc-earth/earth-rotation.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;One site&lt;/a&gt; says something like that the collisions in Universe eventually set the planets rotating, &lt;a href=&quot;http://curious.astro.cornell.edu/about-us/36-our-solar-system/the-earth/general-questions/22-what-makes-the-earth-rotate-intermediate&quot; rel=&quot;nofollow noreferrer&quot;&gt;another site&lt;/a&gt; says that our planet is formed from some collapsing Nebula or whatsoever and Nebula was rotating when it was about to die and that set the Earth rotating at first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question should be probably better specified. Is Earth rotating because of some continuous force, or because at first place there was some one-time force which set it rotating and the effect lasts until now?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the first case, which force is it? As for the second case, that should break our physical laws, because there would be one force, applied a very long time ago which is still in effect and not about to slow down and with a very very long lasting effect. Because thousands or millions of years for rotation is a very long time, isn't it. Or is it slowing down? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW: the infinite movement around orbit is bit mysterious too, because you can say that's because of gravity, but gravity itself attracts the things only to one direction, but orbit movement is like parallel with the gravity emitter, and again it is perpetual.&lt;/p&gt;&#xA;" OwnerUserId="14303" LastEditorUserId="7472" LastEditDate="2017-07-19T10:39:43.520" LastActivityDate="2017-07-19T10:39:43.520" Title="What makes the Earth infinitely rotating around its axis" Tags="&lt;earth&gt;&lt;rotation&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18422" PostTypeId="2" ParentId="18420" CreationDate="2016-09-21T13:05:32.343" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;As for the second case, that should break our physical laws&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is precisely the wrong statement. The fact that the Earth is still spinning matches our physical laws exactly. The physical law I'm specifically referring to is known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion#Newton.27s_first_law&quot; rel=&quot;noreferrer&quot;&gt;Newton's First Law of Motion&lt;/a&gt;, which is loosely paraphrased as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An object at rest remains at rest unless acted upon by a net, outside force. Conversely, an object in motion remains in motion unless acted upon by a net, outside force.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What this means is that once something starts moving (like the Earth spinning) it will keep moving until something else stops it. The Earth long ago started spinning because of the initial angular momentum of the collapsing cloud which eventually made up the Earth. There is nothing in space to stop the Earth from spinning so it is just going to go on spinning&lt;sup&gt;1&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You appear to believe that if you start an object moving by applying some force, it will eventually run down and stop moving of its own accord. This is a common thought, based on everyday experiences. In fact, that was what Aristotle thought (and much of the world up until Isaac Newton's time). But what you're not taking into account is that there's always a force being applied to slow things down. Consider a car rolling along the highway. If you take your foot off the gas pedal, the car starts slowing down. But it isn't because some physical law says it should, it's because a net force is acting on the car to slow it down $-$ that net force being wind resistance on the car. So you see, even common, everyday experiences hold up to Newton's first law.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;Technically there are some &lt;em&gt;small&lt;/em&gt; forces acting on the Earth to slow it down but those are nearly negligible and mostly from the Moon. These forces would take billions of years or longer to make the Earth come to a complete stop (if they ever even could).&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-09-21T13:05:32.343" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18423" PostTypeId="1" AcceptedAnswerId="18424" CreationDate="2016-09-21T14:15:39.950" Score="33" ViewCount="5113" Body="&lt;p&gt;I am writing a science fiction novel, where a ship is stranded in a single star system (a red supergiant). One of the plot points is the star becoming supernova in several hours, so the characters have to fix their ship before that happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have basic knowledge of how it works: Iron generated from nuclear fusion gets accumulated in the core, until it reaches a point when iron fusion starts. As iron fusion is an endothermic reaction, the core is no longer able to generate enough energy to hold against its own gravity and external layers pressure, so it collapses, and explode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read that once the iron fusion starts inside the core, the collapse occurs within minutes, that the collapse itself lasts a few seconds (even less than a second), and that the shockwave takes several hours to reach the surface. Is all that correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is that I need the characters to bee able to predict the explosion in a short term. A few hours or even minutes. It would be great if they could be aware of the core collapse and start a countdown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, are there any external cue of these events, like changes in luminosity or color? Does the star spectrum change when iron fusion starts, or when the core collapses? I know that the core collapse generates a huge amount of neutrinos. Is this amount so intense that it can be easily detectable? (that is, without a huge detector in an underground facility). Can the amount of iron in the core be estimated from tha star spectrum and size, so the aproximate time of the collapse could be predicted?&lt;/p&gt;&#xA;" OwnerUserId="14305" LastActivityDate="2016-09-23T11:01:26.783" Title="Are there observable changes in a star about to become supernova, minutes or hours before the explosion?" Tags="&lt;supernova&gt;&lt;explosion&gt;" AnswerCount="4" CommentCount="15" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18424" PostTypeId="2" ParentId="18423" CreationDate="2016-09-21T15:31:26.793" Score="41" Body="&lt;p&gt;I think your best bet would be detecting neutrinos generated by nuclear burning inside the star (as we do for the Sun). Once the star hits the carbon-burning stage, it's actually putting out more energy in neutrinos than in photons. During the silicon-burning phase, which lasts for a few days and is what creates the degenerate iron core (that collapses once it is massive enough), the neutrino flux increases to about 10&lt;sup&gt;47&lt;/sup&gt; erg/s a few seconds before core collapse. (The peak flux &lt;em&gt;during&lt;/em&gt; core collapse is about 10&lt;sup&gt;52&lt;/sup&gt; to 10&lt;sup&gt;53&lt;/sup&gt; erg/s). &lt;a href=&quot;https://arxiv.org/pdf/1506.01175v4.pdf&quot; rel=&quot;noreferrer&quot;&gt;This paper by Asakura et al.&lt;/a&gt; estimates that the Japanese KamLAND detector could detect the pre-supernova neutrino flux for stars at distances of &lt;em&gt;several hundred parsecs&lt;/em&gt;, and provide advance warning of a core-collapse supernova several hours or even days in advance. Since your characters are &lt;em&gt;in the same system&lt;/em&gt; as the star, they'd hardly need a large underground detector to pick up the neutrinos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This plot shows an example of neutrino luminosity (for anti-electron neutrinos) versus time for a pre-supernova star (from Asakura et al. 2016, based on Odrzywolek &amp;amp; Heger 2010 and Nakazato et al. 2013); core collapse begins at t = 0s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nF9hR.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nF9hR.png&quot; alt=&quot;Figure 1 from Asakura et al. 2016&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By measuring the spectrum of energies for different types of neutrinos and their time evolution, you could probably get a very good idea of how far along the star was, particularly as we can probably assume your characters have much better models for stellar evolution than we currently do. (They'd also want to get accurate measurements of the star's mass, rotation rate, maybe internal structure via astroseismology, etc., in order to fine-tune the stellar-evolution model; these are all things they could do pretty easily.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The core collapse itself would be signalled by the &lt;em&gt;enormous&lt;/em&gt; increase in neutrino flux.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://what-if.xkcd.com/73/&quot; rel=&quot;noreferrer&quot;&gt;This &quot;What If&quot; article&lt;/a&gt; by Randall Munroe estimates that the neutrino flux from a core-collapse supernova would be &lt;em&gt;lethal&lt;/em&gt; to a human being at a distance of around 2 AU. Which, as he points out, could actually be inside of a supergiant star, so your characters would probably be a bit further away than that. But it does show that the neutrino flux would be easily detectable, and that your characters might well get radiation poisoning from it if they were closer than 10 AU. (Of course, you'd want to &lt;em&gt;detect&lt;/em&gt; it more directly than just waiting around till you started to feel sick, since that might take longer than the shock wave takes to reach the surface of the star.) This is just to bring home the fact that they wouldn't have any problem detecting the neutrinos....&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2016-09-23T11:01:26.783" LastActivityDate="2016-09-23T11:01:26.783" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="18425" PostTypeId="2" ParentId="18423" CreationDate="2016-09-21T15:49:13.743" Score="11" Body="&lt;p&gt;As &lt;a href=&quot;https://astronomy.stackexchange.com/questions/18423/are-there-observable-changes-in-a-star-about-to-become-supernova-minutes-or-hou#comment28951_18423&quot;&gt;Dean said&lt;/a&gt;, supernova progenitors typically release neutrinos prior to full core collapse, remnant formation and the ejection of the outer layers of the star. The process - focused here on the neutrinos - goes something like the following:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;At high enough densities (&lt;span class=&quot;math-container&quot;&gt;$\rho\sim10^9\text{ g/cm}^3$&lt;/span&gt;), &lt;a href=&quot;https://en.wikipedia.org/wiki/Electron_capture&quot; rel=&quot;noreferrer&quot;&gt;electron capture&lt;/a&gt; becomes important, where a proton and electron combine to form a neutron and an electron neutrino:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$e^-+p\to n+\nu_e$$&lt;/span&gt;&#xA;Simultaneously, &lt;a href=&quot;https://en.wikipedia.org/wiki/Beta_decay&quot; rel=&quot;noreferrer&quot;&gt;beta decay&lt;/a&gt; may occur, where a neutron decays to a proton, electron and electron antineutrino:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$n\to p+e^-+\bar{\nu}_e$$&lt;/span&gt;&#xA;However, beta decay becomes less important than electron capture at this point.&lt;/li&gt;&#xA;&lt;li&gt;Electron capture reduces electron degeneracy pressure in the core, which leads to accelerated core collapse. Degeneracy pressure is important in the cores of many stars, but in extremely massive stars - red supergiants included - it simply isn't enough to stop the collapse.&lt;/li&gt;&#xA;&lt;li&gt;At densities below &lt;span class=&quot;math-container&quot;&gt;$\sim10^{11}\text{ g/cm}^3$&lt;/span&gt;, neutrinos can carry away energy, and the initial burst leaves the star within about ten seconds. However, core collapse quickly leads to much greater densities, and when &lt;span class=&quot;math-container&quot;&gt;$\rho\sim4\times10^{11}\text{ g/cm}^3$&lt;/span&gt;, neutrinos are trapped. They scatter off nuclei, and transfer energy to electrons. Electron-nuclei scattering is also important, and may be dominant at higher energies.&lt;/li&gt;&#xA;&lt;li&gt;At &lt;span class=&quot;math-container&quot;&gt;$\rho\sim2.5\times10^{14}\text{ g/cm}^3$&lt;/span&gt;, the core undergoes a &amp;quot;bounce&amp;quot;, and the supernova explosion fully begins. A shock wave propagates into the outer core, and more neutrinos are produced via electron capture.&lt;/li&gt;&#xA;&lt;li&gt;Neutrinos still trapped in/by the stellar remnant are released about ten seconds later. Neutrino pair production, too, leads to rapid cooling. Some of these neutrinos may contribute to a revival of the shockwave.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Neutrinos may arrive hours - or possible days, in some circumstances - before the light from the supernova. The former was the case for &lt;a href=&quot;https://en.wikipedia.org/wiki/SN_1987A&quot; rel=&quot;noreferrer&quot;&gt;SN 1987A&lt;/a&gt;, the first supernova from which neutrinos were detected.&lt;/p&gt;&#xA;&lt;h3&gt;References&lt;/h3&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://arxiv.org/abs/1503.08095&quot; rel=&quot;noreferrer&quot;&gt;Balasi et al. (2015)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://arxiv.org/abs/1508.00785&quot; rel=&quot;noreferrer&quot;&gt;Mirizzi et al. (2015)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://arxiv.org/abs/1205.6003&quot; rel=&quot;noreferrer&quot;&gt;Scholberg (2012)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-09-21T15:49:13.743" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18426" PostTypeId="2" ParentId="18322" CreationDate="2016-09-21T16:22:34.383" Score="0" Body="&lt;p&gt;Firstly, observation supports that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;black holes are present in essentially every galaxy that has a bulge&#xA;  component&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, we can say a lot of galaxies have central black holes. But there are some type of galaxies which do not have a bulge, such as dwarf galaxies. For those we can not be sure if there is a black hole in the centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The link between the central super massive black holes (SMBH) and the host galaxies are build because people find a tight correlation between the mass of the SMBHs and the stellar mass of the bulge of the host galaxies. Thus, it's natural to think that there might be an relation in which the SMBHs and the host galaxies are effecting each other while they evolve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, nowadays people find the truth is much more complicated. This is still quite a hot research topic in astrophysics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information, you can check this annual review paper:&#xA;&lt;a href=&quot;https://arxiv.org/abs/1304.7762&quot; rel=&quot;nofollow&quot;&gt;https://arxiv.org/abs/1304.7762&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="14307" LastEditorUserId="14307" LastEditDate="2016-09-21T18:55:10.300" LastActivityDate="2016-09-21T18:55:10.300" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18427" PostTypeId="2" ParentId="18423" CreationDate="2016-09-21T16:33:17.227" Score="17" Body="&lt;p&gt;Other answers are correct; a neutrino pulse is definitely expected as a result of a core-collapse supernova and should occur some hours before a shockwave arrives at the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There essentially would be no visible sign that the star was about to become a supernova and that is because the dynamical timescale of the envelope is relatively long - thus responding slowly with respect to changes in the core. So, even if all support is removed from the centre, the surface can only respond (at best) on a freefall timescale of $\sim (G \rho)^{-1/2}$, where $\rho$ is the average density. If the star is a $10M_{\odot}$ supergiant with a radius of 1 au, then this timescale is tens of &lt;em&gt;days&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility not mentioned so far is gravitational waves. Assuming that a relatively portable gravitational wave detector was available (!) then you would also expect a sharp gravitational wave pulse on the core collapse timescale (a second or less) that would also presage the supernova blast wave some hours later.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-09-21T16:33:17.227" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18429" PostTypeId="1" AcceptedAnswerId="18430" CreationDate="2016-09-21T18:00:31.683" Score="0" ViewCount="121" Body="&lt;p&gt;Can a lunar eclipse completely 'turn off' the mooon, i.e. like a New Moon? Or is it always just a 'shading' of the moon (sometimes red in color)?&lt;/p&gt;&#xA;" OwnerUserId="14311" LastActivityDate="2016-09-21T20:56:43.187" Title="Lunar Eclipse - Total darkness?" Tags="&lt;lunar-eclipse&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18430" PostTypeId="2" ParentId="18429" CreationDate="2016-09-21T18:13:23.763" Score="1" Body="&lt;p&gt;&lt;strong&gt;No.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lunar eclipses are caused when the Moon is in opposition to the Sun. Normally this produces a full moon, but if the Moon is in exact opposition (considering incline of the Moon's orbital plane), all direct sunlight will be blocked from the Moon:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5CKA0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5CKA0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if all the sunlight is blocked, how can we see the Moon? Well, the main reason is that sunlight is often dispersed in the atmosphere, and so it can reach the Moon. In addition, there's  &lt;a href=&quot;https://en.wikipedia.org/wiki/Airglow&quot; rel=&quot;nofollow noreferrer&quot;&gt;airglow&lt;/a&gt;, which is when sunlight hits Earth's upper atmosphere and causes multiple chemical reactions, scattering light throughout the night sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, all this light from Earth, called &lt;a href=&quot;https://en.wikipedia.org/wiki/Planetshine#Earthshine&quot; rel=&quot;nofollow noreferrer&quot;&gt;earthshine&lt;/a&gt;, reflects off the Moon and illuminates it. In addition, Earth removes and blocks parts of the sunlight's spectrum, leaving only the longer wavelengths. This causes the Moon to appear red. Lastly, because the Earth blocks off all the direct sunlight from the Sun (only diffracted sunlight and airglow reach the Moon), we can actually see Earth's shadow on the Moon. &lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-09-21T20:56:43.187" LastActivityDate="2016-09-21T20:56:43.187" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18431" PostTypeId="2" ParentId="18414" CreationDate="2016-09-21T19:21:33.713" Score="0" Body="&lt;p&gt;Space is not a real thing, it is kind of a conceptual picture.  However, there are standard ways to apply that picture that work reasonably well.  One such picture is that all gravitational objects &quot;pull in&quot; space, and keep it as long as the object remains an intact massive body.  Not just black holes, even the Earth.  You can think of the space as being like it is made of rubber and is being contracted whereever there is mass.  This is a poor man's version of the concept of curvature of spacetime, but it works quite well.  Note that if black holes eventually dissipate by Hawking radiation, then they will &quot;let go&quot; of that space.  But it won't re-expand until it rejoins the &quot;Hubble flow&quot; of the Big Bang, and dark energy gets a hold of it-- it seems that empty space is just empty space, it does not have a memory of its own history.  More formally, what we mean by space is just a coordinate system for talking about relationships like distance and motion.  Whether or not space &quot;moves through&quot; an object is really just an issue of reference frame and coordinate system.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2016-09-21T19:31:54.747" LastActivityDate="2016-09-21T19:31:54.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18432" PostTypeId="1" AcceptedAnswerId="18434" CreationDate="2016-09-21T19:22:57.653" Score="1" ViewCount="226" Body="&lt;p&gt;I just had an idea not sure if this would work but if we dropped something into very deep water like the marina trench could the pressure crush it small enough to create a black hole?&lt;/p&gt;&#xA;" OwnerUserId="14312" LastActivityDate="2016-09-24T02:57:11.190" Title="Black hole Creation" Tags="&lt;black-hole&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18433" PostTypeId="2" ParentId="18432" CreationDate="2016-09-21T20:15:21.637" Score="0" Body="&lt;p&gt;No, for a black hole to form the core of a star much larger than our sun must collapse into itself, forming a singularity.  This happens after a star has expended it's fuel no longer being able to offset the inward force of gravity. The result is a rapid release of its outer layers in a supernova. The resulting core then collapses into the singularity. The mass is so great that this singularity is an infinite bend in space time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of note, if we took your object and were able to squeeze it down so small that it's radius reached the Schwarzschild radius, it could become a black hole. The Mariana Trench could not provide enough pressure to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exists little fish and organisms near the bottom of the Mariana Trench. Dropping something like a penny would just cause it be distorted by the pressure above it.&lt;/p&gt;&#xA;" OwnerUserId="13849" LastEditorUserId="13849" LastEditDate="2016-09-21T21:54:30.920" LastActivityDate="2016-09-21T21:54:30.920" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18434" PostTypeId="2" ParentId="18432" CreationDate="2016-09-21T20:19:34.433" Score="8" Body="&lt;p&gt;There already is something at the bottom of Mariana Trench. Rocks and stuff. Luckily they haven't been turned into black holes. So empirically, we can say that the answer is no. If it had, it would have attracted its surroundings, and eventually the rest of Earth would fall into it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To create a black hole, you need to compress a certain mass to within a certain radius. But solids and liquids are really hard to compress. Although the pressure at the bottom of the Mariana Trench is over 1000 atmospheres, water is compressed only by 5%. And solids like rocks and metals are virtually incompressible$^\dagger$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even at the center of the Sun, where the pressure is 250 billion atm, the density is only 150 times that of water (under 1 atm). That is, 1 kg of water placed at the center of the Sun would fill not one liter, but $0.7\,\mathrm{cl}$.&#xA;But to turn a mass $M$ into a black hole, you'd have to compress it to within it so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_radius&quot;&gt;Schwarzschild radius&lt;/a&gt;, which is given by&#xA;$&#xA;r_\mathrm{S} = 2GM/c^2,&#xA;$&#xA;where &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_constant&quot;&gt;$G$&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Speed_of_light&quot;&gt;$c$&lt;/a&gt; are constants. For $M = 1\,\mathrm{kg}$, this equates to $10^{-25}\,\mathrm{cm}$, much smaller than the radius of an atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such conditions arise only in extreme events, such as the core of a dying star collapsing, and even then only the most massive stars. So unfortunately, you won't be able to create a black hole.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;Solids are not &lt;em&gt;entirely&lt;/em&gt; incompressible, since otherwise sounds wouldn't be able to propagate through them.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-09-21T20:51:49.560" LastActivityDate="2016-09-21T20:51:49.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18436" PostTypeId="1" AcceptedAnswerId="18437" CreationDate="2016-09-21T21:20:06.257" Score="2" ViewCount="618" Body="&lt;p&gt;As you know, nebulae such as the Eagle Nebula, the Cat's Eye Nebula, and the Pillars of Creation, have extraordinary shapes.  But have we discovered why that is so?  I've tried researching it myself however all that comes up are theories of a magnetic field, etc.&lt;/p&gt;&#xA;" OwnerUserId="14314" LastActivityDate="2016-09-22T00:55:32.957" Title="Did we discover how nebulae hold their shapes?" Tags="&lt;nebula&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18437" PostTypeId="2" ParentId="18436" CreationDate="2016-09-22T00:55:32.957" Score="6" Body="&lt;p&gt;&amp;quot;Nebula&amp;quot; is quite an imprecise word; over the years, it's been applied to everything from interstellar dust clouds to supernova remnants to the birthplaces of stars to even galaxies. &lt;a href=&quot;https://en.wikipedia.org/wiki/Nebula&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; separates true nebulae into four categories: H II regions, planetary nebulae, supernova remnants, and dark nebulae. In this case, it might be better to make the distinction between nebulae that primarily surround on star/star system and nebulae that exist in interstellar space, potentially encompassing many stars.&lt;/p&gt;&#xA;&lt;h1&gt;Circumstellar nebulae&lt;/h1&gt;&#xA;&lt;p&gt;I'll classify &lt;a href=&quot;https://en.wikipedia.org/wiki/Planetary_nebula&quot; rel=&quot;noreferrer&quot;&gt;planetary nebulae&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Supernova_remnant&quot; rel=&quot;noreferrer&quot;&gt;supernova remnants&lt;/a&gt; as circumstellar nebulae. They're both the result of a star passing into the final phase of its life.&lt;/p&gt;&#xA;&lt;p&gt;Planetary nebulae (which include the Cat's Eye Nebula) are the remains of the outer layers of gas given off by relatively low-mass stars, like the Sun. &lt;a href=&quot;http://www.annualreviews.org/doi/pdf/10.1146/annurev.astro.40.060401.093849&quot; rel=&quot;noreferrer&quot;&gt;Balick &amp;amp; Frank (2002)&lt;/a&gt; subdivide these based on morphology, into four categories:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Round&lt;/li&gt;&#xA;&lt;li&gt;Elliptical&lt;/li&gt;&#xA;&lt;li&gt;Bipolar&lt;/li&gt;&#xA;&lt;li&gt;Irregular&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The names describe the nebulae's characteristics fairly well. However, they show observation bias; for instance, a bipolar nebula may appear to be found if its lobes are lined up perfectly with Earth. The exact reasons for the distinctions aren't well-know; Wikipedia cites &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0501040&quot; rel=&quot;noreferrer&quot;&gt;Jordan et al. (2002)&lt;/a&gt; (which I think you may have already read) in saying that magnetic fields could be the cause, but there are obviously other reasons for a lack of spherical symmetry: intense rotation (possible in an AGB star), violent mass loss (unlikely in stars of this low mass), a strong stellar wind, interactions with other bodies, etc. The latter may have caused the dramatic lobes of &lt;a href=&quot;https://en.wikipedia.org/wiki/M2-9&quot; rel=&quot;noreferrer&quot;&gt;M2-9&lt;/a&gt;, where interactions with a binary partner produce extreme distended streams of gas:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QLyUi.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QLyUi.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;Image courtesy of the ESA under &lt;a href=&quot;https://creativecommons.org/licenses/by/3.0/deed.en&quot; rel=&quot;noreferrer&quot;&gt;the Creative Commons Attribution 3.0 Unported license&lt;/a&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;Supernova remnants fall into three basic categories:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Shell-like, which are simply spherical shells&lt;/li&gt;&#xA;&lt;li&gt;Composite, characterized by radio emission from a pulsar (with potential shell-like structure)&lt;/li&gt;&#xA;&lt;li&gt;Mixed (also sometimes termed &amp;quot;composite&amp;quot;), which has clear characteristics of both of the above, with X-ray and radio emissions&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;NASA &lt;a href=&quot;https://heasarc.gsfc.nasa.gov/docs/objects/snrs/shell-like.html&quot; rel=&quot;noreferrer&quot;&gt;has&lt;/a&gt; &lt;a href=&quot;https://heasarc.gsfc.nasa.gov/docs/objects/snrs/crab-like.html&quot; rel=&quot;noreferrer&quot;&gt;three&lt;/a&gt; &lt;a href=&quot;https://heasarc.gsfc.nasa.gov/docs/objects/snrs/composite.html&quot; rel=&quot;noreferrer&quot;&gt;pages&lt;/a&gt; on the different types, though it uses &amp;quot;composite&amp;quot; in place of &amp;quot;mixed&amp;quot;, designating &amp;quot;composite&amp;quot; nebulae &amp;quot;Crab-like&amp;quot;, after the &lt;a href=&quot;https://en.wikipedia.org/wiki/Crab_Nebula&quot; rel=&quot;noreferrer&quot;&gt;Crab nebula&lt;/a&gt;. The shape of a remnant depends on which type it is; for instance, composite nebulae may be shaped by the stellar wind of the central pulsar.&lt;/p&gt;&#xA;&lt;h1&gt;Interstellar nebulae&lt;/h1&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/H_II_region&quot; rel=&quot;noreferrer&quot;&gt;H II regions&lt;/a&gt; (which include the Eagle Nebula, and of course the Pillars of Creation) and &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_nebula&quot; rel=&quot;noreferrer&quot;&gt;dark nebulae&lt;/a&gt; fall into this category. Both may be part of or interact with &lt;a href=&quot;https://en.wikipedia.org/wiki/Molecular_cloud&quot; rel=&quot;noreferrer&quot;&gt;molecular clouds&lt;/a&gt;, which are the star-forming regions of the galaxy. H II regions, like molecular clouds, are shaped by the stars inside them. Radiation ionizes and even pushes away the gas surrounding young, hot stars, breaking up the nebulae. However, &lt;a href=&quot;https://www.eso.org/public/usa/news/eso1518/&quot; rel=&quot;noreferrer&quot;&gt;variations in dust density&lt;/a&gt; mean that different parts may &amp;quot;evaporate&amp;quot; at different rates, leading to some weird shapes - including the Pillars of Creation.&lt;/p&gt;&#xA;&lt;h1&gt;Evolution and dissipation&lt;/h1&gt;&#xA;&lt;p&gt;When I first read your question, I focused on the phrase (emphasis mine)&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;how nebulae &lt;em&gt;hold&lt;/em&gt; their shapes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;because nebulae are actually often extremely short-lived, on astronomical timescales. While the exact lifetimes vary, they're typically quite short when compared to stellar lifetimes:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Planetary nebulae: On the order of &lt;span class=&quot;math-container&quot;&gt;$\sim10^4$&lt;/span&gt; years, although this may be influenced by the progenitor and the nebula's morphology (see &lt;a href=&quot;http://arxiv.org/abs/1502.01015&quot; rel=&quot;noreferrer&quot;&gt;Badenes et al. (2015)&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;li&gt;Supernova remnants: On the order of &lt;span class=&quot;math-container&quot;&gt;$\sim10^5$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\sim10^6$&lt;/span&gt; years, although this depends on how you define the lifetime of a remnant. It can differ if you define it on X-ray emission, radio emission, or something else entirely.&lt;/li&gt;&#xA;&lt;li&gt;H II regions: On the order of &lt;span class=&quot;math-container&quot;&gt;$\sim10^6$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\sim10^7$&lt;/span&gt; years, depending on the evolution of the star(s) inside the region. Their radiation pressure will dissipate the gas cloud, so its lifetime depends on their characteristics. As &lt;a href=&quot;https://books.google.com/books?id=Xf1eWs_tAeoC&amp;amp;pg=PA254&amp;amp;lpg=PA254&amp;amp;dq=H+ii+region+lifetime&amp;amp;source=bl&amp;amp;ots=-UcBrfmpBP&amp;amp;sig=y24i7KpiOX6KZ73qKKWHEZWIrHc&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwjv5e2R4aHPAhWFFT4KHfUZA_E4ChDoAQhXMAg#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Stars, Nebulae and the Interstellar Medium&lt;/em&gt;&lt;/a&gt; explains, H II regions would live longer if they weren't dissipated by the stars. Otherwise, they could survive for orders of magnitude longer.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-09-22T00:55:32.957" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18438" PostTypeId="1" CreationDate="2016-09-22T03:24:06.237" Score="5" ViewCount="946" Body="&lt;p&gt;Two typical comets that I know are Donati's and Halley's. So comets also have orbits, but are their orbits on the same plane (such as the ecliptic) as the other planets? &lt;/p&gt;&#xA;" OwnerUserId="14254" LastEditorDisplayName="user15217" LastEditDate="2017-01-01T08:43:09.360" LastActivityDate="2017-01-01T08:43:09.360" Title="Do comet orbits lie in the same plane as the other planets?" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;comets&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18439" PostTypeId="2" ParentId="18438" CreationDate="2016-09-22T05:13:23.287" Score="6" Body="&lt;p&gt;Those two comets happen to have large orbital inclinations to the Ecliptic: 117$^\circ$ for&#xA;&lt;a href=&quot;http://www.minorplanetcenter.net/db_search/show_object?utf8=%E2%9C%93&amp;amp;object_id=C%2F1858+L1&quot; rel=&quot;nofollow&quot;&gt;C/1858 L1 (Donati)&lt;/a&gt;&#xA;and 162$^\circ$ for&#xA;&lt;a href=&quot;http://www.minorplanetcenter.net/db_search/show_object?utf8=%E2%9C%93&amp;amp;object_id=1P&quot; rel=&quot;nofollow&quot;&gt;1P/Halley&lt;/a&gt;.&#xA;Long-period comet orbits are oriented essentially at random, suggesting that they originate in the&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Oort_cloud&quot; rel=&quot;nofollow&quot;&gt;Oort cloud&lt;/a&gt;,&#xA;a sphere several thousand AU from the Sun.&#xA;Many &lt;a href=&quot;https://en.wikipedia.org/wiki/Comet#Orbital_characteristics&quot; rel=&quot;nofollow&quot;&gt;short-period&lt;/a&gt;&#xA;comet orbits are inclined only a few degrees to the Ecliptic,&#xA;suggesting origins in the&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Scattered_disc&quot; rel=&quot;nofollow&quot;&gt;scattered disc&lt;/a&gt;,&#xA;a torus several tens to a few hundred AU from the Sun.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2016-09-23T22:43:06.387" LastActivityDate="2016-09-23T22:43:06.387" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18442" PostTypeId="2" ParentId="18052" CreationDate="2016-09-22T12:15:21.393" Score="1" Body="&lt;p&gt;Intuitively you think of quantization as something that discards information. That may be true in the end, but it is not a useful way to look at it. Think the other way around, quantization &lt;em&gt;adds&lt;/em&gt; an error-signal. If you know what this error signal looks like, it gives you opportunity to analyze how digital processing transforms the error and if it ends up interfering with your desired signal (and how much that interference will be).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ALMA is a phased array, it gets its precision from the correlation of phases if multiple receivers (likewise, &lt;em&gt;phase&lt;/em&gt; is typically more important than amplitude in recent modulation schemes). The error function for phase typically is a sawtooth, as the phasor (of a theoretical clean signal) rotates. How the function looks exactly and what the fundamental frequency is, depends on properties of the ADC (and sometimes on AGC settings). The error signal frequency will be n times the received frequency, n=12 or n=8 being typical values. I would have to look into the details of ALMA, I'm not familiar with this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider how this error function is sampled. There is no way to attenuate it before sampling, so aliased images of harmonics of this sawtooth end up in your digital data. You can calculate where these harmonics are and how strong they are. And you can shift them by altering the sampling rate (with a given fixed signal frequency). If you want to observe a certain bandwidth and you do optimize sampling rate, you may find that you have f.e. the 11th harmonic (with amplitude 1/11) somewhere in your signal, but you can avoid all the lower (and stronger) harmonics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Investing in more bits for quantization reduces the amplitude of errors, raising the fundamental freq of the error function at the same time. You may find, that the contribution of quantization errors is already in the magnitude of other noise sources, so there is not much to gain for overall system performance. This is typically the case for direct code spread spectrum applications like GNSS systems.&lt;/p&gt;&#xA;" OwnerUserId="14323" LastActivityDate="2016-09-22T12:15:21.393" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18443" PostTypeId="1" AcceptedAnswerId="18445" CreationDate="2016-09-22T13:52:22.447" Score="5" ViewCount="424" Body="&lt;p&gt;Or are seismic phenomenon named differently when they happen on other celestial bodies? If so, what are they called?&lt;/p&gt;&#xA;" OwnerUserId="9803" LastEditorUserId="37191" LastEditDate="2021-03-07T03:37:27.417" LastActivityDate="2021-03-07T03:37:27.417" Title="If an earthquake happens on Mars, is it still an earthquake?" Tags="&lt;planet&gt;&lt;terminology&gt;&lt;geology&gt;&lt;quake&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18444" PostTypeId="1" AcceptedAnswerId="27041" CreationDate="2016-09-22T14:35:04.560" Score="9" ViewCount="144" Body="&lt;p&gt;&lt;strong&gt;tl;dr&lt;/strong&gt; Has the brown dwarf observation been disproven?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have just started reading about the interesting object V471 Tauri. The first two sentences of the introduction to &lt;a href=&quot;http://arxiv.org/abs/1506.05067&quot; rel=&quot;nofollow noreferrer&quot;&gt;The V471 Tauri System: A Multi-datatype Probe&lt;/a&gt; Vaccaro et al. (2015):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;V471 Tau, a white dwarf-red dwarf eclipsing binary (EB) in the Hyades with orbit period $0^d.52118$, is primarily known for its unique historical role as a stimulus to common envelope evolution theory (Chau et al. 1974; Refsdal 1974; Sparks &amp;amp; Stecher 1974; Ostriker 1976; Paczynski 1976; Alexander et al. 1976; Taam et al. 1978). Other properties include a likely brown dwarf companion to the EB, measured white dwarf spin, mass loss and exchange in a detached binary, differential rotation measured via magnetic spots, spot distributions, accurate white dwarf parameters, and photometric-spectroscopic distance measures that help to pin down the binary’s location within the Hyades.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's the &quot;likely brown dwarf companion&quot; I'd like to ask about. The paper with the catchy title &quot;&lt;a href=&quot;http://arxiv.org/abs/1502.05116&quot; rel=&quot;nofollow noreferrer&quot;&gt;The First Science Results from SPHERE: Disproving the Predicted Brown Dwarf around V471 Tau&lt;/a&gt;&quot; Hardy et al. (2015) &lt;a href=&quot;https://en.wikipedia.org/wiki/Spectro-Polarimetric_High-Contrast_Exoplanet_Research&quot; rel=&quot;nofollow noreferrer&quot;&gt;SPHERE&lt;/a&gt; is a new, advanced adaptive optics system at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Very_Large_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Very Large Telescope&lt;/a&gt; (VLT). The image below (figure 3) is part of the argument, and the suggestion is that if there is no brown dwarf seen in the band between the two white circles, then the predicted brown dwarf does not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be interesting, because an alternative explanation for the slow, periodic drift in the eclipse timing would have to be found. One possibility is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Applegate_mechanism#Mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;Applegate Mechanism&lt;/a&gt; which I don't understand, but may ask a separate question after this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to &lt;a href=&quot;http://arxiv.org/abs/1506.05067&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vaccaro 2015&lt;/a&gt;, section 9, titled &quot;On the Reality of the Third Star&quot; is over six pages of discussion of the underlying assumptions, and if I understand it correctly provides several possible ways in which a suitable brown dwarf could exist but not show up in the SPHERE image. In essence &lt;em&gt;disproving the existence disproof.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I would like to know:&lt;/strong&gt; Is my understanding of the current situation correct? Have there been more recent developments? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/B3HT5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B3HT5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Left panel of Fig. 3 of &lt;a href=&quot;http://arxiv.org/abs/1502.05116&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hardy et al. 2015&lt;/a&gt;: &quot;Figure 3. H-band image of V471 Tau obtained on the SPHERE IRDIS instrument at the VLT. Left panel: Resulting image after angular differential imaging (ADI). The area in-between the white circles denotes the 5 sigma predicted position of the brown dwarf...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/InArY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/InArY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; The monster currently known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Spectro-Polarimetric_High-Contrast_Exoplanet_Research&quot; rel=&quot;nofollow noreferrer&quot;&gt;SPHERE&lt;/a&gt; from &lt;a href=&quot;https://en.wikipedia.org/wiki/File:The_SPHERE_exoplanet_imager_for_the_VLT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2021-04-16T23:23:15.287" LastActivityDate="2021-04-16T23:23:15.287" Title="V471 Tauri's circumbinary brown dwarf non-observation; Applegate, or over-restrictive assumptions?" Tags="&lt;binary-star&gt;&lt;brown-dwarf&gt;&lt;infrared&gt;&lt;adaptive-optics&gt;&lt;eclipsing-star-systems&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18445" PostTypeId="2" ParentId="18443" CreationDate="2016-09-22T14:44:00.340" Score="8" Body="&lt;p&gt;As pointed out in the comments, some do prefer to use different terms for &lt;a href=&quot;https://en.wikipedia.org/wiki/Quake_%28natural_phenomenon%29&quot;&gt;quakes&lt;/a&gt; on other bodies, and you can see &lt;a href=&quot;https://scholar.google.com/scholar?hl=en&amp;amp;q=marsquake&amp;amp;btnG=&amp;amp;as_sdt=1%2C44&amp;amp;as_sdtp=&quot;&gt;here&lt;/a&gt; that &quot;marsquake&quot; does have scholarly usage. That said, there is also scholarly usage for &lt;a href=&quot;https://scholar.google.com/scholar?q=%22earthquake+on+mars%22&amp;amp;btnG=&amp;amp;hl=en&amp;amp;as_sdt=0%2C44&quot;&gt;&quot;earthquake&quot;&lt;/a&gt; in this context. If you think about it, this makes sense given that one definition of &quot;earth&quot; is &lt;a href=&quot;https://en.oxforddictionaries.com/definition/earth&quot;&gt;&quot;the substance of the land surface&quot;&lt;/a&gt;, i.e. the ground. This definition of &quot;earth&quot; is agnostic to which terrestrial planet the land surface is on.&lt;/p&gt;&#xA;" OwnerUserId="6" LastActivityDate="2016-09-22T14:44:00.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18446" PostTypeId="1" CreationDate="2016-09-22T15:36:16.100" Score="1" ViewCount="196" Body="&lt;p&gt;I am a novelist working on a new book, titled Werewolves In The Christmas City, which happens to be Bethlehem, Pa. In 2010 when I was working on a book titled The Christmas City Vampire, a lunar eclipse on the Winter Solstice occurred. As you know it was the first one since 1638. On the 2010 occurrence, I added the resurrection of two werewolves-twins-turned during that eclipse of 1638. My question is: How would my professor character figure out that the event will occur in 2010. The time frame I am in right now is 1947. I would appreciate any help you can give me. Math is certainly not one of my strong suites.&lt;/p&gt;&#xA;" OwnerUserId="14326" LastEditorUserId="13426" LastEditDate="2016-09-22T16:11:31.583" LastActivityDate="2016-09-23T23:16:59.747" Title="Calculating the date and year when a lunar eclipse will fall on the Winter Solstice" Tags="&lt;lunar-eclipse&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18448" PostTypeId="2" ParentId="18446" CreationDate="2016-09-22T20:55:20.157" Score="2" Body="&lt;p&gt;A university library in 1947 might have a copy of T. von Oppolzer,&#xA;&lt;a href=&quot;https://archive.org/details/canonderfinstern00oppo&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Canon der Finsternisse&lt;/em&gt;&lt;/a&gt; (1887),&#xA;which lists those lunar eclipses on pages 368 and 374.&#xA;A reader with a working knowledge of astronomy could make sense of the tables even if not proficient in mathematics or German.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that reference were not available, a text such as R. Buchanan,&#xA;&lt;a href=&quot;https://archive.org/details/mathematicaltheo00buchrich&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;The Mathematical Theory of Eclipses&lt;/em&gt;&lt;/a&gt; (1904),&#xA;would equip the professor to work it out from scratch.&#xA;As lunar eclipses are visible from the entire night hemisphere,&#xA;they are simpler to compute than solar eclipses,&#xA;which are total only along a narrow path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eclipses recur in&#xA;&lt;a href=&quot;http://eclipse.gsfc.nasa.gov/LEsaros/LEperiodicity.html#section104&quot; rel=&quot;nofollow&quot;&gt;saros cycles&lt;/a&gt;&#xA;of 18 years and 11.3 days.&#xA;Given dates of some contemporary eclipses,&#xA;one could trace&#xA;&lt;a href=&quot;http://eclipse.gsfc.nasa.gov/LEsaros/LEsaros119.html&quot; rel=&quot;nofollow&quot;&gt;saros 119&lt;/a&gt;&#xA;back: 1945-06-25, 1927-06-15, ..., 1638-12-21; or trace&#xA;&lt;a href=&quot;http://eclipse.gsfc.nasa.gov/LEsaros/LEsaros125.html&quot; rel=&quot;nofollow&quot;&gt;saros 125&lt;/a&gt;&#xA;forward: 1938-11-07, 1956-11-18, ..., 2010-12-21.&#xA;The saros numbers were not assigned until 1955,&#xA;but the cycles were known to ancient astronomers.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2016-09-23T23:16:59.747" LastActivityDate="2016-09-23T23:16:59.747" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18450" PostTypeId="2" ParentId="18423" CreationDate="2016-09-22T23:48:58.890" Score="1" Body="&lt;p&gt;A superluminous supernova (aka hypernova) can exhibit a double peak to its brightness and some are theorizing that this may be the norm for a superluminous supernova, though as far as I know it's only actually been observed in one case so far (DES14X3taz).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, in (at least) this case there was an initial substantial increase in brightness. Then the brightness dropped (a couple of magnitudes) for a few days, then ramped back up to considerably brighter than the initial &quot;bump&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You probably are going to need to be careful about the distances involved. The initial burst of light is already large enough that unless your people are &lt;em&gt;quite&lt;/em&gt; a ways away, it'll already be enough to fry them to a crisp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one other point that might be interesting for your novel though. After the explosion, what you probably get is a magnetar--which, as you'd guess from the name, is a star with an &lt;em&gt;extremely&lt;/em&gt; strong magnetic field--so strong, in fact, that it's likely to cause all sorts of havoc with anything in the vicinity that depends on anything involving electrical activity--not only electronics, but also probably people's nerves as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an obvious problem here though: a red supergiant is the right type of star as the progenitor for a &quot;normal&quot; supernova. It's probably &lt;em&gt;not&lt;/em&gt; the right type as the progenitor for a superluminous supernova. The progenitor of a supernova is typically something like six or eight solar masses. A superluminous supernova is &lt;em&gt;probably&lt;/em&gt; (only a few are known, so it's hard to generalize) something like a couple hundred solar masses. Given the amount of energy released, it &lt;em&gt;has&lt;/em&gt; to be quite large anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: &lt;a href=&quot;http://arxiv.org/abs/1512.06043&quot; rel=&quot;nofollow&quot;&gt;Smith, et al (2015)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14332" LastEditorUserId="14332" LastEditDate="2016-09-23T00:08:09.243" LastActivityDate="2016-09-23T00:08:09.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18451" PostTypeId="2" ParentId="10360" CreationDate="2016-09-23T03:27:50.270" Score="0" Body="&lt;p&gt;0 K,&quot;absolute zero&quot; (-273.15°C) is the theorized measurement of energy at which all motion stops. Electrons, atoms, and particles stop all movement and it would likely be the lowest possible temperature possible anywhere in the universe. Now that temperature may or may not actually exist in the universe because of all of the radiation and energy from various stars, however it may be possible in the middle of a supervoid or other places.&lt;/p&gt;&#xA;" OwnerUserId="14333" LastEditorUserId="10172" LastEditDate="2016-09-23T06:24:01.907" LastActivityDate="2016-09-23T06:24:01.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18452" PostTypeId="1" AcceptedAnswerId="18458" CreationDate="2016-09-23T03:29:26.573" Score="1" ViewCount="265" Body="&lt;p&gt;Forgive any kind of &quot;dumb&quot; question I may ask as this is a new interest of mine and I know it's purely hypothetical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one were able to surpass light speed and the expansion of space to go beyond the particle horizon, what could be out there? Could it just be an empty void? Could it loop around to the other side of the particle horizon? I'd love to read about any kind of theories or research done on this subject.&lt;/p&gt;&#xA;" OwnerUserId="14333" LastEditorUserId="37191" LastEditDate="2021-02-16T08:24:09.583" LastActivityDate="2021-02-16T08:24:09.583" Title="What could be past the particle horizon?" Tags="&lt;astrophysics&gt;&lt;big-bang-theory&gt;&lt;horizon&gt;&lt;particles&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18454" PostTypeId="2" ParentId="18337" CreationDate="2016-09-23T05:53:29.237" Score="1" Body="&lt;p&gt;The main things to look for are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Decent optics (nearly anything except those with plastic lenses).&lt;/li&gt;&#xA;&lt;li&gt;A steady mount that points where you want to, and moves smoothly. An altitude-azimuth mount is fine because with practice you can guide at high powers.&lt;/li&gt;&#xA;&lt;li&gt;The eyepieces should be 1 1/4 inch size or more expensive 2 inch. The older 0.96 inch are very hard to find. This lets you upgrade eyepieces later.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The larger the aperture the better. Don't worry about magnification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: find your local astronomical society, ask there, go to one of their &quot;star nights&quot;, they may even have telescopes on loan.&lt;/p&gt;&#xA;" OwnerUserId="10761" LastEditorUserId="616" LastEditDate="2016-09-23T15:44:04.717" LastActivityDate="2016-09-23T15:44:04.717" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18455" PostTypeId="1" CreationDate="2016-09-23T13:44:17.147" Score="1" ViewCount="100" Body="&lt;p&gt;Distant galaxies are moving as space expands, not moving through space.  So a photon coming to us from them has to come back through distorted space.  So is not spacetime distorted, so the photon has to go through an area where time slows down.  So would this cause red shift.  So maybe we have a simplistic view of red shift and maybe those galaxies are not going as fast as we think&lt;/p&gt;&#xA;" OwnerUserId="14342" OwnerDisplayName="Roy Proudman" LastEditorUserId="14342" LastEditDate="2016-09-23T15:21:09.760" LastActivityDate="2016-09-25T04:13:58.683" Title="Does red shift need to compensate for space expansion" Tags="&lt;redshift&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18456" PostTypeId="1" CreationDate="2016-09-23T14:59:58.017" Score="5" ViewCount="650" Body="&lt;p&gt;The recent BBC news item &lt;a href=&quot;http://www.bbc.com/news/science-environment-37438458&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gaia clocks speedy cosmic expansion&lt;/a&gt; says that a recent Gaia measurement of 212 Cepheid variables in other galaxies within the local group yields a Hubble constant of ~73 km/s/Mpc which is in agreement with a previous measurement using the Hubble telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The results are reported in: &lt;a href=&quot;https://arxiv.org/abs/1609.05175&quot; rel=&quot;nofollow noreferrer&quot;&gt;A test of Gaia Data Release 1 parallaxes: implications for the local distance scale&lt;/a&gt;. I have included Figure 1 at the end here. If I understand correctly, the x-axis or &quot;Photometric Parallax&quot; represents photometric data which is compared to a model to yield a distance, which is then expressed as an equivalent parallax. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gaia is not a &quot;point and shoot&quot; instrument like the Hubble telescope. It has a slow, extremely steady rotation in order to scan at least 1.5 full revolutions - enough for every start in the scan to pass &lt;a href=&quot;https://space.stackexchange.com/a/18237/12102&quot;&gt;completely across the CCD array through both telescopes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this isn't an optimal observing plan for collecting data on a group of Cepheids. Although there are similarities, the light curves of different stars will be distinct. &lt;em&gt;Am I naive to think that it is necessary to build up a nice, complete light curve with dense points in time in order to use the photometry for precision distance calculations?&lt;/em&gt; Was this actually done within Gaia's survey observing program, or was there a special observing plan for the Cepheid measurements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data also needs to be good enough to calculate the periods of each star with sufficient precision. The curve of absolute magnitude vs period is quite steep at low periods, meaning the error in the calculated distance is strongly affected by an error in the period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And could someone help me understand - what is a TGAS parallax?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QPv0b.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QPv0b.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Figure 1 from &lt;a href=&quot;https://arxiv.org/abs/1609.05175&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/b1Nmm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/b1Nmm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IvKLo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IvKLo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above X2:&lt;/strong&gt; &quot;Figure 4 and 4a: Cepheid light curves (click on images for larger versions) Light curves for the twelve Cepheid variables in M100 that have been observed with Hubble. The absolute magnitude, M, is determined from the period of the Cepheids. Adapted from Freedman et al. (1994).&quot; from the tutorial: &lt;a href=&quot;https://astrosociety.org/edu/publications/tnl/57/realastro3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://astrosociety.org/edu/publications/tnl/57/realastro3.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2016-09-24T00:44:36.450" Title="Did Gaia actually generate complete light curves for 212 Cepheids in other galaxies?" Tags="&lt;distances&gt;&lt;expansion&gt;&lt;cepheids&gt;&lt;gaia&gt;&lt;hubble-constant&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18457" PostTypeId="2" ParentId="18455" CreationDate="2016-09-23T15:26:44.460" Score="7" Body="&lt;p&gt;Redshift &lt;em&gt;is&lt;/em&gt; the &quot;stretching&quot; of the wavelength of light as it travels, over finite time, towards us, due to the ongoing expansion of space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effects of this redshift are indistinguishable from that caused by the doppler effect, so that is why &lt;em&gt;some&lt;/em&gt; elementary treatments choose to discuss it in that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most galaxies only have peculiar velocities with respect to thei comoving volume of a few hundred km/s, and this does perturb the cosmological redshift by that (small) amount. That is why Hubble's law only applies at distances such that cosmological expansion dominates.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-09-23T15:26:44.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18458" PostTypeId="2" ParentId="18452" CreationDate="2016-09-23T16:20:19.357" Score="1" Body="&lt;p&gt;It sounds like your question is about the single most important principle of cosmology, called the cosmological principle.  This asserts that as long as all our observations are consistent with the idea that the universe is the same everywhere at a given age, we will continue to use a model with that attribute.  There are lots of good reasons for adopting that principle, the least of which is not that it allows us to have a model in the first place.  So what this means is, our answer to what is beyond what we can see is that we don't know, but we are going to model it in a way that makes sense to us, so we will use the cosmological principle.   You have to be able to distinguish what we know in the sense of having direct information about, versus what we know about our own best models.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-09-23T16:20:19.357" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18459" PostTypeId="2" ParentId="18456" CreationDate="2016-09-23T16:43:32.693" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Am I naive to think that it is necessary to build up a nice, complete light curve with dense points in time in order to use the photometry for precision distance calculations? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, the more I read about it, the more difficult it seems to be. In principle, if you sample regularly and often, you should eventually get a light curve, unless the period is the same as or a multiple of your sampling period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not easy to find how often GAIA observes a given patch of sky, but &lt;a href=&quot;http://www.esa.int/Our_Activities/Space_Science/Gaia/Gaia_s_first_year_of_scientific_observations&quot;&gt;here&lt;/a&gt; it says that after one year of observing&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Gaia has made an average of roughly 14 measurements of each star on the sky thus far, [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so they seem to have repeat observations a bit more often than once a month. Cepheids have a period of typically 2 to 45 days (&lt;a href=&quot;https://de.wikipedia.org/wiki/Cepheiden&quot;&gt;says german Wikipedia&lt;/a&gt; - didn't find it in the english article). If your sampling is regular, you should be able to get a decent light curve for at least some Cepheids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complete guesswork on my part: Apparently, longer period Cepheids are redder at the minimum. Perhaps they use that to distinguish between the actual period and higher harmonics?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And could someone help me understand - what is a TGAS parallax?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the introduction it says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Tycho-Gaia Astrometric Solution (hereafter TGAS; Michalik et al. 2015)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the abstract of &lt;a href=&quot;http://arxiv.org/abs/1412.8770&quot;&gt;that paper&lt;/a&gt;, it seems to describe their method (they practised with mock GAIA data). Seems like they use the previous Tycho catalogue to improve astrometry and parallax measurements from GAIA data (they mention using Hipparcos parallaxes as a consistency check).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, figure 1 seems to be Cepheid distance vs. Parallax distance.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastActivityDate="2016-09-23T16:43:32.693" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18460" PostTypeId="2" ParentId="18409" CreationDate="2016-09-23T16:45:26.110" Score="2" Body="&lt;p&gt;Addendum - be sure to click here: &lt;a href=&quot;http://www.sankara.net/mom.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.sankara.net/mom.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(choose &quot;Helios&quot; button, top right)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an &quot;incredibly rhetorical&quot; question, heh, can you see any ellipse-ness in the planets orbits?  Your ultimate game result would, I imagine, look something like that webapp.  The pink/orange lines showing the transfer of the two (real) spacecraft Earth to Mars would I guess be something like in your game. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would just reiterate that literally for a game, or any conceivable &quot;display of a planet system&quot;, all you have to do is &quot;draw a circle&quot; and &quot;put a dot on it which moves around in a circle&quot;. Making game-like displays of planet system is &lt;strong&gt;remarkably easy&lt;/strong&gt; - since the scale is so tremendously large, heh!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Again you can simply get the year-length from the trivial formulas linked below, remembering you have to drastically alter timescale.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope it helps!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Just for the record:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To create a moving diagram showing planet(s) orbiting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) as you say, place them at the desired &lt;strong&gt;distance&lt;/strong&gt; and know the &lt;strong&gt;mass&lt;/strong&gt;.  (2) calculate the &quot;year&quot; length of &lt;strong&gt;time&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It is trivial to do that with, as you say, Kepler's equations. Easily found, example: &lt;a href=&quot;http://www.ajdesigner.com/phpgravity/keplers_law_equation_mass.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;plug-in equations&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you'll scale time. (What about 1 Earth-year is one minute?)  (Indeed, you will of course have to radically scale-up the body sizes, or you will simply see absolutely nothing.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had a window open so I just did this in a game engine; it takes 7 clicks and 3 lines of code!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KadDE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KadDE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: you mentioned &quot;simulation&quot; in your original question. That term has a specific technical meaning (especially to anyone on this site), and it is utterly unrelated to what you are asking.  Enjoy!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;FTR I gather from your other question you want to show extreme &quot;video game like&quot; ellipses.  Purely as a matter of interest, for your information the way you literally do that for literally a space-travel-like game (whether app or PC) is either&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) trivially, just have the planet move in and out (whatever percentage you want) each half-year. (simply use a separate transform, so there's a transform for it going around, and a transform for it going in and out) or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) nudge it as outlined for example here, using &lt;a href=&quot;https://stackoverflow.com/questions/30827958/unity-5-moving-planets-in-a-circular-or-elliptical-path-orbits&quot;&gt;sin and cos&lt;/a&gt; (Or, all game engines or graphics/animation apis have a &quot;ping pong&quot; and/or &quot;smoothing-at-ends&quot; call, just use that.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reminder that your display will be drastically altered from reality since you will be presenting 10 or 20 months in, say, 10 or 20 seconds.&lt;/p&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="-1" LastEditDate="2017-05-23T12:37:21.700" LastActivityDate="2016-09-30T13:42:43.570" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18461" PostTypeId="1" AcceptedAnswerId="18463" CreationDate="2016-09-23T17:30:48.617" Score="3" ViewCount="309" Body="&lt;p&gt;After reading this excellent but difficult answer, &lt;a href=&quot;https://astronomy.stackexchange.com/a/18407/13071&quot;&gt;https://astronomy.stackexchange.com/a/18407/13071&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been thinking how difficult it is to get a sense of the distance between and the size of typical items of dust, pebbles, rocks and/or anything else in Saturn's rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, have any photographs been &lt;strong&gt;taken by spacecraft&lt;/strong&gt;, more or less &quot;in&quot; the rings, &lt;strong&gt;close enough to actually see individual&lt;/strong&gt; dust, pebbles and/or rocks?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As I understand it, the rings are on the order of 10m thick.  So I'm asking, have any photos been taken from that order of distance from the rings?  What's simply the closest viewpoint ever taken?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Failing that, is there perhaps an artist's rendition somewhere of &quot;what you'd see from 100m above the rings looking straight down at them?&quot;  What would be in that frustrum - nothing?  Lots of fist-size rocks you could see? Or?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's completely unclear to me if each (say) pebble-size piece is inches apart, meters apart, or for all I know 100s of km apart?&lt;/p&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-09-23T18:20:46.797" Title="Is there, in fact, any close-up photography of Saturn's rings, showing individual pebbles/rocks?" Tags="&lt;saturn&gt;&lt;planetary-ring&gt;&lt;space-probe&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18462" PostTypeId="2" ParentId="18456" CreationDate="2016-09-23T17:58:37.190" Score="3" Body="&lt;p&gt;Casertano et al. used the&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Classical_Cepheid_variable#Period-luminosity_relation&quot; rel=&quot;nofollow&quot;&gt;period-luminosity&lt;/a&gt;&#xA;(P-L) relation of Cepheid variables as a sanity check on Gaia DR1 parallaxes.&#xA;They chose Cepheid variables &lt;em&gt;within&lt;/em&gt; the Milky Way having parallaxes in TGAS (Tycho-Gaia astrometric solution, reusing Hipparcos data for a head start).&#xA;The period and apparent magnitude come from &lt;em&gt;ground-based&lt;/em&gt; photometry&#xA;(&lt;a href=&quot;http://dx.doi.org/10.1111/j.1365-2966.2007.11972.x&quot; rel=&quot;nofollow&quot;&gt;van Leeuwen et al. 2007&lt;/a&gt;, section 2 and appendix A).&#xA;The Cepheid period-luminosity relation gives an absolute magnitude,&#xA;leading to a distance estimate whose reciprocal is an expected parallax.&#xA;The actual parallaxes agree with the photometric estimates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having found the Gaia DR1 parallaxes sane,&#xA;they used those parallaxes to make a tentative estimate of the Hubble constant $H_0$,&#xA;which in turn serves as a sanity check on a recent $H_0$ estimate based on various HST observations&#xA;(&lt;a href=&quot;https://arxiv.org/abs/1604.01424&quot; rel=&quot;nofollow&quot;&gt;Riess et al. 2016&lt;/a&gt;, cited as R16).&#xA;Later they hope to combine HST photometry of Milky Way Cepheids&#xA;with final Gaia parallaxes for an improved $H_0$ estimate.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2016-09-24T00:44:36.450" LastActivityDate="2016-09-24T00:44:36.450" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18463" PostTypeId="2" ParentId="18461" CreationDate="2016-09-23T18:20:46.797" Score="5" Body="&lt;p&gt;The rings are about 3% solid in the densest parts, but this translates to a &lt;a href=&quot;http://classroom.synonym.com/close-rocks-saturns-rings-13152.html&quot; rel=&quot;noreferrer&quot;&gt;separation between 30cm particles of about 1metre&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no images Because approaching the dense part of the rings would be extremely difficult. A probe would be travelling at orbital speeds, and even a small percentage difference in speed would translate to a large absolute difference. As the probe would also be in orbit, it would be extremely difficult to find an orbit that doesn't collide catastrophically with ring particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cassini &lt;a href=&quot;http://www.esa.int/Our_Activities/Space_Science/Cassini-Huygens/Hear_the_passage_through_the_rings&quot; rel=&quot;noreferrer&quot;&gt;did travel through one of the gaps in the ring system&lt;/a&gt; during Saturn orbital insertion. It survived 100000 dust impacts in 5 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Nature&lt;/em&gt; has some &lt;a href=&quot;http://www.nature.com/nature/journal/v440/n7084/fig_tab/440614a_F2.html&quot; rel=&quot;noreferrer&quot;&gt;artist's impressions of the ring particles&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-09-23T18:20:46.797" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18464" PostTypeId="5" CreationDate="2016-09-23T22:33:12.567" Score="0" Body="&lt;p&gt;A refractor telescope uses lenses to focus incoming light. For various practical reasons, refractor telescopes cannot be built as large as mirror telescopes. A great advantage is that the path of incoming light is not obstructed by a secondary mirror.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastEditorUserId="13681" LastEditDate="2016-09-23T23:25:12.257" LastActivityDate="2016-09-23T23:25:12.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18465" PostTypeId="4" CreationDate="2016-09-23T22:33:12.567" Score="0" Body="Questions about telescopes with lenses as their main optical component." OwnerUserId="13681" LastEditorUserId="13681" LastEditDate="2016-09-23T23:25:15.680" LastActivityDate="2016-09-23T23:25:15.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18466" PostTypeId="1" CreationDate="2016-09-24T01:27:47.840" Score="1" ViewCount="333" Body="&lt;p&gt;I'm looking at the Sloan Digital Sky Survey data. I'm primarily using the flux data, from the &lt;a href=&quot;https://data.sdss.org/datamodel/files/BOSS_SPECTRO_REDUX/RUN2D/spectra/PLATE4/spec.html&quot; rel=&quot;nofollow&quot;&gt;documentation&lt;/a&gt; I see the first extend (HDU 1) has &lt;code&gt;flux&lt;/code&gt;, &lt;code&gt;loglam&lt;/code&gt;, &lt;code&gt;ivar&lt;/code&gt;, &lt;code&gt;and_mask&lt;/code&gt;, &lt;code&gt;or_mask&lt;/code&gt;, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's been a lot of confusion here about what the &lt;code&gt;or_mask&lt;/code&gt; is doing exactly, and the documentation hasn't been extremely clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding so far is that the &lt;code&gt;or_mask&lt;/code&gt; tells me when a particular pixel was flagged in one of many ways possible. I've understood from various random sources that this means the pixel is &quot;bad&quot; somehow. I haven't yet figured out how to know exactly &quot;why&quot; it is bad, my working assumption is that the possibilities are numerous, that there are many fields that have flags that might answer that question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My primary question: Is every flux value with a non-zero &lt;code&gt;or_mask&lt;/code&gt; inherently untrustworthy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About 80% of pixels have a non-zero &lt;code&gt;or_mask&lt;/code&gt; value, particularly on the bluer end of the spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is the &lt;code&gt;or_mask&lt;/code&gt; telling me more broadly that there's a flag somewhere, not necessarily even with the flux data, that I might want to consider?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HDU 1 (extname COADD): Coadded Spectrum from spPlate&#xA;&#xA;Binary table with columns:&#xA;                        Required  Columns &#xA;Name      Type      Comment&#xA;flux      float32   coadded calibrated flux [10-17 ergs/s/cm2/Å]&#xA;loglam    float32   log10(wavelength [Å])&#xA;ivar      float32   inverse variance of flux&#xA;and_mask  int32     AND mask&#xA;or_mask   int32     OR mask&#xA;wdisp     float32   wavelength dispersion in pixel=dloglam units&#xA;sky       float32   subtracted sky flux [10-17 ergs/s/cm2/Å]&#xA;model     float32   pipeline best model fit used for classification and redshift&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="14345" LastActivityDate="2016-09-24T01:27:47.840" Title="Sloan Digital Sky Survey (SDSS) HDU `or_mask` - what does it mean exactly?" Tags="&lt;data-analysis&gt;&lt;sky-survey&gt;&lt;sloan-dss&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18468" PostTypeId="2" ParentId="18432" CreationDate="2016-09-24T02:48:45.077" Score="0" Body="&lt;p&gt;Theoretically any object with a mass great enough that its own gravity is able to overcome its molecular movement and provided there is no forces to counteract that force such as atomic fission or fusion can in deed implode to a singularity, however the pressures from the sea are far to weak even at the center of the earth to initiate such a reaction it takes a mass of over 3 times that of our sun to even come close. &#xA;   On the other hand it may be possible hypothetically to compress an Eisenstein Condensate kept at temperatures near 0K (say 1k) at extreme pressure might be able to form a mini hole. But the heat caused by that pressure would disrupt the process and even if successful it would evaporate in pico seconds&lt;/p&gt;&#xA;" OwnerUserId="14346" LastEditorUserId="14346" LastEditDate="2016-09-24T02:57:11.190" LastActivityDate="2016-09-24T02:57:11.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18470" PostTypeId="2" ParentId="488" CreationDate="2016-09-24T04:05:42.310" Score="1" Body="&lt;p&gt;To determine size of a black hole first you need to determine its Schwarzschild Radius using this formula Rs= (2MG)/(c^2). To determine its photo-sphere use the formula 3Rs/2 and extends from the surface of its Rs, to determine its accretion disk (PNR for matter) it is 5.5Rs and measured from the surface of its Rs threshold.&lt;/p&gt;&#xA;" OwnerUserId="14346" LastEditorUserId="14346" LastEditDate="2016-09-24T04:12:18.030" LastActivityDate="2016-09-24T04:12:18.030" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18471" PostTypeId="1" CreationDate="2016-09-24T10:28:18.300" Score="3" ViewCount="359" Body="&lt;p&gt;I am trying to set up 2 Planets in a co-orbital configuration around a Jupiter size planet similarly to Saturn's moons Janus and Epimetheus. I'm trying to do this in Universe Sandbox 2 but when I place these planets they mess with each others orbits too much. Is it possible to have such a system and how would it be done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://abyss.uoregon.edu/~js/ast121/lectures/epimetheus_and_janus.html&quot; rel=&quot;nofollow&quot;&gt;Diagram&lt;/a&gt; showing what Epimetheus and Janus do.&lt;/p&gt;&#xA;" OwnerUserId="14348" LastEditorUserId="7982" LastEditDate="2021-03-08T00:06:54.333" LastActivityDate="2021-03-08T00:06:54.333" Title="2 Earth-like planets in co-orbital configuration" Tags="&lt;orbit&gt;&lt;planet&gt;&lt;universe-sandbox&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18472" PostTypeId="1" CreationDate="2016-09-24T14:25:49.883" Score="1" ViewCount="74" Body="&lt;p&gt;I am planning a game where on could control evolution on an alien world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking red dwarf as per here &lt;a href=&quot;https://astronomy.stackexchange.com/q/10542&quot;&gt;Habitable Planet around Red Dwarf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my understanding three questions arise:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Tidal lock&lt;/li&gt;&#xA;&lt;li&gt;Radiation&lt;/li&gt;&#xA;&lt;li&gt;Sunspots&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Could a parent body (think small gas giant) be tidally locked and a moon then freely rotate? Could the magnetic field of a parent body protect a moon from a sun without the parent radiating too much?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what to say about sunspots though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then about the parent body itself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the formation of the mini-neptune so close to a red dwarf impossible? Would its gravity wreck eveything?&lt;/p&gt;&#xA;" OwnerUserId="14349" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2017-03-07T00:45:10.800" Title="Could a moon of a mini-neptune around a red dwarf harbour life?" Tags="&lt;life&gt;&lt;natural-satellites&gt;&lt;gas-giants&gt;&lt;red-dwarf&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18473" PostTypeId="1" CreationDate="2016-09-24T15:18:15.780" Score="4" ViewCount="381" Body="&lt;p&gt;Is it possible for an observer at Earth's equator to simultaneously observe Polaris and the Southern Cross during an equinox?&lt;/p&gt;&#xA;" OwnerUserId="2757" LastEditorUserId="7695" LastEditDate="2018-08-21T14:19:40.857" LastActivityDate="2018-08-21T14:19:40.857" Title="Is it possible to observe Polaris and the Southern Cross from the equator during an equinox" Tags="&lt;constellations&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="18474" PostTypeId="1" CreationDate="2016-09-24T17:46:41.457" Score="4" ViewCount="385" Body="&lt;p&gt;I'm trying to calculate &lt;strong&gt;the closest path&lt;/strong&gt; of the center line of a given eclipse at a given time by using its besselian elements; However i'm getting an error range up to 30 km (18.6 mile) in the worst case and 5 km (3.1 mile) in the best case (both cases was compared with F. Espenak prediction)&lt;/p&gt;&#xA;&lt;p&gt;Lets use eclipse of August 12, 2026 at 18:00 UTC.&lt;/p&gt;&#xA;&lt;p&gt;F. Espenak &lt;a href=&quot;http://www.eclipsewise.com/solar/SEpath/2001-2100/SE2026Aug12Tpath.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;prediction&lt;/a&gt;: 58°16.1'N   021°34.4'W (58.2683, 21.5733)&lt;/p&gt;&#xA;&lt;p&gt;My prediction: 58°16.8'N 21°47.4'W (58.2795, 21.7901)&lt;/p&gt;&#xA;&lt;p&gt;The difference is close to 13 km (8 mile) which is too big for my application so I need to refine/review the procedure/algorithm I'm using.&lt;/p&gt;&#xA;&lt;h2&gt;Calculus&lt;/h2&gt;&#xA;&lt;p&gt;Given the following besselian elements:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;e20260812 = {&#xA;    t0: 18.00, At: 72.3,&#xA;    x0: 0.475510, x1: 0.518923, x2: -7.727e-05, x3: -8.039e-06,&#xA;    y0: 0.771185, y1: -0.230167, y2: -1.246e-04, y3: 3.767e-06,&#xA;    d0: 14.796683, d1: -0.012065, d2: -3.098e-06, d3: 0.0,&#xA;    l10: 0.537974, l11: 0.000094, l12: -1.212e-05, l13: 0.0,&#xA;    l20: -0.008142, l21: 0.000093, l22: -1.206e-05, l23: 0.0,&#xA;    u0: 88.747810, u1: 15.003090, u2: 1.764e-06, u3: 0.0,&#xA;    tanf1: 0.004614,  tanf2: 0.004591&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;(&lt;code&gt;At&lt;/code&gt; represents &lt;span class=&quot;math-container&quot;&gt;$\Delta T$&lt;/span&gt;)&lt;/p&gt;&#xA;&lt;h3&gt;0 - Reference timestamp.&lt;/h3&gt;&#xA;&lt;p&gt;Where &lt;code&gt;t1 = 18.00&lt;/code&gt;&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;t = t1 - t0 + At/3600.0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;h3&gt;1 - Time dependent calculus&lt;/h3&gt;&#xA;&lt;pre&gt;&lt;code&gt;x = x0 + x1 * t + x2 * t^2 + x3 * t^3&#xA;y = y0 + y1 * t + y2 * t^2 + y3 * t^3&#xA;d = d0 + d1 * t + d2 * t^2 + d3 * t^3&#xA;u = u0 + u1 * t + u2 * t^2 + u3 * t^3 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;h3&gt;2 - Location dependent calculus&lt;/h3&gt;&#xA;&lt;p&gt;Initially I'm using &lt;code&gt;ro=1.0&lt;/code&gt; and following the below sequence to obtain an approximation of &lt;code&gt;phi&lt;/code&gt;, which is the Latitude.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;ro = 1.0&#xA;e = 0.0818192 // earth eccentricity &#xA;sin_d = sin(d * pi / 180.0)&#xA;cos_d = cos(d * pi / 180.0)&#xA;zeta = sqrt(ro^2 + x^2 + y^2)&#xA;cos_theta = (-y * sin_d + zeta * cos_d) / ro&#xA;sin_phi_1 = (y * cos_d + zeta * sen_d) / ro&#xA;cos_phi_1 = sqrt( (x/ro)^2 + cos_theta^2 )&#xA;phi_1 = asin(sin_phi_1) / pi * 180.0&#xA;phi = atan(sin_phi_1 / cos_phi_1 / sqrt(1 - e^2)) / pi * 180.0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;At this point &lt;code&gt;phi = 58.1224&lt;/code&gt; for 18:00 hs UTC.&lt;/p&gt;&#xA;&lt;p&gt;Once &lt;code&gt;phi&lt;/code&gt; was calculated, we recalculate &lt;code&gt;ro&lt;/code&gt; as follow to get a better approximation of &lt;code&gt;phi&lt;/code&gt; and &lt;code&gt;varpi&lt;/code&gt; (not yet calculated)&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;sin_phi = sin(phi * pi / 180.0)&#xA;ro1 = sqrt( (1 - 2e^2 * sin_phi^2 + e^4 * sin_phi^2) / (1 - e^2 * sin_phi^2) )&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Applying the updated &lt;code&gt;ro&lt;/code&gt; (now &lt;code&gt;ro1&lt;/code&gt;), we update &lt;code&gt;sin_d&lt;/code&gt; and &lt;code&gt;cos_d&lt;/code&gt; as follow:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;sin_d1 = sin_d / ro1;&#xA;cos_d1 = sqrt(1-e^2) * (cos_d / ro1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;And update &lt;code&gt;zeta&lt;/code&gt; and &lt;code&gt;cos_theta&lt;/code&gt;&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;zeta1 = sqrt(ro1^2 + x^2 + y^2)&#xA;cos_theta1 = (-y * sin_d1 + zeta1 * cos_d1) / ro1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Finally we calculate &lt;code&gt;vapri&lt;/code&gt; (the Longitud) as follows:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;tan_theta = x / cos_theta1&#xA;theta = atan( tan_theta ) / Math.PI * 180.0&#xA;vapri = u - theta * ro1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;and recalculate &lt;code&gt;phi&lt;/code&gt; by doing:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;cos_phi2 =  cos_theta1 / cos(theta / 180.0 * pi)&#xA;phi1 = acos(cos_phi2) / pi * 180.0 &#xA;phi = atan(tan(phi1 / 180.0 * pi) / sqrt(1 - e^2)) / pi * 180.0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;At this point &lt;code&gt;phi = 58.2795 , vapri = 21.7901&lt;/code&gt; for 18:00 hs UTC (both values MUST be properly signed but I omitted that trivial part) a significant difference compared with the NASA prediction: 58.2683, 21.5733 (13 km or 8 mile)&lt;/p&gt;&#xA;&lt;p&gt;My feeling is that &lt;code&gt;ro1&lt;/code&gt; in not good enough precise, so the final results get shifted, but I have no idea at all about how to getting unblocked.&lt;/p&gt;&#xA;&lt;p&gt;I'm using 64 bit floating point calculator; It is ok for this kind of calculus?&lt;/p&gt;&#xA;&lt;p&gt;Is this method good enough to get an error range near to 500m? if not, could you provide a method/fix to achieve that aim.&lt;/p&gt;&#xA;&lt;p&gt;Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2475" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2021-01-04T12:31:16.390" Title="Closest path of solar eclipse central line" Tags="&lt;solar-eclipse&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="18475" PostTypeId="2" ParentId="18473" CreationDate="2016-09-24T19:58:30.237" Score="6" Body="&lt;p&gt;The March equinox would be a good time to see both; the September equinox would not.&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Crux&quot; rel=&quot;noreferrer&quot;&gt;Crux&lt;/a&gt;&#xA;lies roughly between RA 12h to 13h, Dec -55$^\circ$ to -65$^\circ$.&#xA;For best results, be at latitude 10-15$^\circ$ N so that both Crux and Polaris are well above the horizon, and look between 11:00 and 14:00 local&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sidereal_time&quot; rel=&quot;noreferrer&quot;&gt;sidereal time&lt;/a&gt;&#xA;so that Crux is near the meridian.&#xA;That would be around midnight in March-April, late evening in May, wee hours in February, or daytime in July-December.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-09-24T19:58:30.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18477" PostTypeId="1" AcceptedAnswerId="18486" CreationDate="2016-09-25T02:31:26.420" Score="8" ViewCount="874" Body="&lt;p&gt;Are there orbits with long-term stability (astronomical time scales) around Earth? Low altitude orbits will decay due to friction with Earth's atmosphere. Inside Earth's Roche limit, a Moon could not form anyway due tidal forces. At altitudes closer to that of the Moon, the Moon would presumably destabilize an orbit such that the object would either collide with Earth or the Moon or be ejected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of the other planets in our solar system have multiple natural satellites; where this occurs, they are always vastly smaller than the primary - think of Jupiter, Saturn, and even Mars. On the other hand Earth, and apparently Pluto, each has only one natural satellite, relatively large in comparison to the primary. Is this because a large satellite will tend to clear the entire &quot;sphere of influence&quot; around its primary of any other material?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: does the existence of our Moon effectively prevent the formation or capture of any other natural satellite around Earth? (Considering only true satellites, not &quot;companion&quot; objects or quasi-satellites)&lt;/p&gt;&#xA;" OwnerUserId="1983" LastEditorUserId="1983" LastEditDate="2016-09-26T02:04:52.923" LastActivityDate="2016-10-17T02:44:35.553" Title="Can Earth host another natural satellite than the Moon?" Tags="&lt;the-moon&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18478" PostTypeId="2" ParentId="18455" CreationDate="2016-09-25T04:13:58.683" Score="1" Body="&lt;p&gt;It sounds like the question is asking if there can be two separate contributions due to the expansion, one simply because the distances are increasing so successive wavefronts in the wave train arrive later and later, and a second effect due to more esoteric distortions like time slowing down.  But what of the redshift we attribute to something happening to space, and what we attribute to something happening to time, depend on the coordinates we choose to describe what is going on.  The standard choice is comoving coordinates, where time has the meaning of &quot;age of the universe at the point in question in a frame moving with the average gas.&quot;  In those coordinates, the elapsing time isn't doing anything different as the universe expands, so the only effect you need to worry about is the stretching distances between wavefronts.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-09-25T04:13:58.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18479" PostTypeId="1" CreationDate="2016-09-25T09:12:04.077" Score="3" ViewCount="39" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Very-long-baseline_interferometry&quot; rel=&quot;nofollow&quot;&gt;The Very Long Baseline Array&lt;/a&gt; uses both Arecibo and &lt;a href=&quot;https://en.wikipedia.org/wiki/Spektr-R&quot; rel=&quot;nofollow&quot;&gt;Spectr-R&lt;/a&gt;, the world's largest space telescope with 10 m diameter at Lunar distance, as well as many other antennas. Are there any particular requirements on the design for a radio telescope to be used interferometrically? Could for example the now operating Chinese &lt;a href=&quot;https://en.wikipedia.org/wiki/Five_hundred_meter_Aperture_Spherical_Telescope&quot; rel=&quot;nofollow&quot;&gt;Five Hundred meter Spherical Telescope&lt;/a&gt; be used as a part of the VLBA?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2016-09-25T09:36:23.903" LastActivityDate="2016-09-25T09:36:23.903" Title="Can any radio telescope be hooked up with the Very Long Baseline Array, or other interferometers?" Tags="&lt;radio-telescope&gt;&lt;interferometry&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18480" PostTypeId="1" CreationDate="2016-09-25T09:31:29.180" Score="1" ViewCount="54" Body="&lt;p&gt;I've heard of &lt;em&gt;&quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Interferometric_synthetic_aperture_radar&quot; rel=&quot;nofollow&quot;&gt;interferometric synthetic aperture radar&lt;/a&gt;&quot;&lt;/em&gt; being used for Earth observation from satellites and airplanes. Is active radar used interferometrically for astronomical purposes too? Such as characterizing asteroids or space debris, or identifying volatiles in the polar craters of Mercury.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately for this idea, Arecibo and &lt;a href=&quot;https://en.wikipedia.org/wiki/Five_hundred_meter_Aperture_Spherical_Telescope&quot; rel=&quot;nofollow&quot;&gt;FAST&lt;/a&gt;are located 173 longitudinal degrees apart and have limited fields of view, so they can't observe the same object simultaneously.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-11-23T14:50:56.827" Title="Is radar interferometry used, or feasible, for ground based astronomy?" Tags="&lt;radio-astronomy&gt;&lt;interferometry&gt;&lt;radar&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18482" PostTypeId="1" CreationDate="2016-09-25T17:39:19.910" Score="1" ViewCount="93" Body="&lt;p&gt;Hopefully I'm in the appropriate section as my question is mainly about coding, but it's for astronomy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now I'm trying to figure out how I can take a range of MJD values and convert them into number of days elapsed so I can normalize a plot of two supernova and their absolute magnitudes over time in IDL. Is there a simple conversion for this I'm not seeing or do I need to tediously convert every value to determine the number of days to use for my x-axis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I do not wish to convert to the date, but the actually number of days elapsed as to normalize my two plots for comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="14353" LastActivityDate="2016-09-25T18:57:40.540" Title="Convert MJD date range to number of days elapsed in IDL" Tags="&lt;supernova&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18483" PostTypeId="2" ParentId="10316" CreationDate="2016-09-25T18:28:55.260" Score="1" Body="&lt;p&gt;It is possible. Say Mankind formed 195,000 years ago (oldest humanoid fossil found) and a very massive star closer than 195,000 light years away went supernova or even imploded into a black hole with in 195000 years, there is a chance that some humanoid may have seen it. Historically people have documented seeing supernovas in the past and if any of those would have been massive enough stars, then it may a black hole formation that was witnessed. However no black holes are located in the areas described by the descriptions made by the witnesses but seeing one for is possible, just hope its not very close to our system. The resulting GRB might ruin your day.&lt;/p&gt;&#xA;" OwnerUserId="14346" LastEditorUserId="14346" LastEditDate="2016-09-25T18:55:28.343" LastActivityDate="2016-09-25T18:55:28.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18484" PostTypeId="2" ParentId="18482" CreationDate="2016-09-25T18:52:24.973" Score="2" Body="&lt;p&gt;Like @James K said, the MJD is simply a count, so you can set some arbitrary date to be &lt;code&gt;0&lt;/code&gt; and go from there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, take today's MJD date and say there was one event today:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Today = 57657 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Add three months and 12 days to the next event:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Today + 103 = 57760&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So if you called ten days before the first event &lt;code&gt;0&lt;/code&gt;, then the &quot;dates&quot; of the first two events would simply be the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Start = Event_1 - 10 = 0 &#xA;Event_1 = Start + 10 = 10 &amp;lt;-- Semantics right!&#xA;Event_2 = Event_1 + 103 = 113&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It really is that simple. Just set a variable to equal whatever starting MJD you want to be zero and count earth days between events. &lt;/p&gt;&#xA;" OwnerUserId="2232" LastEditorUserId="2232" LastEditDate="2016-09-25T18:57:40.540" LastActivityDate="2016-09-25T18:57:40.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18485" PostTypeId="2" ParentId="18477" CreationDate="2016-09-26T01:55:23.630" Score="1" Body="&lt;p&gt;A few hundred kilometers beyond geosynchronous is the satellite &lt;a href=&quot;https://en.wikipedia.org/wiki/Graveyard_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;graveyard orbit&lt;/a&gt;, which distance is also a function of satellite size. One could place an asteroid into such a orbit to be a quite visible moon. One can argue that since we do not have such a moon, that it is therefore improbable. However, as below, no moon is forever, and most assuredly in the past we had lots of little moons. Earth more or less has a sort of a second moon, &lt;a href=&quot;http://www.jpl.nasa.gov/news/news.php?feature=6537&quot; rel=&quot;nofollow noreferrer&quot;&gt;and will for centuries&lt;/a&gt; so &quot;can't have&quot; is doubtful.&lt;a href=&quot;https://i.stack.imgur.com/ZLL9v.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZLL9v.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not buy the &quot;I do not like the complicated orbit small temporary natural satellite argument.&quot; Using that argument, &lt;a href=&quot;https://janus.astro.umd.edu/front/pages/links/Phobos1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Phobos&lt;/a&gt; would not be a moon of Mars as its orbit will terminally decay in 30 million years. Our Luna will eventually escape due tidal braking, so, I would prefer to say that no satellite is forever. The next problem is that if 2016HO3 above is not a 'moon' or 'satellite' then Earth is not a 'planet,' as planets 'clear their own orbits' as point 3 of the &lt;a href=&quot;https://en.wikipedia.org/wiki/IAU_definition_of_planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;IAU definition of planet&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UKCow.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UKCow.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hill sphere&lt;/a&gt; is only an approximation, and other forces (such as radiation pressure or the Yarkovsky effect) can eventually perturb an object out of the sphere. This third object should also be of small enough mass that it introduces no additional complications through its own gravity. Detailed numerical calculations show that orbits at or just within the Hill sphere are not stable in the long term; it appears that stable satellite orbits exist only inside 1/2 to 1/3 of the Hill radius. The region of stability for retrograde orbits at a large distance from the primary, is larger than the region for prograde orbits at a large distance from the primary. This was thought to explain the preponderance of retrograde moons around Jupiter; however, Saturn has a more even mix of retrograde/prograde moons so the reasons are more complicated...The Hill sphere of (66391) 1999 KW$_4$, a Mercury-crosser asteroid that has a moon (S/2001 (66391) 1), measures 22 km in radius.&lt;/p&gt;&#xA;" OwnerUserId="14344" LastEditorUserId="14344" LastEditDate="2016-10-17T02:44:35.553" LastActivityDate="2016-10-17T02:44:35.553" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18486" PostTypeId="2" ParentId="18477" CreationDate="2016-09-26T02:59:01.433" Score="8" Body="&lt;p&gt;Short answer:   It might be impossible &quot;long term&quot;.  It's problematic because of the size of the Moon and the proximity to the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth-Moon system in empty space could hold several more moons (like the Pluto-Charon system).    The problem is proximity to the sun in addition to the size of the Moon.    Planets closer to their star have a harder time holding onto moons in part cause the Hill sphere is smaller, in part cause the tidal forces are that much bigger (and those 2 factors may go hand in hand, not in addition to).   But it's not just coincidence that Mercury and Venus don't have moons, they're also less likely to hold onto Moons.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars has a smaller Hill Sphere than Earth due to it's lower mass but it's 2 moons are very tiny.  Also, one of Mars 2 moons might not be there that much longer.  It may crash into Mars in as little as 100 million years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth's proximity to the Sun gives the Earth a &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;noreferrer&quot;&gt;Hill Sphere&lt;/a&gt; of about 1.5 million km, and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere#True_region_of_stability&quot; rel=&quot;noreferrer&quot;&gt;true region of stability&lt;/a&gt; (long term) is about 1/3 to 1/2 of that, so about 500-750,000 km which isn't all that much further out than the Moon is currently.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our Moon's average distance is 384,000 km, but our moon has a slightly eccentric orbit so it's distance varies between about 363,100 and 405,700 km.  &lt;a href=&quot;http://www.space.com/18145-how-far-is-the-moon.html&quot; rel=&quot;noreferrer&quot;&gt;Source.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 2nd factor is the relative mass between the Moon and the planet.   Mars has 2 moons but it's moons are very small and they have very little gravitational effect on each other.    Our Moon is 1/81st the mass of the Earth which is the largest Moon to planet ratio in our solar-system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the simplified hill sphere formula, the cube root of (the ratio of the masses/3).   Basically the cube root of (1/243) gives Earth's moon's a hill sphere of about 16% it's distance to the Earth.   Any Earth orbiting object that passes within the Moon's hill sphere is obviously not going to be in a stable orbit.   In fact the region of instability probably extends well past that 16%.   How far past, I'm not sure, but I can safely say that anything even remotely close to the Moon's orbit couldn't orbit the Earth for any period of time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using 16%, the clearly unstable region is between 300,000 km and 470,000 KM.   It would be impossible for an object to orbit the Earth at a greater distance than the Moon.   Solar and Lunar perturbations would be too big.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's possible for Earth to have a 2nd moon at all, it would need to orbit quite close to the Earth.   Probably well inside 300,000 KM.  The closer to the Earth the more the Earth's gravitation would be dominant, so if I was to guess, I think the Moon would need to be quite close, maybe in the 50,000 km range - but that's just a guess.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparison, Jupiter's 3 moons in orbital resonance; Ganymede, the largest of the 3 has a hill sphere is about 3% it's distance from Jupiter.   The 1:2:4 orbital resonance works out to the 1.5th root of 2 or about a 1.59 to 1 orbital distance ratio.  Those 3 moons also have very nearly circular orbits.    If the Moon's orbit was nearly circular, there might be a stable resonance 1.59 times closer to the Earth, but because of the Moon's orbit is measurably eccentric, I don't believe the 2:1 orbital period ratio would be long term stable because there's too many wobbles in it.   I think your best bet for a stable orbit for a 2nd moon would be very close to the Earth, probably in the 50-100,000 KM range (but that's a rough guess).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth noting that &quot;long term stable&quot; isn't a precise term cause there's no clear cut-off.  A thousand orbits?  A million?   A billion?   Lifespan of the Sun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that &quot;maybe, maybe not&quot; isn't an answer, but I wanted to touch on some of why it's difficult.  A captured satellite couldn't get captured into a relatively circular near earth orbit because the incoming velocity is too great.  In theory, if there was enough debris, a satellite could form close to the Earth and perhaps be stable for relatively long term, but the Moon would to some extent prevent such a formation in a similar way that Jupiter prevents the asteroid belt from coalescing into a tiny planet.   (the low mass of the asteroid belt plays a role in that too, perhaps a bigger role), but Jupiter is a factor. &lt;a href=&quot;http://www.universetoday.com/112464/why-isnt-the-asteroid-belt-a-planet/&quot; rel=&quot;noreferrer&quot;&gt;Article here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-09-26T02:59:01.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18487" PostTypeId="1" AcceptedAnswerId="18489" CreationDate="2016-09-26T10:39:51.107" Score="-4" ViewCount="106" Body="&lt;p&gt;Just messing around and was wondering how stable a system like this would be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dCcbd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dCcbd.jpg&quot; alt=&quot;System&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14348" LastActivityDate="2016-09-26T12:36:57.940" Title="Is this a stable system?" Tags="&lt;orbit&gt;&lt;orbital-resonance&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2016-09-27T14:00:13.543" ContentLicense="CC BY-SA 3.0" />
  <row Id="18488" PostTypeId="1" AcceptedAnswerId="19449" CreationDate="2016-09-26T12:19:45.860" Score="12" ViewCount="243" Body="&lt;p&gt;The gravitational wave detected by the LIGO observatory acted on planet Earth streching and shrinking it a little bit. I assume that not all of this action was fully elastic and that some energy of the gravitational wave was lost by the passage through planet Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there some scientific estimates of how much energy was deposited in planet Earth by this gravitational wave?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there estimates how much energy is in total deposited by gravitational waves in planet Earth per year?&lt;/p&gt;&#xA;" OwnerUserId="11617" LastActivityDate="2016-12-20T10:37:58.793" Title="Energy deposition in planet Earth due to gravitational waves" Tags="&lt;earth&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18489" PostTypeId="2" ParentId="18487" CreationDate="2016-09-26T12:36:57.940" Score="2" Body="&lt;p&gt;Seems like it would be subject to Kozai oscillations at least-- see &lt;a href=&quot;https://www.cfa.harvard.edu/research/ta/kozai-lidov-mechanism&quot; rel=&quot;nofollow&quot;&gt;https://www.cfa.harvard.edu/research/ta/kozai-lidov-mechanism&lt;/a&gt;.  That mechanism tends to swap obliquity for eccentricity, without changing the energies of the orbits, so it is fueled by torque rather than work.  It is easier to get a slow torque from two inclined orbits than it is to get work done, because work requires a consistently oriented force, whereas torque tends to require only an overall average asymmetry in the geometry.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-09-26T12:36:57.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18491" PostTypeId="2" ParentId="18480" CreationDate="2016-09-26T13:57:57.207" Score="2" Body="&lt;p&gt;For obvious reasons you cannot use active radar to observe anything much further away than the Moon. There are plenty of phased-array microwave/radar passive sensing systems that map the energy in those wavelength bands that are emitted from stars, quasars and various other &quot;hot&quot; astronomical bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer one of your examples, measurement of volatiles is usually done by measuring absorption spectra; which is tough to do with Mercury since we can't look at the solar-illuminated side (with special exceptions for certain short-lived orbital positions). Basically, that's why we send off probes to get close to various planets and asteroids, so that we can collect high-resolution visible and microwave images and, in some cases, do some active illumination (radar).  I am not aware of any interferometric active radar on such probes, but someone more knowledgeable may well correct me.  &lt;/p&gt;&#xA;" OwnerUserId="907" LastEditorUserId="2531" LastEditDate="2016-11-23T14:50:56.827" LastActivityDate="2016-11-23T14:50:56.827" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18492" PostTypeId="1" AcceptedAnswerId="18495" CreationDate="2016-09-26T17:15:21.093" Score="8" ViewCount="4738" Body="&lt;p&gt;Does Jupiter help to protect the Earth against comets or asteroids impacts?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some astronomers believe that one reason Earth is habitable is that the gravity of Jupiter does help protect us from some comets. Long-period comets, in particular, enter the solar system from its outer reaches. Jupiter’s gravity is thought to sling most of these fast-moving ice balls out of the solar system before they can get close to Earth. So long-period comets are thought to strike Earth only on very long timescales of millions or tens of millions of years. Without Jupiter nearby, long-period comets would collide with our planet much more frequently.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From: &lt;a href=&quot;http://earthsky.org/space/is-it-true-that-jupiter-protects-earth&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://earthsky.org/space/is-it-true-that-jupiter-protects-earth&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively speaking, it would seem to me that while any massive body in the Solar System would deflect the trajectories of some comets/asteroids that would otherwise pose a threat to Earth, it also must be true that some few ones that were &lt;em&gt;not&lt;/em&gt; in a collision course would then be directed to us after having their trajectories deflected. Why isn't that the case?&lt;/p&gt;&#xA;" OwnerUserId="1447" LastEditorUserId="23486" LastEditDate="2018-06-27T08:49:30.507" LastActivityDate="2018-07-05T07:09:22.430" Title="Does Jupiter help to protect the Earth against comets/asteroids impacts?" Tags="&lt;asteroids&gt;&lt;comets&gt;&lt;jupiter&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="18493" PostTypeId="2" ParentId="10384" CreationDate="2016-09-26T18:30:40.687" Score="2" Body="&lt;p&gt;Brown Dwarfs are bright in radio. They emit plenty of light, just not much visible light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the properties of dark matter is that it also does not obscure light. Interstellar dust (which can be comprised of many larger planet-sized particles) tends to obscure light even if it doesn't emit any. So due to it neither emitting nor obscuring light, we can rule out it being brown dwarfs, dust, or pretty much anything in between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It could be a field of lightweight black holes, but that seems unlikely as black holes are both rare and generally surrounded by evidence of their formation. It is unreasonable to assume that there are huge swarms of lightweight black holes making up a large fraction of the mass in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exotic matter seems strange to us, but it's really not so uncommon. We already know that neutrinos exist and in fact are very abundant. Neutrinos pass through normal matter almost completely uninhibited. They can't be what dark matter is made of, because they travel at the speed of light like photons, while plenty of dark matter is relatively stationary. It seems reasonable to believe that it is some form of exotic matter that doesn't interact with light. It's strange to us, but it seems very consistent with what we know of particle physics.&lt;/p&gt;&#xA;" OwnerUserId="14360" LastActivityDate="2016-09-26T18:30:40.687" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18494" PostTypeId="2" ParentId="14683" CreationDate="2016-09-26T18:58:52.470" Score="3" Body="&lt;p&gt;I built an optical projector to observe the Venus transit with a 10x50 monocular. I mounted the monocular inside a box, leaving a small window to observe the projection. The focusing capability of the monocular was necessary to get a good image. The Sun's image was large and bright and detailed, much better than a pinhole projector. &lt;/p&gt;&#xA;" OwnerUserId="14361" LastActivityDate="2016-09-26T18:58:52.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18495" PostTypeId="2" ParentId="18492" CreationDate="2016-09-27T03:23:05.053" Score="10" Body="&lt;p&gt;Wordy answer, mostly light on Math: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key word here (and the article uses this word) is &quot;long period comets&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First there's Jupiter impacts, but that's a relatively low percentage, cause even large Jupiter is quite small compared to it's orbit.  Even if you extend it out to it's Roche limit where a comet could break apart, it's still a very small target in the grand scheme of things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But any comet that passes near Jupiter (but doesn't hit it, get captured by it or broken apart in it's Roche limit) gets it's orbit changed somewhat by &lt;a href=&quot;http://solarsystem.nasa.gov/basics/grav/primer.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravity assist.&lt;/a&gt;  Jupiter's high mass and relatively high orbital velocity make it the gravity assist king of the solar-system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closer the pass to Jupiter, the bigger the change.  Gravity assists, change both direction and velocity.  They can't change only one.  It's safe to assume that half of those gravity assists add velocity to the comet and half reduce velocity.   Very loosely speaking, the direction of the comet affects it's &lt;a href=&quot;https://en.wikipedia.org/wiki/Perihelion_and_aphelion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Perihelion&lt;/a&gt; and it's velocity and any change in it's velocity affect's it's semi-major axis (explained a bit below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The comet's orbital velocity is obviously dependent on how close the comet is to the sun.   (Kepler's law about equal areas over equal times), but any orbiting object also has an average velocity, which is closely tied to the semi-major axis and orbital period, and it's this average velocity and the velocity at Aphelion (it's slowest velocity) that matters for this discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6&quot; rel=&quot;nofollow noreferrer&quot;&gt;orbital period&lt;/a&gt; of any object orbiting the sun is directly proportional to the 3/2 power of the semi-major axis.  The average velocity is very closely correlated to the square root of the distance between the semi-major axis and the Sun.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you get comets with orbital periods of hundreds or thousands of years, their average orbital velocity, is relatively slow for a celestial object.  Pluto's average orbital velocity is about 4.67 km/s and it's minimum velocity at Aphelion is 3.71 km/s.  &lt;a href=&quot;http://nssdc.gsfc.nasa.gov/planetary/factsheet/plutofact.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For longer orbital periods the orbital velocity drops, roughly by the square root of the relative distance, so a comet with an orbital period of 1,000 years (about 4 times Pluto's), would have an orbital velocity about 1/2 as much and a velocity at Aphelion of maybe 1.8 km/s or a bit less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/vesc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Escape Velocity&lt;/a&gt; for the comet requires just a 41.4% increase on Aphelion orbital velocity, so, with our 1.8 km/s orbital velocity long period comet at Aphelion, if Jupiter gives that comet a push of just about .75 km/s, that would be enough to push that comet out of orbit, where it escapes the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, while it might seem logical that Jupiter could send a comet equally towards earth as away from Earth, the relatively small push that a long period comet needs to escape the solar system, makes that a far more likely scenario than a nearly perfect 100 foot put into the Earth's orbit.   The odds of a long period comet being thrown out of the solar system by Jupiter is relatively high, especially with multiple passes, where, hitting Earth is like hitting a 100 foot put blindfolded.   Earth is a small target.  Out of the solar-system is a large target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For shorter period comets where the added velocity needed to escape the solar system is greater, this grows proportionately less true, but for long period comets which only need a small push to get thrown out of the solar-system, statistically, Jupiter tosses a lot more of those out than it sends towards earth.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick is to be very specific about what we're talking about.   If Jupiter was, for example, to migrate through the asteroid belt, or migrate outwards through the Kuiper belt, then it's size would send a lot of stuff in towards the inner solar-system.   It's not in and of itself, always going to protect Earth and reduce Earth-impacts.  It depends on where it is and what the orbits of the comets are, but specifically for long period comets, Jupiter is a-lot more likely to send them out of the solar-system than towards Earth.  Not all, just statistically more likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth noting that there's a lot we don't know about the oort cloud and how many comets/icy objects are in it.   We don't know what percentage of long period comets are jupiter-crossers vs objects with more distant perihelions.   When a star or large rogue planet passes through Earth's Oort cloud, it can send some of the icy objects out there in towards the Earth.   (Planet 9 - if it exists, probably not so much, as that theoretical planet has probably already mostly cleared out it's orbit), so it's not a key player in sending stuff to the inner solar-systems, but a massive passing object that pass through the oort cloud can do that, at least, that's been theorized, though such events are quite uncommon because space is mostly empty.   When those rare events happens, then there's probably an increase in long period comets and when that happens, Jupiter, for the reasons described above, is pretty good at getting rid of them over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.space.com/28611-star-flew-through-solar-system.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Scholz's star&lt;/a&gt; is thought to have passed within 0.8 light years of our sun some 70,000 years ago and such a pass might be close enough to send a lot of Oort cloud objects in towards the inner solar system (though 0.8 light years it's still a bit far, as the oort cloud is probably quite empty that far out.  For a really good number of new long-period inner solar-system comets you'd probably want a 0.1 lightyear pass, or closer . . . . but massive objects passing that close are extremely rare, and . . .  I digress).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But Scholz's star might have turned a number of Oort cloud objects into long-period comets that pass through the inner solar-system.  (We won't know for a long time, as it'll be perhaps a million years for those new comets to reach us), But assuming Scholz did that, any oort cloud objects entering the inner solar-system from that event would be extremely long period objects and Jupiter would be pretty good at (over time, it wouldn't happen quickly), but over multiple long period orbits which probably takes millions of years, Jupiter would be pretty good at removing most of those from the inner solar-system, but just by chance, it would probably send a few in Earth's direction too.   If there was no Jupiter, it would take a lot longer for those long period comets to get cleared out, so in that sense, it really does protect the Earth from Long period comets.   There would be a lot more of them if not for Jupiter (and to a lesser extent, Saturn) and more comets would mean more Earth impacts.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-07-05T07:09:22.430" LastActivityDate="2018-07-05T07:09:22.430" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="18496" PostTypeId="1" AcceptedAnswerId="18525" CreationDate="2016-09-27T04:43:45.997" Score="6" ViewCount="835" Body="&lt;p&gt;After seeing today's Washington Post's article &lt;a href=&quot;https://www.washingtonpost.com/news/speaking-of-science/wp/2016/09/26/nasa-just-saw-europas-geysers-erupting-into-space-again/&quot; rel=&quot;noreferrer&quot;&gt;NASA just saw Europa’s geysers erupting into space. Again.&lt;/a&gt;, and especially the phrase &quot;geysers erupting into space&quot; I checked and Europa has a surface gravity of about 1.3 $m/s^2$, which is substantial - almost the same as the Moon's at about 1.6 $m/s^2$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The BBC's website article &lt;a href=&quot;http://www.bbc.com/news/science-environment-37473617&quot; rel=&quot;noreferrer&quot;&gt;Europa moon 'spewing water jets'&lt;/a&gt; says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Further evidence has been obtained to show that Jupiter's icy moon Europa throws jets of water out into space.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The suggestion is that the jets reach several hundred kilometres in height before then falling back on to Europa.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On earth, pressure can come from geological hydrostatic forces and by steam produced by geothermal heating, but they rarely rise beyond a few dozen meters. That would be perhaps 100 meters in Europa's gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The images shown of the water above Europa show heights of 100 kilometers and more. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike Earth rock being heavier than water, Europa's ice crust floats. What could be sources of hydrostatic pressure or pressurized steam near the surface that can create such &quot;geysers&quot; or &quot;jets&quot; that can project water to such heights in this gravity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/b1Pco.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/b1Pco.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt;  The geyser Strokkur in Iceland, Earth, from &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Strokkur,_Iceland.jpg&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2016-09-27T05:43:51.247" LastActivityDate="2018-02-10T17:39:23.857" Title="How can &quot;Geysers&quot; on Europa reach heights of 100km?" Tags="&lt;water&gt;&lt;europa&gt;" AnswerCount="3" CommentCount="10" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18497" PostTypeId="1" CreationDate="2016-09-27T06:58:57.083" Score="2" ViewCount="123" Body="&lt;p&gt;Both use the same data. But at least, 90k stars are lost in Tycho-2. &#xA;Why?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="6" LastEditDate="2016-09-27T16:53:40.040" LastActivityDate="2016-09-27T16:53:40.040" Title="Why does the Tycho-2 catalog not include Hipparcos stars?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2016-10-11T19:56:22.197" ContentLicense="CC BY-SA 3.0" />
  <row Id="18498" PostTypeId="2" ParentId="18496" CreationDate="2016-09-27T07:40:38.993" Score="11" Body="&lt;p&gt;Technically those aren't &quot;geysers&quot; on Europa, they're &lt;a href=&quot;https://en.wikipedia.org/wiki/Cryovolcano&quot;&gt;cryovolcanos&lt;/a&gt;.    Though that definition may be a bit imprecise as well, but the 100 km eruptions on Europa probably have more in common with volcanic eruptions on Earth than geysers.   Europa undergoes significant tidal flexing which, in combination to it's solid icy crust, could be compared to explosive volcanic eruptions on Earth and not the cyclic geysers, which are far smaller.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Volcanoes on Earth can shoot material &lt;a href=&quot;http://volcano.oregonstate.edu/how-high-can-explosive-eruptions-go-and-how-far-can-debris-and-ash-be-spread&quot;&gt;high into the atmosphere&lt;/a&gt; on Earth but volcanic ash isn't a good example because it's not so much &quot;shot&quot; into the air as it rides rising hot air to altitudes 30 and 40 KM high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magma fountains on Earth can reach a &lt;a href=&quot;http://volcano.oregonstate.edu/how-high-can-lava-shoot-air&quot;&gt;few hundred meters&lt;/a&gt; and &lt;a href=&quot;http://www.geo.mtu.edu/volcanoes/hazards/primer/tephra.html&quot;&gt;Tephra&lt;/a&gt; (anything larger than Ash, from pebbles to large rocks/boulders) can be found as far as a mile from a volcano and can exit the volcano at &lt;a href=&quot;https://en.wikipedia.org/wiki/Eruption_column#Formation&quot;&gt;several hundred meters per second&lt;/a&gt;.   From the Tephra article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Blocks and bombs as large as 8-30 tons have fallen as far away as 1 km&#xA;  from their source (Bryant, 1991). Small blocks and bombs have been&#xA;  known to travel as far away as 20-80 km (Scott, 1989)! Some of these&#xA;  blocks and bombs can have velocities of 75-200 m/s (Bryant, 1991)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Europa's Tidal-flexing can cause significant pressures within it's icy crust.  See &lt;a href=&quot;http://solarsystem.nasa.gov/europa/tidemovie.cfm&quot;&gt;here&lt;/a&gt;.  and &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_heating&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we draw an Earth-Europa comparison, and assuming my math is right, using 1/2 at^2, all you'd need for material to reach 100 km on Europa is an exit velocity from the ground of a bit over 500 meters per second.   Or, some 20%-30% faster than a 22 caliber bullet.   That's faster than material usually leaves an Earth Volcano, but not all that much faster.   When frozen, Europa's outer icy crust is probably very hard and brittle.  In pressure spots where the pressure exceeds the structural integrity of the icy crust, it's not all that surprising that the results are pretty explosive, leaving the surface at 500 meters per second.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ice is very strong and brittle and when cold enough, where the Earth's crust, while rocks are quite brittle and strong, the crust overall can also be quite malleable, especially a few miles down in hotter temperatures under an erupting volcano.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very cold ice is resistant to cracking, but under enough pressure, when it finally does crack it can release explosive energy.   Ice cracks in water due to &lt;a href=&quot;http://sploid.gizmodo.com/why-do-ice-cubes-crack-when-you-drop-them-in-drinks-1598935375&quot;&gt;differential expansion&lt;/a&gt;, (Silicate rock doesn't expand with temperature nearly as much as frozen ice does), and while the cracking of an ice cube dropped into water might not seem like much, a big enough ice cube dealing with enough tidal heating can have explosive results.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this fun little video might seem unrelated (turn down your volume before watching), it demonstrates the explosive force that rapid warming can have on a hard block of ice.  This isn't steam pressure, as the steam is all escaping.  This is the inside of the block expanding faster than the outside can and the result is impressive.   &lt;a href=&quot;https://www.youtube.com/watch?v=epkRd-w3TGw&quot;&gt;https://www.youtube.com/watch?v=epkRd-w3TGw&lt;/a&gt;   So, with that in mind, explosive energy from a hard, cold 10 mile thick icy surface under the influence of tidal flexing shouldn't be all that surprising.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-09-27T07:40:38.993" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18501" PostTypeId="2" ParentId="18497" CreationDate="2016-09-27T16:37:28.203" Score="3" Body="&lt;p&gt;Could you give a source or example: The Tycho-II catalogue paper by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2000A&amp;amp;A...355L..27H&quot; rel=&quot;nofollow&quot;&gt;Hog et al. (2000)&lt;/a&gt; suggests there are about 19,000 stars in Hipparcos or Tycho-I that are not in Tycho-II. This can be because they are too bright (but that is not very many stars) or for other reasons that are not clearly specified (binarity is an issue I think).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at &lt;a href=&quot;http://www.astro.ku.dk/~erik/Tycho-2/faq.html&quot; rel=&quot;nofollow&quot;&gt;this website&lt;/a&gt;, it suggests 4% of stars did not have proper motions and could not be assigned epoch 2000 positions in Tycho II but I don't think these are &lt;em&gt;missing&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-09-27T16:37:28.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18504" PostTypeId="1" CreationDate="2016-09-27T19:01:10.420" Score="19" ViewCount="4449" Body="&lt;p&gt;Why isn't it possible that the earth formed outside the solar system and got attracted later by the Sun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just need arguments to defeat this hypothesis.&lt;/p&gt;&#xA;" OwnerUserId="14369" LastEditorUserId="11143" LastEditDate="2016-09-27T19:18:41.770" LastActivityDate="2016-09-28T13:26:28.690" Title="Did the earth form outside the solar system?" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;accretion-discs&gt;&lt;origin-of-life&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18505" PostTypeId="2" ParentId="18504" CreationDate="2016-09-27T19:11:59.257" Score="23" Body="&lt;ol&gt;&#xA;&lt;li&gt;The chance happenstance that Earth happened to be floating along and got captured is minuscule. How did Earth wind up floating through space? There's no established mechanism for terrestrial planets to form on their own. As far as we know, they need a host star to form around. So if our Sun captured Earth, it must've formed around another Star, got ejected somehow, and then came under the influence of our Sun where it was captured and nicely placed at its current position. This is all highly, highly unlikely. I'm not going to run through the math, but you probably have a better chance of getting struck by lighting at the exact moment you buy a winning lottery ticket (twice).&lt;/li&gt;&#xA;&lt;li&gt;Aside from a statistical argument, you can look at it from the vantage point that the composition of the Earth properly matches everything else in our solar system. People have made a very close study as to the physical composition of Earth, the Moon, and the other planets and bodies in our solar system. We would have noticed if there was a large disparity in composition or something just didn't match up. As it is, Earth matches up pretty darn well, especially with our Moon. The close resemblance in composition between our Moon and Earth gives strong evidence that the Moon was formed from the Earth. If Earth was captured, so too was the Moon which makes point 1 even less likely.&lt;/li&gt;&#xA;&lt;li&gt;Many people have spent a lot of time simulating the formation of our solar system, from the start up until now. There are a few open questions (such as why Jupiter resides where it does), but by and large, people agree that a planet should form about where Earth is. So if Earth wasn't here during formation and was later captured, why isn't there an extra planet that did form where Earth is? All simulations suggest such a planet should exist. The most likely scenario is that Earth is that planet!&lt;/li&gt;&#xA;&lt;li&gt;It is believed that basic life (i.e., single celled organisms) started on Earth around 3.8 to 4.1 billion years ago. Earth itself has been reliable dated to about 4.53 billion years old. This means life was forming a mere 400-700 million years after Earth first formed. Life requires very very specific conditions (as we know it) to start forming and that can't happen in the dead of space. Earth had to be around a star from the beginning. Any time spent drifting through space until it was captured would have delayed the onset of the formation of life.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;I'm sure there are plenty of arguments that could be added to the list, but that's a good start I think.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: I didn't expect such a spirited debate in response to this answer. First off, I want to say that none of my arguments above are bullet-proof. I'm going to incorporate some of the comments below to help round out this answer. This will be easier than creating a long conversation of comments.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Matthew Whited&lt;/em&gt; - [It] is possible that other rouge planets could have formed outside of their own solar systems. It is also possible that they could form within solar systems and be ejected.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This is certainly true. One cannot soundly reject the hypothesis that Earth formed elsewhere (be it in free space or around another star) and was later captured by our Sun. My arguments are mostly ways to point out that the chance of such an occurrence is diminishingly small due to the unlikely circumstances that must occur for that hypothesis to be true. The betting man would not side with the hypothesis that the Earth formed outside our solar system and was later captured.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Rob Jeffries&lt;/em&gt; - Almost the only way Earth could be captured is during the clustered star formation stage (well established theoretical possibility). In which case it would have the same age and metallicity as if it had formed around the Sun. So age and metallicity are weak arguments.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I don't know if they're as weak as that. If the Earth did come from another stellar system, it would of course come from a nearby one, which would of course have a similar metallicity as our Sun. But I think a point which makes the above arguments stronger is the specific composition of the Earth. If it formed around a star which was slightly different in nature (say an M Dwarf) or at a different distance from the star I think we'd notice that the Earth seemed &amp;quot;out of place&amp;quot; in our solar system. As it is, its iron content agrees quite well with it forming where it currently resides. Again, none of this is bullet-proof, but it cast's doubts on the validity of Earth forming elsewhere.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Rob Jeffries&lt;/em&gt; - Planet capture in the nascent cluster also nullifies arguments 2 (since the Moon would have to be formed after capture) and 4.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;jwenting&lt;/em&gt; - The moon argument isn't a valid one. The moon could have been formed out of the earth after capture. Which doesn't make that capture a more likely scenario of course.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I wouldn't say nullifies. In fact, I would think it makes them a bit stronger. The process of the Earth forming elsewhere, being flung out of its stellar system and captured by ours (and put in its proper place), then forming a Moon and life, requires that these events happened extremely quickly. While not impossible, it is improbable, especially when you add in your argument for the Earth being in a very precise orbit unlikely to have been achieved if it had been captured. If you argue that the Moon/Life formed after capture you're really limiting the time and chances that the Earth didn't form in our system.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-09-28T13:26:28.690" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18507" PostTypeId="2" ParentId="18504" CreationDate="2016-09-27T22:52:30.490" Score="4" Body="&lt;p&gt;It would be really difficult for the Earth to end up in a nearly circular orbit if it came from outside the solar system. Effectively falling from infinity, it would have a hyperbolic orbit and make one swerve around the Sun and depart back into the blackness of interstellar space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What mechanism could get rid of precisely enough energy to keep the Earth from exiting and get it into such a nice orbit? (The eccentricity is .0167, which is too close to a perfect circle to distinguish with the naked eye.) I can't think of anything. At best it should have an orbit that has the outer point way out in the outer solar system and a closest approach near the sun.&lt;/p&gt;&#xA;" OwnerUserId="11458" LastActivityDate="2016-09-27T22:52:30.490" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18508" PostTypeId="2" ParentId="18504" CreationDate="2016-09-27T23:40:46.353" Score="16" Body="&lt;p&gt;I think the Earth's orbit is by far the strongest argument you have that the Earth formed around the Sun. The orbit is nearly circular and almost in the Sun's equatorial plane, similar to the other planets. These facts are naturally accounted for if the Earth formed from material that coagulated in the Sun's primordial disc, where circularisation and collapse to a plane perpendicular to the Sun's rotation axis are expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Capture of planets &lt;em&gt;is&lt;/em&gt; possible and is thought to be reasonably common in the early dense environments of star forming clusters. A captured planet would therefore have a similar age and metallicity to one born around the Sun, since star clusters are empirically almost coeval and chemically homogeneous. The problem is that terrestrial, rocky planets take some tens of millions of years to form. Most dense star clusters have dispersed on much shorter timescales than this. Even if the Sun was born in a rare, long-lived cluster, its primordial disk of gas would have already dispersed within ten million years and thus there would be no mechanism to circularise a captured Earth's orbit or enforce an orbit in the same plane as the other planets.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-09-27T23:40:46.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18509" PostTypeId="2" ParentId="8083" CreationDate="2016-09-28T00:11:29.410" Score="2" Body="&lt;p&gt;This figure comes from numerical simulation by &lt;a href=&quot;https://ui.adsabs.harvard.edu/#abs/2002Natur.419..375L/abstract&quot; rel=&quot;nofollow&quot;&gt;Laskar et al. (2002)&lt;/a&gt;, see Fig. 2. An extended figure from a longer simulation can be found in &lt;a href=&quot;https://ui.adsabs.harvard.edu/#abs/2004Icar..170..343L/abstract&quot; rel=&quot;nofollow&quot;&gt;Laskar et al. (2004)&lt;/a&gt;, Fig 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see more clearly from the extended figure, the simulation doesn't particularly suggest that this is periodic behavior. The question is, then, what happened ~5 million years ago that changed up Mars' obliquity like that? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Laskar mentions the event in the 2002 paper, but doesn't bother to offer any particular physical cause. These are numerical orbital simulations after all, not recorded observations. As unsatisfying as it might be, I think your answer is that evidently the planets aligned, so to speak (albeit probably not in a literal line). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fig. 8 in the 2004 paper gives a tiny bit of context in terms of uncertainty, it shows that the simulation is only precise for the last 10 million years, and then goes chaotic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Laskar mentions that his data are free to anyone who requests it, although I can't say how accessible it would be. In theory, someone could go in there and look for correlations (e.g. maybe there was some kind of strong resonance with the Earth and/or Jupiter around that time). I looked through ADS a bit, and found a newer paper by &lt;a href=&quot;https://ui.adsabs.harvard.edu/#abs/2008AJ....135.1151E/abstract&quot; rel=&quot;nofollow&quot;&gt;Edvardsson and Karlsson (2008)&lt;/a&gt; that pretty much supports Laskar's results in over the past 8 million years, but my brief search didn't turn up anything looking into what exactly caused that anomaly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, from context it appears that &lt;a href=&quot;https://ui.adsabs.harvard.edu/#abs/1993Sci...259.1294T/abstract&quot; rel=&quot;nofollow&quot;&gt;Touma and Wisdom (1993)&lt;/a&gt; were the ones to first discuss that obliquity drop, although their paper unfortunately appears to be hiding behind a Science paywall. &lt;/p&gt;&#xA;" OwnerUserId="14370" LastEditorUserId="14370" LastEditDate="2016-09-28T17:54:51.130" LastActivityDate="2016-09-28T17:54:51.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18510" PostTypeId="1" AcceptedAnswerId="18512" CreationDate="2016-09-28T00:52:04.577" Score="0" ViewCount="79" Body="&lt;p&gt;How to find sub-radar latitude in reference to radar telescopes hitting near earth asteroids?&lt;/p&gt;&#xA;" OwnerUserId="14377" LastEditorUserId="14377" LastEditDate="2016-09-28T15:02:15.537" LastActivityDate="2016-09-28T15:02:15.537" Title="Not clear on how to calculate sub-radar latitude" Tags="&lt;telescope&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ClosedDate="2016-10-02T15:58:26.893" ContentLicense="CC BY-SA 3.0" />
  <row Id="18511" PostTypeId="2" ParentId="18504" CreationDate="2016-09-28T04:45:14.237" Score="3" Body="&lt;p&gt;Occam's razor defeats the hypothesis. There are no astronomical anomalies about earth's obit or position that require the complication of earth's insertion from beyond the solar system. Unless some difficulties in known fact are brought forth &lt;em&gt;requiring&lt;/em&gt; such a hypothesis as their simplest solution, then the idea may simply be rejected as unnecessary.&#xA;The other observations given (orbital circularity, consistency in solar plane, rotation, composition, etc.) lend credibility in that they show how much an insertion hypothesis would need to account for that the born-in-place model already handles.&#xA;NB: This doesn't mean that it isn't possible--only that, scientifically, there's no reason proposed to think it did, and many reasons to think it didn't. Occam's razor isn't a proof; its a methodological principle.&lt;/p&gt;&#xA;" OwnerUserId="14379" LastActivityDate="2016-09-28T04:45:14.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18512" PostTypeId="2" ParentId="18510" CreationDate="2016-09-28T04:58:18.003" Score="5" Body="&lt;p&gt;In the common case asteroids will rotate around a nearly fixed axis, like the Earth. That axis of rotation defines a north pole and south pole for that asteroid. The north pole is the one where, when a viewer is seeing the asteroid from above it, the asteroid is rotating counter-clockwise. Once you have north and south poles, you can define an equator and, thus, a system of latitude. I have no idea what is done for asteroids that are undergoing chaotic tumbling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;sub-radar latitude&quot; is the asteroid's latitude that is directly facing the radar viewer. A good example from the literature that uses the term many times, and includes a short definition, is &lt;A HREF=&quot;http://arxiv.org/abs/1101.3794&quot;&gt;arXiv:1101.3794&lt;/A&gt;. A short excerpt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] implying significant change in the object’s sub-radar latitude (the angle between the asteroid-Earth line and the object’s equatorial plane).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14244" LastActivityDate="2016-09-28T04:58:18.003" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18513" PostTypeId="2" ParentId="18409" CreationDate="2016-09-28T09:14:06.620" Score="1" Body="&lt;p&gt;You don't &lt;em&gt;need&lt;/em&gt; to do a physics simulation to do what you want. Kepler's laws of planetary motion will do just fine and they account for elliptical orbits of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will have to do a bit more Maths if you want your orbits to be elliptical. The Maths you need is summarised &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Position_as_a_function_of_time&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. This is a recipe for calculating $r$ and $\theta$ as a function of time, where $r$ is the helio(astro)centric distance and $\theta$ is the true anomaly, which is the agular coordinate at that point of time. In turn you would have to specify the mass of the star (and planet if it was significant), the orbital period (or semi-major axis), the perihelion(astron) position and the position at $t=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbital inclination is independent of these calculations, since in this basic scheme there is nothing to alter the angular momentum of the orbit and so there is a fifth parameter defining that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculations are reasonably simple though you will have to do something numerical (like the &lt;a href=&quot;http://www.sosmath.com/calculus/diff/der07/der07.html&quot; rel=&quot;nofollow&quot;&gt;Newton-Raphson method&lt;/a&gt;) to get the eccentric anomaly in step 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wish to start taking 3-body (or more) interactions into account then you &lt;em&gt;will&lt;/em&gt; have to think about physics simulations. I think the scheme given by cphyc will work ok so long as you make the time step very small and don't integrate over many orbits. If you are interested in the stability or long term evolution of planetary systems that you setup then you &lt;em&gt;will&lt;/em&gt; have to get into &lt;em&gt;much&lt;/em&gt; more complex schemes, since the one suggested by cphyc is numerically unstable.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-09-28T09:14:06.620" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18514" PostTypeId="2" ParentId="18496" CreationDate="2016-09-28T14:40:20.850" Score="7" Body="&lt;p&gt;I wanted to throw my hat into the mix to flesh out another contributing factor. One reason these &quot;geysers&quot; are so capable of achieving great heights is a lack of atmosphere on Europa to slow them down (and to a lesser extent, weaker gravity). I wrote a basic &quot;physics simulation&quot; in Python 3 which illustrates this purpose, the code for which is below.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;&#xA;# Define properties of Europa&#xA;G = 6.67408E-11     # Gravitational constant, m^3 kg^-1 s^-2&#xA;M = 4.7998E22       # Mass of Europa, kg&#xA;R = 1560000         # Radius of Europa, m&#xA;rho0 = 0.1          # Density of air at surface, kg m^-3&#xA;H = 100             # Scale Height of Atmosphere, m&#xA;&#xA;# Define properties of ejected water droplet&#xA;C_D = 0.5           # Drag coefficient, unitless&#xA;A = np.pi*0.01**2   # Cross-sectional area, m^2&#xA;m = 0.001           # Mass of droplet, kg&#xA;x = 0               # Initial height above surface, m&#xA;v = 500             # Initial velocity, m/s&#xA;t = 0               # Initial time, s&#xA;dT = 1E-3           # Timestep in &quot;simulation&quot;, s&#xA;&#xA;# Run &quot;Simulation&quot;&#xA;while v &amp;gt; 0:&#xA;&#xA;    a = -(G*M*m/(R+x)**2 + 0.5*(rho0*np.exp(-x/H))*v**2*C_D*A) / m&#xA;    v = v + a*dT&#xA;    x = x + v*dT&#xA;&#xA;    t += dT&#xA;&#xA;# Print final results&#xA;print('Time:',round(t,4),'(s)\nHeight:',round(x/1000,4),'(km)\nVelocity:',round(v,5),'(m/s)')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This starts off by defining a droplet with the specified properties (e.g., mass, cross-sectional surface area) and ejects it up with some initial speed from the surface of Europa. As it travels up, the acceleration on it due to gravity and atmospheric drag is constantly calculated and the velocity and subsequently height are updated at every time step. The simulation stops when the droplet reaches the peak of it's upwards journey, indicated by the velocity becoming negative (i.e., the particle has begun falling down again). The final time taken to reach that height, the final height, and the velocity at that height (which should be very nearly zero) are then printed out. The air drag model I used was the basic &lt;a href=&quot;https://en.wikipedia.org/wiki/Drag_equation&quot;&gt;drag equation&lt;/a&gt;. Feel free to substitute something more complex and potentially more realistic. The atmosphere itself, is modeled with a simple exponential model &lt;a href=&quot;https://en.wikipedia.org/wiki/Scale_height#Scale_height_used_in_a_simple_atmospheric_pressure_model&quot;&gt;as described here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to analyze a few results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;No atmosphere&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case just set &lt;code&gt;rho0 = 0&lt;/code&gt;. This will mean only gravity is acting to slow down the droplet. Without an atmosphere, I get the output:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Time: 413.093 (s)&#xA;Height: 101.1158 (km)&#xA;Velocity: -0.00046 (m/s)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To be fair, I chose an ejection velocity which would give me a height of 100 km (which I stole from userLTK's answer). Play around with other ejection velocities as you wish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thin Atmosphere&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two factors here to produce a thin atmosphere. First, setting the density of air at the surface. Without spending the time to dig into more realistic numbers, I'll use &lt;code&gt;rho0 = 0.1&lt;/code&gt; ($kg/m^3$). For reference, on Earth it is $1.225\:kg/m^3$ at the surface. The other factor is the scale height. This number controls how rapidly the atmosphere disappears as you go up. A smaller number means a thinner atmosphere that ends more quickly. A larger number means a persistently thicker atmosphere that extends much farther. Again, I didn't look up what a proper scale height should be, but for reference on Earth the scale height is about $8\:km$. Here I used &lt;code&gt;H = 100&lt;/code&gt; ($\mathrm{meters}$). In this case, I get the output:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Time: 175.804 (s)&#xA;Height: 19.9797 (km)&#xA;Velocity: -0.00028 (m/s)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Even the addition of this thin, weak atmosphere dropped the height achieved by an order of magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thick Atmosphere&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's up the density of the air and make the atmosphere extend more by increasing the scale height. Now &lt;code&gt;rho0 = 0.5&lt;/code&gt; ($kg/m^3$) and &lt;code&gt;H = 1000&lt;/code&gt; ($\mathrm{meters}$). The output is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Time: 7.158 (s)&#xA;Height: 0.1182 (km)&#xA;Velocity: -0.00127 (m/s)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We've dropped the height by three orders of magnitude compared to the no atmosphere model!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the physics model is basic, I don't think it is unrealistic. I'm sure with more investigation, more accurate numbers and models could be used, but I believe the results would be more or less the same. On Earth, it is not gravity which primarily limits the height of our geysers, it the strong atmosphere drag. If I run this simulation on Earth, I find a similar water droplet would reach a height of $40\:m$ in our atmosphere, but without that atmosphere, it would get up to $13\:km$, more than 300 times higher. &lt;em&gt;The reason these geysers can reach such large heights is primarily because of the lack of atmosphere drag&lt;/em&gt;, compared to what we see on Earth. The weaker gravity does help in the geysers reaching higher heights, but to a lesser extent.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2016-09-28T14:45:29.220" LastActivityDate="2016-09-28T14:45:29.220" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18519" PostTypeId="1" AcceptedAnswerId="18523" CreationDate="2016-09-29T11:22:16.370" Score="4" ViewCount="410" Body="&lt;p&gt;After reading &lt;a href=&quot;https://astronomy.stackexchange.com/a/18508/13071&quot;&gt;this interesting answer&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering, do we in fact know &lt;strong&gt;if our Sun&lt;/strong&gt; in particular was created as part of a star-forming-cluster, or, was it more of a solo creation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Or, are all stars created as part of star-forming-clusters, is that the norm and how all stars come to be?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, indeed are we aware of &lt;strong&gt;which stars around us&lt;/strong&gt;, are our sibling star?  (So, there's a list somewhere &quot;stars known to likely be sibling to our Sun from a certain star-forming-cluster&quot;.)  Or have they dispersed over the 4 billion years and it's impossible to know, or is it just the case that &quot;all the stars anywhere nearby&quot; are/were indeed siblings from &quot;our&quot; star-forming-cluster?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, I'm having trouble grasping the order of magnitude of star-forming-clusters. Consider say &quot;all the stars in our arm of the galaxy&quot;: Consider the question &quot;in how many star-forming-clusters did all those stars originate?&quot;  Is the answer like &quot;3&quot;, &quot;a few hundred&quot; or &quot;100 million&quot; ??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps most simply: if sol formed as part of a star-forming-cluster, what's the order of magnitude of &lt;strong&gt;how many stars&lt;/strong&gt; formed with it?&lt;/p&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-09-29T19:47:24.967" Title="Sun's formation as &quot;part of a star-forming-cluster...&quot;?" Tags="&lt;star-formation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18520" PostTypeId="2" ParentId="18519" CreationDate="2016-09-29T12:52:42.220" Score="2" Body="&lt;p&gt;Stars typically do form in clusters, because to get the gravitational instability to initiate, you need a very large mass-- much larger than the mass of a single star.  A typical mass for the gravitational instability is about a million solar masses, though it can vary by an order of magnitude depending on the situation.  But you don't get a single star of that mass because after the instability starts (on the scale of a &quot;giant molecular cloud&quot;), it fragments into smaller bits, which are the star clusters, which fragment into further bits, which are the stars.  There are two basic types of clusters, &quot;open&quot; and &quot;globular&quot;, where the open clusters are not gravitationally bound so the stars wander away from each other after forming (which is presumably what happened to the Sun).  The globular clusters stay bound for very long times and can even wander away from the plane of the spiral galaxy and still be together as clusters, but obviously the Sun is not in one of those and is not old enough to have completely left the plane of the galaxy (though it does oscillate up and down through the plane).  A typical open cluster might be a thousand stars, so it seems likely the Sun might have about a thousand &quot;sister&quot; stars spread around the plane by now, but we'll never know which ones those are since the age cannot be determined precisely enough.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-09-29T12:52:42.220" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="18523" PostTypeId="2" ParentId="18519" CreationDate="2016-09-29T19:47:24.967" Score="5" Body="&lt;p&gt;Fortunately, there is an extensive and highly respected Annual Reviews of Astronomy and Astrophysics article on this very question - Adams (2010), &lt;a href=&quot;https://arxiv.org/abs/1001.5444&quot;&gt;https://arxiv.org/abs/1001.5444&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment it is debated whether most stars form in clusters. They may do, but there is a lot of evidence for hierarchical structure formation that results in lots of stars forming in relative isolation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless, there are a number of strands of evidence that can be used to suggest that &lt;strong&gt;the Sun formed in a cluster of somewhere between a thousand and ten thousand siblings&lt;/strong&gt; - a bit like the Orion Nebula cluster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adams argues for this on the basis of: (1) We have a relatively well-behaved solar system inside the orbit of Neptune. This argues the Sun did not (or rather was not likely to) form in a massive and dense cluster where it suffered lots of very close encounters. On the other hand, the wacky orbits of some more distant objects like Sedna do suggest some early close encounter with another star was required. (2) If the Sun was born in a very massive cluster, it would likely have been born close to some very massive stars. The UV emission from such stars could photoevaporate a circumsolar disk before it formed planets. (3) In tension with this is the probability that the protosolar disk was enriched with radioactive supernova products. This requires the Sun to be born close to moderately massive stars (that would explode on a short timescale), which are only found in moderately massive clusters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adams puts these constraints together to estimate a &lt;em&gt;probability&lt;/em&gt; that the Sun was born in a cluster of a particular size. Figure 7 of the review shows this to be reasonably sharply peaked between a thousand and ten thousand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An order of magnitude uncertainty is at least warranted and some of the assumptions and calculations are open to criticism. But it is a starting point for any further research into the matter.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-09-29T19:47:24.967" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="18524" PostTypeId="1" CreationDate="2016-09-29T20:04:11.617" Score="4" ViewCount="327" Body="&lt;p&gt;How close could two earth sized celestial bodies get in their nearest approach either as planets or moons around a larger planet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Close enough for atmospheric interaction in a stable orbit at nearest approach, or for non-space-fairing species to somehow make the jump without cosmological assistance (aka asteroid impact)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems just from using the roche limit of an earth to earth sized bodies any approach under ~20,000 km would be unstable in the long term, which is orders of magnitude beyond a normal atmosphere size.&lt;/p&gt;&#xA;" OwnerUserId="14406" LastEditorUserId="14406" LastEditDate="2016-09-29T21:28:54.417" LastActivityDate="2017-05-10T18:36:34.803" Title="Could the closest approach of stable orbits of two earth sized planets allow life to jump between?" Tags="&lt;planet&gt;&lt;orbit&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18525" PostTypeId="2" ParentId="18496" CreationDate="2016-09-30T08:38:27.433" Score="1" Body="&lt;p&gt;I've drawn information from the helpful links below &lt;a href=&quot;https://space.stackexchange.com/q/18363/12102&quot;&gt;this question&lt;/a&gt; and this &lt;a href=&quot;https://space.stackexchange.com/a/18365/12102&quot;&gt;partial answer&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think astronomers or planetary scientists expect a geyser of liquid water to be squiring 100 to 200 kilometers above Europa's surface. There are now however at least two published detection of &lt;em&gt;plumes of water vapor&lt;/em&gt; that reach that high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So after doing some reading, asking around, and with helpful conversations in the comments of this questions and the answers here, I think it's pretty clear that there are not geysers on Europa and the term was used incorrectly by the &lt;em&gt;Washington Post&lt;/em&gt;. However, Based on the sub-surface jet model, it might not be incorrect to call them jets, as in the &lt;em&gt;BBC&lt;/em&gt; article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iGGMZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iGGMZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/krLJV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/krLJV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; NASA hubblesite images from &lt;a href=&quot;http://hubblesite.org/newscenter/archive/releases/2016/33/image/b/format/web_print/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://hubblesite.org/newscenter/archive/releases/2016/33/image/c/format/web_print/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image on the left is reminiscent of a water fountain and probably gives the wrong impression. While there is probably some directionality, it might not be a coherent, tight column like that.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Possibility #1 Squirting Water from an Over-Pressured Ocean:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;While it's easy to imagine that stresses and pressure in the thick ice is compressing the liquid below so much that it would squirt out of a hole like a small hole in a water balloon, probably not. Ice floats. The density of ice is about 0.92 to 0.93 (when very cold) compared to cold water at about 1.00. That means if you dug a hole in the ice and reached the ocean, it would rise, but probably not to the surface. For example, if everything is in equilibrium and the ice were 10km deep, it might end up something like 800 meters below the average surface. There could be short term non-equilibrium, but ice deforms and cracks and refreezes, and over time the system would not build up and maintain huge hydrostatic pressure beyond that caused by the weight of the ice. The mechanics and dynamics of the ice a whole topic in itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0TDkw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0TDkw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Image of ice fishing from &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Ministry_of_Defence_of_Russia_-_035.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/2/23/Bogstadvannet_IMG_1611.JPG/665px-Bogstadvannet_IMG_1611.JPG&quot; rel=&quot;nofollow noreferrer&quot;&gt;cf. &quot;I've been waiting all morning - where is the geyser?&quot;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Possibility #2 Squirting Water from an Over-Pressured Trapped Lake:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This one is a bit easier to imagine. A volume of water trapped within the ice and subjected to huge tectonic forces, a bit like a surface spring of subsurface water on Earth. However a volume of trapped gas would really be helpful here. Since water is nearly incompressible, the loss of even a tiny volume will rapidly reduce the pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/do63L.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/do63L.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; &quot;Europa's ice-trapped lake sits above the ocean in an illustration, Illustration courtesy Britney Schmidt and Dead Pixel FX, University of Texas at Austin&quot; from National Geographic's &lt;a href=&quot;http://news.nationalgeographic.com/news/2011/11/111116-lake-europa-jupiter-moons-earth-space-science-nasa/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Great Lakes&quot; Discovered on Jupiter Moon?&lt;/a&gt;  (cropped).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Possibility #3 Sub-surface Vent:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/27p5z.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/27p5z.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Figure 3 from Jared James Berg's thesis &lt;a href=&quot;https://repositories.lib.utexas.edu/handle/2152/34041&quot; rel=&quot;nofollow noreferrer&quot;&gt;Simulating water vapor plumes on Europa&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://repositories.lib.utexas.edu/handle/2152/34041&quot; rel=&quot;nofollow noreferrer&quot;&gt;Berg's thesis&lt;/a&gt; presents &lt;a href=&quot;https://en.wikipedia.org/wiki/Direct_simulation_Monte_Carlo&quot; rel=&quot;nofollow noreferrer&quot;&gt;Direct Simulation Monte Carlo&lt;/a&gt; (DSMC) calculations for a scenario where a reservoir of trapped liquid water below the surface of Europa feeds a &lt;em&gt;volume of water vapor&lt;/em&gt; which continuously vents to the surface. If the cross section of the passage has a constriction or narrowing followed by and expansion area, this forms a sort of nozzle which converts some of the random thermal motion of the atoms into a directed flow. The velocity will be somewhat focused upwards, and the magnitude will be of the same order as the thermal velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This model does not rely on high pressures in the crust, but instead &quot;focuses&quot; the thermal motion upwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simulation includes the dynamics small &quot;grains&quot; of water ice in addition to water vapor molecules. At some point early in the trajectory, the mean free path becomes so large that further condensation stops and molecules are traveling on nearly ballistic trajectories under the influence of gravity.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Comments on Geysers and Atmospheres:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The image of the geyser in the question has initiated some discussion on the effects of atmospheric drag. Geysers on earth reach anywhere from sub 1 meter to as much as 20 meters typically. How much higher would it go if there were no atmosphere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a calculation (python script below) of isolated droplets launched into various density atmospheres with an initial velocity of 25 m/s just as a ballpark estimate of a tall geyser. In each plot the heigh vs time is calculated for five radii geometrically spaced between 0.1 and 10.0 millimeters. It can be a factor of 10 or 20 effect for the smallest droplets, and maybe a factor of 2 or 3 for the largest, but it doesn't get you to 100 kilometers if there's no air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4kZSl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4kZSl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes if Europa had Earths atmosphere, the geysers wouldn't be there, but no, there aren't geysers on Europa because there's no atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this kind of calculation is an &lt;em&gt;overestimate&lt;/em&gt; of the impact of drag. The effect in a real geyser would probably be much smaller than this kind of calculation suggests.  Take a look at an actual geyser! The mass of the water probably dominates the column, and rapidly transfers momentum to the air in the column by the same drag force. The force ($dp/dt$) is equal and opposite, so actually drag may speed up the air much more than it slows down the water. And of course, this is &lt;em&gt;hot air&lt;/em&gt; and steam which are accelerated by buoyancy forces as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7YZhQ.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7YZhQ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The source is the following video - it really gets going around 02:00, and if you keep watching you can see a &lt;a href=&quot;https://www.youtube.com/watch?v=MX0D4oZwCsA&quot; rel=&quot;nofollow noreferrer&quot;&gt;double rainbow&lt;/a&gt; as a bonus! (&lt;a href=&quot;https://youtu.be/OQSNhk5ICTI?t=24)&quot; rel=&quot;nofollow noreferrer&quot;&gt;original&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=y8gLhHzPY5M&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=y8gLhHzPY5M&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def Fdrag(x, v):&#xA;&#xA;    if hscale:&#xA;        rho = rho0 * np.exp(-x/hscale)&#xA;    else:&#xA;        rho = rho0&#xA;&#xA;    return -0.5 * rho * v * abs(v) * CD * area&#xA;&#xA;def deriv(X, t):&#xA;&#xA;    x, v     = X&#xA;    acc_drag = Fdrag(x, v)/mass&#xA;&#xA;    xdot     = v&#xA;    vdot     = -acc_grav + acc_drag&#xA;&#xA;    return np.hstack((xdot, vdot))&#xA;&#xA;class Droplet(object):&#xA;    def __init__(self, r):&#xA;&#xA;        self.r = r&#xA;&#xA;class Atmosphere(object):&#xA;    def __init__(self, rho0, hscale=None):&#xA;&#xA;        self.rho0   = rho0&#xA;        self.hscale = hscale&#xA;&#xA;import numpy as np&#xA;from scipy.integrate import odeint as ODEint&#xA;import matplotlib.pyplot as plt&#xA;import copy&#xA;&#xA;CD_sphere    = 0.47&#xA;H2O_density  = 1E+03        # kg/m^3&#xA;&#xA;hscale_Earth = 1E+04        # meters&#xA;g_Earth      = 9.8          # m/s^2&#xA;rho0_Earth   = 1.3          # kg/m^3&#xA;&#xA;# make some droplets&#xA;&#xA;v0_all = 25.    # m/s&#xA;&#xA;radii = np.logspace(-2, -4, 5)  # meters&#xA;&#xA;droplets = [Droplet(r) for r in radii]    # instantiate&#xA;&#xA;for drop in droplets:     # I love python objects :)&#xA;    drop.area   = np.pi * drop.r**2&#xA;    drop.volume = (4./3.) * np.pi * drop.r**3&#xA;    drop.mass   = H2O_density * drop.volume&#xA;    drop.CD     = CD_sphere&#xA;    drop.v0     = v0_all&#xA;    drop.x0     = 0.0&#xA;&#xA;# make some atmospheres&#xA;&#xA;fractions   = np.array([1E-04, 1E-02, 1])&#xA;names       = ['1E-04', '1E-02', '1.0']&#xA;&#xA;atmospheres = []&#xA;&#xA;for frac, name in zip(fractions, names):&#xA;&#xA;    rho0 = rho0_Earth * frac&#xA;&#xA;    atmosphere = Atmosphere(rho0, hscale=hscale_Earth)&#xA;&#xA;    atmosphere.name = name + &quot; of Earth&quot;&#xA;    atmosphere.droplets = copy.deepcopy(droplets)  # DEEP copy!&#xA;    atmosphere.acc_grav = g_Earth&#xA;&#xA;    atmospheres.append(atmosphere)&#xA;&#xA;tol  = 1E-09&#xA;time = np.linspace(0, 6, 200)&#xA;&#xA;xpts = [2.6, 1.8, 1.2, 0.5, 0.3]&#xA;ypts = [19, 11.8, 6.3, 3.1, 1.2]&#xA;labs = ['0.1', '0.32', '1.0', '3.2', '10'][::-1]&#xA;labs = [n + 'mm' for n in labs]&#xA;&#xA;labzip = zip(xpts, ypts, labs)&#xA;&#xA;for atmosphere in atmospheres:&#xA;&#xA;    for drop in atmosphere.droplets:&#xA;&#xA;        X0       = np.hstack((drop.x0, drop.v0))&#xA;        mass     = drop.mass&#xA;        area     = drop.area&#xA;        CD       = drop.CD&#xA;        rho0     = atmosphere.rho0&#xA;        hscale   = atmosphere.hscale&#xA;        acc_grav = atmosphere.acc_grav&#xA;&#xA;        answer, info = ODEint(deriv, X0, time,&#xA;                              rtol=tol, atol=tol,&#xA;                              full_output=True )&#xA;        drop.answer = answer&#xA;&#xA;plt.figure(figsize=[14,5])&#xA;&#xA;for i, atmosphere in enumerate(atmospheres):&#xA;&#xA;    plt.subplot(1,len(atmospheres),i+1)&#xA;&#xA;    for drop in atmosphere.droplets:&#xA;        x, v = drop.answer.T&#xA;        plt.plot(time, x)&#xA;&#xA;    plt.ylim(0, 34)&#xA;    plt.xlim(0, 5.2)&#xA;&#xA;    title = atmosphere.name + &quot;, v0=&quot; + str(v0_all) + &quot; m/s&quot;&#xA;    plt.title(title, fontsize=14)&#xA;&#xA;    if i == 2:&#xA;        for x, y, s in labzip:&#xA;            plt.text(x, y, s, fontsize=12)&#xA;&#xA;    plt.xlabel('time (sec)', fontsize=14)&#xA;    plt.ylabel('height (m)', fontsize=14)&#xA;&#xA;plt.suptitle('Single, Isolated droplets in air', fontsize=18)&#xA;&#xA;plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-02-10T17:39:23.857" LastActivityDate="2018-02-10T17:39:23.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18526" PostTypeId="1" AcceptedAnswerId="18532" CreationDate="2016-09-30T11:36:26.997" Score="9" ViewCount="679" Body="&lt;p&gt;This question arises from &quot;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/13602/what-planet-is-better-than-earth-to-infer-solar-system-configuration?rq=1&quot;&gt;What planet is better than earth to infer solar system configuration?&lt;/a&gt;&quot;, where one of the answers pointed that it's easier to discover the Moon from Mars than discovering the Galilean satellites of Jupiter from Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then my question is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How easy (or difficult) would be to see the Moon from Mars?&lt;/li&gt;&#xA;&lt;li&gt;Could it be seen with naked eye?&lt;/li&gt;&#xA;&lt;li&gt;Is actually easier to see the Moon from Mars than to see the Galilean moons of Jupiter from Earth?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I know there are favourable factors (Mars is closer to Earth than Earth to Jupiter, and Earth is smaller than Jupiter, so it's expected to obscure less its moon than Jupiter obscures its) and also unfavourable factors (the Moon is closer to the Earth than the Galilean moons to Jupiter, and Earth and Moon are backlit when seen from Mars). What is the overall result?&lt;/p&gt;&#xA;" OwnerUserId="14097" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-09-30T18:20:59.837" Title="How easy would be to see the Moon from Mars?" Tags="&lt;the-moon&gt;&lt;mars&gt;&lt;naked-eye&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18527" PostTypeId="2" ParentId="18526" CreationDate="2016-09-30T13:41:41.663" Score="1" Body="&lt;p&gt;You should be able to answer this easily: just look up the angular subtense of the Galileans from Earth and compare with the angular subtense of our Moon from Mars (and heck, check out the ability of Martians to see their own moons :-) ).  You might also want to check out the orbital radii to ensure you can resolve the moons vs. their home planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Galilean angular separation from Jupiter is &lt;a href=&quot;https://en.wikipedia.org/wiki/Galilean_moons#Visibility&quot; rel=&quot;nofollow&quot;&gt;2 to 10 arc-minutes&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ganymede's diameter is 5300 km; under ideal orbital positioning, it's  &lt;a href=&quot;https://en.wikipedia.org/wiki/Jupiter&quot; rel=&quot;nofollow&quot;&gt;3.95029 AU&lt;/a&gt; , or 590 Gm from Earth, so it subtends about 2 arc-seconds (someone check my math! :-) ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our Moon's diameter is &lt;a href=&quot;https://en.wikipedia.org/wiki/Moon&quot; rel=&quot;nofollow&quot;&gt;3400 km&lt;/a&gt; and its distance from Mars, at closest approach, is &lt;a href=&quot;https://en.wikipedia.org/wiki/Moon&quot; rel=&quot;nofollow&quot;&gt;about 60e6 km&lt;/a&gt;,  for a subtense of about 12 arc-seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That suggests our Moon is easily seen from Mars w/ Galileo's original telescope.&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2016-09-30T13:41:41.663" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18528" PostTypeId="2" ParentId="12068" CreationDate="2016-09-30T14:55:55.217" Score="1" Body="&lt;p&gt;In addition to already present good answers, I would say that if we found life on Mars on a century there is a noticeably probability of it being made from &lt;a href=&quot;https://en.wikipedia.org/wiki/Planetary_protection&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth microbes accidentally taken there by one of our probes&lt;/a&gt; or their descendants. Risk of contamination of other planets is addressed by carefully sterilization of probes but no sterilization is complete, so there is always a small risk.  &lt;/p&gt;&#xA;" OwnerUserId="14097" LastEditorUserId="14097" LastEditDate="2017-09-13T07:40:02.460" LastActivityDate="2017-09-13T07:40:02.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18529" PostTypeId="1" CreationDate="2016-09-30T15:13:20.830" Score="3" ViewCount="4032" Body="&lt;p&gt;I was disappointed to hear Brian Cox's flippant reply to whether or not the LHC could create a black hole big enough to swallow the earth in this video:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=QFy87tFTZwY&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=QFy87tFTZwY&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing that the power used by the LHC is nothing like that used when gravity crushes a star to a singularity, but ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... how big is &quot;nothing like&quot;?&lt;/p&gt;&#xA;" OwnerUserId="14412" LastEditorUserId="14412" LastEditDate="2016-10-01T20:40:23.570" LastActivityDate="2017-02-11T21:37:35.393" Title="Energy needed to create a black hole vs energy needed to run the Large Hadron Collider?" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;high-energy-astrophysics&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18530" PostTypeId="1" AcceptedAnswerId="18536" CreationDate="2016-09-30T15:52:47.503" Score="2" ViewCount="320" Body="&lt;p&gt;To study clustering of galaxies and clusters, one makes use of&#xA;&lt;a href=&quot;https://ned.ipac.caltech.edu/level5/March12/Coil/Coil2.html&quot; rel=&quot;nofollow&quot;&gt;the two-point correlation function&lt;/a&gt;. However, we can see that all &#xA;the estimators of $\xi(r)$ use a random catalogue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the random catalogue used for? &lt;/p&gt;&#xA;" OwnerUserId="2655" LastActivityDate="2016-09-30T19:32:18.793" Title="What is the role of the random catalogue in calculating the two-point correlation function?" Tags="&lt;galaxy&gt;&lt;galaxy-cluster&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18531" PostTypeId="2" ParentId="18529" CreationDate="2016-09-30T18:19:53.933" Score="2" Body="&lt;p&gt;According to our current understanding, there is no lower bound for the energy needed to make a black hole. Any object, no matter how small, if compressed enough, could in theory form a black hole. Make it small enough, and it would not require much energy at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here's the main issue. Black holes have this thing called Hawking radiation: they constantly lose energy (and therefore mass) via radiation, and the smaller the BH is, the faster it loses energy. Super-tiny BHs basically pop out of existence the very moment they are created, that's how quick the process is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a BH to be a threat, it would have to be big enough to be stable long enough to actually start eating surrounding matter, to compensate for loss via radiation. I believe calculations have shown that the BH would then have to be as massive as a big mountain. That's A LOT of energy. The LHC is nowhere near that level.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2016-09-30T18:19:53.933" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18532" PostTypeId="2" ParentId="18526" CreationDate="2016-09-30T18:20:59.837" Score="6" Body="&lt;p&gt;We have pictures of this, thanks to Curiousity!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Q3p2K.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q3p2K.jpg&quot; alt=&quot;The Earth and Moon from the Martian surface&quot;&gt;&lt;/a&gt;&#xA;(&lt;a href=&quot;http://www.universetoday.com/109095/you-are-here-curiositys-1st-photo-of-home-planet-earth-from-mars/&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;“A human observer with normal vision, if standing on Mars, could easily see Earth and the moon as two distinct, bright “evening stars,” said NASA in a statement issued today.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So yes, you can see our moon from Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Extraterrestrial_skies#Earth_from_Mars&quot; rel=&quot;noreferrer&quot;&gt;apparent brightness of our moon from Mars is +0.9&lt;/a&gt;. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Galilean_moons#Visibility&quot; rel=&quot;noreferrer&quot;&gt;apparent brightness of Jupiter's 4 biggest moons is between 4.6 and 5.6 under ideal conditions&lt;/a&gt;.  However, while our moon can be seen visually separate from Earth from Mars, Jupiter's moons can't be distinguished from Jupiter itself with an unaided eye.&lt;/p&gt;&#xA;" OwnerUserId="14255" LastActivityDate="2016-09-30T18:20:59.837" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18533" PostTypeId="1" AcceptedAnswerId="18700" CreationDate="2016-09-30T19:01:35.240" Score="3" ViewCount="1831" Body="&lt;p&gt;I am a student of class 12 and I am studying in India. I want to be an astronomer and I know that I have to get a B.Sc, then M.Sc, and then Ph.D in astronomy. I have also started to learn the python programming language. I came to know that there are less opportunities for research in astronomy, since many organizations like ISRO not only employ scientists, but also employ engineers to make satellites and spacecrafts. I will try my best to become an astronomer, but to have life security (if i am unemployed), can I do aeronautical engineering so that I can even get employed in ISRO? After having a degree in engineering and after doing engineering can i do M.Sc in physics and then i start preparing to be an astronomer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me tell you about myself a bit, I love to study science and I love to learn about stars (I have also watched documentaries on astronomy like &quot;How the universe works&quot; and &quot;Cosmos&quot;). I also love aircraft so aeronautical engineering would also be a job that I would love. And please tell me should I also do some extra courses which would enhance my skills as an astronomer?&lt;/p&gt;&#xA;" OwnerUserId="14413" LastEditorUserId="11143" LastEditDate="2016-09-30T19:09:59.050" LastActivityDate="2016-10-13T20:58:44.670" Title="Can I do M.Sc in physics to become an astronomer after doing B.Tech in aeronautical engineering?" Tags="&lt;astrophysics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18534" PostTypeId="1" AcceptedAnswerId="18568" CreationDate="2016-09-30T19:21:41.617" Score="2" ViewCount="113" Body="&lt;p&gt;As best as is known, what is the latest thinking on in the issue of,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in which constellation was GW150914, and in which constellation was GW151226?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(That is to say, within the large uncertainty, as best as is known in what direction is the likely center of that uncertainty?)&lt;/p&gt;&#xA;" OwnerUserId="13071" LastActivityDate="2016-10-03T22:10:54.913" Title="In which two constellations, were the two gravitational wave events of 2016?" Tags="&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18535" PostTypeId="1" CreationDate="2016-09-30T19:31:11.270" Score="5" ViewCount="1050" Body="&lt;p&gt;There are as far as I know two  fusion reactions by which stars convert hydrogen to helium: The CNO cycle (for carbon–nitrogen–oxygen) and the  proton–proton chain reaction. &#xA;The elements created in the CNO cycle are C,N,O and Fluor. &#xA;But in this video min:13:00 &lt;a href=&quot;https://www.youtube.com/watch?v=pxrPaNDvaQE&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=pxrPaNDvaQE&lt;/a&gt; the prof. talks about Phosphorus and sulfur. Is that true and are those element also created in the CNO cycle or is he talking about something else?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="1056" LastEditDate="2018-03-06T08:35:20.060" LastActivityDate="2018-03-06T08:35:20.060" Title="Are sulfur and phosphorus also created in the nucleosynthesis of big stars?" Tags="&lt;star&gt;&lt;astrochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18536" PostTypeId="2" ParentId="18530" CreationDate="2016-09-30T19:32:18.793" Score="3" Body="&lt;p&gt;The point of the two-point correlation function (pun not intended) is to describe how clustered the galaxies in the universe are. Astronomers want to know if they're all bunched up in tight bundles with huge voids, or maybe they're more or less uniformly distributed.&lt;/p&gt;&#xA;&lt;p&gt;The two-point correlation function is a way of measuring the distribution of these galaxies &lt;em&gt;compared to a random distribution&lt;/em&gt;. In other words, if your distribution of galaxies is not uniformly random (which it isn't for our universe - our universe has a distinct non-random structure of clusters and voids) then your correlation function will indicate the degree to which it isn't random. Let's take a look at the last equation listed in your link.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The most commonly-used estimator is from &lt;a href=&quot;http://cdsads.u-strasbg.fr/abs/1993ApJ...412...64L&quot; rel=&quot;nofollow noreferrer&quot;&gt;Landy &amp;amp; Szalay (1993)&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\xi(r) = \frac{1}{RR(r)}\Big[DD(r)\Big(\frac{n_R}{n_D}\Big)^2 - 2DR(r)\Big(\frac{n_R}{n_D}\Big)+RR(r)\Big]$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The symbols in this equation are defined as follows:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\xi(r)$&lt;/span&gt; - The two-point correlation function.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; - The separation between two galaxies.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$DD$&lt;/span&gt;,&lt;span class=&quot;math-container&quot;&gt;$DR$&lt;/span&gt;,&lt;span class=&quot;math-container&quot;&gt;$RR$&lt;/span&gt; - The counds of pairs of galaxies as a function of separation in the data catalogue, between the data and random catalogue, and in the random catalogue, respectively.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$n_D$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$n_R$&lt;/span&gt; - The mean number density of galaxies in the data and random catalogues, respectively.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;If your data catalogue happens to be completely random, then &lt;span class=&quot;math-container&quot;&gt;$DD=DR=RR$&lt;/span&gt; and you find that &lt;span class=&quot;math-container&quot;&gt;$\xi(r)=0$&lt;/span&gt; (assuming also that &lt;span class=&quot;math-container&quot;&gt;$n_D=n_R$&lt;/span&gt;), showing your data catalogue is indeed indistinguishable from a random distribution. If, however, your data catalogue is not defined with completely random galaxies, you'll find &lt;span class=&quot;math-container&quot;&gt;$\xi(r)\ne0$&lt;/span&gt;, and the degree to which it is non-zero indicates how non-random, or organized your system is.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-09-30T19:32:18.793" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18537" PostTypeId="2" ParentId="18535" CreationDate="2016-09-30T20:13:38.417" Score="5" Body="&lt;p&gt;Yes, they are created through stellar nucleosynthesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sulfur is created from &lt;a href=&quot;https://en.wikipedia.org/wiki/Silicon-burning_process&quot; rel=&quot;nofollow&quot;&gt;silicon&lt;/a&gt; via an &lt;a href=&quot;https://en.wikipedia.org/wiki/Alpha_process&quot; rel=&quot;nofollow&quot;&gt;alpha process&lt;/a&gt;, with the reaction&#xA;$$_{14}^{28}\text{Si }+\text{ } _2^4\text{He}\to _{16}^{32}\text{S }+\gamma$$&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Oxygen-burning_process&quot; rel=&quot;nofollow&quot;&gt;Oxygen burning&lt;/a&gt; can also create $_{16}^{31}\text{S}$, as well as other elements like $_{15}^{31}\text{P}$. Additionally, both oxygen and phosphorus can be formed through supernova nucleosynthesis (see &lt;a href=&quot;https://arxiv.org/abs/1312.3807&quot; rel=&quot;nofollow&quot;&gt;Koo et al. (2013)&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CNO cycle won't produce sulfur or phosphorus. The chain simply isn't set up to create either of them directly.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-10-01T23:06:53.513" LastActivityDate="2016-10-01T23:06:53.513" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18538" PostTypeId="2" ParentId="18535" CreationDate="2016-09-30T21:21:41.330" Score="8" Body="&lt;p&gt;It is not true to say that the CNO cycle produces these elements. It is more like a catalytic chain that aids the conversion of hydrogen to helium. Thus pre-existing CNO nuclei are required and all the reactions do is change the balance of these elements because the reaction steps in the chain have differing timescales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sulphur is mainly produced by alpha capture (fusion) onto. carbon and oxygen via neon, magnesium and silicon. This requires the high temperatures found near the cores of massive stars (more than 8 solar masses) and occurs fairly late in their lives, not long before the type II supernova that casts a lot of the processed material into space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phosphorus is not part of this &quot;alpha chain&quot; (&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Alpha_process&quot;&gt;https://en.m.wikipedia.org/wiki/Alpha_process&lt;/a&gt;) because its mass number is not divisible by 4 and it has an odd atomic number. However, it &lt;em&gt;is&lt;/em&gt; also expected to be produced in massive stars via neutron capture onto isotopes of silicon (see &lt;a href=&quot;https://arxiv.org/abs/1112.3824&quot;&gt;https://arxiv.org/abs/1112.3824&lt;/a&gt; ). This occurs in the neon and oxygen burning shells, late in the lives of massive stars.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-09-30T21:21:41.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18539" PostTypeId="1" AcceptedAnswerId="18542" CreationDate="2016-09-30T21:38:30.177" Score="3" ViewCount="291" Body="&lt;p&gt;Comets lose their mass through water evaporation due to close encounter with sun, so my question is does the loss of mass create an observable change in a comet's orbit? What kind of change that may be shrinking orbit or expanding orbit?&lt;/p&gt;&#xA;" OwnerUserId="14414" LastActivityDate="2016-09-30T23:40:24.597" Title="Does the loss of mass create an observable change in a comet's orbit?" Tags="&lt;orbit&gt;&lt;the-sun&gt;&lt;comets&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18540" PostTypeId="1" AcceptedAnswerId="18550" CreationDate="2016-09-30T22:02:58.033" Score="2" ViewCount="76" Body="&lt;p&gt;Earlier, I was looking for the current distance from Earth to 67P/Churyumov-Gerasimenko, in order to do a &quot;back of envelope&quot; calculation of signal delay. Not important, but I was somewhat puzzled to find quite a large discrepancy between values reported by different sites, and would love to understand why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ESA's Where is Rosetta page currently gives the distance as around 719m km, which, if I'm using it correctly, is backed up by &lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi?find_body=1&amp;amp;body_group=sb&amp;amp;sstr=67P%20[]&quot; rel=&quot;nofollow&quot;&gt;JPL Horizons&lt;/a&gt; tool, which gives me a delta of around 4.81 AU. Which is great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when I look at The Sky Live's &lt;a href=&quot;http://theskylive.com/67p-tracker&quot; rel=&quot;nofollow&quot;&gt;67P live tracker&lt;/a&gt;, it is reporting around 941m km, which their comet info page gives as around 6.3 AU. Am I missing something blindingly obvious?&lt;/p&gt;&#xA;" OwnerUserId="14415" LastActivityDate="2016-10-01T23:03:34.637" Title="Dfferent distances to comet 67P" Tags="&lt;distances&gt;&lt;67p&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18541" PostTypeId="1" AcceptedAnswerId="18928" CreationDate="2016-09-30T22:11:45.123" Score="3" ViewCount="132" Body="&lt;p&gt;I've seen in documentaries that if a Jupiter size planet migrates close to its star then it would remove terrestrial planets along the way. This makes sense and I'm sure most models would predict this, but do we have evidence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, the Kepler survey (taking Kepler 1-447) found 34 Planets&gt;6 R(Earth) less than 0.1 AU from the star and the only one where another planet was found was Kepler 424 which also has a Jupiter size planet at 0.73 AU. Three &quot;warm&quot; Jupiters (0.1- 1.7AU) out of 47 have earth-size planets discovered in same system and all are &quot;mini-solar systems&quot; with the smaller planets far interior. (K68,90 and 407). If you include &quot;super-earths&quot; (R = 1.25-2.0 R(Earth)), then there are 7 additional stars found with &quot;warm Jupiters&quot; in multiple systems. The super-earths are still all inside the Jupiter's (although K18,89 and 118 have Jupiters less than 0.2 AU).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My question is:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Is this finding proof that migrating Jupiters have thrown out&#xA;  terrestrial planets or do their existence make the signals of&#xA;  &quot;exterior&quot; planets hard to discern? And what about smaller planets&#xA;  migrating in &quot;behind&quot; the Jupiter's?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="14097" LastEditDate="2016-10-02T00:17:33.153" LastActivityDate="2016-11-03T18:07:19.710" Title="Hot and &quot;warm&quot; Jupiters expelling terrestrial planets?" Tags="&lt;kepler&gt;&lt;hot-jupiter&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18542" PostTypeId="2" ParentId="18539" CreationDate="2016-09-30T23:40:24.597" Score="2" Body="&lt;p&gt;This problem was studied in &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0256-307X/13/6/021/meta;jsessionid=EEAA7E4E254BA2B9226DD9D56C5C8404.c4.iopscience.cld.iop.org&quot; rel=&quot;nofollow&quot;&gt;Yu &amp;amp; Zheng (1995)&lt;/a&gt;, who evaluated the effects of the change in the Sun's mass over time and the change in a comet's mass over time, for the case of &lt;a href=&quot;https://en.wikipedia.org/wiki/Comet_Shoemaker%E2%80%93Levy_9&quot; rel=&quot;nofollow&quot;&gt;Shoemaker-Levy 9&lt;/a&gt;, which had recently crashed into Jupiter. Given their mass model for the comet (Equation 6), they found that the Sun's mass loss created an increase in semi-major axis of about 8.5 centimeters per year, whereas the comet's mass loss created an increase in semi-major axis of about 10,000 kilometers per year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several things to note:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The comet orbited Jupiter prior to breaking up, so it did not come as close to the Sun as most short-period comets.&lt;/li&gt;&#xA;&lt;li&gt;Mass loss rates can change over time, depending on the distance from the Sun.&lt;/li&gt;&#xA;&lt;li&gt;Shoemaker-Levy 9 should not be considered a normal comet, in any sense, given its orbit and eventual destruction.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;10,000 kilometers a year, though, is nothing to sniff at. Over the course of an orbit, that can be quite a lot - although keep in mind that longer orbits involve much larger semi-major axes - and I'd argue that it should be observable, given the correct calculations of how the orbit should evolve over time.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2016-09-30T23:40:24.597" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18543" PostTypeId="1" CreationDate="2016-10-01T04:27:16.420" Score="2" ViewCount="1950" Body="&lt;p&gt;It takes 8 minutes for the sun's light to reach earth. If the sun were to instantly disappear, what would happen? I imagine that Earth would fly away on a path tangential to its orbit. What would happen to our heat and light during those first 8 minutes after the sun's disappearance? &lt;/p&gt;&#xA;" OwnerUserId="14417" LastActivityDate="2016-10-01T15:41:28.603" Title="Sudden Effects of Sun Disappearing" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;astrophysics&gt;" AnswerCount="3" CommentCount="7" ClosedDate="2018-03-17T03:35:01.607" ContentLicense="CC BY-SA 3.0" />
  <row Id="18544" PostTypeId="2" ParentId="18543" CreationDate="2016-10-01T04:42:50.497" Score="0" Body="&lt;p&gt;It would take the same 8 minutes for the Earth's orbit to change as it does for us to stop receiving light from the Sun.  No signal travels faster than light.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-10-01T04:42:50.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18545" PostTypeId="2" ParentId="18543" CreationDate="2016-10-01T05:11:14.703" Score="0" Body="&lt;p&gt;Since the universal speed limit is the speed of light, for the first 8 minutes nothing would change. Gravity would be the same, our orbit, as you mentioned heat and light would all be the same. Nothing would change until the final signals from the sun hit our planet and ended.&lt;/p&gt;&#xA;" OwnerUserId="14333" LastActivityDate="2016-10-01T05:11:14.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18546" PostTypeId="1" AcceptedAnswerId="18547" CreationDate="2016-10-01T11:30:52.193" Score="2" ViewCount="1039" Body="&lt;p&gt;Including formed &amp;amp; forming planets, will the core of a planet always be very hot, or are there any planets with cold cores? &lt;/p&gt;&#xA;" OwnerUserId="11792" LastEditorUserId="2365" LastEditDate="2016-10-01T13:36:24.103" LastActivityDate="2016-10-01T13:55:32.073" Title="Will a planet's core always be very hot?" Tags="&lt;planet&gt;&lt;core&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18547" PostTypeId="2" ParentId="18546" CreationDate="2016-10-01T13:55:32.073" Score="4" Body="&lt;p&gt;We would expect all large bodies, such as planets, to have a hot interior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interior of a planet is heated through two main mechanisms: Heat released from gravitational energy as the planet is forms, and heat released from the decay of radioactive elements, particularly uranium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outer layers of the planet are a very good insulator, so the any heat is trapped in the interior for a long time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In gas giants, gravitational energy is the main source of heat, Saturn has an internal temperature of about 21000°C, Jupiter is higher, 35000°C in some models. Terrestrial planet's cores are heated by radioactive decay. Earth has a core temperature of about 6000°C. Smaller bodies have cooler temperatures. Dwarf planet Pluto may have a core temperature less than 1000°C, and asteroids would have reached thermal equilibrium, and have roughly the same temperature throughout. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No planet in the universe can be much more than about 10 billion years old, and this is not old enough for a planet like the Earth to have formed a completely cold core.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-10-01T13:55:32.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18548" PostTypeId="2" ParentId="18543" CreationDate="2016-10-01T15:41:28.603" Score="5" Body="&lt;p&gt;Let's assume that every particle in the Sun quantum tunneled far away from us (there's a near 0% chance this will happen, but let's just assume). For the first eight minutes, nothing would change from our point of view. Gravitational waves propagate at lightspeed, so we would continue to orbit the Sun and see it there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After eight minutes, however, the lights would go out. The Moon will no longer shine upon us (it receives its light from the Sun), so our only source of light will be that of the celestial objects. Mercury and Venus will have seemingly disappeared from our optical telescopes — they will no longer be illuminated by the Sun. It would take another 30 minutes to an hour for the reflected light of Jupiter to shut off, and soon all the objects in the Solar System will no longer be visible to optical telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.oocities.org/aa_institute1/ahads_constant_refereed_paper.pdf&quot; rel=&quot;noreferrer&quot;&gt;Ahad (2004)&lt;/a&gt;, the illumination by the rest of the Milky Way would be approximately 1/300 of that of a Full Moon. So we might be able to see around thanks to the night sky and our own man-made technologies. However, most plants will not be able to undergo photosynthesis and will die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, we would be traveling tangentially to our orbit around the Sun, as you mention. We would just keep going into the abyss; it's very unlikely the gravity of any object will pull us in. However, things will get quite cold. The Earth would radiate heat away at a moderate rate, and temperatures would be unsustainable in most parts of the world after a year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luckily, the Earth has a nice and toasty core that could provide life-sustaining heat in some places (like Yellowstone), so humans that migrate over there would likely be better off.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2016-10-01T15:41:28.603" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18549" PostTypeId="2" ParentId="10117" CreationDate="2016-10-01T20:33:49.633" Score="-3" Body="&lt;p&gt;The modern dark matter theory assumes that a spiral galaxy is a system of orbiting objects in the same way that a solar system is a system of orbiting objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is not the case.  The stars in an elliptical galaxy are not &lt;em&gt;in orbit&lt;/em&gt; around the galactic center.  They revolve around it, but they are not in an equilibrium state where they would be both falling into and coasting away from the central object at essentially the same speed.  All the stars are either falling in towards the center -- in a long spiral journey -- or are drifting out towards the intergalactic void in a long spiral journey. You could intuitively draw a circle on a photo of a spiral galaxy demarcating the border separating the ingoing from the outgoing stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has to be remembered that our galaxy, for instance, is roughly a billion times larger in diameter than what might be determined to be the diameter of our solar system.  The motion of the stars, radially and otherwise, is &lt;em&gt;relatively&lt;/em&gt; much, much slower.  Profoundly slower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our solar system, all the mass of the primordial rotating disk has long since either become part of the sun or has coasted outward.  The system is &quot;cleared out&quot; -- all except for a few planets whose very existence is close to being a miracle . A few lucky lumps and orbs, have, in the chaos of motion, somehow fallen into a highly improbable equilibrium state.  The sun has 98.5 per cent of all the mass in the solar system, and most of the rest is Jupiter, so the rest is really just debris.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, it takes a stupendous amount of time for a galaxy to clear out.  The stars would all have gone through their life cycle and would no longer be producing light.  We are watching matter inexorably falling in, and matter inexorably exiting (with maybe just a few stars ultimately ending up in an unlikely equilibrium state).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stars in a galaxy initially got their roughly equivalent momentum from interactive gravitational forces.  But the point is, when we observe a spiral galaxy, we are always looking at something that is just at the beginning of a process that has already run its course in a solar system.  There is no need for 'dark matter' at the periphery of galaxy, because there is nothing being held in orbit.&lt;/p&gt;&#xA;" OwnerUserId="14421" LastEditorUserId="14421" LastEditDate="2016-10-02T12:18:48.017" LastActivityDate="2016-10-02T12:18:48.017" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18550" PostTypeId="2" ParentId="18540" CreationDate="2016-10-01T23:03:34.637" Score="1" Body="&lt;p&gt;I suspect that The Sky Live got their comet IDs mixed up.&#xA;The &lt;a href=&quot;http://theskylive.com/67p-info#elements&quot; rel=&quot;nofollow&quot;&gt;orbital elements&lt;/a&gt; show perihelia in Aug 2015 and Jan 2022,&#xA;aphelion 5.69 AU, period 6.45 years, consistent with other sources e.g.&#xA;&lt;a href=&quot;http://www.minorplanetcenter.net/db_search/show_object?object_id=67P&quot; rel=&quot;nofollow&quot;&gt;Minor Planet Center&lt;/a&gt;.&#xA;However, the &lt;a href=&quot;http://theskylive.com/67p-info#earthdistance&quot; rel=&quot;nofollow&quot;&gt;distance chart&lt;/a&gt;&#xA;shows a perihelion in mid 2018, aphelion about 11 AU, period about 16 years,&#xA;clearly not the same object.&#xA;Report the bug and they'll probably fix it.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-10-01T23:03:34.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18551" PostTypeId="1" AcceptedAnswerId="18555" CreationDate="2016-10-01T23:07:05.633" Score="6" ViewCount="1027" Body="&lt;p&gt;When a comet loses all his volatile material after many orbits around the Sun, it becomes a &amp;quot;dead&amp;quot; comet (like an asteroid, a centaur or whatever), see for example &lt;a href=&quot;https://en.wikipedia.org/wiki/2015_TB145&quot; rel=&quot;nofollow noreferrer&quot;&gt;2015 TB145&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Is there any way to determine the lifetime of a comet (expressed as the number of orbits before it becomes a dead comet)?&lt;/p&gt;&#xA;" OwnerUserId="9949" LastEditorUserId="37191" LastEditorDisplayName="user1569" LastEditDate="2021-02-05T19:46:27.030" LastActivityDate="2021-02-05T19:46:27.030" Title="How long does a comet &quot;live&quot;?" Tags="&lt;orbit&gt;&lt;the-sun&gt;&lt;comets&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18553" PostTypeId="1" AcceptedAnswerId="18557" CreationDate="2016-10-02T03:57:16.367" Score="1" ViewCount="73" Body="&lt;p&gt;I'm reading a paper about a recent x-ray burst from a suspected magnetar (&lt;a href=&quot;https://arxiv.org/pdf/1607.04264v3.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;A. Dai et al 2016&lt;/a&gt;) where they show a light curve of a burst that lasted about 10ms. (Their figure 1). The value of count rate for the peak of the burst from the graph is $6\text{ counts/s}$ over a timespan of less than $0.01s$. This corresponds to about $6*0.01=0.06 \text{ counts}$ for that time period. What does this mean? Is it not counts of photons? What is 0.06 photons? Is there a step in the data analysis I'm missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/S5QCa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S5QCa.png&quot; alt=&quot;Swift/BAT mask-weighted light curves in different energy bands&quot;&gt;&lt;/a&gt;&#xA;(Fig. 1 Swift/BAT mask-weighted light curves in different energy bands)&lt;/p&gt;&#xA;" OwnerUserId="13724" LastActivityDate="2016-10-02T08:15:50.633" Title="What does less than one count from an x-ray detector mean? (Swift BAT detector)" Tags="&lt;photometry&gt;&lt;photons&gt;&lt;x-ray&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18555" PostTypeId="2" ParentId="18551" CreationDate="2016-10-02T05:10:07.043" Score="6" Body="&lt;p&gt;If you don't mind an oversimplified answer.   There is a relatively straight forward relationship.   The life of a comet (in orbits not years) depends on 3 factors.   It's size, it's composition and how close it passes to the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comets are often referred to as dirty snowballs but the ratio of different types of ice and how much &quot;dirt&quot; or silicate material varies from comet to comet.   &lt;a href=&quot;https://en.wikipedia.org/wiki/Comet_nucleus#Size&quot; rel=&quot;nofollow&quot;&gt;Average density is about 0.6&lt;/a&gt; g/cc.  That variation is a factor in how much light gets reflected right back off the comet, and how easily the surface melts, but to make this answer somewhat easier, I'm going to ignore composition and make a simplified assumption that all comets are made of basically ice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that in mind, how much a comet loses per pass around the sun can be simplified into loss of radius.   Think of a block of ice sitting in space, facing the sun.   The rate of sublimation would simply be a product of how close the ice was to the sun.  It's not far off to oversimplify melt rate in terms of loss of radius over a pass near the sun.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Halley's comet, for example &lt;a href=&quot;https://en.wikipedia.org/wiki/Halley&amp;#39;s_Comet#Orbit_and_origin&quot; rel=&quot;nofollow&quot;&gt;has a fairly close pass to the sun&lt;/a&gt;, at about 0.587 astronomical units (closer than Venus).  The closer a comet passes to the sun the faster it sublimates.   Halley's is expected to last another 100-125 orits.  &lt;a href=&quot;http://www.thenakedscientists.com/HTML/questions/question/1000112/&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned above, there's some variation in terms of the density and different materials that make up a comet, but the primary factors are &lt;a href=&quot;https://en.wikipedia.org/wiki/Perihelion_and_aphelion&quot; rel=&quot;nofollow&quot;&gt;Perihelion&lt;/a&gt; and size of the comet.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Halley's has a diameter of about 8 KM. (roughly speaking, it's 16x8x8 km, ellipsoid shape).   8,000 meters over 100-125 passes suggests it loses about 60-80 meters per pass around the sun and if we estimate that the &quot;dead comet&quot; after 100 or so orbits retains 1-3 KM of rocky material, the ballmark melt-rate of Halley's is somewhere in the 30-60 meters per pass around the sun, or a meter every couple-few days by sublimation.   A comet with a perihelion at .8 AU or 1 AU would lose less radius per pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A comet that passes within 0.3 AU instead of 0.6 might lose about 4 times as much per day as it's hit by 4 times the solar energy per square meter, but it spends fewer days due to a faster orbit near the sun.   Somebody might be able to run the math based on highly elliptical orbits, perihelion and solar energy per square meter over the entire pass.   I'm afraid to try myself, but it would be something less than the square of the distance due to the faster orbital speed for smaller perihelion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wit this estimate in mind, Wikipedia says a &lt;a href=&quot;https://en.wikipedia.org/wiki/Sungrazing_comet&quot; rel=&quot;nofollow&quot;&gt;sungrazing&lt;/a&gt; comet with a perihelion of about 0.055 AU and an initial radius of about 2-3 km could survive that pass and be about 1 km.   That would be a loss of 1,000-2,000 KM over the pass, but that's a very close pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that's the start of an answer to this question.  The closer to the sun, the more meters of radius the comet will lose per orbit and calculating the radius losing so many meters every pass gives a rough estimate of the life of the comet.   Sublimation can begin as far away as the orbit of Saturn, but at that distance the sublimation is very slow.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-10-03T06:17:27.763" LastActivityDate="2016-10-03T06:17:27.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18556" PostTypeId="2" ParentId="18529" CreationDate="2016-10-02T07:39:28.073" Score="5" Body="&lt;p&gt;Your question touches on a few points.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, yes, he was flippant, but the risk was super-low.   The simplest way to explain this is that nothing happens in CERN that doesn't happen all over the universe and in the upper atmosphere of the Earth or on the surface of all the moons, planets and stars every day.  For example, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Oh-My-God_Particle&quot; rel=&quot;noreferrer&quot;&gt;Oh My God particle&lt;/a&gt; was traveling with several million times the energy of the particle collisions inside CERN.   At those speeds, accelerating the particle further doesn't add much velocity, what they're basically doing is adding energy or mass to the particle and only slightly increasing it's velocity as it's speed gradually moves closer to the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes CERN useful, isn't that the high speed collisions don't happen anywhere else (they happen all the time all over the Universe).  What makes it useful is the million dollar equipment that zooms in on those collisions, recording what happens with the best cameras in the world.   We can't predict when and where a very fast moving cosmic particle will hit the Earth, but they know when and where the CERN particles are going to collide, because that's planned to happen in a specific place and time.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if those kinds of collisions happen all the time and no planet or star has ever been observed being eaten by a particle collision created black hole, then such events are, at the very least, extremely rare. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add to that, Hawking radiation dictates that black holes that small should evaporate before they can eat anything.   So, when Brian Cox laughs at that question, he has good reason to laugh, because it was never a real risk and it's more of a joke question than a genuine risk.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, when you say &quot;pressure&quot; to create black holes.   The pressure inside a black hole, or, I should say, pressure inside a Neutron Star about to turn into a black hole, or the pressure inside a collapsing core of a very large star about to create a black hole - those pressures are crazy high, but if we could re-create that pressure on earth (We can't), but if we could, it wouldn't be enough to make a black hole.   Pressure alone doesn't do it, you need mass (at last about 3 solar masses) to fit within a certain size and the less mass you have, the greater the required density (and greater pressure) is needed to achieve the necessary mass to schwarzchild radius ratio.  The pressure required inside the forming black hole actually decreases as the mass of the black hole increases.  (based on simple pressure equations and schwarzchild radii).   Equal pressure in a lab on Earth wouldn't accomplish it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes black holes theoretically possible is the small quantum size and possibility of extra dimensions and quantum rules that only apply to the very small, so it's a very different set of circumstances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Collisions in CERN are measured in Electron-volts.   It was theorized that if there were extra dimensions on the quantum level, then &lt;a href=&quot;https://en.wikipedia.org/wiki/Micro_black_hole&quot; rel=&quot;noreferrer&quot;&gt;micro-black holes&lt;/a&gt; might form at the energy of CERN's collisions (this would imply that black holes of this type form in Earth's upper atmosphere all the time, implying they're not dangerous).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But so far, no luck.  No micro-black-holes have been observed.  They should dissipate very rapidly but their decay streams should be recognizable so they should get noticed, but so far, none have been spotted and as a result, there's no evidence of extra dimensions.  That doesn't mean they're not there, and that micro-black-holes can't be created, but so far there's no evidence that they've been created up to about 13 trillion electron volts, the current peak of CERN's collision energy.   See &lt;a href=&quot;https://home.cern/about/physics/extra-dimensions-gravitons-and-tiny-black-holes&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that makes sense.  Corrections welcome.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-10-02T07:39:28.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18557" PostTypeId="2" ParentId="18553" CreationDate="2016-10-02T08:15:50.633" Score="4" Body="&lt;p&gt;I think there is a missing piece of information. The BAT is a coded mask telescope. The imaging is done by photons passing through a mask and falling onto an array of 32768 detectors. &lt;a href=&quot;http://swift.gsfc.nasa.gov/about_swift/bat_desc.html&quot; rel=&quot;nofollow&quot;&gt;http://swift.gsfc.nasa.gov/about_swift/bat_desc.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;mask-weighted&quot; light curve is produced after a complex ray tracing exercise using an estimate of the position of the source. Looking at some of the software specifications (eg &lt;a href=&quot;https://heasarc.gsfc.nasa.gov/ftools/caldb/help/batmasktaglc.html&quot; rel=&quot;nofollow&quot;&gt;https://heasarc.gsfc.nasa.gov/ftools/caldb/help/batmasktaglc.html&lt;/a&gt; ) it seems that the count rate is divided by the number of active detectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this case it looks like the detection rate was more like 200,000 counts/s in the peak 2ms of the event. There would be 400 detected counts, assuming there are no other corrections for sensitivity or vignetting. However,  I think there probably are - the mask must take out about 50% of these counts and there are gaps between the detectors. A background rate has also been subtracted. So overall I guess that the first point in the burst is due to around 100 &lt;em&gt;detected&lt;/em&gt; photons, and so on.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-10-02T08:15:50.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18559" PostTypeId="1" CreationDate="2016-10-02T15:53:22.993" Score="19" ViewCount="452" Body="&lt;p&gt;Thanks to the efforts of the &lt;a href=&quot;https://www.ligo.caltech.edu/page/about-aligo&quot;&gt;aLIGO&lt;/a&gt; team, gravitational wave astronomy is a reality. At the same time, neutrino detectors like &lt;a href=&quot;http://www.hyperk.org/&quot;&gt;Hyperkamiokande&lt;/a&gt; are becoming much more sensitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: what are the prospects for the pseudo-simultaneous detection of gravitational waves &lt;em&gt;and&lt;/em&gt; neutrinos from the same supernovae? What sort of stuff could we learn from such an event, both about supernovae and neutrinos? In particular, what are the prospects for estimating the neutrino mass? &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="6" LastEditDate="2016-10-04T13:32:27.877" LastActivityDate="2018-03-20T21:17:12.000" Title="Multi-messenger astronomy: what is the potential of simultaneous detection of gravitational waves and neutrinos from a supernova?" Tags="&lt;supernova&gt;&lt;gravitational-waves&gt;&lt;neutrinos&gt;&lt;interferometry&gt;" AnswerCount="1" CommentCount="19" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18560" PostTypeId="2" ParentId="13165" CreationDate="2016-10-02T16:56:09.540" Score="4" Body="&lt;p&gt;There are other techniques to construct what you're interested in than making a volume limited sample. What you're trying to construct is called a &quot;luminosity function,&quot; it's the distribution in luminosity normalized so that the area under the curve integrates to the volume density of stars. Constructing a volume limited sample is, perhaps the simplest method of addressing the problem @RobJeffries describes, known as &lt;A HREF=&quot;http://ned.ipac.caltech.edu/level5/March03/Teerikorpi/Teerikorpi2_2.html&quot; rel=&quot;nofollow&quot;&gt;Malmquist bias&lt;/A&gt;. Another technique, known as $1/V_{\mathrm{max}}$, can be summarized as binning by luminosity, then weighting each star by the true volume it could occupy and still be in the luminosity bin. If you have a minimum flux for stars in the sample, $F_{\mathrm{min}}$, no maximum flux, and a maximum distance $d_{\mathrm{max}}$, then the weighting for each star will be:$$w_i  = \frac{3}{\Omega \operatorname{min}\left(d_{\mathrm{max}}, \sqrt{\frac{L_i}{4\pi F_{\mathrm{min}}}}\right)^3\,\Delta L_i},$$ where $\Omega$ is the solid angle on the sky of the survey, $d_{\mathrm{max}}$ is the maximum distance allowed, $L_i$ is the luminosity of the individual star labeled with index $i$, and $\Delta L_i$ is the width of the luminosity bin star $i$ is in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll also need to figure out where your sample comes from. It is well known that the population of stars in the Milky Way &lt;A HREF=&quot;http://www.ifa.hawaii.edu/~barnes/ast626_95/pcmw.html&quot; rel=&quot;nofollow&quot;&gt;varies by location&lt;/A&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At present the galaxy is thought to contain two or three luminous populations (e.g. Wyse 1992). The thin disk and stellar halo correspond to Baade's Pop. I and II, respectively. Still under debate is the existence of a thick disk population which might correspond to the thick disks seen in some other disk galaxies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you limit your study to a single star cluster you can even discover its age. Constructing a Hertzsprung-Russell (HR) diagram, a bivariate distribution on which luminosity is along one axis, and measuring where the stars are pealing away from the main sequence is one way to &lt;A HREF=&quot;http://www.astronomynotes.com/evolutn/s9.htm&quot; rel=&quot;nofollow&quot;&gt;measure the age of a star cluster&lt;/A&gt;.&lt;/p&gt;&#xA;" OwnerUserId="14244" LastActivityDate="2016-10-02T16:56:09.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18561" PostTypeId="1" CreationDate="2016-10-02T18:45:41.450" Score="0" ViewCount="308" Body="&lt;p&gt;I'm looking for the distribution of spectral types in the &quot;giant&quot; star category. This category is defined as basically B and O main sequence stars and any star that is larger than main sequences outside of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first thought was that the M, K, G, F, and A type distribution were the same as they are in the main sequence but then I realized M has to be dropped, and then i'm not quite sure how K, G, F, and A types evolve and usually only Red Giants are mentioned and I couldn't find numbers for the giants so I'm stuck. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be clear, I want to know the percentage of giant stars that are in each of the spectral types. So if there are 1000 stars in a given area and 100 giant stars in that areas I only want to know the percentage of each spectral type of the stars in that 100.&lt;/p&gt;&#xA;" OwnerUserId="14084" LastEditorUserId="14084" LastEditDate="2016-10-02T20:20:13.670" LastActivityDate="2016-10-02T21:32:09.847" Title="Distribution of spectral types of &quot;giant&quot; stars" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18562" PostTypeId="2" ParentId="18561" CreationDate="2016-10-02T21:32:09.847" Score="3" Body="&lt;p&gt;OK, I see what you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I did was take the revised Hipparcos catalogue and select all stars closer than 50pc. At this distance there should be a &lt;em&gt;complete&lt;/em&gt; sample of the types of stars you were interested in. I can plot an absolute magnitude versus colour diagram for these 7096 stars (see below) and use this to select giants and OB main sequence stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do this by using data from chapter 2 of &lt;a href=&quot;http://ads.harvard.edu/books/hsaa/toc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zombeck (1992)&lt;/a&gt;, I define O and B main sequence stars to have $B-V&amp;lt;0$ and $M_V&amp;lt;5$ (to exclude hot white dwarfs). Evolved giants are selected (a bit arbitrarily) to lie significantly above the main sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MbmNu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MbmNu.png&quot; alt=&quot;Stars closer than 50pc&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I can plot a frequency histogram of the B-V values of those stars and label them with spectral types, again according to tables in Zombeck (1992).&#xA;The result is seen below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MsMqy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MsMqy.png&quot; alt=&quot;Distribution of &amp;quot;giant stars&amp;quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of these 235 stars: None are O stars, 29 are B stars, about 120 are G giants, about 80 are K giants and only 2-3 are M giants&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-10-02T21:32:09.847" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="18563" PostTypeId="1" CreationDate="2016-10-03T13:41:21.100" Score="9" ViewCount="9409" Body="&lt;p&gt;It has been some time since I've seen the Moon. What are the conditions that might affect whether or not the Moon is visible within a 24-hour period?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please take into consideration any notable astronomical effects: moon phases, light pollution, other viewing conditions. I want this to be applicable to other cases as well as my own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my particular case:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I live in Wales. The date today is October 3, 2016. There has been no moon visible in the sky for the last 5 days, and I don't know why.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14430" LastEditorUserId="6" LastEditDate="2016-10-06T20:45:01.537" LastActivityDate="2020-10-16T03:41:05.747" Title="Why haven't I seen the Moon?" Tags="&lt;observational-astronomy&gt;&lt;amateur-observing&gt;&lt;the-moon&gt;" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18564" PostTypeId="2" ParentId="12558" CreationDate="2016-10-03T19:50:06.733" Score="8" Body="&lt;p&gt;I'll shamelessly reference &lt;a href=&quot;https://worldbuilding.stackexchange.com/a/9957/627&quot;&gt;an answer I wrote on Worldbuilding&lt;/a&gt; to an almost identical question. &lt;a href=&quot;http://mnras.oxfordjournals.org/content/early/2014/02/17/mnras.stu085.full.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lammer et al. (2014)&lt;/a&gt; suggested that &quot;super-Earths&quot; with masses of $2$-$5M_\oplus$&lt;sup&gt;1&lt;/sup&gt; could retain massive hydrogen/helium envelopes, up to $\sim10^{25}$ kilograms. Above this, up to about $10M_\odot$ or more, &quot;mini-Neptunes&quot; exist, possibly composed of volatiles and having significant gas envelopes&lt;sup&gt;2&lt;/sup&gt;. In other words, there's a transition region between terrestrial planets and gaseous planets (gas dwarfs included) that depends significantly on whether the planet can hold onto an envelope for a significant amount of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of this depends on the environment. For instance, you cite &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler-10c&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kepler-10c&lt;/a&gt; as a possibly upper limit to terrestrial planets. The reason Kepler-10c can't be a gas planet, as &lt;a href=&quot;https://arxiv.org/abs/1405.7881&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dumusque et al. (2014)&lt;/a&gt; explain, is that it's way too close to its parent star to retain a hydrogen envelope. It would likely lose a lot of such an atmosphere via Jeans escape, the same mechanism that leads to &lt;a href=&quot;https://en.wikipedia.org/wiki/Chthonian_planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;chthonian planets&lt;/a&gt;. It's possible, then, that Kepler-10c would have been an ice giant if it had formed farther away from Kepler-10. It has a similar mass and potentially similar composition; it just happens to be much closer to its parent star than Neptune.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is something of an unsatisfying answer, but there simply isn't an exact line that separates terrestrial planets and gas planets, just like there isn't a good line between massive gas giants and, say, &lt;a href=&quot;https://en.wikipedia.org/wiki/Sub-brown_dwarf&quot; rel=&quot;nofollow noreferrer&quot;&gt;sub-brown dwarfs&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1 There are, of course, exceptions, like &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler-138&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kepler-138a&lt;/a&gt;, which has a mass similar to Earth yet is a good gas dwarf candidate.&lt;/sup&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;2 For more information on the transition region, see e.g. &lt;a href=&quot;https://arxiv.org/abs/1311.0329v1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lopez &amp;amp; Fortney (2013)&lt;/a&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:52:24.020" LastActivityDate="2016-10-03T19:50:06.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18565" PostTypeId="2" ParentId="11811" CreationDate="2016-10-03T20:49:23.003" Score="4" Body="&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/1603.01934&quot; rel=&quot;nofollow&quot;&gt;Beichman et al. (2016)&lt;/a&gt; used the Spitzer Space Telescope to view two systems, EPIC 202083828 (one planet) and the aforementioned EPIC 201367065 (three planets), also known as &lt;a href=&quot;https://en.wikipedia.org/wiki/K2-3&quot; rel=&quot;nofollow&quot;&gt;K2-3&lt;/a&gt;. The authors describe several ways in which the observations by &lt;em&gt;Spitzer&lt;/em&gt; complement the &lt;em&gt;Kepler&lt;/em&gt; data:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Spitzer&lt;/em&gt; has better sampling while maintaining the same signal-to-noise ratio, constraining various orbital parameters even more.&lt;/li&gt;&#xA;&lt;li&gt;Observations by &lt;em&gt;Spitzer&lt;/em&gt; occurred over twelve months after the &lt;em&gt;Kepler&lt;/em&gt; observations, allowing for more detailed information on the position and orbits of the planets over time.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Limb_darkening&quot; rel=&quot;nofollow&quot;&gt;Limb darkening&lt;/a&gt; is less of an issue in &lt;em&gt;Spitzer&lt;/em&gt;'s observations than it was in &lt;em&gt;Kepler&lt;/em&gt;'s data.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Spitzer&lt;/em&gt; can reject false transit positives by looking at how the transit depth changes over different wavelengths.&lt;/li&gt;&#xA;&lt;li&gt;There is less noise from the stellar photosphere in infrared bands.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All of this led to much lower uncertainties in parameter observations in &lt;em&gt;Spitzer&lt;/em&gt;'s data alone; combined with &lt;em&gt;Kepler&lt;/em&gt;'s observations, the uncertainties were reduced even further. Period estimates and transit depths, for instance, saw significant improvement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the authors recommend that the system be observed by the James Webb Space Telescope. That won't be launched for another two years, but I suspect its versatility over various wavelengths will better constrain data from EPIC 201367065 and other systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/1501.03798&quot; rel=&quot;nofollow&quot;&gt;Crossfield et al. (2015)&lt;/a&gt; say that the James Webb Space Telescope has the capability to detect spectral signs in the atmospheres of the planets, based on certain models (&lt;em&gt;Hubble&lt;/em&gt; could also observe them). I don't think JWST has been assigned specific targets that include this system, but that's just a guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the first system measured by Beichman et al., EPIC 202083828, ground-based telescopes could also give more information about spectroscopy to rival JWST, according to &lt;a href=&quot;https://arxiv.org/abs/1601.02706&quot; rel=&quot;nofollow&quot;&gt;Schlieder et al. (2016)&lt;/a&gt;. Perhaps EPIC 201367065 could also be observed by these telescopes. &lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-10-10T13:32:52.607" LastActivityDate="2016-10-10T13:32:52.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18566" PostTypeId="2" ParentId="11106" CreationDate="2016-10-03T21:15:49.343" Score="2" Body="&lt;p&gt;I believe &lt;a href=&quot;https://en.wikipedia.org/wiki/WASP-12b&quot; rel=&quot;nofollow noreferrer&quot;&gt;WASP-12b&lt;/a&gt;, a hot Jupiter, is a good example of what you're looking for. &lt;a href=&quot;https://arxiv.org/pdf/1002.4608v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Li et al. (2010)&lt;/a&gt; first predicted that there should be gas transfer - easy, because the planet has a semi-major axis of only about 3 times the radius of WASP-12. Aside from the mass loss, extreme temperatures and distortion from tidal heating make the planet stand out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gas should be transferred into a small disk around the planet, which could reach temperatures of 3,000-4,000 K. Eventually, there should be significant angular momentum transfer between the star/disk and the planet. However, the authors note that certain assumptions of their model mean that this angular momentum transfer may not be as significant. There is also the possibility that a super-Earth in the disk could influence the interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nasa.gov/mission_pages/hubble/science/planet-eater.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Follow-up observations by Hubble&lt;/a&gt; showed that there is in fact mass transfer between WASP-12b and the star, with gas reaching peak velocities of around 100-200 km/s.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The mass loss rate $\dot{M}$ should be approximately&#xA;$$\dot{M}\sim\frac{\dot{E}_tR_p}{2GM_p}$$&#xA;where $R_p$ and $M_p$ are the radius and mass of the planet, and $\dot{E}_t$ is the tidal heating rate, given as&#xA;$$\dot{E}_t\sim\frac{e^2GM_pM_*}{a\tau_e}$$&#xA;for stellar mass $M_*$, semi-major axie $a$ and characteristic eccentricity damping timescale $\tau_e$.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-11-03T17:44:40.387" LastActivityDate="2016-11-03T17:44:40.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18567" PostTypeId="1" CreationDate="2016-10-03T21:40:11.350" Score="8" ViewCount="356" Body="&lt;p&gt;While trying to do some calculations to answer &lt;a href=&quot;https://astronomy.stackexchange.com/q/14475/10172&quot;&gt;this quesion&lt;/a&gt;, I got stuck missing a crucial piece of data: I have no clue how much mass the volcanoes of Io manage to throw out of the gravity well of the Jupiter moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that &lt;em&gt;most&lt;/em&gt; of the erupted material falls back to the surface, and the various velocity estimates I have found are quite a bit lower than the escape velocity (~2.56 km/s). For instance &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Volcanology_of_Io&quot; rel=&quot;nofollow noreferrer&quot;&gt;up to 1 km/s&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One number that is possible to find is how much mass the magnetosphere of Jupiter strips of Io. Most are listing that as 1000kg/s (&lt;a href=&quot;http://www.astro.if.ufrgs.br/solar/io.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;example&lt;/a&gt;). That is about as much as the wheat production of Canada.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the volcanoes directly throwing away a lot more than that? A lot less? Not at all?&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-10-19T14:27:11.307" Title="How much mass do the volcanoes of Io erupt past escape velocity?" Tags="&lt;jupiter&gt;&lt;geology&gt;&lt;volcanism&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18568" PostTypeId="2" ParentId="18534" CreationDate="2016-10-03T22:10:54.913" Score="5" Body="&lt;p&gt;The data is indeed very uncertain. As for GW150914, &lt;a href=&quot;https://www.ligo.caltech.edu/image/ligo20160211b&quot; rel=&quot;noreferrer&quot;&gt;LIGO posted&lt;/a&gt; this image of the probable area of the sky:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/j6EZsm.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j6EZsm.jpg&quot; alt=&quot;Black hole merge event location&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most probable banana-shaped region stretches from Crux to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Large_Magellanic_Cloud&quot; rel=&quot;noreferrer&quot;&gt;Large Magellanic cloud&lt;/a&gt;. Comparing the image with a constellation map, there are several small constellations that may have hosted the event. Those are &lt;a href=&quot;https://en.wikipedia.org/wiki/Mensa_(constellation)&quot; rel=&quot;noreferrer&quot;&gt;Mensa&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Chamaeleon&quot; rel=&quot;noreferrer&quot;&gt;Chamaeleon&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Musca&quot; rel=&quot;noreferrer&quot;&gt;Musca&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Crux&quot; rel=&quot;noreferrer&quot;&gt;Crux&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastActivityDate="2016-10-03T22:10:54.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18569" PostTypeId="2" ParentId="13615" CreationDate="2016-10-04T03:24:55.327" Score="6" Body="&lt;p&gt;The main obstacle with the Oscillating Universe model is the pesky second law of thermodynamics — that entropy always increases within an isolated system, and never the reverse. Thus, as time progresses, entropy increases, until it (as far as we know) reaches its maximum — the heat death or Big Rip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the Big Bang, when entropy was at its minimum, or at least incredibly low. As the Universe evolved, entropy increased, as dictated by the second law. If the Big Crunch were to happen, we'd see a &lt;em&gt;decrease&lt;/em&gt; in entropy as conditions in the Universe recede back to how they were at $t = 0$. As we've stated, the Universe's entropy can never decrease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But say we ignore the whole &quot;Big Crunch problem&quot; and instead focus on the cycles. As stated before, the entropy always increases with time. I'll let &lt;a href=&quot;https://arxiv.org/abs/hep-th/0610213&quot; rel=&quot;nofollow&quot;&gt;Baum &amp;amp; Frampton (2006)&lt;/a&gt; explain the problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...one principal obstacle is the second law of&#xA;  thermodynamics which dictates that the entropy increases from cycle to cycle. If the cycles thereby become longer, extrapolation into the past will lead back to an initial singularity again, thus removing the motivation to consider an oscillatory universe in the first place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As they described, the entropy in the Universe would increase with each cycle. At this point we realize that each cycle in the Universe's history would be different — and if we extrapolated back into the past, we would find that at some point, entropy must have been at its minimum (the Big Bang!) The fact that we would be back to the Big Bang would remove any need for an Oscillating Universe hypothesis — which tries to explain what preceded the Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That problem led many astrophysicists to abandon the Oscillating Universe hypothesis. But nonetheless, the idea that the Universe is reborn is quite attractive, and we still know very little about cosmology — including dark energy, which we think has a vital role in the Universe. Some cosmologists continue to research the Oscillating Universe model in hopes that, as we learn more about dark energy and the Universe, we may be able to avoid the problems with the second law of thermodynamics.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-10-24T15:56:43.873" LastActivityDate="2016-10-24T15:56:43.873" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18570" PostTypeId="2" ParentId="12273" CreationDate="2016-10-04T06:18:16.500" Score="3" Body="&lt;p&gt;You're right, Mars' moons likely did not form with from the same dust mass as it; their spectra, albedo, and density are suspiciously similar to those of C- or D-type asteroids. Thus, Mars presumably captured Phobos and Deimos from the asteroid belt. However, the origin of the Martian moons is quite controversial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1992mars.book.1283B&quot; rel=&quot;nofollow noreferrer&quot;&gt;Burns (1992)&lt;/a&gt; notes that capture requires the loss of energy (and momentum, as you pointed out). So obviously, something would need to dissipate that energy. The immediate candidate is &lt;a href=&quot;https://en.wikipedia.org/wiki/Aerobraking&quot; rel=&quot;nofollow noreferrer&quot;&gt;aerobraking&lt;/a&gt;, but we run into a problem: Mars' atmosphere is too thin to capture a Phobos-sized object through aerobraking. In fact, even if aerobraking could capture Phobos and Deimos, the densities of both moons indicate that they may not survive structurally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several hypotheses have been proposed to answer this. &lt;a href=&quot;https://arxiv.org/pdf/0903.3434.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Landis (2001)&lt;/a&gt; suggests tidal forces are to credit:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The existence of asteroid moons provides a mechanism for the capture of the Martian moons (and the small moons of the outer planets). When a binary asteroid makes a close approach to a planet, tidal forces can strip the moon from the asteroid. Depending on the phasing, the asteroid can then be captured. Clearly, the same process can be used to explain the origin of any of the small moons in the solar system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to Geoffrey Landis, Deimos and Phobos could have each been moons of parent asteroids, and they were separated due to tidal forces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KqtCF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KqtCF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would explain how Phobos and Deimos could survive as moons. I could mention the other hypotheses, but Landis' would answer your question quite well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now, there isn't any clear consensus as to how Phobos and Deimos became moons. More research is still necessary, and we may need to modify our models for the evolution of Mars and the asteroid belt.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-10-04T13:23:47.053" LastActivityDate="2016-10-04T13:23:47.053" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18571" PostTypeId="2" ParentId="16390" CreationDate="2016-10-04T06:39:51.917" Score="3" Body="&lt;p&gt;There doesn't seem to be such a word. Interstellar space within the Solar System is still just interstellar space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There doesn't seem to be a demand for the word you're looking for either. We distinguish space by its contents; the space within the heliosphere is called the interplanetary medium (it contains solar plasma, dust, etc.), while the interstellar medium comprises sparse gaseous molecules from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Interstellar_medium&quot; rel=&quot;nofollow noreferrer&quot;&gt;Local Fluff&lt;/a&gt; and (if you go far enough) the rest of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Local_Bubble&quot; rel=&quot;nofollow noreferrer&quot;&gt;Local Bubble&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas there are about 10 oxygen molecules per cubic centimeter in the interplanetary medium, there is only 1 oxygen molecule per cubic centimeter in the interstellar medium. Seeing that there is no clear boundary for the Milky Way (nor is there any single definition for a boundary for galaxies), the intergalactic medium just vaguely describes the medium within &lt;a href=&quot;https://en.wikipedia.org/wiki/Galaxy_filament&quot; rel=&quot;nofollow noreferrer&quot;&gt;galaxy filaments&lt;/a&gt;; here, the space is so rarified that there is only 1 oxygen molecule per cubic &lt;strong&gt;meter&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2017-01-03T13:41:21.977" LastActivityDate="2017-01-03T13:41:21.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18573" PostTypeId="1" AcceptedAnswerId="18577" CreationDate="2016-10-04T07:42:23.763" Score="29" ViewCount="3491" Body="&lt;p&gt;Can a planet be tectonically inactive and still retain a magnetosphere and protected atmosphere? How does that work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How else could a planet retain a thick atmosphere like Earth's for extensive periods of time? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, I'm not asking on Earth science SE as, well, the question is not about Earth. I was hoping there was some sort of astronomical theory for &lt;em&gt;other&lt;/em&gt; potential habitable planets.&lt;/p&gt;&#xA;" OwnerUserId="13680" LastActivityDate="2016-10-04T21:23:40.927" Title="Can a tectonically inactive planet retain a long-term atmosphere?" Tags="&lt;planetary-atmosphere&gt;&lt;magnetic-field&gt;&lt;earth-like-planet&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18574" PostTypeId="1" AcceptedAnswerId="18582" CreationDate="2016-10-04T08:29:01.680" Score="3" ViewCount="226" Body="&lt;p&gt;So me and my friend were planning a video to explain the Drake Equation (within a time limit of 5 minutes), and we needed some help. This video is aimed at explaining the concept to an age group between 13-18 years, and having gone through loads of articles, we felt that a lot of the common audience would not be able to comprehend the concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So our question was, can we take the Drake Equation and try to explain it with maybe a real life example and on a smaller scale? It is an incredibly interesting equation and we felt like doing a good job of explaining it to a common teenager.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="14435" LastEditorUserId="7982" LastEditDate="2021-02-19T01:25:57.873" LastActivityDate="2021-02-19T01:25:57.873" Title="Explaining the Drake Equation on a smaller scale" Tags="&lt;solar-system&gt;&lt;milky-way&gt;&lt;life&gt;&lt;astrobiology&gt;&lt;drake-equation&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18577" PostTypeId="2" ParentId="18573" CreationDate="2016-10-04T13:26:17.960" Score="42" Body="&lt;h3&gt;Yes, a tectonically inactive planet can retain a long-term atmosphere.&lt;/h3&gt;&#xA;&lt;p&gt;You make the connection that a lack of plate tectonics on a planet indicates a &amp;quot;dead&amp;quot; core and thus said planet has no magnetosphere. As such, I'm going to interpret your question as, can a planet without a magnetosphere retain an atmosphere long-term? As proof, I offer up Venus.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Venus' Magnetic Field&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Venus is a planet without a magnetosphere generated by a core. It is thought that the cause for this is Venus' slow rotation rate (of nearly 243 days) and a lack of convection, allowing for bulk motion in the core. As I'm sure you know, you need moving charge to make magnetic fields and Venus' core just isn't moving. As such, we see that Venus is a tectonically dead planet - it's surface is roughly 500 million years old, whereas Earth's surface gets recycled every 100 million years or less due to our plate tectonics.&lt;/p&gt;&#xA;&lt;p&gt;Now, Venus is not entirely devoid of a magnetic field. Ironically enough, it's lack of a magnetosphere allows for the generation of a magnetic field by it's atmosphere. Because the Sun's radiation is more or less directly hitting the atmosphere, Venus has a strong ionosphere. When you get lots of charged particles moving around in an atmosphere, you get a magnetic field. But on the whole, this field is very, very weak compared to a true magnetosphere such as we have on Earth.&lt;/p&gt;&#xA;&lt;p&gt;I found &lt;a href=&quot;http://www2.ess.ucla.edu/%7Enimmo/website/paper25.pdf&quot; rel=&quot;noreferrer&quot;&gt;this source&lt;/a&gt; which talks a lot about this concept and why Venus doesn't have a magnetosphere. Check it out to get a lot more in depth detail.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Venus' Atmosphere&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;So, Venus has no appreciable magnetosphere (or plate tectonics). Why does it have an atmosphere? And boy does it have an atmosphere. The surface pressure on Venus is estimated to be &lt;span class=&quot;math-container&quot;&gt;$\sim93\:\mathrm{atm}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;In short, the answer is that the inundation of solar wind against an atmosphere is not necessarily the major contributing factor to atmospheric loss. It can be, but not always. For example, Mercury, another planet with a weak (but non-zero) magnetosphere, has no atmosphere (if it ever did) because it is so close to the Sun that the solar wind likely blew away that atmosphere long ago. Venus on the other hand is far enough away that the solar wind just can't strip the atmosphere. Here I'm going to quote &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_escape#Significance_of_solar_winds&quot; rel=&quot;noreferrer&quot;&gt;wikipedia&lt;/a&gt; directly (emphasis mine).&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A lack of magnetic field does not determine the fate of a planet's atmosphere. Venus, for instance, has no powerful magnetic field. Its close proximity to the Sun also increases the speed and number of particles, and would presumably cause the atmosphere to be stripped almost entirely, much like that of Mars. Despite this, the atmosphere of Venus is two orders of magnitudes denser than Earth's. Recent models indicate that stripping by &lt;strong&gt;solar wind accounts for less than 1/3 of total non-thermal loss processes.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Atmospheric Loss&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;If the solar wind isn't the contributing factor in atmospheric loss, what is? The answer to that is a process known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_escape#Thermal_escape_mechanisms&quot; rel=&quot;noreferrer&quot;&gt;Jean's Escape&lt;/a&gt;. To put it simply, for gas particles in the atmosphere to escape into space, they need enough energy to climb out of the planet's gravity well. Some particles will have that energy and thus escape into space. Over time, the atmosphere bleeds off, little by little (this is happening for Earth too!).&lt;/p&gt;&#xA;&lt;p&gt;The factors that contribute to the rate at which a planet loses its atmosphere are such things as the the planet's mass and radius, and the mass of the atmospheric particles. Let's look at Venus. It is comparable in mass and size to Earth and so has a reasonably appreciable gravity well. For something to escape Venus it must be traveling at &lt;span class=&quot;math-container&quot;&gt;$10.4\:\mathrm{km/s}$&lt;/span&gt; (compared to the Earth's &lt;span class=&quot;math-container&quot;&gt;$11.2\:\mathrm{km/s}$&lt;/span&gt;). But, for Venus at least, the important factor there is that the atoms and molecules in its atmosphere are heavy. It is almost entirely (&lt;span class=&quot;math-container&quot;&gt;$\sim97\%$&lt;/span&gt;) carbon dioxide which has a mass of &lt;span class=&quot;math-container&quot;&gt;$\sim44\:\mathrm{amu}$&lt;/span&gt;. That means, the chances of such a massive particle getting the energy to escape is pretty small.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Out-gassing&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Just one more point to add to this. One may argue that possibly the atmosphere is/can be replenished continuously, but that won't work here because we're assuming the planet is tectonically dead. You can't really have out-gassing on a planet with no active surface.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;There are many factors that determine atmospheric escape. Different planets will lose their atmospherics for different reasons. However, it is entirely possible for a planet, under the right conditions, to maintain an atmosphere long-term, despite lacking a global magnetosphere. As we can see from Venus, the conditions are generally that the planet should be sufficiently far from the star, its atmosphere should be sufficiently dense and comprised of heavy particles, and the planet itself should be large enough to have an appreciable gravity well. If these conditions are all met, a planet may retain an atmosphere without having a magnetosphere to protect it.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-10-04T21:23:40.927" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="18579" PostTypeId="2" ParentId="18574" CreationDate="2016-10-04T14:08:33.090" Score="2" Body="&lt;p&gt;As the Drake identity (it's not an equation) is just a trivial exercise in combinatorics, I'd suggest the simplest, most commonly used model in combinatorics: The urn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a number N of balls in an urn. Those represent the stars in the galaxy. Of those only a fraction is green, the rest is red. Green signifies &quot;has planet&quot;, red &quot;doesn't have a planet&quot;.&#xA;You take the fraction of green ones that then host a planet in it's habitable zone and so on, for every characteristic the Drake identity describes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end you just count how many balls with all desired characteristics on them you've taken out of the urn, relative to the total number of balls in the urn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you write the corresponding fractions in your video side-by-side with the undesired balls disappearing, it should increase the understandability further.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2016-10-04T14:08:33.090" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18580" PostTypeId="2" ParentId="13709" CreationDate="2016-10-04T15:04:09.740" Score="6" Body="&lt;p&gt;You are specifically talking about absolute &lt;em&gt;visual&lt;/em&gt; magnitude; how bright an object would appear to our eyes if the object were 10 parsecs away from us. However, you must remember that stars emit a wide spectrum of light — much of it isn't visible to the naked eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luminosity is a measure of the total amount of energy given off by a star (usually as light) in a certain amount of time. Thus, luminosity includes both visible light and invisible light emitted by a star. So there isn't a precise conversion between luminosity and absolute visual magnitude, although there is an approximation we can do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Absolute &lt;em&gt;bolometric&lt;/em&gt; magnitude is an analogous magnitude system considering all the energy emitted by the star. The formula relating absolute bolometric magnitude with luminosity is as follows:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$L_\text{star} = L_0 10^{-0.4 M_\text{Bol}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$L_\text{star}$&lt;/span&gt; is the star's luminosity, &lt;span class=&quot;math-container&quot;&gt;$M_{\text{Bol}}$&lt;/span&gt; is the bolometric magnitude of the star, and &lt;span class=&quot;math-container&quot;&gt;$L_{0}$&lt;/span&gt; is the zero-point luminosity (the luminosity of a star with &lt;span class=&quot;math-container&quot;&gt;$M_{\text{Bol}} = 0$&lt;/span&gt;) arbitrarily defined as &lt;span class=&quot;math-container&quot;&gt;$L_{0}=3.0128×10^{28}$&lt;/span&gt; watts by the &lt;a href=&quot;http://www.iau.org/news/announcements/detail/ann15023/&quot; rel=&quot;nofollow noreferrer&quot;&gt;IAU&lt;/a&gt;. We can approximately convert between the two magnitudes via a &lt;a href=&quot;https://en.wikipedia.org/wiki/Bolometric_correction&quot; rel=&quot;nofollow noreferrer&quot;&gt;bolometric correction&lt;/a&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$M_{\text{bol}}\approx M_V + BC$$&lt;/span&gt;&#xA;The bolometric correction term &lt;span class=&quot;math-container&quot;&gt;$BC$&lt;/span&gt; is empirically determined for the spectral class and evolutionary stage (although it can be modeled; see &lt;a href=&quot;https://arxiv.org/abs/1008.3913v1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Torres (2010)&lt;/a&gt;). A table of values is given on the Wikipedia page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is that stars of the same spectral class and evolutionary stage have a similar distribution of wavelengths, so the discrepancy between all emitted wavelengths (&lt;span class=&quot;math-container&quot;&gt;$M_{\text{bol}}$&lt;/span&gt;) and visible wavelengths (&lt;span class=&quot;math-container&quot;&gt;$M_V$&lt;/span&gt;) should be similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the conversion between luminosity and absolute visual magnitude is approximately&#xA;&lt;span class=&quot;math-container&quot;&gt;$$L_\text{star} \approx L_0 10^{-0.4 (M_V + BC)}$$&lt;/span&gt;&#xA;Of course, this is only an approximation, but it can give us a pretty decent idea of what to expect. Trying it on &lt;a href=&quot;https://en.wikipedia.org/wiki/9_Pegasi&quot; rel=&quot;nofollow noreferrer&quot;&gt;9 Pegasi&lt;/a&gt;, I predicted it would be around &lt;span class=&quot;math-container&quot;&gt;$1000 L_\odot$&lt;/span&gt;, which is still in the ballpark of the actual value &lt;span class=&quot;math-container&quot;&gt;$1950 L_\odot$&lt;/span&gt;. Mind you, the approximation would be better suited for less extreme stars than 9 Pegasi.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2019-04-28T21:29:27.850" LastActivityDate="2019-04-28T21:29:27.850" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18581" PostTypeId="2" ParentId="18529" CreationDate="2016-10-04T15:15:46.430" Score="1" Body="&lt;p&gt;While it's not recommended, your comments cover enough ground that rather than make a long answer, I'll address them here.   If my answer is unsatisfactory, you might want to create a new question, clarifying the specifics.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A star 3 or 5 times the size of our sun will gradually burn up its&#xA;  hydrogen and collapse to form a black hole big enough to swallow the&#xA;  earth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stars don't operate that way.  During the final stages, (red giant, helium flash, etc), stars lose a significant percentage of their matter into space.   &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_evolution#White_and_black_dwarfs&quot; rel=&quot;nofollow noreferrer&quot;&gt;Our sun is expected to lose about half&lt;/a&gt; it's mass by the time it becomes a white dwarf.   larger stars lose an even higher percentage so you need a much more massive star to end up with a black hole. About &lt;a href=&quot;http://hubblesite.org/explore_astronomy/black_holes/encyc_mod3_q12.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;8 solar masses&lt;/a&gt; to end up a neutron star and about 25 solar masses (same link) for a black hole, though a neutron star can accrue mass and become a black hole that way too if it's in a binary system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that's not precisely what you asked, but one of the problems with compression on this scale is you have to ask what you are compressing.   Compressing fissionable materials can lead to &lt;a href=&quot;https://en.wikipedia.org/wiki/Type_Ia_supernova&quot; rel=&quot;nofollow noreferrer&quot;&gt;explosive events&lt;/a&gt; that would resist the compression.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use iron, which won't create any new energy upon being very tightly compacted, then you have a relatively straightforward mass to size problem where you can estimate the pressure. 3 solar masses of iron should be sufficient to form a black hole by its own mass and gravity. It might even be a bit less than that, as it's never been observed and would vary some with speed of rotation and temperature. But &lt;a href=&quot;https://en.wikipedia.org/wiki/Type_Ia_supernova&quot; rel=&quot;nofollow noreferrer&quot;&gt;3.0 solar masses is close enough&lt;/a&gt;.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The force of gravity needed to crush the star to a singularity is $10^x$ joules&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This isn't straightforward because it would vary with the initial mass you wanted to force into a black hole and the math gets very complex, involving equations based on the Pauli exclusion principle and neutron degeneracy pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, in general, with enough mass, no force is needed, because the gravity does all the work itself. There is no energy expenditure with 3 solar masses of iron. There's a huge (enormous) gravitational force, but no energy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With less mass, the force increases because a smaller Schwarzschild radius requires the Neutrons and quarks to get even closer, and the energy of such compression is enormous and hard to calculate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately, there is a way to cheat and get a very simplified answer. The degenerate pressure that keeps the neutron star from becoming a black hole is in balance with the gravitational binding energy and gravitational binding energy can be estimated, assuming uniform density,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U=\frac{3GM^2}{5R}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;https://physics.stackexchange.com/questions/55487/gravitational-potential-of-a-sphere-vs-gravitational-binding-energy-of-a-sphere&quot;&gt;Source&lt;/a&gt;).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's take a millionth of a gram of stuff, something heavy, like a piece of iron dust to avoid any fission energy that would work against black hole formation, and use our handy &lt;a href=&quot;http://xaonon.dyndns.org/hawking/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hawking radiation calculator&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A one millionth of a gram black hole would have a radius of about $1.5 \times 10^{-36}$ meters (calculator above), so the gravitational binding energy, roughly equivalent to the neutron degeneracy pressure at Schwarzchild radius: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{3 \times 6.67 \times 10^{-11} \times 10^{-9}}{5 \times 1.5 \times 10^{-36}} = 2.67 \times 10^{15}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or 26.7 million billion joules, crammed into a space smaller than an atom, just to make a 1 millionth of a gram black hole that should evaporate in a fraction of a fraction of a second.    Now, there's probably a thousand reasons why my estimate is wrong, but it's ballpark (a very big, Texas-sized ballpark), but still in the ballpark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your estimate for CERN is that it uses 800 million watts. Now, remember, CERN accelerates beams of protons, thousands if not millions, and at current peak energy, each individual proton carries a maximum of about 13 trillion electron volts, and there are about $6.24 \times 10^{18}$ electron volts in a joule, so each individual CERN collision, proton onto proton has about 1/48,000 joules, at maximum power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, even with my Texas-sized ballpark estimate, 1/48,000 is a much smaller number than 26.7 million billion. When the numbers are that far off, &quot;bollocks&quot; kinda covers it.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And a millionth of a gram black hole would evaporate anyway. It would be much smaller than an atom and would have a hard time absorbing much of anything. Not to mention the creation of such an object would probably have an initial velocity well above escape velocity and it would simply fly away from the earth (or right through it and away - like a heavy neutrino).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no estimate that creates a dangerous scenario using CERN's 1/48,000 joules per collision energy, and for an individual collision, that's enormous, but it's nowhere close to dangerous, unless you make the mistake of standing in the beam (that happened to somebody once, but...I digress). He lived by the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CERN can't put anywhere close to all 800 megawatts of energy into the acceleration of one proton. If they could, we might discover some very interesting things, but they can't even come close. All that energy powers the magnets and the vacuum and keeps everything cool, but the individual proton energy is a tiny fraction of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, depending on what size black hole you want to create, CERN would need to be, oh, maybe a billion billion billion times more powerful than it currently is, and even so, the black hole created would probably evaporate before it ate anything, or, if Hawking radiation is wrong and it doesn't evaporate it would most likely fly harmlessly through the Earth, or, mostly harmlessly as it might do some harm if it flew through a person.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To create a black hole of low enough temperature to actually be stable, (and theoretically eat the Earth), you'd need, by the Hawking radiation calculator, millions of tons. And clearly we're deep into the impossible if you think CERN can crush millions of tons of iron into a speck the size of an atom. The orders of magnitude of improbability are higher than I can calculate.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, as to the flip side of this question and why it was thought that CERN might be able to create a black hole (so far they haven't), is because quantum black holes (theoretically) obey different rules over very short distances. They wouldn't last long and they wouldn't be dangerous. If quantum black holes can be created in CERN, then they are being created all the time by cosmic rays, on the surface of the moon, in the upper atmosphere of Earth. CERN does nothing that doesn't happen in the upper atmosphere all the time.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to actually turn a speck of matter like a grain of sand into a black hole, the energy required would be billions and billions of times over what CERN is capable of. I can't give you an exact answer because I don't have the training in quantum calculations, and I'm not sure anyone can, or if they can, it will be very complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, &quot;bollocks&quot;, covers it.  But he could have said &quot;yes, that could happen, if CERN were a billion billion billion billion times more powerful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's much easier ways to kill everyone on Earth than making a black hole that eats the Earth. In fact, that might be the hardest way to kill everyone ever thought up.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope you don't mind my narrative, but I think the gravitational binding energy approach isn't a bad way to answer your specific questions on energy required.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-10-04T16:12:21.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18582" PostTypeId="2" ParentId="18574" CreationDate="2016-10-04T16:20:50.803" Score="4" Body="&lt;p&gt;No need to make it complicated: what about this...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just &lt;strong&gt;scribble a rectangle&lt;/strong&gt; on a piece of paper, and say &quot;there are 100 billion stars in our galaxy&quot;....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, color off (let's say) 1/3 of the rectangle, and say &quot;only one third of those are the sort of star that could have life, so that's blah billion&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, color off (say) 9/10ths of &lt;em&gt;that&lt;/em&gt; box, and say &quot;we believe about 90% of those have planets - so that's blah billion&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, color off (say) 1/20th of &lt;em&gt;that&lt;/em&gt; box, and say &quot;of those with planets, it seems that about 1 in 20 have Earth-like planets. Now we're down to blah billion...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: the Drake equation has a number of fairly silly terms relating to &quot;nuclear war!&quot;, which were added as political sops in that era; suggest ignore these unless you want to sound 90 years old!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So just scribble a box or draw a line on a piece of paper ...  or maybe use &quot;a bag of marbles&quot; as the other answer suggests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just BTW there &lt;strong&gt;is in fact&lt;/strong&gt; an entire documentary (I noticed it on &quot;Netflix&quot;) called &quot;The Drake Equation&quot; which does exactly what you say...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dzBJe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dzBJe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.. it is not really very good as I remember.  I think the guy simply draws a line in the ground, to do the &quot;fractions&quot; demo, you know?  (ie, they just erase more and more of the line). It doesn't need to be more complicated than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It's worth noting that the Drake equation simply points out:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) if you multiply those three or four fractions together, you get the number of civilizations in the galaxy. Which is self-evident.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but, the whole point is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) we have &lt;strong&gt;utterly no clue - not even vaguely&lt;/strong&gt; - what most of the fractions are,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could say it's a written formula, which, helps clarify our thinking on, something we are utterly clueless about. So rather than just vaguely saying &quot;we're utterly clueless,&quot; we can speak more clearly about the nature of our cluelessness!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;although interestingly,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(iii) very admirably, the issue of &lt;em&gt;&quot;How many stars have planets?&quot;&lt;/em&gt; ... one could say that issue has been somewhat settled these very years, as we speak - that's great.&lt;/p&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="13071" LastEditDate="2016-10-05T10:22:03.600" LastActivityDate="2016-10-05T10:22:03.600" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18583" PostTypeId="2" ParentId="10909" CreationDate="2016-10-04T18:11:50.010" Score="3" Body="&lt;p&gt;I would argue that simply knowing that a star was about to undergo a supernova could be information enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not always easy to figure out what type of star a given supernova progenitor was. Sometimes, there were no observations of the relevant region of sky prior to the supernova, and so we can't simply look back at prior observations and extrapolate from that. We can definitely look at certain spectral lines and other information from the remnant, but it's rarely as sure as actually observing the progenitor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some classes of stars which have yet to be conclusively identified as supernova progenitors, though there may be evidence suggesting that they should evolve as such. Red supergiants are one group, leading to the so-called &quot;red supergiant problem&quot; (see &lt;a href=&quot;https://arxiv.org/abs/0809.0403&quot; rel=&quot;nofollow noreferrer&quot;&gt;Smartt et al. (2009)&lt;/a&gt; and &lt;a href=&quot;https://astronomy.stackexchange.com/q/16148/2153&quot;&gt;this question&lt;/a&gt;). If we knew that a red supergiant was a supernova progenitor, that would possibly solve the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wolf-Rayet stars, thought to produce Type Ib/c supernovae, are another group (see &lt;a href=&quot;https://arxiv.org/abs/1207.3683&quot; rel=&quot;nofollow noreferrer&quot;&gt;Yoon et al. (2012)&lt;/a&gt;. Evolutionary models and indirect observational evidence predict that they should lead to these supernovae, but no observations of Wolf-Rayet supernova progenitors have been observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eta Carinae A - which I assume is the star you're referring to - appears to be a luminous blue variable (LBV). We do know that LBVs lead to supernovae (see for instance the case of &lt;a href=&quot;https://en.wikipedia.org/wiki/SN_2006jc&quot; rel=&quot;nofollow noreferrer&quot;&gt;SN 2006jc&lt;/a&gt;, which underwent a major outburst two days prior). However, it can be harder to distinguish between supernovae and certain &lt;a href=&quot;https://en.wikipedia.org/wiki/Supernova_impostor&quot; rel=&quot;nofollow noreferrer&quot;&gt;supernova imposter&lt;/a&gt; events, caused by LBVs. I'd speculate that we could try to observe Eta Carinae A in the period leading up to the supernovae, to try to observe the differences between outbursts and the actual supernova, which could give us better information on the nature of the evolution of these stars. Additionally, there is neutrino emission in the periods leading up to supernovae (see &lt;a href=&quot;https://astronomy.stackexchange.com/q/18423/2153&quot;&gt;this question&lt;/a&gt;); perhaps detectors on Earth could monitor the progenitor and see if they can find anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could also, given enough warning time, attempt to study the progenitor in the final phases of stellar evolution. &lt;a href=&quot;http://www.astronomy.ohio-state.edu/~ryden/ast162_4/notes15.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This page&lt;/a&gt; gives a table for the time periods of various fusion phases in the life of a $25M_\odot$ star:&#xA;$$\begin{array}{|c|c|}&#xA;\hline \text{Fusion phase}&amp;amp;\text{Length of phase}\\&#xA;\hline \text{Hydrogen} &amp;amp; \text{7 million years}\\&#xA;\hline \text{Helium} &amp;amp; \text{500,000 years}\\&#xA;\hline \text{Carbon} &amp;amp; \text{600 years}\\&#xA;\hline \text{Neon} &amp;amp; \text{1 year}\\&#xA;\hline \text{Oxygen} &amp;amp; \text{6 months}\\&#xA;\hline \text{Silicon} &amp;amp; \text{1 day}\\&#xA;\hline&#xA;\end{array}$$&#xA;This would let us figure out where the star is in its life and figure out what effects different fusion pathways could be having, if any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I'm assuming that your question means that we know when the light from the supernova reaches Earth, and that this would happen in the near future.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-10-19T17:34:26.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18584" PostTypeId="2" ParentId="13259" CreationDate="2016-10-04T19:05:53.010" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Mass = $\text{atomic mass} \times \text{number of atoms} = 4.4400513610370694 \times 10^{30} \mathrm{\ kg} \approx 2.3M_☉$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Herein lies the problem: your units and exponents are messed up. The mass of your iron isotope is $55.934936 \mathrm{\ u}$, and there are $7.938⋅10^{31}$ atoms. Do the math, and you'd get a total mass of $4.4400513610370694 \times 10^{33}\mathrm{\ u}$. That converts to approximately $7372878 \mathrm{\ kg}$. Nowhere near the mass of the Sun!&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2016-10-04T19:05:53.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18585" PostTypeId="1" AcceptedAnswerId="18655" CreationDate="2016-10-04T19:58:17.057" Score="9" ViewCount="298" Body="&lt;p&gt;Growing up far from any city, I have always been able to view the full night sky. It was just the way how things were. Of course, I knew that light pollution is a thing, and that the sky is a lot less clear if you live in a city.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It just never occurred to how bad it was before talking with people who had experienced it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Inside a city, it is pretty much impossible to view anything.&lt;/li&gt;&#xA;&lt;li&gt;It is not a lot better just outside a city, the sky is still mostly empty.&lt;/li&gt;&#xA;&lt;li&gt;Most of Europe, parts of the US, India, and eastern Asia is so polluted with light that you can not see the Milky way.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Does anyone have a map showing the affected area? How many people have never seen our own galaxy?&lt;/p&gt;&#xA;" OwnerUserId="10172" LastActivityDate="2016-10-09T02:03:51.157" Title="In what parts of the world is it impossible to view the night sky due to light pollution?" Tags="&lt;observational-astronomy&gt;&lt;milky-way&gt;&lt;light-pollution&gt;&lt;night-sky&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18586" PostTypeId="2" ParentId="18585" CreationDate="2016-10-04T20:07:27.423" Score="6" Body="&lt;p&gt;Googling for 'light pollution map' gives a pretty good looking &lt;a href=&quot;http://www.lightpollutionmap.info/&quot; rel=&quot;noreferrer&quot;&gt;result&lt;/a&gt;. Those things usually are created using satellite measurments that are susceptible to scattered light (by measuring the polarity of light) in order to separate direct illumination from actual light pollution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also suggest you visit a city and then compare the light pollution index on this site to what you know as home, in order to get a feeling for the severity of the numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot answer your question about how many people have never seen our galaxy, as this is a function of travel willingness of each individual living in a city and therefore maybe even impossible to answer for our intelligence agencies.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2016-10-04T20:07:27.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18588" PostTypeId="1" CreationDate="2016-10-05T12:33:03.190" Score="8" ViewCount="242" Body="&lt;p&gt;&lt;a href=&quot;http://www.jpl.nasa.gov/news/news.php?feature=4434&quot; rel=&quot;noreferrer&quot;&gt;Saturn has been located&lt;/a&gt; with fantastic precision thanks to Cassini and &lt;a href=&quot;https://en.wikipedia.org/wiki/Very-long-baseline_interferometry&quot; rel=&quot;noreferrer&quot;&gt;VLBI&lt;/a&gt;. The location of the 120,000 km diameter giant planet (when defining the 1 bar pressure level as its surface) is known to within 4 kilometers. This has already been helpful in the (preliminary) discovery and characterization of &quot;planet nine&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose it is the Saturn system's barycenter which has been located gravitationally. At this level of precision I suppose that the moons', especially Titan's tidal effects must be considered. But also Saturn's flattened shape and maybe even weather systems. Saturn is known to be lumpy inside, this has been discovered by the effects it has on the inner rings. Such lumps have similar effects as a moon, so I wonder if these too are being considered at the 4 km precision level, and overall exactly how the point located is defined?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2016-10-05T12:41:54.940" LastActivityDate="2018-06-11T22:21:46.993" Title="Exactly what in Saturn has been located to within 4 km precision?" Tags="&lt;saturn&gt;&lt;gas-giants&gt;&lt;cassini&gt;&lt;ephemeris&gt;&lt;vlbi&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18589" PostTypeId="1" CreationDate="2016-10-05T12:41:35.170" Score="5" ViewCount="53" Body="&lt;p&gt;Juno's low perijove should be great for measuring Jupiter's gravitational effect on Juno's orbit. Cassini together with &lt;a href=&quot;https://en.wikipedia.org/wiki/Very-long-baseline_interferometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;VLBI&lt;/a&gt; has measured Saturn's location to within 4 km. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;Will&lt;/s&gt; Was Juno &lt;s&gt;be&lt;/s&gt; able to be measured by VLBI to determine Jupiter's location and orbit at a similar precision? At a precision useful for mapping the Solar System outside of the Kuiper Belt?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="7982" LastEditDate="2018-11-01T22:55:58.197" LastActivityDate="2018-11-01T22:55:58.197" Title="Did the Juno mission locate Jupiter at the same precision as Saturn (4 km)?" Tags="&lt;jupiter&gt;&lt;ephemeris&gt;&lt;vlbi&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="18590" PostTypeId="1" CreationDate="2016-10-05T14:13:41.433" Score="0" ViewCount="492" Body="&lt;p&gt;Do black holes produce thermal radiation, as expected on theoretical grounds? Does this radiation contain information about their inner structure, as suggested by gauge–gravity duality, or not, as implied by Hawking's original calculation? If not, and black holes can evaporate away, what happens to the information stored in them (since quantum mechanics does not provide for the destruction of information)? Or does the radiation stop at some point leaving black hole remnants? Is there another way to probe their internal structure somehow, if such a structure even exists?&lt;/p&gt;&#xA;" OwnerUserId="14078" LastActivityDate="2016-10-06T06:56:05.877" Title="Do black holes produce thermal radiation?" Tags="&lt;black-hole&gt;&lt;hawking-radiation&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18592" PostTypeId="1" CreationDate="2016-10-05T19:21:41.097" Score="4" ViewCount="2144" Body="&lt;p&gt;I saw the &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1801/why-does-the-darkening-of-the-moon-go-from-left-to-right-during-an-eclipse&quot;&gt;question and answer&lt;/a&gt; for the explanation of the direction of the moon's eclipse from left to right, but what about the direction of the moon as it passes over or between Earth and the sun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the moon always pass from the right to the left or from the left to the right as it covers the sun; if you live above the equator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or does this change? Can it change? August 2017 the USA will have a total eclipse of the sun for the first time in decades.&lt;/p&gt;&#xA;" OwnerUserId="14463" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2021-05-27T07:32:39.793" Title="When the moon eclipses the sun what direction is darkened first: left or right?" Tags="&lt;the-sun&gt;&lt;the-moon&gt;&lt;solar-eclipse&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18593" PostTypeId="2" ParentId="18592" CreationDate="2016-10-05T20:11:37.270" Score="5" Body="&lt;p&gt;Relative to the Sun, stars, and Earth's shadow, the Moon always appears to move from west to east.&#xA;How this looks relative to the horizon depends on your latitude and time of day.&#xA;If a solar eclipse occurs in the middle of the day, the Moon appears to move from right to left if you're in a northern latitude, or from left to right if you're in a southern latitude.&#xA;If you're in a tropical latitude, the Moon appears to move downward if the eclipse occurs in the morning or upward if in the afternoon.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2016-10-06T15:18:28.420" LastActivityDate="2016-10-06T15:18:28.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18595" PostTypeId="2" ParentId="18154" CreationDate="2016-10-05T22:49:33.940" Score="5" Body="&lt;p&gt;The Sun moves in many ways. For one the sun does wobble, as stated above, due to Jupiter's pull, this is actually how astronomers are able to find new planets! They can use physics and Mathematics to figure out a planets size and number of planets orbiting a star just by studying the stars wobble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly the Sun along with the Solar System orbits the center of the Milky Way Galaxy. The length of this process is called a Galactic Year. The Solar System's Galactic year ranges somewhere from 225 to 250 million years.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly our Galaxy and the Sun move as a whole through space, which is what will eventually cause the Milky Way Galaxy to collide with the Andromeda Galaxy.  &lt;/p&gt;&#xA;" OwnerUserId="13617" LastActivityDate="2016-10-05T22:49:33.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18597" PostTypeId="2" ParentId="18590" CreationDate="2016-10-06T06:56:05.877" Score="3" Body="&lt;p&gt;Hawking radiation is the black body radiation that black holes emit. It is a well established theoretical result but it has never been observed.  For all currently known astrophysical black holes it will be totally insignificant and will never be observed.  The reason for this is that the equivalent temperature of the black body is inversely proportional to the mass of e black hole.  For all currently known black holes the temperature would be a tiny fraction of a degree above absolute zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum mechanics is a probabilistic approach to physics, and deals with all kinds of probabilities and uncertainties. There is one such uncertainty of time and energy, given by (Δt) (ΔE) ≥ ℏ/2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of its predictions, which has been experimentally confirmed, is that a particle may take an energy E from “nothing” and yet not violate the Law of Conservation of mass-energy if it “returns” the energy within a time given by (Δt)≥ ℏ/(2E).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of its many implications is that random vacuum fluctuations constantly take place everywhere, i.e. even if a certain region of space has nothing or is purely vacuum, it can still constantly generate and annihilate particles due to the E-t uncertainty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;fast forwarding&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When such a fluctuation causes a particle and its anti-particle to be generated on the edge of the event horizon of a black hole, one of the two enters the black hole and the other escapes. Now for returning the energy to the system, the particle relies on crashing into its anti-particle causing an “annihilation” and conversion of mass to energy. In the scenario I described, this can never take place since the pair of particles have been separated effectively forever. The particle that enters the event horizon has been trapped in an infinite potential gravitational well, so can be thought to have negative energy. So anti-intuitively (wait, what part of all this is even intuitive!), the particle left lurking outside the absorbs some energy from the black hole, (thus conserving mass-energy) gets converted into a “classical” photon and escapes the vicinity of the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus little by little, such events cause a black hole to effectively “evaporate” and diminish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The apparant paradox about the fact that what happens to the information that goes into a black hole is a matter of debate. No one knows exactly how this Black Hole Information Paradox can be solved, however certain speculations are provided by Kip Thorne and Stephen Hawking, which include Firewall theory and Ads/CFT.&lt;/p&gt;&#xA;" OwnerUserId="10576" LastActivityDate="2016-10-06T06:56:05.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18598" PostTypeId="2" ParentId="15078" CreationDate="2016-10-06T07:22:58.063" Score="-1" Body="&lt;p&gt;The Earth's gravity (call it the curvature of space) at the center of the Earth is approximately zero. Similarly, we assume a dense early universe was still &lt;em&gt;infinite&lt;/em&gt; and &lt;em&gt;homogeneous&lt;/em&gt;, so no worries, the gravity canceled out to approximately zero. There is no evidence that it was a lump of matter surrounded by an empty space, like a black hole usually is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, the Big Bang happened right here where you are now, and all the matter became more and more distant from the point where you are. Simply all the distances increased all the time without matter accelerating or moving into some imagined &quot;outer space&quot;.&lt;/p&gt;&#xA;" OwnerUserId="10675" LastActivityDate="2016-10-06T07:22:58.063" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18599" PostTypeId="1" AcceptedAnswerId="18606" CreationDate="2016-10-06T11:01:22.863" Score="4" ViewCount="2198" Body="&lt;p&gt;It is generally accepted that the age of the universe is approximately 12-15 billion years old based on the speed of the expansion of the universe. Since everything is moving &lt;em&gt;very&lt;/em&gt; fast away from us, doesn't relativistic effects from high speeds come into play here? If something is moving quite fast, time slows down, affecting how &quot;old&quot; it is. Is all this taken into account when calculating the age of the universe? I don't know if I'm making any sense to anyone else or not. &lt;/p&gt;&#xA;" OwnerUserId="10117" LastEditorUserId="10172" LastEditDate="2016-10-06T19:34:15.420" LastActivityDate="2016-10-07T00:23:55.543" Title="Does the age of the universe take into account General Relativity / Special Relativity?" Tags="&lt;universe&gt;&lt;observable-universe&gt;&lt;units&gt;&lt;age&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18600" PostTypeId="2" ParentId="18599" CreationDate="2016-10-06T11:12:56.733" Score="11" Body="&lt;p&gt;This is a confusing question - your title mentions GR, but of course the age of the universe is entirely derived as a result of using GR to solve for the dynamics of the expanding universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The text of your question is talking about time dilation and the effects of &lt;em&gt;special&lt;/em&gt; relativity (a subset of GR). Here there is a point to be addressed. It is not a good way to think about the explansion of the universe in terms of things receding from each other at ever increasing velocities. It is better to think about space expanding. It is also not possible to apply special relativity except in local inertial frames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can and has been shown that the redshift-distance relationship &lt;em&gt;cannot&lt;/em&gt; be simply interpreted as the doppler shift due to things moving away from us at ever increasing speeds - since this of course predicts a maximum recession velocity that approaches the speed of light. In fact, beyond redshifts of $z=1.5$, the &quot;recession velocity&quot; in all viable models exceeds the speed of light. What actually happens is that light emitted at some earlier epoch travels across expanding space to get to us and this expansion increases the wavelength of the light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essential reading: &lt;a href=&quot;https://arxiv.org/pdf/astro-ph/0310808.pdf&quot;&gt;Davis &amp;amp; Lineweaver (2003)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-10-06T11:12:56.733" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18603" PostTypeId="1" CreationDate="2016-10-06T12:34:33.733" Score="1" ViewCount="54" Body="&lt;p&gt;Can singularities not hidden behind an event horizon, known as &quot;naked singularities&quot;, arise from realistic initial conditions, or is it possible to prove some version of the &quot;cosmic censorship hypothesis&quot; of Roger Penrose which proposes that this is impossible? Similarly, will the closed timelike curves which arise in some solutions to the equations of general relativity (and which imply the possibility of backwards time travel) be ruled out by a theory of quantum gravity which unites general relativity with quantum mechanics, as suggested by the &quot;chronology protection conjecture&quot; of Stephen Hawking?&lt;/p&gt;&#xA;" OwnerUserId="14078" LastActivityDate="2016-10-06T12:34:33.733" Title="is it possible to prove some version of the &quot;cosmic censorship hypothesis&quot; of Roger Penrose which proposes that this is impossible?" Tags="&lt;general-relativity&gt;&lt;singularity&gt;&lt;quantum-mechanics&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18604" PostTypeId="1" CreationDate="2016-10-06T12:39:06.403" Score="9" ViewCount="2672" Body="&lt;p&gt;What are the percentages of systems that have x number of stars in them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have found thus far is something like:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Single Star Sytems = 69%&lt;/li&gt;&#xA;&lt;li&gt;Double Star Systems = ~10%&lt;/li&gt;&#xA;&lt;li&gt;Triple Star Systems = &amp;lt;20% &lt;/li&gt;&#xA;&lt;li&gt;Quadruple Star Systems = ???&lt;/li&gt;&#xA;&lt;li&gt;Quintuple Star Systems = ???&lt;/li&gt;&#xA;&lt;li&gt;Sextuple Star Systems = ???&lt;/li&gt;&#xA;&lt;li&gt;Septuple Star Systems = ???&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Also... Is there any way to get these percentages for systems that have primaries that are Giants (more than 2 Solar Masses) Major Stars (0.5 to 2 Solar Masses), and Minor Stars (0.5 Solar Masses... red dwarfs)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found an article that says...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Giant Stars = 80% have companions&lt;br&gt;&#xA;Major Stars = 50% have companions&lt;br&gt;&#xA;Minor Stars = 25% have companions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.space.com/1995-astronomers-wrong-stars-single.html&quot;&gt;http://www.space.com/1995-astronomers-wrong-stars-single.html&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://boards.straightdope.com/sdmb/showthread.php?t=499261&quot;&gt;http://boards.straightdope.com/sdmb/showthread.php?t=499261&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.skyandtelescope.com/astronomy-news/surprise-most-star-systems-are-single/&quot;&gt;http://www.skyandtelescope.com/astronomy-news/surprise-most-star-systems-are-single/&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="14084" LastEditorUserId="14084" LastEditDate="2016-10-06T12:59:23.557" LastActivityDate="2016-10-06T21:53:22.020" Title="Multiple Star-system percentages" Tags="&lt;star&gt;&lt;binary-star&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18605" PostTypeId="2" ParentId="18592" CreationDate="2016-10-06T12:59:37.663" Score="4" Body="&lt;p&gt;Coincidentally,  &lt;a href=&quot;https://xkcd.com/1738/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Randall Monroe&lt;/a&gt; posted a drawing covering exactly this issue (with, of course, a few humorous states thrown in).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xKOQC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xKOQC.png&quot; alt=&quot;MoonShapes&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2016-10-06T12:59:37.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18606" PostTypeId="2" ParentId="18599" CreationDate="2016-10-06T14:15:02.467" Score="11" Body="&lt;p&gt;Rob Jeffries gives a good response to this question, but I wanted to go through the basic outline of how the age of the universe is calculated, just so you can see how it works more or less. Be warned though, I'm only giving you the highlights and you'll either have to accept what I'm saying or fill in the blanks yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Friedmann Equation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with anything in physics, when you want to analyze something, you need an equation to represent it. Think of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann_equations&quot;&gt;Friedmann Equation&lt;/a&gt; as the &quot;equation of the universe&quot;. It's kind of crazy (to me) to believe, but yes there is an equation that represents the entire universe. The Friedmann equation is directly derived from the GR field equations, however it is quite possible to derive them from Newtonian mechanics (and a little hand waving). A good source for the Newtonian derivation can be found in Ryden's &lt;a href=&quot;http://atlas.physics.arizona.edu/~kjohns/downloads/lsst/Ryden_IntroCosmo.pdf&quot;&gt;Indroduction to Cosmology&lt;/a&gt;, which is a good undergraduate level cosmology textbook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many forms of the Friedmann equation&lt;sup&gt;1&lt;/sup&gt; and it can be represented numerous ways (especially depending on the universe model), but I think a convenient form is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bigg(\frac{\dot{a}}{a}\bigg)^2 = H_0^2 \Big(\Omega_{m,0}a^{-3} + \Omega_{r,0}a^{-4} + \Omega_{\Lambda,0} + \Omega_{k,0}a^{-2}\Big) \equiv H_0^2E^2(a)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The terms in this equation are defined as follows.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$a(t)$ The size of the universe at time $t$, compared to its size now.&lt;/li&gt;&#xA;&lt;li&gt;$H_0$ The value of Hubble's constant for the current time. Nominally taken to be $\sim68\:\mathrm{km/s/Mpc}$.&lt;/li&gt;&#xA;&lt;li&gt;$\Omega$ The &quot;density&quot; of the various types of matter/energy in our universe. The subscript $0$ indicates that it is the density for the universe at the present time. For these, $m$ represents matter (both normal and dark), $r$ represents radiation (i.e., light), $\Lambda$ represents the cosmological constant (i.e., vacuum energy), and $k$ is the curvature of our universe (i.e. energy in the space-time fabric itself).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This equation may seem scary, but effectively it is telling you is how the size of the universe, has evolved over time based on the current content of the universe. If you know precisely how much matter, radiation, and dark energy you have in the universe right now (or more precisely the density), you can calculate the history of the universe!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;&lt;sub&gt;Technically there is more than one equation, but we'll look at an important form of one of them.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Age of the Universe&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, $a$ is a function of time in the equation above. This means, with a little bit of rearranging and integrating, we can calculate the age of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t=\frac{1}{H_0} \int_{0}^{1} \Big(\Omega_{m,0}a^{-1} + \Omega_{r,0}a^{-2} + \Omega_{\Lambda,0}a^2 + \Omega_{k,0}\Big)^{-1/2} da$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the age of the universe, $t$, you have the non-trivial task of measuring good values for the constants and then doing the integral above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a trivial case, you can assume we live in a flat, matter-only universe - no light, no dark energy, no curvature. In that case $\Omega_{m,0}=1$ and $\Omega_{r,0}=\Omega_{\Lambda,0}=\Omega_{k,0}=0$. You'll wind up with $t=2/(3H_0)$ which amounts to ~10 billion years old. This type of universe is known as an Einstein-de Sitter Universe and is obviously &lt;em&gt;not&lt;/em&gt; the universe we live in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A Note on Dark Energy&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically the equations above are a bit more complicated because I haven't accounted for dark energy. I won't go into the specifics, but one can choose to add components to the equation above which include the dark energy equation of state parameter. Generally this parameter is denoted by $w$. The concepts are more or less the same, the equations just become uglier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With all of the above, I think I can really address your question. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the age of the universe take into account GR/SR?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In short, yes. The Friedmann equation, as I said above, is derived directly from the GR equations. Therefore, we're using an equation to represent the universe which directly results from assuming GR is an accurate theory of gravity in our universe.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since everything is moving very fast away from us, doesn't relativistic effects from high speeds come into play here? If something is moving quite fast, time slows down, affecting how &quot;old&quot; it is. Is all this taken into account when calculating the age of the universe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes and no. As I said, the age of the universe is ultimately derived from GR so anything GR/SR related is accounted for. However, it seems to me that your understanding of how the universe's age is calculated is somewhat simplistic (and there's nothing wrong with that - its a complicated process!). In calculating the age of the universe, we actually don't care about how fast some far-flung galaxy may be moving away from us. We don't even look at those galaxies. What we really look at is the Cosmic Microwave Background (CMB) Radiation. By looking at this radiation, we're able to work out the $\Omega$ parameters more or less (and the dark energy parameter $w$) without every having to concern ourselves with recessional galaxy speeds. Hopefully you can see that the concepts of receding galaxies and time dilation are moot points when it comes to determining the age of the universe.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-10-06T14:15:02.467" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18607" PostTypeId="1" CreationDate="2016-10-06T14:56:02.217" Score="3" ViewCount="69" Body="&lt;p&gt;Neptune was discovered 170 years ago with the help of observations of Uranus' orbit. A great demonstration of the scientific principle for its time. Recently, Earth and cis-Lunar based radio interferometry together with the Cassini orbiter have measured Saturn's orbit, 10 AU from the Sun, with very great precision. This has already helped determine some characteristics of a possible planet Nein 200-700 AU from here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the best way to gravitationally measure the yet unobserved mass concentrations in the Kuiper Belt and beyond, by observing orbits of objects among the known planets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It takes alot to move Saturn to a measurable degree. What about a space probe with the sole purpose of being easy to locate at high precision, maybe in a highly inclined orbit? Or a family of them in different directions. Would its deplacement from the very weak gravity well of a planet Nine be detectable? If it is the history of Saturn that contributes crucially to the precision, what about an asteroid like Ceres which is known since 200+ years? Is there any special property of Saturn that makes it especially useful? Would a second object as well measured as Saturn be extra helpful by some kind of triangulation of distant gravitational influences?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-10-07T09:20:34.443" Title="How can we map masses orbiting the outer Solar System by determining the orbits of planets and smaller objects?" Tags="&lt;gravity&gt;&lt;saturn&gt;&lt;9th-planet&gt;&lt;cassini&gt;&lt;ephemeris&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18608" PostTypeId="2" ParentId="18604" CreationDate="2016-10-06T15:12:52.867" Score="7" Body="&lt;p&gt;Your best bet for finding relevant information on this is to look up actual published papers. I'll walk you through my research process to help in the future, as well as provide the results I found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 1: Google Scholar is your friend&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I start out using &lt;a href=&quot;https://scholar.google.com/&quot;&gt;Google Scholar&lt;/a&gt;. This is much like Google, but rather than returning any old website, it specifically returns published papers and other scholarly articles. Part of this magic is of course knowing what to search. I went with &quot;&lt;a href=&quot;https://scholar.google.com/scholar?q=multiplicity%20of%20stellar%20systems&amp;amp;btnG=&amp;amp;hl=en&amp;amp;as_sdt=0%2C22&quot;&gt;multiplicity of stellar systems&lt;/a&gt;&quot;. Immediately I found two sources which seemed promising.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://iopscience.iop.org/article/10.1086/503158/meta&quot;&gt;Stellar Multiplicity and the Initial Mass Function: Most Stars Are Single (Lada 2006)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mnras.oxfordjournals.org/content/389/2/869.short&quot;&gt;A Catalogue of Multiplicity Among Bright Stellar Systems (Eggleton &amp;amp; Tokovinin 2008)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've lucked out because both articles happen to be free so I can see the full contents!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2: NASA ADS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.adsabs.harvard.edu/&quot;&gt;NASA Astrophysics Data System&lt;/a&gt; is a very large catalogue of just about every astronomy (and some physics) related paper published in any journal. What's even more amazing, very often the papers here are free, even if they aren't free on the journal's actual web page. If you ever run into a paper from a journal that only gives you an abstract, try looking for the paper here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was already free on the journal's webpage, but I was able to look up and find &lt;a href=&quot;http://adsabs.harvard.edu/abs/2006ApJ...640L..63L&quot;&gt;the Lada paper&lt;/a&gt; I cited above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I specifically point you towards ADS for a few reasons.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It has a large collection of free papers.&lt;/li&gt;&#xA;&lt;li&gt;You may have heard of or even use arxiv, which is a great resource, but very often the papers on arxiv are versions which are pre-publication and not the officially peer-reviewed and published papers. ADS usually has the official published papers (and links to arxiv versions!).&lt;/li&gt;&#xA;&lt;li&gt;ADS has another hugely important feature. It lists article citations and articles which have cited that article.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 3: Backtracking and Forwardtracking&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you've found a good source (such as the two papers above), you don't want to stop there. There's likely more or better papers and you can use the papers you have to find new ones. What you want to do now is take a look at all the papers your current source has cited, and look at all the papers which cite your current source. ADS very conveniently gives you all this information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lada 2006 source is 10 years old at this point so I wanted to see if there was anything more recent. I was able to click on the &quot;Citations to the Article&quot; link on ADS for that article, and found the &lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/nph-ref_query?bibcode=2006ApJ...640L..63L&amp;amp;refs=CITATIONS&amp;amp;db_key=AST&quot;&gt;list of 217, more recent papers which cited the Lada paper&lt;/a&gt;. Scanning through this list I found two promising papers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2014AJ....147...28G&quot;&gt;Stellar Multiplicity of the Open Cluster ASCC 113 (Guerrero et al. 2014)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2014ApJ...785...47L&quot;&gt;An Adaptive Optics Multiplicity Census of Young Stars in Upper Scorpius (Lafrenière et al. 2014)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here, just rinse and repeat until you've got the information you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Results&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found four papers which appear to give some or all of the information you want.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Lada 2006 - This paper is more or less a conglomeration of past research on the fraction of stars which are single star systems. It focused on lower mass stars (G to M) and found that the fraction of single star systems ranged from ~43% for G type stars to ~75% for M type stars (which are the most populous type).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Eggleton &amp;amp; Tokovinin 2008 - I think that this is the best source for your particular question. Within their abstract they state &lt;blockquote&gt;We identify 4559 such bright systems (including the Sun), and the frequencies of multiplicities 1, 2,..., 7 are found to be 2718, 1437, 285, 86, 20, 11 and 2.&lt;/blockquote&gt; This implies the fraction of multiplicities is: 59.62%, 31.52%, 6.25%, 1.88%, 0.44%, 0.24%, and 0.04%. However, note they say their measurements have &quot;substantial&quot; uncertainties which they describe.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Guerrero et al. 2014 - This group looked at and around a specific cluster and found &lt;blockquote&gt;a ratio of the number of single to binary stars to be 27:7&lt;/blockquote&gt; within the cluster itself. Including stars around the cluster, the following multiplicities were found (ranging from 1 to 8 companions) &lt;blockquote&gt;125:27:4:1:0:0:0:1&lt;/blockquote&gt; These two sets of ratios indicate that within the cluster they saw 79% single star systems and 21% binary star systems. Overall they saw multiplicity fractions of 79.1%, 17.1%, 2.5%, 0.6%, 0%, 0%, 0%, and 0.6%, respectively.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Lafrenière et al. 2014 - A study that looked at 91 stars and found 57 single stars (63%), 29 binaries (32%), and 5 triple systems (5.5%).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What this suggests is that the multiplicity of stars is highly variable, still somewhat unknown, and that it depends on the environment you look at (and I found a few papers discussing this very point). The Eggleton paper had the largest sample and so may be the most trustworthy in regards to a true average, but make sure to understand what their uncertainties are.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-10-06T15:12:52.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18609" PostTypeId="1" AcceptedAnswerId="18610" CreationDate="2016-10-06T15:46:21.677" Score="10" ViewCount="1710" Body="&lt;p&gt;Andromeda is about 2.5 million ly away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, in this universe, at what &quot;speed&quot; (in km/h) are two objects separating cosmologically - &lt;strong&gt;I mean strictly due to the &quot;expansion of the universe&quot;&lt;/strong&gt; - if they are 2.5 million ly apart?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do understand that local &quot;ordinary&quot; or &quot;peculiar&quot; motion completely swamps this effect.  If I'm not mistaken, the &quot;local&quot; &quot;ordinary&quot; motion of Andromeda per our galaxy happens to be about 400,000 km/h towards us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the &quot;speed&quot; due to the &quot;expansion of the universe&quot; drastically smaller than this?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I assumed that the expansion of the universe (or &quot;of the spacetime metric&quot;) is even everywhere: it's well known that it only affects &quot;the largest structures&quot; but I still assumed that the expansion is the same in my room, my galaxy, my cosmological region.  Perhaps this assumption is totally wrong?&lt;/p&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="13071" LastEditDate="2016-10-06T21:04:37.693" LastActivityDate="2016-10-07T03:03:34.140" Title="In km/h, what actually is the &quot;speed&quot; of Andromeda away from us: cosmologically?" Tags="&lt;expansion&gt;&lt;redshift&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18610" PostTypeId="2" ParentId="18609" CreationDate="2016-10-06T16:08:16.023" Score="20" Body="&lt;p&gt;The rate of expansion, measured in the customary units of (km/s)/Megaparsec&#xA;is not known with great accuracy. Recent measurements include 67.6 (SDSS-III), 73(HST) 67.8 (Plank) 69.3 (WMAP) [&lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble%27s_law&quot;&gt;wikipedia&lt;/a&gt;]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Andromeda_Galaxy&quot;&gt;Andromeda galaxy&lt;/a&gt; is 0.78 Mpc from us, so taking the Hubble constant to be about 70, gives a recession of about 55 km/s. This is not a very great speed: compare with the orbital velocity of the sun around the galaxy at over 200 km/s, or the escape velocity of the galaxy (over 500km/s)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you note, this is pretty much swamped by the proper relative motion of our galaxies. Its blueshift indicates that Andromeda is approaching us at over 100 km/s. For galaxies outside of the the &lt;a href=&quot;https://en.wikipedia.org/wiki/Local_Group&quot;&gt;Local group&lt;/a&gt;, the Hubble flow dominates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the value of 55km/s assumes that space is smooth and homogeneous. This is approximately true on a universal scale, but it is not true on the scale of a galaxy cluster, where local gravitational effects dominate the curvature of spacetime. The general expansion of spacetime has very little effect on the motion of galaxies in the local group, as discussed by &lt;a href=&quot;https://docs.google.com/viewer?url=https://arxiv.org/pdf/1208.1523v4.pdf&quot;&gt;Iorio's paper on the motion of a gravitationally bound binary system&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-10-06T21:30:27.027" LastActivityDate="2016-10-06T21:30:27.027" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="18611" PostTypeId="1" AcceptedAnswerId="18620" CreationDate="2016-10-06T20:03:57.530" Score="11" ViewCount="455" Body="&lt;p&gt;A couple of years ago I started playing a game called Kerbal Space Program (KSP) and my understanding about how orbits actually work increased dramatically. Because of KSP I was looking at a picture of a galaxy with its spiral arms and something struck me as very odd. The stars closer in should orbit the galaxy center faster than the stars farther out and that would smear out these lanes of stars until they were no longer recognizable. Something weird is going on here!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So why do galaxies have arms? What's bunching these stars together like that?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I noticed that the center of galaxies are all smeared out and bulbous like I think that the rest should look like. So what if all that mass rotating around a common center is warping space around it, like the water draining down your sink, and this warping is what catching matter/stars into the lanes.&lt;/p&gt;&#xA;" OwnerUserId="14475" LastEditorUserId="2365" LastEditDate="2016-10-07T05:16:20.217" LastActivityDate="2016-10-07T20:19:44.130" Title="Why do galaxies have arms?" Tags="&lt;orbit&gt;&lt;galaxy&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18613" PostTypeId="1" AcceptedAnswerId="18615" CreationDate="2016-10-06T21:50:39.160" Score="21" ViewCount="1507" Body="&lt;p&gt;When talking about the expansion of the universe, it is said that it can be proven by the red-shifting of light.(As we would need higher than lightspeed to get this redshift by the Doppler effect)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am an amateur, so I am not sure I am correct, but here is what I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Redshifting increases the wavelength of the light. higher wavelength = lower frequency = less energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if my assumptions are correct, where does this energy from the light go?&#xA;If not, where did I make an incorrect assumption?&lt;/p&gt;&#xA;" OwnerUserId="14479" LastEditorUserId="14479" LastEditDate="2016-10-06T22:09:13.053" LastActivityDate="2016-10-16T23:54:23.220" Title="Where does the energy of light go, when it red-shifts?" Tags="&lt;expansion&gt;&lt;redshift&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18614" PostTypeId="2" ParentId="18604" CreationDate="2016-10-06T21:53:22.020" Score="6" Body="&lt;p&gt;The answers are out there. The problem with your question is that the answer is highly mass-dependent. The mass dependence is also somewhat uncertain, with the best empirical knowledge for solar-type stars, with rather more uncertain values and smaller statistical numbers for more and less massive stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best contemporary review of binarity for main sequence stars like the Sun of spectral types F6-K3 (and the one everyone uses) is that of &lt;a href=&quot;https://arxiv.org/abs/1007.0414&quot; rel=&quot;noreferrer&quot;&gt;Raghavan et al (2010)&lt;/a&gt;. Their study deals with binaries over the full range of possible separations and attempts to deal with the complex selection effects of different surveys and observation techniques. The answer you want is directly in the abstract of the paper:  $56\pm 2$% single, $33\pm 2$% binary, $8 \pm 1$% triple, $3 \pm 1$% higher order multiples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best overall review of binary statistics (in my opinion) is that of &lt;a href=&quot;https://arxiv.org/abs/1303.3028&quot; rel=&quot;noreferrer&quot;&gt;Duchene &amp;amp; Kraus (2013)&lt;/a&gt;. This deals quite well with the mass-dependence of the multiplicity frequency, which &lt;em&gt;clearly declines with mass&lt;/em&gt;. The overall multiplicity fraction for solar-type stars is 44% (see above), but this becomes more like 30% for M-dwarfs and perhaps as low as 20% for stars/brown dwarfs with $M&amp;lt;0.1M_{\odot}$, though this latter number has an uncertainty of at least 5%. The fraction of multiple systems that feature $&amp;gt;2$ stars is probably quite similar to solar type stars (i.e. about a quarter of multiple systems are more than binaries).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For higher mass stars the multiplicity is also higher. It is &lt;em&gt;at least&lt;/em&gt; 50% for stars between 1.5-5 solar masses and probably approaches 100% ($&amp;gt;80$%) for O-stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall results are summarised in Table 1 of the review. This is a highly cited and respectable source.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-10-06T21:53:22.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18615" PostTypeId="2" ParentId="18613" CreationDate="2016-10-06T22:10:28.483" Score="11" Body="&lt;p&gt;The problem is that conservation of energy is a slippery concept in General Relativity. There are arguments back and forth but most people accept that conservation of energy is only a local law - it applies only to a local inertial frame and cannot be applied to the universe as a whole. However in an expanding universe it is very difficult to identify &lt;em&gt;any&lt;/em&gt; inertial frames and certainly not ones that encompass a cosmologically significant volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means is that if you make a local &quot;box&quot; small enough that it is not affected by the expansion of the universe, then energy conservation will apply. But of course in such a box, a photon would enter and leave with the same energy because the box is unaffected by the expansion of the universe and so there would be no redshifting of the photon.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-10-06T22:10:28.483" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18616" PostTypeId="1" AcceptedAnswerId="18622" CreationDate="2016-10-06T22:38:19.863" Score="1" ViewCount="146" Body="&lt;p&gt;I currently own an old 1980s Celestron eight inch orange tube schmidt cassegrain telescope with tripod. I am seriously looking to dive into some astrophotography. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also own an olympus pen epl3 camera. It's great for photos during my hikes, but not quite as versatile as an DSLR. Unfortunately, I know of no software out there which can tether this camera to my laptop. I don't have any great ideas on remotely controlling this camera to avoid shaking my telescope.&#xA;I'm thinking of getting a DSLR at some point anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a camera body that anyone would recommend for me?&lt;/p&gt;&#xA;" OwnerUserId="7786" LastEditorUserId="11617" LastEditDate="2016-10-07T14:21:42.510" LastActivityDate="2016-10-07T14:21:42.510" Title="Which camera body is best introduction for beginner amateur astrophotographer?" Tags="&lt;photography&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18618" PostTypeId="1" CreationDate="2016-10-07T01:14:31.863" Score="0" ViewCount="100" Body="&lt;p&gt;I've tried utilizing &quot;Google Sky&quot;, however, once I've found the location of said star and am in the process of zooming in, the site craps out. &#xA;Anyway... if there's someone out there with a little more experience in this field, here are the coordinates of the star i'm searching for: &#xA;10h 10m 11.1s, +4 3' 12.3&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd appreciate the effort, and thank's in advance. &lt;/p&gt;&#xA;" OwnerUserId="14484" LastActivityDate="2016-10-07T07:34:36.807" Title="Need help finding a star" Tags="&lt;star&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18620" PostTypeId="2" ParentId="18611" CreationDate="2016-10-07T02:41:03.433" Score="6" Body="&lt;p&gt;The usual explanation of spiral arms involves &lt;a href=&quot;https://en.wikipedia.org/wiki/Density_wave_theory&quot; rel=&quot;nofollow&quot;&gt;density waves&lt;/a&gt;.&#xA;Nonuniform motion leads to matter alternately bunching up (boosting star formation) and spreading out.&#xA;Like cars on a congested road, stars move through regions of greater or lesser density over time.&#xA;The rotational smearing you anticipate is known as the &quot;winding problem.&quot;&#xA;This &lt;a href=&quot;https://www.scientificamerican.com/article/what-process-creates-and/&quot; rel=&quot;nofollow&quot;&gt;Scientific American article&lt;/a&gt;&#xA;offers three explanations by professional astronomers.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2016-10-07T20:19:44.130" LastActivityDate="2016-10-07T20:19:44.130" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18622" PostTypeId="2" ParentId="18616" CreationDate="2016-10-07T03:10:06.473" Score="2" Body="&lt;p&gt;When I first started out, I used my Olympus pen e-pl1, so you're doing great with the e-pl3. It takes higher resolution photos (330 vs 300 dpi) than my wife's Nikon D5300, so your camera should work just fine. I still whip this one out every now and again just because of its simplicity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one on the right is the adapter that I use regularly. I mounts up easily to each of my telescopes. Works great and has held up nicely the last (almost) decade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ik16u.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ik16u.jpg&quot; alt=&quot;Prime Focus 1.25&amp;quot; micro four thirds adapter.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To avoid the shake, set as long of a timer as you can! I'm not sure if the newer pen cameras have the option for the tethered remote, but mine definitely does not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the software, I wrote some python codes that I use for image processing, and DS9 is the interface that I use regularly, however good old photoshop (pre-everythingisstupidandcloudbased) really works wonders if you're on a budget and have or &quot;find&quot; a copy laying about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p.s. the adapter is from &lt;a href=&quot;http://www.telescopeadapters.com/index.htm?slrkits.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;telescopeadapters.com&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2232" LastActivityDate="2016-10-07T03:10:06.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18623" PostTypeId="1" CreationDate="2016-10-07T04:31:17.767" Score="-4" ViewCount="127" Body="&lt;p&gt;If all of the mass and energy in the universe was in a point smaller than the size of an electron at one time, why did it expand? Did it necessarily expand immediately on formation, or could it have remained in an initial state for some length of time until some phase threshold was reached?&lt;/p&gt;&#xA;" OwnerUserId="9516" LastEditorUserId="9516" LastEditDate="2016-10-07T21:46:13.183" LastActivityDate="2016-10-07T21:46:13.183" Title="Why Did The Universe Expand?" Tags="&lt;black-hole&gt;&lt;cosmology&gt;&lt;singularity&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2016-10-07T19:31:47.570" ContentLicense="CC BY-SA 3.0" />
  <row Id="18625" PostTypeId="2" ParentId="18618" CreationDate="2016-10-07T05:51:16.893" Score="-2" Body="&lt;p&gt;Up to my understanding these are 2 angles, One is vertical one(Where parallel to ground is zero, perpendicular towards sky is 90) and One is horizontal one (Compass, where north will be considered as zero/360). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;10h 10m 11.1s, +4 3' 12.3&quot;(Vertical, Horizontal)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First one , 10h 10m 11.1s. It's hour min sec format. Where 1 hour is 360(total degrees)/24(number of hours in a day)=15 degrees. for 1 minute, 15(degrees per hour)/60 (no of minutes in an hour). same for the second.&#xA;So that this angle will come around ~150+2.5+ (second's part). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;second one, +4 3' 12.3&quot;. This one is degree min sec. Most probably this is compass angle. where north might be zero and clockwise is +ve. representation may vary, but if you'll check the scale you'll be able to figure out that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this will help you.if not feel free to ask more.&lt;/p&gt;&#xA;" OwnerUserId="14491" LastActivityDate="2016-10-07T05:51:16.893" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18626" PostTypeId="2" ParentId="18607" CreationDate="2016-10-07T06:30:52.677" Score="4" Body="&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/1512.05288&quot; rel=&quot;nofollow&quot;&gt;Iorio (Preliminary constraints on the location of Telisto/Planet Nine from planetary orbital dynamics)&lt;/a&gt; has suggested transmissions from the New Horizons probe (currently beyond Pluto) could be used to test for the influence of the proposed new planet, and provide constraints to its possible position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is estimated that the position of New Horizons could be affected by the proposed planet, of the order of 100 metres or more by 2019. It is stressed that this is a preliminary proposal:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Moreover, the present investigation should be considered just as a preliminary sensitivity analysis which, by no means, is meant to replace a full covariance study based on extensive simulations of the ranging data from New Horizons and their reduction with accurate models including also PX itself, which is outside the scopes of this paper. Nonetheless, the possibility of using the telemetry of New Horizons seems appealing, and would deserve further, more detailed and dedicated studies which will, hopefully, prompted by the present one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7491" LastEditorUserId="7491" LastEditDate="2016-10-07T09:20:34.443" LastActivityDate="2016-10-07T09:20:34.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18627" PostTypeId="2" ParentId="18618" CreationDate="2016-10-07T07:34:36.807" Score="1" Body="&lt;p&gt;As far as I can see, this is just a faint star. Here is some data from the USNO B1.0 catalogue:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;USNO B1.0 star&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B1.0 ID: 0940-0181715&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RA: 10h10m11.13s +/- 69 milliarcseconds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;declination: + 4 03' 12.3&quot; +/- 195 milliarcseconds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Epoch: 1977.9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proper motion in RA:  12 +/-       2 milliarcseconds/year&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proper motion in dec: 6 +/-       6 milliarcseconds/year&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photometry:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Band mag   Cal Survey  Emulsi field  S/G   xi    eta&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B1: 14.36   1  POSS-I  103a-O 550     4   -0.05    0.03&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R1: 13.33   1  POSS-I  103a-E 550     1    0.05    0.01&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B2: 14.07   2  POSS-II IIIa-J 782     9    0.09    0.26&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R2: 12.79   2  POSS-II IIIa-F 782     8   -0.07   -0.02&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I:  12.37   2  POSS-II IV-N   781         -0.04   -0.30&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Cal&quot; = calibration: 0 means calibrated via bright standard stars on&#xA;the plate; 1 = via faint standards on the plate; 2 = faint standards on&#xA;adjacent plate; 3 = faint standards two plates away,  etc.&#xA;  &quot;S/G&quot; = star/galaxy index. 11=object looked starlike, to 0=fuzzy/galaxy-like.&#xA;  xi, eta = displacement of the star's position on this plate,  in&#xA;arcseconds,  relative to the mean solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This USNO-B1.0 data was downloaded from a VizieR server.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2016-10-07T07:34:36.807" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18629" PostTypeId="2" ParentId="18613" CreationDate="2016-10-07T14:37:37.687" Score="0" Body="&lt;p&gt;One way to answer this is to say that conservation of energy is a law intended to be applied in a single reference frame-- it is not intended to work if you change reference frames.  It is true that general relativity makes this an even more difficult point, because it's not just changing reference frames there, but a simpler question appears even in special relativity.  If a source of light is moving away from you, and that source is turned on for awhile and then turned off, receivers who are moving away from that source will detect less total light energy than those not moving away.  This is true even if the source is a beam that goes entirely into the detector.  Thus you could equally ask, if the detector moves away, and detects less energy, where did the missing energy go?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is, there is not any missing energy.  The detector moving away always reckoned there was less energy in that beam, and that less energy was conserved the whole time.  Conservation of energy is within a given reference frame-- even the energy of a bullet gets much less if you change to the reference frame of the bullet.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-10-07T14:37:37.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18630" PostTypeId="1" CreationDate="2016-10-07T14:37:45.443" Score="1" ViewCount="774" Body="&lt;p&gt;Solar and lunar eclipses are pretty straightforward. But what about eclipses when you are standing on the Moon? I have been trying to Google the info, but there's some gaming clan that has taken the name &quot;terran eclipse&quot; so it's hard to find the real stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was able to find &lt;a href=&quot;http://blogs.discovermagazine.com/badastronomy/2009/02/19/what-does-a-lunar-eclipse-look-like-from-the-moon/#.V_eysuArJEY&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; on lunar eclipse from the Moon (the Earth eclipsing the sun). But what about the other way - a solar eclipse observed from the Moon? What would Earth look like? Would anything special be happening on the Moon?&lt;/p&gt;&#xA;" OwnerUserId="14497" LastActivityDate="2016-10-13T02:26:43.673" Title="Terran eclipse?" Tags="&lt;solar-eclipse&gt;&lt;lunar-eclipse&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18631" PostTypeId="2" ParentId="18623" CreationDate="2016-10-07T14:46:59.423" Score="0" Body="&lt;p&gt;It's an important question, because it tells us something important about the Big Bang model.  Some think that the Big Bang explains the expansion, but actually, it takes expansion as an initial condition (and it is typical of dynamical models that they require an initial condition that is outside the model) and tracks how it proceeds.  We generally don't explain why our models work, we just test our models.  We would need some new model to test, one that currently does not exist or is highly speculative, to be able to say why the Big Bang model works, and why the universe is expanding.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-10-07T14:46:59.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18632" PostTypeId="2" ParentId="18630" CreationDate="2016-10-07T14:56:46.960" Score="4" Body="&lt;p&gt;A lunar eclipse on earth is a solar eclipse on the moon.&#xA;The earth takes up 2° of sky from the lunar surface (vs 0.5°), so you'd see less corona in your space suit. The moon's speed through earth's shadow is only about 1 km per second, so you'd have &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_eclipse&quot; rel=&quot;nofollow&quot;&gt;up to 107 minutes of totality&lt;/a&gt;. What with earth's atmosphere, I doubt you'd see much in the way of &lt;a href=&quot;https://en.wikipedia.org/wiki/Baily%27s_beads&quot; rel=&quot;nofollow&quot;&gt;Baily's beads&lt;/a&gt;, but there'd likely be a spectacular round the world sunset glow to make up for that.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="2031" LastEditDate="2016-10-07T17:32:48.613" LastActivityDate="2016-10-07T17:32:48.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18633" PostTypeId="1" CreationDate="2016-10-07T15:03:03.040" Score="3" ViewCount="747" Body="&lt;p&gt;When I was a very little girl, my dad talked about the HUGE full moon he saw while in the navy. He made the comment during a full moon we were having in Western NYS. He said, this (harvest) moon is nothing compared to the one he saw. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now years later, I was with my grown son, who's a geologist and who has been all over the world but not during a full moon. He told me my dad was wrong. My son said that the full moon is always about the size of a penny when you hold the penny up to compare the size of the moon to it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have seen pictures of HUGE full moons over cities my whole life and I assumed my dad was correct. Is there a place where I can travel to see a HUGE full moon that would be much bigger than the full moon I see from the shores of Lake Erie? I understand that the horizon makes the moon bigger but the pictures I have seen were not moons on the horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've searched Stacker's data base but I didn't see this question.&lt;/p&gt;&#xA;" OwnerUserId="14463" LastEditorUserId="2365" LastEditDate="2016-10-07T15:49:25.423" LastActivityDate="2016-10-07T16:06:28.113" Title="From our world's perspective, does the full moon look larger in one city than it does in another?" Tags="&lt;the-moon&gt;&lt;night-sky&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18634" PostTypeId="2" ParentId="18633" CreationDate="2016-10-07T15:23:26.593" Score="4" Body="&lt;p&gt;Your son is more or less correct. The moon's size doesn't change appreciably. At best you only see the size vary by about 13% (in angular diameter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variation in the apparent size of the moon comes from the fact that the moon's orbit around us is not perfectly spherical and sometimes it is closer and sometimes farther. The moon will appear largest when a full moon coincides with the moon being at its closest point in its orbit. This is colloquially referred to as a &quot;Super Moon&quot;. The opposite, being at its farthest point in its orbit during a full moon, is known as a &quot;Micro Moon&quot; or &quot;Mini Moon&quot;. These are really the two extremes, but most often the moon is a more average size that doesn't appear to change much. Shown below is an image of both a Super Moon and a Micro Moon. You can see the size difference, which is noticeable, but not hugely so. If you want to see a large moon, find the time when the next Super Moon occurs (a good upcoming Super Moon is on November 14, 2016).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WgVX4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WgVX4.jpg&quot; alt=&quot;Super Moon vs Micro Moon&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also do mention the &lt;a href=&quot;https://en.wikipedia.org/wiki/Moon_illusion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Moon Illusion&lt;/a&gt;, where a Moon on the horizon appears larger than it really is. The cause for this is debatable, but it is a known optical illusion. I believe the reason for this effect is that near the horizon, you have points of reference that allow you to judge the size of the moon different than when it is high in the sky, far from other comparison to other objects. You can attempt to view the Super Moon as it is rising so as to take full effect of this illusion, but try to do so in an area that isn't flat and devoid of structures (e.g., don't watch the Moonrise over a body of water).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your father, possibly he saw a Super Moon near the horizon which maximized both the optical illusion and the moon's natural size variability. Possibly he his memory also became exaggerated over the years so he remembered a Moon which was larger than it really was. &lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-10-07T15:23:26.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18635" PostTypeId="2" ParentId="18633" CreationDate="2016-10-07T16:06:28.113" Score="4" Body="&lt;p&gt;There is essentially no variation in moon size when viewed from different locations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radius of the Earth is about 6500km, but the distance to the moon is about 400000 km: This means that the distance to the moon. and hence its apparent size, varies by less than 2% from place to place. That is not enough to be noticed. There is greater variation due to the non-circular orbit of the moon, which produces &quot;super moons&quot; and &quot;micro moons&quot; at regular intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Images of very large moons over cities are either photoshopped and fake, or taken with a telephoto to make the moon seem large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your son is correct. There are no locations on Earth at which the moon is significantly larger in the sky. The Moon is always about ½ degree. That is significantly smaller than a penny held at arm's length; it's about half the size.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-10-07T16:06:28.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18636" PostTypeId="1" CreationDate="2016-10-07T16:30:30.160" Score="0" ViewCount="201" Body="&lt;p&gt;Given two colorindices LX and LY, such that the color X-Y is given by&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X-Y = -2.5 log(LX/LY)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;is there a way to infer the luminosity in X (or Y)? I'm think not because any information on units has fallen out through this definition, but I could be wrong. I'm just not sure how to compute the luminosity straightforwardly otherwise, as this color index is the only output (along with the distance modulus) from an interpolation scheme of stellar templates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="6965" LastActivityDate="2016-10-07T17:19:29.653" Title="From color index to luminosity" Tags="&lt;galaxy&gt;&lt;luminosity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18637" PostTypeId="2" ParentId="18636" CreationDate="2016-10-07T17:19:29.653" Score="1" Body="&lt;p&gt;You can never tell the luminosity of a star from knowledge of its color X-Y alone, that only tells you something about the surface temperature.  The problem is not solved by knowing the distance modulus, that only tells you how far the star is, but not how big it is.  If you want luminosity from those things, you need to know how bright the star looks, not just its color and distance.  Or you could look at the width of the spectral lines to infer luminosity class, but that brings in line resolution.  But if you know the distance modulus and Lx and Ly, not just Lx/Ly, then you can get the luminosity because you know its color, how bright it looks, and its distance.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-10-07T17:19:29.653" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18640" PostTypeId="1" AcceptedAnswerId="18641" CreationDate="2016-10-07T18:23:19.153" Score="2" ViewCount="245" Body="&lt;p&gt;According to &lt;a href=&quot;http://mnras.oxfordjournals.org/content/387/4/1649.full.pdf&quot; rel=&quot;nofollow&quot;&gt;Begelman et al. (2008)&lt;/a&gt;, one of the most distinguishing features of the hypothetical quasi-star is that it's supported by radiation pressure from the accretion disk of the black hole in its core. The paper suggests that these stars form from Population III stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's more, while almost all of the star's luminosity is from the accretion disk (which is far more luminous than thermonuclear reactions), temperatures around the central black hole are hot enough for fusion:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Although we do not model the non-hydrostatic region of quasi-stars in any detail, in our case high temperatures &lt;em&gt;are&lt;/em&gt; attained in the immediate vicinity of the black hole. However, even in this region the neglect of nuclear reactions is justified, first because black hole accretion is energetically much more efficient than fusion, and secondly because on-going accretion limits the time-scale over which inflowing gas is exposed to high $T$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's all fine, but what surprised me is the footnote on the third page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the most massive quasi-stars, the central temperature may be high enough (a few million K) to initiate lithium burning. This is energetically negligible, and although the presence or absence of lithium does affect the opacity, the effect is small for the photospheric temperatures and densities of interest here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even though this part dismisses lithium burning as an important factor, since it's energy output doesn't compare to that of the accretion disk's, it still mentions that lithium burning can take place. Considering that these form from Population III stars, how is it that much lithium burning can happen in the first place?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although lithium is a relatively light atom, it is considered a metal and lithium burning depletes the amount of lithium in a star. Since Population III stars are supposed to be nearly metal-free, shouldn't only hydrogen or helium fusion be a factor here, even if they only have a small effect?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-10-07T18:33:11.750" LastActivityDate="2016-10-07T19:45:43.590" Title="How could lithium burning take place in a quasi-star?" Tags="&lt;black-hole&gt;&lt;temperature&gt;&lt;early-universe&gt;&lt;metallicity&gt;&lt;hypothetical&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18641" PostTypeId="2" ParentId="18640" CreationDate="2016-10-07T19:30:33.323" Score="5" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Lithium#Astronomical&quot; rel=&quot;nofollow&quot;&gt;Lithium&lt;/a&gt;, along with Hydrogen and Helium, was one of the 3 elements created in the Big Bang. Thus, it should exist to some part in any star that hasn't burnt all of it out, and as mentioned, it's not an easy thing to do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/P/Population+III&quot; rel=&quot;nofollow&quot;&gt;Population III&lt;/a&gt; stars are expected to contain Lithium, and Beryllium as well. The amount, however, is not particularly high.&lt;/p&gt;&#xA;" OwnerUserId="182" LastEditorUserId="182" LastEditDate="2016-10-07T19:45:43.590" LastActivityDate="2016-10-07T19:45:43.590" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18643" PostTypeId="1" CreationDate="2016-10-07T20:18:26.177" Score="-1" ViewCount="223" Body="&lt;p&gt;I know there's no such thing like a perpetual motion machine but I missing something in my reasoning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume there's a gas giant planet where the immense pressure of the core causes it to heat and create a convection current of its gases.&#xA;If there's no atmosphere gases losses to space that means we have a perpetual motion machine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is work being done to move all that gas (or not?). The energy to compress gases is coming from gravity but the total energy of the system remains the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume some heat is being dissipated in space, that means the system is losing energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this possible go forever or will the planet reach a equilibrium where all gases and heat will be distributed in a way the convection cease?&#xA;Or even can possible it loss energy enough to loss mass (and cease the heavy gravity effects)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus:&#xA;Consider we got two kinds of life forms: one breathing XY and exhaling X+Y and another doing the inverse and using the heat from some (goldilock zone) atmosphere layer (some kind of XY + Energy =&gt; X + Y, X + Y + Energy =&gt; XY cycle). Will life perse consume the energy and cooldown the planet (eventually in billions years)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a bit of confusing about this question due to my difficult in express my doubts. Sure it's impossible to achieve total &quot;isolation&quot;. I'm just pondering about an hypothetical scenario.&#xA;I see now the forever convection is just equivalent to a frictionless object spinning foerever in the void.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any energy removed from the system will &quot;slow down&quot; it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The true question is: What if the energy of this isolated system is used to fuel life and this life is inside the system and is part of the system? It still means energy is lost?&lt;/p&gt;&#xA;" OwnerUserId="2951" LastEditorUserId="616" LastEditDate="2021-02-12T21:13:24.223" LastActivityDate="2021-02-12T21:13:24.223" Title="Is a Gas Giant a Perpetual Motion Machine?" Tags="&lt;gas-giants&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18644" PostTypeId="5" CreationDate="2016-10-07T20:28:56.247" Score="0" Body="" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-10-07T20:28:56.247" LastActivityDate="2016-10-07T20:28:56.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18645" PostTypeId="4" CreationDate="2016-10-07T20:28:56.247" Score="0" Body="Questions about hypothetical objects or concepts in astronomy (including wormholes, white holes and quasi-stars). This tag is not for counterfactuals or &quot;what-if&quot; questions." OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-10-07T20:28:56.247" LastActivityDate="2016-10-07T20:28:56.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18646" PostTypeId="2" ParentId="18643" CreationDate="2016-10-07T21:41:35.907" Score="2" Body="&lt;p&gt;You don't have a perpetual motion machine because you are losing energy to space in the form of heat, light, solar wind, etc.  Even if you want to restrict it to the case where you are not expelling gas, you are always going to lose energy to heat radiation.  The Sun loses mass all the time.  It will keep doing this until it cannot generate enough internal heat to counterbalance its tendency to contract due to gravity.  There is a whole progression where it burns its hydrogen, then its helium, etc.  Finally, when it runs out of stuff to burn, it collapses and &quot;blows up&quot; in a nova or supernova, or it could just fizzle out.  Its fate depends upon its mass.&lt;/p&gt;&#xA;" OwnerUserId="9749" LastActivityDate="2016-10-07T21:41:35.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18647" PostTypeId="2" ParentId="1876" CreationDate="2016-10-08T02:06:27.280" Score="1" Body="&lt;p&gt;Consider 2 distant galaxies &quot;moving&quot; apart due to the expansion of space. A photon of light is leaving galaxy A towards galaxy B. The space between galaxy B and galaxy A is expanding. This includes the space between galaxy B and the photon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It takes light about a nanosecond to travel 1 foot. If the space between galaxy B and the photon expands at a rate of more than 1 foot every nanosecond, then the distance between galaxy B and the photon will always be increasing, so the photon will never reach Galaxy B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, the speed of light is constant. The speed is not relative to where the light originated from or to an observer at any arbitrary location. It is measured relative to the light itself. At one moment in time the light is at some location, and a given amount of time later it has moved a distance given by the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You are on a spacecraft traveling in a straight line, very fast, lets say (arbitrarily) at 25,000 kilometers per hour. The particular speed of the spacecraft, or even that it is moving very fast is irrelevant and is mentioned only as an example. &lt;/li&gt;&#xA;&lt;li&gt;You pass by another spacecraft traveling in a straight line, in the exact opposite direction as your spacecraft, also traveling at 25,000 kilometers per hour.&lt;/li&gt;&#xA;&lt;li&gt;The 2 spacecraft are traveling parallel to each other.&lt;/li&gt;&#xA;&lt;li&gt;At the moment you pass the other spacecraft, you project a light out of the front of your spacecraft.&lt;/li&gt;&#xA;&lt;li&gt;At that same moment, a light is projected out of the rear of the other spacecraft.&lt;/li&gt;&#xA;&lt;li&gt;The light from both spacecraft is traveling parallel to each other in the same direction, and at the same speed (&quot;the speed of light&quot;).&lt;/li&gt;&#xA;&lt;li&gt;The light from both spacecraft will reach a distant object at the same time.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14499" LastActivityDate="2016-10-08T02:06:27.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18650" PostTypeId="1" CreationDate="2016-10-08T15:20:56.060" Score="0" ViewCount="205" Body="&lt;p&gt;I am new to this domain and trying to &lt;em&gt;understanding the fundamentals of local sidereal time&lt;/em&gt;.&#xA;Through an online available calculator &lt;a href=&quot;https://www.iiap.res.in/personnel/reks/software/javascript/calclst.php&quot; rel=&quot;nofollow&quot;&gt;https://www.iiap.res.in/personnel/reks/software/javascript/calclst.php&lt;/a&gt; ,I am noting down the &lt;code&gt;local sidereal time&lt;/code&gt; for my &lt;code&gt;longitude&lt;/code&gt; of interest. I have repeated the exercise for same time for five consecutive days. What I observe is that the &lt;code&gt;local sidereal time&lt;/code&gt; is advancing by &lt;code&gt;0.06/07 hours per day&lt;/code&gt; for the same time.Like &lt;code&gt;21.88 Hours&lt;/code&gt; for current date and time , &lt;code&gt;21.94 Hours&lt;/code&gt; for the same time next day ,&lt;code&gt;22.01 Hours&lt;/code&gt; for next to next day and the pattern continues. &lt;em&gt;Kindly let me know how this increment is happening&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="14505" LastActivityDate="2016-10-08T17:37:12.580" Title="Local sidereal time interpretation" Tags="&lt;fundamental-astronomy&gt;&lt;time&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18651" PostTypeId="2" ParentId="18643" CreationDate="2016-10-08T15:27:57.923" Score="5" Body="&lt;p&gt;It is not a perpetual motion machine. Energy is being converted from gravitational potential energy in the gas, to motion, and eventually to heat which is lost as heat radiation. As this happens the planet shrinks and cools. This takes a long time, a lot longer than the few billion years of the the solar system. But ultimately convection will stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All life forms depend on capturing energy. All life processes increase entropy. Life can have a significant effect on a planet, but not by &quot;consuming energy&quot; rather by changing the composition of the atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-10-08T15:27:57.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18652" PostTypeId="2" ParentId="18650" CreationDate="2016-10-08T17:37:12.580" Score="2" Body="&lt;p&gt;If you extrapolate, you'll find that those ~4 minutes per day add up to 24 hours per year.&#xA;The difference is due to the Earth revolving around the Sun at a rate of almost 1 degree per day.&#xA;For every 365$\frac{1}{4}$ solar days, there are 366$\frac{1}{4}$ sidereal days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Civil time is based on mean&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_time&quot; rel=&quot;nofollow&quot;&gt;solar time&lt;/a&gt;,&#xA;in which the Sun crosses the&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Meridian_%28astronomy%29&quot; rel=&quot;nofollow&quot;&gt;celestial meridian&lt;/a&gt;&#xA;every 24h 0m on average.&#xA;In &lt;a href=&quot;https://en.wikipedia.org/wiki/Sidereal_time&quot; rel=&quot;nofollow&quot;&gt;sidereal time&lt;/a&gt;,&#xA;24h 0m is the time between successive meridian transits by any given star, equivalent to 23h 56m 4s of solar time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sidereal time also corresponds to the&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Right_ascension&quot; rel=&quot;nofollow&quot;&gt;right ascension&lt;/a&gt;&#xA;of stars crossing the meridian at that time.&#xA;If a star in Orion has right ascension 5h 30m, then Orion is on the meridian at 05:30 sidereal time, which occurs 2 hours earlier in the solar day per month.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-10-08T17:37:12.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18653" PostTypeId="2" ParentId="18567" CreationDate="2016-10-08T19:30:09.837" Score="4" Body="&lt;p&gt;The velocity required to escape the gravitational attraction of a massive body is given by the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v_{\mathrm{escape}} = \sqrt{\frac{2GM}{R}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $G$ is the gravitational constant ($G = 6.67 \times 10^{-11} \; \mathrm{Nm^{2} {kg}^{-2}}$), $M$ is the mass of the body from which you are escaping, and $R$ is its radius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inputting the values for Io's &lt;a href=&quot;https://en.wikipedia.org/wiki/Io_(moon)&quot; rel=&quot;nofollow&quot;&gt;mass and mean radius&lt;/a&gt;, $M = 0.015 \, M_{\oplus}$ and $R = 0.286 \, R_{\oplus}$,&lt;sup&gt;1&lt;/sup&gt; we get an escape velocity of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v_{\mathrm{escape}} = 2560 \; \mathrm{m/s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the explosive ejecta are ejected from the top of volcanoes, so we should strictly add this to our radius. The tallest volcano on Io is approximately $2.5 \;\mathrm{km}$ above the surface; including this we get a &lt;em&gt;marginally&lt;/em&gt; lower velocity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v_{\mathrm{escape}} = 2559 \; \mathrm{m/s} $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is higher than the $\sim 1000 \; \mathrm{m/s}$ maximum velocity of ejecta calculated in &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0019103583900751&quot; rel=&quot;nofollow&quot;&gt;McEwen &amp;amp; Soderblom (1983)&lt;/a&gt;. Therefore, &lt;strong&gt;no mass is ejected from the surface of Io through volcanic eruptions&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparison, the earth's escape velocity is much higher, $11.2 \; \mathrm{km/s}$.  As discussed in the linked paper, the most extreme ejecta can reach heights of $500 \; \mathrm{km}$ before falling back to the surface.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;&lt;sub&gt;where $M_{\oplus} = 5.972 \times 10^{24} \; \mathrm{kg}$ and $R_{\oplus} = 6371 \;\mathrm{km}$&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="941" LastEditorUserId="941" LastEditDate="2016-10-19T14:27:11.307" LastActivityDate="2016-10-19T14:27:11.307" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18654" PostTypeId="2" ParentId="18154" CreationDate="2016-10-08T21:31:29.807" Score="3" Body="&lt;p&gt;The answer depends on your frame of reference.&#xA;In a &lt;a href=&quot;https://en.wikipedia.org/wiki/Heliocentrism&quot; rel=&quot;nofollow&quot;&gt;heliocentric&lt;/a&gt; frame, the Sun is fixed at the center by definition.&#xA;In a &lt;a href=&quot;https://en.wikipedia.org/wiki/Barycenter&quot; rel=&quot;nofollow&quot;&gt;barycentric&lt;/a&gt; frame,&#xA;defined relative to the Solar system's combined center of mass, the Sun moves in a complex multi-year loop opposite the planets' orbits, never wandering more than its own diameter away from the barycenter as it is the most massive body by far.&#xA;The distinction between these frames may not have been obvious until Newton published his laws of motion and gravitation.&#xA;In a galactic or supergalactic context, the difference in motion between the Sun and planets is negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One frame is not inherently more correct than another, but one may be more useful for a given purpose.&#xA;When astronomers discover an asteroid, they measure its geocentric spherical coordinates.&#xA;Having several such measurements over a span of time, they approximate its current trajectory with an ideal heliocentric ellipse.&#xA;To estimate the asteroid's trajectory 100 years in the future, their N-body gravitational simulation would use barycentric coordinates.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-10-08T21:31:29.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18655" PostTypeId="2" ParentId="18585" CreationDate="2016-10-09T02:03:51.157" Score="6" Body="&lt;p&gt;There's a recent study on this, based on satellite and ground observations around the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://advances.sciencemag.org/content/2/6/e1600377.full&quot;&gt;the paper&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Milky Way is hidden from more than one-third of humanity, including 60% of Europeans and nearly 80% of North Americans.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And there's a map: &lt;a href=&quot;http://cires.colorado.edu/Artificial-light&quot;&gt;The New World Atlas of Artificial Sky Brightness&lt;/a&gt;. This is probably the most accurate and up to date light pollution map we have at the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The percentages are based on population density, I don't know of any numbers for individuals. &lt;/p&gt;&#xA;" OwnerUserId="9682" LastActivityDate="2016-10-09T02:03:51.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18656" PostTypeId="2" ParentId="138" CreationDate="2016-10-09T04:08:37.490" Score="-2" Body="&lt;p&gt;So, far all the answers have shed light on the experimental reasons. But you can say it with full confidence without any astronomical observations that the Sun rotates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logic is that any closed system have a center of mass that is unaltered unless acted upon by some external force. So, given that Earth rotates around Sun then the system forming Earth and Sun must have a center of mass which is not changing. Therefore Sun must rotate to counter the change in center of mass caused by the rotation of Earth. And so it does with any other system belonging to our solar system.&lt;/p&gt;&#xA;" OwnerUserId="712" LastActivityDate="2016-10-09T04:08:37.490" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18657" PostTypeId="2" ParentId="18630" CreationDate="2016-10-09T08:50:49.213" Score="5" Body="&lt;p&gt;As Wayfaring Stranger's answer says, during a lunar eclipse seen from Earth, an observer on the Moon would see a solar eclipse. Earth is about 4 times the apparent diameter of the Sun. You can see a photo of such an eclipse, taken by Japan's Kaguya spacecraft in lunar orbit, &lt;a href=&quot;http://www.universetoday.com/25929/kaguya-captures-eclipse-from-the-moon/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4jZ5w.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4jZ5w.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During a solar eclipse seen from Earth, an observer on the Moon would see the Moon's shadow on the surface of Earth. I don't know of any photos of such a shadow as seen from the Moon, but here's a photo of a solar eclipse seen from the Mir space station in 1999.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CkrNC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CkrNC.jpg&quot; alt=&quot;Looking Back at an Eclipsed Earth&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://apod.nasa.gov/apod/ap110102.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://apod.nasa.gov/apod/ap110102.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="53" LastEditorUserId="53" LastEditDate="2016-10-13T02:26:43.673" LastActivityDate="2016-10-13T02:26:43.673" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18659" PostTypeId="1" CreationDate="2016-10-09T17:17:31.563" Score="1" ViewCount="1365" Body="&lt;p&gt;When a bird is flying or when a copter is hovering or a plane is flying what force is making them rotate with the earth beneath them?&lt;/p&gt;&#xA;" OwnerUserId="14515" LastEditorUserId="13426" LastEditDate="2016-10-09T19:16:52.550" LastActivityDate="2016-10-10T13:24:18.817" Title="What makes a flying object keep moving with the Earth's rotation?" Tags="&lt;rotation&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" ClosedDate="2016-10-12T19:47:28.100" ContentLicense="CC BY-SA 3.0" />
  <row Id="18661" PostTypeId="2" ParentId="18659" CreationDate="2016-10-09T18:16:08.603" Score="0" Body="&lt;p&gt;When something is flying through the air, it keeps moving along with the Earth because it was already moving that way to begin with.  Remember that according to &lt;a href=&quot;https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion#Newton.27s_1st_Law&quot; rel=&quot;nofollow&quot;&gt;Newton's First Law&lt;/a&gt;, things will keep moving in the same direction unless an outside force acts on them.  So when a bird, plane, or helicopter is on the ground, they are rotating along with the Earth.  In the rotating reference frame of the Earth, they are stationary.  From an external frame which sees the Earth as rotating, the object will be rotating in the same way.  Now, when a plane takes off, it is changing its motion relative to the ground.  In that reference frame, it begins moving.  However, from our external reference frame, it was already moving.  So its just changing its already non-zero velocity.&lt;/p&gt;&#xA;" OwnerUserId="13426" LastActivityDate="2016-10-09T18:16:08.603" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="18662" PostTypeId="1" CreationDate="2016-10-09T23:47:08.957" Score="2" ViewCount="61" Body="&lt;p&gt;I'm trying to find out if there is a range of times at night, and/or days during the year, in which the probability of seeing an aurora is higher than average. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aurora section of the British Astronomical Association asserts that there's &#xA;a correlation with season (higher probability near equinoxes), and a correlation with 'magnetic midnight'. &#xA;&lt;a href=&quot;http://www.britastro.org/aurora/&quot; rel=&quot;nofollow&quot;&gt;http://www.britastro.org/aurora/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these assertions correct? If so, what is the definition of magnetic midnight, exactly?&lt;/p&gt;&#xA;" OwnerUserId="5359" LastActivityDate="2017-02-01T11:11:03.903" Title="Auroras: is there any bump in frequency at certain times of night, or times of year?" Tags="&lt;amateur-observing&gt;&lt;aurora&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18664" PostTypeId="1" CreationDate="2016-10-10T03:44:29.763" Score="1" ViewCount="66" Body="&lt;p&gt;Suppose we assume that the gravitational constant is varying with time:&#xA;$$G=G_0 e^{-\frac{t}{t_0}}$$ where $t$ is the time from the Big Bang, and $G_0, t_0$ being some constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming matter-only universe, which one of the expansion equation should we use to obtain the Hubble's parameter $H$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H^2=\left\{\frac{\dot R}{R}\right\}^2=\frac{8\pi G}{3}\rho_0\frac{R_0^3}{R^3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial (H^2R^2)}{\partial (R^2)}=\frac{\ddot R}{R}= -\frac{4\pi G}{3}\rho_0\frac{R_0^3}{R^3}$$&lt;/p&gt;&#xA;" OwnerUserId="4200" LastEditorUserId="9301" LastEditDate="2016-10-27T16:00:50.957" LastActivityDate="2016-10-27T16:00:50.957" Title="Cosmology: about the equations of expansions" Tags="&lt;cosmology&gt;&lt;expansion&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18666" PostTypeId="1" CreationDate="2016-10-10T07:33:56.440" Score="9" ViewCount="354" Body="&lt;p&gt;There are hypotheses about a collision between Earth and an astronomical body the size of Mars, approximately 4.5 billion years ago, about 20 to 100 million years after the solar system coalesced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess this changed the orbital path of earth around the Sun. Do we have any info what was the Earth's orbital path before the collision? (I would prefer some visual explanation.)&lt;/p&gt;&#xA;" OwnerUserId="14522" LastActivityDate="2017-12-30T22:07:08.917" Title="What was the orbital path of earth before collision with a Mars sized planet?" Tags="&lt;impact&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18667" PostTypeId="1" AcceptedAnswerId="18679" CreationDate="2016-10-10T10:14:32.987" Score="5" ViewCount="233" Body="&lt;p&gt;The observation of comet 67P by the Kepler space telescope has been reported in various science news outlets recently. For example, the image below appeared in the recent NASA news article &lt;a href=&quot;http://www.nasa.gov/image-feature/ames/kepler/nasas-kepler-gets-the-big-picture-of-comet-67p&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA's Kepler Gets the 'Big Picture' of Comet 67P&lt;/a&gt; and the same or similar images can be seen elsewhere. The article says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;From Sept. 7 through Sept. 20, the Kepler spacecraft, operating in its K2 mission, fixed its gaze on comet 67P. From the distant vantage point of Kepler, the spacecraft could observe the comet's core and tail. The long-range global view of Kepler complements the close-in view of the Rosetta spacecraft, providing context for the high-resolution investigation Rosetta performed as it descended closer and closer to the comet.&lt;/p&gt;&#xA;&lt;p&gt;During the two-week period of study, Kepler took a picture of the comet every 30 minutes. The animation shows a period of 29.5 hours of observation from Sept. 17 through Sept. 18. The comet is seen passing through Kepler's field of view from top right to bottom left, as outlined by the diagonal strip. The white dots represent stars and other regions in space studied during K2's tenth observing campaign.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I don't really understand this image. It is a GIF assembled from several exposures, but in each frame, individual stars appear as tight clusters of dots for some reason, and the comet moves within a bright, narrow band that crosses the entire image in all frames. Why does this looks so strange?&lt;/p&gt;&#xA;&lt;p&gt;Also, what is the importance of Kepler deviating from it's observations to concentrate on 67P during this period? What is the scientific importance of this series of observations?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/izir3.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/izir3.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Image credit: The Open University/C. Snodgrass and SETI Institute/E. Ryan From &lt;a href=&quot;http://www.nasa.gov/image-feature/ames/kepler/nasas-kepler-gets-the-big-picture-of-comet-67p&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-10-11T23:18:04.623" Title="Why does Kepler's &quot;Big Picture&quot; of comet 67P look so strange? And what is the significance?" Tags="&lt;comets&gt;&lt;photography&gt;&lt;kepler&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18669" PostTypeId="2" ParentId="18659" CreationDate="2016-10-10T13:24:18.817" Score="1" Body="&lt;p&gt;They aren't.  To first order, airborne objects are moving with the Earth because they were moving when on the ground.  However, what they retain is their angular speed, and since the circumference increases with altitude, an object that doesn't accelerate appropriately will appear from the ground to fall behind to the West (the Earth is rotating to the East).  This is a well-known phenomenon whose magnitude is latitude-dependent, falling to zero at the poles. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2016-10-10T13:24:18.817" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18671" PostTypeId="1" AcceptedAnswerId="18674" CreationDate="2016-10-10T15:03:26.683" Score="1" ViewCount="1235" Body="&lt;p&gt;The definition of flux and flux density is fairly straightforward. However, I'm musing over the practical usage of it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one thing, if you observe a source through a filter, how would one go from the total observed energy/number of photons to the flux &lt;em&gt;per unit wavelength&lt;/em&gt;? It's not a matter of simply dividing by the width of the filter -- a filter can have any exotic form of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how is this derived?&#xA;Related, &lt;em&gt;given&lt;/em&gt; a $f_{\nu}$ in some filter, what would be the way to go to total flux within that filter?&lt;/p&gt;&#xA;" OwnerUserId="6965" LastActivityDate="2016-10-10T18:08:13.407" Title="Flux density to flux" Tags="&lt;observational-astronomy&gt;&lt;luminosity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18672" PostTypeId="1" AcceptedAnswerId="18673" CreationDate="2016-10-10T15:53:56.543" Score="1" ViewCount="524" Body="&lt;p&gt;I am a Computer Engineering major and would like to learn more about machine learning methods in Astronomy. What books or other resources could I use?&lt;/p&gt;&#xA;" OwnerUserId="14527" LastEditorUserId="14527" LastEditDate="2016-10-13T21:35:37.793" LastActivityDate="2016-10-13T21:35:37.793" Title="Astronomy &amp; machine learning. Where to start?" Tags="&lt;observational-astronomy&gt;&lt;astrophysics&gt;&lt;radio-astronomy&gt;&lt;data-analysis&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="3" ClosedDate="2016-10-13T16:16:24.707" ContentLicense="CC BY-SA 3.0" />
  <row Id="18673" PostTypeId="2" ParentId="18672" CreationDate="2016-10-10T16:18:18.377" Score="3" Body="&lt;p&gt;This recent &lt;a href=&quot;https://www.youtube.com/watch?v=jAzC1fVksng&quot; rel=&quot;nofollow&quot;&gt;KISSCaltech lecture&lt;/a&gt; by Prof. Rob Fergus at the NYU might be of interest. &lt;em&gt;&quot;An Overview of Machine Learning Approaches: Applications to Exoplanet Detection&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-10-10T16:18:18.377" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18674" PostTypeId="2" ParentId="18671" CreationDate="2016-10-10T17:44:04.620" Score="3" Body="&lt;p&gt;If the normalised filter response function is $R_{\nu}$ then the measured  flux is&#xA;$$ F = \int f_{\nu} R_{\nu}\ d\nu  $$&#xA;The integration is done over the frequency range of the filter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you measure a flux through a filter then the process cannot be inverted exactly. However the &lt;em&gt;average&lt;/em&gt; flux density can be found by dividing the total flux by the effective frequency range&#xA;$$\langle f_{\nu} \rangle = F/ \int R_{\nu} d\nu $$ &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-10-10T18:08:13.407" LastActivityDate="2016-10-10T18:08:13.407" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18675" PostTypeId="2" ParentId="18666" CreationDate="2016-10-10T20:03:01.257" Score="1" Body="&lt;p&gt;The total angular momentum would always be conserved about the Sun (assuming it at rest), as there is no torque about the Sun. Apart from that, I think it is hardly possible to say anything about the orbits. Also, it may be true that the pre-collision orbital plane and the post-collision orbital planes changed due to the collision, as small transverse perturbations must also have resulted from such a collision. However, I would also like to see what other members suggest. &lt;/p&gt;&#xA;" OwnerUserId="13414" LastEditorUserId="1891" LastEditDate="2017-12-30T22:07:08.917" LastActivityDate="2017-12-30T22:07:08.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18676" PostTypeId="1" AcceptedAnswerId="18696" CreationDate="2016-10-11T15:17:25.720" Score="3" ViewCount="85" Body="&lt;p&gt;I &lt;a href=&quot;https://en.wikipedia.org/wiki/Baryon_acoustic_oscillations&quot; rel=&quot;nofollow&quot;&gt;understand from this Wikipedia page&lt;/a&gt; about baryon acoustic oscillations (BAOs). I can also see that BAO measurements were first detected using luminous red galaxy &lt;em&gt;samples&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, many papers use galaxy &lt;em&gt;clusters&lt;/em&gt; to measure the BAO signal. What is the advantage/difference between using galaxy samples and galaxy cluster samples to measure the BAO?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some of the papers using galaxy clusters for BAO detection. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://arxiv.org/abs/0801.3485&quot; rel=&quot;nofollow&quot;&gt;Estrada et al. 2008&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://arxiv.org/abs/1202.0640&quot; rel=&quot;nofollow&quot;&gt;Hong et al. 2012&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://arxiv.org/abs/1311.5895&quot; rel=&quot;nofollow&quot;&gt;Veropalumbo et al. 2014&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2655" LastEditorUserId="2655" LastEditDate="2016-10-13T09:13:31.810" LastActivityDate="2016-10-13T13:53:45.557" Title="Difference in the baryon acoustic oscillation measurements" Tags="&lt;galaxy&gt;&lt;galaxy-cluster&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18677" PostTypeId="1" AcceptedAnswerId="38659" CreationDate="2016-10-11T19:57:33.250" Score="1" ViewCount="95" Body="&lt;p&gt;Are there any citation or documentation showing the measured degradation of the Moon's equatorial rotation velocity (thus affecting its moon's axial rotation)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for the drift of the Moon's axis rotational speed, presumably over 10,000 years or more.&lt;/p&gt;&#xA;" OwnerUserId="14538" LastEditorUserId="14538" LastEditDate="2020-08-26T11:38:25.380" LastActivityDate="2020-08-31T02:58:46.670" Title="The drift of the Moon's equatorial rotation velocity" Tags="&lt;the-moon&gt;&lt;rotation&gt;&lt;tidal-locking&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18679" PostTypeId="2" ParentId="18667" CreationDate="2016-10-11T23:18:04.623" Score="11" Body="&lt;p&gt;One thing to keep in mind is that the Kepler instrument is not a telescope like Hubble.  It is a photometer and though it uses CCDs to look at the sky, it doesn't return a picture in the usual sense.  The way it operates is that you only look at the pixels around the object you're interested in because otherwise you'd never get all those pixels transmitted to the ground because there are so many of them.&lt;/p&gt;&#xA;&lt;p&gt;To observe the comet, they looked at the track it would make across the Kepler field of view.  They chose only the pixels you'd need to look at the comet and its tail, and this turned out to be a 15-pixel-wide swath across the detector.  That band across the image you see is that swath.  You'll notice that most of the picture is black, which are all the pixels that they didn't look at.&lt;/p&gt;&#xA;&lt;p&gt;The stars in the scene where the image isn't back are the stars that Kepler was also looking at.  However, I'm not intimately familiar with raw Kepler data so I can't tell you why the stars are in clusters of four.  Because all the stars are lined up in the same direction, I would surmise that this is because Kepler was tracking the comet and the stars were moving in the background.  From the way they were going to take the measurements, each 36 hour observation was made up of three observations, so I am at a loss as to why there are four dots per star when this would suggest there would be three.&lt;/p&gt;&#xA;&lt;p&gt;The answer to the science question can be found from the &lt;a href=&quot;https://keplerscience.arc.nasa.gov/data/k2-programs/GO10904_Snodgrass.pdf&quot; rel=&quot;noreferrer&quot;&gt;observation proposal&lt;/a&gt;.  To summarize:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Kepler photometry can provide the following key measurements at a time&#xA;when no other platform can observe the comet:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Total brightness of the comet. This relates directly to the amount of dust being released and therefore gives a critical ‘total’ activity&#xA;measurement, which can be compared with the instantaneous (but local)&#xA;measurements from various instruments on Rosetta.&lt;/li&gt;&#xA;&lt;li&gt;Variability of the comet. There are two sources of variability; changing amounts of dust in the coma, and the underlying light-curve&#xA;of the nucleus (its non-spherical shape reflects more or less sunlight&#xA;as it rotates).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;The measurements from Kepler will extend the total activity curve to&#xA;almost the end of the Rosetta mission, meaning that data in the last&#xA;months of Rosetta will not rely on extrapolation of the earlier&#xA;Earth-based observations. As the final months will be some of the most&#xA;exciting – as Rosetta spirals down and makes local measurements from&#xA;incredibly close to the nucleus – this context information is very&#xA;important in this phase.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9749" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-10-11T23:18:04.623" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18680" PostTypeId="1" CreationDate="2016-10-12T13:21:17.280" Score="2" ViewCount="188" Body="&lt;p&gt;I'm well aware of the celestial bodies considered as the oldest visible in our galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question might feel a bit odd: What is the oldest thing in the universe? I am puzzled with this question as the only answer that I have found by myself sounds weird.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the notion of &quot;old&quot; only makes sense in an observable and ellapsing timeline, the most obvious answer to me would be &quot;Time is the oldest thing&quot;. There is no such thing as a measurement of &quot;old&quot; until there is actually mass to allow time, so which predates which? I'm pretty sure the answer will tightly be linked to Planck's constant, but how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong? How could I find relief in an answer that wouldn't be just philosophical, but also logical? :)&lt;/p&gt;&#xA;" OwnerUserId="14550" LastEditorUserId="14550" LastEditDate="2016-10-12T13:56:05.453" LastActivityDate="2016-10-12T23:04:52.447" Title="What is the oldest thing?" Tags="&lt;space-time&gt;&lt;time-dilation&gt;" AnswerCount="3" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="18681" PostTypeId="2" ParentId="18680" CreationDate="2016-10-12T17:30:04.117" Score="8" Body="&lt;p&gt;Pretty much every hydrogen atom that's in a glass of water has a proton that dates from 1 / 1000000 seconds after the big bang.  That's older than the cosmic microwave background, which dates from almost 400,000 years later.&lt;/p&gt;&#xA;" OwnerUserId="14434" LastActivityDate="2016-10-12T17:30:04.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18683" PostTypeId="2" ParentId="18680" CreationDate="2016-10-12T19:41:16.523" Score="4" Body="&lt;p&gt;If we could detect the cosmic neutrino background, then I would class those neutrinos as &quot;things&quot; and there should be lots of them! I guess whether they are detected or not, they are extremely likely to be there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These cosmic background neutrinos were produced when the universe was about 1 second old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree that most protons are almost a second older!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a tiny fraction of a second older than this would be gravitational waves produced at the end of the inflationary epoch (if we can detect them).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-10-12T23:04:52.447" LastActivityDate="2016-10-12T23:04:52.447" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18684" PostTypeId="2" ParentId="18680" CreationDate="2016-10-12T20:14:03.127" Score="2" Body="&lt;p&gt;If &quot;time&quot; is considered to be a valid answer, I would like to propose a more fundamental principle: the &lt;a href=&quot;https://en.wikipedia.org/wiki/Wave_function&quot; rel=&quot;nofollow&quot;&gt;wave function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since a wave function is a description of the quantum state of any physical system, there must necessarily have been a wave function as soon as quantum physics applied to the Universe. Since we generally assume &lt;a href=&quot;https://en.wikipedia.org/wiki/Uniformitarianism&quot; rel=&quot;nofollow&quot;&gt;uniformity in nature&lt;/a&gt;, there is no reason to consider any sort of &quot;pre-quantum&quot; era.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;All right, I can hear some of you complaining already that it says right in the paragraph above that a wave function is a &lt;em&gt;description&lt;/em&gt;, but isn't that what time is after all? It is just a description of a change in the state of the Universe. If you really want to split hairs, you can say it's actually a quantum state--but wait that's just a condition of a quantum system--but wait that's just saying that the first thing is a thing. What is a thing, really?&lt;/p&gt;&#xA;" OwnerUserId="6" LastActivityDate="2016-10-12T20:14:03.127" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18685" PostTypeId="1" CreationDate="2016-10-12T20:26:41.043" Score="8" ViewCount="1279" Body="&lt;p&gt;It is said that because Mars doesn't have a magnetic field its atmosphere disappeared, but how can solar wind cause that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That the protons and electrons can changes molecules in an atmosphere I can understand, but how can solar wind cause molecules to escape from gravity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it that the atmosphere didn't escape in outer space but in contrary the molecules became too heavy, so they got solid and were pulled to the planet's surface?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="10445" LastEditDate="2016-10-12T20:52:26.030" LastActivityDate="2020-03-22T16:36:38.723" Title="How does the solar wind destroy the atmosphere of a planet lacking a magnetic field?" Tags="&lt;planet&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18686" PostTypeId="1" CreationDate="2016-10-12T20:59:28.260" Score="5" ViewCount="210" Body="&lt;p&gt;I have wondered why the &quot;walls and voids&quot; of the cosmological galaxy distribution seem to resemble a &quot;foam&quot;, like rising bread that's ready to fall, with &quot;voids&quot; corresponding to empty bubbles in the foam. James K and Rob Jeffers &lt;a href=&quot;https://astronomy.stackexchange.com/a/18610/14485&quot;&gt;https://astronomy.stackexchange.com/a/18610/14485&lt;/a&gt; seem to be saying local matter tends to relatively &quot;overrule&quot; the cosmological expansion, and that the bubbles (&quot;voids&quot;) get relatively more cosmic expansion. This may explain the &quot;foam&quot; appearance: local large areas of slightly lower relative galaxy density will tend to cosmic expansion more, and will grow (less dense) more than denser regions, like bubbles in the foam, effectively driving all the galaxies into the walls between the bubbles.&lt;/p&gt;&#xA;" OwnerUserId="14485" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-11-03T07:37:58.913" Title="Why are galaxies distributed as walls and voids, or spider webs?" Tags="&lt;gravity&gt;&lt;galaxy&gt;&lt;universe&gt;&lt;cosmology&gt;&lt;expansion&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18687" PostTypeId="2" ParentId="18685" CreationDate="2016-10-12T21:00:39.773" Score="8" Body="&lt;p&gt;What determines how long a planet will keep its atmosphere is the strength of the planet's gravity versus the random thermal motion of the molecules in its atmosphere.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atmosphere is a gas, and all gasses have molecules that move around with velocities that are dependent on the temperature.  High-temperature gasses move faster, and low-temperature gasses move more slowly.  Each molecule of gas will have a velocity $v_{\text{gas}, \text{average}}=\sqrt{\frac{3k_BT}{m_{\text{gas}}}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, each individual molecule is pulled to the Earth by gravity.  Heavier gasses, such as water, carbon dioxide, and oxygen, are pulled more than lighter gasses, such as hydrogen and helium.  The escape velocity of the planet is given by $v_{\text{escape}}=\sqrt{\frac{2GM_{\text{planet}}}{R_{\text{planet}}}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a gas molecule has a high enough thermal energy, then if it is at the top of the atmosphere, it could fly off into space, the thermal motion overcoming gravity's pull.  The lighter it is, the less thermal energy it needs to have.  Of course, this all depends on probability and statistics, but, in general, if $v_{\text{gas}}\ge0.2 \times v_{\text{escape}}$, that species of gas will be reduced to 1/2 its initial amount after 1 billion years.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FnN6w.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FnN6w.gif&quot; alt=&quot;Here is a flowchart describing whether or not a gas will escape.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solar wind consists of charged particles and high energy photons, which are deflected by a planet's magnetic field.  However, in the case of a planet like Mars, without a magnetic field, the particles interact with the atmosphere.  They tend to split molecules apart, resulting in lighter gasses (a process called &quot;photodissociation&quot;).  These gasses, being bound less strongly by gravity, are more free to fly off into space under thermal motion.  As more and more gas leaves the planet, the atmosphere becomes thinner and thinner until, ultimately, only the heavy gasses are left.  In addition, when a molecule is split, the energy imparted to it by the impact may be enough to eject it from the atmosphere (a process called &quot;sputtering&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets (and moons) that are closer to the sun are subject to more intense solar radiation, and also have higher atmospheric temperatures.  This is why distant moons, such as Titan, have thick atmospheres of hydrocarbons.  Planets close to the sun, such as Venus, can retain their atmospheres for a variety of reasons.  Firstly, Venus's atmosphere is primarily carbon dioxide, which is much heavier than Hydrogen, and more difficult to split via photodissociation than water.  Planets like Earth also have hydrogeological cycles which can recycle gas from rocks and water back into the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information on this, read up on &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_escape&quot; rel=&quot;noreferrer&quot;&gt;Jeans Escape&lt;/a&gt;.  A good set of notes for beginners can be found &lt;a href=&quot;http://www.astronomynotes.com/solarsys/s3.htm&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13426" LastEditorUserId="9301" LastEditDate="2016-10-22T21:06:46.850" LastActivityDate="2016-10-22T21:06:46.850" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18688" PostTypeId="2" ParentId="18685" CreationDate="2016-10-12T22:01:07.880" Score="4" Body="&lt;p&gt;At least in the case of Mars, the solar wind carries a magnetic field by the planet.  In the Martian atmosphere are ionized particles created from interactions with UV light, cosmic rays, etc.  The moving magnetic field creates an electric field that accelerates the ions away from the planet.  This effect was &lt;a href=&quot;http://www.nasa.gov/press-release/nasa-mission-reveals-speed-of-solar-wind-stripping-martian-atmosphere&quot; rel=&quot;nofollow&quot;&gt;very recently measured&lt;/a&gt; by the MAVEN spacecraft:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;MAVEN has been examining how solar wind and ultraviolet light strip&#xA;  gas from of the top of the planet's atmosphere. New results indicate&#xA;  that the loss is experienced in three different regions of the Red&#xA;  Planet: down the &quot;tail,&quot; where the solar wind flows behind Mars, above&#xA;  the Martian poles in a &quot;polar plume,&quot; and from an extended cloud of&#xA;  gas surrounding Mars. The science team determined that almost 75&#xA;  percent of the escaping ions come from the tail region, and nearly 25&#xA;  percent are from the plume region, with just a minor contribution from&#xA;  the extended cloud.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The dominant effect could be different other places.  For instance, the moons of Jupiter have to deal with the dynamo generated from the Jovian magnetic field.  And in a case like Io, the atmospheric stripping doesn't happen as quickly as expected because apparently every time Io goes behind Jupiter and loses sunlight, its atmosphere &lt;a href=&quot;http://arstechnica.com/science/2016/08/jupiters-moon-io-loses-its-atmosphere-when-eclipsed/&quot; rel=&quot;nofollow&quot;&gt;condenses out and falls to the surface&lt;/a&gt;!&lt;/p&gt;&#xA;" OwnerUserId="9749" LastActivityDate="2016-10-12T22:01:07.880" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18689" PostTypeId="1" CreationDate="2016-10-12T22:59:56.390" Score="0" ViewCount="90" Body="&lt;p&gt;Is it possible to send data using cosmic rays? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like sending high and low energy cosmic rays, since they can travel farther. And this data can be converted to binary form that can be processed by computers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not an astronomer or studied any thing related to astronomy, I was just imagining if this is possible.&lt;/p&gt;&#xA;" OwnerUserId="14556" LastActivityDate="2016-10-13T13:53:40.507" Title="Can we manipulate cosmic rays?" Tags="&lt;data-analysis&gt;&lt;cosmic-ray&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18690" PostTypeId="1" CreationDate="2016-10-12T23:19:19.987" Score="3" ViewCount="2885" Body="&lt;p&gt;I'm asked to collect data on the orbital properties of the four Galilean satellites of Jupiter and show that they obey the same scaling as in Kepler's 3rd law. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach for moon Io:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Online, I found that the distance from Io to Jupiter is 422,000km ~= 0.00282089577 AU&#xA;which is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2.82\times10^{-3} AU$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Io's orbit around Jupiter is 1.77 Earth days; 1.77/365 ~= $ 4.85\times10^{-3}\, \mathrm{Earth\ years}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Kepler's 3rd law, $P^2(\mathrm{Earth\ years}) = a(AU)^3$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(4.85\times10^{-3})^2 = (2.82\times10^{-3})^3$ which is obviously not true even just by looking at the powers of 10 after distributing the exponent on each side.&lt;/p&gt;&#xA;" OwnerUserId="14557" LastEditorUserId="2365" LastEditDate="2016-10-13T19:54:37.637" LastActivityDate="2016-10-13T19:54:37.637" Title="According to my calculations Jupiter's moons do not follow Kepler's 3rd law - Why is that?" Tags="&lt;jupiter&gt;&lt;kepler&gt;" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18691" PostTypeId="1" CreationDate="2016-10-13T00:10:52.690" Score="1" ViewCount="70" Body="&lt;p&gt;I'm a high school student but have as much interests in Cepheid variable as the professionalists. I'm trying to derive independent distances to Cepheid with the Baade-Wesselink method. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got the angular diameter variations data from the paper &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0311525&quot; rel=&quot;nofollow&quot;&gt;Cepheid distances from infrared long-baseline interferometry&lt;/a&gt; but I'm stuck from the afterward. I want to know how equation 9 in that paper is derived. Where did the number 9.305 suddenly came up? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;θLimb-darkened model angular diameter = amplitude of limb-darkened observed angular diameter + 9.305 (∆D/d)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like I need that equation to derive the distance from the angular diameter values with radial velocity values. &lt;/p&gt;&#xA;" OwnerUserId="13714" LastEditorDisplayName="user1569" LastEditDate="2016-10-31T11:46:48.673" LastActivityDate="2016-10-31T11:46:48.673" Title="Calculating angular diameter amplitude from angular diameter variations of Cepheid variable" Tags="&lt;cepheids&gt;&lt;interferometry&gt;&lt;angular-diameter&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18692" PostTypeId="2" ParentId="18690" CreationDate="2016-10-13T04:56:12.783" Score="2" Body="&lt;p&gt;One must &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law&quot; rel=&quot;nofollow&quot;&gt;consider the mass&lt;/a&gt; of the object being orbited. Kepler's third law is true for all planets orbiting the Sun, and for all moons orbiting Jupiter, but not across different gravity wells. This was not understood until Newton, and must've posed an interesting problem (I've never thought of before) already in Kepler's time since the orbital periods and relative distances of the four Galilean moons were known then. And of course the orbit and relative distance of Earth's own moon. Maybe it was even a key inspiration for Newton's thinking about gravity?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2016-10-13T05:01:51.243" LastActivityDate="2016-10-13T05:01:51.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18694" PostTypeId="2" ParentId="18690" CreationDate="2016-10-13T13:41:03.490" Score="9" Body="&lt;p&gt;Kepler's third law is that $R^3/P^2$ is a constant. However it is not a universal constant; it depends on the mass of the body that is being orbited.&#xA;$$\frac{R^3}{P^2} \simeq \frac{GM}{4\pi^2},$$&#xA;where $M$ is the mass of the orbiting body (assuming that $M\gg$ the mass of the moons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on your level of sophistication, you could try plotting $R^3$ against $P^2$, or better, $\log R$ against $\log P$, to demonstrate consistency with Kepler's third law.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-10-13T13:41:03.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18695" PostTypeId="2" ParentId="18689" CreationDate="2016-10-13T13:53:40.507" Score="2" Body="&lt;p&gt;Most of the communication on Earth and to-fro Space takes place over low frequency visible, infrared, micro and radio wave spectrum of the electromagnetic spectrum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The immediate benefits of using low frequency EM spectrum which I think of are as following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;They are not blocked by Earth's atmosphere and magnetic field. On the other hand, cosmic rays which consists mostly of high energy electrons, protons and ß particles carries electric charge and magnetic field. Hence, they will be hugely affected by Earth's magnetic field. Maybe setting up receivers and transmitters at poles helps a little.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;2.Speed:  EM waves travel at light speed, cosmic rays always travel at lesser speed which brings me to the next point.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Energy needs: The energy required to generate radio waves can be provided by low power solar panels. However, to generate cosmic rays one would need a particle accelerator. Something like a portable LHC, which I don't think is feasible. Moreover, if you want to generate faster cosmic rays then you have to put in more energy. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I can only think of these constraints at this moment. I don't know if sending information via cosmic rays carries any advantage.    &lt;/p&gt;&#xA;" OwnerUserId="10657" LastActivityDate="2016-10-13T13:53:40.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18696" PostTypeId="2" ParentId="18676" CreationDate="2016-10-13T13:53:45.557" Score="2" Body="&lt;p&gt;I think the most important reason is that galaxy clusters more strongly adhere to the BAO shape than individual galaxies do. Galaxies tend to have more dispersion than clusters, making the BAO signal from galaxies smeared out and thus harder to detect. This is alluded to in the Veropalumbo et al. 2014 paper introduction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As tracers of the biggest collapsed structures, they are more strongly clustered than galaxies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another potential reason is that the redshift of a cluster is easier to determine than the redshift of a single galaxy. There are multiple galaxies in a cluster, giving you multiple chances to measure a redshift for that cluster. If you can't measure a redshift for a single galaxy, you can't use it in your analysis. And of course all galaxies in a single cluster are going to have the same redshift to a good approximation. The Estrada et al. 2008 and Hong et al. 2012 both indicate $\sigma_z = 0.01$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that more often than not, these studies use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Photometric_redshift&quot; rel=&quot;nofollow&quot;&gt;photometric redshifts&lt;/a&gt; rather than spectroscopic redshifts. Photometeric redshifts are generally less reliable as they rely on correlations (that isn't to say spectroscopic redshifts are perfect though). If you can measure a few photometric redshifts in a cluster or else measure the photometric redshift for the bright, central galaxy with a high signal to noise, you can reduce your redshift errors, compared to measuring it for a single galaxy.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Lastly, in order to use a particular catalogue to measure the BAO signal, the catalogue needs to be both deep (i.e., measures &lt;em&gt;everything&lt;/em&gt; down to a specific redshift/magnitude limit) and wide (i.e., measures across a large field of the sky). If your catalogue doesn't meet both of these conditions, it can't be used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clusters are easier to spot than individual galaxies. They can be found (as the Estrada et al. 2008 paper describes) by first finding the large, luminous galaxies which generally reside in the center of the cluster and are fairly easy to find. The cluster itself is then identified. Looking at individual galaxies may suffer from magnitude limiting effects or other biases, meaning that your analysis may not include all galaxies that exist at a given redshift. If that is the case, your two-point correlation function will not be accurate and you might miss or inaccurately characterize the BAO signal. I would expect this problem is mitigated by using the easier-to-identify clusters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to point out that cluster samples are likely to be more or less complete, the Estrada et al. 2008 paper is quoted as saying&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Tests on mock catalogs indicate that the MaxBCG sample should be $\gtrsim 90\%$ pure and complete for clusters with $N_{200}\ge 10$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The requirement of a wide catalogue is also pretty restrictive. The BAO is inconveniently large so not many deep galaxy catalogues are going to be wide enough to even see the BAO. The SDSS is one of the few galaxy catalogues that comes to mind that is wide enough and they have &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0501171&quot; rel=&quot;nofollow&quot;&gt;already found the BAO in that&lt;/a&gt;. Cluster catalogues are usually built up not from a specific sky survey as galaxy catalogues usually are, but from individual observations or else conglomerated from many galaxy surveys. As such they can be much wider.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-10-13T13:53:45.557" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18697" PostTypeId="1" AcceptedAnswerId="18699" CreationDate="2016-10-13T14:44:14.933" Score="17" ViewCount="1275" Body="&lt;p&gt;As a child I remember my parents taking camping in California, and upon stepping out of the car I was awestruck at how thick the Milky Way galaxy was and at the number of stars everywhere. I've always wondered how much clearer it could get. What percentage of visible light from stars are absorbed by all of earths atmosphere at the clearest locations on earth? Are we taking about a small percentage or perhaps a significant amount?&lt;/p&gt;&#xA;" OwnerUserId="11589" LastEditorUserId="7982" LastEditDate="2021-02-28T11:27:27.063" LastActivityDate="2021-02-28T11:27:27.063" Title="How much clearer are stars in earths orbit?" Tags="&lt;night-sky&gt;&lt;atmospheric-effects&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18698" PostTypeId="2" ParentId="18690" CreationDate="2016-10-13T14:49:57.250" Score="1" Body="&lt;p&gt;This is what they are looking for. I tried it with numbers and it worked. Thank you for your input, you put me on the right track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M1 + M2 = A3 / P2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consistent units must be used to make this equation work. If the data are not given in a consistent system of units, they must be converted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The masses must be measured in solar masses, where one solar mass is 1.99 X 1033 grams, or 1.99 X 1030 kilograms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The semi-major axis must be measured in Astronomical Units, where 1 AU is 149,600,000 kilometers, or 93,000,000 miles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbital period must be measured in years, where 1 year is 365.25 days. &lt;/p&gt;&#xA;" OwnerUserId="14557" LastEditorUserId="907" LastEditDate="2016-10-13T18:16:38.783" LastActivityDate="2016-10-13T18:16:38.783" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18699" PostTypeId="2" ParentId="18697" CreationDate="2016-10-13T15:05:24.807" Score="22" Body="&lt;p&gt;It varies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best astronomical sites have a visual band extinction of 0.1 mag, which means that only $\sim 10$ per cent of light is absorbed/scattered in the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In dusty, smoggy or polluted sites, this can easily reach one magnitude of extinction, which means that 60 per cent of the light is scattered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These numbers are &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Air_mass_(astronomy)&quot;&gt;per airmass&lt;/a&gt;&lt;/em&gt; - which means looking straight up at zenith. At lower altitudes, the baseline extinction figure would be multiplied by $\sec z$ where $z$ is the angle from the zenith.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This amount of absorption doesn't sound like a huge difference, but what it means is that a large number of extra stars become visible, because roughly-speaking, you get a factor of two increase in the number of stars for every 0.5 magnitudes deeper you can see, and I think that is what is most striking about going to a clear, dark astronomical site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus there wouldn't be &lt;em&gt;much&lt;/em&gt; of an improvement going into space over the best astronomical sites (&lt;em&gt;in the visible band&lt;/em&gt;), at least not due to the decrease in extinction. You do also escape light pollution, which is a different effect. With light pollution it is the fact that the background sky becomes too bright for your eye to pick up the fainter stars. [As Carl Witthoft notes, it is a different ball game at other wavelengths because the atmosphere is significantly more opaque to UV, X-ray and IR wavelengths.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to consider is what you are breathing! Lack of oxygen can hinder visual acuity. &lt;a href=&quot;http://navyaviation.tpub.com/14020/css/Effects-Of-Hypoxia-141.htm&quot;&gt;http://navyaviation.tpub.com/14020/css/Effects-Of-Hypoxia-141.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-10-13T16:26:13.810" LastActivityDate="2016-10-13T16:26:13.810" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18700" PostTypeId="2" ParentId="18533" CreationDate="2016-10-13T15:38:21.030" Score="3" Body="&lt;p&gt;It is good to see that you are interested in astrophysics at an young age. &#xA;No problem with you doing the clichéd BE that everyone in India is forced with, you can still get your PhD in Astrophysics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your 4 years of B.Tech aeronautical is going to be tough, so be prepared for it. In the mean time, keeping in mind your aim of becoming an astrophysicist, prepare yourself equipped with programming languages, especially Python/C++. The reason I say this is because &lt;strong&gt;you don't need too have much of an astronomy background to do M.Sc Astronomy&lt;/strong&gt;, but you definitely &lt;strong&gt;need to have a good knowledge in a programming language.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are websites like &lt;a href=&quot;http://www.courseera.org&quot; rel=&quot;nofollow&quot;&gt;Course Era&lt;/a&gt; or &lt;a href=&quot;http://www.edx.org&quot; rel=&quot;nofollow&quot;&gt;EdX&lt;/a&gt; and plenty more on the internet that can help you with learning python/C++ in an interactive way. Go for it, and trust me it will help you a lot.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;As a final year student doing my PhD in Astrophysics (hailing from India) let me tell you the information I have acquired over the years.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;With regard to the Masters/PhD in Astronomy/Astrophysics, the procedure is quite different in different countries. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Procedure within India&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are plenty of Universities in India that offer you a Masters course in Astrophysics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost all the details required are summarised neatly in &lt;a href=&quot;https://www.iiap.res.in/opportunities/career&quot; rel=&quot;nofollow&quot;&gt;THIS PAGE&lt;/a&gt;, &#xA;so I am not going to go through the list of things required. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Procedure for Europe&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would need:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;All your original mark sheets attested by the college/university (Bachelors, Masters, etc)&lt;/li&gt;&#xA;&lt;li&gt;Two/three recommendation letters, and the number of letters depends on the university you are applying to. &lt;/li&gt;&#xA;&lt;li&gt;A completely filled application form.&lt;/li&gt;&#xA;&lt;li&gt;IELTS scores (depends on country)&lt;/li&gt;&#xA;&lt;li&gt;Statement of purpose, as to why you want to do the course.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is pretty much all you need for Europe. For certain universities, you might be having a Skype interview, wherein they might ask you about the stuff you know. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Procedure for USA:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to the USA, it gets quite complicated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will need:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;All your original mark sheets attested by the college/university (Bachelors, Masters, etc)&lt;/li&gt;&#xA;&lt;li&gt;Two/three recommendation letters, and the number of letters depends on the university you are applying to. For most universities it will be online forms that your professor needs to fill after he/she receives the mail.&lt;/li&gt;&#xA;&lt;li&gt;A completely filled application form. This is going to cost you somewhere from around 50-80 dollars depending on the university.&lt;/li&gt;&#xA;&lt;li&gt;GRE scores (both general GRE and subject (Physics,Chemistry etc.) GRE) that you can send to the university directly online by paying money (25 Dollars for each university). Only a few universities accept you without a subject GRE.&lt;/li&gt;&#xA;&lt;li&gt;IELTS/TOEFL scores. TOEFL scores can be sent online, but if you are taking IELTS you have to go to British council and have them send it. &lt;/li&gt;&#xA;&lt;li&gt;Statement of purpose, as to why you want to do the course. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So it depends on where you want to apply. I would recommend you to apply to both, to increase your chances. &lt;strong&gt;For Europe&lt;/strong&gt;, the chances of you getting through &lt;strong&gt;depends on the country&lt;/strong&gt; you apply. &lt;strong&gt;For USA&lt;/strong&gt;, the chances of you getting through &lt;strong&gt;solely depend on your GRE scores.&lt;/strong&gt; It doesn't matter whether you are a gold medallist in your Bachelors, it all boils down to GRE. If you fail there, your entry into USA will be difficult.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One advantage of Europe is that, mostly, the PhD is only for 3/4 years, depending on the country. Whereas in the USA, it is mandatory for the student to write exams in the first year, teach/be a research assistant in the second year, and only in the third year start the PhD project. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To search for PhDs in Astrophysics depends on the country you are applying for. But as a start, you can have a look at &lt;a href=&quot;http://jobregister.aas.org/&quot; rel=&quot;nofollow&quot;&gt;http://jobregister.aas.org/&lt;/a&gt;. PhD calls, internship calls, post-doc positions, etc. related to astrophysics will be &lt;strong&gt;posted here during the start of every month.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;With regard to your question &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;can I do aeronautical engineering so that I can even get employed in ISRO?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes you can definitely do that. ISRO not only employs aeronautical engineers, but also people who have completed their BE in electrical and electronics and also computer science. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at &lt;a href=&quot;http://www.isro.gov.in/careers&quot; rel=&quot;nofollow&quot;&gt;THE CAREERS WEBSITE OF ISRO&lt;/a&gt; to get more information.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Good luck with your future! &lt;/p&gt;&#xA;" OwnerUserId="2655" LastEditorUserId="2655" LastEditDate="2016-10-13T20:58:44.670" LastActivityDate="2016-10-13T20:58:44.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18701" PostTypeId="1" CreationDate="2016-10-13T15:40:20.110" Score="-2" ViewCount="207" Body="&lt;p&gt;Following on from &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1137/calculate-apsides-without-knowing-eccentricity&quot;&gt;this question&lt;/a&gt;, I wish to know how to find the &lt;em&gt;reverse&lt;/em&gt; - how to find eccentricity given a Semi-minor axis &amp;amp; altitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to use something based on&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b=a\sqrt{1−e^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I have no real way to solve for $e$, for I do not trust my answer,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e=a\sqrt{1}-b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;could anyone please help?&lt;/p&gt;&#xA;" OwnerUserId="14562" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-10-13T17:33:31.700" Title="Calculate eccentricity with altitude and semiminor axes" Tags="&lt;orbital-elements&gt;&lt;orbital-mechanics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18702" PostTypeId="2" ParentId="18701" CreationDate="2016-10-13T17:33:31.700" Score="1" Body="&lt;p&gt;It looks like you did your math wrong.  I'll write out the steps for you:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Square both sides:&#xA;$b^2=a^2(1-e^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide by $a^2$: &#xA;$\frac{b^2}{a^2}=1-e^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isolate $e^2$:&#xA;$e^2=1-\frac{b^2}{a^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the square root results in a solution for eccentricity, based on the length of the semiminor and semimajor axes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e = \sqrt{1-\frac{b^2}{a^2}}$&lt;/p&gt;&#xA;" OwnerUserId="13426" LastActivityDate="2016-10-13T17:33:31.700" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18703" PostTypeId="1" AcceptedAnswerId="20588" CreationDate="2016-10-13T18:15:48.863" Score="5" ViewCount="1032" Body="&lt;p&gt;Though I am active on Physics, I am new to Astronomy. I am an amateur astronomer, actually even worse than amateur. I have just started it.&#xA;I have a 60 mm objective 700mm focal length refractor which I mainly use with a 20 mm eyepiece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a couple of questions regarding the telescope, observations and explanations behind observations. Since they are all interlinked I decided to ask them together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) So, firstly the basics. Regarding the resolution power of a telescope, we know from the Rayleigh criterion that diffraction does happen at the objective. But Diffraction happens when light passes through a narrow slit or bends around a corner (whose dimensions are of the order of the wavelength of light). But there is no slit and probably no bending of light. &lt;strong&gt;&lt;em&gt;How come then do we assume a diffraction pattern to have formed when considering the derivation of the resolution of a telescope?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;strong&gt;How are exit pupil and entry pupil different from the diameter of the eyepiece and objective?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) &lt;em&gt;Why on increasing the magnification does the brightness of the object reduce&lt;/em&gt; (the physics behind it)? &lt;strong&gt;I mean that if I use 4mm eyepiece then, first of all, it becomes very difficult to find the object and secondly its brightness gets reduced when compared to the brightness that I had got using a 20mm eyepiece?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) &lt;em&gt;Secondly why do the Barlow lenses reduce the brightness a lot.&lt;/em&gt; And why on using  Barlow lenses the field of view get smaller and the object passes twice as quickly as without the Barlow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Now the biggest question. &lt;em&gt;How is it possible that every time I point my telescope towards a visible planet or a star (I know it's a star because of the sky map), I see almost the same size, when stars are visible only as small dots.&lt;/em&gt; &lt;strong&gt;The stars appear to be bluish and the planets which I have seen (Mars and Jupiter) appear reddish. That's ok but a star can not be visible as a sphere. Stars appear as dots, don't they?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6)How can I make the telescope give the best results i.e which eyepiece should I use and whether I should use Barlow lenses or not?&lt;/p&gt;&#xA;" OwnerUserId="14565" LastEditorUserId="14565" LastEditDate="2017-05-03T11:42:08.940" LastActivityDate="2017-05-03T11:42:08.940" Title="Question about Telescopes , observations and explanations of the observations" Tags="&lt;telescope&gt;&lt;angular-resolution&gt;&lt;refractor-telescope&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18704" PostTypeId="2" ParentId="17909" CreationDate="2016-10-13T19:57:06.820" Score="1" Body="&lt;p&gt;My first thought would be that the apparent path of a satellite across the sky can appear to turn even when it is straight.  Your brain doesn't easily understand straight lines in all cases.  (Look up lunar terminator illusion)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've several times seen Mir/ISS passes that seem to arrive from one direction, appear to turn, then leave in some unrelated direction.  It's always really moving in a straight line though.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Web sites can show you what path the satellite will make across the starfield, but it can't simulate the illusion you might have on seeing it from the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ground track alone can't tell you the visibility from the ground because it also depends on altitude.  A low-earth-orbit satellite might be visible from a circular patch about 3000km in diameter.  Higher altitude objects would be fainter, but visible across a wider area.    &lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2016-10-13T19:57:06.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18705" PostTypeId="1" CreationDate="2016-10-13T20:35:21.700" Score="1" ViewCount="332" Body="&lt;p&gt;What are the Similarities and Differences between ALMA (Atacama Large Millimeter Array) and FAST (Five hundred meter aperture spherical telescope)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can they see, what's their resolution like? How do they compare?&lt;/p&gt;&#xA;" OwnerUserId="14567" LastActivityDate="2016-10-14T22:44:40.697" Title="What are the Similarities and Differences between ALMA and FAST?" Tags="&lt;radio-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18706" PostTypeId="2" ParentId="18703" CreationDate="2016-10-14T05:49:29.760" Score="3" Body="&lt;p&gt;I'll attempt to answer as many of these questions, to the best of my abilities, as I can at this late hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Lenses have an index of refraction that is different than vacuum, and air, and they are curved. ie: corners when light is concerned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The entrance and exit pupils are the pupil size required to pass an &quot;extreme&quot; ray through an optic. Meaning that if the light ray from the optical axis was not refracted at all as it passed the optical plane, it would still pass within the pupil (entrance) and if reflected perfectly (exit). The diameters of the objective and the eyepiece are subjective, meaning that you can have, within reason, any size objective and use virtually any eyepiece. The restrictions to this would only be focal lengths. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) You are looking through a thicker lens and thereby reducing the light incident on your eye. Think about the shape of the lenses, look at them and it might become apparent just how much more curvature the 4mm lens has than the 20mm lens. This added curvature comes at the cost of a higher angle of incident that a light ray has to take, and thus more optic to pass through. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Barlow lenses reduce the brightness a lot because you are adding additional optics. With each new addition you decrease the amount of light by a factor relative to the thickness and clarity of the optics. The Barlow lens is increasing the magnification and thereby reducing the field of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) With your specific telescope (of which I have one with the exact same size/focal length) you should be able to see Jupiter as a disk, Mars too should be more than a &quot;dot.&quot; Stars are point sources with angular diameters much too small for you to be concerned given the size of your telescope. Also, those planets appear reddish because they are-ish. Many stars will still appear colors other than blue if you can find them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) This is up to your discretion and desired results when observing. Please read your owners manual, or check out the telescope mfr website for recommended setups.&lt;/p&gt;&#xA;" OwnerUserId="2232" LastActivityDate="2016-10-14T05:49:29.760" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18707" PostTypeId="2" ParentId="18703" CreationDate="2016-10-14T06:12:58.450" Score="1" Body="&lt;p&gt;You already have an answer but there are a few interesting Physics things in your question that are linked:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1) Diffraction and the fact that telescopes have no &quot;slit&quot;.&lt;/li&gt;&#xA;&lt;li&gt;5) Stars don't appear as a sphere, they appear as dots.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The telescope has a round aperture (of say 60mm or so) and this behaves as a slit in some ways. It causes diffraction patterns (an &lt;a href=&quot;https://en.wikipedia.org/wiki/Airy_disk&quot; rel=&quot;nofollow&quot;&gt;Airy disc and concentric circles&lt;/a&gt;) when viewing a point source at high magnifications, near the resolution limit of the telescope. Also of slightly out of focus you'll get &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Circular_Aperture_Fresnel_Diffraction_high_res.jpg&quot; rel=&quot;nofollow&quot;&gt;lots of rings&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By the way this the cause of many &quot;weird&quot; unidentified objects that people photograph or film in the sky; especially a DV camera with odd shaped aperture blades. Surprisingly few people want to hear the explanation though.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when viewing stars at high-ish power, on very still nights, they actually look like a small white circle surrounded by a single, very thin, circular ring.&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2016-10-14T06:12:58.450" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="18708" PostTypeId="1" AcceptedAnswerId="18711" CreationDate="2016-10-14T06:40:36.820" Score="21" ViewCount="3089" Body="&lt;p&gt;Nature: &lt;a href=&quot;http://www.nature.com/news/universe-has-ten-times-more-galaxies-than-researchers-thought-1.20809&quot; rel=&quot;noreferrer&quot;&gt;Universe has ten times more galaxies than researchers thought&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NASA feature: &lt;a href=&quot;http://www.nasa.gov/feature/goddard/2016/hubble-reveals-observable-universe-contains-10-times-more-galaxies-than-previously-thought&quot; rel=&quot;noreferrer&quot;&gt;Hubble Reveals Observable Universe Contains 10 Times More Galaxies Than Previously Thought&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Headlines sometimes oversimplify. But if this is really true, that there seems to be about ten times more galaxies than had been generally assumed, does this impact other assumptions? Does this imply the universe is ten times more massive, or just that there is less dark matter than was previously thought. &lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="2531" LastEditDate="2016-10-15T14:15:45.687" LastActivityDate="2016-10-24T01:59:02.010" Title="Does the recent news of &quot;ten times more galaxies&quot; imply that there is correspondingly less dark matter?" Tags="&lt;universe&gt;&lt;cosmology&gt;&lt;dark-matter&gt;&lt;observable-universe&gt;&lt;hubble-telescope&gt;" AnswerCount="1" CommentCount="15" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18710" PostTypeId="1" AcceptedAnswerId="18724" CreationDate="2016-10-14T06:54:59.880" Score="2" ViewCount="112" Body="&lt;p&gt;Astronomers always talk about astrophysical objects being in virial equilibrium (e.g., a stellar system, or a disk of gas within a galaxy, etc.). But I never hear about &lt;strong&gt;thermodynamical&lt;/strong&gt; equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some examples of objects that are not in virial equilibrium?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One example question: suppose that what you thought was a disk of rotating gas within a galaxy (as revealed by spectroscopy) is actually a bipolar outflow of gas. Would the bipolar outflow of gas be in virial equilibrium? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is a radio jet (length of hundreds of kpc) from a supermassive black hole in virial equilibrium? &lt;/p&gt;&#xA;" OwnerUserId="607" LastActivityDate="2016-10-15T00:02:30.957" Title="Examples of astrophysical objects NOT in virial equilibrium?" Tags="&lt;gravity&gt;&lt;galactic-dynamics&gt;&lt;observational-astronomy&gt;&lt;interstellar-medium&gt;&lt;stellar-dynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18711" PostTypeId="2" ParentId="18708" CreationDate="2016-10-14T07:12:02.020" Score="27" Body="&lt;p&gt;All &lt;a href=&quot;https://arxiv.org/abs/1607.03909&quot; rel=&quot;noreferrer&quot;&gt;Conselice et al. (2016)&lt;/a&gt; appear to suggest is that when you look at something like the Hubble deep field, there are many faint (and presumably low mass) galaxies that are not seen. This has absolutely no effect on the need for dark matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main results are: (i) as you look back in time, the overall (co-moving) density of galaxies (more massive than a million times that of the Sun) increases. (ii) But the density of more massive galaxies actually decreases. This is consistent with hierarchical merger picture where small galaxies merge to become larger galaxies.&#xA;This really doesn't have any influence on the need for dark matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the presence of dark matter is inferred from many different observations. Some of these (e.g. galaxy rotation curves) are not influenced at all if there are lots of extra galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the &quot;missing&quot; galaxies are at high redshift, not (or not all) in the present day universe, so they cannot significantly affect a calculation of how much normal matter there is in the universe &lt;em&gt;today&lt;/em&gt;. Presumably, many of these small galaxies then merge to become larger galaxies and the total mass is conserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, just because there are lots of them does not mean they contain much mass anyway. The &quot;mass function&quot; (number density as a function of mass) of galaxies goes roughly as $\phi(M) \propto M^{-1}$ at low masses. This means the mass contained in any interval is&#xA; $$M_{\text{tot}}\propto \int^{M_2}_{M_1} M\phi \ \mathrm{d}M \ = M_2 - M_1$$&#xA;So although, low mass galaxies may be ten times more frequent, they are ten times less massive and so don't change the total mass very much. I will need to read the paper more carefully to see if the authors are suggesting that low mass galaxies are much more common in the early universe than was &lt;em&gt;already&lt;/em&gt; supposed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fourth, primordial nucleosynthesis calculations tell us that only 4 per cent (as a fraction of the critical density) of the energy density of the universe is in the form of baryonic mass. Observations of gravitational lensing, cluster dynamics and the cosmic microwave background tell us that the mass density is actually around 30 per cent of the critical density. Thus most of the dark matter is non-baryonic and cannot be in the form of missing faint galaxies, or any other form of normal baryonic matter. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="9301" LastEditDate="2016-10-24T01:59:02.010" LastActivityDate="2016-10-24T01:59:02.010" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18713" PostTypeId="1" AcceptedAnswerId="18717" CreationDate="2016-10-14T12:24:50.513" Score="2" ViewCount="106" Body="&lt;p&gt;A galaxy cluster consists of somewhere around a few tens to several hundred galaxies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To measure the position of a galaxy cluster, i.e. it's RA and DEC, do astronomers always assume it to be its Brightest Cluster Galaxy BCG's position?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has been shown by &lt;a href=&quot;https://arxiv.org/pdf/0709.1159v1.pdf&quot; rel=&quot;nofollow&quot;&gt;Johnston et al. 2007&lt;/a&gt; that BCGs are not always found in the centre of a cluster/halo. If so, does the assumption of taking the position of a BCG to be the position of the galaxy cluster still hold true?&lt;/p&gt;&#xA;" OwnerUserId="2655" LastEditorUserId="2655" LastEditDate="2016-10-14T13:48:34.317" LastActivityDate="2016-10-14T13:54:12.457" Title="Centre (projected) of a galaxy cluster" Tags="&lt;positional-astronomy&gt;&lt;galaxy-cluster&gt;&lt;declination&gt;&lt;right-ascension&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18714" PostTypeId="1" AcceptedAnswerId="18715" CreationDate="2016-10-14T13:06:57.027" Score="1" ViewCount="1074" Body="&lt;p&gt;For example, how to convert B1950.0 into J2000.0?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we need to know the proper motion of the star and the time it was observed?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="2531" LastEditDate="2016-10-15T09:13:18.117" LastActivityDate="2016-10-15T09:13:18.117" Title="How to convert ra&amp;dec between different epoches?" Tags="&lt;coordinate&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18715" PostTypeId="2" ParentId="18714" CreationDate="2016-10-14T13:26:28.723" Score="4" Body="&lt;p&gt;The difference between B1950.0 AND J2000.0 is not really concerned with proper motion., they refer to the definition of the coordinate system. Thus a star with zero proper motion would still have different B1950 and J2000 coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually (but not always), if you quote a position as equinox 1950 you are implicitly assuming that this is &lt;em&gt;epoch&lt;/em&gt; 1950 (unless otherwise stated). And similarly for J2000, the implicit assumption is that the coordinates refer to epoch 2000 unless you specify otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus there are two things you needs to do. One is to apply a correction for (roughly$^{*}$) 50 years of proper motion (if indeed you only have the epoch 1950 position; or maybe 66 years if you want the coordinates of the star &lt;em&gt;now&lt;/em&gt;). Second is to apply (roughly) 50 years of precession to the coordinates to put them in the J2000 system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://oneau.wordpress.com/2011/10/27/equinox-epoch/&quot; rel=&quot;nofollow&quot;&gt;This description&lt;/a&gt; seems reasonable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^{*}$ There is some detail about B= Besselian years and J = Julian years which you can read about &lt;a href=&quot;https://en.wikipedia.org/wiki/Epoch_(astronomy)&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-10-14T13:45:58.070" LastActivityDate="2016-10-14T13:45:58.070" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18717" PostTypeId="2" ParentId="18713" CreationDate="2016-10-14T13:54:12.457" Score="3" Body="&lt;p&gt;Short answer, no, they don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longer answer, it's complicated. There are, in essence, five different measures for the centre of a galaxy cluster, based on different physical properties of the cluster, and occasionally not in agreement with each other.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;BCG, the brightest cluster galaxy. It should sink to the bottom of the gravitational well, but that's only the case if the cluster had enough time to evolve. Of course, galaxy clusters are, cosmologically speaking, quite young objects and still in the process of forming, so this is not a given.&lt;/li&gt;&#xA;&lt;li&gt;Mean position of the cluster galaxies. Works for spherically symmetrical clusters with lots of galaxies. Problems: Poisson noise due to limited number of galaxies, identification problems (is this galaxy really a part of the cluster), asymmetry.&lt;/li&gt;&#xA;&lt;li&gt;X-ray: the hot intra-cluster gas emits X-ray radiation that we can observe - more and hotter gas means more and harder X-rays and if the gas is in equilibrium, the highest intensity radiation comes from the centre. Unfortunately, the gas is typically not in equilibrium, there are cooling flows, streams, cold cores and other things that make the measurement less straight-forward.&lt;/li&gt;&#xA;&lt;li&gt;(Weak) gravitational lensing. Measures the total mass of the cluster and doesn't care about ugly gas physics. Problems: Shape noise (measurements are based on mean ellipticity of background galaxies), resolution / smoothing issues with the mass map created from averaging over lots of background galaxies.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://ned.ipac.caltech.edu/level5/Sept05/Carlstrom/Carlstrom2.html&quot; rel=&quot;nofollow&quot;&gt;Sunyaev-Zel'dovich&lt;/a&gt; effect. Again the hot cluster gas, but this time we have inverse &lt;a href=&quot;https://en.wikipedia.org/wiki/Compton_scattering&quot; rel=&quot;nofollow&quot;&gt;Compton scattering&lt;/a&gt; of photons in the cosmic microwave background on the electrons in the gas. The problems with gas physics should apply here, too.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, there are other methods to find the centre of a galaxy cluster. &lt;a href=&quot;http://esoads.eso.org/abs/2012MNRAS.419.3547D&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; suggests that ususally the weak lensing centre is offset with respect to both X-ray and BCG. I am not aware of anyone using the mean position of the galaxies as a measure for the cluster centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/1406.4019&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; evaluates the cluster centre offset between measurements in X-ray and the SZ. This is interesting because they both trace the hot intra-cluster gas.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastActivityDate="2016-10-14T13:54:12.457" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18721" PostTypeId="1" CreationDate="2016-10-14T19:44:17.770" Score="2" ViewCount="51" Body="&lt;p&gt;How would Hisenberg's unified field equations be represented if gravity is incorporated. Would it result in a case similar to the Heisenberg-Ivanenkov equation, or am I just confused here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would really appreciate if the Heisenberg-Ivanenkov equation's essence is explained as there is no firm online source to provide more information.&lt;/p&gt;&#xA;" OwnerUserId="14580" LastActivityDate="2016-10-14T19:44:17.770" Title="Modifying Heisenberg's unified field theory with gravity" Tags="&lt;gravity&gt;&lt;quantum-field-theory&gt;" AnswerCount="0" CommentCount="4" FavoriteCount="1" ClosedDate="2016-10-17T16:44:28.170" ContentLicense="CC BY-SA 3.0" />
  <row Id="18722" PostTypeId="2" ParentId="18705" CreationDate="2016-10-14T22:44:40.697" Score="1" Body="&lt;p&gt;This is quite a broad question, since there are far more differences than similarities between the two. This answer will focus on the first part of the question and be about the technical aspects of the two telescopes.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Basics: FAST is a &quot;classic&quot; filled aperture telescope. This means it consists of a single large dish that focuses incoming radio waves on a single receiver (this is simplified). In contrast, ALMA is an interferometer: Many (&gt; 60) dishes each focus incoming radiation on their own receiver. The relative positions of the individual telescopes is known to great precsion, and both amplitude and phase of the incoming radiation measured, which allows astronomers to combine several or all of the telescopes for a single observation. The maximum achievable resolution is then not determined by the diameter of a telescope but by the distance between those furthest apart. (Again, this is greatly simplified; interferometry is complicated.) In principle, FAST could also become part of a huge interferometer called VLBA.&lt;/li&gt;&#xA;&lt;li&gt;Wavelength: FAST is a &quot;classic&quot; radio telescope, observing in the centimeter to meter wavelength range. ALMA is a sub-mm telescope operating at smaller wavelengths of 0.3 mm to about 1 cm.&lt;/li&gt;&#xA;&lt;li&gt;Resolution: since the resolution is $&#xA;\propto \lambda / d$, with $\lambda$ the wavelength and $d$ the diameter of the telescope, ALMA has a much higher resolution - smaller $\lambda$ and greater $d$. ALMAs resolution is given as 10 milliarcseconds, while in the ideal case FASTs is $1.22 \cdot 0.1 / 500 \cdot( 180 / \pi \cdot 3600) = 50$ arcseconds at 10 cm wavelength. For comparison, according to wikipedia the unaided human eye has a resolution of about 60 arcseconds.&lt;/li&gt;&#xA;&lt;li&gt;Sensitivity:&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Atmospheric transmission: FAST observes in a wavelength regime in which our atmosphere is quite transparent. In contrast, the ALMA dishes are located in one of the driest deserts of the world at an altitude of over 5000 m, in order to minimize the effects of water vapour in the atmosphere.&lt;/li&gt;&#xA;&lt;li&gt;Collecting area: FAST has a much larger collecting area. Also (I'm uncertain in this point, please correct this if I'm wrong) you lose some sensitivity when doing interferometry compared to just pooling the collecting areas of the single telescopes in ALMA.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Point_spread_function&quot; rel=&quot;nofollow&quot;&gt;Point Spread Function&lt;/a&gt;: The PSF of a single telescope, even a radio telescope, is much simpler than that of an interferometer. So the data analysis required to turn observations into an image should be much more straightforward for FAST.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are some differences in the required surface accuracy (imperfections significantly smaller in scale than the observed wavelength do not matter) and surface composition - the ALMA dishes are solid metal, whereas the FAST dish is perforated. This works because as long as the holes are smaller than the wavelength, the incoming radio waves see it as solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The receivers used to convert incoming photons to electricity for both telescopes work on the same principles. Practically, it is much more difficult for ALMA due to the high frequencies at the edge of the far-infrared spectrum. ALMAs shortest wavelengths are just a little too long for receivers used in infrared or optical telescopes to work, and almost too short for normal radio receivers to work.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastActivityDate="2016-10-14T22:44:40.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18724" PostTypeId="2" ParentId="18710" CreationDate="2016-10-15T00:02:30.957" Score="4" Body="&lt;p&gt;Not clear what the initial part of your question means. Objects can be in virial equilibrium without being in thermal equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A clear exception to the virial theorem would be any system that is gravitationally unbound. So you couldn't apply it to a supernova explosion or a dissolving cluster of stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The examples you quote are not self-gravitating, bound systems, so no, the virial theorem would not apply to them.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-10-15T00:02:30.957" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18725" PostTypeId="1" AcceptedAnswerId="18733" CreationDate="2016-10-15T01:35:03.930" Score="8" ViewCount="1992" Body="&lt;p&gt;If we assume a geocentric model of the universe (like ancient astronomers did) how could we ever find out (again, like ancient astronomers did) the right distances of the planets? For example, how did they know that Jupiter is&#xA;further than Mars, if Jupiter is brighter?     &lt;/p&gt;&#xA;" OwnerUserId="9949" LastEditorUserId="7472" LastEditDate="2019-05-21T16:10:52.600" LastActivityDate="2019-05-22T00:39:45.587" Title="How did ancient astronomers know to order the planets from the closest to the farthest from the Sun?" Tags="&lt;orbit&gt;&lt;the-sun&gt;&lt;mars&gt;&lt;astrometry&gt;&lt;geocentrism&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="18726" PostTypeId="1" AcceptedAnswerId="18731" CreationDate="2016-10-15T02:16:14.473" Score="7" ViewCount="410" Body="&lt;p&gt;The NY Times article &lt;a href=&quot;http://www.nytimes.com/2016/10/11/science/telescope-earthlike-planet-alpha-centauri-boldlygo-nasa.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Telescope to Seek Earthlike Planet in Alpha Centauri System&lt;/a&gt; describes a proposed effort to put a telescope into orbit fairly quickly with a narrowly defined purpose - to look for planets in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Alpha_Centauri&quot; rel=&quot;nofollow noreferrer&quot;&gt;Alpha Centauri&lt;/a&gt; system, which is about 4.37 lightyears away from Earth.&lt;/p&gt;&#xA;&lt;p&gt;The funding, design, assembly and space deployment fairly quickly is made possible in part by the narrowly defined purpose and the compelling public interest in this particular question.&lt;/p&gt;&#xA;&lt;p&gt;I got out my box of envelope-backs and did the following. Since there is no air, &lt;s&gt;there's no need for adaptive optics so &lt;/s&gt; the standard diffraction limit applies. For an &lt;a href=&quot;https://en.wikipedia.org/wiki/Airy_disk&quot; rel=&quot;nofollow noreferrer&quot;&gt;Airy disk&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \theta \approx 1.22 \frac{\lambda}{d} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The article describes the telescope as washing machine-sized, with an aperture of 20 inches, or about 50 cm.  If I choose a visible wavelength of 500nm, I get a resolution of about 1.22 microradians (&lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt;R). 4.37 lightyears is about 4.13E+13 km or 2.76E+05 AU.&lt;/p&gt;&#xA;&lt;p&gt;That makes the airy disk about 0.34 AU at Alpha Centauri. The actual point spread function of a telescope depends not only on the theoretical diffraction of the aperture, but diffraction from secondaries and supports within the aperture, on aberrations due to imperfectly shaped optical surfaces and scatter and haze due to nano-roughness of mirror surfaces. (see below)&lt;/p&gt;&#xA;&lt;p&gt;Could a 20 inch telescope and associated experiment package be put together quickly that could actually resolve (or &amp;quot;make out&amp;quot; as the article says) an Earth-like planet in the Alpha Centauri system from space? If so, what is it exactly that it could do that can't be done from any existing telescope in orbit or on Earth?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Zu4LE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zu4LE.jpg&quot; alt=&quot;A rendering of the compact exoplanet imaging telescope Project Blue&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; &amp;quot;A rendering of the compact exoplanet imaging telescope Project Blue plans to launch into orbit and aim at Alpha Centauri. Credit Project Blue&amp;quot; from &lt;a href=&quot;http://www.nytimes.com/2016/10/11/science/telescope-earthlike-planet-alpha-centauri-boldlygo-nasa.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;It jumps out that the secondary is mounted off-axis, which can have mechanical and optical advantages (e.g. diffraction) but would probably limit the image quality to a narrow field of view, which might be OK for a single-target telescope if there are suitable calibration targets nearby.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;A nice discussion of the point spread function of the Hubble space telescope can be found here: &lt;a href=&quot;http://www.stsci.edu/institute/org/telescopes/Reports/krist_tinytim_spie.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;20 years of Hubble Space Telescope optical modeling using Tiny Tim&lt;/a&gt; (Krist et al 2011, Proc. of SPIE Vol. 8127 81270J-1), and the reduction of scatter and haze from mirror nano-roughness by using refractive optics (lenses) is addressed by the &lt;a href=&quot;http://www.dunlap.utoronto.ca/instrumentation/dragonfly/&quot; rel=&quot;nofollow noreferrer&quot;&gt;DRAGONFLY&lt;/a&gt; instrument. See also &lt;a href=&quot;https://astronomy.stackexchange.com/q/18074/7982&quot;&gt;What (actually) is the &amp;quot; deprojected half-light radius&amp;quot; of this almost-all-dark-matter Galaxy?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2021-01-05T07:24:10.723" LastActivityDate="2021-01-05T07:24:10.723" Title="How could a 20 inch space telescope &quot;be able to make out Earth-size planets&quot; orbiting Alpha Centauri?" Tags="&lt;exoplanet&gt;&lt;space-telescope&gt;&lt;angular-resolution&gt;&lt;instruments&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="18727" PostTypeId="1" CreationDate="2016-10-15T02:52:06.667" Score="0" ViewCount="245" Body="&lt;p&gt;Where would it be in the sky from Proxima B (planet near Proxima Centauri) for example?&lt;/p&gt;&#xA;" OwnerUserId="14583" LastActivityDate="2016-10-15T03:46:13.837" Title="What constellation would the Sun be in from an exoplanet?" Tags="&lt;star&gt;&lt;amateur-observing&gt;&lt;star-gazing&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18728" PostTypeId="2" ParentId="18727" CreationDate="2016-10-15T03:46:13.837" Score="4" Body="&lt;p&gt;The sun would appear roughly half-way between Capella and the W of Cassiopeia. The constellations are a human invention, and don't correspond to actual groups of stars. Since all the nearby stars would be in somewhat different positions the constellations would be mixed up, for example, Sirius would be close to Betelgeuse. You can't really talk about constellations except from the perspective of Earth, but you could say that the sun would be near some of the stars that form Perseus and Camelopardalis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From more distant exoplanets the constellations would be completely mixed up, so there would be no sense in describing the sun's location in terms of a constellation. The sun would also be too dim to be seen with the naked eye from many exoplanets.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-10-15T03:46:13.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18729" PostTypeId="1" CreationDate="2016-10-15T08:25:00.677" Score="2" ViewCount="130" Body="&lt;p&gt;A planet orbits around its sun on an elliptic orbit, and loses mass slowly due to evaporation. How will the parameters of the orbital ellipse change as a function of time?&#xA;Could we do a generalisation and mathematically model it?&lt;/p&gt;&#xA;" OwnerUserId="14580" LastActivityDate="2016-11-14T12:57:53.977" Title="On orbital mechanics of evaporating planets" Tags="&lt;planet&gt;&lt;astrophysics&gt;&lt;celestial-mechanics&gt;&lt;orbital-mechanics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18731" PostTypeId="2" ParentId="18726" CreationDate="2016-10-15T10:08:53.017" Score="7" Body="&lt;p&gt;I'm not familiar with the design of the ProjectBlue telescope, but I think you have answered your own question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The habitable zones for Alpha Cen A and B, are approximately centred at 1.25au and 0.7au. Both are at a distance of 4.37 light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1au at 4.37 light years, subtends an angle of 0.74 arcseconds. If working at blue wavelengths (the aim appears to be to find a pale blue dot, so let's say 400nm), an Airy disk of 0.74 arcsec is achieved with a telescope with a diameter of just 13cm. To stand a chance of resolving planets in the habitable zone, you would need a telescope a little bigger than this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a bigger problem is separating the signal from the star and planets when they are barely resolved. The Airy disc is just the first minimum in the diffraction pattern. If the star is around 10 billion times as bright as the planet (ball park figure) then it would still swamp a planet that was just resolved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus you need to build a bigger telescope to increase the angular resolution. And you need to build superb optics and a coronagraph to stop the light from the star swamping your detector. But this costs more money, the telescope has more mass, needs a bigger launch vehicle etc. So I guess a 50cm telescope is their compromise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it achievable? I don't know. The best telescopes on Earth achieve almost diffraction-limited angular resolution using adaptive optics. However the contrast they can obtain for exoplanet imaging is compromised by the atmosphere to some extent. If you are looking for a &quot;pale blue dot&quot;, then this can't be done from the ground because adaptive optics systems do not currently work at such wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is quite possible that Earth-like planets in the habitable zone of Alpha Cen could be discovered from the ground in the next few years using improved radial velocity measurements, but not by imaging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting question is why this can't be done now with HST? The proposed telescope will spend two years doing this. Even if HST could cut this to months, I am not convinced that detecting a faint smudge around the nearest star has the necessary science pay off to get the time, especially as there is a fair chance of failure. There may not be planets in the habitable zone or there may be sufficient zodiacal dust in the system to make the measurements impossible due to scattered light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further research suggests that the coronagraph on the HST ACS camera would only give around 7-8 magnitudes (factors of a thousand) contrast at angular separations of an arcsecond. So nowhere near what is required. This new telescope must have radically improved optics and a more advanced coronagraph.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-10-15T11:05:16.480" LastActivityDate="2016-10-15T11:05:16.480" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18733" PostTypeId="2" ParentId="18725" CreationDate="2016-10-15T11:19:41.527" Score="13" Body="&lt;p&gt;According to the Cambridge Concise History of Astronomy (p 33 of my edition), essentially the Greeks took the (not unreasonable) view that the planets that moved more slowly were further away and were orbiting on larger spheres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's obviously not the same as suggesting they knew &quot;the right distances&quot; to the planets, merely the order. They did develop a mathematically sound means of estimating the distance to the Moon, though their observational data let them down on a correct estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They also thought the Sun was in the fourth sphere around the Earth (to explain the observed behaviour of Mercury and Venus).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not all Greeks thought the Sun revolved round the Earth and those who argued for a heliocentric universe said the reasons the stars did not move as the Earth did was because they were very, very far away.&lt;/p&gt;&#xA;" OwnerUserId="823" LastEditorUserId="6980" LastEditDate="2016-10-18T01:22:26.037" LastActivityDate="2016-10-18T01:22:26.037" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18734" PostTypeId="1" CreationDate="2016-10-15T11:54:40.957" Score="2" ViewCount="116" Body="&lt;p&gt;How does the Sun generate its periodically reversing large-scale magnetic field? &lt;/p&gt;&#xA;" OwnerUserId="14078" LastEditorUserId="6" LastEditDate="2016-11-17T20:28:53.527" LastActivityDate="2016-11-17T20:28:53.527" Title="How does the Sun generate its periodically reversing large-scale magnetic field?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;magnetic-field&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2016-10-17T14:13:49.517" ContentLicense="CC BY-SA 3.0" />
  <row Id="18735" PostTypeId="2" ParentId="18729" CreationDate="2016-10-15T12:27:44.230" Score="2" Body="&lt;p&gt;The mass of a planet does not affect its orbit significantly unless it is a very large planet.  On the opposite end of the spectrum would be a very small mass planet that was evaporating, and that is essentially what a comet is.  It's all Keplerian, though comets can have their orbits affected slightly by the tiny push they get from the evaporating gases.  So it's not the change in mass, but how the evaporation actually happens, but this is a small effect.  Even a large planet like Jupiter does not affect the Sun much (the Sun orbits around a point that is not even external to the Sun), so losing mass would not affect the orbit of Jupiter in any significant way either.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-10-15T12:27:44.230" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18737" PostTypeId="1" CreationDate="2016-10-15T15:01:58.170" Score="2" ViewCount="1099" Body="&lt;p&gt;For context: I am visiting Portugal from Australia and after observing that people tend to start their days later, I realised that solar noon also seems to occur at a later time - about 1.20pm at the moment. I don't recall solar noon every occurring at such a late time in Australia. When and where is the latest solar noon on earth?&lt;/p&gt;&#xA;" OwnerUserId="14593" LastEditorUserId="14078" LastEditDate="2016-10-17T16:43:58.870" LastActivityDate="2019-08-08T19:08:48.030" Title="What time and where on earth is the latest solar noon?" Tags="&lt;the-sun&gt;&lt;time&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18738" PostTypeId="1" AcceptedAnswerId="18748" CreationDate="2016-10-15T15:36:31.280" Score="3" ViewCount="208" Body="&lt;p&gt;This may have a simple answer but I can't find it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've come across this term in a paper I was reading, &lt;a href=&quot;http://mnras.oxfordjournals.org/content/419/3/2054.full.pdf&quot; rel=&quot;nofollow&quot;&gt; (Walmswell and Elridge, 2011)&lt;/a&gt; which mentions 'non-detection' of Supernovae, and while I can wrap my head around detection, I was lost at 'non'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried finding a definition but to no avail. Would anyone be able to give one?&lt;/p&gt;&#xA;" OwnerUserId="14594" LastActivityDate="2016-10-16T15:34:29.713" Title="What is 'non-detection' of a Supernova?" Tags="&lt;supernova&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18739" PostTypeId="2" ParentId="18737" CreationDate="2016-10-15T16:45:27.490" Score="6" Body="&lt;p&gt;That should be in western China, since all of China uses Beijing's time zone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: &lt;a href=&quot;https://en.wikipedia.org/wiki/Time_in_China&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Time_in_China&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--- Edit below ---&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In response to the comment by @adrianmcmenamin: I'll leave this as guesswork since I simply don't know the peculiarities of every timezone there is. Here's a back-of-the-envelope calculation for local noon in western China.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun covers 15 degrees in a period of 1 hour. All of China is on Beijing time (UTC + 8 hours). With the above, it's centred on $15 \cdot 8 = 120^{\circ}$ east. As a consistency check, Beijing is at $116^{\circ}$ east. The westernmost part of China is at about $73^\circ$ east (&lt;a href=&quot;https://en.wikipedia.org/wiki/China#Landscape_and_climate&quot; rel=&quot;nofollow&quot;&gt;See here&lt;/a&gt;). Local noon there is delayed by $(120^\circ - 73^\circ)/(15^\circ/h) \simeq 3h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, local noon is a little later than 3pm. This is discounting daylight savings time, which is not currently observed in China.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastEditorUserId="13681" LastEditDate="2016-10-16T09:15:46.213" LastActivityDate="2016-10-16T09:15:46.213" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18740" PostTypeId="1" CreationDate="2016-10-15T16:52:22.717" Score="4" ViewCount="120" Body="&lt;p&gt;I tried to research this and I'm not sure if there is a consensus where the $CO_2$, $N_2$, etc. comes from on terrestrial planets like Venus, Earth and Mars. Possible sources would be accretion from the protoplanetary disk, outgassing (which I would think would require plate tectonics/ molten core) or reactions at the surface caused by stellar radiation. Also, reactions in the atmosphere can alter the composition. Would it be a combination of all of the above and, if so, do we know if one of the processes dominated over the others? Does anybody have a good link that discusses this?&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorDisplayName="user15217" LastEditDate="2016-12-20T06:46:32.113" LastActivityDate="2017-04-29T10:11:52.970" Title="Is there a consensus as to where terrrestrial planet atmospheres in our solar system came from?" Tags="&lt;planet&gt;&lt;planetary-atmosphere&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18741" PostTypeId="1" AcceptedAnswerId="18746" CreationDate="2016-10-15T19:49:36.363" Score="2" ViewCount="342" Body="&lt;p&gt;I understand that the universe after the big bang was a super dense cloud of elementary particles, and because of this high density was &quot;not transparent&quot; to light. &#xA;Then, roughly 400.000 years (380.000 to be exact) after density and temperature lower enough for the first hydrogen atoms to form. Thanks to the lower density, now photons were free to travel. Theoretically the Cosmic Microwave Background Radiation (CMBR) should be this light that &quot;became free&quot; 400.000 years after big bang.&#xA;Let's reason in 2d for simplicity. If the universe was a circle, the maximum radius could have been a bit less than 400.000 light years (it's probably an exaggeration, but let's simply assume that matter can't travel faster than light). So, if we imagine that the matter that eventually generated the earth was on the border of this circle (and somehow we are still close), the most far photon at that time should have been at 800.000 light years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my reasoning is correct, I can not understand how it is possible that we see that photon after 13 billion years. It should have reached us long long time ago.&lt;/p&gt;&#xA;" OwnerUserId="14596" LastEditorUserId="14596" LastEditDate="2016-11-11T18:24:29.253" LastActivityDate="2016-11-11T18:24:29.253" Title="Cosmic microwave background radiation: why did it take so long to get here?" Tags="&lt;astrophysics&gt;&lt;fundamental-astronomy&gt;&lt;big-bang-theory&gt;&lt;radio-astronomy&gt;&lt;cosmic-ray&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18742" PostTypeId="2" ParentId="18737" CreationDate="2016-10-15T19:53:50.627" Score="3" Body="&lt;p&gt;I believe that honor may belong to Adak, Alaska where solar noon does not occur during DST until 2:52 PM, or 1:52 Standard Time. All of Alaska is on Juneau time. You can check it &lt;a href=&quot;http://www.sunrisesunset.com/usa/Alaska.asp&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;br&gt;&lt;br&gt;&lt;b&gt;Edit:&lt;/b&gt;There are maps &lt;a href=&quot;http://www.dailymail.co.uk/sciencetech/article-2572317/Are-YOU-living-sync-Amazing-map-reveals-manmade-timezones-countries-false-sense-sun-rises.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; that show the offset between local and solar time. It's pretty informative. There may be other areas besides western China where solar noon is very late (maybe I'll modify a spreadsheet and see if I can some up with something).&lt;/p&gt;&#xA;" OwnerUserId="7108" LastEditorUserId="7108" LastEditDate="2016-10-17T17:50:29.123" LastActivityDate="2016-10-17T17:50:29.123" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18743" PostTypeId="1" AcceptedAnswerId="18756" CreationDate="2016-10-16T10:12:08.627" Score="5" ViewCount="1196" Body="&lt;p&gt;Galaxies that reside in galaxy clusters are know to have peculiar velocities that cause the &lt;a href=&quot;https://en.wikipedia.org/wiki/Redshift-space_distortions&quot; rel=&quot;noreferrer&quot;&gt;Fingers of God effect&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But what about galaxy clusters themselves?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do they also have peculiar velocites by interacting with other galaxy clusters? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, are they &lt;strong&gt;high enough to be taken into accoun&lt;/strong&gt;t or &lt;strong&gt;low enough to be ignored?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2655" LastEditorUserId="2655" LastEditDate="2016-10-16T13:38:07.360" LastActivityDate="2016-10-17T15:21:55.463" Title="Fingers of God effect for galaxy clusters" Tags="&lt;galaxy&gt;&lt;redshift&gt;&lt;galaxy-cluster&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18744" PostTypeId="2" ParentId="18035" CreationDate="2016-10-16T10:29:06.187" Score="1" Body="&lt;p&gt;First: You can use all galaxies in the line of sight of the cluster to construct their color-magnitude diagram. Those galaxies on the Red Sequence are more likely to be in the cluster (not always true).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second: You should be able to determine their redshift, and compare with the mean cluster redshift. To be able to make cut, see next step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third: You should run some algorithms in order to see whether the redshifts/velocities you have are within the clusters mean velocity dispersion (usually in the clustercentric-velocity space), as used in these papers for example: &#xA;&lt;a href=&quot;https://arxiv.org/pdf/1601.06080.pdf&quot; rel=&quot;nofollow&quot;&gt;https://arxiv.org/pdf/1601.06080.pdf&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://adsabs.harvard.edu/full/1996ApJ...473..670F&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/full/1996ApJ...473..670F&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14592" LastActivityDate="2016-10-16T10:29:06.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18745" PostTypeId="2" ParentId="18035" CreationDate="2016-10-16T15:07:24.097" Score="2" Body="&lt;p&gt;You want to use NED (NASA Extragalactic Database) rather than SIMBAD, since the former has more galaxy- and cluster-specific information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to assume you have a reasonably accurate position for your galaxy as well as its redshift (in km/s). The basic idea is to identify candidate galaxy clusters which are close enough to your galaxy on the sky &lt;em&gt;and&lt;/em&gt; in redshift so that the galaxy could potentially be a member of one of them.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Determine what angular radius $R$ corresponds to 1 or 2 megaparsecs at the redshift of your galaxy (e.g., using &lt;a href=&quot;http://www.astro.ucla.edu/~wright/CosmoCalc.html&quot; rel=&quot;nofollow&quot;&gt;Ned Wright's Cosmology Calculator&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do a search in NED for Galaxy Clusters within &lt;em&gt;R&lt;/em&gt; of the coordinates of your galaxy. If your galaxy has a name, you can skip the coordinate lookup and use the &lt;a href=&quot;http://ned.ipac.caltech.edu/forms/nearname.html&quot; rel=&quot;nofollow&quot;&gt;&quot;Near name&quot; search form&lt;/a&gt;; if you only have coordinates, there's a &lt;a href=&quot;http://ned.ipac.caltech.edu/forms/nearposn.html&quot; rel=&quot;nofollow&quot;&gt;&quot;Near position&quot; search form&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. Put the estimated angular radius $R$ (converted to arc minutes) in the &quot;Search Radius&quot; box;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. Set the &quot;Selection in Redshift&quot; form to &quot;Between&quot; and give it lower and upper limits of $V_{\rm gal} - 2000$ and $V_{\rm gal} + 2000$ in km/s, where $V_{\rm gal}$ is the redshift (in km/s) of your galaxy;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. Under &quot;Selection by Object Type&quot;, select &quot;Galaxy Clusters&quot; in the &quot;Classified Extragalactic Objects&quot; list.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;That should give you a list of galaxy clusters within the specified radius of your galaxy which are at a similar redshift. Note that the same cluster may come up multiple times under different names (i.e., from different cluster catalogs), but it should be clear from the coordinates and the redshifts which are the same clusters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the question is whether your galaxy is plausibly a member of a given candidate cluster or not. Ideally you should find some measurement of each cluster's velocity dispersion $\sigma$; if the galaxy's redshift isn't within $\pm 3 \sigma$ of the cluster's redshift $V_{\rm cluster}$, then you can ignore that cluster, since it's very unlikely that a galaxy with that relative velocity with respect to the cluster is actually gravitationally bound to the cluster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't really do better than a probabilistic assessment: roughly how likely it is that the galaxy is a member of a given cluster. Galaxies which are members of a cluster are more likely to be close to the center (on the sky); member galaxies closer to the center can have a higher range of relative velocities, while member galaxies further away should have velocities closer to the cluster redshift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick example: Is NGC 4889 ($V_{\rm gal} = 6495$ km/s) a member of a cluster? At its redshift, 1 Mpc = 33 arc minutes; if we search on NED for galaxy clusters within 33 arcmin of NGC 4889 with redshift limits of 4500-8500 km/s, we get 6 results, 4 of which are just the Coma Cluster in different catalogs, with $V_{\rm cluster} \sim 6900$ km/s. Since the various distances (&quot;Separ. arcmin&quot;) translate to $\sim$ 100-300 kpc from the cluster center, and since the Coma cluster has a velocity dispersion of $\sim$ 1000 km/s and the difference between NGC 4889 and the cluster is only $\sim$ 400 km/s, it's a very good bet that this galaxy is a member of the Coma Cluster. (It's actually one of the two cD galaxies in the cluster center.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2016-10-16T16:15:40.000" LastActivityDate="2016-10-16T16:15:40.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18746" PostTypeId="2" ParentId="18741" CreationDate="2016-10-16T15:22:37.847" Score="3" Body="&lt;p&gt;Here's a very, very simple explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;why did it take so long to get here?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's two aircraft carriers, A and B, fifty miles apart on the ocean. You've got a helicopter that flies 50 mph.  Of course, it takes one hour to make it from A to B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, imagine A and B are both moving - they're steaming away from each other at a good speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it now takes longer than an hour for the chopper to make it between A and B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the many, many mysteries of the metric expansion of space, this trivial observation is just one thing to bear in mind. Note that &lt;em&gt;of course&lt;/em&gt; this is not a full and complete analogy, and the analogy is indeed utterly wrong / totally irrelevant to many aspects of the broader discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;I thought that the photons were &quot;trapped&quot; in an expanding ball&quot;&lt;/em&gt; sure, the chopper can only move at 50 mph &lt;em&gt;over the water underneath it&lt;/em&gt; meaning that at any given time it can only have moved so far &lt;em&gt;over the water underneath it&lt;/em&gt; (a sort of &quot;circle of achievablity&quot;, if you will).  But you can see with the carriers moving the hell all over the place, it's just an open question whether the chopper reaches B in much less than an hour, much more than an hour, or can never reach B at all.&lt;/p&gt;&#xA;" OwnerUserId="13071" LastActivityDate="2016-10-16T15:22:37.847" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18747" PostTypeId="1" AcceptedAnswerId="18818" CreationDate="2016-10-16T15:29:36.340" Score="12" ViewCount="414" Body="&lt;p&gt;I am not sure if I am doing something wrong, or misunderstanding &lt;a href=&quot;https://arxiv.org/abs/1609.08485&quot; rel=&quot;noreferrer&quot;&gt;Reider and Kenworthy (2016)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just trying to reproduce the orbital velocities listed in Table 1. The second paragraph of Section II list a mass of the primary and semi-major axis for the planet's orbit of 0.9 solar mass and 5.0 AU. From the table the mass of the planet ranges from 20 to 100 Jupiters, which is actually quite sizable, but I'll start without using the reduced mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numerical values I'm using:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$GM_{\odot}=\text{1.327E+20} \ \mathrm{m^3  kg^{-2}}$$&#xA;$$GM=0.9GM_{\odot}$$&#xA;$$\epsilon=0.65$$&#xA;$$1 \ \mathrm{AU} = \text{1.496E+11} \ \mathrm{m}$$&#xA;$$a=5.0 \ \mathrm{AU} \ = \text{7.480E+11} \ \mathrm{m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formulae I'm using:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_{\text{peri}}=a(1-\epsilon)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v^2=GM(2/r-1/a)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_{\text{peri}}=\sqrt{GM(2/r_{\text{peri}}-1/a)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_{\text{peri}}=\text{2.618E+11} \ \mathrm{m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_{\text{peri}}=\text{2.744E+4} \ \mathrm{m/s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is  $27.44 \ \mathrm{km/s}$. But for $\epsilon=0.65$ the table below shows $29.5 \pm 0.4 \ \mathrm{km/s}$. Close but not really close enough, it's off by almost 10%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the mass of the planet (which is quite large) was considered, then the table would have to list a wider range of velocities, wouldn't it?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AfqZF.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AfqZF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2016-10-21T12:00:31.190" LastActivityDate="2016-10-22T01:36:52.967" Title="Orbital velocity of a planet - why is my calculation off by about 10%?" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;exoplanet&gt;&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18748" PostTypeId="2" ParentId="18738" CreationDate="2016-10-16T15:34:29.713" Score="3" Body="&lt;p&gt;From a quick skim of the paper, it appears they using data from a previous paper by &lt;a href=&quot;http://cdsads.u-strasbg.fr/abs/2009MNRAS.395.1409S&quot; rel=&quot;nofollow&quot;&gt;Smartt et al. (2009)&lt;/a&gt;. That paper compiled a sample of observed supernovae (SNe) for which there are good images of the parent galaxies taken before the actual SNe. In some cases, Smartt et al. found plausible progenitor stars for the SNe; those are the &quot;detections&quot;. Cases where they &lt;em&gt;couldn't&lt;/em&gt; find a plausible progenitor on pre-SN images are the &quot;non-detections&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The authors of the paper you are talking about appear to be suggesting that Smartt et al. underestimated the masses of progenitor stars because they failed to account for extra local dust extinction from the more massive red supergiant stars.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2016-10-16T15:34:29.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18749" PostTypeId="2" ParentId="11798" CreationDate="2016-10-16T19:24:32.580" Score="5" Body="&lt;p&gt;Starting from &lt;a href=&quot;https://kepler.nasa.gov/Mission/discoveries/index.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;the index you mentioned&lt;/a&gt;, I clicked through the links for some individual planets, which in turn link to discovery papers or other relevant observations. For planets around Kepler-23, -24, -25, -26, -27, and -28, the relevant papers are &lt;a href=&quot;https://kepler.nasa.gov/Mission/discoveries/index.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ford et al. (2012)&lt;/a&gt; and &lt;a href=&quot;https://kepler.nasa.gov/Mission/discoveries/index.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Steffen et al. (2012)&lt;/a&gt;, two out of a series of papers. Both papers used transit timing variations (TTV) and something called dynamical stability, which gives an upper limit to the mass of the planets such that the relevant n-body system is stable over long timescales. This leads to plots like the following (Fig. 3, Ford et al.):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/h1cnd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h1cnd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arrows indicate simulations where the system was stable over timescales greater than $\sim10^7$ years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, as Ford et al. note,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We caution that the uncertainty in the masses and sizes of the host stars directly translates into uncertainties in the planet masses and sizes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Furthermore, shorter observations by Kepler also lead to larger uncertainties. Thus, the maximum masses of the planets as given in these simulations may very well be much greater than their actual masses. Steffen et al. have similar results, and they use a very similar method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After more reading, it's clear that the same is true for planets around Kepler-32, Kepler-48, Kepler-52, Kepler-54, and Kepler -57 to -60, inclusive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kepler-39b is interesting. The paper NASA cites, &lt;a href=&quot;https://arxiv.org/abs/1106.3225&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bouchy et al. (2011)&lt;/a&gt;, states that the object is &lt;em&gt;larger&lt;/em&gt; than one would expect, even taking into account that there is a large uncertainty in radius. It may be a fast rotator (see &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0004-637X/796/1/67/meta;jsessionid=FAC091F0F12BF9656BDA34DF2AC42F08.c1.iopscience.cld.iop.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zhu et al. (2014)&lt;/a&gt;), which would explain its oblateness, but that still isn't enough to explain the discrepancy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I referred to Kepler-39b as an &quot;object&quot;, not a planet. NASA lists its mass as over 20 Jupiter masses; Bouchy et al. placed it at about 18 Jupiter masses, with a very low uncertainty. This could mean that it's more like a brown dwarf than a massive gas giant, and remember, the distinction is fuzzy when it comes to sub-brown dwarfs. This, then, may be the most likely explanation.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-10-16T19:43:41.333" LastActivityDate="2016-10-16T19:43:41.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18750" PostTypeId="1" CreationDate="2016-10-16T21:53:30.973" Score="1" ViewCount="290" Body="&lt;p&gt;On the molecular level, if an atom, consider helium, were to fall into a black hole, what would be the ultimate fate of its electrons, protons and neutrons?&lt;/p&gt;&#xA;" OwnerUserId="14602" LastEditorUserId="1891" LastEditDate="2019-01-20T16:15:39.567" LastActivityDate="2019-01-21T11:27:39.173" Title="Fate of helium moving toward a singularity" Tags="&lt;black-hole&gt;&lt;helium&gt;&lt;singularity&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18751" PostTypeId="2" ParentId="16239" CreationDate="2016-10-17T13:15:31.627" Score="3" Body="&lt;p&gt;One example is &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler-91b&quot; rel=&quot;nofollow&quot;&gt;Kepler-91b&lt;/a&gt;; early observations included strange light curve behavior. While follow-up observations seem to confirm that it is in fact a planet, rather than some other object, &lt;a href=&quot;https://arxiv.org/abs/1511.01068&quot; rel=&quot;nofollow&quot;&gt;Placek et al. (2015)&lt;/a&gt; suggested that there is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Trojan_(astronomy)&quot; rel=&quot;nofollow&quot;&gt;Trojan&lt;/a&gt; &quot;planet&quot; ($M\sim.025\pm.019M_J$) at one of Kepler-91b's Lagrange points - either L&lt;sub&gt;4&lt;/sub&gt; or L&lt;sub&gt;5&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors believe that this explanation explains the light curve data much better than the single-planet model, as the unexplained dimming could be the result of a &quot;secondary transit&quot; by the Trojan planet, rather than a &quot;primary transit&quot;. However, predictions of the daytime temperature of the secondary planet fall squarely about 5,000 K, which is close to the temperature of the surface of the Sun. Albedo calculations are also higher than expected (0.49).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the authors state that the light curve may therefore be a false positive, and Kepler-91b does not exist - and nor does the hypothetical Trojan planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trojan co-orbital planets have been the main focus of theoretical and experimental research so far; see &lt;a href=&quot;http://mnras.oxfordjournals.org/content/421/1/356.full&quot; rel=&quot;nofollow&quot;&gt;Giuppone (2012)&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-10-17T14:52:11.203" LastActivityDate="2016-10-17T14:52:11.203" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18753" PostTypeId="1" AcceptedAnswerId="18766" CreationDate="2016-10-17T13:33:01.087" Score="3" ViewCount="1009" Body="&lt;p&gt;I am trying to imagine how dark outer space is. Looking at images of cosmonauts in space proved not very useful, as the exposure makes space look completely dark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking about it, I would imagine an object would be quite well illuminated, being bombarded by provided it's not in the shadow of a planet or some other celestial body. Is that correct? How about light scattering? Is cosmic dust abundant enough for the purpose?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; @zephyr made me realize that the quantity of light is a function of position, so let's say the question is with regards to the Milky Way. Still, I would appreciate some comments on what's the situation outside the Milky Way too.&lt;/p&gt;&#xA;" OwnerDisplayName="user14607" LastEditorDisplayName="user14607" LastEditDate="2016-10-17T13:53:43.517" LastActivityDate="2018-10-04T11:29:21.343" Title="Darkness in Outer Space" Tags="&lt;light&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18754" PostTypeId="1" CreationDate="2016-10-17T14:00:56.297" Score="0" ViewCount="136" Body="&lt;p&gt;Do meteors always streak in the same direction? While I know it's not likely to be exactly the same direction, but generally speaking. Would one say meteors always travel, for example, east to west?&lt;/p&gt;&#xA;" OwnerUserId="14608" LastEditorUserId="2153" LastEditDate="2016-10-17T21:02:01.673" LastActivityDate="2016-10-17T21:05:27.553" Title="Direction of Meteors" Tags="&lt;amateur-observing&gt;&lt;meteor&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2016-10-18T13:01:02.937" ContentLicense="CC BY-SA 3.0" />
  <row Id="18756" PostTypeId="2" ParentId="18743" CreationDate="2016-10-17T15:05:05.553" Score="5" Body="&lt;p&gt;I'm going to break your question down into two sections.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do galaxy clusters have measurable peculiar velocities?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer to this depends slightly on what is meant by &quot;peculiar velocity&quot;. In the most general sense, the peculiar velocity of some object is the velocity it has with respect to some standard of rest. &lt;a href=&quot;https://en.wikipedia.org/wiki/Peculiar_velocity#Cosmology&quot; rel=&quot;nofollow noreferrer&quot;&gt;In the cosmological sense&lt;/a&gt;, the peculiar velocity of a galaxy in a cluster is simply any motion that galaxy may have which is not attributable to the Hubble flow. Primarily this peculiar velocity is a result of that galaxy's orbit around the cluster barycenter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that in mind, it should hopefully be easy to reason out that clusters themselves will have peculiar velocities. It is highly unlikely that a cluster will be &quot;fixed&quot; in space and only move according to the Hubble flow. A quick google search revealed &lt;a href=&quot;http://adsabs.harvard.edu/abs/1996ApJ...462L..49B&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bahcall &amp;amp; Oh 1996&lt;/a&gt; which cites an rms one-dimensional galaxy cluster peculiar velocity of $298\pm28\:\mathrm{km/s}$ (based on 22 clusters). While this is smaller than a typical galaxy's peculiar velocity, it is certainly measurable and capable of being distinguished from the Hubble flow (at least for reasonable redshifts).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real difficulty here is not whether the cluster has a non-negligible peculiar velocity, but rather, is that peculiar velocity even measurable due to many of the potential limitations. While the value of the peculiar velocity of a cluster is certainly &quot;high enough to be taken into account&quot; it can often be difficult to measure such a velocity with any reasonable error (a point that Bahcill and Oh discuss).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can these galaxy cluster peculiar velocities produce the &quot;Fingers of God&quot; effect?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To answer this you have to consider a very particular point I made in the last section which was that for galaxies the primary contribution to their peculiar velocity is their orbital motion around the cluster barycenter which can result in peculiar velocities on the order of $10^3\:\mathrm{km/s}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When trying to map out a particular cluster in redshift space, you can often get &quot;Fingers of God&quot; in your maps which are a result of the cluster appearing elongated towards the observer and due entirely to the orbital peculiar motion. An example of such an effect is below, with the red circles added by me to indicate these fingers of god.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6Qekg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6Qekg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxy clusters on the other hand are not gravitationally bound to other galaxy clusters! This means any peculiar motion a cluster has is simply its general motion through space. Since the fingers of god are a result of orbital peculiar motion, it stands to reason that galaxy clusters will not have such a motion and as such clusters won't produce these fingers of god. You'll find that more or less, the peculiar motion of clusters will simply move cause them to be randomly re-positioned without any real &quot;structure&quot; that you see in the fingers of god for galaxies.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-10-17T15:05:05.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18757" PostTypeId="5" CreationDate="2016-10-17T15:06:36.457" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2016-10-17T15:06:36.457" LastActivityDate="2016-10-17T15:06:36.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18758" PostTypeId="4" CreationDate="2016-10-17T15:06:36.457" Score="0" Body="Any element other than hydrogen and helium is considered a metal by astronomers." OwnerUserId="13681" LastEditorUserId="13681" LastEditDate="2016-10-17T15:24:11.443" LastActivityDate="2016-10-17T15:24:11.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18759" PostTypeId="2" ParentId="14842" CreationDate="2016-10-17T15:18:00.683" Score="2" Body="&lt;p&gt;AR Scorpii&lt;sup&gt;1&lt;/sup&gt; appears to be an example. It is a binary system, consisting of a white dwarf and an M-type red dwarf. It was apparently originally classified as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Delta_Scuti_variable&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\delta$ Scuti variable&lt;/a&gt; star, but it was later discovered to be a true close binary, with a period of 3.56 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The white dwarf emits pulses, as discovered by &lt;a href=&quot;http://www.nature.com/nature/journal/v537/n7620/full/nature18620.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Marsh et al. (2016)&lt;/a&gt;, with a period of about two minutes. These are visible in wavelengths from ultraviolet to radio (see Fig. 2):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KaJBj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KaJBj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most noticeable patterns are in the ultraviolet, and the weakest are in the radio band, but the signals are certainly detectable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rational for determining that the compact object is a white dwarf follows from its luminosity, its relatively long spin period (longer than any neutron star pulsar yet discovered), and the fact that the pairing with a red dwarf is &quot;more natural&quot; than the odd alternative involving a low-mass neutron star and high-mass red dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system may form what is known as an &lt;a href=&quot;https://en.wikipedia.org/wiki/Intermediate_polar&quot; rel=&quot;nofollow noreferrer&quot;&gt;intermediate polar&lt;/a&gt;, involving accretion onto a white dwarf influenced by the object's strong magnetic field. The other notable example is &lt;a href=&quot;https://en.wikipedia.org/wiki/AE_Aquarii&quot; rel=&quot;nofollow noreferrer&quot;&gt;AE Aquarii&lt;/a&gt;, as Rob Jeffries mentioned. However, its radio emissions do not involve pulsations. &lt;em&gt;Why&lt;/em&gt; it doesn't emit pulsating signals at radio wavelengths is another question entirely, and not one I can answer right now.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;More recent results (&lt;a href=&quot;http://www.nature.com/articles/s41550-016-0029&quot; rel=&quot;nofollow noreferrer&quot;&gt;Buckley et al. (2017)&lt;/a&gt;) indicate that the radio emissions from the system are polarized, similar to those of neutron stars. It seems that AR Sco shares even more characteristics with pulsars than was previously thought.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1 The &lt;a href=&quot;https://en.wikipedia.org/wiki/AR_Scorpii&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; page, I just discovered, is terrible.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2017-02-08T15:00:04.787" LastActivityDate="2017-02-08T15:00:04.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18760" PostTypeId="2" ParentId="18743" CreationDate="2016-10-17T15:21:55.463" Score="5" Body="&lt;p&gt;I think the key point is how the velocity dispersion in a given structure compares with the velocity dispersion you would expect from the difference in the Hubble flow across the linear extent of the object. Or, equivalently, what range of distances (according to Hubble's law) are implied by the velocity dispersions and how does this compare with the actual size of the structure in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fingers of God effects in clusters are produced because the velocity dispersions ($\sim \pm 1000$ km/s) would imply a front-to-back distance of $\pm 14$ Mpc (for $H_0 = 70$ km s$^{-1}$/Mpc), compared with the actual size of a cluster that typically only has a diameter of $2$ Mpc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxy clusters may themselves be part of superclusters. These superclusters may have sizes of order 100 Mpc or more. However, the peculiar velocities of clusters at similar distances (that can be judged from the average recession velocities of their members) is actually &lt;em&gt;smaller&lt;/em&gt; than the peculiar velocities of individual galaxies. Thus the change in Hubble flow velocity across a supercluster is &lt;em&gt;much&lt;/em&gt; larger than the peculiar velocities of the individual members (andthey are not bound together). So I would not expect a Finger of God effect in the members of a supercluster.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-10-17T15:21:55.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18761" PostTypeId="1" CreationDate="2016-10-17T16:07:28.197" Score="1" ViewCount="5500" Body="&lt;p&gt;I am trying to solve a problem where I need to calculate a satellite's orbit, but first I would like to ask for some clarifications from someone here that might know this stuff. I need to design an orbit so that no maneuvers (one part of the problem) are necessary to maintain it, and the hint is that there are J2 perturbation only and that the orbit is elliptical. I don't really understand this, so can someone explain what they mean with only J2 perturbations and how it is relevant here? (just asking for an explanation) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!  &lt;/p&gt;&#xA;" OwnerUserId="14587" LastActivityDate="2016-10-18T14:12:54.483" Title="J2 perturbations and orbits" Tags="&lt;orbital-mechanics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18763" PostTypeId="2" ParentId="10936" CreationDate="2016-10-17T20:58:31.850" Score="1" Body="&lt;p&gt;As per the above comments, there will still be v=Hd, if v is the rate of change of current distance, and d is the current distance-- as long as the universe continues to obey the cosmological principle.  The nearby version of that will still look like correlating Doppler-shift velocity to distance indicators, and the faraway version will require a dynamical model, albeit a very simple one due to dark energy.  It will lead to a constant H that will not be anything like the inverse of the age of the universe.  And if you go too far into the future, you won't be able to see the CMB, or quasars, so alien cosmology will really struggle.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-10-17T20:58:31.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18764" PostTypeId="2" ParentId="18754" CreationDate="2016-10-17T21:05:27.553" Score="3" Body="&lt;p&gt;Not in general, but sometimes there is a dominant direction.&#xA;When the Earth passes through a stream of debris from a comet or asteroid,&#xA;a&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Meteor_shower&quot; rel=&quot;nofollow&quot;&gt;meteor shower&lt;/a&gt;&#xA;occurs, and those meteors appear to come from a point in the sky called the radiant.&#xA;Since the parent bodies have different orbits, different shower radiants appear in different parts of the sky.&#xA;For example, the Geminid radiant culminates earlier in the night than most, and the Perseid radiant is well north of the Ecliptic.&#xA;As the Earth can encounter meteors from any direction at any time, there are usually a few sporadic meteors unrelated to any active radiant.&#xA;Meteor observers have identified some&#xA;&lt;a href=&quot;http://www.skyandtelescope.com/astronomy-blogs/explore-night-bob-king/random-meteors-arent-random02112015/&quot; rel=&quot;nofollow&quot;&gt;not quite random&lt;/a&gt;&#xA;directions for these.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-10-17T21:05:27.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18765" PostTypeId="2" ParentId="18761" CreationDate="2016-10-18T07:04:24.453" Score="5" Body="&lt;p&gt;Due to the oblateness of the Earth (the Earth's equatorial bulge), a geocentric satellite's orbital plane will precess (rotate) relative to inertial space.&#xA;The rate at which the line of nodes moves owing to this bulge is given by&#xA;$$&#xA;\dot\Omega = -\frac{3}{2}J_2\left(\frac{r_E}{ℓ}\right)^2 n\cos\iota&#xA;$$&lt;br&gt;&#xA;where $J_2$ is the is the zonal harmonic coefficient ($1.08262668\times10^{-3}$ for Earth), $r_E$ is the body's equatorial radius ($6\,378\,137$ m for Earth), $ℓ$ is the orbit parameter (the semi-latus rectum), $n$ is the mean motion, and $\iota$ is the inclination of the orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given orbit parameter ($ℓ$) and mean motion ($n$), the inclination of a geocentric satellite orbit can be selected to obtain, for example, a Sun-synchronous orbit ($\dot\Omega=360^\circ$ per $365.26$ days, or $0.9856$ degrees per day).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Coordinate System&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_elements&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; (and diagram from that article) describes the coordinate system used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tzuPy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tzuPy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbital plane (yellow) intersects a reference plane (gray). For Earth-orbiting satellites, the reference plane is usually the Earth's equatorial plane.  The intersection is called the line of nodes, as it connects the center of mass with the ascending and descending nodes.  This plane, together with the vernal point ($\gamma$), establishes a reference frame.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Two elements that define the size and shape of the elliptical orbit&#xA;are the semi-major axis, $a$, and the eccentricity, $e$. The&#xA;semi-latus rectum is related to $a$ and $e$ by $ℓ=a(1-e^2)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Two elements define the orientation of the orbital plane in which the&#xA;ellipse is embedded: the inclination, $\iota$, and the longitude of the ascending node, $\Omega$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5537" LastEditorUserId="5537" LastEditDate="2016-10-18T14:12:54.483" LastActivityDate="2016-10-18T14:12:54.483" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18766" PostTypeId="2" ParentId="18753" CreationDate="2016-10-18T12:19:04.643" Score="2" Body="&lt;p&gt;Just to put in a simple answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you mean (A) for astronauts just in orbit (at the space station and so on) or on the surface of the moon.  Yes, it is &lt;strong&gt;incredibly bright&lt;/strong&gt;.  Just like on Earth in the Sahara at noon!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A point of confusion is that &lt;strong&gt;&lt;em&gt;THE &gt;&gt; SKY &amp;lt;&amp;lt; IS BLACK&lt;/em&gt;&lt;/strong&gt;, but it is incredibly bright.  On Earth in daytime, the sky happens to be blue, and it is incredibly bright.  On Moon in daytime, the sky happens to be &gt;&gt; black &amp;lt;&amp;lt;, and it is incredibly bright.  (Same deal in orbit as on the Moon.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you are, say, explaining to a young person &quot;is there daytime/nighttime on the moon&quot; that's the salient point to explain: on the moon, confusingly, the sky is black during both daytime and nighttime.  But everything's just as &lt;em&gt;bright&lt;/em&gt; on the moon in daytime as on Earth in daytime, there's the same amount of light, i.e. you can trivially read a book or see the ground, just the same as on Earth.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If your question is about (B) in the solar system - say &quot;on Uranus&quot; or the like.  There is remarkably less light there than on Earth, the sun is much smaller.  Many young people don't realize this; everything's basically dark all the time from about Saturn outwards!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Outstanding page explaining this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astronoo.com/en/children/sun-apparent-size.html&quot; rel=&quot;nofollow&quot;&gt;http://www.astronoo.com/en/children/sun-apparent-size.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Finally if your question is about either (C) inside our galaxy but not near a star, or, (D) midway between the galaxies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's too difficult.&lt;/p&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="13071" LastEditDate="2016-10-18T12:27:06.923" LastActivityDate="2016-10-18T12:27:06.923" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18767" PostTypeId="2" ParentId="18753" CreationDate="2016-10-18T13:00:36.983" Score="0" Body="&lt;p&gt;I think you can actually experience the answer to your question by going out into the deep wilderness late on a night where there is no Moon, make sure no artificial lights are anywhere nearby, and simply observe.  That's how dark it is in a reasonably random location in the Milky Way!  It will be very dark indeed, but you might still be able to see &lt;em&gt;something&lt;/em&gt; (other than the stars, of course).  It's a good question-- can you see your hand in front of your face by starlight alone?  I don't think you can, but I haven't done the experiment!  &lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-10-18T13:00:36.983" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18768" PostTypeId="2" ParentId="16391" CreationDate="2016-10-18T15:19:40.617" Score="9" Body="&lt;p&gt;This is an interesting question and so often interesting questions cannot easily be answered with any current knowledge, but this one can be answered to some degree. I'm going to run through the basics of orbital theory and describe how they can apply to galaxies and how it differs from Keplerian systems. You should have a reasonable understanding of Newtonian physics (after all orbits derive precisely from Newton's laws) and a strong knowledge of math. If you don't have these things, just jump to the end of each section where I'll try to summarize the important points behind the math.&lt;/p&gt;&#xA;&lt;p&gt;A quick note on math notation I'll be using. A dot over a symbol indicates a time derivative (e.g., &lt;span class=&quot;math-container&quot;&gt;$\dot{a}$&lt;/span&gt;) and un-italicised, bold symbols are vector quantities (e.g., &lt;span class=&quot;math-container&quot;&gt;$\bf F$&lt;/span&gt;). Let's get down to business.&lt;/p&gt;&#xA;&lt;h3&gt;The Orbital Equation of Motion&lt;/h3&gt;&#xA;&lt;p&gt;Consider a mass &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; as some position &lt;span class=&quot;math-container&quot;&gt;$\bf r$&lt;/span&gt; and moving with a motion described by &lt;span class=&quot;math-container&quot;&gt;$\bf \dot{r}$&lt;/span&gt;. This mass experiences a force &lt;span class=&quot;math-container&quot;&gt;$\textbf{F}(r)$&lt;/span&gt; which is only a function of radial distance, &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt;, from the center of the coordinate system. The goal here is to determine the equation of motion that can describe the orbit of the mass due to this force. This equation can then be used to solve for &lt;span class=&quot;math-container&quot;&gt;$r(\theta)$&lt;/span&gt;. By Newton's law the equation of motion can be initially defined to be&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\textbf{F}(r) = m\textbf{a} = m\Big(\ddot{r} - r\dot{\theta}^2\Big)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Note that in this case &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; is simply the radial component of &lt;span class=&quot;math-container&quot;&gt;$\bf r$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; is the azimuthal angle of the body in a spherical coordinate system. I'll leave it to you to determine how to break down the acceleration into the two components above, under the appropriate coordinate system. Let's try to remove our &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; dependence so that we have only a function of &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt;. This can be achieved by using angular momentum conservation. The angular momentum per unit mass is given by &lt;span class=&quot;math-container&quot;&gt;$\ell = r^2\dot{\theta}$&lt;/span&gt; so that &lt;span class=&quot;math-container&quot;&gt;$\dot{\theta} = \ell/r^2$&lt;/span&gt;. This gives&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\textbf{F}(r) =  m\big(\ddot{r} - \ell^2/r^3\big)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;This is now a differential equation that allows us to solve for &lt;span class=&quot;math-container&quot;&gt;$r(t)$&lt;/span&gt;, but we want &lt;span class=&quot;math-container&quot;&gt;$r(\theta)$&lt;/span&gt; so we need to do some conversion. Let's re-parameterize by defining &lt;span class=&quot;math-container&quot;&gt;$u \equiv 1/r$&lt;/span&gt; (the reason will become clear in a bit) and determining &lt;span class=&quot;math-container&quot;&gt;$\ddot{r}$&lt;/span&gt; in terms of &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\mathrm{d}}{\mathrm{d}t}(r) = \frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{1}{u}\bigg)=\frac{1}{u^2}\frac{\mathrm{d}u}{\mathrm{d}t}=\frac{1}{u^2}\frac{\mathrm{d}u}{\mathrm{d}\theta}\frac{\mathrm{d}\theta}{\mathrm{d}t}=-\frac{\dot{\theta}}{u^2}\frac{\mathrm{d}u}{\mathrm{d}\theta}=-\ell\frac{\mathrm{d}u}{\mathrm{d}\theta}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Note the substitution of &lt;span class=&quot;math-container&quot;&gt;$\ell=r^2\dot{\theta}=\dot{\theta}/u^2$&lt;/span&gt;. Now differentiate again to determine &lt;span class=&quot;math-container&quot;&gt;$\ddot{r}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\mathrm{d}^2}{\mathrm{d}t^2}(r)=-\ell\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\mathrm{d}u}{\mathrm{d}\theta}\bigg)=-\ell\frac{\mathrm{d}\theta}{\mathrm{d}t}\frac{\mathrm{d}}{\mathrm{d}\theta}\bigg(\frac{\mathrm{d}u}{\mathrm{d}\theta}\bigg) = \ell\dot{\theta}\frac{\mathrm{d}^2u}{\mathrm{d}\theta^2}=-\ell^2 u^2 \frac{\mathrm{d}^2u}{\mathrm{d}\theta^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Putting this into our expression for the equation of motion and making the transformation that &lt;span class=&quot;math-container&quot;&gt;$r=1/u$&lt;/span&gt; finally gives&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\textbf{F}(1/u) =  m\Big(-\ell^2 u^2 \frac{\mathrm{d}^2u}{\mathrm{d}\theta^2} - \ell^2 u^3\Big)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Writing in a more convenient form we finally arrive at&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\boxed{\frac{\mathrm{d}^2u}{\mathrm{d}\theta^2} + u = -\frac{\textbf{F}(1/u)}{m\ell^2u^2}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Remember that, &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; is the mass of the body, &lt;span class=&quot;math-container&quot;&gt;$u(\theta) \equiv 1/r(\theta)$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ell$&lt;/span&gt; is the angular momentum per unit mass, &lt;span class=&quot;math-container&quot;&gt;$\bf F$&lt;/span&gt; is a purely radial force acting on the body, and &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; are the radial and azimuthal coordinate positions of the mass.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Punchline&lt;/strong&gt;: The final result here is a general equation of motion for a body orbiting according to some arbitrary force. This could be gravity, electromagnetic, a spring force, or anything else we decide. It is purposefully derived under general and nonconstrictive assumptions and hopefully you can see that it can be used to understand the orbital motion of a star orbiting in a disk galaxy. The goal with this equation should be to plug in your force (whatever that may be) and solve for &lt;span class=&quot;math-container&quot;&gt;$u(\theta)$&lt;/span&gt;. From there it's easy to determine &lt;span class=&quot;math-container&quot;&gt;$r(\theta)$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h3&gt;Keplerian Motion&lt;/h3&gt;&#xA;&lt;p&gt;Before we go about looking at orbital motion in a galaxy, let's look at standard Keplerian motion so we have something to compare to. Keplerian motion derives from assuming our mass &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; is orbiting a single, point-like mass &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; and under the influence of simple gravity. In that case, we can write our force as &lt;span class=&quot;math-container&quot;&gt;$F(r) = kr^{-2}$&lt;/span&gt; and thus &lt;span class=&quot;math-container&quot;&gt;$F(1/u) = ku^2$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$k\equiv GMm$&lt;/span&gt; is a constant, defined here for mathematical simplicity. Note that &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; is the gravitational constant. The general orbital equation, under this force, now becomes&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\mathrm{d}^2u}{\mathrm{d}\theta^2} + u = -\frac{k}{m\ell^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;This is a standard second-order, nonhomogeneous differential equation with a constant forcing function. If you know your Diff EQ, you should know the solution almost immediately.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$u(\theta) = \frac{k}{m\ell^2} + A\cos (\theta - \theta_0)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;In this equation, &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is some unknown constant, and &lt;span class=&quot;math-container&quot;&gt;$\theta_0$&lt;/span&gt; represents the starting angle of the orbit, which we can arbitrarily choose to be zero. Our ultimate goal is to get &lt;span class=&quot;math-container&quot;&gt;$r(\theta)$&lt;/span&gt; so let's do that. I'm going to do a few steps in one though and leave you to work out the intervening math. I'll fill in for &lt;span class=&quot;math-container&quot;&gt;$k = GMm$&lt;/span&gt; and define the angular momentum as &lt;span class=&quot;math-container&quot;&gt;$L = \ell\mu$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt; is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Reduced_mass&quot; rel=&quot;noreferrer&quot;&gt;reduced mass of our system&lt;/a&gt;. I will also state, without proof, that &lt;span class=&quot;math-container&quot;&gt;$e = A(m\ell^2/k)$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt; is the eccentricity of the orbit.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\boxed{r(\theta) = \frac{L^2/GM\mu^2}{1+e\cos (\theta)}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Punchline&lt;/strong&gt;: We've found a final equation which represents the orbital motion of a mass under the influence of a gravity due to a point-like mass &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt;. If you know your stuff, you'll see that this equation precisely describes the conic sections, depending on the value of &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt;. If &lt;span class=&quot;math-container&quot;&gt;$e = 0$&lt;/span&gt;, you get circular motion (since &lt;span class=&quot;math-container&quot;&gt;$r(\theta)$&lt;/span&gt; becomes a constant). If &lt;span class=&quot;math-container&quot;&gt;$0 &amp;lt; e &amp;lt;1$&lt;/span&gt;, you get elliptical motion, &lt;span class=&quot;math-container&quot;&gt;$e=1$&lt;/span&gt; is parabolic motion and &lt;span class=&quot;math-container&quot;&gt;$e &amp;gt; 1$&lt;/span&gt; is hyperbolic.&lt;/p&gt;&#xA;&lt;p&gt;As expected, Keplerian motion (i.e., having a central force such that &lt;span class=&quot;math-container&quot;&gt;$F\propto r^{-2}$&lt;/span&gt;) resulted in conics, which is precisely Kepler's first law. Kepler's second and third laws more or less derive from the same assumptions. It stands to reason then, that any system where &lt;span class=&quot;math-container&quot;&gt;$F \not \propto r^{-2}$&lt;/span&gt; does not follow any of Kepler's laws. The orbits are not perfect conics (e.g., ellipses, circles, etc.), they do not sweep out equal areas in equal times, and the standard &lt;span class=&quot;math-container&quot;&gt;$P^2 \propto a^3$&lt;/span&gt; certainly does not apply.&lt;/p&gt;&#xA;&lt;h3&gt;Orbital Motion in a Galaxy&lt;/h3&gt;&#xA;&lt;p&gt;Your question correctly describes the situation for stars (or anything really) orbiting in a galaxy. Stars are not orbiting central, point-like masses. They are embedded within both the baryonic and dark matter that comprises the galaxy and are orbiting through it. It is &lt;a href=&quot;https://en.wikipedia.org/wiki/Shell_theorem&quot; rel=&quot;noreferrer&quot;&gt;a well known concept&lt;/a&gt; in physics that spherically symmetric mass distributions do not have a net gravitational pull on objects &lt;em&gt;interior&lt;/em&gt; to that distribution, which means that for stars in a galaxy, the mass affecting its orbit is the mass interior to its radius. If that radius changes, the mass changes!&lt;/p&gt;&#xA;&lt;p&gt;The central force on our star is still going to be gravity, but the mass acting on it will be all the mass interior to some radius, denoted by &lt;span class=&quot;math-container&quot;&gt;$M_r$&lt;/span&gt;. We can see that &lt;span class=&quot;math-container&quot;&gt;$F(r) = GM_r(r)m/r^2$&lt;/span&gt;. If we want to determine the force acting on our star (and thus the exact orbit, via the differential equation above), we first need to figure out what the mass interior to some radius is. This can be achieved by using the mass &lt;a href=&quot;https://en.wikipedia.org/wiki/Continuity_equation&quot; rel=&quot;noreferrer&quot;&gt;continuity equation&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\mathrm{d}M_r}{\mathrm{d}r} = 4\pi r^2 \rho(r)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;You can essentially figure out all the mass interior to &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; by integrating up all the mass density as a function of &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt;. Here of course, you need a good equation for &lt;span class=&quot;math-container&quot;&gt;$\rho(r)$&lt;/span&gt;. A simple, but physically unrealistic density profile is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Singular_isothermal_sphere_profile&quot; rel=&quot;noreferrer&quot;&gt;Singular Isothermal Sphere&lt;/a&gt; (SIS), whereas a more realistic, but mathematically complex equation might be the &lt;a href=&quot;https://en.wikipedia.org/wiki/Navarro%E2%80%93Frenk%E2%80%93White_profile&quot; rel=&quot;noreferrer&quot;&gt;NFW profile&lt;/a&gt; or the &lt;a href=&quot;https://en.wikipedia.org/wiki/Einasto_profile&quot; rel=&quot;noreferrer&quot;&gt;Einasto profile&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;I've now laid out all the steps you need to figure out the orbital motion in a galaxy, but I have to say, it's not pretty. We can look at part of the simplest case though, that of the SIS.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Single Isothermal Sphere&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;For the singlar isothermal sphere, you have that &lt;span class=&quot;math-container&quot;&gt;$\rho(r) = v^2/(4\pi Gr^2)$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; is the rotational velocity of your star. This profile relies on a crucial fact of disk galaxies &lt;span class=&quot;math-container&quot;&gt;$-$&lt;/span&gt; their rotational profile is flat! This was well established, for example by &lt;a href=&quot;http://adsabs.harvard.edu/abs/1978ApJ...225L.107R&quot; rel=&quot;noreferrer&quot;&gt;Ruben et al. 1978&lt;/a&gt;. I've reproduced a figure from this paper below which shows the rotation curve for several galaxies. The important point here is that this shows &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; is constant and does not depend on radius! (Assuming we're not near the galactic bulge or center. That's a whole different beast.)&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OvGBf.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OvGBf.png&quot; alt=&quot;Ruben et al. 1978 Rotation Curves&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;With this crucial piece of information, we can solve for &lt;span class=&quot;math-container&quot;&gt;$M_r$&lt;/span&gt; by integrating &lt;span class=&quot;math-container&quot;&gt;$\rho(r)$&lt;/span&gt; (which I'll leave up to you). The result is that&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M_r = \frac{v^2r}{G}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;This means your force is given by&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F(r) = \frac{v^2rm}{r^2} = \frac{v^2m}{r}\Rightarrow F(1/u) = v^2mu\propto ku$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;You can see here that unlike the Keplerian case, our force is proportional to &lt;span class=&quot;math-container&quot;&gt;$r^{-1}$&lt;/span&gt; rather than &lt;span class=&quot;math-container&quot;&gt;$r^{-2}$&lt;/span&gt;. You can do this process with other density profiles (such as the NFW or Einasto I listed above), but you'll wind up with the same result.&lt;/p&gt;&#xA;&lt;p&gt;If you're so inclined, you could choose to plug this into the orbital equation of motion above and solve it out, but you're now working with a non-linear differential equation and things can get messy fast.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Punchline&lt;/strong&gt;: I'm not sure if this actually answers your question or not. I've led you partially down the rabbit hole, but hopefully you can appreciate how complex it quickly gets. All the above work was using broad assumptions and simplifications. I suppose the short answer to all of this is that stars orbit galaxies in a complex, yet closed orbit which is not easily described precisely (even for our own galaxy) via calculable equations. We can approximate and do our best to work through the math, but in the end its an approximation. In the crudest approximations though, you may as well consider an orbit such as our star to be circular and be done with it.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-10-19T05:07:10.587" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18769" PostTypeId="1" CreationDate="2016-10-18T18:01:55.643" Score="4" ViewCount="88" Body="&lt;p&gt;Why is the sun's &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Magnetic_reconnection&quot; rel=&quot;nofollow noreferrer&quot;&gt;magnetic reconnection&lt;/a&gt; effect many orders of magnitude faster than predicted by standard models?&lt;/p&gt;&#xA;" OwnerUserId="14078" LastEditorUserId="43" LastEditDate="2017-06-10T19:18:46.027" LastActivityDate="2017-06-11T19:49:20.110" Title="Why is the magnetic reconnection(of sun) effect many orders of magnitude faster than predicted by standard models?" Tags="&lt;the-sun&gt;&lt;magnetic-field&gt;&lt;magnitude&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18770" PostTypeId="2" ParentId="10951" CreationDate="2016-10-19T00:37:25.157" Score="2" Body="&lt;p&gt;The effective sunspot number $R_{\text{eff}}$ is calculated through a mix of observations and model calculations. It focuses primarily on some parameter called the frequency of the &lt;a href=&quot;https://en.wikipedia.org/wiki/F_region&quot; rel=&quot;nofollow&quot;&gt;ionosphere F2 layer&lt;/a&gt;, denoted $\text{foF2}$, the highest frequency of electromagnetic radiation which can be reflected off the F2 layer; it can be observed as well as computed based on a certain model (real-time observational data from stations around the world is available &lt;a href=&quot;http://www.ngdc.noaa.gov/stp/IONO/rt-iono/realtime/RealTime_foF2.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;). More information on $\text{foF2}$ can be found on &lt;a href=&quot;https://spawx.nwra.com/spawx/fof2/fof2.html&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mean-squared error between the observational value $\text{foF2}_{\text{obs}}$ and the calculated value $\text{foF2}_{\text{calc}}$ over observations by $n$ stations is&#xA;$$\Delta=\frac{1}{n}\sum_{i=1}^n\left(\text{foF2}_{\text{obs},i}-\text{foF2}_{\text{calc},i}\right)^2$$&#xA;where $_i$ denotes the value at station $i$. $R_\text{eff}$ is defined to be that chosen sunspot number such that $\Delta$ is minimized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just an application of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Least_squares&quot; rel=&quot;nofollow&quot;&gt;least squares&lt;/a&gt; method to fit a model framework to match observations. An example of this using the Simplified Ionospheric Regional Model (SIRM) is &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1029/2003RS002936/pdf&quot; rel=&quot;nofollow&quot;&gt;Zolesi et al. (2004)&lt;/a&gt;, although the procedure for calculating $R_\text{eff}$ was first proposed by &lt;a href=&quot;http://adsabs.harvard.edu/abs/1993JEEEA..13...99H&quot; rel=&quot;nofollow&quot;&gt;Houminer et al. (1993)&lt;/a&gt;. This was introduced to augment or replace something called the 12-month smoothed sunspot number, $R_{12}$, which is entirely based on observations. Zolesi et al. write&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is therefore expected that during periods of relative ionospheric quietness, when the observed $\text{foF2}$ value is much closer to the model predicted median values, the $R_{\text{eff}}$ does not differ much from the $R_{12}$. On the contrary, during periods of ionospheric activity, the $R_{\text{eff}}$ should have large difference form [sic] $R_{12}$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-10-19T15:51:56.137" LastActivityDate="2016-10-19T15:51:56.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18772" PostTypeId="1" CreationDate="2016-10-19T01:29:00.417" Score="3" ViewCount="218" Body="&lt;p&gt;I am building myself a camera that attaches to the back of my Celestron NexStar 4 SE telescope and I am not sure what type of lens to choose to replace the stock lens on the Raspberry Pi camera. I bought &lt;a href=&quot;http://www.m12lenses.com/CS-Mount-Lens-Holder-Metal-p/pt-lh013csm.htm&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; CS Mount and by mistake bought M12 lenses thinking it was a mount for M12 lenses but now I know CS mounts require bigger lenses than M12. So should I instead get a M12 mount and use M12 lenses or should I buy a CS lens and what kind if I am looking to do a high quality video/live stream?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Raspberry Pi Camera: &lt;a href=&quot;https://www.adafruit.com/products/3099&quot; rel=&quot;nofollow&quot;&gt;https://www.adafruit.com/products/3099&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14616" LastActivityDate="2019-09-04T00:11:33.567" Title="What lens should I use for my Raspberry Pi?" Tags="&lt;photography&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18773" PostTypeId="1" AcceptedAnswerId="18780" CreationDate="2016-10-19T03:11:48.863" Score="1" ViewCount="170" Body="&lt;p&gt;I get this question a lot: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where is the center of the Universe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I see this question on the internet all of the time:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where is the center of the Universe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now I have a question regarding all of the answers that I always either give or &lt;a href=&quot;https://astronomy.stackexchange.com/questions/13999/how-does-the-concept-of-a-universe-with-no-center-work&quot;&gt;read&lt;/a&gt;:&#xA;The question linked is the last question and answer that I saw before posing this question here, and is in no way the same question, as this question asks how the concept of a centerless universe can work, whereas the question below asks about how a universe can be perceived to be centerless even when it might not be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is the Universe centerless? Or does it simply appear to be because of our perspective?&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\\\\$&lt;br&gt;&#xA;This is about to be a bad example, but please bear with me.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were a bird floating in the ocean, among a myriad of other birds, and you let time pass such that every bird floated away from the other birds in some random fashion. Would you, at some later time when everybirdy was roughly separated from one another, say that based upon your observations of the present conditions, and some evidence that at one point all of the birds were together, that the ocean is centerless? Or could you assume that because of your position in the flock, that from your perspective you can't tell where the center is, nor do you think that from a different birds position would you be able to tell? Creating the illusion of a centerless ocean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously this requires you not be at or close to the very edge of the flock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Also see examples in KenG's posed answer and comments.&lt;/p&gt;&#xA;" OwnerUserId="2232" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-11-20T04:13:10.490" Title="Centerless expansion or a matter of perspective" Tags="&lt;universe&gt;&lt;expansion&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2016-11-19T16:46:41.753" ContentLicense="CC BY-SA 3.0" />
  <row Id="18774" PostTypeId="1" AcceptedAnswerId="18779" CreationDate="2016-10-19T03:33:19.877" Score="3" ViewCount="462" Body="&lt;p&gt;There is a picture of the moon in National Public Radio's on-line article &lt;a href=&quot;http://www.npr.org/sections/thetwo-way/2016/10/18/498403354/get-ready-for-halloween-by-watching-the-moons-occultation-tonight&quot; rel=&quot;nofollow noreferrer&quot;&gt;Get Ready For Halloween By Watching The Moon's 'Occultation' Tonight&lt;/a&gt;. It looks wrong to me - specifically, the brightness gradation near the terminator - or lack thereof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image is credited &quot;A waning gibbous moon occultation will be visible Tuesday night in parts of the United States. JPL/NASA&quot; but I wonder if this is true. Something just looks &quot;wrong&quot; here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/epKSI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/epKSI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the original article the JPL/NASA contains a link to &lt;a href=&quot;https://nightsky.jpl.nasa.gov/planner.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;, which currently contains &lt;a href=&quot;http://api.usno.navy.mil/imagery/moon.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;this image&lt;/a&gt; also shown below. It looks more like an actual photograph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vkLJd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vkLJd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here is a NASA/JPL image from &lt;a href=&quot;https://svs.gsfc.nasa.gov/4404&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://svs.gsfc.nasa.gov/4404&lt;/a&gt; for 2016-10-19 04:00 UT, which if I uderstand correctly is not actually a photograph, but simulated from LRO data. It also illustrates that the terminator is expected to be graded from light to dark and contain contrast from shadowing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DLwxe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DLwxe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Sn0S0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sn0S0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; terminator from all three images. Realistic terminator on the right show graded intensity and strong shadowing from the highly oblique incident light.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2016-10-19T11:49:18.237" LastActivityDate="2016-10-19T13:07:38.257" Title="Why does the Moon's terminator look &quot;wrong&quot; in this image?" Tags="&lt;photography&gt;&lt;the-moon&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18776" PostTypeId="1" CreationDate="2016-10-19T11:10:36.840" Score="1" ViewCount="93" Body="&lt;p&gt;I have a basic refractor telescope. It's pretty light and so it catches vibrations and they never really settle down. Can I attach some dampers on it to cushion these?&lt;/p&gt;&#xA;" OwnerUserId="14621" LastEditorUserId="9301" LastEditDate="2016-10-20T04:51:38.857" LastActivityDate="2016-10-22T08:17:20.540" Title="Stabilizers/dampers for refractor telescope" Tags="&lt;refractor-telescope&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18777" PostTypeId="1" CreationDate="2016-10-19T11:12:15.100" Score="2" ViewCount="454" Body="&lt;p&gt;I am looking for beginners information on how I can know when the Moon is going to be visible on a particular night ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google-fu says to look for moonrise information about my city from websites such as &lt;a href=&quot;https://www.timeanddate.com/moon/australia/darwin&quot; rel=&quot;nofollow&quot;&gt;https://www.timeanddate.com/moon/australia/darwin&lt;/a&gt;&lt;br&gt;&#xA;So if I followed the above guide for today and started walking on a plane field West-NorthWest direction, at 10.01 PM I should be able to see the Moon. Is my understanding correct? Or is there another way to know when Moon will be visible on a particular (clear) night. &lt;/p&gt;&#xA;" OwnerUserId="14624" LastEditorUserId="2365" LastEditDate="2016-10-22T05:37:07.887" LastActivityDate="2016-10-22T05:37:07.887" Title="How to know when the moon is visible on a particular night?" Tags="&lt;amateur-observing&gt;&lt;the-moon&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18779" PostTypeId="2" ParentId="18774" CreationDate="2016-10-19T13:07:38.257" Score="11" Body="&lt;p&gt;The reason that moon image looks wrong is because it is wrong. It is not a real image of the moon $-$ at least the terminator is not real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.npr.org/sections/thetwo-way/2016/10/18/498403354/get-ready-for-halloween-by-watching-the-moons-occultation-tonight&quot;&gt;original article&lt;/a&gt; you cite has a link just below their image indicating the source of their image of the moon. That source is the &lt;a href=&quot;https://nightsky.jpl.nasa.gov/planner.cfm&quot;&gt;night sky planner&lt;/a&gt;, hosted by JPL. You'll find the same image on that website, albeit slightly darker (it seems the NPR people lightened up the image a bit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do some more digging, you'll see that the night sky planner got its image of that moon from someone else. Within the html code, they have the image defined as:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt;img alt=&quot;the moon&quot; src=&quot;http://api.usno.navy.mil/imagery/moon.png&quot;&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Clearly you can see that the image was taken from the &lt;a href=&quot;http://aa.usno.navy.mil/imagery/moon&quot;&gt;United States National Observatory&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a bit of digging, I found out what exactly is going on here. The purpose of this site is to show you the current phase of the moon. To do this, they take a single image of the full moon and artificially shade out a region to make it appear as the current phase of the moon. You can see their process &lt;a href=&quot;http://tycho.usno.navy.mil/vphase.html&quot;&gt;here&lt;/a&gt; and how it was done by some guy named R. Schmidt. They've broken down the moon phases into 181 images which you can &lt;a href=&quot;http://tycho.usno.navy.mil/gif/Moon.zip&quot;&gt;download here&lt;/a&gt;, if you're interested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the terminator on that image is wrong because it is not a real image of the current phase of the moon, but rather a computer generated &quot;shading&quot; of the full moon to indicate the current moon phase.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-10-19T13:07:38.257" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18780" PostTypeId="2" ParentId="18773" CreationDate="2016-10-19T14:37:19.143" Score="1" Body="&lt;p&gt;The first principle of cosmology is called the &quot;cosmological principle&quot;, which asserts that the universe is more or less the same everywhere spatially, but changes with time.  This is a fundamentally new way to think of the history of the universe, and has revolutionized our ability to understand long-range observations.  We do not, however, know that this principle is true, we only know it is both useful and consistent with all observations.  So we adopt it as a working hypothesis, and all of cosmology follows.  Some day we may discover it is inadequate, but until then, we will continue to use it.  The same could be said for other principles, like conservation of energy-- we don't know that's true either, but it's useful and consistent with observations.  The cosmological principle rules out that the universe has a unique center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question is whether the apparent lack of a unique center stems from our location not being special enough to be able to tell that it does have such a center (say, by encountering a boundary).  Ironically, the cosmological principle turns that question on its ear, and points to the possibility that one's perspective always puts themselves at the apparent center.  So we would say the illusion of &lt;em&gt;having&lt;/em&gt; a center is all about perspective-- not the illusion of &lt;em&gt;not&lt;/em&gt; having a center!&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-10-19T14:37:19.143" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18781" PostTypeId="1" AcceptedAnswerId="18782" CreationDate="2016-10-19T21:50:56.030" Score="2" ViewCount="2224" Body="&lt;p&gt;I know that the equation for the difference in apparent magnitudes can be represented as: &#xA;$\Delta m=-2.5\log\left(\dfrac{I_{1}}{I_{2}}\right)$&lt;br&gt;&#xA;and the apparent brightness of a celestial object is shown as:&lt;br&gt;&#xA;$\ B=\dfrac{L}{4\pi\ D^2}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people in my class told me that$\ B$ and$\ I$ were the same thing but used in different cases. This explanation didn't help me though. While playing around with the equations, I found this nice relation:&lt;br&gt;&#xA;$\ \dfrac{B_{1}}{B_{2}}=\left[\dfrac {\left(\dfrac{L}{4\pi\ D_{2}^2}\right)}{\left(\dfrac{L}{4\pi\ D_{1}^2}\right)} \right]$&lt;br&gt;&#xA;$\ \space\ \space\ \space\ ={\dfrac{4\pi\ D_{1}L}{4\pi\ D_{2}^2L}}$&lt;br&gt;&#xA;$\ \space\ \space\ \space\ ={\dfrac{D_{1}^2}{D_{2}^2}}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if$\ {B_{1}}={I_{1}}$ and$\ B_{2}=I_{2}$holds true, is this following equation also true?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ {\dfrac{D_{1}^2}{D_{2}^2}} \space\ = 10^\left(-{\dfrac{\Delta {m}}{2.5}}\right)$&lt;br&gt;&#xA;$\ \space\ \space\ \space\ \space\ =\dfrac{1}{10^\left({\dfrac{\Delta {m}}{2.5}}\right)}$   &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;However, if$\ B$ and$\ I$ are not the same, what is the difference between the two measurements? And can either one be used to accurately measure how bright a non-luminous celestial body (ie. a planet) appears from Earth?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also... does $\ m$ have an SI unit? I've never used any units when I solve for it&lt;/p&gt;&#xA;" OwnerUserId="14628" LastEditorUserId="14628" LastEditDate="2016-10-20T03:57:41.143" LastActivityDate="2016-10-20T03:57:41.143" Title="What's the difference between apparent brightness and apparent magnitude? Are they the same?" Tags="&lt;apparent-magnitude&gt;&lt;magnitude&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18782" PostTypeId="2" ParentId="18781" CreationDate="2016-10-19T22:08:59.653" Score="6" Body="&lt;p&gt;Yes, $B$ and $I$ are the same things in this context. The $\Delta m$ you derived goes by the name &quot;distance modulus&quot; when one of the distances is set to $10\operatorname{pc}$. The thing to be careful of is that $I$ is also sometimes used for &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Radiative_transfer&quot;&gt;spectral radiance&lt;/A&gt; (or surface brightness) when working with radiative transfer. Spectral radiance doesn't obey the inverse square law, though, so there's no risk that we're confusing them here.&lt;/p&gt;&#xA;" OwnerUserId="14244" LastActivityDate="2016-10-19T22:08:59.653" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18783" PostTypeId="2" ParentId="18777" CreationDate="2016-10-20T04:59:45.467" Score="2" Body="&lt;p&gt;Google-fu is correct... It really depends upon where you are, for an exact time. However, if you understand that there are 8 named phases and they typically have a &quot;standard&quot; rise time, you can know approximately what time and what phase will be visible. The one caveat is that you need to look at the moon to know where to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ecmnW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ecmnW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2232" LastEditorUserId="2232" LastEditDate="2016-10-21T22:32:27.493" LastActivityDate="2016-10-21T22:32:27.493" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18784" PostTypeId="1" AcceptedAnswerId="18821" CreationDate="2016-10-20T07:32:42.447" Score="5" ViewCount="272" Body="&lt;p&gt;The recent NASA feature article &lt;a href=&quot;http://www.nasa.gov/feature/new-horizons-possible-clouds-on-pluto-next-target-is-reddish&quot; rel=&quot;noreferrer&quot;&gt;New Horizons: Possible Clouds on Pluto, Next Target is Reddish&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;MU69 is actually the smallest KBO to have its color measured – and scientists have used that data to confirm the object is part of the so-called cold classical region of the Kuiper Belt, which is believed to contain some of the oldest, most prehistoric material in the solar system.&lt;/p&gt;&#xA;&lt;p&gt;“The reddish color tells us the type of Kuiper Belt object 2014 MU69 is,” said Amanda Zangari, a New Horizons post-doctoral researcher from Southwest Research Institute. “The data confirms that on New Year’s Day 2019, New Horizons will be looking at one of the ancient building blocks of the planets.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://gizmodo.com/new-horizons-spacecraft-is-approaching-a-mysterious-red-1787938286&quot; rel=&quot;noreferrer&quot;&gt;Gizmodo&lt;/a&gt; elaborates on the significance of the red color:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Speaking today at the the American Astronomical Society Division for Planetary Sciences (DPS), mission scientists talked about the Pluto flyby, but they also looked ahead to the Kuiper Belt encounter. Recent observations made by the Hubble Space Telescope suggests MU69 is red—if not redder—than Pluto. MU69 is now the smallest KBO to have its color measured, and it’s our first sense of what this mysterious object actually looks like.&lt;/p&gt;&#xA;&lt;p&gt;The reddish color of MU69, like the iconic red splotches on Pluto and its moon Charon, suggest the presence of tholin, a class of molecules that’s formed through the ultraviolet irradiation of simple organic compounds such as methane and ethane. Tholin doesn’t form naturally on Earth, but it’s abundant on the surface of icy bodies in the outer solar system. Using Hubble, scientists have confirmed that MU69 is part of the “cold classical” region of the Kuiper belt, which contains some of the oldest objects in the Solar System.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;How was the &amp;quot;redness&amp;quot; of 2014 MU69 detected? The second link mentions that it is Hubble data - is it just photometric (images through different filters) or is there also spectroscopy?&lt;/p&gt;&#xA;&lt;p&gt;I'm curious, just how red is it? What exactly are the numerical data that are being interpreted as &amp;quot;redness&amp;quot;? Is this exclusively the assumed visual appearance, or is there excess infrared as well?&lt;/p&gt;&#xA;&lt;p&gt;Also, I wasn't sure if 2014 MU69 is considered a comet or an asteroid. So far everything I've read just calls it a Kuiper Belt Object (KBO). So I've used both tags.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YzQNu.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YzQNu.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; graphic from &lt;a href=&quot;https://blogs.nasa.gov/pluto/2016/06/28/a-world-beyond-pluto-finding-a-new-target-for-new-horizons/&quot; rel=&quot;noreferrer&quot;&gt;NASA blogs&lt;/a&gt;. It looks like they are all red! (joke).&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-06-03T13:35:06.633" Title="Which measurements have shown that 2014 MU69 is red? (New Horizons' next target)" Tags="&lt;asteroids&gt;&lt;comets&gt;&lt;spectra&gt;&lt;kuiper-belt&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18785" PostTypeId="1" AcceptedAnswerId="18786" CreationDate="2016-10-20T07:39:02.743" Score="2" ViewCount="1692" Body="&lt;p&gt;Here is a good example picture of what I mean:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/tFicp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tFicp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, why is this so? I've heard often that in the creation of a planet more the rings have formed, they are most likely just all over the place before they settle to the formation that we know, so why not come to a stable position in such a way. Note that the picture is only an example, rings that horizontal to each other are included too.&lt;/p&gt;&#xA;" OwnerUserId="14614" LastActivityDate="2016-10-20T09:38:48.147" Title="Can a planet have more that one ring?" Tags="&lt;planet&gt;&lt;planetary-ring&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ClosedDate="2016-10-20T10:18:52.640" ContentLicense="CC BY-SA 3.0" />
  <row Id="18786" PostTypeId="2" ParentId="18785" CreationDate="2016-10-20T09:38:48.147" Score="3" Body="&lt;p&gt;Yes, you can have more than one ring. &lt;a href=&quot;http://www.universetoday.com/38097/how-many-rings-does-saturn-have/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Saturn&lt;/a&gt; is an example of a planet with multiple rings on the same axis separated by divisions (gaps between each ring).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About rings that are not on the same axis (similar to your example picture) is addressed in &lt;a href=&quot;https://astronomy.stackexchange.com/questions/16536/could-a-cross-ringed-planet-exist?rq=1&quot;&gt;this previous question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7926" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-10-20T09:38:48.147" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18788" PostTypeId="1" AcceptedAnswerId="18790" CreationDate="2016-10-20T11:29:24.817" Score="8" ViewCount="2573" Body="&lt;p&gt;Is there a way to determine the age of a black hole. Suppose 100 Billion years from now, if two black holes have exactly the same mass(say 30 M☉). One of them formed 10 Billion years from now and other formed 20 billion years from now. At t = t0 + 100 Billion years, looking back into the past, Can we predict how old these black holes are ? Is rate of dissipation of Hawking radiations any different for them? &lt;/p&gt;&#xA;" OwnerUserId="10657" LastActivityDate="2016-10-20T15:42:05.170" Title="Age of a black hole" Tags="&lt;black-hole&gt;&lt;age&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18789" PostTypeId="2" ParentId="16536" CreationDate="2016-10-20T11:40:46.127" Score="6" Body="&lt;p&gt;The answer is a bit more complex than what userLTK presents.&lt;br&gt;&#xA;&lt;strong&gt;Tides&lt;/strong&gt; play a huge role for the orientation of the rings.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the ring system has an inclination relative towards the planet, then &lt;em&gt;inclination damping&lt;/em&gt; will occur.&#xA;This means that the planet, that is flattened due to its own rotation, will exert asymmetric force tugs onto orbiting masses, until the force tugs are symmetric again, which is the case for zero inclination. This works well beyond the Roche-sphere of a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about Uranus ring and moon-system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3VF4Z.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3VF4Z.jpg&quot; alt=&quot;(c) Keck Telescope team&quot;&gt;&lt;/a&gt; (c) Keck Telescope Team &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems implausible from a planetary formation viewpoint that Uranus could have an inclined subdisc from which the rings and moons formed at this inclination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What seems more likely is that the planet was tilted, and then forced rings and moons to follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summarizing&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;Yes&lt;/strong&gt; on very short timescales, but then the rings would look messier and thicker as they would have to be just freshly formed.&lt;br&gt;&#xA;&lt;strong&gt;No&lt;/strong&gt; on long timescales (~10.000 years based on &lt;a href=&quot;https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S1743921314008321&quot; rel=&quot;noreferrer&quot;&gt;a quick google search&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2016-10-20T11:40:46.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18790" PostTypeId="2" ParentId="18788" CreationDate="2016-10-20T12:10:13.880" Score="12" Body="&lt;p&gt;From looking at the Black Hole alone, there is no possibility of determining its age. The state of the Black Hole is fully determined by a few fundamental variables (mass, angular momentum and electric charge). This is the statement of the famous dictum &lt;em&gt;A black hole has no hair.&lt;/em&gt; Hawking radiation in special is only dependent on these variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may be able to determine the age of a Black hole by indirect means (e.g., by looking at its surroundings and see how much it is cleaned from matter).&lt;/p&gt;&#xA;" OwnerUserId="11617" LastActivityDate="2016-10-20T12:10:13.880" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18791" PostTypeId="1" AcceptedAnswerId="31641" CreationDate="2016-10-20T12:45:54.877" Score="6" ViewCount="325" Body="&lt;p&gt;Looking through ALMA on-line documentation and articles, the quantity PWV (precipitable water vapor) is a central theme. I have two linked questions.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What (actually) is precipitable water vapor, and is there a &quot;non-precipitable&quot; component also? &lt;em&gt;Why not just call it &quot;water vapor&quot;?&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;How is it actually measured? It seems to be monitored by radiometers, but I don't understand how that can be used.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I usually see numbers of the order of 5 millimeter to 10 millimeters for very high altitude locations like Atacama. I assume this is the column-integrated amount, expressed in units of liquid, is that right? &lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-04-24T12:29:54.770" LastActivityDate="2019-04-26T15:44:18.703" Title="What is precipitable water vapor in millimeter-wave radioastronomy and how is it measured?" Tags="&lt;radio-astronomy&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="18792" PostTypeId="2" ParentId="18791" CreationDate="2016-10-20T13:11:55.907" Score="4" Body="&lt;p&gt;As defined quite clearly, with equations, &lt;a href=&quot;http://glossary.ametsoc.org/wiki/Precipitable_water&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The total atmospheric water vapor contained in a vertical column of&#xA;unit cross-sectional area extending between any two specified levels,&#xA;commonly expressed in terms of the height to which that water&#xA;substance would stand if completely condensed and collected in a&#xA;vessel of the same unit cross section.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;edit&lt;/h2&gt;&#xA;&lt;p&gt;There's, for example, an &lt;a href=&quot;http://legacy.nrao.edu/alma/memos/html-memos/alma158/memo158.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ALMA paper&lt;/a&gt; which discusses measurements and comparison with meteorological stations.&lt;/p&gt;&#xA;" OwnerUserId="907" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-10-20T13:26:39.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18794" PostTypeId="1" CreationDate="2016-10-20T13:18:26.970" Score="10" ViewCount="2876" Body="&lt;p&gt;From my basic understating, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Momentum is being transfered from the Earth's rotation to the Moon's orbit by tidal friction. The Earth's rotation slows down and the Moon receedes from the Earth as it moves into a higher orbit. This will continue until the Earth's rotational period is equal to the orbital period of the Moon, i.e the Earth is tidally locked to the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming I have the above correct - and please correct me if I don't - will there, realistically, be enough time for tidal locking to  occur before the sun expands and engulfs the Earth? Or is there another reason the Earth will never be locked towards the Moon?&lt;/p&gt;&#xA;" OwnerUserId="14633" LastActivityDate="2016-10-21T15:31:42.233" Title="Will the Earth ever be tidally locked to the Moon?" Tags="&lt;earth&gt;&lt;tidal-forces&gt;&lt;the-moon&gt;&lt;tidal-locking&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18795" PostTypeId="2" ParentId="18794" CreationDate="2016-10-20T13:39:23.853" Score="16" Body="&lt;p&gt;As the moon orbits Earth, tidal forces slow down the Earth's rotation by 2 milliseconds per century. Eventually, in tens of billions of years, the Earth and Moon would achieve a double tidal lock, where both are stuck with one side facing the other as they orbit the Earth-Moon barycenter. In 7.5 billion years, the Sun will expand past the Earth's current orbit, but the Earth may drift out further, preventing it from being vaporized. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is beside the point, because in about one billion years, all of Earth's water will have boiled away, meaning that there would be no more ocean tides, and thus the Earth-Moon system will likely never achieve a double tidal lock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.spaceanswers.com/solar-system/will-the-moon-ever-leave-earths-orbit/&quot; rel=&quot;noreferrer&quot;&gt;Will the Moon ever leave Earth’s orbit? - Giles Sparrow, Space Answers&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://barnyard.syr.edu/eccentric.pdf&quot; rel=&quot;noreferrer&quot;&gt;Harvesting Lunar Eccentricity? - Terry R. McConnell, Syracuse University&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.universetoday.com/128350/will-earth-lock-moon/&quot; rel=&quot;noreferrer&quot;&gt;When Will Earth Lock to the Moon? - Fraser Cain, Universe Today&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.scientificamerican.com/article/the-sun-will-eventually-engulf-earth-maybe/&quot; rel=&quot;noreferrer&quot;&gt;The Sun Will Eventually Engulf Earth--Maybe - David Appell, Scientific American&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://mnras.oxfordjournals.org/content/386/1/155.full&quot; rel=&quot;noreferrer&quot;&gt;Distant future of the Sun and Earth revisited - K.-P. Schröder and Robert Connon Smith, MNRAS&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.sciencedaily.com/releases/2013/12/131216142310.htm&quot; rel=&quot;noreferrer&quot;&gt;When will Earth lose its oceans? - CNRS, Science Daily&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6" LastEditorUserId="6" LastEditDate="2016-10-21T15:31:42.233" LastActivityDate="2016-10-21T15:31:42.233" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18796" PostTypeId="2" ParentId="18784" CreationDate="2016-10-20T15:28:36.173" Score="8" Body="&lt;p&gt;This was a hard one to answer, primarily because of the difficulty in tracking down information.&lt;/p&gt;&#xA;&lt;h3&gt;The Observations of MU69&lt;/h3&gt;&#xA;&lt;p&gt;The extended mission for New Horizons involved adjusting its orbit to do a close fly-by of a Kuiper Belt Object (KBO). The annoying part about this was that they needed to find one first! Ground based observations were unsuccessful due to the faint magnitudes of any potential KBOs and the crowded star field (being that they were looking towards the galactic core). Finally, HST was used and over the course of 194 orbits, between June to September 2014, the entire region of the potentially reachable sky, by New Horizon's designated &lt;span class=&quot;math-container&quot;&gt;$130\:\mathrm{m/s}$&lt;/span&gt; delta-v, was scanned using WFC3 instrument and the &lt;a href=&quot;http://www.stsci.edu/hst/wfc3/documents/handbooks/currentIHB/appendixA12.html&quot; rel=&quot;noreferrer&quot;&gt;F350LP filter&lt;/a&gt;. Image processing revealed 5 potential candidates. This is all laid out in &lt;a href=&quot;http://meetingorganizer.copernicus.org/EPSC2015/EPSC2015-417.pdf&quot; rel=&quot;noreferrer&quot;&gt;a paper&lt;/a&gt; by the lead of the KBO search, &lt;a href=&quot;https://www.boulder.swri.edu/%7Espencer/&quot; rel=&quot;noreferrer&quot;&gt;John Spencer&lt;/a&gt;. You can choose to view any and all images taken during these observations (since it is in the public domain) by downloading them from &lt;a href=&quot;https://hla.stsci.edu/hlaview.html#Images%7CfilterText%3D%24filterTypes%3D%7Cquery_string=0%200%20r%3D180&amp;amp;posfilename=&amp;amp;poslocalname=&amp;amp;posfilecount=&amp;amp;listdelimiter=whitespace&amp;amp;listformat=degrees&amp;amp;RA=0.000000&amp;amp;Dec=0.000000&amp;amp;Radius=180.000000&amp;amp;inst-control=all&amp;amp;inst=ACS&amp;amp;inst=ACSGrism&amp;amp;inst=WFC3&amp;amp;inst=WFPC2&amp;amp;inst=NICMOS&amp;amp;inst=NICGRISM&amp;amp;inst=COS&amp;amp;inst=WFPC2-PC&amp;amp;inst=STIS&amp;amp;inst=FOS&amp;amp;inst=GHRS&amp;amp;imagetype=best&amp;amp;prop_id=13633&amp;amp;spectral_elt=&amp;amp;proprietary=both&amp;amp;preview=1&amp;amp;output_size=256&amp;amp;cutout_size=12.8%7Cra=&amp;amp;dec=&amp;amp;sr=&amp;amp;level=&amp;amp;image=&amp;amp;inst=ACS%2CACSGrism%2CWFC3%2CWFPC2%2CNICMOS%2CNICGRISM%2CCOS%2CWFPC2-PC%2CSTIS%2CFOS%2CGHRS&amp;amp;ds=&quot; rel=&quot;noreferrer&quot;&gt;Hubble's Legacy Archive&lt;/a&gt;. Of course you'll have to do some digging in the 970 total images to find the actual images which contain MU69.&lt;/p&gt;&#xA;&lt;p&gt;Once the entire region was scanned and a set of candidates were found, follow-up observations were made in late October 2014 of each individual candidate to precisely determine orbital parameters and observational characteristics so as to narrow down which candidate was reachable and presented the most scientific advantage. In this case, the same observing group from New Horizons took the images and once again used Hubble's WFC3 and the F350LP filter. Once again, these images can all be found on Hubble's Legacy Archive. An example of a single image which from this run, containing MU39 (although you'd be hard-pressed to pick it out) can be &lt;a href=&quot;http://archive.stsci.edu/cgi-bin/mastpreview?mission=hst&amp;amp;dataid=ICIIJ5YDQ&quot; rel=&quot;noreferrer&quot;&gt;seen here&lt;/a&gt;. Most of this information can be gleaned from the Hubble Legacy Archive and is also described in &lt;a href=&quot;http://www.hou.usra.edu/meetings/lpsc2015/pdf/1301.pdf&quot; rel=&quot;noreferrer&quot;&gt;this paper&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The Minor Planet Catalog (props to AtmosphericPrisonEscape for pointing this out) indicates that further observations of MU69 were taken with the HST in May and July of 2015. I had a bit of trouble tracking these images down, but knowing that John Spencer was the lead in these observations, I went and looked up his 2015 HST proposal which you can &lt;a href=&quot;https://archive.stsci.edu/proposal_search.php?id=14053&amp;amp;mission=hst&quot; rel=&quot;noreferrer&quot;&gt;read here&lt;/a&gt;. With the proper proposal number, I found the set of &lt;a href=&quot;https://hla.stsci.edu/hlaview.html#Images%7CfilterText%3D%24filterTypes%3D%7Cquery_string=0%200%20r%3D180&amp;amp;posfilename=&amp;amp;poslocalname=&amp;amp;posfilecount=&amp;amp;listdelimiter=whitespace&amp;amp;listformat=degrees&amp;amp;RA=280.183117&amp;amp;Dec=-21.029303&amp;amp;Radius=180.00000&amp;amp;inst=WFC3&amp;amp;imagetype=best&amp;amp;prop_id=14053&amp;amp;spectral_elt=&amp;amp;proprietary=both&amp;amp;preview=1&amp;amp;output_size=256&amp;amp;cutout_size=12.8%7Cra=&amp;amp;dec=&amp;amp;sr=&amp;amp;level=&amp;amp;image=&amp;amp;inst=ACS%2CACSGrism%2CWFC3%2CWFPC2%2CNICMOS%2CNICGRISM%2CCOS%2CWFPC2-PC%2CSTIS%2CFOS%2CGHRS&amp;amp;ds=&quot; rel=&quot;noreferrer&quot;&gt;15 images during May and July which were taken of MU69&lt;/a&gt;. Again, the WFC3 instrument was used with the F350LP filter.&lt;/p&gt;&#xA;&lt;h3&gt;The Color of MU69&lt;/h3&gt;&#xA;&lt;p&gt;In total, we have 38 images of MU69, all taken by John Spencer and his team using the HST's WFC3 instrument and the F350LP filter (that peaks at 5845.871 angstroms). The important point here, and the reason I detailed the observations so much in the previous section, was to determine the exact type of images taken of this KBO, so as to ascertain what science data could have been used to determine color.&lt;/p&gt;&#xA;&lt;p&gt;Based on everything I've compiled, it appears to me that the statement of MU69 being &amp;quot;red&amp;quot; is a educated guess. Anyone who knows about telescopes will tell you that imaging an object in a single filter is &lt;em&gt;not&lt;/em&gt; sufficient to reliably measure aspects such as color. You need at least two filters to determine that because you need to understand how the brightness of said object varies with wavelength.&lt;/p&gt;&#xA;&lt;p&gt;However, it could be possible to make a guess at the color of this body from a single filter by making a lot of assumptions. This is purely a guess as to what they did, but you can first measure the apparent magnitude of this object in your filter, then, assuming things like size, albedo, distance, etc., you can create a model of this object's spectrum that you'd expect it to have such that it reproduces your observed image. The &lt;a href=&quot;http://meetingorganizer.copernicus.org/EPSC2015/EPSC2015-417.pdf&quot; rel=&quot;noreferrer&quot;&gt;first paper I linked&lt;/a&gt; details how they did inject false KBOs into their images to ensure their KBO finding algorithms worked which indicates they had some basic ability to model KBO images in their data. If your model can reproduce the observations, you can use that model as a basis for what your KBO is likely like and from that you can say that it appears &amp;quot;red&amp;quot;. Based on experience with KBOs and in partiuclar pluto, we expect they might be red and that the redness corresponds to tholin, a common, red chemical found on KBOs, so the fact that they believe it to be read is not surprising.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-10-20T15:41:43.517" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="18797" PostTypeId="2" ParentId="18788" CreationDate="2016-10-20T15:42:05.170" Score="1" Body="&lt;p&gt;No you can't say anything about their ages and yes, their Hawking radiation is different... not that you could detect the difference. In more detail:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your two black holes began with identical mass, but at different times, the younger hole would have lost less mass than the other through Hawking radiation, with an age difference of only 10 Billion years the mass difference would be unmeasurable (with current technology). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If at some point in the distant future two BHs of different masses are encountered, one could not in general tell whether they had been born at the same time with different masses, or at different times with the same mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intensity of Hawking radiation depends on the temperature of the black hole and as a BH evaporates it effectively heats up, and as it heats up it emits more and more Hawking radiation leading to a final burst before whatever happens right at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is also true that the older black hole, having lost more mass, will be at a slightly (but unmeasurablly so) higher temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have neglected the temperature of the cosmic microwave background, so far. That is currently about 2.7 Kelvin and until the expansion of the universe has lowered it to below that of a black hole, the black hole will actually absorb more energy from the CMB than it emits through Hawking radiation and will actually have increased (immeasurably) in mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why it will take of the order of 10^100 years for all black holes to evaporate, compared to which even your nominal 100 billion year interval is a drop in the (cosmic) ocean.&lt;/p&gt;&#xA;" OwnerUserId="14634" LastActivityDate="2016-10-20T15:42:05.170" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18798" PostTypeId="1" AcceptedAnswerId="18799" CreationDate="2016-10-20T21:38:32.033" Score="2" ViewCount="1854" Body="&lt;p&gt;Title says it all. I'm putting together a presentation on Milankovitch Cycles, and I would like to bring up the difference in distance to the sun (km) at aphelion and perihelion when the earths orbit is at maximum and minimum eccentricity. The numbers I've found are that the maximum eccentricity of Earth's orbit is 0.0679, and the minimum is 0.000055.&lt;/p&gt;&#xA;" OwnerUserId="14639" LastActivityDate="2018-05-08T13:29:23.830" Title="Distance to Sun at Perihilion and Aphelion at max and min Eccentricity" Tags="&lt;orbit&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18799" PostTypeId="2" ParentId="18798" CreationDate="2016-10-20T22:47:50.670" Score="3" Body="&lt;p&gt;I like the math of ellipses cause there's enough parts to keep it interesting, but it's not super complex either.   The answer to your question as you put it, is remarkably simple.  I've bolded the letters that represent distances for easier reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eccentricity or $e = c/a$, where $a$ is the semi major axis and $c$ is the distance between the focal point (sun's location) and the center of the ellipse. $ea$ is the distance between the center point and the focal point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the answer to your question is simply $a$ plus $ea$ and $a$ minus $ea$.   For the Earth and sun, the semi major axis or $a$, is $149,600,000\:\mathrm{km}$, then simply add and subtract $ea$ to both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $e = .0679$, $ea = 10.15\:\mathrm{million\:km}$. So $149,600,000 \pm 10,150,000$. $159,750,000$ and $139,450,000\:\mathrm{km}$ at furthest and closest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $e = .000055$, $ea = .008\:\mathrm{million\:km}$, so $149,680,000$ and $149,520,000$ at most circular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple picture&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/V6UFb.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V6UFb.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More detailed picture&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/995Rd.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/995Rd.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maths and sources of pictures, &lt;a href=&quot;http://facweb.bhc.edu/academics/science/harwoodr/geol102/study/Eccentricity.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/kepler.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;footnote:  I hope this wasn't a homework question.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="11143" LastEditDate="2016-10-21T12:45:29.010" LastActivityDate="2016-10-21T12:45:29.010" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18802" PostTypeId="2" ParentId="16232" CreationDate="2016-10-21T00:28:14.897" Score="5" Body="&lt;p&gt;I can't comment here because I don't have enough reputation, but this is very related to your problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_locking#Timescale&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Tidal_locking#Timescale&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tidal locking will always occur albeit potentially not in a Hubble time. In order for there to be no tidal locking, you would need:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}F}{\mathrm{d}r} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which any extended object will not have. Problems like this always involve timescales and not definite yes/no answers. I hope this helped.&lt;/p&gt;&#xA;" OwnerUserId="9826" LastEditorUserId="9301" LastEditDate="2017-01-04T22:15:24.907" LastActivityDate="2017-01-04T22:15:24.907" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18803" PostTypeId="1" CreationDate="2016-10-21T03:42:59.793" Score="0" ViewCount="192" Body="&lt;p&gt;A neighbor called me to ask about drones flying around. I managed to take several pictures and they don't look like any star, planet or other celestial body I have ever seen, but they seem too steady to be drones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I managed to get &lt;a href=&quot;https://www.flickr.com/photos/25602281@N00/29830323944/in/dateposted-public/&quot; rel=&quot;nofollow noreferrer&quot;&gt;one decent picture&lt;/a&gt;:&#xA;&lt;a href=&quot;https://www.flickr.com/photos/25602281@N00/29830323944/in/dateposted-public/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/A07WP.jpg&quot; alt=&quot;Celestial body or drone?&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any advice for figuring out whether that is a real celestial object or man-made?&lt;/p&gt;&#xA;" OwnerUserId="14641" LastActivityDate="2016-10-21T13:20:46.083" Title="Identifying Celestial Bodies vs Drones" Tags="&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18804" PostTypeId="2" ParentId="18776" CreationDate="2016-10-21T05:46:12.983" Score="2" Body="&lt;p&gt;It depend where vibration are. Often (if you have a good mount for the telescope) all the structure vibrate and there are some good anti-vibration feet that can be useful. If the tripod isn't the problem and the vibration are only near the telescope, probably the problem is the mount that is under-sized for you telescope. In this case the only good way is change the mount with a sturdier one.&lt;/p&gt;&#xA;" OwnerUserId="14600" LastEditorUserId="2365" LastEditDate="2016-10-22T08:17:20.540" LastActivityDate="2016-10-22T08:17:20.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18805" PostTypeId="2" ParentId="18803" CreationDate="2016-10-21T06:13:07.463" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Any advice for figuring out whether that is a real celestial object or man-made?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes - in almost all cases, you can readily identify a celestial object by going out the next clear night, and looking for it in the same position.  (Very few celestial objects move far over a few days.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In response to comments about geostationary satellites:&lt;/strong&gt;&#xA;I doubt it is a geostationary satellite, answers to &lt;a href=&quot;https://space.stackexchange.com/questions/3227/are-any-geosynchronous-satellites-visible-with-the-naked-eye&quot;&gt;this question&lt;/a&gt; indicates these are usually too faint to be seen without special equipment. Also they don't move, whereas nearby stars will move a little over the space of half an hour or more.&lt;/p&gt;&#xA;" OwnerUserId="7491" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2016-10-21T13:20:46.083" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18806" PostTypeId="1" AcceptedAnswerId="18807" CreationDate="2016-10-21T12:50:22.627" Score="4" ViewCount="447" Body="&lt;p&gt;A point spread function (PSF) has many different uses. Consider for instance the following quote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To  extract  the  maximum  information  out  of  an  observation, even the smallest details of the PSF are important. Some examples include: deconvolving the PSF from an observed image to remove the blurring caused by diffraction and reveal fine structure; convolving a model image by the PSF to compare to an observed one;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question about this is the following:&#xA;If we know the PSF of a system, and use this to deconvolve the raw image (purpose 1 from the quote above), why would we &lt;em&gt;convolve&lt;/em&gt; a model with the PSF to compare it to an image (purpose 2 from the quote)? That is, can't we just compare the original, unconvolved model, with the &lt;strong&gt;de&lt;/strong&gt;convolved image? In terms of maximizing the information that seems like the way to go.&lt;/p&gt;&#xA;" OwnerUserId="6965" LastActivityDate="2016-10-21T13:56:29.507" Title="Questions about convolving/deconvolving with a PSF" Tags="&lt;data-analysis&gt;&lt;space-telescope&gt;&lt;hubble-telescope&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18807" PostTypeId="2" ParentId="18806" CreationDate="2016-10-21T13:49:07.023" Score="9" Body="&lt;p&gt;Convolution is not a uniquely invertible process in the presence of random noise in your image. Deconvolving a noisy image can give misleading results, even if you have perfect knowledge of the PSF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, when you are fitting models to data, it is far better to compare the models and data in the observational space of the data, where the uncertainties are reasonably well understood.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-10-21T13:49:07.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18808" PostTypeId="2" ParentId="18806" CreationDate="2016-10-21T13:56:29.507" Score="5" Body="&lt;p&gt;These two uses of PSFs are applicable to different situations. Sometimes you know your PSF well enough to deconvolve your image and get something reasonable out of it, but most of the time you have a lot of assumptions in this process and you're not going to get something perfect, so you might want to go the other way and convolve your model with the best PSF you have to get close to your observations. It also depends who you are - often observational astronomers who collect data from telescopes and theoretical astronomers who build models don't talk as much as we should, so you might not have access to all the information you need to do one of these two processes, so you work with what you've got and do the other one.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2016-10-21T13:56:29.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18809" PostTypeId="1" CreationDate="2016-10-21T13:57:38.493" Score="-5" ViewCount="505" Body="&lt;p&gt;so i was gazing out of my balcony randomly and all of a sudden this red light moves across above, initial thought was that it was a star but it didn't blink, second thoughts was maybe a planet but it was moving fast so i thought maybe an airplane but airplane lights blink too, then i thought maybe i was hallucinating a bit, then when i looked at it again it started to decrease in size and turn dark red and then disappeared, all of  this happened in seconds, fancy how our brains process so fast but yeah help me out with what i just saw please? or else this could be my ufo sighting my friends will be hearing about soon :)&lt;/p&gt;&#xA;" OwnerUserId="14645" LastActivityDate="2016-10-21T16:00:35.413" Title="I think I saw a satellite disappear? or was it a star?" Tags="&lt;space&gt;&lt;natural-satellites&gt;&lt;satellite&gt;&lt;identify-this-object&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18810" PostTypeId="1" AcceptedAnswerId="18811" CreationDate="2016-10-21T14:53:31.590" Score="5" ViewCount="283" Body="&lt;p&gt;I have learned about the various types of supernovae, and both Types Ib and II involve the extremely rapid compaction or explosion of a star's core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the &quot;information&quot; that (respectively) the Chandrasekhar limit or the degeneracy pressure have been reached propagate through the star with enough speed (yet below $c$) that the resulting collapse remains symmetric and happens in a very short time?&lt;/p&gt;&#xA;" OwnerUserId="14646" LastEditorUserId="2153" LastEditDate="2016-10-24T20:03:07.673" LastActivityDate="2016-10-24T20:03:07.673" Title="How do supernova explosions reconcile with light speed constraints?" Tags="&lt;supernova&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18811" PostTypeId="2" ParentId="18810" CreationDate="2016-10-21T15:15:03.167" Score="11" Body="&lt;p&gt;Interestingly enough, the limit on the time needed for such a change to propagate through the star isn't from the speed of light, but from something called the &lt;em&gt;dynamical timescale&lt;/em&gt;, $t_{\text{dyn}}$:&#xA;$$t_{\text{dyn}}\sim\frac{R}{v}\sim\sqrt{\frac{R^3}{GM}}\simeq\frac{1}{\sqrt{G\bar{\rho}}}$$&#xA;where we've assumed that the speed of a pressure wave, $v$ is on the order of the escape velocity of the star. $R$ is the star's radius, $M$ is its mass, $G$ is the gravitational constant, and $\bar{\rho}$ is its mean density (density is not constant throughout a star).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t_{\text{dyn}}$ can vary greatly among different stars, depending on their characteristics. It can be on the order of seconds to minutes to hours to days, and it can determine the time it takes for parts of the supernova to take place. It's a lower limit, though; large dynamical changes can take more time than $t_{\text{dyn}}$, just not less. For more information, see Section II.B of &lt;a href=&quot;http://es.ucsc.edu/~glatz/astr_112/lectures/notes17.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;these notes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to ask ourselves, what precisely is $v$? I waved the specifics away as a &quot;pressure wave&quot;, but is there more to it than that? Unfortunately, we still don't have great models of what goes on inside a supernova - I'm sure you've heard that before - but we do have inklings of the process, which I touched on a bit &lt;a href=&quot;https://astronomy.stackexchange.com/a/18425/2153&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, as the core collapses, matter at the inner regions begins to fall in, reaching extremely high densities. Some strange things happen here, including the production of neutrinos. At first, degeneracy pressure slows the collapse of the inner star, and a shock wave forms. The so-called &quot;supernova bounce&quot; happens, as this wave travels through the outer layers of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shock wave is further energized by the release of neutrinos, created via &lt;a href=&quot;https://en.wikipedia.org/wiki/Electron_capture&quot; rel=&quot;nofollow noreferrer&quot;&gt;electron capture&lt;/a&gt; (note that some have already escaped the outer layers):&#xA;$$e^-+p\to n+\nu_e$$&#xA;This adds to the energy of the shockwave, which eventually blows away the outer layers of the star. Therefore, we can see how the dynamic timescale relates to the shock wave and the time of the resulting explosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact processes causing the bounce and shock wave are not yet fully understood; this is just a basic picture.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-10-21T16:09:59.100" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18812" PostTypeId="1" AcceptedAnswerId="18813" CreationDate="2016-10-21T15:17:58.980" Score="2" ViewCount="505" Body="&lt;p&gt;How long would the orbital period be for an object the mass of the moon orbiting the sun at 1 light year in a circular orbit?&lt;/p&gt;&#xA;" OwnerUserId="6870" LastActivityDate="2016-10-25T18:04:18.797" Title="Orbital period for a 1 light year object?" Tags="&lt;orbit&gt;&lt;kuiper-belt&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18813" PostTypeId="2" ParentId="18812" CreationDate="2016-10-21T15:31:43.967" Score="6" Body="&lt;p&gt;To answer this, let's visit &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law&quot; rel=&quot;nofollow&quot;&gt;Kepler's third law&lt;/a&gt; which tells you the orbital period of a body, if you know how far away it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{P^2}{a^3} = \mathrm{constant}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it turns out, if you use the correct units, for the orbital period, $P$, and the semi-major axis, $a$, namely years and astronomical units (AU) respectively, then that constant is equal to 1! You have our moon orbiting the sun at one light year, so this needs to be converted to AU. Google tells me that $1\:\mathrm{light\:year}$ is $63241.1\:\mathrm{AU}$. For this specific case, you get that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = a^{3/2} = 63241.1^{3/2} = 15,903,734\:\mathrm{years}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that this formula didn't depend on the mass of the moon! (Technically the mass of the moon is in there, but it's pretty negligible and so doesn't matter much.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;sub&gt;You may be scratching your head over the units $-$ how can a timescale squared equal a distance cubed?. Technically there's some trickery going on because that constant isn't unitless and I've implicitly assumed those units are hanging around, without explicitly saying so. Rest assured, the units do make sense for this equation.&lt;/sub&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2016-10-25T18:04:18.797" LastActivityDate="2016-10-25T18:04:18.797" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18814" PostTypeId="2" ParentId="18809" CreationDate="2016-10-21T16:00:35.413" Score="2" Body="&lt;p&gt;If it was about an hour before you posted the question, somewhere near India, then yes it was the International Space Station.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://iss.astroviewer.net/observation.php?lon=77.20902120000005&amp;amp;lat=28.6139391&amp;amp;name=New%20Delhi&quot; rel=&quot;nofollow&quot;&gt;http://iss.astroviewer.net/observation.php?lon=77.20902120000005&amp;amp;lat=28.6139391&amp;amp;name=New%20Delhi&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since that website rolls with time, here is the data cut/pasted from today:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Fri, Oct. 21, 2016 Begin Maximum End &#xA;Time (IST): 6:42:54 PM 6:46:02 PM 6:49:09 PM &#xA;Direction: NW SW SSE &#xA;Altitude: 10° 49° 10° &#xA;Brightness: -2.7 mag  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="10005" LastActivityDate="2016-10-21T16:00:35.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18815" PostTypeId="1" CreationDate="2016-10-21T16:52:35.693" Score="1" ViewCount="65" Body="&lt;p&gt;I’m interested, to give a scientific realism to a project I’m working on, in finding a picture of all stars belonging to a star catalogue such as Tycho 2 or Hipparcos for some very specific sky regions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I have little time available, I’m wondering whether there’s a quicker way (maybe online?) than downloading the whole catalog and browsing through it via some dedicated software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D.&lt;/p&gt;&#xA;" OwnerUserId="14647" LastActivityDate="2016-10-21T16:52:35.693" Title="Stellar catalogues stars in specific sky regions – without downloading the catalogs" Tags="&lt;sky-survey&gt;&lt;star-catalogues&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18816" PostTypeId="2" ParentId="18776" CreationDate="2016-10-21T18:10:35.773" Score="1" Body="&lt;p&gt;Depending on the root cause, vibration damping pads may or may not help.&#xA;Some telescope vendors sell sets of three for $50 or so;&#xA;an Internet search also yields a few do-it-yourself recipes using silicone caulk, sorbothane insoles, etc.&#xA;&lt;a href=&quot;http://www.skyandtelescope.com/astronomy-equipment/tune-up-your-telescope/&quot; rel=&quot;nofollow&quot;&gt;Other home remedies&lt;/a&gt; are possible,&#xA;such as adding weight or shims to the mount, or hanging a chain from the optical tube assembly.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-10-21T18:10:35.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18817" PostTypeId="2" ParentId="14304" CreationDate="2016-10-21T18:20:18.417" Score="11" Body="&lt;p&gt;I was on the targeting team for one of the cameras that discovered most of these new impact craters. The reason for the distribution is simply because it's easiest to find &quot;new&quot; impact craters in the dusty regions of Mars. Often what we see is the dark-toned blast zone created by the impact in lower-resolution data, which has a large areal footprint. Then, we target those dark-toned splotches with the higher resolution cameras---which have MUCH smaller footprints---to confirm whether or not the splotches are craters. See the attached image as an example. The dark splotches really stand out in dusty regions, but in the non-dusty areas there's nothing obvious to pick out to say &quot;hey, there's a new crater&quot; as a human without tediously comparing images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/m7WFZ.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/m7WFZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="14648" LastActivityDate="2016-10-21T18:20:18.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18818" PostTypeId="2" ParentId="18747" CreationDate="2016-10-22T00:13:53.113" Score="13" Body="&lt;p&gt;Well done you. I double checked the calculations and couldn't fault what you had done. So I contacted the lead author of the paper about it and here is the response:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;After checking the numbers in our paper, I found an error: we actually used a mass of 1.0 MSun for J1407 in our simulations, instead of the 0.9 MSun as stated. This accounts for the difference in pericentric velocities (as well as the different semi-major axes, which would be smaller in the 0.9MSun case).&#xA;We will attempt to correct this in the published version, and send a correction to arXiv.&quot;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-10-22T00:13:53.113" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18819" PostTypeId="2" ParentId="18810" CreationDate="2016-10-22T00:41:31.940" Score="3" Body="&lt;p&gt;When the core collapses it is because the pressure falls below that required to support the star. However, at the point where this happens, the sound speed is a reasonable fraction of the speed of light, but decreases rapidly as one moves outward in the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is that the &quot;information&quot; that the core pressure is too low propagates through the core in a very small fraction of a second - and much faster than the dynamical collapse time. However, it takes much longer for sound to propagate into the envelope. The result is that the core collapses and decouples from the envelope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally speaking, the core collapse and supernova are not expected to be perfectly symmetric.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2365" LastEditDate="2016-10-22T03:59:30.013" LastActivityDate="2016-10-22T03:59:30.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18821" PostTypeId="2" ParentId="18784" CreationDate="2016-10-22T13:53:04.687" Score="4" Body="&lt;p&gt;While the answer by @zephyr is extensive, unresolved images of a very distant Kuiper belt object distant body over a short arc of it's orbit through a single filter can not be used to calculate a color. Further, contrary to the claim in that answer, &lt;em&gt;there are indeed&lt;/em&gt; a series of measurements alternating between visible and infrared filters!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sci-News article &lt;a href=&quot;http://www.sci-news.com/space/color-kuiper-belt-objects-04290.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Scientists Determine Color of Kuiper Belt Objects JR1 and MU69&lt;/a&gt; describes the color measurement and shows the image of 2014 MU69. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/58sL3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/58sL3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; &quot;Hubble images of 2014 MU69 taken on June 24, 2014. The images were taken at 10-min intervals. The positions of 2014 MU69 in the images are shown by the green circles. Image credit: NASA / ESA / SwRI / JHU / APL / New Horizons KBO Search Team.&quot; From &lt;a href=&quot;http://www.sci-news.com/space/color-kuiper-belt-objects-04290.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An October 18, 2016 &lt;a href=&quot;http://archive.stsci.edu/proposal_search.php?mission=hst&amp;amp;id=14092&quot; rel=&quot;nofollow noreferrer&quot;&gt;Press Conference&lt;/a&gt; includes slides by Amanda M. Zangari, Post-doctoral Researcher, Southwest Research Institute, titled &quot;2014 MU69 Is a Red Cold Classical KBO&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8ImCd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8ImCd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/obrx4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/obrx4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2hzql.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2hzql.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Slides 4, 5, 6 from (Amanda M. Zangari), October 18, 2016 press conference shown &lt;a href=&quot;http://archive.stsci.edu/proposal_search.php?mission=hst&amp;amp;id=14092&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://arxiv.org/pdf/1605.05376.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Porter et al. 2016&lt;/a&gt; section 2.2 mentions Hubble Space Telescope observing Program #14092, (S. Benecchi PI) which includes 2014 MU69. Part of &lt;a href=&quot;http://archive.stsci.edu/proposal_search.php?mission=hst&amp;amp;id=14092&quot; rel=&quot;nofollow noreferrer&quot;&gt;the proposal abstract&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Fortuitously the New Horizons (NH) spacecraft will fly through this part of the Kuiper Belt after its encounter with the Pluto system providing the opportunity for observing as many as 10 objects from long range. Unfortunately, the NH observations will be panchromatic. We propose to obtain colors and to search for binary companions to 8 of these objects and the NH encounter Kuiper Belt object (KBO; 2 were observed with HST in previous cycles), so that we can link the single filter, and higher resolution, spacecraft data from the distant targets to the color and binary data available in the literature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The data there list a series of HST WFC3 measurements of a variety of KBO objects of potential interest as the next target for [New Horizons][], including 20 measurements tracking the motion of 2014 MU69 of on &lt;code&gt;2016-07-03&lt;/code&gt; between &lt;code&gt;13:48&lt;/code&gt; and &lt;code&gt;19:10&lt;/code&gt;. The measurements alternate between 348 seconds using filter F606W, and 373 seconds using filter F814W, and both using the aperture UVIS2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.stsci.edu/~WFC3/UVIS/SystemThroughput/&quot; rel=&quot;nofollow noreferrer&quot;&gt;tabulated system throughputs&lt;/a&gt; for the aperture/filter combinations are plotted below. Clearly two color measurements are made and the &quot;redness&quot; is established based on green-red vs NIR brightness. In the last slide shown above (#6 from the &lt;a href=&quot;http://archive.stsci.edu/proposal_search.php?mission=hst&amp;amp;id=14092&quot; rel=&quot;nofollow noreferrer&quot;&gt;press release slides&lt;/a&gt; a &quot;redness&quot; scale is devised, &lt;em&gt;with 2014 MU69 falling roughly half-way between Pluto and Mars!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IzF6O.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IzF6O.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; plot of HST system throughputs &lt;a href=&quot;http://www.stsci.edu/~WFC3/UVIS/SystemThroughput/f606w.UVIS2.tab&quot; rel=&quot;nofollow noreferrer&quot;&gt;f606w.UVIS2.tab&lt;/a&gt; (green, solid) and &lt;a href=&quot;http://www.stsci.edu/~WFC3/UVIS/SystemThroughput/f814w.UVIS2.tab&quot; rel=&quot;nofollow noreferrer&quot;&gt;f814w.UVIS2.tab&lt;/a&gt; (red, dashed).&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2016-10-23T12:28:02.510" LastActivityDate="2016-10-23T12:28:02.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18822" PostTypeId="1" CreationDate="2016-10-22T14:42:26.740" Score="2" ViewCount="75" Body="&lt;p&gt;If I take Arcturus for example, its HipID is 32349 and when i look at the raw Hip_Main.dat, it stays that the star's Spectral is K2IIIp whilst Wikipedia says the star's is KOIII.  Why the discrepancy, have I misunderstood the Hip data?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mu Cephei is M2e Ia, in Hip its m2Ia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Achernar is B6 Vep whereas in Hip its B3Vp&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zosma for example, matches, I get A4V in the Hip file and it says A4 V in Wiki.&lt;/p&gt;&#xA;" OwnerUserId="5401" LastActivityDate="2016-10-22T14:42:26.740" Title="Have I misunderstood Hipparcos Data with Arcturus?" Tags="&lt;star&gt;&lt;hipparcos&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18823" PostTypeId="1" AcceptedAnswerId="18858" CreationDate="2016-10-22T14:48:05.963" Score="1" ViewCount="278" Body="&lt;p&gt;I'm a physics graduate student hoping to plot ~10&amp;deg;x10&amp;deg; sections of the sky in right ascension and declination (eventually I will want to plot points on this section at appropriate locations given RA and Dec, preferably in Python). I've seen square plots on the order of arcseconds, where curvature has a small effect, but that won't apply to these larger plots. I'm not familiar with astronomy journals or a module which aids in this sort of plotting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there a commonly accepted method for displaying intermediate-sized sections of a sky-map? How is it implemented?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7480" LastEditorUserId="6980" LastEditDate="2016-10-26T00:42:40.510" LastActivityDate="2016-10-26T00:42:40.510" Title="Journal standard for plotting section of skymap?" Tags="&lt;coordinate&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18826" PostTypeId="1" CreationDate="2016-10-22T18:25:31.697" Score="2" ViewCount="124" Body="&lt;p&gt;I have &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a &lt;a href=&quot;http://www.universetoday.com/19454/the-vixen-ed100sf-refractor-superb-optical-quality/&quot; rel=&quot;nofollow&quot;&gt;Vixen ED100SF ED&lt;/a&gt; refracting telescope and &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.bhphotovideo.com/c/product/246657-REG/Bushnell_191050_10x50_Legend_Binocular.html&quot; rel=&quot;nofollow&quot;&gt;10x50 6.5 Bushnell Legend Porro Prism&lt;/a&gt; binoculars&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and would like to outfit both with filters for solar eclipse viewing. I gather the basic setup simply involves finding an appropriate filter, that fits the equipment, but I'm new to this and the &lt;a href=&quot;http://www.telescope.com/Gift-Center/Sun-Spot/pc/7/483.uts?currentIndex=0&amp;amp;pageSize=22&amp;amp;defaultPageSize=20&amp;amp;mode=viewall&amp;amp;categoryId=7&amp;amp;subCategoryId=483&amp;amp;type=thumbnail2level&quot; rel=&quot;nofollow&quot;&gt;range of options&lt;/a&gt; is bewildering. Ideally I'd also like something that allows me to see solar features, but I gather that these can be quite expensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are my options (or a good reference on options) for viewing solar eclipses and solar features? Would something like an &lt;a href=&quot;http://www.daystarfilters.com/Quark/CALCIUMQuark.shtml&quot; rel=&quot;nofollow&quot;&gt;&quot;eyepiece&quot; filter&lt;/a&gt; work on its own for safe viewing, or would my telescope require something different?&lt;/p&gt;&#xA;" OwnerUserId="2369" LastActivityDate="2016-10-23T18:42:50.960" Title="Filters for solar eclipse and solar feature viewing" Tags="&lt;the-sun&gt;&lt;telescope&gt;&lt;amateur-observing&gt;&lt;solar-eclipse&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18827" PostTypeId="2" ParentId="1350" CreationDate="2016-10-22T19:28:53.447" Score="1" Body="&lt;p&gt;The nebula colours are mostly due to forbidden line emission from oxygen (green) and nitrogen (red) and from the balmer series of hydrogen and occasionally, ionised helium. The key point is that this is usually optically thin emission, meaning that what we see is proportional to the number of excited atoms/ions in the nebula in the line of sight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is important for your question, because in such cases, the surface brightness is &lt;em&gt;independent of distance&lt;/em&gt;. The luminosity increases as one over distance squared, but the area occupied on the sky also increases by the same factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From inside the nebula the surface brightness would be even lower.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-10-22T19:28:53.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18828" PostTypeId="1" CreationDate="2016-10-22T19:57:11.527" Score="1" ViewCount="37" Body="&lt;p&gt;I'm assembling a compact and light travel telescope. I'm going for the small &lt;a href=&quot;https://luntsolarsystems.com/product/lunt-ed-doublet-optical-tube-assemblies/&quot; rel=&quot;nofollow&quot;&gt;Lunt 80mm ed&lt;/a&gt; refractor telescope mounted on a light &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B004XC3GZW&quot; rel=&quot;nofollow&quot;&gt;Dolica tripod&lt;/a&gt;.  For those with this --or similar-- assembly, I would like to ask you how did you attach the Lunt to the Dolica. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;Marcelo&lt;/p&gt;&#xA;" OwnerUserId="14649" LastActivityDate="2016-10-22T19:57:11.527" Title="Travel telescope: attaching a Lunt 80mm ed on a Dolica tripod" Tags="&lt;telescope&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18829" PostTypeId="1" AcceptedAnswerId="18836" CreationDate="2016-10-23T00:44:55.207" Score="1" ViewCount="100" Body="&lt;p&gt;I am taking a class on the space environment and I have a few questions about the solar wind.  A number of sources list the composition as (approximately) 4% He$^{++}$ and 96% H$^{+}$.  What &lt;em&gt;exactly&lt;/em&gt; does this mean?  Does it mean by mass or by number of ions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, when discussing proton flux due to the solar wind at Earth:  What exactly is included in the flux numbers quoted (e.g., one source I have says 2e12 m$^{-2}$s$^{-1}$).  Is that the flux due to only the H$^{+}$ and excluding the He$^{++}$ and other minor constituents?  Or, are they including all the protons from the other ions as well?&lt;/p&gt;&#xA;" OwnerUserId="14657" LastEditorUserId="1203" LastEditDate="2016-10-23T02:40:10.850" LastActivityDate="2016-10-23T15:40:08.627" Title="Composition and Proton Flux from the Solar Wind" Tags="&lt;solar-wind&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18830" PostTypeId="2" ParentId="18826" CreationDate="2016-10-23T02:09:54.160" Score="2" Body="&lt;p&gt;To prevent eye damage, a filter covering the objective (full aperture, energy rejection, safety film, etc.) is mandatory.&#xA;An eyepiece filter also needs an objective filter to protect against overheating.&#xA;Sunspots are visible without an expensive narrowband filter.&#xA;A filter for the hydrogen-alpha (H$\alpha$) line or the calcium line you linked to&#xA;can show more detail within the solar disk or prominences around the edge.&#xA;Sky &amp;amp; Telescope has several &lt;a href=&quot;http://www.skyandtelescope.com/observing/celestial-objects-to-watch/sun/&quot; rel=&quot;nofollow&quot;&gt;articles about observing the Sun&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-10-23T02:09:54.160" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18831" PostTypeId="1" AcceptedAnswerId="18835" CreationDate="2016-10-23T08:47:27.173" Score="2" ViewCount="126" Body="&lt;p&gt;The recently released map of galactic neutral hydrogen density and velocity is really beautiful. The work has been relased by the &lt;a href=&quot;http://www.icrar.org/hi4pi/&quot; rel=&quot;nofollow noreferrer&quot;&gt;HI4PI collaboration&lt;/a&gt; (HI = neutral hydrogen, $4\pi$ = complete spherical coverage).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is fairly simple. Looking at the image near the galactic plane, why do I see near the center a green bump on the left and a blue bump on the right, then near the edges of the image a much stronger blue bump on the left and a green bump on the right. In other words, there is a major and a minor peak in the positive radial velocity gas, and a major and a minor peak in the negative radial velocity gas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this have some simple geometrical explanation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mfmgp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mfmgp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/P9bIg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P9bIg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above x2:&lt;/strong&gt; neutral hydrogen density/velocity map &lt;a href=&quot;http://www.icrar.org/hi4pi/&quot; rel=&quot;nofollow noreferrer&quot;&gt;from here&lt;/a&gt; with inset indicating color scale expanded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/X6sM3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X6sM3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; for those who can not distinguish the green and blue in the image, this plots the average green (solid) and blue (dashed) intensity within the central equatorial band.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a nice article at &lt;a href=&quot;http://phys.org/news/2016-10-scientists-milky.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;phys.org&lt;/a&gt; with a video, which can also be &lt;a href=&quot;https://www.youtube.com/watch?v=Q2mgpsTFuV8&quot; rel=&quot;nofollow noreferrer&quot;&gt;seen in YouTube&lt;/a&gt;. It shows the intensity of HI as a cut in radial velocity is slowly scanned through the data set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=Q2mgpsTFuV8&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=Q2mgpsTFuV8&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the stackexchange YouTube option is not turned on in astronomy stackexchange, here is a GIF of screenshots at -30, -20, -10, 0, 10, 20, and 30 km/sec:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/53pPa.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/53pPa.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2016-10-23T09:09:07.883" LastActivityDate="2016-10-23T12:58:57.857" Title="Why does the neutral hydrogen velocity have this characteristic behavior in the galactic plane?" Tags="&lt;radio-astronomy&gt;&lt;data-analysis&gt;&lt;galactic-dynamics&gt;&lt;velocity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18832" PostTypeId="1" CreationDate="2016-10-21T22:58:12.297" Score="3" ViewCount="1096" Body="&lt;p&gt;I am trying to find out if a planet is in an apparent retrograde motion with respect to the earth at any given point in time. Given the time in Julian days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already have the geocentric and heliocentric coordinates and velocities of the planet using the JPL ephemeris, but not sure of what mathematical formula to use to identify the retrograde motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read this article &quot;&lt;a href=&quot;https://physics.stackexchange.com/questions/249493/mathematically-calculate-if-a-planet-is-in-retrograde&quot;&gt;https://physics.stackexchange.com/questions/249493/mathematically-calculate-if-a-planet-is-in-retrograde&lt;/a&gt;&quot; which had a similar question but I am not sure of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is the answer correct as it has not been marked as correct?&lt;/li&gt;&#xA;&lt;li&gt;How to represent earth in the XY plane only when it has XYZ planes and why do we even need to do that? Can't we just use the heliocentric equatorial plane which I think is the same as the one used by NASA JPL ephemeris.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerDisplayName="ADUser" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-10-24T07:46:00.363" Title="Mathematically calculate if a Planet is in Retrograde" Tags="&lt;orbit&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18833" PostTypeId="2" ParentId="18832" CreationDate="2016-10-23T11:59:59.947" Score="4" Body="&lt;p&gt;Here is a quick answer. Apparent retrograde motion of a planet as seen from Earth means that its ecliptic longitude is decreasing. (Thanks to @barrycarter for correcting my previous statement.) You have to calculate the apparent angular position of the planet from the Earth in ecliptic coordinates (as explained &lt;a href=&quot;https://space.stackexchange.com/a/18647/12102&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://space.stackexchange.com/a/18646/12102&quot;&gt;here&lt;/a&gt;) and find places where the ecliptic longitude is decreasing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can subtract the positions of Earth from the position of the planet in either coordinates. Then you &lt;em&gt;express that vector in spherical coordinates&lt;/em&gt;. It doesn't matter if it's heliocentric or geocentric. What matters is the direction that the axes of the coordinate system is pointing, not where the center is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want RA and dec, use ICRF coordinates, and these can be transformed to ecliptic coordinates as well. An easy way to do this is to use the python package &lt;a href=&quot;http://rhodesmill.org/skyfield/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Skyfield&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a sample script and a plot. This is Mars from 2010.0 to 2020.0 and the thick red line represents retrograde motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MfrlM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MfrlM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;&#xA;from skyfield.api import load&#xA;&#xA;planets = load('de421.bsp')&#xA;&#xA;earth   = planets['earth']&#xA;mars    = planets['mars']&#xA;&#xA;year_zero = 2010&#xA;days = np.linspace(1, 3650, 10000)&#xA;years = year_zero + days / 365.2564&#xA;ts = load.timescale()&#xA;t = ts.utc(year_zero, 1, days)&#xA;&#xA;# thanks to @barrycarter's comment, do it the right way!&#xA;eclat, eclon, ecd = earth.at(t).observe(mars).ecliptic_latlon()&#xA;&#xA;eclondgs      = (180./np.pi) * eclon.radians&#xA;eclondel      = eclondgs[1:] - eclondgs[:-1]&#xA;&#xA;eclondel[eclondel &amp;lt; -300] += 360. # this is a fudge for now&#xA;eclondel[eclondel &amp;gt; +300] -= 360. # this is a fudge for now&#xA;&#xA;prograde   = eclondel &amp;gt; 0.&#xA;&#xA;eclon_prograde   = eclondgs.copy()[:-1]&#xA;eclon_retrograde = eclondgs.copy()[:-1]&#xA;&#xA;eclon_prograde[-prograde]  = np.nan&#xA;eclon_retrograde[prograde] = np.nan&#xA;&#xA;plt.figure()&#xA;plt.plot(years[:-1], eclon_prograde,   '-g', linewidth=1)&#xA;plt.plot(years[:-1], eclon_retrograde, '-r', linewidth=3)&#xA;plt.ylim(0, 360)&#xA;plt.title('Mars Ecliptic longitude (degrees) AD 2010.0 to 2020.0',&#xA;          fontsize=16)&#xA;plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2016-10-24T07:46:00.363" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18835" PostTypeId="2" ParentId="18831" CreationDate="2016-10-23T12:58:57.857" Score="5" Body="&lt;p&gt;The reason is detailed in depth in &lt;a href=&quot;http://euhou.obspm.fr/public/Milky_Way_Rotation_A4.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this pdf&lt;/a&gt;, which contains the following diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MQBLE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MQBLE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some key quantities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$R_0$: Distance from the observer to the center of the Milky Way&lt;/li&gt;&#xA;&lt;li&gt;$R$: Distance from target gas to the center of the Milky Way&lt;/li&gt;&#xA;&lt;li&gt;$V_0$: Velocity of the observer with respect to a certain reference frame&lt;/li&gt;&#xA;&lt;li&gt;$V$: Velocity of target gas with respect to the same reference frame&lt;/li&gt;&#xA;&lt;li&gt;$l$: Angle between $R_0$ and $\mathbf{SM}$, the distance to the gas&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When measuring the radial velocity, we look for the projection of $V$ onto our line of sight. Some geometry results in the relation&#xA;$$V_r=V_0\left(\frac{R_0}{R}-1\right)\sin l$$&#xA;for the radial velocity $V_r$, assuming $V\approx V_0$ (which is not always true!). The $\sin l$ term, in addition to the asymmetrical rotation of the Milky Way, leads to the peculiar data you asked about.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2016-10-23T12:58:57.857" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18836" PostTypeId="2" ParentId="18829" CreationDate="2016-10-23T15:40:08.627" Score="1" Body="&lt;p&gt;That's by number-- you get a lot more helium than that by mass.  Also, when they talk about proton flux, they just count the bare protons, not protons embedded in larger ions like helium.  Those are the kinds of things that should be made clear, but end up being common assumed conventions.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-10-23T15:40:08.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18837" PostTypeId="2" ParentId="18826" CreationDate="2016-10-23T18:42:50.960" Score="3" Body="&lt;p&gt;You mustn't use an eyepiece filter. It is very very dangerous. In my opinion the best is buying an Astrosolar paper, that isn't expensive and you can see very well an eclipse. If you want to see solar features you must spend more and you need an H-alpha filter, but the view with that is amazing&lt;/p&gt;&#xA;" OwnerUserId="14600" LastActivityDate="2016-10-23T18:42:50.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18838" PostTypeId="1" AcceptedAnswerId="18841" CreationDate="2016-10-24T07:29:54.293" Score="-2" ViewCount="6986" Body="&lt;p&gt;i just want to ask in earth what  the difference between Equator &amp;amp; Prime Meridian &amp;amp; also want to know is it same distance Equator &amp;amp; Prime Meridian?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cQqUF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cQqUF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11792" LastEditorUserId="11792" LastEditDate="2016-10-24T07:35:51.120" LastActivityDate="2016-10-24T12:02:59.323" Title="Is it same distance Equator &amp; Prime Meridian?" Tags="&lt;earth&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ClosedDate="2016-10-24T17:08:37.830" ContentLicense="CC BY-SA 3.0" />
  <row Id="18841" PostTypeId="2" ParentId="18838" CreationDate="2016-10-24T12:02:59.323" Score="4" Body="&lt;p&gt;The Earth rotates daily about an axis. This axis of rotation intersects the surface of the Earth at the north and south poles. Imagine a plane that is normal to the axis of rotation and lies halfway between the north and south poles. The equator is the intersection of this plane and the surface of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meridians, or lines of longitude, go from pole to pole along a great circle. The Prime Meridian is simply the meridian that we have deemed to be the zero degrees line of longitude. This was an arbitrarily choice. (The French, for example, argued that the Prime Meridian should pass through the Observatory of Paris.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine going from the south pole to the north pole along the Prime Meridian. The Prime Meridian ends at the north pole. Continuing in the same direction takes you along the 180th degree longitude line. This means that if the Earth was a perfect sphere, the length of the equator would be exactly twice that of the prime meridian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth isn't a perfect sphere. The Earth's rotation gives the Earth a slight equatorial  bulge. The equator (at sea level) is about 21 km further from the center of the Earth than is the north pole. This means the Earth's equatorial circumference (40075 km) is about 67 kilometers longer than the Earth's meridional circumference (40008 km).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That this is a nice round number is not a coincidence. The original definition of the meter was one ten millionth of the distance at sea level between the equator and the north pole. A prototype metre bar was created based on initial estimates of this distance. It turned out that this initial estimate was a bit off, and by then it was too late to redefine the meter to match these better measurements.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2016-10-24T12:02:59.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18842" PostTypeId="2" ParentId="8203" CreationDate="2016-10-24T16:52:33.030" Score="2" Body="&lt;p&gt;&lt;strong&gt;A younger lunar thermal history implies that the Moon was still undergoing a contraction and cooling process,&lt;/strong&gt; which was initiated when it began accreting material together. Evidence has shown that the moon has no tectonic activity, so all geologic features are volcanic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mare are a more recent example of this activity, and have been shown to have fairly young ages, as you indicated. The moon still has some internal heat from its formation that may be leaking out through convection and cooling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though the moon has a relatively small mass, it didn't lose its internal heat as quickly as the earth because there was no tectonic activity to facilitate it fast enough, and convection is a slow process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the moon cools, it contracts, creating sharp ridges and cracks in the surface (see features known as 'scarps', or 'garden'). These contractions contribute to the &quot;deposit&quot; of material formed in the images that paper discussed. As the cooling surface shrinks, it breaks the crust through a &lt;strong&gt;thrust fault&lt;/strong&gt;, which pushes some material higher and shoves other material lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cjrXE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cjrXE.jpg&quot; alt=&quot;Thrust fault causing a lobate scarp&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What may appear to be river channels on the moon are also lava flows. These were initially underground lava channels whose roofs collapsed to expose these subterranean features. More recent incidents of molten material rising to the surface and cooling can contribute to some of the younger ages recorded in the basaltic rocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zqllw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zqllw.jpg&quot; alt=&quot;Cool lava tubes on the moon&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Tldr; Evidence of younger volcanic activity confirms existing theories for the contraction and cooling process of the moon, just that it didn't end as soon as we initially thought&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11684" LastActivityDate="2016-10-24T16:52:33.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18844" PostTypeId="1" AcceptedAnswerId="34373" CreationDate="2016-10-24T20:48:56.183" Score="10" ViewCount="1962" Body="&lt;p&gt;This question may be a little lazy, but can anybody give me a proof of the Hill sphere formula? Acording to &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;noreferrer&quot;&gt;wikipedia&lt;/a&gt;, the formula for the radius, $r$, is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r\approx a(1-e)\left(\frac{m}{3M}\right)^{1/3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where a body of mass $m$ is orbiting a much more massive body of mass $M$ with a semi-major axis $a$ and eccentricty $e$.&lt;/p&gt;&#xA;" OwnerUserId="9949" LastEditorUserId="11143" LastEditDate="2016-10-25T18:39:56.617" LastActivityDate="2020-02-18T20:34:45.180" Title="The deduction of the Hill Sphere formula" Tags="&lt;gravity&gt;&lt;celestial-mechanics&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18845" PostTypeId="1" AcceptedAnswerId="18847" CreationDate="2016-10-24T20:54:02.907" Score="4" ViewCount="132" Body="&lt;p&gt;Cosmic gamma ray bursts are seriously powerful and do a lot of damage. So how did we measure gamma ray frequencies without getting fried?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two possible explanations:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;We never detected them directly but calculated their frequency from secondary radiation from the surrounding gas that was ionized.&lt;/li&gt;&#xA;&lt;li&gt;We get a direct hit but the source is so far that beam divergence will be large enough for only a few photons per square meter to hit us, so there's not enough energy to be destructive.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10910" LastActivityDate="2016-10-25T13:21:04.800" Title="If getting shined on by a gamma ray is an extinction level event how do we detect them?" Tags="&lt;gamma-ray-bursts&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18846" PostTypeId="2" ParentId="18133" CreationDate="2016-10-24T21:14:09.007" Score="3" Body="&lt;p&gt;As Andy pointed out, the spiral galaxy is IC 1296. You're correct that it's not in the Messier Catalog, but it's obviously in the revised Index Catalogue. &lt;a href=&quot;http://ned.ipac.caltech.edu/cgi-bin/nph-objsearch?objname=IC%201296&amp;amp;extend=no&amp;amp;out_csys=Equatorial&amp;amp;out_equinox=J2000.0&amp;amp;obj_sort=RA+or+Longitude&amp;amp;of=pre_text&amp;amp;zv_breaker=30000.0&amp;amp;list_limit=5&amp;amp;img_stamp=YES&quot; rel=&quot;nofollow&quot;&gt;NED&lt;/a&gt; and &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=%402868623&amp;amp;Name=IC+1296&amp;amp;submit=display+all+measurements&quot; rel=&quot;nofollow&quot;&gt;SIMBAD&lt;/a&gt; have some more information on it:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Morphological type:&lt;/strong&gt; SBbc, as classified according to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble_sequence&quot; rel=&quot;nofollow&quot;&gt;Hubble sequence&lt;/a&gt;. It's a spiral galaxy with two clearly visible arms.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Redshift:&lt;/strong&gt; 0.017075 $\pm$ 0.000070, according to &lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/bib_query?1996AJ....112.1803M&quot; rel=&quot;nofollow&quot;&gt;Marzke et al. (1996)&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Distance:&lt;/strong&gt; 66.6 Mpc to 68.9 Mpc after corrections, depending on your distance definition of choice (&lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_diameter_distance&quot; rel=&quot;nofollow&quot;&gt;angular diameter&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Luminosity_distance&quot; rel=&quot;nofollow&quot;&gt;luminosity&lt;/a&gt;, etc.). Large uncertainties arise when calculating the difference from redshift. For comparison, &lt;a href=&quot;https://en.wikipedia.org/wiki/Andromeda_Galaxy&quot; rel=&quot;nofollow&quot;&gt;Andromeda&lt;/a&gt; is approximately 0.78 Mpc away, and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ring_Nebula&quot; rel=&quot;nofollow&quot;&gt;Ring Nebula&lt;/a&gt; is $10^{-3}$ times &lt;em&gt;that&lt;/em&gt; distance.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Apparent magnitude:&lt;/strong&gt; 14.80 $\pm$ 0.30, at visible wavelengths. For comparison, the Ring Nebula's apparent magnitude is 8.8.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2016-10-24T21:14:09.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18847" PostTypeId="2" ParentId="18845" CreationDate="2016-10-24T21:42:11.247" Score="7" Body="&lt;p&gt;The danger from gamma-ray bursts (GRBs) is the potential to reduce atmospheric ozone drastically, allowing harmful ultraviolet radiation to pass through the atmosphere and make life on Earth much more difficult. However, you have to be fairly close for this to happen. &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0211361&quot;&gt;Gehrels et al. (2003)&lt;/a&gt; calculated that a typical GRB from a supernova would have to occur less than 8 parsecs from Earth to cause serious damage to the ozone layer - or, as they put it, to cause levels to drop low enough that they would&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;double the “biologically active” UV ﬂux received at the surface&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously, though, we can detect gamma-rays from distances farther than 8 parsecs! The closest GRB detected so far, &lt;a href=&quot;https://en.wikipedia.org/wiki/GRB_980425&quot;&gt;GRB 980425&lt;/a&gt;, is likely the result of &lt;a href=&quot;https://en.wikipedia.org/wiki/SN_1998bw&quot;&gt;SN 1998bw&lt;/a&gt;, and thus originated 140 million light-years away - over &lt;em&gt;42 million parsecs away&lt;/em&gt;. The radiation from it was nowhere near high enough to harm life on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your second explanation, therefore, is the correct one. The intensity of the signal decreases according to the inverse-square law - that is, $I(r)\propto r^{-2}$ - so at large distances, the flux is small enough that there is negligible impact on Earth's atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phiteros also raised a good point: To cause severe problems from Earth, the gamma-ray emission from a GRB - which is often along an axis from both poles of the progenitor - would have to be pointed nearly directly at Earth. For example, there was once a scare that &lt;a href=&quot;https://en.wikipedia.org/wiki/WR_104#Supernova_explosions&quot;&gt;a supernova by either of two stars in the WR 104 system&lt;/a&gt; could cause problems for Earth, but the odds that the beam would hit Earth full-on are slim.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-10-25T13:21:04.800" LastActivityDate="2016-10-25T13:21:04.800" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18849" PostTypeId="1" AcceptedAnswerId="18850" CreationDate="2016-10-24T23:18:51.300" Score="10" ViewCount="872" Body="&lt;p&gt;I know black holes can be described by only a few parameters. Given the extreme conditions within a neutron star, how much hair does it have, ie. how much of the original star parameters and composition (e.g. metallicity) affects the features of a neutron star?&lt;/p&gt;&#xA;" OwnerUserId="10813" LastActivityDate="2016-10-24T23:37:37.377" Title="No hair and neutron stars" Tags="&lt;neutron-star&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18850" PostTypeId="2" ParentId="18849" CreationDate="2016-10-24T23:37:37.377" Score="10" Body="&lt;p&gt;No complete answer from me, but some obvious examples of &quot;hair&quot; on a neutron star are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature - both internal and surface temperature can be measured (in principle). These are mainly a function of the age of the neutron star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnetic field - young neutron stars have very strong magnetic fields that are probably an inherited property from the composition and rotation of the progenitor. The field decays with time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface composition may vary from neutron star to neutron star. In principle this can be measured with X-ray spectroscopy. The composition may be connected with the progenitor, but could also reflect accreted material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutron stars have a radius! The radius may or may not be a unique function of mass. It could depend on the rotation, composition and age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think though, the gist of your question is could one determine the progenitor of a neutron star from observations of the neutron star? At present the answer to that is no. Even the relationship between neutron star mass and progenitor mass is not understood; it is also unclear why some neutron stars have &lt;em&gt;very&lt;/em&gt; high magnetic fields and so on. But ultimately that would be an aim.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-10-24T23:37:37.377" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18851" PostTypeId="1" CreationDate="2016-10-25T03:20:07.663" Score="-1" ViewCount="192" Body="&lt;p&gt;So, I've been doing research on how no mass could pass the speed of light. So, I've always had a question in mind. What would happen if someone was going at 99.99% the speed of light and they decided to accelerate. What would happen?&lt;/p&gt;&#xA;" OwnerUserId="14662" LastActivityDate="2016-10-25T17:32:35.113" Title="What would happen if a spaceship goes faster than the speed of light?" Tags="&lt;light&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2016-10-25T14:43:32.933" ContentLicense="CC BY-SA 3.0" />
  <row Id="18852" PostTypeId="2" ParentId="18851" CreationDate="2016-10-25T03:55:49.080" Score="3" Body="&lt;p&gt;If someone was going speed $v = \beta c$ where $0 \leq \beta &amp;lt; 1$ and they decided to accelerate, their velocity would increase and approach $c$ but never pass it. This is due to the fact that it would require an infinite amount of energy to accelerate a massive particle &lt;em&gt;to the speed of light in the first place.&lt;/em&gt; You can see this from the energy in special relativity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = \frac{mc^2}{\sqrt{1-\beta^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\beta = v/c$ from above. If $\beta = 1$, or $v = c$, $E \to \infty$. There is a signularity in this equation when $v=c$ or $\beta = 1$, which is interpreted as taking infinite energy just to get to the speed of light. This also restricts massive objects from going the speed of light ever, which means for a massive object $ v &amp;lt; c $ strictly.&lt;/p&gt;&#xA;" OwnerUserId="9826" LastEditorUserId="9301" LastEditDate="2016-10-25T17:32:35.113" LastActivityDate="2016-10-25T17:32:35.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18853" PostTypeId="1" AcceptedAnswerId="18854" CreationDate="2016-10-25T07:16:26.610" Score="9" ViewCount="1528" Body="&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/0808.2813&quot;&gt;Natarajan &amp;amp; Treister (2008)&lt;/a&gt; describes a practical upper limit for black hole masses at $\sim 10^{10} M_\odot$. This is all due to the black hole's interactions with nearby matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, is there a theoretical upper mass limit for black holes in general relativity? More specifically, do any solutions make note of this? Would this depend on whether the black hole described is eternal or time-variant, static or spinning, charged or uncharged, etc.?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, do any metrics make note of lower mass limits? Would it be possible for a black hole with the mass of an electron to exist (at any point in time, putting aside Hawking radiation)?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2018-10-15T14:41:27.293" Title="Are there upper or lower mass limits for black holes?" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;general-relativity&gt;&lt;supermassive-black-hole&gt;&lt;mass&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18854" PostTypeId="2" ParentId="18853" CreationDate="2016-10-25T13:04:30.883" Score="5" Body="&lt;p&gt;In classical General Relativity, black holes can exist at any size (mass) without any problem. The upper limit is given by the available mass of the universe and there is no theoretical lower limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As already noted in the question, quantum effects like Hawking radiation set up lower limits on stable black holes; the ones with too low mass will decay quickly into radiation.&lt;/p&gt;&#xA;" OwnerUserId="11617" LastActivityDate="2016-10-25T13:04:30.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18855" PostTypeId="1" CreationDate="2016-10-25T14:38:47.717" Score="4" ViewCount="60" Body="&lt;p&gt;Most of the spectroscopic surveys like SDSS classify stars into one pure spectral type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a large fraction of stars are in binaries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we know whether a spectral fitting is good enough? Is there something like $\chi^2$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, is it possible we can estimate the probability that a star is in a binary based on one spectrum?&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2016-10-26T19:18:23.843" Title="Does binarity play a role in spectral fitting?" Tags="&lt;binary-star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18856" PostTypeId="1" AcceptedAnswerId="18864" CreationDate="2016-10-25T14:57:15.807" Score="1" ViewCount="167" Body="&lt;p&gt;I saw this &lt;a href=&quot;https://arxiv.org/abs/1610.03031&quot; rel=&quot;nofollow noreferrer&quot;&gt;ArXiv preprint&lt;/a&gt; after reading about it in the &lt;a href=&quot;https://techcrunch.com/2016/10/25/scientists-say-signals-from-certain-stars-are-probably-aliens/&quot; rel=&quot;nofollow noreferrer&quot;&gt;popular media&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;I can not even understand the beginning of the introduction (much less the whole paper):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;INTRODUCTION&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;A Fourier transform analysis of 2.5 million spectra in the Data Release 8 of the Sloan Digital Sky Survey (SDSS) and the SEGUE 2 SDSS was carried out to detect periodic modulations contained in their frequency spectra (Trottier 2012).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;where Trottier 2012 is &lt;em&gt;Recherche de signaux périodiques dans des spectres astronomiques, M.Sc. Thesis, Université Laval&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;The modulation frequency seems to be of order 10&lt;span class=&quot;math-container&quot;&gt;${}^{-12}$&lt;/span&gt; seconds, and the SDSS is (as far as I know) not really a particularly time-resolved data set. 1 micron wavelength light (for general reference) has a frequency of about 3x10&lt;span class=&quot;math-container&quot;&gt;${}^{-14}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;I'm not asking (necessarily) if this result is right or wrong, I'd just like to understand how the SDSS can actually be analyzed to detect or even infer THz rate modulation!&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2020-11-24T01:26:05.887" LastActivityDate="2020-11-24T01:26:05.887" Title="What (actually) does &quot;peculiar periodic spectral modulations&quot; refer to in this preprint?" Tags="&lt;spectroscopy&gt;&lt;spectra&gt;&lt;seti&gt;&lt;sloan-dss&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18858" PostTypeId="2" ParentId="18823" CreationDate="2016-10-25T20:07:09.297" Score="4" Body="&lt;p&gt;I don't think there's a journal standard for this.  An easy and commonly used method is to plot latitude $\delta$ versus longitude $\alpha$, where the aspect ratio is scaled by $\mbox{cos}(\delta_0)$ so the projection is orthonormal at the image center $(\alpha_0, \delta_0)$.  Depending on your region size and on $|\delta_0|$, this may or may not come out looking distorted.   You might be able to pick a coordinate system (i.e. equatorial or galactic) where the latitude is never large enough to give major distortion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another common thing to do, arguably better when you're imaging extended objects, is to use a &lt;a href=&quot;https://lambda.gsfc.nasa.gov/product/iras/coordproj.cfm&quot; rel=&quot;nofollow&quot;&gt;tangent-plane projection&lt;/a&gt;, also called a &lt;a href=&quot;http://mathworld.wolfram.com/GnomonicProjection.html&quot; rel=&quot;nofollow&quot;&gt;gnomonic projection&lt;/a&gt;.  This projection depends on the chosen center, which is usually the image center.  It is included in some plotting and analysis packages, but none that I use regularly.  I recently coded this up for the Astropy package but haven't really tested it yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Projections like Aitoff, Hammer, and Mollweide have been included in many plotting packages, but usually only for whole-sky projections.  You could try to write your own versions but I'm not sure it would be worth the trouble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Whatever you do, make sure longitude increases to the left since you're looking at the sky not the ground...)&lt;/p&gt;&#xA;" OwnerUserId="14668" LastActivityDate="2016-10-25T20:07:09.297" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18859" PostTypeId="1" CreationDate="2016-10-25T20:30:54.737" Score="7" ViewCount="246" Body="&lt;p&gt;The Hubble Space Telescope (HST) is a remarkable device. It produced (produces) images whose resolution and clarity were a step change ahead of anything that could be captured from the ground. Is this still the case, or have adaptive optics and ever larger mirrors caught up with the image quality the HST can achieve? Is it still true that the HST's images are better than anything produced on Earth?&lt;/p&gt;&#xA;" OwnerUserId="14669" LastActivityDate="2018-07-18T19:46:01.780" Title="How good is the HST compared to ground telescopes?" Tags="&lt;hubble-telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18860" PostTypeId="2" ParentId="18859" CreationDate="2016-10-25T23:06:13.323" Score="6" Body="&lt;p&gt;Adaptive optics techniques have only been successfully applied in the far red and near infrared. HST produces images with spatial resolution equivalent to ground based adaptive optics images, but it works at near infrared wavelengths &lt;em&gt;and&lt;/em&gt; visible wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spatial resolution of HST at visible wavelengths has not really been matched with ground based telescopes. &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Lucky_imaging&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lucky imaging&lt;/a&gt;&quot; has shown some promising results, but even here, this is usually applied at far red wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Except now see this &lt;a href=&quot;https://astronomy.stackexchange.com/questions/26989/how-did-vlts-adaptive-optics-obtain-this-resolution-for-neptune-is-it-really-w&quot;&gt;How did VLT&amp;#39;s adaptive optics obtain this resolution for Neptune? Is it really working in visible wavelengths?&lt;/a&gt; Progress occurs! (Though it now seems my statement above should only be slightly modified, since the picture contains no information shortward of 550nm and is largely dominated by light at 600-920 nm).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-07-18T19:46:01.780" LastActivityDate="2018-07-18T19:46:01.780" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="18862" PostTypeId="2" ParentId="16307" CreationDate="2016-10-26T01:05:53.590" Score="1" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;Adding on to answers how the absence of the moon would affect the Earth's axial tilt (aka 'obliquity'):&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Moon were to magically vanish, we can speculate that Earth's obliquity would change over time, similar to Mars's in that Mars's obliquity can vary from 14 to as much as 60 degrees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mAMiF.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mAMiF.gif&quot; alt=&quot;Obliquity of Mars&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars does have two small satellites, but relative to Mars itself, their masses aren't enough to drastically stabilize Mars. Over time this has affected Mars' seasons and its temperature differences at the poles. If the Moon vanished we can assume that Earth's obliquity would vary in the same way, leading to some wacky seasonal changes. If the Earth were tilted nearly 60 degrees, it's poles would be much more exposed to the Sun possibly leading to the melting of the polar ice caps.&lt;/p&gt;&#xA;" OwnerUserId="11684" LastActivityDate="2016-10-26T01:05:53.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18863" PostTypeId="2" ParentId="12654" CreationDate="2016-10-26T01:15:56.867" Score="5" Body="&lt;p&gt;We can identify two distinct realms of planetesimal formation - the inner Solar System and the outer Solar System. The initial group of small bodies&lt;sup&gt;1&lt;/sup&gt; in the inner section of the protoplanetary disk was rather quickly accreted into planets via various interactions; some bodies grew bigger and thus dominated the disk around them. In the outer parts of the disk, many small bodies were accreted into the cores of the giant planets. However, not all were accreted, and there may in fact have been a second generation of planetesimals that formed via gravitational collapse due to disk instabilities.&lt;/p&gt;&#xA;&lt;p&gt;We can compute timescales for this &amp;quot;cleanup&amp;quot; phase, the period during which small bodies were either accreted onto protoplanets or ejected to larger orbits or out of the Solar System entirely. In the inner section of the disk, among the terrestrial planets, it was on the order of &lt;span class=&quot;math-container&quot;&gt;$\sim10^8$&lt;/span&gt; years, while in the outer section, it was on the order of &lt;span class=&quot;math-container&quot;&gt;$\sim10^9$&lt;/span&gt; years. However, the giant planets may actually have finished &lt;em&gt;accreting&lt;/em&gt; matter before the terrestrial planets.&lt;/p&gt;&#xA;&lt;p&gt;This means that the small bodies that were ejected or boosted to larger orbits were largely from the outer Solar System, terrestrial planetary migration effects aside&lt;sup&gt;2&lt;/sup&gt;. Various N-body simulations agree that a total of &lt;span class=&quot;math-container&quot;&gt;$\sim300M_{\oplus}$&lt;/span&gt; of small bodies was ejected from the outer disk, due to interactions with Jupiter and the formation and evolution of Uranus and Neptune. We know that the Oort Cloud's mass is on the order of &lt;span class=&quot;math-container&quot;&gt;$\sim1$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\sim10M_{\oplus}$&lt;/span&gt;, and the Kuiper Belt and scattered disk have masses an order of magnitude or two lower. Therefore, only a small fraction - certainly less than 10% - of the small bodies could have been ejected into the Oort Cloud.&lt;/p&gt;&#xA;&lt;p&gt;If you want to visualize some of this, check out &lt;a href=&quot;https://www.youtube.com/watch?v=VXeOh3xmrQM&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video from a simulation of the Nice Model&lt;/a&gt;&lt;sup&gt;3&lt;/sup&gt;. It shows the orbits of the four giant planets (Jupiter in red, Saturn in orange, Uranus in purple, and Neptune in blue). It shows Neptune and Uranus swap orbits, and also shows the dispersal of a large group of small bodies outside the original orbit of Uranus (which began further out than Neptune). The video shows the simulation over the course of a bit more than one billion years, and you can see a drastic clearing out of the extreme outer disk over this timescale.&lt;/p&gt;&#xA;&lt;p&gt;The above results came before the development of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Nice_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nice Model&lt;/a&gt; and its variants, as well as the possible discovery of Planet Nine and the discoveries of many small objects in the outer Solar System, beyond Neptune. They should therefore not be taken as totally accurate descriptions of the formation of the Solar System - the video in particular is likely incorrect - and I'm guessing I'll get some flak for using them as the basis of this answer.&lt;/p&gt;&#xA;&lt;p&gt;Obviously, many N-body simulations of the Solar System have been done, and over the last decade, a substantial number use some variant of the Nice Model. Some use planetesimal disks - excluding the masses of the eight planets - on the order of &lt;span class=&quot;math-container&quot;&gt;$\sim10$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\sim10^2M_{\oplus}$&lt;/span&gt;; &lt;a href=&quot;https://arxiv.org/abs/1208.2957&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nesvorny &amp;amp; Morbidelli (2012)&lt;/a&gt; (two of the architects of the model) found a sweet spot, so to speak, of &lt;span class=&quot;math-container&quot;&gt;$20$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$50M_{\oplus}$&lt;/span&gt;, depending on the exact result. As I understand it, masses of the surplus disk around these values fit the Nice Model well, and so perhaps we should drop the total mass of the ejected small bodies by a factor or about ten.&lt;/p&gt;&#xA;&lt;h3&gt;Footnotes&lt;/h3&gt;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; I used this term quite a lot here; it refers to bodies with masses in the planetesimal regime and lower. The reason for this is that a substantial amount of the ejection came after the inner planets had formed, and the remaining bodies were small.&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;2&lt;/sup&gt; This should &lt;em&gt;not&lt;/em&gt; be taken to mean that no or very little matter from the inner Solar System was ejected, especially if we assume some movement among the orbits of the inner planets (see &lt;a href=&quot;http://www.eso.org/public/archives/releases/sciencepapers/eso1614/eso1614a.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Meech et al. (2016)&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;3&lt;/sup&gt; Originally brought to my attention through &lt;a href=&quot;https://physics.stackexchange.com/q/160912/56299&quot;&gt;a question on Physics Stack Exchange&lt;/a&gt; by Kyle Oman.&lt;/p&gt;&#xA;&lt;h3&gt;References&lt;/h3&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://authors.library.caltech.edu/37109/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goldreich et al. (2004)&lt;/a&gt; (early N-body simulations)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www2.mps.mpg.de/homes/natalie/PAPERS/stream1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Krivova &amp;amp; Solanki et al. (2003)&lt;/a&gt; (early N-body simulations)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.eso.org/public/archives/releases/sciencepapers/eso1614/eso1614a.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Meech et al. (2016)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://arxiv.org/abs/1208.2957&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nesvorny &amp;amp; Morbidelli (2012)&lt;/a&gt; (more recent results)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-10-26T15:20:08.327" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18864" PostTypeId="2" ParentId="18856" CreationDate="2016-10-26T03:21:01.937" Score="1" Body="&lt;p&gt;The idea of Trottier &amp;amp; Borra is to analyze signals that might be hidden IN the stellar spectrum. I know this sounds weird; it did sound weird to me when I first heard of it too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know how you can decompose light in diferent colors, right? Well, we astronomers like to think of colors in terms of wavelength (with $~4500$ Angstroms being blue and $~7000$ angstroms being red) or in frequency ($1.5\times 10^6$ GHz and $2.3\times 10^6$ GHz respectively. To get these numbers, just divide the wavelengths by $c=3\times 10^8$ m/s, the speed of light). In the range of wavelengths in which the Sloan Digital Sky Survey (SDSS) works, as you can see, it is much more clear to work in wavelength, but people, for different reasons (like in this paper), can also work in frequency-space rather than in wavelength-space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is that the SDSS has spectra for millions of objects, including stars. That is, they have measurements of the flux as a function of wavelength (color) for these objects. Well, the idea in this paper is actually an idea Ermanno Borra published years ago (2012; here is the paper: &lt;a href=&quot;http://adsabs.harvard.edu/abs/2012AJ....144..181B&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2012AJ....144..181B&lt;/a&gt;). In this paper, he proposes that maybe extraterrestials would like to communicate by injecting extra signals &lt;strong&gt;in that spectra&lt;/strong&gt;, i.e., that they would add extra light in different colors to the stellar spectra we observe at Earth. However, he proposes that maybe they would put colors in frequency-space rather than in wavelength-space (because it is easier to modulate that with, e.g., lasers), and that these color modulations will be periodic in this frequency-space spectra. So, basically, this is what they search for in that spectra: periodic modulations &lt;strong&gt;in the colors&lt;/strong&gt;, but in frequency-space rather than in wavelength. This is what they show, for example, in their Figure 4:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gB1HQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gB1HQ.png&quot; alt=&quot;Plot from Trottier &amp;amp; Borra (2016)&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top panel has an added signal with a periodic modulation that is $10^4$ times larger than the one they detected on the real star (lower panel).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, they don't analyze a time series: they analyze the stellar flux in frequency space. Now, I think I have quite a bunch of experience with stellar spectra, and quite frankly I think this result might just be spurious signals due to instrumental problems and/or data reduction issues. I've seen similar modulations before reducing both high-resolution, mid-resolution and even low-resolution spectra (the process of going from the image to the final spectra easily introduces things like this); I wouldn't be surprised if this is the case too.&lt;/p&gt;&#xA;" OwnerUserId="10852" LastEditorUserId="10852" LastEditDate="2016-10-26T03:29:45.783" LastActivityDate="2016-10-26T03:29:45.783" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18865" PostTypeId="1" CreationDate="2016-10-26T04:57:32.050" Score="1" ViewCount="3414" Body="&lt;p&gt;I'd like to add a feature to one of my apps that calculates the current distance between the Earth and Mars (or any other planet, for that matter). My google searches have led me to statistics of average distances (as a planets orbit is elliptical, and therefore the distance can vary greatly), as well as articles about Keplers laws of Planetary motion and more. I have also seen some mentions of &lt;a href=&quot;http://ssd.jpl.nasa.gov/?horizons&quot; rel=&quot;nofollow&quot;&gt;HORIZONS&lt;/a&gt;, but looking over it I did not see anything to fulfill my questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't found any readings explicitly describing how to calculate the current distance between two planets. So, how is this done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please explain your answer and the steps needed, so that I may understand better. Pseudocode is welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="14675" LastActivityDate="2016-10-26T10:11:11.033" Title="How to calculate the current distance between two planets" Tags="&lt;planet&gt;&lt;solar-system&gt;&lt;distances&gt;&lt;ephemeris&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2016-10-26T13:21:23.767" ContentLicense="CC BY-SA 3.0" />
  <row Id="18866" PostTypeId="1" AcceptedAnswerId="18867" CreationDate="2016-10-26T08:42:37.860" Score="12" ViewCount="1766" Body="&lt;p&gt;I know the Sun is orbiting around the Milky Way, but how strong is the attractive force between them (e.g. what is the order of magnitude in terms of newtons)?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastEditorUserId="2153" LastEditDate="2016-10-26T15:03:19.687" LastActivityDate="2016-10-27T08:28:28.817" Title="How strong is the force between the Sun and the centre of the Milky Way?" Tags="&lt;orbit&gt;&lt;the-sun&gt;&lt;milky-way&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18867" PostTypeId="2" ParentId="18866" CreationDate="2016-10-26T09:27:36.067" Score="11" Body="&lt;p&gt;Galactic orbits are not Keplerian: There is not a single massive centre, whose gravity attracts the sun, rather the whole disc, and dark matter halo which surrounds the galaxy. We can't use the inverse square law to calculate the force of gravity without knowing the distribution of mass in the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, the Sun's orbit is roughly circular, so we can use kinematics to get some idea of the forces involved:  For circular motion $a=\frac{v^2}{r}$. The velocity of the sun is about &lt;a href=&quot;http://hypertextbook.com/facts/2001/AngelaChan.shtml&quot; rel=&quot;noreferrer&quot;&gt;225000 m/s&lt;/a&gt;, and we are at a radius of about 2.5e20 m from the centre. The formula above gives a very small centripetal acceleration of 2e-10 m/s²&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the sun is quite massive, 2e30 kg, so using $F=ma$, the force on the sun is of the order 4e20 N.  This is about 0.01 of the force acting on the Earth by the Sun. (3.6e22 N)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-10-27T08:28:28.817" LastActivityDate="2016-10-27T08:28:28.817" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18868" PostTypeId="2" ParentId="18865" CreationDate="2016-10-26T10:11:11.033" Score="3" Body="&lt;p&gt;There are many different ways of doing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way is to use the Keplerian elements and their rates obtained from the &lt;a href=&quot;http://ssd.jpl.nasa.gov/&quot; rel=&quot;nofollow&quot;&gt;JPL Solar System Dynamics&lt;/a&gt; web site to determine the Cartesian coordinates of the planets in the ecliptic plane at a given time. Information and data to allow computation of approximate positions for the planets is provided in the document &lt;a href=&quot;http://ssd.jpl.nasa.gov/txt/aprx_pos_planets.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Keplerian Elements for Approximate Positions of the Major Planets&quot;&lt;/a&gt; by E.M. Standish.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then for each of the planets, $i$, we have the position coordinates in the form&#xA;$$&#xA;\mathbf{r}_i = &#xA;\left[\begin{array}{l}&#xA;x_i \\ y_i \\ z_i&#xA;\end{array}\right]&#xA;$$&#xA;Since these coordinates are all given in the same reference frame, the displacement vector from planet $i$ to planet $j$ can be obtained from&#xA;$$&#xA;\mathbf{r}_{ij} = &#xA;\left[\begin{array}{l}&#xA;x_{ij} \\ y_{ij} \\ z_{ij}&#xA;\end{array}\right] =&#xA;\left[\begin{array}{l}&#xA;x_j-x_i \\ y_j-y_i \\ z_j-z_i&#xA;\end{array}\right]&#xA;$$&#xA;and the distance between planet $i$ and $j$ can be calculated using&#xA;$$&#xA;r_{ij} = \|\mathbf{r}_{ij}\|&#xA;= \sqrt{x_{ij}^2+y_{ij}^2+z_{ij}^2}&#xA;$$&#xA;The method described in the &lt;a href=&quot;http://ssd.jpl.nasa.gov/txt/aprx_pos_planets.pdf&quot; rel=&quot;nofollow&quot;&gt;document&lt;/a&gt; to obtain the ecliptic coordinates are very easy to implement in software.&lt;/p&gt;&#xA;" OwnerUserId="5537" LastActivityDate="2016-10-26T10:11:11.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18870" PostTypeId="1" CreationDate="2016-10-26T16:34:12.593" Score="2" ViewCount="155" Body="&lt;p&gt;I am trying to use &lt;a href=&quot;http://keplerebs.villanova.edu/&quot; rel=&quot;nofollow&quot;&gt;Kepler Data for Eclipsing Binaries&lt;/a&gt; to estimate time period, and then other parameters such as mass, eccentricity, semi-major axis, distance, etc. of the system. I want to write code in MATLAB which will use FFT. The available data has the following columns:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;bjd(date) | phase | raw_flux | raw_err | corr_flux | corr_err | dtr_flux | dtr_err&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the data require any cleaning?&#xA;If I want to take the FFT of the data to find time period, which data should I use?&lt;/p&gt;&#xA;" OwnerUserId="1981" LastEditorUserId="7982" LastEditDate="2021-04-16T23:25:22.017" LastActivityDate="2021-04-16T23:25:22.017" Title="Which data should I use to take the fast Fourier transform (FFT) to find time period for eclipsing binaries?" Tags="&lt;data-analysis&gt;&lt;binary-star&gt;&lt;eclipsing-star-systems&gt;&lt;time-domain-astronomy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18871" PostTypeId="2" ParentId="10419" CreationDate="2016-10-26T17:32:54.217" Score="4" Body="&lt;p&gt;Normally, the two main sources of cosmic rays are considered to be the Sun (&lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_energetic_particles&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar energetic particles&lt;/a&gt;) and galactic or extragalactic sources, although extragalactic sources are more common. Surprisingly, though, there &lt;em&gt;is&lt;/em&gt; a third source, a family of particles known as &lt;a href=&quot;http://solarphysics.livingreviews.org/Articles/lrsp-2013-3/articlese3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;anomalous cosmic rays&lt;/em&gt;&lt;/a&gt; (see &lt;a href=&quot;http://adsabs.harvard.edu/abs/1973ICRC....2.1498H&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hovestadt et al. (1973)&lt;/a&gt; and &lt;a href=&quot;http://adsabs.harvard.edu/abs/1973ApJ...182L..81G&quot; rel=&quot;nofollow noreferrer&quot;&gt;Garcia-Munoz et al. (1973)&lt;/a&gt;). These are particles - generally ionized nuclei of lighter elements - primarily originating from interstellar space with energies on the order of $\sim10$ to $\sim10^2\text{ MeV}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anomalous cosmic rays are generally accelerated at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Heliosphere#Termination_shock&quot; rel=&quot;nofollow noreferrer&quot;&gt;termination shock&lt;/a&gt; of the heliosphere, moving into the heliosheath. Some may then escape into the inner Solar System. Here's an image from &lt;a href=&quot;http://adsabs.harvard.edu/abs/2005Sci...309.2017S&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stone et al. (2005)&lt;/a&gt;, which shows them as a distinct population from galactic cosmic rays:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zqA6B.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zqA6B.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, these particles are technically of interstellar origin, even though they are accelerated to great speeds only at the boundary of the Solar System. However, a related population may have been found in the Kuiper Belt (see &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1029/2002GL015829/abstract;jsessionid=C5D5BFA51F6A411A0ED76B777325DE20.f02t04&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schwadron et al. (2002)&lt;/a&gt;), formed from grains there after collisions between objects. While these are not necessarily - or even likely - the dominant source of anomalous cosmic rays, they do exist, and thus constitute another source of cosmic rays inside the Solar System.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-11-01T23:51:00.540" LastActivityDate="2016-11-01T23:51:00.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18872" PostTypeId="2" ParentId="18855" CreationDate="2016-10-26T19:18:23.843" Score="5" Body="&lt;p&gt;Yes you can, to some extent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the binary has a short period, then the chances are that the two stars have a different radial velocity. In other words, there are two sets of spectral lines in the spectrum. A cross-correlation or direct fitting technique can then pick the two stars out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the two stars are unresolved and have indistinguishable velocities (true for the majority of binaries), then it is more difficult. A fit to a single spectral template may be poor (and you could judge this from the chi-squared of the fit), but there needs to be a significant difference in the spectra of the two stars in order for a two-star model to adequately separate them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is basically a trade-off. You want the stars to be of similar brightness but to have quite different spectra. This is difficult to arrange for main sequence stars and there is normally a sweet-spot where one star is a bit cooler and a bit dimmer, but can still be seen in the combined spectra. Other combinations are easier - a hot white dwarf and an M dwarf for example have very similar luminosity but very different spectra, such that each dominates in different parts of the spectrum. Others are harder - a red giant and a main sequence star of lower mass.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-10-26T19:18:23.843" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18873" PostTypeId="2" ParentId="18686" CreationDate="2016-10-26T19:48:45.047" Score="2" Body="&lt;p&gt;Sorry I have not enough reputation to comment. I am not going to give a full answer as I am on my phone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess there is a (at least seemingly) neat mathematical explanation for the geometry. But it is simply a result of gravitational collapse. The degree of 'formy-ness' depends on cosmology (for example, need more / less time to form structures depending on cosmology). If you have random distribution of matter in infinite 3d with slight density fluctuation, some patch of the space become denser faster than other patches and become even denser and so on. Every matter pulling each other while clustering themselves results in 3d to 2d, 2d to 1d, 1d to 0d structure of overdensity. So most dense regions are point-like structures (super clusters), and next most dense regions are filamentary, then sheets. The rest are voids. Cosmic expansion changes the strength of gravitational pulling but does not really alter the 'shape' or direction of force, hence little effect on the 'formy-ness' of voids. As long as it is a cosmology with collisionless darkmatter, the same formy structure will form given enough time. &#xA; Cosmological N body simulations use fixed(comoving) volume where only the magnitude of gravitation changes due to the cosmic expansion. And in those simulations you can see (and that's where astronomers learned about) the cosmic structures. There are many movies of cosmological simulations you can easily find. &lt;/p&gt;&#xA;" OwnerUserId="13818" LastEditorUserId="13818" LastEditDate="2016-11-03T07:37:58.913" LastActivityDate="2016-11-03T07:37:58.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18874" PostTypeId="2" ParentId="18870" CreationDate="2016-10-27T03:16:31.103" Score="3" Body="&lt;p&gt;For this, you should create a lightcurve, a graph of brightness over time, to view the data. For Kepler data, the bjd(date) column is the time in &lt;a href=&quot;https://en.wikipedia.org/wiki/Barycentric_Julian_Date&quot; rel=&quot;nofollow&quot;&gt;BJD&lt;/a&gt;. The dtr_flux stands for detrended flux, meaning that it should contain &quot;cleaned&quot; data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I would plot the dtr_flux column over bjd(time) to obtain a lightcurve. Depending on the type of eclipsing binary, you should see a minimum and maximum dip occurring periodically. The data also contains a &quot;phase&quot; column, meaning the data has been &quot;folded&quot; to allow you to plot each cycle of the eclipsing binary on top of each other. See &lt;a href=&quot;http://www.southampton.ac.uk/~sdc1g08/BinningFolding.html&quot; rel=&quot;nofollow&quot;&gt;http://www.southampton.ac.uk/~sdc1g08/BinningFolding.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, you can take the FFT if you wish. Remember that since this is an eclipsing binary with both maximum and minumum brightness dips, the FFT will have a spike both at P and P/2.&lt;/p&gt;&#xA;" OwnerUserId="13085" LastActivityDate="2016-10-27T03:16:31.103" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18875" PostTypeId="1" AcceptedAnswerId="18877" CreationDate="2016-10-27T13:52:36.180" Score="12" ViewCount="6084" Body="&lt;p&gt;We're (basically) in the middle of an arm of our galaxy. That is to say, we're sitting in the middle of a dense disc of stars.&lt;/p&gt;&#xA;&lt;p&gt;It would seem to me that. You should see:&lt;/p&gt;&#xA;&lt;p&gt;the thick line of the milky way &lt;strong&gt;all around you&lt;/strong&gt;, i.e. on that plane, in all four directions.&lt;/p&gt;&#xA;&lt;p&gt;(Additionally - sure - in the particular direction of the galaxy center, you would additionally see the huge central bulge.)&lt;/p&gt;&#xA;&lt;p&gt;However: this does &lt;strong&gt;not seem to be the case&lt;/strong&gt;: when you look at the milky way from the vicinity of our solar system, you basically see it &amp;quot;in one direction&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;What am I misunderstanding?  How come the sky object &amp;quot;the milky way&amp;quot; is well-known as only a lump/strip in one direction, rather than, a lump/strip that goes right around us?&lt;/p&gt;&#xA;&lt;p&gt;--&lt;/p&gt;&#xA;&lt;h1&gt;Put it this way...&lt;/h1&gt;&#xA;&lt;p&gt;Quite simply does anyone have any photography of the anti-galactic point?  (Near &amp;quot;Auriga&amp;quot; right?)  Does it show any &amp;quot;milky way band&amp;quot; going through it?&lt;/p&gt;&#xA;&lt;p&gt;If not why not? Looking outwards, we are still looking through ~30k lightyears of the dense disk we are sitting in.&lt;/p&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-11-03T01:50:57.673" Title="Why don't we see &quot;the milky way&quot; in both directions?" Tags="&lt;milky-way&gt;&lt;night-sky&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18877" PostTypeId="2" ParentId="18875" CreationDate="2016-10-27T17:17:36.803" Score="28" Body="&lt;p&gt;I'll turn &lt;a href=&quot;https://astronomy.stackexchange.com/questions/18875/why-dont-we-see-the-milky-way-in-both-directions?noredirect=1#comment30269_18875&quot;&gt;my comment&lt;/a&gt; into a full answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it simply, we actually do see the Milky Way all around us, even in the diametrically opposite direction from the galactic core. You can see this in the image below which is a full sky image I took from &lt;a href=&quot;https://apod.nasa.gov/apod/ap010202.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;APOD&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MG313.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MG313.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the edges of the disk in that image, you're looking at what is actually the edge of our galaxy, opposite in direction from the core. In effect, this is precisely the image you requested because it does contain the part of the sky which contains the anti-core part of the galaxy. It is certainly not as bright, but there are still stars and dust out there. In fact, if you look very closely, you still see lots of dark splotches that mask distance stars and galaxies, indicating there is dust there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the issue you may be having is that you expect many more stars to be within the outer regions of the disk than there actually are. The stellar density profile for our disk is roughly exponential, meaning there are literally exponentially more stars near the core than at the edges. If this means anything to you, the scale length for the exponential radial density profile is ~4 kpc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To really get a good understanding of the stellar distribution, take a look at &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/523619/pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jurić et al. (2008)&lt;/a&gt;. They looked at (~48 million) stars from the SDSS and analyzed the stellar distribution across our galaxy (that is visible to us). You should find figures 10 through 18 of particular interest, however I'll present part of figure 16 here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QDzUS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QDzUS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image shows the (logarithmic) density of stars as a function radius from the galactic core. The varying shades of grey indicate varying heights above the galactic plane (numbered in parsecs). The dashed lines are various exponential decay models with differing scale heights. You can see that these star densities, even within the limited radial ranges covered by the SDSS fall off by an order of magnitude! Hopefully this helps you appreciate the significant difference between the core brightness/visibility and that of the galactic edge.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's a silver of Andromeda spectacularly demonstrating the falloff: even though galaxy photography tends to suggest to the casual eye an evenly-dense plate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5x6WA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5x6WA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This synthetic image &lt;a href=&quot;https://astronomy.stackexchange.com/questions/18875/why-dont-we-see-the-milky-way-in-both-directions/18877?noredirect=1#comment30288_18875&quot;&gt;referenced by Andy&lt;/a&gt;, from the &lt;a href=&quot;https://svs.gsfc.nasa.gov/3442&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tycho Catalog Skymap&lt;/a&gt; also shows the situation clearly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oN2rx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oN2rx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-10-28T15:57:01.643" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18880" PostTypeId="1" AcceptedAnswerId="18882" CreationDate="2016-10-28T00:25:06.627" Score="7" ViewCount="1581" Body="&lt;p&gt;I am just getting into the field of cosmology and was wondering what it means practically to say that the Hubble Constant is decreasing while the expansion rate of the Universe is accelerating. I am not sure how this phenomenon is possible. How do we observe this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is my understanding that the Hubble Constant is the rate at which the Universe is expanding, and it is decreasing with time. (Although it is asymptotic)... Is this conception flawed?&lt;/p&gt;&#xA;" OwnerUserId="14699" LastActivityDate="2020-09-07T06:52:17.927" Title="How is the Universe's Expansion Accelerating if the Hubble Constant is Decreasing?" Tags="&lt;expansion&gt;&lt;dark-energy&gt;&lt;hubble-constant&gt;" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18882" PostTypeId="2" ParentId="18880" CreationDate="2016-10-28T06:35:37.870" Score="8" Body="&lt;p&gt;The Hubble parameter is defined as the rate of change of the distance between two points in the universe, divided by the distance between those two points.&#xA;The Hubble parameter is getting smaller because the denominator is getting bigger more quickly than the numerator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the future, the cosmological constant, $\Lambda$ could result in an exponential expansion with time. A simple piece of maths shows you that the Hubble parameter is a Hubble constant only for an exponential expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose $H(t) = (da(t)/dt)/a(t)$, where $a(t)$ is the distance between two arbitrary points in the universe at epoch $t$. Now let's have an expansion rate $da/dt \propto t$ (ie accelerating). But in this case $a(t) \propto t^2$ and $H(t) \propto t ^{-1}$ (ie decreasing with time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some extra details: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution to the Friedmann equation in a flat universe is&#xA;$$H^2 = \frac{8\pi G}{3}\rho + \frac{\Lambda}{3},$$&#xA;where $\rho$ is the matter density (including dark matter) and $\Lambda$ is the cosmological constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the universe expands, $\rho$ of course decreases as $a(t)^{-3}$, but $\Lambda$ remains constant. Thus the first term on the RHS becomes unimportant if indeed $\Lambda$ is a cosmological constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the Hubble &quot;constant&quot; actually decreases from its current value $H_0$ and asymptotically tends towards $ H = \sqrt{\Lambda/3}$ as time tends towards infinity.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-10-30T22:07:39.913" LastActivityDate="2016-10-30T22:07:39.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18883" PostTypeId="1" AcceptedAnswerId="18888" CreationDate="2016-10-28T14:38:08.810" Score="-1" ViewCount="988" Body="&lt;p&gt;A friend of mine saw a video that claimed the moon appears to be at the feet of the constellation Virgo about once every 17 years with its next occurrence being in September 2017. Is there any reliable sources that back this up? What I mean by a &quot;reliable source&quot; is an organization that speaks with authority on these matters such as NASA. I would prefer that it is available online, however I understand if it is not. If it is, please provide a direct link to it. If it is not online, please quote verbatim what it says about it, preferably in context. Also, here is a picture of what it might mean for the moon to be at the feet of Virgo (do note that the moon might not be to scale):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/75Hf4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/75Hf4.png&quot; alt=&quot;Moon at Virgo&amp;#39;s feet&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14708" LastEditorUserId="14708" LastEditDate="2016-10-28T18:35:01.470" LastActivityDate="2016-10-28T19:52:13.040" Title="Recurrence of Moon at Virgo's &quot;feet&quot;" Tags="&lt;the-moon&gt;&lt;constellations&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18888" PostTypeId="2" ParentId="18883" CreationDate="2016-10-28T19:52:13.040" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CmDbX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CmDbX.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;I think I understand the question, but don't really have an answer.&lt;/p&gt;&#xA;&lt;p&gt;Mu Virgo, the southernmore of Virgo's two feet, has an ecliptic latitude of 9.678 degrees, meaning it's quite far north of the ecliptic.&lt;/p&gt;&#xA;&lt;p&gt;The moon never ventures more than about 5.15 degrees from the ecliptic, so it will never be north of Mu Virgo, or even really that close.&lt;/p&gt;&#xA;&lt;p&gt;The closest interpretation I can give: in some months, the moon's northernmost ecliptic latitude will occur near Mu Virgo, putting it &amp;quot;close&amp;quot; to Virgo's feet.&lt;/p&gt;&#xA;&lt;p&gt;If you run &lt;a href=&quot;http://ssd.jpl.nasa.gov/?horizons&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ssd.jpl.nasa.gov/?horizons&lt;/a&gt; with these settings:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gRyx7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gRyx7.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;you can get a rough idea of where the moon's ecliptic latitude is maximal for each month. Of course, you can refine these settings to get more accuracy.&lt;/p&gt;&#xA;&lt;p&gt;In theory, you can use &lt;a href=&quot;http://wgc.jpl.nasa.gov:8080/webgeocalc/#NewCalculation&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://wgc.jpl.nasa.gov:8080/webgeocalc/#NewCalculation&lt;/a&gt; to get more exact values of when the moon is furthest north of the ecliptic, but I couldn't get this to quite work out (I didn't spend very long on it).&lt;/p&gt;&#xA;&lt;p&gt;I don't see that the ecliptic latitude's maximal position is that much closer to Mu Virgo in September 2017, but didn't look carefully.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-10-28T19:52:13.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18889" PostTypeId="1" CreationDate="2016-10-29T05:11:06.520" Score="0" ViewCount="96" Body="&lt;p&gt;Let's suppose they have similar technical level like us now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In radio band, what is our earth like? What signals can be linked to non-natural behaviour?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about optical band and other methods?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How large a radio-telescope, 1000 pc away, would be needed to detect Earth's ambient human radio emissions?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="2365" LastEditDate="2017-10-01T12:46:16.330" LastActivityDate="2017-10-01T12:46:16.330" Title="What activities of human beings can be detected by aliens 1000 pc away?" Tags="&lt;seti&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2016-10-31T17:27:48.890" ContentLicense="CC BY-SA 3.0" />
  <row Id="18890" PostTypeId="1" AcceptedAnswerId="18891" CreationDate="2016-10-29T12:18:23.000" Score="7" ViewCount="562" Body="&lt;p&gt;Some telescopes in space have been proposed to function as an interferometer. &#xA;Being placed several hundreds or thousands of kilometers apart they could actual image exoplanets directly. To my knowledge none of these space telescopes have yet been deployed.&lt;BR&gt;&#xA;So I was wondering why we don't connect all the several thousand amateur telescopes on Earth together to create a huge interferometer?&lt;BR&gt;&#xA;That sure would require some efforts (especially software to coordinate it all) but compared to sending telescopes into space it seems outright simple.&lt;BR&gt;&#xA;Are amateur telescopes simply not strong enough, even when linked together? What would be the best we could achieve?&lt;/p&gt;&#xA;" OwnerUserId="14278" LastActivityDate="2021-01-16T21:15:02.030" Title="Why don't we use amateur astronomers' telescopes to create a huge interferometer?" Tags="&lt;exoplanet&gt;&lt;amateur-observing&gt;&lt;interferometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18891" PostTypeId="2" ParentId="18890" CreationDate="2016-10-29T12:47:52.830" Score="11" Body="&lt;p&gt;You can't make an interferometer by combining photographs. You need to combine the light sources so that the light waves can &amp;quot;interfere&amp;quot;, which means you need to have not only the intensity of the light, but also the phase of the wavelength. A CCD only detects the intensity.&lt;/p&gt;&#xA;&lt;p&gt;You also need to position your telescopes to sub-wavelength accuracy. For Radio telescopes, with long wavelengths, it is possible to detect and then combine the radio signals from several telescopes electronically. That's not possible with light. Visible wavelength interferometers capture the light and combine it as light. It requires very high precision.&lt;/p&gt;&#xA;&lt;p&gt;Finally, we still have the atmosphere to contend with. Amateur observers know that, beyond a certain size (about 10in), larger telescopes don't have greater resolution, because of the blurring due to the atmosphere&lt;sup&gt;1&lt;/sup&gt;. Bigger mirrors mean more light, mean dimmer objects can be seen, but no more detail is visible. There are ways of getting round this, (Lucky imaging) but these are not suited to interferometery.&lt;/p&gt;&#xA;&lt;p&gt;So since placing amateur telescopes at nanometer accurate positions, and linking their output with single strand optical fibre is beyond the budget of amateurs, and since the results would be limited by the distortion of the atmosphere, this is not normally done.&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;Large observatories like the &lt;a href=&quot;https://en.wikipedia.org/wiki/Very_Large_Telescope&quot; rel=&quot;noreferrer&quot;&gt;Very Large Telescope&lt;/a&gt; use &lt;a href=&quot;https://en.wikipedia.org/wiki/Adaptive_optics&quot; rel=&quot;noreferrer&quot;&gt;Adaptive optics&lt;/a&gt; to obtain dramatically improved resolution through Earth's atmosphere but that kind of technology is sophisticated and out of reach of amateur instruments.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="7982" LastEditDate="2021-01-16T21:15:02.030" LastActivityDate="2021-01-16T21:15:02.030" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="18892" PostTypeId="1" CreationDate="2016-10-29T13:44:46.783" Score="4" ViewCount="281" Body="&lt;p&gt;When the Gaia parallax data (especially for Cepheid stars) become accessible to public? If it is already released for some parts, where can I find it?&lt;/p&gt;&#xA;" OwnerUserId="13714" LastActivityDate="2016-10-29T18:01:17.293" Title="Gaia Parallax data" Tags="&lt;parallax&gt;&lt;gaia&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18893" PostTypeId="1" CreationDate="2016-10-29T14:03:49.177" Score="3" ViewCount="249" Body="&lt;p&gt;I recently ran into this NASA Hubble image of the Crab Nebula in &lt;a href=&quot;http://www.dailygalaxy.com/my_weblog/2016/10/eerie-glow-of-the-crab-nebulas-neutron-star-chinas-guest-star-of-the-year-1050.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; talking about its core. What I am interested to know though is: what are these round dark spots appearing in the image (upper right, lower center and lower left for example)? &quot;Holes&quot; in the Nebula? Objects in front of the Nebula, visible against its illuminated gas? Blemishes in Hubbles lens(es)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CLARIFICATION: I am asking about the small pixel like dots. Not the large &quot;holes&quot; &lt;a href=&quot;https://astronomy.stackexchange.com/questions/18903/what-are-the-large-round-dark-holes-in-this-nasa-hubble-image-of-the-crab-nebu&quot;&gt;from this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CvZvP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CvZvP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14713" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-10-31T11:45:36.330" Title="What are the dark spots in this Hubble image of the Crab Nebula?" Tags="&lt;hubble-telescope&gt;&lt;nebula&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18895" PostTypeId="2" ParentId="13835" CreationDate="2016-10-29T16:24:58.397" Score="1" Body="&lt;p&gt;I believe that black holes are singularities in the sense that matter in it, is not-defined to have any meaningful location or known velocities. I don't think it means singularity in the sense of having zero radius.&lt;/p&gt;&#xA;" OwnerUserId="2880" LastActivityDate="2016-10-29T16:24:58.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18897" PostTypeId="2" ParentId="18892" CreationDate="2016-10-29T18:01:17.293" Score="6" Body="&lt;p&gt;&lt;a href=&quot;http://www.cosmos.esa.int/web/gaia/dr1&quot;&gt;Gaia data release 1&lt;/a&gt; was announced in September 2016. It has parallaxes for 2 million stars previously observed by the Hipparcos mission, a small fraction of the 1.1 billion positions recorded so far. Future releases will of course have more data and smaller errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gaia has ~600 light curves for Cepheids but probably not parallaxes for most of those.&#xA;&lt;a href=&quot;https://arxiv.org/abs/1609.05175&quot;&gt;Casertano et al. 2016&lt;/a&gt; checked Gaia parallax data of some Cepheids against expected values from ground-based photometry.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-10-29T18:01:17.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18899" PostTypeId="1" CreationDate="2016-10-30T11:33:31.633" Score="1" ViewCount="73" Body="&lt;p&gt;How were interstellar reddening values for cepheid first calculated? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean, other than using the derived equation to get extinction-corrected magnitude of Cepheids, I want to know how to drive them independently from measured raw data (maybe V, B-band colors, etc).&lt;/p&gt;&#xA;" OwnerUserId="13714" LastEditorDisplayName="user1569" LastEditDate="2016-10-31T11:47:02.873" LastActivityDate="2016-10-31T11:47:02.873" Title="Interstellar reddening for Cepheids" Tags="&lt;photometry&gt;&lt;cepheids&gt;&lt;extinction&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18900" PostTypeId="1" AcceptedAnswerId="18902" CreationDate="2016-10-30T12:03:44.620" Score="1" ViewCount="107" Body="&lt;p&gt;I just read the following in the NPR website news article &lt;a href=&quot;http://www.npr.org/sections/thetwo-way/2016/10/30/499751470/nasas-new-intruder-alert-system-spots-an-incoming-asteroid&quot; rel=&quot;nofollow&quot;&gt;NASA's New 'Intruder Alert' System Spots An Incoming Asteroid&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;NASA pays for several telescopes around the planet to scan the skies on a nightly basis, looking for these objects. &quot;The NASA surveys are finding something like at least five asteroids every night,&quot; says astronomer Paul Chodas of JPL.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Which telescopes&lt;/em&gt; are the &quot;several telescopes around the planet to scan the skies on a nightly basis&quot; that NASA pays? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, &lt;em&gt;is it really roughly five new actual asteroids identified per day&lt;/em&gt; (almost 2,000 per year), or is that 5 objects detected, and only some small fraction turn out to actually be unique, new objects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 per day sounds like a lot.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorDisplayName="user1569" LastEditDate="2016-10-31T11:46:26.490" LastActivityDate="2016-10-31T11:46:26.490" Title="Which observatories help NASA find 5 new asteroids per day?" Tags="&lt;asteroids&gt;&lt;observatory&gt;&lt;nasa&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18901" PostTypeId="2" ParentId="18893" CreationDate="2016-10-30T14:42:04.613" Score="2" Body="&lt;p&gt;They are most likely dense clumps of dust absorbing whatever wavelength of light this image was captured in. It's hard to say just how big they are as I can't find any scale for this image, but generally nebulae are lightyears massive, with the Crab Nebula according to wiki being 11 light years. So whatever they are, they're still pretty big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Just look at the horizontal band of darker lumps, you can tell that they are between us and the brighter region. The band would be of the same material as the bright green cloud, just less energised as it is further away from the star.&lt;/p&gt;&#xA;" OwnerUserId="11525" LastActivityDate="2016-10-30T14:42:04.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18902" PostTypeId="2" ParentId="18900" CreationDate="2016-10-30T15:01:07.443" Score="7" Body="&lt;p&gt;Given that the article seems to be referring to NASA's &lt;a href=&quot;http://neo.jpl.nasa.gov/index.html&quot; rel=&quot;noreferrer&quot;&gt;Near Earth Object Program&lt;/a&gt;, it appears that there are &lt;a href=&quot;http://neo.jpl.nasa.gov/programs/&quot; rel=&quot;noreferrer&quot;&gt;five subprograms&lt;/a&gt; scanning the skies:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://neo.jpl.nasa.gov/programs/css.html&quot; rel=&quot;noreferrer&quot;&gt;Catalina Sky Survey (CSS)&lt;/a&gt; (four telescopes: 0.50 m, 0.68 m, 1.00 m, 1.50 m)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://neo.jpl.nasa.gov/programs/panstarrs.html&quot; rel=&quot;noreferrer&quot;&gt;Pan-STARRS&lt;/a&gt; (one telescope: 1.80 m)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://neo.jpl.nasa.gov/programs/linear.html&quot; rel=&quot;noreferrer&quot;&gt;LINEAR&lt;/a&gt; (two telescopes: 1.0 m (retired), 3.5 m)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://neo.jpl.nasa.gov/programs/spacewatch.html&quot; rel=&quot;noreferrer&quot;&gt;Spacewatch&lt;/a&gt;  (two telescopes: 0.9 m, 1.8 m)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://neo.jpl.nasa.gov/programs/neowise.html&quot; rel=&quot;noreferrer&quot;&gt;NEOWISE&lt;/a&gt; (one telescope - WISE: 0.4 m)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.jpl.nasa.gov/news/news.php?feature=6664&quot; rel=&quot;noreferrer&quot;&gt;JPL&lt;/a&gt;, CSS and Pan-STARRS are responsible for 90% of new Near Earth Object (NEO) discoveries. LINEAR is responsible for detecting objects one kilometer in diameter or greater, while NEOWISE provides additional measurements from previous &lt;a href=&quot;http://www.nasa.gov/mission_pages/WISE/main/index.html&quot; rel=&quot;noreferrer&quot;&gt;WISE&lt;/a&gt; targets, so it isn't making new detections, merely follow-up observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://neo.jpl.nasa.gov/stats/&quot; rel=&quot;noreferrer&quot;&gt;NASA's statistics&lt;/a&gt; show that over 15,000 Near-Earth Asteroids (NEAs) have been discovered so far, from all programs scanning the sky. In each of the past three years, about 1,500 NEAs have been discovered, most by the five programs listed above. That's roughly consistent with the five-per-day claim. However, it's not yet near 2,000 per year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ReZiY.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ReZiY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-10-30T19:10:58.767" LastActivityDate="2016-10-30T19:10:58.767" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18903" PostTypeId="1" AcceptedAnswerId="18904" CreationDate="2016-10-30T16:38:57.443" Score="14" ViewCount="1237" Body="&lt;p&gt;I came across this image of the Crab Nebula taken from NASA Hubble telescope. What are the large round &quot;holes&quot; and how are they formed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/acNS5.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/acNS5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14713" LastActivityDate="2016-10-30T16:50:38.707" Title="What are the large round dark &quot;holes&quot; in this NASA Hubble image of the Crab Nebula?" Tags="&lt;hubble-telescope&gt;&lt;nebula&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18904" PostTypeId="2" ParentId="18903" CreationDate="2016-10-30T16:50:38.707" Score="12" Body="&lt;p&gt;I think &lt;a href=&quot;https://astronomy.stackexchange.com/a/18898/2153&quot;&gt;my deleted answer&lt;/a&gt; to &lt;a href=&quot;https://astronomy.stackexchange.com/q/18893/2153&quot;&gt;your previous question&lt;/a&gt; covers this well, so I'll add it here.&lt;/p&gt;&#xA;&lt;p&gt;These two spots are known as &lt;strong&gt;the east and west bays&lt;/strong&gt; of the Crab Nebula. They appear to be the result of a torus partially encircling a section of the nebula. The pulsar's magnetic field interacts with the gas and dust in the torus, which blocks &lt;a href=&quot;https://en.wikipedia.org/wiki/Synchrotron_radiation&quot; rel=&quot;noreferrer&quot;&gt;synchrotron radiation&lt;/a&gt; from gases in the nebula. It has been conjectured that the torus is the result of a &lt;span class=&quot;math-container&quot;&gt;$\sim0.1M_{\odot}$&lt;/span&gt; ejection of material from a circumstellar disk prior to the supernova, based on studies of the proper motion of gas on the edges of both bays.&lt;/p&gt;&#xA;&lt;h3&gt;References&lt;/h3&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://adsabs.harvard.edu/full/1992ApJ...399..599F&quot; rel=&quot;noreferrer&quot;&gt;Fesen et al. (1992)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://authors.library.caltech.edu/53632/&quot; rel=&quot;noreferrer&quot;&gt;Hester et al. (1995)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;We can also rule out that this is the result of a defect in Hubble, because these features have been observed in &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Crab_Nebula_in_Multiple_Wavelengths.png&quot; rel=&quot;noreferrer&quot;&gt;different wavelengths&lt;/a&gt;, including infrared, visible, ultraviolet, and x-ray (though not in radio or gamma-ray wavelengths):&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wVPyR.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wVPyR.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;Image courtesy of Wikipedia user Hunster under &lt;a href=&quot;https://creativecommons.org/licenses/by-sa/3.0/deed.en&quot; rel=&quot;noreferrer&quot;&gt;the Creative Commons Attribution-Share Alike 3.0 Unported license&lt;/a&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;The infrared, ultraviolet, and x-ray images come from the Spitzer Space Telescope, the SWIFT observatory, and the Chandra observatory, respectively.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-10-30T16:50:38.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18905" PostTypeId="1" AcceptedAnswerId="18906" CreationDate="2016-10-30T17:59:06.600" Score="8" ViewCount="701" Body="&lt;p&gt;What I'm looking for is a graphic that shows in a general way the best available telescope resolution vs wavelength throughout the entire wavelength spectrum. So for example, there might be two very high resolution peaks aroud &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;millimeter wavelengths (&lt;a href=&quot;https://en.wikipedia.org/wiki/Atacama_Large_Millimeter_Array&quot; rel=&quot;noreferrer&quot;&gt;ALMA&lt;/a&gt;) &lt;/li&gt;&#xA;&lt;li&gt;visible wavelengths (&lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble_Space_Telescope&quot; rel=&quot;noreferrer&quot;&gt;HST&lt;/a&gt; and many ground telescopes with &lt;a href=&quot;https://en.wikipedia.org/wiki/Adaptive_optics&quot; rel=&quot;noreferrer&quot;&gt;Adaptive Optics&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The beautiful image below from &lt;a href=&quot;https://i.stack.imgur.com/wVPyR.jpg&quot; rel=&quot;noreferrer&quot;&gt;this great answer&lt;/a&gt; got me thinking. I have reposted it from there.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wVPyR.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wVPyR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;  &lt;br&gt;&#xA;  &lt;sup&gt;Image courtesy of Wikipedia user Hunster under &lt;a href=&quot;https://creativecommons.org/licenses/by-sa/3.0/deed.en&quot; rel=&quot;noreferrer&quot;&gt;the Creative Commons Attribution-Share Alike 3.0 Unported license&lt;/a&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The infrared, ultraviolet, and x-ray images come from the Spitzer Space Telescope, the SWIFT observatory, and the Chandra observatory, respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2016-10-31T16:41:02.127" Title="Plot of best available resolution vs wavelength - radio through gamma rays?" Tags="&lt;telescope&gt;&lt;radio-astronomy&gt;&lt;angular-resolution&gt;&lt;x-ray&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18906" PostTypeId="2" ParentId="18905" CreationDate="2016-10-30T19:39:45.883" Score="8" Body="&lt;p&gt;After a bit of searching, I found &lt;a href=&quot;http://armchairastronautics.blogspot.com/p/space-observatories.html&quot; rel=&quot;noreferrer&quot;&gt;this blog page&lt;/a&gt;, which has several charts about various observatories, including this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tHg2C.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tHg2C.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;Image courtesy of Olaf Frohn under &lt;a href=&quot;https://creativecommons.org/licenses/by-sa/4.0/&quot; rel=&quot;noreferrer&quot;&gt;the Creative Commons Attribution-Share Alike 4.0 License&lt;/a&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The majority are space-based, although the radio telescopes are largely land-based. They cover existing and future telescopes, at energies from the gamma-ray spectrum to radio waves. You are correct, too, in assuming that adaptive optics can cause dramatic increases in angular resolution; &lt;a href=&quot;https://en.wikipedia.org/wiki/CHARA_array&quot; rel=&quot;noreferrer&quot;&gt;CHARA&lt;/a&gt; and the &lt;a href=&quot;https://en.wikipedia.org/wiki/European_Extremely_Large_Telescope&quot; rel=&quot;noreferrer&quot;&gt;European Extremely Large Telescope&lt;/a&gt; both use adaptive optics, and actually can have better angular resolutions than some space-based telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I annotated the graph to cover in green the smallest angular resolution at various wavelengths:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Rl7rJ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rl7rJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that most of the lines in the radio, microwave, and infrared part of the spectrum are diagonal, with roughly the same slope. This is because they are &lt;a href=&quot;https://en.wikipedia.org/wiki/Diffraction-limited_system&quot; rel=&quot;noreferrer&quot;&gt;limited by diffraction&lt;/a&gt;. In the case of radio waves, this is because the atmosphere has little impact. In the case of infrared- and visible- wavelength telescopes in space - and in space-based telescopes in general, the main thing that stops them is the diffraction limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diffraction limit is&#xA;$$d=\frac{\lambda}{2n\sin\theta}$$&#xA;where $\lambda$ is wavelength and $n\sin\theta$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Numerical_aperture&quot; rel=&quot;noreferrer&quot;&gt;numerical aperture&lt;/a&gt;. On a log-log plot, such as the one above, we have&#xA;$$\log d=\log\lambda-\log(2n\sin\theta)$$&#xA;and&#xA;$$\frac{\mathrm{d}\log d}{\mathrm{d}\log\lambda}=1$$&#xA;for all telescopes limited by the equation. Thus, telescopes restricted by this limit should be described by a diagonal line with a slope of 1 (-1 on this graph).&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-10-31T16:41:02.127" LastActivityDate="2016-10-31T16:41:02.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18907" PostTypeId="2" ParentId="18900" CreationDate="2016-10-30T21:53:33.980" Score="1" Body="&lt;p&gt;Some days there are 10 or more&#xA;&lt;a href=&quot;http://www.minorplanetcenter.net/mpec/RecentMPECs.html&quot; rel=&quot;nofollow&quot;&gt;Minor Planet Electronic Circulars&lt;/a&gt;&#xA;announcing new asteroid discoveries or lost asteroid recoveries.&#xA;The October 30 MPECs included a follow-up&#xA;&lt;a href=&quot;http://www.minorplanetcenter.net/mpec/K16/K16UA8.html&quot; rel=&quot;nofollow&quot;&gt;observation of 2016 UR36&lt;/a&gt;,&#xA;the same object mentioned in the NPR story.&#xA;With an arc of only a few days, its orbit remains highly uncertain.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-10-30T21:53:33.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18908" PostTypeId="1" AcceptedAnswerId="18910" CreationDate="2016-10-31T04:54:58.573" Score="2" ViewCount="1322" Body="&lt;p&gt;Why do main sequence stars more massive than the Sun have lower densities? e.g. Vega, Spica etc.&lt;/p&gt;&#xA;" OwnerUserId="14723" LastActivityDate="2016-10-31T13:47:10.953" Title="Why are main sequence stars more massive than the Sun less dense? e.g. Vega, Spica etc" Tags="&lt;stellar-astrophysics&gt;&lt;main-sequence&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18909" PostTypeId="1" AcceptedAnswerId="18911" CreationDate="2016-10-31T11:39:38.160" Score="-2" ViewCount="187" Body="&lt;p&gt;If a star accretes enough mass it collapses into a black hole (I am not a physicist or an astronomer so I guess this is the simplified version).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what would happen if a forming planet continued to accrete mass (*)?  What transformations would it undergo as it mass increases? Would it become a star? Would it still be a planet until it becomes a black hole? Or something else entirely?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the answer is different depending if the planet is solid or gas, please consider and explain both possibilities.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(*) hypothetically, let's ignore other problems like how would it do that, or if it would be able to maintain orbit or fall into the star, etc.&lt;/p&gt;&#xA;" OwnerUserId="14725" LastEditorUserId="2232" LastEditDate="2016-11-01T05:44:06.023" LastActivityDate="2016-11-01T05:44:06.023" Title="What would happen if a planet would acumulate mass continously?" Tags="&lt;star&gt;&lt;planet&gt;&lt;black-hole&gt;&lt;mass&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18910" PostTypeId="2" ParentId="18908" CreationDate="2016-10-31T12:40:12.450" Score="5" Body="&lt;p&gt;Main sequence stars are defined by being hot enough in the core to fuse hydrogen, so their core is at least about 10 million K, and can get up to 20 million K for the more massive ones (because they are more luminous, so their nuclear fusion has to crank itself up a bit more).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way they get their cores hot enough to fuse hydrogen is via gravitational contraction, but a more massive star does not need to contract as much to reach the necessary temperature.   Since it does not need to contract as much, it is at lower density.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see the scaling approximately, use the virial theorem, which says that the energy per particle in the core (roughly set by the need for fusion) is about $GMm/R$, where $M$ is the mass of the star, and $m$ is the mass of a proton.  Expressed in terms of density $\rho$ instead of radius $R$, that's proportional to $M^{\frac{2}{3}}\rho^{\frac{1}{3}}$.  Ergo, $\rho$ drops when $M$ rises.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="2365" LastEditDate="2016-10-31T13:47:10.953" LastActivityDate="2016-10-31T13:47:10.953" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18911" PostTypeId="2" ParentId="18909" CreationDate="2016-10-31T16:03:28.703" Score="4" Body="&lt;p&gt;It only matters if that mass is iron or not. The state doesn't really matter as it will turn to plasma sooner or later due to gravitational energy transformed to heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a planet continues to accrete mass, if it would be any element before iron in the periodic table, it will - given enough mass - become a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is iron, it will collapse down to a neutron star or black hole if the mass is enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is anything heavier than iron, it either does the same as with iron or will go critical and explode in a nuclear fission reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since most mass in the universe however is either hydrogen or helium, usually the first thing is what happens.&lt;/p&gt;&#xA;" OwnerUserId="14441" LastActivityDate="2016-10-31T16:03:28.703" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18914" PostTypeId="2" ParentId="18880" CreationDate="2016-10-31T19:33:09.433" Score="2" Body="&lt;p&gt;It really comes down to what you define as accelerated expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually accelerated expansion is taken to mean that the first deriviative of the scale factor $a'(t)$ is increasing. However the Hubble parameter is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H \equiv \frac{a'(t)}{a(t)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in expanding Universe $a(t)$ is increasing, it is possible for $a'(t)$ to be increasing, but $H$ to be decreasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose it could be argued that an increasing $H$ (i.e. some kind of phantom energy scenario) is what we should truly reserve the term &quot;accelerated expansion for. E.g. you could argue that a de Sitter Universe where $H$ is a constant, and therefore also any Universe where $H$ is decreasing, shouldn't be described as as &quot;accelerated expansion&quot;, as there is effectively no difference between the expansion at some time $t$ and some later time. However we don't go with this definition.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastActivityDate="2016-10-31T19:33:09.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18918" PostTypeId="1" AcceptedAnswerId="19024" CreationDate="2016-11-01T04:29:48.360" Score="6" ViewCount="330" Body="&lt;p&gt;When we say the sun takes 230M years to orbit the galaxy, what is this in relation to? We measure the earth's rotation relative to the distant stars. What is the reference for our Sun's motion around the galaxy?&lt;/p&gt;&#xA;" OwnerUserId="14723" LastActivityDate="2016-11-11T10:53:17.790" Title="When we say the sun takes 230M years to orbit the galaxy, what is this in relation to?" Tags="&lt;the-sun&gt;&lt;galaxy&gt;&lt;galaxy-center&gt;" AnswerCount="1" CommentCount="13" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18919" PostTypeId="1" AcceptedAnswerId="18920" CreationDate="2016-11-01T06:10:39.100" Score="0" ViewCount="342" Body="&lt;p&gt;Is it possible that a planet originally full of water, and the solar winds decompose the water and hydrogen escapes out, then the oxygen combines with carbon at crust, just like Venus but the whole planets are mostly carbon dioxide and hence become a gas giant?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastActivityDate="2016-11-01T08:23:05.383" Title="Is carbon dioxide gas giants possible?" Tags="&lt;gas-giants&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18920" PostTypeId="2" ParentId="18919" CreationDate="2016-11-01T08:23:05.383" Score="6" Body="&lt;p&gt;While not absolutely impossible this is not a likely situation. The reason is that for a planet to be a &quot;giant&quot;, then it will have enough gravity to hold on to its hydrogen and helium. Conversely, if you remove the H an He from a planet, it won't be &quot;giant&quot;. There are only two gasses that are common enough in the universe to form a massive gas mantle, and those are Hydrogen and Helium.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-11-01T08:23:05.383" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18922" PostTypeId="1" CreationDate="2016-11-01T23:09:09.157" Score="3" ViewCount="149" Body="&lt;p&gt;I'm looking for a function which gives the typical &lt;em&gt;isotropic&lt;/em&gt; spectral power of light emitted by an entire galaxy on the whole electromagnetic spectrum (in watts per frequency unit), as a function of frequency:&#xA;$$\tag{1}&#xA;\mathrm{d}\mathcal{P} = \mathcal{Q}(\omega) \, \mathrm{d}\omega = \; ?&#xA;$$&#xA;The total bolometric power emitted (in watts) is then&#xA;$$\tag{2}&#xA;\mathcal{P}_{\text{tot}} = \int_0^{\infty} \mathcal{Q}(\omega) \, \mathrm{d}\omega.&#xA;$$&#xA;What would be the function $\mathcal{Q}(\omega)$, for an &quot;ideal&quot; (theoretical) spherical galaxy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a candidate, how can we justify a function like the following one, where $\alpha$ and $\omega_0$ are two adjustable positive parameters, and $\mathcal{P}$ is a constant (the total bolometric power)?&#xA;$$\tag{3}&#xA;\mathcal{Q}(\omega) = \frac{\mathcal{P}}{\Gamma(\alpha + 1)} \; \Big( \, \frac{\alpha \, \omega}{\omega_0} \, \Big)^{\alpha} \; e^{-\, \alpha \, \omega / \omega_0} \; \frac{\alpha}{\omega_0}.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="14736" LastEditorUserId="9301" LastEditDate="2016-11-01T23:28:42.930" LastActivityDate="2016-11-01T23:28:42.930" Title="What is the spectral power output of a typical galaxy?" Tags="&lt;galaxy&gt;&lt;cosmology&gt;&lt;astrophysics&gt;&lt;luminosity&gt;&lt;radiation&gt;" AnswerCount="0" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18923" PostTypeId="1" AcceptedAnswerId="18985" CreationDate="2016-11-02T06:02:32.640" Score="3" ViewCount="129" Body="&lt;p&gt;Is it possible that meteoroids, comets, asteroids or other small bodies from around a nearby star (such as Alpha Centauri) could get into the solar system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How low is that possibility?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Oort Cloud is not being gravitationally attracted enough to the Sun to form a disk, I assume this is the same for Alpha Centauri's Oort cloud. Therefore, my question is: can the Sun gravitationally pull a comet from Alpha Centauri's Oort cloud?  &lt;/p&gt;&#xA;" OwnerUserId="11621" LastEditorUserId="6" LastEditDate="2016-11-02T16:19:04.760" LastActivityDate="2016-11-07T22:13:05.673" Title="Are exo-meteoroids possible?" Tags="&lt;asteroids&gt;&lt;comets&gt;&lt;meteoroid&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18924" PostTypeId="1" CreationDate="2016-11-02T06:53:00.587" Score="2" ViewCount="172" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Rogue_planet#Known_or_possible_rogue_planets&quot; rel=&quot;nofollow noreferrer&quot;&gt;The list of known rogue planets&lt;/a&gt; provided by Wikipedia is fairly short. Some are &quot;candidates&quot;; others &quot;may be&quot; red dwarfs. The two confirmed rogue planets are several times the size of Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that rogue planets can only be detected if they glow (which immediately brings the phrase &quot;red dwarf&quot; to mind) or if they cross the path of light emanating from a star. This makes detection extremely difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I ask is I'm not quite certain whether objects the size of Jupiter or smaller can even exist in interstellar space. Maybe dark matter grinds them to dust, or maybe they implode in the absence of any meaningful inertial frame - I don't know. Has anything Earth-sized been detected out there?&lt;/p&gt;&#xA;" OwnerUserId="9510" LastEditorUserId="2365" LastEditDate="2016-11-02T08:15:38.740" LastActivityDate="2016-11-04T12:45:27.337" Title="Has the existence of earth-sized rogue planets been confirmed?" Tags="&lt;rogue-planet&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18925" PostTypeId="2" ParentId="18924" CreationDate="2016-11-02T08:35:05.190" Score="3" Body="&lt;p&gt;The list on Wikipedia looks current. It doesn't list planets that have only been detected as &lt;a href=&quot;http://www.nature.com/news/2011/110518/full/news.2011.303.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;one-off microlensing events.&lt;/a&gt; This can detect Jupiter sized objects (with considerable uncertainty in mass). But observing a microlensing doesn't allow for study, so a microlensing is not considered to be enough evidence for naming or cataloging a known planet. So the short answer is no Earth sized rogue planets are known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is entirely due to the difficulty in detection. An Earth sized rogue planet would cool on its surface to a few degrees above absolute zero, and be undetectable with current telescopes. Microlensing events with small rogue planets would also be smaller and harder to detect above noise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An Earth sized rogue planet has a perfectly good inertial frame, (it is the bound planets that are in a rotating, non-inertial frame), nor does a frame of reference have anything to do with implosions. Dark matter is doesn't grind anything since the main property of Dark matter is that it doesn't interact with regular matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all likelihood, there are billions of rogue planets in the galaxy, and some are smaller, and some are Earth-sized and &quot;rocky&quot;. They could have been formed around a star and then ejected.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-11-02T08:35:05.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18926" PostTypeId="1" CreationDate="2016-11-02T09:15:34.533" Score="1" ViewCount="179" Body="&lt;p&gt;I'm in doubt whether I should put this in the astronomy or rather the coding exchange, but since my interest goes out to the former I will try my luck here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I have an image with a certain pixel scale $x$. I would like to rebin this to a larger pixel scale, $y$, but this is not an integer multiple - so I can't say, for instance, that every group of 4 pixels in the original grid will now be one big pixel in the new grid, and average their flux.&#xA;Rather, I'd like something like i.e. $y=2.3\cdot x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this routinely done in the astronomical community -- and if so, how?&lt;/p&gt;&#xA;" OwnerUserId="6965" LastActivityDate="2016-12-06T19:04:10.793" Title="Rebinning with a non-integer factor" Tags="&lt;data-analysis&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18927" PostTypeId="2" ParentId="18926" CreationDate="2016-11-02T14:45:20.913" Score="3" Body="&lt;p&gt;What you're trying to do is change the pixel scale, but ultimately, you need to decrease the resolution of your image. I'm quite sure this can be done using a variety of image editing tools (even some free ones), but if you insist on doing this yourself, I can show you the general process. Let's say I have the image below, which for convenience is only 6x6 pixels in size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mUEvT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mUEvT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll assume this image has a pixel scale of $p = 1\:\frac{''}{\text{pix}}$ and I want to convert it to an image with a pixel scale of $p'=1.5\:\frac{''}{\text{pix}}=1.5p$. This is effectively saying that I want the resolution of the image to decrease by a factor of $1.5$ and that each pixel in the $p'$ image will be comprised of $1.5^2$ pixels from the $p$ image. This is easily done by defining the new pixel grid over the old one and determining the new pixel &quot;color&quot; as the weighted mean of the old pixels which exist in your new, larger pixel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BcKjk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BcKjk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For convenience, I'm going to define a coordinate system where the pixels are at position $(x,y)$ where $(0,0)$ is the top left pixel and $x$ and $y$ increase to the right and down respectively. This is a common coordinate system for images in programming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at pixel $(0,0)$ in the $p'$ image. This is comprised of pixels $(0,0)$, $(1,0)$, $(0,1)$, and $(1,1)$ from the $p$ image. Specifically, the fractions of these pixels in our new pixel are $1$, $0.5$, $0.5$, and $0.25$ respectively. These fractions will become the weights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you need to do then is look at your four pixels in the $p$ image and determine their color/brightness/property-you're-trying-to-merge. For this simple case, I'll just use a number from zero (white) to one (black) that represents the magnitude of the pixel, but the same process applies no matter what your pixel information contains, be it an RGB value or whatever else. By looking at my four pixels in the $p$ image, I see the magnitudes are $0.05$, $0.05$, $0.25$, and $0.15$ respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The new pixel's magnitude is then the weighted average of these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p'_{00,\text{mag}} = \frac{1\times p_{00,\text{mag}} + 0.5\times p_{10,\text{mag}} + 0.5\times p_{01,\text{mag}} + 0.25\times p_{11,\text{mag}}}{1.5^2} = \frac{1\times 0.05 + 0.5\times 0.05 + 0.5\times 0.25 + 0.25\times 0.15}{1.5^2} = \boxed{0.1056}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process is the same for each pixel in the $p'$ image. Hopefully you can see how this process would generalize for any new pixel scale. The hard part is determining where your new pixels will be and which pixels of your original image will comprise the new pixel. Once you know that, you simply average them together, weighted by the fraction of the original pixels that make up the new pixel. With a little bit of effort, this could all be fairly easily automated by a program and you'd only need to input the new pixel scale and the original image and it could output a new image at the new pixel scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One problem you might run into is that your new image can't be subdivided perfectly. E.g., if, in the above example, you wanted a new scale of $1.4$, you'd still wind up with a new $p'$ image of 4x4 pixels, but you'd only be able to cover 5.6x5.6 pixels of the original image and you'd necessarily have to cut off part of your image, thus losing some information. It'd be up to you to decide what you cut off.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="9301" LastEditDate="2016-12-06T19:04:10.793" LastActivityDate="2016-12-06T19:04:10.793" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18928" PostTypeId="2" ParentId="18541" CreationDate="2016-11-02T15:00:12.847" Score="3" Body="&lt;p&gt;It's not proof that they've ejected other inner planets, because there are plenty of other explanations for why we haven't observed companions. &lt;a href=&quot;https://arxiv.org/abs/1205.2309&quot; rel=&quot;nofollow noreferrer&quot;&gt;Steffen et al. (2012)&lt;/a&gt; analyzed Kepler data - likely some of the same examples you've looked at - and came up with several explanations besides the no-companions-because-of-planet-planet scattering hypothesis:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Inner planets never formed.&lt;/strong&gt; It's possible that mechanisms in systems with hot Jupiters prevent inner planets from forming; the authors are unclear as to what those mechanisms might be.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;The planets are too small to see during transits.&lt;/strong&gt; This is the first explanation based on detection bias. Obviously, &lt;em&gt;Kepler&lt;/em&gt; has its limits, and it's entirely possible that the systems have small bodies it simply can't detect.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;The planets are too low-mass.&lt;/strong&gt; It is possible that there may be planets with masses so low that &lt;a href=&quot;https://en.wikipedia.org/wiki/Transit-timing_variation&quot; rel=&quot;nofollow noreferrer&quot;&gt;transit-timing variations (TTVs)&lt;/a&gt; are too small for &lt;em&gt;Kepler&lt;/em&gt; to see them.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;The planets have high mutual inclinations.&lt;/strong&gt; In other words, there might be planets with orbits highly inclined with respect to the hot Jupiters' orbits; we only detect the ones at inclinations favorable to detection from our angle.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The options the authors suggest rest mainly on experimental biases as opposed to theoretical possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something I was confused about when I read your question was the assumption that hot Jupiters migrate mainly due to planet-planet scattering. &lt;a href=&quot;http://w.astro.berkeley.edu/~kalas/disksite/library/levison06a.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Levison et al.&lt;/a&gt; argue that &lt;em&gt;warm&lt;/em&gt; Jupiters (i.e. giant planets with semi-major axes of about 1 AU) are more likely to have migrated via planet-planet scattering, because it provides a stopping mechanism via damping, assuming the planetesimal number density scales appropriately. This happens when the energy needed to move planetesimals from their orbits is greater than the energy lost from the change in the planet's orbit. Given that planetesimals cannot survive long in orbits less than about twice the stellar radius, warm Jupiters reach semi-major axes no lower than 0.03 to 0.1 AU (see &lt;a href=&quot;http://science.sciencemag.org/content/279/5347/69.full&quot; rel=&quot;nofollow noreferrer&quot;&gt;Murray et al. (1998)&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hot Jupiters, on the other hand, may be driven largely by &lt;a href=&quot;http://www.scholarpedia.org/article/Planetary_formation_and_migration#Gas_disk_migration&quot; rel=&quot;nofollow noreferrer&quot;&gt;Type II gas disk migration&lt;/a&gt;, where giant planets create a gap in a protoplanetary disk that subsequently brings in material; this then brings the planet closer to the star, eventually leading to a hot Jupiter. Here's a visualization, from &lt;a href=&quot;https://blog.planethunters.org/2014/05/09/the-role-of-planetary-migration-in-the-evolution-of-the-solar-system/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planet Hunters&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xLN7a.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xLN7a.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-11-03T18:07:19.710" LastActivityDate="2016-11-03T18:07:19.710" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18930" PostTypeId="2" ParentId="13562" CreationDate="2016-11-02T19:07:54.183" Score="7" Body="&lt;p&gt;Estimates of Planet 9's orbit currently suggest that it doesn't have any major effect on the Kuiper Belt, because its orbit is too far away and on a different plane than the Kuiper belt, and from the inner Oort cloud, at 200 and 1200 aphelion and perihelion, However it does have an effect on Trans-Neptunian objects that transit through the Kuiper belt. It doesn't get closer than 4 times the distance of Pluto and it transits at a different plane to the Kuiper belt.&lt;/p&gt;&#xA;&lt;p&gt;Shepherding is the act of actually creating and directing the asteroid belt by providing many objects with a semi stable orbit with too violent collisions to make a planet (biggest is &lt;a href=&quot;https://en.wikipedia.org/wiki/Ceres_(dwarf_planet)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ceres&lt;/a&gt; at &lt;span class=&quot;math-container&quot;&gt;${\rm 500 km}$&lt;/span&gt; diameter), and having a massive effect on the unstable collection of asteroids, like making large stable groups around Jupiter's Lagrangian points at 60 degrees to Jupiter, and making fast resonant orbits that are not round like stable asteroids but periodically go close to Jupiter and then go straight through the inner asteroid belt and back to towards Jupiter actually in a triangle. &lt;div class=&quot;youtube-embed&quot;&gt;&lt;div&gt;&#xD;&#xA;                &lt;iframe width=&quot;640px&quot; height=&quot;395px&quot; src=&quot;https://www.youtube.com/embed/yt1qPCiOq-8?start=0&quot;&gt;&lt;/iframe&gt;&#xD;&#xA;            &lt;/div&gt;&lt;/div&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WkkcP.png&quot; alt=&quot;https://upload.wikimedia.org/wikipedia/en/timeline/5fb1322f537f8a55d85170976c150191.png&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Neptune's orbit averages 30AU, and the Kuiper belt is generally held to be between Neptune and Pluto's orbits, Pluto orbits at 31-48AU.&lt;/p&gt;&#xA;&lt;p&gt;In that sense, Pluto and Neptune are the two objects which have the biggest effect on the Kuiper belt.&lt;/p&gt;&#xA;&lt;p&gt;The Trans-Neptunian objects all traverse the Kuiper belt, with their Perihelions ranging from 17 to 50 AU, and if Trans-Neptunian objects swing out towards Planet 9, it's a less stable orbit for them, than if they swing out away from planet 9, so all the large Kuiper belt objects swing away from the alledged direction of planet 9, where they can have a stable course without a transiting planet to interfere with them. Trans Neptunians tend to not align themselves with Planet 9's orbit and tend to find a balance away from Planet 9 and away from Jupiter and the majority of them don't get closer than Neptune, about 7/8ths of them are further than Neptune. &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_trans-Neptunian_objects&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/List_of_trans-Neptunian_objects&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The Trans-Neptunian objects are often referred to as Kuiper belt objects, and they didn't start their course to 100ds of AU's distance because of Planet 9, but now they are affected by planet 9 because they are in its vicinity over millions of years. In a million years, they have about 500 chances to approach planet 9, and 500,000 chances in a billion years.&lt;/p&gt;&#xA;&lt;p&gt;The shepherding effect is complex, Jupiter accelerates objects very fast in stable orbits that resonate with it, it prevents planets from forming from the asteroid belt, and it disrupts the otherwise stable asteroid belt, sending objects fast through the asteroid belt and wherever, towards and away from the Sun. Jupiter created and maintains the asteroid belt which is unstable space debris, and it protects Earth from comets that want orbits that through its path.&lt;/p&gt;&#xA;&lt;p&gt;Planet 9's inclination is about 30 degrees, so it can't interact with the inner Oort cloud, and the outer Oort cloud is too far away.&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="37191" LastEditDate="2021-03-03T14:24:08.403" LastActivityDate="2021-03-03T14:24:08.403" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18931" PostTypeId="2" ParentId="18875" CreationDate="2016-11-03T01:50:57.673" Score="1" Body="&lt;p&gt;I suppose that we are so far from the other Arm that we can see only the small part of Milky Way, approximately 27,000 light-years from this center (&lt;a href=&quot;http://solarstory.net/objects/milky-way#visual-idea&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://solarstory.net/objects/milky-way#visual-idea&lt;/a&gt;) and our solar system is so small that we cannot observe all these amazing stars.&lt;/p&gt;&#xA;" OwnerUserId="14746" LastActivityDate="2016-11-03T01:50:57.673" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18932" PostTypeId="1" CreationDate="2016-11-03T02:13:51.450" Score="4" ViewCount="149" Body="&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Within our current knowledge, what is the largest (circular, say) area in the sky which is empty/dark? I'm thinking in terms of angular size.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Motivation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I was wondering what was the largest star as seen from the Earth, other than the Sun of course, in terms of milliarcseconds, say. A search quickly turned up &lt;a href=&quot;https://en.wikipedia.org/wiki/R_Doradus&quot; rel=&quot;nofollow noreferrer&quot;&gt;R Doradus&lt;/a&gt; which is &lt;em&gt;huge&lt;/em&gt; (57 mas as seen from the Earth). But I wondered if it was possible to rule out any (angularly) larger dark objects by finding stars whose light would be blocked if there was such an object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course we should eventually be able to rule out each such area -- as Hubble Deep Field showed, if you look long enough there's a lot to find everywhere -- but relative to our current knowledge, what are the largest gaps?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect that the answer would be in a region which has been missed by a sky survey and which coincidentally doesn't have anything we can see.&lt;/p&gt;&#xA;" OwnerUserId="10111" LastEditorUserId="10111" LastEditDate="2016-11-03T04:48:43.113" LastActivityDate="2016-11-03T04:48:43.113" Title="What is the largest (angular) dark area in the sky?" Tags="&lt;deep-sky-observing&gt;&lt;sky-survey&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18933" PostTypeId="2" ParentId="18932" CreationDate="2016-11-03T03:27:02.280" Score="2" Body="&lt;p&gt;What do you mean by, &quot;relative to our ability to observe it&quot;? If you have a detector sensitive enough you'll see some light no matter what direction you look, but the level of light will depend on the resolution of your detector. When you look at a patch on the sky, all of the light within one resolution element basically gets lumped in together. Think of the analog of looking at the Milky Way with your naked eyes versus using binoculars - with the eyes it looks like a smooth cloud, up the resolution and you can see the stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you up the resolution enough to see past all of the foreground objects (galaxies and stars), you'll see the combination of the foreground milky way clouds and the cosmic microwave background. At a more realistic resolution you need to add in light produced by milky way stars and galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an area of study in astronomy known as the extragalactic background light, here's a &lt;A HREF=&quot;http://rsos.royalsocietypublishing.org/content/3/3/150555&quot; rel=&quot;nofollow noreferrer&quot;&gt;recent review by Asantha Cooray&lt;/A&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the two short answers are, &quot;Nowhere is dark,&quot; or &quot;It depends on your detector's sensitivity and resolution.&quot;&lt;/p&gt;&#xA;" OwnerUserId="14244" LastActivityDate="2016-11-03T03:27:02.280" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18934" PostTypeId="2" ParentId="18932" CreationDate="2016-11-03T04:24:04.807" Score="4" Body="&lt;p&gt;Barnard 68 is the first thing that comes to mind for me, it is a little over 10' across and is opaque in the visible spectrum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cPG9A.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cPG9A.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure there could be something much larger out there, but like I said, this was the first to come to mind. &lt;/p&gt;&#xA;" OwnerUserId="2232" LastActivityDate="2016-11-03T04:24:04.807" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18935" PostTypeId="1" AcceptedAnswerId="18955" CreationDate="2016-11-03T08:36:49.327" Score="2" ViewCount="654" Body="&lt;p&gt;I would like to calculate the Signal-to-Noise ratio (SNR) of long-slit or integral field unit (IFU) observations of H$\alpha$ emission. I can calculate the SNR or each individual spectrum, just fine. However, how can I assign a single SNR value to an observation? Do I just quote the higher SNR or do I calculate the median/mean SNR of all spectra? Is there a convention for this? My data is low-redshift galaxy IFU observations, and thus the SNR is higher in the center of the galaxy and lower at larger radii.&lt;/p&gt;&#xA;" OwnerUserId="1986" LastEditorUserId="11143" LastEditDate="2016-11-03T13:49:56.603" LastActivityDate="2016-12-08T06:25:17.687" Title="How to calculate the S/N of IFU or long-slit galaxy observations" Tags="&lt;observational-astronomy&gt;&lt;spectroscopy&gt;&lt;spectra&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18936" PostTypeId="1" CreationDate="2016-11-03T09:01:45.977" Score="3" ViewCount="710" Body="&lt;p&gt;I'm using the astrometric python module Pyephem, and I would like to get the orbital (keplerian) elements for the  solar system planets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only values I found is the heliocentric latitude, longitude, and the distance to the sun. Is there a way to compte the orbital parameters based on thoses values ? Did I missed a function in Pyephem ? &lt;/p&gt;&#xA;" OwnerUserId="14747" LastActivityDate="2016-11-04T15:31:44.713" Title="Orbital elements of solar system using Pyephem" Tags="&lt;planet&gt;&lt;solar-system&gt;&lt;orbital-elements&gt;&lt;python&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18938" PostTypeId="2" ParentId="18935" CreationDate="2016-11-03T11:26:34.180" Score="2" Body="&lt;p&gt;Unless you in fact find a &quot;standard&quot; definition for the mean or collective SNR, there is no way to answer this.  As you noted, the signal strength varies with view angle (as well as wavelength), so there's really no single value.  If you have a defined algorithm which processes all your spectral data to produce a final image or analysis value, then you can use standard statistical techniques to calculate the SNR of the output as a function of the SNR of all the contributing factors. (Here &quot;factor&quot; means each independent input variable, i.e. spectral line).  I recommend getting a copy of &lt;a href=&quot;http://astro.cornell.edu/academics/courses/astro3310/Books/Bevington_opt.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bevington&lt;/a&gt; to learn how to do this.   Note- that appears to be a not-necessarily-legal-copy; you can purchase the real thing anywhere, e.g. &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0072472278&quot; rel=&quot;nofollow noreferrer&quot;&gt;amazon&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2016-11-03T11:26:34.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18940" PostTypeId="1" CreationDate="2016-11-03T14:26:20.063" Score="1" ViewCount="92" Body="&lt;p&gt;I have a masters degree in computer science and statistics but would like to advance my knowledge in astronomy, ideally in getting some kind of online degree. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody have experience, suggestions or recommendation where I can do what without having to be physically present at a university? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I would like to get view on what specific topics in astronomy it would make sense to focus if my skills are mainly in computer science and data analysis? &lt;/p&gt;&#xA;" OwnerUserId="10557" LastActivityDate="2016-11-03T16:26:37.040" Title="Astronomy online degrees" Tags="&lt;radio-astronomy&gt;&lt;data-analysis&gt;&lt;n-body-simulations&gt;&lt;seti&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18941" PostTypeId="2" ParentId="18940" CreationDate="2016-11-03T15:19:56.230" Score="2" Body="&lt;p&gt;Try &lt;a href=&quot;https://www.coursera.org/courses?query=astronomy&quot; rel=&quot;nofollow&quot;&gt;https://www.coursera.org/courses?query=astronomy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Online courses from several universities. Some are on a schedule with online lectures, etc. Others you can just do yourself at any time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also there are extensive libraries on YouTube, for instance from this guy Michel van Biezen.&#xA;&lt;a href=&quot;http://www.youtube.com/playlist?list=PLX2gX-ftPVXWddG6sGq6H-3cN0IQzfQmG&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/playlist?list=PLX2gX-ftPVXWddG6sGq6H-3cN0IQzfQmG&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That link goes to the first of his playlists about astronomy. His astronomy section has about 20 parts!&lt;/p&gt;&#xA;" OwnerUserId="14752" LastEditorUserId="14752" LastEditDate="2016-11-03T16:26:37.040" LastActivityDate="2016-11-03T16:26:37.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18943" PostTypeId="1" CreationDate="2016-11-03T21:27:31.773" Score="2" ViewCount="83" Body="&lt;p&gt;I am looking for stars in the GALEX data through the GALEX data using GALEXview and plugging in their Henry Draper Numbers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://galex.stsci.edu/galexview/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://galex.stsci.edu/galexview/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the task is tedious by hand. Is there any place that I can look to find the raw text, ascii, or binary data? I want to be able to simply pull the data I need in one go with a program, rather than copying it down by hand. Currently I'm having to look through the data point star by star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://galex.stsci.edu/GR6/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://galex.stsci.edu/GR6/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="13146" LastActivityDate="2016-11-03T21:27:31.773" Title="Where can I find raw text, ascii, or binary data from the GALEX satellite?" Tags="&lt;star&gt;&lt;observational-astronomy&gt;&lt;data-analysis&gt;&lt;space-telescope&gt;&lt;stellar-astrophysics&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18945" PostTypeId="1" AcceptedAnswerId="18959" CreationDate="2016-11-04T10:55:54.030" Score="9" ViewCount="453" Body="&lt;p&gt;Regarding molecular clouds / absorption nebulae / Bok globules / obscuring clouds,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(I'm not sure if there's one generic term encompassing all those and any similar obscuring clouds - I just used &quot;cloud&quot; in the question title),&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is Barnard 68 the only one so close to us that there are no stars between us and it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, what is the list of such?  (Are we observationally confident about this, at this stage?)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I guess a related question would be: &quot;about what is the density of bok globules near us?&quot; ... in a cube of 1000pc on a side, about how many bok globules are we talking? One or two or thousands??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zYSyt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zYSyt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;http://sci.esa.int/gaia/60207-barnard-68/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://sci.esa.int/gaia/60207-barnard-68/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="15415" LastEditDate="2019-04-13T06:17:25.943" LastActivityDate="2019-04-13T06:17:25.943" Title="Is Barnard 68 the only cloud so close to us that there are no stars between us and it?" Tags="&lt;nebula&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="18946" PostTypeId="2" ParentId="18945" CreationDate="2016-11-04T12:10:54.237" Score="9" Body="&lt;p&gt;Barnard 68 has a surface area of around 10 square arcminutes on the sky and is at an estimated distance of around 160 pc. The volume of space encompassed by a sightline to Barnard 68 is therefore about 1 pc$^{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_density&quot; rel=&quot;nofollow noreferrer&quot;&gt;the density of stars&lt;/a&gt; in the solar neighbourhood (hydrogen burning stars with mass above about 0.1$M_{\odot}$) is about 0.1 pc$^{-3}$. So it is not surprising that no stars are seen in front of Barnard 68.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this distance, the brightness of the faintest stars would be around $V \sim 20$ so ought to have been picked up in various observations that have been performed on this cloud. Such stars would be more obvious in the near-infrared, but then you pick up background stars through the cloud and you would need an estimate of the distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not actually aware of any clear statement that there are no faint stars in the foreground, though obviously you can't see any in the widely available visible pictures.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots more &lt;a href=&quot;https://en.wikipedia.org/wiki/Bok_globule&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bok globules&lt;/a&gt; known which appear to have very little in the way of foreground stars. As I outlined in the calculation above - if they are small and nearby, then that is to be expected. You could also look at the &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_dark_nebulae&quot; rel=&quot;nofollow noreferrer&quot;&gt;list of &quot;dark nebulae&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other catalogues include the &lt;a href=&quot;http://cdsarc.u-strasbg.fr/viz-bin/Cat?VII/7A&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lynds catalogue of dark nebulae&lt;/a&gt; or the catalogue of southern dark clouds by &lt;a href=&quot;http://adsabs.harvard.edu/abs/1986A%26AS...63...27H&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hartley et al. (1986)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-11-11T18:11:28.507" LastActivityDate="2016-11-11T18:11:28.507" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18947" PostTypeId="1" AcceptedAnswerId="18949" CreationDate="2016-11-04T12:21:29.320" Score="6" ViewCount="5056" Body="&lt;p&gt;Also, what is the advantage of getting information about the velocity dispersion of a galaxy?&lt;/p&gt;&#xA;" OwnerUserId="14757" LastEditorUserId="9301" LastEditDate="2016-11-04T21:24:44.043" LastActivityDate="2018-01-10T15:30:56.070" Title="What does the velocity dispersion of a galaxy mean?" Tags="&lt;galaxy&gt;&lt;radio-astronomy&gt;&lt;velocity&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18948" PostTypeId="2" ParentId="18924" CreationDate="2016-11-04T12:45:27.337" Score="2" Body="&lt;p&gt;To add to the answer from James K, the ESA program &quot;Gaia&quot;, amongst many other purposes, intends to capture hints predicting microlensing events. There is much to study in that domain, and the amateur astronomer community can help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cosmos.esa.int/web/gaia/IoW_20161027&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.cosmos.esa.int/web/gaia/IoW_20161027&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT :&lt;/strong&gt; forgot to mention the linked possible microlensing event is bound to happen the two first weeks of November 2016, which makes it quite current.&lt;/p&gt;&#xA;" OwnerUserId="1080" LastActivityDate="2016-11-04T12:45:27.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18949" PostTypeId="2" ParentId="18947" CreationDate="2016-11-04T15:17:07.677" Score="12" Body="&lt;strong&gt;Definition of the velocity dispersion&lt;/strong&gt;&#xA;&#xA;&lt;p&gt;From the title of your question, I'm unsure whether you actually know what &quot;dispersion&quot; means: The dispersion of some numbers is the spread around their mean, usually taken to mean their &lt;a href=&quot;https://en.wikipedia.org/wiki/Standard_deviation&quot; rel=&quot;noreferrer&quot;&gt;standard deviation&lt;/a&gt;. If you measure the velocity of several light sources (from the Doppler shift of their spectral lines) that are gravitationally bound, they will have some average velocity$^\dagger$. But since the light sources — which could be stars or gas clouds — move around each other, some move a little faster, and some move a little slower. This will cause their combined emission line to be broadened, instead of being a very narrow line. The central area of this line that emcompasses roughly 68% of the area of the line has a certain width. If the distribution is Gaussian (which is typically the case), then this width, divided by two, is the standard deviation. This is the usual definition, I think, but note that other definitions could be used, e.g. the &lt;a href=&quot;https://en.wikipedia.org/wiki/Full_width_at_half_maximum&quot; rel=&quot;noreferrer&quot;&gt;full width at half maximum&lt;/a&gt; which differs from the standard deviation by a constant factor of $2\sqrt{2\ln2}$ for a Gaussian.&lt;/p&gt;&#xA;&#xA;&lt;strong&gt;A relation between the mass, size, and velocity dispersion&lt;/strong&gt;&#xA;&#xA;&lt;p&gt;The velocity dispersion $\sigma_V$ of gravitationally interacting &quot;particles&quot; such as stars in a galaxy, or galaxies in a cluster, provides a measure of the total mass and its distribution. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Virial_theorem#In_astrophysics&quot; rel=&quot;noreferrer&quot;&gt;virial theorem&lt;/a&gt;, which describes the total energy of a bound system, can be expressed to show that the components of a galaxy of mass $M$ and radius $R$ will have characteristic velocities of the order of the velocity dispersion&#xA;$$&#xA;\sigma_V = \sqrt{\frac{GM}{CR}},&#xA;$$&#xA;where $G$ is the gravitational constant, and the observationally determined factor $C$ depends on the geometry and the actual mass distribution, as well as on whether the system is rotation-dominated (such as disk galaxies) or dispersion-dominated (such as elliptical galaxies) (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2008gady.book.....B&quot; rel=&quot;noreferrer&quot;&gt;Binney &amp;amp; Tremaine 2008&lt;/a&gt;). For dispersion-dominated galaxies, $M$ refers to the total, dynamical mass of the system, and $C \simeq 6.7$ for various galactic mass distributions (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2009ApJ...706.1364F&quot; rel=&quot;noreferrer&quot;&gt;Förster Schreiber et al. 2009&lt;/a&gt;). For rotation-dominated galaxies, the appropriate mass is the mass enclosed within $R$, and $C \simeq 2.25$, again averaged over various galactic mass distribution models (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2009A&amp;amp;A...504..789E&quot; rel=&quot;noreferrer&quot;&gt;Epinat et al. 2009&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The velocity dispersion is rather easy to determine with spectroscopy, since it broadens the spectral lines (other processes also broaden the lines, so one must be careful to disentangle different effects). With photometry, the size of the galaxy can be determined (note though that the size can differ when observed in different bands). Thus, even if you don't know whether a galaxy is rotation- or dispersion-dominated, a moderately precise measure of its mass — one of the main characteristics of a galaxy — can be obtained (e.g. setting $C=5$ won't be too far off, and in astronomy we're usually happy with such small errors).&lt;/p&gt;&#xA;&#xA;&lt;strong&gt;Other relations&lt;/strong&gt;&#xA;&#xA;&lt;p&gt;Because of various scaling relations, $\sigma_V$ can also be used to probe other quantities, or other quantities can be used to probe $\sigma_V$. For instance, the &lt;a href=&quot;https://en.wikipedia.org/wiki/M%E2%80%93sigma_relation&quot; rel=&quot;noreferrer&quot;&gt;M-sigma relation&lt;/a&gt; relates the mass $M_\mathrm{BH}$ of a galaxy's central, supermassive black hole to $\sigma_V$ (through $M_\mathrm{BH} \propto \sigma_V^4$, roughly), and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Faber%E2%80%93Jackson_relation&quot; rel=&quot;noreferrer&quot;&gt;Faber–Jackson relation&lt;/a&gt; relates the luminosity $L$ of a galaxy to $\sigma_V$ (through $L\propto\sigma_V^4$, more roughly).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;In astronomy, this average velocity is called &quot;the systemic velocity&quot;), and is given by the expansion rate of the Universe (the Hubble law), plus in general a small velocity in a random direction (the &quot;peculiar velocity&quot; due to local dynamics).&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-01-10T15:30:56.070" LastActivityDate="2018-01-10T15:30:56.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18950" PostTypeId="2" ParentId="18936" CreationDate="2016-11-04T15:24:45.600" Score="2" Body="&lt;p&gt;There is a website: &lt;a href=&quot;http://orbitsimulator.com/formulas/OrbitalElements.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://orbitsimulator.com/formulas/OrbitalElements.html&lt;/a&gt; that has a javascript program for converting state vectors to orbital elements and back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The source of that website is possible to convert to python: Here is a &lt;code&gt;Body&lt;/code&gt; class that has a method for calculating orbital elements based on that website. The method takes one argument, a body the is the Principal (ie the Sun)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;G = 0.0002946        # in units of seconds, AU and solar masses.&#xA;class Body:&#xA;    &quot;&quot;&quot;A body has attributes r and v, which are its position and &#xA;velocity in cartesian coordinates and a mass. implied units are &#xA;solar masses, AU and seconds.&quot;&quot;&quot; &#xA;    def __init__(self,r,v,mass):&#xA;        self.r = np.array(r,dtype=&quot;float&quot;)&#xA;        self.v = np.array(v,dtype=&quot;float&quot;)&#xA;        self.mass = mass&#xA;        self.GM = self.mass*G&#xA;&#xA;    def orbital_elements(self,principal):&#xA;        '''view-source:http://orbitsimulator.com/formulas/OrbitalElements.html'''&#xA;        mu = G*(principal.mass+self.mass)&#xA;        # calculate relative position,velocity&#xA;        r = self.r - principal.r&#xA;        v = self.v - principal.v&#xA;        try: #catch division by zero&#xA;            R = np.linalg.norm(r)&#xA;            V = np.linalg.norm(v)&#xA;            a = 1/(2/R - V**2/mu)  # semi major axis&#xA;&#xA;            h = np.cross(r,v)&#xA;            H = np.linalg.norm(h)&#xA;&#xA;            P = H**2/mu&#xA;            Q = np.dot(r,v)&#xA;&#xA;            E= np.sqrt(1-P/a)  #eccentricity&#xA;&#xA;            e = [1-R/a,Q/np.sqrt(a*mu)]&#xA;            i = np.arccos(h[2]/H)&#xA;            Omega = 0&#xA;            if i!=0: &#xA;                Omega = np.arctan2(h[0],-h[1]) #Longitude of acending node&#xA;&#xA;            ta = [H**2/(R*mu) -1,H*Q/(R*mu)]&#xA;            TA = np.arctan2(ta[1],ta[0])&#xA;            Cw = (r[0]*np.cos(Omega)+r[1]*np.sin(Omega))/R&#xA;&#xA;&#xA;            if i==0 or i==np.pi:&#xA;                Sw = (r[1]*np.cos(Omega) - r[0]*np.sin(Omega))/R&#xA;            else:&#xA;                Sw = r[2]/(R*np.sin(i))&#xA;&#xA;            omega = np.arctan2(Sw,Cw) - TA  #argument of periapsis&#xA;            if omega&amp;lt;0: omega += 2*np.pi&#xA;&#xA;            u = np.arctan2(e[1],e[0]) &#xA;            M = u - E*np.sin(u) # mean anomaly&#xA;&#xA;            return(a,E,omega,Omega,i,M)&#xA;        except ZeroDivisionError:&#xA;            #meaningless, but avoids crash&#xA;            return(0,0,0,0,0,0)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2016-11-04T15:31:44.713" LastActivityDate="2016-11-04T15:31:44.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18951" PostTypeId="1" AcceptedAnswerId="18952" CreationDate="2016-11-04T17:51:12.100" Score="0" ViewCount="120" Body="&lt;p&gt;Pretend you're in our galaxy near Earth and you're standing upright wrt the flat galactic disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look directly towards the center of the galaxy, that's Sagittarius. If you swivel your head and look directly away from the center, I believe that's around Auriga.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, if you look straight up, about which constellation is that?  And if you look straight down?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If I'm not mistaken, could be wrong, this is equivalent to asking about...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZgB7n.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZgB7n.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia explanation of galactic coordinates&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... galactic coordinates of longitude zero, latitude 90° (up overhead) and -90° (down through your feet).&lt;/p&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="13071" LastEditDate="2016-11-05T01:33:00.653" LastActivityDate="2016-11-05T01:33:00.653" Title="Which constellations are straight up and straight down?" Tags="&lt;coordinate&gt;&lt;constellations&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18952" PostTypeId="2" ParentId="18951" CreationDate="2016-11-04T18:01:31.710" Score="1" Body="&lt;p&gt;Crap: I've just realized the answer is &lt;em&gt;given on that Wikipedia page!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Galactic_coordinate_system&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking up from here: Coma Berenices&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking down from here: Sculptor&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps someone googling.&lt;/p&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="616" LastEditDate="2016-11-04T19:48:30.697" LastActivityDate="2016-11-04T19:48:30.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18953" PostTypeId="1" AcceptedAnswerId="18958" CreationDate="2016-11-04T19:13:55.217" Score="1" ViewCount="70" Body="&lt;p&gt;I recently went to a talk on X-ray emissions from stellar winds of O-type stars, and a paper was explored that in turn referenced &lt;a href=&quot;http://mnras.oxfordjournals.org/content/441/4/3600&quot; rel=&quot;nofollow noreferrer&quot;&gt;ud-Doula et al. (2014)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, O-type stars may show strong x-ray emission due to radial shocks in their stellar winds. In general, the winds are isotropic, and so emission is assumed to be spherically symmetric neglecting rotational effects. However, certain O-type stars (e.g. NGC 1624-2) have strong magnetic fields; this means that outflows are channeled into rather narrow beams within a few stellar radii, which eventually spread out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X-ray emission is characterized by a parameter $X_{T_x}$&#xA;$$X_{T_x}\equiv\rho^2\exp(-T_x/T)$$&#xA;for density $\rho$ and temperature $T$, with a threshold temperature of $T_x$. Simulations place most of the emissions in a rather narrow spot near the star, because while there are indeed high temperatures experienced all along a beam extending outward from the star, the density quickly falls off, as shown in Figure 4:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H5HQa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H5HQa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;From left to right: Logarithmic density, logarithmic temperature, and x-ray emission.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the density drop off so quickly? I would expect that the channeling of more of the wind into a small beam would lead to a rather steady radial density profile along it, as is the case with the temperature graph.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is unrelated to the question, but I'll add that the temperature isn't really constant. 1D and 2D simulations of spherically symmetric show varying behavior due to repeated shocks inside the wind, and I would expect the same to be true here.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2016-11-06T04:09:51.760" Title="Why is there a density dropoff in the stellar winds of magnetic O-type stars?" Tags="&lt;stellar-astrophysics&gt;&lt;density&gt;&lt;solar-wind&gt;&lt;x-ray&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18954" PostTypeId="1" AcceptedAnswerId="18963" CreationDate="2016-11-04T20:10:03.300" Score="9" ViewCount="199" Body="&lt;p&gt;A lead scientist of the Planet Nine theory said that &quot;The period with which it goes around the sun is a rational multiple of the periods of all the furthest Kuiper Belt objects&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To what degree of precision can Konstantin Batygin make that kind of measurement? Has it been backed up by other physicists with a degree of certainty?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He also says that they think it's near it's aphelion &quot;because it hasn't been seen, so it's far away&quot;, not because of the above period coincidence, why can't they say where it is more convincingly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that mean that they have very strong mathematical evidence for Planet Nine?&lt;/p&gt;&#xA;" OwnerUserId="1528" LastActivityDate="2016-11-05T18:04:11.527" Title="Period coincidence of Planet Nine with furthest Kuiper belt objects" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;astrophysics&gt;&lt;asteroids&gt;&lt;comets&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18955" PostTypeId="2" ParentId="18935" CreationDate="2016-11-04T23:32:47.747" Score="1" Body="&lt;p&gt;I'd say quote the central value (and indicate whether this is S/N per Angstrom or whatever spectral binning you're using). If you bin the rest of your IFU data spatially, then you can quote whatever limiting S/N you used to determine the binning (e.g., &quot;we binned the data spatially using Voronoi binning so that each bin had a minimum S/N of $x$&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quoting a mean or median S/N is sensible if you spatially bin your spectra, so that the (small) central bin and the (large) outer bins have roughly similar S/N values. Or you could describe how the S/N varies as you go out from the galaxy center (some people even produce 2D maps of the S/N for their IFU data).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have lots of galaxies to talk about, I'd probably just quote the central spectral S/N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edited to clarify the &quot;mean&quot; S/N idea.]&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2016-11-04T23:40:57.150" LastActivityDate="2016-11-04T23:40:57.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18956" PostTypeId="1" AcceptedAnswerId="18957" CreationDate="2016-11-05T06:54:18.130" Score="-10" ViewCount="370" Body="&lt;p&gt;i always wondered that Is it possible that Fly a Boeing 777 Air Craft on Moon? if cannot how this moon lander's (Apollo) landing in Moon taking off from the moon? &lt;/p&gt;&#xA;" OwnerUserId="11792" LastEditorUserId="11792" LastEditDate="2016-11-05T07:13:15.383" LastActivityDate="2016-11-05T07:13:15.383" Title="Is it possible that Fly a Boeing 777 plane on Moon?" Tags="&lt;the-moon&gt;&lt;planetary-atmosphere&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ClosedDate="2016-11-05T14:03:56.597" ContentLicense="CC BY-SA 3.0" />
  <row Id="18957" PostTypeId="2" ParentId="18956" CreationDate="2016-11-05T07:10:01.593" Score="6" Body="&lt;p&gt;No. The moon is an extremely hard vacuum with an &lt;A HREF=&quot;http://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html&quot;&gt;atmospheric pressure of $3\times10^{-15}\operatorname{bar}$&lt;/a&gt;. The 777 has a &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Boeing_777#Specifications&quot;&gt;service ceiling of $13,100\operatorname{meters}$&lt;/A&gt; where Earth's air pressure is about $1/10^{\mathrm{th}}$ ground level, the air pressure is about $0.1 \operatorname{bar}$. So, there's no oxygen to burn in the engines to generate thrust, no atmosphere to generate lift against the wings. That's why rockets carry both fuel and oxidiser, and don't really bother with wings.&lt;/p&gt;&#xA;" OwnerUserId="14244" LastActivityDate="2016-11-05T07:10:01.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18958" PostTypeId="2" ParentId="18953" CreationDate="2016-11-05T09:31:39.100" Score="3" Body="&lt;p&gt;Conservation of mass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a steady-state wind, the mass loss rate through each shell is the same.&#xA;$$ \frac{dM}{dt} = - 4\pi r^2 \rho(r) v(r),$$&#xA;where $\rho$ and $v$ would be the density and velocity of a spherically symmetric wind, with a fixed mass-loss rate $\dot{M}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus for a fixed wind speed you expect the density to fall as the square of the radius in spherically symmetric wind - and it looks to me that a little outside the closed field loops then spherical symmetry is approximately recovered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This alone, would not account for the sharp drop off in density seen, but the other factor is that a radiation driven wind accelerates with radius when close to the star, which exacerbates the decrease in density. Typically, $v$ would increase by a factor of a few between say 1.5 and 5 times the radius. Multiplying this by a $r^{-2}$ factor would give a factor $\sim 100$ decrease in density overall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The behaviour of the temperature is much more complex, governed by heating and cooling processes and the expansion of the gas.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-11-05T09:31:39.100" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18959" PostTypeId="2" ParentId="18945" CreationDate="2016-11-05T09:34:38.243" Score="6" Body="&lt;p&gt;First query: YES there are many bok clusters without foreground stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For density, That's an unclear field of study, confused with dark matter enigma, and the  estimates of frequency of &quot;Boks&quot; perhaps only have about 1-2 degrees of magnitude precision. We can't say that there is less than 1 bok for every 10 stars, we can't say there is more than 1 for every 1000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance to the nearby and compact globules is difficult to determine. The absence of foreground stars toward globules rules out spectrophotometric distance measurements, and the distance estimation from LSR velocity (kinematic distance) is unreliable. So, listed distances are only values assumed by associating observed globules with neighboring larger molecular clouds whose distance is known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for one thing, you will find that catalogues of Boks, and IRAS, give their galactic coordinates of longitude and lattitude, and not their distance, which they might get wrong by an order of magnitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are dozens of stars within 20ly, there's about 1 star every 280 cubic light years, so within 500 light years from us there are about 2,000,000 stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a catalogue of 380 nearby small compact globules of which 280 are about 1ly across. IRAS Point Source Catalog, Version 2.0 has many thousand of them, they seem to be all in the 4000 parsec range, and stuff that far away becomes difficult do detect as soon as it does have a star in front of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Within 1000 parsecs there are already 100ds of Bok clusters in catalogues, and if there were 10,000 Bok clusters nearby, the opacity of the sky could be measured, there has to be a certain ratio of dark clouds relative to stars to give the current opacity of the solar neighborhood, and perhaps someone else has information about that, it's probably that some studies have been done to measure local opacity, and mass, although that is also complicated by dark matter theories, perhaps there are so many Boks, they make up a large proportion of the local mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a catalogue of actual Bok Globules from 1988, based on globule regularity, roundness, because roundness is used to determine their size and shape rather than precise factors, all their target's did not have a star in front, and they said: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Such a catalog of small molecular clouds was needed for&#xA;our ongoing study of star formation in Bok globules (Clemens&#xA;and Leach 1987; Clemens, Leach, and Barvainis 1988). The&#xA;available catalogs (Lynds 1962; Barnard 1927) are most complete&#xA;for larger clouds (i.e., angular sizes greater than S'-IO').&#xA;However, many smaller clouds do not appear in these catalogs.&#xA;A new search of the Palomar Observatory Sky Survey&#xA;(POSS) was made with the goal of finding the smaller molecular&#xA;clouds.&#xA;We have identified 248 molecular clouds with optical sizes&#xA;smaller than 10'. Most of these have been previously cataloged&#xA;(72%), although the fraction of clouds previously cataloged&#xA;decreases somewhat (to 60%) as the mean size decreases&#xA;to l'-2/&#xA;. A large number of the clouds (60%) have IRAS&#xA;point sources within their boundaries or in their near environs,&#xA;indicating that probing the FIR dust properties of the&#xA;clouds will be productive.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They catalogued 248 easily measured bok globules somewhere nearer than 5000 parsecs and beyond that it's too small to see them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;someone did a survey of SO emissions from the above catalogue, and had images of this kind:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/56DrE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/56DrE.jpg&quot; alt=&quot;SO emission of bok globules&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ChRPi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ChRPi.jpg&quot; alt=&quot;list of bok globules in SO emissions maps study&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2016-11-16T14:25:24.257" LastActivityDate="2016-11-16T14:25:24.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18960" PostTypeId="2" ParentId="13994" CreationDate="2016-11-05T11:03:49.860" Score="2" Body="&lt;p&gt;There will be a complex system of voting to decide on the best name, and the suggestions will come from worldwide sources amidst a media frenzy for a while, the media will sensationalize the naming of the planet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be a 9 year old girl from a school than names a new planet if it is found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planet nine will probably be the place holder name for a while because it sounds ok, unless the media quotes another famous source who suggests a good name. The media will play a large part on publicising all the suggested names and the naming competition, so it's very much an astronomy PR game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the name of another planet would not become official for a good many years after it's discovery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the story will perhaps be similar to the naming of Pluto:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Pluto#Name&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Pluto#Name&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastActivityDate="2016-11-05T11:03:49.860" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18961" PostTypeId="1" AcceptedAnswerId="18966" CreationDate="2016-11-05T14:37:20.540" Score="2" ViewCount="197" Body="&lt;p&gt;It seems all the central compact objects correlated with supernova remnant(SNRs) are isolated, not in binaries, right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find only one exception. A source is claimed to be a compact binary which correlates with SNR G308.3-1.4.&lt;a href=&quot;http://adsabs.harvard.edu/abs/2012ApJ...750....7H&quot; rel=&quot;nofollow noreferrer&quot;&gt;ads link&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, generally, are the compact objects in SNR always isolated? &#xA;And &lt;strong&gt;observationally&lt;/strong&gt;, are the compact objects in SNRs neutron stars?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditDate="2016-11-06T10:33:24.380" LastActivityDate="2016-11-06T10:33:24.380" Title="Are the compact objects in supernova remnants isolated neutron stars?" Tags="&lt;supernova&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18962" PostTypeId="2" ParentId="18961" CreationDate="2016-11-05T15:43:54.967" Score="2" Body="&lt;p&gt;There are plenty of hard-x-ray binaries that involve a compact object with a companion.  Some should be in supernova remnants, but I don't know how many.  Are you sure there is a problem?&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-11-05T15:43:54.967" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18963" PostTypeId="2" ParentId="18954" CreationDate="2016-11-05T15:52:53.517" Score="4" Body="&lt;p&gt;I did a little digging and I found this &lt;a href=&quot;http://www.universetoday.com/127800/could-planet-9-still-exist/&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth noting that the initial publication wasn't about resonances but the fact that all 6 objects came from the same side of the sky, from somewhat similar directions, which, given that they were the 6 most distant at the time, is statistically unusual and it was both similar direction and similar inclination to the plane, so the statistical improbability was high enough to investigate.  Orbital resonance wasn't the primary argument, it was tacked on later as a secondary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Their analysis also offered suggestions as to what kind of resonance&#xA;  the planet has with the KBOs in question. Whereas Sedna’s orbital&#xA;  period would have a &lt;strong&gt;3:2&lt;/strong&gt; resonance with the planet, 2010 GB174&#xA;  would be in a &lt;strong&gt;5:2&lt;/strong&gt; resonance, &lt;em&gt;2994 VN112&lt;/em&gt; in a &lt;strong&gt;3:1&lt;/strong&gt;, &lt;em&gt;2004 VP113&lt;/em&gt;&#xA;  in &lt;strong&gt;4:1&lt;/strong&gt;, and 2013 GP136 in &lt;strong&gt;9:1&lt;/strong&gt;. These sort of resonances are&#xA;  simply not likely without the presence of a larger planet.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“For a resonance to be dynamically meaningful in the outer Solar&#xA;  System, you need one of the objects to have enough mass to have a&#xA;  reasonably strong gravitational effect on the other,” said the&#xA;  research team. “The extreme Kuiper belt objects aren’t really massive&#xA;  enough to be in resonances with each other, but the fact that their&#xA;  orbital periods fall along simple ratios might mean that they each are&#xA;  in resonance with a massive, unseen object.”&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But what is perhaps most exciting is that their findings could help to&#xA;  narrow the range of Planet 9’s possible location. Since each orbital&#xA;  resonance provides a geometric relationship between the bodies&#xA;  involved, the resonant configurations of these KBOs can help point&#xA;  astronomers to the right spot in our Solar System to find it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;now, I want to point out that it's them, not me who did the typos on 2 of the objects, I'm pretty sure (based on Wikipedia and the chart below) that it should be 2004 VN112 (not 2994 VN112) that has the 3:1 and 2012 VP113 (not 2004 VP113) that has the 4:1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This chart below also has a typo, it should be 2010 GB174 not 2012 GB174.  I guess nobody proofreads letter/number assignments.   Correct names will help if anyone wants to run the numbers see how see how close the resonances come to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/giSu7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/giSu7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the degree of statistical certainty?   If you're a good enough statistician, you can run the numbers.   3/2, it seems to me, would be the most common resonance based on Jupiter's abundant Hildas and the Plutinos and only 1 of the 5 potentially resonant objects has a 3/2, but 3/2 is also very far from the sun and hardest to detect, so that's not a deal breaker.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd guess that the degree of certainty from these orbital resonances is OK but not great - but that's entirely a guess.   I also don't know enough about astronomy to know how close numbers have to be to provide some evidence of orbital resonance.  Jupiter's resonant moons, for example, if you check their orbits, are off by 1 or 2 percentage points from each other.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-11-05T18:04:11.527" LastActivityDate="2016-11-05T18:04:11.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18964" PostTypeId="2" ParentId="18750" CreationDate="2016-11-05T17:12:49.373" Score="4" Body="&lt;p&gt;I'll take the reference of an observer falling with the helium atom. At the Event horizon, the tidal force across a helium atom is still minute, &lt;a href=&quot;https://www.google.co.uk/search?q=%5B(1e-12%20*%20c%5E6)%2F(4G%5E2*10%5E60)%5D&amp;amp;oq=%5B(1e-12%20*%20c%5E6)%2F(4G%5E2*10%5E60)%5D&amp;amp;aqs=chrome..69i57j69i64.491j0j4&amp;amp;sourceid=chrome&amp;amp;ie=UTF-8#q=%5B(1e-12%20meters%20*%20c%5E6)%2F(4G%5E2*(5*10%5E30%20kg)%5E2)%5D*(6e-27%20kg)&quot; rel=&quot;nofollow noreferrer&quot;&gt;$10^{-29}N$&lt;/a&gt; much less than the Coulomb forces between the electrons and nucleus, still less than the strong forces inside the nucleus. So the atom will fall through the event horizon without damage. (The region around the black hole can be a pretty violent place, with lots of hard X rays from the superheated matter, which would likely have ionised the atom, but I'll suppose it somehow manages to avoid this)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it falls past the event horizon it is doomed. Think of the singularity less of a place in the middle of the black hole, and more as a point in time. It is inevitable our He atom will reach that point. As it approaches it the tidal forces will grow larger, eventually sufficient pull the atom apart. To describe what happens now would require a good description of quantum gravity, which we don't have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The is no physics that describes what happens &quot;after&quot; the singularity. Think of the singularity as a point in time, but for the atom it is also the &lt;em&gt;end of time&lt;/em&gt; for the atom. It is possible that something beyond our current understanding prevents a singularity from forming, but that is speculation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is explained diagramatically in a &lt;a href=&quot;https://jila.colorado.edu/~ajsh/insidebh/penrose.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Penrose diagram&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SOevH.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SOevH.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-11-05T17:12:49.373" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18965" PostTypeId="2" ParentId="18953" CreationDate="2016-11-06T04:09:51.760" Score="0" Body="&lt;p&gt;The first answer gives a good summary of the overall density falloff you see in all directions, but since you are saying the temperature is fairly constant it sounds like you might be asking about the density in the plane where the winds are colliding.  Is that what you mean?  If so, I think the problem is that none of those figures show the velocity field.  That is not a &quot;beam&quot; you are seeing, it is a cross section of an axially symmetric disk.  The wind is channeled down from the polar regions by the field lines, but the disk circumference increases in proportion to r, and there can be some centrifugal acceleration because the star is rotating.  So those could be the cause of the density drop with radius.  There's also that high density layer that appears very dark, and some if not all of that could end up falling back on the star, so it's not really a beam.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-11-06T04:09:51.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18966" PostTypeId="2" ParentId="18961" CreationDate="2016-11-06T09:58:55.187" Score="2" Body="&lt;p&gt;All type II supernovae are associated with the core collapse of a massive star and are thought to produce some sort of compact star. Supernovae should also always produce an expanding supernova remnant. Almost all massive stars are part of multiple systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These three facts might lead you to believe that most, if not all, supernova remnants should contain an observable compact star in a multiple system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two classes of reason why this is not the case - observational issues and physical mechanisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observationally, we do not see compact multiple systems unless one of the stars is a pulsar, which requires a correct orientation to see it, or if the compact star is accreting mass from its companion. The latter requires a close binary system that thas survived the supernova. Also note that isolated black holes may exist in supernovae remnants but would be unobservable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physically, the supernova explosion may disrupt the binary system. Pulsars are typically travelling at hundreds of km/s thanks to the &quot;kick&quot; they received from their supernova. Supernova remnants are not visible for long ($\sim$ a million years or less), so pulsars that have been kicked out of a binary can still be found within the supernova remnant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short-lived nature of supernovae remnants also biases against finding compact, X-ray accreting binaries. Typically, the compact star's companion must evolve to either fill its Roche lobe (for a low-mass X-ray binary) or to develop high mass loss rates (for a high-mass X-ray binary). This would not normally happen within the short life of the visible supernova remnant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So examples of &lt;em&gt;observable&lt;/em&gt; compact binary systems in supernova remnants are rare. But there are at least two examples known in our Galaxy and several in others. The (at least) two in our Galaxy are &lt;a href=&quot;https://arxiv.org/abs/1312.0632&quot; rel=&quot;nofollow noreferrer&quot;&gt;Circinus X-1&lt;/a&gt; and &#xA;&lt;a href=&quot;https://arxiv.org/abs/1101.3486&quot; rel=&quot;nofollow noreferrer&quot;&gt;SS433&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-11-06T09:58:55.187" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18967" PostTypeId="1" CreationDate="2016-11-06T14:52:40.203" Score="2" ViewCount="113" Body="&lt;p&gt;Two white dwarfs with a total mass $&amp;gt;1.4\:M_{\odot}$ merge. It may trigger a type Ia SN explosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about the three cases below?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Two low mass WDs merge, total mass $&amp;lt;1.4\:M_{\odot}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A low mass WD + a low mass neutron star, total mass $&amp;lt;2\:M_{\odot}$ but $&amp;gt;1.4\:M_{\odot}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A low mass WD + a neutron star, total mass $&amp;gt;2\:M_{\odot}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If any of the above can result in a SN, have we observed this kind of SN?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditDate="2016-11-10T12:37:21.817" LastActivityDate="2016-11-10T12:37:21.817" Title="white-dwarf merge in binaries" Tags="&lt;supernova&gt;&lt;white-dwarf&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18969" PostTypeId="1" CreationDate="2016-11-06T18:58:40.550" Score="1" ViewCount="59" Body="&lt;p&gt;Recently I inherited quite a few old QuickBasic programs which perform various astronomical calculations.  I'm attempting to understand these programs and rewrite some of them in Python.  I do not have a deep background in astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A number of the programs take a parameter file as input, YEAR.DAT.  Below are 5 years of these files (each column represents one file).  I need help in figuring out the various data values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;YEAR.DAT&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;year        2001         2008         2009         2010         2011 &#xA;delta t      66           65           66           66           67 &#xA;tilt         23.43909     23.43818     23.43805     23.43799     23.43786 &#xA;dow           1            2            4            5            6 &#xA;gst           6.71430      6.66860      6.71839      6.702509     6.68659 &#xA;x1          105.690      330.340      310.959      291.631      272.303 &#xA;bs           84           90           88           87           86 &#xA;fs          301          300          298          304          303 &#xA;x2          357.765      356.959      357.689      357.433      357.177 &#xA;x3          354.289      193.159      335.720      105.105      234.489 &#xA;jd      2451910.5    2454466.5    2454832.5    2455197.5    2455562.5 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I believe that all the values which are time dependent are for 0:00 hours on Jan. 1 of the year given.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the values I think I've figured out:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;tilt is the obilquity of the ecliptic&#xA;dow is the day of the week, where Monday is day 1&#xA;bs is the number of the day of the year when British Summer Time (BST) begins&#xA;fs is the number of the day of the year when BST ends&#xA;jd is the Julian day number (of 0:00 hours Jan. 1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Values I'm unsure about:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;delta t is some sort of time delta, but I don't know what&#xA;gst seems to be Greenwich Mean Sidereal Time, but for what moment?&#xA;x1, x2, and x3 I'm clueless about&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here are my questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What might delta t be?  &lt;/li&gt;&#xA;&lt;li&gt;Is gst in fact Greenwich Mean Sidereal Time?  For what moment?  &lt;/li&gt;&#xA;&lt;li&gt;What are x1, x2, and x3?  (This is low-priority question.)&lt;/li&gt;&#xA;&lt;li&gt;How can delta t, gst, and, perhaps other values, be determined for &#xA;    2018, 2019, ...?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Any help will be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roger House&lt;/p&gt;&#xA;" OwnerUserId="14775" LastActivityDate="2016-11-06T18:58:40.550" Title="Attempting to determine astronomy values in a parameter file" Tags="&lt;fundamental-astronomy&gt;&lt;coordinate&gt;&lt;observational-astronomy&gt;&lt;positional-astronomy&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18970" PostTypeId="1" CreationDate="2016-11-07T04:15:15.280" Score="3" ViewCount="1630" Body="&lt;p&gt;This book, &lt;em&gt;Astronomy in Minutes: 200 Key Concepts Explained in an Instant&lt;/em&gt;, by Giles Sparrow, says the photosphere is a visible surface at first, but after that, it says the Sun becomes cool and then transparent, so I am confused.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Our local star is a fiery ball of gas 1.39 million kilometres (863,000 miles) across. Powered by nuclear fusion, it has a three-layered internal structure (see pages 272 and 286). Its visible surface or 'photosphere' marks the zone &lt;strong&gt;where the Sun's hot gases grow sparse and cool enough to become transparent&lt;/strong&gt;, at a temperature of around 5,500°C (9,900°F), but the Sun's sparse atmosphere or corona extends much further outwards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I think the sentence highlighted in &lt;strong&gt;bold&lt;/strong&gt; (not highlighted in the original) represents the definition of chromosphere... Is that right?&lt;/p&gt;&#xA;" OwnerUserId="14782" LastEditorDisplayName="user24157" LastEditDate="2020-07-02T10:11:46.550" LastActivityDate="2020-07-02T10:11:57.387" Title="Photosphere is relatively transparent. Is that right?" Tags="&lt;the-sun&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18972" PostTypeId="2" ParentId="18970" CreationDate="2016-11-07T06:12:04.853" Score="5" Body="&lt;p&gt;See the other &lt;a href=&quot;https://astronomy.stackexchange.com/a/18979/7982&quot;&gt;answer by @RobJeffries&lt;/a&gt; for a more thorough explanation. Here I just address the the question of how something transparent can also be visible.&lt;/p&gt;&#xA;&lt;p&gt;Transparent things &lt;a href=&quot;https://i.stack.imgur.com/gr2fA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;can still emit bright light&lt;/a&gt;. Under some circumstances it can even approximate blackbody radiation. Think about red hot glass, and then imagine it is much bigger and the glass is now gas.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gr2fA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gr2fA.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; from &lt;a href=&quot;http://www.lightlead.com/Glowing_Glass.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Glowing Glass&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aPHo9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aPHo9.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; from Wikipedia&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="16685" LastEditDate="2020-07-01T15:03:09.213" LastActivityDate="2020-07-01T15:03:09.213" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="18974" PostTypeId="1" CreationDate="2016-11-07T12:09:07.513" Score="1" ViewCount="534" Body="&lt;p&gt;I am using Jean Meeus book Astronomical Algorithms and the C software that comes with the book to calculate planetary data. I am comparing two planet's longitude to calculate the time when they are a fixed amount of longitude apart such as 45, 60, 90 degrees apart. These are the standard planetary aspects used in astrology.  I can not calculate the correct time when one or both of the planets is approaching its stationary retrograde motion point. At this point the planets velocity approaches zero and the longitude does not have the decimal point accuracy to make a comparison. The time will be off by anywhere from a few minutes to eleven hours when compared to a commercial ephemeris. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q.) Does anyone know a different method other than simple longitude comparison that can calculate the correct planetary aspect time when planets are near their retrograde motion stationary point?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- Edit is Below --&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example of my problem. My calculations below are made with Astronomical Algorithms software at 1 minute increments for August 1, 2016 at midnight. The calculations show Venus longitude which is moving Direct Motion and Uranus which is moving Retrograde Motion. The American Ephemeris at midnight lists the Venus 120(Trine) Uranus as occurring at 4:10 AM. The Astronomical Algorithms calculations below identify Venus 120 Uranus at 3:56AM which is 16 minutes early. Every time a slow outer planet like Uranus is near a station point and moving very slow my aspect time is always off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q) Is there a different way to calculate the aspect rather than longitude comparison? Maybe use some king of time correction function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geocentric Longitude &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus Direct Motion &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uranus Retrograde Motion &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Date ___ Time _______ Venus _______ Uranus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;08/01/16 03:53 AM   144.5040255637  24.5061221269&lt;/p&gt;&#xA;&#xA;&lt;p&gt;08/01/16 03:54 AM   144.5048790144  24.5061207491&lt;/p&gt;&#xA;&#xA;&lt;p&gt;08/01/16 03:55 AM   144.5057324652  24.5061193708&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;08/01/16 03:56 AM 144.506585916   24.5061179922  Venus 120 Uranus&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;08/01/16 03:57 AM   144.5074393668  24.5061166132&lt;/p&gt;&#xA;&#xA;&lt;p&gt;08/01/16 03:58 AM   144.5082928175  24.5061152338&lt;/p&gt;&#xA;&#xA;&lt;p&gt;08/01/16 03:59 AM   144.5091462683  24.506113854&lt;/p&gt;&#xA;&#xA;&lt;p&gt;08/01/16 04:00 AM   144.509999719   24.5061124738&lt;/p&gt;&#xA;&#xA;&lt;p&gt;08/01/16 04:01 AM   144.5108531697  24.5061110932&lt;/p&gt;&#xA;&#xA;&lt;p&gt;08/01/16 04:02 AM   144.511706621   24.5061097123&lt;/p&gt;&#xA;&#xA;&lt;p&gt;08/01/16 04:03 AM   144.5125600718  24.5061083309&lt;/p&gt;&#xA;&#xA;&lt;p&gt;08/01/16 04:04 AM   144.5134135225  24.5061069491&lt;/p&gt;&#xA;&#xA;&lt;p&gt;08/01/16 04:05 AM   144.5142669732  24.5061055669&lt;/p&gt;&#xA;&#xA;&lt;p&gt;08/01/16 04:06 AM   144.5151204239  24.5061041844&lt;/p&gt;&#xA;&#xA;&lt;p&gt;08/01/16 04:07 AM   144.5159738746  24.5061028014&lt;/p&gt;&#xA;&#xA;&lt;p&gt;08/01/16 04:08 AM   144.5168273253  24.5061014181&lt;/p&gt;&#xA;&#xA;&lt;p&gt;08/01/16 04:09 AM   144.5176807759  24.5061000343&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;08/01/16 04:10 AM 144.5185342266  24.5060986502 American Ephemeris Venus 120 Uranus&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;08/01/16 04:11 AM   144.5193876773  24.5060972657&lt;/p&gt;&#xA;" OwnerUserId="14787" LastEditorUserId="14787" LastEditDate="2016-11-16T21:12:16.237" LastActivityDate="2016-11-16T21:12:16.237" Title="How to calculate planetary aspects at a stationary point?" Tags="&lt;coordinate&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18975" PostTypeId="2" ParentId="13162" CreationDate="2016-11-07T12:33:04.650" Score="3" Body="&lt;p&gt;The book by Dvorak et al. 2006 &quot;Chaos and Stability in Planetary Systems&quot; is a good place to start (as with many titles by Dvorak), but reviewing and studying recent(ish) published work about the topic can greatly enhance understanding of the topic.  For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Malhotra et al. 2011, &lt;a href=&quot;http://www.pnas.org/content/98/22/12342.full&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chaos and stability of the solar system&lt;/a&gt;, &lt;em&gt;PNAS&lt;/em&gt; (full text online) - this article explores the notion that there may be a significant element of chaos in the planetary orbits. A small snippet from the abstract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Over the last two decades, there has come about a recognition that chaotic dynamics is pervasive in the solar system. We now understand that the orbits of small members of the solar system—asteroids, comets, and interplanetary dust—are chaotic and undergo large changes on geological time scales. Are the major planets' orbits also chaotic? The answer is not straightforward, and the subtleties have prompted new questions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In a similar vein, the article: Barnes and Quinn, 2004, &lt;a href=&quot;http://faculty.washington.edu/rkb9/publications/bq04.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;The (In)stability of Planetary Systems&lt;/a&gt;, &lt;em&gt;The Astrophysical Journal&lt;/em&gt; (full PDF online),&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;present results of numerical simulations that examine the dynamical stability of known planetary systems, a&#xA;  star with two or more planets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user14785" LastActivityDate="2016-11-07T12:33:04.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18976" PostTypeId="1" CreationDate="2016-11-07T13:10:52.780" Score="4" ViewCount="150" Body="&lt;p&gt;The planet Mercury has a myriad of impact features - this has been known for some time. However, observe in the following images, a relatively large 'High-Mg region' detected on the planet (below):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fWDsR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fWDsR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image source: &lt;a href=&quot;http://messenger.jhuapl.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Messenger website&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the &lt;a href=&quot;http://messenger.jhuapl.edu/Resources/Press-Conference/20150316.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;group's press conference website&lt;/a&gt;, they state that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The unusual composition of the high-Mg region and the relatively thin crust in this region suggest that it may be the remains of an ancient and highly degraded impact basin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How is the 'unusual composition' of this region suggestive of being the remains of an ancient, highly degraded, impact basin?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user14785" LastEditorUserId="6" LastEditorDisplayName="user15217" LastEditDate="2019-05-02T03:22:36.483" LastActivityDate="2019-05-02T03:22:36.483" Title="Massive degraded impact crater on Mercury - chemical evidence?" Tags="&lt;planet&gt;&lt;fundamental-astronomy&gt;&lt;mercury&gt;&lt;impact&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18977" PostTypeId="2" ParentId="18967" CreationDate="2016-11-07T14:12:16.163" Score="2" Body="&lt;p&gt;I don't think any of those would give a supernova, because the first two don't cross an important mass boundary, and the last one might make a black hole from a neutron star, but that's not going to be much of a supernova because it's already very compact and there's not going to be much to have a core bounce off of, nor is the white dwarf massive enough for a type Ia.  They should all produce some significant emission, depending on how fast the mass is introduced.  Perhaps somewhere between what we would call a supernova, and something more along the lines of a classical nova (for the first one) and quasiperiodic X-ray flaring (for the next two).   They could get pretty bright, and I don't know if the nomenclature has individual names for each one of those, but they don't quite sound like supernovae.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-11-07T14:12:16.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18979" PostTypeId="2" ParentId="18970" CreationDate="2016-11-07T16:36:16.350" Score="5" Body="&lt;p&gt;The Sun is a big ball of relatively dense gas. At most points within the Sun, if a photon is emitted, it will be re-absorbed or scattered on a relatively short length scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The photosphere of the Sun (what we can actually see) is simply defined as that surface at which a photon that is emitted outwards will most likely escape. This is also the definition of transparency!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, what lies &lt;em&gt;above&lt;/em&gt; the photosphere is transparent to the radiation emitted at the photosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This definition is inexact in the sense that one can only define a probability that a photon will escape. By convention, the photosphere is usually defined to be where the &lt;a href=&quot;https://en.wikipedia.org/wiki/Optical_depth&quot; rel=&quot;nofollow noreferrer&quot;&gt;optical depth&lt;/a&gt; is approximately 2/3 and a photon has about a 50% chance of escaping. This in turn is wavelength dependent (since the absorption coefficient of the gas in the Sun's atmosphere varies with wavelength) and this is why we see features and absorption lines in the solar spectrum - we are viewing to different depths (and hence temperatures) into the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that the photosphere is at this particular density/temperature, is that the opacity at the photosphere and just below is dominated by H$^{-}$ ions. These are formed from the copious H atoms combining with free electrons from easily-ionised heavier elements like Na, Fe, Mg etc. These H$^{-}$ ions can then easily be disintegrated by any photon with energy more than about 0.75 eV leading to opaqueness to light with $\lambda &amp;lt; 2.2\,\mu$m. However, as the density/temperature  decreases with height it becomes increasingly difficult to form H$^{-}$ ions because the supply of free electrons dries up, collisonal timescales increase, and the gas becomes transparent to visible light, except at distinct wavelengths corresponding to absorption lines of various chemical species. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you like you can turn this around and ask yourself - if I fired a photon at the Sun, at what position in the Sun's atmosphere would it most likely be absorbed? The answer is the photosphere, the depth at which the Sun ceases to be transparent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB: The chromosphere and corona lie above the photosphere, are hotter but much less dense, and are transparent to visible light of almost all wavelengths. They too emit light, mostly in the form of narrow emission lines with very little continuum. The light from the photosphere below usually swamps our view of the chromosphere, however it can be seen above the limb of the Sun during a solar eclipse and high resolution spectroscopy can detect the chromospheric emission lines, especially in the ultra-violet and EUV regions where it dominates.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-11-07T17:06:52.090" LastActivityDate="2016-11-07T17:06:52.090" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18981" PostTypeId="1" AcceptedAnswerId="18983" CreationDate="2016-11-07T17:52:25.693" Score="3" ViewCount="183" Body="&lt;p&gt;What does &quot;time&quot; mean in the term space-time? I am trying to understand why anything that travels through space, from a proton to massive galaxies, is always referred to as traveling in space-time, not just space. Does this have to do with time being relative?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that 3d space and time together constitute a space-time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone illustrate with a scenario to point out the need to describe things in terms of space-time. &lt;/p&gt;&#xA;" OwnerUserId="14791" LastEditorUserId="6" LastEditDate="2016-11-11T21:21:05.223" LastActivityDate="2017-01-12T20:33:49.050" Title="In layman's terms, why do we put the &quot;time&quot; in space-time?" Tags="&lt;astrophysics&gt;&lt;space-time&gt;&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18983" PostTypeId="2" ParentId="18981" CreationDate="2016-11-07T21:12:37.500" Score="4" Body="&lt;p&gt;To answer your point about why &quot;spacetime&quot; as opposed to space and then time - photons (and other massless particles) travel through spacetime on &quot;null geodesics&quot; - in other words they do not experience time at all. But they are not in some separate class as particles at approach ever closer to the speed of light travel on paths through spacetime that are ever closer to null geodesics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words the four dimensional geometry is inextricably linked and &quot;space&quot; and &quot;time&quot; are a unified whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory humans moving also experience this effect but the speed at which we move is far too slow for this ever to manifest itself to our senses. But we do see its impact in high precision timing - such as with GPS satellites.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2016-11-07T21:12:37.500" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18985" PostTypeId="2" ParentId="18923" CreationDate="2016-11-07T22:13:05.673" Score="4" Body="&lt;p&gt;Most, and perhaps all, stars form in stellar clusters.  Most likely, the sun formed in a open cluster.  In clusters the density of stars is quite high, therefore close encounters are common.  So, I think it would be quite likely for such close encounters to disturb Oort clouds enough to send comets out of their host systems.  Some small fraction of these comets would be recaptured by other stars within the open cluster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, it is thought that in the early period of planetary disk formation, there were many more planets than today, but they often collided with each other, often sending pieces out of the stellar system. Moreove, many small planets and asteroids were thrown out by gravitational interactions with the large planets (see &lt;a href=&quot;https://arxiv.org/abs/1504.01160&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tremaine&lt;/a&gt;).  These 'debris' would be roaming interstellar space.  Occasionally, one could fall into any random stellar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The odds that we have any asteroids or comets from any particular nearby star is low, unless it happened to have been from the Sun's birthplace.  We do not have the ability to determine which of the nearby stars are from the Sun's open cluster, but the Gaia Satellite may make that possible soon.   And, if we explore the solar system long enough, we may find exo-solar asteroids among the hordes of asteroids.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2016-11-07T22:13:05.673" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18986" PostTypeId="1" AcceptedAnswerId="18987" CreationDate="2016-11-07T22:32:50.660" Score="8" ViewCount="3815" Body="&lt;p&gt;While studying photometric redshifts, I came across the ugriz (or u'g'r'i'z) system for classifying magnitudes of galaxies, but I didn't find much information on the internet about how it works. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody explain that? Do the letters &quot;ugriz&quot; have any special meaning, or is its just a nomenclature?&lt;/p&gt;&#xA;" OwnerUserId="14422" LastEditorUserId="2153" LastEditDate="2016-11-08T00:32:27.970" LastActivityDate="2016-11-08T18:31:01.230" Title="What is the ugriz magnitude system?" Tags="&lt;galaxy&gt;&lt;photometry&gt;&lt;magnitude&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18987" PostTypeId="2" ParentId="18986" CreationDate="2016-11-07T22:43:06.383" Score="8" Body="&lt;p&gt;&quot;ugriz&quot; is short for U-band, G-band, R-band, I-band, and Z-band, detailed on &lt;a href=&quot;https://en.wikipedia.org/wiki/Photometric_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Wikipedia article on photometric systems&lt;/a&gt;:&#xA;$$\begin{array}{|c|c|}&#xA;\hline \text{Band}&amp;amp;\text{Effective Wavelength Midpoint }(\lambda_{\text{eff}})\\&#xA;\hline \text{U} &amp;amp; \text{365 nm}\\&#xA;\hline \text{G} &amp;amp; \sim\text{475 nm}\\&#xA;\hline \text{R} &amp;amp; \text{658 nm}\\&#xA;\hline \text{I} &amp;amp; \text{806 nm}\\&#xA;\hline \text{Z} &amp;amp; \text{900 nm}\\&#xA;\hline&#xA;\end{array}$$&#xA;This $ugriz$ system is slightly different from the $u'g'r'i'z'$ system; conversions can be found &lt;a href=&quot;http://classic.sdss.org/dr4/algorithms/jeg_photometric_eq_dr1.html#usno2SDSS&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. The starting point is the same (note that $u(\text{2.5m})=u'$), but there are slight vertical and horizontal shifts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The differences between $ugriz$ and $u'g'r'i'z'$ are much smaller than the differences between $u'g'r'i'z'$ and the $UBV(RI)_C$ system, which was replaced by them both. This older system of filters had large overlaps and did not cover as broad a range. See Fig. 1 from &lt;a href=&quot;http://www.phys.lsu.edu/~jclem/publications/phd.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this thesis&lt;/a&gt; for a comparison:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2PX2z.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2PX2z.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-11-08T18:31:01.230" LastActivityDate="2016-11-08T18:31:01.230" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18988" PostTypeId="1" CreationDate="2016-11-08T00:33:59.493" Score="-1" ViewCount="268" Body="&lt;p&gt;I understood the literal meaning of the red-underlined sentences, but frankly speaking, I don't know the processes. I mean I want to know how Earth's rotation makes a cycle of two high and two low tides in the above context and why, of course how, the Moon experiences only slight tidal forces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in the second sentence that I underlined in red, the phrase 'as its distance from Earth varies throughout each month' means that the Moon has slowly gotten away from the Earth, I think. Am I right? Honestly, I can not understand the process of the second sentence entirely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you. &lt;/p&gt;&#xA;" OwnerUserId="14782" LastEditorUserId="14782" LastEditDate="2016-12-15T04:54:52.677" LastActivityDate="2016-12-15T04:54:52.677" Title="Why and how the Moon experiences only slight tidal forces" Tags="&lt;the-moon&gt;&lt;tidal-forces&gt;" AnswerCount="3" CommentCount="1" ClosedDate="2016-12-15T14:11:48.157" ContentLicense="CC BY-SA 3.0" />
  <row Id="18989" PostTypeId="1" CreationDate="2016-11-08T04:39:27.420" Score="-2" ViewCount="303" Body="&lt;p&gt;&lt;em&gt;By extension, any object in this kind of orbit has also become known as a Trojan, so Mars, Uranus and Neptune all have Trojans of their own. Earth's first Trojan, the asteroid 2010 TK, was identified in 2011. &lt;strong&gt;Saturn's moons Tethys and Dione, meanwhile, share their own orbits with Trojan satellites&lt;/strong&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me understand the last sentence(Saturn's moons Tethys...). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this context, what the Trojan satellites is? &lt;/p&gt;&#xA;" OwnerUserId="14782" LastActivityDate="2016-11-08T12:33:22.803" Title="Trojan satellites" Tags="&lt;trojan-asteroids&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2016-12-18T01:33:32.483" ContentLicense="CC BY-SA 3.0" />
  <row Id="18990" PostTypeId="2" ParentId="4798" CreationDate="2016-11-08T06:14:39.943" Score="3" Body="&lt;p&gt;I don't know what you're talking about. The only one that seems to have mostly flat rocks is Venus. At least based on what little photographs we have from the surface of Venus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iiY9K.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iiY9K.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dNagM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dNagM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14795" LastActivityDate="2016-11-08T06:14:39.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18991" PostTypeId="2" ParentId="18988" CreationDate="2016-11-08T07:02:19.663" Score="2" Body="&lt;p&gt;As stated in the quoted book excerpt, the tidal forces that arise from the Moon's gravity create a bulge in Earth's oceans on the side of the planet facing the Moon and on the side facing opposite the Moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine that you are on a shoreline of some arbitrary continent. As the Earth spins, the shoreline you are on is carried into the oceanic bulge that faces the Moon. This is your first high tide of the day and is the strongest, since this bulge is closest to the Moon and therefore larger. As the Earth continues to spin, you are carried out of this bulge and the water level recedes, reaching your first low tide when the Earth has rotated a further 90 degrees and your shoreline is perpendicular to the tidal bulges. When the Earth rotates another 90 degrees, your shoreline will enter the oceanic bulge opposite the Moon and you will experience your second high tide of the day. Then, just as before, the Earth's rotation will carry you out of the bulge and to a location perpendicular to the tidal bulges, and you will experience your second low tide of the day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that the Moon is slowly receding from the Earth at a rate of about 3 centimeters per year. However, that is not what the second underlined sentence was referring to. It is referring to the slight tidal stretching the Moon experiences during its orbit. The Moon's orbit is not perfectly circular; as with most orbits, it has a slight eccentricity. This causes the Moon to vary in its distance from Earth between 362,600 and 405,400 kilometers over the course of its ~28 day orbit. At its closest approach to Earth (perigee), it feels a stronger pull from Earth's gravity and its crust stretches slightly. Then once it reaches its farthest point away from Earth (apogee) the tidal forces are diminished due to the distance and its crust relaxes, which can lead to moon-quakes.&lt;/p&gt;&#xA;" OwnerUserId="14794" LastActivityDate="2016-11-08T07:02:19.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18992" PostTypeId="2" ParentId="18989" CreationDate="2016-11-08T12:33:22.803" Score="2" Body="&lt;p&gt;I post an answer from the comment. &#xA;Trojan satellites or trojan moon are natural satellites that are in the lagrangian point (L4 and L5) of another bigger (more massive) moon. All the known examples are Saturn's moon. In particular Tethys and Dione have other moon in their lagrangian point L4 and L5&lt;/p&gt;&#xA;" OwnerUserId="14600" LastActivityDate="2016-11-08T12:33:22.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18994" PostTypeId="1" CreationDate="2016-11-08T15:13:18.617" Score="3" ViewCount="275" Body="&lt;p&gt;Why does disorder  (entropy) increase in the same direction of time. as that in which the universe expands?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why can't entropy decrease in the same direction in which it increases ?&lt;/p&gt;&#xA;" OwnerUserId="14078" LastEditorUserId="14078" LastEditDate="2016-11-17T18:08:26.480" LastActivityDate="2016-11-18T09:25:34.473" Title="Why does disorder (entropy) increase in the same direction of time as that in which the universe expands?" Tags="&lt;space-time&gt;&lt;expansion&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18995" PostTypeId="1" CreationDate="2016-11-08T15:14:19.600" Score="9" ViewCount="2590" Body="&lt;p&gt;I know the question seems silly, but as far as I've understood, is this the definition of a dwarf planet:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;circling around a star, and not being a star itself&lt;/li&gt;&#xA;&lt;li&gt;having a round shape (otherwise every comet or asteroid could be a (dwarf) planet)&lt;/li&gt;&#xA;&lt;li&gt;not being able to clear its own orbit (otherwise it's a planet)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If we have a look at Jupiter, then we can easily confirm the two first points, but the third one is an issue: in Jupiter's orbit there are two sets of asteroids, called the Trojans, which are located, together with Jupiter, in some kind of regular triangle, turning around the sun. Following the laws of Kepler, Jupiter and the Trojans have the same speed, so Jupiter can't catch them and although Jupiter's gravity is very strong, it does not stretch to the other side of the sun, which means, following the definition, that Jupiter is a dwarf planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know, Pluto is called a dwarf planet because of the many objects in its orbit, and Pluto's gravity is so small that it can't get rid of those objects. Nevertheless in a scientific you can't use vague wordt like &quot;vicinity&quot;, &quot;close&quot;, ..., so you can't say Pluto is a dwarf while Jupiter isn't because of the vicinity of the mentioned objects (being &quot;too close&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I have a point here or am I missing one? :-)&lt;/p&gt;&#xA;" OwnerUserId="14381" LastActivityDate="2019-04-28T16:49:45.930" Title="Is Jupiter a dwarf planet?" Tags="&lt;planet&gt;&lt;dwarf-planets&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18996" PostTypeId="2" ParentId="18995" CreationDate="2016-11-08T15:26:58.583" Score="13" Body="&lt;p&gt;I suggest taking a look to &lt;a href=&quot;https://en.wikipedia.org/wiki/Clearing_the_neighbourhood&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Clearing_the_neighbourhood&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quoting it,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the end stages of planet formation, a planet (as so defined) will have &quot;cleared the neighbourhood&quot; of its own orbital zone, meaning it has become gravitationally dominant, and there are no other bodies of comparable size other than its satellites or those otherwise under its gravitational influence. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Trojans are textbook example of objects under the influence of a planet (Jupiter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For another example, Pluto is said to not have cleared its neighbourhood because Neptune is in it. Neptune has cleared its neighbourhood although Pluto is in it because Pluto is a lot smaller and Pluto orbit is governed by resonance with Neptune.&lt;/p&gt;&#xA;" OwnerUserId="14097" LastEditorUserId="181" LastEditDate="2017-06-01T14:28:01.927" LastActivityDate="2017-06-01T14:28:01.927" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18997" PostTypeId="2" ParentId="18988" CreationDate="2016-11-08T15:40:01.183" Score="3" Body="&lt;p&gt;I think it's a somewhat poorly written page and Xakarus Alldredge's answer is very good.   I thought I'd add some pictures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Tidal force the Earth exerts on the Moon is much stronger than the one the Moon exerts on the Earth and it's technically not a force either, it's a secondary effect from the force of gravity.   But the tidal &quot;force&quot; on the moon is strong enough that the Moon is tidally locked and the same side of the Moon always faces the Earth.  Any object in a very strong tidal force tends to become tidally locked over time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being tidally locked, the moon has a permanent tidal bulge and that bulge doesn't move very much but it does shift side to side as the moon wobbles slightly and it does grow larger and smaller as the Moon moves closer to and further from the earth.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets that rotate like the Earth have an equatorial bulge.  The Moon, which rotates very slowly, only enough to keep the same size facing the Earth has a bit of an egg shape or tidal bulge, not rounder in the middle like planets with fast rotation, but stretched at 2 points, the part facing the earth and the part facing away from the earth both stretch outwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This drawing is exaggerated but it gives an idea of how the Earth's gravity tidally affects the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZF39q.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZF39q.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://astronomy.nju.edu.cn/~lixd/GA/AT4/AT408/HTML/AT40804.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This diagram is also exaggerated, but, as Xakarus Allderdge points out, when the Moon is closer to the Earth at perigee, the tidal force is stronger and the Moon's tidal bulge grows bigger and at apogee, it's smaller.  The Moon's &lt;a href=&quot;https://community.dur.ac.uk/john.lucey/users/lunar_sid_syn.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;sidereal orbital period&lt;/a&gt;, going from from apogee to perigee back to apogee takes 27.3 days so the rise and fall and slight shift side to side of the lunar bulge happens slowly, much slower than the tides on the Earth which circle the Earth about daily.  The Earth also has a &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth_tide&quot; rel=&quot;nofollow noreferrer&quot;&gt;land bulge&lt;/a&gt; usually called an earth tide, land tide or body tide, caused by the moon, but it's quite a bit smaller than the ocean tides and because there's nothing to measure it against, it can only be observed by very precise satellite measurement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K8NBw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K8NBw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a &lt;a href=&quot;http://www.space.com/26246-lunar-tide-seen-from-space.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;solid article&lt;/a&gt; on the Moon's bulge, Sometimes called a lunar body tide.  It's been measured by satellite.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-11-08T15:40:01.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18998" PostTypeId="2" ParentId="66" CreationDate="2016-11-08T16:37:10.297" Score="2" Body="&lt;p&gt;Yes, the orbits change massively over time.  For example, Earth's eccentricity (how close to a circle the orbit is), its axial tilt (what causes seasons), and precession (which direction the Earth's spin axis points) change on these huge cycles, tens of thousands to hundreds of thousands of years long.  They are caused by the pull of the other planets, mainly Jupiter and Saturn (since they are so huge), as well as the physics of inertial reference frames.  These are known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Milankovitch_cycles&quot; rel=&quot;nofollow noreferrer&quot;&gt;Milankovitch cycles&lt;/a&gt;.  Here is a &lt;a href=&quot;https://www.youtube.com/watch?v=ztninkgZ0ws&quot; rel=&quot;nofollow noreferrer&quot;&gt;neat video&lt;/a&gt; explaining what each of the cycles are in detail.  That video is about Earth, but the same concepts apply to all the other planets too, just on different amounts and timescales.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An asteroid hitting a planet would technically change its orbit, however how much is questionable.  An asteroid a couple km across would barely nudge the Earth since the Earth is bigger by many orders of magnitude.  It would also destroy everything on the planet, but thats another story.  Way back in the solar system's formation, &lt;a href=&quot;https://en.wikipedia.org/wiki/Theia_(planet)&quot; rel=&quot;nofollow noreferrer&quot;&gt;when Thea hit Earth&lt;/a&gt; (theoretically) it was much closer in size than the Earth, so no doubt that it caused a much more significant shift in orbit.&lt;/p&gt;&#xA;" OwnerUserId="14255" LastActivityDate="2016-11-08T16:37:10.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18999" PostTypeId="1" CreationDate="2016-11-08T21:50:51.010" Score="6" ViewCount="158" Body="&lt;p&gt;When a meteor is going through our atmosphere you will see a light beam made by the gas the meteor compressed. But what is the minimum size to see a 'falling star'? Is a 1 mm&lt;sup&gt;3&lt;/sup&gt; meteor visible?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorDisplayName="user1569" LastEditDate="2016-11-09T16:37:56.977" LastActivityDate="2016-11-09T16:37:56.977" Title="What is the minimum meteor size to see a falling star?" Tags="&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19000" PostTypeId="2" ParentId="18999" CreationDate="2016-11-08T23:12:08.610" Score="4" Body="&lt;p&gt;The International Meteor Organization (IMO) defines a meteoroid as &lt;a href=&quot;http://www.imo.net/resources/glossary/#m&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;a solid object moving in interplanetary space, of a size considerably smaller than a asteroid and considerably larger than an atom or molecule.&quot;&lt;/a&gt; This definition is further refined by Rubin and Grossman, who state &quot;a meteoroid is a $10$-$\mathrm{\mu m}$ to $1$-$\mathrm{m}$-size natural solid object moving in interplanetary space.&quot; Additionally, the same authors consider a defintion for a micrometeoroid, which they define as &quot;a meteoroid $10\hspace{2pt}\mathrm{\mu m}$ to $2\hspace{2pt}\mathrm{mm}$ in size.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to astronomer Karen Masters from Cornell, the typical meteor has a mass of &lt;a href=&quot;http://curious.astro.cornell.edu/physics/73-our-solar-system/comets-meteors-and-asteroids/meteors/303-what-is-the-typical-size-of-a-visible-shooting-star-intermediate&quot; rel=&quot;nofollow noreferrer&quot;&gt;a few milligrams&lt;/a&gt;. The American Meteor Society (AMS) also states that the most-commonly observed meteors are roughly the size of a grain of sand. This would seem to imply that many meteors are smaller. In fact, this is also implied by the various statements by many groups, including the AMS, IMO, and IAU, that micrometeoroids commonly cause meteors. Because of this, I would argue that a meteoroid with a volume of $1$ $\mathrm{mm}^3$ would easily cause a relatively bright meteor. Statements surrounding the frequency with which micrometeoroids cause meteors would seem to imply that, on average, any body larger than $10$ $\mathrm{\mu m}$ will be observable as a meteor. These figures will vary largely depending on the entrance velocity of the object.&lt;/p&gt;&#xA;" OwnerDisplayName="user14781" LastActivityDate="2016-11-08T23:12:08.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19001" PostTypeId="1" AcceptedAnswerId="19005" CreationDate="2016-11-08T23:45:41.870" Score="7" ViewCount="166" Body="&lt;p&gt;While reading a paper on the helioseismology of sunspots (&lt;a href=&quot;http://www2.mps.mpg.de/projects/seismo/papers/SLiM.pdf&quot; rel=&quot;noreferrer&quot;&gt;Cameron, et al.&lt;/a&gt;), very brief reference is made to the &quot;vertical displacement of a blob of plasma.&quot; This caused me to wonder if the emissive surface of the sunspot itself were perhaps depressed from the remainder of the photosphere. If this does occur, is there any literature on mechanisms? I would be very interested to learn more. There was one paper I came across around formation of sunspots from a toroidal field (&lt;a href=&quot;http://adsabs.harvard.edu/full/1955ApJ...121..491P&quot; rel=&quot;noreferrer&quot;&gt;Parker&lt;/a&gt;); but, it seemed to focus more on the vertical displacement of magnetic field lines rather than the plasma itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerDisplayName="user14781" LastActivityDate="2016-11-09T23:18:31.530" Title="Are sunspots vertically displaced from the surrounding photospheric plasma?" Tags="&lt;the-sun&gt;&lt;magnetic-field&gt;&lt;plasma-physics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19004" PostTypeId="2" ParentId="18995" CreationDate="2016-11-09T13:12:03.297" Score="11" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Do I have a point here or am I missing one?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are missing several points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point 1: &quot;Clearing the neighborhood&quot; does not mean getting rid of every last spec of mass in the vicinity of a planet. The concept instead addresses the ability to somehow &quot;clear&quot; the vast majority of the mass in the neighborhood of a planet. The total mass of all of Jupiter's Trojans are less than a millionth of Jupiter's mass. By whatever mathematical expression one uses (there are three), that qualifies as clearing the neighborhood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point 2: &quot;Clearing the neighborhood&quot; is as much about the future as it is the present or past. Pluto is not a planet because not only has it not cleared its neighborhood, it will not do so while the Sun is still shining. Jupiter is a planet because by whatever objective measure one uses, it cleared its neighborhood shortly after the solar system formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point 3: There are three competing objective definitions of what qualifies as &quot;clearing the neighborhood&quot; as opposed to the unrealistic concept of having eliminated ever last spec of mass: Alan Stern's and Harold Levison's $\Gamma$, Steven Soter's $\mu$, and Jean-Luc Margot's $\Pi$. For an overview, read the wikipedia article on &lt;a href=&quot;https://en.wikipedia.org/wiki/Clearing_the_neighbourhood&quot; rel=&quot;noreferrer&quot;&gt;clearing the neighborhood&lt;/a&gt;, and follow the references therein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point 4: There are several ways a planet can clear its neighborhood. The most obvious are collisions and ejections from the star system. Less obvious are capturing them as moons or stable pseudo moons. Stable Trojans are an example of the latter.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2016-11-09T13:20:03.710" LastActivityDate="2016-11-09T13:20:03.710" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19005" PostTypeId="2" ParentId="19001" CreationDate="2016-11-09T16:43:07.327" Score="4" Body="&lt;h1&gt;Yes.&lt;/h1&gt;&#xA;&lt;p&gt;The idea that sunspots are depressed slightly came as a possible explanation for the &lt;a href=&quot;https://en.wikipedia.org/wiki/Wilson_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wilson effect&lt;/a&gt;. The Wilson effect was discovered as the shape of sunspots as viewed from Earth changed as the Sun rotates, in a way consistent with the change in perspective looking onto a slightly depressed region. While this isn't the only explanation for the effect, it's certainly the most prevalent.&lt;/p&gt;&#xA;&lt;p&gt;More specifically, as &lt;a href=&quot;http://www2.mps.mpg.de/dokumente/publikationen/solanki/r45.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solanki (2003)&lt;/a&gt; writes, the depressions indicate a lowering of the layer where the optical depth &lt;span class=&quot;math-container&quot;&gt;$\tau=1$&lt;/span&gt; (keep in mind that the bottom of the photosphere is the layer where &lt;span class=&quot;math-container&quot;&gt;$\tau=2/3$&lt;/span&gt;). There are two causes mentioned: lower temperature and magnetic effects.&lt;/p&gt;&#xA;&lt;p&gt;Sunspots are cooler than the surrounding areas, as is well known, and thus appear darker. We see the same thing apparent at the boundaries of &lt;a href=&quot;https://en.wikipedia.org/wiki/Granule_(solar_physics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar granules&lt;/a&gt;: Cooler, darker gas sinks and lets the hotter gas (which is less dense) move upward. Additionally, the opacity &lt;span class=&quot;math-container&quot;&gt;$\kappa$&lt;/span&gt; is temperature-dependent, which may impact how far one can see into the star.&lt;/p&gt;&#xA;&lt;p&gt;Not only is temperature a factor, but so is the magnetic field. Sunspots are, at heart, a magnetic phenomenon, and thus the radial force equation is substantially different. Normally, in a star, the equation of &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrostatic_equilibrium&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrostatic equilibrium&lt;/a&gt; is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\mathrm{d}P}{\mathrm{d}r}=-\rho g$$&lt;/span&gt;&#xA;for pressure &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;, density &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; and gravitational acceleration &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt;. However, when the magnetic field becomes important in a sunspot on the solar surface, the force balance is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\mathrm{d}P}{\mathrm{d}r}=\frac{B_z}{4\pi}\left(\frac{\mathrm{d}B_r}{\mathrm{d}z}-\frac{\mathrm{d}B_z}{\mathrm{d}r}\right)$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; are the radial and vertical coordinates (note the change of coordinate system - &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; is along the surface, and &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; is perpendicular to it!). The force from the magnetic field implies a lower gas pressure and a greater depression.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-11-09T23:18:31.530" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19006" PostTypeId="2" ParentId="6115" CreationDate="2016-11-09T17:13:29.770" Score="2" Body="&lt;p&gt;And here's another way to frame the situation.  Degeneracy is not responsible for the pressure in the electrons in a white dwarf, they would have that same pressure once the star contracts to the same size even if it was all ideal gas (say, if the electrons were distinguishable).  What is meant by &quot;degeneracy pressure&quot; is the perfectly mundane kinetic gas pressure that happens to be reached at the point where the Pauli exclusion principle makes it impossible for the star to lose any more heat, so it cannot contract any further.  Notice this also means that the &quot;degeneracy pressure&quot; limit for a given star is the maximum pressure the star can attain.  This surprises some people, who are inclined to think of degeneracy pressure as a kind of minimum below which the pressure cannot fall, but that would be true for a gas in a box of constant volume, not a self-gravitating contracting star.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2016-11-09T17:26:35.637" LastActivityDate="2016-11-09T17:26:35.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19007" PostTypeId="1" CreationDate="2016-11-09T17:15:36.303" Score="2" ViewCount="61" Body="&lt;p&gt;I'm mainly searching for the best(or good/significant/important) solar data archives via ground-based telescopes. If I'm not being very straightforward... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say if I wanted high quality/resolution images of the sun, I'd use SDO/SOHO's data archives-- they're easily accessible, queryable, and yield impressive images. So I'm looking for mainly visible spectrum images from ground-based telescopes rather than space observatories, although differing spectral images will be used just less-extensively. I've struggled to find any good data archives, however. NSO's archives seem all over the place, although the data it yields is impressive; so, if you are knowledgeable on how to recurringly access and reconstruct their data, I'd be much appreciative. Otherwise, is there any other ground-based observatories' archives that are freely available to the public, easily accessible, and present data online that anyone has used or know's about? Jpeg, png, etc. are all valuable, but if you know of any that use different file formats, they'd also be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind, I'm not looking to discuss/argue what the BEST one is, but rather what commonly used/good ones are. So, any and all input will be greatly appreciated, as my project is best researched with many differing observatories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, the better the telescope's quality in which the data is made available through, the better. So McMath-Pierce and other impressive unit's data should and will be given precedence over telescopes of less accuracy. &lt;/p&gt;&#xA;" OwnerUserId="13808" LastActivityDate="2016-11-09T17:15:36.303" Title="Visible Light Ground-Based Solar Observatory Data Archives?" Tags="&lt;the-sun&gt;&lt;telescope&gt;&lt;amateur-observing&gt;&lt;data-analysis&gt;&lt;observational-astronomy&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19008" PostTypeId="1" CreationDate="2016-11-09T19:50:03.077" Score="20" ViewCount="2190" Body="&lt;p&gt;The diagram below, which I stole from &lt;a href=&quot;https://astronomy.stackexchange.com/questions/18905/plot-of-best-available-resolution-vs-wavelength-radio-through-gamma-rays/18906#18906&quot;&gt;this post&lt;/a&gt; by @HDE226868, shows that angular resolution as a function of wavelength suddenly drops by three orders of magnitudes from visible to UV-light. The resolution of wavelengths shorter than what the Very Large Telescope Interferometer or the European Extremely Large Telescope detect, in the near UV, suddenly cuts off to a factor of a thousand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is obviously because of the properties of Earth's atmosphere. But major space telescopes like the JWST and WFIRST will fill in the far infrared gap. Why aren't there any as ambitious space telescopes planned for UV and shorter wavelengths? (Or is the sudden cut off in that diagram misleading?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because it is more difficult, even from in-space observatories, or is it because the angular resolution of UV and shorter wavelengths are of lesser scientific value?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/toitK.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/toitK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2016-11-09T20:02:39.930" LastActivityDate="2016-11-10T07:32:27.197" Title="Why are wavelengths shorter than visible light neglected by new telescopes?" Tags="&lt;telescope&gt;&lt;observational-astronomy&gt;&lt;spectra&gt;&lt;wavelength&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19009" PostTypeId="2" ParentId="19008" CreationDate="2016-11-09T20:32:57.933" Score="14" Body="&lt;p&gt;You're correct in that the sharp dropoff is simply because there are very few planned major telescopes operating in the UV range, whereas there are a substantial number planned in the infrared range. As I mentioned in my answer you linked to, &lt;a href=&quot;https://en.wikipedia.org/wiki/CHARA_array&quot; rel=&quot;noreferrer&quot;&gt;CHARA&lt;/a&gt; and the &lt;a href=&quot;https://en.wikipedia.org/wiki/European_Extremely_Large_Telescope&quot; rel=&quot;noreferrer&quot;&gt;EELT&lt;/a&gt;, two of the top planned infrared/visible projects, will use new adaptive optics technology, making them far superior to previous telescopes - even though they're ground-based.&lt;/p&gt;&#xA;&lt;p&gt;Obviously, UV telescopes cannot be ground-based, because Earth's atmosphere blocks a substantial amount of UV radiation. Therefore, any substantial improvement in ultraviolet astronomy will require a new space-based mission. The problem is that estimates for even modest increases require much larger mirrors. Proponents of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Advanced_Technology_Large-Aperture_Space_Telescope&quot; rel=&quot;noreferrer&quot;&gt;Advanced Technology Large-Aperture Space Telescope (ATLAST)&lt;/a&gt; proposals say that an 8-meter telescope, at the least, is needed to get good results at 0.11- to 2.5- &lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt;m wavelengths. That's much larger than HST or JWST - and ATLAST could grow to 16 meters!&lt;/p&gt;&#xA;&lt;p&gt;If ATLAST or a similar project is pursued, the angular resolution at UV wavelengths could be on the order of 0.1 arcseconds or, hopefully, lower. That would match and then beat Hubble. But early estimates put the cost at $4.5 billion for the 8-m version, and Hubble and other space-based telescopes have been famously hurt by unforeseen cost increases. Smaller strides may be needed before we can get to 8 meters, and certainly before we can get anywhere near 16. That's going to take a while, probably a decade or more from now.&lt;/p&gt;&#xA;&lt;h3&gt;References&lt;/h3&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.stsci.edu/institute/atlast/documents/mpostman_uvfuture_aip.pdf&quot; rel=&quot;noreferrer&quot;&gt;Postman &amp;amp; Mountain&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://cor.gsfc.nasa.gov/copag/rfi/ATLAST_PAG_white_paper_FINAL.pdf&quot; rel=&quot;noreferrer&quot;&gt;Thronson et al. (2015)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://arxiv.org/ftp/arxiv/papers/0904/0904.0941.pdf&quot; rel=&quot;noreferrer&quot;&gt;Postman et al. (2010)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-11-09T20:38:43.497" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19010" PostTypeId="2" ParentId="19008" CreationDate="2016-11-09T21:44:04.610" Score="15" Body="&lt;p&gt;There &lt;em&gt;are&lt;/em&gt; some technological issues to solve with putting any large telescope into space - and a space telescope is required at UV wavelengths. It is not possible to optimise such an instrument to work at &lt;em&gt;both&lt;/em&gt; UV and IR wavelengths because of issues like cooling, mirror coatings and such-like. The simple angular resolution limit of a telescope goes $\lambda/D$, so on the face of it, to get equivalent resolution to an optical telescope, a UV telescope can be smaller. However, you also have to have optics that are good to a small fraction of a wavelength, so much better than the visible/IR. At even shorter wavelengths then conventional &quot;optics&quot; doesn't work because photons get absorbed and you move to the grazing incidence technologies of X-ray telescopes, which is a whole different game and much harder to achieve a given angular resolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given all that, back in the 80s/90s I would guess that a decision was taken about the wavelength range to be covered by the successor to HST (ie JWST at a cost of approximately 10 billion USD) The real reason that no major UV successor to HST or IUE is ready to go now is simply that it is considered that the most important science priorities are achievable at near and mid-IR wavelengths. These are: observing the high redshift universe (essentially no UV light is detected from galaxies beyond a redshift of 3), observing star and planet formation (mostly in dusty environments where UV light cannot emerge and protoplanetary discs emit mostly at IR wavelengths) and doing exoplanetary science (planets are cooler than stars and emit mostly in the IR).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, I don't think there are any technological showstoppers to a big UV telescope (at least the equivalent of JWST), it just comes down to science prorities.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-11-10T07:32:27.197" LastActivityDate="2016-11-10T07:32:27.197" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19011" PostTypeId="2" ParentId="18988" CreationDate="2016-11-10T02:34:47.427" Score="0" Body="&lt;p&gt;The text in the book leaves something to be desired. It's not true that the moon &lt;em&gt;&quot;experiences only slight tidal forces&quot;&lt;/em&gt; as written. The tidal forces on the Moon due to the Earth are much greater than those on the Earth due to the moon because the Earth is much more massive than the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the existing tidal lock means that those tidal forces are nearly constant in time and so less interesting than the steadily changing effect experienced by most regions of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better statement would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The moon experiences only slight &lt;strong&gt;variations in&lt;/strong&gt; tidal forces&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The moon experiences only slight &lt;strong&gt;periodic distortion due to&lt;/strong&gt; tidal forces&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Both of which are comprehensively explain in the answers by &lt;a href=&quot;https://astronomy.stackexchange.com/a/18991/530&quot;&gt;Xakarus Alldredge&lt;/a&gt; and &lt;a href=&quot;https://astronomy.stackexchange.com/a/18997/530&quot;&gt;userLTK&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="530" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-11-10T02:34:47.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19012" PostTypeId="1" CreationDate="2016-11-10T03:03:45.850" Score="-1" ViewCount="46" Body="&lt;p&gt;I understood the meaning of the sentence literally but that's not enough to understand the process. Could you explain how that works? Thank you! &lt;/p&gt;&#xA;" OwnerUserId="14782" LastEditorUserId="14782" LastEditDate="2016-12-15T04:54:27.543" LastActivityDate="2016-12-15T04:54:27.543" Title="How does the Saturn's low temperature influence the colour of the cloud belts?" Tags="&lt;saturn&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2016-12-15T14:11:55.250" ContentLicense="CC BY-SA 3.0" />
  <row Id="19013" PostTypeId="1" CreationDate="2016-11-10T08:49:52.027" Score="3" ViewCount="66" Body="&lt;p&gt;I occasionally hear claims along the line of ' the mass of the universe is not uniformly distributed but actually is concentrated along a '  line, or some other lower  dimensional manifold &amp;amp; I would like like to know what statistical techniques, if any,  are used to make support these claims.  I'm not interested  in whether the claims have held, just in the what was used for the original analysis.  I have data  whose dimension I would like to estimate, and am familiar with correlation dimension.&lt;/p&gt;&#xA;" OwnerUserId="14806" LastActivityDate="2020-05-06T10:51:10.010" Title="statistical techniques for estimating distribution of mass" Tags="&lt;mass&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19014" PostTypeId="2" ParentId="19012" CreationDate="2016-11-10T11:56:25.290" Score="1" Body="&lt;p&gt;The answer is chemistry in this case.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of Hydrogen and Helium the atmospheres of the gas giants have atmospheres that most likely didn't change much since the formation of the solar system. In heavier elements however the gas giants are enriched (compared to the sun's atmosphere, which we think is the most original gas still around).&lt;br&gt;&#xA;Such a mixture of elements, given at the pressures and temperatures in Saturn's atmosphere (as well as in Jupiter's) will form large amounts of Ammonia naturally by &lt;a href=&quot;https://en.wikipedia.org/wiki/Equilibrium_chemistry&quot; rel=&quot;nofollow noreferrer&quot;&gt;equilibrium chemistry.&lt;/a&gt;&lt;br&gt;&#xA;If then the amount of Ammonia reaches vapour saturation, clouds will form as a result.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2016-11-10T11:56:25.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19015" PostTypeId="1" CreationDate="2016-11-10T13:21:35.617" Score="3" ViewCount="82" Body="&lt;p&gt;I want to watch the ISS with naked eye using this tool NASA gives &lt;a href=&quot;https://spotthestation.nasa.gov/sightings/view.cfm?country=Argentina&amp;amp;region=None&amp;amp;city=Santa_Fe#.WCRzKS3hC1s&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://spotthestation.nasa.gov/sightings/view.cfm?country=Argentina&amp;amp;region=None&amp;amp;city=Santa_Fe#.WCRzKS3hC1s&lt;/a&gt; but I dont know how to read grades in the sense of which direction I have to look to, in order to see the ISS. &lt;/p&gt;&#xA;" OwnerUserId="11562" LastActivityDate="2016-11-10T13:21:35.617" Title="how do I read grades for knowing where I have to look for watching the ISS?" Tags="&lt;artificial-satellite&gt;" AnswerCount="0" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19016" PostTypeId="1" AcceptedAnswerId="19023" CreationDate="2016-11-10T13:28:04.193" Score="2" ViewCount="569" Body="&lt;p&gt;&lt;a href=&quot;http://www.johnstonsarchive.net/astro/browndwarflist.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;There are a large fraction of M dwarfs&lt;/a&gt; which are claimed to be brown dwarfs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do we still use M-type and not create a new stellar type like L, T and Y? The main signature of M is TiO absorption bands. They appear in M-type brown dwarfs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is a spectrum enough to judge whether a celestial object is a brown dwarf or not? Or do we need to know its mass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The earliest type is &lt;a href=&quot;http://www.johnstonsarchive.net/astro/browndwarflist.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;M3.5&lt;/a&gt;. It means M-type objects older than M3.5 are necessarily brown dwarfs? Or is there a safe value, say, M9, is necessarily a brown dwarf? A brown dwarf is necessarily colder than a red dwarf? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There should be more brown dwarfs or red dwarfs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason we do not need to create a new stellar type is the spectra of brown dwarfs are more like red dwarf's spectra? &lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditDate="2016-11-12T09:39:19.793" LastActivityDate="2016-11-12T10:38:54.933" Title="How do I understand a brown dwarf with a M-type spectrum?" Tags="&lt;spectra&gt;&lt;brown-dwarf&gt;&lt;red-dwarf&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19019" PostTypeId="2" ParentId="19016" CreationDate="2016-11-10T18:35:32.930" Score="2" Body="&lt;p&gt;There are a couple different tests you could use to determine whether the object is a brown dwarf of a red dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could try to use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lithium_burning&quot; rel=&quot;nofollow noreferrer&quot;&gt;lithium test&lt;/a&gt; to determine whether the object in question is a brown dwarf or a low-mass red dwarf. Brown dwarfs generally don't reach temperatures high enough for lithium burning to occur at any significant rate, and so in low-mass M-type red dwarfs, there should be no evidence of a lithium line at 670.8 nm. This was originally proposed from models by &lt;a href=&quot;http://adsabs.harvard.edu/abs/1992ApJ...389L..83R&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rebolo et al. (1992)&lt;/a&gt;. The test isn't perfect, as some objects will burn some lithium at different points in their lives, but it &lt;em&gt;can&lt;/em&gt; work, and it's certainly a good starting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lithium isn't the only atmospheric component whose presence is temperature-dependent - titanium oxide, vanadium oxide, carbon monoxide, nitrogen are other examples - so by looking for spectral lines, it might be possible to determine the temperature of the atmosphere and compare the spectra to models of brown dwarf and red dwarf spectra (see &lt;a href=&quot;http://www.iac.es/galeria/nlodieu/media/articles/chapter1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; for more). However, some of these only disappear at temperatures much lower than values for which red dwarfs and brown dwarfs could be confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brown dwarfs also have a very small range of radii. $R\propto M$ for low-mass stars, whereas $R\propto M^{-1/3}$ for brown dwarfs supported by degeneracy pressure (see &lt;a href=&quot;https://arxiv.org/abs/astro-ph/9812091&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oppenheimer et al. (1998)&lt;/a&gt;). If you can measure the effective surface temperature of the object, $T_{\text{eff}}$, you should be able to determine its radius by assuming a black body-like temperature-luminosity relation. $T_{\text{eff}}$ could be determined in a few different ways, including using atmospheric models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I honestly don't see the need for an additional spectral class that would likely overlap the M class. It would create more confusion than already exists, and Morgan-Keenan system already in use is honestly complicated enough, and has been simplified from more complicated and problematic schemes. See the first table &lt;a href=&quot;http://www.stsci.edu/~inr/ldwarf1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, taken from Table 5 of &lt;a href=&quot;http://adsabs.harvard.edu/abs/1999ApJ...519..802K&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kirkpatrick et al. (1999)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="9301" LastEditDate="2016-11-11T13:03:21.797" LastActivityDate="2016-11-11T13:03:21.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19020" PostTypeId="2" ParentId="19013" CreationDate="2016-11-10T19:16:06.897" Score="3" Body="&lt;p&gt;Matter is indeed distributed in knots (clusters), filaments, and sheets (e.g. &lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/1996Natur.380..603B/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bond et al. 1996&lt;/a&gt;). This is predicted from theories and numerical simulations of structure formation under gravitational collapse, and is directly observable in large galaxy surveys by eye, such as &lt;a href=&quot;http://www.2dfgrs.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;The 2dF Galaxy Redshift Survey&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UxMUp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UxMUp.png&quot; alt=&quot;2dfgrs&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Two-point correlation function&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;To quantify this clustering, usually a &lt;em&gt;correlation function&lt;/em&gt; is used, either in 2D (&lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; and &lt;em&gt;y&lt;/em&gt; position on the sky), or 3D (if redshifts are available for the galaxies, giving the &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; dimension). The (two-point) correlation function &lt;span class=&quot;math-container&quot;&gt;$\xi(r)$&lt;/span&gt; gives the excess probability of finding two galaxies at a distance &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; from each other, relative to a random (Poisson) distribution.&#xA;If &lt;span class=&quot;math-container&quot;&gt;$\bar{n}$&lt;/span&gt; is the mean number density of galaxies, then the probability &lt;span class=&quot;math-container&quot;&gt;$dP$&lt;/span&gt; of finding a pair of galaxies at a separation &lt;span class=&quot;math-container&quot;&gt;$r_{12}$&lt;/span&gt; in volume elements &lt;span class=&quot;math-container&quot;&gt;$dV_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$dV_2$&lt;/span&gt; is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;dP = \bar{n}^2 \big(1 + \xi(r_{12})\big) dV_1 dV_2.&#xA;$$&lt;/span&gt;&#xA;The power spectrum &lt;span class=&quot;math-container&quot;&gt;$P(k)$&lt;/span&gt; of the galaxies is related to the two-point correlation function by&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\xi(r) = \frac{1}{2\pi^2} \int dk\,k^2\,P(k)\,\frac{\sin kr}{kr}&#xA;$$&lt;/span&gt;&#xA;Observationally it is found that the two-point correlation function is given by a power-law:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\xi(r) = \left( \frac{r}{r_0} \right)^{-\gamma},&#xA;$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$r_0$&lt;/span&gt; is the correlation length (of the order &lt;span class=&quot;math-container&quot;&gt;$5\,h^{-1}\mathrm{Mpc}$&lt;/span&gt;) and &lt;span class=&quot;math-container&quot;&gt;$\gamma=1.7$&lt;/span&gt;–&lt;span class=&quot;math-container&quot;&gt;$1.8$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the value of &lt;span class=&quot;math-container&quot;&gt;$r_0$&lt;/span&gt; depends on galaxy type; brighter and redder galaxies have a longer correlation length, i.e. they are more strongly clustered (e.g. &lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2008ApJ...687..919W&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wang et al. 2008&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Excursion set theory&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Using a formalism know as &lt;em&gt;excursion set theory&lt;/em&gt;, the mass fraction in knots, filaments, sheets, respectively, can be predicted analytically. The idea of this theory is to model the collapse of structure in an expanding Universe as a stochastic process, where an overdensity fluctuates and forms a structure if it crosses a given threshold. See e.g. &lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2006ApJ...645..783S&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shen et al. (2006)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Dynamical classification&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2009MNRAS.396.1815F&quot; rel=&quot;nofollow noreferrer&quot;&gt;Forero-Romero et al. (2009)&lt;/a&gt; proposed a dynamical classification of the cosmic web, based on evalution of the &lt;em&gt;deformation tensor&lt;/em&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;T_{\alpha\beta} = \frac{\partial^2 \phi}{\partial r_\alpha \partial r_\beta}&#xA;$$&lt;/span&gt;&#xA;of the gravitational potential &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt;, counting the number of eigenvalues above a certain threshold at each grid point in a cosmological N-body simulation, where the case of zero, one, two or three such eigenvalues corresponds to void, sheet, filament or a knot grid point.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Other methods&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I quick &lt;a href=&quot;https://www.google.dk/#q=quantifying+the+cosmic+web&amp;amp;start=10&quot; rel=&quot;nofollow noreferrer&quot;&gt;googling&lt;/a&gt; leads me to other methods I don't really know anything about, e.g. &quot;the Shapefinder diagonistic&quot; (&lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2016IAUS..308..250S/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Prakash 2016&lt;/a&gt;), and large-scale halo ellipticity–ellipticity and ellipticity–direction correlations (&lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2008MNRAS.389.1266L&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lee et al. 2008&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2020-05-06T10:51:10.010" LastActivityDate="2020-05-06T10:51:10.010" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19021" PostTypeId="1" CreationDate="2016-11-11T04:02:45.887" Score="0" ViewCount="198" Body="&lt;p&gt;As we all know the usual distance between earth and moon is 384,400 km.&#xA;But i was thinking how much closer it will be on 14 November 2016, as it will be a supermoon, the brightest and biggest moon in 60 years.&lt;/p&gt;&#xA;" OwnerUserId="14819" LastActivityDate="2016-11-12T16:24:45.580" Title="What would be the difference in distance between earth and moon on supermoon night (14 November 2016)" Tags="&lt;earth&gt;&lt;distances&gt;&lt;the-moon&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19022" PostTypeId="2" ParentId="19021" CreationDate="2016-11-11T07:32:10.283" Score="1" Body="&lt;p&gt;Based on some information from this article&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.space.com/34515-supermoon-guide.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.space.com/34515-supermoon-guide.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Nov 14 2016 supermooon's expected peak of full phase &#xA;is on the morning ov Nov 14 at 8:52 AM EST&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to some quick calculations that I performed&#xA;using PyEphem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and assuming the moon would be viewed by an observer in NYC at 08:52AM EST&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;0.00236372323707 AU&lt;/strong&gt; from observer&#xA;An AU is 1.5 * 10^km&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: I needed a better precision for km to AU&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.iau.org/static/resolutions/IAU2012_English.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.iau.org/static/resolutions/IAU2012_English.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to this, the AU can be more precisely defined as &#xA;149 597 870 700  meters +/- 3 meters&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.00236372323707 * 149 597 870 700 = &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;353 607 963.19 meters&lt;/strong&gt; or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;~ 353,608 km from earth&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here is a quick run through of the inputs I used for the program&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt;&amp;gt;&amp;gt; moon = ephem.Moon()&#xA;&amp;gt;&amp;gt;&amp;gt; nyc = ephem.Observer()&#xA;&amp;gt;&amp;gt;&amp;gt; nyc.long, nyc.lat = '-74.0059', '40.7127'&#xA;&amp;gt;&amp;gt;&amp;gt; nyc.date = '2016/11/14 08:52:00'&#xA;&amp;gt;&amp;gt;&amp;gt; moon.compute(nyc)&#xA;&amp;gt;&amp;gt;&amp;gt; print moon.earth_distance&#xA;&#xA;    0.00236372323707&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Learn more about PyEphem package here&#xA;&lt;a href=&quot;http://rhodesmill.org/pyephem/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://rhodesmill.org/pyephem/index.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note I'm just an amateur and these calculations might not take in certain critical factors. I'm uncertain if PyEphems earth_distance property calculates distance to the moon or the moon's center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it does calculate distance to the center of the moon this number could be about ~1700 km smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given your figure of 384,400 km average, this would put the moon just under&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;~30 792 km closer to the earth or about 91.98 % of it's normal distance&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="14822" LastEditorUserId="14822" LastEditDate="2016-11-12T16:24:45.580" LastActivityDate="2016-11-12T16:24:45.580" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19023" PostTypeId="2" ParentId="19016" CreationDate="2016-11-11T08:53:01.080" Score="5" Body="&lt;p&gt;The spectral type of an object is almost entirely determined by the temperature of its photosphere. ie Saying something is type M3.5 is just a measure of its surface temperature. An M3.5 brown dwarf is at a very similar temperature to an M3.5 star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brown dwarfs begin their lives as hot balls of gas and gradually cool with time. They start off as M-type objects and then cool to become L-type and then eventually, T-type objects. Stars on the other hand, start their lives by cooling, but stabilise their temperatures at a mass-dependent value that corresponds to an M-type classification or even as cool as type L2 for the very lowest mass stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that stars and brown dwarfs appear to have been born throughout Galactic history, then a range of masses are possible for any spectral type. In order to use the surface temperature (or spectral type) as an indicator of whether something is a brown dwarf, then you also need to know its age. In general that is not known, unless the object can be demonstrated to be part of a cluster of stars (of known age).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to be a brown dwarf the mass should be below about $0.075 M_{\odot}$, but unless the object is in a binary system, this cannot be measured. So what people do is trust the models(!) and compare the luminosity (or estimated temperature) with model evolutionary tracks at a given age. This gives an age- and model-dependent mass that can be used to claim that an object is a brown dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that brown dwarfs can have an M-type spectral class when they are young, but then move through L- to T-dwarfs as they get older is not a problem, it is a confirmation that the evolutionary models are (roughly) correct!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A plot might help (from the work of &lt;a href=&quot;http://adsabs.harvard.edu/abs/1997ApJ...491..856B&quot; rel=&quot;nofollow noreferrer&quot;&gt;Burrows et al. (1997)&lt;/a&gt;). This shows the evolution of temperature with time. Each track represents a brown dwarf of a given mass (the tracks from top to bottom) are masses of 0.2 - 0.08$M_{\odot}$, representing &lt;em&gt;stars&lt;/em&gt; (in blue). Then come masses of 0.07 to 0.015$M_{\odot}$ representing brown dwarfs (in green, that will never ignite hydrogen in their cores to any extent). Then below this are the &quot;planetary mass objects&quot; from 14 Jupiter masses to 0.3 Jupiter masses at the very bottom (in blue - the definition here is that these objects don't even ignite deuterium in their cores). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horizontal lines mark the boundaries of the spectral type classifications (M at the top, L in the middle, T at the bottom). &#xA;Notice how the stars level-off in temperature (when they begin H-burning) but can have M-type or even L-type classifications, but the brown dwarfs cool throughout their lives. Even objects as low as $0.01M_{\odot}$ begin their lives as &quot;M-type&quot; objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer some of your more specific queries:&#xA;The warmest spectral type below which you can almost guarantee that an object &lt;em&gt;is&lt;/em&gt; a brown dwarf is $\sim$ L2. Any object warmer than this &lt;em&gt;could&lt;/em&gt; be a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A brown dwarf is always cooler than a low-mass star &lt;em&gt;of the same age&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current observational estimates are that there is about 1 brown dwarf for every 4 low-mass stars - the weighted mean ratio for stars between 0.08 and 1 solar mass to brown dwarfs from observations of several clusters is $4.3 \pm 1.9$, according to &lt;a href=&quot;https://arxiv.org/abs/0807.1354&quot; rel=&quot;nofollow noreferrer&quot;&gt;Andersen et al. (2008)&lt;/a&gt;; the commonly adopted/assumed &quot;Chabrier Initial Mass Function&quot; has 4.9 stars for every brown dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spectrum of a brown dwarf and an M-dwarf star &lt;em&gt;at the same temperature&lt;/em&gt; are very difficult to distinguish. The M-dwarf is only slightly bigger and this might lead to small gravity-dependent differences. Lithium is not burned in brown dwarfs less than $0.06 M_{\odot}$, whereas it &lt;em&gt;is&lt;/em&gt; burned completely in low-mass stars on a mass-dependent timescale. Thus an older (than say 1 billion years) M-dwarf star would certainly have burned its lithum, whereas a younger M-type brown dwarf of lower mass would still have lithium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KEDDE.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KEDDE.gif&quot; alt=&quot;Evolution of brown dwarf temperatures&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-11-12T10:38:54.933" LastActivityDate="2016-11-12T10:38:54.933" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19024" PostTypeId="2" ParentId="18918" CreationDate="2016-11-11T10:53:17.790" Score="7" Body="&lt;p&gt;The Sun's velocity is known (with uncertainties) with respect to the Galactic centre (GC), as is the Sun's distance from the Galactic centre. These measurements have a long tortuous history, which I am not going to attempt to summarise here. However, the motion of the Sun with respect to the GC is established by measuring the proper motion of the Sgr A* source with respect to the assumed zero proper motions of distant radio quasars (see for example &lt;a href=&quot;http://adsabs.harvard.edu/abs/1999ApJ...524..805B&quot; rel=&quot;noreferrer&quot;&gt;Backer &amp;amp; Sramek 1999&lt;/a&gt;; &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0408107&quot; rel=&quot;noreferrer&quot;&gt;Reid &amp;amp; Brundthaler 2004&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The astrometry frame of reference now is known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/International_Celestial_Reference_Frame&quot; rel=&quot;noreferrer&quot;&gt;International Celestial Reference Frame&lt;/a&gt; and is defined by hundreds of compact, extragalactic radio sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you then assume the Sun's orbit is circular and that Sgr A* is at the GC, then the job is done - the relative velocity divided by the distance to the GC gives the angular velocity $\Omega$, and the orbital period is $2\pi/\Omega$ . If you have a model for the Galactic potential then you can do a bit better. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-11-11T10:53:17.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19025" PostTypeId="1" CreationDate="2016-11-11T11:32:10.813" Score="1" ViewCount="217" Body="&lt;p&gt;Many people on the Astronomy stack exchange use the &lt;a href=&quot;http://rhodesmill.org/pyephem/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;PyEphem&lt;/a&gt;  package for the &lt;a href=&quot;https://www.python.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Python&lt;/a&gt;  programmimg language to perform astronomical computations.  Are the underlying numerical routines (from &lt;a href=&quot;http://www.clearskyinstitute.com/xephem/&quot; rel=&quot;nofollow noreferrer&quot;&gt;XEphem&lt;/a&gt;) also available for &lt;a href=&quot;https://www.mathworks.com/products/matlab/&quot; rel=&quot;nofollow noreferrer&quot;&gt;MATLAB&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="5537" LastActivityDate="2016-11-14T01:57:19.637" Title="Is there a MATLAB equivalent of PyEphem?" Tags="&lt;software&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19026" PostTypeId="1" CreationDate="2016-11-11T13:09:44.160" Score="3" ViewCount="51" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Satellite_galaxies_of_the_Milky_Way&quot; rel=&quot;nofollow noreferrer&quot;&gt;Two dwarf satellite galaxies&lt;/a&gt; of our galaxy have large $\frac{\text{diameter}}{\text{distance}}$ values. And one of them, Canis Major Dwarf, is 8kpc away which is &#xA;the same with our distance to the galactic center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we limit the masses(even they are not heavy) of their central black holes now? Like &lt;a href=&quot;http://adsabs.harvard.edu/abs/2009ApJ...692.1075G&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gillessen et al. 2009&lt;/a&gt;, maybe we can find stars orbiting their central black holes. &lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="9301" LastEditDate="2016-11-11T13:31:59.053" LastActivityDate="2016-11-11T13:31:59.053" Title="Can we resolve the centers of Milky way's satellite galaxies now?" Tags="&lt;black-hole&gt;&lt;dwarf-galaxy&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19027" PostTypeId="1" CreationDate="2016-11-11T15:36:31.070" Score="6" ViewCount="150" Body="&lt;p&gt;As reported in some recent observational works works (ref. &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0004-637X/806/1/4&quot; rel=&quot;nofollow noreferrer&quot;&gt;Merten et al. 2015&lt;/a&gt;) the concentration-mass relation of galaxy clusters has a negative slope. This means that at a fixed redshift, haloes with a higher mass have a lower concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some old works say this is a consequence of the fact that the concentration depends only on the density at the assembly time. For instance, &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0004-637X/766/1/32/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bhattacharya et al. 2013&lt;/a&gt;  says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The correlation of halo concentration with mass is based on the idea—as first explicated by NFW—that the concentration is determined by the mean density of the universe when the halo is assembled, with higher concentrations corresponding to higher densities. Thus cluster mass halos, which are assembling today, should have a lower concentration than halos of lower mass that were built up at an earlier epoch, where the mean density was higher. Furthermore, one may expect this trend to flatten out (sufficiently) beyond the nonlinear mass scale $M_∗$, and therefore, since $M_∗$ falls rapidly with redshift, flatten out over an extended range in mass as redshift increases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot understand how this can lead to a negative slope between the concentration and the mass&lt;/p&gt;&#xA;" OwnerUserId="14824" LastEditorUserId="14824" LastEditDate="2016-11-14T12:34:55.910" LastActivityDate="2017-11-20T19:30:54.693" Title="Why does the concentration-mass relation of galaxy clusters have a negative slope?" Tags="&lt;galaxy-cluster&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19028" PostTypeId="1" CreationDate="2016-11-11T18:46:18.380" Score="3" ViewCount="163" Body="&lt;p&gt;In the first version of my answer to &lt;a href=&quot;https://astronomy.stackexchange.com/q/19001/2153&quot;&gt;Are sunspots vertically displaced from the surrounding photospheric plasma?&lt;/a&gt;, I naively used &lt;a href=&quot;https://en.wikipedia.org/wiki/Kramers%27_opacity_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kramer's law&lt;/a&gt; to attempt to show an inverse relationship between opacity $\kappa$ and temperature $T$ due to bound-free or free-free absorption in a region with density $\rho$:&#xA;$$\kappa\propto\rho T^{-7/2}$$&#xA;Rob Jeffries pointed out that this might not be accurate in the photosphere, even though it works well deeper inside most stars (though for high-mass stars, Thomson scattering dominates and $\kappa$ is approximately constant). After some more reading, I concluded that this was right. At lower temperatures, interactions with ionized hydrogen mean that $H^-$ is the major contributor to opacity, and it follows a law&#xA;$$\kappa\propto\rho^{1/2}T^9$$&#xA;I believe the transition of dominance happens around $\sim10^4\text{ K}$, meaning that $H^-$ opacity should be the major contribution in the stellar photosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At even lower temperatures, however ($\sim4000\text{ K}$), molecular scattering can be important (see &lt;a href=&quot;http://www.ucl.ac.uk/phys/amopp/people/jonathan_tennyson/papers/407.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tennyson et al. (2007)&lt;/a&gt;). While this is not necessarily important at all points in the stellar photosphere, I think that it should be important in sunspots, which can have temperatures in this range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is/are the major source(s) of opacity in sunspots? Does molecular scattering dominate, or is $H^-$ still important? If both are important, to varying degrees, how can the mean opacity be calculated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My knowledge of stellar atmospheres and atmospheric models is limited, so I apologize if this comes down to being a basic question.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-11-11T22:41:12.757" Title="What is the main source of opacity in sunspots?" Tags="&lt;the-sun&gt;&lt;opacity&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19029" PostTypeId="2" ParentId="10242" CreationDate="2016-11-11T21:50:38.393" Score="1" Body="&lt;p&gt;Even if an entire volcano worth of smoke and dust particles, like 1000000 tons, and you detonated it near the Sun's surface, it would probably go red and fall into the sun in 5 seconds.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun has turbulent whisps of high energy particles escaping its huge gravity, except that it's invisible and it's called solar wind. If you could see a slice of it, it might look similar to smoke on a large format. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon based smoke ignites at around 400 degrees, and all other smokes decompose into atomic vapor and ion blasted by intense ionic forces at the sun's surface temperature, and some miles form the surface. If you breathed a cigarette at 1 mile from the sun, the smoke would change states into ions because the temperature would excite the molecules and atoms of smoke that they would be a plasma. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun weighs 333,000 times more than here... so the gravity is probably at least 10,000 times as strong at the surface of the sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atmosphere is heavy, and if air is compressed 10,000 times it is closer to being a fluid than something that can handle gas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the Sun doesn't have any atmosphere, because it's too liquid, it bubbles. If the sun were cold, it would perhaps have a bizarre kind of atmosphere, perhaps someone here can say what kind, perhaps only a few meters high, and perhaps 1mm. A human on a cold sun would become a puddle and sink, his bones would flatten under their own weight, to something very fine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temps are too high for visible molecules, and the movement is too violent for smoke, smoke can't exist in a fire, and the Sun is a kind of fire multiplied by 1000. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, what molecules at all can exist at Sun temperatures? Molecules decompose in microseconds and there is perhaps some literature about any &lt;a href=&quot;https://en.wikipedia.org/wiki/Molecules_in_stars&quot; rel=&quot;nofollow noreferrer&quot;&gt;kinds of couplets and triads of atoms that can exist near the surface&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorDisplayName="user5434" LastEditDate="2017-05-06T22:18:09.163" LastActivityDate="2017-05-06T22:18:09.163" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19030" PostTypeId="1" AcceptedAnswerId="19032" CreationDate="2016-11-11T22:40:16.670" Score="2" ViewCount="315" Body="&lt;p&gt;Andromeda and the Milky Way are set to collide in 4bn years. Will the Local Group collide with M81, and what about further away groups? if so, when? &lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2016-11-12T04:28:07.713" LastActivityDate="2016-11-12T16:17:11.933" Title="Is the M81 group bound to the Local group?" Tags="&lt;galaxy&gt;&lt;cosmological-inflation&gt;&lt;local-group&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19031" PostTypeId="2" ParentId="19028" CreationDate="2016-11-11T22:41:12.757" Score="2" Body="&lt;p&gt;In &lt;a href=&quot;https://arxiv.org/pdf/1404.0688.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ignjatovi´c et al. (2014)&lt;/a&gt;, ion-atom absorption processes are discussed. They suggest that these processes are of equal, if not greater, influence on the observed sunspot opacity as absorption processes in $\mathrm{H}(1s)+\mathrm{H}^+$ and $\mathrm{H}(1s)+\mathrm{X}^+$ collisions (where $\mathrm{X}$ is one of the metal atoms $\mathrm{X} = \mathrm{Na, Ca, Mg, Si, Al}$). The main scattering processes discussed within the paper follow the photo-dissociation of $\mathrm{H}_2^+$ and $\mathrm{HX}^+$ by UV photons in the spectral region ($110$ $\mathrm{nm}$ $\lesssim \lambda \lesssim 230$ $\mathrm{nm}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I highly recommend reading through the paper—it is very interesting.&lt;/p&gt;&#xA;" OwnerDisplayName="user14781" LastActivityDate="2016-11-11T22:41:12.757" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19032" PostTypeId="2" ParentId="19030" CreationDate="2016-11-11T23:01:35.670" Score="2" Body="&lt;p&gt;Interestingly enough, although we tend to group galaxy clusters together into superclusters, superclusters are actually &lt;em&gt;not&lt;/em&gt; gravitationally bound. This suggests that, due to &lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/h/hubble+flow&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble flow&lt;/a&gt;, galaxies in separate clusters will never &quot;collide&quot;. In fact, it may be impossible to ever leave one's galaxy cluster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a previous version of this post, I stated the the &lt;a href=&quot;http://www.nasa.gov/mission_pages/hubble/science/great-attractor.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Great Attractor&lt;/a&gt; may be reason to suspect a gravitational binding of Laniakea. In reality, the Great Attractor only reduces the expected relative velocities of these galaxies away from one another by between $200-400$ $\mathrm{km}\cdot\mathrm{s}^{-1}$—I would like to thank eshaya for pointing out my error here. Due to the &lt;a href=&quot;https://arxiv.org/pdf/1607.03143v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;current expansion rate of the universe&lt;/a&gt; of $67.6^{+0.7}_{-0.6}$ $\mathrm{km}\cdot\mathrm{s}^{-1}\cdot\mathrm{Mpc}^{-1}$, some galaxies in Laniakea have recession velocities of up to $30000$ $\mathrm{km}\cdot\mathrm{s}^{-1}$, as given by &lt;a href=&quot;https://arxiv.org/ftp/arxiv/papers/1409/1409.0880.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tully et al. (2014)&lt;/a&gt;—far greater than could ever hope to be counteracted by the Great Attractor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question now becomes one of whether the M81 Group is gravitationally bound to the Local Group. Given the M81 Group's status as being one of the closest to the Local Group, at only $3.6$ $\mathrm{Mpc}$ &lt;a href=&quot;https://arxiv.org/pdf/astro-ph/0410065v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;distant&lt;/a&gt;, I would not be entirely surprised if through some chance it managed to be. Given the current estimate for Hubble's constant, due to Hubble flow the M81 Group would be expected to recede at $\sim243.4$ $\mathrm{km}\cdot\mathrm{s}^{-1}$. This is what would need to be countered through gravitational effects for the two groups to be bound, and thus eventually coalesce.&lt;/p&gt;&#xA;" OwnerDisplayName="user14781" LastEditorDisplayName="user14781" LastEditDate="2016-11-12T16:17:11.933" LastActivityDate="2016-11-12T16:17:11.933" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19033" PostTypeId="1" AcceptedAnswerId="19035" CreationDate="2016-11-11T23:50:55.290" Score="6" ViewCount="1611" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NVSXb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NVSXb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The majority of the stars we see in the sky, like Pollux, are outside of the galactic plane. That means that all those stars we see are not in the galactic disk, and therefore are in the galactic halo, whereas Capella, for example, is where?&lt;/p&gt;&#xA;" OwnerUserId="9949" LastEditorUserId="2153" LastEditDate="2016-11-12T22:33:48.090" LastActivityDate="2016-11-13T14:54:00.380" Title="Are the stars outside of the galactic plane in the galactic halo?" Tags="&lt;star&gt;&lt;galaxy&gt;&lt;milky-way&gt;&lt;stellar-dynamics&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19034" PostTypeId="2" ParentId="19033" CreationDate="2016-11-12T00:02:47.550" Score="7" Body="&lt;p&gt;This is not necessarily true—the galactic disk is not 2-dimensional. In the case of the Milky Way, this disk is approximately $1-3$ $\mathrm{kly}$ in thickness, as per &lt;a href=&quot;https://arxiv.org/pdf/1301.3168v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rix &amp;amp; Bovy (2013)&lt;/a&gt;. This means that stars which do not appear within the galactic plane very well may still be within the disk. In fact, almost &lt;em&gt;all&lt;/em&gt; stars visible from Earth lie within this disk. See &lt;a href=&quot;https://physics.stackexchange.com/questions/45759/what-is-the-farthest-away-star-visible-to-the-naked-eye&quot;&gt;this&lt;/a&gt; question for some further information on distant visible stars.&lt;/p&gt;&#xA;" OwnerDisplayName="user14781" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-11-12T00:02:47.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19035" PostTypeId="2" ParentId="19033" CreationDate="2016-11-12T00:30:01.360" Score="8" Body="&lt;p&gt;The galactic disk, as Riley Jacob wrote, has a definite thickness. It's actually composed of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Thin_disk&quot; rel=&quot;nofollow noreferrer&quot;&gt;thin disk&lt;/a&gt; $\sim0.3\text{ kpc}$ thick and a &lt;a href=&quot;https://en.wikipedia.org/wiki/Thick_disk&quot; rel=&quot;nofollow noreferrer&quot;&gt;thick disk&lt;/a&gt; $\sim1\text{ kpc}$ thick, at least (&lt;a href=&quot;http://mnras.oxfordjournals.org/content/414/3/2446.full.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;McMillan (2011)&lt;/a&gt; has models with &lt;a href=&quot;http://adsabs.harvard.edu/abs/2008ApJ...673..864J&quot; rel=&quot;nofollow noreferrer&quot;&gt;data from the Sloan Digital Sky Survey&lt;/a&gt;). There's also a central bulge that is even thicker, as the following diagram (from &lt;a href=&quot;https://milkyway.cs.rpi.edu/milkyway/science.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;) shows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FCtRc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FCtRc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pollux is $\sim10\text{ pc}$ away from the Solar System, which is about 1% of the thickness of the thick disk (Capella is barely farther). Essentially, it's in the same plane as the Solar System; that's an insignificant distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are stars in the thin disk, thick disk, and halo, which compose &lt;a href=&quot;http://www.ifa.hawaii.edu/~barnes/ast626_95/pcmw.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;different populaitons&lt;/a&gt; based on &lt;a href=&quot;https://en.wikipedia.org/wiki/Metallicity&quot; rel=&quot;nofollow noreferrer&quot;&gt;metallicity&lt;/a&gt;. In the thin disk are Population I stars, which are high in metals. In the halo (and thin disk, to some extent) are Population II stars, which are lower in metals and on average older than Population I stars. The Sun is a Population I star in the disk. There may also be a sub-population of Population II stars in the thick disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Halo stars certainly do not compose the majority of stars in the Milky Way galaxy. Most stars are contained within the thin disk, thick disk, and bulge.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-11-13T14:54:00.380" LastActivityDate="2016-11-13T14:54:00.380" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19036" PostTypeId="2" ParentId="19021" CreationDate="2016-11-12T03:29:58.977" Score="1" Body="&lt;p&gt;From where I am on Vancouver Island,  western Canada, it will be around 356,500 km. &lt;img src=&quot;https://i.stack.imgur.com/ZoQNO.png&quot; alt=&quot;screenshot&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="14820" LastActivityDate="2016-11-12T03:29:58.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19037" PostTypeId="2" ParentId="19025" CreationDate="2016-11-12T04:27:20.163" Score="2" Body="&lt;p&gt;&lt;em&gt;Disclaimer: I've never used this toolbox and so cannot speak to the comparison of this toolbox with PyEphem.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some ephemeral and the like functions within Matlab, for example &lt;a href=&quot;https://www.mathworks.com/help/aerotbx/ug/planetephemeris.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;planetEphemeris&lt;/a&gt;, &lt;a href=&quot;https://www.mathworks.com/help/aerotbx/ug/earthnutation.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;earthNutation&lt;/a&gt;, or &lt;a href=&quot;https://www.mathworks.com/help/aerotbx/ug/earthnutation.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;moonLibration&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately these functions are only accessible within the &lt;a href=&quot;https://www.mathworks.com/help/aerotbx/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aerospace Toolbox&lt;/a&gt; and do not come with a standard Matlab license. You'll need an entirely separate license for the Aerospace Toolbox to use these functions in your Matlab code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is easy for you to obtain this license (be it from some university or organization), then by all means go for it. Otherwise, your best bet may be free packages like PyEphem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also like to echo the &lt;a href=&quot;https://astronomy.stackexchange.com/questions/19025/is-there-a-matlab-equivalent-of-pyephem#comment30692_19025&quot;&gt;comment made by David&lt;/a&gt; in that you should definitely learn Python. It is an amazingly simple, yet robust language that can do just about anything. And its free!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're really dead-set on using Matlab, but the PyEphem package is the only thing that does what you need, you can always consider porting between the two languages. Matlab has several tools and capabilities to interface with other languages. You can read more about using python packages natively in Matlab &lt;a href=&quot;https://www.mathworks.com/help/matlab/getting-started_buik_wp-3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;right here&lt;/a&gt;. I will say, this would probably be much harder than just going with Python outright though.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-11-12T04:27:20.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19038" PostTypeId="2" ParentId="19033" CreationDate="2016-11-12T11:27:13.433" Score="7" Body="&lt;p&gt;I'll turn this around for you. The &lt;em&gt;brightest&lt;/em&gt; star that is classed as a &quot;halo&quot; or metal-poor population II star is &lt;a href=&quot;https://en.m.wikipedia.org/wiki/HD_140283&quot; rel=&quot;noreferrer&quot;&gt;HD 140283&lt;/a&gt;, with a visual magnitude of 7.2 and not even visible to the naked eye. Even this is only a halo object in the sense of being a high velocity object whose trajectory will take it back into the halo in the future - it is presently only 190 light years distant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I completely concur with the other answers, but let me explain &lt;em&gt;why&lt;/em&gt; we don't see halo stars, despite them making up about 1% of the stellar population.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The disc of the Galaxy is roughly 1000 pc thick, so to see a naked eye star actually in the halo, it would be further away than about 500 pc (the Sun is not far from the mid-plane) and have to have an absolute magnitude brighter than -2.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such luminous objects are rare, but not unknown; they would need to be O-type main sequence stars or supergiants. But these are &lt;em&gt;young&lt;/em&gt;, massive stars and recent star formation has only taken place &lt;em&gt;close to the Galactic plane&lt;/em&gt; and so that is where such objects are found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If our eyes were about 3 magnitudes more sensitive then we &lt;em&gt;would&lt;/em&gt; be able to see metal-poor K-type giant stars in the halo.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-11-12T11:27:13.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19039" PostTypeId="1" AcceptedAnswerId="19042" CreationDate="2016-11-12T12:21:15.853" Score="14" ViewCount="1085" Body="&lt;p&gt;Is there any way to avoid the tidal locking of a planet orbiting a red dwarf in the habitable zone?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, could a planet with a 90° obliquity and large moon avoid such a situation?&lt;/p&gt;&#xA;" OwnerUserId="14829" LastEditorUserId="6" LastEditDate="2017-02-03T14:52:24.330" LastActivityDate="2017-02-03T22:14:34.417" Title="Is there any way for a planet orbiting a red dwarf in the habitable zone to not be tidally locked?" Tags="&lt;orbit&gt;&lt;tidal-forces&gt;&lt;celestial-mechanics&gt;&lt;red-dwarf&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19041" PostTypeId="2" ParentId="19039" CreationDate="2016-11-12T16:38:41.330" Score="14" Body="&lt;p&gt;&lt;a href=&quot;http://science.sciencemag.org/content/347/6222/632&quot;&gt;Leconte et al. (2015)&lt;/a&gt; suggested that the presence of an atmosphere could prevent or at least slow tidal locking. The star should exert two separate torques: one on the atmosphere and one on the solid body of the planet:&#xA;$$T_a=-\frac{3}{2}K_ab_a(2\omega-2n),\quad T_g=-\frac{3}{2}K_gb_g(2\omega-2n)$$&#xA;where&#xA;$$K_a\equiv\frac{3M_*R_p^3}{5\bar{\rho}a^3},\quad K_g\equiv\frac{GM_*R_p^5}{a^6}$$&#xA;for stellar mass $M_*$, planetary radius $R_p$, mean density $\rho$, semi-major axis $a$, mean motion $n$, rotation rate $\omega$, and response to torques $b_a$ and $b_g$. The two torques could be equal, and assuming that the atmosphere transfers some angular momentum to the surface of the planet, this could prevent tidal locking. There are several equilibria at which this could occur:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/87Wrk.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/87Wrk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2016-11-12T16:38:41.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19042" PostTypeId="2" ParentId="19039" CreationDate="2016-11-12T21:07:09.217" Score="7" Body="&lt;p&gt;Yes: It has a companion planet or an excessively large moon, with the two bodies orbiting their common center of mass (much like the Earth and the Moon). They could be tidally-locked to each other, but they cannot be tidally-locked to their star.&lt;/p&gt;&#xA;" OwnerUserId="14832" LastEditorUserId="2732" LastEditDate="2017-02-03T22:14:34.417" LastActivityDate="2017-02-03T22:14:34.417" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19043" PostTypeId="1" CreationDate="2016-11-12T23:05:18.127" Score="3" ViewCount="423" Body="&lt;p&gt;The lunar motion can be predicted with basic celestial mechanics, but the perigee and apogee are not always the same, basically because the attraction of the Sun makes some oscilations in the semi-axis of the orbit (&lt;a href=&quot;https://en.wikipedia.org/wiki/Perturbation_(astronomy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Perturbation_(astronomy)&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: how do they know when the full moon will get as close as it got in the recent supermoon? For example, the Dailypost article &lt;a href=&quot;http://www.dailypost.co.uk/whats-on/whats-on-news/biggest-supermoon-68-years-light-12121731&quot; rel=&quot;nofollow noreferrer&quot;&gt;Biggest Supermoon in 68 years will light up the November sky&lt;/a&gt; claims that the full moon won't come this close to Earth until November 25th, 2034.&lt;br /&gt;&#xA;How do they kow that? Are they using the perturbation theory equations to predict this? If so, can anybody give me some source or explanation of how it's done analytically?&lt;/p&gt;&#xA;" OwnerUserId="9949" LastEditorDisplayName="user1569" LastEditDate="2016-11-13T11:15:40.887" LastActivityDate="2016-12-23T02:58:33.967" Title="How can the next supermoon be analytically predicted?" Tags="&lt;orbit&gt;&lt;celestial-mechanics&gt;&lt;the-moon&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19044" PostTypeId="1" AcceptedAnswerId="19958" CreationDate="2016-11-13T10:44:58.223" Score="10" ViewCount="359" Body="&lt;p&gt;The NASA Hubble Space Telescope Science Release &lt;a href=&quot;https://www.spacetelescope.org/news/heic1320/&quot; rel=&quot;noreferrer&quot;&gt;HEIC 1320 - When is a Comet not a Comet? - Hubble astronomers observe bizarre six-tailed asteroid&lt;/a&gt; is self-explanatory, but it doesn't explain what actually happening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes comets will show two tails, but one is usually identified with the path of the comet's orbit, and the other with the direction away from the sun. But this is an asteroid (or so it seems) and there are many tails with complex behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit:&lt;/strong&gt; I've found this ArXiv preprint: &lt;a href=&quot;https://arxiv.org/abs/1311.1483&quot; rel=&quot;noreferrer&quot;&gt;The Extraordinary Multi-Tailed Main-Belt Comet P/2013 P5&lt;/a&gt; where &quot;main belt&quot; refers to the main belt of asteroids, and yet the object is now called a comet.  &lt;a href=&quot;https://en.wikipedia.org/wiki/311P/PANSTARRS&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; lists both names and suggests both designations - asteroid or comet:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;311P/PANSTARRS also known as P/2013 P5 (PANSTARRS) &lt;strong&gt;is an asteroid (or main-belt comet)&lt;/strong&gt; discovered by the Pan-STARRS telescope on 27 August 2013.(4) Observations made by the Hubble Space Telescope revealed that it had six comet-like tails.(5) The tails are suspected to be streams of material ejected by the asteroid as a result of a rubble pile asteroid spinning fast enough to remove material from it.(2)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's been three &lt;em&gt;point five&lt;/em&gt; years now - has this been figured out yet? &lt;em&gt;Is it an asteroid or comet (or both), and how can it produce so many tails?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/A9pFk.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/A9pFk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;strong&gt;above:&lt;/strong&gt; HST image of comet P/2013 P5 from &lt;a href=&quot;https://www.spacetelescope.org/news/heic1320/&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HlcrH.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HlcrH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;strong&gt;above:&lt;/strong&gt; HST image (annotated) of comet P/2013 P5 from &lt;a href=&quot;https://www.spacetelescope.org/images/heic1320b/&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-02-06T15:59:29.380" LastActivityDate="2017-02-06T21:34:19.863" Title="Is this object an asteroid or comet, and how can it produce so many tails?" Tags="&lt;asteroids&gt;&lt;comets&gt;&lt;hubble-telescope&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19045" PostTypeId="5" CreationDate="2016-11-13T10:46:25.627" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2016-11-13T10:46:25.627" LastActivityDate="2016-11-13T10:46:25.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19046" PostTypeId="4" CreationDate="2016-11-13T10:46:25.627" Score="0" Body="Questions about the Hubble Space Telescope itself, and images from it." OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2016-11-13T18:14:13.020" LastActivityDate="2016-11-13T18:14:13.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19047" PostTypeId="2" ParentId="19039" CreationDate="2016-11-13T12:17:14.613" Score="5" Body="&lt;p&gt;The more likely case is actually a spin-orbit resonance that is &lt;em&gt;not&lt;/em&gt; 1:1 but a half odd multiple, like the 3:2 case of our own Mercury.  Having eccentricity in the orbit encourages this situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’ve been meaning to write this up on the Worldbuilding.SE but I have not re-found enough references.  But see &lt;a href=&quot;http://www.seti.org/weeky-lecture/patterns-sunlight-extra-solar-planets&quot;&gt;this video&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7142" LastActivityDate="2016-11-13T12:17:14.613" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19049" PostTypeId="1" CreationDate="2016-11-13T16:52:35.127" Score="3" ViewCount="1077" Body="&lt;p&gt;Twice a year the sun is at a certain angle south (or north) of the equator between The Tropic of Cancer and The Tropic of Capricorn.&#xA;At that time the sun will be exactly overhead for people at the latitude (and longitude).&#xA;Is there a table or - better - a model that takes &#xA;Latitude (and Year) as input&#xA;and responds with&#xA;Date, Time, and Longitude as output.?&#xA;-The 'closest' 'midday' (solar time) (before or after) will be the closest people at that latitude - round the world - will come to experience the sun 'directly' overhead.&#xA;You could call it 'Very High Noon'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;vj&lt;/p&gt;&#xA;" OwnerUserId="14838" LastActivityDate="2016-11-13T21:15:23.033" Title="'Very High Noon' - Sun directly overhead" Tags="&lt;the-sun&gt;&lt;solar-system&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19050" PostTypeId="2" ParentId="19049" CreationDate="2016-11-13T21:15:23.033" Score="3" Body="&lt;p&gt;Let me give a slightly better answer. If you visit&#xA;&lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ssd.jpl.nasa.gov/horizons.cgi&lt;/a&gt; with these parameters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2BvmK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2BvmK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you'll see the Sun's astrometric right ascension and declination for a&#xA;48 hour period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at one of the results and see what it tells us:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;Date__(UT)__HR:MN     R.A._(ICRF/J2000.0)_DEC&#xA;**********************************************&#xA;$$SOE&#xA;2016-Nov-13 00:00     15 13 30.08 -17 57 18.0&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This says the sun's declination is 17$^{\circ}$57'18.0&quot; which means&#xA;the sun is (almost) directly overhead at that latitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right ascension is 15h13m30.08s which means the sun is (almost)&#xA;directly overhead where the local sidereal time is this value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compute local sidereal time we first use&#xA;&lt;a href=&quot;http://aa.usno.navy.mil/faq/docs/GAST.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://aa.usno.navy.mil/faq/docs/GAST.php&lt;/a&gt; to get GMST (I'm using the&#xA;approximation):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;GMST = 18.697374558 + 24.06570982441908 D&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get D, the number of days since the year 2000, I used Unix's date&#xA;command as follows, and used other commands to finish the calculation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;&#xA;: per http://aa.usno.navy.mil/faq/docs/GAST.php we compute the number of&#xA;: days since 2000 January 1, 12h UT (not 0h UT)&#xA;date -d '2000-01-01 12:00 UTC' +%s;: result is 946728000&#xA;date -d '2016-11-13 00:00 UTC' +%s;: result is 1478995200&#xA;: take the difference and convert result to days for formula&#xA;perl -le 'use POSIX; printf(&quot;%.10f\n&quot;, (1478995200-946728000)/86400)'&#xA;: result above is 6160.5 (trailing digits are 0)&#xA;: computing GMST and modding to get result between 0 and 24&#xA;perl -le 'use POSIX; printf(&quot;%.10f\n&quot;, fmod(18.697374558 + 24.06570982441908*6160.5,24))'&#xA;: result above is 3.5027478917&#xA;: convert solar RA to decimal&#xA;perl -le 'use POSIX; printf(&quot;%.10f\n&quot;, 15+13/60+30.08/3600)'&#xA;: result above is 15.2250222222&#xA;: how many hours east of GMST is the Sun (in degrees)? 1h = 15 degrees&#xA;perl -le 'use POSIX; printf(&quot;%.10f\n&quot;, (15.2250222222-3.5027478917)*15)'&#xA;: result above is 175.8341149575&#xA;: converting to minutes and seconds, we first find seconds from fractional part&#xA;perl -le 'use POSIX; printf(&quot;%.10f\n&quot;, 0.8341149575*3600)'&#xA;: result above is 3003 rounded to nearest second&#xA;: and now minutes/seconds from 3003 seconds total&#xA;perl -le 'use POSIX; printf(&quot;%.10f %.10f\n&quot;, floor(3003/60), 3003%60)'&#xA;: result above is 50 and 3&#xA;&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So you would think the Sun is overhead at 17$^{\circ}$57'18.0'' south&#xA;latitude (south because the declination is negative) and longitude&#xA;175$^{\circ}$50'03&quot;E (east because longitude is positive) at 00:00 UTC&#xA;on 2016-Nov-13. Let's use HORIZONS to see if this is true (it turns&#xA;out it won't be, so we'll use a minute on either side of 2016-Nov-13&#xA;00:00 UTC to see what happens):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tqLUu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tqLUu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can confirm the Sun's RA matches local sideral time within 1 second&#xA;of 00:00UTC from these lines in the output:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;Date__(UT)__HR:MN:SC.fff   R.A.        DEC         Azi      Elev    Sid_Time&#xA;2016-Nov-13 00:00:00.000 * 15 13 30.08 -17 57 18.0 104.6639 89.7692 15 13 29.2615&#xA;2016-Nov-13 00:00:01.000 * 15 13 30.08 -17 57 18.0 104.9166 89.7730 15 13 30.2643&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You'll notice the local sidereal time jumps from being 0.82 seconds&#xA;behind the sun's right ascension to 0.18 seconds ahead of the sun's&#xA;right ascension in the one second period (showing the two matched&#xA;sometime within that second, probably at 00:00:00.82 or so).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the Sun's highest position (noted by HORIZONS as 't' for&#xA;transit) occurs a little later:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;Date__(UT)__HR:MN:SC.fff    R.A.        DEC         Azi      Elev    Sid_Time&#xA;2016-Nov-13 00:00:56.000 *  15 13 30.23 -17 57 18.6 178.5844 89.9415 15 14 25.4148&#xA;2016-Nov-13 00:00:57.000 *t 15 13 30.24 -17 57 18.6 182.4651 89.9415 15 14 26.4176&#xA;2016-Nov-13 00:00:58.000 *  15 13 30.24 -17 57 18.6 186.3230 89.9412 15 14 27.4203&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;when the local sideral time is almost a minute ahead of the sun's&#xA;right ascension. What's going on here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Earth is an ellipsoid, and not a sphere, it turns out that&#xA;&quot;up&quot;, the direction perpendicular to a location on Earth, is NOT&#xA;directly away from the center of the Earth (it turns out gravity acts&#xA;in a direction that is neither towards the center of the Earth nor&#xA;perpendicular to the surface, but that's not relevant for this&#xA;question): see &lt;a href=&quot;http://www.oc.nps.edu/oc2902w/c_mtutor/shape/shape4.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.oc.nps.edu/oc2902w/c_mtutor/shape/shape4.htm&lt;/a&gt;&#xA;for details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want really accurate calculations you'll have to use&#xA;&lt;a href=&quot;http://naif.jpl.nasa.gov/naif/tutorials.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://naif.jpl.nasa.gov/naif/tutorials.html&lt;/a&gt; 's&#xA;&lt;a href=&quot;http://naif.jpl.nasa.gov/pub/naif/toolkit_docs/C/cspice/surfnm_c.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://naif.jpl.nasa.gov/pub/naif/toolkit_docs/C/cspice/surfnm_c.html&lt;/a&gt;&#xA;function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although this doesn't cause the discrepancy above, another&#xA;consideration is a location's elevation above the surface of the&#xA;Earth, which can also have slight effect (since &quot;up&quot; isn't always&#xA;towards the zenith). Happily, one thing you can ignore is refraction,&#xA;since refraction at the zenith is effctively zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My &lt;a href=&quot;http://test.barrycarter.info/sunstuff.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://test.barrycarter.info/sunstuff.html&lt;/a&gt; is a very approximate&#xA;representation of where the Sun is overhead and is generated by&#xA;&lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/bc-sun-always-shines.pl&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/bc-sun-always-shines.pl&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2016-11-13T21:15:23.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19051" PostTypeId="1" AcceptedAnswerId="19067" CreationDate="2016-11-13T22:43:32.330" Score="3" ViewCount="227" Body="&lt;p&gt;I can't find information on this for the life of me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do astronomers approximate the mass and semimajor axis of a directly imaged exoplanet? Would several photos need to be taken to find the period and find with Kepler's third law or can it be inferred through the temperature of the planet (or another method)? Also if you have any sources of this that would be great. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="14840" LastEditorUserId="10172" LastEditDate="2016-11-14T00:39:35.773" LastActivityDate="2016-11-15T13:25:42.223" Title="How is the mass and semimajor axis of a directly imaged exoplanet found?" Tags="&lt;exoplanet&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19052" PostTypeId="2" ParentId="19043" CreationDate="2016-11-14T01:13:40.050" Score="2" Body="&lt;p&gt;A slightly more useful answer than my comment above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &lt;a href=&quot;http://wgc.jpl.nasa.gov:8080/webgeocalc/#NewCalculation&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://wgc.jpl.nasa.gov:8080/webgeocalc/#NewCalculation&lt;/a&gt; &quot;Angular&#xA;Separation Finder&quot; feature, we can determine when the angle between&#xA;the Sun and the Moon is at a local maximum. It turns out this isn't&#xA;&lt;em&gt;quite&lt;/em&gt; the definition of Full Moon, but it's very close. I couldn't&#xA;find a way to make webgeocalc compute full moon times exactly, though&#xA;I'm sure there is a way, and, of course, there are plenty of tables of&#xA;full moons online.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting at&#xA;&lt;a href=&quot;http://wgc.jpl.nasa.gov:8080/webgeocalc/#AngularSeparationFinder&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://wgc.jpl.nasa.gov:8080/webgeocalc/#AngularSeparationFinder&lt;/a&gt;&#xA;we fill out the form as so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BU1R9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BU1R9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/KeYl0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KeYl0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/q4p6U.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q4p6U.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and click the &quot;Calculate&quot; button.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This yields a list of all &quot;full&quot; moons. Scroll down and click on the&#xA;&quot;Save All Intervals&quot; button:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/INBLr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/INBLr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now want to find the moon's distance at all of the times we just&#xA;saved (ie, the times of the &quot;full&quot; moons). To do this, start at&#xA;&lt;a href=&quot;http://wgc.jpl.nasa.gov:8080/webgeocalc/#StateVector&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://wgc.jpl.nasa.gov:8080/webgeocalc/#StateVector&lt;/a&gt; and fill out as&#xA;follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8QBrZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8QBrZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/DwZvu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DwZvu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/T3sMU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/T3sMU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: to get the list of intervals above, drag from the saved&#xA;intervals result window into the list of intervals window.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The results, which you can also download, tell you the moon's distance from the Earth at the times of the &quot;full&quot; moons we computed earlier:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KaC4B.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KaC4B.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I downloaded the results in Excel format, imported them into gnumeric&#xA;and sorted by radius (distance from Earth to Moon) and did a little&#xA;additional cleanup to get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FXjFY.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FXjFY.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the full spreadsheet is available at &lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/STACK/StateVectorResults.xls&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/STACK/StateVectorResults.xls&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, yes, tonight's full moon (yellow) is the closest we'll have until&#xA;2034-11-25 (red). It's also the 6th closest moon we'll have this&#xA;century.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see other examples of geocalc in action, see my:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/11899/what-is-the-format-of-the-data-from-the-jpls-horizons-system&quot;&gt;What is the format of the data from the JPL&amp;#39;s HORIZONS system?&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://space.stackexchange.com/questions/18718/calculating-positions-of-moons-ascending-and-descending-nodes-from-ephemeris-fi&quot;&gt;https://space.stackexchange.com/questions/18718/calculating-positions-of-moons-ascending-and-descending-nodes-from-ephemeris-fi&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For more about how NASA calculates these values in the first place: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/2416/where-can-i-find-a-set-of-data-of-the-initial-conditions-of-our-solar-system/2417&quot;&gt;Where can I find a set of data of the initial conditions of our solar system?&lt;/a&gt; which is also now linked to the general &quot;how to compute positions&quot; wiki answer: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/13488/where-can-i-find-visualize-planets-stars-moons-etc-positions/13489#13489&quot;&gt;Where can I find/visualize planets/stars/moons/etc positions?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-11-16T20:49:01.880" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19053" PostTypeId="2" ParentId="19025" CreationDate="2016-11-14T01:57:19.637" Score="2" Body="&lt;p&gt;The canonical reference for all astronomical calculations is the CSPICE library: &lt;a href=&quot;http://naif.jpl.nasa.gov/naif/tutorials.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://naif.jpl.nasa.gov/naif/tutorials.html&lt;/a&gt; and some of the more important files they use are the ephemeris files: &lt;a href=&quot;http://naif.jpl.nasa.gov/pub/naif/generic_kernels/spk/planets/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://naif.jpl.nasa.gov/pub/naif/generic_kernels/spk/planets/&lt;/a&gt; is one example, but you can poke around for others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I helped figure out the format (though it's actually fairly well documented) for Brandon's skyfield project (it's better than pyephem): &lt;a href=&quot;https://github.com/skyfielders/python-skyfield/issues/19&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/skyfielders/python-skyfield/issues/19&lt;/a&gt; even though I'm not a huge fan of Python myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also created Perl and Mathematica programs to read/interpret the ephemeris files:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-xsp2math.pl&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-xsp2math.pl&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-ieee754.pl&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-ieee754.pl&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/STACK/bc-solve-mathematica-105671.m&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/STACK/bc-solve-mathematica-105671.m&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/STACK/bc-solve-mathematica-105671.pl&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/STACK/bc-solve-mathematica-105671.pl&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(these are all fairly ugly, so poke around a bit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, while I don't think there IS a MATLAB equivalent at the moment, it wouldn't be difficult to create one. Of course, using the well-tested CSPICE libraries is a better option unless you happen to loathe C&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2016-11-14T01:57:19.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19055" PostTypeId="1" CreationDate="2016-11-14T08:08:08.973" Score="4" ViewCount="107" Body="&lt;p&gt;I have participated in an exhibition on astronomy and decided to make a project on the topic &quot;weather season and precipitation on other planets&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the knowledge about the weather on earth but I need to know more about that on other planets in the solar system like venus, mars, mercury and Jupiter etc or in other planets in the other galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain me how the precipitation and season  occurs,why do it occur, what is the effect,do they form clouds or something else,how do the temperature rotation and revolution affect the process? You can also suggest me some videos on this topic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="14813" LastEditorUserId="14813" LastEditDate="2016-11-14T08:13:28.127" LastActivityDate="2016-11-14T08:13:28.127" Title="Precipitation and weather and seasons on other planets" Tags="&lt;planet&gt;&lt;rotation&gt;&lt;temperature&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2017-03-16T15:19:12.797" ContentLicense="CC BY-SA 3.0" />
  <row Id="19056" PostTypeId="1" CreationDate="2016-11-14T09:00:51.223" Score="2" ViewCount="93" Body="&lt;p&gt;If primordial black hole do exists today could it be responsible for galaxies maybe even quasars, I mean these ancient monsters gobble up all materials in its wake and grow bigger and eventually evolves into say milky way galaxy etc. What are the odds of such events actually happened?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="6760" LastEditDate="2016-11-14T09:11:35.840" LastActivityDate="2016-11-14T09:11:35.840" Title="Could a super massive black hole actually be primordial black hole in disguise?" Tags="&lt;galaxy&gt;&lt;supermassive-black-hole&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19057" PostTypeId="1" CreationDate="2016-11-14T13:43:05.917" Score="2" ViewCount="106" Body="&lt;p&gt;What are some UK based groups that research astrobiology, especially the chance of extraterrestrial intelligence? &lt;/p&gt;&#xA;" OwnerUserId="14841" LastEditorUserId="2365" LastEditDate="2017-10-01T14:35:54.510" LastActivityDate="2017-10-01T14:35:54.510" Title="UK groups that investigate extraterrestrial life" Tags="&lt;astrobiology&gt;&lt;seti&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19059" PostTypeId="1" AcceptedAnswerId="19061" CreationDate="2016-11-14T18:21:31.283" Score="10" ViewCount="1490" Body="&lt;p&gt;Imagine two major surface observatories, perhaps the &quot;northermost&quot; and &quot;southernmost&quot; such (ideally on a similar longitude).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For nearby stars, could they each take a photo at the same time, and achieve a distance measurement to that star, based on about the size of the Earth as the baseline?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I appreciate that either of the scopes need only wait a few days, for their viewpoint to have moved a much longer baseline distance!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is that distance far too short?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What then is about the minimum baseline we could parallax-measure the nearest stars, with our currently best telescopes?  100,000km, a million? Far more?&lt;/p&gt;&#xA;" OwnerUserId="13071" LastActivityDate="2017-01-07T19:49:14.333" Title="Could we parallax measure stars just based on the Earth's size?" Tags="&lt;telescope&gt;&lt;parallax&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19060" PostTypeId="2" ParentId="19059" CreationDate="2016-11-14T22:07:18.487" Score="2" Body="&lt;p&gt;What you're describing is an interferometer, and in fact we already have an interferometer with a set up as you describe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't happen to know, an interferometer is a set of two or more telescopes, separated by some distance, that work in tandem to take an image of an object. By the basic principles of optics, the effective size of your telescope is governed not by the total, cumulative size of the two or more telescopes, but by the physical separation of the telescopes. That means, if you have one telescope on the north pole and another on the south pole such that they can both observe the same object at the same time, then what you effectively have is a telescope whose aperture is the size of the Earth!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know your optics, you'll know a larger aperture size means better resolution. The already existing, Earth-sized interferometer I alluded to above would be the &lt;a href=&quot;https://en.wikipedia.org/wiki/Very-long-baseline_interferometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Very Long Baseline Interferometer&lt;/a&gt; (VLBI). This telescope can measure at a sub-milli-arcsecond resolution! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;http://www.astro.cornell.edu/research/parallax/&quot; rel=&quot;nofollow noreferrer&quot;&gt;list of 70 pulsars&lt;/a&gt; who have had their parallax measured, a good portion of which have been done using the VLBI.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Some notes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The concept of interferometry is vastly complex and difficult to implement in practice. Because of the physics, the longer the wavelength, the easier it is to having working interferometric systems. As such, most interferometers such as the VLBI, are in the microwave/radio regime. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Navy_Precision_Optical_Interferometer&quot; rel=&quot;nofollow noreferrer&quot;&gt;NPOI&lt;/a&gt; is the only optical interferometer I know of and that only exists because it is funded by the US Military as a necessity for satellites and navigation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Technically, there's a lot more involved that the brief conceptual introduction I gave above but to be frank, you'll need to read an entire textbook to really understand the process and even to me some of it just seems like magic.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In researching the VLBI, you may see references to the VLBA. This is a related, but distinct collection of telescopes. Effectively, the VLBA is the Very Long Baseline Array which is comprised of telescopes all around the world that are &lt;em&gt;owned and operated by the United States&lt;/em&gt;. The VLBI however includes all the telescopes within the VLBA, but also includes other telescopes owned and operated by other countries.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-11-14T22:07:18.487" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19061" PostTypeId="2" ParentId="19059" CreationDate="2016-11-14T22:37:23.793" Score="23" Body="&lt;p&gt;In principle, it's not impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gaia spacecraft, designed primarily for measuring stellar positions, is able to measure parallaxes up to &lt;a href=&quot;http://www.astro.utu.fi/~cflynn/galdyn/lecture10.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;10 kpc away&lt;/a&gt; with 20% uncertainty. Its baseline is 2 AU; $2.3\times10^{4}$ times larger than the diameter of Earth. Thus, placing two Gaias on each side of Earth would be able to measure parallaxes of stars up to a distance of $10\,\mathrm{kpc} \,/\, 2.3\times10^{4} \simeq 0.4\,\mathrm{pc}$, meaning that you would almost be able to measure the distance of our nearest neighboring star, $\alpha$ Centauri, which lies at 1.3 pc. So you would just need improve your Gaias a little bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This ignores small complications such as the atmosphere, but if you're willing to put them outside the atmosphere, you could do it. Of course, it would sort of be a waste of time, since we already know the distances to the nearest stars, but hey, go ahead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wt1Pe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wt1Pe.png&quot; alt=&quot;JoeBlow&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2017-01-07T19:49:14.333" LastActivityDate="2017-01-07T19:49:14.333" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19062" PostTypeId="1" AcceptedAnswerId="19064" CreationDate="2016-11-14T23:23:41.977" Score="0" ViewCount="252" Body="&lt;p&gt;I'm wanting to purchase a telescope for the first time. I've narrowed it down to 2 options based on what I've discovered from researching and my price range. I mainly want do planetary observations and some deep space. The 2 options that I've found are the following&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.amazon.ca/gp/product/B0036GNNCC/ref=ox_sc_act_title_1?smid=A3DWYIK6Y9EEQB&amp;amp;psc=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Celestron 114 LCM Computerized Telescope&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.amazon.ca/Orion-9851-SpaceProbe-Reflector-Telescope/dp/B0000XMSWK&quot; rel=&quot;nofollow noreferrer&quot;&gt;Orion 9851 SpaceProbe 130 EQ Reflector Telescope&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;From what I've gathered, both do have decent specs for what I want to do but I'm still a bit lost when it comes to the aperture and focal length. The Celestron 114LCM has a 1000mm focal length but the aperture is only 114mm where the Orion 130 has a 900mm focal length but a 130mm aperture. Am I right in thinking that the Orion telescope would be better for me? Also, with the Celestron telescope it is computerised so, although the fun would be taken out of messing around with the scope and feeling satisfied with the results, is it really worth it to have this feature for a first time purchase? Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="14851" LastActivityDate="2016-11-15T02:42:42.933" Title="Buying a telescope for the first time. How important is the focal ratio and is a computerised telescope worth it?" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;deep-sky-observing&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19064" PostTypeId="2" ParentId="19062" CreationDate="2016-11-15T02:31:20.170" Score="4" Body="&lt;p&gt;We can compare these scopes on a few grounds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Celestron 114 LCM&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Greater &lt;a href=&quot;https://www.astronomics.com/focal-length_t.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;focal length&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.astronomics.com/altazimuth-mount_t.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;Alt-Az mount&lt;/a&gt; (intuitive)&lt;/li&gt;&#xA;&lt;li&gt;Computerized (easy)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Orion 9851 SpaceProbe 130 EQ&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cheaper&lt;/li&gt;&#xA;&lt;li&gt;Larger aperture&lt;/li&gt;&#xA;&lt;li&gt;Greater &lt;a href=&quot;https://www.astronomics.com/eyepiece-highest-useful-magnification_t.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;highest useful magnification&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Non-computerized (the experience of learning the sky through one's own methods, alone with the telescope, is one which you should not deny yourself, in my humble opinion)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.astronomics.com/equatorial-mount_t.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;EQ mount&lt;/a&gt; (allows for simple star tracking)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These are just some items which stand out to me as being important. The question you have asked is largely one of opinion—there is no &quot;right&quot; answer here. It depends entirely on what &lt;em&gt;you&lt;/em&gt; want. You say you want to focus on planetary observations, a field where extreme magnifications are common and useful. For deep space, larger apertures and shorter focal lengths will aid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://www.astronomics.com/focal-ratio_t.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;focal ratio&lt;/a&gt; is a quick way to judge the &quot;speed&quot; of the telescope's optics. It is found by taking the ratio between the focal length and aperture. The Orion scope in this instance has an f-ratio of f/6.9, while that of the Celestron is f/8.8. This implies that the view through the Orion scope will be slightly brighter, broader, and less-magnified than that through the Celestron, for any given &lt;a href=&quot;https://www.astronomics.com/how-to-pick-an-eyepiece_t.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;eyepiece&lt;/a&gt;. However, these ratios are quite close and are both within a range commonly used both for deep-sky and planetary observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My final piece of advice is that for a beginner, the subtle differences between these scopes in these regards &lt;em&gt;simply does not matter&lt;/em&gt;. You are unlikely to experience any significant difference between these two scopes. Because of that, I would personally recommend going for the Orion telescope—it's cheaper, will give you greater magnifications for planetary observations, and its lack of computerization will force you to truly learn the sky. Should you desire easier set-up, less time navigating through the sky, and so on, then the computerized Celestron may be preferable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Telescopes are beautiful instruments. Whichever you choose will serve you well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clear skies!&lt;/p&gt;&#xA;" OwnerDisplayName="user14781" LastEditorDisplayName="user14781" LastEditDate="2016-11-15T02:42:42.933" LastActivityDate="2016-11-15T02:42:42.933" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19066" PostTypeId="2" ParentId="19051" CreationDate="2016-11-15T07:23:53.263" Score="3" Body="&lt;p&gt;A single direct image of an exoplanets can give an &lt;em&gt;estimate&lt;/em&gt; of the mass and orbital period of an exoplanet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the former, you need a model for how the luminosity of a giant exoplanet (which all the directly imaged ones are) evolves with time, as a function of mass, and an estimate of the age. One simply looks up what mass of planet would have cooled to its current luminosity at the age of the star that hosts the exoplanet. The luminosity measurement needs an accurate distance to the system and benefits from observations at multiple wavelengths. Realistically, these masses are uncertain by a factor of two at least, because of cooling model uncertainties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example of this applied to various directly imaged exoplanets. Evolutionary tracks for various masses are shown in the luminosity versus age plane. Note the size of the age error bars and that is before you factor in uncertainty in the models (two flavours, &quot;hot start&quot; and &quot;cold start&quot; are shown as solid and dashed lines respectively - from &lt;a href=&quot;https://arxiv.org/abs/1308.3859&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bonnefoy et al. 2013&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ApuDK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ApuDK.png&quot; alt=&quot;Luminosity evolution of exoplanets&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The semi-major axis can only be estimated (a lower limit), from a single image, and only then if you have a good distance to the star/planet system. The angular separation is converted to a projected linear separation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A perfect example of the above is the estimate of the mass and orbit for the exoplanet GJ 504b (&lt;a href=&quot;https://arxiv.org/abs/1307.2886&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kuzuhara et al. 2013&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With more images spread over an appreciable fraction of the planet's period one can be more accurate. The projection of the elliptical orbit can be mapped, yielding the orbital inclination and thus a better semi-major axis measurement. The orbital period can then also be roughly estimated and checked for consistency with Kepler's third law and an estimate of the stellar mass. Note though that currently imaged planets have orbital periods of &lt;em&gt;at least decades&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/1607.05272&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wang et al. (2016)&lt;/a&gt; give an example of this approach for beta Pic b. A dozen precise images constrain the semi-major axis to a few percent and give the &lt;em&gt;total&lt;/em&gt; system (star plus planet, but dominated by the star) mass.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-11-15T13:10:36.587" LastActivityDate="2016-11-15T13:10:36.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19067" PostTypeId="2" ParentId="19051" CreationDate="2016-11-15T13:20:26.913" Score="2" Body="&lt;p&gt;I want to expand on Rob Jeffries answer:&lt;br&gt;&#xA;As pointed out in the comments you need to observe a planet over time as it continues in its orbit.&lt;br&gt;&#xA;The reason why you cannot take a simple image to estimate $a$, is that although you observe the planet &lt;em&gt;somewhere&lt;/em&gt; in its orbit, this orbit could be elliptical, and thus the distance where you see it would tell you nothing.&lt;br&gt;&#xA;However observations over time are already possible and have been done, as in the example of the four-planet HR8799 system. Here I just wanna leave that result from &lt;a href=&quot;http://adsabs.harvard.edu/abs/2015ApJ...803...31P&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pueyo et al. 2015 (fig. 9)&lt;/a&gt; with their best-fit orbits:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1CSfR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1CSfR.png&quot; alt=&quot;Pueyo et al. 2015, fig 9&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this article, they collected previous measurments of the planetary positions in HR8799, added some of their own and fitted this data with Kepler ellipses. The existence of those orbits additionally constrains the masses a little bit (they only know for all $m_p &amp;lt; 13 m_{jupiter}$), so that's not too much of a constraint) as the system would have already destabilized itself after the age of $\sim 30 Myr$ if they were higher. Later destabilization is of course still possible.&lt;br&gt;&#xA;Measurments of this type rely on very precise astrometry, which is the determinaton of an object's position. This may sound easier to do than it actually is. In the above figure the error bars of each measurment used are also given, and to suppress them to small values and at the same time make them accurate is the challenge here. One can see that the errors are of the order of 10 milliarcseconds. From the ground this precision is at the moment at the limit of what we can do, better will only Gaia be with errors down to 10-100 microarcseconds.&lt;br&gt;&#xA;But Gaia doesn't have the angular resolution for such systems which has slightly worse than 0.1 arcseconds angular resolution .&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditDate="2016-11-15T13:25:42.223" LastActivityDate="2016-11-15T13:25:42.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19068" PostTypeId="1" CreationDate="2016-11-15T16:24:08.373" Score="1" ViewCount="31" Body="&lt;p&gt;I was given a StarNavigator 90 (DS-2090) with a motorized mount (Single-arm, motorized Alt-Az) which works with 8 AA batteries. Only issue is that it eats the batteries so I was wondering what kind of batteries are recommended for motorized mounts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, can I use a 12v battery (such as those gel batteries, or even car battery) to power the mount?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you,&#xA;Matias.&lt;/p&gt;&#xA;" OwnerUserId="14862" LastActivityDate="2016-11-15T16:24:08.373" Title="StarNavigator 90 motorized battery" Tags="&lt;telescope&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19069" PostTypeId="1" AcceptedAnswerId="19070" CreationDate="2016-11-15T17:43:23.747" Score="8" ViewCount="2167" Body="&lt;p&gt;My question is this : could the Charon-Pluto system becomes in the coming few millions years a full fledged planet, as in the IAU classification ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask this because there are two facts that bother me : &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The system orbit isn't on the ecliptic&lt;/li&gt;&#xA;&lt;li&gt;The system hasn't cleared its neighborhood&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, is it possible that that system isn't stabilized yet, and that 1) its orbit is very slowly changing (for lack of a better word) to the ecliptic, and 2) that the system will slowly collapse, and bring all its moons with it, to clear its neighborhood ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, is Pluto a &quot;proto-planet&quot; (as I would personally name it), or definitely (and for the next millions years) a dwarf planet ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit : I disagree on the fact that this question is a duplicate of &lt;a href=&quot;https://astronomy.stackexchange.com/questions/11084/how-long-will-it-take-pluto-to-grow-to-planet-size&quot;&gt;How long will it take Pluto to grow to planet size?&lt;/a&gt; as this one ask not for the size of Pluto, but rather the other characteristics, the orbit and the clearing the neighborhood.&lt;br&gt;&#xA;Moreover, the question was asked in a broader canvas, to know if it was possible that some elements of the solar system (Pluto, but not only) could be instable enough to reach the level of planet sometime in the futur.&lt;/p&gt;&#xA;" OwnerUserId="14864" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2017-02-03T14:52:03.790" Title="Is Pluto a &quot;proto-planet&quot;?" Tags="&lt;planet&gt;&lt;pluto&gt;&lt;celestial-mechanics&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19070" PostTypeId="2" ParentId="19069" CreationDate="2016-11-15T18:21:05.830" Score="7" Body="&lt;p&gt;The question says a few interesting things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The system orbit isn't on the ecliptic&lt;/li&gt;&#xA;&lt;li&gt;The system hasn't cleared its neighbourhood&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These are not going to change in the next few million years - or ever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/90482_Orcus&quot; rel=&quot;nofollow noreferrer&quot;&gt;Orcus&lt;/a&gt; is an interesting counter-example.  It is in a similar orbit to Pluto - similar aphelion, perihelion and eccentricity, similar orbital period (to within a year or so) and an inclination of around 20 degrees, a little more than Pluto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, in orbital terms Orcus is as much a planet as Pluto is - &lt;strong&gt;except of course&lt;/strong&gt; it's only around half the diameter and has far less mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pluto is never going to displace Orcus from its orbit. In fact they were both pushed into their orbits by Neptune and the &lt;a href=&quot;https://astronomy.stackexchange.com/questions/12228/classification-of-the-kozai-mechanism&quot;&gt;Kozai mechanism&lt;/a&gt;, which causes its inclination. They will never be able to clear their orbits because Neptune dominates the area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the final part of the question - Pluto is officially regarded a dwarf planet by the current definition. Partly because of its orbit as your question states. Whether it's also a protoplanet or not may be debatable. I'd say it's too big to be considered a protoplanet because it's too big. (The larger asteroids like Ceres, Vesta etc. are probably closer to that size range.)&lt;/p&gt;&#xA;" OwnerUserId="7491" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-11-16T08:48:41.457" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19071" PostTypeId="1" AcceptedAnswerId="19077" CreationDate="2016-11-15T22:21:44.730" Score="0" ViewCount="76" Body="&lt;p&gt;Is it possible for a partial lunar eclipse to be a 2 on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Danjon_scale&quot; rel=&quot;nofollow noreferrer&quot;&gt;Danjon scale&lt;/a&gt;? If so, is it possible for &lt;a href=&quot;https://eclipse.gsfc.nasa.gov/5MCLEmap/0001-0100/LE0033-04-03P.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;this lunar eclipse&lt;/a&gt; to have been a 2 on the scale?&lt;/p&gt;&#xA;" OwnerUserId="14708" LastEditorUserId="14708" LastEditDate="2016-11-19T14:39:48.350" LastActivityDate="2016-11-19T14:39:48.350" Title="Partial Blood Moon" Tags="&lt;lunar-eclipse&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19072" PostTypeId="1" AcceptedAnswerId="19080" CreationDate="2016-11-16T04:39:14.597" Score="2" ViewCount="186" Body="&lt;p&gt;I understand that a &quot;regular supermoon&quot; is a full moon when the moon is at perigee (closest to earth in its orbit). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They seem to happen often, &lt;a href=&quot;https://astronomy.stackexchange.com/questions/11457/what-makes-the-moon-a-super-moon&quot;&gt;here's a post from one last year&lt;/a&gt;. But I've seen news articles saying this is the 'superest' super moon since 1948 and we won't see another super moon this bright until 2034.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes the recent super moon so bright?&lt;/p&gt;&#xA;" OwnerUserId="7156" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-11-16T19:10:30.697" Title="What made the recent supermoon so super?" Tags="&lt;orbit&gt;&lt;the-moon&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19073" PostTypeId="1" CreationDate="2016-11-16T07:57:59.600" Score="0" ViewCount="187" Body="&lt;p&gt;For a computer science project  in University; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to find image processing problems on space exploration and Space Telescopes, only related to the image itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have wrote a computer program that is able to separate every pixel of an image according to its pixel intensity (the whitest to the darkest) in 3 (or even more) big groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking that this might be somehow useful in a space image processing situation in a well known challenge of the technology in the Telescopes or machine vision devices around space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Situation can be very specific, just as an example calculating the mass of a star by the look of the pixels which compose it, or something like this.&lt;/p&gt;&#xA;" OwnerUserId="14869" LastActivityDate="2016-12-16T17:06:34.433" Title="Do you know about any Image Processing Challenges known in Space Telescope like Hubble?" Tags="&lt;star-systems&gt;&lt;hubble-telescope&gt;&lt;artificial-satellite&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19074" PostTypeId="1" CreationDate="2016-11-16T10:03:18.770" Score="9" ViewCount="6055" Body="&lt;p&gt;I already googled a face-on galaxy and an edge-on galaxy, but I couldn't find any proper information. Would you explain them to me?&lt;/p&gt;&#xA;" OwnerUserId="14782" LastEditorUserId="14782" LastEditDate="2016-12-15T04:52:57.043" LastActivityDate="2016-12-15T04:52:57.043" Title="face-on galaxy and edge-on galaxy" Tags="&lt;galaxy&gt;&lt;constellations&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19075" PostTypeId="2" ParentId="19074" CreationDate="2016-11-16T10:12:02.670" Score="18" Body="&lt;p&gt;The terms refer to the &lt;em&gt;viewing angle&lt;/em&gt;, i.e. from which direction do we observe a (disk) galaxy. If we happen to be located roughly in its plane of rotation, we see it from the &quot;edge&quot;, whereas if we are more or less above or below its plane, we see its &quot;face&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make a more quantitative statement, the orientation is described by the &lt;em&gt;inclination angle&lt;/em&gt; $i$, which is the angle away from &quot;face-on&quot;. That is, the inclination of a galaxy that is observed face-on is $i=0$, and the inclination of a galaxy observed edge-on is $i = 90º$. For randomly oriented galaxies, the mean inclination angle will be around $i = 60º$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oZavx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oZavx.png&quot; alt=&quot;Inclination&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wrt. your second question, you're right that the text is not correct. The three small constellations lie &lt;em&gt;north&lt;/em&gt; of hydra. &quot;North&quot; means &quot;toward higher values of declination&quot;. Perhaps it's more easily seen in this image (from &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydra_(constellation)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3GY5E.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3GY5E.png&quot; alt=&quot;Sky&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-11-16T11:00:04.663" LastActivityDate="2016-11-16T11:00:04.663" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19076" PostTypeId="2" ParentId="19073" CreationDate="2016-11-16T13:07:20.967" Score="2" Body="&lt;p&gt;First, I'll talk a little bit about &lt;strong&gt;the state of what is being done:&lt;/strong&gt; (according to my knowledge)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Astronomy on-board processing is usually avoided if possible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this is to have the pure data as unspoiled as possible, because instrument effects, cosmic rays, the light from Earth's corona, etc. can introduce many systematic errors into the data. If that would be removed on-board and then sent to Earth, then there would be no control later on what the actual, original data was.&lt;br&gt;&#xA;This became relevant e.e. for the exoplanet-finding &lt;a href=&quot;https://kepler.nasa.gov/Mission/faq/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kepler&lt;/a&gt;(see also the technical data there) space telescope where the photometric data was showing very unexpected behaviour at first, before the reasons for it were realized, obviously long after the spacecraft was designed and built.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the image processing challenges are usually resolved on the ground stations and with the astronomers on their big computers. Also the interpretation of the data, after being cleaned, debiased, etc. is non-trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As however the downlink speed depends on the distance $d$ to Earth like $\sim \frac{1}{d^3}$ so satellites that orbit far away do have to apply some clever ways to reduce data sometimes. The Hubble is not a candidate for this, as it orbits Earth at only some 500 km a.s.l, its data can be transferred to Earth without problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.sea-astronomia.es/drupal/sites/default/files/archivos/proceedings11/instrumentacion/portellj/portellj.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gaia&lt;/a&gt; satellite however, orbits the second Earth-Sun Lagrange point and is thus farther away from Earth, 1.5 million km. This mission is an astrometric mission, meaning it's there to measure positions and velocities of stars to extreme precision. If all the pixels of all onboard systems would continuously be read out and produce data, that would be too much to handle and some would be lost. So Gaia has a pre-detector that tells the main detector &quot;Hey a star is coming&quot;. The main CCD then creates a moving window of 12x12 pixels from which all electrons are read-out and accepted as data. So that's some limited processing there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of data transfer, missions like &lt;a href=&quot;http://pluto.jhuapl.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;New Horizons&lt;/a&gt; (Pluto encounter last year) or the Voyagers, are too far out to simply send their data over. Therefore the &lt;a href=&quot;https://eyes.nasa.gov/dsn/dsn.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Deep Space Network&lt;/a&gt; exists, to look out for radio signals from those probes and actively amplify them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What you could do:&lt;/strong&gt;&lt;br&gt;&#xA;So now that I've spent time explaining that on real spacecraft it's not being done what you want to do, we can bend the rules and just try to think about what's being done on the ground (if you want to):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Image calibration: A detector taking an image wants to be very well understood, before claiming any interesting discoveries or measurments. Part of this process is taking a dark image to characterize what your CCD is doing when nothing is measured. Then a flatfield image is done, an image of an ideally uniformly white surface, in order to characterize pixel-to-pixel variations. &lt;a href=&quot;http://www.astrofriend.eu/astronomy/tutorials/tutorial-pre-calibration-flat-dark/tutorial-pre-calibration-flat-dark.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;More&lt;/a&gt;. For the measured image then, the dark image is substracted and divided through the flatfield image.&lt;/li&gt;&#xA;&lt;li&gt;Cosmic ray removal: Cosmic rays fly through your camera in a random direction at random times (in space as well as on the ground). They produce streaks in CCD cameras, and can be automatically removed by taking the same image three times, and discarding one of them or only a pixel-region if a dramatic intensity streak somewhere is measured.&lt;/li&gt;&#xA;&lt;li&gt;Measuring a point-spread function:  Stars are so far away, that they appear as point sources. However through instrumental effects this ideal point will always be spread out into a finite shape, called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Point_spread_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;PSF&lt;/a&gt;. Because this function can be easily a few pixels wide, it is very important to know the shape of the PSF to be able to assign photons to a star that have been measured in off-center pixels. Knowing then a star's photon output and it's distance one can calculate the luminosity. But even that can't be done on an on-board chip. What you could to is to try to automatically fit an Airy-function to the PSF.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Ok so that was just a few ideas, but I'm not an observer, there may be better ones out there. Also you got keywords not to google further.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditDate="2016-11-16T13:25:58.647" LastActivityDate="2016-11-16T13:25:58.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19077" PostTypeId="2" ParentId="19071" CreationDate="2016-11-16T13:50:02.500" Score="2" Body="&lt;p&gt;If we go with the discussion at &lt;a href=&quot;https://www.timeanddate.com/eclipse/facts-lunar-eclipse.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Time&amp;amp;Date&lt;/a&gt;, it would depend on just how 'red' you want the moon to look.  The term &quot;blood moon&quot; is not an astronomical term, nor is it a specific  full moon, such as &quot;blue,&quot; &quot;harvest,&quot; or even &lt;a href=&quot;http://farmersalmanac.com/full-moon-names/&quot; rel=&quot;nofollow noreferrer&quot;&gt;these&lt;/a&gt; .&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2016-11-16T13:50:02.500" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19078" PostTypeId="2" ParentId="19074" CreationDate="2016-11-16T17:56:52.993" Score="16" Body="&lt;p&gt;I just want to add pictures to augment &lt;a href=&quot;https://astronomy.stackexchange.com/a/19075/14255&quot;&gt;pela's answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just google &quot;edge on galaxy&quot; and &quot;face on galaxy&quot; to find results like these, it is entirely about the viewing angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example of a face on galaxy (Messier 74)&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/KHER5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KHER5.jpg&quot; alt=&quot;Messier 74&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example of an edge on galaxy (NGC 891)&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/HWu12.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HWu12.jpg&quot; alt=&quot;NGC 891&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14255" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-11-16T17:56:52.993" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19079" PostTypeId="1" AcceptedAnswerId="19082" CreationDate="2016-11-16T18:31:39.260" Score="3" ViewCount="164" Body="&lt;p&gt;If something falls into a black hole, but has not yet hit the singularity in the center, can we (outside of the black hole) feel its gravity? On the one hand, I've learned that there's no distance limit to gravity and that everything in the universe feels the gravity of everything else. On the other hand, we would be able to measure the gravitational field of the thing and thus figure out where in the black hole it is, which contradicts the idea that no information can escape a black hole.&lt;/p&gt;&#xA;" OwnerUserId="14878" LastActivityDate="2016-11-17T02:23:16.780" Title="Gravity of other objects in black holes?" Tags="&lt;black-hole&gt;&lt;gravity&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19080" PostTypeId="2" ParentId="19072" CreationDate="2016-11-16T19:10:30.697" Score="7" Body="&lt;p&gt;The thing you have to really understand is that not all supermoons are equal. The official definition of a supermoon, as you state, is that the full moon must coincide with the moon being at perigee.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this never actually happens! The moon being at perigee happens for only a single instance of time - less than a second. The moon being officially &quot;full&quot;, again only happens for a single instance of time (i.e., when the Sun, Earth, and Moon all form a single line). The chances of these two instances of time happening simultaneously are almost non-existent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when someone calls a full moon a supermoon, they really just mean that the exact full moon instance is &quot;pretty close&quot; to the exact moment of reaching perigee. The closer the better and the more super your supermoon will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take a look at the Supermoon which occured on Nov. 14, widely proclaimed to be the &quot;superest&quot; supermoon for a long time, you can see that perigee was reached precisely at 11:23 UTC, while the full moon occurred precisely at 13:52 UTC, two and a half hours later. For pretty much every other supermoon, these two events are close enough for the moon to be called a supermoon, but not two-and-a-half-hours close. The closeness in time between reaching perigee and being truly full made this moon especially big and more important, especially bright.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also another thing which made this supermoon the superest and that was the actual distance of the Moon's perigee. As with anything in our solar system, the Moon's orbit is not perfectly elliptical, but instead has small variations and perturbations over time. This means that the perigee of the Moon's orbit is slightly changing over time. The November supermoon occured during a particularly close perigee that won't be matched until 2034 supermoon which you reference.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-11-16T19:10:30.697" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19081" PostTypeId="2" ParentId="9937" CreationDate="2016-11-17T01:02:09.497" Score="0" Body="&lt;p&gt;believe it or not, we can't expect a doctorate of celestial motion to arrive when we have questions. I'm just someone who has asked this question before, and as far as I can remember.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you gave both the moon and the planet perfectly circular orbits then they would fall out of sync. &#xA;Being in sync would require the gravities to sit at a perfect angle to each other exactly the same as when you balance a spoon and a fork in midair with a matchstick.&#xA;What your question actually sounds like is, &quot;if we round gravities to make it plausible that they could sit in perfectly round orbit, would it stay?&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only then, and even then, only if we artificially created the system, because the universe knows the weight of everything down to the Higgs boson, if it is a single boson asymmetrical in shape or weight of the celestial bodies, then enough would be thrown off. space would have to be a complete vacuum. and something about temperature, radiation, impacts, other celestial bodies. ect ect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much like when you try to balance anything, you are playing with friction. This can be hard to understand, however, every single moon in existence will one day either crash into the planet, get flung away, or otherwise prove wrong anyone claiming a stable orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest possible would be something like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Star A, has Planet A, which has Moon A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moon A is traveling away from Planet A at .00000000000001 mm a year. &#xA;Due to universal frictions and radiation, it is calculated to lose all escape momentum this year. &#xA;However, on average Moon A gets struck with debris that, more often than not, contributes to its escape velocity. Meaning that anytime the moon attains a stable orbit through decay, It loses it to impacts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this example Moon A doesn't get hit every year, some years it decays enough to start inching closer to the Planet, but it inevitably is struck again.&lt;/p&gt;&#xA;" OwnerUserId="14881" LastEditorUserId="14881" LastEditDate="2016-11-17T01:12:05.677" LastActivityDate="2016-11-17T01:12:05.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19082" PostTypeId="2" ParentId="19079" CreationDate="2016-11-17T02:23:16.780" Score="3" Body="&lt;p&gt;Gravity, according to the majority of published and non-published scientists, is able to propagate information from one point to another at the speed of light. This means that as soon as any massive body &quot;physically&quot; interacts with a black hole (enters the Schwartzchild radius) there is no longer a way for anything outside of the black hole to know anything else regarding the status of the body. Therefore the new addition to the black hole simply becomes part of the system and can no longer be considered a separate entity. The body's gravity becomes that of the black hole's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... No&lt;/p&gt;&#xA;" OwnerUserId="2232" LastActivityDate="2016-11-17T02:23:16.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19083" PostTypeId="2" ParentId="9937" CreationDate="2016-11-17T03:29:17.570" Score="3" Body="&lt;p&gt;A couple of points.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A hypothetical simulation starts with a planet called Earth in a&#xA;  perfectly circular orbit around a star called Sol. There is also a&#xA;  moon called Luna in a perfectly circular orbit around Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While a perfectly circular orbit isn't possible, it's true that you can mathematically define a perfectly circular orbit, treating the Earth as a gravitational point at it's center of mass and the Moon the same way.   Orbits are often calculated around point-centers of mass because it's so much simpler and accurate enough, so that part is fine.  You can have a perfectly circular orbit in a mathematical thought experiment.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question: Can Luna's orbit remain perfectly circular or does it become&#xA;  elliptical over time? At what point does it become stable? Never?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Thinking: The distance between Sol and Luna is not constant,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're correct.  While the 2 body system can (mathematically) remain perfectly circular, ignoring tides, gravitational lumpiness, etc, the minute you add a 3rd body, whether the sun or something much smaller, the perfect circle is no longer possible because the 3rd body stretches it.   It creates a gravitational tug on the other 2 bodies.  3 gravitational body mathematics gets enormously complicated.  See &lt;a href=&quot;https://en.wikipedia.org/wiki/Three-body_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;3 body problem&lt;/a&gt;, so I won't get into that.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you've assumed wrong is in thinking that the Sun or 3rd body makes the Moon's orbit elliptical.  The circular orbit is already elliptical because a circle is an ellipse, just one with an eccentricity of zero.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you add the 3rd body, it's gravity bends the shape into a kind of egg shape.  It doesn't turn the circle into an ellipse, it turns the circle into an egg or it turns the ellipse into an elliptical egg or bent or stretched ellipse.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thinking: The distance between Sol and Luna is not constant, it&#xA;  fluctuates as Luna orbits Earth. Therefore the pull sunward also&#xA;  fluctuates. Shouldn't this cause all moon orbits to eventually&#xA;  deteriorate into longer and longer elliptical orbits until they&#xA;  collide with the planet they orbit?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This, again, not really, no.  Orbital deterioration is complicated and it mostly requires more than just a 3rd body tug, even a large tug like the one the sun gives the Moon.  What the large close sun does is it reduces the Earth's &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;hill Sphere&lt;/a&gt;.  Planets closer to the sun for the most part have smaller hill spheres (&lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere#Solar_System&quot; rel=&quot;nofollow noreferrer&quot;&gt;see table&lt;/a&gt;) and the inner 1/3rd or 1/2 of the Hill Sphere is long term stable, so the sun doesn't destabilize Earth's moon at all, it merely jiggles it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbits within the stable part of a planet's hill sphere can get bent by the sun, but they remain stable.   The reason is that the tugs tend to balance out over time.   Sometimes the Sun will bend the Moon's orbit more eccentric and sometimes it will bend it back closer to a circle.  The net effect doesn't accumulate.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mathematics or orbital perturbations is also very complicated.   Earth's Milancovich cycles are orbital perturbations, driven mostly by the influence of Jupiter and Jupiter is not expected to ever destabilize the Earth.  While obviously many collisions have happened, by now, most of the orbits in the solar-system are long-term stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, back to the moon and onto Tidal forces, The Moon is &lt;a href=&quot;http://curious.astro.cornell.edu/about-us/37-our-solar-system/the-moon/the-moon-and-the-earth/111-is-the-moon-moving-away-from-the-earth-when-was-this-discovered-intermediate&quot; rel=&quot;nofollow noreferrer&quot;&gt;moving away from the Earth&lt;/a&gt; at about 3.8cm per year because Earth's tidal forces tug the moon forward in a sense, which speeds it up and as it speeds up it moves into a higher orbit.  Eventually if the Earth's rotation was to stay constant, in about 50 billion years the Earth would eject the Moon out of it's orbit.  Because the Earth's orbit is slowing down and because the sun is going to explode long before then.  For both those reasons, the Earth is not expected to lose the moon.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tidal forces can operate either way on pushing an orbital further out or closer in.  Mars' tidal forces on Phoebos are drawing Phoebos closer to Mars and in time it will crash into the planet or break apart into a ring system and more gradually crash into the planet.  The varying factor is whether the planet rotates faster than the Moon in which case the tidal bulge is ahead of the Moon and pulls it forward and away, or if the Moon orbits faster than the rotation in which case the tidal force is behind the Moon and slows it down, pulling it in.  When objects become tidally locked to each other, the tidal forces are largely minimized.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, back to the point, the tidal forces that push the Moon away from the Earth, also tend to circularize the Moon's orbit, because the forces are greatest when the Moon is closest.   If not for the sun, the Moon's orbit around Earth would be nearly circular.  Tidal forces often circularize orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the Sun does tug on the Moon's orbit in the way you suggest and the Moon speeds up as it moves towards the sun and it slows down as it pulls away, giving it a lumpy orbit around the Earth.   As the Moon moves further from the Earth it's orbit will grow more and more lumpy or chaotic, but it's not expected to ever get chaotic to the point that it will lose it's stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in that sense, the Sun's gravitation is responsible for the Moon's eccentricity around the Earth because if not for the sun, that eccentricity would be nearly zero and nearly a perfect circle so the sun kind of does both. It pulls the Moon's orbit out of an ellipse, but it's also responsible for the Moon's eccentricity being what it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a (perhaps somewhat exaggerated) picture of what the Moon's orbit around the Earth really looks like.  This is because of the stretching and bending the sun does to the Moon's orbit around the Earth.   It's not significant enough to pull it free, but significant enough to jiggle it pretty good. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8mFbX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8mFbX.png&quot; alt=&quot;http://probe-1.com/wp-content/uploads/2013/01/Moons-Orbital-Path-around-Earth-2013-01-16-480x315.png&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pics-about-space.com/moon-orbit-around-earth?p=4&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are 2 pictures of Mercury's &quot;cool&quot; orbit that kind of spirographs around the sun.  This is a more typical 2 body orbit, notice how much neater and closer to an ellipse it is than the ugly Moon orbit above.  That's because it's still essentially a 2 body orbit not a 3 body.  2 body orbits are neat.  3 body orbits can get chaotic, stretched or bent, and the Moon is more than half way towards where it would become unstable and leave the Earth, so it's orbit is more unstable than most &quot;stable&quot; orbits in our solar-system, but it's still considered a stable orbit cause it's well inside the stable region of Earth's Hill Sphere.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9JC5q.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9JC5q.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WZ4f9.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WZ4f9.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an FYI, Mercury's orbit doesn't neatly follow Newton's orbital laws either due to relativity and that was quite the puzzle to scientists for decades.  The non Newtonian gravity accentuates Mercury's Spirograph orbit, pictured above but I used it, cause it gives a good example of a 2 body orbit over multiple orbits compared to our moon's jiggled orbit due to influence of a 3rd body (the sun).  The point of all of this, is that elliptical orbits, (for the most part), are just as mathematically stable as theoretically/mathematically circular ones.  Elliptical orbits aren't unstable nor are they an example of an orbit becoming unstable, they are stable, or at least, not unstable because they are elliptical.   Some orbits are long term stable others are not, but the eccentricity isn't really a measuring stick for instability. The stable region of the hill sphere is a far better measuring stick.   Both the orbits below, provided they are inside the stable region of the hill sphere, would be equally stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sRaxp.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sRaxp.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that wasn't tldr.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-11-17T04:06:33.057" LastActivityDate="2016-11-17T04:06:33.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19084" PostTypeId="1" CreationDate="2016-11-17T06:50:22.917" Score="1" ViewCount="64" Body="&lt;p&gt;According to &lt;a href=&quot;https://astronomy.stackexchange.com/a/19080/7156&quot;&gt;this answer&lt;/a&gt; there is a scale of super moons depending on how close the full moon(henceforth anti-sol) is to perigee. So how close to anti-sol does the moon's perigee need to be to be referred to as a &quot;super moon&quot;?&lt;/p&gt;&#xA;" OwnerUserId="7156" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-11-17T16:58:26.053" Title="How close must a full moon be to become a supermoon?" Tags="&lt;terminology&gt;&lt;the-moon&gt;&lt;definition&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19085" PostTypeId="1" AcceptedAnswerId="19087" CreationDate="2016-11-17T11:21:16.293" Score="8" ViewCount="1600" Body="&lt;p&gt;I can't find any other picture of the Titan surface from the Huygens probe than this one&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any? Are they not disclosed on purpose?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the only one taken, why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://i.picresize.com/images/2016/11/17/uXaDQ.jpg&quot; rel=&quot;noreferrer&quot;&gt;Titan surface http://i.picresize.com/images/2016/11/17/uXaDQ.jpg&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14883" LastActivityDate="2019-05-15T17:00:48.507" Title="Are there other pictures of Titan surface from Huygens?" Tags="&lt;saturn&gt;&lt;titan&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19087" PostTypeId="2" ParentId="19085" CreationDate="2016-11-17T13:22:50.770" Score="16" Body="&lt;p&gt;One second of googling reveals the whole archive:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://esamultimedia.esa.int/docs/titanraw/index.htm&quot; rel=&quot;noreferrer&quot;&gt;http://esamultimedia.esa.int/docs/titanraw/index.htm&lt;/a&gt;&lt;br&gt;&#xA;(note you can click onto the strips to inspect them!)&lt;br&gt;&#xA;The archive depicts the whole decent of the Huygens lander onto Titan, with the first picture being on the ground, and some of them seem to be taken from higher up altitudes, pages 17-27 you can see the flow deltas)&#xA;The three pictures per strip show the views of the three cameras into different directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the location on the planet doesn't change, just the viewing angle around the spacecraft upon descent. This is because Huygens was a lander without any additional propulsion, as it was already considered to be a massive feat if the lander would make it to the surface at all.&lt;br&gt;&#xA;Also a propulsion system would have been more expensive, and more or less useless, as the expected lifetime of the lander in Titan's atmosphere was a few hours, which happened to be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So summa summarum, from the final position in which the lander was fixed, yes this is the only picture of Titan's surface.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditDate="2016-11-17T13:29:44.747" LastActivityDate="2016-11-17T13:29:44.747" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19092" PostTypeId="1" AcceptedAnswerId="26671" CreationDate="2016-11-18T01:57:26.663" Score="5" ViewCount="150" Body="&lt;p&gt;What are the differences between the discoveries of &lt;a href=&quot;http://adsabs.harvard.edu/abs/2004ApJ...617..645B&quot; rel=&quot;noreferrer&quot;&gt;Sedna&lt;/a&gt;, &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014Natur.507..471T&quot; rel=&quot;noreferrer&quot;&gt;2012vp 113&lt;/a&gt;, and the supposed &lt;a href=&quot;http://adsabs.harvard.edu/abs/2016AJ....151...22B&quot; rel=&quot;noreferrer&quot;&gt;Planet 9&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All need deep infrared surveys at the ecliptic plane, but P9 needs a much deeper survey?&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2020-07-11T04:29:54.913" Title="Which kind of strategy should we take to discover Planet 9 and other KBOs?" Tags="&lt;planet&gt;&lt;solar-system&gt;&lt;asteroids&gt;&lt;kuiper-belt&gt;&lt;infrared&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19093" PostTypeId="1" CreationDate="2016-11-18T04:01:51.720" Score="2" ViewCount="507" Body="&lt;p&gt;My question is basically asking how the vernal equinox changes with time, and how we can track where it will be at certain points in time. &lt;/p&gt;&#xA;" OwnerUserId="14890" LastActivityDate="2016-11-21T13:21:51.773" Title="What zodiac constellation will the Vernal Equinox be directed towards in ~6500?" Tags="&lt;coordinate&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19094" PostTypeId="2" ParentId="18994" CreationDate="2016-11-18T09:25:34.473" Score="2" Body="&lt;p&gt;Entropy is the only quantity in the physical sciences that requires a particular direction for time, sometimes called an arrow of time. As one goes &quot;forward&quot; in time, the second law of thermodynamics says, the entropy of an isolated system can increase, but not decrease, Universe can seen as one big isolated system in which total energy is always constant. Hence, from one perspective, &lt;strong&gt;entropy measurement is a way of distinguishing the past from the future.&lt;/strong&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Entropy_(arrow_of_time)&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Entropy_(arrow_of_time)&lt;/a&gt; - wiki for relation between time and entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=870y6GUKbwc&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=870y6GUKbwc&lt;/a&gt; - Basics of entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Highest state of entropy is achieved when the system is in equilibrium, to have this equilibrium, energy entities (mass and other energy radiating sources) should be enough apart from each other that no more energy interaction(give and take) could happen and energy is distributed evenly in the universe, thus expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=xZTb6sfHEX8-&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=xZTb6sfHEX8-&lt;/a&gt; Will the Universe expand forever&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=26_GN-uL4Jc&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=26_GN-uL4Jc&lt;/a&gt; - The Laws of Thermodynamics and                      Fate of The Universe&lt;/p&gt;&#xA;" OwnerUserId="14102" LastActivityDate="2016-11-18T09:25:34.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19095" PostTypeId="2" ParentId="19093" CreationDate="2016-11-18T11:31:35.040" Score="3" Body="&lt;p&gt;Due to Earth's axial precession, the vernal equinox gradually travels westward, completing a cycle in roughly 26,000 years, i.e. about 1º every 72 years. Currently, the vernal equinox crosses the ecliptic in Pisces, near its border with Aquarius. In the year 6500, i.e. in 4500 years, it will have moved a further&#xA;$$&#xA;4500\,\mathrm{yr} \,\, \times \,\, \frac{360º}{26\,000\,\mathrm{yr}} \simeq 62º,&#xA;$$&#xA;that is, to the point that currently has a right ascension of roughly 20 hours. From the map below (from &lt;a href=&quot;https://en.wikipedia.org/wiki/First_Point_of_Aries&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;), you can see that it will just have moved out of Capricornus and into Sagittarius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mTtzv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mTtzv.jpg&quot; alt=&quot;VE&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @barrycarter made me aware, my first drawing was wrong. Today, the vernal equinox in this map is a vertical line in the right-most edge of the map, going from Polaris in Ursa Minor and through Cepheus, Cassiopeia, etc., down to the Octans. However, due to the precession this map will become distorted, and Polaris will no longer be the Northern Star; instead the axis of Earth will travel roughly along the green line, in the year 6500 lying somewhere in the middle of Cepheus. The Vernal Equinox will lie roughly along the red line, going from Cepheus through Cygnus, Vulpecula, Aquila to Sagittarius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I drew this line using &lt;a href=&quot;http://stellarium.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellarium&lt;/a&gt;, setting the date to 6500-01-01, switching on equatorial grid of that date and today, and using the orthographic view. My line on the southern sky may be a somewhat imprecise, but on the northern it should be fine.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-11-21T13:21:51.773" LastActivityDate="2016-11-21T13:21:51.773" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19096" PostTypeId="1" CreationDate="2016-11-19T04:07:06.480" Score="4" ViewCount="1011" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;From a distance of 160,000 light years, the Large Magellanic Cloud looks at first glance like an isolated section of the Milky Way, running across southern Dorado into neighbouring Mensa.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is the Large Magellanic Cloud contained in the Milky Way? I don't think it is, but I'm not entirely sure.&lt;/p&gt;&#xA;" OwnerUserId="14782" LastEditorUserId="9527" LastEditDate="2020-09-25T08:09:58.020" LastActivityDate="2020-09-25T08:09:58.020" Title="Is the Large Magellanic Cloud contained in the Milky Way?" Tags="&lt;milky-way&gt;&lt;magellanic-cloud&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19097" PostTypeId="1" CreationDate="2016-11-19T04:45:32.500" Score="3" ViewCount="611" Body="&lt;p&gt;A tidal day is the MEAN time between successive meridian transits of the Moon. (This is sometimes called called a lunar day, a term that should be deprecated for use in this sense, because it is also used for the length of a day on the Moon and should be reserved for the latter meaning to avoid ambiguity.) Its length exceeds a mean solar day of 24 hours by X minutes, which is also the (mean) time-length by which moonrise (or its meridian transit) is later each day than it was yesterday. (&lt;a href=&quot;https://urldefense.proofpoint.com/v2/url?u=http-3A__oceanservice.noaa.gov_education_kits_tides_media_supp-5Ftide05.html&amp;amp;d=AwMFaQ&amp;amp;c=2X_btuPRWkGwRX26NHIotw&amp;amp;r=uOwu8neHJUHgad2beBfZ56ZZjypo5ad7MWWuBxjkUfQ&amp;amp;m=d_8Uoce1JnY95qK1NFicSrftfWFNbTC-imykJ52c384&amp;amp;s=7TwwObW7gwrUwviHeMjRJVa_7PNTD3qBVVu7_oASD9c&amp;amp;e=&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is an animation by NOAA demonstrating this principle.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact value of X is variously given in different places, ranging from about 48 to 52. However, I cannot find anywhere an authoritative source detailing the &quot;correct&quot; way to calculate X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of an authoritative method of calculating X, and/or can anyone comment on the three methods given below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) I have a personal theory that puts X at about 48.763 minutes, but I am uneasy about how correct the underlying logic of it is, which is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider two successive conjunctions (C1 and C2) of the MEAN sun and MEAN moon. At C1, the Sun's azimuth and the Moon's azimuth will, at ALL places on Earth, be the same, and, although this is no longer true for the rest of the lunation, it will once again be true at C2. This is analogous to two clocks: Clock 1 (the MEAN sun) keeps correct time, while clock 2 (the MEAN moon) loses a constant amount of time (X minutes) each day. Since the two clocks agree at C1 and at C2, clock 2 must have lost exactly 24 hours (1440 minutes) during the intervening synodic lunation, whose length (Ln) is 29.5305891203704 days. This suggests that X (the number of minutes that clock 2 loses per day) = (1440 / Ln) = 48.763.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I feel uneasy about this logic for this reason: If C1 occurs on April 1 at 06:00:00, UTC, and it is then sunrise at Tema, Ghana (on the prime meridian, near the equator), C2 will be on April 30 at 18:44:03, UTC, and that will not be sunrise at Tema; it will be dusk there, and it will be sunrise at east longitude 169°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this negate the above logic suggesting that X = 24 hours / Ln?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Another method for calculating X suggested to me is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is based on the Moon's sidereal period (Ld), 27.321661 days. The MEAN angular distance along its orbit travelled by the Moon each day is (360° / Ld) = 13.17636° (which equates to a mean orbital speed of about 0.55° per hour.) So, each day, the Earth must rotate 360° plus an extra 13.17636° to point the same meridian back at the Moon again. The Earth's rotational speed is (360° / 24 hrs) = 15° per hour. At that rate, the extra time needed by the Earth to rotate those extra 13.17636° is [(13.17636/360) * 24] = 0.878424 hour = 52.7 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) I argue that even if method (2) is right in principle, it requires two amendments, because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) The Earth rotates 360° in a sidereal day, which is nearly 4 minutes shorter than 24 hours, in which time the Earth rotates nearly 361°, or, more precisely, 360.98565°. (The fraction in the latter value is obtained from the average daily portion of the Earth's orbit: 360°/365.2422 days = 0.985647° per day.) (An animation demonstrating why this is so is &lt;a href=&quot;http://moongazer.x10.mx/website/astronomy/downloads.htm#solarVsidereal&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.) So, the extra time needed by the Earth to rotate those extra 13.17636° is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[(13.17636/360.98565)*24] = 0.8760 hours = 52.56 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Furthermore, in those extra 52.56 minutes, the Moon has moved on in its orbit by nearly another half a degree. More precisely, it is (360/27.321661)*(52.56/1440) = 0.48094°. So the calculation should be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[(13.17636+0.48094)/360.98565)*24] = 0.90800 hours = 54.48 minutes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have three different methods of calculating X, yielding values of 48.763, 52.56 or 54.48 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comments anyone? Either endorsements or criticisms of any of these three methods will be gratefully accepted.&lt;/p&gt;&#xA;" OwnerUserId="14889" LastActivityDate="2019-03-02T08:11:55.957" Title="Length of a tidal day or by how much is the Moon late each day?" Tags="&lt;the-moon&gt;&lt;positional-astronomy&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19098" PostTypeId="2" ParentId="19093" CreationDate="2016-11-19T06:21:57.867" Score="2" Body="&lt;p&gt;According to Stellarium, the vernal equinox will indeed be in Sagittarius in the year 6500, right around what we call 20h today (actually 19h53m):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/e7Uvn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/e7Uvn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 0h line, however, will look more like this (blue vertical line in the center of the image), and the celestial north pole will be in Cepheus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qOIPL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qOIPL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the south celestial pole will be in Carina:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/73Xm1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/73Xm1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2016-11-19T06:21:57.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19099" PostTypeId="2" ParentId="19096" CreationDate="2016-11-19T07:06:38.077" Score="7" Body="&lt;p&gt;The short answer is that the Magellanic clouds are not part of the Milky way;&amp;nbsp;they are satellite galaxies of the Milky way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image below shows the Andromeda galaxy. Note the small extra galaxies, marked Messier 32 and Messier 110. They are not part of the Andromeda galaxy, but are satellites of it. They would appear like the Magellanic clouds of the Andromeda galaxy. &lt;a href=&quot;https://i.stack.imgur.com/FVfze.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FVfze.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="7142" LastEditDate="2016-11-20T20:13:50.573" LastActivityDate="2016-11-20T20:13:50.573" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19101" PostTypeId="1" CreationDate="2016-11-19T10:25:12.887" Score="1" ViewCount="87" Body="&lt;p&gt;How much mass per unit volume do known dark matter have, for instance in the Bullet Cluster? Does the density vary in space, for example, do individual dark matter structures have denser cores?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2016-11-19T10:25:12.887" Title="What's the density of dark matter (clouds)?" Tags="&lt;dark-matter&gt;&lt;galaxy-cluster&gt;&lt;density&gt;&lt;intergalactic-space&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19102" PostTypeId="2" ParentId="19057" CreationDate="2016-11-19T11:46:19.573" Score="1" Body="&lt;p&gt;Extraterrestial intelligence is a hard research topic as the search for evidence is not straight forward in any way. The search for intelligent life as far as I know is limited to observations in radio and for that you want to see if any SETI members are UK based.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if you're more into astrobiology in general you might wanna look up the &lt;a href=&quot;http://www.astrobiology.ac.uk/about/%20%22UK%20Centre%20for%20Astrobiology&quot; rel=&quot;nofollow noreferrer&quot;&gt;UK Centre for Astrobiology&lt;/a&gt; which is based in Edinburgh. I don't know any of the faculty but I do know that they are hosting a meeting for early career scientists in 2017. There are also some smaller research groups based in the UK but you might wanna check out the specific research aim for these (e.g. the group at Imperial College in London is focused on Mars) as it might not be the area within astrobiology that interests you the most. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also &lt;a href=&quot;https://www.kent.ac.uk/physical-sciences/staff/profiles/mark-burchell.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mark Burchell&lt;/a&gt; who is a professor at the University of Kent. His group did some exciting work on Panspermia (the idea of life travelling through space) that I happened to read earlier this year. Note that this is focused on microorganisms so this is probably as far from &quot;intelligence&quot; that it gets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you can try to look up departments at different universities. You probably want to look under astronomy or sometimes physical or Earth science as the research aims of some groups put them more in e.g. Earth science than actual astronomy. Sometimes you can also find smaller groups under biology or life science.&lt;/p&gt;&#xA;" OwnerUserId="14899" LastActivityDate="2016-11-19T11:46:19.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19103" PostTypeId="1" AcceptedAnswerId="19106" CreationDate="2016-11-19T12:59:05.197" Score="0" ViewCount="35" Body="&lt;p&gt;I see for example, the L.O. of the Quadrantids is 283.16 and Lyrids is 32.32.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;src: &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_meteor_showers&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/List_of_meteor_showers&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5401" LastActivityDate="2016-11-19T19:42:57.413" Title="In relation to Meteor Showers, what is meant by Lambda Omicron?" Tags="&lt;meteor-shower&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19106" PostTypeId="2" ParentId="19103" CreationDate="2016-11-19T19:42:57.413" Score="1" Body="&lt;p&gt;Not $\lambda \omicron$ or $\Lambda. \mathrm{O}.$ but $\lambda_⊙{}^\circ$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the solar ecliptic longitude, the position of the sun relative to the vernal equinox, in degrees clockwise. This varies from 0 at the spring equinox, over 360 degrees. The Solar longitude is a way of giving the date of a meteor shower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effects of leap days mean that the date of the peak of a meteor shower will vary from year to year by a few days. However, the solar longitude at a meteor's shower's peak should remain the same from year to year. That is any variation in the solar longitude peak represents a real astronomical variation in the shower, not just an artefact of how we structure the calendar. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-11-19T19:42:57.413" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19107" PostTypeId="2" ParentId="19097" CreationDate="2016-11-19T21:32:48.240" Score="2" Body="&lt;ol&gt;&#xA;&lt;li&gt;As you point out, the mean moon conjucts (and actually eclipses) &#xA;the mean Sun every M = 29.5305891203704 days. If the moon's sydonic &#xA;period were exactly 29 days, the gain would be clearly 1440/29 or &#xA;49.6552 minutes. If the moon's sydonic period were exactly 30 days, &#xA;the gain would be clearly 1440/30 or 48 minutes. Thus, your &#xA;calculation seems to be correct here. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;2, 3, 4. I think this calculation is incorrect because you're &#xA;confusing sidereal days with solar days, and your attempt to correct &#xA;for this is in the wrong direction: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Suppose the moon culminates at noon at a given location (which &#xA;technically means the sun also culminates, and we have an eclipse, &#xA;as above). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Twenty four hours later, the sun culminates again, and the moon &#xA;has moved 13.17636 degrees &lt;strong&gt;along the ecliptic&lt;/strong&gt; in its orbit. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt;, the sun has also moved (about 1 degree) along the &#xA;ecliptic in the same direction. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In other words, the moon has only &quot;gained&quot; about 12.17636 degrees &#xA;with respect to the Sun (it's actually a little more, since the Sun &#xA;moves less than one degree along the ecliptic per solar day). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since we measure time by the sun, we only need to rotate an extra &#xA;12.17636 degrees for the moon to culminate. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;That's &lt;strong&gt;less&lt;/strong&gt; time than we'd need otherwise, not more. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I'm too lazy, but if you do the calculations keeping the above in &#xA;mind, you should get the same answer. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2016-11-19T21:32:48.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19108" PostTypeId="1" AcceptedAnswerId="19110" CreationDate="2016-11-19T22:41:09.233" Score="-1" ViewCount="248" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Black holes emit x-rays lights from within their event horizons (I believe this one is unknown because you cannot see what's going on inside event horizon)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A 9 km black hole may be hidden between Saturn and Jupiter (I think it's true because the distance between these planets are larger than 9km).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A black hole in a binary system will suck in its companion (Not sure about this one because I don't think the effect is large in a non-close binary system)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Singularity of black hole has infinite density (I guess it is true but can't find evidence)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Could someone give any suggestion, please?&lt;/p&gt;&#xA;" OwnerUserId="14254" LastActivityDate="2016-11-20T02:01:03.260" Title="Does black hole have the following properties?" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ClosedDate="2016-11-21T14:58:17.003" ContentLicense="CC BY-SA 3.0" />
  <row Id="19110" PostTypeId="2" ParentId="19108" CreationDate="2016-11-19T23:50:38.933" Score="4" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Black Holes do not emit X-Rays from inside the event horizon.  The X-rays are emitted from the black hole's &lt;a href=&quot;https://en.wikipedia.org/wiki/Accretion_disk&quot; rel=&quot;nofollow noreferrer&quot;&gt;accretion disk&lt;/a&gt;.  This is a swirling disk of gas and dust which is spiraling into the black hole.  As it falls inwards, it heats up, giving off energy in the form of x-rays and gamma rays.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A 9km black hole could not lie within the solar system without disrupting the orbits of nearby planets.  A black hole 9km across would have to be about 3 times more massive than our sun.  You can calculate this by using a simplification of the formula for the &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schwarzschild Radius&lt;/a&gt;: $r\approx 2.95 \frac{M}{M_{\odot}}$.  If this were the case, then the orbital motions of the planets would be dominated more by this black hole than by the sun.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A black hole doesn't really &quot;suck&quot; things in.  It only has a gravitational pull relative to its mass.  So if it was a part of a binary system and its neighbor is far away, its partner star will not notice any kind of change when the other star turns into a black hole.  As an example, if the sun were to turn into a black hole right now, the Earth would continue orbiting around it without any kind of change.  However, if the black hole and the other star are &lt;a href=&quot;http://www.astronomy.com/news/2014/01/spanish-researchers-discover-the-first-black-hole-orbiting-a-spinning-star&quot; rel=&quot;nofollow noreferrer&quot;&gt;close enough together&lt;/a&gt;, matter from the star can fall off of it into the black hole.  The black hole's proximity actually disrupts the internal structure of the star, causing matter to be ejected from the star.  A similar situation happens when you have a &lt;a href=&quot;https://en.wikipedia.org/wiki/Type_Ia_supernova&quot; rel=&quot;nofollow noreferrer&quot;&gt;white dwarf orbiting close to another star&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes, you can consider the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_singularity&quot; rel=&quot;nofollow noreferrer&quot;&gt;singularity&lt;/a&gt; as having infinite density.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13426" LastEditorUserId="9301" LastEditDate="2016-11-20T02:01:03.260" LastActivityDate="2016-11-20T02:01:03.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19111" PostTypeId="1" AcceptedAnswerId="19113" CreationDate="2016-11-20T04:34:52.330" Score="2" ViewCount="1266" Body="&lt;p&gt;a) We cannot detect all black holes and therefore don't know the percentage of the galaxy's mass they make up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) We do not know the mass of the Milky Way's central black hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) According to our understanding of stellar evolution, black holes should make up a much lower percentage of the galaxy's mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;d)According to our understanding of stellar evolution, black holes should make up a much higher percentage of the galaxy's mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my own understanding, I feel like it should be a). Since the black holes are forming all the time, I think it is hard to detect all of them. But I am not quite sure, could someone give any insight?&lt;/p&gt;&#xA;" OwnerUserId="14903" LastActivityDate="2016-11-20T18:42:57.090" Title="Why is the statement &quot;black holes make up 1% of the mass of the Milky Way Galaxy&quot; unknown?" Tags="&lt;black-hole&gt;&lt;galaxy&gt;&lt;milky-way&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19113" PostTypeId="2" ParentId="19111" CreationDate="2016-11-20T11:04:30.683" Score="7" Body="&lt;p&gt;(a) We can't detect ALL of &lt;em&gt;anything&lt;/em&gt;, but we can certainly makes estimates of their integral properties. For example, we estimate that there are about 100 billion stars in the galaxy, but we can observe only a tiny fraction of these. We have also observed a tiny fraction of the stellar black holes in our galaxy - those few examples that have revealed themselves through their binary companions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Yes we do. It is just over 4 million solar masses and thus a negligible fraction of the Galaxy's total mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) and (d) It is not just stellar evolution that is involved, it is the distribution of stellar birth masses (the initial mass function) that needs to be known, possibly as a function of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stellar evolution part has a major uncertainty in that although it is reasonably certain that all stars above about 10 solar masses end their lives in supernovae, it is not certain what fraction will leave black hole remnants. This is probably a strong function of mass - higher masses favour a black hole and progenitor metallicity; low metallicity stars lose less mass during their lives and are more likely to directly collapse to a black hole - see &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0212469&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heger et al. (2003)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do a back of the envelope calculation. Suppose the IMF has the Saltpeter form $N(M)\propto M^{-2.35}$ between 0.1 and 100 solar masses. Further assume that: because a 1 solar mass star has a lifetime of 10 billion years, the age of the Galaxy is 10 billion years and the stellar lifetime is a very strong inverse function of mass $(\propto M^{-2.5})$, that the 100 billion stars that we deduce exist in our Galaxy (by counting up the local stars and measuring stellar densities elsewhere and extrapolating) are predominantly all the 0.1 to 1 solar mass stars that have &lt;em&gt;ever&lt;/em&gt; been born. This allows us to then estimate how many stars have been born in any mass range - even if they have long-since died.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So making the above assumptions, I will make the further assumption that all stars from 20-100 solar masses die and leave a black hole remnant.Then if my maths is correct then there are about $10^{11}/1400 \sim 7\times 10^{7}$ black holes in our Galaxy. That is there is 1 black hole for every 1400 stars of 0.1-1 solar mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining parts of the puzzle are what is the mass of a typical black hole and what is the mass of the Galaxy? A good number for the former is about 10 solar masses, since the black hole binaries appear to cluster at this value or a little lower (&lt;a href=&quot;https://arxiv.org/abs/1608.04757&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ozel et al. 2010&lt;/a&gt;). The latter is still a topic of investigation. It appears to be an order of magnitude bigger than all the stars, gas and dust due to some form of dark matter. Let's use the number $7\times 10^{11}$ solar masses (&lt;a href=&quot;https://arxiv.org/abs/1608.04757&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eadie &amp;amp; Harris 2016&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my final number is that black holes form $7\times 10^{7}\times 10/7\times 10^{11} = 0.1$ percent of the Milky Way mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could push this number up by a factor of a few by assuming that the IMF contained more higher mass stars in the past, or that the threshold mass for black hole production was lower in the past for low metallicity stars. Both seem theoretically possible and are active topics of investigation.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-11-20T18:42:57.090" LastActivityDate="2016-11-20T18:42:57.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19115" PostTypeId="1" CreationDate="2016-11-20T18:21:01.677" Score="2" ViewCount="268" Body="&lt;p&gt;Hello I'd like to ask what are steps to get a light curve wfrom Kepler/K2 database. Let's say I'd need to get light curves of 10 random K type stars. How should I access the databases and what should I do next?? &lt;/p&gt;&#xA;" OwnerUserId="14908" LastActivityDate="2016-11-20T19:41:18.590" Title="Getting light curves from Kepler/K2" Tags="&lt;data-analysis&gt;&lt;kepler&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19116" PostTypeId="2" ParentId="19115" CreationDate="2016-11-20T19:41:18.590" Score="4" Body="&lt;p&gt;Go to the Kepler/K2 MAST database &lt;a href=&quot;https://archive.stsci.edu/kepler/data_search/search.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://archive.stsci.edu/kepler/data_search/search.php&lt;/a&gt; or &lt;a href=&quot;https://archive.stsci.edu/k2/epic/search.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://archive.stsci.edu/k2/epic/search.php&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Search on a temperature ($T_{\rm eff}$) range: e.g. for K-stars  4500 .. 5200&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A (big) list of objects will be found and presented to you in a table. Mark the ones you want (all the Kepler objects should have available data, but not all the K2 ones will unless you limit your search to the earlier K2 campaigns) and then click on the &quot;submit marked data for retrieval from STDADS&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will then ask you to confirm that you want to download the data with various options to get at the data and a tick box list of what you want (including light curves).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then once you've got the data you need to look at the FAQ and descriptions of data format etc. that can be found at the main Kepler archive pages &lt;a href=&quot;https://archive.stsci.edu/kepler/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://archive.stsci.edu/kepler/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If on the other hand you just want to look at the light curves then you can do that without downloading the data by clicking on &quot;plot marked light curves&quot; on the search results page.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-11-20T19:41:18.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19117" PostTypeId="1" CreationDate="2016-11-20T19:42:32.420" Score="-4" ViewCount="5787" Body="&lt;p&gt;a) The emission of gravitational waves from merging black holes is predicted by Einstein's general theory of relativity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) The first direct detection of gravitational waves, announced in 2016, came from the LIGO observatory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) Two orbiting neutron stars or black holes will gradually spiral toward each other as a result of energy being carried away by gravitational waves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) The emission of gravitational waves from merging black holes is predicted by Newton's universal law of gravitation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;e) Although gravitational waves are an important theoretical prediction, we do not yet have any observational evidence that they  exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f) Scientists seek to detect gravitational waves by using powerful gamma-ray telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally think e is false because LIGO detected the gravitational wave already. For others, I think they are alright. But I am not sure if I missed anything. Could someone give any suggestion?&lt;/p&gt;&#xA;" OwnerUserId="14254" LastActivityDate="2016-11-21T19:28:00.023" Title="Which of the following statements about gravitational waves are true?" Tags="&lt;orbit&gt;&lt;gravity&gt;&lt;general-relativity&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2016-11-21T13:50:19.737" ContentLicense="CC BY-SA 3.0" />
  <row Id="19118" PostTypeId="1" CreationDate="2016-11-20T19:47:50.837" Score="2" ViewCount="214" Body="&lt;p&gt;I understand that a “large” moon, like ours, keeps the planet’s axis from being peturbed, because the spin axis and the moon’s orbital plane exchange momentum and tend to stay in the same plane, and the &lt;em&gt;orbital&lt;/em&gt; angular momentum is much stronger because of the long lever arm. So, tipping the planet means changing the moon’s orbital plane too, and the peturbations don’t make a significant change to its large value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why is Pluto, along with the orbital plane of its satellites, tipped nearly 90 degrees?&lt;/p&gt;&#xA;" OwnerUserId="7142" LastActivityDate="2016-11-20T20:54:42.960" Title="If a large satellite keeps a planet from wobbling,why is Pluto/Charon on its side?" Tags="&lt;orbit&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19119" PostTypeId="2" ParentId="19117" CreationDate="2016-11-20T20:08:38.383" Score="1" Body="&lt;p&gt;a) This is true.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_wave&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gravitational waves&lt;/a&gt; have been predicted by General Relativity since 1916.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_wave#LIGO_observations&quot; rel=&quot;nofollow noreferrer&quot;&gt;This is true&lt;/a&gt;.  In February 2016, LIGO announced that they had detected gravitational waves resulting from a merger of two black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_wave#Binaries&quot; rel=&quot;nofollow noreferrer&quot;&gt;This is true&lt;/a&gt;.  Two objects orbiting around each other will slowly lose energy due to the production of gravitational waves, causing them to fall inwards towards each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) This is false.  Newton's laws assume that all interactions happen instantaneously.  Gravitational waves and signals from gravity travel at the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e) This is false because (b) is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f) This is false.  There are &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational-wave_observatory&quot; rel=&quot;nofollow noreferrer&quot;&gt;two different ways&lt;/a&gt; to detect gravitational waves.  The first is by using an interferometer, like &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational-wave_observatory&quot; rel=&quot;nofollow noreferrer&quot;&gt;LIGO&lt;/a&gt;, the Laser Interferometer Gravitational-Wave Observatory.  By using a very large interferometer, scientists can detect minute changes in the length of the arms, due to the passage of a gravitational wave.  The second way is by using a &lt;a href=&quot;https://en.wikipedia.org/wiki/Weber_bar&quot; rel=&quot;nofollow noreferrer&quot;&gt;Weber bar&lt;/a&gt;, a large, solid antenna sensitive to changes in strain due to its resonant frequency.  This second method has not yielded any results yet.&lt;/p&gt;&#xA;" OwnerUserId="13426" LastEditorUserId="13426" LastEditDate="2016-11-21T19:28:00.023" LastActivityDate="2016-11-21T19:28:00.023" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19121" PostTypeId="1" AcceptedAnswerId="19122" CreationDate="2016-11-21T00:35:09.430" Score="4" ViewCount="194" Body="&lt;p&gt;I would like to know how large a telescope's  mirror aperture would need to be, that is pointed squarely at Sirius to equal the light intensity at the focus of a 5 inch glass lens of our Sun? I well know that such a telescope is impossible. I am trying to grasp better the true dimness of starlight from a somewhat unorthodox perspective. Thank you.                                                                                                                               &lt;/p&gt;&#xA;" OwnerUserId="14911" LastActivityDate="2016-11-28T15:47:08.733" Title="Light From Sirius" Tags="&lt;light&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19122" PostTypeId="2" ParentId="19121" CreationDate="2016-11-21T09:00:41.497" Score="11" Body="&lt;p&gt;The apparent magnitudes of Sirius and the Sun are $m_\mathrm{Sir} =$ &lt;a href=&quot;https://www.google.dk/search?q=apparent+magnitude+of+Sirius&quot; rel=&quot;nofollow noreferrer&quot;&gt;–1.46&lt;/a&gt; and $m_\odot =$ &lt;a href=&quot;https://www.google.dk/search?q=apparent+magnitude+of+the+Sun&quot; rel=&quot;nofollow noreferrer&quot;&gt;–26.74&lt;/a&gt;, respectively. That means that the Sun is brighter than Sirius by a &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_magnitude#Calculations&quot; rel=&quot;nofollow noreferrer&quot;&gt;factor&lt;/a&gt; of&#xA;$$&#xA;f = \frac{F_\odot}{F_\mathrm{Sir}} = 10^{-(m_\odot-m_\mathrm{Sir}) / 2.5} = 1.3\times10^{10}.&#xA;$$&#xA;The area of your telescope should be larger than your 5&quot; glass lens by the same factor to receive an equal amount of light. Thus, your telescope should have an aperture of $5\!\!&quot; \times \sqrt{f} = 14.5\,\mathrm{km}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extrapolating (linearly in lin-log space) the evolution of telescopes from Galileo's 1620 telescope, the figure demonstrates that Roger H. will have his telescope ready by the year 2555, May 21 9:53 pm$^\dagger$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sqFQY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sqFQY.png&quot; alt=&quot;extrapol&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;± a few minutes.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-11-28T15:47:08.733" LastActivityDate="2016-11-28T15:47:08.733" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19123" PostTypeId="1" AcceptedAnswerId="19125" CreationDate="2016-11-21T12:08:38.250" Score="2" ViewCount="250" Body="&lt;p&gt;When some object is moving, it appears smaller and more massive. What if it's moving so fast it should become a black hole? Some calculations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let observed radius $R_0\gamma$ ($R_0$ - radius at rest) be equal Schwarzschild radius for object with observed mass $M_0\gamma$ ($M_0$ - mass at rest):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_0 / \gamma = \frac{2GM_0\gamma}{c^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\gamma^2=\frac{R_0c^2}{2GM_0}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{v}{c}=\sqrt{1-\frac{2GM_0}{R_0c^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Earth it gives $v=0.999999999304c$ and for Sun $v=0.999997877903c$. Is my reasoning correct? How would we see such objects?&lt;/p&gt;&#xA;" OwnerUserId="14916" LastEditorUserId="9301" LastEditDate="2016-11-21T14:00:32.067" LastActivityDate="2016-11-21T15:34:08.623" Title="Relative black hole?" Tags="&lt;black-hole&gt;&lt;special-relativity&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19125" PostTypeId="2" ParentId="19123" CreationDate="2016-11-21T15:34:08.623" Score="6" Body="&lt;p&gt;To answer quickly, the answer is that &lt;strong&gt;no&lt;/strong&gt;, an object cannot be made to collapse as such. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are multiple reasons as to why. Pela brings up a great argument here, reminding us that if a black hole &lt;em&gt;could&lt;/em&gt; be formed this way, then I could be made to collapse into a black hole just by something moving by me at extreme velocities. Clearly, this cannot occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand why this is, we must first completely throw away the idea you have of &lt;a href=&quot;http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;relativistic mass&lt;/a&gt;. The entire concept of relativistic mass was created to allow for &lt;a href=&quot;https://en.wikipedia.org/wiki/Classical_mechanics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Newtonian mechanics&lt;/a&gt; to still hold in relativistic situations. By modifying the mass of an object at extreme velocities, equations such as $F=ma$ can be made to still hold. To paraphrase &lt;a href=&quot;http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_fast.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Philip Gibbs&lt;/a&gt;, gravity couples not only to mass but also to momentum and itself. Therefore, a description of spacetime in the vicinity of an object will be incomplete without momentum considerations. From this &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/relmom.html#c1&quot; rel=&quot;nofollow noreferrer&quot;&gt;relativistic momentum&lt;/a&gt;, we can compute the &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/releng.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;relativistic energy&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While these quantities are useful for giving detailed descriptions of the body, the answer is actually somewhat simpler. We must remember that an object must behave the same in all inertial reference frames. If the object does not become a black hole in its own frame (as it clearly will not), then it cannot appear to be one in any other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are wondering how to reconcile this with a relativistic mass that approaches infinity, the answer is simply that relativistic mass is not real mass. The metrics which describe spacetime in the vicinity of these objects refer to the &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/mass.html#mas&quot; rel=&quot;nofollow noreferrer&quot;&gt;invariant mass&lt;/a&gt;, which is the rest mass of the body. &lt;/p&gt;&#xA;" OwnerDisplayName="user14781" LastActivityDate="2016-11-21T15:34:08.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19126" PostTypeId="1" AcceptedAnswerId="19137" CreationDate="2016-11-21T22:36:48.383" Score="3" ViewCount="516" Body="&lt;p&gt;Reposting question from &lt;a href=&quot;https://www.reddit.com/r/askscience/comments/5e1zhy/what_does_mercurys_orbit_really_look_like/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.reddit.com/r/askscience/comments/5e1zhy/what_does_mercurys_orbit_really_look_like/&lt;/a&gt; which reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;First the full question: What does the orbit of Mercury really look&#xA;  like visually, without artistic liberties taken to amplify the&#xA;  eccentricity of the orbit, or simplifying it to appear as a perfect&#xA;  circle? What eccentricities are needed for orbits that readily appear&#xA;  elliptical?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I originally did this as a self-answer, but my answer was of such poor quality I've now deleted it. I hope to do a better answer backed by some calculations, but, as always, anyone else is free to answer.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2016-11-22T19:45:12.380" LastActivityDate="2016-11-22T21:44:39.843" Title="Exactly how elliptical is Mercury's orbit, visually, without exaggeration?" Tags="&lt;orbit&gt;&lt;mercury&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19128" PostTypeId="2" ParentId="2221" CreationDate="2016-11-21T22:40:23.187" Score="0" Body="&lt;p&gt;You have slightly misremembered a common (though bad) way of describing Hawking radiation.  Physics popularizers sometimes describe it as a pair of particles being created, one of which is matter and the other of which is negative matter. Or one of which is antimatter and the other of which is negative antimatter.  So your proton escapes and your negative-matter-proton is absorbed. Or your antiproton escapes and your negative-antimatter-antiproton is absorbed. The negative matter (or negative antimatter) shrinks the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although this is a common way to describe things for non-physicists, it's a bad way to describe it. It's confusing because it suggests the exact question that you raised: why doesn't the negative matter fly outward, and shrink the first star or planet it hits? Also, negative matter has never been detected. There is no particular reason to think it's a useful construct for understanding black holes. (Though if it did exist, you might be able to use it to stabilize wormholes, which could be very useful).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's better to describe Hawking radiation as the other answers here have done, without resorting to virtual negative particles.&lt;/p&gt;&#xA;" OwnerUserId="14921" LastActivityDate="2016-11-21T22:40:23.187" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19129" PostTypeId="1" AcceptedAnswerId="32178" CreationDate="2016-11-21T22:42:38.487" Score="9" ViewCount="1315" Body="&lt;p&gt;The cores of iron meteorites exhibit beautiful Widmanstätten patterns, as shown below in the Alvord meteorite (photo by Jim H):&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SUPXH.jpg&quot; alt=&quot;Cutaway view of the Alvord meteorite&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;I understand these patterns form when the molten nickel-iron core of its host body cools. As with normal crystal formation, slower cooling rates lead to longer crystal-formation times, which leads to larger crystals.&lt;/p&gt;&#xA;&lt;h3&gt;Beyond that, what factors influence the formation of the crystal, and in what ways?&lt;/h3&gt;&#xA;&lt;p&gt;I imagine composition and cooling time are the dominant factors....&lt;/p&gt;&#xA;" OwnerUserId="7965" LastEditorUserId="31410" LastEditDate="2021-04-28T22:21:40.373" LastActivityDate="2021-04-28T22:21:40.373" Title="What factors influence the Widmanstätten pattern in iron meteorites?" Tags="&lt;meteor&gt;&lt;formation&gt;&lt;meteorite&gt;&lt;astrochemistry&gt;&lt;meteoroid&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="19130" PostTypeId="1" CreationDate="2016-11-21T22:54:55.100" Score="2" ViewCount="85" Body="&lt;p&gt;Iron-core meteorites display an awesome criss-cross pattern known as the Widmanstätten pattern, shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Nh8lL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nh8lL.jpg&quot; alt=&quot;Cross-section of an iron-core meteorite&quot;&gt;&lt;/a&gt;&#xA;(photo by Dr. Patrick M. Len, taken from &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Widmanstatten_hand.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I TA for an intro-level astronomy/planetary science class, and today we discussed meteorites. A student asked me what the average length of these crystals is for iron meteorites, but I hesitated. The samples we had available (and the image above) made it look like the average length was on the order of a centimeter, but with great variation. I'm curious as to whether there is a more quantitative answer. &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;My question:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;How long is the average length of a given linear crystal in a typical iron meteorite? How much variation is there from sample to sample? &lt;/p&gt;&#xA;" OwnerUserId="7965" LastActivityDate="2016-11-21T22:54:55.100" Title="What is the average Ni-Fe crystal length in iron meteorites?" Tags="&lt;asteroids&gt;&lt;meteor&gt;&lt;meteorite&gt;&lt;meteoroid&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19131" PostTypeId="1" CreationDate="2016-11-22T06:24:37.987" Score="1" ViewCount="55" Body="&lt;p&gt;A directly-imaged object found close to a certain star could be another star in the background or a  companion planet? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would one possibly discern between the two?&#xA;Also would we know it works?&lt;/p&gt;&#xA;" OwnerUserId="14923" LastActivityDate="2016-11-22T21:07:30.330" Title="How to determine what directly imaged objects are?" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19132" PostTypeId="1" CreationDate="2016-11-22T10:39:51.320" Score="0" ViewCount="134" Body="&lt;p&gt;Q2 What does the phrase 'containing several distinct knots of collopsing material' mean? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q3 In this context, does the word 'such clouds' mean protostar? (I'm not an English-native speaker.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q4 In the context, does the word 'a spinning disc' mean an Accretion disk ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q1 And the last... Could you explain 'spiral density wave' to me if you can?  &lt;/p&gt;&#xA;" OwnerUserId="14782" LastEditorUserId="14782" LastEditDate="2016-12-19T07:59:51.107" LastActivityDate="2016-12-19T07:59:51.107" Title="I have some questions about star formation" Tags="&lt;star-formation&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2016-12-04T20:51:56.867" ContentLicense="CC BY-SA 3.0" />
  <row Id="19133" PostTypeId="2" ParentId="19132" CreationDate="2016-11-22T11:34:19.357" Score="3" Body="&lt;p&gt;I'll answer your questions in order 1-4:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The term &quot;spiral density wave&quot; refers to the arms of a spiral galaxy, which have a somewhat higher density than the medium in between (though not &lt;em&gt;much&lt;/em&gt; higher). A gas-forming cloud traveling through such an arm experiences tidal forces, which perturbs the cloud and may cause local regions of the cloud to exceed its &lt;a href=&quot;https://en.wikipedia.org/wiki/Jeans_instability&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jeans mass&lt;/a&gt; and collapse.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The term &quot;knots&quot; means clumps of higher density inside the cloud that has begun to collapse.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Such clouds&quot; refers to the cloud out of which a proto-star is forming. That is, it doesn't refer to the proto-star itself, but to the cloud around it. However, since the proto-star will accrete mass for a while, part of the cloud will eventually become part of the star.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The gas forms a disk around the proto-star, and loses angular momentum through heating+radiation, thus decreasing in distance to the proto-star and eventually accreting onto its surface. So yes, you can call it an accretion disk, but note that accretion disks also exist in other contexts, e.g. the accretion disk around a quasar. Also, once the central tempereture of the proto-star ges hot enough to ignite nuclear fusion and the proto-star becomes a &quot;real&quot; star, radiation pressure will tend to disrupt the disk, blowing the rest of it away (how much depends on the radiation field, which in turn depends on the mass of the star).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2016-11-22T11:34:19.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19134" PostTypeId="1" CreationDate="2016-11-22T12:36:24.940" Score="-1" ViewCount="123" Body="&lt;p&gt;I can not understand the part of the paragraph because I think they are contradiction or irony. Could you explain the process of it? Thank you!&lt;/p&gt;&#xA;" OwnerUserId="14782" LastEditorUserId="14782" LastEditDate="2016-12-19T08:01:02.853" LastActivityDate="2016-12-19T08:01:02.853" Title="Did the 'hot Jupiters' really spiral into the Sun?" Tags="&lt;planet&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2016-12-15T14:15:01.630" ContentLicense="CC BY-SA 3.0" />
  <row Id="19135" PostTypeId="2" ParentId="19134" CreationDate="2016-11-22T14:37:41.483" Score="5" Body="&lt;p&gt;I think you're misinterpreting the passage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gas giants form via a process know as &lt;em&gt;core accretion&lt;/em&gt;. First, a massive core forms from solid material, initially with a small amount of gas. The envelope then grows, producing an object we would recognize as a gas giant - a large gaseous atmosphere, surrounding a comparatively small &quot;rocky&quot; core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with Hot Jupiters is that gas giants, according to current theories of planet formation, cannot form extremely close to a star. Inside a boundary called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Frost_line_%28astrophysics%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;frost line&lt;/em&gt;&lt;/a&gt; (which changes over time), there is not enough solid material for such a massive core to form and accrete enough gas to form a gas giant before the protoplanetary disk dissipates. Therefore, any gas giants that exist inside the snow line should have formed outside it, and then traveled inward via a process known as &lt;a href=&quot;http://www.scholarpedia.org/article/Planetary_formation_and_migration#Planetary_migration&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;migration&lt;/em&gt;&lt;/a&gt;, which can happen through several different mechanisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phrase &quot;spiraled inwards&quot; doesn't mean that the planets will spiral into the star, merely that their semi-major axes will shrink and their orbits will move closer and closer to their parent star, possibly stopping at some finite distance.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Like David Hammen, I'm a bit confused as to whether you think this happened in the Solar System, or just other exoplanetary systems. As far as I know, this sort of formation and migration could not have happened in the Solar System, as per the leading models of its early evolution, because it would have led to a drastically different layout than the one we know today.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-11-22T18:33:10.810" LastActivityDate="2016-11-22T18:33:10.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19136" PostTypeId="1" CreationDate="2016-11-22T15:25:26.190" Score="6" ViewCount="106" Body="&lt;p&gt;BACKGROUND&lt;/p&gt;&#xA;&#xA;&lt;p&gt;May 11, 2013 the Kepler space telescope ended its primary mission. Data captured until then showed that the now famous &quot;Tabby's star&quot; or &quot;Boyajian's star&quot; (KIC 8462852) had been &lt;a href=&quot;https://arxiv.org/pdf/1608.01316v2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;fading throughout&lt;/a&gt; the Kepler mission by about 3% cumulatively. &lt;em&gt;(This independently of other reports of brief very deep dimmings and restorations during the mission.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QUESTION&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today when I post this question, about as much time has passed by since the primary Kepler mission ended, as that mission lasted. So this makes me wonder if other telescopes have observed, or should've been sensitive enough to have observed, a continuation of such a fading trend for that star? Another 3%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DISCUSSION&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, what could the explanation be for why this 3+ year fading trend was suddenly interrupted when the primary Kepler telescope mission was terminated? And how long would it take to determine from ground based observatories whether this particular dimming trend continues or not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0JV7I.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0JV7I.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;em&gt;Fig. 5 from the paper by Montet and Simon 2016.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2016-11-23T11:35:13.717" LastActivityDate="2016-11-23T11:35:13.717" Title="Has star KIC 8462852's 3 year long Kepler-fading continued after the primary Kepler mission ended?" Tags="&lt;observational-astronomy&gt;&lt;kepler&gt;&lt;photometry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19137" PostTypeId="2" ParentId="19126" CreationDate="2016-11-22T20:56:05.733" Score="5" Body="&lt;p&gt;Some things we know about Mercury's orbit:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Semi-major axis:&lt;/strong&gt; 0.387 AU, about 57.9 million km&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Eccentricity:&lt;/strong&gt; 0.205&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We can calculate the semi-&lt;em&gt;minor&lt;/em&gt; axis, $b$ from the equation&#xA;$$e=\sqrt{1-\frac{b^2}{a^2}}\to b=0.379\text{ AU}$$&#xA;We can also calculate the distance to the focus from the center of the ellipse, $f$, as&#xA;$$f=ae=0.078\text{ AU}$$&#xA;I used &lt;em&gt;Mathematica&lt;/em&gt;'s &lt;code&gt;ParametricPlot&lt;/code&gt; to plot Mercury's orbit using these parameters, along with Earth's. Earth's semi-major axis is 1 AU, and with an eccentricity of 0.016, its semi-minor axis is about 1 to three significant figures; I've treated it as a circle. &lt;em&gt;Mathematica&lt;/em&gt; gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mhSuz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mhSuz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ParametricPlot[{{0.387*Sin[x] - 0.0793, 0.379*Cos[x]}, {Sin[x],&#xA;  Cos[x]}}, {x, 0, 2*Pi}, AxesLabel -&amp;gt; {AU, AU}, PlotLegends -&amp;gt; &#xA;  Placed[LineLegend[{Red, Blue}, {&quot;Mercury&quot;, &quot;Earth&quot;}], {1, 0.25}], &#xA;  PlotStyle -&amp;gt; {Red, Blue}]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'd consider the deviation from a perfect circle to be noticeable insofar as the focus is shifted from the point $(0, 0)$, but it's not incredible.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-11-22T21:01:02.670" LastActivityDate="2016-11-22T21:01:02.670" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19138" PostTypeId="2" ParentId="19131" CreationDate="2016-11-22T21:07:30.330" Score="1" Body="&lt;p&gt;Simple really. You observe again and see if the pair are moving together. If the star plus &quot;planet&quot; are quite widely separated (e.g. tens of au or more), then you would expect negligible orbital motion in the course of say a year and their relative position should remain quite constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if the &quot;planet&quot; is a much more distant star then it is quite likely to have a very different proper motion on the sky to the foreground star and so they will show an appreciable difference in their relative positions second time around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More than two observations might be needed to be sure. The bigger the proper motion of the foreground star, the more decisive the test.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-11-22T21:07:30.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19139" PostTypeId="2" ParentId="19126" CreationDate="2016-11-22T21:28:04.020" Score="1" Body="&lt;p&gt;Here you go. Image is from &lt;a href=&quot;https://science.nasa.gov/science-news/science-at-nasa/2000/ast30jun_1m&quot; rel=&quot;nofollow noreferrer&quot;&gt;this NASA website&lt;/a&gt; and shows Earth, Mercury and Mars, with circles for comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RXFrb.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RXFrb.gif&quot; alt=&quot;Orbits of Mercury, Earth and Mars&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-11-22T21:28:04.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19140" PostTypeId="2" ParentId="19126" CreationDate="2016-11-22T21:44:39.843" Score="1" Body="&lt;p&gt;@rob-jeffries and @hde-226868 have both given excellent answers: after remembering that I was the OP, I upvoted both and approved one.This answer is mostly my attempt at redemption:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LRHso.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LRHso.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image above (Sun depicted at 3 times actual diameter) shows the &#xA;orbit of Mercury with the Sun as the center. This agrees with two &#xA;other images that are also meant to show Mercury's orbit to scale: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.reddit.com/r/askscience/comments/5e1zhy/what_does_mercurys_orbit_really_look_like/dab8g7q/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.reddit.com/r/askscience/comments/5e1zhy/what_does_mercurys_orbit_really_look_like/dab8g7q/&lt;/a&gt; which links to: &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9gsnE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9gsnE.png&quot; alt=&quot;http://i.imgur.com/HGXs0Qp.png&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://m.teachastronomy.com/astropedia/article/Mercurys-Orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://m.teachastronomy.com/astropedia/article/Mercurys-Orbit&lt;/a&gt; which embeds: &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/S9RmP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S9RmP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of these look similar to my answer, so I'm reasonably confident &#xA;that I'm correct this &#xA;time. &lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/STACK/bc-mercury.m&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/STACK/bc-mercury.m&lt;/a&gt; &#xA;if anyone wants to check my work &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things to note: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The primary effect of the eccentricity is that the center of &#xA;Mercury's orbit is quite far from the Sun. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The orbit's center is about 0.0796 AU or 11.9 million km from the &#xA;Sun, and the other focus is 23.8 million km from the Sun. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mercury's semiminor axis is 97.86% the length of its semimajor &#xA;axis, so is looks very much like a circle. To see this, we plot &#xA;again, this time using the midpoint of the two foci as the origin: &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/om3no.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/om3no.gif&quot; alt=&quot;[[image37.gif]]&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The x diameter is 587 pixels and the y diameter is 575 pixels. If you &#xA;draw a blue circle with a radius averaging Mercury's semiminor and &#xA;semimajor axes, you get: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oujTx.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oujTx.gif&quot; alt=&quot;[[image38.gif]]&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll leave it to the reader to decide if Mercury's orbit does or does &#xA;not look like a circle. &lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2016-11-22T21:44:39.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19141" PostTypeId="1" CreationDate="2016-11-22T23:06:17.990" Score="1" ViewCount="153" Body="&lt;p&gt;Is having a surname which begins with the letter “A” a big advantage for a researcher in astronomy/astrophysics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have a surname which begins with the letter “M”, and I haven’t published any paper yet, would changing my surname to a name which begins with the letter “A” be a good idea? Or would the advantage gained be too small to make it worth the effort?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Approximately what percentages of research papers in astronomy/astrophysics order their authors …&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;purely alphabetically?&lt;/li&gt;&#xA;&lt;li&gt;half by contribution and half alphabetically?&lt;/li&gt;&#xA;&lt;li&gt;purely by contribution?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6786" LastActivityDate="2016-11-23T14:52:55.653" Title="Is having a surname which begins with the letter “A” a big advantage for a researcher in astronomy/astrophysics? Is changing our surname a good idea?" Tags="&lt;naming&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19142" PostTypeId="2" ParentId="52" CreationDate="2016-11-23T00:06:57.783" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the hottest exoplanet discovered so far?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As far as I can tell, that honor goes to &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler-70b&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Kepler-70b&lt;/strong&gt;&lt;/a&gt;. Estimates for its surface temperature range up to 7,143 K. The other confirmed planet in the system, &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler-70c&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kepler-70c&lt;/a&gt;, may also be extremely hot. The two orbit their parent star - a hot b-type subdwarf - at distances of 0.0060 and 0.0076 AU, respectively, and thus receive quite a lot of radiation. The discovery paper (&lt;a href=&quot;http://www.nature.com/nature/journal/v480/n7378/full/nature10631.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Charpinet et al. (2011)&lt;/a&gt;) reported that the two are likely chthonian planets, having been engulfed in the star's envelope during its red giant phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Habitable Planets Catalog has &lt;a href=&quot;http://phl.upr.edu/projects/habitable-exoplanets-catalog/top10&quot; rel=&quot;nofollow noreferrer&quot;&gt;lists of top exoplanet extremes&lt;/a&gt;. Kepler-70b and Kepler-70c top the list of hottest exoplanets, and are listed as having surface temperatures of 7,143 K and 6,351 K, respectively. I would imagine that planets close in would have much greater surface temperatures, as they are heated more; these two may also have residual heat from the engulfment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several other exoplanets on the lists that top 2,000 K (there are likely more):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Kepler-13 b (3206 K)&lt;/li&gt;&#xA;&lt;li&gt;PSR 1719-14 b (3118 K)&lt;/li&gt;&#xA;&lt;li&gt;WASP-33 b (2505 K)&lt;/li&gt;&#xA;&lt;li&gt;WASP-12 b (2368 K)&lt;/li&gt;&#xA;&lt;li&gt;WASP-103 b (2342 K)&lt;/li&gt;&#xA;&lt;li&gt;KELT-1 b (2296 K)&lt;/li&gt;&#xA;&lt;li&gt;WASP-18 b (2258 K)&lt;/li&gt;&#xA;&lt;li&gt;OGLE2-TR-L9 b (2213 K)&lt;/li&gt;&#xA;&lt;li&gt;WASP-121 b (2202 K)&lt;/li&gt;&#xA;&lt;li&gt;WASP-76 b (2039 K)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The page is a year or two old, and there are clearly other hot exoplanets that have since been discovered. However, it seems that Kepler-70b and Kepler-70c are still the hottest.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-12-26T19:23:03.970" LastActivityDate="2016-12-26T19:23:03.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19143" PostTypeId="2" ParentId="19136" CreationDate="2016-11-23T00:11:53.110" Score="4" Body="&lt;p&gt;Well &quot;if no known astrophysical model can explain it&quot; then nobody told &lt;a href=&quot;https://arxiv.org/abs/1609.03505&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wright &amp;amp; Sigurdsson (2016)&lt;/a&gt; who, cognisant of Montet &amp;amp; Simon's results, explore a number of astrophysical models. They conclude by saying that the most &quot;plausible model&quot; is that of small scale intervening material between us and the star that may be responsible for the short-term dips and the secular dimming. &lt;a href=&quot;https://arxiv.org/abs/1610.03219&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lacki (2016)&lt;/a&gt; appears to agree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wright &amp;amp; Sigurdsson also consider and largely rule out or class as unlikely a number of other possibilities, including artificial structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gaia satellite &lt;a href=&quot;https://arxiv.org/abs/1609.05492&quot; rel=&quot;nofollow noreferrer&quot;&gt;has been observing&lt;/a&gt; this star (and a billion others) since December 2013, but so far the photometry time series has not been released. This should become available towards the end of next year and will probably exceed the accuracy of the Kepler data. In the meantime, it has been claimed by &lt;a href=&quot;https://arxiv.org/abs/1609.09290&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hippke et al. (2016)&lt;/a&gt; that the star was relatively constant between 1934 and 1995 in contradiction to previous claims of a progressive dimming over the last century.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-11-23T00:11:53.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19144" PostTypeId="2" ParentId="19141" CreationDate="2016-11-23T07:44:14.670" Score="5" Body="&lt;p&gt;I would say that it matters not-at-all whether your name is Aardvark or Zygote. Astrophysicists are not stupid, we can see when a large author list has been alphabetised. If anything I would say it is a disadvantage, since a casual glance at an author list might lead someone to think you were the lead author of any paper, even one with just two authors, simply because of your surname.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, the tiny minority of author lists are alphabetised in Astrophysics papers. Off the top of my head I can think of the gravitational wave papers by Abbott et al. and the MACHO papers by Alcock et al. Most many-author papers have a nucleus of authors at the start of the work and then a longer list of names, possibly in alphabetical order, of people who contributed to the overall project or mission. Even these papers are a (growing) minority and it is still true that the majority of Astrophysics papers have a relatively small author list ordered by contribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I will add another personal opinion. The presence of someone's name on many long-author list papers does not necessarily impress me unless they have lead some of them. It is more impressive for a young scientist to have forged their own path and perhaps written a few lead-authored papers with small author lists or even a sole author paper. Unfortunately, much &quot;big science&quot; does involve big consortia and though it can be exciting for an early-career scientist to be part of such an enterprise, it can be much harder to make your mark outside of that consortium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: As a brief, empirical check on my answer, I looked at arXiv for 22nd November 2016. There were 66 new astrophysics papers listed. Of these, 29 were written by three or fewer authors. Only five were written by $&amp;gt;10$ authors and only one paper, from the Pierre Auger collaboration, appeared to have an alphabetised author list. Try hard not to laugh - the paper is by &lt;a href=&quot;https://arxiv.org/abs/1611.06812&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aab, A. et al.&lt;/a&gt; Good luck in your efforts to beat that.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-11-23T14:52:55.653" LastActivityDate="2016-11-23T14:52:55.653" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19145" PostTypeId="1" CreationDate="2016-11-23T11:46:49.257" Score="1" ViewCount="620" Body="&lt;p&gt;I encountered a book discussion that defines the Sun's declination as 23.5 degrees (inclination of Earth's equator to the ecliptic) times the sine of the Sun's longitude. Does this make sense? How do i visualize it?&lt;/p&gt;&#xA;" OwnerUserId="14930" LastEditorUserId="616" LastEditDate="2016-11-23T12:21:12.027" LastActivityDate="2016-11-23T18:16:28.453" Title="Declination of the Sun formula" Tags="&lt;positional-astronomy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19146" PostTypeId="1" CreationDate="2016-11-23T11:57:59.903" Score="4" ViewCount="100" Body="&lt;p&gt;It seems to be common practice to double up observations. For example, shortly before New Horizons flew by Pluto, several of its moons were discovered by telescopes as a concerted effort was put in because of the occasion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have other telescopes made systematic and dedicated observations of the field of view of the Kepler telescope? I imagine speculatively in this paragraph that such observations could've been part of a survey process to determine where to point Kepler out of a set of candidate FOVs to begin with. And once decided upon, that field more closely examined. Then during the Kepler mission persistently observed in order to help rule out noise from passing asteroids and binary and variable stars and to gather complementary data. To characterize transiting planets, it is important to characterize the stars they are orbiting, something which telescopes with other kinds of instruments can be helpful with. Those are just my speculations about why I would think that Kepler's primary mission was &quot;doubled up&quot; by other kinds of telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some of the preparatory, complementary or follow up observation campaigns that have been made of the Kepler field of view? Is Kepler's FOV now one of the most and best observed parts of the sky?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="7982" LastEditDate="2020-10-20T01:17:12.480" LastActivityDate="2020-10-20T01:17:12.480" Title="What complementary observation campaigns have looked at Kepler's field of view?" Tags="&lt;observational-astronomy&gt;&lt;kepler&gt;&lt;sky-survey&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19147" PostTypeId="1" CreationDate="2016-11-23T12:09:51.333" Score="0" ViewCount="91" Body="&lt;p&gt;Orbits of planets and stars decay due to gravitational wave radiation. An elliptical orbit would become more circular with time. This is especially observed well in binary systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the example of binaries, we can compute the rate of change of eccentricity of the orbit and the rate of change of the semi-major axis as a function of time. If we consider a highly elliptic orbit of a binary, with eccentricity, $\epsilon\rightarrow{1}$, we can prove that with time the orbit would eventually become a circular one, as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The power emitted by gravitational waves is given by:&#xA;  $$P_{GW}= \frac{c^5}{G}\left(\frac{GM}{c^5l}\right)^5$$&#xA;  Very compact binaries will lose energy rapidly by GW radiation.&#xA;  If we assume that the two bodies making up the binary lie in the $x-y$ plane and their &lt;strong&gt;orbits are circular&lt;/strong&gt; ($\epsilon=0$), then only non-vanishing components of quadrupole tensors are:&#xA;  $$\boxed{Q_{xx}=-Q_{yy}=\frac{1}{2}(\mu)a^{2}\cos2\Omega t}$$ and &#xA;  $$\boxed{Q_{xy}=Q_{ya}=\frac{1}{2}(\mu)a^{2}\sin2\Omega t}$$&#xA;  Where $\Omega$ is the orbital velocity, $\mu=\dfrac{m_{1}m_{2}}{m}$ is the reduced mass and where $m=m_{1}+m_{2}$&#xA;  The luminosity of the system can be deduced as:&#xA;  $$L^{GM}=\frac{32}{5}\frac{G}{c^5}\mu^2 a^4\Omega^6=\frac{32}{5}\frac{G^4}{c^5}\frac{M^3 \mu^2}{a^5}$$&#xA;  The latter part is obtained from Kepler third law: $\Omega^2=\frac{GM}{a^3}$&#xA;  As the gravitating system loses energy by emitting radiation, the distance between the 2 bodies shrinks at a rate:&#xA;  $$\boxed{\frac{da}{dt}=\frac{64}{5}\frac{G^{3}M\mu}{c^5 a^3}}$$&#xA;  The binary would hence colase at a time :&#xA;  $$\tau=\frac{5}{256}\frac{c^5}{G^3}\frac{a_{0}^4}{\mu M^4}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;By applying a similar treatment to elliptical orbits , the following can be computed:&#xA;  $$\left&amp;lt;\frac{da}{dt}\right&amp;gt;=-\frac{64}{5}\frac{G^{3}m_{1}m_{2}(m_{1}+m_{2})}{c^{5}a^{4}(1-e^{2})^{\frac{7}{2}}}\left(1+\frac{73e^2}{24}+\frac{37e^4}{96}\right)$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\left&amp;lt;\frac{de}{dt}\right&amp;gt;=-\frac{304}{15}\frac{G^{3}m_{1}m_{2}(m_{1}+m_{2})}{c^{5}a^{4}(1-e^{2})^{\frac{5}{2}}}\left(1+\frac{121e^2}{304}\right)$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Solving this system of ODE's ultimately results in the equation:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$T(a_{0},e_{0})=\frac{12(c_{0}^4)}{19\gamma}\int_{0}^{e_0}{\frac{e^{29/19}[1+(121/304)e^2]^{1181/2299}}{(1-e^2)^{3/2}}}de\tag1$$&#xA;  Where $$\gamma=\frac{64G^3}{5c^5}m_{1}m_{2}(m_{1}+m_{2})$$&#xA;  Upon solving this one can find the time taken for the orbit to decay into a circle from an ellipse due to gravitational wave radiation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A similar treatment can be done for the orbital decay of the planets of our solar system. Now my question is if we apply time reversal symmetry, the orbits become more and more elliptical and eventually tend to become a straight line(an elliptical orbit with $\epsilon\rightarrow{1}$) with a chunk of mass(moving in the straight line) from which all the planets and the Sun  of our solar system arose. Now the Milky Way &quot;orbits&quot; the &lt;a href=&quot;http://en.wikipedia.org/wiki/Local_Group&quot; rel=&quot;nofollow noreferrer&quot;&gt;Local Group&lt;/a&gt;, which in turn &quot;orbits&quot; the &lt;a href=&quot;http://en.wikipedia.org/wiki/Virgo_Supercluster&quot; rel=&quot;nofollow noreferrer&quot;&gt;Virgo Supercluster&lt;/a&gt;.&#xA;Beyond that, the expansion of the universe starts to dominate over gravitation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So reversing time to a space after the Big Bang, there must have been many galaxies and solar systems that were forming; so according to the above result the mass chunks from which galaxies, stars and planets arose must have had a highly elliptic orbits. Thus, the spread of matter in space must have been along approximate lines(which were orbits for the massive mass chunk a that formed them). This moves towards a linear arrangement of galaxies, contrary to what is observed. Can anyone please resolve my doubt. Any help is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="14580" LastEditorUserId="14580" LastEditDate="2016-11-24T19:29:21.250" LastActivityDate="2020-07-11T14:52:20.907" Title="On gravitational wave radiation and arrangement of galaxies post- big bang" Tags="&lt;big-bang-theory&gt;&lt;gravitational-waves&gt;&lt;orbital-migration&gt;&lt;eccentric-orbit&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19148" PostTypeId="2" ParentId="19145" CreationDate="2016-11-23T12:49:00.260" Score="1" Body="&lt;p&gt;Yes, it is correct. For better precision:&#xA;Declination =23.5*sin(longitude)&#xA;This is true only if you don't take into account the eccentricity of Earth's orbit and other movement. It isn't easy to understand why, but I'll try ti explain it.&#xA;The different declination of the sun during the year is generated by the different angle of inclination of the Earth (the north pole point always in the same direction, so during the earth revolution the poles don't point always in the sun direction and the protection of the inclination on the sun-earth plane change during the year) &#xA;So, what wee see of the sun is the projection of the circular movement of the earth on a line of sight, so an harmonic motion.&lt;/p&gt;&#xA;" OwnerUserId="14600" LastActivityDate="2016-11-23T12:49:00.260" CommentCount="1" CommunityOwnedDate="2016-11-23T12:49:00.260" ContentLicense="CC BY-SA 3.0" />
  <row Id="19149" PostTypeId="2" ParentId="19145" CreationDate="2016-11-23T13:06:02.250" Score="1" Body="&lt;p&gt;Yes, see &lt;a href=&quot;https://www.desmos.com/calculator/vq7djgkyir&quot; rel=&quot;nofollow noreferrer&quot;&gt;graph&lt;/a&gt;.&#xA;At the March equinox, the Sun's ecliptic longitude $\lambda_\odot$ is zero, and the Sun's declination $\delta_\odot$ is zero and increasing.&#xA;At the June solstice, $\lambda_\odot$ is 90$^\circ$ and $\delta_\odot$ is at a maximum, 23.5$^\circ$ north of the equator.&#xA;At the September equinox, $\lambda_\odot$ is 180$^\circ$ and $\delta_\odot$ is zero and decreasing.&#xA;At the December solstice, $\lambda_\odot$ is 270$^\circ$ and $\delta_\odot$ is at a minimum of -23.5$^\circ$.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2016-11-23T18:16:28.453" LastActivityDate="2016-11-23T18:16:28.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19150" PostTypeId="1" AcceptedAnswerId="40155" CreationDate="2016-11-23T13:40:09.073" Score="11" ViewCount="292" Body="&lt;p&gt;I've read that in Mars' poles, the winds can be as fast as 400 km/h, when the poles are exposed to sunlight because the frozen $CO_2$ sublimes. I know that the Martian atmosphere is much thinner than Earth's atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, by knowing the wind speeds on Mars, is there any way to get an idea of its intensity, or in other words, the intensity of a wind of x speed in Mars, to which speed of wind of Earth is comparable, for them to have the same intensity?&lt;/p&gt;&#xA;" OwnerUserId="11562" LastEditorUserId="11562" LastEditDate="2017-12-13T17:09:34.877" LastActivityDate="2020-12-01T23:20:22.220" Title="Is the wind's intensity on Mars similar to Earth?" Tags="&lt;planetary-atmosphere&gt;&lt;mars&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19151" PostTypeId="1" AcceptedAnswerId="19152" CreationDate="2016-11-23T14:47:36.330" Score="1" ViewCount="332" Body="&lt;p&gt;How is a galaxy formed? I know that the center of our galaxy is considered to be Sagittarius A* and it's surrounded by a lot of stars (also a lot of neutron stars).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what makes this happen? I mean, why are all those stars in that right formation, like other galaxies, rather than just being random clouds of stars in the universe?&lt;/p&gt;&#xA;" OwnerUserId="14936" LastEditorUserId="2365" LastEditDate="2016-11-23T21:41:00.657" LastActivityDate="2016-12-26T13:49:33.290" Title="How is a Galaxy formed?" Tags="&lt;galaxy&gt;&lt;universe&gt;&lt;milky-way&gt;&lt;space&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19152" PostTypeId="2" ParentId="19151" CreationDate="2016-11-23T19:43:56.747" Score="2" Body="&lt;p&gt;Well this is not a question that can be answered in a few sentences!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We still do not know, how exactly the galaxies formed and have the shapes they possess. The large-scale structure (LSS) in the Universe as we see them today are a consequence of tiny &lt;a href=&quot;https://en.wikipedia.org/wiki/Primordial_fluctuations&quot; rel=&quot;nofollow noreferrer&quot;&gt;primordial density fluctuations&lt;/a&gt; that arose right after the &#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Bang&quot; rel=&quot;nofollow noreferrer&quot;&gt;Big Bang&lt;/a&gt;. &#xA;The reason for these tiny density fluctuations is believed to be quantum in nature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, these tiny fluctuations in the early past lead to agglomeration of gas and dust clouds, leading to certain areas becoming denser. These denser areas slowed down the &lt;a href=&quot;https://en.wikipedia.org/wiki/Accelerating_expansion_of_the_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;expansion of the Universe&lt;/a&gt;, allowing the gas to accumalate into small protogalactic clouds. Gravity in these clouds casued the gas and dust to collapse, and in turn form stars. These stars burned quickly and became globular clusters (while gravity was still collapsing the dust and gas). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QcGkm.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QcGkm.gif&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, according to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lambda-CDM_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\Lambda$CDM model&lt;/a&gt;, the structures form in a &quot;bottom-up&quot; fashion, i.e. small structures forming first (stars and galaxies) followed by large structures (galaxy clusters). This is exactly what we are observing today thanks to surveys that are probing high redshift ranges. &lt;/p&gt;&#xA;" OwnerUserId="2655" LastActivityDate="2016-11-23T19:43:56.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19155" PostTypeId="1" AcceptedAnswerId="19156" CreationDate="2016-11-24T02:35:07.477" Score="10" ViewCount="708" Body="&lt;p&gt;The planets in our system are most often grouped into two categories:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Terrestrial:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mercury&lt;/li&gt;&#xA;&lt;li&gt;Venus&lt;/li&gt;&#xA;&lt;li&gt;Earth&lt;/li&gt;&#xA;&lt;li&gt;Mars&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Gas Giants:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Jupiter&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Saturn&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Uranus&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Neptune&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Why is it that we don't see &quot;in-between&quot; planets in our system - large, rocky worlds larger than Earth but with thick atmospheres somewhat similar to those of gas giants? Are they possible? Did they not occur by random chance?&lt;/p&gt;&#xA;" OwnerUserId="14918" LastActivityDate="2020-05-19T14:24:44.740" Title="Why don't we have in-between planets?" Tags="&lt;planet&gt;&lt;gas-giants&gt;&lt;planetary-formation&gt;&lt;terrestrial-planets&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19156" PostTypeId="2" ParentId="19155" CreationDate="2016-11-24T02:47:35.177" Score="10" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Super-Earth&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Super-Earths&lt;/strong&gt;&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Super-Earth&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Mini-Neptunes&lt;/strong&gt;&lt;/a&gt; are the &quot;in-between&quot; types of exoplanets you're looking for. A sweeping generalization would put most in the range of $\sim1$-$10M_{\oplus}$ (Earth masses), with some outliers a bit above that. They may have significant quantities of hydrogen and helium in their atmospheres, as well as water, in liquid or vapor form. The latter should also have what are referred to as &lt;a href=&quot;https://en.wikipedia.org/wiki/Volatiles&quot; rel=&quot;nofollow noreferrer&quot;&gt;volatiles&lt;/a&gt; (colloquially called &quot;ices&quot;) - compounds such as ammonia and water that are present in the atmospheres of proper ice giants like Uranus and Neptune.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One reason we don't have any of these in the Solar System is simply that no worlds with the proper mass formed. I explained &lt;a href=&quot;https://astronomy.stackexchange.com/a/18564/2153&quot;&gt;here&lt;/a&gt; that there is a transition region of sorts, but most super-Earths require minimum masses of $1.5$-$2M_{\oplus}$ to keep their large hydrogen/helium envelopes (which terrestrial planets may accrete early in their lives but later lose). There would have to be a planet just in that mass range &lt;em&gt;and&lt;/em&gt; at the right distance from the Sun (see &lt;a href=&quot;https://arxiv.org/abs/1606.08088&quot; rel=&quot;nofollow noreferrer&quot;&gt;D'Angelo &amp;amp; Bodenheimer (2016)&lt;/a&gt; for general constraints on semi-major axes). The environment in the protoplanetary disk may simply not have been suitable to produce the right body at the right time.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-11-24T02:47:35.177" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19158" PostTypeId="1" CreationDate="2016-11-24T05:01:24.013" Score="1" ViewCount="146" Body="&lt;p&gt;do most or only a minority of all planetary systems to show transits?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like its minority due to the random orientation of their orbits?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the thing is I seem to find conflicting information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone explain which answer is correct because now I am confused?&lt;/p&gt;&#xA;" OwnerUserId="14941" LastActivityDate="2016-11-26T14:22:17.963" Title="does a minority of all planetary systems show transits?" Tags="&lt;planet&gt;&lt;fundamental-astronomy&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19160" PostTypeId="1" CreationDate="2016-11-24T08:01:48.183" Score="0" ViewCount="83" Body="&lt;p&gt;Supposing that another planetary system is in one plane like our solar system, if let's say that the transit of one planet is visible then is it safe to also assume that all planets in the system must transit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Logically I feel like that this makes sense because since they are all in the same plane then they therefore must all transit as well. But then I feel like the transit probability equation could make this false? Could someone please clarify my doubts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also suppose that we are in another system where each planet doesn't necessarily orbit in the same plane.  Is it easier to find transits if the planets are orbiting in the same or random planes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that you have a better chance detecting a planet if they are in random orbits. but i am a bit lost in regards to detecting multiple transits.&lt;/p&gt;&#xA;" OwnerUserId="14941" LastEditorDisplayName="user5434" LastEditDate="2017-05-06T22:18:00.573" LastActivityDate="2017-05-06T22:18:00.573" Title="Frequency of transit in planar and non-planar planetary systems" Tags="&lt;observational-astronomy&gt;&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19161" PostTypeId="1" CreationDate="2016-11-24T09:59:26.057" Score="2" ViewCount="143" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Mj8EN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mj8EN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is radio-emitting gas? (I think it refers to the Magellanic Stream, but I can't be convinced.)&lt;/p&gt;&#xA;" OwnerUserId="14782" LastEditorUserId="2365" LastEditDate="2016-12-15T06:42:13.687" LastActivityDate="2016-12-15T14:56:16.390" Title="What is radio-emitting gas?" Tags="&lt;milky-way&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2016-12-15T14:13:41.347" ContentLicense="CC BY-SA 3.0" />
  <row Id="19162" PostTypeId="2" ParentId="19161" CreationDate="2016-11-24T12:41:20.660" Score="2" Body="&lt;p&gt;One type of ratio-emitting gas is ionized gas, which is penetrated by a magnetic field. Also known as a plasma, the charged particles that make up this gas will start rotating around the magnetic field lines.&lt;br&gt;&#xA;Their paths will have a local radius of curvature equal to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gyroradius&quot; rel=&quot;nofollow noreferrer&quot;&gt;gyroradius&lt;/a&gt;, which is the natural movement radius of charged particles in a magnetic field.&lt;br&gt;&#xA;Then, because circular motion is accelerated motion, and accelerated charges in general emit electromagnetic radiation (EMR), the gyrating charges &lt;a href=&quot;https://physics.stackexchange.com/questions/65339/how-and-why-does-accelerating-charges-radiate-electromagnetic-radiation&quot;&gt;will emit EMR&lt;/a&gt;.&lt;br&gt;&#xA;One can &lt;a href=&quot;http://www.mathpages.com/home/kmath528/kmath528.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;calculate&lt;/a&gt; that for typical magnetic field strengths in interstellar space this radiation will then be emitted in the radio-range.&lt;br&gt;&#xA;The details depend a lot on where the charged particles come from, what their energies are, viewing geometry of the observer etc. but essentially it's that.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-11-24T16:59:22.960" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19163" PostTypeId="2" ParentId="19160" CreationDate="2016-11-24T12:50:41.353" Score="2" Body="&lt;p&gt;This is unfortunately not true.&lt;br&gt;&#xA;If you already look at &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_inclination&quot; rel=&quot;nofollow noreferrer&quot;&gt;our solar system&lt;/a&gt;, even this is not co-planar.&lt;br&gt;&#xA;The few degrees inclination that the planets have to each other would already be enough to make some of them undetectable, when others are detectable by transits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to your question therefore is not so much about the transit probability, but more about what the mutual inclinations are of the planets. Those can be stirred up by close-planet encounters or resonances after the gas disc dissipates in which the planets form.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2016-11-24T12:50:41.353" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19165" PostTypeId="1" AcceptedAnswerId="19167" CreationDate="2016-11-24T16:26:47.037" Score="2" ViewCount="143" Body="&lt;p&gt;The energy-momentum relation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E^2 = m^2c^4 +p^2c^2,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;lets us derive the momentum of a massless particle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p = \frac{E}{c} = \frac{h\nu}{c}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the expansion of the Universe redshifts light. This should decrease the momentum of photons. Where would the momentum go, in order for conservation of momentum to hold?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2016-11-26T13:07:14.640" Title="Does the redshifting of photons from the Universe's expansion violate conservation of momentum?" Tags="&lt;light&gt;&lt;cosmology&gt;&lt;general-relativity&gt;&lt;expansion&gt;&lt;redshift&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19166" PostTypeId="2" ParentId="19161" CreationDate="2016-11-24T16:31:40.323" Score="5" Body="&lt;p&gt;Referring &lt;em&gt;only&lt;/em&gt; to the image you present, the radio part of the image (I think this is a stacked optical and radio image) shows something called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Magellanic_Stream&quot; rel=&quot;nofollow noreferrer&quot;&gt;Magellanic stream&lt;/a&gt;. Despite what the wikipedia page says, I believe it was discovered by &lt;a href=&quot;http://adsabs.harvard.edu/abs/1971A&amp;amp;A....12...59D&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dieter (1971)&lt;/a&gt; using 21 cm observations of &lt;em&gt;neutral&lt;/em&gt; hydrogen &lt;em&gt;atoms&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radio emission in the case of the picture you show arises from the so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrogen_line&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;spin flip&quot;&lt;/a&gt; transition associated with changes in the relative spin of the proton and electron in a hydrogen atom. A configuration where the spins are anti-parallel has a lower energy than where they are parallel. (Quantum mechanics only allows these two possibilities). The energy difference is small - only $5.87 \times 10^{-6}$ eV. In hydrogen gas in space it is possible for the atom to make a transition from the higher level state to the lower level, in the course of which it emits radiation with a wavelength of 21 cm (i.e. radio waves). This is not seen in terrestrial laboratory samples because the gas is collisionally de-excited before it has the chance to emit radiation. Hence 21cm radiation is an example of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Forbidden_mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;forbidden line&lt;/a&gt;, with a long radiative lifetime and which is only observed from the rarefied neutral hydrogen gas found in the interstellar medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intrinsic width of the emitted spectral line is very narrow. This means that the observed wavelength of this 21 cm line is an excellent probe of the motion of the emitting gas, since the observed radio waves will be doppler shifted from the 21 cm wavelength at which they were emitted according to the line-of-sight velocity of the gas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gas in question here has been stripped from the two satellite galaxies of the Milky Way, known as the Magellanic clouds. Although the picture also shows copious 21 cm radiation arising from neutral hydrogen gas in the plane of our Milky Way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many other ways that gas can be a source of radio emission. In addition to atomic and molecular transitions that give radio emission at discrete wavelengths (including the radio equivalent of lasers, called &lt;a href=&quot;https://en.wikipedia.org/wiki/Maser&quot; rel=&quot;nofollow noreferrer&quot;&gt;masers&lt;/a&gt;), there can be continuum emission caused by the acceleration of electrons in the electric fields of partially and completely ionised ions - so called &lt;a href=&quot;https://en.wikipedia.org/wiki/Bremsstrahlung&quot; rel=&quot;nofollow noreferrer&quot;&gt;bremsstrahlung&lt;/a&gt; or free-free emission. There is also recombination radiation (free-bound) radiation and continuum radiation emitted by electrons that are accelerated in electromagnetic fields - &lt;a href=&quot;https://en.wikipedia.org/wiki/Cyclotron_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;cyclotron radiation&lt;/a&gt;; &lt;a href=&quot;https://en.wikipedia.org/wiki/Synchrotron_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;synchrotron radiation&lt;/a&gt; (if the electrons are relativistic); curvature radiation (a variety of synchrotron radiation) or even just the Rayleigh-Jeans tail from cold blackbody emission (e.g. in the cosmic microwave background).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A decent introductory summary can be found &lt;a href=&quot;http://astronomy.swinburne.edu.au/sao/downloads/HET608-M03A02.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-12-15T14:56:16.390" LastActivityDate="2016-12-15T14:56:16.390" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19167" PostTypeId="2" ParentId="19165" CreationDate="2016-11-24T17:47:31.733" Score="5" Body="&lt;p&gt;In relativity you can think of a single conservation law that unites conservation of energy and momentum -- conservation of &lt;a href=&quot;https://en.wikipedia.org/wiki/Four-momentum#Conservation_of_4-momentum&quot; rel=&quot;nofollow noreferrer&quot;&gt;four-momentum&lt;/a&gt;. Energy and momentum are the zeroth and the first to third components of the four-momentum respectively. Such conservation laws arise from invariance of the Lagrangian with respect to a translation in space-time coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In General Relativity these conservation laws are local concepts that (most people think) can only be applied in local, inertial (flat) frames of reference. In particular, they cannot be applied in changing space-times and so cannot be applied to situations involving the expansion of the universe.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-11-24T17:47:31.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19169" PostTypeId="1" CreationDate="2016-11-25T08:54:07.643" Score="0" ViewCount="130" Body="&lt;p&gt;How could visible light be in pitch-black? Can it be?&lt;/p&gt;&#xA;" OwnerUserId="14782" LastEditorUserId="14782" LastEditDate="2016-12-15T04:54:09.783" LastActivityDate="2016-12-15T04:54:09.783" Title="How could visible light be in pitch-black? Can it be?" Tags="&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2016-12-15T14:14:04.563" ContentLicense="CC BY-SA 3.0" />
  <row Id="19170" PostTypeId="1" CreationDate="2016-11-25T09:26:31.763" Score="1" ViewCount="3457" Body="&lt;p&gt;I have the moon's position in equatorial coordinates (RA and Dec) and now I want to calculate same moon's position in terms of earth centered coordinates i.e. Geocentric (latitude &amp;amp; longitude). My main goal is to establish a c++-code that gives me this transformation from RA,Dec to Lat,Lon and vice versa. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT&lt;/em&gt;: More precisely, I have moon’s position in equatorial geocentric coordinates with following details:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Origin: Center of Earth&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;Lat   : Declination With fundamental Plane: Celestial Equator&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;Lon   : R.A.        With Primary direction: Vernal Equinox&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I am able to obtain the moon’s position corresponding to a specific Modified Julian Date (MJD). I am also able to calculate moon’s position in xyz-coordinates by following calculations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X= EarthMoon*sin(PI/2.-dec_moon*deg2rad)*cos(ra_moon*deg2rad);$&#xA;$Y= EarthMoon*sin(PI/2.-dec_moon*deg2rad)*sin(ra_moon*deg2rad);$&#xA;$Z= EarthMoon*cos(PI/2.-dec_moon*deg2rad);$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where EarthMoon is the distance between Earth and Moon. Further, I am converting them into Horizon Coordinates (topocentric) w.r.t to an observer at a specific location by using my subroutine and this part done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, I am initiating my simulated CR-particles from observer site. E.g. vCR[3]= {R,lat,lon}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;vCR[3]={Rearth+2300, 31.30, 55.74}&lt;/em&gt; //Geocentric CS (earth-based)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And converting them into Geocentric Cartesian coordinates(x,y,z). Finally, I have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;vM[3]={X,Y,Z};&lt;/em&gt; in Equatorial geocentric coordinate system&lt;br&gt;&#xA;&lt;em&gt;vCR[3]={x,y,z};&lt;/em&gt; in Geocentric Coordinate System i.e. Earth based  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is my particular question in which I want to make vM[moon position] and vCR[CR position] to lie in same coordinates to get my final result. i.e. transformation from Geocentric (R,lat,lon) into Equatorial Geocentric (h,R.A,Dec).&lt;br&gt;&#xA;&lt;strong&gt;Note&lt;/strong&gt;: For vCR[3]: zero-longitude is the Prime Meridian, Greenwich whereas for vM[3] it’s vernal equinox in celestial sphere there is the difference.&lt;/p&gt;&#xA;" OwnerUserId="14951" LastEditorUserId="14951" LastEditDate="2016-12-04T07:43:37.147" LastActivityDate="2016-12-04T07:43:37.147" Title="Transformation from Geocentric Coordinates into Equatorial Coordinates?" Tags="&lt;earth&gt;&lt;coordinate&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19171" PostTypeId="2" ParentId="19147" CreationDate="2016-11-25T10:29:28.823" Score="1" Body="&lt;p&gt;When you go back in time, take collisions into account. A very highly elliptical orbit primarily leads to chunks colliding at the very first periapsis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if they are not chunks but very dense and small objects (point-like), if you extrapolate back in time so far as to the straight line &quot;orbit&quot;, what does it mean? Extreme case of objects with zero relative initial velocity going against each other on a collision course - ups, this &quot;orbit&quot; does not have a chance to &quot;decay due to gravitational wave radiation&quot;. Two smaller chunks collide and form a larger chunk, which has now a non-zero velocity (probably, if the chunks weren't &lt;em&gt;exactly&lt;/em&gt; equal momentum). The chunk won't enter a straight line &quot;orbit&quot; around (towards) anything, but either an elliptical or hyperbolic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your extrapolation does suggest that initial &quot;chunks&quot; hadn't ideally equal momenta (masses*velocities); that's all.&lt;/p&gt;&#xA;" OwnerUserId="10675" LastActivityDate="2016-11-25T10:29:28.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19172" PostTypeId="1" CreationDate="2016-11-25T11:03:16.483" Score="0" ViewCount="783" Body="&lt;p&gt;I'm currently developing a javascript application in which I want to calculate the approximate position of the sun. This works quite fine but requires the value for deltaT (TT-UT) to be set depending on the year for which I want to calculate the solar position. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, I'm using a default value of 67 for my calculation. However, since I want to calculate the solar position for several years I'm looking for a convenient way to obtain the deltaT value for each year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To all of you, that have some experience with programming:&#xA;Is there any interface (API) that provides me with the desired values?&#xA;Of course, it would also be sufficient to get the universal and terrestrial time so that I can calculate deltaT on my own.&lt;/p&gt;&#xA;" OwnerUserId="14953" LastActivityDate="2017-01-26T23:04:46.163" Title="Obtaining deltaT for use in software" Tags="&lt;the-sun&gt;&lt;software&gt;&lt;time&gt;&lt;azimuth&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19173" PostTypeId="2" ParentId="19170" CreationDate="2016-11-25T12:58:05.397" Score="1" Body="&lt;p&gt;I think, you are mixing two different concepts. Equatorial coordinates are coordinates in the equatorial coordinate system (as opposed to the ecliptical, galactic, super galactic, or horizontal coordinate systems). The coordinates also refer to a specific origin. This origin might be heliocentric (centre of the Sun), baryocentric (centre of mass of the Solar System), geocentric (centre of the Earth), or topocentric (a specific location on the Earth's ellipsoid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you may have heliocentric equatorial coordinates, geocentric equatorial coordinates, or topocentric equatorial coordinates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From your question I infer that you have &lt;em&gt;geocentric&lt;/em&gt; coordinates of the Moon and you want to convert them to &lt;em&gt;topocentric&lt;/em&gt; (not geocentric!) coordinates for an observer at a specific location (latitude and longitude) on the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way to go about this would be to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First, convert you geocentric equatorial coordinates to its rectangular equivalent (i.e. find the $X_M$, $Y_M$, and $Z_M$ equatorial coordinates).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Second calculate the $X_o$, $Y_o$, and $Z_o$ coordinates of the observer on Earth in the equatorial system.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Third, subtract the $X_o$, $Y_o$, and $Z_o$ coordinates of the observer from the $X_M$, $Y_M$, and $Z_M$ coordinates of the Moon.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, convert the coordinates back to spherical coordinates, i.e. the $X$, $Y$, and $Z$ coordinates back to right ascension, declination, and distance.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you have the geocentric right ascension $\alpha_M$, declination $\delta_M$, and distance $d_M$, then you can get the rectangular coordinates from:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ X_M = d_M \cos(\alpha_M) \cos(\delta_M) $$&#xA;$$ Y_M = d_M \sin(\alpha_M) \cos(\delta_M) $$&#xA;$$ Z_M = d_M \sin(\delta_M)  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $X$-axis is pointed towards the vernal equinox, and the $Z$-axis is pointed towards the celestial north pole (i.e. parallel with the Earth's rotation axis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the rectangular equatorial coordinates of the observer, you need the geographical latitude ($\phi$) and longitude ($L$), the elevation ($H$) and the sidereal time ($\theta$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the rectangular coordinates of the observer is a bit more difficult as the Earth's surface is (approximately) an ellipsoid. The following is taken from &lt;a href=&quot;http://www.willbell.com/math/mc1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jean Meeus book, Astronomical Algorithms&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b' = 0.99664719$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is a measure of the ellipsoid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ u = \arctan(b' \tan(\phi)) $$&#xA;$$ \rho\sin(\phi) = R_E b' \sin(u) + H \sin(\phi)) $$&#xA;$$ \rho\cos(\phi) = R_E \cos(u) + H \cos(\phi) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R_E = 6378140.0 \textrm{m}$ is the Earth's radius.&#xA;We can then calculate the rectangular geocentric equatorial coordinates of the observer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ X_o =  \rho \cos(\phi) \cos(\theta)$$&#xA;$$ Y_o =  \rho \cos(\phi) \sin(\theta)$$&#xA;$$ Z_o =  \rho \sin(\phi) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The topocentric rectangular equatorial coordinates of the Moon are then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ X = X_M - X_o $$ &#xA;$$ Y = Y_M - Y_o $$ &#xA;$$ Z = Z_M - Z_o $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then convert back to spherical coordinates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ d' = \sqrt{X^2 + Y^2 + Z^2} $$&#xA;$$ \alpha' = \arctan \left(\frac{Y}{X} \right) $$&#xA;$$ \delta' = \arcsin \left(\frac{Z}{d'} \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then have the topocentric equatorial coordinates of the Moon: right ascension $\alpha'$, declination $\delta'$ and the distance $d'$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will have to make sure that $\alpha'$ is in the correct quadrant. Programmatically this can be easily achieved with the &lt;code&gt;atan2&lt;/code&gt; function, i.e. &lt;code&gt;alpha = atan2(y,x)&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't tested these formulae so I hope that they are correct.&lt;/p&gt;&#xA;" OwnerUserId="864" LastEditorUserId="864" LastEditDate="2016-11-25T13:41:17.640" LastActivityDate="2016-11-25T13:41:17.640" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19174" PostTypeId="2" ParentId="19172" CreationDate="2016-11-25T13:57:44.400" Score="0" Body="&lt;p&gt;I'm not aware of any APIs that provide $\Delta T$, but you may be able to parse &lt;a href=&quot;https://datacenter.iers.org/eop/-/somos/5Rgv/latest/16&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://datacenter.iers.org/eop/-/somos/5Rgv/latest/16&lt;/a&gt; for the value $\Delta T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course if you want to calculate the &lt;em&gt;approximate&lt;/em&gt; position of the Sun, a few second more or less should not matter too much ;-)&lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2016-11-25T13:57:44.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19175" PostTypeId="1" AcceptedAnswerId="19177" CreationDate="2016-11-25T14:46:51.617" Score="11" ViewCount="933" Body="&lt;p&gt;Here's our galaxy as best we know, courtesy of this &lt;a href=&quot;https://en.wikipedia.org/wiki/Milky_Way&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J5AKf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J5AKf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that as best we know, each arm &lt;strong&gt;is a compressed or brighter area (rather like a wave)&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Looking at that image, do we know/guess if the arm-patterns are rotating? Or do they just sit there statically, unchanging from that diagram? &lt;strong&gt;Which way&lt;/strong&gt; does it rotate, cw or acw per that diagram? (If we know.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the pattern is rotating - how long does it take it to rotate 360 degrees? As far as we know?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And hence ..&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Indeed, is the arm-pattern catching us and moving through us ... or is the arm pattern slower than us and we're moving through it?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="13071" LastEditDate="2016-11-28T12:07:37.557" LastActivityDate="2016-11-28T13:13:01.087" Title="Do we know if we are overtaking or falling behind our arm?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;spiral-arms&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19177" PostTypeId="2" ParentId="19175" CreationDate="2016-11-25T15:47:18.497" Score="10" Body="&lt;p&gt;The answer to the question depends on precisely where you are in the Milky Way. According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Density_wave_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;density wave theory&lt;/a&gt;, which appears to describe the formation and evolution of spiral arms quite well, the arms move relative to some static inertial frame at a global pattern speed $\Omega_{gp}$. At some defining radius $R_c$, known as the corotation radius, stars move at $\Omega_{gp}$. Inside $R_c$, stars move quicker and thus move ahead of the arms; outside $R_c$, stars move slower and thus &quot;fall behind&quot; the arms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is uncertainty surrounding the values of $R_c$ and $\Omega_{gp}$ in the Milky Way in all its spiral arms - especially since local stellar kinematics differ in each of the arms, and so too may $R_c$ and $\Omega_{gp}$ - but measurements place the Solar System slightly inside the corotation radius, meaning that we are moving ahead of the arms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have data from a review by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2011MSAIS..18..185G&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gerhard (2011)&lt;/a&gt;. Here are some basic average parameters, for the Milky Way as a whole:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Bar&lt;/em&gt; corotation radius ($R_{CR}$): $3.5$-$4.5\text{ kpc}$&lt;/li&gt;&#xA;&lt;li&gt;Spiral arm corotation radius: $8.1$-$8.4\text{ kpc}$&lt;/li&gt;&#xA;&lt;li&gt;Bar pattern speed ($\Omega_b$): $50$-$60\text{ km}\text{ s}^{-1}\text{ kpc}^{-1}$&lt;/li&gt;&#xA;&lt;li&gt;Spiral arm pattern speed ($\Omega_{sp}$): $17$-$28\text{ km}\text{ s}^{-1}\text{ kpc}^{-1}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This also goes along with a local mean speed of stars in the solar neighborhood - the Sun included - of $\sim220\text{ km}\text{ s}^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hence for the three questions: (1) The pattern is rotating clockwise as seen in that diagram; (2) It rotates about once every 300 million years; (3) and at our radius, we're moving a little faster than the pattern so we are overtaking it.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-11-28T13:13:01.087" LastActivityDate="2016-11-28T13:13:01.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19178" PostTypeId="2" ParentId="19172" CreationDate="2016-11-25T21:11:59.380" Score="1" Body="&lt;p&gt;See &lt;a href=&quot;https://astronomy.stackexchange.com/questions/13488/where-can-i-find-visualize-planets-stars-moons-etc-positions/13489#13489&quot;&gt;Where can I find/visualize planets/stars/moons/etc positions?&lt;/a&gt; for an extremely general answer on how to compute positions, but the files you're looking for specifically are the &quot;leap second kernels&quot; at &lt;a href=&quot;http://naif.jpl.nasa.gov/pub/naif/generic_kernels/lsk/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://naif.jpl.nasa.gov/pub/naif/generic_kernels/lsk/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was a mild kerfuffle on spice-discussion lists when NASA failed to update the kernels after the end-of-2016 leap second was announced, but they have updated it now.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-11-25T21:11:59.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19180" PostTypeId="2" ParentId="19169" CreationDate="2016-11-26T05:55:47.110" Score="3" Body="&lt;p&gt;Black is not a color; it is the absence of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you interpret the sentence as &quot;the light is black&quot;, but it's just a way of saying &quot;there is no light (at optical wavelengths, but there is at radio wavelengths)&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion this is not a good way of explaining the CMB. The sky is &lt;em&gt;not&lt;/em&gt; black at any wavelengths, but has a background radiation at all wavelengths. In the optical, this background originates (primarily) from the integrated light from stars; in the microwave it originates (primarily) from the afterglow of Big Bang; in the infrared from dust, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the figure in &lt;a href=&quot;https://physics.stackexchange.com/a/196373/70207&quot;&gt;this answer&lt;/a&gt; to a question about the number density of CMB photons, to see how large the background is in various wavelength regions.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-11-26T05:55:47.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19181" PostTypeId="1" CreationDate="2016-11-26T07:10:31.443" Score="0" ViewCount="861" Body="&lt;p&gt;For 26 Nov 2016:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider Tehran with 06:51 sunrise and 16:52 sunset.&#xA;Another city is Rafsanjan in 850KM southeast of Tehran, sunrise is 6:21 and 16:45 for sunset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time zone of two cities is +3.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why difference of sunrise is 30 minutes while sunset is 7 minutes?&lt;/p&gt;&#xA;" OwnerUserId="14967" LastActivityDate="2016-11-26T14:12:58.427" Title="Difference of sunset and sunrise of two cities are different" Tags="&lt;time&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19182" PostTypeId="2" ParentId="19181" CreationDate="2016-11-26T12:47:17.190" Score="0" Body="&lt;p&gt;I would have commented but I don't have enough reputation. &#xA;Sunrise and sunset times may depend on a lot of factors. For example if your city had a big mountain in the west, the sunset would be earlier than what it would have been with the sea instead of the mountain. These things may depends on a lot of factors and you should know the morphology of the zone where the cities are to answer properly. For example there may be mountains in the east of Tehran and that would explain the late sunrise. I don't know the details but I think morphology of the earth is the answer. On a big scale you can say it's a sphere, locally it's everything but smooth.&lt;/p&gt;&#xA;" OwnerUserId="13436" LastActivityDate="2016-11-26T12:47:17.190" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19184" PostTypeId="2" ParentId="1068" CreationDate="2016-11-26T13:56:33.030" Score="7" Body="&lt;p&gt;I've read lichens have been shown to survive in a Mars-like atmosphere, tested by NASA. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This other article says they can survive vacuum and radiation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astrobio.net/extreme-life/lichen-on-mars/&quot; rel=&quot;noreferrer&quot;&gt;http://www.astrobio.net/extreme-life/lichen-on-mars/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11562" LastEditorUserId="11562" LastEditDate="2017-12-13T17:10:15.807" LastActivityDate="2017-12-13T17:10:15.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19185" PostTypeId="2" ParentId="19181" CreationDate="2016-11-26T14:12:58.427" Score="1" Body="&lt;p&gt;Tehran (35.7$^\circ$ N) has 23 fewer minutes of daylight than Rafsanjan (30.4$^\circ$ N) on Nov 26 because the Sun is 21$^\circ$ south of the celestial equator and Tehran is 5.3$^\circ$ farther north than Rafsanjan.&#xA;For the same reason, the north pole has 24-hour night and the south pole has 24-hour daylight at this time of year.&#xA;On May 26, when the Sun's declination is 21$^\circ$ north, Tehran has 25 more minutes of daylight than Rafsanjan does, and the north pole has 24 more hours of daylight than the south pole does.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-11-26T14:12:58.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19186" PostTypeId="1" AcceptedAnswerId="19188" CreationDate="2016-11-26T14:16:37.613" Score="1" ViewCount="522" Body="&lt;p&gt;I was showing a recent video about Ceres to a friend and we looked up asteroids. Ceres has attained a fairly spherical shape, but Vesta still has a way to go. From &lt;a href=&quot;https://en.wikipedia.org/wiki/4_Vesta#Physical_characteristics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Vesta's shape is close to a gravitationally relaxed oblate spheroid, but the large concavity and protrusion at the southern pole (see 'Surface features' below) combined with a mass less than $5×10^{20}$ kg precluded Vesta from automatically being considered a dwarf planet under International Astronomical Union (IAU) Resolution XXVI 5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know the proper term to use here. Perhaps it is &quot;hydrostatically relaxed&quot; or something similar – I'm looking for the term that would suggest a body has reached a fairly round shape without major excursions, under the influence of its own gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think Ceres meets this criterion, and Vesta doesn't. Is Vesta the largest such body (by dimension or volume, not sure which would be the most reasonable measure)?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="14006" LastEditDate="2018-02-04T17:36:05.690" LastActivityDate="2021-01-18T12:20:21.470" Title="Ceres is spherical, Vesta isn't – is it the largest non-spherical solid body in the solar system?" Tags="&lt;asteroids&gt;&lt;planetary-formation&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19187" PostTypeId="2" ParentId="19158" CreationDate="2016-11-26T14:22:17.963" Score="1" Body="&lt;p&gt;Consider a circular orbit and approximate the planet as a point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance from Earth to a star is so large compared to the diameter of the star, which we'll call d, that we approximate the lines joining the two poles of the star with the observer as parallel to each other and to the x-axis that we choose. Thanks to symmetry, we choose our coordinates so that the plane of the planet's orbit passes through the tangent plane of the celestial sphere at the star in two places on the y-axis. By symmetry again we ignore half the cases for rotation and also ignore the direction of the orbit and only have to consider how the orbit is inclined at between 0 and pi/2 radians from the x-axis (Edit for clarity: in the XZ plane). Call that angle a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the radius r of the planet's orbit is such that d/r = sin(a), the planet will just barely touch the limb of the star when it is closest to the Earth. If sin(a) &gt; d/r it will never happen and if sin(a) &amp;lt; d/r there will be a transit of finite length. Since there is no preference for the value of a, we conclude that the probability of the planet transiting must be $$P(sin(a) &amp;lt; d/r) = P{ \left( a &amp;lt; {arcsin \left( { d \over r } \right)  } \right) } = { 2 arcsin \left( { d \over r } \right) \over \pi } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will notice that for a star-grazing planet, the probability of a transit approaches certainty. Overall probability will depend on the probability distribution of semi-major axes and possibly eccentricity which I ignored here.&lt;/p&gt;&#xA;" OwnerUserId="14969" LastActivityDate="2016-11-26T14:22:17.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19188" PostTypeId="2" ParentId="19186" CreationDate="2016-11-26T15:21:11.530" Score="2" Body="&lt;p&gt;Among the asteroids 4 Vesta is the largest non-spherical object in the Solar system. But far out there in the Kuiper Belt is 136108 Haumea, and maybe yet undiscovered objects, that are very elongated. Haumea is 1,960 × 1,518 × 996 km. That's larger than the 960 km diameter Ceres and hardly spherical with 2:1 proportions. Its fast rotation and composition of malleable volatiles makes this possible. Maybe Haumea is &quot;hydrostatically relaxed&quot; without being spherical?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Haumea images below from &lt;a href=&quot;https://en.wikipedia.org/wiki/Haumea&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/58wJy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/58wJy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/9Y6K5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9Y6K5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2016-11-26T15:27:42.257" LastActivityDate="2016-11-26T15:27:42.257" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19190" PostTypeId="1" CreationDate="2016-11-26T15:50:01.123" Score="9" ViewCount="2573" Body="&lt;p&gt;First of all, the Sun's surface temperature is only about 6000 K. Years ago I was working at at a melt work with furnace temperatures 0f about 2-3000 K; of course you could not be near a furnace without protection, when it was open, but you could still stay some 20-30 metres away from it. Likewise, a light bulb has no long reach, till it fades away. But how about the Sun? Why can experience it at 150 000 000 km away? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides, I wonder why space is not heated up, because the heat radiating from all the stars and galaxies is so enormous, and actually the matter density in space is so low. How, actually, can heat be absorbed in space? What temperature is measured in space - is it just the wavelengths of radiations, from which temperature is calculated?&lt;/p&gt;&#xA;" OwnerUserId="14971" LastEditorUserId="2153" LastEditDate="2016-11-26T23:06:35.627" LastActivityDate="2019-10-11T11:39:22.523" Title="Why is the Sun so bright, but you can feel it far away?" Tags="&lt;the-sun&gt;&lt;temperature&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19191" PostTypeId="1" CreationDate="2016-11-26T16:32:29.880" Score="-3" ViewCount="142" Body="&lt;p&gt;I can't understand the last sentence in this text, entirely. &#xA;Please, help me.&#xA;Thank you &lt;/p&gt;&#xA;" OwnerUserId="14782" LastEditorUserId="14782" LastEditDate="2016-12-19T08:00:31.730" LastActivityDate="2016-12-19T08:00:31.730" Title="What does a solar magnetosphere &quot;short circuit&quot; mean?" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2016-12-15T14:16:18.727" ContentLicense="CC BY-SA 3.0" />
  <row Id="19192" PostTypeId="2" ParentId="19190" CreationDate="2016-11-26T17:06:18.627" Score="5" Body="&lt;p&gt;Your question is far more complex than you think. You need matter to have temperature, cause temperature is nothing else than the measure of the mean energy of a group of particles (this definition is quite accurate for our purposes here). So you need particles and something that can give them energy to heat the matter up. The photons from the stars heat the matter and you measure a temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what's the temperature in space? It depends, it depends where you are and what's there. In a galaxy for example you have some matter between stars, it's called the interstellar medium (ISM), it's mainly gas, but there is a very important (for astronomers) fraction of (usually cold) dust. So what are the temperature of this medium? We can divide it in categories:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hot ionized medium: composed of ionized particles. Its temperature ranges from one million kelvin to 10 millions kelvin. You can find it in the bulge of our galaxy &lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Warm ionized medium: temperatures around 10000 kelvins. You can find it in the spiral arms, around the hottest stars (O, B type: the blue ones).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Warm neutral medium: Its temperature ranges from 1000 to 10000 kelvins. It's diffused in the disk of our galaxy and it extends way farther than the zone where the stars in the galaxy are.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Cold neutral medium: Its temperature ranges from 10 to 100 kelvins, it's mainly composed of molecules of hydrogen and it's very important for the star formations, so you can find it in the spiral arms.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then you have the dust which get heated up by stars radiation and its temperature can vary depending on its composition, location and dimensions of grains.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What temperature is measured in space? -Is it just wavelengths of radiations, from which temperature is calculated?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Basically yes, we measure the radiation that arrive to us, and we know the processes that can make this radiation possible and from this knowledge we can calculate the temperature those particle had to have in order to emit like that.&lt;/p&gt;&#xA;" OwnerUserId="13436" LastEditorUserId="13436" LastEditDate="2019-10-11T11:13:37.557" LastActivityDate="2019-10-11T11:13:37.557" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="19197" PostTypeId="1" CreationDate="2016-11-26T22:27:55.610" Score="-6" ViewCount="137" Body="&lt;p&gt;I have two simple questions about Neptunian system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understood the text just iterally, but I can't grasp the mechanism of them. Could you explain them? &lt;/p&gt;&#xA;" OwnerUserId="14782" LastEditorUserId="14782" LastEditDate="2016-12-19T07:58:40.383" LastActivityDate="2016-12-19T07:58:40.383" Title="I have simple questions about neptunian system" Tags="&lt;neptune&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2016-11-28T05:33:50.770" ContentLicense="CC BY-SA 3.0" />
  <row Id="19198" PostTypeId="2" ParentId="19190" CreationDate="2016-11-26T22:43:28.943" Score="9" Body="&lt;p&gt;The answer to your first question has to do with &lt;a href=&quot;https://en.wikipedia.org/wiki/Luminosity&quot; rel=&quot;nofollow noreferrer&quot;&gt;luminosity&lt;/a&gt;. It's a measure of power, the energy given off by an object in a certain amount of time, which you can think of as brightness. The more luminous the object, the brighter it appears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can treat the Sun as an idealized object called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Black_body&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;black body&lt;/em&gt;&lt;/a&gt;, which emits thermal radiation according to something called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stefan-Boltzmann law&lt;/a&gt;. Assuming the Sun is spherical, with a radius $R$ and temperature $T$, then its luminosity $L$ is&#xA;$$L=4\pi\sigma R^2T^4$$&#xA;where $\sigma$ is the Stefan-Boltzmann constant. The important thing here is that $L\propto R^2$. If the Sun was the size of one of your furnaces, it would not be very bright; if the furnace was spherical and could be treated as a black body, the two would have the same luminosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun, however, has a radius on the order of about $7\times10^8$ meters. I don't know how big the furnace is, but I'm assuming it can't be larger than ten meters, to an order of magnitude. First, that's a size difference of about $10^7$. Second, we square the radius when determining luminosity, meaning that there's a difference of a factor of $10^{14}$. To make this &lt;a href=&quot;https://xkcd.com/558/&quot; rel=&quot;nofollow noreferrer&quot;&gt;a bit more meaningful&lt;/a&gt;, the Sun is then essentially one hundred million million times more luminous than the furnace, even though the temperatures are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like your second question may be a duplicate of &lt;a href=&quot;https://astronomy.stackexchange.com/q/2001/2153&quot;&gt;What is the temperature of outerspace?&lt;/a&gt; (see also the questions linked therein), and &lt;a href=&quot;https://astronomy.stackexchange.com/a/19192/2153&quot;&gt;Run like hell&lt;/a&gt; has already tried to address it. &lt;a href=&quot;https://astronomy.stackexchange.com/questions/19190/suns-extreme-brightness#comment31057_19190&quot;&gt;barrycarter's comment&lt;/a&gt;, though, is spot-on. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Flux&quot; rel=&quot;nofollow noreferrer&quot;&gt;flux&lt;/a&gt; from the source, $F$, drops off via the inverse-square law - that is, at a distance $r$ from the source,&#xA;$$F\propto\frac{1}{r^2}$$&#xA;Remember how we just established the power (pun intended) of squaring a large quantity? Interstellar distances are enormous; the nearest star system, Alpha Centauri, is roughly $10^{16}$ meters away, or ten thousand million million meters. Now square that distance. That's a &lt;em&gt;tiny&lt;/em&gt; factor - even when multiplied by the luminosity of all three stars, which combined is still on the same order of magnitude as the Sun.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-11-26T22:43:28.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19199" PostTypeId="1" CreationDate="2016-11-27T06:27:37.120" Score="0" ViewCount="117" Body="&lt;p&gt;Why do planets revolve around the sun? I understand that gravity plays a huge role in allowing it to move around. But what caused it to revolve in the first place? One possible solution is explained with the theory of relativity but is there a classical mechanics approach to this question? &lt;/p&gt;&#xA;" OwnerUserId="9394" LastActivityDate="2016-11-27T07:08:09.547" Title="Revolution of Planets" Tags="&lt;gravity&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2016-11-27T17:30:09.417" ContentLicense="CC BY-SA 3.0" />
  <row Id="19200" PostTypeId="2" ParentId="19172" CreationDate="2016-11-27T06:43:33.303" Score="0" Body="&lt;p&gt;A historical table from 1961 to the present of TAI-UTC is maintained here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;ftp://hpiers.obspm.fr/iers/bul/bulc/UTC-TAI.history&quot; rel=&quot;nofollow noreferrer&quot;&gt;ftp://hpiers.obspm.fr/iers/bul/bulc/UTC-TAI.history&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Delta T can be calculated by adding 32.184s, the difference between TT and TAI, to the value (TAI-UTC) in the table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So currently Delta T is about 68 and will likely be 69 in a few years. It's increasing by one every 3 years or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, values of UTC are adjusted with leap seconds so we can use precision clocks that aren't continuously tweaked. UT1 is the precise measure of &quot;Earth&quot; time. You can modify this to reference UT1 using the DUT1 value which is distributed with NIST time signals, WWV etc. It's value is published here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hpiers.obspm.fr/eoppc/bul/buld/bulletind.131&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://hpiers.obspm.fr/eoppc/bul/buld/bulletind.131&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14973" LastEditorUserId="14973" LastEditDate="2016-11-27T07:14:50.583" LastActivityDate="2016-11-27T07:14:50.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19201" PostTypeId="2" ParentId="19199" CreationDate="2016-11-27T07:08:09.547" Score="0" Body="&lt;p&gt;Classical mechanics approach to this would be &lt;strong&gt;gravitational force&lt;/strong&gt; as we know &lt;code&gt;F=GMm/r2&lt;/code&gt; and &lt;strong&gt;Centripetal force&lt;/strong&gt; &lt;code&gt;F=mv2/r&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In free space we when two things are kept some distance apart they accelarate towards each other (along straight line from center of on body to center of another body) due to force of gravity but if they are under some influence with other bodies they does not get accelarate exactly towards each other(deviation along the path), it causes the dirrection of bodies to change. In this case the planets(cause of relatively small mass). &#xA;If this deviation is large body(planet) fly by without any effect. If deviation is small, bigger body continuously change smaller one's direction, thus orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://curious.astro.cornell.edu/about-us/57-our-solar-system/planets-and-dwarf-planets/orbits/243-why-do-the-planets-orbit-the-sun-beginner&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://curious.astro.cornell.edu/about-us/57-our-solar-system/planets-and-dwarf-planets/orbits/243-why-do-the-planets-orbit-the-sun-beginner&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.windows2universe.org/our_solar_system/formation.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.windows2universe.org/our_solar_system/formation.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;fun Q:why does astronauts on ISS feel 0 gravity when ISS is revolving under earth's gravity? &#xA;because they experience gravity but the revolution causes them to free fall, which they feel like 0 gravity.&lt;/p&gt;&#xA;" OwnerUserId="14102" LastActivityDate="2016-11-27T07:08:09.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19202" PostTypeId="1" AcceptedAnswerId="19207" CreationDate="2016-11-27T07:35:16.960" Score="1" ViewCount="103" Body="&lt;p&gt;An &lt;a href=&quot;https://astronomy.stackexchange.com/q/19186/7982&quot;&gt;intriguing comment&lt;/a&gt; below &lt;a href=&quot;https://astronomy.stackexchange.com/questions/19186/#comment31073_19186&quot;&gt;this question&lt;/a&gt; links to &lt;a href=&quot;https://www.quora.com/How-and-why-are-planets-spherical-What-makes-them-round#!n=96&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Quora question&lt;/a&gt; and points to the answer by Robert Walker. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sentence that particularly intrigued me is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It turns out that there are two possible solutions, as the spin rate increases. You can get an oblate spheroid, or a triaxial ellipsoid - the solution &quot;bifurcates&quot;. But the triaxial ellipsoid is the most stable of the two as Jacobi found out in his paper published in 1834.  Figures of Equilibrium - Historical Account* - Chandrasekar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems to refer both to an 1834 paper by Jacobi and it's reference in the 1964 paper &lt;em&gt;Ellipsoidal figures of equilibrium—an historical account&lt;/em&gt; by S. Chandrasekhar, Communications on Pure and Applied Mathematics, 20 (2), May 1967, 251–265. While the latter is paywalled, there seems to be a readable version &lt;a href=&quot;https://books.google.com.tw/books?id=h_9pDQAAQBAJ&amp;amp;lpg=PA1145&amp;amp;ots=w9oKES1OYs&amp;amp;dq=Figures%20of%20Equilibrium%20-%20Historical%20Account&amp;amp;pg=PA1138#v=onepage&amp;amp;q=Figures%20of%20Equilibrium%20-%20Historical%20Account&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;here in google&lt;/a&gt; in the book &lt;em&gt;A Quest for Perspectives: Selected Works of S. Chandrasekhar : with Commentary By Subrahmanyan Chandrasekhar&lt;/em&gt;, Kameshwar C. Wali, Volume 1. Imperial College Press, 2001.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematical bifurcation is a fascinating topic and there is even an &lt;a href=&quot;http://www.worldscientific.com/worldscinet/ijbc&quot; rel=&quot;nofollow noreferrer&quot;&gt;entire mathematics journal&lt;/a&gt; dedicated to the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, &lt;em&gt;where exactly in the calculation of hydrostatic equillibrium of rotating bodies does this bifurcation&lt;/em&gt; (mentioned in the Quora answer) &lt;em&gt;occur?&lt;/em&gt; I'm wondering if it simply means that for a given volume and density, below a certain rotation rate the hydrostatic equilibrium shape is an oblate sphere, but above it, the hydrostatic equilibrium shape will be a triaxial ellipsoid? Or is the bifurcation part of the evolution of the shape over time, involving viscocity or external effects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the triaxial ellipsoide is always(?) the most stable, then is a oblate spheroid actually an intermediate shape, and not really the shape of hydrostatic equillibrium?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-11-27T20:18:13.770" Title="What is meant by the bifurcation of hydrostatic equilibrium shapes in planetary formation?" Tags="&lt;planetary-formation&gt;&lt;mathematics&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19203" PostTypeId="1" AcceptedAnswerId="19204" CreationDate="2016-11-27T08:46:27.780" Score="2" ViewCount="309" Body="&lt;p&gt;At the beginning of a video on youtube titled &quot;history of the world in 2 hours&quot;, it is said that &quot;every thing begins with this tiny bundle of energy smaller than an atom&quot;. I want to know what it means.  atom is matter.  energy is different from matter and is measured in joules where as size is measured in volume units. How can one say energy smaller than an atom?&lt;/p&gt;&#xA;" OwnerUserId="14975" LastEditorUserId="14975" LastEditDate="2016-11-27T09:56:43.643" LastActivityDate="2016-11-28T16:10:46.040" Title="What is the meaning tiny bundle of energy smaller than an atom" Tags="&lt;early-universe&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19204" PostTypeId="2" ParentId="19203" CreationDate="2016-11-27T11:40:33.227" Score="5" Body="&lt;p&gt;What the video is clumsily alluding to is that the universe was very much smaller than it is today. However, if the universe is infinite, then it was still infinite at all times in the past. All one can say is that the density was much higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes commentators are more precise. One can define a size scale for the currently observable universe and ask how that volume of space evolved with time. This of course assumes an accurate knowledge of cosmology and the relevant cosmological parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ball park number (see &lt;a href=&quot;https://physics.stackexchange.com/questions/32917/size-of-universe-after-inflation&quot;&gt;here&lt;/a&gt;) is that the universe has got bigger by a factor of $10^{30}$ since the end of the inflationary epoch (at $t \sim 10^{-32}$ s; which is not fully understood yet). Thus the current observable universe, of size $10^{27}$ m, was contained within a diameter of 1 mm at the end of the inflationary epoch. Prior to the inflationary epoch it must have been compacted into a &lt;em&gt;much&lt;/em&gt; smaller volume than this - and I think this is the origin of the statement in your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that your source talks about energy is because that is in what form the density of the universe was at that epoch. It was only later that this energy started to &quot;freeze out&quot; into various forms of matter. It is now commonly supposed that the &lt;a href=&quot;https://en.wikipedia.org/wiki/Higgs_mechanism#Consequences_for_fermions&quot; rel=&quot;nofollow noreferrer&quot;&gt;Higgs mechanism&lt;/a&gt; or some form of it, is what gives elementary particles the property of a rest mass that might serve as a definition of &quot;matter&quot;. The Higgs mechanism kicks in at the epoch of &lt;a href=&quot;https://en.wikipedia.org/wiki/Spontaneous_symmetry_breaking#Higgs_mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;electroweak symmetry breaking&lt;/a&gt; (at temperatures of about $10^{15}$ K), giving paticles mass. Prior to this the density of the universe was in the form of massless entities that you might refer to as &quot;energy&quot;. This symmetry breaking occurs &lt;em&gt;after&lt;/em&gt; the end of the inflationary epoch at $t \sim 10^{-12}$s, and therefore when the currently observable universe was larger than 1 mm. So prior to this, and certainly when the currently observable universe was &quot;smaller than an atom&quot;, the contents of the universe could be described in term of energy (without rest mass).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-11-28T16:10:46.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19205" PostTypeId="2" ParentId="19155" CreationDate="2016-11-27T15:35:06.533" Score="4" Body="&lt;p&gt;This is not a characteristic of the solar system. It is a characteristic of the definitions of the names you used. Neptune and Uranus are the bodies you believe to be missing. In fact, with the mass of Earth at 6*10^24 kg, Uranus at 9*10^25 kg and Jupiter 2*10^27 kg, you'll notice that Uranus is only ~15 times the mass of Earth while Jupiter is ~20 times the mass of Uranus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial growth rate of planets was limited by how much solid dust there was that could clump together to form them. Planets that formed in colder conditions outside the ice line were able to grow faster because of ices (especially water but also ammonia, etc. depending on how far out) that were prevalent in those parts of the nebula from which the solar system formed. Once they reached a critical mass and were able to hold on to hydrogen and helium gas gravitationally, they would grow much faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter grew fast due to its location just outside the ice line, where it had a comparatively dense supply of ice, and could then quickly gobble up a large supply of gas. Neptune and Uranus did not grow as quickly and had only collected a small amount of gas by the time the solar wind blew the nebula away. This is why they are composed of a much larger percentage ices than Jupiter and Saturn; in fact they are often called the &quot;ice giants&quot; which is the correct intermediate classification. (The &quot;ice&quot; in &quot;ice giant&quot; refers to the substance, which is of course no longer generally solid in the atmospheres of these two bodies.)&lt;/p&gt;&#xA;" OwnerUserId="14969" LastActivityDate="2016-11-27T15:35:06.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19207" PostTypeId="2" ParentId="19202" CreationDate="2016-11-27T20:09:52.727" Score="2" Body="&lt;p&gt;I think all this means is that up to some critical rotation rate, the equilibrium shape of a rotating, self-gravitating fluid is an oblate spheroid. That is, this shape defines a unique global minimum of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above this threshold there are two possible equilibrium solutions. Apparently one is an oblate spheroid, whereas the other is a triaxial ellipsoid. These will define &lt;em&gt;local&lt;/em&gt; energy minima, but the triaxial ellipsoid is the global minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gory mathematical details are no doubt contained in Chandrasekhar's textbook &lt;a href=&quot;http://ukads.nottingham.ac.uk/abs/1969efe..book.....C&quot; rel=&quot;nofollow noreferrer&quot;&gt;on equilibrium ellipsoidal figures&lt;/a&gt;. Pages 3-5 of &lt;a href=&quot;https://arxiv.org/abs/1409.3858&quot; rel=&quot;nofollow noreferrer&quot;&gt;Iurato (2014)&lt;/a&gt; appear to summarise the historical development of this bifurcation in the equilibrium solutions.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-11-27T20:18:13.770" LastActivityDate="2016-11-27T20:18:13.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19209" PostTypeId="2" ParentId="19197" CreationDate="2016-11-28T02:12:38.377" Score="1" Body="&lt;p&gt;The sections of your book are brief and I understand your confusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most important aspect is the difference between the 3 kinds of moons.   Impact moons, Moons that formed with the planet at the time of formation and captured moons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Impact moons&lt;/strong&gt; (like our moon, or Pluto's Charon) can be quite large compared to the planet.   Charon at 1/11th Pluto's mass and our moon at 1/81st the Earth's mass are the 2 largest ratios of satellite to central object in our solar-system (ignoring small binary meteor/asteroid/comet systems).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gas giant planets are unlikely to have impact moons because of their gaseous atmosphere.  It's much less likely to blow debris off a gaseous atmosphere in such a way as it forms into a new moon.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moons that form with the planet (&lt;strong&gt;Formation moons&lt;/strong&gt;) - I'm not sure that's a term or not, but I like the term.    Many of the Moons in our solar system likely formed with the planet.   Jupiter's Galilean Moons, Many of Saturn's and Uranus' Moons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nature of a spiraling disk of matter is that it all doesn't end up in the center, but a percentage of the material, due to angular momentum ends up in far smaller, separate units orbiting around the center-mass.   Explaining why this happens is very mathematical and above my ability, but all coalescing bodies, whether stars or planets, have an angular momentum.  (see fun video &lt;a href=&quot;https://www.youtube.com/watch?v=tmNXKqeUtJM&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;).   Not all the material gets caught in the center of mass, but some form into orbiting satellites.  The ratio of mass between planet and moon of formation tends to be very planet dominant, hundreds or thousands to one.   &quot;Formation&quot; moons need to orbit in the same direction that the planet rotates, though, theoretically, a giant impact could reverse a planet's rotation, so there may be rare cases where the planet rotates in the opposite direction, but in that case, the Moons would still orbit in the &quot;right&quot; way.   Wrong way moons are rare and are thought to only be possible via capture (or giant impact), not formation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Captured Moons&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's difficult for a planet to capture a moon.  It might seem easy that a massive body would capture moons left and right but the mathematics of gravity doesn't work that way.   As an object falls towards a massive body like a planet it gains velocity and the gained velocity helps it accelerate away from the planet.   In a simple 2 body Newtonian system, capturing a body into orbit isn't possible.  The object is either already in orbit, or it can't be captured because there's no force slowing it down.   A capture is only possible if there's 3 bodies and 2 of the orbiting bodies interact, making the capture possible, or if there's some kind of atmospheric drag or orbital debris that slows down the new object as it passes by the planet.   There's also, I suppose, relativistic effects that make capture possible, but lets not go there.   None of that is really important.  What is important is that capture of an asteroid is very difficult and it happens rarely and when capture happens, it's usually with a very eccentric orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Triton, because it's a wrong-way moon, capture is the most likely scenario because a formation moon can't be a wrong-way moon and an impact moon, like a formation moon, also needs to orbit in the direction that the planet rotates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Triton was captured (most likely scenario), it's relatively high mass to most moons in the solar system would make it dominant and as it's highly eccentric orbit was circularized over time due to tidal forces with Neptune.  It now has a nearly circular orbit, but it's wrong way orbit and it's eccentric to circular orbital change over time, wrecked havoc on the other moons of Neptune and cast out a lot of the.   It's the reason Neptune has comparatively few moons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any system with multiple orbitals around a central body, there's greater stability if all the larger orbitals move in the same direction and in roughly the same plane and if they are mostly circular.   A large body (Triton is very large for a moon), orbiting in the wrong direction with presumably a very elliptical orbit would cause chaos among the Moons around the planet.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another property of a highly elliptical orbit is strong tidal variation that stretch and squeeze the moon, heating it's insides.  Triton no longer has those tidal forces because it's orbit is nearly circular, but when it was captured and it had to have a very elliptic orbit, it would have experienced tidal heating and volcanic eruptions as a result.   Over time as it's orbit circularized this tidal heating decreased to the point now where it's very low.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth pointing out that it might be difficult to tell if a moon is captured or was there at formation if it orbits in the expected direction due to orbits circularizing over time.  Triton's wrong way motion is strong evidence that it was captured.  Mars' 2 tiny moons are thought to be captured and several of the smaller gas giant moons were probably captured.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the basic explanation anyway.   Hope that helps &amp;amp; corrections are welcome if I made any errors.   I'll try to add some links later to make this answer a bit better and clear up some of the wordiness.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-11-29T03:02:01.207" LastActivityDate="2016-11-29T03:02:01.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19210" PostTypeId="2" ParentId="7891" CreationDate="2016-11-28T04:58:16.577" Score="2" Body="&lt;p&gt;But note that high-mass stars begin helium fusion when the electrons are still an ideal gas, but lower-mass stars like the Sun begin helium fusion when the electrons are degenerate.  So that produces a &quot;helium flash,&quot; which the question is not about, but after the flash stabilizes (when the electrons are once again acting like an ideal gas), you get normal helium fusion in low-mass stars.  Typically, the density will be much higher in a low-mass star, perhaps an order of magnitude higher than what is quoted for the high-mass stars.  So the stellar mass will matter in answering the question, but you get the basic idea.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-11-28T04:58:16.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19211" PostTypeId="2" ParentId="7891" CreationDate="2016-11-28T07:44:16.860" Score="3" Body="&lt;p&gt;Your question inspired me to have a look at the data that is used in plots produced by the stellar evolution code &lt;a href=&quot;http://mesa.sourceforge.net&quot; rel=&quot;nofollow noreferrer&quot;&gt;MESA&lt;/a&gt;. Here's a $\rho$-$T$ diagram showing boundaries of roughly where various nuclear reaction chains kick in as well as a solar model (black dashed line) and the a line indicating the onset of electron degeneracy (red line).&#xA;&lt;a href=&quot;https://i.stack.imgur.com/suVDn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/suVDn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;To answer the question, the point at which the &quot;He burn&quot; line intersects $10^8\,\mathrm{K}$ is around $4000\,\mathrm{g.cm}^{-3}$ but you can also see how the boundary varies somewhat as a function of temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a disclaimer, I don't actually know where these numbers come from but I'm asking around about it. If I find out, I'll update this answer.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2016-11-28T07:44:16.860" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19212" PostTypeId="2" ParentId="8102" CreationDate="2016-11-28T07:50:26.827" Score="1" Body="&lt;p&gt;The source &lt;em&gt;is&lt;/em&gt; Sco X-1. It is a point source of X-rays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll have get back to you on why it looks that way in chromoscope, but it looks incorrect to me.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-11-28T07:55:39.183" LastActivityDate="2016-11-28T07:55:39.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19213" PostTypeId="1" AcceptedAnswerId="19238" CreationDate="2016-11-28T09:03:41.617" Score="1" ViewCount="117" Body="&lt;p&gt;I've asked earlierin this &lt;a href=&quot;https://astronomy.stackexchange.com/q/16262/7982&quot;&gt;unanswered question&lt;/a&gt; about an &quot;extinct species&quot; of meteorite, now I've just read in Phys.org about Barbarian asteroids:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Named after the first asteroid of this type discovered, Barbara (234), Barbarians are a key to understanding how the solar system first formed. Barbarians are extremely rare and ancient, and were present before the Earth was created. Only 13 Barbarians have ever been discovered.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Are these two different types of extremely rare materials, or are the &quot;extinct&quot; meteorites actually products of the &quot;rare Barbarian&quot; asteroids?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From: &lt;a href=&quot;http://m.phys.org/news/2016-11-uwa-zadko-telescope-reconstruct-barbarian.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;UWA Zadko Telescope helps reconstruct 'Barbarian' asteroids&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-11-29T19:28:43.017" Title="Are the &quot;extinct species&quot; of meteorites originally from the &quot;Barbarian&quot; asteroids?" Tags="&lt;solar-system&gt;&lt;asteroids&gt;&lt;planetary-formation&gt;&lt;meteorite&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19214" PostTypeId="1" CreationDate="2016-11-28T13:22:41.593" Score="0" ViewCount="201" Body="&lt;p&gt;Maybe this sounds like a silly question (and hopefully is legitimate), but I will elaborate a little more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I seem to love astronomy, every single thing about it, I find it brilliant, interesting, mysterious, insanely amazing and mind boggling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is only so many documentaries you can watch, read articles on the web etc before you start to see repetition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just wondered if there was some advice of what I could do to keep that interesting fire burning...because I absolutely love it!! &lt;/p&gt;&#xA;" OwnerUserId="14060" LastEditorUserId="14060" LastEditDate="2016-11-28T13:51:54.530" LastActivityDate="2016-11-28T18:18:53.393" Title="What can the average Astronomy Enthusiast do to learn more" Tags="&lt;planet&gt;&lt;the-sun&gt;&lt;solar-system&gt;&lt;black-hole&gt;&lt;galaxy&gt;" AnswerCount="3" CommentCount="4" ClosedDate="2016-11-29T20:25:22.310" ContentLicense="CC BY-SA 3.0" />
  <row Id="19216" PostTypeId="1" AcceptedAnswerId="19273" CreationDate="2016-11-28T14:39:28.777" Score="6" ViewCount="632" Body="&lt;p&gt;What is the density difference in our galaxy (or in a typical spiral),&lt;/p&gt;&#xA;&lt;p&gt;at A, B, C versus at a, b, c?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pA2B1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pA2B1.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Consider the yellow track.&lt;/p&gt;&#xA;&lt;p&gt;Do astronomers have a graph like this...&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FisGn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FisGn.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;...which shows the shape of the density through and between the arms?&lt;/p&gt;&#xA;&lt;p&gt;In short,&lt;/p&gt;&#xA;&lt;h1&gt;Do we roughly know the density between-arms versus in-arms ... is it 10%, 50% or 90%?&lt;/h1&gt;&#xA;&lt;p&gt;(Perhaps at the &amp;quot;corotation radius&amp;quot;, which is about where we are.)&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Further issues arising:&lt;/p&gt;&#xA;&lt;p&gt;= I understand that one of the reasons the arms &amp;quot;appear real&amp;quot; is just that there are more bright young stars forming in the arms (due to the higher pressure). Perhaps the actual density difference is very low?&lt;/p&gt;&#xA;&lt;p&gt;= Perhaps the density variation is dependent on the type of star, or other material - what about things like gasses?&lt;/p&gt;&#xA;&lt;p&gt;= Let's say the percentage I am asking about is X% - at our radius. Is it X% at all radii?  How does the &amp;quot;strengthiness&amp;quot; percentage vary with radius? This is not immediately intuitive from looking at images.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Milky_Way&quot; rel=&quot;nofollow noreferrer&quot;&gt;(original galaxy arm pattern diagram courtesy wikipedia)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-12-03T18:17:13.073" Title="Star density in- versus out- of the arms of the galactic pattern?" Tags="&lt;galaxy&gt;&lt;galactic-dynamics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19217" PostTypeId="2" ParentId="19191" CreationDate="2016-11-28T14:53:10.963" Score="2" Body="&lt;p&gt;To understand what's intended, you need to be comfortable with the concept of electromagnetic fields, and the difference between field strength and field gradients.  I think the author is trying to describe the collapse of a magnetic field over a large area into a field over a very small area but with steeper gradients.  At some point in this process, the field gives up energy (which means photons are generated), some of which forces physical particles free of the sun's gravity well. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2016-11-28T14:53:10.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19218" PostTypeId="2" ParentId="19214" CreationDate="2016-11-28T17:49:34.640" Score="0" Body="&lt;p&gt;If you want a sip from the fire hose, try reading a few preprints via &lt;a href=&quot;https://arxiver.wordpress.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;arXiver&lt;/a&gt;.&#xA;Some of those papers get published and some don't; some will interest you personally and some won't.&#xA;If something goes over your head at first, look it up, and if it still doesn't make sense, ask about it here.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-11-28T17:49:34.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19219" PostTypeId="2" ParentId="19214" CreationDate="2016-11-28T17:55:30.500" Score="2" Body="&lt;p&gt;I recommend that you learn your way around the night sky, so start by buying a book of star charts.  Next try a book that covers interesting objects in each constellation, such as Burnham's Celestial Handbook, or The Constellations by Motz &amp;amp; Nathanson.  This should keep you busy for a while.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2016-11-28T17:55:30.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19220" PostTypeId="1" CreationDate="2016-11-28T18:06:07.763" Score="14" ViewCount="1432" Body="&lt;p&gt;Here is a graphic of cloud layers of Jupiter (&lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Jupiter#/media/File:Structure_of_Jovian_atmosphere.png&quot; rel=&quot;noreferrer&quot;&gt;source: Wikipedia&lt;/a&gt;):&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UdH0A.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UdH0A.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;There are three distinct cloud layers of ammonia, ammonium hydrosulfide, and water. The temperature and pressure conditions seem to be surprisingly earth-like; temperatures between 200 and 300 K, pressures about 1 to 10 atm, gravity around 1.3g.&lt;/p&gt;&#xA;&lt;p&gt;Clouds (of water) form on earth because they solar energy causes them to evaporate from a solid surface, rise a few km, then condense to form water droplets (or solid crystal snow). But Jupiter does not have a solid surface, or nearly as much solar energy as the Earth.&lt;/p&gt;&#xA;&lt;p&gt;All three of the cloud forming compounds should be liquid at the conditions of their cloud layer. Given the density of those liquids (between 0.7 and 1.2 g/cm&lt;span class=&quot;math-container&quot;&gt;$^3$&lt;/span&gt;) and the density of the bulk of the hydrogen and helium atmosphere, &lt;strong&gt;how do the clouds not fall as precipitation into Jupiter's interior and never re-emerge?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="14989" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-09-10T11:18:07.297" Title="How can clouds form in Jupiter's atmosphere of Hydrogen and Helium?" Tags="&lt;planet&gt;&lt;solar-system&gt;&lt;jupiter&gt;&lt;planetary-atmosphere&gt;&lt;gas-giants&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="19221" PostTypeId="2" ParentId="19214" CreationDate="2016-11-28T18:18:53.393" Score="2" Body="&lt;p&gt;You could take a more serious course. Formal education is always an option if you have the time and money and there are many good MOOCs in astronomy. For example, &lt;a href=&quot;https://extend.duke.edu/courses/course-v1:DukeExtend+introastro+2016-2017/about&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Introduction to Astronomy&lt;/a&gt; MOOC started today.&lt;/p&gt;&#xA;" OwnerUserId="14969" LastActivityDate="2016-11-28T18:18:53.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19222" PostTypeId="1" AcceptedAnswerId="19234" CreationDate="2016-11-28T21:06:26.327" Score="1" ViewCount="551" Body="&lt;p&gt;Suppose I want to calculate the diameter of the crater. I apply the small angle formula $$diameter=\frac{\theta_{diameter}}{206265''}\times D$$ where $\theta_{diameter}$ is the angular size of the crater, in arcseconds, and $D$ is the average surface-to-surface distance from the Earth to the Moon, which is around $375,900$ km. We know that the moon is about $\frac12$ degree, which is $1860$ arcseconds. But how do we estimate the angular size of the crater without using lens, with only  eyes? I was thinking using ruler and applying scale but that way, I think, is inaccurate because the Moon has a round surface. &lt;/p&gt;&#xA;" OwnerUserId="14254" LastEditorUserId="11143" LastEditDate="2016-11-29T16:12:22.527" LastActivityDate="2016-11-29T16:12:22.527" Title="How do we get the angular size of a crater without lens?" Tags="&lt;the-moon&gt;&lt;size&gt;&lt;surface&gt;&lt;angular-diameter&gt;&lt;crater&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19223" PostTypeId="1" AcceptedAnswerId="19236" CreationDate="2016-11-28T23:15:35.120" Score="3" ViewCount="995" Body="&lt;p&gt;In their 1966 book, &lt;em&gt;Intelligent Life in the Universe&lt;/em&gt;, Carl Sagan and Iosif Shklovsky state that &quot;a natural satellite cannot be a hollow object.&quot; How do we know this is absolutely true?&lt;/p&gt;&#xA;" OwnerUserId="14994" LastActivityDate="2016-11-29T17:03:04.857" Title="How do we know that a natural satellite cannot be a hollow object?" Tags="&lt;natural-satellites&gt;&lt;the-moon&gt;&lt;artificial-satellite&gt;&lt;phobos&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19225" PostTypeId="1" AcceptedAnswerId="19227" CreationDate="2016-11-29T02:23:42.993" Score="2" ViewCount="132" Body="&lt;p&gt;Do they expect to see wave modulations like the ones we use in our technology? what do they expect the waves to be like exactly?&lt;/p&gt;&#xA;" OwnerUserId="11272" LastActivityDate="2016-11-29T11:28:58.243" Title="How do astronomers at SETI know if they received intelligent radio signals from other planets?" Tags="&lt;radio-astronomy&gt;&lt;observational-astronomy&gt;&lt;radio-telescope&gt;&lt;radio&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19226" PostTypeId="2" ParentId="19220" CreationDate="2016-11-29T06:46:07.373" Score="10" Body="&lt;p&gt;First, it's a great question.  Mostly the answer is straight forward, so I can answer it, but it's still a great question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I'll add a similar, but slightly more detailed picture to the one you posted.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FES8t.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FES8t.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.windows2universe.org/jupiter/atmosphere/J_clouds_decks.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're right that there is a clear difference between Earth's surface where liquid water can exist, evaporate, make clouds, rain and repeat.   Earth's water cycle in theory, could go on indefinitely so long as Earth's atmosphere and the solar input are maintained (and lost hydrogen is replaced), but it's a circular system that only needs solar input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter is different because over time, the heavier gases in Jupiter probably will sink deeper towards the center and Jupiter's cloud forming gases should decrease given enough time.   Some of Jupiter's &quot;rain&quot; probably does fall too deep in its swirling mix of gases, and leaves Jupiter's cloud cycle permanently, similar to water seeping underground and leaving Earth's water-cloud cycle.    So, in 100 billion or a trillion years or so, Jupiter could lose its clouds and cloud forming gases in its upper atmosphere for the reasons you suspect.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this hasn't happened yet is simply mixing.  While density of gas tends towards layers of increasing density, the internal heat inside Jupiter also wants to even out, so there's enormous convection going on pretty much all the way through the planet.   This keeps some heavier gases in Jupiter's upper atmosphere.   Jupiter's much too turbulent to have just hydrogen and helium in its upper atmosphere.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, once we begin with the observation that Jupiter's upper atmosphere is (about) 90% hydrogen, 9% helium, 1% other gases and the mixing maintains the 1% of other, after that it's just &lt;a href=&quot;https://en.wikipedia.org/wiki/Cloud_physics&quot; rel=&quot;nofollow noreferrer&quot;&gt;cloud physics&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clouds look like puffy collections of water vapor (tiny droplets of ice or water, as water vapor is actually transparent).  They look like objects with shapes, but that's not entirely accurate.   If you are up close to a cloud (flying in a plane for example), the clear edges disappear.  A cloud isn't an object so much, it's a visible phase change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atmosphere on Earth is about 78% nitrogen, 21% oxygen, 0.9% argon and (usually not listed because it's so variable), about 0.4% water vapor on average, as high as 1% with high temperature and high humidity and close to 0% in cold temperatures or dry-deserts.   When you take warm surface air that's 0.6-0.8% water vapor, and that air rises (as hot air does), it's the phase change that creates the clouds.   The cloud forms in the hot rising air as it cools.  There's some electrostatic attraction, but mostly it's just a block of similar air undergoing cooling and the cloud looks like it has solid edges, but it doesn't.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same exact thing happens on Jupiter, different gases phase change at different temperatures/pressures, but the process is the same.  And, just as on Earth, once the droplets or &quot;icelets&quot; form, they are denser and they begin to fall, but falling droplets are very small so they fall very slowly and for the most part, they are falling through rising atmosphere.  Also, as they are a phase change, new cloud is being formed and old cloud is being disbursed or returned to gas all the time, kind of like sea ice.   The clouds have the appearance of semi-permanence, but clouds are dynamic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my explanation doesn't work for you, here's an &lt;a href=&quot;https://www.reddit.com/r/askscience/comments/h4rmp/what_holds_clouds_together/&quot; rel=&quot;nofollow noreferrer&quot;&gt;explanation&lt;/a&gt; on clouds and how they really aren't bound together even though they look that way.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that's the gist of it, mixing keeps the upper atmosphere of Jupiter from being pure hydrogen and helium (or pure hydrogen), and after that, the cloud formation is pretty much the same as on Earth, just without a surface.   Some of the heavier gases probably do get lost from the cycle, but the loss is slow enough that Jupiter still has some heavy clouds forming gases in its upper atmosphere and will for probably billions of years to come.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The greater density variation between H/He and other gases likely plays a role in how the clouds behave, as the density variation is greater, but the wind speeds are also higher on Jupiter.   All that's really needed is mixing.  After that, with gases that can become liquid or ice under temperature/pressure variations, the phase changes create the clouds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also possible that Jupiter's cloud forming gases, from time to time, are replenished by asteroid and comet impacts.   Shoemaker-Levy 9 was about 5 km in diameter and a fair percentage of that was probably ammonia and water ice.   That's a lot of cloud forming gas added to Jupiter's upper atmosphere.   Jupiter's faint ring system, which might have been much larger millions of years ago, but since rained onto Jupiter, and eruptions from Io might also play roles in keeping Jupiter's upper atmosphere rich enough in cloud making elements like water and ammonia.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="7472" LastEditDate="2018-09-10T11:18:07.297" LastActivityDate="2018-09-10T11:18:07.297" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="19227" PostTypeId="2" ParentId="19225" CreationDate="2016-11-29T08:55:10.350" Score="5" Body="&lt;p&gt;I'm not sure what you mean by &quot;wave modulations&quot;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Loosely speaking, there's 2 things to look for regarding alien intelligence/technology.   One is a signal of some kind and the other is a very large or very energetic structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most obvious place to look for a signal is in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Water_hole_(radio)&quot; rel=&quot;nofollow noreferrer&quot;&gt;water-hole&lt;/a&gt;.  Good article on that &lt;a href=&quot;http://astronomynow.com/news/n1004/26seti5/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.  Because of the relatively low interference in that bandwidth, it's a good region to send a signal from one solar-system and hope it gets picked up by another.   Stellar distances are so large that this still requires a very large output and very sensitive receiving devices.   People have been listening for water-hole signals for decades, with no luck, outside of the extremely inconclusive &lt;a href=&quot;https://en.wikipedia.org/wiki/Wow!_signal&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Wow&quot; signal&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thing they can look for is mega-structures but that's much harder.   We have a hard enough time seeing entire planets in other solar-systems and the idea of an alien technology building something as large as a planet, but distinguishable . . . a bit of a stretch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets are mostly those are observed by seeing wobbles or changes in the star's orbit or brightness as the planet passes in-front of them.   To see an alien structure and recognize it as something manufactured, not natural, it would need to be very large.  See &lt;a href=&quot;http://www.dailymail.co.uk/sciencetech/article-3726229/Alien-megastructure-mystery-deepens-Dyson-sphere-star-dimming-dramatically-knows-why.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;silly article&lt;/a&gt;.  Outside of a structure as large as a dyson-sphere, we don't have the means of seeing an alien structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, and to your question.  There is no technological &quot;wavelength&quot; to look for, light is light.  We transmit our cell-phone signals at about 2 gigahertz but an alien species might use 5 or 8.  There are ranges that are best for traveling between atmosphere and satellite and regions of comparatively low interference that are best for communication across a solar-system, but at the distance of several light-year's it's difficult to overcome the background radiation.   We could recognize an intelligence signal if we saw one.  Any information or language carrying signal would have a different pattern than natural signals, but the hard part is finding the signal in the first place, over such vast distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, Seti is limited to looking for the extremely large and so far, (at least as far as I know), nothing even remotely close to definitive has ever been observed.   They don't have the means to observe most technology, assuming there is technology to be seen.  Only the most enormous mega-structure could maybe be observed, and only under the right conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As our telescopes get better they might find something.  The atmospheric signature of a technological alien society might become visible to the &lt;a href=&quot;http://www.jwst.nasa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;James Webb telescope&lt;/a&gt;, but only for planets that transit their stars, not all planets and that might only offer possibilities, nothing definitive.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Signals like our television, cell phone and satellite broadcasts, those would be extremely hard to pick up from light-years away.  The signals are too weak and over light-years of distance, the background radiation of space would grow too loud by comparison.   Likewise the visible light from the night-side of Earth, while we can see that from the space station orbiting the Earth, observing that kind of light from light years away would require a huge telescope, many times larger than anything we have.   I don't have the knowledge to give you all the specifics on how hard to see or how big a telescope but much bigger than anything we have.   An Alien race with a telescope a few miles across might be able to observe the light from our cities on the night side of Earth.   We don't have the ability to observe anything like that ourselves.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2016-11-29T08:55:10.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19228" PostTypeId="2" ParentId="19170" CreationDate="2016-11-29T09:28:16.707" Score="2" Body="&lt;p&gt;If I understand your question correctly than your problem is that the primary direction of &lt;code&gt;vCR[3]&lt;/code&gt; is the intersection between the prime meridian and the Earth's equator, instead of the vernal equinox. The geographical latitude is the same as the declination, but to get the right ascension you will need to rotate your coordinates around the z-axis by an amount that is equal to the sidereal time at Greenwich ($\theta_G$). This is done, simply by replacing the the longitude (lon) with the sidereal time at your location $\theta = \theta_G-lon$.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x =  R* cos(lat) * cos(theta);&#xA;y=   R* cos(lat) * sin(theta);&#xA;z =  R* sin(lat);&#xA;&#xA;vCR[3]={x,y,z};&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where &lt;code&gt;theta&lt;/code&gt; is the sidereal time $\theta$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sidereal time is not a time but actually an angle that gives the right ascension of objects that are at that moment crossing the local meridian (i.e. the line between South and the Celestial north pole, for observers on the northern hemisphere). It is in a sense the longitude of your position on Earth if the geographical coordinate system were to rotate along with the celestial sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: To calculate the Sidereal time, you can use the method explained in &lt;a href=&quot;http://www.willbell.com/math/mc1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomical Algorithms by Jean Meeus&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T = \frac{JD-2451545.0}{3625}$$&#xA;$$\theta_0 = 100.46061837 + 36000.770053608 T + 0.000387933 T^2 - \frac{T^3}{38710000}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sidereal time is expressed in degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then express the UTC time as an angle: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t = 15h+0.25m+\frac{0.25}{60}s$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $h$ is the number of hours, $m$, the number of minutes, and $s$ the number of seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sidereal time at Greenwich at the specified time is then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta_G = \theta + 1.00273790935*t $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the local sidereal time is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta = \theta_G - L$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $L$ is the geographical longitude (positive west from Greenwich).&lt;/p&gt;&#xA;" OwnerUserId="864" LastEditorUserId="864" LastEditDate="2016-11-29T18:45:48.040" LastActivityDate="2016-11-29T18:45:48.040" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19229" PostTypeId="2" ParentId="19225" CreationDate="2016-11-29T11:28:58.243" Score="1" Body="&lt;p&gt;Well, they don't. You can only disprove that it's aliens.&lt;br&gt;&#xA;All remaining signals that are leftover after natural signals are substracted are simply of unknown origin.&lt;br&gt;&#xA;Then, depending how hard a particular scientist or journalist wants to sell out, he or she declares 'maybe aliens' are found. But it's not like we will ever be able to prove such a thing, unless we really get some crazy stuff like the Fibonacci sequence sent to us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true for photometric signals and the recent unscientific popular discussion about &lt;a href=&quot;https://en.wikipedia.org/wiki/KIC_8462852&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tabby's star&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2016-11-29T11:28:58.243" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19230" PostTypeId="1" AcceptedAnswerId="19510" CreationDate="2016-11-29T11:42:29.670" Score="2" ViewCount="521" Body="&lt;p&gt;Is an ice planet made only by ice? &#xA;What causes the water to conglomerate and form ice in a planet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if the planet is not made entirely by ice but has an high % of ice (e.g. 50% ice), how can it be explained?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I heard about ice planets with surface temperature way above the melting point.&#xA;Can it be explained with surface pressure?&#xA;&lt;strong&gt;Edit: This is probably wrong but feel free to point it out in a full answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="14936" LastEditorUserId="14936" LastEditDate="2016-11-29T15:09:06.867" LastActivityDate="2016-12-25T11:26:57.220" Title="Is an ice planet made only by Ice?" Tags="&lt;planet&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19234" PostTypeId="2" ParentId="19222" CreationDate="2016-11-29T15:49:30.657" Score="1" Body="&lt;p&gt;Most lunar craters are too small to resolve with the naked eye; I would measure a published photograph. A crater near the limb appears as an ellipse whose minor axis is foreshortened but whose major axis measures about the same as if the crater were centrally located. Surface curvature has little effect on this measurement because most craters are small compared to the lunar radius.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2016-11-29T15:49:30.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19236" PostTypeId="2" ParentId="19223" CreationDate="2016-11-29T17:03:04.857" Score="4" Body="&lt;p&gt;I'd caution that absolute statements are not good things in science. When Sagan says a natural satellite cannot be hallow, it is an absolute statement, but there's no way to 100% prove it true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, everything we know about planet/satellite formation tells us that the chances of a satellite forming as a hollow structure (or becoming hollow afterwards) is pretty much nil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is still much debate on the specifics of planetary formation, but the general process seems to be that small rocks and particles start to form, then they coalesce into bigger chunks and slowly that planetesimal grows in size. This type of formation process does not allow for the center to be hollow since it is the center which forms first! You cannot form the &quot;edges&quot; of your hollow planet and have those edges come together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moons and satellites can potentially have a different formation path. Look at our own moon for example. It is hypothesized that a Mars-sized proto-planet crashed into the Earth, flinging out a large amount of material. Much of this material then coalesced into our current Moon. Again, this process doesn't really allow for making something which is hollow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from the difficulty in forming a hollow satellite, you also have the challenge of maintaining the structure. Gravity constantly pulls everything to the center of the body. If there's nothing at the center, there's nothing preventing the outer shell from collapsing. It is technically possible that the satellite could not collapse, but that would require it to be fairly small and be perfectly structured such that the structural integrity of the rock forming the shell could withstand the crushing gravitational forces. Really though, such an object would have to be intelligently designed as the chances of it forming perfectly like that are non-existent. There's no possible way an object like Phobos could maintain itself without collapsing if it were hollow.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-11-29T17:03:04.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19237" PostTypeId="1" CreationDate="2016-11-29T18:57:05.113" Score="0" ViewCount="995" Body="&lt;p&gt;Because a star's temperature is an indication of the energy passing through each unit of its surface, it follows that for cool red and hot blue stars of identical luminosities, the cooler red star must be considerably smaller so that the energy escapes through a reduced surface area and has a far greater &lt;strong&gt;&lt;em&gt;heating effect&lt;/strong&gt;.&lt;/em&gt; Hence, knowledge of a star's colour and luminosity can reveal its size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, I can't understand the relation between energy and the heating effect. Of course, I searched for the word &quot;the heating effect&quot; in my own language. Also, I Googled in English. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this context, what is the heating effect? Could you explain to me? If you give me any help, it will be very useful to me.  &lt;/p&gt;&#xA;" OwnerUserId="14782" LastEditorUserId="2153" LastEditDate="2016-11-29T20:23:53.850" LastActivityDate="2018-02-13T15:59:46.120" Title="The relation between the energy of star and heating effect" Tags="&lt;star&gt;&lt;temperature&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19238" PostTypeId="2" ParentId="19213" CreationDate="2016-11-29T19:28:43.017" Score="2" Body="&lt;p&gt;The answer is in the part of the abstract that you quote in your other question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;...This may be the first documented example of an ‘extinct’ meteorite, that is, a meteorite type that does not fall on Earth today because its parent body has been consumed by collisions.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What they're saying here is that there used to exist some asteroids of some particular type (here they mean formed from the pieces of a single body). After some time all asteroids of that type were destroyed in collisions, fell on planets or into the sun or were ejected from the solar system. Some of the pieces fell on Earth. As far as we can tell there are no pieces left in solar orbit and so none of these type of meteorite fall on Earth any more, hence it is called &quot;extinct&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Barbarian asteroids currently exist, they are by definition not extinct.&lt;/p&gt;&#xA;" OwnerUserId="14969" LastActivityDate="2016-11-29T19:28:43.017" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19239" PostTypeId="2" ParentId="19237" CreationDate="2016-11-29T20:16:21.060" Score="5" Body="&lt;p&gt;I'm not sure what you mean by &quot;heating effect&quot;. The amount another body is heated by a star with luminosity $L$ is quantified by its &lt;a href=&quot;https://en.wikipedia.org/wiki/Effective_temperature&quot; rel=&quot;nofollow noreferrer&quot;&gt;effective temperature&lt;/a&gt;:&#xA;$$T_{eff}=\left(\frac{L(1-a)}{16\pi\sigma D^2}\right)^{1/4}$$&#xA;where $a$ is the albedo, $D$ is the distance to the other body, and $\sigma$ is the Stefan-Boltzmann constant. If the luminosity is given, then there is no direct dependence on the size of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing that matters is the luminosity. You stated yourself that the luminosities are the same, which means that the red star in question is likely a &lt;a href=&quot;https://en.wikipedia.org/wiki/Red_supergiant&quot; rel=&quot;nofollow noreferrer&quot;&gt;red supergiant&lt;/a&gt; - a member of a group containing many of the largest stars in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why, mathematically, is this the case? Well, stellar models can tell us that this is the case, but we can also figure this out via the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stefan-Boltzmann law&lt;/a&gt; by assuming that the stars are &lt;a href=&quot;https://en.wikipedia.org/wiki/Black_body&quot; rel=&quot;nofollow noreferrer&quot;&gt;black bodies&lt;/a&gt;. For an object of radius $R$ and temperature $T$, the luminosity is approximately&#xA;$$L=4\pi\sigma R^2T^4$$&#xA;If star 1 is a blue supergiant and star 2 is a red supergiant, then, setting the luminosities equal, we have&#xA;$$L_1=L_2\to R_1^2T_1^4=R_2^2T_2^4$$&#xA;We know that $T_2$ is much less than $T_1$ - possibly by an order of magnitude - so $R_2$ must be much larger for the luminosities to be the same. This matches what we observe and what models predict.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2016-12-10T15:29:44.563" LastActivityDate="2016-12-10T15:29:44.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19240" PostTypeId="2" ParentId="19220" CreationDate="2016-11-29T20:21:13.173" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How do the clouds not fall as precipitation into Jupiter's interior and never re-emerge?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gases in a planet's troposphere don't differentiate chemically; the turbulence driven by heating and planet rotation keep the atmosphere well-mixed. We can see this in our own atmosphere. Carbon dioxide and argon are considerably more dense than are the nitrogen and oxygen that form the bulk of the atmosphere. Yet we don't have a layer of carbon dioxide at the bottom of the atmosphere. The turbopause marks where an atmosphere shifts from being dominated by turbulent mixing to being dominated by diffusion. Chemical differentiation by atomic mass does occur above the turbopause, but even there, it's gradual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about rain? The answer is simple: It evaporates. That happens here on Earth, particularly in arid regions. Clouds form, and rain falls from those clouds, but the rain sometimes evaporates before it reaches the ground. This is called virga.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Temperatures rise inside Jupiter due to compressional heating, at a rate of about 1.85 K per kilometer of increasing depth. That means the temperature reaches the water's critical temperature (647 K) about 240 kilometers below the 1 bar pressure level. So even if rain water could fall that far as rain before evaporating (which is dubious), it would cease to be a liquid.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2016-11-29T20:21:13.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19241" PostTypeId="2" ParentId="2092" CreationDate="2016-11-30T05:11:15.763" Score="3" Body="&lt;p&gt;It collapses onto itself, becoming more spherical. This process is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_collapse&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gravitational collapse&lt;/a&gt;, and for a cloud of dust, will happen when the dust cloud is greater than the &lt;a href=&quot;https://en.wikipedia.org/wiki/Jeans_mass&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jeans mass&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Megan Whewell, Education Team Presenter for the National Space Centre, &lt;a href=&quot;https://www.spaceanswers.com/deep-space/what-is-the-minimum-size-a-celestial-body-can-become-a-sphere/&quot; rel=&quot;nofollow noreferrer&quot;&gt;writes of other radiuses&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[F]or bodies made mainly of rock, the minimum size to become a self-gravitating sphere is about 600km diameter; but, for bodies mainly made of ice, the minimum size is about 400km diameter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously, some level of collapse can happen before that point, non-solid objects would need to be larger, and objects made out of stronger stuff like steel will also need to be larger, but that gives some idea of the necessary scale for solids at least.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It’s worth noting that a hollow sphere won’t collapse under its own gravity, because the net force of gravity at any point inside a sphere is zero, the greater mass of the more distant part of the shell counteracting the lesser mass of the closer part.&lt;/p&gt;&#xA;" OwnerUserId="6600" LastActivityDate="2016-11-30T05:11:15.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19243" PostTypeId="1" AcceptedAnswerId="19252" CreationDate="2016-11-30T08:56:52.500" Score="5" ViewCount="132" Body="&lt;p&gt;Is it possible to find out how old is a planet, especially &lt;a href=&quot;https://en.wikipedia.org/wiki/Rogue_planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;a rogue planet&lt;/a&gt;? I know that people measured the radioactive decays to determine Earth's age with some amazing accuracy, but what about interstellar planets be, they gas or terrestrial?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="2153" LastEditDate="2016-11-30T15:42:23.873" LastActivityDate="2016-11-30T20:11:04.173" Title="How can we tell the age of a rogue planet?" Tags="&lt;gas-giants&gt;&lt;rogue-planet&gt;&lt;terrestrial-planets&gt;&lt;age&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19244" PostTypeId="1" AcceptedAnswerId="19461" CreationDate="2016-11-30T09:57:16.423" Score="0" ViewCount="922" Body="&lt;p&gt;I want to see Mars polar caps by myself. Is it feasible to make a picture of mars like this&lt;a href=&quot;https://i.stack.imgur.com/osxMN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/osxMN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; at home or to achive this very expensive telescope is required?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Maybe someone can give me a link, where it is shown how polar caps are changing during the seasons? Many pictures or animation.&lt;/p&gt;&#xA;" OwnerUserId="15004" LastActivityDate="2016-12-22T10:46:13.043" Title="Question about mars polar caps" Tags="&lt;telescope&gt;&lt;mars&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19245" PostTypeId="1" CreationDate="2016-11-30T13:23:33.200" Score="2" ViewCount="1463" Body="&lt;p&gt;I've always wondered how our planet Earth or any other large spheroid-shaped thing in our universe has gravity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it have a small black hole on its core causing it to have gravitational force?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take our planet as an example: what causes it to have gravity?&lt;/p&gt;&#xA;" OwnerUserId="15007" LastEditorUserId="6" LastEditDate="2016-11-30T16:42:27.633" LastActivityDate="2021-05-21T19:42:50.477" Title="What causes the Earth to have gravity?" Tags="&lt;gravity&gt;&lt;earth&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19246" PostTypeId="2" ParentId="19245" CreationDate="2016-11-30T13:44:24.300" Score="8" Body="&lt;p&gt;Any massive body (i.e., a body having mass) like the stars, planets, moons, and any atom in the universe have, what is called, a gravitational potential which results in its gravity.  This is described by &lt;a href=&quot;https://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation&quot; rel=&quot;noreferrer&quot;&gt;Newton's Law of Gravity&lt;/a&gt;. &lt;em&gt;Why&lt;/em&gt; masses cause gravity, we do not know.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in short, the earth's mass causes it to have gravity.&lt;/p&gt;&#xA;" OwnerUserId="5537" LastEditorUserId="1203" LastEditDate="2016-11-30T13:59:53.913" LastActivityDate="2016-11-30T13:59:53.913" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19248" PostTypeId="1" AcceptedAnswerId="19250" CreationDate="2016-11-30T14:22:57.920" Score="12" ViewCount="7831" Body="&lt;p&gt;I have a general understanding of how and why a body in space can be tidally locked to it's planet or sun and I'm aware that our moon is in such a state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, if our moon once had a rotation, has it slowed down to a point where across the course of our normal lifetimes we can't observe it's spin but across the span of hundreds, if not thousands, of years we could observe its current rotation?  In other words, if it has a rotation and if I looked at the moon today and jumped into the future, how far would I have to go to see a noticeable difference?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the question is to relative then I'll ask it like this-- at the moons current rate of rotational speed and deceleration, how long would it take for the moon to rotate, say, 15 degrees on it's current axis?  I assume this would be enough to make the moon look &quot;different&quot; to the naked eye so I'll go with that concrete figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...or is the moon in either a state of near-equilibrium or is it &quot;wobbling&quot; due to other forces outside the gravitational pull of the earth (e.g. pull of the sun, asteroid bombardment, comets passing by, etc.) having minor effects on its rotational state and thus making it's rotation inconsistent and unobservable?&lt;/p&gt;&#xA;" OwnerUserId="9434" LastActivityDate="2020-12-12T18:41:32.317" Title="Is the moon &quot;perfectly&quot; tidally locked and, if not, how long would it take us to observe it's rotation?" Tags="&lt;the-moon&gt;&lt;tidal-locking&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19250" PostTypeId="2" ParentId="19248" CreationDate="2016-11-30T14:53:13.713" Score="11" Body="&lt;p&gt;The question is interesting, but I suspect the answer is that the Moon will never show its &amp;quot;far side&amp;quot; to the Earth, because there are differences between the side that faces us and the far side that suggest there is something quite permanent about its orientation.  So while the rotation was locking, it either settled into a state of minimum potential that it has maintained ever since, or it built up the differences between the side we see and the side we don't after it tidally locked, but either way, this means it has showed us the same side for billions of years, so will likely continue to do that.  I don't think anything would likely knock it out, but we can't rule out chaotic orbital effects so I don't really know.  The Moon's orbit varies a lot with time, so maybe it's possible that the orbit itself could change, showing us the other side of the Moon.  Or, an asteroid hit or some such thing might occur.  A billion years is a long time, but I'll bet humanity will always see the same side from Earth.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="36644" LastEditDate="2020-12-12T18:41:32.317" LastActivityDate="2020-12-12T18:41:32.317" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="19251" PostTypeId="2" ParentId="19245" CreationDate="2016-11-30T17:07:54.287" Score="8" Body="&lt;p&gt;Anything that has mass will have gravity ( Exhibit a phenomemon called gravitation) not just black holes. &#xA;One way of understanding gravity was put forth by Newton as his law of gravitation. As @christo said, there is no clear knowledge of why it has gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more generic way to put it would be to say, anything that has mass curves the space around it. The force or gravity you would feel can be considered as the way you are being forced to follow the curved lines instead of your intended path. This is how Einstein's general theory of relativity explains gravitation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum mechanics, It is speculated that gravity is caused by exchanging particles called gravitons, think of it like the way you would play catch with your friends. Multiple objects having some mass throwing small balls called gravitons at each other. There is no evidence for the existence of gravitons yet, but it explains a lot of things in the quantum realm.&lt;/p&gt;&#xA;" OwnerUserId="15012" LastActivityDate="2016-11-30T17:07:54.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19252" PostTypeId="2" ParentId="19243" CreationDate="2016-11-30T20:11:04.173" Score="4" Body="&lt;p&gt;At the moment there is basically only one way. That is to associate the planetary-sized object with a cluster of stars or moving group of stars of known age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's basically it. If the planetary-sized object really can't be associated with another object, then only limits can be placed on its age by comparing it's luminosity to theoretical planet cooling models. But to use these models you need to know the mass!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at the moment, the only &quot;free-floating&quot; planetary mass objects known (they could all be low-mass brown dwarfs actually) &lt;em&gt;are&lt;/em&gt; those for which an age is estimated by association and therefore a (planetary) mass has been deduced from cooling models.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-11-30T20:11:04.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19253" PostTypeId="1" AcceptedAnswerId="19255" CreationDate="2016-11-30T21:47:21.333" Score="7" ViewCount="861" Body="&lt;p&gt;According to modern cosmology, space is expanding, causing proper distances (but not comoving distances) to increase between galaxies. In the Big Crunch hypothesis, gravity halts and reverses the expansion of the Universe, causing all matter to collide and eventually form a single black hole. This gives way to other oscillating universe hypotheses, which generally propose that the conditions in a compressed Universe would be the same as those during the Big Bang, leading a cycle of expanding and contracting universes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring the &lt;a href=&quot;https://astronomy.stackexchange.com/questions/13615/are-there-any-studies-that-prove-disprove-the-oscillating-universe-theory/18569#18569&quot;&gt;problems with entropy&lt;/a&gt; in reverting the Universe to Big Bang conditions, how can gravity be the cause of a Big Crunch in the first place? Specifically, gravity (to my knowledge) only curves space; the idea that it can revert the Universe to the conditions of the Big Bang seems to imply that gravity actually can contract space. Is this actually the case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, the gravitating objects should be moving through a comoving coordinate system, so space itself would not actually be contracting. As far as I can tell, we would have all the Universe's matter compressed within a single point in space, instead of space itself contracting. This should be completely unlike the Big Bang, when space was far less dilated than it is now. If this is in fact what the Big Crunch hypothesis describes, than I'm utterly confused as to how an oscillating universe could work in such a situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I mistaken, or do the Big Crunch and oscillating universe hypotheses imply that gravity actually contracts space (as in, the comoving distances of gravitationally attracting objects would not change)? If not, how could gravity possibly lead to these scenarios?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2016-12-08T04:29:01.163" LastActivityDate="2016-12-26T07:42:50.870" Title="How can gravity lead to the Big Crunch scenario?" Tags="&lt;gravity&gt;&lt;cosmology&gt;&lt;general-relativity&gt;&lt;expansion&gt;&lt;fate-of-universe&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19254" PostTypeId="2" ParentId="19248" CreationDate="2016-12-01T04:37:55.763" Score="0" Body="&lt;p&gt;If The moon's movement relative to Earth is due to it's residual rotation rather than due to the effect of the pacific, the residual rotation would be decreasing similar to a pendulum thats almost still, imagine you are swinging a pendulum around, so the moon's movement is probably a flat ellipse, a figure of 8 or a resonating spirograph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the moon formed from an accretion disk, it's material was less spherical and more fluid and under the influence of our gravity. If it did manage to keep a rotation at that stage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps there isn't a measurement of the moon's rotation, because it's difficult to measure, and it may in fact be a slow rocking motion due to the pacific rather than residual motion from it's formation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To measure the movement of the moon is technically challenging, perhaps you would have to place a camera on the moon and take a photo of a place on earth at the same time every day and then see if the variation is due to the pacific or the residual rotation. perhaps the cheapes way to measure it's movement is to measure the length of the flagpole, or to send a digital flagpole there. The shadows of the moon shift so much, it's not practical to use it's topography for reference. &lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2016-12-03T02:54:44.847" LastActivityDate="2016-12-03T02:54:44.847" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19255" PostTypeId="2" ParentId="19253" CreationDate="2016-12-01T08:03:09.887" Score="5" Body="&lt;p&gt;The amount of matter in the universe is directly related to the curvature of space itself.  We can look at the Friedmann Equations to see how this works:&#xA;$$&#xA;H(t)^2 = \frac{R'(t)^2}{R(t)^2} = \frac{8\pi G}{3}(\rho_{m} + \rho_{r}) + \frac{1}{3}\Lambda - \frac{c^2}{R^2\mathscr{R}^2}&#xA;$$&#xA;This is the equation for determining the scale factor for finding distances in the universe at any given time.  Here, $H$ is the Hubble Constant, $R$ is the scale factor, $G$ is the gravitational constant, $\rho_m$ is the density of matter (both dark matter and baryonic matter), $\rho_r$ is the density of radiation (photons), $\Lambda$ is the cosmological constant due to dark energy, and $\mathscr{R}$ is the radius of curvature of space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we define a quantity we call the &quot;critical density&quot; ($\rho_c$).  This is the density of matter needed to make the universe go from an &quot;open&quot;, hyperbolic geometry to a &quot;closed&quot;, spherical geometry.  $\rho_c=\frac{3H^2}{8\pi G}$. We create a value $\Omega=\frac{\rho}{\rho_c}=\frac{8\pi G \rho}{3H^2}$ to take into account this critical density, making our equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;H^2 = \Omega_m + \Omega_r + \Omega_\Lambda - \frac{c^2}{R^2\mathscr{R}^2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here, we input our present-day values for everything.  We define $R$ to be 1 and $H$ to be $H_0$ at the present day.  We also choose to neglect the density of radiation and dark energy.  Simplifying the equation a little gives us:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{c^2}{\mathscr{R}^2} = H_0^2(\Omega_{m,0} - 1)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here, we can see that the curvature of the universe is dependent on $\Omega_{m,0}$, which is directly related to the density of matter in the universe.  Specifically, for $\Omega_{m,0}\gt1$, $\mathscr{R}$ will be positive, meaning this will be a spherical, closed universe.  If $\Omega_{m,0}\lt1$, $\mathscr{R}$ will be negative, meaning the universe will expand forever, having hyperbolic curvature.  If $\Omega_{m,0}=1$, then $\mathscr{R}=\infty$, which is a flat universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we can see that having a dense enough universe, gravity will curve space and make it spherical.  You can also use the Friedmann Equation to calculate the deceleration parameter of space - how quickly the expansion is slowing or speeding up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;q_0=\frac{\Omega_{m,0}}{2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, we can see that in a positively curved universe, $q_0&amp;gt;0$, meaning that the universe's expansion will be decelerating.  Eventually, the expansion rate will be negative, and then it will begin collapsing in on itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing to keep in mind when thinking about this is that gravity can, in a way, be considered a 'fictitious force'. Gravity is the force objects feel as they attempt to travel in straight lines through curved spacetime.  Matter is what is responsible for the curving of spacetime, so by increasing the density of matter in the universe, you increase the curvature, and thus make objects want to move closer together, which will increase the density, which increases the curvature.  Thus you have a feedback cycle, where you will ultimately collect all matter at a single point, and $\mathscr{R}=0$, so spacetime will have a 0 curvature radius, meaning spacetime has collapsed as well.&lt;/p&gt;&#xA;" OwnerUserId="13426" LastEditorUserId="2365" LastEditDate="2016-12-26T07:42:50.870" LastActivityDate="2016-12-26T07:42:50.870" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="19256" PostTypeId="1" AcceptedAnswerId="19264" CreationDate="2016-12-01T08:30:18.280" Score="4" ViewCount="75" Body="&lt;p&gt;I know by observing the dimness of a star it is possible to calculate an exoplanet's distance from the star and its mass by how much the star wobbles. However, is there any way to accurately determine if said planet has any magnetic field through observation alone directly?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="2153" LastEditDate="2016-12-01T18:21:02.620" LastActivityDate="2016-12-01T19:08:47.893" Title="Is it possible to tell if a certain extrasolar planet produces its own magnetic field?" Tags="&lt;exoplanet&gt;&lt;magnetic-field&gt;&lt;planetary-transits&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19257" PostTypeId="1" AcceptedAnswerId="19258" CreationDate="2016-12-01T08:50:55.260" Score="2" ViewCount="257" Body="&lt;p&gt;All gas giants in the solar systems have two things in common: abundant hydrogen gas and a magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering: could there be a gas planet that somehow can't produce any magnetic field on its own?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I learned there many kinds of bizarre and unfamiliar worlds from reading up terrestrial exoplanets and moons, so I thought maybe there is a unique gas planet that doesn't produce any magnetic field - or maybe the magnetic field disappeared after a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW I'm aware of hydrogen becoming metallic under a specific condition.&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="2365" LastEditDate="2016-12-27T22:24:58.043" LastActivityDate="2016-12-27T22:24:58.043" Title="Are all gas planets guaranteed to produce their own magnetic field?" Tags="&lt;planet&gt;&lt;magnetic-field&gt;&lt;gas-giants&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19258" PostTypeId="2" ParentId="19257" CreationDate="2016-12-01T12:50:11.600" Score="3" Body="&lt;p&gt;It's not technically impossible, but it does seem incredibly unlikely that a gas planet would lack a magnetic field. This is because hydrogen (by far the most abundant element) attains metallic properties when put under extreme pressure. And a spinning metal/conductor generates magnetic fields via the dynamo effect. This is why earth has a magnetic field, except instead of metallic hydrogen spinning in our core, its our liquid iron outer core that does the trick. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any collection of gas large enough to form a planet would certainly create the needed pressure to transform hydrogen in its metallic form. For a gas planet to have a magnetic field, it really only needs to meet 2 conditions: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The planet needs abundant free hydrogen (which is extremely&#xA;likely given that almost all matter is hydrogen).&lt;/li&gt;&#xA;&lt;li&gt;The planet needs to be spinning (also extremely likely because a&#xA;planet retains the net angular momentum of all the stuff that fell&#xA;onto it while it was forming... the only way a planet avoids&#xA;spinning is by tidal locking, such as the moon always facing one&#xA;side towards the earth... and even then it's still spinning&#xA;slowly... it's just spinning once every time it completes an orbit.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Ultimately, given the vast and possibly infinite size of the universe, and the great abundance of planets, I'd bet there is a planet somewhere that happens to be made of gasses other than hydrogen, or happens to not be rotating, and thus there would be a gas giant without a magnetosphere. But I'd be very surprised if we find one in my lifetime, or even in the the next 1000 years!&lt;/p&gt;&#xA;" OwnerUserId="9610" LastEditorDisplayName="user15217" LastEditDate="2016-12-27T04:38:54.727" LastActivityDate="2016-12-27T04:38:54.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19259" PostTypeId="1" AcceptedAnswerId="19262" CreationDate="2016-12-01T13:15:44.080" Score="5" ViewCount="492" Body="&lt;p&gt;Supposing a planet is made entirely of gas, is it possible for the planet to become tidally locked with the star it orbits? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know when a terrestrial planet orbits a star in a different amount of time than the length of its day, the planet warps continuously, resulting in tidal heating. This heating takes energy from the planet's rotation until it stabilizes in a tide-locked formation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe a purely gaseous planet would also have some sort of internal heating for the same reason. But because the &quot;surface&quot; of a gas planet is fluid, it must also obey thermodynamic principles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a basic example, the side of the planet facing a star gets hot, and the side facing away gets cold. The hot gas expands and the cold gas contracts and you get convection, Hadley cycles, prevailing winds, and chaotic weather patterns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it seems like there's a battle between gravity, which wants the gas to stop moving and become tide-locked, and thermodynamics, which wants the surface to roil with convection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does one process dominate over the other? Does one process slowly take over as the other fades away? Can it go either way?&lt;/p&gt;&#xA;" OwnerUserId="9610" LastEditorUserId="9610" LastEditDate="2016-12-04T03:13:36.153" LastActivityDate="2016-12-04T03:13:36.153" Title="How can a gas planet become tidally locked with a star?" Tags="&lt;star&gt;&lt;planet&gt;&lt;orbit&gt;&lt;gravity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19260" PostTypeId="1" AcceptedAnswerId="19263" CreationDate="2016-12-01T13:38:14.503" Score="0" ViewCount="1243" Body="&lt;p&gt;A star's nearby environment may give clue to its age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different stellar type have different spectral features. If we just have its spectrum, how do we know its age?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a star may be born with a M3-type(I know it is more difficult for M-type compared with OBAFGK) and stay here for a long time.&#xA;How do we know it was born with a M3-type? &#xA;How do we know its age? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is better to talk about different stellar types and give reference papers.&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditDate="2016-12-01T14:05:57.103" LastActivityDate="2016-12-01T15:57:28.340" Title="How do we know a star's age based on its spectrum?" Tags="&lt;star&gt;&lt;stellar-evolution&gt;&lt;star-formation&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2016-12-03T01:36:59.280" ContentLicense="CC BY-SA 3.0" />
  <row Id="19261" PostTypeId="2" ParentId="19253" CreationDate="2016-12-01T14:37:05.100" Score="1" Body="&lt;p&gt;In less mathy words, the Big Crunch scenario occurs if the ratio of the total density of the Universe to its expansion rate is sufficiently large$^\dagger$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand your question, you're basically asking, &quot;&lt;em&gt;Why would a sufficiently dense universe not simply contract into a clump? Why does it have to pull space itself with it?&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you basically answer that question yourself: &lt;strong&gt;Yes, space is indeed &quot;tied&quot; to matter&lt;/strong&gt;. This is the essence of the Friedmann equation, and of general relativity in general. To my knowledge, there's no &quot;proof&quot; of this other than, well, it's one of the foundations of GR, which has so far proved an immensely successful theory. In moderately overdense regions (galaxy clusters) space expands slower than in underdense regions (voids). In very overdense regions (galaxies, stars, cats, etc.), it doesn't expand at all. And in extremely overdense regions (black holes), space contracts. &#xA;In the case of a black hole, space only contracts locally, but in principle the whole of the Universe could do the same. Only expansion prevents this, and seems to be able to prevent it forever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We believe that the Universe is homogeneous and isotropic; if this is indeed the case, matter wouldn't be able to contract to a point within the Universe and make one giant black hole in an otherwise expanding Universe, since every bit of matter is attracted the same in all directions. You could perhaps envision clumps of matter contract on very large, yet sub-Universal, scales, to create many supersupermassive black holes, but as it turns out, the expansion rate was simply too large in the early Universe for this to happen, and now it's too late.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;This is analogous to a rock thrown upwards falling back again if the ratio of the gravitational attraction between Earth and rock to the speed with which it is thrown is suffiently large.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2016-12-01T15:14:37.413" LastActivityDate="2016-12-01T15:14:37.413" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19262" PostTypeId="2" ParentId="19259" CreationDate="2016-12-01T15:44:11.210" Score="2" Body="&lt;p&gt;Gaseous planets would tidally lock, just as binary stars do.  That would occur in addition to convection-- tidal effects are weak and operate on very long timescales, convection is strong and operates on much shorter timescales.  It is true that a gaseous planet conforms more easily to the tidal potential even if it is rotating, so there is less of a turning of the tidal bulges, but the tidal bulges are larger because they fill the equipotential, whereas solid surfaces don't.  So the rapid response of gas could make the locking process either faster or slower, depending on which effect dominates.  But you are probably right to focus on energy dissipation as the key requirement, and gas is probably less dissipative, so I would guess the locking time is longer for gas.  Still, there is energy dissipation, there is a delay as the gas flows, and there is a tidal locking effect, all going on on timescales much slower than convection.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-12-01T15:44:11.210" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19263" PostTypeId="2" ParentId="19260" CreationDate="2016-12-01T15:57:28.340" Score="1" Body="&lt;p&gt;There are two separate issues when you talk about a star's age.  Are you talking about its absolute age, in years, or its age in terms of how far along is it within its own evolution?  The difference is, massive stars go through their evolutionary phases very quickly, and low-mass stars very slowly, so essentially all very low-mass stars are older in years than essentially every very high-mass star.   Also, you can know more than the &quot;spectral type&quot; from a spectrum, you can also know the &quot;luminosity class&quot;-- that's the roman numeral that tells you if the star is, for example, a main-sequence dwarf or not.  So if you pick an M spectral type main-sequence star, and an O spectral type main-sequence star, as two stars at random, it is extremely likely that the M star is much older than the O star, and you can know that purely from the spectral type and luminosity class.  But if you want to know the age of the M star, or the O star, in years within their evolutionary stages, then it gets more difficult because there is rather little main-sequence evolution in the spectrum.  There is some evolution though, especially late in the main sequence, so you could tell its age if there is evidence of main-sequence evolution in the spectrum (for one thing, the luminosity rises with age, within a spectral type, but that's pretty hard to tell from the spectrum).  If the star is both low-mass and early in its main-sequence evolution, the best way to tell its age is to look at the rotational broadening of its lines, as low-mass stars spin down rapidly as they age (due to strong magnetic fields and weak winds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another good way to tell the age of stars is if they are in clusters that all formed together, you can look at what is called the &quot;main sequence turnoff&quot;, which basically means you look for the lowest mass star that has gone supernova, and you know how long that takes for that mass of star.  You tell this from the spectrum because the turnoff is in the &quot;Hertsprung-Russell&quot; diagram, which you make from the spectra, but it requires a lot of stars in the cluster, so it's not really a property of a single spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the bottom line is, you need a suite of different techniques for assessing the ages of stars, it is not always easy or straightforward but you can take each case individually and look for ways to infer age.  Much easier is determining which evolutionary class a star is in, because stars change dramatically in one phase or another, and the spectrum alone can tell you if you have a dwarf or a giant, etc.  (Basically, the lines get much narrower as a star puffs out in radius, because the pressure drops.)&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-12-01T15:57:28.340" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19264" PostTypeId="2" ParentId="19256" CreationDate="2016-12-01T19:02:42.380" Score="8" Body="&lt;p&gt;There are three approaches with which people have looked for this, and not yet been too successful:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A transiting planet planet with a strong magnetic dipole and/or very strong host star winds might produce a visible signal when the magnetospheric bow-shock passes in front of the star as well. The idea is that at the bow-shock the streaming hydrogen would be heated up, ionize and produce a detectable ultraviolet photometry signal upon recombination at a characteristic distance (the stand-off distance) from the planet.&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/HD_189733_b&quot; rel=&quot;noreferrer&quot;&gt;HD189733b&lt;/a&gt; was for a time the first planet to have measured its magnetic field in that way, however the discovery was retracted because the ultraviolet opacities that the group used to determine the bow-shock position were wrong.&lt;br&gt;&#xA;It turns out there is no signal from the bow-shock.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The second idea is to look at a hot Jupiter that has an evaporating atmosphere. The shape of the ionized hydrogen tail in the magnetosphere will be influenced by the strength of the planetary dipole. But then, the amount of hydrogen that is in the planet's shadow is also a function of the magnetic field strength.&lt;br&gt;&#xA;Hydrogen that comes into the planet's shadow will recombine with magnetospheric electrons due to the lack of direct irradiation from the star and produce a signal in the Lyman-$\alpha$ recombination line.&lt;br&gt;&#xA;Going further, only the particles that stream away from the planet, and towards us will be part of this process. Thus the blue-shifted side of the Lyman-$\alpha$ line can inform us about the strength of the magnetic field.&lt;br&gt;&#xA;So far the theory. The practice is that our last remaining UV-spectrometer in space that could perform such a thing (mounted on the Hubble) is even only barely capable of observing this process. But it has been done for the planet &lt;a href=&quot;https://en.wikipedia.org/wiki/HD_209458_b&quot; rel=&quot;noreferrer&quot;&gt;HD209458b&lt;/a&gt;&#xA;The data quality is lousy, and the interpretation that has been done in &lt;a href=&quot;http://science.sciencemag.org/content/346/6212/981&quot; rel=&quot;noreferrer&quot;&gt;Kislyakova et al. 2015&lt;/a&gt; has more parameter than datapoints and is therefore questionable.&lt;br&gt;&#xA;But if you choose to believe the authors, then this planet has a magnetic moment of $0.1$ Jupiter magnetic moments.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The third possibility that could yield real results are radio observations with the &lt;a href=&quot;https://en.wikipedia.org/wiki/LOFAR&quot; rel=&quot;noreferrer&quot;&gt;LOFAR&lt;/a&gt; array. LOFAR observes at the edge of observable frequencies from Earth (at very low frequencies the ionosphere interferes with radio observations) at the for the moment highest achievable spatial resolution. The idea here is to look for analogues of &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetosphere_of_Jupiter#/media/File:Jupiter_radio.jpg&quot; rel=&quot;noreferrer&quot;&gt;Jupiters decametric radiation&lt;/a&gt;, which is simply synchrotron radiation produced when ionized particles start gyrating around a magnetic field line.&lt;br&gt;&#xA;This already has been tried at higher frequencies / different observatories with the planets around $\epsilon$ Andromedae  and $\tau$ Bootes, as yet to no avail.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditDate="2016-12-01T19:08:47.893" LastActivityDate="2016-12-01T19:08:47.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19265" PostTypeId="2" ParentId="19248" CreationDate="2016-12-01T23:55:45.790" Score="11" Body="&lt;p&gt;The Moon indeed &quot;wobbles&quot; about in its orbit because it goes around the Earth in an ellipse and not a circle.  From our point of view, it wobbles a little back and forth such that over a lunar cycle, we end up seeing about 59% of its surface.  The effect is referred to as &lt;em&gt;lunar libration&lt;/em&gt;.  You can find a number of videos that &lt;a href=&quot;https://www.youtube.com/watch?v=LdQXtTDzB_c&quot; rel=&quot;noreferrer&quot;&gt;show how the Moon looks to us&lt;/a&gt; over the course of a cycle.&lt;/p&gt;&#xA;" OwnerUserId="9749" LastActivityDate="2016-12-01T23:55:45.790" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19266" PostTypeId="1" AcceptedAnswerId="19269" CreationDate="2016-12-02T04:13:50.970" Score="1" ViewCount="75" Body="&lt;p&gt;Over a year, the ratio of daylight changes, e.g. in Montreal there is more daylight in the summer than in the winter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know if, given a bad data sample, e.g. five years, one might notice a trend of increasing or decreasing daylight ratios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's why I ask. I'm doing research on the Dead Sea scrolls, and some of the fragments contain ratios that a scholar has argued represent daylight. What I'd like to know is a trend that the community of Qumran might have noticed in their dataset. They had a theory that darkness was increasing. I doubt their data was exhaustive, but I wonder if around Jerusalem there might have been a period of decreasing daylight ratios over a period of a few years roughly around 100 B.C.E.&lt;/p&gt;&#xA;" OwnerUserId="15024" LastActivityDate="2016-12-02T13:06:13.010" Title="Does the ratio of daylight change from year to year?" Tags="&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="19269" PostTypeId="2" ParentId="19266" CreationDate="2016-12-02T13:06:13.010" Score="2" Body="&lt;p&gt;I don't think it's possible to have a change in the ratio of day to night when averaged on a yearly basis.  To do so would require either a nonlinear change in the Earth's rotation or a significant shift in the polar axis tilt within a year.&lt;br&gt;&#xA;If the rotation were simply slowing down over time, the year-averaged ratio would be the same every year (only the number of days in a year would change).  Similarly, the only way a polar tilt could cause a change would be if the change in angle occurred at some (sinusoidal) rate not an integral ratio of the orbital period (year).&lt;br&gt;&#xA;Either of those putative mechanisms can only change the per-year ratio by producing a day/night ratio in &quot;summer&quot; that is not the inverse of the &quot;winter&quot; ratio, &lt;strong&gt;and&lt;/strong&gt; producing a &lt;strong&gt;different&lt;/strong&gt; non-inverse the following year.  (I'm oversimplifying a bit, since the elliptic orbit produces a repeatable asymmetry in the Sun's zenith pattern).&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2016-12-02T13:06:13.010" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19270" PostTypeId="1" CreationDate="2016-12-03T02:46:11.017" Score="3" ViewCount="263" Body="&lt;p&gt;I took a photo of SuperMoon 2016 and it was very shiny, the light around it showed seven colours. Was it my camera's effect or was it the moonlight's true colours. &lt;a href=&quot;https://i.stack.imgur.com/4rhCI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4rhCI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15038" OwnerDisplayName="preeti" LastEditorUserId="1528" LastEditDate="2016-12-03T08:35:03.713" LastActivityDate="2016-12-03T23:35:02.430" Title="Does supermoon's white light splits into seven colours" Tags="&lt;the-moon&gt;&lt;photography&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19271" PostTypeId="2" ParentId="19270" CreationDate="2016-12-03T04:06:12.707" Score="2" Body="&lt;p&gt;The moon doesnt have really have 7 colors, you are having a slight prismatic effect from the lens. If it was an effect of the atmosphere you would have seen it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun's spectrum peaks in greenish colors, and the moon peaks in yellow and red colors. That explains why the moon is more yellow than the sun, although they both can look red when they are rising and setting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The colors on your photo are more blue, if you look into the lens of your camera near a window you will probably see a blue window and a white window, it is the lens flare coating and a lens artefact. The flare on the moon is also blueish green, same as your lens. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are photos of lens flare of the sun and there is a lot of information on avoiding lens flare and different types. This pic of the sun also has violet lens flare:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/TsZCl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TsZCl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.fr/search?q=flare+and+ghosting&amp;amp;num=100&amp;amp;espv=2&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ved=0ahUKEwiBvJ-S89fQAhXBtxQKHTDKAggQ_AUICCgB&amp;amp;biw=896&amp;amp;bih=542&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.google.fr/search?q=flare+and+ghosting&amp;amp;num=100&amp;amp;espv=2&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ved=0ahUKEwiBvJ-S89fQAhXBtxQKHTDKAggQ_AUICCgB&amp;amp;biw=896&amp;amp;bih=542&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here are the Sun and the Moon's spectra/colors if you have a good prism: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;sun&#xA;&lt;a href=&quot;https://i.stack.imgur.com/scHZs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/scHZs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;moon&#xA;&lt;a href=&quot;https://i.stack.imgur.com/dA15P.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dA15P.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2016-12-03T11:34:19.657" LastActivityDate="2016-12-03T11:34:19.657" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19272" PostTypeId="2" ParentId="19270" CreationDate="2016-12-03T06:32:49.393" Score="6" Body="&lt;p&gt;The answer by @comprehensible nicely states that whatever the effect might be, it is related to your camera. Of course the moon's light has a spectrum all the time, just like the sun, but we don't see it until we look through some dispersive material or object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought I would just add a little back-up information as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took your photo and rotated, cropped, and flipped it left-right on my laptop, then made 1D plots of the three color channels (R, G, B) normalized to local (R+G+B) by projecting down to the new x-axis. Note that this is &lt;strong&gt;very unscientifically done&lt;/strong&gt;, but it's still interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the dispersion were diffraction, then moving away from the moon you'd expect to see the order B, G, R, B... since diffraction peaks are at larger angles for longer wavelengths. It would repeat if there were multiple orders. Of course this is messy geometry and the source (the moon) is very wide, so this is only guessing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surprisingly, this is the order that appears! So it's possible that there is some periodic pattern somewhere on the lens or in the camera. Since the characteristic size of one cycle is of the order of the diameter of the moon or about 0.01 radian, if there scratches or streaks on the lens the period owould be of the order of 100 wavelengths. There is an obvious periodicity in the camera's CCD (assuming it's not a film camera) but then the ray tracing would be more complicated (multiple reflections) and this simple analysis wouldn't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just speculation, and thin film interferece can also produce color effects also, and non-uniform thickness (like a soap bubble) can make color gradients as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note, our perception of color is a complex subject, and while our eye and brain together try to reconstruct a rich color behavior, the simple up and down wiggles of the intensity in the three color channels of your camera's sensor is all that is actually in the photo. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The color modulation is quite strong, so I am guessing that whatever is producing the 1 directional flair is also the thing that has the periodicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why don't you try a similar experiment and point at strongly overexposed LEDs or maybe street lights (Sodium or Mercury vapor) and see if you get a clearer effect. This would be just for fun. This is just an artifact of your camera and has nothing to do with the moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MkNMT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MkNMT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2N8Nb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2N8Nb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the Python script I used to plot the image after I rotated/cropped/flipped it on my laptop:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;&#xA;fname = &quot;Moon Spectrum.png&quot;&#xA;&#xA;img = plt.imread(fname)&#xA;&#xA;img1D = img.sum(axis=0).T[:3]&#xA;&#xA;img1Dnorm = img1D/img1D.sum(axis=0)&#xA;&#xA;plt.figure()&#xA;&#xA;plt.subplot(2,1,1)&#xA;plt.imshow(img)&#xA;&#xA;plt.subplot(2,1,2)&#xA;for thing, color in zip(img1Dnorm, ['red', 'green', 'blue']):&#xA;    plt.plot(thing, color=color)&#xA;&#xA;plt.xlim(0, img1Dnorm.shape[1])&#xA;&#xA;for lab, x, y in [('B',  58, 0.35), ('G',  80, 0.39), ('R',  97, 0.31),&#xA;                  ('B', 126, 0.36), ('G', 141, 0.46), ('R', 166, 0.40) ]:        plt.text(x, y, lab, fontsize=14)&#xA;&#xA;plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2016-12-03T23:35:02.430" LastActivityDate="2016-12-03T23:35:02.430" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19273" PostTypeId="2" ParentId="19216" CreationDate="2016-12-03T10:34:24.733" Score="2" Body="&lt;p&gt;I'm afraid there is no answer, yet... The problem is that you would have to know the star densities in the different regions (in the spiral arms and outside of the spiral arms). While it is certainly possible to have an accurate estimate of the density of massive stars because of their brightness, the density of low mass stars in different regions of the galaxy is much more difficult to estimate. And since low mass stars make up the vast majority of stars (see for instance the &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/a/ac/Imf.svg&quot; rel=&quot;nofollow&quot;&gt;initial mass function&lt;/a&gt; which is a measure of the distribution of stars over the different masses), estimating the density of stars in different regions becomes very uncertain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, at the moment the &lt;a href=&quot;http://sci.esa.int/gaia/&quot; rel=&quot;nofollow&quot;&gt;Gaia&lt;/a&gt; satellite is measuring the position and distances of about a billion stars and other objects. A much more detailed three dimensional map will be created with that data, including distance data for dim stars whose distances we were as yet unable to measure. So there is hope ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB. The first data release of Gaia has been published in September. I haven't found any papers that were able to answer your question (my search was not extensive though), but they might be forthcoming. Of course if you have some programming skills, you might determine the density differences from the data (&lt;a href=&quot;http://gea.esac.esa.int/archive/&quot; rel=&quot;nofollow&quot;&gt;available here&lt;/a&gt;) yourself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT - Just took a look at the data, the parallax (and therefore the distance) is still lacking for many stars. Hopefully this will change with future data releases.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;= I understand that one of the reasons the arms &quot;appear real&quot; is just that there are more bright young stars forming in the arms (due to the&#xA;  higher pressure). Perhaps the actual density difference is very low?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would expect that the density difference of stars is quite small. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;= Perhaps the density variation is dependent on the type of star, or other material - what about things like gasses?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The density variation is certainly different between different types of stars. High mass stars only occur near star forming regions (i.e. in spiral arms). Their lifespan is so short that they do not have the time to get very far from the region in which they were formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd expect that the density variation for gas and dust will be higher (but I have no knowledge to support this).&lt;/p&gt;&#xA;" OwnerUserId="864" LastEditorUserId="864" LastEditDate="2016-12-03T18:17:13.073" LastActivityDate="2016-12-03T18:17:13.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19274" PostTypeId="2" ParentId="19270" CreationDate="2016-12-03T12:02:51.613" Score="1" Body="&lt;p&gt;As others have mentioned, it is a prismatic effect due to your lens and/or the periodic nature of the way your camera's sensor absorbs light. Such effects are present in pretty much &lt;em&gt;all&lt;/em&gt; photographs of large, strong light sources taken with cameras and lenses intended for artistic photography. When properly exposed the effects are generally controlled well enough to not be blatantly visible in the resulting photograph. When grossly overexposed, however, as is the case in this photo then the effects are more pronounced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a good example of how exposure differences affects lens flare and other effects caused by light reflecting off the surface of optical elements of a lens or sensor, please see &lt;a href=&quot;https://photo.stackexchange.com/a/41724/15871&quot;&gt;this answer&lt;/a&gt; to this question at Photo.stackexchange: &lt;a href=&quot;https://photo.stackexchange.com/questions/41722/can-you-photograph-the-milky-way-with-a-full-moon-out/41724#41724&quot;&gt;Can you photograph the milky way with a full moon out?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For how to get best results when photographing the moon at night, please see &lt;a href=&quot;https://photo.stackexchange.com/questions/459/how-do-i-set-the-proper-exposure-for-nighttime-moon-photos/34084#34084&quot;&gt;How do I set proper exposure for night time Moon photos?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15040" LastEditorUserId="-1" LastEditDate="2017-04-13T12:44:06.413" LastActivityDate="2016-12-03T12:30:50.503" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19275" PostTypeId="1" AcceptedAnswerId="19278" CreationDate="2016-12-03T13:53:11.607" Score="1" ViewCount="114" Body="&lt;p&gt;I came to know that dark energy is constant.As mentioned in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Friedmann equations&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can a &lt;code&gt;constant energy&lt;/code&gt; cause an &lt;code&gt;accelerating rate of expansion&lt;/code&gt; in the universe?&lt;/p&gt;&#xA;" OwnerUserId="15032" LastEditorUserId="11143" LastEditDate="2016-12-03T15:30:45.870" LastActivityDate="2016-12-04T01:21:43.913" Title="How can space expand in accelerating speed when dark energy is constant?" Tags="&lt;universe&gt;&lt;cosmology&gt;&lt;expansion&gt;&lt;dark-energy&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19276" PostTypeId="1" CreationDate="2016-12-03T16:50:24.563" Score="2" ViewCount="880" Body="&lt;p&gt;Okay, so the speed of revolution of earth around sun is 30 km/s[1], speed of revolution of sun around milky way's center is ~200 km/s[2], speed of rotation of milky way is 828,000 km/h[3] i.e. 230000 m/s. Since speed of light is 300000000 m/s, so on an even bigger scale (bigger than clusters) are we already moving faster than the speed of light? Or am I totally wrong?&lt;/p&gt;&#xA;" OwnerUserId="15042" LastActivityDate="2016-12-05T23:31:47.977" Title="Are we already moving faster than light?" Tags="&lt;light&gt;&lt;space&gt;&lt;speed&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19277" PostTypeId="2" ParentId="19276" CreationDate="2016-12-03T17:02:46.493" Score="4" Body="&lt;p&gt;&lt;em&gt;Short Answer&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;NO.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Long Answer -&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The Earth &quot;wobbles&quot; on its polar axis.  This motion is not relevant (IMO) to the calculations needed to return to earth from interstellar space as it is an &quot;in situ&quot; motion, but it exists.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Earth revolves {spins} on its axis. For these calculations we will use the equator with a circumference of approx. 25,000 miles. One rotation of Earth is approx. 24 hours. Again this is &quot;in situ&quot; motion so not relevant to space travel, but adds to our &quot;stationary chair&quot; model &#xA;25000/24 =1041.7 MPH&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Earth is orbiting the Sun once a year. The circumference of the Earth's orbit is approx.  607.6  million miles {or 940 million kilometers}. One year is approx. 365 days &#xA;365 days X 24 = 8760 hours &#xA;607,600,000/8760 = 69,360.73 MPH&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So far we have basically 3 Directions of Motion {Wobble, Spin and&#xA;  Orbit} and a combined speed of 1041.70 + 69,360.73 = 70,402.43 MPH for&#xA;  a person sitting in a chair at the Equator.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;The sun {and hence the solar system} is moving towards the constellation Hercules, namely to the star Lambda Herculis at 12 miles per second {or 20 kilometers per second} which is 43,200 MPH&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Solar system is also moving upwards, at 90 degrees to the plane of the Milky Way, at 4.34 miles per second or 15,624 MPH. But we are actually leaving the Galaxy, out about 50 light years now and will be moving out to 250 light years before it reverses. Details of the mechanics of this are explained in the link below. We also crossed the Galactic plane 2 million years ago.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Solar system is orbiting around the Galaxy at an &quot;estimated&quot; speed of 124 miles per second {or 200 kilometers per second} which is 446,400 MPH. The way that figure has been calculated can be found at the link below. &#xA;So our &quot;armchair Astronaut&quot; is now moving through 6 different directions and a combined speed of approximately 574,585 MPH &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;69,361 MPH Spin and Orbit    43,200 MPH Towards Lambda Herculis&lt;br&gt;&#xA;  15,624 MPH Perpendicular to Galactic Plane  446,400 MPH Orbiting the&#xA;  Galactic Center {or Galactic Spin Rate}&lt;/strong&gt;  &lt;strong&gt;574,585 MPH  or 256862.478 m/s Speed of Earth&#xA;  within Our Galaxy&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;light's= velocity = 3*10^8 m/s&#xA;So the ratio = 256862.478 / 3*10^8 = 0.00085620826&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="15032" LastEditorUserId="15032" LastEditDate="2016-12-04T05:52:44.460" LastActivityDate="2016-12-04T05:52:44.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19278" PostTypeId="2" ParentId="19275" CreationDate="2016-12-03T17:29:06.520" Score="2" Body="&lt;p&gt;This is a very good question and it comes down to what is meant by &quot;accelerating expansion&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In cosmology we define something called the scale factor $a$. By definition at the present time $a=1$, billions of years ago when linear distances on a cosmic scale (e.g. the distance to a faraway galaxy) were half what they are now $a$ was equal to $0.5$ and many years in the future when cosmic linear distances have increased to twice their current size the $a$ will equal $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a$ is a function of cosmological time and we also define its 1st derivative and 2nd derivatives, with respect to cosmological time, $\dot{a}$ and $\ddot{a}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{a}$ is a measure of the Universe's rate of expansion and so when $\dot{a}$ is increasing with time we say that the expansion of the Universe is accelerating. $\ddot{a}$ is a measure of the change of the rate of the Universe's expansion so equivalently we say if  $\ddot{a}$ is positive then the expansion of the Universe is accelerating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However note we have defined $a$ at the present time to be $1$, but the present time is entirely arbitrary. This means that $a$, $\dot{a}$ and $\ddot{a}$ depend on when we define $a$ to be equal to $1$. To get round this can instead think of the Hubble Parameter $H = \frac{\dot{a}}{a}$ as a better measure of the rate of expansion as it does not depend on when $a=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a de Sitter Universe (spatially flat, empty, positive cosmological constant) the Friedmann equations reduce to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\ddot{a}}{a} = H^2 = \frac{\Lambda c^2}{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this we can see that $H$ and $\ddot{a}$ are both positive constants. So in one sense (the sense of $H$) the rate of expansion of the Universe is constant, but in another sense (in the sense of $\dot{a}$) the rate of expansion is increasing. So we can see a link between constant energy desnity (the energy density of the cosmological constant is constant) and a constant rate of expansion as long as we define the rate of expansion in a certain way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the currently-favoured cosmological model dark energy is a cosmological constant and we are currently in an era when dark energy dominates the dynamics of the Universe. (Looking at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;full Friedmann equations&lt;/a&gt;) the contribution of matter reduces $\ddot{a}$, but as the cosmological constant dominates it is still positive and we say that the Universe's acceleration is expanding. Looking at $H$ however we say that, because the energy density of matter decreases with time asymptotically to zero, $H$ is decreasing asymptotically to the value given above for the de Sitter Universe.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastEditorUserId="2092" LastEditDate="2016-12-04T01:21:43.913" LastActivityDate="2016-12-04T01:21:43.913" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19279" PostTypeId="2" ParentId="19276" CreationDate="2016-12-03T23:46:06.370" Score="2" Body="&lt;p&gt;Short answer: &lt;strong&gt;no&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever you mention a speed you must state what you are measuring the speed relative to. When a traffic sign says maximum speed 50kph it means relative to the Earth's surface. In measuring speed in space there is no such obvious frame of reference. The theory of relativity does not allow objects to move through space relative to each other, at speeds greater than the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For objects inside the galaxy, the force of galaxy dominates their motion. We are in orbit in a galaxy, but our speed, relative to every other part of the galaxy is well below the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For greater distances, the Hubble flow dominates motion: distant galaxies are moving away from us. The more distant galaxies are moving away from us faster. Galaxies more than about 4 billion parsecs (or 13billion light years) would be moving away faster than the speed of light. This doesn't break the theory of relativity as spacetime is being stretched between them and us. Objects can't move through space at speeds greater than the speed of light, but if the spacetime between objects is expanding, then they can appear to be moving faster than light, depending on your point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of this answer should be taken as contradicting the short answer above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;No&lt;/strong&gt;, we are not moving through space faster than the speed of light relative to any object in our frame of reference.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-12-03T23:46:06.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19280" PostTypeId="1" CreationDate="2016-12-04T02:13:42.337" Score="3" ViewCount="128" Body="&lt;p&gt;What would the inner and outer boundaries of the “habitable zone” be for planets with an albedo that differs from that of the Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example a planet with an albedo of 0.07 or 0.90 (the values for Mercury and Venus, respectively). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to compare these values to the distances in AU's of the rocky inner planets.&lt;/p&gt;&#xA;" OwnerUserId="15044" LastEditorUserId="2365" LastEditDate="2016-12-04T20:45:31.090" LastActivityDate="2016-12-04T20:45:31.090" Title="Estimating inner and outer boundaries of habitable zone for Sun" Tags="&lt;habitable-zone&gt;&lt;albedo&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19281" PostTypeId="1" CreationDate="2016-12-04T08:08:52.363" Score="1" ViewCount="193" Body="&lt;p&gt;Why is cubic spline interpolation used to analyze light curve of pulsating stars? Is there any scientific justification of such usages? &#xA;One example of using the cubic spline interpolation is &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/340567/fulltext/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, section 4.&lt;/p&gt;&#xA;" OwnerUserId="13714" LastActivityDate="2016-12-05T03:06:58.657" Title="Cubic spline interpolation in light curve of pulsating stars" Tags="&lt;photometry&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19282" PostTypeId="1" AcceptedAnswerId="19311" CreationDate="2016-12-04T12:24:27.370" Score="2" ViewCount="103" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Transiting_Exoplanet_Survey_Satellite&quot; rel=&quot;nofollow noreferrer&quot;&gt;TESS&lt;/a&gt; is designed to perform an all-sky transiting exoplanet survey.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;TESS's survey will focus on nearby G, K and M type stars with apparent&#xA;  magnitudes brighter than magnitude 12. Approximately 500,000 stars will&#xA;  be studied, including the 1,000 closest red dwarfs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since its targets are very bright and its sensitivity is relatively worse than Kepler, I wonder whether it can be replaced by gound-based-small-telescope arrays. There can be no gaps in the time series too.&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2016-12-08T09:11:23.637" Title="Is TESS replaceable by gound-based-small-telescope arrays?" Tags="&lt;exoplanet&gt;&lt;photometry&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19284" PostTypeId="1" CreationDate="2016-12-04T16:55:15.903" Score="1" ViewCount="874" Body="&lt;p&gt;The sun is used to determine and measuring time in our solar system. What if we were to travel outside the solar system i.e. to interstellar space, or other galaxies and other celestial bodies (out of our solar system). How would we keep a tab on time then?&lt;/p&gt;&#xA;" OwnerUserId="7478" LastEditorUserId="2365" LastEditDate="2016-12-04T19:59:15.200" LastActivityDate="2016-12-06T07:09:32.127" Title="Time measurement outside solar system?" Tags="&lt;general-relativity&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19285" PostTypeId="2" ParentId="19284" CreationDate="2016-12-04T19:07:31.200" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Units of time are actually quite arbitrary.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Their relationship to astronomical features is historical:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;it's what people used when counting rotations and revolutions of the planet were the best tools to hand. Nowadays, they are legacy values: the &quot;true&quot; units are other things that happen to be feasible to measure using modern tools (wavelengths of light, resonances of atoms), with particular values chosen and scaled to match the old values for continuity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And &lt;em&gt;&lt;code&gt;other units&lt;/code&gt;&lt;/em&gt; derived from them were purely matters of arithmetical convenience. Factors of 12 and 60 were common because they make it easy to do arithmetic in thirds and quarters, and when doing fractions, they reduce easily. So a value like a &quot;second&quot; is absolutely and utterly arbitrary; it applies to outer space exactly as well as it does here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;The second's arbitrary value is the fundamental unit of the metric system,&lt;/code&gt; that that's what scientists use to communicate with each other. The numbers can be scaled for their convenience: as long as they're labeled, one system is as good as any other. (Especially with computers to do the work.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomers try to measure rotations and revolutions of distant bodies, and when speaking of them in public they might use &quot;day&quot; or &quot;year&quot; to help people visualize it. But when speaking to each other, they know what the technical terms mean, and avoid the ambiguity of layperson speak.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So in the end, it's all just &lt;code&gt;&quot;seconds&quot;&lt;/code&gt;, scaled up or down. It may be&#xA;  converted to something else for public consumption or even with each&#xA;  other, but when it comes down to brass tacks they find it's easiest to&#xA;  all use the same unit, even if it's completely arbitrary.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(It is worth noting that some physicists use a rather unfamiliar set of units call &quot;natural units&quot;, under which important constants like the speed of light and the pull of gravity work out to precisely 1. These are convenient when you're pushing around equations, but require a fair bit of effort to translate into clocks and rulers when you want to build an actual experiment.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Another approach can be &lt;code&gt;Through an atomic clock.&lt;/code&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Modern atomic clocks use atoms to determine the passing of time. So you would measure it using that. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For example 1 second is defined as --- the duration of 9192631770&#xA;  periods of the radiation corresponding to the transition between the&#xA;  two hyperfine levels of the ground state of the caesium 133 atom This&#xA;  mean that if you can make a laser shine on a ceasium 133 atom and by&#xA;  counting the periods of the light emitted you will now how many&#xA;  seconds have passed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a interesting side note sins our orbit around the sun is not constant it is sometimes necessary by the scientific community to prolong a year by a second or so every now and then to make clocks match the moment of the earth. There called leap seconds&lt;/p&gt;&#xA;" OwnerUserId="15032" LastActivityDate="2016-12-04T19:07:31.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19287" PostTypeId="1" CreationDate="2016-12-04T19:43:33.907" Score="-1" ViewCount="1286" Body="&lt;p&gt;I'd like to find the angle in radians between two  equatorial coordinate i.e from RA dec of point 1 and RA and dec of point 2.&lt;/p&gt;&#xA;" OwnerUserId="15051" LastEditorUserId="2365" LastEditDate="2016-12-04T19:54:02.267" LastActivityDate="2019-02-24T23:35:18.983" Title="Angle from 2 point on celestial sphere" Tags="&lt;coordinate&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19288" PostTypeId="2" ParentId="19284" CreationDate="2016-12-04T20:10:01.427" Score="3" Body="&lt;p&gt;We would use a clock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, we don't use the position of the sun for determining time. The apparent position of the sun doesn't change with enough regularity. Events on the Earth, such as Earthquakes, can change the length of a &quot;day&quot;. The spinning of the Earth is a good clock, but not the best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best clocks use the frequency of light emitted by an atom (often a caesium atom). This frequency is very stable, far more stable than the Earth. If we were far from the sun, we could still use a clock to provide accurate timekeeping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clock would record local time, which would not keep in sync with Earth time if the clock were moving relative to Earth, due to relativistic effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether we would still use &quot;days&quot;, &quot;months&quot;, &quot;years&quot; etc is speculation. (and hence off topic) I'll note that our internal body clocks work best in a cycle of about 24 hours.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-12-04T20:10:01.427" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19290" PostTypeId="2" ParentId="19281" CreationDate="2016-12-04T20:24:42.980" Score="0" Body="&lt;p&gt;I can't speak for the general case; I'm not aware that cubic splines are always used for modelling light curves. However, a cubic spline is the lowest order of spline that allows for inflection points, thus offering a good combination of flexibility with simplicity. A cubic spline could fit an arbitrary smooth curve with only a few control points. It is also offered in many data analysis and drawing packages. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the paper, the authors are fitting a light curve by eye. Cubic splines are an &quot;obvious&quot; way of doing this.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-12-04T20:24:42.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19292" PostTypeId="2" ParentId="19287" CreationDate="2016-12-04T22:58:20.823" Score="7" Body="&lt;p&gt;There are multiple formulae for how to calculate this angle. The simplest is to construct the unit vectors:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\hat{n}_i &amp;amp; = \left[\begin{array}{c}&#xA;  \cos \delta_i \cos \alpha_i \\&#xA;  \cos \delta_i \sin \alpha_i \\&#xA;  \sin \delta_i&#xA;  \end{array}\right]&#xA;\end{align}$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\delta$&lt;/span&gt; is the declination and &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; is the right ascension, then take the dot product of the vectors for the two positions, and finally use &lt;span class=&quot;math-container&quot;&gt;$\hat{n}_1 \cdot \hat{n}_2 = \cos \theta_{12}$&lt;/span&gt;. The dot product results in the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\cos(\theta_{12}) = \sin(\delta_1)\sin(\delta_2) + \cos(\delta_1)\cos(\delta_2)\cos(\alpha_1 - \alpha_2)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with that approach is that it loses numerical accuracy very quickly when &lt;span class=&quot;math-container&quot;&gt;$\theta_{12} \ll 1$&lt;/span&gt; radian. Thus the &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Haversine_formula&quot; rel=&quot;nofollow noreferrer&quot;&gt;Haversine formula&lt;/A&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\theta_{12} = 2\operatorname{arcsin} \left(\sqrt{\sin^2\left(\frac{\delta_2 - \delta_1}{2}\right) + \cos \delta_1 \cos \delta_2 \sin^2 \left(\frac{\alpha_2 - \alpha_1}{2}\right)}\right),$$&lt;/span&gt;&#xA;is often preferable. There is an algorithm that covers all cases, &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Vincenty%27s_formulae&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vincenty's formulae&lt;/A&gt;, but it is much more complicated to implement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find a good discussion of practical differences at this &lt;A HREF=&quot;http://gis.stackexchange.com/questions/4906/why-is-law-of-cosines-more-preferable-than-haversine-when-calculating-distance-b&quot;&gt;Geographic Information Systems Stack Exchange Question&lt;/A&gt;.&lt;/p&gt;&#xA;" OwnerUserId="14244" LastEditorUserId="14244" LastEditDate="2019-02-24T23:35:18.983" LastActivityDate="2019-02-24T23:35:18.983" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="19293" PostTypeId="2" ParentId="7869" CreationDate="2016-12-04T23:00:11.317" Score="0" Body="&lt;p&gt;Often stars can form when a supernova compresses surrounding clouds into higher densities with interstellar shock waves, and it can also send the stars outwards from the point of the supernova. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two days after seeing this topic, I read a &lt;a href=&quot;https://twin-cities.umn.edu/news-events/researchers-propose-low-mass-supernova-triggered-formation-solar-system&quot; rel=&quot;nofollow noreferrer&quot;&gt;report&lt;/a&gt; that identifies the imprints of a low mass supernova in meteorites of the solar system, by measuring which radioactive isotopes are in meteorites:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Identifying these 'fingerprints' of the final supernova is what we needed to help us understand how the formation of the solar system was initiated,&quot; Professor Heger said.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;The fingerprints uniquely point to a low-mass supernova as the trigger.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;The findings in this paper have opened up a whole new direction of research focusing on low-mass supernovae,&quot; he said.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In addition to explaining the abundance of Beryllium-10, this low-mass supernova model would also explain the short-lived nuclei Calcium-41, Palladium-107, and a few others found in meteorites.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The speed at which star clusters diverge is relatively high, Teegarden's star for example is 11,502 Million years old, 12 ly away, and it's group's movement has been traced back roughly to a zone of origin, and the group's stars are twice as distant from each other as they were at their formation, the group spans at least 50 light years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't want to get in over my head here, but after 4 billion years, stars of a group can be millions of light years apart. High velocity stars can move at 200 light years per million years (70km/s), so that's 200.000 light years per billion  years. Teegarden's star is moving that fast and it's 0.2 times the mass of the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Teegarden's group are distancing themselves by about .5 light years per million years, and they are currently 10-60 light years away from the sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will be a steep task to find where the Sun came from and where its neighbors are. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proxima Centauri currently moves toward Earth at a rate of 22.4 km/s. Ηowever, after 26,700 years, when it will come as close as 3.11 light-years, it will begin to move farther away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why I am surprised by the journal suggesting a stellar neighbour at 110 ly away, seeing as some of its group can potentially have travelled 200,000 ly although not in a straight line. &#xA;There are proper motion simulations on the web which are interesting to see.&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorDisplayName="user5434" LastEditDate="2017-05-06T22:15:46.943" LastActivityDate="2017-05-06T22:15:46.943" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19294" PostTypeId="2" ParentId="18052" CreationDate="2016-12-04T23:47:38.237" Score="4" Body="&lt;p&gt;It is wasteful to sample with many bits because the signal to noise ratio at the ADC of a radio telescope is typically &amp;lt;&amp;lt; 1, so using many bits would just be resolving noise. (An exception to this is when there is strong radio-frequency interference that needs to be resolved, but this is not a big problem for ALMA due to its location and observing frequencies).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;High dynamic range measurements arise after averaging together many samples (or correlations of samples), which boosts the SNR to a meaningful level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using very few bits at the ADC does introduce quantization noise that reduces the efficiency of the instrument, but 3 bits is enough to achieve 96% efficiency [1].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1029/2006RS003585/full&quot; rel=&quot;nofollow noreferrer&quot;&gt;[1] Convenient formulas for quantization efficiency&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15053" LastActivityDate="2016-12-04T23:47:38.237" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19295" PostTypeId="2" ParentId="19281" CreationDate="2016-12-05T02:32:10.560" Score="1" Body="&lt;p&gt;The main thing a cubic spine lets you do is have a continuous second derivative.  If all you need is a continuous curve, connecting the dots with straight lines suffices, with no overshoot, but a very discontinuous derivative.  So if you want to think about the derivative of the light curve, you'd have to go at least to a quadratic spline, which allows a continuous first derivative but introduces the potential for overshooting (a curve that goes outside the range of the points themselves).  If you actually want to think about the second derivative also, then you'll need that to be continuous, so that will require a cubic spline.  But the potential for overshoot keeps getting worse when you add more continuous derivatives.  Personally, I find cubic splines to be very unreliable, the overshoot can be terrible unless you have a very nicely resolved curve with not much noise, but that may be just what they do have.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-12-05T02:32:10.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19296" PostTypeId="2" ParentId="19281" CreationDate="2016-12-05T03:06:58.657" Score="0" Body="&lt;p&gt;Cubic spline interpolations are used because if you have a good pipeline you only need to adjust 2 variables to get a pretty good fit. Granted, a cubic spline have numerous limitations, and offers a ton of inflection points that won't allow for best fitting minimization. That being said, it is fast, dirty, easy, and widely accepted as good enough. Plus, it oscillates just like a pulsating or variable star, so that's another good excuse to go the cubic spline route.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are literally innumerable scientists and papers that base all of their results on what the cubic spline gives them, in fields from photometry as you've linked, to temperatures derived from Ly$\alpha$ absorption profiles as seen &lt;a href=&quot;http://adsabs.harvard.edu/abs/2011MNRAS.410.1096B&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a &lt;a href=&quot;http://www.geos.ed.ac.uk/~yliu23/docs/lect_spline.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; to a reference on spline interpolation I used in my undergraduate thesis. You might find it helpful.&lt;/p&gt;&#xA;" OwnerUserId="2232" LastActivityDate="2016-12-05T03:06:58.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19297" PostTypeId="1" AcceptedAnswerId="19313" CreationDate="2016-12-05T04:34:02.547" Score="1" ViewCount="68" Body="&lt;p&gt;There are two slits and two slots on EIS, are they just four types of grating?&lt;/p&gt;&#xA;" OwnerUserId="14686" LastActivityDate="2016-12-06T17:53:34.297" Title="What's the slit and slot on EIS(EUV Imaging Spectrometer) instrument of Hinode?" Tags="&lt;the-sun&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19298" PostTypeId="1" AcceptedAnswerId="19299" CreationDate="2016-12-05T05:27:47.850" Score="1" ViewCount="780" Body="&lt;p&gt;mostly, we use the one-dimensional spectrum. But sometimes we use two-dimensional spectrum, what's the differences between them?&lt;/p&gt;&#xA;" OwnerUserId="14686" LastActivityDate="2016-12-05T16:40:13.033" Title="What's the differences between one-dimensional spectrum and two-dimensional spectrum?" Tags="&lt;spectroscopy&gt;&lt;spectra&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19299" PostTypeId="2" ParentId="19298" CreationDate="2016-12-05T07:30:22.183" Score="3" Body="&lt;p&gt;When you place a spectrograph slit on a source, the spectrum recorded can be thought of as lots of images of the slit at different wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ordinarily, you would sum up this spectrum along the direction of the slit images to give you a one dimensional spectrum. If however, you leave the image as recorded, then you have a two dimensional spectrum - intensity as a function of wavelength along one axis and as a function of position along the slit in the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two dimensional spectra are used when we expect the spectrum to vary with position along the slit. Examples might include a spectrum recorded across a galaxy, or a spectrum of a binary star with the slit placed across both components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example is shown below. The inset image shows a broad-band image of V458 Vul - a classical nova that is surrounded by (not visible) shells of ionised material. The authors of this &lt;a href=&quot;http://www.vgoranskij.net/v458/v458atel2.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;particular study&lt;/a&gt; lined up a spectrograph slit as shown in the inset and then obtained the two two-dimensional spectra shown in the main image. What you have to imagine is that each position on the slit produces a horizontal spectrum at a vertical position that corresponds to its position on the slit. Therefore we see a bright spectrum across the middle corresponding to the central source, but there are then &quot;knots&quot; of emission at particular wavelengths that are some distance away from the central star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BZTkR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BZTkR.jpg&quot; alt=&quot;A two dimensional spectrum&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slitless two dimensional spectroscopy is also possible using integral field spectrographs. Fibers record spectra over a two dimensional area. This can also be referred to as two dimensional spectroscopy.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-12-05T16:40:13.033" LastActivityDate="2016-12-05T16:40:13.033" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19301" PostTypeId="1" AcceptedAnswerId="19306" CreationDate="2016-12-05T14:26:24.337" Score="2" ViewCount="745" Body="&lt;p&gt;What is the period of unique horoscopes? In other words how long does it take before any configuration of the classical planets repeats?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The seven classical &quot;wandering stars&quot; are: Sun, Moon, Saturn, Jupiter, Mars, Venus, and Mercury.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume a horoscope is created by dividing the ecliptic into 12 equal regions and then for a particular day noting which region each of the 7 &quot;wandering stars&quot; is occupying relative to the earth. For example, a horoscope could be: Sun in Aries, Moon in Capricorn, Saturn in Taurus, Jupiter in Libra, Mars in Pisces, Venus in Aries, Mercury in Aries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Saturn revolves around the sun once every 29 earth years, Jupiter every 12 years, and Mars 1.88 or so, multiplying together we have 654. So, relative to each, they would seem to repeat their configuration every 654 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is potentially a complicated question which might require a simulation to determine the answer. For example, today (5 December 2016), the following is true for the tropical zodiac:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sun in Sagittarius&lt;br&gt;&#xA;Moon in Aquarius&lt;br&gt;&#xA;Mercury in Capricorn&lt;br&gt;&#xA;Venus in Capricorn&lt;br&gt;&#xA;Mars in Aquarius&lt;br&gt;&#xA;Jupiter in Libra&lt;br&gt;&#xA;Saturn in Sagittarius&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, one instance of this problem would be to determine the next date at which this same configuration will next recur.&lt;/p&gt;&#xA;" OwnerUserId="2800" LastEditorUserId="2800" LastEditDate="2016-12-05T17:14:47.967" LastActivityDate="2020-12-14T01:03:04.313" Title="Period of unique horoscopes?" Tags="&lt;solar-system&gt;" AnswerCount="1" CommentCount="13" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19302" PostTypeId="2" ParentId="19282" CreationDate="2016-12-05T18:21:53.840" Score="2" Body="&lt;p&gt;Depends on the wavelength ranges to be studied as well.  Quoting from the link you provided:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The sole instrument on TESS is a package of four wide-field-of-view&#xA;  CCD cameras. Each camera features a low-noise, low-power 16.8&#xA;  megapixel CCD detector created by the MIT Lincoln Laboratory. Each has&#xA;  a 24° × 24° field of view, a 100 mm (4 in) effective pupil diameter, a&#xA;  lens assembly with seven optical elements, and a bandpass range of 600&#xA;  to 1000 nm&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Atmospheric absorption has some serious peaks near 750 and 920 nm, which is one reason to place these cameras in orbit above the atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2016-12-05T18:21:53.840" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19303" PostTypeId="1" CreationDate="2016-12-05T21:02:24.327" Score="0" ViewCount="524" Body="&lt;p&gt;I have read that Chinese months start on the day on which the new moon is observed from the 120th meridian east.  From this I am assuming that the new moon is observable from there each time and that it doesn't matter where on the meridian you are (although I suppose it's implied you're in the northern hemisphere).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate the time at which the moon is new from a particular meridian?&lt;/p&gt;&#xA;" OwnerUserId="15059" LastActivityDate="2016-12-07T11:48:14.083" Title="How can I calculate when the new moon will be observed from a particular meridian?" Tags="&lt;the-moon&gt;&lt;moon-phases&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19304" PostTypeId="1" AcceptedAnswerId="19305" CreationDate="2016-12-05T21:42:40.277" Score="1" ViewCount="61" Body="&lt;p&gt;Do high velocity stars decelerate if they transit at 300ly per million years? What kind of forces can decelerate them and in what kind of time frame?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about ordinary stars which travel at 70 light years per million, i.e alpha centauri? How often do they change direction and do they end 1000ds of light years away after a billion years?&lt;/p&gt;&#xA;" OwnerUserId="1528" LastActivityDate="2016-12-05T22:35:56.043" Title="how far away can close stars diverge with time?" Tags="&lt;star&gt;&lt;stellar-dynamics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19305" PostTypeId="2" ParentId="19304" CreationDate="2016-12-05T22:35:56.043" Score="5" Body="&lt;p&gt;Only one force counts on galactic scales: gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars like Alpha Centauri are orbiting in the galaxy, as is the sun. Both stars are moving in a similar direction and at roughly the same speed, relative to the galactic centre. The distance to such stars will vary slowly, but both stars will remain within the galaxy, and the stars will stay fairly near each other for some time yet. The direction of stars doesn't change suddenly, but the gravity of the galaxy will pull the star's path into a roughly circular orbit within the galactic disc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;High-velocity stars are still affected by the gravitational pull of the galaxy. The galactic escape velocity is over 500km/s (1800 ly per million years) Stars can have sufficient speed to escape from the galaxy's gravitational pull. Such stars can escape from the galaxy into intergalactic space. They will be decelerated by the gravity of the galaxy, but if their speed is enough, then gravity won't be enough to pull them back. Stars that are moving more slowly than that will pass into the galactic halo, before being pulled back by the gravity of the galaxy. They will then usually orbit in the galactic halo, with an orbital period measured in hundreds of millions of years.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-12-05T22:35:56.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19306" PostTypeId="2" ParentId="19301" CreationDate="2016-12-05T22:37:25.070" Score="4" Body="&lt;p&gt;I wrote &lt;a href=&quot;https://github.com/barrycarter/bcapps/tree/a944a8d8315fd6169907f73e47698f030befe3bb/ASTRO/bc-zodiac.c&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/tree/a944a8d8315fd6169907f73e47698f030befe3bb/ASTRO/bc-zodiac.c&lt;/a&gt; to solve this. I may've tweaked bc-zodiac.c since then, so you'll need the exact link above (not the most recent version of bc-zodiac.c) to confirm/replicate my results. In the directory &lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The file &quot;houses.txt.bz2&quot; contains a list of all house changes for now +- ~15000 years in a mostly-English-readable format.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The file &quot;house-states.txt.bz2&quot; is a list of all &quot;horoscopes&quot;(house states for all planets) in machine readable format for the same period. Please see &lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-zodiac.txt&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-zodiac.txt&lt;/a&gt; (which also contains the text of this answer) for details on how to read this file.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Important caveats below, but first, some notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You'll notice that lines like:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;BCE 13200-JUL-12 09:48 SUN ENTERS ARIES PROGRADE -479626193452 S0+&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;suggest the vernal equinox was on July 12th about 15000 years ago (of course, no one actually used our calendar back then), which seems a bit off. Further, you'll notice:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;CE 17191-FEB-23 00:27 SUN ENTERS ARIES PROGRADE 479386211293 S0+&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;suggesting the vernal equinox will occur on February 23rd (about a month early) 15000 years from now (assuming they continue using our calendar).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://astronomy.stackexchange.com/a/13009/21&quot;&gt;https://astronomy.stackexchange.com/a/13009/21&lt;/a&gt; for why this happens (my answer there has slightly different dates/times, because I used a less accurate model).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In the 11,100,017 days (about 30,391 years) covered, there are 6,359,874 horoscope changes, for an average horoscope change roughly every 1.75 days. Broken down by planet:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The moon changes house 4,875,276 times, about once every 2.3 days.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mercury changes house 450,523 times (including retrograde house changes), about once every 25 days.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Venus changes house 384,971 times (including retrograde), about once every 29 days.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Sun comes in third with 364,691 changes, about once every 30 days. A more exact number is once every 30.4367697585 days. When you multiply this by 12, you get 365.2412371020 days, which, as expected, is a good estimate of the length of the tropical year (the exact value given by &lt;a href=&quot;http://hpiers.obspm.fr/eop-pc/models/constants.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://hpiers.obspm.fr/eop-pc/models/constants.html&lt;/a&gt; is 365.242190402, which is 82 seconds different).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mars changes house (including retrograde) 209,501 times, about once every 53 days.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Jupiter (w/ retrograde) 49,215 times, once every ~225.5 days.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Saturn (w/ retrograde) 25,697 times, ~432 days.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As expected, most of these house changes (6,283,141) are in the forward or &quot;prograde&quot; direction and very few (76,733) are in the retrograde direction. Of course, the Sun and Moon always move prograde, another reason the prograde number is so large.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since there are 12 houses and 7 planets, it would seem there are 12^7 = 35,831,808 possible horoscopes. However, there are actually only 3,732,480:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Since Mercury's greatest elongation is 28 degrees, it must remain within one house of the Sun. Given the Sun's house, there are only 3 possibilities for Mercury's house (either the Sun's house or one of the two adjacent houses).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since Venus' greatest elongation is 47 degrees, it must remain within two houses of the Sun, so there are only 5 possibilities given the Sun's house.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Combining these, we see there are 12*12*3*5*12*12*12 or 3,732,480 possible horoscopes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Of these possibilities there are 2,565,570 (about 69% of the total possible) actual horoscopes. Of these,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;969,067 horoscopes occur only once&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The horoscope &quot;4045424&quot; (translated below) occurs 18 times (most recently in 1799 and next in 5391), but two others occur 17 times, 10 appear 16 times, and many appear 15 times, so it's only the most frequent by coincidence (probably just because of where DE431 happens to start and end).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;4045424&quot; translated is Sun in Leo, Moon in Aries, Mercury in Leo, Venus in Virgo, Mars in Leo, Jupiter in Gemini, Saturn in Leo.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that this horoscope is probably only the most frequent because of how I chose the start and end dates (based on the DE431 ephemeris).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Important caveats:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;My standard disclaimer: please do not rely on anything I do/post anywhere. I can and do make mistakes. Feel free to check my work and alert me to mistakes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you actually want to check my results, you will need to download kernels from NASA. See &lt;a href=&quot;http://naif.jpl.nasa.gov/naif/tutorials.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://naif.jpl.nasa.gov/naif/tutorials.html&lt;/a&gt; for more details on how this all works.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Note that NASA can't predict planet positions (or even the Earth's own orientation) into the far future: they solve differential equations to predict planetary positions and then publish an approximation of those solutions. The further away we get from today, the less accurate these predictions are.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We have a pretty good approximation of how the Earth precesses, but we can't even predict THAT super accurately more than a few months in advance. Note that &lt;a href=&quot;https://naif.jpl.nasa.gov/pub/naif/generic_kernels/pck/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://naif.jpl.nasa.gov/pub/naif/generic_kernels/pck/&lt;/a&gt; shows the high precision precession kernel is only good through 6 Mar 2017, and even THAT is updated twice a week (see &lt;a href=&quot;https://naif.jpl.nasa.gov/pub/naif/generic_kernels/pck/aareadme.txt&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://naif.jpl.nasa.gov/pub/naif/generic_kernels/pck/aareadme.txt&lt;/a&gt; for details). Since precession affects the location of the first point of Aries, this adds another uncertainty.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Precession doesn't change the ecliptic itself, but nutation does. We don't have a super-accurate model of nutation, and I am using the undocumented SPICE functions zzmobliq_ and zzwahr_ as per a private reply to my &lt;a href=&quot;https://naif.jpl.nasa.gov/pipermail/spice_discussion/2016-January/000423.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://naif.jpl.nasa.gov/pipermail/spice_discussion/2016-January/000423.html&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The calendar on &lt;a href=&quot;https://cafeastrology.com/astrologyof2017horoscopes.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://cafeastrology.com/astrologyof2017horoscopes.html&lt;/a&gt; (and the 2016 equivalent) should agree with my results within +- 1 minutes. Remember to convert between ET on their site and UTC in my file.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://aa.usno.navy.mil/data/docs/EarthSeasons.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://aa.usno.navy.mil/data/docs/EarthSeasons.php&lt;/a&gt; can also be used to check some of my calculations (within +- 1 minute), recalling that, in the northern hemisphere:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;SUN ENTERS ARIES&quot; (as in &quot;first point of Aries&quot;) marks the vernal equinox.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;SUN ENTERS CANCER&quot; (as in &quot;Tropic of Cancer&quot;) marks the summer solstice.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;SUN ENTERS LIBRA&quot; marks the autumnal equinox. As nearly as I can tell, there is no well-known associated phrase with this point in the zodiac.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;SUN ENTERS CAPRICORN&quot; (as in &quot;Tropic of Capricorn&quot;) marks the winter solstice.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you use HORIZONS (&lt;a href=&quot;http://ssd.jpl.nasa.gov/?horizons&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ssd.jpl.nasa.gov/?horizons&lt;/a&gt;) to check my work, you'll notice my results are accurate to the minute for about +- 1000 years from now, but drift in accuracy after that. I'm not exactly sure why this is.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I used a &quot;stepsize&quot; of 86400 seconds (1 day), so if a planet changed house and then changed back in under a day, my program may miss it. Someone w/ more computing power may wish to re-run my program with a smaller stepsize.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The moon's position varies with your location on Earth. I used a geocentric model, giving the moon's position for a theoretical observer located at the center of the Earth.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I actually computed planetary barycenters (the center of mass of the planet and its satellites) because NASA doesn't provide planet position information for the entire 30,000 year period I computed (see &lt;a href=&quot;http://ssd.jpl.nasa.gov/eph_spans.cgi?id=A&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ssd.jpl.nasa.gov/eph_spans.cgi?id=A&lt;/a&gt; for details).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For Mercury and Venus, this no effect (no moons), and the Mars' system barycenter is within a few inches of Mars' own center (Phobos and Deimos have very little mass). Jupiter is fairly massive, so its system's barycenter is fairly close to its own center, and even the effect Titan has on Saturn isn't that great.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The human-readable time in houses.txt.bz2 is the output of the C function &lt;a href=&quot;http://naif.jpl.nasa.gov/pub/naif/toolkit_docs/C/cspice/timout_c.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://naif.jpl.nasa.gov/pub/naif/toolkit_docs/C/cspice/timout_c.html&lt;/a&gt; (format &quot;&quot;ERAYYYY##-MON-DD HR:MN ::MCAL ::RND&quot; modified slightly by Perl). They are reasonable for current dates, but not necessarily for past dates because the Gregorian Reformation of the calendar occurred at different times for different countries, and the BCE/CE time system didn't exist prior to 1 CE (and probably not even then). For precision work, see bc-zodiac.txt for details.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As above, I use &quot;ARIES&quot; to mean the &quot;1st house&quot;, ie, the constellation that starts at the vernal equinox (&quot;the first point of Aries&quot;, where the ascending ecliptic meets the celestial equator) and ends 30 ecliptic degrees of ecliptic longitude later.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This was reasonably accurate in 56 BCE, but is certainly not true today as the vernal equinox is in Pisces and drifts towards Aquarius, heralding the &quot;Age of Aquarius&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 30,000+ year period covered by my program, the first point of Aries makes a complete circuit (and a bit more) of the Zodiac, even ending up in Ophiuchus for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the constellation boundaries I mention above are those defined by the IAU. I don't know how (or if) the ancients drew their constellation boundaries.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I am leaving my old answer below, because NASA is a trusted site for astronomers, and WebGeoCalc and HORIZONS are important resources. The step-by-step method I show below can help people here make their own calculations or help others. I am in no way affiliated with NASA or the government of the United States, except as a citizen.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT: As I should have guessed, people have already done this: &lt;a href=&quot;https://cafeastrology.com/astrologyof2017horoscopes.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://cafeastrology.com/astrologyof2017horoscopes.html&lt;/a&gt; (and I'm guessing there are many many other sites too, just google &quot;Mercury enters Aries&quot; &quot;Venus enters Taurus&quot; type stuff).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope to provide a better answer later, but for now:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Visit &lt;a href=&quot;http://wgc.jpl.nasa.gov:8080/webgeocalc/#CoordinateFinder&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://wgc.jpl.nasa.gov:8080/webgeocalc/#CoordinateFinder&lt;/a&gt; and do this:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2PXLm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2PXLm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/YqHiV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YqHiV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/HMquw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HMquw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and click &quot;Calculate&quot;. This will tell you every time Mercury has an ecliptic longitude of 30 degrees, the cusp between astrological Aries and Taurus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do this for all 7 planets and every multiple of 30 degrees (12*7 = 84 times), you can use the results to see when the combination of constellations change. However, this is has several problems:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;It's extremely inefficient&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you hover over &quot;Solar System Kernels kernel set&quot; above, it says the kernels are valid from 1950 to 2050. However, with the options I chose, you can only get results from 1966 to 2034, possibly because the ECLIPDATE frame is only defined in that interval (when I choose years outside this range, I get an unhelpful and inaccurate error-- you may get different results for other planets).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The results are in calendar times; however, this may be my fault. You may be able to get the results in a more numerical and useful format.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The &quot;correct&quot; way to do this is to use CSPICE (&lt;a href=&quot;http://naif.jpl.nasa.gov/naif/tutorials.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://naif.jpl.nasa.gov/naif/tutorials.html&lt;/a&gt;) which lets you make calculations +- 10000 years from now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leaving that aside, let's check our results. If you scroll down a bit in the results window, you'll see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/k0AVl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k0AVl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's check the result for 2017-04-20 by visiting &lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ssd.jpl.nasa.gov/horizons.cgi&lt;/a&gt; and entering these parameters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kOVA0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kOVA0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to get these results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eXrCp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eXrCp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, Mercury's ecliptic longitude crosses 30 degrees between 2017-04-20 and 2017-04-21, so our results are reasonable. Interestingly, Mercury is in retrograde at this time, so it actually crosses from Taurus into Aries, not vice versa.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-12-14T07:58:04.683" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19307" PostTypeId="2" ParentId="19276" CreationDate="2016-12-05T23:03:07.403" Score="1" Body="&lt;p&gt;You're totally wrong, but it's a good question.  Leonard Susskind actually uses something like the scenario that you're referring to in order to demonstrate how to add velocities when things like special relativity (which is a little different from what you're talking about, but similar enough for this discussion) are of concern.  Here are some links to the math behind adding velocities in special relativity (again, not technically what you're talking about but similar enough).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html&lt;/a&gt;&#xA;&lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/einvel.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/einvel.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing in your reference frame (pretend you're an observer for whom the sun is moving at the very, very, very limit of the speed of light) can travel at a speed faster than the speed of light (in a vacuum).  A lot of people can describe the consequences thereof very well (well enough to have actually made sure that it's a solid theory anyway), but I don't think anybody knows exactly &lt;em&gt;why&lt;/em&gt; the speed of light acts like a cosmic speed limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:  Here's the Susskind lecture: &lt;a href=&quot;https://www.youtube.com/watch?v=qfTJP7Soto4&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=qfTJP7Soto4&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15060" LastEditorUserId="15060" LastEditDate="2016-12-05T23:31:47.977" LastActivityDate="2016-12-05T23:31:47.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19310" PostTypeId="2" ParentId="19284" CreationDate="2016-12-06T07:09:32.127" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The sun is used to determine and measuring time in our solar system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Correction: The Sun (or more precisely, the Earth's rotation) &lt;strong&gt;was&lt;/strong&gt; used to determine and measuring time in our solar system. Over two thousand years ago, keeping time via the appearance of the Sun was known to be problematic. The Babylonians, Egyptians, and other ancient cultures knew this. This became problematic with the invention of accurate clocks about four hundred years ago. Kepler argued that a clock rather than the position of the Sun was the proper way to keep time. Clocks and sundials disagree by as much as 16.5 minutes due to the equation of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over a hundred years ago, the Earth's rotation rate was found to be less even less rock solid than thought. The second was redefined to be based on the Earth's orbit about the Sun as opposed to the Earth's average rotation rate. This, too, proved unreliable as we developed ever more accurate timekeeping devices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays we keep time with atomic clocks. To keep time in sync with the rotating Earth, people worldwide observe the difference between the smoothed Earth rotation rate and what atomic clocks tell us. We occasionally have to introduce a leap second to maintain the fiction of a solar-based concept of time. For example, the last minute of 2016 will be 61 seconds long. (Leap seconds are a plague upon humanity. Leap minutes or leap hours would have been a better solution. Ask google. Or ask me. I'm currently dealing with a leap second problem in real life. It is a massive PITA.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even atomic clocks are a bit problematic. A minor problem is that some atomic clocks are more accurate than others. Even more problematic is that even the most accurate of atomic clocks won't keep time with one another due to the ever changing shape of the Earth. While our last ice age ended about twelve thousand years ago, the Earth has not quite recovered from the effects of the kilometer-plus thick sheets of ice that formerly covered parts of the Earth. The ongoing post glacial rebound affects the rate at which different atomic clocks tick.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2016-12-06T07:09:32.127" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19311" PostTypeId="2" ParentId="19282" CreationDate="2016-12-06T14:46:44.787" Score="1" Body="&lt;p&gt;There's several reasons why we might want a space-based exoplanet telescope over a ground-based telescope (or array of telescopes).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;As Carl Witthoft already said, atmospheric absorption limits what we can do from the ground. The atmosphere absorbs at important wavelengths and, more importantly, atmospheric disturbances introduce unwanted noise in your data. In a field where clean, noiseless data is paramount, putting something in space to remove those problems is necessary.&lt;/li&gt;&#xA;&lt;li&gt;Earth-based, exoplanet-hunting telescopes are limited by the sky they can see in a few ways. For one thing, no single Earth-based location can view the entire sky so if you really want full coverage, you need these telescopes in multiple locations around the world and that can get expensive. One example of a project like this is the &lt;a href=&quot;https://en.wikipedia.org/wiki/MEarth_Project&quot; rel=&quot;nofollow noreferrer&quot;&gt;MEarth Project&lt;/a&gt;, but there are plenty of other ones. On the whole, a single, multi-camera telescope in space can cover more of the sky, and do so more easily, than an array of telescopes scattered across the planet. Secondly, the idea of being &quot;limited by the sky&quot; also comes into play when you consider that ground-based telescopes are limited by weather and light pollution. A space-based telescope doesn't have these limitations and can more or less view any star at any time (so long as it is not occluded by the Moon/Earth or too close to the Sun).&lt;/li&gt;&#xA;&lt;li&gt;Ground based telescopes are also limited by another factor (that ties into the previous two) which is that they fairly magnitude limited. Most ground-based exoplanet-hunting telescopes are groups of small telescopes which either (a) specifically target nearby, bright stars, or (b) scan the sky, extracting data for only the brightest stars. Neither of these situations is ideal for really being able to look at 500,000 stars as TESS is proposing. For example the MEarth Project I cited above was originally only targeting 1,976 stars! That's far less than TESS's 500,000, and just like MEarth, I'm sure TESS will continue to expand it's operational parameters.&lt;/li&gt;&#xA;&lt;li&gt;Another factor to consider is that there's already numerous ground-based exoplanet-hunting telescopes. However, there's not nearly as many space-based telescopes dedicated to searching for exoplanets. Why limit ourselves and choose to do one or other when we could do both?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2016-12-07T16:12:55.553" LastActivityDate="2016-12-07T16:12:55.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19313" PostTypeId="2" ParentId="19297" CreationDate="2016-12-06T17:53:34.297" Score="1" Body="&lt;p&gt;You can go back and reference the paper published on this telescope. Specifically it is &lt;a href=&quot;https://www.osapublishing.org/ao/fulltext.cfm?uri=ao-45-34-8674&amp;amp;id=118233#articleBody&quot; rel=&quot;nofollow noreferrer&quot;&gt;Culhane et al. 2007&lt;/a&gt;. This paper talks about the overall telescope, however the paper describing the optics is &lt;a href=&quot;https://www.osapublishing.org/ao/fulltext.cfm?uri=ao-45-34-8674&amp;amp;id=118233#articleBody&quot; rel=&quot;nofollow noreferrer&quot;&gt;Korendyke et al. (2006)&lt;/a&gt;. Unfortunately this optics paper is behind a paywall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Within the Culhane paper, in section 3, the describe the slit mechanism as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Exposure times are controlled by a rotating shutter while a slit exchange mechanism can allow selection of four possible apertures — two spectral slits and two spectral imaging slots.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's about as much detail as they give, except to note in table 1 that the slit and slot widths are 1, 2, 40, and 266 arcsec.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Korendyke optics paper, they give a bit more detail.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Light passing through the slit is dispersed and stigmatically reimaged by the toroidal grating onto two 1024 x 2048 pixel CCD detectors. The focus of the grating can be remotely adjusted with a mechanism. In flight, the mirror will be rotated in ~0.125 arcsec increments about the Y axis [solar north–south (N-S)] to sample different structures of the solar surface with the slit. High-resolution spectroheliograms (raster images) are formed by steadily moving the solar image in fine increments on the spectrograph slit and taking re-peated exposures. An interchange mechanism allows selection among two slits (1 and 2 arc sec wide) and two slots (40 or 266 arc sec wide). The slot observations of the solar disk produce spectrally pure images&#xA;  of large areas of the disk in bright solar emission lines with each exposure. The slot images exhibit a modest spatial blur along the dispersion direction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For completion, they also include the diagram below of the optics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xHKlk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xHKlk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From reading all of this, it appears to me that the various slit/slot widths are designed for taking different types of images/spectra, either of very focused, single-instance areas (such as with the slits) or else of large regions (using the slots).&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-12-06T17:53:34.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19316" PostTypeId="1" CreationDate="2016-12-06T20:01:08.060" Score="1" ViewCount="887" Body="&lt;p&gt;I have a fits file of the galaxy m87 produced by SCUBA 2 and I am wondering how I obtain the fluxes at 450 and 850 microns. the output image gives pixel magnitudes in mJy/beam where I want to find eh flux for a certain part of the galaxy. I am as far as summit the values of all pixels within an aperture but I am not sure where to go from there to obtain the flux as there are more than one pixel within the FWHM. say I have a an aperture of radius 1.5 pixels that gives a sum of 6985.04885203 mJy/beam with a pixel size of 3.99999999 arc seconds and a beam size of 13 arc seconds how do I obtain the flux for that aperture at that wavelength?&lt;/p&gt;&#xA;" OwnerUserId="15070" LastActivityDate="2018-03-12T03:52:45.547" Title="How do I get flux density when I have mJy/beam , beam size and pixel size in an astronomical image" Tags="&lt;galaxy&gt;&lt;photometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19317" PostTypeId="1" CreationDate="2016-12-06T20:53:28.897" Score="2" ViewCount="6370" Body="&lt;p&gt;I have a photo where objects and their shadows (e.g. trees, people) are visible, the Sun is not visible directly, but it is clearly neither dusk nor dawn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume that coordinates of this place are known more or less precisely so that I can find it on google maps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know the date when this photo was taken but don't know &lt;em&gt;the time of day&lt;/em&gt;. Is it possible to estimate it from the photo?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a debate with a friend of mine about this.&#xA;My friend states that a reasonable enough estimate is to calculate the Sun's elevation above the horizon by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\text{elevation} = \arctan \frac{\text{object's height}}{\text{shadow's length}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then use one the myriad websites on the internets to look up the time for that date and location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I doubt it because&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The length of a shadow on the photograph depends on the relative position of the Sun, the object &lt;em&gt;and the camera&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Most importantly, the slope of the ground can't be estimated neither from the photo, nor from google maps. &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/0/04/Apollo14Shadows.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;This picture&lt;/a&gt; is a good example.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Who's right? Am I missing something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it would be interesting to understand the role of uncertainties in this calculation. Assuming that the surface is reasonably smooth and the slope is close to zero, what uncertainties (location, object's height, shadow's length, the slope, camera position) would play the major role? Does it depend on the time of year?&lt;/p&gt;&#xA;" OwnerUserId="15069" LastEditorUserId="9301" LastEditDate="2016-12-07T00:28:45.460" LastActivityDate="2017-12-28T13:27:57.367" Title="Is it possible to tell the time of day by shadows on the photo" Tags="&lt;observational-astronomy&gt;&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19318" PostTypeId="2" ParentId="19317" CreationDate="2016-12-06T22:54:34.350" Score="6" Body="&lt;p&gt;In short: If you know length of shadow and object height (assuming vertical object casting shadow of an horizontal surface), you can tell time if you know date and you can tell date if you know time. If you know direction of shadow (e.g. relative to north) you can tell date and time without knowing any of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most tricky part is to measure objects and shadows from a photograph - because you need their real length and perspective distorts object and shadow lengths -. To do this you need to make some assumptions about the geometry of the objects in the photograph. In fact, if you know the geometry of the settings, after some geometrical calculation you can find the Sun position and therefore date and time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, with an image like &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Aerial_view_of_Washington_Monument_04799v.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt; and a detailed map, you can find date, time, position of camera and even focal length of lens - with a lot of measurement and geometry calculation, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the only objects casting shadows in your image are trees and people, measuring their exact position and size would be impossible and date and time estimations will be quite rough, specially if the image doesn't show any other identifiable and measurable features - and even worse if you can't tell the north from features in the photo. In general, finding date and time from shadows works better with buildings, specially with sharp shaped buildings with known orientation.  &lt;/p&gt;&#xA;" OwnerUserId="14097" LastEditorUserId="14097" LastEditDate="2017-12-28T13:27:57.367" LastActivityDate="2017-12-28T13:27:57.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19320" PostTypeId="2" ParentId="19146" CreationDate="2016-12-07T02:35:48.943" Score="2" Body="&lt;p&gt;The Kepler Field had been covered in the near IR by the 2MASS catalog since that covered the entire northern sky.  There was also a Smithsonian Institute program called the &lt;a href=&quot;http://adsabs.harvard.edu/abs/2005AAS...20711013L&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kepler Input Catalog program&lt;/a&gt; that observed the Kepler field with the 48-inch telescope at the Whipple Observatory on Mount Hopkins, Arizona in the SDSS g, r, i, and z bands plus a custom filter centered on the Mg b lines.  Also many individual objects studied by Kepler were observed in follow up studies.  So, yes this is a well studied region of the sky now.  Now, however, the Pan-Starrs survey has imaged most of the whole sky with several optical filters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the reaction wheel failure on Kepler, it was repurpose as the K2 project.  In this project, it observes a different part of the ecliptic plane every 3 months.  Catalogs of stars and galaxies in each of these locations are created ahead of time with the &lt;a href=&quot;http://adsabs.harvard.edu/abs/2016ApJS..224....2H&quot; rel=&quot;nofollow noreferrer&quot;&gt;EPIC survey&lt;/a&gt;.  In addition, there is a SN/AGN search ongoing with K2 that includes simultaneous groundbased observations using various wide field imaging telescopes around the world such as the SkyMapper in Australia.  A &lt;a href=&quot;https://keplerscience.arc.nasa.gov/k2-supernova-experiment-workshop-on-feb-14-16-2017.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;workshop on this&lt;/a&gt; will be held in February.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2016-12-07T02:35:48.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19322" PostTypeId="1" CreationDate="2016-12-07T05:29:52.460" Score="1" ViewCount="735" Body="&lt;p&gt;I am doing a project on &quot;how astronomers measure distance&quot; where i came about with stellar parallax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i realized how it  works with trigonometry but i am having a slight doubt&#xA;tan p=distance between sun and earth/distance between sun and star&#xA;where i can find out distance between two star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but why do we have to wait for six months?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can measure the distance simply staying in june&lt;/p&gt;&#xA;&#xA;&lt;p&gt;why do we need to go to december again??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can only work with triangle ACD&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iAuCX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iAuCX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14813" LastActivityDate="2016-12-07T07:52:38.893" Title="Stellar parallax" Tags="&lt;star&gt;&lt;observational-astronomy&gt;&lt;parallax&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19324" PostTypeId="2" ParentId="19322" CreationDate="2016-12-07T07:52:38.893" Score="2" Body="&lt;p&gt;There is obviously the practical problem that we cannot observe from the Sun! But of course there would be a time of year when the Earth was on the projection of the line CD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cartoon is just that. In practice, measurements are taken at different points in the orbit on multiple occasions over several years. Two measurements would not be enough to separate the apparent motion of the star from parallax from its proper motion on the sky. The cartoon is shown in that way because making observations at those points for that star, will give the largest measurable difference in position and therefore the most precise estimate of the parallax.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-12-07T07:52:38.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19325" PostTypeId="1" CreationDate="2016-12-07T10:48:32.323" Score="0" ViewCount="1197" Body="&lt;p&gt;The present northern direction to the nearest 0.5 degrees with degree zero being the start of January and measured counterclockwise (same as Earth's orbit relative to celestial north pole). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the assumptions below do not allow a precision of 0.5 degrees then I imagine this question is unanswerable without specifying a specific date and time, in which case the specific date and time would be 1 January 2017, 00:00:00 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumptions: (1) That the Earth's orbit around the sun is a perfect circle (I know it's very slightly elliptical but intuitively imagine that won't significantly affect the precision I need). (2) That the Sun's center is at the center of Earth's circular orbit (I know the Sun's center isn't exactly but don't imagine that will affect the accuracy I need either).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 1: I've found three general answers from three different sites, but, two of them appear to me to conflict and the other is too imprecise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 2: This is my first post to this site and I did not plow through the hundreds of questions related to these tags. So if this question has already &#xA;been answered I apologize and please be merciful and point me in right direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="15080" LastEditorUserId="15080" LastEditDate="2016-12-07T16:20:19.270" LastActivityDate="2016-12-10T05:09:49.240" Title="What is direction of Sun's axis of rotation relative to the Earth's orbital plane around the Sun?" Tags="&lt;the-sun&gt;&lt;solar-system&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19326" PostTypeId="2" ParentId="19325" CreationDate="2016-12-07T11:26:18.263" Score="2" Body="&lt;p&gt;Well, Wikipedia says that it's 7.25°to the ecliptic and gives a &lt;a href=&quot;http://nssdc.gsfc.nasa.gov/planetary/factsheet/sunfact.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Looking there clarifies the value in 3 dimensions:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;North Pole of Rotation&lt;/p&gt;&#xA;&lt;p&gt;Right Ascension: 286.13&lt;br /&gt;&#xA;Declination    :  63.87&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Since those coordinates are &lt;a href=&quot;https://en.wikipedia.org/wiki/Right_ascension&quot; rel=&quot;nofollow noreferrer&quot;&gt;relative to the Earth’s axis&lt;/a&gt;, being a &lt;em&gt;practical&lt;/em&gt; measurement system for Earth ground observers to use in pointing their telescopes, you need to combine that with Earth's orbit.&lt;/p&gt;&#xA;&lt;p&gt;As a useful visualization, &lt;em&gt;mark that point&lt;/em&gt; on a model of the celesial sphere.  Compare that with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ecliptic&quot; rel=&quot;nofollow noreferrer&quot;&gt;ecliptic&lt;/a&gt; pole: the difference between thise points is what you want to know.&lt;/p&gt;&#xA;&lt;p&gt;I expect you’ll find they are separated by 7.25°, but by marking a model sphere you’ll understand the actual direction.&lt;/p&gt;&#xA;" OwnerUserId="7142" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-12-07T11:26:18.263" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19327" PostTypeId="2" ParentId="19303" CreationDate="2016-12-07T11:48:14.083" Score="1" Body="&lt;p&gt;You need an &lt;a href=&quot;https://en.wikipedia.org/wiki/Orrery&quot; rel=&quot;nofollow noreferrer&quot;&gt;Orrery&lt;/a&gt;. Specifically, one that includes Earth’s rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My noting the meridian on the model Earth, you can &lt;em&gt;see&lt;/em&gt; how the Earth may be rotated some amount each time the Moon passes on a line between the Earth and Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/New_moon#Determining_new_moons:_an_approximate_formula&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt; page gives a formula.  Take the result (given in UT) and apply the timezone offset for the desired longitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same page discusses Lunar calindars and may have the info you are really interested in.&lt;/p&gt;&#xA;" OwnerUserId="7142" LastActivityDate="2016-12-07T11:48:14.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19328" PostTypeId="1" CreationDate="2016-12-07T12:54:04.790" Score="2" ViewCount="35" Body="&lt;p&gt;I'm looking to get some optical equipment that will be used for both terrestrial and astronomical viewing (ratio 70/30).  It will also have to be portable to a degree although I realise that with a proper mount it wont be able to be carried for long distances.  I've a long held interest in astronomy but never had anything more powerful that 10x50 binoculars.  I've no great leaning to what I want to view but I do understand that planetary viewing might be limited with the items above compared to more dedicated gear but as there does have to be fair degree of terrestrial use then that's always going to be an issue&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've narrowed it down to 3 choices which all come within my budget of ~ 350 GBP(including tripod / mount).  These are:&#xA;Helios 4x100 binoculars&#xA;Celestron 25x100 spotting scope&#xA;Startravel 120 OTA refractor (which has a correcting prism)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I appreciate that each have their plusses and minuses and that there's always going to be compromises so would be grateful for any advice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in anticipation&lt;/p&gt;&#xA;" OwnerUserId="15081" LastActivityDate="2016-12-07T12:54:04.790" Title="Equipment question" Tags="&lt;binoculars&gt;&lt;refractor-telescope&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19330" PostTypeId="1" CreationDate="2016-12-07T15:04:30.737" Score="2" ViewCount="105" Body="&lt;p&gt;If alliptical galaxies seem to a line with their associated galactic filament and that orientation is related to connections between the mass of said galaxy and the dominating central object, which evolves that orientation. Is there a relation between these filaments and the supermassive central object as a start or destination point? &lt;/p&gt;&#xA;" OwnerUserId="15083" LastActivityDate="2016-12-07T15:04:30.737" Title="Galactic Filaments and their connection to Galaxy centre" Tags="&lt;galaxy&gt;&lt;cosmology&gt;&lt;supermassive-black-hole&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19331" PostTypeId="2" ParentId="14885" CreationDate="2016-12-07T20:28:59.367" Score="0" Body="&lt;p&gt;This is one of the most essential questions in cosmology.  The first gedankenexperiment probably. If the universe were finite what could be at the boundary? If there were some sort of wall, what would be on the other side?  Can space just end?  What would stop one from pushing past the end?  If you extend your arm out past the end does your arm cease to exist?  It is because we cannot provide any logically consistent answers to these questions that we accept that there are only two possibilities for the boundaries of the universe: 1) the universe is infinite and has no boundaries or 2) the universe is finite, but bends around and closes in on itself like the surface of a sphere and has no boundaries.  That is, there are no boundaries. Alternative 2 then forces one to contemplate that there is at least one more dimension to space then the known 3.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2016-12-07T20:28:59.367" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19332" PostTypeId="2" ParentId="19282" CreationDate="2016-12-07T20:38:40.073" Score="1" Body="&lt;p&gt;I think the biggest issue is the precision of the photometry.  The atmospheric turbulence limits ground-based photometric precision to 1%.  In space, Kepler achieved about 0.001% on bright stars over a few hours.  This precision is needed to detect an earth-sized planet transiting a sun-sized star.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2016-12-08T01:24:47.007" LastActivityDate="2016-12-08T01:24:47.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19333" PostTypeId="1" AcceptedAnswerId="19334" CreationDate="2016-12-07T21:40:18.487" Score="5" ViewCount="1183" Body="&lt;p&gt;I want to make clock that shows current time on different solar bodies. For that I need to know how to count time on those planets. Is there any official definition of times on other planets in our solar system? Brief google search did not help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am capable of implementing library for this myself but I wonder about current &lt;em&gt;state of the art&lt;/em&gt; of time telling on other planets. So basically I want to know these things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is there any official time standard?&lt;/li&gt;&#xA;&lt;li&gt;If so, where can I find it?&lt;/li&gt;&#xA;&lt;li&gt;If not, how should I approach this problem? Should I make something like Unix epoch at some time in the past for every planet?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks for any tips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Ok, so on Mars, seconds, minutes and hours were made longer to keep 24-hour day. Is this viable to other planets as well? Venus aside for now :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was also thinking about keeping the hours the same, having days as long as many whole hours fits in them plus smaller compensation hours which would have length of (total_minutes_in_solar_day % 60). This concept is from David Weber novels, if you are familiar with them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT2: Based on discussion in the comments, I'll try to make some basic model that would be if possible applicable to any planetary body (thinking ahead :)). This is obviously not going to have any serious real world application but I would still like it to have some sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your input, I will leave the question open for some more time if anyone wants to add their opinion.&lt;/p&gt;&#xA;" OwnerUserId="15089" LastEditorUserId="15089" LastEditDate="2016-12-07T22:26:26.067" LastActivityDate="2020-08-20T12:40:03.403" Title="&quot;Official&quot; time for other planets" Tags="&lt;planet&gt;&lt;solar-system&gt;&lt;time&gt;" AnswerCount="2" CommentCount="19" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19334" PostTypeId="2" ParentId="19333" CreationDate="2016-12-07T21:49:39.973" Score="5" Body="&lt;p&gt;Time on Mars is measured in &lt;a href=&quot;https://en.wikipedia.org/wiki/Timekeeping_on_Mars&quot; rel=&quot;nofollow noreferrer&quot;&gt;sols&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://space.stackexchange.com/q/18655/58&quot;&gt;Venus&lt;/a&gt; could be a little trickier, depending on your application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not aware of any timekeeping system for any other planets. There hasn't been a need for one yet. So, again depending on your application, you could come up with various different possible schemes.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Speculation (based on the OP's input regarding application):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mercury - because of the length of the day, I doubt would-be colonists would measure time in days. They would probably use Earth time (because of communications) and also keep track of Mercury years.&lt;/li&gt;&#xA;&lt;li&gt;Jupiter, Saturn, Uranus, Neptune - no colonization, no timekeeping&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2016-12-07T22:21:05.810" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19335" PostTypeId="2" ParentId="19282" CreationDate="2016-12-08T00:28:10.893" Score="1" Body="&lt;p&gt;The raison d'etre for TESS is to do a precise and complete survey for exoplanets, that are &lt;em&gt;smaller&lt;/em&gt; than Jupiter, around bright stars in a short period of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason to go to space, is to do more precise photometry than is possible (in an efficient way) from the ground looking through the Earth's atmosphere. This allows the detection of Neptune and even Earth-sized planets, which are currently not detected by the all-sky ground-based experiments like WASP and HATnet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The predicted yield and improvement is shown below (from the &lt;a href=&quot;https://tess.gsfc.nasa.gov/whytess.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA TESS&lt;/a&gt; site.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Nwhvt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nwhvt.jpg&quot; alt=&quot;Predicted TESS yield&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best prospect from the ground now appears to be &lt;a href=&quot;http://www.ngtransits.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;NGTS&lt;/a&gt;, which will probably have the precision to get down to the sub-Neptune sized planets, but which will be limited to the southern hemisphere. NGTS started last year, so as long as it can get results out very soon, then it will steal a march on TESS.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-12-08T09:11:23.637" LastActivityDate="2016-12-08T09:11:23.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19336" PostTypeId="1" AcceptedAnswerId="19337" CreationDate="2016-12-08T01:10:37.810" Score="5" ViewCount="233" Body="&lt;p&gt;From what I've read, given the speed of stars rotating within a galaxy and the known mass, the stars' speed is too fast to stay in orbit. Dark matter was proposed to explain this but is it possible that there are undiscovered smaller black holes with no orbiting stars?&lt;/p&gt;&#xA;" OwnerUserId="15091" LastActivityDate="2016-12-09T20:04:55.087" Title="Could undiscovered smaller black holes within galaxies be an explanation for dark matter?" Tags="&lt;black-hole&gt;&lt;galaxy&gt;&lt;dark-matter&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19337" PostTypeId="2" ParentId="19336" CreationDate="2016-12-08T02:08:42.793" Score="9" Body="&lt;p&gt;MACHOs (massive compact halo objects) are thought to make up a portion of the dark mass and portion of these are thought to be black holes, but there are several very solid reasons why black holes not generally seen as a serious alternative explanation to non-baryonic dark matter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Observed elemental abundances are not consistent with the amount of baryonic matter being produced in the early Universe to explain the dark mass&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Microlensing searches for MACHOs of the size of black holes formed by stellar evolution rules them out as making a significant amount of dark matter in the Milky Way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Black holes are observable through accretion. For black holes to make a significant amount of dark matter they would have to be in isolated locations within the galaxy so as not to be visible. However most dark matter must be distributed evenly throughout the Milk Way to explain the rotation curve.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastEditorUserId="2092" LastEditDate="2016-12-08T19:07:05.513" LastActivityDate="2016-12-08T19:07:05.513" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19339" PostTypeId="2" ParentId="18935" CreationDate="2016-12-08T06:25:17.687" Score="2" Body="&lt;p&gt;There are different types of signal to noise ratios ($S/N$) that you might need to consider. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Read Noise dominated $S/N$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Extremely faint sources &lt;em&gt;and&lt;/em&gt; background&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Or a really poor detector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you are RN dominated&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$N\approx\sigma_{\rm RN}$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;which means that your $S/N$ is proportionate to the exposure time $t$.&#xA;$$\frac{S}{N}\Bigg|_{\rm RN}=\frac{f_{\lambda}A_{\rm eff}\Delta\lambda\eta_{\lambda}/h\nu)t}{\sigma_{\rm RN}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f$ is the specific flux, $A$ is the area of the primary mirror area, $\Delta\lambda$ is the wavelength range of your filter/optics, $\eta$ is the end-to-end efficiency of the system, $h\nu$ is your energy, $t$ is the exposure time, and $\sigma_{\rm RN}$ is the read noise error.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Source dominated $S/N$&lt;/strong&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Source count rate significantly exceeds that of all others&lt;/li&gt;&#xA;&lt;li&gt;Typically for bright sources&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$N\approx\sigma_{\rm obj}$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Here your $S/N$ is proportional to $(f_{\lambda}t)^{1/2}$&#xA;$$\frac{S}{N}\Bigg|_{\rm obj}=\sqrt{\frac{f_{\lambda}A_{\rm eff}\Delta\lambda\eta_{\lambda}t}{h\nu}} $$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Background dominated $S/N$&lt;/strong&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Common for faint sources or very bright backgrounds&lt;/li&gt;&#xA;&lt;li&gt;$N\propto\sigma_{\rm sky}$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;like the previous, $S/N$ is proportional to $t^{1/2}$&#xA;$$\frac{S}{N}\Bigg|_{\rm obj}=\frac{f_{\lambda}t^{1/2}\sqrt{A_{\rm eff}\Delta\lambda\eta_{\lambda}}}{\sqrt{\mu_{\lambda}A_B}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\mu$ is surface brightness and $A_B$ is the background area.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another option, which might be a TON easier, depending upon the reason you're trying to calculate the $S/N$... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;a href=&quot;http://www.stsci.edu/hst/stis/software/planning/etc/&quot; rel=&quot;nofollow noreferrer&quot;&gt;HST&lt;/a&gt; and &lt;a href=&quot;https://www2.keck.hawaii.edu/inst/esi/etc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Keck&lt;/a&gt;, there is software on their respective sites, that estimates the $S/N$ for an observation, is intended to capture all of the factors for a given instrument+telescope (ie: $\eta_{\lambda}$, $A_{\rm eff}$).&lt;/p&gt;&#xA;" OwnerUserId="2232" LastActivityDate="2016-12-08T06:25:17.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19342" PostTypeId="2" ParentId="14885" CreationDate="2016-12-08T13:50:04.530" Score="3" Body="&lt;p&gt;One important concept that has not been mentioned yet is the &quot;cosmological principle.&quot;  This is the key simplifying idea for our cosmological models, it says that the universe is the same everywhere on the largest scales at a given age, so alien astronomers 50 billion light years from us that also conclude the universe is 13.8 billion years old will be observing pretty much all the same things we are.  This principle is not proven by data, but the data is consistent with it, most notably the Hubble law and the homogeneity of the cosmic microwave background.  It allows us to understand the past of our own part of the universe by seeing the past of distant parts (a constraint we are stuck with given the speed of light).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that's why a universe model that is finite in size must curve back on itself-- nothing else would satisfy the cosmological principle.  This doesn't mean the principle is true, it means we are not going to part with it unless we have to.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-12-08T13:50:04.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19343" PostTypeId="1" AcceptedAnswerId="19347" CreationDate="2016-12-08T14:00:35.213" Score="3" ViewCount="322" Body="&lt;p&gt;Ignoring the turbulent beginnings of our solar system I wonder if there is a meaningful way to calculate the relative position of the planets of our solar system 100 million years into the future using the standard parameters &#xA;N /  i  /  w /  a /  e   /  M / +- pertubations , {assuming no unannounced black hole or neutron star zips through our solar system in the meantime.} &#xA;How far can the orbital calculations be projected into the future before error accumulation make these calculations completely meaningless? &lt;/p&gt;&#xA;" OwnerUserId="15095" LastEditorUserId="15095" LastEditDate="2016-12-09T00:26:56.097" LastActivityDate="2016-12-09T17:05:40.203" Title="How to calculate planetary position over vast time spans" Tags="&lt;solar-system&gt;&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19345" PostTypeId="1" CreationDate="2016-12-09T05:20:01.440" Score="6" ViewCount="682" Body="&lt;p&gt;JW Space Telescope is designed for a 5-10 year mission duration. Seeing as Hubble and other space missions have paved the way for JWST for reliability issues, It even has zero friction gyro's with 100+ years mean time before failure. Is it reasonable to think that JWST can last longer than Hubble?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;JWST will orbit a Lagrange point in earth's shadow and will be 4 times further than the moon, around 10 days of travel away, and NASA states that JWST won't be serviceable. Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are JWST's fuel limitations so short? Are there no provisions for refueling? even a refueling craft? How many gallons of dinitrogen tetroxide ($N_2O_4$) as oxidizer and hydrazine ($N_2H_4$) will it contain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely NASA are secretly hoping for a 20 year mission?&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorDisplayName="user15217" LastEditDate="2016-12-20T04:24:40.910" LastActivityDate="2016-12-20T04:24:40.910" Title="Will JWST be as durable as the hubble telescope?" Tags="&lt;telescope&gt;&lt;satellite&gt;&lt;lagrange-point&gt;&lt;nasa&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19346" PostTypeId="2" ParentId="19345" CreationDate="2016-12-09T05:49:30.363" Score="15" Body="&lt;p&gt;Hubble was in low earth orbit, and was always intended to be serviceable. In fact, the original plan for Hubble was to have the space shuttle carry it down from orbit and take it back up, but they decided that was too risky compared to servicing in orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;JWST, on the other hand, will be at the Earth/Sun L2 Lagrange point, like &lt;A HREF=&quot;http://map.gsfc.nasa.gov/mission/observatory_l2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;WMAP and Planck&lt;/A&gt; before it. The distance to that point is 0.01 astronomical units, which is about 1.5 million kilometers from Earth's center. The Moon's orbit around the Earth is &lt;A HREF=&quot;http://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;about $1/4^{\mathrm{th}}$&lt;/A&gt; of that. So, a manned servicing mission would be an unprecedented project (beyond Apollo), and an unmanned mission, also unprecedented, would face automation challenges from the 5 second light travel time (10 second round trip).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the L2 Lagrange point is only meta-stable (stable to offsets in two directions, unstable along the Earth-Sun line), JWST will have to expend fuel on station-keeping just to stay there. It needs to stay there because it can only be as cool as it needs to be by using the Earth as a sun-shade - those tennis court sized shades built in are actually to shade JWST from the thermal radiation of Earth &amp;amp; the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put simply, it would be cheaper to build and launch another JWST than to service the existing one.&lt;/p&gt;&#xA;" OwnerUserId="14244" LastEditorUserId="14244" LastEditDate="2016-12-13T22:28:39.210" LastActivityDate="2016-12-13T22:28:39.210" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19347" PostTypeId="2" ParentId="19343" CreationDate="2016-12-09T13:13:36.967" Score="4" Body="&lt;p&gt;Jacques Laskar, at IMCCE, has been working on describing planetary motions on long term time scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quoting and translating &lt;a href=&quot;http://www.bourbaphy.fr/laskar.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;a paper by him&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Motion in the solar system is chaotic, meaning that trajectories diverge exponentially: an error on the location of a planet gets multiplied by 10 every 10 million years&lt;/li&gt;&#xA;&lt;li&gt;This means that a 15 meter error on the initial location of the Earth yields an error of 150 meters after 10 million years, and 150 million km afer 100 million years. Long term predictions involving a single trajectory are impossible.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For predictions over longer periods, it becomes necessary to study multiple possible trajectories. Laskar and Gastineau have &lt;a href=&quot;https://www.imcce.fr/fr/presentation/equipes/ASD/person/Laskar/papers/2009/nature08096%26s1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;published results of a simulation&lt;/a&gt; of the solar system involving 2501 variations of the location of Mercury. In this simulation, &quot;initial conditions  differ by only 0.38 mm in the semi-major axis of Mercury&quot;. Some of those trajectories involve &quot;possible collisions of Mercury, Mars or Venus with the Earth&quot;.&lt;/p&gt;&#xA;" OwnerDisplayName="user15104" LastEditorDisplayName="user15104" LastEditDate="2016-12-09T17:05:40.203" LastActivityDate="2016-12-09T17:05:40.203" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19348" PostTypeId="1" AcceptedAnswerId="19349" CreationDate="2016-12-09T14:08:30.950" Score="1" ViewCount="460" Body="&lt;p&gt;Before it was widely accepted that the Earth was round, how did astronomers describe the yearly movement of the sun to south in the Northern Winter that produced the seasons? Did the Earth wobble or did the Sun just not have a perfectly circular trajectory around the Earth? Did this perspective change when we realized the Earth was a globe? What about before Heliocentrism was the norm?&lt;/p&gt;&#xA;" OwnerUserId="15106" LastEditorUserId="864" LastEditDate="2016-12-09T15:52:49.437" LastActivityDate="2016-12-09T17:13:58.350" Title="Did Earth &quot;tilt&quot; seasonally before it was accepted that the world was not flat?" Tags="&lt;the-sun&gt;&lt;history&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19349" PostTypeId="2" ParentId="19348" CreationDate="2016-12-09T16:54:54.357" Score="2" Body="&lt;p&gt;This is not really a physics or astronomy question, but I reckon that when you model the Earth as flat, the stars and the Sun are merely minor objects and their motion is governed by heavenly spirits rather than regular orbits. So, rather then the flat Earth tilting, the Sun's trajectory changed with the seasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Btw, what I find more concerning is how the (infinitely extended) flat Earth can be reconciled with the Sun rising in the East after setting in the West: how did it get there?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Btw2, I don't think that elites in classic times (Greek/Egyptian/Middle East Cultures ~1000Bc and more recent) seriously considered the Earth to be flat. All the constructions of epicyles etc for explaining the data assume a spherical Earth in the centre of the cosmos. Flat is much more arcane than Geocentric.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2016-12-09T17:13:58.350" LastActivityDate="2016-12-09T17:13:58.350" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19358" PostTypeId="2" ParentId="19325" CreationDate="2016-12-10T05:09:49.240" Score="1" Body="&lt;p&gt;This isn't really a great answer, but it has an image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lESX5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lESX5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planetarium programs like Stellarium let you view the sky from different planets, including the Sun. As above, the Sun's &quot;north pole&quot; is in Draco, very close to NGC 6552 (note it's declination puts it about 2.5 minutes away from the Sun's north pole).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although Stellarium is fairly accurate, this answer shouldn't be regarded as canonical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;REMINDER: Do not observe directly from the surface of the Sun. You will be burnt to a crisp ;) &lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2016-12-10T05:09:49.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19359" PostTypeId="1" CreationDate="2016-12-10T10:10:15.530" Score="0" ViewCount="114" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rCWoI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rCWoI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If hypothetically, the vantage point would move 2AU along an orbit around the star being measured, would that make it seem as if the other star moved twice the distance over the back-drop?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="15119" LastActivityDate="2017-08-15T23:58:50.503" Title="If a star orbited another star, how would that affect parallax measurements of the other star from the vantage point of the first?" Tags="&lt;star&gt;&lt;solar-system&gt;&lt;universe&gt;&lt;data-analysis&gt;&lt;parallax&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19360" PostTypeId="1" AcceptedAnswerId="19361" CreationDate="2016-12-10T10:57:44.513" Score="1" ViewCount="182" Body="&lt;p&gt;When I'm looking at the inside of a mirror telescope:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U76M9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U76M9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering why the secondary mirror does not block half of the incoming light? Is it &quot;transparent&quot; in this direction?&lt;/p&gt;&#xA;" OwnerUserId="15120" LastActivityDate="2016-12-10T14:23:19.290" Title="Mirror Telescope" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19361" PostTypeId="2" ParentId="19360" CreationDate="2016-12-10T14:23:19.290" Score="2" Body="&lt;p&gt;In a Newtonian reflector, as pictured, the secondary mirror does block some of the light, but maybe less than you think.  Even if the secondary were half the diameter of the primary, it would only block 1/4 of the light ($\ (1/2)^2$).  In a more typical case the secondary would be somewhat smaller - perhaps a quarter of the size of the primary (or less).  Hence a $1/16th$ or less is blocked, which is not too bad.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2016-12-10T14:23:19.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19362" PostTypeId="1" AcceptedAnswerId="19364" CreationDate="2016-12-10T15:43:21.490" Score="2" ViewCount="117" Body="&lt;p&gt;I'm trying to decode the Variable Types of stars in the Hipparcos Library.  I see there are a number of stars for example :-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EA/AR - Does that mean the star is both a EA(Eclipsing Binary Star) and AR (Detached System of AR Lacertae) type variable star ? e.g. HIP112009&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whats the difference between EA (Delta Orionis) and EA: (Omicron Puppis) type stars ?  What does the colon signify?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edited&lt;/em&gt; - So on line, 2575 of hip_va_1.dat, the following line has a Variable Type of EA/AR, what does the slash signify.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;112009| |F5V         |P|EA/AR | | 9.016| | 9.65 |  -1.7|   2.96850  |  -4.1|8501.4698|4|A| |W_Gru       |   2.96850|2430132.15| 9.40|10.00|P|R&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What does the (+) sign signify in a Variable Type e.g. SR+ZA (HIP36623)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really like to understand what the significance of the +, / and : mean &lt;/p&gt;&#xA;" OwnerUserId="5401" LastEditorUserId="7982" LastEditDate="2021-04-16T23:20:05.010" LastActivityDate="2021-04-16T23:20:05.010" Title="Variable Types of Stars" Tags="&lt;variable-star&gt;&lt;hipparcos&gt;&lt;eclipsing-star-systems&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19364" PostTypeId="2" ParentId="19362" CreationDate="2016-12-10T19:59:22.480" Score="4" Body="&lt;p&gt;It seems to me that you're mostly asking about star naming conventions, which is unfortunately a difficult thing to master because there are many many conventions. What makes this process difficult is that the conventions you seem to be referencing don't come from Hipparcos at all. Likely you're seeing the names of these stars as they're referred to by some other, more popular convention. I'll try to go through the things you have listed and describe how those conventions work.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Hipparcos Names&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;A star in the Hipparcos catalogue is given a number. That number is preceded by HIP and you cite a good example of &lt;em&gt;HIP 36623&lt;/em&gt;. The number itself is more or less meaningless for the star. Some catalogues provide numbers to stars based on their RA and/or Dec, but for the Hipparcos catalogue the number is the same as the that in the Hipparcos Input Catalog (all HIP stars were pre-chosen for observation and defined in an Input Catalog). The star entries are, with a few exceptions, ordered by increasing HIP number, which basically follows the order of the object's right ascension (Equinox J2000) independent of declination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find a pretty good list of Hipparcos variable conventions on &lt;a href=&quot;https://heasarc.gsfc.nasa.gov/W3Browse/all/hipparcos.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Bayer Convention&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;You also cite the names Delta Orionis and Omicron Puppis. These names come from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_naming_conventions#Bayer_designation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bayer naming convention&lt;/a&gt;. The general scheme for this convention is to group stars by the constellation they're in. Then you rank all of them by brightness. You give each star a name which includes a Greek letter then the Latin name of the constellation it exists in. The brightest star is alpha (equivalent to A), the next brightest beta (equivalent to B), then gamma (equivalent to C), and so on. There is a convention about what to do once you've run out of Greek letters, but honestly, I've never seen anyone use the Bayer convention for such stars. It's generally only used for the brightest stars in a constellation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the examples you gave, Delta Orionis is the 4th brightest star in the Orion Constellation (since delta is the fourth letter in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Greek_alphabet&quot; rel=&quot;nofollow noreferrer&quot;&gt;Greek alphabet&lt;/a&gt;). For Omicron Puppis, that's the 15th brightest star in the Puppis constellation. In this convention, you can sometimes see the greek letter written as a Greek letter, rather than spelled out, and occasionally the constellation name is shortened. Thus, for your examples, you might see them written as $\mathrm{\delta}$ Ori and $\mathrm{\omicron}$ Pup. Note that lower-case Greek letters are used (first), not upper case.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Variable Star Type Designations&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Variable stars are complex and come in many many different flavors. As such, there's a whole slew of different names and schemes to be used when referring to a variable star. A good reference site is the &lt;a href=&quot;https://www.aavso.org/vsx/index.php?view=about.vartypes&quot; rel=&quot;nofollow noreferrer&quot;&gt;International Variable Star Index page&lt;/a&gt;. You can refer here to answer most of your questions, but I'll post the descriptions of the non-alphabetic characters.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A pipe character ( $|$ ) between two different types signifies a logical OR; the classification is uncertain and all possible types are indicated. An example of this is ELL|DSCT, where the star may be an ellipsoidal binary system or a DSCT-type pulsating variable with half the given period.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A plus character ( $+$ ) signifies a logical AND; two different variability types are seen in the same star or system. An example of this would be ELL+DSCT, where one of the components of an ellipsoidal binary system is a DSCT-type pulsating variable.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A slash character ( $/$ ) indicates a subtype. In the case of binary systems (eclipsing, ellipsoidal or reflection variables) it is used to help describe either the physical properties of the system (E/PN or EA/RS), the luminosity class of the components (EA/DM), or the degree of filling of their inner Roche lobes (EA/SD). This is the GCVS classification system. In cataclysmic variables, slash characters are used to indicate some properties of the system, as in the degree of polarization (NA/DQ) or the nature of their components (UG/IBWD).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This site doesn't specifically indicate the meaning of the colon, but research elsewhere indicates that the colon is used to &quot;indicate the uncertain nature of the classification&quot;. It looks to me that within variable star designations, and particularly by the General Catalogue of Variable Stars, a colon used anywhere (not just on the variable type designation) implies uncertainty.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2016-12-10T19:59:22.480" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19366" PostTypeId="1" CreationDate="2016-12-11T01:04:10.210" Score="2" ViewCount="384" Body="&lt;p&gt;For the following graph, I have some confusions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How do galaxies rank by mass? We know that on main-sequence, more luminous stars are also more massive, so I guess it also applies to galaxies.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do red sequence galaxies split into blue galaxies, or do blue galaxies merge into bigger red galaxies?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can galaxies ever be born as red sequence members? I doubt it because all galaxies should start from blue. But I am not 100% sure.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do red sequence galaxies tend to become redder over time?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do large galaxies use up star forming gas more rapidly than smaller galaxies?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/F77mD.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F77mD.jpg&quot; alt=&quot;color-luminosity chart for galaxies&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14254" LastActivityDate="2016-12-11T01:04:10.210" Title="What characteristics or patterns do galaxies have on color-luminosity chart?" Tags="&lt;galaxy&gt;&lt;star-formation&gt;&lt;luminosity&gt;&lt;main-sequence&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ClosedDate="2016-12-12T17:04:04.403" ContentLicense="CC BY-SA 3.0" />
  <row Id="19367" PostTypeId="1" AcceptedAnswerId="19372" CreationDate="2016-12-11T03:01:08.503" Score="8" ViewCount="267" Body="&lt;p&gt;What events led to the nearly universal acceptance of the letter &quot;&lt;em&gt;z&lt;/em&gt;&quot; as the denotation of redshift? What did the letter originally stand for?&lt;/p&gt;&#xA;" OwnerUserId="15129" LastActivityDate="2016-12-11T23:04:57.047" Title="Why is the letter *z* customarily used to denote redshift?" Tags="&lt;redshift&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19369" PostTypeId="1" CreationDate="2016-12-11T05:09:47.717" Score="1" ViewCount="82" Body="&lt;p&gt;Just received an old &lt;a href=&quot;https://cl.ly/3j0s2m2V351d/29049_6-98.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Orion Deep Space Explorer 6&quot; Dobsonian Reflecting Telescope&lt;/a&gt;.  I'm hoping to gift it to my kids, but I want to add a twist...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a microscope with an adapter for an iPhone that has been great fun with allowing time lapse photography and very nice images on large monitors, so I wanted to accomplish something similar with this device.  Telescopes, in the past, have always been a dud due to the lack of multiple kids and family members wanting to be able to see the display simultaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone had any experience at retrofitting a unit such as this with an adapter that has given good results?  Any gotchas to be aware of?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm eying the &quot;&lt;a href=&quot;http://carson.com/products/digiscoping-hookupz2-is-200/&quot; rel=&quot;nofollow noreferrer&quot;&gt;HookUpz 2.0 Universal Smartphone Optics Digiscoping Adapter&lt;/a&gt;&quot;, but a little worried as I've never retrofitted such a device before and would like to hit this spot on since the timeframe is tight for me.&lt;/p&gt;&#xA;" OwnerUserId="15132" LastActivityDate="2016-12-11T05:09:47.717" Title="Retrofitting a Dobsonian Reflecting Telescope for smart phones?" Tags="&lt;telescope&gt;&lt;orion&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19371" PostTypeId="1" CreationDate="2016-12-11T19:55:59.337" Score="3" ViewCount="356" Body="&lt;p&gt;The rate of expansion of the universe after the Big Bang is described as Inflation. It is stated as being initially very fast then slowed. But the expansion is still continuing.  Some say expansion is currently powered by Dark Energy. The equations unlike gravity say that the rate appears to be linearly proportional to distance, like Hooke's Law for springs. Initially the universe was very small, so if the spring constant is dependent on the size of the universe, could this explain the inflation? What experiment could be devised to test this?&lt;/p&gt;&#xA;" OwnerUserId="15118" LastEditorUserId="13426" LastEditDate="2016-12-11T20:37:38.903" LastActivityDate="2016-12-31T11:24:07.007" Title="Is the Big Bang inflation caused by Dark Energy" Tags="&lt;cosmological-inflation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19372" PostTypeId="2" ParentId="19367" CreationDate="2016-12-11T20:34:42.350" Score="5" Body="&lt;p&gt;It may be that someone will come along with a better answer, but the first example I can find for the use of &quot;z&quot; explicitly for redshift is is &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1934BAN.....7..205D&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;William de Sitter's 1934 paper&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm tempted to think this is one of the earliest uses if not the first use, certainly looking at &lt;a href=&quot;https://catalog.hathitrust.org/Record/001477670&quot; rel=&quot;nofollow noreferrer&quot;&gt;de Sitter's 1933 book&lt;/a&gt; where he touches on this briefly, he does not use the letter &quot;z&quot;. However he did use &quot;z&quot; in &lt;a href=&quot;http://www.pnas.org/content/16/7/474.full.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;a 1930 paper&lt;/a&gt; to represent the ratio between the radius of the Universe and initial radius, which in this context is actually the same quantity as 1 + z and so why he would've used &quot;z&quot; in his 1934 paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I had to hazard a guess, there is no particular reason for the choice of &quot;z&quot; for redshift, it was just a convenient letter to use for the variable.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastEditorUserId="2092" LastEditDate="2016-12-11T23:04:57.047" LastActivityDate="2016-12-11T23:04:57.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19373" PostTypeId="2" ParentId="19371" CreationDate="2016-12-11T20:50:44.097" Score="5" Body="&lt;p&gt;The expansion caused by a cosmological constant (a particularly simple version of dark energy) is not exactly Hooke's law, it is an expansion that maintains a fixed Hubble constant. So this means $\mathrm{d}a/\mathrm{d}t = Ha$, for scale parameter $a$ and constant $H$. The solution is exponentially growing in time.  Hooke's law (without the minus sign) would have the acceleration $\mathrm{d}^2 a/\mathrm{d}t^2$ being proportional to $a$, which permits more complicated solutions, but you're right it certainly has similarities  if one picks out the exponentially growing solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's more, you are right that some treatments of inflation look like that of a cosmological constant, in that they are also exponential in time, but with a much much larger H constant (like 50 orders of magnitude larger).  So if it's the same thing, then for some reason the cosmological constant must have dropped vastly in value.  It's thought to be for a rather different reason, though this is all speculation.  The common speculation is that the &quot;false vacuum&quot; went through a phase change during inflation, leading to the vacuum we have now, and the vacuum we have now has a very different cosmological constant.  Maybe you can think of it as a phase change in the cosmological constant itself, I don't know.  So there could be some connection, but they can't be just the same.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="9301" LastEditDate="2016-12-31T11:24:07.007" LastActivityDate="2016-12-31T11:24:07.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19374" PostTypeId="2" ParentId="19371" CreationDate="2016-12-11T20:52:01.677" Score="3" Body="&lt;p&gt;The big similarity between inflation and dark energy is that the preferred models for both can be described by scalar fields, however there is a huge difference between the energy scale, with the energy scale of inflation being 27 orders of magnitude bigger than the energy scale of dark energy!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could they be related? Possibly, but the difference between the energy scales is not easy to explain. There are a number of papers proposing a relationship, but nothing that has really gone beyond speculation. Any test of a connection between the two would depend on the model used.&lt;/p&gt;&#xA;" OwnerUserId="2092" LastActivityDate="2016-12-11T20:52:01.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19377" PostTypeId="1" AcceptedAnswerId="19378" CreationDate="2016-12-12T15:24:11.493" Score="14" ViewCount="964" Body="&lt;p&gt;I was wondering if the constellations visible from Earth would also be visible from Mars.  Would they look the same but just shifted slightly in the sky due to the larger circumference of the orbit?&lt;/p&gt;&#xA;" OwnerUserId="15139" LastActivityDate="2016-12-12T16:38:28.060" Title="Martian Constellations" Tags="&lt;mars&gt;&lt;constellations&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19378" PostTypeId="2" ParentId="19377" CreationDate="2016-12-12T16:13:18.557" Score="23" Body="&lt;p&gt;The stars are so immensely far away that to the human eye there would be no noticeable difference. The nearest stars are moving roughly 1.5 arcsecond wrt. the background when viewed from Earth's position with half a year's interval (i.e. half a revolution around the Sun). Mars' orbit is only 50% larger, but the human eye cannot resolve better than roughly one arcminute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the sky would look different for another reason: Although stars are point sources, on Earth stars are smeared out due to the atmosphere. On Mars, there's almost no atmosphere, but on the other hand there's lots of dust whirling around. Dust tends to absorb blue light more than red and thus &lt;em&gt;redden&lt;/em&gt; objects. However, Martian dust happens to scatter red more than blue (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1997JGR...102.9039O&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ockert-Bell et al. 1997&lt;/a&gt;), so the stars should appear more blue. Also, the dust will scatter the Sun's light so much that no stars are visible during the day (as opposed to e.g. the Moon).&#xA;Note also that the north pole of Mars points not toward Polaris, but toward Deneb in Cygnus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since a shift in position of 1 AU (the distance from Earth to the Sun) corresponds to a shift in position of 1 arcsec for a star which is 1 pc (3.26 lightyears) away, if you want to go to a place where the nearest star (Proxima Centari; 1.3 pc away) has moved a full moon diameter, i.e. ½ a degree — you would need to travel roughly 2340 AU away, i.e. much farther away than Pluto, but still inside the Solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most stars are much farther away, however. The brightest stars that give the constellations their appearance are tens to hundreds of pc away, so to really change the constellation, you should go 10 or 100 times farther away, i.e. roughly 1 pc. In other words, the constellation &lt;em&gt;will&lt;/em&gt; look (slightly, but conspicuously) different from out nearest neighbor stars.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="10437" LastEditDate="2016-12-12T16:38:28.060" LastActivityDate="2016-12-12T16:38:28.060" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19379" PostTypeId="1" AcceptedAnswerId="19380" CreationDate="2016-12-12T17:36:55.063" Score="3" ViewCount="273" Body="&lt;p&gt;The Fomalhaut system is a trinary system, with Fomalhaut A (1.9 M), Fomalhaut B (0.7M) and Fomalhaut C (0.2M). Fomalhaut C is 2.5 light years from Fomalhaut A, and 3.2 light years away from Fomalhaut B. The tidal radius of the Fomalhaut system is 6.2 light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sirius system (Sirius A, 2M, Sirius B, 0.9M) is roughly 1.1x more massive than Fomalhaut A and B. What would be some ways to calculate a tidal radius of the Sirius sytem, to get an estimate of at what distance other stars could orbit the Sirius system?&lt;/p&gt;&#xA;" OwnerUserId="15140" LastActivityDate="2019-11-27T16:43:49.000" Title="What is the tidal radius of the Sirius system?" Tags="&lt;astrophysics&gt;&lt;binary-star&gt;&lt;star-systems&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19380" PostTypeId="2" ParentId="19379" CreationDate="2016-12-12T21:03:58.770" Score="5" Body="&lt;p&gt;The tidal radius for a star or collection of stars near the Sun and on a circular orbit (matches Sirius reasonably well) is given by (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/1998MNRAS.299..955P&quot; rel=&quot;noreferrer&quot;&gt;Pinfield et al. 1998&lt;/a&gt;)&#xA;$$R_T \simeq \left( \frac{GM}{2(A-B)^2} \right)^{1/3} ,$$&#xA;where $A$ and $B$ are the &lt;a href=&quot;https://en.wikipedia.org/wiki/Oort_constants&quot; rel=&quot;noreferrer&quot;&gt;Oort constants&lt;/a&gt; and $M$ is the total mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the value of $A-B$ from &lt;a href=&quot;https://arxiv.org/abs/astro-ph/9706293&quot; rel=&quot;noreferrer&quot;&gt;Feast et al. (1997)&lt;/a&gt;, this reduces to&#xA;$$R_T = 1.43 \left( \frac{M}{M_{\odot}} \right)^{1/3}\ \ {\rm pc}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this formula, I get that the tidal radius for Fomalhaut is 2.01 pc (in reasonable agreement with whatever source you are using). As the mass of the Sirius system is only marginally larger then the tidal radius will be almost the same, since it depends on the cube root of mass.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-12-12T21:14:34.123" LastActivityDate="2016-12-12T21:14:34.123" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19381" PostTypeId="1" AcceptedAnswerId="19382" CreationDate="2016-12-12T21:23:47.213" Score="6" ViewCount="251" Body="&lt;p&gt;If I look at the night sky, which easily identifiable stars have known exoplanets, and what are the names of those planets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the recent discovery of so many exoplanets, it is a nice idea to look at the stars with this new knowledge. They will no longer seem like a bunch of anonymous white dots. &lt;/p&gt;&#xA;" OwnerUserId="15141" LastEditorUserId="6" LastEditDate="2016-12-12T22:00:06.663" LastActivityDate="2016-12-12T22:00:06.663" Title="Which stars, identifiable with the naked eye, have exoplanets?" Tags="&lt;exoplanet&gt;&lt;constellations&gt;&lt;star-gazing&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19382" PostTypeId="2" ParentId="19381" CreationDate="2016-12-12T21:42:54.737" Score="5" Body="&lt;p&gt;Stars with an apparent magnitude of 7.0 or greater are typically not visible to the naked eye. So if you take a list of exoplanetary host stars, such as &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_exoplanetary_host_stars&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;, and you sort by apparent magnitude, in ascending order, then you will see a list of identifiable stars known to have exoplanets, until you reach an apparent magnitude of 7.0 on the list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding the star can be trickier, but Wikipedia will usually show you what constellation you can find the star in. With a little work, and maybe the help of some other posts on this site, you could probably identify which stars are visible in your area and time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most exoplanets have not been given formal names themselves, but instead a lowercase letter is appended to the end of the name or designation of the star, for example &quot;Beta Geminorum b (abbreviated β Geminorum b, β Gem b) and HD 62509 b&quot; for Pollux b, the first planet discovered around Pollux. Incidentally in 2014, the International Astronomical Union did release a process for giving proper names to exoplanets, and Pollux b was given the name Thestias. If you look up other exoplanets on Wikipedia, you can find out whether they have been given proper names as well.&lt;/p&gt;&#xA;" OwnerUserId="6" LastActivityDate="2016-12-12T21:42:54.737" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19384" PostTypeId="1" AcceptedAnswerId="19386" CreationDate="2016-12-13T02:35:14.697" Score="30" ViewCount="5806" Body="&lt;p&gt;Shortly after the Big Bang, temperatures cooled from the Planck temperature. Once temperatures lowered to 116 gigakelvins, nucleosynthesis took place and helium, lithium and trace amounts of other elements were created.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the temperatures were so high shortly after the Big Bang, why weren't much heavier elements produced? 116 gigakelvins is obviously far above the temperature required for elements like carbon and oxygen to fuse. In addition, shouldn't most of the protons at those temperatures have fused, leaving the Universe with mostly heavier elements?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2016-12-14T09:52:36.753" Title="Why didn't the Big Bang produce heavier elements?" Tags="&lt;cosmology&gt;&lt;big-bang-theory&gt;&lt;temperature&gt;&lt;early-universe&gt;&lt;nucleosynthesis&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19385" PostTypeId="1" AcceptedAnswerId="19387" CreationDate="2016-12-13T02:55:36.727" Score="8" ViewCount="668" Body="&lt;p&gt;It stands to reason that some of the hydrogen and helium that formed directly as a product of the big bang might never have fallen into a star to re-ejected when that star explodes.  My question is, given the best theory, what percentage of that matter has managed to drift without being sucked into a star.  Do we have any idea?&lt;/p&gt;&#xA;" OwnerUserId="15146" LastActivityDate="2016-12-13T09:32:13.537" Title="What percentage of the hydrogen today has never been in a star" Tags="&lt;cosmology&gt;&lt;matter&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19386" PostTypeId="2" ParentId="19384" CreationDate="2016-12-13T03:24:09.400" Score="38" Body="&lt;p&gt;I think that your thought process is flawed in that you assume that by drastically increasing the temperature you are guaranteed to get heavy elements. As odd as this may sound, this isn't the case (especially during the Big Bang Nucleosynthesis (BBN)) for a few reasons. In fact, if you took a hydrogen-only star and made it go supernova, you wouldn't get heavy elements like you see in current stars going supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BBN Timescale&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One major point to consider is that the BBN era is calculated to be only ~20 min long. That isn't really much time to form elements. Sure, supernovae happen in an instantaneous flash, but there are other things going on there, which I'll get to in a second. The main point here is that fusing takes time and 20 min isn't that much time to form heavy elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Deuterium&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get heavy elements, you need to build up to them. You can't just smash together 50 protons and 50 neutrons and get tin. So the first step is to smash together a proton and neutron to get deuterium, but here you already run into a problem known as the deuterium bottleneck. As it turns out, the huge temperatures actually (and somewhat counterintuitively) impede the creation of deuterium. This is mainly because the deuteron will end up having so much energy that it will be able to overcome the binding energy (and deuterium has pretty low binding energy being that its only two nucleons) and will likely break apart again. Of course, given the density and temperature you can still get a good amount of deuterium simply by force of will, but not as much and not at the rate you'd expect otherwise. Another point that makes deuterium form less frequently that you'd naïvely expect is that the proton to neutron ratio before BBN was about 7:1 due to the proton being more favorable to be created since it has a slightly lower mass. So 6 out of 7 protons didn't have a corresponding neutron to combine with and had to wait for deuterium to form first before it could combine with anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tritium, Helium, Lithium, Oh My!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deuterium is then the catalyst for forming all the next stages of particles in your soup. From here you can throw them together with various other things to get $^3\mathrm{He}$, $^3\mathrm{H}$, and $^4\mathrm{He}$. Once you've got a good amount of deuterium, tritium, and helium isotopes floating around, you can start making lithium and if you're lucky a bit of beryllium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To Boron and Beyond&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now, once again you run into a bottleneck, and one more severe than the deuterium bottleneck. You can't easily jump to heavier elements with what you have on hand. The next fusion chain, and the way stars do it, is the triple-alpha process which helps to form carbon but to perform this chain and build up enough carbon you need a &lt;em&gt;lot&lt;/em&gt; of time. And we only have 20 minutes! There just isn't time to form the carbon we need to progress along the fusion cycle. As I hinted at the beginning, pure hydrogen stars also wouldn't produce heavy elements upon supernova for this reason. They're able to produce heavy elements now because they've had billions of years before their SN event to build up a base amount of carbon, nitrogen, oxygen, etc. that can aid in the heavy element fusion processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you don't have the time to follow the triple alpha process and make carbon — what about other processes? Surely the temperatures are high enough that you can do different fusion methods not seen in stars. Well... no. You can't even smash together lots of $\mathrm{He}$ or $\mathrm{Li}$ to get really heavy elements because of the fact that heavy nuclei are only stable if they have way more neutrons than protons. And we already said there was a big neutron deficiency early on so the chance you have enough neutrons hanging around to smash together to get, say $^{112}\mathrm{Sn}$ (that's tin with 62 neutrons), is pretty small. What's more, you can't even try to either skip carbon by making something slightly heavier or form something intermediary between lithium and carbon. Again, this is because of stability issues. So with no other options, you have to shoot for carbon after lithium, and as stated above, you just don't have time for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, BBN is limited to getting only to lithium because of limited time, proton to neutron abundance ratios, and fusion bottlenecks that slow things down. All these come together to produce ~75% $^1\mathrm{H}$, ~25% $^4\mathrm{He}$, ~0.01% $^2\mathrm{H}$ and $^3\mathrm{He}$, and trace amounts of $\mathrm{Li}$.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="9301" LastEditDate="2016-12-13T03:59:43.327" LastActivityDate="2016-12-13T03:59:43.327" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19387" PostTypeId="2" ParentId="19385" CreationDate="2016-12-13T08:55:07.577" Score="6" Body="&lt;p&gt;About 70% of the baryonic matter in the universe is hydrogen, with a mean density of about $4\times 10^{-29}$ kg/m$^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the stars that have ever been born are still alive, since an average star is only about $0.25 M_{\odot}$ and has a lifetime much longer than the age of the universe (so very little material has actually been recycled). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume there are $10^{22}$ stars of $0.25 M_{\odot}$ in an observable universe of radius 47 billion light years, that are 70% H by mass, the &quot;stellar&quot; hydrogen is only one part in 73.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there is only one hydrogen nucleus (a proton) in a star for every 73 in the universe. This ratio would have been smaller in the past (E.g. when theSun was born). But as I mentioned, most of this hydrogen (about 90%) is in stars that are longer lived than the universe. Therefore my very rough estimate is that around 1 hydrogen atom in every 1000 on Earth has been inside a star. This contrasts markedly with say that 100% of carbon and oxygen atoms have been inside a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: To be fair, this calculation hinges a lot on how many stars there are in the observable universe. This number is very uncertain and could be higher - perhaps $10^{23}$ (see &lt;a href=&quot;https://physics.stackexchange.com/questions/137341/how-do-we-estimate-1023-stars-in-the-observable-universe&quot;&gt;here&lt;/a&gt;), in which case my numbers are somewhat pessimistic and it might be more like 1 H atom in 7 is inside a star and 1 H atom in about 100 on Earth were inside a star. However, I don't think there is any argument that the majority of hydrogen in the universe is not, and has never been, in a star, but whether that is 90% or 99% is still moot.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-12-13T09:32:13.537" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19390" PostTypeId="1" CreationDate="2016-12-13T14:32:03.393" Score="3" ViewCount="251" Body="&lt;p&gt;Some stars, if observed repeatedly over time, show a perturbation or &quot;wobble&quot; in their proper motion. If this is a periodic occurrence we can infer that the perturbation occurs due to the gravitational influence of an unseen companion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Sirius Matters, Noah Brosch describes a secondary wiggle,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The trajectory described by the two components of Sirius shows an additional component of motion as tiny wiggles super-posed on the wider modulations of the proper motion by the orbital motion of the Sirius pair. The wiggles have a period of one year. &quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What causes this second wiggle?&lt;/p&gt;&#xA;" OwnerUserId="15140" LastActivityDate="2016-12-29T13:24:18.797" Title="Why does Sirius proper motion &quot;wiggle&quot; with a 1 year period, when Sirius A and Bs binary orbit is a 50 year period?" Tags="&lt;astrophysics&gt;&lt;binary-star&gt;&lt;star-systems&gt;&lt;astrometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19393" PostTypeId="1" AcceptedAnswerId="19395" CreationDate="2016-12-13T17:15:49.980" Score="3" ViewCount="759" Body="&lt;p&gt;I'm looking for the latest values (with uncertainties) of the four main cosmological density parameters $\Omega_i$ :&#xA;\begin{align}\tag{1}&#xA;\Omega_{\text{mat}} &amp;amp;={} ?,&#xA;&amp;amp;\Omega_{\text{rad}} &amp;amp;={} ?,&#xA;&amp;amp;\Omega_{\Lambda} &amp;amp;={} ?,&#xA;&amp;amp;\Omega_{k} &amp;amp;={} ?.&#xA;\end{align}&#xA;I know that $\Omega_{\text{mat}} \approx 0.30$, $\Omega_{\text{rad}} \approx 0.00$, $\Omega_{\Lambda} \approx 0.70$ and $\Omega_{k} \approx 0.00$, but I would like to have more precise values (with uncertainties, if possible).  Take note that these parameters are constrained by the following relation :&#xA;\begin{equation}\tag{2}&#xA;\Omega_{\text{mat}} + \Omega_{\text{rad}} + \Omega_{\Lambda} + \Omega_{k} \equiv 1.&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, I checked &lt;em&gt;Wikipedia&lt;/em&gt; but I don't trust it very much :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Lambda-CDM_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Lambda-CDM_model&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also checked on &lt;em&gt;arXiv&lt;/em&gt;.  For example :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/1502.01589&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/1502.01589&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I don't find clear final and consensual values in this paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help would be appreciated.  Please state your sources.&lt;/p&gt;&#xA;" OwnerUserId="14736" LastActivityDate="2018-05-29T12:33:52.310" Title="Latest cosmological parameters" Tags="&lt;universe&gt;&lt;cosmology&gt;&lt;general-relativity&gt;&lt;dark-matter&gt;&lt;hubble-constant&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19394" PostTypeId="1" CreationDate="2016-12-13T17:29:39.960" Score="1" ViewCount="100" Body="&lt;p&gt;The seventeenth century saw a revolution in astronomy. The invention of the telescope and the acknowledgement of the heliocentric system triggered a race amongst astronomers to measure the parallax of stars - the annual displacement of stellar positions due to Earth's motion around the Sun. In the late 1830s these measurements enabled astronomers to determine the distances to a handful of stars for the first time. From the 1850s onwards, the application of photography to astronomical observations transformed the practice of charting the sky, allowing the compilation of larger and larger catalogues of stellar positions and distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That robust marvel of trigonometry applied to observations of other stars, lets me look up into the sky for 6 months, measure the apparent motion, then divide 1 by that, and voila, the distance to the star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the distance to Alpha Centauri, &lt;a href=&quot;https://www.wolframalpha.com/input/?i=1+%2F+.77&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.wolframalpha.com/input/?i=1+%2F+.77&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These distances are used in all textbooks, all databases, all conversations, and does not correct for any space-time curvature whatsoever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To draw a line on the sky, and use trigonometry, seems almost a bit too simple. Anyone have any thoughts on if space-time might curve in weird ways sort of like how a fish eye lens on Instagram distorts an image?&lt;/p&gt;&#xA;" OwnerUserId="15140" LastActivityDate="2016-12-14T05:58:10.880" Title="Could space time curve from the vantage point of earth and out into the universe, and would d = 1 / P in that case give false estimates?" Tags="&lt;amateur-observing&gt;&lt;distances&gt;&lt;space-time&gt;&lt;fundamental-astronomy&gt;&lt;dark-matter&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19395" PostTypeId="2" ParentId="19393" CreationDate="2016-12-13T19:33:39.813" Score="12" Body="&lt;p&gt;Cosmological parameters are measured in a variety of ways, and their values will depend on which measurements you trust the most. The paper you link to (&lt;a href=&quot;https://arxiv.org/abs/1502.01589&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck Collaboration et al. 2016&lt;/a&gt;) with the 2015 results from the Planck observations of the cosmic microwave background is probably the one that most people will accept, but even in that paper you will find different values, depending on which observables you combine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will find the values in their Table 4. I think that &lt;strong&gt;most people use the values in the column called &quot;TT+lowP+lensing&quot;&lt;/strong&gt; (e.g. &lt;a href=&quot;https://arxiv.org/abs/1512.00564&quot; rel=&quot;nofollow noreferrer&quot;&gt;Geil et al. 2016&lt;/a&gt;, &lt;a href=&quot;https://arxiv.org/abs/1607.04291&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ricotti et al. 2016&lt;/a&gt;, and &lt;a href=&quot;https://arxiv.org/abs/1608.00819&quot; rel=&quot;nofollow noreferrer&quot;&gt;Liu et a. 2016&lt;/a&gt;), which is the &quot;conservative&quot; choice. However, you'll also find some (e.g. &lt;a href=&quot;https://arxiv.org/abs/1603.03037&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chevallard &amp;amp; Charlot 2016&lt;/a&gt; and &lt;a href=&quot;https://arxiv.org/abs/1611.09846&quot; rel=&quot;nofollow noreferrer&quot;&gt;Silk 2016&lt;/a&gt;) who use the values in the last column, called &quot;TT,TE,EE+lowP+lensing+ext&quot;. These values takes into account external data (baryonic acoustic oscillations and supernovae data), which reduce the uncertainties, arguably to unnaturally small values. The TT, TE, TT, and lowP refer to the polarization maps used, and &quot;lensing&quot; to the weak gravitational lensing measurements by Planck.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Standard cosmological parameters&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The table below is a modified version from the Planck paper where I show only the most used parameters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VvN7V.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VvN7V.png&quot; alt=&quot;PlanckData&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $n_s$ is the slope of the primordial power spectrum, $H_0$ is the Hubble constant in km s&lt;sup&gt;–1&lt;/sup&gt; Mpc&lt;sup&gt;–1&lt;/sup&gt;, $\Omega_\Lambda$ and $\Omega_m$ are the density parameters of dark energy and total (dark+baryonic) matter, $\Omega_\mathrm{b}h^2$ and $\Omega_\mathrm{c}h^2$ are the density parameters of baryonic and dark matter, multiplied by the factor $h \equiv H_0/100$ (squared), $\sigma_8$ is the matter density fluctuations on scales of 8 (comoving) Mpc, $z_\mathrm{re}$ is the redshift at which the Universe was reionized (assuming instant reionization), and the last row shows the inferred age of the Universe in billion years.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Curvature density&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The constraints on the curvature parameter $\Omega_K$ is given in Table 5, which has somewhat different combinations of data. All in all, Planck constrains the curvature to $|\Omega_K| &amp;lt; 0.005$, but you will rarely offend anyone by simply setting the curvature to zero.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Temperature and radiation density&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The radiation density is a bit more convoluted. It has a contribution from both photons and neutrinos, and their densities are related as&#xA;$$&#xA;\rho_\nu = N_\mathrm{eff} \frac{7}{8} \left(\frac{4}{11}\right)^{4/3}\rho_\gamma,&#xA;$$&#xA;where $N_\mathrm{eff} = 3.046$ is the &lt;em&gt;effective&lt;/em&gt; number of neutrino species. Following the procedure by &lt;a href=&quot;https://physics.stackexchange.com/users/24142/pulsar&quot;&gt;Pulsar&lt;/a&gt; in &lt;a href=&quot;https://physics.stackexchange.com/a/94221/70207&quot;&gt;this answer&lt;/a&gt;, but with updated parameters (i.e. the $N_\mathrm{eff}$ given above, and the average CMB temperature of $T_0 = 2.722\pm0.027$ (Eq. 83a)), I get that&#xA;$$&#xA;\begin{array}{rcl}&#xA;\Omega_\mathrm{rad}h^2 &amp;amp; = &amp;amp; \Omega_\nu h^2+ \Omega_\gamma h^2 \\&#xA;                       &amp;amp; = &amp;amp; (1.7018 + 4.6213) \times 10^{-5} \\&#xA;                       &amp;amp; = &amp;amp; 4.1620\times10^{-5},&#xA;\end{array}&#xA;$$&#xA;that is, with $h = 0.6781$,&#xA;$$&#xA;\Omega_\mathrm{rad} = 9.0513\times10^{-5}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Recap&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;So, answering your question is a little difficult, as there's no single answer, and since the curvature is given with 95% confidence (&quot;2$\sigma$&quot;), rather than 68% (&quot;1$\sigma$&quot;). For the Hubble constant and the matter and dark energy, I'd recommend $H_0 = 67.81\pm0.92 \,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$, $\Omega_\mathrm{m} = 0.308\pm0.012$ and $\Omega_\Lambda = 0.692\pm0.012$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For curvature, I would use 0 (especially because your calculations probably would need to switch between ordinary and hyperbolic trigonometry depending on the sign),  but if you do want to include uncertainty, you can say $\Omega_K=0\pm0.005$ (95%). Or, you could simply use $\Omega_K = 1 - \Omega_\mathrm{m} - \Omega_\Lambda - \Omega_\mathrm{rad}$ and propagate the errors to get $\Omega_K = 0\pm0.017$, which will give you a more conservative value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For radiation, to propagate uncertainties you would have to know the covariance matrix of the input parameters, but since the total error is dominated by that of the Hubble constant, standard error propagation $(\sigma_{\Omega_\mathrm{rad}}/\Omega_\mathrm{rad} \simeq 2 \sigma_h/h)$ yields a value of $\Omega_\mathrm{rad} = (9.0513\pm0.2456)\times10^{-5}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to be explicit my recommendation is:&#xA;$$&#xA;\{\Omega_\mathrm{m}, \Omega_\Lambda, \Omega_\mathrm{rad}, \Omega_K\}&#xA;= \{ 0.308, 0.692, 9.05\times10^{-5},0\}&#xA;\pm \{0.012, 0.012, 2.46\times10^{-6}, 0 \}.&#xA;$$&#xA;But I think the most important thing is to state where you take the parameters from. People rarely state &lt;em&gt;why&lt;/em&gt; the choose a particular set of parameters, and although Planck gives very small error bars, other probes give error bars so small that they're basically mutually incompatible. That's why you can still easily get away with $\{\Omega_\mathrm{m}, \Omega_\Lambda, \Omega_\mathrm{rad}, \Omega_K\} = \{ 0.3,0.7,0,0\} \pm \{0,0,0,0\}$.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-05-29T12:33:52.310" LastActivityDate="2018-05-29T12:33:52.310" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="19396" PostTypeId="1" AcceptedAnswerId="19401" CreationDate="2016-12-13T23:22:39.607" Score="0" ViewCount="1464" Body="&lt;p&gt;I had this question for a long time. But otherwise the question basically sums the whole thing up.&lt;/p&gt;&#xA;" OwnerUserId="15167" LastActivityDate="2016-12-14T14:12:23.093" Title="Can virtual particles be the source of dark matter and dark energy?" Tags="&lt;dark-matter&gt;&lt;dark-energy&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2016-12-14T15:01:01.543" ContentLicense="CC BY-SA 3.0" />
  <row Id="19398" PostTypeId="2" ParentId="19394" CreationDate="2016-12-14T04:46:23.130" Score="1" Body="&lt;p&gt;Yes, but the amount of curvature is too small in scale proportionally to affect such things, especially when one thinks of how away from us the objects are that we observe.  Even the moon is a quarter of a million miles away, and the Earth's gravitational well is quite small.&lt;/p&gt;&#xA;" OwnerUserId="15171" LastActivityDate="2016-12-14T04:46:23.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19399" PostTypeId="2" ParentId="19394" CreationDate="2016-12-14T05:51:39.060" Score="1" Body="&lt;p&gt;The real answer to your question is ... No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is that when measuring the distance to a star, any star, you are either looking at a star within our own galaxy, or you're doing something wrong. This means that you're not looking at objects far enough away for any possible space-time curvature to be detectable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the reason &quot;simple&quot; trigonometry works, and why these distances are universally used. &lt;/p&gt;&#xA;" OwnerUserId="2232" LastActivityDate="2016-12-14T05:51:39.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19400" PostTypeId="2" ParentId="19394" CreationDate="2016-12-14T05:58:10.880" Score="1" Body="&lt;p&gt;We have a very good idea of what causes spacetime to curve. It is described by general relativity, a theory that has been demonstrated to be an excellent model on numerous occasions, both in near and far from the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space is curved by mass (or more precisely by mass and momentum as described by the stress-energy tensor), and we know that the space between us and other stars are empty, and so in GR the space is approximately flat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If spacetime around the Earth was being curved by some unseen mass, the result wouldn't be incorrect distance calculations, instead there would be very obvious distortions: Einstein rings or double images. We don't see this, and so we can be confident that the curvature of spacetime has is negligible when calculating distance by parallax.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-12-14T05:58:10.880" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19401" PostTypeId="2" ParentId="19396" CreationDate="2016-12-14T14:12:23.093" Score="1" Body="&lt;p&gt;The problem with associating virtual particles with either dark matter or dark energy is that, quite literally, virtual particles aren't real.  By that I mean, they are intended to change the effects of energy, like how the energy is transported and where it goes, but they shouldn't alter the &lt;em&gt;amount&lt;/em&gt; of energy.  As such, virtual particles are just a kind of picture, a way to view the terms in a particular way of writing the equations.  Many physicists reject that picture, they think it gets taken too literally, while others embrace it wholeheartedly, but it shouldn't appear in the stress energy that is used in general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course, the problem is that general relativity may be wrong, especially since it does not seem very consistent with quantum mechanics.  So the idea is, maybe there is some way to correct general relativity where virtual particles do have a more direct role.  Then the people who tend to dismiss their relevance would have to reconsider.  But so far it's not at all clear that this will be the right way to fix GR, because you run into significant problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problems you run into with dark matter being virtual particles is that most of the effects of dark matter are seen on completely Newtonian scales, like the gravity in a galaxy.  On that scale, one should not need GR, and no one is suggesting that we need to fix Newtonian gravity by including virtual particles, that would not be consistent with everything Newton did.  For example, we see no effects of dark matter within our own solar system, so why would we need to fix Newtonian gravity on the galactic scale but not the solar system scale?  We would need to do more than include virtual particles, we'd need a whole new kind of gravity (sometimes called &quot;MOND&quot;, modifications to Newtonian dynamics&quot;).  If you are going to do MOND, it opens a full range of possibilities, there's really no reason to be constrained by the virtual particle concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problems you run into if you try to associate virtual particles with dark energy is that the answer you get, if you proceed in the most straightforward way possible, is that the dark energy should either be zero (if you don't think virtual particles count in the energy of the vacuum), or something more than 100 orders of magnitude too large (if you think they do count).  So how do you take that virtual particle concept and whittle down its effects from something more than 100 magnitudes too large to something tiny but not zero?  It's not at all clear this is the best starting point!&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-12-14T14:12:23.093" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19402" PostTypeId="1" AcceptedAnswerId="19408" CreationDate="2016-12-14T14:23:00.077" Score="6" ViewCount="171" Body="&lt;p&gt;Is it possible for a black-hole to easily stretch or turn any hard object like (Diamond , Gold bar , Platinum bar , Brass) into spaghettification state in just a seconds or minute once it gets near nor reached the black-hole's event horizon??&lt;/p&gt;&#xA;" OwnerUserId="15007" LastActivityDate="2019-01-17T12:40:14.243" Title="Spaghettification of hard-things/Elements" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19404" PostTypeId="1" CreationDate="2016-12-14T15:25:43.463" Score="1" ViewCount="546" Body="&lt;p&gt;Space is curved by mass (or more precisely by mass and momentum as described by the stress-energy tensor). The milky way system has a lot of mass and a lot of energy. In what ways is space-time curved within it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=-m3SddsTSAY&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X441R.png&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15140" LastActivityDate="2020-07-09T10:58:02.480" Title="In what ways does the mass and energy of the milky way curve space time?" Tags="&lt;astrophysics&gt;&lt;space-time&gt;&lt;general-relativity&gt;&lt;dark-matter&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19405" PostTypeId="1" CreationDate="2016-12-14T18:26:07.353" Score="3" ViewCount="2183" Body="&lt;p&gt;I know what a jansky is, but I don't understand what beam (in &lt;span class=&quot;math-container&quot;&gt;$\frac{1 Jy}{beam}$&lt;/span&gt;) is.&#xA;If jansky is a unit of flux density, what kind of unit is &lt;span class=&quot;math-container&quot;&gt;$\frac{1 Jy}{beam}$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="15180" LastEditorUserId="25994" LastEditDate="2019-03-06T15:32:11.647" LastActivityDate="2019-07-02T07:34:01.137" Title="Why levels of radio contours maps are given in mJy/beam and what does it mean?" Tags="&lt;radio-astronomy&gt;&lt;units&gt;&lt;radio&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="19406" PostTypeId="1" AcceptedAnswerId="19411" CreationDate="2016-12-14T18:58:21.173" Score="3" ViewCount="2196" Body="&lt;p&gt;In the Nova episode &quot;Origins: Earth is Born&quot;, Neil DeGrasse Tyson states, &quot;But Mars is just a fraction the size of the Earth, so it cooled more rapidly. And as it cooled, its molten iron core hardened. As a result, Mars stopped generating its magnetic shield.&quot; (February 28, 2004)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;see: &lt;a href=&quot;http://www.pbs.org/wgbh/nova/space/origins-series-overview.html#origins-earth-born&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.pbs.org/wgbh/nova/space/origins-series-overview.html#origins-earth-born&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This comment from Tyson is consistent with what I have long believed. But I wonder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The melting point of iron (Fe) is around 1538 °C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;see: &lt;a href=&quot;https://en.wikipedia.org/wiki/Melting_points_of_the_elements_(data_page)&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Melting_points_of_the_elements_(data_page)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The the geothermal gradient for Mars is estimated to be around 6 °C per km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;see: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/14875/what-is-the-temperature-55-km-beneath-the-surface-of-mars&quot;&gt;What is the temperature 55 km beneath the surface of Mars?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the estimate of 6 °C per km is the ballpark, then I would expect that iron is molten as a mere 256 km of depth. At the core of Mars (over 3000 km in depth) I would expect the temperatures to be far higher than 1538 °C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that it is impossible for iron to be in solid form at Mars' core.&lt;/p&gt;&#xA;" OwnerUserId="15179" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2016-12-18T07:31:34.217" Title="Has the iron core of Mars really solidified?" Tags="&lt;mars&gt;&lt;core&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19407" PostTypeId="2" ParentId="19406" CreationDate="2016-12-14T19:56:26.493" Score="5" Body="&lt;p&gt;You're mostly missing an important point to do with the geothermal gradient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The geothermal gradient used in that question, and applied to a $55\:\mathrm{km}$ depth is &lt;em&gt;not&lt;/em&gt; consistent all the way through Mar's interior. They cite the Earth's geothermal gradient (see image below) and from that page you can see how the temperature profile changes with depth. There is a &lt;em&gt;much&lt;/em&gt; stronger gradient towards the surface. Thus you can't naively apply the surface geothermal gradient through to the core. Rather, to get the core temperature, you'd need to integrate the gradient function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Earth's Temperature Profile&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/41r0N.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/41r0N.jpg&quot; alt=&quot;Earth&amp;#39;s Temperature Profile&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2016-12-14T20:54:27.530" LastActivityDate="2016-12-14T20:54:27.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19408" PostTypeId="2" ParentId="19402" CreationDate="2016-12-14T20:56:04.740" Score="11" Body="&lt;p&gt;It depends on the size of the black hole and the size of the bit of material you are trying to pull apart, as well as what it is made of. What is important is the &lt;em&gt;difference&lt;/em&gt; in gravitational pull between one end of your sample and the other. This difference force grows as the inverse of distance &lt;em&gt;cubed&lt;/em&gt; as one moves in towards the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, nothing special marks crossing event horizon and if we were talking about a &lt;em&gt;supermassive&lt;/em&gt; black hole (a million solar masses or more), you might not even notice it had happened, since the force difference, sometimes called the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Tidal_force&quot; rel=&quot;nofollow noreferrer&quot;&gt;tidal force&lt;/a&gt;, would be quite small. At the event horizon of a &lt;em&gt;smaller&lt;/em&gt; black hole, the tidal force would be much larger. To see this, we note that the difference in force between two ends of a bar of length &lt;span class=&quot;math-container&quot;&gt;$\Delta r$&lt;/span&gt; is approximately&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ F = \frac{8GM}{r^3}\Delta r,$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is the black hole mass. At the event horizon &lt;span class=&quot;math-container&quot;&gt;$r = 2GM/c^2$&lt;/span&gt;, so&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ F = \frac{c^6}{G^2M^2}\Delta r.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spaghettification force continues to increase as the object heads towards &lt;span class=&quot;math-container&quot;&gt;$r=0$&lt;/span&gt;, which will occur in a finite proper time for the falling object. So it will always increase sufficiently to rip something apart, no matter what it is made of; the question is merely at what radius does this happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is of interest however to calculate whether it would happen to an object &lt;em&gt;outside&lt;/em&gt; the event horizon, simply because then an external observer (at a safe distance) might witness it. This in turn depends on the material making up the object, it's size and the mass of the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a material of tensile strength &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;, size &lt;span class=&quot;math-container&quot;&gt;$\Delta r$&lt;/span&gt; and cross-sectional area &lt;span class=&quot;math-container&quot;&gt;$\sim (\Delta r)^2$&lt;/span&gt;. For this to break up &lt;em&gt;before&lt;/em&gt; reaching the event horizon then the tidal force at the event horizon must exceed &lt;span class=&quot;math-container&quot;&gt;$T (\Delta r)^2$&lt;/span&gt; and thus&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{c^6}{G^2M^2}\Delta r &amp;gt; T (\Delta r)^2 $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ M &amp;lt; 2\times10^5 \left({T \Delta r}\right)^{-1/2}\ M_{\odot}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diamond has &lt;span class=&quot;math-container&quot;&gt;$T\sim 10^{11}$&lt;/span&gt; N/m&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt;. If it was of size 1 cm, then a black hole would need to be &lt;em&gt;larger&lt;/em&gt; than 6 solar masses for this chunk of diamond to survive being ripped apart &lt;em&gt;outside&lt;/em&gt; the event horizon. Other materials or larger bits of material are easier to break and would not survive outside the horizon of even more massive black holes. e.g. The tensile strength of bone is about &lt;span class=&quot;math-container&quot;&gt;$10^{8}$&lt;/span&gt; N/m&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt;, so a 0.1m piece of bone would be pulled apart prior to falling through the event horizon if &lt;span class=&quot;math-container&quot;&gt;$M&amp;lt; 60M_{\odot}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-01-17T12:40:14.243" LastActivityDate="2019-01-17T12:40:14.243" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19409" PostTypeId="1" CreationDate="2016-12-14T22:53:05.090" Score="1" ViewCount="198" Body="&lt;p&gt;I have used the (matlab/octave) SOFA routine/function iauEpv00 to get the barycentric position and velocity of the Earth, but now need to find barycentric position of the sun. Can't seem to find the right routine/function.&lt;/p&gt;&#xA;" OwnerUserId="15181" LastActivityDate="2016-12-14T22:53:05.090" Title="How do I calculate the barycentric position of the sun at a given julian date?" Tags="&lt;the-sun&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19410" PostTypeId="2" ParentId="19406" CreationDate="2016-12-15T02:02:03.450" Score="2" Body="&lt;p&gt;The absence of a magnetic field is not sufficient evidence to conclude that a planet has a solid core. It may be that the interior of Mars is liquid but lacks convective currents vigorous enough to generate a magnetic field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;see: &lt;a href=&quot;http://adsabs.harvard.edu/abs/2015EGUGA..1715240D&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2015EGUGA..1715240D&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NASA has scheduled a mission to Mars (for May of 2018) which is designed to give us new information about the deep interior. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;see: &lt;a href=&quot;http://insight.jpl.nasa.gov/overview.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://insight.jpl.nasa.gov/overview.cfm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in conceptualizing a geothermal temperature gradient for Mars (or any other planetary body), the relationship between depth and temperature is not linear. The curve will also have variances according to the composition of the layers, which is currently not well known. Furthermore, eutectic properties of the Martian lithosphere could result in unexpected melting points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, knowing the radius of a planetary body is also not necessarily sufficient evidence to conclude that it has a molten core.&lt;/p&gt;&#xA;" OwnerUserId="15179" LastActivityDate="2016-12-15T02:02:03.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19411" PostTypeId="2" ParentId="19406" CreationDate="2016-12-15T02:13:20.780" Score="5" Body="&lt;p&gt;Mars core may or may not be solid, I'm of the opinion that it's not solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See recent NASA press release  &lt;a href=&quot;http://newatlas.com/mars-gravity-map/42430/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and a much earlier study &lt;a href=&quot;https://www.newscientist.com/article/dn11962-lab-study-indicates-mars-has-a-molten-core/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It has been known since 2003 that at least part of Mars’ interior is&#xA;  molten, based on how easily the Sun’s gravity distorts the planet’s&#xA;  shape, but no one knew whether it is completely liquid, or whether&#xA;  there is a solid inner core like Earth’s.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and Nasa's 2003 announcement &lt;a href=&quot;https://www.nasa.gov/home/hqnews/2003/mar/HP_news_03094.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;on Neil deGrasse Tyson, link &lt;a href=&quot;https://gregbentall.com/tag/neil-degrasse-tyson/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, but I'm not sure it's from him, so . . I might delete later.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard (but should provide a source) it said that Mars core has solidified.  I remember NdGT saying that, but if he says it on a show, it's really the show saying, not him.   Anyway, I'll try to get that point worked out.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, these seemingly conflicting statements might still have some relation (I think).  A basic explanation is that the difference between liquid and solid under the pressures you get inside a planet are somewhat inexact.   The other thing is, to generate a magnetic field, you need more than just a liquid metal around a solid metal, but you need dynamic spiraling.  See &lt;a href=&quot;https://en.wikipedia.org/wiki/Dynamo_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://abyss.uoregon.edu/~js/glossary/dynamo_effect.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus rotates very slowly, this rotation means that Venus has essentially zero Coriolis effect, which might help it generate a permanent magnetic field.   Earth's relatively rapid rotation and Coriolis effect, and perhaps it's very gradually slowing rotation due to the Moon, help Earth maintain it's magnetic field.  A very slow variation in the rate of rotation of the solid core and outer solid mantle could generate enough spin to maintain the magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.physics.org/article-questions.asp?id=64&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; on the Coriolis force and &lt;a href=&quot;https://astronomynow.com/2016/04/01/moon-thought-to-play-major-role-in-maintaining-earths-magnetic-field/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; on the moon's possible role on maintaining the Earth's magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point of all this, planetary magnetic fields are complicated but they likely require the right type of permanent flow within the liquid part of the core, probably around a solid iron core.  Earth maintains that perhaps in part due to higher core temperatures which enable more easy flowing as well as the release of heat through condensation as the outer core solidifies onto the inner core and perhaps also, due to the Moon.    Mars at one time in it's past probably had a magnetic field similar to or perhaps even stronger than Earth, evidenced by some powerful crustal fields that remain.  See &lt;a href=&quot;http://www.space.dtu.dk/english/Research/Universe_and_Solar_System/magnetic_field&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know why he uses the word &quot;solidified&quot; when other research says liquid.   That might be a good question to ask him, but, Neil deGrass Tyson may be right that the cooling of the smaller Mars may have caused it to lose it's magnetic field, but based on article's I've read, he's incorrect to imply it solidified, but the liquid core may have become solid enough that the dynamics that could have once caused a magnetic field could have stopped, at least, I would speculate that this is possible but I don't think anyone knows for 100% certain.  The history of planetary magnetic fields as well as the precise makeup of planetary cores and what turns on or off a magnetic field inside a planet isn't well understood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and Zepher's answer is also important that the temperature of a liquid iron core at pressures inside of a planet are very different than the melting temperature at 1 ATM.   Earth's inner core is hotter than it's outer core but the greater pressure keeps it's inner core solid while it's outer core is liquid.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2016-12-15T13:16:08.287" LastActivityDate="2016-12-15T13:16:08.287" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19412" PostTypeId="1" AcceptedAnswerId="20885" CreationDate="2016-12-15T02:35:57.997" Score="3" ViewCount="1933" Body="&lt;p&gt;I have a 130mm (5&quot;) diameter Newtonian reflector with a 650mm focal length (f/5). I have collimated my telescope already but when I'm looking at Mars or Jupiter (not a star I'm sure), all I see is a white dot. Even if I use my 6mm with a 2x barlow, Mars still looks like a white dot but with a liiittle bit of a gray shade like a 3/4 moon. Of course I focused my eyepiece before. Why can't I see Mars or any other planet better than that ?&lt;/p&gt;&#xA;" OwnerUserId="15182" LastEditorUserId="11143" LastEditDate="2016-12-15T13:34:03.373" LastActivityDate="2017-04-28T10:09:15.557" Title="Why does any planet look like a (not so) big white ball through my telescope?" Tags="&lt;telescope&gt;&lt;newtonian-telescope&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19413" PostTypeId="2" ParentId="19406" CreationDate="2016-12-15T09:11:36.627" Score="2" Body="&lt;p&gt;Scientists cannot theorize clearly if minerals and elements are arranged in plumes and strata inside mars, and the diameters that make up the martian condensed/fluid inner/outer strata and plumes, they don't know if there is something fluid and metallic [mobile electrically charged plumes of (probably) heavy elements (perhaps including iron, the 4th most common element on our world's surface), and heat from radioactive decay]. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They will know more if there is any thermally driven geological movement inside the planet, including molten iron, if they can find any small temporal fluctuations in the magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the moment, a molten core only gives way to theories on &quot;paleomagnetism, a latent magnetic crust of 10-125Km&quot; deep stating that all the magnetism is left over from 4 billion years ago, and that certain zones have been de-magnetized by impact collision. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meteorites de-magnetize the crust and it may be possible to date the chronology of the weakening molten core of mars from billions of years ago, to know how long ago the molten core was magnetic, and how fast it ceased being as fluid as it was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yt is not possible to determine detailed vertical variations of the magnetization of the Martian crust on the basis of magnetic data analysis alone, it is possible to estimate the thickness of the magnetic part of the Martian crust using other independent observations. The magnetic layer is bounded at the bottom by the depth to Curie isotherm of its major magnetic minerals, and at the top by the depth to the base of a near surface zone that has been demagnetized by impact-induced shock waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some studies have demonstrated that the secondary magnetization acquired by the&#xA;lower crust in the absence of the core dynamo has little contribution to the observed magnetic anomalies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The measurements of the magnetic crust are the only witness of the inner core's activity, and of it's spherical geometry:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/eRm0V.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eRm0V.jpg&quot; alt=&quot;Irregular magnetic field on mars&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Difficult measures are needed: 3D maps of the magnetic field, seismic data from multiple stations and meteorites, Magnetic basalt/igneous maps of 1/2/3 billion  years chronology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xhmuq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xhmuq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/X6kQ9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X6kQ9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Some elements like Tg are twice as heavy as iron, melt at 2.5 times higher temps, and are conductive, so they would be much more likely than iron to be solid and to sink inwards if they are not melted into core alloys of metals. It's difficult to know the alloys and stratification of molten/solid materials that happen inside a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Maximum pressure at the center of mars is about &amp;lt;50 GigaPascals, because earth's core pressure is 330-360Gpa maximum, and mars is 10 times lighter. Here are images where you will see that H20 at the center of mars would need over 800'C to change from ice to water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron needs to be over 3000K to melt inside mars at the very center , so we have no idea, all we know is that the magnetic field of earth is 40 times stronger than that of mars, which is a fairly good measure of the quantity and fluidity of metals inside mars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scientists have more questions than answers about earth and mars's core physics, even after many years of seismology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For our planet, they have been able to determine that the center of the earth is solid iron/nickel, surrounded by a viscous and fluid outer core of similar material. If there is a lot of friction at the core of our planet, in a conductive medium, it is channeled in a N/S arrangement precisely matched with the rotation of the Earth. The Martian core is completely different, because it's magnetosphere is very irregular and lop-sided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some interesting facts about martian geology: &lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/2008/1710.html?referrer=https://www.google.fr/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.planetary.org/blogs/emily-lakdawalla/2008/1710.html?referrer=https://www.google.fr/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/k34Wa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k34Wa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2016-12-18T07:31:34.217" LastActivityDate="2016-12-18T07:31:34.217" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19416" PostTypeId="1" CreationDate="2016-12-15T13:40:30.620" Score="2" ViewCount="657" Body="&lt;p&gt;I get the below from &lt;a href=&quot;https://en.wikipedia.org/wiki/Neptune&quot; rel=&quot;nofollow noreferrer&quot;&gt;Neptune&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Pluto&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pluto&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In 1821, Alexis Bouvard published astronomical tables of the orbit of Neptune's neighbour Uranus.[22] Subsequent observations revealed substantial deviations from the tables, leading Bouvard to hypothesise that an unknown body was perturbing the orbit through gravitational interaction.[23] In 1843, John Couch Adams began work on the orbit of Uranus using the data he had. Via Cambridge Observatory director James Challis, he requested extra data from Sir George Airy, the Astronomer Royal, who supplied it in February 1844. Adams continued to work in 1845–46 and produced several different estimates of a new planet.&lt;/p&gt;&#xA;&lt;p&gt;In the 1840s, Urbain Le Verrier used Newtonian mechanics to predict the position of the then-undiscovered planet Neptune after analysing perturbations in the orbit of Uranus.[26] Subsequent observations of Neptune in the late 19th century led astronomers to speculate that Uranus's orbit was being disturbed by another planet besides Neptune.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Does anyone know how did people calculate their orbits in 19th century? It is not a three-body problem, right?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2016-12-15T13:47:28.513" Title="How to calculate the orbits of Neptune and Pluto based on the orbits of Uranus and Neptune respectively?" Tags="&lt;planet&gt;&lt;solar-system&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19419" PostTypeId="2" ParentId="703" CreationDate="2016-12-15T17:53:56.627" Score="0" Body="&lt;p&gt;Isn't this star almost certainly Capella?  As Vega begins to set early in the west, Capella is rising in the east.  It looks yellowish-white to me, and is in the constellation Auriga.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sirius is more south I think, and it would be easy to tell if the star is Sirius since it will be below Orion.&lt;/p&gt;&#xA;" OwnerUserId="15193" LastActivityDate="2016-12-15T17:53:56.627" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19420" PostTypeId="1" AcceptedAnswerId="19425" CreationDate="2016-12-15T21:12:12.490" Score="3" ViewCount="139" Body="&lt;p&gt;Have we detected continuum optical emission from any rotating neutron stars that do not have an accretion disk dominating the light? I ask because I know we have observed Doppler broadening of spectral lines from the rotation of even ordinary stars rotating at ordinary rates, but it seems to me that only the rapid rotation of a neutron star would be fast enough to make the Doppler broadening of the thermal continuum peak to be observable.&lt;/p&gt;&#xA;" OwnerUserId="14244" LastActivityDate="2016-12-16T09:26:55.673" Title="Have we Observed Continuum Emission from Neutron Stars?" Tags="&lt;spectroscopy&gt;&lt;neutron-star&gt;&lt;special-relativity&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19421" PostTypeId="1" CreationDate="2016-12-16T00:00:24.053" Score="8" ViewCount="874" Body="&lt;p&gt;What would happen to a neutrino that, traveling parallel to the black hole, crossed it's event horizon?  Would it come through the other side unaffected? Or, A neutrino and a black hole just happen to be traveling in the same direction but the neutrino get close enough to cross the event horizon.  Is it now trapped inside the event horizon even though technically it only interacted with the black hole on a gravitational level?&lt;/p&gt;&#xA;" OwnerUserId="15171" LastEditorUserId="15171" LastEditDate="2016-12-18T18:23:07.517" LastActivityDate="2018-04-12T01:49:12.180" Title="Neutrinos passing through black holes" Tags="&lt;black-hole&gt;&lt;neutrinos&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19422" PostTypeId="2" ParentId="19421" CreationDate="2016-12-16T00:37:04.137" Score="13" Body="&lt;p&gt;In general, don't think of black holes as material objects like stars or planets. We can ascribe properties like mass, charge and angular momentum to them, but if you're near a black hole and reach out to touch its event horizon, you won't feel anything (assuming you're still alive). Your hand will simply pass through empty space, and then into the black hole. Think of black holes instead as regions of spacetime, with the curious property that if you enter one, you can never exit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing your question is motivated by the fact that neutrinos interact extremely weakly with normal matter. They have very little mass and no electric charge, and mainly influence other particles through the weak nuclear force - which is, of course, weak, although not as weak as gravity. The problem with taking this assumption further and concluding that a neutrino might pass through a black hole unharmed is that black holes aren't &quot;normal matter&quot;. They're sections of spacetime, and while they may contain matter, that matter is not at all important when interacting with other particles. The only thing here that is important is that they are sections of space which have only one-way openings. Once inside, nothing can leave.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2016-12-16T00:37:04.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19425" PostTypeId="2" ParentId="19420" CreationDate="2016-12-16T09:26:55.673" Score="5" Body="&lt;p&gt;Yes - quite a few isolated neutron stars have been observed, where any magnetospheric emission or accretion-related emission is either negligible or has been otherwise separated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you suspect, this emission is thermal in nature. Neutron stars are roughly approximated by black bodies but, like &quot;normal&quot; stars, they do have atmospheres and strong magnetic fields that modify the black body spectrum considerably. If the neutron star is isolated then the only reason that you might be able to detect thermal emission from the surface is that it is very young (less than about a million years after formation) and still hot. The difficulty is in ruling out any residual, pulsed magnetospheric component or a contribution from accretion of the interstellar medium itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of examples to look at in the optical part of the spectrum would be &lt;a href=&quot;http://arxiv.org/abs/astro-ph/9803024&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kulkarni &amp;amp; van Kerwijk (1998)&lt;/a&gt; who detected optical radiation from the counterpart to RX0720.4-3125; and &lt;a href=&quot;http://www.nature.com/nature/journal/v389/n6649/abs/389358a0.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Walter &amp;amp; Matthews (1997)&lt;/a&gt; who claimed a detection of thermal optical emission detection from the isolated neutron star RXJ185635-3754 (and see picture below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several more examples of thermal emission detected at X-ray wavelengths from young, isolated neutron stars - the so-called X-ray dim neutron stars (see review by &lt;a href=&quot;https://arxiv.org/abs/1008.2891&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mereghetti 2010&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course if you are extending your definition of continuum emission to include non-thermal (but not accretion-related) processes, then continuum emission is detected from all isolated pulsars. Synchrotron radiation is a continuum process and has been detected right across the electromagnetic spectrum in many pulsars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A visible band HST image of the isolated neutron star RXJ185635-3754, attributable to Walter &amp;amp; Matthews 1997.&lt;/em&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/pGeqn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pGeqn.jpg&quot; alt=&quot;HST image of RXJ185635-3754&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-12-16T09:26:55.673" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19426" PostTypeId="1" CreationDate="2016-12-16T11:55:48.093" Score="5" ViewCount="1133" Body="&lt;p&gt;Generally, the parallax is contributed to the motion of the earth around the sun. However, at the same time, the sun is moving too. It carries the earth with it. Presumably, the shift of the sun is bigger than the earth-sun distance. So, that will also cause apparent shift of a star in the background. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not so familiar with astronomy. Is my argument right? &lt;/p&gt;&#xA;" OwnerUserId="1905" LastActivityDate="2016-12-16T12:15:12.283" Title="Parallax due to the motion of the sun" Tags="&lt;parallax&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19427" PostTypeId="2" ParentId="19426" CreationDate="2016-12-16T12:06:48.800" Score="12" Body="&lt;p&gt;You are referring to what is known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Proper_motion&quot; rel=&quot;nofollow noreferrer&quot;&gt;proper motion&lt;/a&gt;, which is the apparent motion on the sky due to the relative motion of the solar system and the stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To measure an accurate parallax you have to observe for long enough that you can disentangle the parallax motion, which is of course modulated on a 1 year cycle, from the proper motion of a star, which produces an approximately linear, monotonic drift of the star's position with time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is perfectly illustrated in this graphic produced by HST/NASA, which shows the track on the sky taken by Proxima Centauri (a nearby star with a large parallax &lt;em&gt;and&lt;/em&gt; a large proper motion). The small, periodic wiggles are caused by the parallax shift, whilst the linear trend in the path is due to proper motion. The two are easily separated if you observe for at least 18 months with several precise position measurements during that time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1nVKe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1nVKe.jpg&quot; alt=&quot;The track of Proxima Centauri&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-12-16T12:15:12.283" LastActivityDate="2016-12-16T12:15:12.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19428" PostTypeId="1" CreationDate="2016-12-16T13:21:16.543" Score="3" ViewCount="60" Body="&lt;p&gt;As far as I understand, the effect of parallax exhibits itself in the shift of a star relative to the background stars on photographs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the aberration of light, how is it reflected? It would not cause the relative shift of a star in a background, right? Because all the stars (the foreground one and the background ones) will be rotated as a whole, right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what was Bradley's observation?    &lt;/p&gt;&#xA;" OwnerUserId="1905" LastActivityDate="2016-12-16T13:43:44.997" Title="What was Bradley's observation concerning abevation of light?" Tags="&lt;observational-astronomy&gt;&lt;parallax&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19429" PostTypeId="2" ParentId="19428" CreationDate="2016-12-16T13:43:44.997" Score="4" Body="&lt;p&gt;You are right that what you must use as a referent is different-- for parallax, your referent will be a much more distant object, because it shows little parallax.  In aberration, it happens to the whole visual field.  But what is shifted is the angle relative to the ground at which the light arrives, so if you set your telescope to see a given star, in principle you would need to correct for both parallax and aberration as the Earth orbits. But aberration is usually much larger than parallax because it scales with v/c rather than D/d (where D is 1 AU and d is the distance to the star).  For the Earth's orbit, v/c is about $10^{-4}$, but stars are much farther than $10^4$ AU.  As for Bradley's observation, a good version is given at &lt;a href=&quot;http://cseligman.com/text/history/bradley.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://cseligman.com/text/history/bradley.htm&lt;/a&gt;.  He was looking for the parallax effect, and stumbled into the much larger aberration effect.  He could tell it was something different because the sign was opposite and it did not depend on distance to the source.  This was 50 years after Romer, so the speed of light was already known.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-12-16T13:43:44.997" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19430" PostTypeId="1" AcceptedAnswerId="19434" CreationDate="2016-12-16T16:43:14.493" Score="6" ViewCount="221" Body="&lt;p&gt;Recently I read about comet &lt;a href=&quot;https://en.wikipedia.org/wiki/C/2014_Q2_(Lovejoy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;C/2014 Q2 (Lovejoy)&lt;/a&gt;.The thing got me amazed was that scientists found it was dumping ethyl alcohol all along its way, at a rate of around 300-500 bottles per second. So I want to ask how common is alcohol in outer space, and how did it form? I usually imagine space as an inert mixture of dry dust and light gases but it turns out there are sometimes alcoholic compounds.&lt;/p&gt;&#xA;" OwnerUserId="15201" LastEditorUserId="2153" LastEditDate="2016-12-16T17:48:21.973" LastActivityDate="2016-12-17T12:11:58.183" Title="How does this comet have alcohol in it?" Tags="&lt;comets&gt;&lt;astrochemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19431" PostTypeId="2" ParentId="19073" CreationDate="2016-12-16T17:06:34.433" Score="0" Body="&lt;p&gt;Based on the original question... &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;which I feel like is a little late now, as most universities are finished until next year. &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;isn't as involved or difficult as it has been presented. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all you're looking for are image processing problems, then you're going to need to look into the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Bias levels&lt;/li&gt;&#xA;&lt;li&gt;Dark levels&lt;/li&gt;&#xA;&lt;li&gt;Background levels&lt;/li&gt;&#xA;&lt;li&gt;Cosmic Rays&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Anything else regarding the images ie: stacking of filter combinations, are secondary and rarely required for any science case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The funny thing is, these are common for any CCD image with a long exposure, so the space telescopes bit only adds more chance for cosmic ray interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enjoy&lt;/p&gt;&#xA;" OwnerUserId="2232" LastActivityDate="2016-12-16T17:06:34.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19433" PostTypeId="2" ParentId="19404" CreationDate="2016-12-17T08:25:26.640" Score="3" Body="&lt;p&gt;In short, not by much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out by us the Earth the gravitational acceleration towards the centre of the galaxy is $v^2/r = \frac{(2\times2\times 10^{20}\times \pi\  \mathrm{metres} / 230000000\  \mathrm{years})^2 }{ 2\times 10^{20} \mathrm{metres}}$, which is of the order of $10^{-10} ms^{-2}$ : in other words, it is very small. This is because spacetime is not highly curved except close to massive bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This spacetime is approximately flat. It can be well approximated by Minkowski space. If you need to include in your model the distribution of mass in the galaxy, then Newtonian mechanics is sufficient, for example &lt;a href=&quot;https://physics.stackexchange.com/questions/62637/the-potential-and-the-intensity-of-the-gravitational-field-in-the-axis-of-a-circ&quot;&gt;https://physics.stackexchange.com/questions/62637/the-potential-and-the-intensity-of-the-gravitational-field-in-the-axis-of-a-circ&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the centre of the galaxy is a black hole, and it is in the neighbourhood of the black hole that GR really comes into play. Near the black hole, the Kerr solution of GR is a good model (the Kerr solution describes a symmetric uncharged spinning mass)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now on the massive scale, a galaxy, or a cluster of galaxies can have a significant effect on light, given enough time, and enough distance. Light, travelling for many millions of years, past a galaxy, can have its path significantly bent by gravity. (The Schwartzchild solution is a first approximation to the shape of spacetime around a galaxy, though for a better approximation, numerical relativity can be used) This causes gravitational lensing. For light traveling inside a galaxy this is not a significant factor, There isn't time in a few years for light to be bent significantly by the general gravitational field of the galaxy. However light from a star passes close to another star, then there can be measurable bending. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The measurement of the bending of light of stars behind the sun during a solar eclipse was one of the first experimental confirmations of General Relativity (1919) &lt;a href=&quot;https://www.wired.com/2009/05/dayintech_0529/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.wired.com/2009/05/dayintech_0529/&lt;/a&gt; Note that even light that passes right over the Sun's surface would only be deflected by less that 2 arcseconds.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-12-17T14:12:35.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19434" PostTypeId="2" ParentId="19430" CreationDate="2016-12-17T12:11:58.183" Score="8" Body="&lt;p&gt;Getting atoms together into molecules requires two basic things - getting the reactants together and them providing the means to overcome the &quot;activation energy&quot; that is due to the mutual repulsion of the electrons that surround them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astrophysical chemistry is a rich and complex field of study. Most molecules are produced by a wide variety of processes. For ethanol in solar system objects, a cartoon of the process might be this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Grains of dust (basic silicates and soot) form in the protosolar nebula; cosmic rays and high energy radiation from the protosun ionise atoms (remove outer electrons), which makes them &quot;sticky&quot;; these ionised atoms are concentrated onto the surfaces of dust grains, increasing their densities to the extent that reactions between them become likely; hydrogen ions are able to hydrogenate carbon and oxygen to produce CH and OH radicals; these then encounter each other in the presence of more hydrogen to form ethanol; dust grains with ethanol are then incorporated into solar system bodies like comets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as I said, for any individual ethanol molecule, other pathways are possible.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-12-17T12:11:58.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19435" PostTypeId="1" CreationDate="2016-12-18T01:31:42.287" Score="2" ViewCount="416" Body="&lt;p&gt;How did Edwin Hubble determine the Universe was expanding?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was he looking in a particular direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And where does this Galaxy sit in the expansion.&lt;/p&gt;&#xA;" OwnerUserId="15211" LastEditorUserId="2365" LastEditDate="2016-12-18T13:12:19.437" LastActivityDate="2016-12-18T13:23:32.687" Title="Understanding the theory of an expanding Universe" Tags="&lt;solar-system&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19436" PostTypeId="2" ParentId="12768" CreationDate="2016-12-18T02:36:54.907" Score="0" Body="&lt;p&gt;It was just a century ago that we learned through the careful observations of astronomer Edwin Hubble, that the universe extends far beyond our own Milky Way galaxy. Mysterious cosmic clouds, then categorized as 'nebulae', are actually galaxies like our own; gravitationally bound collections of stars, dust, gas, and collapsed matter. At the center of these structures lies a supermassive black hole, an object millions to billions of times the mass of our own star, the Sun. Today, we know quasars (quasi-stars) are really just galaxies which are actively consuming matter at its core, and we call them 'active galaxies'. Since most known active galaxies date back to the beginnings of the universe, we can surmise that most of the supermassive black hole growth in the universe occurred long ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gamma ray bursts, of varying durations, are electromagnetic indications of intense nuclear interactions. These are thought to coincide with major stellar events like the final collapse and explosion after the nuclear fuel has been spent (supernova or hyper nova), or the collision/merger of neutron stars and/or black holes. And these phenomena have been seen right back to the beginnings of the universe, one example being GRB 090423 observed by the Swift mission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in terms of scale, active galaxies might be consuming anywhere from 10 to 1000 solar masses of matter per year, while a GRB may only represent a one time event of seconds to hours for the conversion of matter, but the rate of that matter conversion is much higher. A neutron star may weigh anywhere from 1 to 3 solar masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, both phenomena have been observed right back near the beginning of the universe, today estimated to be about 13.8 billion years ago.&lt;/p&gt;&#xA;" OwnerUserId="15209" LastActivityDate="2016-12-18T02:36:54.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19437" PostTypeId="2" ParentId="19435" CreationDate="2016-12-18T13:23:32.687" Score="2" Body="&lt;p&gt;Hubble discovered that he could measure the approximate distance to some galaxies by observing a type of variable star. The absolute brightness of these stars is known, so by measuring their apparent brightness Hubble could find how far away a galaxy was. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was also discovered that most galaxies were &quot;red-shifted&quot;, which was interpreted as meaning that the other galaxies were moving away from the Milkyway. Hubble plotted a scatter graph with distance on one axis and speed on the other, and noted that the more distant galaxies were moving faster, and the relationship was approximately a straight line. This meant that all galaxies were moving away from all other galaxies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This expansion could be predicted by general relativity, which allowed for  the space between galaxies to expand. It also suggests that the galaxies were closer together in the past, suggesting that the early universe was hot and dense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Milkyway does not have a special position in this expansion, all galaxies are moving away from each other. There is no implication that the Milkyway is central to the universe.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-12-18T13:23:32.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19439" PostTypeId="1" AcceptedAnswerId="19468" CreationDate="2016-12-19T00:46:37.393" Score="5" ViewCount="764" Body="&lt;p&gt;Reading through questions about my favourite planet, Venus, I came across &lt;a href=&quot;https://astronomy.stackexchange.com/a/57/15217&quot;&gt;this answer&lt;/a&gt; to the question &lt;a href=&quot;https://astronomy.stackexchange.com/q/26/15217&quot;&gt;What is the current accepted theory as to why Venus has a slow retrograde rotation?&lt;/a&gt;, specifically where the answer suggests one of the possible reasons for the slow retrograde motion as being&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The spin slowed to a standstill and then reversed, caused by the sun's gravity, the dense atmosphere and friction between core and mantle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on a fairly recent study published in the Nature article &quot;&lt;a href=&quot;http://www.nature.com/nature/journal/v411/n6839/abs/411767a0.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The four final rotation states of Venus&lt;/a&gt; (Alexandre C. M. Correia &amp;amp; Jacques Laskar, 2001).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the relatively high density of the Venusian atmosphere, this leads to the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How does the dense atmosphere of Venus affect the planet's rotation?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15217" LastEditorUserId="-1" LastEditorDisplayName="user15217" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-12-21T05:58:57.283" Title="How does the dense atmosphere of Venus affect the planet's rotation?" Tags="&lt;planet&gt;&lt;rotation&gt;&lt;planetary-atmosphere&gt;&lt;venus&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19440" PostTypeId="1" CreationDate="2016-12-19T03:36:51.930" Score="10" ViewCount="202" Body="&lt;p&gt;In a recent observation reported in the ScienceAlert article '&lt;a href=&quot;http://www.sciencealert.com/boron-has-been-detected-on-mars-for-the-first-time&quot; rel=&quot;noreferrer&quot;&gt;Boron has been detected on Mars for the first time&lt;/a&gt;&quot; reporting that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Curiosity rover has found boron on the surface of Mars, indicating that, at some point, the Red Planet had long-term habitable groundwater.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Making the analogy that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Boron here on Earth is associated with sites where there was once lots of water, but it's since evaporated away - like California's Death Valley.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, they state that this may not necessarily be the case for Mars, so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What could be the possible origin of boron found in the Martian soil if not from ancient groundwater?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15217" LastEditorUserId="6" LastEditorDisplayName="user15217" LastEditDate="2019-05-02T03:22:28.550" LastActivityDate="2019-05-02T03:22:28.550" Title="What could be the possible origin of boron found in the Martian soil if not from ancient groundwater?" Tags="&lt;planet&gt;&lt;mars&gt;&lt;planetary-formation&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19441" PostTypeId="2" ParentId="19439" CreationDate="2016-12-19T04:20:08.620" Score="1" Body="&lt;p&gt;The idea must have to do with tidal locking of Venus' rotation with its orbit.  That also happened to Mercury, except that Mercury has maintained its elliptical orbit so it actually switches the face toward the Sun each time it passes perihelion.  Venus' orbit is circular (probably because of the stronger tidal deformations of its malleable atmosphere), so tidal bulges on its atmosphere would tend to slow the rotation to match the orbital period and keep the same side facing the Sun.  That goes a long way toward stopping the rotation, though what &lt;em&gt;reverses&lt;/em&gt; it a little must be something else, a final step that has to do with the mantle rotating differently from the atmosphere.  That last bit must be complicated.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-12-19T04:20:08.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19443" PostTypeId="2" ParentId="14540" CreationDate="2016-12-19T05:48:16.390" Score="7" Body="&lt;p&gt;According to the paper &quot;&lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1029/2003JA009961/pdf&quot; rel=&quot;noreferrer&quot;&gt;Simultaneous ionospheric variability on Earth and Mars&lt;/a&gt;&quot; (Mendillo et al. 2003), the overall structure of any ionised region of a planetary atmosphere depends on&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a blend of in situ production and loss&#xA;  processes, plus effects of transport of ionization into or out&#xA;  of the local region of interest&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Specifically, according to Mendillo et al., the only significant influence is from the photon flux from the parent star - so, the further the planet is from the star, the less the incoming photon flux.  The other mechanisms are suggested to be planetary-based, including (from the article):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;planetary rotation rate&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;orbital obliquity&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;which both also affect the photon flux reaching that particular part of the planetary atmosphere. Other mechanisms are to do with the planetary atmospheres themselves, with factors including:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the thermal structure, chemistry and dynamics of the planetary &lt;a href=&quot;http://link.springer.com/referenceworkentry/10.1007%2F978-3-642-11274-4_1050&quot; rel=&quot;noreferrer&quot;&gt;'neutral' atmosphere&lt;/a&gt; itself.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;by how much energetic particles (from the stellar wind and/or magnetosphere) affects the planetary atmosphere.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;diffusion and electrodynamics from coupling from above the layer; and tides, waves and electrodynamics from coupling with below the layer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A further study in the paper &quot;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0032063310002187&quot; rel=&quot;noreferrer&quot;&gt;Ionospheric photoelectrons: Comparing Venus, Earth, Mars and Titan&lt;/a&gt;&quot; (Coates et al. 2011) compared the sunlit portion of planetary ionospheres, which is sustained by &lt;a href=&quot;https://en.wikipedia.org/wiki/Photoionization&quot; rel=&quot;noreferrer&quot;&gt;photoionization&lt;/a&gt; of primarily neutral atmospheric constituents by solar EUV (Extreme UV), a process originally identified for Earth, but recently:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the Mars Express, Venus Express and Cassini-Huygens missions have revealed the importance of this process at Mars, Venus and Titan, respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It should be noted, that for planets without magnetic fields (Venus and Mars for example), the Max Planck Institute for Solar System Research page &quot;&lt;a href=&quot;http://www.mps.mpg.de/planetary-science/planetary-plasma-environment&quot; rel=&quot;noreferrer&quot;&gt;Research about the planetary plasma environment&lt;/a&gt;&quot;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The ionosphere is assumed to display two distinct states, &quot;magnetized&quot; and &quot;unmagnetized&quot;, &lt;strong&gt;depending on the penetration depth of the solar wind magnetic field&lt;/strong&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One more factor is suggested - dust, theorised from observations reported in the paper &quot;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0032063311001693&quot; rel=&quot;noreferrer&quot;&gt;On the role of dust in the lunar ionosphere&lt;/a&gt;&quot; (Stubbs et al. 2011). According to the NASA page summarising the paper &quot;&lt;a href=&quot;https://science.nasa.gov/science-news/science-at-nasa/2011/14nov_lunarionosphere&quot; rel=&quot;noreferrer&quot;&gt;Mystery of the Lunar Ionosphere&lt;/a&gt;&quot;, ubiquitous grains of dust from the Lunar surface may be the source of the Lunar ionosphere.  The research from Stubbs et al. is summarised as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;UV rays from the sun hit the grains and ionize them.   According to their calculations, this process produces enough charge (positive grains surrounded by negative electrons) to create the observed ionosphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The process with the lunar ionosphere is illustrated below (from the NASA page linked):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Y9g3v.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y9g3v.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15217" LastEditorDisplayName="user15217" LastEditDate="2016-12-19T20:19:25.043" LastActivityDate="2016-12-19T20:19:25.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19445" PostTypeId="1" CreationDate="2016-12-19T06:57:50.197" Score="5" ViewCount="740" Body="&lt;p&gt;A full disclosure to begin with: I'm a PhD student in mathematics and while I understand most of standard cosmological-astronomical terms and I've followed a one semester course on cosmology, I don't really have a deep comprehension of the well known material neither on cosmology or astronomy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've recently heard about the possibility of a &quot;redshift quantization&quot;. As I understood this is not a mainstream result, which means that as 2007 was still quite debated. Anyway I couldn't understand if it really was ruled out by some more recent observations or if it's a still ongoing idea. But to be more specific what I really don't understand are the implications of this result and if this result would invalidate some current cosmological theory and why. I've heard it might have some implication on Quasar formation and change the current timeline of the Big-bang but didn't get how nor why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody give me some hints?&lt;/p&gt;&#xA;" OwnerUserId="9860" LastEditorUserId="9860" LastEditDate="2016-12-19T07:14:31.190" LastActivityDate="2020-05-30T16:33:38.183" Title="Redshift quantization" Tags="&lt;big-bang-theory&gt;&lt;redshift&gt;&lt;quasars&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19446" PostTypeId="1" CreationDate="2016-12-19T09:33:06.120" Score="45" ViewCount="11991" Body="&lt;p&gt;Scientists have known about the star &lt;em&gt;HD 140283&lt;/em&gt;, informally nicknamed the Methuselah star, for more than 100 years, since it cruises across the sky at a relatively rapid clip. The star moves at about 800,000 mph (1.3 million km/h) and travels the width of the full moon in the sky every 1,500 years or so, researchers said. Its apparent magnitude is 7.223.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.space.com/20112-oldest-known-star-universe.html&quot; rel=&quot;noreferrer&quot;&gt;Previous research&lt;/a&gt; had estimated that the Milky Way galaxy's so-called&#xA;&quot;Methuselah star&quot; is up to &lt;strong&gt;16 billion years old&lt;/strong&gt;. That's a problem,&#xA;since most researchers agree that the Big Bang that created the&#xA;universe occurred about &lt;strong&gt;13.8 billion years ago&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.nasa.gov/mission_pages/hubble/science/hd140283.html&quot; rel=&quot;noreferrer&quot;&gt;Later estimates&lt;/a&gt; show that the star could be as old as &lt;strong&gt;14.5 billion years&lt;/strong&gt; (± 0.8 billion years), which is still it older than the universe's calculated age of about &lt;strong&gt;13.8 billion years&lt;/strong&gt;. This is an obvious dilemma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can a star be older than the universe itself?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="15219" LastEditorUserId="17315" LastEditDate="2017-07-11T17:57:25.210" LastActivityDate="2019-08-12T01:47:37.540" Title="How can 'HD 140283' be older than the universe?" Tags="&lt;star&gt;&lt;universe&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="19447" PostTypeId="2" ParentId="19446" CreationDate="2016-12-19T11:34:52.613" Score="40" Body="&lt;p&gt;It's not - the current age is an estimate. The discrepancy comes from astronomer's attempts to fully and correctly measure aspects of the star, particularly its distance.  From the &lt;a href=&quot;https://www.nasa.gov/mission_pages/hubble/science/hd140283.html&quot; rel=&quot;noreferrer&quot;&gt;NASA&lt;/a&gt; page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The new Hubble age estimates reduce the range of measurement uncertainty, so that &lt;strong&gt;the star's age overlaps with the universe's age&lt;/strong&gt; — as independently determined by the rate of expansion of space, an analysis of the microwave background from the big bang, and measurements of radioactive decay.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Currently, if you take the uncertainty into account, the current &lt;em&gt;estimate&lt;/em&gt; for the star's age is 14.5 +/- 0.8 billion years  or between 15.3 and 13.7 billion years - the latter end of the estimate overlaps with the current model of the age of the Universe - so a more correct way of describing this age is that &lt;strong&gt;HD 140283 is at least 13.7 billion years old by this estimation&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another paper based on this star in Nature - &quot;&lt;a href=&quot;http://www.nature.com/news/nearby-star-is-almost-as-old-as-the-universe-1.12196&quot; rel=&quot;noreferrer&quot;&gt;Nearby star is almost as old as the Universe&lt;/a&gt;&quot;, a team has narrowed down the age to 13.9 +/- 0.7 billion years, stating that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Taking into account that experimental error, the age does not conflict with the age of the Universe, 13.77 billion years. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;more concisely stating that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The star's age is therefore at least 13.2 billion years&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No doubt further work is being performed to further refine the age of the star.&lt;/p&gt;&#xA;" OwnerDisplayName="user15217" LastEditorDisplayName="user15217" LastEditDate="2017-01-06T22:29:12.430" LastActivityDate="2017-01-06T22:29:12.430" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="19448" PostTypeId="2" ParentId="19446" CreationDate="2016-12-19T12:00:21.817" Score="16" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Later estimates shows that the star could be as old as 14.5 billion years (± 0.8 billion years), which is still it older than the universe's calculated age of about 13.8 billion years, an obvious dilemma.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no dilemma. That ± 0.8 billion years is important. Subtract 0.8 billion years from that 14.5 billion year figure (&lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0004-637X/792/2/110/meta&quot; rel=&quot;noreferrer&quot;&gt;later revised to 14.27 billion years&lt;/a&gt;) and you get 13.7 billion years, less than the age of the universe. Also note that that 0.8 billion years uncertainty is the standard error, or roughly one standard deviation. That revised age of 14.27 billion years makes the dilemma essentially non-existent. That's older than the estimate age of the universe by 0.4 to 0.5 million years, and this is not significant even at the one sigma level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Particle physicists generally use the five sigma level as distinguishing between &quot;almost certain&quot; and &quot;maybe&quot; (and particle physicists are reluctant to publish &quot;maybes&quot;). Astronomers generally use a three sigma level. Even social scientists would balk at a deviation that is less than one sigma as being significant.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="10434" LastEditDate="2016-12-19T17:11:06.783" LastActivityDate="2016-12-19T17:11:06.783" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19449" PostTypeId="2" ParentId="18488" CreationDate="2016-12-19T14:12:59.540" Score="7" Body="&lt;p&gt;Don't know about the energy transfer but from a rough assumption we can calculate the energy amount that actually passed through Earth depending on the mass converted and the distance of the event. The energy absorbed by Earth would be far less than that obviously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The measured event was two black holes merging with a total mass loss of $3 M_{\odot}$. While the distance for that event was calculated to $1.3*10^9 ly$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With comparing the total energy released (assuming full conversion of mass to gravitational wave energy) over the released area:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;E_{Earth} = \pi r_{Earth}^2*\frac{m_{conv}c^2}{4\pi r_{dist}^2}&#xA;$$&#xA;$$&#xA;E_{Earth} \approx (6378km)^2*\frac{3M_{\odot}*c^2}{4*(1.3*10^9years*c)^2}&#xA;$$&#xA;$$&#xA;E_{Earth} \approx 4.0*10^{13}m^2*\frac{5.7*10^{30}kg}{6.72*10^{33}s^2}&#xA;$$&#xA;$$&#xA;E_{Earth} \approx 3.39 * 10^{10}J&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the maximum amount of energy passing through earth would have been roughly $34 GJ$. That is about the kinetic energy of three Boeing 747 in flight - or neglectable on a planetary scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even IF the total energy would have been absorbed by Earth, the amount would still have been - not much.&lt;/p&gt;&#xA;" OwnerUserId="14441" LastEditorUserId="14441" LastEditDate="2016-12-20T10:37:58.793" LastActivityDate="2016-12-20T10:37:58.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19450" PostTypeId="2" ParentId="19445" CreationDate="2016-12-19T15:07:46.643" Score="4" Body="&lt;p&gt;There's a pretty comprehensive &lt;a href=&quot;https://en.wikipedia.org/wiki/Redshift_quantization&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia page&lt;/a&gt; on the subject. In summary....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Throughout the 2000's conflicting results were published using the same data. The conflict seems to center on the selection effects of the surveys, and the corrections applied to account for them. I did a quick search for 'redshift quantisation' on the Arxiv and found no papers since 2009, which suggests the supporting data has been comprehensibly discredited, or explained away by natural mechanisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two explanations for the quantization that do not require exotic cosmologies. The first is the large scale clustering of galaxies that traces the cosmic web. The walls and filaments formed in this web leads to groupings of galaxies around specific redshifts along the line of sight. The second explanation is the quantisation of emission from Hydrogen, known as the Lyman series. Since photons can only be released with certain energies, the measured redshifts are also quantised (this is only true in low pressure and temperature environments, otherwise pressure and thermal broadening occur).&lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2016-12-19T15:07:46.643" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19452" PostTypeId="1" CreationDate="2016-12-19T18:54:45.843" Score="1" ViewCount="142" Body="&lt;p&gt;Lets assume I have a source and receive $8 \times 10^{-20}$ ${\rm Wm}^{-2}$ in (400-750) nm filter. How do you convert this received flux to the unit Jansky, is there a formula? (Ultimately, I want to express this flux in magnitude AB.)&lt;/p&gt;&#xA;" OwnerUserId="15233" LastEditorUserId="2232" LastEditDate="2016-12-22T03:29:39.073" LastActivityDate="2016-12-22T03:29:39.073" Title="${\rm W/m}^2$ to Jansky" Tags="&lt;light&gt;&lt;observational-astronomy&gt;&lt;magnitude&gt;&lt;units&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2016-12-23T00:28:40.217" ContentLicense="CC BY-SA 3.0" />
  <row Id="19453" PostTypeId="1" AcceptedAnswerId="19458" CreationDate="2016-12-19T21:54:48.867" Score="12" ViewCount="593" Body="&lt;p&gt;I would like to demonstrate the existence of another planet (Neptune) by studying the discrepancy between the observation of the Uranus orbit and the mathematical prediction, this work was made from Le Verrier and I would like to understand his method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read the Chapter 2, &quot;The Discovery of Neptune (1845-1846),&quot; in the biography Le Verrier -- Magnificent and Detestable Astronomer,&#xA;but it is too much in-depth and I did not understand very well his work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am studying the three-body problem (Sun, Uranus, Neptune) via Matlab and the two body problem (Sun, Uranus) taking the initial condition from here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://nssdc.gsfc.nasa.gov/planetary/factsheet/uranusfact.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://nssdc.gsfc.nasa.gov/planetary/factsheet/uranusfact.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried this method:&#xA;I put Uranus in the Perihelion with the Max. orbital velocity and I calculate the semi-major axis, and it is more accurate than the one obtained from putting Uranus and Neptune in the Perihelion with their respective Max. orbital velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here a cool pic made with Matlab:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/gxLML.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gxLML.jpg&quot; alt=&quot;Here a cool pic&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody help me? what I have to do and with what data I have to compare my prediction? Even a simple link could be helpful.&lt;/p&gt;&#xA;" OwnerUserId="15240" LastEditorUserId="2365" LastEditDate="2016-12-23T23:55:54.683" LastActivityDate="2016-12-23T23:55:54.683" Title="How to discover Neptune from the Uranus orbit (by computer simulation)" Tags="&lt;orbit&gt;&lt;fundamental-astronomy&gt;&lt;n-body-simulations&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19454" PostTypeId="2" ParentId="19453" CreationDate="2016-12-20T07:27:57.253" Score="5" Body="&lt;p&gt;If I understand correctly, you're modeling Uranus' orbit as an ellipse and want to compare it to Uranus' actual orbit as perturbed by Neptune? I don't have an answer, but &lt;a href=&quot;https://astronomy.stackexchange.com/questions/13488/where-can-i-find-visualize-planets-stars-moons-etc-positions&quot;&gt;Where can I find/visualize planets/stars/moons/etc positions?&lt;/a&gt; explains how to use SPICE, HORIZONS, and other tools to find Uranus' true position at a given time +-15000 years from now, including the best fit elliptical parameters (using HORIZONS &quot;orbital elements&quot; feature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, anything you do will be &quot;circular&quot; in some sense, since HORIZONS computed position of Uranus in the past already includes Neptune's perturbations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could find tables of Uranus position predictions or something from the past, you might have something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, feel free to contact me (see profile for details) if this project extends beyond a stackexchange question.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-12-20T07:27:57.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19456" PostTypeId="2" ParentId="10395" CreationDate="2016-12-20T09:00:45.917" Score="11" Body="&lt;p&gt;A very well documented example of a small celestial object with a gas giant is the collision between Comet Shoemaker-Levy 9 with Jupiter.  As it was a comet that was tracked, the whole process of the collision was well documented and observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though the collision occurred in July, 1994 - the first effects occurred 2 years earlier, according to the NASA Technical Report &quot;&lt;a href=&quot;https://ntrs.nasa.gov/search.jsp?R=19940030900&quot;&gt;Shoemaker-Levy 9 and the tidal disruption of comets&lt;/a&gt;&quot;, when the &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;break-up of Periodic Comet Shoemaker-Levy 9 into multiple pieces following its grazing encounter with Jupiter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This represents one of the first stages of the collision that could occur as the celestial object approached the large planet.  In this case, this breakup resulted in Comet Shoemaker-Levy 9 to form a train of fragments to eventually collide into Jupiter - the full process at this stage is shown in the diagram below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tYO6M.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tYO6M.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image Source: &lt;a href=&quot;http://ase.tufts.edu/cosmos/print_images.asp?id=16&quot;&gt;Tufts University&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The train of fragments was captured by the Hubble Space Telescope (shown below):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cH4Qg.gif&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cH4Qg.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image Source: &lt;a href=&quot;http://www2.jpl.nasa.gov/sl9/image2.html&quot;&gt;NASA&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each fragment was designated a letter (e.g. Fragment A etc).  According to the article &quot;&lt;a href=&quot;https://www.ncbi.nlm.nih.gov/pubmed/11539477&quot;&gt;The collision of Jupiter and Comet Shoemaker-Levy 9.&lt;/a&gt;&quot; (Zahnle and Mac Low, 1994), these fragments were in the order of about 100 m to 5 km in size, heading straight towards Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Then to the main event - the collision of the comet fragments with Jupiter.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example was from Fragment G, one of the larger fragments, when it collided with Jupiter (image below):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1CjRr.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1CjRr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image source: &lt;a href=&quot;http://stardust.jpl.nasa.gov/comets/sl9.html&quot;&gt;NASA&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the explosion was spectacular - about 21 fragments impacted into Jupiter over a number of days, given Jupiter's relatively fast rotation, the impacts became quite distributed on the top of the clouds of Jupiter, as shown in the ultraviolet image below (the circle shadow below is the moon Io):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mCcWz.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mCcWz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image Source &lt;a href=&quot;https://airandspace.si.edu/multimedia-gallery/web12139-2011hjpg&quot;&gt;Air and Space Museum&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been several impacts into Jupiter (and presumably into the other gas/ice giants as well), another one that underwent scientific scrutiny was an asteroid 'the size of the Titanic' colliding with Jupiter in 2009, reported in the article &quot;&lt;a href=&quot;http://phys.org/news/2011-01-evidence-asteroid-comet-struck-jupiter.html&quot;&gt;New evidence that asteroid, not comet, struck Jupiter in 2009&lt;/a&gt;&quot;, describes a comparison between the 2009 asteroid and Comet Shoemaker-Levy 9 are described as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The dark debris, the heated atmosphere and upwelling of ammonia were similar for this impact and Shoemaker-Levy, but the debris plume in this case didn't reach such high altitudes, didn't heat the high stratosphere, and contained signatures for hydrocarbons, silicates and silicas that weren't seen before. The presence of hydrocarbons, and the absence of carbon monoxide, provide strong evidence for a water-depleted impactor in 2009.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Overall, the article mentioned the likely process during any collisions of a celestial body with a gaseous planet, using the example from Jupiter:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Plunging through Jupiter's atmosphere, the object created a channel of super-heated atmospheric gases and debris. An explosion deep below the clouds – probably releasing at least around 200 trillion trillion ergs of energy, or more than 5 gigatons of TNT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the celestial object itself - vaporised, totally destroyed in the resulting explosion.&lt;/p&gt;&#xA;" OwnerDisplayName="user15217" LastEditorDisplayName="user15217" LastEditDate="2016-12-20T09:14:15.130" LastActivityDate="2016-12-20T09:14:15.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19458" PostTypeId="2" ParentId="19453" CreationDate="2016-12-20T16:14:36.600" Score="12" Body="&lt;p&gt;Here is what I did: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Based on their masses, it is safest to initially consider Jupiter and Saturn as well as Uranus. It might also be fruitful to include the Earth in the analysis, to get relative positions, observation angles, etc. So, I will be considering: &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sun&lt;/li&gt;&#xA;&lt;li&gt;Earth&lt;/li&gt;&#xA;&lt;li&gt;Jupiter&lt;/li&gt;&#xA;&lt;li&gt;Saturn &lt;/li&gt;&#xA;&lt;li&gt;Uranus&lt;/li&gt;&#xA;&lt;li&gt;Neptune&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Get the &lt;a href=&quot;https://en.wikipedia.org/wiki/Standard_gravitational_parameter&quot; rel=&quot;noreferrer&quot;&gt;standard gravitational parameters&lt;/a&gt; (μ) for all of them&lt;/li&gt;&#xA;&lt;li&gt;Get initial positions and velocities via &lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;noreferrer&quot;&gt;JPL/HORIZONS&lt;/a&gt; for all these planets. I had some data lying around from J2000.5, so I just used the state vectors from the 1st of January, 2000, at noon. &lt;/li&gt;&#xA;&lt;li&gt;Write an N-body integrator with built-in MATLAB tools. Integrate this incomplete Solar system once without Neptune, and once with Neptune included. &lt;/li&gt;&#xA;&lt;li&gt;Analyze and compare!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, here's my data, and N-body integrator: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function [t, yout_noNeptune, yout_withNeptune] = discover_Neptune()&#xA;&#xA;    % Time of integration (in years)&#xA;    tspan = [0 97] * 365.25 * 86400;&#xA;&#xA;    % std. gravitational parameters [km/s²/kg]&#xA;    mus_noNeptune = [1.32712439940e11; % Sun&#xA;                     398600.4415       % Earth&#xA;                     1.26686534e8      % Jupiter&#xA;                     3.7931187e7       % Saturn&#xA;                     5.793939e6];      % Uranus&#xA;&#xA;    mus_withNeptune = [mus_noNeptune&#xA;                       6.836529e6]; % Neptune&#xA;&#xA;    % Initial positions [km] and velocities [km/s] on 2000/Jan/1, 00:00&#xA;    % These positions describe the barycenter of the associated system,&#xA;    % e.g., sJupiter equals the statevector of the Jovian system barycenter.&#xA;    % Coordinates are expressed in ICRF, Solar system barycenter&#xA;    sSun     = [0 0 0 0 0 0].';&#xA;    sEarth   = [-2.519628815461580E+07  1.449304809540383E+08 -6.175201582312584E+02,...&#xA;                -2.984033716426881E+01 -5.204660244783900E+00  6.043671763866776E-05].';&#xA;    sJupiter = [ 5.989286428194381E+08  4.390950273441353E+08 -1.523283183395675E+07,...&#xA;                -7.900977458946710E+00  1.116263478937066E+01  1.306377465321731E-01].';&#xA;    sSaturn  = [ 9.587405702749230E+08  9.825345942920649E+08 -5.522129405702555E+07,...&#xA;                -7.429660072417541E+00  6.738335806405299E+00  1.781138895399632E-01].';&#xA;    sUranus  = [ 2.158728913593440E+09 -2.054869688179662E+09 -3.562250313222718E+07,...&#xA;                 4.637622471852293E+00  4.627114800383241E+00 -4.290473194118749E-02].';&#xA;    sNeptune = [ 2.514787652167830E+09 -3.738894534538290E+09  1.904284739289832E+07,...&#xA;                 4.466005624145428E+00  3.075618250100339E+00 -1.666451179600835E-01].';&#xA;&#xA;    y0_noNeptune   = [sSun; sEarth; sJupiter; sSaturn; sUranus];&#xA;    y0_withNeptune = [y0_noNeptune; sNeptune];&#xA;&#xA;    % Integrate the partial Solar system &#xA;    % once with Neptune, and once without&#xA;    options = odeset('AbsTol', 1e-8,...&#xA;                     'RelTol', 1e-10);&#xA;&#xA;    [t, yout_noNeptune]   = ode113(@(t,y) odefcn(t,y,mus_noNeptune)  , tspan, y0_noNeptune  , options);&#xA;    [~, yout_withNeptune] = ode113(@(t,y) odefcn(t,y,mus_withNeptune),     t, y0_withNeptune, options);&#xA;&#xA;end&#xA;&#xA;% The differential equation &#xA;%&#xA;%    dy/dt = d/dt [r₀ v₀ r₁ v₁ r₂ v₂ ... rₙ vₙ]    &#xA;%          = [v₀ a₀ v₁ a₁ v₂ a₂ ... vₙ aₙ]    &#xA;%&#xA;%  with &#xA;%&#xA;%    aₓ = Σₘ -G·mₘ/|rₘ-rₓ|² · (rₘ-rₓ) / |rₘ-rₓ| &#xA;%       = Σₘ -μₘ·(rₘ-rₓ)/|rₘ-rₓ|³  &#xA;%&#xA;function dydt = odefcn(~, y, mus)&#xA;&#xA;    % Split up position and velocity&#xA;    rs = y([1:6:end; 2:6:end; 3:6:end]);&#xA;    vs = y([4:6:end; 5:6:end; 6:6:end]);&#xA;&#xA;     % Number of celestial bodies&#xA;    N = size(rs,2);&#xA;&#xA;    % Compute interplanetary distances to the power -3/2&#xA;    df  = bsxfun(@minus, permute(rs, [1 3 2]), rs);&#xA;    D32 = permute(sum(df.^2), [3 2 1]).^(-3/2);&#xA;    D32(1:N+1:end) = 0; % (remove infs)&#xA;&#xA;    % Compute all accelerations     &#xA;    as = -bsxfun(@times, mus.', D32);              % (magnitudes)    &#xA;    as = bsxfun(@times, df, permute(as, [3 2 1])); % (directions)    &#xA;    as = reshape(sum(as,2), [],1);                 % (total)&#xA;&#xA;    % Output derivatives of the state vectors&#xA;    dydt = y;&#xA;    dydt([1:6:end; 2:6:end; 3:6:end]) = vs;&#xA;    dydt([4:6:end; 5:6:end; 6:6:end]) = as;&#xA;&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here is the driver script I used to get some nice plots out: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;clc&#xA;close all&#xA;&#xA;% Get coordinates from N-body simulation&#xA;[t, yout_noNeptune, yout_withNeptune] = discover_Neptune();&#xA;&#xA;% For plot titles etc.&#xA;bodies = {'Sun'&#xA;          'Earth'&#xA;          'Jupiter'&#xA;          'Saturn'&#xA;          'Uranus'&#xA;          'Neptune'};&#xA;&#xA;&#xA;% Extract positions&#xA;rs_noNeptune   = yout_noNeptune  (:, [1:6:end; 2:6:end; 3:6:end]);&#xA;rs_withNeptune = yout_withNeptune(:, [1:6:end; 2:6:end; 3:6:end]);&#xA;&#xA;&#xA;&#xA;% Figure of the whole Solar sysetm, just to check&#xA;% whether everything went OK&#xA;figure, clf, hold on&#xA;for ii = 1:numel(bodies)&#xA;    plot3(rs_withNeptune(:,3*(ii-1)+1),...&#xA;          rs_withNeptune(:,3*(ii-1)+2),...&#xA;          rs_withNeptune(:,3*(ii-1)+3),...&#xA;          'color', rand(1,3));&#xA;end&#xA;&#xA;axis equal&#xA;legend(bodies);&#xA;xlabel('X [km]');&#xA;ylabel('Y [km]');&#xA;title('Just the Solar system, nothing to see here');&#xA;&#xA;&#xA;% Compare positions of Uranus with and without Neptune&#xA;rs_Uranus_noNeptune   = rs_noNeptune  (:, 13:15);&#xA;rs_Uranus_withNeptune = rs_withNeptune(:, 13:15);&#xA;&#xA;figure, clf, hold on&#xA;&#xA;plot3(rs_Uranus_noNeptune(:,1),...&#xA;      rs_Uranus_noNeptune(:,2),...&#xA;      rs_Uranus_noNeptune(:,3),...&#xA;      'b.');&#xA;&#xA;plot3(rs_Uranus_withNeptune(:,1),...&#xA;      rs_Uranus_withNeptune(:,2),...&#xA;      rs_Uranus_withNeptune(:,3),...&#xA;      'r.');&#xA;&#xA;axis equal&#xA;xlabel('X [km]');&#xA;ylabel('Y [km]');&#xA;legend('Uranus, no Neptune',...&#xA;       'Uranus, with Neptune');&#xA;&#xA;&#xA;% Norm of the difference over time&#xA;figure, clf, hold on&#xA;&#xA;rescaled_t = t/365.25/86400;&#xA;&#xA;dx = sqrt(sum((rs_Uranus_noNeptune - rs_Uranus_withNeptune).^2,2));&#xA;plot(rescaled_t,dx);&#xA;xlabel('Time [years]');&#xA;ylabel('Absolute offset [km]');&#xA;title({'Euclidian distance between'&#xA;       'the two Uranuses'});&#xA;&#xA;&#xA;% Angles from Earth&#xA;figure, clf, hold on&#xA;&#xA;rs_Earth_noNeptune   = rs_noNeptune  (:, 4:6);&#xA;rs_Earth_withNeptune = rs_withNeptune(:, 4:6);&#xA;&#xA;v0 = rs_Uranus_noNeptune   - rs_Earth_noNeptune;&#xA;v1 = rs_Uranus_withNeptune - rs_Earth_withNeptune;&#xA;&#xA;nv0 = sqrt(sum(v0.^2,2));&#xA;nv1 = sqrt(sum(v1.^2,2));&#xA;&#xA;dPhi = 180/pi * 3600 * acos(min(1,max(0, sum(v0.*v1,2) ./ (nv0.*nv1) )));&#xA;plot(rescaled_t, dPhi);&#xA;&#xA;xlabel('Time [years]');&#xA;ylabel('Separation [arcsec]')&#xA;title({'Angular separation between the two'&#xA;       'Uranuses when observed from Earth'});&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which I'll describe here step by step. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, a plot of the Solar system to check that the N-body integrator works as it should: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XcDzN.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XcDzN.png&quot; alt=&quot;the Solar system&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nice! Next, I wanted to see the difference between the positions of Uranus with and without the influence of Neptune. So, I extracted just the positions of those two Uranuses, and plotted them: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/erO5x.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/erO5x.png&quot; alt=&quot;Two Uranuses, with and without Neptune&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...that's hardly useful. Even when zooming in greatly and rotating the heck out of it, this is just not a useful plot. So I looked at the evolution of the absolute Euclidian distance between the two Uranuses: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oL7G7.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oL7G7.png&quot; alt=&quot;Time evolution of the Euclidian distance between the two Uranuses&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's starting to look more like it! Roughly 80 years after the start of our analysis, the two Uranuses are almost 6 million km apart! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Large as that may sound, in the grander scale of things this might drown in the noise when we take measurements here on Earth. Plus, it still does not tell the whole story, as we'll see in a moment. So next, let's look at the angular difference between the observation vectors, from Earth towards the two Uranuses to see how large that angle is, and if it can stand out above the observational error thresholds: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xwZlg.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xwZlg.png&quot; alt=&quot;Angular separation between the two Uranuses&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...whoa! Well over 300 arcseconds difference, plus all sorts of wobbly bobbley timey wimey rippling going on. That seems well within the observational capabilities of the time (although I can't find a reliable source on this so quickly; anyone?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for good measure, I also produced that last plot leaving Jupiter and Saturn out of the picture. Although some perturbation theory had been developed in the 17&lt;sup&gt;th&lt;/sup&gt; and 18&lt;sup&gt;th&lt;/sup&gt; centuries, it was not very well developed and I doubt even Le Verrier took Jupiter into consideration (but again, I could be wrong; please correct me if you know more). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, here is the last plot without Jupiter and Saturn: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GaLrT.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GaLrT.png&quot; alt=&quot;Angular separation between the two Uranuses, leaving Jupiter and Saturn out of the equation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although there are differences, they are minute, and most importantly irrelevant for discovering Neptune. &lt;/p&gt;&#xA;" OwnerUserId="599" LastEditorUserId="599" LastEditDate="2016-12-20T18:57:43.217" LastActivityDate="2016-12-20T18:57:43.217" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19459" PostTypeId="1" AcceptedAnswerId="19460" CreationDate="2016-12-20T17:22:21.787" Score="11" ViewCount="2261" Body="&lt;p&gt;This question arose to me when I saw a SciFi movie where they shot with laser guns and you clearly could see the dashes of light beams travelling from the shooter to the target. Nonsense of course, reality is much more boring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I wondered if you would just need more distance: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Given a limited light beam traveling from A to B, where &quot;limited&quot; means that there is a start and / or end of the emission source causing that beam.&lt;/li&gt;&#xA;&lt;li&gt;Given enough distance from us to A, B and the line between them&lt;/li&gt;&#xA;&lt;li&gt;And add some kind of nebular smoke in between just enough to disperse as much light as needed so we can see it here but not as much to prevent it from reaching B.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Couldn't we observe the starting and / or the ending of that light beam, progressing on it's way from A to B, like a jet trail in the sky?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is possible, where can this be observed? Or do the laws of physics forbid such a thing?&lt;/p&gt;&#xA;" OwnerUserId="10921" LastEditorUserId="10921" LastEditDate="2016-12-20T17:36:30.943" LastActivityDate="2016-12-21T06:43:22.220" Title="Can you see the starting and the ending of a light beam passing in the distance?" Tags="&lt;light&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19460" PostTypeId="2" ParentId="19459" CreationDate="2016-12-20T18:51:53.277" Score="18" Body="&lt;p&gt;Nothing forbids this, and it is actually observed astronomically. You need a very bright source of light: such as a supernova (which isn't a beam, but radiates in all directions) and very large distances. The flash of light can be seen spreading out from supernova in a circle, as it illuminates the dust and gas ejected by the supernova progenitor star in the years before&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect is known as a light echo. Over time the &quot;beams&quot; of light move further out, and so the echo widens, as shown in this series of images (from an Australian telescope so you have to read from the bottom up, 507 days means 507 days after the supernova in 1987.) For scale, 5 arcmin is quite small: The moon is about 30 arcmin. &lt;a href=&quot;http://oldweb.aao.gov.au/images/captions/aat066.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BSkxZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BSkxZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-12-20T18:51:53.277" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19461" PostTypeId="2" ParentId="19244" CreationDate="2016-12-20T19:24:02.220" Score="1" Body="&lt;h2&gt;Regarding the original picture from the question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;No, it is not feasible to get a picture like this. According to &lt;a href=&quot;http://nssdc.gsfc.nasa.gov/photo_gallery/photogallery-mars.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this webpage&lt;/a&gt; the photograph you posted was taken with the Hubble Space Telescope's &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Wide_Field_and_Planetary_Camera_2&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wide-Field Planetary Camera 2&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The telescope itself has a 2.4 m (7.9 ft) mirror and the focal length of 57.6 m (189 ft) and is placed outside the Earth's atmosphere. You'd need a large instrument with adaptive optics to get near this quality (assuming you can't afford to place a telescope in orbit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, it is possible for you to capture the polar ice caps on Mars using an amateur telescope. The quality of these will largely depend on the equipment you have and the location (&lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_seeing#Overcoming_atmospheric_seeing&quot; rel=&quot;nofollow noreferrer&quot;&gt;atmospheric turbulence is very unforgiving&lt;/a&gt; if you use large magnifications) as well as the imaging technique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a &lt;a href=&quot;http://www.skyandtelescope.com/online-gallery/planet-pictures/&quot; rel=&quot;nofollow noreferrer&quot;&gt;gallery of planet pictures taken by amateurs&lt;/a&gt;. You can find some photos of Mars in which the polar caps are visible.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Regarding the Mars pictures you found on the Sky &amp;amp; Telescope website&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;To answer your questions about &lt;a href=&quot;http://www.skyandtelescope.com/online-gallery/rotation-mars-0507-06062016/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this image&lt;/a&gt;, I see no reason to think it wasn't taken with an 8&quot; telescope. However, there must have been more to it than just snapping a picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the author, the photographs were taken using a &lt;a href=&quot;http://www.celestron.com/browse-shop/astronomy/telescopes/advance-vx/advanced-vx-8-schmidt-cassegrain-telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Celestron 8&quot; SCT&lt;/a&gt; telescope with a Meade 3x Telextender and Expanse 120 Camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're talking about a focal length of 2000 mm with 3x extender, which gives us 6000 mm. Not bad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Expanse 120 is a CCD camera. Now, I'm not sure that's the case with these specific pictures but one way to reduce the impact of atmospheric turbulence is to record a video of a planet and stack the individual frames. Many of them will look very bad due to turbulence but occasionally, you'll get a better quality image. Selecting the best frames from a video and combining them into a single image can help you get a lot more detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at &lt;a href=&quot;https://nightskyinfocus.com/equipment/imaging-planets-using-webcams/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this guide&lt;/a&gt; explaining how consumer-grade computer webcams can be used. It contains a neat comparison of a single frame and a stacked image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that I'm merely speculating as to which techiniques could be used to obtain such images. I can see that the author posted their email on the S&amp;amp;T site and that there's a comments section as well. Feel free to ask them yourself.&lt;/p&gt;&#xA;" OwnerUserId="211" LastEditorUserId="211" LastEditDate="2016-12-22T10:46:13.043" LastActivityDate="2016-12-22T10:46:13.043" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19462" PostTypeId="2" ParentId="19459" CreationDate="2016-12-20T19:36:42.130" Score="8" Body="&lt;p&gt;What, like this? Resolved astrophysical jets are not unusual. In this example, showing the region near the centre of the active galactic nucleus of M87, relativistic particles are emitted in a collimated beam from bear the central black hole. The particles both produce radiation and interact with material to produce more radiation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whilst originating from a jet of particles, there is certainly an intense beam of radiation emitted along the path you can see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PZCNp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PZCNp.jpg&quot; alt=&quot;Jet from M87&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-12-20T19:36:42.130" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19463" PostTypeId="2" ParentId="669" CreationDate="2016-12-20T19:46:05.690" Score="2" Body="&lt;p&gt;That's not actually how explosions work. When nitroglycerin detonates, it does not leave a hole in the centre. Just like an explosion, the big bang does not work that way either. In any valid frame of reference, the universe started expanding at the speed of light without leaving a hole in the centre and the centre is not a special place. Because of strange laws of the universe, there isn't only one valid frame of reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe follows general relativity which simplifies to special relativity in the absence of a gravitational field and in the absence of objects with an escape velocity that's a significant fraction of the speed of light, very closely follows a version of special relativity where gravity is a real force that does not bend space-time. See &lt;a href=&quot;https://physics.stackexchange.com/questions/19937/time-dilation-as-an-observer-in-special-relativity/384547#384547&quot;&gt;https://physics.stackexchange.com/questions/19937/time-dilation-as-an-observer-in-special-relativity/384547#384547&lt;/a&gt; to learn how special relativity works. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to special relativity, the universe has no centre. Any nonrotating object travelling at any constant velocity slower than the speed of light is a valid frame of reference and in its frame of reference, the centre of the universe is the place where the big bang has occurred. There is no timelike line that all observers agree is the centre of the universe. In any frame of reference, the centre of the universe in that frame of reference can't be a special place because it's not the centre in another frame of reference. When we look at galaxies near the edge of the universe, we see ones similar to those that occured near the beginning of the universe but we're really only looking back at galaxies from when when they were about half the age of our universe in our frame of reference. They're like much younger galaxies only because of their own time dilation and in their own frame of reference, are actually much younger. In any frame of reference, what happens if you're near the edge of the universe and stationary? You see yourself as being near the edge. In another frame of reference, you're in the center of the universe and moving and the aberration of light you observe makes you perceive yourself as not being in the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's just what special relativity predicts but in reality, the universe doesn't follow special relativity but some of the results I already mentioned are still true. The universe is accelerating so galaxies will eventually recede from us faster than light because space itself is dragging them away faster than light. We probably live in a De Sitter universe. Our cosmic horizon, the region of space that's moving away from us at the speed of light in our frame of reference behaves just like a black hole in the sense that we will see galaxies exponentially approach the cosmic horizon without ever quite reaching it and getting more red shifted without bound as it gets closer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/De_Sitter_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/De_Sitter_universe&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13685" LastEditorUserId="13685" LastEditDate="2018-04-06T01:13:41.640" LastActivityDate="2018-04-06T01:13:41.640" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19465" PostTypeId="2" ParentId="19459" CreationDate="2016-12-20T21:42:05.733" Score="3" Body="&lt;p&gt;I think the problem/contention on all the answers (and the reason the question is interesting) is the issue of seeing light.  I'm pretty sure you can only see what is reflected towards you, so if light is passing through completely empty space (and not pointed toward you) then you can't see it at all.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were passing through an area filled with dust, asteroids or other objects then you can see the light as it's reflected from those objects--and in this case  you would obviously see it progress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps in the case of the laser weapon it might vaporize all the dust/material it comes across which might actually make for a really cool visual effect.&lt;/p&gt;&#xA;" OwnerUserId="15256" LastActivityDate="2016-12-20T21:42:05.733" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19467" PostTypeId="1" AcceptedAnswerId="19470" CreationDate="2016-12-21T01:07:01.980" Score="2" ViewCount="146" Body="&lt;p&gt;I have 2 simulated IFU data cubes of the same observation, let's say A and B. The 3rd dimension of cube A is in wavelengths. In order to create cube B I just rebinned cube A in log-wavelengths and then converted that to velocity (km/s). The rebinning to log-wavelength was necessary in order to convert the 3rd dimension of cube B in km/s. For an explanation see this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/16455/converting-%C3%A5ngstr%C3%B6m-spectral-dimension-to-galaxy-speed-km-s&quot;&gt;Converting &amp;#229;ngstr&amp;#246;m spectral dimension to galaxy speed (km/s)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to apply instrumental broadening on both cubes. In order to do that, for each cube, I have to create a Gaussian profile with FWHM equal to the resolution of the instrument (or Line Spread Function, or LSF) and then convolve that with every spectrum of the cube. Of course the LSF of cube A will be in wavelengths while the LSF of cube B in km/s. Here is my question now:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Since all pixel shifts in cube A do not correspond to the same km/s value, one of the two LSFs won't be a perfect Gaussian, right? Shall I assume that would be the velocity LSF?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the correct way of creating the velocity LSF? Do I just create a wavelength LSF and then rebin it to km/s, the same way I rebinned cube A to cube B?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Does it worth the effort though? Will it make a difference? I have seen peer reviewed papers using velocity LSFs which are perfect Gaussians. Is this because the difference is negligible or my whole thinking/question is totally wrong?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1986" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-12-21T09:26:48.680" Title="wavelength-based IFU data cubes VS velocity-based IFU data cubes" Tags="&lt;spectroscopy&gt;&lt;spectra&gt;&lt;units&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19468" PostTypeId="2" ParentId="19439" CreationDate="2016-12-21T05:50:50.537" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How does the dense atmosphere of Venus affect the planet's rotation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know all the details, but I can give a partial answer, with links to further reading if interested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tidal locking, and orbits slowing down due to a tidal bulge and tidal forces and/or a denser or greater gravitational gradient on one side of the tidally locked object are all pretty straight forward, and I don't think that part needs to be explained too much.  The tidal bulge has mass and that mass creates drag on the rotation (or speeds it up if the bulge is behind the object causing the tides).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it's worth adding that rocky worlds are gravitationally lumpy, where the heavier side (gravitational gradient), tends to, over time and in tidal locking, face the object it orbits.   The near side of the Moon, for example, is slightly denser on average than the far side.  See &lt;a href=&quot;https://physics.stackexchange.com/questions/4116/why-does-the-moon-face-earth-with-the-same-side&quot;&gt;here&lt;/a&gt; for a better explanation and &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitation_of_the_Moon&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for gravitational map of the moon (the Earth side is on the left in the picture below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6Ygqp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6Ygqp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus' atmosphere, particularly it's upper atmosphere, behaves very differently.   There's still an atmospheric tidal bulge, but because of the heat from the sun, the hot gas on the sun side of Venus upper atmosphere has lower density, so the tidal bulge may have less mass, not more mass, even with greater size.   Atmosphere doesn't behave like a typical tidal bulge orbiting close to a star because the hot atmosphere is less dense and the night-side of Venus' upper atmosphere is more dense.  The (by mass) true tidal bulge is on the far side of Venus, away from the sun, not the near side.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another effect is that the heat from the sun causes something like a jet-stream on Venus, a very powerful jet stream.   The wind speed on Venus' upper atmosphere is faster than any hurricane on Earth and the upper atmosphere tends to rotate in the same direction around Venus.   This solar driven rotation (may be) a factor in Venus' surface having a slow retrograde rotation where, without an atmosphere it might be tidally locked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's difficult to imagine an upper atmosphere wind could &quot;spin&quot; a planet, and it would only do so very very slowly, (Venus' atmosphere is about 1/10,000th it's total mass, not too far off from the ratio of Earth's oceans to Earth's total mass) and the tidal slowing of a planet is very gradual, at Venus distance taking perhaps 100 million years to slow a rocky body of that size to tidal locking.     So the rotation may be a balance between Venus' solar driven jet stream and it's gradual tidal locking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least, that's what I came up with doing some reading on this.    From &lt;a href=&quot;https://en.wikipedia.org/wiki/Venus#Orbit_and_rotation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikpedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The rotation period of Venus may represent an equilibrium state&#xA;  between tidal locking to the Sun's gravitation, which tends to slow&#xA;  rotation, and an atmospheric tide created by solar heating of the&#xA;  thick Venusian atmosphere&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and link to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Venus#cite_note-102&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;, which is in French but English translation is available.   (see footnote 100 and 101)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Venus &quot;jet-stream&quot; like upper atmosphere, note that while the jet-stream on Earth is driven by surface temperature variation perhaps more than solar and by Earth's Coriolis effect, and it varies by seasons.   Venus doesn't have surface temperature variation to speak of and essentially no Coriolis effect due to it's slow rotation, so it's jet-stream tends to be much more along latitude lines and much less curved than Earth's.    That's not to say that there's no variation.   Venus' wind seems to operate on a 5 day cycle and it appears to have grown stronger in the years since we've been observing Venus.  Neither of those is well understood.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.universetoday.com/36816/winds-on-venus/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Article here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At the very top of the cloud layers on Venus, wind speeds reach 355&#xA;  km/hour (or 100 meters/second). This is the same the jet stream here&#xA;  on Earth. As you descend through the cloud layers, though, the wind&#xA;  speeds pick up. In the middle layer, the winds can reach speeds of&#xA;  more than 700 km/hour. That’s faster than the fastest tornado speed&#xA;  ever recorded on Earth.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But then as you descend further down through the clouds, the&#xA;  thickening atmosphere slows the winds down, so that they act more like&#xA;  currents in the ocean than winds in the atmosphere. Down at the&#xA;  surface, the winds only move at a few km/hour. That’s not much, but&#xA;  the thick atmosphere can still kick up dust and push around small&#xA;  rocks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The winds on Venus travel in a westerly direction, the same backwards&#xA;  direction that Venus rotates. Seen from above, Venus rotates in a&#xA;  clockwise direction. This is backwards from the other 7 planets, which&#xA;  rotate counter-clockwise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;another &lt;a href=&quot;http://www.universetoday.com/103001/venus-winds-are-mysteriously-speeding-up/&quot; rel=&quot;nofollow noreferrer&quot;&gt;article here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and, related, quote from &lt;a href=&quot;https://en.wikipedia.org/wiki/Venus#Orbit_and_rotation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; and source &lt;a href=&quot;http://www.esa.int/Our_Activities/Space_Science/Venus_Express/Could_Venus_be_shifting_gear&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Venus's rotation has slowed down by 6.5 min per Venusian sidereal day&#xA;  in the 16 years between the Magellan spacecraft and Venus Express&#xA;  visits&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, short summary, Venus atmosphere may play a key role in it's reverse rotation, or, maybe it's rotation was reversed by an impact a long time ago and it's still slowing down from that impact.   The impact theory might make the most sense, but impacts large enough to spin a planet are quite rare so, it shouldn't be assumed.   I don't believe it's known 100% whether Venus' atmosphere caused it's backwards rotation, but it seems reasonable and possible.   My personal favorite idea, though it may not be likely, is the gyroscope flipping hypothesis.  (Posting video cause it's really cool, but it has nothing to do with your question).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=1n-HMSCDYtM&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=1n-HMSCDYtM&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2016-12-21T05:58:57.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19469" PostTypeId="2" ParentId="19459" CreationDate="2016-12-21T06:43:22.220" Score="4" Body="&lt;p&gt;Answers above will explain seeing with naked eyes.&lt;br&gt;&#xA;But you can film light beam for sure&lt;br&gt;&#xA;See this amazing TED video of a light beam, filmed at trillion frames per second.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=Y_9vd4HWlVA&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=Y_9vd4HWlVA&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15264" LastActivityDate="2016-12-21T06:43:22.220" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19470" PostTypeId="2" ParentId="19467" CreationDate="2016-12-21T09:26:48.680" Score="3" Body="&lt;p&gt;Your logic seems entirely correct. Whatever you do is an approximation, the difference between a Gaussian profile in cube B and what it &quot;should&quot; look like will be small for any sensible resolution. By sensible, I mean if your wavelength/velocity axis covers many resolution elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is a &quot;correct&quot; way. Generate a Gaussian in wavelength space $f(\lambda)$ (if you think that's what your LSD is, where $\lambda$ here is short hand for $\lambda - \lambda_0$, the separation from the central wavelength) and then convert to velocity space in a flux-conserving manner: &#xA;$$f(\lambda)\ d\lambda = g(\log \lambda)\ d\log \lambda$$&#xA;$$ g(\log \lambda) = \lambda f(\lambda)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe a better way to do this is to get some data, like an arc calibration and see how unresolved lines look in your observational space?&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-12-21T09:26:48.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19471" PostTypeId="1" AcceptedAnswerId="19474" CreationDate="2016-12-21T16:33:36.940" Score="1" ViewCount="3142" Body="&lt;p&gt;From my home I can see a very bright star (or a planet ?), appearing before 5pm local time, when I look toward South. It's really much more brighter than any other star, it's very often the only one you can see because of ambient light in the neighbourhood. It's even visible when it's a little bit cloudy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Details:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;My Location: N 50° 43.640' E 005° 41.430' (East of Belgium)&#xA;Time : 5pm (4pm UTC), December 21st, 2016&#xA;Direction : Almost straight on South (190°)&#xA;Altitude : 35°&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that angles are rough estimates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what's that star ?&lt;/p&gt;&#xA;" OwnerUserId="15269" LastEditorUserId="15269" LastEditDate="2016-12-22T08:44:29.843" LastActivityDate="2016-12-22T08:44:29.843" Title="What's that very bright star (south-oriented, appearing at 5 p.m.)?" Tags="&lt;star&gt;&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19472" PostTypeId="2" ParentId="19471" CreationDate="2016-12-21T17:58:21.463" Score="3" Body="&lt;p&gt;You could check out Night Vision by Brian Simpson, its a Free application that you can download and install.  I use it.  You can then match what you see with what on screen and be able to identify (&lt;a href=&quot;http://www.nvastro.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.nvastro.com/&lt;/a&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess it might not be the answer you're looking for but if no one is able to answer, this could be the way to go.  There's also Microsoft WWT and Google Earth too.&lt;/p&gt;&#xA;" OwnerUserId="5401" LastActivityDate="2016-12-21T17:58:21.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19474" PostTypeId="2" ParentId="19471" CreationDate="2016-12-21T18:26:48.107" Score="6" Body="&lt;p&gt;That star is probably the planet Venus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is always close to the Sun, and &lt;a href=&quot;http://www.stellarium.org/&quot; rel=&quot;noreferrer&quot;&gt;Stellarium&lt;/a&gt; shows it to be in the south and well above the horizon at around 5pm local time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check, right above it should be Mars (also a planet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if you see a very bright star, it is very likely that it is a planet, and most likely to be Venus, Mars, or Jupiter. The latter two can appear at any time during the night, but Venus is always close to the Sun, so it can be seen just after sunset or right before sunrise. All planets appear close to the track of the Sun across the sky.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastEditorUserId="13681" LastEditDate="2016-12-21T18:39:35.590" LastActivityDate="2016-12-21T18:39:35.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19476" PostTypeId="1" CreationDate="2016-12-21T21:51:18.607" Score="8" ViewCount="454" Body="&lt;p&gt;I'm studying a particular case of the restricted three-body problem. It's been found that some objects follow a horseshoe orbit pattern, and I'm trying to sort out something through an integration code in C. I'm following some advice in the article &lt;a href=&quot;http://www.aanda.org/articles/aa/abs/2005/12/aa1483/aa1483.html&quot; rel=&quot;noreferrer&quot;&gt;Families of periodic horseshoe orbits in the restricted three-body problem&lt;/a&gt;, which gives me ideal initial conditions and the equations in the centre of mass system. (m is the mass of Earth, and consequent position of the sun in the center of mass reference system, (x,y) are the coordinates of the third body, assumed massless (as the restricted problem requires).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ O=(x^2+y^2)/2+\dfrac{(1-m)}{r1}+\dfrac{m}{r2}+\dfrac{(1-m)m}{2} $$&#xA;$$r1^2=(x-m)^2+y^2$$&lt;br&gt;&#xA;$$r2^2=(x-m+1)^2+y^2$$&#xA;$$ a(x)=\dfrac{\mathrm{d}{O}}{\mathrm{d}x} + 2v(y)$$&#xA;$$a(y)=\dfrac{\mathrm{d}{O}}{\mathrm{d}y}-2v(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The positions of the &quot;sun&quot; and &quot;earth&quot; are fixed to (m,0) and (m-1,0), in the same reference system. (rotating reference system, assuming the earth has a circular orbit.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From all this I've calculated the equations to describe the system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a(x)=x+\dfrac{(m-1)(x-m)}{((x-m)^2+y^2)^1.5}-\dfrac{2m(x-m+1)}{((x-m+1)^2+y^2)^1.5} + 2v(y)$$&#xA;$$a(y)=y-\dfrac{y(1-m)}{((x-m)^2+y^2)^1.5} -\dfrac{2ym}{((x-m+1)^2+y^2)^1.5} -2v(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've used the algorithm of Runge-Kutta 4 to integrate those equations.&#xA;(I know the code is quite mind-twisting, but I just can't use pointers and I use structs everywhere).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#include&amp;lt;stdio.h&amp;gt;&#xA;#include&amp;lt;stdlib.h&amp;gt;&#xA;#include&amp;lt;math.h&amp;gt;&#xA;&#xA;#define dt 0.0001&#xA;#define N 100&#xA;&#xA;typedef struct{&#xA;    long double x,y;&#xA;}vec;&#xA;&#xA;typedef struct{&#xA;    vec k1,k2,k3,k4;&#xA;}runge;&#xA;&#xA;typedef struct{&#xA;    runge r,v;&#xA;}big;&#xA;&#xA;double dS,dE,m;&#xA;&#xA;double accx(double,double,double);&#xA;double accy(double,double,double);&#xA;void rad(vec);&#xA;big rungekutta(vec,vec);&#xA;vec moto(vec,runge);&#xA;double jacobi(vec);&#xA;&#xA;int main(){&#xA;  vec r,v;&#xA;    big f;&#xA;    double J,t;&#xA;    int i,Num;&#xA;    FILE* s1;&#xA;    s1=fopen(&quot;HorseShoe.dat&quot;,&quot;w&quot;);&#xA;&#xA;    Num=(int)N/dt;&#xA;    scanf(&quot;%Lf&quot;,&amp;amp;r.x);&#xA;    scanf(&quot;%Lf&quot;,&amp;amp;r.y);&#xA;    scanf(&quot;%Lf&quot;,&amp;amp;v.x);&#xA;    scanf(&quot;%Lf&quot;,&amp;amp;v.y);&#xA;    scanf(&quot;%lf&quot;,&amp;amp;m);&#xA;&#xA;    for(i=0;i&amp;lt;Num;i++){&#xA;        t=(i+1)*dt;&#xA;        rad(r);&#xA;        f=rungekutta(r,v);&#xA;        r=moto(r,f.r);&#xA;        v=moto(v,f.v);&#xA;        J=jacobi(r);&#xA;        fprintf(s1,&quot;%lf\t%Lf\t%Lf\t%Lf\t%Lf\t%lf\n&quot;,t,r.x,r.y,v.x,v.y,J);&#xA;    }&#xA;return 0;&#xA;}&#xA;&#xA;void rad(vec r){&#xA;    dS=pow(r.x-m,2)+pow(r.y,2);&#xA;    dE=pow(r.x-m+1,2)+pow(r.y,2);&#xA;}&#xA;&#xA;double jacobi(vec r){&#xA;    return pow(r.x,2)+pow(r.y,2)+2*(1-m)/dS+2*m/dE+m*(1-m);&#xA;}&#xA;&#xA;double accx(double x,double y,double v){&#xA;    return x-(x-m)*(1-m)/pow(pow(x-m,2)+pow(y,2),1.5)-m*(x-m+1)/pow(pow(x-m+1,2)+pow(y,2),1.5)+2*v;&#xA;}&#xA;&#xA;double accy(double x,double y,double v){&#xA;    return y-(1-m)*y/pow(pow(y,2)+pow(x-m,2),1.5)-m*y/pow(pow(y,2)+pow(x-m+1,2),1.5)-2*v;&#xA;}&#xA;&#xA;big rungekutta(vec r,vec v){&#xA;    big f;&#xA;    f.r.k1.x=v.x;&#xA;    f.r.k1.y=v.y;&#xA;    f.v.k1.x=accx(r.x,r.y,v.y);&#xA;    f.v.k1.y=accy(r.x,r.y,v.x);&#xA;    f.r.k2.x=v.x+f.v.k1.x*dt/2;&#xA;    f.r.k2.y=v.y+f.v.k1.y*dt/2;&#xA;    f.v.k2.x=accx(r.x+f.r.k1.x*dt/2,r.y+f.r.k1.y*dt/2,v.y+f.v.k1.y*dt/2);&#xA;    f.v.k2.y=accy(r.x+f.r.k1.x*dt/2,r.y+f.r.k1.y*dt/2,v.x+f.v.k1.x*dt/2);&#xA;    f.r.k3.x=v.x+f.v.k2.x*dt/2;&#xA;    f.r.k3.y=v.y+f.v.k2.y*dt/2;&#xA;    f.v.k3.x=accx(r.x+f.r.k2.x*dt/2,r.y+f.r.k2.y*dt/2,v.y+f.v.k2.y*dt/2);&#xA;    f.v.k3.y=accy(r.x+f.r.k2.x*dt/2,r.y+f.r.k2.y*dt/2,v.x+f.v.k2.x*dt/2);&#xA;    f.r.k4.x=v.x+f.v.k3.x*dt;&#xA;    f.r.k4.y=v.y+f.v.k3.y*dt;&#xA;    f.v.k4.x=accx(r.x+f.r.k3.x*dt,r.y+f.r.k3.y*dt,v.y+f.v.k3.y*dt);&#xA;    f.v.k4.y=accy(r.x+f.r.k3.x*dt,r.y+f.r.k3.y*dt,v.x+f.v.k3.x*dt);&#xA;    return f;&#xA;}&#xA;&#xA;vec moto(vec r,runge rk){&#xA;    r.x+=(rk.k1.x+2*rk.k2.x+2*rk.k3.x+rk.k4.x)*dt/6;&#xA;    r.y+=(rk.k1.y+2*rk.k2.y+2*rk.k3.y+rk.k4.y)*dt/6;&#xA;    return r;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Plotting the results I just get a spiral, while using the given inputs I should get a horseshoe orbit. I've tried many different inputs, (m=0.0001 and m=0.000003, the latter is in scale with the actual values of Earth and Sun masses (sun mass is 1-m)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just can't see what's wrong (probably everything :D), can anyone please help me?&lt;/p&gt;&#xA;" OwnerUserId="15272" LastEditorUserId="2365" LastEditDate="2016-12-23T14:46:55.977" LastActivityDate="2019-07-15T07:31:23.930" Title="Horseshoe orbits and integration in C" Tags="&lt;gravity&gt;&lt;n-body-simulations&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19477" PostTypeId="2" ParentId="669" CreationDate="2016-12-22T06:49:47.510" Score="3" Body="&lt;p&gt;The amorphous geometry of the Universe is currently being studied, and the large scale distribution of the galaxies is similar to a sponge. The measure in the middle of the image represents 1.5 billion light years. light travels in every direction, and at the time of the big bang, there was no light to travel anywhere, and early in the theory of the big bang, there were no 3D directions that we can conceive, no definition of straightness and edge, no distance in between anything in a know geometry, in 3D, 4D, 5D, 12D superstring theory. So to find the geometry you need, Mathematics can become 12D/28D and is confusing to us, the notion of center is different in 12/20 dimensions. The Big Bang high temperature predates atoms, light, subatomic particles, matter, gravity, it predates the existence of known geometry, its contents exceeds any geometrical or finite measure, the only focal point is time, so to measure it you need to invent many new dimensions and geometry models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2eHYr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2eHYr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;The number of voids in the sponge could be well over trillion times more numerous than the number of atoms in the ocean. There could be a Googolplex MPC's as a mote of the total. So where is the center of that? When will time end?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big bang was amorphous from our viewpoint, and in that sense you could say it is &quot;amassive&quot; It's cosmic, space and physical properties are incommensurate (it's a nice word to say unmeasurable/unrelated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you imagine that our view of the cosmic background radiation(13.8bn LY) has the diameter of an atom in the sea. The big bang perhaps also happened in another atom on the other side of the sea, so the geometry doesn't have a gradation of measurement that can be defined within observation. If the big universe has a different appearance a Googolplex cubed trillion light years away, you will have a hard time finding out about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An object without symmetry or measurement and without a boundary cannot have a center. It has a cubic googolplex measurement rather than a single center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are therefore asking a geometric question similar to &quot;where is the center on the surface of a sphere, and a hoop&quot;? &lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorDisplayName="user5434" LastEditDate="2017-05-06T22:18:14.830" LastActivityDate="2017-05-06T22:18:14.830" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19479" PostTypeId="1" AcceptedAnswerId="19481" CreationDate="2016-12-22T11:36:57.493" Score="5" ViewCount="164" Body="&lt;p&gt;Is normal matter always accompanied by dark matter and dark matter by normal matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was ever observed effect of gravity lensing without normal matter?&lt;/p&gt;&#xA;" OwnerUserId="15282" LastEditorUserId="15282" LastEditDate="2016-12-22T11:49:30.367" LastActivityDate="2016-12-22T13:12:16.407" Title="Is normal matter always accompanied by dark matter and vice versa?" Tags="&lt;dark-matter&gt;&lt;matter&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19480" PostTypeId="2" ParentId="19479" CreationDate="2016-12-22T12:31:34.283" Score="-2" Body="&lt;p&gt;Mass attracts mass, and so dark matter will attract normal matter and vice versa. However, people have looked for &quot;dark lenses&quot;, so far without success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://arxiv.org/abs/astro-ph/0307381&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/astro-ph/0307381&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://arxiv.org/abs/astro-ph/0203061&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/astro-ph/0203061&lt;/a&gt; (&quot;dark&quot; only in X-ray?)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://arxiv.org/abs/astro-ph/9804136&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/astro-ph/9804136&lt;/a&gt; (mostly &quot;not a lens&quot; rather than a dark one)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://arxiv.org/abs/astro-ph/0202360&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/astro-ph/0202360&lt;/a&gt; discusses what a dark lens should look like&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;On the other side of the question, normal matter has a chance to cool and thus clump together more densely than is possible for dark matter. Hence, small-scale dense environments are dominated by normal matter.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastActivityDate="2016-12-22T12:31:34.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19481" PostTypeId="2" ParentId="19479" CreationDate="2016-12-22T13:12:16.407" Score="14" Body="&lt;p&gt;We lack the precision to say that there aren't regions where there is matter without dark matter or vice-versa. But what &lt;em&gt;is&lt;/em&gt; clear is that the ratio of dark matter to normal matter, which is (or needs to be) around 5 on average to explain the flatness of the universe, varies by orders of magnitude from place to place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this is that matter interacts with matter in a completely different way to that in which dark matter interacts (weakly) with normal matter or itself by any other means than gravity. Gravitational interactions between normal/dark matter are not dissipative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By that I mean that the sum of gravitational and kinetic energy is preserved. When normal matter interacts with itself there are normally dissipative consequences. For example, interacting matter may become hot and radiate away energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is why the normal matter in our Milky Way galaxy is concentrated into a plane. Dissipative interactions during the gravitational collapse of our Galaxy, combined with the requirement for conservation of angular momentum result in disc formation. The same is &lt;em&gt;not&lt;/em&gt; true for the dark matter associated with the Milky Way, which is expected to have a much more spherical distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, when we look at the required density profiles of normal and dark matter in order to explain the kinematics of stars and gas in our Milky Way, we see that the normal matter is much more centrally concentrated than is the dark matter - hence we talk about a &quot;dark halo&quot;. Again, dissipative interactions in normal matter are the reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another famous example of the separation between dark and normal matter is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Bullet_Cluster&quot; rel=&quot;nofollow noreferrer&quot;&gt;bullet cluster&lt;/a&gt;. The image below shows where the hot gas is (normal matter, observed by an X-ray telescope) compared to where it is inferred that the dark matter is (shown by contours) on the basis of gravitational lensing. This pair of galaxy clusters have recently interacted, passing through each other. The hot, intracluster gas has been stripped out of the pair and shock-heated, so that it appears concentrated &lt;em&gt;between&lt;/em&gt; the two galaxy clusters. The dark matter content of each cluster has sailed on oblivious and remains concentrated around the visible cluster galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XBVki.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XBVki.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another example would be &lt;a href=&quot;https://en.wikipedia.org/wiki/Dwarf_spheroidal_galaxy&quot; rel=&quot;nofollow noreferrer&quot;&gt;dwarf spheroidal galaxies&lt;/a&gt;, these are relatively small galaxies that are nevertheless almost &lt;a href=&quot;https://arxiv.org/abs/0709.1510&quot; rel=&quot;nofollow noreferrer&quot;&gt;totally dominated by dark matter&lt;/a&gt; - they have mass-to-light ratios of $\sim$1000, compared with a &quot;normal&quot; galaxy, where this ratio might be $\sim$ 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also examples of &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_galaxy&quot; rel=&quot;nofollow noreferrer&quot;&gt;dark galaxies&lt;/a&gt;. These may not be galaxies made of non-baryonic dark matter, but dark matter in the sense that for whatever reason, these galaxies do not contain many bright stars. Thus although they may be massive, they do not emit much light and contain a lot of &quot;inert&quot;, non-starforming gas.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-12-22T13:12:16.407" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19482" PostTypeId="1" CreationDate="2016-12-22T14:01:11.173" Score="3" ViewCount="437" Body="&lt;p&gt;Solar radiation is strong. Following problems for &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar telescopes&lt;/a&gt; may be telescope site selection and temperature control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder the telescopes themselves and their instruments are different? &lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditDate="2016-12-22T14:09:18.497" LastActivityDate="2018-07-20T05:33:07.013" Title="What's the difference between optical telescopes and solar telescopes?" Tags="&lt;the-sun&gt;&lt;telescope&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19483" PostTypeId="1" CreationDate="2016-12-22T14:08:35.043" Score="1" ViewCount="258" Body="&lt;p&gt;Fast-moving objects near our earth have different paths in the sky. They are different with stars far away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For telescopes, when we do not know an object's orbit, catching up with its orbit and not losing it is a problem ? How to figure out a new object's orbit parameters quickly before we lose it? For an old object with an irregular orbit, can a telescope track it?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The typical speed in the asteroid belt may be smaller than 1&quot;/min? But there may be faster objects. Then the paths are not so simple and regular.&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditDate="2016-12-23T12:20:48.333" LastActivityDate="2016-12-23T12:20:48.333" Title="How to track asteroids or NEOs?" Tags="&lt;telescope&gt;&lt;asteroids&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19484" PostTypeId="2" ParentId="19483" CreationDate="2016-12-22T14:44:07.537" Score="2" Body="&lt;p&gt;Specialty hardware will be able to do it, as might clever programming if you have a commercial, computerized mount and feel like hacking away. Either way, an ordinary mount will be able to follow targets in the asteroid belt easily -- their motion across our sky isn't actually very fast. What I'm about to write really only applies to NEOs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With NEOs, the problem isn't just the apparent speed of the objects across our sky, it's that the objects are usually in irregular orbits while moving at that speed. This is in addition to the apparent motion due to the rotation of the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To track these objects you'd have to project the position of the object onto our sky, using the object's ephemeris data as your starting point, then add the apparent motion due to the rotation of the earth. That position could be calculated by the mount or calculated by a computer tethered to it, which would send updated pointing directions to the mount. The specifics on how to do this depend on your hardware. It's not a small project.&lt;/p&gt;&#xA;" OwnerUserId="7154" LastEditorUserId="7154" LastEditDate="2016-12-22T14:58:56.860" LastActivityDate="2016-12-22T14:58:56.860" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="19485" PostTypeId="1" CreationDate="2016-12-22T15:02:40.707" Score="6" ViewCount="272" Body="&lt;p&gt;Is the ultimate reason that the neutron star loses angular momentum that the &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetic_moment&quot; rel=&quot;noreferrer&quot;&gt;magnetic-dipole radiation&lt;/a&gt; is not spherically symmetric and photons have momentum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for the loss of angular momentum, particles emitted from the neutron star surface are not necessary, right?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="2531" LastEditDate="2016-12-22T21:24:30.563" LastActivityDate="2016-12-22T21:24:30.563" Title="How does a neutron star emitting magnetic-dipole radiation lose angular momentum?" Tags="&lt;neutron-star&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19486" PostTypeId="2" ParentId="373" CreationDate="2016-12-22T15:22:57.380" Score="2" Body="&lt;p&gt;The orbit is around L2 at a vast distance from it, perhas  &gt; 10,000 km around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a map of the stability of the lagrangian points:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/whDsq.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/whDsq.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;L2 is unstable, like balancing a pencil on it's tip, but l4 and l5 have a force restoring a deprating object back onto L4.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/tHyoc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tHyoc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a lot les space debris at L2 that orbiting the earth, JWST has less chance of a hit than the space station. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://space.stackexchange.com/questions/6642/collision-with-space-junk-where-is-it-safe-where-is-it-most-dangerous&quot;&gt;https://space.stackexchange.com/questions/6642/collision-with-space-junk-where-is-it-safe-where-is-it-most-dangerous&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2016-12-22T15:22:57.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19490" PostTypeId="2" ParentId="19485" CreationDate="2016-12-22T21:20:01.187" Score="9" Body="&lt;p&gt;Radiation carries a momentum &lt;em&gt;per unit volume&lt;/em&gt; given by&#xA;$$ \vec{p} = \epsilon_0 \vec{E} \times \vec{B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It therefore carries an angular momentum of&#xA;$$ \vec{L} = \vec{r}\times \vec{p} = \epsilon_0 \vec{r} \times (\vec{E}\times \vec{B})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For plane-polarised transverse electromagnetic waves propagating in the $\hat{r}$ direction, then $\vec{L}=0$. However, if the axis of a magnetic dipole is inclined to the rotation axis then the radiation is not a single pure transverse wave; the dipole moment is the sum of two oscillating magnetic dipole moments that are perpendicular and oscillate $\pi/2$ out of phase. These produce two transverse waves, with arbitrary amplitudes, perpendicular polarisations and a $\pi/2$ phase shift between them - a.k.a. &lt;a href=&quot;https://en.wikipedia.org/wiki/Elliptical_polarization&quot; rel=&quot;noreferrer&quot;&gt;elliptically polarised light&lt;/a&gt;. Elliptically polarised light carries &lt;a href=&quot;https://en.wikipedia.org/wiki/Spin_angular_momentum_of_light&quot; rel=&quot;noreferrer&quot;&gt;angular momentum&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-12-22T21:20:01.187" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19491" PostTypeId="2" ParentId="18976" CreationDate="2016-12-23T03:28:52.183" Score="1" Body="&lt;p&gt;There is still some research to do to definitively answer this question. Having said that though, according to research published in the paper &quot;&lt;a href=&quot;http://home.dtm.ciw.edu/users/lrn/preprints/weider_epsl2015.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Evidence for geochemical terranes on Mercury: Global mapping&#xA;of major elements with MESSENGER’s X-Ray Spectrometer&lt;/a&gt; (Weider et al. 2015), the high Mg region (HMR) could be due to the consequences of an impact event - the authors from the article suggest that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;therefore, that the HMR may also be an impact related&#xA;  feature, and &lt;strong&gt;that its high Mg/Si ratio is evidence of&#xA;  mantle excavation during a large impact event&lt;/strong&gt; early in Mercury’s&#xA;  history (&gt;4.1 Ga).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and later, partly resurfaced by subsequent volcanism and smaller impact events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors elaborate further that the mechanism behind it could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If an ancient large impact at the HMR site penetrated into a&#xA;  vertically differentiated mantle (with additional lateral variations&#xA;  possible) and reached a deep, otherwise unsampled layer (i.e., the&#xA;  lherzolitic source proposed by Charlier et al. 2013*), the HMR composition&#xA;  would be the signature of high-degree partial melting of&#xA;  that mantle source.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Additional resource cited:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Charlier et al. 2013, &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0012821X12007078&quot; rel=&quot;nofollow noreferrer&quot;&gt;Phase equilibria of ultramafic compositions on Mercury and the origin of the compositional dichotomy&lt;/a&gt;, &lt;em&gt;Earth and Planetary Science Letters&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15217" LastActivityDate="2016-12-23T03:28:52.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19492" PostTypeId="1" AcceptedAnswerId="19493" CreationDate="2016-12-23T12:34:37.800" Score="1" ViewCount="301" Body="&lt;p&gt;I was reading up on compton radiation, and a question came up which I could not quite answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compton scattering is known to be less efficient at high energies, due to the fact that the Klein-Nishina cross-section falls off at the high energy end. This means that the Universe is more transparent to Compton radiation at high energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What other processes could stop high energy photons from propogating through the Universe (i.e., a process that preferentially stops photons at the high-energy end)?&lt;/p&gt;&#xA;" OwnerUserId="6965" LastActivityDate="2016-12-23T16:35:36.613" Title="Compton scattering of high energy radiation" Tags="&lt;radiation&gt;&lt;high-energy-astrophysics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19493" PostTypeId="2" ParentId="19492" CreationDate="2016-12-23T16:28:35.903" Score="3" Body="&lt;p&gt;The answer is &lt;a href=&quot;https://en.wikipedia.org/wiki/Pair_production&quot; rel=&quot;nofollow noreferrer&quot;&gt;pair production&lt;/a&gt;. Once photon energies exceed 1.02 MeV it is possible to spontaneously create an electron-positron pair in the presence of an atomic nucleus (to conserve momentum). In general for high energy photon interactions with matter you need to consider the photoelectric effect, Compton scattering and pair production. The former is more important at lower energies, the latter at higher energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture below shows the mass attenuation coefficients for Aluminium and Iron as a function of energy, with the contributions of these three processed identified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NFbG6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NFbG6.png&quot; alt=&quot;Mass attenuation coefficients&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-12-23T16:35:36.613" LastActivityDate="2016-12-23T16:35:36.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19494" PostTypeId="2" ParentId="19492" CreationDate="2016-12-23T16:29:34.290" Score="-2" Body="&lt;p&gt;The Compton cross section starts to get reduced significantly only when the photon energy is significant compared to the electron rest energy.  So at that energy scale, for a more isotropic radiation field, you would have to start worrying about some of the photons having enough energy to combine with another photon and create an electron-positron pair.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-12-23T16:29:34.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19498" PostTypeId="1" CreationDate="2016-12-24T06:23:00.340" Score="8" ViewCount="537" Body="&lt;p&gt;The physical nature of &lt;strong&gt;Fast Radio Bursts&lt;/strong&gt; (&lt;strong&gt;&lt;em&gt;FRBs&lt;/em&gt;&lt;/strong&gt;), a new type of &lt;strong&gt;&lt;em&gt;cosmological transients&lt;/em&gt;&lt;/strong&gt; discovered recently, is not known. It has been suggested that &lt;strong&gt;&lt;em&gt;FRBs&lt;/em&gt;&lt;/strong&gt; can be produced when a &lt;strong&gt;&lt;em&gt;spinning supra-massive neutron star loses centrifugal support&lt;/em&gt;&lt;/strong&gt; and &lt;strong&gt;&lt;em&gt;collapses to a black hole&lt;/em&gt;&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we suggest that such implosions can happen in supra-massive neutron stars shortly (hundreds to thousands of seconds) after their births, and an &lt;strong&gt;&lt;em&gt;observational signature of such implosions&lt;/em&gt;&lt;/strong&gt; may have been observed in the &lt;strong&gt;&lt;em&gt;X-ray afterglows&lt;/em&gt;&lt;/strong&gt; of some long and short &lt;strong&gt;Gamma-Ray Bursts&lt;/strong&gt; (&lt;strong&gt;&lt;em&gt;GRBs&lt;/em&gt;&lt;/strong&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under this picture, a small fraction of &lt;em&gt;FRBs&lt;/em&gt; would be physically connected to &lt;em&gt;GRBs&lt;/em&gt;. Is there any possible connection between &lt;strong&gt;&lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Gamma-ray_burst&quot; rel=&quot;noreferrer&quot;&gt;Gamma Ray Bursts&lt;/a&gt;&lt;/em&gt;&lt;/strong&gt; and &lt;strong&gt;&lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Fast_radio_burst&quot; rel=&quot;noreferrer&quot;&gt;Fast Radio Bursts&lt;/a&gt;&lt;/em&gt;&lt;/strong&gt; ??&lt;/p&gt;&#xA;" OwnerUserId="15219" LastEditorUserId="15219" LastEditDate="2016-12-24T19:01:16.603" LastActivityDate="2017-08-04T11:29:31.613" Title="Is there any possible connection between Gamma Ray Burst and Fast Radio Bursts?" Tags="&lt;gamma-ray-bursts&gt;&lt;fast-radio-bursts&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19501" PostTypeId="2" ParentId="19482" CreationDate="2016-12-24T14:38:05.360" Score="1" Body="&lt;p&gt;Solar telescopes are optical telescopes, and it is quite possible to use an optical telescope to observe the sun. The usual way is to use the telescope to project an image of the sun, by placing a card at the focus of the eyepiece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A professional telescope is a complex piece of engineering, it is designed for a particular task, such as observing very faint objects. It isn't designed for solar observation, and would most likely be rapidly damaged if it were pointed at the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solar telescopes have plenty of light to observe. The technical challenges involve controlling that light, so that it doesn't damage the equipment, and filter it, so particular aspects of the sun can be observed. The engineering problems of a solar and &quot;nocturnal&quot; telescope are different, so you can't use the same professional telescope for both. Analogy: an F1 race car and a rally car are similar in some ways but have been designed to overcome different technical challenges. An F1 car would never win a rally, nor would a rally car win at F1.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-12-24T14:38:05.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19502" PostTypeId="1" CreationDate="2016-12-24T18:17:08.510" Score="1" ViewCount="45" Body="&lt;p&gt;At about 4.15 am this morning a light travelling in a southerly direction passed over Adelaide South Australia. It was travelling 10 - 15 times the speed of an Airliner. The light was constant and moving in a straight line&lt;/p&gt;&#xA;" OwnerUserId="15293" LastActivityDate="2016-12-24T19:59:03.233" Title="Fast moving southward tracking light" Tags="&lt;light&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2016-12-26T21:58:59.960" ContentLicense="CC BY-SA 3.0" />
  <row Id="19506" PostTypeId="1" CreationDate="2016-12-25T02:01:12.320" Score="1" ViewCount="325" Body="&lt;p&gt;I have a star project, and I couldn't find anything on the amount of time it takes from nebula to neutron star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How long does it take, from a nebula, to form a neutron star?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="15297" LastActivityDate="2016-12-26T14:52:09.200" Title="What is the average time between a nebula formation and a neutron star?" Tags="&lt;neutron-star&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19507" PostTypeId="2" ParentId="19506" CreationDate="2016-12-25T03:56:23.273" Score="1" Body="&lt;p&gt;&lt;strong&gt;Approximately greater than or equal to $2\times10^7$ years.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approximate number comes from &lt;em&gt;Stellar Structure and Evolution&lt;/em&gt; by Dina Prialnik&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The contraction from nebula phase takes about $10^5$ years, and the red giant envelope dispersion takes about another $10^5$ years. Neither of which impact the timescale due to the two orders of magnitude longer MS lifetime.&lt;/p&gt;&#xA;" OwnerUserId="2232" LastActivityDate="2016-12-25T03:56:23.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19509" PostTypeId="2" ParentId="19506" CreationDate="2016-12-25T10:06:04.897" Score="2" Body="&lt;p&gt;Short answer: It depends on the progenitor mass, but somewhere between 5 million and 30 million years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I presume you are talking about the sum of the star formation timescale and the time it takes for the star to evolve through the various phases of its life and then explode as a supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As LaserYeti has pointed out, the formation timescale is extremely short compared to the rest of a massive star's life. In fact, for the massive stars that will eventually yield supernovae, there really isn't a pre-main sequence phase at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lifetime of the star then almost entirely depends on its mass, and to a lesser extent on its rotation and metallicity. The types of star that produce supernovae and neutron stars are probably in the range 9 to 40 solar-masses. Less massive progenitors produce white dwarfs, more massive progenitors probably produce black holes and there may not even be a supernova. Stars at 9 solar masses are about 30 times more commonly born than 40 solar-metallicity stars. To find their lifetimes we must appeal to theoretical models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ukads.nottingham.ac.uk/abs/1992A%26AS...96..269S&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schaller et al. (1992)&lt;/a&gt; provide a well-respected set of &quot;vanilla&quot; calculations for non-rotating, solar-metallicity stars. Their table 45, summing up the H-burning and He-burning stages (subsequent evolution happens comparatively instantaneously), suggests the evolutionary lifetimes range from about 30 million years for 9 solar-mass star, to as short as 5 million years for a 40 solar-mass star. These would be the numbers I would use, though rapid rotation can increase these lifetimes by perhaps 30%. Metallicity is a smaller effect.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2016-12-26T14:52:09.200" LastActivityDate="2016-12-26T14:52:09.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19510" PostTypeId="2" ParentId="19230" CreationDate="2016-12-25T10:40:14.437" Score="6" Body="&lt;p&gt;There are several compounds referred to as 'ices' in the solar system, these are mostly hydrogen-based compounds such as water, ammonia and methane (and others) accreted around what is believed to be a rocky core, these compounds are important, as they comprise a great proportion of the two 'Ice Giants' observed in our Solar System (Uranus and Neptune), as seen in the diagram below, compared to the Gas Giants - Jupiter and Saturn:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IwKCL.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IwKCL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Ice_giant&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer will just mainly focus on the Ice Giants. as discussion of these answer both your questions regarding the ice planets and what is sometimes described as ices existing at high temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of planets with proportionally large amounts of ice, as mentioned in the comments, there is a dividing line when the solar system formed known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Frost_line_(astrophysics)&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;Frost Line&lt;/strong&gt;&lt;/a&gt;, illustrated in the diagram below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/S91P8.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S91P8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;a href=&quot;http://lasp.colorado.edu/education/outerplanets/solsys_planets.php&quot; rel=&quot;noreferrer&quot;&gt;Image Source&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside the Frost Line, the hydrogen compounds remain vapourised, beyond the line, hydrogen compound ices can condense and solidify, as seen for Uranus and Neptune (and an array of dwarf planets, moons and other celestial objects), these 'ices' comprise the mantle of both planets. An important distinction about what these &quot;ices&quot; are is explained in the paper &quot;&lt;a href=&quot;https://www.ucolick.org/~jfortney/papers/FortneySSR09.pdf&quot; rel=&quot;noreferrer&quot;&gt;The Interior Structure, Composition, and Evolution of Giant Planets&lt;/a&gt;&quot; (Fortney and Nettelmann, 2009) as being:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The label ice refers to a mixture of $H_2O$, $CH_4$, and $NH_3$ that are supposed to have been in an ice phase during protoplanetary core formation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The compounds in the mantle in the Ice Giants are under immense pressure (in excess of 10 GPa), far in excess of those existing on Earth. Greater pressures change the properties of the hydrogen-based ices - so it is one of the reasons that it is sometimes referred to as 'hot ice', according to the AstroTech Gallery page '&lt;a href=&quot;http://cronodon.com/SpaceTech/IceGiant.html&quot; rel=&quot;noreferrer&quot;&gt;Ice Giant&lt;/a&gt;', describe a model of the Ice Giant planets as being:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Much of this mass (perhaps about 65%) is made up of hot ices, which despite being very hot are probably &#xA;  kept at least partially solid by the enormous pressures within these massive planets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The effect of pressure on water (as an example) can be seen in the example phase diagram below (of course there would be differences depending on the actual nature of the mixtures and compounds present):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DjBqq.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DjBqq.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image Source: &lt;a href=&quot;http://www.phy.duke.edu/~hsg/363/table-images/water-phase-diagram.html&quot; rel=&quot;noreferrer&quot;&gt;Duke University&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, according to the article &lt;a href=&quot;http://phys.org/news/2015-09-gas-ice-giant-neptune.html&quot; rel=&quot;noreferrer&quot;&gt;The gas (and ice) giant Neptune&lt;/a&gt;, the present icy (or 'hot ice') mantle of Neptune is described as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The mantle is equivalent to 10 – 15 Earth masses and is rich in water, ammonia and methane. This mixture is referred to as icy even though it is a hot, dense fluid, and is sometimes called a &quot;water-ammonia ocean&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A similar situation is hypothesised to exist for Uranus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;It is known that ice exist on Mercury, Earth, Moon and Mars, but these comprise a small proportion compared to the planet itself, so not included in this answer.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15217" LastEditorDisplayName="user15217" LastEditDate="2016-12-25T11:26:57.220" LastActivityDate="2016-12-25T11:26:57.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19511" PostTypeId="2" ParentId="12327" CreationDate="2016-12-25T13:35:24.247" Score="2" Body="&lt;p&gt;The reason you would want to cover most of your aperture is so you can point directly at a huge light source (i.e. a star), but ignore most of the light coming from it. This makes it easier to directly detect faint features around the source that would ordinarily be washed out by the light from the source itself (i.e. planets and the like). I believe this is what LocalFluff was getting at with &quot;advanced coronagraphy,&quot; and I don't believe it has anything to do with interferometery.&lt;/p&gt;&#xA;" OwnerUserId="15299" LastActivityDate="2016-12-25T13:35:24.247" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19514" PostTypeId="2" ParentId="19151" CreationDate="2016-12-26T13:49:33.290" Score="0" Body="&lt;p&gt;You are closer to the answer than you think with this question. That revolves around the connection between the SMBH central object and the mass of the Galaxy. For instance; all stars rotate around this thing, the Galactic Plane is about its 'equator'(if it ever associates to jetting - thats at right angles to the Galactic plane), a connection in mass and also the density of stars increases, across the galaxy, up to this object.&#xA;That connection is not really explained by merger, which is unlikely to result in stars orbiting v close to the centre. Matter is 'burping' or spewing from this object in outflow that forms stars locally and that is their origin. Just like in Dwarf Galaxies and Globular Clusters, which seem to have a correlation in general characteristics and morphology, show signs of star forming in the centre with, sometimes, large numbers of young stars in the middle.&lt;/p&gt;&#xA;" OwnerUserId="15083" LastActivityDate="2016-12-26T13:49:33.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19515" PostTypeId="1" CreationDate="2016-12-26T14:50:26.590" Score="2" ViewCount="156" Body="&lt;p&gt;According to the ScienceAlert article &lt;a href=&quot;http://www.sciencealert.com/mysterious-repeating-radio-signals-have-been-detected-coming-from-outside-our-galaxy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mysterious repeating radio signals have been detected coming from outside our galaxy&lt;/a&gt;, based on the Letter to Nature &quot;&lt;a href=&quot;http://www.nature.com/nature/journal/v531/n7593/full/nature17168.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;A repeating fast radio burst&lt;/a&gt;&quot;, there has been a series of signals detected that the article reports that scientists have not seen before.  From the article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Astronomers have detected repeating blasts of radio signals coming from deep space. These short-lived signals are known as 'fast radio bursts' (FRBs), and although we've heard them before, they were always thought to be one-off events coming from random locations. But for the first time ever, researchers have now heard repeating signals, &lt;strong&gt;all emanating from a single unknown source outside our galaxy&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Emphasis mine&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Letter to Nature, it is hypothesised that the source &lt;em&gt;could&lt;/em&gt; be that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;these repeat bursts with high dispersion measure and variable spectra specifically seen from the direction of FRB 121102 support an origin in a young, highly magnetized, extragalactic neutron star&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This leads me to the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How could a neutron star cause repeated 'fast radio bursts' detected coming from outside of the galaxy?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15217" LastActivityDate="2016-12-26T19:50:49.907" Title="How could a neutron star cause repeated 'fast radio bursts' detected coming from outside of the galaxy?" Tags="&lt;neutron-star&gt;&lt;fast-radio-bursts&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19516" PostTypeId="2" ParentId="19515" CreationDate="2016-12-26T19:50:49.907" Score="4" Body="&lt;p&gt;The progenitors of FRBs have been tied to &lt;a href=&quot;https://en.wikipedia.org/wiki/Soft_gamma_repeater&quot; rel=&quot;nofollow noreferrer&quot;&gt;soft gamma-ray repeaters (SGRs)&lt;/a&gt;, and are likely young magnetars displaying irregular atmospheric and magnetic activity. The earliest connection between the two - as far as I know - was made by &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/345493/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lyutikov (2002)&lt;/a&gt;, who compared the outbursts to solar flares (the magnetar bursts are sometimes referred to as &quot;giant pulses&quot;), based on models by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2002ApJ...574..332T&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thompson et al. (2002)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory is that currents dissipate into the neutron star's magnetosphere, at rates coupled to the star's crust. Slow deformation can allow dissipation to proceed at a constant rate, while &quot;fractures&quot; in the crust can lead to quick outbursts. The release of magnetic energy through &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetic_reconnection&quot; rel=&quot;nofollow noreferrer&quot;&gt;reconnection&lt;/a&gt; may then be responsible for the radio outbursts. The evolution of the field is similar to the behavior of the Sun's magnetic field in the corona, leading to the analogy to solar flares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/0710.2006&quot; rel=&quot;nofollow noreferrer&quot;&gt;Popov &amp;amp; Postner (2007)&lt;/a&gt; state that Lyutikov's model's predictions agree with several observed bursts. While &lt;a href=&quot;https://arxiv.org/abs/1603.00581&quot; rel=&quot;nofollow noreferrer&quot;&gt;the arXiv version&lt;/a&gt; of the Letter concerning FRB 121102  does not specifically delve into the model, it seems to support it, although there are doubts as to the feasibility of repeated giant flares.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2016-12-26T19:50:49.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19518" PostTypeId="1" CreationDate="2016-12-26T21:35:54.573" Score="9" ViewCount="820" Body="&lt;p&gt;In 1977, Uranus occulted a star, and the starlight flickered 40 minutes before and after the actual occultation, revealing Uranus' rings.&lt;br /&gt;This is common knowledge to astronomers, but what I can't find out is &lt;em&gt;which star&lt;/em&gt; Uranus occulted. Can someone give me the name of this star?&lt;/p&gt;&#xA;" OwnerUserId="15129" LastEditorDisplayName="user1569" LastEditDate="2016-12-27T13:53:52.123" LastActivityDate="2019-01-26T00:26:19.893" Title="Which star did Uranus occult in 1977 when the planet's rings were discovered?" Tags="&lt;history&gt;&lt;planetary-ring&gt;&lt;uranus&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19519" PostTypeId="2" ParentId="19518" CreationDate="2016-12-26T22:17:56.020" Score="13" Body="&lt;p&gt;From wikipedia, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Rings_of_Uranus&quot; rel=&quot;noreferrer&quot;&gt;rings of uranus&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The definitive discovery of the Uranian Rings [...] use[d] the occultation of the star SAO 158687[...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The star &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-basic?Ident=SAO%20158687&amp;amp;submit=SIMBAD%20search&quot; rel=&quot;noreferrer&quot;&gt;SAO 158687&lt;/a&gt;, also cataloged as &lt;a href=&quot;http://news.sky-map.org/starview?object_type=1&amp;amp;object_id=1405322&amp;amp;object_name=HD%20128598&amp;amp;locale=SV&quot; rel=&quot;noreferrer&quot;&gt;HD 128598&lt;/a&gt; is a magnitude 8.7, orange dwarf star in Libra. There seems to be nothing special about the star, except that one day, a planet passed in front of it.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-12-26T22:17:56.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19520" PostTypeId="1" AcceptedAnswerId="19960" CreationDate="2016-12-27T06:00:54.497" Score="0" ViewCount="131" Body="&lt;p&gt;In around late June, I saw two star-like bodies in the sky. At first they appeared to be twinkling, but when I looked closely they appeared to change color; blue, green, a slight red, then yellow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It freaked me out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm wondering, did I see dying stars or was it some sort of rare phenomenon occurring because of the earth's atmosphere or was it something else?&lt;/p&gt;&#xA;" OwnerUserId="15312" LastEditorUserId="2092" LastEditDate="2016-12-27T07:24:17.063" LastActivityDate="2017-03-09T06:18:32.817" Title="What object could I have seen at late June on the Sky?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="19525" PostTypeId="1" CreationDate="2016-12-27T11:25:15.620" Score="3" ViewCount="1326" Body="&lt;p&gt;For a data visualization project, I want to extract galaxy data (in x y z coordinates) from galaxy catalogue of within 150 mpc and project each of those galaxies as dots and visualize them in a 2D projection of a spherical volume. Need your help in getting this data set and how to achieve it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would be grateful for any help in any way!&lt;/p&gt;&#xA;" OwnerUserId="15314" LastEditorUserId="10196" LastEditDate="2016-12-27T19:30:27.040" LastActivityDate="2017-03-28T17:36:03.063" Title="How to project galaxy data in x y z coordinates?" Tags="&lt;galaxy&gt;&lt;coordinate&gt;&lt;space-geometry&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19526" PostTypeId="1" AcceptedAnswerId="19527" CreationDate="2016-12-27T11:59:14.563" Score="-5" ViewCount="240" Body="&lt;p&gt;can a life exist outside of universe. but we know that in our universe many galaxies have life  without water and oxygen and it is possible that life exist outside the universe so help me to find the right answer??and what is the possibility of this??&lt;/p&gt;&#xA;" OwnerUserId="15316" LastEditorUserId="15316" LastEditDate="2016-12-28T13:22:40.530" LastActivityDate="2016-12-28T13:22:40.530" Title="can a life exist outside the universe?" Tags="&lt;galaxy&gt;&lt;exoplanet&gt;&lt;cosmology&gt;&lt;life&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" ClosedDate="2016-12-27T16:47:58.500" ContentLicense="CC BY-SA 3.0" />
  <row Id="19527" PostTypeId="2" ParentId="19526" CreationDate="2016-12-27T14:38:06.377" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;can a life exist outside of universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's no way to define &quot;outside of universe&quot; I know of.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;we all know that life can not exist without water and oxygen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We don't know that.  It's an assumption, and not one I would make.  I would not even make it about Earth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;but we know that in our universe many galaxies have life without water and oxygen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We certainly do not know that.  It contradicts your original statement, but there's been no evidence found of life anywhere except on Earth ( to date ).  There are many theories saying life probably exists ( in some basic form, like cells ) outside of Earth, but they are simply theories at this point.  Science is not about excepting theories without evidence, it's about theories fitting evidence.  At this point we have no theories on life outside Earth that we have evidence to back.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and it is possible that life exist outside the universe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, we cannot define &quot;outside the universe&quot; so that has no real meaning.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;so help me to find the right answer??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Perhaps try the PBS Space Time channel on YouTube, as it discusses these things at a reachable level.  It recently discussed life ( advanced technological civilizations ) and why we cannot see any evidence of them in several videos.  Worth a look and they also make great introductions to advanced current thinking in space-time physics, including ideas like the multiverse, which often mislead people.  They usually supply links for people who want to get more serious about study as well as the video overview.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2016-12-27T14:38:06.377" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19528" PostTypeId="1" AcceptedAnswerId="19530" CreationDate="2016-12-27T17:35:30.050" Score="9" ViewCount="160" Body="&lt;p&gt;Wikipedia's section on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbit_of_the_Moon#Inclination&quot; rel=&quot;noreferrer&quot;&gt;Moon's inclination&lt;/a&gt; says it is always 1.5&amp;deg; to the ecliptic. Does that mean that as seen from its surface, over the year the sun at zenith would move 1.5&amp;deg; to the north, then to the south, varying by 3&amp;deg; overall?&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2016-12-27T22:23:36.293" Title="How does the sun move over the year in the sky of the Moon?" Tags="&lt;the-moon&gt;&lt;inclination&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19530" PostTypeId="2" ParentId="19528" CreationDate="2016-12-27T22:23:36.293" Score="10" Body="&lt;p&gt;That's right. Unlike the Earth, which has a substantial inclination of 23.4 degrees to the ecliptic, the Moon is inclined by only 1.5 degrees. This means that there are no substantial seasons. The moon also spins much more slowly than the Earth. Each &quot;day&quot; lasts for one Earth month. It also means that at the poles there are craters that are completely shaded from the sun. In these craters, there may be useful amounts of water ice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now as noted by the Wikipedia site, this is due to two motions being linked. The moon's orbit is inclined by 5.1 degrees to the ecliptic, and the moon's equator is inclined by 6.7 degrees to its orbit, however the direction of the pole moves to keep these two angles always in opposite directions to each other (There can be no pole star on the moon) The result is that the inclination of the moon to the ecliptic is fixed at 1.5 degrees.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-12-27T22:23:36.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19532" PostTypeId="2" ParentId="19525" CreationDate="2016-12-28T08:35:52.530" Score="3" Body="&lt;p&gt;You could try the &lt;a href=&quot;http://leda.univ-lyon1.fr&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hyperleda&lt;/a&gt; database. You will have to learn to use the database, which possibly means learning a little bit of SQL. For starters you could select all items with redshift $cz &amp;lt; H_0\times 150$ km/s, using redshift as a distance indicator. This should be roughly ok at 150 Mpc (though I don't see any viable alternative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The returned catalogue should have (at least) RA, Dec, redshift, possibly a size and morphology and, for a small subset of the galaxies, a redshift-independent distance estimate. Given that very nearby galaxies (in the local group or a $d &amp;lt; 50 $Mpc) are not fully part of the &quot;Hubble flow&quot; you might choose to use these in preference to a redshift-distance where available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X,Y,Z coordinates (based in the equatorial system) can be calculated using &lt;a href=&quot;http://www.astronexus.com/a-a/positions&quot; rel=&quot;nofollow noreferrer&quot;&gt;this prescription&lt;/a&gt; or in Galactic coordinates (with respect to the centre and orientation of our Galaxy can then be found using the operations defined in &lt;a href=&quot;http://adsabs.harvard.edu/full/1987AJ.....93..864J&quot; rel=&quot;nofollow noreferrer&quot;&gt;Johnson &amp;amp; Soderblom (1987)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea how complete this database will be. Certainly there will be an issue with extinction near the Galactic plane.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-03-28T17:36:03.063" LastActivityDate="2017-03-28T17:36:03.063" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19533" PostTypeId="1" AcceptedAnswerId="19535" CreationDate="2016-12-28T09:19:17.647" Score="2" ViewCount="122" Body="&lt;p&gt;Just a little something whilst a &lt;a href=&quot;http://edition.cnn.com/2016/12/27/us/vera-rubin-dark-matter-astronomy-obit-trnd/&quot; rel=&quot;nofollow noreferrer&quot;&gt;dark matter pioneer&lt;/a&gt; moves onto a better place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This commentator &lt;a href=&quot;https://news.ycombinator.com/item?id=12703911&quot; rel=&quot;nofollow noreferrer&quot;&gt;writes&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My understanding is that dark matter was &quot;discovered&quot; when trying to find out why stars in the outer edge of a galaxy rotates with the same speed as the inner. A ten-fold increase in number of galaxies doesn't seem to affect this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When reading &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_matter#History&quot; rel=&quot;nofollow noreferrer&quot;&gt;the history&lt;/a&gt; - this simple explanation doesn't quite come across. Perhaps the statement above is oversimplified. For example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;mass in the galactic plane must be greater than what was observed&lt;/em&gt; (doesn't quite talk about comparative rotational velocity)&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;estimated its mass based on the motions of galaxies near its edge&lt;/em&gt; (doesn't quite talk about comparative rotational velocity)&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;some unseen matter provided the mass and associated gravitation attraction to hold the cluster together&lt;/em&gt; (doesn't quite talk about comparative rotational velocity)&lt;/li&gt;&#xA;&lt;li&gt;measurements of galaxy rotation curves (what is a galaxy rotational curve? Shouldn't it be a straight line if they're rotating at the same angular velocity?)&lt;/li&gt;&#xA;&lt;li&gt;gravitational lensing of background objects by galaxy clusters (not talking about comparative rotational velocity?)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;strong&gt;Is it true dark matter was discovered whilst looking at the rotational speed of galaxies?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2894" LastEditorUserId="2894" LastEditDate="2016-12-28T10:58:24.547" LastActivityDate="2016-12-28T15:20:14.273" Title="Is it true dark matter was discovered whilst looking at the rotational speed of galaxies?" Tags="&lt;galaxy&gt;&lt;rotation&gt;&lt;dark-matter&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19534" PostTypeId="2" ParentId="10374" CreationDate="2016-12-28T12:39:40.173" Score="-2" Body="&lt;p&gt;The apparent density of the universe is inconsistent with available observations.  Because of the relatively slow speed of light, observations do not reflect actual distributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxies are now being observed in positions and size billions of years ago.  While our own Milky Way has a computational time dispersion over millions of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add to this that galaxies are observably moving apart from each other at greater than the speed of light, means that there are galaxies moving away from us at speeds greater than the speed of light and are likely not detected because of their weakly dispersed signature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The observable universe is likely skewed with sufficient variability in trustworthy evidence, that makes measuring universal density impractical, and untrustworthy.&lt;/p&gt;&#xA;" OwnerUserId="15322" LastActivityDate="2016-12-28T12:39:40.173" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19535" PostTypeId="2" ParentId="19533" CreationDate="2016-12-28T15:12:51.283" Score="1" Body="&lt;p&gt;What you're possibly missing here, judging from your questions, is the link between the mass distribution of a galaxy and the rotation velocities.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest way to obtain a prediction for a tangential velocity component is to look at the Kepler problem, i.e. a central mass, dominating another smaller mass, orbiting it.&lt;br&gt;&#xA;Newton's force-law $\vec F = m \dot{\vec{v}}$ relates acceleration and acting force, which is given by gravity.&lt;br&gt;&#xA;All acting forces then sum up to change the velocity of a test-particle $\dot{\vec{v}} = \sum_i \frac{\vec F_i}{m}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we get a number of cases, depending on how many summands there are, and how they are distributed spatially:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Kepler problem: Only one central $F_i$ exists, which dominates the system's behaviour. For a circular orbit one gets the well-known tangential velocity $v_t = \sqrt{\frac{GM}{r}}$, with $M$ being the mass of the central object and $r$ the distance from it.&lt;/li&gt;&#xA;&lt;li&gt;A Galaxy: In a galaxy the visible mass distribution is very different from a stellar system. It is spaced out more equidistantly and thus would give rise to a peak in $v_t$ and then a decrease again, after one gets far away from most of the mass. This is however not observed, and calculating the measured velocities of visible mass backward to the true mass distribution leads to the postulation of dark matter.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A related problem is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Virial_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;virialisation&lt;/a&gt; of galaxy clusters. When a bunch of mass particles interact with each other gravitationally for long enough, in this case galaxies, their averaged kinetic energy $&amp;lt;T&amp;gt; = &amp;lt;\frac{1}{2}mv^2&amp;gt;$ becomes half the available gravitational potential energy $&amp;lt;V&amp;gt;$. Thus, under the assumption of virial equilibrium one can measure the total mass in a system by measuring the velocity dispersion. Because galaxy clusters are very old objects, it is usually assumed they display this equilibrium. Departures from this are still ongoing research and for reasons I don't remember the milky way is usually not assumed to be virialised.&lt;br&gt;&#xA;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1933AcHPh...6..110Z&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zwicky 1933&lt;/a&gt; realised that the Coma cluster then needs to contain as much as 400 times the observed mass, to explain it's observed velocity dispersion.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditDate="2016-12-28T15:20:14.273" LastActivityDate="2016-12-28T15:20:14.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19536" PostTypeId="1" AcceptedAnswerId="19538" CreationDate="2016-12-28T16:54:00.013" Score="4" ViewCount="309" Body="&lt;p&gt;Dark Matter's observational evidence is &lt;a href=&quot;http://www.nytimes.com/2016/12/27/science/vera-rubin-astronomist-who-made-the-case-for-dark-matter-dies-at-88.html&quot; rel=&quot;noreferrer&quot;&gt;widely attributed&lt;/a&gt; to Vera Rubin based on her &lt;a href=&quot;http://adsabs.harvard.edu/doi/10.1086/150317&quot; rel=&quot;noreferrer&quot;&gt;observation&lt;/a&gt; of galaxy rotation curves with Kent Ford in 1970s, which was later &lt;a href=&quot;http://ned.ipac.caltech.edu/level5/March05/Bosma/frames.html&quot; rel=&quot;noreferrer&quot;&gt;confirmed&lt;/a&gt; by Albert Bosma in 1978. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it was in the late 1930s that Fritz Zwicky inferred the presence of dark matter based on his &lt;a href=&quot;http://spiff.rit.edu/classes/phys440/lectures/gal_clus/gal_clus.html&quot; rel=&quot;noreferrer&quot;&gt;observations&lt;/a&gt; of Coma galaxy cluster using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Virial_theorem&quot; rel=&quot;noreferrer&quot;&gt;virial theorem&lt;/a&gt;. Herein lies my confusion. Why does the scientific community and people, in general, give credits to Vera Rubin instead of Zwicky for inferring/observing the evidence for dark matter? Or, is the recognition given because she independently strengthened Zwicky's observation thereby confirming the existence of dark matter?&lt;/p&gt;&#xA;" OwnerUserId="15327" LastActivityDate="2016-12-28T18:00:34.750" Title="Who found the existence of Dark Matter?" Tags="&lt;dark-matter&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19537" PostTypeId="2" ParentId="10374" CreationDate="2016-12-28T17:39:40.010" Score="-2" Body="&lt;p&gt;There remains a question of semantics, which is simply, what is meant by the &quot;observable universe.&quot;  The fact is, different people are going to mean different things by that very phrase.   Indeed, the Wiki on the &quot;observable universe&quot; contradicts itself in its very first paragraph, stating first that &quot;The observable universe is a spherical region of the Universe comprising all matter that may be observed from Earth at the present time, because light and other signals from these objects have had time to reach Earth since the beginning of the cosmological expansion&quot;, but then four sentences later it changes its meaning, saying &quot;Every location in the Universe has its own observable universe, which may or may not overlap with the one centered on Earth.&quot;  So they seem to pick a particular universal age for their meaning, but not necessarily a particular vantage point.  But note that in this meaning, there is no way to answer how it changes with time, as it exists only at a single time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This meaning gives us various choices for how to extend the &quot;observable universe&quot; forward and backward in time.  For example, we could take all the stuff in today's &quot;observable universe&quot; from Earth, and ask where that stuff will be in the future, and where it was in the past.  Then we can use language like &quot;when the observable universe was the size of a grapefruit,&quot; etc., but notice the ambiguity:  when applied to the future, like &quot;what will the observable universe be at such-and-such a time,&quot; we invariably imagine updating what could be seen by the beings of the day, but when applied to the past, we generally don't imagine there are any beings at all, so we don't update what their observable universe would be, we take our own and just shrink it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So quite frankly, the term is really something of a mess, and so to answer your question, we would need to clarify which meaning you are taking.  Let's assume you mean the &quot;obserable universe&quot; that is constantly updating what hypothetical beings could observe had they existed on Earth at the time, then we do have a time-dependent mass.  As the Earth ages, there will be more time for light to reach us, so the observable universe will increase in size, but it will not necessarily increase in mass.  Assuming the acceleration continues as expected, the observable universe mass will increase by something like a factor of 2, reach a maximum, and then begin to decrease.  Its size will always be increasing with time, but its mass will then be decreasing.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-12-28T17:39:40.010" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19538" PostTypeId="2" ParentId="19536" CreationDate="2016-12-28T17:53:13.520" Score="3" Body="&lt;p&gt;You have a clear view of the history, and it wasn't an independent confirmation, it was simply the best evidence.  Whenever one talks about the discovery of dark matter, one should start with Zwicky, and move to Rubin.  One might do something similar with the Big Bang, starting with Hubble, and moving to Penzias and Wilson.  Normally we give credit both to the initial discovery, and to the confirmation, but it is the most conclusive evidence that generally gets associated the largest credit.  After all, it was Edgar Allan Poe who first described the Big Bang model, but he gets no astronomical credit because he provided no evidence.  Zwicky certainly did better than that, so does deserve (and should receive) much credit, but there were still many doubters in his day, much as with Hubble and the Big Bang.  Perhaps Zwicky's role contributed to the reason that Rubin never received the Nobel prize, I hope it wasn't simply the unspeakable reason, but of course it's too late now for both of them.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2016-12-28T18:00:34.750" LastActivityDate="2016-12-28T18:00:34.750" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19539" PostTypeId="1" CreationDate="2016-12-28T19:41:45.313" Score="4" ViewCount="154" Body="&lt;p&gt;I watched an interesting excerpt from a &lt;a href=&quot;https://www.youtube.com/watch?v=LpjbdH1y_ds&quot; rel=&quot;nofollow noreferrer&quot;&gt;BBC show on the work of the McDonald Observatory&lt;/a&gt; in its laser measurements of the distance to the moon which are apparently accurate to within a few centimeters. The observatory used to make a measurement every day and by this means a plot of the Moon's orbit around the earth was obtained over a period of approximately 40 years from about 1970 to about 2010.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of the excerpt Dr. Peter Shelus says that the measurements show that the shape of the moon's orbit is not quite as expected by &quot;Newton's&quot; theories, which I interpreted to mean that the orbit is non-Keplerian, but possibly he meant something else. It is hard to know because the excerpt just cuts off mid-sentence, so it is unclear what he was saying exactly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In what way is the orbit of the moon irregular according to the MacDonald measurement series?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as an aside, I notice that the &lt;a href=&quot;https://www.theguardian.com/technology/2009/jun/21/mcdonald-observatory-space-laser-funding&quot; rel=&quot;nofollow noreferrer&quot;&gt;observatory's funding was cut in 2009&lt;/a&gt; so that these measurements are no longer being made, but two other observatories are making new series. Are these other observatories making daily measurements the way the MacDonald Observatory used to do?&lt;/p&gt;&#xA;" OwnerUserId="2800" LastEditorUserId="2618" LastEditDate="2017-03-06T11:33:02.187" LastActivityDate="2017-03-06T11:33:02.187" Title="Non-Newtonian motion of the moon" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19540" PostTypeId="1" CreationDate="2016-12-28T20:51:44.953" Score="2" ViewCount="110" Body="&lt;p&gt;In the recently released movie &lt;a href=&quot;https://en.wikipedia.org/wiki/Passengers_(2016_film)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Passengers&lt;/a&gt; the ship hits a chunk of rock about the size of a softball that punches through multiple layers of hull and creates a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that a spaceship is traveling away from Earth at near light speed through interstellar space for a distance of 2 to 10 light years is there a significant probability that it would hit a rock that could do serious damage?&lt;/p&gt;&#xA;" OwnerUserId="2800" LastEditorUserId="13426" LastEditDate="2016-12-29T00:00:03.780" LastActivityDate="2016-12-29T00:00:03.780" Title="Do interstellar rocks pose a threat to interstellar travel?" Tags="&lt;interstellar-travel&gt;&lt;space-travel&gt;" AnswerCount="0" CommentCount="16" ClosedDate="2016-12-29T16:05:01.673" ContentLicense="CC BY-SA 3.0" />
  <row Id="19541" PostTypeId="1" AcceptedAnswerId="19543" CreationDate="2016-12-28T20:59:09.677" Score="2" ViewCount="689" Body="&lt;p&gt;I'm curious about the difference between low redshift and high redshift universe. Is there any defined limit of redshift beyond which we call things high redshifted?&lt;/p&gt;&#xA;" OwnerUserId="11679" LastActivityDate="2016-12-28T23:15:58.667" Title="What is high redshift?" Tags="&lt;universe&gt;&lt;redshift&gt;&lt;quasars&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19542" PostTypeId="2" ParentId="19539" CreationDate="2016-12-28T21:44:23.977" Score="2" Body="&lt;p&gt;The motion of the moon, as measured by the Lunar Laser Ranging data is not as predicted by Newton. As the scientist being interviewed says, Newton's model was correct to the accuracy of the data available to him, and was good enough to land the Eagle on the moon in '69. However the Lunar Laser Ranging data can position the moon to an accuracy of 3cm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relativistic effects cause the moon to deviate from the orbit predicted by Newtonian mechanics by about 1m, mostly due to special relativistic effects such as Lorentz contraction. General Relativity accounts for a variation of 10cm. This is much less than the perturbation due to Jupiter (about 1km). However, general relativity does fully account for the motion of the moon, to the accuracy of the experiment.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;All tests are in agreement with Einstein's General Relativity which is used for the numerical integration of ephemerides. No variation of the gravitational constant is discernible. &lt;a href=&quot;http://ilrs.gsfc.nasa.gov/docs/williams_lw13.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lunar ranging is ongoing at the &lt;a href=&quot;http://physics.ucsd.edu/~tmurphy/apollo/apollo.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;A&lt;/strong&gt;pache &lt;strong&gt;P&lt;/strong&gt;oint &lt;strong&gt;O&lt;/strong&gt;bservatory &lt;strong&gt;L&lt;/strong&gt;unar &lt;strong&gt;L&lt;/strong&gt;aser-ranging &lt;strong&gt;O&lt;/strong&gt;peration.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-12-28T21:44:23.977" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19543" PostTypeId="2" ParentId="19541" CreationDate="2016-12-28T23:15:58.667" Score="2" Body="&lt;p&gt;There definitely is no definition of what is low and what is high redshift — it is always seen in relation to something. An astronomer working in, say, the interstellar medium of the Milky Way ($z\sim0$) might talk about the ISM of a $z\sim2$ galaxy as being high redshift, whereas another astronomer working the reionization of the Universe (which took place around $z\sim10$) would consider $z\sim2$ to be low redshift. A cosmologist working in the CMB or inflation, however, would probably call $z\sim10$ low redshift. It all depends on the context. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless I think that &lt;em&gt;most&lt;/em&gt; people would consider $z\lesssim0.1$ to be low redshift. &lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2016-12-28T23:15:58.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19547" PostTypeId="1" AcceptedAnswerId="19549" CreationDate="2016-12-29T10:30:33.250" Score="1" ViewCount="93" Body="&lt;p&gt;Is the CMB that we see the same, unchanged energy from this moment? Does the radiation of the CMB exist as Photons and are we only witnessing the photons that have not been absorbed or destroyed? Will there be a time we can no longer witness the CMB because it has all been absorbed? How does red-shift factor in to this?&lt;/p&gt;&#xA;" OwnerUserId="13936" LastActivityDate="2016-12-29T13:13:09.417" Title="Is the CMB the photons that were created at the birth of the atom?" Tags="&lt;redshift&gt;&lt;cmb&gt;&lt;photons&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19548" PostTypeId="1" AcceptedAnswerId="19565" CreationDate="2016-12-29T11:16:44.277" Score="0" ViewCount="116" Body="&lt;p&gt;Does the distribution of mass or density of mass inside the event horizon matter for a black hole to exist as we currently theorize? Could the mass inside the event horizon be distributed in any way at all, or have any density so long as it is dense enough to locate 100% of the mass inside the Schwarzschild radius?&lt;/p&gt;&#xA;" OwnerUserId="13936" LastActivityDate="2016-12-30T21:47:46.020" Title="Does our current understanding of the mass of black hole only allow for it to be located at the singularity?" Tags="&lt;black-hole&gt;&lt;mass&gt;&lt;density&gt;&lt;singularity&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2016-12-30T22:56:53.423" ContentLicense="CC BY-SA 3.0" />
  <row Id="19549" PostTypeId="2" ParentId="19547" CreationDate="2016-12-29T13:13:09.417" Score="3" Body="&lt;p&gt;The CMB consists of photons that were emitted very shortly before and during the recombination of electrons with hydrogen and helium nuclei. [Side note - &lt;em&gt;why is&lt;/em&gt; it called the epoch of recombination when the atoms were never &quot;combined&quot; to begin with?]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those photons were then able to travel, unabsorbed across the universe until they encountered our microwave telescopes 13.7 billion years later. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the photons are unabsorbed, the expansion of the universe causes a redshift as they travel. Their current wavelengths are approximately 1100 times larger than when they were emitted and the universe is &lt;em&gt;almost&lt;/em&gt; completely transparent to photons of this wavelength - that is there is very little chance of a photon being &lt;em&gt;absorbed&lt;/em&gt; even if it travels across the diameter of the observable universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;em&gt;scattering&lt;/em&gt; processes that can deviate the photons or even increase their energy (and decrease their wavelength) via inverse Compton scattering (the &lt;a href=&quot;https://en.wikipedia.org/wiki/Sunyaev%E2%80%93Zel&amp;#39;dovich_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sunyaev-Zel'dovich effect&lt;/a&gt;), but this does not affect the CMB spectrum a lot and is difficult to observe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our current ideas about the future of the universe are that it will continue to expand at an accelerating rate. The CMB photons will become even more redshifted - to radio wavelengths and beyond - and the universe will become even more transparent to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/16508/future-redshifting-and-effect-on-the-pitch-of-cmb-radiation?rq=1&quot;&gt;Future redshifting and effect on the &amp;#39;pitch&amp;#39; of CMB radiation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-12-29T13:13:09.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19550" PostTypeId="2" ParentId="19390" CreationDate="2016-12-29T13:24:18.797" Score="3" Body="&lt;p&gt;Your vantage point changes throughout the year as the Earth moves around the solar-system barycentre (close to the centre of the Sun). That means the positions of nearby stars &quot;wiggle&quot; with respect to more distant background stars, with a period of one year; the amplitude of this wiggle is how you determine the &lt;a href=&quot;https://en.wikipedia.org/wiki/Parallax#Parallax_in_astronomy&quot; rel=&quot;nofollow noreferrer&quot;&gt;parallax distance&lt;/a&gt; to the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of Sirius AB, this wiggle is superimposed on a motion caused by the binary orbit of the two components in the Sirius AB system and by the motion of the Sirius system with respect to the Sun projected onto the plane of the sky - the &lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/P/Proper+motion&quot; rel=&quot;nofollow noreferrer&quot;&gt;proper motion&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-12-29T13:24:18.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19551" PostTypeId="2" ParentId="13131" CreationDate="2016-12-29T13:32:24.697" Score="2" Body="&lt;p&gt;Roughly: the mass loss rate for a spherically symmetric wind is given by&#xA;$$\dot{M} = 4\pi r^2 \rho v_w,$$&#xA;where $\rho$ is the density in the wind, $r$ is the radius at which that density is measured and $v_w$ is the wind velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you have $v_w$ and have the mass of material within two concentric radii - so you can work out the average density at some average radius in the shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The age of the shell is not relevant to the mass-loss &lt;em&gt;rate&lt;/em&gt;, but assuming the wind parameters are unchanged over that time, they do tell you ho much mass has been lost in total $\simeq \dot{M} \tau$.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-12-29T13:32:24.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19552" PostTypeId="2" ParentId="13131" CreationDate="2016-12-29T18:12:07.113" Score="2" Body="&lt;p&gt;There is a lot of additional information you would need to get a mass-loss rate.  What you have is a detection of some dust mass, and a time interval over which that mass was lost, so you could get something that looks like a mass-loss rate by taking the mass and dividing by the time over which it was lost, so 0.1 M_sun divided by 26,000 years.  That's already a mass-loss rate of .0004 M_sun/year, which is quite high, but there are several problems that could lead to this being either an overestimate or an underestimate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) You presumably want an average mass-loss rate over very long time intervals, not just 26,000 years.  So you need to know if this represents a steady process, or a kind of special event.  Stars related to Wolf-Rayet stars, called &quot;luminous blue variables,&quot; are often observed to lose mass in discrete episodes that happen an unpredictable intervals.  So is the dust surrounding your star a typical feature that should be expected to always be there, or was the star selected because it has such a feature?  If the latter, then the 26,000 year timescale is not a characteristic time for that object.  A related issue is that 60 km/s is a very slow speed for a Wolf-Rayet wind, so that also suggests you had some kind of mass release incident when the star was very puffed out, not a more steady kind of wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) An even worse problem is that all you know about is the mass in dust.  That is probably only a small fraction of the total mass lost, because depending on the evolutionary state of the Wolf-Rayet star, there could be a lot of hydrogen, and very likely a lot of helium, that do not participate in dust-making.  Also, the reason dust is not detected inside 10000 AU could be that the dust hasn't formed yet when it is so close to the ionizing radiation from the Wolf-Rayet star.  If so, you cannot even use the dust mass to characterize the mass of the material that does form dust, without also knowing the fraction that has made dust at each particular radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Also, if the outer radius of the feature is at 60000 AU, and the feature has taken 26,000 years to form, then the average speed of the outer edge of the feature is the ratio of those numbers, yielding 2.3 AU/year.  That translates into 11 km/s.  So if the dust is moving at 60 km/s, but the outer edge of the feature is only moving at 11 km/s, that means the wind is piling into material in the interstellar medium.  So it's not at all clear that the dust mass you are seeing is all from the Wolf-Rayet wind, some of it could be forming out of gas that was previously in the interstellar medium because something like 5/6 of the gas mass in that feature will have to have come from the interstellar medium.  The Wolf-Rayet wind will be enriched in dust-forming material, but if it's only 5 times more so enriched, there will still be dust coming from the ISM.  So that also depends on the evolutionary state of the Wolf-Rayet star.  (The fact that you see so much dust suggests that this might be a &quot;WC&quot; type star, which are very enriched in carbon, so this last issue might not be a concern.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it would take some more thinking and more research on the object, with additional quantitative constraints, to get a longer term mass-loss rate estimate.  But at least you have .0004 M_sun/yr as a kind of benchmark to begin that conversation.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-12-29T18:12:07.113" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19553" PostTypeId="1" CreationDate="2016-12-29T20:56:58.440" Score="2" ViewCount="53" Body="&lt;p&gt;When analyzing an image I'm used to thinking of the pixels as a grid of boxes the photons have fallen into. I assume that there's zero space between the boxes. That seems obviously wrong, since there needs to be some separation on the detector to prevent electron bleed between pixels. So, what's the real geometry of pixels in a detector like (especially for: CCDs and infrared arrays)? Is it closer to a grid of points than a grid of boxes?&lt;/p&gt;&#xA;" OwnerUserId="14244" LastActivityDate="2016-12-30T05:11:55.213" Title="Detector Pixel Geometry?" Tags="&lt;data-analysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19555" PostTypeId="2" ParentId="19553" CreationDate="2016-12-30T05:06:08.253" Score="1" Body="&lt;p&gt;Usually the separation exists, but is negligible compared to the size of the pixels.  Hence, the loss due to photons &quot;falling through the cracks&quot; is usually very small, and can be well accounted for using either Monte Carlo-based methods or experimentally using a focused source of known brightness. For example, attached is the user manual for the WIYN One-Degree Imager (ODI), a CCD detector that I have done some work on. There is a section on detector geometry that will probably help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wiyn.org/ODI/ODIUserManual.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.wiyn.org/ODI/ODIUserManual.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15299" LastEditorUserId="15299" LastEditDate="2016-12-30T05:11:55.213" LastActivityDate="2016-12-30T05:11:55.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19556" PostTypeId="2" ParentId="16215" CreationDate="2016-12-30T09:04:19.093" Score="1" Body="&lt;p&gt;&lt;em&gt;Converting answer in comment to community wiki&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlikely. Neptune is warmer than Uranus, more internal heat means more turbulent atmosphere. The reason why this is so isn't known, but it's possible that Uranus had a more turbulent past and it emitted heat from its formation more quickly. It's also possible that Neptune had a collision more recently that may have added to its internal heat. Triton is too small to have the effect you describe and the retrograde motion isn't a factor. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-12-30T09:04:19.093" CommentCount="0" CommunityOwnedDate="2016-12-30T09:04:19.093" ContentLicense="CC BY-SA 3.0" />
  <row Id="19557" PostTypeId="2" ParentId="15036" CreationDate="2016-12-30T09:43:17.867" Score="2" Body="&lt;p&gt;Most galaxies have enough dust to hide their cores from investigation in visible light. But the jet from an AGN is composed of large amounts of relativistic particles, and is easily powerful enough to clear any dust out of its path. We see in M87 a jet punch 5000 light years from an active nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dust that hides Sagittarius A* is not localised around the black hole, it is spread through the bulge, bar and disk. You would not want to be anywhere near an active galactic nucleus. Large amounts of radiation are produced by the accretion disc surrounding the black hole. Distance and dust may provide some protection to a star orbiting an active galaxy at the distance of the sun from Sag A*, but not if the star is directly in the jet.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-12-30T09:43:17.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19559" PostTypeId="1" CreationDate="2016-12-30T12:00:41.853" Score="1" ViewCount="146" Body="&lt;p&gt;This is an addition to this question which is closed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/1443/by-putting-a-mirror-in-space-would-we-be-able-to-see-into-the-past/8721#8721&quot;&gt;By putting a mirror in space, would we be able to see into the past?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine there is a supernova a hundred years ago. Lots of people see it but nobody has a decent telescope. A hundred years later we point a powerful telescope in the opposite direction, and find a huge alien mirror a hundred light years away, that allows us to watch the supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what is the limit of such a technique?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could we see right back to the big bang if the mirror was large enough and far enough away?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Please note this is not a question about whether the telescope or mirror are realistically possible.&#xA;Assume we are using some superior alien technology.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="15340" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2016-12-30T18:20:25.177" Title="(Thought experiment) if we put a huge mirror into space, could we see back in time to the big bang?" Tags="&lt;telescope&gt;&lt;light&gt;&lt;big-bang-theory&gt;&lt;mirror&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2016-12-31T00:21:41.100" ContentLicense="CC BY-SA 3.0" />
  <row Id="19560" PostTypeId="2" ParentId="19559" CreationDate="2016-12-30T12:55:44.793" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Imagine there is a supernova a hundred years ago. Lots of people see it but nobody has a decent telescope. A hundred years later we point a powerful telescope in the opposite direction, and find a huge alien mirror a hundred light years away, that allows us to watch the supernova.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Light travels at a fixed speed so a hundred years later when we build our telescope we would see nothing in the opposite direction, because the light reflected from your alien mirror would only have reached that mirror, and has another hundred years to reach our telescope.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My question is, what is the limit of such a technique ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's a thought experiment, not a technique.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Could we see right back to the big bang if the mirror was large enough and far enough away?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Light takes time to reach us, so when we look at very distant objects they're already telling us about the distant past.  This is how we have built up an understanding of what happened in the distant past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to explore the subject of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cosmic Microwave Background&lt;/a&gt; here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle it would be easier to detect any radiation from the big bang directly (without a mirror) in a similar way.  But the universe is not a quiet place and radiation from the big bang has literally been through everything and won't be as easily detected (if it's possible at all).  We don't even have a proper theory to know what to expect to be able to detect, if anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the mirror does is complicate things.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastEditorUserId="13681" LastEditDate="2016-12-30T18:20:25.177" LastActivityDate="2016-12-30T18:20:25.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19561" PostTypeId="1" CreationDate="2016-12-30T14:31:06.817" Score="0" ViewCount="607" Body="&lt;p&gt;My book says on the topic 'Formation of stars' that a galaxy starts forming by the accumulation of hydrogen gas in the form of very large clouds called nebulae. Then local lumps of gas are formed. These lumps start getting even denser and eventually form gaseous bodies called stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That seems a fair explanation but in the topic 'Our solar system' my book says &quot;The nebula from which our Solar system is supposed to have been formed, started its collapse and core formation some time 5-5.6 billion years ago and the planets were formed about 4.6 billion years ago.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is the nebula being talked about in the topic 'Formation of stars' same as the nebula being talked about in the topic 'Our solar system'?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question came to my mind because the nebula being talked about in 'formation of stars' topic might be very large, but when talking about our solar system it is eventually formed into the Sun, so how can they talk about the nebula while describing about our solar system, as the solar system is something within our galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any picture helping to visualize will be very nice. &lt;/p&gt;&#xA;" OwnerUserId="15313" LastEditorUserId="2365" LastEditDate="2017-01-28T14:12:57.233" LastActivityDate="2017-01-28T14:12:57.233" Title="Is the galaxy made of a nebula or the solar system?" Tags="&lt;solar-system&gt;&lt;galaxy&gt;&lt;nebula&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19562" PostTypeId="1" AcceptedAnswerId="19574" CreationDate="2016-12-30T16:07:15.633" Score="5" ViewCount="188" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Pulsar_kick&quot; rel=&quot;noreferrer&quot;&gt;Pulsar kicks&lt;/a&gt; originating from supernovae can impart neutron stars with speeds of 100-500 km/s, often at or close to the escape velocity of the Milky Way. Even if a pulsar fails to escape the galaxy, it can be launched into unusual orbits. Assuming that the velocity distribution is isotropic, a decent fraction of such pulsars should have large orbital inclinations after the kicks, such that they rise above the galactic plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two questions regarding this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is there a large - i.e. noticeable - population of such neutron stars, distinguishable by modern telescopes, or is the fraction of total neutron stars relatively small?&lt;/li&gt;&#xA;&lt;li&gt;Does observing a pulsar - at any wavelength - in such an orbit have any advantages over observing an identical pulsar firmly in the galactic disk? I would assume that some kicks could propel them high enough that scattering along the line of sight could be lessened, leading to a smaller dispersion measure smearing, and could also isolate the pulsars from ambient radio sources in the disk, such as a the galactic center.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I understand that &quot;noticeable&quot;, at the least, is a bit subjective, but I can't come up with any non-arbitrary criteria to change that.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2016-12-31T16:42:54.863" Title="Is there a significant population of neutron stars outside the galactic plane?" Tags="&lt;neutron-star&gt;&lt;observational-astronomy&gt;&lt;pulsar&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19563" PostTypeId="2" ParentId="19561" CreationDate="2016-12-30T17:32:00.260" Score="4" Body="&lt;p&gt;No, those are two different nebulae.  There are stars in our galaxy much older than 5.6 billion years.  The problem is that &quot;nebula&quot; is a very generic term, so it can be a whole galaxy or even the remains of a single star.  It's not even clear what is meant by the nebula from which our Sun formed, it could be the giant molecular cloud if our Sun formed in a huge cluster, or it could be something called a &quot;proplyd&quot; (which is much smaller) if the Sun formed in relative isolation.  There are even proplyds that break off from giant molecular clouds in a kind of combined scenario where the GMC forms the proplyd and then the proplyd forms the star.  In any event, it is important to realize that the galaxy formed from a huge nebula some 10 billion years ago, and the Sun formed from a much smaller nebula much later.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2016-12-30T17:32:00.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19565" PostTypeId="2" ParentId="19548" CreationDate="2016-12-30T21:37:17.720" Score="1" Body="&lt;p&gt;As a star or neutron star collapses to form a blackhole there will, of course, be a moment when the matter is distributed everywhere within the event horizon.  So, there may not be a singularity within a full fledged blackhole, at first.  But, the inward forces within the event horizon are irresistible (considering that escape velocity is greater than the speed of light) and so everything within the event horizon must fall into a singularity.  How long this takes is complicated by the issue of which reference frame you chose to time it.  &lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2016-12-30T21:47:46.020" LastActivityDate="2016-12-30T21:47:46.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19566" PostTypeId="1" AcceptedAnswerId="19567" CreationDate="2016-12-30T22:21:19.007" Score="2" ViewCount="2038" Body="&lt;p&gt;When looking into the sky at night using my bare eyes, I see that stars appear in different colors. From my understanding this is caused by different chemical compositions of those stars which show up in their spectral lines and therefore defining their color. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does blue and red shift visibly change the color perception when looked at the stars with your naked eye?&lt;/p&gt;&#xA;" OwnerUserId="11911" LastEditorUserId="2365" LastEditDate="2016-12-30T22:49:55.447" LastActivityDate="2016-12-31T08:25:17.653" Title="Are blue and red shift visible?" Tags="&lt;redshift&gt;&lt;spectroscopy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19567" PostTypeId="2" ParentId="19566" CreationDate="2016-12-30T22:49:22.167" Score="6" Body="&lt;p&gt;No, the blue and red shift of stars is not possible to detect with the naked eye. There are a couple of reasons for this. First the effect is slight: even for an object moving at thousands of km/s (way faster than stars in the milky way move), it is only a tiny change in hue. Next, since the light of stars is a mixture of many different colours (a black body spectrum) if the light from a star is red shifted, then ultra-violet is shifted down and takes the place of blue, the blue moves towards green and so on. The end result is a mix of all the colours, and so very little change in the perceived colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The colour of stars has relatively little to do with their composition, and a lot to do with their temperature. Cool stars (either small, or large, old and with puffed up cool outer layers) look red or orange. Hot stars (large and powerful stars) are white or blue-white. Stars have a spectrum that approximates a blackbody, and so the actual composition has relatively little effect on the colour. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The composition of the atmosphere of a star will absorb particular wavelengths of light, and this results in a dark line in the spectrum (not enough is absorbed to change the apparent colour). It is by measuring the position of these lines in the spectrum that red- and blueshift is observed.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-12-30T22:49:22.167" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19568" PostTypeId="1" CreationDate="2016-12-30T13:13:48.787" Score="3" ViewCount="733" Body="&lt;p&gt;Just got a Celestron Astromaster 114eq as a christmas present and I have tried setting up the Star Pointer in the daylight. I did not use the red dot method as in the instruction manual, instead, I tried to do this by viewing a object through the main scope (placing the object in the center of the main scope) and then I made some slight adjustments to the Star Pointer finder scope aligning both circle and dots on the finder scope using the two adjustment screws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to work fine and finds the targets quite well,but when it turns dark and I turn on the the red dot in the finder scope and aim at a star I have yet to manage to focus in on it. Should the method I used to set up the scope finder or do I need to set up the scope finder in the dark?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerDisplayName="bigal" LastEditorUserId="38" LastEditDate="2017-01-01T14:29:28.010" LastActivityDate="2017-01-02T05:29:05.110" Title="Celestron Astromaster 114eq Star pointer finder scope help required" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19570" PostTypeId="2" ParentId="370" CreationDate="2016-12-31T07:21:44.463" Score="1" Body="&lt;p&gt;Since gas giants have a low density, they normally weigh less than a rocky planet of the same size would. This also means that gravity is very low and it could be unable to clear it's neighbourhood. Thus it is not defined as a planet.&lt;/p&gt;&#xA;" OwnerUserId="15345" LastEditorUserId="1891" LastEditDate="2016-12-31T10:46:36.630" LastActivityDate="2016-12-31T10:46:36.630" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19571" PostTypeId="2" ParentId="19566" CreationDate="2016-12-31T08:25:17.653" Score="2" Body="&lt;p&gt;As James K says, it is not possible with the naked eye. But in fact, it almost is. The quasar &lt;a href=&quot;https://en.wikipedia.org/wiki/CTA-102&quot; rel=&quot;nofollow noreferrer&quot;&gt;CTA 102&lt;/a&gt; lies at a redshift of $z\simeq1$ and usually has an apparent magnitude of +17. Very recently (within the last few months) it had a bright outburst, increasing its magnitude to +11.5, i.e. more than two orders of magnitude in brightness! This is still not visible with the naked eye, but a colleague of mine just photographed it with a regular camera. A 10 sec exposure with no tracking makes it barely visible, but a 30 sec exposure, piggybacking on a telescope for tracking the sky, displays a beautiful reddish dot.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2016-12-31T08:25:17.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19572" PostTypeId="1" CreationDate="2016-12-31T13:32:39.050" Score="8" ViewCount="332" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble_sequence&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble sequence&lt;/a&gt; sorts and characterizes or classifies galaxies by their shape. It was suggested by Edwin Hubble back in 1926.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One drawback, or criticism, of this scheme is that Hubble had a limited data set of galaxies at hand, compared to today's data. He &quot;only&quot; had a few hundred galaxies in visible light from which he could build the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, adding today's knowledge of way more galaxies imaged in all different kinds of spectra how correct is Hubble's scheme? Is it skewed because of this originally limited selection of data? Are there other shapes missing in the sequence?&lt;/p&gt;&#xA;" OwnerUserId="7807" LastEditorUserId="-1" LastEditDate="2016-12-31T19:39:41.913" LastActivityDate="2016-12-31T19:39:41.913" Title="How accurate and complete is the Hubble sequence diagram?" Tags="&lt;galaxy&gt;&lt;classification&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19573" PostTypeId="2" ParentId="19572" CreationDate="2016-12-31T15:02:05.700" Score="2" Body="&lt;p&gt;The only way the Hubble sequence is truly incomplete is that it assumes such a classification applied to all stages of the universe. We now have telescopes that can see much further in space&amp;mdash;and thus in time&amp;mdash;and so we can see many more galaxies at different periods of time in the universe. Observations show that the Hubble sequence is still valid with one modification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &lt;a href=&quot;https://www.spacetelescope.org/images/heic1002a/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Hubble Space Telescope&lt;/a&gt; article explains, &lt;a href=&quot;https://en.wikipedia.org/wiki/Peculiar_galaxy&quot; rel=&quot;nofollow noreferrer&quot;&gt;peculiar galaxies&lt;/a&gt;&amp;mdash;so named because they exhibit odd features, including sizes and shapes&amp;mdash;appear to constitute over 50% of distant galaxies observed by &lt;em&gt;Hubble&lt;/em&gt;, while they make up only about 10% of local galaxies. Elliptical and lenticular galaxies made up roughly the same fractions (3-4% and 13-15%) in both periods, while spiral galaxies make up 72% of local galaxies and only 31% of distant galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hubble sequence classification still seems to organize galaxies well in this earlier period, with the glaring exception of peculiar galaxies. The conclusion the team has drawn is that many of the peculiar galaxies, one way or another, became spiral galaxies through collisions or other processes.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2016-12-31T19:39:19.757" LastActivityDate="2016-12-31T19:39:19.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19574" PostTypeId="2" ParentId="19562" CreationDate="2016-12-31T16:42:54.863" Score="3" Body="&lt;p&gt;I'll partly answer your question with the following plot from the &lt;a href=&quot;http://www.atnf.csiro.au/people/pulsar/psrcat/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ATNF pulsar database&lt;/a&gt; (which is easy to play around with). The plot shows the Galactic z coordinate of the pulsars (distance from the Galactic plane in kpc) versus their estimated age (from their spin down rates, in years).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic picture is that young pulsars are all near the Galactic plane, because they are born from high mass stars that are born, live and die near the plane and even if the neutron star got a kick, there has been insufficient time to move far from the plane. A typical kick might be 500-1000 km/s, roughly 0.5-1 kpc in a million years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once pulsars reach a million years old, then indeed some reach a kpc above/below the Galactic plane and then a few even older examples reach 10 kpc.&#xA;Most normal pulsars only show pulsar activity for 10-100 million years (the older ones on the plot are probably a different population of pulsars in binaries), so they aren't observed at much greater distances (though there will be old neutron stars out there).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WIuSt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WIuSt.png&quot; alt=&quot;STAND pulsars&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2016-12-31T16:42:54.863" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19575" PostTypeId="1" CreationDate="2016-12-31T17:13:33.913" Score="7" ViewCount="1139" Body="&lt;p&gt;There is a gravitational force between every pair of objects, because of which planets revolve around the Sun in elliptical orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this: As planets move in elliptical orbits with the Sun at a focus so do the stars in galaxies also orbit, and if then what are the foci of their orbits. Since the gravity of black holes is very high are black holes at any of the focus of a star's orbit?&lt;/p&gt;&#xA;" OwnerUserId="15347" LastEditorUserId="38" LastEditDate="2017-07-14T14:46:20.350" LastActivityDate="2020-07-09T08:58:20.177" Title="Do Stars in Galaxies also move in elliptical orbits?" Tags="&lt;orbit&gt;&lt;galaxy&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19577" PostTypeId="1" AcceptedAnswerId="19578" CreationDate="2016-12-31T18:19:28.433" Score="9" ViewCount="6607" Body="&lt;p&gt;It takes one year for the earth to go around the sun, but one year was already this long when they didn't even know that the earth revolves around the sun at all. So how did they figure it out?&lt;/p&gt;&#xA;" OwnerUserId="15348" LastEditorDisplayName="user2216" LastEditDate="2017-01-04T19:28:17.903" LastActivityDate="2017-01-04T19:28:17.903" Title="How did they decide the length of one year even though it wasn't known that earth revolves around the sun?" Tags="&lt;the-sun&gt;&lt;solar-system&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19578" PostTypeId="2" ParentId="19577" CreationDate="2016-12-31T18:40:07.933" Score="13" Body="&lt;p&gt;Originally a year would be noticed as the time between two winters (or two harvests, two annual floods etc), although the exact number of days might not be known with any accuracy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As people start to observe the sun more carefully, it would be noticed that the position of sunrise and sunset move regularly. Stonehenge and other ancient monuments seem aligned to midwinter sunset or midsummer sunrise. Since this would allow one to predict the seasons, it would have been very useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More careful observations led to the geocentric model, in which the sun takes just over 365 days to move around the Earth once relative to the background stars. The length of the year was known to be about 365.25 days since the times of the ancient Egyptians.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-12-31T18:40:07.933" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19579" PostTypeId="2" ParentId="19575" CreationDate="2016-12-31T18:58:14.930" Score="8" Body="&lt;p&gt;Planets orbit the sun in approximately elliptical orbits. The orbits of the planets are perturbed by each other, particularly by Jupiter. So the shape of orbits are very nearly, but not exactly elliptical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple elliptical orbit is found when there is a single dominating mass, for example in the solar system the Sun is much more massive than the rest of the solar system put together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars in galaxies do move. However, they don't move in simple elliptic orbits for the simple reason that the mass of a galaxy is not concetrated at a single point. There is a supermassive black hole at the centre of the galaxy, but although it is massive, its mass is dwarfed by the mass of the rest of the galaxy (it has a mass of about 4 million suns, the galaxy has a mass of 600000 million suns.) The stars move in orbits around the centre of mass of the galaxy, to which the black hole contributes only a very small amount. The shape of the orbit is a &quot;wavy circle&quot;. It is not elliptical, and so doesn't have foci. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2016-12-31T18:58:14.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19580" PostTypeId="1" AcceptedAnswerId="19584" CreationDate="2016-12-31T19:55:15.117" Score="8" ViewCount="372" Body="&lt;p&gt;Consider all of the naturally-occurring objects in the solar system : in the reference frame of the object's center (ie, ignoring the orbital velocity around the sun or, in the case of satellites, their planet) and considering an average point on their surface : &lt;strong&gt;which object has the fastest tangential velocity?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note - I am &lt;em&gt;not&lt;/em&gt; asking about the &lt;em&gt;rotational velocity&lt;/em&gt; ($\omega$), but the tangential velocity, which for a solid spherical body would be simply $\vec v=\vec \omega\times r $. Since the solar system is typically not made of up homogeneous solid spheres, I can't just look up the tables of rotational velocities and radii of the objects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking to be surprised, so everything goes : planets, their satellites and any of the other known objects (comets, asteroids, etc) - the only condition should be that this velocity should be more or less constant (I'm not sure if bodies in highly-elliptical orbits have severe changes in rotational velocity as they approach the sun).&lt;/p&gt;&#xA;" OwnerUserId="5506" LastActivityDate="2017-01-04T16:37:39.323" Title="Which celestial object in the solar system has the fastest tangential velocity at its surface?" Tags="&lt;solar-system&gt;&lt;rotation&gt;&lt;natural-satellites&gt;&lt;velocity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19581" PostTypeId="2" ParentId="19577" CreationDate="2016-12-31T20:05:28.740" Score="4" Body="&lt;p&gt;There are several ways to measure a year. The time it takes for Earth to complete one orbit around the sun is called a sidereal year or 365.256363 days, and though it is a valid definition of a year, it is not the one our calendars measure, which is a tropical year. A tropical year is defined &quot;as the period of time for the mean ecliptic longitude of the Sun to increase by 360 degrees,&quot; and is 365.24219 days in length. One way to think of the tropical year is the length of time it takes for the sun to complete its north-south journey through our sky.&lt;br&gt;&lt;br&gt;There is a decent discussion of the various definitions of a year &lt;a href=&quot;https://en.wikipedia.org/wiki/Year&quot; rel=&quot;nofollow noreferrer&quot;&gt;at Wikipedia.&lt;/a&gt; It's worth a read if you're interested.&lt;/p&gt;&#xA;" OwnerUserId="7108" LastEditorUserId="7108" LastEditDate="2017-01-01T00:54:52.337" LastActivityDate="2017-01-01T00:54:52.337" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19582" PostTypeId="2" ParentId="19575" CreationDate="2016-12-31T22:18:27.680" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Since the gravity of black holes is very high&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I feel there's a slight misconception floating around this expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes have a quite ordinary gravitational fields at stellar distances.  They're no different from any star in this sense.  Their field is governed by their mass, which is not particularly larger than stars for typical black holes ( the majority of which are the dead remains of an even larger star ).  This reputation for a very high field is only relevant when you're close to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black holes at the center of many galaxies are exceptionally large and that means they are proportionately stronger, but just being a black hole doesn't make an object suddenly gain a stronger field than it's mass would dictate.  In some ways it could be said to be safer - it's harder to collide with a black hole than a sun simply because it's smaller for the same mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black hole thought to be at the center of our galaxy is estimated to be about 41 light-seconds across which makes it about 9 times the diameter of our Sun.  But our Sun is a pretty mediocre star and there are many stars larger than that.  The largest we know of are estimated to be over the 1,500 times the Sun's radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as we start the New Year, can we take a vow to stop portraying harmless black holes as monsters of high gravity ? :-)&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2016-12-31T22:18:27.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19583" PostTypeId="2" ParentId="19539" CreationDate="2017-01-01T00:11:50.707" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;At the end of the excerpt Dr. Peter Shelus says that the measurements show that the shape of the moon's orbit is not quite as expected by &quot;Newton's&quot; theories, which I interpreted to mean that the orbit is non-Keplerian, but possibly he meant something else.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's a bad assumption. He undoubtably meant something else. Even Kepler knew the Moon's orbit about the Earth was not Keplerian. Newton tried to explain this non-Keplerian behavior in terms of perturbations from the Sun. He did manage to explain some of the key lunar anomalies, but the math of his time was not quite up to the challenge. The full Newtonian explanation would have to wait a couple of hundred years, culminating in the work of Ernest William Brown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To develop his (classical) lunar theory, Brown had to account for gravitational perturbations from a number of different objects: The Sun, Venus, Jupiter (and to a lesser extent, the other planets), and also the non-spherical nature of the Earth's and the Moon's gravity fields. Even with that, Brown had to introduce a fudge factor to account for a 10 arc second fluctuation in the Moon's angular position. It turned out that the need for this fudge was the transfer of angular momentum from the Earth's rotation to the Earth's orbit. This slows down the Earth's rotation rate (making time based on the the Earth's rotation a less than stellar idea) and makes the Moon recede from the Earth. Brown eventually added these concepts to his lunar theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brown's student Wallace J. Eckert later extended Brown's lunar theory. Eckert was the first person to apply digital computers (as opposed to human computers) to the problem of predicting the Moon's orbit. He was also the person who developed the lunar ephemerides used by the Apollo program. Eckert's work was still classical as opposed to relativistic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The retroreflectors placed on the Moon by the U.S. and the Soviet Union during the 1960s and 1970s finally necessitated the use of general relativity to explain the Moon's orbit. The effects are extremely subtle, but they are there. In particular, those retroreflectors serve as one of the highest precision tests of a key precept of general relativity, the strong equivalence principle. The equivalence principle comes in a number of forms:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Galilean (or Newtonian) equivalence principle, which addresses the equivalence of inertial and gravitational mass, and hence the universality of free fall.&lt;/li&gt;&#xA;&lt;li&gt;The weak equivalence principle, which says the motion of test bodies with negligible self-gravity is independent of their properties (this essentially restates the Galilean equivalence principle in relativistic terms),&lt;/li&gt;&#xA;&lt;li&gt;The Einstein equivalence principle, which says that a state of rest in homogeneous gravitational field is physcially equivalent to a state of uniform acceleration in gravity-free space,&lt;/li&gt;&#xA;&lt;li&gt;The strong equivalence principle, which says that at any point of spacetime, physics is locally described by special relativity and is not affected by the presence of a gravitational field.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Given sufficiently accurate measurements, the Moon's weird composition forms an excellent basis for testing both the various forms of the equivalence principle. The Moon's far side has a much thicker crust than does its near side. This makes for a two kilometer offset between the Moon's center of mass and center of figure. The thick crust on the far side (mostly silicon, oxygen, and aluminum) the offset core on the near side (mostly iron and nickel) would subtly affect the Moon's orbit if weak form of the equivalence principle didn't apply. In fact, the Moon's orbit as measured by lunar laser ranging experiments provides an excellent basis for testing even the strong equivalence principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that Newtonian mechanics disagrees with all but the Galilean equivalence principle. Newtonian mechanics also fails to match the observed orbit of Mercury, and to a lesser extent, the observed orbit of the Moon. Viable alternatives to general relativity agree with the weak form of the equivalence principle, and most agree with the Einstein form of the equivalence principle. General relativity is consistent with the strong equivalence principle (as are a very small number of alternatives to general relativity).&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-01-01T00:11:50.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19584" PostTypeId="2" ParentId="19580" CreationDate="2017-01-01T01:48:45.377" Score="8" Body="&lt;p&gt;Hands down it's Jupiter.  It has both the largest radius and faster rotational velocity of the planets, and smaller objects will not be able to compete with that huge radius.  Jupiter's rotational velocity is approximately 13.4 km/s (with some variation due to wind currents, as it is a gas), which can be determined both from Doppler shifts of reflected sunlight, and also from watching the motion of equatorial cloud features.  Saturn is only 10.9 km/s, as it has a longer rotation period and a smaller radius.  None of the other planets can compete with Jupiter, and smaller objects would need prohibitively short rotation periods to have any chance.  For example, the Earth would need to have a day that was about half an hour long to compete with Jupiter.  Even a molecule that spins millions of times a second would need a size on the order of a cm to compete with Jupiter.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2017-01-04T16:37:39.323" LastActivityDate="2017-01-04T16:37:39.323" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19586" PostTypeId="1" AcceptedAnswerId="19587" CreationDate="2017-01-01T14:17:13.437" Score="5" ViewCount="345" Body="&lt;p&gt;I'm watching a BBC special on the formation of solar systems, and they're covering the topic of &quot;hot Jupiters&quot; that exist near their host stars. The prevailing theory, according to this 2015 special, is that the planets form farther out, then migrate in. This addresses the issue of their gas being driven off by the host star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't also possible that these planets formed shortly before the ignition of the host star? The mass that will form the star is already mostly collected, and could hold bodies in orbit, but the heat has yet to begin, and wouldn't drive off gas. &lt;/p&gt;&#xA;" OwnerUserId="11388" LastActivityDate="2017-01-01T21:36:12.807" Title="Formation of &quot;hot Jupiters&quot; before star ignition" Tags="&lt;planetary-formation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19587" PostTypeId="2" ParentId="19586" CreationDate="2017-01-01T14:59:58.443" Score="4" Body="&lt;p&gt;Hot Jupiters &lt;em&gt;can and do&lt;/em&gt; form before the hot star &quot;ignites&quot;, as you put it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conventional core-accretion theory, by which gas-giant planets form, requires them to accrete from a disk of gas and dust for about 10 million years. Coincidentally (or perhaps not a coincidence?), the empirically determined lifetimes of circumstellar gas and dust disks around young stars is also about 10 million years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, the pre-main-sequence lifetimes of newly-born stars (which I define here as the length of time it takes to contract from a large ball of gas down to a much more compact configuration, with a central temperature high enough to initiate hydrogen fusion), range from about 10 million years for stars a little more massive than the Sun to 100 million years or more for stars that are less massive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think, though, that the information that you are missing is that there really is no period of time where the &quot;heat has yet to begin&quot; whilst the protostar is surrounded by a gas/dust disk. A protostar is actually &lt;em&gt;more&lt;/em&gt; luminous when it is younger and becomes fainter as it contracts towards its main sequence configuration. The energy that powers this luminosity is derived from gravitational potential energy. The inner portions of the gas/dust disk are always too hot to form hot Jupiters &lt;em&gt;in situ&lt;/em&gt; and so all current models focus on how hot Jupiters can be &quot;moved&quot; from where they were formed to where they are found now.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="4155" LastEditDate="2017-01-01T21:36:12.807" LastActivityDate="2017-01-01T21:36:12.807" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19588" PostTypeId="1" CreationDate="2017-01-01T20:48:43.840" Score="1" ViewCount="346" Body="&lt;p&gt;Tonight high up in the night sky, it was 8.20 precisely, I saw a bright orange light floating slowly across the sky ,occasionally slowing as if looking for something and flickering, it was too high up and far away to see an exact shape, but it looked like a star, so was it a star, do we have bright orange stars&lt;/p&gt;&#xA;" OwnerUserId="15358" LastActivityDate="2017-01-01T23:07:48.437" Title="Orange star like ight" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2017-01-02T01:01:13.930" ContentLicense="CC BY-SA 3.0" />
  <row Id="19589" PostTypeId="2" ParentId="19588" CreationDate="2017-01-01T23:07:48.437" Score="2" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;I saw a bright orange light floating slowly across the sky ,occasionally slowing as if looking for something&lt;/p&gt;&#xA;&lt;p&gt;it looked like a star, so was it a star,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;For the sake of teaching some scientific reasoning skills, I'll point out that just because you think it looks like a star does not mean it is a star.&lt;/p&gt;&#xA;&lt;p&gt;In fact the motion of the object suggests it was not a star.  Stars do not move except as part of a general motion of the night sky dues to Earth's rotation.  What you describe is not that.&lt;/p&gt;&#xA;&lt;p&gt;When we try and work out what something is, we use all the evidence.&lt;/p&gt;&#xA;&lt;p&gt;What would it be like ?  Perhaps a helicopter with lights.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-01-01T23:07:48.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19590" PostTypeId="1" CreationDate="2017-01-02T01:52:27.193" Score="0" ViewCount="398" Body="&lt;p&gt;I have recently purchased an Orion AstroView 90mm Equatorial Refractor telescope, also this is my first time having a telescope. The telescope came with a 25mm and a 10mm eyepiece, I also bought a Celestron OMNI 2x Barlow lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just the other night I was looking through my telescope at the moon, and with all my eyepieces and Barlow lens the image of the moon comes out really clearly. But when I go to look at mars or venus (also in the western sky) the image of the two planets come out really blurry. The image of the planets through my telescope come out as a circle with a bunch of horizontal lines through the circle also a couple of black dots on the outer edge of the circle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that these two circles that I was looking at were planets, and I know that they were not stars. I am really confused about the whole thing of me being able to see the moon clearly but not see any of the planets clearly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please help me figure this out. I have also tried looking through the telescope without the star diagonal, to be honest I have tried all possibilities with my lenses and all of my parts. &lt;/p&gt;&#xA;" OwnerUserId="15362" LastEditorUserId="2365" LastEditDate="2017-01-02T08:57:50.510" LastActivityDate="2018-03-12T13:22:37.467" Title="Planets not visible" Tags="&lt;telescope&gt;&lt;observational-astronomy&gt;" AnswerCount="0" CommentCount="11" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19592" PostTypeId="1" AcceptedAnswerId="19594" CreationDate="2017-01-02T05:14:17.197" Score="9" ViewCount="1158" Body="&lt;p&gt;Does the earth's orbital plane change with time relative to the distant stars? Does the Sun follow the same path through the zodiac (ignoring proper motions of the zodiacal stars)?&lt;/p&gt;&#xA;" OwnerUserId="14723" LastActivityDate="2020-03-29T21:15:38.023" Title="How does the earth's plane of orbit (ecliptic) vary over time?" Tags="&lt;the-sun&gt;&lt;ecliptic&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19593" PostTypeId="2" ParentId="19568" CreationDate="2017-01-02T05:29:05.110" Score="1" Body="&lt;p&gt;For an astronomical object, a 1000 mm focal length objective forms an image 1000 mm away. However, the image of a terrestrial object focuses farther out: 1010 mm for a target 100 m away, 1020 mm for a target 50 m away. When switching from a terrestrial target to an astronomical one, the eyepiece must move closer in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aligning the finder on a terrestrial target is bound to produce some parallax error. After focusing the telescope on a bright planet or the Moon, aim the telescope at Polaris and fine-tune the finder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, even with good finder alignment, sometimes you will have to sweep to locate your target. It gets easier with practice.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2017-01-02T05:29:05.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19594" PostTypeId="2" ParentId="19592" CreationDate="2017-01-02T08:55:02.560" Score="7" Body="&lt;p&gt;The position of the ecliptic does change, due to perturbation of other planets, but only very slowly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The invariant plane is the plane of angular momentum of the whole solar system. This is dominated by the orbital momentum of the planets, particularly Jupiter, and since other stars are much too far away to perturb the solar system, it is invariant with respect to distant stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ecliptic does vary. It is currently decreasing by about 1.4 degrees per 10000 years, as indicated by &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Ecliptic_inclination_dziobek.PNG&quot; rel=&quot;noreferrer&quot;&gt;this graph, (from Wikipedia)&lt;/a&gt;. This variation is one hundredth of the variation in the Earth's rotational axis.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-01-02T08:55:02.560" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19595" PostTypeId="2" ParentId="18662" CreationDate="2017-01-02T10:35:03.110" Score="1" Body="&lt;p&gt;Magnetic midnight occurs when the sun, the observer and the magnetic poles are in the same plane. This occurs around local midnight: In the USA it is about 1 hour before local midnight, in the UK it is about 2 hours before local midnight (local time is the time as shown on a sundial, with no adjustments for daylight saving etc)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aurora season is the winter, mostly because at the high latitudes where aurora are visible, the sky is not fully dark between April and September.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aurora are most likely during solar maximum, which occurs every 11 years (with a good deal of variability). Solar activity remains high for about 4 years. There have been maxima in 2000 and 2014 (the last cycle has been long and weak) The next is due around 2023-2025.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-01-02T10:35:03.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19597" PostTypeId="2" ParentId="14213" CreationDate="2017-01-02T10:58:07.140" Score="3" Body="&lt;p&gt;Both are solar observatories, and so have different requirements from conventional telescopes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sloping sides may allow for smoother air flow around the building, as heating during the day causes air to rise, which affects the resolution of the image. Also the pyramid is structurally strong (useful when you have a building on top of a mountain) and attractive, echoing she shape of the mountains. The Nepali observatory uses the flat walls to mount photovoltaic cells.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-01-02T10:58:07.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19598" PostTypeId="2" ParentId="13672" CreationDate="2017-01-02T11:07:03.673" Score="1" Body="&lt;p&gt;Ionised gas will emit radiation at particular frequencies, Nebulae that generate their own light this way are called &quot;Emission nebulae&quot; and they have spectra that are characterised by a collection of bright lines. They contrast with reflection nebulae that are lit by bright stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars also have part of their spectrum that exceeds what would be expected from a blackbody due to emission from gas in their photosphere's &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-01-02T11:07:03.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19599" PostTypeId="1" AcceptedAnswerId="19601" CreationDate="2017-01-02T13:46:35.353" Score="3" ViewCount="131" Body="&lt;p&gt;Motivated by this question &lt;a href=&quot;https://worldbuilding.stackexchange.com/questions/66468/how-long-could-an-operational-virginia-class-sub-survive-in-the-jurassic-sea/66491#66491&quot;&gt;about a stranded submarine in the Jurassic&lt;/a&gt; I would like to know:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything remotely left of our current skyline if we go back or forward millions of years so that the stars rotating around the milky way covered a measurable distance ? I would expect that the proper motion of a star causes an irreversible drift out of the original position. Are there any stabilizing factors which may allow that some star configurations are still discernible ?&lt;/p&gt;&#xA;" OwnerUserId="7695" LastEditorUserId="-1" LastEditDate="2017-04-13T12:52:24.020" LastActivityDate="2017-01-05T13:30:55.197" Title="Sky in the very far future / past" Tags="&lt;star-formation&gt;&lt;celestial-mechanics&gt;&lt;night-sky&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19600" PostTypeId="2" ParentId="13672" CreationDate="2017-01-02T14:06:25.113" Score="3" Body="&lt;p&gt;The key thing about lines is that they have a higher opacity than the continuum, so they are generally formed in layers that are above where the continuum formed.  Thus there are two general ways to get emission lines (i.e., lines that are brighter than the continuum), one is to make the higher regions hotter than what is below, and the other is to make them physically larger.  The first means that at line wavelengths, the object looks brighter, and the second means that at line wavelengths, the object looks larger.  The first happens in what is called a &quot;chromosphere&quot; of the star (a shell that is hotter because of some additional heating going on above the photosphere of the star), and the second happens if the star has a wind or anything else that physically extends it in ways that are transparent in the continuum but can be seen in lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third way to get emission lines is if the continuum is not optically thick (i.e., the object is transparent in the continuum), as happens with various low-density nebulae, but your question seems to stipulate that there is a bright continuum there, so there is an opaque object emitting something akin to blackbody radiation.  &lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2017-01-02T14:06:25.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19601" PostTypeId="2" ParentId="19599" CreationDate="2017-01-02T14:27:07.587" Score="5" Body="&lt;p&gt;The sun still rises in the East and sets in the West. So you quickly identify the cardinal directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By observing the point around which the stars rotate each night you can find the altitude of the pole, which gives you your latitude. You can't find your absolute longitude, but with good time keeping you can find your longitude relative to your starting point. The planets are still in the same orbits. You might find that Saturn doesn't have rings (it's not certain how old the rings are) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Galaxies would still be in the same positions, The Milkyway, Andromeda and the Magellanic clouds are all visible. Many more with a moderate telescope, The Milkyway would look different, in detail, the LMC and SMC would be in different positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the stars in the background have all changed. There are no constellations that have not undergone radical changes. There would be no stars that you recognise: Stars are visible either because they are close (and so have a large proper motion) or powerful and short-lived. All the stars you know would either have moved, or not be born yet.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-01-02T14:27:07.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19602" PostTypeId="1" AcceptedAnswerId="19603" CreationDate="2017-01-02T15:50:07.260" Score="11" ViewCount="351" Body="&lt;p&gt;How do scientists know the positions of black holes?&lt;br&gt;&#xA;Observe the highlighted area in the image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/I1aq0.png&quot; alt=&quot;Observe the Image&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15347" LastEditorUserId="15347" LastEditDate="2017-02-07T16:09:37.130" LastActivityDate="2017-02-07T16:09:37.130" Title="How the position of this black hole known?" Tags="&lt;black-hole&gt;&lt;observational-astronomy&gt;&lt;positional-astronomy&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19603" PostTypeId="2" ParentId="19602" CreationDate="2017-01-02T17:24:10.523" Score="13" Body="&lt;p&gt;The article is referring to the black hole known as &lt;a href=&quot;https://en.wikipedia.org/wiki/V4641_Sagittarii&quot; rel=&quot;nofollow noreferrer&quot;&gt;V4641 Sagittarii&lt;/a&gt;. It was originally thought to be the closest known black hole to Earth, at 1,600 light-years (0.5 kiloparsecs). It's a binary system which emits X-rays. Original distance measurements were made based on observations during and following an X-ray outburst in 1999. Analysis of the jets emitted and of the inclination of the system led to initial estimates of $0.4$-$1.7\text{ kpc}$ (see &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/317255;jsessionid=5787B4D414742976CD20AAEBA53E7239.c1.iopscience.cld.iop.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hjellming et al. (2000)&lt;/a&gt;), with lower estimates of around $\sim0.5\text{ kpc}$ being favored - the distance that was then commonly quoted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later spectroscopic measurements in more detail appear to show these early observations to be incorrect, as &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0103045v1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Orosz et al. (2000)&lt;/a&gt; found. By observing the orbits of the two objects in the system, their masses were determined, which led to an estimate of the companion star's radius. The temperature of the star was determined by examining its spectrum, and its luminosity was computed from this and the radius. By using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Color_index&quot; rel=&quot;nofollow noreferrer&quot;&gt;color index&lt;/a&gt; and taking into account &lt;a href=&quot;https://en.wikipedia.org/wiki/Extinction_%28astronomy%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;interstellar extinction&lt;/a&gt;, the group was able to calculate the distance to the companion star, and thus to the black hole. They found $7.40\leq d\leq12.31\text{ kpc}$, values closer to the currently accepted distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An absorption study (&lt;a href=&quot;http://adsabs.harvard.edu/full/2001ESASP.459..479C&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chaty et al. (2001)&lt;/a&gt;) found values somewhere in the middle, with $3\leq d\leq8\text{ kpc}$. However, all groups agree that the actual distance to V4641 Sagittarii is much further than 1,600 light-years. The closest known black hole is now thought to be &lt;a href=&quot;https://en.wikipedia.org/wiki/A0620-00&quot; rel=&quot;nofollow noreferrer&quot;&gt;A0620-00&lt;/a&gt;, also known as V616 Monocerotis, at around 2,800 light-years (about $0.85\text{ kpc}$) from Earth.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;General black hole detection is discussed in the answers to &lt;a href=&quot;https://astronomy.stackexchange.com/q/24/2153&quot;&gt;How are black holes found?&lt;/a&gt;. The takeaway for this case - a &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_black_hole&quot; rel=&quot;nofollow noreferrer&quot;&gt;stellar-mass black hole&lt;/a&gt; in a binary system - is that effects from the companion star must be observed and measured. Fortunately, there are typically plenty of ways to choose from.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2017-02-02T16:31:57.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19605" PostTypeId="2" ParentId="19440" CreationDate="2017-01-03T01:49:37.420" Score="2" Body="&lt;p&gt;Research published in the &lt;em&gt;PLoS One&lt;/em&gt; paper &quot;&lt;a href=&quot;http://journals.plos.org/plosone/article?id=10.1371/journal.pone.0064624&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boron Enrichment in Martian Clay&lt;/a&gt;&quot; (Stephenson et al. 2013) suggest a volcanic origin for the boron, in the chemical analysis of a Martian meteorite (found in Antarctica), they explain:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A straightforward geochemical interpretation of our results is that boron, a relatively volatile and soluble element, was &lt;strong&gt;first concentrated in the fluid dregs of lava&lt;/strong&gt; (4–7 ppm boron has been detected in the late stage mesostasis of other &lt;a href=&quot;https://en.wikipedia.org/wiki/Nakhlite&quot; rel=&quot;nofollow noreferrer&quot;&gt;nakhlites&lt;/a&gt;) and &lt;strong&gt;then became further concentrated by any groundwater or hydrothermal fluids&lt;/strong&gt; that came into contact with the rock. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Further modelling reported in the paper &quot;&lt;a href=&quot;https://www.repository.cam.ac.uk/handle/1810/246646&quot; rel=&quot;nofollow noreferrer&quot;&gt;Partitioning of light lithophile elements during basalt eruptions on Earth and application to Martian shergottites&lt;/a&gt;&quot; (Edmonds, 2014), also point towards a volcanic origin. Their modelling is based on analyses of &lt;a href=&quot;https://en.wikipedia.org/wiki/Shergotty_meteorite&quot; rel=&quot;nofollow noreferrer&quot;&gt;shergottite&lt;/a&gt; pyroxenes (Martian meteorites), stating that observed elevated boron alongside other light lithophile elements or LLE are to do with the earlier mantle chemical processes, extrapolated from those observed on Earth:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The range in LLE and trace elements observed in shergottite pyroxenes are instead consistent with concurrent mixing and fractionation of heterogeneous melts from the mantle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, a volcanic origin is suggested in the Goldschmidt Conference Abstract &quot;&lt;a href=&quot;https://goldschmidtabstracts.info/2008/889.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boron isotopes in the nakhlites: Implications for crustal fluids on Mars&lt;/a&gt;&quot; (Spivak-Berndorf et al. 2008), also based on analyses of nakhlites:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The lack of isotopic fractionation between the igneous and&#xA;  secondary alteration phases in the nakhlites indicates that (1)&#xA;  the B in the fluid that deposited the &lt;a href=&quot;https://en.wikipedia.org/wiki/Iddingsite&quot; rel=&quot;nofollow noreferrer&quot;&gt;iddingsite&lt;/a&gt; was derived&#xA;  from the primary igneous minerals and (2) there was minimal&#xA;  B isotope fractionation between the fluid and the iddingsite&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user15217" LastActivityDate="2017-01-03T01:49:37.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19606" PostTypeId="2" ParentId="10354" CreationDate="2017-01-03T09:48:50.460" Score="1" Body="&lt;p&gt;&lt;em&gt;Convert answer in comment to community wiki&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would need to do statistical tests. i.e. Is the data you have (given the known noise properties) consistent with a featureless spectrum? If not, then it has features. Of course, the lower the signal-to-noise or the weaker the features might be, then the harder it would be to rule out a featureless continuum.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-01-03T09:48:50.460" CommentCount="0" CommunityOwnedDate="2017-01-03T09:48:50.460" ContentLicense="CC BY-SA 3.0" />
  <row Id="19607" PostTypeId="2" ParentId="10247" CreationDate="2017-01-03T11:38:52.237" Score="3" Body="&lt;p&gt;The mass of Vesta was determined by observing its effect on 197 Arete. The model used is that Vesta will cause the eccentricity and semi-major axis to change. By observing the asteroid before and after their encounter, and determining the asteroid's orbital elements before and after, one can find the angle of deflection. Model the motion as a hyperbolic encounter and one can find the combined mass. Since 197 Arete is much smaller than Vesta, this gives a good estimate of the mass of Vesta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of asteroids can also be found by their perturbations of Mars, providing there are sufficiently accurate observations. Viking radar data (with an accuracy of 7m) allowed for good estimates of the mass of the major asteroids. The Dawn mission to Vesta and Ceres made very high accuracy measurements of the gravity field around them and so their mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The math involved is essentially just Newtonian mechanics. You know the positions and velocities before the encounter. There is an energy conserving encounter with a force of $$F = \frac{m_1 m_2 G}{r^2}$$ Finding the angle of perturbation allows you to find the reduced mass $$\frac{m_1 m_2}{m_1+m_2}$$&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-01-03T11:38:52.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19608" PostTypeId="2" ParentId="16390" CreationDate="2017-01-03T15:45:14.310" Score="1" Body="&lt;p&gt;The Sun's Hill Sphere may be the term you're looking for. Or perhaps &quot;within the Oort Cloud diameter&quot; would do it for you. According to Wikipedia, the outer gravitational zero velocity distance (Hill Sphere) isn't known - it's thought to lie between 50,000 AU and 200,000 AU - or up to over 3 light years.&lt;/p&gt;&#xA;" OwnerUserId="15374" LastActivityDate="2017-01-03T15:45:14.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19609" PostTypeId="1" AcceptedAnswerId="19610" CreationDate="2017-01-03T16:23:34.890" Score="-2" ViewCount="69" Body="&lt;p&gt;Would inertia not grow if we took out the reference frame?&lt;/p&gt;&#xA;" OwnerUserId="7178" LastActivityDate="2017-01-03T16:38:58.960" Title="If velocity is relative, why mass (inertia) is increasing absolutely?" Tags="&lt;mass&gt;&lt;velocity&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2017-01-03T16:49:28.507" ContentLicense="CC BY-SA 3.0" />
  <row Id="19610" PostTypeId="2" ParentId="19609" CreationDate="2017-01-03T16:38:58.960" Score="1" Body="&lt;p&gt;First a conceptual thing: There's no 'taking out' the reference frame. Only changing it.&lt;br&gt;&#xA;Next: Velocity $\dot x$ is the change of position $x$, thus this is a coordinate-dependent quantity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then mass is a bit more complicated, as there are two concepts of mass: &lt;strong&gt;Inertial&lt;/strong&gt; mass and &lt;strong&gt;gravitational&lt;/strong&gt; mass.&lt;br&gt;&#xA;Inertial mass is the proportionality constant between an acting force $\vec F$ on a body and it's change of velocity $\ddot x$, thus we get Newton's 2nd law that states $\vec F = m \ddot x$. Thus inertial mass is coordinate-independent by construction. Similar for the gravitational mass.&lt;br&gt;&#xA;Then experiments show that inertial and gravitational mass are equal, also there are &lt;a href=&quot;https://en.wikipedia.org/wiki/Equivalence_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;theoretical reasons to believe this.&lt;/a&gt; Thus we can show that mass in total is coordinate independent.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2017-01-03T16:38:58.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19613" PostTypeId="1" CreationDate="2017-01-03T19:11:27.420" Score="3" ViewCount="478" Body="&lt;p&gt;I'm new here and searched for solution but did not found one. I have found algorithms in order to calculate sunrise and sunset times for an observer on the surface of Earth with explanations of derivations of these formulae, but I want to improve it by including the altitude of the observer. I found &lt;a href=&quot;https://en.wikipedia.org/wiki/Sunrise_equation#Hour_angle&quot; rel=&quot;nofollow noreferrer&quot;&gt;an article&lt;/a&gt; in Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could one explain to me, from where is this &quot;2.076 degrees&quot; in the formula $-2.076^{\mathrm{o}} \sqrt{\mathrm{elevation\:of\:observer\:in\:meters}}/60^{\mathrm{o}}$ derived and calculated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OM8lD.png&quot; alt=&quot;Screenshot of equation in question&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15378" LastEditorUserId="11143" LastEditDate="2017-01-04T16:11:26.713" LastActivityDate="2017-09-02T02:25:48.927" Title="About the formula for taking in count of the observer altitude in sunrise/sunset formula" Tags="&lt;the-sun&gt;&lt;positional-astronomy&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19614" PostTypeId="1" AcceptedAnswerId="19616" CreationDate="2017-01-03T20:11:13.817" Score="31" ViewCount="6056" Body="&lt;p&gt;I've read various articles and books (like &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0824707095&quot; rel=&quot;noreferrer&quot;&gt;this one&lt;/a&gt;) stating that we are not certain about the geometry of the universe, but there were experiments on-going or planned that would help us find out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently though, I've watched a lecture by &lt;a href=&quot;https://en.wikipedia.org/wiki/Lawrence_M._Krauss&quot; rel=&quot;noreferrer&quot;&gt;cosmologist Lawrence Krauss&lt;/a&gt; where he seems to categorically assert that the universe has been proven to be flat by the &lt;a href=&quot;https://en.wikipedia.org/wiki/BOOMERanG_experiment&quot; rel=&quot;noreferrer&quot;&gt;BOOMERanG experiment&lt;/a&gt;. &lt;a href=&quot;https://youtu.be/veU6hK3jMH4?t=12m25s&quot; rel=&quot;noreferrer&quot;&gt;Here's the relevant portion of the talk&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've looked around and there are still articles stating that we still don't know the answer to this question, like &lt;a href=&quot;http://curious.astro.cornell.edu/103-the-universe/cosmology-and-the-big-bang/geometry-of-space-time/604-what-is-the-shape-of-the-universe-intermediate&quot; rel=&quot;noreferrer&quot;&gt;this one&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is two-fold:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Am I mixing concepts and talking about different things?&lt;/li&gt;&#xA;&lt;li&gt;If not then is this evidence not widely accepted by some reason? What reason would that be?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15379" LastActivityDate="2019-07-31T21:36:02.197" Title="Is the universe considered to be flat?" Tags="&lt;cosmology&gt;&lt;space-geometry&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="19615" PostTypeId="2" ParentId="19614" CreationDate="2017-01-03T21:15:01.260" Score="2" Body="&lt;p&gt;Yes, it's considered to be spatially flat on the largest scales we can observe, but we must remember that scientific measurements come with uncertainties, and our models can be replaced by better ones.  At the present time, we have observations that say the universe is spatially flat to within a high degree of accuracy, but there is still some wiggle room there for it to be slightly curved that we cannot rule out.  Also, we can only observe the part of the universe we can see, we cannot know that the rest of the universe has the same curvature as our portion does.  We have a theoretical understanding that it would be very difficult for the universe to be close to flat, without being extremely close to flat, so we expect it is extremely close to flat.  But theories can be replaced and usually are, and even if the theoretical expectation continues to pass tests way into the future, we still can never know the universe is &lt;em&gt;exactly&lt;/em&gt; flat.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the bottom line is, we have both very good observations, and a good theory (the theory of inflation, and the fact that flatness is unstable with age under general relativity), that agree the universe on the largest scales we can observe is &lt;em&gt;very close to&lt;/em&gt; spatially flat.  Hence, we can create a model in which is it is flat, and use that model successfully.  That's all you ever get in science.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2017-01-03T21:21:57.637" LastActivityDate="2017-01-03T21:21:57.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19616" PostTypeId="2" ParentId="19614" CreationDate="2017-01-03T21:17:02.477" Score="34" Body="&lt;p&gt;I think the reason you're suffering from conflicting sources is that you're mixing both new and old, out-of-date pieces of information. First off, the book you cited was published in 2001 - 15 years ago - and the other article you cite was published in 1999 - 17 years ago. There's been a &lt;em&gt;lot&lt;/em&gt; of work done in the past 15 years, often under the term &quot;precision cosmology&quot;, in an attempt to really nail down the precise content, shape, size, etc. of our Universe. By the early 2000's we pretty much knew the science behind everything (we knew about dark matter, dark energy, had well-developed theories on the Big Bang, etc.) but what we didn't have, were good, solid, believable numbers to put into these theories, explaining why the flatness of the universe was still contested in your sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll direct you to two incredibly important observatories which have been paramount in achieving our goal of having &quot;good numbers&quot;. The first is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Wilkinson_Microwave_Anisotropy_Probe&quot; rel=&quot;noreferrer&quot;&gt;Wilkinson Microwave Anistropy Probe (WMAP)&lt;/a&gt;, launched in 2001, and the second is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Planck_(spacecraft)&quot; rel=&quot;noreferrer&quot;&gt;Planck satellite&lt;/a&gt;, launched in 2009. Both missions were designed to stare intently at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;noreferrer&quot;&gt;Cosmic Microwave Background (CMB)&lt;/a&gt; radiation and try to sort out the treasure trove of information which can be gleaned from it. In this vein, you might also come upon the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_Background_Explorer&quot; rel=&quot;noreferrer&quot;&gt;Cosmic Background Explorer (COBE)&lt;/a&gt;, launched in 1989. This satellite had a similar purpose as the other two, but was not nearly as precise as the later two missions as to provide us with good numbers and definitive statements by the early 2000's. For that reason I'll mostly focus on what WMAP and Planck have told us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WMAP was a hugely successful mission which stared at the CMB for 9 years and created the most detailed and comprehensive map of its day. With 9 years of data, scientists were really able to reduce the observational errors on various cosmological quantities, including the flatness of the universe. You can see a table of their &lt;a href=&quot;https://lambda.gsfc.nasa.gov/product/map/current/params/lcdm_wmap9.cfm&quot; rel=&quot;noreferrer&quot;&gt;final cosmological parameters here&lt;/a&gt;. For the flatness, what you want to do is add up $\Omega_b$ (the baryonic matter density), $\Omega_d$ (the dark matter density), and $\Omega_\Lambda$ (the dark energy density). This will give you the &lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/D/Density+Parameter&quot; rel=&quot;noreferrer&quot;&gt;overall density parameter&lt;/a&gt;, $\Omega_0$, which tells you the flatness of our universe. As I'm sure you know from your sources, if $\Omega_0 &amp;lt; 1$ we have a hyperbolic universe, if $\Omega_0 = 1$ our universe is flat, and $\Omega_0 &amp;gt; 1$ implies a spherical universe. From the results of WMAP, we have that $\Omega_0 = 1.000 \pm 0.049$ (someone can check my math) which is very close to one, indicating a flat universe. As far as I know, WMAP was the first instrument to give a truly precise measurement of $\Omega_0$, allowing us to say definitively that our universe appears flat. As you say, the BOOMERanG experiment also provided good evidence for this, but I don't think the results were nearly as powerful as WMAP's was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other important satellite here is Planck. Launched in 2009, this satellite has provided us with the best high precision measurements of the CMB to-date. I'll let you dig through their results in &lt;a href=&quot;https://arxiv.org/pdf/1502.01589v3.pdf&quot; rel=&quot;noreferrer&quot;&gt;their paper&lt;/a&gt;, but the punchline is that they measure the flatness of our universe to be $\Omega_0 = 0.9986 \pm 0.0314$ (calculated from this &lt;a href=&quot;https://en.wikipedia.org/wiki/Lambda-CDM_model#Parameters&quot; rel=&quot;noreferrer&quot;&gt;result table&lt;/a&gt;), again extremely close to one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, recent results (within the past 15 years) allow us to definitively state that our Universe appears flat. I don't think, at this time, anyone contests that or believes it is still uncertain. As it usually goes with science, answering one question has only resulted in more questions. Now that we know $\Omega_0 \simeq 1$, we have to ask why is it one? Current theory suggests it shouldn't be - that it should be either enormously small or enormously large. This is known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Flatness_problem&quot; rel=&quot;noreferrer&quot;&gt;Flatness Problem&lt;/a&gt;. That in turn delves into the &lt;a href=&quot;https://en.wikipedia.org/wiki/Anthropic_principle&quot; rel=&quot;noreferrer&quot;&gt;Anthropic Principle&lt;/a&gt; as an attempted answer, but then, I'm getting out of the scope of this question.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-01-03T21:17:02.477" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19617" PostTypeId="2" ParentId="19614" CreationDate="2017-01-03T21:19:31.633" Score="11" Body="&lt;p&gt;The basic assumptions of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmological_principle&quot; rel=&quot;noreferrer&quot;&gt;cosmological principle&lt;/a&gt; mean that space can only have constant scalar curvature. This can be positive, negative or zero and a flat Universe is one where the curvature is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The curvature of space is something that can be measured and the current value &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0201137&quot; rel=&quot;noreferrer&quot;&gt;is known to be close to zero&lt;/a&gt;, not just from BOOMERanG, &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0603449&quot; rel=&quot;noreferrer&quot;&gt;but from subsequent observations&lt;/a&gt;. Vanilla FLRW cosmology has difficulty explaining this and it is known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Flatness_problem&quot; rel=&quot;noreferrer&quot;&gt;the flatness problem&lt;/a&gt;. However the conventional view is that &lt;a href=&quot;https://en.wikipedia.org/wiki/Inflation_(cosmology)&quot; rel=&quot;noreferrer&quot;&gt;cosmic inflation&lt;/a&gt; does a very neat job in solving this problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However a truly flat Universe must have a spatial curvature of exactly zero on a large scale, so to truly determine if the Universe is flat, even using a number of reasonable assumptions requires an exact measurement, which is impossible. So observation can never rule out the possibility that the Universe might have a very small positive or negative curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, if you slightly relax the cosmological principle from its strictest interpretation, the scalar curvature does not completely determine the topology of the Universe, opening the door to so-called exotic topologies. For example a flat Universe could have a toroidal topology and be compact (of finite spatial volume).&lt;/p&gt;&#xA;" OwnerUserId="2092" LastActivityDate="2017-01-03T21:19:31.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19619" PostTypeId="1" AcceptedAnswerId="19621" CreationDate="2017-01-04T09:25:29.270" Score="2" ViewCount="160" Body="&lt;p&gt;Each time I look at a shadow cast by a sunlit object it seems as if the shadow doesn't move at all. But if I look at it after say 5 minutes, I can notice that the shadow has actually moved a bit. This allows to say &quot;yes, the Sun has moved in the sky&quot;. But this doesn't make it all that obvious that &quot;the Sun is continuously moving all the time&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'd like to make some optical configuration which would allow me to watch a spot of sunlight or a shadow moving much faster — say, 1 mm/s — still being from the Sun, not from an artificial light source. I've tried doing this with a cylindrical silvery surface like that of a kitchen rail system, sticking black tape on it and leaving only a lateral strip of reflecting surface, then then letting Sun rays to hit it almost tangentially, and to observe the reflection. But either this doesn't give good enough magnification to make Sun's motion apparent, or the reflected spot just becomes too faint for the naked eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what can I do to effectively magnify angular motion of the Sun to observe it (almost) directly at the scale of seconds? I's OK if it could be visible for e.g. a minute and required me to adjust my instrument to view the next piece of motion. But I'd like to do it as simply as possible, without too expensive hardware.&lt;/p&gt;&#xA;" OwnerUserId="735" LastActivityDate="2017-01-04T13:43:38.603" Title="How to make motion of the Sun more apparent at seconds scale?" Tags="&lt;the-sun&gt;&lt;amateur-observing&gt;&lt;apparent-motion&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19620" PostTypeId="2" ParentId="19614" CreationDate="2017-01-04T10:00:39.217" Score="1" Body="&lt;p&gt;Just to add to @zephyr 's answer, LISA fired 3 lasers in space to form a triangle in order to measure the flatness of space: if the sum of the 3 angles exactly equals 180 degrees, then the space is flat; deviation from 180 degrees tells you how much the space is curved and the orientation of the curvature. But if the size of the space is too small, then the angles will sum to exactly 180 degrees; this is like looking at the surface of the Earth and thinking it looks flat when it is actually round. LISA measured exactly 180 degrees, so either space is actually flat or we can constrain the curvature of space on larger scales with error-bars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: It was LISA, not WMAP, that did the laser experiment. Thanks to @zephyr for the correction. &lt;/p&gt;&#xA;" OwnerDisplayName="user15317" LastEditorUserId="6" LastEditDate="2017-01-04T14:42:10.120" LastActivityDate="2017-01-04T14:42:10.120" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19621" PostTypeId="2" ParentId="19619" CreationDate="2017-01-04T13:43:38.603" Score="1" Body="&lt;p&gt;The obvious answer is to use a telephoto lens -- or any other magnification scheme.  As an extreme example, if you had a post 1 km tall, the shadow of its top (think sundial) would move pretty clearly.  If you re-image a short post, or just re-image the sun itself, which enough magnification, you'll see the motion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is kind of the same as trying to observe a celestical object with a high-power telescope. If you don't have an equatorial mount, everything in your field of view will move quite noticeablly.&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-01-04T13:43:38.603" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19623" PostTypeId="2" ParentId="19613" CreationDate="2017-01-04T17:04:25.447" Score="2" Body="&lt;p&gt;It's simple geometry, you are calculating the downward angle you need to look relative to the horizontal, from a given elevation, so see the horizon. To lowest order, that angle simply gets added to the angle of rotation of the Earth to make the limb of the Sun reach the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate that angle, let's measure all distances relative to Earth radius, and say you are at altitude $x$ above the surface. Draw a straight line to the horizon, where that line makes a right angle to the radius of the Earth (it is tangent to the horizon at that point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now you have a right triangle with hypotenuse $1+x$, and a side of length $1$, so the angle between the direction you are looking, and the direction to the center of Earth, has a sine of $1/(1+x)$. That means the downward angle we want is however much less that angle is from a right angle. Let's call the downward angle we want $y$, so that $y \ll 1$ if measured in radians, and then what we have from the above is $\sin(\pi/2 - y) = 1/(1+x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a trigonometric identity then says $\cos(y) = 1/(1+x)$. But since $x$ and $y$ are $\ll 1$, $\cos(y)$ is close to $1 - y^2/2$ and $1/(1+x)$ is close to $1-x$, so we can say $y$ is close to the $\sqrt{2x}$. That's where the formula comes from, the rest is just unit conversion from radians to degrees and from the radius of the Earth to the altitude in meters. &lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="9301" LastEditDate="2017-05-04T23:42:32.107" LastActivityDate="2017-05-04T23:42:32.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19624" PostTypeId="1" CreationDate="2017-01-04T18:22:31.677" Score="0" ViewCount="159" Body="&lt;p&gt;I'll take number per average star, multiple by appropriate numbers , star/Galaxy, galaxies/universe, 14 billion years, and try to figure their location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was looking for a rough number and a distribution function.&lt;/p&gt;&#xA;" OwnerUserId="15394" LastEditorUserId="1203" LastEditDate="2017-01-05T18:49:57.477" LastActivityDate="2017-02-06T23:32:26.343" Title="Number of photons in universe? I'll take number from a star in unit time and extrapolate" Tags="&lt;galaxy&gt;&lt;cosmology&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19626" PostTypeId="1" CreationDate="2017-01-04T21:19:28.097" Score="3" ViewCount="885" Body="&lt;p&gt;I have a Skywatcher 130/900 and I found that if I do an accurate polar alignment I'll need only one motor to track a celestial body, so I decided to build my motor. Right now I'm facing a problem with the mating of the spur gear that drive the worm gear&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YvhQf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YvhQf.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Using a caliper I found that the gear has the outer diameter of 50mm, and a tooth is 1mm in height, so the reference diameter is 49mm; the gear has 140 teeth, and so the module (reference_diameter / n_of_teeth) is 0.35.&lt;/p&gt;&#xA;&lt;p&gt;After this I used Solidworks to model a spur gear that could match it, and printed this&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FNcju.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FNcju.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;When I tried to check the mating, I found this&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KBH17.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KBH17.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The teeth seems right, but they're too short and can't mate.&lt;/p&gt;&#xA;&lt;p&gt;I used a caliper and counted the teeth to obtain the gear specs, and I think is kinda stupid as approach, but I could not think of something better to mate my gear. Is there a better approach to this?&lt;/p&gt;&#xA;&lt;p&gt;Then, the idea was to try and remove the gear (removing the screw) and replace it with a 3d printed gear, but it does not seems to work (and I think it could break the telescope mount). Did someone try to do it?&lt;/p&gt;&#xA;&lt;p&gt;Searching online I found that (obviously) there are some tracking motors and another question came to my mind: how can they mate the gear? Is there some kind of standard? Does it mean that every telescope's mount has some kind of standard gear? If yes, where can I found them?&lt;/p&gt;&#xA;&lt;p&gt;EDIT: That's the situation: on the right there is the spur gear that I'm trying to mate (or eventually remove, if possible) and in the center there is the worm gear that move the telescope on the RA&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UC9sK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UC9sK.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;EDIT2: Here's the project of an involute gear with Solidworks. (I followed &lt;a href=&quot;https://grabcad.com/tutorials/tutorial-how-to-model-involute-gears-in-solidworks-and-show-design-intent&quot; rel=&quot;nofollow noreferrer&quot;&gt;this tutorial&lt;/a&gt;) but as you can see, the teeth overlap.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZV11A.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZV11A.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/HxB8R.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HxB8R.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;As a reminder, I have the following parameters&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Pitch diameter (PD): 49mm (caliper measurement)&lt;br /&gt;&#xA;Number of teeth (Z): 140&lt;br /&gt;&#xA;Module (PD/Z): 0.35&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Am I doing something wrong (or am I missing something)?&lt;/p&gt;&#xA;" OwnerUserId="15397" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2021-02-24T01:09:18.927" Title="DIY tracking motor - gear mating" Tags="&lt;telescope&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19627" PostTypeId="1" AcceptedAnswerId="19640" CreationDate="2017-01-04T21:28:35.590" Score="3" ViewCount="367" Body="&lt;p&gt;Reading up on rogue planets (i. e. planets not orbiting a star), it turns out that a rogue Super-Earth with a sufficiently dense hydrogen atmosphere could retain its internal heat and have oceans/lakes of liquid water. What I'd like to know is whether those oceans could contain dissolved oxygen in equilibrium. Please advise if this should be best posted to Chemistry...&lt;/p&gt;&#xA;" OwnerUserId="15398" LastEditorUserId="6" LastEditDate="2019-05-02T03:22:54.450" LastActivityDate="2019-05-02T03:22:54.450" Title="Can a planet with a hydrogen atmosphere have a water ocean with dissolved oxygen?" Tags="&lt;planetary-atmosphere&gt;&lt;water&gt;&lt;rogue-planet&gt;" AnswerCount="2" CommentCount="5" ClosedDate="2017-01-08T13:10:21.663" ContentLicense="CC BY-SA 3.0" />
  <row Id="19629" PostTypeId="2" ParentId="19626" CreationDate="2017-01-05T00:04:08.770" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Gear&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gear&lt;/a&gt; teeth usually have a special shape which is either an epicycloid or involute curve. Making worm gears is even more complicated because the teeth have a helical curve in addition to having an involute profile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making good-quality gears requires specialized tools, knowledge and equipment.&lt;/p&gt;&#xA;" OwnerUserId="2800" LastActivityDate="2017-01-05T00:04:08.770" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19630" PostTypeId="1" AcceptedAnswerId="19638" CreationDate="2017-01-05T00:21:29.920" Score="2" ViewCount="109" Body="&lt;p&gt;Every 24 hour day, there is a point as the earth rotates about its axis, where a location on earth is the closest to the ecliptic and where it's the furthest. Do those two points have a name?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I mean is on the plane of the ecliptic, as the earth rotates around its axis, a point on the earth at 45 deg north is going to hit a low-point perpendicular to that plane, and will hit a high point perpendicular to that plane. &lt;/p&gt;&#xA;" OwnerUserId="15366" LastEditorUserId="15366" LastEditDate="2017-01-05T21:23:01.203" LastActivityDate="2017-01-05T21:23:01.203" Title="What's the name of the point where you are closest or furthest from the ecliptic?" Tags="&lt;ecliptic&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19631" PostTypeId="2" ParentId="19627" CreationDate="2017-01-05T01:14:21.983" Score="1" Body="&lt;p&gt;If it has internal heat and that the main source of energy in the ocean, then it won't ever be in equilibrium.  The internal heat must slowly decrease anyway and the ocean will be warmed from the bottom, setting up a bottom-to-top circulation.  It would be a bit like a giant pot on a stove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in principle any body of water can have a dissolved oxygen content.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how the circulation of water would affect it.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-01-05T01:14:21.983" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19632" PostTypeId="1" AcceptedAnswerId="19634" CreationDate="2017-01-05T06:18:54.437" Score="3" ViewCount="83" Body="&lt;p&gt;We quote that the Earth is about 4 billion years old. However earth started with billions of cigarette smoke sized particles. What event or size or something do we use to say this is the start of Earth? Can we know how long the formation from dust to current size took?&lt;/p&gt;&#xA;" OwnerUserId="15403" LastEditorUserId="15403" LastEditDate="2017-01-05T11:02:58.660" LastActivityDate="2017-01-05T11:12:09.860" Title="Age of a planetary body" Tags="&lt;earth&gt;&lt;planetary-formation&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19633" PostTypeId="2" ParentId="19614" CreationDate="2017-01-05T10:09:27.710" Score="3" Body="&lt;p&gt;You ask &quot;Am I mixing concepts and talking about different things?&quot;  I have no way of knowing whether you are or are not, but the title of your post and the first sentence are somewhat at odds.  Your question &quot;Is the Universe considered to be flat?&quot; concerns curvature, which, by itself, does not completely determine geometry, while the statement &quot;we are not certain about the geometry of the universe, but there were experiments on-going or planned that would help us find out&quot; may be talking about something more general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first of your links is to Jeffrey Weeks' book &lt;em&gt;The Shape of Space&lt;/em&gt;, which focuses a lot of attention on the topology of space.  Table 19.1 on page 186 lists some possible topologies for the cases of positively curved, flat, and negatively curved space.  The same page contains the surprising statement &quot;When the first edition of this book appeared in 1985, many cosmologists were completely unaware of closed manifolds with flat or hyperbolic geometry.&quot;  I am curious about whether that is a fair characterization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the preceding page of that book (page 185), the evidence, as of 2001, for a flat geometry is briefly outlined.  In particular, there is the statement that &quot;New data (coming from studies of distant supernovas and the cosmic microwave background radiation) make a strong case that the visible universe is not hyperbolic, but flat.&quot;  The same page contains the question &quot;Is the universe closed or open?  In other words, is space finite or infinite?&quot; and the answer &quot;Put briefly, we don't know.&quot;  The last two chapters of the book discuss &quot;Cosmic crystallography&quot; and &quot;Circles in the sky&quot;, two proposed observational approaches to the topology of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently work in the topology of the universe continues to be active.  Scholarpedia contains a recent &lt;a href=&quot;http://www.scholarpedia.org/article/Cosmic_Topology&quot; rel=&quot;nofollow noreferrer&quot;&gt;review&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="15388" LastActivityDate="2017-01-05T10:09:27.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19634" PostTypeId="2" ParentId="19632" CreationDate="2017-01-05T10:58:45.647" Score="3" Body="&lt;p&gt;Radioactive dating works based on the assumption that once a particular rock (in this case, a tiny zircon crystal) has cooled and solidified, the material within it will not be exchanged with material outside it. Since the rock contains trace (known) amounts of radioactive isotopes at its formation, measuring the amount of radioactivity today gives the amount of time since that rock solidified and cooled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The USGS claims that the age of the zircon crystals is roughly the age of the Earth. They are doing so based on the assumption that the zircon crystals solidified shortly after the Earth was formed. Considering that Earth formed when a diffuse cloud of tiny particles collapsed into a comparatively tiny ball, releasing a bunch of potential energy in doing so, it is reasonable to suspect that Earth, just after it formed, was extremely hot, certainly hot enough to melt zircon at 1900 degrees C. As discussed above, melting something &quot;resets the clock&quot; with respect to isotope composition, as new radioactive isotopes readily diffuse in to replace the old ones. Therefore, their assumption seems to be well grounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the last question, it depends on what you mean by &quot;know&quot;. Obviously we don't have planet-sized clouds of dust to play around with in laboratories, and we also don't really have enough observation time to see these things happen (even though I said they occur &quot;quickly&quot;, not many processes on the scale of planets take less than a few thousand years to happen). So we will probably never see more than a snapshot of this evolution process directly. Then again, if you trust the simulations that people are doing nowadays (considering how much thought is generally put into those things, it's reasonable to do so), we seem to have a general idea of how long the collapse &lt;em&gt;should&lt;/em&gt; take, which is on the order of a million years, starting from the phase where the dust has already collapsed into 1 km-wide bits: &lt;a href=&quot;https://www.cfa.harvard.edu/~kenyon/pf/terra/t0/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.cfa.harvard.edu/~kenyon/pf/terra/t0/index.html&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="15299" LastEditorUserId="15299" LastEditDate="2017-01-05T11:12:09.860" LastActivityDate="2017-01-05T11:12:09.860" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19637" PostTypeId="2" ParentId="6014" CreationDate="2017-01-05T16:29:53.013" Score="0" Body="&lt;p&gt;You can check out &lt;a href=&quot;https://en.wikipedia.org/wiki/Olbers&amp;#39;_paradox&quot; rel=&quot;nofollow noreferrer&quot;&gt;Olbers' paradox&lt;/a&gt;, also known as &lt;em&gt;dark night sky paradox&lt;/em&gt;.&#xA;In short, it argues that the universe can't be a static one (ie. no time component = no beginning and no ending, infinite in size, etc), because if it were that way, than in this infinite universe, &lt;strong&gt;there should be an infinite amount of stars&lt;/strong&gt; and the light of those stars would had an infinite amount of time to reach us. This would mean, that we should see the light of all (infinite) stars at all times, which is definitely not the case if you go out at night and look up.&lt;/p&gt;&#xA;" OwnerUserId="10611" LastActivityDate="2017-01-05T16:29:53.013" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19638" PostTypeId="2" ParentId="19630" CreationDate="2017-01-05T17:12:15.183" Score="1" Body="&lt;p&gt;I don't know of a standard term for that quantity, but I'd call it the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ecliptic_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;ecliptic latitude&lt;/a&gt; of the zenith.&#xA;With north positive and south negative, this would reach a minimum at 6:00 and a maximum at 18:00 local &lt;a href=&quot;https://en.wikipedia.org/wiki/Sidereal_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;sidereal time&lt;/a&gt;.&#xA;These correspond to local solar noon or midnight at the solstices and occur 2 hours earlier per month.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vectors perpendicular to the ecliptic are called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ecliptic_pole&quot; rel=&quot;nofollow noreferrer&quot;&gt;ecliptic poles&lt;/a&gt;. The zenith only coincides with one of these on the Arctic or Antarctic circle, at sunrise (winter-spring) or sunset (summer-fall).&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2017-01-05T17:12:15.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19639" PostTypeId="1" AcceptedAnswerId="19641" CreationDate="2017-01-05T19:19:32.887" Score="5" ViewCount="336" Body="&lt;p&gt;Triton is very special, since its orbit is retrograde respect Neptune, its obliquity is near 0° and its orbit is almost circular (according to &lt;a href=&quot;http://adsabs.harvard.edu/abs/1984NASCP2330..357H&quot; rel=&quot;nofollow noreferrer&quot;&gt;Harris (1984)&lt;/a&gt; is $5 \times 10^{-4}$). I'm familiar with the theory that claims that Triton was several Myr ago a KBO that was captured by the gravity of Neptune, but I didn't find an actual &lt;strong&gt;explanation&lt;/strong&gt; of why the orbit is almost circular. Can anybody give me some theories that actually explain the cause of this? I've heard about tidal circularization, but I don't know if that explains it at all...&lt;/p&gt;&#xA;" OwnerUserId="9949" LastEditorUserId="9949" LastEditDate="2017-01-07T17:14:34.347" LastActivityDate="2017-01-07T17:14:34.347" Title="Why is the eccentricity of Triton so low?" Tags="&lt;orbit&gt;&lt;gravity&gt;&lt;neptune&gt;&lt;tidal-forces&gt;&lt;triton&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19640" PostTypeId="2" ParentId="19627" CreationDate="2017-01-05T20:21:31.833" Score="2" Body="&lt;p&gt;I'm going to say no, or at least, very unlikely.   Short answer, no sunlight, no oxygen formation.    Longer answer below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A water ocean on a large Rogue planet with a thick enough atmosphere is possible, at least for while.  If there's enough heat coming from the surface or under the oceans through volcanism, perhaps with it being a a younger planet with higher internal heat and more radioactive elements, and/or if the atmosphere was thick enough then liquid oceans are certainly possible.  You run into problems with too think an atmosphere, because the pressure and perhaps temperature would be enormous where liquid water could form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we assume somewhat Earth like, with extensive volcanic activity to maintain enough surface heat, Oxygen is more unlikely for a few reasons.   The hydrogen atmosphere would prevent significant amounts of free Oxygen from ever being present and without oxygen in the atmosphere, disolved oxygen in the oceans would be temporary if it was formed there, not equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxygen is also very reactive.  It readily binds or reacts with all kinds of common elements like hydrogen, methane, Iron, Silicon, Magnesium, Aluminum, Sulfur and inside the Earth, trapped Oxygen wouldn't escape as a gas but it would likely bind with silicates pretty much permanently.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxygen in it's pure form is rare on planets because a whole lot of it needs to be produced before it's abundant enough to begin collecting in the atmosphere (or in liquid form if the planet is cold enough . . . but that's not relevant with a liquid water question).  Photosynthesis is a great way to produce Oxygen from water and CO2 but it only works with sunlight as the process is very energy intensive.   On a rogue planet without sunlight, I can't think of any way to produce any statistically relevant amounts of oxygen inside the oceans.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plants on Earth produce oxygen in abundance because of free energy from sunlight and easy access to CO2 and H2O, but without all that free energy, I don't see it happening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Extremophile&quot; rel=&quot;nofollow noreferrer&quot;&gt;Extremophiles&lt;/a&gt; come in may types, so I don't want to say it's impossible, but life that we know of relies on chemical energy and the laws of chemical energy are pretty straight forward and the production of free oxygen as a by product is difficult without a significant supply of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extremophiles, loosely speaking, eek out a living on not very abundant chemicals from volcanic gas escaping into the ocean floor or in other difficult situations.    There's &lt;a href=&quot;https://en.wikipedia.org/wiki/Extremophile#Classifications&quot; rel=&quot;nofollow noreferrer&quot;&gt;many different kinds&lt;/a&gt; and more possibilities if you factor in &lt;a href=&quot;https://en.wikipedia.org/wiki/Astrobiology&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astrobiology&lt;/a&gt; theories, but I have a tough time seeing a method where oxygen is released.   There's plenty of heat in ocean vents, but heat isn't easily converted into usable energy by bacteria.  Chemical energy and exothermic chemical reactions to create stored energy, or photosynthesis is generally how the life we know creates and stores it's energy.   The chemical energy required to release oxygen in significant amounts seems very impractical without significant sunlight.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, granted, the biology and chemistry behind what extremophiles can produce, gets pretty complicated and is closer to biology and perhaps chemistry than astronomy (though some might say astrobiology is part of astronomy), but in addition to that, exploring what chemical processes might be possible would get very long and above my paygrade so, there's a chance it's possible, but I feel pretty confident in saying your scenario is unlikely. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="43" LastEditDate="2017-01-06T11:37:30.717" LastActivityDate="2017-01-06T11:37:30.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19641" PostTypeId="2" ParentId="19639" CreationDate="2017-01-05T21:26:49.747" Score="5" Body="&lt;p&gt;Tidal circularization is indeed a plausible mechanism for why Triton's orbital eccentricity is so low, and I believe it is in fact the prevailing theory, at the moment. In short, tidal forces from Neptune dissipate energy in Triton's interior, essentially &quot;squeezing&quot; the moon. This results in a loss of orbital kinetic energy. The exact calculations that lead to the decrease in eccentricity are a little complicated - and described in the paper you linked to - but in a nutshell, Triton's orbit could have circularized on a timescale of about $\sim10^7$ to $\sim10^8$ years (&lt;a href=&quot;http://onlinelibrary.wiley.com/wol1/doi/10.1029/GL017i010p01749/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ross &amp;amp; Schubert (1990)&lt;/a&gt;, for instance, computed the latter order-of-magnitude timescale).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tidal circularization is not the only possible explanation for Triton's low eccentricity. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2005DPS....37.2511C&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cuk &amp;amp; Gladman (2005)&lt;/a&gt; hypothesized that there was once a debris disk around Neptune, possibly caused by the collision of two or more other satellites after gravitational interactions with Triton. Interactions with the disk then contributed to energy loss from the moon in the same way that tidal circularization might have, thus leading to a smaller eccentricity. Interestingly enough, though, this could have happened within only $\sim10^5$ years! The authors have suggested that a hybrid of the two models may be the best explanation.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2017-01-05T21:26:49.747" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19642" PostTypeId="1" CreationDate="2017-01-05T22:00:47.290" Score="2" ViewCount="445" Body="&lt;p&gt;Is there any way through which we could determine the ROTATION period of an exoplanet based on the data we observed?(from transit)&lt;/p&gt;&#xA;" OwnerUserId="15036" LastActivityDate="2017-01-06T11:43:58.297" Title="Exoplanet rotation period" Tags="&lt;exoplanet&gt;&lt;planetary-transits&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19643" PostTypeId="1" AcceptedAnswerId="19645" CreationDate="2017-01-06T03:06:27.773" Score="3" ViewCount="281" Body="&lt;p&gt;Can a satellite maintain an equatorial orbit around Earth near the Moon's orbit to receive partial gravitational boosts by gaining momentum as the satellite passes though the Moon's gravity well? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not asking about hanging out in the Lagrange points, but passing between the L2 point and the Moon or the Moon and Earth aligned, in a Moon assisted orbit.&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="14094" LastEditDate="2018-12-18T22:03:16.590" LastActivityDate="2018-12-18T22:03:16.590" Title="Can the Moon provide momentum to an object in Earth's orbit? Gravity Assisted Boost" Tags="&lt;the-moon&gt;&lt;natural-satellites&gt;&lt;orbital-mechanics&gt;&lt;satellite&gt;&lt;eccentric-orbit&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ClosedDate="2017-01-09T14:48:35.690" ContentLicense="CC BY-SA 4.0" />
  <row Id="19644" PostTypeId="2" ParentId="6014" CreationDate="2017-01-06T03:25:49.867" Score="-1" Body="&lt;p&gt;There is an uncountable amount of stars from our perspective, but space does have a limit of size which is the distance which subatomic particle and matter as traveled since the big bang and with that limit of stars. Although stars die and form beyond our ability to count so observably we can never have an exact count of stars at anytime as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would bet on answer B because The universe does have a total mass of about 25 billion galaxies. We can mathematically calculate the mass of our universe.  &lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2017-04-16T18:59:16.663" LastActivityDate="2017-04-16T18:59:16.663" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19645" PostTypeId="2" ParentId="19643" CreationDate="2017-01-06T03:35:31.370" Score="4" Body="&lt;p&gt;The Moon has a &quot;Roche lobe&quot;, where the Moon's gravity dominates, and all you need for the situation you describe is that the Roche lobe lies physically outside the Moon.  That will be true, because the Lagrange points are on the outside edge of that Roche lobe, and it's outside the Moon.  The Moon's Roche lobe is shown here: &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/lagpt.html&quot; rel=&quot;noreferrer&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/lagpt.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2017-01-06T03:35:31.370" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19646" PostTypeId="1" AcceptedAnswerId="19650" CreationDate="2017-01-06T03:47:37.497" Score="3" ViewCount="266" Body="&lt;p&gt;I read &lt;a href=&quot;http://www.iflscience.com/space/a-star-is-on-a-collision-course-with-the-edge-of-our-solar-system/&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;this article&quot;&gt;this article&lt;/a&gt; on &lt;a href=&quot;https://en.wikipedia.org/wiki/Gliese_710&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gliese 710&lt;/a&gt; that stated it would come within about 1.1 ly (+/- 0.577 ly) of earth in 1.35 million years. Wikipedia &lt;a href=&quot;https://en.wikipedia.org/wiki/Gliese_710&quot; rel=&quot;nofollow noreferrer&quot;&gt;states&lt;/a&gt; it would have an apparent magnitude of 0.5, and the &lt;a href=&quot;http://www.aanda.org/articles/aa/abs/2016/11/aa29835-16/aa29835-16.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;original paper abstract&lt;/a&gt; states it would have a -2.7 apparent magnitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From reading the &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_brightest_stars&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;brightest stars&quot;&gt;list of brightest stars&lt;/a&gt;, apparent magnitude of 0.5 would make it about the 10th brightest star, and -2.7 would make it the brightest (besides the sun). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious about the discrepancy between these sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How bright would Gliese 710 appear if it were at the near end (0.523 ly) and far end (1.677 ly) of the expected closest approach to us?&lt;/p&gt;&#xA;" OwnerUserId="6870" LastEditorUserId="1030" LastEditDate="2017-01-06T12:22:52.920" LastActivityDate="2017-01-06T16:12:56.200" Title="How bright could Gliese 710 appear in 1.35 million years?" Tags="&lt;star&gt;&lt;apparent-magnitude&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19647" PostTypeId="1" CreationDate="2017-01-06T11:05:30.263" Score="-2" ViewCount="275" Body="&lt;p&gt;Just curious i heard mention of the Hubble being retired soon...not sure if it was 2017...but the same article said there was a more updated telescope ready to replace it? &lt;/p&gt;&#xA;" OwnerUserId="15419" LastActivityDate="2017-01-19T21:41:11.167" Title="Is the Hubble Telescope due to be retired and replaced this year?" Tags="&lt;hubble-telescope&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19648" PostTypeId="2" ParentId="19642" CreationDate="2017-01-06T11:43:58.297" Score="1" Body="&lt;p&gt;Do you mean the time it takes for the exoplanet to rotate around its axis? The transit method is used to detect the presence of an exoplanet as it passes in front of the star (as seen from Earth). After several transits you can get the &lt;em&gt;orbital&lt;/em&gt; period, i. e. the time it takes for the planet to orbit around the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are methods to measure the rotation period of astronomical objects (exoplanets included). You can use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Doppler_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Doppler effect&lt;/a&gt; to measure the change of wavelengths of the planet's light as it rotates. This technique has been used to &lt;a href=&quot;https://www.eso.org/public/news/eso1414/&quot; rel=&quot;nofollow noreferrer&quot;&gt;measure the rotation period of exoplanet Beta Pictoris b&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the planet (as seen from our point of view) rotates from right to left. If so the right-hand edge of the planet's disc will be moving towards us, while the left-hand edge will be moving away from us. Therefore, the wavelength of the light coming from the right will be slightly shorter (i. e. it will be &lt;em&gt;blue-shifted&lt;/em&gt;) than that coming from the left (which will be &lt;em&gt;red-shifted&lt;/em&gt;). The greater the difference in wavelengths, the faster the planet is rotating. The method cited in the article above seems to be a bit more complicated, but that's the basics of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if you can image the planet with enough resolution, you might be able to spot different terrains or patterns on its surface, and by measuring the time it takes for those patterns to appear again in the same apparent position you might be able to calculate the rotation period. But this is probably unfeasible for most exoplanets, at least with the instruments we have right now.&lt;/p&gt;&#xA;" OwnerUserId="15398" LastActivityDate="2017-01-06T11:43:58.297" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19649" PostTypeId="1" AcceptedAnswerId="19651" CreationDate="2017-01-06T15:49:26.670" Score="0" ViewCount="88" Body="&lt;p&gt;Many images of the Andromeda galaxy depict a yellow-red-ish bright center and then those brown &quot;dust layers&quot; around the center. As far as I know the bright light comes from stars, but what is this &quot;brown dust&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/9/98/Andromeda_Galaxy_%28with_h-alpha%29.jpg&quot; width=&quot;500&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="15424" LastActivityDate="2017-01-06T17:44:31.483" Title="Brown bands in Andromeda galaxy?" Tags="&lt;galaxy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19650" PostTypeId="2" ParentId="19646" CreationDate="2017-01-06T16:12:56.200" Score="2" Body="&lt;p&gt;WolframAlpha says (to the query &quot;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=apparent%20magnitude%20of%20star%20with%20absolute%20magnitude%20%3D%208.23%20and%20distance%20%3D%201.1%20ly&quot; rel=&quot;nofollow noreferrer&quot;&gt;apparent magnitude of star with absolute magnitude = 8.23 and distance = 1.1 ly&lt;/a&gt;&quot;) that the apparent magnitude of Gliese 170 would be about 0.87 on average. I took the value of the absolute magnitude (8.23) from Wikipedia. You can plug in different values for the distance in the formula and see what comes out. The formula used by WolframAlpha is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D_l = 10^{{m-M \over 5} +1} pc$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $D_l$ = luminosity distance, $m$ = apparent magnitude, $M$ = absolute magnitude, $pc$ = parsec.&lt;/p&gt;&#xA;" OwnerUserId="15398" LastActivityDate="2017-01-06T16:12:56.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19651" PostTypeId="2" ParentId="19649" CreationDate="2017-01-06T17:44:31.483" Score="6" Body="&lt;p&gt;It is the remains of dead stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Large stars are able to fuse light elements (like helium) to make heavier ones (like Carbon, Oxygen and Silicon, but also all the elements found on Earth). When these stars die, some of these heavier elements are ejected into space, where they condense to form dust particles. The dust particles are small, typically a few micrometres long (similar in size to a bacteria). The dust particles are mixed with gas and ices and form nebulae. If the gas and dust can become compressed enough, then clumps of dust and gas can collapse to form new stars, and planets. Every carbon atom in your body started out in interstellar dust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dust is therefore composed of normal matter, and in fairly familiar forms, such as &quot;silicon oxide&quot; (which is what much of the rock on Earth is made of) There are also carbon grains, and metals: aluminium, magnesium and iron (usually as oxides).&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-01-06T17:44:31.483" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19652" PostTypeId="1" CreationDate="2017-01-06T18:16:08.077" Score="3" ViewCount="231" Body="&lt;p&gt;Having a look on &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble&amp;#39;s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; I noticed the following values (of the last 6 years) for the Hubble constant:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;67.6±0.7 SDSS-III Baryon Oscillation Spectroscopic Survey    &lt;/li&gt;&#xA;&lt;li&gt;73.00±1.75   Hubble Space Telescope&lt;/li&gt;&#xA;&lt;li&gt;67.80±0.77   Planck Mission&lt;/li&gt;&#xA;&lt;li&gt;69.32±0.80   WMAP (9-years)&lt;/li&gt;&#xA;&lt;li&gt;70.4±1.3     WMAP (7-years)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Most of experiments exclude the values of the others so that WMAP excludes the values of SDSS and from Hubble, while Hubble excludes Plank Mission's values, SDSS and WMAP 9.&#xA;Can anybody tell what's the problem here and which one is the more reliable source for $H_0$?&lt;/p&gt;&#xA;" OwnerUserId="9860" LastEditorUserId="9860" LastEditDate="2017-01-06T22:22:04.107" LastActivityDate="2017-01-06T22:22:04.107" Title="Why the Hubble constant is so inconsistent?" Tags="&lt;astrophysics&gt;&lt;fundamental-astronomy&gt;&lt;hubble-constant&gt;" AnswerCount="0" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19654" PostTypeId="1" CreationDate="2017-01-07T05:04:27.943" Score="5" ViewCount="967" Body="&lt;p&gt;Why are &lt;a href=&quot;https://en.wikipedia.org/wiki/Comet&quot; rel=&quot;nofollow noreferrer&quot;&gt;comets&lt;/a&gt; able to cross planetary orbits without colliding with the planets - &lt;a href=&quot;https://www.youtube.com/watch?v=C8zV1xiGqf4&quot; rel=&quot;nofollow noreferrer&quot;&gt;Why is the comets' orbits elliptical, intersecting all the other planets' orbit.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why don't they get attracted by the planets when they are in their orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, why doesn't the force of gravitation from the planets not attract or compel the &lt;a href=&quot;http://www.space.com/53-comets-formation-discovery-and-exploration.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;comet&lt;/a&gt; to rotate around them.&lt;/p&gt;&#xA;" OwnerUserId="15347" LastEditorUserId="2365" LastEditDate="2017-01-12T21:50:01.250" LastActivityDate="2017-01-12T21:53:02.770" Title="Why the comets can cross the orbits of planets?" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;gravity&gt;&lt;comets&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19655" PostTypeId="2" ParentId="8541" CreationDate="2017-01-07T05:25:54.373" Score="2" Body="&lt;p&gt;This was likely the international space station.  It can take &lt;a href=&quot;http://www.space.com/32387-satellite-space-station-skywatching-mobile-apps.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;up to 5 minutes&lt;/a&gt; for the ISS to pass from horizon to horizon.  However, later in the night, that time gets shorter as the suns relative angle to the ISS and you changes.  It will be quickly back into the shadow of the earth and lower on the horizon, staying low throughout.  It can reach &lt;a href=&quot;https://space.stackexchange.com/questions/5592/why-does-the-international-space-station-have-a-downward-facing-light&quot;&gt;up to -8 magnitude&lt;/a&gt; on occasion.  This is very easy to see, however near midnight it would likely be closer to 0 or 1.&lt;/p&gt;&#xA;" OwnerUserId="11343" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2017-01-07T05:25:54.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19656" PostTypeId="1" AcceptedAnswerId="19663" CreationDate="2017-01-07T06:51:54.207" Score="6" ViewCount="1544" Body="&lt;p&gt;Looking at spectra of emission nebulae like the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lagoon_Nebula&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lagoon Nebula&lt;/a&gt;, the $[\text{O III}]$ lines are prominent in intensity. However, the abundance of oxygen is minuscule compared to hydrogen. How then are the $[\text{O III}]$ spectral lines of such high intensity?&lt;/p&gt;&#xA;" OwnerUserId="14723" LastEditorUserId="2153" LastEditDate="2017-01-07T19:54:01.677" LastActivityDate="2017-01-08T19:35:39.840" Title="Why are O III lines so prominent in the spectra of emission nebulae when the amount of oxygen relative to hydrogen is a million times smaller?" Tags="&lt;spectra&gt;&lt;nebula&gt;&lt;elemental-abundances&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19657" PostTypeId="2" ParentId="19654" CreationDate="2017-01-07T08:42:07.370" Score="3" Body="&lt;p&gt;There are several reasons, the main ones are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The orbit of some comets may appear to 'cross' a planetary orbit when viewing the solar system in 2-dimensions, but their orbit does not lie in the ecliptic plane (that the planet orbits follow).  An example is Comet Hale-Bopp:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aRObs.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aRObs.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analogous to why &lt;a href=&quot;https://astronomy.stackexchange.com/a/11783/15217&quot;&gt;Pluto won't hit Neptune despite their orbits appear to 'cross' in 2D representations&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image source: &lt;a href=&quot;http://spiff.rit.edu/classes/phys235/comets/comets.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://spiff.rit.edu/classes/phys235/comets/comets.html&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Some comets do have orbits on the ecliptic, particularly some short-period comets, such as Comet Encke:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dP8FQ.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dP8FQ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Same image source as above&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is important to note that the chances for the comet and planet to be in enough of a vicinity to each other to result in the comet being 'captured' by the planet is low - so if the planet is not nearby, the comet can just keep on going on its normal orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though we have witnessed if a comet gets too close to a planet, when &lt;a href=&quot;https://astronomy.stackexchange.com/a/19456/15217&quot;&gt;Comet Shoemaker-Levy 9 got caught in Jupiter's gravitational field, tidally fragmented and eventually collided in 1994&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in terms of whether a comet could be captured in orbit around a planet, this has been addressed as answers to the question &quot;&lt;a href=&quot;https://astronomy.stackexchange.com/q/15051/15217&quot;&gt;Can a comet orbit a planet?&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;" OwnerDisplayName="user15217" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2017-01-12T21:53:02.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19658" PostTypeId="1" AcceptedAnswerId="19726" CreationDate="2017-01-07T09:33:27.420" Score="5" ViewCount="6824" Body="&lt;p&gt;While looking for a way to prolong an orbit via moon power I was faced with the parameters of what would be a stable orbit by length of time an object stayed in orbit, What is the definition of a stable orbit? Is there a such an orbit since all orbits eventually degrade?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/19643/can-the-moons-gravity-provide-enough-force-to-keep-an-object-in-earths-orbit&quot;&gt;Can the moon&amp;#39;s gravity provide enough force to keep an object in Earth&amp;#39;s orbit?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-11-04T03:12:26.740" LastActivityDate="2018-11-04T03:12:26.740" Title="What defines a stable orbit?" Tags="&lt;orbit&gt;&lt;orbital-mechanics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19659" PostTypeId="2" ParentId="19654" CreationDate="2017-01-07T12:01:00.967" Score="2" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Why does the comets are able to cross the planets orbit without touching the planet&lt;/p&gt;&#xA;&lt;p&gt;Why not they get attracted by the planets when they are in their orbit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;They are attracted to the planets, but unless they get very close to a planet, the Sun's attraction is dominant by quite a margin and the planets only have very small effects on the comets' paths.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Touch&amp;quot; would be the same as &amp;quot;collide&amp;quot; in this context.  The distances are simply huge and in general they are not going to be that close.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Why the force of gravitation could not attract or compel the COMET to rotate around them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The Sun's the big force and you need to be much closer to a planet to be significantly affected by its gravitational attraction compared to the Sun's.  Jupiter is the next most important (which is why Jupiter has, over huge timescales, built up a large number of small moons).&lt;/p&gt;&#xA;" OwnerUserId="10196" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-01-07T15:10:54.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19661" PostTypeId="1" AcceptedAnswerId="19667" CreationDate="2017-01-07T16:25:12.280" Score="2" ViewCount="99" Body="&lt;p&gt;The curve that separates the bright and dark parts of the moon's disc as viewed from the earth is called the inner &lt;a href=&quot;https://en.wikipedia.org/wiki/Terminator_(solar)#Lunar_terminator&quot; rel=&quot;nofollow noreferrer&quot;&gt;lunar terminator&lt;/a&gt;, which is semi-elliptical in shape. We can model the disc as a unit circle centred on the origin, in which the inner terminator runs from (0,1) to (0,-1) and its major semi-axis always has length 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does its minor semi-axis vary during the lunar cycle? Please assume that orbits are circular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, if the point halfway along the inner terminator is X, what is the equation for the motion of X back and forth along the line-segment from (-1,0) to (1,0)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the shape of the visible disc is not a &lt;a href=&quot;https://en.wikipedia.org/wiki/Lune_(geometry)&quot; rel=&quot;nofollow noreferrer&quot;&gt;lune&lt;/a&gt;, except in the two trivial cases.&lt;/p&gt;&#xA;" OwnerDisplayName="user15430" LastEditorDisplayName="user15430" LastEditDate="2017-01-07T18:40:00.440" LastActivityDate="2017-01-08T12:43:35.010" Title="How does the length of the minor semi-axis of the moon's disc's inner terminator vary through the cycle?" Tags="&lt;the-moon&gt;&lt;moon-phases&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19662" PostTypeId="1" CreationDate="2017-01-07T16:26:45.597" Score="3" ViewCount="73" Body="&lt;p&gt;How can an extrasolar comet be detected - this is based on an article regarding the influence on a extra solar gas giant on these comets causing changes in their orbit and leading to them being swallowed by the star ..&#xA;Still stuck on understanding how we detect them .. they certainly have no measurable effect on the movement of the planet or the star .. and the chemical make-up of the comet seems irrelevant given the size of the star and observations we can make ? Just missing something Simple I think ?&lt;/p&gt;&#xA;" OwnerUserId="14603" LastActivityDate="2017-01-07T17:34:48.627" Title="extrasolar comets - basic question" Tags="&lt;comets&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19663" PostTypeId="2" ParentId="19656" CreationDate="2017-01-07T17:25:39.520" Score="4" Body="&lt;p&gt;This is an excellent question. Think about the way in which emission happens. &lt;a href=&quot;https://en.wikipedia.org/wiki/H-alpha&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\text{H}\alpha$ emission&lt;/a&gt; happens when an electron makes a transition from the third energy level to the second, emitting a photon in the process with the energy equivalent to the difference of the energies of the two states, roughly $1.9\text{ eV}$. There's a lot of hydrogen in emission nebulae - and in nebulae in general - and so it plays an important role in the emission spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Emission from $[\text{O III}]$ comes from a different process: collisions. Free electrons with energies of $\sim1\text{ eV}$ - about the amount of the oxygen's low excitation potential - collide with oxygen atoms. This produces &lt;a href=&quot;https://en.wikipedia.org/wiki/Forbidden_mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;forbidden lines&lt;/a&gt; that would otherwise have been unlikely to be created, as the ionizing photons typically would not have the right energies. Therefore, doubly-ionized oxygen (in particular) becomes extremely important in spectra, generally second only in intensity to $\text{H}\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that $[\text{O III}]$ emission is collision-based, it makes sense that there is a temperature dependence - which there is, as is the case with $[\text{N II}]$ lines and is responsible for their high intensities. Conditions in many hot nebulae (in this case, $&amp;gt;10,000\text{ K}$) are right for such collisions and thus forbidden line emission. Hot host stars can sometimes help achieve high electron temperatures.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2017-01-08T19:35:39.840" LastActivityDate="2017-01-08T19:35:39.840" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19664" PostTypeId="2" ParentId="19662" CreationDate="2017-01-07T17:34:48.627" Score="2" Body="&lt;p&gt;The chemical make-up of the comet is indeed relevant. From Wikipedia (&lt;a href=&quot;https://en.wikipedia.org/wiki/Exocomet&quot; rel=&quot;nofollow noreferrer&quot;&gt;Exocomet&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The exocomets can be detected by spectroscopy as they transit their host stars. The transits of exocomets, like the transits of exoplanets, produce variations in the light received from the star. Changes are observed in the absorption lines of the stellar spectrum: the occultation of the star by the gas cloud coming from the exocomet produces additional absorption features beyond those normally seen in that star, like those observed in the ionized calcium lines. As the comet comes close enough to the star, cometary gas is evolved from the evaporation of volatile ices and dust with it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is not to say that you can determine anything else (besides their existence and chemical composition) about the comets. &lt;/p&gt;&#xA;" OwnerUserId="15398" LastActivityDate="2017-01-07T17:34:48.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19665" PostTypeId="2" ParentId="10506" CreationDate="2017-01-07T18:31:10.830" Score="0" Body="&lt;p&gt;This may seem like a cheat, but why not just look up the star in a catalogue, or using one of the several free astronomy planetarium type apps - eg Stellarium, Cartes du Ciel etc.  The star in question, nor any one is likely to find, is not unknown.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-01-07T18:31:10.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19666" PostTypeId="1" AcceptedAnswerId="20372" CreationDate="2017-01-07T18:54:45.593" Score="12" ViewCount="3879" Body="&lt;p&gt;I'm wondering what the farthest object we know of, as measured only by parallax methods, from either earth or satellite based telescopes, and at either visible light or radio wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, if we had no other way of measuring distances,&#xA; what would be the farthest object we knew about? I'm guessing this measurement would probably be one from &lt;em&gt;Gaia&lt;/em&gt;, it being the latest telescope to measure such things TTBOMK. &lt;/p&gt;&#xA;" OwnerUserId="11222" LastActivityDate="2017-03-13T11:53:32.513" Title="What's the farthest object as determined only by parallax?" Tags="&lt;distances&gt;&lt;parallax&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19667" PostTypeId="2" ParentId="19661" CreationDate="2017-01-07T22:28:46.913" Score="1" Body="&lt;p&gt;The moon is rotating wrt to the sun. As the moon rotates, the longitude of the terminator increases linearly. Treating the moon as being at &quot;great distance, so the moon appears as an orthographic projection, the position of the middle of the terminator varies sinusoidally, it moves fastest when the moon is at first or third quarter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the orthographic projection, the longitude and latitude $(\lambda,\varphi)$ are mapped to (x,y) by&#xA;$$&#xA;\begin{align}&#xA;x &amp;amp;= R\,\cos\varphi \sin\left(\lambda - \lambda_0\right) \\&#xA;y &amp;amp;= R\big(\cos\varphi_0 \sin\varphi - \sin\varphi_0 \cos\varphi \cos\left(\lambda - \lambda_0\right)\big)&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For points on the equator, and treating the central longitude $\lambda_0$ as the lunar prime meridian, and a scale of R=1, that simplifies to&#xA;$x = \sin(\lambda)$. As the terminator moves, its longitude increases linearly, so x varies sinusoidally. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Correcting for elliptical orbits, librations of the moon and a general perspective view would only marginally alter this.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-01-07T23:31:13.620" LastActivityDate="2017-01-07T23:31:13.620" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19668" PostTypeId="2" ParentId="19624" CreationDate="2017-01-07T23:23:10.463" Score="1" Body="&lt;p&gt;&lt;em&gt;Extending Zephyrs comment as community wiki&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://physics.stackexchange.com/questions/196366/number-density-of-cmb-photons&quot;&gt;https://physics.stackexchange.com/questions/196366/number-density-of-cmb-photons&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this response, the number of photons per cubic meter is estimated as $n_\lambda = 10^8$. Since the observable universe has a volume of the order $10^{80}\ m^3$, there are about $10^{88}$ photons in the observable universe.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2017-01-07T23:23:10.463" CommentCount="0" CommunityOwnedDate="2017-01-07T23:23:10.463" ContentLicense="CC BY-SA 3.0" />
  <row Id="19670" PostTypeId="2" ParentId="19661" CreationDate="2017-01-08T04:58:02.597" Score="1" Body="&lt;p&gt;As this &lt;a href=&quot;https://svs.gsfc.nasa.gov/4537#67907&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA visualization video&lt;/a&gt; shows, the terminator moves &lt;em&gt;almost&lt;/em&gt; sinusoidally, always from lunar east to west, never from west to east. We see the sunrise half of the terminator in the waxing phases and the sunset half in the waning phases. At the new and full phases, our view of the terminator switches from one half to the other, making the position vs. time function a &lt;em&gt;discontinuous&lt;/em&gt;, piecewise-sinusoidal sawtooth wave.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2017-01-08T12:43:35.010" LastActivityDate="2017-01-08T12:43:35.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19672" PostTypeId="2" ParentId="1238" CreationDate="2017-01-08T16:04:18.317" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/1005.1663&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carter &amp;amp; Winn (2010)&lt;/a&gt; suggest that the most promising means of detecting exoplanet obliquity would be through miniscule signatures imprinted on the transit light signal at ingress and egress (~ 200 parts per million for a planet as oblate as Saturn). &lt;a href=&quot;https://arxiv.org/abs/1410.0361&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zhu et al. (2014)&lt;/a&gt; use this technique to make the first tentative detection of exoplanet obliquity from a Kepler object, the 18 Jupiter mass brown dwarf Kepler 39b (KOI-423.01). They measure an oblateness of 0.22±0.11. They also place some upper constraints on the oblateness of other planets in the Kepler catalogue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ST2LK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ST2LK.png&quot; alt=&quot;Transit signal for KOI-423.01 over 12 orbits. The residuals of two models, one with oblateness and one without, are plotted at bottom. The Oblateness model is a better fit to the data.&quot;&gt;&lt;/a&gt;&#xA;&lt;em&gt;Transit signal for KOI-423.01 over 12 orbits. The residuals of two models, one with oblateness and one without, are plotted at bottom. The Oblateness model is a better fit to the data.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oblateness variations are thought to be conducive to exoplanet habitability by &lt;a href=&quot;https://arxiv.org/abs/1404.3686&quot; rel=&quot;nofollow noreferrer&quot;&gt;regulating temperature modulations&lt;/a&gt;, so I expect that measurements of this property over time will be a priority in future exoplanet observations for astrobiology and SETI studies.&lt;/p&gt;&#xA;" OwnerUserId="941" LastEditorUserId="941" LastEditDate="2017-01-08T16:09:37.207" LastActivityDate="2017-01-08T16:09:37.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19673" PostTypeId="1" CreationDate="2017-01-08T18:14:15.320" Score="3" ViewCount="197" Body="&lt;p&gt;Are there any mechanisms which could increase the density of the moon's atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon's existing atmosphere is exceeding rare, with about 1 000 000 molecules of gas per cubic centimeter at the surface, ten trillion times less than the Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The source of this gas is the impact of solar wind on the moon's surface, and some outgassing from the interior of the moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this density vary significantly? Could it reach a density comparable to Earth's? Are there natural processes which could cause the atmospheric density to increase, either in the short term, or over a longer time span.&lt;/p&gt;&#xA;" OwnerUserId="14603" LastEditorUserId="2365" LastEditDate="2017-01-14T18:08:24.453" LastActivityDate="2017-01-16T17:41:50.607" Title="Can lunar atmospheric density, as it exists today, increase naturally?" Tags="&lt;the-moon&gt;&lt;planetary-atmosphere&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19674" PostTypeId="1" CreationDate="2017-01-08T19:05:48.127" Score="2" ViewCount="35" Body="&lt;p&gt;I would like to know what conclusions one could draw from mathematically analyzing the depth of various craters across the surface of a given planet, and then relating that to their position on said planet. Note that I am not looking for immediate conclusions, but rather conclusions that could actually benefit manned exploration in that region. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="15435" LastActivityDate="2017-01-08T19:05:48.127" Title="Any benefit for analyzing the depth of craters across the surface of a planet?" Tags="&lt;data-analysis&gt;&lt;crater&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19675" PostTypeId="2" ParentId="14406" CreationDate="2017-01-08T19:31:42.013" Score="1" Body="&lt;p&gt;&lt;em&gt;Converting comment to community wiki&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Aircraft lights (but only if non-blinking); &lt;/li&gt;&#xA;&lt;li&gt;genuine satellites/old launch stages not in the database; &lt;/li&gt;&#xA;&lt;li&gt;random (&quot;sporadic&quot;) meteors not part of an established shower; &lt;/li&gt;&#xA;&lt;li&gt;birds (yes, sometimes light-coloured migrating birds reflect city light from their bellies). &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-01-08T19:31:42.013" CommentCount="0" CommunityOwnedDate="2017-01-08T19:31:42.013" ContentLicense="CC BY-SA 3.0" />
  <row Id="19676" PostTypeId="2" ParentId="14388" CreationDate="2017-01-08T19:45:11.163" Score="1" Body="&lt;p&gt;As an example consider the prime meridian for Vesta. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on Hubble images, the IAU had established a prime meridian based on the observation of the Obler's regio (a dark region). The prime meridian was defined as passes through the middle of this region. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the Dawn mission arrived at Vesta, it was discovered that the asteroid's pole was not in the same place that it had been assumed to be (based on Hubble data) and that Obler's regio was not a convenient basis for a prime meridian, as it was not even discernable from up close. Instead they chose a small crater, Claudia, that was close to the equator, and defined the prime meridian based on that (the prime meridian is defined as a passing 4 degrees from the centre of Claudia) This is now the generally accepted coordinate system, although it has not been &quot;officially&quot; adopted by the IAU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if you are the first to map a new world, you get to choose where do draw your meridian.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-01-08T19:45:11.163" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19677" PostTypeId="2" ParentId="19673" CreationDate="2017-01-09T00:30:46.870" Score="4" Body="&lt;p&gt;There's not enough gassible (is gassible a word?   I'm guessing not), OK, there's not enough of the right elements to form enough gas on or in the moon to create any kind of atmosphere that would measurably slow down falling objects (say, enough to land a probe with a parachute, like they do on Mars).   Atmospheres are heavy.  They may not seem heavy but they are.   Mars' &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Mars&quot; rel=&quot;nofollow noreferrer&quot;&gt;atmosphere&lt;/a&gt; has a mass of about 25 trillion metric tonnes and that's a very thin atmosphere compared to Earth.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon, with roughly 1/4th Mars' surface area, would need about 1/4 of that mass in it's atmosphere to have an atmosphere with similar air resistance to Mars, somewhere in the 6-7 trillion metric tonne range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's some ice on the moon in craters that never get direct sunlight, but not nearly enough ice to make that kind of an atmosphere.   The Moon also has very infrequent volcanic eruptions, but while some gas is likely released, again it wouldn't be nearly enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible scenario would be a large asteroid or comet impact that could both release some trapped gas from below the surface and release gas from the impactor.   The problem with such a scenario is that the heat from the impact would likely send much of the gas away from the planet at above escape velocity shortly after impact.    The escape velocity of the Moon is 2.38 km/s and at the top of this theoretical atmosphere it would be a little bit less than that.  A large impact from a comet is the best way I can think of, but that wouldn't be very efficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you found a way to create that kind of atmosphere on the Moon, it wouldn't last very long.  The Moon is simply too small, with too little gravity and it's too close to the sun to maintain an atmosphere, unless that atmosphere was Xenon or if you want to be a bit more daring, sulfur-dioxide (acidic when it comes in contact with water and stinky) or sulfur-hexaflouride, which isn't toxic in and of itself but becomes very toxic when exposed to UV light from the sun, so, lets not use either of those, but Xenon is safe, so we can use that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;see &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_escape&quot; rel=&quot;nofollow noreferrer&quot;&gt;atmospheric escape velocity&lt;/a&gt;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jkhOt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jkhOt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to your question is yes, the Moon could be given a thick enough atmosphere to slow falling objects down, but it would probably need to be done artificially, as there's no natural process I can see that would do that, and ideally, you'd use a very heavy and non toxic gas like Xenon.    All the Xenon in the Earth's atmosphere wouldn't be enough.   A rough calculation, Earth's atmospheric Xenon comes to less than a billion metric tons, which is still several thousand times too little mass to create a mars like atmosphere on the Moon.   Now if you harvested all the Xenon from Jupiter and/or Saturn and moved that to the moon, then you might be able to create a somewhat long lasting atmosphere on the moon that would provide some measurable air resistance.   It wouldn't be breathable, but it wouldn't be toxic, which is a plus and with enough Xenon, you might be able to get around on the moon with just an oxygen tank and no pressure suit.  Buildings with Oxygen/Nitrogen would need to be air-tight but not vacuum resistant, which is far easier to build.   Enough Xenon would make living on the moon quite a bit easier.   Still wouldn't be easy, but definitely easier.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-01-09T00:30:46.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19678" PostTypeId="2" ParentId="18563" CreationDate="2017-01-09T01:21:53.010" Score="4" Body="&lt;p&gt;According to the NASA page about &lt;a href=&quot;http://moon.nasa.gov/newsdisplay.cfm?Subsite_News_ID=51124&amp;amp;SiteID=6&amp;amp;iSiteID=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;news and events related to the moon&lt;/a&gt;, the start of October, 2016 saw the moon enter a New Moon phase - where it is not visible, as explained on the site:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the new moon isn't visible, because it's between Earth and the sun, and the unlit side is facing Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The 'unlit side' is often referred to as being the 'dark side' of the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and as shown in the terribly-but-necessarily-not-to-scale diagram below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VQ3GW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VQ3GW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image source and further information: &lt;a href=&quot;http://www.sems.und.edu/index_MoonPhases.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;SEMS: Moon Phases&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lunar cycle is roughly 29 days (the approximate time between consecutive new moon cycles as an example).  There are many online 'lunar calendars' are available, such as from &lt;a href=&quot;http://www.moonconnection.com/moon_phases_calendar.phtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;Moon Connection&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is just as easy to observe the lunar cycles throughout the nights (weather permitting) - when the lunar cycles are heading towards a 'new moon', then it will appear to be 'waning', 'waxing' refers to the image of the moon 'growing' towards a full moon.  Examples of how the moon would appear is shown in the NASA image below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/B3f8G.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B3f8G.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Image from the NASA page linked above) - the cycle progresses anti-clockwise in the diagram.&lt;/p&gt;&#xA;" OwnerDisplayName="user15217" LastActivityDate="2017-01-09T01:21:53.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19680" PostTypeId="1" CreationDate="2017-01-09T05:06:37.947" Score="4" ViewCount="500" Body="&lt;p&gt;I have the orbital elements of an asteroid's heliocentric motion (the orbit has non-zero inclination and is elliptical). But I want to simulate the motion of the sun around the asteroid. So without having the sun's ephemeris, and given only the asteroid's orbital elements, how can I determine an equivalent orbital elements for the sun's motion around the asteroid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;I realize that the question wasn't clear enough based on the comment. So, what I am trying to do is to numerically solve the two body problem equation in the inertial frame. In the classic two-body problem 2nd order ODE, in my case, the position vector is defined from the centre of the asteroid to the sun, which is absolutely valid and upon integration gives me the position an velocity for the sun with respect to an inertial frame centred at the asteroid. My question is concerned with the initial conditions to begin the numerical integration. As of now I determine the initial state vector for the Sun from some random orbital elements. But to perform a more realistic simulation, Is it possible to determine the real orbital elements for the asteroid-centric motion of the Sun from a given set of orbital elements of a heliocentric-motion of the asteroid? I hope I was clear and not made this more confusing. &lt;/p&gt;&#xA;" OwnerUserId="15440" LastEditorUserId="15440" LastEditDate="2017-01-10T21:20:53.460" LastActivityDate="2017-02-11T04:27:43.190" Title="Sun's orbital elements" Tags="&lt;the-sun&gt;&lt;orbital-elements&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19681" PostTypeId="2" ParentId="19680" CreationDate="2017-01-09T13:16:27.180" Score="3" Body="&lt;p&gt;Strictly speaking the Sun's motion is going to be dominated by the gas giants and the planets and one asteroid is going to have negligible effect on it compared to those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if we assume there are no other items we can get an approximate idea of the effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the similarity of the orbits.  The Sun ( in this simplified model ) follows a complimentary orbit around the center of mass of the two bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distances you have for the size of the obit scale like this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_1 = \frac \mu m_1 r$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_2 = \frac \mu m_2 r$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mu$ is the reduced mass :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu = \frac { m_1 m_2 } { m_1 + m_2 }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as you can see $r = r_1 + r_2$ is the distance between the two bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now as you can see from those equations, the most massive body moves the least.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mathematics comes from the way the standard &lt;a href=&quot;https://en.wikipedia.org/wiki/Two-body_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;two body problem&lt;/a&gt; is solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be instructive for you to compare the effect of your asteroid to that of Jupiter or Earth.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-01-09T13:16:27.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19682" PostTypeId="2" ParentId="19666" CreationDate="2017-01-09T14:18:13.303" Score="7" Body="&lt;p&gt;The first data release from GAIA was in 2016 so fairly new. If you want to be certain you should take a look at it because as you say, GAIA will no doubt be the source of the most accurate parallax measurements in the near future. Speaking of accuracy, I believe that this is the essence of your question because &lt;strong&gt;it basically depends on how uncertain measurements you are willing to accept.&lt;/strong&gt; &#xA;Basically, modern values of parallaxes are very small numbers, e.g. 0.1 milliarcsec which means a distance of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;d(pc) = 1/(parallax in arc-seconds)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;around 10,000 pc (one parsec is 3.26 light years). But the important point is the accuracy of these estimates. Clearly if your measurements is 0.1 milliarcsec, but the accuracy is 1 milliarcsec, it would be going out on a limb to postulate that you have measured a distance of 10,000 pc. So the real answer to your question is another question &quot;to what accuracy?&quot; So if you want something like 10 or 20 % accuracy you are more or less limited to our neighbourhood, say some 200 pc. If you check out &lt;a href=&quot;http://sci.esa.int/education/35616-stellar-distances/?fbodylongid=1661&quot; rel=&quot;noreferrer&quot;&gt;ESA's website on stellar parallax&lt;/a&gt;, they will go up to &lt;strong&gt;500 pc&lt;/strong&gt; but it really depends on what level of accuracy your work requires.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you check out the GAIA satellite then the aim is actually to go down to an accuracy of 24 micro-arc seconds so whatever object is measured to be farthest (with a given accuracy) is sure to replaced as the data from GAIA is analysed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For fun, you might wanna look at this &lt;a href=&quot;https://i.stack.imgur.com/pmH1a.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pmH1a.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also found at the &lt;a href=&quot;http://sci.esa.int/gaia/27100-gaia-an-ambitious-space-observatory-in-astronomy/&quot; rel=&quot;noreferrer&quot;&gt;GAIA website.&lt;/a&gt;. I think it provides a nice visual guideline to what distances we can expect (and it also nicely shows a ring at 500 pc - the limit of fairly accurate measurements from Hipparchos!)&lt;/p&gt;&#xA;" OwnerUserId="14899" LastActivityDate="2017-01-09T14:18:13.303" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19683" PostTypeId="1" AcceptedAnswerId="19685" CreationDate="2017-01-09T16:25:25.877" Score="-2" ViewCount="507" Body="&lt;p&gt;Why does the traslation movement of the earth last the same time as the four seasons? The seasons could take 100 days or 200 or 20 years (aprox. like in GoT)... but exactly (exactly???) 1 year. Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I think that I need to redefine the question.&lt;br&gt;&#xA;I define a year as the full traslation movement of the earth around the sun.&lt;br&gt;&#xA;And the four seasons as the full movement of the perpendicular axis from 0 to 23.5 to 0 again to -23.5 to 0 again.&lt;br&gt;&#xA;Why do these two movements take the same time?&lt;/p&gt;&#xA;" OwnerUserId="15451" LastEditorUserId="15451" LastEditDate="2017-01-10T15:30:14.163" LastActivityDate="2017-01-10T15:30:14.163" Title="Why does the traslation movement of the earth last the same time as the four seasons?" Tags="&lt;earth&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19684" PostTypeId="1" CreationDate="2017-01-09T16:38:47.480" Score="1" ViewCount="71" Body="&lt;p&gt;When checking the internal consistency of the Tycho-2 catalogue, I came across 254 entries in the main catalogue file that are also present in supplement 1 (the Hipparcos/Tycho-1 stars not included in Tycho-2). These entries have the same TYC1/2/3 identifiers, and the same HIP number, but they refer to different CCDM components. For all but a few cases, the A component is found in the main catalog file, while the B component is found in supplement 1.&#xA;From the description Tycho-2 catalogue guide, I would expect these stars to have at least a different TYC3 identifier. Does anybody know why these stars were not given a unique TYC identifier?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brightest example: HIP 84606 (TYC 2604-1775-1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I loaded the Tycho-2 catalogue into a MySQL database, the following query selects all the duplicate entries, ordered by increasing magnitude:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;SELECT tyc, MIN(HIP) AS hip, MIN(VTmag) AS mag FROM&#xA;(&#xA;    SELECT CONCAT(TYC1, '-', TYC2, '-', TYC3) AS tyc, HIP, VTmag FROM tyc2&#xA;    UNION ALL&#xA;    SELECT CONCAT(TYC1, '-', TYC2, '-', TYC3) AS tyc, HIP, VTmag FROM suppl_1&#xA;    UNION ALL&#xA;    SELECT CONCAT(TYC1, '-', TYC2, '-', TYC3) AS tyc, HIP, VTmag FROM suppl_2&#xA;) AS t1&#xA;GROUP BY tyc HAVING COUNT(tyc)&amp;gt;1 ORDER BY mag ASC&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="15443" LastActivityDate="2017-01-09T16:38:47.480" Title="Duplicate entries in Tycho-2 main and supplement 1" Tags="&lt;star-catalogues&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19685" PostTypeId="2" ParentId="19683" CreationDate="2017-01-09T16:42:53.523" Score="6" Body="&lt;p&gt;I suspect that this question may get closed as too trivial, but to put it simply a year is more or less defined as the time that it takes for the Earth to complete an orbit around the Sun. Because the &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_tilt#Earth&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth rotates on an axis that is tilted relative to its orbit around the Sun&lt;/a&gt;, rather than perpendicular, the Northern and Southern hemispheres alternate being inclined toward or away from the Sun. The hemisphere that is facing the Sun experiences Summer and the hemisphere that is facing away from the Sun experiences Winter. So a cycle of seasons is roughly equivalent to the time it takes for the Earth to complete its orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qbxdb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qbxdb.jpg&quot; alt=&quot;Earth&amp;#39;s orbit plus axial tilt creates seasons&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Image Credit: &lt;a href=&quot;http://scijinks.jpl.nasa.gov/review/solstice/seasons-earth-orbit_large.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="6" LastActivityDate="2017-01-09T16:42:53.523" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19686" PostTypeId="2" ParentId="19683" CreationDate="2017-01-09T18:57:50.010" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does a year last the same time as the four seasons ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Long before we had any serious science of astronomy, we had seasons.  We defined a year as being four seasons.  All you need to do that is long term memory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you can count and record events  in some way, you start counting days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we accumulated knowledge of the stars, we noticed that the changes of positions of the stars in the heavens followed very regular cycles.  That gave us a way to relate a year to an exact number of days ( and in fact to a fraction of a day ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we developed physics enough to be able to associate the yearly cycle with an orbital cycle we changed our definition of year to be exactly that cycle ( and we even allow for slight changes in that's cycle's length ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we invented the approximate year first and then refined it to match our understanding of physics.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The seasons could take 100 days or 200 or 20 years (aprox. like in GoT)... but exactly (exactly???) 1 year. Why?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because a year happens to be the period of time that matches the changes in seasons and heavens we first noticed and first gave that name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've just refined the definition a bit more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some thought I've edited my original to include a reference to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Egyptian_calendar&quot; rel=&quot;nofollow noreferrer&quot;&gt;Egyptian Calendar&lt;/a&gt; which has three seasons ( winter, summer and inundation ), so four seasons doesn't have to be the number.  It is notable that the approximate year length seems to be the same.  I dare say other similar examples are out there and unknown to me.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastEditorUserId="10196" LastEditDate="2017-01-09T19:40:01.947" LastActivityDate="2017-01-09T19:40:01.947" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19687" PostTypeId="1" AcceptedAnswerId="19690" CreationDate="2017-01-09T20:04:44.847" Score="19" ViewCount="9528" Body="&lt;p&gt;In my research on exoplanets, I have heard many people talk &quot;forward modeling of exoplanet atmospheres&quot;.  I don't know what the &quot;forward&quot; means in &quot;forward modeling&quot; and how it compares with &quot;reverse modeling&quot;, if that's even a thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is forward modeling, and why is it so special that it needs to be distinguished from just plain ol' regular modeling?&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2019-10-02T15:00:46.000" Title="What does forward modeling mean?" Tags="&lt;exoplanet&gt;&lt;planetary-atmosphere&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19689" PostTypeId="2" ParentId="19687" CreationDate="2017-01-09T22:07:26.033" Score="8" Body="&lt;p&gt;Forward modeling is the use of a model in order to simulate an outcome. The problem of getting the model to produce data from the input is called the &lt;em&gt;forward problem&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The forward model takes certain parameters and produces data that can then be compared with the actual observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forward modeling seems to be in common use in the Earth sciences, referring e. g. to models of global climate, seismic events, etc.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Forward problem&lt;/strong&gt; (direct problem, normal problem): The problem of&#xA;  calculating what should be observed for a particular model, e.g.&#xA;  calculating the gravity anomaly that would be observed for a given&#xA;  model of a salt dome. (&lt;a href=&quot;http://www.encyclopedia.com/science/dictionaries-thesauruses-pictures-and-press-releases/forward-problem&quot; rel=&quot;noreferrer&quot;&gt;A Dictionary of Earth Sciences&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The opposite procedure is called the &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Inverse_problem&quot; rel=&quot;noreferrer&quot;&gt;inverse problem&lt;/a&gt;&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An inverse problem in science is the process of calculating from a set&#xA;  of observations the causal factors that produced them: for example,&#xA;  calculating an image in computer tomography, source reconstructing in&#xA;  acoustics, or calculating the density of the Earth from measurements&#xA;  of its gravity field.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It is called an inverse problem because it&#xA;  starts with the results and then calculates the causes. This is the&#xA;  inverse of a forward problem, which starts with the causes and then&#xA;  calculates the results.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solving an inverse problem then means, given a set of observations, constructing a model that accounts for them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose it's to be expected that exoplanet atmospheres are studied through forward modeling, because we already have adequate atmospheric models for Earth and the understanding to adjust them to other planets, while we don't yet have an adequate characterization of exoplanet atmospheres.&lt;/p&gt;&#xA;" OwnerUserId="15398" LastActivityDate="2017-01-09T22:07:26.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19690" PostTypeId="2" ParentId="19687" CreationDate="2017-01-09T22:07:41.907" Score="13" Body="&lt;p&gt;There are different ways to model something. From what you're asking, there are two main types of modeling: forward modeling and inverse modeling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Forward Modeling&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this type of modeling, you have a specific model that defines the &quot;current&quot; state of your system. In the case of exoplanet atmospheres, it'd likely be something that defines the molecular content, ionization level, density, etc. of your exoplanet atmosphere. Then, you use the known physics/math of your system to decide how it will behave. In this setup, what you've created is a system for predicting system states from a predetermined physics model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such an example would be someone creating their own atmosphere of an exoplanet in a model and then saying, okay what happens when I shine light through this atmosphere. What observations might I record?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Inverse Modeling&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some sense this is the opposite of forward modeling, albeit it doesn't really mean you're running a model to see into the past. Instead, what happens with this setup is you know a particular state or result, and you want to construct a model of your system which can produce said state. Essentially, you want your model to arrive at a certain state when it is done calculating. If it does, you have a reasonable confidence that your model was some indication of what your system is actually like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this situation, you'd measure components of the atmosphere, e.g. the radius of the planet as a function of wavelength, and then create a model of the atmosphere which can hopefully reproduce your observations. If you can, then the hope is that the model accurately represents what your system is.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-01-09T22:07:41.907" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19691" PostTypeId="1" CreationDate="2017-01-10T00:50:57.290" Score="-1" ViewCount="552" Body="&lt;p&gt;For every location on earth, there are parts of the sky that the sun can never be in (e.g., directly overhead anywhere outside the tropics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the sun can be in about 40% of the sky (a hemisphere).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll withhold my reasoning/calculations for now, because I'd like the opinion(s) of others more-expert (most are) than me in this subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please weigh in. I believe this &quot;simple&quot; geometric question will have been answered centuries ago, but I have not found it as an answer, nor even as a subject.&lt;/p&gt;&#xA;" OwnerUserId="15450" LastActivityDate="2017-01-10T05:58:45.530" Title="What percentage of the sky could the sun be in?" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19692" PostTypeId="5" CreationDate="2017-01-10T01:50:46.183" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2017-01-10T01:50:46.183" LastActivityDate="2017-01-10T01:50:46.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19693" PostTypeId="4" CreationDate="2017-01-10T01:50:46.183" Score="0" Body="For questions pertaining to the structure, dynamics and interactions with Earth's natural satellite." OwnerDisplayName="user15217" LastEditorDisplayName="user15217" LastEditDate="2017-01-10T11:52:26.627" LastActivityDate="2017-01-10T11:52:26.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19694" PostTypeId="1" CreationDate="2017-01-10T03:07:57.790" Score="7" ViewCount="136" Body="&lt;p&gt;Several missions to the Moon have revealed a &lt;strong&gt;lunar horizon glow&lt;/strong&gt;, such as the ones seen below by the spacecraft Surveyor 7 in 1968:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6jFE7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6jFE7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image source: &lt;a href=&quot;https://www.nasa.gov/centers/goddard/news/features/2010/lhg.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the NASA article &quot;&lt;a href=&quot;https://www.nasa.gov/centers/goddard/news/features/2010/lhg.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Model Helps Search for Moon Dust Fountains&lt;/a&gt;&quot; written in 2010 suggests the cause of the phenomenon is due to sunlight shining through a dust layer suspended in the very thin atmosphere (exosphere) of the moon.  From the NASA webpage:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;It has been suggested that electrostatic forces play a role in the ejection of dust from the lunar surface, and its dynamics in the atmosphere, but we &lt;strong&gt;really don’t understand how it gets there in such high abundances&lt;/strong&gt;.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given that there is no wind or precipitation in the lunar sky, my question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What causes the abundance of dust to 'float' in the thin lunar atmosphere?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15217" LastEditorUserId="2365" LastEditDate="2018-12-07T17:53:04.047" LastActivityDate="2018-12-08T10:35:32.150" Title="What causes the abundance of dust to 'float' in the thin lunar atmosphere?" Tags="&lt;the-moon&gt;&lt;planetary-atmosphere&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19695" PostTypeId="2" ParentId="19694" CreationDate="2017-01-10T03:07:57.790" Score="7" Body="&lt;p&gt;A recent study in Geophysical Research Letters — &quot;&lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/2016GL069491/full&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dust charging and transport on airless planetary bodies&lt;/a&gt;&quot; (Wang et al. 2016) and reported in the ScienceAlert article &quot;&lt;a href=&quot;http://www.sciencealert.com/nasa-just-explained-why-moon-dust-is-levitating-above-the-lunar-surface&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA just explained why Moon dust is 'levitating' above the lunar surface&lt;/a&gt;&quot; — seems to explain the phenomenon through laboratory simulation experiments which observe what occurs when micron-sized dust is exposed to ultraviolet (UV) radiation or electrically-charged gases known as plasmas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From these experiments, the researchers observed that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In both cases, the dust particles would leap several centimetres above the surface.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The experiment observations can be seen below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zKW0Y.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zKW0Y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image source: Science Alert article linked above&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, one can conclude that on the Moon light would scatter through these levitating particles to create the horizon glow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism behind the 'levitation' is stated as being:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This is due to the reaction [&lt;em&gt;with the plasma and UV light&lt;/em&gt;] causing an emission and re-absorption of electrons inside 'micro-cavities' formed between neighbouring particles, which can generate unexpectedly large electrical charges and intense repulsive forces.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;These forces cause the dust particles or clusters to lift off the&#xA;  surface, or 'levitate'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although the effects are observed with the lunar horizon glow, the researchers expect that effect would cause a suspension of dust across the entire lunar surface to some degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A significant implication of this research is stated in the NASA article &quot;&lt;a href=&quot;https://www.nasa.gov/feature/ames/dust-floats-above-lunar-surface-electrostatic-dust-transport-reshapes-surfaces-of-airless&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dust “Floats” Above Lunar Surface: Electrostatic Dust Transport Reshapes Surfaces of Airless Planetary Bodies&lt;/a&gt;&quot; as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The laboratory observations also showed dusty surfaces becoming smooth as a consequence of dust mobilization. These electrostatic dust processes could help to explain the formation of the “dust ponds” on asteroid Eros and comet 67P, and the unexpectedly smooth surface on Saturn’s icy satellite Atlas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user15217" LastEditorUserId="7411" LastEditDate="2018-12-08T10:35:32.150" LastActivityDate="2018-12-08T10:35:32.150" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19696" PostTypeId="2" ParentId="19691" CreationDate="2017-01-10T05:58:45.530" Score="0" Body="&lt;p&gt;Yes, you're right. The current inclination of the earth's axis of rotation to the plane of orbit around sun is 23.43708° degrees. &#xA;The sun can effectively cover the area from latitudes +23.43708° to -23.43708° on a sphere. Ignoring the night side of the celestial sphere gives the required area as Sin(23.43708°) = 39.774% ~40%.&lt;/p&gt;&#xA;" OwnerUserId="14723" LastActivityDate="2017-01-10T05:58:45.530" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19697" PostTypeId="1" CreationDate="2017-01-10T09:38:14.817" Score="4" ViewCount="129" Body="&lt;p&gt;The question says it all: are there observations of something like &lt;a href=&quot;http://www.dtic.mil/dtic/tr/fulltext/u2/a464576.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;non thermal filaments&lt;/a&gt; in other Galaxies other than our?&lt;/p&gt;&#xA;" OwnerUserId="9860" LastEditorUserId="9860" LastEditDate="2017-01-10T14:10:26.337" LastActivityDate="2017-01-10T16:36:21.797" Title="Are there non thermal filaments outside our Galaxy?" Tags="&lt;galaxy&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19698" PostTypeId="2" ParentId="18563" CreationDate="2017-01-10T11:03:14.990" Score="3" Body="&lt;p&gt;Just to add to the great answers already posted: They are all concerned with the lunar cycle and rise/set time because these are mostly what governs whether you see the Moon or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;You mention that you also want light pollution and other viewing conditions taken into consideration and these are all near insignificant compared to the lunar cycle and rise/set time (i.e. position)&lt;/strong&gt;. While light pollution can easily prevent you from observing stars it doesn't really matter when it comes to the Moon. It is so bright that you can see it even in the crescent phases - just like you can often see the Moon during daylight. &lt;a href=&quot;https://en.wikipedia.org/wiki/Limiting_magnitude&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; puts the limiting apparent magnitude in Manhattan at 2, while the Moon doesn't go above 0 (it's at around -12 at full). Actually, the best viewing conditions for astronomers are when the Moon is not visible because it is a big source of light pollution itself.&#xA;Other than that it might be cloudy but often you can catch a glimpse of the Moon or at least you can see some of the light from it through the clouds. Clouds also tend to disappear eventually so really, the most important aspects are the ones already mentioned in answers/comments. If you live near a mountain or tall buildings the Moon could certainly hide behind them but I don't think that counts.&lt;/p&gt;&#xA;" OwnerUserId="14899" LastActivityDate="2017-01-10T11:03:14.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19699" PostTypeId="2" ParentId="19683" CreationDate="2017-01-10T13:31:40.727" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does a year last the same time as the four seasons?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because that is how a year is defined. Our year is based on the time from one March equinox to the next. The yearly cycle of equinoxes and solstices is in turn driven by the Earth's axial tilt. This axial tilt is what drives the seasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiple ancient societies independently discovered the relationship between how these cycles and when was the best time to plant and time to harvest. They also discovered that one year is a bit more than 365 days long, and curiously, that the stars do not quite follow the same cycle. Over the course of 70.5 years, the date on which a star rose at a certain location shifted by one day. We base our year on the Sun, not on the stars, and because we base it on the Sun, a year stays in sync with the seasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an issue here, which is that a year is a bit more than 365 days long. Our civil calendar is based on days rather than years. The Romans addressed this issue by adding a leap day every fourth year. This made a year 365.25 days long, on average. The correct value needed to stay in sync with the seasons is a year that is 365.2421875 days long, on average.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This became a problem (rather than just an issue) during the Renaissance. The discrepancy between 365.25 days and 365.2421875 days had made the seasons fall out of sync with the calendar by 10 days at the time of Pope Gregory XIII. Most importantly (to the Roman Catholic church, that is), properly calculating the date on which Easter fell was in jeopardy. The Julian calendar had 100 leap years for every 400 years. The Gregorian reform made 10 days vanish (or more, depending on when a country adopted the Gregorian calendar), and made only 97 years out of 400 be leap years. This makes the Gregorian year be 365.2425 days long, on average. This is considerably closer to the ideal of 365.2421875 days than is the 365.25 days used in Roman times.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-01-10T13:31:40.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19700" PostTypeId="2" ParentId="19697" CreationDate="2017-01-10T16:36:21.797" Score="2" Body="&lt;p&gt;You have two questions here, which are somewhat mutually exclusive. I'll start with the second.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Are there observations of something like non thermal filaments in other galaxies other than our?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.dtic.mil/dtic/tr/fulltext/u2/a464576.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper you linked&lt;/a&gt; pretty plainly spells out that only a handful of nonthermal filaments (NTF) have been found, and all of them have been found near our galactic core. To quote the paper:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Observationally, the NTFs are unique to the GC&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Specifically they mention that there were only 9 known NTFs previous to this paper:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;nine previously identified isolated NTFs—the most recent NTF discovery is described by Reich (2003)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;and this paper reports that they found another 20:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;New high-resolution, wide-field 90 cm VLA observations of the Galactic Center (GC) region by Nord and coworkers have revealed 20 nonthermal filament (NTF) candidates.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So it seems the answer to your question, at least according to this paper, is that no, there are no observations of NTFs in other galaxies at this point in time.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Are there non thermal filaments outside our Galaxy?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;I interpret this to mean, &lt;em&gt;could there exist&lt;/em&gt; NTFs in other galaxies, and not so much have we actually seen NTFs in other galaxies. I think the answer to this, despite being a total guess with no proof to back it up, is a resounding yes. There's no reason to think our galaxy and it's structure/evolution is unique. Very likely NTFs exist in many other galaxies which formed under similar processes.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nrao.edu/pr/2004/filaments/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article by the NRAO&lt;/a&gt; details an attempt to explain where these NTFs originate from and the proposed solution is that they are a result of starburst activity. To quote from the article:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Yusef-Zadeh determined that a number of the non-thermal filaments seemed to connect to concentrated areas of thermal emission, which identify pockets of star formation. Galatic Center Combined radio image from the Very Large Array and Green Bank Telescope. The linear filaments near the top are some of the nonthermal radio filaments (NRFs) studied by the researchers. Other features, such as supernova remnants (SNRs) and the area surrounding our Galaxy's supermassive black hole (Sgr A) are shown.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;What this showed us is that two seemingly disparate processes, thermal and non-thermal radio emission, can be created by the very same phenomenon,&amp;quot; said Yusef-Zadeh. &amp;quot;In this case, that phenomenon is pockets of starburst activity.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If this is indeed the source of NTFs, then any galaxy which has or did have starburst formation would likely have NTFs as well.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-01-10T16:36:21.797" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19703" PostTypeId="1" CreationDate="2017-01-10T19:14:15.430" Score="3" ViewCount="109" Body="&lt;p&gt;According to &lt;a href=&quot;http://www.universetoday.com/36816/winds-on-venus/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.universetoday.com/36816/winds-on-venus/&lt;/a&gt;, the high altitude winds on Venus travel at about 100 m/s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think frictional resistance is really what's limiting the speed of the winds. Here's a possible cause of resistance to the winds moving even faster. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Venus has a single convection cell per hemisphere. &lt;/li&gt;&#xA;&lt;li&gt;The atmosphere cools by adiabatic expansion as it rises at the equator, &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;then because it's colder than the low altitude air, it emits radiation at a lower rate than the low altitude air &lt;/li&gt;&#xA;&lt;li&gt;so it heat's by absorbing radiation from the low altitude air faster than it cools by emitting radiation &lt;/li&gt;&#xA;&lt;li&gt;so the temperature drops less rapidly with altitude than that from adiabatic expansion. &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;As the air at the equator rises, it cools &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;but then it's colder than the rest of the air at that altitude &lt;/li&gt;&#xA;&lt;li&gt;so it resists being accelerated upward by the high pressure of the air under it. &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Also because of that resistance, it rises slowly enough that it heats up from absorption fast enough to not drop further in temperature as it goes higher preventing it from exerting a stronger force of resistance to the flow of the convection cells than the driving force for the convection.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13685" LastEditorUserId="6" LastEditDate="2017-01-11T13:21:38.303" LastActivityDate="2017-01-11T13:27:38.507" Title="What mechanisms limit the top speed of high altitude winds on Venus?" Tags="&lt;planet&gt;&lt;weather&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19706" PostTypeId="2" ParentId="19703" CreationDate="2017-01-11T13:27:38.507" Score="4" Body="&lt;p&gt;I found some recent papers which, if nothing else, discuss apparent mechanisms.&#xA;Quoting from &lt;a href=&quot;http://lasp.colorado.edu/~espoclass/ASTR_5835_2015_Readings_Notes/Schubert_Et_Al-EVTP.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schubert, 2016&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The large-scale circulation of the upper atmosphere from ~90 to ~200&#xA;  km altitude (upper mesosphere and thermosphere1 ) is a combination of&#xA;  two distinct flow patterns: (1) a relatively stable&#xA;  subsolar-to-antisolar (SS-AS) circulation cell driven by solar (EUV-&#xA;  UV) and IR heating, and (2) a highly variable retrograde superrotating&#xA;  zonal (RSZ) flow, in part a continuation of the lower-atmosphere RSZ&#xA;  flow discussed above&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, &lt;a href=&quot;http://sci.esa.int/venus-express/54065-4-super-rotation-is-speeding-up/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ESA&lt;/a&gt; says, in part,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 2006, average cloud-top wind speeds between latitudes 50° on either&#xA;  side of the equator were clocked at roughly 300 km/h. However,&#xA;  detailed cloud tracking studies revealed that these already remarkably&#xA;  rapid winds are becoming even faster, increasing to 400 km/h over the&#xA;  course of the mission. &lt;strong&gt;The reason for this dramatic increase is&#xA;  unknown&lt;/strong&gt;.&#xA;  [emphasis mine]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternatively,  &lt;a href=&quot;https://arxiv.org/pdf/1609.06549.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;these guys&lt;/a&gt; think they have a working model:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The atmospheric circulation in Venus is well known to exhibit strong&#xA;  super-rotation. However, the atmospheric mechanisms responsible for&#xA;  the formation of this super-rotation are still not fully understood.&#xA;  In this work, we developed a new Venus general circulation model to&#xA;  study the most likely mechanisms driving the atmosphere to the current&#xA;  observed circulation. Our model includes a new radiative transfer,&#xA;  convection and suitably adapted boundary layer schemes and a dynamical&#xA;  core that takes into account the dependence of the heat capacity at&#xA;  constant pressure with temperature. The new Venus model is able to&#xA;  simulate a super-rotation phenomenon in the cloud region&#xA;  quantitatively similar to the one observed. The mechanisms maintaining&#xA;  the strong winds in the cloud region were found in the model results&#xA;  to be a combination of zonal mean circulation, thermal tides and&#xA;  transient waves. In this process, the semi-diurnal tide excited in the&#xA;  upper clouds has a key contribution in transporting axial angular&#xA;  momentum mainly from the upper atmosphere towards the cloud region.&#xA;  The magnitude of the super-rotation in the cloud region is sensitive&#xA;  to various radiative parameters such as the amount of solar radiative&#xA;  energy absorbed by the surface, which controls the static stability&#xA;  near the surface. In this work, we also discuss the main difficulties&#xA;  in representing the flow below the cloud base in Venus atmospheric&#xA;  models&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-01-11T13:27:38.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19707" PostTypeId="1" CreationDate="2017-01-11T17:13:43.363" Score="4" ViewCount="54" Body="&lt;p&gt;A black hole's gravity prevents light from escaping, how can a black hole emit X-rays? Visible light and X-rays are both electromagnetic radiation, shouldn't the black hole's gravity prevent X-rays from escaping?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about &lt;a href=&quot;https://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hawking radiation&lt;/a&gt;  that is predicted to be released by &lt;a href=&quot;https://en.wikipedia.org/wiki/Black_hole&quot; rel=&quot;nofollow noreferrer&quot;&gt;black holes&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="15347" LastActivityDate="2017-01-11T17:13:43.363" Title="How can X-rays be emitted from a black hole?" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;hawking-radiation&gt;&lt;x-ray&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2017-01-11T18:44:53.277" ContentLicense="CC BY-SA 3.0" />
  <row Id="19708" PostTypeId="2" ParentId="15008" CreationDate="2017-01-11T19:16:14.223" Score="1" Body="&lt;p&gt;&lt;em&gt;Convert commments to community wiki&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a standard problem with using an arctan function (inverse tan function returns a value between -90 and +90). A lot of programming languages have two functions the simple one like you appear to have used and a more complex one that takes two parameters for when you are dividing (before the arctan function) as in this case. (This is described in further detail at (&lt;a href=&quot;http://youtu.be/_JbkTK17MrY&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://youtu.be/_JbkTK17MrY&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Java you can use &lt;code&gt;α=atan2((cos ε∗sin λ−sin ε∗tan β), cos λ)&lt;/code&gt; you may need to add 360 if the angle returned is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative is &lt;code&gt;α=atan2(cosβsinλcosε−sinβsinε,cosβcosλ)&lt;/code&gt;. This is equivalent in most cases but avoids an infinite term &lt;code&gt;tanβtan⁡β&lt;/code&gt; at the ecliptic poles, costing one more trig function call. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-01-11T19:16:14.223" CommentCount="0" CommunityOwnedDate="2017-01-11T19:16:14.223" ContentLicense="CC BY-SA 3.0" />
  <row Id="19709" PostTypeId="1" AcceptedAnswerId="19722" CreationDate="2017-01-11T20:43:58.213" Score="1" ViewCount="2691" Body="&lt;p&gt;how large, massive, far away (from earth), and fast would an object (planet-like) have to be to block out the Sun for 3 hours and not disturb Venus, Earth, or the Moon?&lt;/p&gt;&#xA;" OwnerUserId="15471" LastActivityDate="2020-02-04T16:21:11.977" Title="Could an object block out the Sun from Earth's view for 3 hours?" Tags="&lt;earth&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="19711" PostTypeId="1" CreationDate="2017-01-12T00:29:39.907" Score="2" ViewCount="651" Body="&lt;p&gt;So i want to know why world changed from the empty state ( no materials, energy and ...) from to the it's staring state in the big bang shape?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;based of this answer's to this question:&#xA;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/2342/big-bang-thoery-what-triggered-it-and-why?rq=1&quot;&gt;big-bang-thoery-what-triggered-it-and-why&lt;/a&gt;&#xA;they said:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You may think of the pre-Planck and the Planck epoch as a state of the&#xA;  universe, which is dominated by quantum fluctuations. They don't need&#xA;  a reason or a cause to occur. They are a consequence of Heisenberg's&#xA;  uncertainty principle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So,Did any law like uncertainty principle caused this to be happened or any somethings? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one meaningless question?!!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot.&lt;/p&gt;&#xA;" OwnerUserId="15472" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2017-02-06T11:23:49.413" Title="Did uncertainty principle triggered big bang?" Tags="&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ClosedDate="2017-01-13T15:30:14.983" ContentLicense="CC BY-SA 3.0" />
  <row Id="19713" PostTypeId="1" AcceptedAnswerId="19715" CreationDate="2017-01-12T08:05:42.847" Score="9" ViewCount="181" Body="&lt;p&gt;There are between &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Stable_nuclide&quot; rel=&quot;nofollow noreferrer&quot;&gt;90 and 254 stable nuclei&lt;/A&gt; all the way up to element number 82. In discussions and graphs about &lt;A HREF=&quot;http://www.astro.ucla.edu/~wright/BBNS.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;big bang nucleosynthesis&lt;/A&gt; nothing above lithium is even mentioned. It's a pretty safe bet that none of the heavier elements have been observed in reasonably primordial gases, since the implications would be profound enough to require mention in even popular level books about the big bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, has anyone bothered to perform searches for all of the heavier elements? As in, do we have explicit experimental upper limits on the concentrations of some or all of these heavier elements, or just a more qualitative, &quot;No evidence seen for anything else&quot;?&lt;/p&gt;&#xA;" OwnerUserId="14244" LastEditorUserId="14244" LastEditDate="2017-01-12T08:39:36.417" LastActivityDate="2017-01-14T20:50:29.750" Title="How Good Are the Upper Limits on Heavy Elements?" Tags="&lt;cosmology&gt;&lt;big-bang-theory&gt;&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19715" PostTypeId="2" ParentId="19713" CreationDate="2017-01-12T08:56:44.920" Score="6" Body="&lt;p&gt;Good question! Normally, the lack of heavy elements is taken as an indicator that something was made from (close-to-primordial) gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So an answer could be to ask what are the lowest abundances ever measured with respect to hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not completely up to date with the current record holders, but stars with iron abundances that are 5 orders of magnitude less than in the Sun have been found (&lt;a href=&quot;https://inis.iaea.org/search/search.aspx?orig_q=RN:44121795&quot; rel=&quot;noreferrer&quot;&gt;Norris et al 2013&lt;/a&gt;). This corresponds to A(Fe)$\sim 2$ on the usual logarithmic scale where hydrogen has A(H)=12.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constraints on other iron-peak elements are similar. Alpha elements like O, Mg are usually enhanced in very metal poor stars, so the constraints are an order of magnitude higher.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-01-12T08:56:44.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19717" PostTypeId="2" ParentId="19713" CreationDate="2017-01-12T19:48:24.177" Score="0" Body="&lt;p&gt;After astronomers observed the collision of two neutron stars back in June of 2013, a theory emerged that posits that most (all?) of the elements heavier than iron were synthesized in neutron star-neutron star or black hole-neutron star collisions. There is an article in &lt;a href=&quot;http://www.smithsonianmag.com/science-nature/all-the-gold-in-the-universe-could-come-from-the-collisions-of-neutron-stars-13474145/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Smithsonian&lt;/a&gt; that has a decent explanation.&lt;br&gt;&lt;br&gt;It's my understanding that this theory has yet to gain widespread acceptance, though the authors seem to make a pretty convincing argument in favor of it. Time and further observations may seal the deal or not.&lt;br&gt;&lt;br&gt;&lt;b&gt;Edit:&lt;/b&gt; Adding some addition references: from &lt;a href=&quot;https://phys.org/news/2011-09-cosmic-forging-gold.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Physics.org&lt;/a&gt;. And here is the paper the &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/2041-8205/738/2/L32/meta;jsessionid=5B0AF854DFCF7FEBB63068FD10E827C1.c2.iopscience.cld.iop.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;article cites&lt;/a&gt;. And from &lt;a href=&quot;http://aasnova.org/2016/09/21/colliding-neutron-stars-as-the-source-of-heavy-elements/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nova&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="7108" LastEditorUserId="7108" LastEditDate="2017-01-14T20:50:29.750" LastActivityDate="2017-01-14T20:50:29.750" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19718" PostTypeId="2" ParentId="19666" CreationDate="2017-01-12T20:12:03.117" Score="0" Body="&lt;p&gt;The most distant parallax measurements are made with very long baseline radio interferometry, mapping the apparent motions of clumps of water vapor masers.  I think the distance record is still held&lt;sup&gt;1&lt;/sup&gt; by NGC 4258, but the method is applied to many of the galaxies in the Local Group, as well as to a number of Galactic water masers at distances beyond 10 kpc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;[1] See a recent &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0004-637X/775/1/13/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;2013 paper&lt;/a&gt; giving ~7.6 Mpc or 25 million light years away.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="11023" LastEditorUserId="11222" LastEditDate="2017-03-13T11:53:32.513" LastActivityDate="2017-03-13T11:53:32.513" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19719" PostTypeId="2" ParentId="18031" CreationDate="2017-01-12T20:32:45.373" Score="2" Body="&lt;p&gt;&lt;em&gt;Comments formed into community wiki&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Satellites can and have been seen through telescopes. Possibly it is just a very small or faint satellite which requires a telescope to see and is too difficult to detect by the naked eye. The description of a &quot;moderate speed in a straight line and eventually dimming out&quot; sounds like every satellite. A piece of space junk is also possible. Lots of second stage boosters and other rocket parts floating around up there that are small enough that you would only notice them through a telescope &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be able to identify it by going to the &lt;a href=&quot;http://heavens-above.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heavens Above&lt;/a&gt; site and providing your location, the date &amp;amp; time.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-01-12T20:32:45.373" CommentCount="0" CommunityOwnedDate="2017-01-12T20:32:45.373" ContentLicense="CC BY-SA 3.0" />
  <row Id="19722" PostTypeId="2" ParentId="19709" CreationDate="2017-01-12T21:51:48.297" Score="5" Body="&lt;p&gt;Is it possible?   Not really.  Not with your criteria.   Could some mad scientists make it happen, like if we make the Moon bigger?   Er, maybe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.timeanddate.com/eclipse/eclipse-information.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Types of eclipses&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/p8VCD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p8VCD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.timeanddate.com/eclipse/eclipse-information.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nasa's eclipse page&lt;/a&gt; (scroll down for maps over 20 year periods).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2001-2020 map below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iSg9W.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iSg9W.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The map has a notation for hybrid eclipses (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_eclipse#Types&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;), not all that relevant to the question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total solar eclipses (blue lines on the map), are quite small because the &lt;a href=&quot;https://en.wikipedia.org/wiki/Umbra,_penumbra_and_antumbra#Umbra&quot; rel=&quot;nofollow noreferrer&quot;&gt;umbra&lt;/a&gt; or fully shaded region that the Moon casts on Earth is small, maybe 150 km in diameter on average, 267 km max.  All planets and objects that orbit the sun cast an Umbra that extends behind them.   The Moon (diameter 3,474 km), is always casting an umbra somewhere in space, being about 1/400th the diameter of the sun, it's umbra is about 1/399th the distance between the sun and the moon, or roughly 375,000 km on average.   Eclipses happen on Earth when the Earth is lined up below the Moon's Umbra.  Mostly they don't line up which is why eclipses are rare, but the Umbra is always somewhere in space, not all that far from the Earth, following the moon's path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CXQN9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CXQN9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rogue Planet&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your scenario mentioned planet-like.  Now, if it's a planet from our solar-system, that would imply disturbing the orbits of Earth/Venus/Moon, which you specified you didn't want.   There's also real danger if you were to have a planet like object in a near Earth orbit that passes from time to time, close enough to cast it's Umbra over Earth.   For safety reasons &quot;just passing through&quot; is the way to go.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rogue planet on a &lt;a href=&quot;https://en.wikipedia.org/wiki/Hyperbolic_trajectory&quot; rel=&quot;nofollow noreferrer&quot;&gt;hyperbolic trajectory&lt;/a&gt; past the sun that, by very rare coincidence, passes between the Earth and the Sun, but close enough to cast it's Umbra over Earth would be very rare to say the least, but a theoretically possible scenario to cast a total eclipse over Earth, but it would need to pass very close and at just the right angle to block the 1/2 of 1 degree of arc that the Sun takes up in the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How close?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars is roughly twice the diameter of the Moon, so would need to pass within about 750,000 km to the earth, and line up just right, to cast a tiny umbra that moved across the Earth.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus, about 3.5 times the diameter, about 1.3 million km&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter, about 41 times the diameter of the moon, would need to pass within about 15 million KM to cast an umbra on the Earth of similar size to the typical Moon's eclipse Umbra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And those are the minimum distances.  You'd need the theoretical planet to be pretty well inside those limits to have an Umbra of significant size and for a 3 hour eclipse, bigger is better. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What would a rogue planet passing that close look like?&lt;/strong&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider that it's passing close enough that it looks larger than the Moon, at least for a 3-plus hours.   How close it would need to be depends on the size, as noted above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimum velocity for an object on a hyperbolic trajectory just passing through the solar system but near Earth distance would be Earth's escape velocity, or the orbital velocity of the Earth (108,000 kph) times the square root of 2, and if it approached on the same plane and in the same direction as the Earth, the relative velocity between this Rogue planet and Earth, the minimum relative velocity would be about 45,000 kph, give or take.   That's absolute bare minimum as a theoretical rogue planet passing inside the Earth's orbit would usually be traveling a fair bit faster than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for a planet like that to remain between the Earth and the Sun for 3 hours, it would need to be about 3 times 45,000 km in diameter, or, nearly the size of Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small rogue planet couldn't cast an Umbra over Earth anywhere close to 3 hours because it would pass over the Earth much too quickly.   A Jupiter-sized Rogue planet could, theoretically, if it was closer than closer than 15 million km, cast it's umbra over Earth for 3 hours or so, but that would be close enough to change the Earth's orbit and, at the very least, give the Earth significant climate change by pushing the Earth a few million km either closer or further from the sun.    Not an ideal scenario.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there's no scenario that does what you ask.  You'd need a planet the size of Jupiter, with the mass of Neptune or perhaps less to leave the Earth/Moon undisturbed.    A Neptune mass planet with a large ring system . . . just maybe.   But given how rare your scenario is to begin with, I'm going to say basically no.   You could do it, but it would change the Earth's orbit, or you could do it with a smaller planet but for much less than 3 hours.  In any case, it would be a preposterously rare scenario no matter what.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What if we move the Moon?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With rogue planets, I just looked at the relative velocity between the planets, which is close enough for a near-sighted umpire.  When you look at the Moon's umbra, the Moon's motion is much slower, so you also have to take into account the Earth's rotation too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon (Synodic period), takes 59 minutes to cross the sun, that's tip to full cover, and another 59 minutes back to tip.  (based on a 29.5 day synodic orbit, 29.5 days to cover 360 degrees, the sun, roughly 1/2 of 1 degree in the sky).   So, 158 minutes of partial eclipse, and a record, of 7 minutes total eclipse in one spot on Earth (7 minutes, the record for a total eclipse is possible when the Moon is slightly larger than the sun, close to it's &lt;a href=&quot;https://en.wikipedia.org/wiki/Apsis&quot; rel=&quot;nofollow noreferrer&quot;&gt;perigee&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if our mad scientists pull the moon closer to the Earth, the period of time for a total eclipse increases but the rate that the Moon moves across the sun also increases, so you're effectively increasing the 7 minutes of total but decreasing the 158 minutes of partial.   If you pull the Moon all the way into a geostationary orbit, some 10.5 times closer, the Moon would span an enormous 5.25% of the sky, but it wouldn't move in the sky while the sun would and the sun, which crosses about 1 degree of sky in 4 minutes would pass behind the enormous geo-stationary moon in just 21 minutes.   You can't create a 3 hour eclipse by pulling the Moon closer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where the Earth' rotation speed comes in.  The Moon moves over the Earth in it's orbit at about 3,683 kph, and relative to the sun, but because the Earth is in orbit to, the effective velocity relative to the sun (&lt;a href=&quot;https://community.dur.ac.uk/john.lucey/users/lunar_sid_syn.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Synodic&lt;/a&gt; not sidereal) is about 8% less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon's umbra effectively moves with the Moon's relative velocity - at least, close enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth rotates in the same direction that the Moon moves, peak velocity at the equator of a bit over 1,600 kph.   In effect, the rotation of the Earth tries to keep up with the movement of the Umbra during an eclipse, but the Umbra moves West to East about twice as fast as the Earth turns West to East.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the solution for your 3 hour eclipse?   Make the Moon bigger and push it further away so it orbits the Earth more slowly and it's Umbra moves closer to the Earth's rotation speed, making a 3 hour eclipse, at mid day, close to the equator . . . possible.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we replace the Moon with Venus (3.5 times the diameter of the Moon) and we push it some 3 times further away (which would be pushing the limits of the stability of the Earth's hill sphere), then you'd be a lot closer, as our new &quot;Venus Moon&quot; would move across the sky and across the sun still to fast, but it would move more slowly and closer to Earth's rotational velocity leading to eclipses that stayed in the sky much longer.  If you did that, you might get your 3 hours of total eclipse.  You wouldn't want to push Venus too far, cause you'd want it a big bigger than the Sun in the sky, but 3 times the Moons distance, it would be about 15% larger and close to the limit of orbital stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other silly ideas.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you were to Move the Planet Uranus into Earth's L1 Trojan point, which . . . well, the Trojan point would be further away due to the larger planet's mass dominating so . . . that wouldn't work and because L1 points aren't stable, it would also be highly dangerous.   Moving on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to move Uranus to where the Earth is and make the Earth a satellite of Uranus . . . we might get longer than 3 hour eclipses every time we were on the far side of Uranus making an orbit.   Seems a lot of work though just for an eclipse. . . . but that would work too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, I'll stop now.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-04-26T05:38:54.903" LastActivityDate="2018-04-26T05:38:54.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19723" PostTypeId="2" ParentId="19658" CreationDate="2017-01-12T22:18:09.923" Score="4" Body="&lt;p&gt;The orbit of the moon is stable on a timescale of billions of years. Low Earth Orbit is not stable in the long term. However, Geostationary orbits are very long lived. The satellites in geostationary orbit will eventually be perturbed (some to collide with the Earth, some probably to escape) But this will take much longer than human lifespans. For practical purposes the geostationary orbits are stable. Geostationary satellites use thrusters for station keeping, not to prevent their orbits from decaying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond the Earth, it is believed that the planet's orbits are stable. There is a very small chance that a resonance between Mercury and Jupiter could develop, and that could cause Mercury's orbit to change substantially (or even collide with another planet), over the course of about 3 billion years. More generally the solar system is known to be technically &lt;em&gt;chaotic&lt;/em&gt;. While the orbits of the planets won't change, it is not possible to predict the position of the planets after about 50-100 million years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stability of the solar system and Geostationary orbit is in contrast to truly unstable orbital configurations. LEO is unstable due to atmospheric drag. Many 3-body configurations are unstable, and result in the ejection of one member of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A three body system may be periodic but unstable, in the sense that a perturbation of the system results in chaos, and the ejection of one of the members of the system. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-01-12T22:18:09.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19724" PostTypeId="1" CreationDate="2017-01-12T22:35:53.890" Score="1" ViewCount="228" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ewodE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ewodE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have this image from a treasure hunt, I'm fairily certain it's a real piece of the sky, but I've been unable to identify it, can anyone help?&lt;/p&gt;&#xA;" OwnerUserId="15485" LastEditorUserId="11143" LastEditDate="2017-01-13T15:25:41.953" LastActivityDate="2020-10-13T19:04:05.450" Title="Can someone help me to identify this section of the night sky?" Tags="&lt;amateur-observing&gt;&lt;constellations&gt;&lt;night-sky&gt;" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19725" PostTypeId="2" ParentId="19711" CreationDate="2017-01-12T23:07:02.547" Score="1" Body="&lt;p&gt;I think there is a real question in there, just not the one that you think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gerald isn't saying that the uncertainty principle played a role in the big bang.   He's saying it played a role in and shortly after the Planck era which came after the big bang.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the big bang - nobody knows.   Really, that's the answer.  &quot;Was the big bang this&quot; answer &quot;We don't know&quot;.   Period.   End of discussion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't know what came before the big bang, We don't know if space was empty or not empty.  We don't know what physical laws may have applied.  Nobody knows.   The initial, basic theory was that the big bang happened at a point or a singularity, which if you think about it, doesn't make much sense cause gravity wins at a point that dense, but the big bang at a point idea does resemble, kinda/sorta a really crazy particle that, by something kinda like quantum tunneling, could become a universe - er, maybe.   String Theory brings other dimensions into this particle becomes a universe - idea, and again, maybe, but nobody knows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more modern model, but still, we don't know, is that the universe was infinite at the time of the big bang expansion, and it grew, not from a point but it expanded from infinite into infinite, which works mathematically and it makes (I think) a bit more sense than the super-particle model.  Still the answer is &quot;We don't know&quot;.   The answer is always &quot;We don't know&quot;.   There's also 4 dimensional models or two time dimension models (those are my favorite when you have x-axis and y-axis time), which are creative ways to avoid the singularity but . . . I'm repeating myself.  Nobody knows.  The answer to what happened at the big bang is always nobody knows.  We don't know if it was empty or not, we don't know if there was some 4th dimensional crossing of &lt;a href=&quot;https://en.wikipedia.org/wiki/M-theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;membranes&lt;/a&gt; which caused the 3d universe we live in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, on modeling back to the big bang, the experts can run the model backwards and apply known laws and have a theory that works to a time shortly after the big bang, which still isn't certainty, but it's a reasonable hypothesis using the laws of physics as they work today.  But it doesn't work all the way back to the big bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the paragraph that you quote, I think there's a solid question in there, but the quote doesn't mean what you think. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You may think of the pre-Planck and the Planck epoch as a state of the universe, which is dominated by quantum fluctuations. They don't need a reason or a cause to occur. They are a consequence of Heisenberg's uncertainty principle.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;So,Did any law like uncertainty principle caused this to be happened or any somethings?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gerald is talking about a period AFTER the big bang and what the models tell us about that period when the universe was, (theoretically), very dense and very hot and governed by quantum mechanics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps a better question, which I'm not qualified to answer but might be worth asking, is, &quot;&lt;em&gt;how did the uncertainty principle apply during and shortly after the Planck Era&lt;/em&gt;&quot; - at least, that's the question I take away from the paragraph you quoted.   Gerald didn't say the uncertainty principal applied to the big bang, only to the Planck Era.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="11264" LastEditDate="2017-02-06T11:23:49.413" LastActivityDate="2017-02-06T11:23:49.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19726" PostTypeId="2" ParentId="19658" CreationDate="2017-01-13T00:17:25.887" Score="6" Body="&lt;p&gt;There's no clearly defined definition of what a stable orbit is.  Orbits can last for hundreds or thousands of years, millions, billions or even trillions of years.  There's no neat and tidy definition between stable and unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hubble for example, orbits the Earth every 97 minutes, and in a few years, primarily slowed by the very slight air resistance, it is expected to fall to the Earth.   Many satellites, because their equipment doesn't last forever, are designed to gradually slow by air resistance and fall to the Earth after a few decades of operation.   Satellites can be positioned where they stay in orbit much longer simply by moving them a bit further out, but there's little benefit to that.   It's advantageous to have satellites fall to the Earth after they've served their usefulness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbits around the Moon are more unstable than orbits around the Earth for 2 reasons.   One, because the Moon is gravitationally lumpy so low orbits around the Moon don't stay in place very long.  See &lt;a href=&quot;https://science.nasa.gov/science-news/science-at-nasa/2006/06nov_loworbit&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and two, because the Earth is gravitationally dominant, so the Moon doesn't have a very big hill sphere.  Celestial mechanics doesn't like satellites to have satellites.   See &lt;a href=&quot;http://www.universetoday.com/109666/can-moons-have-moons/&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://curious.astro.cornell.edu/about-us/44-our-solar-system/the-moon/general-questions/104-can-moons-have-moons-intermediate&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.  Our moon could have a satellite, but it likely wouldn't be stable for very long, at least, not long in a celestial time-scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trojan points, L4 and L5 are generally stable in a System where the larger body (Earth) is more than 25 times as massive as the smaller body (moon), but our Moon's orbit is too eccentric and too chaotic for lunar Trojan points to be stable for any period of time.     Earth's L4 and L5 points around the sun are more stable and there are a few small asteroids in Earth's Trojan points.  The Moon's L4 and L5 aren't stable cause the Moon wobbles too much in it's orbit.  Eccentricity and perturbations tend to destabilize L4 and L5 points.   That said, the Moon's L4 and L5 are more stable than L1, L2 and L3.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars moon Phobos is in a &lt;a href=&quot;http://www.universetoday.com/14258/phobos-might-only-have-10-million-years-to-live/&quot; rel=&quot;noreferrer&quot;&gt;decaying orbit&lt;/a&gt; that may only last 10 million years or so.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our Moon, however, even though it's slowly moving away from Earth, is sufficiently stable that it will outlast our sun.  It's stable on the scale of many billions of years.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your question &quot;prolong an orbit via moon power&quot;, The moon tends to destabilize, not stabilize orbits around the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But while L1 and L2 points aren't gravitationally stable, they are still useful places for satellites because they require less adjustment.   Satalites with orbits that move around lagrange points are often used as useful locations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take this bizar orbit around the Moon, where the NASA satellite transfers from L2 to L1 and could, in theory, be transferred back and forth without expending too much energy.  This isn't a &quot;stable&quot; orbit by any means but it's an example of an unstable but useful orbit for the life of a satellite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8SqXF.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8SqXF.jpg&quot; alt=&quot;http://www.berkeley.edu/news2/2010/10/artemis_diagram.jpg&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://news.berkeley.edu/2010/10/27/artemis_moon_mission/&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The James Webb Telescope, when it's launched will orbit around Earth's L2 Lagrange point, which isn't stable, but it requires relatively small adjustments to stay in that order.  It's also useful because it will be nearly entirely shaded from the sun by the Earth and it works best in cold space, away from sunlight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rMqLq.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rMqLq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.nasa.gov/topics/universe/features/webb-l2.html&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those are 2 examples of useful but unstable orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter's orbit around the sun, for example, will probably be stable for trillions of years.   Jupiter's moons around Jupiter?  Hard to say, with 4 Galilean moons, it's difficult to predict their orbits over long periods of time.&lt;br&gt;&#xA;.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-01-13T00:17:25.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19727" PostTypeId="1" AcceptedAnswerId="19739" CreationDate="2017-01-13T01:16:10.783" Score="2" ViewCount="1291" Body="&lt;p&gt;Okay ... so, I woke up from a dream last night wherein I was (apparently) attempting to devise a time-keeping system for interstellar travel (similar to the stardate system used in Star Trek). In the dream, I was attempting to correlate our current calendar, based on Earth's movements relative to the sun, with this fictitious calendar using the positions of intergalactic stellar bodies in comparison to Milky Way. (i.e., What if we could tell time based on the way other galaxies moved in relation to our own galaxy?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in the dream, I was understanding that Milky Way is a spiral galaxy which rotates around a galactic center. That Earth is toward the middle-end of one arm of Milky Way's spirals. That almost every star we see in the night sky is &lt;em&gt;inside&lt;/em&gt; Milky Way with us, and that the galaxy Andromeda is a great distance beyond Milky Way and can be seen in the night sky over Earth in the same location (between constellations Cassiopeia and Pegasus) every night since (at least) 10th century C.E.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming Milky Way was graphed on a flat sheet of paper with the Galactic Center plotted at X=0,Y=0, and the arm which contains Earth was located at X=0,Y=-δ, and also assuming the galaxy Andromeda can be seen in earth's sky from this position ... why then, when Milky Way rotates around it's galactic center so that the arm which contains Earth is located at X=+δ,Y=0 (a rotation of 90°), or even X=0,Y=+δ (a rotation of 180°) is the galaxy Andromeda still visible in the same location in the sky?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless the galaxy Andromeda is in some sort of galactic-synchronous orbit with Milky Way (which seems unlikely since most of the posts referencing Andromeda here deal with the collision of the two galaxies), then, in the in either of the second rotational positions listed above, Andromeda should no longer be visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing? Why does Andromeda, which is not a part of Milky Way, only move in the night sky in relation to Earth's seasons and not in relation to any other galactic movement?&lt;/p&gt;&#xA;" OwnerUserId="15486" LastActivityDate="2017-01-16T17:08:22.207" Title="Why can I see the Andromeda Galaxy despite Milky Way's rotation?" Tags="&lt;orbit&gt;&lt;milky-way&gt;&lt;galaxy-center&gt;&lt;night-sky&gt;&lt;intergalactic-space&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19729" PostTypeId="2" ParentId="19724" CreationDate="2017-01-13T02:30:11.000" Score="0" Body="&lt;p&gt;I do not believe that is the real sky. The rings of Saturn are not that large with the naked eye. I am not sure what the other formation is either. Perhaps some one else could shed better information about the picture. &lt;/p&gt;&#xA;" OwnerUserId="15488" LastActivityDate="2017-01-13T02:30:11.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19731" PostTypeId="1" AcceptedAnswerId="19760" CreationDate="2017-01-13T07:12:58.110" Score="5" ViewCount="132" Body="&lt;p&gt;Let's say we're mapping Callisto's northern celestial hemisphere in a way similar to &lt;a href=&quot;https://i.imgur.com/FPiEE3l.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this map of Earth's&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Along with the stars (which let's just ignore for this question), this map will identify the paths of four objects:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The Sun&lt;/li&gt;&#xA;&lt;li&gt;Jupiter&lt;/li&gt;&#xA;&lt;li&gt;Ganymede (closer to Jupiter)&lt;/li&gt;&#xA;&lt;li&gt;Himalia (farther from Jupiter)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Where on the map would these objects' paths be? Or at least how would I find the exact answers with math/software/etc.?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My assumptions/guesses:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Jupiter's path is relatively similar to the ecliptic but Callisto's orbital inclination or maybe Jupiter's might separate the two&lt;/li&gt;&#xA;&lt;li&gt;Jupiter's moons are within a similar orbital plane to Callisto and would follow near Jupiter's path&lt;/li&gt;&#xA;&lt;li&gt;the distance Callisto travels away from the sun in its own orbit is insignificant in determining the shape of the ecliptic&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I have no idea if the axial tilt of Callisto or Jupiter would affect anything but it probably does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it helps, I've made this incredibly stylish and oh-so-fabulous template in case anyone wants to sketch it in Paint or something. The black dot at the center is the celestial north pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/W4q5A.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W4q5A.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your time!&lt;/p&gt;&#xA;" OwnerUserId="15489" LastActivityDate="2017-01-14T16:12:31.150" Title="How would a celestial map for Callisto depict the paths of Jupiter, the Sun, etc.?" Tags="&lt;orbit&gt;&lt;solar-system&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19733" PostTypeId="1" CreationDate="2017-01-13T09:09:44.890" Score="9" ViewCount="1013" Body="&lt;p&gt;In Sobel's famous book 'Longitude', she described the story of Harrison. I still have one question. What if it is cloudy on the sea? In this case, the sailors have nothing to compare with, right? This still means in many many cases, they have no means to determine their position, neither latitude nor longitude, right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that the problem is not so satisfactorily solved. &lt;/p&gt;&#xA;" OwnerUserId="11754" LastActivityDate="2017-01-14T05:51:32.387" Title="Determining latitude and longitude in bad weather" Tags="&lt;earth&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19734" PostTypeId="1" AcceptedAnswerId="19740" CreationDate="2017-01-13T09:33:32.803" Score="10" ViewCount="1143" Body="&lt;p&gt;I know that even light can't escape from a black hole's gravity and the velocity of light and gravitational waves are the same. How can only gravitational waves escape from its gravity?&lt;/p&gt;&#xA;" OwnerUserId="14078" LastEditorUserId="11143" LastEditDate="2017-01-13T13:55:04.110" LastActivityDate="2017-01-13T14:36:38.000" Title="How can gravitational waves escape from the gravity of black hole?" Tags="&lt;black-hole&gt;&lt;gravity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19735" PostTypeId="2" ParentId="19733" CreationDate="2017-01-13T10:58:05.800" Score="5" Body="&lt;p&gt;When it's cloudy at sea, the one measurement one still has is approximate sunrise or sunset, which can be used to determine longitude, but the uncertainty is very large, in particular at higher latitudes where sunrise and -set are slow.  Having both (length of day) one can estimate latitude as well.  Although one cannot determine those to the same precision and accuracy as in clear conditions, an experienced person should still be able to do a pretty good job.  If they are 10 minutes off in their estimate, that translates to around 2.5° of longitude.  And of course it means they can only make 2 measurements per 24 hours.  Between those, they'd have to extrapolate.&lt;/p&gt;&#xA;" OwnerUserId="1462" LastEditorUserId="1462" LastEditDate="2017-01-13T17:03:20.573" LastActivityDate="2017-01-13T17:03:20.573" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19736" PostTypeId="2" ParentId="19733" CreationDate="2017-01-13T12:08:20.947" Score="7" Body="&lt;p&gt;&lt;a href=&quot;http://www.livescience.com/27696-viking-sunstone-shipwreck.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sunstones&lt;/a&gt; are believed to have been used by vikings to determine the direction to the Sun on cloudy days. That helps you point the ship in the right direction, and by experience you can estimate the side ways drift in the water. But the longitude will still be very uncertain without precision time keeping. The news paper article I linked to says that a Sunstone was find at a wreckage in the English Channel from as late as 1592.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2017-01-13T12:08:20.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19737" PostTypeId="2" ParentId="19734" CreationDate="2017-01-13T12:32:33.743" Score="4" Body="&lt;p&gt;Gravitational waves are a distortion of space-time outside the black hole.  They don't have to escape, because they're already outside.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-01-13T12:32:33.743" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19738" PostTypeId="1" AcceptedAnswerId="19744" CreationDate="2017-01-13T13:24:43.630" Score="6" ViewCount="369" Body="&lt;p&gt;Suppose an alien race fired a laser beam from the centre of the Milky Way, it went through the Sun and then off further out of the galaxy.  What Constellation would the light beam head towards, closest star also if possible?  Am not interested about how much devastation the light beam may or may not cause, just the journey and destination.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would that direction be the quickest/shortest way out of the galaxy if we had a spaceship that could travel that fast?&lt;/p&gt;&#xA;" OwnerUserId="5401" LastActivityDate="2017-01-14T16:58:45.763" Title="Opposite Constellation of Sagittarius A*" Tags="&lt;milky-way&gt;&lt;constellations&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19739" PostTypeId="2" ParentId="19727" CreationDate="2017-01-13T14:01:05.260" Score="5" Body="&lt;p&gt;As barrycarter already pointed out, it is due to the huge time it takes for the milky way to rotate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For once, as he pointed out, the stars in an orbit similar like the sun take about 250 million years for one full revolution around the center of the galaxy. If you put that in a relation to an earth year and take our civilization of roughy 6000 years of keeping records in comparison, we exist roughy 12 minutes of our galactic year. If you only take the last 150 years of the modern era, it's down to 18 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you compare that, how much does Earth move around the sun in 18 seconds? That is about how much Andromeda has apparently moved relative to the constellations in the night sky since we can capture visual images on film.&lt;/p&gt;&#xA;" OwnerUserId="14441" LastEditorUserId="14441" LastEditDate="2017-01-13T14:06:49.953" LastActivityDate="2017-01-13T14:06:49.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19740" PostTypeId="2" ParentId="19734" CreationDate="2017-01-13T14:26:16.350" Score="7" Body="&lt;p&gt;I see this phrase all the time, and I have to say I've come to greatly dislike it because it's a very bad misnomer. Nine times out of ten, when someone is talking about a black hole, they describe it as an object with such strong gravity that &quot;not even light can escape&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this unqualified statement presents a strong misconception as to what black holes actually are and how they work and accomplishes nothing but confusing innocent bystanders such as yourself. The gravity of a black hole is no more or less strong than any other object in the universe. Black holes are not cosmic vacuums which use their powerful gravitational forces to suck up all nearby matter, light, etc. In fact, if you replaced our Sun with a black hole of exactly the same mass, all the planets in our system would go about orbiting exactly the same way and wouldn't notice a difference at all (aside from the mass extinction on Earth due to no longer receiving any energy from the Sun).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, let's paint a better picture of what a black hole is and how it works. A black hole is a clump of mass which has become so massive that the gravitational force of that mass on itself, trying to pull it together, actually collapses the mass into a singularity. The singularity is a point-like region of space where all the mass becomes contained. Slightly outside this singularity, the physics gets weird. For example, if you're right next to this singularity, and you calculate the speed needed to get away from that singularity (e.g., you need to travel ~11 km/s to get away from Earth) you find a speed which is much greater than the speed of light. That's the origin of the phrase &quot;not even light can escape&quot;. &lt;strong&gt;But&lt;/strong&gt;, if you start farther away from the singularity, you need less speed to escape it because you feel less gravitational pull from it (gravity decreases with distance). This means, at some distance from the singularity, the speed of light actually is fast enough to escape the black hole. This distance is so important that scientists have given it a special name, the &lt;em&gt;event horizon&lt;/em&gt;. It can get a lot more complicated than the simple picture I've painted above, but that's the general idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you put all that together, then that tells you that any light which is outside the event horizon has no trouble escape the black hole. It is only the light &lt;em&gt;inside&lt;/em&gt; this event horizon which cannot escape. Likewise, any gravitational waves outside the event horizon can escape just as easily. This is what the answer by StephenG meant by saying they were &quot;outside&quot; the black hole. By outside, he meant outside the event horizon. And it is true that as long as the creation of the gravitational wave happens outside the event horizon it will escape the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And just for reference of size, the supermassive black hole in the center of our galaxy, which is 4,000,000 times more massive than our Sun, has an event horizon which only extends ~10,000,000 km. That's barely out to the orbit of Mercury if it were in the position of our Sun. So you can see, it's not very hard to be outside the event horizon as the event horizon isn't that big in astronomical terms.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2017-01-13T14:36:38.000" LastActivityDate="2017-01-13T14:36:38.000" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19741" PostTypeId="1" CreationDate="2017-01-13T15:50:41.317" Score="7" ViewCount="971" Body="&lt;p&gt;Let there be an accelerating force acting on earth over an arbitrarily long time to change its orbit around the sun (e.g. by &lt;a href=&quot;https://www.theguardian.com/environment/2001/jun/10/globalwarming.climatechange&quot; rel=&quot;noreferrer&quot;&gt;asteroids repeatedly passing by close to earth (intentional or not)&lt;/a&gt; or whatever). For this question consider only the effect of earth's gravity on the moon (and vice versa), not other influences (e.g. not the immediate effect the asteroids passing by have on the moon's orbit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If such perturbations are weak enough yet of sufficient duration to change earth's orbit around the sun, how does it change the moon's orbit around earth? My considerations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4Fqvh.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4Fqvh.png&quot; alt=&quot;Sketch of effects of a force on earth on the moon&amp;#39;s orbit&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the moon remains in orbit around earth even as the earth goes around the sun there obviously is some lower limit to how fast earth can accelerate. However, that force wanders around earth once a year, hence its effect on the moon's orbit is evening out - and the effects from the sun are supposed to be ignored here, anyway, so let's have the force go in a constant direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force acting on earth can be decomposed into $F_\perp$ perpendicular to and $F_=$ coplanar with the moon's orbital plane. $F_\perp$ moves earth out of the plane of the moon's current orbit, so the moon's orbit will adjust to find a new plane: if the force is continuously applied in the same direction the moon's orbit will change until $F_\perp = 0$, so this factor can be neglected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_=$ will cause the moon's orbit to become very elliptical, the moon passing closer on the side toward which earth is accelerated and being further away on the other side. If the force is indefinitely sustained something will happen to the moon, but under which circumstances will which scenario happen?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the moon remains in orbit,&lt;/li&gt;&#xA;&lt;li&gt;the moon gets tidally torn apart,&lt;/li&gt;&#xA;&lt;li&gt;the moon crashes into earth,&lt;/li&gt;&#xA;&lt;li&gt;the moon leaves orbit of earth (though the moon will probably say that it's earth which ran away...),&lt;/li&gt;&#xA;&lt;li&gt;anything else interesting I forgot.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15494" LastEditorUserId="15494" LastEditDate="2017-01-13T18:48:50.837" LastActivityDate="2017-01-14T23:09:54.473" Title="If earth changed its orbit, what would happen to the moon?" Tags="&lt;gravity&gt;&lt;earth&gt;&lt;the-moon&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19742" PostTypeId="2" ParentId="19724" CreationDate="2017-01-13T16:09:10.687" Score="0" Body="&lt;p&gt;It doesn't seem to be real, moreover, as the other user has said, the rings of Saturn are not visible with the naked eyes.  It is merely seen a dot. I think you need a minimum of 30x magnification to see the rings. &lt;/p&gt;&#xA;" OwnerUserId="15476" LastActivityDate="2017-01-13T16:09:10.687" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19743" PostTypeId="1" CreationDate="2017-01-13T16:50:40.410" Score="6" ViewCount="492" Body="&lt;p&gt;Why doesn't the earth have a &lt;a href=&quot;https://en.wikipedia.org/wiki/Ring_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;ring&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If solid material such as dust and &lt;a href=&quot;https://en.wikipedia.org/wiki/Moonlet&quot; rel=&quot;nofollow noreferrer&quot;&gt;moonlets&lt;/a&gt;, and common components of &lt;a href=&quot;https://en.wikipedia.org/wiki/Satellite_system_(astronomy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;satellite systems&lt;/a&gt; are brought nearer to the Earth, then what will happen? Will it take the form of a ring or fall down on the Earth?&lt;/p&gt;&#xA;" OwnerUserId="15347" LastEditorUserId="2365" LastEditDate="2017-01-13T19:54:07.577" LastActivityDate="2017-01-13T19:54:07.577" Title="Can Earth have a ring?" Tags="&lt;gravity&gt;&lt;earth&gt;&lt;planetary-ring&gt;&lt;near-earth-object&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19744" PostTypeId="2" ParentId="19738" CreationDate="2017-01-13T16:59:12.713" Score="2" Body="&lt;p&gt;Per &lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_Center&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Galactic_Center&lt;/a&gt;, the galactic center &#xA;is at RA 17h 45m 40.04s, Dec -29° 00' 28.1&quot; in the J2000 epoch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point opposite this would be RA 5h 45m 40.04s, Dec +29° 00' 28.1&quot; &#xA;(J2000). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nearest star I could find to this position is a relatively &#xA;uninteresting 16th magnitude star: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6aA4T.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6aA4T.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nearest &quot;interesting&quot; star is HIP 27088, which is an 8th magnitude &#xA;double star:   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iwW2j.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iwW2j.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My calculations show this point is in Auriga, not Gemini, albeit just barely: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EMlxO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EMlxO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A slightly more zoomed out view: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EkrdY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EkrdY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @zephyr notes, the fastest way out of our galaxy would be &quot;straight up&quot; (perpendicular to the plane of the galaxy), not away from the center, since the Milky Way's &quot;height&quot; is much smaller than its length/width.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone wants to replicate what I did on Stellarium and use RA/DEC grids, be sure to set the date to 1999 Dec 31 at noon GMT: the grids Stellarium displays are based on the current date, not J2000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: &lt;a href=&quot;http://www.usno.navy.mil/USNO/astrometry/optical-IR-prod/nomad&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.usno.navy.mil/USNO/astrometry/optical-IR-prod/nomad&lt;/a&gt; lists even more stars than Stellarium (Stellarium uses only part of this catalog), so you might be able to find an even closer star to this point, but I was too lazy to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Newer versions of Stellarium let you overlay galactic coordinates. I'm too lazy to do this myself, but if someone else wants to, it would be a better answer.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2017-01-14T16:58:45.763" LastActivityDate="2017-01-14T16:58:45.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19745" PostTypeId="2" ParentId="12925" CreationDate="2017-01-13T18:01:29.363" Score="5" Body="&lt;p&gt;Darkness is not a thing. It is just the absence of light. It doesn't have speed. Try observe the image:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/c58Xf.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c58Xf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this image this is shown that suppose there is darkness on any many light year far object if someone switch the torch in the direction of object and after some instant raise his hand and cover the light then before the light the darkness will reach on object which is completely wrong. Since the darkness which is assumed to be reach before light is the darkness there before the light reach on the object.&lt;/p&gt;&#xA;" OwnerUserId="15347" LastActivityDate="2017-01-13T18:01:29.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19746" PostTypeId="2" ParentId="19733" CreationDate="2017-01-13T18:09:56.800" Score="2" Body="&lt;p&gt;First off, there is evidence to suggest that sunstones were used to find the direction of the sun on cloudy days.  However the compass has completely eliminated the need for a sunstone as it works in cloudy weather.  You are correct that taking star, planet and moon sights with a sextant in cloudy conditions is problematic.  However the sky is not always completely occluded.  If even a single navigational star appears for long enough than a sight can be taken.  Without seeing other stars and constellations one can make educated guesses as to which navigational star this is.  For example with a star finder and the most general idea of where one is on earth the name of the star could be deducted.  I believe that just such a sight during Ernest Shackleton's historic voyage in Antarctica enabled him to navigate to a whaling station to get assistance.&lt;/p&gt;&#xA;" OwnerUserId="15497" LastActivityDate="2017-01-13T18:09:56.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19747" PostTypeId="2" ParentId="19743" CreationDate="2017-01-13T18:16:51.100" Score="6" Body="&lt;p&gt;Most planets don't have rings.   The ring region is inside the &lt;a href=&quot;https://en.wikipedia.org/wiki/Roche_limit&quot; rel=&quot;noreferrer&quot;&gt;Roche limit&lt;/a&gt; which is quite close to the planet.  A ring system outside the Roche limit needs to be either very faint or, it would over time coalesce into a moon or possibly pair of moons at Trojan points to each other.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's different Roche limits for rocky bodies vs icy, but an icy body couldn't exist in Earth's orbit for long, so it's not really relevant for the Earth.  Saturn's rings are mostly ice, which extends about twice as far as the rocky body Roche limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lSSrd.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lSSrd.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How long a ring system lasts depends on a few factors.   How big it is, whether it's given new material regularly or not, what it's made of and how vulnerable it is to the sun's rays.   Saturn's rings, for example, are losing material, &lt;a href=&quot;http://www.universetoday.com/101374/rain-is-falling-from-saturns-rings/&quot; rel=&quot;noreferrer&quot;&gt;raining down on Saturn&lt;/a&gt; as well as &lt;a href=&quot;http://saturn-archive.jpl.nasa.gov/science/moons/enceladus/enceladusfeedring/&quot; rel=&quot;noreferrer&quot;&gt;slowly adding material&lt;/a&gt; from Enceladus, one of it's small and close moon's that gives up ring material by tidal-volcanic eruptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth's large moon could (not saying it would but it might) play a role in disrupting and Earth-Ring system more quickly than it would with no moon.   A 2 body system with the 2nd body as large as the moon isn't gravitationally friendly to other orbiting objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many man made satellites orbit well within the Roche limit but they are small enough to be held together and not be affected by tidal forces.   Larger objects, just a few miles across, perhaps less, are more vulnerable to tidal forces and tend to break up inside the Roche Limit.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-01-13T18:16:51.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19748" PostTypeId="1" CreationDate="2017-01-13T21:14:41.937" Score="3" ViewCount="40" Body="&lt;p&gt;Why do red galaxies tend to be more massive than blue galaxies? Do we have an explanation for this observation?&lt;/p&gt;&#xA;" OwnerUserId="15126" LastEditorUserId="2365" LastEditDate="2017-01-13T21:44:40.443" LastActivityDate="2017-01-13T23:43:02.383" Title="Bimodal Galaxy Distribution" Tags="&lt;galaxy&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19750" PostTypeId="2" ParentId="13367" CreationDate="2017-01-13T22:10:02.113" Score="1" Body="&lt;p&gt;The best inertial frame is that which is comoving with the cosmic microwave background. There is a notable dipole in the microwave background, the result of the Solar system's motion relative to it. The invariance of this dipole provides evidence that the frame of the barycentre of the solar system is also a very good inertial frame. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is of no significance to navigation, either by the stars, or among the planets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Within the solar system, the sun's gravitation influence is dominant. While the sun is orbiting with the galaxy, a space probe is also orbiting the galaxy, and the variation in gravitation due to the galaxy is extremely small. Galactic tides are several orders less significant than photon pressure, various quadrupole effects or the influence of other bodies.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-01-13T22:10:02.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19751" PostTypeId="2" ParentId="13069" CreationDate="2017-01-13T22:30:43.590" Score="1" Body="&lt;p&gt;In principal, Laplace's method (and Gauss's method) should be very accurate for determining orbits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course &quot;garbage in garbage out&quot; applies, and if the observations were inaccurate the results may also be inaccurate. However, you should not get great discrepancies. I would consider careful checking of your calculation, and perhaps comparing your results with those of an independent orbital determiner, to check for a mistake in the maths.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-01-13T22:30:43.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19752" PostTypeId="1" AcceptedAnswerId="19754" CreationDate="2017-01-13T22:55:41.403" Score="4" ViewCount="102" Body="&lt;p&gt;In his popular youtube channel, Scott Manely addressed a question of whether an astronaut can orbit a space station. The provided answer is negative since such an orbit would require the space station to be impractically dense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This got me wondering if similar arguments can limit the number of nested orbits in general. For example, say you are free to choose any solid bodies A, B, C,... s.t. A orbits B, B orbits C, etc. Is there a maximal number of such bodies and do we know it?&lt;/p&gt;&#xA;" OwnerUserId="14092" LastActivityDate="2017-01-13T23:48:00.593" Title="Is there a non-trivial limit to a number of nested orbits?" Tags="&lt;orbit&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19753" PostTypeId="2" ParentId="19748" CreationDate="2017-01-13T23:43:02.383" Score="3" Body="&lt;p&gt;My understanding is that the matter is not settled, definitively. There is a chain of logic that goes something like this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Why is any galaxy red? Because it isn't forming any new, high mass, short lived, blue stars.&lt;/li&gt;&#xA;&lt;li&gt;Why would a galaxy not form stars? Because the gas in the galaxy is either not dense enough or not cold enough to gravitationally collapse into new stars.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are two main explanations put forward for the inability of the gas in really large galaxies to collapse and form stars. First, the most massive galaxies tend to be found in clusters of galaxies, and the gas in the galaxy clusters is too hot to collapse and form stars. Second, when large galaxies of comparable mass merge simulations suggest that the merger process heats up the gas in the galaxies, cutting off star formation. &lt;/p&gt;&#xA;" OwnerUserId="14244" LastActivityDate="2017-01-13T23:43:02.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19754" PostTypeId="2" ParentId="19752" CreationDate="2017-01-13T23:48:00.593" Score="5" Body="&lt;p&gt;There is no theoretical maximum. It would be possible for the solar system to orbit a sufficiently large mass, (while the planets' orbits remain stable). And it would be possible for that large mass to orbit another, even larger mass. The process could, in theory, be repeated without limit. The masses of the central bodies would have to be exponentially larger with each embedding, and so could not be conventional stars, beyond a certain point. (They could be black holes)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the absence of theoretical limit imposed by the theory of gravitation. In practice, multiple levels of orbits don't exist, since there does not exist a mechanism that can create the massive central bodies in the right orbits.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-01-13T23:48:00.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19756" PostTypeId="1" AcceptedAnswerId="19758" CreationDate="2017-01-14T00:56:34.957" Score="10" ViewCount="503" Body="&lt;p&gt;Is there a place in the universe where, with the naked eye, you wouldn't see any stars, galaxies or light emitting phenomena? From what I understand at some point into the far future in Earth's location, we wouldn't be able to see anything. Is there a place like that could be like that now?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p.s. Not sure what the appropriate tags to put on this question. Feel free to add/change tags as needed. &lt;/p&gt;&#xA;" OwnerUserId="10117" LastEditorUserId="2531" LastEditDate="2017-01-17T14:45:26.723" LastActivityDate="2017-01-26T01:03:57.530" Title="Is there a place in the known universe that there would be no visible light from other stars/galaxies?" Tags="&lt;light&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19758" PostTypeId="2" ParentId="19756" CreationDate="2017-01-14T11:10:11.450" Score="12" Body="&lt;p&gt;If you don't want to look at the stars and galaxies, you can do two things: You can travel so far that they're too far away to see, or you can block your vision.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Go far away&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;On large scales, i.e. above a billion lightyears or so, the Universe is observed to be roughly homogeneous. On smaller scales, however, matter is distributed in the so-called &lt;em&gt;cosmic web&lt;/em&gt;, where galaxies lie in filaments and sheet, meeting each other in knots (where the largest clusters are found). In between these overdensities, there are underdensities known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Void_(astronomy)&quot;&gt;voids&lt;/a&gt;. These voids are virtually free of any luminous matter, and since they're up to several million lightyears across, you could place yourself in the middle of such a void and only see darkness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, consider a void of 100 Mpc (~300 million lightyears) in radius, corresponding to a distance modulus&lt;sup&gt;1&lt;/sup&gt; of $\mu = 35$. If a Milky Way-like galaxy (absolute magnitude&lt;sup&gt;2&lt;/sup&gt; $M=-20.5$) were at the border of the void, its apparent magnitude&lt;sup&gt;3&lt;/sup&gt; would be $m = M+\mu = 14.5$, invisible to the naked eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small telescope, however, &lt;em&gt;would&lt;/em&gt; enable you to see galaxies from here.&#xA;The limiting magnitude&lt;sup&gt;4&lt;/sup&gt; of the naked eye is roughly 6–7 (though some claim to be able to see $m=8$ objects), so the Milky Way-like galaxy would be roughly 8 magnitudes too faint to see. The gain in limiting magnitude when using a telescope is roughly $g = 5\log(D_\mathrm{tel}/D_\mathrm{pupil})$, so with your 6 mm pupil, a telescope with a diameter of&#xA;$$&#xA;D_\mathrm{tel} = D_\mathrm{pupil} 10^{g/5} \\&#xA;= 6\,\mathrm{mm}\times10^{8/5} \\&#xA;\simeq 200\mathrm{-}250\,\mathrm{mm},&#xA;$$&#xA;would enable you to see Milky Way-like galaxies from the center of the void.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note though that no matter how far you are from galaxies, there will always be some radiation around, if nothing else then at least the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot;&gt;CMB&lt;/a&gt;. Of course, being microwaves, this is not visible to the naked eye.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Block your vision&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Apart from going down in your basement and turn off the light, could there be &quot;astronomical&quot; places where you would be unable to see anything? As David Hammen comments, at the surface of a cloud-enshrouded planet or moon, you could face total darkness (at least on the night-side of the planet). But perhaps you could also, as Wayfaring Stranger comments, go inside a dense, interstellar cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Bok_globule&quot;&gt;Bok globules&lt;/a&gt; are small ($R\sim10^4\,\mathrm{AU}$), dense ($n\sim10^{4\mathrm{-}6}\,\mathrm{cm}^{-3}$) nebulae of gas and dust. A sightline from the center of such a cloud would result in the apparent magnitude being extincted (i.e. increased) by up to several tens in visual light. The largest extinction through the center of a Bok globule &lt;em&gt;that I've been able to find&lt;/em&gt; — but I'm not expert on this — is &quot;FeSt 1-457&quot; with a visual extinction of $A_V = 41$ (&lt;a href=&quot;https://arxiv.org/abs/astro-ph/0506205&quot;&gt;Kandori et al. 2005&lt;/a&gt;), so from the center and out is roughly $A_V\simeq20$. That means that the fraction of light from outside sources that make it into the center is&#xA;$$&#xA;f = 10^{-A_V/2.5} \sim 10^{-8},&#xA;$$&#xA;which is not a lot. However, Bok globules are found in the vicinity of young stars which tend to be bright. An O star has an absolute magnitude of $M\simeq-4$. Such a star located right outside the cloud (say, at a distance of $d=2\times10^4\,\mathrm{AU}=0.1\,\mathrm{pc}$) would have an apparent magnitude of $m=M+\mu+A_V\simeq6$, and thus be just visible to the naked eye, albeit very faint. And if it were just a little farther away, you wouldn't be able to see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that you're human, but if you are in fact a cichlid, you should be able to see infrared light (&lt;a href=&quot;http://link.springer.com/article/10.1007%2Fs00114-012-0980-7&quot;&gt;Meuthen et al. 2012&lt;/a&gt;). In the IR, the extinction is much smaller than in the visible, so hiding in a Bok globule won't help you.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;&lt;sub&gt;The &lt;em&gt;distance modulus&lt;/em&gt; $\mu$ is a logarithmic way of expressing distances: $\mu \equiv 5\log(d/10\mathrm{pc})$.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2&lt;/sup&gt;&lt;sub&gt;The &lt;em&gt;absolute magnitude&lt;/em&gt; $M$ is a (logarithmic) measure of the luminosity $L$ of an object: $M=-2.5\log(L)+\mathrm{constant}$. Note the minus sign; the brighter an object is, the smaller $M$ is.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;3&lt;/sup&gt;&lt;sub&gt;The &lt;em&gt;apparent magnitude&lt;/em&gt; $m$ is a logarithmic measure of how bright an object looks to an observer, and so depends on the distance to the object: $m = M + \mu$. Stars visible to the naked have roughly $m=0$–$6$. Again, the smaller the number, the brighter the object is. An extremely distant galaxy may have $m=30$, while the Sun has $m=-27$. &lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;4&lt;/sup&gt;&lt;sub&gt;The &lt;em&gt;limiting magnitude&lt;/em&gt; is the largest magnitude (i.e. faintest object) visible.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2017-01-17T11:52:31.993" LastActivityDate="2017-01-17T11:52:31.993" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="19760" PostTypeId="2" ParentId="19731" CreationDate="2017-01-14T16:12:31.150" Score="3" Body="&lt;p&gt;I'm not sure this answers your question, but I used Stellarium (free) to generate this image from Callisto's north pole (so its celestial north pole, in Draco, is overhead) with orbit trails turned on:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GHOJY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GHOJY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the image is not only messy, but may be inaccurate, since Stellarium doesn't do a great job of plotting orbits long-term: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/11617/stellarium-0-10-4-planet-orbits-change-over-time&quot;&gt;Stellarium 0.10.4: planet orbits change over time?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I wrote &lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/STELLARIUM/callisto.ssc&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/STELLARIUM/callisto.ssc&lt;/a&gt; to create a video of the view from Callisto's north pole, but the results are fairly uninteresting: &lt;a href=&quot;https://youtu.be/O2GHJM54uhQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/O2GHJM54uhQ&lt;/a&gt; and come with several caveats:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I assumed that Stellarium is accurate, not just in where it displays the planet, but also in how it determines Callisto's rotational orientation, and any other details required to depict Callisto's celestial northern hemisphere.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The video is in &quot;starchart mode&quot;, which means the stars don't move. In reality, the stars would make a complete rotation about every 16.69 days. More specifically, in the video, Jupiter moves through the constellations. In reality, Jupiter would remain fixed (Callisto is tidally locked), and the constellations would move.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The one year (+ about 2 months) period I chose is not &quot;representative&quot; and does not repeat. There may be quite a few more interesting things happening after the period the video represents. In particular, note that the Sun takes about 12 years (Jupiter's orbital period) to travel Callisto's ecliptic, and only a fraction of that travel is show in the video.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user21" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2017-01-14T16:12:31.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19761" PostTypeId="1" AcceptedAnswerId="19764" CreationDate="2017-01-14T17:18:40.647" Score="2" ViewCount="340" Body="&lt;p&gt;I have a question that I don't know how to solve, any help is appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Rewrite the equation for the virial mass of a cluster so that if velocities are entered in km/sec and dimensions in Mpc, the resulting mass is in solar mass&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well the equation is: $M=\frac{5(v_r)^2R}{G}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I have to use dimensional analysis but I really don't know how to get there&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[M]=\mathrm{\frac{(\frac{km}{s})^2 km}{\frac{m^3}{kg \cdot s^2}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with one $\mathrm{mpc}$ being $3.09 \times 10^{19} \ \mathrm{km}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance :)&lt;/p&gt;&#xA;" OwnerUserId="15386" LastEditorUserId="9301" LastEditDate="2017-01-15T03:43:39.033" LastActivityDate="2017-01-15T03:43:39.033" Title="Rewriting the equation for the virial mass of a cluster" Tags="&lt;mass&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19762" PostTypeId="1" CreationDate="2017-01-14T18:12:28.907" Score="6" ViewCount="303" Body="&lt;p&gt;I couldn't find this answer anywhere.  Before 1998, was there a notable physicist who conjectured that the universe is accelerating and perhaps even submitted some equations to support it (and the existence of dark energy).  Such a hypothesis would likely have been rejected outright and fallen into oblivion, which is why I have to ask this question to get the answer.  And I don't mean Einstein's cosmological constant since he was trying to make the universe static rather than accelerating.  Someone who actually believed that the universe's expanding rate was increasing even before there was any evidence of it.&lt;/p&gt;&#xA;" OwnerUserId="15502" LastActivityDate="2017-01-14T19:43:19.060" Title="Who first hypothesized that the universe is accelerating?" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19763" PostTypeId="2" ParentId="19762" CreationDate="2017-01-14T19:43:19.060" Score="8" Body="&lt;p&gt;Science isn't concerned with belief. The possibility of a dark energy term and therefore an accelerating universe is implicit in GR and in the consequent Friedmann-Lemaitre equations that govern the dynamics of the universe. When the supernova cosmology teams set out to measure the deceleration of the universe, the models that they fitted to their data explicitly allowed for the possibility that the universe might actually be accelerating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, such hypothetical universes had certainly been explored before. The &lt;a href=&quot;https://en.wikipedia.org/wiki/De_Sitter_universe&quot; rel=&quot;noreferrer&quot;&gt;De Sitter universe&lt;/a&gt; (proposed by Willem de Sitter in 1917) is an accelerating universe consisting only of a cosmological constant. Our own universe appears to be headed asymptotically towards being described as a De Sitter universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea of $\Lambda$ as a form of &quot;dark energy&quot; that had the &lt;em&gt;potential&lt;/em&gt; to cause an accelerating universe was already put forward as an idea by &lt;a href=&quot;https://en.wikipedia.org/wiki/Georges_Lema%C3%AEtre&quot; rel=&quot;noreferrer&quot;&gt;Lemaitre&lt;/a&gt; in the late 1920s and early 30s. The picture below shows the evolution of the scale factor for families of solutions and was apparently made by Lemaitre in 1927 (see &lt;a href=&quot;http://blogs.futura-sciences.com/e-luminet/2015/06/21/the-rise-of-big-bang-models-4-lemaitre/&quot; rel=&quot;noreferrer&quot;&gt;this science history blog about Lemaitre&lt;/a&gt;). The upper 5 curves correspond to a universe that is accelerating today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WEpIw.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WEpIw.jpg&quot; alt=&quot;Scale factor in Lemaitre universe&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-01-14T19:43:19.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19764" PostTypeId="2" ParentId="19761" CreationDate="2017-01-14T20:09:13.110" Score="3" Body="&lt;p&gt;This is something that crops up a lot in Astrophysics and Astronomy, so here is the general approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Get your equation into SI units. Thus&#xA;$$M = \frac{5v^2R}{G} = 7.5\times10^{10} v^2 R,$$&#xA;with $M$ in kg, $v$ in m/s and $R$ in m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you wish to replace $v^2$ by $(v/1000)^2$ (which is velocity expressed in km/s), you would simultaneously multiply the numerical constant at the front by $1000^2$. Similarly, to replace $R$ by $R/3.09\times 10^{22}$ (which is $R$ expressed in Mpc), then you have to multiply the numerical constant by $3.09\times10^{22}$ to leave everything unchanged. This then leaves your result still expressed in kg. To express it in solar masses, simply divide by a solar mass of $2\times 10^{30}$ kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&#xA;$$\frac{M}{M_{\odot}} = \frac{7.5\times 10^{10}\times 10^6 \times 3.09\times10^{22}}{2.0\times 10^{30}} \left(\frac{v}{1 {\rm km/s}}\right)^2 \left(\frac{R}{1 {\rm Mpc}}\right)$$&#xA;$$\frac{M}{M_{\odot}} =  1.16\times10^{9} \left(\frac{v}{1 {\rm km/s}}\right)^2 \left(\frac{R}{1 {\rm Mpc}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, check your result makes sense. Galaxy clusters are of order a Mpc in size, have velocity dispersions of hundreds of km/s and consists of hundreds of galaxies with total mass of $\sim 10^{14} M_{\odot}$ or so. This seems to check out - usually if a mistake has been made you will be many, many orders of magnitude out.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-01-14T20:09:13.110" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19765" PostTypeId="2" ParentId="19741" CreationDate="2017-01-14T23:09:54.473" Score="1" Body="&lt;p&gt;Aside from the fact that it's 81 times easier to move the Moon than the Earth, so, in such a scenario, the Moon could be moved too with an extra 1.23% of time/money/energy, so, it really would make sense to move them both, not just move the Earth and risk losing the moon or worse, crashing into it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But (lets credit the need to save 1.23% to a congressional budget committee) and just move the Earth, as you propose.   Would the Earth lose the Moon?  . . . It depends, but not necessarily.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first, idea to get out of the way is that there's no gravitational string connecting the Earth and Moon, so if the Earth was accelerated fast enough (say, launching millions of very high velocity rockets off the same side of the earth at the same time) there's no attachment that would pull the Moon with the Earth.   This is one of those scenarios where the &lt;a href=&quot;http://theory.uwinnipeg.ca/users/gabor/black_holes/slide5.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;rubber sheet analogy&lt;/a&gt; works quite well.  Accelerate the Earth and the corresponding gravity well moves away from the Moon and the Moon moves corresponding to the decrease (or increase) in the slope of Earth's &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/2/21/Gravity_well_plot.svg/600px-Gravity_well_plot.svg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravity well&lt;/a&gt;, so the Moon's velocity carries it relative to the new &lt;em&gt;slope&lt;/em&gt;.  But that's the argument for a single acceleration.   For multiple gravity assists, you have to account for the moon being in different positions so the relative movement wouldn't all be in the same direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Asteroid Gravity Assist - how close to the Earth?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A passing asteroid would affect both the Moon and the Earth, but for optimal Earth movement, you'd want it pass within, say, a few thousand KM of Earth's surface, outside of the atmosphere but much closer to the Earth than the Moon, where it's gravitational tug on the Earth would be over 1,000 times the tug on the Moon at closest pass.  The entire path wouldn't be a 1000 to 1 ratio, but with a close enough pass we can effectively say the Asteroid is giving the Earth a much larger gravitational tug, where the tug on the Moon is somewhat negligible.  In effect, moving the Earth's orbit but not moving the moon's orbit nearly as much.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asteroids passing that close would be inside the Roche limit and probably would begin to break apart, but that wouldn't change the gravity assist or be a problem for the Earth if the velocity was sufficient to pass through and not remain in a near-Earth orbit.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it was a single asteroid of high mass with a near-Earth pass that effectively moved the Earth but not the Moon, Then any of your scenarios are possible.  The Earth could be moved sufficiently where it could be pulled away from the Moon (a change in velocity of about 1,500 km/hour in the right direction would do that) or the Earth could be pulled into the Moon (a change in velocity of about 3,600 km/hour), or the orbit could be sufficiently jiggled where one of those two scenarios would happen eventually if not right away.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To change the Earth's velocity that much, with a single asteroid pass, you'd need that asteroid to be, (if my math is right), a few times the mass of the Moon, and an object with that mass passing within a few thousand KM over the Earth's surface, lets just say that would be a bad idea for a number of reasons. - so, it's better to do it the way you suggested.  Many asteroids, over a long time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Many asteroids - what would happen to the Earth-Moon system&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each asteroid, passing within a few thousand km of Earth would move the Earth slightly, but as the Moon moves around the Earth, many of those movements would cancel out.   They would move the Earth towards the Moon when the Moon was on the same side of the Earth as the asteroid, and away from it when the Moon was on the opposite side.  Similarly, eccentricity could be increased or decreased.   There's no one answer as to whether the Earth would lose the moon without adding up the position of the moon and averaging it out.     Moving the Earth could be timed, where, in effect the Moon would be carried along with it.  This works both ways too.   We could move the Earth's orbit by properly timed thrusters or explosions on the surface of the Moon, sometimes moving the Moon away from the Earth, sometimes closer, but gravity assists are probably easier with less energy involved, as Jupiter could be used to accelerate the asteroids, which would reduce the energy input needed.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net effect is that you can move the Moon with the Earth, or the Earth with the Moon, but pushing on just one of them.   That's possible with gravitationally bound objects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, depending on the timing of the asteroid gravity assists, you could create any of your four scenarios, but without specific timing, there's at least a pretty good chance the Moon stay with the Earth.  The 2nd scenario, the Moon getting tidally torn apart, that would only happen if it got very close to the Earth.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-01-14T23:09:54.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19766" PostTypeId="1" CreationDate="2017-01-14T23:12:45.753" Score="2" ViewCount="639" Body="&lt;p&gt;As the Sun nears the end of its life and expands, would the Earths orbit get bigger as it moves out or would it stay in the same orbital plane and not move out?  Therefore in the far off future, we'd have 512 days instead of 365 days?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the Moon is moving out at a distance of 3.8 cms a year (ref: &lt;a href=&quot;http://curious.astro.cornell.edu/about-us/37-our-solar-system/the-moon/the-moon-and-the-earth/111-is-the-moon-moving-away-from-the-earth-when-was-this-discovered-intermediate&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://curious.astro.cornell.edu/about-us/37-our-solar-system/the-moon/the-moon-and-the-earth/111-is-the-moon-moving-away-from-the-earth-when-was-this-discovered-intermediate&lt;/a&gt;), are we moving out or in towards the Sun or is this not possible to work out because there is nothing to measure against?&lt;/p&gt;&#xA;" OwnerUserId="5401" LastEditorUserId="2531" LastEditDate="2017-01-18T20:57:55.410" LastActivityDate="2017-01-25T18:23:01.920" Title="As the Sun dies, does the Earth's orbit change?" Tags="&lt;orbit&gt;&lt;the-sun&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19767" PostTypeId="2" ParentId="11442" CreationDate="2017-01-15T20:28:07.930" Score="2" Body="&lt;p&gt;&lt;em&gt;Comments converted to community wiki&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ceres' density (2.08 g/cc) suggests it has abundant water/ice below its presumably mostly dry surface. That and the shiny spot (salts left after subsurface water has sublimated) suggests it wouldn't too difficult. Low gravity, keep the dig in shade to prevent the ice from sublimating in direct sunlight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could not do it now, but with improved technology and continued space exploration, it should be very possible. Ceres should be a comparatively easy place to build a space elevator too. It seems a pretty good mining option. Only potential issue might be the occasional asteroid impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ask the asteroid mining companies &quot;Planetary Resources&quot; and &quot;Deep Space Industries&quot;. Actual and serious companies. See the websites www.planetaryresources.com and www.deepspaceindustries.com&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-01-15T20:28:07.930" CommentCount="0" CommunityOwnedDate="2017-01-15T20:28:07.930" ContentLicense="CC BY-SA 3.0" />
  <row Id="19768" PostTypeId="2" ParentId="11418" CreationDate="2017-01-15T20:36:17.967" Score="2" Body="&lt;p&gt;Comets are part of the meteor calculation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space junk is a minor concern. Most satellite and rocket bodies are fairly flimsy and are destroyed by the atmosphere. Sometimes chunks do come down. But by the time they reach Earth they are not falling fast enough to form a significant crater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However being directly hit could be fatal. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;According to Mark Matney, a scientist in the Orbital Debris Program Office at NASA's Johnson Space Center in Houston, the odds that any of the 7 billion people on Earth will be struck by a piece of the soon-to-fall satellite is 1 in 3,200. &quot;The odds that you will be hit are 1 in several trillion &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In '97 a piece of debris from a Delta rocket hit Lottie Williams in Tulsa. It caused no significant injury. &lt;a href=&quot;http://www.livescience.com/33511-falling-nasa-satellite-uars-risk.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.livescience.com/33511-falling-nasa-satellite-uars-risk.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-01-15T20:36:17.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19769" PostTypeId="2" ParentId="11414" CreationDate="2017-01-15T20:40:54.767" Score="4" Body="&lt;p&gt;As per the &lt;a href=&quot;https://www.google.com/search?q=uranus&amp;amp;biw=1357&amp;amp;bih=807&amp;amp;source=lnt&amp;amp;tbs=cdr%3A1%2Ccd_min%3A1%2F1%2F1900%2Ccd_max%3A12%2F31%2F1985&amp;amp;tbm=bks#tbs=cdr:1%2Ccd_min:1%2F1%2F1900%2Ccd_max:12%2F31%2F1985&amp;amp;tbm=bks&amp;amp;q=uranus+picture&quot; rel=&quot;nofollow noreferrer&quot;&gt;books referred to in comments&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both planets could be resolved as a disk, but no surface features could be observed on either. Spectrograms had been taken, so the general colour of the planets was known. The extreme axial tilt of Uranus was known about. But the relative warmth of Neptune was not.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-01-15T20:40:54.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19770" PostTypeId="2" ParentId="11102" CreationDate="2017-01-15T20:47:37.353" Score="0" Body="&lt;p&gt;&lt;em&gt;comments converted to community wiki&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The day of the week by itself has no use in astronomy unless you also know the corresponding calendar date. And since the day of the week is computable from the calendar date, it's really just superfluous information. Nevertheless, most astronomy apps such as Sky Safari seem to use the first three letters (in English) abbreviating the names for the days of the week. This use seems to conform with the &lt;a href=&quot;https://en.wikipedia.org/wiki/ISO_week_date&quot; rel=&quot;nofollow noreferrer&quot;&gt;ISO week date standard&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-01-15T20:47:37.353" CommentCount="3" CommunityOwnedDate="2017-01-15T20:47:37.353" ContentLicense="CC BY-SA 3.0" />
  <row Id="19772" PostTypeId="1" AcceptedAnswerId="19878" CreationDate="2017-01-16T06:57:34.207" Score="5" ViewCount="193" Body="&lt;p&gt;In &lt;a href=&quot;https://space.stackexchange.com/a/8061/12102&quot;&gt;this answer&lt;/a&gt; I have just read about The University of Texas Rio Grande Valley's &lt;a href=&quot;http://www.utrgv.edu/cos/programs/stargate/index.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Spacecraft Tracking and Astronomical Research into Giga-hertz Astrophysical Transient Emission&lt;/a&gt; (STARGATE).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While &lt;a href=&quot;https://en.wikipedia.org/wiki/Fast_radio_burst&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fast Radio Bursts&lt;/a&gt; (FRBs) seem to fit that description spot-on, I am wondering if the phrase was chosen to be in some way broader or more general than FRB to include other phenomenon, and if so, have there been any Giga-hertz Astrophysical Transient Emissions observations &lt;em&gt;that are not also considered FRBs&lt;/em&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{4cm}$&lt;img src=&quot;https://i.stack.imgur.com/HaokK.jpg&quot; alt=&quot;The University of Texas Rio Grande Valley STARGATE&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Logo of &lt;a href=&quot;http://www.utrgv.edu/cos/programs/stargate/index.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;The University of Texas Rio Grande Valley STARGATE program&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-11-13T15:13:32.160" LastActivityDate="2017-11-13T15:13:32.160" Title="Is &quot;Giga-hertz Astrophysical Transient Emission&quot; distinct in any way from &quot;Fast Radio Burst&quot;?" Tags="&lt;radio-astronomy&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19774" PostTypeId="2" ParentId="19727" CreationDate="2017-01-16T17:03:20.900" Score="2" Body="&lt;p&gt;It is certainly true that the galactic orbit is very slow.  Yet it should be noted that even if the Sun/Earth system orbited the galaxy once every century, the Andromeda galaxy would still be seen in essentially the &quot;same part of the sky&quot; at all times.  This is because the only way objects appear to move in the sky, due to the motion of the Earth, is an effect called parallax, which depends on the ratio of how far the observer moves divided by how far away is the object.  Andromeda is so far away that this ratio is 0.01, so that's not enough to make Andromeda move significantly to a different part of the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the Sun orbited the galaxy every century, then all the stars in our galaxy would also be in rapid orbits at different orbital rates, so the stars and constellations would be scrambled over human history, so much more rapidly than they already are.  But that really doesn't relate to where we would see Andromeda, only what background of stars we would see it against, so that doesn't seem to be the issue here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main point is, as objects go around in an orbit, one must avoid the tendency to imagine that the various axes of orientation rotate as well.  This is a common mistake for understanding seasons, for the explanation would not work if the Earth's axis of rotation also re-aligned in synch with the Earth's position around the Sun, as if connected rigidly to the Earth-Sun line.  So we should not think of an orbit as a global rotation, we should think of it as a circular translation of just the object and not any of its axes.  It's true that we can go into an orbiting frame by using a rotation, but then we also have to rotate the distant objects in that frame, which is confusing if we want to understand where they will appear in the sky. Restricting to a circular translation shows that the only effect is parallax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, we should constrast the general situation of how parallax alters location in the sky, with what happens in the special case of the Earth's orbit around the Sun over the seasons.  As pointed out in the question, objects do appear to change locations in the sky in different seasons, even though the parallax is very small.  But that's only because what is actually changing is the location of the Sun in the sky, not the other objects, and the parallax for the Sun is very large.  Since the Sun is so bright, its location affects not the location of other things, but when we can see them, as its brightness blots out our view.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2017-01-16T17:08:22.207" LastActivityDate="2017-01-16T17:08:22.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19775" PostTypeId="1" CreationDate="2017-01-16T18:20:51.563" Score="1" ViewCount="49" Body="&lt;p&gt;It is assumed that particles and anti-particles pop out in vacuum (vacuum energy). Could this have an effect on the paths of photons when they interact with these particles? The amount of photons going straight from the source would then gradually diminish. The supernovae would then appear dimmer than without this effect.&lt;/p&gt;&#xA;" OwnerUserId="15519" LastActivityDate="2017-01-17T07:18:34.990" Title="Could vacuum energy dim standard candles?" Tags="&lt;light&gt;&lt;vacuum&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19777" PostTypeId="2" ParentId="19775" CreationDate="2017-01-16T20:13:04.903" Score="3" Body="&lt;p&gt;Locally, energy and momentum must be conserved. So no, a photon cannot lose energy or change momentum by interacting with virtual particle pairs that have a net energy and momentum of zero.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-01-17T07:18:34.990" LastActivityDate="2017-01-17T07:18:34.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19778" PostTypeId="1" AcceptedAnswerId="19780" CreationDate="2017-01-16T20:20:46.923" Score="4" ViewCount="166" Body="&lt;p&gt;Cancri is what to Cancer?  or Leonis Minoris to Leo Minor?   Arietis to Aries ?  or Lyncis to Lynx.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are they the Flamesteed names, the Latin names, Greek names?&lt;br&gt;&#xA;What is their name?  &lt;/p&gt;&#xA;" OwnerUserId="5401" LastEditorUserId="2365" LastEditDate="2017-01-16T21:59:35.123" LastActivityDate="2017-01-17T21:41:50.487" Title="Why do constellation names change (eg from Cancer to Cancri) when used as part of a star's name?" Tags="&lt;constellations&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19779" PostTypeId="1" AcceptedAnswerId="19782" CreationDate="2017-01-16T21:50:11.977" Score="6" ViewCount="234" Body="&lt;p&gt;When Kepler went to Tycho Brahe, he learned that Tycho Brahe was using the mean position of the sun to note down the position of Mars. Well, Kepler needed to convert these datas of Brahe to do his own calculations for the orbit of Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this link, the difference between the real and mean position is shown.&#xA;&lt;a href=&quot;http://www.keplersdiscovery.com/MeanSun.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.keplersdiscovery.com/MeanSun.html&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.keplersdiscovery.com/CorrectedTable.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.keplersdiscovery.com/CorrectedTable.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how did Kepler calculate the real position of the Sun? I don't get it. And due to my first language is not English it might be possible that I didn't understand some important aspects of the given links.&lt;/p&gt;&#xA;" OwnerUserId="15474" LastEditorUserId="35433" LastEditDate="2020-10-06T17:22:42.967" LastActivityDate="2020-10-06T17:22:42.967" Title="Kepler and the Sun position" Tags="&lt;the-sun&gt;&lt;planet&gt;&lt;astrophysics&gt;&lt;mars&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19780" PostTypeId="2" ParentId="19778" CreationDate="2017-01-16T21:58:01.980" Score="8" Body="&lt;p&gt;These are the Latin word endings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Latin nouns change their endings according to how the word is being used. English doesn't have this grammar. But note how the word &quot;I&quot; changes to &quot;my&quot;. In Latin you would call &quot;my&quot; the genitive form of &quot;I&quot; The good people at &lt;a href=&quot;https://latin.stackexchange.com/&quot;&gt;https://latin.stackexchange.com/&lt;/a&gt; can give more explanation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;em&gt;&quot;Leo&quot;&lt;/em&gt; is a third declension noun and its genitive &quot;Of the lion&quot; is &lt;em&gt;Leonis&lt;/em&gt;. Third declension nouns have a genitive that ends in -is.  On the other hand &lt;em&gt;&quot;Cancer&quot;&lt;/em&gt; is a second declension noun with Genitive &lt;em&gt;&quot;Cancri&quot;&lt;/em&gt;. A Latin student spends much of their first year learning all these word endings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you write &lt;em&gt;&quot;Alpha Cancri&quot;&lt;/em&gt; you mean &quot;the alpha star of the Crab&quot;, because &quot;of the crab&quot; is &lt;em&gt;cancri&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="-1" LastEditDate="2017-04-13T13:01:09.390" LastActivityDate="2017-01-17T21:41:50.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19782" PostTypeId="2" ParentId="19779" CreationDate="2017-01-17T07:55:03.533" Score="7" Body="&lt;p&gt;It depends what you mean by &quot;real position&quot;.    Also, just to clarify something that should be gotten out of the way, the Earth rotates, obviously, and the night sky appears to rotate around the Earth a complete 360 degrees in a 24 hours.  That's why the geocentric model lasted as long as it did.  That's what it looks like, and that's what everyone believed prior to Copernicus, but that's not really relevant to your question.   I mention it, because what is relevant is that the stars are so far away that they appear fixed.   They rotate together, they don't move past each other so the stars operate as a fixed background against which the sun, moon and planets move.   So, when we talk about &quot;movement&quot; in the sky or position, we're talking about against the fixed stars in the background, not the rotation of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the fixed stars, the Sun is the simplest object in the sky, it moves a little bit against the background every day and it returns to the same place every year, the Sun's position against the fixed background stars was known long before Kepler, so Kepler always knew the position or direction of the Sun, even at night because the Sun held it's position against the background stars, moving only a little bit each day.   He didn't know the distance to the sun (more on that later), but he knew the position.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kepler also knew that Mars' orbit around the sun was 687 days.   How did he know that?   Well (these words aren't my own, but it's discussed &lt;a href=&quot;https://www.physicsforums.com/threads/how-did-kepler-know-that-the-time-of-mars-orbit-was-687-days.506262/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It's a simple matter to measure that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Synodic_period&quot; rel=&quot;nofollow noreferrer&quot;&gt;synodic period&lt;/a&gt; of Mars is&#xA;  about 780 days. This is the time between successive oppositions of&#xA;  Mars, meaning that Mars is due South at midnight. Given the&#xA;  heliocentric model, it is easy to see, as shown in the Wikipedia&#xA;  article, that 1/Synodic Period = 1/Earth's orbital period - 1/Mars&#xA;  orbital period. So, once the heliocentric model was understood, it all&#xA;  becomes easy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Kepler was a brilliant Mathematician.   Brahe wasn't so much.   Now, I'm not saying Brahe was dumb, cause that's not accurate either, but he wasn't skilled in math the way Kepler was.   Brahe was very patient and dedicated and he made the most accurate observations to date, some 10 times more accurate than anyone before him.   He had the best equipment for making accurate observations in history, so to a person like Kepler, Brahe's notes were pure gold, containing decades of great accuracy.    There was even a rumor that Kepler killed Brahe to get access to his observatory and his work, but most people think that's just a rumor, that it didn't actually happen, but there was enough uncertainty that Brahe's body was exhumed and tested for poison.  (none was found).    But I digress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 3rd bit of information is that by using parallax and the maximum angle that Mars varies in the sky, you can get a rough estimate of the distance to the planets relative to each other.   Copernicus did that so Kepler didn't have to.   A bit more on that &lt;a href=&quot;https://physics.stackexchange.com/questions/36458/how-did-copernicus-establish-the-relative-distance-to-the-superior-planets&quot;&gt;here&lt;/a&gt;.  and &lt;a href=&quot;http://astro.unl.edu/naap/ssm/ssm_advanced.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Kepler had a lot of information at his fingertips when he got access to Brahe's notes.   He knew that Mars was about 1.5 times as far from the Sun as the Earth (from Copernicus) and he knew that Mars orbit was 687 days (not difficult given the heliocentric model), which means, every year and 322 days, Mars is in the same spot relative to the Sun, but the Earth has moved. - So you have 2 still objects and one moving object that's some &quot;43 days&quot; earlier in it's orbit each time Mars is in the same place and by taking that measurement every 687 days, you can learn a lot about the relative position of Mars, Earth and the Sun.   And that's what Kepler did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His two major breakthroughs (in my opinion) were observing that Mars wasn't on the same plane as the Earth (He did that observing Copernicus' observations, not Brahe's - I read that, I can't find the notation right now).    And his other breakthrough was to figure out that the Sun wasn't at the center of the Earth's orbit.   He did that by measuring the angle between the Earth and Sun and Earth and Mars every 687 days.   More on that &lt;a href=&quot;http://science.larouchepac.com/kepler/newastronomy/part3/24/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His measurements telling him that the sun wasn't at the center of the Earth's orbit led him to his ellipse model.   After that it was all clarification and more detailed measurement that led him to his 3 laws, but I think, recognizing that the Sun wasn't at the center of the Earth's orbit was his big &quot;aha&quot; moment that and his discovery of Mars being on a different orbital plane.   Kepler's measurement was quite accurate.   Mars' inclination to Earth is 1.87 degrees.   Kepler had it measured at 1 degree, 50 minutes, or 1.83 degrees.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2017-01-18T23:20:42.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19783" PostTypeId="1" CreationDate="2017-01-17T10:59:30.093" Score="4" ViewCount="16098" Body="&lt;p&gt;I have got a Celestron Astromaster 130EQ telescope, which lens should I use to see the various planets and nebula? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I tried to see Mars it was only a small red dot on scope. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As per the technical description, it's got 6&quot; mirror with 650 mm focal length.&#xA;Kindly guide me to which lens of what mm should be used to see Mars, Saturn, Jupiter, Venus, and nebula.&lt;/p&gt;&#xA;" OwnerUserId="15523" LastEditorUserId="11143" LastEditDate="2017-01-17T13:27:44.040" LastActivityDate="2019-03-11T22:33:32.897" Title="Which planets, nebula, and other objects can be seen using Celestron Astromaster 130EQ" Tags="&lt;telescope&gt;&lt;terrestrial-planets&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19784" PostTypeId="1" AcceptedAnswerId="19787" CreationDate="2017-01-17T15:59:13.737" Score="30" ViewCount="4155" Body="&lt;p&gt;We measure movement, position, and many other things of an object in space because of its light and what we can measure with it. But as far as I know there's supposed to be a HUGE amount of dark matter in space whose mass and size we don't know, and because it has mass it has gravity and it can bend light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know physics can take into account the gravity of stars and huge stuff in space, but how can they be certain of any of their measurements (specially the position) if they don't know how that light has deviated from a straight line?  &lt;/p&gt;&#xA;" OwnerUserId="15526" LastEditorUserId="2153" LastEditDate="2017-01-17T16:49:09.913" LastActivityDate="2017-01-18T05:19:19.263" Title="If dark matter bends light, how do we know the stuff in the sky is where we think it is?" Tags="&lt;gravity&gt;&lt;light&gt;&lt;dark-matter&gt;&lt;positional-astronomy&gt;&lt;gravitational-lensing&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19785" PostTypeId="2" ParentId="14507" CreationDate="2017-01-17T16:05:14.933" Score="1" Body="&lt;p&gt;There is no net internal magnetic field to perturb, so how could one induce currents?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect arises on Earth because some transient event (e.g., &lt;a href=&quot;https://en.wikipedia.org/wiki/Coronal_mass_ejection&quot; rel=&quot;nofollow noreferrer&quot;&gt;coronal mass ejection&lt;/a&gt; or CME) hits Earth's &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;magnetosphere&lt;/a&gt; and causes a sudden $\tfrac{dB}{dt}$ which, from Faraday's law, results in induced electric fields.  These induced electric fields then drive currents in things like power grids (typically long-distance wires running east-west at high latitudes), oil pipelines, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only changing magnetic field would be that of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_wind&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar wind&lt;/a&gt; and the small, localized &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetic_anomaly&quot; rel=&quot;nofollow noreferrer&quot;&gt;magnetic anomalies&lt;/a&gt; on the surface of Mars.  However, these fields are incredibly small so the $\tfrac{dB}{dt}$ would be even smaller.&lt;/p&gt;&#xA;" OwnerUserId="13663" LastActivityDate="2017-01-17T16:05:14.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19786" PostTypeId="1" AcceptedAnswerId="19792" CreationDate="2017-01-17T16:12:10.590" Score="2" ViewCount="145" Body="&lt;p&gt;The recent approval of a NASA mission to study the asteroid 16 Psyche brought to my attention that Psyche (about 200 km in diameter) is thought to be the exposed iron core left over from a protoplanet (theorized as being about 500 km in diameter). My question is: what is the mechanism by which a body of that size can accrete, differentiate as thoroughly as Psyche apparently did, and then completely lose its mantle? I have a basic understanding of planetary differentiation and I had the impression that the migration of iron and nickel to the core took a long time, and that by that time major impacts capable of shattering large bodies must have been rare.&lt;/p&gt;&#xA;" OwnerUserId="15398" LastActivityDate="2017-01-17T19:59:14.023" Title="How can 16 Psyche be the core of a protoplanet?" Tags="&lt;solar-system&gt;&lt;asteroids&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19787" PostTypeId="2" ParentId="19784" CreationDate="2017-01-17T16:24:22.007" Score="25" Body="&lt;p&gt;The local dark matter density is actually quite tiny, on the order of $\rho\sim10^{-19}\text{ g/cm}^3$ (see e.g. &lt;a href=&quot;https://arxiv.org/abs/1205.4033&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bovy &amp;amp; Tremaine (2012)&lt;/a&gt;). This means that there is roughly $0.001$-$0.01M_{\odot}$ of dark matter per cubic parsec - a staggeringly small amount. 1000 cubic parsecs would contain about one solar mass of dark matter - and that's a cube 10 parsecs in length on each side! Now, the distribution of dark matter in galaxies is not homogeneous - it follows, roughly, a &lt;a href=&quot;https://en.wikipedia.org/wiki/Navarro%E2%80%93Frenk%E2%80%93White_profile&quot; rel=&quot;nofollow noreferrer&quot;&gt;Navarro-Frenk-White profile&lt;/a&gt;, decreasing in density from the center of the galaxy - but on the scale of parsecs (and certainly &lt;a href=&quot;https://physics.stackexchange.com/q/194107/56299&quot;&gt;in the Solar System&lt;/a&gt;), we can consider it to have roughly uniform density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On small scales, then, we have approximate homogeneity and low density. This means that any gravitational lensing effects from dark matter should be extremely low or self-cancelling, arising only from inhomogeneities containing large clumps of dark matter. However, such clumps are unlikely to form solely through dark matter's interaction with itself (if we discount the &lt;a href=&quot;https://en.wikipedia.org/wiki/Massive_compact_halo_object&quot; rel=&quot;nofollow noreferrer&quot;&gt;MACHO&lt;/a&gt; hypothesis, which, as far as I know, is not currently favored).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On intergalactic scales, however, dark matter &lt;em&gt;can&lt;/em&gt; have some effects. &lt;a href=&quot;https://en.wikipedia.org/wiki/Weak_gravitational_lensing&quot; rel=&quot;nofollow noreferrer&quot;&gt;Weak lensing&lt;/a&gt; is a commonly-observed phenomenon in galaxy clusters, which may have extremely high fractions of dark matter. There are several techniques currently used to model the mass distribution of the lensing galaxy (see the &lt;a href=&quot;http://www.astro.ipm.ir/ISGLT08/Lecture3.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;KSB+ method&lt;/a&gt;) and to reconstruct the image and position of the original galaxy via &lt;a href=&quot;https://en.wikipedia.org/wiki/Deconvolution&quot; rel=&quot;nofollow noreferrer&quot;&gt;deconvolution&lt;/a&gt; (see &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0612094&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chantry &amp;amp; Magain&lt;/a&gt;; a visual example is given &lt;a href=&quot;http://www.nature.com/nature/journal/v481/n7381/fig_tab/nature10669_F1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;). I'm not too familiar with either technique, though, so I can't give you a good overview.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even large-scale lensing has large mass requirements. zephyr pointed out that the foreground object that created the &lt;a href=&quot;https://en.wikipedia.org/wiki/Einstein_Cross&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein cross&lt;/a&gt; contained $\sim10^{10}M_{\odot}$ of dark matter (&lt;a href=&quot;https://arxiv.org/pdf/0807.4175v2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;van de Ven et al. (2010)&lt;/a&gt;). That is &lt;em&gt;enormous&lt;/em&gt;!&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2017-01-18T05:19:19.263" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19788" PostTypeId="1" AcceptedAnswerId="19790" CreationDate="2017-01-17T17:17:37.577" Score="8" ViewCount="261" Body="&lt;p&gt;A &lt;a href=&quot;https://arxiv.org/abs/1411.0115&quot;&gt;review by Martin Kilbinger, 2015&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cosmic shear is the distortion of images of distant galaxies due to weak gravitational lensing by the large-scale structure in the Universe. Such images are coherently deformed by the tidal field of matter inhomogeneities along the line of sight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;As a rough analogy&lt;/em&gt;: imagine looking at an image full of galaxies through a rippled piece of glass. While a circular lens might make a galaxy look smaller or larger, a ridge would have a preferential direction of magnification, and the galaxies under that ridge would have a statistical tendency to have slightly different sizes on average parallel vs perpendicular to the ridge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is such a phenomenon generally agreed to have been observed yet? Are there any plots (of any kind) that show such experimental data?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If I take this &lt;em&gt;rough analogy&lt;/em&gt; one step further, while the peaks and troughs of the ripple magnify and demagnify, the midpoints between would be more prismatic - they would tend to deflect the centers, or locations of objects systematically, rather than affect the shapes. Applying this to cosmic shear, wouldn't the existence of shear essentially suggest that - to put it simply - stuff is, in fact, &quot;not quite&quot; where it appears to be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gUigV.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gUigV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; from &lt;a href=&quot;https://www.behance.net/gallery/16447189/Free-Ripple-and-Frosted-Glass-Overlays&quot;&gt;Free Ripple and Frosted Glass Overlays, Bojan Živković&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-01-17T17:33:57.867" LastActivityDate="2017-01-17T18:02:56.890" Title="Is cosmic shear generally agreed to have been observed?" Tags="&lt;gravitational-lensing&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19790" PostTypeId="2" ParentId="19788" CreationDate="2017-01-17T18:02:56.890" Score="6" Body="&lt;p&gt;Yes it has been observed. There are numerous papers covering this concept. Before I get into weak-lensing (because that is what you reference in your question) I'll note that cosmic shear is simply a specific type of distortion of an object by a lensing event. It is not limited to the weak lensing regime and cases of strong lensing exhibit cosmic shear as well. So images like &lt;a href=&quot;https://apod.nasa.gov/apod/ap151127.html&quot;&gt;this one&lt;/a&gt; show cosmic shear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as cosmic shear by weak lensing goes, the first published and recorded case of a weak lensing measurment was in 1996 and published by &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1996ApJ...466..623B&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot;&gt;Brainerd, Blandford, &amp;amp; Smail (1996)&lt;/a&gt;. The cosmic shear you speak of is referred to as polarization in that paper, but they're essentially trying to measure cosmic shear. Note also, while this is weak lensing, most of that lensing occurs by one galaxy lensing another. As such it is also termed galaxy-galaxy lensing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the first detection, there's been steady, but limited work on galaxy-galaxy lensing and shear measurments by papers such as &lt;a href=&quot;https://arxiv.org/pdf/astro-ph/0507108v2.pdf&quot;&gt;Mandelbaum et al. (2006)&lt;/a&gt;, &lt;a href=&quot;http://www.aanda.org/articles/aa/pdf/2011/10/aa17308-11.pdf&quot;&gt;van Uitert et al. (2011)&lt;/a&gt;, and &lt;a href=&quot;https://arxiv.org/pdf/1304.4265v2.pdf&quot;&gt;Velander et al. (2013)&lt;/a&gt;. I think the Velander paper is the latest and greatest measurements you can get and they have a lot of good material in there. Because you requested an image, I'll pull out one from this paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G7bRP.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G7bRP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Caption&lt;/strong&gt;: Galaxy-galaxy lensing signal around red lenses which have been split into luminosity bins according to Table 1, and modeled using the halo model described in Section 3.2. The black dots are the measured differential surface density, $\Delta \Sigma$ , and the black line is the best-fit halo model with the separate components displayed using the same convention as  in Figure 3. Grey triangles represent negative points that are included  unaltered in the model fitting procedure, but that have here been moved up to positive value s as a reference. The dotted error bars are the unaltered error bars belonging to the negative points. The grey squares represent distance bins containing no objects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that in this paper, they're ultimately trying to measure the tangential component of the cosmic shear (you can't measure the radial component), denoted by $\gamma_t$ and by equation (1) in their paper, they relate this to the surface mass density $\Delta \Sigma$. So in their plot above, when they say they're measuring $\Delta \Sigma$, they're really measuring $\gamma_t$ and converting through a multiplicative constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limiting factor in progressing further in this field is the lack of detailed observations required to do a true cosmic shear measurement. Since this measurement can only be observed statistically and not directly, you need observations of the sky which are both deep and wide. It is hoped that the &lt;a href=&quot;https://en.wikipedia.org/wiki/Wide_Field_Infrared_Survey_Telescope&quot;&gt;WFIRST satellite&lt;/a&gt; will provide these measurements and thereby expand ten-fold the weak lensing work astronomers can do which in turn may some day allow us to make important statements about the content, distribution, and properties of Dark Matter.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-01-17T18:02:56.890" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19792" PostTypeId="2" ParentId="19786" CreationDate="2017-01-17T19:59:14.023" Score="2" Body="&lt;p&gt;If this hypothesis is correct, the body from which 16 Psyche formed was a 500km planetesimal (about half the diameter of Ceres) It would have been destroyed in a collision with another body. The other body would have had to have been large enough (300km) to complete fragment the asteroid. &lt;a href=&quot;http://adsabs.harvard.edu/abs/1999Icar..137..140D&quot; rel=&quot;nofollow noreferrer&quot;&gt;The problem with this hypothesis&lt;/a&gt; is that no other pieces of this planetesimal have been found (they should still have similar orbits as Psyche).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another hypothesis is that Psyche suffered from a series of smaller impact, that fractured and disrupted the parent body, but did not destroy it. This would give Psyche a composition similar to the stony-iron mesosiderite meteorites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Psyche mission is intended to answer these questions. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-01-17T19:59:14.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19794" PostTypeId="2" ParentId="12131" CreationDate="2017-01-17T20:24:58.567" Score="1" Body="&lt;p&gt;As of 2016, the cause of the fluctuations in not known. There were significant dips in brightness in 2011, and 750 days later in 2013. It was not observed during the next potential dip in 2015. If the dips are caused by an orbiting body or bodies, the next dip will be in May 2017.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-01-17T20:24:58.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19795" PostTypeId="1" CreationDate="2017-01-17T20:46:12.550" Score="3" ViewCount="110" Body="&lt;p&gt;Spiral arms in galaxies can be modeled by potentials of the form&#xA;$$\Phi_{\text{spiral}}(R,\theta,z,t)=\sum_mA_m(R,z)\cos(m\theta-m\theta_0-\phi_m(R)-\Omega_pt)$$&#xA;where $A_m(R,z)$ is the $m$th-order coefficient based on some potential - such as a Miyamoto-Nagai potential - $\phi_m(R)$ is a function typically of the form&#xA;$$\phi_m(R)=\alpha_m\ln (R/R_{0,m})$$&#xA;for some characteristic radius $R$, and $\Omega_p$ is some pattern speed. I've read that typically the $m=0$ component dominates, followed by the $m=2$ term. There may also be an $m=4$ term, though it is typically weaker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This model describes the disk component of a galaxy with two main spiral arms. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Milky_Way#Spiral_arms&quot; rel=&quot;nofollow noreferrer&quot;&gt;spiral arms of the Milky Way&lt;/a&gt; are more complicated; some consider it to have two major arms and two more subordinate arms, while others consider it to have four major arms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have any potential or density models of the Milky Way been created using these Fourier components, and if so, what are the parameters used (the $m$s, $A_m$s, $\phi_m$s, and $R_{0,m}$s)?&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2017-01-18T14:25:32.123" LastActivityDate="2017-02-07T19:56:56.180" Title="Have parameters been determined for the Fourier components of a Milky Way spiral density-potential model?" Tags="&lt;milky-way&gt;&lt;spiral-arms&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19796" PostTypeId="2" ParentId="19795" CreationDate="2017-01-17T20:46:12.550" Score="2" Body="&lt;h1&gt;Yes.&lt;/h1&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://arxiv.org/abs/1106.1170&quot; rel=&quot;nofollow noreferrer&quot;&gt;Antoja et al. (2011)&lt;/a&gt;&lt;/strong&gt; created two models: The Tight-Winding Approximation model (TWA) and the PERLAS model. They decided to use just the &lt;span class=&quot;math-container&quot;&gt;$m=2$&lt;/span&gt; term for their TWA, based on data from &lt;a href=&quot;https://en.wikipedia.org/wiki/Spitzer_Space_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Spitzer Space Telescope&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The amplitude &lt;span class=&quot;math-container&quot;&gt;$A(R)$&lt;/span&gt; was&#xA;&lt;span class=&quot;math-container&quot;&gt;$$A(R)=A_{sp}Re^{-R/R_{\Sigma}}$$&lt;/span&gt;&#xA;Their &lt;span class=&quot;math-container&quot;&gt;$\phi_2(R)$&lt;/span&gt; (denoted &lt;span class=&quot;math-container&quot;&gt;$g(R)$&lt;/span&gt; in the paper) was of the form&#xA;&lt;span class=&quot;math-container&quot;&gt;$$g(R)=\left(\frac{2}{N\tan i}\right)\ln\left(1+\left(\frac{R}{R_{sp}}\right)^N\right)$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; measures, in their words,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;how sharply the change from a bar-like to spiral-like occurs in the inner regions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;They approximated using &lt;span class=&quot;math-container&quot;&gt;$N=100$&lt;/span&gt;, although &lt;span class=&quot;math-container&quot;&gt;$g(R)$&lt;/span&gt; should properly use the limit &lt;span class=&quot;math-container&quot;&gt;$N\to\infty$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; is the pitch angle of the arms, which they measured to be &lt;span class=&quot;math-container&quot;&gt;$15.5$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$12.8^{\circ}$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$R_{sp}$&lt;/span&gt;, the beginning of the spiral, was found to be &lt;span class=&quot;math-container&quot;&gt;$2.6$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$3.6\text{ kpc}$&lt;/span&gt;. For the other constants, &lt;span class=&quot;math-container&quot;&gt;$A_{sp}$&lt;/span&gt; was taken to be &lt;span class=&quot;math-container&quot;&gt;$650$&lt;/span&gt;-&lt;span class=&quot;math-container&quot;&gt;$1100$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$850$&lt;/span&gt;-&lt;span class=&quot;math-container&quot;&gt;$1300[\text{km s}^{-1}]^2\text{ kpc}^{-1}$&lt;/span&gt;, depending on the exact arm shape, and the scale length &lt;span class=&quot;math-container&quot;&gt;$R_{\Sigma}$&lt;/span&gt; was set to &lt;span class=&quot;math-container&quot;&gt;$2.5\text{ kpc}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The pattern speed &lt;span class=&quot;math-container&quot;&gt;$\Omega_p$&lt;/span&gt; was set in the range &lt;span class=&quot;math-container&quot;&gt;$15$&lt;/span&gt;-&lt;span class=&quot;math-container&quot;&gt;$30\text{ km/s}$&lt;/span&gt;, which fits nicely with &lt;a href=&quot;https://astronomy.stackexchange.com/a/19177/2153&quot;&gt;other values I've seen&lt;/a&gt;. &lt;span class=&quot;math-container&quot;&gt;$\theta_0$&lt;/span&gt; (denoted &lt;span class=&quot;math-container&quot;&gt;$\phi_0$&lt;/span&gt; in the paper) was set to &lt;span class=&quot;math-container&quot;&gt;$88$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$60^{\circ}$&lt;/span&gt; - a little bigger than the spiral arm pitch angle. putting all this together, we get a potential of the form&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\Phi_\text{spiral}(R,\theta,t)=A_{sp}Re^{-R/(2.5\text{ kpc})}\times\cos\left(2(\theta-88\text{ to }60)+\left(\frac{2}{100\tan (15.5\text{ to }12.8^{\circ})}\right)\ln\left(1+\left(\frac{R}{2.6\text{ to }3.6\text{ kpc}}\right)^{100}\right)-\Omega_{p}t\right)$$&lt;/span&gt;&#xA;A &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;-component could be multiplied to get a three-dimensional potential, likely exponentially decaying with height.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://academic.oup.com/mnras/article/433/3/2511/1235707/Fitting-the-Lin-Shu-type-density-wave-theory-for&quot; rel=&quot;nofollow noreferrer&quot;&gt;Griv et al. (2013)&lt;/a&gt;&lt;/strong&gt; also used density-wave theory. Their potential was&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\Phi(R,\theta,t)=\Phi_{\text{mean}}(R)+\tilde\Phi\sqrt{\frac{R_0}{R}}\cos\phi$$&lt;/span&gt;&#xA;where&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\phi=\phi_0+m\left[\Omega_pt-\varphi+(1/\tan p)\ln(R/R_0)\right]$$&lt;/span&gt;&#xA;which is essentially the same potential as discussed in the question, with&#xA;&lt;span class=&quot;math-container&quot;&gt;$$A(R)=\tilde{\Phi}\sqrt{\frac{R_0}{R}}$$&lt;/span&gt;&#xA;The authors chose &lt;span class=&quot;math-container&quot;&gt;$m=1,2,3$&lt;/span&gt; components, with different parameters for each. &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; (in degrees) ranged from &lt;span class=&quot;math-container&quot;&gt;$-6$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$-20$&lt;/span&gt;, while &lt;span class=&quot;math-container&quot;&gt;$\phi_0$&lt;/span&gt; (also in degrees) ranged from &lt;span class=&quot;math-container&quot;&gt;$239$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$247$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$\Omega_p$&lt;/span&gt; was from &lt;span class=&quot;math-container&quot;&gt;$32.4$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$37.6$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$\text{ km/s}$&lt;/span&gt;), also consistent with other models. &lt;span class=&quot;math-container&quot;&gt;$\tilde{\Phi}$&lt;/span&gt; saw the most variation, ranging from &lt;span class=&quot;math-container&quot;&gt;$-65$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$-390$&lt;/span&gt; (in &lt;span class=&quot;math-container&quot;&gt;$\text{kpc}^{-2}\text{s}^{-2}$&lt;/span&gt;). &lt;span class=&quot;math-container&quot;&gt;$R_0$&lt;/span&gt; was set to &lt;span class=&quot;math-container&quot;&gt;$8\text{ kpc}$&lt;/span&gt;, I believe. Again, an exponentially decaying function should account for the &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;-dependence.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-02-07T19:56:56.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19797" PostTypeId="2" ParentId="19359" CreationDate="2017-01-17T21:38:32.007" Score="1" Body="&lt;p&gt;The other star would have a significant proper motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over the course of one orbit, it would appear to travel right around the sky once. The variation in position due to the Earth's orbit of the sun would be superimposed on top of this motion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Except that the proper motion of the star, and its parallax would be much greater than for any real star, this is not really different to the way in which the parallax motion of a star is superimposed on its proper motion.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-01-17T21:38:32.007" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19798" PostTypeId="1" AcceptedAnswerId="19809" CreationDate="2017-01-17T22:36:28.230" Score="1" ViewCount="134" Body="&lt;p&gt;The widely known phenomenon termed &lt;em&gt;gravitational lensing&lt;/em&gt; is believed to curve space, thereby affecting the path followed by light and other electromagnetic waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This effect is associated principally with galaxies and galaxy clusters, as it requires considerable mass. But in theory any star or star cluster, or black hole, can have this effect to some degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can we tell for sure about the position of external galaxies, if we are uncertain that their light is actually coming to us along a straight path: i.e. if light follows the curvature of space, and that curvature is not uniform, how can we be sure we are not observing, for example, the light from a single source appearing to us as more than one object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can a galaxy appear to be in more than one place in the sky?&lt;/p&gt;&#xA;" OwnerUserId="15531" LastActivityDate="2017-01-19T10:27:31.983" Title="Given that gravity bends the path of light, how confident can we be that galaxies are where they seem to be?" Tags="&lt;gravitational-lensing&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19800" PostTypeId="1" CreationDate="2017-01-18T02:53:50.160" Score="2" ViewCount="131" Body="&lt;p&gt;I have taken photos at different exposures from different places of the Orion constellation. I wish to find the effect of skyglow on the limiting magnitude of stars in the Orion constellation. How can I estimate the limiting value of any star from its RGB pixel value in the photograph?&lt;/p&gt;&#xA;" OwnerUserId="15512" LastActivityDate="2017-01-18T02:53:50.160" Title="Convert RGB pixel value of an image to limiting magnitude" Tags="&lt;photography&gt;&lt;photometry&gt;&lt;light-pollution&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19801" PostTypeId="2" ParentId="279" CreationDate="2017-01-18T12:40:13.287" Score="4" Body="&lt;p&gt;A recent study reported in the Phys.org article &quot;&lt;a href=&quot;https://phys.org/news/2017-01-main-moon-formation-theory.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Study crashes main Moon-formation theory&lt;/a&gt;&quot; , based on the Nature Geoscience paper &quot;&lt;a href=&quot;http://www.nature.com/ngeo/journal/vaop/ncurrent/full/ngeo2866.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;A multiple-impact origin for the Moon&lt;/a&gt;&quot; performed numerical modelling that suggests that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the Moon could instead be the product of a succession of a variety of smaller collisions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The specific strength of such a theory is that &quot;Numerous &quot;impactors&quot; would have excavated more Earth material than a single one&quot; according to the authors, allowing for a stronger composition-based correlation with the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors also contend that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;sub-lunar moonlets are a common result of impacts expected onto the proto-Earth in the early Solar System and find that the planetary rotation is limited by impact angular momentum drain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The process is simplified as follows (from the article):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xwI66.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xwI66.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting consequence of this model, according to the authors is that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Building the Moon in this way takes many millions of years, implying that the Moon's formation overlapped with a considerable portion of Earth's growth,&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user15217" LastActivityDate="2017-01-18T12:40:13.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19802" PostTypeId="2" ParentId="19766" CreationDate="2017-01-18T22:15:38.153" Score="6" Body="&lt;p&gt;The answer is yes. As the Sun ages, it will become a red giant and the mass loss rate from its surface will increase. This effect will increase (dramatically) further when the Sun enters the asymptotic giant branch phase, where thermal pulsations drives a cool wind that may carry away a millionth of a solar mass per year, eventually leaving a burned-out core in the form of a white dwarf with about half a solar mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any point in this evolution we can model the evolution of the Earth's orbit using some simple approximations - that the wind from the Sun escapes to infinity, that a negligible proportion is actually accreted by the Earth and nor does it exert a torque,  that the mass loss takes place on a timescale much longer than the Earth's orbit and that the mass of the Earth $m$ is always much less than the time-dependent mass of the Sun $M(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In which case we consider the orbital angular momentum of the Earth:&#xA;$$ m a \omega^2 \simeq G\frac{M m}{a^2},$$&#xA;where $a$ is the semi major axis. So the angular momentum $J = m a^2 \omega$ is given by&#xA;$$ J^2 = m^2 a^4 \frac{G M m}{m a^3} \propto M a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the angular momentum of the Earth's orbit is conserved, the $M(t) a(t)$ is constant and as the Sun loses mass, the semi major axis increases by the same factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coming to the specifics - when the Sun is a half solar mass white dwarf, the semi-major axis will be 2 au (assuming the giant Sun did not quite engulf it - it will be a close-run thing) and Kepler's third law $(P^2 \propto a^3/M)$ can be used to estimate an orbital period of 4 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tidal effects of the Sun on the Earth's orbit are quite negligible compared with these mass loss effects.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-01-18T22:24:44.507" LastActivityDate="2017-01-18T22:24:44.507" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19804" PostTypeId="1" CreationDate="2017-01-19T07:10:28.277" Score="2" ViewCount="68" Body="&lt;p&gt;Is the total energy of the CMBR diminishing with time? Since it's existence the volume of the Universe (at least the visible part) has grown by a factor three, which is to say, the number of photons per unit volume is diminished by a factor 27. But the wavelength of these photons has grown by a much bigger factor. So you should think the total energy of all the CMBR photons has become less over time. &lt;/p&gt;&#xA;" OwnerUserId="10659" LastActivityDate="2017-01-19T07:17:03.433" Title="Is the total energy of the CMBR diminishing with time?" Tags="&lt;cmb&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19805" PostTypeId="2" ParentId="19804" CreationDate="2017-01-19T07:17:03.433" Score="1" Body="&lt;p&gt;Yes. The energy density in the radiation field scales like $a^{-4}$, and the volume scales like $a^{3}$. Since the total energy is the density times the volume, the total energy scales like $a^{-1}$. Note that $a$ is called the &quot;scale factor&quot; and $a = \frac{1}{1+z}$, where $z$ is the redshift, assuming $z=0$ now.&lt;/p&gt;&#xA;" OwnerUserId="14244" LastActivityDate="2017-01-19T07:17:03.433" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19806" PostTypeId="1" CreationDate="2017-01-19T08:43:37.763" Score="4" ViewCount="112" Body="&lt;p&gt;I was just told the CMBR energy diminishes with time. Can it be that the lost energy is somehow transferred to spacetime, causing the expansion to accelerate?&lt;/p&gt;&#xA;" OwnerUserId="10659" LastActivityDate="2017-01-19T09:42:31.977" Title="Can the diminishing energy of the CMBR be the source of dark energy?" Tags="&lt;expansion&gt;&lt;dark-energy&gt;&lt;cmb&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19808" PostTypeId="2" ParentId="19806" CreationDate="2017-01-19T09:42:31.977" Score="8" Body="&lt;p&gt;No - the decreasing energy in the CMB is already well modeled in the &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Friedmann_equations&quot;&gt;Friedmann equations&lt;/A&gt;. The term in the density parameter that is proportional to $a^{-4}$ is the contribution of radiation energy density to the evolution of the universe, the term proportional to $a^{-3}$ is matter density (mostly dark, but includes ordinary matter), $a^{-2}$ is the contribution of the curvature of space-time itself, and the term without any factors of $a$ is the contribution of dark energy. The size of the radiation density, today, is already a small fraction of the matter density (about 0.03% of the matter density, 0.01% of the density of the universe overall - ordinary matter is about 5% overall). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last time the energy density in the radiation fields was the same size as what's in the matter fields was &lt;A HREF=&quot;https://lambda.gsfc.nasa.gov/product/map/current/params/lcdm_wmap9.cfm&quot;&gt;around $z=3,300$&lt;/A&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also disagree with @J.Chomel's answer - the energy stored in the radiation field is decreasing. Then energy density in the radiation field scales like $a^{-4}$, and the volume scales like $a^3$. Since the total energy is the energy density times the volume, the total energy scales like $a^{-1}$, just as you would expect with the number of photons being fixed, but the energy in each photon scaling as $a^{-1}$ as the wavelength increases. &lt;/p&gt;&#xA;" OwnerUserId="14244" LastActivityDate="2017-01-19T09:42:31.977" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="19809" PostTypeId="2" ParentId="19798" CreationDate="2017-01-19T10:27:31.983" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The widely known phenomenon termed gravitational lensing is believed&#xA;  to curve space, thereby affecting the path followed by light and other&#xA;  electromagnetic waves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and gravitational waves! In fact, the paths followed by light and other massless particles (for the lack of a better word) are called &lt;em&gt;null geodesics&lt;/em&gt; and they have in common that $$\mathrm ds^2 = 0, $$ i.e. the 4-distance they experience is zero.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This effect is associated principally with galaxies and galaxy&#xA;  clusters, as it requires considerable mass. But in theory any star or&#xA;  star cluster, or black hole, can have this effect to some degree.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One could argue that the Shapiro delay is of the same nature, only temporal rather than spatial. You can observe the Shapiro delay from the sun using radar probes!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What can we tell for sure about the position of external galaxies, if&#xA;  we are uncertain that their light is actually coming to us along a&#xA;  straight path: &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is by definition! The path that light takes &lt;em&gt;is&lt;/em&gt; straight, even if it seems bent to us in 3-dimensional space. Therefore, galaxies appear where they are because that's where they are. It's just that we project our idea of Euclidean (i.e. unbent) space onto the sky!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know what you want to ask though: How can we be sure that the sky isn't curved in some insane way somewhere and we're actually looking somewhere else? We &lt;em&gt;do&lt;/em&gt; assume that on the whole, space is Euclidean. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, on the small scale this might be the case: We might be looking at stuff whilst it is &quot;really&quot; somewhere else (note that it is really there, the &quot;really somewhere else part&quot; is just us imposing our coordinates on the sky!!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we know beyond reasonable doubt that the Universe, as a whole, is more or less Euclidean. At least up to redshift z~1000, i.e. about 300,000 years after the Big Bang. We can observe the Cosmic Microwave Background, and it tells us something about the curvature in the sky. We know that there's structures there which have a certain angular size, as we know that they had 300,000 years to form and we know the speed at which they formed. So we can predict their typical size, and it turns out that the angle under which they appear would be bigger if the universe were oblate/hyperbolic and smaller if the universe were closed. They are just right for a flat universe. So we know that on big scales, space is indeed Euclidean. It might not be the case for individual stars, galaxies, etc. but on average, it works. And cosmology / gravitational lensing usually works with averages and statistics.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can a galaxy appear to be in more than one place in the sky?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Totally. There's a plethora of cases of &lt;a href=&quot;http://www.roe.ac.uk/~heymans/website_images/148928main_image_feature_575_ys_4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;multi image lens-source systems&lt;/a&gt;. You can usually sort them out by looking at their spectra and checking their time-variability.&lt;/p&gt;&#xA;" OwnerUserId="15548" LastActivityDate="2017-01-19T10:27:31.983" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19810" PostTypeId="1" AcceptedAnswerId="19811" CreationDate="2017-01-19T12:23:53.537" Score="2" ViewCount="347" Body="&lt;p&gt;If I'm on a life boat, in the middle of the ocean, radioing another ship to come pick me up - I'd say something like &quot;I'm 40 kilometers directly east of you.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In space, it's not that easy. How do you give a direction?&lt;/p&gt;&#xA;" OwnerUserId="15550" LastActivityDate="2017-01-19T12:58:56.673" Title="How do we give directions in space?" Tags="&lt;navigation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19811" PostTypeId="2" ParentId="19810" CreationDate="2017-01-19T12:58:56.673" Score="4" Body="&lt;p&gt;You need to define an agreed standard for orientation of axis and origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What co-ordinate system you choose as a standard depends on your purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some examples and links : &lt;a href=&quot;https://en.wikipedia.org/wiki/Celestial_coordinate_system&quot;&gt;Celestial coordinate systems&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical reference points are defined using very distant objects whose position can be measured accurately ( typically quasars ).&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-01-19T12:58:56.673" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19813" PostTypeId="1" AcceptedAnswerId="19815" CreationDate="2017-01-19T19:28:35.820" Score="13" ViewCount="1047" Body="&lt;p&gt;What is the range of percentage mass of parent star left in a stellar black hole directly after its formation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What factors determine this number for a specific case?&lt;/p&gt;&#xA;" OwnerUserId="15555" LastActivityDate="2017-01-19T20:35:27.883" Title="Mass of black holes compared to parent star" Tags="&lt;star&gt;&lt;black-hole&gt;&lt;mass&gt;&lt;stellar-evolution&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19814" PostTypeId="2" ParentId="19813" CreationDate="2017-01-19T20:31:20.810" Score="5" Body="&lt;p&gt;Your question regards the formation of stellar-mass black holes, which form as the result of a Type II or Type Ib supernova explosion. This occurs when a massive star's core collapses from its own self-gravity, driving a rapid release of energy through nuclear reactions. This imparts a tremendous amount of energy in the form of photons and neutrinos to the rest of the star, which, as a result, blows the star up. This core region either becomes a neutron star or, when the mass of this core region is high enough, collapses directly into a black hole. While stars that can explode through this channel are rare in the Milky Way, i.e., compared to stars like our Sun, there are likely ~billions of neutron stars and stellar-mass black holes that have formed through this process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars that explode as supernova are indeed massive, weighing in with masses at least ~8 times the mass of the Sun. Those that produce black holes in the center are even higher, usually above ~20 solar masses or so (this number is disputed... some of the nuclear physics in these extreme environments is uncertain). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure 2 of &lt;a href=&quot;https://arxiv.org/pdf/astro-ph/9510136v1.pdf&quot;&gt;this paper&lt;/a&gt; might shed some light (...) on your question. This paper ran a set of stellar evolution models to track how much mass was expelled during the explosion and how much mass remained post explosion. The horizontal axis gives the original mass of the star (in units of the Sun's mass, e.g., a value of 10 means 10 times the mass of the Sun), and the solid circles identify the final mass of the leftover remnant--which is either a neutron star or black hole. The vertical axis gives the mass of the remnant. Sadly they decided to use logarithmic space for the vertical axis, even though the range is only over a single order of magnitude. So to get the actual amount of mass, you have to undo the base-10 logarithm. For example, if a black dot had a value of 0.3 on the vertical axis, the mass of the remnant would be 10^(0.3) = 2.0 times the mass of the Sun. A value of 0.6 would be 10^(0.6) = 3.98 times the mass of the Sun, etc. They considered several different mechanisms for the explosion at higher masses (remember, things get more uncertain the bigger the star gets), which is why some horizontal values have multiple black dots. If you're curious, weaker explosions can allow some of the material to fall back down onto the remnant, which results in a black dot that is higher up on the plot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless, you can see that, for example, a 20 solar mass star creates a 10^(0.3)=2 solar mass remnant. A 30 solar mass star might create a remnant that is between 2 and 4 times the mass of the Sun. In all cases, the majority of the star's original mass is lost. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also glance at the plots of &lt;a href=&quot;https://arxiv.org/pdf/0904.2784v2.pdf&quot;&gt;this paper&lt;/a&gt; as well. This paper looks like it did a slightly more careful job. Either paper still gives you the basic picture, however. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Aside: Figure 2 is for 'solar metallicity' stars, which means 'stars that you might find in the Milky Way.' Figure 1 is for stars that would have formed in the early Universe, before a considerable amount of elements beyond helium had been formed.)&lt;/p&gt;&#xA;" OwnerUserId="15557" LastActivityDate="2017-01-19T20:31:20.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19815" PostTypeId="2" ParentId="19813" CreationDate="2017-01-19T20:35:27.883" Score="9" Body="&lt;p&gt;There is no general consensus on this. Different evolutionary models give different results. The factors (in addition to the initial mass of the star) that effect the final black hole mass would be the rotation rate of the progenitor, its composition (or metallicity) and whether it was in a binary system or not and whether that binary system was able to transfer mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotation is thought to be important because it affects internal mixing and therefore the rate at which fuel is supplied to the core and the rat at which processed material gets to the surface, affecting the atmospheric composition. It can also enhance mass loss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The composition is important because mass loss is driven by radiation and radiative opacities are higher for high metallicity compositions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set of calculations by &lt;a href=&quot;http://cds.cern.ch/record/589942/files/0211062.pdf&quot;&gt;Heger et al. (2003)&lt;/a&gt; are one of the canonical works on this subject. Below is a plot of initial mass versus remnant mass for stars with big bang primordial abundance (zero initial metallicity) and then the same again for stars of solar metallicity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio of the red line to dotted &quot;no mass loss&quot; line gives the fraction you are after. &#xA;In zero metallicity (primordial) stars it increases from 10-40% for initial masses of 25-100 solar masses and is perhaps even higher for supermassive population III stars. (I stress that these are &lt;em&gt;theoretical results&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For solar metallicity stars the results are a bit different. The ratio of the red line to the dotted line varies from 10-25% for 25-40 solar masses, but then it is not clear whether black holes can even form at even higher masses because of the much higher mass loss rates (see the difference between the dotted line and the blue curve).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Xydf3.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xydf3.png&quot; alt=&quot;Zero metallicity relation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eh8HQ.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eh8HQ.png&quot; alt=&quot;Solar metallicity relation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-01-19T20:35:27.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19816" PostTypeId="1" CreationDate="2017-01-19T21:06:38.617" Score="-2" ViewCount="229" Body="&lt;p&gt;i just don't really understand this topic.&lt;/p&gt;&#xA;" OwnerUserId="15558" LastActivityDate="2017-01-20T14:31:03.363" Title="Does making a ground based telescope larger always give you a better resolution?? why?? or why not?" Tags="&lt;telescope&gt;&lt;newtonian-telescope&gt;&lt;telescope-lens&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2017-01-23T14:58:06.627" ContentLicense="CC BY-SA 3.0" />
  <row Id="19817" PostTypeId="2" ParentId="19647" CreationDate="2017-01-19T21:41:11.167" Score="3" Body="&lt;p&gt;The Hubble Space Telescope is not expected to be retired unless a failure of one or more components dramatically effects its capabilities and reduces the science return.  There is no provision currently for repairing Hubble, which depended on NASA's Space Shuttle program, no longer operating.  Lots more information about &lt;em&gt;HST&lt;/em&gt; is at &lt;a href=&quot;http://hubblesite.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://hubblesite.org/&lt;/a&gt; and elsewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NASA's next flagship space science mission is the James Webb Space Telescope, expected to be launched in late 2018.  This is a follow-on to Hubble's science mission and will be able to answer questions that Hubble is not designed to do. Among many other things it is expected to study some of the first stars and galaxies that formed not long after the beginning of the Universe, known as the Big Bang. &#xA;Get much more information about &lt;em&gt;JWST&lt;/em&gt; at &lt;a href=&quot;http://webbtelescope.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://webbtelescope.org/&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="15559" LastActivityDate="2017-01-19T21:41:11.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19818" PostTypeId="2" ParentId="19783" CreationDate="2017-01-20T00:05:23.713" Score="2" Body="&lt;p&gt;By lens in this case I presume you mean eyepiece.&lt;br /&gt;&#xA;The magnification you get from an eyepiece is given by dividing the focal length of the mirror (650) by the focal length of the eyepiece. You did not say what eyepieces you have but if, for instance, you had a 10mm eyepiece you would get a magnification of 650/10=65x. A 20mm eyepiece would give a magnification of 33x and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars is notoriously hard to see any detail on, it is small and never gets very close to earth, a small red dot is about all you are ever likely to see apart from every couple of years around opposition when it is close to us and you may see a white area at one pole or the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus at the moment is easily visible in the Western sky before around sunset, it shows phases like the moon and you should be able to see those, it is currently half full.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nebulae, galaxies and the like require low power eyepieces (in general) and dark skies, the aperture of the main mirror (6&quot; for you) is much more important here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try the scope out on the moon, the Orion nebula and maybe Saturn to get the feel for what you can see, a small scope like this will show plenty, just don't expect to see anything like the photographs you see in magazines and books, visual astronomy is hard and mostly black and white!&lt;/p&gt;&#xA;" OwnerUserId="15560" LastEditorDisplayName="user1569" LastEditDate="2017-01-20T14:41:46.800" LastActivityDate="2017-01-20T14:41:46.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19820" PostTypeId="1" AcceptedAnswerId="19822" CreationDate="2017-01-20T10:59:21.643" Score="8" ViewCount="190" Body="&lt;p&gt;I have read in relation to Big Bang theory that after 300,000 years of big Bang temperature was reduced to 4500 Kelvin and this gave rise to Atomic matter.So,my question is why reduction in temperature makes it suitable for atom formation?&lt;/p&gt;&#xA;" OwnerUserId="15313" LastActivityDate="2017-01-20T14:07:20.627" Title="What made cooler temperatures suitable for atom formation?" Tags="&lt;big-bang-theory&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19821" PostTypeId="2" ParentId="19820" CreationDate="2017-01-20T13:09:07.180" Score="0" Body="&lt;p&gt;Keeping it as simple as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more energy there is, the harder it is for the (relatively) weak electromagnetic force to bind electrons to nuclei ( and an atom is a nucleus with electrons bound to it ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When there was more energy the electrons and nuclei had too much energy to be bound together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simplified way of looking at this is that the electrons and nuclei were simply moving to fast when they were hot ( temperature is related to average energy which is related to motion ).&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-01-20T13:09:07.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19822" PostTypeId="2" ParentId="19820" CreationDate="2017-01-20T13:12:50.490" Score="9" Body="&lt;p&gt;The temperature of a gas is a measure of the kinetic energy of the particles. For molecules you can have rotational and vibrational energy, while for single atoms you just have translational energy, or &quot;thermal motion&quot;. At a given temperature, the particles don't have exactly the same energy, but a &lt;em&gt;distribution&lt;/em&gt; of energies, and hence velocities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most (&gt;90%) of the gas in the Universe is hydrogen. The energy needed to knock the electron off (i.e. ionize) a hydrogen atom is 13.6 eV. For a gas of $T \gtrsim 3000\,\mathrm{K}$, the fraction of particles with sufficient energy to ionize hydrogen is so high, that the majority of the atoms are ionized&lt;sup&gt;$\dagger$&lt;/sup&gt;, i.e. split up in protons and electrons. This was the case in the beginning of the history of the Universe. Everytime a proton and electron met and recombined to form a neutral atom, the electron would almost immediately be knocked off again by a high-energy particle (usually an electron, but it could also be a proton or photons, since all particles were in &quot;thermodynamic equilibrium&quot;, i.e. shared the same distribution of energies).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the Universe expanded, the gas cooled. At some point, 380,000 years after Big Bang, the temperature had decreased enough that it was no longer possible to keep the atoms ionized, so over a rather short period of time ($\sim10^4$ years), they all recombined. This epoch is hence called &lt;em&gt;the epoch of recombination&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until this point, all photons kept scattering on the free electrons. With the electrons &quot;trapped&quot; in atoms, they could now stream freely, and &quot;decouple&quot;. They have been traveling freely ever since, but since they travel through an expanding Universe, they become redshifted along the way. Since then, the Universe has expanded by a factor of ~1100, and so have the wavelength of the photons, so that today they have temperature of $3000\,\mathrm{K}/1100\simeq2.7\,\mathrm{K}$. This is what we see as the &lt;em&gt;cosmic microwave background&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;$\dagger$&lt;/sup&gt;In which case they're in principle not &quot;atoms&quot;, but a plasma. However, in astronomy it's quite normal to call it atoms anyway.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2017-01-20T13:12:50.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19823" PostTypeId="2" ParentId="19820" CreationDate="2017-01-20T14:07:20.627" Score="0" Body="&lt;p&gt;Atoms are formed by nuclei surrounded by electrons, bound by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Electromagnetism&quot; rel=&quot;nofollow noreferrer&quot;&gt;electromagnetic force&lt;/a&gt;; nuclei are composed of protons and neutrons, bound by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Strong_interaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;strong nuclear force&lt;/a&gt;; protons and neutrons are in turn composed of (different types and amounts of) &lt;a href=&quot;https://en.wikipedia.org/wiki/Quark&quot; rel=&quot;nofollow noreferrer&quot;&gt;quarks&lt;/a&gt;, also bound by the strong force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the very beginning of the universe it is theorized that all forces (those mentioned plus gravity and the weak nuclear force) were one; as the temperature dropped they started becoming distinct. Temperature, for particles, means energy. You cannot have atomic nuclei until quarks become bound as protons and nucleons and these in turn bound to each other, that is, you need the strong nuclear force to be distinct and overcome the tendency of energetic particles to just go away at random.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the temperature drops, protons and neutrons form and then become bound in nuclei. Electrons are still scattered about, even though the electromagnetic force has become distinct as well, because they are still very energetic and they are hit all the time by other energetic particles, and the electromagnetic force is very, very weak compared to the strong nuclear force. At this stage the universe is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Plasma_(physics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;plasma&lt;/a&gt;, that is, a soup of nuclei and free electrons. (It still is, for the most part and excluding dark matter and dark energy, though that's not the case on our tiny corner of it, i. e. Earth.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the temperature drops some more and the electromagnetic force, which attracts positively-charged nuclei to negatively-charged electrons, begins to be felt. At this point &quot;regular&quot; atoms can form (and once the temperature is low enough, they can also bond to each other forming molecules).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can easily force nuclei and electrons to become detached again by using high temperature (or other forms of energy). It's rather more costly to &lt;a href=&quot;https://en.wikipedia.org/wiki/Nuclear_fission&quot; rel=&quot;nofollow noreferrer&quot;&gt;dismantle nuclei&lt;/a&gt; (unless they are inherently unstable, i. e. radioactive, as in the case of uranium), and &lt;em&gt;extremely&lt;/em&gt; difficult to &lt;a href=&quot;https://en.wikipedia.org/wiki/Quark%E2%80%93gluon_plasma&quot; rel=&quot;nofollow noreferrer&quot;&gt;dismantle protons or neutrons&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="15398" LastActivityDate="2017-01-20T14:07:20.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19824" PostTypeId="2" ParentId="19816" CreationDate="2017-01-20T14:31:03.363" Score="2" Body="&lt;p&gt;The diffraction limit of telescope optics, for apertures large compared to the wavelength of light, is inversely proportional to the diameter of the aperture. See the relevant &lt;a href=&quot;https://en.wikipedia.org/wiki/Diffraction-limited_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For moderate to large telescopes with simple optics this resolution limit is usually not realisable due to seeing. The seeing limit is due to thermal inhomogeneity and turbulence in the atmosphere between the telescope and the target. See this &lt;a href=&quot;http://www.skyatnightmagazine.com/feature/general-guide/guide-seeing-and-atmospheric-transparency&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sky at Night page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To some extent the seeing limit can be overcome using adaptive optics. See the relevant &lt;a href=&quot;https://en.wikipedia.org/wiki/Adaptive_optics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2017-01-20T14:31:03.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19826" PostTypeId="1" CreationDate="2017-01-20T18:07:46.823" Score="2" ViewCount="2568" Body="&lt;p&gt;Today I try to watch Venus through a mirror telescope. I am no professional. No matter what lens I take, I see this image:&lt;a href=&quot;https://i.stack.imgur.com/N3ZeI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/N3ZeI.png&quot; alt=&quot;strange&quot;&gt;&lt;/a&gt; What am I doing wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: What happens when I refocus:&lt;a href=&quot;https://i.stack.imgur.com/ZPmGu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZPmGu.png&quot; alt=&quot;focus 1&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZXYC1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZXYC1.png&quot; alt=&quot;focus 2&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/yPnXL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yPnXL.jpg&quot; alt=&quot;focus 3&quot;&gt;&lt;/a&gt;Sorry for the bad quality, I didn't manage to make better images. :-(&lt;/p&gt;&#xA;" OwnerUserId="15572" LastEditorUserId="15572" LastEditDate="2017-01-20T19:01:54.960" LastActivityDate="2017-03-24T17:15:53.290" Title="Mirror telescope: blocked sight" Tags="&lt;telescope&gt;" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="19827" PostTypeId="1" AcceptedAnswerId="19828" CreationDate="2017-01-20T21:12:59.307" Score="4" ViewCount="892" Body="&lt;p&gt;I'm working on my assessment for Physics in the IB, and I'm planning to investigate stars in spiral galaxies and determine their ages. For this I'm going to use data from the SDSS (Sloan digital sky survey &lt;a href=&quot;http://skyserver.sdss.org/dr13/en/tools/chart/navi.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://skyserver.sdss.org/dr13/en/tools/chart/navi.aspx&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to get the luminosity of a star from the absolute bolometric magnitude. However the SDSS only gives magnitude in 5 values, u g r i and z. I've tried to understand what to do with these and what they mean but I can't figure it out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I get one number the absolute bolometric magnitude from 5 different values, and what do they mean?&lt;/p&gt;&#xA;" OwnerUserId="15573" LastEditorUserId="2365" LastEditDate="2017-01-21T23:24:00.363" LastActivityDate="2017-04-20T11:55:37.720" Title="Finding luminosity of a star with magnitude" Tags="&lt;luminosity&gt;&lt;magnitude&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19828" PostTypeId="2" ParentId="19827" CreationDate="2017-01-20T22:28:48.103" Score="3" Body="&lt;p&gt;The ugriz values are just measures of the stellar brightness taken in 5 bands from the near ultraviolet u through to the near infrared z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need the distance to the star to convert the apparent SDSS magnitudes to absolute magnitudes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you need to find a relationship (or relationships) between bolometric correction in a particular band and the colour of the star (by colour I mean something like g-i). Possibly instead you might find relationships between bolometric correction and temperature and will need to first estimate the temperatures of the stars from their colours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have absolute magnitudes and a bolometric correction, then the absolute bolometric magnitude follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle you could try this with a number of different colours and see if the results agree. Alternatively, you would fit a library of stellar models to the spectral energy distribution defined by the 5 magnitude values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A table of bolometric corrections for the SDSS filters was calculated by &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0404358&quot; rel=&quot;nofollow noreferrer&quot;&gt;Girardi et al. (2004)&lt;/a&gt; and can be found &lt;a href=&quot;http://pleiadi.oapd.inaf.it/isoc_photsys.01/isoc_sloan/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. (Other tables/formulations are available, there is no universally accepted version).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some bolometric correction tables can be found in &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/677403/pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schmidt et al. (2014)&lt;/a&gt; for very cool objects, which are more problematic.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-04-20T11:55:37.720" LastActivityDate="2017-04-20T11:55:37.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19829" PostTypeId="2" ParentId="19783" CreationDate="2017-01-21T00:09:12.430" Score="6" Body="&lt;p&gt;Okay, so that's a 130 mm newtonian with an f/5 focal ratio. Per &lt;a href=&quot;http://www.celestron.com/browse-shop/astronomy/telescopes/astromaster-telescopes/astromaster-130eq-telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;the manufacturer's website&lt;/a&gt;, it comes with two eyepieces, 10 mm and 20 mm, giving 66x and 33x magnifications respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, make sure your scope is collimated. The procedure is described in their generic Telescope Maintenance document, and also more specifically in the Astromaster Manual. See links on the site. The performance of your scope when observing planets depends a lot on collimation. I made it a habit to check collimation every time before I use the scope; I'm so used to it that it only takes 5 minutes now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The topic of collimation is huge, just google &quot;telescope collimation&quot; and learn. I'm still learning new things after years in this hobby.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the scope is collimated, you could try to push magnification even higher, by using a 2x ... 3x barlow with either of the eyepieces - that would double ... triple magnification. Please be aware that more magnification is not always better; for any observation there's an optimal magnification, not too big, not too small. But I doubt your scope would do well above 200x, even though the theoretical limit is 260x. Probably not worth trying anything beyond the high 1XX values - that is, for your eyepieces, a 3x barlow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the &lt;strong&gt;Moon&lt;/strong&gt;, any magnification works. Low magnification gives you broad images with lots of stuff to see. High magnification zooms into particular details. Good collimation is critical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For planets, it depends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Saturn&lt;/strong&gt; looks good at any magnification. If you can see the rings, the shadow of the rings on the surface of the planet, and the main split inside the rings (&lt;a href=&quot;https://en.wikipedia.org/wiki/Rings_of_Saturn#Cassini_Division&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Cassini division&lt;/a&gt;) then you're doing fine. It doesn't have a lot of detail to justify very high magnification, unless you're looking for the finer ring divisions, but your scope can only see Cassini.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Jupiter&lt;/strong&gt; is a bit tricky. It has enough fine detail to justify high magnification, but it has low contrast - and then as you increase magnification it gets more washed out, more dark, and hard to see. &quot;Medium-high&quot; magnification works best. Just try different eyepieces, with and without barlow, and see what works best for you. It also helps to keep your eye at the eyepiece, keep tracking the planet, and let your eye adapt - in time your brain/eye system will start pulling more detail out of the image. It takes me at least 2 ... 3 minutes to start seeing the really fine stuff there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mars&lt;/strong&gt; - don't even bother unless it's at opposition. It only looks good when it's closest to Earth. &lt;a href=&quot;http://www.wolframalpha.com/input/?i=next%20Mars%20opposition&quot; rel=&quot;nofollow noreferrer&quot;&gt;The next Mars opposition&lt;/a&gt; is in July 2018, so there's a bit of wait until then (will start to look good a few months before that). Even then, you need several things to see Mars well:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;very high magnification&lt;/li&gt;&#xA;&lt;li&gt;excellent collimation&lt;/li&gt;&#xA;&lt;li&gt;a good instrument&lt;/li&gt;&#xA;&lt;li&gt;excellent seeing (the inverse of turbulence)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Otherwise it just looks like a red dot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_seeing&quot; rel=&quot;nofollow noreferrer&quot;&gt;Seeing&lt;/a&gt; will affect all solar system objects (Moon and planets). When seeing is bad (when atmosphere is turbulent), any object will look bad. It depends on the weather - not just weather at ground level (what you think of as &quot;weather&quot;) but weather in the whole thickness of the atmosphere, all the way up (the stuff only airplane pilots usually care about). If you live in North America, &lt;a href=&quot;http://www.cleardarksky.com/csk/&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Clear Dark Sky site&lt;/a&gt; can give you a prediction of a few parameters including seeing. When seeing is bad, high magnification is worthless (the image is fuzzy and keeps shaking like jello).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing that helps: if there's a large temperature difference between inside and outside, leave the scope outside for a while to catch up with the air temperature. This will reduce internal turbulence and improve image quality on planets. I always leave my scope at &quot;outside&quot; temperature for about 1 hour before observing. When there is no temperature difference between inside and outside, then don't worry about this. This is like seeing (turbulence), but inside your scope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that light pollution has absolutely no effect on solar system objects (see below re: light pollution). Even when light pollution is very high, the Moon and the planets still look the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark adaptation does not matter for solar system objects (see below re: dark adaptation). You don't have to be in a dark place when observing them. I've looked at Saturn from a sidewalk under the bright lights downtown (city center), with cars and people passing by - no problem.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Deep sky objects (nebulae, star clusters, and galaxies), a.k.a. DSOs, are different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are objects with low brightness. Because of that, your eye cannot see details or colors. Collimation is less important. Seeing also doesn't matter much. Collimation and seeing affect fine details, but you can't see fine detail in DSOs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does affect them is light pollution - the light from all sources around you, glowing in the dust, water drops, etc in the atmosphere. This is why it's hard to see DSOs from the city. If you can drive 1 hour away from the city, to a place far from sources of light (cities, factories), then DSOs are easier to see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google &quot;light pollution map&quot; and you will find several sites that give you light pollution intensities in your area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even from the city, there are many DSOs that you can see. M31 (the Andromeda galaxy), the Orion nebula, the M13 cluster, the Dumbbell nebula - all these are clearly visible in an instrument like yours even from very light-polluted places. I've watched M13 many times from the middle of the Silicon Valley.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick is how to find these objects. Get a book called &quot;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521153972&quot; rel=&quot;nofollow noreferrer&quot;&gt;Turn left at Orion&lt;/a&gt;&quot; - it's a guide for beginners showing how to find many DSOs. Get a good planisphere (star chart) like &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1928771017&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt; and keep it next to the scope when you're observing DSOs. Start with something easy like the Orion nebula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you could google &quot;how to find the Orion nebula&quot; (or M13, etc) and you'll find online guides which are sometimes helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually you don't need high magnification for DSOs. Low and medium magnification work well - but it depends; e.g. I've watched M13 in high-ish magnification and it looks good. What really does help here is eyepieces with a wide field of view, because it makes it easier to find the objects. I'm guessing your eyepieces are of the Plossl design, with a 52 degree FoV, which is fine, but more is better for DSOs. Now don't go ahead and buy some fancy expensive eyepieces - keep them proportional in price with the scope, and the original ones should be fine for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing that really helps with DSOs is telescope size - bigger is always better. That's also true for solar system objects, but it's even more important here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, make sure your eyes are dark adapted before looking for DSOs. Observe from a dark place, and stay in the dark for at least 15 ... 20 minutes before observing. Do not use anything with a screen (smartphone, computer) because that destroys your dark adaptation. Do not use flashlights. There are flashlights for astronomers, with a red LED and a dial button, and you need to dial down the light to the least amount that allows you to see and read the star chart. Very weak red light will not destroy your dark adaptation. You could also improvise a red flashlight, but make it as weak as possible (or put a dial button on it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clear skies to you and good luck!&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2017-01-24T19:03:13.887" LastActivityDate="2017-01-24T19:03:13.887" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19830" PostTypeId="2" ParentId="19826" CreationDate="2017-01-21T00:21:19.133" Score="2" Body="&lt;p&gt;I think you just need to keep winding the focusser knob, you are way past focus in the first picture but you can see the image get smaller as you approach focus. The last image looks like you may be there or thereabouts, I think the colored diffraction patterns are caused by the camera optics.&lt;/p&gt;&#xA;" OwnerUserId="15560" LastActivityDate="2017-01-21T00:21:19.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19832" PostTypeId="1" AcceptedAnswerId="19838" CreationDate="2017-01-21T19:29:49.787" Score="3" ViewCount="94" Body="&lt;p&gt;Aurorae are caused by energetic particles interacting with planetary atmospheres. Part of this flux comes from the Sun (solar wind) while the rest is cosmic radiation. My main question is: could a planetary body located far away from the Sun (or any other star), with a suitable atmosphere and magnetosphere, have noticeable* aurorae, just from being exposed to cosmic rays? A related question would be: do we know how much of the aurora is caused by the solar wind, and how much by cosmic rays? And thirdly: is there any other way that a planetary body could produce aurorae, such as its own electromagnetic emissions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*I know &quot;noticeable&quot; doesn't really work as a limit, but for the time being, I'm talking of electromagnetic radiation that can be distinguished clearly from the background of a planet's disc using instruments not more advanced than our current telescopes at interplanetary distances.&lt;/p&gt;&#xA;" OwnerUserId="15398" LastActivityDate="2017-01-22T22:33:22.723" Title="Can cosmic rays alone produce noticeable aurorae in bodies far away from the Sun?" Tags="&lt;solar-wind&gt;&lt;cosmic-ray&gt;&lt;aurora&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19833" PostTypeId="1" CreationDate="2017-01-22T05:59:20.210" Score="5" ViewCount="3639" Body="&lt;p&gt;The Earth is about 4.543 billion years old.&lt;br&gt;&#xA;However, the &lt;em&gt;length of a year&lt;/em&gt; can be changed by anything from an impact to a natural cycle of orbit changes to a slow drift toward or away from the sun to the tug of a nearby body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while the Earth is about 4.543 billion &lt;em&gt;modern years old&lt;/em&gt;, it may have revolved around the sun a significantly different number of times.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is a reasonable estimate for how many revolutions - factoring in known geologic events - that the Earth has made around the Sun?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/11243/how-has-the-earths-orbit-changed-over-hundreds-of-millions-or-billion-of-years&quot;&gt;This question&lt;/a&gt; touches on the topic of orbital changes but never really answers what I'm asking here.  &lt;/li&gt;&#xA;&lt;li&gt;The second answer on &lt;a href=&quot;https://www.quora.com/How-does-the-length-of-a-year-change-over-time&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Quora question&lt;/a&gt; seems to answer my question but I don't know if that source is reputable, and the actual calculations and reasoning are not given&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.reddit.com/r/askscience/comments/2ct3yo/has_the_length_of_earths_year_changed_appreciably/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This reddit discussion&lt;/a&gt; touches on the subject but has differing / inconclusive answers&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14918" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2017-01-25T14:10:47.637" Title="About how many revolutions has the Earth made around the sun?" Tags="&lt;orbit&gt;&lt;earth&gt;&lt;orbital-migration&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19834" PostTypeId="2" ParentId="19833" CreationDate="2017-01-22T14:19:06.047" Score="-2" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Back-of-the-envelope_calculation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Back of the envelope&lt;/a&gt; time :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking a rough estimate of 35 days per half billion years as the change in orbital period over time (from that Reddit page), we get the orbital period of about 365+4.5*2*35 days 4.53 billion years ago ( about 680 days ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average period over that time is about 522 days, so over 4.5 billion years (with those years being our current definition of year ) that's about 3.1 billion orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's a pretty useless number, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth noting that we could not accurately simulate such a large number of orbits with any accuracy.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-01-22T14:19:06.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19835" PostTypeId="2" ParentId="19833" CreationDate="2017-01-22T16:00:43.990" Score="7" Body="&lt;p&gt;Since the very early solar system, there have not been large-scale movements of the planets.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the early solar system, while the planets were still embedded in a protoplanetary disk, there were large movements. (notably the hypothesised &quot;Grand Tack&quot; of Jupiter) However once Jupiter reached it's current position of 5.2AU it remained there, and the Earth formed then too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have not been any significant changes to the Earth's orbit since the Earth formed, otherwise there would have been significant heating or cooling. So the average year length was and is about 30 000 000 seconds, and Earth has made about 4.5 billion orbits of the sun since then.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-01-22T21:46:57.127" LastActivityDate="2017-01-22T21:46:57.127" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19836" PostTypeId="2" ParentId="19826" CreationDate="2017-01-22T19:23:57.123" Score="2" Body="&lt;p&gt;The images in the question show a progression from severely out of focus (large disc with shadow of secondary mirror) to nearly in focus (six-way diffraction pattern). There are a few things you can try:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In daylight, make sure that the the secondary mirror support vanes are parallel to the optical tube axis. This may reduce the diffraction artifacts.&lt;/li&gt;&#xA;&lt;li&gt;The light from a very bright source will show weaknesses in all but the finest optical systems. Use a neutral-density &quot;moon&quot; filter to reduce the overall brightness.&lt;/li&gt;&#xA;&lt;li&gt;Point sources make it easier to find the point of best focus. Focus on a nearby star faintly visible to the naked eye, then aim the telescope at the planet and wait for steady air.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2017-01-23T17:04:46.743" LastActivityDate="2017-01-23T17:04:46.743" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19837" PostTypeId="1" CreationDate="2017-01-22T22:24:50.323" Score="2" ViewCount="168" Body="&lt;p&gt;If magnetism requires some energy to exert change of motion to attract ferrous metals, then, does gravity require energy to exert attraction, or is gravity energy itself, creating attraction through centrifugal force from the earth?&lt;/p&gt;&#xA;" OwnerUserId="15582" LastActivityDate="2017-03-26T19:22:58.143" Title="Is energy required for gravity to have effect, or is it an energy itself?" Tags="&lt;gravity&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19838" PostTypeId="2" ParentId="19832" CreationDate="2017-01-22T22:33:22.723" Score="1" Body="&lt;p&gt;You may be interested to learn that Jupiter experiences aurora on its poles nearly (if not totally) &lt;a href=&quot;https://www.nasa.gov/feature/goddard/2016/hubble-captures-vivid-auroras-in-jupiter-s-atmosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;non-stop&lt;/a&gt;. These aurora never cease, because they do not rely entirely on the sun as their source of charged particles, instead receiving particles from the volcanic ejections of its moon Io, as well as from cosmic rays. Granted, these aurora are occurring in the ultra-violet, but they are still 'noticeable'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, yes aurora can be caused by non-solar sources, and conceivably cosmic rays at that. As to your other question concerning 'how much' of the aurora is due to which sources... I couldn't tell you how much of the auroras visibility comes from the sun, and I doubt there is much (or any) information on that subject, since determining the source of such convoluted light is likely outside the realms of both interest and empiricism. Perhaps you could pioneer that research? In any case, an intriguing subject to be sure.&lt;/p&gt;&#xA;" OwnerUserId="11585" LastActivityDate="2017-01-22T22:33:22.723" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19839" PostTypeId="1" AcceptedAnswerId="19841" CreationDate="2017-01-23T08:06:20.813" Score="4" ViewCount="514" Body="&lt;p&gt;Is there a particular galaxy cataloged by astronomers which is considered to most closely resemble what the structure of the Milky Way is believed to look like, when viewed from a similar distance?&lt;/p&gt;&#xA;" OwnerUserId="15587" LastActivityDate="2017-01-23T11:43:18.953" Title="What galaxy looks the most like the Milky Way?" Tags="&lt;galaxy&gt;&lt;observational-astronomy&gt;&lt;ngc&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19840" PostTypeId="1" CreationDate="2017-01-23T11:39:54.753" Score="4" ViewCount="376" Body="&lt;p&gt;&lt;strong&gt;What existed before the big bang?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What all thing are there inside big bang object ?.&lt;/p&gt;&#xA;" OwnerUserId="14868" LastActivityDate="2017-01-24T23:03:30.970" Title="What existed before the big bang" Tags="&lt;universe&gt;&lt;big-bang-theory&gt;&lt;early-universe&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2017-01-23T14:57:40.053" ContentLicense="CC BY-SA 3.0" />
  <row Id="19841" PostTypeId="2" ParentId="19839" CreationDate="2017-01-23T11:43:18.953" Score="6" Body="&lt;p&gt;According to &lt;a href=&quot;http://www.space.com/11841-milky-spiral-galaxy-twins-photo.html&quot; rel=&quot;noreferrer&quot;&gt;this article&lt;/a&gt; on space.com the most similar looking galaxy to our own is NGC 6744. The source of this article comes from a statement by the ESO (European Southern Observatory) which says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If we had the technology to escape the Milky Way and could look down on it from intergalactic space, this view is close to the one we would see — striking spiral arms wrapping around a dense, elongated nucleus and a dusty disc.&lt;/p&gt;&#xA;&lt;p&gt;There is even a distorted companion galaxy — NGC 6744A, seen here as a smudge to the lower right of NGC 6744, which is reminiscent of one of the Milky Way’s neighboring Magellanic Clouds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Included below is an image of the galaxy (NGC 6744) in question, taken with the Wide Field Imager on the MPG/ESO 2.2-meter telescope.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vMzBT.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vMzBT.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;However it is also noted that while the Milky Way and NGC 6744 are similar in appearance, NGC 6744 is almost twice the diameter of the Milky Way.&lt;/p&gt;&#xA;" OwnerUserId="10437" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-01-23T11:43:18.953" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19842" PostTypeId="2" ParentId="19840" CreationDate="2017-01-23T13:00:34.130" Score="5" Body="&lt;p&gt;We don't have theories that can describe the big bang (or whatever happened) properly so we can't say anything definite about the big bang itself, let alone if/what was before it.  We would need (at a minimum) something like a quantum field theory for gravity and we don't have one yet.  We've no reason to think that's all we'd need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not even possible (as I understand it) to define &quot;before&quot; the big bang.  Time, in essence, started at the big bang.  It has no common sense meaning beyond that.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastEditorUserId="10196" LastEditDate="2017-01-24T23:03:30.970" LastActivityDate="2017-01-24T23:03:30.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19843" PostTypeId="2" ParentId="19833" CreationDate="2017-01-23T13:33:47.173" Score="0" Body="&lt;p&gt;Phil Plait recently wrote a &lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2017/01/01/how_do_we_define_the_year.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;relevant article&lt;/a&gt; which, while partially tongue-in-cheek, does point out the difficulty in defining a year, let alone counting how many there have been. The problem is that Sol is moving relative to the stellar background so there's no absolute reference for getting &quot;back where we started.&quot;  It gets worse when you try to map years to days, since even for a static origin, a body which rotates while orbiting executes a different number of rotations depending on whether you observe from the origin or the stellar background. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-01-23T13:33:47.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19844" PostTypeId="2" ParentId="12049" CreationDate="2017-01-23T20:39:54.980" Score="2" Body="&lt;p&gt;Yes, they can been seen together. I have added screenshot of sky at 24th Jan 2 am IST from Sky Safari app,  &lt;a href=&quot;https://i.stack.imgur.com/7odBV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7odBV.jpg&quot; alt=&quot;Sky on 24th Jan&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15592" LastActivityDate="2017-01-23T20:39:54.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19845" PostTypeId="2" ParentId="19833" CreationDate="2017-01-25T08:27:13.780" Score="5" Body="&lt;p&gt;The configuration of the solar system is thought to have been more or less settled after the first 10-20 million years or so. However, what governs the Earth's orbital period is it's orbital angular momentum and the mass of the Sun. Two events have certainly changed the Earth's orbital period (a) whatever collision formed the Moon and (b) the continuous process of mass loss from the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that (a) probably happened sometime in the first tens of millions of years and likely did not alter the Earth's angular momentum greatly  - it depends on the speed, mass and direction of the impactor and the amount of mass lost from the Earth-Moon system - I will ignore it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know a little more about (b). It seems that the mass loss from the early Sun was much greater than it is now. A review by &lt;a href=&quot;https://arxiv.org/abs/0712.1763&quot; rel=&quot;nofollow noreferrer&quot;&gt;Guedel (2007)&lt;/a&gt; suggests a mass loss rate over the last 4.5 billion years that increases as $t^{-2.3}$ (with considerable uncertainty on the power law index), where $t$ is time since birth, and an initial solar mass between 1% and 7% more than it is now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conservation of angular momentum and Kepler's third law means that $a \propto M^{-1}$ and $P \propto M^{-2}$. Therefore the Earth's orbital period was 2-14% shorter in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this considerable uncertainty means is that your question cannot be answered to three significant figures, or perhaps even two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the solar wind power law time dependence is very steep, then most of the mass loss occurred early, but the total mass loss would have been greater. On the other hand, a lower total mass loss implies a shallower mass loss and the earth spending a longer time in smaller orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have time I will run through a calculation, &lt;strong&gt;but it seems safe to assume that the earth has executed more orbits than its age in years, but probably not by more than a few percent.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;A further consideration could be the tidal torque exerted by the Sun on the Earth-Sun orbit, which would increase the orbital separation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantifying this is difficult. The tidal torque on a planet from a the Sun is&#xA;$$ T = \frac{3}{2} \frac{k_E}{Q} \frac{GM_{\odot}^{2} R_{E}^{5}}{a^6},$$&#xA;where $R_E$ is an Earth radius and $k_E/Q$ is the ratio of the tidal &lt;a href=&quot;https://en.wikipedia.org/wiki/Love_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;Love number&lt;/a&gt; and $Q$ a tidal dissipation factor (see &lt;a href=&quot;https://arxiv.org/abs/1206.0334&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sasaki et al. (2012)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://astro.cornell.edu/academics/courses/astro6570/Tidal_evolution.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;These lecture notes&lt;/a&gt; suggest values of $k_E/Q\sim 0.1$ for the Earth and therefore a tidal torque of $4\times 10^{16}$ Nm. Given that the orbital angular momentum of the Earth is $\sim 3\times 10^{40}$ kgm$^2$s$^{-1}$, then the timescale to change the Earth's angular momentum (and therefore $a$ and $P$) is $&amp;gt;10^{16}$ years and thus this effect is &lt;em&gt;negligible&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-01-25T14:10:47.637" LastActivityDate="2017-01-25T14:10:47.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19848" PostTypeId="1" AcceptedAnswerId="19863" CreationDate="2017-01-25T14:18:33.497" Score="2" ViewCount="3741" Body="&lt;p&gt;If the Earth as-is were moved to Venus's orbit (and Venus removed), would the Earth remain habitable, would it eventually turn into what Venus currently is, or would it turn into something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm mainly wondering which characteristics of Venus (hot, no magnetic shielding, etc) are the result of how it is composed compared to it's proximity to the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar question: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1812/if-venus-and-mars-changed-places-would-we-then-have-3-habitable-planets&quot;&gt;If Venus and Mars changed places, would we then have 3 habitable planets?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15608" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2017-01-27T02:10:32.690" Title="If the Earth was in Venus's orbit, would it be habitable?" Tags="&lt;venus&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19850" PostTypeId="2" ParentId="19848" CreationDate="2017-01-25T14:45:53.527" Score="7" Body="&lt;p&gt;A planet's environment is a complex interplay of factors.  If you dramatically change the insolation of a planet, it becomes very difficult to predict how the other factors which influence the planet's environment will respond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus, for example, seems locked into its high-CO2 environment (which makes its surface hot) because its surface is too hot to allow for the formation of carbonates.  Even if it had water (also needed for the formation of carbonates), it would be continuously boiled away from the surface.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All planets - Earth, Mars, and Venus included - have their temperatures influenced by their albedo.  Albedos of planets are often strongly correlated with cloud cover.  Cloud cover generally increases as a wet planet warms, reflecting more light.  But atmospheric composition can overcome this balance, as in the case with Venus, which has a very high albedo but is still swelteringly hot on the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you get rid of the feedback factors like water and carbon dioxide, the planets have very different nitrogen levels - Venus at about 2 1/2 times that of Earth (enough to cause mild nitrogen narcosis on its own), and Mars almost nothing.  Really, Mars in Venus's position would be in a terrible state - it's been so stripped of its atmosphere even as far from the sun as it is, it would be all the more stripped being much closer to the sun.  Over geological timescales it'd become highly water stripped as well, as it'd be much easier for water to enter a gas phase, and the solar wind can very effectively strip water planets even as large as Venus, let alone Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus in Mars' position would probably fare much better.  But it's difficult to say exactly how much.  Given that we can't say with great confidence exactly what set in motion the path that led to its current state today (thicker, more viscous lithosphere preventing crustal overturn? cause or consequence?), it's hard to say whether that would have happened had it been in Mars' orbit.  One curiosity if you took today's Venus and moved it to Mars' position: you may well cause an atmospheric collapse to a supercritical state. It's been theorized that even in its current orbit that's happened before (temperature runs counter to the formation of supercritical fluids).  &lt;/p&gt;&#xA;" OwnerUserId="15606" LastActivityDate="2017-01-25T14:45:53.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19851" PostTypeId="1" AcceptedAnswerId="19852" CreationDate="2017-01-25T14:55:33.087" Score="5" ViewCount="748" Body="&lt;p&gt;I have read before that Jupiter is named after the great Roman god because of how bright it is. I can't help but wonder, why wasn't the name given to Venus instead? Isn't Venus much brighter than Jupiter in the night-sky, or am I mistaken?&lt;/p&gt;&#xA;" OwnerUserId="15609" LastActivityDate="2017-01-25T18:00:52.630" Title="Why is Jupiter called so?" Tags="&lt;jupiter&gt;&lt;venus&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19852" PostTypeId="2" ParentId="19851" CreationDate="2017-01-25T16:06:23.250" Score="6" Body="&lt;p&gt;I don't think this will be a great answer because these are just speculations. I cannot find concrete evidence or sources discussing why Jupiter (the planet) became associated with the &quot;king of gods&quot; and not Venus, but here's my go at citing some potential reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To start, I want to point out that ultimately, we have to trace this back to the Greeks and Babylonians. Sure Jupiter (the planet) is named after the Roman god Jupiter, but really the Roman mythology traces pretty distinctly to Greek mythology and Greek mythology has strong ties to Babylonian mythology. The Greeks were the ones who associated Zeus with the planet Jupiter (and they were likely taking cues from the Babylonians) and the Romans just tagged along with that idea, but changed the god's name. So I think your question then transforms into, &lt;em&gt;Why didn't the Greeks/Babylonians decide to associate the planet Venus with their chief god (being that it is the brightest) and instead chose the planet Jupiter?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Brightness of Venus and Jupiter&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let's just confirm that Venus is indeed brighter. If you look up the &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_magnitude&quot;&gt;Apparent Magnitude&lt;/a&gt; system of Wikipedia, you see a description of how astronomers measure visible brightnesses of astronomical objects (and coincidentally enough, this system was created by a Greek). In this system, smaller and more negative numbers indicate brighter objects. From the table on that page, we can see that the apparent magnitude of Venus varies from $-4.89$ at its brightest to $-3.82$ at its dimmest (this variation is caused by the different configurations of Earth, Venus, Sun positions). For Jupiter, we see that at its brightest it has an apparent magnitude of $-2.94$ and a minimum of $-1.61$. Clearly, even at its brightest, Jupiter is far outshone by Venus at its dimmest. Note that none of this takes into account light pollution or other atmospheric effects which would have been negligible to the Greeks and Babylonians anyway. So we can say emphatically yes that Venus is distinctly brighter than Jupiter, but importantly, we note that Jupiter is still brighter than every other star in the sky $-$ the brightest being Sirius with a magnitude of $-1.47$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Confusion about Venus&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, the Greeks and Babylonians were sitting around looking at these planets and wondering what they were. Likely they started thinking they were big, great deities and started attributing gods to them. I can't say this is a certainty, but one possible reason Venus didn't get the top god was that initially the Greeks and Babylonians weren't even sure Venus was a planet like the rest of the planets were. Because Venus is so close to the Sun, we only ever see it at sunrise and sunset. Otherwise, it is in the sky during the day and dwarfed by the brightness of the Sun. This meant that it took the ancient peoples a long time before they even realized the bright object they saw in the evening was the same as the bright object they saw in the morning. To quote &lt;a href=&quot;https://en.wikipedia.org/wiki/Ancient_Greek_astronomy&quot;&gt;Wikipedia&lt;/a&gt; on this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Observations of Venus are not straightforward. Early Greeks thought that the evening and morning appearances of Venus represented two different objects, calling it Hesperus (&quot;evening star&quot;) when it appeared in the western evening sky and Phosphorus (&quot;light-bringer&quot;) when it appeared in the eastern morning sky. They eventually came to recognize that both objects were the same planet. Pythagoras is given credit for this realization.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Potentially, the ancient Greeks didn't even consider Venus to be a planet initially. The word planet is Greek and means &quot;wandering star&quot;. The ancient Greeks identified objects in the night sky which were different and seemingly more important that the normal stars. However, initially, they don't seem to have grouped Jupiter and Mars in with the same type of object as Venus (and possibly Mercury). Jupiter's motion around the sky may have seemed more grandiose and full of purpose, hence why they attributed it to being their biggest and most important god.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Love and Beauty&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said, the Babylonians, as well as the Greeks before they figured it out, didn't consider Venus to be a planet, but rather two separate and strange &quot;stars&quot;. However, once they figured out the two &quot;stars&quot; were really one and the same and Venus was a planet just like Jupiter, they decided to attribute a proper god or goddess from the pantheon of gods to it. Jupiter was already taken so they had to pick someone else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite that, I don't think the argument that Venus is brighter automatically means it should be the most important god. Note that Venus is named after the Roman goddess Venus (of course) who is also the Greek goddess Aphrodite. This is the Greek goddess of love and beauty and she herself was supposed to be exceedingly beautiful. Venus, the planet, is a beautiful sight to behold and it seems appropriate that the most radiant and beautiful planet be named after the most radiant and beautiful Goddess.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2017-01-25T16:51:31.383" LastActivityDate="2017-01-25T16:51:31.383" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19853" PostTypeId="2" ParentId="19851" CreationDate="2017-01-25T18:00:52.630" Score="5" Body="&lt;p&gt;The ancient Mesopotamian astrologers associated each of the known moving lights in the sky (the five planets, sun and moon) with one of their gods. You can see their names, in the case of Babylon, &lt;a href=&quot;https://en.wikipedia.org/wiki/Babylonian_astrology#Planets_and_gods&quot;&gt;here&lt;/a&gt;. The days of the week are all named after these lights, although in English they often use the name of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Interpretatio_graeca#Interpretatio_germanica&quot;&gt;closest Germanic god&lt;/a&gt;. (This practice of associating gods across cultures will become a theme.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ancient Greeks learned much from Babylon, but were very much not a culture of religious freedom. &lt;a href=&quot;https://en.wikipedia.org/wiki/Impiety&quot;&gt;Impiety&lt;/a&gt; was a crime for which one could be executed in Athens, for example, and importing foreign religion was not allowed in at least some cases. It is therefore unsurprising that when the Greeks borrowed the Babylonian knowledge of the planets, they justified it by replacing each god with whichever Greek god seemed most similar. &lt;a href=&quot;https://en.wikipedia.org/wiki/Nergal&quot;&gt;Nergal&lt;/a&gt;, a god associated with war and plague, was for example replaced with the Greek war god Ares. &lt;a href=&quot;https://en.wikipedia.org/wiki/Ishtar&quot;&gt;Ishtar&lt;/a&gt;, a goddes of love, sex and war, was replaced with Aphrodite. The resident god of Babylon and chief god at the time, &lt;a href=&quot;https://en.wikipedia.org/wiki/Marduk&quot;&gt;Marduk&lt;/a&gt;, became Zeus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Romans, on the other hand, did generally have religious freedom. The ancient Romans in general held the belief that, since the gods they worship are real, similar gods of other cultures must simply be the same gods by other names. It should be stressed that despite this the practices and myths were different, even between Greece and Rome. The Romans therefore thought nothing of calling the planets by the name of whichever god of theirs they felt the Greek name referred to, Venus in the case of Aphrodite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having gotten that far, we now need to establish why Ishtar (Venus) is the brightest planet and not Marduk (Jupiter), the chief god of Babylon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, Marduk was only in later Babylonian times the chief deity. In earlier periods when Ur was in its prime, the chief god was &lt;a href=&quot;https://en.wikipedia.org/wiki/Sin_(mythology)&quot;&gt;Sin&lt;/a&gt;, the moon god, &lt;strong&gt;whose daughter was Ishtar&lt;/strong&gt; according to Wikipedia. (My copy of Gilgamesh lists Anu as the father of Ishtar. This is likely to have been changed by Babylonian times, in keeping with the local god of Babylon, Marduk, displacing Anu who was father of the gods in Sumerian times.) Marduk was a minor god at the time when the Epic of Gilgamesh was written. In fact, every one of the gods we call the &quot;chief god&quot; was really just the chief god when his city reigned supreme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ishtar was also a very fickle goddess. Consider this partial summary of Tablet VI of the Standard Version of the Epic of Gilgamesh, taken from The Epic of Gilgamesh: A New Translation by Andrew George (1999, Penguin, ISBN 0-14-044721-0, p. 47):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Back in Uruk Gilgamesh's beauty provokes the desire of the goddess Ishtar and she proposes to him. Gilgamesh scorns her, &lt;strong&gt;reminding her of the fates suffered by her former conquests&lt;/strong&gt;...persuades Anu, her father, to give her...the constellation Taurus...The Bull of Heaven causes havoc in Uruk...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Furthermore, (p. 223.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sometimes she is a mature woman, sometimes an impetuous young virgin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This fits perfectly with the fickleness of Venus in the sky. It took a long time for humanity to realise that the morning star and the evening star are even the same object, which appears briefly in twilight and even then is sometimes gone (behind the sun). It was very well known in ancient times, for example see Isaiah 14:12 in the Bible, which mentions the fickleness of the morning star. (And is probably the source of the name &quot;Lucifer&quot; for the devil, Lucifer or &quot;Bringer of Light&quot; being an alternative Roman name for the planet Venus.)&lt;/p&gt;&#xA;" OwnerUserId="14969" LastActivityDate="2017-01-25T18:00:52.630" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19855" PostTypeId="2" ParentId="19766" CreationDate="2017-01-25T18:23:01.920" Score="1" Body="&lt;p&gt;Rob Jeffries has already answered the question, but has left out a detail that should be included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass that the sun loses to solar wind will cause the Earth to migrate outward. Increased drag from moving through a denser solar wind will slow the Earth and partly counteract the effect. Initially not by much, but as the sun expands the Earth will find itself moving through an increasingly dense medium. According to &lt;a href=&quot;http://sro.sussex.ac.uk/1800/2/DoomsdayResub.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;, drag in the lower chromosphere will eventually be high enough for the planet to fall into the sun.&lt;/p&gt;&#xA;" OwnerUserId="14969" LastActivityDate="2017-01-25T18:23:01.920" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19856" PostTypeId="2" ParentId="19837" CreationDate="2017-01-25T18:42:58.173" Score="1" Body="&lt;p&gt;Gravity is an effect of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor&quot; rel=&quot;nofollow noreferrer&quot;&gt;stress-energy tensor&lt;/a&gt;. Put differently, energy and momentum (they're related concepts) are the &lt;em&gt;cause&lt;/em&gt; of gravity in general relativity. A fully charged electric battery has a very slightly stronger gravitational field than a depleted one. You may have heard that $E = mc^2$ or in as the terms are more descriptively defined in modern usage, &lt;a href=&quot;https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation&quot; rel=&quot;nofollow noreferrer&quot;&gt;$E^2 = m^2c^4 + p^2c^2$&lt;/a&gt; . Just thinking about these equations from special relativity, you can see that more energy means more mass, which means more gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond that, gravitational waves carry energy. You may remember the recent &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_wave#LIGO_observations&quot; rel=&quot;nofollow noreferrer&quot;&gt;discovery of gravitational waves from a black hole collision&lt;/a&gt;, in which about three solar masses worth of energy were calculated to have been given off as gravitational waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting back to plain old Newtonian mechanics, when masses move around in a gravitational field energy is converted from one form to another as Dean noted in their comment on the question. For example when a comet is near the sun on its highly elliptic orbit, it has high kinetic energy and low potential energy. It loses kinetic energy as it moves outward, being slowed by gravitational energy, but gains gravitational potential energy. This potential is, quite simply, a representation of the fact that it can get its &quot;lost&quot; kinetic energy back when it falls in again on the inbound leg of its orbit, being helped along by gravity instead of hindered when inbound.&lt;/p&gt;&#xA;" OwnerUserId="14969" LastActivityDate="2017-01-25T18:42:58.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19857" PostTypeId="2" ParentId="11243" CreationDate="2017-01-25T19:36:20.920" Score="11" Body="&lt;p&gt;What governs the Earth's orbital period is its orbital angular momentum and the mass of the Sun. Two events have certainly changed the Earth's orbital period (a) whatever collision formed the Moon and (b) the continuous process of mass loss from the Sun. A third possibility (c) is that tidal torques from the Sun have increased the angular momentum of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that (a) probably happened sometime in the first tens of millions of years and likely did not alter the Earth's angular momentum greatly  - it depends on the speed, mass and direction of the impactor and the amount of mass lost from the Earth-Moon system - I will ignore it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b). It seems, from observations of younger solar analogues, that the mass loss from the early Sun was much greater thanthe modest rate at which it loses mass now via the solar wind. A review by &lt;a href=&quot;https://arxiv.org/abs/0712.1763&quot; rel=&quot;noreferrer&quot;&gt;Guedel (2007)&lt;/a&gt; suggests a mass loss rate over the last 4.5 billion years that increases as $t^{-2.3}$ (with considerable uncertainty on the power law index), where $t$ is time since birth, and suggests an initial solar mass between 1% and 7% larger than it is now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conservation of angular momentum and Kepler's third law means that $a \propto M^{-1}$ and $P \propto M^{-2}$. Therefore the Earth's orbital period was 2-14% shorter in the past due to solar mass loss, but has been close to its current value for the last 2-3 billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the solar wind power law time dependence is very steep, then most of the mass loss occurred early, but the total mass loss would have been greater. On the other hand, a lower total mass loss implies a shallower mass loss and the earth spending a longer time in a smaller orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c)&#xA;The tidal torque exerted by the Sun on the Earth-Sun orbit increases the orbital separation, because the Sun's rotation period is shorter than the Earth's orbital period. The Sun's tidal &quot;bulge&quot; induced by the Earth applies a torque that increases the orbital angular momentum, much like the effect of the Earth on the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantifying this is difficult. The tidal torque on a planet from a the Sun is&#xA;$$ T = \frac{3}{2} \frac{k_E}{Q} \frac{GM_{\odot}^{2} R_{E}^{5}}{a^6},$$&#xA;where $R_E$ is an Earth radius and $k_E/Q$ is the ratio of the tidal &lt;a href=&quot;https://en.wikipedia.org/wiki/Love_number&quot; rel=&quot;noreferrer&quot;&gt;Love number&lt;/a&gt; and $Q$ a tidal dissipation factor (see &lt;a href=&quot;https://arxiv.org/abs/1206.0334&quot; rel=&quot;noreferrer&quot;&gt;Sasaki et al. (2012)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://astro.cornell.edu/academics/courses/astro6570/Tidal_evolution.pdf&quot; rel=&quot;noreferrer&quot;&gt;These lecture notes&lt;/a&gt; suggest values of $k_E/Q\sim 0.1$ for the Earth and therefore a tidal torque of $4\times 10^{16}$ Nm. Given that the orbital angular momentum of the Earth is $\sim 3\times 10^{40}$ kgm$^2$s$^{-1}$, then the timescale to change the Earth's angular momentum (and therefore $a$ and $P$) is $&amp;gt;10^{16}$ years and thus this effect is &lt;em&gt;negligible&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-01-25T19:36:20.920" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19858" PostTypeId="2" ParentId="19756" CreationDate="2017-01-26T01:03:57.530" Score="3" Body="&lt;p&gt;Here's a simplified answer. &quot;About&quot; the farthest thing we can see with the naked eye is the Andromeda galaxy at about 2 million light years. Therefore, I would say that you just have to be in a void with no galaxies closer than about 2 or 3 million light years. I could be wrong, but I'm thinking that this is most of the known universe. &#xA;   As far as future Earth goes, it won't be about distance, but about when the last star close enough to see &quot;burns out, which will be a VERY long time from now. There may not even be an Earth then, if it is engulfed by the sun going red giant.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2017-01-26T01:03:57.530" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19859" PostTypeId="1" AcceptedAnswerId="19860" CreationDate="2017-01-26T15:49:59.480" Score="2" ViewCount="272" Body="&lt;p&gt;It sounds like dark matter makes up one quarter of the universe's mass-energy.  And it's density is fairly uniform.  But could it be simultaneously both uniform and moving - like currents in a lake.  Does it have a standing wave?  A propagating wave?  A rotation around some universal central point?  Local rotations around massive objects like low pressure systems on a weather map?&lt;/p&gt;&#xA;" OwnerUserId="14085" LastActivityDate="2017-01-28T08:12:47.990" Title="Is Dark Matter moving?" Tags="&lt;dark-matter&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19860" PostTypeId="2" ParentId="19859" CreationDate="2017-01-26T16:25:59.587" Score="4" Body="&lt;p&gt;Yes and no. In general, Dark Matter tends to exist in the same places as galaxies. It isn't really distributed throughout all of space uniformly. It has voids and filaments just like normal matter does. In fact, on the whole, normal matter is a pretty good tracer of where Dark Matter is. In some cases Dark Matter can separate from normal matter, and it's believed by some that Dark Matter can clump in areas where normal matter doesn't, but more or less, on a cosmological scale, it is distributed and acts like regular matter. On short timescales, Dark Matter most just sits there. There isn't mass motion or propagating waves. On long timescales and in very large views, you can see actual motion and interesting structures (e.g., &lt;a href=&quot;https://en.wikipedia.org/wiki/Baryon_acoustic_oscillations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Baryonic Acoustic Oscillations&lt;/a&gt; - technically we observe these in normal matter, but Dark Matter would be affected in the same way).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give you an idea of this, check out the &lt;a href=&quot;https://www.youtube.com/watch?v=UC5pDPY5Nz4&quot; rel=&quot;nofollow noreferrer&quot;&gt;Millennium Simulation&lt;/a&gt; (that's a YouTube video which moves around and views the results of the simulation). This was a set of simulations of the universe as a whole, in an attempt to test out current theories about our universe and see if those theories could produce a universe that looks like ours. Suffice to say, they were pretty successful. An image of the results of the simulation is shown below. This represents the universe at a very very large scale and you can see all the dark matter and how it is distributed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Uh3TW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uh3TW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also want to check out the &lt;a href=&quot;https://www.youtube.com/watch?v=2qeT4DkEX-w&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aquarius Simulation&lt;/a&gt; which was very similar in nature to the Millennium Simulation. However, unlike the video for the Millennium Simulation, this video shows the evolution of the universe. This really gives you an idea of the flow and motion of dark matter over the entire history of the Universe.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2017-01-27T13:52:19.487" LastActivityDate="2017-01-27T13:52:19.487" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19861" PostTypeId="1" CreationDate="2017-01-26T18:20:57.710" Score="1" ViewCount="312" Body="&lt;p&gt;I am trying to write an app to determine planetary positions, using Jean Meeus' &lt;em&gt;Astronomical Algorithms&lt;/em&gt; as a reference. His guidance on using the VSOP87D data is straightforward, as are many other sources that I googled. But all of these talk only about the &lt;strong&gt;last three columns&lt;/strong&gt; of that data. What is the purpose of the preceding columns? Some are self explanatory (e.g., the simple numbering of each row), but others seem to have no explanation either in the file itself or the 'read me' documents that accompany it on the FTP site. I want to ensure I fully understand the data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example of the third row of VSOP87D for the planet Jupiter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4510, 3, 0, 0, 0, 0, 2, -5, 0, 0, 0, 0, 0, 0, -0.00566632795, -0.00089195704, 0.00573610145, 1.44406205976, 7.11354700080&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where I have replaced the various spaces with commas to save space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will note that the term .00573610145, for example, differs by by several orders of magnitude from that listed in Meeus' table, as do some other terms. But this does not seem to correspond to any of the numbers in the other columns, unless I'm missing the obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data is taken from the FTP website: &lt;a href=&quot;ftp://cdsarc.u-strasbg.fr/pub/cats/VI%2F81/VSOP87D.jup&quot; rel=&quot;nofollow noreferrer&quot;&gt;ftp://cdsarc.u-strasbg.fr/pub/cats/VI%2F81/VSOP87D.jup&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="15619" LastActivityDate="2017-05-26T19:12:44.370" Title="VSOP87D: What are the purpose of the initial columns in the VSOP87D data?" Tags="&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19862" PostTypeId="2" ParentId="19172" CreationDate="2017-01-26T18:41:53.047" Score="1" Body="&lt;p&gt;The actual equations used by NASA are located here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://eclipse.gsfc.nasa.gov/SEcat5/deltatpoly.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://eclipse.gsfc.nasa.gov/SEcat5/deltatpoly.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I failed to find any pre-written code and consequently wrote my own in Swift. The equations are fairly straightforward and a list of the possible errors these equations may produce is linked to that page as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the polynomials:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the ΔT values derived from the historical record and from direct observations (see: Table 1 and Table 2 ), a series of polynomial expressions have been created to simplify the evaluation of ΔT for any time during the interval -1999 to +3000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define the decimal year &quot;y&quot; as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    y = year + (month - 0.5)/12&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This gives &quot;y&quot; for the middle of the month, which is accurate enough given the precision in the known values of ΔT. The following polynomial expressions can be used calculate the value of ΔT (in seconds) over the time period covered by of the Five Millennium Canon of Solar Eclipses: -1999 to +3000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before the year -500, calculate:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    ΔT = -20 + 32 * u^2&#xA;    where:  u = (y-1820)/100&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Between years -500 and +500, we use the data from Table 1, except that for the year -500 we changed the value 17190 to 17203.7 in order to avoid a discontinuity with the previous formula at that epoch. The value for ΔT is given by a polynomial of the 6th degree, which reproduces the values in Table 1 with an error not larger than 4 seconds:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ΔT = 10583.6 - 1014.41 * u + 33.78311 * u^2 - 5.952053 * u^3&#xA;    - 0.1798452 * u^4 + 0.022174192 * u^5 + 0.0090316521 * u^6 &#xA;where: u = y/100&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Between years +500 and +1600, we again use the data from Table 1 to derive a polynomial of the 6th degree.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ΔT = 1574.2 - 556.01 * u + 71.23472 * u^2 + 0.319781 * u^3&#xA;    - 0.8503463 * u^4 - 0.005050998 * u^5 + 0.0083572073 * u^6&#xA;where: u = (y-1000)/100&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Between years +1600 and +1700, calculate:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ΔT = 120 - 0.9808 * t - 0.01532 * t^2 + t^3 / 7129&#xA;where:  t = y - 1600&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Between years +1700 and +1800, calculate:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ΔT = 8.83 + 0.1603 * t - 0.0059285 * t^2 + 0.00013336 * t^3 - t^4 / 1174000&#xA;where: t = y - 1700&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Between years +1800 and +1860, calculate:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ΔT = 13.72 - 0.332447 * t + 0.0068612 * t^2 + 0.0041116 * t^3 - 0.00037436 * t^4 &#xA;    + 0.0000121272 * t^5 - 0.0000001699 * t^6 + 0.000000000875 * t^7&#xA;where: t = y - 1800&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Between years 1860 and 1900, calculate:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ΔT = 7.62 + 0.5737 * t - 0.251754 * t^2 + 0.01680668 * t^3&#xA;    -0.0004473624 * t^4 + t^5 / 233174&#xA;where: t = y - 1860&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Between years 1900 and 1920, calculate:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ΔT = -2.79 + 1.494119 * t - 0.0598939 * t^2 + 0.0061966 * t^3 - 0.000197 * t^4&#xA;where: t = y - 1900&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Between years 1920 and 1941, calculate:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ΔT = 21.20 + 0.84493*t - 0.076100 * t^2 + 0.0020936 * t^3&#xA;where: t = y - 1920&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Between years 1941 and 1961, calculate:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ΔT = 29.07 + 0.407*t - t^2/233 + t^3 / 2547&#xA;where: t = y - 1950&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Between years 1961 and 1986, calculate:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ΔT = 45.45 + 1.067*t - t^2/260 - t^3 / 718&#xA;where: t = y - 1975&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Between years 1986 and 2005, calculate:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ΔT = 63.86 + 0.3345 * t - 0.060374 * t^2 + 0.0017275 * t^3 + 0.000651814 * t^4 &#xA;    + 0.00002373599 * t^5&#xA;where: t = y - 2000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Between years 2005 and 2050, calculate:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ΔT = 62.92 + 0.32217 * t + 0.005589 * t^2&#xA;where: t = y - 2000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This expression is derived from estimated values of ΔT in the years 2010 and 2050. The value for 2010 (66.9 seconds) is based on a linearly extrapolation from 2005 using 0.39 seconds/year (average from 1995 to 2005). The value for 2050 (93 seconds) is linearly extrapolated from 2010 using 0.66 seconds/year (average rate from 1901 to 2000).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Between years 2050 and 2150, calculate:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ΔT = -20 + 32 * ((y-1820)/100)^2 - 0.5628 * (2150 - y)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The last term is introduced to eliminate the discontinuity at 2050.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After 2150, calculate:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ΔT = -20 + 32 * u^2&#xA;where:  u = (y-1820)/100&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;All values of ΔT based on Morrison and Stephenson [2004] assume a value for the Moon's secular acceleration of -26 arcsec/cy^2. However, the ELP-2000/82 lunar ephemeris employed in the Canon uses a slightly different value of -25.858 arcsec/cy^2. Thus, a small correction &quot;c&quot; must be added to the values derived from the polynomial expressions for ΔT before they can be used in the Canon&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;c = -0.000012932 * (y - 1955)^2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since the values of ΔT for the interval 1955 to 2005 were derived independent of any lunar ephemeris, no correction is needed for this period.&lt;/p&gt;&#xA;" OwnerUserId="15619" LastEditorUserId="15619" LastEditDate="2017-01-26T23:04:46.163" LastActivityDate="2017-01-26T23:04:46.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19863" PostTypeId="2" ParentId="19848" CreationDate="2017-01-27T02:02:21.217" Score="4" Body="&lt;p&gt;KarenRei's answer is very good as there is a significant degree of uncertainty, that said, I think the answer to this question is a clear no, the Earth would not be habitable at Venus' distance to the Sun.  Not unless you're a waterbear living in a very deep cave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus' average distance to the sun is .722 AU.   &lt;a href=&quot;http://www.universetoday.com/15462/how-far-are-the-planets-from-the-sun/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.  That means the surface of the sun is about 38.5% closer (A bit more than that since distance is measured to the center of the sun, not the surface) and because illumination drops by the square of the distance, Venus gets hit by 92% more solar energy per square meter than the Earth does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's thought that life on the Earth's surface could get difficult with just a small percent luminosity increase from the sun in as little as 150 million years or so and with a 10%-15% luminosity increase range (1 to 1.5 billion years), Earth could begin to have surface water vaporization, leading to a runaway greenhouse effect.    &lt;a href=&quot;http://www.sciencemag.org/news/2014/01/earth-wont-die-soon-thought&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.   Some estimates say even sooner, 600-700 million years, or a 6-7% increase in luminosity.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These estimates are difficult to model and should be taken with a grain of salt, but if a 15% increase begins vaporizing lakes on land, and with water being a greenhouse gas, a 92% increase would be catastrophic, almost certainly leading to a very hot, Venus like surface temperature on Earth with a significant increase in atmospheric water vapor and perhaps Methane trapped below frozen tundra and CO2 from dying planets as well.   I agree with KarenRei that there's considerable uncertainty in the outcome and cloud formation and albedo are difficult to predict, but Earth would still likely be hot enough to cook a pot-roast on the surface in that orbit.   I don't see any way it wouldn't be, and possibly not enough to melt lead like the surface of Venus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps, in the deep oceans (if the oceans survive, and I'm not sure they would), or in caves or permanently shaded caverns with access to underground water, some rugged life might survive.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to Earth's magnetic field, it's not all that relevant to your question.  Without a magnetic field, a planet is more likely to lose it's atmosphere, but that takes millions, if not tens or hundreds of millions of years.   Venus has already been stripped of much of it's lighter gases.   If you put Earth in Venus' orbit, that process would probably still happen, but it would take a long time and Earth's magnetic field would likely make the process slower.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason Earth has a magnetic field is somewhat complicated and there's many good threads on Earth's magnetic field in &lt;a href=&quot;https://earthscience.stackexchange.com/&quot;&gt;Earth Science Stack Exchange&lt;/a&gt;, but in a nutshell, Earth's solid inner core, liquid outer core and Earth's relatively fast rotation period are key factors.  Earth's moon &lt;em&gt;might&lt;/em&gt; play a role in maintaining it too.   Earth's magnetic field wouldn't go away if it was in Venus's orbit.   Venus lack of a magnetic field may be due to it's inner core not being solid and it could also be due to it's very slow rotation or a combination of both.   But Venus' atmosphere creates an &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Venus#Induced_magnetosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;induced magnetic field&lt;/a&gt;, which helps it retain it's atmosphere.  But that's only relevant to a planet's climate over the very long term as atmosphere gets stripped by the solar wind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 3rd consequence of Earth in Venus' orbit is the Moon would be much closer to an unstable orbital region.   It would still probably remain in the Earth's orbit but it could be touch and go over thousands maybe millions of orbits, and the Moon would be significantly more perturbed and probably more eccentric at times and there's a chance Earth could lose the moon in this scenario, turning it into a dangerous near earth object that we would basically share an orbit with, and possibly crash into the Earth, or get tossed away by gravity assist.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:29.267" LastActivityDate="2017-01-27T02:10:32.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19864" PostTypeId="1" AcceptedAnswerId="19869" CreationDate="2017-01-27T11:36:41.763" Score="3" ViewCount="501" Body="&lt;p&gt;Based on this question: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/12854/how-loud-would-the-sun-be&quot;&gt;How loud would the Sun be?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If instead of a vacuum we had air between the sun and earth (let's say of the same composition we have here on earth) how warm would our planet get?&lt;/p&gt;&#xA;" OwnerUserId="15468" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2017-01-27T17:25:39.747" Title="If there was air between the Sun and Earth, how warm would we get?" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;heat&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19866" PostTypeId="1" AcceptedAnswerId="19885" CreationDate="2017-01-27T14:43:29.253" Score="2" ViewCount="99" Body="&lt;p&gt;I'm quite novice about all this, only have read some vulgarization books and sites, but I wonder: could all the observable universe be falling outwards on some non-observable, more compact kind of barrier - I'll call it &quot;edge&quot;? Could this be a part of the explanation of cosmological accelerated expansion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like say we would be in a kind of bubble, or egg-shape thing? This thing could be inflating, and the closer matter would be from the edges, the quicker it would fall? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good question &quot;why would we be at the center of it?&quot;... I know already there could be &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_flow&quot; rel=&quot;nofollow noreferrer&quot;&gt;a kind of &quot;flow&quot; of matter at large scales&lt;/a&gt; which is already kind of violation of universe isotropy. Could we remove this issue with some kind &lt;strong&gt;gravity propagation&lt;/strong&gt; itself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any sense to anybody it makes?&lt;/p&gt;&#xA;" OwnerUserId="11264" LastEditorUserId="11264" LastEditDate="2017-01-30T08:14:30.087" LastActivityDate="2017-01-30T08:14:30.087" Title="Could the accelerating expansion of universe inidicate we are surrounded by &quot;denser&quot; space?" Tags="&lt;cosmology&gt;&lt;space-time&gt;&lt;cosmological-inflation&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ClosedDate="2017-01-29T11:25:37.667" ContentLicense="CC BY-SA 3.0" />
  <row Id="19868" PostTypeId="2" ParentId="19864" CreationDate="2017-01-27T16:07:37.610" Score="4" Body="&lt;p&gt;Sorta depends -- you can't have a uniform density atmosphere, so to reach something close to one standard atm. near Earth's orbit (which, BTW would wreak havoc with orbital stability in the first place due to drag), the density would have to be something horrible nearer to the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be an excellent xkcd what-if question if phrased more along the lines of &quot;what density profile&quot; ;  my careful, detailed calculation (i.e. pulled out of an impolite orifice) suggests that such an atmosphere would rapidly collapse into the Sun proper due to gravitational attraction.  I hesitate to speculate on just how hot the outer reaches would get due to optical absorption  -- supposing that this atmosphere suddenly popped into place at an overall temperature in the 200-300 K range and immediately started both heating and collapsing. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-01-27T16:07:37.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19869" PostTypeId="2" ParentId="19864" CreationDate="2017-01-27T17:25:39.747" Score="10" Body="&lt;p&gt;This is a really hard question to answer as Carl Witthoft alludes to. The main problem is that there is no way you can finagle putting at atmosphere around the Sun and have it extend to the Earth. It would be wildly unstable and no matter what happens, the general result will be an explosion of energy that kills the Earth. And you can't really say you're going to wave some magic wand and prevent all of that because you're then fundamentally altering physics and you won't end up answering the question you wanted to ask to begin with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your only way of answering this question is to look at the current atmosphere of the Sun and try to extrapolate. Take a look at the figure below. This represents both the temperature (solid line) and density (dashed line) of the Sun vs height from the surface (what the &quot;surface&quot; of the Sun means is a whole other question). The temperature at the surface of the Sun is a measly $5800\:\mathrm{Kelvin}$. As you increase in height, this temperature (i.e., the temperature of the plasma atmosphere above the &quot;surface&quot; of the Sun) doesn't change much. This region of the atmosphere is known as the chromosphere. But then, at a particular height known as the transition region, there's a &lt;em&gt;drastic&lt;/em&gt; increase in the temperature. It goes from a few thousand degrees to a couple million degrees. This area of the atmosphere is the Corona. This graph cuts off at $10,000\:km$ but the Sun's atmosphere extends far out beyond Pluto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OBpND.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OBpND.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully you can fully appreciate the point I'm getting at which is that Earth actually &lt;em&gt;is&lt;/em&gt; inside the atmosphere of the Sun and that atmosphere is really hot! The reason we of course aren't boiled to death is that the atmosphere is incredibly sparse. Only a handful of particles per cubic centimeter. So what you're really proposing is to increase the density of this atmosphere by an insane amount. You want to make it approximately $10^{20}$ times more dense! If we just extrapolate properties, that's going to turn the already hot atmosphere of the Sun (around the Earth) into something incredibly hot. And the huge density means we'll really feel the effects. The Earth will likely be vaporized in a flash, before the cataclysmic explosion which I alluded to above can even happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, long answer short: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/19864/if-there-was-air-between-the-sun-and-earth-how-warm-would-we-get#comment32627_19864&quot;&gt;About a billion degrees&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2017-01-27T17:25:39.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19870" PostTypeId="1" CreationDate="2017-01-27T17:45:18.000" Score="0" ViewCount="108" Body="&lt;p&gt;They say that the observable universe is only 5% of the entire equation. The rest of it consists of dark matter and energy. I believe that they are trying to find a way to see something that cannot be seen due to being outside of the observable universe, sort of like how galaxies redshift and slowly face away of out existence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;tl;dr  does dark matter/energy increase every year?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What comes after red shift? Black? If so, then does that mean the universe is mostly black because 95% of matter has &quot;Blackshifted&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did I post a question correctly?&lt;/p&gt;&#xA;" OwnerUserId="15624" LastEditorUserId="15624" LastEditDate="2017-01-27T18:04:15.757" LastActivityDate="2017-01-27T22:19:09.737" Title="How do you measure something that has already left the observable universe?" Tags="&lt;dark-energy&gt;&lt;observable-universe&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19871" PostTypeId="2" ParentId="19870" CreationDate="2017-01-27T19:03:55.607" Score="3" Body="&lt;p&gt;First off, you need to understand what &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Redshift&quot; rel=&quot;nofollow noreferrer&quot;&gt;Redshift&lt;/a&gt;&quot; is.  Light that is redshifted means that its wavelength has increased, making it appear more red.  This can come from a variety of sources, including Doppler shifting.  What we're more concerned with here is &quot;cosmological redshift&quot;.  As light travels through space, its wavelength increases due to the expansion of space.  We can use a measurement of how much the light has redshifted to calculate the distance to its source.  There is nothing &quot;beyond&quot; redshift, because redshift is just a lengthening of the wavelength.  And nothing is happening to the objects themselves, they just appear to get redder to us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, the definition of the &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Observable_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;observable universe&lt;/a&gt;&quot;.  Because light travels at a finite speed, it takes a certain amount of time to reach us.  The universe had a beginning in the big bang, meaning that light has only had a certain amount of time to travel.  Therefore we can only see things out to a certain distance.  Light coming from objects further than that has not had enough time to reach us yet, so we can't see them.  That's what we mean by the &quot;observable universe&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxies do not &quot;phase out of existence&quot; because of redshift.  As we look outwards, we're looking at older and older objects.  Because the light takes a certain amount of time to reach us, we're essentially looking back in time.  As time passes, however, we can see more and more because light from further away objects has more time to reach us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_matter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dark matter&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;dark energy&lt;/a&gt; cannot be directly observed because they do not interact with light.  They do not absorb it, emit it, reflect it, or anything else.  However, they are still within the observable universe, because the observable universe is basically just the size of the universe.  There are many different theories as to what dark matter and dark energy actually are, but they are not ordinary matter that has been &quot;blackshifted&quot;.&lt;/p&gt;&#xA;" OwnerUserId="13426" LastEditorUserId="13426" LastEditDate="2017-01-27T21:16:16.523" LastActivityDate="2017-01-27T21:16:16.523" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19872" PostTypeId="2" ParentId="19870" CreationDate="2017-01-27T21:54:48.810" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;They say that the observable universe is only 5% of the entire&#xA;  equation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The observable universe and regular matter are different things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Regular matter&lt;/strong&gt; is the stuff that you and I are made of. Besides regular matter, the Universe is made of dark matter and dark energy. According to some estimates, regular matter is indeed only a few percentage points of the &quot;stuff&quot; in the observable universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dark matter&lt;/strong&gt; - it's not clear what it is, but it appears to be matter-like and it interacts gravitationally. It does not seem to interact in any other way; it could be shooting right now through your head and you wouldn't know it. It does not make clumps like regular matter, so it tends to stay in a diffuse state. That's all we know about it, everything else is speculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dark energy&lt;/strong&gt; is something that drives and accelerates the expansion of the universe but doesn't seem to do anything else. It is not related to dark matter despite both being called &quot;dark&quot;. That's all we know about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The observable universe&lt;/strong&gt; is the chunk of the Universe that we can see: regular matter, dark matter, dark energy, and the space in between. By definition, the observable universe is everything that we could (theoretically) observe or detect. It's basically everything within a radius of a few dozen billion light years measured from the chair you're sitting on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything placed further than that is outside the observable universe and by definition cannot be observed. Not only that, but it cannot influence us in any way whatsoever - if it could, it would be observable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;sort of like how galaxies redshift and slowly face away of out&#xA;  existence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's a tricky statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe expands. Things close to us (on a cosmic scale) are drifting away from us slowly. Things far away from us are drifting apart quickly. Things very far are drifting away very quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep going like this and there comes a place where things are drifting away from us at light speed. Beyond this threshold there's nothing for us to see anymore. That's the theoretical outer limit of the observable universe. Anything beyond it is outside the observable universe. In practice you can't quite see that far anyway, to be honest, because at those huge distances it's hard to detect anything, and because at those huge distances you only see very old objects which may not be bright enough yet to be seen (before stars are formed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that once something is out of the observable universe, it's basically as if it doesn't exist anymore. But that's only true from your frame of reference. From their frame of reference they continue to exist just fine. Both frames are correct, BTW. Disagreements like this one are common in relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So don't say they're fading out of existence, because they may get angry at you. :) Simply say they exit the observable universe.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;does dark matter/energy increase every year?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Please be aware that anything that gets out of our observable universe, be it regular matter, dark matter, or dark energy, is lost to us forever. The event horizon (outer limit) of the observable universe does not apply only to regular matter, it applies to everything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, within the observable universe, we don't really know if the relative percentages of regular matter, dark matter, and dark energy are changing a whole lot.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What comes after red shift? Black?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Red shift does not have a limit. The name seems to suggest it stops at red - but really it doesn't. It's just not a perfect name, that's all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As things move away from you faster and faster and faster, light coming in from there gets &quot;red shifted&quot; - it simply means the wavelength becomes bigger and bigger and bigger. If light starts out in the color blue, &quot;red shift&quot; just means it becomes green, then yellow, orange, red, infrared, microwave, radio waves... just getting stretched out more and more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some point the wavelength gets so big it becomes very hard to detect anything.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do you measure something that has already left the observable&#xA;  universe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You don't. It's out of OU, it's gone. Completely and forever. Just let it go cause it ain't coming back.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;does that mean the universe is mostly black because 95% of matter has&#xA;  &quot;Blackshifted&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The universe is mostly black because there's not enough regular matter (stars, more specifically) in the observable universe to light it up fully. It's just a whole lot of empty space - hence dark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that in the outer parts of the observable universe objects are so red shifted that it's hard to see them anymore. That's one thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thing is - once objects are so far that they are moving away from us faster than light speed, it is 100% impossible for us to see them, forever. They are completely outside the observable universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the night sky is dark because the observable universe is not infinite, and because it's mostly empty. This is the solution to &lt;a href=&quot;https://en.wikipedia.org/wiki/Olbers&amp;#39;_paradox&quot; rel=&quot;nofollow noreferrer&quot;&gt;Olbers' paradox&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2017-01-27T22:19:09.737" LastActivityDate="2017-01-27T22:19:09.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19873" PostTypeId="2" ParentId="11242" CreationDate="2017-01-28T02:25:49.960" Score="1" Body="&lt;p&gt;A blink comparator was a viewing apparatus used by astronomers to find differences between two photographs of the night sky. It permitted rapidly switching from viewing one photograph to viewing the other, &quot;blinking&quot; back and forth between the two taken of the same area of the sky at different times. This allowed the user to more easily spot objects in the night sky that changed position. It was also sometimes known as a blink microscope. It was invented by physicist Carl Pulfrich at Carl Zeiss AG, then constituted as Carl-Zeiss-Stiftung, in 1904&lt;/p&gt;&#xA;" OwnerUserId="15626" LastActivityDate="2017-01-28T02:25:49.960" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19874" PostTypeId="2" ParentId="11242" CreationDate="2017-01-28T05:32:17.903" Score="2" Body="&lt;p&gt;I don't know if the loss of an eye (mentioned in &lt;a href=&quot;https://astronomy.stackexchange.com/a/11254/7982&quot;&gt;this answer&lt;/a&gt;) is related, but there is a one-eyed embodiment described in the 1909 patent &lt;a href=&quot;https://www.google.com/patents/US941812&quot; rel=&quot;nofollow noreferrer&quot;&gt; US941812, Telemeter for two successive observations&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Another arrangement of the telemeter, in which the two images are produced successively in a common image field filling the entire ocular field, seems to be still more advantageous.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WG9md.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WG9md.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus it seems the comment at the bottom of the the first page in &lt;a href=&quot;http://www.math.harvard.edu/~knill/history/wolf/bea_proofs_wolf.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; might not be 100% correct. I don't know the date, but the one-eyed blink comparator in Wikipedia looks pretty darn old!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Tragically, Pulfrich lost one eye in 1906,&#xA;  preventing him from using the stereographic tool from then on.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The full quote can be found in the &lt;a href=&quot;https://astronomy.stackexchange.com/a/11254/7982&quot;&gt;other answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oM7Ie.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oM7Ie.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; &quot;English: Blink comparator in the Meridian hall of the Helsinki observatory. Made by Carl Zeiss in early 20th century.&quot; from &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Blink_comparator_1.JPG&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/14U0c.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/14U0c.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; from &lt;a href=&quot;https://airandspace.si.edu/stories/editorial/finding-pluto-blink-comparator&quot; rel=&quot;nofollow noreferrer&quot;&gt;Finding Pluto with a Blink Comparator&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2017-01-28T05:51:23.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19875" PostTypeId="1" AcceptedAnswerId="19876" CreationDate="2017-01-28T06:09:01.960" Score="3" ViewCount="864" Body="&lt;p&gt;In &lt;a href=&quot;https://space.stackexchange.com/q/19936/12102&quot;&gt;this question&lt;/a&gt; I mention the GIF in EarthSky.org's Moon Phases update &lt;a href=&quot;http://earthsky.org/astronomy-essentials/lengths-of-lunar-months-in-2017&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lengths of lunar months in 2017&lt;/a&gt;. After watching it a while, I've noticed a pattern of parallel nearly vertical lines on the Moon image that's used to generate this simulation. I've shown it below. However, if I look at the moon image in Wikipedia, which is much larger, higher resolution, and has perhaps higher contrast, I don't see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It could be there, but less noticable due to the enhanced level of detail i the large image, but I am wondering - is it real, or not. If not, how did it get into this image?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wstX2.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wstX2.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Animation used in &lt;a href=&quot;https://space.stackexchange.com/q/19936/12102&quot;&gt;EarthSky.org&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AXqTC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AXqTC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Frame 29 from the GIF Animation used in &lt;a href=&quot;https://space.stackexchange.com/q/19936/12102&quot;&gt;EarthSky.org&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NGil0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NGil0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; A casual analysis of the cropped ROI. The traces are color: top, middle, and bottom third, column averages, black: entire column average. Short vertical lines are drawn by hand to try to point out the pattern that I'm seeing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vanMZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vanMZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; &quot;Full Moon photograph taken 10-22-2010 from Madison, Alabama, USA. Photographed with a Celestron 9.25 Schmidt-Cassegrain telescope...&quot; from &lt;a href=&quot;https://en.wikipedia.org/wiki/File:FullMoon2010.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.210" LastActivityDate="2017-01-28T11:13:40.197" Title="Are these parallel lines I'm seeing in this image a real pattern on the Moon?" Tags="&lt;the-moon&gt;&lt;surface&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19876" PostTypeId="2" ParentId="19875" CreationDate="2017-01-28T07:27:10.857" Score="11" Body="&lt;p&gt;They are not real. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The images used in the animation use a mosaic of photographs taken by an orbiting satellite, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Clementine_(spacecraft)&quot;&gt;Clementine orbiter&lt;/a&gt;. This satellite has a polar orbit, and so returns images in strips that run roughly north-south. Due to variations in lighting between orbits, when the strips are joined together, there are artefacts at the joins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole moon image from a ground-based telescope doesn't have such artefacts. The lunar orbiter view was used as it covers the whole moon, including the far-side. As the moon orbits Earth, it appears to wobble (an effect known as libration.) By starting with a whole moon map, and then projecting it to a sphere, this libration can be simulated. You couldn't do that with a single ground-based image.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-01-28T11:13:40.197" LastActivityDate="2017-01-28T11:13:40.197" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19877" PostTypeId="2" ParentId="19859" CreationDate="2017-01-28T08:12:47.990" Score="3" Body="&lt;p&gt;Hypothesised, weakly interacting, cold dark matter is affected by gravity, but behaves differently to normal matter, in terms of its spatial distribution and kinematics, because it is dissipationless. That is, it cannot lose kinetic energy through interactions and radiate heat away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distribution and motion of dark matter can be looked at on a number of scales. On the scale of galaxy clusters and superclusters, dark matter is arranged into voids and filaments, as shown by Zephyr's answer. It is decidedly non-uniform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the scale of a galaxy, our Milky way galaxy for instance, dark matter is thought to be distributed in a smoothly spherical, centrally concentrated way, and extends well beyond where the normal matter is found. The dark matter &lt;em&gt;is&lt;/em&gt; moving; it orbits in the overall Galactic gravitational potential like everything else, but the orbits will be much more radial, rather than the circular orbits followed by stars in the disk of our Galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means for dark matter on solar system scales, is that we expect there is a dark matter &quot;wind&quot;, because the Sun is orbiting at around 200 km/s with respect to the Galaxy, and so the dark matter will have this large number superimposed on its relative velocity with respect to the solar system. The wind at the Earth will also get stronger and weaker by 30 km/s as the Earth travels in its orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spatial distribution of dark matter on solar system scales should be quite uniform, and be less than one hundredth the density of normal matter in interplanetary space. There should however be a small (of order 1% density contrast) gravitational focusing effect due to the Sun as the &quot;wind&quot; passes by it.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-01-28T08:12:47.990" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19878" PostTypeId="2" ParentId="19772" CreationDate="2017-01-28T09:16:05.307" Score="1" Body="&lt;p&gt;I am going to float this as a tentative answer. I'll appreciate feedback, either validating it or letting me know it's wrong, in which case I'll delete it. If it is part of a better answer, please run with it and I'll also delete this. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;X-ray navigation &lt;a href=&quot;https://en.wikipedia.org/wiki/X-ray_pulsar-based_navigation&quot; rel=&quot;nofollow noreferrer&quot;&gt;XNAV&lt;/a&gt; is being look at as a source of navigational data for spacecraft in deep space that does not require measurements and telemetry from a deep space facility on Earth. While a spacecraft could not easily receive radio pulses from a pulsar without a very large antenna to overcome the thermal noise in the receiver's front-end, a suitable X-ray detector could potentially build up timing data from several pulsars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since a spacecraft is usually moving without thrust under gravitational forces from known bodies, an instantaneous fix is not needed and data could be collected for several minutes or even longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While radio pulses from pulsars are generally periodic, there are transient phenomena that can be related to infall of ionized material interracting with the magnetic field causing bumps and wiggles of individual radio pulses, which may have a bearing on X-ray time structure as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so I am speculating that these transient deviations in a Pulsar's otherwise regular behavior would be of relevance to technology development for both deep space exploration and business development.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, while intermittent download of data from spacecraft could be considered Transient GHz Emissions, it would not be considered Astrophysical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; The variability is in the production of electromagnetic radiation in a pulsar's incredibly intense magnetic field (radio) and at the surface (X-ray) produced during the infall of matter along the field lines towards the magnetic poles. The underlying rotation of the neutron star and it's field are still quite regular. &lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-01-28T11:21:09.300" LastActivityDate="2017-01-28T11:21:09.300" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19880" PostTypeId="1" CreationDate="2017-01-28T18:20:27.830" Score="0" ViewCount="1807" Body="&lt;p&gt;I want to calculate the Luminosity of a star e.g. Vega but can be any star and came across the formula for Luminosity at the University of Oregon (&lt;a href=&quot;http://pages.uoregon.edu/soper/Light/luminosity.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://pages.uoregon.edu/soper/Light/luminosity.html&lt;/a&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not an expert on Maths, could anyone help me with the figures to put into the formula.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L = (4\pi d^2)b$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d$ is the distance but the page doesn't say Parsecs or Light Years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b$ is the apparent brightness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L = ((4\times3.141) \times (25.04^2) ) \times 0.026 = 204.81 \mathrm{(light\ years)}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L = ((4\times3.141) \times (7.68^2)) \times 0.026 = 19.26 \mathrm{(parsecs)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which gives me 204.81 which is different to Luminosity which is quoted (40.12) on the Wikipedia page.  Is this the right formula for what I'm after.  Please no scientific equations like what you see on the page linked above.  Any calculations would need to be the type you would put into a non-scientific calculator.  &lt;/p&gt;&#xA;" OwnerUserId="5401" LastEditorUserId="2365" LastEditDate="2017-01-28T19:55:08.140" LastActivityDate="2017-01-29T12:17:57.840" Title="Calculate Luminosity of Vega" Tags="&lt;star&gt;&lt;luminosity&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19881" PostTypeId="1" CreationDate="2017-01-28T21:22:20.353" Score="-5" ViewCount="350" Body="&lt;p&gt;When people say that 95% of the universe is dark matter and dark energy, are they talking about it being in the observable universe or the unobservable universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because surely all matter in the universe must have a gravitational pull because it has mass, just because we dont see it anymore doesn't mean it's force has disappeared from the observable universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what if the matter we can't see outside the observable universe shows itself by imposing its gravitational forces on everything we can see within the observable universe?&lt;/p&gt;&#xA;" OwnerUserId="15624" LastEditorUserId="10437" LastEditDate="2017-01-30T11:50:45.040" LastActivityDate="2017-02-28T08:52:33.767" Title="Are they talking about the observable universe or the unobservable universe?" Tags="&lt;dark-matter&gt;&lt;dark-energy&gt;&lt;observable-universe&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2017-02-11T18:48:33.727" ContentLicense="CC BY-SA 3.0" />
  <row Id="19882" PostTypeId="2" ParentId="19881" CreationDate="2017-01-28T21:54:46.743" Score="3" Body="&lt;p&gt;We can't know anything certain about the unobservable universe. It is too distant for information from it to have reached us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is generally believed that the universe is much larger than the part which we can observe. It may be infinitely large. There is no reason to suppose that the unobservable part of the universe is fundamentally any different from the part which we can observe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark energy and dark matter are nothing to do with the observable and unobservable universe. There is dark matter in the observable universe. We can observe it, as it has a gravitational effect. But we can't see it. It probably is some as-yet-undiscovered particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark energy is also in the observable universe, but what exactly it is is very uncertain. It may just be a number in a formula: not actually &quot;energy&quot; at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When people say that only 5% of the universe is normal matter, they are speaking about the observable universe, however, there is no reason to suppose the rest of the universe is any different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matter that is outside the observable universe is matter. And dark matter is dark matter. But matter that is currently outside our observable universe is too far away for any information about it to reach us. This means that the gravity of that galaxy doesn't affect us. In some sense, its gravitational force has left our observable universe. (always assuming that the observable universe isn't somehow special). Remember that gravitation isn't instant. Changes in gravitational pull propagate at the speed of light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the ascii art below, the S represents a galaxy, and u represents you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The galaxy is &quot;now&quot; outside our observable universe (the region in the / \ is observable) It has no gravitational or other effects, however it is quite reasonable to believe the ratio of matter to dark matter is the same for this galaxy as for any other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't observe the galaxy &quot;now&quot;, but you can observe its past form, since, as you see the early galaxy is in our observable part of the universe.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;          S    u     ---Now----&#xA; ^        S   / \&#xA; |        S  /   \&#xA; T        S /     \&#xA; I        S/       \&#xA; M        S         \&#xA; E       /S          \&#xA; ----Space----------------&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-01-28T22:25:53.413" LastActivityDate="2017-01-28T22:25:53.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19883" PostTypeId="2" ParentId="19881" CreationDate="2017-01-28T22:42:48.117" Score="1" Body="&lt;p&gt;Your numbers are approximately those measured for the observable universe and predicted for the entire Universe, assuming the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lambda-CDM_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;ΛCDM model&lt;/a&gt; (aka the standard model of cosmology) is correct. The ΛCDM model has proven to be a very precise approximation for the entire Universe, and it indicates that a geometrically flat universe which follows the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmological_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;cosmological principle&lt;/a&gt; and expands at the same rate as our Universe would have an ordinary matter density of 4.86%, a dark matter density of 25.89% and a dark energy density of 69.11%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 2006, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Wilkinson_Microwave_Anisotropy_Probe&quot; rel=&quot;nofollow noreferrer&quot;&gt;WMAP spacecraft&lt;/a&gt; measured the spatial geometry of the Universe to be nearly flat and determined that the observable universe consists of 4.628% ordinary matter, 24.02% dark matter and 71.35% dark energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other methods have been used to confirm the accuracy of the ΛCDM model in terms of the universe's matter-energy distribution. &lt;a href=&quot;https://arxiv.org/abs/0804.4142&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kowalski et al. (2008)&lt;/a&gt; show that supernova observations indicate 71.3% of the observable universe is composed of dark energy, whereas the other 28.7% is the total matter content.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more advanced &lt;a href=&quot;https://en.wikipedia.org/wiki/Planck_(spacecraft)&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Planck&lt;/em&gt; spacecraft&lt;/a&gt; gave a better estimation in 2013, with its data indicating the observable universe is composed of 4.9% ordinary matter, 26.8% dark matter and 68.3% dark energy, almost exactly what the ΛCDM model predicts for the entire (approximated) Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, our measurements of the observable universe strongly line up with our models for the entire Universe. Keep in mind the ΛCDM model is not entirely perfect, but it approximates the Universe relatively well. While we cannot know for certain the mass-energy distribution of the entire Universe, we can predict its distribution based on the precision of our models and how closely the data for the observable universe matches up with them.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2017-02-28T08:52:33.767" LastActivityDate="2017-02-28T08:52:33.767" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19885" PostTypeId="2" ParentId="19866" CreationDate="2017-01-29T00:35:29.987" Score="6" Body="&lt;p&gt;It's a pretty clever idea and a solid question that unseen mass might cause gravity outside the observable that tugs on the universe and might be the cause of dark energy as opposed to some unknown energy that seems to permeate all space.  It's not a new idea, but it's worth answering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's 2, possibly 3 (or 4) pretty big problems to that approach.   The first is that Dark Energy expansion is very uniform.  For it to be explained by gravity, that gravity would need to be close to uniform around the observable universe, with Earth close to the center.    A giant &quot;edge&quot; as you called it, or maybe we're in a low mass bubble surrounded by higher mass.    Not impossible, but somewhat improbable that the Earth is in the center of this theoretical gravity field pulling the observable universe apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 2nd problem is time.  Gravity travels at the speed of light, so this gravity field would need to have existed prior to the big bang or be very close to the edge of the observable universe.  Still not impossible, but it's a difficult theory to work with as it requires a few leaps of faith.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A possible 3rd problem is that measurements show that the &lt;a href=&quot;http://www.space.com/34928-the-universe-is-flat-now-what.html&quot; rel=&quot;noreferrer&quot;&gt;universe is very flat&lt;/a&gt;.  I don't know if a gravity field of the sort you suggest would be flat or not, I suspect not, unless it was enormously large.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given enough time, maybe they can find a test to detect gravitons (if they exist) or, find dark energy and answer this one with better certainty.  The question is asked here on &lt;a href=&quot;https://www.quora.com/Could-dark-energy-just-be-the-gravity-of-the-galaxies-and-stars-beyond-the-observable-universe&quot; rel=&quot;noreferrer&quot;&gt;Quora&lt;/a&gt; if interested.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll add a 4th problem.   The velocity of observed young galaxies at the edge of the observable universe is close to the speed of light away from us in all directions.  Relative velocity that high is difficult to explain by high mass outside of the observable universe cause that mass would need to be enormous, measurably greater than the mass of the observable universe.   All the factors taken into account, it's an improbable explanation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If I've missed anything obvious or broken any laws of physics with the answer above, corrections are always welcome.&lt;/em&gt;  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-01-29T04:45:47.640" LastActivityDate="2017-01-29T04:45:47.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19886" PostTypeId="2" ParentId="19880" CreationDate="2017-01-29T04:48:49.033" Score="1" Body="&lt;p&gt;The $b$ in your equation, brightness, is what we call a &quot;Flux&quot;.  That's the amount of energy per unit area at a certain distance.  Flux can be calculated using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stefan-Boltzmann law&lt;/a&gt;.&lt;br&gt;&#xA;$$&#xA;J=\sigma T^4&#xA;$$&#xA;where $\sigma=5.670373\times10^{-8}\space W\space m^{-2}\space K^{-4}$is the Stefan-Boltzmann constant, This is measured in Watts per square meter.  That means that we will need our distance to be in meters, and we will get a luminosity in Watts.  We can go a little further and divide our answer by $3.828\times10^{26} \space W$ to get our final value in solar luminosities (L$_\odot$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That will make our final equation:&#xA;$$&#xA;L=\frac{4\pi R^2 \sigma T^4}{L_\odot}&#xA;$$&#xA;Because we're using the Stefan-Boltzmann equation, instead of the distance to the star, we have to use its radius.  Vega's radius is $2.362 \space R_\odot$, which is $1.64\times10^9$ meters. Its surface temperature is 9,600 K.  Plugging in those numbers yields a luminosity of:&#xA;$$&#xA;L=42.7L_\odot&#xA;$$&#xA;This value is slightly different from the value quoted by Wikipedia, due to rounding, but is well within the acceptable range.&lt;/p&gt;&#xA;" OwnerUserId="13426" LastActivityDate="2017-01-29T04:48:49.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19887" PostTypeId="2" ParentId="19880" CreationDate="2017-01-29T09:48:41.850" Score="1" Body="&lt;p&gt;This calculation attempts to estimate the luminosity from the distance and apparent brightness, instead of modelling the star as a black body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I notice that you have &quot;b=0.026&quot;, this is not the brightness, but instead the apparent magnitude of Vega. If you are trying to estimate the luminosity from distance and apparent magnitude, you can proceed as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absolute magnitude $M$  of a star can be found from the apparent magnitude $m$ and distance $d$ (in parsecs).&#xA;$$M = m - 5 \left(\log_{10}d-1\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The luminosity is related to the absolute bolometric magnitude, which considers all the energy at all wavelengths (including infrared and ultraviolet). Bolometric magnitude differs most from visible magnitude when the star is very cool or very hot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vega has a absolute visible magnitude of 0.57, and a bolometric magnitude only slightly brighter (ie smaller) than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The luminosity, L is found from the bolometric magnitude as:&#xA;$$L=10^{\frac{71.197-M_\mathrm{bol}}{2.5}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting the value 0.55 into this formula give a luminosity of $1.8\times10^{28}$Watts, or about $46L_\odot$, which differs slightly from the published value, due I think to not properly adjusting for bolometric magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining these gives&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;L = (10^26.5 * d^2)/2.51^m / 3.28e26&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Formulae from wikipedia: &lt;a href=&quot;https://en.wikipedia.org/wiki/Absolute_magnitude&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Absolute_magnitude&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-01-29T12:17:57.840" LastActivityDate="2017-01-29T12:17:57.840" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19888" PostTypeId="2" ParentId="19880" CreationDate="2017-01-29T10:03:48.510" Score="1" Body="&lt;p&gt;The root of your problem is not the maths, it is choosing the correct units. The relevant equation &lt;em&gt;is&lt;/em&gt; &#xA;$$L = 4\pi d^2 b,$$&#xA;Where $b$ is the power per unit area received at Earth from Vega, and $4\pi d^2$ is the surface area of an enormous sphere constructed around Vega, with radius equal to the distance to the earth and which would catch &lt;em&gt;all&lt;/em&gt; the light emitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a luminosity in Watts, we need to express $b$ in Watts per square metre and the area in square metres. It helps to know that $d =19.26 \times 3.08\times 10^{16}$ m. The value of $b$ you have used is the &lt;em&gt;apparent visual magnitude&lt;/em&gt; of Vega and I'm afraid that is not the power per unit area received at the Earth. A bit of reverse engineering tells me that actually $b = 3.5\times 10^{-9}$ W/m$^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To obtain a final answer in terms of a multiple of solar luminosities, you divide your result by the solar luminosity in Watts, which is $L_{\odot}= 3.83\times 10^{26}$ W.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I think you &lt;em&gt;want&lt;/em&gt; to be able to calculate the stellar luminosity from the distance and apparent magnitude. In which case, I refer you to James K's answer, noting that the bolometric correction depends on what type of star you are looking at.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-01-29T10:20:59.003" LastActivityDate="2017-01-29T10:20:59.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19889" PostTypeId="1" AcceptedAnswerId="19890" CreationDate="2017-01-29T13:36:53.280" Score="2" ViewCount="68" Body="&lt;p&gt;Why do gaseous planets spin faster than telluric planets? I've seen thats because of primitive nebula but I didn't understand why? Could someone help me? I would appreciate a lot!&lt;/p&gt;&#xA;" OwnerUserId="15636" LastActivityDate="2017-01-29T13:57:55.903" Title="Speed gaseous planets" Tags="&lt;planet&gt;&lt;speed&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19890" PostTypeId="2" ParentId="19889" CreationDate="2017-01-29T13:57:55.903" Score="4" Body="&lt;p&gt;There isn't such a great difference between Earth and Mars (at about 25 hours) and the gas/ice giants (between 10 and 20 hours). Earth had a day of about 6 hours in the early Solar system, but has been slowed by lunar tides. Small planets will not accrete gas in their early formation, and so will not pick up angular momentum from accretion discs, which can explain why Jupiter and Saturn have rapid rotation rates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury and Venus are slowed by Solar tides. In the case of Mercury, it is in a resonance between its orbit and the solar tide. In the case of Venus, there are significant atmospheric tides, that slow it down (It actually spins backwards, and the full explanation is incomplete.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is three-fold: Distant gas giants pick up lots of momentum in their formation, they are not much affected by Solar tides, and the differences between them and the rocky planets are not actually so great anyway. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-01-29T13:57:55.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19891" PostTypeId="1" CreationDate="2017-01-29T16:07:13.007" Score="-1" ViewCount="1842" Body="&lt;p&gt;What is the difference between Absolute Magnitude and Absolute Bolometric Magnitude, is it a case of multiply one by a constant to get the other?  Can you give an example for a star, say Vega?&lt;/p&gt;&#xA;" OwnerUserId="5401" LastActivityDate="2017-05-31T05:40:08.850" Title="Absolute Magnitude and Absolute Bolometric Magnitude" Tags="&lt;star&gt;&lt;magnitude&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19893" PostTypeId="1" AcceptedAnswerId="19895" CreationDate="2017-01-30T02:51:55.260" Score="9" ViewCount="361" Body="&lt;p&gt;If you were flying along in your spacecraft somewhere in our solar system and then a large, stable wormhole suddenly opened and you flew into it and were whisked away to some distant location in the galaxy, would you be able to figure out where you were? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, a large stable wormhole is exceedingly unlikely, and then there's the matter of the power of your onboard computer and telescope(s). But all that aside, I'm wondering if it would be even theoretically possible to figure out where you were, since the constellations would probably be unrecognizable and you wouldn't have an understandable flight path.&lt;/p&gt;&#xA;" OwnerUserId="15637" LastActivityDate="2017-03-13T22:02:43.037" Title="Is it possible to figure out your location in the Milky Way if you are suddenly wormholed to a random, distant location in the galaxy?" Tags="&lt;galaxy&gt;&lt;telescope&gt;&lt;milky-way&gt;&lt;positional-astronomy&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19894" PostTypeId="2" ParentId="19893" CreationDate="2017-01-30T05:19:09.083" Score="1" Body="&lt;p&gt;If a spaceship travels through a space warp into an unknown region of space or gets lost some other way the navigators should look for familiar objects. Fortunately space is transparent. Very transparent. Very, very, very transparent. The thinnest and most transparent vacuum made on Earth is like a sheet of lead compared to the densest, dustiest part of interstellar space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a clear spot you can see forever, so to speak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the navigators should look for bright objects in the visible and other parts of the electromagnetic spectrum. They should take the spectra of the 10 or 100 objects that are brightest in each of various bands of visible light, radio, infrared, ultraviolet, X Rays, gamma rays etc., and compare them to the spectra of the 10 or 100 brightest objects in those bands as seen from Earth or their home world wherever it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hope will be that you haven't traveled too far from known space and some at least of the objects observed in the new location will match those on the list as seen from the home world. All you need to do is observe three known objects and then from differences in their apparent magnitudes and the angles between them as seen in your new location and as seen from your home world you can calculate your new position. Modern astronomical instruments that are vital for interstellar navigation can measure such angles to incredible precision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the 10 or 100 brightest objects visible in any band of the spectrum from any location will usually be at a great variety of distances. Some just a few light years away, some hundreds, some tens of thousands, some millions of light years away, some hundreds of millions, etc. The ones that are far enough away will still be among the brightest objects in that band no matter where you go within our galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can be Lost in space in the sense of not having a way to get home, but so long as you don't jump millions of light years or something you can't get Lost in space in the sense of not being able to figure out where you are.&lt;/p&gt;&#xA;" OwnerUserId="15639" LastActivityDate="2017-01-30T05:19:09.083" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19895" PostTypeId="2" ParentId="19893" CreationDate="2017-01-30T07:27:36.757" Score="7" Body="&lt;p&gt;Assuming you had access to the relevant astronomical catalogues and data, then yes it would.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at constellations would of course be hopeless and most stellar catalogues only contain star positions for a relatively small volume in our Galaxy. However, there are classes of object inside and outside the Galaxy that would do the job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A displacement in our own Galaxy would produce a notable displacement in the positions of local group galaxies, like Andromeda and M33, with respect to distant quasars that define the ICRS coordinate reference frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Providing one had the means (i.e. telescopes) to make the necessary observations, then it should be relatively straightforward to reconstruct your position in the Galaxy from the new apparent positions of recognisable local group galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further possibilities are offered by globular clusters and pulsars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The colour-magnitude diagrams and chemical compositions of globular clusters are all subtley different, which should enable their identification. They are also spread in a roughly spherical distribution around the Galaxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radio pulsars have unique pulse shapes and timing properties that make them readily identifiable. A possible/probable problem is that many pulsars would be rendered invisible by the change in viewing angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both these cases, a map of the revised spatial locations with respect to the ICRS should enable you to work out where you are compared with the position of the Earth now. I am unsure what precision could be achieved, but then we are unsure of our present location with respect to distance to the Galactic centre by a few percent.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-03-13T22:02:43.037" LastActivityDate="2017-03-13T22:02:43.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19896" PostTypeId="1" AcceptedAnswerId="19897" CreationDate="2017-01-30T08:58:13.433" Score="5" ViewCount="522" Body="&lt;p&gt;There are a finite number of currently observable galaxies due to the finite age of the universe and the speed of light. What fraction of these galaxies have we actually observed (by eye, telescope, etc.)?&lt;/p&gt;&#xA;" OwnerUserId="941" LastEditorUserId="941" LastEditDate="2017-01-31T15:03:35.857" LastActivityDate="2017-09-26T18:00:44.553" Title="What fraction of galaxies in the observable universe have we actually observed?" Tags="&lt;galaxy&gt;&lt;observable-universe&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19897" PostTypeId="2" ParentId="19896" CreationDate="2017-01-30T08:58:13.433" Score="7" Body="&lt;p&gt;There are around 2 trillion galaxies in the currently observable universe according to the &lt;a href=&quot;http://www.nature.com/news/universe-has-ten-times-more-galaxies-than-researchers-thought-1.20809#/b1&quot; rel=&quot;noreferrer&quot;&gt;latest&lt;/a&gt; &lt;a href=&quot;https://www.spacetelescope.org/news/heic1620/&quot; rel=&quot;noreferrer&quot;&gt;estimates&lt;/a&gt;, obtained by &lt;a href=&quot;https://arxiv.org/abs/1607.03909&quot; rel=&quot;noreferrer&quot;&gt;integrating theoretical galaxy stellar mass functions&lt;/a&gt; above $10^{6} M_{\odot}$ between $0 \leqslant z \leqslant 8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's difficult to get a precise number for the total observed galaxies as the results from new surveys are being released all the time, and updated analysis of legacy data is revealing more galaxies. For example, the &lt;a href=&quot;https://www.sdss3.org/dr13/&quot; rel=&quot;noreferrer&quot;&gt;latest SDSS data release&lt;/a&gt; identified close to &lt;a href=&quot;http://www.sdss.org/dr13/scope/&quot; rel=&quot;noreferrer&quot;&gt;200 million galaxies&lt;/a&gt;, whilst the ongoing Dark Energy Survey seeks to identify around &lt;a href=&quot;http://www.space.com/33766-dark-energy-survey.html&quot; rel=&quot;noreferrer&quot;&gt;300 million galaxies&lt;/a&gt;. &lt;a href=&quot;https://www.lsst.org/&quot; rel=&quot;noreferrer&quot;&gt;LSST&lt;/a&gt;, in Chile, will observe close to &lt;a href=&quot;http://www.dailygalaxy.com/my_weblog/2015/05/-observing-20-billion-galaxies-the-large-synoptic-survey-telescope-lsst.html&quot; rel=&quot;noreferrer&quot;&gt;20 billion&lt;/a&gt; galaxies, an order of magnitude greater than anything before it. However, all of these surveys operate up to relatively low redshifts. Probing to higher redshifts requires bigger, preferably space based telescopes, and much longer exposure times. As such, the high redshift universe has been poorly documented. Our understanding of galaxy formation at high redshift is not as complete as at low redshift, and so we may be under- or over-estimating the number of galaxies at this epoch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A vague upper limit: we've observed less than 0.01% of all currently observable galaxies. &lt;/p&gt;&#xA;" OwnerUserId="941" LastEditorUserId="941" LastEditDate="2017-09-26T18:00:44.553" LastActivityDate="2017-09-26T18:00:44.553" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="19898" PostTypeId="1" CreationDate="2017-01-30T14:01:18.520" Score="2" ViewCount="116" Body="&lt;p&gt;I have a problem: If I would dig a hole through the earth, and jump right in, disregarding the rotation of the earth, I should come back out on the other side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, orbital mechanics say otherwise: Specific Orbital Energy tells me:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;v^2/2 - µ/r = -µ/2a&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, on the planet, my relative velocity is 0 m/s, so v^2/2 = 0. The term simplifies to:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-µ/r = -µ/2a | / (-µ)&#xA;1/r = 1/2a | ^(-1)&#xA;r = 2a | /2&#xA;a = r/2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;My apoapsis is 3000 km by this equation. But it should be 6000 km, since I come out of the other side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this, and did I do a wrong assumption?&lt;/p&gt;&#xA;" OwnerUserId="15644" LastActivityDate="2017-01-30T16:01:14.187" Title="Hole through the earth" Tags="&lt;orbit&gt;&lt;earth&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2017-01-31T15:04:52.607" ContentLicense="CC BY-SA 3.0" />
  <row Id="19899" PostTypeId="2" ParentId="19891" CreationDate="2017-01-30T14:03:51.400" Score="1" Body="&lt;p&gt;From the page titled, &quot;Absolute Magnitude&quot; and with the disclaimer on that page &quot;not about the magazine,&quot;  [emphasis mine to aid in finding the info]&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Absolute magnitude is the measure of intrinsic brightness of a&#xA;  celestial object. It is the hypothetical apparent magnitude of an&#xA;  object at a standard distance of exactly 10 parsecs (32.6 light years)&#xA;  from the observer, assuming no astronomical extinction of starlight.&#xA;  This places the objects on a common basis and allows the true energy&#xA;  output of astronomical objects to be compared without the distortion&#xA;  introduced by distance. As with all astronomical magnitudes, the&#xA;  absolute magnitude can be specified for different wavelength&#xA;  intervals; for stars the most commonly quoted absolute magnitude is&#xA;  the absolute visual magnitude, which uses only the visual (V) band of&#xA;  the spectrum (UBV system). Also commonly used is the &lt;strong&gt;absolute&#xA;  bolometric magnitude&lt;/strong&gt;, which is the total luminosity expressed in&#xA;  magnitude units that takes into account energy radiated at all&#xA;  wavelengths, whether visible or not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So in short, AbsMag is Visual Luminosity and AbsBol is Total Luminosity, Visual and Otherwise. (from comment by MiscellaneousUser)&lt;/p&gt;&#xA;" OwnerUserId="907" LastEditorUserId="2365" LastEditDate="2017-05-31T05:40:08.850" LastActivityDate="2017-05-31T05:40:08.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19900" PostTypeId="2" ParentId="13242" CreationDate="2017-01-30T15:36:32.043" Score="0" Body="&lt;p&gt;Let's take the case we enter $\mathrm{BH}_\mathrm{Sgr\,A^*}$ (the super-massive black hole at the center of the way) backwards; from what I understood &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10467/time-according-to-the-gravity-of-sagittarius-a&quot;&gt;here&lt;/a&gt;, an external observer should experience you slowing down, whereas you will see is time accelerate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had the possibility to &quot;hover&quot; at 1 meter from the surface, you would see time be accelerated a million time for ordinary mater in low-gravity fields  (see calculation at earlier reference), and more and more the closer you get to the boundary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then once you pass the border, I recon you should see yourself in the first place, because photons you emit will be dragged back to the black hole by its gravity!! I guess it requires you enter the black hole with a specific trajectory to compensate rotational effects. But since other light should also be entering the BH, it should superpose to yours...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, there is a limit to the amount of light that enters the BH: its the rate at which BH grows: if light entering BH would be infinite, then BH mass should become infinite as soon as the BH becomes a BH. &lt;/p&gt;&#xA;" OwnerUserId="11264" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2017-01-30T15:36:32.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19901" PostTypeId="2" ParentId="19898" CreationDate="2017-01-30T15:55:41.303" Score="5" Body="&lt;p&gt;The formula you are using is for a point mass, but a hole through the Earth experiences less gravity than a point mass, as the comment points out.  One can then ask, what if the Earth's mass were all at the center, why wouldn't the mass fly through and go to a distance r on the other side?  It's because a mass like that is a singularity, one that doesn't matter unless you fall right through it.  If the Earth's mass was really concentrated at a point, then falling from r would still tend to just miss that point, since it's just a point anyway, and the huge forces would turn the orbit around, such that the object fell from r, went to zero, turned around, and flew back out to r.  That's a very squeezed ellipse, with a=r/2, as desired.  The motion that goes straight through and out to the opposite r, on the other hand, is not an ellipse with the mass at one focus, so it violates Kepler's laws.  Again, the violation of Kepler's laws is that the mass is singular in a way that matters if you fall through a hole in it.  Interestingly, it doesn't matter how concentrated the mass is toward the center-- if there's a hole in it that the object falls through, then the object does not obey Kepler's laws.  It will only obey Kepler's laws if the orbiting object does not enter the mass.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2017-01-30T16:01:14.187" LastActivityDate="2017-01-30T16:01:14.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19902" PostTypeId="1" CreationDate="2017-01-30T18:36:50.030" Score="1" ViewCount="1275" Body="&lt;p&gt;I'm a right that the difference is that precession indicates the direction of the inclination of the Earth'axis and the tilt is the quantity of this inclination? So that when there is no tilt there can't be any precession?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And are the inclination degrees of the precession and the tilt related or the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that the precession causes different possible polarstar, but is the tilt also have effect on this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps this above it not right, so what else is the difference? Is perhaps the precession just the changing of the tilt?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastActivityDate="2017-01-31T22:10:03.613" Title="What is the difference between precession and axial tilt?" Tags="&lt;planet&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19903" PostTypeId="2" ParentId="19902" CreationDate="2017-01-30T20:56:46.913" Score="2" Body="&lt;p&gt;Tilt is the angle between the earth's rotation axis, and its orbital plane around the sun.  Precession is the phenomenon whereby the direction of the tilt changes, which is why Polaris won't always be the pole star (until it comes round again).  The amount of the tilt also varies, but only by a few degrees.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-01-30T20:56:46.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19904" PostTypeId="2" ParentId="19902" CreationDate="2017-01-30T21:00:20.683" Score="3" Body="&lt;p&gt;The Earth orbits the sun. The plane in which it orbits is called the Ecliptic plane, and perpendicular to the plane is a direction, known as the ecliptic pole. The direction of the Ecliptic pole is towards the constellation &quot;Draco&quot;. The Earth's North pole doesn't point in the same direction. It points towards the pole star &quot;Polaris&quot; in Ursa Minor. The difference between these two poles is the axial tilt (it is about 23.5 degrees)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over time the direction of the North Pole changes, it moves in a circle around the Ecliptic pole, taking about 26000 years to complete one circuit. This is precession. The amount of tilt doesn't change much since the North Pole moves in roughly a circle. The tilt varies over a few degrees with a period of about 41000 years. The variation in tilt is not precession.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So axial tilt and precession are not the same thing. One is the angle between the ecliptic pole and north pole, the other is the motion of the north pole relative to the ecliptic.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-01-31T22:10:03.613" LastActivityDate="2017-01-31T22:10:03.613" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19905" PostTypeId="1" AcceptedAnswerId="19908" CreationDate="2017-01-30T21:57:51.767" Score="4" ViewCount="602" Body="&lt;p&gt;As far as I know, the Big Rip occurs when the scale factor reaches infinity in a finite time. This will only happen in a universe dominated by phantom energy (i.e. a universe with an equation of state of $&amp;lt; -1$). However, according to &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Rip#Expansion&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A universe dominated by phantom energy is an accelerating universe, expanding at an ever-increasing rate. However, this implies that the size of the observable universe is continually shrinking; the distance to the edge of the observable universe which is moving away at the speed of light from any point moves ever closer. When the size of the observable universe becomes smaller than any particular structure, no interaction by any of the fundamental forces can occur between the most remote parts of the structure. When these interactions become impossible, the structure is &quot;ripped apart&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On Wikipedia, there is a &quot;[why]&quot; tag, indicating that this wasn't explained very well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This raises two questions. In a universe dominated by phantom energy, why would the observable universe not continue growing and approaching the cosmic event horizon? After all, its comoving distance is just $\int_0^t \frac{\mathrm{d}t}{a(t)}$, so shouldn't it just monotonically increase?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More confusingly, Wikipedia implied that the Big Rip occurs when the observable universe's radius becomes incredibly small and the fundamental forces can no longer hold any structure together, rather than when the scale factor reaches infinity. Am I right in assuming these are separate events, or do they occur at the same time?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2017-01-30T22:27:49.893" LastActivityDate="2017-02-02T01:30:18.827" Title="How can the observable universe shrink in a Big Rip?" Tags="&lt;universe&gt;&lt;cosmology&gt;&lt;expansion&gt;&lt;observable-universe&gt;&lt;fate-of-universe&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19906" PostTypeId="1" CreationDate="2017-01-30T22:20:25.493" Score="1" ViewCount="107" Body="&lt;p&gt;The star is located at a distance of 163,000 light years in the constellation of Circinus.  As the Milky Way is 100,000 light years, this would then imply the star is outside the Milky Way?  Would it be a data anomoly if that is the case or is it in another galaxy like r136a?  Would it be on the far side of the galaxy, other side of the Galactic Centre?  &lt;/p&gt;&#xA;" OwnerUserId="5401" LastActivityDate="2017-01-30T22:20:25.493" Title="Location of star HD 135591" Tags="&lt;star&gt;" AnswerCount="0" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19908" PostTypeId="2" ParentId="19905" CreationDate="2017-01-31T14:06:08.927" Score="2" Body="&lt;p&gt;I've found the answer to this is actually very interesting and (to me, at least) surprising — it is not just the light cone that shrinks, but the cosmic event horizon as well! To explain this, let us first look at a de Sitter universe. The scale factor $a$ of a de Sitter universe can be expressed as a function of time with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a(t) = e^{kt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $k$ is some constant, depending on the specifics of our universe (in a de Sitter universe, $k = H$, the Hubble parameter). A universe dominated by phantom energy is actually pretty similar to a de Sitter universe, but the expansion rate gets exponentially faster. Thus, in a Big Rip universe $k$ is time dependent ($\dot{k} &amp;gt; 0$), so the scale factor can be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a(t) = e^{k(t)t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this relevant? Well, the comoving distance to the event horizon is actually inversely proportional to $a$ and $k$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_{eh}(t)\propto\frac{1}{a(t)k(t)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a de Sitter universe (and our own universe, assuming $w = -1$), $k$ is equal to the Hubble constant, so the event horizon’s distance will remain constant (in a de Sitter universe, the $\dot{H} = 0$ because the deceleration parameter $q=-1$). In a Rip universe, $k$ is time dependent; as $k$ increases, the distance to the horizon decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this formula, we can see that the event horizon’s distance is inversely proportional to the scale factor; thus, if $a(t)$ reaches infinity, the event horizon’s distance will reach zero. This means that the Big Rip can be defined as either the scale factor reaching infinity, or the observable universe’s diameter reaching zero — these two events happen at the same time.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2017-02-02T01:30:18.827" LastActivityDate="2017-02-02T01:30:18.827" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19909" PostTypeId="1" AcceptedAnswerId="19910" CreationDate="2017-01-31T15:45:59.363" Score="11" ViewCount="2405" Body="&lt;p&gt;Does the fact that universe is currently expanding at an accelerating rate means that far far away objects (&lt;em&gt;FAO&lt;/em&gt;) might start disappearing with time, set aside the &quot;red shift&quot;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm saying this because from what I understand until now:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;we receive photons from &lt;em&gt;FAO&lt;/em&gt; that were emitted long long ago&lt;/li&gt;&#xA;&lt;li&gt;now due to space expansion, some of those objects are now moving away at speed greater than the speed of light, but at the time the photons we receive &lt;em&gt;today&lt;/em&gt; were emitted, these &lt;em&gt;FAO&lt;/em&gt; where not moving away faster than light&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So there must be a point in time at which these far away objects disappear from view. How is this concept called?, or is it only a mistake I'm doing in my thinking? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a simple drawing to illustrate:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;at t=-11 billion years(By), photons are emitted in a slow expanding universe, and today we can view the object as it was&lt;/li&gt;&#xA;&lt;li&gt;at t=-2 By, the object moves faster than speed of light, so photon emitted at that time will never reach us, even if we wait the 9 by + &quot;expansion correction&quot; needed...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VEZjt.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VEZjt.png&quot; alt=&quot;light from fast receding objects will never reach us&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't be convinced by the answer I found &lt;a href=&quot;https://astronomy.stackexchange.com/a/14338/11264&quot;&gt;here&lt;/a&gt;, saying nothing disappears out of horizon. I don't see how it complies with &lt;a href=&quot;https://astronomy.stackexchange.com/a/6035/11264&quot;&gt;that answer&lt;/a&gt; were event horizon approaches us.&lt;/p&gt;&#xA;" OwnerUserId="11264" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2017-01-31T22:40:54.040" Title="Will we start seeing galaxies disappear due to Universe expansion?" Tags="&lt;universe&gt;&lt;expansion&gt;&lt;fate-of-universe&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="6" ClosedDate="2017-10-17T16:35:07.857" ContentLicense="CC BY-SA 3.0" />
  <row Id="19910" PostTypeId="2" ParentId="19909" CreationDate="2017-01-31T16:55:51.953" Score="16" Body="&lt;p&gt;It is a common misconception that galaxies receding faster than light cannot be observed. There are two versions of this misconception:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Galaxies that are &lt;em&gt;now&lt;/em&gt; receding faster than light cannot be seen.&lt;/li&gt;&#xA;&lt;li&gt;If we observe a galaxy &lt;em&gt;today&lt;/em&gt;, it may recede faster than light &lt;em&gt;now&lt;/em&gt;, but when it emitted the light we see now, it did not.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Both are incorrect.&lt;/p&gt;&#xA;&lt;strong&gt;Intuitive explanation&lt;/strong&gt;&#xA;&lt;p&gt;This is most easily seen in a spacetime diagram. But before showing that, a somewhat intuitive way of understanding this apparent paradox is to consider the journey of the photon from the reference frame of the emitting galaxy: In this frame, the photon leaves the emitter with a velocity &lt;span class=&quot;math-container&quot;&gt;$v=c$&lt;/span&gt; (as it should), but since its velocity is always &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; &lt;em&gt;locally&lt;/em&gt;, then in the frame of the emitter, its velocity increases without bounds. This is not a proof, but may help to show that &amp;quot;light may travel faster the the speed of light&amp;quot;.&lt;/p&gt;&#xA;&lt;strong&gt;Spacetime diagram&lt;/strong&gt;&#xA;&lt;p&gt;Now to the spacetime diagram. My favorite version is from the classic paper by &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0310808&quot; rel=&quot;noreferrer&quot;&gt;Davis &amp;amp; Lineweaver (2004)&lt;/a&gt;, but I actually like better this version from &lt;a href=&quot;https://physics.stackexchange.com/a/63780/70207&quot;&gt;Pulsar's excellent (!) post&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SR9EP.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SR9EP.png&quot; alt=&quot;spacetime&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;strong&gt;Real example of a galaxy that has always been, and always will be, receding with superluminal velocity&lt;/strong&gt;&#xA;&lt;p&gt;Such a diagram shows age (or, equivalently, size) of the Universe, as a function of the corresponding comoving distance&lt;span class=&quot;math-container&quot;&gt;$^1$&lt;/span&gt; of events. A lot of things are going on in this diagram, and I'll refer to Pulsar's answer for an interesting discussion of all this. For our purposes, consider the only thing the I have added to the figure: the world line — i.e. the &amp;quot;path&amp;quot; taken through spacetime of some object — of the galaxy &lt;a href=&quot;https://en.wikipedia.org/wiki/GN-z11&quot; rel=&quot;noreferrer&quot;&gt;GN-z11&lt;/a&gt;, seen as a vertical dashed line at a comoving distance &lt;span class=&quot;math-container&quot;&gt;$d_C \simeq 32.2\,\mathrm{Gly}$&lt;/span&gt;. It is vertical, because it doesn't move in comoving coordinates. We observe it at a redshift of &lt;span class=&quot;math-container&quot;&gt;$z=11.09$&lt;/span&gt;, which means that the light we see (moving at 45º toward &lt;em&gt;our&lt;/em&gt; worldline which is at &lt;span class=&quot;math-container&quot;&gt;$d_C=0$&lt;/span&gt;) was emitted when the Universe was roughly 400 Myr old. At that time, GN-z11 was receding at a velocity &lt;span class=&quot;math-container&quot;&gt;$v\simeq4c$&lt;/span&gt;!&lt;/p&gt;&#xA;&lt;strong&gt;Hubble sphere and event horizon&lt;/strong&gt;&#xA;&lt;p&gt;The innermost green bubble (labeled &amp;quot;&lt;span class=&quot;math-container&quot;&gt;$v_\mathrm{rec}=c$&lt;/span&gt;&amp;quot;) is the &lt;em&gt;Hubble sphere&lt;/em&gt;, defined such that at any point in time, everything outside recedes faster than &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;. Meanwhile, the red line shows the &lt;em&gt;event horizon&lt;/em&gt;, which defines the boundary between everything that &lt;em&gt;can&lt;/em&gt; eventually be observed (light red region), and everything that cannot. A subset of this is out &lt;em&gt;past lightcone&lt;/em&gt; (orange region) which is stuff that we have already observed (or at least have had the chance to).&lt;/p&gt;&#xA;&lt;p&gt;Everything lying in between the Hubble sphere and the event horizon consists of events that we have been, or will eventually be, able to see, but which recede faster than the speed of light when it emits the light we see.&lt;/p&gt;&#xA;&lt;p&gt;As you can see the region becomes narrower with time&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt;, meaning that the accelerated expansion of the Universe makes it harder and harder for photons emitted far away to reach us. We will never be able to see galaxies that today are farther away than roughly 17 Gly. But galaxies that are nearer &lt;em&gt;will&lt;/em&gt; be seen (if we look), and a subset of these — namely the ones with &lt;span class=&quot;math-container&quot;&gt;$14\,\mathrm{Gly} \lesssim d_C \lesssim 17\,\mathrm{Gly}$&lt;/span&gt; — are currently receding faster than the speed of light.&lt;/p&gt;&#xA;&lt;strong&gt;The size of the observable Universe always increases&lt;/strong&gt;&#xA;&lt;p&gt;The observable Universe is defined as the region within which light has had the time to reach us. The &amp;quot;edge&amp;quot; of this is called the &lt;em&gt;particle horizon&lt;/em&gt;, and as seen from the blue line, it always increases its distance from us (in comoving coordinates, and thus even more so in physical coordinates). In comoving coodinates, it reach a maximum size of roughly 63 Gly. That means that more and more galaxies will enter our observable Universe, but as time goes they will do so at an ever-decreasing pace, asymptotically reaching a finite value (of 5 trillion galaxies, more or less). Once inside the particle horizon, a galaxy will never leave. It will, however, keep increasing its distance from us in physical coordinates, as well as its speed, and thus redshift without bounds and fade to black.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^1$&lt;/span&gt;&lt;sub&gt;Comoving coordinates are defined so that they &amp;quot;expand with the expansion of the Universe&amp;quot;. That means that in these coordinates, galaxies lie approximately still wrt. each other. To get the &lt;em&gt;physical&lt;/em&gt; coordinates — i.e. what you would measure if you froze the Universe and laid out measuring rods — multiply the comoving coordinates by the scale factor &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; seen on the right &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; axis. Since today &lt;span class=&quot;math-container&quot;&gt;$a\equiv1$&lt;/span&gt;, comoving and physical coordinates coincide today.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt;&lt;sub&gt;The fact that the diagram is in comoving coordinates makes it a bit hard to see, but this is still the case in physical coordinates.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-01-31T22:40:54.040" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="19911" PostTypeId="1" AcceptedAnswerId="19912" CreationDate="2017-01-31T22:16:38.577" Score="5" ViewCount="297" Body="&lt;p&gt;Imagine this scenario - I'm in a space station with one of those huge torus shaped constructions, that you spin to simulate gravity through centripetal force. But it's turned off right now, so I'm floating in the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gMvjN.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gMvjN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When happens when they turn it on and it starts spinning?&#xA;I can imagine what would happen if I was grabbing on to something. I'd be pulled around, and the angular velocity would drag me to the outer part of the torus. But if I'm just floating there, with no friction to the ground to catch me, do I stay put, with the thing just whizzing past me?&lt;/p&gt;&#xA;" OwnerUserId="15550" LastActivityDate="2017-02-01T19:27:31.157" Title="In zero gravity - I'm floating inside a torus as they spin it up, do I float do the &quot;ground&quot;?" Tags="&lt;gravity&gt;&lt;space-travel&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19912" PostTypeId="2" ParentId="19911" CreationDate="2017-01-31T23:06:19.157" Score="7" Body="&lt;p&gt;Yes...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the scenario you describe, you (and the air) will remain at rest, while the torus speeds up.  One hopes there are no radial partitions, or, if there are, one comes along to pick you up fairly early in the session.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the rotating torus will impart its velocity to the air in contact with all the irregularities in the wall/floor/ceiling.  This would then be transferred to the rest of the air.  Floating there, you would experience a very slight breeze blowing you in the direction of rotation of the station.  This in turn would cause you to slowly drift radially outward, until you hit the floor and rolled along until you were at rest with respect to the floor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if the station is not running up from zero rotation rate, you can experience some strange effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the station is big enough, you can travel with low energy.  Simply run &lt;em&gt;against&lt;/em&gt; the rotation of the station.  When you are running at the rotation speed, lift your feet, and drift in place until your destination comes along.  Put down you feet, start running, and slow to a walk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be careful, though.  If you face in the wrong direction and run at that same speed, your apparent weight will quadruple as you crash to the floor...&lt;/p&gt;&#xA;" OwnerUserId="420" LastEditorUserId="420" LastEditDate="2017-02-01T19:27:31.157" LastActivityDate="2017-02-01T19:27:31.157" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19916" PostTypeId="1" CreationDate="2017-02-01T22:05:26.397" Score="1" ViewCount="1121" Body="&lt;p&gt;Basically, I've been seeing these graphs that say: Cepheid Period-Luminosity Relation and the Period axis is supposed to be &quot;logarithmic&quot;;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this graph work? Since the luminosity of a star doesn't just increase, as we know, but it increases and decreases periodically, why does the magnitude keep raising?&#xA;How do I find the pulsating period of such a star through a graph like this?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/4mMpN.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4mMpN.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15377" LastEditorDisplayName="user1569" LastEditDate="2017-02-02T14:43:05.100" LastActivityDate="2017-02-02T14:43:05.100" Title="How to find a Cepheid's pulsation period using its graph?" Tags="&lt;distances&gt;&lt;luminosity&gt;&lt;cepheids&gt;&lt;absolute-magnitude&gt;&lt;variable-star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19917" PostTypeId="2" ParentId="19916" CreationDate="2017-02-01T22:44:07.953" Score="3" Body="&lt;p&gt;This graph is not the light curve of a cepheid. It is used to find the average brightness of the star, and from that, estimate its distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate at which cepheids pulse is related to their &lt;em&gt;average&lt;/em&gt; luminosity (and so their average absolute magnitude). This graph relates the period to their average absolute magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use the graph, observe a cepheid for long enough to establish a period (say 30 days, for example). Then read the graph (an absolute magnitude of just under -5.5 in the example)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You now know how intrinsically bright the star is on average. By looking at its average apparent brightness, you can calculate an estimate of its distance. See &lt;a href=&quot;https://astronomy.stackexchange.com/questions/17826/how-does-the-cepheid-variables-help-in-finding-distance?rq=1&quot;&gt;How does the cepheid variables help in finding distance&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2017-02-01T22:44:07.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19918" PostTypeId="1" CreationDate="2017-02-01T22:47:30.513" Score="0" ViewCount="395" Body="&lt;p&gt;All we know that a star's altitude at its upper culmination is 38 degrees and 52 arcminutes and that its altitude at its lower culmination is 12 degrees and 24 arcminutes (so it never really sets at that place).&#xA;How can we, through this information, find that place's latitude?&lt;/p&gt;&#xA;" OwnerUserId="15377" LastActivityDate="2017-02-09T18:15:13.493" Title="How to find a place's latitude when you know a star's minimum and maximum altitude?" Tags="&lt;star&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19919" PostTypeId="1" AcceptedAnswerId="19936" CreationDate="2017-02-02T01:27:38.390" Score="1" ViewCount="189" Body="&lt;p&gt;I saw a very bright star in the night sky on Monday (Jan 30, 2017). The star was in the Western sky and was directly above the crescent moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My location is Toronto, Ontario. Canada. It was about 8pm at night, but by 10pm the star had disappeared over the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure this was a planet as it shined brighter then any of the other stars, but I couldn't guess which one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you identify this planet for me?&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2017-02-03T22:11:47.413" Title="Can you identify this planet seen in the night sky?" Tags="&lt;star&gt;&lt;planet&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19922" PostTypeId="2" ParentId="10354" CreationDate="2017-02-02T13:57:37.093" Score="1" Body="&lt;p&gt;Both spectra you show clearly have real features. There are perturbations in the spectra at certain wavelengths that are far larger than the rms noise that can be assessed empirically (by eye, or calculation) from looking at other parts of the spectrum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would form that basis of an empirical test for the reality of spectral features. On the other hand, if your spectrum was full of (small) real features, then you would have to rely on a theoretical estimate of what the rms noise in your spectrum &lt;em&gt;should&lt;/em&gt; be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, your questions as posed cannot really be answered in any detail. If you have a spectrum that is consistent with being featureless, there will be some amplitude of real feature below which it would be undetectable. This amplitude upper limit will increase as the signal-to-noise decreases - i.e. the better your data, the smaller the features you can rule out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, you can never be sure that you have a &quot;featureless continuum&quot;, only that any features are smaller than some threshold, that must be established by calculation or simulation.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-02-02T13:57:37.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19923" PostTypeId="1" CreationDate="2017-02-02T20:07:58.543" Score="1" ViewCount="734" Body="&lt;p&gt;From what I can see in a physics textbook page (on redshift), it seems to imply that all planets/galaxies emit red, rather than blue waves (for red-shift).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wouldn't this mean that everything is moving &lt;em&gt;away from us&lt;/em&gt; and that we're the centre of the Big Bang? (See quote from the book below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a similar note, does everything move at the same speed away from the centre of the Universe? - If not, then wouldn't that mean that we'd see a few planets/galaxies that emit blue, as they are traveling with a greater velocity from the centre than we are (with it being closer to the centre) and that they are 'catching up' with us?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not a &lt;em&gt;big&lt;/em&gt; deal and I don't &lt;strong&gt;need&lt;/strong&gt; to know it, but I would like to know, if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;~StratusGold&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quoting the textbook: &quot;When we look at light from distant galaxies, we find that the wavelength has increased. The wavelengths are longer than they should be - they're shifted towards the red end of the spectrum. This is called red-shift. &lt;em&gt;(...omitted...)&lt;/em&gt; Measurements of the red-shift indicate that these galaxies are moving away from us (receding) very quickly - and it's the same result whichever direction you look in&quot; - CGP GCSE AQA 9-1 Physics Revision Guide, Topic 8 'Space Physics', Pg 102&lt;/p&gt;&#xA;" OwnerDisplayName="user15682" LastEditorDisplayName="user15682" LastEditDate="2017-09-01T22:51:35.280" LastActivityDate="2017-09-01T22:51:35.280" Title="Are all planets/galaxies moving away from *us*?" Tags="&lt;redshift&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2017-02-03T13:36:59.763" ContentLicense="CC BY-SA 3.0" />
  <row Id="19924" PostTypeId="2" ParentId="19923" CreationDate="2017-02-02T23:17:50.000" Score="3" Body="&lt;p&gt;No, not all of them. Only the ones very far away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you walk your dog, the dog is tied to you via the leash. It does not recede into the distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets in the Solar System are all tied up to the Sun via gravity. They don't recede into the distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All stars in our galaxy are bound together in a big clump (the galaxy) and so they stay together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, all galaxies in our neighborhood are also bound by gravity, so they stay together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond that range, things are not bound by gravity, so they keep running away from each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity is like a leash. If things are close enough, they stay together.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that the universe does not have a center. It's infinite (as far as we can tell), and infinite things don't have centers. Rather, if you take any two distant galaxies, they all recede from each other. But the expansion does not go back in time to a common center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a comparison. Imagine you're a bi-dimensional creature; instead of having 3 dimensions, you only have 2. Like a stick figure drawn on a paper. You only exist in the 2 dimensions of the paper, and can't even imagine the 3rd dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine a balloon. You draw a bunch of stick figures on it. Now blow air into the balloon. All stick figures recede from each other, but for them there's no center for that expansion - because all they know is the surface of the balloon. They don't know anything that's not on the surface. So they all see each other running away, but there doesn't seem to be a center for that expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's just an analogy, because the universe is not a balloon, but it will get you started.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Finally, red-shifted galaxies don't emit only red light. They emit the usual radiation like everything else. But because us and them are running away from each other, radiation from one side reaches the other side a little bit stretched out. For the visible part of the spectrum, it's like taking the spectrum and shifting it towards the red part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's all there is to it. It's still the usual radiation, but all wavelengths are a little bit blown up.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2017-02-02T23:17:50.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19926" PostTypeId="1" CreationDate="2017-02-03T04:33:53.300" Score="2" ViewCount="3058" Body="&lt;p&gt;My question is simple:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the gravitational relationship between Earth and the moon, how wide is Earth's roche limit?  How close does the moon have to orbit before getting disintegrated by Earth's gravitational pull, and how far does it have to orbit before being free from Earth's orbit entirely?&lt;/p&gt;&#xA;" OwnerUserId="10085" LastActivityDate="2017-02-03T20:38:10.047" Title="How Wide is Earth's Roche Limit?" Tags="&lt;orbit&gt;&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19927" PostTypeId="1" AcceptedAnswerId="19928" CreationDate="2017-02-03T10:21:58.090" Score="13" ViewCount="2466" Body="&lt;p&gt;I know that some of them are pulsars and pulsars spin very fast, but do all of the neutron stars spin? I would think that they would because of conservation of momentum, but I'm really just not sure.&lt;/p&gt;&#xA;" OwnerUserId="6783" LastActivityDate="2017-02-03T16:31:19.150" Title="Do all neutron stars spin?" Tags="&lt;neutron-star&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19928" PostTypeId="2" ParentId="19927" CreationDate="2017-02-03T10:45:35.830" Score="18" Body="&lt;p&gt;I think it is absolutely safe to say that all neutron stars spin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conservation of angular momentum ensures that as they collapse from a massive stellar core the size (roughly) of the Earth, to something with a 10km radius, their angular velocity increases roughly as the square of the decrease in their radius (i.e. a factor of $\sim 4\times 10^5$. Thus, even if the stellar core had the slightest spin to begin with, then the neutron star will be spinning very fast indeed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Young pulsars appears to be born with rotation periods that vary from about 0.01 s to perhaps a second or so. They then lose angular momentum as they get older.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty in answering your question without any equivocation is that once neutron stars spin down to periods of longer than 1-10 seconds or so (depending on the pulsar magnetic field strength - see picture below from &lt;a href=&quot;http://www.aanda.org/articles/aa/abs/2011/02/aa15190-10/aa15190-10.html&quot;&gt;Wang et al. 2011&lt;/a&gt;) then the pulsar mechanism switches off and the neutron star, to all intents and purposes, becomes invisible (the so-called &quot;pulsar deathline&quot;, &lt;a href=&quot;http://zmtt.bao.ac.cn/xian/proceedings/ZhangB/fin/ZhangB.pdf&quot;&gt;Zhang 2003&lt;/a&gt;) .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0KHef.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0KHef.jpg&quot; alt=&quot;Pulsar period distribution&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is expected that the spin down will continue, even after the pulsar has switched off, but the spin down rate itself depends on the rotation speed, so the neutron star will never come to a halt, even after tens of billions of years.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-02-03T16:31:19.150" LastActivityDate="2017-02-03T16:31:19.150" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19929" PostTypeId="2" ParentId="19927" CreationDate="2017-02-03T10:53:56.390" Score="7" Body="&lt;p&gt;I'd say every stellar object spin. It comes from the accretion process that cannot be purely radial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you get the unlikely chance that a stellar object receive the right forces to stop spin in the local place you look it from, next time a small force is applied on it, it will start spinning again (even very slowly).&lt;/p&gt;&#xA;" OwnerUserId="11264" LastActivityDate="2017-02-03T10:53:56.390" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19930" PostTypeId="1" AcceptedAnswerId="19932" CreationDate="2017-02-03T14:27:46.570" Score="2" ViewCount="541" Body="&lt;p&gt;For the last week or so, I have been teaching myself orbital mechanics within the context of &lt;a href=&quot;http://www.braeunig.us/space/index.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Braeunig's Rocket and Space Technology&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I noticed a symbol, $\phi '$, and was wondering what context that was used in? I think it is an angle measurement, but I know that it is almost equal to the semi-major axis on earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the box labeled &lt;strong&gt;Geodetic Latitude, Geocentric Latitude, and Declination&lt;/strong&gt;, found about half way down after selecting &lt;strong&gt;Orbital Mechanics&lt;/strong&gt; &lt;a href=&quot;http://www.braeunig.us/space/index.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OEYWY.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OEYWY.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15687" LastEditorUserId="7982" LastEditDate="2017-02-04T13:36:43.030" LastActivityDate="2017-02-04T13:36:43.030" Title="What is $\phi '$ in orbital mechanics?" Tags="&lt;orbital-mechanics&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19932" PostTypeId="2" ParentId="19930" CreationDate="2017-02-03T14:46:39.080" Score="5" Body="&lt;p&gt;This appears to be the &lt;a href=&quot;https://en.wikipedia.org/wiki/Latitude#Geocentric_latitude&quot; rel=&quot;nofollow noreferrer&quot;&gt;geocentric latitude&lt;/a&gt; ($\psi$), which is the angle between the equatorial plane and the point on the surface of the ellipse. It can be calculated from the geodetic latitude (also known as the geographic latitude) ($\phi$) and the eccentricity of the ellipse ($e$):&#xA;$$\psi(\phi)=\tan^{-1}\left(\left(1-e^2\right)\tan\phi\right)$$&#xA;If $e=0$, then the two lattitudes are the same, because the focus is at the center of the ellipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/0/06/Two-types-of-latitude.png&quot; alt=&quot;&quot;&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;Image courtesy of Wikipedia user Wmc824 under &lt;a href=&quot;https://creativecommons.org/licenses/by/3.0/deed.en&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Creative Commons Attribution 3.0 Unported license&lt;/a&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might make more sense to use this latitude to describe the position of an orbiting object, but the vector from the focus to the object's position $(r,\theta)$ is perpendicular to the tangential component of the object's velocity vector, meaning that the geodetic latitude may be better.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2017-02-03T15:00:28.830" LastActivityDate="2017-02-03T15:00:28.830" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19933" PostTypeId="2" ParentId="19926" CreationDate="2017-02-03T18:30:14.730" Score="5" Body="&lt;p&gt;A little reading and effort is recommended and the Wikipedia page is straight forward enough, if you're not completely math-phobic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you want an answer, Rob Jeffries 9,500 km is right, however, that's center point to center point.   Surface to surface, that would only be less than 2,000 km, which would be crazy close.   That close the Moon would take up about (guessing) close to 50% of the sky, maybe 70 degrees or so, when directly overhead.    But that 2000 km number doesn't take into account tidal stretching or rotation, as the Moon would almost certainly be tidally locked and rotating in sync with it's orbit like it does now and the Earth would be similarly tidally locked and have a tidal and gravitational bulge in the direction of the Moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking into account rotation, that pushes it out to the cube root of 3, not 2 or about 14% further, but that's center to center.  The surface to surface distance would be increased to about 3,000 KM.   Still insanely close.   Any orbital eccentricity would increase instability though, so for a safe distance, Earth to Moon where the Moon wouldn't break apart at all, but it would still stretch measurably, I wouldn't want it any closer than about 5,000 or 6,000 km from the Earth's surface.   That's a ballpark guess with tidal forces, tidal locking, solar perturbations and orbital eccentricity all in mind and even there, some loose moon dust and rocks might regularly blow off and shower onto the Earth.   6,000 KM (surface to surface) is less than twice the Moon's diameter, so it would still be enormous in the sky, but I think that's about the distance where the Moon would hold together in orbit.   The math says even closer, but that doesn't take into effect various issues like eccentricity, rigidity, etc.   It gets rather hard to calculate with rock solid accuracy taking into account all the details.  But, point is, the Moon would need to be stupid/gonzo/cray close before it begins to break apart.   Some 50-60 times closer (surface to surface) than it is now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now as far as escape, the math of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;noreferrer&quot;&gt;Hill Sphere&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/Sphere_of_influence_(astrodynamics)&quot; rel=&quot;noreferrer&quot;&gt;Sphere of Influence&lt;/a&gt;  is pretty straight forward, but the long-term stable zone is about 1/2 to 1/3rd of the Hill Sphere calculation, some 500,000 to 750,000 km.   Somewhere in there or perhaps just beyond that range the Moon probably becomes unstable and escapes the Earth's orbit becoming a very large and potentially very dangerous near earth object.   That doesn't mean that the Moon would just fly away past 750,000 km.  It would remain in orbit for multiple orbits, hundreds, maybe thousands even past that distance, but past the stable region it would probably destabilize over time and escape eventually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon is perhaps as much as 70% of the way towards where it could eventually escape the Earth, but that final 30% is a LOT.  The moon isn't going anywhere anytime soon.  It's still comfortably inside the long-term stable region and should remain orbiting the Earth for billions of years.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-02-03T20:38:10.047" LastActivityDate="2017-02-03T20:38:10.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19934" PostTypeId="1" AcceptedAnswerId="19935" CreationDate="2017-02-03T18:39:43.867" Score="5" ViewCount="635" Body="&lt;p&gt;First of all, I'm from Syria. In Syria we have 70% of the night without electricity. So I love hanging out at this period of time in the night, looking up just as Prof. Stephen Hawking said to the sky, observing the beauty of the night sky. Tonight, when I was looking arbitrarily I saw a star. Suddenly its brightness increased gradually, and then this star released two big sparkles. They were more like distorted yellow or golden rings formed or more like waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how to express this phenomenon! Then those two sparkles disappeared, and the brightness of this star decreased very very fast until it just disappeared. It was the first time in my life I saw something like this! I just don't know what to say! This phenomenon lasted for a couple seconds.&#xA;I was shocked as much as you can imagine. I was unable to speak, totally impressed! I just cannot express my feeling! Maybe this is not a supernova! But what is that? Why did this star do that? &lt;/p&gt;&#xA;" OwnerUserId="15693" LastEditorUserId="2153" LastEditDate="2017-02-03T19:27:33.957" LastActivityDate="2018-09-05T09:04:21.417" Title="Have I just observed a supernova?" Tags="&lt;amateur-observing&gt;&lt;supernova&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19935" PostTypeId="2" ParentId="19934" CreationDate="2017-02-03T19:25:13.127" Score="10" Body="&lt;p&gt;The odds of a supernova being visible to the naked eye is slim. &lt;a href=&quot;https://arxiv.org/abs/1306.0559&quot;&gt;Adams et al. (2013)&lt;/a&gt; estimated that there is a 20% chance that we will be able to see such a supernova within the next 50 years. That should give you an idea of how unlikely this is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vast majority of supernovae that have been discovered lately have magnitudes ranging from +20 to +16 (check out &lt;a href=&quot;http://www.rochesterastronomy.org/snimages/sndate.html&quot;&gt;this page&lt;/a&gt;, updated daily). The naked eye can see objects of magnitude +6 or less, meaning that a supernova of magnitude 16 is $10^{10}$ times fainter. Powerful telescopes are needed to observe these events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supernovae also aren't that quick. They fade over days or weeks or months or even years. You would not see a bright, quick flash of light. I don't want to say that it's impossible . . . but it's impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this &lt;em&gt;was&lt;/em&gt; a supernova, it's even more unlikely that you could have seen such detail. When I say that a supernova is visible to the naked eye, that just means that you can see the light from it. It doesn't mean that the structure of any gas or dust being expelled can be observed. To make out that level of detail, it would need to be quite close to Earth - and so it would definitely have been observed by astronomers and amateur stargazers in quite a few places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So no, I don't think this was a supernova. I have no idea what you saw, but it was not what you think it is.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2017-02-04T23:52:48.907" LastActivityDate="2017-02-04T23:52:48.907" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="19936" PostTypeId="2" ParentId="19919" CreationDate="2017-02-03T22:11:47.413" Score="3" Body="&lt;p&gt;&lt;sub&gt;This Community Wiki answer was created to preserve knowledge while other answer(s) are under review.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://astronomy.stackexchange.com/users/11822/dr-chuck&quot;&gt;Dr Chuck&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It was Venus. It should have been 10 - 15 degrees above the moon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://astronomy.stackexchange.com/users/2031/wayfaring-stranger&quot;&gt;Wayfaring Stranger&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Venus is whiter than Jupiter, Saturn, or Mars, which have yellowish, yellow, and reddish tints instead. It's also usually much brighter than the other planets, including Mercury. Mercury always hangs closer to the sun than did Venus in early Feb 2017.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2017-02-03T22:11:47.413" CommentCount="0" CommunityOwnedDate="2017-02-03T22:11:47.413" ContentLicense="CC BY-SA 3.0" />
  <row Id="19937" PostTypeId="2" ParentId="7778" CreationDate="2017-02-03T22:25:56.613" Score="1" Body="&lt;p&gt;The outer edge of a neutron star contains very tightly packed neutrons, protons and electrons.   I'd try shooting the Neutron star with positrons to collide with the electrons creating heat and growing positive charge.   The combination of heat and positive charge and localized matter-anti-matter explosion (just might) gradually shed some mass, a few protons here and there, reaching escape velocity.   It would take a long time, but it just might work.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But remember to stand back when the star gets sufficiently lighter and reaches the critical inverse of the Tolman–Oppenheimer–Volkoff limit when it could and probably will rapidly de-Neutron and rapidly expand.      I think this might be the best way to do it (though I also like the spin it very rapidly answer).  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-02-03T22:25:56.613" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19939" PostTypeId="1" CreationDate="2017-02-04T10:23:06.853" Score="3" ViewCount="275" Body="&lt;p&gt;&lt;strong&gt;Background, Europa:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an earlier question &lt;a href=&quot;https://astronomy.stackexchange.com/q/18496/7982&quot;&gt;How can “Geysers” on Europa reach heights of 100km?&lt;/a&gt; I brought up the use of the word &quot;geyser&quot; in association with the reported confirming observations of water vapor plumes on Europa. Europa size and surface gravity are nearly as large as those of Earth's moon. With an escape velocity of about &lt;a href=&quot;https://en.wikipedia.org/wiki/Europa_(moon)&quot; rel=&quot;nofollow noreferrer&quot;&gt;2,025 m/s&lt;/a&gt; it was hard for me to understand how an &lt;em&gt;actual geyser&lt;/em&gt; could exist within the ice, a geyser so powerful that it could be shooting liquid water from the surface at 2 kilometers per second. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of the problem here is that I still can't be absolutely sure there is a clear definition of what is or isn't an &lt;em&gt;actual geyser&lt;/em&gt; on solar system bodies other than Earth. On Earth, while a geyser is usually powered by geothermal conversion of water into steam, it pushes additional liquid water up to and above the surface as liquid. I believe that other sources of pressure may still result in proper Earth geysers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most scientific references I found referred to the vapor as plumes or jets, the word geyser was used in more popular and general interested articles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems one likely explanation is a venting of water vapor (gas) from a particularly shaped vent can accelerate the molecules to an average velocity that is several times the speed of sound by converting random thermal motion to directed vertical motion, as described in a &lt;a href=&quot;https://repositories.lib.utexas.edu/handle/2152/34041&quot; rel=&quot;nofollow noreferrer&quot;&gt;U.T. Austin thesis by Jared James Berg&lt;/a&gt;. Simulations there showed that this jet phenomenon can result in some water vapor reaching a substantial fraction of escape velocity, and rise to 100 to 200km altitudes. The model also addresses the formation of solids as ice crystals within the plume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;About Enceladus:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have just come across the paper &lt;a href=&quot;http://ciclops.org/media/sp/2014/7927_18911_0.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;How the Geysers, Tidal Stresses, and Thermal Emission Across the South Polar Terrain of Enceladus are Related&lt;/a&gt;, (Porco, DiNino and Nimmo (2014), Ast. J. 148 (3), 45), an in depth study of sources of water vapor from Enceladus. I saw that both geysers and jets were mentioned in the abstract, but at the end of the second paragraph in the introduction I ran into the very interesting sentence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(Throughout this paper, we use the terms “jets” and “geysers” interchangeably.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The escape velocity on Enceladus is only about &lt;a href=&quot;https://en.wikipedia.org/wiki/Enceladus&quot; rel=&quot;nofollow noreferrer&quot;&gt;240 m/s&lt;/a&gt;, &lt;em&gt;nearly an order of magnitude lower&lt;/em&gt; than that on Europa, which means much less sub-surface pressure would be required to push liquid water high above the surface into near space. So I can understand that the term geyser might be more likely to be appropriate for these plumes than it would be for the plumes on higher surface gravity Europa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would know why the words &quot;geysers&quot; and &quot;jets&quot;/&quot;plumes&quot; would be used interchangeably. Is this because it hasn't been confirmed yet what they are, and the authors don't wish to show favoritism by choosing one or the other, or are they synonymous and can always be used interchangeably, or si there some other reason?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZbCYQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZbCYQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Lxqt9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lxqt9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above x2:&lt;/strong&gt; Figures 3a and 3b from &lt;a href=&quot;http://ciclops.org/media/sp/2014/7927_18911_0.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Porco, DiNino and Nimmo (2014)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-02-10T17:40:20.943" LastActivityDate="2018-02-10T17:40:20.943" Title="Enceladus; why use the words &quot;geysers&quot;, &quot;jets&quot;, and &quot;plumes&quot; interchangeably?" Tags="&lt;water&gt;&lt;europa&gt;&lt;enceladus&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19940" PostTypeId="1" CreationDate="2017-02-04T16:51:27.743" Score="4" ViewCount="391" Body="&lt;p&gt;Within a few billion years our Sun would become a red giant and destroying Mercury and Venus and perhaps even the Earth. But nevertheless our solar system still have some planets like Jupiter and Neptune orbitting around the sun, because his mass wouldn't be very different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After another while our Sun would turn into a white dwarf. Probably his mass changed a bit, but would it be enough to lose for example Neptune?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, is there a prediction at what time, if ever, our 'sun' will lose (some of) his planets for the first time?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastActivityDate="2017-02-04T22:28:56.627" Title="Is there a prediction when our solar system would fall apart?" Tags="&lt;solar-system&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19941" PostTypeId="2" ParentId="19940" CreationDate="2017-02-04T22:28:56.627" Score="1" Body="&lt;p&gt;The process whereby the Sun will become a red giant and then turn into a white dwarf is neither short nor simple. It's absolutely not the case that it will expand, blow away a chunk of its mass, contract again, and that will be it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sun#Life_phases&quot; rel=&quot;nofollow noreferrer&quot;&gt;According to Wikipedia&lt;/a&gt; (emphasis mine):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For the Sun, four thermal pulses are predicted before it completely loses its outer envelope and starts to make a planetary nebula. By the end of that phase – lasting approximately 500,000 years – the Sun will only have &lt;strong&gt;about half of its current mass&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's a good lengthy coverage of the last days of the Sun &lt;a href=&quot;https://www.quora.com/Will-the-Sun-have-enough-gravity-to-keep-the-planets-in-the-orbit-after-it-becomes-a-white-dwarf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. It cites a study and then concludes that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The bottom line is that the orbits are unpredictable and where the&#xA;  planets find themselves once the star dies is dependent on when, how&#xA;  much and how quickly the sun shed half its mass. This makes a major&#xA;  difference in how the solar system will look in the end and how easily&#xA;  the planets may be ejected from their orbits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the simple answer to your question is: as the Sun loses mass, the orbits of the outer planets will become larger. When they become wide enough, they will escape into interstellar space. If the hypothesized &lt;a href=&quot;https://en.wikipedia.org/wiki/Planet_Nine&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planet Nine&lt;/a&gt; exists, it could help kick out Uranus and Neptune. Jupiter (and Mars, I guess) may or may not remain. Earth will in all likelihood be dragged into the Sun during the red giant phase as the Sun's outer layers expand to fill its orbit.&lt;/p&gt;&#xA;" OwnerUserId="15398" LastActivityDate="2017-02-04T22:28:56.627" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19942" PostTypeId="2" ParentId="19939" CreationDate="2017-02-04T23:21:55.433" Score="3" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Question:&lt;/p&gt;&#xA;&lt;p&gt;I would know why the words &amp;quot;geysers&amp;quot; and &amp;quot;jets&amp;quot; would be used interchangeably. Is this because it hasn't been confirmed yet what they are, and the authors don't wish to show favoritism by choosing one or the other, or are they synonymous and can always be used interchangeably, or is there some other reason ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A jet is a generic term to describe the ejection of material.  You'll see it used all over the place in physics (and astronomy).  Jet's can be any form of matter ( solid, liquid, gas, plasma ).&lt;/p&gt;&#xA;&lt;p&gt;A geyser is a specific type of geological process and, as I understand it, means (at the least) a liquid is being vented (typically water).  The exact process is, of course, not known for certain for non-terrestrial bodies.  I don't think using the term &amp;quot;geyser&amp;quot; is a statement of certainty, but would be used to indicate the general type of event.&lt;/p&gt;&#xA;&lt;p&gt;I think you're more likely to use &amp;quot;jets&amp;quot; if it's not properly understood and &amp;quot;geysers&amp;quot; when you're either proposing a mechanism that's geyser-like or it's generally accepted that they're geyser-like already.  But I don't think that's written in stone.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-02-04T23:21:55.433" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19943" PostTypeId="1" CreationDate="2017-02-05T02:16:25.967" Score="5" ViewCount="216" Body="&lt;p&gt;Is there an established distinction between a geyser and a cryovolcano in the context of cold solar-system bodies such as the moons of Jupiter and Saturn, or Kuiper belt objects (e.g. Pluto)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both can lead to spreading of new water across the surface, and both can result in plumes of water vapor above the surface. Do they have distinctly different mechanisms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it unclear or in a state of flux, a bit like the diagram in @zephyr's &lt;a href=&quot;https://astronomy.stackexchange.com/a/19973/7982&quot;&gt;answer&lt;/a&gt; to &lt;em&gt;Do astronomers generally agree that the distinction between comets and astroids is not so clear?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-02-05T01:13:01.720" LastActivityDate="2019-02-05T21:20:32.340" Title="Is there an established distinction between a geyser and a cryovolcano in solar system bodies?" Tags="&lt;kuiper-belt&gt;&lt;geysers&gt;&lt;planetary-science&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19944" PostTypeId="1" AcceptedAnswerId="19949" CreationDate="2017-02-05T02:23:49.817" Score="0" ViewCount="191" Body="&lt;p&gt;What is the probability that we are the first ever intelligent life developed all over the universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything starts somewhere, Is there any probability that we are the beginners to be part of intelligent life all over universe?!&lt;/p&gt;&#xA;" OwnerUserId="15717" LastEditorUserId="2531" LastEditDate="2017-02-05T18:48:35.983" LastActivityDate="2017-02-05T18:48:35.983" Title="What is the probability that we are the first ever intelligent life in the whole universe?" Tags="&lt;life&gt;&lt;early-universe&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2017-02-06T15:54:08.960" ContentLicense="CC BY-SA 3.0" />
  <row Id="19945" PostTypeId="1" AcceptedAnswerId="19951" CreationDate="2017-02-05T03:14:12.170" Score="0" ViewCount="1756" Body="&lt;p&gt;Is the Sun really burning and getting smaller? If so, because there is no air in space, what can be keeping it burning and what materials get burnt? If no air is required, what can speed up or can stop the burning process?&lt;/p&gt;&#xA;" OwnerUserId="15717" LastEditorUserId="9301" LastEditDate="2017-02-06T00:18:31.597" LastActivityDate="2017-05-05T04:26:06.327" Title="Is the Sun really burning and getting smaller?" Tags="&lt;star&gt;&lt;the-sun&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19946" PostTypeId="2" ParentId="19945" CreationDate="2017-02-05T03:41:06.850" Score="6" Body="&lt;p&gt;The Sun is not &lt;em&gt;burning&lt;/em&gt;, but undergoing &lt;a href=&quot;http://earthguide.ucsd.edu/virtualmuseum/ita/07_2.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;nuclear fusion&lt;/a&gt; at the core of the star. There, under great pressure, and high temperature, Hydrogen is converted into Helium and energy (light).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you recall Einstein's equation $E = mc^2$, energy is equivalent to mass, meaning a small amount of mass can be converted into a large amount of energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual process of converting the Hydrogen to Helium &lt;a href=&quot;http://www.emc2-explained.info/Emc2/Fusion.htm#.WJad6hIrLXE&quot; rel=&quot;nofollow noreferrer&quot;&gt;is complicated&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6870" LastEditorUserId="-1" LastEditDate="2017-02-05T23:40:16.367" LastActivityDate="2017-02-05T23:40:16.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19947" PostTypeId="1" CreationDate="2017-02-05T06:24:17.580" Score="9" ViewCount="866" Body="&lt;p&gt;Uranus and Pluto have their axes of rotation almost 90 degrees towards the ecliptic. But why do their moons tilt the same way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it mean that their tilts were caused by passing near some external mass that affected the planet and its moons equally, rather than some planet specific process such as a comet impact or tectonics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars' rotational axis tilts over time, but its moons today are spot on over its temporary equator. Earth's moon is aligned with the ecliptic and doesn't seem to care about our planetary tilt, although (or because of?) oceanic masses tugging on its tides. Exoplanets have been found that don't orbit in alignment with their star's rotation, and why would they do so billions of years after formation?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="2618" LastEditDate="2017-03-07T00:40:14.990" LastActivityDate="2017-03-07T00:40:14.990" Title="Why do Uranus' and Pluto's moons orbit the equator? What makes a moons' inclination tilt with the rotational axis of their planet?" Tags="&lt;orbit&gt;&lt;rotation&gt;&lt;natural-satellites&gt;&lt;inclination&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19948" PostTypeId="2" ParentId="19945" CreationDate="2017-02-05T06:50:13.660" Score="14" Body="&lt;p&gt;I won't bother answering how the Sun &quot;burns&quot; its fuel — that is already answered in &lt;a href=&quot;https://astronomy.stackexchange.com/questions/2302/how-can-the-sun-burn-without-oxygen&quot;&gt;this&lt;/a&gt; question. Your second question is whether the Sun is getting smaller; the answer depends on what you mean by &quot;smaller&quot;. We'll analyze this in two ways:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Radius&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we define the radius of the Sun by distance between the core and photosphere, then the Sun is not getting smaller. In fact, it is &lt;strong&gt;growing over time&lt;/strong&gt;, as demonstrated in this graph:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/po7NV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/po7NV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Image courtesy of &lt;a href=&quot;https://arxiv.org/abs/0911.4872&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ribas (2009)&lt;/a&gt;, colored by user RJHall on Wikipedia under the &lt;a href=&quot;https://creativecommons.org/licenses/by-sa/3.0/deed.en&quot; rel=&quot;nofollow noreferrer&quot;&gt;Creative Commons Attribution-Share Alike 3.0 Unported&lt;/a&gt; license&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The height of the photosphere is increasing ever so slightly, even in the Sun's main sequence stage; specifically, Table 3 of &lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1993ApJ...418..457S&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sackmann et al. (1993)&lt;/a&gt; states that at the Sun's zero-age main sequence, it was $0.9R_\odot$ in radius, and at its terminal age main sequence it will be $1.6R_\odot$. That means that, over the course of about 10.9 billion years, the Sun's radius will have increased by $0.7R_\odot$ — about 70% its current radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we do the math $\left (0.7R_\odot \over 10.9 \ \text{Gyr} \right )$, we could find that the Sun's average increase in radius (from beginning to end of main sequence) is about 4.5 centimeters each year (1.77 inches). This calculation shouldn't be taken too seriously, since the growth rate is not linear, but it demonstrates how slowly the Sun has been growing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun will continue to grow at a very slow rate until it leaves its main sequence stage. Once it enters the red giant branch, it will grow many times its current size, to approximately 1 AU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mass&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun is losing mass for two reasons: the nuclear fusion in its core converts some mass into energy, and high energy protons and electrons are constantly ejected from the Sun's corona as solar winds. &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/340797&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wood et al. (2002)&lt;/a&gt; shows that the Sun's mass-loss rate $\dot{M}$ is (approximately) inversely proportional to the square of its age $t$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{M} \propto t^{-2.00 \pm 0.52}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As of now, $\dot{M} = 10^{-14}\frac{M_\odot}{\mathrm{yr}}$. After the Sun leaves the main sequence, it will lose its mass at a much faster rate. As stated in Sackermann et al., the rate at which it a post-main sequence star loses mass can be generalized with &lt;a href=&quot;http://adsabs.harvard.edu/abs/1975MSRSL...8..369R&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reimers' law&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{M} = -\eta (4 \times 10^{-13}) \frac{LR}{M}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\eta$ is a value depending on the star (derived to be ~0.6 for the Sun), $M$, $L$ and $R$ are in solar units, and $\dot{M}$ is in solar masses per year. This is simplified, but gives us an idea of how the Sun's phase affects its mass-loss rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the time the Sun enters the red giant branch, it will have only 72.5% its current mass, and its luminosity and radius will reach $2300L_\odot$ and $170R_\odot$, respectively. Plugging those values in, we find that the rate of mass loss will be $1.3 \times 10^{-7} \frac{M_\odot}{\mathrm{yr}}$. It will lose mass even faster in the asymptotic giant branch, when its mass will become ~54.1% of what it is now, meaning $\dot{M}$ will be $2.5 \times 10^{-7} \frac{M_\odot}{\mathrm{yr}}$.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2017-05-05T04:26:06.327" LastActivityDate="2017-05-05T04:26:06.327" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19949" PostTypeId="2" ParentId="19944" CreationDate="2017-02-05T10:25:14.157" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the probability of we are the first ever intelligent life developed all over the universe ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thinking on this varies over time and, in view of the subject, and sometimes because of entirely non-scientific thinking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that we think life (e.g. microorganisms) ought to be pretty common, and intelligent life fairly common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Hitchhiker%27s_Guide_to_the_Galaxy&quot; rel=&quot;nofollow noreferrer&quot;&gt;school of thought&lt;/a&gt; that, including dolphins, there's only one intelligent life form on Earth.  YMMV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a good discussion on this subject on the &lt;a href=&quot;https://www.youtube.com/watch?v=BAoEnV3Zt_Q&quot; rel=&quot;nofollow noreferrer&quot;&gt;PBS Space Time&lt;/a&gt; channel on YouTube.  They had some related discussions &lt;a href=&quot;https://www.youtube.com/watch?v=4H55wybU3rI&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://www.youtube.com/watch?v=Ttwl_zH_DZ8&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think overall there is a general view that we expect there to be other intelligent life and (relatively speaking) lots of it.  The questions of why we can't detect it or it doesn't seem interested in us if such life exists are dealt with to some extent in those PBS Space Time videos.  I'd describe the whole problem as an open one at this stage.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Everything starts somewhere, Is there any probability that we are the beginners to be part of intelligent life all over universe?!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is a probability we're the first, a probability we're the last, a probability we're the insert-any-number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the general scientific view at the moment is that it would be surprising if we were the first intelligent civilization in the galaxy, let alone the entire universe.  But that's not to say we might not be the only intelligent civilization, just to say that there's always a possibility.  From a scientific point of view there's no reason (at present) to suppose we're unique or the first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's not much to say beyond speculation and back-of-the-envelope calculations at present.  And those calculations change wildly depending on who you ask.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-02-05T10:25:14.157" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19950" PostTypeId="2" ParentId="19947" CreationDate="2017-02-05T14:06:30.727" Score="5" Body="&lt;p&gt;I have managed to find two hypotheses for Uranus's tilt that explain why its moons are also orbiting on a tilted plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;&lt;a href=&quot;http://io9.gizmodo.com/5847495/how-uranus-got-knocked-on-its-side&quot; rel=&quot;noreferrer&quot;&gt;multiple impacts&lt;/a&gt;&lt;/strong&gt; hypotheses lies on the fact that, if Uranus had become tilted by the force of a single large impact (as was commonly believed), then the moons should have stayed on their original plane. A corrected version of this hypothesis said that, if the accretion disc from which Uranus was forming was still around, an impact would have destroyed it and recreated it on the same place as the new equator. The moons would then have accreted from the disc on that plane. This, however, would have produced moons with retrograde motion (which is not the case for most). So the latest idea is that Uranus suffered multiple impacts, which, according to the simulations, works to produce the planetary system as we see it today. (There's another problem with this hypothesis.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;&lt;a href=&quot;https://www.technologyreview.com/s/416515/collision-free-theory-explains-why-uranus-is-lying-on-its-side/&quot; rel=&quot;noreferrer&quot;&gt;collision-free theory&lt;/a&gt;&lt;/strong&gt; (or &lt;a href=&quot;https://arxiv.org/abs/0912.0181&quot; rel=&quot;noreferrer&quot;&gt;collissionless scenario&lt;/a&gt; as described in the paper) proposes that Uranus could have become tilted progressively as it formed without the need to postulate an impact (or series of impacts), if it had an additional satellite and started with a large inclination (which is allowed by current models); the extra moon could have been ejected later. The simulations worked for an initial inclination of more 17 degrees and a moon of 0.01 Uranian mass and at 50 Uranian radii.&lt;/p&gt;&#xA;" OwnerUserId="15398" LastActivityDate="2017-02-05T14:06:30.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19951" PostTypeId="2" ParentId="19945" CreationDate="2017-02-05T14:24:27.457" Score="2" Body="&lt;p&gt;The only other detail I can add to what has already been said above is about the actual fusion process inside he Sun - its engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4H -&gt; He + neutrinos + gamma-ray photons&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass of 4H = 6.693e-27 Kg&#xA;Mass of He = 6.645e-27 Kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass lost per single fusion reaction = 0.048e-27 Kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This generates 4.3e-12 J&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 1Kg of H is converted into He, a mass of 0.007 kg (0.7% of total initial mass) is lost in the process, generated 6.3e14 J. This is equivalent to 20e6 kg of coal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the Sun's luminosity is 3.9e26 J/s which when divided by the energy/kg defined above gives us the mass of H converted per second: 600e12 Kg. &lt;/p&gt;&#xA;" OwnerUserId="15568" LastActivityDate="2017-02-05T14:24:27.457" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19952" PostTypeId="2" ParentId="19947" CreationDate="2017-02-05T16:39:58.687" Score="3" Body="&lt;p&gt;Just as tidal forces can push a moon outwards and over time, circularize the orbit, the planet's equatorial bulge has a similar force that draws the Moon into an orbit over the equator.   This is true given sufficient time and enough rotation speed for the planet to have a sufficient equatorial bulge.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a formation moon, there should be some similarity in the angular momentum during formation naturally.   Similar to an impact moon like our own where the impact sets the planet rotating.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A captured moon can orbit in either direction with or against the planet's rotation at any angle to the planet's equatorial planet, at least, initially but the equatorial bulge has a tidal effect that over time, draws the moon over it.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth's tidal bulge, that moves the moon a few cm away every year is only a couple meters high over the oceans and a few inches over land.   Earth's &lt;a href=&quot;https://en.wikipedia.org/wiki/Equatorial_bulge&quot; rel=&quot;nofollow noreferrer&quot;&gt;equatorial bulge&lt;/a&gt; is 42.77 km.  Some 4 orders of magnitude more significant than the tidal bulge.  It's not hard to see how a bulge of that size, over many orbits, would influence the moon into an orbit over the equator.  The same is true for ring systems, which form over a planet's equator.    &lt;a href=&quot;http://curious.astro.cornell.edu/about-us/59-our-solar-system/planets-and-dwarf-planets/moon/240-why-do-most-moons-orbit-their-planets-at-the-equator-intermediate&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Ask an Astronomer&quot; says&lt;/a&gt; basically the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exceptions would be if the moon was recently captured, or if the planet rotated very slowly (Venus for example), or if the moon was close to the unstable region of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere#True_region_of_stability&quot; rel=&quot;nofollow noreferrer&quot;&gt;hill sphere&lt;/a&gt; where the object it orbits (planet) and the object that limits the planet's hill sphere (sun) both have an effect on the moon.  I suspect (but I don't know how to do the math), that our moon's orbit being 5 and change degrees off Earth's equator is due to the proximity to the sun and Moon being no too far from the unstable part of the Hill Sphere.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other moons you mention are much further inside their planet's Hill Spheres and much more governed by the planet's equatorial bulge.  see &lt;a href=&quot;http://www.windows2universe.org/our_solar_system/moons_table.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;table&lt;/a&gt; and chart below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XxJk6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XxJk6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-02-05T16:39:58.687" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19953" PostTypeId="2" ParentId="19944" CreationDate="2017-02-05T18:46:28.587" Score="3" Body="&lt;p&gt;We know that intelligent life exists on one planet. We do not fully understand the processes by which life (a) starts and (b) becomes intelligent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the probability of these things happening elsewhere ranges from 1 to as close-to-zero as needs to be to ensure we are the first and perhaps only intelligent life in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the answer to your question ranges anywhere from zero to one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more sensible, informed calculation could arise if we ever create life in the lab or if life with an independent genesis is found elsewhere (in the solar system or beyond).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-02-05T18:46:28.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19956" PostTypeId="1" CreationDate="2017-02-06T17:26:41.760" Score="-1" ViewCount="120" Body="&lt;p&gt;Until recently, it was thought that should the size of Jupiter exceed 86,881.4 miles in diameter or the mass 318 times greater than Earth's, it would become a dwarf star, not a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WASP-17b is twice as wide as Jupiter but only half as dense, and that's likely because of its gravitational proximity to its star, which is close enough to be categorized as a &quot;Hot Jupiter&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A really mind-boggling discovery is HD 106906b.  Its mass exceeds Jupiter's by eleven times.  Which raises the question:  &lt;strong&gt;How can we be sure that HD 106906b is a planet as opposed to a wandering dwarf star?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10085" LastActivityDate="2017-02-06T18:13:52.713" Title="What IS HD 106906b?" Tags="&lt;exoplanet&gt;&lt;mass&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19957" PostTypeId="2" ParentId="19956" CreationDate="2017-02-06T18:13:52.713" Score="1" Body="&lt;p&gt;Judging from the distance to its parent star, this could easily be a brown dwarf that formed together with the more massive star in a progenitor cloud. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However as this object is directly imaged, it's orbital properties are unknown. A high eccentricity together with semimajor-axis of a few tens to 100 AU, if found, would speak for the co-formation scenario. The &lt;a href=&quot;https://arxiv.org/abs/1312.1265&quot; rel=&quot;nofollow noreferrer&quot;&gt;discovery paper&lt;/a&gt; mentions that formation is unlikely to have happened from the protoplanetary disc, as well that it is co-moving with the host star, a.k.a. physically associated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also remember, that the $11\pm2 M_J$ mass of the planet is not measured directly, but derived from a model, that associates the observed luminosity to a mass, assuming a formation and atmospheric structure mechanism. Thus, it might have to be corrected upwards or downwards, emphasizing its origin as fragment of the original parent cloud.&lt;br&gt;&#xA;This means there's a possibility to not have been formed in the way normal planets are formed.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2017-02-06T18:13:52.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19958" PostTypeId="2" ParentId="19044" CreationDate="2017-02-06T21:34:19.863" Score="2" Body="&lt;p&gt;A paper titled &quot;&lt;a href=&quot;http://old.iaa.es/content/monte-carlo-models-dust-environment-sample-comets-oort-cloud-outer-main-asteroid-belt&quot; rel=&quot;nofollow noreferrer&quot;&gt;Monte Carlo Models of the dust environment of a sample of comets from the Oort Cloud to the outer Main Asteroid Belt&lt;/a&gt;&quot; (PDF &lt;a href=&quot;http://digital.csic.es/bitstream/10261/137982/1/Tesis_Fran_Pozuelos.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;also here&lt;/a&gt;) (2014) studied the behaviour of P/2013 P5 using models and concluded that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;strong&gt;The object has been subjected to an intermittent dust mass loss, most likely associated with a rotational disruption.&lt;/strong&gt; This is confirmed&#xA;  from the analysis of both HST and GTC images. The total dust mass&#xA;  released was of the order of $10^7\ kg$, for particle density of&#xA;  $1000\ kg\ m^{−3}$ and geometric albedo $p_v = 0.04$.&lt;/li&gt;&#xA;  &lt;li&gt;The model of rotational disruption, based on simulations of an object that loses mass from its equatorial region, and whose&#xA;  rotational axis is perpendicular to its orbit plane, reproduces to the&#xA;  last detail the observed complex brightness pattern at four different&#xA;  epochs of HST and GTC observations. For obliquities different from 0°&#xA;  or 180° , the fits get much worse. On the other hand, an isotropic&#xA;  ejection model does not fit the HST data because it produces much more&#xA;  diffuse tails than observed.&lt;/li&gt;&#xA;  &lt;li&gt;The ejection velocities are very low, of the order of $0.02–0.05\ m s^{−1}$. This places the limit on the size of the object to be in the&#xA;  range $30–134\ m$ for assumed densities of $3000–1000\ kg m^{−3}$.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This seems to confirm what the original question cited as a possible explanation, i. e. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The tails are suspected to be streams of material ejected by the&#xA;  asteroid as a result of a rubble pile asteroid spinning fast enough to&#xA;  remove material from it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The asteroid is loosely bound (a &quot;rubble pile&quot;) and spinning fast, so everytime some of the boulders which form it shift a bit, some material may escape into space and form a small tail.&lt;/p&gt;&#xA;" OwnerUserId="15398" LastActivityDate="2017-02-06T21:34:19.863" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19960" PostTypeId="2" ParentId="19520" CreationDate="2017-02-07T05:41:32.730" Score="2" Body="&lt;p&gt;Checked on Sky Map on my phone for late June, 2016 in Kolkata - and Saturn and Antares were close to each other around then in the Eastern sky, so these may be the two objects you saw.  The colour change sounds like atmospheric effects affecting the &lt;em&gt;seeing&lt;/em&gt; that night.  I don't know if June in Kolkata is particularly worse than usual for atmospheric conditions, but even on a seemingly clear night there can be a layer of turbulent air above you that will diffract the light from the sources and produce unsteady images.  This is usually worse closer to the horizon as well (as these two objects were).&#xA;Planets don't usually twinkle, so if Saturn was one of the objects, then it must have been a strong effect.&#xA;Nothing to freak out about, and keep looking up! :) &lt;/p&gt;&#xA;" OwnerUserId="15741" LastActivityDate="2017-02-07T05:41:32.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19961" PostTypeId="2" ParentId="13509" CreationDate="2017-02-07T06:33:16.653" Score="2" Body="&lt;p&gt;Conservation of angular momentum.  The spin of the protoplanetary disc will be randomly determined when it initially forms, but then it becomes the dominant factor.  The matter in the disc is then orbiting the centre of mass in the same direction even as it groups into asteroids and then protoplanets. Even though objects have their own individual spin, they all have the larger effect of the disc influencing them. &#xA;So all the planets rotate the same direction, except Uranus and Venus.  I think the hypothesis for those is still protoplanetary collision which has knocked Uranus on it's side and Venus right over.   &lt;/p&gt;&#xA;" OwnerUserId="15741" LastActivityDate="2017-02-07T06:33:16.653" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19962" PostTypeId="2" ParentId="13509" CreationDate="2017-02-07T08:47:17.017" Score="3" Body="&lt;p&gt;Remember that in a protoplanetary disc the rotation velocity $v_r$, which is keplerian, as a distance of star r varies like&lt;br&gt;&#xA;$$v_r(r) = \sqrt{\frac{GM}{r}} ~ ~~~~~~~~~~~~   (1)$$&#xA;This should serve to illustrate part of the point: at any $r&amp;lt;r_0$ we have $v_r&amp;gt;v_r(r_0)$ and the other way around. Thus viewed from the planet's position gas and dust 'left' of it is systematically flowing faster, and 'right' of it systematically flowing slower than the planet.&lt;br&gt;&#xA;Thus, if you would accrete a substantial fraction of your total, final mass and thus angular momentum from this flow, you'd induce a systematic spin automatically.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when is this relevant?&lt;br&gt;&#xA;The region from which a protoplanet or asteroid can accrete is maximally its gravitational sphere of influence, also Hill-sphere with radius &#xA;$$r_H = r_0 \sqrt[3]{\frac{m_{planet}}{3 m_{star}}} ~~~~~~~~~(2) $$&#xA;where $r_0$ as above, is a semi-major axis distance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if this $r_H$ is too small to feel the velocity-gradients in (1), or said differently, if the accreting object is not massive enough for $r_H$ to extend significantly into the protostellar disc, then accretion will accumulate random momenta.&lt;br&gt;&#xA;If the protoplanet manages to grow to a substantial Hill-sphere, it starts accreting gas and solids with a huge velocity difference $v_r(r)-v_r(r_0)$, which is always systematic, instead of random.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR&lt;/strong&gt; &#xA;Small objects, roughly below the size of asteroids, accrete random momenta pushes. Massive objects, protoplanetary and above, accrete systematic velocity differences, thus giving them a net angular momentum.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2017-02-07T08:47:17.017" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19963" PostTypeId="1" AcceptedAnswerId="19973" CreationDate="2017-02-07T10:39:36.560" Score="8" ViewCount="449" Body="&lt;p&gt;&lt;strong&gt;edit:&lt;/strong&gt; I just saw &lt;a href=&quot;https://twitter.com/Yeqzids/status/852903116244746240&quot; rel=&quot;nofollow noreferrer&quot;&gt;this tweet&lt;/a&gt; and find it incredibly relevant :)&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mKh1a.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mKh1a.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/admcn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/admcn.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;begin question:&lt;/strong&gt; See &lt;a href=&quot;https://astronomy.stackexchange.com/a/19958/7982&quot;&gt;this answer&lt;/a&gt; and then consider if there are known or likely solar system bodies that might be identified as both asteroids and comets.&lt;/p&gt;&#xA;&lt;p&gt;If so (there seems to be an example there) would this be because there is insufficient information known about the body, and the ambiguity could be cleared up with enough data (say a visit by a spacecraft)?  Or are the categories of asteroids and comets actually overlapping?&lt;/p&gt;&#xA;&lt;p&gt;If the latter, is there work underway to improve or update these definitions?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="15835" LastEditDate="2020-12-09T04:14:40.667" LastActivityDate="2020-12-09T04:14:40.667" Title="Do astronomers generally agree that the distinction between comets and asteroids is not so clear?" Tags="&lt;solar-system&gt;&lt;asteroids&gt;&lt;comets&gt;&lt;terminology&gt;&lt;definition&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="19964" PostTypeId="1" CreationDate="2017-02-07T10:41:07.407" Score="1" ViewCount="38" Body="&lt;p&gt;I have been reading the &lt;a href=&quot;http://omniweb.gsfc.nasa.gov/html/ow_data.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;OMNIWeb data documentation&lt;/a&gt; in order to clarify one particular question without success. I need to associate each measurement in the OMNI dataset to a moment in time at the sun (at least to narrow it to a time interval). Therefore, I need to understand how long it takes for the solar wind to get from the sun to the point where it is measured. I understand that the solar wind velocity is variable, so this time interval is constantly changing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you please help me understand this question? Can the solar wind measurements be used to trace back into the moment at which it left the sun? I would be very grateful if you could explain or point me to some sufficient references.&lt;/p&gt;&#xA;" OwnerUserId="15743" LastActivityDate="2017-02-07T10:41:07.407" Title="Tracing back OMNI data to the sun" Tags="&lt;the-sun&gt;&lt;data-analysis&gt;&lt;solar-wind&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19966" PostTypeId="1" AcceptedAnswerId="19967" CreationDate="2017-02-07T13:27:06.483" Score="13" ViewCount="3058" Body="&lt;p&gt;I was trying to describe how vast the largest known star is to someone and felt I wasn't quite able to relate the scale difference. I know it's roughly 1500 times larger than the sun. Anyone know of a good analogy? &lt;/p&gt;&#xA;" OwnerUserId="15745" LastEditorUserId="2365" LastEditDate="2017-02-07T23:34:27.617" LastActivityDate="2018-08-21T15:56:51.737" Title="Relatable comparison of VY Canis Majoris to the Sun?" Tags="&lt;star&gt;&lt;size&gt;" AnswerCount="7" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19967" PostTypeId="2" ParentId="19966" CreationDate="2017-02-07T13:57:39.160" Score="26" Body="&lt;p&gt;After some playing around with wolfram alpha and google my best comparison has been&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The sun compared to VY Canis Majoris is like a donut compared to the London Eye.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The London Eye is about 120m in diameter, this divided by 1500 is about 8cm which is roughly the diameter of a ring donut. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jl6l5.jpg&quot; alt=&quot;VY Canis Majoris&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15745" LastEditorUserId="10589" LastEditDate="2018-08-21T15:56:51.737" LastActivityDate="2018-08-21T15:56:51.737" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="19968" PostTypeId="2" ParentId="19966" CreationDate="2017-02-07T14:15:00.173" Score="5" Body="&lt;p&gt;Depends a bit on whether you want to compare diameters or volume (3D). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For radius,  consider that it's approximately 6.6 A.U., so if you dropped it into the Solar System it'd extend out to well past Jupiter's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or it's the ratio of a standard tenpin bowling ball to the point of a &lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20110303132601AAOmiN7&quot; rel=&quot;noreferrer&quot;&gt;sewing needle&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The volumetric ratio is 1420^2 or about 2 million to one. That's roughly one person out of the entire population of Houston, TX.   Or the ratio of 100 liters of water to a single drop from an eyedropper.&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-02-07T14:15:00.173" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19970" PostTypeId="2" ParentId="19966" CreationDate="2017-02-07T15:46:33.727" Score="22" Body="&lt;h2&gt;On the banana scale, it's an ore freighter.&lt;/h2&gt;&#xA;&lt;p&gt;Every physicist worth their salt knows that the most important scale in the galaxy is the banana. Now, your average banana is &lt;a href=&quot;http://healthyeating.sfgate.com/portion-size-bananas-7664.html&quot; rel=&quot;noreferrer&quot;&gt;between 7-8 inches in length&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Approximately the same shape are the ore freighters that go on the Great Lakes. They're called 1000 footers because, go figure, they're roughly 1000 feet long. The banana is to the freighter as the sun is to YV Canis Major.&lt;/p&gt;&#xA;&lt;p&gt;Here's a picture of one of those freighters. Now imagine one of the workers is having lunch, eating a banana.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bpJOd.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bpJOd.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2239" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-02-07T15:46:33.727" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19972" PostTypeId="2" ParentId="19918" CreationDate="2017-02-07T18:26:08.633" Score="1" Body="&lt;p&gt;As barrycarter mentions in the comments, if you take the average of the two altitudes, you'll get the altitude of the celestial pole—north if you're in the northern hemisphere, south if you're in the southern hemisphere.  In this case, the average is just the average in degrees plus the average in minutes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{38+12}{2} = 25 \text{ degrees}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{52+24}{2} = 38 \text{ minutes}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equal to the latitude of the observation point.  One can also obtain the declination of the star in question, by halving the &lt;em&gt;difference&lt;/em&gt; between the two altitudes, rather than their sum, and then subtracting the result from $90$ degrees:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{38-12}{2} = 13 \text{ degrees}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{52-24}{2} = 14 \text{ minutes}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you subtract that from $90$ degrees, you get a declination of $76$ degrees, $46$ minutes.  I know of no bright star particularly near that declination, but perhaps this is just an academic exercise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The foregoing neglects atmospheric refraction.  This turns out to be not entirely negligible at the level of precision provided in the question.  At an altitude of $39$ degrees, atmospheric refraction produces an additional elevation of about $1.2$ minutes (barring local and daily effects), meaning that if a star is observed at $38$ degrees $52$ minutes, it's &quot;really&quot; at $38$ degrees $51$ minutes or so.  At an altitude of $12$ degrees, the additional elevation is about $4.5$ minutes, meaning that the star observed at $12$ degrees $24$ minutes is &quot;really&quot; at $12$ degrees $19–20$ minutes.  If the altitude figures given are as observed, then the corrected figures should be used instead.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;ETA: Here's a page with information on the extent of &lt;a href=&quot;http://www.mysundial.ca/tsp/refraction_of_light.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;atmospheric refraction&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="15753" LastEditorUserId="15753" LastEditDate="2017-02-09T18:15:13.493" LastActivityDate="2017-02-09T18:15:13.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19973" PostTypeId="2" ParentId="19963" CreationDate="2017-02-07T19:27:02.613" Score="10" Body="&lt;p&gt;If we're going to get technical, Asteroids are not really an &lt;em&gt;official&lt;/em&gt; name anymore. In 2006, when the IAU redefined what a planet was (and thus demoted Pluto), they also decided to more formally define other terms to identify objects in our solar system. You can see a diagram of all the official terms and how they relate below. Notice the important factor that asteroids no longer made the cut. There is the collection of &amp;quot;Small Solar System Bodies&amp;quot; (SSSB) and within that falls comets and centaurs, but asteroids are not exclusively called out.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/owHbc.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/owHbc.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;So if you want to get technical, asteroids are really SSSBs now and comets are SSSBs as well, but also happen to fit into the sub-category of comets. What defines an SSSB as a comet is its ability to form a coma based on the sublimation of icy surface volatiles. So you might distinguish an &amp;quot;asteroid&amp;quot; as an object which cannot form a coma (and orbits inside Neptune), but that still leads to an ambiguity because what happens when a comet runs out of surface volatiles? Does it now become an asteroid? Is it still considered a comet? There are no clear answers.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-02-07T19:27:02.613" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="19974" PostTypeId="1" AcceptedAnswerId="19975" CreationDate="2017-02-07T20:30:52.743" Score="2" ViewCount="230" Body="&lt;p&gt;I recently observed the beautiful sight of Venus quite close to the crescent moon, and was wondering, what the closest (angular distance as seen from earth) they come.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there on-line tables or calculators for finding when such nearest events occur and are visible?&lt;/p&gt;&#xA;" OwnerUserId="11222" LastActivityDate="2017-02-08T17:58:31.550" Title="What's the closest Venus gets to the moon?" Tags="&lt;observational-astronomy&gt;&lt;venus&gt;&lt;the-moon&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19975" PostTypeId="2" ParentId="19974" CreationDate="2017-02-07T20:44:07.470" Score="5" Body="&lt;p&gt;Venus is, on occasion, occulted by the Moon (that is to say, the Moon passes in front of Venus as seen from our point of view).  Exactly when that happens is dependent on your location on the Earth, since the location of the Moon in the skies varies by about a degree or so depending on your location on the Earth, and of course, it might be on the other side of the Earth when it happens (it only lasts about an hour or less).  From North America, it happened in December 2015; it will happen again in September 2017, but only from Australia and neighboring countries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you google &quot;Venus occultation&quot; you should get something like what you want.  ETA: Here's one link: &lt;a href=&quot;http://www.lunar-occultations.com/iota/planets/planets.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.lunar-occultations.com/iota/planets/planets.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15753" LastEditorUserId="15753" LastEditDate="2017-02-08T17:58:31.550" LastActivityDate="2017-02-08T17:58:31.550" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19978" PostTypeId="2" ParentId="19966" CreationDate="2017-02-08T02:37:28.077" Score="8" Body="&lt;p&gt;&lt;b&gt;An airplane flying along the surface of the sun would take about 6.6 months to circle it once.&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The same airplane would take 787 years to complete one trip around VY Canis Majoris.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;hr&gt;&#xA;Aircraft's speed: &lt;b&gt;900 km / h&lt;/b&gt;&#xA;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sun's radius: &lt;b&gt;696,000 km&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One circle around the sun: &lt;b&gt;4,373,096.97 km&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time for aircraft to complete this circle: &lt;b&gt;4858 hours ≈ 6.6 months &lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&#xA;VY Canis Majoris radius: &lt;b&gt;988,320,000 km&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One circle around VY Canis Majoris: &lt;b&gt;6,209,797,703 km&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time for aircraft to complete this circle: &lt;b&gt;6899775.225 hours&#xA;≈ 787.645 years! &lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;sub&gt;PS: Looks like the math done in one of the videos shared in the comments above is incorrect :)&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="10589" LastActivityDate="2017-02-08T02:37:28.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19980" PostTypeId="1" CreationDate="2017-02-08T04:23:03.430" Score="1" ViewCount="253" Body="&lt;p&gt;&lt;strong&gt;note:&lt;/strong&gt; This question is asking for a representation of the &lt;em&gt;astronomical classification scheme now used for solar system bodies&lt;/em&gt; as established (presumably) by the Working Group for Planetary System Nomenclature (WGPSN) of the &lt;a href=&quot;https://en.wikipedia.org/wiki/International_Astronomical_Union&quot; rel=&quot;nofollow noreferrer&quot;&gt;International Astronomical Union&lt;/a&gt; or IAU. While graphical layout is mentioned as part of the question's motivation, the question is not about layout or graphics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler_diagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;Euler diagram&lt;/a&gt; shown in &lt;a href=&quot;https://astronomy.stackexchange.com/a/19973/7982&quot;&gt;this helpful answer&lt;/a&gt; is hard for me to understand, I think mostly due to graphical layout. &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Euler_diagram_of_solar_system_bodies.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;The diagram shows&lt;/a&gt; the mathematical relationship between various sets (in this case solar system body classifications) so it could be drawn with widely differing choices of layout and still retain exactly the same mathematical relationships.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering if there is some attempt here to indicate relative numbers or ratios by the degree of overlap? Look closely at the small overlap of Minor Planets with Comets. Is that meant to reflect in some way a relatively small overlap quantitatively? Did you notice it took a while to even make sure you were seeing the overlap correctly? This style of layout requires the use of color perception to help identify which label is associated with which contour while simultaneously trying to follow one line in a groups of perfectly parallel lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have other versions of this Euler diagram representation of solar system body classification been published? A link to one or even inclusion of the diagram in the answer itself would be greatly appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/owHbc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/owHbc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Example of an Euler diagram in &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler_diagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Wikipedia article&lt;/a&gt;, from &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Euler_diagram_of_solar_system_bodies.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qB5w4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qB5w4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Example of a somewhat easier to use and understand Euler diagram, from &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:British_Isles_Euler_Diagram_13.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, used in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler_diagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;same Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2017-02-09T16:27:42.227" Title="Is there an alternative (potentially easier to understand) Euler diagram of the new solar system body classification?" Tags="&lt;asteroids&gt;&lt;comets&gt;&lt;classification&gt;&lt;visualization&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19981" PostTypeId="1" AcceptedAnswerId="19984" CreationDate="2017-02-08T09:11:21.070" Score="2" ViewCount="191" Body="&lt;p&gt;I searched the internet for images taken by the &lt;a href=&quot;https://en.wikipedia.org/wiki/South_Pole_Telescope#The_telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;South Pole Telescope&lt;/a&gt; but without successes in finding it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:Some images&#xA;&lt;a href=&quot;https://i.stack.imgur.com/j31Rn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j31Rn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10763" LastEditorUserId="10763" LastEditDate="2017-02-09T11:42:02.490" LastActivityDate="2017-02-09T11:42:02.490" Title="Is there available space images taken by the south pole Gregorian telescope?" Tags="&lt;telescope&gt;&lt;space&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19982" PostTypeId="2" ParentId="19966" CreationDate="2017-02-08T11:49:48.127" Score="0" Body="&lt;p&gt;I think both the ore freighter / banana and donut / London Eye comparisons have scales that are relatable but lack &quot;universality&quot; in the sense that people may not know the size of the London Eye or an ore freighter. I'd suggest thumbnail vs bus, as both exist all over the world.&lt;/p&gt;&#xA;" OwnerUserId="15154" LastActivityDate="2017-02-08T11:49:48.127" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19983" PostTypeId="1" AcceptedAnswerId="20040" CreationDate="2017-02-08T15:29:38.057" Score="2" ViewCount="126" Body="&lt;p&gt;We know that gravitational waves stretch space-time so light travelling through that space should be stretched as well. So my question is how do we know anything is stretching if the effect is experienced by every object in the universe. &lt;/p&gt;&#xA;" OwnerUserId="15616" LastActivityDate="2017-02-13T12:31:45.830" Title="Gravitational waves" Tags="&lt;gravity&gt;&lt;astrophysics&gt;&lt;space-time&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19984" PostTypeId="2" ParentId="19981" CreationDate="2017-02-08T16:07:19.477" Score="3" Body="&lt;p&gt;The South Pole Telescope (SPT) was funded, in part, by the NSF. This means their data must be made public as that is almost always a condition of funding by the NSF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you go to the SPT's homepage, they link to their &lt;a href=&quot;http://pole.uchicago.edu/spt/hep_summary_dmp/Summary%20for%20HEP%20of%20Data%20Management%20Principles%20(SPT).pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Data Management Policy&lt;/a&gt;. Within this policy they state:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The collaboration – in particular the P.I. and data managers at the University of Chicago – bear the responsibility for making data public. Public data products are hosted on University of Chicago machines and on the NASA Legacy Archive for Microwave Background Data Analysis (LAMBDA) server.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I went over to NASA's LAMBDA server and found that SPT data was hosted &lt;a href=&quot;https://lambda.gsfc.nasa.gov/product/spt/&quot; rel=&quot;nofollow noreferrer&quot;&gt;right here&lt;/a&gt;. Look to the left-hand side menu to see the links to raw data for the SPT-SZ and SPTpol instruments, individually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The University of Chicago hosts their data &lt;a href=&quot;http://%20pole.uchicago.edu/public/publications.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;right here&lt;/a&gt; (props to called2voyage). It seems that they provide publications and include any data used in that publication rather than just hosting all the RAW data like the LAMBDA server does. Use whichever is most convenient for you.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2017-02-08T17:00:20.987" LastActivityDate="2017-02-08T17:00:20.987" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19985" PostTypeId="1" CreationDate="2017-02-08T17:02:33.967" Score="1" ViewCount="465" Body="&lt;p&gt;Sometimes, you can see the Moon high in the night sky, which is illuminated from the side where Sun goes down but slightly from above. It is paradoxical when Sun hangs at the lower level above the horizon than the Moon. Sun can even go down behind the horizon, yet, the Moon is illuminated slightly from above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paradox is defined &lt;a href=&quot;https://en.wikipedia.org/wiki/Terminator_(solar)#Lunar_terminator_illusion&quot; rel=&quot;nofollow noreferrer&quot;&gt;in Wikipedia&lt;/a&gt; and captured in &lt;a href=&quot;https://youtu.be/bhEDxh5moq4?t=53&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a video from &quot;Vsauce&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://youtu.be/Y2gTSjoEExc?t=280&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://youtu.be/Y2gTSjoEExc?t=280&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which pretends to explain the phenomenon but fails, IMO. From his explanation I only get that light beam should deviate from the direct route to move along the orbital around the Earth, like if it is the Sun or Moon orbit seen from the Earth. But why? The nebular objects orbit because they have mass and gravitationally bound. Why should light beam travelling from Sun to the Moon bend the same way as if it orbits around the Earth? I do not get the explanations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The popular explanation of the illusion does not hold the water&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://youtube/watch?v=Rsz2QNPprB8&amp;amp;lc=z134gjggimumexwgr04cdxbrnlattlwp5n40k&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://youtube/watch?v=Rsz2QNPprB8&amp;amp;lc=z134gjggimumexwgr04cdxbrnlattlwp5n40k&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15769" LastEditorUserId="2365" LastEditDate="2017-04-11T19:58:59.640" LastActivityDate="2017-06-10T20:16:33.933" Title="Lunar terminator illusion" Tags="&lt;optics&gt;&lt;the-moon&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19986" PostTypeId="1" AcceptedAnswerId="19992" CreationDate="2017-02-08T17:29:07.563" Score="0" ViewCount="118" Body="&lt;p&gt;On the following page, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.uni.edu/morgans/astro/course/Notes/section2/spectraltemps.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.uni.edu/morgans/astro/course/Notes/section2/spectraltemps.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It mentions Main, Giants and SuperGiants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However for Dwarf Stars (e.g. D, Sd) , do I apply Main Sequence to the solar type?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For other types, IV, II, do I apply Main Sequence / Subgiant / Giant to the type?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the linked webpage, you'll see it mentions Types (V,III,I) but for instance where would Procyon (F5IV-V), Sargas (F1II) fit or should there be a different table for types IV, II?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would groups (D, sd, IV,m II) have a different table?  Rigil Kentaurus (G2V) would clear fit into Main Sequence.&lt;/p&gt;&#xA;" OwnerUserId="5401" LastEditorUserId="5401" LastEditDate="2017-02-09T07:06:56.043" LastActivityDate="2017-03-11T07:25:00.063" Title="Solar Spectral Types and Dwarf Stars" Tags="&lt;dwarf-star&gt;&lt;spectral-type&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19990" PostTypeId="2" ParentId="19966" CreationDate="2017-02-08T19:02:32.260" Score="0" Body="&lt;p&gt;Well, taking the scale of the solar system as an example, you can say that VY Canis Majoris is around 70% the size of Saturn's orbit.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference numbers:&lt;br/&gt;&#xA;Saturn's average distance from the Sun = 1,433,449,370 km.&lt;br/&gt;&#xA;VY Canis Majoris radius = 988,300,000 km.&lt;/p&gt;&#xA;" OwnerUserId="15774" LastActivityDate="2017-02-08T19:02:32.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19991" PostTypeId="2" ParentId="19966" CreationDate="2017-02-08T21:11:32.740" Score="1" Body="&lt;p&gt;You should look at &lt;a href=&quot;https://www.youtube.com/watch?v=pR5VJo5ifdE&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mark Rober's video&lt;/a&gt;, he uses a size comparaison known by &lt;strike&gt;everybody&lt;/strike&gt;every American: a football field!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the Sun is a normal soccer ball...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Earth is a pinhead at 26 yards (23m) away from the Sun/ball.&lt;/li&gt;&#xA;&lt;li&gt;Jupiter is a grape at 135 yards (123m) away from the Sun/ball, almost the size of the field.&lt;/li&gt;&#xA;&lt;li&gt;Saturn is a smaller grape at &quot;2,5 football fields away from the Sun/ball&quot;, which is almost 300 yards (274m)&lt;/li&gt;&#xA;&lt;li&gt;At 6:02 of the video, &quot;the largest star known&quot; would be the size of the Empire State Building: 381m (416 yards)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15771" LastActivityDate="2017-02-08T21:11:32.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19992" PostTypeId="2" ParentId="19986" CreationDate="2017-02-09T02:46:07.557" Score="0" Body="&lt;p&gt;The modern system of stellar classification is 2-dimensional, with one axis conveying spectral class, and the other conveying luminosity class. The spectral types are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;O-type, &quot;blue&quot;, $\geq 30000$ $\mathrm{K}$&lt;/li&gt;&#xA;&lt;li&gt;B-type, &quot;blue-white&quot;, $10000-30000$ $\mathrm{K}$&lt;/li&gt;&#xA;&lt;li&gt;A-type, &quot;white&quot;, $7500-10000$ $\mathrm{K}$&lt;/li&gt;&#xA;&lt;li&gt;F-type, &quot;yellow-white&quot;, $6000-7500$ $\mathrm{K}$&lt;/li&gt;&#xA;&lt;li&gt;G-type, &quot;yellow&quot;, $5200-6000$ $\mathrm{K}$&lt;/li&gt;&#xA;&lt;li&gt;K-type, &quot;orange&quot;, $3700-5200$ $\mathrm{K}$&lt;/li&gt;&#xA;&lt;li&gt;M-type, &quot;red&quot;, $2400-3700$ $\mathrm{K}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are also some extended spectral types, including D for white dwarfs, C for carbon stars, and several more for brown drawfs. The above are the predominant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnitude axis consists of the following groupings:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;0-type, &quot;hypergiants&quot;&lt;/li&gt;&#xA;&lt;li&gt;I-type, &quot;supergiants&quot;&lt;/li&gt;&#xA;&lt;li&gt;II-type, &quot;bright giants&quot;&lt;/li&gt;&#xA;&lt;li&gt;III-type, &quot;giants&quot;&lt;/li&gt;&#xA;&lt;li&gt;IV-type, &quot;subgiants&quot;&lt;/li&gt;&#xA;&lt;li&gt;V-type, &quot;dwarfs&quot; (these are the main-sequence stars)&lt;/li&gt;&#xA;&lt;li&gt;VI-type, &quot;subdwarfs&quot;&lt;/li&gt;&#xA;&lt;li&gt;VII-type, &quot;white dwarfs&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When these systems are used, a spectral class is combined with a number ranging from 0-9 which symbolizes the location between adjacent spectral classes, and the luminosity class is then appended. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the Sun is a type G2V star. This means it is a G-type star on the main sequence. It could then informally also be referred to as a &quot;yellow dwarf&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system is more complex than this, and there are many additional suffixes and other symbols to further explain the nuances of individual stars. This does cover the basics quite well, however. &lt;/p&gt;&#xA;" OwnerDisplayName="user14781" LastActivityDate="2017-02-09T02:46:07.557" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19993" PostTypeId="1" CreationDate="2017-02-09T02:55:05.780" Score="1" ViewCount="549" Body="&lt;p&gt;I just enrolled in an introductory Astronomy course, and today in lecture the professor asked this question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Give 4 ways in which the day/night cycle and the seasons would behave on Earth if our rotational axis was in our orbital plane (Uranus is close to this). What about if the rotational axis was perpendicular to the orbital plane (so the ecliptic and Celestial Equator lay on top of each other), but the Earth's orbit was quite eccentric instead of almost circular?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the first part, it means the Earth is rotating by North/South, right? Would this mean that the day/nights would be &quot;flipped around,&quot; i.e. it would be night in a city when it would normally be day. I don't think the seasons would change, since the orbit remains the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really know what the 2nd part is saying. Is it simply asking what would change if the orbit was eccentric rather than what it is now?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help or clarification on this would be greatly helpful. Thank you!&lt;/p&gt;&#xA;" OwnerUserId="15781" LastActivityDate="2017-02-09T03:12:56.440" Title="Day/Night Cycles and Seasons if the Earth's Rotational Axis and Orbit Changed" Tags="&lt;orbit&gt;&lt;rotation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19994" PostTypeId="2" ParentId="19993" CreationDate="2017-02-09T03:12:56.440" Score="1" Body="&lt;p&gt;The Earth's rotational axis is essentially the line connecting the North and South poles.  This axis is slightly tilted, making the Earth's equator 23.5 degrees offset from the orbital plane (the ecliptic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GIDl5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GIDl5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;From &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth&amp;#39;s_rotation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first part of the question is asking what would happen if we were to make the rotational axis line up with the orbital plane.  That would essentially mean tilting it so that the Earth's equator would be 90 degrees offset from the ecliptic.  One way to do this would be to point the North or South pole directly at the sun, or to point it in the same direction as Earth's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second part of the question is asking what would happen if the rotation axis was perfectly up and down.  This would mean that the equator would not be offset from the ecliptic at all.  However, for this part of the question, the Earth's orbit would also be highly eccentric, meaning that our distance to the sun would change a great amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I won't post the answer here, since this looks like it is probably a homework question, but to find the answer, first read up about what causes the &lt;a href=&quot;https://en.wikipedia.org/wiki/Season&quot; rel=&quot;nofollow noreferrer&quot;&gt;seasons&lt;/a&gt;.  To talk about the day/night cycle, think about how the &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth&amp;#39;s_rotation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth's rotation&lt;/a&gt; causes day and night, by moving the sun across the sky.  If we changed how the Earth was tilted, how would that affect the seasons?  How would that affect where the sun appears in the sky?&lt;/p&gt;&#xA;" OwnerUserId="13426" LastActivityDate="2017-02-09T03:12:56.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19995" PostTypeId="1" CreationDate="2017-02-09T04:56:03.637" Score="0" ViewCount="320" Body="&lt;p&gt;If the Earth's rotational axis was perpendicular to the orbital plane, but the orbit also highly eccentric (sun at center), how would that affect the day/night cycles? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I think is that during the time the Earth's closer to the sun, because the gravitational pull is stronger, it would make the Earth spin faster than thereby creating much shorter days than if the Earth was near the focii of the orbit (on which the days/nights would be longer). Is this line of thinking correct?&lt;/p&gt;&#xA;" OwnerUserId="15784" LastActivityDate="2017-02-10T14:31:24.000" Title="Eccentric Earth Orbit and Day/Night Cycles" Tags="&lt;orbit&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19997" PostTypeId="2" ParentId="19995" CreationDate="2017-02-09T13:51:24.830" Score="1" Body="&lt;p&gt;Your thinking is incorrect.    High gravity, or, more specifically, strong tidal forces lead to tidal locking not faster spinning.   Moons are a good example of this.   The Earth's gravity keeps the same side of the Mon facing the Earth at all times.   The Moon still rotates but it's rotation is in sync with it's orbit around the Earth, 1:1, so the same size of the Moon (the heavier/denser side) always points towards Earth.   Many moons are tidally locked to their planets as a result of the strong tidal force.  Pluto and Charon are tidally locked to each other.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, tidal locking takes a long time.  Just passing close to a massive object wouldn't stop a planet from rotating.   It takes many orbits to create the tidal locking, usually millions or billions.  Because planets tend to be pretty uniform in density, gravity doesn't affect their rotation very much. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A highly eccentric orbit, the sun isn't close to the center, either.   A low eccentricity orbit, the sun is close to the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yVpQ5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yVpQ5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A highly eccentric orbit would affect seasons.  With no Axial tilt, the seasons would be entirely governed by the distance to the sun, not the position of the sun in the sky which governs seasons on Earth as it is.   To clarify, it's the Axial tilt that causes the seasons, not the distance to the sun.  But as you specified with zero axial tilt, then the distance would govern the seasons, but at Earth's current eccentricity, that effect would be small.   The greater the eccentricity the greater the seasonal variation in this hypothetical and the seasons would be planet wide, not alternating between the two hemispheres.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summer would also be shorter than winter due to the Earth moving faster around the sun at the closest point,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The days themselves wouldn't be affected much.   The length of a typical day (&lt;a href=&quot;http://astro.unl.edu/naap/motion3/sidereal_synodic.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Synodic&lt;/a&gt;) would vary by a minute or two based on the Earth's orbital speed.  It wouldn't be very noticeable but with a perfect 0 degree axial tilt, the 12 hours of sunshine would vary slightly based on orbital speed and position in the eccentric orbit. . . and the size of the sun in the sky and atmospheric refraction. . . . but only a couple minutes here or there with the Earth's current rotation speed. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-02-10T14:31:24.000" LastActivityDate="2017-02-10T14:31:24.000" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19998" PostTypeId="2" ParentId="19980" CreationDate="2017-02-09T16:27:42.227" Score="1" Body="&lt;p&gt;Per the IAU (Resolution B5, adopted by the 2006 IAU General Assembly), there are five supposedly mutually exclusive categories of things in the solar system:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Sun (only one), which is distinguished by the ability to fuse hydrogen into helium in its core.&lt;/li&gt;&#xA;&lt;li&gt;The planets (eight, might possibly become nine if Michael Brown is correct), which are distinguished by having the ability to &quot;clear their orbit&quot; (a well-defined concept with numerical meaning) before the Sun turns into a red giant.&lt;/li&gt;&#xA;&lt;li&gt;The dwarf planets (unknown number, most likely well over 100), which are distinguished as being bodies that don't qualify as the sun or as planets, but are sufficiently massive to more or less have pulled themselves into a roundish shape. &lt;/li&gt;&#xA;&lt;li&gt;Small solar system bodies (unknown number, well into the millions) which are distinguished by not being distinguishable as any of the other categories.&lt;/li&gt;&#xA;&lt;li&gt;The satellites of planets, dwarf planets, and small solar system bodies (unknown number, the number of which change over time as satellites are captured or rejected) which are distinguished by being in orbit about a planet / dwarf planet / SSSB rather than the Sun.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Contrary to that resolution, these boundaries are fuzzy rather than sharp, particularly when one looks outside of our solar system. (Note well: The IAU definitions pertain to our solar system only. There are currently eight planets in the entire universe, with the possibility of growing to nine if the conjectures regarding a largish planet beyond Pluto turn out to be correct.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the first boundary, that which separates planets from stars. This becomes fuzzy with the smallest red dwarf and the largest super-Jupiter. There's no clear-cut boundary here. In fact, there's a good amount of debate over this very issue. This boundary is quite sharp in our solar system, but it almost certainly is anything but sharp in star systems beyond our own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second boundary, the one between planets and dwarf planets (which, despite the name, are not planets) also has the potential for being fuzzy. In our solar system, that boundary is once again rather sharp. Any time you can come up with a boundary that is separated by five orders of magnitude you have a nice sharp classification boundary. However, given the hugeness of the universe, there is almost certainly some star system out there, somewhere, where this boundary is as fuzzy as can be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third boundary, the one between dwarf planets and small solar system bodies, is fuzzy in our own solar system. Objects less than 100 meters in diameter are, as a rule, anything but &quot;round&quot;. Objects over a kilometer in diameter are, as a rule, quite &quot;round&quot; (when the shape is resolvable). Astronomers have found lots and lots of objects between 200 and 600 meters in diameter. Squint hard enough at the smaller ones and you might be able to convince yourself that they are &quot;roundish&quot;. Focus intensely on the larger ones and you might be able to convince yourself that they are not quite round. There's no sharp boundary here, even in our own solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fifth category, satellites vs non-satellites, opens yet another can of worms. Planets (particularly the giant planets) can temporarily capture objects as satellites, only to lose them as satellites after a short time (tens or even hundreds of thousands of years is a &quot;short time&quot; in this context).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;Note well: I'm not saying Pluto should once again be promoted to the status of &quot;planethood&quot;. That very sharp boundary (at least in our solar system) says that it most definitely is not a &quot;planet&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about asteroids versus comets, which is what started this series of questions? That boundary is ridiculously fuzzy. Fuzzy logic (a real and serious concept) doesn't come close to addressing how &quot;fuzzy&quot; this is. Centaurs and formerly active comets are the boundary cases. There is no boundary. There instead is a soft melding from one to another.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-02-09T16:27:42.227" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19999" PostTypeId="2" ParentId="19985" CreationDate="2017-02-09T17:12:34.737" Score="1" Body="&lt;p&gt;The light ray does not bend (in fact it does a little bit, but that's small and doesn't matter now). However, what you see is not a path of a ray of light, but a projection of this path on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Celestial_sphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;celestial sphere&lt;/a&gt; around you. This projection is an arc of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Great_circle&quot; rel=&quot;nofollow noreferrer&quot;&gt;great circle&lt;/a&gt; with you in the centre.&lt;/p&gt;&#xA;" OwnerUserId="15069" LastActivityDate="2017-02-09T17:12:34.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20000" PostTypeId="1" CreationDate="2017-02-09T19:30:29.530" Score="3" ViewCount="135" Body="&lt;p&gt;In our Milkyway all the stars are orbitting around the black hole in the center. But what is the cause it this rotation and are there galaxies where the stars aren't orbitting a center?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="10445" LastEditDate="2017-02-10T11:27:02.397" LastActivityDate="2017-02-10T11:27:02.397" Title="Are the stars in all galaxies orbitting around a middle?" Tags="&lt;galaxy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20002" PostTypeId="2" ParentId="20000" CreationDate="2017-02-09T20:51:47.913" Score="0" Body="&lt;p&gt;The cause of the rotation is that the hydrogen cloud that the galaxy condensed out of had some angular momentum which was conserved as the cloud collapsed into a galaxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stars in a galaxy have to be in orbit around something. (Even if it's just the center of mass of the galaxy.) If they were at rest with respect to the center of mass, they'd fall into it!&lt;/p&gt;&#xA;" OwnerUserId="11458" LastActivityDate="2017-02-09T20:51:47.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20003" PostTypeId="2" ParentId="20000" CreationDate="2017-02-09T21:11:02.990" Score="4" Body="&lt;p&gt;Stars aren't really orbiting the &lt;em&gt;center&lt;/em&gt; of a galaxy, as much as they're orbiting the gravitational potential of the galaxy. The galaxy doesn't need to have a black hole in the center for the stars to orbit, and even if it does (which in fact it does most of the time), the black hole does not dominate the gravitational potential, except in the most central part (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Sphere_of_influence_(black_hole)&quot;&gt;sphere of influence of a black hole&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All matter in the galaxy contributes to the gravitational potential, but since dark matter comprises 5/6 of the mass, this is the most important contributor. Only in the hypothetical case of a perfectly symmetrical distribution of masses, would the black hole and the center of rotation coincide. And since the paths of the stars are ellipses and not perfect circles, the center of the potential will not lie in the center of their orbits, but in one of the focal points. Furthermore, galaxies are dynamical and the potential continually changes, in turn changing the exact orbit of a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxies form due to the collapse of a dark matter halo, and the gas that follows  along. Such a halo will in general have a non-zero angular momentum, which is partially conserved during the collapse (they do lose a significant amount, for instance through minor and major merging; see e.g. &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0602005&quot;&gt;D'Onghia et al. 2006&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in the case of a forming star, a forming planet, an ice skater, or you on an office chair, reducing the radius means increasing orbital speed. This is the origin of the rotation.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2017-02-09T21:11:02.990" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20004" PostTypeId="1" CreationDate="2017-02-10T00:16:08.763" Score="2" ViewCount="89" Body="&lt;p&gt;What is the mathematical formula used to do that? Is there a website/app that does the calculation for me?&lt;/p&gt;&#xA;" OwnerUserId="15790" LastActivityDate="2017-04-28T14:13:28.300" Title="How to calculate what is the best date and latitude to observe a certain object in space?" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20006" PostTypeId="2" ParentId="19963" CreationDate="2017-02-10T02:28:43.600" Score="1" Body="&lt;p&gt;This is supplemental information to further clarify the answer. The drawing is confusing to me an likely to others. I've written a color filter in Python to help isolate only four of the boundaries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gJe2N.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gJe2N.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;&#xA;colorz = dict()&#xA;&#xA;colorz['red']    = 1.0,   0.0,   0.0&#xA;colorz['green']  = 0.0,   0.502, 0.0&#xA;colorz['blue']   = 0.0,   0.0,   1.0&#xA;colorz['purple'] = 0.435, 0.192, 0.596&#xA;colorz['brown']  = 0.612, 0.353, 0.235&#xA;colorz['pink']   = 1.0,   0.0,   1.0&#xA;colorz['gray']   = 0.471, 0.471, 0.471&#xA;colorz['orange'] = 1.0,   0.494, 0.0&#xA;colorz['yellow'] = 1.0,   0.8,   0.0&#xA;&#xA;namez = dict()&#xA;&#xA;namez['red']    = &quot;Satellites (natural)&quot;&#xA;namez['green']  = &quot;Dwarf planets&quot;&#xA;namez['blue']   = &quot;Planets&quot;&#xA;namez['purple'] = &quot;Minor planets&quot;&#xA;namez['brown']  = &quot;Trans-Neptunian Objects&quot;&#xA;namez['pink']   = &quot;Planetoiods&quot;&#xA;namez['gray']   = &quot;Small Solar System bodies&quot;&#xA;namez['orange'] = &quot;Comets&quot;&#xA;namez['yellow'] = &quot;Centaurs&quot;&#xA;&#xA;fname = &quot;Euler_diagram_of_solar_system_bodies.svg.png&quot;&#xA;&#xA;img = plt.imread(fname)&#xA;&#xA;def fakeimg(img, colors, hw):&#xA;&#xA;    img3 = img[..., :3]&#xA;&#xA;    imgnew = np.ones_like(img3)&#xA;&#xA;    for color in colors:&#xA;&#xA;        col = np.array(color)[:3]&#xA;&#xA;        mask = (np.abs(img3-col) &amp;lt; hw).all(axis=-1)&#xA;&#xA;        imgnew[mask] = col&#xA;&#xA;    return imgnew&#xA;&#xA;cs = [colorz[c] for c in ['orange', 'green', 'purple', 'gray']]&#xA;&#xA;imnew = fakeimg(img, cs, 0.01)&#xA;&#xA;if 1 == 1:&#xA;    plt.figure()&#xA;    plt.imshow(imnew)&#xA;    plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2017-02-10T02:28:43.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20009" PostTypeId="1" AcceptedAnswerId="20012" CreationDate="2017-02-10T13:23:10.773" Score="2" ViewCount="300" Body="&lt;p&gt;If planetary rings are maintained by staying within the Roche limit, how is it that  J1407b's rings extend so far? Surely the gravity and tidal forces should be low enough to allow a moon to form.&lt;/p&gt;&#xA;" OwnerUserId="11525" LastEditorUserId="2153" LastEditDate="2017-02-10T19:18:22.513" LastActivityDate="2017-02-10T19:18:22.513" Title="How are J1407b's rings possible?" Tags="&lt;planetary-ring&gt;&lt;roche-limit&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20010" PostTypeId="1" AcceptedAnswerId="21613" CreationDate="2017-02-10T14:15:26.357" Score="1" ViewCount="407" Body="&lt;p&gt;I'm playing around with Sérsic profiles, and would like to retrieve the effective (half-light) radius that I put in by integration of the profile. I'm not managing to, so perhaps someone can help me out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I create the profile using astropy:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from astropy.modeling.functional_models import Sersic1D&#xA;from scipy.integrate import simps&#xA;Sersic = Sersic1D(amplitude=1,r_eff=1,n=1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I can then make a light profile:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x = np.arange(1000)&#xA;I = Sersic(x)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then we can perform an integral over this I(r):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;integral = [simps(I[0:i],x[0:i]) for i in np.arange(len(I)-1)+1]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From which one would finally find the radius at which the enclosed flux is half that of the total:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from scipy.interpolate import interp1d&#xA;&#xA;f_int = interp1d(integral,np.arange(len(I)-1)+1)&#xA;print 'Half radius: ', f_int(integral[-1]/2.)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However, this does not give me the desired result -- the input radius of 1. Why not? Is this a conceptual error or a numerical one - or both?&lt;/p&gt;&#xA;" OwnerUserId="6965" LastEditorUserId="13681" LastEditDate="2017-02-10T19:47:01.957" LastActivityDate="2017-08-11T21:35:52.217" Title="Retrieving input effective radius from Sérsic profile" Tags="&lt;size&gt;&lt;astropy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20012" PostTypeId="2" ParentId="20009" CreationDate="2017-02-10T16:00:43.277" Score="2" Body="&lt;p&gt;There are a few ways that a gigantic ring system outside the Roche limit might be possible.   Ring systems can extend well past the Roche limit if they are very faint or very young.   A faint/light ring system doesn't have enough mass to coalesce and a young system might not have enough time.   Young small moons with lots internal heat of formation can be volcanically active too, feeding a ring system.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn's E ring is well outside it's Roche limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/f/f7/Saturn%27s_Rings_PIA03550.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://upload.wikimedia.org/wikipedia/commons/f/f7/Saturn%27s_Rings_PIA03550.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most likely explanation for it's enormous and very dense rings, however is that J1407b is very young and a lot of it's ring system will likely to coalesce into moons in time.  The breaks in it's ring system suggests that some moons are already clearing paths in it's rings.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's mass is also very significant.   While difficult to estimate, it's thought to be about 10-40 Jupiters in mass.  &lt;a href=&quot;http://www.rochester.edu/newscenter/gigantic-ring-system-around-j1407b/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.  That would give it a mass in the range of 33 to 130 Saturns and a Roche limit some 3 to 5 times (cube root of the mass), distance to Saturn's center, perhaps 6-10 times the distance from the surface surface.   Because it's ring system is about 200 times the size of Saturn's rings, the mass alone wouldn't explains part of it.    Mostly likely it's a combination of very massive planet/perhaps a brown dwarf star and young plant/star with a satellite system still in formation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There might also be a very powerful magnetic field at play that helps prevent the ring debris from forming into satellites, but that's just speculation on my part.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-02-10T16:00:43.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20014" PostTypeId="2" ParentId="20009" CreationDate="2017-02-10T19:16:48.077" Score="2" Body="&lt;p&gt;There are actually two main ways to form rings, which I'll call the bottom-up and top-down approaches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the bottom-up approach, a moon  or other body moves inside the planet's Roche limit, and is torn apart by tidal forces. The resulting particles then settle down into a plane and spread around the planet until a ring is formed, of roughly uniform density. Over time, the ring may dissipate and dissolve, or it may grow larger if more satellites are broken up close to the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top-down approach involves material left over from the beginning of the planetary system. Planets, moons and other bodies form from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Protoplanetary_disk&quot; rel=&quot;nofollow noreferrer&quot;&gt;protoplanetary disk&lt;/a&gt; early in a system's lifetime, and coalesce from smaller fragments. More massive objects rich in volatiles grow bigger and become gas giants. They then further dominate their part of the planetary system, and may accrete more matter after they have finished forming. This matter can create a large ring around the planet, without any need for satellites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the current hypothesis about how J1407b's rings formed. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2015ApJ...800..126K&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kenworthy &amp;amp; Mamajek (2015)&lt;/a&gt; write&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This eclipse and model imply that we are seeing a circumplanetary disk undergoing a dynamic transition to an exosatellite-sculpted ring structure, which is one of the ﬁrst seen outside our solar system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This circumplanetary disk is likely a leftover from the system's original protoplanetary disk. It may also be leading to the formation of an exomoon.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2017-02-10T19:16:48.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20015" PostTypeId="2" ParentId="20010" CreationDate="2017-02-10T19:22:23.693" Score="1" Body="&lt;p&gt;There are a number of conceptual and numeric problems with your code. They contribute, but I haven't found the conceptual problem yet. I will edit this answer if / when I find it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To start, you set the half-light radius for an exponential profile at 1, create a profile out to 1000, and with a resolution of 1. You need more resolution at the centre, and the last 990 or so bins (probably more than that) are useless. I changed your &lt;code&gt;x&lt;/code&gt; to&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x = np.arange(0,10,0.01) # tried with 100, too -- no change&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The half-light radius doesn't change if you go from an upper end of 1000 to 100 or even to 10. It does change quite dramatically, though, if you change the resolution from 1 to 0.1 to 0.01. If we increase it to 0.001, it starts to slowly stabilize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This brings us to the next problem: These are the results I get when I increase the resolution:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1   : 1.518&lt;/li&gt;&#xA;&lt;li&gt;0.1 : 5.139&lt;/li&gt;&#xA;&lt;li&gt;0.01: 42.30&lt;/li&gt;&#xA;&lt;li&gt;0.001:414.0&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You can see the problem: the interpolation counts bins and assumes a width of 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, I did a small test to see what I'd get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;cs = np.cumsum(I) # Array of length I, with cs[n] = sum(I[:n])&#xA;total = cs[-1] # Total flux count&#xA;halflight = np.where(cs &amp;gt; (total / 2.))[0][0] # First entry where I &amp;gt; total / 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Result is 412, consistent with yours. So, your integration and interpolation works. Right now, I think the problem might be in the definition of the half-light radius (missing pre-factor?). You should be able to find that in the astropy documentation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quick check: Changing &lt;code&gt;r_eff&lt;/code&gt; to 10, and integrating out to 100 gives a factor 10 larger half-light radius.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit before submitting: it looks like you're missing a factor of $(\pi / 2)^2$.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastActivityDate="2017-02-10T19:22:23.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20016" PostTypeId="1" CreationDate="2017-02-10T20:15:56.963" Score="3" ViewCount="112" Body="&lt;p&gt;I'm considering a problem involving star S, planet A, and a comet B. The orbits of the planet and comet are eccentric, with eccentricities $e_A$ and $e_B$. I'm trying to calculate the average change in position that A should cause in B. I'm assuming $M_A \gg M_B$, so B should have negligible effect on A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I start using a coordinate system and assume S is at the origin, and A and B have semimajor axes $a_A$ and $a_B$ then A's orbit has polar equation $$r_A(\alpha) = \frac{a_A(1-e_A^2)}{1 - e_A\cos(\alpha)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if the angle between the major axes of A and B (the axes connecting their foci) at S is $\theta$ then B's orbit has polar equation $$r_B(\beta) = \frac{a_B(1-e_B^2)}{1 - e_B\cos(\beta-\theta)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Newton I know that the acceleration on B toward A is given by $$\text{acc}(\alpha, \beta) := \frac{GM_A (\vec{x}_A(\alpha) - \vec{x}_B(\beta))}{\mid\mid\vec{x}_A(\alpha) - \vec{x}_B(\beta)\mid\mid^3}$$ where $\vec{x}_P(\phi)$ is my shorthand for the position of object P, $(r_P(\phi), \phi)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial thought was the position change is given by $$\int_0^{2\pi}\int_0^{2\pi}\text{acc}(\alpha, \beta) \; \mathrm{d}\alpha  \; \mathrm{d}\beta$$ and so the average position change would be $$\frac{1}{4\pi^2}\int_0^{2\pi}\int_0^{2\pi}\text{acc}(\alpha, \beta) \; \mathrm{d}\alpha  \; \mathrm{d}\beta$$ however a colleague suggested that this assumes that the distribution of A and B in their orbits is uniform, which it isn't. So I did some calculations and worked out that object P has probability distribution $$\mu_{P}(\theta) = \frac{(1 - e_{P}^{2})^{3/2}}{2\pi (1-e\cos(\theta))^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Armed with this knowledge I now think that the average effect of A on B is the change in position given by $$\frac{1}{4\pi^2}\int_0^{2\pi}\int_0^{2\pi}\text{acc}(\alpha, \beta)\mu_A(\alpha)\mu_B(\beta) \; \mathrm{d}\alpha  \; \mathrm{d}\beta.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said I'm not sure if I'm missing anything or if my whole approach is off, and so I thought I'd ask here if my reasoning and ultimately my concluding integral seem correct. I'm a pure mathematician so I'm not very comfortable sorting out whether an application of theory has been properly applied to a real world situation. Any input is most welcome.&lt;/p&gt;&#xA;" OwnerUserId="15798" LastEditorUserId="9301" LastEditDate="2017-02-10T20:55:56.597" LastActivityDate="2017-02-10T20:55:56.597" Title="Average effect of a planet on a comet" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;comets&gt;&lt;eccentric-orbit&gt;&lt;mathematics&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20017" PostTypeId="1" CreationDate="2017-02-10T21:32:03.410" Score="2" ViewCount="92" Body="&lt;p&gt;I'm aware of hard radiations from space can wipe out life but I believe that would also causes the planet to be inhabitable for a period of time. Is there any other kind of energetic astronomical event that could destroy life through a transient pulse of radiation (particles or photons) without any &quot;cool down&quot; needed for new life to develop?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The event could be known or theoretical or just plausible that spans entire galaxy if necessary.&lt;/p&gt;&#xA;" OwnerUserId="15800" LastEditorUserId="7982" LastEditDate="2017-02-11T03:28:08.883" LastActivityDate="2017-02-13T16:17:39.740" Title="Is there any plausible astronomical source of energy that could wipe out humanity without rendering the planet inhabitable?" Tags="&lt;galaxy&gt;&lt;space&gt;&lt;life&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20018" PostTypeId="1" CreationDate="2017-02-10T22:56:46.547" Score="4" ViewCount="293" Body="&lt;p&gt;I'm interested lately in Stellarium and the functionalities which it provides especially scripting. I want to make a simulation that shows the celestial sphere with some planets like Venus, Mars, and the Pole-star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would make a kind of animation that starts from a departure point (France) and arrives in (Canada), but in this animation, I couldn't figure out how to synchronise between time and position (Supposing that I'm travelling to Canada at a fixed speed like in a vessel). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've already looked for some examples in the scripts folder and I read the documentation and I found that I can use &lt;code&gt;setObserverLocation()&lt;/code&gt; and maybe &lt;code&gt;setDate()&lt;/code&gt; but I don't know to synchronise between both of them to make the animation realistic.&lt;/p&gt;&#xA;" OwnerUserId="15799" LastEditorUserId="2365" LastEditDate="2017-02-11T00:15:50.210" LastActivityDate="2017-09-08T06:01:40.153" Title="Stellarium simulation" Tags="&lt;earth&gt;&lt;observational-astronomy&gt;&lt;positional-astronomy&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20019" PostTypeId="2" ParentId="19680" CreationDate="2017-02-11T04:27:43.190" Score="1" Body="&lt;p&gt;This is a really good question. Any attempt at a good answer requires breaking the news that in a realistic solar system with some big planets &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_elements&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Keplerian orbit is merely an idealized, mathematical approximation at a particular time&lt;/a&gt;&quot;. If you are writing a computer program and not using ink, feathers, and parchment, it's best to just integrate the appropriate ODEs for the six element state vectors ($x, y, z, v_x, v_y, v_z$) of each body in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several ways to do this. You can use an ephemeris to get the state vectors at one moment in time (set up initial conditions) and integrate yourself, or you can get tables of positions of everything besides your asteroid and interpolate them. Two examples of resources for these are NASA JPL's &lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;Horizons&lt;/a&gt; and the python package &lt;a href=&quot;http://rhodesmill.org/skyfield/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Skyfield&lt;/a&gt; but there are several other options. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do some looking around here and in the Space Exploration stackexchange sites for tags for those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun's motion relative to the solar system's center of mass is not a simple orbit, but reflects the motion of (primarily) Jupiter, Saturn, Uranus and Neptune, and the only way to get it right is to either calculate all of them yourself, or use a pre-calculated ephemeris.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are necessary both to get the position of the sun correctly, and to calculate the gravitational effects of those planets on your asteroid!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember though, that gravity travels at the speed of light, so if you want to do precise calculations you have to at least make a first order approximation at each time step to where the planet &lt;em&gt;would have been in the past&lt;/em&gt; such that the gravity &quot;signal&quot; will have just arrived at your asteroid at the time reflected in your time step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example of the sun's motion with python. Dashed lines represent the equatorial radius of the sun - the motion regularly moves beyond one solar radius from the solar system center of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NLHi6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NLHi6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4gEXM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4gEXM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;from skyfield.api import Loader&#xA;&#xA;load = Loader('~/Documents/fishing/SkyData')&#xA;&#xA;data = load('de405.bsp')  # this one is big, de421 is smaller but more limited years&#xA;ts   = load.timescale()&#xA;&#xA;years  = np.arange(1980, 2030, 0.1, dtype=float)&#xA;&#xA;times  = ts.utc(years, 1, 1)&#xA;&#xA;sun    = data['sun']&#xA;&#xA;sunpos = sun.at(times).ecliptic_position().km&#xA;&#xA;sunlims = 695700. * np.array([-1, 1])[:, None] * np.ones_like(years)&#xA;&#xA;if 1 == 1:&#xA;    plt.figure()&#xA;    for thing in sunpos:&#xA;        plt.plot(years, 1E-06*thing)&#xA;    plt.ylabel('millions of km', fontsize=16)&#xA;    plt.xlabel('year', fontsize=16)&#xA;    for sunlim in sunlims:&#xA;        plt.plot(years, 1E-06*sunlim, '--k')&#xA;    plt.title('J2000 ecliptic position of the Sun (x, y, z)', fontsize=16)&#xA;    plt.show()&#xA;&#xA;&quot;&quot;&quot;&#xA;# you can save as a numpy or json file:&#xA;&#xA;np.save('sunpos', sunpos)  # You can save as .npy if you want to&#xA;&#xA;with open('sunpos.json', 'w') as outfile:&#xA;    json.dump(sunpos.round(1).tolist(), outfile)&#xA;&quot;&quot;&quot;   &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2017-02-11T04:27:43.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20023" PostTypeId="1" CreationDate="2017-02-11T22:53:59.217" Score="1" ViewCount="161" Body="&lt;p&gt;A few weeks ago my daughter phoned me to say she was looking at a very large bright object in the sky in the direction of the Preseli hills. She wondered if it was a planet. I went out to have a look and sure enough this was much brighter and much larger than a star or planet. I looked at it for several minutes and quite suddenly it just, for want of a better phrase, went out. Just completely gone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last night I saw something very similar. Again it was much larger than a star but this time appeared yellowish in colour. I kept looking as I was doing a few small outdoor jobs and then, although I didn't see it happen this time, it suddenly disappeared. I am mystified as to what these lights could have been. Looked on internet and hoped someone else may have seen them but no luck. &lt;/p&gt;&#xA;" OwnerUserId="15807" LastEditorUserId="10437" LastEditDate="2017-02-13T10:18:59.203" LastActivityDate="2017-02-13T10:18:59.203" Title="Strange light in the sky" Tags="&lt;light&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2017-02-15T22:19:10.413" ContentLicense="CC BY-SA 3.0" />
  <row Id="20024" PostTypeId="2" ParentId="16630" CreationDate="2017-02-12T12:26:21.920" Score="3" Body="&lt;p&gt;Another way to put it is to think of why planets or satellites orbiting their parent don't fall into them. In a similar fashion chunks of matter swirl around the &lt;em&gt;black&lt;/em&gt; hole. In process due to the high energy of their inertia, they emit some energy when opposed by other particles. &#xA;So they shed some mass in the form of light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important that we realize that it is energy with the right momentum (both direction and speed) that prepares a particle to escape the clutches of gravity. So if a &lt;em&gt;particle of light&lt;/em&gt; has enough energy and is headed in the &lt;em&gt;right direction&lt;/em&gt;, it &lt;strong&gt;will&lt;/strong&gt; escape from its outer orbit beyond the event horizon, usually in jets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The light emitted from such jets is of immense energy and is usually observed in the gamma spectrum. If you are interested in how these emitted jets can be used to study the black hole or in general high energy particle physics, look up, Very High Energy Gamma Ray Astronomy. There are whole classes of objects that emit VHE gamma photons and other intriguing stuff.&lt;/p&gt;&#xA;" OwnerUserId="15012" LastActivityDate="2017-02-12T12:26:21.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20027" PostTypeId="1" AcceptedAnswerId="20036" CreationDate="2017-02-12T15:28:22.880" Score="1" ViewCount="413" Body="&lt;p&gt;I'm writing an SF story in which a main belt asteroid is diverted into a collision course trajectory with earth (long story), and unknown to us here on earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its H magnitude is 12. As it nears earth, I would assume it gets brighter in V--apparent magnitude. &lt;strong&gt;Considering all the eyes on the sky these days, how bright, apparent mag V, would it get before being discovered here on earth?&lt;/strong&gt; 9 or 10 seem reasonable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if you could give its distance in AU, at discovery, that would be super (trying to find out how much time we have left ;) )!&lt;/p&gt;&#xA;" OwnerUserId="15814" LastEditorUserId="15814" LastEditDate="2017-02-12T16:37:54.577" LastActivityDate="2017-02-13T07:37:03.437" Title="Asteroid apparent magnitude" Tags="&lt;asteroids&gt;&lt;apparent-magnitude&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20029" PostTypeId="1" CreationDate="2017-02-12T20:31:00.210" Score="2" ViewCount="54" Body="&lt;p&gt;When speaking about energy transport, the textbook, physics-101 always lists three possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;radiation&lt;/li&gt;&#xA;&lt;li&gt;convection&lt;/li&gt;&#xA;&lt;li&gt;conduction&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now when studying radiation transport in (stellar) atmospheres and of course astronomy &amp;amp; astrophysics in general, we are always concerned about the photon mean-free path $\lambda_{mfp}$. When $\lambda_{mfp}$ becomes too small compared to a scattering or absorption mean-free path, the equations of radiation transport turn into a diffusion equation for the photon intensity $I_{\nu}$ at frequency $\nu$, so that we get&#xA;$$ \partial_t I_{\nu} \sim \nabla (D \nabla I_{\nu})$$&#xA;where the diffusion coefficient $D$ in the optically thick limit is proportional to density $\rho$ and opacity $\kappa_{\nu}$, $D \sim (\rho \kappa_{\nu})^{-1}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to understand differences of this process to conduction, which is similarly modeled with a diffusion equation, but different underlying transport physics. In solid state physics lectures one often derives the heat and transport of crystals, which then is given by the properties of how well phonons can transport momentum inside the crystal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So finally, my question:&#xA;As we go deep into a star, a white dwarf or a Jupiter-like planet, the efficiency of optically thick photon transport becomes less and less.&lt;br&gt;&#xA;So in our transport equations, do we need to replace photon transport terms at some point brute-force by phonon/sound wave energy transport terms, or do they arise naturally when photons are scattered off phonons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll add a &lt;strong&gt;TLDR&lt;/strong&gt;:&#xA;Is optically thick phonon transport and conduction like in solid state physics sometimes the same?&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditDate="2017-02-13T20:31:16.230" LastActivityDate="2017-02-13T20:31:16.230" Title="To which extent is optically thick photon transport related to conduction?" Tags="&lt;star&gt;&lt;stellar-astrophysics&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20030" PostTypeId="1" AcceptedAnswerId="20031" CreationDate="2017-02-12T22:51:08.113" Score="8" ViewCount="469" Body="&lt;p&gt;From what my non-scientifically-trained mind understands, matter typically enters a black hole via getting caught up in the accretion disc going faster and closer to the black hole until it passes the event horizon, the point of no return. Using Earth as an analogy for a black hole, I assume the accretion disc would orbit around the equator, similar to Saturn's rings, and the matter might eventually crash into Brazil or somewhere along the equator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My main question is, can matter simply do a straight shot into the black hole at a higher latitude, say straight into New York City, avoiding the disc altogether?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can probably tell by my simplistic analogy, I am a complete layperson. While fancy jargon and artistic looking equations might be expected (which is fine), please also dumb it down to an average intelligent business administration major's level of scientific understanding. &lt;/p&gt;&#xA;" OwnerUserId="10117" LastActivityDate="2017-02-13T19:44:03.723" Title="Can matter fly directly into a black hole and avoid the accretion disc?" Tags="&lt;black-hole&gt;&lt;accretion-discs&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20031" PostTypeId="2" ParentId="20030" CreationDate="2017-02-12T23:13:10.380" Score="4" Body="&lt;p&gt;In your analogy: A satellite stays in orbit because it has angular momentum. This always balances the force of gravity, thus when being in the satellite (or in the ISS) it seems as if you're weightless, because force balance. This is very important to understand, because it is the reason that in space all orbital motions can -in principle- go on forever. Because angular momentum is conserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in order to crash into the Earth/BH you need to have very low angular momentum (which is equivalent to having only linear momentum straight down into the BH) or you have to loose the angular momentum somehow.&lt;br&gt;&#xA;Near Earth, &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_station-keeping&quot; rel=&quot;noreferrer&quot;&gt;this happens to the ISS&lt;/a&gt;, because it has a low-earth-orbit between 350-450 km and at this height there are still enough atmospheric particles around to produce drag.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Around black holes it is a very similar thing.&lt;br&gt;&#xA;I will define the Schwarzschild-radius $R_s = \frac{GM}{c^2}$ as a relevant length-scale. Then, as long as we are outside $3 R_s$ distance from the BH, we need, as on Earth, friction to dissipate our angular momentum. In the case of accretion discs that form around BH's this can be accomplished through turbulent dissipation as laid out in the famous paper by &lt;a href=&quot;http://adsabs.harvard.edu/abs/1973A%26A....24..337S&quot; rel=&quot;noreferrer&quot;&gt;Shakura and Sunyaev&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, only very close to the BH, namely at a distance $&amp;lt; 3R_s$, space-time distortions help us to fall into the BH without the need for any additional source of friction. As you see, this is a region very close to the Black Hole and it is misrepresented in popular accounts of BHs that all matter orbiting it will magically fall into it.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2017-02-12T23:13:10.380" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20032" PostTypeId="1" CreationDate="2017-02-13T01:57:07.383" Score="3" ViewCount="2677" Body="&lt;p&gt;I've looked this up, and was wondering the differences between them. Luminosity is the total energy emitted by a source, in watts. However, I'm confused about luminous flux and radiant flux. Since radiant flux is the total energy emitted by an object, would that make luminosity the same as radiant flux? Then how does luminous flux relate to luminosity? Is luminous flux just radiant flux, but only showing the visible light? And this website, &lt;a href=&quot;https://www.ifa.hawaii.edu/~barnes/ASTR110L_S03/inversesquare.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.ifa.hawaii.edu/~barnes/ASTR110L_S03/inversesquare.html&lt;/a&gt;, says lumens are a unit of luminosity, but how if lumens are candela times steradians?&lt;/p&gt;&#xA;" OwnerUserId="15819" LastActivityDate="2019-04-21T15:17:16.463" Title="Difference between luminosity, luminous flux, and radiant flux?" Tags="&lt;star&gt;&lt;light&gt;&lt;luminosity&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20033" PostTypeId="1" CreationDate="2017-02-13T02:06:32.497" Score="0" ViewCount="157" Body="&lt;p&gt;Yesterday while coming out of a restaurant my wife and I noticed what appeared be a satellite passing quite high in the sky. Based on the location, time, and brightness this appears to have been &lt;a href=&quot;http://heavens-above.com/passdetails.aspx?lat=52.2681&amp;amp;lng=-113.8112&amp;amp;loc=Red%20Deer&amp;amp;alt=859&amp;amp;tz=MST&amp;amp;satid=7735&amp;amp;mjd=57796.1104311148&quot; rel=&quot;nofollow noreferrer&quot;&gt;a passing of the GEOS 3 Rocket&lt;/a&gt;. What made this unusual is that when it got low (perhaps 25-30 degrees off the horizon) it suddenly appeared to dart side to side (over 15 degree sweeps) multiple times over perhaps a second, while continuing its general western trend, before vanishing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this all happened rather quickly I had to estimate the distances after the fact using my hands and what few reference points were available. It is certainly possible that it was lower in the sky than my estimate as that is always a tricky part to judge. I'm quite certain of the sweep distance but an answer that could explain a smaller sweep could still be very useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could cause a satellite to appear to move side to side in this fashion?&lt;/p&gt;&#xA;" OwnerUserId="10953" LastEditorUserId="10953" LastEditDate="2017-02-13T16:57:19.610" LastActivityDate="2017-03-15T16:57:58.917" Title="What could have made a satellite appear to move side to side while near the horizon?" Tags="&lt;artificial-satellite&gt;&lt;optics&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20034" PostTypeId="2" ParentId="20023" CreationDate="2017-02-13T04:21:04.803" Score="1" Body="&lt;p&gt;If you have a date and time you can go to &lt;a href=&quot;https://www.flightradar24.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;Flightradar24&lt;/a&gt; and recreate all of the planes that flew over the area. An oncoming large body jet can look a lot like a yellow star if it has its &quot;headlights&quot; turned on, and if it is coming in your direction and descending, it can look surprisingly motionless. Since the light is focused forward, it can suddenly dim if the plane changes direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lovely NATS video &lt;a href=&quot;https://www.youtube.com/watch?v=gmtejbrFte0&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Day Over the UK&lt;/a&gt; shows a substantial amount of East/West air traffic in that area, including several flights off of the main corridor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if you watched it for 20 minutes, this can obviously be ruled out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus and Mars were too close to the horizon at 21:00 GMT in the UK, but I recommend you play with &lt;a href=&quot;https://in-the-sky.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;in-the-sky.org&lt;/a&gt; a bit. Choose &lt;code&gt;Planetarium&lt;/code&gt;. It should know your approximate location based on your internet access so as long as the location is roughly correct the only think you need to do is choose a date and time to recreate, and enjoy! I hear the site works on phones nicely as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UvOaI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UvOaI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Screen shot fro &lt;a href=&quot;https://in-the-sky.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;in-the-sky.org&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2017-02-13T04:21:04.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20035" PostTypeId="2" ParentId="20033" CreationDate="2017-02-13T05:09:09.460" Score="1" Body="&lt;p&gt;Your link shows estimated magnitude, and it gets quite dim when it gets close to the horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ej43B.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ej43B.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; from &lt;a href=&quot;http://heavens-above.com/passdetails.aspx?lat=52.2681&amp;amp;lng=-113.8112&amp;amp;loc=Red%20Deer&amp;amp;alt=859&amp;amp;tz=MST&amp;amp;satid=7735&amp;amp;mjd=57796.1104311148&quot; rel=&quot;nofollow noreferrer&quot;&gt;heavens-above&lt;/a&gt;, (from url: lat=52.2681&amp;amp;lng=-113.8112)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a spacecraft in low earth orbit appears to approach the horizon, the line-of-sight distance from it to you can change from a few hundred to a few thousands kilometers, which is why the brightness in your table changes so much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The path in the sky shown in the link is going to be almost vertical (reaches 70 degrees). Wikipedia says &lt;a href=&quot;https://celestrak.com/satcat/search.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;GEOS-3&lt;/a&gt; is a satellite put into orbit in 1975. These days there is a big effort to de-orbit the &quot;leftovers&quot; so they burn up in the atmosphere (space junk mitigation) but back then it was common for each launch to leave other stuff up there for decades or longer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia says the COSPAR-ID for the actual GOES-3 satellite is 1975-027A. That means it was the 27th internationally recognized launch in 1975. &quot;A&quot; is assigned to the GEOS-3 satellite itself, but if you look up &lt;code&gt;1975-027&lt;/code&gt; in the satellite catalog (SatCat) at &lt;a href=&quot;https://celestrak.com/satcat/search.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;Celestrak.org&lt;/a&gt; you can see there are were four other objects associated with the launch being tracked. One of them has burned up, but the others are still there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can double check if 1975-027B is the one that your link is talking about. Anyway, since they are so old, none of them is capable of a spectacularly sudden engine burn and change in orbit. So I'd have to say that whatever you did see, it isn't associated with the GEOS-3 launch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point to consider; if you just coming out of a restaurant, the chances are low that you could see a 5th magnitude object. Near the horizon the apparent speed would be much slower than that of a satellite overhead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Bbq05.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bbq05.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2017-02-13T05:09:09.460" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="20036" PostTypeId="2" ParentId="20027" CreationDate="2017-02-13T07:37:03.437" Score="2" Body="&lt;p&gt;That's a big asteroid!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recent discoveries of objects of this size are rare, one example was 2013UQ4. We believe that all objects of this size in the main asteroid belt have been discovered. (The discovery of 2013US10, which appeared to be main belt object turned out to be an error, the combination of observations of two different objects) This object has a highly inclined and retrograde orbit. It turned out to be a comet not an asteroid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For context US10 it was at magnitude 19 when discovered. It is unlikely that a large body would remain undiscovered until it was at mag. 10 or 9, especially if it was in the main belt. A very small object that is rapidly brightening could reach these magnitudes before discovery, but not a 10-20km sized object that your story implies. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-02-13T07:37:03.437" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20037" PostTypeId="1" CreationDate="2017-02-13T08:12:55.947" Score="1" ViewCount="50" Body="&lt;p&gt;In this case I'm using a CCD camera mounted telescope pointed at the clouds to take solar spectra and was wondering how the telescope being pointed at the clouds through a double gazing (two layer) glass window would affect the spectra, if at all? &lt;/p&gt;&#xA;" OwnerUserId="15822" LastActivityDate="2017-02-13T16:23:02.633" Title="How does glass affect taking solar spectra?" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20038" PostTypeId="1" CreationDate="2017-02-13T09:57:01.347" Score="1" ViewCount="105" Body="&lt;p&gt;It is my understanding that the Earth's axial precession is such that the direction of the North pole pointing to the celestial sphere traces out a circle with a period of roughly 26,000 years. My first question on reading this was, where is the centre of that circle? In other words, what axis is the Earth precessing about? I had assumed this would be somewhat arbitrary but have since learned that it points towards the north ecliptic pole. Am I then correct to assume that planets naturally tend to rotate in the same plane as their orbits, but the effects of precession caused by other massive bodies cause the axis of rotation to oscillate around the north ecliptic instead?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, then what is going on with Uranus and it's unusually high axial tilt? Is this another phenomenon or does Uranus' rotation axis also precess about it's own ecliptic north pole? Albeit with a much larger amplitude.&lt;/p&gt;&#xA;" OwnerUserId="15823" LastActivityDate="2017-02-13T16:21:44.093" Title="Is the centre of axial precession perfectly perpendicular to the ecliptic?" Tags="&lt;solar-system&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20039" PostTypeId="2" ParentId="19983" CreationDate="2017-02-13T11:35:16.080" Score="0" Body="&lt;p&gt;Geometry of gravitational waves is such, that it stretches and shrinks spacetime differently in perpendicular directions. So the detector notices the difference in flight time between two photons going along different axes of the detector, as the velocity of light is constant. &quot;Stretching&quot; of a photon would mean, that it changes color (wavelength) but not flight time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're interested in the math have a look at &lt;a href=&quot;http://bcc.impan.pl/13Gravitational/uploads/Whelan_geometry.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Geometry of Gravitational Wave Detection&lt;/a&gt; by John T Wheelan.&lt;/p&gt;&#xA;" OwnerUserId="15824" LastActivityDate="2017-02-13T11:35:16.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20040" PostTypeId="2" ParentId="19983" CreationDate="2017-02-13T12:31:45.830" Score="3" Body="&lt;p&gt;I understand your question and wrestled with it myself for some time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that the Michelson-type interferometer setup works is that the travel time of the photons in each arm is &lt;em&gt;much&lt;/em&gt; shorter than the time period of the gravitational wave. Thus the photons are not really stretched by the gravitational wave at all - they are simply emitted into an arm that has changed its apparent length. The effect is different in the two perpendicular arms (one is stretched whilst the other is contracted) and this produces a phase difference in the signal when the beams in the two arms are recombined. The phase difference is modulated with the frequency of the gravitational wave (which is &lt;em&gt;much&lt;/em&gt; lower than the frequency of the laser light).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the LIGO arms are effectively 1120 km long, because the light is bounced backwards and forwards 280 times. Light spends 3.7 micro-seconds in the apparatus. The gravitational waves that are being detected have frequencies below a kHz, so the arm length only changes significantly on times longer than a milli-second. Thus the photons &quot;see&quot; arms of fixed length.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-02-13T12:31:45.830" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20041" PostTypeId="1" CreationDate="2017-02-13T13:46:53.480" Score="-2" ViewCount="1859" Body="&lt;p&gt;Please put the light year also. Its for our report so I need some examples. I've already searched on some books but I don't find anything. &lt;/p&gt;&#xA;" OwnerUserId="15825" LastActivityDate="2017-02-13T15:23:12.860" Title="What are the other examples of medium-sized stars?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;solar-system&gt;&lt;galaxy&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2017-02-15T12:59:56.960" ContentLicense="CC BY-SA 3.0" />
  <row Id="20042" PostTypeId="2" ParentId="20041" CreationDate="2017-02-13T13:56:10.627" Score="0" Body="&lt;p&gt;I think you're looking for something like the &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_analog&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solar Analog&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My impression, by the way, is that the Sun itself is not a medium sized star, but is smaller than average for stars in this galaxy.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-02-13T13:56:10.627" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20043" PostTypeId="2" ParentId="20041" CreationDate="2017-02-13T15:23:12.860" Score="1" Body="&lt;p&gt;1.) I think it's unclear what you're asking. Especially if you ask about the distance in light-years, then ask about the distance in light-years. Not 'light-years'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) As soon as you're clear on what you're asking, you can try to look e.e. &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_nearest_stars_and_brown_dwarfs&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. They don't list masses or sizes, but remember that most of those stars are on the main sequence, so the stellar class informs you about the mass or radius. Distances in Ly are given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) To address @StephenG's comment: Yes the sun is a low-mass star, as e.e. classified on the x-axis of fig. 1 of &lt;a href=&quot;https://www.cambridge.org/core/journals/publications-of-the-astronomical-society-of-australia/article/div-classtitlethe-dawes-review-2-nucleosynthesis-and-stellar-yields-of-low-and-intermediate-mass-single-starsdiv/59DFD9C086C70D153511319ACF918A18&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2017-02-13T15:23:12.860" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20044" PostTypeId="2" ParentId="20017" CreationDate="2017-02-13T16:17:39.740" Score="1" Body="&lt;p&gt;The cockroach and the tardigrade can withstand just about anything.  Next, any radiation blast consisting of wavelengths reasonably attenuated with water wouldn't affect, directly at least, all the deep-sea creatures.&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-02-13T16:17:39.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20045" PostTypeId="2" ParentId="20038" CreationDate="2017-02-13T16:21:44.093" Score="1" Body="&lt;p&gt;The answer lies in the cause of the precession. Spinning objects don't simply precess of their own accord, it is the effect of other bodies acting on the rotation object to induce a torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the precession of the equinoxes, the precession is caused by the other bodies in the solar system, most significantly, the sun and the moon. Because the Earth is slightly oblique (the equatorial radius is greater than the polar radius), there is an equatorial bulge. Further, because of the tilt of the Earth's axis, this bulge is tilted with respect to the ecliptic. The gravitational pull of the sun acts more strongly on the side of the bulge closer to it and creates a small torque. Similar effects take place due to the presence of the moon, the combined effect is known as lunisolar precession. Other near by massive bodies also play a role.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A complete understanding of how/why this torque produces the effects of precession requires an understanding of angular momentum in general and is relatively trivial given that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the question explicitly, the centre of axial precession points roughly along the ecliptic pole because the effect is caused by the sun (which is always on the ecliptic) and the moon (which is very close to the ecliptic).&lt;/p&gt;&#xA;" OwnerUserId="15823" LastActivityDate="2017-02-13T16:21:44.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20046" PostTypeId="2" ParentId="20037" CreationDate="2017-02-13T16:23:02.633" Score="2" Body="&lt;p&gt;you'll need to find the spectral transmissivity of each piece of glass.  Most likely they all block UV and large sections of the IR bands.  Further,  unless you've removed the internal IR filter, your camera won't record any IR.  CCDs also cut off in the blue, so no UV will be recorded.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, how are you separating the wavelengths at the camera?  Typically one uses a (expensive) grating spectrometer.&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-02-13T16:23:02.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20047" PostTypeId="2" ParentId="20030" CreationDate="2017-02-13T19:44:03.723" Score="-1" Body="&lt;p&gt;I suppose its certainly possible, however it's highly unlikely. First of all, you'd need your matter to fall in directly, meaning it would have to be traveling directly at the BH. Any motion slightly to the side, and it will start orbiting and spiral into the black hole. That's a pretty small target so that would already make this scenario not very plausible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, your main problem is going to be the fact that (1) black holes have magnetic fields usually and (2) black holes also usually spin. What this means is that anything trying to get into the black hole will tend to get swept up and taken along with the rotating magnetic field, causing it to start spiraling in. You could of course say that your infalling material is neutral and thus doesn't interact with the field, but there's no way it will stay neutral all the way in. At some point friction from falling in will make it a plasma and it will then be &quot;bound&quot; to the magnetic field. If you're knowledgeable about your magnetic fields, you may suggest making it so this matter could simply fall into the magnetic poles as it wouldn't then wouldn't be subjected to being dragged by the magnetic field. Unfortunately, the poles tend to be the regions where material escapes from the BH so you're unlikely to get infalling material through that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, it &lt;em&gt;might&lt;/em&gt; be possible to contrive some scenario where matter is just &quot;dropped&quot; onto the black hole but the circumstances for such an event to occur are exceedingly rare and will likely be ruined if the black hole is either spinning or has a magnetic field (which almost all do).&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-02-13T19:44:03.723" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20048" PostTypeId="1" CreationDate="2017-02-13T20:15:52.507" Score="2" ViewCount="82" Body="&lt;p&gt;Yesterday night I have tried find comet 45P, that should be still easily visible with my equipment at hand, however I was unable to find it.&#xA;I used data provided by &lt;a href=&quot;https://theskylive.com/45p-info&quot; rel=&quot;nofollow noreferrer&quot;&gt;this website&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today I've checked the website's data against the Minor Planets Center's data loaded into Stellarium (the program) and found the predicted positions of the comet to differ by more than 7°. That is totally unacceptable when trying to find the comet with a FOV of a few tens of arcminutes (the finder presumably is too light-weak).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So does anybody know a definite, reliable source for the position data of 45P? I'm gonna try again today anyway with the positions provided by Stellarium, but it might be good to have another alternative.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2017-02-15T18:03:44.380" Title="Observing Comet 45P - reliable sources of position data" Tags="&lt;amateur-observing&gt;&lt;comets&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20050" PostTypeId="2" ParentId="20048" CreationDate="2017-02-13T23:15:43.100" Score="0" Body="&lt;p&gt;You can try this site&lt;br&gt;&#xA;&lt;a href=&quot;http://www.heavens-above.com/comet.aspx?cid=45P&amp;amp;lat=0&amp;amp;lng=0&amp;amp;loc=Unspecified&amp;amp;alt=0&amp;amp;tz=UCT&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.heavens-above.com/comet.aspx?cid=45P&amp;amp;lat=0&amp;amp;lng=0&amp;amp;loc=Unspecified&amp;amp;alt=0&amp;amp;tz=UCT&lt;/a&gt;&#xA;which agrees with the planetarium program I use.  I think Stellarium is reliable, and there is Cartes du Ciel, which is also free.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-02-13T23:15:43.100" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20051" PostTypeId="1" CreationDate="2017-02-14T00:12:13.270" Score="2" ViewCount="71" Body="&lt;p&gt;I have read that one drawback of Adaptive Optics (AO) IFU observations is that they have lower SNR (compared natural seeing observations) and as a result it is harder to observe faint structures. For example, in a galaxy observation, while AO can give great resolution in the inner parts of the galaxy, the faint outskirts or the disk are more visible with natural seeing observations. Is that true? Why AO are not good with faint structures? I assume there is some loss of light during the whole process involved with AO which results in lower SNR? Can someone give me more details please?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this paper: &lt;a href=&quot;http://adsabs.harvard.edu/abs/2011A%26A...528A..88G&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2011A%26A...528A..88G&lt;/a&gt; The authors say: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the galaxies from the LSD sample we find more complex morphologies&#xA;  which are likely the consequence of the AO-assisted observations&#xA;  providing higher spatial resolution (∼0.2''FWHM) which allows us to&#xA;  spatially resolve more complex structures, but lower sensitivity to&#xA;  extended sources given the higher surface brightness detection&#xA;  threshold.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which implies that the AO are causing lower sensitivity. Maybe the lower sensitivity has something to do with the brightness of the laser used in the AO observations?&lt;/p&gt;&#xA;" OwnerUserId="1986" LastEditorUserId="1986" LastEditDate="2017-02-23T12:16:58.760" LastActivityDate="2017-02-23T12:16:58.760" Title="Adaptive Optics observations worse than natural seeing observations for faint structures?" Tags="&lt;galaxy&gt;&lt;adaptive-optics&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20052" PostTypeId="1" AcceptedAnswerId="20057" CreationDate="2017-02-14T02:02:31.010" Score="13" ViewCount="289" Body="&lt;p&gt;Are black holes expected to contain the same ratio of dark matter to regular matter as the rest of the universe? I've heard that dark matter is distributed in halos around galaxies. Does that make it less likely to be ingested into a black hole?&lt;/p&gt;&#xA;" OwnerUserId="6870" LastActivityDate="2017-02-19T23:04:17.133" Title="Are black holes expected to contain the same ratio of dark matter to regular matter as the rest of the universe?" Tags="&lt;black-hole&gt;&lt;dark-matter&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20053" PostTypeId="1" AcceptedAnswerId="20108" CreationDate="2017-02-14T02:52:16.693" Score="3" ViewCount="322" Body="&lt;p&gt;Suppose I wanted to construct the spectral energy distribution (SED) of an object. Further suppose I observe this object through the 9 broad-band filters using the Subaru Suprimecam (just for an example) here: &lt;a href=&quot;http://www.naoj.org/Observing/Instruments/SCam/sensitivity.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.naoj.org/Observing/Instruments/SCam/sensitivity.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the statement below the table of broad-band filters says, the Johnson-Cousins filters (capital letters) are on the Vega magnitude system whereas the SDSS filters (lowercase letters) are on the AB magnitude system. I understand what these two magnitude systems are and how to convert between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: what is the point of the two sets of filters being on different magnitude systems, one Vega and the other AB? It's not as if the Johnson-Cousins filter &lt;em&gt;hardware&lt;/em&gt; physically has the Vega spectrum built into it, right...? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I do aperture photometry on an I-band image and an i'-band image, will the first one return Vega magnitude of the object and the second return AB magnitude of the object, by default? And if I wanted to do SED fitting, then I need to be consistent such that all of my data points (even from other telescopes) need to be on the same magnitude system (e.g., convert the I-band Vega magnitude into an I-band AB magnitude)? (Or do that conversion first, and only &lt;em&gt;then&lt;/em&gt; convert to flux units via $\nu F_{\nu}$ or $\lambda F_{\lambda}$).&lt;/p&gt;&#xA;" OwnerUserId="607" LastActivityDate="2017-02-20T09:55:15.183" Title="Why do different instrumental filters use different magnitude systems (Vega vs AB)?" Tags="&lt;spectroscopy&gt;&lt;luminosity&gt;&lt;photometry&gt;&lt;apparent-magnitude&gt;&lt;magnitude&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20054" PostTypeId="1" AcceptedAnswerId="20056" CreationDate="2017-02-14T06:41:20.757" Score="9" ViewCount="220" Body="&lt;p&gt;The phys.org article &lt;a href=&quot;https://phys.org/news/2017-02-scientists-huge-dataset-nearby-stars.html&quot;&gt;Scientists make huge dataset of nearby stars available to public&lt;/a&gt; describes the release of a publicly accessible database of Echelle radial velocity measurements; &lt;a href=&quot;https://arxiv.org/abs/1702.03571&quot;&gt;The LCES HIRES/KECK Precision Radial Velocity Exoplanet Survey&lt;/a&gt;. See also Keck's &lt;a href=&quot;https://www2.keck.hawaii.edu/inst/hires/&quot;&gt;HIRES home page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For two decades, these scientists have pointed HIRES at more than 1,600 &quot;neighborhood&quot; stars, all within a relatively close 100 parsecs, or 325 light years, from Earth. The instrument has recorded almost 61,000 observations, each lasting anywhere from 30 seconds to 20 minutes, depending on how precise the measurements needed to be. With all these data compiled, any given star in the dataset can have several days', years', ore even more than a decade's worth of observations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This part caught my interest especially:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;strong&gt;We recently discovered a six-planet system orbiting a star&lt;/strong&gt;, which is a big number,&quot; Burt says. &quot;We don't often detect systems with more than three to four planets, but we could successfully map out all six in this system because we had over 18 years of data on the host star.&quot; (emphasis added)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For very simple cases of one, or maybe two planets with minimal interplanetary gravitational interaction, a Fourier transform of a nice, long, continuous radial velocity measurement would show two main peaks, and possibly other artifacts. If the stellar motion induced by each planet were similar magnitude, the analysis might be fairly simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for the six planet case mentioned in the quote (I don't know which one it is), and patchy time coverage (it's a survey) how was this analysis done? Peaks alone? Or just throw it to a supercomputer simulation of every possible combination and let simulated annealing run for a month?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or was there some 'detective work' involved as well - assumptions, limitations of the fitting space, or even inclusion of other data from outside the study?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2017-02-14T09:34:50.707" Title="Was the discovery of six exoplanets around one star as &quot;easy&quot; as counting six peaks in the FT?" Tags="&lt;exoplanet&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20056" PostTypeId="2" ParentId="20054" CreationDate="2017-02-14T09:34:50.707" Score="9" Body="&lt;p&gt;I suspect that the record holder (as of 14/2/2017) is &lt;a href=&quot;https://en.wikipedia.org/wiki/HD_10180&quot;&gt;HD 10180&lt;/a&gt; which has at least 7 planets and possible evidence for as many as 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/1011.4994&quot;&gt;Lovis et al. (2011)&lt;/a&gt; announced the initial discovery based on 190 radial velocity measurements taken over 6 years. The precision of the measurements was 0.3-0.9 m/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Section 4 of that paper describes how they go about finding the planets in the data. It is a hybrid of fourier and fitting methods. As each peak in the periodogram is found it is added to a model which is then iterated towards a best solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Successive planets reduce the rms scatter around the predicted radial velocity curve. Ultimately a judgement has to be made whether the improvement in the fitting statistic justifies the addition of another planet (and more free parameters) to the model. The final rms after adding 7 planets was just over 1 m/s, which is worse than the expected precision, but which Lovis et al. attribute to (probable) radial velocity jitter due to stellar activity. The orbital model is then refined to include the effects of planet-planet interactions and tidal forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radial velocity amplitudes attributable to each planet range from 0.8-4.5 m/s. The most marginal detection has the smallest amplitude, but the shortest period (more cycles and hence easier to detect smaller amplitudes). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A later paper by &lt;a href=&quot;https://arxiv.org/abs/1204.1254v1&quot;&gt;Tuomi (2012)&lt;/a&gt; used a more conventional Monte Carlo Markov Chain in a Bayesian framework to fit a model of $n$ non-interacting planets to the radial velocity data. Again, there is lots of discussion (see section 3.3 and 3.4) of exactly how many planets are &lt;em&gt;required&lt;/em&gt; to fit the data. Tuomi claims there is strong evidence for an 8th and 9th planet in their analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of important assumptions made in these types of analysis. The main one is that you have to assume some sort of model for the background noise and it is often assumed to be Gaussian and non-periodic.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-02-14T09:34:50.707" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20057" PostTypeId="2" ParentId="20052" CreationDate="2017-02-14T18:59:39.120" Score="13" Body="&lt;p&gt;(Short answer: No, scroll to the last point.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is irrelevant to an external observer whether the matter that fell into the black hole was dark matter or baryonic, by the no hair theorem. The only properties of a black hole from our point of view are mass, electric charge and angular momentum. (But of course we don't understand quantum gravity.)&lt;/li&gt;&#xA;&lt;li&gt;From the point of view of matter which has fallen into the black hole, nothing special happens upon crossing the event horizon. This means that dark matter stays dark and baryonic matter stays baryonic when viewed from inside the black hole.&lt;/li&gt;&#xA;&lt;li&gt;There is some controversy about how much dark matter exists in the universe. &lt;a href=&quot;https://arxiv.org/abs/0903.1962&quot; rel=&quot;noreferrer&quot;&gt;This recent article&lt;/a&gt;, for example, indicates in the abstract that more accurate modelling of galactic rotation curves could eliminate a large percentage of the expected non-baryonic dark matter. (Note as @pela indicated in the comments, that this author's papers have not been peer reviewed and could be suspect.) Obviously, the amount of dark matter in the universe would greatly affect the question's answer. I should note that the controversy is mostly composed of a small number of vocal scientists who appear disproportionately in the media. Following the mainstream news science sections, I get the impression that the death of dark matter seems to be announced once a month or so.&lt;/li&gt;&#xA;&lt;li&gt;The formation of supermassive black holes is poorly understood. One hypothesis is that they may form by successive merger of stellar mass black holes. As there have recently been &lt;a href=&quot;http://www.ligo.org/science/Publication-GW150914/&quot; rel=&quot;noreferrer&quot;&gt;gravitational wave observations of such mergers&lt;/a&gt;, and as candidates for intermediate mass black holes &lt;a href=&quot;http://www.nature.com/nature/journal/v542/n7640/full/nature21361.html&quot; rel=&quot;noreferrer&quot;&gt;have also been observed recently&lt;/a&gt;, I will assume here that this is how they form and that supermassive black holes are therefore made of roughly the same stuff as stellar mass black holes.&lt;/li&gt;&#xA;&lt;li&gt;Black holes lose most of their mass during the formation process. It is important to always keep in mind whether we are talking about the mass of the stellar core which collapsed to form the black hole (this is often the &quot;mass&quot; of a black hole that is referred to when speaking about e.g. the minimum size black hole that can form from core collapse) or the mass of the black hole as seen by a distant observer after the supernova.&lt;/li&gt;&#xA;&lt;li&gt;Dark matter particles can not lose much orbital energy by interacting with other matter nor by radiation, therefore will remain in orbit around a black hole rather than falling in, unless they happen by unlikely chance to hit it near the event horizon. &lt;a href=&quot;https://arxiv.org/abs/0802.2041&quot; rel=&quot;noreferrer&quot;&gt;This paper&lt;/a&gt; indicates that simulated supermassive black holes derive no more than about 10% of their mass from dark matter.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, it must be said that some scientists suspect that dark matter &lt;a href=&quot;https://www.nasa.gov/feature/goddard/2016/nasa-scientist-suggests-possible-link-between-primordial-black-holes-and-dark-matter&quot; rel=&quot;noreferrer&quot;&gt;is made of primordial black holes in the first place.&lt;/a&gt; There is also the theory of MACHOs (&lt;a href=&quot;https://en.wikipedia.org/wiki/Massive_compact_halo_object&quot; rel=&quot;noreferrer&quot;&gt;Massive Compact Halo Objects&lt;/a&gt;), that dark matter is composed of large compact bodies such as black holes, but it is believed by most that this theory can not account for the dark matter in the universe.&lt;/p&gt;&#xA;" OwnerUserId="14969" LastEditorUserId="14969" LastEditDate="2017-02-15T05:09:15.537" LastActivityDate="2017-02-15T05:09:15.537" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20058" PostTypeId="1" AcceptedAnswerId="24194" CreationDate="2017-02-14T23:05:58.510" Score="4" ViewCount="244" Body="&lt;p&gt;Consider a loose pile of rocks in an early solar system, say a tiny asteroid. As it accretes more and more material, the pressure at its center eventually becomes great enough to collapse the gaps between the rocks, which further increases the density, which increases the gravitational force, ... which leads to a runaway process that compacts a loose pile of rubble into a near-spherical body of almost solid rock.  (And some ices and liquids.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But where do the rocks &lt;em&gt;themselves&lt;/em&gt; come from?  Surely at least that much pressure is needed to fuse dust into macroscopic rocks?  Why isn't this a chicken-and-egg problem?&lt;/p&gt;&#xA;" OwnerUserId="15835" LastActivityDate="2017-12-30T21:03:51.980" Title="Where do rocks come from?" Tags="&lt;gravity&gt;&lt;asteroids&gt;&lt;accretion&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20059" PostTypeId="1" AcceptedAnswerId="20060" CreationDate="2017-02-15T00:12:45.777" Score="2" ViewCount="54" Body="&lt;p&gt;Its rotation period is 45 hours. Its satellite is at the distance of 15000 km. Given this data, and assuming the satellite's orbital period the same as the main body's rotation period, 45 hours, what is the mass of the main body?&lt;/p&gt;&#xA;" OwnerUserId="2541" LastEditorUserId="2365" LastEditDate="2017-02-17T01:39:58.170" LastActivityDate="2017-02-17T01:39:58.170" Title="Assuming 2007 OR10 is in tidal lock with its satellite, what is its mass?" Tags="&lt;rotation&gt;&lt;dwarf-planets&gt;&lt;tno&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20060" PostTypeId="2" ParentId="20059" CreationDate="2017-02-15T01:16:13.500" Score="2" Body="&lt;p&gt;We can use Kepler's Third Law, the Law of Periods, to solve this problem. The law can be stated mathematically as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T^2 = \frac{4\pi^2}{G(M+m)} a^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T$ is the orbital period,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G$ is the gravitational constant,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M$ is the mass of the primary,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m$ is the mass of the secondary, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a$ is the semimajor axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $(M + m)$ with the given values returns a &lt;strong&gt;system mass&lt;/strong&gt; of approximately $8 \times 10^{22}$ $\mathrm{kg}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other data are necessary to isolate the mass of the satellite from the primary. &lt;/p&gt;&#xA;" OwnerDisplayName="user14781" LastActivityDate="2017-02-15T01:16:13.500" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20061" PostTypeId="1" CreationDate="2017-02-15T01:34:53.593" Score="2" ViewCount="2008" Body="&lt;p&gt;Stars convert mass into energy. Even converting the slightest mass into energy is immense because the speed of light is so great and $E = mc^2$. This means that stars have to emit large amounts of energy. So can they emit so much energy in a form of electromagnetic radiation that is not visible light? Resulting in a star that cannot be detected using our senses(+ a telescope). If so what problems could this star pose?&lt;/p&gt;&#xA;" OwnerUserId="15836" LastEditorUserId="15836" LastEditDate="2017-02-21T19:40:13.590" LastActivityDate="2017-02-21T19:58:31.567" Title="Can some stars not emit any energy in the visible spectrum?" Tags="&lt;light&gt;&lt;fundamental-astronomy&gt;&lt;space&gt;&lt;spectroscopy&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20062" PostTypeId="2" ParentId="20061" CreationDate="2017-02-15T02:05:29.620" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Light that is not light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's meaningless.  All light is electromagnetic radiation.  A finite part of the infinitely large range of the electromagnetic spectrum is visible light.  So you should talk about EM radiation and to discuss the visible spectrum just say visible spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars emit lots of energy at frequencies that are outside the visible range.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dark Matter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Dark matter does &lt;em&gt;not&lt;/em&gt; mean that it's just not radiating in the visible part of the spectrum.  It means it's not radiating in any part of the EM spectrum much.  If dark matter exists (and that's not certain) it won't interact with most matter much at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If dark matter were in star-like objects, it sounds very unlikely that there would not be any EM radiation as a result of the interactions that would result in a star-like object.  We'd detect it if it were there.  So I don't think that's likely.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-02-15T02:05:29.620" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20063" PostTypeId="1" CreationDate="2017-02-15T02:35:13.310" Score="1" ViewCount="47" Body="&lt;p&gt;I'm just starting a project to analyze some of the images obtained from the Dark Energy Survey (DES), and I'd like to get some of the COADD images from the SV data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, this paper: &lt;a href=&quot;https://arxiv.org/pdf/1512.03062v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/pdf/1512.03062v1.pdf&lt;/a&gt; shows a series of images that are apparently from the 'DESDM database of coadd objects'.  Just looking at image (a) on page 3 of the paper, I can find images in the DES Y1 data from NOAO Science Archive at &lt;a href=&quot;http://archive.noao.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://archive.noao.edu/&lt;/a&gt; that show the objects in image (a) from the paper, but they are individual exposures - I can't find any COADDed images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone tell me how to get these COADD images?&lt;/p&gt;&#xA;" OwnerUserId="15838" LastActivityDate="2017-02-15T02:35:13.310" Title="DESDM COADD image access" Tags="&lt;cosmology&gt;&lt;astrophysics&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20064" PostTypeId="1" AcceptedAnswerId="20066" CreationDate="2017-02-15T04:07:31.977" Score="2" ViewCount="294" Body="&lt;p&gt;Below &lt;a href=&quot;https://astronomy.stackexchange.com/q/20058/7982&quot;&gt;this question&lt;/a&gt; there is a &lt;a href=&quot;https://astronomy.stackexchange.com/q/20058/#comment33190_20058&quot;&gt;comment&lt;/a&gt; that caught my interest:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;First, there's a misconception. Compressing a body does not increase the force of gravity on outer-lying masses. That's Gauss's famous divergence theorem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I understand that one can use the divergence theorem to address (derive) the &lt;a href=&quot;https://en.wikipedia.org/wiki/Shell_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;shell theorem&lt;/a&gt; - one consequence of quick is that a particle inside a spherically symmetric shell experiences no net gravitational force from the shell, but this seems different as it refers &lt;em&gt;changes in average density&lt;/em&gt; of a body and gravitational forces on &lt;em&gt;outer-lying masses&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'd like to understand better: &lt;em&gt;What does Gauss' divergence theorem say about compression of a body under self-gravitation?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference, &lt;a href=&quot;https://astronomy.stackexchange.com/a/8032/7982&quot;&gt;this answer&lt;/a&gt; covers the &lt;a href=&quot;https://astronomy.stackexchange.com/a/8032/7982&quot;&gt;shell theorem&lt;/a&gt; nicely.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2017-02-15T13:27:08.713" Title="What does Gauss' divergence theorem say about compression of a body under self-gravitation?" Tags="&lt;gravity&gt;&lt;planetary-formation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20065" PostTypeId="2" ParentId="20061" CreationDate="2017-02-15T10:25:59.440" Score="4" Body="&lt;p&gt;You mix up a lot of things in your assumptions.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Stars convert mass into energy.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$E = mc^2$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The way stars produce energy is nuclear fusion. One main process there is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Triple-alpha_process&quot; rel=&quot;nofollow noreferrer&quot;&gt;tripple-alpha process&lt;/a&gt;, which releases a net energy of about &lt;span class=&quot;math-container&quot;&gt;$E \sim 7$&lt;/span&gt; MeV (that is not &lt;em&gt;that&lt;/em&gt; much)&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;So can they emit so much energy in a form of light that is not light?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;By &lt;em&gt;light&lt;/em&gt; you are speaking of &lt;strong&gt;visible light&lt;/strong&gt;. But radiation goes over all frequencies and wavelength, not just the one the human eye can see (which is roughly &lt;span class=&quot;math-container&quot;&gt;$\lambda = 400 - 700$&lt;/span&gt; nm).&#xA;Here you want to check out the theory behind &lt;a href=&quot;https://en.wikipedia.org/wiki/Black-body_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;black body radiation&lt;/a&gt;, which describes - more or less - the intensity of the light at different wavelengths/frequencies at a given temperature.&lt;/p&gt;&#xA;&lt;p&gt;For an overwiev, look at this figure. The sun's surface temperature ist about 6000 Kelvin, so you can infer, that it's radiation maximum is in the range, which the human eye would perceive as green.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/bWXBI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bWXBI.png&quot; alt=&quot;Planck's law at different temperatures [from wikipedia]&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10611" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-02-15T10:25:59.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20066" PostTypeId="2" ParentId="20064" CreationDate="2017-02-15T13:27:08.713" Score="5" Body="&lt;p&gt;Gauss's divergence theorem applied to the gravitational field $\vec{g}$ is that&#xA;$$ \oint \vec{g} \cdot d\vec{A} = \int \nabla \cdot \vec{g}\ dV,$$&#xA;where the left hand side is the flux of gravitational field into/out of a closed surface and the right hand side is the integral of the divergence of that field over the volume enclose by the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fundamental definition of how mass produces gravitational field is that&#xA;$$ \nabla \cdot \vec{g} = -4\pi G \rho,$$&#xA;where $\rho$ is the mass density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are at a point &lt;em&gt;outside&lt;/em&gt; the mass then the right hand side of the divergence theorem  becomes a &lt;em&gt;constant&lt;/em&gt;.&#xA;$$ \oint \vec{g} \cdot d\vec{A} = -4\pi G \int \rho\ dV = -4\pi GM.$$&#xA;Therefore the left hand side is also constant when calculated over any surface that encloses &lt;em&gt;all&lt;/em&gt; the mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you allow the mass to contract &lt;em&gt;in a spherically symmetric fashion&lt;/em&gt;, then as the gravitational field on the left hand side is then always in the radial direction then $\vec{g}$ must stay the same at any position in space &lt;em&gt;outside&lt;/em&gt; the mass distribution.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-02-15T13:27:08.713" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20067" PostTypeId="1" CreationDate="2017-02-15T13:47:33.057" Score="9" ViewCount="107" Body="&lt;p&gt;Has the rate of neutrino production in the universe varied over time? Are there more or less being created now against the early days?&lt;/p&gt;&#xA;" OwnerUserId="7579" LastActivityDate="2017-02-15T23:56:05.377" Title="Neutrino producton rate over time" Tags="&lt;cosmology&gt;&lt;stellar-evolution&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20068" PostTypeId="2" ParentId="20032" CreationDate="2017-02-15T15:23:04.320" Score="4" Body="&lt;p&gt;Photometric units can often seem complex at first sight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Luminosity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luminosity refers to the &lt;em&gt;total energy released by an object per unit time&lt;/em&gt;. This quantity is dimensionally equivalent to power, and is therefore measured in joules per second, or watts $(\mathrm{W})$. Another common unit is the solar luminosity $\mathrm{L}_{\odot} = 3.846 \times 10^{26}$ $\mathrm{W}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Luminous Intensity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luminous intensity refers to the &lt;em&gt;wavelength-weighted&lt;/em&gt; power emitted by a source &lt;em&gt;in a particular direction&lt;/em&gt;. Wavelength-weighted is important here. Because the human eye is more sensitive to some wavelengths than others, this weighting is designed to set a standard for perception. This is set at $555$ $\mathrm{nm}$, the wavelength to which the human eye is most sensitive. This unit is called the candela $(\mathrm{cd})$, and is equivalent to a source emitting at $555$ $\mathrm{nm}$ with a power of $\frac{1}{683}$ $\mathrm{W}\cdot \mathrm{sr}^{-1}$. Essentially this is taking into account the fact that concentrating the same amount of light into a smaller solid angle will cause it to appear brighter. Finding the luminous intensity of a source with multiple wavelengths is a little more complicated. Let us define the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I_v$ is the luminous intensity in candelas,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I_e$ is the radiant intensity in watts per steradian, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bar{y}(\lambda)$ is the standard luminosity function, which does the wavelength-weighting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we can define the luminosity intensity of a multi-wavelength source to be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I_v = 683 \int_{0}^{\infty} \bar{y}(\lambda) \cdot \frac{dI_e(\lambda)}{d\lambda}d\lambda$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Luminous Flux&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The luminous flux is equivalent to the total wavelength-weighted power emitted by a source. It is similar to luminosity. It is measured in the lumen $\mathrm{lm}$, which is equivalent to the candela-steradian $(\mathrm{cd}\cdot\mathrm{sr})$. One can imagine this as the human-useful amount of light emitted by a source. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Radiometric Units&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radiometric units are the same, except they refer to the true emitted power, with no wavelength-weighting. &lt;/p&gt;&#xA;" OwnerDisplayName="user14781" LastActivityDate="2017-02-15T15:23:04.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20069" PostTypeId="2" ParentId="20048" CreationDate="2017-02-15T15:50:39.067" Score="1" Body="&lt;p&gt;The best way to get ephemeris of solar system bodies is to use JPL HORIZONS:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ssd.jpl.nasa.gov/horizons.cgi&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You put &quot;Target Body&quot; as 45P, your &quot;Location&quot;, &quot;Time&quot;, and click &quot;Generate Ephemeris&quot;. This will give the RA/DEC information for time by default.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want some advanced data other than the listed, e.g., phase angle, V-magnitude, RA/DEC change rate, etc, you can tune &quot;Table Settings&quot;. I usually skip daytime and airmass &amp;lt; 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can be some mismatch between the ephemeris and real observation. This is not because JPL HORIZONS gives wrong calculation, but because the reported observation data is not perfect. It calculates the orbit of the target from previously reported observational data to IAU MPC. IAU MPC provides the orbital elements inferred from that reported data, and these elements are used to generate ephemeris from JPL HORIZONS. So if the report is slightly imperfect and thus the inferred orbital element differs from the true value by a slight amount, it may cause the mismatch at a time far from the original observation. This is the reason we need more and more data continuously and update the catalog.&lt;/p&gt;&#xA;" OwnerUserId="15846" LastEditorUserId="2365" LastEditDate="2017-02-15T18:03:44.380" LastActivityDate="2017-02-15T18:03:44.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20070" PostTypeId="1" AcceptedAnswerId="20078" CreationDate="2017-02-15T16:32:30.033" Score="7" ViewCount="144" Body="&lt;p&gt;Years ago I came upon a scientific text which mentioned different types of radiation and how they are perceived. One such form of radiation was described (from our perspective and understanding of space-time) to originate from the future and travel backwards through time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot for the life of me remember what this type of radiation was called and I would love to know it.&lt;/p&gt;&#xA;" OwnerUserId="15847" LastEditorUserId="2365" LastEditDate="2017-02-15T18:02:24.000" LastActivityDate="2017-02-16T14:05:40.470" Title="What Types of Radiation Emanate in the Future and are Perceived in the Present?" Tags="&lt;space-time&gt;&lt;radiation&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20071" PostTypeId="2" ParentId="20070" CreationDate="2017-02-15T18:01:54.457" Score="5" Body="&lt;p&gt;Nothing that is known to exist&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relativity does not allow for a massive particle to travel at the speed of light, but it doesn't prevent a particle from travelling faster than light. Such a particle has been called a &lt;em&gt;Tachyon&lt;/em&gt;. No such particle has ever been observed. There are good reasons for believing that they don't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a particle would be extremely strange. The mass of such a particle would not be positive, it would be imaginary! However you could not use such a particle to send a message to the past. The particles can't be localised, which means that you can't detect them as being &quot;at&quot; a particular place at a particular time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, such particles have never been detected, probably don't exist, and couldn't be used to send messages.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;read more at &lt;a href=&quot;http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/tachyons.html&quot; rel=&quot;noreferrer&quot;&gt;http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/tachyons.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-02-15T18:01:54.457" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20072" PostTypeId="1" CreationDate="2017-02-15T18:04:16.150" Score="2" ViewCount="189" Body="&lt;p&gt;It seems a bunch of people, including myself, have got it into our heads to create a 3D visualisation of the stars around us - one where you get to fly around them rather than view them from the viewpoint of Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means, obviously, that I'm limited to stars that have parallax values, which in the case of the &lt;a href=&quot;https://gea.esac.esa.int/archive/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gaia archive&lt;/a&gt; is about 2 million whereas in &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Simbad&lt;/a&gt; 100 000. The reason why I'd like to use both is that Simbad appears to have more useful information, including names and types of star. Also, amazingly, the highest parallax (in milliarcseconds) in gaia is in the 200s, which suggests that alpha centauri is not in the archive! My hunch is that Simbad is the place to go for nearby and named stars, whereas Gaia has higher numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine (perhaps incorrectly) that there is a way to identify a Simbad star on the Gaia database and vice versa. Gaia has a bunch of id values (solution_id, source_id, random_index) whereas on Simbad there's just the one (Identifier). I have the suspicion that Simbad may have more columns I'm not aware of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, my plan would be to query both archives, and weed out multiples based on an identifier of some sort. Maybe there's a third database that links the two? Maybe I'm missing something obvious? Perhaps there's the perfect page out there that tells me precisely how to do this which has eluded my googling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="15848" LastActivityDate="2017-02-15T18:04:16.150" Title="A question about using both the Simbad and the Gaia archive" Tags="&lt;gaia&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20074" PostTypeId="2" ParentId="20067" CreationDate="2017-02-15T22:28:10.353" Score="6" Body="&lt;p&gt;Likely more Neurinos are produced billions of years ago than today because there were more very large stars in young galaxies.   I'm not going to touch the first few seconds after the big bang.   Someone else can if they want, but post star formation, it's a surprisingly simple problem to work out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutrinos are produced primarily in nuclear reactions in stars.   Large stars produce far more than smaller stars.  A star with 10 solar masses, would, very roughly, produce about 10,000 more neutrinos.   Stars that size are uncommon because they only live about 10-15 million years.   Only about 1 in 800 stars are B-type or greater in terms of neutrino production and most of those are only a few solar masses, producing between maybe 50 and 500 times the number of neutrinos our sun produces.   Very high neutrino producing stars are sufficiently rare that they don't throw off the numbers.  (I'm ignoring supernovas, but I'll get to that).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we estimate 100 billion stars in the milky way, the larger ones give off far more neutrinos than our sun and the smaller ones, far fewer.   And our sun gives off 1.79 x 10^38 neutrinos per second.   Doing some very bad estimating, all the stars in the Milky way together give off probably in the range of 1 x 10^49 neutrinos per second.  Figuring our sun, being on the large size, gives off more neutrino to mass ratio than average, but not enormously more).   Again, we're ignoring supernovas.  You'll see why soon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, bad estimate, 1 x 10^49 per second from all the stars in the galaxy, or 3 x 10^56 per year.   Or 1 x 10^58 in about 30 years.   (Why do I care about 10^58 you ask?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because that's the number of neutrinos that the type II supernova observed in 1987 released.  &lt;a href=&quot;https://profmattstrassler.com/2011/09/20/supernovas-and-neutrinos/&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;.    So one type II nova produces the same number of neutrinos as all the stars in the milky way currently produce in about 30 years, give or take.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The milkyway has an estimated &lt;a href=&quot;https://www.nasa.gov/mission_pages/GLAST/science/neutron_stars.html&quot; rel=&quot;noreferrer&quot;&gt;one billion neutron stars&lt;/a&gt; and about &lt;a href=&quot;http://hubblesite.org/explore_astronomy/black_holes/encyc_mod3_q7.html&quot; rel=&quot;noreferrer&quot;&gt;100 million stellar mass black holes&lt;/a&gt;.   Each of those can only form by type 2 supernovae.   Given the age of the Milkyway, about 13 billion years, if we average it out, that's one type II supernovae every 12 years or so.   The current rate is much slower, so the young milky way had significantly more type 2 supernovae than it currently averages, and, it would be consistant that there were significantly more large stars.   Even with significantly fewer smaller and main sequence stars, the large stars and type II nova are the clear winners in neutrino production.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current supernova rate (Type 1 and II) in the milky way is &lt;a href=&quot;http://www.skyandtelescope.com/astronomy-news/milky-way-supernova-rate-confirmed/&quot; rel=&quot;noreferrer&quot;&gt;one every 50 years&lt;/a&gt;, so the rate had to be several times higher when the galaxy was young.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Peak Neutrino production would coincide with peak type II supernova explosions, which would only be 5-15 million years or so after peak very large star formation.   I don't mention Type 1 supernova, because, while more common now, they (I think) produce far fewer neutrinos.  Still a great deal, but well fewer than 10^58, though I had some difficulty finding a specific number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very likely that young spiral galaxies like our milky way produce (very bad guess), maybe 10 times more neutrinos than they produce some 13 billion years after formation, so that would be the most likely answer.   Once large star formation in galaxies was well underway, Neutrino production probably peaked, whether that was when the universe was a few hundred million years old or a couple billion, I don't know, but neutrino production likely peaked early with the formation of many large stars and began a steady gradual decline after the peak. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corrections welcome.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-02-15T23:56:05.377" LastActivityDate="2017-02-15T23:56:05.377" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20076" PostTypeId="2" ParentId="20061" CreationDate="2017-02-16T02:43:39.157" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So can they emit so much energy in a form of light that is not light? Resulting in a star that cannot be detected using our senses(+ a telescope).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Presumably by &quot;light that is not light&quot; you mean light outside of the visible spectrum. This is why the fields of gamma ray astronomy, x-ray astronomy, ultraviolet astronomy, infrared astronomy, microwave astronomy, and radio astronomy exist. Astronomy is not limited to the visible spectrum, and there's lots of interesting stuff to &quot;see&quot; from using a telescope that senses light outside of the visible spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hottest of stars emit most of their radiation as x-rays and ultraviolet light, but they also emit a lot of visible light. These very hot stars are quite large, short-lived, and very rare. The coolest of stars emit most of their radiation in the infrared, but they also emit some light in the visible range. These very cool stars are quite small, very long-lived, and extremely numerous. Their cool temperatures makes them rather hard to see, even in the infrared.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If so what problems could this star pose? Could this be dark matter?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The very smallest of those cool stars, along with black holes, cool neutron stars, brown dwarfs, and rogue planets collectively form a set of candidates for dark matter called &quot;massive astrophysical compact halo objects&quot;, or MACHOs for short. However, theoretical calculations along with observations says that MACHOs can account for at most about 20% of dark matter.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-02-16T02:43:39.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20077" PostTypeId="1" AcceptedAnswerId="20080" CreationDate="2017-02-16T04:17:35.320" Score="10" ViewCount="309" Body="&lt;p&gt;In the interstellar medium, there are several different diffuse phases of gas, distinguished by their density and temperature. Specifically, the cold neutral medium has temperatures from ~50-100 K and the warm neutral medium ranges from 6,000-10,000 K. According to my professor, there have been few gas clouds found with temperatures between these two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know why such a large gap exists?&lt;/p&gt;&#xA;" OwnerUserId="13426" LastEditorUserId="2153" LastEditDate="2017-02-16T21:57:57.073" LastActivityDate="2017-02-17T14:52:58.293" Title="Why are there no ISM clouds with temperatures between 100 and 6,000K?" Tags="&lt;temperature&gt;&lt;interstellar-medium&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20078" PostTypeId="2" ParentId="20070" CreationDate="2017-02-16T14:05:40.470" Score="2" Body="&lt;p&gt;One concept of antiparticles (see Feynman) is that they're regular particles travelling backward in time.  If you go with that, then by extending the meaning of &quot;radiation&quot; to include physical particles -- which is common usage, e.g. $\alpha$ and $\beta $ particles -- then antiparticles such as positrons came from the future.&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-02-16T14:05:40.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20080" PostTypeId="2" ParentId="20077" CreationDate="2017-02-16T20:22:31.650" Score="10" Body="&lt;p&gt;Let $n$, $T$, and $x_i$ be the number density of hydrogen, the temperature of the gas, and $n_i/n$, where $n_i$ is the number density of the $i$th component of the interstellar medium. We can then write the criteria for thermal equilibrium as&#xA;$$n^2\Lambda(n,T,x_i)-n\Gamma(n,T,x_i)\equiv n^2\mathcal{L}=0$$&#xA;where $\Lambda$ and $\Gamma$ and the heating and cooling functions, respectively, and $\mathcal{L}$ is defined through those functions and $n$. If the equilibrium is unstable,&#xA;$$\left(\frac{\partial\mathcal{L}}{\partial S}\right)&amp;lt;0$$&#xA;for entropy $S$. This leads to different instability conditions, termed the &lt;em&gt;isochoric&lt;/em&gt; and &lt;em&gt;isobaric&lt;/em&gt; instabilities (&lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/bib_query?1965ApJ...142..531F&quot; rel=&quot;noreferrer&quot;&gt;Field (1965)&lt;/a&gt;, $\text{Eq } 4a,4b$). These can be determined from the temperature, pressure, and density of the gas (also assuming that the gas can be approximated as an ideal gas).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, $\Lambda$ and $\Gamma$ are complicated to determine, although combinations of power-law and exponentially decaying factors may often be sufficient. An example curve that seems to pop up a lot as an example was calculated by &lt;a href=&quot;http://adsabs.harvard.edu/abs/1972ARA%26A..10..375D&quot; rel=&quot;noreferrer&quot;&gt;Dalgarno &amp;amp; McCray (1972)&lt;/a&gt;, Figure 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tFGfT.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tFGfT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More accurate measurements have been made since then, but the general shape is still applicable. The warm neutral medium occupies the area near the sharp change at around $\sim10,000\text{ K}$, and the cold neutral medium occupies the area in the left of the diagram. Another way to visualize this is on a $\log P/\log n$ diagram like this one (from &lt;a href=&quot;http://w.astro.berkeley.edu/~ay216/08/NOTES/Lecture11-08.pdf&quot; rel=&quot;noreferrer&quot;&gt;these slides&lt;/a&gt;, annotated from &lt;a href=&quot;http://adsabs.harvard.edu/abs/1995ApJ...443..152W&quot; rel=&quot;noreferrer&quot;&gt;Wolfire et al. (1995)&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7lPas.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7lPas.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, the two-phase model is an oversimplification, and the ISM has more distinct components. However, the isochoric/isobaric instabilities still limit the range in which clouds can exist in stable equilibria, and explain the dearth of gas in the relevant temperature range.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let me elaborate on the terms &lt;em&gt;isobaric&lt;/em&gt; and &lt;em&gt;isochoric&lt;/em&gt;. In thermodynamics, it is sometimes convenient to assume that some thermodynamic variables remain constant in a certain situation. &lt;a href=&quot;https://en.wikipedia.org/wiki/Isothermal_process&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Isothermal&lt;/em&gt; processes&lt;/a&gt; occur at constant temperature; likewise, &lt;a href=&quot;https://en.wikipedia.org/wiki/Isobaric_process&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;isobaric&lt;/em&gt; processes&lt;/a&gt; occur at constant pressure and &lt;a href=&quot;https://en.wikipedia.org/wiki/Isochoric_process&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;isochoric&lt;/em&gt; processes&lt;/a&gt; occur at constant volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations for the two instabilities are&#xA;$$\left(\frac{\partial\mathcal{L}}{\partial T}\right)_{\rho}&amp;lt;0\tag{Isochoric}$$&#xA;$$\left(\frac{\partial\mathcal{L}}{\partial T}\right)_{p}=\left(\frac{\partial\mathcal{L}}{\partial T}\right)_{\rho}-\frac{\rho_0}{T_0}\left(\frac{\partial\mathcal{L}}{\partial\rho}\right)_T&amp;lt;0\tag{Isobaric}$$&#xA;In the first, we assume that the cloud is at constant volume, and since the total amount of matter in the system is constant, the (mean) density must also be constant. In the second case, we assume that the cloud is at constant pressure. Perturbations that lead to instabilities thus arise from perturbations of other thermodynamic variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a final note on notation,&#xA;$$\left(\frac{\partial\mathcal{L}}{\partial A}\right)_B$$&#xA;means that we take the partial derivative of $\mathcal{L}$ with respect to $A$ while keeping $B$ constant. It's a common thermodynamical convention.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2017-02-17T14:52:58.293" LastActivityDate="2017-02-17T14:52:58.293" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20081" PostTypeId="1" AcceptedAnswerId="20093" CreationDate="2017-02-17T01:12:06.713" Score="6" ViewCount="152" Body="&lt;p&gt;Los Angeles Times's article &lt;a href=&quot;http://www.latimes.com/science/sciencenow/la-sci-sn-life-asteroid-ceres-20170216-story.html&quot; rel=&quot;noreferrer&quot;&gt;NASA's Dawn mission finds life's building blocks on dwarf planet Ceres&lt;/a&gt; says;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;While the scientists aren’t sure exactly what the compounds are, the fingerprint is characteristic of material containing carbon-hydrogen bonds, and may include components like methyl and methylene.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've located the mentioned Science article &lt;a href=&quot;http://science.sciencemag.org/content/355/6326/719&quot; rel=&quot;noreferrer&quot;&gt;Localized aliphatic organic material on the surface of Ceres&lt;/a&gt; but it is paywalled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was found? Is it some spectroscopic feature? &lt;a href=&quot;https://en.wikipedia.org/wiki/Aliphatic_compound&quot; rel=&quot;noreferrer&quot;&gt;Aliphatic&lt;/a&gt; compounds are hydrocarbons that are not &lt;a href=&quot;https://en.wikipedia.org/wiki/Aromaticity&quot; rel=&quot;noreferrer&quot;&gt;aromatic&lt;/a&gt;. Is there a spectral feature that says &quot;I am not aromatic&quot; specifically, or does this just mean that aromaticity was not detected?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2017-02-18T05:56:58.380" Title="What aliphatic compounds were found on Ceres, and how where they identified?" Tags="&lt;astrochemistry&gt;&lt;ceres&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20082" PostTypeId="1" AcceptedAnswerId="20086" CreationDate="2017-02-17T03:57:52.567" Score="4" ViewCount="469" Body="&lt;p&gt;The BBC article &lt;a href=&quot;http://www.bbc.com/news/science-environment-38937141&quot; rel=&quot;nofollow noreferrer&quot;&gt;Event Horizon Telescope ready to image black hole&lt;/a&gt; describes the &lt;a href=&quot;https://en.wikipedia.org/wiki/Event_Horizon_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Event Horizon Telescope&lt;/a&gt;, a coordinated observing technique with several radio telescope arrays across the globe forming a synthetic aperture with an Earth-sized baseline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\lambda}{r_{Earth}} \sim \frac{r_{Sag A*}}{D_{Sag A*}} \sim  10^{-11}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...when plugging in 1 millimeter for $\lambda$, and with $r_{Sag A*}$ and $D_{Sag A*}$ the radius of, and distance from Earth to Sagittarius A*, the black hole in the center of the Milky Way Galaxy of 20 million km and 26,000 light years, respectively. (values taken from the article). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation I've written shows that millimeter wavelength interferometry with an Earth-sized baseline has the possibility to resolve the existence of some structure with scale of the black hole's event horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is &lt;em&gt;How does the Event Horizon Telescope implement the interferometry?&lt;/em&gt; It would certainly be impossible to bring all signals together to a central site and perform the interference there in real time as down-converted analog signals, and quite difficult/expensive to do it with dedicated, synchronized digital optical fiber lines. Is the massive amounts of data sent as IP packets over the internet to a central correlator (numerical interferometer)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article mentions atomic clocks and lots of hard drives, and I have a hunch these have something to do with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ndfbs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ndfbs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; &quot;The eventual EHT array will have 12 widely spaced participating radio facilities&quot;. From &lt;a href=&quot;http://www.bbc.com/news/science-environment-38937141&quot; rel=&quot;nofollow noreferrer&quot;&gt;BBC&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-03-14T05:58:48.363" LastActivityDate="2019-07-20T00:32:30.537" Title="How does the Event Horizon Telescope implement the interferometry?" Tags="&lt;black-hole&gt;&lt;radio-astronomy&gt;&lt;sagittarius-a&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20083" PostTypeId="1" CreationDate="2017-02-17T05:15:57.127" Score="1" ViewCount="39" Body="&lt;p&gt;The question &lt;a href=&quot;https://astronomy.stackexchange.com/q/20077/7982&quot;&gt;Why are there no ISM clouds with temperatures between 100 and 6,000K?&lt;/a&gt; has an accepted answer of the form:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $n$, $T$, and $x_i$ be the number density of hydrogen, the temperature of the gas, and $n_i/n$, where $n_i$ is the number density of the $i$th component of the interstellar medium. We can then write the criteria for thermal equilibrium as&#xA;  $$n^2\Lambda(n,T,x_i)-n\Gamma(n,T,x_i)\equiv n^2\mathcal{L}=0$$&#xA;  where $\Lambda$ and $\Gamma$ and the heating and cooling functions, respectively, and $\mathcal{L}$ is defined through those functions and $n$. If the equilibrium is unstable,&#xA;  $$\left(\frac{\partial\mathcal{L}}{\partial S}\right)&amp;lt;0$$&#xA;  for entropy $S$.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This leads to different instability conditions, termed the &lt;em&gt;isochoric&lt;/em&gt; and &lt;em&gt;isobaric&lt;/em&gt; instabilities (&lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/bib_query?1965ApJ...142..531F&quot; rel=&quot;nofollow noreferrer&quot;&gt;Field (1965)&lt;/a&gt;, $\text{Eq } 4a,4b$). These can be determined from the temperature, pressure, and density of the gas (also assuming that the gas can be approximated as an ideal gas).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is not helping me &lt;em&gt;understand why there aren't any&lt;/em&gt; beyond saying simply &quot;because they are unstable.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the key is to understand what the &quot;different instability conditions, termed the &lt;em&gt;isochoric&lt;/em&gt; and &lt;em&gt;isobaric&lt;/em&gt; instabilities&quot; refer to - what those terms mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to provide an approximate explanation of these two instability conditions in words that would not presuppose the reader has substantial depth in statistical physics? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There may in fact be an explanation in &lt;a href=&quot;https://physics.stackexchange.com/a/310190/83380&quot;&gt;this answer&lt;/a&gt;, but I need something unwound a bit further. Also possible help in the several answers to &lt;a href=&quot;https://astronomy.stackexchange.com/q/15086/7982&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit:&lt;/strong&gt; I've looked at &lt;a href=&quot;http://w.astro.berkeley.edu/~ay216/05/NOTES/Lecture10.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://crescent.astro.illinois.edu/yhchu/ISM/Lectures/ISM_L14_Pogge.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; as well, and I know there is a region when the line labeled &quot;heating rate&quot; has a negative slope, but I don't understand what this actually means. Is there something like a gravitational effect (triggered by density) or a phase change or chemical reaction that 'triggers' this instability? &lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2017-02-17T09:15:38.467" Title="Explanation of the isochoric and isobaric instability conditions of interstellar medium gas clouds" Tags="&lt;interstellar-medium&gt;&lt;thermodynamics&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2017-02-19T01:56:52.180" ContentLicense="CC BY-SA 3.0" />
  <row Id="20084" PostTypeId="1" AcceptedAnswerId="20085" CreationDate="2017-02-17T05:39:06.677" Score="4" ViewCount="81" Body="&lt;p&gt;If all trojans from the Greek camp and all the trojans of the Trojan camp were combined into two bodies respectively, what sizes and masses these bodies would have? Would they qualify as dwarf planets? Would they be greater than Ceres?&lt;/p&gt;&#xA;" OwnerUserId="2541" LastActivityDate="2017-02-18T12:13:06.613" Title="What would be the sizes of the bodies created from Jupiter's trojans?" Tags="&lt;dwarf-planets&gt;&lt;trojan-asteroids&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20085" PostTypeId="2" ParentId="20084" CreationDate="2017-02-17T11:12:36.963" Score="8" Body="&lt;p&gt;The mass of the Trojans (both camps as far as I know) is estimated at around 0.0001 Earth masses (&lt;a href=&quot;https://arxiv.org/abs/astro-ph/0004117&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference&lt;/a&gt;) (Note that this number is debated). The density of them vary but the highest found is around 2.5 g/cm^3 (&lt;a href=&quot;http://home.dtm.ciw.edu/users/sheppard/pub/Sheppard04JupChapter.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference&lt;/a&gt;) Assuming that the composition is mainly rocky/icy with very little gas, the act of combining all of them should make a rocky/icy object with that mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of Ceres is estimated to be 0.00015 Earth masses with a density of some 2.1 g/cm^3 (&lt;a href=&quot;https://nesf2015.arc.nasa.gov/sites/default/files/downloads/pdf/05.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume that the object made from the Trojans has the same density then they should roughly make an object the same size or slightly smaller than Ceres (about 1 % the mass of the Moon). Make that into two equal bodies and you have half a Ceres per group (so each smaller than Ceres) (assuming that there are the same mass of bodies in each group originally).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that there is much uncertainty associated with this, I would bet that the mass and size would be smaller than this.&#xA;Whether they would qualify as dwarf planets, I'm not sure. But they would have to be captured into the orbit rather than be formed there as I'm pretty sure the gravitational influence of Jupiter would prevent any coalescence and gravitational stability of the Trojans (though I could be wrong on this).&lt;/p&gt;&#xA;" OwnerUserId="14899" LastEditorUserId="14899" LastEditDate="2017-02-18T12:13:06.613" LastActivityDate="2017-02-18T12:13:06.613" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20086" PostTypeId="2" ParentId="20082" CreationDate="2017-02-17T14:46:24.187" Score="4" Body="&lt;p&gt;Assuming I understand your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in actually mentioned in the article, all data collected are stored on hard drives, timestamped with an atomic clock, and then flown to a central location where the interferometry actually takes place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further reading:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://iopscience.iop.org/article/10.1086/684513&quot; rel=&quot;nofollow noreferrer&quot;&gt;R2DBE: A Wideband Digital Backend for the Event Horizon Telescop&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://eventhorizontelescope.org/publications&quot; rel=&quot;nofollow noreferrer&quot;&gt;eventhorizontelescope.org/publications&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user14781" LastEditorUserId="7982" LastEditDate="2019-03-27T07:19:14.093" LastActivityDate="2019-03-27T07:19:14.093" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="20087" PostTypeId="1" CreationDate="2017-02-17T17:27:40.607" Score="9" ViewCount="89" Body="&lt;p&gt;&lt;a href=&quot;http://www3.mpifr-bonn.mpg.de/staff/sbritzen/maser.pdf&quot; rel=&quot;noreferrer&quot;&gt;I was reading that:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, unlike Galactic sources such as W3(OH), the emission is unpolarized and the 1667 MHz line is stronger than the 1665 MHz line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but how is this possible? Does not the 1665 MHz line and 1667 MHz line correspond to a particular transition and dipole moment? Won't emitted photons naturally have a polarization? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, they state:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The characteristics of the λ18 cm OH mega-maser emission differ from those of Galactic maser sources, such as the main line intensity ratio (T1667MHz/T1665 MHz &gt; 1), large linewidth (&gt;100 km s^−1), and unpolarized emission.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But why is unpolarized emission expected from extragalactic masers (arising near AGN) and not galactic ones (arising in circumstellar and interstellar environments)?&lt;/p&gt;&#xA;" OwnerUserId="15865" LastActivityDate="2017-02-17T17:27:40.607" Title="How can maser emission be unpolarized?" Tags="&lt;radio-astronomy&gt;&lt;polarization&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20088" PostTypeId="1" AcceptedAnswerId="20090" CreationDate="2017-02-17T22:09:35.973" Score="3" ViewCount="157" Body="&lt;p&gt;I'm asked first to calculate the critical energy density $\rho_{crit}$ of the universe given certain constants. The value I got was $5500\,\mathrm{MeV}\,\mathrm{m}^{-3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I'm asked to calculate various number densities of particles assuming the universe is made up of a single type. To do this, I'm told I need to set the universe's matter density equal to $\rho_{crit}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to do this since they are in different units. My expression to find $\rho_{crit}$ had a $c^2$ in it, and if I take that out I would have units of matter density. Is that what I'm supposed to do?&lt;/p&gt;&#xA;" OwnerUserId="15870" LastActivityDate="2017-02-18T02:00:30.960" Title="Equating critical energy density to matter density" Tags="&lt;cosmology&gt;&lt;density&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20089" PostTypeId="1" CreationDate="2017-02-17T23:00:57.400" Score="1" ViewCount="219" Body="&lt;p&gt;I have recently gotten into observing the night skies and learning about the stars/ star systems that are visible to me using an app on my phone. For the last month or so Venus and Mars appear in the early evening followed by some of the regular constellations (Orion, the two Canis'....) and later in the night skies Jupiter followed closely by the moon with Saturn peering out before daybreak. Yesterday night at around 20:15hrs I was looking up at the clear skies as I happened to be outside. I took a peek at the bright stars Sirius, Canopus and Rigel before concentrating on the Pleiades. As I was looking at that, one of the stars seemed to disappear in front my eyes. I thought it an illusion and carried on looking and about 6-7secs later a star appeared to the right of where I thought I saw one disappear initially. The light stayed for about 2secs and off it went again. I felt like someone was flashing a torch or even space station personnel having a call and response (baffled). One thing I was sure of was that it wasn't a plane as I am an Aeronautics man by profession and am familiar with all aircraft strobes. Another 6-7secs later it reappeared even further to the right than the second time round. For the next 11mins or so it came on and off at those regular intervals and yet still traveling to the right until it was out of my sight. tried to record a video of it with my phone but it wouldn't pick it up nor could it the other brighter stars. I was almost convinced by the end that it was some celestial matter but for the flashing light at exact intervals. Is it one of satellites in orbit? If so, is it normal for it to be equipped with flashing light/ strobe?&lt;/p&gt;&#xA;" OwnerUserId="15871" LastActivityDate="2017-02-19T10:45:01.943" Title="Travelling, flashing Star?" Tags="&lt;star&gt;&lt;satellite&gt;&lt;flare&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" ClosedDate="2017-02-22T13:45:36.227" ContentLicense="CC BY-SA 3.0" />
  <row Id="20090" PostTypeId="2" ParentId="20088" CreationDate="2017-02-18T02:00:30.960" Score="2" Body="&lt;p&gt;By the mass-energy equivalence: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E = mc^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $E = 5500$ $\mathrm{MeV}$, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m = \frac{E}{c^2} = 9.8 \times 10^{-27}$ $\mathrm{kg}$. &lt;/p&gt;&#xA;" OwnerDisplayName="user14781" LastActivityDate="2017-02-18T02:00:30.960" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20091" PostTypeId="1" AcceptedAnswerId="20094" CreationDate="2017-02-18T02:59:12.943" Score="3" ViewCount="265" Body="&lt;p&gt;According to Einsteins theory of special relativity, when something is travelling close to light speed, travel path is contracted. When the traveller is light itself, does it see the travel path contracted to zero? In another word, does light see length of the whole universe is zero across its travel path? How?&lt;/p&gt;&#xA;" OwnerUserId="15770" LastEditorUserId="15770" LastEditDate="2020-04-28T00:24:54.720" LastActivityDate="2020-04-28T00:24:54.720" Title="What is length of the universe on lights perspective" Tags="&lt;light&gt;&lt;space-time&gt;&lt;general-relativity&gt;&lt;special-relativity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="20092" PostTypeId="1" AcceptedAnswerId="20097" CreationDate="2017-02-18T03:55:39.497" Score="1" ViewCount="497" Body="&lt;p&gt;I was wondering how I could get the sun's azimuth and elevation values to shade a particular window with a roller shutter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first I thought azimuth is good enough; of course, it wasn't, as I found out in practice.&#xA;While there are calculators to find a and e for a specific location at a certain date, I lack the maths to figure out when the sun will hit a western window.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The window faces 270 deg, I know how far the eaves stick out, and the top and bottom edges of the window (and of course location).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get the idea the maths will be complicated... however, (it won't matter much) as I want to program this into a controller to operate the shutter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints appreciated.&lt;/p&gt;&#xA;" OwnerUserId="15872" LastActivityDate="2017-02-18T09:22:21.167" Title="Determine sun azimuth and elevation for shading a window throughout the year" Tags="&lt;the-sun&gt;&lt;azimuth&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20093" PostTypeId="2" ParentId="20081" CreationDate="2017-02-18T05:40:53.113" Score="5" Body="&lt;p&gt;OK, I did a bit of research on this.   The &lt;a href=&quot;http://www.latimes.com/science/sciencenow/la-sci-sn-life-asteroid-ceres-20170216-story.html&quot; rel=&quot;noreferrer&quot;&gt;LA Times article&lt;/a&gt; you quoted says something a little bit useful.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;While the scientists aren’t sure exactly what the compounds are, the&#xA;  fingerprint is characteristic of material containing carbon-hydrogen&#xA;  bonds, and may include components like methyl and methylene.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Methyl means a carbon with at least 3 hydrogen bonded to it.  Because carbon has 4 bonds, that 4th bond can be part of a chain, so a methyl group offers a very large body of molecules to choose from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YIlOE.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YIlOE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://study.com/academy/lesson/methyl-group-structure-formula.html&quot; rel=&quot;noreferrer&quot;&gt;More info&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Methylene_group&quot; rel=&quot;noreferrer&quot;&gt;Methlyene&lt;/a&gt; is a carbon with 2 hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://theconversation.com/organic-molecules-found-on-giant-asteroid-ceres-why-thats-such-a-huge-deal-73147&quot; rel=&quot;noreferrer&quot;&gt;article&lt;/a&gt;, which is a pretty good summary, says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The organic molecules that have been detected on Ceres are complex&#xA;  aliphatic compounds that seem to be almost tar-like in nature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tar is a pretty nonspecific term, tar-like even more so, but tar is made by cooking coal or wood without oxygen, so I would assume this implies it's not Oxygen rich.  More &quot;simple&quot; complex hydrocarbon chains (and because &quot;simple&quot; complex is somewhat confusing, another way to say that it's probably mostly Cs and Hs in a fairly big molecule, probably 5, maybe 6 carbons minimum I would think, otherwise it would be too quickly sublimated into gas.  Tar like aliphatic compounds would still be vulnerable to photo-disintegration by UV light from the sun, which is common enough on Ceres Surface, but would be high enough melting points to not evaporate somewhat rapidly like shorter chains would.   That these compounds were found locally probably implies somewhat recent, localized hydro-thermal activity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Aromatic C-H bond and the Aliphatic C-H bond have different spectroscopy.   &lt;a href=&quot;http://www.800mainstreet.com/irsp/eir.html&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a weak alkene CH stretch above 3000 cm-1. This comes from the&#xA;  C-H bonds on carbons 1 and 2, the two carbons that are held&#xA;  together by the double bond. The strong CH stretch bands below 3000&#xA;  cm-1 come from carbon-hydrogen bonds in the CH2 and CH3 groups.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The IR spectrum for benzene, C6H6, has only four prominent bands&#xA;  because it is a very symmetric molecule. Every carbon has a single&#xA;  bond to a hydrogen. Each carbon is bonded to two other carbons and the&#xA;  carbon-carbon bonds are alike for all six carbons. The molecule is&#xA;  planar. The aromatic CH stretch appears at 3100-3000 cm-1 There are&#xA;  aromatic CC stretch bands (for the carbon-carbon bonds in the aromatic&#xA;  ring) at about 1500 cm-1. Two bands are caused by bending motions&#xA;  involving carbon-hydrogen bonds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, less than 3,000 cm wavelength implies what the LA-Times article said, Methyl and Methylene or CH3 and CH2 groups on fairly long chain carbon molecules, probably tar-like.   But that's hundreds, perhaps thousands of molecules that fit those criteria, some with single bonds between carbons, some with double bonds, some maybe more complex and from all the articles I've read, they don't know which ones.  They can't tell which hydrocarbons by orbiting spectroscopy, they can only see the signature of the CH3 and/or CH2 groups at the end of the chains.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To know which molecules, they'll need to test a sample.    So, a mars-like rover is probably going to go to Ceres at some point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why this is interesting is that these molecules aren't that long lived on the surface of Ceres, because UV light form the sun destroys them and they're location implies they came from inside Ceres, probably by hydro-thermal activity.   There's also ammonia rich clay and carbonate rich clay and those are 3 key ingredients to life.    None of this is proof that there's life underground in Ceres, but the ingredients are all there (or mostly there).  Warm water (below the surface - and Ceres has a lot of internal water based on it's density), and the basic building blocks.   It's still a far cry from proof of life as nothing has been found on the surface of Ceres that doesn't exist in comets, other than the implication of underground liquid water.   But having all the ingredients in the same place is interesting.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small point to add, but the &lt;a href=&quot;http://theconversation.com/organic-molecules-found-on-giant-asteroid-ceres-why-thats-such-a-huge-deal-73147&quot; rel=&quot;noreferrer&quot;&gt;article&lt;/a&gt; above mention Ammoniated Clay and Carbonates.  Both of those are building blocks.  Neither is organic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This is strengthened by the fact that the molecules are found together&#xA;  with carbonates and clays containing ammonia. These have been observed&#xA;  in many regions of Ceres&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.reference.com/science/carbonates-a2c20e20f893c085&quot; rel=&quot;noreferrer&quot;&gt;Carbonates&lt;/a&gt; are simple.   A molecule or molecules bound to a CO3(2-) is a carbonate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QONGY.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QONGY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ammoniated clay took some looking up, but I found &lt;a href=&quot;http://news.nationalgeographic.com/2015/11/151110-ceres-dawn-ammonia-outer-solar-system-nasa/&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The orbiting Dawn spacecraft has a more ideal perch to observe how&#xA;  molecules on the dwarf planet’s surface reflect various wavelengths of&#xA;  light. It was in those wavelengths that Dawn investigator Carle&#xA;  Pieters and her colleagues spied the signature of &lt;strong&gt;ammoniated&lt;/strong&gt;&#xA;  &lt;strong&gt;phyllosilicates&lt;/strong&gt;, which are minerals similar to clays on Earth, mixed&#xA;  in with other materials.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.tulane.edu/~sanelson/eens211/phyllosilicates.htm&quot; rel=&quot;noreferrer&quot;&gt;Phyllosilicates&lt;/a&gt; are just clay or dissolved rock, with one or more OH- molecules.   Ammoniated Phylosilicates, instead of having an OH- (&lt;a href=&quot;https://en.wikipedia.org/wiki/Hydroxy_group#Occurrence&quot; rel=&quot;noreferrer&quot;&gt;Hydroxy group&lt;/a&gt;) have one or more NH2- (&lt;a href=&quot;https://chem.libretexts.org/Core/Organic_Chemistry/Amides&quot; rel=&quot;noreferrer&quot;&gt;amides&lt;/a&gt;) or perhaps NH4+ (&lt;a href=&quot;https://en.wikipedia.org/wiki/Ammonium#Ammonium_salts&quot; rel=&quot;noreferrer&quot;&gt;ammonium&lt;/a&gt;), which is a different charge so that wouldn't replace the OH but might still bond with silicates with hydro-thermal mixing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this &lt;a href=&quot;https://en.wikipedia.org/wiki/Ammonium#Ammonium_salts&quot; rel=&quot;noreferrer&quot;&gt;article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The researchers also identified a mixture of minerals on the surface&#xA;  of Ceres, which they think are ammonia-bearing clay minerals and&#xA;  magnesium carbonate. The clay minerals could have been produced by&#xA;  silicates reacting with ammonia ice. However, if Ceres had formed&#xA;  where it is now, it would not have been able to pick up any ammonia&#xA;  ice to enable such a reaction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ceres ammonia clay implies ammonia rich waters under it's surface, so Ceres may have formed further away from the sun than it's current position, perhaps in the Kuipebelt where ammonia ice can exist and be part of a planet or moon's formation.  Ceres lower density than most asteroid belt objects is also evidence of that.   While it's important that Ceres has abundant ammonia, where it formed isn't relevant to this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somewhat complex Chemistry happening inside a planet with heat, pressure and liquid water isn't surprising.   None of these discoveries are even close to proof that Ceres has life, but they indicate that Ceres has all the basic building blocks of life.  Bound Ammonia, hydrocarbons, carbonates and liquid water.   It's still a huge step from building blocks to life, but these ingredients suggest it's worth taking a closer look.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corrections are welcome.   That's all I could deduce from reading the articles. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-02-18T05:56:58.380" LastActivityDate="2017-02-18T05:56:58.380" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20094" PostTypeId="2" ParentId="20091" CreationDate="2017-02-18T06:05:54.647" Score="6" Body="&lt;p&gt;Sort of.  The Lorentz factor is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whereupon a stationary object in the stationary reference frame of length $L$ has a length of $L' = \frac{L}{\gamma} = L\sqrt{1-\frac{v^2}{c^2}}$ in the moving reference frame.  As the velocity $v$ increases toward $c$, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\lim_{v \to c} L' = L \lim_{v \to c} \sqrt{1-\frac{v^2}{c^2}} = 0&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is to say, if a spaceship were to move very rapidly from here to the center of the galaxy and back again, a clock aboard that spaceship would advance much less than an initially synchronized clock here at home.  From our point of view, the reason is that the clock on board the spaceship slowed down (time dilation).  From the point of view of someone on the spaceship, the reason is that the distance to the center of the galaxy was shortened (length contraction).  And in the limit (which cannot actually be reached by anything with mass), the clock on the spaceship becomes arbitrarily slow from our point of view, while the distance to the center of the galaxy becomes arbitrarily short from their point of view.  Of course, we can't actually ask the photon how it experiences the trip (even in principle), since the trip is instantaneous for it.  So the way to express it is to say what the limiting behavior is, as $v$ approaches $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia has a useful article on &lt;a href=&quot;https://en.wikipedia.org/wiki/Special_relativity&quot; rel=&quot;noreferrer&quot;&gt;special relativity&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="15753" LastActivityDate="2017-02-18T06:05:54.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20097" PostTypeId="2" ParentId="20092" CreationDate="2017-02-18T09:22:21.167" Score="1" Body="&lt;p&gt;You can use a package such as the pyephem package for python. It can calculate the position of any astronomical object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However for the sun since it moves along the ecliptic, roughly at a uniform rate, you can approximate the position by doing coordinate transformations. The sun's longitude moves by 365.24/360 degrees a day from 0 at the vernal equinox and has an ecliptic latitude of 90 degrees. This can be converted into equatorial coordinates (right ascension and declension), and then, given the hour, the alt and az positions can be found. The formulae can all be found on &lt;a href=&quot;https://en.wikipedia.org/wiki/Celestial_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Celestial_coordinate_system&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-02-18T09:22:21.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20098" PostTypeId="1" CreationDate="2017-02-18T13:42:31.060" Score="1" ViewCount="68" Body="&lt;p&gt;I'm interested in the Alfonsine tables published in Paris in the 1400s. I know there is an English translation of what survived of earlier Alfonsine tables from Castile, but that's not what I'm interested in. If you know of software that implements the algorithms implied by the table, that would be good too.&lt;/p&gt;&#xA;" OwnerUserId="7385" LastActivityDate="2017-02-18T13:42:31.060" Title="Is there an English translation of the Parisian Alfonsine tables" Tags="&lt;coordinate&gt;&lt;ephemeris&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20099" PostTypeId="2" ParentId="14258" CreationDate="2017-02-18T14:21:01.513" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;1) In the early universe 13 billion years ago when the star first emitted the photon, there was no Hubble Telescope&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No thats not the case. Theoretically, in photons perspective there will be no present, past and future just like there will be no space across the journey. All the events what we call present, past and future in our inertial reference frame perspective exist altogether in reality. So Hubbles telescope is already there. :)&lt;/p&gt;&#xA;" OwnerUserId="15770" LastEditorUserId="15770" LastEditDate="2020-04-30T10:38:46.837" LastActivityDate="2020-04-30T10:38:46.837" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="20100" PostTypeId="2" ParentId="7781" CreationDate="2017-02-18T15:01:52.643" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If you consider only GR. &lt;strong&gt;Yes, A have to wait 13by to actually see B&lt;/strong&gt;. But optionally, we know the universe is expanding. Hence, when we consider SR though the age of the universe is 13.8by, its expected radius is about 40b+ light year right now. So, an observer at A will have to wait actually more than 13by.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the universe, all the events and spaces, what we call present, past and future exist altogether. Time is just an abstract label we put for our convenience in the fabric of cosmos. For example, from lights perspective, there is no start and end point of across a travel path, because each and every start and end point across the journey is contracted essentialy at a same point.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you are interested more, you can watch &lt;a href=&quot;https://www.youtube.com/watch?v=Zuedaptw73w&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video&lt;/a&gt; by Brian Greene to understand the spooky reality.&lt;/p&gt;&#xA;" OwnerUserId="15770" LastEditorUserId="15770" LastEditDate="2020-04-28T00:18:41.937" LastActivityDate="2020-04-28T00:18:41.937" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="20102" PostTypeId="2" ParentId="20089" CreationDate="2017-02-19T10:45:01.943" Score="1" Body="&lt;p&gt;It is highly likely that what you have seen is an Iridium communications satellite. Because of the material covered, it can reflect sun light and appear very bright and sometimes flashing. &lt;/p&gt;&#xA;" OwnerUserId="14307" LastActivityDate="2017-02-19T10:45:01.943" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20107" PostTypeId="2" ParentId="20052" CreationDate="2017-02-19T23:04:17.133" Score="2" Body="&lt;p&gt;Dark matter is (thought to be) in halos which extend both to the centers of galaxies &lt;em&gt;and&lt;/em&gt; outside most of the normal matter in galaxies (gas, stars, dust). So a black hole inside a galaxy &lt;em&gt;could&lt;/em&gt; and undoubtedly will ingest some dark matter. However:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stellar-mass black holes form from the core-collapse of a massive star. Since stars are almost entirely of regular matter, the initially formed BH remnant would itself have been made almost entirely of regular matter. Such BHs might later grow by accreting gas (e.g., from a close binary companion star), in which case they're gaining mass in the form of regular matter. There would inevitably be &lt;em&gt;some&lt;/em&gt; dark matter swallowed by the BH as it orbited within its parent galaxy -- just as the BH would swallow some interstellar dust, for example. But it would still be overwhelmingly formed out of regular matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supermassive black holes in galaxy centers would probably start out from some kind of early-universe collapse of a gas cloud or very massive star, which would again be mostly regular matter. Subsequent growth of supermassive BHs comes primarily from interstellar gas feeding an accretion disk around the BH, plus the occasional star that wanders too close -- so once again it's mostly regular matter that falls in the black hole. (The central regions of galaxies &lt;em&gt;do&lt;/em&gt; have some dark matter, but they're dominated by regular matter. Plus, regular matter in the form of gas clouds can easily lose energy via cloud-cloud collisions and sink to the center of the galaxy, where it could feed a supermassive BH; dark matter can't do this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, as user25972 points out, it's largely irrelevant to outsiders like us &lt;em&gt;what&lt;/em&gt; kind of matter goes into making a BH. A black hole formed out of dark matter would behave identically to one formed out of regular matter.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2017-02-19T23:04:17.133" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20108" PostTypeId="2" ParentId="20053" CreationDate="2017-02-19T23:59:52.650" Score="2" Body="&lt;p&gt;I'd guess it's largely a matter of historical accident and inertia, mixed with a certain amount of taste.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vega magnitudes are the traditional system, based on how bright the star Vega is within the filter (+ detector response, etc.). The Johnson and Cousins filter systems were defined back in the early 1950s when Vega magnitudes were the default, and were to some extent meant to reproduce earlier photographic magnitudes. The AB system was proposed a bit later (1960s) and was originally used for narrow-band photometry and spectrophotometry, not broad-band photometry like the Johnson-Cousins and similar filters. Some more modern filter systems -- such as the SDSS filter system -- have been defined using AB instead, because the creators preferred it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some filter systems have both definitions. For example, the near/mid-IR filters of the IRAC camera on the &lt;em&gt;Spitzer Space Telescope&lt;/em&gt; have both Vega and AB definitions; some papers use one, some the other. (Sometimes papers that are part of the same overall project.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If I do aperture photometry on an I-band image and an i'-band image, will the first one return Vega magnitude of the object and the second return AB magnitude of the object, by default?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By default, no. Doing aperture photometry implies choosing a calibration: that is, a conversion between observed counts and magnitudes. (Such a calibration might already be provided in an image header, e.g., from SDSS.) It's certainly possible to do aperture photometry on an I-band image &lt;em&gt;calibrated&lt;/em&gt; to i-band, and vice-versa. I have on occasion done things like aperture photometry on &lt;em&gt;R&lt;/em&gt;-band images calibrated to &lt;em&gt;K&lt;/em&gt;-band, though I certainly wasn't expecting things to be very accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;if I wanted to do SED fitting ...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the answer depends on what form your model SED is in. If for some reason your SED is in AB magnitudes, then, yes, you'd want to convert all your observed data to AB magnitudes. If, on the other hand, your model is in flux units, then you'd want to convert your data to flux units (either Vega mag to flux units or AB mag to flux units).&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2017-02-20T09:55:15.183" LastActivityDate="2017-02-20T09:55:15.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20109" PostTypeId="1" AcceptedAnswerId="20110" CreationDate="2017-02-20T02:07:44.770" Score="1" ViewCount="915" Body="&lt;p&gt;I'm working with galaxy spectral templates (e.g., Bruzual &amp;amp; Charlot 2003) which seem to always come with y-axis units of $L_{\odot}$/A and x-axis units of Angstroms. Thus the y-axis is a &lt;em&gt;luminosity density&lt;/em&gt; instead of a flux density. In contrast, observationally, we tend to always work with spectra that have y-axis units of flux density ($F_{\lambda}$ in erg/s/cm$^2$/A or $F_{\nu}$ in erg/s/cm$^2$/Hz). Similarly, spectral energy distributions (SEDs) from photometry tend to have $\lambda F_{\lambda}$ or $\nu F_{\nu}$ such that the y-axis is flux, not flux density. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How do I convert a theoretical template spectrum from units of luminosity density ($L_{\odot}$/A) to flux density (erg/s/cm$^2$/A)?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For context, I want to fit spectral templates to an observed SED. The observed SED is for an object at a redshift $z$, so I think I can either convert the templates to flux density units, or I can convert my observed SED to luminosity density units. I feel like working in flux density units is more natural -- plus I'm not sure if multiplying the observed SED y-axis values by $4\pi D^2$ (D is distance of object) and x-axis (wavelengths) by $1/(1+z)$ would be sufficient (e.g., normalization concerns).&lt;/p&gt;&#xA;" OwnerUserId="607" LastActivityDate="2017-02-20T03:19:56.913" Title="How to convert theoretical template spectrum from luminosity density to flux density units?" Tags="&lt;galaxy&gt;&lt;redshift&gt;&lt;spectroscopy&gt;&lt;spectra&gt;&lt;luminosity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20110" PostTypeId="2" ParentId="20109" CreationDate="2017-02-20T03:19:56.913" Score="1" Body="&lt;p&gt;Use this equation:&#xA;$$F_\nu = \frac{L_\nu}{4\pi D_L^2}.$$&#xA;This is the relationship that defines the luminosity distance, $D_L$, in a static Euclidean universe (i.e. not ours). Hogg has a good review on &lt;A HREF=&quot;https://arxiv.org/abs/astro-ph/9905116&quot; rel=&quot;nofollow noreferrer&quot;&gt;the arXiv&lt;/A&gt; of how to handle the relationship in an expanding universe, including when you need to perform something called a &quot;$K$-correction&quot; for spectral densities like the question asks.&lt;/p&gt;&#xA;" OwnerUserId="14244" LastActivityDate="2017-02-20T03:19:56.913" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20111" PostTypeId="1" CreationDate="2017-02-20T06:39:09.337" Score="2" ViewCount="211" Body="&lt;p&gt;Am new to this field. My son is 9 years old and extremely interested in viewing the Solar system. I do not want to buy a low quality one. I have no idea what to buy, I will have to buy it from Dubai as good quality telescopes are not available where I live.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please can you help me to choose the right tool. He wants to view as many planets as possible. What would be the right amount to pay in US dollars ?&lt;/p&gt;&#xA;" OwnerUserId="15886" LastEditorUserId="2365" LastEditDate="2017-05-01T14:06:09.667" LastActivityDate="2017-05-31T14:29:13.600" Title="Telescope Specs" Tags="&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20113" PostTypeId="1" CreationDate="2017-02-20T18:50:15.957" Score="1" ViewCount="684" Body="&lt;p&gt;I'm trying to get an equation for the energy density of matter of the universe $\rho(t)$, assuming the number of particles is conserved and rest mass energy is much greater than kinetic energy. $\rho(t)$ should be a function of the scale factor of the universe $a(t)$, mass of particles $m$, and the number density of the particles $n(t_0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to approach this since $\rho(t)$ is in units of $\frac{Mass}{Length^3}$, $m$ is in units of $\frac{Mass}{Particle}$, $n(t_0)$ is in units of $\frac{Particle}{Length^3}$, and $a(t)$ is in units of $Length$. I guess there could be a factor of $\frac{a(t)}{a(t_0)}$ or $\frac{a(t_0)}{a(t)}$, but I don't know where I'd get that from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT! I forgot that $a(t)$ is dimensionless.&lt;/p&gt;&#xA;" OwnerUserId="15870" LastEditorUserId="15870" LastEditDate="2017-02-20T19:52:12.043" LastActivityDate="2017-02-20T22:13:12.707" Title="Finding equation for energy density of matter" Tags="&lt;universe&gt;&lt;cosmology&gt;&lt;matter&gt;&lt;density&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20114" PostTypeId="2" ParentId="20113" CreationDate="2017-02-20T22:13:12.707" Score="2" Body="&lt;p&gt;Energy density $\rho \sim nmc^2$ if kinetic energy is unimportant (and you are ignoring dark energy and other forms of energy density).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the number &lt;em&gt;density&lt;/em&gt; of particles varies as $a^{-3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&#xA;$$ \frac{\rho}{\rho_0} = a^{-3},$$&#xA;where $\rho_0 = n(0)mc^2$ is the present day energy density and the present day scale factor is 1.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-02-20T22:13:12.707" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20115" PostTypeId="1" AcceptedAnswerId="21460" CreationDate="2017-02-20T23:53:28.413" Score="9" ViewCount="7536" Body="&lt;p&gt;In Disney's &quot;Moana&quot;, we get a hint of a Polynesian wayfinding technique of holding your hand to &quot;measure the stars&quot; (note, you're not giving the sky a high-five). We get to see two different constellations measured: Crux (the Southern Cross), and Orion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By matching the positions of the stars either in real life or on Stellarium, I was able to figure out something of how the Southern Cross is used. You place your fingers against the four brightest stars (two pointers plus the two brightest of the cross itself), and then the south celestial pole is right in the curve of your outstretched thumb. &lt;em&gt;(Aside: It makes an awesome demo in Stellarium to place your hand on the screen, do the measurements, and then turn on the grid, highlighting the south pole beautifully.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I haven't managed to do, though, is match the corresponding measurement in Orion. The fingers are placed near Orion's belt and heel, but I can't figure out what is being measured. Can anyone provide either certain information, or plausible guesses?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I can provide a screencap if there's an appropriate way to include it in the question.)&lt;/p&gt;&#xA;" OwnerUserId="15894" LastActivityDate="2017-06-29T22:58:14.610" Title="Polynesian wayfinding techniques - how does this work?" Tags="&lt;constellations&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20116" PostTypeId="1" AcceptedAnswerId="20119" CreationDate="2017-02-21T03:19:18.047" Score="2" ViewCount="130" Body="&lt;p&gt;What is the name and unit of the quantity in which star brightness due to skyglow is measured? How can it be measured from RAW image files of stars?&lt;/p&gt;&#xA;" OwnerUserId="15512" LastEditorUserId="2365" LastEditDate="2017-02-21T06:18:35.520" LastActivityDate="2017-02-21T18:11:55.787" Title="What is the unit used for star brightness due to skyglow?" Tags="&lt;star&gt;&lt;astrophysics&gt;&lt;photometry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20117" PostTypeId="2" ParentId="20116" CreationDate="2017-02-21T06:42:17.970" Score="4" Body="&lt;p&gt;The brightness of a diffuse area is it's luminance, and can be quoted in units of $cd\,m^{-2}$. Measuring it would require knowledge of the properties of the camera: the sensitivity (at various wavelengths) of the ccd, the aperture, length of exposure and so on. This could probably be found using calibration images, in which a known luminance is imaged and then the brightness of the skyglow is compared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The brightness of a star is usually given as a magnitude which is related to the amount of electromagnetic energy received from the star per second, in a particular range of wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's unclear what you mean by &quot;brightness due to skyglow&quot;. The brightness of stars is not due to skyglow, but excessive skyglow can mask dimmer stars and diffuse objects.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-02-21T17:13:03.173" LastActivityDate="2017-02-21T17:13:03.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20119" PostTypeId="2" ParentId="20116" CreationDate="2017-02-21T17:45:57.870" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Surface_brightness&quot; rel=&quot;nofollow noreferrer&quot;&gt;Surface brightness&lt;/a&gt;&#xA;of galaxies and nebulae is typically expressed as visual magnitude per square arcsecond.&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Skyglow&quot; rel=&quot;nofollow noreferrer&quot;&gt;Skyglow&lt;/a&gt;&#xA;is often quantified in the same way.&#xA;Given the angular dimensions of the frame and the magnitudes of a few stars in it, photometry software could compute sky brightness by excluding all visible stars and dividing the total brightness of the remainder by its angular area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an&#xA;&lt;a href=&quot;http://www.unihedron.com/projects/darksky/magconv.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;online tool to convert&lt;/a&gt;&#xA;between $\mathrm{mag} / \mathrm{arcsec}^2$ and $\mathrm{cd} / \mathrm{m}^2$.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2017-02-21T18:11:55.787" LastActivityDate="2017-02-21T18:11:55.787" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20120" PostTypeId="2" ParentId="20061" CreationDate="2017-02-21T19:58:31.567" Score="3" Body="&lt;p&gt;I will just point out two things that haven't really been discussed above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, stars generally fall into two categories. 'Hot stars' are those that convert mass into energy by nuclear fusion, and 'cold stars' are those that have ended the nuclear burning phase, e.g. white dwarfs and neutron stars. These still have a finite temperature and therefore will emit radiation. As you can imagine, however, since they no longer produce energy, eventually they will become actually cold, hence not radiating (or radiating very little). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second point to note is that radiations from stars far away are redshifted as they reach us, and they are likely to be attenuated by dust. Therefore, even if the radiation itself when emitted is visible light, by the time it gets to us it may no longer be. However, there are still many other different spectroscopic techniques that will allow us to 'see' it.&lt;/p&gt;&#xA;" OwnerUserId="15906" LastActivityDate="2017-02-21T19:58:31.567" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20122" PostTypeId="1" AcceptedAnswerId="20124" CreationDate="2017-02-22T05:41:57.793" Score="13" ViewCount="3457" Body="&lt;p&gt;Sirius B is a massive white dwarf of 1 Solar mass, orbiting at about 25 AU distance from the 2 Solar mass Sirius A. As it evolves and expands, will the A star start shedding matter to the white dwarf, and when will this start happening? Will the Sun be at a safe distance when/if it does happen, or is Sirius our doom?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2019-09-03T11:44:23.940" Title="Will Sirius B start accreting from A and become a supernova type Ia?" Tags="&lt;supernova&gt;&lt;binary-star&gt;&lt;white-dwarf&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20124" PostTypeId="2" ParentId="20122" CreationDate="2017-02-22T08:44:33.700" Score="17" Body="&lt;p&gt;Will Sirius B start accreting? Yes, it is doing so now. Sirius A will have a wind and some of that wind will be captured by the white dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effectiveness of wind capture is a strong function of relative wind speed. An analytic approximation to the accretion rate, known as &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Bondi_accretion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bondi-Hoyle accretion&lt;/a&gt;, goes as the inverse cube of the relative speed. In its present evolutionary state, the mass loss from Sirius A will be relatively weak (like the Sun) and relatively fast (like the Sun). This disfavours any significant accretion by the white dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the later stages of its life, Sirius A will swell to become an asymptotic giant star. The envelopes of such stars are gradually (on timescales of millions of years) blown away quite slowly by a dusty wind. If Sirius A is around 2 solar masses now, it will lose about 1.4 solar masses during this phase at speeds of only 10-20 km/s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only &lt;em&gt;a fraction&lt;/em&gt; of this mass may be accreted by the white dwarf, because the separation between the stars is still large at 25 AU (and will become larger as mass is lost from the system) compared with the likely terminal size of Sirius A (probably of order 2 au). If you look at the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Roche_lobe&quot; rel=&quot;nofollow noreferrer&quot;&gt;likely Roche lobe size&lt;/a&gt;, then Roche lobe overflow would require A to reach about 40% of the separation, which isn't going to happen. Exactly what fraction &lt;em&gt;is&lt;/em&gt; captured by the less efficient wind accretion process (the majority will likely disappear into space &lt;em&gt;and&lt;/em&gt; widen the orbit) depends strongly on the wind speed, which is hard to predict. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if Sirius B could accrete the 0.35-0.4 solar masses (I think that is unlikely, but lack the wherewithal to do the hydrodynamic simulation) it needs to become unstable, it is not clear if that mass will &quot;stick&quot;. A build up of hydrogen-rich material can ignite and explode in a nova (not supernova) on the surface of a white dwarf, causing mass loss!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, when will this happen? Well Sirius is probably about 300 million years old now and has another perhaps 500 million years before it starts to evolve in the way I described. It will be nowhere near the Sun then.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-09-03T11:44:23.940" LastActivityDate="2019-09-03T11:44:23.940" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="20125" PostTypeId="1" AcceptedAnswerId="20145" CreationDate="2017-02-22T12:27:53.550" Score="0" ViewCount="554" Body="&lt;p&gt;I have zero clue in astronomy and I need astronomical data to write a program. What I learned so far:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;celestial coordinates are usually measured relatively to vernal equinox aka meridian of First point of Aries (which is not located in Aries anymore)&lt;/li&gt;&#xA;&lt;li&gt;the GHA of first point of Aries changes constantly&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I was searching for vernal equinox GHA data and it seems to me that &lt;a href=&quot;http://sv-inua.net/the-nautical-almanac&quot; rel=&quot;nofollow noreferrer&quot;&gt;this personal, generated Almanac version&lt;/a&gt; lists it as &quot;Aries&quot; (it was misleading me).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expected that&lt;br&gt;&#xA;( GHA(object) - GHA(v. equinox) ) mod 24h = RA (object)&lt;br&gt;&#xA;as per &lt;a href=&quot;https://en.wikipedia.org/wiki/Right_ascension&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt;. However I have found &lt;a href=&quot;http://www.moshier.net/aadoc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;a program&lt;/a&gt; which calculates declination and right ascension and I have found out that the relation is sort of inverse. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://sv-inua.net/wp-content/uploads/2016/12/almanac2020.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://sv-inua.net/wp-content/uploads/2016/12/almanac2020.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the program output:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;               The Moon   &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &#xA;  &lt;p&gt;JD 2458849.50,  2019 December 31 Tuesday 23h 58m 49.174s  UT 2020&lt;br&gt;&#xA;  &lt;strong&gt;January 1 Wednesday  0h 00m 00.000s  TDT&lt;/strong&gt; nutation dRA -1.043s dDec&lt;br&gt;&#xA;  -6.12&quot; Geometric lon -13.866 deg, lat -4.894 deg, rad 2.6996e-003 au   Apparent geocentric longitude 346.129 deg   latitude -4.894 deg&lt;br&gt;&#xA;  Distance 63.319 Earth-radii Horizontal parallax    0d 54' 17.67&quot;&lt;br&gt;&#xA;  Semidiameter    0d 14' 47.60&quot; Elongation from sun 66.20 deg,&lt;br&gt;&#xA;  Illuminated fraction 0.30 Phase 2.0 days before First Quarter&lt;br&gt;&#xA;      &lt;strong&gt;Apparent:  R.A. 23h 16m 38.283s  Declination -  9d 58' 41.62&quot;&lt;/strong&gt;&lt;br&gt;&#xA;  Local apparent sidereal time   1h 54m 46.005s diurnal aberration dRA&lt;br&gt;&#xA;  0.012s dDec -0.03&quot; diurnal parallax dRA -104.977s dDec -2485.74&quot;      atmospheric refraction 0.033 deg  dRA 4.296s dDec 102.06&quot; Topocentric:&lt;br&gt;&#xA;  Altitude 25.678 deg, Azimuth 224.447 deg Topocentric: R.A. 23h 14m&lt;br&gt;&#xA;  57.615s   Dec. - 10d 38' 25.32&quot; local meridian transit 2019 December 31 Tuesday 21h 16m 02.396s  UT rises 2019 December 31 Tuesday 15h 49m&lt;br&gt;&#xA;  26.475s  UT sets 2020 January 1 Wednesday  2h 50m 11.075s  UT Visible hours 11.0124      &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So:&#xA;110 deg 56,9 sec - 100 deg 7,1 sec = 10 deg 49,8 sec&lt;br&gt;&#xA;which is inverse of R.A. which program lists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I have a misconception?&lt;/p&gt;&#xA;" OwnerUserId="15914" LastActivityDate="2017-02-23T12:47:02.377" Title="How do GHA of vernal equinox, GHA of astronomical object and RA relate?" Tags="&lt;coordinate&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20126" PostTypeId="1" CreationDate="2017-02-22T18:40:55.550" Score="1" ViewCount="82" Body="&lt;p&gt;One of the tools being developed for the study of exoplanets is called &lt;A HREF=&quot;http://www.exoclimes.com/topics/transmission-spectroscopy/&quot; rel=&quot;nofollow noreferrer&quot;&gt;transmission spectroscopy&lt;/A&gt; - basically looking at how passing through the exoplanet's atmosphere changes the spectrum of the star. The problem, though, is that spectroscopy is hard, and requires a big telescope, probably space-based. This makes the process slow. It seems that it would be very useful if we would triage the data using broad-band photometry to more quickly limit the presence of compounds like water, $\mathrm{CO}_2$, and $\mathrm{O}_3$. Looking at a plot of &lt;A HREF=&quot;https://upload.wikimedia.org/wikipedia/commons/7/7c/Atmospheric_Transmission.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth's atmospheric transmission&lt;/A&gt;, with a breakdown by chemical compound, it seems like looking at the color of the transit depths in and between the windows in the wavelength range from 1 to 20 microns, might provide a plausible alternative to quickly get information on the atmospheres of planets and brown dwarf stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the exoplanets, specifically, the wavelengths blocked by an atmosphere make the planet, effectively, bigger at some wavelengths than at others, depending on the opacity at that wavelength, the thickness (density) of the atmosphere, and vertical extent of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: is this practical? Specifically, with a 1 to 2 meter class space telescope that simultaneously images in 2 to 8 filters?&lt;/p&gt;&#xA;" OwnerUserId="14244" LastActivityDate="2017-02-22T22:08:33.940" Title="Transit Depth Color Versus Transmission Spectroscopy" Tags="&lt;exoplanet&gt;&lt;space-telescope&gt;&lt;photometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20127" PostTypeId="1" AcceptedAnswerId="20129" CreationDate="2017-02-22T19:18:15.963" Score="22" ViewCount="4215" Body="&lt;p&gt;The &lt;a href=&quot;https://exoplanets.nasa.gov/news/1419/nasa-telescope-reveals-largest-batch-of-earth-size-habitable-zone-planets-around-single-star/&quot; rel=&quot;noreferrer&quot;&gt;TRAPPIST-1 system&lt;/a&gt; is around an ultra-cool dwarf star. I went looking for more information on that kind of star, and found very little. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Ultra-cool_dwarf&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; on it lengthened from a minimal stub to one paragraph in the 15 minutes from when I began the search, after the press conference on the discovery of 7 Earth-sized planets in orbit around TRAPPIST-1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of star is this?&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2153" LastEditDate="2017-02-23T14:33:17.903" LastActivityDate="2017-02-24T20:09:00.337" Title="So, what exactly is an 'ultra-cool' dwarf star?" Tags="&lt;star&gt;&lt;terminology&gt;&lt;dwarf-star&gt;&lt;trappist-1&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20128" PostTypeId="2" ParentId="20127" CreationDate="2017-02-22T19:30:37.797" Score="8" Body="&lt;p&gt;According to &lt;a href=&quot;http://adsabs.harvard.edu/full/1998ASPC..134..405K&quot; rel=&quot;noreferrer&quot;&gt;Spectroscopic Properties of Ultra-cool Dwarfs and Brown Dwarf&lt;/a&gt; by J.D. Kirkpatrick, the working definition of ultra-cool dwarf is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;dwarfs having classifications of M7 or later.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is referring to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_classification&quot; rel=&quot;noreferrer&quot;&gt;stellar classification&lt;/a&gt; system which categorizes stars by their spectral characteristics. Class M stars have a surface temperature less than 4000 K, and &quot;M7 or later&quot; means surface temperature less than 2500 K. Though it should be noted that this is a working definition, and the Wikipedia article identifies ultra-cool dwarfs as objects with surface temperature less than 2700 K. It is possible that the term is still used somewhat loosely in scientific papers.&lt;/p&gt;&#xA;" OwnerUserId="6" LastActivityDate="2017-02-22T19:30:37.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20129" PostTypeId="2" ParentId="20127" CreationDate="2017-02-22T19:48:10.733" Score="22" Body="&lt;p&gt;Breaking the phrase down: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Dwarf_star&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Dwarf star&lt;/em&gt;&lt;/a&gt; - a term I will never understand - is used to describe relatively small, dim stars. Unfortunately, this encompasses most main-sequence stars, which are indeed dwarfs compared to large giants and supergiants.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Ultra-cool&lt;/em&gt;, as &lt;a href=&quot;https://astronomy.stackexchange.com/a/20128/2153&quot;&gt;called2voyage&lt;/a&gt; already discussed, &lt;a href=&quot;http://www.eso.org/public/archives/releases/sciencepapers/eso1615/eso1615a.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;means that&lt;/a&gt; the star has an&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;effective temperatures of less than 2,700 kelvin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is about half of the Sun's surface temperature. Spectroscopically, it means that these stars are of class &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/510132/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;M7-L8&lt;/a&gt;, i.e. really, really cool stars that just barely reach the threshold for nuclear fusion. Actually, some L dwarfs will never fuse hydrogen as normal stars do, and will become brown dwarfs. Therefore, &quot;ultra-cool dwarf&quot; doesn't necessarily refer to just main sequence stars.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Other characteristics in some cases (from &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/510132/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cruz et al&lt;/a&gt;.):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Lithium in some L-type objects, indicative of low temperatures, low masses, and young ages in those particular objects.&lt;/li&gt;&#xA;&lt;li&gt;Very low surface gravities in late M-type dwarfs (though likely not as much in an M7 dwarf like TRAPPIST-1).&lt;/li&gt;&#xA;&lt;li&gt;Very low metallicity in certain other L-type objects.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;TRAPPIST-1 actually appears to be a rather massive ultra-cool dwarf, then, in comparison to some of these other objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an annotated Hertzsprung-Russell diagram, with a box for ultra-cool dwarfs and a circle for where TRAPPIST-1 is, approximately:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZOlPP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZOlPP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;Image from &lt;a href=&quot;https://en.wikipedia.org/wiki/File:HRDiagram.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; courtesy of user Saibo, under &lt;a href=&quot;https://creativecommons.org/licenses/by-sa/2.5/deed.en&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Creative Commons Attribution-Share Alike 2.5 Generic license&lt;/a&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2017-02-24T20:09:00.337" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20130" PostTypeId="1" AcceptedAnswerId="20131" CreationDate="2017-02-22T20:57:24.117" Score="7" ViewCount="245" Body="&lt;p&gt;With the NASA announcement today regarding the discovery of a system containing seven earth-sized planets (3 within the habitable zone), I wondered about the seemingly crowded conditions.  What principles guide the formation of planets and the distances between them?  Are there laws governing this that are well established?  How close have we observed two planets form?&lt;/p&gt;&#xA;" OwnerUserId="15139" LastActivityDate="2017-02-22T21:38:37.787" Title="How close can planets form to one another?" Tags="&lt;planet&gt;&lt;exoplanet&gt;&lt;planetary-formation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20131" PostTypeId="2" ParentId="20130" CreationDate="2017-02-22T21:22:28.377" Score="7" Body="&lt;p&gt;If you're thinking about how close planets can be, you should probably consider each planet's &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;noreferrer&quot;&gt;Hill sphere&lt;/a&gt;, the region in which it can retain satellites. &lt;a href=&quot;https://arxiv.org/abs/1302.7190&quot; rel=&quot;noreferrer&quot;&gt;Fang &amp;amp; Margot (2013)&lt;/a&gt; did an analysis of &lt;em&gt;Kepler&lt;/em&gt; data and found that planets had mean values of $\Delta = 21.7$, where $\Delta$ is a parameter given for two adjacent planets by&#xA;$$\Delta=\frac{a_2-a_1}{R_{H1,2}}$$&#xA;where the $a$s are the semi-major axes and $R_{H1,2}$ is the mutual Hill radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One system the authors consider is &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler-11&quot; rel=&quot;noreferrer&quot;&gt;Kepler-11&lt;/a&gt;, which has 6 planets, all with semi-major axes $\leq0.466\text{ AU}$ and with only one semi-major axis greater than $0.25\text{ AU}$. The smallest $\Delta$ there is approximately $5.7$, although all the other $\Delta$s are quite small. &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler-36&quot; rel=&quot;noreferrer&quot;&gt;Kepler-36&lt;/a&gt;, with only two planets, still has a $\Delta$ of $4.7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.nature.com/nature/journal/v542/n7642/full/nature21360.html&quot; rel=&quot;noreferrer&quot;&gt;the &lt;em&gt;Nature&lt;/em&gt; paper&lt;/a&gt; about TRAPPIST-1, all seven planets have semi-major axes within $\sim0.063\text{ AU}$. They have mean $\Delta$s of $10.5\pm1.9$ - not much different from the Kepler-11 planets, because they have smaller Hill spheres. They may be closer together, but they can be much closer together without having stability problems. Additionally, they are in a &quot;near-&lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_resonance&quot; rel=&quot;noreferrer&quot;&gt;resonant&lt;/a&gt;&quot; configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How close planets can be depends strongly on their masses, then, which in turn determines their mutual Hill radii, which determines stability.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;All that said, the authors believe that the TRAPPIST-1 planets may have migrated in from further out, thus entering the resonances. Without more information, we can't know whether this is the case, but if so, it is not, then, an example of planets forming close to each other.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2017-02-22T21:38:37.787" LastActivityDate="2017-02-22T21:38:37.787" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20132" PostTypeId="2" ParentId="20126" CreationDate="2017-02-22T22:08:33.940" Score="1" Body="&lt;p&gt;But this is already possible. Spitzer has performed &quot;low-resolution&quot; transit spectrophotometry of several exoplanets in its near infrared bands - see for example &lt;a href=&quot;http://ukads.nottingham.ac.uk/abs/2007Natur.448..169T&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tinetti et al. (2007)&lt;/a&gt;;  &lt;a href=&quot;https://arxiv.org/abs/1104.2901&quot; rel=&quot;nofollow noreferrer&quot;&gt;Knutson et al. (2011)&lt;/a&gt;; &lt;a href=&quot;https://arxiv.org/abs/1008.2481&quot; rel=&quot;nofollow noreferrer&quot;&gt;Desert et al. (2011)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I dare say you could get better data on more planets with a dedicated telescope with a larger mirror and the ability to image in several bands at once. All that is required is money.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-02-22T22:08:33.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20134" PostTypeId="1" CreationDate="2017-02-22T23:12:49.387" Score="4" ViewCount="1405" Body="&lt;p&gt;Not sure if this is the best place for this, maybe physics would be a better fit, but I'm trying to build a periscope and am running into an issue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traditional mirror periscopes work well, but when I look into the one end of the periscope, I see the reflected image at the end of a long tube. I figure that using some combination of lenses, you should be able to make it look as if the position of your eye/head was actually &lt;strong&gt;at the top&lt;/strong&gt; of the tube looking out at the world rather than looking &lt;strong&gt;through&lt;/strong&gt; a tube. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thinking here is that when you look through binoculars or a telescope, you don't see the inside of the tube, just the enlarged image. So shouldn't you be able to do the same thing but without the magnification?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I purchased some lenses from Amazon with focal lengths of 200mm and -100mm and based on some optics simulators assumed(incorrectly apparently) that I could arrange them 10cm apart and then another set reversed on the other side like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;`````()````````````````)(```````````)(``````````````()&lt;br/&gt;&#xA;200mm````````-100mm``-100mm```````200mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;() = convex lens&lt;br/&gt;&#xA;)( = concave lens&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thinking here was that one set would condense the image down so that the walls of the tube were no longer visible within that viewing angle, and then the other reversed set would reverse that and the image would be back at normal size to my eye, it would just look as though my eye were actually at the opposite end of the tube. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This didn't work. The image is close to life size, but the tube looks even longer: a foot long tube has turned into a 3 foot long tube, and the circle I can see the world through is very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a pipe dream? I feel like it must be possible if they used periscope in military applications for years before electronics and fiber-optic cables were invented. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where did I go wrong? Any guidance would be hugely appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: some follow up questions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I stuck having a small field of view out of the end? Or would using a lower focal point(wide angle) lens allow me to see a wider angle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is there any way to have a variable length tube? For instance if I'm looking around a wall, is there a way to install the lenses so that the tube could be &quot;telescoped&quot; out to look around a thicker wall?&lt;/p&gt;&#xA;" OwnerUserId="15937" LastEditorUserId="15937" LastEditDate="2017-02-23T19:05:07.607" LastActivityDate="2020-02-04T10:15:22.160" Title="Constructing a periscope/telescope - trouble with lenses" Tags="&lt;telescope&gt;&lt;light&gt;&lt;optics&gt;&lt;telescope-lens&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20135" PostTypeId="1" AcceptedAnswerId="20174" CreationDate="2017-02-23T00:26:26.460" Score="3" ViewCount="144" Body="&lt;p&gt;When will the moon be far enough away from the Earth that we would no longer consider it a 'solar eclipse', but rather a 'lunar transit' when the moon travels between the Earth and Sun?&lt;/p&gt;&#xA;" OwnerUserId="9438" LastActivityDate="2017-02-24T14:54:22.370" Title="When will solar eclipses become lunar transits?" Tags="&lt;solar-eclipse&gt;&lt;the-moon&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20136" PostTypeId="2" ParentId="20134" CreationDate="2017-02-23T04:53:40.033" Score="7" Body="&lt;p&gt;You're close!  What you built is a back-to-back set of Galilean telescopes (with an overall magnification of 1:1).  The reason you're getting a smaller circle, but with a life-size image, is that the light bundles in the middle are diverging faster, and so you lose the edges if the Galilean assemblies are too far apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Military perisopes use optical relays to get the image down the tube without blocking out any of the view.  In simple terms, they have one relatively short-focusing  group of achromatic/apochromatic lenses (called an Objective) to focus the light to a crosshair, like in a rifle scope.  From there, the image is picked up by another long-focusing lens group called a relay.  The relay re-collimates the light (but at low angles).  The light is focused and collimated several times through the tube, until it reaches a final image plane, which is then looked at with an eyepiece (like you might have in a telescope).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, you can think of a periscope as a telescope/monocular/riflescope with relatively low magnification with several pairs of relays in between.  A point of note, you need an even number of image planes (0, 2, etc) to avoid having an &#xA;upside-down image.  I can go into a lot more detail or provide some resources, but I think this most directly answers your question.  Please let me know if there's any other questions on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: I design optics for periscopes for naval applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit for additional questions:&lt;/strong&gt;&#xA;First, the most comprehensive resource I found in my lazy search is this page: &lt;a href=&quot;http://taylortechassoc.com/?page_id=97&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://taylortechassoc.com/?page_id=97&lt;/a&gt;, which is actually written by a previous engineer at my company.  There's a ton of information there, mostly intended for people who already know about optical design, but there's probably a good amount you can get from it as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To the question in your comment:&lt;/strong&gt; in your preliminary design, the positive and negative lenses are taking collimated light input, and outputting collimated light.  However, since they are acting as a telescope with magnification power, the angle of the rays coming out is wider than the angle going in (by a factor equal to your magnification).  The trick to a relay lens design (used in periscopes and some kinds of endoscopes, as well as other applications) is to get the collimated spaces short enough that you don't lose that light before it spreads out.  Or more accurately, that you get the angles of the collimated light to be small enough that you can spread out your lenses quite a bit while still maintaining this condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Regarding field of view:&lt;/strong&gt; you're absolutely not stuck with a small FOV.  That depends entirely on the focal length of your objective.  A short focal length lens as your objective will give you a wider field of view (so long as your light isn't clipped by your top mirror).  You could also have an objective that changes focal length (a zoom lens) to change your field of view as desired.  For your application, an SLR camera lens could work quite well for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Regarding tube length:&lt;/strong&gt; You could vary the spacing in the sections where the light is collimated.  This will impact how much light is being lost from it spreading out too much (typically limited by whichever section is longest).  The result will be a dropoff in the relative illumination (how bright the edge of the FOV is compared to the center).  This is an effect known as vignetting, and can be mitigated, but not without tradeoffs.&lt;/p&gt;&#xA;" OwnerUserId="10820" LastEditorUserId="10820" LastEditDate="2017-02-25T03:24:15.427" LastActivityDate="2017-02-25T03:24:15.427" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20138" PostTypeId="1" AcceptedAnswerId="20149" CreationDate="2017-02-23T06:48:25.313" Score="5" ViewCount="127" Body="&lt;p&gt;Which rotational frame does the universe prefer?&#xA;We say that the earth rotates once per 24 hours.&#xA;But is this the master frame?&#xA;What is the solution to &lt;a href=&quot;https://en.wikipedia.org/wiki/Bucket_argument&quot; rel=&quot;nofollow noreferrer&quot;&gt;Newtons Bucket Problem&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like if the answer is a detailed explanation.&#xA;I know about high school math and science and participated in a astronomy olympiad.&lt;/p&gt;&#xA;" OwnerUserId="15901" LastEditorUserId="15901" LastEditDate="2017-02-23T14:43:47.890" LastActivityDate="2017-02-23T15:17:47.297" Title="Which rotational frame does the universe prefer?" Tags="&lt;frame-of-reference&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20140" PostTypeId="1" AcceptedAnswerId="20153" CreationDate="2017-02-23T08:51:03.850" Score="4" ViewCount="256" Body="&lt;p&gt;Most of the news about extraterrestrial search focuses on possibility for existence of water and temperatures like earth. Why do scientists feel this would lead them to extraterrestrial life? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;After all, we are searching for something that doesn't exist on earth. Why do we assume it would have behavior like us? Maybe the matrix inside their cells is of acetone or something else. Water may not play the role in their lives as it does for us. Maybe they can sustain high temperatures, maybe they can live in extreme cold. Why should we look for earth-like planets, isn't there a baseless (of course it's not baseless, there must be some logic which I want to know) assumption that the aliens are human-like (or any species on earth)?&lt;/p&gt;&#xA;" OwnerUserId="15928" LastEditorUserId="7472" LastEditDate="2017-10-15T19:16:16.807" LastActivityDate="2017-10-15T19:16:16.807" Title="Why search for water?" Tags="&lt;astrobiology&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20141" PostTypeId="1" CreationDate="2017-02-23T09:03:02.623" Score="-1" ViewCount="105" Body="&lt;p&gt;me and my friends likes to stargaze at night and every time we see this star like light that glides in one direction. they sometimes even cross in the middle then disappear sometime. i observed that it cant be seen if its too dark. we wondered what it could possibly be, we even make jokes that it was an alien craft that monitors our actions. we even nicknamed it &quot;the fifth wave&quot;(after the movie &lt;em&gt;The Fifth Wave&lt;/em&gt;). its just something to wonder upon, really, but if you have something to share i'd be willing to read it and share it with my friends.&lt;/p&gt;&#xA;" OwnerUserId="15927" LastActivityDate="2017-02-23T14:22:31.123" Title="Star like lights at night" Tags="&lt;star&gt;&lt;light&gt;&lt;night-sky&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2017-02-23T17:55:13.517" ContentLicense="CC BY-SA 3.0" />
  <row Id="20142" PostTypeId="1" AcceptedAnswerId="20144" CreationDate="2017-02-23T09:19:14.810" Score="54" ViewCount="7424" Body="&lt;p&gt;According to the &lt;a href=&quot;https://exoplanets.nasa.gov/trappist1/&quot; rel=&quot;noreferrer&quot;&gt;NASA announcement&lt;/a&gt; and the &lt;a href=&quot;https://en.wikipedia.org/wiki/TRAPPIST-1&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article on TRAPPIST-1&lt;/a&gt;, the planets around the TRAPPIST-1 star are named TRAPPIST-1 b through h. What happened to &quot;TRAPPIST-1 a&quot;?&lt;/p&gt;&#xA;" OwnerUserId="15930" LastEditorUserId="2153" LastEditDate="2017-02-23T14:24:30.827" LastActivityDate="2019-04-19T00:23:33.890" Title="Where did TRAPPIST-1a go?" Tags="&lt;exoplanet&gt;&lt;naming&gt;&lt;trappist-1&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20143" PostTypeId="1" AcceptedAnswerId="20150" CreationDate="2017-02-23T09:52:51.903" Score="2" ViewCount="998" Body="&lt;p&gt;After hearing that it has been confirmed that we found seven similar exoplanets like earth. I thought exoplanets were dead stars. After doing a little bit of research I found out that exoplanets are planets that are orbiting a star, NOT the sun. But isn't a star like the Sun? Can someone please explain this to me?&lt;/p&gt;&#xA;" OwnerDisplayName="user15932" LastEditorUserId="2365" LastEditDate="2017-02-23T19:35:37.767" LastActivityDate="2017-02-23T19:35:37.767" Title="Exo Planets and Planets difference?" Tags="&lt;planet&gt;&lt;exoplanet&gt;&lt;space&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20144" PostTypeId="2" ParentId="20142" CreationDate="2017-02-23T10:38:30.617" Score="39" Body="&lt;p&gt;The convention for planetary naming is that the closest planet to the star (if multiple planets are found at the same time) is named &quot;star&quot;b, then &quot;star&quot;c and so on. As correctly pointed out by Zephyr, if the discoveries are more haphazard, the order of discovery takes precedence over distance from the star).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there is no Trappist-1a. Or you can think of it being the star itself if you like, although that becomes confusing if the star is itself part of a multiple star system, where it might be known as &quot;star&quot;A or &quot;star&quot;B etc. (note, capital letters), depending on where it is in the hierarchy.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-02-24T08:55:40.173" LastActivityDate="2017-02-24T08:55:40.173" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20145" PostTypeId="2" ParentId="20125" CreationDate="2017-02-23T12:47:02.377" Score="1" Body="&lt;p&gt;I'm using the &lt;em&gt;Explanatory Supplement to the Astronomical Almanac&lt;/em&gt; 3rd ed. as a reference. The glossary tells us that Greenwich Apparent Sidereal Time (GAST) is the GHA of the true equinox of date, that is, &quot;GHA(v. equinox)&quot; in your equation. On page 80 equation 3.14 states&lt;/p&gt;&#xA;&#xA;&lt;p&gt;local hour angle = local apparent sidereal time - apparent right ascension (3.14)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one is located on the Greenwich meridian, then that equation becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GHA = GAST - apparent right ascension (3.14a)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;solving for apparent RA,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;apparent RA = GAST - GHA (3.14b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing your equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;( GHA(object) - GHA(v. equinox) ) mod 24h = RA (object)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your sign seems to be inverted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking up values using UT1 and the Multiyear Computer Interactive Almanac, rounding to nearest second,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;apparent RA 23 h 16 m 38 s&#xA;GAST         6 h 40 m 29 s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting into eq. 3.14a, GHA = 7 h 23 m 51 s&lt;/p&gt;&#xA;" OwnerUserId="7385" LastActivityDate="2017-02-23T12:47:02.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20146" PostTypeId="2" ParentId="20141" CreationDate="2017-02-23T14:22:31.123" Score="3" Body="&lt;p&gt;Generally, these will be low orbit satellites. They are visible when still lit by the sun (visible over the horizon from behind the edge of Earth at their altitude) and they disappear when the enter Earth shadow (&quot;night&quot;) and are no longer highlit by the sun. Obviously, that happens only during evening and dawn - in the middle of the night they are in full shadow.&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2017-02-23T14:22:31.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20147" PostTypeId="2" ParentId="20142" CreationDate="2017-02-23T14:24:16.077" Score="53" Body="&lt;p&gt;To add to Rob's answer, I wanted to expand on where this naming convention comes from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The International Astronomical Union (IAU) is the organization which generally sets conventions and definitions. They're the ones who demoted Pluto to being a dwarf planet in 2006. Anyway, before any exoplanets were found, there existed a convention for naming multiple-star systems. The rule was that you gave the system a name, for example Alpha Centauri, and then the brightest star in that system is designated by &quot;A&quot;. Any other objects are then given the letters &quot;B&quot;, &quot;C&quot;, etc. The Alpha Centauri system has three stars in it and so they're named Alpha Centauri A (The brightest of the three), Alpha Centauri B, and Alpha Centauri C (a.k.a. Proxima Centauri).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This convention then continued on to naming exoplanets. The star, according to this convention is &quot;A&quot;, whereas all the exoplanets are given the letters &quot;b&quot; and onwards. Note that generally the exoplanet letters are lowercase, whereas they'd be uppercase for stars. Also note that if you have a star (say TRAPPIST-1) around which planets are found, the star's name doesn't then become TRAPPIST-1 A. The &quot;A&quot; is implicit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to point out another point of interest with this naming convention, the letters of the planets don't always signify their distance from the star. That is, the &quot;b&quot; planet isn't necessarily the closest, followed by &quot;c&quot;, etc. The planets are labeled at their time of discovery. Someone could find a planet, label it as &quot;b&quot; and then later find a second, closer planet and label it &quot;c&quot;. It just so happens that the lettering of TRAPPIST 1 is in order because the first three were found at the same time and labeled according to distance, then the last 4 were found at the same time and labelled according to distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read more about the full set of rules concerning this on Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Exoplanet_naming_convention&quot; rel=&quot;noreferrer&quot;&gt;exoplanet naming convention&lt;/a&gt; page. There are a lot of caveats such as what to do when your planet is orbiting a binary system.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-02-23T14:24:16.077" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20149" PostTypeId="2" ParentId="20138" CreationDate="2017-02-23T15:17:47.297" Score="7" Body="&lt;p&gt;There's a lot of loosely connected questions here, so let's go through them one by one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which rotational frame does the universe prefer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The universe doesn't prefer any frame. There is no &quot;universal&quot; reference frame. This is one of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Special_relativity#Lack_of_an_absolute_reference_frame&quot; rel=&quot;nofollow noreferrer&quot;&gt;principle axioms of Special Relativity&lt;/a&gt;. You cannot pick a frame and say this is the &quot;correct&quot; frame. The reason being that you have no means of establishing what &quot;correct&quot; even means. One person may claim their chosen frame is correct, but another person might choose something different and claim theirs is correct. Neither person would be able to show their frame is better or more &quot;correct&quot; than the other's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, I think its important to point out that while no true universal reference frame exists, scientists sometimes use a de facto &quot;psuedo-universal&quot; reference frame. That frame is the one which is at rest with respect to the Cosmic Microwave Background (CMB) radiation. It is not a true universal reference frame, but because of the importance of the CMB, the frame at rest with respect to the CMB has been given more (arbitrary) importance by scientists, hence the term psuedo-universal. You can read more about this at this great &lt;a href=&quot;https://physics.stackexchange.com/questions/25928/is-the-cmb-rest-frame-special-where-does-it-come-from&quot;&gt;physics Stack Exchange question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We say that the earth rotates once per 24 hours. But is this the master frame?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a good question. Anytime someone mentions motion, you should always ask, motion with respect to what? If I'm inside a car, I would say the car isn't moving at all because my motion, &lt;em&gt;with respect to the car&lt;/em&gt; is zero. A person standing on the side of the road would say the car is traveling at 25 mph. So you can see that defining motion requires defining a frame to measure that motion in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to talking about the rotation of the Earth, there are actually a few different frames that scientists use. They're all very similar and produce similar answers (all about 24 hours). I'll restrict myself to talking about the frame which defines the &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solar Day&lt;/a&gt; (another example would be the &lt;a href=&quot;https://en.wikipedia.org/wiki/Sidereal_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sidereal Day&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Solar Day is the time it takes the Sun to return to the same position in the sky as from the previous day. As you can hopefully see, this measure of a day is fixed to a frame with respect to the Sun. It would be the same regardless of the motion of our solar system through the universe and thus wouldn't be considered a &quot;master&quot; frame (not that there is one).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the solution to &lt;a href=&quot;https://en.wikipedia.org/wiki/Bucket_argument&quot; rel=&quot;nofollow noreferrer&quot;&gt;Newtons Bucket Problem&lt;/a&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This, I would say, is really outside the realm of the Astronomy Stack Exchange and really should be its own question, preferably on the Physics Stack Exchange. For that reason, I won't answer this question here, but I will point you to this &lt;a href=&quot;http://www-groups.dcs.st-and.ac.uk/history/HistTopics/Newton_bucket.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;great source&lt;/a&gt; which talks about the Bucket Problem is detail.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2017-02-23T15:17:47.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20150" PostTypeId="2" ParentId="20143" CreationDate="2017-02-23T15:27:47.107" Score="5" Body="&lt;p&gt;Not all planets are exoplanets, but all exoplanets are planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we differentiate the two, we use &quot;planets&quot; to refer to the eight planets revolving around our Sun. The Greek prefix &quot;exo&quot; (outside) should indicate that exoplanets orbit other stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, you can call exoplanets &quot;planets&quot;, and you'll often hear &quot;Astronomers have found x planets around star y&quot;. The word &quot;exoplanet&quot; is used interchangeably with planets from other planetary systems.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2017-02-23T15:27:47.107" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20151" PostTypeId="1" AcceptedAnswerId="20156" CreationDate="2017-02-23T15:55:13.423" Score="19" ViewCount="1582" Body="&lt;p&gt;I would like to know if besides &lt;a href=&quot;https://en.wikipedia.org/wiki/TRAPPIST-1&quot; rel=&quot;noreferrer&quot;&gt;these planets&lt;/a&gt;, astronomers have discovered a gas planet like Jupiter in this system and if on these planets could exist life without a comet protector like Jupiter does for the Earth. &lt;/p&gt;&#xA;" OwnerUserId="828" LastEditorUserId="2153" LastEditDate="2017-02-23T16:18:47.837" LastActivityDate="2017-02-23T17:48:05.017" Title="Is there a gas giant orbiting TRAPPIST-1?" Tags="&lt;exoplanet&gt;&lt;gas-giants&gt;&lt;trappist-1&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20153" PostTypeId="2" ParentId="20140" CreationDate="2017-02-23T16:21:05.030" Score="5" Body="&lt;p&gt;A lot of different &quot;alternative biologies&quot; have been considered, but if you analyze their required chemistry, this always falls apart at some point, some mechanisms necessary for life are simply impossible. Replace carbon with silicon, increase ambient temperature, and you're getting a lot of interesting reactions that could be conductive to life. But instead of CO2, which is a gas, you have SIO2 which is a crystal. You can't breathe crystal. It's awfully insoluble, so no simple hemoglobin counterpart. And so on. Won't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water and carbon based biology is one &lt;em&gt;most plausible&lt;/em&gt;. There are odd variants, like &lt;a href=&quot;https://phys.org/news/2017-02-year-old-microbes-mexican-nasa-scientist.html&quot; rel=&quot;noreferrer&quot;&gt;bacteria thriving on sulfite, manganese and copper oxide&lt;/a&gt;, but they still require water to live, and still use carbon as the fundamental building block.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While others might exist, chance for them to form is much lower than for one similar to ours.  And since we haven't designed even one fully plausible alternate biology that doesn't involve water we don't even know what sort of detectable signatures it would produce. So we search for something we know might yield results; hint at possibility of life somewhat similar to ours - instead of a wild goose chase after the unknown.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Actually, there exist one &quot;alternate biology&quot; which is fully plausible even if its early origins would be Earth-like: Silicon based, artificial intelligence, robotic physiology, electrical power, proliferation through fully automated manufacturing processes. It's unlikely to produce any detectable chemical signature (at least distinguishable from half of the natural celestial bodies out there), but we are fairly sure its electromagnetic spectrum - radio waves signature - would be quite specific. Well, The &lt;a href=&quot;https://www.seti.org/&quot; rel=&quot;noreferrer&quot;&gt;SETI&lt;/a&gt; project is on it.&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2017-02-23T16:21:05.030" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20154" PostTypeId="1" AcceptedAnswerId="20157" CreationDate="2017-02-23T16:27:56.233" Score="4" ViewCount="143" Body="&lt;p&gt;It is possible to search for life from here, how?&#xA;What methods could be used?&#xA;It is possible to look for organic matter?&#xA;Could we know if there is life according to the components of the planets atmosphere?&lt;/p&gt;&#xA;" OwnerUserId="15948" LastEditorUserId="11143" LastEditDate="2017-02-23T16:59:38.717" LastActivityDate="2017-02-23T16:59:38.717" Title="How could we search for life in TRAPPIST-1" Tags="&lt;observational-astronomy&gt;&lt;astrobiology&gt;&lt;trappist-1&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2017-02-27T14:17:25.743" ContentLicense="CC BY-SA 3.0" />
  <row Id="20155" PostTypeId="2" ParentId="20151" CreationDate="2017-02-23T16:33:46.320" Score="5" Body="&lt;p&gt;As &lt;a href=&quot;http://www.eso.org/public/archives/releases/sciencepapers/eso1615/eso1615a.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;the paper on the first three planets discovered around the star&lt;/a&gt; states, even though firm constraints have not yet been put on the masses of the planets,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The results of planetary thermal evolution models — and the intense extreme ultraviolet (1−1,000 Å) emission of low-mass stars18 during their early lives — make it unlikely that such small planets would have thick envelopes of hydrogen and/or helium gases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no evidence for planets beyond TRAPPIST-1h.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The evolutionary history of the system is unclear. It is thought that stars like TRAPPIST-1 - termed &quot;ultracool dwarfs&quot; - could have rocky planets around them, but they would have to have formed beyond the &lt;a href=&quot;https://en.wikipedia.org/wiki/Frost_line&quot; rel=&quot;nofollow noreferrer&quot;&gt;frost line&lt;/a&gt;, in the region where volatiles exist. They would then have migrated inwards, falling into &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_resonance&quot; rel=&quot;nofollow noreferrer&quot;&gt;orbital resonances&lt;/a&gt;. Any putative gas giant would need to have an orbital history consistent with such an evolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomers have not observed any other objects in the system - exomoons or exocomets included - so we do not have a good idea of what small bodies may exist in the system, and thus how they might impact life on the planets.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The team used the &lt;a href=&quot;https://en.wikipedia.org/wiki/Transit-timing_variation&quot; rel=&quot;nofollow noreferrer&quot;&gt;transit-timing variation (TTV)&lt;/a&gt; method to detect the planets. Essentially, it looks for perturbations in transits of planets to figure out if there are other planets in the system. Models can then be created that attempt to reproduce the results. They found that a 6-planet model with the data for 6 planets; the seventh planet - with poorly constrained data - can still be consistently included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are instability issues. Over one million years, they determined that the system has a 25% chance of an instability; over one billion years, there is only an &lt;em&gt;8.1%&lt;/em&gt; chance it will survive with little or no changes. In other words, the systems is not particularly stable over long periods of time, and it remains it be seen how a gas giant could play into that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If&lt;/em&gt; there is a gas giant, it could interact with the planets and might further throw the system into havoc, meaning that it would have been hard to survive even 500 million years, the age of the system. Add to that the fact that the planets likely formed beyond the frost line and so would have been near where the gas giant formed, and you have a recipe for disaster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the authors note that there are poor constraints on many orbital parameters and masses, and it is &lt;em&gt;possible&lt;/em&gt; that one or more extra planets could stabilize the system. However, they haven't seen anything else, which is troubling - and a gas giant would have a good chance of showing up via the TTV method.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2017-02-23T16:55:38.693" LastActivityDate="2017-02-23T16:55:38.693" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20156" PostTypeId="2" ParentId="20151" CreationDate="2017-02-23T16:42:10.293" Score="19" Body="&lt;p&gt;No such planet has been announced as having been discovered. The paper only shows evidence for the 7 (really 6 because the 7th can't be officially confirmed with only 1 observation) terrestrial planets and does not make the case for any other planets. The paper doesn't indicate that more planets could exist, but does remark that there are large error bars on some of their data, leaving room for uncertainty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, I think we can cross off a gas giant existing in this system for a few reasons.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We don't see transit timing variations due to a gas giant. A gas giant would have noticeable gravitational influences on the inner 7 planets and we would see this influence by the small variations in the orbital periods of the inner planets. The authors of the paper did see transit timing variations, but they were able to explain all of the variations as due to the gravitational influences of the 7 planets only. They never had to invoke an 8th unseen planet to explain what they say.&lt;/li&gt;&#xA;&lt;li&gt;We don't observe such a (easily-detectable) transit. There are only two reasons we wouldn't see the transit of this planet. Either the gas giant is orbiting in a different plane than the rest of the planets (which are all remarkably close to the same plane, as pointed out by the authors), or else the orbital period is so long, it was never captured by any of the previous observations (which spans a number of years). Neither situation seems likely to occur.&lt;/li&gt;&#xA;&lt;li&gt;The mass of the central star is only 8% of the Sun's mass. Smaller stars tend to form smaller planets. Its really hard for gas giants to form around small stars, primarily due to lack of material. From what we understand about planet formation, the chances of a gas giant even being able to form around this star is pretty small.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of course, only continued observation will really be able to convince us that a gas giant doesn't exist.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On these planets could exist life without a comet protector like Jupiter does for the Earth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a great question. I think the answer is, we can't be sure. Jupiter does do a great job shepherding comets and protecting the Earth. Possibly this system is rife with comets that constantly bombard the planets. However, that's just one small part of the puzzle. Our Moon does a phenomenal job in protecting us as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think, when it comes to these planets, your primary concern for whether life can exist or not is the central star. It is a low-mass, ultra-cool dwarf star. These stars tend to be very volatile, much more so than our generally quiescent Sun. This means these planets are likely to receive much more radiation and be hit by many more solar storms than we are. Furthermore, these planets are so close to TRAPPIST-1 that they're all tidally locked - one face is always towards the star and one is always away. This could make one side inhospitably hot and the other inhospitably cold. The climate/weather on such a planet would likely be unsuitable for life (but who knows for sure). The tidal locking could potentially be good though as it means only the outward-facing side would generally be hit by comets, thus protecting any life on the Sun facing side (assuming its not too hot on that side).&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2017-02-23T17:48:05.017" LastActivityDate="2017-02-23T17:48:05.017" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20157" PostTypeId="2" ParentId="20154" CreationDate="2017-02-23T16:59:05.380" Score="3" Body="&lt;p&gt;As far as I know, and aside from direct communication, there's only one way to &quot;find life&quot; on another planet. That method is to look for molecular oxygen in the atmosphere. Molecular oxygen hates existing by itself and loves to combine with other atoms and molecules through oxidation. The Earth's atmosphere is currently ~30% $\mathrm{O_2}$, but that's only because we have a huge amount of plant life (primarily algae) constantly producing it. If you removed all life from the Earth, the oxygen in the atmosphere would quickly oxidize out and you'd be left with an atmosphere primarily of nitrogen and a few smaller constituents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means is that if we do see a substantial amount of oxygen on a planet, the best bet is that that oxygen was artificially produced by life. It's not ironclad evidence, but its a very strong indicator. We can search for oxygen through a few techniques that already exist. They're still being developed, but the technology is very quickly coming to fruition. Ultimately, you'd need to look at the spectrum of the planet's atmosphere. This can be achieved by either looking at the star's light as it passes through the planet's atmosphere, or else observing at numerous wavelengths and seeing how the transiting planet's radius changes (due to changing atmospheric opacity at different wavelengths).&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-02-23T16:59:05.380" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20160" PostTypeId="1" AcceptedAnswerId="20162" CreationDate="2017-02-23T22:23:00.083" Score="12" ViewCount="409" Body="&lt;p&gt;The Trappist-1 planets all orbit very close to each other.  During &lt;a href=&quot;https://www.youtube.com/watch?v=v5Xr-WkW5JM&quot; rel=&quot;noreferrer&quot;&gt;NASA's press release&lt;/a&gt;, they mentioned that these planets are close enough to disturb each others orbits.  Is this system stable over a long time scale?  Or could we perhaps just have imaged this system before one or more planets is ejected or destroyed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/v5Xr-WkW5JM?t=33m17s&quot; rel=&quot;noreferrer&quot;&gt;Later into the press conference&lt;/a&gt;, one of the scientists says &quot;these planets should have formed further out and migrated inwards&quot;.  Could they possibly still be moving inwards, or are their orbits now stable?&lt;/p&gt;&#xA;" OwnerUserId="13426" LastEditorUserId="13426" LastEditDate="2017-02-23T22:37:50.670" LastActivityDate="2017-02-24T14:43:25.207" Title="Are the Trappist-1 planets in stable orbits?" Tags="&lt;orbital-mechanics&gt;&lt;trappist-1&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20161" PostTypeId="1" AcceptedAnswerId="20180" CreationDate="2017-02-23T22:41:16.370" Score="8" ViewCount="528" Body="&lt;p&gt;According to the &lt;a href=&quot;https://youtu.be/v5Xr-WkW5JM?t=23m39s&quot; rel=&quot;noreferrer&quot;&gt;NASA press release&lt;/a&gt;, the Trappist planets are close enough (only a few million kilometers) that &quot;the tidal forces between the planets are not negligible&quot;.  The speaker says that this could cause ocean tides on the planets.  Are they close enough that the tidal forces could heat the interior of the planets?&lt;/p&gt;&#xA;" OwnerUserId="13426" LastActivityDate="2019-04-09T22:26:31.163" Title="Are the Trappist-1 planets close enough to be tidally heated?" Tags="&lt;tidal-forces&gt;&lt;trappist-1&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20162" PostTypeId="2" ParentId="20160" CreationDate="2017-02-24T00:26:32.013" Score="10" Body="&lt;h3&gt;We don't know.&lt;/h3&gt;&#xA;&lt;p&gt;Much of what we do know - or at least think we know - about the planets' orbital parameters comes from simulations the team ran via &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;-body methods&lt;sup&gt;&lt;a href=&quot;http://www.nature.com/nature/journal/v542/n7642/full/nature21360.html&quot; rel=&quot;noreferrer&quot;&gt;[1]&lt;/a&gt;&lt;/sup&gt;. Some methods of integration led to short-term disruptions, on the order of less than &lt;span class=&quot;math-container&quot;&gt;$\sim10^6$&lt;/span&gt; years. That said, the system is at least &lt;span class=&quot;math-container&quot;&gt;$5\times10^8$&lt;/span&gt; years, and it would be odd if the observations came just before the final instability, in the authors' opinions.&lt;/p&gt;&#xA;&lt;p&gt;Fortunately, using a different (statistical) method, the team got different results. The found that the system has&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;a 25% chance of suffering an instability over 1 Myr, and an 8.1% chance of surviving for 1 billion years&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This does not seem great for long- &lt;em&gt;or&lt;/em&gt; short- term instabilities. However, the authors were &lt;em&gt;extremely&lt;/em&gt; cautious, emphasizing that there is plenty of uncertainty. There are many other factors that need to be taken into account:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Tidal interactions between the planets and the star and between each other&lt;/li&gt;&#xA;&lt;li&gt;Possible other planets in the system (although &lt;a href=&quot;https://astronomy.stackexchange.com/q/20151/2153&quot;&gt;giant planets may be ruled out&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;The masses and orbits are not known to great accuracy, and so interactions can not be modeled as ideally as one would like.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_resonance&quot; rel=&quot;noreferrer&quot;&gt;Resonances&lt;/a&gt; could be important, and the planets in the system fall into near-integer resonances, which is interesting.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Whether or not the planets are still changing their orbits is an interesting question. The authors believe that the planets moved inwards via gas disk migration, which would require interactions with the initial protoplanetary disk. However, as the disk has clearly dissipated, it would seem that the migration has stopped, and has not happened for some time.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-02-24T14:43:25.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20163" PostTypeId="2" ParentId="20140" CreationDate="2017-02-24T00:36:14.770" Score="3" Body="&lt;p&gt;To copy &lt;a href=&quot;https://space.stackexchange.com/questions/13624/why-do-we-search-for-extra-terrestrial-water/13628#13628&quot;&gt;my answer from Space.SE&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Virtually all chemical reactions take place in a liquid or gaseous phase. Uncontained gasses aren't a likely candidate for life, so we're looking for solvents for life-related reactions to take place in.&lt;/p&gt;&#xA;&lt;p&gt;A good solvent is (a) common, (b) dissolves a wide variety of substances, and (c) is liquid at a wide range of temperatures. Very few liquids meet even two of these criteria, much less all three. The likely solvents for life are water and ammonia, and maybe methane.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1559" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-02-24T00:36:14.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20164" PostTypeId="1" AcceptedAnswerId="20167" CreationDate="2017-02-24T03:01:12.637" Score="21" ViewCount="4816" Body="&lt;p&gt;I was going through all the questions in this community related to &lt;strong&gt;TRAPPIST-1&lt;/strong&gt; in order to know how planets TRAPPIST-1b to TRAPPIST-1h were discovered, but there aren't any. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How were they discovered?&lt;/p&gt;&#xA;" OwnerUserId="13247" LastEditorDisplayName="user1569" LastEditDate="2019-03-07T14:46:58.080" LastActivityDate="2019-03-07T14:46:58.080" Title="How was Trappist-1 discovered?" Tags="&lt;exoplanet&gt;&lt;observational-astronomy&gt;&lt;nasa&gt;&lt;trappist-1&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="20165" PostTypeId="2" ParentId="20164" CreationDate="2017-02-24T04:00:50.163" Score="6" Body="&lt;p&gt;The dwarf star 2MASS J23062928-0502285 was first catalogued in 1999, if I've got that right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In May last year (2016) the Transiting Planets and Planetesimals Small Telescope–South (TRAPPIST) facility (its a .6 m automated scope in Chile) published their observations of the dwarf star and announced they had found 3 exo-planets orbiting it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their observations were then followed up by the VLT and the Spitzer Space Telescope (and others), and the 500 hours of observation from the SST has resulted in this announcement of the extra 4 exo-planets being identified, and further they are able to use that data to measure the sizes and masses of 6 of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wiki doth provide:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/TRAPPIST&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/TRAPPIST&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/TRAPPIST-1&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/TRAPPIST-1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=2MASS+J23062928-0502285#lab_notes&quot; rel=&quot;noreferrer&quot;&gt;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=2MASS+J23062928-0502285#lab_notes&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15741" LastActivityDate="2017-02-24T04:00:50.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20166" PostTypeId="2" ParentId="20164" CreationDate="2017-02-24T04:09:48.650" Score="3" Body="&lt;p&gt;The discovery, reported in the journal Nature, was made by astronomers using Nasa's exoplanet-hunting &lt;strong&gt;&lt;em&gt;Spitzer Space Telescope&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The telescope operates at the infrared wavelengths which glow brightest from &lt;strong&gt;&lt;em&gt;TRAPPIST-1&lt;/em&gt;&lt;/strong&gt;, and can detect the tiny dimming that occurs when a passing or &quot;transiting&quot; planet blocks out light from its star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spitzer's data allowed the team to measure precisely the sizes of the seven planets and estimate the masses and densities of six of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Spitzer was launched in 2003, and was never meant to continue in space for this long but the telescope is still making discoveries beyond what was imagined. It follows the earth's orbit around the sun, but travels slightly slower, so over time it gets further away from the earth. It is now in its &quot;final&quot; phase, which lasts until 2018.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further details:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;a href=&quot;https://www.theguardian.com/science/2017/feb/22/thrilling-discovery-of-seven-earth-sized-planets-discovered-orbiting-trappist-1-star&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.theguardian.com/science/2017/feb/22/thrilling-discovery-of-seven-earth-sized-planets-discovered-orbiting-trappist-1-star&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;a href=&quot;https://www.nasa.gov/press-release/nasa-telescope-reveals-largest-batch-of-earth-size-habitable-zone-planets-around/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.nasa.gov/press-release/nasa-telescope-reveals-largest-batch-of-earth-size-habitable-zone-planets-around/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15219" LastActivityDate="2017-02-24T04:09:48.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20167" PostTypeId="2" ParentId="20164" CreationDate="2017-02-24T04:18:46.673" Score="28" Body="&lt;p&gt;The star at the center of TRAPPIST-1 is called &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-basic?Ident=2MASS%20J23062928-0502285&quot; rel=&quot;nofollow noreferrer&quot;&gt;2MASS J23062928-0502285&lt;/a&gt;.  It was discovered by the &lt;a href=&quot;https://en.wikipedia.org/wiki/2MASS&quot; rel=&quot;nofollow noreferrer&quot;&gt;Two Micron All-Sky Survey&lt;/a&gt; (2MASS), which imaged the whole sky in the infrared between 1997 and 2001.  This resulted in a catalog of over 300 million objects.  TRAPPIST-1 itself was cataloged in 1999.  The name is actually its coordinates in right ascension and declination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The planets of the TRAPPIST-1 were discovered by the method of &lt;a href=&quot;https://en.wikipedia.org/wiki/Methods_of_detecting_exoplanets#Transit_photometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;transit photometry&lt;/a&gt;.  The way this works is a telescope watches a star for a period of time and records the amount of light coming from the star.  They chart how much light is coming from the star as a function of time, creating a &lt;a href=&quot;https://imagine.gsfc.nasa.gov/science/toolbox/timing1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;light curve&lt;/a&gt;.  If they see periodic dips in the intensity from the star, there is a high likelihood that that star has a &lt;a href=&quot;http://www.planetary.org/explore/space-topics/exoplanets/transit-photometry.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;planet in orbit around it&lt;/a&gt;.  The planet blocks the light from the star every time it passes between us and the star.  This causes the dips in the light curve.  One advantage to this method is that you can scan multiple stars in the same field of view, analyzing them all for planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By measuring how long it takes planets to pass in front of the star, how much light it blocks, and how often they orbit, scientists can calculate the masses of these planets and how far away from the star they are by using &lt;a href=&quot;https://www.nasa.gov/ames/kepler/measuring-the-mass-of-a-mars-size-exoplanet&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kepler's laws of motion&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;TRAPPIST-1 was initially determined to have planets orbiting it by the Transiting Planets and Planetesimals Small Telescope - South team.  From their data they determined that it had at least 3 planets.  One of these planets was in the habitable zone of the star.  They published their results in the journal &lt;a href=&quot;http://www.nature.com/nature/journal/v533/n7602/full/nature17448.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Nature&lt;/em&gt;&lt;/a&gt; in May 2016.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once TRAPPIST had determined that the system had planets around it, NASA trained the Spitzer Space Telescope on it.  Ground-based observations of Trappist-1 are difficult because it is so dim.  Spitzer, an infrared telescope, made more precise measurements of the light curves and determined that there were at least &lt;a href=&quot;https://www.nasa.gov/press-release/nasa-telescope-reveals-largest-batch-of-earth-size-habitable-zone-planets-around&quot; rel=&quot;nofollow noreferrer&quot;&gt;7 planets&lt;/a&gt; in orbit around it, 3 of which were in the habitable zone.  Additional observations were made by numerous other telescopes including the Very Large Telescope, UKIRT, the Liverpool Telescope, and the William Herschel Telescope.  The results were also published in &lt;a href=&quot;http://www.nature.com/nature/journal/v542/n7642/full/nature21360.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Nature&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JNH3s.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JNH3s.png&quot; alt=&quot;TRAPPIST-1&amp;#39;s light curve&quot;&gt;&lt;/a&gt;&#xA;Here's an image showing the light curve of the TRAPPIST-1 system &lt;a href=&quot;http://www.spitzer.caltech.edu/images/6282-ssc2017-01e-500-Hours-of-Spitzer-Observations-of-TRAPPIST-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;as measured by Spitzer&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13426" LastEditorUserId="13426" LastEditDate="2017-02-28T07:02:19.367" LastActivityDate="2017-02-28T07:02:19.367" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="20168" PostTypeId="2" ParentId="20164" CreationDate="2017-02-24T10:07:13.933" Score="18" Body="&lt;p&gt;Trappist-1 was first catalogued by the 2MASS survey about 17 years ago and has the catalogue number 2MASS J23062928-0502285.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was identified as an ultra- low-mass star with a spectral type of M7.5 by &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0004361&quot; rel=&quot;noreferrer&quot;&gt;Gizis et al. (2000)&lt;/a&gt; and &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0307429&quot; rel=&quot;noreferrer&quot;&gt;Cruz et al. (2003)&lt;/a&gt;, using a combination of 2MASS and proper motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason it was monitored by the Trappist telescope is that is was found to be reasonably close ($12.2\pm 0.4$ pc) by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2006AJ....132.1234C&quot; rel=&quot;noreferrer&quot;&gt;Costa et al. (2006)&lt;/a&gt; (who assigned it a spectral type M8) and is therefore quite bright for a star of its type at $V=18.8$.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-02-24T11:59:38.393" LastActivityDate="2017-02-24T11:59:38.393" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20169" PostTypeId="1" AcceptedAnswerId="20173" CreationDate="2017-02-24T10:37:46.520" Score="5" ViewCount="954" Body="&lt;p&gt;If I understand correctly, all the current exoplanet search is based on planets transitioning the distant star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the plane of ecliptic of that star happens to be perpendicular to line of sight from Earth to that star - or sufficiently inclined, so that orbits of all its planets as seen from Earth &quot;circle&quot; the star without ever transitioning it - does that mean we're completely unable to detect them?&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="83" LastEditDate="2017-02-24T10:47:26.650" LastActivityDate="2017-02-24T17:06:10.640" Title="Can we detect exoplanets with perpendicular plane of ecliptic?" Tags="&lt;exoplanet&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20173" PostTypeId="2" ParentId="20169" CreationDate="2017-02-24T13:42:18.517" Score="10" Body="&lt;p&gt;There are (at least) three main methods to detect planets that orbit so that the orbital plane is at right angles (or thereabouts) to our line of sight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first of these is direct imaging. This is already responsible for the discovery of a number of planets. However, in order to be visible against the glare of their host star, the exoplanets need to be in wide orbits and be massive. It also favours the detection of giant planets around &lt;em&gt;young&lt;/em&gt; stars, since these planets are still contracting, are still warm, and emitting their own infrared light. A list of about 20 directly imaged exoplanets can be found &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_directly_imaged_exoplanets&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;; arguably some of these should more rightly be called brown dwarfs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second method is the so-called astrometric detection of exoplanets. This works because the star and planet orbit their common centre of mass. This means that the star executes an orbit with the same period as the exoplanet, but with a semi-major axis that is smaller by the ratio of the planet mass to the star mass. This motion of the star can be detected, in a similar way to which parallax is measured, by precise measurement of position with respect to background objects. The method is most sensitive to wider binaries and massive stars, but is limited by the requirement to see appreciable motion of the star within the duration of the observational dataset. In practice then, this means it is difficult to detect planets with orbital periods of decades, unless one is prepared to wait for some time!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may only be one accepted exoplanet discovery by this method so far (&lt;a href=&quot;http://exoplanet.eu/catalog/hd_176051_b/&quot; rel=&quot;nofollow noreferrer&quot;&gt;HD 176051b&lt;/a&gt;), but the Gaia satellite should be capable of detecting many thousands of exoplanets with masses at or above that of Jupiter and with orbital radii of about 1--10 au (see &lt;a href=&quot;https://arxiv.org/abs/1411.1173&quot; rel=&quot;nofollow noreferrer&quot;&gt;Perryman et al. 2004&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thirdly, gravitational microlensing is quite insensitive to the plane of the orbit. This works when a foreground object gravitationally lenses and magnifies the light from a background star. An exoplanet is marked by a &quot;blip&quot; in the lensed light curve. There are also about &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_exoplanets_detected_by_microlensing&quot; rel=&quot;nofollow noreferrer&quot;&gt;20 planets detected by microlensing&lt;/a&gt;. The disadvantage of this technique is that the event is a non-repeatable one-off, so not much more can be learned about these systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notably, all three of these techniques are less sensitive to planets that have small mass (as indeed are transit experiments and the radial velocity technique). However, whilst small planets (in terms of mass) &lt;em&gt;have&lt;/em&gt; been found by microlensing, it seems unlikely that the other two techniques will discover Earth-like planets for quite some time.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-02-24T17:06:10.640" LastActivityDate="2017-02-24T17:06:10.640" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20174" PostTypeId="2" ParentId="20135" CreationDate="2017-02-24T14:28:40.330" Score="1" Body="&lt;p&gt;When looking at definitions of [transit] and [eclipse] or more general [occultations] it is noteworthy that the degree of covering is not defined, so technically speaking each eclipse is also a transit and an annular eclipse even more so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Details for next eclipses can be found at [this NASA page]. The next annular eclipse, err, transit &lt;a href=&quot;https://eclipse.gsfc.nasa.gov/SEplot/SEplot2001/SE2017Feb26A.GIF&quot; rel=&quot;nofollow noreferrer&quot;&gt;happens next Sunday 2017-02-26&lt;/a&gt; and is visible in South America, south Atlantic and Afrika.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you were referring to the [increasing distance between earth and moon] (+3.8 cm /yr), it then depends on an exact definition on when an eclipse is not regarded as an eclipse anymore. Assume for simplicity that this is the case, when the diameter of the moon has halved (one quarter the area), this means that the distance had to be doubled. &#xA;This would take [380 Gm] / 3.8 cm/yr = 10 Tyr (using SI prefixes). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that one might need to take other effects into account, too, for example the [increasing distance of earth and sun] (~15 cm / a) and that the [sun is becoming a red giant] in a couple of Gyr. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, would have loved to post more links, but are not allowed to currently. &lt;/p&gt;&#xA;" OwnerUserId="15824" LastEditorUserId="15824" LastEditDate="2017-02-24T14:54:22.370" LastActivityDate="2017-02-24T14:54:22.370" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20175" PostTypeId="1" CreationDate="2017-02-24T14:52:45.320" Score="3" ViewCount="194" Body="&lt;p&gt;The recent discovery of &lt;a href=&quot;http://www.nature.com/nature/journal/v542/n7642/full/nature21360.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;new planets around the star TRAPPIST-1&lt;/a&gt; happened by observations from 7 ground-based telescopes, as well as the Spitzer Space Telescope. The team of astronomers responsible has noted that there are many relevant parameters which still have large uncertainties, including orbital information and planetary mass. There's also some excitement about possible atmospheric composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any &lt;em&gt;definite&lt;/em&gt; planned future observations of the system? I assume that at some point it will be targeted by the &lt;a href=&quot;https://en.wikipedia.org/wiki/James_Webb_Space_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;James Webb Space Telescope&lt;/a&gt;, but that won't launch for another year and a half, and it has plenty of other targets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not looking for detailed information - I know that cannot be easily gathered - but if there are any general mentions, &lt;a href=&quot;http://subarutelescope.org/Topics/2016/01/21/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;like this one about the Subaru Telescope and Planet Nine&lt;/a&gt;, that would be fantastic.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2017-02-24T19:24:23.770" LastActivityDate="2017-05-26T18:03:00.890" Title="Are there any planned future observations of TRAPPIST-1 and its planets?" Tags="&lt;exoplanet&gt;&lt;observational-astronomy&gt;&lt;trappist-1&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20176" PostTypeId="1" AcceptedAnswerId="20177" CreationDate="2017-02-24T15:25:45.077" Score="4" ViewCount="127" Body="&lt;p&gt;Using transit photometry astronomers were able to discover Trappist b through h. Based on &lt;a href=&quot;https://astronomy.stackexchange.com/questions/20142/where-did-trappist-1a-go&quot;&gt;this&lt;/a&gt; question we learn that the naming of planets is based on their distance from their star (b being the closest).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do we actually determine the distance from the star? From my understanding with transit photometry we basically just see a shadow against the backdrop of the star. That can tell us the planet exists, but how do we know the distance between the planet and the star?&lt;/p&gt;&#xA;" OwnerUserId="1549" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2017-02-24T15:46:23.940" Title="How do we know the order of the new Trappist-1 planets?" Tags="&lt;exoplanet&gt;&lt;planetary-transits&gt;&lt;trappist-1&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20177" PostTypeId="2" ParentId="20176" CreationDate="2017-02-24T15:46:23.940" Score="4" Body="&lt;h3&gt;Using Kepler's laws.&lt;/h3&gt;&#xA;&lt;p&gt;We can determine the orbital period, &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;, each planet has by just looking at the time between transits for a given planet. Each planet has a distinct transit so its easy to distinguish which planet is transiting and we can calculate a period pretty easily.&lt;/p&gt;&#xA;&lt;p&gt;Kepler's law then tells us that if you're orbiting with a specific period, you &lt;strong&gt;have&lt;/strong&gt; to be orbiting at a specific distance (based on parameters of the system). The general equation is given by&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P^2 = \frac{4\pi}{GM_{star}}a^3$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; is just the gravitational constant, &lt;span class=&quot;math-container&quot;&gt;$M_{star}$&lt;/span&gt; is of course the mass of the star and fairly well known, and &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is the radius at which the planet orbits. Simply calculate &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; for each of your planets and put them in the correct order. You could even do this yourself using data from the paper.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-02-24T15:46:23.940" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20178" PostTypeId="1" AcceptedAnswerId="20186" CreationDate="2017-02-24T16:17:03.473" Score="12" ViewCount="3267" Body="&lt;p&gt;I know you need oxygen for the ignition, but presumably, if the bullet is impervious (or water-tight) and if there is a little air enclosed in the bullet case, wouldn't that be sufficient to fire it ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, giving the speed, would that be enough/too much to send it into orbit around the moon ? Would you need to fire it at a certain angle ?(which will presumably make it an elliptic orbit)&lt;/p&gt;&#xA;" OwnerUserId="10836" LastEditorUserId="1378" LastEditDate="2017-02-25T00:29:41.567" LastActivityDate="2020-06-02T14:53:51.020" Title="Can a bullet be fired on the moon and sent it into orbit?" Tags="&lt;gravity&gt;&lt;the-moon&gt;&lt;orbital-mechanics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20180" PostTypeId="2" ParentId="20161" CreationDate="2017-02-24T16:45:24.570" Score="5" Body="&lt;p&gt;This is a complicated question that would really require a full physics simulation and better knowledge of the system to accurately answer. But let's try a few back of the envelope calculations to see what we get.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Calculating tidal forces from TRAPPIST-1c on TRAPPIST-1b&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I'm going to calculate the tidal effects of TRAPPIST-1c on TRAPPIST-1b (simply because, a priori, this seems likely to be where the strongest tidal heating will be induced). See the figure below which describes the parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/g1qj4.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g1qj4.png&quot; alt=&quot;Diagram of setup&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tidal force of 1c on 1b is defined as the differential force of gravity across 1b, that is, the difference of the force of gravity on the side of 1b facing towards 1c and the force of gravity on the side of 1b facing away from 1c. Mathematically, we get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{tide,c-b} = F_{g,-R_b} - F_{g,+R_b} = \frac{GM_bM_c}{(D-R_b)^2} - \frac{GM_bM_c}{(D+R_b)^2} = 4GM_bM_c\frac{DR_b}{(D^2-R_b^2)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can presume that $R_b &amp;lt;&amp;lt; D$ (for this case $R_b/D = 1\%$) and reduce this to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{tide,c-b}(D) \approx 4GM_bM_c\frac{R_b}{D^3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this isn't enough to determine the amount of tidal heating that may occur. Tidal heating only occurs when the tidal force &lt;em&gt;changes&lt;/em&gt;. It is this constantly changing tidal force which results in tidal flexing of the planet and thus creating heat through tidal friction. Fortunately, for these two planets, the tidal force will be changing since $D$ will be constantly changing. So let's calculate $F_{tide}$ for the two extremes where these planets are as close as possible and as far as possible and difference them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta F_{tide,c-b} = F_{tide}(0.004\:\mathrm{AU}) - F_{tide}(0.026\:\mathrm{AU})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I plug in numbers to this, I find that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta F_{tide,c-b} \approx 3.7\times10^{20}\:\mathrm{N}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, but what do we do with this number? It's somehow a metric of the change in tidal forcing that 1c imparts on 1b, but is it negligible? To determine this, we have to compare it to something. Let's compare this to the tidal forcing that TRAPPIST-1b would receive from the star.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Calculating tidal forces from TRAPPIST-1 on TRAPPIST-1b&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I've already set up the math, so we don't need to go over that again. But first, let me discuss where this tidal forcing is actually coming from. Quoting an &lt;a href=&quot;http://www.space.com/35790-seven-earth-size-planets-trappist-1-discovery.html&quot; rel=&quot;noreferrer&quot;&gt;article from space.com&lt;/a&gt;, the paper's author, Gillon, states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because the seven alien worlds orbit so tightly, they're probably all tidally locked, Gillon said. That is, they likely always show the same face to their host star, just as Earth's moon only shows the &quot;near side&quot; to us. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As I said above, the only way to produce tidal heating is to have changing tidal forces. These planets are likely synchronized and always present the same side to the star. It's marginally possible these planets aren't perfectly tidally locked, but rather have some higher spin-orbit resonance. That is, their spin orbit resonance might not be 1:1 (as it would be if they were tidally locked) but instead could be something like 3:2 (&lt;a href=&quot;https://en.wikipedia.org/wiki/Mercury_(planet)#Spin.E2.80.93orbit_resonance&quot; rel=&quot;noreferrer&quot;&gt;which is what Mercury has&lt;/a&gt;). I'll ignore that complication though and just assume 1:1 resonance. So if they're tidally locked, they can't experience differing tidal forces through their own rotation. Instead, the differential tidal forcing comes from the orbit's ellipticity. Sometimes the planet will be closer and sometimes it will be farther, causing a differential tidal force on TRAPPIST-1b from the star as it orbits. This is exactly what occurs in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_heating_of_Io&quot; rel=&quot;noreferrer&quot;&gt;tidal heating of Io&lt;/a&gt;. Let's calculate $\Delta F_{tide,*-b}$ by using the varying distances TRAPPIST-1b will have from the star. I found that TRAPPIST-1b will orbit between $0.0101\:\mathrm{AU}$ and $0.0119\:\mathrm{AU}$&lt;sup&gt;1&lt;/sup&gt;. This means the differential tidal force is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta F_{tide,*-b} \approx 4GM_bM_*R_b\left(\frac{1}{(0.0119\:\mathrm{AU})^3} - \frac{1}{(0.0101\:\mathrm{AU})^3}\right) = 1.8\times10^{23}\:\mathrm{N}$$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Is planetary tidal heating non-negligible?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The back of the envelope calculations show that the differential tidal force on TRAPPIST-1b from TRAPPIST-1c is about $0.2\%$ of the differential tidal force due to the star. Whether you consider this negligible or not is up to you. I'd personally consider it a pretty small effect and say that most of the tidal heating these planets experience comes from the star itself.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Could the inter-planetary tidal heating still contribute to the tidal heating of the planets enough to heat the interior?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This is a remarkably hard question to answer and I can't even really do a back of the envelope calculation without making wild, unjustifiable assumptions. The calculations above simply determined the maximum tidal force variation over time. That doesn't tell us anything about how much tidal heating this may induce though. That requires knowing more about the planet itself, in particular the planet's &lt;a href=&quot;https://en.wikipedia.org/wiki/Love_number&quot; rel=&quot;noreferrer&quot;&gt;Love numbers&lt;/a&gt; which define the rigidity of the body and thus how easy it is to stretch via differential tidal forces. You can vary your tidal forcing as much as you want, but if your planet is pure iron (and thus very rigid) you're unlikely to have as much of an effect as if it were primarily silicate (and thus much less rigid). The paper produces the plot below which defines the potential constituents of each planet. This would be a first step in determining the planet rigidities, but as you can see from the error bars, it would be highly uncertain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IaIUl.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IaIUl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Overall, and this is entirely opinion based and from my calculations above, but I'd say the chances of the inter-planetary tidal heating having significant effects on the interior heat of of these planets is negligible. More than likely the biggest contributing factor is radioactive decay, followed by tidal heating from the star (but this is amplified by the eccentric orbits induced by planetary gravitational perturbations)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1 Note that this calculation involves using the eccentricity and the paper only provides an upper limit. These distances then represent an upper limit as well and the final answer will also be an upper limit. It may be less.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&#xA;Values used in calculations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$G = 6.67\times10^{-11}\:\mathrm{m^3 kg^{-1} s^{-2}}$&lt;/li&gt;&#xA;&lt;li&gt;$M_b = 5.075\times10^{24}\:\mathrm{kg}$&lt;/li&gt;&#xA;&lt;li&gt;$M_c = 8.239\times10^{24}\:\mathrm{kg}$&lt;/li&gt;&#xA;&lt;li&gt;$M_* = 1.604\times10^{29}\:\mathrm{kg}$&lt;/li&gt;&#xA;&lt;li&gt;$R_b = 7.34\times10^{6}\:\mathrm{m}$&#xA;&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-02-24T16:45:24.570" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20181" PostTypeId="2" ParentId="20175" CreationDate="2017-02-24T16:57:13.400" Score="2" Body="&lt;p&gt;A website &lt;a href=&quot;http://www.trappist.one/&quot; rel=&quot;nofollow noreferrer&quot;&gt;trappist.one&lt;/a&gt; has been created to detail information about this discovery. In the &amp;quot;Future&amp;quot; section, they specifically state:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In the short term, photometric follow-up using the repurposed Kepler satellite (named K2) along with with newer observations using Spitzer ought to reveal the period of planet 1h. We will also search for additional planets. Theses lightcurves, combined with ground-based measurements will increase the number of transit timing measurements for each of the planets. This will give us more accurate masses and orbital eccentricities. This will confirm whether the planets are mostly rocky or whether they contain a certain amount of volatiles, like water.&lt;/p&gt;&#xA;&lt;p&gt;In the medium term, we can expect the first attempts at detecting the atmospheres of the TRAPPIST-1 planets, using Hubble, followed by deeper investigations thanks to the James Webb. The James Webb could in principle measure the temperature of the planets, and detect the chemical composition of their atmospheres. It will do so by collecting dozens of eclipses of the TRAPPIST-1 planets. The advantage of having seven planets in one system is that we will be able to compare them to one another.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It seems they're currently using &lt;em&gt;Kepler&lt;/em&gt; and &lt;em&gt;Spitzer&lt;/em&gt; and plan to use &lt;em&gt;Hubble&lt;/em&gt; in the near future as well as &lt;em&gt;JWST&lt;/em&gt; after it launches.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-02-24T16:57:13.400" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20182" PostTypeId="1" AcceptedAnswerId="20183" CreationDate="2017-02-24T17:07:32.737" Score="2" ViewCount="242" Body="&lt;p&gt;As I understood tidal heating, it comes from tidal force acting upon a body as it spins, distorting it; the wave of distortion travels along the surface (along with apparent travel of the other body on the sky) and continued friction as the matter is strained into the traveling wave is the source of tidal heating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the two bodies are in a tidal lock, the distortion remains constant - it doesn't move. There's no new work done as the bodies remain immobile relative to each other. The tidal heating should be a flat zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, Io has immense volcanic activity attributed to tidal heating - despite it being tidally locked to Jupiter. While it still heats Jupiter, dragging its own tidal wave around it, Jupiter shouldn't contribute any heat to Io as its distortion remains constant over time, a stable equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that just residual heat from times when Io was spinning or am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2017-02-24T17:43:52.850" Title="What is the level of tidal heating between bodies that are already in mutual tidal lock?" Tags="&lt;tidal-forces&gt;&lt;tidal-locking&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20183" PostTypeId="2" ParentId="20182" CreationDate="2017-02-24T17:43:52.850" Score="1" Body="&lt;p&gt;There are actually several components of tidal forces which serve to distort a planet or moon - diurnal, nonsynchronous rotation, ice-shell thickening, orbit obliquity, and polar wander.  A moon could be stressed by any combination of these mechanisms, leading to frictional forces and heating.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Diurnal Stress&lt;/strong&gt; - Since orbits are ellipses, not circles, the moon will experience a differential gravitational field.  When the moon is closer to the planet, the tidal stresses will be slightly larger than when it is closer away, since the gradient will be steeper. In addition, since Kepler's 2nd Law informs us that a body moves faster when it is closer to its primary, this means that the tidal locking is imperfect.  When the moon is close to its primary, it is moving slightly faster than it is rotating.  Likewise, when it is further away it is moving slightly slower than it is rotating.  This leads to the moon seeing its primary as oscillating slightly in the sky.  &lt;a href=&quot;http://adsabs.harvard.edu/full/1997LPI....28..457G&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Nonsynchronous Rotation Stress&lt;/strong&gt; - If the moon's crust is decoupled from its core by a liquid layer (either liquid rock or liquid water), the crust can rotate freely over the core.  The core will stay tidally locked to the primary, but the shell can move around.  Since the core will have a tidal bulge, when the crust moves around it will become stressed.  The crust feels a &lt;a href=&quot;http://s3.amazonaws.com/academia.edu.documents/43068116/Thermal_state_of_an_ice_shell_on_Europa20160225-9471-11o5zx5.pdf?AWSAccessKeyId=AKIAIWOWYYGZ2Y53UL3A&amp;amp;Expires=1487960820&amp;amp;Signature=uI2xzaZtmBSa244Cjyw0%2FvqN0yI%3D&amp;amp;response-content-disposition=inline%3B%20filename%3DThermal_state_of_an_ice_shell_on_Europa.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;torque&lt;/a&gt; because its thickness varies across the moon's surface.  &lt;a href=&quot;http://planetary.brown.edu/pdfs/ab13.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Ice-Shell Thickening&lt;/strong&gt; - Icy moons, such as Europa, can experience stresses caused by their icy outer shells freezing and thickening.  As the moon loses heat, the water at the bottom of the shell will freeze.  This will increase its volume, creating extensional stresses.  At the surface, the cooling of the ice will contract it, causing compressional stresses.  While this is not a tidal stress, it is still a source of stressing which can act on these moons, so I thought I'd throw it in.  &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1029/2004JE002347/full&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Orbit Obliquity&lt;/strong&gt; - Most moons do not rotate exactly perpendicular to their orbital plane.  Rather, their rotational axis has some amount of obliquity.  This obliquity changes the latitudinal orientation of the tidal bulge as the moon orbits the planet.  This creates additional stresses as the tidal bulge gets pulled on.  &lt;a href=&quot;https://www.researchgate.net/profile/Hermes_Jara_Orue/publication/253804323_Effects_of_low-viscous_layers_and_a_non-zero_obliquity_on_surface_stresses_induced_by_diurnal_tides_and_non-synchronous_rotation_The_case_of_Europa/links/5817629508ae90acb2428cfe.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Polar Wander&lt;/strong&gt; - Large impacts can cause a moon's lithosphere to rotate and reorient itself relative to its rotational axis.  When this rotation changes the apparent location of the rotational poles, it is called &quot;polar wander&quot;.  Polar wander causes stress in a similar manner to nonsynchronous rotation.  The lithosphere rotates over the core's bulge, pushing on the crust.  &lt;a href=&quot;https://websites.pmc.ucsc.edu/~fnimmo/website/matsuyama_stress.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There may be other stress mechanisms that I am not aware of, but these are the main ones.  If you want more information about the topic, or want to see some visualizations of what these various stresses look like, check out &lt;a href=&quot;https://github.com/SatStressGUI/SatStressGUI&quot; rel=&quot;nofollow noreferrer&quot;&gt;SatStressGUI&lt;/a&gt;, a program I helped developed which models stresses on icy moons.&lt;/p&gt;&#xA;" OwnerUserId="13426" LastActivityDate="2017-02-24T17:43:52.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20184" PostTypeId="1" AcceptedAnswerId="20192" CreationDate="2017-02-24T18:24:12.187" Score="9" ViewCount="291" Body="&lt;p&gt;The planets orbiting TRAPPIST-1 are in orbits around their star much closer than Earth orbits Sol. However, since TRAPPIST-1 is a cool brown dwarf, some of the planets are in an &quot;inhabitable zone&quot;, where the temperature would not be too extreme for life as we know it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I wonder is, how long have those planets been in an inhabitable zone? The &lt;a href=&quot;https://en.wikipedia.org/wiki/TRAPPIST-1&quot; rel=&quot;noreferrer&quot;&gt;wikipedia page&lt;/a&gt; says that TRAPPIST-1 is older than 1 Gyr. But, how much of that time has it been a cool brown dwarf?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, since the abiogenesis of life and the evolution of living organisms has dependency on time, how much time has life had a chance to develop in the star's habitable zone?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this question, I'm not concerned about the chemistry or other characteristics of the planets themselves.&lt;/p&gt;&#xA;" OwnerUserId="10878" LastEditorUserId="10878" LastEditDate="2017-02-25T19:56:05.267" LastActivityDate="2017-02-25T22:40:12.673" Title="How long have the TRAPPIST-1 planets been in a habitable zone?" Tags="&lt;exoplanet&gt;&lt;trappist-1&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20186" PostTypeId="2" ParentId="20178" CreationDate="2017-02-24T19:01:32.693" Score="16" Body="&lt;p&gt;Almost. (&lt;strong&gt;EDIT:&lt;/strong&gt; actually yes, see below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed of an object in orbit depends on the radius of the orbit, and the mass of the body being orbited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WF8nb.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WF8nb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lowest orbit possible is where the bullet is just grazing the surface, so r = the radius of the Moon. M is the mass of the Moon, and G is the gravitational constant. Let's do the math:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=sqrt(gravitational%20constant%20*%20mass%20of%20moon%20%2F%20radius%20of%20moon)&quot; rel=&quot;noreferrer&quot;&gt;wolfram alpha calculation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is approx 1.7 km/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/.50_BMG&quot; rel=&quot;noreferrer&quot;&gt;.50 BMG&lt;/a&gt; Saboted Light Armor Penetrator (SLAP) reaches 1.2 km/s muzzle velocity. So even a very powerful round such as this one moves a bit too slowly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is conceivable that a tricked-out .50 round, with an extra load of powder, and possibly shot from a specially prepared rifle, could reach the 1.7 km/s speed needed; that would be a 40% increase in velocity over the SLAP round.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; According to &lt;a href=&quot;https://www.thefirearmblog.com/blog/2019/05/03/one-mile-per-second-or-5280-fps/&quot; rel=&quot;noreferrer&quot;&gt;this article&lt;/a&gt; saboted .224 caliber bullets loaded into .300 RUM (.300 Remington Ultra Magnum) cases can reach a speed of 1.7 km/s (the average they measured over the whole test was a little less than that, but the highest values exceeded 1.7).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case, if fired at the horizon, it would do a circular orbit, would go round the Moon and hit the shooter in the back 1 hour and 47 minutes later, provided it doesn't hit a mountain or something en route, and provided the orbit is along the equator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're not shooting along the equator, it will still do an orbit, but its motion relative to the surface of the Moon will be a weird spiral and it will not return to the starting point (the orbit itself will still be a closed loop, a circle, but the Moon spins underneath it, so seen from the Moon it appears like a spiral).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The circular orbit (standing on a hill, shooting at the horizon) requires the smallest muzzle speed to complete the loop. Any other orbit would require a greater muzzle velocity from the rifle. The more elongated the orbit, the higher the velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plus, if you shoot higher than the horizon, the bullet would hit the ground before it reaches you. You have to stand on an elevated point (a hill), and make sure the rifle does not point at the ground either forward or backward. This is because the return trajectory of the bullet after completing one loop is aligned with the rifle (the rifle's barrel is tangent to the trajectory), and if the butt of your rifle points at the ground, on the return trajectory the bullet would have to come out of the ground, which is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A laser mounted on the rifle must not point at the ground no matter whether the laser is pointing forward or backward, otherwise the orbit gets clipped by the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TLDR:&lt;/strong&gt; Stand on a tall hill or a mountain, shoot while the rifle is exactly horizontal. Use a top-tier extremely high velocity round, custom-loaded with extra powder, with a high caliber sniper rifle. Anything less would not work.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I know you need oxygen for the ignition&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You don't. Gun powder contains its own oxygen. You can fire guns in a vacuum just fine. Primers also go off just fine in a vacuum.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It should be noted that the important number here, 1.7 km/s, is the orbital velocity near ground level on the Moon. It must not be confused with the Moon's escape velocity (the speed at which an object breaks away from orbit and keeps going away into space forever, never to return), which is higher.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2020-05-31T23:40:56.190" LastActivityDate="2020-05-31T23:40:56.190" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="20187" PostTypeId="1" AcceptedAnswerId="20188" CreationDate="2017-02-24T19:23:01.680" Score="1" ViewCount="135" Body="&lt;p&gt;Assuming modern and near future technology, could we build a telescope powerful enough to actually see TRAPPIST-1's planets?&lt;/p&gt;&#xA;" OwnerUserId="15984" LastActivityDate="2017-02-24T19:44:59.363" Title="Is it realistic to build a telescope powerful enough to take a photo of TRAPPIST-1's planets?" Tags="&lt;telescope&gt;&lt;trappist-1&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20188" PostTypeId="2" ParentId="20187" CreationDate="2017-02-24T19:44:59.363" Score="1" Body="&lt;p&gt;It's not out of the question, but it would be very difficult to use &lt;a href=&quot;https://en.wikipedia.org/wiki/Methods_of_detecting_exoplanets#Direct_imaging&quot; rel=&quot;nofollow noreferrer&quot;&gt;direct imaging&lt;/a&gt; for any of the exoplanets around TRAPPIST-1:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The exoplanets are small, all not much bigger than Earth. Ideally, you want a target the size of Jupiter or bigger. Likewise, they are very low-mass, and are thus small. Once more, this is a problem. Direct imaging has only had much success with large planets, of many Jupiter masses.&lt;/li&gt;&#xA;&lt;li&gt;The exoplanets are close to the star, with the furthest one out - admittedly with a poorly-constrained orbit - having a semi-major axis of 0.063 AU (and a year of 20 days), give or take. That is tiny. Mercury's semi-major axis is more than six times that. If the exoplanets are close to the star, it makes it much harder to distinguish them from the star's glare. Granted, there are methods that do this very well, but they're not meant for planets this close.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To see what I'm talking about, look at &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_directly_imaged_exoplanets&quot; rel=&quot;nofollow noreferrer&quot;&gt;some examples of directly-imaged exoplanets&lt;/a&gt;. Almost all in that list have many times Jupiter's mass and at least its radius. Their semi-major axes are many, many - sometimes thousands - times that of Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't speculate as to what technology we'll have decades from now, but if you want to find out more about the exoplanets around TRAPPIST-1, direct imaging is not the way to go, and I suspect it will not be a very popular choice among astronomers.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2017-02-24T19:44:59.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20189" PostTypeId="1" CreationDate="2017-02-25T03:03:55.453" Score="4" ViewCount="149" Body="&lt;p&gt;Is the albedo of any of the Trappist-1 planets know? Is it high enough so if you are on the surface of one of the planets, you can see the others? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked up the distances from the star and size of the planets and it seems a few of them will appear larger than the Moon in the night sky, unless the albedo is really low that is.&lt;/p&gt;&#xA;" OwnerUserId="14988" LastActivityDate="2020-09-11T18:04:50.490" Title="Trappist-1 planet albedo" Tags="&lt;exoplanet&gt;&lt;trappist-1&gt;&lt;albedo&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20190" PostTypeId="1" AcceptedAnswerId="20191" CreationDate="2017-02-25T05:35:49.130" Score="2" ViewCount="358" Body="&lt;p&gt;Has the Roche limit affect been observed having an effect on an constructed object gravitational wise to cause damage? Could the Roche limit be used to slow an object?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-01-18T21:16:57.770" LastActivityDate="2018-01-18T21:16:57.770" Title="Is the Roche limit used this way?" Tags="&lt;gravity&gt;&lt;observational-astronomy&gt;&lt;roche-limit&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20191" PostTypeId="2" ParentId="20190" CreationDate="2017-02-25T10:24:06.027" Score="8" Body="&lt;p&gt;&lt;strong&gt;No&lt;/strong&gt;. The Roche limit refers to the radius at which a satellite which is held together &lt;strong&gt;only by its own self gravity&lt;/strong&gt; is disrupted by tidal forces. Spacecraft are not such satellites, but are much harder to tear apart.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2017-02-25T10:24:06.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20192" PostTypeId="2" ParentId="20184" CreationDate="2017-02-25T10:50:04.880" Score="8" Body="&lt;p&gt;The luminosity of Trappist-1 is estimated to be $5.25\times 10^{-4}\ L_{\odot}$, but it has not always been like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The luminosity of a brown dwarf decreases with time and it this measured luminosity (along with the spectral type) that allows an &lt;em&gt;estimate&lt;/em&gt; of the mass and a &lt;em&gt;lower limit&lt;/em&gt; to the age using stellar evolutionary models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I look at the &lt;a href=&quot;https://arxiv.org/abs/1503.04107&quot; rel=&quot;nofollow noreferrer&quot;&gt;Baraffe et al. (2015)&lt;/a&gt; low-mass evolutionary models and look at the locus of luminosity versus time for a $0.08\ M_{\odot}$ star like Trappist-1, you can see that the current luminosity implies an age of $\sim 500$ million years. But if you go back in time, the star was more luminous and for that reason, planets that are &lt;em&gt;currently&lt;/em&gt; in the habitable zone (said to be planets e,f,g) were not so in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The details of a habitable zone (HZ) calculation can be complex, but basically the radius of the habitable zone scales as the square root of the luminosity. If planets d and h are &lt;em&gt;not&lt;/em&gt; currently in the HZ then we can use these as a conservative definition of the HZ boundary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this (and using the published orbital radii of the planets), I can see that if the luminosity is increased by a factor of 9, then &lt;em&gt;none&lt;/em&gt; of the planets b-g are in the HZ, it is larger than all of their orbits. Trappist-1 had a luminosity that was 9 times larger when it was younger than 27 million years old. On the other hand, if I want to move the HZ just outside the orbit of planet e (and simultaneously include planet h inside the HZ), then this would occur when Trappist-1 was at an age of 206 million years. As a final thought, you can see from this particular model that Trappist-1 may fade by a further factor of two as it gets older. This &lt;em&gt;decreases&lt;/em&gt; the HZ radius by a factor of 1.41 and would mean that g (and possibly f) would fall outside the HZ, whilst d (and possibly c) would be brought into the HZ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted however that: different models give slightly different results, these loci are mass dependent and the mass is not known, it is inferred from the same models using a temperature estimate (which is also uncertain). So whilst my qualitative conclusions about the &lt;em&gt;past&lt;/em&gt; location of the HZ are likely to be correct (though the detailed age numbers are model-dependent), the future behaviour of the HZ is more uncertain because Trappist-1 may be slightly more massive than assumed and &lt;em&gt;already&lt;/em&gt; reached its minimum luminosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/g9a7q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g9a7q.png&quot; alt=&quot;Luminosity evolution of Trappist-1&quot;&gt;&lt;/a&gt;&#xA;&lt;em&gt;The luminosity evolution of Trappist-1, assuming a mass of $0.08M_{\odot}$ and the models of Baraffe et al. (2015). The horizontal dashed line marks the best estimate of its current luminosity, for which planets e-g are said to be in the HZ. If we go back in time, the rightmost vertical dashed line marks the age below which the luminosity increases to the point that e becomes too hot to be habitable. Then further back, the leftmost vertical dashed line marks the point where all the currently known planets (b-h) become uninhabitable.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to your question is quite uncertain and depends critically on the age of Trappist-1 now and of course &lt;strong&gt;whether the planets were always at the orbital radii they are now&lt;/strong&gt;$^{*}$. As you can see from the plot above (note the logarithmic scale on the x-axis), the luminosity evolution referred to above takes place early-on. If Trappist-1 could be as young as 500 million years, then life on planet e may only have been possible for 300 million years. However, if the star is slightly more massive and is 10 billion years old, then life has had 9.8 billion years to get going.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are talking about planet f, then it has had slightly longer ($\sim +100$ million years) inside the HZ, and planet g slightly longer ($\sim +70$ million years) again. Planet h will have spent comparatively little time (in the past), less than a few hundred million years, inside the HZ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^{*}$ The abstract of the discovery paper by &lt;a href=&quot;http://www.eso.org/public/archives/releases/sciencepapers/eso1706/eso1706a.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gillon et al. (2017)&lt;/a&gt; briefly discusses the possibility that the planets migrated &lt;em&gt;inwards&lt;/em&gt; after formation, through a process of &quot;disc-driven migration&quot;. If so, then this will &lt;em&gt;not&lt;/em&gt; alter the discussion above. Disc around very low-mass stars may be longer-lived than those around higher mass stars, but have essentially dispersed after $\sim 10-20$ million years (&lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0004-637X/695/2/1210/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kennedy &amp;amp; Kenyon 2009&lt;/a&gt;; &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013MNRAS.429..903D&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dawson et al. 2013&lt;/a&gt;; &lt;a href=&quot;https://arxiv.org/abs/1611.07416&quot; rel=&quot;nofollow noreferrer&quot;&gt;Binks &amp;amp; Jeffries 2017&lt;/a&gt;), and the planetary configuration would have to be settled where it is now by the time the disc has gone.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-02-25T22:40:12.673" LastActivityDate="2017-02-25T22:40:12.673" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20193" PostTypeId="2" ParentId="20190" CreationDate="2017-02-25T14:18:54.910" Score="5" Body="&lt;p&gt;You have asked two different questions, Muze. The title of your question asks whether the Roche limit is dangerous to spacecraft, but the body asks whether the Roche limit has been observed having an effect on an object. The answers to these two very different questions are &quot;no&quot; and &quot;yes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main question first: The Roche limit applies to objects that are weakly bound by self gravitation. A good example was Comet Shoemaker–Levy 9, which broke apart into a number of pieces before colliding with Jupiter in 1994. A prior close approach to Jupiter in 1992 tore the comet into pieces thanks to tidal effects. The collision in 1994 was a &lt;a href=&quot;http://www2.jpl.nasa.gov/sl9/back1.html&quot; rel=&quot;noreferrer&quot;&gt;string of pearls&lt;/a&gt; hitting Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spacecraft are held together via welds, nuts and bolts, screws, and inter-atomic and inter-molecular forces. Each of these is many orders of magnitude stronger than is gravitation. The Roche limit doesn't apply to spacecraft as spacecraft aren't held together by gravitation.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2017-02-25T14:24:04.257" LastActivityDate="2017-02-25T14:24:04.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20194" PostTypeId="2" ParentId="20164" CreationDate="2017-02-25T14:43:19.733" Score="2" Body="&lt;p&gt;Another reason for the interest. Because the star is so dim and small, planetary signatures in the infrared stand out much better than they would with a sunlike star. The star has been described as an 'ultracool brown dwarf', which would imply that it doesn't have much nuclear fusion in its interior. The planets are incredibly close to their star (much closer than Mercury in our system) which is why they are relatively warm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, for the planets to be found at all, there is a freakish alignment where the planetary orbits are all aligned so that they eclipse their parent star from our vantage point - they are all moving in the ecliptic - the 'dinner plate' formed by their circles around their parent star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither of these would be true if we were looking at our own solar system from afar - the Sun would drown out the signatures of Earth-sized planets with current telescope technology, and only one or two of the planets in our system would pass in front of the Sun, due to orbits in our solar system being inclined above and below the ecliptic. So this is extreme luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Discussing the planets as 'earthlike' is a HUGE stretch. They aren't gas giants like Jupiter, and their size indicates they are probably rocky. But Earth and Venus would look the same from this distance - and the surface of Venus is near 1000F with atmospheric pressure 100x that of Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as visiting - the most advanced plans for interstellar spacecraft involve &quot;ships&quot; weighting a few grams moving a few percent of the speed of light. It would take several hundred years for such microprobes to reach this system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big excitement is that, having such a small and dim star, space telescopes in the near term will be able to collect infrared signatures from the planets, and thereby get atmospheric composition - not possible with other &quot;earthlike&quot; planets to date. And with 7 examples, we'll have our first real statistics for &quot;earthlike&quot; exoplanet features.&lt;/p&gt;&#xA;" OwnerUserId="16001" LastActivityDate="2017-02-25T14:43:19.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20195" PostTypeId="2" ParentId="20175" CreationDate="2017-02-25T16:47:56.290" Score="0" Body="&lt;p&gt;The video shared by NASA specifically mentions that JWST will be used to observe the TRAPPIST 1 system. Infact that's going to be one of the prime targets.&lt;/p&gt;&#xA;" OwnerUserId="15901" LastActivityDate="2017-02-25T16:47:56.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20196" PostTypeId="1" AcceptedAnswerId="20197" CreationDate="2017-02-25T17:22:36.057" Score="4" ViewCount="259" Body="&lt;p&gt;According to NASA, we've discovered 3,453 exoplanets so far&lt;a href=&quot;http://exoplanetarchive.ipac.caltech.edu/&quot; rel=&quot;noreferrer&quot;&gt; [1]&lt;/a&gt;. With some of the farthest being at 13,000 light years away&lt;a href=&quot;http://www.space.com/29120-alien-planet-among-farthest-known.html&quot; rel=&quot;noreferrer&quot;&gt; [2]&lt;/a&gt;. This would suggest today scientists have a really advanced capability for this kind of discoveries. However, the existence of a ninth planet in the solar system (which has not only been proposed but its characteristics are already predicted with great precision&lt;a href=&quot;https://en.wikipedia.org/wiki/Planet_Nine&quot; rel=&quot;noreferrer&quot;&gt;[3]&lt;/a&gt;) has not yet been discovered. If a ninth planet existed in our solar system, shouldn't it have already been discovered? How can exoplanets thousands of light years away be discovered weekly, while a giant planet right in our backyard still lays undiscovered?&lt;/p&gt;&#xA;" OwnerUserId="16007" LastActivityDate="2017-02-25T19:58:15.313" Title="If thousands of exoplanets have been discovered, why haven't we discovered planet 9 yet?" Tags="&lt;planet&gt;&lt;telescope&gt;&lt;exoplanet&gt;&lt;9th-planet&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2017-02-26T13:55:26.430" ContentLicense="CC BY-SA 3.0" />
  <row Id="20197" PostTypeId="2" ParentId="20196" CreationDate="2017-02-25T18:38:50.197" Score="6" Body="&lt;p&gt;There are two primary reasons &lt;a href=&quot;https://en.wikipedia.org/wiki/Planet_Nine&quot; rel=&quot;noreferrer&quot;&gt;Planet Nine&lt;/a&gt; (if it exists) is extremely hard to detect: it's very dim and its position is not well constrained.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Brightness&lt;/strong&gt; - We are able to see planets in our solar system because they reflect light from the sun.  The amount of light they reflect depends on how large the object is, how reflective it is, and (most importantly) the amount of sunlight that hits it.  The hypothesized distance at which Planet Nine is orbiting is so far away that it will receive only a tiny amount of sunlight. At this distance, its apparent magnitude would be greater than 22.  In comparison, &lt;a href=&quot;https://en.wikipedia.org/wiki/Pluto&quot; rel=&quot;noreferrer&quot;&gt;Pluto's apparent magnitude&lt;/a&gt; is about 15.  Magnitude is a logarithmic scale, so that means that Planet Nine would be $631$ times dimmer than Pluto, making it incredibly hard to see.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Unknown Position&lt;/strong&gt; - The estimates of Planet Nine's characteristics are based on how it gravitationally disrupts the orbits of some trans-Neptunian objects.  Based on these observations, scientists can calculate where it is likely to be, but these calculations depend on its size.  &lt;a href=&quot;https://www.washingtonpost.com/news/speaking-of-science/wp/2016/01/20/new-evidence-suggests-a-ninth-planet-lurking-at-the-edge-of-the-solar-system/&quot; rel=&quot;noreferrer&quot;&gt;Estimates of its mass&lt;/a&gt; range from 2 to 4 times the mass of the Earth.  And all of these are just estimates, giving us a relatively large patch of sky to search to find Planet Nine.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Combining both of these reasons, it is very difficult to search for Planet Nine.  We're looking for an incredibly dim object in a large area of the sky.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, when we look for exoplanets, things are actually a lot easier.  To &lt;a href=&quot;https://en.wikipedia.org/wiki/Methods_of_detecting_exoplanets&quot; rel=&quot;noreferrer&quot;&gt;detect exoplanets&lt;/a&gt; we hardly ever look for the planets directly.  We've only detected about &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_directly_imaged_exoplanets&quot; rel=&quot;noreferrer&quot;&gt;20 planets&lt;/a&gt; via direct imaging.  Instead, we look at the light from the star.  The stars are many orders of magnitude brighter than the planets, and are very easy to see.  All we have to do is see how the light from the star changes over time.  An early method detect shifts in the radial velocity of the star (its motion towards or away from us) using &lt;a href=&quot;https://en.wikipedia.org/wiki/Doppler_spectroscopy&quot; rel=&quot;noreferrer&quot;&gt;Doppler spectroscopy&lt;/a&gt;.  The more prevalent method today, the one that the Kepler mission used, is &lt;a href=&quot;https://en.wikipedia.org/wiki/Methods_of_detecting_exoplanets#Transit_photometry&quot; rel=&quot;noreferrer&quot;&gt;transit photometry&lt;/a&gt;.  We measure the light output of a star, and look for dips in its intensity, caused by the planet orbiting in front of the star and blocking some of the light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because we're looking at a star, we don't have the problems we have when looking for Planet Nine.  Stars are very bright, their positions are known very accurately, and we don't have to wait too long to make a detection.  Plus, surveys like &lt;a href=&quot;https://www.nasa.gov/mission_pages/kepler/overview/index.html&quot; rel=&quot;noreferrer&quot;&gt;Kepler&lt;/a&gt; are able to scan many, many stars at once.&lt;/p&gt;&#xA;" OwnerUserId="13426" LastEditorUserId="13426" LastEditDate="2017-02-25T19:58:15.313" LastActivityDate="2017-02-25T19:58:15.313" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20199" PostTypeId="1" AcceptedAnswerId="20204" CreationDate="2017-02-26T00:42:31.777" Score="3" ViewCount="578" Body="&lt;p&gt;The Moon is tidally locked to the Earth, the Four Galilean are tidally locked as well, and the recently found planetary sistem TRAPPIST-1 has seven tidally locked planets, but Venus or Mercury are not. Why? Is there any conditions or analytic formula which says if a small body will be eventually tidally locked to the major body?&lt;/p&gt;&#xA;" OwnerUserId="9949" LastEditorUserId="9949" LastEditDate="2017-02-26T00:51:52.000" LastActivityDate="2019-05-21T14:38:25.430" Title="Requirements for a satellite/planet to be tidally locked to a planet/star" Tags="&lt;gravity&gt;&lt;tidal-forces&gt;&lt;celestial-mechanics&gt;&lt;tidal-locking&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20200" PostTypeId="1" AcceptedAnswerId="33180" CreationDate="2017-02-26T01:23:48.623" Score="5" ViewCount="698" Body="&lt;p&gt;There is a this poster on the &lt;a href=&quot;https://exoplanets.nasa.gov/trappist1/&quot; rel=&quot;noreferrer&quot;&gt;NASA site&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;https://i.stack.imgur.com/wcOhu.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wcOhu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that irks me. Of course its an artist's fantasy, but since it is on a NASA site, I see people considering it as scientifically accurate. So, beyond the two obvious &quot;In which direction(s) is the sun&quot;, and &quot;OMG these two planets are crashing into each other! Before we crash in the debris!&quot;, what else is wrong? Tides? Roche's limit? Atmospheres ripped out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, since 1e is an intermediate planet, it is possible to have the other six planets all conveniently placed in the same area of the sky?&lt;/p&gt;&#xA;" OwnerUserId="16013" LastActivityDate="2019-08-28T10:28:57.673" Title="How many things are wrong in this &quot;artist view&quot; of the TRAPPIST-1 system?" Tags="&lt;trappist-1&gt;&lt;planetary-science&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20201" PostTypeId="1" CreationDate="2017-02-26T01:52:40.247" Score="0" ViewCount="200" Body="&lt;p&gt;Are there any galaxies that are partially with the visible universe? It seems unlikely as most of the universe is empty space however our circumference makes it seem likely as well. It would be a rather interesting thing to know about.&lt;/p&gt;&#xA;" OwnerUserId="16015" LastActivityDate="2017-02-27T15:23:45.493" Title="Are any galaxies partially in the observable universe?" Tags="&lt;galaxy&gt;&lt;observable-universe&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20202" PostTypeId="2" ParentId="20200" CreationDate="2017-02-26T05:05:27.757" Score="5" Body="&lt;p&gt;The 2 big planets are probably f &amp;amp; g and they don't look up to scale to me. While f is the size of Earth, or almost 4 times bigger than the Moon, but it's also further away, ~ 1.3 million kilometers, as opposed to 400,000km for the Moon. So I would say f will be somewhat larger than the moon but not anywhere near what the poster shows. G will appear even smaller than f because it's further away.&#xA;I haven't looked at the orbits in details but I think that sort of event should be possible. Trappist-1e being intermediate planet makes no difference. &#xA;As for &quot;where is the sun&quot; - it appears to be somewhere to the right, probably below the horizon. That makes sense because any base would probably be along the terminator.&lt;/p&gt;&#xA;" OwnerUserId="14988" LastActivityDate="2017-02-26T05:05:27.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20203" PostTypeId="2" ParentId="20200" CreationDate="2017-02-26T07:01:38.377" Score="4" Body="&lt;p&gt;&lt;strong&gt;Hominids in the neighborhood&lt;/strong&gt; is probably the biggest mistake made by the artist. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry to be a little less optimistic than people at NASA, but a journey this far would probably take hundreds of thousands of years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we're not in the best condition to prepare for it. The technology needed for such a trip should take time to develop. Too much time, too much ressources... Our time there is limited, and we can difficulty move to another planet to live.&lt;/p&gt;&#xA;" OwnerUserId="11264" LastActivityDate="2017-02-26T07:01:38.377" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20204" PostTypeId="2" ParentId="20199" CreationDate="2017-02-26T08:13:09.647" Score="4" Body="&lt;p&gt;Wikipedia gives the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$t_{\text{lock}} \approx \frac{\omega a^6 I Q}{3 G m_p^2 k_2 R^5}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; is the initial spin rate expressed in radians per second,&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is the semi-major axis of the motion of the satellite around the central body (given by the average of the periapsis and apoapsis distances),&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$I\approx 0.4 m_s R^2$&lt;/span&gt; is the moment of inertia of the satellite, where &lt;span class=&quot;math-container&quot;&gt;$m_s$&lt;/span&gt; is the mass of the satellite and &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; is the mean radius of the satellite,&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt; is the dissipation function of the satellite,&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; is the gravitational constant,&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$m_p$&lt;/span&gt; is the mass of the central body, and&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$k_2$&lt;/span&gt; is the tidal Love number of the satellite.&quot; &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_locking&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Notice that the semi-major-axis is the power of 6. A small change in the orbital distance can have a very large effect on whether the body will become locked. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note the terms &lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt; (a measure of how elastic the body is, and hence how much energy is lost in distortion &lt;a href=&quot;https://websites.pmc.ucsc.edu/~pkoch/EART_206/09-0127/Goldreich%20&amp;amp;%20Soter%2066%20Icarus%205-375.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;) and &lt;span class=&quot;math-container&quot;&gt;$k_2$&lt;/span&gt; (a measure of the rigidity of the body: how much it is distorted by tides &lt;a href=&quot;https://en.wikipedia.org/wiki/Love_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;), which are both hard to measure. The 7 detected planets in the Trappist-1 system are all much closer to their star than any of the sun's planets and for reasonable values of &lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$k_2$&lt;/span&gt;, the planets will be tidally locked within a few million years, much as the moons of Jupiter are tidally locked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our solar system, mercury is asynchronously locked with the sun (on a 3:2 resonance) The Earth is not locked: it is too far from the sun, and the moon has a big effect. Mars is much too far out, and Venus is &lt;a href=&quot;https://space.stackexchange.com/questions/2289/why-is-the-rotation-rate-of-venus-so-slow&quot;&gt;odd&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="1987" LastEditDate="2019-05-21T14:38:25.430" LastActivityDate="2019-05-21T14:38:25.430" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="20205" PostTypeId="1" CreationDate="2017-02-26T14:13:24.693" Score="8" ViewCount="281" Body="&lt;p&gt;What are future steps that can be taken in order to learn more about the inner workings of a Neutron Star such as &quot;nuclear pasta&quot; and the materials in the inner core?&lt;/p&gt;&#xA;" OwnerUserId="16019" LastActivityDate="2017-02-26T15:22:53.993" Title="Inner Workings of Neutron Stars" Tags="&lt;neutron-star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20206" PostTypeId="2" ParentId="20205" CreationDate="2017-02-26T15:22:53.993" Score="9" Body="&lt;p&gt;An incomplete shortlist would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) If you can measure the maximum possible mass of a neutron star (the most massive neutron star gives a lower limit to that), then it tells you how &quot;hard&quot; or &quot;soft&quot; the equation of state (the relationship between pressure, density and composition) is in the interior (especially at the core), and therefore tells you something about the matter composition and the nature of the strong nuclear force at small separations. Nuclear pasta is unlikely to have a dramatic effect on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) Measuring the masses &lt;em&gt;and&lt;/em&gt; radii of neutrons stars would directly tell you about the equation of state at different depths inside a neutron star. Again, I think nuclear pasta does not have a major bearing on these diagnostics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(iii) Measuring the rate at which neutron stars cool, tells you about the interior composition. Quark matter and boson condensations in the core would allow for faster cooling. Abrupt changes to the cooling rate can tell you about superfluid phase transitions. Cooling can also be accelerated by nuclear pasta phases by allowing the URCA process to operate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(iv) Measuring the response of pulsars to glitches tells you about superfluidity in the interior and crust and the coupling of crust and core will depend on the properties of what is in between - i.e. nuclear pasta.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-02-26T15:22:53.993" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20207" PostTypeId="1" CreationDate="2017-02-26T19:13:02.523" Score="4" ViewCount="91" Body="&lt;p&gt;In my assignment I have to deal with the 2D spacetime metric $$ds^2=-\rho^2d\alpha^2+d\rho^2.$$ During this assignment we have shown that the constant $$l=\rho^2\frac{d\alpha}{d\tau}$$ is a constant along the geodesics in that space. After this we found an expression for $\rho$ in terms of $\alpha$, for which $\rho(\alpha=0)=l$, namely: $$\rho(\alpha)=l\frac{2e^\alpha}{e^{2\alpha}+1}.$$&#xA;We plotted this function for various values of l, which gave:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/6SEfy.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6SEfy.jpg&quot; alt=&quot;Expression for $\rho(\alpha)$ plotted for various values of l.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we showed that in the close neighbourhood of the Schwarschild radius, i.e. $0&amp;lt;r-r_s\ll 1$, the metric $ds^2=-\rho^2 d\alpha^2+d\rho^2$ corresponds to the metric $ds^2=-(1-\frac{r_s}{r})dt^2+(1-\frac{r_2}{r})^{-1}dr^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what we need to do is, use the result that the metric is approximately the schwarzschild metric at the schwarzschild radius to interpret the plots we found in the figure shown above and give a possible interpretation for l. And I really have no idea how to interpret these figures or what l might be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas are welcome!&lt;/p&gt;&#xA;" OwnerUserId="9713" LastActivityDate="2017-02-26T19:13:02.523" Title="On the geodesics of the metric $ds^2=-\rho^2d\alpha^2+d\rho^2$ and the constant $l=\rho^2\frac{d\alpha}{d\tau}$" Tags="&lt;general-relativity&gt;&lt;mathematics&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20210" PostTypeId="1" AcceptedAnswerId="20213" CreationDate="2017-02-27T11:05:42.550" Score="5" ViewCount="1337" Body="&lt;p&gt;As far as my (somewhat basic) knowledge of astrophysics goes in general the closer to a star your orbit gets smaller (because you travel less distance) and faster (because you're deeper in the gravity well and need to go quicker to avoid falling in.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what happens on bodies orbiting close to black holes and subject to time dilation as far as an external observer is concerned?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question isn't about interstellar but it's a good point of reference. Take the water planet they drop onto where gravitational time dilation is enough to make 20 or 30 years pass to the external observers while only a few hours pass for them. (I don't know if this is a realistic rate(?) of time dilation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would the apparent orbit of their planet be to the external orbit and what would his orbit appear to be for them? Even though he is further from the star would he complete more orbits in the elapsed time despite being further from the centre of the system? And would they appear to be orbiting &quot;slower&quot; than him from his point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out of interest, though perhaps a second question is more appropriate, how would the apparent difference in relative orbital speed affect rendezvous manoeuvres if at all?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example would the standard decelerate and fall into a lower orbit still apply but the trajectory plotted from a static observer look weird as time dilation set in? Or would the manoeuvre have to be different to account for the apparent velocity difference? What would a non-circular orbit look like in this system when observed from a circular orbit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also is the velocity required for orbit so close to a black hole high enough that speed based relativity also comes into play?&lt;/p&gt;&#xA;" OwnerUserId="16022" LastActivityDate="2017-02-28T21:51:05.513" Title="What effect does time dilation have on bodies orbiting close to Black Holes?" Tags="&lt;black-hole&gt;&lt;space-time&gt;&lt;orbital-mechanics&gt;&lt;time-dilation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20211" PostTypeId="2" ParentId="20201" CreationDate="2017-02-27T14:56:45.380" Score="0" Body="&lt;p&gt;The observable universe, by definition, is the area of the universe from which light has reached us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this boundary could lie in the middle of a galaxy, some of the oldest light (at the edge of the observable universe) could be old enough that it originates from a time before galactic formation. In this case, the cutoff could exist on a semi-congealed mass of gas or whatever structure was present at the time, but there would be no galaxies to speak of.&lt;/p&gt;&#xA;" OwnerUserId="15904" LastActivityDate="2017-02-27T14:56:45.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20212" PostTypeId="2" ParentId="20201" CreationDate="2017-02-27T15:23:45.493" Score="2" Body="&lt;p&gt;Perhaps, on rare occasions this will occur.  Quoting from Phil Plait in an  &lt;a href=&quot;http://www.astronomycast.com/2014/05/ep-343-the-universe-is-trying-to-kill-you/&quot; rel=&quot;nofollow noreferrer&quot;&gt;astronomycast&lt;/a&gt; , &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dr. Plait:    Yeah, because the universe is expanding ever faster every&#xA;  day, basically the distance we will be able to see, away from the&#xA;  earth, is going to get closer and closer to us. Anything past the&#xA;  distance will be moving away from us too rapidly to see and so&#xA;  eventually, all that is going to be left is some sort of super galaxy&#xA;  that is us and the Andromeda galaxy and all of the little dwarf&#xA;  galaxy’s, will have all combined into, sort of, one thing. Of course,&#xA;  all of the stars will be dead and then that’s it.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;So, you know, if you are a civilization, somehow, that evolves a&#xA;  hundred trillion years from now, the universe is going to look way&#xA;  different than it does now. You are not going to see any other&#xA;  galaxies; it’s just going to be you.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Frazier:  You are not going to see any stars, because it will all be&#xA;  dark.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Dr. Plait:    That’s right and for a long period of time that is all&#xA;  there is going to be, it’s just going to be that. Even on the scale of&#xA;  what we are talking about, we are talking about a long period of time,&#xA;  now instead of an octillion years, which is what, ten to the twenty&#xA;  seven, now we are starting to talk about ten to the fifty and sixty&#xA;  years, so far into the future, that the present age of the universe,&#xA;  13 billion years, is so small, that I don’t even have, ten to the&#xA;  sixty divided by ten to the thirteen is ten to the forty seven, that&#xA;  is such a huge number, that there is no ratio that exists today.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On those expansion scales, a galaxy is small (!!) enough that it'll pretty much disappear all at once.&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-02-27T15:23:45.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20213" PostTypeId="2" ParentId="20210" CreationDate="2017-02-28T06:53:20.533" Score="3" Body="&lt;p&gt;Trajectories &lt;em&gt;close&lt;/em&gt; to a black hole can't be approximated by Newtonian mechanics, General Relativity is needed. The black hole doesn't have a surface, but the Schwarzschild radius is point of no return. However it is not possible to orbit just above the Schwartzchild radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are less than 1.5 radii from the event horizon, there are no orbits. The orbital velocity closer to a black hole than 1.5 radii would be greater than the speed of light. Inside this radius, the faster you try to go, the less the centrifugal force you experience becomes, and so the faster you would fall towards the blackhole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact you won't find a stable orbit closer than 3 radii from the black hole. If the black hole is spinning (as most younger ones are expected to do) the space around the black hole is dragged along by the spinning black hole. It is possible to gain energy by entering this region of dragged space around a black hole. This all means that a space ship can't remain in the region of ultra-intense gravity for many hours by orbiting, It can't experience (from the point of view of a distant observer) extended gravitation time dilation by orbiting a black hole only a few km above its Schwarzschild radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to approach closer than 1.5 radii. An outside observer would see the spaceship approach the black-hole in a spiral at very high speed, but not cross the event horizon. The orbit would not appear to be elliptical and it would appear time dilated. Outside of 1.5 radii the dilation would be less. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trajectories close to the black hole would have to be calculated using GR. Newtonian mechanics would not be a close approximation. The calculation in GR would be computationally intensive, since the GR equations can't be integrated exactly, even for a two body problem, so a numerical technique would have to be used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Provided you stay rather further from the black hole, then the usual Newtonian approximation is valid. There's no certain boundary at which Newtonian mechanics is a good approximation. It depends on the accuracy needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember also that the region close to a black hole is a violent region of space. It would be dangerous for a ship to be so close to a black hole due to the intense radiation produced by infalling matter, and tidal forces.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-02-28T21:51:05.513" LastActivityDate="2017-02-28T21:51:05.513" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20214" PostTypeId="1" AcceptedAnswerId="20244" CreationDate="2017-02-28T07:17:19.400" Score="9" ViewCount="261" Body="&lt;p&gt;A Gamma Ray Burst was detected 0.4 seconds after the gravitational wave event, &lt;a href=&quot;https://en.wikipedia.org/wiki/First_observation_of_gravitational_waves&quot; rel=&quot;nofollow noreferrer&quot;&gt;GW150914&lt;/a&gt;, caused by a black hole merger, and it was in the same part of the sky. It is uncertain whether &lt;a href=&quot;https://en.wikipedia.org/wiki/First_observation_of_gravitational_waves#Coincident_gamma-ray_observation&quot; rel=&quot;nofollow noreferrer&quot;&gt;that Gamma Ray Burst was associated with the black hole merger&lt;/a&gt;. The odds of a GRB being coincident (or just background noise) is 0.22%. That implies a 99.78% chance the black hole merger was related to the GRB. Later analysis suggests the GRB was just a background event that occurred in the same place in the sky at just 0.4 seconds after the black hole merger, and therefore not related.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While a GRB may have relativistic jets beaming out from it in opposite directions, research &quot;excludes the possibility that the event is associated with substantial gamma-ray radiation, directed towards the observer.&quot; I interpret that to mean this GRB did &lt;strong&gt;not&lt;/strong&gt; have relativistic jets, but was omnidirectional. (Not sure how astronomers came to that conclusion.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow, whether it was coincident or not, I am asking about how the energy output of that GRB compares to other GRB's. The Wikipedia article in the link says the &quot;energy emission in gamma-rays and hard X-rays from the event was less than one millionth of the energy emitted as gravitational waves.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much energy does a typical GRB emit? What was the spectrum spread of a typical GRB? Over what amount of time does a typical GRB emit that energy? (How many seconds?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How do the energy levels, spectrums, and time durations of other GRB's compare to the GRB associated with the black hole merger?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is similar to other GRB's, that would support the hypothesis this GRB was just a coincident event at nearly the same time and location in the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this GRB has different energy emissions and time duration than other GRB's, that would support the hypothesis it is truly associated with the black hole merger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;When you answer, please provide data, citations, or quotes from original research. I am not looking for unsupported speculation, but real analysis supported by real data.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11175" LastEditorUserId="11264" LastEditDate="2017-03-01T13:22:43.220" LastActivityDate="2017-03-06T10:52:56.347" Title="How does the Gamma Ray Burst that occurred when 2 black holes merged compare to other GRB's?" Tags="&lt;black-hole&gt;&lt;gravitational-waves&gt;&lt;gamma-ray-bursts&gt;&lt;high-energy-astrophysics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20216" PostTypeId="1" CreationDate="2017-02-28T13:25:23.320" Score="10" ViewCount="687" Body="&lt;p&gt;How could Hawking get the conclusion &quot;[...] &lt;strong&gt;and black holes have positive energy&lt;/strong&gt;. That’s why empty space is stable. Bodies such as stars or black holes cannot just appear out of nothing. But a whole universe can.&quot;? It would make sense to me, if a black hole has negative energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quote from book &lt;em&gt;The Grand Design&lt;/em&gt;, page 180:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the total energy of the universe must always remain zero, and it costs energy to create a body, how can a whole universe be created from nothing? That is why there must be a law like gravity. Because gravity is attractive, gravitational energy is negative: One has to do work to separate a gravitationally bound system, such as the earth and moon. This negative energy can balance the positive energy needed to create matter, but it’s not quite that simple. The negative gravitational energy of the earth, for example, is less than a billionth of the positive energy of the matter particles the earth is made of. A body such as a star will have more negative gravitational energy, and the smaller it is (the closer the different parts of it are to each other), the greater this negative gravitational energy will be. But before it can become greater than the positive energy of the matter, the star will collapse to a black hole, and black holes have positive energy. That’s why empty space is stable. Bodies such as stars or black holes cannot just appear out of nothing. But a whole universe can.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16036" LastEditorUserId="11143" LastEditDate="2017-02-28T13:46:35.653" LastActivityDate="2019-02-10T09:35:44.497" Title="Why black holes have positive energy?" Tags="&lt;black-hole&gt;&lt;gravity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20217" PostTypeId="2" ParentId="20216" CreationDate="2017-02-28T16:21:56.833" Score="6" Body="&lt;p&gt;To understand this, you have to understand what it means to have negative or positive energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following the example given by the text, the gravitational attraction of the Earth and Moon is a system with negative energy. This is because it requires an energy &lt;em&gt;input&lt;/em&gt; to undo the system. One must exert energy to separate them, thus requiring energy input to the system to break it apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matter comprising the Earth however is all positive energy. You can get an energy &lt;em&gt;output&lt;/em&gt; from this matter and thus it must be positive. An example of energy output might be a &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravity_assist&quot; rel=&quot;noreferrer&quot;&gt;gravity assist&lt;/a&gt; around the Earth, radioactive decay of nuclear materials in the Earth, or matter-antimatter annihilation of Earth particles. Any process that involves a matter interaction producing energy in the form of heat/light/whatever takes that energy out the Earth's matter and represents an energy output. Note that energy in the matter may take many forms such as heat, motion, or simply the intrinsic energy comprised in matter (via the famous $E=mc^2$ equation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black hole, much like the Earth, is simply comprised of matter (albeit in some unknown form at the singularity) and as such energy can be extracted from this matter, making it a system with positive energy. &lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-02-28T16:21:56.833" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20219" PostTypeId="1" AcceptedAnswerId="20241" CreationDate="2017-02-28T20:08:27.020" Score="2" ViewCount="1306" Body="&lt;p&gt;I have found in numerous places such as this website: &lt;a href=&quot;http://burro.astr.cwru.edu/Academics/Astr222/Galaxy/Structure/metals.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://burro.astr.cwru.edu/Academics/Astr222/Galaxy/Structure/metals.html&lt;/a&gt; or in &quot;Introduction to Stellar Astrophysics&quot; by B. Carroll, that state:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Population I: &lt;br&gt;&#xA;  metal rich [Fe/H] &gt; -1 &lt;br&gt;&#xA;  disk stars &lt;br&gt;&#xA;  open clusters &lt;br&gt;&#xA;  metal rich makes stars redder&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Population II&lt;br&gt;&#xA;  metal poor [Fe/H] &amp;lt; -1  &lt;br&gt;&#xA;  halo stars&lt;br&gt;&#xA;  globular clusters &lt;br&gt;&#xA;  metal poor makes stars bluer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I also found (from: &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/Starlog/pop12.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/hbase/Starlog/pop12.html&lt;/a&gt;): &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Population I stars include the sun and tend to be luminous, hot and young, concentrated in the disks of spiral galaxies. They are particularly found in the spiral arms. With the model of heavy element formation in supernovae, this suggests that the gas from which they formed had been seeded with the heavy elements formed from previous giant stars. About 2% of the total belong to Population I.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Population II stars tend to be found in globular clusters and the nucleus of a galaxy. They tend to be older, less luminous and cooler than Population I stars. They have fewer heavy elements, either by being older or being in regions where no heavy-element producing predecessors would be found. Astronomers often describe this condition by saying that they are &quot;metal poor&quot;, and the &quot;metallicity&quot; is used as an indication of age.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;If Population I stars are metal rich and therefore appear red, how can they be luminous, hot, and young, and concentrated in disk (the region of star formation)? Shouldn't they appear blue? How does color relate to metallicity?&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hot stars should be blue, shouldn't they? As stated here for example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A star’s color is critical in identifying the star, because it tells us the star’s surface temperature in the black body radiation scale. The sun has a surface temperature of 5,500 K, typical for a yellow star. Red stars are cooler than the sun, with surface temperatures of 3,500 K for a bright red star and 2,500 K for a dark red star. The hottest stars are blue, with their surface temperatures falling anywhere between 10,000 K and 50,000 K. (&lt;a href=&quot;http://www.webexhibits.org/causesofcolor/18B.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.webexhibits.org/causesofcolor/18B.html&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, this question is a little similar but not quite: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/11057/metal-rich-or-metal-poor&quot;&gt;Metal rich or metal poor?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a graph: &lt;a href=&quot;http://science.psu.edu/alert/images/SDSSmetals.jpg/image_view_fullscreen&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://science.psu.edu/alert/images/SDSSmetals.jpg/image_view_fullscreen&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if the colors there correspond to the colors of the stars, but they used blue to refer to old stars. Why?&lt;/p&gt;&#xA;" OwnerUserId="2913" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2017-03-02T09:20:45.320" Title="Relationship between metallicity and color? Should Pop. I stars be blue?" Tags="&lt;star&gt;&lt;stellar-evolution&gt;&lt;metallicity&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20220" PostTypeId="1" CreationDate="2017-03-01T01:39:00.303" Score="10" ViewCount="1892" Body="&lt;p&gt;Could there be such thing as anti-dark matter? If there is anti regular matter, why not anti-dark matter? If there is no such thing as anti-dark matter, why?&lt;/p&gt;&#xA;" OwnerUserId="16049" LastActivityDate="2017-03-01T14:15:23.587" Title="Is Anti-Dark Matter a thing?" Tags="&lt;dark-matter&gt;&lt;antimatter&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20221" PostTypeId="1" CreationDate="2017-03-01T02:55:12.553" Score="4" ViewCount="160" Body="&lt;p&gt;When I was reading &lt;a href=&quot;http://www.space.com/17262-quasar-definition.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;information about QSOs&lt;/a&gt; on a website, it said that Seyfert Galaxies emit less energy than Quasars and QSOs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do Seyfert Galaxies emit less energy than quasars, and how is this energy measured?&lt;/p&gt;&#xA;" OwnerUserId="16050" LastEditorUserId="37191" LastEditDate="2021-04-15T10:55:28.290" LastActivityDate="2021-04-15T10:55:28.290" Title="Why do Seyfert Galaxies emit less energy than Quasars and Blazars, and how is the emitted energy measured?" Tags="&lt;radio-astronomy&gt;&lt;supermassive-black-hole&gt;&lt;quasars&gt;&lt;seyfert-galaxy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20222" PostTypeId="1" AcceptedAnswerId="20243" CreationDate="2017-03-01T03:47:40.160" Score="0" ViewCount="137" Body="&lt;p&gt;There's a lot of interest with the recent discovery of seven earth sized planets around &lt;a href=&quot;http://www.trappist.one/&quot; rel=&quot;nofollow noreferrer&quot;&gt;TRAPPIST-1&lt;/a&gt;. I saw &lt;a href=&quot;https://exoplanets.nasa.gov/resources/2159/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this poster&lt;/a&gt;, and the &lt;a href=&quot;https://www.nasa.gov/press-release/nasa-telescope-reveals-largest-batch-of-earth-size-habitable-zone-planets-around&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA video&lt;/a&gt; says that in the system you could see a lot of other planets whipping around in their orbits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It got me wondering:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Obviously that poster is exaggerated, but how exaggerated? &lt;/li&gt;&#xA;&lt;li&gt;What would you actually see with the naked eye, standing on TRAPPIST-1g, and looking back toward the other planets? &lt;/li&gt;&#xA;&lt;li&gt;How big would they appear?&lt;/li&gt;&#xA;&lt;li&gt;What might TRAPPIST-1 look like?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6870" LastActivityDate="2017-03-03T06:19:15.800" Title="What would the view be list from TRAPPIST-1g, looking back towards TRAPPIST-1?" Tags="&lt;trappist-1&gt;&lt;dwarf-star&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2017-03-02T19:32:45.437" ContentLicense="CC BY-SA 3.0" />
  <row Id="20225" PostTypeId="1" CreationDate="2017-03-01T10:21:14.460" Score="2" ViewCount="50" Body="&lt;p&gt;In the recent press conference about the planets orbiting around Trappist-1, one of the astronomers mentioned that the planets are close enough to each other that the tidal forces between planets are non-negligible. Supposing there were oceans on 1, 2 or all 3 of the planets in the habitable zone. What would the tides on these planets look like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the dynamics of planet-planet interactions would be quite different from a planet-moon interaction, the tides would be far more complicated than a sinusoidal low-tide/high-tide at any given point on the planet. The abstract of the Nature paper on these planets states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The six inner planets form a near-resonant chain, such that their orbital periods (1.51, 2.42, 4.04, 6.06, 9.1 and 12.35 days) are near-ratios of small integers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What effect would this have on the tides? How powerful would the tides be?&lt;/p&gt;&#xA;" OwnerUserId="13085" LastActivityDate="2017-03-01T10:21:14.460" Title="What would the tides on the Trappist-1 planets be like?" Tags="&lt;exoplanet&gt;&lt;trappist-1&gt;&lt;tidal-forces&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2017-03-07T17:27:19.490" ContentLicense="CC BY-SA 3.0" />
  <row Id="20226" PostTypeId="2" ParentId="20220" CreationDate="2017-03-01T11:15:30.020" Score="0" Body="&lt;p&gt;It depends on the composition of the dark matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the description I read it is clear that MACHOs (composed of normal baryonic matters) can have anti-objects, and there are also &lt;a href=&quot;http://heasarc.gsfc.nasa.gov/docs/objects/heapow/archive/normal_galaxies/fermi_m31.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;anti-WIMPs&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure about the Axion, it seem to be a different beast (on par with the photon that is its own anti-particle). &lt;/p&gt;&#xA;" OwnerUserId="11617" LastEditorUserId="11617" LastEditDate="2017-03-01T14:15:23.587" LastActivityDate="2017-03-01T14:15:23.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20227" PostTypeId="2" ParentId="20220" CreationDate="2017-03-01T12:37:30.737" Score="11" Body="&lt;p&gt;Yes, it is a thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weakly Interacting Massive Particles (aka &lt;a href=&quot;https://en.wikipedia.org/wiki/Weakly_interacting_massive_particles&quot; rel=&quot;noreferrer&quot;&gt;WIMPS&lt;/a&gt;) are thought to come in matter and anti-matter forms and have a self-annihilation cross-section in order to produce the &quot;correct&quot; amount of dark matter (in relation to photons) that we measure today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the opening paragraph of the wikipedia article on WIMPS states - there are a number of experiments searching for the photons associated with Annihilation of WIMP anti-WIMP pairs, from which it appears that there is expected to be a &lt;a href=&quot;https://astrobites.org/2013/03/02/the-search-for-wimp-annihilation-in-the-sun/&quot; rel=&quot;noreferrer&quot;&gt;neutrino and a gamma ray signature&lt;/a&gt; or even the presence of anti-matter (positrons) that might be detected in cosmic ray experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the other dark matter candidates - whether &lt;a href=&quot;https://en.wikipedia.org/wiki/Sterile_neutrino&quot; rel=&quot;noreferrer&quot;&gt;sterile neutrinos&lt;/a&gt; have an anti-particle depends on whether they are &lt;a href=&quot;https://en.wikipedia.org/wiki/Majorana_fermion&quot; rel=&quot;noreferrer&quot;&gt;Majorana fermions&lt;/a&gt; or not. The hypothetical &lt;a href=&quot;https://en.wikipedia.org/wiki/Axion&quot; rel=&quot;noreferrer&quot;&gt;axion&lt;/a&gt; I think is its own anti-particle (a chargeless boson) and there are no annihilation-type experiments for their detection that I can find.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-03-01T12:37:30.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20229" PostTypeId="1" AcceptedAnswerId="20242" CreationDate="2017-03-01T16:51:05.450" Score="3" ViewCount="789" Body="&lt;p&gt;Looking at sunrise and sunset data around 2017 vernal equinox (because of GMT timezone I took London) &lt;a href=&quot;https://www.timeanddate.com/sun/uk/london&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.timeanddate.com/sun/uk/london&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Date  Sunrise      Sunset        Day Length  Solar Noon&#xA;16    06:12 (92°)  18:07 (269°)  11:54:55    12:09 (37.0°)&#xA;17    06:09 (91°)  18:08 (269°)  11:58:53    12:08 (37.4°)&#xA;18    06:07 (90°)  18:10 (270°)  12:02:52    12:08 (37.8°)&#xA;19    06:05 (90°)  18:12 (271°)  12:06:50    12:08 (38.2°)&#xA;20    06:03 (89°)  18:13 (271°)  12:10:48    12:07 (38.6°)&#xA;21    06:00 (88°)  18:15 (272°)  12:14:47    12:07 (39.0°)&#xA;22    05:58 (88°)  18:17 (273°)  12:18:45    12:07 (39.3°)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We see that 12-hours day happens around 17-18 of March, however the Vernal Equinox takes place on 20 March 2017, 10:29 GMT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that due to refraction in the Earth's atmosphere and due to the fact that sunrises and sunsets are defined by the top of the Sun's disk, sunrises are &quot;earlier&quot; and sunsets are &quot;later&quot;, so the 12-hours day indeed should happen before the equinox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why the directions of sunrise and sunset are east--west on that same day? The above factors should not influence &lt;em&gt;directions&lt;/em&gt;, should they?&lt;/p&gt;&#xA;" OwnerUserId="15069" LastActivityDate="2017-03-02T10:37:01.333" Title="Equinox sunrise sunset direction" Tags="&lt;the-sun&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20230" PostTypeId="2" ParentId="20111" CreationDate="2017-03-01T18:36:55.420" Score="0" Body="&lt;ul&gt;&#xA;&lt;li&gt;Magnification. Since your wish is to view planets, you will want sufficient magnification to see them somewhat clearly. I'm afraid that what you will see looks nothing like the spectacular photos in the media (and on telescope packaging). Your son may be disappointed and lose interest in astronomy. I recommend getting a plan of other things to view, such as nebulae. Magnification is mostly important for planets and deep field observations; the latter will exceed your budget. Perhaps mapping features on the moon could maintain his interest for a while (I don't recommend the full moon; it's very bright unless you do something about that).&lt;/li&gt;&#xA;&lt;li&gt;Aperture, aperture and aperture. Your eyes care more about how bright something is than how big it is. Also remember that magnification reduces makes the object fainter, so you need more aperture for more magnification.&lt;/li&gt;&#xA;&lt;li&gt;Mounting. This is very, very important and usually overlooked by beginners. If you are not able to keep the telescope still and pointed at an object, you will not be able to make out any details. Unfortunately you will not be able to view the telescope before buying. Get a reputable brand name and ask around about the specific models that you're interested in. A polar mount is harder to set up to start viewing but can track stars more easily than an alt-azimuth mount.&lt;/li&gt;&#xA;&lt;li&gt;Filters. You can get filters that assist with viewing certain features in the sky. For example light pollution filters will reduce the glare from older-type artificial lights when you're near cities.&lt;/li&gt;&#xA;&lt;li&gt;Optical quality. Once again, you will not be able to check in advance if the scope was properly built. The same factory in China may produce anything from garbage to perfect scopes which is why it is a good idea to buy from a reputable brand (good resellers do most of their own quality control).&lt;/li&gt;&#xA;&lt;li&gt;Size. Despite the importance of aperture, the scope must be small enough that you can move it to where you will use it regularly enough or you will just waste money on a white elephant.&lt;/li&gt;&#xA;&lt;li&gt;I cannot give a good answer in dollars as I'm not up to date and have no idea of your budget.&lt;/li&gt;&#xA;&lt;li&gt;Personally I recommend a good binocular as a starting point before you get a telescope. Stay away from fancy coloured coatings on the lenses, get one of the simpler ones where the objective lenses are not in line with the eyepieces (Porro prism, not roof prism) and also stay away from things that zoom.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14969" LastActivityDate="2017-03-01T18:36:55.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20236" PostTypeId="1" CreationDate="2017-03-02T01:11:58.023" Score="4" ViewCount="777" Body="&lt;p&gt;According to general relativity, as I understand it, the space around a black hole's event horizon is distorted by gravity, such that the distance to a point approaching the event horizon from an observational point further out from the EH approaches infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the distance to any mass inside the EH appears to be infinite, how does this mass assert gravitational effects outside the EH?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps a different question, what is the difference between the gravitational effects of a mass inside the EH of a black hole and a similar mass that is outside the cosmic event horizon of an observational point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(edit as follows)&#xA;My original intent for this question is rethink how we think about black holes.  The distance to a point on the event horizon of Sag A is as far to the point on an observer's cosmic event horizon.  You can travel no farther than such a point on the CEH than you could to a point on Sag A's event horizon, at least using our 4D coordinate system.  BTW, I think all talk about a black hole having a singularity at its center is bunk, first of all, we can never get there to confirm or deny. Secondly, the density of Sag A if it were uniform, would mathematically would be around the density of liquid water as it exists on earth, but we would never be able to know, just as we would we would ever be able to know the density of the universe outside of our cosmic event horizons.  Any answers to those two questions would be meaningless anyway, as long as there is no away around the speed of causality rule enforced by our universe.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I believe I have read that Hawking Radiation has been shown to exist.  However, does that mean the part of whatever is outside our cosmic event horizon is evaporating as well? &lt;/p&gt;&#xA;" OwnerUserId="4122" LastEditorUserId="4122" LastEditDate="2020-04-19T20:28:42.690" LastActivityDate="2020-04-19T20:28:42.690" Title="How does gravity have an effect from the inside the event horizon of a black hole with the rest of the universe?" Tags="&lt;black-hole&gt;&lt;gravity&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="20237" PostTypeId="2" ParentId="13749" CreationDate="2017-03-02T04:51:58.940" Score="-1" Body="&lt;p&gt;particle and anti-particle pair - basically it has a particle that is attracted to the black hole and one that is repelled by it. Since these particles contain mass, the particle that is repelled is left as radiation, and the one that is attracted is added mass, and by adding this mass it will decrease the total density, since it isn't just an addition problem, there is lost mass increase that makes the density and the mass of the black hole a inverse relationship. This states that the more mass the less dense which means that is will shrink and decrease in length.&lt;/p&gt;&#xA;" OwnerUserId="16060" LastActivityDate="2017-03-02T04:51:58.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20238" PostTypeId="2" ParentId="20236" CreationDate="2017-03-02T05:10:20.163" Score="-1" Body="&lt;p&gt;Inside the event horizon, or what is thought of as the mass of the black hole is the section in which you can tell the &quot;strength&quot; of the pull from the singularity. The larger the event horizon (which is initially based on the star it originated from, r= 2GM/c^2, this gives the event horizon radius where G is the gravity constant and c is the speed of light) the stronger the singularity is, and the more dense the black hole is. By adding mass, or having the particle anti-particle pairs forming and then adding mass to the singularity, then the density decreases which decreases its strength therefore it decreases the radius of the event horizon. Where the event horizon is what we &quot;see&quot; or detect through equipment it just takes up more space, and by taking up more space, it has the greater pull on things around it. So basically by the size of the black holes event horizon, it will have a greater pull on things outside in space.&lt;/p&gt;&#xA;" OwnerUserId="16060" LastActivityDate="2017-03-02T05:10:20.163" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20241" PostTypeId="2" ParentId="20219" CreationDate="2017-03-02T09:20:45.320" Score="2" Body="&lt;p&gt;I think the misunderstanding (which I've been asked about before) is because Population II stars are redder &lt;em&gt;as a population&lt;/em&gt;, even though for a given mass, a main-sequence Pop II (i.e. metal-poor) star would be bluer.  That is, if you compare representative samples of Pop I and Pop II stars, the Pop II stars will on average be redder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This because Pop II stars are generally old so the blue stars have already died as supernovae or evolved into compact remnants. So looking at the whole population, you'll see more of the longer-lived, lower-mass stars (including red giants), which are redder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same argument applies to galaxies that aren't forming stars appearing redder than those galaxies that are still forming stars.  Since new stars are being born, some are blue, but they live fast and die young, so unless the galaxy keeps forming stars, it will become red overall.  Again, this isn't because of the stars being redder for a given property but because the population lacks young, hot, blue stars.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2017-03-02T09:20:45.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20242" PostTypeId="2" ParentId="20229" CreationDate="2017-03-02T10:37:01.333" Score="4" Body="&lt;p&gt;Due to the atmospheric refraction the Sun rises earlier and sets later as you correctly wrote. Refraction however is perpendicular to the horizon. So if the apparent sun rise is earlier, than the true (actual) position of the Sun will be more to the north east (smaller azimuth) than when the true sun rise occurs. As refraction is perpendicular the apparent sun rise will also be more to the north east and hence the azimuth will have a lower value (88-89° instead of 90°).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB. This is of course only true from the northern hemisphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB2. The Sun is not to scale in the figure below.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/nfk0B.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nfk0B.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="864" LastActivityDate="2017-03-02T10:37:01.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20243" PostTypeId="2" ParentId="20222" CreationDate="2017-03-02T14:18:12.507" Score="1" Body="&lt;p&gt;Let's take a look at the star and the planet's characteristics first. We have the star TRAPPIST-1:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$M = 0.082 M_{\odot}$&lt;/li&gt;&#xA;&lt;li&gt;$R = 0.117 R_{\odot}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The planets are all roughly in the scale of $0.7 R_{\oplus}$ to $1.2 R_{\oplus}$ with their semi-major axes of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;b: $1.66 \times 10^6 \; \mathrm{km}$&lt;/li&gt;&#xA;&lt;li&gt;c: $2.88 \times 10^6 \; \mathrm{km}$&lt;/li&gt;&#xA;&lt;li&gt;d: $3.14 \times 10^6 \; \mathrm{km}$&lt;/li&gt;&#xA;&lt;li&gt;e: $4.19 \times 10^6 \; \mathrm{km}$&lt;/li&gt;&#xA;&lt;li&gt;f: $5.57 \times 10^6 \; \mathrm{km}$&lt;/li&gt;&#xA;&lt;li&gt;g: $6.73 \times 10^6 \; \mathrm{km}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So if we would stand on the surface of TRAPPIST-1g and look towards the star, we would be at $1/25$ the distance to the star compared to the sun. The star however is roughly $1/8 R_{\odot}$. That would make the star appear about at about $3$ times the diameter compared to our Sun from the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TRAPPIST-1f would every few days pass TRAPPIST-1g with a closest approach of $1.16 \times 10^6 \; \mathrm{km} \approx 3 \times r_{\mathrm{Earth-Moon}}$ which is roughly three times the distance to the Moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $R_{\mathrm{TRAPPIST-1f}} \approx R_{\oplus}$ and $R_{\oplus}/R_{\mathrm{Moon}} = 3.67$ it would seem that TRAPPIST-1f would appear roughly the size of the Moon on closest approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other planets are similar in radius but their distances are $2$ to $5$ times further away and would thus appear smaller by that factor.&lt;/p&gt;&#xA;" OwnerUserId="14441" LastEditorUserId="14441" LastEditDate="2017-03-03T06:19:15.800" LastActivityDate="2017-03-03T06:19:15.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20244" PostTypeId="2" ParentId="20214" CreationDate="2017-03-02T15:13:53.433" Score="3" Body="&lt;p&gt;The interpretation you suggest in the second paragraph is incorrect. It is understandable, since &lt;strong&gt;there is a debate in the literature - different papers come to potentially contradicting conclusions&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Excluding a possibility that the event is associated with substantial gamma-ray radiation, directed towards the observer&quot; simply means that &lt;strong&gt;no observable GRB is found, hinting that may be the original detection was caused by instrumental background&lt;/strong&gt;: each detector has it's own instrumental backgrounds, only real events should be seen by all sufficiently sensitive instruments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, it is always possible that the merger generated beamed emission, directed somewhere else. No instrument should have seen it directly, and there is no simple way to know if it happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This upper limit is derived from the observation of another satellite (INTEGRAL/SPI-ACS, &lt;a href=&quot;https://arxiv.org/abs/1602.04180&quot; rel=&quot;nofollow noreferrer&quot;&gt;Savchenko et al 2016&lt;/a&gt;) than the original detection (Fermi/GBM, &lt;a href=&quot;https://arxiv.org/abs/1602.03920&quot; rel=&quot;nofollow noreferrer&quot;&gt;Connaughton et al 2016&lt;/a&gt;).&#xA;Also an alternative analysis of the Fermi/GBM data (&lt;a href=&quot;https://arxiv.org/abs/1606.00314&quot; rel=&quot;nofollow noreferrer&quot;&gt;Greiner et al 2016&lt;/a&gt;) suggested that no event can be found in the GBM data - their opinion is that it was &lt;strong&gt;a background fluctuation of some kind&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Right now, the teams who reported these conflicting results are working together, trying to come up with a consistent picture, which might be, in principle, a GRB with this or that properties associated with the GW150914, an unrelated GRB with some properties, or no detection whatsoever&lt;/strong&gt;. This work is centered on cross-calibrating and comparing instruments, and is also useful to avoid these kind of uncertainties in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could try to characterize the spectral properties of this event, following the approach of the original Fermi/GBM team. But unfortunately, the measurement appeared to be in very unlucky conditions for Fermi/GBM (in bad direction). Which is why &lt;strong&gt;the signal was very weak&lt;/strong&gt; (below what would be usually reported for a real GRB, though recently the attempts were made to decrease these thresholds, see &lt;a href=&quot;https://arxiv.org/abs/1612.02395&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goldstein et al 2017&lt;/a&gt; ), and &lt;strong&gt;spectral characterization is loose&lt;/strong&gt;. You can look for some details to &lt;a href=&quot;https://arxiv.org/abs/1607.02616&quot; rel=&quot;nofollow noreferrer&quot;&gt;Veres et al 2016&lt;/a&gt;.&#xA;&lt;strong&gt;With these large uncertainties the spectrum is compatible to that of known short GRBs.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;luminosity estimate depends on the spectrum&lt;/strong&gt;, but is seems to be at the &lt;strong&gt;lower end of the short GRB sample&lt;/strong&gt; (see e.g. &lt;a href=&quot;https://arxiv.org/abs/1405.5878&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wanderman et al 2015&lt;/a&gt;)&lt;br&gt;&#xA;But because the uncertainties are large, the event, if real, might be unusual as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The INTEGRAL observation, non-detection, would imply much softer (perhaps, unusual for a short GRB) or/and a weaker burst, possibly incompatible even with the &lt;strong&gt;highly uncertain Fermi/GBM data&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Duration&lt;/strong&gt; of this possible GRB possibly associated with GW event is the easy part and is about 1 s long, &lt;strong&gt;typical for a short GRB&lt;/strong&gt; (&lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1993ApJ...413L.101K&amp;amp;db_key=AST&amp;amp;high=403acf95b220083&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kouveliotou  et al 1993&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="16065" LastEditorUserId="16065" LastEditDate="2017-03-06T10:52:56.347" LastActivityDate="2017-03-06T10:52:56.347" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20246" PostTypeId="1" CreationDate="2017-03-02T16:36:40.143" Score="1" ViewCount="163" Body="&lt;p&gt;So is the effect between the passing planets different whether they encounter in the same or opposite direction in their orbits with the same relative speed difference?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="10445" LastEditDate="2017-03-11T09:51:18.500" LastActivityDate="2017-03-11T09:51:18.500" Title="Is the effect when two planets passing each other different, when they pass in the same or opposite direction?" Tags="&lt;orbit&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20248" PostTypeId="2" ParentId="20221" CreationDate="2017-03-02T20:51:06.263" Score="2" Body="&lt;p&gt;The linked website seems to summarise generally accepted theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seyfert galaxies appear less energetic than quasars or blazars. The energy is measured by observations of their brightness in the sky (at different wavelengths of light) combined with their distance (inferred from their red-shift). There are also Radio-galaxies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We think that active galactic nuclei don't release the same amount of energy in all directions, instead they have a &quot;beam&quot;. They also have an accretion disc, the inner parts of which are extremely hot If we happen to be looking directly down the beam, they appear exceptionally powerful, and we call them &quot;blazars&quot; If we see them off centre we can see into the extremely hot central part of the accretion disk and they will be a quasar or (if less powerful) a Seyfert galaxy, and if we see them side-on they appear less powerful, but the two lobes of high-energy gas and particles emitted along the beam will be visible, and we will observe a two-lobed radio galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See these related answers: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/2461/quasars-and-smbh/2492&quot;&gt;Quasars and SMBH&lt;/a&gt; and &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6099/difference-between-quasar-and-active-galactic-nuclei&quot;&gt;Difference between quasar and Active Galactic Nuclei?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astronomy.ohio-state.edu/~ryden/ast162_9/notes37.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.astronomy.ohio-state.edu/~ryden/ast162_9/notes37.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2017-03-02T20:51:06.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20249" PostTypeId="2" ParentId="20236" CreationDate="2017-03-03T03:11:11.197" Score="0" Body="&lt;p&gt;Here's my understanding as a backyard astronomer... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, gravity is more accurately defined in Einstein's STR as energy that bends space/time (because energy and mass are linked by E=mc^2)(1).  So let go of Newtonian mass attracting mass.  A black hole then is a huge energy source that is warping space/time initially incredibly deeply, but then tapering off according to the inverse-square law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The event horizon (EH) is defined as the boundary where anything with mass, which includes light, has to travel more than the speed of light (c) to attain escape velocity.  As c is the absolute upper limit on speed, nothing can escape and the 'hole' is 'black'. (2) Trying to calculate what is going on within the EH is fruitless, as current equations tend to return &quot;Divide By 0&quot;! ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, with the mass of the singularity warping space/time causing what we call 'gravity', gravity itself doesn't have to escape anything - it just is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus a black hole can be observed exerting a gravitational influence on objects around it, such as the observations of stars and stellar clouds in orbits around Sagittarius A* (3)&lt;br&gt;&#xA;As per any gravitational system, an object may be attracted towards a black hole and then ejected, it may be in a stable orbit, or it may be spiralling inwards to eventually be torn apart by tidal forces and join the accretion disc around the EH. Here it may fall in to add to the mass of the singularity or may be radiated by the energy of the accretion disc and jets (4).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a system consists of binary black holes, they might orbit one another until eventually spiralling into each other and merging in a cataclysmic event that the LIGO team recently detected (5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps a bit!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) &lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/M/Mass&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://astronomy.swin.edu.au/cosmos/M/Mass&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) &lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/E/Event+Horizon&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://astronomy.swin.edu.au/cosmos/E/Event+Horizon&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) &lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/C/Centre+Of+The+Milky+Way&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://astronomy.swin.edu.au/cosmos/C/Centre+Of+The+Milky+Way&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) &lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/J/Jets&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://astronomy.swin.edu.au/cosmos/J/Jets&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) &lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/G/Gravitational+Waves&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://astronomy.swin.edu.au/cosmos/G/Gravitational+Waves&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15741" LastActivityDate="2017-03-03T03:11:11.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20250" PostTypeId="1" AcceptedAnswerId="20251" CreationDate="2017-03-03T06:40:28.480" Score="2" ViewCount="207" Body="&lt;p&gt;At first I assumed the amplification factor meant an increase in brightness. But since the equation: $$A(u)=\frac{u^2+2}{u\sqrt{u^2+4}}$$ is independent of mass, this make no sense to me. Wouldn't everything be infinity magnified (if its mass independent empty space should work, right)? Even if its simply not valid for a zero mass, everything would be amplified by an arbitrarily high amount any time even a neutrino passes arbitrarily close to the right $u$?&#xA;But if its not brightness, I don't know what signal is being amplified.&lt;/p&gt;&#xA;" OwnerUserId="16069" LastActivityDate="2017-03-04T08:07:55.617" Title="What does the 'amplification factor' of gravitational microlensing mean?" Tags="&lt;gravitational-lensing&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20251" PostTypeId="2" ParentId="20250" CreationDate="2017-03-03T08:28:47.313" Score="6" Body="&lt;p&gt;Gravitational lensing is just geometry, as is optical lensing. The amplification factor describes the increase of the &lt;strong&gt;area&lt;/strong&gt; of the image (at constant surface brightness). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, the amplification factor &lt;em&gt;does&lt;/em&gt; depend on the mass $M$ of the lens, since $u=\theta/\theta_E$ and $\theta_E$ depends on $M$. Here $\theta$ and $\theta_E$ are the relative angular distance between source and lens and the radius (angle) of the Einstein ring for the lens, respectively.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2017-03-04T08:07:55.617" LastActivityDate="2017-03-04T08:07:55.617" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20253" PostTypeId="2" ParentId="20236" CreationDate="2017-03-03T21:39:14.403" Score="4" Body="&lt;p&gt;As is the case even in Newtonian gravity, the gravitational attraction from a spherical body exterted on a far-away object is identical the that exterted by a point mass located at the the center of the body. This is the so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Shell_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;shell theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To an outside observer, due to the infinite time dilation at the event horizon of a black hole (BH), all mass that has ever fallen into the BH has actually never reached the event horizon, but is instead placed in a shell around the BH, still falling ever-so-slowly toward the BH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, what we feel is in fact the gravitation not from a point mass, but from a shell of matter, but there is no way for us to tell the difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you may ask, &quot;What if we drop a massive object, e.g. a planet, into a symmetric black hole? Since it never reaches the BH, won't the effective gravitational potential be asymmetric?&quot; What happens is that the way that space is &quot;warped&quot; around the black, the gravitational field is bent along with it in such a way that sufficiently far from the BH+planet, the gravitational field looks symmetric. Sort of like the following drawing shows close to (&lt;em&gt;left&lt;/em&gt;) and far from (&lt;em&gt;right&lt;/em&gt;) the BH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eWR6W.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eWR6W.png&quot; alt=&quot;BH&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2017-03-03T21:39:14.403" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20254" PostTypeId="1" AcceptedAnswerId="20260" CreationDate="2017-03-04T00:34:32.017" Score="1" ViewCount="98" Body="&lt;p&gt;I have found oddly different answers through my multiple google searches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found one article that mentioned this did not alter the estimated number of stars in the universe. Link: &lt;a href=&quot;http://www.universetoday.com/30305/how-many-galaxies-in-the-universe/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.universetoday.com/30305/how-many-galaxies-in-the-universe/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet other article gave the estimate of stars an increase. Link: &lt;a href=&quot;http://gizmodo.com/we-were-very-wrong-about-the-number-of-galaxies-in-the-1787750693&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://gizmodo.com/we-were-very-wrong-about-the-number-of-galaxies-in-the-1787750693&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There were other examples but they also had similar contradictory answers.&#xA;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="16080" LastActivityDate="2017-03-04T11:31:44.613" Title="The # of Estimated Galaxies increased 10 fold. Where does that leave our # of stars estimate?" Tags="&lt;universe&gt;&lt;observable-universe&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20256" PostTypeId="2" ParentId="20236" CreationDate="2017-03-04T01:30:47.103" Score="2" Body="&lt;p&gt;The physical distance does not tend to infinity. What does tend to infinity is the 'coordinate distance', which is unphysical. Think about it: there are probably numerous black holes in our observable universe - how can they all be infinitely far away?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physics of black holes is described by general relativity. It cannot be understood using any Newtonian formula, such as F = GM1M2/r^2. &lt;/p&gt;&#xA;" OwnerUserId="15906" LastActivityDate="2017-03-04T01:30:47.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20257" PostTypeId="2" ParentId="20216" CreationDate="2017-03-04T01:35:44.920" Score="0" Body="&lt;p&gt;The reason is because of the positive energy theorem in GR. See, for a brief qualitative account, wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's probably a little too advanced to explain in 'The Grand Design'.&lt;/p&gt;&#xA;" OwnerUserId="15906" LastActivityDate="2017-03-04T01:35:44.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20258" PostTypeId="2" ParentId="20254" CreationDate="2017-03-04T07:14:06.307" Score="3" Body="&lt;p&gt;The result from the Christopher Conselice of the University of Nottingham is based on the number of galaxies that could be visible to us. This is not the same as the number of galaxies that exist &quot;today&quot;, since when we look into the deep space, we also look into the past. We can think of a &quot;light cone&quot; that stretches out in space-time. The number of galaxies means the number of galaxies on our light-cone. This makes sense because the value 200 billion, or 2 trillion, is the number of galaxies in the observable universe. (The entire Universe may be infinite)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conclusion is that in the Early universe, galaxies were smaller, dimmer, and more numerous. Over time these small galaxies have merged to form the large galaxies that we see in our local neighbourhood. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it should be noted that this is a model based conclusion. Conselice has looked at the distribution of galaxies at different distances, and hence different times, and used a model of galaxy formation to infer that at great distances (and early times) there are a great many galaxies that can't be seen. These distant, early, small and dim galaxies greatly outnumber those that can be seen. The conclusion is that there are many more galaxies on our light-cone, but most are invisible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since these galaxies are small and dim they contain fewer stars. There is no change to the estimated number of stars or to the estimated mass of matter in the universe, only to the distribution of these stars in the early universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.nasa.gov/feature/goddard/2016/hubble-reveals-observable-universe-contains-10-times-more-galaxies-than-previously-thought&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.nasa.gov/feature/goddard/2016/hubble-reveals-observable-universe-contains-10-times-more-galaxies-than-previously-thought&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-03-04T07:14:06.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20259" PostTypeId="1" CreationDate="2017-03-04T07:15:11.613" Score="1" ViewCount="72" Body="&lt;p&gt;Suppose I have an astronomical image in some bandpass and I do photometry of an extended source (i.e., a galaxy). I measure an enclosed magnitude within my aperture or galaxy profile. What is a straightforward recipe for obtaining a realistic uncertainty on that magnitude? Can I just do something like 1/(S/N) or is that only for point source photometry?&lt;/p&gt;&#xA;" OwnerUserId="607" LastActivityDate="2017-03-04T07:15:11.613" Title="How to derive uncertainty on photometric magnitudes of extended objects?" Tags="&lt;galaxy&gt;&lt;photometry&gt;&lt;magnitude&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20260" PostTypeId="2" ParentId="20254" CreationDate="2017-03-04T11:31:44.613" Score="4" Body="&lt;p&gt;The sensationalist headlines mask a rather more prosaic result that has absolutely no effect on the number of stars we might estimate are present in the observable universe &lt;em&gt;today&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The study in question has discovered (actually it was always supposed and suggested that this was the case), that the distribution of galaxy masses was much more skewed to lower mass galaxies in the past. That is, there were many more small galaxies in the past. But this happens at the expense of the large galaxies - there were fewer large galaxies in the past. To first order, the overall mass contained within galaxies stays the same and so long as the ratio of stellar mass to luminosity (since this is actually what is measured) remains roughly constant, than any estimate of stellar numbers stays the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, a better estimate for the number of stars in the observable universe uses the present day galactic mass function and then extrapolates that to the size of the observable universe today. This is a far more honest estimate, since we have a much better understanding of how many stars there are in our local part of the universe and can then invoke the cosmological principle to say that other places (but not times) in the universe are similar. This cannot possibly be affected by the new study since that is dealing with the numbers of galaxies &lt;em&gt;in the past&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-03-04T11:31:44.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20261" PostTypeId="1" AcceptedAnswerId="20264" CreationDate="2017-03-04T15:10:37.753" Score="4" ViewCount="579" Body="&lt;p&gt;I just read &lt;a href=&quot;http://www.seeker.com/pluto-could-be-made-a-planet-again-along-with-102-other-celestial-bodi-2295672083.html&quot; rel=&quot;noreferrer&quot;&gt;this article&lt;/a&gt;. I don't know how credible it is, since I couldn't find any link to the original paper. But it says that pluto will be considered a planet again, along with more than 100 other celestial objects including our moon. So, if the definition given here is applied, then would there be anything as dwarf planet or satellite? I think even comets and asteriouds would come under this defintion. I am talking about this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The paper, published in Lunar and Planetary Science, declares a planet as a round body that's never undergone fusion, and has less mass than a star. That's it. This means, Pluto would be planet again-- But it would also bring along Ceres, Eris, Makemake and Haumea... as well as 97 other objects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10741" LastActivityDate="2017-03-04T18:37:36.020" Title="So is there nothing like satellite anymore?" Tags="&lt;planet&gt;&lt;natural-satellites&gt;&lt;definition&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20262" PostTypeId="2" ParentId="20261" CreationDate="2017-03-04T15:53:53.030" Score="2" Body="&lt;p&gt;Small objects like comets and asteroids are generaly not round.&#xA;The &lt;em&gt;roundness&lt;/em&gt; of bigger bodies is a consequence of the body own gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to answer your question, no, not all celestial bodies would become a planet with this definition as small objects like comets and asteroids doesn't have a sufficent mass to become round.&lt;/p&gt;&#xA;" OwnerUserId="11521" LastActivityDate="2017-03-04T15:53:53.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20263" PostTypeId="2" ParentId="20261" CreationDate="2017-03-04T17:23:19.377" Score="0" Body="&lt;p&gt;I doubt if the definition given above will actually be applied ever, but if it is applied then all the current natural satellites (moons) will not be considered as a planet. So is the case with dwarf planets, comets, asteroids, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asteroids and comets are not round in shape (even if some are, they are not round as a result of their own gravity), so they will not be considered as planets. Most of the moons in the solar system are not round in shape. For example, moons of Mars, Phobos and Demos, are not in round shape. They will continue to be considered as moons or natural satellites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in short, all the natural satellites, dwarf planets, etc., will not become planets.&lt;/p&gt;&#xA;" OwnerUserId="15703" LastActivityDate="2017-03-04T17:23:19.377" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20264" PostTypeId="2" ParentId="20261" CreationDate="2017-03-04T18:37:36.020" Score="5" Body="&lt;p&gt;This is really a question of Language rather than Astronomy. The meaning of words is defined by their use. The IAU has proposed one way of using the word &quot;planet&quot;, this group of scientists have proposed another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone can propose a definition of any word. This group proposes that a &quot;planet&quot; is any astronomical body which 1) Does not have, and never has had sustained nuclear fusion in its core, and 2) Is large enough for gravity to pull it into an ellipsoidal shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intention of this definition is to make a planet defined entirely by &quot;what&quot; a body is, not by &quot;where&quot; it is. Under this definition the classic 8 planets, Ceres, Pluto, Charon, the Moon, the larger moons of Jupiter, Saturn Uranus and Neptune, Eris, Sedna, and the larger Kuiper belt objects would all be &quot;planets&quot; - about 100 known in total. The exoplanets would also be &quot;planets&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asteroids (except Ceres), comets and other small solar system bodies are too small, and so would not be planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantage of this definition is that the usually where something is located is not part of the definition of what it is. However, it contradicts several hundred years of use by defining the larger moons as planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike the 8 planet definition, this does not have the support of a large international body of astronomers (the IAU). If you wish to use this definition you are free to. I doubt that without the support of the IAU this definition is ever going to be widely used. Most people will either use the 8 planet definition or the 8+Pluto definition.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-03-04T18:37:36.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20265" PostTypeId="1" CreationDate="2017-03-04T19:44:48.367" Score="-1" ViewCount="179" Body="&lt;p&gt;&lt;a href=&quot;https://worldbuilding.stackexchange.com/questions/17244/ftl-travel-is-impossible-how-is-interstellar-trade-possible/71030?noredirect=1#comment211727_71030&quot;&gt;Some suggest&lt;/a&gt; that there might be currents of charged particles moving at 0.98c in interstellar space.  Since this is tainted by association with the crack “Electric Universe” cosmology, my first impression is to be dismissive of the possibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, proponents point to current observations and ideas, so I thought I’d check on this claim in and of itself.  Given what we do know, are such currents possible or excluded?  And, if possible, could they be massive enough to propell a space craft bearing a large sail?&lt;/p&gt;&#xA;" OwnerUserId="7142" LastEditorUserId="-1" LastEditDate="2017-04-13T12:52:24.020" LastActivityDate="2017-03-04T22:00:40.860" Title="Could there be massive charged “rivers” in interstellar space?" Tags="&lt;cosmology&gt;&lt;interstellar-medium&gt;&lt;interstellar-travel&gt;&lt;plasma-physics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20266" PostTypeId="2" ParentId="20265" CreationDate="2017-03-04T22:00:40.860" Score="2" Body="&lt;p&gt;Define &quot;current&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one particle makes a current, there are lots of particles that hit the Earth at energies much more than 0.98c, most famously the &quot;&lt;a href=&quot;http://www.cosmic-ray.org/reading/flyseye.html#SEC10&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oh my god particle&lt;/a&gt;&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have many particles travelling at these speeds you don't have a river, more a beam, or a jet or particles. These are produced by active accretion disks around black holes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dQjwT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dQjwT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this combination image from Nasa (the background is a visible light image from hubble, the pink lobes are from the VLA, a radio telescope), an active radio galaxy is producing two jets, that are much longer than the size of the galaxy. These consist of particles that have been accelerated by the black hole's gravity and shot into space. A ship would find it hard to use the energy in these jets. Particles travelling at close to the speed of light are high energy radiation. A space ship unlucky enough to pass unprotected through the beams would be irradiated. I haven't done the maths, but the dose would probably be enough to kill anyone on board. It's hard to extract useful energy from anything so violent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if you are looking for random currents of charge, curving between the stars, and interstellar sailors tacking and jibing betweent them, you may be out of luck.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-03-04T22:00:40.860" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="20267" PostTypeId="1" AcceptedAnswerId="20271" CreationDate="2017-03-05T14:05:04.293" Score="1" ViewCount="123" Body="&lt;p&gt;For example, pisces birthdays are from Febuary 19th-March 20th. Would the sun be near or at the pisces constellation during Mid febuary to mid march? Basically, I was wondering whether the astrology we use now is related to the sun's position relative to the zodaic constellations. &lt;/p&gt;&#xA;" OwnerUserId="15910" LastActivityDate="2017-03-05T19:02:10.340" Title="Does the astrology of constellations relate to its position during the year?" Tags="&lt;the-sun&gt;&lt;constellations&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20268" PostTypeId="1" AcceptedAnswerId="20269" CreationDate="2017-03-05T14:50:47.933" Score="4" ViewCount="1499" Body="&lt;p&gt;Venus has a greenhouse affect because its atmosphere is 96 percent carbon dioxide. However, although mercury is closer to the sun it does not have the greenhouse effect because it has no atmosphere and only an exosphere. I was wondering why doesn't mars have a greenhouse effect like Venus? They both have around 95 percent carbon dioxide in their atmosphere and consist of an atmosphere. In addition, I was also wondering about what are the requirements to have a greenhouse effect on a planet?&lt;/p&gt;&#xA;" OwnerUserId="15910" LastActivityDate="2017-07-21T06:36:36.600" Title="Greenhouse effect on planets" Tags="&lt;planet&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20269" PostTypeId="2" ParentId="20268" CreationDate="2017-03-05T15:29:07.450" Score="7" Body="&lt;p&gt;For there to be a significant greenhouse effect there has to be a lot of greenhouse gas in the atmosphere. Not just a large percentage, but a large amount of gas&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a greenhouse effect on Mars, however, as its atmosphere is very thin. The surface pressure on Mars is only 0.6% of the pressure on Earth. As such there very little of anything to trap heat. Even though a large percentage of the atmosphere is CO2, the greenhouse effect only raises temperatures by about 5 Kelvin. &lt;a href=&quot;http://nova.stanford.edu/projects/mod-x/id-green.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://nova.stanford.edu/projects/mod-x/id-green.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It wasn't always this way. During a dust storm, more heat can be trapped, and temporarily raise temperatures. Early in Mars's past the atmosphere was thicker at temperatures were higher. It is even possible that at one point, a runaway greenhouse effect heated Mars, until the atmosphere thinned and the planet cooled &lt;a href=&quot;http://www.astrobio.net/mars/did-ancient-mars-have-a-runaway-greenhouse/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.astrobio.net/mars/did-ancient-mars-have-a-runaway-greenhouse/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only requirement for a planet to be warmed by the greenhouse effect is for its atmosphere to contain greenhouse gases. Mars does have a greenhouse effect, as other bodies with atmospheres: Earth, Titan and Venus. &lt;a href=&quot;https://www.nasa.gov/home/hqnews/1991/91-143.txt&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.nasa.gov/home/hqnews/1991/91-143.txt&lt;/a&gt; Other rocky bodies don't have a thick enough atmosphere for a greenhouse effect to be measured.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="6841" LastEditDate="2017-07-21T06:36:36.600" LastActivityDate="2017-07-21T06:36:36.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20270" PostTypeId="1" CreationDate="2017-03-05T15:32:03.603" Score="1" ViewCount="73" Body="&lt;p&gt;Electrons move to higher energy levels and drop back (electron migration). When they drop levels, photons are emitted and the wavelength/frequency emitted is based on how many levels are dropped. My question is, how does this work with elements with only one shell, such as hydrogen? Also, what determines how many energy levels are dropped?&lt;/p&gt;&#xA;" OwnerUserId="16090" LastEditorUserId="2531" LastEditDate="2017-03-05T16:04:01.137" LastActivityDate="2017-03-05T16:04:01.137" Title="How do elements with only one shell emit light?" Tags="&lt;light&gt;&lt;spectroscopy&gt;&lt;spectra&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2017-03-06T19:53:26.003" ContentLicense="CC BY-SA 3.0" />
  <row Id="20271" PostTypeId="2" ParentId="20267" CreationDate="2017-03-05T15:35:50.253" Score="3" Body="&lt;p&gt;Yes, but no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic idea behind the zodiac is as you stated what constellation the Sun is located in during the year. But the time assigned to each constellation (1/12 of the year) is not accurate, for several reasons. Firstly, because of modern constellation borders, some constellations are larger than others. This has also caused the ecliptic to pass through Ophiuchus, not in the zodiac. Secondly, &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;axial precession&lt;/a&gt; has rotated the zodiac by almost an entire constellation over the last 2000 years. That is not accounted for in astrological dates.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="10172" LastEditDate="2017-03-05T19:02:10.340" LastActivityDate="2017-03-05T19:02:10.340" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20272" PostTypeId="2" ParentId="20270" CreationDate="2017-03-05T15:54:19.227" Score="4" Body="&lt;p&gt;Actually, hydrogen can have as many electron levels as you want. However, they only exist at specific energy levels, with the potential energy given as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E=-\frac{R_H}{n^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $R_h$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Rydberg_constant&quot; rel=&quot;noreferrer&quot;&gt;Rydberg constant&lt;/a&gt;, and $n$ is any integer. When electrons are exited to a higher energy lever, they will fall back after some time and emit photons (not protons, by the way). So how many levels do they drop? Well, in the case of hydrogen, it will eventually go back to the basic energy level, but this is not guaranteed to happen in one jump. This may or may not take multiple jumps, that depends on probability. For a simple atom like hydrogen, those probabilities may be possible to calculate analytically. For more complex atoms however, this becomes an extremely difficult task, usually approached by simulations.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastActivityDate="2017-03-05T15:54:19.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20274" PostTypeId="2" ParentId="20246" CreationDate="2017-03-05T19:57:52.403" Score="3" Body="&lt;p&gt;From the any of the planet's point of view, the encounter is exactly the same regardless of direction if the relative speed and force is equal. This should be pretty clear for symmetry reasons. For any encounter, the law is $v_{in} = v_{out}$. However, if the encounter takes a long time, you would also have to consider the influence of the central body of the system. The problem then becomes a three-body problem and this case is not restricted enough for an analytical solutions. We can make a few observations though: As the velocity vectors of both objects before the encounter have no radial component, both orbits are going to get a higher eccentricity. Any other information depends on knowing the distance of the encounter and the relative masses.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastActivityDate="2017-03-05T19:57:52.403" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20275" PostTypeId="1" CreationDate="2017-03-05T20:44:21.493" Score="1" ViewCount="126" Body="&lt;p&gt;I got this question while watching the COSMOS TV series. I couldn't find the right answer in google.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read about Stellar classification and it states that we can classify the stars based on their spectral characteristics. My question is how can we be sure of which particular star the light is coming from. Since Earth is getting light from so many stars How the the light is classified?  Even if we can get different spectra then how can find their exact sources?  &lt;/p&gt;&#xA;" OwnerUserId="16094" LastEditorUserId="2365" LastEditDate="2017-03-05T23:09:00.257" LastActivityDate="2017-03-05T23:09:00.257" Title="How did we differentiate lights from stars?" Tags="&lt;light&gt;&lt;spectra&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20276" PostTypeId="1" AcceptedAnswerId="20292" CreationDate="2017-03-05T20:54:27.530" Score="18" ViewCount="9200" Body="&lt;p&gt;I have just been watching a podcast called &quot;Deep Astronomy&quot; and the discussion was about a super fast spinning black hole discovered with the NuSTAR space observatory. This black hole was modeled with high confidence to be spinning at about 99% of the maximum spin rate. They stopped short of saying that the tangential velocity of this spin rate is &quot;c&quot; (and how can a singularity have a &quot;tangential velocity&quot;?) They did say that the event horizon at maximum spin of a stellar black hole is about 1-1/2 km. and that if a black hole were to spin faster then the result would be a &quot;naked black hole&quot; that would defy the laws of physics (GR). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, shouldn't all black holes spin extremely fast (conservation of angular momentum) or would a retrograde accretion disk slow it down. Could somebody clarify this whole &quot;black hole spin thing&quot; without getting too complicated??&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2019-06-28T07:30:17.083" Title="Maximum spin rate of a black hole?" Tags="&lt;black-hole&gt;&lt;general-relativity&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="20277" PostTypeId="2" ParentId="20275" CreationDate="2017-03-05T21:36:20.333" Score="2" Body="&lt;p&gt;If two lights are shining in your room, one red and one blue, how do you know which one is red and which is blue?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the two lights, the lens in your eye will focus the red light at a different place from the blue light, so you can see two lights in different places. Even though the light that reaches your eye is mixed, the lens can separate the light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Telescopes have lenses (or curved mirrors that do the same thing). The light from different stars is focused by the telescope, to form an image, and so the light from different stars is separated. The light from individual stars can then be split into colours. This is the star's spectrum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The job of the telescope is two-fold: collect a lot of light and then focus it so that individual objects can be seen and analysed individually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many ways of classifying a spectrum, Stars have continuous spectra crossed by dark lines. The temperature of the star changes the spectrum, the stars are classified in order &lt;code&gt;O B A F G K M&lt;/code&gt; from hottest to coolest. Nebula sometimes have emmission spectra which looks mostly dark, crossed by bright lines.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-03-05T21:36:20.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20278" PostTypeId="1" CreationDate="2017-03-06T01:13:17.763" Score="1" ViewCount="462" Body="&lt;p&gt;We have been calling the star by looking at which we can make out the direction of the North Pole the North Star. This was the only method at the disposal of the sailors during the days when they did not have any compass with them. When the axis of the earth would point toward Denab, Denab would become the North Star. When it would point toward Vega, Vega would become the North Star. Likewise, when the axis would point toward Thuban, Thuban would become the North Star. When we look at the Polaris from the earth, the Big Bear appears to be revolving around it. So I am curious to know which constellations would appear to be revolving around the Denab, Vega and Thuban when the axis of the earth would be pointing toward them due to wobbling of the earth.   &lt;/p&gt;&#xA;" OwnerUserId="13871" LastEditorUserId="13871" LastEditDate="2017-03-24T22:55:14.857" LastActivityDate="2017-03-25T04:27:28.790" Title="Which constellation would appear to revolve around Denab,Vega or Thuban when axis of the earth would point toward them?" Tags="&lt;star&gt;&lt;earth&gt;&lt;big-bang-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20280" PostTypeId="2" ParentId="20161" CreationDate="2017-03-06T03:18:32.533" Score="1" Body="&lt;p&gt;What drove the comment that &quot;the tidal forces between the planets are not negligible&quot; was the Jovian moons. The three innermost of the Galilean moons of Jupiter, Io, Europa, and Ganymede, are in a 4:2:1 orbital resonance. Io would not exhibit any tidal heating if its orbit was circular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Io's orbit is not circular, thanks to those orbital resonances. One of the consequences of these resonances is that Europa and Ganymede act to pull Io's orbit out-of-round; i.e., more elliptical. The elliptical nature of Io's orbit results in time-varying tidal stresses on Io, which makes Io geologically active. Those tidal stresses by Jupiter in turn act to circularize Io's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tidal stresses become less severe as Io's orbit becomes closer to circular. Io cools down, resulting in an increase in its tidal qualify factor Q. This makes it less susceptible to further circularization. The competing forces from Europa and Ganymede can then make Io's orbit more eccentric. Tidal stresses eventually come into play again, warming up Io and decrasing its tidal quality factor. Now Jupiter is the driver. This makes for a rather nice hysteresis loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;What prompted that comment that the TRAPPIST-1 planets may be subject to tidal heating is that some of those planets appear to be in an orbital resonance, with periods being very close to small integer multiples of one another.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-03-06T03:18:32.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20281" PostTypeId="1" CreationDate="2017-03-06T04:08:59.243" Score="0" ViewCount="242" Body="&lt;p&gt;Nothing in the universe stands still. It's always on its way from somewhere to somewhere. A black hole follows a trajectory through the universe, moving from one locus in spacetime to another locus.&#xA;&lt;/p&gt;&lt;p&gt;Deep inside each black hole is a point that we've never been able to characterize.&lt;/p&gt;&lt;p&gt;My question is, does the singularity, and possibly the powerful gravitational fields around the singularity have any measurable effect on the area of spacetime it passed through? One imagines an undeformed area of spacetime. After some time, a black hole passes through this spacetime, then moves on. Does the spacetime it passed through simply go back to being everyday, garden variety undeformed spacetime?&lt;/p&gt;&#xA;" OwnerUserId="9516" LastActivityDate="2017-03-06T11:33:23.253" Title="As a black hole passes through spacetime, does it affect it?" Tags="&lt;black-hole&gt;&lt;space-time&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20282" PostTypeId="1" AcceptedAnswerId="20285" CreationDate="2017-03-06T05:37:11.317" Score="4" ViewCount="218" Body="&lt;p&gt;I am of course not an expert on astronomy. But from the literature, I have the impression that the lunar motion is very complicated. It seems that Kepler can make sense of the motion of the planets but cannot do that for the moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why is the moon's motion more complicated than that of planets like Mars? &lt;/p&gt;&#xA;" OwnerUserId="16097" LastEditorUserId="2618" LastEditDate="2017-03-06T11:20:29.360" LastActivityDate="2017-03-06T11:20:29.360" Title="Why is the moon's motion more complicated than that of those planets?" Tags="&lt;orbit&gt;&lt;gravity&gt;&lt;the-moon&gt;&lt;data-analysis&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20283" PostTypeId="2" ParentId="20281" CreationDate="2017-03-06T10:10:24.497" Score="3" Body="&lt;p&gt;No, a black hole, or any other gravitationally attracting object for that matter, does not change spacetime permanently, but only affects spacetime around it. The easiest way to see this, is to view the BH in its own frame of reference. Here, the BH is stationary, and spacetime around it is described by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_metric&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schwarzschild metric&lt;/a&gt;. If a BH came by you and left a physical imprint on spacetime after it (like a wake), then from the BH's frame that would mean that you would fly by it, and it would start producing a wake in the direction that you moved. You don't have such an effect on BHs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RsS4J.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RsS4J.jpg&quot; alt=&quot;BHwake&quot;&gt;&lt;/a&gt;&#xA;&lt;sup&gt;Wrong scenario of observer and black hole passing each other, as viewed from the observer's (&lt;em&gt;left&lt;/em&gt;) and the black hole's (&lt;em&gt;right&lt;/em&gt;) reference frames.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2017-03-06T10:10:24.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20284" PostTypeId="1" CreationDate="2017-03-06T11:02:08.407" Score="3" ViewCount="413" Body="&lt;p&gt;I am using the following R script to plot the moon phase and the moon's illuminated fraction:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;library(oce)&#xA;&#xA;X11()&#xA;&#xA;t_start &amp;lt;- as.POSIXct(&quot;2017-02-23&quot;, tz='UTC')&#xA;t_end   &amp;lt;- t_start + 75 * 24 * 3600  # 35 days&#xA;&#xA;tm &amp;lt;- seq(from = t_start,&#xA;           to  = t_end,&#xA;           by  = 10000&#xA;        )&#xA;&#xA;moon     &amp;lt;- moonAngle(t=tm, longitude=-8, latitude=47)&#xA;fraction &amp;lt;- moon$illuminatedFraction&#xA;phase    &amp;lt;- moon$phase - floor(moon$phase)&#xA;&#xA;plot (tm, fraction, type=&quot;l&quot;, col='blue')&#xA;lines(tm, phase   , type=&quot;l&quot;, col='red' )&#xA;&#xA;grid()&#xA;&#xA;# wait for mouse click or enter pressed&#xA;locator(1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is plotted:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/WezOk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WezOk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red lines indicate the phase (0 = new moon, 1/4 = first quarter, 1/2 = full moon etc.), the blue curve indicates the how much of the moon is illuminated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have expected the illuminated fraction to be zero exactly then when also the phase is zero. However, this is not the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is there something I am missing, am I doing something wrong in my script or is the implementation of &lt;code&gt;oce&lt;/code&gt; inaccurate?&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="16462" LastEditDate="2017-04-06T06:00:04.227" LastActivityDate="2017-10-03T10:25:29.767" Title="Why is the moon phase not in sync with the moon's illuminated fraction?" Tags="&lt;the-moon&gt;&lt;software&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20285" PostTypeId="2" ParentId="20282" CreationDate="2017-03-06T11:03:19.843" Score="4" Body="&lt;p&gt;Keplerian motion would be a correct way to explain the Moon's orbit about the Earth if the Sun, Venus, Jupiter, and all the other planets weren't present, if the Moon wasn't slowly receding from the Earth due to tidal acceleration, and if Newtonian gravity properly described the underlying physics. None of these is the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of modeling the behaviors of the major objects that comprise the solar system (i.e., the Sun and the eight planets), perturbations involving planets interacting with one another, particularly with Jupiter, make the orbits of the planets slightly non-Keplerian. That the universe is relativistic rather than Newtonian famously makes Mercury's orbit precess in a way that even accounting for planetary perturbations cannot explain. That said, Keplerian motion is a fairly good approximation with regard to the Sun and the eight planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is not true for the Moon. It's orbit is notably non-Keplerian, even over the course of just a month. Ptolemy (after the fact) and Brahe (also after the fact) both noticed irregularities in the Moon's motion that cannot be explained in terms of Keplerian elliptical orbits in which area swept out per unit time is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton explained some of these effects by noticing that the Sun is a huge perturber of the Moon's orbit about the Earth. Newton was not quite successful in this endeavor; the mathematical tools needed to model the Moon's motion hadn't been developed. The full development of classical lunar theory took two hundred years after Newton's time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another complicating factor is that the Moon is gradually receding from the Earth due to tidal acceleration. That the Moon is slowly receding from the Earth was first hypothesized by Halley, one of Newton's contemporaries. A century after the first publication of Newton's &lt;em&gt;Principia&lt;/em&gt;, Laplace erroneously attributed this lunar recession solely to perturbations from the Sun. In the next hundred years, others corrected Laplace's calculations and found that solar perturbations are unable to explain this recession. George Darwin (Charles' son) explained the recession in terms of the tides. The full explanation? People are still writing papers on this subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relativity makes for yet another complication. Relativistic effects on the Moon are very hard to see given the many orders of magnitude larger complications due to perturbations by the Sun, Venus, Jupiter, ..., and tidal acceleration.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-03-06T11:03:19.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20286" PostTypeId="2" ParentId="20281" CreationDate="2017-03-06T11:33:23.253" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Nothing in the universe stands still.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is not quite correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every object is still (i.e. at rest) in it's own reference frame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A black hole follows a trajectory through the universe, moving from one locus in spacetime to another locus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Except that the black hole is at rest in it's own frame of reference.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Deep inside each black hole is a point that we've never been able to characterize.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We do characterize it as a singularity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My question is, does the singularity, and possibly the powerful gravitational fields around the singularity have any measurable effect on the area of spacetime it passed through ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The singularity is a property of the space-time distortion inside a black hole, i.e. a result of the gravitational field of the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational field has effects on space-time both inside and outside the event horizon.  The event horizon is what defines the black hole's extent.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One imagines an undeformed area of spacetime. After some time, a black hole passes through this spacetime, then moves on. Does the spacetime it passed through simply go back to being everyday, garden variety undeformed spacetime?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It becomes more and more distorted and then less and less distorted.  As the range of the gravitational field is infinite, it never actually is undistorted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no permanent effect on space-time (e.g. nothing like a track &quot;goughed&quot; through space-time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as the space-time is distorted there is a kind of permanent effect in the sense that e.g. an observer could see an object passing through such a gravitational field as experiencing a time dilation relative to the (distant) observer.  That would be permanent in the sense that the difference in the time observer and object calculate is not going to be undone.  The rate at which time changes will return to normal, but not the accumulated time they each record, which will remain different (in the absence of the observer passing through a similar distortion in space-time).&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-03-06T11:33:23.253" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20287" PostTypeId="1" AcceptedAnswerId="20294" CreationDate="2017-03-06T12:47:53.250" Score="0" ViewCount="383" Body="&lt;p&gt;I was watching &lt;strong&gt;Passengers (2016)&lt;/strong&gt;. In this movie, a swimming pool exists in the space station. In middle of the movie, Artificial Gravity is turned off and the water rises, with the swimmer suffocating as they try to get out of the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suddenly Artificial Gravity turned itself on again and the swimmer survives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=uP0_3MstCAA&quot; rel=&quot;nofollow noreferrer&quot;&gt;A clip of this scene is found on YouTube.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, water is an anti-ferromagnetic material - it can't attract to magnetic bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How then, can artificial gravity hold the water in swimming pool?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is totally different from this &lt;a href=&quot;https://astronomy.stackexchange.com/questions/2097/how-would-water-waves-behave-in-partial-gravity&quot;&gt;question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16100" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2017-03-06T17:15:50.207" Title="How can artificial gravity hold the water in swimming pool?" Tags="&lt;gravity&gt;&lt;water&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2017-03-06T19:56:13.983" ContentLicense="CC BY-SA 3.0" />
  <row Id="20288" PostTypeId="2" ParentId="20276" CreationDate="2017-03-06T13:33:04.473" Score="2" Body="&lt;p&gt;From a quick drive around the InformationSuperHighway, I'd say the answer will remain a complicated mess :-) .  I did find a reasonably non-mathematical discussion at &lt;a href=&quot;http://www.universetoday.com/109308/how-fast-do-black-holes-spin/&quot; rel=&quot;nofollow noreferrer&quot;&gt;universetoday&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The speed limit is set by the event horizon, eventually, at a high&#xA;  enough spin, reaches the singularity. You can’t have what’s called a&#xA;  naked singularity. You can’t have a singularity exposed to the rest of&#xA;  the Universe. That would mean that the singularity itself could emit&#xA;  energy or light and somebody outside could actually see it. And that&#xA;  can’t happen. That’s the physical limitation of how fast it can spin.&#xA;  Physicists use units for angular momentum that are cast in terms of&#xA;  mass, which is a curious thing, and the speed limit can be described&#xA;  as the angular momentum equals the mass of the black hole, and that&#xA;  sets the speed limit.”&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Just imagine. The black hole spins up to the point that it’s just&#xA;  about to reveal itself. But that’s impossible. The laws of physics&#xA;  won’t let it spin any faster. And here’s the amazing part. Astronomers&#xA;  have actually detected supermassive black holes spinning at the limits&#xA;  predicted by these theories.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;One black hole, at the heart of galaxy NGC 1365 is turning at 84% the&#xA;  speed of light. It has reached the cosmic speed limit, and can’t spin&#xA;  any faster without revealing its singularity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-03-06T13:33:04.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20289" PostTypeId="1" CreationDate="2017-03-06T14:02:15.747" Score="1" ViewCount="574" Body="&lt;p&gt;If declination is the degree measurement from the celestial equator roughly the Earth equator unless I've misunderstand plane.  &quot;The celestial equator is a great circle on the imaginary celestial sphere, in the same plane as the Earth's equator.&quot;  &lt;a href=&quot;https://en.wikipedia.org/wiki/Celestial_equator&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Celestial_equator&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is zero degrees for Right Ascension, is it the GMT Line or International Date Line or some other line?&lt;/p&gt;&#xA;" OwnerUserId="5401" LastActivityDate="2017-03-06T19:24:34.497" Title="Where is 0 R.A.?" Tags="&lt;declination&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20290" PostTypeId="2" ParentId="20289" CreationDate="2017-03-06T14:51:35.583" Score="3" Body="&lt;p&gt;0 RA is the direction from the center of the Earth to the vernal equinox. You've already imagined the plane in space created by projecting the equator into space. Now consider the plane formed by the mean orbit of the Earth; this is the ecliptic. The two planes intersect in a line. The ray from the center of the Earth toward where the Sun is in March is the vernal equinox; the ray from the center of the Earth toward where the Sun is in September is the autumnal equinox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big difference is that the location of the vernal equinox among the stars changes very slowly, taking about 26,000 years to sweep all around the ecliptic. The plane that passes through the north and south pole and the Royal Observatory, Greenwich, sweeps through all the stars every day.&lt;/p&gt;&#xA;" OwnerUserId="7385" LastEditorUserId="7385" LastEditDate="2017-03-06T19:24:34.497" LastActivityDate="2017-03-06T19:24:34.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20292" PostTypeId="2" ParentId="20276" CreationDate="2017-03-06T15:59:03.203" Score="26" Body="&lt;p&gt;Since I like math, let's throw some math into this. I'll try to keep it as simple as possible though.&lt;/p&gt;&#xA;&lt;h3&gt;Kerr Black Holes&lt;/h3&gt;&#xA;&lt;p&gt;A rotating black hole is known as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Kerr_metric&quot; rel=&quot;noreferrer&quot;&gt;Kerr Black Hole&lt;/a&gt; (named after &lt;a href=&quot;https://en.wikipedia.org/wiki/Roy_Kerr&quot; rel=&quot;noreferrer&quot;&gt;Roy Kerr&lt;/a&gt; who found the numerical solution to GR equations for rotating black holes). In the case of a rotating black hole, there are two important parameters used to describe the black hole. The first is of course the mass of the black hole &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt;. The second is the spin &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt;. Really &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; isn't the spin itself &lt;span class=&quot;math-container&quot;&gt;$-$&lt;/span&gt; it's defined by &lt;span class=&quot;math-container&quot;&gt;$a=J/M$&lt;/span&gt; &lt;sub&gt;(see footnote)&lt;/sub&gt; where &lt;span class=&quot;math-container&quot;&gt;$J$&lt;/span&gt; is the angular momentum of the black hole &lt;span class=&quot;math-container&quot;&gt;$-$&lt;/span&gt; but it's a good proxy for spin so often you'll see scientists get lazy and just call it the spin of the black hole. The mathematics will tell you that Kerr black holes have the limitation that&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$0 \le a/M \le 1$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Black Hole Event Horizon&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The important parameter we want to calculate is the black hole's radius. If you run through the math, you find that this radius is given by&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$r_e = M+(M^2-a^2)^{1/2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;In the case when &lt;span class=&quot;math-container&quot;&gt;$a/M=0$&lt;/span&gt; (and thus &lt;span class=&quot;math-container&quot;&gt;$a=0$&lt;/span&gt;), this reduces to just &lt;span class=&quot;math-container&quot;&gt;$r_e=2M$&lt;/span&gt;, or in regular units (instead of geometrized units) &lt;span class=&quot;math-container&quot;&gt;$r_e=2GM/c^2$&lt;/span&gt;. Hopefully you can see that this just reduces to the normal &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;noreferrer&quot;&gt;Schwarzchild radius&lt;/a&gt; for a non-rotating black hole and thus the equation above is a generalization to account for spin. Let's look at the other limit when &lt;span class=&quot;math-container&quot;&gt;$a/M=1$&lt;/span&gt; (and thus &lt;span class=&quot;math-container&quot;&gt;$a=M$&lt;/span&gt;). In this case, you find that the radius is &lt;span class=&quot;math-container&quot;&gt;$r_e=M$&lt;/span&gt;. When &lt;span class=&quot;math-container&quot;&gt;$a/M=1$&lt;/span&gt;, you have a &lt;em&gt;maximally rotating&lt;/em&gt; black hole, and your radius is half the normal Schwarzchild radius of a non-rotating black hole. This equation defines the radius of the Event Horizon, the point after which there is no returning from the black hole.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Ergosphere&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;As it turns out, when you define your equation to calculate the radius of the black hole, there are actually multiple solutions! The section above shows one such solution, but there's another important solution as well. This radius, sometimes called the &lt;em&gt;static limit&lt;/em&gt; is given by the equation&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$r_{s} = M+\left(M-a^2\cos^2(\theta)\right)^{1/2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Notice that this is almost exactly the same as above, except for that extra &lt;span class=&quot;math-container&quot;&gt;$\cos^2(\theta)$&lt;/span&gt;. This defines a different, slightly larger and somewhat &amp;quot;pumpkin-shaped&amp;quot; horizon which encompasses the inner Event Horizon defined above. The region between this outer horizon and the inner horizon is known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ergosphere&quot; rel=&quot;noreferrer&quot;&gt;Ergosphere&lt;/a&gt;. Without getting into the nitty gritty details, I'll just say that one important point about the Ergosphere is that anything inside it (that is, &lt;span class=&quot;math-container&quot;&gt;$r_e&amp;lt;r&amp;lt;r_s$&lt;/span&gt;) must rotate exactly with the black hole - it is physically impossible to stay still here!&lt;/p&gt;&#xA;&lt;h3&gt;Answers&lt;/h3&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;They stopped short of saying that the tangential velocity of this spin rate is &amp;quot;c&amp;quot; (and how can a singularity have a &amp;quot;tangential velocity&amp;quot;?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;When you talk about the tangential velocity, there are multiple components of this black hole you/they may be talking about. One such tangential velocity is the tangential velocity of the event horizon (defined by &lt;span class=&quot;math-container&quot;&gt;$r_e$&lt;/span&gt; above). We can take a look at the case of a maximally rotating black hole and say that the angular momentum, based on the equations above, of such a black hole is given by&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$J_{max} = a_{max}Mc = M^2c$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Note that I've dropped the geometrized units just to be completely explicit. This has introduced an extra &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; now. Remember that &lt;span class=&quot;math-container&quot;&gt;$a_{max}$&lt;/span&gt; is achieved when &lt;span class=&quot;math-container&quot;&gt;$a/M=1$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;We can also define the angular momentum using the standard equation from physics 101, &lt;span class=&quot;math-container&quot;&gt;$J=rMv_\perp$&lt;/span&gt;, where of course &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; is the radius of your object, and &lt;span class=&quot;math-container&quot;&gt;$v_\perp$&lt;/span&gt; is the perpendicular, or else tangential, velocity of your spinning object. Recall from above that for a maximally rotating black hole, &lt;span class=&quot;math-container&quot;&gt;$r_e=M$&lt;/span&gt; so we also have that&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$J_{max} = r_eMv_{\perp} = M^2v_\perp$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;You can see that these two equations for &lt;span class=&quot;math-container&quot;&gt;$J_{max}$&lt;/span&gt; only equal each other if the tangential velocity &lt;span class=&quot;math-container&quot;&gt;$v_\perp$&lt;/span&gt; is equal to the speed of light &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;. So yes, you are correct to presume that at the fastest possible rotations, the black hole's event horizon is rotating at the speed of light!&lt;/p&gt;&#xA;&lt;p&gt;I said though that there are multiple components you could talk about when discussing rotating black holes. The other, as you allude to, is the rotating singularity. You correctly point out - &amp;quot;how can a singularity have a tangential velocity&amp;quot;? As it turns out, Kerr black holes don't have point singularities, they have &lt;a href=&quot;https://en.wikipedia.org/wiki/Ring_singularity&quot; rel=&quot;noreferrer&quot;&gt;ring singularities&lt;/a&gt;. These are &amp;quot;rings&amp;quot; of mass with zero width but some finite radius. Almost like a disk of no height. These rings of course can then have a tangential velocity. You were correct to be suspicious of a point singularity having tangential velocity though. That is not possible.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;They did say that the event horizon at maximum spin of a stellar black hole is about 1-1/2 km. and that if a black hole were to spin faster then the result would be a &amp;quot;naked black hole&amp;quot; that would defy the laws of physics (GR).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;We know the equation exactly, since I defined it above. The radius of a stellar black hole (that is a black hole with mass exactly equal to the mass of the Sun, &lt;span class=&quot;math-container&quot;&gt;$M_\odot$&lt;/span&gt;) is given by&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$r=\frac{GM_{\odot}}{c}=1.48\:km$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So yes, they were correct on their radius. They also state that spinning any faster results in a naked singularity. This is entirely true. To see this, go back to the equation for the event horizon. Remember that our upper spin limit is that &lt;span class=&quot;math-container&quot;&gt;$a=M$&lt;/span&gt;. What happens to our event horizon radius when &lt;span class=&quot;math-container&quot;&gt;$a&amp;gt;M$&lt;/span&gt; (and thus &lt;span class=&quot;math-container&quot;&gt;$a/M&amp;gt;1$&lt;/span&gt;)? For arguments say, lets say &lt;span class=&quot;math-container&quot;&gt;$a=2M$&lt;/span&gt;. Then our event horizon radius becomes&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$r_e=M-(M^2-a^2)^{1/2}=M-(M^2-4M^2)^{1/2}=M-(-3M^2)^{1/2}=M-i\sqrt{3}M$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Suddenly our radius is complex and has an imaginary component! That means it is not physical and thus &lt;em&gt;cannot exist&lt;/em&gt;. Now that we don't have an event horizon, our singularity can't hide behind it and is &amp;quot;naked&amp;quot;, exposed to the universe for anyone to see. GR tells us such an event shouldn't be allowed to happen because it results in all sorts of violations of physics. So somehow, something has to prevent black holes from spinning faster than a maximal black hole.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Shouldn't all black holes spin extremely fast (conservation of angular momentum) or would a retrograde accretion disk slow it down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Yes, that is true in general. All black holes should spin extremely fast, simply because of conservation of angular momentum. In fact, I don't think I can come up with a case where a black hole was found to be not spinning. Shown below is a plot from &lt;a href=&quot;http://www.nature.com/news/spin-rate-of-black-holes-pinned-down-1.13512&quot; rel=&quot;noreferrer&quot;&gt;this Nature paper&lt;/a&gt; which shows the measured spin of 19 supermassive black holes. They're all spinning pretty fast with some of them almost at the speed of light. None of them are even close to not spinning.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4JYOz.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4JYOz.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;sub&gt;Footnote: In GR, to make the math easier, often scientists adopt special units known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Geometrized_unit_system&quot; rel=&quot;noreferrer&quot;&gt;geometrized units&lt;/a&gt;. These are units chosen in just such a way that the gravitational constant, &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt;, and the speed of light, &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;, are equal to one. There's infinitely many units which allow this. Essentially this means no GR equations have &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; in them, but they're implicitly there, they're just equal to one and so not shown.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-03-07T13:35:08.187" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20293" PostTypeId="2" ParentId="20284" CreationDate="2017-03-06T16:00:31.750" Score="1" Body="&lt;p&gt;This is a partial answer since your question is still unclear.  I would recommmend finding an astronomy package for tools like this rather than using what's basically a &quot;bonus feature&quot; from an oceanographic package. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the help pages for package &lt;code&gt;oce&lt;/code&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;illuminatedFraction&lt;/strong&gt;&lt;br&gt;&#xA;   fraction of moon’s visible disk that is&#xA;   illuminated&lt;br&gt;&#xA;  &lt;strong&gt;phase&lt;/strong&gt;&lt;br&gt;&#xA;   phase of the moon, defined in equation 32.3 of Meeus&#xA;   [1982]. The fractional part of which is 0 for new moon, 1/4 for first&#xA;   quarter, 1/2 for full moon, and 3/4 for last quarter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Perhaps you are missing the fact that marginally more than 50% of the moon is illuminated, and thus at first/last quarter marginally more than 50% of the visible disk is illuminated?  I would also recommend looking up the referenced &quot;equation 32.3&quot; to see what the definition is and how it applies to a zero-altitude viewer.&lt;/p&gt;&#xA;" OwnerUserId="907" LastEditorUserId="907" LastEditDate="2017-03-06T16:14:10.937" LastActivityDate="2017-03-06T16:14:10.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20294" PostTypeId="2" ParentId="20287" CreationDate="2017-03-06T16:39:27.960" Score="2" Body="&lt;p&gt;&quot;Artificial gravity&quot; should act like gravity. So just as gravity has a direction on earth and holds water in a pool, so should  &quot;artificial gravity&quot; hold water in a pool on a space station with  &quot;artificial gravity.&quot; Notice that bubbles rise in the space station pool just as they would on earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If gravity were loss on a space station with a pool, then the surface tension of the water would want to pull the water into a sphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll also point out that water is an anti-ferromagnetic material in earth swimming pools, and gravity works just fine. So using magnetic boots on a steel deck is very different that having gravity which would work with an aluminum deck. &lt;/p&gt;&#xA;" OwnerUserId="9570" LastEditorUserId="9570" LastEditDate="2017-03-06T17:15:50.207" LastActivityDate="2017-03-06T17:15:50.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20296" PostTypeId="1" CreationDate="2017-03-06T20:26:34.453" Score="1" ViewCount="3636" Body="&lt;p&gt;There are many calculators online to convert local time to local sidereal time, but I can't find any resources to convert the other way.  Can you convert back by doing the same conversion backwards?  I found equations on &lt;a href=&quot;http://aa.usno.navy.mil/faq/docs/GAST.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;this website&lt;/a&gt; to convert to LST.  Would it be valid to use this equation to convert back:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;UT = \frac{LST - \frac{L}{15} - 6.697374558 - 2400.051336*T - 0.000025862*T^2)}{1.0027379093}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $UT$ is the time in Universal Coordinated Time, $LST$ is the Local Sidereal Time, $L$ is the longitude, and $T$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;T = \frac{JD-2451545.0}{36525.0}&#xA;$$&#xA;where JD is the Julian Date at noon UTC.  Once I've calculated the time in UT, I then add or subtract 24 until the result is within 0 - 24 hours.  Then I can convert from UT to local time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would prefer to have this as accurate as possible.  I did a calculation using this, but according to Stellarium I was off by about an hour.  Could this be due to Daylight Savings time or something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the Python code I made to do the calculation.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;LST = 8 + ((37 + (5.642/60))/60)&#xA;JD0 = 2457813.5&#xA;L = - 106.6056&#xA;GMST = LST - L/(365/(23 + ((56 + (4.1/60))/60)))&#xA;T = (JD0 - 2451545)/36525&#xA;UT = (GMST - 6.697374558 - 2400.051336*T - 0.000025862*(T*T))/1.0027379093&#xA;&#xA;&#xA;while UT &amp;lt;0:&#xA;    UT += 24&#xA;&#xA;&#xA;LT = UT - 7&#xA;&#xA;while LT &amp;lt; 0:&#xA;    LT += 24&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="13426" LastEditorUserId="13426" LastEditDate="2017-12-29T20:50:43.180" LastActivityDate="2017-12-29T20:50:43.180" Title="How do you calculate local time from local sidereal time?" Tags="&lt;time&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20298" PostTypeId="1" CreationDate="2017-03-06T22:01:39.967" Score="1" ViewCount="683" Body="&lt;p&gt;I want to be able to predict how long the summer solstice lasts at any part of world given the maximum tilt angle $φ = 0.4101524 \; \mathrm{rad}$ and the latitude $θ \; \mathrm{rad}$. I specifically want to do this using a geometric approach. The problem is, I never had experience with spherical trig before, and the values I got are somewhat off from their real values. So I want to know if that's because I didn't do the math correctly, and/or because of physical reasons (e.g. the Earth isn't perfectly spherical, the tilt changes slightly etc.).&#xA;Here is how I did it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the Earth be represented as a perfect sphere of radius $R$ moving at a constant angular velocity. Let the Earth's rotational axis be in the z direction, and the axis of the circle forming the dark/light boundary be tilted by angle $φ$ relative to the Earth's axis. During the summer solstice, our location of interest (say, in the northern hemisphere) will trace out a circular path of radius $r &amp;lt; R$ perpendicular to the z-axis. These three circles, $C_{\text{rot}}$, $C_{\text{bound}}$ and $C_{\text{path}}$ create a spherical triangle. Using the arc length $s$ of this triangle along $C_{\text{path}}$, we can calculate the length of the day in hours using: $$24 \times \frac{πr + 2s}{2πr}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point where $C_{\text{rot}}$ and $C_{\text{path}}$ meet is located at $$P(\text{rot/path}) = (R\cosθ, 0, R\sinθ)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the point where $C_{\text{bound}}$ and $C_{\text{path}}$ meet, we first observe that $$y = \sqrt{R^2 - x^2}\sinφ$$ and $$z = \sqrt{R^2 - x^2}\cosφ$$&#xA;Which we can solve for $x$ by setting $z$ equal to the $z$ from the previous point&#xA;$$x = R\sqrt{1 - \frac{\sin^2θ}{\cos^2φ}}$$&#xA;Plugging this back into $y$ and $z$, we finally get&#xA;$$P(\text{bound/path}) = \left (R\sqrt{1 - \frac{\sin^2θ}{\cos^2φ}},R\sinθ\tanφ,R\sinθ \right )$$&#xA;Now these two points are on the same plane in the z-axis, which is the plane of $P_{\text{path}}$. The distance $d$ between these points is &#xA;$$d = \sqrt{\left (R\sqrt{1 - \frac{\sin^2θ}{\cos^2φ}} - R\cosθ \right)^2 + R^2\sinθ^2\tan^2φ}$$&#xA;From this we can get $s$ by using the arc length formula:&#xA;$$s = 2\arcsin \left (\frac{d}{2r} \right )r$$&#xA;I used Mathematica to compare the calculated and expected values in 2017:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;New York: $θ = 0.7105724077$; length (calculated) = 14 hours 56 minutes; length (expected) = 15 hours, 5 minutes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;London: $θ = 0.8989737191$; length (calculated) = 14 hours 25 minutes; length (expected) = 16 hours, 38 minutes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, there is a noticeable difference (10+ minutes).&lt;/p&gt;&#xA;" OwnerUserId="16111" LastEditorUserId="9301" LastEditDate="2017-03-06T22:40:17.077" LastActivityDate="2017-12-09T14:16:58.147" Title="Calculating length of summer solstice day using tilt and latitude" Tags="&lt;earth&gt;&lt;rotation&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20301" PostTypeId="1" CreationDate="2017-03-06T23:13:28.230" Score="6" ViewCount="837" Body="&lt;p&gt;How do we know degenerate matter exists on white dwarfs? Is it purely hypothetical or have we observed it before? Have we ever created a form of degenerate matter on earth?&lt;/p&gt;&#xA;" OwnerUserId="15902" LastActivityDate="2017-03-07T16:25:07.027" Title="White Dwarf and Degenerate Matter" Tags="&lt;white-dwarf&gt;&lt;degenerate-matter&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20302" PostTypeId="2" ParentId="20301" CreationDate="2017-03-07T00:39:02.693" Score="10" Body="&lt;p&gt;The density of white dwarfs is not hypothetical, it can be measured. The short answer is that the density is so high that a stable star can &lt;em&gt;only&lt;/em&gt; be supported by electron degeneracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sirius B is an example. The radius can be estimated by combining the luminosity of the white dwarf with its temperature estimated from spectroscopy. The mass can then be determined from the binary motion (with Sirius A), or gravitational redshift or from estimating the surface gravity from the width of absorption lines in its spectrum. These tell us that Sirius B has a radius of $0.0084 R_{\odot}$ and a mass of about $1 M_{\odot}$ (e.g. &lt;a href=&quot;https://arxiv.org/abs/1201.3822&quot; rel=&quot;noreferrer&quot;&gt;Holberg et al. 2012&lt;/a&gt;). This leads to an estimate of the &lt;em&gt;average&lt;/em&gt; density of $2.4 \times 10^{9}$ kg/m$^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we know that for such an object to find any kind of equilibrium that the pressure must be greatest in the middle. Since for any plausible equation of state, the pressure increases with density, then the white dwarf must be &lt;em&gt;even&lt;/em&gt; denser than the average at its centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there it is a standard piece of bookwork to show why the white dwarf &lt;em&gt;must&lt;/em&gt; be supported by electron degeneracy pressure and I will repeat it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An electron gas can only escape degeneracy &lt;em&gt;if&lt;/em&gt; $E_F - m_e c^2 &amp;lt; k_{b}T$, where $E_F - m_e c^2$ is the kinetic energy associated with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Fermi_energy&quot; rel=&quot;noreferrer&quot;&gt;Fermi energy&lt;/a&gt; of the electrons, $m_e$ is the electron mass and $T$ is the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Fermi energy can be derived from the following formula (again, standard bookwork, this is not exotic physics, it is the same physics that describes how conduction takes place in metals).&#xA;$$ E_F = \left[ \left(\frac{3h^3n_e}{8\pi}\right)^{2/3}c^2 + m_e^2 c^4 \right]^{1/2},$$&#xA;where $n_e$ is the electron number density. Now it turns out that the temperatures we are going to need to avoid degeneracy (see below) are easily going to be sufficient to completely ionise the contents of the white dwarf. If we assume the white dwarf is made out of carbon or oxygen, then the relationship between mass density and $n_e$ is just $\rho = 2 n_e m_u$, where $m_u$ is an atomic mass unit (the 2 just comes from the fact that there are 2 mass units per free electron). We could mess about with the composition &lt;em&gt;a little&lt;/em&gt; to change this relationship, but it doesn't matter very much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making the appropriate substitutions and using $\rho = 2.4 \times 10^{9}$ kg/m$^3$, we obtain $E_F = 0.75$ MeV (or $1.2\times10^{-13}$ J). If we now use the test&#xA;$$ T &amp;gt; \frac{E_F - m_e ,c^2}{k_b},$$&#xA;we find that to avoid electron degeneracy then the temperature inside the white dwarf must exceed $2.75 \times 10^{9}$ K (and be much higher in the centre of the white dwarf where the density and Fermi energy are larger). If temperatures were this high inside a white dwarf and it was made of something like carbon or oxygen then nuclear fusion would take place (rapidly) and the luminosity of the white dwarf would not be as low as it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if it were made of iron? There would be no nuclear fusion, even at these high temperatures. OK, leaving aside that if the interior really were that hot, it could not be hidden very easily by an atmosphere, then this would not be a stable situation. The star would be losing copious heat via neutrino losses. If the gas were non-degenerate, the star would rapidly contract and become even hotter and even denser, and if you are not going to allow for degeneracy (i.e. you postulate that quantum mechanics does not exist), then rather quickly the white dwarf ends up being a black hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A back of the envelope estimate would be to divide the gravitational potential energy by the current luminosity. Taking Sirius B as an example, the timescale is&#xA;$$ \tau \sim \frac{GM^2}{RL},$$&#xA;where $L$ is the luminosity. Taking the current luminosity from photons at the surface this timescale would be a few $10^{10}$ years (plausibly stable). But if the interior were at $\sim 10^{10}$ K, the neutrino losses would be many orders of magnitude greater and the timescale would be just millions of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that this isn't happening to white dwarfs because we know they are stable and long-lived. We see them in clusters of stars of known age and associated with stars that are much older than millions of years. For instance we know that stars like Sirius A are hundreds of millions of years old.&#xA;We can calculate using the theory of electron degenerate structures that the faintest white dwarfs we see have been cooling for 10 billion years without collapsing - precisely the age expected for the galactic disk. The cooling ages of white dwarfs in clusters match the ages of those clusters very well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Degenerate matter is very common on earth. The &lt;a href=&quot;http://farside.ph.utexas.edu/teaching/sm1/lectures/node86.html&quot; rel=&quot;noreferrer&quot;&gt;conduction electrons&lt;/a&gt; in a metal form a degenerate gas. If you mean the kind of degenerate matter in white dwarfs, then no, at least not in any stable state. The pressures required to constrain such energy densities are too large to create in the lab.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="11143" LastEditDate="2017-03-07T16:25:07.027" LastActivityDate="2017-03-07T16:25:07.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20304" PostTypeId="1" CreationDate="2017-03-07T01:50:49.370" Score="6" ViewCount="570" Body="&lt;p&gt;How do magnetars create/sustain such strong magnetic fields? If a dynamo effect did create a magnetar, then what creates the magnetic field? The dynamo effect requires an electrically conductive fluid, but magnetars are purely composed of neutrons. In Earth, the dynamo effect is supported with evidence, but how does it apply to magnetars?&lt;/p&gt;&#xA;" OwnerUserId="16113" LastActivityDate="2017-03-07T17:05:36.100" Title="Magnetars and the Dynamo Effect" Tags="&lt;neutron-star&gt;&lt;magnetic-field&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20305" PostTypeId="1" AcceptedAnswerId="20308" CreationDate="2017-03-07T02:44:15.830" Score="3" ViewCount="145" Body="&lt;p&gt;According to the big bang theory, the universe started from a small intial point and is essentially expanding. However, my question is that if the universe is expanding how is it possible for galactic collisons to occur? Are the galacies moving away from a relative position but not moving away from each other? If so, how do we know this?&lt;/p&gt;&#xA;" OwnerUserId="15910" LastActivityDate="2017-03-07T07:52:02.480" Title="Collision of Galaxies" Tags="&lt;galaxy&gt;&lt;big-bang-theory&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20306" PostTypeId="2" ParentId="20305" CreationDate="2017-03-07T04:22:09.473" Score="7" Body="&lt;p&gt;The expansion of space is something that happens on the largest scales.  At small scales, such as distances between nearby galaxies, other forces, such as gravity, dominate.  Galaxy clusters are held together by the attractive force of gravity between these galaxies.  Space in these regions is still expanding, but the gravity pulls on these galaxies much more than space's expansion moves them apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxies that are close enough together, such as the Milky Way and Andromeda, are pulled together by the force of gravity between them.  &lt;/p&gt;&#xA;" OwnerUserId="13426" LastActivityDate="2017-03-07T04:22:09.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20307" PostTypeId="1" CreationDate="2017-03-07T06:34:05.317" Score="2" ViewCount="7716" Body="&lt;p&gt;The ratio for a planet to be tidally locked has to be 1:1, but the ratio for Mercury is 3:2. How is Mercury tidally locked if the ratio is not 1:1?&lt;/p&gt;&#xA;" OwnerUserId="16110" LastActivityDate="2019-08-06T14:24:50.260" Title="How is Mercury tidally locked if the ratio is not 1:1?" Tags="&lt;tidal-locking&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20308" PostTypeId="2" ParentId="20305" CreationDate="2017-03-07T07:52:02.480" Score="2" Body="&lt;p&gt;Even more concise: &lt;em&gt;the universal expansion of space does not affect the space within gravitationally bound structures&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pair of colliding galaxies clearly fall into this category.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general expansion of the universe is only apparent at the largest scales, where the universe can be treated as an isotropic, homogeneous fluid, where there is no net gravitational force acting on any &quot;particle&quot;. At smaller scales, this isn't true. The universe at small scales is messy and inhomogeneous. The anisotropic gravitational forces between bound objects completely dominate the expansion effect.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-03-07T07:52:02.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20309" PostTypeId="2" ParentId="19498" CreationDate="2017-03-07T08:50:27.463" Score="1" Body="&lt;p&gt;There is a tentative detection of a GRB simultaneously with an FRB,  &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/2041-8205/832/1/L1/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;DeLaunay et al 2016&lt;/a&gt;. If it is true, it would mean that at least some FRBs have the same origin as some GRBs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically many possibilities were suggested before and after this tentative, detection. For now they all remain quite speculative, and more observations are required (and planned)&#xA;For some selection of models, it could be interesting to look into &lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/nph-ref_query?bibcode=2016ApJ...832L...1D&amp;amp;refs=CITATIONS&amp;amp;db_key=AST&quot; rel=&quot;nofollow noreferrer&quot;&gt;citations&lt;/a&gt; to this paper or its &lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/nph-ref_query?bibcode=2016ApJ...832L...1D&amp;amp;refs=REFERENCES&amp;amp;db_key=AST&quot; rel=&quot;nofollow noreferrer&quot;&gt;references&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16065" LastActivityDate="2017-03-07T08:50:27.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20310" PostTypeId="1" CreationDate="2017-03-07T11:42:48.603" Score="5" ViewCount="10066" Body="&lt;p&gt;Why is it that Auroras only occur in the North and South poles. Why don't they occur in the equator or anywhere for that matter?&lt;/p&gt;&#xA;" OwnerUserId="16075" LastActivityDate="2017-03-07T14:30:07.123" Title="Why do Auroras only occur in polar regions?" Tags="&lt;aurora&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20311" PostTypeId="2" ParentId="20310" CreationDate="2017-03-07T13:57:35.617" Score="7" Body="&lt;p&gt;The reason is because of Earth's global magnetic field. Take a look at the image below which shows a representation of what the magnetic field lines around Earth look like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lQNc2.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lQNc2.jpg&quot; alt=&quot;Earth&amp;#39;s magnetic field&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you see above is what's known as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetic_dipole&quot; rel=&quot;noreferrer&quot;&gt;Magnetic Dipole&lt;/a&gt; field. The reason being that there are two (Di means two) magnetic poles, namely the north pole and the south pole. This is completely analogous to a giant bar magnet you may have played with in science class. You can see here that the field lines leave the magnetic south pole and travel around, going up into the magnetic north pole. The field lines don't really leave from or enter the Earth's equator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why does this mean aurora's happen at the poles? Because, according to the theory of electromagnetism, charged particles are forced to travel only along magnetic field lines. They cannot travel across them or break free from them. Once they get stuck, they're stuck (this isn't 100% true but for the purposes of this answer, it is). Note that the particles can travel both forward and backward on the field lines. They're not constrained to going only in the direction of the field line. This means that the primary location in which these charged particles can enter the Earth's atmosphere is where the field lines tend to hit the atmosphere and we already noted that that location was the magnetic north and south pole (which coincidentally is near the physical north and south pole).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Auroras themselves are just emissions from our atmosphere in response to being bombarded by high energy particles. To put this all together, here's what happens: The Sun spews out huge amounts of energy and tons of charged particles. Some of those charged particles hit the Earth's magnetic field and get trapped in it, being forced to travel along the magnetic field lines. They are then constrained to primarily hit the Earth's atmosphere at the magnetic poles where their extreme energy excites the atmosphere, causing it to emit light in shimmering red and green sheets. The auroras can't occur at the equator simply because not enough particles can hit the atmosphere there to cause auroras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day though, we may see auroras at the equator. The Earth's magnetic field is not fixed and moves with time. About every 450,000 years the field flips around completely. Its been about &lt;a href=&quot;http://www.astronoo.com/en/articles/magnetosphere-earth.html&quot; rel=&quot;noreferrer&quot;&gt;750,000 years&lt;/a&gt; since the last reversal so we're due for one potentially soon (and by soon I mean in the next few thousand or tens of thousands of years). During the flip, while the magnetic poles are over the equator, you'll see auroras there.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2017-03-07T14:30:07.123" LastActivityDate="2017-03-07T14:30:07.123" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20312" PostTypeId="2" ParentId="20304" CreationDate="2017-03-07T14:57:35.120" Score="6" Body="&lt;p&gt;Magnetars (and neutron stars in general) don't need a dynamo to create their magnetic fields. Their magnetic fields are &quot;frozen in&quot; at the time of their formation. To really see why this is, you have to understand a lot about electromagnetism, but I can boil it down to the basics. Keep in mind that neutron stars are highly mysterious objects and we don't have a huge amount of good, observational evidence behind our theories since they're so hard to find and observe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutron stars (and magnetars) are superconductors (or at least &lt;a href=&quot;http://physicsworld.com/cws/article/news/2011/mar/02/neutron-star-has-superfluid-core&quot; rel=&quot;nofollow noreferrer&quot;&gt;most theory and objective evidence suggests that is the case&lt;/a&gt;). What this means is that there is literally no resistance to the motion of electric charges moving throughout the star. To point out, even if the entire star was all neutrons (which &lt;a href=&quot;https://astronomy.stackexchange.com/questions/20304/magnetars-and-the-dynamo-effect#comment33689_20304&quot;&gt;zibadawa timmy points out is not the case&lt;/a&gt;) you'd still have charges since neutrons themselves are composed of charged particles. In any case, since there's no resistance, there cannot exist any electric fields inside the star. Any field which arose would induce a force on the charged particles which would be able to immediately move to cancel said field out. All electric fields destroy themselves immediately in a superconductor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we venture into electromagnetic theory equations, you'll find the very useful equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \times \textbf{E} = -\frac{\partial \textbf{B}}{\partial t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without getting into the details of this equation, the general idea behind it is that electric fields result in time varying magnetic fields and vice versa. But we just said above that our neutron star was superconducting and thus has no electric fields. This means the left hand side of that equation is zero. The right side represents the change in magnetic field over time, but we know that has to be zero now so we're forced to conclude &lt;strong&gt;magnetic fields in neutron stars (and magnetars) cannot change. They're frozen in at the point of creation (when the star becomes superconducting).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can venture one step further and say that the magnetic fields of neutron stars and magnetars is so strong simply because of magnetic flux conservation. That is, you have a huge star of a few solar masses which has a massive field in its own right. That star then collapses into a neutron star but in the process, the magnetic field flux through its surface has to stay conserved. Magnetic field flux is a function of both the field strength and the radius of the star. Since the radius is decreasing so immensely, &lt;strong&gt;the magnetic field strength has to increase in proportion to the radius decrease (squared)&lt;/strong&gt; to compensate and keep the overall flux the same, at which point that magnetic field is locked into place due to the superconducting nature of the star.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2017-03-07T16:20:32.920" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20313" PostTypeId="1" CreationDate="2017-03-07T16:10:57.543" Score="1" ViewCount="33" Body="&lt;p&gt;I am working on a Bayesian analysis project for class using data I gathered some data from NASA's Sloan Atlas (&lt;a href=&quot;http://www.nsatlas.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.nsatlas.org/&lt;/a&gt;). One of the variables I am dealing with is the Sérsic Index (calculated by &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0004-6256/142/1/31/pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://iopscience.iop.org/article/10.1088/0004-6256/142/1/31/pdf&lt;/a&gt;).&#xA;I am wondering if there is some error associated with this value.  I hope to model each galaxies Sersic index as a gaussian, but I can´t wrap my head around what to write as the error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your time!&lt;/p&gt;&#xA;" OwnerUserId="16117" LastActivityDate="2017-03-07T16:10:57.543" Title="Error associated with Sersic Profile calculations?" Tags="&lt;galaxy&gt;&lt;data-analysis&gt;&lt;sloan-dss&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20314" PostTypeId="1" CreationDate="2017-03-07T16:37:58.927" Score="1" ViewCount="3422" Body="&lt;p&gt;There are times when the sun goes down and it takes a while for the moon to rise, because the moon obviously doesn't rise with the sunset whole year. I guess this happens only when the moon is full. Also, in some days, moon sets in the midnight and there is no moon on the sky for the rest of the night. So, I was wondering what that time period of the night is called when the moon has gone down, or has yet to rise? Can we call it 'moondown'? I am not talking about the new moon when there is no moon the whole night. Please help, and thank you in advance!&lt;/p&gt;&#xA;" OwnerUserId="16118" LastActivityDate="2017-04-06T22:16:23.070" Title="What is the specific term for time of the night when there is no moon?" Tags="&lt;the-sun&gt;&lt;the-moon&gt;&lt;night-sky&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20315" PostTypeId="1" CreationDate="2017-03-07T16:48:01.233" Score="4" ViewCount="1309" Body="&lt;p&gt;I've just begun learning of Astronomy and I can't figure out why any stars would begin their life with such elements if nuclear fusion hasn't created them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't all stars begin life as Hydrogen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the presence of these elements can change the luminosity of the given star. Is this also correct?&lt;/p&gt;&#xA;" OwnerUserId="16119" LastActivityDate="2017-03-10T16:10:50.127" Title="How much of carbon, sodium, silicon, and magnesium does the Sun have?" Tags="&lt;star&gt;&lt;stellar-evolution&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20316" PostTypeId="2" ParentId="20304" CreationDate="2017-03-07T17:05:36.100" Score="6" Body="&lt;p&gt;There are, in general, two classes of explanations for neutron star magnetic fields: fossilized magnetic fields and active magnetic fields (see &lt;a href=&quot;http://dspace.rri.res.in/bitstream/2289/3315/20/Chapter%203.pdf&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; for an early overview on some of the internal battery models).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;fossilized&quot; field theory - which is well-accepted, as far as I know - states that neutron star magnetic fields are leftovers from the magnetic fields of the progenitor stars. This seems plausible, and some (e.g. &lt;a href=&quot;https://arxiv.org/abs/0711.3650&quot; rel=&quot;noreferrer&quot;&gt;Spruit (2008)&lt;/a&gt;) have suggested that the supernova that formed the neutron stars may have endowed them with exceptionally strong fields during core collapse, leaving behind magnetars. This is what zephyr means by &quot;frozen in&quot;: the magnetic fields remains the same after the star becomes a neutron star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;active&quot; theories - and I'm using &quot;active&quot; as a non-technical term - posit that neutron stars continue to generate magnetic fields. This makes it possible for the magnetic fields to grow in strength, which can explain why magnetars have exceptionally strong fields; fossilized fields may not be sufficient to explain this in all cases. There have been several suggestions over the years for changes in magnetic fields, some of which are no longer accepted but some of which are still possible:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;The battery model.&lt;/strong&gt; This was originally proposed as a mechanism for generating magnetic fields in normal stars. It holds that electrons inside a star drift outward slightly relative to the ions, due to different effects from the gravitational field and any centrifugal force. The resulting pressures cause the electrons to move in ways similar to a battery, which generates a magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As zephyr mentioned, the important equation is&#xA;$$-\frac{\partial\mathbf{B}}{\partial t}=\nabla\times\mathbf{E}$$&#xA;In the battery model, $\nabla\times\mathbf{E}$ turns out to be non-zero, meaning that the magnetic field can, in fact, change, thanks to thermal effects.  I believe, however, that the battery model has been ruled out for neutron stars. Taking degeneracy pressure into account does, in fact, lead to a vanishing $\nabla\times\mathbf{E}$, and therefore there is no time-varying field.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;The thermoelectric mechanism.&lt;/strong&gt; This is actually a variation on the pure battery model, and is applicable only in the neutron star's crust. If there is a non-zero radial heat gradient &lt;em&gt;and&lt;/em&gt; a small magnetic field, electrons will create a small pressure gradient, which in turn causes an opposing thermoelectric field to arise - which leads to a non-zero $\nabla&#xA;\times\mathbf{E}$! The precise equation is&#xA;$$\frac{\partial\mathbf{B}}{\partial t}=\overbrace{\nabla\times\left(\mathbf{V}\times\mathbf{B}\right)}^{\text{Field convection term}}-\overbrace{\nabla Q_0\times\nabla T}^{\text{Battery term}}-\overbrace{\nabla\times\left[\frac{\nabla\times\mathbf{B}}{4\pi\sigma_0}\right]}^{\text{Ohmic decay term}}$$&#xA;The thermoelectric model is much better than the traditional battery model, and does allow for the magnetic field to grow.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Accretion from a companion.&lt;/strong&gt; This is a slightly newer idea, which has gained traction after observations of binary systems (see a discussion by &lt;a href=&quot;http://adsabs.harvard.edu/full/1999ptgr.conf..235B&quot; rel=&quot;noreferrer&quot;&gt;Bhattacharya (1999)&lt;/a&gt;). Matter from a companion star follows the neutron star's pre-existing magnetic field lines. Pressure causes the matter to &quot;drag&quot; the field lines along the neutron star's surface until &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetic_reconnection&quot; rel=&quot;noreferrer&quot;&gt;magnetic reconnection&lt;/a&gt; occurs, &quot;screening&quot; the field underneath the accreted matter. This actually weakens the field, making it decay over time - which does match observations of some neutron stars. However, instabilities make it a difficult possibility.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2017-03-07T17:05:36.100" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20318" PostTypeId="2" ParentId="20315" CreationDate="2017-03-07T18:59:39.647" Score="6" Body="&lt;p&gt;Stars are mostly hydrogen and helium. The presence of small amounts of other elements does change their luminosity, and their spectra. The heavy elements were formed in other stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the big bang there was a short period in which the conditions were right for hydrogen to fuse into helium. The conditions were not suitable for any heavier elements to form, the state of the universe right after this time was about 3/4 hydrogen, 1/4 helium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some time clouds of hydrogen and helium began to form (around clumps of dark matter) and inside some of these clumps, smaller clouds collapsed to form the first generation of stars. These were large and they had no elements other than hydrogen and helium. These stars produced in their cores heavier elements like Carbon and Oxygen, and when they had reached the end of their lives, being big stars, they exploded in supernovae, and these elements were returned to the interstellar space, along with other heavier elements that were formed by the supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next generation of stars formed from a mixture of hydrogen, helium and the heavier elements formed in the first generation. These stars still had less heavy elements than modern stars, but were more like normal stars. Some of these stars are still around, they are known as Population 2 stars. They are found mostly in globular clusters &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/Starlog/pop12.html&quot; rel=&quot;noreferrer&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/hbase/Starlog/pop12.html&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the population 2 stars had lived their lives and died, either in a supernova or in a planetary nebula, elements that had been made in their cores was returned to the intersellar space, futher enriching it with heavy elements. It was from one of these clouds, seeded with elements from stars that had gone before it, that the sun and the solar system formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why Carl Sagan says &lt;a href=&quot;https://www.google.co.uk/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=1&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0ahUKEwiJwM2QicXSAhWlI8AKHSmTDI8QyCkIHDAA&amp;amp;url=https%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3DtLPkpBN6bEI&amp;amp;usg=AFQjCNHOIJ3CRxlExzJHhEBtoCq5XKPX1g&amp;amp;sig2=fHb_SQYFjaGznPVf_28k4w&amp;amp;bvm=bv.148747831,d.d24&quot; rel=&quot;noreferrer&quot;&gt;We are star stuff&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-03-07T18:59:39.647" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20319" PostTypeId="2" ParentId="20307" CreationDate="2017-03-07T19:59:35.977" Score="3" Body="&lt;p&gt;The simple answer to your question is that Mercury is not tidally locked. You may have seen old books (before 1965) that said it was tidally locked, because it was once assumed to be so. Alternatively, as zephyr said, your source may have been referring to the 3:2 resonance, but that is also not really the same thing.&lt;/p&gt;&#xA;" OwnerUserId="7752" LastEditorUserId="7752" LastEditDate="2017-03-07T20:33:41.970" LastActivityDate="2017-03-07T20:33:41.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20321" PostTypeId="2" ParentId="20298" CreationDate="2017-03-07T22:00:04.290" Score="3" Body="&lt;p&gt;If you are reducing everything to plain geometry, you can compute the fraction of the &quot;sunny&quot; arc of the circle of latitude for a given location. (Hint: draw a diagram showing the tropic of Cancer, the circle of latitude, the terminator and the axis, all as straight lines.) The height of this segment of a circle of latitude (at $\lambda$) is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h = \frac{\cos(\lambda) + \sin(\lambda)\times \tan(\epsilon)}{\cos(\lambda)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle of this segment is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a = 2\times \arccos(1 - h)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;New York: 14.9297092807291&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;London: 16.4195095586004&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;which is fairly close to the &quot;expected&quot; numbers you have provided. The difference is very likely due to Mathematica using a geoid rather than a sphere and including atmospheric refraction.&lt;/p&gt;&#xA;" OwnerUserId="16122" LastEditorUserId="9301" LastEditDate="2017-03-07T22:29:34.823" LastActivityDate="2017-03-07T22:29:34.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20322" PostTypeId="2" ParentId="20314" CreationDate="2017-03-07T22:01:00.020" Score="1" Body="&lt;p&gt;You are right that the moon only rises as the sun sets when it is full. At other times there is some time each night when both the moon and the sun are both set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The terms moonrise and moonset are well established, by analogy with sunrise and sunset. The time when the moon is at its highest point in the sky is known as lunar culmination. You can use &quot;lunar day&quot; for the 24.84 hour period between lunar culminations. I know of no specific term for the period when the moon is set, or for when it is risen, instead you can say &quot;after moonrise&quot; or &quot;before moonset&quot;, as appropriate.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-03-07T22:01:00.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20323" PostTypeId="2" ParentId="20307" CreationDate="2017-03-07T22:14:08.060" Score="2" Body="&lt;p&gt;The reason that Mercury is not locked into a 1:1 spin orbit resonance is its eccentric orbit. In fact, the 3:2 resonance is the result of weak tides in conjunction with the orbital eccentricity. This is in fact mentioned on the Wikipedia page.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2017-03-07T22:14:08.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20324" PostTypeId="1" CreationDate="2017-03-07T22:15:38.933" Score="2" ViewCount="3457" Body="&lt;p&gt;I searched Google for a while and of so many available cameras, I couldn't find a moon live stream video camera. Does anyone know any free website that offers this? Where can you see live the moon right now?&lt;/p&gt;&#xA;" OwnerUserId="16125" LastActivityDate="2019-08-07T16:26:55.773" Title="Any live moon stream?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20326" PostTypeId="2" ParentId="20315" CreationDate="2017-03-08T08:55:56.200" Score="4" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/20318/1577&quot;&gt;James' answer&lt;/a&gt; already covers the fact that since helium was formed in the Big Bang (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Bang_nucleosynthesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Big Bang Nucleosynthesis&lt;/a&gt;), stars always formed with some helium (about 1/4 by mass). Those first stars fused the hydrogen and helium into heavier elements, and some of them eventually spewed these products into their neighbourhoods, so the next generation of stars was born with those &quot;metals&quot; (all elements heavier than helium) from the start. Each subsequent &quot;generation&quot; of stars was born with more and more of the detritus of the previous generations, so the Sun was born with a wide range of metals. All that said, the metal content in your average stars is still low as a fraction of mass. e.g. around 2% in the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the title question, however, I suggest starting with a &lt;a href=&quot;http://adsabs.harvard.edu/abs/2009ARA%26A..47..481A&quot; rel=&quot;nofollow noreferrer&quot;&gt;widely-cited review article&lt;/a&gt; from 2009.  Their Table 1 gives the logarithm of the number densities relative to hydrogen, plus 12. i.e.&#xA;$$\log_{10}\epsilon_X=\log_{10}(n_X/n_\mathrm{H})+12$$&#xA;The values for carbon, sodium, silicon, and magnesium are 8.43 ± 0.05, 6.24 ± 0.04, 7.51 ± 0.03 and 7.60 ± 0.04, respectively. If you follow the paper trail, you'll find other estimates too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, the metal abundance of the Sun is actually a contemporary problem. Though this oversimplifies the details substantially, we generally measure the abundances by comparing spectral lines with models of the Sun's outer layers atmosphere. In the late 1990's and early 2000's, those models leaped forward from generally one-dimensional, hydrostatic models, to three-dimensional hydrodynamic models.  Although the 3D models fit the spectral data better, they do so with lower abundances than we inferred with the 1D models.  This revision isn't itself a problem but it means that our models of the Sun are quite a lot worse than we used to think (see Fig. 8 in the review), so there's a fair amount of effort going into trying to find what's missing from the models to explain the remaining discrepancy.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2017-03-08T08:55:56.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20327" PostTypeId="2" ParentId="13953" CreationDate="2017-03-08T13:23:37.570" Score="0" Body="&lt;p&gt;Apparently there is a way of testing a convex mirror by making it from optical grade glass and testing it through the back. The refractive part of the light path creates a situation that if you test it as a sphere or paraboloid in that way (I forget which) the actual curve you get is a hyperboloid.  &lt;/p&gt;&#xA;" OwnerUserId="16130" LastActivityDate="2017-03-08T13:23:37.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20328" PostTypeId="2" ParentId="20324" CreationDate="2017-03-08T16:41:19.950" Score="3" Body="&lt;p&gt;Since the moon isn't visible all the time from any single camera location on Earth, such a website would have to be an aggregate of many different cameras around the world, displaying a single camera based on current time and weather conditions. Even in this scenario, there's a chance all the cameras will be obscured by clouds and there will be no view of the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a website doesn't exist &lt;em&gt;that I know of&lt;/em&gt;. Perhaps it would be an interesting project if you're interested in such a thing.&lt;/p&gt;&#xA;" OwnerUserId="15904" LastActivityDate="2017-03-08T16:41:19.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20329" PostTypeId="1" CreationDate="2017-03-08T17:46:32.193" Score="1" ViewCount="262" Body="&lt;p&gt;I have a question concerning how Hawking radiation works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand, a pair of particle-antiparticle may pop to existence near the event horizon (from a virtual particle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then it might happen that the negative particle (antimatter) gets captured by the black hole whereas the positive particle (matter) escapes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This results in a net loss of the black hole mass. If you repeat this process an astonishing number of times the Black Hole might end up completely evaporating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right so far? (Even if the explanation is not very mathematical)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question then is:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This assumes that when a pair particle-antiparticle is created, a bias towards antiparticles being captured and particles escaping must exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, the mass of the black hole might fluctuate, but in average for big numbers of pairs, equal percentage of particles and antiparticles would be captured/escape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is there really a bias needed for the black hole to loose mass, or is it something wrong about my understanding of Hawking radiation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="16132" LastActivityDate="2017-05-08T22:12:41.370" Title="Symmetry in Hawking radiation?" Tags="&lt;black-hole&gt;&lt;antimatter&gt;&lt;hawking-radiation&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20333" PostTypeId="1" AcceptedAnswerId="20334" CreationDate="2017-03-09T08:39:23.807" Score="3" ViewCount="1248" Body="&lt;p&gt;I'm trying to write a Python program that will do the following: take an image in fits format and a reg file with a list of regions in the sky, and determine which regions in the list are inside the fits image and which ones are outside. This can be done visually with ds9, but I have no idea of how I can access the fits file and determine the coordinates of the boundary of the image. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any advice? I don't know much about the fits format, but it won't open with a usual text editor like gedit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Now I have a more specific problem. I have four points in the sky defined by their declination and right ascension. Is there a simple way to check if these coordinates belong to the image coded in the fits file? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2:&lt;/strong&gt; This is what I see as the header of my fits file:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/kX26H.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kX26H.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm no expert in physics, just a mathematician, so I'm not sure how to proceed from here. Any help would be appreciated...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 3:&lt;/strong&gt; Something seems to change when I put the terminal in full screen mode. Is there any useful info here?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/9VsAk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9VsAk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16142" LastEditorUserId="16142" LastEditDate="2017-03-09T17:40:00.097" LastActivityDate="2017-09-24T21:35:19.003" Title="Python with image in fits format" Tags="&lt;data-analysis&gt;&lt;python&gt;" AnswerCount="3" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="20334" PostTypeId="2" ParentId="20333" CreationDate="2017-03-09T09:34:39.793" Score="4" Body="&lt;p&gt;I personally use &lt;a href=&quot;http://www.astropy.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astropy&lt;/a&gt;, specifically &lt;code&gt;astropy.io.fits&lt;/code&gt;, although I'm not a seasoned user of FITS files and I don't really know their layout.  As an example snippet of code, I often load data from FITS files using&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from astropy.io import fits&#xA;data = fits.open('data_file.fits')[0].data&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You'll find more information in the &lt;a href=&quot;http://docs.astropy.org/en/stable/io/fits/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;documentation&lt;/a&gt; on the FITS module.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2017-03-09T09:34:39.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20338" PostTypeId="2" ParentId="20333" CreationDate="2017-03-09T13:56:29.577" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;There's no clear answer to this, it would all depend on your specific file, which of course we know nothing about. The information you need is likely going to exist in the fits header though (if it exists at all). I'm assuming you're reading your fits file using the &lt;code&gt;astropy&lt;/code&gt; package since you're using Python (if not, definitely check it out). If that is the case, you can read all about looking at the fits file headers &lt;a href=&quot;http://docs.astropy.org/en/stable/io/fits/#working-with-fits-headers&quot; rel=&quot;nofollow noreferrer&quot;&gt;in the AstroPy documentation&lt;/a&gt;. You'll have to look at the header files and scan through to see if any information included with the fits file tells you precisely where in the sky the image is.&lt;/li&gt;&#xA;&lt;li&gt;No image is going to be &quot;circular&quot;. Images are taken on square CCDs and thus produce square images of the sky. A &quot;regular&quot; image will be square.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-03-09T13:56:29.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20339" PostTypeId="1" CreationDate="2017-03-09T17:07:10.187" Score="1" ViewCount="38" Body="&lt;p&gt;I'd like to know if it's possible to create a strong enough onboard magnetic field, similar to Earths' that would protect deep space spacecrafts from the solar wind?&lt;/p&gt;&#xA;" OwnerUserId="16150" LastActivityDate="2017-03-09T17:07:10.187" Title="magnetic shield for spacecraft outside of Earths shield" Tags="&lt;astrophysics&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20340" PostTypeId="1" AcceptedAnswerId="20341" CreationDate="2017-03-09T17:27:41.630" Score="9" ViewCount="2648" Body="&lt;p&gt;My question is about the equivalence about having an event horizon and having a singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one side the implication looks pretty obvious: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A singularity implies having an event horizon and therefore a black hole. Since the mass is compressed in a zero volume space, if you get close enough there will be a point where the escape velocity gets bigger than the speed of light so you will get a black hole by definition.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But what about the opposite? Does having an event horizon imply the existence of a singularity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it be that you have a neutron star massive enough to reach a escape velocity equal to the speed of light but not strong enough to make the matter collapse?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if such star can not exist because the strong force collapses before reaching an event horizon, this doesn't mean an equivalence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It just means that for some specific value of the maximum strong force this is not possible, but image now an imaginary exotic matter that has a way bigger strong force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For such &quot;science fiction&quot; matter, it would be possible to reach an event horizon without collapsing to a singularity, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it really an equivalence between this two concepts, such that no matter how resistant matter is to collapse it will never reach an event horizon?&lt;/p&gt;&#xA;" OwnerUserId="16132" LastEditorUserId="11143" LastEditDate="2017-03-09T17:33:02.157" LastActivityDate="2021-02-20T21:28:29.960" Title="Black hole without singularity?" Tags="&lt;black-hole&gt;&lt;singularity&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20341" PostTypeId="2" ParentId="20340" CreationDate="2017-03-09T17:45:17.730" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does having an event horizon imply the existence of a singularity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An event horizon is not an inherent component of any given object. It's not like once a star turns into a black hole, it suddenly gets an event horizon. The event horizon is merely a mathematical boundary which defines the distance from a mass $M$ where the escape velocity equals the speed of light. I can calculate such a boundary for a black hole, for the Sun, the Earth, or even you. So I guess the answer here is &lt;strong&gt;no, having an event horizon does not imply the existence of a singularity&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Could it be that you have a neutron star massive enough to reach a escape velocity equal to the speed of light but not strong enough to make the matter collapse?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer here, technically, is no. The reason being that once it requires a speed greater than or equal to the speed of light to escape your object, it is necessarily a black hole. That is the definition of a black hole. So that means this neutron star you propose is actually a black hole. Another equivalent definition of a black hole is any object whose mass is concentrated inside that object's event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you might still ask, could you have a black hole where the mass inside the event horizon is not a singularity. This would require some sort of support to prevent the matter from collapsing down to the singularity. The answer to this is that it is currently unknown. The problem is that inside event horizons, suddenly you need to work with both GR and quantum field theory but those two theories don't play nice. Instead you should be using a Quantum Gravity theory but this theory hasn't been developed. So ultimately any answer to this would be a guess until this theory gets fully fleshed out.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-03-09T17:45:17.730" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20342" PostTypeId="2" ParentId="20329" CreationDate="2017-03-09T21:18:59.700" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;This assumes that when a pair particle-antiparticle is created, a bias towards antiparticles being captured and particles escaping must exist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No such assumption or bias exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as we know the universe treats anti-particles just like particles with different properties (e.g. charge is opposite compared to the corresponding particle).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Otherwise, the mass of the black hole might fluctuate, but in average for big numbers of pairs, equal percentage of particles and antiparticles would be captured/escape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You seem to be implying that anti-matter has negative mass.  All particles (including anti-matter) have either zero mass (like photons) or positive mass (like electrons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So regardless of whether the anti-particle or particle escapes, the black hole looses mass (more precisely it looses energy).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So, is there really a bias needed for the black hole to loose mass, or is it something wrong about my understanding of Hawking radiation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your understanding of the mass of anti-particles is likely the problem.  They have positive mass, just like particles, because they are particles.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-03-09T21:18:59.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20343" PostTypeId="1" AcceptedAnswerId="20351" CreationDate="2017-03-09T22:33:50.283" Score="0" ViewCount="1464" Body="&lt;p&gt;I'm a software engineer (a moderately good one). I like programming, mathematics and I love everything about astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be best way to go ahead?&lt;/p&gt;&#xA;" OwnerUserId="16158" LastEditorUserId="16158" LastEditDate="2017-03-15T17:31:13.610" LastActivityDate="2017-03-15T17:31:13.610" Title="Astronomy as a profession for moderately good software engineer" Tags="&lt;software&gt;" AnswerCount="2" CommentCount="12" FavoriteCount="3" ClosedDate="2017-03-15T12:56:28.933" ContentLicense="CC BY-SA 3.0" />
  <row Id="20344" PostTypeId="2" ParentId="20340" CreationDate="2017-03-10T02:01:12.987" Score="3" Body="&lt;p&gt;A real neutron star would start to collapse when the strength of its gravity exceeds the strength of the neutron degeneracy pressure, before it has an event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you approach the event horizon, the force required to stop a stationary mass from falling in approaches infinity. So I don't think any finite force, fictional or otherwise, can keep a star star-shaped after it reaches the critical density at which it has an event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, predicting exactly what happens at or inside the event horizon would likely require a quantum theory of gravity, and I don't have one of those.&lt;/p&gt;&#xA;" OwnerUserId="16160" LastActivityDate="2017-03-10T02:01:12.987" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20345" PostTypeId="2" ParentId="20340" CreationDate="2017-03-10T02:16:33.490" Score="2" Body="&lt;p&gt;zephyr is right that you would need quantum gravity to really understand what happens inside event horizons. But the traditional description of what would happen inside the event horizon of a black hole (more or less ignoring quantum mechanics) is that there is no force that can stop matter from forming a singularity. The coordinate system inside the event horizon is such that, speaking crudely, the future direction points towards the center. So you can't just have a pile of matter dense enough to fit inside an event horizon, and at the same time strong enough not to collapse to a singularity. &lt;/p&gt;&#xA;" OwnerUserId="7752" LastActivityDate="2017-03-10T02:16:33.490" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20346" PostTypeId="2" ParentId="20343" CreationDate="2017-03-10T07:10:31.127" Score="-1" Body="&lt;p&gt;Considering your conditions and what you've already achieved. You could most likely start by doing a masters degree and then doing a PhD in astronomy if you want to do research.&#xA;The reason why I say to start out on a masters is because I assume you have a firm understanding of physics considering what you've done as I said. If you don't have enough background in physics, you could probably do a couple bridging courses and then do your masters. &#xA;That's a simple way to put it.&#xA;Regards, Ethan  &lt;/p&gt;&#xA;" OwnerUserId="16168" LastActivityDate="2017-03-10T07:10:31.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20347" PostTypeId="2" ParentId="20315" CreationDate="2017-03-10T07:12:00.343" Score="1" Body="&lt;p&gt;Nuclear fusion is the process in which atomic nuclei are fused together. You may remember that like charges repel and opposite charges attract. In order for atomic nuclei to undergo nuclear fusion, they need to overcome the Coulomb Barrier, which is basically a minimum threshold of energy needed for like charges to overcome their repulsion. The only place this happens is inside the cores of stars due to the high temperature needed to overcome this threshold. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although oversimplified, the main idea is that fusion of atomic nuclei combines tiny masses of particles and releases energy (via $\Delta E = \Delta m * c^2$). A very small amount of mass can be converted into a lot of energy. But the combined mass of fusing atomic nuclei is slightly less than the addition of the individual masses of atomic nuclei; it is this tiny mass difference that is converted into energy that we receive from the Sun in the form of heat and light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a cup of water that you drop a rock into. Because the density of the rock is greater than the density of the water, the rock will sink to the bottom. Similarly, elements born via nuclear fusion are heavier than their &quot;parents&quot;. In this way, nuclear fusion will keep building up heavier and heavier elements in the core of the star while the byproducts of the lighter elements &quot;float&quot; towards the top. But now that there are heavier elements in the core, the amount of energy required for nuclear fusion increases. Once the nuclear fusion &lt;a href=&quot;https://i.imgur.com/fvKTTcH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;chain reaches iron&lt;/a&gt;, the energy transfer is no longer cost-effective; the output energy from nuclear fusion will be less than the input energy needed for nuclear fusion, so the temperature is no longer sufficient to act as the input energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the star dies and goes nova, its elements (via fusion) are scattered in all directions, which can &lt;a href=&quot;https://i.imgur.com/kVzcJzq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;partially explain the galactic abundance of elements&lt;/a&gt;. You may notice from an &lt;a href=&quot;https://i.imgur.com/VqZovef.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;HR-diagram&lt;/a&gt; that the characteristics of a star (like luminosity, mass, temperature, and size) change over time. If you look specifically at the sharp turns in the diagram throughout the lifespan of a star (like our Sun), you will notice that they occur when new fusion changes begin, such as the CNO cycle (wh&lt;/p&gt;&#xA;" OwnerDisplayName="user15317" LastEditorUserId="2153" LastEditDate="2017-03-10T16:10:50.127" LastActivityDate="2017-03-10T16:10:50.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20349" PostTypeId="2" ParentId="20340" CreationDate="2017-03-10T14:24:10.647" Score="4" Body="&lt;p&gt;Singularity means &quot;my theory doesn't work here&quot;. In other words, GR is unable to predict what happens at the point, so it calls this point a singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most important thing is not to mistake map for the territory. GR is the map, a real black hole is the territory. GR is the map which allows us to predict what we will find in the territory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the map says &quot;don't really know about this point&quot; you really shouldn't expect that when you go into the territory you will see an Unmeasurable Infinite Thing there. It's very much against our historical experience. To date, time after time we observed normal finite things in the territory, but we never have seen Unmeasurable Infinite Things. In every case when an old map said we would see infinity, we found that measurements of a territory were finite and so &lt;em&gt;falsified&lt;/em&gt; that map (that theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it seems we should expect singularity as a word to only refer to the map. You really &lt;em&gt;shouldn't&lt;/em&gt; expect to observe a singularity (a map thing) when your actual ship enters a black hole (a territory thing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might turn out that GR is approximately right about an event horizon, but we already know it is not good enough to describe what is at the center.&lt;/p&gt;&#xA;" OwnerUserId="10675" LastActivityDate="2017-03-10T14:24:10.647" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20350" PostTypeId="2" ParentId="18524" CreationDate="2017-03-10T22:46:59.337" Score="1" Body="&lt;p&gt;&lt;em&gt;From comments&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;no, you can't have orbits anything like that close. Nor can you have a large moon anything like as close as you are thinking. 2 earth sized planets near each other's Roche limits would cause catastrophic earthquakes. Independent orbits would need, probably a few million miles apart for stability. Orbiting each other you could work them to perhaps, earth-Moon distance or even closer assuming they were tidally locked to each other to avoid too much disruptive tidal forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BUT if this is for a story: don't forget the other direction. One could say it is perhaps generally agreed today that &quot;panspermia&quot; can happen: life - microbes - via asteroids/etc can travel large astronomical distances. If we learned today that Mars has life - and indeed it's the &quot;same&quot; as Earth life - we'd assume that it travelled via ejecta - the Mars-Earth distance is no problem for such a putative story&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Comment converted to community wiki&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-03-10T22:56:00.233" LastActivityDate="2017-03-10T22:56:00.233" CommentCount="0" CommunityOwnedDate="2017-03-10T22:46:59.337" ContentLicense="CC BY-SA 3.0" />
  <row Id="20351" PostTypeId="2" ParentId="20343" CreationDate="2017-03-10T23:47:41.263" Score="4" Body="&lt;p&gt;A profession in astronomy can entail various things.  Unlike what other people are saying, you don't have to have a PhD in astronomy to do work in this field, to be an astronomer.  There are other options, but you will have to work hard.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Study as much physics and mathematics as possible.&lt;/strong&gt;  Astronomy is essentially a branch of physics.  If you're still in school, then take classes in these areas.  If you're not in school anymore, then there are plenty of &lt;a href=&quot;https://www.susanjfowler.com/blog/2016/8/13/so-you-want-to-learn-physics&quot; rel=&quot;noreferrer&quot;&gt;online resources&lt;/a&gt; to continue your education.  This will allow you to understand what people are talking about when they discuss technical things.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Decide what exactly you're interested in.&lt;/strong&gt;  There are lots of different areas of study within astronomy.  Decide what appeals to you the most.  Do you like planets?  Galaxies?  Cosmology?  All of these different focuses require more specialized knowledge.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Your skills are already marketable.&lt;/strong&gt;  Being an astronomer can mean lots of different things.  There are some astronomers who hardly ever look through telescopes, and spend most of their time reducing data.  A large part of astronomy is computer programming.  As a software engineer, you may already have the skills you need to work with astronomy research.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Keep an open mind.&lt;/strong&gt;  There are lots of jobs in astronomy research where you might not be called an &quot;astronomer&quot;.  For example, as a software engineer, you could find work with NASA supporting space research.  Your job title might not be &quot;Astronomer&quot;, but you'll still be in the field.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Be prepared for hard work.&lt;/strong&gt;  Being an astronomer can be difficult, not just because of the subject matter, but also because of the effort you have to put in to get a job.  It can be relatively difficult to find an employer looking for people with your particular skillset.  My recommendation would be to keep an eye on the &lt;a href=&quot;https://nasajobs.nasa.gov/&quot; rel=&quot;noreferrer&quot;&gt;NASA Jobs website&lt;/a&gt; to see when they're hiring.  You can also search for internships on &lt;a href=&quot;https://intern.nasa.gov/ossi/web/public/main/&quot; rel=&quot;noreferrer&quot;&gt;NASA's OSSI site&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Try some amateur astronomy.&lt;/strong&gt; There are lots of ways you can get involved in astronomy without having it as a job.  A quick Google search for &quot;Amateur astronomy projects&quot; yields many different results.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Join a local astronomy group.&lt;/strong&gt; If you live in a reasonably sized city, there should be an astronomy group near you.  Find them and join them.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Don't give up!&lt;/strong&gt; Having astronomy as a profession can be difficult, but it is also rewarding.  If you truly want to do this, you'll need to work at it.  Just don't give up!&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13426" LastActivityDate="2017-03-10T23:47:41.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20352" PostTypeId="1" CreationDate="2017-03-11T00:08:28.730" Score="4" ViewCount="545" Body="&lt;p&gt;If I had 2 really strong magnets, and I threw one into a black hole, while holding one outisde it at a safe distance, what would happen? Would the magnets affect each other in any way? Would one push/pull the other?&lt;/p&gt;&#xA;" OwnerUserId="16177" LastActivityDate="2017-03-12T00:04:44.170" Title="Does a magnet outside a black hole affect one inside it?" Tags="&lt;black-hole&gt;&lt;magnetic-field&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20353" PostTypeId="1" CreationDate="2017-03-11T00:11:53.917" Score="3" ViewCount="530" Body="&lt;p&gt;In Duncan and Lorimer's &lt;em&gt;Handbook of Pulsar Astronomy&lt;/em&gt;, the equation for the flux density using a radio telescope is given as (for a top-hat function):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S=\frac{\beta\,\mathrm{SNR}\, T_{sys}}{G\sqrt{n_{p}\Delta f\,T}}\sqrt{\frac{W_{eq}}{P-W_{eq}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $\beta$ is a correction factor, $\mathrm{SNR}$ is the signal-to-noise ratio, $T_{sys}$ is the system temperature, $G$ is the telescope gain, $n_{p}$ is the number of polarizations, $\Delta f$ is the observing bandwidth, $T$ is the integration time, $W$ is the equivalent width of the pulse, and $P$ is the period of the pulse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one go from this to the so-called &quot;modified radiometer equation&quot; that is often used (often for an arbitrarily shaped pulse):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S=\frac{\beta\,A\, T_{sys}}{G\,N_{bin}\,\sigma_{off}\sqrt{n_{p}\Delta f\,t_{obs}}}$$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ is the area under the pulse, $N_{bin}$ is the total number of phase bins in the profile, $\sigma_{off}$ is the RMS of the noise in the pulsar time series, and $t_{obs}$ is the total integration time of each phase bin.&lt;/p&gt;&#xA;" OwnerUserId="16178" LastActivityDate="2017-03-11T00:11:53.917" Title="Radiometer Equation for Flux Density" Tags="&lt;telescope&gt;&lt;radio-astronomy&gt;&lt;radiation&gt;&lt;radio-telescope&gt;&lt;radio&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20354" PostTypeId="1" AcceptedAnswerId="20356" CreationDate="2017-03-11T02:07:53.720" Score="5" ViewCount="655" Body="&lt;p&gt;I am wondering whether it is possible for a guy in a deep well to see the stars in the daytime. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the background is dark. So might the stars be visible? &lt;/p&gt;&#xA;" OwnerUserId="16181" LastActivityDate="2020-10-20T17:18:18.140" Title="Can someone in a deep well see the stars in daytime?" Tags="&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20355" PostTypeId="1" AcceptedAnswerId="20357" CreationDate="2017-03-11T06:20:50.990" Score="3" ViewCount="266" Body="&lt;p&gt;I recently came across &lt;a href=&quot;https://www.youtube.com/watch?v=gai8dMA19Sw&quot; rel=&quot;noreferrer&quot;&gt;this youtube video&lt;/a&gt;, in which a few people are singing about the history of exoplanet discovery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It suggests that, in the 1990s, exoplanet hunters were largely ignored or ridiculed. Is there any truth to this?&lt;/p&gt;&#xA;" OwnerUserId="16185" LastActivityDate="2017-03-13T00:06:50.827" Title="Exoplanet hunters ignored or ridiculed in the 1980s or 1990s?" Tags="&lt;exoplanet&gt;&lt;history&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20356" PostTypeId="2" ParentId="20354" CreationDate="2017-03-11T08:40:49.223" Score="10" Body="&lt;p&gt;No, because being in a deep well doesn't make the background dark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason we can't see stars in the daytime is the atmosphere, which scatters sunlight in all directions such that along every sightline from the ground toward the sky you receive much more photons from the Sun than from a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the photons that makes it impossible to see Vega, for instance, are the photons from the Sun that make their last scattering event at a molecule lying between you and Vega. Being in a well blocks the star- and sunlight from other directions, but lets through both star- and sunlight from Vega's direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5lURG.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5lURG.png&quot; alt=&quot;well&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2017-03-11T08:40:49.223" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20357" PostTypeId="2" ParentId="20355" CreationDate="2017-03-11T12:10:34.617" Score="8" Body="&lt;p&gt;Everyone expected exoplanets to exist - there was no reason to suppose that the conditions that existed when the sun formed were unique, and we knew that multiple star systems were common. Newton wrote &quot;And if the fixed stars are the centres of similar systems, they will all be constructed according to a similar design&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observations in the 80s were not ignored, they just lacked sufficient evidence. I.e, they were considered and properly rejected. Better observations have since confirmed some of the planets that were &quot;discovered&quot; in the 80s, but denied others. Scientific scepticism means that you don't accept something without evidence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We knew that finding planets would be hard. The technology of sensitive ccd devices that could be used for constant monitoring of multiple stars didn't exist in a form that could be put on a space telescope, and there were other priorities for space science: solar system exploration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing that planets would be impossible to observe with contempory technology meant that &quot;planet hunting&quot; wasn't a thing. The first planet to be detected was around a pulsar, as the detection depends on changes in the timing of the pulsar, not on measurements of variations in light intensity during a transit, or doppler shift as the star is wobbled. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-03-11T12:10:34.617" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20359" PostTypeId="2" ParentId="20352" CreationDate="2017-03-11T23:46:57.730" Score="2" Body="&lt;p&gt;The tidal forces emerging from the black hole would at first break the magnet flung towards it into two pieces. Those two pieces would themselves undergo deformation and splitting, as the process of &lt;a href=&quot;https://en.wikipedia.org/wiki/Spaghettification&quot; rel=&quot;nofollow noreferrer&quot;&gt;spaghettification&lt;/a&gt; will continue beyond the molecular, atomic and even sub-atomic levels, reducing everything that crosses the &lt;a href=&quot;https://en.wikipedia.org/wiki/Event_horizon&quot; rel=&quot;nofollow noreferrer&quot;&gt;Event Horizon&lt;/a&gt; to a &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_singularity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Singularity&lt;/a&gt;​. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Note:&lt;/b&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You won't actually see the magnet thrown towards the black hole crossing the event horizon, but that's a different matter altogether - please see &lt;a href=&quot;https://physics.stackexchange.com/q/21319/139130&quot;&gt;this&lt;/a&gt; if it might interest you.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, magnetic attraction is essentially just an Electromagnetic Force, the range (not to be misunderstood as strength) of which is infinite - see &lt;a href=&quot;https://en.wikipedia.org/wiki/Fundamental_interaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of the scenario stated in the question, the fate of attraction between a pair of magnets, one of which is thrown into a black hole, could be visualised by the following analogy: imagine holding a magnet in your hand and the other attached to a train which is accelerating downhill into a gorge. The two possible results are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If the attraction is strong enough to pull the magnets together before the train falls into the gorge, then they both crash into the abyss together (assuming that the magnet attached to the train would easily be able to pull the one that's being held by hand). In case of a black hole, this would mean that both the objects (magnets) would be crushed to singularity. However, we might safely say that they both were &quot;attached&quot; &lt;b&gt;before&lt;/b&gt; crossing the Event Horizon.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Or, the acceleration of the train &lt;b&gt;reduces&lt;/b&gt; the attractive force infinitesimally, with the magnet in hand &lt;b&gt;barely&lt;/b&gt; having any effect on the other magnet that crashed (and vice-versa). In a classical sense, this means that the force of attraction would still exist, even if it's reduced infinitesimally, to almost zero. However, with regards to a black hole, Event Horizon ensures that anything inside the horizon can no longer communicate with or influence anything outside the horizon (and vice-versa). This means, if the magnet somehow crosses the event horizon as a single piece, before losing its existence to singularity, its magnetic attraction would &lt;b&gt;still&lt;/b&gt; never be able to cross the &lt;b&gt;boundary&lt;/b&gt; defined by the horizon, even though the range of influence of the electromagnetic field is infinity. Ultimately, this would only last for a very brief moment and the magnet would be unable to prevent being pulverised to singularity, either by itself or by any attempts from any other entities outside of the Event Horizon.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="10589" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2017-03-12T00:04:44.170" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20361" PostTypeId="1" AcceptedAnswerId="20366" CreationDate="2017-03-12T08:12:30.010" Score="6" ViewCount="121" Body="&lt;p&gt;From &lt;a href=&quot;https://en.wikipedia.org/wiki/File:S74-15583skylabsunview.jpg&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; and &lt;a href=&quot;https://spaceflight.nasa.gov/gallery/images/skylab/skylab3/html/s74-15583.html&quot; rel=&quot;noreferrer&quot;&gt;NASA&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Caption:&lt;/strong&gt; NASA: S74-15583 &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(July 1973) --- A huge solar eruption can be seen in this Spectroheliogram obtained during the Skylab 3 mission by the Extreme Ultraviolet Spectrograph/Spectroheliograph SO82A Experiment aboard the Skylab space station in Earth orbit. SO82 is one of the Apollo Telescope Mount experiments. The SO82 A instrument covers the wavelength region from 150-650 angstroms (EUV regions). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The caption may have been written in the 1970's. Is there still a mystery about the forces, or are things somewhat better understood now? The only other (fundamental) force that could be relevant is electric, but I wonder if that was meant to be implicit (as electromagnetic). See statement highlighted in bold below, as the caption continues:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The magnitude of the eruption can be visualized by comparing it with the small white dot that represents the size of Earth. This photograph reveals for the first time that helium erupting from the sun can stay together to altitudes of up to 500,000 miles. After being ejected from the sun, the gas clouds seem to have come to a standstill, as though blocked by an unseen wall. Some materials appear to have been directed back toward the sun as a rain, distinguished by fine threads. &lt;strong&gt;At present it is a challenge to explain this mystery--what forces expelled these huge clouds, then blocked its further progress, yet allowed the cloud to maintain its threads. Both magnetic fields and gravity must play a part, but these curious forms seem to defy explanation based on magnetic and gravitational fields alone&lt;/strong&gt;. The EUV spectroheliograph was designed and constructed by the U.S. Naval Research Laboratory and the Ball Brothers Research Corporation under the direction of Dr. R. Tousey, the principal investigator for this NASA experiment. On the left may be seen the sun's image in emission from iron atoms which have lost 14 electrons by collision in the sun's million-degree coronal plasma gas.  (emphasis added)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I understand correctily, this is a telescopic view with a diffraction grating in the optical path - the clear image and its &quot;echoes&quot; are the result of each strong, narrow emission line producing an image at a different angle of the grating's dispersion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PcZdr.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PcZdr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-03-12T08:24:08.993" LastActivityDate="2017-03-12T16:17:07.100" Title="What forces expelled these huge clouds, then blocked further progress, yet allowed it to maintain its threads?" Tags="&lt;the-sun&gt;&lt;spectroscopy&gt;&lt;magnetic-field&gt;&lt;solar-flare&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20362" PostTypeId="1" AcceptedAnswerId="20363" CreationDate="2017-03-12T11:12:51.310" Score="16" ViewCount="46932" Body="&lt;p&gt;I recently saw this image, but somebody told me that it is fake, possibly photoshopped. The moon and the sun appear to be on the same perpendicular from the horizon. Is there any way to tell if this shot was actually captured?&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/8Rwu1.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8Rwu1.jpg&quot; alt=&quot;Image of the sun near the horizon with the moon above it&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16193" LastEditorUserId="-1" LastEditDate="2017-03-12T22:14:56.633" LastActivityDate="2020-01-28T07:57:26.813" Title="Can this picture of the moon directly above the sun from the horizon be real?" Tags="&lt;the-sun&gt;&lt;the-moon&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20363" PostTypeId="2" ParentId="20362" CreationDate="2017-03-12T13:39:55.043" Score="33" Body="&lt;p&gt;No, it's not real.  Given that the moon appears to be nearly full, it would be near to the horizon opposite the sun, which is clearly not the case.  Also the framing of the sun and moon between the trees means that the sun and moon are on the same side of the trees, which is not possible when the moon is full.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-03-12T13:39:55.043" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20364" PostTypeId="1" CreationDate="2017-03-12T13:52:06.757" Score="0" ViewCount="105" Body="&lt;p&gt;I recently bought a telescope, but it doesn't have a very clear image.  Where can I buy a better mirror?&lt;/p&gt;&#xA;" OwnerUserId="16194" LastEditorUserId="13426" LastEditDate="2017-03-12T19:36:58.107" LastActivityDate="2017-03-13T19:00:52.713" Title="Where do I find a better mirror?" Tags="&lt;telescope&gt;&lt;mirror&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2017-03-13T20:49:44.620" ContentLicense="CC BY-SA 3.0" />
  <row Id="20365" PostTypeId="2" ParentId="20362" CreationDate="2017-03-12T15:14:31.287" Score="11" Body="&lt;p&gt;No, neither is true. The moon itself does not emit light at optical wavelengths. It only reflects the sunlight, although in rare cases you can see it reflecting light from our earth (in which case the light is also from the sun).  The side being illuminated should face the sun. So, if you see the sun aligned with the moon as in the photo, it must be faked.  But you need the moon and the sun to be in the same line of sight in this case, e.g. a solar/moon eclipse.&lt;/p&gt;&#xA;" OwnerUserId="14307" LastEditorUserId="14307" LastEditDate="2017-03-18T09:04:50.130" LastActivityDate="2017-03-18T09:04:50.130" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="20366" PostTypeId="2" ParentId="20361" CreationDate="2017-03-12T16:17:07.100" Score="6" Body="&lt;p&gt;This was a coronal mass ejection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those 1973 astronomers weren't looking at the picture correctly. They didn't have the tools at that time to look at the picture correctly. Coronal mass ejections (the term used now) were only discovered a couple of years prior to that picture taken from Skylab, via the Orbiting Solar Observatory 7 satellite. Those early 1970s astronomers called such events &quot;coronal transients&quot;. OSO-7 took 45 minutes to transmit a single 256x256 image back to Earth. A proper look at the picture took a few more years to develop. Those coronal mass ejections (the term used now) eject massive amounts of solar material into space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When looked at via a single still image, it does indeed appear that material is looping up only to fall back to the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bK1wz.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bK1wz.jpg&quot; alt=&quot;A still image of a coronal mass ejection, apparently showing lines of material that appear to move away from the Sun only to later fall back onto the Sun.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Image credit: SOHO (ESA &amp;amp; NASA)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better view of a coronal mass ejection became available once those satellites had the ability to send more than an image per hour back to Earth. The image below portrays a coronal mass ejection that occurred in 2001.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bK1wz.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f3yLb.gif&quot; alt=&quot;A time lapsed image of a coronal mass ejection, clearly showing large volumes of material being ejected from the Sun&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Image credit: SOHO (ESA &amp;amp; NASA)&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-03-12T16:17:07.100" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20367" PostTypeId="1" CreationDate="2017-03-12T18:13:28.993" Score="-6" ViewCount="323" Body="&lt;p&gt;This may seem like a trivia question or a question for skeptics, but I really would like to know, &lt;em&gt;what is the significance of the asteroid belt?&lt;/em&gt; The only reason I've seen from my research is &quot;because it's there&quot;, but I'd like to know a more scientific reason. Evolutionists state the reason is &quot;because the Belt will give us more clues to the formation of the Solar System&quot;; but I don't buy into that theory. Some dreamers say that asteroids may one day be mined for precious minerals; but this is only a far-off dream.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theses three reasons given are all unsatisfactory; could a professional or knowledgeable amateur astronomer help me out here?&lt;/p&gt;&#xA;" OwnerUserId="15129" LastEditorUserId="15129" LastEditDate="2017-03-13T01:41:18.563" LastActivityDate="2017-03-18T19:17:00.857" Title="What is the significance of the asteroid belt?" Tags="&lt;asteroids&gt;&lt;asteroid-belt&gt;" AnswerCount="1" CommentCount="9" ClosedDate="2017-03-13T13:34:03.863" ContentLicense="CC BY-SA 3.0" />
  <row Id="20368" PostTypeId="2" ParentId="20367" CreationDate="2017-03-12T20:25:03.723" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The only reason I've seen from my research is &quot;because it's there&quot;, but I'd like to know a more scientific reason.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Because it's there&quot; &lt;strong&gt;is&lt;/strong&gt; a scientific reason. Couple that with the fact that we know there's a whole lot of stuff we don't know about the asteroids makes studying the asteroids interesting from a pure scientific perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That studying asteroids yields insights into the formation of the solar system makes such studies relevant. That asteroids have caused significant damage to the Earth (e.g., the Chicxulub impact event) and can do so again makes such studies even more relevant. That asteroids might well be a trove of wealth makes such studies more relevant still.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2017-03-18T19:17:00.857" LastActivityDate="2017-03-18T19:17:00.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20369" PostTypeId="1" AcceptedAnswerId="20370" CreationDate="2017-03-12T20:39:41.397" Score="0" ViewCount="89" Body="&lt;p&gt;Gegenschein is a &quot;faint brightening of the night sky&quot; at the anti-solar point.  A naked eye limiting magnitude of about 7.6 might enable an observer to make out gegenschein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia article on gegenschein suggests that interplanetary dust at the Sun-Earth L2 point might be responsible for gegenschein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The James Webb Space Telescope will also be at the Sun-Earth L2 point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this imply that the same interplanetary dust that may be responsible for gegenschein will affect the JWST?&lt;/p&gt;&#xA;" OwnerUserId="12061" LastActivityDate="2017-03-12T22:12:34.093" Title="Will the JWST be affected by dust at L2 (gegenschein?)" Tags="&lt;space-telescope&gt;&lt;lagrange-point&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20370" PostTypeId="2" ParentId="20369" CreationDate="2017-03-12T22:03:23.823" Score="3" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Gegenschein&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; cites &quot;Zdeněk (1962)&quot; for the statement that the dust responsible for the Gegenschein has a possible concentration at L2. I haven't been able to obtain that paper, but I can't really see why that would be the case, since L2 is not dynamically stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the dust consists of millimeter-sized grain (see e.g. &lt;a href=&quot;https://apod.nasa.gov/apod/ap140114.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this APOD image&lt;/a&gt;), which is quite large for dust. Such large grains probably has a highly &lt;em&gt;an&lt;/em&gt;isotropic &quot;phase function&quot;, with a high preference for backscattering. Since when you look from Earth toward L2 you have the Sun in your back, you will thus see an increased brightness compared to other directions, even though the density of that dust is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the case, then placing JWST in L2 is no problem (except when it looks exactly away from the Sun, but that would be a problem anywhere in the ecliptic).&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2017-03-12T22:03:23.823" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20371" PostTypeId="2" ParentId="20369" CreationDate="2017-03-12T22:12:34.093" Score="2" Body="&lt;p&gt;Not likely. The sun shade will always be pointed towards Earth and the sun. Thinking about the design, that means the telescope will never be pointed in a direction where it will have gegeschein in its field of view, because otherwise the heating from IR radiation from Earth would cause it problems. The term for the angle between where you're looking and where the sun is is &quot;solar elongation&quot;. I don't know what the ranges of solar elongation JWST will be limited to, but I'm sure it excludes a substantial angle around $180^\circ$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;Spitzer&lt;/em&gt; space telescope is similarly limited in the parts of the sky it can view at any one time by the need to keep the sun behind it's shade.&lt;/p&gt;&#xA;" OwnerUserId="14244" LastActivityDate="2017-03-12T22:12:34.093" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20372" PostTypeId="2" ParentId="19666" CreationDate="2017-03-12T23:27:05.687" Score="5" Body="&lt;p&gt;It depends what parallax uncertainty you are prepared to tolerate. Very Long Baseline Interferometry (VLBI) at long wavelengths currently provides the most precise parallaxes. Parallaxes to bright radio sources measured in this way can have precisions of around 10 microarcseconds (see for example &lt;a href=&quot;https://arxiv.org/abs/1409.2453&quot; rel=&quot;noreferrer&quot;&gt;Reid et al. (2014)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the review by &lt;a href=&quot;https://arxiv.org/abs/1312.2871&quot; rel=&quot;noreferrer&quot;&gt;Reid &amp;amp; Honma (2014)&lt;/a&gt;, the most distant source with a VLBI-based trigonometric parallax is the star forming region W49N. The source has a parallax of $90\pm 6$ microarcseconds and a corresponding distance of $11.1 \pm 0.8$ kpc (&lt;a href=&quot;https://arxiv.org/abs/1312.3856&quot; rel=&quot;noreferrer&quot;&gt;Zhang et al. 2013&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The precision of these parallaxes is similar to what is likely to be possible with Gaia &lt;em&gt;for the brightest stars&lt;/em&gt; (e.g 5-16 microarcseconds according to &lt;a href=&quot;https://www.cosmos.esa.int/web/gaia/science-performance&quot; rel=&quot;noreferrer&quot;&gt;https://www.cosmos.esa.int/web/gaia/science-performance&lt;/a&gt; ). Only the most luminous giants will be this bright at distances of $&amp;gt;10$ kpc.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-03-12T23:27:05.687" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20373" PostTypeId="2" ParentId="20355" CreationDate="2017-03-12T23:55:15.547" Score="6" Body="&lt;p&gt;The problem in the 80s and 90s for exoplanet hunters was that nobody expected that they would be able to detect any planets, even if they were there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I was a postgraduate/postdoc at Birmingham University in the early 1990s, there was a solar and stellar seismology group that used a Potassium magneto-optical filter that achieved radial velocity precisions of some 10s of m/s. From the introduction of &lt;a href=&quot;http://adsabs.harvard.edu/abs/1991MNRAS.249..643I&quot; rel=&quot;noreferrer&quot;&gt;Innis et al. (1991)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On much longer time-scales, accurate velocity&#xA;  measurements of a star to the 1-m s-1 level will allow a&#xA;  search to be made for possible planetary companions. For&#xA;  example, the reflex motion of the Sun, due to the orbit of&#xA;  Jupiter, has a semi amplitude of around 12.5 ms&quot;1&#xA;  , with aperiod of 12 yr. A very long series of observations (decades&#xA;  to centuries) could reveal the presence of similar motions of&#xA;  solar-like stars. A further paper (in preparation) will discuss&#xA;  the long-term (‘planetary search’) aspects of the velocity&#xA;  measurements of this star.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This paragraph illustrates that the potential and the ideas were there, but it also illustrates why it was difficult to get funding and access to telescopes to pursue these ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expected timescales for these small variations in radial velocities were years and decades and the expected signals were about at the limits of the precisions of the instruments at hand. Other astronomers were not convinced that such absolute accuracy could be maintained over such long durations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What changed everything was the discovery of &quot;hot Jupiters&quot; in orbits of only a few days and which caused radial velocity variations that were bigger than the precisions that spectrographs were capable of - especially the newly developed iodine cells (see &lt;a href=&quot;http://adsabs.harvard.edu/abs/1992PASP..104..270M&quot; rel=&quot;noreferrer&quot;&gt;Marcy &amp;amp; Butler 1992&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I do not think it is the case that exoplanet hunters were &quot;ignored and ridiculed&quot;, it is simply that &lt;strong&gt;planetary formation scenarios did not permit the formation of &quot;hot Jupiters&quot;&lt;/strong&gt; and thus nobody expected that planets could be detected with the available technology without observing stars for years, if not decades. It thus required time on telescopes that was dedicated to this purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it turns out, and with the benefit of hindsight, anyone with a CCD camera and a modest sized telescope can identify the transits of hot Jupiters. But even if someone had gone out on a limb and searched for such a thing systematically, they would have been unable to confirm their discovery (as opposed to the vast numbers of possible false positive transit signals) without spectroscopic backup to confirm the planetary mass of the companion. The first transiting planet was found by searching for transits among hot Jupiters identified by the doppler method. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-03-13T00:06:50.827" LastActivityDate="2017-03-13T00:06:50.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20374" PostTypeId="1" CreationDate="2017-03-13T00:23:54.807" Score="7" ViewCount="743" Body="&lt;p&gt;In class, we read a Scientific American article, &quot;Cloudy with a Chance of Stars&quot; that explained how the cores and the dust around them formed stars(he called them eggs for a chicken/egg analogy, where cores are eggs and I suppose novae are the chickens because the novae cause the core to start to collapse), but we don't know how the very first stars were actually formed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the leading hypotheses in this area, or am I misinterpreting the article? If you have a Scientific American account, &lt;a href=&quot;https://www.scientificamerican.com/article/cloudy-with-a-chance-of-stars/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here's&lt;/a&gt; the article I am referring to.&lt;/p&gt;&#xA;" OwnerUserId="16026" LastEditorUserId="11264" LastEditDate="2017-08-25T13:01:56.217" LastActivityDate="2018-01-15T19:50:31.393" Title="How were the first stars formed?" Tags="&lt;star&gt;&lt;star-formation&gt;&lt;early-universe&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20375" PostTypeId="1" CreationDate="2017-03-13T06:12:27.670" Score="3" ViewCount="55" Body="&lt;p&gt;For the moment, the TRAPPIST-1 worlds are just known by their official designations, like &quot;TRAPPIST-1a&quot; etc. NASA made a casual tweet asking people what they would name the worlds, and there has been some talk of naming the worlds after Belgian beers due to the origin of the &quot;Trappist&quot; name, but I haven't heard any official talk about what names might be given to the new exoplanets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering, is there a timeline (either official or traditional) for the naming of these worlds? Based on previous names given to astronomical bodies, are they likely to have names within the year, or is it likely to take at least five years or something, or is it completely unpredictable?&lt;/p&gt;&#xA;" OwnerUserId="529" LastActivityDate="2017-03-13T06:12:27.670" Title="Is there a timeline for the naming of the TRAPPIST-1 worlds?" Tags="&lt;naming&gt;&lt;trappist-1&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20376" PostTypeId="1" CreationDate="2017-03-13T07:28:40.417" Score="4" ViewCount="55" Body="&lt;p&gt;In Stargate they talk about using the stars on other planets to figure out their location. I know that's fiction, but it made me wonder if such an algorithm (with arguably little usefulness right now) exists.&lt;/p&gt;&#xA;" OwnerUserId="16208" LastActivityDate="2017-03-13T20:51:23.243" Title="Does an algorithm exist that takes an image of the night sky and tell you where in the milky way that perspective likely originates?" Tags="&lt;fundamental-astronomy&gt;&lt;observational-astronomy&gt;&lt;positional-astronomy&gt;&lt;algorithm&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2017-03-15T12:59:07.390" ContentLicense="CC BY-SA 3.0" />
  <row Id="20377" PostTypeId="1" AcceptedAnswerId="20379" CreationDate="2017-03-13T08:47:14.780" Score="4" ViewCount="171" Body="&lt;p&gt;I'm currently trying to write a C++ library that makes accessing binary data from a JPL ephemeris file easier. So far I sifted through the web, trying to find a good and thorough documentation about what is actually stored in there and how to work with the data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know of a PDF/Book or something similar, that explains anything of that sort?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="16209" LastActivityDate="2017-03-14T06:18:21.843" Title="JPL DE Documentation" Tags="&lt;solar-system&gt;&lt;ephemeris&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20378" PostTypeId="1" AcceptedAnswerId="20380" CreationDate="2017-03-13T11:16:48.317" Score="5" ViewCount="849" Body="&lt;p&gt;I'm an Astronomy noob, but I'm interested in learning how stars and planets work. Everything started with the Trappist-1. I read about it, and it's fascinating. However, it's impossible for me to localize it with my current state of knowledge. Therefore, I would like to ask some general questions here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, during high school, I have learned that we are in the solar system and there are some planets orbiting the sun. Easy to remember and imagine, but there are stars, planets, constellations, etc. which I can't manage to relate to this basic system that I learned about. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what confused me a lot are these two pictures: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nrcOMm.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nrcOMm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kyJNX.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kyJNX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this definition of an exoplanet:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An exoplanet or extrasolar planet[...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the pictures, I can see that Trappist is in fact in the solar system. Is this right? And if this is right, why in the world did it take us so many years to find it? We have one Voyager that is almost out from the solar system. Couldn't it find it before? And is it possible to find more planets in our solar system? Is it because Voyager went only in the x-y direction, and not &quot;perpendicular&quot; to the orbits?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry for asking these stupid questions, but I simply can't build new knowledge on my current low-level understanding of the solar system. This is my current knowledge: &#xA;&lt;a href=&quot;https://i.stack.imgur.com/z97A8.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z97A8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please help me to answer the above questions?&lt;/p&gt;&#xA;" OwnerUserId="16210" LastActivityDate="2017-09-20T17:18:52.020" Title="Are the planets Trappist-1 in the solar system?" Tags="&lt;solar-system&gt;&lt;exoplanet&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20379" PostTypeId="2" ParentId="20377" CreationDate="2017-03-13T11:19:48.750" Score="3" Body="&lt;p&gt;Be aware there are differing formats. SPK is the newer format - the older text format seems to be deprecated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth having a look at the documentation on &lt;a href=&quot;https://pypi.python.org/pypi/jplephem&quot; rel=&quot;nofollow noreferrer&quot;&gt;this jplephem python library&lt;/a&gt; as it explains the format reasonably well. If you do use SPK, you'll need to read all of &lt;a href=&quot;https://naif.jpl.nasa.gov/pub/naif/toolkit_docs/FORTRAN/req/spk.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;the documentation in NASA's SPK repository&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if this is from the same author as &lt;a href=&quot;http://cow.physics.wisc.edu/~craigm/idl/ephem.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this library&lt;/a&gt; - worth a read as it discusses the FITS format.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2017-03-13T11:19:48.750" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20380" PostTypeId="2" ParentId="20378" CreationDate="2017-03-13T13:04:59.600" Score="15" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;During high school, I have learned that we are in the solar system and there are some planets orbiting the Sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes that is correct. According to current definitions of what counts as a planet, there are 8 planets orbiting our Sun $-$ Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune. There are also a collection of smaller &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Dwarf_planet&quot; rel=&quot;noreferrer&quot;&gt;Dwarf Planets&lt;/a&gt;&quot; which include (in no particular order): Pluto, Ceres, Eris, and Makemake. There's also a lot of other minor solar system bodies out there like asteroids, comets, etc.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;but there are stars, planets, constellations&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, there are many other stars out there besides our own (as you can see by simply looking up at the night sky). Each of those stars has the potential for having its own planets, much like ours does. Any planet outside our own solar system is referred to as an exoplanet (&lt;a href=&quot;http://www.dictionary.com/browse/exo-?s=t&quot; rel=&quot;noreferrer&quot;&gt;exo-&lt;/a&gt; meaning outside or external). Within the past 20 or so years we've developed the technology to find exoplanets around other stars and this TRAPPIST-1 discovery was simply the latest and most sensational discovery. To date we've discovered &lt;a href=&quot;https://exoplanets.nasa.gov/&quot; rel=&quot;noreferrer&quot;&gt;~3500 exoplanets&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What confused me a lot are these two pictures. [...] And this definition of an exoplanet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those pictures, while seeming to show the the TRAPPIST-1 planets in our solar system, are not indicating that those planets actually reside in our solar system. Instead, what they're trying to do is compare the new TRAPPIST-1 system to our own so you can see the relative sizes. In your first image, you see Jupiter along with the orbits and distances of a few of its Moons. Next to it, you see the TRAPPIST-1 system with the seven planets we know of around it. Next to that is the Sun where you can just begin to see the orbit of Mercury (with the other planets beyond the bounds of the image. The point here is just to compare sizes of orbits. Clearly this TRAPPIST-1 system is much smaller than our own since all 7 planets orbit closer to the main star that even Mercury does for us. However, it is still bigger than Jupiter and its moons. They're not trying to make the point that these systems reside in the same place, just putting them together for comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the second image you have, what they're trying to show here (I assume) is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Circumstellar_habitable_zone&quot; rel=&quot;noreferrer&quot;&gt;habitability zone&lt;/a&gt;. Around every star, depending on the star's size and temperature, there will be a good distance to be from the star that is optimal for life to exist. You don't want your planet to be too hot (by being too close) or too cold (by being too far). This image appears to be showing you the regions where a planet can be at a good distance from TRAPPIST-1 and still be at a good temperature. There is the Inner B line which I take to be the Inner &lt;em&gt;Boundary&lt;/em&gt; of this zone. In other words, if a planet is any closer than this, it will be too hot. There's also the Outer B (or Outer Boundary) where beyond that, any planet would be too cold for life. Finally they have the Earth B. I can't be certain but I believe this point is actually the distance for the &quot;optimal&quot; temperature that we have here on Earth. In other words, the best chance a planet has of being at a good temperature for life is to orbit between Inner B and Outer B and ideally be at the distance of Earth B. This image also shows that the orbit of the planet TRAPPIST-1e is close to Earth B and still inside Outer B, meaning it is likely to be at a hospitable temperature for life (which is good news for us!). I won't go into what an AU (aside from mentioning it is a unit of distance) or the frost line, since they're not pertinent to the point of the image. Feel free to ask other questions for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, the definition of an exoplanet you found describes it an &quot;extrasolar planet&quot;. That just means it is beyond our solar system $-$ &quot;&lt;a href=&quot;http://www.dictionary.com/browse/extra-?r=75&amp;amp;src=ref&amp;amp;ch=dic&quot; rel=&quot;noreferrer&quot;&gt;extra-&lt;/a&gt;&quot; meaning &quot;outside&quot; or &quot;beyond&quot; and &quot;&lt;a href=&quot;http://www.dictionary.com/browse/solar?s=t&quot; rel=&quot;noreferrer&quot;&gt;solar&lt;/a&gt;&quot; referring to our own Sun.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-03-13T13:04:59.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20381" PostTypeId="2" ParentId="20362" CreationDate="2017-03-13T13:27:34.047" Score="8" Body="&lt;p&gt;The position of the moon and the sun in this picture are possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's more feasible in between &lt;a href=&quot;https://en.wikipedia.org/wiki/Tropics&quot; rel=&quot;nofollow noreferrer&quot;&gt;the tropics&lt;/a&gt; and depends on the direction of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbit_of_the_Moon#Inclination&quot; rel=&quot;nofollow noreferrer&quot;&gt;moon orbit inclination&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue in this picture is the homogeneous brightness of the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's look like a full moon, which happen when the observer (here we will talk about the earth) is between the moon and sun, which is not the case in this picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what the moon should look like ?&#xA;In this case the visible part of the moon will be lighted by the sun in a small crescent.&#xA;And the rest of the visible part of the moon will be lighted by an almost full earth. That it, a &lt;a href=&quot;http://traduction.sensagent.com/earthlight/en-en/&quot; rel=&quot;nofollow noreferrer&quot;&gt;earthlight&lt;/a&gt;.&#xA;The brightness contrast is gigantic but you can see it by your eyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here a picture of it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VyKon.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VyKon.jpg&quot; alt=&quot;http://blogs.futura-sciences.com/feldmann/wp-content/uploads/sites/9/2015/12/cendree2.jpg&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And another one with tree to have a brightness reference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lInk7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lInk7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can't see such a contrast on the original image. And the picture doesn't look like a &lt;a href=&quot;https://en.wikipedia.org/wiki/High-dynamic-range_imaging&quot; rel=&quot;nofollow noreferrer&quot;&gt;HDR image&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes. This is a photomontage.&lt;/p&gt;&#xA;" OwnerUserId="9327" LastEditorUserId="9327" LastEditDate="2020-01-28T07:57:26.813" LastActivityDate="2020-01-28T07:57:26.813" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="20382" PostTypeId="2" ParentId="20378" CreationDate="2017-03-13T13:58:36.030" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;From the pictures, I can see that Trappist is in fact in the solar system. Is this right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The intent of those pictures is to give an idea of the small size of the TRAPPIST-1 star system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TRAPPIST-1 is not in the solar system. It is instead a star system (a very small star system) that is about 40 light years from our own solar system. By comparison, the distance between the Sun and Proxima Centauri, the nearest star other than the Sun, is about 4.22 light years, and the distance between the Sun and Voyager 1 is currently about 0.0022 light years.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-03-13T13:58:36.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20383" PostTypeId="1" AcceptedAnswerId="20385" CreationDate="2017-03-13T16:41:03.553" Score="39" ViewCount="14206" Body="&lt;p&gt;It seems to me that the Moon has appeared in many locations - apparent locations - in the 'dome of the sky'. I know that the Sun follows an arc from east to west and that the arc is smaller in winter than in summer. Therefore the Sun cannot appear in every location in the 'dome of the sky'.&lt;br&gt;&#xA;My question is: can the moon appear in any (every?) location in the 'dome of the sky'?&lt;/p&gt;&#xA;" OwnerUserId="16216" LastActivityDate="2017-03-15T17:08:09.103" Title="Can the moon appear to be in any position in the sky?" Tags="&lt;the-moon&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="20384" PostTypeId="2" ParentId="20383" CreationDate="2017-03-13T17:01:31.837" Score="7" Body="&lt;p&gt;No, it cannot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me elaborate. If the Moon was orbiting around the Earth in the plane of the ecliptic (the plane containing the orbit of the Earth around the Sun), then it would be obvious that the Sun and Moon would follow the same path. However, both planes are tilted by approximately 5°. This means that the Moon's location is constrained to a small band on both sides of the ecliptic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that due to a phenomenon called the diurnal parallax (the Sun and Moon are not at infinity), their location on the sky depends slightly on the location of the observer on Earth.&lt;/p&gt;&#xA;" OwnerDisplayName="user15104" LastActivityDate="2017-03-13T17:01:31.837" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20385" PostTypeId="2" ParentId="20383" CreationDate="2017-03-13T17:45:21.317" Score="44" Body="&lt;p&gt;Like the sun, the moon, from our perspective on the surface of the Earth, rises in the East and sets in the West. However, it does not rise exactly due East and set exactly due West. If you were to track the position of the moonrise along the horizon over the course of a year, you would notice that is varies considerably. &lt;em&gt;It cannot, however, be located everywhere in the night sky, even over the course of a year.&lt;/em&gt; To understand why, we need to understand some celestial geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://qph.ec.quoracdn.net/main-qimg-ef4ba644d7e18a6c9b3cf25fa496f1e2-c&quot; alt=&quot;Ecliptic Plane&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The ecliptic plane is the plane which a hypothetical line between the Earth and the sun sweeps out over the course of a year. In the image above, you can see &#xA;how the Earth's orbit exists within this plane. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Also notice that the line from the Earth's geographic south pole through its north pole (its rotational axis) is not perpendicular to this ecliptic plane. In fact it is 23.5 degrees off this perpendicular direction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Notice that the Earth's rotational axis points in the same direction regardless of which side of the Sun the Earth is currently on. This rotational axis always points towards the star Polaris. This means that on different parts of the year, the sun will rise at varying points along the horizon. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Earth's equatorial plane is a plane defined to extend out from the Earth's equator in all directions. This plane is inclined relative to the ecliptic plane by 23.5 degrees (the same amount as the rotational axis is tilted relative to the perpendicular direction of the ecliptic plane).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/4/43/Earth-Moon.PNG/300px-Earth-Moon.PNG&quot; alt=&quot;Earth Moon System&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the Earth moon system. In order to determine how far North or South the moon can appear to rise from, you need to consider the angle between the Earth's equatorial plane and the moon's orbital plane. Since the Earth's rotational axis is inclined 23.5 degrees relative to the ecliptic, and the Moon's orbit is inclined 5.14 degrees relative to the ecliptic, the highest the Moon's orbit can be relative to the equatorial plane of the Earth is 28.64 degrees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the Moon cannot be directly overhead of any point on the surface of the Earth if that point is 28.64 degrees above or (by an symmetric argument) below the equator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the Moon cannot appear in every location in the sky for a given location on Earth. There are parts of the sky that the geometry of the Earth moon system simply will not permit the Moon to exist.&lt;/p&gt;&#xA;" OwnerUserId="14937" LastEditorUserId="1056" LastEditDate="2017-03-13T22:12:52.553" LastActivityDate="2017-03-13T22:12:52.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20386" PostTypeId="2" ParentId="20364" CreationDate="2017-03-13T18:54:52.087" Score="2" Body="&lt;p&gt;Well, if you buy a car and then you go &quot;this thing doesn't drive well, where do I find a better engine&quot; - then folks are going to ask you questions such as: have you driven stick shift before, do you understand how the clutch operates, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many reasons why a telescope does not provide a satisfactory image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the expectations were too high. An amateur telescope can provide quite amazing views, but it's never going to be equal to a large professional telescope in orbit around Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also in line with the expectations thing, all else being equal, a large telescope will always perform better than a small one. A 6&quot; scope and a 24&quot; scope will provide very different levels of performance. If you have the 6&quot; scope, but you expect 24&quot;-level performance, that's never going to happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps seeing (atmospheric turbulence) was bad during the time when you tested the instrument. For objects in the solar system, seeing is one of the major impediments - it reduces resolution and muddies up the image. It changes in time; sometimes it's good, other times it's bad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps light pollution was the problem. DSOs (deep space objects, which have low brightness) will easily get masked by the regular light pollution in cities. It changes with the place; it's bad in cities, it's good far from cities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was the telescope collimated? Some telescopes, such as dobsonians (and most scopes with a newtonian OTA) require regular collimation. Many SCTs do too. Performing collimation on such a scope is like changing the oil on your car - you just have to do it periodically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was the telescope stored in a warm place (inside the house), then suddenly taken outside in the cold night? The temperature differential will be similar to bad seeing for up to one hour, until the scope and the air take the same temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was the scope properly focused? In many cases, finding perfect focus is a little tricky. A small turn of the knob could make a big difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you even reach focus? It's surprising how often people have weird setups where the eyepiece cannot even reach the focal plane of the primary mirror. In that case, no amount of twiddling the focuser will provide a good image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the eyepiece bad? Sometimes mass-produced scopes come with eyepieces that are just not very good. It happens. Trying a known good eyepiece instead should provide proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the optics very dirty? A layer of dust on the primary will not affect performance in any way that you can perceive, so leave the dust alone. However, if there's a big fat oily smudge on the back side of the eyepiece, then of course performance will be degraded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the optics seated properly in their mirror cells, etc? A primary mirror not sitting flat in its cell will mean the scope will never perform well; this could happen during transport with some cell designs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the frame or the OTA intact? Has it been bumped or otherwise deformed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, yes, in some cases mass-produced scopes have bad optics. In that case there's nothing you can do except replace something - either the defective mirror or the whole instrument. Keep in mind that a custom mirror will cost more than your whole instrument, if it's high quality.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TLDR&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're not providing any details, so it's hard to say what's going on. Based on experience, I'd say the most likely candidates are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your scope is a dobsonian or an SCT, I would start with &lt;strong&gt;collimation&lt;/strong&gt;. Learn how to collimate the scope, and do it regularly. It has a very, very large impact on observations of the Moon and the planets. Do this before you try anything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're observing DSOs (galaxies, nebulae), maybe &lt;strong&gt;light pollution&lt;/strong&gt; is the reason. Load up the scope in your car and drive about 1 hour away from the city, and try again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Learn how to &lt;strong&gt;focus&lt;/strong&gt; the instrument. It could be trickier than it seems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, go to a local astronomy club meetup and try and use an instrument similar to yours. Maybe your &lt;strong&gt;expectations&lt;/strong&gt; are not calibrated properly.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2017-03-13T19:00:52.713" LastActivityDate="2017-03-13T19:00:52.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20389" PostTypeId="1" CreationDate="2017-03-13T15:28:05.070" Score="4" ViewCount="1858" Body="&lt;p&gt;We know that a planet is formed from the nebula of gas and dust. Gravity pulls the particles together in a core. So we all know that the gas cloud is a cloud of hydrogen.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how did water form? From where did oxygen come? How did it combine to make $\text{H}_2\text{O}$?&lt;/p&gt;&#xA;" OwnerUserId="16270" OwnerDisplayName="avito009" LastEditorUserId="2153" LastEditDate="2017-03-13T23:41:38.193" LastActivityDate="2017-03-18T22:49:18.133" Title="How was water formed?" Tags="&lt;water&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20390" PostTypeId="2" ParentId="20377" CreationDate="2017-03-14T06:18:21.843" Score="0" Body="&lt;p&gt;The Python package &lt;a href=&quot;http://rhodesmill.org/skyfield/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Skyfield&lt;/a&gt; was written fairly recently, from scratch, by the &lt;a href=&quot;https://stackoverflow.com/users/85360/brandon-rhodes&quot;&gt;same person&lt;/a&gt; who maintains &lt;a href=&quot;http://rhodesmill.org/pyephem/&quot; rel=&quot;nofollow noreferrer&quot;&gt;PyEphem&lt;/a&gt;. You may find helpful information inside there as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had some difficulty reading a JPL kernel with a variant format, and was pointed to a home-brew python package that solved my problems. The whole thing &lt;a href=&quot;https://space.stackexchange.com/a/19263/12102&quot;&gt;is described here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the candid comments here: &lt;a href=&quot;https://github.com/whiskie14142/spktype01&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/whiskie14142/spktype01&lt;/a&gt; you'll notice that the task is not always transparent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the helpful &lt;a href=&quot;https://naif.jpl.nasa.gov/pub/naif/toolkit_docs/C/req/spk.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;SPK Required Reading&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2017-05-23T12:37:21.700" LastActivityDate="2017-03-14T06:18:21.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20391" PostTypeId="2" ParentId="19405" CreationDate="2017-03-14T10:31:10.523" Score="4" Body="&lt;p&gt;Jansky is defined as  &lt;span class=&quot;math-container&quot;&gt;$1 Jy= 10^{-23}erg/s/cm^2/Hz$&lt;/span&gt;. So all the energy coming from a given solid angle per time, per frequency bin and per detector area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is convenient for sources of small angular extent (smaller than observing beam), e.g. for point sources, because the flux remains constant for varying beam sizes. For extended sources, the surface brightness is often described with units of Jy per solid angle. Radio contour maps, e.g. of interstellar clouds, usually show objects that are larger than the beam, so that the observed fluxes are a function of the beam size. To clean the beam influence from the date Jy/beam (beam area in steradian) is used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is a surface brightness. For a flux of 1 Jy measured with a beam of 1 sr solid angle we find:   &lt;span class=&quot;math-container&quot;&gt;$1 Jy/beam= 10^{-23} erg/s/cm^2/Hz/sr$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, that 1 sr is already a huge portion of the sky. Beams are much smaller, on the order of arcsec to arcmin diameters.&lt;/p&gt;&#xA;" OwnerUserId="1988" LastEditorUserId="1988" LastEditDate="2019-07-02T07:34:01.137" LastActivityDate="2019-07-02T07:34:01.137" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="20392" PostTypeId="2" ParentId="20383" CreationDate="2017-03-14T11:23:24.597" Score="7" Body="&lt;p&gt;A lot of the debate in the answers and comments arises from the ambiguity in the OP's phrase &lt;em&gt;the Dome of the Sky&lt;/em&gt;. It depends whether you think this means &lt;em&gt;the Sky Above Your Head&lt;/em&gt; or &lt;em&gt;the Celestial Dome That is Fixed to the Stars&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;Sky Above Your Head&lt;/em&gt; depends on where you are in the world. If you go to the equator, you will, on occasion, see the Moon directly overhead, at the zenith. If you go the North Pole, you will never see it there. If you follow this interpretation, the answer is certainly &lt;em&gt;yes&lt;/em&gt; since you are taking the Moon's possible track over time and painting it over the whole sky as you move about. This seems to me a peculiarly pedantic position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;Celestial Dome&lt;/em&gt; doesn't care where you are and the question becomes &lt;em&gt;can the Moon occlude every star in the sky?&lt;/em&gt; Here the answer is obviously &lt;em&gt;no&lt;/em&gt;. There are broad circumpolar regions of the Northern and Southern sky where the Moon cannot go. The Moon can never appear in &lt;em&gt;Ursa Major&lt;/em&gt;, for example.&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="2938" LastEditDate="2017-03-14T15:36:05.227" LastActivityDate="2017-03-14T15:36:05.227" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20394" PostTypeId="2" ParentId="20378" CreationDate="2017-03-14T13:42:44.460" Score="2" Body="&lt;p&gt;You're not stupid for asking this. I'll give you what i think will help you. You that all of these stars are part of the milky way galaxy, that means that they are all a collection of planets, stars, and other objects that are in a cluster by gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the TRAPPIST-1 system is around forty light years away, which means if we were traveling at the speed of light, it would take 40 years to get there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, in order for the TRAPPIST-! system to be part of our own system, it needs to be orbiting the sun, or the star in the TRAPPIST-1 system needs to be orbiting our sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When two stars orbit each other, it is called a binary system. There have been some planets found in binary systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the TRAPPIST-1 system was part of the solar system, its planets and its star would most likely be orbiting us. But the distance from the sun makes the planets orbit TRAPPIST-1 (their star), and not out star (the Sun).&lt;/p&gt;&#xA;" OwnerUserId="16048" LastActivityDate="2017-03-14T13:42:44.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20395" PostTypeId="2" ParentId="20142" CreationDate="2017-03-14T13:46:53.333" Score="5" Body="&lt;p&gt;The reason is because exoplanets orbiting a star are named in the order they were discovered, starting with the letter b. Sometimes, in a system with two stars, say Alpha Centauri, there are two stars, Alpha Centauri A and Alpha Centauri B, no planes have been found in the system so far. If there were planets orbiting either one of those two stars, say for example a planet orbited Alpha Centauri A, the first planet discovered would be named Alpha Centauri Ab, and so on for Alpha Centauri B. Upper case letters are usually only used in the stars' names.&lt;/p&gt;&#xA;" OwnerUserId="16048" LastActivityDate="2017-03-14T13:46:53.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20397" PostTypeId="2" ParentId="20383" CreationDate="2017-03-14T16:44:02.223" Score="12" Body="&lt;p&gt;The Moon can appear in only about &lt;strong&gt;48% of the sky&lt;/strong&gt;&lt;sup&gt;*&lt;/sup&gt; of a stationary Earth observer. The moon must be at least 61.4 degrees away from the celestial poles, such as the one near the North Star.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/c6K9q.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c6K9q.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compared with the sun, which must always be more than 66.6 degrees from the poles (40% of the sky), the moon has just a little more room to travel in the 'dome' of your sky, a little higher than the summer solar arc, a little lower than winter's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To visualize what portion of your sky can have a moon in it, start by visualizing the &lt;strong&gt;celestial equator&lt;/strong&gt;. That's the arc the sun appears to travel on the equinoxes. Here's that arc in the sky of someone at middle north latitudes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/c6K9q.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UpBkk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon must be within about 28.6 degrees north or south of that equatorial arc.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 28.6 figure adds the &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_tilt&quot; rel=&quot;noreferrer&quot;&gt;Earth's axial tilt&lt;/a&gt; - the famous angle between Earth's spin and orbit, 23.437 degrees - to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbit_of_the_Moon&quot; rel=&quot;noreferrer&quot;&gt;angle between the Moon and Earth's orbits&lt;/a&gt;, 5.145 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So imagine a &lt;strong&gt;57.2 degree band across your sky&lt;/strong&gt; (28.6 doubled), along the celestial equator. The sun is confined to a similar 46.9 degree band. Here's a visual for this kind of segment of your celestial sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/c6K9q.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GglqQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this figure which I also just stole off the interwebs, the sphere does not represent Earth, it represents your sky, without the Earth in the way.  O is you, call it the Observer, and the angle between the red lines is 57.2 degrees.  And you need to imagine this tilted to the same extent as the equator in your sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that at latitudes higher than 61.4 degrees (90 minus 28.6), the moon could appear anywhere on your horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;*48% is the sine of 28.6 degrees. That &lt;a href=&quot;https://www.quora.com/Why-is-the-area-of-a-sphere-equal-to-the-lateral-area-of-the-circumscribed-open-cylinder&quot; rel=&quot;noreferrer&quot;&gt;surprisingly simple&lt;/a&gt; calculation is the portion of the sky taken up by an equatorial band, based on the &lt;a href=&quot;http://mathworld.wolfram.com/SphericalSegment.html&quot; rel=&quot;noreferrer&quot;&gt;area of a spherical segment&lt;/a&gt;, $2\pi R h$ divided by the &lt;a href=&quot;http://mathcentral.uregina.ca/QQ/database/QQ.09.07/h/nicholas4.html&quot; rel=&quot;noreferrer&quot;&gt;area of a hemisphere&lt;/a&gt; $2 \pi R^2$, which is $h / R$ or the sine of the angle. That's for a pole observer,&#xA; but whether at the pole, the equator, or anywhere in between, you can see about half the band and half the sky.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="71" LastEditDate="2017-03-15T17:08:09.103" LastActivityDate="2017-03-15T17:08:09.103" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20399" PostTypeId="2" ParentId="6240" CreationDate="2017-03-14T20:01:24.513" Score="5" Body="&lt;p&gt;This is basically @agtoever comment, converted to an answer as a community wiki.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an online service &lt;a href=&quot;http://nova.astrometry.net/&quot; rel=&quot;noreferrer&quot;&gt;http://nova.astrometry.net/&lt;/a&gt; which can identify star fields in photographs, and so convert from pixel position to ra/dec. It can export to FITS and generally seems to work well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, a bot on reddit.com/r/Astronomy can automatically annotate them images posted there, if you don't mind others seeing them.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-03-14T20:01:24.513" CommentCount="0" CommunityOwnedDate="2017-03-14T20:01:24.513" ContentLicense="CC BY-SA 3.0" />
  <row Id="20400" PostTypeId="2" ParentId="17844" CreationDate="2017-03-14T21:38:07.490" Score="0" Body="&lt;p&gt;No. It would be really hard because although the coolest Brown dwarfs might be able to sustain water, they are made nearly completely of gas and that might by a problem. It would be very hard for water to exist on a brown dwarf, because if you were in the BD's &quot;surface&quot; the water would probably be in a gas phase due to the brown dwarf's low density there and pressure. But if it could exist as a pressurized water it probably still would be extremely hot, and unless the life form could survive that kind of temperatures it wouldn't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, this is also very similar to life in gas giants. If it was to exist, it would need to be very small and not very dense. That way, it could still be part of the gases in the atmosphere and not sink.&lt;/p&gt;&#xA;" OwnerUserId="16048" LastActivityDate="2017-03-14T21:38:07.490" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20401" PostTypeId="2" ParentId="20307" CreationDate="2017-03-14T21:53:02.310" Score="3" Body="&lt;p&gt;It's not tidally locked like the moon is because it is in a 3:2 resonance with the sun. It rotates three times for every two orbits it makes. So it isn't considered a tidal lock because it means they &lt;em&gt;usually&lt;/em&gt; need to be in a 1:1 resonance. I think you were referring to Wikipedia, where it said Mercury was in a tidal lock with the sun. A 3:2 resonance would not be considered regular tidal locking, but elliptical tidal locking. Elliptical tidal locking means when a body is in a stable resonance that is not 1:1, so Mercury wouldn't be the best example of a tidal lock, but it would be a good example of elliptical tidal locking.&lt;/p&gt;&#xA;" OwnerUserId="16048" LastActivityDate="2017-03-14T21:53:02.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20402" PostTypeId="1" AcceptedAnswerId="20413" CreationDate="2017-03-14T21:56:20.253" Score="2" ViewCount="1469" Body="&lt;p&gt;I know that a a planet or moon's rotation will slow down or speed up until it is tidally locked with a body, is Pluto's rotation being affected by the sun or is it being forced by Charon just to to be tidally locked with Charon?&lt;/p&gt;&#xA;" OwnerUserId="16048" LastActivityDate="2017-03-15T22:43:07.640" Title="If Pluto is tidally locked with Charon, is its rotation still slowing down?" Tags="&lt;pluto&gt;&lt;tidal-locking&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20404" PostTypeId="2" ParentId="17844" CreationDate="2017-03-14T23:08:06.890" Score="2" Body="&lt;p&gt;Although you might think that Y-dwarfs (brown dwarfs with temperatures $&amp;lt;500$ K) might become cool enough to support liquid water. In fact that does not happen. As discussed by &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0304226&quot; rel=&quot;nofollow noreferrer&quot;&gt;Burrows et al. (2003)&lt;/a&gt; and &lt;a href=&quot;https://arxiv.org/abs/1404.0005&quot; rel=&quot;nofollow noreferrer&quot;&gt;Morley et al. (2014)&lt;/a&gt;, in Y-dwarf atmospheres, when the temperature becomes low enough for water to condense out of the gas phase ($&amp;lt;375$K), it goes straight to water ice particles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, although water vapour &lt;em&gt;will&lt;/em&gt; condense  in clouds in a cold brown dwarf, it will be in the form of ice particles, and there is not expected to be liquid water in a brown dwarf atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-03-16T07:13:09.180" LastActivityDate="2017-03-16T07:13:09.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20405" PostTypeId="1" AcceptedAnswerId="20406" CreationDate="2017-03-15T04:41:30.913" Score="3" ViewCount="472" Body="&lt;p&gt;So this may be an incorrect assumption, but from my knowledge solar systems are heliocentric and there are always suns. (I realize to be a solar system there must be a sun semantically.) But it seems to me that planets and stars are just bodies of mass, so are there cases of solar systems where the star is just one of the bodies of the system and orbits around a super massive planet? And similarly are there 'solar systems' without stars, where there are a group of planets orbiting a massive planet at the center of the system?&lt;/p&gt;&#xA;" OwnerUserId="16230" LastActivityDate="2017-03-15T14:21:18.923" Title="Why are solar systems heliocentric" Tags="&lt;star&gt;&lt;planet&gt;&lt;orbit&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20406" PostTypeId="2" ParentId="20405" CreationDate="2017-03-15T04:53:48.197" Score="8" Body="&lt;p&gt;Generally, astronomical bodies rotate around their &lt;em&gt;common center of mass&lt;/em&gt;, not the one or the other body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one of the bodies is much heavier, the center of mass is much nearer to its center, sometimes even &lt;em&gt;inside&lt;/em&gt; the larger body. For example, Earth and Sun both rotate around a point that is much nearer to the Sun's center than to the Earth's center, as the sun is much heavier. Looking from the outside, it seems as Earth rotates around the Sun, while the Sun just wobbles a bit in sync. But the reason for that is exclusively the masses of the bodies, not their being-a-star or not-being-a-star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your theoretical example of a super-heavy planet and a lighter star &lt;em&gt;would&lt;/em&gt; result in the star seemingly rotating around the planet; however, here you need to look at why a star is a star:&lt;br&gt;&#xA;Stars are created by gravity that is compressing the mass to a level that starts fusion, and thereby a star. In your example, either the 'light star' wouldn't make it into a star (and be a light body/planet), or the 'super-heavy planet' becoming a much brighter star (or both of it). That reverses the roles effectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at it this way, the bodies in a system rotate always around the center of the mass, which is near the largest mass, which will be the first star of the system (if any).&lt;br&gt;&#xA;So your first constellation is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second is well possible - if neither of the bodies has enough mass to become a star, they will be still rotating around each other the same way, with the heaviest body being nearest the center. Such a system would be dark and basically invisible from some distance. We wouldn't be able to find them.&lt;/p&gt;&#xA;" OwnerUserId="10084" LastActivityDate="2017-03-15T04:53:48.197" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20408" PostTypeId="2" ParentId="1860" CreationDate="2017-03-15T10:10:36.617" Score="0" Body="&lt;p&gt;The observable universe is 879,873,000,000,000,000,000,000 kilometers across. Using the mean measured Hubble Constant of 75 kilometers per second per megaparsec (30,800,000,000,000,000,000 km) for the expansion of space, you can then calculate the rate of expansion for the entire universe and that number is 2,113,636 kilometers per second. That says the expansion of the universe across the entire diameter is expanding at 7.05 (+/-2.33) times the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Diameter of the universe 93,000,000,000 LY&#xA;Megaparsec 3,300,000 LY&lt;/li&gt;&#xA;&lt;li&gt;((Universe Diameter / Megaparsec = 28,182) x 75 kps )=2,113,636 kps or 7.05 x the speed of light.&lt;/li&gt;&#xA;&lt;li&gt;This means the universe expands 0.0000000000000000000008 % every earth year.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Locally this works out to ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Milky way should experience 3.46 kilometers per second expansion or 108,988,052 kilometers per year. (using 75 kps - within margin of error)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few papers now reporting an indirect but measurable increase in space locally. One peer reviewed paper published in 2015 in &lt;em&gt;Gravitation and Cosmology periodical&lt;/em&gt; states the measured effect on &lt;strong&gt;&lt;em&gt;Earths orbit is about 5 meters per year&lt;/em&gt;&lt;/strong&gt; (about 1/2 of the hubble constant) Manifestations of dark energy in the solar system &lt;a href=&quot;https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S1743921309993012&quot; rel=&quot;nofollow noreferrer&quot;&gt;Manifestations of dark energy in the solar system&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16234" LastActivityDate="2017-03-15T10:10:36.617" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20410" PostTypeId="2" ParentId="20405" CreationDate="2017-03-15T14:21:18.923" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;From my knowledge solar systems are heliocentric and there are always suns. (I realize to be a solar system there must be a sun semantically.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just a semantic note, systems of stars and planets different from our own are referred to as &lt;em&gt;stellar&lt;/em&gt; systems. The solar system refers specifically to our own stellar system since solar means &quot;of or relating to our own Sun&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To remark on this point though, yes this is (somewhat) true. All stellar systems are heliocentric in that all bodies orbit the central star (or stars) of that system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is technically more true is that all systems are &lt;em&gt;barycentric&lt;/em&gt;, in that all bodies (including the star) orbit the common center of mass, known as the barycenter. It just so happens that the star tends to have $&amp;gt;99\%$ of the mass of any stellar system and so the barycenter is very very close to the center of the star. Our own Sun does actually have a measurable orbit around this barycenter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where it gets really interesting is when you have a stellar system with more than one star. In this case, the barycenter is no longer near the center of any particular star and so you have noticeable orbits of the stars.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But it seems to me that planets and stars are just bodies of mass, so are there cases of solar systems where the star is just one of the bodies of the system and orbits around a super massive planet?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As stated above, all bodies orbit the barycenter. The location of the barycenter depends on the masses of all the bodies in the system and will be closest to the most massive body. Think of the barycenter as a weighted mean of the positions of all the bodies, where the weight is the respective body's mass (because that's exactly what it is). That's just how the math works out. It is impossible then for the objects in a stellar system to orbit a less massive body as the barycenter could never be closest to that body.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And similarly are there 'solar systems' without stars, where there are a group of planets orbiting a massive planet at the center of the system?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only possible case would be a system which doesn't technically have a star. It is quite possible that a system might try to form but just not have enough matter to get very big. The central object would try to become a star, but might not actually be big enough to start shining. We call these objects brown dwarfs. There are almost assuredly systems which have no official stars and the central body is a brown dwarf that every other body in the system orbits around. Whether you want to consider this brown dwarf a &quot;massive planet&quot; is a bit hazy.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-03-15T14:21:18.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20412" PostTypeId="1" CreationDate="2017-03-15T19:55:38.337" Score="6" ViewCount="653" Body="&lt;p&gt;I am currently working on a simple program to randomly generate and display rocky exoplanets (for a space based strategy game), but I am having some problems figuring out what colour the rocks of rocky planets may or may not have.&lt;/p&gt;&#xA;&lt;p&gt;Specifically, I am thinking of what natural colours may be so dominantly found on the surface of a planet or moon, that the whole planet or moon, seen from space would seem to be (completely or partially) that colour, &lt;strong&gt;which is not made out of ice, and which neither has oceans nor life&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;As far as I can tell from looking at pictures of the not lifebearing rocky planets, dwarf planets and moons in our solar system, it seems that red and yellow colors are dominant on larger bodies such as Mars, Venus or Io (maybe because their atmosphere cause oxidation of metals on the surface, or in Io's case volcanic activities as pointed out by Phiteros):&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CpwOo.jpg&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Io looks red or yellow&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CpwOo.jpg&quot; alt=&quot;Io looks red/yellow-ish&quot; title=&quot;Io looks red or yellow&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;All while smaller bodies such as the Moon or Ceres are completely grey:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NnGiA.jpg&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Ceres is very grey&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NnGiA.jpg&quot; alt=&quot;very grey Ceres&quot; title=&quot;Ceres is very grey&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Therefore my current approach is to make sure the red component of the RGB colour of the planets is the biggest, the green the second biggest and the blue the smallest.&lt;/p&gt;&#xA;&lt;p&gt;However just the fact that there aren't any rocky planets or moons in our solar system made of blue and green rock, doesn't mean that they can't possibly exist.&lt;/p&gt;&#xA;&lt;p&gt;Therefore my question is: is my assumption that rocky planets tend to be mostly red or yellow correct, or are there any good logical arguments for that predominantly blue, green or purple rocky planets could exist.&lt;/p&gt;&#xA;&lt;h1&gt;Edit&lt;/h1&gt;&#xA;&lt;p&gt;Maybe i was a little to vargue on defining colour; i was thinking about the »diffuse colour« (sometimes called »albedo colour«) input, used in the openGL shader that i am using to display the planets; this colour is however the same as something not very reflective (like planets) has when under white light.&lt;/p&gt;&#xA;" OwnerUserId="16247" OwnerDisplayName="Nikolaj" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-03-17T16:22:12.837" Title="What colours can rocky planets (or moons) have" Tags="&lt;planet&gt;&lt;natural-satellites&gt;&lt;surface&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20413" PostTypeId="2" ParentId="20402" CreationDate="2017-03-15T21:01:57.330" Score="2" Body="&lt;p&gt;Short answer.  Pluto's rotation is 100% governed by Charon, or, more accurately, the Pluto-Charon system.  The sun is irrelevant to it's rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small point to add.  Pluto's rotation is pretty much governed by the Pluto-Charon orbit.  If Charon moves closer to Pluto, Pluto's rotation will correspondingly increase.   If Charon moves further away, Pluto's rotation will decrease.  That's how tidal locking works.   I answered the question in regards to the orbit changing, not the rotation changing, but in a tidally locked system the rotation period follows the orbital period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I know that a a planet or moon's rotation will slow down or speed up&#xA;  until it is tidally locked with a body,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tidal locking doesn't always happen.  For distant orbits the tidal force that makes tidal locking happen is often very weak and very slow, often requiring more time than the lifespan of the star in that solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, for high eccentricity orbits, classic tidal locking is impossible though resonant orbits like Mercury's 3:2 around the sun are probably fairly common.   And many moons will escape their planet or crash into their planet long before mutual tidal locking occurs, though it's not uncommon for moons themselves to be tidally locked.   Also small moons don't have sufficient gravity to pull the object they orbit into mutual tidal locking.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For mutual tidal locking, like Pluto-Charon, there needs to be respectible mass ratio or, perhaps an initial formation where orbital speed and rotation are very close, so it doesn't take much force to lead to mutual tidal locking.  Our moon, at 1/81st the mass of the Earth, would eventually cause the Earth to tidal lock to it, but estimates say it will take longer than it will take the sun to go red-giant for that to happen, and the expected outcome of our moon is that it will crash into the Earth before there is mutual tidal locking.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, mutual tidal locking like pluto-charon is actually pretty rare because you need the smaller object to have a significant percentage of mass to the larger one.   One-way tidal locking is quite common, but your question addresses mutual tidal locking.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is Pluto's rotation being affected by the sun or is it being forced by&#xA;  Charon just to to be tidally locked with Charon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a quick approximation, the tidal force can roughly/bad approximation be estimated by the size of the object in the sky times it's density.   There's other factors, ofcourse, like the size of the object receiving the tidal force, but for a rough approximation, the size in the sky gives an estimate.   For example, the sun and moon each take up about 1/2 a degree of arc when viewed from the surface of the Earth, so they have similar tidal forces due to having similar size.   The major difference being that the Moon is about 2.4 times more dense than the sun so it's tides are correspondingly about 2.4 times as strong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This visual size approximation doesn't work for very-near-earth orbits, as it's the size that would appear from the center of the Earth, not the surface that should be applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Charon from the surface of Pluto is huge.   It spans several degrees of arc, over 10 times the size the Moon appears from the Earth.   The sun, from Pluto is tiny about 1/30th the size it appears from Earth, which is only about 50% larger than Venus appears in the sky at it's largest, or, about 1 minute of arc, but even at that distance the sun is still bright enough to cast visible shadows and would probably damage your retina if you stared at it.   It would be the most impressive light in the sky from Pluto, but still tiny, so the tidal effect on Pluto-Charon would be quite small.  (I'd prefer not to do the math, my math tends to run into errors and I'm not good with mathjax).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the sun's tidal influence on the Pluto-Charon system is close to negligible, It can be calculated and it's not zero, but it's very small.   Over a very very very long time, the Sun's tidal force will sap energy from the Pluto-Charon system and draw them closer towards each other, but at that distance, billions maybe trillions of years.  Pluto is probably more likely to meet a different fate well before then, such as crashing into Neptune or being thrown out of the solar-system before the Pluto-Charon system becomes unstable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be different if Pluto-Charon orbited at Mercury or Venus distance (ignoring the obvious melting of ices and outgassing in a warmer environment).  That close, the sun becomes a major player in the future of the mutually tidally-locked systems, likely causing them to crash into each other, but at Pluto distance, there's virtually no effect.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;When a system is mutually tidally locked with a nearly circular orbit, then it's very stable.  If there's no tidal locking or one way tidal locking like Earth-Moon, then there's a tidal bulge that leads or follows the moon and that creates a force that can slow down or speed up the planet and push the moon away or draw it in closer.   With tidal locking, there's no leading or following with the tidal bulge, so the orbit is less prone to change.   There's always some orbital variation and some orbital energy converted to heat due to no orbits being perfectly circular, but mutual tidal locking is still a very stable orbital system with less change than most, as long as there isn't a 3rd party forcing change, such as a debris cloud creating drag or outgassing or orbiting close to a massive body that saps energy from the tidally locked orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No orbits are stable forever, and considering all the factors, Pluto and Charon are close enough that they exchange some atmosphere and as the sun grows more luminous over time, that exchange will probably increase, and likely both are losing mass slowly over time due to the solar wind.    This very gradual loss of mass, I would think, would cause the orbit to expand, but very slowly.   With no loss of gas by the solar wind, the Pluto-Charon orbit should very slowly contract due to tidal forces from the sun and some orbital energy being converted into heat, but both of those effects are very slow.   Mutual tidal locking is for the most part a very stable system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that wasn't too long, but that's the gist of it.   My answer could probably be improved with some math and some linked references but that would make it even longer.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-03-15T22:43:07.640" LastActivityDate="2017-03-15T22:43:07.640" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="20414" PostTypeId="2" ParentId="20402" CreationDate="2017-03-15T21:12:07.943" Score="1" Body="&lt;p&gt;The specifics of the Pluto Charon system are complicated, so I will answer more generally for now. A single body orbiting the sun can be tidally locked with the sun if its rotational period (its day) is equal to its orbital period (its year). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon is tidally locked to the Earth because the period of one orbit of the moon around the Earth takes the same amount of time as one rotation of the moon about its axis. (Note that the Earth is not tidally locked with the Moon. One Earth day does not correspond to one lunar orbit.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A binary (dwarf) planet system (like Pluto and Charon) is considered to be tidally locked if their respective orbital periods &lt;em&gt;about the common barycenter of mass&lt;/em&gt; are equal to their respective rotational periods, and all of these periods, for both objects, are equal to each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You specifically asked about the tidal effects of the Sun on Pluto. Unfortunately, the complete answer depends on the orbital parameters of Charon and Pluto's orbits around the barycenter of mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will assume the orbits of Pluto and Charon are such that their orbital plane about the barycenter is not perpendicular to a radial vector drawn from the Sun to the barycenter of mass for the following explanation. (i.e. one body is slightly closer to the Sun at a given instant in the same way the Moon gets closer and farther from the Sun as it orbits the Earth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this assumption, tidal effects from the Sun should slow the rotation of the Pluto Charon system about the barycenter (the orbital period of each body about the barycenter of mass should decrease over time). This slowing would occur because the radially inward (toward the Sun) side of the closer body would have a greater gravitational pull on it that the farther side of that body. Thus, the Sun's gravitational pull would attempt to slow the rotational period of that body (equivalently: increase the length of that body's &quot;day&quot;). This process I am describing is the same as the process that caused the Moon to become tidally locked to the Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Pluto Charon system introduces complications to this explanation because of the highly inclined and atypical orbital parameters of the Pluto-Charon-[Other Plutonian Bodies] system, the varying masses of the system bodies, and the system's extreme distance from the Sun. The large orbital distance of the system from the Sun means that any tidal effects on Pluto and Charon are mild at their strongest. Thus, measurable changes due to tidal effects are next to impossible to measure over human time scales.&lt;/p&gt;&#xA;" OwnerUserId="14937" LastEditorUserId="14937" LastEditDate="2017-03-15T21:38:06.450" LastActivityDate="2017-03-15T21:38:06.450" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20415" PostTypeId="1" AcceptedAnswerId="26222" CreationDate="2017-03-15T23:28:26.117" Score="6" ViewCount="1894" Body="&lt;p&gt;I'm new to astronomy and recently got myself a telescope with an equatorial mount. I've found lots of resources online showing how to align the telescope but they all operate on the assumption that you live in the Northern Hemisphere and align to True North or Polaris.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the procedure for those of us who live on the other half of Earth? Pointing my telescope north, I still suffer problems finding and tracking objects in the sky.&lt;/p&gt;&#xA;" OwnerUserId="16250" LastActivityDate="2018-05-09T15:06:15.860" Title="How to align equatorial mount in the Southern Hemisphere?" Tags="&lt;telescope&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20417" PostTypeId="2" ParentId="20415" CreationDate="2017-03-16T05:59:08.767" Score="1" Body="&lt;p&gt;Okay so firstly your telescope would never work if you point in to the north(as you are in the southern hemisphere)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why dont you look these links up? They completely answer your question and more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=35E9Lqjypz8&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=35E9Lqjypz8&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=35E9Lqjypz8&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=35E9Lqjypz8&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.celestron.com/support/knowledgebase/articles/how-do-i-polar-align-my-telescope-if-im-in-the-southern-hemisphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.celestron.com/support/knowledgebase/articles/how-do-i-polar-align-my-telescope-if-im-in-the-southern-hemisphere&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here I assume a basic level of telescope familiarity, if you still need more help, pls feel free to comment or downvote:)&lt;/p&gt;&#xA;" OwnerUserId="15901" LastActivityDate="2017-03-16T05:59:08.767" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20418" PostTypeId="2" ParentId="20389" CreationDate="2017-03-16T06:20:47.807" Score="2" Body="&lt;p&gt;Stars, especially larger stars, are hot enough at their core to produce significant amounts of heavier elements: Carbon, Nitrogen and Oxygen. When the stars die, some of these elements are returned to the interstellar medium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nebula from which stars form contains not only Hydrogen, but also all the other elements found on Earth. The nebula contains the Carbon that is the basis of life, the Nitrogen that forms the atmosphere, the Silicon that makes rocks, and the oxygen that reacts with hydrogen to form water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the water on Earth was actually brought to the Earth on asteroids after the Earth had formed, see &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10277/how-did-water-get-on-earth&quot;&gt;How did water get on Earth&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2017-03-16T06:20:47.807" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20419" PostTypeId="1" CreationDate="2017-03-16T10:05:31.753" Score="1" ViewCount="1171" Body="&lt;p&gt;I just wanted to know if what we can only see out-side of our planet is just the past of other objects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 1 Light Year is equals to 365.25 Days, then I saw an object outside our planet that is 1 Light Year away from us, does it means I just saw its past 365.25 days ago?&lt;/p&gt;&#xA;" OwnerUserId="16255" LastActivityDate="2017-03-16T15:42:28.107" Title="Does Light Year is their past?" Tags="&lt;space-time&gt;" AnswerCount="2" CommentCount="0" ClosedDate="2017-03-16T15:18:40.697" ContentLicense="CC BY-SA 3.0" />
  <row Id="20420" PostTypeId="2" ParentId="14525" CreationDate="2017-03-16T10:22:45.467" Score="3" Body="&lt;p&gt;A common technique to estimate the beam position on the sky are so called pointing measurements. Ideally you try to use a radio point source with a known position on the sky. Scanning this point source with your radio beam will give maximum signal strength once the beam is centered on the object. By doing this for example as a cross-scan you are trying to establish the detailed angular position of the max. signal strength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5aTIh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5aTIh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some problems involved. First of all, there are not too many radio point sources on the sky (depending on the frequency you are measuring). Maser emission is usually a good candidate. But for higher frequencies the list becomes very short. A commonly used pointing source for carbon line emission receivers is the carbon star IRC 10216. Also planets are used as pointing sources, because they are effectively point sources for most of the non too big telescopes. Depending on the frequency they are suffieciently radio bright to be used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also point on extended objects where you know the shape exactly, e.g.&lt;br&gt;&#xA;cross pointing at the stellar or lunar disk is also used. &#xA;Usually  these pointing measurements have to be performed at multiple elevations to establish a good model of the pointing. This is for example because the radia antennas suffer from gravitational deformation depending on the elevation they are looking at (they are not infinitely stiff structures). The also suffer from thermal expasion, wind pressure, etc. (See for example : &lt;a href=&quot;http://ipnpr.jpl.nasa.gov/progress_report/42-159/159A.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ipnpr.jpl.nasa.gov/progress_report/42-159/159A.pdf&lt;/a&gt; , and &lt;a href=&quot;http://lss.fnal.gov/archive/other1/iram-298.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://lss.fnal.gov/archive/other1/iram-298.pdf&lt;/a&gt; ) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these techniques depend on the visibility of the pointing sources.&#xA;An alternative is the usage of an optical guiding telescope that is attached to the radio telescope. Here you use optical star positions to determine your pointing. Then you just have to establish the  relative alignment between the optical axis of the optical telescope and the radio beam. This of course requires the visibility of stars, i.e. you can use it only during night observations ( while sun pointing is used during day time). The Herschel satellite used an optical guide telesope (pointing in the opposite direction) to point the telescope. ( see here: &lt;a href=&quot;http://herschel.esac.esa.int/Docs/Herschel/html/ch02s04.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://herschel.esac.esa.int/Docs/Herschel/html/ch02s04.html&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="1988" LastActivityDate="2017-03-16T10:22:45.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20421" PostTypeId="2" ParentId="20419" CreationDate="2017-03-16T11:25:05.253" Score="1" Body="&lt;p&gt;Light year is a measure of distance. Light in vacuum travels at a (large) finite value, $3 \cdot 10^5$ kilometers per second to be almost exact. We can exploit this speed as a function of time. If an object is 5 light years away, it means light must travel at that large speed for 5 years to reach Earth from where it left. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun is about 8 light minutes away. It takes light 8 minutes to reach Earth from the Sun. The sunlight you see is from 8 minutes ago, as it was when it left its source — the Sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there were a telescope powerful enough to clearly see objects hundreds of millions of light years away, and if some hypothetical advanced alien race used such a telescope right now from hundreds of millions of light years away, they would see dinosaurs on Earth since that light had enough time to reach its destination. We aren't visible to them yet because our light is still traveling on its way to destination. &lt;/p&gt;&#xA;" OwnerDisplayName="user15317" LastEditorUserId="9301" LastEditDate="2017-03-16T15:42:28.107" LastActivityDate="2017-03-16T15:42:28.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20424" PostTypeId="1" CreationDate="2017-03-16T14:42:10.553" Score="1" ViewCount="69" Body="&lt;p&gt;I am having trouble finding a solidifying definition of accretion when it comes to galaxy formation models and galaxy evolutions. Where as, in the context of satellite galaxies, dwarf galaxies, or gasses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, how do interpret the sentence &quot;Even&#xA;dwarf and Milky Way-sized halos do accrete new mass at low&#xA;redshifts&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that they &quot;gain&quot; new mass or does this mean they &quot;leak&quot; new mass? &lt;/p&gt;&#xA;" OwnerUserId="16260" LastActivityDate="2017-03-16T14:42:10.553" Title="How does one interpret the definition of &quot;accretion&quot; in galaxy formation?" Tags="&lt;galaxy&gt;&lt;astrophysics&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20425" PostTypeId="2" ParentId="20419" CreationDate="2017-03-16T14:47:32.360" Score="1" Body="&lt;p&gt;Yes, you are correct. Since Alpha Centauri is four light years away, you see it as it was four years ago. For example, if the sun was to suddenly disappear it would take eight minutes for us to see it disappear. The reason is that light takes eight minutes to travel from the sun to us. Also, the Moon is nearly one light second away, so were are looking at as it was one second ago.&lt;/p&gt;&#xA;" OwnerUserId="16048" LastActivityDate="2017-03-16T14:47:32.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20426" PostTypeId="1" CreationDate="2017-03-16T17:18:38.200" Score="2" ViewCount="60" Body="&lt;p&gt;If we assume that mass was initially uniformly distributed in an area, then why does it flatten out after long time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider our solar system or most galaxies out there.&lt;/p&gt;&#xA;" OwnerUserId="16134" LastEditorUserId="16134" LastEditDate="2017-03-16T18:16:00.523" LastActivityDate="2017-03-16T18:16:00.523" Title="Why do most galaxies or planetary systems have disc shape?" Tags="&lt;solar-system&gt;&lt;galaxy&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2017-03-17T02:32:51.793" ContentLicense="CC BY-SA 3.0" />
  <row Id="20427" PostTypeId="2" ParentId="20412" CreationDate="2017-03-16T17:46:12.387" Score="0" Body="&lt;p&gt;You've got two problems (yeah, I know that's the start of a bad software joke).    First, you have to decide if you want the planets/moons/asteroids to appear as they would under strong solar illumination, or from a red dwarf (or a rgb scanning laser on your spaceship).&lt;br&gt;&#xA;Second, you have to decide what elements and simple compounds make up the surface. I'd say the comments have provided some realistic color choices, but you could always posit a pure-amethyst rock that was formed as ejecta from some violent explosion, for example.  &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-03-16T17:46:12.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20428" PostTypeId="1" AcceptedAnswerId="20446" CreationDate="2017-03-17T10:42:07.123" Score="4" ViewCount="326" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_energy&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The density of dark energy ($~ 7 × 10^{−30} g/cm^3$) is very low, much less than the density of ordinary matter or dark matter within galaxies. However, it comes to dominate the mass–energy of the universe because it is uniform across space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other sources I found also give the density either as percentage of mass of the universe, or just as mass density like above, $kg/m^3$ or similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In about all other applications and contexts of usage of energy density, I'd find &lt;a href=&quot;https://en.wikipedia.org/wiki/Energy_density&quot; rel=&quot;noreferrer&quot;&gt;MJ/L&lt;/a&gt;, or other units of &lt;em&gt;energy per volume&lt;/em&gt;. Why is energy of density given as &lt;em&gt;mass per volume&lt;/em&gt;, and how do I convert it to more familiar units? Using $E=mc^2$ ?&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2017-03-18T10:53:05.890" Title="Why is dark energy density given in grams per cm$^3?$" Tags="&lt;dark-energy&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20429" PostTypeId="2" ParentId="20428" CreationDate="2017-03-17T14:06:08.670" Score="4" Body="&lt;p&gt;This is a case of using a sort of &lt;a href=&quot;https://en.wikipedia.org/wiki/Natural_units#Systems_of_natural_units&quot; rel=&quot;nofollow noreferrer&quot;&gt;natural units&lt;/a&gt; (similar to &lt;a href=&quot;https://en.wikipedia.org/wiki/Geometrized_unit_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;geometerized units&lt;/a&gt;), which are often convenient in relativity (special and general). Take, for instance, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Minkowski_space#Minkowski_metric&quot; rel=&quot;nofollow noreferrer&quot;&gt;Minkowski metric&lt;/a&gt; of special relativity. In dimensional units, it is&#xA;$$\mathrm{d}s^2=-c^2\mathrm{d}t^2+\mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2$$&#xA;Setting $c=1$ makes it a bit more convenient, and gives time and space the same units - which is more intuitive, if you think of time as another dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also brought up $E=mc^2$. This is a special case of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation&quot; rel=&quot;nofollow noreferrer&quot;&gt;energy-momentum relation&lt;/a&gt;:&#xA;$$E^2=p^2c^2+m_0^2c^4$$&#xA;Setting $c=1$, we have the much more convenient form&#xA;$$E=p^2+m_0^2$$&#xA;This comes up again and again. Add to this the &lt;a href=&quot;https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation&quot; rel=&quot;nofollow noreferrer&quot;&gt;mass-energy equivalence&lt;/a&gt; that is also important in relativity (especially the right-hand side of the Einstein field equations), and it simply makes sense to treat energy and mass as the same thing. In fact, it becomes perhaps even more useful than in special relativity, because we also often set $G=1$ - and that comes up a &lt;em&gt;lot&lt;/em&gt; in general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the units of density under this system are $\text{Length}^{-2}$, and the density above is clearly not in those units. I'm guessing that this just isn't a case of setting $G=1$ - and that makes sense, because in cosmology and astronomy, many types of densities are measured in g/cm&lt;sup&gt;3&lt;/sup&gt;: dark matter, interstellar gas and dust, and sometimes dark energy. Keeping it in $\text{Mass}\cdot\text{Length}^{-3}$ makes it easy to compare these, working in a system that is more commonly used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a nutshell, then: Cosmologists sometimes like $c=1$ (and sometimes $G=1$) because it makes a lot of equations a lot easier to write out. Here, it becomes really easy to compare mass densities and energy densities.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2017-03-17T14:06:08.670" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20430" PostTypeId="2" ParentId="20428" CreationDate="2017-03-17T15:45:46.283" Score="0" Body="&lt;p&gt;Aside from cosmology,  the advantage of reporting density in $\frac{g}{cc}$ is that density and &lt;a href=&quot;https://en.wikipedia.org/wiki/Specific_gravity&quot; rel=&quot;nofollow noreferrer&quot;&gt;specific gravity &lt;/a&gt;  are identical in this case (unless some wiseguy doesn't use water as the reference material).&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-03-17T15:45:46.283" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20431" PostTypeId="2" ParentId="20412" CreationDate="2017-03-17T16:22:12.837" Score="4" Body="&lt;p&gt;I'll begin by saying that for exoplanets, the answer to your question is an active area of research, and currently there are no observational constraints on what rocky exoplanets would look like to the human eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, this may be a lot more complicated an answer than what you want for a computer game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to hunt around for this in the astronomical or planetary science literature, what you're trying to do is determine the reflectance, or albedo spectrum, of the planet. That's what tells you how much of the incoming light is bounced back off the surface for you to see as function of wavelength. Note that this assumes the planet is not hot enough to thermally emit in the optical, which will only start to matter if it's &gt;3000K or so. This albedo spectrum is determined by the surface properties of the planet: what it's made out of, how shiny it is, the scattering properties, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have an albedo spectrum, you then can multiply it be the incoming radiation from whatever primary star you've chosen. That's what will make a planet seem slightly more orange around an M-dwarf, and more &quot;normal&quot; around a Sun-like star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get to how a person would see it, you'd then need to then multiply that resulting reflected-light spectrum by the spectral sensitivity of the human eye, and then convert that final observed spectrum into a color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the planetary albedo spectrum, there aren't a lot of sources in the astronomical literature. The best I can find is &lt;a href=&quot;https://arxiv.org/abs/1204.1544&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;. You'll probably have more luck looking through Solar System planetary stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the stellar spectra, you can try &lt;a href=&quot;http://www.eso.org/sci/facilities/paranal/decommissioned/isaac/tools/lib.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, but that presumes you're comfortable with astronomical FITS files. For stars hotter than &gt;4500K (the Sun is 5800K), you can probably just assume that the incoming radiation into the planet's surface is blackbody of whatever temperature you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I am not familiar with the spectral sensitivity of the human eye, or how to go from an input observed spectrum to a single color value.&lt;/p&gt;&#xA;" OwnerUserId="16268" LastActivityDate="2017-03-17T16:22:12.837" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20432" PostTypeId="1" AcceptedAnswerId="20438" CreationDate="2017-03-17T18:38:06.480" Score="1" ViewCount="55" Body="&lt;p&gt;Is it possible to derive Metallicity/Age/Radial Velocity/Rotational Velocity/Surface Gravity of a star from Hipparcos data files or are do they require individual studying?&lt;/p&gt;&#xA;" OwnerUserId="5401" LastEditorUserId="37191" LastEditDate="2021-02-15T15:14:01.110" LastActivityDate="2021-02-15T15:14:01.110" Title="Stellar Information from Hipparcos" Tags="&lt;star&gt;&lt;star-catalogues&gt;&lt;hipparcos&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20433" PostTypeId="1" CreationDate="2017-03-17T18:42:39.180" Score="13" ViewCount="1852" Body="&lt;p&gt;I don't know how common it is for a system to have two stars (or perhaps even more) but how do they arise? Is that due to the stellar accretion disc, or the composition of the stellar nebula? &#xA;Or are most of them created by a collision of stars?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="2153" LastEditDate="2017-03-17T19:06:04.207" LastActivityDate="2017-03-17T21:14:27.243" Title="How are binary star systems created?" Tags="&lt;star&gt;&lt;star-formation&gt;&lt;binary-star&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20434" PostTypeId="2" ParentId="20433" CreationDate="2017-03-17T19:05:26.737" Score="12" Body="&lt;p&gt;There are two main theories for the formation of binary stars - one accepted, and one mainly deprecated.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;http://www.phys.lsu.edu/astro/nap98/bf.final.html&quot; rel=&quot;noreferrer&quot;&gt;The fission hypothesis.&lt;/a&gt;&lt;/strong&gt; The fission hypothesis states that the binary system forms after the collapse of the original gas cloud into a protostar. Angular momentum is conserved, so as the extremely large cloud slowly contracts, it spins faster. After enough time, the protostar could form a dumbbell shape, one part of which eventually breaks off. There are now two blobs, and thus two protostars, which will evolve into a binary star system.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1988ComAp..12..169B&amp;amp;defaultprint=YES&amp;amp;filetype=.pdf&quot; rel=&quot;noreferrer&quot;&gt;The fragmentation hypothesis.&lt;/a&gt;&lt;/strong&gt; The fragmentation hypothesis states that the gas cloud &lt;em&gt;fragments&lt;/em&gt; earlier on, due to some instability or cooling/heating effects. Each of the two fragments then evolves separately, forming a binary star system.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The fission hypothesis is no longer favored. It cannot explain the existence of systems with certain mass ratios or separations. The fragmentation hypothesis, however, can, and &lt;em&gt;is&lt;/em&gt; accepted. It is also possible for a binary system to form through the capture of one star by another, but that requires a third body, and probably cannot explain the high fraction of binary systems in the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, then, the main theory to date is that the protostellar nebula breaks up, and then both fragments collapse independently.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2017-03-17T19:05:26.737" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20437" PostTypeId="2" ParentId="20433" CreationDate="2017-03-17T21:14:27.243" Score="3" Body="&lt;p&gt;That's a really good question. I'll explain as best as I can, there are about a number of ways this can happen.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Fragmentation, as the protostar forms, the disk may fragment under its own gravity, or like the method listed below, it may rotate enough until some of the mass breaks off so another star may form.&lt;/li&gt;&#xA;&lt;li&gt;Accretion, another object in the system may &quot;steal&quot; matter from the forming star, and when it's mass reaches a high enough point, a second star may start to form.&lt;/li&gt;&#xA;&lt;li&gt;Breaking up, if the star rotates fast enough, it may split apart into two separate stars. But like HDE 226868 said, it is no longer favored.&lt;/li&gt;&#xA;&lt;li&gt;Normal formation, when the stars are forming, both may acquire enough mass without any of these effects to form a binary system.&lt;/li&gt;&#xA;&lt;li&gt;Capture, a nearby star may capture another so that they orbit each other.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that there are probably more ways for binary stars to form. Also, a collision isn't very likely to form these systems.&lt;/p&gt;&#xA;" OwnerUserId="16048" LastActivityDate="2017-03-17T21:14:27.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20438" PostTypeId="2" ParentId="20432" CreationDate="2017-03-17T21:16:05.033" Score="2" Body="&lt;p&gt;Hipparcos measured the parallax (distance) of stars, along with their colors in two wavelength regions (B_Tycho, &quot;blue&quot; and V_Tycho, &quot;green&quot;). Using the color information you can make a reasonable estimate for the stellar temperature, and coupled with the parallactic distance, this would allow you to get an estimate for the stellar radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But Hipparcos does not provide us with any good information about stellar metallicities, ages*, radial velocity, rotation, or surface gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* - aside from stars that you can group in position and velocity space to identify as belonging to stellar moving groups with defined ages.&lt;/p&gt;&#xA;" OwnerUserId="16268" LastActivityDate="2017-03-17T21:16:05.033" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20440" PostTypeId="2" ParentId="11102" CreationDate="2017-03-17T23:33:15.790" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Template:WEEKDAYNAME&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia uses&lt;/a&gt;: mon, tue, wed, thu, sat, sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/ISO_week_date&quot; rel=&quot;nofollow noreferrer&quot;&gt;ISO 8601&lt;/a&gt; defines an exchange standard for dates based on weeks and days, which looks like this: 2017-W01-3 or this 2017W011 (Year, ISO week number, number of day of week starting with Monday). Not so surprisingly it does not standardize names of days as far as I can tell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you could claim to use ISO-8601 conventions and simply use 1-7 (starting with Monday).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A comprehensive listing of abbreviations of day names for European countries is given by the European Medicines Agency for &lt;a href=&quot;http://www.ema.europa.eu/docs/en_GB/document_library/Other/2012/04/WC500124995.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;blistered drugs&lt;/a&gt; for the curious.&lt;/p&gt;&#xA;" OwnerUserId="15824" LastActivityDate="2017-03-17T23:33:15.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20441" PostTypeId="1" CreationDate="2017-03-18T01:30:11.910" Score="6" ViewCount="2688" Body="&lt;p&gt;What is the currently accepted estimated range of the amount of energy of the Big Bang event?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In joules at some estimated size, so a temperature may be calculated.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For context, I wonder if the temperature of the Big Bang would make its heat radiation wavelength shorter than the Planck length.&lt;/p&gt;&#xA;" OwnerUserId="9433" LastActivityDate="2020-01-10T20:41:19.543" Title="Amount of energy of the Big Bang" Tags="&lt;big-bang-theory&gt;&lt;high-energy-astrophysics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20442" PostTypeId="2" ParentId="20441" CreationDate="2017-03-18T02:23:59.223" Score="7" Body="&lt;p&gt;Let's start by making some points clear:&lt;/p&gt;&#xA;&lt;h1&gt;1. We don't know what the Big Bang was.&lt;/h1&gt;&#xA;&lt;p&gt;Rather, we know that the Universe is expanding. If you extrapolate backwards, you'd expect the Universe to be denser and denser. More specifically, we talk about this as a change in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Scale_factor_(cosmology)&quot; rel=&quot;nofollow noreferrer&quot;&gt;scale factor&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt;, and this gets smaller and smaller as we look further back in time. According to general relativity (our modern theory of gravity), 13.8 billion years ago, &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; should have been &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt;; however, you can't have a metric with &lt;span class=&quot;math-container&quot;&gt;$a = 0$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Thus, we know that general relativity is necessarily incomplete. It breaks down at the conditions of the early universe, so we currently have no physical model to explain that time. Rather, we know that the early universe expanded, and the Big Bang is the time that perplexes cosmologists. Some theories, like &lt;a href=&quot;https://en.wikipedia.org/wiki/Quantum_gravity&quot; rel=&quot;nofollow noreferrer&quot;&gt;quantum gravity&lt;/a&gt;, have emerged in an effort to explain the Big Bang; however, we currently have little understanding of what it actually was.&lt;/p&gt;&#xA;&lt;p&gt;So no, we can't tell you what the energy output of the event was, since we don't know what actually happened.&lt;/p&gt;&#xA;&lt;h1&gt;2. The temperature of the early Universe was high&lt;/h1&gt;&#xA;&lt;p&gt;Our theories break down at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Timeline_of_the_formation_of_the_Universe#Planck_epoch&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck epoch&lt;/a&gt; of the Universe. The Planck epoch was the earliest epoch of the Universe and lasted until &lt;span class=&quot;math-container&quot;&gt;$10^{-42}$&lt;/span&gt; seconds after the Big Bang — that's 200 &lt;a href=&quot;https://en.wikipedia.org/wiki/Planck_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck times&lt;/a&gt;, which are the shortest meaningful measurement of time.&lt;/p&gt;&#xA;&lt;p&gt;During this epoch, the entire Universe was at &lt;span class=&quot;math-container&quot;&gt;$1.417×10^{32} \; \mathrm{K}$&lt;/span&gt;, which is the Planck temperature. This is the hottest possible temperature; an object at this temperature will emit photons with wavelengths of a Planck length (you can read more about this in my answer &lt;a href=&quot;https://astronomy.stackexchange.com/questions/16206/is-there-a-limit-to-how-hot-an-object-an-get/16208#16208&quot;&gt;here&lt;/a&gt;). The point is that there is no meaningful distance smaller than a Planck length, so the Universe couldn't be hotter than the Planck temperature.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-05-04T12:28:27.517" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="20444" PostTypeId="1" CreationDate="2017-03-18T07:24:49.710" Score="1" ViewCount="136" Body="&lt;p&gt;The center of the galaxy is densely packed with stars and obscured by a whole lot of dust between us and it. For those reasons, groups that study the motion of stars around the super-massive black-hole in the center of the Milky way need to use big telescopes at near infrared wavelengths. Questions of dust obscuration aside, what resolution would a telescope need to have to pick out a background quasar looking through the center of the Milky Way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SDSS QSOs start at around 8th magnitude (Vega) in 2MASS photometry ($H$ and $K_s$ bands), and rise like a power law as flux goes down Here's a quick and dirty plot of SDSS QSO survey made from SDSS data release 10 made using &lt;A HREF=&quot;http://www.star.bris.ac.uk/%7Embt/topcat/&quot; rel=&quot;nofollow noreferrer&quot;&gt;topcat&lt;/A&gt;. I would convert the graph to counts per solid angle, but I'm too lazy to look up the survey area of the SDSS spectroscopic survey right now. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/PXVvp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PXVvp.png&quot; alt=&quot;Histogram of Quasar Fluxes&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a quick and dirty example of the effect I'm describing in Figure 2 of the &lt;A HREF=&quot;https://arxiv.org/abs/0808.2870&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ghez et al. (2008)&lt;/A&gt; paper Rob Jeffries linked to below. The difference between the red points and blue points is both exposure time (more photons) and higher &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Strehl_ratio&quot; rel=&quot;nofollow noreferrer&quot;&gt;Strehl ratios&lt;/A&gt; (a measure of how close to diffraction limited the image is and, hence, a proxy for resolution given the same optical system).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generically, you could say that the question is: how do we determine the confusion limit in images with point sources? This is just specifically applied to the field of view on the sky with the highest density of resolvable point sources available. Concretely, say we wanted to resolve $22^{\mathrm{nd}}$ magnitude sources in views of the galactic center, what resolution is required to do so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting to the full answer to the more concrete question &quot;how deep would we have to look to see a background quasar&quot; would require additional information: size of field of view, and dust extinction along the sight line. So, for simplicity, I'm assuming that if you're resolving $22^{\mathrm{nd}}$ magnitude sources in $K$ band, you're probably seeing a quasar. Thus we only need the projected density on the sky of sources brighter than that cutoff, roughly, and some description of how resolved sources need to be from the background brightness to detect them to answer the question.&lt;/p&gt;&#xA;" OwnerUserId="14244" LastEditorUserId="14244" LastEditDate="2017-03-18T21:26:04.237" LastActivityDate="2017-03-18T21:26:04.237" Title="What resolution is needed to see a quasar through the galactic center?" Tags="&lt;milky-way&gt;&lt;angular-resolution&gt;&lt;astrometry&gt;&lt;quasars&gt;&lt;galactic-center&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20445" PostTypeId="2" ParentId="20444" CreationDate="2017-03-18T10:31:06.910" Score="1" Body="&lt;p&gt;Whilst awaiting clarification, I'm going to assume that your question is asking about calibrating the reference frame within which the motions of stars at the Galactic centre are measured. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process is described by &lt;a href=&quot;https://arxiv.org/abs/0808.2870&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ghez et al. (2008)&lt;/a&gt;; does not involve observing quasars at infrared wavelengths; and the study does not suffer as a result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Galactic centre adaptive optics images are 5-10 arcseconds across, but contain thousands of stars. Only a fraction of these are close to the Galactic centre. The ensemble can be used, iteratively, to define a relative reference frame by minimising the displacements of thousands of stars, excluding those very close to Sgr A* and those with large motions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is really all that is required to do the orbital analyses of the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put the coordinates in the International Celestial Reference System, a wider field is observed that contains a few infrared-bright giant stars, that are also bright maser sources at radio wavelengths. The positions of these are used to bootstrap the coordinates onto the ICRS, and this is important if, for example, we are interested in the relative motion of the Sun around the Galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maser sources have ICRS positions by virtue of their &lt;em&gt;radio&lt;/em&gt; source positions versus the &lt;em&gt;radio&lt;/em&gt; coordinates of distant (but still bright and point-like) background quasars.The issue of background sources, extinction and confusion does not arise at radio wavelengths.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-03-18T10:31:06.910" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20446" PostTypeId="2" ParentId="20428" CreationDate="2017-03-18T10:53:05.890" Score="3" Body="&lt;p&gt;Yes. Just convert the mass to its energy equivalent via $c^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$7\times10^{-30}$ g/cc is entirely equivalent to $7\times 10^{-27}$ kg/m$^3$ or $6.3\times 10^{-10}$ J/m$^3$ or 3.94 GeV/m$^3$ or 3.94 keV/cc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is quoted as a mass density so that its importance in terms of the dynamics of the universe, can be compared with that of normal and dark matter. In fact in most applications it is more normal to quote and compare in terms of energy density, since that takes account of the fact that matter has kinetic energy as well as rest-mass energy and more neatly includes radiation energy into the comparison.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-03-18T10:53:05.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20447" PostTypeId="1" AcceptedAnswerId="20452" CreationDate="2017-03-18T13:00:57.797" Score="3" ViewCount="508" Body="&lt;p&gt;I asked a question previous similar to this, but I'm wondering, can a star make a moon move closer to its planet or further away? How?&lt;/p&gt;&#xA;" OwnerUserId="16048" LastActivityDate="2017-03-21T03:12:22.403" Title="How do stars affect the orbits of moons?" Tags="&lt;star&gt;&lt;planet&gt;&lt;tidal-forces&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20448" PostTypeId="1" AcceptedAnswerId="20449" CreationDate="2017-03-18T16:47:41.910" Score="39" ViewCount="3752" Body="&lt;p&gt;How large can a ball of water be without fusion starting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Peculiar question: some explanation might be necessary. My young son is into ‘space’ and astronomy. One of his posters says that Saturn could float, if a sufficiently large ocean could be found. Obviously that wouldn’t work: Saturn’s atmosphere would peel off and join or become the atmosphere of the larger body, and then Saturn’s dense core would sink.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But could such an ocean even exist without fusion starting?&lt;/p&gt;&#xA;" OwnerUserId="16283" LastActivityDate="2020-08-20T10:06:25.347" Title="How large can a ball of water be without fusion starting?" Tags="&lt;gravity&gt;&lt;saturn&gt;&lt;brown-dwarf&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20449" PostTypeId="2" ParentId="20448" CreationDate="2017-03-18T20:00:32.187" Score="45" Body="&lt;p&gt;You really need a full-blown stellar evolution model to answer this precisely and I'm not sure anyone would ever have done this with an oxygen-dominated star.&lt;/p&gt;&#xA;&lt;p&gt;To &lt;em&gt;zeroth&lt;/em&gt; order the answer will be the similar to a metal-rich star - i.e. about 0.075 times the mass of the Sun. Any less than this and the brown dwarf (for that is what we call a star that never gets hot enough at its centre to initiate significant fusion) can be supported by electron degeneracy pressure.&lt;/p&gt;&#xA;&lt;p&gt;A star/brown dwarf with the composition you suggest would be a bit different.  The composition would be thoroughly and homogeneously mixed by convection. Note that other than a thin layer near the surface, the water would be completely dissociated and the hydrogen and oxygen atoms completely ionised. Hence the density of protons in the core would be lower for the same mass density than in a &amp;quot;normal star&amp;quot;.  However, the temperature dependence is so steep I think this would be a minor factor and nuclear fusion would be significant at a similar temperature.&lt;/p&gt;&#xA;&lt;p&gt;Of much greater importance is that there would be fewer electrons and fewer particles at the same density. This decreases both the electron degeneracy pressure and normal gas pressure at a given mass density.&#xA;The star is therefore able to contract to much smaller radii before degeneracy pressure becomes important and can thus reach higher temperatures for the same mass as a result.&lt;/p&gt;&#xA;&lt;p&gt;For that reason I think that the minimum mass for hydrogen fusion of a &amp;quot;water star&amp;quot; would be &lt;em&gt;smaller&lt;/em&gt; than for a star made mainly of hydrogen.&lt;/p&gt;&#xA;&lt;p&gt;But how much smaller? Back-of-the-envelope time!&lt;/p&gt;&#xA;&lt;p&gt;Use the virial theorem to get a relationship between perfect gas pressure and the temperature, mass and radius of a star. Let gravitational potential energy be &lt;span class=&quot;math-container&quot;&gt;$\Omega$&lt;/span&gt;, then the virial theorem says&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \Omega = -3 \int P \ dV$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;If we only have a perfect gas then &lt;span class=&quot;math-container&quot;&gt;$P = \rho kT/\mu m_u$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is the temperature, &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; the mass density, &lt;span class=&quot;math-container&quot;&gt;$m_u$&lt;/span&gt; an atomic mass unit and &lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt; the average number of mass units per particle in the gas.&lt;/p&gt;&#xA;&lt;p&gt;Assuming a constant density star (back of the envelope) then &lt;span class=&quot;math-container&quot;&gt;$dV = dM/\rho$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$dM$&lt;/span&gt; is a mass shell and &lt;span class=&quot;math-container&quot;&gt;$\Omega = -3GM^2/5R$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; is the &amp;quot;stellar&amp;quot; radius. Thus&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{GM^2}{5R} = \frac{kT}{\mu m_u} \int dM$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ T = \frac{GM \mu m_u}{5k R}$$&lt;/span&gt;&#xA;and so the central temperature &lt;span class=&quot;math-container&quot;&gt;$T \propto \mu MR^{-1}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Now what we do is say that the star contracts until at this temperature, the phase space occupied by its electrons is &lt;span class=&quot;math-container&quot;&gt;$\sim h^3$&lt;/span&gt; and electron degeneracy becomes important.&lt;/p&gt;&#xA;&lt;p&gt;A standard treatment of this is to say that the physical volume occupied by an electron is &lt;span class=&quot;math-container&quot;&gt;$1/n_e$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$n_e$&lt;/span&gt; is the electron number density and that the momentum volume occupied is &lt;span class=&quot;math-container&quot;&gt;$\sim (6m_e kT)^{3/2}$&lt;/span&gt;. The electron number density is related to the mass density by &lt;span class=&quot;math-container&quot;&gt;$n_e = \rho /\mu_e m_u$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\mu_e$&lt;/span&gt; is the number of mass units per electron. For ionised hydrogen &lt;span class=&quot;math-container&quot;&gt;$\mu_e=1$&lt;/span&gt;, but for oxygen &lt;span class=&quot;math-container&quot;&gt;$\mu_e=2$&lt;/span&gt; (all the gas would be ionised near the temperatures for nuclear fusion). The average density &lt;span class=&quot;math-container&quot;&gt;$\rho = 3M/4\pi R^3$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Putting these things together we get&#xA;&lt;span class=&quot;math-container&quot;&gt;$$h^3 = \frac{ (6m_e kT)^{3/2}}{n_e} = \frac{4\pi \mu_e}{3}\left(\frac{6 \mu}{5}\right)^{3/2} (Gm_e R)^{3/2} m_u^{5/2} M^{1/2}$$&lt;/span&gt;&lt;br /&gt;&#xA;Thus the radius to which the star contracts in order for degeneracy pressure to be important is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ R \propto \mu_e^{-2/3} \mu^{-1} M^{-1/3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;If we now substitute this into the expression for central temperature, we find&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ T \propto \mu M \mu_e^{2/3} \mu M^{1/3} \propto \mu^2 \mu_e^{2/3} M^{4/3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Finally, if we argue that the temperature for fusion is the same in a &amp;quot;normal&amp;quot; star and our &amp;quot;water star&amp;quot;, then the mass at which fusion will occur is given by the proportionality&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ M \propto \mu^{-3/2} \mu_e^{-1/2}$$&lt;/span&gt; .&lt;/p&gt;&#xA;&lt;p&gt;For a normal star with a hydrogen/helium mass ratio of 75:25, then &lt;span class=&quot;math-container&quot;&gt;$\mu \simeq 16/27$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mu_e \simeq 8/7$&lt;/span&gt;.  For a &amp;quot;water star&amp;quot;,  &lt;span class=&quot;math-container&quot;&gt;$\mu = 18/11$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mu_e= 9/5$&lt;/span&gt;. Thus if the former set of parameters leads to a minimum mass for fusion of &lt;span class=&quot;math-container&quot;&gt;$0.075 M_{\odot}$&lt;/span&gt;, then by increasing &lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mu_e$&lt;/span&gt; this becomes smaller by the appropriate factor &lt;span class=&quot;math-container&quot;&gt;$(18\times 27/11\times 16)^{-3/2} (9\times 7/5\times 8)^{-1/2} = 0.173$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Thus a water star would undergo H fusion at &lt;span class=&quot;math-container&quot;&gt;$0.013 M_{\odot}$&lt;/span&gt; or about 13 times the mass of Jupiter!&lt;/p&gt;&#xA;&lt;p&gt;NB This only deals with hydrogen fusion. The small amount of deuterium would fuse at lower temperatures. A similar analysis would give a minimum mass for this to occur of about 3 Jupiter masses.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2020-08-20T10:06:25.347" LastActivityDate="2020-08-20T10:06:25.347" CommentCount="15" ContentLicense="CC BY-SA 4.0" />
  <row Id="20450" PostTypeId="2" ParentId="20389" CreationDate="2017-03-18T22:49:18.133" Score="2" Body="&lt;p&gt;Usually, when a star dies, it will usually form a planetary nebula or supernova and most of this matter will reach the interstellar medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it collapses due to some disturbance, a protoplanetary disk forms, it will cool off eventually. Once it is low enough temperature to form compounds, usually molecular hydrogen (H2) will form first. Here is something that may help you: &#xA;No, H2O cannot exist in stars, but H and O separately can. Hydrogen is the basic building material of the universe, created in the Big Bang. Oxygen is created by nuclear reactions in stars. If you put H and O together in the cold of space, you get H2O. There are enormous amounts of water in space. In fact, nearly all of the oxygen in space is in the form of water or carbon monoxide. Similarly, most the carbon and nitrogen in space are also in their most hydrogenated forms: methane (CH4) and ammonia (NH3).&#xA;&lt;a href=&quot;http://curious.astro.cornell.edu/about-us/101-the-universe/cosmology-and-the-big-bang/general-questions/580-where-did-all-the-water-in-the-universe-come-from-intermediate&quot; rel=&quot;nofollow noreferrer&quot;&gt;(Source)&lt;/a&gt;&#xA;Since water is H2O, it is likely to form because nebulae contain so much hydrogen and some oxygen. The Earth was most likely hit by comets and asteroids that contained water, and when our planet cooled off, we got our oceans.&lt;/p&gt;&#xA;" OwnerUserId="16048" LastActivityDate="2017-03-18T22:49:18.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20451" PostTypeId="1" AcceptedAnswerId="20453" CreationDate="2017-03-18T23:10:54.760" Score="8" ViewCount="624" Body="&lt;p&gt;I have binge watched a lot of documentaries and movies about meteors hitting the earth. The one pattern I have seen, even in the science-fact documentaries like Nova, is that the animations show meteors hitting the earth on the day side. By that, I mean at a near 90° angle in the center of the day zone. As if it were to hit the earth at high noon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What seems unlikely to me is &lt;em&gt;where&lt;/em&gt; the meteor hits. It seems intuitive to me that statistics would favor a meteor strike on the night side of the planet rather than the day side. Or, possibly from the transition between day and night if the meteor came in from the side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my suspicion correct? Is it unlikely for a meteor to hit the earth on the day side and far more likely to hit the night side?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify, I'm thinking of significant meteor strike, like the Chicxulub impact  that killed off the dinosaurs. If something that massive came in from the day side, wouldn't it more likely to have been captured or redirected by the sun because of the sun's gravity?&lt;/p&gt;&#xA;" OwnerUserId="16286" LastEditorUserId="16286" LastEditDate="2017-03-18T23:24:06.647" LastActivityDate="2017-03-22T14:50:34.203" Title="Is it unlikely for a meteor to hit the day side of the earth?" Tags="&lt;meteor&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20452" PostTypeId="2" ParentId="20447" CreationDate="2017-03-18T23:44:55.383" Score="4" Body="&lt;p&gt;Short answer : yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer : it's complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a classic &lt;a href=&quot;https://en.wikipedia.org/wiki/Three-body_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;three body problem&lt;/a&gt;, and a real example would be the Sun-Earth-Moon system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately there is no general solution of the 3-body Newtonian gravity problem, so we have to rely on special methods for special cases and even these turn out to be quite complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best answer I could find to your specific question applied to the Sun-Earth-Moon is &lt;a href=&quot;https://ca.answers.yahoo.com/question/index?qid=20071222002428AAxq0TI&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one on Yahoo&lt;/a&gt;.  As you can see the resulting effect is quite a complex set of periodic effects.  It's small but not negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can intuitively visualize the effect in such a system by noting that typically to a good approximation the planet-moon system will orbit the star at a considerably larger distance than the planet-moon distance.  The planet and moon themselves orbit around their center of mass.  The relative difference in gravitational forces between the position where the moon is furthest away and closest will give you a rough guide to the order of magnitude of the effect.  A back of the envelope calculation for Earth makes it about a 1% order of magnitude for the Sun-Earth-Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other systems will be different, but there will certainly be an effect.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-03-18T23:44:55.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20453" PostTypeId="2" ParentId="20451" CreationDate="2017-03-19T07:13:05.660" Score="10" Body="&lt;p&gt;Remember that the Earth, and the meteoroid, are both in orbit around the sun. As the meteoroid approaches Earth, it will also curve due to Earth's gravity. If you imagine the Earth to be stationary and the meteoroid to be moving in a straight line your intuition is likely to be wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result of the complex dynamics is that there is no time or direction from which it is impossible for a meteoroid to have come from. On average the direction of impact is 45 degrees, and the timing is somewhat more likely in the morning when the rotation of Earth combines with its movement through space. A direct impact at noon, from the apparent direction of the sun, is unlikely but possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you look at lists of meteors, many are at night. Except the brightest fireballs, a meteor is much more noticeable at night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nasa hs a database of boloids: &lt;a href=&quot;http://neo.jpl.nasa.gov/fireballs/&quot; rel=&quot;noreferrer&quot;&gt;http://neo.jpl.nasa.gov/fireballs/&lt;/a&gt; By combining the UT time with the latitude information it is possible to get an estimate of the local solar time of the fireball:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Ue4fu.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ue4fu.png&quot; alt=&quot;Local solar time of boloids since 2007&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This histogram of time of impact of about 250 fireballs since 2007 shows no particular decrease around midday. The documentary graphics may show an apparently fully lit earth, and a perpendicular impact as it is easier to recognise the Day side of the Earth, and it looks more dramatic. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-03-19T08:41:11.233" LastActivityDate="2017-03-19T08:41:11.233" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20455" PostTypeId="1" AcceptedAnswerId="20600" CreationDate="2017-03-19T13:33:54.357" Score="2" ViewCount="485" Body="&lt;p&gt;I heard our universe is flat. Then one question is puzzling in my mind. If our universe is really flat, why we measure distance from any point across in a spherical way. In other words, why we say, the comoving distance (radius) is now about 46.6 billion light years. If this radius is only treated for observable universe, then how does science find that the universe is flat knowing nothing beyond the limit of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1299/what-is-the-physical-geometric-shape-of-the-universe&quot;&gt;questions&lt;/a&gt; in SO, however, they do not assimilate what I intended to know. In &lt;a href=&quot;https://astronomy.stackexchange.com/q/1299/15770&quot;&gt;this&lt;/a&gt; question, for instance, OP asked about whats the shape of the universe. And also there are some terms related to black holes, 3D shape, ball etc. Apart from those complexity, my intention is simply to be clarified, why we call our universe flat? And by word &lt;a href=&quot;https://en.oxforddictionaries.com/definition/flat&quot; rel=&quot;nofollow noreferrer&quot;&gt;flat&lt;/a&gt;, what I meant &quot;having a level surface; without raised areas or indentations&quot;. Am i missing any alrernative meaning of flat here?&lt;/p&gt;&#xA;" OwnerUserId="15770" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2017-04-01T20:36:14.620" Title="Is our universe flat?" Tags="&lt;universe&gt;&lt;distances&gt;&lt;observable-universe&gt;&lt;age&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20456" PostTypeId="2" ParentId="9990" CreationDate="2017-03-19T13:53:06.923" Score="2" Body="&lt;p&gt;Answer is &lt;strong&gt;Yes&lt;/strong&gt;. Not only to sun and moon, we are seeing each and every distant object we are ever experiencing with certain delay. For example, if you are reading this text on a computer screen at 50 centimetres distant from your eyes, you are actually experiencing the past version of &lt;code&gt;1.66777852e-9&lt;/code&gt; &lt;a href=&quot;https://www.google.com/search?sxsrf=ALeKk01ZqsGKNWEEzI-fA7t4dzsqtdRBqg%3A1588034587721&amp;amp;ei=G3ynXoXRK7ea4-EPubqx6A0&amp;amp;q=50%20%2F2.998e%2B10&amp;amp;oq=50%20%2F2.998e%2B10&amp;amp;gs_lcp=CgZwc3ktYWIQAzIECAAQRzIECAAQRzIECAAQRzIECAAQRzIECAAQRzIECAAQRzIECAAQRzIECAAQR1AAWABgyMsRaABwAngAgAEAiAEAkgEAmAEAqgEHZ3dzLXdpeg&amp;amp;sclient=psy-ab&amp;amp;ved=0ahUKEwjFyvfr8YnpAhU3zTgGHTldDN0Q4dUDCAw&amp;amp;uact=5&quot; rel=&quot;nofollow noreferrer&quot;&gt;seconds&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One interesting thing to know is, in reality past, present and future exist all together as spacetime in the fabric of cosmos. For example, when you say, we are seeing 10 years past version of a star which is 10 light years away from our planet, similarly an alien right now is watching 10 years past version of our earth seating on that star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An excellent explanation can be found &lt;a href=&quot;https://youtu.be/vrqmMoI0wks&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; presented by Brian Greene to understand how mind boggling reality we are living in.&lt;/p&gt;&#xA;" OwnerUserId="15770" LastEditorUserId="15770" LastEditDate="2020-04-29T07:08:49.600" LastActivityDate="2020-04-29T07:08:49.600" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="20457" PostTypeId="1" AcceptedAnswerId="20458" CreationDate="2017-03-19T15:31:11.290" Score="5" ViewCount="1843" Body="&lt;p&gt;I understand that a black hole happens after a star exhaustions elements capable of undergoing fusion.  Without the energy produced by fusion the remaining mass of the star collapses on itself and forms a black hole.  I conclude black holes must be deficient in hydrogen, helium and elements that undergo fusion in stars.  I understand that hydrogen and the like undergoes fusion in stars in part (or completely?) because of the gravitational pressure which occurs in stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose a young star replete with hydrogen merged with a black hole.  My question: would the result be a larger black hole, or a radiant star with a very massive heart?  Could a massive infusion of (fusible?) elements reboot the black hole back into a star?&lt;/p&gt;&#xA;" OwnerUserId="16293" LastActivityDate="2017-03-19T16:30:22.993" Title="Can a black hole turn back into a star?" Tags="&lt;star&gt;&lt;black-hole&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2017-03-21T13:34:23.627" ContentLicense="CC BY-SA 3.0" />
  <row Id="20458" PostTypeId="2" ParentId="20457" CreationDate="2017-03-19T16:30:22.993" Score="9" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A black hole is characterised by the formation of an event horizon. Matter which travels past an event horizon cannot return to the space outside the event horizon, to do so would require the matter to travel faster than light, and so would require infinite energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is nothing like a regular star &quot;inside&quot; a black hole. The general relativity model of a black hole contains a singularity. Matter which enters a black hole is certain to reach this singularity in a (usually short) amount of time. Adding hydrogen to a black hole would just make the black hole more massive. There is a theorem in mathematical physics that the only properties that a black hole can have are mass, angular momentum and charge. In other words, a black hole made of collapsed hydrogen would be identical to one made of collapsed iron. There is no matter in a black hole, only mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formation of a black hole normally occurs when a core in a massive star has a certain amount of iron, which saps the core of energy causing collapse. If the resulting object has more than about 3 solar masses, then nothing in the universe can stop it from collapsing all the way to a singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with much about general relativity, your intuition is a very bad guide to how black holes work. However General relativity is a very good model for how black holes should form and behave &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-03-19T16:30:22.993" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20460" PostTypeId="2" ParentId="20455" CreationDate="2017-03-19T17:49:46.070" Score="2" Body="&lt;p&gt;The aspect of the universe described as &quot;flat&quot; is the spacetime curvature. In General Relativity gravity isn't treated as a force but as a curvature of the spacetime manifold. At large scales the universe is expanding: negative curvature like the top of a hill, where things roll downhill away from each other. But mass causes a positive curvature like a bowl, where things roll together, toward the bottom of the bowl. On average the two regimes of curvature appear to be balanced - i.e. the universe is, on average, flat. &lt;/p&gt;&#xA;" OwnerUserId="4042" LastActivityDate="2017-03-19T17:49:46.070" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20463" PostTypeId="1" AcceptedAnswerId="20464" CreationDate="2017-03-20T03:46:05.257" Score="-4" ViewCount="252" Body="&lt;p&gt;How far in miles does the Moon have to be for the Earth not to feel its effects?&lt;/p&gt;&#xA;" OwnerUserId="16295" LastEditorUserId="6" LastEditDate="2017-03-20T14:27:07.463" LastActivityDate="2017-03-20T14:27:07.463" Title="How far away does the Moon have to be until Earth doesn't feel its effects?" Tags="&lt;the-moon&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2017-03-20T14:52:49.143" ContentLicense="CC BY-SA 3.0" />
  <row Id="20464" PostTypeId="2" ParentId="20463" CreationDate="2017-03-20T04:13:12.550" Score="2" Body="&lt;p&gt;We can use Newton's law of gravity to calculate the force the Earth feels from the moon:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;F_g = \frac{GM_{Earth}M_{Moon}}{r^2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where G is the universal gravitational constant, $6.62*10^{-11}$ $m^3 kg^{-1} s^{-2}$, $M_{Earth}$ and $M_{Moon}$ are the masses of the Earth and Moon in kilograms, and $r$ is the distance between their center s in meters.  We can rearrange this equation to solve for a distance like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;r = \sqrt{\frac{GM_{Earth}M_{Moon}}{F_g}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can input any force we like to see how far away the Moon would have to be in order for us to feel that force.  If we want the force of gravity to go to zero, we can take the limit as $F_g\rightarrow 0$.  When we do this, we see that for the Earth to feel zero gravitational force from the Moon, it would have to be infinitely far away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, infinity is not a very useful number; why don't we look for something a little more practical?  Let's say that we want the Earth to feel exactly 0.01 Newton of gravitational force from the Moon.  That's about the weight of 1 gram of matter at the Earth's surface.  In that case, we will have $r=5.411*10^{19}$ meters or $3.4*10^{14}$ miles, about $1.4*10^{11}$ times further away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind, though, that this is still not zero force.  It's small enough that we probably wouldn't be able to detect it, but it is nonzero.&lt;/p&gt;&#xA;" OwnerUserId="13426" LastActivityDate="2017-03-20T04:13:12.550" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20466" PostTypeId="1" AcceptedAnswerId="20469" CreationDate="2017-03-20T15:38:30.147" Score="80" ViewCount="20871" Body="&lt;p&gt;In astronomy distances are generally expressed in non-metric units like: light-years, astronomical units (AU), parsecs, etc. Why don't they use meters (or multiples thereof) to measure distances, as these are the SI unit for distance? Since the meter is already used in particle physics to measure the size of atoms, why couldn't it be used in astrophysics to measure the large distances in the Universe? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The ISS orbits about 400 km above Earth.&lt;/li&gt;&#xA;&lt;li&gt;The diameter of the Sun is 1.39 Gm (gigameters).&lt;/li&gt;&#xA;&lt;li&gt;The distance to the Andromeda Galaxy is 23 Zm (zettameters).&lt;/li&gt;&#xA;&lt;li&gt;At its furthest point, Pluto is 5.83 Tm (terameters) from the Sun.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Edit: some have answered that meters are too small and therefore not intuitive for measuring large distances, however there are plenty of situations where this is not a problem, for example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bytes are used for measuring gigantic amounts of data, for example terabytes (1e+12) or petabytes (1e+15)&lt;/li&gt;&#xA;&lt;li&gt;The energy released by large explosions is usually expressed in megatons, which is based on grams (1e+12)&lt;/li&gt;&#xA;&lt;li&gt;The SI unit Hertz is often expressed in gigahertz (1e+9) or terahertz (1e+12) for measuring network frequencies or processor clock speeds.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If the main reason for not using meters is historical, is it reasonable to expect that SI-unites will become the standard in astronomy, like most of the world switched from native to SI-units for everyday measurements? &lt;/p&gt;&#xA;" OwnerUserId="16304" LastEditorUserId="16304" LastEditDate="2017-04-04T12:47:49.947" LastActivityDate="2020-02-18T17:01:42.763" Title="Why don't astronomers use meters to measure astronomical distances?" Tags="&lt;distances&gt;&lt;units&gt;" AnswerCount="12" CommentCount="21" FavoriteCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="20469" PostTypeId="2" ParentId="20466" CreationDate="2017-03-20T17:09:52.617" Score="84" Body="&lt;p&gt;In addition to the answer provided by @HDE226868, there are historical reasons. Before the advent of using radar ranging to find distances in the solar system, we had to use other clever methods for finding the distance from the Earth to the sun; for example, &lt;A HREF=&quot;https://science.nasa.gov/science-news/science-at-nasa/2004/28may_cook/&quot; rel=&quot;noreferrer&quot;&gt;measuring the transit of Venus across the surface of the sun&lt;/A&gt;. These methods are not as super accurate as what is available today, so it makes sense to specify distances, that are all based on measuring parallaxes, in terms of the uncertain, but fixed, Earth-Sun distance. That way, if future measurements change the conversion value from AU to meters, you don't have to change as many papers and textbooks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not to mention that such calibration uncertainties introduce correlated errors into an analysis that aren't defeatable using large sample sizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't speak authoritatively on the actual history, but solar system measurements were all initially done in terms of the Earth/sun distance. For example, a little geometry shows that it's pretty straightforward to back out the size of Venus's and Mercury's orbit in AU from their maximum solar elongation. I don't know how they worked out the orbital radii of Mars, etc, but they were almost certainly done in AU long before the AU was known, and all of that before the MKS system existed, let alone became standardized. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For stars, the base of what is known as the &quot;cosmological distance ladder&quot; (that is &quot;all distance measures&quot; in astronomy) rests on measuring the parallax angle:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\tan \pi_{\mathrm{angle}} = \frac{1 AU}{D}.$$&lt;/span&gt;&#xA;To measure &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; in 'parsecs' is to setup the equation so that the angle being measured in arcseconds fits the small angle approximation. That is:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{D}{1\, \mathrm{parsec}} = \frac{\frac{\pi}{180\times60\times60}}{\tan\left(\pi_{\mathrm{angle}} \frac{\pi\, \mathrm{radians}}{180\times60\times60 \, \mathrm{arcsec}}\right)}.$$&lt;/span&gt;&#xA;In other words, &lt;span class=&quot;math-container&quot;&gt;$1\operatorname{parsec} = \frac{180\times 3600}{\pi} \operatorname{AU}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomers also have a marked preference for the close cousin of mks/SI units, known as &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Centimetre%E2%80%93gram%E2%80%93second_system_of_units&quot; rel=&quot;noreferrer&quot;&gt;cgs&lt;/A&gt;. As far as I can tell, this is due to the influence of spectroscopists who liked the &quot;Gaussian units&quot; part of it for electromagnetism because it set Coulomb's constant to 1, simplifying calculations.&lt;/p&gt;&#xA;" OwnerUserId="14244" LastEditorUserId="14244" LastEditDate="2019-09-03T11:52:07.463" LastActivityDate="2019-09-03T11:52:07.463" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="20471" PostTypeId="2" ParentId="20466" CreationDate="2017-03-20T19:07:19.887" Score="24" Body="&lt;p&gt;I would suggest it also makes the material more reachable for the human mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just can't work with insanely large or small numbers.  They convey no meaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But 1 AU is easy, even if I don;t know exactly what that is in meters, I know what it &lt;em&gt;means&lt;/em&gt; and it is a convenient scale for the mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise when we talk about stellar distances, what use is the distance in meters (or AU) ?  It makes more sense to work with light years.  Again most people know what that means even if they don't know exactly what it is in meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And when we go cosmic you're also talking about colossal &lt;em&gt;times&lt;/em&gt; in the past, so light years do convey a double meaning here.  If I told you the distance in meters, that doesn't instantly tell you how far back in time it is as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think it's a matter on convenience and comprehension.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-03-20T19:07:19.887" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="20473" PostTypeId="1" AcceptedAnswerId="20474" CreationDate="2017-03-21T00:27:15.700" Score="6" ViewCount="431" Body="&lt;p&gt;When a star system forms from a cloud of gas and dust, why doesn't the dust migrate towards the centre of gravity with the gas to form a star? Why does dust in the accretion disk aggregate to form rocky planets, rather than fall into the proto-star?&lt;/p&gt;&#xA;" OwnerUserId="13887" LastActivityDate="2017-03-21T01:55:15.510" Title="Why do rocky planets form?" Tags="&lt;planetary-formation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20474" PostTypeId="2" ParentId="20473" CreationDate="2017-03-21T01:37:48.950" Score="6" Body="&lt;p&gt;The nebula from which the sun and the solar system formed contained mostly hydrogen and helium, but also small amounts of heavier elements, including &quot;dust&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any dust that ended up in the centre of the accretion disc became part of the sun. The sun contains 0.01% silicon, for example. This is only a small proportion, but is still about $2\times10^{26}kg$, 100 times the mass of the Earth. When you add in the Iron, Nickel, Oxygen, Carbon and other elements that make up the rocks, you find that there is enough matter in the sun to make up thousands of rocky planets. (&lt;a href=&quot;http://www.space.com/17170-what-is-the-sun-made-of.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.space.com/17170-what-is-the-sun-made-of.html&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is only that part of the dust that happens not to fall into the sun that goes on to form rocky planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The details of the process of planetary formation are complex (and not completely understood) however after the formation of the sun, solar wind cleared out the inner solar system of gas and volatiles, leaving only the rocky planets which were too small to hold on a large hydrogen-helium envelope. Further out, beyond the ice-line, water and other volatiles are solid, and much larger planets can form, that gravitationally attract hydrogen and helium mantles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;see &lt;a href=&quot;https://astronomy.stackexchange.com/questions/16140/how-did-the-earth-come-to-be-in-orbit-around-the-sun&quot;&gt;How did the Earth come to be in orbit around the sun?&lt;/a&gt; for more on planetary formation.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2017-03-21T01:55:15.510" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20476" PostTypeId="2" ParentId="20447" CreationDate="2017-03-21T03:12:22.403" Score="4" Body="&lt;p&gt;Just to add to StephenG's answer, and because I think a key point wasn't covered, I'll give this one a go.  Warning - answer is probably too long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Yes, it's complicated and the math gets a little intense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2), As StevenG points out it's a 3-body problem and The 3-body problem has no solution but it can be run on a supercomputer in many cases for millions or tens of millions of orbits, and millions or tens of millions of years into the future with pretty good accuracy.   The margin of error doubles with time, so there are limits to such calculations, but within those limits, models of the future of 3-body orbits can be made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) and my main point.  &lt;strong&gt;There's a difference between a stable 3 body orbit&lt;/strong&gt; (Moon orbits the Earth, Earth orbits the sun), &lt;strong&gt;and the unstable model&lt;/strong&gt; that's typically referred in a 3 body problem like the image below from the &lt;a href=&quot;https://en.wikipedia.org/wiki/N-body_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;n-body problem in Wikipedia&lt;/a&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kiR9d.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kiR9d.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the Moon is comfortably inside the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere#True_region_of_stability&quot; rel=&quot;nofollow noreferrer&quot;&gt;true region of stability&lt;/a&gt; inside Earth's Hill Sphere, it's orbit can be considered stable.   Ultimately all orbits are unstable, so, when I say it's stable, I mean, stable for hundreds of millions or billions of years.  That's a nice round number.  Phobos, for example, is expected to crash into Mars or break apart inside Mars' roche limit in about 10 million years.   Some people call that an unstable orbit, but 10 million years is a long time.  There's no precise dividing line between stable and unstable orbits, it's a term of convenience, more than accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Kepler's laws and the Earth-Moon-Sun as an example, The Earth-Moon barycenter orbits the sun in basically a Kepler orbit.   Pretty close anyway.     If you want to be a stickler, it orbits the 3 body barycenter, but with the sun's mass ratio to the Earth/Moon being about 330,000 to 1, it's pretty close to just orbiting the sun.  Kepler's laws are pretty close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon does not orbit the Earth in a neat Kepler orbit because solar tides are significant, making the Moon's elliptical orbit pretty stretched and wobbled.   The same sun-tides that affect oceans on earth affect the Lunar orbit around the Earth.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This diagrams is exaggerated, so take with a grain of salt, but this is the gist of what the sun does to the Moon's orbit.   It's still ellipse-ish, but visibly stretched.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Orbit_of_the_Moon#Rotation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Apsidal precession&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hwCS1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hwCS1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon's apsidal precession completes a full 360 cycle every 8.85 years or roughly 109 synodic orbits, a bit over 3 degrees off where it should be by kepler's laws per orbit.   That's over 1000 times the variation of &lt;a href=&quot;https://en.wikipedia.org/wiki/Tests_of_general_relativity#Perihelion_precession_of_Mercury&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mercury's precession&lt;/a&gt;, which confused astronomers for decades.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This significant lunar precession is almost entirely due to solar tides and it's an example of how solar tides affect the Moon's orbit around the Earth.   If Venus or Mercury had moons, those moons would probably have even bigger precessions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, long term changes are far slower as the precession tends to balance out over time.  Even with the Moon's significant precession it's still considered a stable orbit and it's comfortably inside the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere#True_region_of_stability&quot; rel=&quot;nofollow noreferrer&quot;&gt;true region of stability&lt;/a&gt; inside Earth's Hill Sphere.  There's also very little &lt;a href=&quot;https://en.wikipedia.org/wiki/Chaos_theory#History&quot; rel=&quot;nofollow noreferrer&quot;&gt;chaotic unpredictability&lt;/a&gt; in the Moon's orbit that is typical of a standard 3-body problem.   That's not to say there's none, there's still some, but there's also a long term constancy to the Moon's orbit.   See here, for a &lt;a href=&quot;http://w.askamathematician.com/2011/10/q-what-is-the-three-body-problem/&quot; rel=&quot;nofollow noreferrer&quot;&gt;very nice explanation of the 3-body problem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, long story short, what happens with a structured orbit (Sun-Planet-Moon), where all orbits are inside true regions of stability, becomes somewhat predictable.  Basically the sun steals or saps energy from the planet-moon orbit as an effect of the tidal force.   How this happens is tricky to explain and maybe there's an expert who can explain it better or give cool graphics.  I don't have those skills.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now &quot;stealing&quot; is perhaps a bad word. The planet orbits the star and the moon, orbits the planet but that orbit is stretched and squeezed by the stars tidal force.   If you think of the Moon as a &quot;bulge&quot; on the planet the solar gravity tugs on that bulge, slowing it down.   But the effect is quite small as what it does is both speed it up over 1/2 the orbit and slow it down over the 2nd half, but the de-circularizaiton of the orbit when added up, works out to a decrease in orbital energy over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun is doing this to the Earth-Moon system now, but it's measurably smaller than the Earth-Moon dynamic that &lt;a href=&quot;http://curious.astro.cornell.edu/about-us/37-our-solar-system/the-moon/the-moon-and-the-earth/111-is-the-moon-moving-away-from-the-earth-when-was-this-discovered-intermediate&quot; rel=&quot;nofollow noreferrer&quot;&gt;pushes the Moon away from the Earth&lt;/a&gt; about 3.8 cm per year.  Over time, as the Earth loses it's oceans, and the Moon moves slightly further out, the solar effect will become the primary force, slowing drawing the Moon back towards the Earth, But that's probably billions of years away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternate way to think about this, instead of Sun-Planet-Moon is Planet-Moon-sub satellite or, planet-moon-sub moon.  This model we have actual observed data for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moons as a rule, don't have their own moons and the reason for this is the same - tidal forces.   And on average, the tidal force a moon receives from it's planet are significantly greater than the force a planet gets from it's star, but while the tides are many times stronger, the principal is still the same.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.universetoday.com/109666/can-moons-have-moons/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Can moons have moons&quot;&lt;/a&gt;.   Dr. Fraser Cain explains what happens, he doesn't get into the specifics of why (math drives away readers).   But he explains what happens.  From his article&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;No satellite we’ve sent to the Moon has ever orbited for longer than a&#xA;  few years before crashing down into the lunar surface. In theory, you&#xA;  could probably get a satellite to last a few hundred years around the&#xA;  Moon.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But why? How come we can’t make moons for our moon to have a moon of&#xA;  it’s own for all time? It all comes down to gravity and tidal forces.&#xA;  Every object in the Universe is surrounded by an invisible sphere of&#xA;  gravity. Anything within this volume, which astronomers call the “Hill&#xA;  Sphere”, will tend to orbit the object.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;So, if you had the Moon out in the middle of space, without any&#xA;  interactions, it could easily have multiple moons orbiting around it.&#xA;  But you get problems when you have these overlapping spheres of&#xA;  influence. The strength of gravity from the Earth tangles with the&#xA;  force of gravity from the Moon.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Although a spacecraft can orbit the Moon for a while, it’s just not&#xA;  stable. The tidal forces will cause the spacecraft’s orbit to decay&#xA;  until it crashes. But further out in the Solar System, there are tiny&#xA;  asteroids with even tinier moons. This is possible because they’re so&#xA;  far away from the Sun. Bring these asteroids closer to the Sun, and&#xA;  someone’s losing a moon.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The object with the largest Hill Sphere in the Solar System is&#xA;  Neptune. Because it’s so far away from the Sun, and it’s so massive,&#xA;  it can truly influence its environment. You could imagine a massive&#xA;  moon distantly orbiting Neptune, and around that moon, there could be&#xA;  a moon of its own. But this doesn’t appear to be the case.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;NASA is considering a mission to capture an asteroid and put it into&#xA;  orbit around the Moon. This would be safer than having it orbit the&#xA;  Earth, but still keep it close enough to extract resources. But&#xA;  without any kind of orbital boost, those tidal forces will eventually&#xA;  crash it onto the Moon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Earth's tidal forces will cause an object in Lunar orbit to crash into the Moon.  (the Moon's &lt;a href=&quot;https://science.nasa.gov/science-news/science-at-nasa/2006/06nov_loworbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravitational lumpiness&lt;/a&gt; doesn't help either - for an entirely different reason, both are factors.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the Moon's permanent tidal lock on the Earth makes a lunar orbit difficult.  In a sense, there's a permanent heavy side facing the earth and a permanent light side of the moon always facing away from the earth. See article &lt;a href=&quot;http://curious.astro.cornell.edu/about-us/44-our-solar-system/the-moon/general-questions/104-can-moons-have-moons-intermediate&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.  That imbalance helps create unstable orbits around the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more massive a planet, on average, the more gravitationally spherical it becomes (ignoring rotational bulging, but since most orbits are equatorial, bulging isn't an issue).  Moons and smaller planets (like Mars) can be more gravitationally lumpy, which makes orbits somewhat less stable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to your question, The solar tidal force does draw moons closer to their planets, but outside of Mercury and Venus, the solar tides are small enough that this isn't much of a factor.    When we get better looks at exoplanets, and red-dwarf stars with close orbiting planets, we might find that moons are rare on those planets because of the comparatively larger solar tidal forces, but such observations are probably a ways away.   I'm not sure even the James Webb Telescope will identify exo-moons.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other solar effects, like evaporation/transpiration off a planet's surface would make the planet lighter, so in that sense, a sun can cause a planet to lose a moon if the planet is losing mass due to solar wind.   (this effect is quite slow - Titan comes to mind as the &lt;a href=&quot;http://link.springer.com/chapter/10.1007%2F978-1-4020-9215-2_15#page-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;best example of a solar-system object that's losing mass&lt;/a&gt; due to the solar wind, but even for Titan, this loss is very slow.    If Titan had a moon, however, the solar wind would cause it's moon to slowly move away (but Saturn's gravity would draw the moon closer to Titan).  In your specific question about Pluto, this is probably a bigger effect than tides, so the sun is probably causing Charon and Pluto to slowly move apart due to outgassing, as it's called. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as a final example of solar effects gaseous pressure from an expanding star would cause a lunar orbit to slow down and cause the moon to crash into it's planet, but gaseous pressure like that isn't expected until our sun goes red giant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the answer I think you're looking for is that solar tides will sap energy from a planet-moon orbit around that star over time and quite slowly.  As I said above.   If somebody can explain the math behind that or show with graphics, please feel free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(too long?)&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-03-21T03:12:22.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20477" PostTypeId="1" AcceptedAnswerId="20484" CreationDate="2017-03-21T05:23:33.473" Score="6" ViewCount="1504" Body="&lt;p&gt;The sun lines up with this fountain, which is at the Salk Institute, during the equinoxes. Will precession eventually throw the alignment out of whack? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aH3vK.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aH3vK.jpg&quot; alt=&quot;Salk Institute fountain&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16317" LastActivityDate="2020-08-25T00:24:14.847" Title="Precession and the location of the sun on the horizon at equinox" Tags="&lt;the-sun&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20478" PostTypeId="2" ParentId="20466" CreationDate="2017-03-21T06:05:47.050" Score="5" Body="&lt;p&gt;Perhaps one needs to go back in time and think about why the cubit (length of forearm), league (distance walked in one hour), foot, (metre - one ten-millionth of a quadrant of the Earth?? and so perhaps should not be i this list) etc were chosen as the units of distance?&lt;br&gt;&#xA;They were easily understood and reproducible  whilst at the same time being of a scale comparable with distances to be measured.&lt;br&gt;&#xA;So in the modern world people have chosen further units of distance which initially had those characteristics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once these new units gain favour and papers, textbooks etc are written it is difficult to get rid of them and some would say - &quot;Why bother?&quot;.  &lt;/p&gt;&#xA;" OwnerUserId="16303" LastEditorUserId="16303" LastEditDate="2017-03-21T09:53:26.943" LastActivityDate="2017-03-21T09:53:26.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20479" PostTypeId="1" CreationDate="2017-03-21T07:23:54.107" Score="5" ViewCount="141" Body="&lt;p&gt;Last 65 days of Kepler's light curve for the infamous KIC 8462852 star is shown below from the published paper's graph. Skip the dips here, I wonder about two other features:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Day 1559, during a period which is calm but also the very brightest, the light curve instantly and permanently drops by half of a percent.&lt;/li&gt;&#xA;&lt;li&gt;Days 1581 to 1587 the light curve is exactly flat (also days 803-809&#xA;not shown here). Is the real data for those periods lost?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Are such features commonplace or another particularity for KIC 8462852?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How wide would a confidence interval typically be for a finished Kepler light curve when all adjustments for telescope movements and whatnot have been made?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tLYb4.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tLYb4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2017-03-21T07:48:45.537" LastActivityDate="2017-03-21T13:40:45.263" Title="What caused instant jumps and exactly flat periods in Kepler's light curves?" Tags="&lt;data-analysis&gt;&lt;kepler&gt;&lt;light-curve&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20480" PostTypeId="2" ParentId="20477" CreationDate="2017-03-21T08:02:08.980" Score="3" Body="&lt;p&gt;The alignment should be maintained.  In general, at the equinoxes, the sun rises due east and sets due west.  As East and West are determined relative to the Earth's axis of rotation, it doesn't matter that the direction of the axis changes due to precession.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, Stonehenge is aligned to the midsummer sunrise (OK, not the equinox, but it makes the point), and has been for 4000 years (I haven't checked when the relevant part is thought to have been built).&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-03-21T08:02:08.980" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20481" PostTypeId="2" ParentId="20466" CreationDate="2017-03-21T10:51:24.707" Score="1" Body="&lt;p&gt;It's tough to relate something like a terameter to &quot;real lengths&quot;, because of the lack of knowledge of physical objects to compare them to.  Also, because after a while, these units become just &quot;so many more zeros&quot;.  So I would suggest the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space Marginal Unit (SMU): 1,000,000 meters, or roughly the distance of from one end of France to the other.  The minimum distance two spacecraft would have to be from each other before they would have to coordinate trajectories or go into docking manuvers.  (Give me a bit of suspension of disbelief here folks.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Length of Earth Orbit (LEO): 1,000,000,000,000 meters, the distance the Earth travels in one year.  (The distance is actually about 6% less than that, but the LEO is something that can be visualized.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kaid: 1,000,000,000,000,000,000 meters.  That's a bit more than the distance from here to the star Alkaid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above readily lend themselves to everyday conversation -- if we ever get to a point where we talk about such things everyday!&lt;/p&gt;&#xA;" OwnerUserId="16324" LastActivityDate="2017-03-21T10:51:24.707" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20482" PostTypeId="2" ParentId="20466" CreationDate="2017-03-21T12:51:46.003" Score="7" Body="&lt;p&gt;Another not yet mentioned reason:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;There were no usable SI prefixes for such distances.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to use an unit, you need something which allows to express&#xA;a specific quantity without too many leading or trailing zeroes.&#xA;I do not express human height as 1 670 000 µm or the size of a bacteria as 0.000 02 m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look up the &lt;a href=&quot;https://en.wikipedia.org/wiki/Metric_prefix&quot; rel=&quot;noreferrer&quot;&gt;table of prefixes&lt;/a&gt; you see that giga and tera&#xA;was &lt;em&gt;defined&lt;/em&gt; the first time 1960. But definition does not include&#xA;usage and those definitions were exactly as exotic as &lt;em&gt;octillion&lt;/em&gt;; sure&#xA;it exists as definition, but noone uses it or knows of its existence.&#xA;During academic studies in physics in the 90s (!) &lt;strike&gt;it still was not widely known, 30 years after introduction. Still many scientists do not use&#xA;giga- or tera- at all.&lt;/strike&gt; Hint by gerrit: Physicists used frequencies with the giga-/tera- prefix, I forgot that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 AU is then 150 gigameter or 0.15 terameter. If you are using light years,&#xA;1 light year is already 9500 terameter which is not a convenient unit. Thirty years later they finally introduced some usable metric prefixes, but I still&#xA;have to find someone who uses exa-, peta-, yotta- or zetta-.&lt;/p&gt;&#xA;" OwnerUserId="7695" LastEditorUserId="7695" LastEditDate="2017-03-21T15:34:24.463" LastActivityDate="2017-03-21T15:34:24.463" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20483" PostTypeId="2" ParentId="20479" CreationDate="2017-03-21T13:40:45.263" Score="6" Body="&lt;p&gt;There are two separate points of interest you're looking at so I'll separate this into sections.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Sudden Drop at Day 1559&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;As near as I can tell, this is the result of a quarterly roll of the satellite, specifically the end of quarter 17. Every 90 days, which NASA calls a quarter, the space craft does a 90 degree roll to optimize the solar panel efficiency. What this means for the data is that when this roll occurs at the end of the quarter, the photons from the star now fall on a different CCD channel which is calibrated slightly differently, causing a slight discontinuity in the data. You can read about &lt;a href=&quot;https://keplerscience.arc.nasa.gov/PyKEprimerStitching.shtml&quot; rel=&quot;noreferrer&quot;&gt;&quot;Quarter Stitching&quot; here&lt;/a&gt;. I've extracted some relevant text and an image for posterity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;After each roll maneuver, most Kepler targets fall on a different CCD channel and the target’s point-spread function will be distributed differently across neighboring pixels. Naturally this redistribution requires a new computation of the target mask and optimal aperture size within the Kepler pipeline, taking into account the point-spread function, new CCD characteristics, and new estimates of nearby source crowding. The operational outcome is often a different mask shape, with differing amounts of flux within the optimal aperture from both the target and contaminating sources.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RbGei.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RbGei.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;sup&gt;&lt;em&gt;Light curves extracted from all single pixels within the quarter 4, 5 and 6 Target Pixel Files for KIC 9603833 (the symbiotic star StHA 169). Gray pixels comprise the optimal apertures that yield the archived light curves. The target point-spread function is distributed across neighboring pixels differently from quarter to quarter and hence the optimal aperture varies in size and shape from quarter to quarter.&lt;/em&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Flat line between days 1581 to 1587&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I'm less certain about this one. I believe the flat line is just missing data. To keep the line continuous, they just connected the data points they had with a straight line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why the data is missing, I can't be sure. I know that Quarter 17 ended with the failure of Kepler's second reaction wheel so my guess is that this missing data is in some way related to these reaction wheel troubles. I do know they tried,in the past, to put the spacecraft in 10-day &quot;wheel rest&quot; mode in hopes that it would give the reaction wheels a break, let them get lubed up again, and be better when they restarted. Perhaps, by this stage, they had already suspected a second reaction wheel problem and were trying out a wheel rest mode. I can't find any evidence a wheel rest occurred during Quarter 17 though, so this is just my guess.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-03-21T13:40:45.263" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20484" PostTypeId="2" ParentId="20477" CreationDate="2017-03-21T13:54:57.900" Score="6" Body="&lt;p&gt;The fountain will always line up with the Sun during the equinox. The precession of the Earth's axis will not change this fact. What it will change is exactly when the equinox happens. However, the fountain doesn't care about when the equinox occurs, only that it does and when it does happen, it will be pointed at the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equinox is simply the precise point in time when the Earth's axis is neither leaning towards or away from the Sun, but rather leaning along it's direction of travel. There are many analgous definitions of the equinox, but the main point here is that the equinox is simply a specific orientation of the Earth's tilt with respect to the Sun. So long as the Earth is in that specific state, the equinox will occur and the fountain will line up with the Sun. As I said though, the precession may cause the date of the equinox to drift over time.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-03-21T13:54:57.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20486" PostTypeId="2" ParentId="20466" CreationDate="2017-03-21T16:00:14.827" Score="2" Body="&lt;p&gt;Units like metres are simply too small to be used when measuring distances on an astronomical scale. While one could, in theory, use metres in conjunction with scientific notation, it is unnecessarily difficult. One Astronomical Unit is the distance between the Earth and the Sun, this acts as a sort of a cosmic metre stick.&lt;/p&gt;&#xA;" OwnerUserId="16335" LastActivityDate="2017-03-21T16:00:14.827" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20487" PostTypeId="1" AcceptedAnswerId="20489" CreationDate="2017-03-21T17:04:06.600" Score="3" ViewCount="1622" Body="&lt;p&gt;Imagine an astronaut standing on an asteroid 2.5 A.U. distant from the Sun. He is gazing, naked-eye, towards the Sun.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Bt7TX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bt7TX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the apparent magnitude of the Sun at that distance, and is that bright enough to blot out any of the inner planets? No light blockers and solely viewing with the Eyeball Mark I?&lt;/p&gt;&#xA;" OwnerUserId="15814" LastEditorUserId="15814" LastEditDate="2017-03-21T20:00:20.103" LastActivityDate="2017-03-21T20:18:23.370" Title="Sun's apparent magnitude at 2.5 AU?" Tags="&lt;the-sun&gt;&lt;apparent-magnitude&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20488" PostTypeId="2" ParentId="20477" CreationDate="2017-03-21T18:29:55.907" Score="0" Body="&lt;p&gt;Just to add some detail to the above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you set the 21st of March and September as the equinox dates, which is mostly not done, but still &lt;a href=&quot;https://en.wikipedia.org/wiki/March_equinox#Calendars&quot; rel=&quot;nofollow noreferrer&quot;&gt;occasionally&lt;/a&gt;, and, also, not astrology but more, cultural, so probably not worth mentioning, but by the 21st or 22nd dates, then there's some shifting.   As others have said, the date and time shifts, the sun passing over the equinox point always happens, every year, pretty close to March and Sept 21.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Variation on the specific date and time changes not just because of orbital cycles but also, time-zones and leap-years. see &lt;a href=&quot;https://www.timeanddate.com/calendar/varying-march-equinox-date.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;on the Milankovich cycles&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changes in the Earth's &lt;a href=&quot;https://en.wikipedia.org/wiki/Milankovitch_cycles#Orbital_shape_.28eccentricity.29&quot; rel=&quot;nofollow noreferrer&quot;&gt;eccentricity cycle&lt;/a&gt; affects the inequality of length of the 4 seasons.    For now, Winter lasts 88.99 days, Spring, 92.75 days, Summer, 93.66 days and fall 89.85 days.  A year remains pretty much constant at 365.242 days but the length of the seasons varies slowly over the periods of two Milankovich cycles, eccentricity and axial precession.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The eccentricity cycle is the most inconsistent of the Milankovich cycles, having periods of 95,000, 125,000 and 400,000 years and it won't peak for another 150,000 or 200,000 years or so.   See &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/6/66/MilankovitchCyclesOrbitandCoresRecaptioned.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;chart&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a curious sidebar, the lengths of the months are somewhat adjusted to match this seasonal variation.   Spring and Summer both have 2 months with 31 days.   Winter has 2 but also 28 day February.  Fall has only 1 month with 31 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Milankovitch_cycles#Axial_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;Axial precession cycle&lt;/a&gt; that you asked about, the 25,771 year cycle, often abbreviated to 26,000, right now it's closely lined up so that if lands only a couple weeks off the winter solstice.  In 13,000 years or so it will line up near the summer solstice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means, 13,000 years from now, winter will be the longest season and summer the shortest.  The Spring Equinox will &quot;spring forward&quot; a few days and the fall equinox will fall back a few days.  This happens gradually, not all at once, but in 13,000 years, we might see the Spring equinox around March 25-27 and the fall around September 16-18, as a ballpark guess.    This variation will be even more pronounced as the Earth's eccentricity grows. which right now it's not, but in 180,000 or 200,000 years, there's a chance, we'll get spring equinox in April.   (Not making a guarantee, just saying there's a chance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as others have said, the equinoxes will always line up perfectly with the structures, in as much as the structures are correctly built.   But the dates when they line up will change, over Milankovich time, then change back.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tried to clean that up and remove the earlier errors.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-03-22T05:17:19.690" LastActivityDate="2017-03-22T05:17:19.690" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20489" PostTypeId="2" ParentId="20487" CreationDate="2017-03-21T20:12:26.107" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the apparent magnitude of the Sun at that distance?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fortunately, there's a formula known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Distance_modulus&quot; rel=&quot;noreferrer&quot;&gt;Distance Modulus&lt;/a&gt; which is defined to calculate precisely this information. The equation is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m-M=5\log_{10}(d_{pc}) - 5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_magnitude&quot; rel=&quot;noreferrer&quot;&gt;apparent magnitude&lt;/a&gt; of your star, $M$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Absolute_magnitude&quot; rel=&quot;noreferrer&quot;&gt;absolute magnitude&lt;/a&gt;, and $d$ is the distance from the star, in units of $\mathrm{parsecs}$. The absolute magnitude of the Sun is easily looked up and &lt;a href=&quot;https://www.google.com/search?q=absolute%20magnitude%20sun&amp;amp;ie=utf-8&amp;amp;oe=utf-8&quot; rel=&quot;noreferrer&quot;&gt;found to be&lt;/a&gt; $M_\odot = 4.83$. The distance, you provide as $2.5\:\mathrm{AU} = 1.2120342\times10^{-5}\:\mathrm{pc}$. Now its simple to solve for $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = 5\log_{10}(1.2120342\times10^{-5}) - 5 + 4.83$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = -24.75$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as apparent magnitudes go, this is still pretty bright. You can see a table of typical apparent magnitudes for comparison &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_magnitude#Table_of_notable_celestial_objects&quot; rel=&quot;noreferrer&quot;&gt;right here&lt;/a&gt;. Interestingly, this table lists $-25$ as the &quot;minimum brightness that causes the typical eye slight pain to look at&quot;. At $m=-24.75$ (remember higher value = less bright) you'd be just above this threshold and so the Sun would be safe to look at without causing eye strain or pain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is that bright enough to blot out any of the inner planets?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In order to determine the the brightness of the inner planets, we have to do some more math. Unfortunately, people don't really calculate the absolute magnitudes of planets so the form of the distance modulus listed above isn't entirely useful. Instead, we'll use a different form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_1 - m_2 = -5\log_{10}\left(\frac{d_2}{d_1}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation relates an object with an apparent magnitude of $m_1$ at a distance $d_1$ to its apparent magnitude $m_2$ at a distance $d_2$. If we look up the apparent magnitude of the various inner systems from Earth, and figure out their distance from Earth at those times, we can figure out the apparent magnitude at the distance of your asteroid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll work this out for Venus since Venus will likely be one of the brightest of the inner planets (given its high &lt;a href=&quot;https://en.wikipedia.org/wiki/Albedo&quot; rel=&quot;noreferrer&quot;&gt;albedo&lt;/a&gt;). Venus as viewed from Earth, at its brightest, has an apparent magnitude of $m_1 = −4.89$. This occurs when it is an the point of greatest elongation. As determined from this &lt;a href=&quot;https://www.fourmilab.ch/cgi-bin/Solar&quot; rel=&quot;noreferrer&quot;&gt;solar system orrery&lt;/a&gt;, the distance is approximately $d_1 = 0.75\:\mathrm{AU} = 0.36\times10^{-5}\:\mathrm{pc}$. On your asteroid, you'll be about $d_2 = 2.17\:\mathrm{AU} = 1.05\times10^{-5}\:\mathrm{pc}$ from Venus at the same point of maximum brightness (based on a little trig). From this we can calculate $m_2$ to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_2 = −4.89 + 5\log_{10}\left(\frac{1.05\times10^{-5}\:\mathrm{pc}}{0.36\times10^{-5}\:\mathrm{pc}}\right) = -2.57$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the comparison table listed above, this would make Venus, at its brightest, about as dim as the &quot;minimum brightness of the Moon&quot;. This is a rough calculation, but I don't expect the actual answer to vary too much from this value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind, that's the maximum brightness of one of the brightest inner planets. Any other time Venus will be dimmer and all other planets will never be brighter than this. So all in all, the inner planets will likely be outshined by the much brighter Sun. There's a chance, with the right conditions, if you managed to properly block the light from the Sun using some sort of occulting disk or &lt;a href=&quot;https://exoplanets.nasa.gov/resources/1015/&quot; rel=&quot;noreferrer&quot;&gt;flower starshade&lt;/a&gt;, that you could see the inner planets with the naked eye as they still would be pretty bright in their own right. You just need to get rid of the sunlight outshining them. &lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-03-21T20:12:26.107" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20490" PostTypeId="2" ParentId="20487" CreationDate="2017-03-21T20:18:23.370" Score="4" Body="&lt;p&gt;It's too bright to look at. Your astronaut has permanent retinal damage&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The apparent magnitude of the sun from Earth is -26.7. From the asteroid belt it is -24.7 (&lt;a href=&quot;http://www.1728.org/magntudj.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.1728.org/magntudj.htm&lt;/a&gt;). So it is dimmer, but it is still far too bright for her to look at without eye protection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since she is standing in the vacuum of space, there is no atmosphere to protect her eyes from ultraviolet. Her eyes would be damaged by the intensity of the light. Within a few days her vision would become fuzzy, from the production of free radicals from ultraviolet exposure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If she shaded her eyes from the sun, then the lack of atmosphere would help. With no atmosphere to scatter the sunlight, the sky around the sun would be dark, and the planets would be pretty bright. However, you specify &quot;no light blockers&quot; which means that our astronaut's eyes are basically toast.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-03-21T20:18:23.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20491" PostTypeId="1" AcceptedAnswerId="20492" CreationDate="2017-03-21T20:47:16.977" Score="1" ViewCount="73" Body="&lt;p&gt;I have a question about purchasing another lens that's right for my situation. I own the Orion Skyquest XT10 Dobsonian, a classic case of buying a scope that is WAY to big to be carrying around everywhere, and yet I manage to get it out periodically throughout the year. I am very amateur obviously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I enjoy using my stock 25mm but find the stock 10mm just too blurry. I do admit, I rarely get out of the neighborhood, the seeing is not great in my valley, there's city light, and I must admit I long for those hubble images popping out of my own lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that said, I just feel like I should be able to get some more clarity with a new lens. I'd really like to be able to see jupiter with some more detail, right now the best I get is a small orange blur with some minor bands visible if you use your imagination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What wouuld you recommend as the next lens in the $50-$200 dollar range that will provide a bit more detail than my 25mm but with good clarity?&lt;/p&gt;&#xA;" OwnerUserId="13272" LastActivityDate="2017-03-22T00:19:31.263" Title="What is the next lens I should buy?" Tags="&lt;telescope-lens&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2017-03-23T21:23:56.073" ContentLicense="CC BY-SA 3.0" />
  <row Id="20492" PostTypeId="2" ParentId="20491" CreationDate="2017-03-21T21:40:05.417" Score="4" Body="&lt;p&gt;It's a pretty common tendency to fix problems by buying more eyepieces, but that's rarely how it works in practice. There are a few steps you need to take before buying more glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Collimation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check collimation first. This should be your first item on the list, full stop. The performance of the scope decreases rapidly when it's out of collimation (for solar system objects). Check the documentation that came with your scope. Also check these links:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.skyandtelescope.com/astronomy-resources/how-to-align-your-newtonian-reflector-telescope/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.skyandtelescope.com/astronomy-resources/how-to-align-your-newtonian-reflector-telescope/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://garyseronik.com/a-beginners-guide-to-collimation/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://garyseronik.com/a-beginners-guide-to-collimation/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the scope is not in good collimation, a new eyepiece is a waste of money. All the eyepiece does is magnify the real image generated by the scope in the primary focal plane - so if that image is out of whack, then no eyepiece will give you the performance you're looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check collimation every time you use the scope. With practice, it will end up taking only a couple minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Seeing&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It varies with time and location. Sometimes it's better, other times it's bad. Go on the CDS site, search for your location and use the Seeing forecast; good seeing is represented by dark blue, bad seeing by light blue or white.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cleardarksky.com/csk/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.cleardarksky.com/csk/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plan your observations according to the forecast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing only affects fine-detail objects like the Moon or the planets. It does not affect faint DSOs (galaxies, nebulae, etc) because your eye does not resolve fine detail on DSOs anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thermal equilibrium&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the primary mirror is not at the same temperature as the air, &quot;local seeing&quot; conditions (vortices of air on the primary) will blur the image. Take the scope outside 1 hour before you start observing and let it acclimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a scope 8&quot; or bigger it's often a good idea to install a fan on the back of the primary to accelerate the process - but get everything else done first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-1/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-1/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Light pollution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only affects DSOs (galaxies, nebulae, clusters). It does not affect solar system objects such as planets. The only remedy is to load the scope up in a car and drive away from the city.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, all Messier objects are visible in most amateur telescopes even from very light-polluted places like major cities. They just don't look as spectacular as from a dark sky place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Finding focus&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure the scope is actually focused properly. This might be more tricky than it seems. Use a star nearby and adjust until the star is as sharp as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also make or buy a Bahtinov mask, which will give you proof positive of perfect focus (adjust focus with the mask on, until all 3 lines intersect each other in the center).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Bahtinov_mask&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Bahtinov_mask&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.astrozap.com/scripts/prodList.asp?idCategory=74&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.astrozap.com/scripts/prodList.asp?idCategory=74&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;With all of the above taken care of you should definitely see Jupiter's equatorial belts. They become visible even in 2&quot; of aperture, and a slightly larger instrument should show them pretty clearly, even in a modest eyepiece. Under good seeing, with a perfectly collimated instrument, you can even start noticing the curls and the vortices in the belts even with an aperture as low as 4&quot; ... 6&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter getting more and more blurry as you increase magnification is normal. It's not a sign that the eyepiece is bad. What's not normal is not seeing the belts in a 10&quot; scope. That indicates that either collimation is bad, or seeing is exceptionally bad, or both; typically it's collimation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your eyepieces should actually give decent performance in the center of the field of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need to let your brain-eye system adapt to the image. You don't have to observe from a very dark spot, Jupiter is bright enough to not require perfect dark adaptation. But your brain and your eye need a bit of time until they can extract all details from the image. Let it drift through the field, stay relaxed, follow the image and just take it in slowly. You will start seeing details after 3 or 4 minutes that were not visible at first (assuming good collimation, seeing, etc).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now on to the eyepiece. Your instrument is a f/4.7 newtonian, 1200 mm focal length. A 10 mm eyepiece will give you 120x magnification. This should not be too much even under bad seeing - provided the scope is collimated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've observed Jupiter a lot at 1200 mm f.l. in a smaller instrument. I've found that a 8.8 mm eyepiece (136x) was a reliable workhorse, never needed anything with less magnification. When seeing got better, I could push it up to 6.7 mm (180x) or even 4.7mm (255x). But speaking in general, Jupiter is low contrast and does not benefit from pushing the magnification too hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars OTOH does very well under very high magnification. But that assumes your collimation is top-notch and seeing is very good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Provided that you take care of collimation, etc first, you could try and get better eyepieces. The ones that came with your scope are not that bad actually, they do fine in the center, it's the edge performance that's weaker. But if you do get new eyepieces, you might as well get something better. I've found that the Explore Scientific 82 degree series provides a good bang for the buck - not too far from the performance of top glass, but much cheaper than top prices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://explorescientificusa.com/collections/82-series-eyepiece&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://explorescientificusa.com/collections/82-series-eyepiece&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the options here are very numerous, so do your own research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also benefit from having a barlow, maybe a 2x or a 2.5x. However, the much higher magnification generated with the barlow &lt;strong&gt;will not work&lt;/strong&gt; unless collimation, seeing, etc are all taken care of!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a little difficult when you juggle collimation and seeing at the same time, one you can control (collimation), the other you can't (seeing), but realize that this will take some time to figure out, so go one step at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also understand that this is not the kind of problem that is solved by simply throwing money at it.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2017-03-22T00:19:31.263" LastActivityDate="2017-03-22T00:19:31.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20493" PostTypeId="2" ParentId="20466" CreationDate="2017-03-22T03:27:50.873" Score="4" Body="&lt;p&gt;I do not know how it is in your country, but here in Russia, astronomical articles and news very often report astronomical distances in kilometers, million kilometers, billion kilometers, trillion kilometers etc. It is just we do not use units like gigameters, petameters and the like, but kilometer is the standard unit in astronomy.&lt;/p&gt;&#xA;" OwnerUserId="2541" LastEditorUserId="2541" LastEditDate="2017-03-23T12:16:02.280" LastActivityDate="2017-03-23T12:16:02.280" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20494" PostTypeId="1" AcceptedAnswerId="20495" CreationDate="2017-03-22T04:29:13.830" Score="12" ViewCount="9280" Body="&lt;p&gt;My understanding is $\sim$ generally means &quot;on the order of magnitude of&quot; e.g. $T \sim 10^5$ K&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\approx$ is obviously &quot;approximately equal to&quot; so for example one might write $d \approx 400$ pc rather than $d=4 \times 10^2$ pc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\simeq$ and $\cong$, are where I am more confused as their usage is rarer and less consistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The below IAU resolution, for example, seems to use $\simeq$ for truncation $m_{\mbox{bol}} = -26.83199\ldots \simeq -26.832$, but I have also seen $\simeq$ used as &quot;approximately equal to&quot; in other works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/pdf/1510.06262&quot; rel=&quot;noreferrer&quot;&gt;http://arxiv.org/pdf/1510.06262&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should one be using $\sim$, $\approx$, $\simeq$, and $\cong$?&lt;/p&gt;&#xA;" OwnerUserId="16345" LastEditorUserId="16345" LastEditDate="2017-03-22T06:07:04.323" LastActivityDate="2017-03-24T17:14:23.697" Title="Usage of $\sim$, $\approx$, $\simeq$, and $\cong$ in observational astronomy?" Tags="&lt;terminology&gt;&lt;notation&gt;&lt;symbols&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20495" PostTypeId="2" ParentId="20494" CreationDate="2017-03-22T05:57:29.057" Score="12" Body="&lt;p&gt;$\simeq$ and $\approx$ both mean &quot;approximately equal to&quot;. I don't think $\cong$ is used so often, but if I read it, I would interpret is as the same as the two others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sim$ &lt;em&gt;in principle&lt;/em&gt; means &quot;of the order of&quot;, i.e. correct to within an order of magnitude. However, astronomers tend to be rather slobby sometimes (we call all elements except H and He for &quot;metals&quot;, we call it &quot;gas&quot; even when it's ionized and should be called &quot;plasma&quot;, we omit factors of order unity if we can get away with it, we mix up terms like &quot;flux&quot;, &quot;flux density&quot;, and &quot;intensity&quot;, and so on), and thus $\sim$ is very often used to mean &quot;approximately equal to&quot;. It is even sometimes used as &quot;proportional to&quot;, but (usually) only when there's no risk of misunderstanding (e.g. $L_\mathrm{UV}\!~\sim\!~\mathrm{SFR}$, meaning that the UV luminosity is proportional to the star formation rate, omitting a factor of $10^{27}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Journals usually have rather strict standards for style and nomenclature, so they might not accept a $\sim$ for a $\propto$, or even for a $\simeq$. But if you read a conference proceeding or lecture notes or some other non-refereed text (even a paper on arXiv waiting to be refereed), you best be prepared for anything.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2017-03-22T05:57:29.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20496" PostTypeId="1" AcceptedAnswerId="20506" CreationDate="2017-03-22T08:28:58.240" Score="11" ViewCount="789" Body="&lt;p&gt;I'm writing a piece on Hawking radiation, and find I have something of a problem. The &quot;given&quot; explanation which I find on &lt;a href=&quot;https://en.wikipedia.org/wiki/Hawking_radiation#Overview&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; and elsewhere is unsatisfactory: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Physical insight into the process may be gained by imagining that particle–antiparticle radiation is emitted from just beyond the event horizon. This radiation does not come directly from the black hole itself, but rather is a result of virtual particles being &quot;boosted&quot; by the black hole's gravitation into becoming real particles&lt;sup&gt;[10]&lt;/sup&gt;. As the particle–antiparticle pair was produced by the black hole's gravitational energy, the escape of one of the particles lowers the mass of the black hole&lt;sup&gt;[11]&lt;/sup&gt;. An alternative view of the process is that vacuum fluctuations cause a particle–antiparticle pair to appear close to the event horizon of a black hole. One of the pair falls into the black hole while the other escapes. In order to preserve total energy, the particle that fell into the black hole must have had a negative energy...&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It relies on virtual particles and an negative-energy particles. However vacuum fluctuations are not the same thing as virtual particles, which &lt;a href=&quot;https://physics.stackexchange.com/a/185139/76162&quot;&gt;only exist in the mathematics of the model&lt;/a&gt;, and we know of know negative-energy particles. So I'm looking for a better explanation. The Wikpedia article also says this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;In another model, the process is a quantum tunnelling effect, whereby particle–antiparticle pairs will form from the vacuum, and one will tunnel outside the event horizon&lt;sup&gt;[10]&lt;/sup&gt;.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However that suggests pair production is occurring inside the event horizon, which seems to disregard the infinite gravitational time dilation, and that one of them a) appears outside of the event horizon and b) escapes as Hawking radiation when pair production typically involves the creation of an electron and a positron. Again it's unsatisfactory. So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there a better explanation of Hawking radiation?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9772" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2019-12-08T01:46:19.063" Title="Is there a better explanation of Hawking radiation?" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;general-relativity&gt;&lt;hawking-radiation&gt;&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="12" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20498" PostTypeId="2" ParentId="20494" CreationDate="2017-03-22T10:49:50.007" Score="1" Body="&lt;p&gt;Wikipedia (where would we be without it) has a &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_mathematical_symbols&quot; rel=&quot;nofollow noreferrer&quot;&gt;handy list of mathematical symbols&lt;/a&gt; which is pretty useful in this context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in particular :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\sim$ will crop up often as meaning &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Asymptotic_analysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;asymptotically equivalent&lt;/a&gt;&lt;/em&gt; which has a very specific meaning and should &lt;strong&gt;not&lt;/strong&gt; be used for as an approximate symbol - they're quite different and some mathematicians will lynch you if you call it approximately.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Last time I checked &lt;a href=&quot;https://www.youtube.com/watch?v=LYNOGk3ZjFM&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carl Bender's lectures on mathematical physics&lt;/a&gt; were available on YouTube if it's of interest to anyone and these include asymptotic analysis.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-03-22T10:49:50.007" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20499" PostTypeId="1" AcceptedAnswerId="20503" CreationDate="2017-03-22T12:21:39.960" Score="45" ViewCount="16063" Body="&lt;p&gt;Is there a stable geostationary orbit around the moon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My feeling is, that the orbit would collide with earth, because of the moon's slow rotation.&lt;/p&gt;&#xA;" OwnerUserId="16354" LastEditorUserId="16354" LastEditDate="2017-03-23T21:14:04.043" LastActivityDate="2019-09-25T03:39:13.567" Title="Is it possible to achieve a stable &quot;lunarstationary&quot; orbit around the moon?" Tags="&lt;orbit&gt;&lt;the-moon&gt;" AnswerCount="2" CommentCount="20" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20502" PostTypeId="2" ParentId="20499" CreationDate="2017-03-22T15:13:34.770" Score="56" Body="&lt;p&gt;As the answer by zephyr describes very well, there are very few stable orbits around the moon, and none of them are stationary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the moon is tidally locked to Earth. That means that all of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lagrangian_point&quot; rel=&quot;noreferrer&quot;&gt;Lagrangian points&lt;/a&gt; of the Earth-Moon system are stationary relative to the Moon surface.&lt;/p&gt;&#xA;" OwnerUserId="7261" LastEditorUserId="7261" LastEditDate="2017-03-22T15:19:25.153" LastActivityDate="2017-03-22T15:19:25.153" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="20503" PostTypeId="2" ParentId="20499" CreationDate="2017-03-22T15:14:32.860" Score="70" Body="&lt;p&gt;First off, such an orbit wouldn't be a &lt;em&gt;geostationary&lt;/em&gt; orbit since &lt;em&gt;geo-&lt;/em&gt; refers to the Earth. A more appropriate name would be &lt;em&gt;lunarstationary&lt;/em&gt; or &lt;em&gt;selenostationary&lt;/em&gt;. I'm not sure if there is an officially accepted term since you rarely hear people talk about such an orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can calculate the orbital distance of a selenostationary orbit using Kepler's law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a = \left(\frac{P^2GM_{\text{Moon}}}{4\pi^2}\right)^{1/3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, $a$ is your orbital distance of interest, $P$ is the orbital period (which we know to be 27.321 days or 2360534 seconds), $G$ is just the gravitational constant, and hopefully it is obvious that $M_{\text{Moon}}$ is the mass of the Moon. All we have to do is plug in numbers. I find that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a = 88,417\:\mathrm{km}=0.23\:\mathrm{Earth\mathit{-}Moon\:Distance}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I at least match your calculation pretty well. I think you were just relying on Wolfram Alpha a bit too much to get the units right. The units do work out right though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to determine if this orbit can exist however, you need to do a bit more work. As a first step, calculate the Moon's &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;noreferrer&quot;&gt;Hill Sphere&lt;/a&gt;. This is the radius at which the Moon still maintains control over it's satellite, without the Earth causing problems. The equation for this radius is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r \approx a_{\text{Moon}}(1-e_{\text{Moon}})\sqrt[3]{\frac{M_{\text{Moon}}}{3M_{\text{Earth}}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this equation, $a_{\text{Moon}} = 348,399\:\mathrm{km}$ is the Moon's semi-major axis around the Earth and $e_{\text{Moon}} = 0.0549$ is the Moon's orbital eccentricity. I'm sure you can figure out that the $M$'s are the masses of the respective bodies. Just plug and chug and you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r \approx 52,700\:\mathrm{km}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://articles.adsabs.harvard.edu//full/1964SvA.....7..618C/0000622.000.html&quot; rel=&quot;noreferrer&quot;&gt;more careful calculation&lt;/a&gt;, including the effects of the Sun is slightly more optimistic and provides a Hill radius of $r = 58,050\:\mathrm{km}$. In either case though, hopefully you can see that the radius for a selenostationary orbit is much farther than the Hill radius, meaning that no stable orbit can be achieved as it would be too much perturbed by the Earth and/or the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One final, semi-related point. It turns out almost no orbits around the Moon are stable, even if they're inside the Hill radius. This is primarily to do with &lt;a href=&quot;https://en.wikipedia.org/wiki/Mass_concentration_(astronomy)&quot; rel=&quot;noreferrer&quot;&gt;mass concentrations&lt;/a&gt; (or mascons) in the Moon's crust and mantle which make the gravitational field non-uniform and act to degrade orbits. There are only a handful of &quot;stable&quot; orbits and these are only achieved by orbiting in such a way as to miss passing over these mascons.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2017-03-22T15:56:03.350" LastActivityDate="2017-03-22T15:56:03.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20505" PostTypeId="1" AcceptedAnswerId="21431" CreationDate="2017-03-22T18:12:23.977" Score="3" ViewCount="853" Body="&lt;p&gt;In preperation for the upcoming solar eclipse I'd like to practice my photography to be able to take some (hopefully) good photos of the corona during totallity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the brightness of the sun's corona? How bright is it compared to a full moon (or other celestial phenomena)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At what point during the &quot;diamond ring&quot; does it become too bright to look at safely without safety equipment, and what is the relative brightness of the diamond?&lt;/p&gt;&#xA;" OwnerUserId="16360" LastActivityDate="2017-09-06T15:22:12.253" Title="Relative brightness of the sun's corona" Tags="&lt;the-sun&gt;&lt;solar-eclipse&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20506" PostTypeId="2" ParentId="20496" CreationDate="2017-03-22T19:19:05.243" Score="7" Body="&lt;p&gt;Andy Gould proposed a classical derivation of Hawking radiation in a somewhat &lt;a href=&quot;http://inspirehep.net/record/231344/files/slac-pub-4061.pdf&quot; rel=&quot;noreferrer&quot;&gt;obscure paper from 1987&lt;/a&gt;.  The essential argument is that a black hole must have a finite, non-zero entropy (otherwise you could violate the second law of thermodynamics with a black hole).  Moreover, the entropy of the black hole must depend only on its area (otherwise you could change the area of a black hole via the Penrose process and lower its entropy and make a perpetual motion machine).  If a black hole has an entropy and a mass, then it has a temperature.  If it has a temperature, then it must radiate thermally (otherwise you could again violate the second law of thermodynamics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you look at the Hawking radiation temperature, there's a Planck's constant in there, so it has to know something about quantum mechanics, right?  But it turns out that it's actually thermodynamics in general that knows about quantum mechanics, not general relativity --- Planck's constant is only needed to keep entropies finite (and therefore temperatures non-zero).  This is true of black holes and blackbodies alike.&lt;/p&gt;&#xA;" OwnerUserId="10371" LastActivityDate="2017-03-22T19:19:05.243" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20509" PostTypeId="1" AcceptedAnswerId="20511" CreationDate="2017-03-23T01:18:49.537" Score="0" ViewCount="423" Body="&lt;p&gt;I'm preparing for the solar eclipse coming in August of this year, and want to make a decision on whether it will be worth my while to travel north from San Francisco.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As observed from San Francisco, when compared to the May 2012 eclipse, will there be more of the sun obscured by the moon in 2017, or less?&lt;/p&gt;&#xA;" OwnerUserId="16366" LastActivityDate="2017-03-23T18:09:45.563" Title="Eclipse comparison: May 20 2012 vs Aug 21 2017 as observed from San Francisco, CA" Tags="&lt;solar-eclipse&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20510" PostTypeId="2" ParentId="20509" CreationDate="2017-03-23T06:52:47.987" Score="2" Body="&lt;p&gt;This is going to be the eclipse of the century for the USA. To make the most of it, you need to be on the path of totality (see map) to see a total eclipse. If you stay in San Francisco, you will only see a partial eclipse. You need to make arrangements &lt;strong&gt;now&lt;/strong&gt; if you need somewhere to stay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gray band indicates the path of totality. Inside it, you will see a total eclipse, and will be able to see the Sun's corona in all its glory. If you are outside the path of totality, you will only see a partial eclipse (and no corona). The closer you are to the blue centerline, the longer totality will last. The black dot near Nashville indicates the location where totality will last the longest (2 minutes and 40 seconds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During totality (but only totality), it is safe to observe the Sun's corona with the naked eye. At all other times, you will need to use a protective (solar) filter if you want to look at the Sun directly, or project an image of the Sun onto a screen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hoeGb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hoeGb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.eclipse2017.org/2017/maps.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;www.eclipse2017.org&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13887" LastEditorUserId="13887" LastEditDate="2017-03-23T18:09:45.563" LastActivityDate="2017-03-23T18:09:45.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20511" PostTypeId="2" ParentId="20509" CreationDate="2017-03-23T07:00:47.250" Score="1" Body="&lt;p&gt;Less of the sun will be obscured, the eclipse will be less total, from San Fransisco in 2017 than in 2012. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simulations can be found at &lt;a href=&quot;https://www.timeanddate.com/eclipse/in/usa/san-francisco&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.timeanddate.com/eclipse/in/usa/san-francisco&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to see the totality, you will need to head North or East.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-03-23T07:00:47.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20512" PostTypeId="2" ParentId="20466" CreationDate="2017-03-23T11:07:21.180" Score="9" Body="&lt;p&gt;Along with the other answers, there is one other reason, specifically when measuring the distances to other galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When stating the distance to other galaxies, Astronomers rarely ever state the distance in any unit of length, they tend to use redshifts (&lt;em&gt;z&lt;/em&gt;).  This unit is &lt;strong&gt;not&lt;/strong&gt; actually a unit of length (it is a dimensionless ratio of wavelengths), nor does it linearly convert to a distance (&lt;em&gt;z=2&lt;/em&gt; is &lt;strong&gt;not&lt;/strong&gt; twice as far as &lt;em&gt;z=1&lt;/em&gt;), nor is there an excepted conversion between redshift and distance (it depends on what model of the universe you assume).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Redshift is used because it can be very accurately measured.  There are features in a star or a galaxies spectra that we know the exact wavelength that they are emitted at and so the redshift can be calculated exactly by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;z=\frac{\lambda_{obs}}{\lambda_{em}}-1&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an observed, exact (within experimental error) property.  Converting this to a distance is confusing: are you talking about the distance the object is away from us &lt;em&gt;instantaneously now&lt;/em&gt;, or &lt;em&gt;instantaneously when the photon that you see was emitted&lt;/em&gt;, or &lt;em&gt;the distance the photon you see travelled&lt;/em&gt;?  Do you wish to take into consideration local movement as well as Hubble (universe) expansion?  Add on to this the shape of the universe, the rate of expansion of the universe, the rate of change of the expansion of the universe (dark energy/Hubble constants/other effects), and you see that any conversion to an actual distance is problematic and would require that you define exactly what type of conversion and with what assumptions.  It is easier to stay with the well-defined easy-to-measure redshift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good (degree-level) work that summarises all the different types of cosmological distances and their calculations is &lt;a href=&quot;https://arxiv.org/abs/astro-ph/9905116&quot; rel=&quot;noreferrer&quot;&gt;Hogg 2000&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16369" LastActivityDate="2017-03-23T11:07:21.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20514" PostTypeId="1" AcceptedAnswerId="20523" CreationDate="2017-03-23T20:17:36.160" Score="5" ViewCount="223" Body="&lt;p&gt;We think, more or less, that our Moon was formed from an accretion disk caused by a Mars-sized impactor. Cool. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tidal forces can break apart a moon, causing rings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ashes to ashes, dust to dust: the moon came from an orbiting debris disk and may return to an orbiting debris disk someday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What determines whether an accretion disk or ring system will form moons vs. remain as stable rings?&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="7965" LastActivityDate="2017-03-24T12:31:02.010" Title="What stabilizes rings or accretion disks?" Tags="&lt;the-moon&gt;&lt;planetary-formation&gt;&lt;planetary-ring&gt;&lt;orbital-mechanics&gt;&lt;accretion-discs&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20515" PostTypeId="1" CreationDate="2017-03-23T21:30:57.923" Score="2" ViewCount="1002" Body="&lt;p&gt;I have problems to understand the point of &quot;mean sun&quot; and &quot;true sun&quot;.&#xA;As I understood the mean sun is set in the center of the earth orbit (which is a circle). So Ptolemy, Copernicus and Brahe used the &quot;mean sun&quot;, however they knew that this isn't the real position of the sun? Did they do that due to the reason it would fit with the mars oppositions?*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*Three of them used circles to describe orbits of planets. But if they want to link that together with the observations it wouldn't fit.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that right? So even Ptolemy knew that the sun isn't in the middle of the orbit of the earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a website which I think describes it: &lt;a href=&quot;http://science.larouchepac.com/kepler/newastronomy/part1/5/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://science.larouchepac.com/kepler/newastronomy/part1/5/index.html&lt;/a&gt;&#xA;&lt;a href=&quot;http://science.larouchepac.com/kepler/newastronomy/part1/6/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://science.larouchepac.com/kepler/newastronomy/part1/6/index.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to my first language is not English it might be possible that I didn't understand some important aspects of the given links.&lt;/p&gt;&#xA;" OwnerUserId="15474" LastActivityDate="2017-05-24T00:04:45.477" Title="Mean and true Sun" Tags="&lt;the-sun&gt;&lt;orbit&gt;&lt;amateur-observing&gt;&lt;mars&gt;&lt;kepler&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20516" PostTypeId="2" ParentId="20515" CreationDate="2017-03-23T21:48:01.927" Score="2" Body="&lt;p&gt;Yes, Ptolemy adopted Hipparchus's idea that epicycles and eccentric orbits could explain the irregular motion of the Sun. From David McClung's biography of Hipparchus:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Hipparchus also created the first reasonably accurate model of the motion of the sun and moon. Using the idea of epicycles, probably introduced by the mathematician Apollonius of Perga, he was able to create a model that accurately predicted some key positions of the moon's motion. He created a similar model to predict the position of the sun. (See the Ptolemy page for more information on epicycles and its subsequent use in the planetary models.) His model didn't work for all cases, but it marked the real beginning of trying to create a predictive model that matched data.&lt;/p&gt;&#xA;&lt;p&gt;Even though the orbits are circular and going around the earth, the earth is not at the exact center of the orbits of the sun and moon. (They were eccentric orbits.) Hipparchus needed to use the epicycles and eccentric orbits because the observed motion of the sun around the ecliptic was not at a constant rate - in the winter the sun moved a little bit faster and in the summer the sun moved a little bit slower. Using an eccentric orbit for the sun and moon, and using small epicycles, Hipparchus gives the foundations of the classic Greek model of the solar system that approximately predicts the positions of the moon and sun that matched positions and predicted key dates and events pretty well. &lt;strong&gt;Eventually, Ptolemy takes this Hipparchan model and fully applies it to the motion of the planets - and eventually produces a full, working model of the solar system.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;sup&gt;emphasis added&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;An epicycle is like an orbit along an orbit:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/P82K2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P82K2.png&quot; alt=&quot;Epicycle&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;An eccentric orbit is an orbit where the inner body does not occupy the center point of the outer body's orbit.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.themcclungs.net/astronomy/people/hipparchus.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Historical Astronomy: Ancient Greeks: Hipparchus&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-03-24T22:43:57.043" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20517" PostTypeId="1" CreationDate="2017-03-23T21:54:39.717" Score="2" ViewCount="109" Body="&lt;p&gt;I'm asked to verify an expression for the ionization fraction of helium in a universe in which helium dominates baryonic matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm given that the ionization fraction $X = \frac{n_{He^+}}{n_{He^+}+n_{He}}$ with a given ionization energy $Q_{He}$, and number densities by the formula $n_i=g_i(\frac{m_ik_BT}{2\pi \hbar})^{3/2}\exp(-\frac{m_ic^2}{k_BT})$. Also the baryon-photon ratio in this case is $\eta=\frac{4(n_{He^+}+n_{He})}{n_\gamma}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to show that by putting all this together I get something with the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln(\frac{1-X}{X^2})=A+\ln(\eta)-B\,\ln(\frac{Q_{He}}{k_BT})+\frac{Q_{He}}{k_BT}$. where $A$ and $B$ are constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I plug everything in, I got $\frac{1-X}{X^2}=\frac{n_\gamma}{8}\eta \, \exp(-\frac{Q_{He}}{K_BT})$, which gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln(\frac{1-X}{X^2})=\ln(\frac{n_\gamma}{8})+\ln(\eta)+\frac{Q_{He}}{k_BT}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas of where I might have gone wrong?&lt;/p&gt;&#xA;" OwnerUserId="15870" LastEditorUserId="9301" LastEditDate="2017-03-25T06:23:30.783" LastActivityDate="2017-03-25T09:30:52.970" Title="Ionization Fraction of Helium in alternate, He dominated early universe" Tags="&lt;cosmology&gt;&lt;early-universe&gt;&lt;nucleosynthesis&gt;&lt;helium&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20519" PostTypeId="2" ParentId="20494" CreationDate="2017-03-23T23:44:01.853" Score="2" Body="&lt;p&gt;Recently, some in the astronomy literature are adopting $\sim$ in the statistics sense to mean &quot;is sampled from&quot; or &quot;has the distribution of&quot; (&lt;a href=&quot;http://mathworld.wolfram.com/Tilde.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;the last definition on mathworld&lt;/a&gt;). For example, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x \sim p(x)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;would read &quot;$x$ is sampled from the distribution $p(x)$&quot;.&lt;/p&gt;&#xA;" OwnerUserId="215" LastActivityDate="2017-03-23T23:44:01.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20520" PostTypeId="1" AcceptedAnswerId="20522" CreationDate="2017-03-24T04:19:25.127" Score="5" ViewCount="130" Body="&lt;p&gt;Lord Kelvin's prediction of heat death indicates the eventual end to star formation in the observable universe. Have any astrophysicists made any predictions for the endurance of star formation in the observable universe? &lt;/p&gt;&#xA;" OwnerUserId="14131" LastEditorUserId="14131" LastEditDate="2017-03-24T11:47:22.823" LastActivityDate="2017-03-24T11:47:22.823" Title="How long will star formation endure before the eventual heat death of the observable universe?" Tags="&lt;star-formation&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20521" PostTypeId="1" CreationDate="2017-03-24T11:00:40.857" Score="5" ViewCount="166" Body="&lt;p&gt;In Chapter 58 in Astronomia Nova, Kepler made his final step to finally create the ellipse (for the orbit of planets) &#xA;&lt;a href=&quot;http://science.larouchepac.com/kepler/newastronomy/part4/58/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://science.larouchepac.com/kepler/newastronomy/part4/58/index.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the given source there is written: &#xA;&quot;Dissatisfied with so large an error, Kepler writes that he &lt;strong&gt;rejected his hypothesis of chapter 56&lt;/strong&gt;, which &lt;em&gt;was supported by the distances of chapters 51&lt;/em&gt; and 53 as well as the physical arguments of chapter 57&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For his hypothesis of chapter 56 he made use of the distances of chapters 51 and 53. And now, he rejects this hypothesis. Does that mean the distances he calculated were wrong? Or did he just use these (correct) datas for a wrong suggestion for orbits?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you can see the determination of the distances:&#xA;&lt;a href=&quot;http://science.larouchepac.com/kepler/newastronomy/part4/51/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://science.larouchepac.com/kepler/newastronomy/part4/51/index.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15474" LastEditorUserId="9527" LastEditDate="2020-05-26T06:51:13.923" LastActivityDate="2020-05-26T06:51:13.923" Title="Kepler - Creating the Ellipse (Astronomia Nova)" Tags="&lt;orbit&gt;&lt;planet&gt;&lt;mars&gt;&lt;history&gt;&lt;johannes-kepler&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20522" PostTypeId="2" ParentId="20520" CreationDate="2017-03-24T11:45:22.510" Score="6" Body="&lt;p&gt;This is actually very well researched. Wikipedia &lt;a href=&quot;https://en.wikipedia.org/wiki/Future_of_an_expanding_universe&quot; rel=&quot;noreferrer&quot;&gt;has quite a good summary&lt;/a&gt;, stelliferous era ends after about $10^{14}$ years, assuming an expanding universe.&lt;/p&gt;&#xA;" OwnerUserId="16392" LastActivityDate="2017-03-24T11:45:22.510" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20523" PostTypeId="2" ParentId="20514" CreationDate="2017-03-24T12:31:02.010" Score="4" Body="&lt;p&gt;I believe the two biggest things that determine whether or not a ring system forms into a moon depends on are the parent planet's &lt;a href=&quot;https://en.wikipedia.org/wiki/Roche_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Roche Limit&lt;/a&gt; and whether or not there are other moons that can disrupt the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Roche_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;roche limit&lt;/a&gt; of a planet is how close a large body can orbit the planet without being torn apart by the planet's gravitational pull. The article I've referenced can show how this is calculated. Small moons like Jupiter's Metis can orbit within the roche limit as they are so small, the difference between the tidal forces acting on their near and far sides is not large enough to tug the moon apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The presence of other moons - or in the case of stellar disks, other planets - will play a large role. Like &lt;a href=&quot;http://www.cosmosup.com/how-did-the-asteroid-belt-form/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; explains about our asteroid belt, the gravitational perturbations of Jupiter prevented a planet forming between it and Mars. Smaller bodies cannot easily gather together when a beast like Jupiter swings by to scatter them all!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also means though that most asteroids within the belt are confined between Mars and Jupiter. Around Saturn's F ring, there are two small moons that slow down / speed up debris within the ring so that it remains gathered in a small space. This does keep rings and disks stable, but it also stops larger bodies forming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long story short; if the disk if far enough away from its parent, and nothing else large is orbiting nearby, you should get a moon / planet. If it's too close to the parent and being shepherded by another large body, it's more likely to remain a stable disk.&lt;/p&gt;&#xA;" OwnerDisplayName="user10106" LastActivityDate="2017-03-24T12:31:02.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20524" PostTypeId="2" ParentId="20494" CreationDate="2017-03-24T17:14:23.697" Score="1" Body="&lt;p&gt;As an (ex) professional astronomer, I can tell you that there are no rules that are followed.  There are very very few exact measurements in Astronomy, &lt;em&gt;x&lt;/em&gt; rarely technically &lt;em&gt;equals&lt;/em&gt;  anything.  In general, $\sim$ means less accurate than $\approx$ which means less accurate than $=$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematicians (and Chemists) frequently complain about our loose use of mathematics (and chemistry).&lt;/p&gt;&#xA;" OwnerUserId="16369" LastActivityDate="2017-03-24T17:14:23.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20526" PostTypeId="2" ParentId="20278" CreationDate="2017-03-24T23:46:31.540" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;When we look at the Polaris from the earth, the Big Bear appears to be revolving around it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ursa Major is relatively close to Polaris in angular distance, so this motion is easier to see.  But in fact &lt;em&gt;all&lt;/em&gt; the constellations appear to revolve around the axis.  The closer they are, the tighter the arc and the easier this motion is to see.  If they are far away and you are not at very high latitudes, they will be obscured at some point by the horizon and the motion is hidden.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for your question, look at a starmap and the closer a constellation is to the future axis, the more it will behave like Ursa Major today.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2017-03-24T23:46:31.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20528" PostTypeId="2" ParentId="20278" CreationDate="2017-03-25T04:27:28.790" Score="2" Body="&lt;p&gt;You can easily determine this by looking at a star chart or by using free software like &lt;strong&gt;Stellarium&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are two pictures that I obtained from Stellarium....they are centered on Thuban and Vega. &lt;a href=&quot;https://i.stack.imgur.com/fcSiV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fcSiV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z9aOx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z9aOx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16405" LastActivityDate="2017-03-25T04:27:28.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20529" PostTypeId="2" ParentId="20517" CreationDate="2017-03-25T09:30:52.970" Score="1" Body="&lt;p&gt;My answer is incomplete (but still possibly helpful, only posting because I don't have enough &quot;points&quot; to comment).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did a &lt;a href=&quot;https://i.imgur.com/bcDMOq8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;bit of the math involved&lt;/a&gt; but did not substitute terms with n's and eta's. However, from the result at the bottom of that page, it looks to me like a &lt;a href=&quot;https://www.wolframalpha.com/input/?i=binomial+expansion+ln(+a%2Fb+%2B+(a%2Fb)%5E2)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Binomial Expansion&lt;/a&gt; (truncated Taylor Expansion). The link shows the highest order for your terms (which is okay since higher order terms can usually be regarded as negligible) with the same +/- pattern you expect in your answer.&lt;/p&gt;&#xA;" OwnerDisplayName="user15317" LastActivityDate="2017-03-25T09:30:52.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20530" PostTypeId="1" CreationDate="2017-03-25T10:35:44.650" Score="7" ViewCount="255" Body="&lt;p&gt;The supermassive black hole in the center of the Milky Way is blocked by dust and gas clouds as seen from where we are. Which wavelengths are blocked and which are most undisturbed for observation of Sagittarius A* and its immediate neighborhood? Does the matter between us change for example polarization at some wavelengths or cause disturbances other than blocking the light? Could galactic cosmic rays, heavy ions, from it be detected if they were emitted, or would they be deflected by magnetic fields?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, at what distance from us is the most dust located? Is it concentrated in one region or are there several dust clouds in the way?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="2153" LastEditDate="2017-03-25T18:01:23.137" LastActivityDate="2018-12-07T17:49:32.060" Title="At what wavelengths can black hole Sagittarius A* be observed from Earth?" Tags="&lt;observational-astronomy&gt;&lt;dust&gt;&lt;wavelength&gt;&lt;extinction&gt;&lt;sagittarius-a&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20531" PostTypeId="1" AcceptedAnswerId="20538" CreationDate="2017-03-25T13:07:31.077" Score="4" ViewCount="206" Body="&lt;p&gt;Suppose those people engage in similar activity, walking on surface etc.&lt;/p&gt;&#xA;" OwnerUserId="2541" LastActivityDate="2017-03-27T20:23:25.657" Title="If Earth had no atmosphere, how many people would be killed by meteors every year?" Tags="&lt;meteor&gt;&lt;meteor-shower&gt;&lt;meteoroid&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2017-03-28T13:40:06.607" ContentLicense="CC BY-SA 3.0" />
  <row Id="20532" PostTypeId="1" CreationDate="2017-03-25T13:19:56.577" Score="1" ViewCount="152" Body="&lt;p&gt;Dwarf stars have terrestrial sized planets orbiting in habitable zones very close to them. These exoplanets are often said to be tidally locked to their star, like the Moon is to Earth, and that they thus have one hot hemisphere and one cold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the Solar system there's only one tidally locked planet, and that is Mercury which is asynchronously locked in a 2:3 relation. It does rotate relative to the sun anyway. And only one known planet which practically doesn't rotate at all, Venus. But Venus has a super-rotating atmosphere that distributes heat from insolation all around anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there reasons to believe that asynchronous tidal locking and super-rotating atmospheres are less common for planets with a dwarf stars than in a planetary systems with a Solar like star?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2017-03-26T23:58:17.897" Title="Are exoplanets at dwarf stars less likely to have super-rotating atmospheres or asynchronous tidal locking?" Tags="&lt;exoplanet&gt;&lt;rotation&gt;&lt;planetary-atmosphere&gt;&lt;tidal-locking&gt;&lt;dwarf-star&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20533" PostTypeId="2" ParentId="20530" CreationDate="2017-03-25T14:34:45.513" Score="5" Body="&lt;p&gt;From &lt;a href=&quot;http://adsabs.harvard.edu/abs/2010RvMP...82.3121G&quot; rel=&quot;nofollow noreferrer&quot;&gt;Genzel et al. (2010)&lt;/a&gt;, here's part of Fig. 7.7.1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qyVgM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qyVgM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is part of the spectral energy distribution of Sagittarius A*, a flot of &lt;span class=&quot;math-container&quot;&gt;$\nu$&lt;/span&gt; (frequency) vs. &lt;span class=&quot;math-container&quot;&gt;$\nu L_{\nu}$&lt;/span&gt; (frequency times luminosity). For comparison, visible light is in wavelengths from &lt;span class=&quot;math-container&quot;&gt;$\sim4\times10^{14}\text{ Hz}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\sim8\times10^{14}\text{ Hz}$&lt;/span&gt;, which happens to be around the bottom of the trough of nonthermal electron emission. Right away, this makes submillimeter and millimeter wavelengths good candidates, leading to &lt;a href=&quot;http://www.eventhorizontelescope.org/docs/AUS%20261%20271F.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;studies using Very Long Baseline Interferometry&lt;/a&gt;. Likewise, infrared emissions are a good target, and so &lt;a href=&quot;http://www.spitzer.caltech.edu/news/1892-feature16-11-Seeing-the-Milky-Way-s-Giant-Black-Hole-with-New-Eyes&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Spitzer Space Telescope, for instance, has been used&lt;/a&gt;. X-ray flares of up to &lt;span class=&quot;math-container&quot;&gt;$\sim10^{36}\text{ erg/s}$&lt;/span&gt; also occur from time to time,&lt;sup&gt;1&lt;/sup&gt; so that part of the spectrum is sometimes used to observe this activity. Finally, of course, Sagittarius A* is a very strong radio source, and it was initially observed in radio wavelengths (and it still is!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &lt;a href=&quot;http://adsabs.harvard.edu/abs/2010IAUS..261..271F&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fish &amp;amp; Doeleman 2010&lt;/a&gt; write,&lt;sup&gt;2&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Interstellar scattering, which varies as &lt;span class=&quot;math-container&quot;&gt;$\lambda^2$&lt;/span&gt;, dominates over intrinsic source structure at longer wavelengths, and the emission from Sgr A* transitions from optically thick to optically thin near &lt;span class=&quot;math-container&quot;&gt;$\lambda = 1\text{ mm}$&lt;/span&gt; (Doeleman et al. 2001). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This means that the visible part of the spectrum is dimmed even more. Combine that with relatively low emission at those wavelengths, and you have a pretty poor target for optical telescopes, and a pretty good target (all other factors considering) at other wavelengths, especially near the emission peak.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; Flares also are visible in other wavelengths, but I mention X-rays here because Sagittarius A* is typically much less luminous in X-rays.&#xA;&lt;br&gt;&#xA;&lt;sup&gt;2&lt;/sup&gt; I &lt;em&gt;believe&lt;/em&gt; they mean &lt;span class=&quot;math-container&quot;&gt;$\lambda^{-2}$&lt;/span&gt;. Power laws for scattering in the area vary, but they are generally given between as &lt;span class=&quot;math-container&quot;&gt;$\lambda^{-1.5}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\lambda^{-2}$&lt;/span&gt;. At any rate, the index may differ over different regimes.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2018-12-07T17:49:32.060" LastActivityDate="2018-12-07T17:49:32.060" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="20534" PostTypeId="2" ParentId="20466" CreationDate="2017-03-25T18:39:25.687" Score="2" Body="&lt;p&gt;&lt;strong&gt;Astronomers don't and cannot &lt;em&gt;measure&lt;/em&gt; distances.&lt;/strong&gt; Distances are merely inferred from what actually has been measured, such as an angle, a relative luminosity, a time period, etc.. Most astronomical distance determinations ultimately hinge on the Earth-Sun distance (astronomical unit), which therefore is of fundamental importance (and only in modern time is known with good accuracy). For nearby stars, the parallax angle is directly related to the distance, but the distance inferred from that is not a proper measured  distance: its uncertainty is not normally distributed (think about a negative parallax measurement).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomers know, of course, how many meters a parsec is, and know that using meters for galactic distances is only confusing, because you have to make sure you get the correct number of 0000 all the time (or the correct power of ten).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, unlike particle physics, astronomy as a science predates the meter system, at least its wider use. Changing from a well working system to something else only for the sake of conformity with SI, but for the price of inconvenience and confusion seems a stupid idea.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2017-03-25T18:46:00.967" LastActivityDate="2017-03-25T18:46:00.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20535" PostTypeId="2" ParentId="20532" CreationDate="2017-03-25T18:53:13.837" Score="0" Body="&lt;p&gt;Mercury is in a spin-orbit resonance other than 1:1, hence is  &lt;strong&gt;not tidally locked&lt;/strong&gt;. This can occur in eccentric orbits and when the tides are weak (so that the orbit remains eccentric). The recently discovered system of planets (Trappist-1) has 7 planets in orbits with very small eccentricity, so the Situation as for Mercury will not occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; See also &lt;a href=&quot;https://astronomy.stackexchange.com/questions/20307/how-is-mercury-tidally-locked-if-the-ratio-is-not-11/20323#20323&quot;&gt;this recent question&lt;/a&gt; regarding the spin-orbit locking. The eccentricity and spin-orbit evolution are closely coupled (because of conservation of total angular momentum), but the time scale for the latter is much shorter. Therefore, the ratio between spin and orbital frequency quickly reaches an equilibrium (which depends on the eccentricity), but if the tides are weak the eccentricity may not change and the system hardly changes. This is the situation for Mercury.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2017-03-26T15:34:24.613" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20538" PostTypeId="2" ParentId="20531" CreationDate="2017-03-25T21:27:14.803" Score="10" Body="&lt;p&gt;Three millions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this answer to be any fun, let's first assume that before removing the atmosphere, we have equipped all humans with oxygen supplies. In return, they are not allowed to go indoors or otherwise protect themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/I2fig.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I2fig.png&quot; alt=&quot;oxygen&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of meteors — as well as most other things in the Universe — has a distribution that is well described by a power-law. The larger the meteor, the rarer they are.&#xA;Small meteors can be lethal only by a direct hit, while large meteors &lt;em&gt;in the real world&lt;/em&gt; can be lethal due to the effect they have on the environment; dust can cover the sky for years and prevent the Sun from keeping us warm. However, since we have removed the atmosphere, the debris will fall down immediately, and the only way to kill us is by the explosion. But, as will be shown below, the dangerous ones are the small ones because the rate at which large ones falls off so quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number $N$ of meteors entering Earth's atmosphere every year with a diameter $D$ (measured in meters) or larger is given by (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2002Natur.420..294B&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brown et al. 2002&lt;/a&gt;)&#xA;$$&#xA;\frac{N(&amp;gt;D)}{\mathrm{year}} \simeq 37 \left( \frac{D}{\mathrm{m}} \right)^{-2.7}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;strong&gt;Minimum size needed to be lethal&lt;/strong&gt;&#xA;&#xA;&lt;p&gt;What makes a meteor lethal is really outside astronomy, and someone may be able to come up with a better answer, but for the sake of the argument, let's assume that the energy needed to kill a human by a direct hit is equal to the energy of a bullet fired from the world's smallest firearm, the &lt;a href=&quot;https://en.wikipedia.org/wiki/2mm_Kolibri&quot; rel=&quot;nofollow noreferrer&quot;&gt;2mm Kolibri&lt;/a&gt;, which has a muzzle energy of $E_\mathrm{lethal} \simeq 4\,\mathrm{J}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth orbits the Sun at 30 km/s. Depending on the direction from where the meteor arrives, it will have different speeds, but it will at least enter with the escape velocity of Earth, i.e. 11 km/s, and a typical speed is $v_\mathrm{impact} = 17\,\mathrm{km/s}$ (&lt;a href=&quot;http://impact.ese.ic.ac.uk/ImpactEffects/effects.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Marcus et al. 2010&lt;/a&gt;). Thus, the minimum mass to kill by a direct hit is as small as&#xA;$$&#xA;m_\mathrm{lethal} = 2 \, E_\mathrm{lethal} \, / \, v_\mathrm{impact}^2 \simeq 30 \,\mu\mathrm{g}.&#xA;$$&#xA;At an &lt;a href=&quot;http://meteorites.wustl.edu/id/density.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;average density of 4 g/cm$^3$&lt;/a&gt;, this means that a small grain of a diameter of 0.2 mm will kill you. According to the formula above, the number of such grains (or larger) that hit Earth every year is $2\times10^{11}$.&#xA;Since the surface area of Earth is 510 million $\mathrm{km}^2$, that's $n_\mathrm{hits}\sim5\times10^{-4}\,\mathrm{hits}\,\mathrm{m}^{-2}\,\mathrm{year}^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some 7 billion people live on Earth. They each occupy roughly $1\,\mathrm{m}^2$ (and weren't allowed to protect themselves, remember), so in total they occupy an area of $A_\mathrm{folks,tot} = 7 \times 10^9\, \mathrm{m}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the number of people being killed by direct hit meteors every year is&#xA;$$&#xA;N_\mathrm{kill} = n_\mathrm{hits} \, \times \, A_\mathrm{folks,tot}&#xA; = \mathbf{3\,million\,year^{-1}},&#xA;$$&#xA;or one person every 10 seconds.&lt;/p&gt;&#xA;&#xA;&lt;strong&gt;Death by explosion&lt;/strong&gt;&#xA;&#xA;&lt;p&gt;How about non-direct hits? Let's assume that the minimum size to kill you would have an energy equal to that of a hand grenade holding 100 g explosives, and that you're killed if it hits within a few meters of you, so that you &quot;occupy&quot; 10 m$^2$. Similar to above, you can work out that this corresponds to a meteor of ~1 cm being lethal, and that the rate at which such meteors is only $2\times10^4$ per year, so only 2-3 people would be killed in this way. Even larger explosions are even more rare. Thus, the probability of killed by a explosion is negligible compared to direct hits, so the final answer would be the above, i.e. 3 million per year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conclusion: Let's not remove the atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2017-03-27T20:23:25.657" LastActivityDate="2017-03-27T20:23:25.657" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20541" PostTypeId="2" ParentId="20532" CreationDate="2017-03-26T12:23:49.217" Score="3" Body="&lt;p&gt;I'm not exactly sure what you're asking, cause you touch on a few related points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason planets around red-dwarf stars are thought to be tidally locked is because the tidal force is comparatively much greater for the habitable zone.   Take, for example, a star with half the mass of our sun.   It would have, roughly speaking, 1/16th the solar output, so a planet, to get the same amount of heat, would need to be 4 times closer or 1/4 AU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tidal force increases by the cube of the distance, or, 4^3, or 64, and the mass, 1/2, means that for just 1/2 the solar mass, the tidal force on a planet getting equal heat is 32 times greater.   That's a ballpark ratio to the 5th power, which is highly significant.    When you get into solar masses 25% the mass of the sun or less, the tidal force for a habitable zone planet can be thousands of times greater than the tidal force for the habitable zone in our solar system.   Because of the high tidal forces, planets in the habitable zone of small stars are very likely tidally locked, at least for any near-circular orbits.  That's pretty darn close to certain when the star is below a certain size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some theories that thermal heating of a planet's atmosphere, which takes time, so it doesn't happen precisely at mid-day, but a couple hours after that, there's expected to be a tidal torque in the upper atmosphere and an equilibrium where the constant rotation of the atmosphere drags on the surface and keeps the planet rotating, so the planet never becomes tidally locked.    Venus might be an example of that.    But the articles below suggest that this is predicted to only happen in the larger red-dwarfs, 50% the mass of the sun or larger.  Article &lt;a href=&quot;http://physicsworld.com/cws/article/news/2015/jan/15/exoplanets-could-avoid-tidal-locking-if-they-have-atmospheres&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://arstechnica.com/science/2015/01/thin-atmosphere-is-enough-to-keep-many-exoplanets-spinning/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for more details.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, for smaller red dwarf stars, tidal locking is probably very common within the habitable zone.   For larger red dwarf stars, you might get a mix, growing less common for further out planets.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotations maintained by wind torque are probably, at least, to my thinking, not very rapidly rotating, similar to Venus, a slow rotation makes super-rotating trade winds easy.  (I say trade winds because I don't think weather-wind or local wind is considered super-rotating), so I think we should just consider relatively permanent wind speed and direction).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for any slowly rotating planet, Super-rotating wind should be quite common.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the very smallest stars, like Trappist 1, which at .08 solar masses isn't far off from the minimum size for a red dwarf, it's inner most planet has a period of just 1.5 days, it's corresponding rotation (Sidereal, not solar) has a 1.5 day period too.   That's probably a fast enough rotation to generate a significant Coriolis effect and some interesting weather (provided the planet hasn't lost it's atmosphere - which is also possible with close orbits around small stars).   That's an equatorial surface velocity of over 600 km/h, so for very close orbit planet around the smallest red dwarfs, super-rotating winds may not happen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the &lt;a href=&quot;https://en.wikipedia.org/wiki/TRAPPIST-1#Planetary_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;planets orbital periods around Trappist 1&lt;/a&gt;.  The inner 2-3 planets might rotate too fast to have super-rotating winds.   The 5th one, for example, with an orbital period of 9 days, corresponding to a sidereal rotating speed at the equator of not much more than 100 km/h, it's probably likely to have super-rotating winds.  The longer the orbital period, the more likely super-rotating winds are to happen.  Any tidally locked planet with an orbital period over 10 days, should be statistically likely to have super-rotating winds (compared to sidereal rotation).   Solar rotation, well, by that measurement, all tidally locked planets with atmosphere have super-rotating winds.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-03-26T23:58:17.897" LastActivityDate="2017-03-26T23:58:17.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20542" PostTypeId="1" AcceptedAnswerId="20545" CreationDate="2017-03-26T23:42:25.253" Score="5" ViewCount="155" Body="&lt;p&gt;I'm not sure for how many planetary systems we have a fairly good idea of the inclination and &quot;argument of periapsis&quot; of the orbital plane. I'm assuming there are a number of them that we can estimate due to co-planar assumptions in a multi-planet system or some other means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asking if there are enough &quot;known&quot; orbital planes that we can have some statistical idea whether or not the orbital planes are somewhat aligned with the galactic plane or whether local forces would dominate galactic orbit angular momentum. (I'm thinking the latter is true.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would have to keep in mind of course that a system &quot;directly above&quot; us and parallel to the galactic plane would be &quot;face on&quot;.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2017-03-27T14:02:48.637" Title="Is there a preference for orbital planes of planetary systems to be aligned with the galactic disk?" Tags="&lt;orbital-elements&gt;&lt;planetary-systems&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20545" PostTypeId="2" ParentId="20542" CreationDate="2017-03-27T14:02:48.637" Score="4" Body="&lt;p&gt;Unless I'm missing something we know very little about the orbital planes of individual systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most exoplanetary systems (all of the transiting ones, whereas it is just a strong bias for the doppler-discovered planets) are discovered because they have high orbital inclination. This means that the planet's orbit takes it across the stellar disc as we view it. However, the entire arrangement can be rotated around the line of sight by any angle and this would give exactly the same observational signature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being the case, the only test I can immediately suggest is that &lt;em&gt;if&lt;/em&gt; the planes of planetary systems were preferentially aligned with the galactic plane (ours isn't), then it would result in a lower planetary detection rate for stars in directions perpendicular to the Galactic plane in comparison with stars lying in the Galactic plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be some mileage in pursuing this, though I don't know of any results. For example the main Kepler field was directed just out of the Galactic plane, but the K2 fields are at a variety of Galactic latitudes along the ecliptic plane. If one could normalise somehow for the types of stars observed in each field and the differing data qualities then it might be possible to look at the &quot;planetary occurrence rate&quot; as a function of Galactic latitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;all-sky&quot; ground-based surveys are less suitable for this because they tend to avoid the Galactic plane because it is too crowded with targets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question can be more broadly addressed by looking at &lt;em&gt;visual&lt;/em&gt; binary systems - i.e. those where you can resolve the components and study the orbit. There, it &lt;em&gt;is&lt;/em&gt; possible to say something about the orbital planes of individual systems. A paper by &lt;a href=&quot;https://arxiv.org/abs/1411.4919&quot; rel=&quot;nofollow noreferrer&quot;&gt;Agati et al. (2015)&lt;/a&gt; has looked at this. They collected 51 systems with the requisite information, that are close to the Sun. They found only weak evidence for any anisotropy of the orbital poles and this weak alignment was closer to the ecliptic pole than the Galactic pole. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-03-27T14:02:48.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20549" PostTypeId="1" CreationDate="2017-03-28T03:29:49.023" Score="8" ViewCount="510" Body="&lt;p&gt;The evolutionary path of the Sun has been described in some detail, and, subtle differences aside, it's been described as such for decades - the Earth-engulfing red giant stage, the helium flash, etc.. It would seem that the knowledge necessary to make such a prediction has been around since the 60s, if not earlier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we know it will happen, and on the timescales described? If we make some measurements/estimations of the mass and composition of the Sun, and use our understanding of nuclear physics - energy of fusion and so forth, and an understanding of Newtonian physics (or do we have to invoke Einstein?), do the predictions just inevitably fall out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could the predictions be wrong? Is there any significant measure of uncertainty, perhaps due to sensitivity of certain calculations to initial conditions, etc?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the red giant future of our Sun a firm conclusion, or just a likely possibility among a range of others and commonly talked about merely because it seems the most likely one?&lt;/p&gt;&#xA;" OwnerUserId="1983" LastActivityDate="2017-03-29T01:34:24.313" Title="How do we know how the Sun (or any star, for that matter) will evolve?" Tags="&lt;the-sun&gt;&lt;stellar-evolution&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20550" PostTypeId="1" AcceptedAnswerId="20581" CreationDate="2017-03-28T06:48:11.787" Score="3" ViewCount="219" Body="&lt;p&gt;Problem: There is an satellite that orbits the Earth. Its orbit is circular and its orbital plane is perpendicular to Earth's Equatorial Plane. It orbits is such that a person who is on equator see this satellite on Zenith every 12Hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. Find Satellite's Period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. What is distance of satellite to person when it Sets on person's horizon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem with this question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first part: somewhere said period is 12hours. Somewhere else said the person on equator also moves (Earth Rotation) so 12 hours is P/2 and P= 24hours. What is the right answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part: Which of the following figures show the position of satellite when it sets? (Figure 1 or 2?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HOBaw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HOBaw.png&quot; alt=&quot;figure 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AkzjR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AkzjR.png&quot; alt=&quot;figure 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16441" LastActivityDate="2017-07-02T06:43:02.470" Title="Problem about orbiting of a satelite" Tags="&lt;orbit&gt;&lt;earth&gt;&lt;satellite&gt;&lt;orbital-mechanics&gt;&lt;artificial-satellite&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20551" PostTypeId="2" ParentId="20549" CreationDate="2017-03-28T08:53:33.113" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the red giant future of our Sun a firm conclusion, or just a likely possibility among a range of others and commonly talked about merely because it seems the most likely one?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ignoring the minute possibility of something like a neutron star colliding with our Sun sometime during the next few billion years, yes, it's an inevitability. There's some debate about borderline cases such as low metallicity stars (&quot;metal&quot; to an astronomer means anything but hydrogen or helium), and low or high mass stars. Our Sun does not fall into these borderline cases.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-03-28T08:53:33.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20553" PostTypeId="2" ParentId="20549" CreationDate="2017-03-28T13:00:27.080" Score="10" Body="&lt;p&gt;The thing you are not taking into account is that our predictions for the future evolution of the Sun are not based on our understanding and observations of the Sun alone. They're based on our understanding and observations of all the stars we've ever looked at. The thing is, stars of the same mass and composition must necessarily evolve in the same way. So we can look out at other stars and find some which are more or less the same as our own, but might be farther along the evolutionary track. We can study these stars and see what is happening to them. By conglomerating a lot of data about many stars at many different points in their evolutionary life cycle, we can put a coherent picture together for our own Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scientists have spent a &lt;em&gt;lot&lt;/em&gt; of effort to create stellar evolutionary models. To do this, they put together the multitudinous information gleaned from the tens of millions of stars we've observed, add in a lot of math and physics representing the physical processes occurring in these stars and what you find is that the &quot;predictions just inevitably fall out&quot;. There are many stellar evolutionary code bases out there you can find. For example, &lt;a href=&quot;http://mesa.sourceforge.net/pdfs/pols_98.pdf&quot; rel=&quot;noreferrer&quot;&gt;Pols et al. 1998&lt;/a&gt; applied stellar evolutionary code to clusters of stars. A figure pulled from their paper is shown below. It shows a Color-Magnitude diagram of the M67 open cluster of stars, along with predictions of the evolutionary track of those stars. There's of course some variance as every star is not exactly the same, but overall, you can see that the stars follow the predicted evolutionary track pretty well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dt3dx.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dt3dx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have a well calibrated stellar evolutionary program, it's not too difficult to apply that program to our Sun. There will be some unknowns and some variance, e.g., we might not know the physical extent of the red giant phases' expansion, but we know it will happen.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-03-28T13:00:27.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20554" PostTypeId="2" ParentId="20549" CreationDate="2017-03-28T13:15:20.933" Score="6" Body="&lt;h1&gt;From theory&lt;/h1&gt;&#xA;&lt;p&gt;It is possible to &lt;a href=&quot;http://www.tifr.res.in/%7Eantia/wball_notes.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;numerically integrate the equations of stellar structure over time&lt;/a&gt; to figure out how a star will evolve. Some basic informed &amp;quot;guesses&amp;quot; about the central and surface properties of the model star need to be made, but as the model is created, future iterations can use different values for these, making the model more and more accurate until it is self-consistent.&lt;/p&gt;&#xA;&lt;p&gt;The equations are based on several key assumptions (for which there is a good amount of evidence). Here are a few:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Stars are in approximate &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrostatic_equilibrium&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrostatic equilibrium&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Energy is (for the most part) generated via well-explored nuclear fusion pathways&lt;/li&gt;&#xA;&lt;li&gt;Energy is conserved&lt;/li&gt;&#xA;&lt;li&gt;There is a relation - an &lt;a href=&quot;https://en.wikipedia.org/wiki/Equation_of_state&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;equation of state&lt;/em&gt;&lt;/a&gt; - that describes the mathematical relationship via many of the key quantities (e.g. pressure, density, temperature, etc.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h1&gt;From experiment&lt;/h1&gt;&#xA;&lt;p&gt;There are hundreds of billions of stars in the Milky Way alone. Clearly, we don't know all of them - and &lt;a href=&quot;http://sci.esa.int/gaia/&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Gaia spacecraft will only observe (still astonishingly many) one billion of them&lt;/a&gt; - but we do have observations for a number of &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_analog&quot; rel=&quot;nofollow noreferrer&quot;&gt;stars like the Sun&lt;/a&gt; and stars that are somewhere along evolutionary tracks very similar to the one it is predicted the Sun will take. Data from these stars confirms many models (while showing that others need modification).&lt;/p&gt;&#xA;&lt;p&gt;Thanks to repeated comparisons of theory and simulation to real life, the models have been refined over time, and more and more evidence has been garnered in their favor. There is always a strong interplay between theory and experiment, and the model predictions match what we see in the sky, while the things we see in the sky give us better and better data for making newer, even more accurate, models.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;In a nutshell . . .&lt;/h1&gt;&#xA;&lt;p&gt;. . . we have good theories for the processes that make stars possible, one of which is nuclear fusion. We know very well how nuclear fusion should change elements over time and thus how the composition of any sample of material should change. We also know that fusion is the process the powers stars; therefore, we can create models that show how fusion in a star should cause it to change and evolve over time. Those models then match empirical data.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-03-29T01:34:24.313" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20556" PostTypeId="1" AcceptedAnswerId="20558" CreationDate="2017-03-28T17:25:33.810" Score="3" ViewCount="150" Body="&lt;p&gt;I'm happy that the universe is expanding, because we have good scientific evidence in the form of galactic redshift. The Wikipedia &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble's law&lt;/a&gt; article gives what I think is a fair explanation of that. It redirects from cosmological redshift and maybe gives too much emphasis to Doppler shift, but no matter. Because like I said I'm happy that the universe is expanding. Even if there was no evidence, my understanding of general relativity would be telling me that the universe just &lt;em&gt;has&lt;/em&gt; to expand. For the life of me, I still don't know why Einstein didn't predict it.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what I'm not happy about is inflation. I take note of articles like &lt;a href=&quot;https://blogs.scientificamerican.com/cross-check/physicist-slams-cosmic-theory-he-helped-conceive/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Physicist Slams Cosmic Theory He Helped Conceive&lt;/a&gt; where Paul Steinhardt is critical of inflation. He says things like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The whole point of inflation was to get rid of fine-tuning – to explain features of the original big bang model that must be fine-tuned to match observations. The fact that we had to introduce one fine-tuning to remove another was worrisome. This problem has never been resolved&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I empathize. I don't empathize with his cyclic universe, but I share his view that we have not explained any feature of the universe by introducing &lt;a href=&quot;https://en.wikipedia.org/wiki/Inflation_(cosmology)&quot; rel=&quot;nofollow noreferrer&quot;&gt;inflation&lt;/a&gt;. In fact, I would dispute the claims that inflation explains why the universe is isotropic, homogeneous, flat, and bereft of magnetic monopoles. I'd even say that I think inflation is a solution to a problem that doesn't exist. But as the Wikipedia article says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The basic inflationary paradigm is accepted by most scientists, who believe a number of predictions have been confirmed by observation;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't. I'm just some IT guy, but I'm like &quot;the substantial minority of scientists who dissent from this position&quot;. Steinhardt is mentioned, as are John Earman and Jesus Mosterin. I can read their paper on &lt;a href=&quot;http://sci-hub.cc/10.2307/188736&quot; rel=&quot;nofollow noreferrer&quot;&gt;SciHub&lt;/a&gt;. But what I'm struggling to find is any big-bang cosmological model or models that don't feature inflation. It seems to be included in all big bang literature and images, including this &quot;wine flute&quot; public domain image courtesy of NASA:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lAeFc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lAeFc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody point some out to me? In other words my question is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are there expanding-universe cosmological models that do not feature inflation?&lt;/strong&gt;  &lt;/p&gt;&#xA;" OwnerUserId="9772" LastEditorUserId="2365" LastEditDate="2017-03-28T18:50:33.790" LastActivityDate="2017-03-28T18:50:33.790" Title="Are there expanding-universe cosmological models that do not include inflation?" Tags="&lt;cosmological-inflation&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20557" PostTypeId="1" AcceptedAnswerId="20567" CreationDate="2017-03-28T17:35:11.357" Score="6" ViewCount="387" Body="&lt;p&gt;I understand that space was compressed to a single point and that during the big bang all points within that expanded away from each other at phenomenal speeds. I also have heard that during this the universe wasn't the same everywhere, as in small quantum fluctuations that expanded to macroscopic scales. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this mean that higher density than normal areas could create massive black holes? Maybe not as big as the ones today but perhaps large enough to solve some of the problems with super massive black holes being observed in a short amount of time after the big bang&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not asking &quot;why didn't the universe turn into a singularity?&quot; I am asking &quot;Could the variable density afterwards create many singularities which could explain for the problems associated with super massive blackholes?&quot;.&lt;/p&gt;&#xA;" OwnerUserId="16449" LastEditorDisplayName="user5434" LastEditDate="2017-11-25T21:54:35.420" LastActivityDate="2017-11-25T21:54:35.420" Title="Could the big bang have created super massive black holes?" Tags="&lt;big-bang-theory&gt;&lt;supermassive-black-hole&gt;&lt;early-universe&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20558" PostTypeId="2" ParentId="20556" CreationDate="2017-03-28T17:49:43.223" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Variable_speed_of_light#Modern_VSL_theories_as_an_alternative_to_cosmic_inflation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Variable speed of light (VSL)&lt;/a&gt; theories (for instance &lt;a href=&quot;http://www.bigravitytheory.com/pdf/modern_physics_letters_a1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Petit (1988)&lt;/a&gt; and &lt;a href=&quot;https://arxiv.org/abs/gr-qc/9211020v2&quot; rel=&quot;nofollow noreferrer&quot;&gt;Moffat (1993)&lt;/a&gt;) have been introduced in the past to solve the &lt;a href=&quot;https://en.wikipedia.org/wiki/Horizon_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;horizon problem&lt;/a&gt; you mentioned, i.e. that the observable universe appears to be isotorpic and homogeneous. If certain universal &quot;constants&quot; (including $c$, the speed of light) are allowed to change, then it is possible for interactions between extremely distant sections of the universe to occur, solving the horizon problem. These modifications still fit in with general relativity and the idea of an expanding universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;VSL theories, though, have not gained much traction, and inflation is still the prevailing theory in the scientific community today. It solves the horizon problem, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Flatness_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;flatness problem&lt;/a&gt; (dealing with how fine-tuned the initial density of the universe appears to have been), and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Inflation_(cosmology)#Magnetic-monopole_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;magnetic monopole problem&lt;/a&gt; (i.e. why we don't see magnetic monopoles). Inflation isn't the &lt;em&gt;only&lt;/em&gt; possible solution to these problems, but it works quite well.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2017-03-28T17:49:43.223" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20559" PostTypeId="1" AcceptedAnswerId="20569" CreationDate="2017-03-29T00:56:09.363" Score="5" ViewCount="708" Body="&lt;p&gt;When I see an animation of two colliding black holes anywhere, it looks like two merging beads of water. But why would the event horizon form a bridge during collision? There is no material flowing in-between. Wouldn't the event horizons &quot;push&quot; each-other away as they merge?(the effect not the black hole itself) For what I understand the gravity would poorly, but surely, be canceled out. Wouldn't this &quot;push&quot; only disappear from view as they got significantly closer, where it still exists but is trapped in a much larger one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to make sure, as I see the similar questions: I don't mean what happens as they collide, but rather why isn't this portrayed if it does this effect at all?&lt;/p&gt;&#xA;" OwnerUserId="16449" LastActivityDate="2017-03-29T22:27:28.013" Title="Why do &quot;they&quot; portray colliding black holes like that?" Tags="&lt;black-hole&gt;&lt;event-horizon&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20560" PostTypeId="1" CreationDate="2017-03-29T07:48:48.573" Score="3" ViewCount="5556" Body="&lt;p&gt;I have a world map, and I want to know, given a timestamp, where is the sun located right now, in terms of long/lat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a website that does that:&#xA;&lt;a href=&quot;https://www.timeanddate.com/worldclock/sunearth.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.timeanddate.com/worldclock/sunearth.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But they do not share the calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate the sun's position in earth gps terms given a timestamp?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: some context&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a plane that is determined by 2 points on earth, and that plain is perpendicular to the earth in those points. I want to know where is the sun relative to that plane, so I can know if the sun is to the right of it, or to the left of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of working in 3 dimensions, I decided to work in 2. I already have that plane in 2D (just a line on the map), now I need to know where the sun is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit2&lt;/strong&gt;: Using the wiki page &lt;a href=&quot;https://en.wikipedia.org/wiki/Position_of_the_Sun&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, this is what I got so far&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The interesting part is &lt;code&gt;getCurrentPosition&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;class Sun {&#xA;    private julianTime(timestamp: number) {&#xA;        // Source: http://stackoverflow.com/questions/466321/convert-unix-timestamp-to-julian&#xA;        return timestamp / 86400 + 2440587.5;&#xA;    }&#xA;&#xA;    private convertEclipticEquatorial(lambda: number, beta: number, R: number, n) {&#xA;        const epsilon = 23.439 - 0.0000004 * n; // Obliquity of the ecliptic&#xA;        console.log(&quot;epsilon&quot;, epsilon);&#xA;        const alpha = Math.atan2(Degrees.cos(epsilon) * Degrees.sin(lambda), Degrees.cos(lambda));&#xA;        console.log(&quot;alpha&quot;, alpha);&#xA;        const delta = Math.asin(Degrees.sin(epsilon) * Degrees.sin(lambda)); // declination&#xA;        console.log(&quot;delta&quot;, delta);&#xA;    }&#xA;&#xA;    getCurrentPosition(timestamp: number) {&#xA;        const JD = this.julianTime(timestamp);&#xA;        console.log(&quot;JD&quot;, JD);&#xA;&#xA;        // Source: https://en.wikipedia.org/wiki/Position_of_the_Sun&#xA;        const n = JD - 2451545;&#xA;        console.log(&quot;n&quot;, n);&#xA;        const L = (280.460 + 0.9856474 * n) % 360;&#xA;        console.log(&quot;L&quot;, L);&#xA;        const g = (357.528 + 0.9856003 * n) % 360;&#xA;        console.log(&quot;g&quot;, g);&#xA;        const lambda = (L + 1.915 * Degrees.sin(g) + 0.020 * Degrees.sin(2 * g)) % 360;&#xA;        console.log(&quot;lambda&quot;, lambda);&#xA;        const R = 1.00014 - 0.01671 * Degrees.cos(g) - 0.00014 * Degrees.cos(2 * g); // Distance&#xA;        console.log(&quot;R&quot;, R);&#xA;&#xA;        this.convertEclipticEquatorial(lambda, 0, R, n);&#xA;    }&#xA;}&#xA;&#xA;class Degrees {&#xA;    static sin(angle: number) {&#xA;        return Math.sin(angle / 180 * Math.PI);&#xA;    }&#xA;    static cos(angle: number) {&#xA;        return Math.cos(angle / 180 * Math.PI);&#xA;    }&#xA;}&#xA;&#xA;console.log(new Sun().getCurrentPosition(new Date().getTime() / 1000));&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Running this now, gives:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;JD&lt;/code&gt; 2457844.130512928 (can confirm, this is correct)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;n&lt;/code&gt; 6299.1305129281245 (simple math)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;L&lt;/code&gt; 9.181612328272422 (simple math)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;g&lt;/code&gt; 85.95292328111373 (simple math)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;lambda&lt;/code&gt; 11.09465308568091&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;R&lt;/code&gt; 0.9990992788207762&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;epsilon&lt;/code&gt; 23.436480347794827 (can confirm, that is roughly correct)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;alpha&lt;/code&gt; 0.17801325468993906&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;delta&lt;/code&gt; 0.0766106706512887&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16453" LastEditorUserId="16453" LastEditDate="2017-03-31T15:19:09.500" LastActivityDate="2020-05-13T20:35:56.743" Title="How to calculate the position of the sun in long/lat?" Tags="&lt;the-sun&gt;&lt;mathematics&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20561" PostTypeId="1" AcceptedAnswerId="20563" CreationDate="2017-03-29T08:17:10.297" Score="0" ViewCount="208" Body="&lt;p&gt;I'm new to astronomy and have to make a star map that displays the stars above a current location. From the information I've pieced together I understand that declination is equal to the current latitude of the observer, however I am slightly confused about right ascension. My two quesions are;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If I work out GMST using the equations here; &lt;a href=&quot;http://www2.arnes.si/~gljsentvid10/sidereal.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www2.arnes.si/~gljsentvid10/sidereal.htm&lt;/a&gt; will the answer be the current right ascension above Greenwich or are there more steps I've missed?&lt;/li&gt;&#xA;&lt;li&gt;If I want to find the sidereal time of, for example, Sydney, Australia, do I need to account for daylight savings at either location or do I simply add 11 hours to GMST?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thank you for your help.&lt;/p&gt;&#xA;" OwnerUserId="16454" LastActivityDate="2017-03-29T09:20:07.840" Title="Sidereal time and right ascension" Tags="&lt;right-ascension&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20562" PostTypeId="1" AcceptedAnswerId="20566" CreationDate="2017-03-29T08:46:20.027" Score="2" ViewCount="313" Body="&lt;p&gt;We know that we can use Doppler effect to determine equatorial velocity of a planet. We send a electromagnetic wave with wavelength $\lambda_0$ to an equatorial edge of planet and we get wavelength $\lambda$. So we can use Doppler Effect formula, (assuming $v \ll c$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\Delta\lambda}{\lambda_0} = \frac{v_r}{c}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;such that $v_r$ is radial velocity. But $v_r$ = $2*v_p$. And $v_p$ is equatorial velocity of planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know where does the $``2&quot;$ comes from. Someone said it is because of receiving the signal and then reflecting it. I want to know the math behind this not just to say receiving and reflecting. Why do we need to multiply the velocity by 2 to get the radial velocity?&lt;/p&gt;&#xA;" OwnerUserId="16441" LastEditorUserId="16441" LastEditDate="2017-03-29T14:16:03.567" LastActivityDate="2017-03-29T15:05:35.033" Title="Question about doppler effect" Tags="&lt;velocity&gt;&lt;wavelength&gt;&lt;doppler-effect&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20563" PostTypeId="2" ParentId="20561" CreationDate="2017-03-29T09:20:07.840" Score="0" Body="&lt;p&gt;I can't vouch for the formula, but you can try it out and check with some internet source, eg USNO, to see if it works.  If the formula is correct, that's it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the sidereal time in Sydney, you need the difference in longitude between Greenwich and Sydney, and use the conversion rate of 15 degrees longitude is equivalent to 1 hour. &lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-03-29T09:20:07.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20564" PostTypeId="1" AcceptedAnswerId="20579" CreationDate="2017-03-29T10:46:54.917" Score="17" ViewCount="587" Body="&lt;p&gt;Over in the Physics SE a question was posted asking about the difference in the time dilation of the Earth between perihelion and aphelion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/321910/does-earth-experience-any-significant-measurable-time-dilation-at-perihelion&quot;&gt;Does Earth experience any significant, measurable time dilation at perihelion?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather to my surprise it turns out that because of the changes in the Earth-Sun distance and the Earth's orbital velocity there is a difference of about $60\mu$s per day between the two extremes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A commentator pointed out that pulsars can be measured accurately enough to detect this difference. However I have never heard of a pulsar measurement having to be corrected for the time of year, and Googling has found me nothing related. I would be interested to know if this is something that needs to be considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference is slightly over one part in $10^9$, so presumably it depends on whether pulsars can be timed this accurately.&lt;/p&gt;&#xA;" OwnerUserId="10687" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2020-05-09T17:31:08.270" Title="Does the observed period of a pulsar change with the time of year?" Tags="&lt;pulsar&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20565" PostTypeId="2" ParentId="14808" CreationDate="2017-03-29T13:20:01.463" Score="2" Body="&lt;p&gt;Fresnel lenses are being considered for high energy waves, like X-rays. Problem with X-rays is that they are not reflected by mirrors of traditional telescopes : they just pass through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a new type of telescopes have been designed for X-ray astronomy, &quot;Grazing Incidence Telescopes&quot;. In them the X_rays are reflected at a very small angle, to just change their direction slightly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SAIZh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SAIZh.jpg&quot; alt=&quot;Grazing Incidence&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so new difficulties arise form using reflection for X-ray and gamma rays, that Fresnel telescopes are being considered. It could be worth to try to overcome disadvantages as chromatic aberration instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is some interesting link :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/astro-ph/0308445&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fresnel lenses for X-ray and Gamma-ray Astronomy&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16455" LastActivityDate="2017-03-29T13:20:01.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20566" PostTypeId="2" ParentId="20562" CreationDate="2017-03-29T15:05:35.033" Score="4" Body="&lt;p&gt;Before I start, let me just say that this topic is vastly more complicated than you've presented and what I will be showing. The trouble here is that ultimately, everything you've done and I will do (to a lesser extent) uses approximations and assumptions. Because of this, it can sometimes be hard to understand the true underlying physics when looking at only the &quot;simple&quot; versions of these equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say that the you've heard that the source of the two is because of receiving the signal and reflecting it, which is entirely true. Let's consider the simpler case of a police officer using radar to track a car's speed (just so we don't have to drag in the more complicated concepts of orbital motions of satellites, dispersion relations, etc.) You have a police officer standing by the side of the road with a radar gun, shooting radar waves of some frequency $f_o$ at the car as it recedes from him (note I'm going to use frequency but it is completely analogous to wavelength, assuming a simple dispersion relation). That wave travels to the car and the first peak of the wave hits the car. The car then has time to move slightly forward by the time the next peak hits the car. From the car's perspective, because of his relative motion to the police officer, the frequency of the wave hitting the car has been Doppler shifted to a new frequency, $f_1$. This new frequency is given by the Doppler equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{f_1}{f_o} = \left(1+\frac{v}{c}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this equation is the same as what you presented, except that it uses frequency instead of wavlength. So now, our car has been hit by these frequency shifted waves. It will now reflect those waves back at the police officer. The process of reflection effectively makes the car an emitter of the frequency $f_1$ but since it's obviously moving with respect to the police office, the police officer receives a Doppler shifted frequency, which I'll call $f_2$. Since the speed between the car and the police officer is still $v$, we know that the Doppler equation for the frequency shift of the reflection is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{f_2}{f_1} = \left(1+\frac{v}{c}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the police officer receives back a frequency $f_2$, but he originally emitted $f_o$, so he wants to know the velocity which caused the shift from $f_o$ to $f_2$. Fortunately, this can be determined by combining the two shifts above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{f_2}{f_o} = \frac{f_2}{f_1}\frac{f_1}{f_o} = \left(1+\frac{v}{c}\right)\left(1+\frac{v}{c}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{f_2}{f_o} = \left(1+\frac{v}{c}\right)^2 = 1 + 2\frac{v}{c} + \frac{v^2}{c^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this stage, we make the assumption that $v^2/c^2 \ll v/c$ and this second order term can be safely ignored. We then wind up with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{f_2-f_o}{f_o} = \frac{\Delta f}{f_o} = \frac{2v}{c}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the police officer measures by using the radar gun and applying the Doppler equation is $2v$, but of course we know this to be twice the speed of the car, so of course to get the actual speed of the car, the speed measured by the radar gun must be divided by two. As you can hopefully see, this 2 comes from the double receive and reflect of the radio waves by the car. The same exact principle holds whether we're talking about radar guns and cars or satellites and planets.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Note: As I said at the start, thing's are vastly more complicated than I've presented. For example, we assumed that $v^2/c^2 \ll v/c$. In reality, to get a more correct answer, you wouldn't want to drop this term and you'd find that $v_r \ne 2v_p$ as you've indicated in your answer, and there's actually more to it than that. This is all further exacerbated by the fact that the doppler equation I started with already had some assumptions built in. The fully correct equation is more complicated than that.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-03-29T15:05:35.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20567" PostTypeId="2" ParentId="20557" CreationDate="2017-03-29T21:30:26.573" Score="1" Body="&lt;p&gt;Arxiv prodives Haiman 2012 paper &lt;a href=&quot;https://arxiv.org/pdf/1203.6075.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;The Formation of the First Massive Black Holes&lt;/em&gt;&lt;/a&gt;. The leading theory, based on WMAP, is that the first perturbations resulted in dark matter halos, which were the first gravitational concentrations of matter.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the first nonlinear objects in the universe were born inside ∼ 10^5 M⊙&#xA;  dark matter halos at redshifts of z ∼ 20−30&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The non-linear objects do not refer precisely to black holes. The key part for us:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] question of how the  first SMBHs  were  assembled. It is  worth&#xA;  emphasizing that this is an unsolved problem – indeed, it is not&#xA;  entirely clear even whether the first nonlinear objects in the&#xA;  universe were stars or black holes, and whether galaxies or their&#xA;  central black holes formed first.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The article proceeds to tell a very wide range of theories, but the summary is - we are quite clear about simulating the early dark matter halos but we really don't know how to clearly show whether/how it led to SMBH formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to emphasize what is (maybe) obvious: SMBH requires a very concentrated gravity. We assume that a very early universe despite being dense didn't have enough gravity. All theories are generally about how density dropped far down and in what ways could it start to grow again, but now only inside the concentrations (i.e. when the gravity started to affect universe, the density was much lower that inside stars or black holes).&lt;/p&gt;&#xA;" OwnerUserId="10675" LastEditorUserId="10675" LastEditDate="2017-03-29T21:58:41.340" LastActivityDate="2017-03-29T21:58:41.340" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20568" PostTypeId="1" AcceptedAnswerId="20587" CreationDate="2017-03-29T22:16:32.187" Score="0" ViewCount="287" Body="&lt;p&gt;When planets are forming, does the densest material sink the core and the less dense material &quot;float&quot; on top? What I'm asking is that are the more dense materials more likely to appear in the core rather than the surface. I'm also wondering, can a planet have an iron surface and still have rock beneath the surface in its crust and core?&lt;/p&gt;&#xA;" OwnerUserId="16048" LastActivityDate="2017-03-31T17:35:36.470" Title="Are planets more likely to have dense materials at their cores?" Tags="&lt;planet&gt;&lt;planetary-formation&gt;&lt;core&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20569" PostTypeId="2" ParentId="20559" CreationDate="2017-03-29T22:27:28.013" Score="6" Body="&lt;p&gt;The event horizon is not a physical structure. Rather it is the boundary of a region of spacetime from which no information can escape. If you were (unfortunate enough to be) between two black holes, you may be in a position in which there is no net force, and yet you could not travel from your location to the point distant from the two black holes, no matter how fast you were able to move. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were in such a position, you would be inside the event horizon of the two black holes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding the event horizon means solving the Field equations of General relativity. Exact solutions are known in several situations: A single mass concentrated at a point, a mass with a charge and a rotating mass. If there are two masses, then the Field equations have no exact solution, but can be solved numerically (with a fast enough supercomputer) It is the results of these calculations that you are seeing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the shape you see is the event horizon, The event horizon of two nearby black holes can be calculated by a computer. Since you may be trapped between two black holes the event horizon will be deformed towards another black hole's event horizon, giving the &quot;teardrop-&gt; bridge-&gt;merger&quot; appearance of the final moments of two back holes.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-03-29T22:27:28.013" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20570" PostTypeId="1" AcceptedAnswerId="20575" CreationDate="2017-03-29T22:45:29.443" Score="2" ViewCount="252" Body="&lt;p&gt;When we talk about Milky Way kinematics, we define it as rotating clockwise from top.  In the rest of physics, rotating objects are defined as rotating counterclockwise from top (following the right-hand rule).  Why do we do the opposite when talking about the Milky Way?&lt;/p&gt;&#xA;" OwnerUserId="13426" LastActivityDate="2017-03-30T21:31:52.410" Title="Why do we define the Milky Way's rotation in a left-hand coordinate system?" Tags="&lt;milky-way&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20571" PostTypeId="1" CreationDate="2017-03-29T22:49:02.973" Score="1" ViewCount="155" Body="&lt;p&gt;And subsequently how long will it be until we can never see them again?&lt;/p&gt;&#xA;" OwnerUserId="16345" LastActivityDate="2017-03-31T04:13:44.687" Title="How long will it be until the nearby groups of galaxies are receding at speeds faster than light?" Tags="&lt;expansion&gt;&lt;redshift&gt;&lt;galaxy-cluster&gt;&lt;fate-of-universe&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20573" PostTypeId="1" AcceptedAnswerId="20574" CreationDate="2017-03-30T07:26:36.103" Score="1" ViewCount="215" Body="&lt;p&gt;A star in Pleiades cluster has apparent magnitude of +12 ($m_{star} = +12$) determine it's distance from the sun? (We know the Sun properties such apparent and absolute magnitudes, distance from earth, luminosity, brightness (energy flux) and others but nothing is given about the other star expect that apparent magnitude).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is that we don't have luminosity of the star. We can say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_{sun} - m_{star} = -2.5 * \log_{10}{\frac{b_{sun}} {b_{star}}} \Rightarrow 10^{(-26.8 - 12)/(-2.5)} \approx 3.31131 * 10^{15} = \frac{b_{sun}} {b_{star}} \approx \frac{1370 Wm^-2}{L_{star} / (4  \pi  d^2)} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we can see, we don't have luminosity of the star. Can this question be solved with this information?&lt;/p&gt;&#xA;" OwnerUserId="16441" LastActivityDate="2017-03-30T10:35:23.107" Title="Problem about finding distance with magnitude given" Tags="&lt;distances&gt;&lt;luminosity&gt;&lt;apparent-magnitude&gt;&lt;magnitude&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20574" PostTypeId="2" ParentId="20573" CreationDate="2017-03-30T10:35:23.107" Score="2" Body="&lt;p&gt;Short answer: no, you can't solve the question if all you know is its apparent magnitude. It could be a faint star that's nearby, or a very luminous star that's far away. (The Sun's parameters are irrelevant.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm ignoring the bit about it being &quot;a star in the Pleiades&quot;, which implies that it's a physical member of the Pleiades cluster, in which you &lt;em&gt;already know its distance&lt;/em&gt;: it's the same as the distance of the Pleiades cluster. Maybe you just meant &quot;it's in the same part of the sky&quot;.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2017-03-30T10:35:23.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20575" PostTypeId="2" ParentId="20570" CreationDate="2017-03-30T10:46:00.410" Score="7" Body="&lt;p&gt;The problem, as far as I can tell, is that the North and South Galactic Poles were defined well before we had any idea of which way the Galaxy rotated. It turned out that the Galaxy's angular momentum vector (in the standard right-hand sense) actually points in the same direction as the previously defined South Galactic Pole. Since there's a separate convention for having &quot;north&quot; point &quot;up&quot;, you end up with the backwards rotation definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;William Herschel seems to have started this back in the late 18th Century; I assume his idea was that whichever Galactic pole was in the North &lt;em&gt;Celestial&lt;/em&gt; Hemisphere should be called the &quot;northern&quot; pole. (No one back then had any clear sense that the Galaxy necessarily rotated at all.) By the time the actual direction of Galactic rotation was figured out in the 20th Century, the traditional orientation was too well entrenched to be worth upending.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2017-03-30T21:31:52.410" LastActivityDate="2017-03-30T21:31:52.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20576" PostTypeId="2" ParentId="20568" CreationDate="2017-03-30T12:41:37.307" Score="0" Body="&lt;p&gt;Apparently you don't fully appreciate the results of  gravity.  Try to model a fluid sphere (for our purposes here, a collection of rock, liquid, and gas is semifluidic) such that somehow there's less gravitational pressure near the center than near the outermost shell.  Can't happen. &#xA;Next, think about trying to make chunks of lead, or granite, float on top of a pond (of liquid water).  Won't happen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your last question is kind of pointless - there are &quot;rocks&quot; with densities both greater and less than the density of pure iron. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-03-30T12:41:37.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20577" PostTypeId="2" ParentId="20571" CreationDate="2017-03-30T12:57:37.973" Score="1" Body="&lt;p&gt;This will take about &lt;strong&gt;2 trillion years&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given everything we know now, it's not too hard to predict the eventual fate of the universe. Of course there may be some changes as our knowledge advances, but I think the general course of events will occur as we expect them to. Based on our current knowledge, if you run through the math, you find that after about 2 trillion years from now, the expansion of the universe will have made it so that all by the closest galaxies will no longer be visible to us. You do ask two separate questions, but they're more or less the same question. They might not occur precisely at the same time, but predictions of the future aren't precise enough to distinguish such a small time difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, you can read more about the future of the universe on &lt;a href=&quot;https://en.wikipedia.org/wiki/Future_of_an_expanding_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, and specifically &lt;a href=&quot;https://en.wikipedia.org/wiki/Future_of_an_expanding_universe#Galaxies_outside_the_Local_Supercluster_are_no_longer_detectable&quot; rel=&quot;nofollow noreferrer&quot;&gt;this section&lt;/a&gt; as it relates to your question.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-03-30T12:57:37.973" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20578" PostTypeId="2" ParentId="20557" CreationDate="2017-03-30T19:12:32.670" Score="1" Body="&lt;p&gt;The problem of early SMBH formation arises because of the Eddington limit: accreting matter heats up, radiates, and the radiation pressure limits further accretion. The problem is particularly acute if the SMBH is spinning fast, because then the innermost stable circular orbit (ISCO), which sets the inner edge of the accretion disc, is closer to the hole and the accreting matter heats up much more before passing the event horizon. However, SMBHs may actually not spin much (as a result of randomly orientated infall events), when the formation of observed SMBHs at high redshifts are not a problem.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2017-03-30T19:12:32.670" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20579" PostTypeId="2" ParentId="20564" CreationDate="2017-03-30T20:53:20.340" Score="8" Body="&lt;p&gt;Yes. In terms of pulsar timing measurements, this is a massive effect! A +/- 30 km/s doppler shift changes the pulsar frequency by +/- 1 part in 10000. This sounds small, but the accumulated phase shift over many periods is readily apparent. In addition, the light travel time across the solar system has to be taken into account, as well as the Earth's rotation and some other smaller effects - such as Shapiro delay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the question refers to the specific annually varying difference in clock rates caused by the different gravitational potential experienced by an Earthbound telescope on an elliptical (as opposed to circular) orbit -the answer is still yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is item 4 in the list of applied corrections given on p.52 of &quot;Pulsar Astronomy&quot; by Lyne et al. The maximum effect is a rate change of $3\times 10^{-9}$ leading to a maximum lead or lag of 1.7 ms.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-03-30T22:32:52.407" LastActivityDate="2017-03-30T22:32:52.407" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20581" PostTypeId="2" ParentId="20550" CreationDate="2017-03-31T13:19:36.303" Score="4" Body="&lt;p&gt;I'm treating this as a homework question. You have made a clear attempt, so it is not off topic, but you should clarify the source of the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will first need to establish if the rotation of the Earth is to be considered (it makes a big difference, but in a homework, sometimes the Earth stops spinning!) You have then correctly found the two possible answers, either 24 or 12 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also consider the motion of the Earth around the sun (as the Earth only takes 23hr, 56min to spin once, relative to the stars). What this means is that, after 12 hours, the Earth has turned more than 180 degrees, and there is no orbital plane that passes over the person, and through the origin, and is perpendicular to the equator. However the error is only about 1 degree, so probably can be neglected for the purposes of homework.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part, the first picture is certainly wrong, as the line of sight (in green) passes through the Earth. The second shows the line of sight (now in red) being tangent to the Earth's surface. You will need to use orbital mechanics to find the distance from the Earth's centre to the satellite, and find out the distance from the Earth's centre to its surface (green lines), giving two sides of a right-angled triangle. And the rest is maths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few details that may affect the answer: you should consider if the motion of the person will affect the answer to part (b) (and if not, why not). You should know that the Earth is not spherical. You should consider whether the optical refraction of light by the Earth's atmosphere should be considered.  However, in a homework problem, such issues can often be ignored, as their effects are small.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-04-02T12:25:23.767" LastActivityDate="2017-04-02T12:25:23.767" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20583" PostTypeId="1" CreationDate="2017-03-31T14:26:58.747" Score="0" ViewCount="110" Body="&lt;p&gt;If we convert the mass of all the stars and all the planets which must be weighing trillions and trillions of tonnes by the equation E = m.c^2 it gets converted into an unimaginable amount of energy. Would so much of energy not have gone into the evolution of the Universe? I am at a great suspense whether the Big Bang theory or any other theory at all answers from where so much of energy would have come that would have got converted into so many trillions and trillions of tonnes of the mass of all the galaxies put together. How we can believe in the big-bang-theory unless we know from where so much of energy would have arrived?&lt;/p&gt;&#xA;" OwnerUserId="13871" LastEditorUserId="13871" LastEditDate="2018-04-02T06:56:38.597" LastActivityDate="2018-04-02T06:56:38.597" Title="From where so much of energy would have arrived as was required for such a giant universe to have come into existence?" Tags="&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2018-04-01T18:56:23.403" ContentLicense="CC BY-SA 3.0" />
  <row Id="20584" PostTypeId="2" ParentId="20564" CreationDate="2017-03-31T14:37:16.223" Score="5" Body="&lt;p&gt;Short answer is: yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longer answer is: correcting for the time dilation effects of Earth moving around the Sun's gravitational potential is actually relatively standard in almost all branches of astronomy. To the point where running that correction is a sentence in a paper (sometimes less), and is probably why you had trouble Googling for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'll caveat all this by saying I'm mostly familiar with exoplanet transit and RV timing issues, but they should be the same as what the pulsar folks have to deal with).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As some background, the base time-keeping system used around the world is International Atomic Time (TAI), which is a weighted average of over 300 atomic clocks determined by the International Bureau of Weights and Measures outside of Paris. Importantly, TAI is strictly continuous: there are no leap seconds added. This is important if you care about sub-second timing precision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we use as normal &quot;clock&quot; time is Coordinated Universal Time (UTC), which is TAI with leap seconds subtracted off. Those leap seconds are present to deal with the fact that 86,400 SI seconds are 1 to 3 milliseconds less than one mean Solar day, and so ensure that our clock time is linked to the position of the Sun. The most recent leap second was added just this past New Year's, making UTC = TAI - 37 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even further down the time-keeping rabbit hole is Barycentric Dynamic Time (TDB), which accounts for the variable relativistic time dilation over the course of a year that you asked about. TDB has a fixed offset from TAI of 32.184 seconds due to how the zero-points of the the two systems were defined, and otherwise stays within 1.6 milliseconds of TAI - depending upon where Earth is in its orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Effectively all precise times reported by astronomers these days are the barycentric Julian date in the Barycentric Dynamic Time system (BJD_TDB). This is the Julian date an event would appear to happen for an observer located at the Solar System's barycenter using TDB as their timekeeping system. Note that the fact this is at the SS barycenter does matter, since observations on Earth will see similar events up to ~16 minutes apart over the course of the year due to light-travel time delay (Roemer Delay, for the aficionados) across the Earth's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yeah, this all has to be accounted for all the time. As I said, these days the transformation is standard enough that you usually just list a time as &quot;BJD_TDB&quot; and don't have to explicitly discuss the transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more reading about astronomical timekeeping, see &lt;a href=&quot;http://adsabs.harvard.edu/abs/2010PASP..122..935E&quot; rel=&quot;noreferrer&quot;&gt;Eastman et al. (2010)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS - In case you're wondering why Barycentric Dynamic Time is abbreviated TDB and Coordinated Universal Time is UTC, it's because we all use the French abbreviations.&lt;/p&gt;&#xA;" OwnerUserId="16268" LastActivityDate="2017-03-31T14:37:16.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20585" PostTypeId="2" ParentId="20560" CreationDate="2017-03-31T14:54:54.883" Score="1" Body="&lt;p&gt;If you don't need this precisely, the easy way to figure this out would be the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can assume that at noon UTC on one of the equinoxes, the Sun will be directly over (0N, 0E). The longitude of the sub-Solar point will then progress evenly over the course of a day: 4 hours later it's at 60W, 12 hours later at 180W, 18 hours later 90E, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latitude of the sub-Solar point will either increase or decrease (depending upon the starting equinox) as a sine curve with a period of year and an amplitude equal to the axial tilt of the Earth (23.43705 degrees).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That will be good to ~5 degrees over the course of the year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a more precise answer, you'd need to calculate the Analemma over the course of the year, and position the sub-Solar point accordingly. Formally, this would mean calculating the &lt;a href=&quot;https://en.wikipedia.org/wiki/Equation_of_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;Equation of Time&lt;/a&gt; and transforming the resulting time offset into a position offset for the sub-Solar point. E.g., &lt;a href=&quot;https://en.wikipedia.org/wiki/File:EquationofTimeandAnalemma.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;what's happening in this image&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not familiar with how to do that, so hopefully someone else can help - or the simple way above works.&lt;/p&gt;&#xA;" OwnerUserId="16268" LastActivityDate="2017-03-31T14:54:54.883" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20587" PostTypeId="2" ParentId="20568" CreationDate="2017-03-31T17:35:36.470" Score="1" Body="&lt;p&gt;Essentially, a planet can be considered as a self-gravitating fluid. As was already known to &lt;strong&gt;Archimedes&lt;/strong&gt;, under gravity low-density fluids are &lt;strong&gt;buoyant&lt;/strong&gt; and rise, while high-density fluids sink. So, with some local exceptions, planets are always well structured from the densest material in the core to lower-density material further up.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2017-03-31T17:35:36.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20588" PostTypeId="2" ParentId="18703" CreationDate="2017-03-31T18:30:13.840" Score="3" Body="&lt;p&gt;&lt;strong&gt;1. Resolving power and diffraction&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diffraction happens anywhere there's an edge. &lt;em&gt;&quot;It is defined as the bending of light around the corners of an obstacle or aperture into the region of geometrical shadow of the obstacle.&quot;&lt;/em&gt; (&lt;a href=&quot;https://en.wikipedia.org/wiki/Diffraction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;) So all you need for diffraction is an obstacle of any shape - at the edge of the obstacle, the light will bend a little.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/q1Ob9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q1Ob9.png&quot; alt=&quot;diffraction&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In telescopes, diffraction takes place all around the edge of the objective, be that a lens or a mirror. With a smaller objective, with a more strongly curved edge, and with opposite sides of the edge closer together, you get stronger diffraction (larger angle of deviation). With a larger objective, more straight edge, opposite sides of the edge further apart, you get less diffraction (smaller angle of deviation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why larger telescopes have smaller &lt;a href=&quot;https://en.wikipedia.org/wiki/Airy_disk&quot; rel=&quot;nofollow noreferrer&quot;&gt;Airy disks&lt;/a&gt;. It's similar to how a narrow slit makes a wider diffraction figure, whereas a wider slit makes a tighter figure. In the case of a telescope, the diffraction figure is the Airy disk - instead of having parallel diffraction zones, it's circular. A larger aperture makes a smaller Airy disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qdYLm.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qdYLm.gif&quot; alt=&quot;Airy disk&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Entrance/exit pupils and apertures&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a lot of modern amateur telescopes, the entrance pupil is actually the aperture of the telescope, or the visible diameter of the objective. In some cases the instrument has an aperture stop that reduces the size of the visible part, but that's rare. In most cases diameter of mirror = entrance pupil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exit pupil is simply due to the way the rays of light converge after they exit the eyepiece - they converge until they're bundled in a smallest area (the exit pupil) then diverge again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PHzEC.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PHzEC.gif&quot; alt=&quot;pupils&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the exit pupil is actually smaller than the diameter of the final lens in the eyepiece, because the rays keep converging for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, you want to keep your eye at the eyepiece so that the exit pupil coincides with the pupil in your eye, so as to capture the greatest amount of light coming out of the scope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The exact place where you keep your eye does not change the size of the image. The size of the virtual image made by the telescope depends only on object size and magnification - nothing else matters. Changing the eye position only changes the amount of light that enters your visual system. (Explaining why this is would probably require a whole 'nother topic on this forum.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3. Magnification and brightness&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of light captured by the scope stays the same. It is given by the size of the aperture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But with a larger image (higher magnification) that same amount of light coming from the object is spread out over a bigger solid angle. You see the same number of photons, but now they are spread out over a larger apparent surface. Of course it will look more dim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same amount of butter spread over a larger slice of bread.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4. Barlow, magnification, and field of view&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Barlow increases magnification. As shown in paragraph #3 above, more magnification = reduced apparent brightness. Same thing would happen without a Barlow, but just using a much stronger eyepiece instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's all about magnification. The light absorption in the glass in the Barlow is negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the object seems to pass more quickly through the field of view because the apparent field of view stays the same, but the whole image is magnified. Because the image is magnified, any motion within it must appear &quot;faster&quot;. So the object will seem to pass more quickly through the same field of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5. Stars and planets appear the same&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They shouldn't. It's a performance issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars should appear as tiny dots at low to medium magnification. They should appear as Airy disks (but still not large) at high magnification, fluctuating due to seeing (air turbulence).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most planets (Venus, Mars, Jupiter, Saturn) should exhibit disks that are clearly bigger than the stars. Even Uranus and Neptune (which are far away) should still show disks that, upon close examination, are revealed to be bigger than Airy disks, in an instrument as small as 150 mm aperture (perhaps even smaller).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could be a number of reasons for this. At 60 mm aperture, your scope should have enough resolving power to distinguish between the largest planetary disks (Jupiter) and any stellar Airy disks, and the rings of Saturn should start to become visible although very tiny. If that doesn't happen, then perhaps:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It's a collimation issue. All telescopes except most refractors require periodic collimation, otherwise performance decreases&lt;/li&gt;&#xA;&lt;li&gt;Quality of optics. Not sure what optics you have, but if they have performance issues then the images will be bloated&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As long as there are lingering issues with the instrument, higher magnification will not help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Stars are so far away that they are point-like objects to us. So their images in a scope &quot;ought to be&quot; also point-like. But they're not. This is due to diffraction and aberrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diffraction I've explained above. The Airy disk is the smallest possible image of a star you could ever have in a telescope. You cannot make a stellar image smaller than that, due to diffraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there's aberrations. There are monochromatic aberrations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;coma: the further away from the center the image is, the less point-like it becomes&lt;/li&gt;&#xA;&lt;li&gt;spherical mirrors or lenses DO NOT make perfect point-like images of stars. The point-like image is just an approximation. Spheres make imperfect images by their very design and shape.&lt;/li&gt;&#xA;&lt;li&gt;astigmatism - like coma, images near the edge of the field of view are distorted. The difference is that astigmatism makes a symmetrical distortion, whereas coma is weird shaped (like a seagull)&lt;/li&gt;&#xA;&lt;li&gt;field curvature - the focal plane of the primary mirror is not flat, but it's curved. So if the scope is in perfect focus in the center, it's not in perfect focus at the edge, and viceversa.&lt;/li&gt;&#xA;&lt;li&gt;distortion - a square grid is bloated so it looks like a curved cushion. Doesn't matter much for astronomical instruments.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then there are chromatic aberrations, which happen with refractors because the refraction index of the objective is not the same for all color of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the hallmark of a good telescope that its aberrations are less than the size of the Airy disk, at least for most of the field of view (perhaps excluding the zone close to the edge). That instrument is said to be &quot;diffraction limited&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If stellar images are very bloated, the instrument is likely not diffraction limited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://starizona.com/acb/basics/equip_optics101_aberrations.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://starizona.com/acb/basics/equip_optics101_aberrations.aspx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.telescope-optics.net/aberrations.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.telescope-optics.net/aberrations.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;6. How to optimize an instrument&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could write entire books on this topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Barlow is not magic. It only gives you more magnification, but magnification is the most misunderstood parameter of a scope. Unless everything else is perfect, more magnification does not help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a lot of reading you could do, and there is a lot of experimentation. Here are some older topics on this stack:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/8788/how-much-magnification-is-needed-to-see-planets-of-solar-system&quot;&gt;How much magnification is needed to see planets of solar system?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/2196/first-night-on-a-telescope-questions&quot;&gt;First night on a telescope questions&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/2193/best-telescope-for-the-viewing-of-nebulae-stars-and-planets&quot;&gt;Best telescope for the viewing of Nebulae, Stars and Planets&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/2223/please-guide-me-to-buy-my-first-telescope&quot;&gt;Please Guide me to buy my first Telescope&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2017-04-21T17:57:04.453" LastActivityDate="2017-04-21T17:57:04.453" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="20589" PostTypeId="2" ParentId="20560" CreationDate="2017-03-31T23:34:28.973" Score="3" Body="&lt;p&gt;By &lt;strong&gt;latitude&lt;/strong&gt; and &lt;strong&gt;longitude&lt;/strong&gt; of the sun, I think you mean &lt;strong&gt;latitude&lt;/strong&gt; and &lt;strong&gt;longitude&lt;/strong&gt; of a point on the earth's surface where the rays of the sun fall perpendicular to the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In astronavigation, this point is called the &lt;strong&gt;Geographical Position&lt;/strong&gt; (GP) of the sun and terminology to describe this point is slightly different:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Latitude&lt;/strong&gt; of the GP is called &lt;strong&gt;Declination&lt;/strong&gt;, and it is numerically equal to what one might call &quot;Latitude&quot;. It is measured in degrees North or South the equator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Longitude&lt;/strong&gt; of the GP is called &lt;strong&gt;Greenwich Hour Angle&lt;/strong&gt; (GHA). Just like longitude, GHA is measured East or West of Greenwich Meridian. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Describing the sun's position in this way is called the Equatorial Coordinate System.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia page you have used gives the additional formulae below to convert from Ecliptic Coordinates (the steps you followed) to Equatorial Coordinates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4anh0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4anh0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Follow these steps to determine Declination &lt;span class=&quot;math-container&quot;&gt;$\delta$&lt;/span&gt; (Equivalent to Latitude).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the sun's GHA, you need to first calculate &lt;span class=&quot;math-container&quot;&gt;$RA$&lt;/span&gt; (Right Ascension) as shown above and then convert this to GHA by these= rules...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$SHA° = 360 – 15 · RA[h] $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$GHA_{Aries}° = 15 · GST[h] $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$GHA = SHA + GHA_{Aries}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second step, you will need &lt;span class=&quot;math-container&quot;&gt;$GST$&lt;/span&gt; (Greenwich Sidereal Time) which is you can calculate if you know UT (UT can be obtained from the time shown by a clock). Use &lt;a href=&quot;http://www.astro.umd.edu/~jph/GST_eqn.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this link to convert directly from UT to GST&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a simple, concise way to be able to calculate the positions of all manner of celestial bodies, the sun, moon, planets, stars, then I highly recommended the book &lt;em&gt;Astronomical Formulae for Calculators&lt;/em&gt; by &lt;em&gt;Jean Meeus&lt;/em&gt;. It is the simplest, most comprehensive collection of astronomical formulae you are likely find anywhere on earth.&lt;/p&gt;&#xA;" OwnerUserId="16406" LastEditorUserId="7472" LastEditDate="2020-05-13T20:35:56.743" LastActivityDate="2020-05-13T20:35:56.743" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="20590" PostTypeId="2" ParentId="20550" CreationDate="2017-04-01T00:24:03.667" Score="3" Body="&lt;p&gt;Part (1). Due to the earth's rotation, an observer on the equator moves 180 deg every 12 hours. Your satellite is moving perpendicular to the equatorial plane and it meets the observer on the other side of its orbit after 12 hours. Therefore it too has travelled 180 degrees. To revolve a further 180 degrees (and complete a full 360 degree revolution), in time to meet the observer again, the satellite would take another 12 hours. Therefore the time period of the satellite would have to be 12h + 12h = &lt;strong&gt;24h&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) A satellite with a time period of 24h is called a geosynchronous satellite. The radius $r$ of the orbit of this type of satellite is approximately 36,000 km, can be determined by working further into this formula...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{v^2}{r} = \frac{GM}{r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in your second diagram, we can see that the radius of the satellite is 36,000 km, and we know the radius of the earth is 6400km, so by Pythagoras rule, the distance $d$ to the satellite when it sets can be obtained by,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d = \sqrt{36000^2 - 6400^2}$$&lt;/p&gt;&#xA;" OwnerUserId="16406" LastActivityDate="2017-04-01T00:24:03.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20591" PostTypeId="2" ParentId="6089" CreationDate="2017-04-01T00:35:54.853" Score="0" Body="&lt;p&gt;(1) By the law of universal gravitation, All satellites MUST orbit around the centre of the earth. The plane of the Tropic of Cancer does not pass through the earth's centre and therefore cannot be the orbital plane for &lt;strong&gt;ANY&lt;/strong&gt; satellite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Since geostationary satellites must be located directly over the equator, their footprint does not extend to extreme North and South latitudes. To overcome this, an extremely eccentric orbital pattern called a Molniya orbit is used by countries like Russia (extreme North latitudes). The orbit is designed so that when the satellite is over Russian cities, the satellite is moving very slowly (in accordance with Kepler's second law) and remains &quot;visible&quot; to radio receivers for maximum time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6EDEb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6EDEb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16406" LastActivityDate="2017-04-01T00:35:54.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20592" PostTypeId="2" ParentId="6089" CreationDate="2017-04-01T01:04:51.933" Score="0" Body="&lt;p&gt;Your statement , &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Any geostationary satellite can be placed so that a designated city is always in line of sight&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is incorrect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Geostationary&lt;/em&gt; orbit simply implies a &lt;em&gt;geosynchronous&lt;/em&gt; orbit where the orbital plane and the equatorial plane are the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could place a &lt;em&gt;geosynchronous&lt;/em&gt; satellite over London, but it would not be visible to London all the time. It's orbit would have a period of 24h, and if were exactly above London at 12 Noon, it will be exactly over London at 12 Noon tomorrow, but between these two times, it would not remain above London. &lt;/p&gt;&#xA;" OwnerUserId="16406" LastActivityDate="2017-04-01T01:04:51.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20593" PostTypeId="1" AcceptedAnswerId="20596" CreationDate="2017-04-01T02:56:40.160" Score="3" ViewCount="4729" Body="&lt;p&gt;I'm obviously a novice. I see a star in the eastern sky every night. It is quite large and impressive. How can I identify it. Thank you. &lt;/p&gt;&#xA;" OwnerUserId="16474" LastActivityDate="2017-04-01T21:21:43.073" Title="How can I identify a star?" Tags="&lt;star&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20595" PostTypeId="2" ParentId="20583" CreationDate="2017-04-01T05:46:57.453" Score="4" Body="&lt;p&gt;I think it'd be helpful to explain what you mean by &quot;have gone into the evolution of the Universe&quot;. Also, please stay assured that a lot of attention is going towards the shortcomings of big bang theory and other cosmological theories already :) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If by evolution of the universe you mean its expansion and accelerated expansion, then it does not require the energy equivalent of the galaxies. The reason is that it is already happening, without using their energy equivalent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the energy came from: Probably one of the biggest physics and philosophy open questions, if not the biggest one. So, no answer can be given right now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last but not least, the energy equivalent of the galaxies is currently considered to be only a fraction of the total mass/energy in the universe. The rest being the unknown factor named dark matter and mainly the unknown factor called dark energy. The latter is actually related to the evolution of the universe (&lt;em&gt;after&lt;/em&gt; the big bang of course). Its nature not known yet, but it clearly is not in the form of matter, unlike galaxies. &lt;/p&gt;&#xA;" OwnerUserId="16462" LastActivityDate="2017-04-01T05:46:57.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20596" PostTypeId="2" ParentId="20593" CreationDate="2017-04-01T06:29:11.717" Score="8" Body="&lt;p&gt;If an astronomical object appears to be very bright, and does not appear to 'twinkle' very much, then it is probably a planet, rather than a star. Jupiter is almost on the opposite side of the Earth to the Sun right now, so it will rise in the East when the Sun is setting in the West. Since Jupiter does look like a very bright star to the naked eye, and since it is in the part of the sky that you describe, then the object that you are seeing is almost certainly Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for recognising actual stars, you really need to learn the layout of the constellations to some extent. This will allow you to find your way around the night sky. A useful tool for this is a program like &lt;a href=&quot;http://www.stellarium.org&quot; rel=&quot;noreferrer&quot;&gt;Stellarium&lt;/a&gt;. This will show you the positions of stars and other astronomical objects, including the planets.&lt;/p&gt;&#xA;" OwnerUserId="13887" LastEditorUserId="13887" LastEditDate="2017-04-01T06:43:10.690" LastActivityDate="2017-04-01T06:43:10.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20597" PostTypeId="1" AcceptedAnswerId="20643" CreationDate="2017-04-01T09:25:53.833" Score="3" ViewCount="854" Body="&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt; The &lt;a href=&quot;http://www.ianww.com/meteor-showers/&quot; rel=&quot;nofollow noreferrer&quot;&gt;on-line simulation&lt;/a&gt; seems to be working beautifully now! I'd recommend anyone to go back and take another look to enjoy both the mathematical beauty of orbital mechanics, and the aesthetics of the solar system visualization!&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;New York Times article &lt;a href=&quot;https://www.nytimes.com/2017/03/24/science/visualizing-the-cosmic-streams-that-spew-meteor-showers.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Visualizing the Cosmic Streams That Spew Meteor Showers&lt;/a&gt; links to a &lt;a href=&quot;http://www.ianww.com/meteor-showers/&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar system viewer&lt;/a&gt; that allows one to visualize aspects of meteor-shower-inducing comet orbits by Ian Webb.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Now, astronomers and engineers have created &lt;a href=&quot;http://www.ianww.com/meteor-showers/#Orionids&quot; rel=&quot;nofollow noreferrer&quot;&gt;an animation&lt;/a&gt; that lets you witness the entire journey. Using data from the &lt;a href=&quot;http://cams.seti.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cameras for Allsky Meteor Surveillance&lt;/a&gt;, a network of about 60 cameras pointed at the sky above San Francisco Bay, researchers have recorded more than 300,000 meteoroid trajectories since 2010. They plan to use the data to confirm more than 300 potential meteor showers that scientists have observed, but not verified.&lt;/p&gt;&#xA;&lt;p&gt;“Each dot that you see is a shooting star that was captured by one of our cameras,” said &lt;a href=&quot;https://www.seti.org/users/peter-jenniskens&quot; rel=&quot;nofollow noreferrer&quot;&gt;Peter Jenniskens&lt;/a&gt;, an astronomer at the SETI Institute and NASA Ames Research Center in Silicon Valley, Calif., who runs CAMS. His interactive transforms meteor showers like the &lt;a href=&quot;http://www.ianww.com/meteor-showers/#Geminids&quot; rel=&quot;nofollow noreferrer&quot;&gt;Geminids&lt;/a&gt; and the &lt;a href=&quot;https://www.nytimes.com/2016/10/21/science/orionid-meteor-shower-halleys-comet.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Orionids&lt;/a&gt; into shimmering rivers of space rocks. Viewers can pinpoint the moment the &lt;a href=&quot;https://www.nytimes.com/2016/05/03/science/if-you-blink-you-might-miss-the-eta-aquarids-meteor-shower.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lyrids&lt;/a&gt; or &lt;a href=&quot;http://www.ianww.com/meteor-showers/#all&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eta Aquarids&lt;/a&gt; light up the night by watching when their streams intersect with the Earth’s orbit, shown in blue. There is even an option to see all the meteor showers at once, making it look like a &lt;a href=&quot;http://www.ianww.com/meteor-showers/#all&quot; rel=&quot;nofollow noreferrer&quot;&gt;meteor hurricane&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I need some help understanding what I'm looking at. When I first open it up I believe I see a number of objects following the same general hyperbolic (or very elliptical) orbit. What bothers me is that some seem to zip past very fast, some seem to crawl slowly, and some are moving at intermediate speed. If they are all associated with one primary comet, and follow similar paths in this large (roughly 10 AU) view, shouldn't they have at least roughly similar velocities?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I've snapped screenshots from &lt;code&gt;30-10-2017&lt;/code&gt; to &lt;code&gt;04-11-2017&lt;/code&gt; at a speed setting of &lt;code&gt;0.005&lt;/code&gt; and made two GIFs below The second one is annotated with a red, green, and blue arrow indicating three objects (slow, medium, and fast) following nearly the same orbit with at least a factor of 10 different speeds between them.&lt;/p&gt;&#xA;&lt;p&gt;I have replayed viewing from different angles, they follow nearly the same orbit in 3D, all the way around the sun and back into space - it is not related to a particular view.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;I believe this to be completely unphysical!&lt;/em&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/t8KUf.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t8KUf.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/I32b5.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I32b5.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt; Red, Green Blue (slow to fast)&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3hyCt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3hyCt.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/Yy7MP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Yy7MP.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-04-20T01:00:40.227" Title="Why are these objects moving at Vastly Different Speeds along the same orbit?" Tags="&lt;orbit&gt;&lt;comets&gt;&lt;meteor&gt;&lt;orbital-mechanics&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20598" PostTypeId="1" AcceptedAnswerId="20609" CreationDate="2017-04-01T09:45:03.650" Score="7" ViewCount="204" Body="&lt;p&gt;What is the difference in the creation process of solar flares and CME (coronal mass ejections)? How do they relate to sunspots and coronal holes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/87/what-are-the-main-differences-between-solar-flares-and-coronal-mass-ejections/88#88&quot;&gt;Here&lt;/a&gt; there are some good explanations about the flares' and CME's differences. But my question focuses on how each of them is created, and what are the respective conditions at the corona. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;In an effort to make the questions more clear, I could add that as far as I can tell both effects result from magnetic reconnection, but in one case we have only energetic particles emitted while in the other we have whole chunks of plasma. Do we know which is the determining factor about which of the two will happen? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I understand that flares are mostly related to sunspots while ejections are to coronal holes. Is this correct? If yes then it sounds like a pretty big difference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Please be reminded that specifying that the answer is not known yet &lt;em&gt;is&lt;/em&gt; an answer.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit #2:&lt;/strong&gt; Please see &lt;a href=&quot;https://astronomy.stackexchange.com/questions/20598/difference-in-creation-of-solar-flares-and-cme#comment34597_20609&quot;&gt;this comment&lt;/a&gt; for further clarification of what the question is about. &lt;/p&gt;&#xA;" OwnerUserId="16462" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2017-05-06T03:24:20.643" Title="Difference in creation of solar flares and CME?" Tags="&lt;the-sun&gt;&lt;solar-flare&gt;&lt;coronal-mass-ejection&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20599" PostTypeId="2" ParentId="20597" CreationDate="2017-04-01T09:55:56.877" Score="0" Body="&lt;p&gt;Over time, small objects relatively close together at the start will be pulled into quite different orbits by the perturbations of larger objects (like Jupiter) and even Earth if they approach closely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relatively small changes at close approach can have quite large effects when the objects reach their furthest approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbits are approximately elliptical (some may be hyperbolic) and the orbital velocity will vary in such an orbit (unlike a circular orbit).  So you'll see objects drifting along at the furthest reach of their orbit and speeding up as they &quot;fall&quot; in to the closest point of their orbit.  They'll again slow down as they &quot;rise up&quot; away from the closest approach (exactly as a stone you throw into the air).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the combination of these things means that the objects can separate significantly over time into quite distinct orbits that have quite significant differences in orbital velocities.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-04-01T09:55:56.877" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20600" PostTypeId="2" ParentId="20455" CreationDate="2017-04-01T14:54:31.797" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;And also there are some terms related to black holes, 3D shape, ball&#xA;  etc. Apart from those complexity, my intention is simply to be&#xA;  clarified, why we call our universe flat? And by word flat, what I&#xA;  meant &quot;having a level surface; without raised areas or indentations&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Imho in order to accept the answer, you have to realize the contradiction in the sentence above: &lt;em&gt;You mean&lt;/em&gt; level surface, but &lt;em&gt;we call&lt;/em&gt; the universe flat :) &quot;Flat&quot; in general relativity does not refer to &quot;having a level surface; without raised areas or indentations&quot;. Please accept this first. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's think of a ball (yes, a ball; it's not complexity). Would you call its surface flat or curved? I'd call it curved. Let's think of a pan. I'd call its surface flat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both the surface of a ball and a pan have two dimensions. The &lt;em&gt;ball&lt;/em&gt; itself has three dimensions obviously, but its &lt;em&gt;surface&lt;/em&gt; has two dimensions. I hope this is straightforward. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, everything described above about surfaces also applies to volumes. Volumes can be flat, and they can be curved. The issue is that humans cannot visualize things with three dimensions bending, so they cannot visualize a curved volume. But this is okay: We can still grasp the meaning of flat and curved, just by seeing it on surfaces, and understand that it can be extended to volumes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it's not just visualization and understanding: Flat and curved volumes (i.e. spaces) have very different physical and mathematical properties. A usual example is that in a curved space the sum of the angles in a triangle is not 180 degrees. There are also other properties and those enable astrophysicists to measure the actual curvature of the universe. I hope this was helpful. &lt;/p&gt;&#xA;" OwnerUserId="16462" LastActivityDate="2017-04-01T14:54:31.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20601" PostTypeId="2" ParentId="20455" CreationDate="2017-04-01T17:49:18.807" Score="3" Body="&lt;p&gt;&lt;em&gt;Flat&lt;/em&gt; here means &quot;has Euklidean geometry&quot; (on large scales), as explained by Helen's answer. Observationally, the universe appears flat, i.e. any deviations are within the uncertainties. Of course, if the universe is infinite, no statements on the whole universe can ever be made, but only on our local observable patch of it.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2017-04-01T20:36:14.620" LastActivityDate="2017-04-01T20:36:14.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20602" PostTypeId="2" ParentId="19896" CreationDate="2017-04-01T20:54:17.493" Score="4" Body="&lt;p&gt;The problem is that we don't really know how many small and diffuse galaxies there are. Even in our own cosmic backyard, the local group, we are still discovering new galaxies. Since these &lt;em&gt;dwarf spheroidal galaxies&lt;/em&gt; are by far the most common, we have essentially observed a &lt;strong&gt;negligible fraction&lt;/strong&gt; of all galaxies in the observable universe, and will never be able do increase this significantly.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2017-04-01T20:54:17.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20603" PostTypeId="2" ParentId="20593" CreationDate="2017-04-01T21:21:43.073" Score="5" Body="&lt;p&gt;The best way to tell the difference between planets and stars is:&#xA;(From &lt;a href=&quot;http://curious.astro.cornell.edu/physics/58-our-solar-system/planets-and-dwarf-planets/planet-watching/255-what-is-the-observational-difference-between-a-star-and-a-planet-beginner&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;)&#xA;Early astronomers were able to tell the difference between planets and stars because planets in our Solar System appear to move in complicated paths across the sky, but stars don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, if you observe the sky night after night, the stars will all appear in fixed positions with respect to each other. They will rise and set a few minutes earlier each night (an effect that is due to the Earth's motion around the Sun), but otherwise nothing will change. This is why the background stars are sometimes referred to as the &quot;celestial sphere&quot; -- from our point of view, it looks like the stars are &quot;painted&quot; onto a gigantic sphere that surrounds Earth and therefore are unable to move with respect to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets, on the other hand, are observed to move in very complicated paths with respect to the background stars, sometimes even appearing to go &quot;against the grain&quot; and reverse their directions. Therefore, they are easily distinguishable from stars if you look at the sky night after night. Although ancient astronomers did not have a correct explanation for this phenomenon, we now know that the complicated motion is just a projection effect -- it is due to the fact that Earth and the other planets are physically moving in orbits around the Sun, so the planets' relative positions as seen from Earth (with respect to the fixed background stars) change as time goes on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other observational differences between planets and stars too, by the way -- such as the fact that planets almost never twinkle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like the text said, if an object doesn't twinkle, it's most likely to be a planet. If you observe it every night, and it appears to change its position more than stars do. Although stars also change their positions, they don't change their positions in the sky as much as planets throughout the year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars maintain positions relative to one another, constellations are a great to tell the difference between planets in stars. What I mean by maintaining their positions is that each night they are usually in the same pattern and usually the same distance apart in the night sky.&lt;/p&gt;&#xA;" OwnerUserId="16048" LastActivityDate="2017-04-01T21:21:43.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20604" PostTypeId="2" ParentId="20550" CreationDate="2017-04-02T12:07:02.553" Score="1" Body="&lt;p&gt;The question is wrong: such a satellite orbit does not exist. Due to angular-momentum conservation, the orbital plane of the satellite is fixed in space, but the position of the observer rotates by about $\left( 360+\frac{360}{365.25} \right)^{\circ}$ every $24$ hours. This is simply because the Earth rotates more than once in an average day, i.e. the time between consecutive noons (Sun in zenith on the equator).&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="10825" LastEditDate="2017-07-02T06:43:02.470" LastActivityDate="2017-07-02T06:43:02.470" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20605" PostTypeId="1" AcceptedAnswerId="20617" CreationDate="2017-04-02T12:34:50.230" Score="2" ViewCount="281" Body="&lt;p&gt;Telescopes that use adaptive optics now use flexible mirrors to adjust for distortions in the atmosphere.  The mirrors are flexed in real time to account for the distortions.  I am wondering if it would be feasible to instead use fixed mirrors and use software to make the needed corrections to the image?&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2017-04-05T02:14:13.257" Title="Instead of using flexible mirrors, could software be used for adaptive optics?" Tags="&lt;telescope&gt;&lt;photography&gt;&lt;software&gt;&lt;adaptive-optics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20606" PostTypeId="1" AcceptedAnswerId="20608" CreationDate="2017-04-02T16:23:25.390" Score="1" ViewCount="92" Body="&lt;p&gt;Often I look at NASA's &lt;a href=&quot;https://eclipse.gsfc.nasa.gov/eclipse.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eclipse Web Site&lt;/a&gt;. More specifically, I often look at the &lt;a href=&quot;https://eclipse.gsfc.nasa.gov/LEcat5/LEcatalog.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Index to Five Millennium Catalog of Lunar Eclipse&lt;/a&gt;. How can I tell if an eclipse results in a &quot;blood moon&quot;? I know not all lunar eclipses result in a blood moon, but I do know that some do. So how can I know if one is from looking at it? A blood moon is the phenomenon whereby the moon in total eclipse appears reddish in color as it is illuminated by sunlight filtered and refracted by the earth's atmosphere. (&lt;a href=&quot;https://www.google.com/webhp?sourceid=chrome-instant&amp;amp;ion=1&amp;amp;espv=2&amp;amp;ie=UTF-8#newwindow=1&amp;amp;safe=active&amp;amp;q=blood%20moon&amp;amp;*&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;James K says this question may be a duplicate of another question. I want to know how to identify &quot;blood moons&quot; from looking at NASA's reports. The other question is more general. However the answer may be the same.&lt;/p&gt;&#xA;" OwnerUserId="14708" LastEditorUserId="14708" LastEditDate="2017-04-03T12:28:00.530" LastActivityDate="2017-04-03T12:28:00.530" Title="Blood Moon Identification in NASA reports" Tags="&lt;lunar-eclipse&gt;&lt;nasa&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2017-04-05T15:31:41.683" ContentLicense="CC BY-SA 3.0" />
  <row Id="20607" PostTypeId="1" AcceptedAnswerId="20673" CreationDate="2017-04-02T16:31:09.543" Score="5" ViewCount="126" Body="&lt;p&gt;I have seen the term superficial gravity used and it seems to be equivalent to surface gravity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;seen, e.g., here &lt;a href=&quot;http://arxiv.org/pdf/1701.02295&quot; rel=&quot;noreferrer&quot;&gt;http://arxiv.org/pdf/1701.02295&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any difference between superficial and surface gravity? When should one be using one or the other?&lt;/p&gt;&#xA;" OwnerUserId="16345" LastActivityDate="2017-04-08T23:17:33.420" Title="What is &quot;superficial gravity&quot;" Tags="&lt;gravity&gt;&lt;terminology&gt;&lt;white-dwarf&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20608" PostTypeId="2" ParentId="20606" CreationDate="2017-04-02T17:11:50.877" Score="2" Body="&lt;p&gt;&quot;Blood moon&quot; is not a term used scientifically. Therefore look for total lunar eclipses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As some light is diffracted by earth's atmosphere, the moon is not entirely black but appears red. (Blue light is scattered, see why is the sky blue).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All total lunar eclipses will show a reddish colour. The brightness may vary according to the state of the atmosphere but the moon will never be black, or any other colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(James K contributed equally to this answer)&lt;/p&gt;&#xA;" OwnerUserId="15824" LastEditorUserId="15824" LastEditDate="2017-04-03T07:14:37.370" LastActivityDate="2017-04-03T07:14:37.370" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20609" PostTypeId="2" ParentId="20598" CreationDate="2017-04-02T21:46:38.327" Score="5" Body="&lt;p&gt;The point of my comment was that the connection between CMEs and flares would make for a good 20 page research paper in a college course.  The connections are still only understood at a kind of cartoon level, after decades of concerted research.  But I can tell you what the basic cartoon is, and you can think of that as a kind of operating hypothesis that helps organize many of the observations, but is still a long way from a predictive model.  The basic idea is that the energy for flares and for CMEs both seem to come from magnetic free energy stored in the corona due to the magnetic and turbulent nature of the convection zone.  The magnetic field is capable of trapping pockets of hot gas that has been energized by this magnetic activity, but then when the field lines get significantly twisted up they reach a point where they can release that energy and relax to a smoother configuration.  This relaxation does two things-- it releases energy, and it reconfigures the field such that it can release these pockets of trapped plasma.  The former is the solar flare, the latter is the CME, and that's why the two often come together.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there seems to be a full range of weak to strong flares, and weak ones would not be associated with CMEs.  On top of all that, there may be other processes that heat the corona, and other processes that lead to the solar wind, so one big question is if the heating of the corona is mostly from lots and lots of small flares, and the solar wind is lots and lots of small CMEs, or if the general heating of the corona and general driving of the wind are separate processes from flares and CMEs.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you mention closed and open field regions, so these questions can be asked separately in those two domains.  But CMEs would normally come from closed field regions that open up when the field reconfiguration occurs, I believe open field regions tend to be rather sedate and not associated with either strong flares or CMEs, but rather a more steady wind.  In the closed field regions, CMEs are often associated with features called  &quot;helmet streamers,&quot; which are regions of trapped plasma that are kind of combed out at their highest point by the solar wind, creating an interface between closed and open fields that is a good setup for getting field reconfiguration that could launch out some plasma.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2017-05-06T03:24:20.643" LastActivityDate="2017-05-06T03:24:20.643" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20610" PostTypeId="1" AcceptedAnswerId="20611" CreationDate="2017-04-03T00:07:29.137" Score="11" ViewCount="3591" Body="&lt;p&gt;Or would the planet(s) need to be too far away for life to exist?&lt;/p&gt;&#xA;" OwnerUserId="16490" LastActivityDate="2018-11-27T02:26:33.953" Title="Do blue giants have a habitable zone?" Tags="&lt;planet&gt;&lt;habitable-zone&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20611" PostTypeId="2" ParentId="20610" CreationDate="2017-04-03T01:52:12.230" Score="15" Body="&lt;p&gt;Just because a planet is in the &quot;Habitable zone&quot; doesn't mean it is habitable. A planet is said to be in the habitable zone if it is possible for liquid water to exist on the surface. A planet may be in the habitable zone, and yet not be habitable if, for instance, it's atmospheric pressure is too low (like Mars) or too high (like Venus), both of which are in the habitable zone of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A star may be unsuitable for life for other reasons: it may be prone to extreme flares, for example. A blue giant is a star that burns bright and dies young, only lasting a few million years. Planets around such a star would have only just formed (they may still have molten surfaces). Blue giants tend to be unstable, being variable over a range of timescales, and ejecting large amounts of matter into space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while there would be a region which is &quot;habitable&quot; in the sense that the surface temperature of a planet could be in the range 0-100, you wouldn't expect such a planet to actually be suitable for life.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-04-03T01:52:12.230" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20613" PostTypeId="1" CreationDate="2017-04-03T04:13:49.810" Score="1" ViewCount="104" Body="&lt;p&gt;I am working on a code for galactic dynamics simulation. Initial conditions (positions and velocities) are based on a model with a massive dark halo and a thin exponential disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, from the following paper &lt;a href=&quot;https://arxiv.org/pdf/astro-ph/9807320.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Tidal tails in CDM cosmologies&quot;&lt;/a&gt;, I am studying the response of the dark matter profile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a capture of the article part that confuses me :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8s2bY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8s2bY.png&quot; alt=&quot;Response of dark matter profile&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, I don't understand how one gets equation (16) from the fact that angular momentum of individual dark matter orbits is&#xA;conserved, i.e :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;r_{i}\,M(r_{i})=r_{f}\,M_{f}(r_{f})\quad\quad\text{(eq 16)}       &#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot; where $r_{i}$ and $r_{f}$ are the initial and final radii of some dark matter mass shell, $M(r)$ gives the initial NFW mass&#xA;profile, and $M_{f}(r)$ is the final cumulative mass profile after the disk is formed.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1) Is (equation 16) got by the momentum theorem ? ($\dfrac{\text{d}\vec{L}}{\text{d}t} = \vec{OM} \times \vec{F}$)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;And how to prove it, i.e what's the demonstration to get (equation 16) ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand what's the difference between $M(r)$ and $M_{f}(r)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After, it is said that $M_{f}(r)$ is the sum of the cumulative mass of the disk and the dark mass inside the initial radius&#xA;and author writes (&lt;strong&gt;equation 17 on paper&lt;/strong&gt;) :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;M_{f}\,(r_{f})=M_{d}\,(r_{f})+(1 - m_{d})\,M(r_{i})\quad\quad\text{(eq 17)}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to connect this equation with the total mass of galaxy which could be written (according to me) as :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;M_{total}\,(r) = M_{d}\,(r) + M_{DM}\,(r) = m_{d}\, M_{DM}\,(r) + M_{DM}\,(r) = (1+m_d)\, M_{DM}\,(r)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $m_{d}$ the ratio between disk mass and halo mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2) Could you help me to connect the total mass with this equation (17) ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After with equation (18), author of paper writes that the final profile $M_{h}(r)$ of the dark matter halo is then given by :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;M_{h}\,(r) = M_{f}\,(r) - M_{d}\,(r)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 3) So should I conclude that $M_{f}\,(r)$ is the total mass of galaxy (disk + halo) as a function of $r$ ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be great,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE 1 :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have contacted the author of paper &lt;a href=&quot;https://arxiv.org/pdf/astro-ph/9807320.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Tidal tails in CDM cosmologies&quot;&lt;/a&gt;, i.e Volker Springel and here are its comments :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question 1*) Answer: For simplicity, you can consider that all particles are on circular&#xA;  orbits. The initial circular velocity of a particle is $$v_i = \sqrt{(G M(r_i)/r_i)}$$, the final one is $$v_f = \sqrt{(G M(r_f)/r_f)}$$. The&#xA;  corresponding angular momenta are $$L_i = m r_i v_i$$ and $$L_f = m r_f v_f$$.&#xA;  Now, the angular momentum is an adiabatic invariant when the halo is&#xA;  slowly compressed, hence $$L_i = L_f$$. The latter equation then gives eqn&#xA;  (16) when one plugs in the velocities.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Question 2*) &quot;I don't understand what's the difference between $M(r)$ and&#xA;  $M_f(r)$ ?&quot; &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Answer: &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$M_i(r)$ is the initial and $M_f(r)$ the final cumulative mass profile.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Question 3*)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I would like to connect this equation with the total mass of galaxy which could be written as :&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$ M_{total}(r) = M_d(r) + M_DM(r) = m_d M_DM(r) + M_DM(r) = (1+m_d) M_DM(r)$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Answer : This can only be written for the initial distribution in&#xA;  this form, where the assumption is that the mass that will later make&#xA;  up the disk is distributed proportionally to the dark matter.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Question 4*)&#xA;  Could you help me to connect the total mass with this equation (17) ?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;After with equation (18), you write that the final profile M_{h}(r)$ of the dark matter halo is then given by :&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$M_h (r) = M_f(r) - M_d(r)$$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;So should I conclude that M_f(r) is the total mass of galaxy (disk + halo) as a function of r ?&#xA;  &amp;lt; Answer : &quot;Yes.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In conclusion, I can solve my question 1*) but not question 2*) nor question 3*)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now, my goal is to demonstrate how to get  :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{equation}&#xA;M_{f}\,(r_{f})=M_{d}\,(r_{f})+(1 - m_{d})\,M(r_{i})\quad\quad\text{(eq 17)}&#xA;\end{equation}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help to acheive this is welcome, thanks&lt;/p&gt;&#xA;" OwnerUserId="16492" LastEditorUserId="16492" LastEditDate="2017-04-05T10:07:55.010" LastActivityDate="2017-04-05T10:07:55.010" Title="Need help on Springel model for galaxy simulation" Tags="&lt;galaxy&gt;&lt;dark-matter&gt;&lt;galactic-dynamics&gt;&lt;disk&gt;" AnswerCount="0" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20614" PostTypeId="1" CreationDate="2017-04-03T05:11:31.123" Score="3" ViewCount="2598" Body="&lt;p&gt;It is believed that there is a black hole at the centre of the Milky Way galaxy. And the sun revolves around the centre of the milky way.&#xA;So does it mean our sun is revolving around the centre of the galaxy due to this black hole? &lt;/p&gt;&#xA;" OwnerUserId="16495" LastEditorUserId="16495" LastEditDate="2017-04-06T14:07:04.207" LastActivityDate="2019-08-11T20:53:12.983" Title="Revolution of the sun" Tags="&lt;the-sun&gt;&lt;galaxy-center&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20615" PostTypeId="1" CreationDate="2017-04-03T06:05:43.393" Score="-1" ViewCount="86" Body="&lt;p&gt;When the sun reaches close to the centre of the Milky Way galaxy while revolving around it (like the earth closes down on the sun while revolving around it, perihelion), will the gravitational effect of the relatively nearby stars and that of the black hole itself be affecting the planets of the solar system?&lt;/p&gt;&#xA;" OwnerUserId="16495" LastEditorUserId="16495" LastEditDate="2017-04-04T06:45:23.313" LastActivityDate="2017-04-04T06:45:23.313" Title="Effect of gravitation" Tags="&lt;the-sun&gt;&lt;gravity&gt;&lt;galaxy&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2017-04-04T06:29:38.277" ContentLicense="CC BY-SA 3.0" />
  <row Id="20616" PostTypeId="2" ParentId="20614" CreationDate="2017-04-03T07:04:02.237" Score="9" Body="&lt;p&gt;The Sun orbits in the entire Galactic gravitational potential, which is very approximately similar to that provided by considering that &lt;em&gt;all&lt;/em&gt; the mass interior to its Galactic orbit (about 100 billion solar masses) were concentrated at the Galactic centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black hole at the centre has a mass of 4 million Suns, so contributes less than 0.01% of the mass required to keep the Sun in its present Galactic orbit of about 200 million years. Or to put it another way, if the Sun just orbited the black hole with a period of 200 million years, we would need to be about 800 light years from the Galactic centre, rather than the 30,000 light years we presently are.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-04-03T10:42:17.013" LastActivityDate="2017-04-03T10:42:17.013" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20617" PostTypeId="2" ParentId="20605" CreationDate="2017-04-03T13:27:04.707" Score="6" Body="&lt;p&gt;Short answer: no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why not?  because the mirror is placed at a position in the optical train such that it controls the wavefront phase and tilt of each subsection, &quot;subaperture&quot; of the incoming beam.  Once an optical system has formed an image, all phase information is lost.  There  are some software methods which can correct for simple aberrations such as astigmatism or coma, but those require a near-constant type of error across the input field.  The problem with atmospheric turbulence is that the errors are essentially of a localized and random nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that even a plenoptic camera which, by tricky use of sublenses at the focal plane, record enough information to change &lt;strong&gt;focus&lt;/strong&gt;, cannot correct for general phase aberrations.&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-04-03T13:27:04.707" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20618" PostTypeId="2" ParentId="20614" CreationDate="2017-04-03T16:49:24.027" Score="0" Body="&lt;p&gt;&lt;strong&gt;Yes, of course&lt;/strong&gt;. In the same way as the answer to &lt;em&gt;If the Moon revolves around Earth, does it mean it revolves around you?&lt;/em&gt; is Yes!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does not mean, though, that you determine or even significantly affect the orbit of the Moon, and so it is with the supermassive black hole (SMBH) at the center of the Milky Way and the orbit of the Sun, as Rob explains (but fails to answer your question ;-).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational tug of the Milky Way at the Sun is the sum of the tugs of all its components, not just the SMBH. This is dominated by the vast number of stars (and dark matter), which together weight much more than the SMBH.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2017-04-03T16:49:24.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20619" PostTypeId="1" CreationDate="2017-04-03T17:02:59.987" Score="3" ViewCount="124" Body="&lt;p&gt;I made an illustration to explain what i mean : If we assume we have two similar planet/star systems (similars in size/mass/period..) but tilted differently relative to us.&#xA;How can we predict that differences in dimming measurements (time &amp;amp; intensity) are caused by the tilting of orbital plane? (and not by the diameter of the exoplanet)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KFZWV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KFZWV.png&quot; alt=&quot;two different tiltings&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16500" LastActivityDate="2017-04-03T23:14:04.897" Title="What's the influence of a tilted orbital plane, when observing an exoplanet transit?" Tags="&lt;exoplanet&gt;&lt;planetary-transits&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20620" PostTypeId="1" CreationDate="2017-04-03T17:17:28.843" Score="4" ViewCount="1556" Body="&lt;p&gt;I have noticed that the inclination of the moon at a given location varies throughout the night. From the UK, the terminator in a half-moon is clearly sloping from top left to bottom right early in the evening, but later on, as the moon is setting, it is clearly sloping from top right to bottom left. I know that the shape of the moon changes from night to night, but I just can't see why the inclination would change during a single night. I know that the appearance of the moon is due to a combination of the tilt of the earth's axis and the latitude of my position (the moon slopes oppositely in the southern hemisphere), but I just can't visualise in three dimensions exactly what is going on. Please can you help me to do this.&lt;/p&gt;&#xA;" OwnerUserId="16501" LastActivityDate="2019-02-21T23:12:07.210" Title="Changing Appearance of the Moon During a Single Night" Tags="&lt;moon-phases&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20621" PostTypeId="2" ParentId="20619" CreationDate="2017-04-03T20:52:37.467" Score="4" Body="&lt;p&gt;The &lt;em&gt;depth&lt;/em&gt; of the transit tells you the size of the planet (relative to the star) in &lt;em&gt;both&lt;/em&gt; cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference in total eclipse width tells you the impact parameter (how far from the centre of the disk, the transit crosses). This in turn depends on the orbital inclination and the ratio of the stellar radius to the radius of the orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second constraint on inclination and the ratio of stellar radius to orbital radius comes from the ratio of total transit duration to the time when the planet is fully eclipsing the star. A more inclined transit has a shorter period of total eclipse but longer ingress and egress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a full solution then requires some estimate of the stellar mass, which leads via Kepler's third law to the orbital radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also some important additional complications due to limb darkening and eccentricity that mean, in practice, this problem is not solved analytically, but by fitting a detailed model to the light curve. For instance, the transit depth becomes smaller for transits that don't cross the centre of the stellar disk, because of limb darkening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best picture I can find is copyrighted material, but you can find it here from &lt;a href=&quot;https://www.paulanthonywilson.com/wp-content/uploads/2014/08/lc_impact-700x355.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carole Haswell's book on Transiting Exoplanets&lt;/a&gt;. The inclinations range from 90 degrees to 80 degrees.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-04-03T23:14:04.897" LastActivityDate="2017-04-03T23:14:04.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20622" PostTypeId="2" ParentId="20620" CreationDate="2017-04-03T21:47:35.670" Score="5" Body="&lt;p&gt;If you take a line from the middle of the terminator, but at 90 degrees to the termination, it more or less points towards the sun.  Now consider the moon at first quarter: as the moon rises, the sun is due south, so imagine a line from the moon to the sun, and the terminator will be at 90 degrees to this, so pointing from bottom right to top left.  At mid-day, the sun is in the west, and the terminator will be sloping bottom left to top right.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-04-03T21:47:35.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20623" PostTypeId="2" ParentId="20615" CreationDate="2017-04-03T23:14:47.087" Score="1" Body="&lt;p&gt;You might want to consult this &lt;a href=&quot;https://en.wikipedia.org/wiki/Formation_and_evolution_of_the_Solar_System#Galactic_collision_and_planetary_disruption&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page on the formation and future of our solar system and galaxy&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two events to note :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In about 4 billion years our galaxy and the Andromeda galaxy will collide (although not with a bang and they'll basically merge).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Sometime around 5-7 billion years from now the Sun will become a red giant.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;No mention of us heading into the center of the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's estimated we &quot;orbit&quot; the galaxy about once every 240 million years note that even if we were falling in it would take an minimum of that kind of time scale to reach the center of the galaxy, something I've heard no evidence for.  It's only 16 galactic orbits until the Milky Way - Andromeda merger is under way, and that's probably not enough for the solar system to &quot;fall&quot; into our galaxy's center even if we thought it might.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So most likely scenario is that we end up in the merged galaxy long before we end up reaching our solar system.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the gravitational effect of the relatively nearby stars and that of the black hole itself will be affecting the planets of the solar system?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On this point note that the black hole and the rest of the galaxy already effect us gravitationally, and have done for about 20 galactic orbits already with no obvious effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black hole at the galaxy's center is large, but actually not very much of a problem - it's gravitational influence is pretty relaxed until you're very, very, very close to it - think light year.  This is not an issue we're going to have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational forces between individual stars are pretty small on average, and we'd expect no major effect from those unless we had a relatively close approach.  The most &quot;worrying&quot; of these is the one coming in about 1.3 million years with &lt;a href=&quot;https://en.wikipedia.org/wiki/Gliese_710&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gliese 710&lt;/a&gt;.  I personally expect to be dead by then, but good luck to the rest of you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another possible close approach by a star in about 250,000 years, but that's not as widely accepted an estimate.  Again, I'll be dead by then anyway, but there's plenty of time to build a shelter in the basement if ya want to.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-04-03T23:14:47.087" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20625" PostTypeId="2" ParentId="20620" CreationDate="2017-04-04T11:47:21.220" Score="2" Body="&lt;p&gt;Adding to Chuck's answer, you may be interested in looking into the parallactic angle. Peter Meadows discusses it in a &lt;a href=&quot;http://www.petermeadows.com/html/parallactic.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;page on his site&lt;/a&gt;. There are a couple of things to keep in mind:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;that page explains it in the context of observations of the Sun rather than the Moon,&lt;/li&gt;&#xA;&lt;li&gt;the Moon's terminator doesn't point at the celestial poles.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aQ5s2.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aQ5s2.gif&quot; alt=&quot;PeterMeadows.com dawn&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/FGqJI.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FGqJI.gif&quot; alt=&quot;PeterMeadows.com dusk&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.petermeadows.com/html/parallactic.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The naked eye view of the Sun just after sunrise and just before sunset at one the equinoxes and for an observer at at London (51.5°N, 0.0°E) . The two rectangles show the field of view for an alt/azi and equatorial telescope mounting.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user15104" LastEditorUserId="7982" LastEditDate="2019-02-21T23:12:07.210" LastActivityDate="2019-02-21T23:12:07.210" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="20626" PostTypeId="1" CreationDate="2017-04-04T15:39:22.267" Score="0" ViewCount="275" Body="&lt;p&gt;I see this question in &amp;quot;An introduction to modern cosmology - Andrew Liddle - Wiley Publication&amp;quot;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In the real Universe the expansion is not completely uniform. Rather,&#xA;galaxies exhibit some random motion relative to the overall Hubble&#xA;expansion, known as their pecuLiar veLocity and caused by the&#xA;gravitational pull of their near neighbours. Supposing that a typical&#xA;(eg root mean square) galaxy peculiar velocity is 600 km/s, how far&#xA;away would a galaxy have to be before it could be used to determine&#xA;the Hubble constant to ten per cent accuracy, supposing&lt;/p&gt;&#xA;&lt;p&gt;(a) The true&#xA;value of the Hubble constant is 100 km/s Mpc?&lt;/p&gt;&#xA;&lt;p&gt;(b) The true value&#xA;of the Hubble constant is 50 km/sMpc ?&lt;/p&gt;&#xA;&lt;p&gt;Assume in your&#xA;calculation that the galaxy distance and red-shift could be measured&#xA;exactly. Unfortunately, that is not true of real observations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I really don't know how to start solving this. What do we need to find Hubble constant. I know &lt;span class=&quot;math-container&quot;&gt;$v=H.r$&lt;/span&gt; but I don't know how to calculate the effect of peculiar velocity. we don't know &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; in Hubble formula, how to find &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="16441" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-04-04T16:07:55.780" Title="Question about Hubble parameter (Hubble constant) and measuring it" Tags="&lt;cosmology&gt;&lt;redshift&gt;&lt;doppler-effect&gt;&lt;hubble-constant&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20627" PostTypeId="2" ParentId="20626" CreationDate="2017-04-04T16:07:55.780" Score="1" Body="&lt;p&gt;You can start by working out how fast a galaxy must be receding if its peculiar velocity (given as about 600 km/s) is less than or equal to 10% of its expansion velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not going to spell out the next step as you said you don't know where to start.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-04-04T16:07:55.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20629" PostTypeId="1" AcceptedAnswerId="20630" CreationDate="2017-04-04T18:54:43.827" Score="3" ViewCount="578" Body="&lt;p&gt;After doing research it said that nebulas are made of gas and dust. Are they actually gases and solids, or are they plasma?&lt;/p&gt;&#xA;" OwnerUserId="16048" LastActivityDate="2017-04-04T22:18:55.500" Title="What state of matter are nebulae in?" Tags="&lt;nebula&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20630" PostTypeId="2" ParentId="20629" CreationDate="2017-04-04T19:15:37.963" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Nebula&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nebulas&lt;/a&gt; are not static objects - they evolve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially a nebula might contain mostly (as an example) hydrogen, helium and some other elements in a cloud, but over time that material will clump together to form larger and/or denser material.  Eventually parts of some nebula can even condense enough to form a star and it's planets.  Our solar system was (probably) once part of a nebula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there's no single answer to your question.  They can contain most forms of matter.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-04-04T19:15:37.963" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20635" PostTypeId="1" CreationDate="2017-04-05T02:21:24.980" Score="0" ViewCount="1228" Body="&lt;p&gt;I don't understand and I haven't understood for a while.&#xA;How is it that all the galaxies in the universe are moving, how they are moving and how fast?&lt;/p&gt;&#xA;" OwnerUserId="16516" LastActivityDate="2017-04-05T08:53:53.810" Title="How are galaxies moving so much and why are they moving?" Tags="&lt;galaxy&gt;&lt;movement&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2017-04-05T15:32:11.753" ContentLicense="CC BY-SA 3.0" />
  <row Id="20636" PostTypeId="1" AcceptedAnswerId="20638" CreationDate="2017-04-05T07:26:03.670" Score="2" ViewCount="133" Body="&lt;p&gt;For example, a geostationary satellite travels slower than the ISS. What is going on?&lt;/p&gt;&#xA;" OwnerUserId="16519" LastActivityDate="2017-04-05T08:47:19.763" Title="Why do satellites with greater altitude travel at a slower speed?" Tags="&lt;satellite&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20638" PostTypeId="2" ParentId="20636" CreationDate="2017-04-05T08:47:19.763" Score="4" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law&quot; rel=&quot;noreferrer&quot;&gt;Kepler's third law&lt;/a&gt; in this situation tells us that&#xA;$$ \frac{R^3}{P^2} = \frac{GM}{4\pi^2}, $$&#xA;where $R$ is the orbital radius (or semi-major axis for an elliptical orbit), $P$ is the orbital period and $M$ is the mass of the Earth (actually, it is the mas of the Earth plus the mass of the satellite, but we can neglect the latter). Thus the orbital period $P \propto R^{3/2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the orbital speed, we divide the orbital circumference by the orbital period&#xA;$$ v = \frac{2\pi R}{P}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then substituting for $P$ from Kepler's third law, we get&#xA;$$ v = \sqrt{\frac{GM}{R}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the larger the orbital radius, the slower the orbital speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can just use Newton's second law and equate the centripetal acceleration to the gravitational force:&#xA;$$ m \frac{v^2}{R} = G\frac{Mm}{R^2},$$&#xA;where $m$ is the mass of the satellite, which leads directly to the same result.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-04-05T08:47:19.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20640" PostTypeId="2" ParentId="20635" CreationDate="2017-04-05T08:53:53.810" Score="2" Body="&lt;p&gt;There are several sources of movement for galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Galaxies may rotate&lt;/strong&gt;.  Galaxies are formed essentially from matter collapsing under gravity into &quot;clumps&quot;.  This process isn't uniform due to gravity from other clumps of matter and so it is very rare that all the particles in a single clump &quot;fall&quot; directly into it's centre.  This creates rotation (Angular Momentum) and so the eventual galaxy will rotate.  Note that some galaxies show very little overall rotation, with the individual stars orbiting the centre of the galaxy in a multiple directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Galaxies interact with each other&lt;/strong&gt;. Most galaxies reside within a &lt;a href=&quot;https://en.wikipedia.org/wiki/Galaxy_cluster&quot; rel=&quot;nofollow noreferrer&quot;&gt;group or cluster&lt;/a&gt; of galaxies, our galaxy is within a group called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Local_Group&quot; rel=&quot;nofollow noreferrer&quot;&gt;Local Group&lt;/a&gt; with Andromeda and several ($\geq$52) dwarf galaxies.  These dwarf galaxies are usually part of a &quot;satellite-subsystem&quot; where they orbit a large galaxy under gravity.  These orbits are affected by the other local galaxies and occasionally some systems can be &quot;expelled&quot; due to the gravitational forces happening to speed the body up past the velocity required to escape the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The major galaxies in a group or cluster interact with each other, orbiting a centre of gravity.  This is due to the same mechanism as galaxy rotation.  This is usually a very dense area of galaxies, or a massive &lt;a href=&quot;https://en.wikipedia.org/wiki/Brightest_cluster_galaxy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brightest Cluster Galaxy (BCG)&lt;/a&gt;.  In the local group, the two main galaxies (our Milky Way and the Andromeda Galaxy) are getting closer together and will &lt;a href=&quot;https://en.wikipedia.org/wiki/Andromeda%E2%80%93Milky_Way_collision&quot; rel=&quot;nofollow noreferrer&quot;&gt;collide&lt;/a&gt; in approximately 4 billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The universe is expanding&lt;/strong&gt;. The universe is expanding and so galaxies appear to be moving away from us in every direction.  We see this as &lt;a href=&quot;https://en.wikipedia.org/wiki/Redshift&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;redshift&lt;/em&gt;&lt;/a&gt;, a change in the wavelength of light as it reaches us due to the difference in speed between the origin an us.  This is not really anything to do with galaxies moving, and more to do with the fabric of the universe.  Whether this will continue, or whether the universe will stop expanding and collapse is a current &lt;a href=&quot;https://en.wikipedia.org/wiki/Ultimate_fate_of_the_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;active area of research&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16369" LastActivityDate="2017-04-05T08:53:53.810" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20642" PostTypeId="1" AcceptedAnswerId="20723" CreationDate="2017-04-05T09:43:43.880" Score="0" ViewCount="257" Body="&lt;p&gt;I read in Sofue's articles (1999) that he used a method to investigate the bulge of galaxies to obtain rotation curve of this part of galaxies. So my question is how could Vera Rubin et al. obtained the rotation curve in their article (1978), especially the steep increase in the begenning of the curve since in Sofue's article he seems to say it was rather difficult, at that time, to investigate galactic bulges ?&lt;/p&gt;&#xA;" OwnerUserId="16521" LastActivityDate="2017-05-14T18:38:25.947" Title="rotation curve in the innermost part of galaxies" Tags="&lt;rotation&gt;&lt;dark-matter&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20643" PostTypeId="2" ParentId="20597" CreationDate="2017-04-05T15:59:34.290" Score="3" Body="&lt;p&gt;Looking at the simulation, and the gifs attached here, some things are clear. The dots cannot represent actual meteoroids in their orbits, as the 6 orbital elements fully determine the state vectors of a body. You can't be in the same orbit, at the same place, at the same time and have very different velocity. Thus the dots don't obey Kepler's third law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not clear that they obey the second law either (that equal areas are swept out in equal times) They do seem to move faster at perihelion but perhaps not fast enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graphic doesn't seem to show the &quot;lumpiness&quot; of some meteor streams. Notably the Leonids, which have a very intense storm, roughly every 33 years. Other streams are also lumpy which doesn't seem to be apparent in this graphic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What it does show is the variation in orbit of the meteoroids that form a meteor shower. This is where (I guess) Nasa was involved. From the observation of a meteor (ideally several observations of the same meteor to triangulate its position) it should be possible to work out its orbit before it fell into Earth's gravitational field and hit the atmosphere. Observe enough meteors and you can get an idea of the distribution of orbits: the mean and standard deviation in the inclination, eccentricity, semimajor axis, ascending node and angle of perihelion. From these, you can get a distribution of elliptical orbits. There may be some modelling work do be done: we can only observe those meteorids that hit the Earth, but we can suppose the they are distributed all around the parent body. But from our sample of meteoroids that hit the Earth, we can model the distribution of those that never will.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dots are drawn on these Keplerian orbits, but the speed at which the move on the orbit doesn't seem to represent the actual speed of meteoroids in space. Rather it is a way of illustrating the variation in elliptical orbits, without drawing a solid ring for each orbit.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-04-05T16:14:17.887" LastActivityDate="2017-04-05T16:14:17.887" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20644" PostTypeId="1" CreationDate="2017-04-05T21:04:04.630" Score="2" ViewCount="103" Body="&lt;p&gt;Given a nearly infinite timeline, would the expansion of space eventually cause individual particles to become &quot;unobservable&quot; from one another due to the vast distance and increasing rate of expansion with that distance.&lt;/p&gt;&#xA;" OwnerUserId="16529" LastActivityDate="2017-04-05T22:58:06.247" Title="Expansion of Space in the Very Distant Future" Tags="&lt;expansion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20645" PostTypeId="1" CreationDate="2017-04-05T21:21:42.977" Score="1" ViewCount="475" Body="&lt;p&gt;Using data from Hipparcos is it possible to calculate Radial Velocity.  If we take 19 Leonis as example, RV is -3.70.  Is it calculated from Proper Motion?  Can I ask that any formulas be done in long hand please, i.e. how you would enter it into a standard spreadsheet rather than with symbols etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-basic?Ident=19+Leonis&amp;amp;submit=SIMBAD+search&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://simbad.u-strasbg.fr/simbad/sim-basic?Ident=19+Leonis&amp;amp;submit=SIMBAD+search&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Took the advice and found CRVAD2.dat ( &lt;a href=&quot;http://cdsarc.u-strasbg.fr/viz-bin/Cat?III/254&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://cdsarc.u-strasbg.fr/viz-bin/Cat?III/254&lt;/a&gt; ).  The following line comes from CRVAD2.dat file and when I looked up HipId 40829 which is below, the RV value is -3.70 which matches the link above.  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9.790554129  11.56817726   0.8   0.1   11.06   1.00   -48.63    -3.39   1.24   0.69  6.709  6.436 0.005 0.005 0.007   D   AY D A3n          A 7 VN                48029  84722  948942 4 ABA    -3.70  -9.99 B   9 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If I look up Sirius (32349), it says the RV is -7.60 but when you look up Sirius at Simbad, it says -5.50 ( &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-basic?Ident=Sirius&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://simbad.u-strasbg.fr/simbad/sim-basic?Ident=Sirius&lt;/a&gt; ) &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;6.752569403 -16.71314306   1.2   0.1  379.20   1.58  -552.63 -1206.40   0.50   0.50 -1.430 -1.121 0.011 0.002 0.039   U     OD A1V+A5wd     A1V+A5wd              32349  48915 1447791 4 AB     -7.60  -9.99 A   0 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Whats the reason for the discrepancy, CRVAD2.dat is an old version like there's two Hip data files (97 and 07)?&lt;/p&gt;&#xA;" OwnerUserId="5401" LastEditorUserId="5401" LastEditDate="2017-04-07T21:24:10.970" LastActivityDate="2017-04-07T21:24:10.970" Title="Calculating Radial Velocity from Hipparcos" Tags="&lt;star&gt;&lt;hipparcos&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20646" PostTypeId="2" ParentId="20645" CreationDate="2017-04-05T22:10:38.253" Score="3" Body="&lt;p&gt;You cannot calculate the radial velocity from a proper motion. The proper motion is the motion you see the star move on the sky, so perpendicular to your line of sight towards the star. The radial velocity is the velocity of the star along this line of sight. Perhaps this picture will help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/F1xwY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F1xwY.jpg&quot; alt=&quot;Proper motion picture&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Proper_motion&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Proper_motion&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radial velocities are actually calculated by measuring the Doppler shift of the spectra of stars, but this is not very easy, and you need the raw spectrum data. You are probably best off just taking the values of the radial velocity from that site for example.&lt;/p&gt;&#xA;" OwnerUserId="16531" LastActivityDate="2017-04-05T22:10:38.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20647" PostTypeId="2" ParentId="20644" CreationDate="2017-04-05T22:58:06.247" Score="3" Body="&lt;p&gt;Maybe someone can give chapter and verse on this, but my understanding is that bound objects are unaffected by the expansion for the case of dark energy represented by a simple cosmological constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Rip&quot; rel=&quot;nofollow noreferrer&quot;&gt;big rip&lt;/a&gt;&quot; will occur in a finite time if the dark energy equation of state $P = w \rho$ is characterised by $w &amp;lt; -1$. In this scenario, the expansion accelerates so quickly that the scale factor becomes infinite in a finite time and structures must disintegrate when the size of the observable universe becomes smaller than that structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The present state of play is that $w$ appears to be very close to -1 (a cosmological constant). But given that dark energy is not understood, who can really say whether it might change in the future...&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-04-05T22:58:06.247" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20648" PostTypeId="1" CreationDate="2017-04-05T23:50:40.733" Score="0" ViewCount="626" Body="&lt;p&gt;Is gravitational lensing responsible for obscuring (figuratively) the supposed supermassive black holes at the center of our own and other galaxies? By &quot;see&quot; I mean relative to the measuring tools we are using. A radio telescope &quot;sees&quot; in radio waves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Hubble photos of galaxies, no black holes appear to be present in galaxy centers, at least not visually. How can this be explained?&lt;/p&gt;&#xA;" OwnerUserId="16533" LastEditorUserId="2365" LastEditDate="2017-04-07T13:33:07.140" LastActivityDate="2017-04-07T13:33:07.140" Title="Why can't we see the supermassive black holes supposedly at the center of galaxies?" Tags="&lt;astrophysics&gt;&lt;supermassive-black-hole&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20649" PostTypeId="2" ParentId="20645" CreationDate="2017-04-06T00:52:21.667" Score="1" Body="&lt;p&gt;As user2998 correctly states, one cannot measure the RV from proper motion, simply because they are perpendicular to each other.&lt;br&gt;&#xA;What needs to be done to get the RV is to measure the &lt;a href=&quot;https://en.wikipedia.org/wiki/Doppler_shift&quot; rel=&quot;nofollow noreferrer&quot;&gt;Doppler shift&lt;/a&gt; of one or several spectral lines relative to laboratory lines in the atmosphere of a given star. This is usually done from the ground in follow-up surveys. This means that once a star is identified and has measured proper motion, there are people on the ground that take the star catalogues from e.e. Hipparcos, find it again (due to known proper motion) and measure its RV.&lt;br&gt;&#xA;You can find links to the paper where your value of $-3.70 km/s$ originates from on simbad, directly below this number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there is the follow up mission for Hipparcos, which is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Gaia_(spacecraft)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gaia&lt;/a&gt;, which itself has limited RV capabilities. But also for this satellite there are already ground follow-ups in motion.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2017-04-06T00:52:21.667" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20650" PostTypeId="2" ParentId="20648" CreationDate="2017-04-06T04:20:03.057" Score="5" Body="&lt;p&gt;The main reason supermassive black holes aren't visible in images of our galaxy is because of all the dust and stuff in between the center of the galaxy and us.  The Milky Way's bulge is ~16000 light-years thick.  There are millions of stars, and huge clouds of dust in between us and the black hole, which block visible light.  Most of our measurements of the center of our galaxy are made in radio wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, compared to galaxies, black holes are small.  A supermassive black hole &lt;a href=&quot;https://en.wikipedia.org/wiki/Black_hole#Physical_properties&quot; rel=&quot;nofollow noreferrer&quot;&gt;might have a radius&lt;/a&gt; of up to 400 AU (for the really large ones).  The nearest spiral galaxy to us, &lt;a href=&quot;https://en.wikipedia.org/wiki/Andromeda_Galaxy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Andromeda&lt;/a&gt;, is 778,000 parsecs away.  This corresponds to an angular size of 0.0005 arcseconds.  Even the &lt;a href=&quot;https://www.spacetelescope.org/about/faq/#21&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble Space Telescope&lt;/a&gt; only has a resolution of 0.043 arcseconds in the optical.  That means that Hubble can only see things that are 100 times this size.  So we wouldn't even be able to see any black holes in another galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, we can make observations of black holes using indirect means.  For example, one of the ways we are able to precisely measure the black hole in our galaxy is by &lt;a href=&quot;https://phys.org/news/2016-06-star-black-hole.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;watching stars orbit around it&lt;/a&gt;.  We take pictures of these stars in the infrared and radio wavelengths, which pass through the dust.  By measuring &lt;a href=&quot;https://www.youtube.com/watch?v=AZhUQl-wmq0&quot; rel=&quot;nofollow noreferrer&quot;&gt;how fast these stars are moving&lt;/a&gt;, we can calculate the mass of the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other galaxies, we can observe other effects of black holes, such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Active_galactic_nucleus&quot; rel=&quot;nofollow noreferrer&quot;&gt;active galactic nuclei&lt;/a&gt;, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Quasar&quot; rel=&quot;nofollow noreferrer&quot;&gt;quasars&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So despite being unable to get an optical picture of a black hole, we still know that they're there.&lt;/p&gt;&#xA;" OwnerUserId="13426" LastEditorUserId="13426" LastEditDate="2017-04-06T20:44:45.407" LastActivityDate="2017-04-06T20:44:45.407" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20651" PostTypeId="2" ParentId="20645" CreationDate="2017-04-06T08:03:54.987" Score="2" Body="&lt;p&gt;For your purpose, I would suggest you follow the advice of the other answers - proper motion measures the component of velocity perpendicular to a line joining the Earth and the star and won't ordinarily tell you anything about radial velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, I wanted to point out to the interested reader, that it has long been known that radial velocities &lt;em&gt;can&lt;/em&gt; be estimated from sufficiently accurate astrometric (parallax and proper motion) in at least three ways. In the forthcoming Gaia era, these might be useful, and the third of these methods &lt;em&gt;has&lt;/em&gt; been used with Hipparcos data to find radial velocities in nearby clusters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These geometric (as opposed to spectroscopic) methods are described by &lt;a href=&quot;https://arxiv.org/abs/astro-ph/9907145&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dravins et al. 1999&lt;/a&gt;, and are of great interest because they allow the study of phenomena that affect spectroscopic methods (e.g. General Relativistic redshift). The methods are - and I summarise here; the full details are in Dravins et al.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method 1: Look for a change in parallax with time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method 2: If a star is moving away as well as at right angles to us, then the proper motion will change with time. This is both as a consequence of the change in distance, but also due to the change in perspective as the apparent position of the star changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you might imagine, the first two effects are small on the timescales of the few years of a space mission. They may become possible to apply to nearby stars when we have the full Gaia astrometry dataset in a few years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method 3: If a star belongs to a known moving group or cluster of stars, then its radial velocity can be deduced from the total velocity of the cluster, the proper motion and the parallax, since the vector sum of the radial velocity and tangential velocity must equal the total velocity. This method does not measure a rate of change and requires less precision. It can and has been used for nearby, large clusters and kinematic groups of stars. Obviously it can only be applied to stars that are known or suspected group members.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-04-06T08:03:54.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20652" PostTypeId="1" CreationDate="2017-04-06T08:05:41.097" Score="3" ViewCount="146" Body="&lt;p&gt;Do IRAS sources refer to star forming regions? I searched google and I came to know that it stands for IR astronomical satellite.But what do they search for? Please help me&lt;/p&gt;&#xA;" OwnerUserId="11434" LastActivityDate="2017-04-06T09:13:46.957" Title="What are IRAS sources?" Tags="&lt;astrophysics&gt;&lt;observational-astronomy&gt;&lt;star-formation&gt;&lt;space-telescope&gt;&lt;infrared&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20653" PostTypeId="2" ParentId="20652" CreationDate="2017-04-06T09:13:46.957" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://heasarc.gsfc.nasa.gov/W3Browse/iras/iraspsc.html&quot; rel=&quot;noreferrer&quot;&gt;The IRAS Point Source Catalog&lt;/a&gt;, Version 2.0, is a catalog of some 250,000 well-confirmed infrared point sources observed by the Infrared Astronomical Satellite (IRAS), i.e., sources with angular extents less than approximately 0.5, 0.5, 1.0, and 2.0 arcminutes in the in-scan direction at 12, 25, 60, and 100 microns (um), respectively. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This includes some comets, planetary disks, and even minor planets. Some sources are likely to be star forming, but an IR source meeting the spec above could be many things.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2017-04-06T09:13:46.957" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20654" PostTypeId="1" AcceptedAnswerId="20657" CreationDate="2017-04-06T14:28:46.463" Score="4" ViewCount="748" Body="&lt;p&gt;In a lecture which I attended at my institution on gravitational lensing, the lecturer linked gravitational lensing and dark matter research. Can anyone explain what is the relation of both?&lt;/p&gt;&#xA;" OwnerUserId="14301" LastEditorUserId="7472" LastEditDate="2021-03-01T09:38:16.423" LastActivityDate="2021-03-01T09:38:16.423" Title="How are gravitational lensing and dark matter related?" Tags="&lt;dark-matter&gt;&lt;gravitational-lensing&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20655" PostTypeId="2" ParentId="20654" CreationDate="2017-04-06T15:02:03.443" Score="3" Body="&lt;p&gt;Microlensing by massive compact halo objects (MACHOs)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it is generally believed that most dark matter is in the form of exotic particles that don't interact with normal matter, a certain proportion of the dark matter in the universe is made up of ordinary matter that is simply dark. This matter could be in the form of brown dwarfs, rogue planets, or even small black holes. The common property is that the mass in these objects is formed of normal atoms, not from anything exotic. But these objects are invisible because they are too cool to shine, and too small to image directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a MACHO were to pass in front of a star, the mass of the MACHO would cause light from the star to be slightly bent, and become focussed. The result would be the star would briefly brighten. &lt;a href=&quot;https://i.stack.imgur.com/wiDR5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wiDR5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;(image by Jan Skowron, under cc-sa-2.5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The frequency of such events places limits on how many such MACHOs there are in space. The conclusion is that there are not nearly enough to account for all the mass in the galaxy, and so there must be something else, something exotic, a particle that we haven't yet detected, a weakly interacting massive particle, or WIMP. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So gravitational lensing is used to detect a specific type of dark matter, the MACHO.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-04-06T15:02:03.443" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20656" PostTypeId="1" CreationDate="2017-04-06T15:45:05.957" Score="1" ViewCount="750" Body="&lt;p&gt;What would give a gas giant a red color (dark or light)? &lt;/p&gt;&#xA;" OwnerUserId="16490" LastActivityDate="2017-04-08T17:22:35.537" Title="What elements and compounds would be present in a red gas giant?" Tags="&lt;exoplanet&gt;&lt;gas-giants&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20657" PostTypeId="2" ParentId="20654" CreationDate="2017-04-06T15:53:36.117" Score="7" Body="&lt;p&gt;Gravitational lensing of background galaxies and quasars is used to probe the amount of gravitating matter in foreground objects such as clusters of galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example, taken with the Hubble Space Telescope.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/k7bzA.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k7bzA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The foreground object is a cluster of galaxies and the lensed background objects, some of which are marked, are galaxies and quasars that are much further away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shapes of these lensed objects can be used to deduce the gravitational potential in the foreground cluster. This potential is due both to the galaxies that can be seen, hot intra-cluster gas (that can be seen in X-rays), but chiefly is due to dark matter that pervades the cluster. Thus a combination of optical, X-ray and lensing observations can map the distribution of dark matter and hot gas in the cluster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example is shown below for Abell 1689, where the diffuse, blobby emission is a false-colour representation of the density distribution of dark matter inferred to be in this cluster from the many lensed images of background galaxies that can be seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IkssI.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IkssI.jpg&quot; alt=&quot;Abell 1689&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;http://hubblesite.org/news_release/news/2010-37&quot; rel=&quot;noreferrer&quot;&gt;http://hubblesite.org/news_release/news/2010-37&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A particularly important example that is a landmark in dark matter research is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Bullet_Cluster&quot; rel=&quot;noreferrer&quot;&gt;Bullet Cluster&lt;/a&gt;. This is actually two clusters that appear to have collided and passed through each other about 150 million years ago. The distribution of galaxies in the clusters can be seen in optical images; the hot gas can be traced by X-ray observations and the dark matter is traced by looking at the lensed images of distant background galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key image is shown below - it illustrates the distribution of optical galaxies, the X-ray-emitting hot gas (pink/red) and the inferred distribution of dark matter (blue).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the clusters move through each other, the hot gas interacts and piles up into bow-shocks that are displaced from the distribution of optical galaxies. On the other hand, the dark matter, which is thought to consist of non-baryonic, non-interacting particles remains centred on the visible galaxies and each dark matter &quot;cloud&quot; has apparently passed through the other with no interaction occuring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The displacement between the hot gas and the majority of the gravitating matter, that is dark, is very strong evidence for a non-baryonic particle description of dark matter, rather than some of the alternatives to dark matter like &lt;a href=&quot;https://en.wikipedia.org/wiki/Modified_Newtonian_dynamics&quot; rel=&quot;noreferrer&quot;&gt;Modified Newtonian Dynamics&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XCMF1.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XCMF1.jpg&quot; alt=&quot;The Bullet Cluster&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-04-06T16:22:00.563" LastActivityDate="2017-04-06T16:22:00.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20666" PostTypeId="1" AcceptedAnswerId="20667" CreationDate="2017-04-07T16:45:58.420" Score="0" ViewCount="71" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sugar_Grove_Station&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sugar Grove Station&lt;/a&gt; was an outpost designed to listen in on Soviet radio signals reflected off the moon.  How would this work?  Wouldn't the radio signals be distorted by the imperfect surface of the moon?  And wouldn't they also be extremely weak since they wouldn't be beamed directly at the moon?&lt;/p&gt;&#xA;" OwnerUserId="13426" LastActivityDate="2017-04-07T18:32:36.740" Title="How was Sugar Grove Station supposed to work?" Tags="&lt;radio-astronomy&gt;&lt;communication&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20667" PostTypeId="2" ParentId="20666" CreationDate="2017-04-07T18:32:36.740" Score="3" Body="&lt;p&gt;I can't comment in detail but &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth%E2%80%93Moon%E2%80%93Earth_communication&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth-Moon-Earth&lt;/a&gt; communication at radio wavelengths is well established.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At wavelengths longer than a few cm (frequencies below 10 GHz), the Moon appears smooth and acts like a reflective sphere as far as radio waves are concerned.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-04-07T18:32:36.740" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20668" PostTypeId="2" ParentId="20656" CreationDate="2017-04-07T19:35:26.013" Score="1" Body="&lt;p&gt;It's a more complicated question than it appears.   Also, a couple points are in order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you're thinking about gas giant color is wrong.  It's not the ingredients as much as the temperature.   The ingredients don't change all that much.  They change some, but gas giant color has more to do with temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two, color is a very narrow perspective.  We see, only a narrow band of the spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/L1tI8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L1tI8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NASA brings a variety of cameras on their missions, Infra Red, Radio, UV as well as visible light.   See &lt;a href=&quot;http://www.nasa.gov/mission_pages/newhorizons/spacecraft/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, many photos that we see of planets are doctored to make the variations stand out more visibly, and I think it's great that NASA does this.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And three, the planet's star can affect it's color.   Color is different under a blue star compared to a red dwarf, so there's 2 colors to consider, color under a white light and color under the planet's star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that said, color's still interesting and fun and it tells us something.   Using visible light only, many gases are largely transparent, but that doesn't mean they're colorless.  Any gas will reflect and diffract some sunlight and gas giants have atmospheres thousands of miles thick.  There's no such thing as a transparent gas giant planet, with an earth like atmosphere that you can see through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen and Helium are both highly transparent to visible light but bluish when thick enough.  (Neptune gets it's blue color from a combination of it's hydrogen and helium and trace amounts of Methane ice).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upper-atmosphere temperature is also very important.    Saturn, Uranus and Neptune are cold enough that mostly their color is defined by ice, which tends to be more reflective, giving them lighter and generally considered &quot;prettier&quot; colors.   Uranus and Neptune's methane ice give them their blue/green and more pure blue colors.    Saturn is a little too warm for Methane ice, but it has ammonia ice giving it a lighter color and other gases in it's upper atmosphere giving it some darker yellow and orange colors.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter's bands can be color-differentiated by where there is abundant ice high in it's atmosphere and where there is more gas.   The darker bands are warmer, generally circulating upwards from deeper in the planet, like Hadley cells on Earth, driven in part by Jupiter's fast rotation and strong Coriolis effect and significant internal heat.   It's lighter bands are colder, where ice reflection is the primary factor in it's color.   Jupiter also has some complex hydrocarbons in it's upper atmosphere due to chemistry, intense lightning and atmospheric mixing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there's quite a few difficulties in answering this because a gas giant planet's color depends on many factors, How well it's upper atmosphere circulates, it's temperature and what ices, if any, can form in it's upper atmosphere.   Any trace elements that might carry color, even a 1 part in tens of thousands, can be enough for a dominant color if the other gases are mostly transparent.   Take Earth for example - clouds are a very small part of the atmosphere by mass but they change the color of the sky (viewed from space) significantly.  This makes predicting a gas giant planet's color very tricky and borderline guess-work, because one trace gas can be enough to change it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gas giants are mostly hydrogen and helium in their upper atmospheres, but, as noted above, it's often the other gases, ices or trace gases, not the hydrogen and helium that will give them their color.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I always thought Jupiter's &quot;great red spot&quot; was misnamed.  It looks more like an orange-brown to me.    But if that counts as red, then red might be relatively common.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be UV rays and a trace gas, ammonium hydro-sulfide, that gives Jupiter's spot it's color.  See article &lt;a href=&quot;http://www.sciencemag.org/news/2016/03/cosmic-rays-help-give-jupiter-s-great-red-spot-its-color&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.   I gather there's still some debate on that question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a &lt;a href=&quot;http://Astronomy.stackexchange.com/questions/14032/color-of-planets&quot;&gt;question&lt;/a&gt; with a nice answer on the colors of the planets.   Venus is unusually bright of the reflective SO2 high in it's atmosphere.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-04-08T17:22:35.537" LastActivityDate="2017-04-08T17:22:35.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20669" PostTypeId="1" CreationDate="2017-04-07T23:35:50.717" Score="-1" ViewCount="289" Body="&lt;p&gt;Is there a practical way to determine the galactic coordinates directly from the equatorial system? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, I can't find the $(l,b)$ for the Pleiades star cluster (M45) anywhere. I am trying to determine if they reside in the thick disk or in the thin disk. &#xA;(I am assuming these are the possible options, since the halo mainly contains globular clusters and the bulge is too messy so that the Pleiades wouldn't be so noticeable.)&lt;/p&gt;&#xA;" OwnerUserId="14422" LastActivityDate="2017-04-08T06:25:58.760" Title="Pleiades galactic coordinates" Tags="&lt;coordinate&gt;&lt;star-cluster&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20670" PostTypeId="2" ParentId="20669" CreationDate="2017-04-08T06:25:58.760" Score="1" Body="&lt;p&gt;The galactic coordinate system is centred on the Earth (not the galactic centre) with 0 degrees longitude in the direction of the galactic centre. (as measured by observation of the distribution of neutral hydrogen). The physical centre, marked by Sagittarius A* is offset by less than 0.1 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coordinate transform from equatorial (right ascension, declension) to galactic is just spherical trigonometry&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formulae, copied from &lt;a href=&quot;https://en.wikipedia.org/wiki/Celestial_coordinate_system#Converting_coordinates&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$l = 303^\circ - \arctan\left({\sin(192.25^\circ - \alpha) \over \cos(192.25^\circ - \alpha) \sin 27.4^\circ - \tan\delta \cos 27.4^\circ}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin b = \sin\delta \sin 27.4^\circ + \cos\delta \cos 27.4^\circ \cos (192.25^\circ - \alpha)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This uses the B1950 coordinates, which differs slightly from the J2000 coordinates, due to precession.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine the 3D position of a star cluster relative to the disk, you would also need distance information.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-04-08T06:25:58.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20673" PostTypeId="2" ParentId="20607" CreationDate="2017-04-08T23:17:33.420" Score="3" Body="&lt;p&gt;The term &quot;superficial&quot; means &quot;on the surface,&quot; so &quot;superficial gravity&quot; is the same as surface gravity. As @pela pointed out in the comments, the author of the article is Portuguese and in that language the word for surface is 'superficie.'&lt;/p&gt;&#xA;" OwnerUserId="4042" LastActivityDate="2017-04-08T23:17:33.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20674" PostTypeId="1" AcceptedAnswerId="20675" CreationDate="2017-04-09T05:56:08.887" Score="2" ViewCount="143" Body="&lt;p&gt;I was looking at a star recently and some other object that appeared to be stationary in relation to the star was occluding the light to only one of my eyes. This is why it caught my attention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was able to move my head around this shadow and see the star but there was a definite tiny area where the star was clearly hidden. The object hiding the star also appeared to be a similar size to the tiny point of light of the star. Is this common to observe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the last few days I have been asking myself how many objects could be hidden like this (though most with shadows that engulf the earth); How thick are asteroid belts etc, that are between our eyes and the rest of the stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is most of it hidden?&lt;/p&gt;&#xA;" OwnerUserId="16561" LastActivityDate="2017-04-09T13:03:47.897" Title="Do objects that are invisible to the naked eye occlude much of the stars to Earth based observers?" Tags="&lt;star&gt;&lt;observational-astronomy&gt;&lt;amateur-observing&gt;&lt;asteroids&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20675" PostTypeId="2" ParentId="20674" CreationDate="2017-04-09T07:13:33.587" Score="4" Body="&lt;p&gt;From your description, an object that blocks out light from a star in only one location cannot be an astronomical object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Occultations by asteroids are fairly common. You can see a list at  &lt;a href=&quot;http://www.asteroidoccultation.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.asteroidoccultation.com/&lt;/a&gt;. When a star is being occluded by an asteroid there will be a fairly narrow strip from which the star will be hidden. But it is still about 50km wide. The star appears to rapidly fade, and it remains hidden for at most a few seconds, then reappears. Your description of &quot;only hidden to one of my eyes&quot; doesn't fit an asteroid occultation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The object hiding the star must, therefore, have been local, and not moving. It would be speculation to suggest what it could be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are may objects that are hidden in visible light, not by asteroids, but by interstellar dust clouds. We can see them by using other wavelengths.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-04-09T13:03:47.897" LastActivityDate="2017-04-09T13:03:47.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20676" PostTypeId="1" CreationDate="2017-04-09T13:31:16.110" Score="1" ViewCount="99" Body="&lt;p&gt;I observed a greenish blue patch in the red sky after sunset, and it stayed for not more than 5 minutes. It didn't curve or appear distorted; it was just like a carpet. Why is it so? Could it have been an aurora? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;location:&lt;/strong&gt; 9° N, 78° E; Temperature 35 deg C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CXH6Z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CXH6Z.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11237" LastEditorUserId="7982" LastEditDate="2017-04-10T06:02:44.747" LastActivityDate="2017-04-10T06:02:44.747" Title="A greenish blue patch in the sky during sunset" Tags="&lt;observational-astronomy&gt;&lt;amateur-observing&gt;" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20677" PostTypeId="1" CreationDate="2017-04-09T15:35:56.860" Score="3" ViewCount="125" Body="&lt;p&gt;Why are Fabry-Perot Interferometers (FPIs) and Fourier Transform Spectrographs (FTSs less common than Integral Field Spectrographs (IFS) nowadays? My understanding is that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;With FPIs and FTSs you have to scan your spectral range over time. This (A) Takes more telescope time, and (B) Is affected by seeing variations over time (i.e., different wavelengths end up having different PSFs).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Limited spectral range. As a result, you have limited information about the observed object.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I was also under the impression that reducing FPI or FTS data is more complex. However I do not have a reference to support this claim. It this true? If it is, can someone explain why?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For some reason FPIs use Image Photon Counting System (IPCS) instead of CCDs. Why is that? Does this somehow make them &quot;worse&quot; than IFSs?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What other advantages does IFS have over FPI and FTS?&lt;/p&gt;&#xA;" OwnerUserId="1986" LastEditorUserId="7411" LastEditDate="2017-04-21T13:55:48.727" LastActivityDate="2017-05-21T15:13:04.237" Title="Integral Field Spectroscopy vs Fabry-Perot Interferometry" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20678" PostTypeId="1" CreationDate="2017-04-09T17:00:14.540" Score="2" ViewCount="863" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;We have a telescope that can see stars with about +14 magnitude. how to find its lens (or mirror) diameter? (I mean the $D$ in formula $\theta = 1.22 \lambda / D $)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know whether I can assume that the temperature of star is equal to temperature of the sun or not. but if we think that they are equal we get this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_{\mathrm{sun}} - m_{\mathrm{star}} = -2.5 \log_{10} (\frac{b_{\mathrm{sun}}}{b_{\mathrm{star}}}) = -2.5 \log (\frac{\sigma t^4 \times 4 \pi R_{\mathrm{sun}}^2 /(4 \pi r^2) }{\sigma t^4 \times 4\pi R_{\mathrm{star}}^2 / (4\pi r_{\mathrm{star}}^2)})$ if we solve this with what we know about magnitude of sun (-26.83 or something close to it between -25 to -27) and another things like radii we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($R/r = 3.16605 \times 10^{-11} \mathrm{rad} = \tan \theta/2 = \theta/2$ and from that and formula&#xA;$\theta = 1.22 \lambda / D $ and $\lambda = 550 nm$ we get something like D = $10596.8 \mathrm{m}$. which is very big for a telescope. what's the problem? how to solve this? Is this answer correct?&lt;/p&gt;&#xA;" OwnerUserId="16441" LastEditorUserId="9301" LastEditDate="2017-04-09T19:07:38.137" LastActivityDate="2017-04-09T20:29:14.593" Title="Finding the lens diameter of a telescope by magnitude" Tags="&lt;telescope&gt;&lt;magnitude&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20679" PostTypeId="2" ParentId="20678" CreationDate="2017-04-09T18:36:16.580" Score="3" Body="&lt;p&gt;The unaided eye can typically see mag 6 objects. With your telescope, you can see an additional 8 magnitudes. This requires a factor of $100^{8/5}$ of additional light gathering power (since 5 magnitudes equals 100 times the brightness). I.e. about 1580 times larger aperture than the pupil of the eye.  Assuming the pupil has a diameter of 7mm, then the diameter of your telescope would have to be $7\times 1580^{1/2}$, which is about 280 mm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is fairly close I think - perhaps a bit on the high side but not much.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastEditorUserId="2365" LastEditDate="2017-04-09T20:29:14.593" LastActivityDate="2017-04-09T20:29:14.593" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20680" PostTypeId="1" CreationDate="2017-04-09T19:59:20.583" Score="0" ViewCount="168" Body="&lt;p&gt;I've come across &lt;a href=&quot;http://cdsarc.u-strasbg.fr/viz-bin/Cat?I/311#sRM2.1&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://cdsarc.u-strasbg.fr/viz-bin/Cat?I/311#sRM2.1&lt;/a&gt;  and am trying to decipher the hip2.dat file to work out whether a star is a solo, binary or multiple star.  Ideally would be interested in knowing how many stars are in the system but I'd be happy with just knowing if its (SBM) solo, binary or multi.  What I don't understand is which item in the line identifies its companionship (SBM), I believe it its 8-10 but the value for 32349 is 0 when its widely known that Sirius is a double star system so I think I'm wrong there.  I've deliberately excluded the names of the star.  I'm not expecting the answers to everyone, I just need the explanation of how to decipher one line and i'll do the rest because I can't understand what they mean.  Do the values under - d and -s refer to bitwise values of the value in 8-10?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know 49669 has 4 stars in its system, 32349 has 2 and 79374 has an alleged 7.  The other descriptions of the other items in the list are about Parallax and Coordinates, nothing that appears related to the information what I'm after. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the stars I looked up are all either binary or multiple stars, I couldn't find a solo star, apart from the Sun (debatable).  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 49669   5 0 1  2.6545337495  0.2088669583   41.13  -248.73     5.59   0.32   0.21   0.35   0.35   0.21  75  9.03  0    0.0    0  1.3232 0.0006 0.005 0 -0.087 0.015 -0.100   1.23   4.74   1.02   0.34   0.56   1.00   3.14   0.67  -0.35   1.11   2.94   1.16  -0.22   4.20   1.00&#xA; 91262   1 0 1  4.8735545728  0.6768909262  130.23   200.94   286.23   0.26   0.28   0.36   0.32   0.40  98 -0.14  0    1.8    0  0.0868 0.0021 0.014 2 -0.001 0.005 -0.010   1.01   0.00   1.00   0.24   0.08   1.11  -0.02  -0.09   1.14   1.00  -0.09  -0.18   0.76  -0.08   1.00&#xA; 79374  15 3 2  4.2411312043 -0.3396523621    6.88    -7.65   -23.71   0.64   0.46   0.76   0.71   0.47  46 11.32  0    0.0    0  4.0076 0.0011 0.007 0  0.076 0.044  0.140   1.05  -0.37   1.19   0.83   1.54   1.08  -0.70   0.29  -0.40   1.00   0.51  -1.45   2.12  -0.13   1.00&#xA; 50583  17 3 2  2.7051266664  0.3463057904   25.07   304.30  -154.28   0.58   0.32   0.52   0.71   0.36  77 15.43  1    0.0  608  2.1684 0.0011 0.006 2  1.128 0.029  1.170   1.31   1.31   1.09   0.33   0.78   1.07   0.72   1.28  -0.91   1.04   2.13  -0.56   0.89   0.88   1.07&#xA; 32349   0 4 1  1.7678185359 -0.2916993748  379.21  -546.01 -1223.07   1.21   1.04   1.58   1.33   1.24  47  0.00  0    0.0    0 -1.0876 0.0024 0.040 0  0.009 0.007 -0.020   0.83  -0.04   0.96  -0.21   0.32   0.68   0.46  -0.12   0.13   0.90  -0.20   0.62  -0.14  -0.16   1.00&#xA; 11767   5 0 1  0.6622851337  1.5579531082    7.54    44.48   -11.85   0.09   0.10   0.11   0.11   0.13 127  1.08  2    0.0    0  2.1077 0.0021 0.014 1  0.636 0.003  0.700   1.02   1.23   1.04  -0.42  -2.11   1.01   1.67  -0.26  -0.93   1.06  -0.06   1.42   0.64   2.87   1.00&#xA;&#xA;&#xA;Bytes Format Units    Label   Explanations&#xA;&#xA;1-  6  I6    ---      HIP     Hipparcos identifier&#xA;8- 10  I3    ---      Sn      [0,159] Solution type new reduction (1)&#xA;12  I1    ---      So      [0,5] Solution type old reduction (2)&#xA;14  I1    ---      Nc      Number of components&#xA;&#xA;Note (1): Solution type.&#xA;The solution type is a number 10xd+s consisting of two parts d and s:&#xA;- s describes the type of solution adopted:&#xA;  1 = stochastic solution (dispersion is given in the 'var' column)&#xA;  3 = VIM solution (additional parameters in file hipvim.dat)&#xA;  5 = 5-parameter solution (this file)&#xA;  7 = 7-parameter solution (additional parameters in hip7p.dat)&#xA;  9 = 9-parameter solution (additional parameters in hip9p.dat)&#xA;- d describes peculiarities, as a combination of values:&#xA;  0 = single star&#xA;  1 = double star&#xA;  2 = variable in the system with amplitude &amp;gt; 0.2mag&#xA;  4 = astrometry refers to the photocenter&#xA;  8 = measurements concern the secondary (fainter) in the double system&#xA;Note (2): as follows:&#xA;  0 = standard 5-parameter solution&#xA;  1 = 7- or 9-parameter solution&#xA;  2 = stochastic solution&#xA;  3 = double and multiple stars&#xA;  4 = orbital binary as resolved in the published catalog&#xA;  5 = VIM (variability-induced mover) solution&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Edit 2. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 09474+1134 A          +001.89+0005.5 10                02  6.4 A7*-0064 0000+12 2095.0+11 1142.6  84722          09420N1202A  48029&#xA; 09474+1134 B* MAL  34                   1980 206   0.1 12                                                        09420N1202B  48029&#xA; 10083+1159 A          +004.30-0058.1 10                04  1.3 B8*-0248+0003+12 2149.0   98967.8  87901  A 7654  10030N1227A  49669&#xA; 10083+1159 B  STF5006 -005.13+0048.7 10 1836 307 176.9 24  7.6 G *-0225+0006+12 2147.0+12 1198.6  87884  A 7654N 10030N1227B       &#xA; 10083+1159BC  HDO 127                   1943 086   2.5 12 13.1                                           A 7654  10030N1227C       &#xA; 10083+1159 D  HDO 127                   1924 274 217.0  1                                                A 7654  10030N1227D       &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5401" LastEditorUserId="5401" LastEditDate="2017-04-10T18:45:13.667" LastActivityDate="2017-04-17T15:22:40.507" Title="Hip2.dat and Solo, Binary and Multiple Star Systems" Tags="&lt;star&gt;&lt;hipparcos&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20681" PostTypeId="2" ParentId="20680" CreationDate="2017-04-09T20:29:27.397" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;All the stars I looked up are all either binary or multiple stars, I couldn't find a solo star, apart from the Sun (debatable).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;HIP2.dat (the Hipparcos parallax and proper motion catalogue) is &lt;em&gt;not&lt;/em&gt; a comprehensive catalogue of multiple stars. In the excerpt you have shown only HIP79374 and HIP50583 needed to be treated as multiple stars in order to get a solution for their proper motion and parallax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, all the flags are telling you about is whether the star needed to be treated as a binary system in order to get an accurate model for its motion on the sky. For many multiple systems this would not be necessary, for example where the components had either a very small or very large angular separation or a very large brightness contrast. I imagine Sirius AB falls into this latter category. See &lt;a href=&quot;https://www.cosmos.esa.int/documents/532822/546798/oral01_04.pdf/a32e3f88-4a5c-4179-a8f7-9fe8a145b757&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lindegren (1997).&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-04-17T15:22:40.507" LastActivityDate="2017-04-17T15:22:40.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20682" PostTypeId="1" CreationDate="2017-04-09T22:14:32.320" Score="5" ViewCount="611" Body="&lt;p&gt;I was looking at the Jupiter for the first time today and I have noticed a smudge near it. Could someone explain what it is, please? I was looking online but did not find anything. So I thought that some kind of defect in my setup but I couldn't get rid of it. I mean the thing below its moons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit1&lt;/strong&gt;:&#xA;My telescope is a Celestron AstroMaster 76 ESQ. The smudge was there no matter what eyepiece I have used. I was able to see it even without a camera. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit2&lt;/strong&gt;:&#xA;I was watching it for about 10 minutes, and the smudge was moving along with the rest of the objects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took just this one image for the StackExchange purposes. I have put the camera in front of an eyepiece to take the picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uvEKu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16564" LastEditorUserId="16564" LastEditDate="2017-04-10T06:43:26.607" LastActivityDate="2020-07-26T11:44:25.017" Title="A smudge near Jupiter" Tags="&lt;telescope&gt;&lt;jupiter&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20683" PostTypeId="2" ParentId="20682" CreationDate="2017-04-09T23:37:47.810" Score="7" Body="&lt;p&gt;The faint green smudge looks to me like a &lt;a href=&quot;https://en.wikipedia.org/wiki/Lens_flare&quot; rel=&quot;nofollow noreferrer&quot;&gt;flare&lt;/a&gt; (that is, an optical defect called a flare).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I'd expect a flare to be symmetric with the bright object it relates to about the center of the field of view.  This does imply either the image is off center (e.g. the camera is off center or the image is a crop off center) or there's a misalignment of your optics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question on Physics SE is about &lt;a href=&quot;https://physics.stackexchange.com/questions/25637/how-to-correct-flares-and-sharply-focus-on-my-telescope-when-viewing-planets&quot;&gt;correcting flare in a telescope&lt;/a&gt; and may be relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The similar question &lt;a href=&quot;https://astronomy.stackexchange.com/q/32940/7982&quot;&gt;Does this smartphone photo show Mars just below the Sun?&lt;/a&gt; and its answers contain some other examples. &lt;/p&gt;&#xA;" OwnerUserId="10196" LastEditorUserId="7982" LastEditDate="2019-12-24T01:52:08.677" LastActivityDate="2019-12-24T01:52:08.677" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="20684" PostTypeId="1" AcceptedAnswerId="20686" CreationDate="2017-04-10T06:23:26.263" Score="1" ViewCount="77" Body="&lt;p&gt;This recent HST image of Jupiter near opposition is &lt;a href=&quot;http://hubblesite.org/image/4013/news_release/2017-15&quot; rel=&quot;nofollow noreferrer&quot;&gt;described here&lt;/a&gt; and &lt;a href=&quot;https://svs.gsfc.nasa.gov/cgi-bin/details.cgi?aid=12570&amp;amp;button=recent&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and was made under &lt;a href=&quot;http://archive.stsci.edu/proposal_search.php?mission=hst&amp;amp;id=14756&quot; rel=&quot;nofollow noreferrer&quot;&gt;HST proposal 14756&lt;/a&gt;. It was taken with WFC3/UVIS. It is made from three images using narrow-band filters at 395, 502, and 631 nanometers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering if there were additional images recorded with filters at other wavelengths. If so, how could I go about finding them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WQpjc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WQpjc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2017-04-10T15:27:29.543" Title="Are there more images with different filters available for the April 6 Hubble image of Jupiter?" Tags="&lt;solar-system&gt;&lt;jupiter&gt;&lt;data-analysis&gt;&lt;hubble-telescope&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20686" PostTypeId="2" ParentId="20684" CreationDate="2017-04-10T12:13:49.427" Score="2" Body="&lt;p&gt;As those observations are from only a few days ago, I think the data will still be embargoed, usually one year, until it's publicly available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To confirm this, I've looked at the &lt;a href=&quot;https://hla.stsci.edu/hlaview.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble Legacy archive&lt;/a&gt;, where Hubble data is made publicly available. When you look there, the last public data for Jupiter is from February 2016.&lt;br&gt;&#xA;Although I couldn't use the retrieved data from 2016, as it's not the FITS format, but it downloaded something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So either you'll have to wait it seems, or you contact the project PI and ask nicely ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the HST proposal page from the MAST archive (that links to the legacy archive) you see the photos they've taken listed, but the data is not available, I guess due to embargo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;However I'm not sure if the legacy archive has all the data, as there's one dataset from 2011 with which I've worked that should be there, and I can't find it. Weird.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditDate="2017-04-10T15:27:29.543" LastActivityDate="2017-04-10T15:27:29.543" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20687" PostTypeId="1" AcceptedAnswerId="20688" CreationDate="2017-04-10T15:22:42.040" Score="1" ViewCount="858" Body="&lt;p&gt;All I could find on the internet was about how stars vary in brightness depending on their distance to Earth, temperature, type of star...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my question is, can a star can change its brightness.&lt;/p&gt;&#xA;" OwnerUserId="16569" LastEditorUserId="2365" LastEditDate="2017-04-10T17:33:05.937" LastActivityDate="2017-04-13T12:13:26.027" Title="Do stars vary their own brightness?" Tags="&lt;variable-star&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20688" PostTypeId="2" ParentId="20687" CreationDate="2017-04-10T17:22:43.540" Score="4" Body="&lt;p&gt;They can, and some do. These stars are called &lt;a href=&quot;https://en.wikipedia.org/wiki/Variable_star&quot; rel=&quot;noreferrer&quot;&gt;variable stars&lt;/a&gt;, because their luminosities &lt;em&gt;as observed from Earth&lt;/em&gt; vary over time, often (though not always) in a regular period. Here are some broad categories:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_pulsations&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;Pulsating stars&lt;/strong&gt;&lt;/a&gt;, where fluctuations lead to increases and decreases in size or temperature, which in turn produce changes in the star's brightness. Example: The oft-cited &lt;a href=&quot;https://en.wikipedia.org/wiki/Cepheid_variable&quot; rel=&quot;noreferrer&quot;&gt;Cepheid variables&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Binary_star#Eclipsing_binaries&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;Eclipsing binary stars&lt;/strong&gt;&lt;/a&gt; involve two stars orbiting each other. When one passes between the other and Earth, the combined luminosity of the system appears to decrease, even though the stars' intrinsic luminosities are probably constant. Example: &lt;a href=&quot;https://en.wikipedia.org/wiki/Algol&quot; rel=&quot;noreferrer&quot;&gt;Algol&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;So-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Variable_star#Eruptive_variable_stars&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;eruptive variables&lt;/strong&gt;&lt;/a&gt; (often lumped together with &lt;a href=&quot;https://en.wikipedia.org/wiki/Cataclysmic_variable_star&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;cataclysmic variables&lt;/strong&gt;&lt;/a&gt;, which are different) may have irregular outbursts caused by flares or other phenomena. Example: &lt;a href=&quot;https://en.wikipedia.org/wiki/Luminous_blue_variable&quot; rel=&quot;noreferrer&quot;&gt;Luminous blue variables (LBVs)&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The change in brightness and the length of the variations depend on the type of variable star in question. The luminosity can vary by anywhere from a fraction of a magnitude to many magnitudes. Likewise, the period (if there is one) can range from hours to years.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2017-04-10T17:22:43.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20689" PostTypeId="1" CreationDate="2017-04-10T18:02:03.813" Score="1" ViewCount="83" Body="&lt;p&gt;I'm participating in the Zooniverse Exoplanet-explorers project (transit method), which provides only minimal guidance on interpreting their light curve and transit data graphics, as visible here:&#xA; &lt;a href=&quot;https://www.zooniverse.org/projects/ianc2/exoplanet-explorers/classify&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.zooniverse.org/projects/ianc2/exoplanet-explorers/classify&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone point me to source(s) that will help me read these graphs in detail, pls?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Re own comment below, suggest view &lt;a href=&quot;https://www.zooniverse.org/projects/ianc2/exoplanet-explorers/talk/820/277985?comment=457728&amp;amp;page=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Weird example&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: Just located &lt;a href=&quot;http://exoplanetarchive.ipac.caltech.edu/applications/ETSS/Kepler_index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA Exo-PLanet Archive&lt;/a&gt; which looks like it might hold what iSeek. But further recommendations would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="16570" LastEditorUserId="16570" LastEditDate="2017-04-11T11:17:15.883" LastActivityDate="2017-04-11T11:17:15.883" Title="Where can I find detail on interpreting the exo-planet transit-data graphics used in the Zooniverse Project?" Tags="&lt;exoplanet&gt;&lt;kepler&gt;&lt;light-curve&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20690" PostTypeId="2" ParentId="20687" CreationDate="2017-04-10T18:35:13.450" Score="3" Body="&lt;p&gt;Looking at the question longer-term, a star's brightness also changes as it evolves:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;starting possibly with some brightness of its initial accretion disc,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;then as nuclear burning begins it becomes a true star (check out the Hertzsprung Russell diagram showing stellar evolution)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;later in its life, depending mainly on its mass it can either evolve &#xA;into a near-dead dwarf, which will be increasingly dim&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;alternatively, heavier stars go supernova and become so bright as to &#xA;briefly outshine galaxies before becoming dimmer again as neutron stars or black holes.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So to answer your question broadly - stars during their lifetimes do vary in brightness on a vast scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If interested further, I suggest you look up: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Hertzsprung%E2%80%93Russell_diagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Hertzsprung%E2%80%93Russell_diagram&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_evolution&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Stellar_evolution&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16570" LastEditorUserId="16570" LastEditDate="2017-04-13T12:13:26.027" LastActivityDate="2017-04-13T12:13:26.027" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20691" PostTypeId="1" AcceptedAnswerId="20692" CreationDate="2017-04-11T05:45:12.440" Score="3" ViewCount="154" Body="&lt;p&gt;I read the paper &quot;Observation of Gravitational Waves from a Binary Black Hole Merger&quot;&#xA;&lt;a href=&quot;https://physics.aps.org/featured-article-pdf/10.1103/PhysRevLett.116.061102&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://physics.aps.org/featured-article-pdf/10.1103/PhysRevLett.116.061102&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to understand all the graphs and data observed in the paper.&#xA;I am curious how these data and graphs conclude that the gravitational waves observed were because of binary black holes merger? Also what calculations and equations led to the same?&#xA;Also how will the calculations and equations look if neutron stars were there instead of black hole in the same merger?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(graphs and data can be seen in the observation section in the paper)&lt;/p&gt;&#xA;" OwnerUserId="14301" LastActivityDate="2017-04-11T09:52:55.177" Title="Observation of Gravitional waves by LIGO from binary black hole merger" Tags="&lt;observational-astronomy&gt;&lt;data-analysis&gt;&lt;gravitational-waves&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20692" PostTypeId="2" ParentId="20691" CreationDate="2017-04-11T07:42:22.517" Score="4" Body="&lt;p&gt;Neutron stars are less massive but larger than black holes. The frequency of the gravitational waves is due to the orbital period of the objects as they spiral together. A neutron star merger would have a different orbital speed, so the frequency would be different and lower. It would rise at a different rate. The signal from neutron stars is also much weaker, so couldn't be detected from the same range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prior to the detection of the signal, numerical simulations of merging black holes of various sizes were done, so the theoretical shape of the waveform expected from the merger of black holes (and neutron stars) was known. It was possible from this theoretical work to estimate the mass of the two black holes taking part in the merger by matching the observed signal to a theoretical one. You can see the theoretical signal for black holes with 36 and 29 in the diagrams labelled as &quot;numerical relativity&quot; &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-04-11T07:42:22.517" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20694" PostTypeId="2" ParentId="20691" CreationDate="2017-04-11T09:52:55.177" Score="3" Body="&lt;p&gt;This is to supplement &lt;a href=&quot;https://astronomy.stackexchange.com/a/20692/7982&quot;&gt;@JamesK's very clear answer&lt;/a&gt;. From the oscillations shown in Figure 1, one can &lt;em&gt;estimate&lt;/em&gt; a period of the &lt;em&gt;of the measured gravitational wave&lt;/em&gt; of about 5 milliseconds, which would correspond to an &lt;em&gt;estimated frequency of the wave&lt;/em&gt; of about 200 Hz. Because the phenomenon is binary - a &quot;wave&quot; will come from each of the two objects, one can &lt;em&gt;estimate&lt;/em&gt; the instantaneous orbital frequency at this moment to be half that, or about 100 Hz. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discussion at the end of Section II mentions that in this case an orbital frequency reaching 75 Hz &lt;em&gt;before contact&lt;/em&gt; could only be produced by two compact massive objects — black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The annotations indicate an estimation from the figures only. An accurate analysis would require modeling the original data set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://physics.aps.org/featured-article-pdf/10.1103/PhysRevLett.116.061102&quot; rel=&quot;nofollow noreferrer&quot;&gt;P. B. Abbott et al. PRL 116, 061102 (2016)&lt;/a&gt;, section II and Figure 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To reach an orbital frequency of 75 Hz (half the gravitational-wave frequency) the objects must have been very close and very compact; equal Newtonian point masses orbiting at this frequency would be only ≃350 km apart. A pair of neutron stars, while compact, would not have the required&#xA;  mass, while a black hole neutron star binary with the deduced chirp mass would have a very large total mass, and would thus merge at much lower frequency. This leaves black holes as the only known objects compact&#xA;  enough to reach an orbital frequency of 75 Hz without contact. Furthermore, the decay of the waveform after it peaks is consistent with the damped oscillations of a black hole relaxing to a final stationary Kerr configuration. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7QD96.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7QD96.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YgB0M.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YgB0M.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2017-04-11T09:52:55.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20695" PostTypeId="1" AcceptedAnswerId="20711" CreationDate="2017-04-11T13:10:27.897" Score="0" ViewCount="3402" Body="&lt;p&gt;I've been reading this article&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Celestial_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Celestial_coordinate_system&lt;/a&gt;&#xA;and a question came to my mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimum units are seconds ( ″ ). I picture an angle of 1 second as being like a &quot;cone&quot; going out from Earth. It's really small at the beginning but as you go further its width increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How wide does this cone get as you get further into the universe? How big is an arcsecond on different astronomical bodies?&lt;/p&gt;&#xA;" OwnerUserId="16569" LastEditorUserId="2365" LastEditDate="2017-04-13T08:16:24.430" LastActivityDate="2017-04-14T09:43:08.040" Title="How big is one arcsecond at various distances?" Tags="&lt;coordinate&gt;&lt;size&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20697" PostTypeId="2" ParentId="10047" CreationDate="2017-04-11T14:32:06.217" Score="2" Body="&lt;p&gt;This is a back-of-the-envelope analysis to double-check @RobJeffries' self-described &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10047/#comment13163_10050&quot;&gt;case-closed&lt;/a&gt; answer &lt;a href=&quot;https://astronomy.stackexchange.com/a/10050/7982&quot;&gt;posted (above)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light received by Jupiter and Ganymede:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_{Jup} = r_{Jup}^2$$&#xA;$$I_{Gan} = r_{Gan}^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fraction of Jupiter's received light received by Ganymede:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f = a_{Jup}\frac{1}{2}\frac{r_{Gan}^2}{4 R_{Gan}^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\frac{1}{2}$ because Ganymede is lit by Jupiter from the side, half is invisible from Earth. I'm likewise surprised that it's so small, until realizing that Jupiter is only 7.5 degrees wide in Ganymede's sky. It's those darn NASA images with the satellites dwarfed by Jupiter that's throwing us off perhaps &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{f \times I_{Jup}}{I_{Gan}} = \frac{a_{Jup}}{8}\left( \frac{r_{Jup}}{R_{Gan}} \right)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a_{Jup}}{8}\left( \frac{r_{Jup}}{R_{Gan}} \right)^2 =  \frac{0.34}{8}\left( \frac{70,000}{1,070,400} \right)^2 \approx \text{1.8E-03}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frighteningly close to @RobJeffries' answer. Yay for envelope backs!&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:07.017" LastActivityDate="2017-04-11T14:37:16.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20698" PostTypeId="1" CreationDate="2017-04-11T14:35:39.937" Score="1" ViewCount="627" Body="&lt;p&gt;Both low-mass PMS (pre-main-sequence) stars and young brown dwarfs can fuse lithium in their cores and the lithium can be depleted throughout the star/brown dwarf very quickly. &lt;a href=&quot;https://en.wikipedia.org/wiki/Lithium_burning&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wiki&lt;/a&gt;. Then the Li I 6708A absorption line disappears from the object's spectrum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why and how can the Lithium test be used as an indicator to distinguish between young brown dwarfs and young low-mass stars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody summarise in what cases we can tell whether an object is a brown dwarf or a low-mass star?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="2531" LastEditDate="2017-04-17T13:10:57.590" LastActivityDate="2018-07-14T18:41:08.053" Title="Concerning the &quot;Lithium test&quot; for Brown Dwarfs" Tags="&lt;star&gt;&lt;stellar-evolution&gt;&lt;brown-dwarf&gt;&lt;elemental-abundances&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20700" PostTypeId="1" AcceptedAnswerId="20724" CreationDate="2017-04-11T14:51:24.340" Score="3" ViewCount="160" Body="&lt;p&gt;Red spectra of M stars are dominated by TiO(Titanium oxide) bands, which have become the primary means of establishing spectral types.see &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/305587/fulltext/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ApJ&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is TiO special? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There should be much more Silicon elements than Titanium if we suppose the element abundance of M stars are similar to our sun.(Solar abundance is &lt;a href=&quot;https://www.thoughtco.com/element-composition-of-sun-607581&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;). If similar，there should be much more SiO2 than TiO?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditDate="2017-04-11T15:07:53.987" LastActivityDate="2017-04-13T10:56:00.200" Title="Why is TiO special?" Tags="&lt;spectral-type&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20702" PostTypeId="1" CreationDate="2017-04-11T15:14:32.027" Score="3" ViewCount="4193" Body="&lt;p&gt;In an attempt to visualise nearby stars, I've tried to get the star radius (measured in solar radii) using the equation provided on the following site:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cas.sdss.org/dr4/en/proj/advanced/hr/radius1.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://cas.sdss.org/dr4/en/proj/advanced/hr/radius1.asp&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now though the equation seems to work fine for a star like Sirius, when I plug in the values for absolute magnitude and temperature for Barnard's star (according to wikipedia, 13.21 and 3134 K respectively) I get a radius of 0.0722. The radius provided by wikipedia is more than twice that value, at 0.196.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many potential culprits. Could it be floating point errors in the php code? The equation looks like this by the way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{radius = pow(5800/3134, 2)} \times \text{pow(pow(2.512, (4.83 - 13.21)), 0.5})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is that the equation is too simplistic, and there are other variables I need to take into account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any insights would be greatly appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Michael&lt;/p&gt;&#xA;" OwnerUserId="15848" LastEditorUserId="9301" LastEditDate="2017-04-11T16:19:35.903" LastActivityDate="2020-05-01T19:37:25.987" Title="Calculating star radius from temperature and absolute magnitude - what am I missing?" Tags="&lt;star&gt;&lt;temperature&gt;&lt;absolute-magnitude&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20703" PostTypeId="2" ParentId="20702" CreationDate="2017-04-11T22:10:00.580" Score="-1" Body="&lt;p&gt;In order to calculate this, you can use the Stefan-Boltzmann law to calculate the star's surface flux and its absolute magnitude to get the luminosity.  Once you know the surface flux and luminosity, you can find the radius of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stefan-Boltzmann Law:&#xA;$$&#xA;F=\sigma T^4&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Absolute Magnitude - Luminosity calculation:&#xA;$$&#xA;\frac{L}{L_\odot} = 10^{0.4(M_\odot - M)}&#xA;$$&#xA;where $L_\odot$ is solar luminosity and $M_\odot$ is the absolute magnitude of the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, you use the formula for intensity and solve for distance:&#xA;$$&#xA;r = \sqrt{\frac{L}{4\pi F}}&#xA;$$&#xA;Compiling all this into one equation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;r = \sqrt{\frac{L_\odot * 10^{0.4(M\odot - M)}}{4\pi \sigma T^4}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as you use SI units for $L_\odot$, $\sigma$, and $T$, you'll get $r$ in meters.&lt;/p&gt;&#xA;" OwnerUserId="13426" LastActivityDate="2017-04-11T22:10:00.580" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20704" PostTypeId="1" AcceptedAnswerId="20705" CreationDate="2017-04-12T02:15:38.510" Score="1" ViewCount="1746" Body="&lt;p&gt;I'm reading some &lt;a href=&quot;https://en.wikipedia.org/wiki/Atacama_Large_Millimeter_Array&quot; rel=&quot;nofollow noreferrer&quot;&gt;ALMA&lt;/a&gt; proposals and I am often seeing a conversion from Jy to Kelvin when, for example, people quote noise levels or source flux.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example on their sensitivity calculator&#xA;(&lt;a href=&quot;https://almascience.nrao.edu/proposing/sensitivity-calculator&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://almascience.nrao.edu/proposing/sensitivity-calculator&lt;/a&gt;) 1 mJy is equivalent to around 1 K when the angular resolution is 1&quot;. I can tell it somehow depends on the resolution because it won't give me a conversion unless I enter a resolution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's happening under the hood here? And why does their documentation treat sensitivities entered in Kelvin differently from when they're entered in Jy? Is this an interferometry thing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="13724" LastActivityDate="2017-04-12T02:34:57.077" Title="How are people converting intensities in Janskys to Kelvin?" Tags="&lt;observational-astronomy&gt;&lt;units&gt;&lt;interferometry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20705" PostTypeId="2" ParentId="20704" CreationDate="2017-04-12T02:34:57.077" Score="6" Body="&lt;p&gt;This is, indeed, a result of how we measure things in radio astronomy.  It's not just interferometry, but radio astronomy in general.  The thing they're referring to is a concept called &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Brightness_temperature&quot; rel=&quot;noreferrer&quot;&gt;brightness temperature&lt;/a&gt;&quot;.  In the low frequency limit (valid for radio astronomy), we can use the Rayleigh-Jeans approximation, which gives us the expression &#xA;$$&#xA;T_B = \frac{S_\nu \lambda^2}{2k \theta_s^2}&#xA;$$&#xA;where $S_\nu$ is the flux density, $\lambda$ is the wavelength, $k$ is Boltzmann's constant, and $\theta_s$ is the size of the source (or the resolution in this case).  As long as your source and wavelength are constant, brightness temperature and flux density can be interchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason astronomers talk about temperatures instead of fluxes sometimes is twofold.  First, a large number of objects examined by radio astronomers are not blackbodies.  Therefore we cannot calculate their temperatures like we normally would.  Instead, we calculate what their temperature would be if they were blackbodies.  This at least gives us a value to compare them to other things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, when discussing radio telescopes and antennae, we usually say that various parts of the antenna contribute different 'temperatures'.  We can quantify system temperatures to help us understand noise.&lt;/p&gt;&#xA;" OwnerUserId="13426" LastActivityDate="2017-04-12T02:34:57.077" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20706" PostTypeId="1" CreationDate="2017-04-12T10:39:58.337" Score="1" ViewCount="169" Body="&lt;p&gt;Consider the vector positions of the center of mass of Moon, Sun and Earth, $\vec{r}_{\rm M}$, $\vec{r}_{\rm S}$, $\vec{r}_{\rm E}$, respectively, in a given reference frame. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the angle between the vector $\vec{r}_{\rm S} - \vec{r}_{\rm E}$ and $\vec{r}_{\rm M} - \vec{r}_{\rm E}$,&lt;/li&gt;&#xA;&lt;li&gt;the angle between the vector $\vec{r}_{\rm E} - \vec{r}_{\rm S}$ and $\vec{r}_{\rm M} - \vec{r}_{\rm S}$,&lt;/li&gt;&#xA;&lt;li&gt;the angle between the vector $\vec{r}_{\rm E} - \vec{r}_{\rm M}$ and $\vec{r}_{\rm S} - \vec{r}_{\rm M}$.   &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Does any of these angles have a name in astronomy, or can be important for some specific reasons? If so, may you please give me some general indications of why this is the case, and point to some references? &lt;/p&gt;&#xA;" OwnerUserId="16581" LastActivityDate="2017-04-12T10:39:58.337" Title="Angles in Moon, Earth and Sun system" Tags="&lt;the-moon&gt;&lt;natural-satellites&gt;&lt;celestial-mechanics&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20708" PostTypeId="2" ParentId="12198" CreationDate="2017-04-12T12:23:48.727" Score="1" Body="&lt;p&gt;The OP's &quot;simple (and obvious) explanation for the dimming of KIC 8462852?&quot; is unlikely to work as there very likely is no &lt;em&gt;single&lt;/em&gt; simple explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The May 2017 Scientific American (pp. 30-35) has an article on KIC 8462852 (also known as Boyajian's Star or Tabby's Star). which reviews recent thinking on a variety of explanations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the article concludes that more information is required, I wonder why we should expect a complex phenomenon to have one explanation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my view: the simpl&lt;em&gt;est&lt;/em&gt;, and to me most obvious explanation, is that it's a &lt;em&gt;combination&lt;/em&gt; of a number of mechanisms, some, like the stars dimming, may be intrinsic, others may rely on some inhomogeneous materials (at possibly more than one intervening distance) to be irregularly occluding light from the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: Looking at the other entry in StackExchange on this topic &lt;a href=&quot;https://astronomy.stackexchange.com/questions/19136/has-star-kic-8462852s-3-year-long-kepler-fading-continued-after-the-primary-kep&quot;&gt;post 19136&lt;/a&gt; it appears that there's an arXiv post from 2016, not referenced in the Scientific American article, by Jason T. Wright, Steinn Sigurdsson &lt;a href=&quot;https://arxiv.org/abs/1609.03505&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Families of Plausible Solutions to the Puzzle of Boyajian's Star&quot;&lt;/a&gt;, which appears to have anticipated the (admittedly obvious but perhaps understated) idea of multiple solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another useful summary is &lt;a href=&quot;https://en.wikipedia.org/wiki/KIC_8462852&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16570" LastEditorUserId="-1" LastEditDate="2017-04-13T12:59:06.910" LastActivityDate="2017-04-12T12:47:25.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20711" PostTypeId="2" ParentId="20695" CreationDate="2017-04-12T15:11:33.010" Score="9" Body="&lt;p&gt;&lt;strong&gt;How big is one arcsecond at various distances?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An arcsecond is a small angle, 1/3600 of a degree or about 5 millionths of a radian ($4.85\times10^{-6}$). To estimate the size of something that appears 1 arcsecond across you can use the small angle approximation to trigonometry: Multiply the distance to the object by $4.85\times10^{-6}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One arcsecond on the moon is 1.87km&lt;/li&gt;&#xA;&lt;li&gt;One arcsecond on the sun is 727km&lt;/li&gt;&#xA;&lt;li&gt;One arcsecond on Mars (when it is closest to Earth) is 237km&lt;/li&gt;&#xA;&lt;li&gt;One arcsecond at distance of one parsec is one astronomical unit (AU), by definition&lt;/li&gt;&#xA;&lt;li&gt;One arcsecond on Alpha Centauri is 200 million km&lt;/li&gt;&#xA;&lt;li&gt;One arcsecond on the Andromeda galaxy is 100 trillion km&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The atmosphere limits how much detail you can see. Typically the smallest detail you can see is about 3 or 4 arcseconds, though professional telescopes do better by being built on the top of mountains and using various tricks, like adaptive optics.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorDisplayName="user15104" LastEditDate="2017-04-13T12:38:16.223" LastActivityDate="2017-04-13T12:38:16.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20712" PostTypeId="1" AcceptedAnswerId="20713" CreationDate="2017-04-12T16:40:05.870" Score="0" ViewCount="169" Body="&lt;p&gt;Will a laser collimator like the Gosky 1.25 Metal Laser Collimator, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B01M4IVUYG&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amazon.com/gp/product/B01M4IVUYG/ref=crt_ewc_title_dp_1?ie=UTF8&amp;amp;psc=1&amp;amp;smid=AHQ6VCR020F8X&lt;/a&gt;, work for a dobsonian even though it says it's for a newtonian?&lt;/p&gt;&#xA;" OwnerUserId="13272" LastActivityDate="2017-04-12T17:30:38.203" Title="Laser Collimator For Dobsonian vs Newtonian" Tags="&lt;newtonian-telescope&gt;&lt;dobsonian-telescope&gt;&lt;collimation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20713" PostTypeId="2" ParentId="20712" CreationDate="2017-04-12T17:30:38.203" Score="1" Body="&lt;p&gt;A Dob is a Newtonian, so yes it will work.  Strictly speaking the Dob bit is the mount, but it's nearly always a newtonian that's mounted.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-04-12T17:30:38.203" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20714" PostTypeId="1" AcceptedAnswerId="20734" CreationDate="2017-04-12T20:10:45.787" Score="2" ViewCount="356" Body="&lt;p&gt;A spaceship with mass $m = 25\:kg$ is moving toward a red giant with $M = M_{sun}$ and $L = 1000\:L_{sun}$ (L: Luminosity). It is moving on a line (so this is not an orbital question). When it gets too close to the red giant, it's sensor open its solar sails (they get and then reflect the light completely), hence the radiation pressure help it to escape from the red giant (on the same line). Find the area of solar sail. (It is from National Olympiad of an Asian Country)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is with the radiation pressure formula. for solving this problem, somewhere said the radiation pressure is: ($c$: light speed, $\sigma$: Stefan-Boltzman Constant)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P= \frac{4\sigma T^4}{3c} = \frac{4}{3}\frac{L}{4\pi D^2}$$&#xA;Where $T$ is the equivalent Temprature in distance $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somewhere else said:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P= \frac{4\sigma T^4}{c} = 4 \frac{L}{4\pi D^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In another book it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P=2 \frac{4\sigma T^4}{c} =  2\times 4 \frac{L}{4\pi D^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 is because it gets the light and then reflecting it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the final:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P\times A = \frac{GmM}{D^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know what is the right solution for calculating radiation pressure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must say that the below picture is also with the question but in the books, it was used for another part of the question which is not my problem and I didn't mentioned it above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PqyIq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PqyIq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16441" LastEditorUserId="11143" LastEditDate="2017-04-13T13:11:56.640" LastActivityDate="2017-04-14T18:47:25.517" Title="Question about radiation pressure" Tags="&lt;luminosity&gt;&lt;radiation&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20715" PostTypeId="1" AcceptedAnswerId="20718" CreationDate="2017-04-12T21:15:31.843" Score="2" ViewCount="223" Body="&lt;p&gt;This may sound a little bit confusing, but I will try to explain. (I have done research on this topic already) &#xA;I have looked at the temperatures of planets and I was wondering, is there a point where a planet is at perigee and it's temperature cannot reach any higher? Let's just say the planet is tidally locked, and is currently at perigee. Would the planet be at a point where it's temperature can no longer be increased at the current distance from its star?&lt;/p&gt;&#xA;" OwnerUserId="16048" LastActivityDate="2017-04-13T10:59:12.050" Title="Is there a maximum temperature a planet can have?" Tags="&lt;planet&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20716" PostTypeId="1" AcceptedAnswerId="20722" CreationDate="2017-04-12T21:53:02.647" Score="1" ViewCount="248" Body="&lt;p&gt;I was recently asked a question for a lab I am doing for an astronomy class where we collected data from the M44 cluster (through Google Sky) and used information about the Hyades cluster to estimate the age and distance of M44. Anyways, I was asked whether there was more scatter in my M44 data or the provided Hyades data. There was definitely more scatter (see below) and I'm not sure how to explain it. I believe it is because of these reasons: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There is measurement error. &lt;/li&gt;&#xA;&lt;li&gt;And I might be picking up background/foreground stars This would definitely lead to scatter. If I measure a foreground star and assume it is as far as the M44 star cluster, it would appear brighter than if it were truly in the M44 cluster.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Am I on the right track or am I missing something? Any help would be great. Here is my data, the green is the collected M44 data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZWVZ3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZWVZ3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16583" LastEditorUserId="2365" LastEditDate="2017-04-13T08:42:13.563" LastActivityDate="2017-04-14T02:12:52.737" Title="Understanding Scatter in Data collected for HR Plot" Tags="&lt;star&gt;&lt;amateur-observing&gt;&lt;star-cluster&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20717" PostTypeId="1" AcceptedAnswerId="20745" CreationDate="2017-04-12T23:35:40.107" Score="1" ViewCount="76" Body="&lt;p&gt;I study Computer Science and I need to understand perfectly how is the process of publishing data to the virtual observatory, since I will start a project that seeks to develop another platform to enter data into the virtual observatory (in a more intuitive way if it's possible)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand correctly I believe that this is currently done by means of &lt;a href=&quot;http://cdsweb.u-strasbg.fr/publication-support#yourDataInVizieR&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://cdsweb.u-strasbg.fr/publication-support#yourDataInVizieR&lt;/a&gt; and trough other web pages like &lt;a href=&quot;http://www.cadc-ccda.hia-iha.nrc-cnrc.gc.ca/en/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.cadc-ccda.hia-iha.nrc-cnrc.gc.ca/en/&lt;/a&gt; or programs like &lt;a href=&quot;http://aladin.u-strasbg.fr/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://aladin.u-strasbg.fr/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone understand the topic or where can I find documentation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel a bit overwhelmed because in these web pages I see there are many concepts that I do not understand well catalogs, VOTable, I do not even understand very well what a virtual observatory, I do not want to ask my &quot;thesis&quot; adviser every 5 minutes about basic concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help will be well received.&lt;/p&gt;&#xA;" OwnerUserId="16585" LastActivityDate="2017-04-16T01:20:19.850" Title="Publish data in the virtual observatory" Tags="&lt;observatory&gt;&lt;resource-request&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2017-04-16T09:39:01.560" ContentLicense="CC BY-SA 3.0" />
  <row Id="20718" PostTypeId="2" ParentId="20715" CreationDate="2017-04-13T01:00:28.497" Score="4" Body="&lt;p&gt;A planet in a stable orbit can only receive a finite maximum power from it's star and likewise can only generate a finite energy from internal heat (like our cooling core, which is a tiny power output compared to what we get from the Sun).  There's a finite limit to available power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that power is not radiated out (which is thermal radiation) the planet's temperate would rise without limit.  This never happens because hot things radiate heat (as EM radiation) to try and reach thermal equilibrium with their environment (which in the case of a planet is the very cold space around it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there must be a finite limit to the temperature any planet in a stable orbit can reach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @Cody indicated the actual point of maximum temperature may not be at periapsis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atmospheres can act in complex ways to raise the temperature of the surface significantly.  Venus is a prime example as &lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20101007094209AAmddCT&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page demonstrates&lt;/a&gt;.  Without it's pressure cooker-like atmosphere we'd expect a temperature of around 240 K, whereas it's actually a rather staggering 740 K.  But all of that is solar powered and can't rise without limit.  But it illustrates the importance of atmospheric composition (and indeed the Greenhouse effect) on planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt; : I came across another page on the &lt;a href=&quot;http://www.dangermouse.net/gurps/science/temps.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;basic theory of planetary temperature&lt;/a&gt; while looking into something else. It may be of use so I've added it in this edit.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastEditorUserId="10196" LastEditDate="2017-04-13T10:59:12.050" LastActivityDate="2017-04-13T10:59:12.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20719" PostTypeId="1" CreationDate="2017-04-13T02:53:49.903" Score="1" ViewCount="199" Body="&lt;p&gt;So assume we have a planet of temperature T, and mass M, We can calculate it's internal energy/heat content by the formula $U=cMT$ where $c$ is the specific heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if we also know the gravitational energy of the planet via $\frac{GM^2}{R}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;how do I approach the following question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the release of gravitational energy during planet formation in principle (is there enough) explain this much heat&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="15892" LastActivityDate="2017-04-15T15:14:07.940" Title="gravitational energy and internal energy" Tags="&lt;planet&gt;&lt;heat&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20720" PostTypeId="1" AcceptedAnswerId="20726" CreationDate="2017-04-13T07:29:23.270" Score="3" ViewCount="230" Body="&lt;p&gt;In the mid 1970's, Franklin and O'Meara saw persistent &quot;radial spoke-like features&quot; in the rings of Saturn, that should not have existed due to the differential rotation of the rings.  A publication on this observation was rejected by a journal apparently on the grounds that the phenomenon was considered to be illusory (c.f Sciparelli/Lowell's Martian canals?).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ckUgL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ckUgL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;From &quot;Seeing in the Dark: How&#xA;  Amateur Astronomers Are Discovering the Wonder&quot; By Timothy Ferris.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question is, how reliably were these features visible from the ground-based telescopes of the day?  Could sufficiently strong evidence for the existence been obtained before images from space probes?  Was the rejection of the paper reasonable in the historical context in which the paper was submitted?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2017-04-13T13:37:42.173" Title="Are the radial spokes in Saturn's rings reliably visible via ground-based telescopes" Tags="&lt;amateur-observing&gt;&lt;saturn&gt;&lt;planetary-ring&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20721" PostTypeId="2" ParentId="20695" CreationDate="2017-04-13T08:35:51.903" Score="4" Body="&lt;p&gt;James K's answer is right on, and probably what you're after. I just wanted to mention an effect which comes into play if you look really, really far away:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because light moves at a finite speed, we see galaxies (and other things) as they were in the past. And because the Universe expands, everything was closer together in the past, so we see distant galaxies as they were when they were closer and hence spanned a larger angle on the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we have two competing effects: On the one hand, galaxies become smaller and smaller with distance, as expected and as decribed in James K's answer. On the other hand, we look further and further into the past, and hence see galaxies at a time when they looked larger and larger. The result — which is obtained by integrating the &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Friedmann equation&lt;/a&gt; — is that galaxies become smaller and smaller out to a certain distance (roughly 15 billion lightyears), after which the second effect starts to dominate and they start to grow in size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, in other words, an arcseconds spans a larger and larger physical size (as described by James K) out to ~15 Gly (where it spans roughly 28,000 lightyears), after which it spans a smaller and smaller size. The relation looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NMPox.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NMPox.png&quot; alt=&quot;scale&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2017-04-13T20:01:40.733" LastActivityDate="2017-04-13T20:01:40.733" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20722" PostTypeId="2" ParentId="20716" CreationDate="2017-04-13T09:24:04.650" Score="3" Body="&lt;p&gt;So, scatter in observational measurements of clusters in the HR diagram is mainly, in my experience, caused by 3 things: observational errors, binaries and contamination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you mentioned, there are observational errors, which are contributed to by the instrument, the data reduction, the weather conditions, etc. This error usually gets worse as you get fainter toward the limits of the telescope and looks like a &quot;widening&quot; of the CMD (seen here notably in magnitudes &gt; 19).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://www.researchgate.net/profile/Puragra_Guhathakurta/publication/231129963/figure/fig2/AS:300350830596099@1448620626576/Fig-4-A-B-vs-B-V-color-magnitude-diagram-of-all-9507-stars-in-the-WFPC2-image-of-M30.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next problem is contamination. When you point your telescope at an object like M44 or the Hyades, it's not obvious which stars belong to the cluster and which stars just happen to be in the field of view, so you will get contamination from normal milky way stars (which will not follow that isochronic relationship that cluster members will). This looks something like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://picload.org/image/rcaacarr/screenshot2017-04-14at10.11.35.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final problem that I can think of is a binary population. A &quot;photometric binary&quot; star is what happens when two stars overlap in an observation. This happens quite often in clusters, since there are &quot;intrinsic binaries&quot; which are stars that are very close to, and orbit each other. They are usually different masses, and hence different colors. This is usually seen as another line, next to the main isochrone line, that is a little redder than all of the other stars and would look something like this image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://picload.org/image/rcaacdwl/fig2.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this scatter looks far too big to be measurement error alone, modern telescopes will have magnitude errors of much much less than a magnitude (think milli-magnitudes, depending on specifics -- I'm unfamiliar with Google Sky, but any telescope is going to do better than a magnitude error, for reference the magnitude system is based on eyesight &quot;just-noticable-difference&quot;, so your eyes have an approximate error of one magnitude).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And since the scatter doesn't follow some other systematic relationship, I would discard the binaries idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most reasonable explanation to me is, like you said, foreground contamination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.researchgate.net/figure/231129963_fig2_Fig-4-A-B-vs-B-V-color-magnitude-diagram-of-all-9507-stars-in-the-WFPC2-image-of-M30&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.researchgate.net/figure/231129963_fig2_Fig-4-A-B-vs-B-V-color-magnitude-diagram-of-all-9507-stars-in-the-WFPC2-image-of-M30&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://home.strw.leidenuniv.nl/~piladiez/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://home.strw.leidenuniv.nl/~piladiez/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://inspirehep.net/record/1217369/plots&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://inspirehep.net/record/1217369/plots&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16592" LastEditorUserId="16592" LastEditDate="2017-04-14T02:12:52.737" LastActivityDate="2017-04-14T02:12:52.737" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20723" PostTypeId="2" ParentId="20642" CreationDate="2017-04-13T09:57:19.370" Score="2" Body="&lt;p&gt;So they are talking about different scales. Rubin is finding the rotation curve on kpc scales. And Sofue is talking about how hard it is to get the rotation curve in the central less than a kiloparsec. In the center you cannot use the same techniques as Rubin did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotation curves of galaxies are most often measured by the gas speed. The speed of the stars can be used, but owing to dynamical heating over time, the stellar rotation speed does not indicate the true rotation speed of the galaxy as well as the gas speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When looking for the speed of something, astronomers will look for the redshift or blueshift of the stars or the gas which they are observing. So, they need to identify a gas, which will be present in all galaxies, which has a well known and easy to observe &quot;line&quot; in the spectra which will allow a velocity determination by seeing how red or blueshifted it is compared to that emission line in a laboratory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two such lines are H$\alpha$ and molecular CO (the one mentioned in that &lt;a href=&quot;https://arxiv.org/pdf/astro-ph/9910004v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper you've linked&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The H$\alpha$ line is an optical line at 656.28 nm. The CO lines are radio waves at 2.6 and 1.3 mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like the &lt;a href=&quot;http://cdsads.u-strasbg.fr/abs/1978ApJ...225L.107R&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rubin+1978&lt;/a&gt; paper took spectra of the H$\alpha$ lines, and from Figure 2, it looks as if they've collected spectra from various radii from the center. If you do it to many different points inside the galaxy at different distances from the nucleus, you can see how fast they are moving relative to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That steep increase at the beginning of the curve usually occurs for around the extent of the bulge, a few kiloparsecs (as in this image from Rubin+1978).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/polcH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/polcH.png&quot; alt=&quot;curves&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like they had a resolution of about a kpc, which would not tell you much about the center, except that the rotation curve is very high, very quickly (you assume the very center is 0 rotation):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7IkAQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7IkAQ.png&quot; alt=&quot;curve&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sofue+1999 seems to be talking about even more central than that, like the &quot;central few hundred pc region&quot; which will require a new technique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A large limitation, as mentioned in the Sofue+1999 paper, is that HI is often obscured in the very centers (&amp;lt;0.5 kpc) of these galaxies by dust. Galactic dust, in general extincts shorter wavelength light much more effectively than long wavelength light. In this way, looking at the galaxy in longer wavelengths allows you to see through the dust. The HI line is much more easily erased by galactic extinction than the CO line at a longer wavelength. So CO makes it easier to look into the galactic central hundred parsecs than H$\alpha$. You can see this in the galactic extinction curve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/E1SDT.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E1SDT.gif&quot; alt=&quot;Extinction Curve&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CO line is in the radio band (mm range), a type of astronomy that was just picking up speed in the 1960s and 1970s. Having been predicted in the 1800s by Maxwell’s equations, scientists were unable to observe any radio signals from the Galaxy until a serendipitous observation of Sagittarius A* by Karl Jansky at Bell Labs in the 1930s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, radio astronomy comes with a problem of resolution, the resolving power of a telescope is basically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Theta = 1.22 * \frac{\lambda}{D}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where lambda is the wavelength of the observation and D is the diameter of the telescope. And, for reference, the distances to two galaxies from that rotation curve above are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NGC4594 = 9.45 Mpc&lt;br&gt;&#xA;NGC7217 = 15 Mpc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the distances of these galaxies, we need to be looking in the right wavelength to penetrate, and we need enough resolution to get the rotation speed at maybe 10 parsec scales. This resolution, at 15 Mpc, is maybe 0.1 arcseconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3600 * arctan(\frac{10}{15e6}) * \frac{180}{\pi} = 0.1375 arcseconds$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get that resolution observing at the 2.6 mm you need for the CO line, you need a:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1.22 * 3600 * \frac{180}{\pi} * \frac{2.5e-3}{0.1375} = 4581.48$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.5 km telescope… Hmmmmm. A bit of a problem considering the unfortunate tale of the 300 foot telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KJPFn.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KJPFn.gif&quot; alt=&quot;before&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/KXNNo.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXNNo.gif&quot; alt=&quot;after&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, computers to the rescue. &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_interferometer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Interferometry&lt;/a&gt; is an idea whereby one can use two telescopes to increase their resolving power. When combining their signals, they have an effective diameter of their separation. Of course, they are not equivalent to a single dish telescope of that diameter, because they collect far less total starlight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qAzug.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qAzug.jpg&quot; alt=&quot;dish array&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The technique was successfully utilized scientifically as early as the 1920s, to resolve the star Betelgeuse. But it was difficult to use this technique in it’s modern form, combining tens or hundreds of telescopes at once without the advent of computers which could handle the calculations in the 1960s and 1970s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now, the final piece. Interferometry grows much &lt;a href=&quot;https://en.wikipedia.org/wiki/Aperture_synthesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;more difficult the smaller your wavelength&lt;/a&gt;. Interferometry is only possible if amplitude and phase of the incoming wave is recorded and recombined precisely. So an optical interferopeter needs: more precisely polished mirrors, more precisely measured separations, stable signal information transport from the dishes (fiberoptics), and more computing power to run more fourier transforms. While radio interferometry was performed in the 1920s, it took all the way until the 1990s to do optical interferometry, and it's still far from routine (see the &lt;a href=&quot;https://en.wikipedia.org/wiki/Large_Binocular_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Large Binocular Telescope&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to summarize:&lt;br&gt;&#xA;1) To get a rotation curve of a galaxy, we need a molecular line that we can get a redshift or blueshift from. HI (optical) and CO (mm) are good choices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The very centers of galaxies are obscured by dust, which destroys optical signals, but allows mm to pass, so CO is the molecule of choice for rotation curves of the nuclei of these galaxies, as H$\alpha$ signal is poor in the centers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) However, for a mm telescope to be able to spatially resolve small parts of the galaxies at the large distances of 10+ MPC in this wavelength, we need a telescope a few kilometers wide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) We can solve this with interferometry, but only recently has the technology to do interferometry at this wavelength been accomplished. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first interferometer observing at this wavelength, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Submillimeter_Array&quot; rel=&quot;nofollow noreferrer&quot;&gt;SubMillimeter Array&lt;/a&gt; (SMA; eight, 6-meter telescopes), opened to guest astronomers for the first time in 1990. The SMA was still being expanded at the time of the paper you linked. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2CFzf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2CFzf.jpg&quot; alt=&quot;sma&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These days, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Atacama_Large_Millimeter_Array&quot; rel=&quot;nofollow noreferrer&quot;&gt;Atacama Large Millimeter Array&lt;/a&gt; (ALMA) is just now nearing completion, and is focusing on the same wavelength range only with huge resolution and surface area (66 telescopes from 7-12 meters). They are collecting CO rotation curves to such precision that they claim to be able to estimate the central black hole masses of galaxies to 5%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xfjsF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xfjsF.jpg&quot; alt=&quot;alma&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stuff:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_interferometer&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Astronomical_interferometer&lt;/a&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Aperture_synthesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Aperture_synthesis&lt;/a&gt;&#xA;&lt;a href=&quot;https://ned.ipac.caltech.edu/level5/Fitzpatrick/Figures/figure1.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ned.ipac.caltech.edu/level5/Fitzpatrick/Figures/figure1.gif&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.nrao.edu/whatisra/images/300ft-before.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.nrao.edu/whatisra/images/300ft-before.gif&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.nrao.edu/whatisra/images/300ft-after.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.nrao.edu/whatisra/images/300ft-after.gif&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.atnf.csiro.au/outreach//images/wallpapers/atcamorningsml.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.atnf.csiro.au/outreach//images/wallpapers/atcamorningsml.jpg&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.almaobservatory.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.almaobservatory.org/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16592" LastEditorUserId="16592" LastEditDate="2017-04-14T18:31:33.443" LastActivityDate="2017-04-14T18:31:33.443" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20724" PostTypeId="2" ParentId="20700" CreationDate="2017-04-13T10:24:51.193" Score="2" Body="&lt;p&gt;TiO is used as a sunscreen, it has high absorption and emittance rates and a high radiation potential. SiO2 is used for fiber optic cables and probably deflects and transmits photons a lot more, and it is very chemically inert, that's why the desert is made of quartz, it's all that is left after the rest has been weathered, it's orbitals are very stable and difficult to change. That's why silicon life forms probably don't exist, especially in a water world, they can't really bond with H and they absorb oxygen very tenaciously. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a molecule absorbs a wavelength of light, an electron is excited from a full orbital into an empty anti-bonding orbital. Each jump takes energy from the light, and a big jump obviously needs more energy than a small one. Each wavelength of light has a particular energy associated with it. If that particular amount of energy is just right for making one of these energy jumps, then that wavelength will be absorbed - its energy will have been used in promoting an electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/7437/why-glass-is-transparent&quot;&gt;https://physics.stackexchange.com/questions/7437/why-glass-is-transparent&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:15.377" LastActivityDate="2017-04-13T10:56:00.200" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20725" PostTypeId="2" ParentId="20719" CreationDate="2017-04-13T10:56:30.937" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can the release of gravitational energy during planet formation in principle (is there enough) explain this much heat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Planets in general can have internal heat sources (like radioactive breakdown or a complex hot core).  During planetary formation there are thought to be multiple collisions with large proto-planets before things settle down to a more-or-less stable set of orbits and planets.  These collisions add energy to the planets as they develop and increase temperature (a lot), as does ongoing bombardment from boring-old asteroids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now all of this energy does basically come from gravitational sources, but as you can see it's a complex (messy) business and there are many significant events that can alter temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be instructive for you to read about existing theories on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Formation_and_evolution_of_the_Solar_System&quot; rel=&quot;nofollow noreferrer&quot;&gt;formation of Solar system&lt;/a&gt; as well as material related to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Structure_of_the_Earth#Core&quot; rel=&quot;nofollow noreferrer&quot;&gt;structure of the Earth&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-04-13T10:56:30.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20726" PostTypeId="2" ParentId="20720" CreationDate="2017-04-13T13:37:42.173" Score="5" Body="&lt;p&gt;&lt;a href=&quot;http://www.narit.or.th/en/files/2007JAHHvol10/2007JAHH...10..148B.pdf&quot; rel=&quot;noreferrer&quot;&gt;Bryan (2007)&lt;/a&gt; gives a number of reasons why O'Meara's discovery was largely discounted:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;As you stated, the behavior was entirely inconsistent with Keplerian predictions of motion.&lt;/li&gt;&#xA;&lt;li&gt;While O'Meara was able to reproduce his findings, no other independent observers could.&lt;/li&gt;&#xA;&lt;li&gt;The detections were done entirely visually, rather than with numerical measurements.&lt;/li&gt;&#xA;&lt;li&gt;There had been previous observations (i.e. in the 19th century), but in different places in the rings (the A and C rings, not the B ring). These have never been confirmed, and it is still believed that these were &lt;em&gt;actually&lt;/em&gt; illusions.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Reliable observations from ground-based telescopes, especially those of amateurs, have arisen &lt;a href=&quot;http://www.astrosurf.com/delcroix/doc/EPSC-DPS2011/Delcroix%20M.%20et%20al.%202011-First%20ground%20observations%20of%20Saturn%27s%20spokes%20around%202009%20equinox%20(abstract)-EPSC-DPS2011-336.pdf&quot; rel=&quot;noreferrer&quot;&gt;within the past decade or so&lt;/a&gt;. Since 2007, 67 &quot;candidate&quot; observations have been made of the spokes, many during the 2009 Saturnian equinox (keep in mind that the spokes may be a seasonal phenomenon). One additional factor that made observations difficult was that the spokes seemed to have vanished, even from space probes, in the years after the &lt;em&gt;Voyager&lt;/em&gt; observations, making it impossible for ground-based telescopes to see them.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2017-04-13T13:37:42.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20728" PostTypeId="2" ParentId="20695" CreationDate="2017-04-13T17:51:42.277" Score="0" Body="&lt;p&gt;You can do your own calculations easily, no need to ask anyone. Just change the distance units in the examples below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=tan(1%20arcsec)%20*%2010%20km&quot; rel=&quot;nofollow noreferrer&quot;&gt;tan(1 arcsec) * 10 km&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=tan(1%20arcsec)%20*%204%20light%20years&quot; rel=&quot;nofollow noreferrer&quot;&gt;tan(1 arcsec) * 4 light years&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2017-04-13T17:51:42.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20729" PostTypeId="1" CreationDate="2017-04-13T19:22:09.037" Score="2" ViewCount="89" Body="&lt;p&gt;In the celestial sphere below, what is the $\omega$ and $\tau$ physically? I mean, In the normal world, how do we see them? for example, we can see the altitude of the sun (a) by examining the shadow of a pillar and comparing it with its height, How do we see $\omega$ and $\tau$ in real world?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S is the sun and W is the west.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also  I want to know how do they effect on the shadow of wall which is built in direction of &quot;$SCP$ to $NCP$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7hBDn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7hBDn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16441" LastActivityDate="2017-04-14T14:41:18.263" Title="What is $\omega$ and $\tau$ in this celestial sphere" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20730" PostTypeId="1" AcceptedAnswerId="20732" CreationDate="2017-04-14T05:19:40.603" Score="1" ViewCount="82" Body="&lt;p&gt;I know that there has been conflicts between scientists and the native population. Astronomers want to build new telescopes on Mauna Kea, however, several sacred sites would have to be built on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, are there any current plans to move forward with the observatories, or have they reached a gridlock?&lt;/p&gt;&#xA;" OwnerUserId="16597" LastActivityDate="2017-04-14T11:35:43.873" Title="Status of new telescopes on Mauna Kea?" Tags="&lt;telescope&gt;&lt;observational-astronomy&gt;&lt;observatory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20732" PostTypeId="2" ParentId="20730" CreationDate="2017-04-14T11:35:43.873" Score="2" Body="&lt;p&gt;You're talking about the construction of the '&lt;a href=&quot;https://en.wikipedia.org/wiki/Thirty_Meter_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thirty Meter Telescope&lt;/a&gt;'.&lt;br&gt;&#xA;Currently the construction has been halted by the State Supreme Court of Hawaii as a reaction to the violation of native sacred land.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Negotiations are going on about moving the construction site to the canary Islands, which has been deemed an acceptable alternative in the northern hemisphere, even if only barely.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2017-04-14T11:35:43.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20733" PostTypeId="2" ParentId="20729" CreationDate="2017-04-14T14:41:18.263" Score="3" Body="&lt;p&gt;There seems to be very little to go on here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Zenith&quot; and &quot;Horizon&quot; are easy to understand, it suggests that this is a projection of the sky (and ground) In that context &quot;W&quot; would appear to be &quot;West&quot;, and you give that &quot;S&quot; is &quot;sun&quot;. The horizon is curved. If you project the entire hemisphere the horizon would go straight across this isn't a complete hemispheric projection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we are looking West, then North would be to the right, and South to the left. The point P seems to be identified with NCP. The first guess would be that N of NCP means North, but this contradicts the reasoning that North is on the right. You mention SCP in the text, but it isn't in the diagram. The sun appears to be West, or just North of West, which is not an unusual but not impossible position. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lines SW, WP and PS are curved. This may be to indicate that they are great circles on the celestial sphere. If so then $\omega$ is the spherical angle SWP and $\tau$ is the great circle distance SW, perhaps given as an angle with vertex at the centre of the sphere. Given three points on a sphere (in alt-az coordinates), it is a small exercise in spherical geometry to find the spherical distance and angles between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure why you are being coy about the source of this. If you could link to the question from which this is taken the meanings of $\delta$ and H could be clearer. I guess the solution is an exercise in spherical angles, and projections from a circle to the plane&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-04-14T14:41:18.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20734" PostTypeId="2" ParentId="20714" CreationDate="2017-04-14T14:51:11.350" Score="3" Body="&lt;p&gt;I might venture into some math which is slightly above what you need, but I wanted to show you the reason behind the correct equation to use. To start with, the pressure due to radiation, in the most general sense, is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{\mathrm{absorption}} = \frac{\langle S\rangle}{c} \cos(\theta)$$&#xA;$$P_{\mathrm{reflection}} = 2\frac{\langle S\rangle}{c} \cos^2(\theta)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top equation represents the case where some light hits your object and the light gets absorbed, and the bottom equation represents the case where light hits and bounces off your object (really it's re-emitted). Your case is the bottom equation so I'll work with that from now on and drop the reflection subscript. Without needing to explain exactly what it is, just accept that $\langle S\rangle$ is the average energy of the photon being absorbed or emitted. Now, that's for a single photon. If we have a lot of photons, generally the energy distribution of all the photons together is said to be described by some function, such as $I_\lambda$ (the subscript refers to wavelength). We can define the full pressure then by not only integrating over all wavelengths of all the photons, but importantly &lt;em&gt;all the directions that the photons may hit or be emitted&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = \frac{2}{c}\int_0^\infty\int_0^{2\pi}\int_0^{\pi/2} I_\lambda(\lambda) \cos^2(\theta)\sin(\theta)\:d\theta\:d\phi\:d\lambda$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note now that we've replaced $\langle S\rangle$ with $I_\lambda(\lambda)$. Also, since we're assuming this is pressure on a flat surface, we're &lt;em&gt;not&lt;/em&gt; integrating over every angle, but only half of them $-$ any photon, in order to hit the solar sail, has to come from the direction the sail is facing, and can only reflect in the direction it came from, i.e., we're discluding photons &quot;behind&quot; the sail. For that reason, the $d\theta$ integral only goes to $\pi/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't understand the math at this point, that's fine but here's the important takeaway: If we only integrated over all wavelengths, you'd get the equation &lt;em&gt;without the three&lt;/em&gt; that you listed. What that means is that the equation without the three doesn't take into account all the possible angles and orientations of the incoming photons. It simply assumes all your photons hit your surface perpendicularly, and then bounce off perpendicularly. This is of course not physically accurate as the photons can come in from a variety of angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do the angle integrals out, you find that you get the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = \frac{2}{c}\int_0^{2\pi}\:d\phi \int_0^{\pi/2} \cos^2(\theta)\sin(\theta)\:d\theta\int_0^\infty I_\lambda(\lambda)\:d\lambda$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = \frac{2}{c}\frac{2\pi}{3}\int_0^\infty I_\lambda(\lambda)\:d\lambda$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see there in the last line that the three popped out of those integrals over all the possible angles (specifically the $d\theta$ integral). Now, you get the rest of this by assuming some sort of function for $I_\lambda(\lambda)$ which describes the energy distribution of the photons as a function of wavelength. Since your light comes from a star, you know that's just going to be the &lt;a href=&quot;https://en.wikipedia.org/wiki/Planck%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck distribution&lt;/a&gt; (specifically the equation for $B_\lambda(\lambda,T)$). This integral is non-trivial, but if you again run through the math, you find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = \frac{2}{c}\frac{2\pi}{3}\int_0^\infty \frac{2hc^2}{\lambda^5}\frac{1}{e^{hc/\lambda k_BT}-1}\:d\lambda = \frac{2}{c}\frac{2\pi}{3}\frac{\sigma T^4}{\pi}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\boxed{P = \frac{4\sigma T^4}{3c}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the true equation for radiation pressure on the solar sail, assuming reflection is occurring. This includes the three because we need to account for the photons hitting and bouncing off at all possible angles (for the solar sail). Note one important differences here is that my equation already has the extra &quot;2&quot; due to reflection factored in. Your last equation differs from mine in that it throws in that extra &quot;2&quot; for reflection. So either it erroneously included the factor of &quot;2&quot; twice, or (more likely) &lt;em&gt;its account for reflection at every angle, not just all possible angles&lt;/em&gt;. Remember above, we only integrated the $d\theta$ angle up to $\pi/2$ for the reason explained there. If we integrated up to $\pi$ to include every angle, then we'd get an extra &quot;2&quot; as well and my equation would match your last one. Your solar sails though, don't receive or re-emit photons at every angle, and so my equation differs by that factor of &quot;2&quot;.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2017-04-14T18:30:02.613" LastActivityDate="2017-04-14T18:30:02.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20735" PostTypeId="2" ParentId="20714" CreationDate="2017-04-14T18:40:30.383" Score="2" Body="&lt;p&gt;I actually think you are supposed to assume the solar sail is unfurled while the spaceship is still quite far from the red giant (despite &quot;too close&quot;), because otherwise the problem is extremely difficult.  At least if you are far from the red giant, you can assume all the light is incident along the normal to the sail, and if it's a mirror, it will also reflect back the same way.  In that situation (only), you can simply take the radiative flux density (L / 4pi D^2), divide by c (to get a momentum flux), multiply by the area A, and double it (to account for straight-back reflection), resulting in an expression that is like zephyr's except without the 1/3 (which comes from assuming we are only talking about one instant and it is right at the surface of a blackbody with no limb darkening, none of which we should expect from a red giant), and my expression also has a 1/D^2 which is actually important in what follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I think this is what is intended is because only then is D not essential to treat in detail.  It then becomes easy to figure out what the area of the sail needs to be, we can do it with energy considerations.  At large D, the radiative force will be a constant multiple of the gravity, and that multiple needs to be &quot;2&quot; in order to get the spaceship to go back to infinity where it came from (assuming either the original speed at infinity was small, or that your goal is to return so whatever that original speed was, as these are the only ways you don't need to know that original speed).  The gravitational force is GMm/D^2, so P times A needs to be twice that, where P is given above.  The answer comes out A = 4pi G M m c / L, which is independent of the D where the sail is unfurled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason it's much harder if you don't make D large is that the incident flux is not along the normal any more, so you need an r-dependent factor that ranges from 1/3 right at the surface, to 1 farther out.  You'd have to integrate the role of that factor to get the necessary A, whereas my answer doesn't need any of that.  What's more, if you unfurl the sail right at the surface, the spacecraft will enter the star, and the radiative environment becomes even more complicated, not to mention possibly destroying the spacecraft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I don't think any of those expressions are what you want, because you can't just get the force at the surface, you need to know the force that would succeed in getting the spacecraft to turn around, so it must exceed gravity by a factor that would be very hard to compute unless you take D large.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2017-04-14T18:47:25.517" LastActivityDate="2017-04-14T18:47:25.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20736" PostTypeId="1" CreationDate="2017-04-14T20:43:32.760" Score="14" ViewCount="3762" Body="&lt;p&gt;So I was watching something that said&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;if we compressed Earth into the size of a peanut: we would get a black hole;&lt;/p&gt;&#xA;&lt;p&gt;if we compressed Mount Everest into a few nano-meters; we would get a black hole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Can I make a black hole with one or two atoms? If yes, would it become larger and turn into normal-sized black hole?&lt;/p&gt;&#xA;" OwnerUserId="16599" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-04-19T08:21:15.667" Title="Can I make a black hole with one or two atoms?" Tags="&lt;black-hole&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20737" PostTypeId="2" ParentId="20736" CreationDate="2017-04-14T22:18:52.503" Score="23" Body="&lt;p&gt;There are two answers: yes and no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Yes&lt;/em&gt; because every mass M has a Schwarzschild radius given by $\frac{2GM}{c^2}$ (where G is the gravitational constant (about $6.7\times10^{-11}$ and c is the speed of light (about 300 000 000 $\mathrm{m/s}$). If something is compressed to its Schwarzschild radius it becomes a black hole. You can do this for an atom. An atom of carbon (for example) has a mass of $2\times10^{-26}\mathrm{kg}$ so its Schwartzschild radius is &#xA;$$ \frac{2\times (6.7\times10^{-11})\times (2\times10^{-26})}{300 000 000^2}\approx 3\times10^{-53}\ \mathrm{metres}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the actual answer is &lt;em&gt;no&lt;/em&gt; as there is no feasible way of compressing an atom to this size. Of significance here is the fact that this size is so small that objects this small don't behave like small balls but as quantum mechanical objects. But a black hole is a gravitational object modeled by General Relativity, and Relativity and quantum mechanics don't work well together. In other words, we don't have a scientific model for describing how an atomic mass black hole would behave. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stephen Hawking has shown that small black holes are unstable, so an atomic mass black hole would be very unstable, evaporating in a very short time.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="7142" LastEditDate="2017-04-16T06:29:42.303" LastActivityDate="2017-04-16T06:29:42.303" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20738" PostTypeId="2" ParentId="20736" CreationDate="2017-04-14T23:53:28.127" Score="15" Body="&lt;p&gt;I think the answer is &lt;strong&gt;No&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we try and compress these atoms, we end up (eventually) with the nuclei close enough to be forced to fuse.  Fusion would mean we've formed a single nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This stage is unavoidable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your two atoms question now reduces to &lt;em&gt;whether a single nucleus can form a black hole ?&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nucleus is a kind of complex quark-gluon mix and if we compress it more we end up with a very dense version of that which we basically don't have physics to model properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's extremely unlikely that conventional general relativity can be applied to something that will be so small it's actually smaller than we think we can apply quantum theory.  And the energy density involved at that point would be so high our current theories don't make sense any more.  We need a quantum theory of gravity to do this and we don't have one that works well enough.  In fact we're not even sure a quantum theory of gravity would allow us to go to such small, high energy scales - even that is unknown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we're in uncharted waters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So why &quot;no&quot; ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, to force such a compression of a nucleus we'd have to apply energies to a very small region of space - smaller than we think it's possible to do, because of the consequences of the uncertainty principle.  Put simplistically, beyond some point we'd not be able to simultaneously say where the nucleus is and how fast it's moving.  It would be impossible to confine to a smaller region.  This would happen long before we reach the Schwarzschild radius, at around the &lt;a href=&quot;https://en.wikipedia.org/wiki/Planck_length&quot; rel=&quot;noreferrer&quot;&gt;Planck length&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you'll see from the answer by @James-K , the Schwarzschild radius is about 10&lt;sup&gt;−53&lt;/sup&gt; m, but the Planck length is &lt;em&gt;18 orders of magnitude larger&lt;/em&gt; at about 10&lt;sup&gt;−35&lt;/sup&gt; m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we could not realistically confine and compress our nucleus into a small enough space to ever reach its black hole size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can make a generic catch-all statement that a new theory might provide some loophole that lets us get around that, but it does seem unlikely as we'd expect a new theory to reproduce most of what we already know at those limits.  It's hard to imagine the uncertainty principle &quot;going away&quot; so I don't see a way around that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;There's an unproven possibility of a yes.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quantum theory of gravity that works &lt;em&gt;might&lt;/em&gt; (repeat &lt;em&gt;might&lt;/em&gt; or &lt;em&gt;might not&lt;/em&gt;) find that gravity at that scale changes its character and allows it to form event horizons at larger sizes than we'd currently expect for such mass-energy ranges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we lack any evidence to support that idea, and I won't convert a &quot;no&quot; to a &quot;maybe yes&quot; simply to allow room for any wild idea.  That's science fiction, not science.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastEditorUserId="7142" LastEditDate="2017-04-16T08:22:54.993" LastActivityDate="2017-04-16T08:22:54.993" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20739" PostTypeId="2" ParentId="20736" CreationDate="2017-04-15T00:33:41.487" Score="9" Body="&lt;p&gt;A small addition to the answers above (I like the Planck length answer).  It was thought that it might be possible to make very small black holes at CERN, theoretically anyway, but that theory &lt;a href=&quot;https://home.cern/about/physics/extra-dimensions-gravitons-and-tiny-black-holes&quot; rel=&quot;nofollow noreferrer&quot;&gt;required extra dimensions&lt;/a&gt; to exist.   Because no black holes were observed, the extra dimensions (on very small scales) theory took a hit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if those black holes could be created, they are predicted to evaporate very very quickly.  (billionth of a billionth of a billionth of a second), but even that rate of decay should be noticeable.    None were noticed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also worth asking, if CERN smashes two protons together really really fast, and, if that makes a black hole (in theory), as in, pretend it's possible . . . Would this theoretical black hole really be made up of two protons or is it made up of two protons and 14 TeV plus kinetic energy?  I think it's more accurate to say that such a black hole is really created out the kinetic energy not the atoms themselves.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some might call that splitting hairs on Schrodinger's cat, but I think it's an important point.  The enormous kinetic energy of a near light speed collision, might just be able to create a micro black hole, and in that case, it's the kinetic energy that should be called the primary ingredient not the atoms.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="10128" LastEditDate="2017-04-16T16:33:31.807" LastActivityDate="2017-04-16T16:33:31.807" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20740" PostTypeId="2" ParentId="20698" CreationDate="2017-04-15T08:14:55.410" Score="2" Body="&lt;p&gt;The &quot;lithium test&quot; for a brown dwarf involves measuring two things - the lithium content and the spectral type (a proxy for surface temperature) or luminosity. But you may also need to know (or assume) something about the age of the object. The method is useful because the tracks of low-mass stars and brown dwarfs run together very closely in the usual luminosity versus Teff plane of the Hertzsprung-Russell diagram. In addition to the considerable uncertainty in the placement of these tracks, one really needs to know the &lt;em&gt;age&lt;/em&gt; of an object in order for its luminosity or Teff to lead to even a crude mass estimate and hence the Li-test is pursued.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Li is completely and rapidly burned in a fully convective, low-mass object, once the core temperature reaches about 3 million degrees. The time it takes for a contracting pre main sequence star (or some brown dwarfs) to reach this temperature is mass-dependent. It takes about 20 million years at 0.2 solar masses, about 120 million years at 0.075 solar masses (roughly the boundary between stars, and brown dwarfs that will never become hot enough to fuse &lt;em&gt;hydrogen&lt;/em&gt;), and if the object is below 0.06 solar masses, it &lt;em&gt;never&lt;/em&gt; gets rid of its Li. The physics of this process is well understood and hence it has become a very useful and trusted tool for mass and age estimation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same time as this happens, the surface temperature and luminosity are also diagnostic of mass, but in an age dependent way. An M7 spectral type is on the brown dwarf limit at about 120 Myr, but a similar spectral type in an older and more luminous object would mean it was more massive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Historically then, the Li test was applied to low-mass objects in the Pleiades cluster (see &lt;a href=&quot;https://arxiv.org/abs/astro-ph/9607002&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rebolo et al. 1996&lt;/a&gt;), which has an age of about 120 Myr. Objects with spectral types cooler than M7 &lt;em&gt;at that age&lt;/em&gt; should be brown dwarfs. This was confirmed by finding Li in their atmospheres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Li test is not a completely unambiguous indicator of whether an object of unknown age is a star or a brown dwarf. Both stars and brown dwarfs can have spectral types M6.5-L2, but a star would be older at the same spectral type. If you have measured the luminosity (or spectral type) of an object and know whether it has lithium or not, then this provides a constraint on both mass and age. The diagram below, from &lt;a href=&quot;http://adsabs.harvard.edu/abs/1998ASPC..134..394B&quot; rel=&quot;nofollow noreferrer&quot;&gt;Basri et al. (1998)&lt;/a&gt;, is I think as clear as it can be made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GKXLd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GKXLd.png&quot; alt=&quot;Li-temperature relation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more-or-less diagonal curves are the tracks along which low-mass stars and brown dwarfs of the labelled mass evolve (getting cooler with age). The diagonal dashed region shows the area of the temperature versus age plot where Li is expected to be depleted. I have labelled the equivalent spectral types. The stippled (dotted) area marks where the presence of Li &lt;em&gt;guarantees&lt;/em&gt; that you have a brown dwarf with mass $&amp;lt;$ 0.075 solar masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus an M6.5 object with Li could either be a brown dwarf or a star younger than about 140 Myr, but an M6.5 object without Li is a star. An M8 dwarf with Li is a brown dwarf with an age less than 200 Myr, but an M8 dwarf without Li could either be a more massive brown dwarf with age just over 200 Myr or an older low-mass star. At spectral type L0-L2 then an object with Li will be a brown dwarf if its age is less than 500-1000 Myr, but could still be a brown dwarf with mass above 0.06 solar masses with an age just a little older than this, or an even older low-mass star. Objects cooler than L2 are always brown dwarfs and always have Li - but this test may fail at much cooler temperatures (cooler than abut L8) because the Li atoms form LiCl and LiOH molecules, the atomic Li I line at 6708A disappears from the spectrum, and the molecular features are too weak/confused to be identified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar diagram can be made with luminosity on the y-axis and actually, that is more secure, since Li depletion as a function of luminosity is much easier to &#xA;predict than the behaviour of effective temperature, which depends a lot on the details of the atmosphere. Of course to measure a luminosity you need the distance to the object though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays, the presence of Li in a low mass objects atmosphere is more likely to be used as an age, rather than mass indicator. For example, if one sees Li in an M7 object, then it is likely to be younger than 120 Myr (and vice versa). The prediction of Li depletion, particularly as a function of luminosity or spectral type is thought to be more secure than predicting the position of such an object in the HR diagram. This forms the basis of the&#xA;&lt;a href=&quot;http://www.astro.keele.ac.uk/~rdj/pages/ldb.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;lithium depletion boundary&lt;/a&gt; age determination method for clusters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To sum up&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;An object with a spectral type of M8 or cooler &lt;em&gt;with Li&lt;/em&gt; is a substellar brown dwarf. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An object with Li but a warmer spectral type &lt;em&gt;may&lt;/em&gt; be a brown dwarf if it is old enough.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An object younger than $\sim 150$ Myr &lt;em&gt;without&lt;/em&gt; Li is a star.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An object cooler than about spectral type L2 is a brown dwarf.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-07-14T18:41:08.053" LastActivityDate="2018-07-14T18:41:08.053" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="20742" PostTypeId="2" ParentId="20702" CreationDate="2017-04-15T08:33:08.910" Score="3" Body="&lt;p&gt;The page you quote is guilty of a gross simplification (or error). The luminosity ratio &lt;em&gt;cannot&lt;/em&gt; be estimated from the difference in visual magnitudes for stars of different spectral type. You need to use the difference in bolometric magnitude, which indicates the luminosity integrated over all wavelengths. This will involve making &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Bolometric_correction&quot; rel=&quot;nofollow noreferrer&quot;&gt;bolometric corrections&lt;/a&gt; to the visual absolute magnitudes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A star like Proxima Cen radiates much of its light outside the visible band and has a much different and larger bolometric correction than the Sun or Sirius. This accounts for your problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct calculation uses a table of bolometric corrections vs temperature. This rapidly changes for M-dwarfs in the visual band (you would be more precise using K magnitude), but is about 2.6 mag for a star at 3100 K (see &lt;a href=&quot;http://ads.harvard.edu/cgi-bin/bbrowse?book=hsaa&amp;amp;chap=2&amp;amp;page=0069&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zombeck 1992&lt;/a&gt;). That is, the bolometric magnitude is smaller than $M_V$ by 2.6 mag. The equivalent correction for the Sun is only 0.1 mag.&#xA;$$ M_{\rm bol}= M_V - BC$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus there is a 2.5 mag &lt;em&gt;less&lt;/em&gt; difference in the &lt;em&gt;bolometric&lt;/em&gt; magnitudes in your final equation than the difference in absolute visual magnitudes, and I get a radius for Barnard's star of $0.23\ R_{\odot}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probable reason for the discrepancy between this and the radius quoted in wikipedia (actually it is the radius determined by an interferometric angular diameter measurement plus the parallax by Lane et al. 2001; a more recent measurement by &lt;a href=&quot;https://arxiv.org/abs/1208.2431&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boyajian et al. 2012&lt;/a&gt; gives $0.187\pm 0.001\ R_{\odot}$) is that (a) the assumed temperature is too low or (b) that the bolometric correction scale for V-band magnitudes is not very well defined for very cool stars. Both (a) and (b) are possible, especially as Barnard' star probably has a significantly subsolar metallicity.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-07-16T17:16:58.427" LastActivityDate="2017-07-16T17:16:58.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20743" PostTypeId="2" ParentId="8518" CreationDate="2017-04-15T14:04:29.967" Score="0" Body="&lt;p&gt;Your latest information is not a clear image of H2O economy of the evolving earth. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/bYdmQ.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bYdmQ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it takes a lot of hot and cold and solar wind to dry a planet into a rock, a solar system planet's natural state is to have a lot of ices, if it wasn't for atmospheric loss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that there are vast amounts of H and O inside the sun and solar system, and the earth would have constantly received large volumes of water throughout it's formation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solar radiation and freezing conditions evaporate/sublimate water and volatile gaseous chemistries, so inner planets would have more water if it were not for solar radiation and dry frost, because it sends the masses of water back out into space. H and O bond well with other minerals and they exist in many acids and minerals more prevalently than in water, i.e. there is more water 100ds of kilometers inside the planet than in the oceans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen and Oxygen are some of the lightest elements and most likely to escape to the high atmosphere, so the end balance of the planets has been an equation between massive constant arrival of water and massive constant evaporation, where only the earth has had ambient temperature, high planet gravity, and the magnetic field to keep water as oceans, rather than H in ices and acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atmospheric losses can be more of light Hydrogen than Oxygen, Venus has masses of atmospheric oxygen and almost no hydrogen.&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2017-04-15T14:12:11.280" LastActivityDate="2017-04-15T14:12:11.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20745" PostTypeId="2" ParentId="20717" CreationDate="2017-04-15T14:37:39.907" Score="1" Body="&lt;p&gt;&quot;Virtual Observatory&quot; is a catchall phrase basically for a place any scientist can publish their data (ie. tables in their papers) or for hosting databases (such as the latest releases of major survey measurements) which comply with TAP protocols and so can be easily communicated with using a variety of astronomical data analysis programs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many many virtual observatories, here, for example, is the German Virtual Observatory, which is run by the excellent people in Heidelberg: &lt;a href=&quot;http://www.g-vo.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.g-vo.org/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have a link describing the data publishing process here: &lt;a href=&quot;http://www.g-vo.org/pmwiki/About/PublishYourData&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.g-vo.org/pmwiki/About/PublishYourData&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually what this entails is putting your data, such as tables from your paper, into a universal format, like VOTable. This type of table is readily accessed by astronomy data analysis tools such as TopCat (or Aladin, like you mentioned): &lt;a href=&quot;http://www.star.bris.ac.uk/~mbt/topcat/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.star.bris.ac.uk/~mbt/topcat/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Programs such as TopCat can also pull data from these virtual observatories using the TAP protocol which is described a bit here in reference to the GAVO: &lt;a href=&quot;http://dc.zah.uni-heidelberg.de/__system__/tap/run/info&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://dc.zah.uni-heidelberg.de/&lt;strong&gt;system&lt;/strong&gt;/tap/run/info&lt;/a&gt; (note: this link is displayed incorrectly) and a bit here: &lt;a href=&quot;http://www.star.bris.ac.uk/~mbt/topcat/sun253/TapTableLoadDialog.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.star.bris.ac.uk/~mbt/topcat/sun253/TapTableLoadDialog.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main purpose of the VO type databases is to allow people to query them while referencing data they have on their computer by allowing SQL (ADQL) queries (crossmatching, selection cuts, etc.) The TAP protocol also allows programs such as Aladin and TopCat to communitcate, so you can do a crossmatch of your data in TopCat with a catalog on the VO and the data will be instantly displayed visually in Aladin as well. Vizier, as far as I know, is just a repository of databases and does not allow this type of querying and interaction.&lt;/p&gt;&#xA;" OwnerUserId="16592" LastEditorUserId="16592" LastEditDate="2017-04-16T01:20:19.850" LastActivityDate="2017-04-16T01:20:19.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20746" PostTypeId="2" ParentId="20719" CreationDate="2017-04-15T15:14:07.940" Score="1" Body="&lt;p&gt;It's important not to confuse the amount of heat the Earth has radiated in its lifetime (a quantity you have no access to here) with its current heat content.  This question is asking you to neglect all the heat radiated (which could be affected by things like radioactivity), and just look at the current heat content, and ask if that is much more, or much less, than the gravitational energy GM^2/R.  So find some way to get the interior T of the Earth (not the surface T, that has nothing to do with this question), and you should find that the total heat content is way &lt;em&gt;less&lt;/em&gt; than the gravitational energy released.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The virial theorem was mentioned above, so if you know about that, you might have thought the heat content would be half the gravitational energy.  That's only true of normal stars, which are held up by the gas pressure of largely noninteracting particles, whereas the Earth is held up by forces within the rocks that are not included in the usual virial theorem and have little to do with heat content.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, the gravitational energy gives the energy that was released in forming the Earth, so when you find how little the heat content is in comparison, it gives you a good idea of how much the Earth has cooled since forming.  It also gives you a lower bound on how much net heat it has radiated (over and above the heat of sunlight it has absorbed).  It's only a lower bound because there are also internal radioactive heat sources, but I believe their sum total effect is still fairly small compared to GM^2/R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the answer to your question is going to turn out to be &quot;yes&quot; when you do that comparison.  (I wouldn't give that spoiler except that some of the answers above might have led you to think the answer should be &quot;no.&quot;)&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2017-04-15T15:14:07.940" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20747" PostTypeId="1" CreationDate="2017-04-15T15:20:51.333" Score="2" ViewCount="3904" Body="&lt;p&gt;If we know a star's magnitudes from several bands(say u,g,r,i,z,J,H,k) and the star's distance, how to calculate Lbol?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's suppose the star has a black body spectrum. Is there any manual about the calculation? It's better to have a lot of technical details.&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="13587" LastEditDate="2017-08-04T12:00:44.643" LastActivityDate="2017-08-04T12:00:44.643" Title="How to convert magnitudes to bolometric luminosity?" Tags="&lt;luminosity&gt;" AnswerCount="1" CommentCount="7" ClosedDate="2017-04-20T10:35:34.607" ContentLicense="CC BY-SA 3.0" />
  <row Id="20751" PostTypeId="1" CreationDate="2017-04-16T09:00:18.603" Score="1" ViewCount="54" Body="&lt;p&gt;vernal equinox for a very high tower ($h_1&amp;gt;3000m$) in a city ($\lambda , \phi$) in altitude $h_1$ from the sea level is 8:00 Local time. Does this time change if we go to another altitude in that same position ($\lambda , \phi$), for example altitude $h_2$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other way, does the time of vernal equinox change with altitude?&lt;/p&gt;&#xA;" OwnerUserId="16441" LastActivityDate="2017-04-18T17:33:31.753" Title="Change of time in vernal equinox by altitude from the Sea level" Tags="&lt;the-sun&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20753" PostTypeId="2" ParentId="20751" CreationDate="2017-04-16T14:24:07.750" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In other way, does the time of vernal equinox change with altitude?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equinoxes are defined as the points in time at which the Earth's equatorial plane intersects the center of the Sun. From an Earth-centric perspective, the Sun appears to crosses the equatorial plane from south to north at the March equinox, and from north to south at the September equinox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Altitude, latitude, and longitude, and length of day do not come into play. In fact, days are slightly longer than are nights almost everywhere on the Earth on the dates of the equinoxes. This is a result of the Sun's angular diameter and atmospheric refraction, both of which mean we can see the Sun when the center of the Sun is a bit (0.83 degrees) below the horizon.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2017-04-18T17:33:31.753" LastActivityDate="2017-04-18T17:33:31.753" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20754" PostTypeId="1" CreationDate="2017-04-16T15:04:55.993" Score="0" ViewCount="482" Body="&lt;p&gt;Let universe be completely made from hydrogen. And also we have redshift $z= 6$. with Hubble constant $H_{0} = 2.1941747572815535\times 10^{-18}\:\mathrm{s}^{-1}$. We also know that density of the universe $\rho =\frac{3H^2}{8\pi G}$ ($G= 6.67\times 10^{-11}\:\mathrm{m^3\ kg^{-1}\ s^{-2}}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is the following approach right to find the number density of hydrogen in universe?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we calculate density $\rho =\frac{3H^2}{8\pi G}$, then we have&#xA;$1+z = \sqrt{\frac{1+v/c}{1-v/c}}$. From that we get $v = H\times R$ and from this we calculate $R$, the radius of observable universe. With that, we calculate volume of universe $V = (4\pi/3)R^3$ and we get Mass $M=\rho V$. Then we calculate number of Hydrogen atoms with $N = M/m_{\mathrm{hydrogen}}$. Now we have number density of hydrogen $\rho = N/V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this solution right? Or we must solve it another way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;The source of question is National Olympiad of Iran which is not available in English so the above is my translation for the question.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="16441" LastEditorUserId="11143" LastEditDate="2017-04-17T13:13:06.230" LastActivityDate="2018-06-25T18:34:12.503" Title="Approximating density of hydrogen in [observable] universe" Tags="&lt;cosmology&gt;&lt;redshift&gt;&lt;density&gt;&lt;hubble-constant&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20756" PostTypeId="1" AcceptedAnswerId="20758" CreationDate="2017-04-16T19:19:33.407" Score="0" ViewCount="89" Body="&lt;p&gt;I was looking at the Wikipedia article &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_potentially_habitable_exoplanets&quot; rel=&quot;nofollow noreferrer&quot;&gt;List of potentially habitable exoplanets&lt;/a&gt;&quot; and I noticed that many of the closest planets listed(tens or hundreds instead of thousands of light-years away) orbited M-class stars. I have also read that scientists are more certain about the possibility of habitable planets around G and K class stars, because the habitability of red dwarf systems is still debated. I was wondering whether there is something that makes the larger stars be farther away from each other, or whether it is just a coincidence. So is it possible to have many G and K class stars tens of light years near each other?&lt;/p&gt;&#xA;" OwnerUserId="16623" LastEditorUserId="2365" LastEditDate="2017-04-16T20:54:50.343" LastActivityDate="2017-04-16T22:03:51.440" Title="Spacing of G and K class stars" Tags="&lt;star&gt;&lt;distances&gt;&lt;star-formation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20758" PostTypeId="2" ParentId="20756" CreationDate="2017-04-16T21:52:18.967" Score="2" Body="&lt;p&gt;The solution to your mystery about why habitable planets are found around nearby low-mass stars, but distant G/K stars is all to do with observational selection effects and biases. There are many G/K stars within 100 light years of the Sun, but almost none have been examined in the detail required to reveal small, habitable planets around them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost all the small &quot;Earthlike&quot; planets in habitable zones have been found by the transit technique. It is much easier to find such planets around small stars for two reasons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transit technique yields a signal that depends on $(R_p/R_{*})^2$. Thus small planet transits are easier to see around small stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of a transit depends on the ratio of the star size to orbital radius. This sounds like it works against finding planets around small stars; and it does at a fixed orbital radius. But, because the habitable zone is much closer-in to a smaller, less luminous star, then the habitable planets around such stars are &lt;em&gt;more&lt;/em&gt; likely to transit. In addition, their orbital periods will be tens of days rather than the months to a year around G/K stars and this makes it much easier to observe repeated transits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the combination of these two selection effects makes it possible to find habitable zone planets around small stars all over the sky from &lt;em&gt;ground-based&lt;/em&gt; observatories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The downside is that small stars are low luminosity, so the planets that can be found are found around nearby examples at tens of light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, habitable zone planets around G/K stars are mostly found in months-year orbits by Kepler, which stared at a single small field for 4 years. The majority of the stars targeted by Kepler for exoplanet searches were 10th-15th magnitude G/K stars (there are few brighter examples in such a limited patch of sky), which are at distances of hundreds to thousands of light years.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-04-16T22:03:51.440" LastActivityDate="2017-04-16T22:03:51.440" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20760" PostTypeId="1" AcceptedAnswerId="20764" CreationDate="2017-04-17T05:07:30.807" Score="0" ViewCount="295" Body="&lt;p&gt;Say we have a closed universe that follows the cosmological principle. As a result, mass is distributed equally and, if as far as I know, gravity should have no net effect on the movements of galaxies (as in, there are no peculiar velocities).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the Big Crunch, the Universe stops expanding and gravity causes the scale factor to decrease. How is this possible? Shouldn't gravity have no effect on a homogenous and isotropic universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On that note, if gravity actually caused a collapse, would gravity cause the comoving coordinates of the galaxies to change?&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2019-03-28T20:41:11.310" Title="Can gravity cause a Big Crunch, if the cosmological principle holds?" Tags="&lt;gravity&gt;&lt;cosmology&gt;&lt;general-relativity&gt;&lt;fate-of-universe&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20761" PostTypeId="1" CreationDate="2017-04-17T06:12:10.837" Score="4" ViewCount="616" Body="&lt;p&gt;We know Eddington Limit (Eddington Luminosity) is calculated with this formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L_{Ed} \approx 3 * 10^4 * L_{sun} * \frac{M}{M_{sun}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: For a black hole with $10^4 M_{sun}$, we get&#xA;$$L_{Ed} = 1.16 * 10^{35} W$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is what does it really show? What do we get by calculating this Number? Is this formula really used to calculate the luminosity or it is used for calculating mass?&lt;/p&gt;&#xA;" OwnerUserId="16441" LastEditorUserId="7982" LastEditDate="2020-07-06T03:21:31.340" LastActivityDate="2020-07-06T03:21:31.340" Title="What does Eddington Limit (Eddington Luminosity) Show?" Tags="&lt;astrophysics&gt;&lt;luminosity&gt;&lt;radiation&gt;&lt;eddington-luminosity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20762" PostTypeId="2" ParentId="20761" CreationDate="2017-04-17T06:38:41.580" Score="5" Body="&lt;p&gt;The Eddington limit represents the maximum luminosity that can be achieved by a body (such as the star) when there is hydrostatic equilibrium (&lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/H/Hydrostatic+Equilibrium&quot; rel=&quot;noreferrer&quot;&gt;http://astronomy.swin.edu.au/cosmos/H/Hydrostatic+Equilibrium&lt;/a&gt;). For luminosity greater than Eddington limit, the radiative force of the luminosity on matter exceeds the gravitational force on the matter. If the luminosity radiated by an accretion disk exceeds the Eddington limit, the matter falling towards the supermassive black hole can be blown away. &lt;/p&gt;&#xA;" OwnerUserId="16628" LastActivityDate="2017-04-17T06:38:41.580" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20764" PostTypeId="2" ParentId="20760" CreationDate="2017-04-17T12:49:47.420" Score="3" Body="&lt;p&gt;You are thinking about gravity and how it affects the universe as a whole in the wrong way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dynamics of the universe are governed by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Friedmann equations&lt;/a&gt;, which can be solved to say how the scale factor of a homogeneous, isotropic universe behaves with time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is enough gravitating matter/energy density in the universe then the scale factor may eventually decrease to zero in a finite time. The simplest case is a closed universe with no cosmological constant and a mass density greater than the critical density (which is not the universe we currently inhabit: our universe appears to be (a) flat and (b) expanding at an accelerating rate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interpretation of this is that the space &lt;em&gt;between&lt;/em&gt; the galaxies is contracting. It is the reversal of what we currently observe in the expanding universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way that you do not interpret the universal expansion of a homogeneous, isotropic universe as some sort of repulsion between galaxies; neither would the contraction be interpreted as some sort of attractive force between galaxies. As you rightly point out, in a strictly Newtonian sense, there would be no net gravitational force on any one galaxy and so its co-moving coordinates would be unchanged during the contraction.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-04-17T12:49:47.420" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20765" PostTypeId="2" ParentId="20747" CreationDate="2017-04-17T13:00:33.850" Score="0" Body="&lt;p&gt;Quoting from Wikipedia,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The bolometric magnitude Mbol, takes into account electromagnetic&#xA;  radiation at all wavelengths. It includes those unobserved due to&#xA;  instrumental pass-band, the Earth's atmospheric absorption, and&#xA;  extinction by interstellar dust. It is defined based on the luminosity&#xA;  of the stars. In the case of stars with few observations, it must be&#xA;  computed assuming an effective temperature. Classically, the&#xA;  difference in bolometric magnitude is related to the luminosity ratio&#xA;  according to:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$ M_{bol,*} - M_{bol,sun} = -2.5log_{10}(\frac{L_*}{L_{sun}})$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In August 2015, the International Astronomical Union passed Resolution&#xA;  B2[7] defining the zero points of the absolute and apparent bolometric&#xA;  magnitude scales in SI units for power (watts) and irradiance (W/m2),&#xA;  respectively. Although bolometric magnitudes had been used by&#xA;  astronomers for many decades, there had been systematic differences in&#xA;  the absolute magnitude-luminosity scales presented in various&#xA;  astronomical references, and no international standardization. This&#xA;  led to systematic differences in bolometric corrections scales, which&#xA;  when combined with incorrect assumed absolute bolometric magnitudes&#xA;  for the Sun could lead to systematic errors in estimated stellar&#xA;  luminosities (and stellar properties calculated which rely on stellar&#xA;  luminosity, such as radii, ages, and so on).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[leading to the accepted definition of]&#xA;$ M_{bol} = -2.5log_{10}(L_*) + 71.1974... $ , where the constant term is the zero-point luminosity $L_0$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dunno if this helps, other than that you have to determine the spectral luminosity of the star in question. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-04-17T13:00:33.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20766" PostTypeId="1" AcceptedAnswerId="20770" CreationDate="2017-04-17T15:42:57.043" Score="-2" ViewCount="136" Body="&lt;p&gt;I'm not scientist at all so the answer might be obvious but I was pondering about the big bang. I started wondering how is it possible that the earth had to accelerate at some point in history to the speed it currently has but not continue accelerating past it? Isn't acceleration infinite out in space?&lt;/p&gt;&#xA;" OwnerUserId="16631" LastActivityDate="2017-04-17T22:06:19.057" Title="How does the earth not continue accelerating?" Tags="&lt;big-bang-theory&gt;" AnswerCount="2" CommentCount="5" ClosedDate="2017-04-20T10:34:29.517" ContentLicense="CC BY-SA 3.0" />
  <row Id="20767" PostTypeId="2" ParentId="20766" CreationDate="2017-04-17T16:19:12.530" Score="-2" Body="&lt;p&gt;If we assume that the big bang creates 'all', and the expansion is not of matter in a pre-existing space, but of matter in a co-existing space, created with it, then no acceleration exists at all...&lt;/p&gt;&#xA;" OwnerUserId="16238" LastEditorUserId="16238" LastEditDate="2017-04-17T20:30:41.763" LastActivityDate="2017-04-17T20:30:41.763" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20770" PostTypeId="2" ParentId="20766" CreationDate="2017-04-17T22:06:19.057" Score="1" Body="&lt;p&gt;The solar system and the Earth have formed billions of years ago from a pile of stuff (mostly hydrogen gas and rocks, with traces of other things) that collapsed together due to gravity pulling all those parts together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current orbital speeds and spin motions of planets are just residuals from that collapse. In other words, a pile of rocks and stuff collapsed and made the Earth, and the result happened to spin at whatever rate back then - and that's all there is to it. Other &quot;piles of stuff&quot; (planets) spin at different rates, because that's how they collapsed from the cloud of gas and rubble; there's no special reason for it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Isn't acceleration infinite out in space?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Acceleration only happens when there's an external force accelerating the object along its direction of motion. I.e. when something is actively pushing it.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2017-04-17T22:06:19.057" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20771" PostTypeId="1" AcceptedAnswerId="20772" CreationDate="2017-04-18T20:45:55.793" Score="8" ViewCount="638" Body="&lt;p&gt;A recent paper in Nature &lt;em&gt;&amp;quot;Planetary science: Reckless orbiting in the Solar System&amp;quot;&lt;/em&gt; (Morais &amp;amp; Namouni, 2017) presents the following series of four co-orbital states:&lt;/p&gt;&#xA;&lt;img src=&quot;https://media.springernature.com/full/springer-static/image/art%3A10.1038%2F543635a/MediaObjects/41586_2017_Article_BF543635a_Fig1_HTML.jpg?as=webp&quot;&gt;&#xA;&lt;p&gt;While I understand the development of the &lt;em&gt;quasi-satellite&lt;/em&gt; and &lt;em&gt;trisectrix&lt;/em&gt; states, I am unable to inuit the development of the &lt;em&gt;tadpole&lt;/em&gt; and &lt;em&gt;horseshoe&lt;/em&gt; co-orbital states.&lt;/p&gt;&#xA;&lt;p&gt;From explanation provided in the article:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The tadpole and horseshoe shapes arise because the planet's&#xA;gravitational attraction alters the body's orbital path — the body&#xA;goes through a cycle of catching up with the planet and falling&#xA;behind, seeming to change direction from the perspective of the&#xA;planet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;How, when approaching the planet, does the body &amp;quot;fall behind&amp;quot; instead of continuing to accelerate toward the planet?&lt;/p&gt;&#xA;&lt;p&gt;The only means that comes to mind is if the planet orbits faster than the body, but then I imagine it would&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;No longer be considered co-orbital (because it would have a different &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_period&quot; rel=&quot;nofollow noreferrer&quot;&gt;orbital period&lt;/a&gt;), and&lt;/li&gt;&#xA;&lt;li&gt;Would eventually be overtaken by the planet.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;How, then, do these first two co-orbital states work?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Citation:&lt;/strong&gt; &lt;a href=&quot;http://www.nature.com/nature/journal/v543/n7647/full/543635a.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Nature&lt;/em&gt; 543, 635–636 (30 March 2017)&#xA;&lt;code&gt;doi:10.1038/543635a&lt;/code&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14439" LastEditorUserId="14439" LastEditDate="2021-05-24T14:29:08.023" LastActivityDate="2021-05-24T14:29:08.023" Title="How do tadpole, horseshoe co-orbital states arise and how are they stable?" Tags="&lt;orbital-mechanics&gt;&lt;orbital-resonance&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="20772" PostTypeId="2" ParentId="20771" CreationDate="2017-04-18T21:27:10.543" Score="9" Body="&lt;p&gt;If the body is in front of the planet (relative to the planet's orbital motion) and a little further from the sun, it will orbit the sun slightly slower than the planet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it is slower, the planet will slowly catch it up. (It takes many &quot;years&quot; for the planet to get close to the body.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the planet catches up with the body, the gravitational effect of the planet will tend to pull back on the body, this causes the body to lose energy and begin to fall slighly towards the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slight fall towards the sun cause the body to move to a lower orbit (relative to the sun) and so speed up. It appears paradoxical, but it the fact that the gravitational drag of the planet on the body actually causes it to speed up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The body speeds up and so begins to move away from the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the body is makes an angle of more than 60 degrees from the planet, the gravitational pull of the planet now has the opposite effect. It combines with the pull of the sun to cause the body to accelerate, and (apparently paradoxically) this acceleration causes the body to move to a more distant orbit, and slow down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two things key to understanding this. First remember that the diagrams you show are drawn with a fixed planet. I.e. the &quot;camera&quot; is rotating at the same rate as the planet takes to orbit the sun, so the planet appears fixed. If you don't rotates the camera, you would see that the body actually has a nearly elliptical orbit, relative to the sun. Secondly you need to grasp the idea that, in orbit, drag doesn't slow you down. Drag causes you to fall towards the sun and speed up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did bodies get into such orbits? Basically by luck. They happened to have orbits that were close to Earth's orbit, and by a lucky effect of the gravity of the other planets they were dislodged from their regular orbit of the Sun to a tadpole orbit. Once there, they can remain in their orbit for tens fo thousands of years. Even so, only one object is known to exist in a tadpole orbit relative to Earth.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-04-18T21:27:10.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20773" PostTypeId="1" AcceptedAnswerId="20776" CreationDate="2017-04-18T21:57:16.913" Score="1" ViewCount="106" Body="&lt;p&gt;I'm new to all this and trying to implement galaxy counting. For the initial stage &lt;strong&gt;I need raw image data, that is images with galaxies and stars along with a count of how many galaxies there are in the image.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came across this which seems to have a way to label galaxies &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://skyserver.sdss.org/dr13/en/tools/chart/navi.aspx?ra=179.68929&amp;amp;dec=-0.45438&amp;amp;opt=&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://skyserver.sdss.org/dr13/en/tools/chart/navi.aspx?ra=179.68929&amp;amp;dec=-0.45438&amp;amp;opt=&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;but I cannot figure it out at all&lt;/strong&gt;, I used this page &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://skyserver.sdss.org/dr13/en/tools/chart/chartinfo.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://skyserver.sdss.org/dr13/en/tools/chart/chartinfo.aspx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to try to get it to label only galaxies are seen but I got nothing (put G in opt section, did not make a difference).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know how to operate the above or &lt;strong&gt;where I could get images with lots of galaxies (any number would do) in them along with a count of how many galaxies there are in the image?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found &lt;a href=&quot;https://data.galaxyzoo.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://data.galaxyzoo.org/&lt;/a&gt; but traversing it seems impossible&lt;/p&gt;&#xA;" OwnerUserId="16435" LastEditorUserId="16435" LastEditDate="2017-04-19T21:33:28.337" LastActivityDate="2017-04-19T21:33:28.337" Title="I want a galaxy count database, also how do I use SDSS skyserver to fetch galaxies?" Tags="&lt;galaxy&gt;&lt;data-analysis&gt;&lt;photography&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20774" PostTypeId="2" ParentId="20771" CreationDate="2017-04-19T00:50:08.283" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How, when approaching the planet, does the body &quot;fall behind&quot; instead&#xA;  of continuing to accelerate toward the planet?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is fundamentally the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravity_assist&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravity assist&lt;/a&gt; problem.   In the 2 body system, the smaller object falls towards the Earth, accelerates, misses, then flies away from the earth giving back the velocity in flying away that it added flying towards.   The net velocity change relative to the Earth is close to zero during the pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 3 body system, relative to the sun, that all changes.  The asteroid effectively adds or removes velocity relative to the sun.   If it slows down, it falls closer to the sun, if it speeds up, it moves away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horseshoe orbit is a relatively perfect dance where it does this back and forth, crossing the Earth's orbital path for dozens or hundreds, maybe even thousands or oribts, looking like a horseshoe from Earth's frame of reference.   From the sun's frame of reference, the asteroid can't make up it's mind if it wants to be closer than the Earth or further than the Earth.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-04-19T00:50:08.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20775" PostTypeId="1" AcceptedAnswerId="20780" CreationDate="2017-04-19T03:43:04.413" Score="0" ViewCount="231" Body="&lt;p&gt;Say I have an arbitary image of stars and galaxies without any extra information (no other spectra info other than image itself). Can you help me distinguish them? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any specific features / properties I could look for? Say for eg the R G or B value or maybe values in some other color space (YUV etc.) be more or less for stars than galaxies, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I dont have any &quot;special&quot; data (no sensor or spectral data), I only have data from the internet (common images). The point is to detect stars and galaxies from images commonly found on the internet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of them is this: &lt;a href=&quot;https://cdn.spacetelescope.org/archives/images/large/heic0916a.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://cdn.spacetelescope.org/archives/images/large/heic0916a.jpg&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16435" LastEditorUserId="16435" LastEditDate="2017-04-19T12:45:57.053" LastActivityDate="2017-04-19T15:48:55.383" Title="Image processing techniques to distinguish between stars and galaxies in an image?" Tags="&lt;star&gt;&lt;galaxy&gt;&lt;data-analysis&gt;&lt;photography&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2017-04-23T21:39:44.670" ContentLicense="CC BY-SA 3.0" />
  <row Id="20776" PostTypeId="2" ParentId="20773" CreationDate="2017-04-19T04:56:18.473" Score="0" Body="&lt;p&gt;You might find the &lt;a href=&quot;https://ned.ipac.caltech.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;NED&lt;/a&gt; database and &lt;a href=&quot;http://astrometry.net&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astrometry.net&lt;/a&gt; useful. NED is NASA's extragalactic database. You can search by a position in the sky with a radius and it will give you a list of known objects in that field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astrometry.net lets you upload images and will resolve what you're looking at, also giving you a list of known objects in your image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like to use &lt;a href=&quot;http://cds.u-strasbg.fr/&quot; rel=&quot;nofollow noreferrer&quot;&gt;CDS&lt;/a&gt; for a quick look up on objects. This will let you browse images from different surveys easily and links you to other sources on that object as well as publications about your object. &lt;/p&gt;&#xA;" OwnerUserId="13724" LastActivityDate="2017-04-19T04:56:18.473" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="20777" PostTypeId="1" AcceptedAnswerId="20779" CreationDate="2017-04-19T08:19:20.380" Score="6" ViewCount="670" Body="&lt;p&gt;Recently I've explored an area on Mars' surface which one looks like huge hatch cap &amp;amp; hole, about 10 kilometers, between tons of circular craters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formation is very unique and strange to me. Somehow it has moved from original location and broke into two pieces then half part disappeared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you think about it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.com/maps/space/mars/@25.3717187,166.6337854,206134m/data=!3m1!1e3&quot; rel=&quot;noreferrer&quot;&gt;Google Maps link&lt;/a&gt;(25.3717187,166.6337854)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SB09V.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SB09V.png&quot; alt=&quot;strange formation on Mars&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16643" LastEditorUserId="10172" LastEditDate="2017-04-19T13:29:00.317" LastActivityDate="2017-04-19T17:56:41.070" Title="Strange square-like formation on Mars" Tags="&lt;mars&gt;&lt;surface&gt;&lt;crater&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20779" PostTypeId="2" ParentId="20777" CreationDate="2017-04-19T13:56:11.300" Score="12" Body="&lt;p&gt;I looked at the image and annotated it based on what it most likely is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xJ3nf.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xJ3nf.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In black is the formation itself.&lt;/li&gt;&#xA;&lt;li&gt;In blue is a depression underneath it.&lt;/li&gt;&#xA;&lt;li&gt;In grey is part of the boundary of what you believed to be the disappeared half.&lt;/li&gt;&#xA;&lt;li&gt;In green is a ridge to the north.&lt;/li&gt;&#xA;&lt;li&gt;In red are several craters.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Here's a more likely hypothesis for what this actually is.&lt;/p&gt;&#xA;&lt;h1&gt;The depression&lt;/h1&gt;&#xA;&lt;p&gt;The blue oval contains a slight depression. It could be a number of things, including an old crater, a basin, or the result of the collapse of part of the surface. I'm actually a little bothered by the idea of it as an impact crater. We can see from the shadows on the western walls that it is only a little shallow, especially compared to other similarly-sized impact craters in the area. It's possible that it's a crater that was filled in with sediment.&lt;/p&gt;&#xA;&lt;p&gt;If it's a natural basin, it could once have been a &lt;a href=&quot;https://en.wikipedia.org/wiki/Lakes_on_Mars&quot; rel=&quot;noreferrer&quot;&gt;Martian lake&lt;/a&gt; - although this is quite speculative. It could also be the remains of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Pit_crater&quot; rel=&quot;noreferrer&quot;&gt;pit crater&lt;/a&gt; (there are many &lt;a href=&quot;https://en.wikipedia.org/wiki/Elysium_quadrangle#Fossae.2Fpit_craters&quot; rel=&quot;noreferrer&quot;&gt;pit craters in the area&lt;/a&gt;), formed possibly from the cave-in of a series of &lt;a href=&quot;https://en.wikipedia.org/wiki/Martian_lava_tube&quot; rel=&quot;noreferrer&quot;&gt;lava tubes&lt;/a&gt;. Again, this is also speculative, and the depression is quite large, about 10 miles long on its major axis.&lt;/p&gt;&#xA;&lt;p&gt;If one of these options is correct, the depression is not very mysterious. You've identified the grey part of it as the location of the other half of the formation, partly because it appears to be roughly the same size. I find this unlikely, for a couple reasons:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The ridges look similar in height to the edge of the depression on the other side of the formation, indicating that it is actually a part of that feature.&lt;/li&gt;&#xA;&lt;li&gt;Half of an outcropping does not simply vanish by itself. I know of no mechanism for this to happen. Erosion could take its toll over time, and perhaps that caused weathering of the formation, but it couldn't move it.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;I think it would be clearer that this is actually part of the depression if the ridge enclosed in green were not there.&lt;/p&gt;&#xA;&lt;h1&gt;The mesa&lt;/h1&gt;&#xA;&lt;p&gt;So if we assume that the outcropping is all that there even was of the object, we simply have to explain why it's so square. This really isn't that mysterious. There are plenty of other mesas in this cartographical area of Mars, designated the &lt;a href=&quot;https://en.wikipedia.org/wiki/Elysium_quadrangle#Mesas&quot; rel=&quot;noreferrer&quot;&gt;Elysium quadrangle&lt;/a&gt; by the USGS for mapping purposes. Here are &lt;a href=&quot;https://en.wikipedia.org/wiki/File:45016_2080mesas.jpg&quot; rel=&quot;noreferrer&quot;&gt;some&lt;/a&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Cerberus_Palus.JPG&quot; rel=&quot;noreferrer&quot;&gt;examples&lt;/a&gt;, which are similarly shaped:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/4/46/45016_2080mesas.jpg&quot; alt=&quot;&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/c/c4/Cerberus_Palus.JPG&quot; alt=&quot;&quot; /&gt;&lt;/p&gt;&#xA;&lt;h1&gt;The craters&lt;/h1&gt;&#xA;&lt;p&gt;Finally, we have the three large craters. They aren't too mysterious; craters are normal on Mars, and there are plenty, of varying sizes, in the Elysium quadrangle. The one reason the smallest stands out is because it is near the center of the northeast edge of the mesa. I'm reasonably certain this is just a coincidence.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Here's my hypothetical and likely not entirely correct timeline for this formation:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The large basin forms, from either a crater or some other process.&lt;/li&gt;&#xA;&lt;li&gt;The mesa and ridge form, possibly as a result of the erosion of some larger feature.&lt;/li&gt;&#xA;&lt;li&gt;Craters form through impacts. One happens to be on the mesa itself.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;I think your identification is simply a case of &lt;a href=&quot;https://en.wikipedia.org/wiki/Pareidolia&quot; rel=&quot;noreferrer&quot;&gt;pareidolia&lt;/a&gt;, which is the human tendency to find shapes and patterns where none really exist. A well-known Martian example is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cydonia_(region_of_Mars)#.22Face_on_Mars.22&quot; rel=&quot;noreferrer&quot;&gt;Face of Mars&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-04-19T17:56:41.070" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20780" PostTypeId="2" ParentId="20775" CreationDate="2017-04-19T15:18:31.707" Score="3" Body="&lt;p&gt;If you have the CCD data and instrument profile, then a star may be treated as a &quot;point source&quot; and a galaxy as an &quot;extended source&quot;. This means that, under ideal conditions a star will have it's light spread through the &quot;point spread function&quot; of the telescope, usually these objects have a Gaussian(ish) intensity profile with one pixel of the CCD claiming most of the light and adjacent pixels falling off in intensity, while a galaxy will have an extended, flattened profile with 2 or 3 or 20 pixels claiming most of the light and then falling off in adjacent pixels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have the full CCD information, this is the method of choice. This breaks down when galaxies are so far away that they are resolved smaller than the &quot;point source&quot; stars (larger telescopes can resolve farther away galaxies as extended than smaller, back yard telescopes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just have catalog data, such as colors, then you want to look for items falling off of the &quot;stellar color locus.&quot; Stars, in color space tend to sit on a &quot;ribbon&quot; in color-color space (see this from &lt;a href=&quot;http://adsabs.harvard.edu/abs/2000ApJ...540..825Y&quot; rel=&quot;nofollow noreferrer&quot;&gt;Yanny+2000&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tCQ71.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tCQ71.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objects with colors which do not lie on this &quot;ribbon&quot; are special objects. For example the box labeled &quot;Q&quot; is Quasars, which is a type of galaxy. Not shown in this image is a &quot;cloud&quot; of objects which are galaxies (in this image, the cloud will start around that quasar box and extend much redder in both color axes). If you take all of your objects from a telescope and plot them in any(ish) color-color space you will find a tight band, which are stars, and a cloud near this band (which are galaxies). This is because stars operate as a blackbody, while galaxies operate as a dual blackbody (with a lot of light emitted by the stars, following the same blackbody radiation profile, and a lot of gas absorbing this light and re-emitting it in cooler, infrared blackbody colors).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method is &lt;em&gt;not&lt;/em&gt; perfect, this is a statistical method for selecting say 90% galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Rob is absolutely correct that you cannot do any of these with jpeg images. A lot of jpegs, especially promotional images have had their colors doctored, or even include false colors (for example to represent ultraviolet or infrared light; this is also why you can't do this with random jpegs, each camera will have it's own way of representing color, so my polaroid camera picture will give different colors than a modern random DLSR camera... you can only do this if all the images are collected from identical cameras). jpegs also do not include the &quot;full CCD information&quot; ie. the intensity of light falling onto each pixel of the CCD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 2: Yes you can estimate the PSF for a telescope with an image, but only a CCD image which shows the photon count in each pixel. A CCD image is a 3d image, which shows how many photons fall into each bin, a jpeg does not have this information, so it is... impossible I think, unless the jpeg is produced to show the photon count and not the color.&lt;/p&gt;&#xA;" OwnerUserId="16592" LastEditorUserId="16592" LastEditDate="2017-04-19T15:48:55.383" LastActivityDate="2017-04-19T15:48:55.383" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20783" PostTypeId="1" AcceptedAnswerId="20788" CreationDate="2017-04-19T17:18:47.717" Score="1" ViewCount="100" Body="&lt;p&gt;What is the smallest and largest sun an Earth like planet could have in its sky? What kinds of stars are those?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw &lt;a href=&quot;https://www.youtube.com/watch?v=mUYrCul8A9M&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video&lt;/a&gt; and thought, yeah, some of those stars are really big in absolute terms, but Earth would have to be orbiting them proportionally further to keep the climate, making them smaller in the sky, so the video is not accurate.&lt;/p&gt;&#xA;" OwnerUserId="16646" LastActivityDate="2017-04-20T02:30:01.300" Title="What apparent sizes of the sun an Earth like planet can have?" Tags="&lt;earth-like-planet&gt;&lt;apparent-magnitude&gt;&lt;habitable-zone&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20785" PostTypeId="1" AcceptedAnswerId="20787" CreationDate="2017-04-19T20:37:34.690" Score="8" ViewCount="341" Body="&lt;p&gt;I've read a few articles written in $2008$ that some stars which have enough mass just collapse into black holes without a supernova, is this proven?&lt;/p&gt;&#xA;" OwnerUserId="16649" LastActivityDate="2017-04-20T13:17:38.453" Title="Do all massive stars explode?" Tags="&lt;star&gt;&lt;black-hole&gt;&lt;supernova&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20787" PostTypeId="2" ParentId="20785" CreationDate="2017-04-20T00:14:26.173" Score="10" Body="&lt;p&gt;There are a variety of models of stars collapsing into black holes without proper supernova explosions. These are often called &lt;a href=&quot;https://en.wikipedia.org/wiki/Failed_supernova&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;failed supernovae&lt;/em&gt;&lt;/a&gt; or &lt;em&gt;direct collapse black holes&lt;/em&gt; - although the former seems more common. Failed supernovae can happen if the initial shock wave rebounding from a collapse loses enough energy, causing it to fizzle out. Neutrinos produced in the core often supply this energy, but a small enough burst may not be enough to keep the shock going. There might then be a simple collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimum mass for the progenitor of a failed supernova is unknown (and as Rob Jeffries pointed out, this is metallicity-dependent). Some early models (e.g. &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/307647/fulltext/&quot; rel=&quot;noreferrer&quot;&gt;Fryer (1999)&lt;/a&gt;) found that stars above $40M_{\odot}$ could produced failed supernovae, while newer ones suggest that this could be as low as $25M_{\odot}$. This lower limit would include a sizable portion of red supergiants - which could be a possible solution to &lt;a href=&quot;https://astronomy.stackexchange.com/q/16148/2153&quot;&gt;the red supergiant problem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A number of searches, largely targeting supergiants, have been carried out. Just a few are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://arxiv.org/abs/1609.01283&quot; rel=&quot;noreferrer&quot;&gt;Adams et al. (2016)&lt;/a&gt;, using the Hubble Space Telescope to examine &lt;a href=&quot;https://en.wikipedia.org/wiki/N6946-BH1&quot; rel=&quot;noreferrer&quot;&gt;N6946-BH1&lt;/a&gt;, a disappeared red supergiant of $25M_{\odot}$.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://arxiv.org/abs/1507.05823&quot; rel=&quot;noreferrer&quot;&gt;Reynolds et al. (2015)&lt;/a&gt;, looking at past Hubble data. They found one possible candidate, a $25\text{-}30M_{\odot}$ red supergiant in &lt;a href=&quot;https://en.wikipedia.org/wiki/NGC_3021&quot; rel=&quot;noreferrer&quot;&gt;NGC 3021&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So far, we have no definite confirmation that these stars were failed supernovae. However, they do seem plausible candidates, and there are many more possibilities out there. &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/590053/meta&quot; rel=&quot;noreferrer&quot;&gt;One group proposed monitoring one million supergiants&lt;/a&gt; in an attempt to catch failed supernovae. Perhaps they'll be successful.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2017-04-20T13:17:38.453" LastActivityDate="2017-04-20T13:17:38.453" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20788" PostTypeId="2" ParentId="20783" CreationDate="2017-04-20T02:12:25.253" Score="2" Body="&lt;p&gt;I'll make a very simple dimensional calculation here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total power from of a star with radius $R$ and temperature $T$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P_{tot} \ = \ R^2 \ T^4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flux $I$ (power per unit area) at an orbital distance $r$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I \ = \ \frac{R^2 \ T^4}{r^2} \ = \ T^4 \ \left( \frac{R}{r} \right)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparent size (radians, small angle approx):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta \ \approx \frac{R}{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta \ \approx \ \frac{\sqrt{I}}{T^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surprisingly, for a fixed incident flux of light (ignoring different planetary warming effects of different wavelengths) the apparent size of a sun viewed from a star varies roughly as the inverse square of its temperature, and does not depend on the actual size of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your milage may vary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &lt;a href=&quot;https://en.wikipedia.org/wiki/Wien%27s_displacement_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wein's dispacement law&lt;/a&gt; which relates the wavelength of maximum spectral radiance of a blackbody $\lambda_{max}$ with its temperature $T$ using Wien's displacement constant $b$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda_{max} \ = \ \frac{b}{T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The apparent size scales with the square of the peak wavelength of a star's blackbody spectrum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta \ \approx \ \lambda_{max}^2 \ \sqrt{I}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yay! Reddish stars will appear much larger than bluish stars - the result we were all expecting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if you'd like do some landscaping, the 2008 Scientific American article &lt;a href=&quot;https://www.ebscohost.com/uploads/imported/thisTopic-dbTopic-1033.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Color of Plants on Other Worlds&lt;/a&gt; is a good read in stellar spectra, atmospheres, and photosynthesis.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-04-20T02:30:01.300" LastActivityDate="2017-04-20T02:30:01.300" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20790" PostTypeId="1" AcceptedAnswerId="20791" CreationDate="2017-04-20T05:14:32.593" Score="4" ViewCount="427" Body="&lt;p&gt;After origin of Earth ( &lt;a href=&quot;https://en.wikipedia.org/wiki/History_of_Earth&quot; rel=&quot;nofollow noreferrer&quot;&gt;4.54 billion years ago&lt;/a&gt; ) to today; what are the changes  through-which the sun undergone with time? , ( if seen from Earth. )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm doing some palaeobiology-landscape rendering; so I'm thinking if I want to add the sun in such pictures; &lt;strong&gt;should I draw it just like present day&lt;/strong&gt;? Or &lt;em&gt;the look/ appearance&lt;/em&gt; (colour, size, flames, position, halo etc) &lt;strong&gt;&lt;em&gt;varied in observable amount&lt;/strong&gt; through the geologic time-scale&lt;/em&gt;(precambrian-cambrian-ordovician-silurian-devonian-carboniferous-permian-triassic-jurassic-cretaceous-palaeogene-neogene etc). (the timescale terms provided just to hint the answerer the level of details may NOT to go )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; Here I attach a  small version of &lt;a href=&quot;https://en.wikipedia.org/wiki/Geologic_time_scale&quot; rel=&quot;nofollow noreferrer&quot;&gt;geologic time table&lt;/a&gt; to indicate time ranges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Vp3fn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vp3fn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;from &lt;a href=&quot;http://paleo.cortland.edu/tutorial/Timescale/timescale.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see the entire chart; &lt;a href=&quot;http://www.stratigraphy.org/index.php/ics-chart-timescale&quot; rel=&quot;nofollow noreferrer&quot;&gt;see international commission of stratigraphy&lt;/a&gt; charts&lt;/p&gt;&#xA;" OwnerUserId="16657" LastEditorUserId="16657" LastEditDate="2017-04-20T07:41:30.810" LastActivityDate="2017-04-21T07:42:40.823" Title="Appearance of sun at geologic past" Tags="&lt;the-sun&gt;&lt;stellar-evolution&gt;&lt;geology&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20791" PostTypeId="2" ParentId="20790" CreationDate="2017-04-20T09:39:10.617" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm doing some palaeobiology-landscape rendering; so I'm thinking if I want to add the sun in such pictures; should I draw it just like present day ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although the Sun was less luminous 4.5 billion years ago (about 70% of it's current luminosity, it is thought), and while this would affect it's color, making it perhaps a little more orange than yellow, the effect is small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun is still going to be a blazing near-white object in the sky of roughly the same size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although there was probably a color shift to a slightly more orange tint, the &lt;a href=&quot;http://www.ux1.eiu.edu/~cfjps/1400/atmos_origin.html&quot; rel=&quot;noreferrer&quot;&gt;composition of the Earth's atmosphere&lt;/a&gt; was radically different (see for example this page on &lt;a href=&quot;https://en.wikipedia.org/wiki/Paleoclimatology&quot; rel=&quot;noreferrer&quot;&gt;Paleoclimatology&lt;/a&gt;) and this would have had a much more significant effect on the appearance of the Sun (and the lighting of the landscape) than any minor change in the Sun's appearance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some issues relating to the development of life and this change in the Sun's luminosity (like the &lt;a href=&quot;https://en.wikipedia.org/wiki/Faint_young_Sun_paradox&quot; rel=&quot;noreferrer&quot;&gt;Faint Young Sun Problem&lt;/a&gt;) but these won't make much difference to your rendering, I think, compared to the atmospheric and climate issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the climate changes are not fully understood and it's rather speculative to try and make specific claims about what the overall effect would have been on the illumination levels and color of light reaching the surface in these periods.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-04-20T09:39:10.617" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20792" PostTypeId="2" ParentId="20790" CreationDate="2017-04-20T10:23:09.323" Score="5" Body="&lt;p&gt;You can divide this into two astronomical epochs, which are most easily understood in terms of the Hertzsprung-Russell diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Pre-main-sequence_star&quot; rel=&quot;nofollow noreferrer&quot;&gt;pre-main-sequence&lt;/a&gt; (PMS) phase, where the protosun contracts towards its main sequence, hydrogen burning, configuration. During this phase the (proto-)Sun declines in luminosity at roughly constant temperature for several million years, then heats up, with a slightly rising luminosity, before settling onto the main sequence with a slightly lower luminosity (see diagram below) after about 20 million years. The diagram gives you a rough idea of the &quot;colour&quot; that the protosun would have during this phase. The Sun is of course changing in brightness accordingly, but the radius of the Sun also decreases throughout this phase, so would cover a progressively smaller angle on the sky as seen from the Earth.&#xA;$$ R = \left(\frac{L}{4\pi \sigma T^4}\right)^{1/2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WHIwQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WHIwQ.jpg&quot; alt=&quot;PMS track&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A plot showing the paths (solid lines) of stars of different masses from birth (the Sun is still $\simeq 1M_{\odot}$ at birth) in luminosity and temperature, until they begin the H-burning main sequence. The dashed lines are &quot;isochrones&quot; at the indicated ages. The Sun &quot;begins&quot; its life (or at least emerges from its natal gas and dust less than a million years after birth) at the top of the $1M_{\odot}$ track.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second phase is of course much longer, but the changes are far less dramatic. For the next 4.45 billion years the Sun just gets a little brighter and a little larger, but stays more-or-less at the same temperature. Over that time the Sun has got about 30% more luminous and 15% larger at a roughly constant rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need more detail than that, then you need to access a proper stellar evolutionary model. There are lots of those available. e.g. the &lt;a href=&quot;https://obswww.unige.ch/Recherche/evol/Geneva-grids-of-stellar-evolution&quot; rel=&quot;nofollow noreferrer&quot;&gt;Geneva models&lt;/a&gt; (the Schaller et al. 1992 set are appropriate for the Sun).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-04-21T07:42:40.823" LastActivityDate="2017-04-21T07:42:40.823" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="20793" PostTypeId="1" CreationDate="2017-04-20T23:02:31.780" Score="2" ViewCount="51" Body="&lt;p&gt;This &lt;a href=&quot;https://earthscience.stackexchange.com/questions/10158/which-physical-parameters-can-we-actually-measure-to-confirm-the-existence-of-th&quot;&gt;question&lt;/a&gt; got me thinking about this.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that we measure the Earth's surface temperature by satellite (perhaps somewhat inaccurately, but it's done all the same).   Using Venus as an example, can we see see evidence of Venus' 800 degree surface from space or is Venus' temperature deduced more by modeling and the Russian ship that landed there.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All things with heat give off a thermal signature and that signature isn't a single wavelength but a range of wavelength.   I assume, by measuring the overall color emitted and/or the peak wavelength, it's a fairly straight forward calculation to work out the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when you have an atmosphere, and this is the main part of of my question.   I assume, an atmosphere is to some degree transparent, so you'd be seeing in a sense, heat through a medium and inside as well as on the upper surface.  But the heat you see diminishes depending on how opaque/transparent that atmosphere is to the specific wavelength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any good ballpark estimates to how thick (or how dense or pressurized), we can see temperature signatures through an atmosphere?     In the case of Venus, does any of it's 800 degree surface temperature generated IR make it through it's atmosphere, where it would be observable by spacecraft, giving a good measurement of it's surface temperature, or was it's surface temperature deduced by different types of models and that mistreated Soviet spacecraft that didn't last long once it had landed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine some atmospheric gases are more transparent than others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to clarify - a specific &quot;yes/no we can detect IR from Venus' surface&quot; and a more general answer as to how far we can see heat through an atmosphere, are both fine.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="907" LastEditDate="2017-04-21T18:32:38.507" LastActivityDate="2017-04-23T06:52:45.730" Title="Can IR sensitive cameras read signals from Venus's surface emissions?" Tags="&lt;planetary-atmosphere&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20794" PostTypeId="1" AcceptedAnswerId="20796" CreationDate="2017-04-20T23:04:42.217" Score="2" ViewCount="170" Body="&lt;p&gt;Could dark energy (the accelerating expansion of the universe) be the result of gravity from the dense past universe &quot;pulling&quot; outward from all directions? The universe theoretically started out extremely dense and hot, and we can see the light reaching us as the cosmic microwave background. I wonder if that same region and denser portions beyond are exerting a gravitational effect that is stretching out everything within toward it, effectively being a cosmic gravity background. If so, is this the explanation for dark energy?&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="2153" LastEditDate="2017-04-21T13:45:07.390" LastActivityDate="2017-04-21T15:47:52.823" Title="Could Dark Energy be a &quot;Cosmic Gravity Background&quot;" Tags="&lt;gravity&gt;&lt;expansion&gt;&lt;cosmological-inflation&gt;&lt;dark-energy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20795" PostTypeId="1" AcceptedAnswerId="20803" CreationDate="2017-04-20T23:23:28.367" Score="9" ViewCount="1572" Body="&lt;p&gt;Enceladus, one of Saturn's moons, has a geometric albedo of 1.38 and a bond albedo of 0.81. How can the geometric albedo of Enceladus be greater than one? What are the similarities or differences between a geometric albedo and a bond albedo?&lt;/p&gt;&#xA;" OwnerUserId="16660" LastActivityDate="2017-10-10T16:07:34.433" Title="Why is Enceladus's albedo greater than 1?" Tags="&lt;saturn&gt;&lt;albedo&gt;&lt;enceladus&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20796" PostTypeId="2" ParentId="20794" CreationDate="2017-04-21T00:17:44.900" Score="5" Body="&lt;p&gt;There is in fact a cosmic &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_wave_background&quot; rel=&quot;noreferrer&quot;&gt;gravitational wave background&lt;/a&gt;. These waves are expected to be &lt;a href=&quot;http://www.ligo.org/science/GW-Stochastic.php&quot; rel=&quot;noreferrer&quot;&gt;stochastic&lt;/a&gt;, having originated in the early universe (much earlier than the cosmic microwave background). Random fluctuations were subsequently stretched during &lt;a href=&quot;https://en.wikipedia.org/wiki/Inflation_(cosmology)&quot; rel=&quot;noreferrer&quot;&gt;inflation&lt;/a&gt;, making them observable over many wavelengths. A good and reasonably up-to-date introduction I read is &lt;a href=&quot;https://journals.aps.org/prx/abstract/10.1103/PhysRevX.6.011035&quot; rel=&quot;noreferrer&quot;&gt;Lasky et al. (2016)&lt;/a&gt;, which gives some of the most recent constraints on the parameters of the frequency spectrum $\Omega_{\text{GW}}(f)$. The great thing about the wide spectrum is that a number of different tools (&lt;a href=&quot;https://en.wikipedia.org/wiki/Pulsar_timing_array&quot; rel=&quot;noreferrer&quot;&gt;pulsar timing arrays&lt;/a&gt;, interferometers like LIGO, etc.) can be used to constrain different parts of it, both directly and indirectly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, dark energy is almost certainly not a result of the gravitational wave background. First, there's no theoretical mechanism to explain how such an effect would arise. Second, the background is probably not strong enough to have an influence on the current expansion of the universe (see the graph &lt;a href=&quot;http://www.tapir.caltech.edu/~teviet/Waves/gwave_spectrum.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; to see how weak it is). Finally, dark energy is a relatively recent phenomenon, arising several billion years ago. The background was around long before then, and yet had no effects.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2017-04-21T00:17:44.900" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20797" PostTypeId="2" ParentId="12267" CreationDate="2017-04-21T00:59:45.533" Score="5" Body="&lt;p&gt;Short answer: &lt;span class=&quot;math-container&quot;&gt;$t \lesssim 10^5\ \mathrm{years}$&lt;/span&gt; &lt;em&gt;(maybe)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An &quot;overcontact binary&quot; is just another way of saying &quot;common envelope binary&quot;. The two phrases are exactly the same and it's frustrating that the authors on the VFTS 352 paper decided to create their own convention - as if astrophysical classifications weren't confusing enough!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A contact binary exists on timescales predominantly dependent on stellar evolution, so figuring out how long a contact binary will exist is heavily dependent on the mass, metallicity, and rotation of the primary star among other things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Deriving the timescale:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's keep the scope to systems like VFTS 352, where the primary is massive and the binary has an orbital period less than 4 years (2.5 AU separation). In order to have a common envelope event, the stars must have overflown their Roche lobes. &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1983ApJ...268..368E&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;The radius for the Roche lobe of two point masses is&lt;/a&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{equation*}&#xA;r_L = \frac{0.49 q^{\frac{2}{3}}}{0.6q^{\frac{2}{3}}+\mathrm{ln}(1+q^{\frac{1}{3}})}a&#xA;\end{equation*}&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is the separation. For close binaries, the general observed trend is a high mass ratio &lt;span class=&quot;math-container&quot;&gt;$q=M_2/M_1$&lt;/span&gt;. So, if we assume &lt;span class=&quot;math-container&quot;&gt;$q=1$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$r_L = 0.38a$&lt;/span&gt;. Hence, for a binary with &lt;span class=&quot;math-container&quot;&gt;$a&amp;lt;2.5$&lt;/span&gt; AU,&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align*}&#xA;r_L &amp;amp;\lesssim 1\ \mathrm{AU}\\&#xA;r_L &amp;amp;\lesssim 215\ R_{\odot}&#xA;\end{align*}&lt;/span&gt;&#xA;since &lt;span class=&quot;math-container&quot;&gt;$q=1$&lt;/span&gt; is an upper bound on the Roche lobe radius. Now, performing some trivial rearrangement of the blackbody luminosity equation &lt;span class=&quot;math-container&quot;&gt;$L=4\pi\sigma_{SB}R^2T^4$&lt;/span&gt;, we find that&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{equation*}&#xA;R \approx 3.31\times10^{7} \bigg(\frac{L}{L_{\odot}}\bigg)^{\frac{1}{2}}\bigg(\frac{1\ \mathrm{K}}{T}\bigg)^2 \ R_{\odot}.&#xA;\end{equation*}&lt;/span&gt;&#xA;Massive stars typically have roughly constant luminosity, so we will choose &lt;span class=&quot;math-container&quot;&gt;$L\approx10^5\ L_{\odot}$&lt;/span&gt;. Hence,&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{equation*}&#xA;R\approx 1\times10^{10}\bigg(\frac{1\ \mathrm{K}}{T}\bigg)^2 \ R_{\odot}&#xA;\end{equation*}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The massive star needs to evolve until its radius is equal to that of the Roche lobe radius, so we find that the star reaches the common envelope phase for&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{equation*}&#xA;T \gtrsim 7000\ \mathrm{K}&#xA;\end{equation*}&lt;/span&gt;&#xA;Taking a peek at an HR diagram, this star varies from about &lt;span class=&quot;math-container&quot;&gt;$30000\ \mathrm{K}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$4000\ \mathrm{K}$&lt;/span&gt; from ZAMS to end of main-sequence. Thus the primary spends roughly 3/4 of its time on the main-sequence not in the common envelope phase. Hence, this binary's common envelope phase lasts for, at most, 1/4 the primary's total lifetime, which is on the order of &lt;span class=&quot;math-container&quot;&gt;$10^6$&lt;/span&gt; years. Thus, the upper bound for the timescale of a common envelope event &lt;strong&gt;with massive stars&lt;/strong&gt; with negligible rotation is &lt;span class=&quot;math-container&quot;&gt;$\sim10^5$&lt;/span&gt; years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that this derivation does not take into account the bulging effect that occurs as the separation decreases. This will certainly lower this upper bound, but by how much I'm not sure. It could lower it by 1 year, or &lt;span class=&quot;math-container&quot;&gt;$10^5\ \mathrm{years}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lower bounds to this timescale are entirely ambiguous and not particularly helpful in any physical context. The stars could be spinning really fast, have high or low metallicity, the binary could have a different mass ratio, there could be another binary close by, and there may be magnetic interaction (?). The list goes on! I'm sure there's something I left out.&lt;/p&gt;&#xA;" OwnerUserId="16661" LastEditorUserId="2365" LastEditDate="2019-04-20T19:12:23.167" LastActivityDate="2019-04-20T19:12:23.167" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="20799" PostTypeId="2" ParentId="20677" CreationDate="2017-04-21T14:32:41.123" Score="1" Body="&lt;p&gt;I think you've got it pretty well summarized. In particular, your first three points are dead-on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm going to use &quot;IFU&quot; = integral field unit instead of &quot;IFS&quot; = integral field spectrograph, but that's just because I'm more used to the former term; they're basically synonyms.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me combine the &quot;data reduction&quot; question with one other key advantage of IFUs that you may have missed, which is that they are (almost) plug-in substitutions for traditional long-slit spectrographs, which means they can take direct advantage of existing spectrograph-design expertise and data-reduction software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, one common form of IFU is a closely packed array of small input apertures attached to fiber-optic cables. Away from the IFU, the cables can be re-arranged so that their output ends form a 1D row, very much like the output of a long slit in a traditional spectrograph. (And even more like a multi-object spectrograph design, where multiple individual input apertures are scattered across the field of view rather than being organized into a contiguous 2D array.) This means that the &lt;em&gt;rest&lt;/em&gt; of the spectrograph design -- gratings or prisms, collimation optics, imaging onto a 2D CCD detector, etc. -- is pretty much the same as traditional spectrograph. There are instruments on some telescopes which allow you to switch between long-slit, multi-object, and IFU modes, with the same set of gratings, filters, and detector being used for all modes (for example, the &lt;a href=&quot;http://www.gemini.edu/sciops/instruments/gmos/&quot; rel=&quot;nofollow noreferrer&quot;&gt;GMOS instruments&lt;/a&gt; on the Gemini North and South Telescopes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This carries over into ease of data reduction: the spatial+spectral image on the CCD detector (or near-IR detector array) produced by an IFU is very similar to that produced by a traditional long-slit spectrograph, so the same data-reduction techniques can (with a little modification) be used. In essence, you just chop up the big 2D data image you get from the CCD into a bunch of little sub-images, one for each fiber that came from the IFU, and you can treat each one as though it were a standard single-aperture or single-slit dataset. This contrasts quite strongly with the very different (unique) form of data and data reduction that F-P requires.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2017-04-21T14:32:41.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20800" PostTypeId="1" CreationDate="2017-04-21T15:04:34.523" Score="1" ViewCount="140" Body="&lt;p&gt;This question is raised by someone I've discussed with, after watching the movie &quot;Interstellar&quot;, in which a clip showing Cooper and Amelia landing on Miller's planet near a super-massive blackhole and they experienced quite amount of time dilation due to the gravitational effect. They spent like 20 minutes there and when they backed to the station they found out that 23 years have passed on station (where Romilly waited them for 23 years to return).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's okay for this conception to be implemented here I believe most people will understand. We will just think that yeah 23 years has passed but they just feel normal, just like a 20-minutes park walking. BUT, no one has ever experienced such a strong time dilation, how could you be sure Cooper &quot;just feel normal&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember reading a book which says if you are rushing directly to center of a blackhole, when you are very very close to the Schwartzchild radius, you may experience a whole lifespan of universe IN A SECOND, in other words, you are rushing to the real end of the world. I really doubt any life could 'feel' that second passing. Anyway, this is an extreme example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question: will biological process behaves normal under certain gravitational condition? If it does, can we feel differently? Does our conscious get used to the normal time flow all life on earth are living with? more crudely, will we die due to the extreme time dilation caused by gravity (provided that gravity affect homogeneously upon us) and if so, how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting extra: If Romilly could observe Cooper from station and sees a near still 'slow-motion' picture of him, will Cooper conversely see a 'fast-motion Romilly' from the ground of Miller's planet?&lt;/p&gt;&#xA;" OwnerUserId="16663" LastActivityDate="2017-04-22T07:42:02.757" Title="Will biological process be disrupted under strong gravitational time dilation?" Tags="&lt;general-relativity&gt;&lt;supermassive-black-hole&gt;&lt;time-dilation&gt;&lt;astrobiology&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20801" PostTypeId="2" ParentId="20800" CreationDate="2017-04-21T15:35:58.893" Score="3" Body="&lt;p&gt;Remember this is a theory of &quot;relativity&quot;. Now, time dilation due to gravitation effects is rather outside our normal experience. But there are relativistic effects that that you experience all the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sitting on a train. Relative to my computer I'm not moving. Relative to the track I'm travelling at 100km/h. Relative to the sun I'm moving about 30 km/s, as the Earth orbits. Relative to a distant galaxy I'm moving close to the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in &lt;em&gt;my&lt;/em&gt; frame of reference, I'm not moving at all. So, even though I'm moving at nearly the speed of light, relative to some distant galaxy, I don't notice any slowing down of my watch. My watch is moving with me, in my frame of reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be the same if you were experiencing intense gravitational time dilation. For example, if you were falling around a black hole. Your watch would be in the same frame of reference as you, so there would not be any observable difference, locally. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you tried to stop yourself falling, the force required would crush you. Small black holes have such extreme variation in gravity that your head and feet would be pulled apart by the tidal difference, and if you passed the event horizon, you can never come back. So you would be dead if you tried the experiment in practice. However &quot;gravity&quot; itself can't hurt you, only variation in gravity or your own attempts to act against gravity.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-04-21T15:35:58.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20802" PostTypeId="2" ParentId="20794" CreationDate="2017-04-21T15:47:52.823" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Could dark energy (the accelerating expansion of the universe) be the result of gravity from the dense past universe &quot;pulling&quot; outward from all directions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, because gravity alters the motion of light and matter through space. It doesn't make space fall down in some Chicken-Little fashion. You may hear that it does, as per &lt;a href=&quot;https://www.google.co.uk/#q=black+hole+waterfall+analogy&amp;amp;spf=382&quot; rel=&quot;nofollow noreferrer&quot;&gt;the waterfall analogy&lt;/a&gt;, but it doesn't. In similar vein the &quot;reversed&quot; gravity that you're suggesting doesn't make space fall up, as it were.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The universe theoretically started out extremely dense and hot, and we can see the light reaching us as the cosmic microwave background. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think most people are fairly happy with the CMBR. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I wonder if that same region and denser portions beyond are exerting a gravitational effect that is stretching out everything within toward it, effectively being a cosmic gravity background. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's good to think for yourself, and it's good to wonder. But I have to say I think this idea is a non-starter. We have galaxies receding faster than light (see &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0310808&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;). That means it's space expanding, not gravity pulling.    &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If so, is this the explanation for dark energy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sorry, no. If I were you I'd have a look at vacuum energy. &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2017-04-21T15:47:52.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20803" PostTypeId="2" ParentId="20795" CreationDate="2017-04-21T16:40:12.230" Score="12" Body="&lt;p&gt;To answer this, one really has to understand how the geometric and bond albedos are defined. Let's start with the bond albedo since its simpler.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Bond Albedo&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Bond_albedo&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bond Albedo&lt;/a&gt; is just the fraction of energy hitting a surface that gets reflected. To simplify the process, let's say I shoot 100 photons of all the same energy at Enceladus. Of those 100 photons that hit the surface, 81 of them will be reflected back and 19 of them will be absorbed. That means the bond albedo will be $81/100 = 0.81$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple right? As such, the bond albedo can never be greater than 1 because you obviously can't reflect back more photons than you received. This means that a bond albedo of zero implies no light was reflected and all of it was absorbed, whereas a bond albedo of one implies all light was reflected and no light was absorbed.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Geometric Albedo&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The definition of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Geometric_albedo&quot; rel=&quot;nofollow noreferrer&quot;&gt;geometric albedo&lt;/a&gt; is a bit more complicated. The first thing you have to realize though is that it is not a straight zero to one scale. A geometric albedo of one doesn't mean all the light got reflected like it did for the bond albedo. So let's try to define the geometric albedo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two important points about the geometric albedo that make it differ from the bond albedo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the bond albedo, we talked solely about the total amount of incident light and the total amount of reflected light. This says nothing about an observer of said light. We just have some magical way of knowing about every photon that hits and is reflected by your surface and that allows us to calculate the albedo according to the bond albedo definition. The geometric albedo on the other hand specifically takes into account your ability to observe the light from a specific vantage point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second major point is that the geometric albedo is a measure of how well your particular surface reflected the light back to you &lt;em&gt;compared to a reference surface&lt;/em&gt;. Imagine now you have two surfaces, one is the surface of Enceladus, the other is your &quot;reference surface&quot;. This reference surface is an &quot;idealized&quot; reflector which means that it reflects every photon that hits it (making it have a bond albedo of 1). The key here though, is that it reflects light isotropically. What I mean by that is that the light doesn't get reflected in any preferred direction, but rather gets reflected in all directions equally. So now you have your 100 photons hitting your reference surface and all 100 photons get reflected, but because they're all reflected in random directions, only 10 photons actually get to your camera/eye/detector. What can happen with Enceladus though, is that the surface has just the right properties such that 100 photons hit the surface, and 14 photons get reflected to your detector because the surface &lt;em&gt;preferentially reflects light in a specific direction&lt;/em&gt;. It's sending more photons to your detector than the reference surface. The geometric albedo is the ratio of how much light gets reflected to your detector by Enceladus over how much light gets reflected to your detector by the reference surface. In this case, $14/10 = 1.4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One small, additional point, is that this definition relies on your detector being in the same direction as your light source. In other words, the geometric albedo is a measure of how much your surface can &lt;a href=&quot;https://en.wikipedia.org/wiki/Retroreflector&quot; rel=&quot;nofollow noreferrer&quot;&gt;retro-reflect&lt;/a&gt; (i.e., reflect back to the source of the photons) compared to the reference surface. Technically, the geometric albedo attains a maximum value when your surface is made up of retroreflectors. &lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2017-10-10T16:07:34.433" LastActivityDate="2017-10-10T16:07:34.433" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20804" PostTypeId="1" CreationDate="2017-04-21T19:18:16.373" Score="2" ViewCount="96" Body="&lt;p&gt;Has the Lyman-alpha forest ever been used to test Halton Arp's theory that quasars are not cosmological but instead are ejected from relatively near-by galaxies?  If Arp was right, then the spectra of the quasar and the 'parent galaxy' should both show the same Lyman-alpha forest; if it only appears in the quasar spectra, then the quasar must be much further away.&lt;/p&gt;&#xA;" OwnerUserId="16667" LastActivityDate="2017-04-21T23:05:03.570" Title="Has the Lyman-alpha forest ever been used to test Arp's theory?" Tags="&lt;distances&gt;&lt;quasars&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20805" PostTypeId="1" AcceptedAnswerId="20806" CreationDate="2017-04-21T20:25:30.580" Score="4" ViewCount="4293" Body="&lt;p&gt;I gather that the impact speed depends on the radius of the orbit of the asteroid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the orbit of asteroids in the same direction around the sun, or can they move in the &quot;opposite&quot; direction?&lt;/p&gt;&#xA;" OwnerUserId="14150" LastActivityDate="2020-12-15T02:16:48.820" Title="What are the maximum and minimum impact speeds for an asteroid that would strike the Earth?" Tags="&lt;asteroids&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20806" PostTypeId="2" ParentId="20805" CreationDate="2017-04-21T21:54:39.140" Score="9" Body="&lt;p&gt;Although most objects orbit the Sun in the same direction — having emerged out of the same rotating gas cloud that spawned the Solar System — some asteroids and other &lt;a href=&quot;https://en.wikipedia.org/wiki/Minor_planet&quot; rel=&quot;noreferrer&quot;&gt;minor planets&lt;/a&gt; do move in opposite, or &lt;em&gt;retrograde&lt;/em&gt;, orbits (see &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_exceptional_asteroids#Retrograde&quot; rel=&quot;noreferrer&quot;&gt;this Wikipedia article&lt;/a&gt; for a list of such objects).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimum speed for an asteroid is achieved if it has more or less the same velocity around the Sun as the Earth. In this case the gravitational attraction of Earth will accelerate the object to the escape velocity of Earth, i.e.&#xA;$$&#xA;v_\mathrm{min} = v_{\mathrm{esc,}\oplus} = \sqrt{\frac{2GM_\oplus}{R_\oplus}} \simeq 11\,\mathrm{km}\,\mathrm{s}^{-1}.&#xA;$$&#xA;Here, $G$, $M_\oplus$, and $R_\oplus$ are the gravitational constant and the mass and radius of Earth, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum speed is achieved at a &quot;head-on&quot; collision. Earth's speed around the Sun is&#xA;$$&#xA;v_{\mathrm{orb,}\oplus} = \sqrt{\frac{GM_\odot}{d}} \simeq 30\,\mathrm{km}\,\mathrm{s}^{-1},&#xA;$$&#xA;where $M_\odot$ and $d$ are the mass of the Sun and the distance from Sun to Earth (1 AU).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the asteroid travels on the same orbit, but in the opposite direction, the impact will then be at 60 km/s. However, if the asteroid comes from far away (e.g. the Oort Cloud), it will be accelerated by the Sun and achieve a velocity equal to &lt;em&gt;the escape velocity from the Sun at the location of Earth&lt;/em&gt;. As is seen from the two equations above, the orbital speed and the escape velocity differ by a factor of $\sqrt{2}$. That is, an object falling from infinity toward the Sun, will have a speed equal to $30\,\mathrm{km}\,\mathrm{s}^{-1}\times\sqrt{2}=42\,\mathrm{km}\,\mathrm{s}^{-1}$ when it reaches Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the maximum impact velocity is&#xA;$$&#xA;v_\mathrm{max} = 30\,\mathrm{km}\,\mathrm{s}^{-1} + 42\,\mathrm{km}\,\mathrm{s}^{-1} = 72\,\mathrm{km}\,\mathrm{s}^{-1}.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2017-04-22T15:26:54.617" LastActivityDate="2017-04-22T15:26:54.617" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="20807" PostTypeId="2" ParentId="20804" CreationDate="2017-04-21T23:05:03.570" Score="2" Body="&lt;p&gt;All possible tests of the Arp model have been failed. High redshift quasars have Lyman alpha forests, low redshift galaxies do not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Arp model suggests that the Lyman alpha forests are somehow intrinsic to the quasar. There are many problems with this idea.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-04-21T23:05:03.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20808" PostTypeId="1" CreationDate="2017-04-21T23:32:00.610" Score="3" ViewCount="273" Body="&lt;p&gt;From &lt;em&gt;Wikipedia&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;The spectral classes O through M, as well as other more specialized classes discussed later, are &lt;strong&gt;subdivided by Arabic numerals (0–9), where 0 denotes the hottest stars of a given class&lt;/strong&gt;. For example, A0 denotes the hottest stars in the A class and A9 denotes the coolest ones. Fractional numbers are allowed; for example, the star &lt;strong&gt;Mu Normae is classified as O9.7&lt;/strong&gt;. The Sun is classified as G2.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So our Sun is a G2V yellow dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The G identifies its temperature as within the range of 5,200-6,000K.&#xA;The V tells us it is on its main sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That leaves us with the 2. The 2 tells us that the sun is warmer than a G3 star and cooler than a G1. I get that. &lt;strong&gt;My problem: I just can't seem to figure out how the ranges of these specific numerical subclasses are determined&lt;/strong&gt;. Are they explicitly defined? Logarithmic? Is there a way to calculate it exactly? Why is Mu Normae a O9.7?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My best guess is to take the range of the stellar class and divide it into 10 subsections but I can't find any source to confirm that.&lt;/p&gt;&#xA;" OwnerUserId="16668" LastActivityDate="2017-04-21T23:32:00.610" Title="How are the numerical stellar subclassifications determined?" Tags="&lt;star&gt;&lt;classification&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20809" PostTypeId="1" AcceptedAnswerId="20810" CreationDate="2017-04-22T02:39:11.783" Score="5" ViewCount="483" Body="&lt;p&gt;I don't know if this would fall under the other questions but I have always been confused by this... Information has left the black hole; its mass, rotation, location, and a few other properties. My problem has been that a true singularity should collapse the space around it and become separate from the universe as being able to observe the effects of a singularity breaks the fact that it is a singularity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I see a black hole I can already tell its mass and be effected by the gravity of an object which I can receive no other information from. This would break things too! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's just say I am an all powerful being who can do almost impossible things, but I still have to follow the laws of the universe. I decide to create a gravity observation station in orbit which uses lasers to see very small changes in the black hole. It also has a teleporter of sorts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then enter the black hole and construct another station, this one however uses small black holes and impossibly powerful thrusters and materials to do certain things. Its purpose is to manipulate and sustain a small black hole &quot;orbiting&quot; the inside of the bigger black hole (being sustained, barely, by the station) it then changes the direction this smaller black hole orbits in a kind of binary way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then step into a &quot;teleporter&quot; converted to binary instructions and encoded into these gravitational and rotational information. (Let's just say the station has no lifespan) and this station over a huge amount of time begins to piece together this code, and reconstructs me in the &quot;teleporter&quot; on board. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have I just circumvented the event horizon using disclosed information available to me? Why would or wouldn't this work because for what I see being able to describe a black hole violates the properties of the black hole.&lt;/p&gt;&#xA;" OwnerUserId="16449" LastEditorUserId="16449" LastEditDate="2017-04-27T05:30:44.577" LastActivityDate="2017-04-27T05:30:44.577" Title="Whats the deal with black holes and &quot;no information from inside the event horizon can leave&quot;?" Tags="&lt;black-hole&gt;&lt;event-horizon&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20810" PostTypeId="2" ParentId="20809" CreationDate="2017-04-22T05:57:04.320" Score="8" Body="&lt;p&gt;You are right in that, if you could send a message of any kind from inside a black hole then there is no magic required rebuild you from the information about you. If you could send even a single bit of information, then the black hole wouldn't be so black. You want to do this using the rotation of the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't work, for various reasons.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Information has left the black hole; its mass, rotation, location, and a few other properties&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The mass etc is a property of the black hole, not information travelling from inside the black hole. The reason is that you can't manipulate them in the way you describe, so they can't be used to send a message. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I then enter the black hole and construct another station&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is your first problem. From the frame of reference outside the black hole, there is infinite time dilation at the event horizon. Hence you never quite reach the event horizon, nor enter the black hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the frame of reference of the infalling person, the event horizon is crossed. So let's somehow pretend that you are now inside the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Its purpose is to manipulate and sustain a small black hole &quot;orbiting&quot; the inside of the bigger black hole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can't orbit, you can only fall. The orbital speed is higher than the speed of light, so you will reach the singularity, as surely as you will reach next Tuesday. Your time has now been infinitely dilated, so events that take happen in finite time for you, take an infinite amount of time in the frame of the outside observer, so that makes sending a message using variation in rotation impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, from inside the black hole you can't change the total mass or rotation. The angular momentum of a closed system is conserved, so if you use your thrusters to change the direction of the small black hole you are carrying, the total angular momentum of the black hole remains constant. You can't change the total angular momentum with thrusters if you include both the angular momentum of the ship and the angular momentum of the gas ejected by the thrusters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly you can't change the total mass or location of the black hole from inside the black hole. And you can't do anything in a finite time from the perspective of an outside observer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus no information can be encoded in the observed properties of the black hole.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-04-22T06:05:09.383" LastActivityDate="2017-04-22T06:05:09.383" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20811" PostTypeId="2" ParentId="20800" CreationDate="2017-04-22T07:42:02.757" Score="0" Body="&lt;p&gt;One of the fundamentals of General Relativity is that an observer travelling freely in a gravitational field cannot do any local experiment that would give different results to the same experiment performed outside a gravitational field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbiting a black hole falls into this category of an inertial frame if there are no external forces besides gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the extent that GR is a good theory, then no, there is no &quot;local experiment&quot; (including how you feel) that would differ if you were travelling in a strong gravitational field.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-04-22T07:42:02.757" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20812" PostTypeId="1" CreationDate="2017-04-22T09:38:00.950" Score="7" ViewCount="523" Body="&lt;p&gt;I just re-read a popular astronomy book that was published when I was a child, about 1973. It starts with an interview to Harold C. Urey where the interviewer asks about a hypothetical planet in our solar system called Biga.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has a hypothetical planet ever been called &quot;Biga&quot; by its  proponents, in the same way there was a proposed planet &quot;Vulcan&quot;, or is this &quot;Biga&quot; name a printing or translation error?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's a mistake and there never was a &quot;Biga&quot; proposal, which is the proposed planet with the most similar name?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.librosmaravillosos.com/elsistemasolar/&quot; rel=&quot;noreferrer&quot;&gt;This link contains such interview&lt;/a&gt;, in Spanish. I will translate and quote its most relevant text:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Interviewer:&lt;/strong&gt; At present, do we know all the planets of the Solar&#xA;  System, or is there the possibility of still unknown planets in our&#xA;  system?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;H.C. Urey:&lt;/strong&gt; We certainly know the nine usual planets. Recently there&#xA;  has been much talk of another object moving in space and supposed to&#xA;  be close to the Solar System, but I am not sure that it is a real&#xA;  hypothesis, and personally I think it is better to wait before&#xA;  deciding whether it is Not of a planet. I consider it sufficient to&#xA;  say that there are Mercury, Venus, Earth, Mars, Jupiter, Saturn,&#xA;  Uranus, Neptune, Pluto and the asteroidal belt.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Interviewer:&lt;/strong&gt; When talking about this object, are you referring to the&#xA;  so-called planet Biga?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;H.C. Urey:&lt;/strong&gt; Yes, that's what I meant, but I'm not sure of its&#xA;  existence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Please, remember:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- This interview mas made about 45 yeras ago. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- I'm not asking if this hypothesis is correct or even possible. Only want to know more about this hypothesis which was well known by H. C. Urey in the 70's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- I'm not asking if the name &quot;Biga&quot; is a correct proposal of a name for a planet in our Solar System accordind to current naming rules.&lt;/p&gt;&#xA;" OwnerUserId="11487" LastEditorUserId="11487" LastEditDate="2017-04-22T13:00:47.080" LastActivityDate="2018-02-21T19:45:05.563" Title="Was there ever a hypothesis about a planet called Biga?" Tags="&lt;planet&gt;&lt;solar-system&gt;&lt;history&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20813" PostTypeId="2" ParentId="20812" CreationDate="2017-04-22T10:00:21.213" Score="3" Body="&lt;p&gt;If another planet in our solar system was to be named, I think that Biga wouldn't be the first name to come up, all of the known planets in our solar system are named after Roman deities and Biga is not a deity but rather a cart on which mostly Luna is depicted. I have never heard of this proposal and if the 'rules' are used for naming, it wouldn't be the first candidate.&lt;/p&gt;&#xA;" OwnerUserId="16670" LastActivityDate="2017-04-22T10:00:21.213" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20814" PostTypeId="2" ParentId="8139" CreationDate="2017-04-22T15:00:26.163" Score="-3" Body="&lt;p&gt;You asked:&#xA;For example, on the ISS during the night (the Earth that cover the Sun) looking at the deep sky, how many stars can you see?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is 0 (at ISS-night) and 1 (at ISS-day) which is called the Sun. Well, that is what NASA, the visitors to the ISS, the Apollo astronauts, the videos of the ISS space walks and other pictures produced by NASA say. There are no pictures still or video taken from the ISS that show more than 1 star, that well known nearest one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly there was one Russian astronaut (Not on ISS but early Russian space craft) who saw something different when he was up in space. He saw the sky full of stars and the light from them.(I'll try to find you a link later).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is clearly one obvious conclusion. You know what it is and it has nothing to do with optical effects, the physics of light or the way your eyeball works.&lt;/p&gt;&#xA;" OwnerUserId="16672" LastActivityDate="2017-04-22T15:00:26.163" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20816" PostTypeId="2" ParentId="8139" CreationDate="2017-04-22T19:19:46.557" Score="2" Body="&lt;p&gt;Stars can be seen from space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronauts, such as &lt;a href=&quot;https://www.youtube.com/watch?v=LaUCMzgidvs&amp;amp;feature=youtu.be&quot; rel=&quot;nofollow noreferrer&quot;&gt;James Reilly Describes Seeing Stars in Space&lt;/a&gt;. In the video he notes that the stars in space don't twinkle, and that more stars are visible compared with Earth (he actually describes seeing &quot;literally millions&quot;, but that is obviously hyperbole). He notes that it can be difficult to pick out the constellations as many of the fainter stars are visible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, several thousand stars are visible, but it depends on the acuity of your vision, and your skill at observing, so no exact number can be given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are used to typical light polluted skies, then seeing the stars in a truly dark sky is pretty awe inspiring.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-04-22T19:25:21.230" LastActivityDate="2017-04-22T19:25:21.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20817" PostTypeId="2" ParentId="20793" CreationDate="2017-04-23T06:52:45.730" Score="2" Body="&lt;p&gt;Yes. A t&lt;a href=&quot;http://www.planetary-astronomy-and-imaging.com/en/image-surface-venus/&quot; rel=&quot;nofollow noreferrer&quot;&gt;hermal image from the surface can be detected&lt;/a&gt;, even by amateurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ground is heated to about 700K, and thermal emission at a wavelength of 1 micron is just about observable by skilled amateurs, with the right equipment. &lt;a href=&quot;http://www.danielegasparri.com/venus_nightside_gasparri_.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mountains, which are cooler by about 30K, can be seen at this wavelength&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From orbit, more details can be seen. &lt;a href=&quot;http://www.planetary.org/multimedia/space-images/venus/akatsuki-sees-venus-surface.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Akatsuki used the 1 micron band to observe the surface in 2015&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At other wavelengths, cloud patterns are revealed. &lt;a href=&quot;https://www.jpl.nasa.gov/spaceimages/details.php?id=PIA00112&quot; rel=&quot;nofollow noreferrer&quot;&gt;Galileo took images at 2.3 microns during its flyby&lt;/a&gt;. They show sulphuric acid clouds at 50km above the surface (10km below the visible cloud layer) and at about 270K, with gaps exposing the lower atmosphere, glowing at 500K. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-04-23T06:52:45.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20821" PostTypeId="1" CreationDate="2017-04-23T16:44:06.560" Score="1" ViewCount="152" Body="&lt;p&gt;Can the Keck Observatory resolution be improved by adding a third telescope? I imagine that that would be cheaper than building a completely new telescope?&lt;/p&gt;&#xA;" OwnerUserId="11740" LastActivityDate="2017-04-28T02:45:39.770" Title="Can the Keck Observatory be improved with a third telescope?" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20822" PostTypeId="2" ParentId="20821" CreationDate="2017-04-23T17:16:22.437" Score="3" Body="&lt;h1&gt;Original Overview&lt;/h1&gt;&#xA;&lt;p&gt;The Keck telescopes mainly work independently of one another. Therefore, adding a third telescope wouldn't really have much of a point (aside from the benefits of having a telescope on top of a mountain with low light pollution and excellent atmospheric conditions). In the past, this might have been a slightly more appealing idea, because &lt;a href=&quot;https://science.nasa.gov/missions/keck&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Keck Interferometer (KI) ran from 2003 to 2012&lt;/a&gt;, and was a combination of the two telescopes (an effective increase from two 10-meter telescopes to one 85-meter telescope).&lt;/p&gt;&#xA;&lt;p&gt;KI had some problems, though, that prevented it from living up to its full potential. For instance, &lt;a href=&quot;http://www.skyandtelescope.com/astronomy-news/closure-looms-for-keck-interferometer/&quot; rel=&quot;nofollow noreferrer&quot;&gt;four secondary &amp;quot;outrigger&amp;quot; telescopes to improve the interferometer were never built&lt;/a&gt;, even though they were an important part of &lt;a href=&quot;http://www2.keck.hawaii.edu/1stLight/keck-spie-00-annot1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;early plans&lt;/a&gt;. It was eventually &amp;quot;mothballed&amp;quot; after several of the main scientific objectives were completed. Obviously, the interferometer could be reactivated, but you'd have to make a strong case for doing so. If you could get that to happen, and get permission to build another telescopes, many might choose to use the funds for these outriggers, rather than a whole new instrument.&lt;/p&gt;&#xA;&lt;p&gt;Finally, &lt;a href=&quot;https://en.wikipedia.org/wiki/Opposition_to_the_Mauna_Kea_Observatories&quot; rel=&quot;nofollow noreferrer&quot;&gt;building telescopes on Mauna Kea is controversial&lt;/a&gt; because of possible environmental impacts, as well as arguments that it endangers Hawaiian cultural and religious sites. There are objections enough over the planned &lt;a href=&quot;https://en.wikipedia.org/wiki/Thirty_Meter_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thirty Meter Telescope&lt;/a&gt;, never mind another Keck (which would, admittedly, probably be cheaper, although the two would use different wavelengths).&lt;/p&gt;&#xA;&lt;p&gt;In short, a third telescope would help only if the interferometer was brought back on line - and even then, you'd have to make a case for both building a new telescope and for choosing it over the original outriggers.&lt;/p&gt;&#xA;&lt;h1&gt;Resolution&lt;/h1&gt;&#xA;&lt;p&gt;The angular resolution for a single telescope, to an order of magnitude, is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\theta\sim\frac{\lambda}{D}\tag{1}$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; is the wavelength of observations and &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; is the diameter of the dish (&lt;a href=&quot;https://en.wikipedia.org/wiki/Adaptive_optics&quot; rel=&quot;nofollow noreferrer&quot;&gt;adaptive optics&lt;/a&gt;, by the way, can make things a bit better). Each Keck telescope has a diameter of &lt;span class=&quot;math-container&quot;&gt;$10\text{ m}$&lt;/span&gt; and observes at the optical and near-infrared ranges - so about &lt;span class=&quot;math-container&quot;&gt;$4\times10^{-7}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$4\times10^{-6}\text{ m}$&lt;/span&gt;. This approximation yields angular resolutions of about &lt;span class=&quot;math-container&quot;&gt;$0.0083$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$0.083$&lt;/span&gt; arcseconds. This approximation is accurate to within an order of magnitude.&lt;/p&gt;&#xA;&lt;p&gt;Likewise, for an interferometer with two telescopes separated by a baseline of distance &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt;, the angular resolution is approximately&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\theta\sim\frac{\lambda}{b}$$&lt;/span&gt;&#xA;The Keck interferometer had a baseline of &lt;span class=&quot;math-container&quot;&gt;$85\text{ m}$&lt;/span&gt; and observed at wavelengths of &lt;span class=&quot;math-container&quot;&gt;$2.2\times10^{-6}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$10^{-5}\text{ m}$&lt;/span&gt;. The approximation gives us resolutions of &lt;span class=&quot;math-container&quot;&gt;$0.0053$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$0.024$&lt;/span&gt; arcseconds - pretty close to the actual values.&lt;/p&gt;&#xA;&lt;p&gt;What if we add a third telescope, identical to the other two and placed in a line next to them, &lt;span class=&quot;math-container&quot;&gt;$85\text{ m}$&lt;/span&gt; from the closer one and &lt;span class=&quot;math-container&quot;&gt;$170\text{ m}$&lt;/span&gt; from the farther one? The longest baseline is now &lt;span class=&quot;math-container&quot;&gt;$2\times85=170\text{ m}$&lt;/span&gt;, twice the baseline of the two-dish setup. We've now made the angular resolution twice as small as before. That's excellent. At &lt;span class=&quot;math-container&quot;&gt;$2.2\times10^{-6}\text{ m}$&lt;/span&gt;, the resolution is &lt;span class=&quot;math-container&quot;&gt;$1.3\times10^{-8}$&lt;/span&gt; radians.&lt;/p&gt;&#xA;&lt;p&gt;Let's think about observations at &lt;span class=&quot;math-container&quot;&gt;$2.2\times10^{-6}\text{ m}$&lt;/span&gt;, at the shortest wavelengths the interferometer used. Assume we want to build another telescope with the same angular resolution. Rearranging &lt;span class=&quot;math-container&quot;&gt;$(1)$&lt;/span&gt; gives us&#xA;&lt;span class=&quot;math-container&quot;&gt;$$D\sim\frac{\lambda}{\theta}$$&lt;/span&gt;&#xA;Plugging in our numbers (&lt;span class=&quot;math-container&quot;&gt;$\lambda=2.2\times10^{-6}\text{ m}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\theta=1.3\times10^{-8}$&lt;/span&gt; radians), we find that &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; should be about &lt;span class=&quot;math-container&quot;&gt;$170\text{ m}$&lt;/span&gt; - which is of course the longest baseline of the three-dish interferometer. This is simply implausible at these sort of wavelengths. The largest single-dish optical telescope - observing in shorter wavelengths than this by a factor of about three - is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gran_Telescopio_Canarias&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gran Telescopio Canarias&lt;/a&gt;, at &lt;span class=&quot;math-container&quot;&gt;$10.4\text{ m}$&lt;/span&gt; (although the &lt;a href=&quot;https://en.wikipedia.org/wiki/Large_Binocular_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Large Binocular Telescope&lt;/a&gt; is slightly better - &lt;span class=&quot;math-container&quot;&gt;$11.8\text{ m}$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;p&gt;If you want to use the interferometer at optical wavelengths, you've still looking at a &lt;span class=&quot;math-container&quot;&gt;$\sim55\text{ m}$&lt;/span&gt; dish, and that's currently maybe feasible. There are several &lt;a href=&quot;https://en.wikipedia.org/wiki/Extremely_large_telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;extremely large telescopes&lt;/a&gt; being planned, notably, the Thirty Meter telescope I mentioned earlier and the &lt;a href=&quot;https://en.wikipedia.org/wiki/European_Extremely_Large_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;European Extremely Large Telescope (E-ELT)&lt;/a&gt;, coming in at around &lt;span class=&quot;math-container&quot;&gt;$40\text{ m}$&lt;/span&gt;. The E-ELT should be online as soon as 2-24 (!), which would be fantastic, and would have a comparable resolution to this three-dish interferometer.&lt;/p&gt;&#xA;&lt;h1&gt;Costs&lt;/h1&gt;&#xA;&lt;p&gt;Let's look at how much all of this might cost. The E-ELT is expected to cost &lt;a href=&quot;https://www.eso.org/public/about-eso/faq/faq-elt/#14&quot; rel=&quot;nofollow noreferrer&quot;&gt;a little over one billion euros&lt;/a&gt; (1.083 billion, to be precise), which comes out to around \$1-\$1.25 billion, in US dollars (that sounds like a lot, but space-based telescopes can be up to ten times that - though often with higher-quality images). Let's assume that our putative &lt;span class=&quot;math-container&quot;&gt;$\sim55\text{ m}$&lt;/span&gt; optical telescope costs \$1.5-\$2 billion to build.&lt;/p&gt;&#xA;&lt;p&gt;The cost of the instruments for the Keck Observatory ran about &lt;a href=&quot;http://ast.noao.edu/system/tsip/more-info/time-calc-keck&quot; rel=&quot;nofollow noreferrer&quot;&gt;\$80 million&lt;/a&gt;. This includes the telescopes proper, as well as the cameras and other equipment needed to actually capture the images. Making improvements and other developments has run the total project to about \$200 million (not counting observing costs) We can be generous and say that building a third telescope - identical to the other two - and getting the interferometer system up and running might cost \$50-\$60 million. Maybe that's off, but it certainly can't be wrong by over an order of magnitude. At that point, the entirety of this new Keck Observatory has cost only about one tenth of what a single dish would cost.&lt;/p&gt;&#xA;&lt;p&gt;Yes, it's likely that I'm wrong somewhere above - that I've made an error in the size of the single-dish telescope, or in estimating the total cost. But the discrepancy is &lt;em&gt;still&lt;/em&gt; extraordinary.&lt;/p&gt;&#xA;&lt;p&gt;Here's the problem, though: Nightly observing costs are high. &lt;a href=&quot;http://ast.noao.edu/system/tsip/more-info/time-calc-keck&quot; rel=&quot;nofollow noreferrer&quot;&gt;One night using on Keck telescope&lt;/a&gt; costs \$54,000. The interferometer uses &lt;em&gt;both&lt;/em&gt; telescopes, and, as &lt;a href=&quot;http://www.skyandtelescope.com/astronomy-news/closure-looms-for-keck-interferometer/&quot; rel=&quot;nofollow noreferrer&quot;&gt;an article I cited earlier&lt;/a&gt; notes,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;it's complicated and expensive to link the giant eyes through a system of optical pathways for just a few dozen nights each year.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Interferometry isn't as simply as looking at an object with two different telescopes. It's complicated and costly. If we assume that operating the interferometer would cost around \$200,000 per night (is that gratuitous?), we find yearly operating costs of &lt;span class=&quot;math-container&quot;&gt;$\sim$&lt;/span&gt;\$73 million - assuming that it's used every night, which isn't likely. But even accounting for all the nights when the interferometer wouldn't be in use, that's probably still more than the &lt;a href=&quot;http://www.photoxels.com/worlds-largest-optical-near-infrared-telescope-e-elt-will-be-built-with-new-advanced-powerful-cameras/&quot; rel=&quot;nofollow noreferrer&quot;&gt;estimated yearly costs for the E-ELT&lt;/a&gt; (50 million euros, or \$54.33 million). The single-dish telescope - again, operating a shorter wavelengths than Keck - would probably cost less per night than the new Keck Interferometer.&lt;/p&gt;&#xA;&lt;p&gt;Perhaps the relatively low construction costs of the third telescope would offset all of this in the short-term (it likely would).&lt;/p&gt;&#xA;&lt;p&gt;Here's the really obvious thing, however, that I've been ignoring: More dish means more light. An interferometer with an &lt;span class=&quot;math-container&quot;&gt;$170\text{ m}$&lt;/span&gt; baseline is &lt;em&gt;not&lt;/em&gt; the same as a single dish telescope of the same diameter. The three dishes are much, much smaller than the single one. While the resolutions are the same, the size of the dishes really does matter.&lt;/p&gt;&#xA;&lt;p&gt;A dish with a diameter of &lt;span class=&quot;math-container&quot;&gt;$10\text{ m}$&lt;/span&gt; has an area of &lt;span class=&quot;math-container&quot;&gt;$25\pi\text{ m}^2$&lt;/span&gt;. A dish with a diameter of &lt;span class=&quot;math-container&quot;&gt;$170\text{ m}$&lt;/span&gt; has an area of &lt;span class=&quot;math-container&quot;&gt;$7225\pi\text{ m}^2$&lt;/span&gt; - a difference of a factor of 289! If you want to make true comparisons between the new interferometer and a large telescope, you'll need more than three dishes. And &lt;em&gt;that&lt;/em&gt; will make costs skyrocket.&lt;/p&gt;&#xA;&lt;p&gt;Just for fun, imagine that we end up having 50 total dishes, with a baseline. If each costs \$30 to \$40 million to build (and here I'm being generously low), we quickly find that building 48 more dishes runs us to &lt;em&gt;\$1.9 billion&lt;/em&gt;, or at least the cost of building our &lt;span class=&quot;math-container&quot;&gt;$55\text{ m}$&lt;/span&gt; optical telescope.&lt;/p&gt;&#xA;&lt;p&gt;Okay, so let's again assume that I made an error somewhere, and this figure is too high. Also, if you ended up building this interferometer (Where would you put it, by the way? Mauna Kea's crowded enough!), you could probably make each dish simpler than Keck I or Keck II. You still have to deal with high nightly operating costs - &lt;em&gt;really&lt;/em&gt; high operating costs.&lt;/p&gt;&#xA;&lt;h1&gt;Science!&lt;/h1&gt;&#xA;&lt;p&gt;So, why isn't the interferometer up and running? According to &lt;a href=&quot;https://science.nasa.gov/missions/keck&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA&lt;/a&gt;, it's because the interferometer's main target, observing certain circumstellar disks, is done:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;NASA’s primary goal in developing and operating the two 10-m telescope KI was the characterization of faint dust disks around nearby main sequence stars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;For an overview, see &lt;a href=&quot;http://lanl.arxiv.org/abs/1104.1382&quot; rel=&quot;nofollow noreferrer&quot;&gt;Millan-Gabet (2011)&lt;/a&gt;. As the authors note in section 3, the interferometer's primary observing wavelengths (about &lt;span class=&quot;math-container&quot;&gt;$10^{-6}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$10^{-7}\text{ m}$&lt;/span&gt;, the micrometer range) encompass the peak emission wavelengths of circumstellar dust. Additionally, the resolution at these wavelengths can resolve features of circumstellar disks.&lt;/p&gt;&#xA;&lt;p&gt;As we determined before, it's hard to build a telescope of comparable resolution operating in the near-infrared. Remember, our single-dish telescope operates largely in the optical range, although it would of course have near-infrared capabilities, to an extent. Therefore, an interferometer like the Keck Interferometer is our best choice - and perhaps our only choice, in the foreseeable future. However, it's certainly possible to build extremely large telescopes operating at slightly shorter wavelengths, with roughly the same angular resolution (to an order of magnitude, really). Plus, they'll gather more light, and - as I found shortly before posting this, in &lt;a href=&quot;https://astronomy.stackexchange.com/a/14470/2153&quot;&gt;an excellent answer by Rob Jeffries&lt;/a&gt; - limitations of interferometer instruments and atmospheric issues makes them less versatile than large, single-dish telescopes.&lt;/p&gt;&#xA;&lt;h1&gt;TL;DR&lt;/h1&gt;&#xA;&lt;p&gt;There are a few takeaways here:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Adding a third telescope would give you a resolution unmatched at those wavelengths by any other land-based telescopes.&lt;/li&gt;&#xA;&lt;li&gt;This would likely be a lot cheaper.&lt;/li&gt;&#xA;&lt;li&gt;However, you'd gather only a small fraction of the light that a much larger telescope would.&lt;/li&gt;&#xA;&lt;li&gt;To gather more light, costs would skyrocket - as would technical problems.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;So, economically and technically, it's really not feasible to add a third disk to Keck - at least, compared to building a single extremely large telescope.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;This is a pretty long answer, and I'm guessing I made an error somewhere. I'll go back and do one more check in the next day or so. If anyone finds one, do let me know!&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-04-28T02:45:39.770" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20823" PostTypeId="1" AcceptedAnswerId="20824" CreationDate="2017-04-23T18:33:29.133" Score="16" ViewCount="2586" Body="&lt;p&gt;I've heard that the more an object spins, the less of a true sphere it is.&#xA;Using this logic most of neutron stars would be far from spherical,in general what shape are most neutron stars?&lt;/p&gt;&#xA;" OwnerUserId="16649" LastActivityDate="2017-04-24T19:45:54.300" Title="Shape of neutron stars" Tags="&lt;rotation&gt;&lt;neutron-star&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20824" PostTypeId="2" ParentId="20823" CreationDate="2017-04-23T18:57:47.980" Score="20" Body="&lt;p&gt;I don't think you'll find a single agreed shape for a rotating neutron star, not least because we don't have an agreed single model for the equation of state of the material in a neutron star (which is more complex than the name suggests).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found one &lt;a href=&quot;https://arxiv.org/pdf/1303.3259.pdf&quot; rel=&quot;noreferrer&quot;&gt;openly available paper&lt;/a&gt; (I'm sure there are more) which will give you a rough flavor for the complexity of modeling the shape of neutron stars.  As you'll see the difficulty of no single model for an equation of state (EOS is the shorthand typically used) is just one issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think &quot;ellipsoid&quot; should be considered as an approximation, although it's not something I'd consider written in stone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that to be useful a paper has to provide not just a model for what the shape might be, but also someone has to provide a way to measure this, which is challenging.  I think one of the hopes for the new era of gravitational wave astronomy is to be able (eventually) to make more useful and measurements that help us investigate the interior of neutron stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is an open question, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;@Rob-Jeffries asked a question in comment about typical numbers for the deformation, and I answered in comment but comments can be removed by the system, so I'm adding that information as an edit :&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first section of the paper I linked to they do quote fractional deformations as being typically $10^{−5}$, perhaps $10^{−4}$ in special cases and in extreme cases up to $10^{−3}$. However another paper gives an analysis based on crustal rigidity and a very small deformation for a particular neutron star. The paper I initially liked to describes an upper limit based on gravitational wave considerations, I think, rather than a general analysis.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastEditorUserId="-1" LastEditDate="2017-04-24T14:58:00.783" LastActivityDate="2017-04-24T14:58:00.783" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20825" PostTypeId="1" AcceptedAnswerId="20827" CreationDate="2017-04-23T18:59:14.227" Score="0" ViewCount="268" Body="&lt;p&gt;How much gravitational effect do we experience (e.g. maybe -.00001 G or smaller) from the edge of the visible universe?  By edge of the visible universe, I am talking about the region of the cosmic microwave background and beyond.  In the past, the universe was very hot and dense, and we are seeing the light from this time period as the Cosmic Microwave Background.  It took slightly less than the age of the universe for that light to reach us, and it has become extremely redshifted with the expansion of space.  Since gravity apparently travels at the speed of light, we should be experiencing a gravitational effect from this shell around us as well.  As I understand, the universe was extremely dense back then, and it is around us in every direction, which would intensify the gravity greatly.  However, it is extremely distant, which would weaken the gravity greatly.  I don't know if redshift would affect gravity as it has the light from the cosmic microwave background, but if so, this would weaken the effect greatly as well.  With all this in mind, I am curious, what would be the effect of gravity from the edge of the universe on us?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I would expect this to be a very small outward effect in all directions, something like -.00000001 g or something like that, but not 0.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="1386" LastEditDate="2017-04-24T20:38:02.927" LastActivityDate="2017-04-24T20:38:02.927" Title="How strong is the gravitational stretch we experience from the edge of the universe?" Tags="&lt;gravity&gt;&lt;dark-energy&gt;&lt;cosmological-horizon&gt;&lt;doppler-effect&gt;&lt;hubble-constant&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20826" PostTypeId="2" ParentId="20823" CreationDate="2017-04-23T19:19:47.607" Score="-1" Body="&lt;p&gt;Logically speaking, they should be spherical as things with higher gravity tend to collapse into spheres.  Neutron stars are extremely dense, and have high gravity.  However, they are as far as we know rotating extremely fast as well (e.g. Pulsars).  It should be that the faster they rotate, the more disc-like they would become (like an ellipse or slight possibility of more of a disc in extreme cases).  So, depending on rotation speed, sphere for no to fairly high rotation speed, ellipse for high rotation speed, or possibly even a disc for extremely high rotation speed.  There is room for debate here, but this is how I logically see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: By ellipse, I mean a 3 dimensional ellipse, like an egg, but &quot;squished the other way&quot;.  Basically a sphere that has been stretched out on its equator.  The faster it spins, the more it should be deformed (stretched along the equator).  Dantopia's answer shows the shape I am describing.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="1386" LastEditDate="2017-04-24T19:45:54.300" LastActivityDate="2017-04-24T19:45:54.300" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20827" PostTypeId="2" ParentId="20825" CreationDate="2017-04-24T00:20:35.863" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In the past, the universe was very hot and dense, and we are seeing the light from this time period as the Cosmic Microwave Background.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This statement is (partially) correct. &lt;br&gt;&#xA;However....&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since gravity apparently travels at the speed of light, we should be experiencing a gravitational effect from this &lt;b&gt;shell&lt;/b&gt; around us as well. As I understand, the universe was extremely dense back then, and &lt;b&gt;it is&lt;/b&gt; around us in every direction, which would &lt;b&gt;intensify&lt;/b&gt; the gravity greatly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This statement perhaps indicates that there are some misconceptions about the present state of the universe and also about what Cosmic Microwave Background Radiation really is (emphases added to identify the problem areas). So, in order to give a reasonable answer to the question asked, it would require addressing these misconceptions too.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The Big Bang model suggests that everything we see in our Observable Universe was once all concentrated in an infinitesimally small, hot and dense region of space. However, the &lt;a href=&quot;https://physics.stackexchange.com/q/136860/139130&quot;&gt;Big Bang happened everywhere&lt;/a&gt;, which was followed by a super-luminal expansion of the space by a process called &lt;a href=&quot;https://en.wikipedia.org/wiki/Inflation_(cosmology)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Inflation&lt;/a&gt;. The universe remained too hot still for any photons to roam around freely. It wasn't until the next 380,000 years that it continued to expand and eventually got cooled enough for the first hydrogen atoms to form. This is when the light first became visible. The remnant of this light is what we now now call &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cosmic Microwave Background Radiation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, it's important to note that the space was expanding in all possible directions continuously when the first light was emitted everywhere. In other words, the &quot;dense&quot; region from which the spacetime had begun, had now become comparatively much less dense overall. All through these years, this &lt;a href=&quot;https://en.wikipedia.org/wiki/Metric_expansion_of_space&quot; rel=&quot;nofollow noreferrer&quot;&gt;expansion hasn't stopped&lt;/a&gt; and &lt;b&gt;every region where the light was first emitted, has now receded to great distances (46 billion light years in any direction)&lt;/b&gt;. All the regions have cooled down further still to evolve into stars and planets, galaxies, clusters and any other celestial objects imaginable. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, to answer the questions asked (for simplicity, we'll stick to CMB as the &quot;edge&quot;).... &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;There is &lt;b&gt;no&lt;/b&gt; space outside our Observable Universe which is surrounding us as a &lt;i&gt;&lt;b&gt;concrete shell&lt;/b&gt;&lt;/i&gt;. To be precise, it is merely a limit to how far back into the past and how deep we can peek into the space with our current technologies. Also, this has got nothing to do with the present state of the universe there. &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmological_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;The universe is homogeneous and isotropic&lt;/a&gt; — meaning, you could go to any other place in the universe and it would still appear the same as it does from here — stars and galaxies everywhere. &lt;b&gt; Your vantage point would determine which galaxy appears as a distant blob to you and which region of space would merely appear as a CMB radiation. &lt;/b&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The regions which we now see as CMB, 46 billion light years away, are indeed &lt;i&gt;one of the farthest reaches of space&lt;/i&gt; from where anything, including gravity, would have had any effects on us. By now, everything that we see as CMB would have coalesced to form celestial objects. But, there's a catch.... Remember, if the Sun was to disappear from the Solar System now, it would still take approximately 8 minutes for the Earth to be set free from its orbit. Meaning, the &lt;b&gt;state of the Sun 8 minutes ago&lt;/b&gt; determines what gravity we feel because of it &lt;b&gt;now&lt;/b&gt;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;So, to compute the gravitational tug from CMB (46 billion light years away), we need to look at the state of the matter from 13.8 billion years ago minus 380,000 years (when the CMB was emitted). We can roughly assume that, back then, it was just Hydrogen atoms all over. The overall mass of all those atoms combined would have been equivalent to the &lt;b&gt;mass of the Observable Universe&lt;/b&gt;, i.e. $10^{53}kg$.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;But again, the &lt;b&gt;CMB was emitted everywhere&lt;/b&gt; and the &lt;a href=&quot;http://www.universeadventure.org/big_bang/cmb-origins.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;isotropic nature of CMB&lt;/a&gt; tells us that &lt;a href=&quot;http://www.astro.ucla.edu/~wright/CMB-DT.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;departures from uniformity are only up to seven parts in a million&lt;/a&gt;! Meaning, when the CMB was emitted 13.8 billion years minus 380,000 years ago, the mass of the Observable Universe was distributed almost perfectly uniformly in every direction possible. From the distance of 46 billion light years then, this translates into &lt;b&gt;almost&lt;/b&gt; equal gravitational pull from everywhere that light (and gravity) has reached us since then. Hence, the gravitational effects of the &lt;b&gt;matter&lt;/b&gt; that is (i.e. was) spread uniformly in a radius of 46 billion light years from us, would be equivalent to what one would feel if they were to somehow reach the center of the earth — i.e. experiencing a nearly perfect equilibrium, with a net force of zero. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;PS: Should there be any discrepancies above, I'd wholeheartedly welcome the suggestions on correcting or improving the answer.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="10589" LastActivityDate="2017-04-24T00:20:35.863" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="20828" PostTypeId="1" AcceptedAnswerId="20829" CreationDate="2017-04-24T00:57:17.523" Score="0" ViewCount="667" Body="&lt;p&gt;Take a star of a given mass (say $1.0\ \mathrm{M_{\odot}}$ or $1.1\ \mathrm{M_{\odot}}$), what affects the star's luminosity as a function of time and how much? (metallicity?, rotation?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that stellar evolution is basically set by the laws of physics; if we know certain parameters (e.g., mass). Of course, that's the trick since we can't directly measure the mass of a star with no companion(s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have run across a few issues with published values being in conflict. For example, &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/591785/pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this study&lt;/a&gt; gives the age of Zeta Tucanae as being about $2.1-3.8\ \mathrm{Gyr}$. However, mass/radius/luminosity is given as 0.99/1.08/1.26 relative to sun (Wikipedia sources). If you look at an evolutionary plot of the Sun something is wrong here. (I'm guessing mass.) By the way [Fe/H] is given as -0.18.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="69" LastEditDate="2017-04-25T12:27:26.550" LastActivityDate="2017-04-25T12:27:26.550" Title="What affects the evolution curve of a star's luminosity as a function of time?" Tags="&lt;stellar-evolution&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20829" PostTypeId="2" ParentId="20828" CreationDate="2017-04-24T06:46:39.710" Score="4" Body="&lt;p&gt;I think this question is too broad, but I'll take a stab at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Russell-Vogt (or sometimes Vogt-Russell) theorem is that the position of a star in the HR diagram is determined by its mass and composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The luminosity is determined mostly by its central temperature and composition. In turn, the central temperature depends on mass and radius and the radius depends on the luninosity and effective temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the question you ask fills textbooks. But to first order.&#xA;The time dependence of luminosity is set by the time dependence of mass - ie mass loss (or gain) and the rate of change of composition, particularly in the nuclear burning regions of a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a star like the Sun, mass loss is relatively unimportant, so it is the rate at which hydrogen is turned into helium in the core that sets the timescale for luminosity evolution. Other processes that alter the core composition like mixing (due to convection, rotational mixing or diffusion) are thought to be second order effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Restricting myself to the main sequence as an example. The luminosity of the Sun is set by the core temperature. The the nuclear reactions become faster as the mean particle weight increases and the core contracts and the temperature rises to maintain hydrostatic equilibrium. The rate at which this occurs is given by the luminosity of the star divided by the amount of H available (proportional to the mass of the star).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall chemical composition drives this as a second order effect. A lower metallicity star is less opaque to radiation. The radiation escapes more efficiently and the star is smaller and hotter at the surface for the same luminosity. Smaller stars are hotter in the middle for the same mass and thus more luminous. Thus the luminosity evolution of low metallicity stars is faster than those of higher metallicity. See for example Figs 1 and 2 of &lt;a href=&quot;http://adsabs.harvard.edu/abs/1990ApJ...354..644B&quot; rel=&quot;noreferrer&quot;&gt;Bazan &amp;amp; Mathews (1990)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-04-24T08:15:50.640" LastActivityDate="2017-04-24T08:15:50.640" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20830" PostTypeId="2" ParentId="14613" CreationDate="2017-04-24T07:00:54.830" Score="1" Body="&lt;p&gt;I'm assuming you're talking about the visible universe, not the observable universe. If so, the comoving distance we could see is about 45.7 Billion light years. Even though the universe is only 13.8 Billion years old, due to expansion it took longer than that to reach us. It'd be like walking up an escalator thats going down vs walking up stairs, I'll get there but it'll just take much longer. A COMOVING distance would relate better to the person on the stairs, it factors out expansion, thus making calculations much easier. Using the comoving distance, figuring out it's volume is simple math, unless of course you plan on figuring it's volume  without mass, just the space. Then of course you'd have to find each masses volume, subtract and blah, blah, and impossible, since we don't know every object in the visible universe. Using V=4/3πr^3, V = approximately 3.99795E+32ly^3. Or, 399 nonillion Ly^3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine the escalator you're on starts running just as fast as you, eventually faster; You'll never reach the top, in fact you'll move away from the top despite running towards it. This is what is happening to our visible universe, it's slowly receding faster than light due to expansion. So the light will continue to travel towards us but will never reach us once it's emitted beyond about 4500 megaparsecs away, it's event horizon, sometimes called particle horizon. &#xA;Point being, the visible universe volume is constantly shrinking. &lt;/p&gt;&#xA;" OwnerUserId="16686" LastActivityDate="2017-04-24T07:00:54.830" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20831" PostTypeId="1" AcceptedAnswerId="20842" CreationDate="2017-04-24T11:15:29.933" Score="2" ViewCount="140" Body="&lt;p&gt;After a supernova neutrinos get released by electron capture,and they escape in a flood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is because they have tiny masses and get momentum from momentum conservation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, the huge number of neutrinos it emits carry away so much energy that the temperature of an isolated neutron star falls within a few years to around $10^6$ kelvin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reference : &lt;a href=&quot;https://en.wikipedia.org/wiki/Neutron_star#Mass_and_temperature&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;,&lt;a href=&quot;http://www.astro.umd.edu/~miller/nstar.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Original&lt;/a&gt;(wikipedia reference)&#xA;Wouldn't most of the neutrinos scatter after supernova almost instantaneously before a formation of a neutron star?&lt;/p&gt;&#xA;" OwnerUserId="16649" LastEditorUserId="16649" LastEditDate="2017-04-24T18:41:40.313" LastActivityDate="2017-04-24T20:18:23.607" Title="Why do neutrinos take time to scatter?" Tags="&lt;supernova&gt;&lt;neutron-star&gt;&lt;neutrinos&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20833" PostTypeId="1" AcceptedAnswerId="20834" CreationDate="2017-04-24T15:02:15.950" Score="2" ViewCount="726" Body="&lt;p&gt;For example, the average Sun-Earth distance is 1 AU. Is this&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the distance between the center of the Sun and the center of the Earth&lt;/li&gt;&#xA;&lt;li&gt;the distance bewteen the surface of the Sun and the surface of the Earth,&lt;/li&gt;&#xA;&lt;li&gt;or something else?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13473" LastActivityDate="2017-04-25T15:16:54.203" Title="Are the distances between two celestial bodies the distance of their centers?" Tags="&lt;distances&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20834" PostTypeId="2" ParentId="20833" CreationDate="2017-04-24T15:49:26.080" Score="5" Body="&lt;p&gt;Distance is an ambiguous term that can mean both centre-centre and surface-surface depending on the context it is used.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In cases where the size of the objects is trivial compared to the distance, 'distance' is almost exclusively used for the &quot;distance between centres&quot; of the two objects. This is the case for Sun-Earth, Earth-Moon and almost any other pair of significant solar system object. The alternative meaning must be specified if intended, like: &quot;the distance &lt;em&gt;from the surface&lt;/em&gt; of the Sun to Earth&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It may however also mean the &quot;distance between surfaces&quot;, especially when the two objects are very close to each other. This is often the case with &lt;a href=&quot;https://en.wikipedia.org/wiki/Binary_asteroid&quot; rel=&quot;nofollow noreferrer&quot;&gt;binary asteroids&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For  an orbital spacecraft, the word 'distance' is mostly not used when referring to the central object. The terms 'altitude' or 'orbital radius' are preferred. Altitude is the distance to the surface (or rather, to a theoretical sealevel. It may also refer to the actual surface though, so be careful). Orbital radius is the distance to the centre of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note &quot;average distance&quot; in an orbit is referring to the semi-major axis as that is the number used for calculating the orbital period and different velocities.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastEditorUserId="10172" LastEditDate="2017-04-25T15:16:54.203" LastActivityDate="2017-04-25T15:16:54.203" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20835" PostTypeId="1" AcceptedAnswerId="20864" CreationDate="2017-04-24T16:02:46.230" Score="2" ViewCount="119" Body="&lt;p&gt;Let us suppose a solar system with a Hot/Warm Jupiter HJ (with a semi-major axis somewhere from 0.5 AU to 0.0001 AU) and an Earth-Like planet EL in it (with a semi-major axis of about 1 AU). Let their orbital inclinations be a couple of degrees apart. Let the star be a G2V or so, like our Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://arxiv.org/abs/1003.2206&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; and other sources, we can tell that Hot Jupiters are &quot;bleeding&quot; atmosphere due to their proximity to their sun. Some of it, I presume, gets pushed out with the solar wind farther out into the solar system, at least as far as the EL is.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What effect will this bled atmosphere from HJ have on EL, if any?&lt;/li&gt;&#xA;&lt;li&gt;If there is an effect, when does it happen? On all of EL's orbit, despite the different inclinations? Only when EL crosses a transit node? Only when there's a transit (or rather, when EL passes through the transit node a certain amount of time after HJ does, to account for the velocity of the atmospheric particles through space)?&lt;/li&gt;&#xA;&lt;li&gt;Will this bled atmosphere be (naked-eye) visible from EL? What would it look like?&lt;/li&gt;&#xA;&lt;li&gt;Would all of this have any effect on life/civilization on the EL?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14931" LastEditorUserId="14931" LastEditDate="2017-04-24T19:27:05.563" LastActivityDate="2017-04-26T23:40:00.173" Title="Effect of Hot/Warm Jupiter atmosphere loss on other planets" Tags="&lt;planetary-atmosphere&gt;&lt;hot-jupiter&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20836" PostTypeId="2" ParentId="20823" CreationDate="2017-04-24T17:20:53.063" Score="1" Body="&lt;p&gt;Current understanding is that star is an &lt;em&gt;oblate spheroid&lt;/em&gt;. An extreme example is shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U3CA5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U3CA5.png&quot; alt=&quot;oblate spheroid&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a neutron star, the difference between the polar diameter and the equatorial diameter is around 10% and would look more like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J8Gar.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J8Gar.png&quot; alt=&quot;neutron&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16684" LastActivityDate="2017-04-24T17:20:53.063" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20837" PostTypeId="2" ParentId="20831" CreationDate="2017-04-24T17:31:28.283" Score="1" Body="&lt;p&gt;Simple answer: the matter density is incredibly high. The energy density may permit coherent scattering of the neutrinos among the nucleons. There is a lot of matter, and it is not transparent to neutrinos. The mean free path for scattering is very small, and the neutrinos are bouncing back and forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.int.washington.edu/talks/WorkShops/RIA/People/Liebendoerfer_M/liebendoerfer.RIA.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Neutrino Transport in Core Collapse Supernovae&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://arxiv.org/pdf/nucl-th/0211003.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Neutrino mean free path and in-medium nuclear interaction&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16684" LastEditorUserId="907" LastEditDate="2017-04-24T18:08:44.457" LastActivityDate="2017-04-24T18:08:44.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20839" PostTypeId="1" CreationDate="2017-04-24T18:29:19.967" Score="1" ViewCount="249" Body="&lt;p&gt;I see that there was already about Pluto's orbital inclination, but I would like to ask a more general question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do planets have orbital inclinations?  And is there any particular reason, other than chance, for Mercury's inclination being so small?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="16696" LastActivityDate="2017-04-24T19:44:30.163" Title="Reasons for orbital inclinations" Tags="&lt;inclination&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20840" PostTypeId="1" AcceptedAnswerId="26328" CreationDate="2017-04-24T18:34:20.110" Score="2" ViewCount="230" Body="&lt;p&gt;I understand that the kappa mechanism (that leads to star variability) causes an increase of opacity with increasing temperature in partial ionization zones.  However, I'm not sure I understand what causes the increasing opacity with temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until recently, I thought that the increasing opacity was due to increasing Thompson opacity due to the increasing ionization fraction with temperature in partial ionization zones.  However, some things I read suggested that that was not the cause.  So I wanted to ask, what is the cause of increasing opacity with temperature in partial ionization zones?&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2018-05-19T03:16:42.730" Title="What leads to increase in opacity in kappa mechanism?" Tags="&lt;variable-star&gt;&lt;opacity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20841" PostTypeId="2" ParentId="20839" CreationDate="2017-04-24T19:44:30.163" Score="1" Body="&lt;p&gt;The planets formed from a disc of dust and gas. Since the orbiting the proto-sun formed into a disc the planets, which formed from this matter are in a disc too, but not a perfect disc. There was some scattering of the matter in the primordial disc, and so the planets are also slightly scattered. Interactions between the planets have also scattered the planets out from the disc too. The formation of the planets caused them to have small, but not exactly zero inclination&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth has the greatest inclination (except pluto) however, by convention, we measure inclination relative to the Earth, all the planets have a positive inclination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is no special reason that the planets have slight inclinations. It is just what would be expected due to random effects during their formation. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-04-24T19:44:30.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20842" PostTypeId="2" ParentId="20831" CreationDate="2017-04-24T19:49:20.367" Score="6" Body="&lt;p&gt;You are barking up the wrong tree. The passage you quote is talking about how neutron stars cool &lt;em&gt;after&lt;/em&gt; they have formed. Neutrinos are emitted in a burst lasting a few seconds after core collapse but then continue to be produced whilst the neutron star is hot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutrino production is indeed fearsomely high in the first fraction of a second after core collapse. However, the hot, dense material is opaque to neutrinos (mean free paths of 10-100 m) and they diffuse outward in much the same way as photons from the Sun. But the timescale for this is only about 10 seconds. Once the neutron star has cooled to about $10^{10}$ K, the fermions (neutrons, protons, electrons) in the neutron star become degenerate and the neutron star becomes transparent to neutrinos because only the small fraction of fermions within $kT$ of their Fermi surfaces can interact with thermal neutrinos and the neutrino mean free paths increase as $\sim T^{-2}$ and the production rate of neutrinos falls as $T^6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, the neutron star would then radiate away all of its thermal energy within a further 10 seconds, but the main neutrino-generating reactions - cycles of beta and inverse beta decay (aka the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Urca_process&quot; rel=&quot;noreferrer&quot;&gt;Urca process&lt;/a&gt;) become blocked (except perhaps right in the core at the highest densities) by an inability to simultaneously conserve energy and momentum in the degenerate gas. Instead, neutrinos continue to be generated by the modified Urca process at a &lt;em&gt;much&lt;/em&gt; slower rate, using &quot;bystander&quot; baryons to conserve momentum.&#xA;$$ n + n \rightarrow n + p + e + \bar{\nu}_e$$&#xA;$$ n + p + e \rightarrow n + n + \nu_e$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though the efficiency of this process is low, so is the thermal energy contained within degenerate gases. Thus rapid cooling still occurs. I am not clear where Wikipedia gets its information from, but cooling to even a surface temperature of $10^{6}$K (the interiors are 1-2 orders of magnitude hotter) is likely to take longer than a few years - somewhere between 100 and $10^4$ years, unless direct URCA processes are somehow permitted (e.g. in quark matter, see &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0402143&quot; rel=&quot;noreferrer&quot;&gt;Yakovlev &amp;amp; Pethick 2004, for a thorough review of neutron star cooling&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-04-24T20:18:23.607" LastActivityDate="2017-04-24T20:18:23.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20844" PostTypeId="1" AcceptedAnswerId="20852" CreationDate="2017-04-24T23:46:20.660" Score="2" ViewCount="141" Body="&lt;p&gt;I would like to know, how strong would the gravity be from the distant universe if we were somehow able to remove half of it.  For example, if everything behind us did not exist, how strongly would the stuff in front of us be pulling?  By distant, let us consider everything beyond 10 billion light years (e.g. remove all the closer stuff).  How many G's acceleration would I experience (discounting the fact that I would be in free fall and wouldn't feel it)?&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2017-04-25T13:58:53.327" Title="How strong is the gravity from half of the distant universe?" Tags="&lt;gravity&gt;&lt;dark-energy&gt;&lt;cosmological-horizon&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20845" PostTypeId="1" AcceptedAnswerId="20849" CreationDate="2017-04-25T07:05:35.413" Score="3" ViewCount="1817" Body="&lt;p&gt;It's a fairly simple question (at least in my layman's mind). Can a radio telescope that has no mirrors or lenses to capture visible light (to human eyes) still produce imagery like an optical telescope such as the Hubble? I'm talking just ROYGBIV light (no infrared or UV).&lt;/p&gt;&#xA;" OwnerUserId="10117" LastActivityDate="2017-04-29T09:12:22.687" Title="Can radio-based telescopes produce visible light images?" Tags="&lt;telescope&gt;&lt;hubble-telescope&gt;&lt;radio-telescope&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20846" PostTypeId="2" ParentId="20845" CreationDate="2017-04-25T07:41:26.450" Score="0" Body="&lt;p&gt;Here are images produced with radio waves in RGB, but they are frequency-inaccurate. YES, Radio telescopes can produce imagery... NO, radios are not in the visible range, otherwise a talky walky areal would glow like a lamp when it transmits information. No, Radio antenna don't capture photons electrically and it is their voltage change which is measured, they can only get hot and malfunction from light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is mystifying... You must read about electromagnetic spectrum energy!!! it's simple!! it is the source of all collected astronomy data and studies. It's like subtraction and division in maths, electromagnetic spectrum is fundamental. If you mix up photons and radio waves, it's like you are mixing subtractions and divisions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bubbles here are supernova remnants, on the right is a radio galaxy.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/7Y83i.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Y83i.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The colors here are red to green based on real radio wave length according to the researcher, however i think it's a compilation of multiple images of different radio periods. source is 70-230mhz = 1.2meters to 4meters&#xA;&lt;a href=&quot;https://i.stack.imgur.com/s38MB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/s38MB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.ted.com/talks/natasha_hurley_walker_how_radio_telescopes_show_us_unseen_galaxies&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.ted.com/talks/natasha_hurley_walker_how_radio_telescopes_show_us_unseen_galaxies&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2017-04-29T09:12:22.687" LastActivityDate="2017-04-29T09:12:22.687" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20847" PostTypeId="1" CreationDate="2017-04-25T10:04:26.117" Score="3" ViewCount="569" Body="&lt;p&gt;According to cosmological constant, dark energy should have a constant density but since the universe is expanding so is the volume of the universe and so should the mass of dark energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how does dark energy gain mass or energy?&lt;/p&gt;&#xA;" OwnerUserId="16649" LastEditorUserId="1038" LastEditDate="2017-04-25T11:36:42.317" LastActivityDate="2017-04-26T10:20:06.643" Title="How does dark energy remain constant?" Tags="&lt;universe&gt;&lt;expansion&gt;&lt;dark-energy&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20848" PostTypeId="1" CreationDate="2017-04-25T10:06:24.813" Score="1" ViewCount="39" Body="&lt;p&gt;The cycle of solar maximums and solar minimums are each about 11 years, which is close to the orbital period of Jupiter. I thought it was a numerical coincidence, until google turned up some results of a possible connection. A lot of it is beyond my head though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was under the impression that there are gaps in our knowledge regarding Jupiter's core, which would likely be the driver of its magnetic field. And since solar cycles are related to sunspot activity, which in turn is related to the magnetic field of the Sun, I am struggling to find even a basic understanding of the cause-effect relationship (if there is one). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a connection between Jupiter's orbital period and the Sun's solar cycles? Or is it blind speculation?&lt;/p&gt;&#xA;" OwnerDisplayName="user15317" LastActivityDate="2017-04-25T10:06:24.813" Title="Is there a connection between Jupiter's orbital period and the Sun's solar cycles?" Tags="&lt;the-sun&gt;&lt;jupiter&gt;&lt;magnetic-field&gt;&lt;orbital-mechanics&gt;&lt;core&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2017-04-25T19:21:41.600" ContentLicense="CC BY-SA 3.0" />
  <row Id="20849" PostTypeId="2" ParentId="20845" CreationDate="2017-04-25T10:10:37.160" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can a radio telescope that has no mirrors or lenses to capture visible light (to human eyes) still produce imagery like an optical telescope such as the Hubble?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course, either via false color or pseudocolor. A false color radio image maps various frequencies (typically three) in a multispectral radio image to visible colors. A pseudocolor radio image maps various intensities in a single spectrum gray scale radio image to visible colors. The same techniques are also used for gamma, X-ray, UV, IR, and microwave imagery. We can't see in those frequencies, so something needs to be done to the captured imagery to make them visible to us.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If the question is asking whether a radio telescope can be used to capture visible light, the answer is no. I'll start with the receiver. It's designed to capture long wavelength radiation. Photons in the visible spectrum will have minimal, if any, impact on a radio telescope receiver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An even bigger problem with some radio telescopes is depicted below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FRLBV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FRLBV.jpg&quot; alt=&quot;Radar telescope at the Haystack Observatory. The telescope is hidden from view, inside the large radome.&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sup&gt;Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Haystack_Observatory&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Haystack_Observatory&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is the Haystack Radio Telescope at the Haystack Observatory in Westford, Massachusetts. The radome protects the telescope against weather, but also hides the antenna from optical light. A couple of other radio telescopes at the Haystack Observatory show yet another problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DmaHl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DmaHl.jpg&quot; alt=&quot;Two more radar telescope antennae at the Haystack Observatory. These are not protected by a radome. They however are see-thru in the visible portion of the spectrum. The sky can be seen through the movable antenna, and trees and hills can be seen through the fixed zenith antenna.&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sup&gt;Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Haystack_Observatory&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Haystack_Observatory&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fixed zenith telescope antenna in the foreground and the fully movable telescope antenna in the background are made of a wire mesh, making them see-thru in the visible portion of the spectrum. This wire mesh construction is quite common for long wavelength radio antennae because it allows the wind to blow through it. Shorter wavelength radio antennae are more likely to be solid, but those are oftentimes painted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A wire mesh with wires separated by centimeters is very smooth to a meter long radio wave, as solid surface smooth at the millimeter level to a centimeter long radio wave. A surface with millimeter-scale roughness would make for an extremely lousy optical mirror. There is no reason to make a radio telescope antenna smooth to the visible portion of the spectrum, and there are many millions of reasons (i.e., many millions of dollars) not to do so.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2017-04-25T15:01:40.747" LastActivityDate="2017-04-25T15:01:40.747" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20850" PostTypeId="2" ParentId="20847" CreationDate="2017-04-25T10:19:19.453" Score="2" Body="&lt;p&gt;Not all models of dark energy assume it constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when this is assumed true, dark energy is usually compared to the energy density of the vacuum. In other words, while the Universe expands, the dark energy increases proportionally, so its ratio with the Universe volume (the density) stays constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, better, we generally believe those models where the dark energy is a constant because they fit better our observations of the Universe, in particular the Big Bang model, which describes how our Universe evolves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you like some math, &lt;a href=&quot;https://ned.ipac.caltech.edu/level5/March01/Carroll/Carroll1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;you can have it here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastActivityDate="2017-04-25T10:19:19.453" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20851" PostTypeId="2" ParentId="20847" CreationDate="2017-04-25T13:53:28.277" Score="0" Body="&lt;p&gt;You're missing the point:  it's called &quot;dark energy&quot; not because it's dark or because it's energy.  The name is the cosmological equivalent of the &quot;black box&quot; in a laboratory setting.  Until we figure out what the heck it is, there's no reason for it to follow our known laws. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-04-25T13:53:28.277" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20852" PostTypeId="2" ParentId="20844" CreationDate="2017-04-25T13:53:28.803" Score="5" Body="&lt;p&gt;Not very strong at all. I get a rough figure of $3.725\times 10^{-9} m/s^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To perform that calculation I made a few simplifying assumptions.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Assume that we can ignore everything outside the &lt;a href=&quot;https://en.wikipedia.org/wiki/Observable_universe&quot; rel=&quot;noreferrer&quot;&gt;observable&#xA;universe&lt;/a&gt;.   &lt;/li&gt;&#xA;&lt;li&gt;Assume that the observable universe is a static homogeneous sphere&#xA;with a radius of 46.6 billion light-years, and a mean density of&#xA;$\rho = 0.85\times 10^{-26} kg/m^3$, that figure is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Observable_universe#Estimates_based_on_critical_density&quot; rel=&quot;noreferrer&quot;&gt;critical&#xA;density&lt;/a&gt;&#xA;needed for the global curvature of the universe to be zero (i.e., on&#xA;the very large scale spacetime is flat).   &lt;/li&gt;&#xA;&lt;li&gt;Assume that we can use Newtonian physics to calculate the&#xA;acceleration due to gravity.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I didn't bother removing all the stuff closer than 10 billion light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're standing at the centre of the flat surface of a hemisphere of radius $r$ and we want to know how strong the gravity is. We can pretend that all the mass of the hemisphere is concentrated at its centre of mass, which is $\frac{3r}{8}$ below us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Newtonian formula for gravitational acceleration is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a = \frac{GM}{r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass is volume times density, i.e., $M=V \rho$ and the volume of a hemisphere is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V = \frac{2}{3}\pi r^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining everything, not forgetting to use $\frac{3r}{8}$ as the distance term in the denominator of the acceleration formula, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a = G \rho \frac{128}{27} \pi r$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging in the values I mentioned earlier for $\rho$ and $r$, the &lt;a href=&quot;https://www.google.com/#q=G*0.85E-26(kg/m%5E3)*128/27*pi*(46.6E9+light-years)&quot; rel=&quot;noreferrer&quot;&gt;Google Calculator&lt;/a&gt; says $3.72484086\times 10^{-9} m/s^2$. That's about $3.8\times 10^{-10}$ standard gravities.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="16685" LastEditDate="2017-04-25T13:58:53.327" LastActivityDate="2017-04-25T13:58:53.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20853" PostTypeId="2" ParentId="20845" CreationDate="2017-04-25T13:57:34.513" Score="-2" Body="&lt;p&gt;The answer is either &quot;no, of course not,&quot;  or &quot;yes, in a pathological sense.&quot;  A radio &quot;telescope&quot; captures photons which are capable of forcing electromagnetic responses in metal -- which is to say, the induced voltage caused by an E-M field. The statistical odds of a visible-range photon causing an induced signal of detectable level in a metal antenna are ridiculuously small.  &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-04-25T13:57:34.513" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20854" PostTypeId="2" ParentId="13178" CreationDate="2017-04-25T17:01:17.257" Score="1" Body="&lt;p&gt;As another answer says, there &lt;em&gt;is&lt;/em&gt; a tendency to do a lot of algebra involving planets' periods. Another approach however is examining the timing of actual solar events. In &lt;a href=&quot;https://arxiv.org/abs/1702.00641&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; case, solar flares and their time distribution is related to the relative motion of Jupiter and Saturn, providing predictions. &lt;/p&gt;&#xA;" OwnerUserId="16462" LastActivityDate="2017-04-25T17:01:17.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20855" PostTypeId="2" ParentId="2439" CreationDate="2017-04-25T17:31:01.380" Score="3" Body="&lt;p&gt;There exists various theories, some stating that &lt;a href=&quot;https://www.insidescience.org/news/every-black-hole-contains-new-universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;every black hole contains a new universe&lt;/a&gt;... In fact as a once very dense object, our universe could also be seen as a black hole itself!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About what matter's enthropy becomes, Hawking proposes that it stays on the event horizon as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Holographic_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;holographic structure&lt;/a&gt;, also compared to some &lt;a href=&quot;http://www.livescience.com/53363-stephen-hawking-black-holes-have-hair.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;kind of electric hairs&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16702" LastActivityDate="2017-04-25T17:31:01.380" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20856" PostTypeId="1" AcceptedAnswerId="20861" CreationDate="2017-04-26T06:50:56.787" Score="0" ViewCount="774" Body="&lt;p&gt;My wife just bought me an &lt;a href=&quot;http://www.celestron.com/browse-shop/astronomy/telescopes/astromaster-telescopes/astromaster-114eq-telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;AstroMaster 114EQ&lt;/a&gt; which came with an &quot;upgraded starpointer finderscope&quot; -- it looks different than the one depicted in that link. I have tried following the instructions in their documentation to calibrate the finderscope with the primary telescope image and, unfortunately, the small red dot in the finderscope simply will not align with the main image of the telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In attempting to calibrate it, I pointed the scope at a building about a mile and a half away and got it clearly focused in the main scope. I then looked through the starpointer finderscope and tried to adjust it but the range of adjustment would not allow me to line up the red dot with this building -- the dot was high and to the right and the adjustment knobs simply would not turn any more -- they were at the end of their adjustable range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The supplementary documentation for the &lt;a href=&quot;http://www.celestron.com/browse-shop/astronomy/accessories/navigation/star-pointer-finderscope&quot; rel=&quot;nofollow noreferrer&quot;&gt;starpointer finderscope&lt;/a&gt; that came with the scope doesn't tell you which end points toward your target and which toward your face. I cannot really tell from the images in their website which orientation is correct. The mount doesn't seem to enforce any particular orientation. I assumed that the lense should be at the far end of the scope toward your objective and the power/brightness knob and the up/down adjustment knob, should be close to your face.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone suggest how I might remedy this? I did search here first and this &lt;a href=&quot;https://astronomy.stackexchange.com/questions/19568/celestron-astromaster-114eq-star-pointer-finder-scope-help-required&quot;&gt;other post&lt;/a&gt; did not seem to have the information I was after. I'm supposed to go camping the day after tomorrow and would love to be able to locate Jupiter. I'm concerned that the scope may have been incorrectly assembled...i.e., the mount is off or something.&lt;/p&gt;&#xA;" OwnerUserId="16705" LastActivityDate="2017-04-26T17:27:59.190" Title="Cannot calibrate “upgraded starpointer finderscope” for Celestron AstroMaster 114EQ" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;newtonian-telescope&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20857" PostTypeId="2" ParentId="20847" CreationDate="2017-04-26T10:20:06.643" Score="0" Body="&lt;p&gt;The energy of universe increases, that energy is contributed to dark energy.&#xA;In order for accelerated expansion of universe, dark energy would have to have negative pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to first law of thermodynamics $\mathrm{d}U=\mathrm{d}Q-P\mathrm{d}V$,since the universe is an isolated system and the heat doesn't change $\mathrm{d}Q=0$, so it's expected by thermodynamics that the universe gains energy which is attributed to dark energy.&lt;/p&gt;&#xA;" OwnerUserId="16649" LastActivityDate="2017-04-26T10:20:06.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20858" PostTypeId="1" AcceptedAnswerId="20863" CreationDate="2017-04-26T10:56:11.777" Score="11" ViewCount="4410" Body="&lt;p&gt;So correct me if I'm wrong, but when searching for extra terrestrial life, scientists usually search for signs they would expect for life on earth to thrive in. We look for conditions on planets that meet certain criteria e.g close enough to a sun for water, environments for carbon based life forms to exist. I was wondering; why cant an alien be a sulphur based life form that for the sake of argument drinks lead and breathes in chlorine like we do oxygen.&lt;/p&gt;&#xA;" OwnerUserId="16708" LastEditorUserId="2365" LastEditDate="2017-10-01T12:49:36.680" LastActivityDate="2019-09-30T23:22:46.230" Title="Why do aliens have to be carbon based lifeforms?" Tags="&lt;astrobiology&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20860" PostTypeId="1" CreationDate="2017-04-26T17:26:06.507" Score="0" ViewCount="81" Body="&lt;p&gt;We might expect to not find life around halo stars (“Population II” stars) because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a)&#xA;they are out of the plane of the galaxy and thus exposed to more cosmic rays (energetic&#xA;nuclei)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b)&#xA;these stars are all too hot and thus will expose any planets to too much ultraviolet radiation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c)&#xA;these stars lack the heavy elements (“metals”) out of which life can form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d)&#xA;these stars have existed too short a time for life to have evolved&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e)&#xA;the stars all pass too close to the central supermassive black hole in the Milky Way&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am stumped on this question, any ideas?&lt;/p&gt;&#xA;" OwnerUserId="16712" LastEditorUserId="2365" LastEditDate="2017-04-26T17:46:12.653" LastActivityDate="2017-04-26T17:53:48.503" Title="Help with homwork - Astronomy" Tags="&lt;radio-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20861" PostTypeId="2" ParentId="20856" CreationDate="2017-04-26T17:27:59.190" Score="1" Body="&lt;p&gt;I do not have one of these scopes, so my answer reflects general knowledge of astronomical telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, the lens should be at the far end, away from your eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the alignment question, you can probably get a bit more movement in &quot;left/right&quot; direction by loosening the screws that hold the base of the finder to the scope, and twisting the mount and then re-tighten it.  For the up/down adjustment, you may need to use a shim (a thin bit of metal) to adjust the angle, but the adjustment built into the finder is normally adequate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I doubt if there is a problem with the scope - if you can get a good focus, that's fine.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-04-26T17:27:59.190" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20862" PostTypeId="2" ParentId="20860" CreationDate="2017-04-26T17:53:48.503" Score="4" Body="&lt;p&gt;You will need to find out what &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_population#Population_II_stars&quot; rel=&quot;nofollow noreferrer&quot;&gt;population II &#xA; stars&lt;/a&gt; are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you understand what a population II star is, ask yourself two questions about each statement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example a) &quot;Are population II stars out of the plane of the galaxy, and do they receive more cosmic rays&quot;, and &quot;Would receiving more cosmic rays actually prevent life from forming&quot;. If you can answer yes to both questions, you have a potential answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you judge that several answers are possible, then re-read the definition of a population II star, and ask which statement is most pertinent to the definition of such a star. Good luck.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-04-26T17:53:48.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20863" PostTypeId="2" ParentId="20858" CreationDate="2017-04-26T21:08:49.180" Score="20" Body="&lt;p&gt;&lt;sub&gt;I want to start with a disclaimer that I'm not a chemist and hopefully I don't screw any of the chemistry up. If I do, please let me know.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think &lt;a href=&quot;https://astronomy.stackexchange.com/questions/20858/why-do-aliens-have-to-be-carbon-based-lifeforms#comment35226_20858&quot;&gt;Bill Oertell&lt;/a&gt; hit the nail on the head. Just because you can imagine a species which can drink lead doesn't mean it's actually physically possible. Only specific atoms can form the basis of the (extremely) complex organic molecules needed to form life and Carbon happens to be the atom which can do that. The primary reason for this is that Carbon has four valence electrons and its outer shell can hold eight, allowing it to form strong and varied bonds that most other atoms just can't form. As such, Carbon is the most logical element to form the backbone of organic molecules.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Other ___-based Lifeforms&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Moving down the periodic table, you can see that Silicon has the same properties (4 valence elections, 8 possible) and you might be inclined to think that you could form some sort of silicon based life. This topic has been &lt;a href=&quot;https://en.wikipedia.org/wiki/Hypothetical_types_of_biochemistry#Silicon_biochemistry&quot; rel=&quot;noreferrer&quot;&gt;widely considered&lt;/a&gt;, but you run into a number of problems that make Silicon-based life unfeasible. The primary reason is that Silicon is just much bigger and more massive than Carbon (having a nucleus with 28 nucleons compared to Carbon's 12). This makes it harder for Silicon to form the necessary bonds that Carbon can readily form and you're out of luck right form the get-go. Similar issues occur with other &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Carbon_group&quot; rel=&quot;noreferrer&quot;&gt;Carbon group&lt;/a&gt;&quot; elements, except they're even further exacerbated by even higher masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many other &quot;organic&quot; configurations chemists have considered such as boron-based life or sulfur-based life but they all have some sort of problem that makes it hard for it to exist. Some are more possible than others, but Carbon-based life, as we know it, just seems to be the best solution. If you gave someone all the elements in the periodic table and said, make organic life, they'd probably come back with Carbon-based life because its the easiest. You don't have to overcome as many obstacles.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Observations&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Aside from the impracticality of non-Carbon based life, you have to consider that everything we've seen in the universe, as far as organic chemistry goes, points to the fact that the Universe is filled with Carbon-based chemistry. Astronomers have been able to look out at comets, interstellar dust clouds, and all sorts of places and time after time, they see complex, Carbon-based, organic molecules. For example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.astrochem.org/docs/IAU251_Salama.pdf&quot; rel=&quot;noreferrer&quot;&gt;We've found Polycyclic Aromatic Hydrocarbons (PAHs) all over the solar system and galaxy&lt;/a&gt;,&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.bbc.com/news/science-environment-29368984&quot; rel=&quot;noreferrer&quot;&gt;Astronomers have also found things like i-propyl cyanide&lt;/a&gt; in meteorites and in interstellar dust $-$ an organic molecule which is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Isopropyl_cyanide#Importance_in_life.27s_origin&quot; rel=&quot;noreferrer&quot;&gt;necessary ingredient in amino acids&lt;/a&gt;,&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.popularmechanics.com/space/solar-system/news/a25235/ceres-organics/&quot; rel=&quot;noreferrer&quot;&gt;Organic molecules have been found (and believed to have formed) on Ceres.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The point I'm driving at here is that the universe seems ready-made to start forming Carbon-based life everywhere. Everywhere we look, we see organic molecules which have Carbon as the backbone of their structure. Nowhere do we see complex silicon based molecules (or something even crazier). This is highly suggestive that Carbon is the way to go if you want to quickly and easily form life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are probably a 100 other reasons to favor Carbon over something else, but that delves into more complicated chemistry than I've touched on here and I'm not as knowledgeable about that. I would think this question would get a more in-depth chemistry answer on the &lt;a href=&quot;https://chemistry.stackexchange.com/&quot;&gt;chemistry stack exchange&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-04-26T21:08:49.180" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20864" PostTypeId="2" ParentId="20835" CreationDate="2017-04-26T23:40:00.173" Score="1" Body="&lt;p&gt;The primary effects wouldn't be the things you ask about.   First, an object the size of Jupiter would change the day and night sky drastically.    The sky now is basically, Sun, Moon and a bunch of yellow dots (and one reddish dot) mostly fixed, 4 move around visibly to the naked eye (Venus, Mars, Jupiter, Saturn).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus stands out a little as larger and brighter, but it looks pretty much like an abnormally bright star.   All that changes if you put a Jupiter sized planet in the orbits around where Mercury or Venus are.   Jupiter is about 1/10th the diameter of the Sun.   If you put it about midway between Earth and the Sun it would be about 1/5th the diameter and transits over the sun would be obvious, unlike &lt;a href=&quot;https://en.wikipedia.org/wiki/Transit_of_Venus&quot; rel=&quot;nofollow noreferrer&quot;&gt;Venus transits&lt;/a&gt; which are observable (not a good idea without protective lenses), but easily missed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter in an orbit between the Earth and the Sun would have apparent phases like the Moon and visible size variation.  It's likely that in such a scenario, the geocentric model would never have been adapted, cause Jupiter would raise too many questions in this hypothetical.   Aristotle and Ptolomy were smart.   This situation would have been enough to perhaps look at things differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also the 3 body gravitation problem.   Jupiter inside the Earth's orbit might wreck havoc with the Earth's orbit.  Earth's eccentricity might vary a lot more perhaps leading to more extreme climate change periods.   There's a good chance an Earth like planet's relatively stable orbit would be subject too a lot of undesirable changes with a hot Jupiter between it and the sun.  Comet impacts could be a problem too.   A distant Jupiter changing the orbit of passing comets helps the Earth avoid comet impacts.  A Jupiter that close to the sun could do the opposite.    There's also the migration problem.   If Jupiter forms out far from the sun and migrates inwards, that would wreck havoc with any planet orbits it passes.    So there's a huge gravity problem with your scenario.   If you're looking for Earth like planets, solar-systems with hot Jupiters are probably not the best place to look.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;that said, onto your questions.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What effect will this bled atmosphere from HJ have on EL, if any?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If there is an effect, when does it happen? On all of EL's orbit,&#xA;  despite the different inclinations? Only when EL crosses a transit&#xA;  node? Only when there's a transit (or rather, when EL passes through&#xA;  the transit node a certain amount of time after HJ does, to account&#xA;  for the velocity of the atmospheric particles through space)?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Will this bled atmosphere be (naked-eye) visible from EL? What would&#xA;  it look like?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's difficult to say with precision.   Certainly if the bleeding was significant enough it might be visible, but on average I think probably not.   As a bad comparison, we can't see Coronal Mass Ejections.   Most of the &quot;bleeding&quot; from a hot Jupiter would be surface gas molecules, primarily hydrogen and helium, most of it uncharged.  Those gas molecules, spread out in space would be essentially transparent, unless it was an absolutely gonzo insane amount.    Very sensitive cameras might pick them up, but they should be largely invisible from Earth.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comet tails are visible, even occasionally visible during the day, but comet tails are different.  They're full of dust that breaks off the comet and the escape velocity off a comet is low, so the dust that breaks off from the comet tends to follow in a similar orbit, hence the relative density and apparent cohesion of a comet tail.   With a hot Jupiter and gas escaping in a variety of directions presumably due to heat escape or Jeans escape primarily, there would be very little cohesion.   Escaped gas molecules would quickly spread out in space.   The charged particles that escape from this theoretical Jupiter might be enough to create a faint Aurora, but should be significantly smaller than the charged particles escaping the sun.   The uncharged particles, a small percentage would collide with Earth, most would miss.   There's no scenario I can imagine that would be visibly impressive from such bleeding unless the hot Jupiter was super-close to the sun, say closer than Mercury and the bleeding was extreme.   Even then, the primary visual would be the stream of gas falling into the sun, not the spread of gas moving towards and past the Earth's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something not too different happened in the Earth's distant past.  Venus used to have oceans and it bled probably at least an Atlantic ocean of water, perhaps a Pacific ocean of water into space that drifted mostly past the Earth.  Likely there was very little effect to Earth drifting through a very disburse amount of water-vapor in it's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a ballpark answer anyway.  Corrections are welcome.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-04-26T23:40:00.173" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20865" PostTypeId="1" AcceptedAnswerId="20874" CreationDate="2017-04-27T08:35:00.807" Score="3" ViewCount="232" Body="&lt;p&gt;In the Los Angeles Times news item &lt;a href=&quot;http://www.latimes.com/science/sciencenow/la-sci-sn-supernova-gravitational-lensing-20170420-story.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Scientists get a rare view of a type Ia supernova magnified 50 times&lt;/a&gt; &lt;em&gt;what exactly is magnified 50 times?&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This supernova is really very far away. Is it somehow imaged by the gravitational lensing and being resolved — the actual supernova itself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit:&lt;/strong&gt; I believe this is it, but it's behind a paywall (will confirm in due course): &lt;a href=&quot;http://science.sciencemag.org/content/356/6335/291&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://science.sciencemag.org/content/356/6335/291&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FfdWV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FfdWV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; From the &lt;a href=&quot;http://www.latimes.com/science/sciencenow/la-sci-sn-supernova-gravitational-lensing-20170420-story.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Los Angeles Times&lt;/a&gt;. Photo credit: Joel Johansson&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PSXKk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PSXKk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; &quot;This schematic image represents how light from a distant galaxy is distorted by the gravitational effects of a nearer foreground galaxy, which acts like a lens and makes the distant source appear distorted, but brighter, forming characteristic rings of light, known as Einstein rings. An analysis of the distortion has revealed that some of the distant star-forming galaxies are as bright as 40 trillion Suns, and have been magnified by the gravitational lens by up to 22 times. (Credit: ALMA ESO/NRAO/NAOJ, L. Calçada (ESO), Y. Hezaveh et al., edited and modified by Joel Johansson)&quot; From the &lt;a href=&quot;http://www.latimes.com/science/sciencenow/la-sci-sn-supernova-gravitational-lensing-20170420-story.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Los Angeles Times&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-04-28T11:43:18.530" LastActivityDate="2017-04-28T11:43:18.530" Title="What exactly is it that is being magnified 50 times in this gravitational lensing observation?" Tags="&lt;supernova&gt;&lt;gravitational-lensing&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20866" PostTypeId="2" ParentId="20865" CreationDate="2017-04-27T13:24:50.833" Score="-3" Body="&lt;p&gt;It appears to just be a regular photo. It could've been taken by Hubble, and someone just magnified the photo. It doesn't appear to be gravitational lensing. If it was you would be able to see space being bent a bit in the photo, &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_lens&quot; rel=&quot;nofollow noreferrer&quot;&gt;read here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16723" LastActivityDate="2017-04-27T13:24:50.833" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20867" PostTypeId="2" ParentId="20865" CreationDate="2017-04-27T13:44:33.733" Score="3" Body="&lt;p&gt;Reading through &lt;a href=&quot;http://science.sciencemag.org/content/sci/356/6335/291.full.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;the full article&lt;/a&gt;, I see that the authors write&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;we estimated the lensing amplification to be $\mu\sim52$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This refers to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_lensing_formalism#Magnification&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;magnification factor&lt;/em&gt;&lt;/a&gt;. Essentially, it describes the solid angle of the image, as related to the solid angle of the source&lt;sup&gt;&lt;a href=&quot;http://www.pa.msu.edu/~abdo/GravitationalLensing.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;&lt;/sup&gt;:&#xA;$$\mu\equiv\frac{\theta}{\beta}\frac{d\theta}{d\beta}$$&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/File:Gravitational-lensing-angles.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's&lt;/a&gt; a diagram of the magnitudes of the angles:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/e/e4/Gravitational-lensing-angles.png&quot; alt=&quot;&quot;&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;Image courtesy of Wikipedia user Falcorian under &lt;a href=&quot;https://creativecommons.org/licenses/by-sa/3.0/deed.en&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Creative Commons Attribution-Share Alike 3.0 Unported license&lt;/a&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2017-04-27T13:44:33.733" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20868" PostTypeId="1" CreationDate="2017-04-27T13:55:31.353" Score="5" ViewCount="230" Body="&lt;p&gt;A rather stupid question but I was just curious if the radiation could be virtual.  From the instrument....sort of like virtual particles. In other words if you just build a sensitive microwave detector and stick it in a South African Durban Deep gold mine 5 miles under the earth would it do the same thing? &lt;/p&gt;&#xA;" OwnerUserId="16724" LastActivityDate="2018-11-11T19:08:34.047" Title="Has the microwave background radiation ever been blocked in order to prove it's really from the sky?" Tags="&lt;radio-astronomy&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20869" PostTypeId="2" ParentId="20858" CreationDate="2017-04-27T14:38:46.030" Score="2" Body="&lt;p&gt;Carbon can hold one of the most stable 4 bond connections out of the elements. We see how life is shaped on earth and based on the properties of life here, we make educated guesses on what life must be based around in other parts of the universe, given that there is other life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the second part of the second, the reason we breathe in oxygen is to allow certain cellular functions to take place, the most important reason is respiration. C6H12O6 + O2 yields CO2 + H2O + 36-38 ATP. ATP is the molecule the body uses as energy. Oxygen is essential for us because it's used in the chemical reaction that gives our body energy. If an alien were to be breathing chlorine, it would need to have a chemical process that uses the chlorine in their body. &#xA;As for there being sulphur-based life, I don't know whether sulfur is stable enough to hold the same bonds as carbon. From what scientists know, carbon is the best universal organic base, because of its stable 4 bonds. &lt;/p&gt;&#xA;" OwnerUserId="16723" LastEditorUserId="16723" LastEditDate="2017-05-03T11:56:28.587" LastActivityDate="2017-05-03T11:56:28.587" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20870" PostTypeId="2" ParentId="20868" CreationDate="2017-04-27T15:57:25.920" Score="4" Body="&lt;p&gt;I think this is a good question, not necessarily doubting the experimental results, but asking if a simple sanity check that one would do with other kinds of measurements could be done with CMB. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's really difficult. For light you could wave any convenient-sized blocker in front, even your hand for small apertures. But for this test you'd have to wave something colder than the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow noreferrer&quot;&gt;CMB&lt;/a&gt;, which is only about 2.73 degrees above absolute zero. That particular test would be hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could cool something to 4K with boiling liquid helium at 1 atmosphere, then hold it in front and see a &lt;em&gt;uniform&lt;/em&gt; temperature appear, and that probably was done during the testing phase before satellites were launched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the CMB has both coarse and fine structure, and so one can point the detector to different places and and see different patterns. On the rotating Earth, you'd have to keep moving the detector to see the same pattern. The pattern correlates only with the direction with respect to the celestial sphere, not with respect to the telescope or Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's not the same test as blocking the aperture and seeing the signal going away, but it makes a strong case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;note:&lt;/em&gt; current CMB imaging is done with imaging telescopes such as the &lt;a href=&quot;https://en.wikipedia.org/wiki/South_Pole_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;South Pole Telescope&lt;/a&gt; for example (see &lt;a href=&quot;https://astronomy.stackexchange.com/a/19984/7982&quot;&gt;@zephyr's answer&lt;/a&gt;), unlike the early measurements that were done by measuring a few points at a time (per wavelength band), for example &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_Background_Explorer&quot; rel=&quot;nofollow noreferrer&quot;&gt;COBE&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Wilkinson_Microwave_Anisotropy_Probe&quot; rel=&quot;nofollow noreferrer&quot;&gt;WMAP&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Planck_(spacecraft)&quot; rel=&quot;nofollow noreferrer&quot;&gt;PLANCK&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-04-28T07:45:46.057" LastActivityDate="2017-04-28T07:45:46.057" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20871" PostTypeId="1" AcceptedAnswerId="20872" CreationDate="2017-04-27T18:03:22.883" Score="4" ViewCount="483" Body="&lt;p&gt;According to Wikipedia, re Europa's tidal flexing:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the tidal flexing kneads Europa's interior and gives it a source of&#xA;  heat, possibly allowing its ocean to stay liquid while driving&#xA;  subsurface geological processes.The ultimate source of this energy is&#xA;  Jupiter's rotation, which is tapped by Io through the tides it raises&#xA;  on Jupiter and is transferred to Europa and Ganymede by the orbital&#xA;  resonance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Europa_(moon)&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Europa_(moon)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this imply that Jupiter's rotation is slowing over time, since it is losing energy due to raising tides on Io?  (Or just generally, does raising tides cause rotation to slow?)&lt;/p&gt;&#xA;" OwnerUserId="16727" LastActivityDate="2017-04-30T04:54:06.057" Title="Is Jupiter's rotation slowing?" Tags="&lt;jupiter&gt;&lt;tidal-forces&gt;&lt;europa&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20872" PostTypeId="2" ParentId="20871" CreationDate="2017-04-27T18:56:44.490" Score="6" Body="&lt;p&gt;&lt;strong&gt;Yes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_locking&quot; rel=&quot;noreferrer&quot;&gt;Tidal locking&lt;/a&gt; is a mutual process, and both the satellite and the parent body will eventually &#xA;permanently face each other. However, while tidal locking of a satellite can happen relatively quickly (that is, within the life span of the solar system), tidal locking of the parent body is extremely slow and will in a practical sense never happen (the current age of the universe is not enough in most cases).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relative mass of the satellite and the parent body is one of the most important factors:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Charon and Pluto with a mass ratio of just 8.1 are already mutually locked.&lt;/li&gt;&#xA;&lt;li&gt;The Moon and Earth has a relatively high mass ratio of 81. This causes a measurable effect on the Earth.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Compare that to the mass ratio of Juptier and Europa of 39,600 (or Io: 21300, Ganymede 12,800, Callisto: 17,600). While the time required for a locking of Jupiter itself is difficult to measure due to the fact that we don't have the necessary data about the moons, this is not a noticeable effect before the Sun dies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tidal locking of the parent body can most likely be justified by conservation of energy, but conservation of momentum gives a more intuitive view: As the satellite slows down and moves away from the parent body, its angular momentum increases (angular momentum is proportional to distance times velocity, and the orbital velocity is proportional to the square root of distance). This angular momentum must be supplied from somewhere. The only source for that is the rotation of the parent body. Even if all the moons of Jupiter were to completely escape, the angular momentum drain is not enough to stop the rotation of Jupiter.&lt;/p&gt;&#xA;" OwnerUserId="10172" LastActivityDate="2017-04-27T18:56:44.490" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20873" PostTypeId="1" AcceptedAnswerId="20902" CreationDate="2017-04-27T19:24:01.047" Score="2" ViewCount="2919" Body="&lt;p&gt;How much ultraviolet light comes from the moon at night and from the Sun during the day? I am interested in using a UV camera on a robot to see the UV from the sun/moon to do robot localization.&lt;/p&gt;&#xA;" OwnerUserId="16728" LastActivityDate="2017-07-29T17:50:41.100" Title="Moon and Sun UV Light" Tags="&lt;the-sun&gt;&lt;the-moon&gt;&lt;light&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20874" PostTypeId="2" ParentId="20865" CreationDate="2017-04-27T19:57:15.170" Score="3" Body="&lt;p&gt;The magnification refers to the increase in angular extent (expressed as a &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Solid_angle&quot; rel=&quot;nofollow noreferrer&quot;&gt;solid angle&lt;/a&gt;) of the background source, but is also the factor by which its total brightness is increased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this is that the flux received &lt;em&gt;per unit solid angle&lt;/em&gt; is unchanged by gravitational lensing. So if the source area increases by a factor of $M$, then so does the overall brightness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This increase in total flux observed applies whether the magnified source is resolved by the telescope or not, since that is merely an instrumental issue.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-04-27T20:04:49.723" LastActivityDate="2017-04-27T20:04:49.723" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20875" PostTypeId="1" CreationDate="2017-04-27T20:40:01.183" Score="-4" ViewCount="97" Body="&lt;p&gt;Nobody has been able to find an answer how anybody may be able to see the inner organs of the human body as Natalya Demkina is able to see. In this context, I want to mention that the same way as we are able to see things in the light of an electric lamp or the sun-rays even though our eyes don't emit the visible band of the electromagnetic rays may it not be possible for her to see the inner organs of other people because the sun-rays contain X-rays also. According to me, it is wrong to say that since human eyes can't emit X-rays she may not be able to see the inner organs of other persons the same way as we are able to see everything even though our eyes don't emit the visible band of the electromagnetic waves. I want to know whether it may be possible that her brain's circuitry may be allowing her to see the inner organs through the X-ray band of the sun-rays unlike the rest of us or not. The chances of her having some special brain-circuitry of this type can't be ruled out according to me.    &lt;/p&gt;&#xA;" OwnerUserId="13871" LastActivityDate="2017-04-27T21:04:42.693" Title="If sun-rays also contain any X-rays why we can't see the inner organs of the human body through these rays?" Tags="&lt;x-ray&gt;&lt;sun-rays&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2017-04-28T05:10:00.093" ContentLicense="CC BY-SA 3.0" />
  <row Id="20876" PostTypeId="2" ParentId="20875" CreationDate="2017-04-27T20:49:46.717" Score="2" Body="&lt;p&gt;X-Rays from the sun don't reach the Earth's surface, as our atmosphere is opaque to this kind of electromagnetic radiation.&lt;br&gt;&#xA;This is why X-Ray observatories like the &lt;a href=&quot;https://en.wikipedia.org/wiki/Chandra_X-ray_Observatory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chandra telescope&lt;/a&gt; have to be placed in space. Thus it is excluded, that anybody having an ability to see X-Rays would see people's inner organs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other issue is that in general, human physiology doesn't allow to see electromagnetic radiation this energetic. Our abilities to see high-energetic radiation ends at the violet colour, which is where UV radiation has it's name from. X-Rays are even more energetic than UV.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditDate="2017-04-27T21:04:42.693" LastActivityDate="2017-04-27T21:04:42.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20877" PostTypeId="1" AcceptedAnswerId="20882" CreationDate="2017-04-27T23:26:27.937" Score="2" ViewCount="264" Body="&lt;p&gt;In NASA fact sheets, such as &lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/venusfact.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one for Venus&lt;/a&gt;, a radius of the planet is given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that radius the radius of the surface of the planet, or of the envelope of atmosphere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put another way, if calculating the angular diameter of a planet/moon from a location, do I only have to look up this radius, or do I have to look up this radius along with the atmosphere height to determine the angular diameter?&lt;/p&gt;&#xA;" OwnerUserId="14931" LastActivityDate="2017-04-28T05:59:42.790" Title="Does the given radius of a planet/moon include its atmosphere?" Tags="&lt;planetary-atmosphere&gt;&lt;angular-diameter&gt;&lt;mathematics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20878" PostTypeId="1" AcceptedAnswerId="20879" CreationDate="2017-04-28T00:23:17.110" Score="1" ViewCount="358" Body="&lt;p&gt;I'm viewing the 2017 Solar Eclipse in Graham County, NC.  How important is it to be in the center(blue line)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is almost entirely national forest there so it would be fairly easy to get to assuming I can park somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://xjubier.free.fr/en/site_pages/solar_eclipses/TSE_2017_GoogleMapFull.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://xjubier.free.fr/en/site_pages/solar_eclipses/TSE_2017_GoogleMapFull.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14226" LastActivityDate="2017-04-29T18:53:52.567" Title="Viewing 2017 Solar Eclipse - How important is it to be in the center(blue line)?" Tags="&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20879" PostTypeId="2" ParentId="20878" CreationDate="2017-04-28T00:46:48.350" Score="2" Body="&lt;p&gt;If you want to see a total solar eclipse, you just need to be between the two red lines on that map.  As long as you're within there, what we call the &quot;path of totality&quot;, you will be able to see a total solar eclipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are outside of the red lines, you will only see a partial solar eclipse.  &lt;a href=&quot;https://www.greatamericaneclipse.com/nation/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This map&lt;/a&gt; shows what percentage of the sun will be covered the further north or south you are of the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graham County is entirely within the path of totality, so you should see a total solar eclipse.  If you're planning to travel, however, it might be too late.  Most campgrounds and hotels within the path are already full.  If you're already within the path of totality, my recommendation would be to stay put.  Travelling on that day, or even that week, will be difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The center of the path of totality will offer the longest viewing for the eclipse.  The further from the center, the shorter it will last.  &lt;a href=&quot;https://www.greatamericaneclipse.com/nation/&quot; rel=&quot;nofollow noreferrer&quot;&gt;The aforementioned map&lt;/a&gt; shows timings at different distances from the center.  Notice that it also helps to be close longitudinally as well as latitudinally (that may not be a word).  &lt;/p&gt;&#xA;" OwnerUserId="13426" LastEditorUserId="13426" LastEditDate="2017-04-28T18:44:25.037" LastActivityDate="2017-04-28T18:44:25.037" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20880" PostTypeId="1" AcceptedAnswerId="20881" CreationDate="2017-04-28T01:21:12.903" Score="4" ViewCount="358" Body="&lt;p&gt;This is a question I asked years ago in an astronomy course, but to which I never got a straight answer. Please feel free to correct me if any of my assumptions here stray from facts. It goes like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To any observer, at large enough scale the universe appears to be receding in accordance with Hubble's constant, which causes the light that reaches them from objects such as distant galaxies to experience a red shift. Technically, though, if the observer is standing on a planet, they are in orbit moving in a particular direction at any given time, and so the distant galaxies in the direction of that should have a(n extremely) slightly smaller red shift while those in the opposite direction should have one slightly more severe. This means that, relatively, the galaxies behind are receding at a faster rate from the observer than those in front. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the observer were to possess instruments of sufficient precision to determine the discrepancy in the red shifts and the means to adjust their velocity in any necessary way, would it be possible for them to equalize the red shift in every direction and thus be motionless in respect to the expansion of the universe itself? The cosmic background radiation might be a good reference for measuring red shifts in every direction, since it should be equally distant everywhere and very, very red shifted.&lt;/p&gt;&#xA;" OwnerUserId="16730" LastEditorUserId="16730" LastEditDate="2017-04-29T13:45:59.980" LastActivityDate="2017-04-29T13:45:59.980" Title="&quot;True&quot; motionlessness - red shifts and CMB" Tags="&lt;redshift&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20881" PostTypeId="2" ParentId="20880" CreationDate="2017-04-28T01:28:33.197" Score="7" Body="&lt;p&gt;Yes. This is is actually something that is done in the analysis of redshift data. If no correction to the redshifts are made, then the redshifts are known as &quot;geocentric.&quot; When you correct for the motion of the Earth around the Sun, the redshifts are known as &quot;heliocentric&quot; (see, for example, the description of data in the &lt;A HREF=&quot;http://www-wfau.roe.ac.uk/6dFGS/spectra.html&quot; rel=&quot;noreferrer&quot;&gt;6dFGS database&lt;/A&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Less common is to correct redshifts for the motion of the Milky Way all the way to the cosmic microwave background's (CMB) frame. This is done by applying a shift to a frame where the &lt;A HREF=&quot;http://www.astro.ucla.edu/~wright/CMB-dipole-history.html&quot; rel=&quot;noreferrer&quot;&gt;CMB dipole&lt;/A&gt; is zero. There are two reasons why this is less common. First, the CMB dipole was only well measured relatively recently by the WMAP and Planck missions. Second, correcting for the solar system's motion relative to the CMB fixes only half of the problem. The target galaxy will also be moving with respect to the CMB at speeds that are probably comparable to the Milky Way. This fact leads to distortions in when we try to use redshift as a proxy of distance, some of which are called &quot;&lt;A HREF=&quot;http://w.astro.berkeley.edu/~louis/astro228/redshift.html&quot; rel=&quot;noreferrer&quot;&gt;Finger-of-God&lt;/A&gt;&quot; for their tendency to make plots of galaxies appear to contain filaments that point at the observer.&lt;/p&gt;&#xA;" OwnerUserId="14244" LastEditorUserId="14244" LastEditDate="2017-04-28T10:59:08.277" LastActivityDate="2017-04-28T10:59:08.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20882" PostTypeId="2" ParentId="20877" CreationDate="2017-04-28T05:59:42.790" Score="3" Body="&lt;p&gt;I looked for a source without luck, so I'll just post cause I've read this answer before.   The diameter is the solid surface, not the atmosphere for rocky planets.   For gas giants it's the atmosphere up to 1 bar.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine Venus' angular diameter, you'd need to add something for it's atmosphere.   It's difficult to say how high it's atmosphere stops being reflective (and if anyone knows the details, please answer), but I would guess about 50 km where &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Venus#Troposphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;it's clouds reach to&lt;/a&gt; on it's sunny side.   You'd need to add more if you look at it through infra-red.   But I think it would be an error to add the full 250 km listed as the thickness of it's atmosphere and it's reflectivity probably trails off well before then.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A curious sidebar, &lt;a href=&quot;https://www.quora.com/Which-moon-is-bigger-Titan-or-Ganymede&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ganymede is listed as the largest moon by diameter&lt;/a&gt;, but because of it's atmosphere, Titan appears larger.  It's visibly larger on accurate moon charts.   Because of ice in it's upper atmosphere and low gravity, Titan's atmosphere remains reflective for a significant distance.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-04-28T05:59:42.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20883" PostTypeId="2" ParentId="20871" CreationDate="2017-04-28T06:21:41.893" Score="4" Body="&lt;p&gt;&lt;strong&gt;Nope&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry, the other answer beginning with yes, I wanted to begin one with no), but both answers are correct from a certain point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter's rotation is speeding up because it's contracting.   See &lt;a href=&quot;https://www.quora.com/Is-Jupiter-shrinking-If-yes-why&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.     Jupiter does slow down due to it's tidal bulge and tugging from it's moons, but the bigger factor is that Jupiter loses more heat than it gets from the sun.  As Jupiter loses heat and as it gravitationally coalesces, it gets smaller, which speeds up it's rotation.   That's the larger of the two factors, though running the numbers makes me cringe a little, but I'm fairly certain.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate of rotation of a gas giant is somewhat difficult to measure with accuracy, I'm fairly confident that overall the shrinking is the larger of the two factors and Jupiter's rotation is slowly increasing.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your 2nd question, a tidal bulge can slow down or speed up a planet, it depends on whether the moon moves ahead of or behind the tidal bulge.   If Venus, which rotates very slowly, had a moon, the moon would likely slowly increase Venus' rate of rotation and the Moon would move towards Venus, not away from Venus.   For moons around more rapidly rotating planets, the planet slows down by the tidal interaction and the Moons move away.     In Jupiter's case, it's dense ionosphere may create some resistance that counteracts any tidal force moving it's moons away.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.space.com/16419-io-facts-about-jupiters-volcanic-moon.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Io, Jupiter's innermost moon&lt;/a&gt;, by one article, loses about a ton a second to volcanic activity and tidal flexing and some of that material spirals into Jupiter creating it's permanent Auroras.   It's difficult to say what the net effect on rotation from that would be, but it could have an effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a few different factors too look at beyond just the tidal interaction.   Io, if it was alone with Jupiter, it could well slowly drift closer, but the 4 moon resonance might stabilize it's orbit somewhat.   The entire picture of the motions of Jupiter's moons is above my pay-grade, but Jupiter's shrinking is still likely the single biggest factor, which means it's rotation rate is probably slowly increasing not decreasing.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-04-30T04:54:06.057" LastActivityDate="2017-04-30T04:54:06.057" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20884" PostTypeId="2" ParentId="20004" CreationDate="2017-04-28T09:02:51.700" Score="1" Body="&lt;p&gt;It's a very general question, as a rule of thumb the higher an object is on the sky the light is going to be less dispersed as the atmosphere is thinner. So, the best point and moment to observe an object is when the object is in zenith (directly over your head). Usually it is not possible to travel around the globe to make your observations, so we are going to see better the objects that happen to be near zenith, or they that are crossing the meridian line (with circumpolar objects, it's better wait until the object the meridian line nearest to the zenith)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have here my books so I can't give you a proper formula, but if you want, I can try to give you that formulae.&lt;/p&gt;&#xA;" OwnerUserId="16061" LastEditorUserId="10172" LastEditDate="2017-04-28T14:13:28.300" LastActivityDate="2017-04-28T14:13:28.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20885" PostTypeId="2" ParentId="19412" CreationDate="2017-04-28T10:09:15.557" Score="2" Body="&lt;p&gt;It seems that you haven't align properly your finder (or) laser. You can align at day with a chimney, but at night you must start with the moon and after that you are sure that you are pointing correctly move to a well known feature (like Pleiades cluster in Taurus) and try to find it with the scope (after that you can move to a star (maybe Mizar in Ursa Mayoris or Albireo in Cygnus if you are in the northern hemisphere) after that you must be sure that the finder and laser are pointing correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other question is that (probably) you are using a very high magnification with that telescope. I have myself a RTF newtonian 114 / 500 and I usually works with 20 mm eyepice (that is 25x), you are using about 216x that is almost the maximum theorically magnification for that aperture. Maybe you should use the 6 mm eyepiece without the barlow (that is 108x more or less) this should be enough but with cheap optics that could be a very high magnification also. I think that you could try with a bigger eyepiece... maybe 15mm plössl (30x and about 1.5º of size)&lt;/p&gt;&#xA;" OwnerUserId="16061" LastActivityDate="2017-04-28T10:09:15.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20886" PostTypeId="2" ParentId="2284" CreationDate="2017-04-28T10:16:49.800" Score="2" Body="&lt;p&gt;A planet's magnetic field extends furthest in the direction away from the sun.  This longest part of the magnetic field is called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetosphere#Magnetotail&quot; rel=&quot;nofollow noreferrer&quot;&gt;magnetotail&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetosphere_of_Jupiter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jupiter's magnetic field&lt;/a&gt; is huge.   It's tail extends up to 7,000 Jupiter radii, or about 490 million km per Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As closest pass, &lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/jupiterfact.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jupiter's aphelion&lt;/a&gt; is about 816 million km from the sun and &lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/saturnfact.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Saturn's perihelion&lt;/a&gt;, is about 1,352 million km from the sun, so they get as close as 546 million km.  The distance Saturn's magnetosphere reaches forward would be a handful of Jupiter radii, but what the math suggests is that mostly Jupiter's tail doesn't reach that far, but it does reach most of the way to Saturn.   There's some room for error based on how accurate the 7,000 Jupiter radii figure is.  At the very least it nearly reaches Saturn.   The fields might occasionally touch and exchange particles but probably rarely and not a lot of particles, given that it's the tail end of Jupiter's magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a sidebar, the &lt;a href=&quot;https://www.nasa.gov/topics/moonmars/features/magnetotail_080416.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Moon passes through the Earth's magnetotail&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="1322" LastEditDate="2017-06-03T13:42:24.310" LastActivityDate="2017-06-03T13:42:24.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20887" PostTypeId="1" AcceptedAnswerId="20889" CreationDate="2017-04-28T14:05:51.570" Score="4" ViewCount="101" Body="&lt;p&gt;Edwin Hubble, in &lt;a href=&quot;https://ned.ipac.caltech.edu/level5/Sept04/Hubble/paper.pdf&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;The Observational Approach to Cosmology&lt;/em&gt;&lt;/a&gt;, wrote&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thus the density of the nebular distribution increases outwards,&#xA;  symmetrically in all directions, leaving the observer in a unique position. Such a favoured position, of course, is intolerable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first question is: do more recent observations still show a symmetrical nebulae density increase? Or did Hubble just do a miscalculation, misinterpretation of an observation etc.?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, Hubble can solve this problem with &lt;em&gt;spatial curvature&lt;/em&gt;. Further below, he writes&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Therefore, in order to restore homogeneity, and to escape the horror of a unique position, the departures from uniformity, which are introduced by the recession factors, must be compensated by the second term representing effects of spatial curvature. - There seems to be no other escape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I understand how an expanding Universe explains the symmetrical redshift, but I fail to understand how it explains the symmetrical increase of nebulae density.&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="11143" LastEditDate="2017-04-28T19:45:25.693" LastActivityDate="2017-11-11T23:19:26.443" Title="If the density of nebular distribution increases symmetrically in all directions, how does space curvature explain it?" Tags="&lt;space-time&gt;&lt;space&gt;&lt;expansion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20889" PostTypeId="2" ParentId="20887" CreationDate="2017-04-28T20:33:47.787" Score="7" Body="&lt;p&gt;Let me preface my answer by saying that you're reading a book from 1937 which references the scientific knowledge from the early 1900's. In case you're unfamiliar with astronomical history, I'll say that we've come a &lt;em&gt;looong&lt;/em&gt; way since the early 1900's. For example, this book discuses &quot;Nebula as Island Universes&quot;. At this point, astronomers had only just discovered that these mysterious &quot;nebulae&quot; were actually entire galaxies that were millions of light years away. Hubble is still referring to them as nebula though, rather than galaxies. That being said, it may actually be kind of difficult to &quot;translate&quot; some of this science from almost 100 years ago into modern theories and terminology. You may even find things which are flat out wrong. Hubble did make a lot of important discoveries, but there were many details he was mistaken on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Do more recent observations still show a symmetrical nebulae density increase?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, nebulae here refers to galaxies, not what we've come to consider as &lt;a href=&quot;https://apod.nasa.gov/cgi-bin/apod/apod_search&quot; rel=&quot;noreferrer&quot;&gt;nebulae&lt;/a&gt;. What you're asking here, and what Hubble was referring to, before it had an official name, is about the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmological_principle&quot; rel=&quot;noreferrer&quot;&gt;Cosmological Principle&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the appropriate answer to this question though, is to point out that I believe you're misinterpreting the text. In the &lt;em&gt;Comparison of Observations with Theory&lt;/em&gt; section, Hubble is showing that observations of nebulae (sic, galaxies) indicate they are uniform through space and that his uniformity is easily explained by a static, non-expanding universe.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If both of the extra terms (for recession and for curvature) were absent, the surveys would clearly fit the formula because the situation would be precisely that in a stationary universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &quot;formula&quot; he's talking about here is the one which indicates&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;nebulae are uniformly distributed through a non-expanding universe in which red-shifts are not primarily velocity-shifts, [and] then the numbers should be proportional to the volumes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, he also knows that the Universe is expanding (he was the one to discover this after all) so he took the standard formula and threw in universal expansion. What he finds though, is that the equations now predict that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the numbers of nebulae increase faster than the volume of &#xA;  space through which they are scattered&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The important point here is that an expanding universe (with no curvature) &lt;em&gt;predicts&lt;/em&gt; this is the case, not that it actually is the case. His previous sections and accompanying plot indicate that observations show the universe is homogeneous. Hubble makes the point (and you reference it) that this is discrepancy can be solved by including the curvature of the universe into the mix as well in which case the predictions now match up with the basic &quot;formula&quot; which didn't include expansion or curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as an addendum, I'll expand on this and add that current observations do in fact show that galaxies are isotropic and homogeneous on large scales. One just survey showing this is the &lt;a href=&quot;http://www.roe.ac.uk/~jap/2df/&quot; rel=&quot;noreferrer&quot;&gt;2dF Galaxy Redshift Survey&lt;/a&gt;. Shown below is a plot from their results. While full statistical analysis is needed to says something conclusive, you can see that to the naked eye, the galaxy distribution does appear to be homogeneous and isotropic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4XNaW.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4XNaW.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;sub&gt;Note that this only covers part of the sky because the rest of it is blocked by the disk of our galaxy.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I understand how an expanding Universe explains the symmetrical redshift, but I fail to understand how it explains the symmetrical increase of nebulae density.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interpreting this to mean, you're wondering how the &quot;formula&quot; with expansion included can produce the result that &quot;the numbers of nebulae increase faster than the volume of space through which they are scattered&quot;. This I think is more or less a math question. If you're really interested in seeing how it was derived, feel free to look at Hubble's original paper: &lt;a href=&quot;http://adsabs.harvard.edu/abs/1935ApJ....82..302H&quot; rel=&quot;noreferrer&quot;&gt;Hubble, E., Tolman, R. C., &lt;em&gt;Astrophysical Journal&lt;/em&gt;, 82, 302, 1935&lt;/a&gt;. I think you'll find the summary equations in Section 12 (page 328 in the paper, page 27 in the pdf) with lengthy derivations before that.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-04-28T20:33:47.787" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20891" PostTypeId="2" ParentId="20880" CreationDate="2017-04-29T07:46:51.077" Score="2" Body="&lt;p&gt;You can define an observer to be motionless if they observe no dipole for the CMB. Galaxies are much less useful (than CMB), suffering from shot noise and random motions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, even the CMB dipole is slightly distorted (from that in a smooth expanding universe), but it's the best you can get.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2017-04-29T07:46:51.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20892" PostTypeId="2" ParentId="20868" CreationDate="2017-04-29T09:39:55.030" Score="2" Body="&lt;p&gt;It has been tested as being from the sky, by blocking the signal from all other directions than that being measured, by comparing results of directions and by triangulating output form different receivers. When testing a directional signal, changing the direction of the receiver is in effect blocking the signal from the entire sky except for one aperture. They then compare 1.5 years of results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different observations have correlated the same patterns in the same zones of the sky, so it's the same as pointing two cameras at the same object and questioning if the object exists or if it's an artifact of the camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what is interesting is to wonder what the sea is like, if our own CMB is only one atom of the sea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The receiver equipment was also carefully calibrated using this kind of terminology:&#xA;&lt;a href=&quot;https://arxiv.org/abs/1505.08022&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Planck 2015 results. V. LFI calibration(150 additional authors not shown)&lt;/em&gt;&lt;/a&gt;&#xA;We present a description of the pipeline used to calibrate the Planck Low Frequency Instrument (LFI) timelines into thermodynamic temperatures for the Planck 2015 data release, covering four years of uninterrupted operations. As in the 2013 data release, our calibrator is provided by the spin-synchronous modulation of the cosmic microwave background dipole, but we now use the orbital component, rather than adopting the Wilkinson Microwave Anisotropy Probe (WMAP) solar dipole. This allows our 2015 LFI analysis to provide an independent Solar dipole estimate, which is in excellent agreement with that of HFI and within 1σ (0.3% in amplitude) of the WMAP value. This 0.3% shift in the peak-to-peak dipole temperature from WMAP and a global overhaul of the iterative calibration code increases the overall level of the LFI maps by 0.45% (30 GHz), 0.64% (44 GHz), and 0.82% (70 GHz) in temperature with respect to the 2013 Planck data release, thus reducing the discrepancy with the power spectrum measured by WMAP. We estimate that the LFI calibration uncertainty is now at the level of 0.20% for the 70 GHz map, 0.26% for the 44 GHz map, and 0.35% for the 30 GHz map. We provide a detailed description of the impact of all the changes implemented in the calibration since the previous data release.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/glNrR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/glNrR.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastActivityDate="2017-04-29T09:39:55.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20893" PostTypeId="2" ParentId="18740" CreationDate="2017-04-29T10:11:52.970" Score="2" Body="&lt;p&gt;&lt;em&gt;Three decent answers in comments converted to a community wiki.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quite recent study about the origin and development of terrestrial planet atmospheres is the 2015 paper &quot;&lt;a href=&quot;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20140017487.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Atmospheres of the Terrestrial Planets: Clues to the Origins and Early Evolution of Venus, Earth, and Mars&lt;/a&gt;&quot; (Baines et al.) includes a fairly comprehensive analysis of chemical evidence from Earth and from Martian meteorites and observations of the atmosphere of Venus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general consensus is that much of the atmospheric gases were brought through c&lt;a href=&quot;https://www.sciencedaily.com/releases/2009/12/091210153538.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;omet and asteroid impacts&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pre-planetary disk would have been centrifugally sorted. Many iron, stone, carbonacious meteorites, watery, ice-iron-rock agglomerates would have brought a very diverse amalgam of different stones, ices, and room temperature gas elements, throughout the formation. CO2 and N2 objects are a bit less stable in direct solar radiation at 1 AU. CO2 and N2 would have pounded into the world in considerable quantities throughout the accretion process, with a wide range of particulate sizes from microns to kilometers in diameter without a marked difference in element type from start to end.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-04-29T10:11:52.970" CommentCount="0" CommunityOwnedDate="2017-04-29T10:11:52.970" ContentLicense="CC BY-SA 3.0" />
  <row Id="20895" PostTypeId="2" ParentId="14935" CreationDate="2017-04-29T11:29:43.730" Score="3" Body="&lt;p&gt;It would be good business for the news industry: sky lights, plane crashes, exploding grid transformers, city blackouts, satellite damage, long, electricity-carrying wires would spark, start fires and and send signals when they are switched off. Burning buildings, dead people, everything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;big flares occur about every 100 years. the last was the Carrington event in 1859. It triggering telegraph malfunctions (&lt;a href=&quot;https://science.nasa.gov/science-news/science-at-nasa/2008/06may_carringtonflare/&quot; rel=&quot;nofollow noreferrer&quot;&gt;even giving operators electrical shocks&lt;/a&gt;) and caused some telegraph pylons to suddenly spark and catch fire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A surge in solar wind can blow out power transformers by melting the coating that insulates copper windings and by melting the copper itself, especially in highly interconnected regions (i.e. the East Coast), transformer failures can trigger cascading effects, spreading power outages over wide areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a flash of light is seen from the sun 17 hours in advance of the magnetic storm, and sunspots are monitored for activity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a massive CME were spotted, 3-day forecasts give us some lead time: there are some measures electric utilities could take to protect their equipment, such as quickly disconnecting transformers. Polar flights, which travel at the highest altitudes, could be rerouted to avoid contact with damaging solar particles, and some satellites could be switched into a safe mode to minimize damage. Here on Earth, at the very least, we’d have some time to prepare for potential power blackouts and other problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the scope of our current technological infrastructure, it’s been estimated that the next Carrington-scale event (which happens roughly once every 100 years) will cause upwards of $2.6 trillion in damage across the globe, and require four to ten years for complete recovery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scientists studying tree rings found a sharp increase in the amount of radioactive carbon-14 recorded in the rings of ancient Japanese cedar trees between 774 and 775. Carbon-14 can be created by cosmic ray particles arriving from space, and the 774 incident could have been a solar flare. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;there is a list of big solar storms here: &#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_solar_storms&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/List_of_solar_storms&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2017-04-29T11:36:33.570" LastActivityDate="2017-04-29T11:36:33.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20900" PostTypeId="2" ParentId="20858" CreationDate="2017-04-29T19:31:19.887" Score="4" Body="&lt;p&gt;Carbon is like lego-technic. It connects water into organic compounds which have every chemical and physical property from volatile solvent to syrup to oil to tar and rock. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silicon is like lego, try building a car out of it, it's difficult to find a good solvent for it. it can't even bond H. Silane which is the equivalent of methane doesn't occur naturally on our planet, it's too unstable. since silicon doesn't form OH compounds, it doesn't have the malleability of water, it's always rocky and unstable except at -200 degrees where it could be more versatile than carbon. silicon compounds under -180 degrees C are interesting and worth studying, they would need a kryogenic space suit to keep them cool, and if you shoot them they would spray liquid oxygen everywhere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sulphur is even less versatile than silicon. it only forms linear chains and is hyper reactive with hydrogen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'd need a solvent other than water which is used for carbon based. Ammonia, formaldehyde, sulphuric acid, liquid nitrogen, hydrogen fluoride, methane, hydrocarbons, supercritical fluids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps complex robotic life forms are some of the easiest to make up. Life forms that can be designed by artificial intelligence, to use most commonly available resources at ambient temperatures are more likely than complex life forms using other than carbon. That's because a large computer can design transistors/mechanics that work with any metals and conductors in very cold temperatures (aluminium can structurally handle -130 kelvin easily), and robotic life forms would prefer to be in a stable environment without biology, clouds, rain, water, in cold environments, it would be easier for them. &lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2017-04-29T23:17:44.503" LastActivityDate="2017-04-29T23:17:44.503" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20901" PostTypeId="2" ParentId="16450" CreationDate="2017-04-30T16:15:44.987" Score="2" Body="&lt;p&gt;The light from Earth into space has a blueish colour, and when that light reflects from the Moon and is seen by someone on Earth it is a bit less blue because the lunar surface reflects yellow light more than blue. The net effect is to give us earthshine that is blueish, or turquoise.&lt;/p&gt;&#xA;" OwnerUserId="16745" LastActivityDate="2017-04-30T16:15:44.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20902" PostTypeId="2" ParentId="20873" CreationDate="2017-04-30T17:04:10.687" Score="1" Body="&lt;p&gt;&quot;A little&quot; (compared to how much visible light there is)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/myJVk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/myJVk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;(from a &lt;a href=&quot;https://physics.stackexchange.com/questions/203411/explanation-for-the-spectrum-of-the-sun&quot;&gt;question on physics se&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun produces relatively little UV radiation, most of its light is visible, or infrared, and anything shorter than about 290nm is completely blocked by the atmosphere (remember to thank the atmosphere today)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UVa (in the range 400-320) and UVb (320-290nm) do get through. The sun is so very bright, that it is by far the brightest natural source of UV on the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon is about 400,000 times dimmer than the sun. Ultraviolet reflected off the moon could reach the Earth. It would be very dim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At night UV camera systems use active illumination rather than depending on the moon.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-04-30T17:04:10.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20905" PostTypeId="1" AcceptedAnswerId="20908" CreationDate="2017-04-30T18:02:20.033" Score="1" ViewCount="256" Body="&lt;p&gt;(Background: Trying to learn something about the mathematical and computational part of astronomy I have acquired a copy of Meeus' &quot;Astronomical Algorithms&quot; and implemented the material up to Chapter 33 in Java, with only one major sidestep: VSOP87, for which I obtained the full data set and &quot;translated&quot; the FORTRAN program.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;VSOP87 exhibits the reason for my question: you can compute rectangular/spherical heliocentric positions either for the reference frame J2000 or the one of the given date. Clearly, if I want to observe some planets tonight, I'll compute the data for today's equinox. On the other hand, there is the possibility of a transformation between equinoxes of different dates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why are these methods (here and for other methods, too) provided for both reference frames? Was it the (then) weaker computer performance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only a little later, iauPlan94 provides the result only for J2000 although Meeus in the chapter dedicated to the paper of J.L.Simon e.al. provides tables with data from their theory for J2000 and the equinox of the date. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be good reasons for today and the years to come to refer a computation to J2000? &lt;/p&gt;&#xA;" OwnerUserId="16122" LastEditorUserId="13426" LastEditDate="2017-04-30T18:18:12.823" LastActivityDate="2017-05-01T06:59:46.000" Title="What are the reasons for referring to equinox J2000 or the equinox of the date?" Tags="&lt;algorithm&gt;&lt;frame-of-reference&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20906" PostTypeId="1" AcceptedAnswerId="20907" CreationDate="2017-04-30T18:39:58.500" Score="2" ViewCount="1790" Body="&lt;p&gt;I need a resource that can tell me the stars that make up a constellation (pattern), preferably by Henry Draper or some catalogue number? All i can find is lists of every star in a constellation (area) or incomplete lists of stars by name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this doesn't exist can you suggest a way to work it out myself? &lt;/p&gt;&#xA;" OwnerUserId="16454" LastActivityDate="2020-04-26T08:42:45.573" Title="Stars in constellation patterns by catalogue number?" Tags="&lt;star&gt;&lt;constellations&gt;" AnswerCount="5" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20907" PostTypeId="2" ParentId="20906" CreationDate="2017-04-30T20:03:28.037" Score="3" Body="&lt;p&gt;What you ask for does not exist. There is no standard for what the patterns are. In modern astronomy, the constellations are areas of the sky. You can find lists of the stars in each area, but it seems that you want only the stars in the &quot;pattern&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you see a star map, lines are sometimes drawn between the brighter stars. These help people to compare the stars to the map, our brains like to look for connections, and so linking the stars together helps us to make sense of the sky. However different stellar-cartographers will draw the lines in different ways. Consider Gemini:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/5/58/Gemini_constellation_map.png/271px-Gemini_constellation_map.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/5/58/Gemini_constellation_map.png/271px-Gemini_constellation_map.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/VE0Ww.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VE0Ww.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same constellation is represented in different ways. Neither presentation is official, it is an artistic choice which stars to include and how to link them. It is therefore impossible to say which stars &quot;make up a pattern&quot; and which do not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you look at the catalogues, and see which stars have received a &lt;a href=&quot;https://en.wikipedia.org/wiki/Bayer_designation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bayer designation&lt;/a&gt; you will have a &lt;a href=&quot;https://en.wikipedia.org/wiki/Table_of_stars_with_Bayer_designations&quot; rel=&quot;nofollow noreferrer&quot;&gt;list of the visible stars&lt;/a&gt; in each constellation. This won't, of course, tell you how the stars may be linked.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-04-30T20:03:28.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20908" PostTypeId="2" ParentId="20905" CreationDate="2017-05-01T06:59:46.000" Score="2" Body="&lt;p&gt;Maybe you are interested in watching the occultation of a star by a solar system object. In which case you will need to compare their positions. Catalogues of stellar positions are most commonly given with J2000 positions.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-05-01T06:59:46.000" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20909" PostTypeId="2" ParentId="20111" CreationDate="2017-05-01T14:28:15.827" Score="1" Body="&lt;p&gt;$0 (at first)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using an astronomical telescope is hard work, and the results are nearly always disappointing. Planets are very small targets, and through a small telescope they look like a little circle. You are lucky to see any detail at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First just get used to watching the night sky, learning the constellations and asterisms. Watching the motion of the planets against the stars. See Venus brighten then disappear behind the setting sun, only to re-appear as a morning object. Look for pleasing alignments and syzygy. Visit some dark sky locations and see the milky way. Point a camera at the sky and let it record star trails around Polaris. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then a pair of binocular or a spotting scope is useful to see some detail on the moon and deeper into star fields. These are general purpose instruments, good for birdwatching as well as stargazing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sign up with the Bradford Robotic telescope, then you will have shared use of a 34-inch photographic telescope, in a dark location. You will get better photos of the planet than you could ever hope with backyard equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Join up with a local astronomical society, you will see what is possible with an amateur 'scope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And when you find this equipment limiting you can look to purchase a decent sized reflector, but by then you will know what you get for your money, and how much you are willing to spend.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-05-01T14:28:15.827" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20912" PostTypeId="2" ParentId="16450" CreationDate="2017-05-01T20:05:43.377" Score="2" Body="&lt;p&gt;You must put it in a spectral histogram program. same color as a shark/manatee according to one crazy program. I found this from basic &lt;a href=&quot;http://www.wisegeek.com/how-can-i-find-the-average-color-in-a-photograph.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;blur pic&lt;/a&gt; from a true color source:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/dFDOc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dFDOc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/gma5t.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gma5t.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/1UDa4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1UDa4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/LYTDH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LYTDH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;and for true color of all planets there was this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/RME65.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RME65.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2017-05-01T20:11:56.777" LastActivityDate="2017-05-01T20:11:56.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20914" PostTypeId="2" ParentId="7679" CreationDate="2017-05-02T16:37:36.787" Score="0" Body="&lt;p&gt;Life here on earth developed on earth as far as scientists are aware. It is believed that life began in nutrient-rich puddles of a mud-like substance that is known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Primordial_soup&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;primordial soup&quot;&lt;/a&gt;. This was the primitive ocean of the earth. Heat was also needed to develop life, and it's believed that life developed in the mid regions of Africa due to it being the hottest areas of the earth. I don't know about the formation of the cell under these conditions so that's an area I won't be able to explain. But from there, the theory of evolution took over. Cells start forming, mutating, developing some degree of genetic variation. Eventually, after millions of years, multi-celled organisms formed, like plants and basic animals. It's debated whether it was jellyfish or sea sponges that came first. Personally, I believe it's the sea sponge simply because it seems a more simple creature, it's just a bunch of identical cells working together collectively. Then evolution just continues on, creating more and more complex life. &lt;/p&gt;&#xA;" OwnerUserId="16723" LastActivityDate="2017-05-02T16:37:36.787" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20915" PostTypeId="1" CreationDate="2017-05-02T20:58:20.747" Score="7" ViewCount="603" Body="&lt;p&gt;It's always made intuitive sense to me that a higher surface gravity leads to broader spectral lines from a star, but yesterday I realized that I don't have a physical explanation for why this is so.  Why is this so?&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2017-05-03T07:24:56.847" Title="Why does higher surface gravity broaden spectral lines?" Tags="&lt;star&gt;&lt;spectra&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20917" PostTypeId="1" AcceptedAnswerId="20923" CreationDate="2017-05-03T05:11:35.337" Score="1" ViewCount="311" Body="&lt;p&gt;Pick a any notion of year and any type of day measurement.  Outside of history and the common cycle of day an night, why are these viewed as integrally or even fractionally commensurate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why all the stunts with leap days and leap seconds, etc... when we now have atomic clocks?  Why not decouple the two?&lt;/p&gt;&#xA;" OwnerUserId="16758" LastEditorUserId="16758" LastEditDate="2020-05-10T03:05:48.527" LastActivityDate="2020-05-10T06:50:42.773" Title="Why the continued obsession in measuring the year as a multiple of days?" Tags="&lt;history&gt;&lt;time&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="20918" PostTypeId="1" AcceptedAnswerId="20919" CreationDate="2017-05-03T06:21:45.247" Score="26" ViewCount="5710" Body="&lt;p&gt;I always hear the narrator of documentaries say that a star exploded because it ran out of fuel.&#xA;Usually things explode when they have too much fuel, not when they run out of fuel.&#xA;Please explain...&lt;/p&gt;&#xA;" OwnerUserId="15605" LastActivityDate="2017-09-19T14:06:46.513" Title="Why do stars explode?" Tags="&lt;star&gt;&lt;explosion&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20919" PostTypeId="2" ParentId="20918" CreationDate="2017-05-03T07:06:48.830" Score="33" Body="&lt;p&gt;&lt;strong&gt;Short answer:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A tiny fraction of the gravitational potential energy released by the very rapid collapse of the inert iron core gets transferred to the outer layers and this is sufficient to power the observed explosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In more detail:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the energetics of an idealised model star. It has a &quot;core&quot; of mass $M$ and initial radius $R_0$ and an outer envelope of mass $m$ and radius $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose the core collapses to a much smaller radius $R \ll R_0$ on such a short timescale that it decouples from the envelope. The amount of gravitational potential energy released will​ be $\sim GM^2/R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fraction of this released energy can be transferred to the envelope in the form of outward moving shocks and radiation. If the transferred energy exceeds the gravitational binding energy of the envelope $\sim Gm^2/r$ then the envelope can be blown into space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an exploding star (a type II core collapse supernovae) $R_0\sim 10^4$ km, $R\sim 10$ km and $r \sim 10^8$ km. The core mass is $M \sim 1.2M_{\odot}$ and  the envelope mass is $m \sim 10M_{\odot}$. The dense core is mostly made of iron and supported by &lt;a href=&quot;https://en.wikipedia.org/wiki/Electron_degeneracy_pressure&quot; rel=&quot;noreferrer&quot;&gt;electron degeneracy pressure&lt;/a&gt;. The star is said to have &quot;run out of fuel&quot; because fusion reactions with iron nuclei do not release significant amounts of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The collapse is triggered because nuclear burning continues around the core and so the core mass is gradually increased and as it does so it gradually &lt;em&gt;shrinks&lt;/em&gt; (a peculiarity of structures supported by degeneracy pressure), the density increases and then an instability is introduced either by electron capture reactions or photodisintegration of iron nuclei. Either way, electrons (which are what is providing the support for the core) are mopped up by protons to form neutrons and the core collapses on a free fall timescale of $\sim 1$ s!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The collapse is halted by the strong nuclear force and neutron degeneracy pressure. The core bounces; a shock wave travels outwards; most of the gravitational energy is stored in neutrinos and a fraction of this is transferred to the shock before the neutrinos escape, driving away the outer envelope. An excellent descriptive account of this and the previous paragraph can be read in &lt;a href=&quot;https://arxiv.org/pdf/astro-ph/0601261.pdf&quot; rel=&quot;noreferrer&quot;&gt;Woosley &amp;amp; Janka (2005)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting in some numbers.&#xA;$$GM^2/R = 4\times 10^{46}\ {\rm J}$$&#xA;$$Gm^2/r = 3\times 10^{44}\ {\rm J}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one only needs to transfer of order 1% of the collapsing core's released potential energy to the envelope in order to drive the supernova explosion. This is actually not yet understood in detail, though somehow supernovae find a way to do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A key point&lt;/strong&gt; is that the rapid collapse takes place &lt;em&gt;only&lt;/em&gt; in the core of the star. If the entire star collapsed as one, then most of the gravitational potential energy would escape as radiation and neutrinos and there would be insufficient energy even to reverse the collapse. In the &lt;em&gt;core&lt;/em&gt; collapse model, most (90%+) of the released gravitational energy &lt;em&gt;is&lt;/em&gt; lost as neutrinos, but what remains is still easily sufficient to unbind the uncollapsed &lt;em&gt;envelope&lt;/em&gt;. The collapsed core remains bound and becomes either a neutron star or black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second way to cause a star (a white dwarf) to explode is a thermonuclear reaction. If the carbon and oxygen can be ignited in nuclear fusion reactions then enough energy is released to exceed the gravitational binding energy of the white dwarf. These are type Ia supernovae.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-05-03T12:15:10.537" LastActivityDate="2017-05-03T12:15:10.537" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20920" PostTypeId="2" ParentId="20915" CreationDate="2017-05-03T07:24:56.847" Score="11" Body="&lt;p&gt;In order to be in hydrostatic equilibrium, high surface gravity atmospheres have higher pressures at the same sort of temperatures. That means the density of atoms and ions is higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the particle density is higher then there is more chance of interactions between the particles. In particular, collisions between atoms perturb their energy levels and reduce the lifetimes of atoms in particular states. The natural frequency width is proportional to the reciprocal of the radiative lifetime. If this lifetime is truncated due to frequent collisional (de)excitation, then the lifetime is shortened and the frequency width increased. This is known as &lt;a href=&quot;http://www.fas.harvard.edu/~scidemos/QuantumRelativity/CollisionalBroadening/CollisionalBroadening.html&quot; rel=&quot;noreferrer&quot;&gt;collisional broadening&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-05-03T07:24:56.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20921" PostTypeId="2" ParentId="20918" CreationDate="2017-05-03T09:37:59.243" Score="4" Body="&lt;p&gt;To give an answer in more simple turns. (Yes very simplified, but it should introduce the basic concept).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Star &quot;burns&quot; by nuclear fusion between lighter elements such as Hydrogen turning to Helium. The heat and energy of that burning constantly pushes on the matter inside the star holding it up. The fusing hydrogen generates enough energy to stop it from being able to collapse down into the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the star starts running out of fuel that &quot;fire&quot; gets colder, and the pushing out gets weaker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually the push isn't enough to keep the star apart and it all rushes back together. That collapse releases a huge amount of energy which causes the explosion.&lt;/p&gt;&#xA;" OwnerUserId="16762" LastActivityDate="2017-05-03T09:37:59.243" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20922" PostTypeId="2" ParentId="20917" CreationDate="2017-05-03T12:08:43.537" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why all the stunts with leap days and leap seconds, etc... when we now have atomic clocks? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Science exists to serve mankind, not to rule it. Calendars were one of the first concepts developed by humankind, arguably predating writing. Timekeeping is even more important to modern society than it was to the ancients. You're not going to get rid of calendars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One key purpose of the Gregorian calendar is to keep the calendar in sync with the seasons. Every solar calendar requires some intercalation mechanism. The Gregorian calendar uses leap days. The Gregorian scheme (years divisible by 4 have an extra day, except for years that are multiples of 100 that aren't divisible by 400) will fall out of sync with the seasons at at the very slow rate of one day per 3030 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leap seconds are a rather different beast. They are inserted, unpredictably, to keep the clock in sync with the day/night cycle. Without leap seconds, counting 86400 atomic clock seconds = one day would already have the clock out of sync by about half a minute. Leap seconds are problematic for business. There is an ongoing effort to remove leap seconds and just live with a day of 86400 seconds going out of sync with the day, at least until it truly becomes noticeable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why not decouple the two and say that day X belongs to year Y if at 12:00 noon of that day is &quot;in&quot; year Y?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What's this &quot;year Y&quot; thing you're writing about? How do you know what year it is without a calendar?&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-05-03T12:08:43.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20923" PostTypeId="2" ParentId="20917" CreationDate="2017-05-03T16:56:37.073" Score="7" Body="&lt;p&gt;The year is not defined as a multiple of days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The year is the time for the Earth to orbit the sun once. This is not a constant amount of time. So it is not defined in seconds or days or any other length of time. However it only varies a little.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The (synodic) day is the amount of time for the Earth to rotate once relative to the Sun. This is also not a fixed amount of time. But like the year only varies a little.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time is measured in seconds. one second is 9,192,631,770 cycles of the radiation produced by the transition between two levels of the caesium 133 atom. This is among the most stable clocks we know. The second can be redefined if a more stable clock exists. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the practical aspect of calendar forming it is convenient to have a simple rule that anyone can apply and can decide, without the need to make careful observations, which day is in which year. To this end the Gregorian Calendar gives a fairly simple but accurate approximation to the true astronomical year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The disadvantages of using astronomical observations to decide the first day of the year are significant. Politically, who makes the observations? Russia? China? Japan? Botswana? In theory it would mean that I could not know which day a year was in, until after that day had been reached, and the observations done. In practice there would be no difference from the Gregorian calandar for many centuries; so why bother.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarise: The astronomical year is approximated by the Gregorian year. This is simple to use. It is accurate for most calendar applications. It allows for the calendar to be extended into the future. The Gregorian calendar has wide acceptance among many countries. If precise timekeeping is needed then one doesn't use &quot;years&quot; instead one uses the SI unit of time, the second.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2020-05-10T06:50:42.773" LastActivityDate="2020-05-10T06:50:42.773" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="20925" PostTypeId="1" AcceptedAnswerId="20926" CreationDate="2017-05-03T22:52:05.947" Score="7" ViewCount="300" Body="&lt;p&gt;How are the world's biggest telescopes, (such as the Very Large Telescope for example) cleaned?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My research includes what I have read about CO&lt;sub&gt;2&lt;/sub&gt; cleaning &lt;a href=&quot;https://www.quora.com/How-do-they-clean-those-larger-telescope-mirrors&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. Are there any other ways to cleaning the mirrors? What if we were talking about the (cancelled) &lt;a href=&quot;https://en.wikipedia.org/wiki/Overwhelmingly_Large_Telescope&quot; rel=&quot;noreferrer&quot;&gt;Overwhelmingly Large Telescope&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="16776" LastActivityDate="2017-05-03T23:34:53.270" Title="How are the world's biggest telescopes cleaned?" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20926" PostTypeId="2" ParentId="20925" CreationDate="2017-05-03T23:34:53.270" Score="5" Body="&lt;p&gt;One of the ways they clean telescopes is to replace the metal coating on their surface. See, for telescopes, unlike bathroom mirrors, the metal is on the front of the glass. The reason for this difference is that having glass on the front makes the mirror more durable, but it also reduces the mirror's efficiency at creating an image. Because of this every few years the mirror needs to be taken off of the telescope, the old metal removed from the glass by an acid or some other chemical means, and then a new coating of metal is applied. On the tour of the &lt;A HREF=&quot;http://www.astro.caltech.edu/palomar/homepage.html&quot; rel=&quot;noreferrer&quot;&gt;Palomar Observatory&lt;/A&gt; they'll tell you that the 5.1 meter Hale telescope uses about the same amount of aluminum to cover it as is in a quarter (not sure if they meant volume or mass - also not completely sure it wasn't a nickle, or 2 quarters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of telescopes with segmented mirrors, like Keck, they can do this one segment at a time, which makes the process more convenient and spreads it out over time. It's also possible, though I don't know if they do this, to keep segments in backup to put in place of removed segments, because the telescope is made from a relatively small number of segments that have the same curvature and are, therefore, interchangeable.&lt;/p&gt;&#xA;" OwnerUserId="14244" LastActivityDate="2017-05-03T23:34:53.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20927" PostTypeId="2" ParentId="20917" CreationDate="2017-05-04T03:08:05.267" Score="0" Body="&lt;p&gt;I think that sometimes we do things for convenience, not just scientific rigor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, it may be much more computationally efficient for humans to choose a new number system based on 12 numeric symbols, because 12 has factors 2,3,4,6 rather than a decimal number system which has factors only 2,5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we stick to the decimal system because its simpler to count on our 10 fingers.&lt;/p&gt;&#xA;" OwnerUserId="16406" LastActivityDate="2017-05-04T03:08:05.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20928" PostTypeId="1" AcceptedAnswerId="20933" CreationDate="2017-05-04T06:20:55.927" Score="5" ViewCount="8287" Body="&lt;p&gt;Last weekend (29/30 April) the sky in my area (Central Europe) was pretty clear and the Moon was nicely visible. Funnily, one could not only see the sunlit area of the Moon, but also the dark side&lt;sup&gt;*&lt;/sup&gt; could be easily recognised throughout the whole evening from blue hour until dark night. I attached two photos of the Moon that should illustrate the effect. The pictures (sorry for the poor quality) needed to get a little overexposed to capture the illuminated dark side, but with plain eye the effect could be well seen.&#xA;We were speculating about the reasons why the non-sunlit area of the Moon could be seen.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;diffraction of sunlight through the atmosphere can be neglected as the Moon doesn't have an atmosphere. &lt;/li&gt;&#xA;&lt;li&gt;reflection of light that in turn was reflected from earth (Sun -&gt; Earth -&gt; Moon -&gt; observer) sounded a little unlikely to me&lt;/li&gt;&#xA;&lt;li&gt;a psychological effect that kind of mentally completes the shape of the Moon, because we know it should be circular, sounded ok to me. But could be debunked as the dark side stays visible if one covers the bright side on the photo? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So I guess the  sunlight has to reach the dark side of the Moon somehow differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could be provide enough light to make the dark side visible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;* not the far side of the Moon, but the non-sunlit area of the Moon&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VjIpK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VjIpK.jpg&quot; alt=&quot;visible dark side of the Moon in the evening (sorry for the blurred picture)&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ViNRE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ViNRE.jpg&quot; alt=&quot;visible dark side of the Moon later at night&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16756" LastEditorDisplayName="user1569" LastEditDate="2019-09-19T08:57:56.657" LastActivityDate="2019-09-19T08:57:56.657" Title="Why can I see the dark side of the moon?" Tags="&lt;the-moon&gt;&lt;moon-phases&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="20929" PostTypeId="2" ParentId="20928" CreationDate="2017-05-04T08:21:42.720" Score="2" Body="&lt;p&gt;A full Earth on the Moon is much brighter than a full Moon is on Earth, because the Earth is significantly bigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Btw, &quot;dark side of the Moon&quot; is not really a technical term (it's rather a song of Pink Floyd). On the Moon, there is (as on Earth) day and night (though the day is as long as month on Earth) and no spot remains dark for longer (except for the pole caps as on Earth). There is, of course, the far side of the Moon, most of which is invisible from Earth (because the Moon is tidally locked to spin with the same rate as it rotates around Earth).&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2017-05-04T08:21:42.720" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20932" PostTypeId="1" CreationDate="2017-05-04T15:36:19.837" Score="2" ViewCount="348" Body="&lt;p&gt;During a solar eclipse Sir Arthur Eddington measured the deflection of star light by the Sun. Einstein predicted a deflection of 1.74 arc seconds, but according to Newton the deflection should have been 0.86 arc seconds. So is it a coincidence that the prediction of Einstein is twice as large or is there a kind of law involved which makes Einstein's outcome of his GR twice as big as that according to Newton?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="2531" LastEditDate="2019-04-19T10:37:06.197" LastActivityDate="2019-04-19T10:37:06.197" Title="Is it coincidence that Einstein predicted exactly twice the outcome of Newton for the gravitational deflection of light?" Tags="&lt;general-relativity&gt;" AnswerCount="2" CommentCount="0" ClosedDate="2017-05-06T16:57:59.967" ContentLicense="CC BY-SA 4.0" />
  <row Id="20933" PostTypeId="2" ParentId="20928" CreationDate="2017-05-04T18:08:52.933" Score="8" Body="&lt;p&gt;You can generally see the unlit side of the moon when a considerable amount of sunlight is reflected off the earth. This reflected sunlight illuminates the unlit side of the moon. This is referred to as earthshine, and a decent explanation can be found at &lt;a href=&quot;https://www.timeanddate.com/astronomy/earthshine.html&quot; rel=&quot;noreferrer&quot;&gt;timeanddate.com&lt;/a&gt;. I seem to recall reading somewhere (but now can't find a reference) that it is more prominent when it is very cloudy on earth near where the moon is overhead.&lt;/p&gt;&#xA;" OwnerUserId="7108" LastActivityDate="2017-05-04T18:08:52.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20934" PostTypeId="2" ParentId="20932" CreationDate="2017-05-04T19:08:04.820" Score="4" Body="&lt;p&gt;As far as I know Newtonian prediction is based on Soldner's work in 1801 and Einstein 1911 also predicted around $0.87$arcsec (he was also accused for plagiarizing Soldner's work the article:&lt;a href=&quot;http://articles.adsabs.harvard.edu//full/1923PASP...35..185T/0000185.000.html&quot; rel=&quot;noreferrer&quot;&gt;Historical Note on the Problem of Light Deflection in the Sun's Gravitational Field&lt;/a&gt; explains why the accusations were wrong).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Newtonian approach accounted for the space curvature, as did special relativity which Einstein first used, but it missed the part that not only space is curved but space time is curved. So the factor of two was because of the spatial curvature, that's because by &lt;a href=&quot;https://en.wikipedia.org/wiki/Metric_tensor_(general_relativity)#Flat_spacetime&quot; rel=&quot;noreferrer&quot;&gt;Minkowski metric&lt;/a&gt; we consider that light travels flat so that $\rm{d}s^2=0$.The correct answer would be to use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_metric&quot; rel=&quot;noreferrer&quot;&gt;Schwarzschild metric&lt;/a&gt; and get the correct answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This paper gives more insight on the formulas and the correct approach. &lt;a href=&quot;http://home.fnal.gov/~syphers/Education/Notes/lightbend.pdf&quot; rel=&quot;noreferrer&quot;&gt;Gravitational Bending of Light&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16649" LastActivityDate="2017-05-04T19:08:04.820" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20936" PostTypeId="2" ParentId="6205" CreationDate="2017-05-04T21:42:26.043" Score="4" Body="&lt;p&gt;I think we can work this our pretty well with a few approximations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To start, let's say that the observable limit of the naked eye under completely dark skies is 6th magnitude in $V$. Let's also say that the human eye has an angular resolution of 1 arcminute, which will subtend $\pi \cdot 1^2 \approx 3$ square arcminutes on the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actual dark skies on Earth still have some measurable sky brightness - which is one reason we can't see arbitrarily faint stars - of about $21.8 \ \mathrm{mag} \over \text{arcsec}^2$ in V-band. &lt;a href=&quot;http://www.eso.org/~fpatat/science/skybright/twilight.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;See Figure 1 here.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integrated brightness, $m$, of the sky background, $S$, will be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = S - 2.5 \cdot \log(A)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ is the angular area. Note that the integrated sky brightness will just scale along with $S$, the surface brightness of the sky, since $A$ remains constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if for $S = {21 \ \mathrm{mag} \over \text{arcsec}^2}$ the observable limit is 6th mag in $V$, the limit for arbitrary $S$ will be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_{\text{limit}} = 6 - [21 - S(z)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $S(z)$ is the sky brightness when the Sun is at some zenith angle $z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From that reference I linked to, $S = 21.8$ at about $z = 105^{\circ}$ (Fig 5), and it goes up pretty much linearly to $S = 10$ in $V$ at $z = 94^{\circ}$. So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S(z) = 1.07 \cdot z - 90.5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the limiting visible $V$ magnitude for a star, as function of the Sun's zenith angle ought to be roughly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_{\text{limit}}(z) = 1.07 \cdot z - 105.5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you wanted the altitude of the Sun below the horizon, not away from the zenith, so subtract $90^{\circ}$ from $z$ in the above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_{\text{limit}} = 1.07 \cdot \theta - 10$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\theta$ is how far the Sun is below the horizon in degrees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Approximately, at any rate.&lt;/p&gt;&#xA;" OwnerUserId="16268" LastEditorUserId="9301" LastEditDate="2017-05-04T23:57:54.117" LastActivityDate="2017-05-04T23:57:54.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20937" PostTypeId="1" CreationDate="2017-05-05T00:08:18.737" Score="3" ViewCount="528" Body="&lt;p&gt;I've heard that KIC 9832227 is a contact binary star system that is expected to merge and explode as a &quot;red nova&quot; sometime in the next ~5 years or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sciencemag.org/news/2017/01/colliding-stars-will-light-night-sky-2022&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.sciencemag.org/news/2017/01/colliding-stars-will-light-night-sky-2022&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before that happens, I'd like to see it.  Where would I need to be / what sort of equipment would I need in order to be able to view this before it goes nova?&lt;/p&gt;&#xA;" OwnerUserId="16802" LastActivityDate="2018-06-22T12:53:51.230" Title="How Can I View KIC 9832227?" Tags="&lt;star&gt;&lt;binary-star&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20938" PostTypeId="2" ParentId="20937" CreationDate="2017-05-05T05:54:42.213" Score="3" Body="&lt;p&gt;Right now it doesn't look like much. It has a magnitude of about 12. And appears as a single point of light. So it requires the upper end of amateur equipment to observe. If you want to measure the variation, which is only 0.2 magnitudes, you would need to take images. So you would need a proper photo-imaging system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words it is a tricky amateur target. You could try using an online telescope. The star is within the range of the Bradford robotic telescope.  (Except this is offline and doesn't have an expected date to be back in service :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Digital sky survey (using professional equipment), the star appears in the &lt;a href=&quot;http://www.wikisky.org/?ra=19.487762&amp;amp;de=46.62219199999999&amp;amp;zoom=11&amp;amp;show_grid=1&amp;amp;show_constellation_lines=1&amp;amp;show_constellation_boundaries=1&amp;amp;show_const_names=0&amp;amp;show_galaxies=1&amp;amp;show_box=1&amp;amp;box_ra=19.487762&amp;amp;box_de=46.622192&amp;amp;box_width=50&amp;amp;box_height=50&amp;amp;img_source=DSS2&quot; rel=&quot;nofollow noreferrer&quot;&gt;centre of this image&lt;/a&gt;. This has very high magnification, this image is about 4 arcminutes across, about 1/8 the apparent size of the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/trSO5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/trSO5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is hard to give an exact spec. There is a formula: $m=\log_{10}(D)\times 5 + 7$ which calculates the limiting magnitude $m$, from the diameter of the mirror in cm $D$. This suggests you need at least 13cm mirror. However it assumes clear and stable skies, so in practice you would need something clearly in excess of that.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-05-05T18:05:50.027" LastActivityDate="2017-05-05T18:05:50.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20940" PostTypeId="2" ParentId="872" CreationDate="2017-05-05T10:01:43.390" Score="3" Body="&lt;p&gt;I am not feeling completely satisfied with Alexey Bobrick's argument: &quot;interesting things will start happening when the total mass of the planets will start being comparable to the mass of the star. So the limit would certainly exist.&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us consider a self-similar hierarchical planetary system, where planet number $p$ has semi-major axis $a_p$ and where $a_{p+1}&amp;gt;&amp;gt;a_p$ (say, as in a geometric progression). For planet $p$, all the mass inside its orbit is that &quot;of the star&quot;. In other words, the effective mass of the star depends on the planet we are considering and it has no limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see any argument going against the stability of such a system.&lt;/p&gt;&#xA;" OwnerDisplayName="user15104" LastActivityDate="2017-05-05T10:01:43.390" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20941" PostTypeId="2" ParentId="20932" CreationDate="2017-05-05T12:36:18.660" Score="-2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Einstein predicted a declination of 1.74 arc seconds, but according to Newton the declination should have been 0.86 arc seconds. So is it a coincidence that the prediction of Einstein is twice as large... ? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. Einstein's figure is twice Newton's because of the wave nature of matter. Check out Ned Wright's &lt;a href=&quot;http://www.astro.ucla.edu/~wright/deflection-delay.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;deflection and delay&lt;/a&gt; article:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ndyzV.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ndyzV.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He says this: &lt;em&gt;&quot;Newtonian mechanics predicts that a particle traveling at the speed of light which just grazes the edge of the Sun will be deflected by 0.875 seconds of arc&quot;&lt;/em&gt;. But that's a particle such as an electron, which we might have created from light in &lt;a href=&quot;https://en.wikipedia.org/wiki/Pair_production&quot; rel=&quot;nofollow noreferrer&quot;&gt;pair production&lt;/a&gt;. Pair production &lt;em&gt;&quot;often refers specifically to a photon creating an electron-positron pair&quot;.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously an electron isn't the same thing as a photon. One essential difference is mass, another is charge, and another is the way it's deflected by gravity. If you threw an electron fast past the Sun, nearly as fast as light, its deflection would be half the deflection of a photon.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can understand why by thinking what would happen if you trapped the photon in a gedanken mirror-box such that it was going round and round a square path. Then if you put it in a gravitational field the light would be deflected like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OZdQ3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OZdQ3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horizontals curve down, and because of this the position changes. The photon in its gedanken mirror-box &lt;em&gt;falls down&lt;/em&gt;. Then if you were to throw it fast past the Sun you'd say it was deflected - it's moving fast from left to right, but it falls down too. But note that the verticals are not affected. Only half the light path is curved by the Sun. The electron is affected in a similar fashion because of the &lt;a href=&quot;https://www.google.co.uk/?gfe_rd=cr&amp;amp;ei=EUM_VY2-Juiq8wfv_oCgDA&amp;amp;gws_rd=ssl#q=wave+nature+of+matter&amp;amp;spf=406&quot; rel=&quot;nofollow noreferrer&quot;&gt;wave nature of matter&lt;/a&gt;. This is why the deflection of matter is half the deflection of light.    &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2017-05-05T12:36:18.660" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20946" PostTypeId="2" ParentId="20284" CreationDate="2017-05-06T08:34:41.667" Score="1" Body="&lt;p&gt;The blue &lt;code&gt;fraction&lt;/code&gt; function reaches its maximum at full moon time, 12. März in your diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At that time, the fractional part of the red &lt;code&gt;phase&lt;/code&gt;, i.e., &lt;code&gt;moon$phase - floor(moon$phase)&lt;/code&gt; reaches 0.5, because a full Moon occurs at half time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see no &quot;out of sync&quot; here.&lt;/p&gt;&#xA;" OwnerUserId="16122" LastActivityDate="2017-05-06T08:34:41.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20947" PostTypeId="1" CreationDate="2017-05-06T09:30:56.773" Score="4" ViewCount="11634" Body="&lt;p&gt;When the moon is full, fully lit by the sun, will it appear full everywhere on the planet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. if I where to set the date for the next three full moons for say: Sweden, Bali and Seattle how different would those dates be?&lt;/p&gt;&#xA;" OwnerUserId="16813" LastEditorUserId="37191" LastEditDate="2021-03-07T03:28:45.370" LastActivityDate="2021-03-07T03:28:45.370" Title="Is the full moon full everywhere on the planet?" Tags="&lt;the-moon&gt;&lt;moon-phases&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20948" PostTypeId="1" AcceptedAnswerId="20950" CreationDate="2017-05-06T10:59:25.273" Score="1" ViewCount="154" Body="&lt;p&gt;I am interested in learning about big bang nucleosynthesis, nuclear fusion up to iron in stellar cores and beyond iron in supernovas, and into the lithium problem (galactic abundance anomoly for atomic numbers 3,4,5). As these topics are interlinked, I have been hoping to find good resources on the subject. Can anyone recommend a good source for self-study? In case it's relevant (regarding difficulty level), I'm a physics grad student lacking exposure to related topics. &lt;/p&gt;&#xA;" OwnerDisplayName="user15317" LastActivityDate="2017-05-25T00:25:53.480" Title="Can you recommend a book about big bang nucleosynthesis and chemical abundances?" Tags="&lt;supernova&gt;&lt;core&gt;&lt;elemental-abundances&gt;&lt;nucleosynthesis&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20949" PostTypeId="2" ParentId="20947" CreationDate="2017-05-06T11:03:05.700" Score="2" Body="&lt;p&gt;The points in time where the Moon is new/1st quarter/full/last quarter are defined to be from a &lt;em&gt;geocentric&lt;/em&gt; point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plane defined by Sun, Earth/Moon barycenter, Moon separates those that can see a full Moon - i.e., the ones on that plane and the ones closer to the Sun - from those who have to wait a little.&lt;/p&gt;&#xA;" OwnerUserId="16122" LastActivityDate="2017-05-06T11:03:05.700" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20950" PostTypeId="2" ParentId="20948" CreationDate="2017-05-06T11:41:43.177" Score="3" Body="&lt;p&gt;I highly recommend &lt;a href=&quot;https://www.amazon.co.uk/Nucleosynthesis-Chemical-Evolution.../dp/0521840309&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nucleosynthesis and Chemical Evolution of Galaxies&lt;/a&gt; by Bernard Pagel. It contains the basics of nuclear reactions andstellar evolution, chapters on big bang nucleosynthesis and light element production, as well as covering the broad swathe of stellar nucleoynthesis and how these link together into predicting the chemical evolution of galaxies.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-05-06T11:41:43.177" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20951" PostTypeId="2" ParentId="20947" CreationDate="2017-05-06T14:34:43.410" Score="5" Body="&lt;p&gt;The moon phase is corresponding to the position of the moon round the earth. The peak of full moon is the same time. At this time, one side of the world that can see the peak of full moon and another side not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the phase change is much slower than earth rotation, another side of the world can see near full moon at the moon-set or at the moon-rise, a few hours before or after the peak of full moon, depending on location. While it is not at the peak, by comparing to previous and next day, it is still perceived as full moon on this day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, whether Sweden, Bali and Seattle can see this peak at the same time, it varies on each full moon. But their longitude varies too much, it is impossible for three places to see this peak at the same time. At least one of them is facing away from the moon at the peak of full moon.&lt;/p&gt;&#xA;" OwnerUserId="2074" LastActivityDate="2017-05-06T14:34:43.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20952" PostTypeId="1" AcceptedAnswerId="20956" CreationDate="2017-05-06T17:37:50.897" Score="3" ViewCount="610" Body="&lt;p&gt;I know that light is considered as energy, and as such has a mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is certainly taken into account in universes models and calculations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm wondering: How much of the total mass of the universe does light and electromagnetic waves accounts for? And how much does it contributes to its expansion? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it positive or negative? Since it is considered as mass only, it should decrease expansion somehow. Am I right about this?&lt;/p&gt;&#xA;" OwnerUserId="15907" LastEditorUserId="4085" LastEditDate="2017-05-07T05:01:08.293" LastActivityDate="2017-05-07T14:58:18.053" Title="Total mass of light?" Tags="&lt;universe&gt;&lt;light&gt;&lt;space-time&gt;&lt;expansion&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ClosedDate="2017-05-08T13:19:20.123" ContentLicense="CC BY-SA 3.0" />
  <row Id="20954" PostTypeId="1" AcceptedAnswerId="20958" CreationDate="2017-05-06T22:14:03.320" Score="4" ViewCount="231" Body="&lt;p&gt;(Edited at James K's recommendation to stay on-topic for this SE.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://astronomy.stackexchange.com/questions/856/do-moons-have-moons&quot;&gt;my&lt;/a&gt; &lt;a href=&quot;https://astronomy.stackexchange.com/questions/18179/how-massive-must-a-rocky-body-be-to-liquify-adequately-to-eliminate-internal-mas&quot;&gt;reading&lt;/a&gt; on this SE and &lt;a href=&quot;http://curious.astro.cornell.edu/about-us/44-our-solar-system/the-moon/general-questions/104-can-moons-have-moons-intermediate&quot; rel=&quot;nofollow noreferrer&quot;&gt;from&lt;/a&gt; &lt;a href=&quot;https://www.universetoday.com/109666/can-moons-have-moons/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Googling&lt;/a&gt; I've gotten conflicting reports on the viability of subsatellites this close to the Sun, but a naive application of the Hill Sphere equation suggests that if it got captured at all, a large asteroid could theoretically orbit the moon in a stable orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't know is this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;For how long? Longer than a couple hundred thousand years, such that humans on Earth wouldn't have known anything else? Longer than a million years? Probably not longer than a billion, due to the tidal effects decaying the asteroid's orbit, but is that a bad assumption?&lt;/li&gt;&#xA;&lt;li&gt;How eccentric would the asteroid's orbit be? Would it be unlikely for it to orbit with negligible eccentricity, whether on the ecliptic or not? If so, what kind of eccentricity/orbital inclination would be reasonable?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16816" LastEditorUserId="16816" LastEditDate="2017-05-07T19:07:08.857" LastActivityDate="2017-05-07T19:07:08.857" Title="Could the Moon naturally capture a large asteroid and keep it as a subsatellite? For how long?" Tags="&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20956" PostTypeId="2" ParentId="20952" CreationDate="2017-05-07T14:58:18.053" Score="2" Body="&lt;h1&gt;First, some clarifications:&lt;/h1&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I know that light is considered as energy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, light is a group of photons and those have energy&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So it is certainly taken into account in universes models and calculations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The energy of radiation is minimal, reaching about $0.01\%$ (yes, what you &lt;a href=&quot;https://astronomy.stackexchange.com/questions/20952/total-mass-of-light#comment35426_20952&quot;&gt;said is correct, kingW3&lt;/a&gt;) of the energy of the Universe, and that's because of two phenomena causing it to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Decrease in absolute amounts: because of redshift.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Decrease in relative amounts: because of more dark energy, because it doesn't &quot;dilute&quot; with the expansion:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Think of it in this way: even if there is more space (because of expansion) the ratio between the dark energy and volume of the observable universe is constant &lt;sub&gt;(almost, this is more like a metaphor)&lt;/sub&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Another way to think about (and &lt;strong&gt;was&lt;/strong&gt; one of the possibilities of what is dark energy) is to think of it as a property of space. So you would see that the ratio remains constant.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it positive or negative?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If it were negative, we would be testing some kind of light propulsion engine or using it for building an Alcubierre warp drive. You don't see that happening. Also, light would be a tachyon (which are bad: if you ever see a theory produces a tachyon, &lt;em&gt;throw the theory out of the window&lt;/em&gt;, as it's equivalent to saying it doesn't work)&lt;/li&gt;&#xA;&lt;li&gt;If it were positive it would have mass, we wouldn't have discovered a lot of things, such as Special Relativity because of it traveling at subluminal speeds, and possibly &lt;strike&gt;it would hurt&lt;/strike&gt; we could &quot;feel it&quot; (because of touch) it depending on the exact value of its mass.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;it should decrease expansion somehow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It does, but $0.01\%$ of the energy of the Universe clearly is negligible, and as I said, that number decreases over time.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;The mass of light is equal to...?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Light is massless, meaning that its rest mass ($m_0$) is zero. But that doesn't mean it can't interact gravitationally: the reason objects have gravity is that they have mass (and momentum, if not you would have what I think is about $92\%$ less mass), and in this case, mass is &lt;em&gt;equivalent&lt;/em&gt; to energy in the sense that it curves Spacetime similarly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other way to phrase that last sentence is as &lt;a href=&quot;https://astronomy.stackexchange.com/questions/20952/total-mass-of-light#comment35430_20952&quot;&gt;StephenG&lt;/a&gt; put it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] Energy (i.e. non-mass energy) can contribute to the distortion of space-time, but that's not the same as it having mass.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Summary of what I said, answering other questions:&lt;/h1&gt;&#xA;&#xA;&lt;h3&gt;How much of the total mass of the universe does light and electromagnetic waves accounts for?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;About $0.01\%$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;And how much does it [energy of radiation] contributes to its [the Universe's] expansion?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Almost nothing.&lt;/p&gt;&#xA;" OwnerUserId="16776" LastActivityDate="2017-05-07T14:58:18.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20958" PostTypeId="2" ParentId="20954" CreationDate="2017-05-07T15:26:48.170" Score="8" Body="&lt;p&gt;I'm going to discuss only our moon, to keep this on topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sort of orbit could be stable on the sort of timescales that you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objects in Low Lunar Orbit will likely suffer from the irregular gravity of the moon. There are &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Mass_concentration_(astronomy)&quot; rel=&quot;noreferrer&quot;&gt;mass concentrations&lt;/a&gt;&quot; as a result of ancient meteorite impacts. These &quot;mascons&quot; distort the orbits of objects in low orbit, and cause them to collide with the moon in a relatively short time. However at 24000km the gravity field is smoothed by your distance from the surface. You should not have problems with mascons at this distance, at least not over 100,000 years or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://curious.astro.cornell.edu/about-us/44-our-solar-system/the-moon/general-questions/104-can-moons-have-moons-intermediate&quot; rel=&quot;noreferrer&quot;&gt;Hill sphere of the moon extends to about 60000km&lt;/a&gt;, so your orbit is well within the Hill radius, and so can be stable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon spins slowly, so tidal effects would tend to cause your asteroid to lose altitude, and eventually hit the moon. The timescale of this would be longer than a few 100,000 years &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting your asteroid in orbit is very tricky. You would need to hypothesise double asteroids transferring momentum between each other and the moon, and one being lost while the other ends up in lunar orbit. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-05-07T15:26:48.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20959" PostTypeId="1" CreationDate="2017-05-07T18:30:02.953" Score="-1" ViewCount="216" Body="&lt;p&gt;If Saturn is 83x bigger that Earth by surface area why isn't the volume also 83x bigger than earth by volume? Is there a geometric or astronomical reason?&lt;/p&gt;&#xA;" OwnerUserId="16821" LastEditorUserId="2365" LastEditDate="2017-05-07T19:57:03.570" LastActivityDate="2017-05-09T14:40:07.187" Title="Why is the volume of Saturn is 764 times bigger than Earth, but its surface area only 83 times bigger?" Tags="&lt;saturn&gt;&lt;space-geometry&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2017-05-09T18:03:43.770" ContentLicense="CC BY-SA 3.0" />
  <row Id="20960" PostTypeId="2" ParentId="20959" CreationDate="2017-05-07T19:21:42.367" Score="2" Body="&lt;p&gt;The surface area of a sphere is $A=4\pi r^2$ and the volume is $V=\frac 4 3 \pi r^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the ratio $a = \frac {A_{saturn}} {A_{earth}} = (\frac {r_{saturn}} {r_{earth}})^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the volume ratio is $b = \frac {V_{saturn}} {V_{earth}} = (\frac {r_{saturn}} {r_{earth}})^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting ratio is $a = b ^{\frac 2 3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;User @Alex added that in your example, $764^{\frac 2 3}\approx83.6$, which I'd neglected to mention.  Thank's @Alex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And clearly that won't be equal to unless the two planets had the same radius.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastEditorUserId="10196" LastEditDate="2017-05-09T14:40:07.187" LastActivityDate="2017-05-09T14:40:07.187" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20961" PostTypeId="2" ParentId="14910" CreationDate="2017-05-07T19:27:11.430" Score="0" Body="&lt;p&gt;Yes.  The Tele Vue NP101is telescope, Finger Lakes Instrumentation Microline&#xA;ML8051 camera and Software Bisque MyT Paramount with matching tripod is &quot;all&quot; of the 2017 off-the-shelf amateur astronomy equipment you need to measure starlight-bending general relativity effects during a total solar eclipse ....plus a LOT of prep work and a LOT of data analysis!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's some articles of interest with details of how to measure GR light bending during a total eclipse:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.skyandtelescope.com/wp-content/uploads/Bruns-SAS2016-paper-v7.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.skyandtelescope.com/wp-content/uploads/Bruns-SAS2016-paper-v7.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://w.astro.berkeley.edu/~kalas/labs/documents/kennefick_phystoday_09.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://w.astro.berkeley.edu/~kalas/labs/documents/kennefick_phystoday_09.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16822" LastEditorUserId="16822" LastEditDate="2017-05-07T20:47:32.393" LastActivityDate="2017-05-07T20:47:32.393" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20970" PostTypeId="2" ParentId="20947" CreationDate="2017-05-08T12:55:20.710" Score="6" Body="&lt;p&gt;As others have said, the full moon (date and time) is set by when the center of the Moon passes &lt;em&gt;over or under&lt;/em&gt; a line that passes from the center of the sun through the center of the Earth.    That gives the full moon a precise time, to the minute, that's the same everywhere on earth, provided you adjust for timezone.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;over or under&quot; is important, cause mostly the Moon doesn't orbit in the same plane as the Earth's orbit, so it's usually above or below that line drawn from the sun through the center of the Earth.   If it's near the Spring or Fall Equinox and they are more lined up, and the Moon passes very close to the line, then it passes through the Earth's shadow.   In the sense of being a full circle in the sky, the Moon is never that, because that would require the Moon be in the Earth's shadow.    That's why full moon's often appear wider than they are tall.   A full moon often appears like a squashed ball, not a circle.   Because by definition, a full Moon is full East to West, but not North to South.   It remains within about 6 degrees of full North to South all the time, so that's generally ignored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, unlike the new-years countdown, which happens one hour later for every timezone, the full moon moment happens the same time everywhere on Earth, day or night and whether the moon is visible in the sky or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike sunsets, most of us see the full moon as lasting the entire night and perhaps, even back to back nights.  The specific minute of the full moon doesn't strike most of us as particularly relevant.  &quot;The moon is full tonight&quot; is still a perfectly good thing to say, and &quot;The moon was full 10 minutes ago, dang we missed it&quot;, would frankly be a strange thing to say, unless you wanted to catch it at Moonrise or Moonset - which can make for some great low to the horizon moon watching.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, if we look at the full moon as the night when it's fullest, then you have up to a 12 hour difference, depending on longitude, because night begins 12 hours later on one side of the globe, than on the other.   (And the moon moves across the sky, so this takes a bit longer than 12 hours, but lets not get into that).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you wanted to get silly-fussy and you wanted to know when the Moon would appears fullest to you.   That's not an official fullest time, but that does change slightly depending on your longitude.  It takes about 3 1/2 hours for the Moon to cross the Earth's diameter, so when it appears fullest varies by about that much time, depending on where you are on the earth and if the moon is visible.  Our eyes can't detect a change in fullness that occurs over a 3.5 hour period.  It's too small, so this is kind of an irrelevant point and not astronomically relevant, but if you wanted to define the moment when the Moon is visually fullest, not just the moment when it passes over or under the line through the center of the Earth, that would give you about 3.5 hours of variation from one side of the Earth to the other.    I've never heard of anyone who felt that distinction was of particular use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct answer is that the astronomical full moon happens the same time everywhere on Earth.   The point of view answer can give you variations of up to 12 hours if you consider full moon starting on the rise of the full moon during the night of the full-moon, or 3.5 hours if you fuss over the precise longitude of Earth as the measuring point of when the Moon appears 100% fullest.  But those are point of view answers and not astronomical answers.   Officially The Moon is full everywhere on Earth at the same precise minute.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-05-08T12:55:20.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20971" PostTypeId="1" AcceptedAnswerId="20972" CreationDate="2017-05-08T18:53:56.463" Score="2" ViewCount="232" Body="&lt;p&gt;Would it be possible to measure the distance to a black hole of known mass by measuring the radius of the most redshifted radiation that is detectable? The idea is that since the point where that radiation vanishes is related to the gravitational potential difference between the observer and the point of emission of that radiation, we should be able to get an estimate of how far we are by observing how far into the black hole we can see (assuming that for a given mass and an observer at infinity that radius is fixed).&lt;/p&gt;&#xA;" OwnerUserId="9838" LastActivityDate="2017-05-08T19:10:46.257" Title="Theoretical measure of distance to black hole?" Tags="&lt;black-hole&gt;&lt;event-horizon&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20972" PostTypeId="2" ParentId="20971" CreationDate="2017-05-08T19:10:46.257" Score="3" Body="&lt;p&gt;Yes. If you could spatially resolve a black hole and measure the redshift of radiation at a given angular radius, then that would yield the distance to the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works because the amount of redshift depends on the ratio of radius to Schwarzschild radius, but the latter is known if the mass is known. Thus the redshift yields the radius. The distance is then given by the radius divided by the apparent angular radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There will be complications due to gravitational lensing and any orbital velocity around the black hole, but this does not stop it working in principle.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-05-08T19:10:46.257" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20973" PostTypeId="1" CreationDate="2017-05-08T19:50:15.750" Score="5" ViewCount="6158" Body="&lt;p&gt;Although &lt;a href=&quot;https://astronomy.stackexchange.com/questions/914/how-can-i-calculate-moon-phase-and-height&quot;&gt;this question&lt;/a&gt; apparently addresses the topic, it does not tell appropriate factors involving the moon phase.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found two distinct algorithms. &lt;a href=&quot;https://gist.github.com/miklb/ed145757971096565723&quot; rel=&quot;noreferrer&quot;&gt;this one&lt;/a&gt; and also &lt;a href=&quot;https://www.daniweb.com/programming/software-development/code/453788/moon-phase-at-a-given-date-python&quot; rel=&quot;noreferrer&quot;&gt;this one&lt;/a&gt;. I am less than a newbie, less than a layman, to astronomy - but I am not to programming, but I would like to skip testing the algorithm for a couple of years before choosing either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After unsuccessfully finding a comparison of different algorithms, I'd like to know of either of these algorithms is less error-prone, or if there is someone less error prone-to these (I know there is no stuff like a perfect algorithm, but I'd like to have a good one), or having appropriate ways to judge an algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my &lt;strong&gt;question&lt;/strong&gt; comes like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do these methods have a small error, enough to accurately calculate the moon phase?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you give me the most precise algorithm -or concepts / underlying theory I should know- astronomers would use as of today to calculate the moon phase?&lt;/p&gt;&#xA;" OwnerUserId="16834" LastActivityDate="2021-04-25T10:41:19.193" Title="What should I consider in order to accurately calculate the moon phase?" Tags="&lt;moon-phases&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20974" PostTypeId="1" AcceptedAnswerId="20977" CreationDate="2017-05-08T19:56:54.207" Score="12" ViewCount="1092" Body="&lt;p&gt;I have Googled &quot;iridium content of comets&quot; and can't find anything that suggests comets contain iridium in any detectable quantities. Yet, whenever I read an article about the Chicxulub impactor, the article invariably says something to the effect that it may have been created by an asteroid &lt;i&gt;or a comet&lt;/i&gt;. But Alvarez and company first offered the impact hypothesis for the K-T extinction by observing an iridium layer in global sedimentary deposits, which could only have come from an asteroid impact, since iridium is extremely rare in the Earth's crust. Later the Chicxulub structure was identified as the probable site of the asteroid impact.&lt;br&gt;&lt;br&gt; While a comet could have caused the Chicxulub crater, it could not have been responsible for the global iridium layer. So why do science writers offer the comet alternative? If Chicxulub was created by a comet, then another site must have been the source of the iridium layer.&lt;/p&gt;&#xA;" OwnerUserId="7108" LastEditorUserId="2365" LastEditDate="2018-08-01T10:18:06.340" LastActivityDate="2019-12-17T13:48:38.270" Title="Do comets contain any significant amounts of iridium?" Tags="&lt;asteroids&gt;&lt;comets&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20976" PostTypeId="1" CreationDate="2017-05-08T21:01:22.570" Score="1" ViewCount="1007" Body="&lt;p&gt;If I observe the sun and measure the local azimuth and elevation, and I also have my latitude, can I calculate the local sunrise and local sunset times? If so, how would I calculate it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, can the same calculations be used for the moonrise and moonset?&lt;/p&gt;&#xA;" OwnerUserId="16805" LastActivityDate="2017-05-18T20:49:03.487" Title="Determining sunrise and sunset times based on azimuth and elevation" Tags="&lt;the-sun&gt;&lt;the-moon&gt;&lt;azimuth&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20977" PostTypeId="2" ParentId="20974" CreationDate="2017-05-08T22:00:05.427" Score="14" Body="&lt;p&gt;I also googled &quot;iridium content of comets&quot;, and the first result was &lt;a href=&quot;https://news.dartmouth.edu/news/2013/04/dartmouth-researchers-say-comet-killed-dinosaurs&quot; rel=&quot;noreferrer&quot;&gt;https://news.dartmouth.edu/news/2013/04/dartmouth-researchers-say-comet-killed-dinosaurs&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there is currently no consensus on the nature of the Chicxulub impactor. The observations of iridium and osmium suggest an asteroid. A minority opinion is that a comet may be responsible. The authors of the article in the linked news report suggest that measurements of iridium concentrations from ocean cores may be erroneously high, due to sorting in marine sediments. If these cores are excluded, the amount of iridium measured is lower, and consistent with some models of the composition of comets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, comets do contain some iridium. The amount of iridium is not certain. Some models of comet nuclei contain more dust and less ice, and so have more iridium. The exact amount of iridium in the KT boundary layer is also uncertain. The low end of estimates would be consistent with a cometary impact. The nature of the impactor is not yet settled science, but the balance of opinion currently favours a stony asteroid.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-05-09T06:00:42.570" LastActivityDate="2017-05-09T06:00:42.570" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20978" PostTypeId="2" ParentId="20973" CreationDate="2017-05-09T10:12:39.813" Score="6" Body="&lt;p&gt;The moon's phase isn't easy to calculate accurately. Fortunately, the 4 main phases (new moon, first quarter, full moon, last quarter) are a bit more stable than the intermediate phases. Good main phase algorithms use a dozen or so sine calculations for the new or full moon, and a similar number for the first or last quarter. Those two algorithms you linked use very simple approximations so their results are rather crude.&lt;/p&gt;&#xA;&lt;p&gt;The period of the moon's phase cycle (the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_month#Synodic_month&quot; rel=&quot;nofollow noreferrer&quot;&gt;synodic month&lt;/a&gt;) is stable over the long term, but it drifts considerably over the short term. So algorithms like those in your links are usable over a rather long time span, and they can be useful first approximations to feed to better algorithms, but by themselves they only yield precision results by accident. ;)&lt;/p&gt;&#xA;&lt;p&gt;The guru of astronomical calculations for bodies in the solar system is &lt;a href=&quot;https://en.wikipedia.org/wiki/Jean_Meeus&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jean Meeus&lt;/a&gt;, so if you want quality algorithms make sure they reference Meuss. :)&lt;/p&gt;&#xA;&lt;p&gt;For seriously accurate ephemerides, especially if you need to cover a wide time span, consider using the work of JPL. You can access their ephemerides at the &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;Horizons&lt;/a&gt; Web page; a telnet and an email-based interface are also provided.&lt;/p&gt;&#xA;&lt;p&gt;If you want to do precise astronomical calculations (eg for eclipse prediction) you'll need to learn about the various time systems used in astronomy, in particular &lt;a href=&quot;https://en.wikipedia.org/wiki/Terrestrial_Time&quot; rel=&quot;nofollow noreferrer&quot;&gt;Terrestrial Time&lt;/a&gt; and its forerunner, Ephemeris Time.&lt;/p&gt;&#xA;&lt;p&gt;I see from your SO profile that you're familiar with Python. Good stand-alone Python moon phase code isn't easy to find online, but eventually I did find keturn's &lt;a href=&quot;http://bazaar.launchpad.net/%7Eketurn/py-moon-phase/trunk/annotate/head:/moon.py&quot; rel=&quot;nofollow noreferrer&quot;&gt;py-moon-phase&lt;/a&gt;, which is a Python port of the core code of John Walker's rather ancient &lt;a href=&quot;http://www.fourmilab.ch/moontoolw&quot; rel=&quot;nofollow noreferrer&quot;&gt;Moontool&lt;/a&gt;, which uses algorithms from Meeus and also from Peter Duffett-Smith. I haven't actually tested John Walker / keturn's code, but it does look rather good (although it uses UTC instead of Ephemeris Time or Terrestrial Time). I have very similar (but rather messy) C code that's generally accurate to half a minute or better over the 20th century.&lt;/p&gt;&#xA;&lt;p&gt;However, there's a package called &lt;a href=&quot;http://rhodesmill.org/pyephem/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;PyEphem&lt;/a&gt; that can do all sorts of astronomical calculations. James K notes in the comments that PyEphem is now discouraged for new projects. The PyEphem developers recommend &lt;a href=&quot;https://rhodesmill.org/skyfield/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Skyfield&lt;/a&gt; instead. &lt;a href=&quot;https://www.astropy.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astropy&lt;/a&gt; can also do a lot of what PyEphem does, and &lt;em&gt;much&lt;/em&gt; more.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="16685" LastEditDate="2021-04-25T10:41:19.193" LastActivityDate="2021-04-25T10:41:19.193" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="20979" PostTypeId="1" AcceptedAnswerId="20985" CreationDate="2017-05-09T11:24:34.340" Score="-1" ViewCount="84" Body="&lt;p&gt;Will that be able to answer questions such as our origins or why we exist? Will it unlock some sort of new perception towards anything?  &lt;/p&gt;&#xA;" OwnerUserId="16838" LastActivityDate="2017-05-09T17:56:31.397" Title="What questions will we be able to answer if we find life on other celestial bodies?" Tags="&lt;universe&gt;&lt;life&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2017-05-09T18:03:25.623" ContentLicense="CC BY-SA 3.0" />
  <row Id="20980" PostTypeId="2" ParentId="20974" CreationDate="2017-05-09T12:41:53.883" Score="6" Body="&lt;p&gt;It occurs to me that part of your confusion may be in defining the difference between a comet and an asteroid in the first place.  Here are a couple claims.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://www.universetoday.com/33006/what-is-the-difference-between-asteroids-and-comets/&quot; rel=&quot;noreferrer&quot;&gt;UniverseToday&lt;/a&gt;, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Asteroids and comets have a few things in common. They are both&#xA;  celestial bodies orbiting our Sun, and they both can have unusual&#xA;  orbits, sometimes straying close to Earth or the other planets. They&#xA;  are both “leftovers” — made from materials from the formation of our&#xA;  Solar System 4.5 billion years ago. But there are a few notable&#xA;  differences between these two objects, as well. The biggest difference&#xA;  between comets and asteroids, however, is what they are made of.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;While asteroids consist of metals and rocky material, comets are made&#xA;  up of ice, dust, rocky materials and organic compounds. When comets&#xA;  get closer to the Sun, they lose material with each orbit because some&#xA;  of their ice melts and vaporizes. Asteroids typically remain solid,&#xA;  even when near the Sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://coolcosmos.ipac.caltech.edu/ask/181-What-is-the-difference-between-an-asteroid-and-a-comet-&quot; rel=&quot;noreferrer&quot;&gt;CalTech&lt;/a&gt;, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The main difference between asteroids and comets is their composition,&#xA;  as in, what they are made of. Asteroids are made up of metals and&#xA;  rocky material, while comets are made up of ice, dust and rocky&#xA;  material. Both asteroids and comets were formed early in the history&#xA;  of the solar system about 4.5 billion years ago. Asteroids formed much&#xA;  closer to the Sun, where it was too warm for ices to remain solid.&#xA;  Comets formed farther from the Sun where ices would not melt. Comets&#xA;  which approach the Sun lose material with each orbit because some of&#xA;  their ice melts and vaporizes to form a tail.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As JamesK noted, it is possible for &quot;rocky material&quot; in either a comet or an asteroid to contain iridium -- or not contain iridium.  I think the reason iridium is tied to the KT event is that there appears to be a world-wide (ish) layer with iridium, and that layer consistently dates to approximately the same epoch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. note also &lt;a href=&quot;https://physics.stackexchange.com/questions/25645/how-do-the-terms-comet-asteroid-meteoroid-meteor-and-meteorite-differ&quot;&gt;Physics.SE&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-05-09T12:41:53.883" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20981" PostTypeId="1" CreationDate="2017-05-09T14:34:12.170" Score="3" ViewCount="2406" Body="&lt;p&gt;Some of the spotting scopes which are e.g. 70mm claim to magnify terrestrial objects at 75x. But yet they cannot produce sharp and clear images at even 30x-35x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While 90mm scopes can do a better job than them but still they stop  producing clear view at higher magnification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I want to know is there a formula from which I can &lt;em&gt;estimate&lt;/em&gt; (if not accurately tell) their useful magnification at which they are able to focus objects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came across this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Magnification = Focal length / Eye piece focal length&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But not sure if that works for what I am asking.&lt;/p&gt;&#xA;" OwnerUserId="16841" LastEditorUserId="1378" LastEditDate="2017-08-07T22:23:36.687" LastActivityDate="2019-01-26T00:54:51.263" Title="What's the useful magnification for a scope with 70mm objective lens?" Tags="&lt;telescope&gt;&lt;fundamental-astronomy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20982" PostTypeId="2" ParentId="20981" CreationDate="2017-05-09T15:07:09.787" Score="1" Body="&lt;p&gt;You are correct that magnification is the Telescope Focal length / Eye piece focal length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an F/10 70 mm refractor, the focal length is 700 mm.  A common 20 mm eyepiece will get 35x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With excellent optics, a good rule of thumb is the maximum useful magnification is about 1.5x the telescope's aperture in mm.  So an excellent 70mm F/10 refractor might be useful up to 105x, which corresponds to a 7mm eyepiece.  If you have a fast, F/5 spotting scope, the maximum magnification would come with a 3.5mm eyepiece; eyepieces with such short focal lengths tend to be hard to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very cheap telescopes may have misaligned or badly figured optics, however, and the maximum useful magnification may be much lower.  To get maximum performance, the eyepieces have to be high-quality as well.  Spotting scopes tend to have lower focal ratios so they can be more compact; this requires more strongly curved surfaces on the lenses and makes them harder to figure correctly.  All telescopes, regardless of price, need to be correctly aligned to get good performance.&lt;/p&gt;&#xA;" OwnerUserId="14434" LastActivityDate="2017-05-09T15:07:09.787" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20984" PostTypeId="1" CreationDate="2017-05-09T17:09:53.693" Score="1" ViewCount="86" Body="&lt;p&gt;So, I've received this question to solve, it's not a very advanced one, though advanced enough to cause headaches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation is as follows: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Say, you are observing with a telescope with an effective area of 44000 $cm^2$. Assume that the total efficiency through the atmosphere, the telescope, optics and the filter is 50%. How long would it take to collect 80 photons from a source with V=20?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have little to no idea how to tackle this problem. Can someone help? I've had lectures about spectral flux density, and I'd assume you'll have to use it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adam.&lt;/p&gt;&#xA;" OwnerUserId="16845" LastActivityDate="2017-05-09T17:09:53.693" Title="How long would it take to collect a certain amount of photons?" Tags="&lt;fundamental-astronomy&gt;&lt;spectroscopy&gt;&lt;photons&gt;" AnswerCount="0" CommentCount="6" ClosedDate="2017-05-15T13:02:11.657" ContentLicense="CC BY-SA 3.0" />
  <row Id="20985" PostTypeId="2" ParentId="20979" CreationDate="2017-05-09T17:56:31.397" Score="2" Body="&lt;p&gt;The question is strange, but I think, answerable to a point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest question in astronomy today, hands down, is whether there is life out there.    Answering that question alone would be answering the single biggest question.    This is something that maybe the James Webb Space Telescope will be able to answer.   It's a huge question.   Answering it, IMHO, would be the top scientific discovery in my lifetime.  Bigger than Higgs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answering a key question tends to raise other questions.  That's the nature of discovery.   You should probably ask what questions it would raise not what questions it would answer, and even that might get closed for being too general and too dependent on what exactly is discovered.   What we'll be able to answer after would depend on how good a look we have at what we discovered and if we have a sample.   This is too generalized to give a real answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding life on other planets would raise questions such as &quot;what kind of life?&quot;, &quot;How advanced?&quot;, &quot;How did it form?&quot; &quot;How long has it been there?&quot;  &quot;How similar to live on Earth?&quot;  &quot;Warm or cold blooded?&quot;   &quot;How common is life on other planets&quot;, &quot;How abundant on that planet?&quot; &quot;Can we talk to them?/Do they have language?&quot;   &quot;Should we talk to them?&quot;   &quot;Will they try to eat us if we get too close?&quot;  &quot;How can we study them further?&quot;.   &quot;Do they have technology and if so, how advanced?&quot;  &quot;Are they friendly?&quot;  and probably several more.   Discoveries lead to more questions as a general rule of science.   Answers to the new questions can be answered in their own time.  Depending on how easy the study is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To your specific questions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Will that be able to answer questions such as our origins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Probably not without a sample.  That would require a close look at the chemistry to see if it's a similar enough pattern to have a shared origin.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;or why we exist?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This &quot;Why&quot; question is a whole different ballgame.  Orders of magnitude harder. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Will it unlock some sort of new perception towards anything?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It could.  It would depend what was discovered about the life.   This would depend on a lot of things, but discovering life on other planets certainly could change our perception in a number scientific fields and perhaps more generally than that, could change people's perception overall.   This is a very vague question though and not good for this board.    I've heard it said that the moon landings and pictures of the Earth from space changed our overall perception of the Earth as one planet, not divided nations.   I was to young during the moon landings to have that impression personally, so I don't know if that's a valid interpretation or not.   Might depend on the individual.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-05-09T17:56:31.397" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20987" PostTypeId="2" ParentId="18524" CreationDate="2017-05-10T15:53:16.920" Score="0" Body="&lt;p&gt;If it was a one time pass and the life exchanged was bacterial, then yes, it's possible.    That close, both planets would experience enormous tidal flexing and probably extinction level events, but the good news is that if both orbited the sun, passing that close they would change each other's orbits, so it could be a once time thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Events like that would be rare and potentially dangerous as a collision is also possible.   But in theory, a near pass that exchanges some primitive bacterial life is just possible, making it possible for one planet to give life to the other.     There's easier ways for that to happen like a large asteroid impact and debris knocked into space.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not possible for the two planets to do this periodically.   That would be too unstable and it's not possible for birds to jump from planet to planet, unless you don't mind them dying in the process.    But a one time pass and bacteria.   A very rare occurrence but possible.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-05-10T15:53:16.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20988" PostTypeId="1" CreationDate="2017-05-10T16:28:33.117" Score="1" ViewCount="225" Body="&lt;p&gt;&lt;strong&gt;Disclaimer:&lt;/strong&gt; I have a Machine Learning/Computer Vision background, &lt;em&gt;not&lt;/em&gt; Astrophysics, so please be a bit patient with me. If you feel like this question is too far out there it would be great if you could direct me to meta or some bulletin board/mailing list that you think is a better fit. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Important:&lt;/strong&gt; I am talking about &quot;datasets&quot; in a machine learning / computer vision manner. I do &lt;em&gt;not&lt;/em&gt; mean &quot;the entirety of SDSS data&quot; but rather a common set of inputs and (for supervised learning) associated outputs for a specific problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been looking for a while now to tinker with some problems in Astronomy/Astrophysics and I keep finding the same kind of advice when it comes to datasets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go to SDSS/GalaxyZoo, they provide downloads for their data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, while that is true, none of these datasets are anywhere near as accessible as datasets in Computer Vision. &quot;Over here&quot; a dataset is mostly a bunch of (huge) tar files with jpeg images and some &lt;code&gt;.mat&lt;/code&gt; or &lt;code&gt;.csv&lt;/code&gt; for annotations. The process of getting started is pretty straigth-forward: unzip, load annotations into memory, start sciencing (or usually, applying wild heuristics). Obviously I simplify slightly, but you get the idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to stay with the example of GalaxyZoo, the object coordinates, annotations and computed features are conveniently &lt;a href=&quot;https://data.galaxyzoo.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;presented&lt;/a&gt; as a &lt;code&gt;csv&lt;/code&gt;, &lt;code&gt;fits&lt;/code&gt; or &lt;code&gt;voTable&lt;/code&gt; file. But to get the actual visual data (if I understand this correctly) I have to, for each annotated galaxy...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;determine a bunch of &lt;code&gt;.fits&lt;/code&gt; urls from the coordinates in the annotations (one per band and galaxy) and download them (~12MB/file)&lt;/li&gt;&#xA;&lt;li&gt;download the containing &quot;tiles&quot; from the SDSS servers (in each of the bands I'd like to work with)&lt;/li&gt;&#xA;&lt;li&gt;extract the patch containing the galaxies (with the aptly named sExtractor)&lt;/li&gt;&#xA;&lt;li&gt;possibly apply some non-trivial corrections/transforms to the image patch&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I tried following along this procedure with a semi-randomly chosen paper from arXiv for which there is code available &lt;a href=&quot;https://github.com/EdwardJKim/dl4astro&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and without wanting to make any statement about said paper/code, this process is not comparable to the simple workflow I described above, and I would probably download some TB of data, were I to use the entire dataset, of which only a tiny fraction would actually be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting dataset only occupies a few GB of space, and has a very simple nature: &lt;strong&gt;ca 50000 images, with associated labels -1 or 1, for stars or galaxies&lt;/strong&gt;. The data preparation pipeline, while one might also want to optimize it for this task, has next to nothing to do with the &quot;real&quot; work being done in the paper, and so having the reader go through those steps makes little sense to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Did I understand the workflow correctly?&lt;/strong&gt; Maybe I'm just following the wrong kind of paper, or I really just overlooked a big red &quot;Download Here!&quot; button? Otherwise:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Did anyone go through these steps and make the resulting dataset available?&lt;/strong&gt; Or are there good reasons not to do so? From the size of the sExtractor manual I gather that the choice of parameters in this processing step might significantly influence the outcome of any downstream method. So...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Are there any sane/agreed-upon default settings for the preprocessing pipeline?&lt;/strong&gt; While for any given method, applying extra &quot;magic&quot; in the preprocessing might make sense, having a default set of parameters would make dataset creation and a bunch of other things much easier, no?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I am sorry if this is a naive newcomer question. And for the long post. :P&#xA;As a meta, I would add tags &quot;dataset&quot; and &quot;programming&quot; or some such, but I lack the reputation.&lt;/p&gt;&#xA;" OwnerUserId="14358" LastEditorUserId="14358" LastEditDate="2017-05-14T12:42:47.587" LastActivityDate="2017-05-14T12:42:47.587" Title="Are there any ready-made machine-learning datasets of photometric data that are easy to use?" Tags="&lt;data-analysis&gt;" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20989" PostTypeId="2" ParentId="18524" CreationDate="2017-05-10T18:36:34.803" Score="0" Body="&lt;p&gt;If you're thinking of occasional close encounters, this is very unlikely to work. The approach would be catastrophical for both planets. Their plate tectonics would get completely screwed up. It would kill a whole lot of life on both planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it could be possible that two planet-size bodies might coexist very close to each other, so that the tops of their atmospheres would not be separated by a very large gap. It's not clear whether this is stable long term from a gravitational p.o.v. Fluids would definitely migrate towards the stronger gravitational field over the very long term, unless there's some active mechanism (perhaps powered somehow by heat from the star?) pumping them back occasionally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last scenario is explored in the sci-fi book Rocheworld by Robert Forward. It's sci-fi but it does not stray too far from rigorous science.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2017-05-10T18:36:34.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20992" PostTypeId="2" ParentId="20018" CreationDate="2017-05-11T03:11:24.963" Score="2" Body="&lt;p&gt;This doesn't really answer your question, but may help you get started. If you run &lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/STACK/bc-stell-travel.pl&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/STACK/bc-stell-travel.pl&lt;/a&gt; and follow the instructions at &lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/STACK/README&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/STACK/README&lt;/a&gt; you can create a video similar to &lt;a href=&quot;https://www.youtube.com/watch?v=_4ej5rfYm4A&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=_4ej5rfYm4A&lt;/a&gt; but higher quality (youtube lowers the quality of uploaded videos).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the default settings, it's a ~6 hour Mach 1 flight from Menton, France to Rampart House, YT Canada starting at 2017-02-10 at 00:00:00 GMT at approximately 120x regular speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only vaguely interesting things about the default flight:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Since the flight goes fairly far north, you can see Polaris rising in the Northern sky for most of the flight.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since you traveling westwards fairly quickly, once you get far north enough, the stars go &quot;backwards&quot;: they rise in the west and set in the east.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2017-05-11T03:11:24.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20996" PostTypeId="1" CreationDate="2017-05-11T12:47:33.903" Score="3" ViewCount="72" Body="&lt;p&gt;I'm working with images from LASCO C3 coronagraph&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm able to download pictures from LASCO C3 using this form&#xA;&lt;a href=&quot;https://sohodata.nascom.nasa.gov/cgi-bin/data_query&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://sohodata.nascom.nasa.gov/cgi-bin/data_query&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, I want to do some analysis of CMEs on this pictures and determine some of their parameters, such as direction and speed (i know that is better to use STEREO A\B for that purposes), but also I need exact coordinates of SOHO station to determine the direction of the ejection in space&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I get the coordinates of the station for these pictures?&lt;/p&gt;&#xA;" OwnerUserId="16859" LastEditorDisplayName="user1569" LastEditDate="2018-09-30T16:53:28.483" LastActivityDate="2018-09-30T16:53:28.483" Title="How to get SOHO Satellite images' coordinates?" Tags="&lt;the-sun&gt;&lt;coordinate&gt;&lt;soho&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20998" PostTypeId="1" CreationDate="2017-05-11T22:07:44.180" Score="3" ViewCount="123" Body="&lt;p&gt;What i mean, why we still get dark if all the stars are sending their light to earth.&lt;/p&gt;&#xA;" OwnerUserId="16862" LastActivityDate="2017-05-12T01:27:08.110" Title="Why at night the light of all stars get in small amount to earth?" Tags="&lt;star&gt;&lt;light&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2017-05-14T21:56:45.257" ContentLicense="CC BY-SA 3.0" />
  <row Id="21000" PostTypeId="2" ParentId="20906" CreationDate="2017-05-12T03:18:08.083" Score="6" Body="&lt;p&gt;If you visit &lt;a href=&quot;https://github.com/Stellarium/stellarium/tree/master/skycultures&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/Stellarium/stellarium/tree/master/skycultures&lt;/a&gt; you see several sky cultures that include various constellation shapes (you also get this directory when you download stellarium).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, lets look at &lt;a href=&quot;https://github.com/Stellarium/stellarium/tree/master/skycultures/western_rey&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/Stellarium/stellarium/tree/master/skycultures/western_rey&lt;/a&gt;, in particular the file &lt;a href=&quot;https://github.com/Stellarium/stellarium/blob/master/skycultures/western_rey/constellationship.fab&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/Stellarium/stellarium/blob/master/skycultures/western_rey/constellationship.fab&lt;/a&gt; and line number 127 which reads:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Aql 10 98036 97649 97649 97278 97278 96229 96229 95501 95501 97804 99473 97804 95501 93747 93747 93244 95501 93805 93805 93429&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;Aql&quot; stands for the for the constellation Aquila, the Eagle (&lt;a href=&quot;https://github.com/Stellarium/stellarium/blob/master/skycultures/western_rey/constellation_names.eng.fab&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/Stellarium/stellarium/blob/master/skycultures/western_rey/constellation_names.eng.fab&lt;/a&gt; verifies this)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;10&quot; means there are 10 lines that make up the markings for Aquila. Note that these lines are specific to the H.A. Rey skyculture. Different cultures will have different lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next 20 number represent the lines themselves. For example &quot;98036 97649&quot; is a line from star 98036 to star 97649, and &quot;97649 97278&quot; is a line from star 97649 to star 97278. Since these two lines share star 97649, they are connected, but that's not always the case with adjacent lines. For example &quot;95501 97804 99473 97804&quot; represents a line from star 95501 to star 97804 and another unconnected line from star 99473 to star 97804.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next question: what exactly IS star 98036?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at &lt;a href=&quot;https://github.com/Stellarium/stellarium/blob/master/stars/default/name.fab&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/Stellarium/stellarium/blob/master/stars/default/name.fab&lt;/a&gt; you'll see it's Beta Aquilae, but not all stars have names in this file, and the name isn't particularly helpful in finding the star itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stellarium itself stores this data in the &quot;catalog files&quot; at &lt;a href=&quot;https://github.com/Stellarium/stellarium/tree/master/stars/default&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/Stellarium/stellarium/tree/master/stars/default&lt;/a&gt; with the format described at &lt;a href=&quot;http://stellarium.sourceforge.net/wiki/index.php/Star_Catalogue_Format&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://stellarium.sourceforge.net/wiki/index.php/Star_Catalogue_Format&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this can be ugly. Since Stellarium uses the standard Hipparcos catalog, you can visit &lt;a href=&quot;ftp://cdsarc.u-strasbg.fr/pub/cats/I/239/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ftp://cdsarc.u-strasbg.fr/pub/cats/I/239/&lt;/a&gt; and download &quot;hip_main.dat.gz&quot;. The format is described at &lt;a href=&quot;http://cdsarc.u-strasbg.fr/viz-bin/Cat?I/239&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://cdsarc.u-strasbg.fr/viz-bin/Cat?I/239&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The line relating to star 98036 starts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;H|       98036| |19 55 18.77|+06 24 28.6| 3.71|...&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which tells us star 98036 is at J2000 right ascension 19h55m18.77s, declination +06 degrees, 24'28.6&quot; declination, and has a magnitude of 3.71. This agrees almost perfectly with &lt;a href=&quot;https://en.wikipedia.org/wiki/Beta_Aquilae&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Beta_Aquilae&lt;/a&gt; (the declination is slightly off, but this may be because Beta Aquilae is a double star, and Hipparcos references the brighter star and Wikipedia references the midpoint of the double star-- or something).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other files in the stellarium github (eg, culture-specific star names) that also use the Hipparcos numbering designation.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorUserId="33520" LastEditDate="2020-04-26T08:42:45.573" LastActivityDate="2020-04-26T08:42:45.573" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21001" PostTypeId="2" ParentId="20947" CreationDate="2017-05-12T03:45:48.757" Score="8" Body="&lt;p&gt;[This is intended to supplement @userLTK's answer and reaches the same conclusion.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition, the Moon is full when its ecliptic longitude is exactly 180 degrees greater than the Sun's ecliptic longitude for a theoretical geocentric observer (first paragraph of &lt;a href=&quot;https://en.wikipedia.org/wiki/Full_moon&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Full_moon&lt;/a&gt; and definition of &quot;full moon&quot; at &lt;a href=&quot;http://www.ncsm.city.nagoya.jp/astro/astrotool/glossary.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.ncsm.city.nagoya.jp/astro/astrotool/glossary.html&lt;/a&gt; but couldn't find a 100% reliable resource).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By this definition, your location on Earth doesn't matter, since the full moon is defined by a theoretical observer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Realistically, the moon's position for an observer on the surface of the Earth can differ by as much as a degree from the geocentric position. (This is the arctangent of 1/60, since the moon's distance is about 60 times the Earth's radius.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the moon circles the ecliptic once every 27.321662 days or so (its &quot;sidereal period&quot;), its ecliptic longitude increases by about 0.55 degrees per hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, roughly speaking, the &quot;full moon&quot; at a given point on Earth can differ by as much as ±1.8 hours from the true full moon. By this token, it takes the &quot;full moon&quot; about 3.6 hours to cross the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, all of this refers to the moon's geometric position. If you allow for refraction, the results could be slightly different, especially since both the sun and moon experience over 0.5 degree refraction when rising/setting and the full moon rises right about when the sun sets and vice versa.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorUserId="18497" LastEditDate="2017-10-17T03:42:30.567" LastActivityDate="2017-10-17T03:42:30.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21002" PostTypeId="1" AcceptedAnswerId="21008" CreationDate="2017-05-12T15:44:43.553" Score="2" ViewCount="8203" Body="&lt;p&gt;I'm writing a program that converts position in the ECI (CIS, EPOCH JD2000.0) coordinate system to WGS 84 (CTS, ECEF) coordinates. To do so, I'm following what's described in &lt;a href=&quot;http://earth-info.nga.mil/GandG/publications/tr8350.2/tr8350.2-a/Appendix.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this document&lt;/a&gt; and implementing C++ code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My knowledge of astrophysics is very limited, so I find it hard to understand some terms. Especially, I can't seem to understand properly how the Greenwich Mean Sideral Time is computed, and why I always get different values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance (assuming time is &lt;em&gt;&lt;code&gt;2019-01-01 08:00:00&lt;/code&gt;&lt;/em&gt; and Julian Date is &lt;code&gt;2458484.833333&lt;/code&gt;):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If I use this on-line &lt;a href=&quot;http://dc.zah.uni-heidelberg.de/apfs/times/q/form&quot; rel=&quot;nofollow noreferrer&quot;&gt;GMST calculator&lt;/a&gt;, I find that GMST is &lt;em&gt;&lt;code&gt;14:42:45.3766&lt;/code&gt;&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If I use the expression used in &lt;a href=&quot;http://es.mathworks.com/matlabcentral/fileexchange/28233-convert-eci-to-ecef-coordinates?focused=5234019&amp;amp;tab=function&quot; rel=&quot;nofollow noreferrer&quot;&gt;this MATLAB script&lt;/a&gt; (showing the value in time, not degrees) I also get &lt;em&gt;&lt;code&gt;14:42:45.350&lt;/code&gt;&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;However, if I calculate it myself with the expression given in the above appendix, I get &lt;em&gt;&lt;code&gt;06:42:45.379163&lt;/code&gt;&lt;/em&gt; (with a clear offset of &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;According to the equations in page A-26, GMST depends upon the number of Julian centuries (Tu). H0 expression is exactly the same than that of &lt;a href=&quot;https://www.celestrak.com/columns/v02n01/&quot; rel=&quot;nofollow noreferrer&quot;&gt;CelesTrak&lt;/a&gt; for θ(0h). However, I can't really understand what the &lt;strong&gt;du&lt;/strong&gt; truly stands for, or more specifically, why the document shows &lt;code&gt;du → ±0.5, 1.5, 2.5...&lt;/code&gt;. Moreover, to find Λ (which is stated to be in &lt;em&gt;seconds of time&lt;/em&gt;) the expressions use values in radians and seconds of time indistinctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could somebody shed light into this?&lt;/p&gt;&#xA;" OwnerUserId="16864" LastActivityDate="2019-03-06T10:56:51.667" Title="How to find Greenwich Mean Sideral Time?" Tags="&lt;coordinate&gt;&lt;time&gt;&lt;orbital-mechanics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21003" PostTypeId="2" ParentId="8133" CreationDate="2017-05-12T16:03:42.683" Score="5" Body="&lt;p&gt;&lt;a href=&quot;https://medium.com/starts-with-a-bang/ask-ethan-how-bright-is-the-earth-as-seen-from-the-moon-627eb1554ca9&quot; rel=&quot;noreferrer&quot;&gt;Ask Ethan: How bright is the Earth as seen from the Moon?&lt;/a&gt; has some detailed explanations, including discussions of lunar eclipses as seen from the moon. It doesn't include magnitude calculations, but it concludes that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a “full Earth” as seen from the Moon is about 43 times brighter than the full Moon is as seen from Earth. When the icecaps are larger and the cloud cover is greater — and also when the deserts are visible in the Sun — the Earth appears at its brightest, up to approximately 55 times brighter than the Moon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;log(43) to the base 2.512 is 4.1, which when subtracted from the full moon's mean magnitude of -12.7 yields magnitude -16.8. Using the &quot;55 times brighter&quot; figure would lead to a maximum brightness of -12.7 - 4.4 = -17.1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't seem to cover the effects of the distance between the Earth and Moon, so it might get even brighter at perigee.&lt;/p&gt;&#xA;" OwnerUserId="16634" LastActivityDate="2017-05-12T16:03:42.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21004" PostTypeId="2" ParentId="21002" CreationDate="2017-05-12T17:25:38.623" Score="2" Body="&lt;p&gt;I worked out the GMST using the method described in &lt;a href=&quot;http://aa.usno.navy.mil/faq/docs/GAST.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://aa.usno.navy.mil/faq/docs/GAST.php&lt;/a&gt;. This should be an authoritative source, and claims 0.1 second accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In pseudo-code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;midnight = floor(2458484.833333) + 0.5            // J0 = 2458484.5&#xA;days_since_midnight = 2458484.833333 - 2458484.5  // = 0.333333&#xA;hours_since_midnight = 0.333333 * 24              // H = 8.0&#xA;days_since_epoch = 2458484.833333 - 2451545.0     // D = 6939.833333&#xA;centuries_since_epoch = days_since_epoch / 36525  // T = 0.190002&#xA;whole_days_since_epoch = 2458484.5 - 2451545.0    // D0 = 6939.5&#xA;&#xA;GMST = 6.697374558 &#xA;     + 0.06570982441908 * whole_days_since_epoch &#xA;     + 1.00273790935 * hours_since_midnight&#xA;     + 0.000026 * centuries_since_epoch**2        //=470.712605328&#xA;&#xA;GMST_hours = 470 % 24 // = 14&#xA;GMST_minutes =  floor(0.712605328 * 60) //=42(.7563197)&#xA;GMST_seconds =  0.7563197*60 // =45.38&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This calculation agrees with the online calculator. The method here (or as described at the link) should be easy to implement.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-11-05T20:55:25.280" LastActivityDate="2018-11-05T20:55:25.280" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21006" PostTypeId="1" AcceptedAnswerId="21017" CreationDate="2017-05-13T10:04:12.467" Score="3" ViewCount="1328" Body="&lt;p&gt;I am trying to get a better understanding of cosmological distances, in particular the angular diameter distance which I have also seen referred to as angular size distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am looking for is a more detailed explanation of why the angular diameter distance &quot;turns over&quot; in the following graph.&lt;a href=&quot;https://i.stack.imgur.com/byFEx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/byFEx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia states in their article on angular diameter distance that objects beyond a certain redshift have a smaller angular diameter and as such appear larger on the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have searched around and have been unable to find a complete explanation of this behaviour and have only found brief answers which state the expansion of the universe or the size of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like if possible for a more detailed explanation as well as the physical significance if any at all of what it means for an object to be at this redshift when they emitted light we are now observing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for any feedback or and discussion or suggestions where to look. I would also like to include as it may be of some help that I am currently in my final year of undergraduate studies in physics so please do not hold back any details or technicalities!&lt;/p&gt;&#xA;" OwnerUserId="16872" LastEditorUserId="2365" LastEditDate="2017-05-13T11:50:43.117" LastActivityDate="2017-05-14T12:40:23.927" Title="Understanding The Turnover Point of Angular Diameter Distance" Tags="&lt;cosmology&gt;&lt;distances&gt;&lt;redshift&gt;&lt;angular-diameter&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21007" PostTypeId="1" AcceptedAnswerId="21014" CreationDate="2017-05-13T10:18:42.187" Score="-1" ViewCount="773" Body="&lt;p&gt;If the planetary geocentric position of the moon is known (say, longitude &#xA;&quot;254°31'35&quot;&quot;) is it possible to calculate such position for north/south (ascending/descending) moon nodes? Or what else needs to be known?&lt;/p&gt;&#xA;" OwnerUserId="16871" LastEditorUserId="2365" LastEditDate="2017-05-13T11:51:56.470" LastActivityDate="2017-05-13T21:43:25.927" Title="How moon north/south nodes can be calculated" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21008" PostTypeId="2" ParentId="21002" CreationDate="2017-05-13T15:58:21.740" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;To do so, I'm following what's described in &lt;a href=&quot;http://earth-info.nga.mil/GandG/publications/tr8350.2/tr8350.2-a/Appendix.pdf&quot; rel=&quot;noreferrer&quot;&gt;this document&lt;/a&gt; and implementing C++ code.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is a very old document you are using. It's using the 1982 IAU precession model. There have been multiple updates since then. The latest set of changes are working toward eliminating the concept of Greenwich Sidereal Time. (In software terms, GMST and GAST are deprecated concepts. While still currently supported, support will be dropped in some future release of the Earth rotation model.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Other than using an outdated version of a deprecated concept, what did you do wrong? The expression you are using to compute GMST is at the top of page A-26:&#xA;$$\Lambda = H_0 + \Delta H + \omega^{\ast}\,(t-\Delta t)$$&#xA;The $\Delta H$ term takes one from GMST to GAST. Dropping that term, the expression for GMST is&#xA;$$H = H_0 + \omega^{\ast}\,(t-\Delta t)\tag1$$&#xA;where&#xA;$$\begin{align}&#xA;H_0 = &amp;amp;\,24110.54841 + 8640184.812866 T_u + 0.093104 {T_u}^2 - 6.2\times10^{-6} {T_u}^3\tag 2 \\&#xA;&amp;amp;\,\text{is the Greenwich sidereal time at midnight, in seconds of time} \\&#xA;\omega^{\ast} = &amp;amp;\,7.2921158553\times10^{-5} + 4.3\times10^{-15} T_u\tag3 \\&#xA;&amp;amp;\,\text{is the Earth's sidereal rotation rate, in radians}/\text{second} \\&#xA;t\phantom{\Delta}\quad&amp;amp;\text{is UTC time of day, in seconds, and} \\&#xA;\Delta t\quad&amp;amp;\text{is the difference between UT1 and UTC.}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there's a units mismatch in equation (1). Equation (2) yields a result in units of sidereal seconds while equation (3), when multiplied by time of day, yields a result in units of radians. The latter needs to be converted to sidereal seconds. To do this, simply multiple both terms in equation (3) by $\frac{86400}{2\pi}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\omega^{\ast} = &amp;amp;\,1.00273790935 + 5.9\times10^{-11} T_u\tag4 \\&#xA;&amp;amp;\,\text{is the Earth's sidereal rotation rate, in sidereal seconds}/\text{UT second}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying this to 2019 Jan 01 08:00:00 UTC yields the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Julian date at midnight UTC 2019 Jan 01 is $2458484.5$, or $2451545+6939.5$. See footnote 1 for an explanation of this convention.&lt;/li&gt;&#xA;&lt;li&gt;Dividing the residual 6939.5 by 36525 yields $T_u$: $T_u = 6939.5 / 36525 = 0.1899931553730322$.&lt;/li&gt;&#xA;&lt;li&gt;Applying equation (2) yields the Greenwich Mean Sidereal Time at midnight: $H_0 = 1665686.527373333$ sidereal seconds.&lt;/li&gt;&#xA;&lt;li&gt;Applying equation (4) and multiplying by $t=8*3600$ seconds yields another 28878.85178960284 sidereal seconds to the above. What about $\Delta t$, aka dUT1? I've ignored it (i.e., treated it as if it was zero.) See footnote 2.&lt;/li&gt;&#xA;&lt;li&gt;1665686.527373333 + 28878.85178960284 = 52965.37916293584 (modulo 86400). This is the mean sidereal time, in seconds. Converting to hours, minutes and seconds results in 14:42:45.4. The 0.4 seconds is bogus because I ignored dUT1.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Footnotes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;There's a problem with using Julian dates to express time when high precision is of concern. The problem is that adding 2.45 million and change to the time since noon on 1 Jan 2000 results in potential truncation problems. The next representable Julian date after midnight 1 Jan 2000 is 40 microseconds after midnight. I myself try to stay a couple of orders of magnitude away from the quantization level, which means problems for dates at the millisecond level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost every scientific package that supports Julian dates either uses a pair of double precision numbers or uses some sort of modified Julian date, such as time since 12 Noon 16 Nov 1858 (omits the leading 2.4 million in Julian date), time since 12 noon 23 May 1968 (omits the leading 2.44 million), or time since J2000.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The term $\Delta t$ in equations (2) and (4) is better known as dUT1. This is the difference between UT1 (time using the mean rotating Earth as a clock) and UTC (time that runs at the same rate as an atomic clock, but with leap seconds occasionally inserted to keep UT1 and UTC in sync). Those leap seconds keep $|\text{UT1}-\text{UTC}| &amp;lt; 0.9\,\text{seconds}$. Simply ignore the $\Delta t$ term if that's the case -- but remember that fractions of a second are meaningless.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-05-13T15:58:21.740" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21009" PostTypeId="5" CreationDate="2017-05-13T17:32:00.193" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2017-05-13T17:32:00.193" LastActivityDate="2017-05-13T17:32:00.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21010" PostTypeId="4" CreationDate="2017-05-13T17:32:00.193" Score="0" Body="Questions about using the Python programming language to solve astronomical problems. General questions about Python are off-topic but might be asked on Stack Overflow." OwnerUserId="13681" LastEditorUserId="7472" LastEditDate="2019-04-27T18:34:32.247" LastActivityDate="2019-04-27T18:34:32.247" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21011" PostTypeId="1" CreationDate="2017-05-13T17:34:54.200" Score="4" ViewCount="201" Body="&lt;p&gt;Is &lt;a href=&quot;https://en.wikipedia.org/wiki/Inflation_(cosmology)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cosmological inflation&lt;/a&gt; still considered as hypothesis?&#xA;Is CMBR evidence not enough?&lt;/p&gt;&#xA;" OwnerUserId="16877" LastEditorUserId="10589" LastEditDate="2017-05-14T03:57:18.587" LastActivityDate="2017-05-14T03:57:18.587" Title="Is cosmic inflation still a hypothesis?" Tags="&lt;cosmological-inflation&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21012" PostTypeId="1" AcceptedAnswerId="21013" CreationDate="2017-05-13T17:40:57.787" Score="2" ViewCount="122" Body="&lt;p&gt;I just read that Cassini discovered titan had &quot;active Weather patterns&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does that mean?&lt;/p&gt;&#xA;" OwnerUserId="16878" LastEditorUserId="2365" LastEditDate="2017-05-13T22:05:24.233" LastActivityDate="2017-05-13T22:05:24.233" Title="What does &quot;Active Weather Patterns&quot; mean?" Tags="&lt;saturn&gt;&lt;titan&gt;&lt;weather&gt;&lt;climate&gt;&lt;cassini&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21013" PostTypeId="2" ParentId="21012" CreationDate="2017-05-13T19:23:53.260" Score="4" Body="&lt;p&gt;It means that Titan has weather (driven by methane rather than water) and that it's weather changes with the seasons. Cassini has been observing Titan for almost half of a Titan year, which is 29.457 times longer than is our year. Titan's weather patterns (where it's cloudy, where it rains) has visibly changed over this time, just as weather patterns here change from summer to winter / winter to summer.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-05-13T19:23:53.260" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21014" PostTypeId="2" ParentId="21007" CreationDate="2017-05-13T21:43:25.927" Score="1" Body="&lt;p&gt;In short, no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lunar nodes vary in a roughly uniform fashion, taking about 18 years to vary by 360 degrees. The longitude of the moon at a given time is not part of the calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The details are complex, since they involve the dynamics of a three body system. Nasa has a system called &quot;SPICE&quot; that lets you calculate positions of orbiting bodies. There is &lt;a href=&quot;http://wgc.jpl.nasa.gov:8080/webgeocalc/#NewCalculation&quot; rel=&quot;nofollow noreferrer&quot;&gt;a web interface to spice&lt;/a&gt; available. And the details of using this to calculate lunar nodes is described on the &lt;a href=&quot;https://space.stackexchange.com/questions/18718/calculating-positions-of-moons-ascending-and-descending-nodes-from-ephemeris-fi&quot;&gt;space exploration stack exchange&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rough calculation can be done by approximating by a linear function of time.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-05-13T21:43:25.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21015" PostTypeId="1" CreationDate="2017-05-14T10:35:43.687" Score="1" ViewCount="103" Body="&lt;p&gt;I've read, that on Enceladus scientists observed plumes containing water vapor. Can I use plumes and geysers interchangeable? if not, What's the difference between Plumes and Geysers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks!&lt;/p&gt;&#xA;" OwnerUserId="16878" LastActivityDate="2017-05-14T11:02:27.150" Title="Plumes vs Geyser" Tags="&lt;saturn&gt;&lt;enceladus&gt;&lt;geysers&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" ClosedDate="2017-05-19T12:11:46.653" ContentLicense="CC BY-SA 3.0" />
  <row Id="21016" PostTypeId="2" ParentId="19939" CreationDate="2017-05-14T11:02:27.150" Score="3" Body="&lt;p&gt;A &lt;em&gt;geyser&lt;/em&gt; is a water volcano, they are familiar and fairly common on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By analogy, a water volcano on other planets can also be called a geyser, although the mechanism may be rather different. Geysers on Earth are caused by heating of sub-surface water by magma chambers. On Enceladus, there is no magma. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy of the volcanos of Enceladus is carried purely by the water, and so they are sometimes called &lt;em&gt;cryovolcanos&lt;/em&gt; They function in a similar way to volcanos on Earth, but with water instead of silicates, and at much lower temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a geyser on Earth erupts, the water all falls back to the ground or evaporates and is trapped by the atmosphere. Enceladus has much lower gravity and no significant atmosphere, so the water from geysers can escape into space, appearing as &lt;em&gt;plumes&lt;/em&gt; of water vapour. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A geyser is a water volcano. On Enceladus (but not Earth), the geysers are a form of cryovolcano. A plume appears when water from a geyser escapes into space&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-05-14T11:02:27.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21017" PostTypeId="2" ParentId="21006" CreationDate="2017-05-14T12:40:23.927" Score="0" Body="&lt;p&gt;On the one hand an object spans a smaller angle the farther away it is, as expected. On the other hand, due to the expansion of the Universe &lt;em&gt;and&lt;/em&gt; the finite speed of light, very distant objects were closer to us when they emitted the light we see today. At that time they spanned a larger angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following paragraph is hard to understand without a drawing , so have a look at &lt;a href=&quot;https://astronomy.stackexchange.com/a/8688/5264&quot;&gt;the drawing in this answer&lt;/a&gt;, which may help. The light from the edges of a galaxy that happened to travel toward where we were at the time it was emitted did &lt;em&gt;not&lt;/em&gt; reach us, because by the time it arrived to that place, we were no longer there. But the light that happened to be emitted toward some point behind us reached that point at the same time as we did, so that's what we see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact turnover, the threshold between &quot;looking smaller because far away&quot; and &quot;looking larger because closer in the past&quot; depends on the expansion rate history of the Universe, as well as on the way that light propagates in the Universe, which in turn depends on the densities od the various constituents of the Universe. The interrelationship of these quantities is given by the Friedmann equation.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2017-05-14T12:40:23.927" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21018" PostTypeId="1" AcceptedAnswerId="21025" CreationDate="2017-05-14T17:48:08.650" Score="-3" ViewCount="149" Body="&lt;p&gt;How can we say that universe is expanding as we r sure about that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;energy is neither created nor destroyed&#xA;  then from where does this more energy come from&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16884" LastEditorUserId="10271" LastEditDate="2017-05-15T13:54:05.813" LastActivityDate="2017-05-15T13:54:05.813" Title="How can we say that the universe is expanding" Tags="&lt;universe&gt;&lt;fate-of-universe&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2017-05-15T13:55:09.677" ContentLicense="CC BY-SA 3.0" />
  <row Id="21020" PostTypeId="1" CreationDate="2017-05-14T21:29:18.380" Score="1" ViewCount="1807" Body="&lt;p&gt;I'm from the Philippines, this is the first time I've seen such a bright star. I don't normally see stars at all cause of my poor eyesight but this star is very bright and really noticeable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took the first picture around 4:45 am, then now the little white which is still the star. The second I took it around 5:15 am when the sun already rising. The star and the sun are both in my east. The star is like going up while the sun is rising. &lt;img src=&quot;https://i.stack.imgur.com/1VYlT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ieD6s.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16887" LastEditorUserId="2365" LastEditDate="2017-05-14T22:26:53.370" LastActivityDate="2017-05-18T01:52:20.277" Title="very bright white star before sunrise" Tags="&lt;amateur-observing&gt;&lt;venus&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21021" PostTypeId="2" ParentId="21020" CreationDate="2017-05-14T22:22:10.697" Score="6" Body="&lt;p&gt;An exceptionally bright star in the morning or evening is usually the planet Venus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In May 2017, Venus is rising about two hours before the sun, and is by far the brightest star-like object in the morning sky. As it moves around the sun, it will get less bright. In a telescope, you would see it change from a crescent at the beginning of the month, to a semi-circle by the end. #&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planetarium software, such as Stellarium, can show the position of stars and planets at a given time and so help identify stars and planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look out for Venus on the Morning of May 22, when it will be close to the crescent moon.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-05-14T22:22:10.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21022" PostTypeId="1" AcceptedAnswerId="21024" CreationDate="2017-05-14T23:44:19.537" Score="5" ViewCount="4686" Body="&lt;p&gt;If I were to travel 1000 light-years, and witness only 1 year elapse in my frame of reference, how much time would have passed in the frame of reference of the Earth?&lt;/p&gt;&#xA;" OwnerUserId="16889" LastEditorUserId="13426" LastEditDate="2017-05-15T15:13:55.387" LastActivityDate="2021-02-19T00:39:24.433" Title="How old would I be if I travelled 1000 light years in one year" Tags="&lt;special-relativity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21024" PostTypeId="2" ParentId="21022" CreationDate="2017-05-15T09:09:00.663" Score="20" Body="&lt;p&gt;The answer is sort of trivial. If you travel 1000 ly so fast that in your own reference frame it takes one year, then you will have aged by one year in your own reference frame. To do so, you will need a speed of almost the speed of light, so in the reference frame of Earth, you will have spent just a tad more that 1000 yr to travel 1000 ly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the time dilation is given by the Lorentz factor $\gamma = 1/\sqrt{1-v^2/c^2}$ so to be exact, your speed must be&#xA;$$&#xA;1000^2 = \frac{1}{1 - v^2/c^2} ⇒ \\&#xA;     v = (1-10^{-6})^{1/2}\,c = 0.9999995c&#xA;$$&#xA;so your journey will take&#xA;$$&#xA;t = \frac{d}{v} = 1000.0005\,\mathrm{yr},&#xA;$$&#xA;i.e. 1000 years, 4 hours, and 23 minutes in Earth's reference frame.&lt;/p&gt;&#xA;&#xA;&lt;strong&gt;&quot;Realistic&quot; acceleration&lt;/strong&gt;&#xA;&#xA;&lt;p&gt;As David Hammen comments below, this assumes that your spaceship accelerates instantaneously to $v$. There are infinitely many ways to achieve that speed. &#xA;The &lt;em&gt;proper&lt;/em&gt; time $\tau$ (i.e. the time experienced by the traveler) to reach a distance $d$ when traveling at a constant acceleration $a$ is&#xA;$$&#xA;\tau = \frac{c}{a} \cosh^{-1} \left( \frac{ad}{c^2} +1 \right).&#xA;$$&#xA;Solving for $a$ yields the acceleration needed. The most pleasant way would arguably be accelarating at $a\simeq 19.2\,\mathrm{G}$ for half a year, and the decelerate at the same amount for the rest of the journey. A more pleasant way in the beginning, but less pleasant in the end, would be to accelerate at $a\simeq9.6\,\mathrm{G}$ for one year, and then crash into planet WASP-142b, which lies at a distance of roughly 1000 ly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your journey, as measured by people on Earth, would then take&#xA;$$&#xA;t(\tau) = \frac{c}{a} \sinh \left( \frac{a \tau}{c} \right),&#xA;$$&#xA;which works out to roughly 18 and 36 days more, respectively, than in the instantaneous case. The reason that it doesn't differ that much is that you actually reach relativistic speeds pretty fast at this acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest G-forces can be with endured in the &quot;forward direction&quot;, i.e. corresponding to lying on your back and accelerating upwards (accelerating along the direction of your spine tends to break it, and accelerating &quot;backward&quot; makes your eyes pop out). According to Wikipedia, &quot;acceleration pioneer&quot; &lt;a href=&quot;https://en.wikipedia.org/wiki/John_Stapp&quot; rel=&quot;noreferrer&quot;&gt;John Stapp&lt;/a&gt; withstood 25 G for 1.1 seconds, so you might want to send him instead of yourself.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2017-05-15T15:30:21.283" LastActivityDate="2017-05-15T15:30:21.283" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="21025" PostTypeId="2" ParentId="21018" CreationDate="2017-05-15T10:29:37.380" Score="0" Body="&lt;p&gt;The theory of the expanding universe comes from Hubble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hubble studied galaxies and plotted their red shift vs. their estimated distance. From that he found that there seems to be a correlation between distance and redshift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And since redshift is caused by something flying away from us, stronger redshift means its flying away faster. So since the further away a galaxy is, the stronger its redshift, implies that the further away something is from earth, the faster it flies away from us (thats where the Hubble constant comes from, it basically says how much the redshift increases if an object is a certain distance from earth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this the model formed, that the reason for them to be flying faster away from us, the further away they are, is because the universe is expanding. This model fits all of our current evidence and checks out with other implications as far as I know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are however other explentations on why stuff that is further away is more redshifted. At the moment the &quot;Expanding Universe&quot; model is the most accepted one in the community of scientists.&lt;/p&gt;&#xA;" OwnerUserId="10271" LastActivityDate="2017-05-15T10:29:37.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21026" PostTypeId="1" AcceptedAnswerId="21305" CreationDate="2017-05-15T15:30:35.727" Score="2" ViewCount="139" Body="&lt;p&gt;Angular momentum, and in particular the conservation of angular momentum, is of course very relevant in many processes in astrophysics, such as e.g. star- and planet formation, and disk formation in galaxies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept of objects spinning up upon collapse is straightforward enough to understand, but this begs the question: where does the primordial angular momentum come from? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tidal interactions between haloes, I suppose? &lt;/p&gt;&#xA;" OwnerUserId="6965" LastActivityDate="2017-06-17T09:27:24.983" Title="Primordial angular momentum?" Tags="&lt;astrophysics&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21028" PostTypeId="1" CreationDate="2017-05-15T16:56:31.040" Score="0" ViewCount="364" Body="&lt;p&gt;What is the maximum size for a gas giant? &lt;/p&gt;&#xA;" OwnerUserId="16894" LastActivityDate="2017-05-16T06:26:51.383" Title="Why could a gas planet not get much larger than Jupiter?" Tags="&lt;jupiter&gt;&lt;gas-giants&gt;&lt;size&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2017-05-16T13:16:52.793" ContentLicense="CC BY-SA 3.0" />
  <row Id="21029" PostTypeId="1" AcceptedAnswerId="21037" CreationDate="2017-05-15T17:59:32.997" Score="3" ViewCount="286" Body="&lt;p&gt;Night sky is dark because light from distant stars has yet not reached our earth. Light emitted by stars formed after the instant of creation i.e. 13.7 billion light years ago is on the way to reach us. As light from these distant stars reach us, night sky must keep getting brighter.&lt;br&gt;&#xA;How true is this? And to what extent? Can one day night sky become as bright as a day?&lt;/p&gt;&#xA;" OwnerUserId="16895" LastEditorUserId="16895" LastEditDate="2017-05-16T06:18:02.300" LastActivityDate="2017-05-17T01:04:31.557" Title="Is night sky getting brighter day by day?" Tags="&lt;star&gt;&lt;night-sky&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21030" PostTypeId="2" ParentId="21028" CreationDate="2017-05-15T18:14:24.010" Score="-1" Body="&lt;p&gt;Usually they can get is &lt;strong&gt;Almost twice&lt;/strong&gt;.  Here's the evidence:&lt;br&gt;&#xA; It is difficult for them to become even bigger than that as once they reach 15 Jupiter masses or so there is enough pressure in the core to ignite deuterium fusion, so those are considered “brown dwarfs” rather than planets.&lt;br&gt;&#xA;An example is TrES-4, which is located 1,400 light years away in the constellation Hercules. The planet has been measured to be 1.4 times the size of Jupiter, but it only has 0.84 times Jupiter’s mass. With such a low density, it is dubbed &lt;strong&gt;the puffy planet&lt;/strong&gt;.&lt;br&gt;&#xA;It was discovered in 2006 orbiting a brightish star 1500 light years from Earth. Its diameter of about 1.8 times Jupiter’s makes it the largest accurately measured planet. Strangely, though, TrES-4 is very light for its size. It has only 88 per cent of Jupiter’s mass, giving it a density of roughly 0.2 grams per cubic centimetre, less than that of cork. Just how a planet can be as fluffy as TrES-4 remains a mystery.&lt;/p&gt;&#xA;" OwnerUserId="16895" LastEditorUserId="16895" LastEditDate="2017-05-16T06:26:51.383" LastActivityDate="2017-05-16T06:26:51.383" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21031" PostTypeId="2" ParentId="1592" CreationDate="2017-05-16T02:01:56.047" Score="4" Body="&lt;p&gt;To supplement Gerald's answer, which focuses on the physical survival of the comet, I'll note that the orbit is chaotic over a similar time period.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://arxiv.org/abs/1409.7762&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Chaotic Dynamics of Comet 1P/Halley; Lyapunov Exponent and Survival Time Expectancy&lt;/em&gt; M. A. Muñoz-Gutiérrez, M. Reyes-Ruiz, B. Pichardo, arXiv:1409.7762, 2014&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On average, it is unstable on a timescale of hundreds of thousands of years. The chaotic nature of Halley's present day orbit implies that a precise determination of its motion, at the level of the present day observational uncertainty, is difficult to predict on a timescale of approximately 100 years. Furthermore, we also find that the ejection of Halley from the solar system or its collision with another body could occur on a timescale as short as 10,000 years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16634" LastActivityDate="2017-05-16T02:01:56.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21032" PostTypeId="1" CreationDate="2017-05-16T02:16:58.590" Score="3" ViewCount="55" Body="&lt;p&gt;The effects of outgassing on comet orbits can be noticeable. Yeomans, in  &lt;a href=&quot;http://lpi.usra.edu/books/CometsII/7009.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cometary orbit determination and nongravitational forces, 2004&lt;/a&gt;, wrote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As the astrometric datasets improve and lengthen and as the modeling of the cometary nongravitational effects becomes more realistic, there remains the strong possibility that some physical characteristics of comets will soon be accurately determined from the orbit-determination process alone. We are already beginning to see signs that this is the case.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How much does the orbital period of comet Halley vary due to these nongravitational forces? Hours? Days?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which comets have the largest perturbations due to these forces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: a more sophisticated question along these lines is &lt;a href=&quot;https://space.stackexchange.com/questions/18474/did-rosetta-improve-on-models-of-non-gravitational-effects-on-comet-67ps-orbit&quot;&gt;Did Rosetta improve on models of non-gravitational effects on comet 67P's orbit? - Space Exploration Stack Exchange&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16634" LastActivityDate="2017-05-16T02:16:58.590" Title="How large are the effects of nongravitational forces on comet Halley?" Tags="&lt;comets&gt;&lt;orbital-mechanics&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21034" PostTypeId="1" CreationDate="2017-05-16T15:13:22.160" Score="1" ViewCount="97" Body="&lt;p&gt;In this image where would the antumbra be?:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://img.tfd.com/hm/JPG/A5umbra.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://img.tfd.com/hm/JPG/A5umbra.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16886" LastEditorUserId="2365" LastEditDate="2017-05-16T19:42:35.837" LastActivityDate="2017-05-16T19:42:35.837" Title="In this image, where would the antumbra be?" Tags="&lt;the-moon&gt;&lt;space&gt;&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21035" PostTypeId="2" ParentId="21034" CreationDate="2017-05-16T16:32:02.203" Score="6" Body="&lt;p&gt;There is no antumbra in this image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An antumbra only exists when the moon is further from the Earth than in that image, so that the moon appears smaller in the sky than the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case, when the moon is perfectly aligned with sun, a ring of sun is visible behind the moon. The region in which the ring is visible is called the antumbra. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the image, the moon is closer to the Earth, so when the moon and sun are aligned, the moon entirely blocks out the sun, and so there is a region of umbra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That image shows a total eclipse. An antumbra is present only in an annular eclipse. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-05-16T16:32:02.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21036" PostTypeId="1" CreationDate="2017-05-16T23:55:27.577" Score="3" ViewCount="1159" Body="&lt;p&gt;What's the orbital period of an asteroid which has synodic period of 429 days?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the formula of orbital period for inner planets is $$P_{orb} = \frac{P_{syn}}{P_{syn} + 1}$$ and for outer planets is $$P_{orb} = \frac{P_{syn}}{P_{syn} - 1}$$ where $P_{syn}$ is the synodic period of the planet measured in years. However, I don't know whether the asteroid should be counted as an inner planet or outer planet.&lt;/p&gt;&#xA;" OwnerUserId="16906" LastActivityDate="2017-05-17T02:53:02.807" Title="Orbital Period of an Asteroid" Tags="&lt;asteroids&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21037" PostTypeId="2" ParentId="21029" CreationDate="2017-05-17T01:04:31.557" Score="4" Body="&lt;p&gt;The rate of star formation in the universe &lt;a href=&quot;https://astrobites.org/2016/11/11/the-changing-star-formation-rate-of-the-universe/&quot; rel=&quot;nofollow noreferrer&quot;&gt;peaked about 3.5 billion years after the Big Bang&lt;/a&gt;, or &lt;a href=&quot;http://www.space.com/18370-universe-star-formation-rate-decline.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;2.7 billion years after it&lt;/a&gt;, according to a different study—in any case, for most of the universe's lifetime fewer and fewer stars have been forming each year. Though I don't have precise data on the death rate, on the whole &lt;a href=&quot;https://www.theguardian.com/science/2015/aug/10/universe-slowly-dying-gama-old-stars-fade-faster-than-new-ones-born&quot; rel=&quot;nofollow noreferrer&quot;&gt;the number of stars is decreasing&lt;/a&gt; as more old stars die than new ones form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, the light from old stars is being &lt;a href=&quot;https://en.wikipedia.org/wiki/Redshift&quot; rel=&quot;nofollow noreferrer&quot;&gt;redshifted&lt;/a&gt; out of existence by the expansion of the universe, which is itself accelerating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure there's a mathematical model that can quantify this properly, but I think those two facts, combined, must be the answer to why the sky isn't getting brighter.&lt;/p&gt;&#xA;" OwnerUserId="15398" LastActivityDate="2017-05-17T01:04:31.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21038" PostTypeId="2" ParentId="21036" CreationDate="2017-05-17T02:53:02.807" Score="1" Body="&lt;p&gt;Just to clarify, inner planets usually refers to Mercury, Venus, Earth, Mars.    But in relation to your question, the &quot;inner&quot; planets are just Mercury and Venus and the outer planets are Mars and out, because the +1 / -1 formula is in relation to whether the Earth is passing the planet as it orbits the sun or being passed by the planet.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone else can do the math if they like, the gist is explained &lt;a href=&quot;http://astronomyonline.org/Science/SiderealSynodicPeriod.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.  Because it's the motion relative to the stars in the background that you're tracking, anything that orbits closer to the sun and faster than Earth would move in a different direction and require a different formula than orbits further out and slower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your asteroid remains outside the Earth's orbit, then it would obey the outer planet formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your asteroid is an &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_Earth-crossing_minor_planets&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth crosser&lt;/a&gt;, which isn't uncommon, then it would have a different type of orbital period that would follow neither formula, and I'm not sure what formula would apply for an earth-crossing asteroid.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-05-17T02:53:02.807" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21039" PostTypeId="1" CreationDate="2017-05-17T07:25:15.917" Score="7" ViewCount="214" Body="&lt;p&gt;Specifically, excluding the Sun and the Moon:&lt;br/&gt;&#xA;How much energy, as light and other frequencies reaches a point in space on the average, if you add up all the sources like distant stars, the Milky Way, distant galaxies, nebulae etc.? How much brighter is the space than &quot;perfect blackness&quot; on the average?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say, we've developed a magical solar battery that has 100% efficiency, can absorb all electromagnetic frequencies, is spherical, fully omnidirectional and has &lt;span class=&quot;math-container&quot;&gt;$1 m^2$&lt;/span&gt; of cross-section surface (56cm radius);  we sent it out halfway to Proxima Centauri and we measure how much power it produces - how many watts we'd get out of the battery utilizing all the starlight available?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; If &quot;all frequencies&quot; is not easily obtainable, the luminous counterpart to radiant values will be okay (only taking visible spectrum into account). &lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="25994" LastEditDate="2019-05-15T18:22:07.290" LastActivityDate="2019-05-16T13:57:08.183" Title="What is the radiant intensity of the celestial sphere?" Tags="&lt;luminosity&gt;&lt;radiation&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21040" PostTypeId="1" AcceptedAnswerId="21044" CreationDate="2017-05-17T10:46:10.007" Score="3" ViewCount="939" Body="&lt;p&gt;I'm trying to understand following schema of the wikipedia page &lt;a href=&quot;https://en.wikipedia.org/wiki/Umbra,_penumbra_and_antumbra&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OMjGg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OMjGg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image subtitle is &lt;em&gt;&quot;Umbra, penumbra, and antumbra, for a lunar eclipse&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distances and sizes are not to scale. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If distance Earth-Sun is about 150 Gm (gigameters) and Sun diameter is 1.4 Gm, $\arctan 1.4/150 = 0.57º$ (degrees). Thus, the Sun is seen from Earth with an angle of approx. 0.57º (0.01 radians). The Blue and red lines in schema should be near to parallel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means the point where umbra and antumbra touch is (being 12 Mm the approx diameter of Earth) at a distance of about $\frac{150}{1.4\times0.012}=1.3 \mathrm{Gm}$ from Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As distance Earth-Moon is only 0.400 Gm, it seems impossible that Moon was never in the antumbra zone of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm doing a basic error somewhere, but I do not know where. &lt;/p&gt;&#xA;" OwnerUserId="16912" LastEditorUserId="16912" LastEditDate="2017-05-17T17:38:55.803" LastActivityDate="2017-05-17T21:16:47.780" Title="Umbra, penumbra and antumbra" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;solar-eclipse&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21041" PostTypeId="1" CreationDate="2017-05-17T15:15:35.843" Score="1" ViewCount="61" Body="&lt;p&gt;Today, at about 23:40 in Sungkyungwan University, South Korea, a friend and I spotted a bright light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was brighter than nearby stars. Since I was looking at South, and at that time, the brightest star in the Southern sky was Saturn, with an apparent magnitude of about 0.9, I'm safely assuming that its apparent magnitude was brighter than 0.9. The thing was even brighter than that (I think that it was as bright as a full moon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its initial altitude was about 60degrees then it fell below the horizon very quickly (in about 3~4 seconds). At first, I thought that it was a comet of some kind, but It wasn`t leaving any traces in the sky, and also it was moving in sort of a random way, going from left to right to left to right and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was a point, but it was so bright that it appeared larger than Vega(I'm not talking about absolute sizes, as stars don't have sizes. I'm just saying that, just as brighter stars seem larger, this moving thing was brighter and therefore, seemed larger.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My friend and I saw it together, and wondered what it was. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, after about 10 minutes, it appeared again. But this time, it came from the place where it disappeared at its first sight. It moved up to about 10degrees (altitude). When it was going up, it was going up in a parabolic path. Then, it came down but along the way, its path changed from parabola to zig zag. Going from up to down took about 2 seconds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could this be? This cannot have been an optical illusion because both my friend and I witnessed it. Also, it could not have been a satellite as it was too bright(almost as bright as the moon) and moved in a non-linear path (also it moved very very fast). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could this be? &lt;/p&gt;&#xA;" OwnerUserId="16913" LastEditorUserId="2365" LastEditDate="2017-05-17T17:03:46.180" LastActivityDate="2017-05-17T17:03:46.180" Title="I saw a light moving fast what could it be?" Tags="&lt;star&gt;&lt;light&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2017-05-17T20:23:08.283" ContentLicense="CC BY-SA 3.0" />
  <row Id="21042" PostTypeId="2" ParentId="21026" CreationDate="2017-05-17T16:49:08.197" Score="0" Body="&lt;p&gt;Let's say I'm floating in space with two toy tops. I spin them in opposite directions at the same speed. Each of them has angular momentum but the total angular momentum of the system is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a larger scale the same thing could happen universe-wide. As stars and galaxies formed they would end up spinning every which way, but the total angular momentum would be unchanged. &lt;/p&gt;&#xA;" OwnerUserId="4042" LastActivityDate="2017-05-17T16:49:08.197" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21043" PostTypeId="2" ParentId="21040" CreationDate="2017-05-17T16:49:26.080" Score="2" Body="&lt;p&gt;The calculations seem reasonable. The conclusion is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earths antumbra starts 1.3 Gm from Earth. At less than 1.3 Gm there is umbra, penumbra and non-shadow. The moon is 0.4 Gm from Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0.4 &amp;lt; 1.3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence it is impossible for the moon to be in the Earth's antumbra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the Moon's antumbra forms at almost exactly one lunar distance. We notice that the moon and the sun are almost exactly the same size in the sky. When the moon is further from the Earth, an annular eclipse can form, as a small part of the Earth will then be in the moon's antumbra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've made a scale model of the sun-earth-moon shadows, at &lt;a href=&quot;https://ggbm.at/CbbvmBFq&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ggbm.at/CbbvmBFq&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-05-17T16:49:26.080" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21044" PostTypeId="2" ParentId="21040" CreationDate="2017-05-17T21:16:47.780" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;As distance Earth-Moon is only 0.400 Gm, it seems impossible that Moon was never in the antumbra zone of the Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is correct. You were reading yet another poorly written wikipedia page. The information on that page isn't quite incorrect, but it isn't quite correct, either. There is no such thing as an annular lunar eclipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That page would have been much more meaningful had it used the Moon's shadow as the example. The Moon's umbra trails from 368000 km to 381000 km behind the Moon, depending on how close the Moon is to the Sun. This is very similar to the distance between the center of the Moon and the closest point on the surface of the Earth to the Moon, which varies from 362000 km to 400000 km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the Moon is directly between the Earth and the Sun, the resulting solar eclipse can be total or annular. When the Earth is directly between the Moon and the Sun, the resulting lunar eclipse is always total.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-05-17T21:16:47.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21045" PostTypeId="2" ParentId="21020" CreationDate="2017-05-18T01:52:20.277" Score="4" Body="&lt;p&gt;Similarly to James K, I suspected that such an object may be Venus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To confirm this, I loaded up Stellarium, and set the location to Victoria, Philippines. I took your time and date as well. A screenshot of the view seems to confirm that your object is in fact Venus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GGRRA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GGRRA.jpg&quot; alt=&quot;A view of Venus in the Eastern sky on the morning of May-14 2017, from Victoria, Philippines.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I highly recommend downloading Stellarium for yourself. It is free, and allows you to check the night sky from any location on Earth at any time.&lt;/p&gt;&#xA;" OwnerDisplayName="user14781" LastActivityDate="2017-05-18T01:52:20.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21046" PostTypeId="1" AcceptedAnswerId="21056" CreationDate="2017-05-18T06:44:27.320" Score="1" ViewCount="360" Body="&lt;p&gt;While reading about a White holes, I stumbled upon a question &quot;Where does the matter emitted by a White hole come from?&quot; There seems to be 2 possible candidates:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It was the matter engulfed by a Black hole.  &lt;/li&gt;&#xA;&lt;li&gt;Matter was created inside the White hole.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My search on wiki stated that:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Penrose Diagram of Schwarzschild radius, objects first leave the White hole and then may enter into a Black hole. But, due to Hawking radiation, a Black hole can achieve thermal equilibrium with gas of radiation. This equilibrium state is reversal invariant, implying that time reversal of a Black hole is again a Black hole. This means that a Black hole and a White hole are same objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My 2nd candidate suggests that the big-bang was a white hole. I thought of ruling this out because it requires the existence of a white hole before the big-bang.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This forced me to draw the conclusion that a white hole must either be inside a black hole or behind it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct (provided that we haven't yet proven the existence of a White hole)? Where am I wrong?&lt;/p&gt;&#xA;" OwnerUserId="16895" LastEditorUserId="2365" LastEditDate="2017-05-18T19:15:58.177" LastActivityDate="2017-05-19T12:28:26.413" Title="Does a White hole lies behind a Black hole?" Tags="&lt;black-hole&gt;&lt;big-bang-theory&gt;&lt;white-hole&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21047" PostTypeId="1" AcceptedAnswerId="21052" CreationDate="2017-05-18T10:26:04.627" Score="6" ViewCount="98" Body="&lt;p&gt;When I first learned about Positional Astronomy (50 years ago), everything was described with respect to the Equator and Ecliptic (true or mean, of date or of a specific epoch). Then 'they' invented the International Celestial Reference System (ICRS). An inertial reference system, so true and mean no longer apply and neither does it require an epoch to be given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I see things are referred to the International Celestial Reference System Version 2 (ICRS2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What changed?&lt;/p&gt;&#xA;" OwnerUserId="16920" LastActivityDate="2017-05-18T17:41:27.717" Title="What is the difference between ICRS and ICRS2?" Tags="&lt;positional-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21048" PostTypeId="1" CreationDate="2017-05-18T08:23:23.747" Score="1" ViewCount="335" Body="&lt;p&gt;Like every planet they are in orbit of something or another planet like our solar system every planet revolves around the sun, my question is: what's the name of the planet or object which the sun revolves around; if the sun is stationary or if our galaxy is in motion, around what it is revolving.&lt;/p&gt;&#xA;" OwnerDisplayName="hemanjosko" LastEditorUserId="7472" LastEditDate="2018-02-02T00:06:50.433" LastActivityDate="2018-02-02T00:06:50.433" Title="The Sun or Our Galaxy revolves around what?" Tags="&lt;the-sun&gt;&lt;milky-way&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2017-05-18T19:53:28.013" ContentLicense="CC BY-SA 3.0" />
  <row Id="21049" PostTypeId="2" ParentId="21048" CreationDate="2017-05-18T10:50:37.483" Score="4" Body="&lt;p&gt;As a previous comment states; it is complicated. But in principle the Sun is just one object in a large collection of objects which we call the Milky Way Galaxy. All the objects in this Galaxy orbit the barycentre (or barycenter, if you prefer the American spelling). The barycentre is, essentially, the common centre of mass of the whole galaxy. In the same way the planets in our solar system don't actually orbit the Sun, they orbit the Solar System Barycentre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the Milky Way Galaxy, that too is a member of a larger group of galaxies, and they all orbit their common barycentre.&lt;/p&gt;&#xA;" OwnerUserId="16920" LastActivityDate="2017-05-18T10:50:37.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21050" PostTypeId="2" ParentId="21048" CreationDate="2017-05-18T16:16:55.490" Score="1" Body="&lt;p&gt;The Sun, like all other stars in the Milky Way Galaxy, is rotating around the &lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_Center&quot; rel=&quot;nofollow noreferrer&quot;&gt;Galactic Center&lt;/a&gt;. The Galactic Center is around 25,000 light years away, and it may contain a supermassive black hole. There is a very intense radio source in the center of the galaxy, called Saggitarius A*, which is thought to be the same as the black hole. &lt;/p&gt;&#xA;" OwnerUserId="14989" LastActivityDate="2017-05-18T16:16:55.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21051" PostTypeId="1" AcceptedAnswerId="21060" CreationDate="2017-05-18T16:59:33.903" Score="4" ViewCount="400" Body="&lt;p&gt;I am making a small orbit simulator. I figured out Kepler's laws and know how to work with them to &quot;update&quot; an orbit based on time. So now I want to know how I can describe orbits with newton laws. Given a position vector and a velocity vector of a satellite, What I currently do is this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Calculate gravitational acceleration: (g*mass)/distance^2&lt;/li&gt;&#xA;&lt;li&gt;Add this number to the velocity vector.&lt;/li&gt;&#xA;&lt;li&gt;Multiply the new velocity vector with the DeltaTime&lt;/li&gt;&#xA;&lt;li&gt;Add it to the position vector&lt;/li&gt;&#xA;&lt;li&gt;Go back to 1.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However when converting this to Keplerian Orbit Elements doesn't really make them constant.(Calculations from here: &lt;a href=&quot;http://ccar.colorado.edu/asen5070/handouts/cart2kep2002.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cartesian To Kepler&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the orbit isn't really stable, and also wobbles around. So can anyone help me fix this? Or am I doing something wrong?&lt;/p&gt;&#xA;" OwnerUserId="16923" LastEditorUserId="2365" LastEditDate="2017-05-18T18:05:08.497" LastActivityDate="2017-05-22T10:56:08.637" Title="Orbits using Newtons laws" Tags="&lt;orbit&gt;&lt;orbital-elements&gt;&lt;orbital-mechanics&gt;&lt;newtonian-gravity&gt;" AnswerCount="1" CommentCount="13" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21052" PostTypeId="2" ParentId="21047" CreationDate="2017-05-18T17:41:27.717" Score="5" Body="&lt;p&gt;Not much has changed. Each update is consistent with its predecessor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original ICRF is defined by 212 extragalactic sources. The ICRF2 is defined by 295 sources, 97 of which are shared by ICRF1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for the change was greater accuracy. The 295 sources in ICRF2 allow for locations accurate to less than 40 micro-arcseconds (µas) compared with 250 µas for ICRF1. The ICRF2 is consistent with ICRF1, to within the latter's precision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both reference frames agree with the J2000.0 frame, to within the precision of the J2000.0 frame. The ICRF is several orders of magnitude more precise than can be achieved by optical astronomy.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-05-18T17:41:27.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21054" PostTypeId="2" ParentId="20976" CreationDate="2017-05-18T20:43:26.247" Score="5" Body="&lt;p&gt;$ &#xA;\cos ^{-1}\left(-\frac{\tan (\lambda ) (\cos (\lambda ) \cos (\phi ) \cos &#xA;(Z)+\sin (\lambda ) \sin (Z))}{\sqrt{(\cos (\lambda ) \sin (Z)-\sin (\lambda) &#xA;\cos (\phi ) \cos (Z))^2+\sin ^2(\phi ) \cos ^2(Z)}}\right)-\tan^{-1} &#xA;(\cos (\lambda ) \sin (Z)-\sin (\lambda ) \cos (\phi ) \cos (Z),\sin &#xA;(\phi ) (-\cos (Z))) &#xA;$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the amount of time from now (see important notes below) a celestial object will set, where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\phi$ is the azimuth of the object&lt;/li&gt;&#xA;&lt;li&gt;$Z$ is the altitude of the object above the horizon&lt;/li&gt;&#xA;&lt;li&gt;$\lambda$ is the latitude of the observer&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note the two-argument form of arctangent is required for accuracy: &lt;a href=&quot;https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Two-argument_variant_of_arctangent&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Two-argument_variant_of_arctangent&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Important notes (MUST READ!)&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The result is in radians. To convert to sidereal hours, multiple by $\frac{12}{\pi }$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the object has a fixed right ascension and declination, divide sidereal hours by 1.002737909350795 to get clock hours.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For the Sun (which &lt;em&gt;doesn't&lt;/em&gt; have fixed right ascension/declination), do NOT divide as above. By not dividing, you compensate for the Sun's change in right ascension.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For the Moon, see &quot;&lt;strong&gt;The Moon&lt;/strong&gt;&quot; section below.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The time computed above is for &lt;strong&gt;geometric midpoint&lt;/strong&gt; setting. In reality, refraction near the horizon means the object will set later. Additionally, for objects that have an angular diameter (eg, the Sun), setting usually means when the top edge disappears over the horizon, which will make the set time even later. Accounting for both effects should be possible, but might require numerical methods instead of a closed form formula as above.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the quantity inside the arccosine is greater than 1, the object is always in the sky and never sets or rises.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the quantity inside the arccosine is less than -1, the object is never in the sky and thus also never sets or rises.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I did only minimal testing. As always, do not rely on my answers for anything important.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Moon&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The Moon's right ascension and declination change rapidly, so this calculation does not work well for the moon.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You could compensate for the change in right ascension (and thus hour angle) by approximating the increase as 24 hours every 27.32158 days (its sidereal period) and do an iterative calculation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An even better compensation for the change in right ascension would be to approximate the moon's movement in ecliptic longitude (which is more constant than its movement its right ascension) as 360 degrees per 27.32158 days and then project the ecliptic longitude back to right ascension, and then iterate.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Compensating for the moon's change in declination is more difficult. The moon's ecliptic latitude (which can be converted to declination) varies sinusoidally, but the equation $\sin (x)=a$ normally has two solutions. Unless you know whether the moon's ecliptic latitude is increasing or decreasing (ie, whether it's between ascending and descending nodes or vice versa), you won't know the direction of declination change.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Less important notes (optional):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;See &lt;a href=&quot;https://astronomy.stackexchange.com/questions/14492&quot;&gt;Need Simple equation for Rise, Transit, and Set time&lt;/a&gt; for more general equations on when an object rises/sets/etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Calculations for this answer at: &lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/STACK/bc-object-riset-from-az-elt.m&quot; rel=&quot;noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/STACK/bc-object-riset-from-az-elt.m&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Related calculations: &lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/STACK/bc-rst.m&quot; rel=&quot;noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/STACK/bc-rst.m&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mathematica was unable to find a simpler form for the &quot;time to set&quot; above, though I sense a simpler form does exist (I could be wrong).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you know the Sun's declination (which you can get from its azimuth and elevation as above), you can &lt;em&gt;almost&lt;/em&gt; determine the date. However, the sun reaches a given declination twice a year (example: it reaches 0 degrees declination on both equinoxes, by definition), so you can only know it's one of two days.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution notes:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I learned quite a bit answering this question, and thought it was only solvable numerically until I figured out the shortcut:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;To convert from the azimuth/elevation sphere to the hour angle/declination sphere, you just rotate around the y axis by $\frac{\pi }{2}-\lambda$ (90 degrees minus the latitude) and then rotate $pi$ (180 degrees) around the z axis.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Once you have the declination, computing the hour angle when an object sets is easy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You then subtract the setting time from the current hour angle to get the answer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;TODO: I was hoping to add some graphs/visualizations to this answer but couldn't quite get things working.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2017-05-18T20:49:03.487" LastActivityDate="2017-05-18T20:49:03.487" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21056" PostTypeId="2" ParentId="21046" CreationDate="2017-05-19T12:28:26.413" Score="2" Body="&lt;p&gt;No, there is probably no such thing as a white hole in real life. A white hole is basically a time-reversed version of a black hole. Just as a black hole cannot be destroyed in classical General Relativity, a white hole can never be formed in GR. So while a white hole is technically a thing that could exist, there is no way to form one unless it was there at the beginning of the universe. We don't have any reason to think white holes existed at the Big Bang, so we think they don't exist (and never will).&lt;/p&gt;&#xA;" OwnerUserId="16932" LastActivityDate="2017-05-19T12:28:26.413" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21059" PostTypeId="1" CreationDate="2017-05-19T14:58:15.307" Score="2" ViewCount="1404" Body="&lt;p&gt;I'm looking for a simple algorithm to compare astronomical images (of the same sky region) against each other, compute their movement and rotation, to finally stack them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment I'm already having a more or less working algorithm. First I extract all the stars out of an image (including information like brightness and FWHM), and then I walk through all the resulting &quot;points&quot; and create triangles out of the current point and those two other stars that have the shortest distance to this star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This list of triangles is created for every image. After this I take one image as a reference and then I walk through the list of triangles in the reference image to find a triangle in the other image with the same length of each side of the triangle (I also &quot;allow&quot; some tolerance due minimal relative differences of the star positions in each image). For this matches I calculate movement and rotation relative to the reference image. The last step is to find the matched triangles that have to same relative movement and rotation like the other matches. This is done by calculating the standard deviation, sorting out triangles that are not within 1 or 2 sigma and repeat this process until I have a very small standard deviation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last part, finding &quot;valid&quot; triangles with the same movement/rotation, is working fine. The problem is that sometimes I have only like 2 or 3 &quot;valid&quot; triangles out of 300 initial triangles. All other triangles have side lengths different to those of the reference image. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I assume it's the way I generate my initial triangles which causes the problem. Sortings stars by their brightness and using this data to generate the triangles also doesn't work. So is there a better way to create the initial triangles in all the images?&lt;/p&gt;&#xA;" OwnerUserId="1000" LastEditorUserId="2365" LastEditDate="2017-05-19T15:10:58.160" LastActivityDate="2017-05-21T09:07:32.870" Title="Algorithm to stack astronomical images" Tags="&lt;star&gt;&lt;photography&gt;&lt;algorithm&gt;&lt;alignment&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21060" PostTypeId="2" ParentId="21051" CreationDate="2017-05-19T19:50:12.943" Score="7" Body="&lt;p&gt;I like to classify solutions of the problem of the time evolution of the&#xA;complete initial state of a set of objects at some epoch time, where the&#xA;objects are subject to Newtonian gravitation into two main groups.&#xA;One approach is to use orbital elements of some sort.&#xA;The other is to use a numerical initial value problem solver,&#xA;aka a numerical integrator. The latter is the primary subject of this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note well: This classification isn't quite perfect as hybrid approaches are&#xA;also possible, wherein one uses a numerical integrator to integrate&#xA;time-varying orbital elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;Keplerian orbital elements work quite nicely in the case of two point&#xA;masses, or more generally, two objects with a spherical mass distribution.&#xA;The anomaly is the only Keplerian orbital element that changes over time.&#xA;Keplerian elements can be used in situations where the underlying&#xA;assumptions are approximately correct by developing a model of how&#xA;those supposedly unchanging elements vary with time. One way of doing&#xA;this is to use Lagrange's planetary equations. (There are other related&#xA;approaches such as Gauss' planetary equations, Delaunay's planetary&#xA;equations, etc.)&#xA;Lagrange's planetary equations yield expressions for how Keplerian orbital&#xA;elements vary over time given a set of perturbing forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another approach is to use something akin to those Keplerian elements&#xA;such as Delaunay elements), coupled with planetary equations for those&#xA;alternative elements. Yet another approach is to use orbital elements&#xA;(e.g., Brouwer-Lyddane elements, SGP4 elements) in which the planetary&#xA;equations are embedded in the orbital element to Cartesian state&#xA;transformation algorithm. This final approach is used to this day to&#xA;describe vehicles in Earth orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;The other approach is to use numerical integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll start with a discussion of how to solve for the value of a scalar&#xA;function $x(t)$ at some time $t_1$ given an initial value $x(t_0) = x_0$&#xA;and some well-behaved (continuous and bounded) derivative function&#xA;$f(x(t),t) = dx(t)/dt$ that describes the time evolution of $x$.&#xA;This falls in the very broad category of initial value problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the ordinary differential equation cannot be solved analytically&#xA;and cannot be expressed terms of a useful power series. This doesn't mean&#xA;nothing can be done. There are a number of techniques for solving this&#xA;problem numerically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the dependence of the derivative function on the dependent variable $x$.&#xA;This becomes the much simpler problem of numerical quadrature if the&#xA;derivative function can be expressed independent of $x$.&#xA;The discussion that follows assumes that the derivative function $f$ does&#xA;indeed depend on $x$.&#xA;Note well: Newtonian gravitation falls in this category.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also assumes the derivative function is well-behaved.&#xA;Numerically integrating across a discontinuity is a bad idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The foundation of the integration-based techniques for a scalar function&#xA;is the mean value theorem, which says that at some time $t_c$ between&#xA;$t_0$ and $t_1$, the value at $t_1$ is exactly&#xA;$x(t_1) = x(t_0) + (t_1-t_0)\,f(x(t_c),t_c)$. If only we could find that&#xA;magical $t_c$ and the derivative at that point. There's a chicken and&#xA;egg problem here: that magical point in time is not known. Even if it was,&#xA;the derivative function depends on state, and that too isn't known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very simple approach around this problem is to assume that this magical&#xA;point is the initial point:&#xA;$$x(t_1) = x(t_0) + (t_1-t_0)f(x(t_0),t0)$$&#xA;This works quite nicely for values of $t_1$ that are very close to $t_0$.&#xA;It doesn't work very well at all where $|t_1 - t_0|$ is not small.&#xA;This suggests splitting the interval $(t_0, t_1)$ into a number of smaller&#xA;intervals. This results in Euler's method: Apply the above to advance&#xA;state to time $t_0+\Delta t$, then to $t_0+2\Delta t$, and so on, eventually&#xA;reaching the desired time. Euler's method is rather lousy, even for a simple&#xA;first order scalar ODE. We can do much better than this. The key reason for&#xA;discussing Euler's method is that it is the basis for many other integration&#xA;techniques. Learn how it works, then toss it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One approach to approving on Euler's method is to somehow correct the&#xA;result from Euler's method. For example, take an Euler step and compute&#xA;the derivative at the end point. Then use the average of those two&#xA;derivative values (the original value used to make the Euler step, and&#xA;the other from the end of the Euler step) to recompute the step from&#xA;$t$ to $t+\Delta t$. THis is Heun's method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another approach is to guess that the magical point $t_c$ lies&#xA;somewhere between $t$ and $t+\Delta t$. Perhaps the middle? We can use&#xA;Euler's method to advance state to the midpoint, and then use the&#xA;derivative at that point to advance state from $t$ to $t+\Delta t$.&#xA;This is the midpoint method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both Heun's method and the midpoint method appear to be steps backwards,&#xA;computationally. While Euler's method requires but one evaluation of the&#xA;derivative function per time step, these improved methods require two.&#xA;However, the error growth is in general so much smaller with either&#xA;Heun's method or the midpoint method compared to that from Euler's method.&#xA;This means that those &quot;improvements&quot; most definitely are improvements.&#xA;The expense of calling the derivative function twice per step is more than&#xA;offset by the fact that imprpovements enables take steps that are orders&#xA;of magnitude larger than one can make with Euler's method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both Heun's method and the midpoint method are simple improvements.&#xA;This problem has been studied in many guises. There are many more advanced&#xA;techniques. One is the class of Runge-Kutta integration techniques.&#xA;Both Heun's method and the midpoint method fall into this class.&#xA;The most popular of these, classical Runge-Kutta 4, is a significant&#xA;improvement on those two methods. There are even higher order Runge-Kutta&#xA;integrators than RK4.&#xA;Heun's method also falls into the broad class of predictor-correctors,&#xA;wherein one method (the predictor) advances state to the end of the interval&#xA;and another method (the corrector) uses the derivative at this approximate&#xA;endpoint to correct the guess made by the predictor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above focused solely on first order ODEs involving a scalar function.&#xA;What if the problem is multidimensional or involves higher order derivatives?&#xA;The mathematics of the techniques described above can easily handle&#xA;multidimensional data: Simply use the vector-valued time derivative.&#xA;Since a higher order ODE can be converted to a first order ODE via an&#xA;augmented vector-valued state, the same approaches used to address&#xA;multidimensional data can also be employed to address higher order ODEs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a problem with doing this: It throws out geometry. For example,&#xA;consider the rather simple first order ODE $\dot x = -y, \dot y = x$.&#xA;The solution to this multivariate ODE is uniform circular motion.&#xA;Applying Euler's method to this results in&#xA;$$\begin{aligned}&#xA;x(t+\Delta t) &amp;amp;= x(t) - \Delta t\,y(t) \\&#xA;y(t+\Delta t) &amp;amp;= y(t) + \Delta t\,x(t)&#xA;\end{aligned}$$&#xA;The square magnitude of this new vector is&#xA;$(x(t)^2+y(t)^2)(1+\Delta t^2)$, which is always greater than magnitude&#xA;of the vector at the start of the step. This is not uniform circular motion.&#xA;The solution obtained via Euler's method instead spirals out. Other techniques&#xA;spiral inward. A geometric integrator on the other hand will somehow&#xA;maintain the constraint that $x^2+y^2$ is a constant of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above example showed why we don't want to toss geometry in a very&#xA;simple problem. The geometry of Newtonian gravitation, along with much&#xA;classical mechanics in general, is symplectic geometry.&#xA;This is why symplectic integrators are of great concern. A simple example&#xA;again, with Euler's method: Suppose the second derivative of position&#xA;is given by some function $\ddot {\vec x}(t) = \vec f(x(t),t)$.&#xA;Applying the basic Euler method against 3+3 dimensional phase space&#xA;dictates that&#xA;$$\begin{aligned}&#xA;\vec x(t+\Delta t) &amp;amp;= \vec x(t) + \Delta t \vec v(t) \\&#xA;\vec v(t+\Delta t) &amp;amp;= \vec v(t) + \Delta t \vec f(\vec x(t),t)&#xA;\end{aligned}$$&#xA;A simple change makes this symplectic:&#xA;$$\begin{aligned}&#xA;\vec v(t+\Delta t) &amp;amp;= \vec v(t) + \Delta t \vec f(\vec x(t),t) \\&#xA;\vec x(t+\Delta t) &amp;amp;= \vec x(t) + \Delta t \vec v(t+\Delta t)&#xA;\end{aligned}$$&#xA;As is the case with the scalar techniques discussed at the start,&#xA;Euler's method is a starting point rather than the end with&#xA;regard to symplectic integration techniques. Symplectic Euler's method&#xA;is rather lousy. But at least orbits don't spiral outward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In making an N-body gravitation simulation, the size of N (the number of&#xA;bodies) is a key concern. Simulating a galaxy is a very different concern&#xA;from simulating a star system. The techniques used in simulating the&#xA;formation of a galaxy are very different from those used to develop a&#xA;solar system ephemeris. Galactic scale simulations cannot afford to&#xA;calculate all of the N&lt;sup&gt;2&lt;/sup&gt; gravitational interactions amongst all&#xA;the particles, and because N is so large, it cannot afford anything&#xA;more complex than very simple integrators. A star system model that does&#xA;not calculate all N&lt;sup&gt;2&lt;/sup&gt; of the gravitational interactions&#xA;or that uses very a simple integrator will by viewed in disdain.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2017-05-22T10:56:08.637" LastActivityDate="2017-05-22T10:56:08.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21062" PostTypeId="2" ParentId="21059" CreationDate="2017-05-20T18:12:07.493" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://pixinsight.com/doc/tools/StarAlignment/StarAlignment.html#description_003&quot; rel=&quot;nofollow noreferrer&quot;&gt;This page about a commercial product&lt;/a&gt; goes into some detail about their algorithm.  It does the triangle matching you describe, with something like simulated annealing to get a more optimal solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The accepted answer to &lt;a href=&quot;https://astronomy.stackexchange.com/q/14551/14434&quot;&gt;this closely related question&lt;/a&gt; recommends &lt;a href=&quot;http://hugin.sourceforge.net&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hugin panorama software&lt;/a&gt;; it's open source, so you should be able to glean the algorithms used.&lt;/p&gt;&#xA;" OwnerUserId="14434" LastActivityDate="2017-05-20T18:12:07.493" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21063" PostTypeId="2" ParentId="21059" CreationDate="2017-05-21T09:07:32.870" Score="3" Body="&lt;p&gt;Have a look at &lt;a href=&quot;https://www.astromatic.net/software/scamp&quot; rel=&quot;nofollow noreferrer&quot;&gt;SCAMP&lt;/a&gt; for astrometry and &lt;a href=&quot;https://www.astromatic.net/software/swarp&quot; rel=&quot;nofollow noreferrer&quot;&gt;SWarp&lt;/a&gt; for stacking. Like the software mentioned in the other answer, both are open source, so you can check what algorithms they use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SCAMP documentation is &lt;a href=&quot;https://www.astromatic.net/pubsvn/software/scamp/trunk/doc/scamp.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, with an explanation of the algorithm in chapter 6.7 (page 25). There's also a short &lt;a href=&quot;http://adsabs.harvard.edu/abs/2006ASPC..351..112B&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper&lt;/a&gt;, but the manual seems more thorough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the software is written with wide-field multi-CCD mosaic detectors in mind, so what they do might be overkill for what you have in mind.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastActivityDate="2017-05-21T09:07:32.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21065" PostTypeId="1" AcceptedAnswerId="21066" CreationDate="2017-05-22T01:58:13.730" Score="0" ViewCount="360" Body="&lt;p&gt;Assuming two similarly sized, tidally locked planets orbit a common barycenter, how would the gravity of the barycenter affect the surface gravity? For example, on the inner sides of the planets, what would be the effect of the multiple gravity centers (the planets' and the barycenter), especially to a being on the surface? &lt;/p&gt;&#xA;" OwnerUserId="16777" LastActivityDate="2017-05-22T07:06:49.673" Title="How would the barycenter of a double-planet system affect the two planets' surface gravity?" Tags="&lt;planet&gt;&lt;gravity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21066" PostTypeId="2" ParentId="21065" CreationDate="2017-05-22T07:06:49.673" Score="2" Body="&lt;p&gt;There's basically no gravitational pull towards the barycenter.  The two objects are in free-fall around each other and objects in free-fall experience (almost) no gravity from the object they are in free-fall around.  Anyone standing on either planet would basically only feel the gravitation of that object alone, as if it were sitting still in space (though nothing actually &quot;sits still&quot; in space), everything in space is flying through it's local gravitational field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can be small variations to the local objects gravity, generated by any rotation.  Tidal locking might be common with a dual-planet system and Tidal locking is usually a relatively slow rotation unless the objects are very close, but while it's a slow rotation, it's still greater than zero, which creates a tiny tidal bulge and a very slight &lt;a href=&quot;http://www.calctool.org/CALC/phys/newtonian/centrifugal.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;centrifugal force&lt;/a&gt; which would effectively lower the gravity.   This has nothing to do with the barycener, it is a product of the objects rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the objects get closer there is a tidal bulging which has some effect on gravity.   Parts of the planet that bulge outwards experience lower gravity and parts that are effectively stretched in closer to the center experience an increase in gravity.   This effect is pretty small and it's an indirect effect, caused by tidal forces from the other object, not caused by the barycenter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the &lt;a href=&quot;https://en.wikipedia.org/wiki/Roche_limit#Explanation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Roche Limit&lt;/a&gt;, gravity is effectively cancelled out by the tidal force and objects can drift off the surface both on the near and far side of the smaller of the two tightly orbiting objects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of gravitational variation would be small, unless the orbit was very close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to post a map of the change in gravity on Pluto's Moon Charon, or Jupiter's innermost Moon Io due to the tidal forces and tidal bulging, but I couldn't find any written results on that and I'm reluctant to try to do the math myself.  Even under very strong tidal forces like those two, the effect is likely still quite small, perhaps in the range of 1% or less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a &lt;a href=&quot;https://space.stackexchange.com/questions/3977/would-it-be-possible-to-fall-off-phobos&quot;&gt;question&lt;/a&gt; where the tidal force and it's effect on Phobos' gravity is discussed.  You need very strong tidal forces before the gravity gets significantly effected.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-05-22T07:06:49.673" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21067" PostTypeId="2" ParentId="11332" CreationDate="2017-05-22T15:50:28.867" Score="3" Body="&lt;p&gt;I was fortunate enough to spend the past week at a workshop held at the Green Bank Observatory (GBO), so I can give you a partial answer based on how the Breakthrough Initiatives affects it in particular.&lt;/p&gt;&#xA;&lt;h2&gt;Telescope time&lt;/h2&gt;&#xA;&lt;p&gt;The GBO actually operates a selection of telescopes, of which the most widely-known (and largest) is the Green Bank Telescope (GBT). The GBT is the main instrument used by Breakthrough Listen at the observatory. During &lt;a href=&quot;https://science.nrao.edu/observing/proposal-types/tac-reports/16b-tac-report&quot; rel=&quot;nofollow noreferrer&quot;&gt;the second half of 2016&lt;/a&gt;, the GBT was oversubscribed by a factor of 2.6, meaning that only about one in every 2.6 proposals was approved outright (“filler” proposals will be accepted if short periods of time are available). This doesn’t mean that for every hour of telescope time, 2.6 hours worth of observations were proposed, although the ratio is close (2.3), but it does mean that the GBT is very much in demand - and those numbers have increased this year. The telescope is not in use on only two days annually, Thanksgiving and Christmas, although there are blocks set out each week for maintenance.&lt;/p&gt;&#xA;&lt;p&gt;The answer to your first question, then, is yes - there is approximately indeed a zero-sum redistribution, because the telescope is extraordinarily busy. The actual data-gathering proportion may be slightly different because overhead times vary by the type of observations, and things like changing receivers or pointing at a different source do take time. However, it is true that Breakthrough Listen, like any other set of observations, takes time away from other scientists - &lt;a href=&quot;http://greenbankobservatory.org/science/breakthrough-listen/&quot; rel=&quot;nofollow noreferrer&quot;&gt;about 20% of the GBT’s annual total observing time&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;(As an aside, Breakthrough Listen operates a “backend” in the GBO computer room - a set of GPUs that analyze incoming data. There are plenty of backends for various receivers and instruments, and they do take up space - and power. We were told that the room the Breakthrough Listen computers are in wasn’t actually designed to hold that many computers, which did cause some initial cooling issues (which were quickly rectified) when these new units were installed. So it’s important to realize that the project takes up more resources than just observing time!)&lt;/p&gt;&#xA;&lt;h2&gt;Money&lt;/h2&gt;&#xA;&lt;p&gt;The financial system at Green Bank is actually a little complicated. In 2012, it was recommended that the GBT and the &lt;a href=&quot;https://www.lbo.us/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Long Baseline Array&lt;/a&gt; be removed from the NRAO, and &lt;a href=&quot;http://www.aui.edu/news/nrao-structural-changes-announcing-the-separation-of-the-green-bank-observatory-and-the-long-baseline-observatory/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this was implemented in 2016&lt;/a&gt;. In some ways, the transition is still happening - much official GBO information is still on the NRAO’s website - but the relevant point here is that in the 2017, the GBO &lt;a href=&quot;http://greenbankobservatory.org/about/green-bank-facts/#funding&quot; rel=&quot;nofollow noreferrer&quot;&gt;only received about 66% of its funds from the National Science Foundation (NSF)&lt;/a&gt;. The current agreement is that the NSF will gradually defund the GBO over the next few years.&lt;/p&gt;&#xA;&lt;p&gt;This is why Breakthrough Listen is so financially crucial, and why it is getting so much publicity at the GBO. Without sources (and currently, without &lt;em&gt;non-governmental&lt;/em&gt; sources) of funding, the observatory simply cannot operate. That said, the GBO is not beholden to people like Yuri Milner (the man behind the Breakthrough Initiatives) who have specific projects in mind - and indeed, that type of donor is quite the outlier. It does seek out donors and philanthropists who are willing to give money to simply ensure that the facility keeps going.&lt;sup&gt;1&lt;/sup&gt; If it can be guaranteed that these sources will continue giving without setting limits on what types of observing should be done, then things can progress as close to normal as possible.&lt;/p&gt;&#xA;&lt;p&gt;The exact details of funding for the GBO are not (as far as I know) public, and I don’t want to speculate too much. However, I can say that the observatory is almost certainly not going to push ordinary astronomers out of the way for the sake of big projects - in the short-term. Long-term funding is uncertain, true, but that does not mean that it will be renamed the Breakthrough Listen Observatory any time soon. The number of people who have operated the GBT number in the thousands, and they don’t pay to use it. It was built using public funds and has operated largely using public funds for quite some time now. I think the GBO intends to keep operating like this, and will not turn to the alternative - subject to the interests of only a select few projects - unless matters become extraordinarily desperate. It’s possible, but not something to be immediately worried about.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; One day last week, we noticed a helicopter landing at the airstrip. It turns out that the passenger was a VIP who was given a full tour of the telescope, guided by several observatory higher-ups. He was, as you can image, a prospective donor.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-05-22T17:54:39.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21068" PostTypeId="2" ParentId="20812" CreationDate="2017-05-22T17:59:04.123" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_hypothetical_Solar_System_objects&quot; rel=&quot;nofollow noreferrer&quot;&gt;This Wikipedia page&lt;/a&gt; doesn't say anything about a planet called Biga, and I couldn't find anything on the internet related to Harold Urey, and a possible tenth planet (netheir on Google nor on Wikipedia).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Planets_beyond_Neptune&quot; rel=&quot;nofollow noreferrer&quot;&gt;This page&lt;/a&gt; says that at that time, there was a theorized planet X, which might be what you are looking for, but lacking any references for Biga on the internet, it is hard to say anything.&lt;/p&gt;&#xA;" OwnerUserId="13473" LastEditorUserId="13473" LastEditDate="2018-02-21T19:45:05.563" LastActivityDate="2018-02-21T19:45:05.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21070" PostTypeId="1" CreationDate="2017-05-23T07:59:09.790" Score="6" ViewCount="4976" Body="&lt;p&gt;I'm watching &lt;a href=&quot;https://www.youtube.com/watch?v=MTY1Kje0yLg&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; YouTube video, and in it, the lecturer explains a model of planetary orbit. At &lt;a href=&quot;http://www.youtube.com/watch?v=MTY1Kje0yLg&amp;amp;t=1m55s&quot; rel=&quot;noreferrer&quot;&gt;1:55&lt;/a&gt; he shows how a marble will naturally orbit the center object, and he briefly claims that planets will not lose energy like the marble does. But I cant see how that is the case. Surely there must be meteors, asteroids, comets, including general space debris, that will over the course of millennia, diminish this energy/momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How then, can planets retain their energy, when there are almost only forces that would slow it down? Slowly increasing the energy wouldn't help either, as it would offset the balance. How can there be stability, especially over the course of millions of years?&lt;/p&gt;&#xA;" OwnerUserId="16961" LastActivityDate="2017-05-27T06:56:20.503" Title="How do planets retain momentum?" Tags="&lt;orbit&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21071" PostTypeId="2" ParentId="21070" CreationDate="2017-05-23T08:36:03.150" Score="29" Body="&lt;p&gt;Such forces do exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are utterly negligible in most circumstances. If you imagine a planet &quot;running into&quot; matter whilst travelling in its orbit. Let's say that on average that matter is at rest with respect to the star the planet is orbiting. Then in order to make a very significant change to the linear momentum of the planet, then it must accrete a significant fraction of its own mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a thing might be important in the very early history of a planetary system when the system is full of gas and debris (e.g. the formation of the Moon), but later on (and bear in mind that the &lt;a href=&quot;https://www.iau.org/static/resolutions/Resolution_GA26-5-6.pdf&quot; rel=&quot;noreferrer&quot;&gt;IAU definition&lt;/a&gt; of a planet includes the specification that it has &quot;cleared the neighbourhood around its orbit&quot;) this just isn't an issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://curious.astro.cornell.edu/about-us/75-our-solar-system/comets-meteors-and-asteroids/meteorites/313-how-many-meteorites-hit-earth-each-year-intermediate&quot; rel=&quot;noreferrer&quot;&gt;This&lt;/a&gt; popular science article suggests that 37,000-78,000 tonnes of material hits the Earth every year. Sounds a lot, but in comparison to the mass of the Earth ($6\times 10^{21}$ tonnes) it is very small. If the material was at rest with respect to the Earth's orbit, then an order of magnitude estimate for how long such impacts would take to &lt;em&gt;significantly&lt;/em&gt; affect the Earth's orbit would be $\sim 10^{16}$ years.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="13753" LastEditDate="2017-05-24T04:53:03.283" LastActivityDate="2017-05-24T04:53:03.283" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21072" PostTypeId="1" AcceptedAnswerId="21076" CreationDate="2017-05-23T10:03:31.387" Score="1" ViewCount="173" Body="&lt;p&gt;First of all sorry for my immense ignorance in this matter and for all the stupid things I'll say here.&lt;/p&gt;&#xA;&lt;p&gt;Yesterday, May 22nd 2017, I removed the telescope from its dusty box&lt;sup&gt;1&lt;/sup&gt; and decided to watch Jupiter. After about 2 minutes of observation I spotted something across the sky &lt;em&gt;[X-files music in background]&lt;/em&gt;, it was a little less bright as Jupiter's &lt;em&gt;&amp;quot;Galilean&amp;quot;&lt;/em&gt; moons&lt;sup&gt;2&lt;/sup&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Below you can see an accurate &lt;em&gt;[Windows Paint]&lt;/em&gt; reconstruction of the fact I saw in my telescope. The red line indicates the direction of the thing (from right to left outside of the telescope).&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.imgur.com/SxzHjGh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/SxzHjGh.png&quot; alt=&quot;Click to enlarge&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Available data to help identify the object&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;Time: May 22nd, 2017 @ 22:12 UTC&#xA;My position: {&#xA;    Lat: 41.720194062015096&#xA;    Long: 12.713427287843729&#xA;    Sea level: 540 m&#xA;    }&#xA;Direction: Probably South // The direction of Jupiter from that location at that time (didn't have a compass)&#xA;Telescope: {&#xA;    Focal length: 700 mm&#xA;    Diameter: 60 mm&#xA;    Lens: H 20 mm&#xA;    }&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;I only have data of the ISS and Iridium insights on an Android app&lt;sup&gt;3&lt;/sup&gt; so I am here to ask if there is some tool with which I can identify by myself as many satellites as possible across the sky.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Notes&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Winter is not nice with observers from here&lt;/li&gt;&#xA;&lt;li&gt;Not visible at naked eye&lt;/li&gt;&#xA;&lt;li&gt;I can almost certainly say that the &amp;quot;thing&amp;quot; was not the ISS or one of the Iridium satellites&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;strong&gt;P.s&lt;/strong&gt;: I will post here what the object was (with reference if possible) as soon as I identify it for curious people like me.&lt;/p&gt;&#xA;" OwnerUserId="16962" LastEditorUserId="1082" LastEditDate="2021-01-07T23:28:03.307" LastActivityDate="2021-01-08T00:58:01.660" Title="&quot;What was that spot in the sky?&quot; - How to identify satellites" Tags="&lt;amateur-observing&gt;&lt;jupiter&gt;&lt;artificial-satellite&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21073" PostTypeId="1" AcceptedAnswerId="21079" CreationDate="2017-05-23T10:42:27.933" Score="2" ViewCount="1683" Body="&lt;p&gt;Imagine standing (somewhere) on the Moon's surface. How big would the Earth look from the Moon, compared to the image of the Moon we see from Earth? What would be the ratio between the diameters of each body? How would this ratio be calculated? Any help with answering such a calculation would be highly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="7872" LastEditorUserId="23940" LastEditDate="2018-11-17T11:02:16.397" LastActivityDate="2019-10-26T01:35:13.140" Title="How big would the Earth look from the Moon?" Tags="&lt;the-moon&gt;&lt;distances&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21074" PostTypeId="1" CreationDate="2017-05-23T11:15:03.953" Score="1" ViewCount="91" Body="&lt;p&gt;I recall when I was a child (many years ago) my father taking us out to gaze at the starts. I remember being able to see the Milky Way with our naked eyes (North Central New Jersey). Nowadays living down here in Maryland I can count the stars on my hand because of all of the light pollution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I ask, is there some sort of listing that shows where the best places are to gaze without a telescope up at the stars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to see the Milky Way again and not the white glow of the city lights.&lt;/p&gt;&#xA;" OwnerUserId="16964" LastActivityDate="2017-05-23T19:11:11.830" Title="Is there any official star gazing location list?" Tags="&lt;milky-way&gt;&lt;star-gazing&gt;&lt;light-pollution&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21075" PostTypeId="2" ParentId="21074" CreationDate="2017-05-23T13:00:25.547" Score="3" Body="&lt;p&gt;You may be out of luck. If you live pretty much anywhere in the North East, there are very few locations where you can see a truly dark sky. You might be able to get somewhere where you can see the faint glow of the milky way band, but to see a really dark sky, you pretty much have to be out west in the middle of no where.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One resource to help you out is to look up the &lt;a href=&quot;https://en.wikipedia.org/wiki/Bortle_scale&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bortle Scale&lt;/a&gt;. This is a metric used to describe how dark the sky is in a given area and thus what you're likely able to see with the unaided eye. There are numerous sites which attempt to predict or inform what your Bortle scale number is at your specific location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general though, you can get a feel of the light pollution in a particular area by using light pollution maps like &lt;a href=&quot;https://www.lightpollutionmap.info/#zoom=4&amp;amp;lat=5123902&amp;amp;lon=-10909018&amp;amp;layers=B0TFFFFF&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt; or &lt;a href=&quot;http://darksitefinder.com/maps/world.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;. Attached below is an image from the first link showing the general light pollution level across the US.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7UjlV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7UjlV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-05-23T13:00:25.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21076" PostTypeId="2" ParentId="21072" CreationDate="2017-05-23T17:54:49.290" Score="2" Body="&lt;p&gt;This will be hard to track down. I'd suggest looking at something like &lt;a href=&quot;http://www.satview.org/tm.php?group=master_tle&amp;amp;lang=us&amp;amp;dmy=22052017&amp;amp;hms=2210&quot; rel=&quot;nofollow noreferrer&quot;&gt;SATVIEW Time Machine&lt;/a&gt;. I've entered the location and time you provided (I had to deviate by a few minutes to find something so maybe your time is not 100% accurate) and I can see there were a few satellites flying overhead. With enough work, you may be able to figure out which one was most likely the one you saw based on the Alt-Elev information detailed in the table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's plenty of other online satellite tracking websites out there, but keep in mind, most of them aren't going to have everything. And possibly, what you saw is in no database because it's a classified satellite in which case you probably won't find any record of it.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-05-23T17:54:49.290" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21077" PostTypeId="2" ParentId="21074" CreationDate="2017-05-23T19:07:51.793" Score="1" Body="&lt;p&gt;Have a look at the &lt;a href=&quot;http://darksky.org/idsp/reserves/&quot; rel=&quot;nofollow noreferrer&quot;&gt;the list of Dark Sky reserves&lt;/a&gt; from the international dark-sky association. This lists areas with almost 'unaffected' sky, where   arrangements are in place to conserve the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://darksky.org/idsp/finder/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; they offer a map of places with 'better' conditions.&lt;/p&gt;&#xA;" OwnerUserId="15824" LastActivityDate="2017-05-23T19:07:51.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21078" PostTypeId="2" ParentId="21074" CreationDate="2017-05-23T19:11:11.830" Score="1" Body="&lt;p&gt;There's a state park in Pennsylvania that has some of the darkest skies on the east coast. It's so dark at night that the Milky Way casts a shadow. It's probably the best place you can go on the east coast to see the galaxy. I'm not sure of the distance from Maryland, but from New York City it's about 5 hours away, so it's a weekend trip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's called cherry springs state park, and here's a link to the Wikipedia page: &#xA;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Cherry_Springs_State_Park&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Cherry_Springs_State_Park&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16600" LastActivityDate="2017-05-23T19:11:11.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21079" PostTypeId="2" ParentId="21073" CreationDate="2017-05-24T00:59:33.687" Score="7" Body="&lt;p&gt;There are a couple questions here. The ones you have asked are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What would be the ratio between the respective diameters?&lt;/li&gt;&#xA;&lt;li&gt;How would this ratio be calculated?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But I think you meant to ask:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;How much larger does the Earth appear when standing on the Moon than the Moon appears when standing on the Earth? (and the calculation)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For the answer to #1 (and #2), it is the &lt;em&gt;ratio between the numbers (of the Earth's diameter of 12,742 km and the Moon's diameter of 3,474 km)&lt;/em&gt; which is &lt;span class=&quot;math-container&quot;&gt;$Earth:Moon = 3.668:1$&lt;/span&gt; as &lt;a href=&quot;https://astronomy.stackexchange.com/questions/21073/how-big-would-the-earth-look-from-the-moon/21079#comment35690_21073&quot;&gt;@pela's comment&lt;/a&gt; mentions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the answer to #3, it is the same! The more interesting part is why. We know that the distance from the Earth to the Moon is the same as the Moon to the Earth, so it follows that when we are on one of them looking at the other we are gazing across the same distance. Next, we know that an object (approximately) appears larger in the same proportion as it is larger, so an object that is twice as wide or tall will appear twice as wide or tall. Since the Earth is 3.668 times as wide as the Moon, it will appear that much larger when viewed from the same distance, and we've already discussed why it is the same distance. So, that's it really. It is the ratio of the diameters.&lt;/p&gt;&#xA;" OwnerUserId="16975" LastEditorUserId="7982" LastEditDate="2019-10-26T01:35:13.140" LastActivityDate="2019-10-26T01:35:13.140" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21081" PostTypeId="1" CreationDate="2017-05-24T08:46:28.377" Score="2" ViewCount="182" Body="&lt;p&gt;I've studied that the major axis of a planets orbit = ( radius of perihelion + radius of aphelion) . But then  how can we calculate the value of the minor axis if we are provided with the values of radius of perihelion and radius of aphelion only ?&lt;/p&gt;&#xA;" OwnerUserId="16980" LastActivityDate="2017-05-24T11:10:54.640" Title="Property of elliptical orbits" Tags="&lt;planet&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21083" PostTypeId="2" ParentId="21081" CreationDate="2017-05-24T11:10:54.640" Score="6" Body="&lt;p&gt;If $r_p$ is the perihelion distance (the closest distance of the planet in its elliptical orbit from the Sun), and if $r_a$ is the aphelion distance (the farthest distance from the Sun), the major axis is&lt;br&gt;&#xA;$$&#xA;2a = r_a+r_p&#xA;$$&#xA;where $a$ is the &lt;em&gt;semi-major axis&lt;/em&gt; (half of the major axis).&#xA;The &lt;em&gt;eccentricity&lt;/em&gt; of the orbit can be calculated from the perihelion and aphelion distances using&#xA;$$&#xA;e = \frac{r_a-r_p}{r_a+r_p}&#xA;$$&#xA;Once the eccentricity is known, the &lt;em&gt;semi-minor axis&lt;/em&gt; can be calculated from&#xA;$$&#xA;b = a\sqrt{1-e^2}&#xA;$$&#xA;The minor axis, of course, is twice the semi-minor axis, or $2b$.&lt;/p&gt;&#xA;" OwnerUserId="5537" LastActivityDate="2017-05-24T11:10:54.640" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21086" PostTypeId="2" ParentId="20948" CreationDate="2017-05-25T00:25:53.480" Score="1" Body="&lt;p&gt;My copy of &quot;Principles of Stellar Evolution and Nucleosynthesis&quot; by Donald D. Clayton is from 1968, but it covers what you're interested in and is intended for a graduate-level course.  It's also well-written.&lt;/p&gt;&#xA;" OwnerUserId="16986" LastActivityDate="2017-05-25T00:25:53.480" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21088" PostTypeId="2" ParentId="2302" CreationDate="2017-05-25T03:44:18.973" Score="2" Body="&lt;p&gt;The sun is a big ball of hydrogen atoms, that get compressed together due to a huge gravitational force.  We start off with 4 H atoms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P-e, P-e, P-e, P-e (note: P=proton, e=electron)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when a proton (P) and electron (e) fuse together, they simply combine into a neutron (note: neutron=N). So, if two of those P-e pairs fuse into a neutron, we now have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N, N, P-e, P-e&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two neutrons, two protons, two electrons. The 4 hydrogen atoms have become a single helium atom. But wait... the mass went on a diet! If we &quot;weighed&quot; the 4 hydrogen atoms, their total mass was = 4(1.004u) = 4.016u. But the end result, a single helium atom, is only 4.003u.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, from 4 hydrogens to 1 helium, some &quot;weight&quot; was lost, or more exactly, some of the mass converted into energy. How much energy? This much energy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e=mc^2&lt;/p&gt;&#xA;" OwnerUserId="16988" LastActivityDate="2017-05-25T03:44:18.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21089" PostTypeId="2" ParentId="872" CreationDate="2017-05-25T08:11:21.973" Score="3" Body="&lt;p&gt;Lets start with some basics and, before I continue, this is a criteria based answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short answer:   30.   (OK that sounds nuts, but hear me out).  That's about the upper, upper, gonzo, bananas limit for planet definition and long-term stable orbits.  I'm tempted to say 25 as an upper limit just because 30 seems too improbable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gist of the problem, is that a star and protoplanetary disk is unlikely to form the maximum possible number of planets.   Gravity tends to clump around the larger objects.  Planetary perturbations and migration make the maximum possible stable number unlikely to be reached, but with luck of a &quot;just right&quot; formation and some planet capture, I reached a ballpark estimate of about 30.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer: let's assume we're talking about only stable planetary orbits by the definition of having cleared out their orbital path and don't cross each other's orbits.   This eliminates any Trojan planets and doesn't eliminate, but makes highly elliptical orbits problematic because they span a greater orbital range.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And lets dismiss any large &lt;a href=&quot;https://en.wikipedia.org/wiki/Planetesimal&quot; rel=&quot;nofollow noreferrer&quot;&gt;planetesimals&lt;/a&gt; that might be planet sized and any planet sized dwarf planets that cross other planet's orbits.    We're only counting orbit dominating planet definition planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets also eliminate any binary or trinary systems, and only use single star systems, but the star could have some very massive planets that are borderline brown dwarf stars if you like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using our solar-system as a guideline and quoting from the planetesimals article above:   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is generally thought that about 3.8 billion years ago, after a&#xA;  period known as the Late Heavy Bombardment, most of the planetesimals&#xA;  within the Solar System had either been ejected from the Solar System&#xA;  entirely, into distant eccentric orbits such as the Oort cloud, or had&#xA;  collided with larger objects due to the regular gravitational nudges&#xA;  from the giant planets&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd also like to set some kind of time-limit because young solar-systems can have hundreds of large planitesimals.    By about 700 million years of age, our solar-system had, for the most part, settled down into the 8, &lt;a href=&quot;https://en.wikipedia.org/wiki/Planet_Nine&quot; rel=&quot;nofollow noreferrer&quot;&gt;maybe soon to be 9&lt;/a&gt;, planets that are currently known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A larger star probably has the potential for a good deal more than 9.   But if it takes 700 million years (give or take) for a protoplanetary disk to work itself out into planets with stable, semi-permanent orbits, that puts a limit on the size of the star.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 40 solar-mass star has a lifespan of only a million years or so before it goes Supernova.  That's far too short a lifespan for planetary systems to form.   Even a 10 solar mass star lasts just 30 million years or so.   Again, too short.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 4 solar mass star has a lifespan some 30 times shorter than our sun (&lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/M/Main+Sequence+Lifetime&quot; rel=&quot;nofollow noreferrer&quot;&gt;using the 2.5 power rule&lt;/a&gt;, which I've also seen as a 3 power rule, but all this is pretty ballpark.   Point is, a star with 4 solar masses has less than 400 million years for it's planetary system.   5 solar masses, as little as 200 million years.    That's pretty close to what I'd call the minimum amount of time for a planetary system to have relevance, so I'm going to go with a 4 solar mass upper limit.   The romantic notion of a star 20 times the mass of our sun, with 100 planets might make good science fiction, but it's not realistic.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 2nd factor to consider is the mass and size of the planetary debris field.  Our sun is about 99.8% of the mass of the solar-system, leaving 0.2% of the solar-system's mass to form all the planets and other stuff.   There was probably more mass in the debris field originally, some of which was lost as rogue planets, rogue comets and asteroids, so the original planetary debris field might have been higher, but not all that much higher.   Larger objects can cast out smaller ones.   The ratio of lost debris to remaining debris shouldn't be all that high.   (if anyone knows, feel free to post a comment).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The highest percentage of mass in a forming solar-system is difficult to calculate and it depends on the total angular momentum of the debris field that collapses into the spiraling disk of matter, but it's improbable that the % of mass gets too high.   1%-3% might be on the upper limit.  If we go with 3% of the mass of a 4 solar mass star in the planetary disk that's about 40,000 Earth masses or about 125 Jupiter masses.  That's obviously ballpark, perhaps too ballpark, but it helps to have a sense of how much stuff we have to work with.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of a debris field is important too.  &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0004-637X/808/1/69&quot; rel=&quot;nofollow noreferrer&quot;&gt;By this article&lt;/a&gt;, the largest debris field ever observed is about 1,000 AU in diameter (500 AU in radius) with a debris field mass of about 3.1 += .6 Jupiter masses and a central star perhaps less massive than our sun.   Whether such a system could form planets as far out as 500 AU is hard to say, but I'm inclined to think that the outermost planet would form comfortably inside that debris field, not at the observed edge.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth pointing out that planetary formation is a chaotic mess.   A young protoplanetary disk, especially one with some 125 jupiter masses worth of material could easily form over 100 planet sized objects early in formation, but it wouldn't retain that many.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets perturb each other's orbits and they need space.  You'd get collisions like the collection that formed our Moon and larger planets can send smaller planets any which way.   No system could keep 100 planets.  It's too many and would be much too unstable.   There would be far fewer when  a mostly stable formation is reached.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter, for example, is believed to have migrated towards the sun when our solar system was young, them migrated back outwards, called &lt;a href=&quot;https://en.wikipedia.org/wiki/Planetary_migration#Type_II_migration&quot; rel=&quot;nofollow noreferrer&quot;&gt;type II migration&lt;/a&gt;.   Migrating Jupiters are both good and bad if you want a lot of planets.    Jupiter's migration is believed to be the reason why there's no planets and so much empty space between Mars and Jupiter and why Mars is so small.   Jupiter's migration may have also sent Uranus, Neptune out to their current distant orbits, so gas giant migration can move planets around, but it can also cast them completely out of a solar-system.   The larger the gas giant, the greater a kick it can give to smaller planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very massive planets are bad if you want the highest number of planets because they cause greater perturbations and demand the greatest space around them.   With a lot of debris in a planetary disk, very large planets are likely to form so more debris isn't always better.  What you probably want is a larger, more spread out disk, where you don't get any super massive planets, but some massive enough to push some young forming planets outwards to create more planets at greater distances.  Planets are unlikely to form at very great distances, but they can be tossed out there by larger planets to very distant orbits.  By tossing a number of fledgling planets outwards early in formation, the total number of planets in a solar-system could increase.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How close can the planets be to each other?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets don't like to be too close to each other.   While we can't see small planets very well, Kepler observations seem to confirm this that very close planets are rare.  When they're too close, there's orbital instability.  Earth and Venus are the closest planets by multiple, where Earth is 1.38 times the distance from the sun as Venus.  By this &lt;a href=&quot;http://curious.astro.cornell.edu/about-us/57-our-solar-system/planets-and-dwarf-planets/orbits/237-what-is-the-pattern-to-the-distances-between-each-planet-and-the-sun-intermediate&quot; rel=&quot;nofollow noreferrer&quot;&gt;short article&lt;/a&gt;, a multiple of 1.4 to 1.8 times the distance between planets is suggested.  Observations of exo-solar-systems find very few planets closer than 1.4 times their nearest observed neighbor, so for an entire system, a 1.4 to 1.8 multiple seems about right on average.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets around small stars, like &lt;a href=&quot;https://en.wikipedia.org/wiki/TRAPPIST-1#Planetary_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Trappist 1&lt;/a&gt; can get very close to each other, close enough that they can appear about moon sized from their closest neighbors, but those systems are almost entirely around small red-dwarf stars with very tight orbits, often with orbital resonance and even with very close orbiting planets, they still average out to about the 1.4 multiple or greater.   Planets in a 3/2 orbital resonance that corresponds to a 1.31 distance multiple, and such resonances depend on the interactive tidal force that are only possible at close distances around smaller stars.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler-36&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kepler 36&lt;/a&gt; is an oddball with two very close planets with a 7:6 orbital resonance, but building an entire solar system from planets that close seems enormously improbable.   So, a key criteria to my estimate is the 1.4 distance multiple, and that's probably conservative over an entire system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How close can the closest planets be to the star?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat of a 4 solar-mass star is a problem for very close planets.   A 4 solar mass star, (while the luminosity changes over it's lifetime), is over 100 times more luminous than our sun, so the innermost rocky planet should probably begin at roughly about 10 times the distance Mercury is from our sun.  Much closer than that and the planet would be in danger of being vaporized.   So for a 4 solar mass star, 3 AU might be a good starting point.  Applying the 1.4 multiple to a 3 AU starting point.   A hot Jupiter might survive closer than that, but a hot Jupiter couldn't form that close, so that would probably require too much migration for our goal of highest number of planets.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so, if we start at 3 AU, and we do a 1.4 distance multiple, then our 4 solar mass star can have up to 30 planets within an orbits less than one light year, and just 32 within 2 light years, so you don't add much by doubling the distance, at least, using the 1.4 multiple.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An obvious question that follows might be, well, maybe the 1.4 multiple no longer applies at larger distances, but planets would need to grow fairly large to effectively clear out their orbit and have an effect on near-by asteroids and comets, like Neptune does and Planet 9 is believed to, so as the distance grows, you can't have mercury sized planets and define them as planets, and as the distance grows, the planets gravitational effect on each other remains consistent, so the 1.4 multiple rule should still apply even at very distant orbits.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury for example, is massive enough to be a planet where it is, but if it was out past Neptune, it would be perhaps too small to clear out it's orbit.   Here's a &lt;a href=&quot;https://astronomy.stackexchange.com/questions/11399/how-much-more-massive-would-pluto-have-to-be-to-clear-its-neighborhood&quot;&gt;question that discusses this in more detail&lt;/a&gt; and it raises the problem that if Pluto was some 15-20 times more massive, the minimum mass it would need and assuming it wasn't crossing Neptune's orbit, That theoretical object would still need a billion years to clear it's orbit and that's over twice our star's lifetime and the necessary minimum size grows a larger at greater distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we go with our one-light year proposal, an object orbiting around a 4 solar mass star at 1 light year distance has an orbital period of about 8 million years and an orbital velocity of about .23 km/s and it would have a required minimum mass to clear out it's orbit of at least several Earths.  Planet 9, for comparison, is thought to have an orbital period between 10,000 and 20,000 years and an orbital velocity in the .5-.7 km/s range and a semi-major axis of about 600-800 AU or about 1/90th of a lightyear.  Those numbers are all ballpark and just posted for comparison.   But it points out the difficulty in recognizing a planet in a very distant orbit.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for a planet to get that distant, it would need to be thrown out there by a larger planet, presumably undergoing type II migration or, perhaps captured from a passing star.   I think you'd probably want some of both to maximize the number of planets.   A star with a very large very distant planet could be effective in &lt;a href=&quot;https://phys.org/news/2012-04-stars-capture-rogue-planets.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;helping capture planets&lt;/a&gt; and/or debris from nearby stars that pass too close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases, the planet cast out very far or captured planets would initially have a very eccentric orbit and it would take some time for any such planets to circularize and you'd need the orbits to circularize, because a handful of eccentric orbits don't meet the planet criteria if they cross other planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, using our solar-system as a model, the outer planets, Uranus, Neptune and Planet 9 (if it exists) are all thought to have formed quite a bit closer to the sun than where they are now and migrated outwards, presumably by Jupiter.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A large star could have upwards of 100 Mercury or maybe even Earth sized objects in it's orbit, but no where close to that many that would meet the planet criteria.   30 is pushing it.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A large star capturing planets whether rogue, or capturing planets off a smaller star is certainly possible.  3 body dynamics does make planet capture possible, but there's still the problem of eccentricity and orbits crossing other orbits not meeting the criteria of a planet.  If you dismiss that standard orbital criteria or a planet, then the number goes up.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, using the criteria for a large star (4 solar masses) an innermost planet (3 AU) an outermost (1 light year - a bit of a stretch), and distance multiple (1.4 - also probably on the low side), a 4 solar mass star could have a maximum of 30 planets.   If you run different criteria, you get different numbers, but I think that's a pretty good upper benchmark, perhaps on the generous side.    Such a system could have a lot more objects that meet the dwarf-planet criteria, some of them even what we think of as planet sized, but meeting the complete &lt;a href=&quot;https://www.loc.gov/rr/scitech/mysteries/pluto.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;planet criteria&lt;/a&gt;, 30 seems a pretty good gonzo upper limit.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something interesting happens if you make the star smaller.    If we make the star 2 solar masses instead of 4 and put the outermost planet at the inverse square law or .707 light years, not 1 light year.   A 2 solar mass planet is about 12-16 times as luminous as our sun and 12-16 times less luminous than a 4 solar mass star, so the outermost planet that wouldn't get vaporized is now about 1 AU, not 3 AU.   So the inner part of the planet region is 3 times closer and just 1.4 times close on the outside, so curiously a 2 solar mass star could perhaps maybe hold more planets than the 4 solar mass star.   It wouldn't capture as many, on average, but the upper limit still goes up, using the same criteria to 32 or 33 for a 2 solar mass star and continues to grow as the star gets smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same time, as stars get smaller, the upper end mass of the planetary debris field grows smaller too and the ability to capture planets drops, so I don't small stars are good candidates for the most planets, but interestingly, smaller stars with smaller protoplanetary disks could still, on average have as many planets as their larger neighbors.  When James Webb starts to take a look, maybe we'll get an answer on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously if you had all no criteria, and a star a few million light years from the nearest galaxy or massive object, you could design something with many more planets, but I'm thinking formation within a galaxy and I'm thinking that both planet capture and the right set of circumstances during formation would both play a role in maximizing the number of planets.  A star that far from other stars would be unlikely to capture any planets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that's not too world-building an an answer or too long.   I'll try to check it for typos tomorrow.  (kinda late now). &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-05-25T08:29:53.360" LastActivityDate="2017-05-25T08:29:53.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21093" PostTypeId="1" CreationDate="2017-05-25T14:05:48.840" Score="1" ViewCount="54" Body="&lt;p&gt;I have no formal training in physics, just a casual interest, so if this question doesn't make sense i would love to understand why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know from casual physics talks etc that pairs of particles pop in and out of existence due to fluctuations in the underlying relevant field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that mean that for every instant there is an extra amount of mass per unit of space equal to the massive 'fluctuation' particles that still did not pop out of existence/annihilate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that holds true has anyone measured it or calculated it?&lt;/p&gt;&#xA;" OwnerUserId="16993" LastActivityDate="2017-05-25T14:05:48.840" Title="Mass of quantum fluctuations in space?" Tags="&lt;space&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21095" PostTypeId="1" CreationDate="2017-05-25T15:33:58.060" Score="6" ViewCount="330" Body="&lt;p&gt;August 21 is going to see a total eclipse across most of the US, and I'm planning a trip to what I hope is an &lt;a href=&quot;http://www.stjomo.com/eclipse2017/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ideal viewing location&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are good ways to safely view the eclipse?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't want to make a big trip out of it, only to be looking into a poor quality pin-hole shoebox, or $2 mirrored cardboard glasses.  How can get I get a good, detailed clear view of it while staying safe?  What about ways to photograph it effectively?&lt;/p&gt;&#xA;" OwnerUserId="2234" LastEditorUserId="2234" LastEditDate="2017-05-25T18:36:43.760" LastActivityDate="2017-08-02T19:39:09.877" Title="Good ways to safely view an Eclipse?" Tags="&lt;solar-eclipse&gt;&lt;photography&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21096" PostTypeId="2" ParentId="20620" CreationDate="2017-05-25T17:09:43.350" Score="3" Body="&lt;p&gt;A relatively simple way to understand why the moon axis seems to rotate is to observe that the stars do the same thing. Watch over several hours as &lt;a href=&quot;https://en.wikipedia.org/wiki/Cassiopeia_(constellation)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cassiopeia&lt;/a&gt; - located near the North Star moves and you'll notice the orientation changes. Cassiopeia, for example, is shaped like a &quot;W&quot; when lowest in the sky, a &quot;3&quot; as it rises, an &quot;M&quot; as it reaches its highest point overhead, a Σ (Sigma) as it is setting. Below is the constellation as it appears lowest in the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;img src=&quot;https://i.stack.imgur.com/3PvCQ.png&quot; alt=&quot;Cassiopea at its lowest point&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image from IAU and &lt;a href=&quot;http://www.iau.org/public/themes/constellations/#cas&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sky &amp;amp; Telescope magazine (Roger Sinnott &amp;amp; Rick Fienberg)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because Cassiopeia is close to the north celestial pole observers in the Northern  hemisphere can watch the constellation rotate through all 360 degrees through the year.  Orion, however is near the celestial &lt;em&gt;equator&lt;/em&gt; and the amount of apparent rotation relative to the ground is always less than 180 degrees and as little as 45 degrees - depending on your latitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;moon&lt;/strong&gt; can be found anywhere from +28.5° to −28.5 (above or below) the celestial equator - its position relative to your observing latitude dictates the maximum amount of apparent rotation.&lt;/p&gt;&#xA;" OwnerUserId="16846" LastActivityDate="2017-05-25T17:09:43.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21097" PostTypeId="2" ParentId="21095" CreationDate="2017-05-25T17:42:02.037" Score="3" Body="&lt;p&gt;I don't think you can get much better safety instructions for viewing than from &lt;a href=&quot;https://eclipse.gsfc.nasa.gov/SEhelp/safety.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA&lt;/a&gt;. Welder's classes are a pretty good choice. As for photography, I think &lt;a href=&quot;http://www.nikonusa.com/en/learn-and-explore/a/tips-and-techniques/how-to-photograph-a-solar-eclipse.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nikon&lt;/a&gt; would be a knowledgeable source, and they recommend a solar filter.&lt;/p&gt;&#xA;" OwnerUserId="14261" LastActivityDate="2017-05-25T17:42:02.037" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21098" PostTypeId="2" ParentId="19861" CreationDate="2017-05-25T18:06:38.140" Score="1" Body="&lt;p&gt;First, the data in most of these columns is not required for the computation of planetary positions using the theory if the 'second' form equation is used. The terms are, using the order of the example of Jupiter VSOP 87D's third line ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4510 ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;4&quot; means the 'version code', i.e., this is the &quot;D&quot; version of the theory which allows computation of the heliocentric rectangular coordinates of the planet for the ecliptic of date. A &quot;3&quot; would be for the spherical heliocentric coordinates for the ecliptic of date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;5&quot; is the 'body' code, Jupiter being the FIFTH planet. Mercury would be planet '1', for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;1&quot; is the 'index of coordinate', meaning in this example that this line is part of the 'longitude' coordinate calculation. Later lines show a '2' for the B (or latitude) coordinate or '3' for the R (radial or distance) coordinate)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;0&quot; is the 'degree' or power of the time variable. Here, terms are multiplied by T to the ZERO power (or 1). Later terms are multiplied by successively higher powers of T.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 &#xA;... is simply the sequential number of this line in this series. Jupiter has 5 series (each multiplied by a higher power of T), so this number 'resets' for each series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next 12 terms, mostly zeros in this example, are the 'coefficients of mean longitude'. As near as I can tell, these coefficients are used to correct mean longitude quantities for secular perturbations. They are used only in the first form of the equation as shown below with the next two terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next two terms are amplitudes S and K, used in the first form of the positional equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T**alpha * (S sin phi + K cos phi)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where phi = sum[i, 1 thru 12] [a(i) * lambda(i)] and lambda(i) values are given for each solar system body in the explanatory text.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final 3 terms are Amplitude A, Amplitude B and Phase C for use in the second form of the equations to calculate position:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T**alpha * A * cos(B + CT)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... with T being time in Julian centuries, and alpha the 'power' of T.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My comment in the original question about the terms being 'off' by several orders of magnitude is moot ... for convenience either in calculating or printing, Meeus multiplied the terms by 10**8, then corrects for it later. This probably facilitated better computation on machines of the time it was written and also made for neater printing in the book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The document provided by barrycarter, along with some other study of planetary motion allowed me to finally understand this, thanks!&lt;/p&gt;&#xA;" OwnerUserId="15619" LastEditorUserId="15619" LastEditDate="2017-05-26T19:12:44.370" LastActivityDate="2017-05-26T19:12:44.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21099" PostTypeId="2" ParentId="21095" CreationDate="2017-05-25T19:34:38.930" Score="5" Body="&lt;p&gt;Projection is a safe way to view the partial phase of the eclipse. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pinhole produces a small image of the sun, but doesn't require any special equipment. A larger image can be obtained by projecting through a telescope:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point the scope at the sun, but never look down a scope that is pointing anywhere near the sun, neither look along the tube. Use its shadow to line it up with the sun.  Hold a white card half a metre or so behind the eyepiece, and focus the telescope to make a sharp image of the sun. You will want to provide some shade for the image. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iQIp6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iQIp6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image from &lt;a href=&quot;https://www.bakersfieldcollege.edu/planetarium/observing-the-sun-safely&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bakersfield College &lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works well with binoculars, you need some way to mount, or prop up the binoculars, and you will need to cover up one of the lenses (or you will get two overlapping images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a larger telescope, you will want to cut down the aperture, to prevent overheating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will be able to see sunspots on the disc of the sun and the growing bite of the sun hidden behind the moon (which will be otherwise unseen). You can practice on any sunny day, You should be able to see some sunspots, and perhaps faculae on any day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other alternatives are specially designed solar viewing glasses. They must be unscratched and meet ISO 12312-2:2015 standard. Welders glass is also possible if rated &quot;14&quot; (the darkest possible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not possible to use any form of glasses with a telescope. It is possible to get an aperture filter, which may be useful if you have the right telescope, but the projection method is better if you want more than one person to view at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any kind of improvised solar filter is unsafe. This includes: stacks of sunglasses, pop-tarts wrappers, exposed film and viewing reflections in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During totality, the sun is hidden by the moon, and so it is not dangerous to look at the solar corona during totality with the naked eye. If you had been wearing solar viewing glasses, be ready to replace them. You MUST put them back on before totality ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photography is hard. The images of the solar eclipse are made with professional telephoto lenses and good deal of skill. Instead you can photograph the scene. A timelapse of the eclipse can be a very effective way to record the event, especially if you have some interesting foreground and background. The eclipse itself will be a tiny dot you can still document how the scenery and shadows change during the event.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2773" LastEditDate="2017-08-02T19:39:09.877" LastActivityDate="2017-08-02T19:39:09.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21100" PostTypeId="2" ParentId="21095" CreationDate="2017-05-25T20:04:42.400" Score="0" Body="&lt;p&gt;Astro solar filter is good.  Baader Planetarium make this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.baader-planetarium.com/en/solar-viewer-astrosolar-silver-gold-(1pc-10pc-25pc-100pc).html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.baader-planetarium.com/en/solar-viewer-astrosolar-silver-gold-(1pc-10pc-25pc-100pc).html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have used this filter in front of a 4 inch refractor with no adverse effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know where you are based but this product should be easily obtainable in the UK, USA and Germany.  I have no affiliation with the company.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-05-25T20:04:42.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21101" PostTypeId="1" CreationDate="2017-05-25T21:49:24.770" Score="3" ViewCount="466" Body="&lt;p&gt;I was reading &lt;a href=&quot;http://newatlas.com/brown-dwarf-is-a-planet/49445/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; article and it said something that didn't sit right with me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;SIMP0136 still gives off light mostly in the infrared wavelengths, as its temperature is now approximately 830° C (1,526° F),&quot; Gagné told us. &quot;It will take it approximately &lt;strong&gt;10 billion years more before it reaches 0° C&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SIMP0136 was classified as a brown dwarf star but when they just recently figured out its mass to be only 12.7 times that of Jupiter they downgraded it to a gas giant planet.  The threshold to qualify as a brown dwarf is 13 Jupiter masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like gas giant planets, brown dwarf stars do not gave enough mass for fusion to occur so the term star is a bit misleading.  Since there is no fusion, then it should eventually cool down.  The 10 billion year cool down time to go from its present 830 deg C to 0 deg C seems way too long.  How do they calculate this? &lt;/p&gt;&#xA;" OwnerUserId="16998" LastActivityDate="2017-05-28T19:38:16.677" Title="How do we calculate heat flow or cooling rate for no fusion brown dwarf star?" Tags="&lt;astrophysics&gt;&lt;star-formation&gt;&lt;gas-giants&gt;&lt;brown-dwarf&gt;&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21102" PostTypeId="2" ParentId="9917" CreationDate="2017-05-25T22:42:06.910" Score="1" Body="&lt;p&gt;10^19 words have been spoken in all of human history. This accounts for 109 billion people who have ever lived and each speaking 309 billion words in their life time. The amount of stars in the universe in a very rough estimate is somewhere around 10^24. For every word spoken in human existence there are over 100,000 stars in the universe. In conclusion, there are more stars in the universe than for every word ever spoken by humankind as well as every grain of sand on all the beaches on earth and many many more.&lt;/p&gt;&#xA;" OwnerUserId="16999" LastActivityDate="2017-05-25T22:42:06.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21104" PostTypeId="1" CreationDate="2017-05-26T03:49:02.240" Score="3" ViewCount="351" Body="&lt;p&gt;I've a data file with 2 columns, each containing time sampled voltage output (a sequence of real numbers) from a radio telescope, so there is a pair of telescope which are measuring the signal here. The source of the data is a pulsar and I've to find its period and dispersion measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm an amateur trying to understand what is really happening in the method. The data is time sampled at the Nyquist sampling frequency (2B) and the signal is measured at central frequency f and broadband B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I studied the cross correlation between the two voltages V1 and V2 from essential radio astronomy, but didn't quite understand the method. Do we have to evaluate the cross correlation function of each reading and then plot it with time to get the pulse profile? Or do we have to take fourier transform of each pair of voltage(how?) and then plot it to get the pulse profile? We find period of a pulsar from its average pulse profile, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also written that to find pulsar timing, we've to fold data from many pulses modulo instantaneous pulse period (how can we do that if we don't know the pulse period in the first place?). This sentence is everywhere and I'm not able to understand it each time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confused about the appropriate series of steps to be done here to get the pulsar period and dispersion measure. Could you give me some insight?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="16991" LastActivityDate="2018-01-17T15:56:54.683" Title="How to find period of a pulsar from a given time sampled voltage data?" Tags="&lt;radio-astronomy&gt;&lt;data-analysis&gt;&lt;observational-astronomy&gt;&lt;pulsar&gt;&lt;radio-telescope&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21105" PostTypeId="1" CreationDate="2017-05-26T08:17:26.860" Score="0" ViewCount="129" Body="&lt;p&gt;Looking at &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1350/are-nebulae-only-visible-from-a-distance/1354&quot;&gt;answers to this question&lt;/a&gt;, people seem pretty convinced that human naked eye should be able to see the colour of ionised gas/dust cloud when they are inside an emission nebula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we know &lt;em&gt;Solar system&lt;/em&gt; is surrounded by &lt;a href=&quot;https://en.wikipedia.org/wiki/Interplanetary_dust_cloud&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Interplanetary dust cloud&lt;/em&gt;&lt;/a&gt;, and possibly inside an &lt;a href=&quot;https://en.wikipedia.org/wiki/Local_Interstellar_Cloud&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Interstellar cloud&lt;/em&gt;&lt;/a&gt;. &lt;strong&gt;Why don't we see a colorised space backdrop?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it have to do with density of gas cloud? or how ionised they are? Does camera with long exposure see some background colour our naked eye can't?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Excuse me for any wrong terminology.)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="17002" LastActivityDate="2020-06-20T04:03:05.140" Title="Why isn't our local interstellar space colorised?" Tags="&lt;space&gt;&lt;nebula&gt;&lt;naked-eye&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21106" PostTypeId="2" ParentId="21070" CreationDate="2017-05-26T08:26:23.107" Score="0" Body="&lt;p&gt;Obviously, there are objects floating in space. Consider the space a fabric (Here, a Stretched Piece Of Cloth). If you consider the sun as 10Kg metal ball and put it into the center of the Fabric, you will see how deep the fabric is at the center and as you go far away its height increases. Now, consider the earth as a 50 gram ball and put it reasonably away from the metal ball. You will notice that the 'earth' moves towards the 'sun'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what this implies. The force of attraction is due to gravity. Gravity of a large object &lt;strong&gt;bends&lt;/strong&gt; the space (Fabric) around it. This is why nothing escapes a black hole (infinite mass gives it infinite Gravity). Now, gravity causes the &lt;strong&gt;centripetal force&lt;/strong&gt; and forces an object to move around it in nearly circular orbits. The direction of movement on any object in a circular path is given by the tangent at is point. Therefore, the earth is always moving away from the sun at the same time  getting pulled towards the sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies the moving in the orbit, the earth maintains a constant velocity. Since the space is a vacuum, there are no resistance offered and earth doesn't lose its energy. Obviously, if we keep 0.001 gram objects in the path of a 50 gram ball that's under the gravity of a 10kg ball, there will be no effect on the motion of the 'earth'. Hope you got it!&lt;/p&gt;&#xA;" OwnerUserId="17003" LastActivityDate="2017-05-26T08:26:23.107" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21107" PostTypeId="1" AcceptedAnswerId="21126" CreationDate="2017-05-26T08:36:15.357" Score="2" ViewCount="224" Body="&lt;p&gt;This should be a straightforward exercise, but I'm struggling. Help is much appreciated&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to parameterize a Navarro-Frenk-White profile with one parameter, using the correlation between total mass and the concentration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a starting point, I have the density in terms of two parameters: the total dark matter mass inside the virial radius $r_{vir}$, and the concentration parameter $c=r_{vir}/r_{s}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho_{DM}(r) = \frac{M_{DM}}{4\pi A(c)}\frac{1}{r(r_{s}+r)^{2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the function $A(c)$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A(c) = \ln{(1+c)} - \frac{c}{1+c} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the correlation between $M_{DM}$ and $c$ (from e.g. Napolitano et al. 2005)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c(M_{DM}) \approx 9.195 \left(\frac{M_{DM}}{h^{-1}10^{12} M_{\odot}} \right)^{-0.094}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it should then in principle be possible to parameterize $\rho_{DM}$ as a function of $M_{DM}$ only. This is how they do it in e.g. Williams et al. 2009.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm not managing to reproduce that. Sure, given the correlation above, $M_{DM}$ determines $A(c)$, but won't $\rho_{DM}$ still depend on $r_{s}$? &lt;/p&gt;&#xA;" OwnerUserId="6965" LastActivityDate="2017-05-27T13:24:41.320" Title="1-parameter NFW dark matter profile" Tags="&lt;dark-matter&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21108" PostTypeId="2" ParentId="21105" CreationDate="2017-05-26T11:56:48.047" Score="1" Body="&lt;p&gt;This is really more of a biology question, I think, so: The Mark-I Eyeball is far less sensitive to color than to intensity.  There are rods and cones in the retina. The neural signals from the rods are used to report intensity only. There are different spectral sensitivity curves for different cones (typically three types in humans), so the relative outputs from all cones are used to generate color information.  However, cones are less sensitive than rods, so for dim light such as that reaching us from nebulas, only the rods produce a neural output. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-05-26T11:56:48.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21109" PostTypeId="2" ParentId="21101" CreationDate="2017-05-26T12:08:12.913" Score="0" Body="&lt;p&gt;The surface temperature of a brown dwarf (or a star, for that matter) is considerably less than the core temperature.  As &lt;a href=&quot;http://www.scholarpedia.org/article/Brown_dwarfs&quot; rel=&quot;nofollow noreferrer&quot;&gt;scholarpedia&lt;/a&gt; writes, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Depending on the mass of the brown dwarf, core temperatures can reach&#xA;  10^4 to 6×10^6 K, while core densities range from 10 to 10^3 g/cm^3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's a lot more energy in the body than if the entire dwarf were at 1100 K. Even after downgrading to planet status, the core temperature is maybe a decade greater than the surface temperature.&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-05-26T12:08:12.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21110" PostTypeId="1" AcceptedAnswerId="21113" CreationDate="2017-05-26T14:02:31.880" Score="1" ViewCount="1618" Body="&lt;p&gt;I'm looking for some realistic data about the size (in &lt;em&gt;meters&lt;/em&gt;) and mass (in &lt;em&gt;kg&lt;/em&gt;) of a typical small asteroids that impact the Earth, without disintegrating in the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestion ?&lt;/p&gt;&#xA;" OwnerUserId="14736" LastActivityDate="2020-12-01T16:05:59.593" Title="Size and Mass of a typical &quot;small&quot; asteroid that impacts the Earth?" Tags="&lt;earth&gt;&lt;asteroids&gt;&lt;mass&gt;&lt;impact&gt;&lt;near-earth-object&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21111" PostTypeId="2" ParentId="21101" CreationDate="2017-05-26T14:13:53.297" Score="3" Body="&lt;p&gt;Brown dwarves usually exhibit deuterium burning in their core, which is where this omnious limit of 13 Jupiter masses comes from. It is the limit at which the core is dense and hot enough (as others have pointed out) enough to make deuterium nuclei fuse together. Deuterium has a relative abundance compared with Hydrogen of $10^{-4}$, but is easier to fuse than regular hydrogen. This is why brown dwarves can exist, after all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy flux they generate in the core is slowly diffused outwards until it reaches the shell from which it can escape. The virial theorem then tells us, that exactly half of their luminosity we see on the surface is retained in the dwarf. Thus, is we know an energy content $E$ and we know the energy loss rate $\dot E$ (which is the luminosity) astronomers usually go ahead to construct an estimate of the e-folding cooling &amp;amp; contraction time as $t_{\text{cool}} = E/\dot E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be used as a first estimate of the future fate of the brown dwarf. The estimate will now change based on whether the object is actually a brown dwarf that fuses deuterium, or not. This is easily understandable because above the deuterium burning limit the energy loss rate will be something like $\dot E = \dot E_{\text{contraction}} + \dot E_{\text{deuterium burning}}$, while below that limit the deuterium burning term is zero, and thus the estimate for $t_{\text{cool}}$ increases.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="9301" LastEditDate="2017-05-28T05:33:44.650" LastActivityDate="2017-05-28T05:33:44.650" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21113" PostTypeId="2" ParentId="21110" CreationDate="2017-05-26T15:07:56.563" Score="3" Body="&lt;p&gt;For the asteroid not to break up it needs to either be big, or tough. &#xA;You can experiment with the &lt;a href=&quot;http://impact.ese.ic.ac.uk/ImpactEffects/&quot; rel=&quot;nofollow noreferrer&quot;&gt;impact effects calculator&lt;/a&gt; but you should notice that a rocky asteroid with a diameter of less than about 1km will partially break up on impact with the atmosphere, and there may be multiple craters formed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the asteroid is less than about 100m in diameter than the breakup will be complete enough that the resulting meteorites will fall at terminal velocity, and won't form craters in rock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron meteorites are tougher, but even these are likely to break up if they are less than 500m in diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Needless to say, these are not small events, and we have never actually witnessed any object hitting the Earth that is large enough to make it through our atmosphere in one piece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The direct answer to the question is that there are no typical small impacts of bodies that don't break up. The Barringer &quot;meteor crator&quot; impactor was an iron asteroid, with a diameter of about 40 or 50m, that broke up on impact with the atmosphere, and fell in a circle of about 150m in diameter, forming the 1km diameter crater we see now. The mass of the impactor was about 1 to 2 million tonnes. It hit the atmosphere at about 15 to 17 km/s but by the time it reached Earth, it had slowed to less than 12 km/s. It exploded with the energy of about 10 Megatonnes of TNT. The dispersal of the fragments was not enough to from several craters. See &lt;a href=&quot;http://impact.ese.ic.ac.uk/cgi-bin/crater.cgi?dist=100&amp;amp;distanceUnits=1&amp;amp;diam=40&amp;amp;diameterUnits=1&amp;amp;pdens=&amp;amp;pdens_select=8000&amp;amp;vel=17&amp;amp;velocityUnits=1&amp;amp;theta=60&amp;amp;wdepth=&amp;amp;wdepthUnits=1&amp;amp;tdens=2500&quot; rel=&quot;nofollow noreferrer&quot;&gt;a barringer type impact&lt;/a&gt; Or read &lt;a href=&quot;http://www.space.com/834-mystery-arizona-meteor-crater-solved.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;about the impact&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-05-26T15:22:31.707" LastActivityDate="2017-05-26T15:22:31.707" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21115" PostTypeId="1" CreationDate="2017-05-26T18:05:58.850" Score="0" ViewCount="1406" Body="&lt;p&gt;My query is basically, say for example if the new moon is seen in the Middle east region, does it mean that it was the day of new moon in sub continental country as well ? You can't have a new moon in India for example the next day scientifically, right ? Please help me understand this with diagrams so I can explain it to others. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me it doesn't make sense that we will have two consecutive new moons in the same region of the planet. New moon in middle east but new moon in India is on the next day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware since it's a broad topic so timezone can change the answer so in this question. I am restricting my question to the sighting of the new moon in the same region as in say Moon is sighted in Dubai, doesn't it make sense that India will also have the first day of moon on the same day given that it all falls in the same plane.&lt;/p&gt;&#xA;" OwnerUserId="17007" LastEditorUserId="633" LastEditDate="2017-05-27T06:23:36.073" LastActivityDate="2017-05-27T07:00:00.410" Title="Does the new moon appear at the same time around the world or in one region last" Tags="&lt;the-moon&gt;&lt;moon-phases&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2017-06-04T10:44:24.440" ContentLicense="CC BY-SA 3.0" />
  <row Id="21116" PostTypeId="1" CreationDate="2017-05-26T19:18:23.553" Score="1" ViewCount="1621" Body="&lt;p&gt;Is there a formula to convert the B-V index to U-B index or are they totally different and separate observations and can't be convert like you can with Fahrenheit-&gt;Celsius?  (Updated)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know F-&gt;C is not astronomy related directly but there is a formula to convert between the two measurements.&lt;/p&gt;&#xA;" OwnerUserId="5401" LastEditorUserId="5401" LastEditDate="2017-05-26T20:09:43.077" LastActivityDate="2018-08-17T14:00:28.850" Title="B-V to U-B colour index" Tags="&lt;star&gt;&lt;spectral-type&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21117" PostTypeId="2" ParentId="21115" CreationDate="2017-05-26T19:27:51.490" Score="0" Body="&lt;p&gt;If the new moon is measured by moonrise then it's possible that it could land on a different day in different longitudes.   Or, if it's not measured by day, but by astronomical position, it's still possible that it would fall on a different day in different timezones.    180 degrees of longitude should equal roughly a 50% chance of landing on one day before or after (2 days apart isn't possible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;30 degrees of longitude should be roughly 1 chance in 12 of landing on a different day.    More detail is in the possible duplicate question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Map of &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/e/e8/Standard_World_Time_Zones.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;timezones&lt;/a&gt;.  While the lines aren't precisely straight, each line should correspond to about 15 degrees longitude.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-05-27T07:00:00.410" LastActivityDate="2017-05-27T07:00:00.410" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21118" PostTypeId="2" ParentId="21110" CreationDate="2017-05-26T19:57:01.317" Score="1" Body="&lt;p&gt;It is best summed up by this graph of meteorite diameter versus frequency of impact, divided by the probability of atmosphere traversal.&lt;/p&gt;&#xA;&lt;p&gt;Micrometeorites (particles normally less than 1 mm in&#xA;size) constitute the main part of the flux of extraterrestrial&#xA;matter accreting on Earth (1–3). (from arctic samples)&lt;/p&gt;&#xA;&lt;p&gt;Over the whole surface area of Earth, there's 18,000 to 84,000 meteorites bigger than 10 grams per year. (study  by P. A. Bland and was published in Monthly Notices of the Royal Astronomical Society.)&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QW7jS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QW7jS.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2020-12-01T16:05:59.593" LastActivityDate="2020-12-01T16:05:59.593" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21119" PostTypeId="1" CreationDate="2017-05-26T21:13:45.157" Score="2" ViewCount="65" Body="&lt;p&gt;From my research it doesn't appear to be from tidal locking with the sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there somewhere I can start reading about the probable causes? &lt;/p&gt;&#xA;" OwnerUserId="1582" LastActivityDate="2017-05-27T15:46:24.803" Title="Why does Venus spin so slowly?" Tags="&lt;venus&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2017-05-29T17:47:38.080" ContentLicense="CC BY-SA 3.0" />
  <row Id="21120" PostTypeId="1" CreationDate="2017-05-26T18:59:42.753" Score="0" ViewCount="1101" Body="&lt;p&gt;Have this telescope and tripod that is missing the 'tripod head platform' to mount to a binoculars or telescope. Can I buy it somewhere? [Photos attached]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to connect the tripod to the D70 telescope, and also this Celestron binoculars that comes with a tripod mount piece that has two screws in it, you could see in the picture on the right of the binoculars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: added additional photos&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photos from the Celestron manual, showing the tripod with all parts:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/NOoy4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NOoy4.jpg&quot; alt=&quot;telescope manual descriptions&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/3JgLK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3JgLK.jpg&quot; alt=&quot;manual2&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/oiOIg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oiOIg.jpg&quot; alt=&quot;manual3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My tripod, missing the tripod head:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/0MIcX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0MIcX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Telescope:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/SGkeZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SGkeZ.jpg&quot; alt=&quot;telescope&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ZKFDh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZKFDh.jpg&quot; alt=&quot;binoculars&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17024" OwnerDisplayName="larry909" LastEditorUserId="17024" LastEditDate="2017-05-29T05:11:18.383" LastActivityDate="2017-05-29T13:55:22.517" Title="How do you mount these binoculars and telescope on this tripod?" Tags="&lt;binoculars&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21121" PostTypeId="1" AcceptedAnswerId="21442" CreationDate="2017-05-27T01:33:34.240" Score="1" ViewCount="220" Body="&lt;p&gt;Sky &amp;amp; Telescope's &lt;a href=&quot;http://www.skyandtelescope.com/astronomy-news/askap-joins-the-hunt-for-mysterious-bursts/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ASKAP Joins the Hunt for Mysterious Bursts&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A new telescope, the Australian Square Kilometre Array Pathfinder (ASKAP), has joined the search for energetic and elusive fast radio bursts. And in just a few days of looking, it’s already had success!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later on, the article goes into some detail about ASKAP:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A new player is now on the scene, however, and it’s already had huge success. ASKAP is a wide-field radio telescope made up of an array of 12-meter antennas. Using phased-array-feed technology, ASKAP is able to instantaneously observe an effective area of 160 square degrees — an enormous field compared to Parkes’ 0.6 square degrees! This capability significantly increases our chances of being able to detect fast radio bursts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The article links to &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/2041-8213/aa71ff&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Detection of an Extremely Bright Fast Radio Burst in a Phased Array Feed Survey&lt;/a&gt; &lt;em&gt;which is not paywalled&lt;/em&gt; and contains some discussion of the array.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sounds really interesting! It is a phased array of dishes, but at the focal plane of each dish there is also a phased array of receivers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is different than just adding an array of feed horns and independent receivers as I've discussed in my surprisingly robustly unanswered question &lt;a href=&quot;https://astronomy.stackexchange.com/q/16237/7982&quot;&gt;What is the highest granularity focal-plane array on a dish radio telescope? Or is this the ONLY ONE?&lt;/a&gt; where receivers measure intensity as pixels, in a similar way to a camera (thus the counterintuitively placed &lt;code&gt;astrophotography&lt;/code&gt; tag there).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, each focal plane is equipped with a true phased array, where if I understand correctly, phase relationships between elements in the focal plane are maintained and recorded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Australian_Square_Kilometre_Array_Pathfinder&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; and CSIRO's &lt;a href=&quot;http://www.ska.gov.au/multimedia/Documents/Publications/Factsheets/ASKAP_FastFacts.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Australian Square Kilometre Array&#xA;Pathfinder – Fast Facts&lt;/a&gt; there are 36 separate 12-meter dishes in the entire array, and the focal plane of each dish is equipped with a 188 element focal-plane phased-array. 188 $\times$ 36 = 6768 channels, not counting polarizations because I don't know if they are separate channels in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That would require quite a hefty correlater if the analysis was done flat, without any heierarchy in the calculation. This is a pathfinder for the SKA so pushing limits is important, but I'd like to know how this large number of signals is handled. Is there pre-processing?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How do the phased array feeds on each dish of the ASKAP interact with the entire array phasing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure of the &quot;focal plane array&quot; is located precisely at the focal plane or not. If it were, then a given source in the sky would produce signals in only one, or a small group of receivers, &lt;em&gt;and not the whole focal plane array!&lt;/em&gt; One normally uses a phased array &lt;em&gt;instead of an imaging system&lt;/em&gt; so there must be something more interesting going on here, perhaps only nearest and next-nearest neighbor correlations within the focal plane array but full correlation between the dishes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this thing really work?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below x2:&lt;/strong&gt; Cropped from &lt;a href=&quot;https://en.wikipedia.org/wiki/File:CSIRO_ScienceImage_2161_Close_up_of_a_radio_astronomy_telescope_with_several_more_in_the_background.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;CSIRO ScienceImage 2161 Close up of a radio astronomy telescope with several more in the background&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Rqq4X.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rqq4X.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/03IcN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/03IcN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2020-12-02T01:47:33.320" LastActivityDate="2020-12-02T01:47:33.320" Title="How do ASKAP's focal plane phased array feeds interact with the entire array phasing?" Tags="&lt;observational-astronomy&gt;&lt;radio-astronomy&gt;&lt;radio-telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21123" PostTypeId="2" ParentId="21070" CreationDate="2017-05-27T06:56:20.503" Score="1" Body="&lt;p&gt;Actually, the planets do &lt;strong&gt;not&lt;/strong&gt; retain their energy or angular momentum. However, this is not (or hardly as explained by Rob) due to collisions with meteroids or any other dust, but due to their mutual gravitational interactions. These are far more important. The gravitational interaction energy of the Earth with Jupiter is about $10^{-4}$ times that with the Sun, $\sim10^{11}$ times more important than the 'dust'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational interactions are conservative: energy and angular momentum are not lost, only exchanged, but this does not mean that the Solar system is stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, tidal forces result in the dissipation of energy (but not the total angular momentum including spin), for example the Moon's orbit around Earth increases in radius by about 1inch/yr due the tidal interaction with Earth. For the dynamics of the Solar system these tides are the most important non-conservative forces.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2017-05-27T06:56:20.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21124" PostTypeId="1" AcceptedAnswerId="21135" CreationDate="2017-05-27T07:33:44.277" Score="2" ViewCount="1261" Body="&lt;p&gt;I know that the Hubble telescope was named after Edwin Hubble, due to his astronomical discoveries of astronomy, but who designed it? Which people were most directly responsible for the inspiring and planning the HST project.&lt;/p&gt;&#xA;" OwnerUserId="17010" LastEditorUserId="2365" LastEditDate="2017-05-28T19:26:43.383" LastActivityDate="2019-01-07T03:26:22.787" Title="who invented the Hubble telescope" Tags="&lt;hubble-telescope&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21126" PostTypeId="2" ParentId="21107" CreationDate="2017-05-27T11:12:05.150" Score="4" Body="&lt;p&gt;You are missing one last ingredient: the correlation between the virial radius $r_{\mathrm{vir}}$ and the virial mass $M_{\mathrm{vir}}$. The idea is that the orbital time scale of a particle orbiting in the halo becomes roughly equal to the age of the universe at $r_{\mathrm{vir}}$. This implies a relation between the (mean) density of the halo and of the universe:&#xA;$$&#xA;    \bar{\rho}_{\mathrm{vir}} \equiv \frac{3M_{\mathrm{vir}}}{4\pi r_{\mathrm{vir}}^3} = \Delta \rho_{\mathrm{crit}}(z)&#xA;$$&#xA;where $\bar{\rho}_{\mathrm{vir}}$ is the halo's mean density interior of $r_{\mathrm{vir}}$, $\rho_{\mathrm{crit}}$ is the critical density of the universe (as function of redshift $z$), and $\Delta$ a parameter that depends on the cosmology. Often $\Delta=200$ is used and the corresponding quantities labeled $r_{200}$ and $M_{200}$ instead. See also &lt;a href=&quot;https://en.wikipedia.org/wiki/Virial_mass&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Wikipedia page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the way to construct a halo is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Pick $M_{\mathrm{vir}}$ (or $M_{200}$) and redshift $z$ (and your favorite cosmology)&lt;/li&gt;&#xA;&lt;li&gt;Use above relation to find $r_{\mathrm{vir}}$ (or $r_{200}$)&lt;/li&gt;&#xA;&lt;li&gt;Use the relation from Napolitano et al. (2005), or equivalent, to find $c$ and hence $r_s$&lt;/li&gt;&#xA;&lt;li&gt;Obtain $\rho_{\mathrm{DM}}$ (if you still need it)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2017-05-27T13:24:41.320" LastActivityDate="2017-05-27T13:24:41.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21127" PostTypeId="1" CreationDate="2017-05-27T13:39:51.610" Score="4" ViewCount="426" Body="&lt;p&gt;I was out taking some milky way pictures last night, and saw a much quicker shooting star type thing, only with a much more random movement - any ideas what this could be - this is the RAW file from my camera - converted to JPEG for upload&#xA;&lt;a href=&quot;https://i.stack.imgur.com/LgZLb.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LgZLb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17011" LastEditorUserId="17011" LastEditDate="2017-05-27T13:45:22.823" LastActivityDate="2017-06-08T11:46:56.570" Title="What have I seen in the night sky" Tags="&lt;observational-astronomy&gt;&lt;identify-this-object&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21128" PostTypeId="2" ParentId="21127" CreationDate="2017-05-27T15:09:11.510" Score="9" Body="&lt;p&gt;I took the liberty of looking a bit closer at your image and annotate several features.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PMaXy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PMaXy.jpg&quot; alt=&quot;Image from OP, just with annotated features&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look closely, you see that several of the brighter objects in your image do the exact same random walk. Those are e.e. the objects A, B and C. This already strongly hints to some shaking of the camera, rather than A being a UFO.&#xA;If you look closer at your own image (which you always should), you see those lines all over the image from other weaker stars as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This automatically begs the question: How long was your exposure? Or was it composed from several single frames? Because it looks like one of those options  for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To substantiate this, let's look at some other features: Feature D doesn't show any spots, thus must have been taken while a shake of the camera was happening, thus must have moved quickly, thus this must be a shooting star. Of course a shooting star would appear like this as well on a non-shaking camera.&lt;br&gt;&#xA;For contrast, feature E is also linear, but shows spots, which means while this line was putting light into the camera it must have stopped shaking on regular time intervals. This is a bit weird. But because the angle of the line, just as D, doesn't match any of the streaks that we follow in the figure A, this must also be a self-moving object. Now a self moving object of much lower speed than a shooting star (D), can be a satellite or an airplane.&lt;br&gt;&#xA;Then there's feature F which clearly is also slow-moving, but does a weird rectangle that I cannot match to the 'master-shake' of feature A. It could mean that it's also self-moving taken together with one of the corners of the master shake. But then I don't understand why F is doing the cornery thing, but E does not. Another possibility in my opinion is, that E and F are both satellites, which faded in and out of brightness at different moments when the picture was taken, thus a different 'shaking' was imprinted on them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yeah, &lt;strong&gt;summarizing&lt;/strong&gt;&lt;br&gt;&#xA;Your image does some weird things, most of them clearly caused by camera-shaking. I don't understand all the details in your picture for that matter, but that's far from a reliable UFO claim. In particular because essentially EVERY UFO claim is based on shaky/nonsharp images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.&lt;br&gt;&#xA;Also note how the grass in the lower left is shifted from really dark, to slightly dark. There are two central features in the grass that can be brought to roughly, but not exactly match when moving them over each other in photoshop. This indicates that this is the same patch of grass, just being photographed at two different times, after the grass was moving around a bit.&lt;br&gt;&#xA;I suggest a more stable mount for your camera, if you're sure you didn't bounce into it by yourself. &lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditDate="2017-06-08T11:46:56.570" LastActivityDate="2017-06-08T11:46:56.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21130" PostTypeId="1" CreationDate="2017-05-27T16:48:39.700" Score="4" ViewCount="170" Body="&lt;p&gt;I'm currently studying the star system of &lt;a href=&quot;https://en.wikipedia.org/wiki/Gliese_570&quot; rel=&quot;noreferrer&quot;&gt;Gliese 570 &lt;/a&gt; and I´m trying to calculate the habitable zone of the component A from the luminosity value in relation of the Sun´s. I´ve read that the needed value of luminosity is the value of the &lt;strong&gt;bolometric luminosity&lt;/strong&gt; (here 26%), but the calculations of the HZ that I found is about 0.40 AU, which seems to be calculated from the &lt;strong&gt;visual luminosity&lt;/strong&gt; (about 16%). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the questions are: &lt;strong&gt;Which of these 2 values do I have to use to calculate the habitable zone? Does the stellar class of the star affect the selection of it?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="17013" LastActivityDate="2017-06-01T16:56:42.700" Title="Luminosity value to use for the calculation of the habitable zone?" Tags="&lt;star&gt;&lt;exoplanet&gt;&lt;habitable-zone&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21131" PostTypeId="2" ParentId="21120" CreationDate="2017-05-27T21:21:47.993" Score="1" Body="&lt;p&gt;What you want to attach the telescope to a tripod is something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.firstlightoptics.com/dovetail-bars/baader-vixen-style-dovetail-clamp.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.firstlightoptics.com/dovetail-bars/baader-vixen-style-dovetail-clamp.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is a clamp that holds the standard vixen dovetail used by most small telescopes.  The important thing with the one linked to is that it has a threaded hole to allow you to attach it to a standard 1/4&quot; photo tripod screw. (Not all do - some of them are designed to bolt to a mount, so only have countersunk holes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the tripod adapter that came with the binoculars should have a threaded hole on the bottom part to attach it to the tripod. The captive screw in the adapter attaches to a threaded socket on the front of the&#xA;binocular hinge - there's likely to be a circular cover that unscrews or clips off to reveal the socket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, judging from the picture of the tripod, it looks as though you may be missing the camera platform bit. Normal photo tripod heads will either have a platform with a captive 1/4&quot; whitworth tripod thread thumbscrew, or a platform with a quick release holder for a removable plate which has the screw (the quick release versions make it easier to quickly attach or remove the camera (with the QR plate attached) from the tripod.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the picture, it looks to me as though that platform part is missing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the tripod head was designed to be removable, it will usually be screwed onto a 3/8&quot; bolt attached to the leg section. There may also be some extra&#xA;smaller screws on the underside of the platform the head is attached to to&#xA;prevent it from unscrewing by accident. If it is removable, you may be able to replace the entire tripod head unit (This is fairly common with more expensive photo tripods, where you can buy the legs and heads separately).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you might find it cheaper to look for a secondhand tripod.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For supporting a telescope, you usually want a fairly solid and stable tripod. That can often mean heavy, large and expensive. Most of the smalller /  cheaper tripods are designed for lightweight cameras and can be shaky with&#xA;a heavier load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with using a photo tripod for normal binoculars is that when you're observing near the zenith (looking straight up) the tripod gets in the&#xA;way. You can also get things called parallelogram mounts that put the binoculars out on a pivoting arm with a counterweight on the other side - since the binoculars aren't over the tripod you can point them straight up without the tripod getting in the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit - now that you've added the manual pictures, it looks as though you may just be able to attach the scope directly to a standard tripod, without needing a dovetail clamp, assuming at least one of the two holes in the bottom of the telescope dovetail is threaded. Some tripod heads have a rectractible or spring loaded pin, as well as the main screw, mainly designed for use with movie/video cameras. If the spacing on that matches up with the other hole in the scope dovetail, then the pin helps prevent the scope accidentally unscrewing (since it stops the scope turning round on the platform).]&lt;/p&gt;&#xA;" OwnerUserId="17016" LastEditorUserId="17016" LastEditDate="2017-05-29T13:55:22.517" LastActivityDate="2017-05-29T13:55:22.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21133" PostTypeId="2" ParentId="12901" CreationDate="2017-05-27T22:14:40.500" Score="0" Body="&lt;p&gt;Please see &lt;a href=&quot;http://www.pbs.org/wgbh/nova/blogs/physics/2012/01/the-chandrasekhar-limit-the-threshold-that-makes-life-possible/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; that describes the Chandrasekhar limit. Any star remnant more massive than the Chandrasekhar limit, 1.4x sun mass, will become neutron star or black hole. Otherwise it will become white dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This mass limit is the threshold at which gravitational will overcome &lt;a href=&quot;https://en.wikipedia.org/wiki/Electron_degeneracy_pressure&quot; rel=&quot;nofollow noreferrer&quot;&gt;electron degeneration pressure&lt;/a&gt;, causing the electrons to join the nucleus. The protons in the nucleus combine with the electrons to become neutrons. Thus the term neutron star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the mass of the neutron star surpasses the level at which the &lt;a href=&quot;https://en.wikipedia.org/wiki/Escape_velocity&quot; rel=&quot;nofollow noreferrer&quot;&gt;escape velocity&lt;/a&gt; for the neutron star exceeds the speed of light, then even light cannot escape its gravitation, thus becoming a black hole.&lt;/p&gt;&#xA;" OwnerUserId="16998" LastEditorUserId="16998" LastEditDate="2017-05-28T16:56:49.217" LastActivityDate="2017-05-28T16:56:49.217" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21134" PostTypeId="2" ParentId="21101" CreationDate="2017-05-28T17:09:18.400" Score="5" Body="&lt;p&gt;Whether or not the object is just above or below the threshold (about 13 Jupiter masses) at which the core will become hot enough to burn deuterium is not really relevant to the calculation you wish to understand. The object has cooled beyond the point at which D burning has ceased. It has also pretty much contracted to a minimum sized configuration where electron degeneracy pressure supports it. Therefore very little gravitational potential energy can contribute to subsequent thermal evolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that is left is to calculate the thermal energy content (roughly $3kT/2 \times N_i$ - the number of hydrogen and helium ions, where $T$ is the &lt;em&gt;internal&lt;/em&gt; temperature; the degenerate electrons contribute very little to the heat capacity) and then divide this by the luminosity radiated from the surface. The calculation is a little more complex than that because the luminosity fades as the brown dwarf cools.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for kicks - here is a back of the envelope calculation. The central temperature of a contracting ball of gas is given by the virial theorem. The brown dwarf you are talking about will contract until D burning (nearly) starts at about $10^{6}$ K - let's assume that is an upper limit to the current core temperature. The average temperature in the interior will be a bit lower than that (let's divide by 2), so $\langle T \rangle \sim 5\times 10^{5}$ K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the mass is about $0.012 M_{\odot}$ and it is made almost totally of hydrogen, then the number of protons is about $N_i \sim 1.4 \times 10^{55}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the total thermal energy is $E \sim 3N_i k\langle T \rangle/2 \sim 1.5\times 10^{38}$ J.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;current&lt;/em&gt; luminosity $L \sim 4 \pi \sigma R^2 T_s^{4}$, where $T_s \simeq 1100$ K is the &lt;em&gt;surface&lt;/em&gt; temperature and $R$ is the radius, which will be about the same as Jupiter, i.e. $R\sim 70,000$ km. Thus $L \sim 5 \times 10^{21}$ W. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current cooling timescale is $E/L \sim 10^{9}$ years - but this is a &lt;em&gt;lower limit&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not surprised that this is somewhat lower than your estimate because the cooling time gets much longer as the object cools. This is clearly seen when looking at a proper model calculation; for example see the picture below taken from &lt;a href=&quot;https://arxiv.org/abs/astro-ph/9705201&quot; rel=&quot;nofollow noreferrer&quot;&gt;Burrows et al. (1997)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lines show the cooling tracks of surface temperature versus (log) time. The red curves are &quot;planets&quot;, the green are &quot;brown dwarfs&quot; - the dividing line is the D-burning threshold - seen as a plateau in the the green brown dwarf curves at early times. Your object is around this threshold. I have marked two points on the plot - the first corresponding to a 13 Jupiter mass object with a surface temperature of 1100K (you can see it must already be at least a hundred million years old) , the second marks the same object when it has cooled to about 273 K (zero Celsius). It does indeed occur after about 10 billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eRXBi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eRXBi.jpg&quot; alt=&quot;Burrows cooling curves&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-05-28T19:38:16.677" LastActivityDate="2017-05-28T19:38:16.677" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21135" PostTypeId="2" ParentId="21124" CreationDate="2017-05-28T19:23:00.917" Score="6" Body="&lt;p&gt;There was no single designer of the Hubble telescope. It was designed and build by NASA, with contributions from the European Space Agency (ESA).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantages of a telescope in space have been clear for over 100 years. These ideas were crystallized in &lt;strong&gt;Lyman Spitzer's&lt;/strong&gt; 1946 paper. Spitzer continued to push for a telescope through the '50s and '60s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Congress eventually funded the telescope in '78, the task of design was given to the &lt;strong&gt;Marshal Space Flight Center&lt;/strong&gt;. The head of the space telescope project was &lt;strong&gt;Dr. Bob O'Dell&lt;/strong&gt;, chair of the Astronomy dept. at the University of Chicago. In a video from NASA, &lt;a href=&quot;http://spacetelescope.org/videos/hst3_odell/&quot; rel=&quot;noreferrer&quot;&gt;Dr O'Dell talks about the early years of HST development&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over the years others have worked as Project Manager. There never was a single &quot;inventor&quot;. Hundreds of scientists have been involved in its design and planning. Something as complex as the HST could never be designed by a solo inventor.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-05-28T19:23:00.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21136" PostTypeId="1" CreationDate="2017-05-29T00:21:35.910" Score="1" ViewCount="314" Body="&lt;p&gt;I am just wondering: is there a potential galactic centre out there that's very close to us? I know that CMDG is irregular, but can it still contain a black hole at the centre?&lt;/p&gt;&#xA;" OwnerUserId="16849" LastEditorUserId="9527" LastEditDate="2020-05-22T07:58:02.907" LastActivityDate="2020-05-22T07:58:02.907" Title="Does the Canis Major Dwarf Galaxy have a black hole at its centre?" Tags="&lt;black-hole&gt;&lt;observational-astronomy&gt;&lt;galactic-center&gt;&lt;dwarf-galaxy&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21138" PostTypeId="1" AcceptedAnswerId="21150" CreationDate="2017-05-29T05:01:05.043" Score="4" ViewCount="1258" Body="&lt;p&gt;Suppose that I have an array of stellar masses of tens of thousands of galaxies, and that I also already know what the total &quot;survey volume&quot; is (in Mpc^3) within which all of those galaxies are contained. For context, my data is from a simulation of galaxy formation at redshift ~ 0, rather than from an observational simple -- so in principle I don't think I need to worry about incompleteness corrections at the faint (low-mass) end. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given my array of stellar masses and the total survey volume in Mpc^3, how do I create and plot the &quot;galaxy stellar mass function&quot; that I always see people talking about? Isn't it just basically a histogram of the stellar mass array, and then divide the # of galaxies in every bin by the survey volume (it's almost just like a normalization constant)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course this question could equally apply to galaxy luminosity functions.) &lt;/p&gt;&#xA;" OwnerUserId="607" LastActivityDate="2019-03-26T19:24:30.640" Title="How do I create a galaxy stellar mass function?" Tags="&lt;galaxy&gt;&lt;cosmology&gt;&lt;mass&gt;&lt;python&gt;&lt;sky-survey&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21139" PostTypeId="1" AcceptedAnswerId="21141" CreationDate="2017-05-29T07:24:26.427" Score="3" ViewCount="3512" Body="&lt;p&gt;I'm trying to see how far can our star reaches with its gravity. I'm asking if anyone could give info as to what's our star's limit or the furthest object found in our solar system.&lt;/p&gt;&#xA;" OwnerUserId="16990" LastEditorUserId="2365" LastEditDate="2017-05-30T14:17:41.550" LastActivityDate="2020-12-03T00:14:36.830" Title="How far could be an object from the Sun and still be under the influence of its gravitational field?" Tags="&lt;the-sun&gt;&lt;solar-system&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21140" PostTypeId="1" AcceptedAnswerId="21189" CreationDate="2017-05-29T07:54:25.057" Score="-3" ViewCount="589" Body="&lt;p&gt;What is the powers the telescope is used at? &lt;/p&gt;&#xA;" OwnerUserId="17024" LastEditorUserId="17024" LastEditDate="2017-06-02T18:39:10.343" LastActivityDate="2017-06-02T18:39:10.343" Title="What is the magnification of the Keck telescopes in Hawaii?" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2017-06-04T10:42:10.183" ContentLicense="CC BY-SA 3.0" />
  <row Id="21141" PostTypeId="2" ParentId="21139" CreationDate="2017-05-29T08:03:57.840" Score="11" Body="&lt;p&gt;The Sun's gravity extends infinitely, but eventually solar objects would be unstable due to the influence of other stars. The minor planet &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/90377_Sedna&quot; rel=&quot;noreferrer&quot;&gt;Sedna&lt;/a&gt;&quot; has an orbit which takes it nearly 1000 AU (0.016 light years) from the sun at its furthest point (but now it is a lot closer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also thought that billions of comets must orbit in the outer part of the solar system, out to 50000AU, or 0.8 light years, (or possibly further) forming the &lt;a href=&quot;https://en.wikipedia.org/wiki/Oort_cloud&quot; rel=&quot;noreferrer&quot;&gt;Oort Cloud&lt;/a&gt;. However, at such distances, they could not be directly observed. This marks the greatest distance at which orbiting solar system bodies can be found.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-05-30T16:38:40.130" LastActivityDate="2017-05-30T16:38:40.130" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21142" PostTypeId="1" AcceptedAnswerId="21144" CreationDate="2017-05-29T09:09:23.110" Score="0" ViewCount="426" Body="&lt;p&gt;What is the maximum planet mass for human habitation? How many Gs can humans withstand on a constant basis?&lt;/p&gt;&#xA;" OwnerUserId="17026" LastActivityDate="2017-05-29T14:14:35.323" Title="What is the maximum planet mass for human habitation?" Tags="&lt;exoplanet&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2017-05-29T17:47:19.770" ContentLicense="CC BY-SA 3.0" />
  <row Id="21143" PostTypeId="2" ParentId="21139" CreationDate="2017-05-29T10:30:05.313" Score="7" Body="&lt;p&gt;There's no straight forward answer.   In the solar-system, which is well ordered, objects that are in stable orbits, and not too elliptical, have well defined &lt;a href=&quot;https://en.wikipedia.org/wiki/Sphere_of_influence_(astrodynamics)#Table_of_selected_SOI_radii&quot; rel=&quot;noreferrer&quot;&gt;spheres of influence&lt;/a&gt;.   Planet 9, if/when it's discovered, will probably have the largest sphere of influence for known solar-system objects.   Currently, Neptune has the largest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the stars near the sun were static relative to each other, the Sun's sphere of influence could be calculated and it would probably extend between 2 and 3 light years.   But because the stars are not static, the sphere of influence is constantly changing and stars (probably) exchange outer, loosely orbiting debris fairly frequently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.space.com/16401-oort-cloud-the-outer-solar-system-s-icy-shell.html&quot; rel=&quot;noreferrer&quot;&gt;The Oort cloud&lt;/a&gt; by this article is thought to extend to almost 2 light years, so that's one possible answer to your question.  If you want to know the most distant &lt;a href=&quot;https://en.wikipedia.org/wiki/Perihelion_and_aphelion&quot; rel=&quot;noreferrer&quot;&gt;aphelion&lt;/a&gt; of an object currently orbiting the sun, James Ks answer is good, but I think the outer most aphelion is a bit further than the 0.8 light years that he suggests.   At least 2 light-years, possibly even 3.  The problem is, an orbit that distant, such an object has a good chance of being deflected before it reaches it's perihelion, a journey that takes over 10 million years.  Orbits that distant are likely not very stable.  A lot depends on how close other stars get to our sun.  A star that passes too close would likely throw everything in the vicinity that it passes through out of wack.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://1.bp.blogspot.com/-vf2qvNTzHB8/UWbIkBQ71PI/AAAAAAAAFlo/MKyPTbJGY48/s1600/Post+-+April+2013+(4)+-+1.jpg&quot; rel=&quot;noreferrer&quot;&gt;chart&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_nearest_stars_and_brown_dwarfs#Future_and_past&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tiny &lt;a href=&quot;https://en.wikipedia.org/wiki/Scholz%27s_star&quot; rel=&quot;noreferrer&quot;&gt;Scholz's star&lt;/a&gt; is thought to have passed within 1 light year of our sun about 70,000 years ago.   Stars passing that close are quite rare, but, from the link above&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A star is expected to pass through the Oort Cloud every 100,000 years&#xA;  or so. An approach as close or closer than 52,000 AU is expected to&#xA;  occur about every 9 million years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This does make defining an outermost orbit somewhat difficult, as the most distant orbits take millions of years to reach their closest point from their most distant point, and they run a pretty good chance of being perturbed within a single orbit.   Stars likely play Frisbee with their outer-most orbiting objects all the time.   Picking an outermost stable orbit is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A curious sidebar on Scholz, is that, it may have sent a bunch of outer comets and oort cloud objects heading towards the inner solar system.   We wont find out how many for another 2 million years or so.   That's how long it will take any objects that were sent towards the inner solar-system to reach it.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-05-29T10:30:05.313" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21144" PostTypeId="2" ParentId="21142" CreationDate="2017-05-29T14:14:35.323" Score="3" Body="&lt;p&gt;When astronomers discuss the &quot;habitable zone&quot; around a star they're really concerned with someone likely to provide liquid water and the right amount of radiation for basic life (cells) as we understand them.  But this doesn't mean they're suitable for humans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maximum mass is not an issue, it's gravity that matters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need a few things from gravity to be human habitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly we're &quot;designed&quot; (evolved) to operate in a one-gee environment.  As far as I know there have been no studies on long term high gee effects on humans (low gee, yes, but not high gee).  However, considering the &quot;plumbing&quot; and structure involved in a human, it seems unlikely to me we'd adapt long term to more than a small percentage larger than one-gee.  Much more on a sustained basis would seem likely to increase physical stress on the bones, muscles, tendons and so on and lead to more rapid development of related illnesses and disabilities.  The plumbing (fluids are being moved all over your body) can cope with quite high gees for short periods, but it's simply not designed to cope with sustained high gee.  The strain on the circulatory system strikes me as the greatest vulnerability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Movement in high gee is also an issue.  I'd be surprised if people could adapt to long term requirements without serious health side effects.  We'd also be more vulnerable to injury from falls and slips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider how long you could walk in 1.5g.  It would be the equivalent of walking while carrying something weighing half as much as you do yourself. If you were 100 kilos, how long could you walk while carrying 50 kilos ?   If you had to do that all day, everyday, how long before the strain caused serious injuries and other health problems ?  Not long, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So just on this basis I'd suggest a maximum sustained gee of about 1.1 is ballpark.  Walking around with ten kilos when you are 100 kilos sounds like a lot all day everyday, but it's probably doable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But gravity does more than provide a way to nail us to the ground.  It provides a way to nail the atmosphere to ground at an atmospheric pressure high enough for us to use, and not too high for us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it would appear from human habitations in the deep sea, that we can sustain quite high atmospheric pressures, so that would not seem likely to be a problem.  What tends to hurt humans is shifting quickly from high to low pressures.  As I've argued that we can't sustain high gees for long and need something close to one gee for health reasons, the atmosphere doesn't seem to be a problem in that case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what does 1.1g mean in terms of planet size and density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational pull is proportional to planet average density and radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we need a planet that's around the 1.1g mark meaning :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1.1 \rho_{earth} r_{earth} \approx \rho_{new-earth} r_{new-earth}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\rho$ is density and $r$ is radius and this represents the limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this may make it easier to find suitable planets.  We aren't restricted to specific mass ranges, but can look for planets that are less dense and larger or more dense and smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this does make me wonder if the notion of human colonization of planets ever made any sense.  Building our own large habitats in space is starting to make a lot more sense to me as it restricts us less.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-05-29T14:14:35.323" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21145" PostTypeId="1" CreationDate="2017-05-29T18:57:39.227" Score="6" ViewCount="108" Body="&lt;p&gt;Are there any cases in which the gravitational lensing of a quasar has been observed in both visible light and in radio waves, and if so, is the radio-wave resolution good enough to show that the lensing at these two different wavelengths is the same? &lt;/p&gt;&#xA;" OwnerUserId="16667" LastActivityDate="2017-05-30T14:14:38.393" Title="Gravitational lensing of quasars" Tags="&lt;gravitational-lensing&gt;&lt;quasars&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21146" PostTypeId="1" AcceptedAnswerId="21147" CreationDate="2017-05-29T20:46:00.733" Score="1" ViewCount="277" Body="&lt;p&gt;I am just curious about photos taken by space telescopes &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of type of camera systems are used in space telescopes like HUBBLE and JWST? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are those cameras systems specially built for that purpose? &#xA;Is there a significant difference between cameras we use and of those? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What type of lenses are used? &lt;/p&gt;&#xA;" OwnerUserId="11951" LastEditorUserId="2365" LastEditDate="2017-05-30T05:44:33.137" LastActivityDate="2017-05-30T06:06:47.627" Title="What type of camera instruments are used to record optical images in space telescopes like HUBBLE &amp; JWST.?" Tags="&lt;space-telescope&gt;&lt;photography&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21147" PostTypeId="2" ParentId="21146" CreationDate="2017-05-29T22:14:32.643" Score="1" Body="&lt;p&gt;The HST and JWST are both mirror telescopes. There are many instruments on each, but the best known is the wide field and planetary camera (wfpc1), and its successors (wfpc2 and wfc3). These are all CCD cameras, and as such are based on the same basic technology that is found in many digital cameras, but there the similarity ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The camera is placed at the primary focus of the telescope. The telescope, therefore, acts as the main &quot;lens&quot;. There are secondary optics, particularly in wfpc2 and 3 there are optics to correct the miscurvature of the main mirror. These optics are relay mirrors, not lenses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each camera is a one-off. Designed and built in-house by NASA. Compared to modern cameras, the resolution of the CCD is not so high. The original WFPC had an 800x800 pixel CCD, the WFPC2 had 4 800x800 pixel CCD sensors (at different pixel densities) If these seem low, remember that Hubble was designed and built in the '80s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cameras have to deal with a lot of things that the cameras we use don't. There are high-energy cosmic rays that can damage CCDs, and servicing required a shuttle launch. The CCDs are sensitive in a wide range of frequencies, from infrared to ultraviolet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latest camera, installed in 2009, has two light paths as shown, separating the visible from the IR imaging. As noted by Sean Lake in a comment, CCDs are read out by moving charge across the detector, infrared arrays read out individual pixels. IR arrays have more problems with stuck pixels or variable sensitivity across the array.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/v2MyM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/v2MyM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Schematic from the &lt;a href=&quot;http://www.stsci.edu/hst/wfc3/design/at_a_glance/&quot; rel=&quot;nofollow noreferrer&quot;&gt;space telescope science institute&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the cameras on Hubble, and the cameras that will be placed on JWST are distant relations of commercial digital cameras. They have no lenses, can see in a wider range of frequencies. The camera and its housing are also about 2m wide. However, the basic technology of converting an image into a digital file is shared between commercial digital cameras and the HST&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-05-30T06:06:47.627" LastActivityDate="2017-05-30T06:06:47.627" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21148" PostTypeId="2" ParentId="21145" CreationDate="2017-05-29T22:19:39.183" Score="4" Body="&lt;p&gt;Yes, there are many. One of the first, if not the first was the radio lensed quasar B1938+666, in which a near-IR Einstein ring was reported by &lt;a href=&quot;https://arxiv.org/abs/astro-ph/9710171&quot; rel=&quot;nofollow noreferrer&quot;&gt;King et al. (1998)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to your real question though is not as straightforward as that. Although lensing should be wavelength independent (according to GR), it is not independent of the size and position of the source. The radio waves and optical emission arise from different source sizes and locations. In addition, if you are interested in testing whether the surface brightness magnification is the same, well that relies on knowing the intrinsic source properties; and the radio to optical brightness of quasars varies wildly.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-05-30T14:14:38.393" LastActivityDate="2017-05-30T14:14:38.393" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21150" PostTypeId="2" ParentId="21138" CreationDate="2017-05-30T08:47:01.983" Score="8" Body="&lt;p&gt;Your approach is completely correct, just note three things:&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Logarithmic distribution&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;First, since the distribution of masses is logarithmic in nature (as is most other things), be sure to bin them logarithmically. Otherwise you will oversample (undersample) the bins at the low-(high-)mass end.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Comoving densities&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Second, to be able to compare mass functions at different redshifts, ones uses the &lt;em&gt;comoving&lt;/em&gt; volume rather than the physical volume, such that the expansion of the Universe is factored out. The two are related as &lt;span class=&quot;math-container&quot;&gt;$V_\mathrm{com} = V_\mathrm{phys}(1+z)^3$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Damn you, little &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt;!&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Finally, observers and modellers tend to use a slightly different definition of the unit volume. Whereas observers usually use &lt;span class=&quot;math-container&quot;&gt;$\mathrm{Mpc}$&lt;/span&gt; for distances, and hence &lt;span class=&quot;math-container&quot;&gt;$\mathrm{Mpc}^{-3}$&lt;/span&gt; for number densities, if your galaxies come from a cosmological simulation where the cosmological parameters can when tuned at will, it is custom to factor out the Hubble constant &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt;. In simulations, masses and distances are then measured in &lt;span class=&quot;math-container&quot;&gt;$h^{-1}M_\odot$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$h^{-1}\mathrm{Mpc}$&lt;/span&gt;, respectively, so number densities are measured in &lt;span class=&quot;math-container&quot;&gt;$h^3  \mathrm{Mpc}^{-3}$&lt;/span&gt;. Here &lt;span class=&quot;math-container&quot;&gt;$h\equiv H_0\,/\,(100\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1})$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is probably a reminiscence of the time when the Hubble constant was rather uncertain. Nowadays, in my opinion there is no need to do this, but since everybody does, it's difficult to go against the stream. For a discussion of this issue, see &lt;a href=&quot;https://arxiv.org/abs/1308.4150&quot; rel=&quot;nofollow noreferrer&quot;&gt;Croton (2013)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Python code&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Since you've tagged the question with &lt;a href=&quot;/questions/tagged/python&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;python&amp;#39;&quot; rel=&quot;tag&quot;&gt;python&lt;/a&gt;, I wrote this little snippet that should do the work (I randomly chose &lt;span class=&quot;math-container&quot;&gt;$10^5\,\mathrm{Mpc}$&lt;/span&gt; as your survey volume; note also that in this example I don't factor out &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt;):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;&#xA;M     = np.loadtxt('Mstar.dat')         #Read stellar masses in Msun&#xA;logM  = np.log10(M)                     #Take logarithm&#xA;nbins = 10                              #Number of bins to divide data into&#xA;V     = 1e5                             #Survey volume in Mpc3&#xA;Phi,edg = np.histogram(logM,bins=nbins) #Unnormalized histogram and bin edges&#xA;dM    = edg[1] - edg[0]                 #Bin size&#xA;Max   = edg[0:-1] + dM/2.               #Mass axis&#xA;Phi   = Phi / V / dM                    #Normalize to volume and bin size&#xA;&#xA;plt.clf()&#xA;plt.yscale('log')&#xA;plt.xlabel(r'&lt;span class=&quot;math-container&quot;&gt;$\log(M_\star\,/\,M_\odot)$&lt;/span&gt;')&#xA;plt.ylabel(r'&lt;span class=&quot;math-container&quot;&gt;$\Phi\,/\,\mathrm{dex}^{-1}\,\mathrm{Mpc}^{-3}$&lt;/span&gt;')&#xA;plt.plot(Max,Phi,ls='steps-post')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2019-03-26T19:24:30.640" LastActivityDate="2019-03-26T19:24:30.640" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21152" PostTypeId="1" AcceptedAnswerId="21153" CreationDate="2017-05-30T10:12:13.487" Score="2" ViewCount="269" Body="&lt;p&gt;Dark matter is not distributed evenly. It forms clouds, it concentrates in galaxies, it forms &lt;a href=&quot;https://www.nasa.gov/feature/jpl/earth-might-have-hairy-dark-matter&quot; rel=&quot;nofollow noreferrer&quot;&gt;hairs&lt;/a&gt; around celestial bodies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A google search for dark energy distribution floods me with dark matter distribution results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is dark energy distributed? Does it cover the space uniformly, at even density, or does it create areas of concentration or other irregular forms?&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2017-05-30T10:16:27.943" Title="What is the distribution of dark energy?" Tags="&lt;dark-energy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21153" PostTypeId="2" ParentId="21152" CreationDate="2017-05-30T10:16:27.943" Score="1" Body="&lt;p&gt;The actual model says that dark energy is distributed uniformly around all the universe and as you probably know it expands the universe with him so it stays distributing uniformly, there are no clouds or irregular forms because that wouldn't explain the actual universe model. Maybe in the future that change.&lt;/p&gt;&#xA;" OwnerUserId="16990" LastActivityDate="2017-05-30T10:16:27.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21157" PostTypeId="1" CreationDate="2017-05-31T05:51:57.817" Score="5" ViewCount="1588" Body="&lt;p&gt;A simple, but interesting question I think.&#xA;I understand that after the hydrogen exhaustion on the main sequence the post main sequence starts with the SGB where they burn hydrogen in a shell around the Helium core.&#xA;We also know that stars under $ 2 M_{\odot }$ experience a Helium flash and start Helium core fusion after the Red giant tip. Whereas stars greater than $2 M_{\odot }$ do not experience this Helium flash and starting the He-core burning immediately after the SGB without a helium flash (very massive stars burn Helium on the main sequence).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is now what specific physical process happens in a sub giant, which makes it a red giant afterwards?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance (I am very open for mathematical/physics explanations as well, since I study physics myself).&lt;/p&gt;&#xA;" OwnerUserId="16625" LastEditorUserId="16625" LastEditDate="2017-06-01T07:53:40.453" LastActivityDate="2021-01-27T20:01:16.123" Title="When exactly does a sub giant become a red giant?" Tags="&lt;star&gt;&lt;red-giant&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21158" PostTypeId="1" AcceptedAnswerId="21160" CreationDate="2017-05-31T07:50:40.877" Score="0" ViewCount="589" Body="&lt;p&gt;I'm new to Observational Astronomy, and I just started observations with a telescope and read about RA and Declination(celestial co-ordinates). Although I think I understood the concept, I'm confused about what RA/Dec is visible from the location at present(Summer Sky).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I have understood, since I'm in Sendai, Japan(~38.26N, 140.86E), the +38 degree declination is at my zenith at midnight. So almost -30 Declination will be at my horizion. That should mean that I can see stars from -30 Declination to 90 Declination at my location. Is that correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about RA though, what range of RAs will it be possible to view while its dark(2000 hours to 0300 hours)  &lt;/p&gt;&#xA;" OwnerUserId="17050" LastActivityDate="2017-05-31T12:45:26.393" Title="What Right Ascension and Declinations are observable by a telescope from a particular location?" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21159" PostTypeId="2" ParentId="21157" CreationDate="2017-05-31T08:40:34.003" Score="0" Body="&lt;p&gt;I don't know if this answer your question, but both stars suffer the same process, what happens to the subgiants is that the free electrons cloud stops a bit the procces, but in the massive stars this cloud hasn't been altered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Degenerated electron cloud:  non-interacting particles with pressure and other physical characteristics determined by quantum mechanical effects. It is the analogue of an ideal gas in classical mechanics. The degenerate state of matter, insofar as it deviates from an ideal gas, arises at extraordinarily high density (in compact stars) or at extremely low temperatures in laboratories. It occurs for matter particles such as electrons, neutrons, protons, and fermions in general and in which case it is referred to as electron-degenerate matter, neutron-degenerate matter, and so on. In a mixture of particles, such as ions and electrons in white dwarfs or metals, the electrons may be degenerate, while the ions are not.&lt;/p&gt;&#xA;" OwnerUserId="16990" LastEditorUserId="16990" LastEditDate="2017-05-31T09:47:15.937" LastActivityDate="2017-05-31T09:47:15.937" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21160" PostTypeId="2" ParentId="21158" CreationDate="2017-05-31T12:45:26.393" Score="1" Body="&lt;p&gt;The declination limit, looking south, is about -52 degrees.  This is basically your latitude minus 90.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 8pm, due south will be about RA 12hrs, and at 3 am it will be about 19hrs.  The eastern horizon will be about 6 hrs ahead, and that increases as you look further north, until you hit the circumpolar region, when all RAs are visible.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-05-31T12:45:26.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21161" PostTypeId="1" AcceptedAnswerId="21165" CreationDate="2017-05-31T14:39:00.383" Score="4" ViewCount="248" Body="&lt;p&gt;Since I'm not a physicist, I want to know how that system works. &#xA;For instance, if there was a planet, very similar to the Earth, but with two moons, and the moons where in a horseshoe orbit, how does it work? Is it stable?&lt;br&gt;&#xA;I need it explained in an easy way, and also how would the moons look like from the Earth?&#xA;Thanks! &lt;/p&gt;&#xA;" OwnerUserId="17056" LastEditorUserId="6" LastEditDate="2019-01-29T14:45:16.840" LastActivityDate="2019-01-29T14:45:16.840" Title="Wondering about a horseshoe orbit" Tags="&lt;orbit&gt;&lt;natural-satellites&gt;&lt;earth-like-planet&gt;&lt;planetary-science&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21162" PostTypeId="2" ParentId="21157" CreationDate="2017-05-31T14:40:14.920" Score="2" Body="&lt;p&gt;Let's walk through the stages of post-main sequence evolution. For reference, the images, and much of the content is being pulled from &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0805304029&quot; rel=&quot;nofollow noreferrer&quot;&gt;An Introduction to Modern Astrophysics by Carroll and Ostlie&lt;/a&gt;. I'll break this into low mass stars ($\sim1\:\mathrm{M_\odot}$) and intermediate mass stars (($\sim5\:\mathrm{M_\odot}$)&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Low Mass Evolution&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tqgQI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tqgQI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above is an &lt;a href=&quot;https://en.wikipedia.org/wiki/Hertzsprung%E2%80%93Russell_diagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;H-R Diagram&lt;/a&gt; representing the path of a low mass star through the luminosity-temperature phase space. The star leaves the Zero-Age Main Sequence (ZAMS) once the core of the star has run out of hydrogen and fusion of hydrogen into helium ceases. As you stated, you begin to have a slight core contraction and then hydrogen in a shell around the core begins to burn, producing a helium shell. This hydrogen shell burning causes the core to continue to grow in size and at some point, you reach the &lt;a href=&quot;https://en.wikipedia.org/wiki/Sch%C3%B6nberg%E2%80%93Chandrasekhar_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schonberg-Chandrasekhar limit&lt;/a&gt;, essentially the mass limit for the core which is capable of supporting the gravitational pressure of itself and the envelope. Once the core becomes too massive, it rapidly contracts, releasing a lot of gravitational potential energy which is dumped into the envelope, causing it to heat up. At this point, the star is at the end of the Sub-Giant Branch (SGB) portion of the path and just about to enter the Red Giant Branch (RGB).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a collapsing core (and continued burning of the hydrogen shell) producing huge amounts of energy and an envelope which is heating up, the star needs to reach an equillibrium. This is achieved by the envelope expanding drastically. This, I would argue, is when the star becomes a red giant. After the core contraction energy has been dumped into the envelope and the envelope has expanded due to being heated up. A lot more stuff happens after this point, including a dredge-up from the core and the helium flash, but that is outside the scope of your question.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Intermediate Mass Evolution&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7J6im.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7J6im.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the H-R Diagram representing the path of an intermediate mass star as it leaves the ZAMS. Most of the description I provided above is applicable here, in that the intermediate mass star follows the same process. There are only a few differences in the details. The major difference you may pick out is that as the star moves along the SGB, it actually decreases in luminosity towards the end before venturing up the RGB. This is simply due to the fact that after the core contracts and releases energy into the envelope, the envelope is so much larger than the low mass case, that it can absorb a lot more energy before needing to re-equillibrate. For a time, the core contracts, dumping energy into the envelope, and the overall star luminosity decreases because the core's energy has not found its way out of the star yet, but the star is now smaller. Once the envelope reaches its peak energy consumption it then expands, just as in the low mass state $-$ it simply takes more time and energy to reach this point.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-05-31T14:40:14.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21163" PostTypeId="2" ParentId="21157" CreationDate="2017-05-31T15:02:28.287" Score="4" Body="&lt;p&gt;The straightforward definition is in terms of where a star lies on its evolutionary track in the HR diagram (see below). The subgiant branch stars are those which have exhausted their hydrogen core and which are burning hydrogen in a shell &lt;em&gt;but their He cores have not begun to contract significantly&lt;/em&gt;. The distinct upturn in luminosity marks the beginning of the red giant branch. This occurs when the core grows significantly more massive, cannot support itself hydrostatically and begins to contract. At the same time the envelope expands and becomes convective and the H-burning shell moves inwards and increases in temperature and luminosity.&lt;/p&gt;&#xA;&lt;p&gt;The tip of the red giant branch is where He is ignited. This takes place &amp;quot;explosively&amp;quot; in a degenerate core if the star is below about &lt;span class=&quot;math-container&quot;&gt;$2M_{\odot}$&lt;/span&gt; (not &lt;span class=&quot;math-container&quot;&gt;$5M_{\odot}$&lt;/span&gt;), but begins smoothly in higher mass stars. This causes the core to expand, pushes the H-burning shell out and reduces the luminosity.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xXNBP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xXNBP.png&quot; alt=&quot;HR diagram&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;HR diagram showing the location of the subgiant and red giant branches on the evolutionary tracks for stars of different mass. The tracks begin at the main sequence and evolve to the termination of the main sequence, across the subgiant branch, turn up the red giant branch, which terminates when core He burning commences and the stars become less luminous and hotter.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="17043" LastEditDate="2021-01-27T20:01:16.123" LastActivityDate="2021-01-27T20:01:16.123" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21164" PostTypeId="1" CreationDate="2017-05-31T17:14:04.283" Score="1" ViewCount="143" Body="&lt;p&gt;Title says it, really. As I understand, our total peculiar motion with respect to the CMB is 631 (+/-20) km/h, in the direction of 10.5h RA; -24 deg Dec (or rather away from 22.5h RA; +24 deg Dec, right? That's close to the direction of Lambda Pegasi).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we draw a line from the Moon to Lambda Pegasi, does Earth ever cross it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the answer is no, but don't have enough of a model to prove it to myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inclination of the Moon's orbit is 5.14 degrees, and the Earth subtends about 1.2 degrees from the moon, but the Moon's path extends off at 24 degrees above the ecliptic. So the Earth won't cross it, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all assuming that the vector info above is the total, and already accounts for the motion of the sun around the milky way. If that's not the case, please let me know.&lt;/p&gt;&#xA;" OwnerUserId="17058" LastEditorUserId="13426" LastEditDate="2017-05-31T18:23:05.177" LastActivityDate="2017-07-04T11:26:13.327" Title="Does the Earth ever cross the Moon's path with respect to the CMB?" Tags="&lt;the-moon&gt;&lt;cmb&gt;&lt;ecliptic&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21165" PostTypeId="2" ParentId="21161" CreationDate="2017-05-31T21:10:06.250" Score="5" Body="&lt;p&gt;A basic orbital rule is Kepler's third law, which states that the closer the moon is to the planet, the quicker is its orbit. A body that is close to the Earth will move in orbit faster than one which is distant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fundamental weirdness: If a body is in orbit, and you give it a &lt;em&gt;push&lt;/em&gt; in the direction it is moving, it will &lt;em&gt;gain&lt;/em&gt; energy, move &lt;em&gt;away&lt;/em&gt; from the planet, and &lt;strong&gt;slow down&lt;/strong&gt;. Contrariwise, If you give it a pull back, it will drop to a lower orbit and speed up. This is fundamentally weird because we are used to the idea that if you push something it will speed up, for bodies in orbit the opposite happens. Make sure you understand why this is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose we have two moons, a big one and a little one. The big one is in a circular orbit (and is big enough that the orbit won't change) The little one is on the opposite side of the planet, and slightly closer to the planet than the big moon (only slightly closer). The little one is also in a circular orbit, in the same direction as the big moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the little one is closer to the planet, it orbits faster than the big moon, so will slowly catch up the big moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it catches up the big moon, the pull of the big moon's gravity will pull the little moon forward. Now remember the fundamental weirdness: if the little moon is pulled forward, it will move away from the planet and &lt;em&gt;slow down&lt;/em&gt;. The big moon is also pulled, but we'll suppose the big moon is big enough for the gravity of the little moon to be negligible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the little moon has slowed down. It is still going in the same direction as the big moon, but now it is going slower, so it falls behind the big moon. Eventually it falls so far behind that the big moon begins to catch it up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the gravity of the big moon is pulling the little moon back, and because of the fundamental weirdness, this makes the little moon drop towards the planet, and speed up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process can now repeat itself, with the little moon repeatedly catching up, then falling back from the big moon. An observer on the Big moon would see the little moon slowly approach then fall back, go behind the planet, approach on the other side, then fall back again. It would appear to move in a horseshoe shape, relative to the big moon. Relative to the planet, both moons move in circles or near circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How stable it is depends on all sorts of things: tides, other planets, the effect of the little moon on the big. However such orbit can be stable over long periods of time. But perhaps not over the 5 billion years of the solar system.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-05-31T21:10:06.250" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21166" PostTypeId="1" CreationDate="2017-06-01T02:25:57.437" Score="4" ViewCount="511" Body="&lt;p&gt;So I'm in year 11 and have an aspiration to be an observational astronomer but I'm worried that if I study astrophysics I won't be able to be an astronomer in particular. It probably sounds like a really dumb question. I am open to doing any job in astronomy/astrophysics but I just need to know. Thanks :)&lt;/p&gt;&#xA;" OwnerUserId="17060" LastActivityDate="2017-06-01T15:02:42.780" Title="Can I have a career in observational astronomy if I study astrophysics or does it have to be particularly astronomy?" Tags="&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21167" PostTypeId="1" AcceptedAnswerId="21168" CreationDate="2017-06-01T02:53:40.590" Score="1" ViewCount="246" Body="&lt;p&gt;Consider an Earth-like planet, with two moons. I guess the two moons would have to be smaller than our own, for the system to be stable. Is that stable? I did some research, and apparently Janus and Epimetheus (Saturn moons that are in a horseshoe orbit) are like that, but I don't know how their cycles work, or how they look from Saturn. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence,: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Would a moon have to be smaller that the other one, or would both have to be the same size? &lt;/li&gt;&#xA;&lt;li&gt;Would both have the same cycle? &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;How could nights be seen from the Earth?&lt;/strong&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm looking for that sort of answer. &#xA;Thanks in advance! :) &lt;/p&gt;&#xA;" OwnerUserId="17056" LastActivityDate="2017-06-01T04:26:54.027" Title="If I had a horseshoe orbit of two moons..." Tags="&lt;orbit&gt;&lt;natural-satellites&gt;&lt;earth-like-planet&gt;&lt;orbital-mechanics&gt;&lt;moon-phases&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21168" PostTypeId="2" ParentId="21167" CreationDate="2017-06-01T04:26:54.027" Score="2" Body="&lt;p&gt;It's not the size of the Moons so much as the proximity to the sun that makes a 2 moon system (and especially a 2 moon horse-shoe orbit) unstable.   Size matters too, but the proximity to the sun matters more.    The sun's tidal force perturbs the Moon's orbit around the earth quite a bit.   If Earth was much further from the sun, it could have several moons, not just one and some a fair bit more distant and the big moon's orbit would be nearly circular due the inverse tidal forces, which tend to circularize an orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making the moon's smaller helps some, but what you really need, if you want that orbit to be relatively long-lasting, is to push the Earth away from the sun at least a few more astronomical units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we ignore the solar-tidal effects, then Earth could have 2 moons in a horse-shoe orbit to each other.    As to what it would look like, the first thing to consider is that everything in the sky rises in the east and sets in the west and that's because the Earth rotates.    If you watch each of the two horseshoe moons individually, they would move much like our moon, rising in the east, moving slowly against the fixed stars, and setting in the west.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second thing to consider is that the Moons, the planets and the sun move against the fixed background stars, but this movement is slow unless the Moon is very close to Earth.  Our Moon, moves a full 360 degrees across the fixed stars once every 27 days (&lt;a href=&quot;https://community.dur.ac.uk/john.lucey/users/lunar_sid_syn.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;sidereal orbit&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motion of the two moons individually would be very similar to our moon now in terms of new, waxing, full, waning, and repeat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the horseshoe orbit would do is you'd see one moon slowly catching up to the other, perhaps touching it from the point of view of earth, but never passing it.  it would get close, then it would reverse direction relative to the other moon and in a couple/few weeks, it would approach from the other direction, slow down and reverse.    Each moon individual would behave much like our moon but relative to each other there would be this weird game of attraction-repulsion between them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe (but don't quote me on this), from Earth you'd never see the Moons' cross, but you'd perhaps see them touch.   The nature of the horeshoe orbit prevents crossing and passing, you'd just see them move closer, then move further away from each other.   If you had a really good eye, you might be able to notice the smaller moon change size as it moves closer and further from the Earth as the moon's get close, but that difference would be subtle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the best diagram I've seen for what happens as the moons get close.  Unfortunately it's not in English, but it's not hard to follow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/9/99/JanusEpimetheus.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://upload.wikimedia.org/wikipedia/commons/9/99/JanusEpimetheus.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horseshoe back and forth, assuming some degree of stability, it should have a regular period, just as the full-moons would have a period.   The optimal time for moon watching would be when those 2 periods coincide and you have a double full moon.   The horseshoe period and the full moon period (currently 29 days - &lt;a href=&quot;https://community.dur.ac.uk/john.lucey/users/lunar_sid_syn.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;sidereal orbit&lt;/a&gt;) should have nothing to do with each other but they could in theory be coincidentally in some kind of resonance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed out in the other thread, even if the Earth was far enough away from the sun to have two moons in this pattern, it's probably not truly long-term stable.   It might last for a few hundred orbits, or a few thousand.   It's not what is generally considered a stable orbit.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Would both have the same cycle?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a bit of a trick question.   On average, yes.  The orbital periods would need to be the same for the horseshoe to work when averaged out, but individual orbits would need to vary by a little bit, so the individual orbits wouldn't have the same cycle.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corrections are welcome if I got any of that not quite right.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-06-01T04:26:54.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21169" PostTypeId="1" AcceptedAnswerId="21171" CreationDate="2017-06-01T06:24:34.043" Score="4" ViewCount="913" Body="&lt;p&gt;So my question is that, is not a lie that we are flying away from the sun slowly, that is increasing by time because the sun loss mass, so when the scientist usually talk that we will get swallow by Sun when it get red giant, do they count on this? I mean, we can get far enough to survive or they just calculate with the actual orbit?&lt;/p&gt;&#xA;" OwnerUserId="16990" LastEditorUserId="1891" LastEditDate="2019-04-14T17:44:54.707" LastActivityDate="2019-04-14T17:44:54.707" Title="Could the Earth survive a red giant Sun?" Tags="&lt;the-sun&gt;&lt;red-giant&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21170" PostTypeId="2" ParentId="21169" CreationDate="2017-06-01T06:42:49.833" Score="2" Body="&lt;p&gt;The time frames on these two phenomenon are quite distinct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will take approximately 5 billion years for the sun to reach this red giant phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now at the speed of earth drifts away, and taken into account the &lt;a href=&quot;http://www.live-counter.com/sun/&quot; rel=&quot;nofollow noreferrer&quot;&gt;mass loss&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;By fusion, the sun &quot;burns&quot; about 564 million tons hydrogen per second, resulting in 559.7 million tons of helium. The loss of mass, about 4.3 million tons per second, is transformed into energy. But don't worry, it's only 0.0000000000000000002 percent of the sun's entire mass. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Plus it's kind of negligible how the earth drifts away in the present years, even if it could change drastically, but we know nothing about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Wikipedia: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 1989, Jacques Laskar's work indicated that the Earth's orbit (as well as the orbits of all the inner planets) can become chaotic and that an error as small as 15 meters in measuring the initial position of the Earth today would make it impossible to predict where the Earth would be in its orbit in just over 100 million years' time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;there are possibilities&lt;/strong&gt; the earth will escape its fate: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if &lt;strong&gt;things stay roughly the same it won't&lt;/strong&gt; (orbit go on expanding the way it does now, with sun mass loss taken into account)&lt;/li&gt;&#xA;&lt;li&gt;its possible future civilizations will continue to exist, see a need to keep it and have sufficient technology to do so, so they will &quot;move&quot; it in the right place to avoid it.&lt;/li&gt;&#xA;&lt;li&gt;its possible gravitational instability will do the job and sling the earth out of the sun's outer layers. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11264" LastEditorUserId="7472" LastEditDate="2017-06-01T14:43:34.910" LastActivityDate="2017-06-01T14:43:34.910" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="21171" PostTypeId="2" ParentId="21169" CreationDate="2017-06-01T09:28:06.573" Score="8" Body="&lt;p&gt;The answer is yes, there are many authors who do take mass loss from stars into account when trying to work out the fate of their planetary systems. Examples include &lt;a href=&quot;https://arxiv.org/abs/0801.4031&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schroder &amp;amp; Smith 2008&lt;/a&gt;; &lt;a href=&quot;https://arxiv.org/abs/1303.3841&quot; rel=&quot;nofollow noreferrer&quot;&gt;Adams et al. (2013)&lt;/a&gt;; &lt;a href=&quot;https://arxiv.org/abs/1310.2577&quot; rel=&quot;nofollow noreferrer&quot;&gt;Adams &amp;amp; Bloch (2013)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is thought that (e.g. &lt;a href=&quot;https://arxiv.org/abs/0706.3894&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kalirai et al. 2008&lt;/a&gt;) that the Sun will lose half its mass by the time it ends up as a white dwarf. Almost all of this mass loss occurs in the final hundred million years or so (in just over 7.5 billion years time) whilst the Sun is on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Red-giant_branch&quot; rel=&quot;nofollow noreferrer&quot;&gt;red giant branch&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Asymptotic_giant_branch&quot; rel=&quot;nofollow noreferrer&quot;&gt;asymptotic giant branch&lt;/a&gt;. The mass loss that occurs before then is small in comparison - about $10^{-13}$ of a solar mass every year (see &lt;a href=&quot;https://arxiv.org/abs/0801.3807&quot; rel=&quot;nofollow noreferrer&quot;&gt;Noerdlinger 2008&lt;/a&gt;), so losing less than 1% of its mass prior to the RGB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consideration of conservation of angular momentum alone leads to the idea that $M(t)R$ is constant, where $M(t)$ is the mass of the Sun as a function of time and $R$ is the semi-major axis of a planetary orbit. Thus &lt;em&gt;prior&lt;/em&gt; to the RGB phase, the Earth's orbit expands by maybe 1.4 cm per year, or by a total of $7\times 10^{-4}$ au prior to the RGB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, during and after the RGB we see that the Earth ends up with an orbital radius of $\sim 1.5$ au (or in solar units, about 290 $R_{\odot}$) after the RGB and maybe 430 $R_{\odot}$ after the AGB. So, whilst the Sun is in the RGB/AGB phase and losing mass, the Earth moves outwards from about 215 to 430$ R_{\odot}$. Meanwhile, as it ascends the RGB, the Sun will likely increase in size to about $200 R_{\odot}$. Therefore it is a close-run thing as to whether the Earth will be engulfed within the outer envelope of the evolved Sun.&#xA;Major uncertainties include just exactly how big the Sun will get, how much mass it will lose on the RGB and then on the AGB, and it turns out that the time-profile of the mass loss (which will happen in pulses, rather than continuously) may also play a role.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plot below (from Schroder &amp;amp; Smith 2008) may help to make some sense of this. It shows the (log) of the radius of the Sun versus time in the RGB and AGB phases as it loses mass. Also shown with a dashed line is the orbital radius of the Earth. Fractions written on the plot show what fraction of the Sun's present mass remains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kLbdh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kLbdh.png&quot; alt=&quot;Evolution of the Sun&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, even if the Earth escapes immediate engulfment, the Sun would be several thousand times more luminous than it is today, so the temperatures at the surface of the Earth could reach $&amp;gt;1000$ Celsius even if it stays an astronomical unit away from the Sun's surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A further issue is the effects of tidal dissipation. The Earth will be dragged back by tidal bulges induced in the Sun's extended envelope. The rate at which angular momentum is lost is however exquisitely sensitive to the exact radius of the Sun and the radius of the Earth's orbit, to the extent that their orders of magnitude uncertainty in whether the Earth will be dragged into the Sun's envelope even before it reaches the tip of the RGB (this is what Schroder &amp;amp; Smith 2008 claim) or whether the loss rates will be too low to have much effect. It all depends on how much mass loss there is (Schroder &amp;amp; Smith use a slightly lower estimate than above) and therefore what radius the Earth reaches and what the final radius of the Sun becomes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment I would say there is sufficient uncertainty that the fate of the Earth is undecided, but it will unarguably be uninhabitable.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-06-01T16:02:04.023" LastActivityDate="2017-06-01T16:02:04.023" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21172" PostTypeId="1" CreationDate="2017-06-01T10:10:42.317" Score="4" ViewCount="955" Body="&lt;p&gt;For astrophotography using 'medium size' land telescope at good location,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;How many time longer exposure&lt;/strong&gt;, in view of boosting the system's ability to photograph even darker objects, can one gets from cooling down digital camera sensor chip?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there a '&lt;strong&gt;sweet spot&lt;/strong&gt;' temperature where most thermal related noise is suppressed and further lowering of temperature yields litter extra benefit.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the &lt;strong&gt;ball park temperature&lt;/strong&gt; for low (amateur), medium and high (professional) grade cooled camera?  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Web search shows that Peltier electric cooled camera works at around &quot;&lt;strong&gt;minus a few tens degrees centigrade&lt;/strong&gt;&quot;. Does the really professional one cool to more extreme level, somewhere close to liquid nitrogen temperature? &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Does CCD, CMOS or other &lt;strong&gt;sensor&lt;/strong&gt; have different 'sweet spots' temperature?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13130" LastEditorUserId="13130" LastEditDate="2017-06-02T15:45:54.137" LastActivityDate="2017-12-04T03:43:57.680" Title="Cooling of astrophotography camera sensor chip" Tags="&lt;photography&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21173" PostTypeId="2" ParentId="21166" CreationDate="2017-06-01T10:51:23.447" Score="5" Body="&lt;p&gt;I don't think there really is a difference between &quot;astronomer&quot; and &quot;astrophysicist&quot;. Both are physicists using their knowledge to learn about the Universe. Their is perhaps a tendency to call yourself an astronomer if you're doing observational stuff, and an astrophysicist if you're more theoretical/numerical. If people ask me what I do, I use &quot;astronomer&quot; if I want to sound romantic, and &quot;astrophysicist&quot; if I want to sound clever. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you are interested in doing observations, rather than doing simulations or pure theory. But in all cases, you start by gaining basic knowledge of both observational astronomy, &quot;normal&quot; physics, theory, and coding. Depending on which country you study in, at some point you will probably have the opportunity to have more influence on the courses you take, so that you could skip courses of, say, particle physics and general relativity, in favor of data analysis and, even better, observing summer schools (which I would recommend even for students planning to do theory, because it's a fantastic experience). Finally, your Master's thesis (to a small degree) and eventually your Ph.D. (to a large degree) will have an impact on the path you choose for your career. But many astro*s have been known to switch.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2017-06-01T10:51:23.447" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21174" PostTypeId="2" ParentId="21166" CreationDate="2017-06-01T15:02:42.780" Score="4" Body="&lt;p&gt;This is mostly semantics, but I would say that Astronomy is the umbrella term for both space physics and astrophysics. Space physics essentially involves studying things like plasmas, heliophysics, magnetospheres, heliospheres, planetary atmospheres, ionospheres, aurora, space weather, etc. Astrophysics involves studying things like black holes, stars, the Universe (e.g., cosmology), exoplanets, galaxies, interstellar dust, etc. Observational astronomy could fall under either space physics or astrophysics, depending on what you want to observe. But honestly, it doesn't matter much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During your undergraduate education, you'll likely have a career path nearly indistinguishable from a physics student, with the exception that you may take a handful of more astronomy oriented classes. You'll still need all the basic physics courses. During graduate school is really when you focus on what your future career will be, and even then, not until you start your PhD dissertation. You'll have a good one to two years of classes at the graduate level that will be just more advanced astronomy/astrophysics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For now, my advice is study physics as best you can and point your education towards astronomy, but don't get bogged down in future career paths just yet. You still have a good 5 to 7 years before you really have to start tightening your scope.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-06-01T15:02:42.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21177" PostTypeId="2" ParentId="21130" CreationDate="2017-06-01T16:56:42.700" Score="2" Body="&lt;p&gt;The number you need is the bolometric luminosity. This tells you how many Watts per square metre are incident upon the planet. That is true for the Sun and it is true for GL 570.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond that, you will need to make assumptions about an atmosphere and an albedo if you are trying to calculate a realistic habitable zone. If you just assume the planet is like the Earth then using the same albedo value will probably do. But, because Gl 570 has a spectrum that peaks at 50% longer wavelengths than the Sun, then the albedo values at longer wavelengths are more important for a planet orbiting Gl 570. Whether that is bigger or smaller than the Earth would depend on exactly what sort of atmosphere/clouds/terrain were on the planet.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-06-01T16:56:42.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21178" PostTypeId="1" AcceptedAnswerId="21199" CreationDate="2017-06-01T20:15:34.127" Score="2" ViewCount="102" Body="&lt;p&gt;I remember reading an article on the subject highlighting the periodicity of this event throughout history supposing it could be caused by the rotation of the sun around the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Do we have any clues that can confirm this hypothesis? What should we expect to see? Shouldn't it leave any remains?&lt;/p&gt;&#xA;" OwnerDisplayName="user17066" LastEditorUserId="2365" LastEditDate="2018-08-01T10:17:28.190" LastActivityDate="2018-08-01T10:17:28.190" Title="Should we expect clues from a gamma ray burst causing massive extinctions on Earth?" Tags="&lt;gamma-ray-bursts&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21181" PostTypeId="1" AcceptedAnswerId="21186" CreationDate="2017-06-02T01:41:59.710" Score="1" ViewCount="117" Body="&lt;p&gt;Okay, so imagine that there is an Earth-like planet, with the following conditions: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It's further from it's sun than the Earth is from the Sun (not much, but enough for weather to be a little colder than the Earth). &lt;/li&gt;&#xA;&lt;li&gt;It's more or less the same size as the Earth (same proportion of land and water, too). &lt;/li&gt;&#xA;&lt;li&gt;Since it's further from it's sun, it takes more time to complete an orbit that the Earth (for instance, 400 Earth-days, instead of 365). &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Given those conditions, I would like to know which of the following would be more stable for this Earth-like planet to have a similar day/night and moon cycle to the Earth: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A moon that's the same size as the Earth's moon, only closer. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;OR&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A moon that's as far away as the Moon is to Earth, only that the moon of this planet is bigger (let's say, twice as big as our moon). &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Sorry if it's kind of complicated. Please tell me if it's not understandable and I'll try to edit it so it's as clearly as possible!&#xA;Thanks in advance :) &lt;/p&gt;&#xA;" OwnerUserId="17056" LastActivityDate="2017-06-02T10:09:35.273" Title="What of the following moon-situations would be better?" Tags="&lt;the-moon&gt;&lt;natural-satellites&gt;&lt;positional-astronomy&gt;&lt;earth-like-planet&gt;&lt;night-sky&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21182" PostTypeId="1" AcceptedAnswerId="21185" CreationDate="2017-06-02T02:58:32.663" Score="5" ViewCount="3314" Body="&lt;p&gt;This morning I was interested in finding out how many days are in one year on average, to several decimal places (because I am a mathematician, so I like to be accurate). Unfortunately, Googling the phrase &quot;days in a year&quot; results in several answers that differ as soon as the third decimal place (taking into account rounding).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second question in addition to the title question: Does the duration of earth's orbit increase and decrease, or change only monotonically over time?&lt;/p&gt;&#xA;" OwnerUserId="17068" LastEditorUserId="17068" LastEditDate="2017-06-02T07:41:12.497" LastActivityDate="2020-09-28T06:16:37.017" Title="To several decimal places, how many days are in one year?" Tags="&lt;orbit&gt;" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="21183" PostTypeId="2" ParentId="21181" CreationDate="2017-06-02T06:11:50.127" Score="1" Body="&lt;p&gt;A closer moon would bring higher gravitational pull and it orbits faster than our moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A moon that's as far away as the Moon is to Earth, only that the moon of this planet is bigger (let's say, twice as big as our moon).&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case too the moon cycle won't be similar to ours, since again the size of object matters in gravitation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but the rotation of That Earth like planet gives day/night cycles which you haven't touched of.&lt;/p&gt;&#xA;" OwnerUserId="17069" LastEditorUserId="17069" LastEditDate="2017-06-02T08:24:55.387" LastActivityDate="2017-06-02T08:24:55.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21184" PostTypeId="2" ParentId="21181" CreationDate="2017-06-02T07:24:30.027" Score="1" Body="&lt;p&gt;It looks like you are mixing things a little: If earth was farther away from the sun, it will mean nothing to the day/night cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Day/night cycle only depends on the way the earth rotates on itself, not around the sun. So your question is the same for an planet with same orbit as ours as it is for a planet with an orbit a little larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[for it] to have a similar day/night and moon cycle to the Earth: &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Both are possible. Day/night cycle greatly depends on initial planet's spin. What changes here is &lt;strong&gt;the mass of the earth-like planet&lt;/strong&gt; we are talking about, and this changes the speed at which the moon rotates around. the gravity of the moon will affect the planets rotation speed over very long time range.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;first case, planet is heavier to have a moon the size our moon (I assume same mass as ours) rotating faster around - moon cycle is different, but you can have same day/night cycles as ours&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;second case, planet is lighter, the big moon rotates slower, but you can also have same day/night cycles&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11264" LastActivityDate="2017-06-02T07:24:30.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21185" PostTypeId="2" ParentId="21182" CreationDate="2017-06-02T09:27:40.833" Score="8" Body="&lt;p&gt;A mathematician would probably find the imprecision of precise orbital time a little frustrating.&lt;/p&gt;&#xA;&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Second&quot; rel=&quot;nofollow noreferrer&quot;&gt;second can be defined as&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;the duration of 9 192 631 770 periods of the radiation corresponding&#xA;to the transition between the two hyperfine levels of the ground state&#xA;of the caesium 133 atom&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A 24 hour day is 86 400 of those, well, more or less.&lt;/p&gt;&#xA;&lt;p&gt;A problem that early time-keepers ran into was that the length of the day varied based on the time of year.   Earth does a complete rotation in about 23 hours, 56 minutes, so, the rotation really doesn't factor into it directly, but because Earth orbits the sun about 1 degree of arc every 24 hours, an apparent solar day is has been defined as 24 hours long for centuries.    Days are shorter than that when the earth is closer to the sun, &lt;a href=&quot;https://en.wikipedia.org/wiki/Perihelion_and_aphelion&quot; rel=&quot;nofollow noreferrer&quot;&gt;at perihelion&lt;/a&gt;, (&lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_time#Mean_solar_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;by about 20 seconds&lt;/a&gt;) and days are longer than the 24 hour average by about 30 seconds at aphelion.   Not enough for anyone to notice unless they were measuring solar time very precisely.&lt;/p&gt;&#xA;&lt;p&gt;So a day has to be averaged out, called a mean solar day, and that's what 24 hours was based on.   The problem is, Earth's rotation is slowing down.  The Moon is very gradually slowing down the rotation of the Earth by tidal interaction, so &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_time#Introduction&quot; rel=&quot;nofollow noreferrer&quot;&gt;an average day is currently about 86 400.002 seconds&lt;/a&gt;.   Leap seconds are added almost every year and in the future, leap seconds will be added more often.   Officially a day is still 86 400 seconds, and when a leap-second is added, the &amp;quot;leap-day&amp;quot; is 86 401 seconds.   But you could use either number for number of days in a year, average day (currently 86 400.002) or &lt;a href=&quot;https://en.wikipedia.org/wiki/Day#International_System_of_Units_.28SI.29&quot; rel=&quot;nofollow noreferrer&quot;&gt;SI day&lt;/a&gt;* (86 400).  As I understand it, a year is usually measured in 86 400 second days or SI days, not &amp;quot;average&amp;quot; days, but it's important to be specific about which one you're using.  I've seen the occasional article that says that 200 million years ago a year was about 400 days long.  That's clearly using average says, not SI days.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;And if you live in high altitude, your clock will run even faster and a day will run a tiny bit longer thanks to relativity but . . . lets not even go there for the calculations).&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;A solar year, based on the position of the sun and the precise time of the solstice of your choice, which is a precise and momentary alignment.   The length of a solar year changes a bit on a year-to-year basis.   The position of the moon and gravitational perturbations from other planets, affect Earth's period of rotation a little bit, so like the day, the year should be averaged out also.   Modern &lt;a href=&quot;https://www.timeanddate.com/date/perfect-calendar.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;leap-year calendars&lt;/a&gt; adjust days to years quote well, but not to the accuracy you're asking about.&lt;/p&gt;&#xA;&lt;p&gt;Looking at the time of &lt;a href=&quot;https://en.wikipedia.org/wiki/Summer_solstice&quot; rel=&quot;nofollow noreferrer&quot;&gt;recent summer solstices&lt;/a&gt; (I've copied them down below for the June solstice).&lt;/p&gt;&#xA;&lt;p&gt;2010-06-21 11:28&lt;br /&gt;&#xA;2011-06-21 17:16&lt;br /&gt;&#xA;2012-06-20 23:09&lt;br /&gt;&#xA;2013-06-21 05:04&lt;br /&gt;&#xA;2014-06-21 10:51&lt;br /&gt;&#xA;2015-06-21 16:38&lt;br /&gt;&#xA;2016-06-20 22:34&lt;br /&gt;&#xA;2017-06-21 04:24&lt;br /&gt;&#xA;2018-06-21 10:07&lt;br /&gt;&#xA;2019-06-21 15:54&lt;br /&gt;&#xA;2020-06-20 21:44&lt;/p&gt;&#xA;&lt;p&gt;There's as much as 15 minutes variation in the length of a year over this 11 year sample.  That's not because Earth's orbit is permanently changing.  These are mostly fluctuations due to the Moon and nearby planets.&lt;/p&gt;&#xA;&lt;p&gt;An &amp;quot;average year&amp;quot;, or &lt;a href=&quot;https://en.wikipedia.org/wiki/Tropical_year#Length_of_tropical_year&quot; rel=&quot;nofollow noreferrer&quot;&gt;mean tropical year&lt;/a&gt;, is currently about 365.2422 86 400 second days and &lt;a href=&quot;https://en.wikipedia.org/wiki/Tropical_year#Mean_tropical_year_current_value&quot; rel=&quot;nofollow noreferrer&quot;&gt;if you want to get more accurate&lt;/a&gt;, 365.2421897 days on 1 January 2000.  Formula in the link for adjusting to different years.&lt;/p&gt;&#xA;&lt;p&gt;I suspect, but I'm not 100% certain that the Earth's mean tropical year moves back and forth with some of Earth's Milankovich cycles.   It's not slowing consistently in the sense that a day is slowing due to the Moon's gravity and inverse tidal force.&lt;/p&gt;&#xA;&lt;p&gt;According to Kepler's laws, the orbital period is based mostly on Earth's semi-major axis.  The problem is that Kepler's laws aren't 100% accurate.   Relativity is a factor when you try to pinpoint a year precisely to the second and variation in eccentricity, which affects Earth's orbital speed and the precise timing of an average year.&lt;/p&gt;&#xA;&lt;p&gt;What I know about orbital stability, is that Earth's semi-major axis changes very little, even during the Milankovich cycles, and orbital projections say, that it will probably change little even over time-frames of millions of years.&lt;/p&gt;&#xA;&lt;p&gt;The Sun is also losing mass, which should slow the Earth's orbit over time, but that effect is also very tiny.   Over millions of years, it's thought that the length of a year doesn't change very much.  The length of a day, over time-periods that long, changes quite a bit more rapidly.  An unanticipated orbital change is always possible, but not thought to be likely.   Orbits don't leave much in the way of track-able footprints, so they're hard to pinpoint over long periods of time and are mostly estimated by mathematical models so, take with a grain of salt, but models do suggest that a year is pretty much constant even over periods of millions of years.&lt;/p&gt;&#xA;&lt;p&gt;For further reading, maybe start here:   &lt;a href=&quot;https://en.wikipedia.org/wiki/Equation_of_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Equation_of_time&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Hope that helps.  Some of the math gets kinda tricky.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Thanks zephyr.  I wasn't familiar with that term.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="35314" LastEditDate="2020-09-28T06:16:37.017" LastActivityDate="2020-09-28T06:16:37.017" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="21186" PostTypeId="2" ParentId="21181" CreationDate="2017-06-02T10:09:35.273" Score="2" Body="&lt;p&gt;Both would be stable.   A single moon orbiting a planet is generally stable provided it's comfortably inside the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere#True_region_of_stability&quot; rel=&quot;nofollow noreferrer&quot;&gt;true region of stability&lt;/a&gt;, in the Hill Sphere and provided that it's not so large that tidal forces become a stability issue or not so close that it dips inside the roche limit and might break apart. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way you might run into instability in your scenarios, is if you have the smaller moon very close, inside the &lt;a href=&quot;https://en.wikipedia.org/wiki/Geosynchronous_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Geosynchronous&lt;/a&gt; orbital distance, where the Moon would move ahead of the tidal bulge it creates and that would cause the Moon to be drawn towards the planet and in time, crash into it, similar to how Phobos is &lt;a href=&quot;https://www.universetoday.com/14258/phobos-might-only-have-10-million-years-to-live/&quot; rel=&quot;nofollow noreferrer&quot;&gt;expected to crash into/break into a ring system around Mars&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the planet moves away from the sun it's stable region expands.  The geosynchronous orbit is unrelated to the sun, that's between the masses of the planet and moon only, but as long as your moon is orbiting between geosynchronous orbit and inside the stable region of the hill sphere, it should be fine.   (unless you give the planet or moons a retrograde rotation, then you need to re-examine the long term tidal effects).    For most orbits, the planet rotates ahead of the Moon so the tidal bulge pushes the Moon slowly outwards, but this effect gets weaker as the Moon gets further, so it's generally long term stable over billions of years, unless the planet is very close to the sun and it has a tiny hill-sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only two well known moons in our solar-system orbit ahead of their tidal bulges that leads to orbital decay (a handful of tiny captured moons do too, but those are so small and their tidal forces are largely irrelevant).   Only Phobos and Neptune's wrong way moon Triton orbit ahead of their tidal bulges, resulting in them being pulled towards their planets, not pushed away.   Triton is sufficiently distant that it's timing to &lt;a href=&quot;https://www.quora.com/When-will-Triton-crash-to-Neptune&quot; rel=&quot;nofollow noreferrer&quot;&gt;crash into/break into a ring system around&lt;/a&gt; Neptune is billions of years off.   All the other moons, for tidal reasons, should be moving away from their planets, mostly very very gradually.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 4 Jovian moons, however, are more complicated as Jupiter has a gonzo magnetic field, a magnetosphere that's almost wind like and it's moons are tidally locked to each other and it's innermost moon is losing mass to Jupiter's magnetosphere - so . . . all bets are off when it comes to Jupiter's 4 inner moons.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 or more large moons can sometimes interact with each other, creating some instability.  One moon systems are, more often than not, long term stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(too long?)&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-06-02T10:09:35.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21189" PostTypeId="2" ParentId="21140" CreationDate="2017-06-02T16:01:27.160" Score="1" Body="&lt;p&gt;It really doesn't make much sense to talk about magnification unless you are actually looking into an eyepiece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have a &quot;backyard&quot; telescope, for direct observation (not photographic) the magnification is a sensible thing. The light recording surface is fixed (your eye). The primary focus of the telescope is a small image of the sky, so an eyepiece is used to magnify the primary focus. The overall magnification of the sky depends focal length of the main mirror, and the focal length of the eyepiece. $M=F_O/F_e$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a small telescope, the maximum useful magnification is limited by the size of the main mirror. A larger mirror would allow for greater resolution, and so a more powerful eyepiece can be used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the Keck is a different beast. The limit on resolution is determined not by the size of the mirror, but by the blurring due to the atmosphere. It doesn't matter how big the mirror is, there is no more information that can be gained by greater magnification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nobody is sticking an eyepiece at the primary focus and looking down it. Instead the light at the primary focus is passes into various scientific instruments. Most of the are spectrometers of one sort of another, and so they don't produce magnified images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have an image, the concept of magnification is unclear. If I have a photo of Jupiter, and I move my head closer to the image, it gets bigger. So what is the magnification? It depends on the distance of the observer to the image. It really doesn't make much sense to talk about magnification unless you are actually looking into an eyepiece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some cameras that are used, such as the &lt;a href=&quot;https://www2.keck.hawaii.edu/inst/nirc2/genspecs.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NIRC2&lt;/a&gt;. It has a narrow view mode with 0.009942 arcsec/pixel. On a typical 100 pixels per inch display, Jupiter (at 50 arcseconds) would be 50 inches across.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-06-02T16:01:27.160" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21190" PostTypeId="1" CreationDate="2017-06-02T19:07:46.130" Score="1" ViewCount="77" Body="&lt;p&gt;I'm calculating the total angular momentum of a bunch of theoretical stars of a few solar masses, separated by a few light years, moving at velocities of about a few dozens of &lt;strong&gt;km / sec&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mechanical angular momentum relative to an arbitrary origin (the cluster's center of mass is a good choice) is this :&#xA;\begin{equation}&#xA;\vec{L} = \sum_i \vec{r}_i \times m_i \, \vec{v}_i,&#xA;\end{equation}&#xA;so the &lt;em&gt;SI&lt;/em&gt; units are $\mathrm{kg} \cdot \mathrm{m}^2 / \mathrm{sec}$.  But in real astronomy, what should be the most simple/natural units for star clusters ?  $M_{\odot} \cdot (\mathrm{ly})^2 / \mathrm{year} \equiv M_{\odot} \, c^2 \cdot \mathrm{year}$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the angular momentum units used by astronomers ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what about energy units for a cluster of stars ?  I'm interested in kinematical energy and gravitationnal potential energy of the whole cluster.  Certainly not &lt;strong&gt;joules&lt;/strong&gt; !?  And $M_{\odot} \, c^2$ would certainly be too much, I suppose !&lt;/p&gt;&#xA;" OwnerUserId="14736" LastActivityDate="2017-09-01T14:25:38.037" Title="Natural units of star clusters angular momentum and energy" Tags="&lt;star&gt;&lt;newtonian-gravity&gt;&lt;units&gt;&lt;star-cluster&gt;&lt;globular-clusters&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21191" PostTypeId="1" CreationDate="2017-06-03T06:09:52.440" Score="3" ViewCount="120" Body="&lt;p&gt;One method used for detecting exo-planets is to look for a slight dip in the parent star's luminosity as the planet transits the stellar disc. Intuitively, it seems to me that if planetary systems in our galactic neighborhood are randomly oriented, there would have to be a very large proportion of them in which transits can never happen from Earth's viewpoint. Perhaps, however, the assumption of random orientation is incorrect, and there is some alignment of the axes of rotation of planetary systems, which would facilitate detection of planets in some preferred plane (the galactic plane?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In popular presentations concerning the search for exo-planets, I have never seen this issue addressed. What observations and/or assumptions are used in arriving at a realistic estimate of the number of exo-planets in our region of the galaxy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There are related questions in this forum, but I haven't found one that asks about the possible alignment of the axes of rotation.)&lt;/p&gt;&#xA;" OwnerUserId="17076" LastActivityDate="2017-06-03T20:12:56.750" Title="Detection of exo-planets" Tags="&lt;exoplanet&gt;&lt;planetary-transits&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21192" PostTypeId="2" ParentId="21191" CreationDate="2017-06-03T06:52:57.220" Score="2" Body="&lt;p&gt;The assumption of random orientations is a reasonable one. One reason that exoplanets weren't detected in the 1980s was the expectation that most solar systems would be like ours, with large planets at a great distance, making transits rare, infrequent and hard to detect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hot Jupiters changed that. Most of the planets that Kepler detects are very close to their host star. This means that no great coincidence is required for the inclination of the axis of rotation relative to the solar system. An axial inclination of between 80 and 90 degrees would allow for a transit in many of the systems discovered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is taken into account when estimating the number of stars with planets, with the conclusion that nearly all sun-like stars have planetary systems. Kepler can only detect a fraction of these, but it surveys so many stars that it has found a good number of planetary systems. But most of the stars observed haven't shown a transit. Extrapolating from its discoveries, we have to conclude that the main reason that we don't detect planets around the other stars is due to the inclination of the exoplanetary systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For analysis of the probabilities involved in transiting exoplanets, you can consult &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0004-637X/712/2/1433/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;&#xA;Transit Probabilities for Stars with Stellar Inclination Constraints&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-06-03T20:12:56.750" LastActivityDate="2017-06-03T20:12:56.750" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21193" PostTypeId="2" ParentId="21191" CreationDate="2017-06-03T10:09:09.373" Score="3" Body="&lt;p&gt;It isn't usually an issue because most experiments are simply concerned with finding exoplanets. They are rarely designed in such a way that it is easy to estimate population statistics because of all sorts of biases that go into selecting the targets. Unfortunately the search for exoplanets has turned into a sport where discovery is everything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one assumes random orientation of orbits (and that is all it is, an assumption) then the probability of a transit scales roughly as &#xA;$$P \simeq \frac{R_p+ R_s}{a}$$&#xA;where $R_p$ and $R_s$ are the radius of the planet and hot star respectively and $a$ the planet's orbital radius (with small modifications for non-circular orbits). The larger this is, the more likely a transit is to occur. Hence large exoplanets orbiting close to large stars are more likely to transit. In principle then, this effect can be corrected for when calculating the statistics and frequency of exoplanets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how good is the random orbital inclination assumption? I honestly think nobody knows at the moment. I have done work on the possible alignment of spin axes within the low-mass stars of clusters (&lt;a href=&quot;https://arxiv.org/abs/0911.1075&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jackson &amp;amp; Jeffries 2010&lt;/a&gt;) finding consistency with the random hypothesis. More recent work using asteroseismology suggests that there may be alignment for more massive stars (&lt;a href=&quot;https://arxiv.org/abs/1703.05588&quot; rel=&quot;nofollow noreferrer&quot;&gt;Corsaro et al. 2017&lt;/a&gt;). However, even if the spin axes (and therefore presumably the majority of planet orbits) of stars in clusters line up, there is no obvious reason why each cluster should have the same angular momentum vector When the clusters eventually disperse into the field then they would, presumably, form a pseudo-random distribution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Except, &lt;em&gt;what if&lt;/em&gt; the Galactic tides or a large-scale Galactic magnetic field played a role in shaping the angular momentum direction of the clouds that formed the clusters. Might it be possible for some alignment to persist to old age? Corsaro et al. argue that interactions within a cluster are not sufficient to &quot;scramble&quot; the angular momenta after star formation has finished. Close interactions between stars become much less likely after they emerge from a cluster into the field. An intriguing piece of work by &lt;a href=&quot;https://arxiv.org/abs/1307.5711&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rees &amp;amp; Zijlstra (2013)&lt;/a&gt; found that there was evidence for a non-random distribution of orientation for bipolar planetary nebulae towards the Galactic bulge. This suggested that the orbital angular momenta of binary systems responsible for the bipolar shape of the nebulae were oriented &lt;em&gt;in&lt;/em&gt; the Galactic plane. The result is highly statistically significant but as far as I know has not been followed up despite its obvious implications for estimations of transit yields from exoplanetary surveys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that there will be a much better answer to this question once we have all-sky exoplanet searches of the quality of the Kepler satellite (the main Kepler survey was in one particular direction). It should become very obvious if there are changes in the planet yields as a function of position of the sky (although you also have to control for the types of star being observed) associated with any large-scale alignment. Maybe there is enough information in the Kepler K2 fields that are taken at positions around the ecliptic - I have not seen any analysis. However, such data will surely become available with the launch of NASA's all-sky TESS satellite in 2018.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-06-03T10:09:09.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21194" PostTypeId="1" CreationDate="2017-06-03T11:06:27.040" Score="0" ViewCount="74" Body="&lt;p&gt;The Force of attraction between two bodies is proportional to the product of their masses. And considering the equation $F=ma$. Acceleration due to gravity will be more for objects lower mass. This contradicts the established theory that objects with different masses take the same time to reach the ground if dropped from the same height. Please explain&lt;/p&gt;&#xA;" OwnerUserId="17077" LastEditorUserId="2365" LastEditDate="2017-06-03T17:24:28.610" LastActivityDate="2017-06-04T12:47:08.780" Title="Newtons law of Gravitation" Tags="&lt;near-earth-object&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21195" PostTypeId="2" ParentId="21194" CreationDate="2017-06-03T11:51:05.053" Score="6" Body="&lt;p&gt;For gravity the force is :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F=\frac {G M_1 M_2} {r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the acceleration :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a=\frac F {M_2} =\frac {G M_1 } {r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this is the same for all bodies as it depends only on the primary's mass ($M_1$).&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-06-03T11:51:05.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21196" PostTypeId="2" ParentId="21190" CreationDate="2017-06-03T12:33:22.907" Score="1" Body="&lt;p&gt;Using unit analysis, I've found that a natural unit of angular momentum should be&#xA;\begin{equation}\tag{1}&#xA;L_0 = \frac{G M_{\odot}}{c} \approx 8.816 \times 10^{41} \, \text{kg} \, \text{m}^2/\text{s}.&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a &lt;em&gt;bounded&lt;/em&gt; system of two equal masses, circularly moving around their center of mass, the total angular momentum (relative to the center of mass) is&#xA;\begin{equation}\tag{2}&#xA;L_{\text{bound}} = 2 M \, v \, r = \frac{2M \, v^2 \, r}{v}.&#xA;\end{equation}&#xA;Now, using Newton's equation and the gravitationnal force, we have&#xA;\begin{equation}&#xA;M \, v^2 = \frac{G M^2}{4 \, r}.&#xA;\end{equation}&#xA;Thus, equation (2) becomes :&#xA;\begin{equation}\tag{3}&#xA;L_{\text{bound}} = \frac{G M^2}{2 \, v}.&#xA;\end{equation}&#xA;The classical theory cannot be trusted when $v \sim c$, and since $v &amp;lt; c$ (forgetting the 2 factor) :&#xA;\begin{equation}\tag{4}&#xA;L_{\text{bound}} \approx \frac{G M^2}{v} &amp;gt; \frac{G M^2}{c} \equiv L_0.&#xA;\end{equation}&#xA;So $L_0$ appears to be a good unit.  When $L \sim L_0$, we then need relativity.  We're in a classical regime when $L \gg L_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system isn't bounded when $L \gg L_{\text{bound}}$, which implies $L \ggg L_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if this is a good way to introduce the $L_0$ unit defined by (1).  Any opinion/suggestions/comments on this ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do astronomers use $G M^2 / c$ as natural unit of angular momentum, when they study star clusters and globulars ?&lt;/p&gt;&#xA;" OwnerUserId="14736" LastEditorUserId="14736" LastEditDate="2017-06-03T12:47:38.460" LastActivityDate="2017-06-03T12:47:38.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21197" PostTypeId="1" AcceptedAnswerId="21283" CreationDate="2017-06-03T12:58:07.613" Score="3" ViewCount="185" Body="&lt;p&gt;The StarShot project from &lt;a href=&quot;https://breakthroughinitiatives.org/Initiative/3&quot; rel=&quot;nofollow noreferrer&quot;&gt;Breakthrough Initiatives&lt;/a&gt; aims to send a paper light spacecraft to alpha centauri using pressure from a laser located on earth at 20,000kms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the options for a transmitter that can be received from 1-4 lyrs away, if the transmitter weighs the same as a few jellybeans (&amp;lt;10 grams)? What wavelength and power range can they work with for that distance using that weight of equipment?&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2017-06-06T09:59:55.017" LastActivityDate="2017-06-14T13:57:04.510" Title="Power requirement for transmitting communications further than 1 ly?" Tags="&lt;star&gt;&lt;communication&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21198" PostTypeId="1" AcceptedAnswerId="21200" CreationDate="2017-06-03T13:50:53.747" Score="1" ViewCount="204" Body="&lt;p&gt;In the excellent chemistry series of &lt;a href=&quot;https://www.youtube.com/user/periodicvideos/videos&quot; rel=&quot;nofollow noreferrer&quot;&gt;Periodic Videos&lt;/a&gt;' episode &lt;a href=&quot;https://youtu.be/4ESW_NTIhBM&quot; rel=&quot;nofollow noreferrer&quot;&gt;Amazing piece of metal (speculum)&lt;/a&gt; there is an interesting showing and discussion of Isaac Newton's first telescope. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At &lt;code&gt;02:36&lt;/code&gt; there is a short cut to a view of a very large, modern telescope mirror. I'm confused by the image because it seems to show what looks like a small 45 degree elliptical, diagonal secondary mirror that you might see near the top end of a long focal length Newtonian telescope, except that it seems to be sitting in front of the primary, facing towards the stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone identify this telescope, and help me understand the basics of how this particular optical system is configured?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit:&lt;/strong&gt; I was so focused on the small mirror in the center that I neglected to read the caption. This &quot;very large telescope&quot; may in fact be &lt;em&gt;THE Very Large Telescope&lt;/em&gt;, one of it's components at least. But because of all the shiny metal an reflections in mirrors, I still don't really understand what I'm looking at as far as the optical system is configured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vrzFI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vrzFI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=4ESW_NTIhBM&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=4ESW_NTIhBM&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2017-06-03T15:37:52.910" Title="What telescope is this, and how does it work?" Tags="&lt;telescope&gt;&lt;optics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21199" PostTypeId="2" ParentId="21178" CreationDate="2017-06-03T15:14:18.103" Score="2" Body="&lt;p&gt;A gamma ray burst could cause an extinction event. There is no evidence that this has ever happened to the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gamma ray bursts are rare events, we don't know exactly how rare, but to cause a mass extinction a GRB would need to be near (6000 light years or less is near in this context) and pointed right at Earth. GRBs are rare enough for it to be possible that this has never happened. It is possible that one or two of the extinctions in the last billion years were due to GRBs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass extinction that is most likely to have been caused by a GRB is the &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0309415&quot; rel=&quot;nofollow noreferrer&quot;&gt;Orodvician-Silurian Event&lt;/a&gt;, 444 million years ago. There is no positive evidence of a GRB, but it can't be excluded. The best that can be said is that the pattern of extinction is consistent with a GRB. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A gamma ray burst would leave few signs that could be used to prove it occurred. The radiation would strip the ozone layer, leaving the surface vulnerable to UV radiation. But it wouldn't leave a crater, or a geological layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gamma ray bursts can occur when a hypernova occurs. There is no reason to think that the Earth is more vulnerable to a hypernova when it is in the galatic plane. The evidence of a 26 million year periodicity to mass extinctions is weak, and there is no evidence that closely ties this to the sun's orbit around the milky way. There is good reason to tie other extinctions to particular impacts, for example there is no need to hypothesise that a GRB occurred 65 million years ago, at the K-T mass extinction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gamma ray bursts are one of the less likely causes of a mass extinction. Researcher David Thompson likens them to the danger of finding a polar bear in his closet. It &lt;em&gt;could&lt;/em&gt; happen; it would be dangerous, but the chance of it happening is so low that it is not a risk worth worrying about.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-06-03T15:14:18.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21200" PostTypeId="2" ParentId="21198" CreationDate="2017-06-03T15:22:42.137" Score="4" Body="&lt;p&gt;This is indeed the VLT, or rather one of the four 8.2m Unit Telescopes (UT in ESO parlance). Looking at the &lt;a href=&quot;https://www.eso.org/sci/facilities/paranal/telescopes/ut/optlao6.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;optical layout&lt;/a&gt; (found &lt;a href=&quot;https://www.eso.org/sci/facilities/paranal/telescopes/ut.html#par_title&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; along with a description of the telescope), the small mirror in the centre of your screenshot is M3, the tertiary mirror. M1 is the primary (in the background), and M2 is out of the frame towards the top- the white supports you see in the foreground hold it above M1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One advantage of this design is that you can have two large instrument platforms on the vertical tilt axis of the telescope - this way, you don't have to dangle several tons of instrument from the support structure (Nasmyth focus in the image).&lt;/p&gt;&#xA;" OwnerUserId="13681" LastEditorUserId="13681" LastEditDate="2017-06-03T15:37:52.910" LastActivityDate="2017-06-03T15:37:52.910" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21202" PostTypeId="1" CreationDate="2017-06-04T00:13:18.173" Score="1" ViewCount="98" Body="&lt;p&gt;If the first stars in the universe were formed 400 million years after the big bang, and the universe is expanding, should that stars at that time not been in our cosmic backyard?  And should their light not have reached earth much sooner?&lt;/p&gt;&#xA;" OwnerUserId="17081" LastActivityDate="2017-06-04T09:04:05.933" Title="Expanding universe" Tags="&lt;early-universe&gt;" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="21204" PostTypeId="1" CreationDate="2017-06-04T04:40:00.920" Score="0" ViewCount="2541" Body="&lt;p&gt;I looked outside my window one night, and the moon was shining brightly, and it wasn't a full moon. But thats not the point. Really close to it was something that looked liked a star. It was really bright, and it gave off a white light, but the more I looked at it, the more I realized that it wasn't flickering or twinkling the way stars do. A few minutes later, I looked at it again, and this time it was smaller, less bright, and farther away from the moon. I know its not a satalite because I've seen those before and they move faster, and they give off the same amount of light all the time. If anyone has any idea what this thing could be, please tell me, because I really want to know.&lt;/p&gt;&#xA;" OwnerUserId="17082" LastActivityDate="2017-06-12T21:57:20.333" Title="I see a non flickering star. How is this possible?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21205" PostTypeId="2" ParentId="21202" CreationDate="2017-06-04T07:46:50.710" Score="4" Body="&lt;p&gt;Yes, and indeed it did. For some stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some first stars were close to us, some were far away. The light from the ones that were very far away has yet to reach us, while the light from the ones that were close to us, reached us in the past; if we look &lt;em&gt;today&lt;/em&gt; at their location, we no longer see first stars, but instead see evolved galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the light from the ones that happened to lie at the particular distance of roughly 2.7 billion lightyears (Gly) when the Universe was 400 Myr old reaches us today. An example of this is the galaxy &lt;a href=&quot;https://en.wikipedia.org/wiki/GN-z11&quot; rel=&quot;noreferrer&quot;&gt;GN-z11&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;2.7 Gly?&lt;/em&gt;&quot;, you say, &quot;&lt;em&gt;Shouldn't that light have reached us after 2.7 Gyr, then?&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, because the Universe expands, it takes longer. In fact at that time, the part of the Universe that was at the distance of GN-z11 expanded at four times the speed of light. That means that even though light traveled in our direction (locally at the speed of light), the distance from us$^\dagger$ to the photons increased. Eventually, however, as the photons reached regions that expanded less rapidly, they &quot;overtook&quot; the expansion of the Universe, and slowly started decreasing their distance to us.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;The hypothetical &quot;us&quot;; our Sun wasn't even born at that time, and probably the Milky Way was just a small density fluctuation in the background density.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2017-06-04T07:46:50.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21206" PostTypeId="2" ParentId="21202" CreationDate="2017-06-04T09:04:05.933" Score="4" Body="&lt;p&gt;The very first stars were probably, massive, ultraluminous, and very short-lived. So if they were formed 400 million years after the big bang, then they would have ceased to exist only a few million years later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that, if we wish to observe the first stars, then we have to look at the light from galaxies that has been travelling for 13.3 billion years to get to us. Hence we must look at very, very distant galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider very old, but second generation stars (known as &quot;Population II stars) - those born from material that incorporates the detritus of the first stars, then we &lt;em&gt;can&lt;/em&gt; see those closer to home. That is because some of those stars were long lived and survive to the present day in our own and other nearby galaxies.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-06-04T09:04:05.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21207" PostTypeId="1" AcceptedAnswerId="21346" CreationDate="2017-06-04T09:06:21.530" Score="2" ViewCount="310" Body="&lt;p&gt;I was looking at the comparison of 2007 and 2015 HST images of the &amp;quot;disappearing&amp;quot; star N6946-BH1 in Phys.org's article &lt;a href=&quot;https://phys.org/news/2017-05-collapsing-star-birth-black-hole.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Collapsing star gives birth to a black hole&lt;/a&gt;. The 2007 images is from the WFPC2 while the one from 2015 is from WFC3.&lt;/p&gt;&#xA;&lt;p&gt;I wondered why the WFPC2 image is substantially fuzzier than that from WFC3. I looked at the &lt;a href=&quot;http://documents.stsci.edu/hst/wfpc2/documents/handbooks/cycle17/wfpc2_ihb.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;WFPC2 manual for cycle 17&lt;/a&gt; and found it has the &amp;quot;L-shaped&amp;quot; detector. Three quadrants have CCD elements with pixel spacing of 0.1 arcsec, the fourth &amp;quot;planetary&amp;quot; array has a pixel spacing of 0.046. However it seems more complicated because - if I understand correctly - there are different focal ratios used for the two.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The WFPC2 field-of-view is divided into four cameras by a four-faceted&#xA;pyramid mirror near the HST focal plane. Each of the four cameras&#xA;contains an 800x800 pixel Loral CCD detector. Three cameras operate at&#xA;an image scale of 0.1″ per pixel (F/12.9) and comprise the Wide Field&#xA;Camera (WFC) with an “L” shaped field-of-view. The fourth camera&#xA;operates at 0.046″ per pixel (F/28.3) and is referred to as the Planetary&#xA;Camera (PC).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;However, while the &lt;a href=&quot;http://documents.stsci.edu/hst/wfc3/documents/handbooks/cycle23/wfc3_ihb.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;WFC3 manual for cycle 23&lt;/a&gt; gives the pixel spacing as&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Table 5.1: WFC3 Detector Characteristics&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;0.04″ per pixel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I didn't find a focal ratio.&lt;/p&gt;&#xA;&lt;p&gt;I'd like to understand why the 2007 image is so &amp;quot;gaussian fuzzy&amp;quot;. If it were only because of larger pixel spacing it would look more blockish or pixellated. Does this mean it was using a different optical path with lower optical resolution? Is the 2007 image from the WFC part at F/12.9? Does the WFC3 use F/28.3?&lt;/p&gt;&#xA;&lt;p&gt;I also don't understand the optical reason why different paths would have different optical resolutions. This would not come from different Airy disks from simple diffraction - the angular size would be unchanged.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AOdsV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AOdsV.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/DeqRC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DeqRC.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; cropped sections from the same image shown below.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MGJxL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MGJxL.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; From &lt;a href=&quot;https://phys.org/news/2017-05-collapsing-star-birth-black-hole.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-06-26T07:39:36.690" Title="Why is this HST WFPC2 image so much fuzzier than the WFC3 image of the same field?" Tags="&lt;optics&gt;&lt;angular-resolution&gt;&lt;hubble-telescope&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21208" PostTypeId="2" ParentId="21164" CreationDate="2017-06-04T10:27:47.663" Score="2" Body="&lt;p&gt;Not within plus/minus a few hundred years at least! The Moon's orbital plane around the Earth (viewed in an inertial frame) rocks back and forth a bit (&lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_precession#Nodal_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lunar Nodal Precession&lt;/a&gt; with a period of about 18.6 years), but that's not going to help here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can play with the numbers in this program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DyMxe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DyMxe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pbmLJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pbmLJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ARF5P.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ARF5P.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1Rwsn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1Rwsn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;from skyfield.api import Loader&#xA;&#xA;degs = 180./np.pi&#xA;&#xA;load = Loader('~/Documents/yournamehere/SkyData')&#xA;data = load('de405.bsp')&#xA;ts   = load.timescale()&#xA;&#xA;timez = []&#xA;dayz  = np.arange(1, 31, 0.1)&#xA;dayz  = np.arange(1, 366, 0.1)&#xA;&#xA;#for year in range(1700, 2200):&#xA;for year in range(2017, 2028):&#xA;#for year in range(2000, 2030):&#xA;&#xA;    #time    = ts.utc(year, 7,  dayz)  # July&#xA;    time    = ts.utc(year, 1, dayz)   # January&#xA;    timez.append(time)&#xA;&#xA;earth = data['earth']&#xA;moon  = data['moon']&#xA;&#xA;RAdegs, Decdegs = [], []&#xA;&#xA;for time in timez:&#xA;&#xA;    RA, Dec, dist = earth.at(time).observe(moon).radec()&#xA;    RAdeg, Decdeg = degs*RA.radians, degs*Dec.radians&#xA;&#xA;    # lazy way to block the wrapparound&#xA;    dRAdeg = RAdeg[1:] - RAdeg[:-1]&#xA;    dropit = np.abs(dRAdeg) &amp;gt; 100&#xA;    RAdeg[:-1][dropit]  = np.nan&#xA;    Decdeg[:-1][dropit] = np.nan&#xA;&#xA;    RAdegs.append(RAdeg)&#xA;    Decdegs.append(Decdeg)&#xA;&#xA;RAdegpts  = [157.5, 157.5+180]  # coordinates from the question&#xA;Decdegpts = [-24.0, +24.0] &#xA;&#xA;if 1 == 1:&#xA;    plt.figure()&#xA;    for RAdeg, Decdeg in zip(RAdegs, Decdegs): #[::5]:  # plot every 5th year&#xA;        plt.plot(RAdeg, Decdeg)&#xA;    plt.scatter(RAdegpts, Decdegpts)&#xA;    plt.title(&quot;2017 throuth 2027 continuously&quot;, fontsize=20)&#xA;    plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-06-04T10:36:15.257" LastActivityDate="2017-06-04T10:36:15.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21209" PostTypeId="2" ParentId="21194" CreationDate="2017-06-04T12:47:08.780" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The force of attraction between two bodies is proportional to the product of their masses. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Correct. StephenG gave the expression: $F=\frac {G M_1 M_2} {r^2}$. If the masses are greater the force is even greater. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This contradicts the established theory that objects with different masses take the same time to reach the ground if dropped from the same height. Please explain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's simpler than you think. Imagine you have two Earth-sized objects A and B. Object A will &quot;fall&quot; towards object B at 9.8m/s². In addition object B will &quot;fall&quot; towards object A at 9.8m/s². So their closing acceleration is 19.6m/s². It's more than 9.8m/s². The net effect is that the two objects fall together faster. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if your two objects are apple-sized and Earth-sized, object A will &quot;fall&quot; towards object B at 9.8m/s², but object B experiences no measurable &quot;fall&quot; towards object B. An apple is just too small. It's the same if you have a melon-sized object and an Earth-sized object. It's also the same if you have a melon+apple-sized object and an Earth-sized object. You measure an acceleration of 9.8m/s². Your melon+apple needs to be very &lt;em&gt;very&lt;/em&gt; big before you measure a closing acceleration of more than 9.8m/s².&lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2017-06-04T12:47:08.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21210" PostTypeId="1" CreationDate="2017-06-04T16:10:54.660" Score="3" ViewCount="114" Body="&lt;p&gt;The Abell 1689 is a large group of galaxies, containing a lot of elliptical galaxies :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Abell_1689&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Abell_1689&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, an elliptical galaxy don't have large angular momentum around its center of mass (spin angular momentum, if you prefer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is about the galactic group, globally as a whole.  Does it have angular momentum around its center of mass, or should it be negligible ?&lt;/p&gt;&#xA;" OwnerUserId="14736" LastActivityDate="2017-06-13T07:33:00.643" Title="Does the Abell 1689 supercluster have angular momentum?" Tags="&lt;galaxy&gt;&lt;rotation&gt;&lt;galactic-dynamics&gt;&lt;galaxy-cluster&gt;&lt;kinematics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21211" PostTypeId="1" CreationDate="2017-06-04T16:35:00.213" Score="5" ViewCount="353" Body="&lt;p&gt;When you look at effective temperature ranges and escape velocities of planets, there is a &quot;population&quot; where hydrogen escapes, but helium doesn't. Is it feasible that there may be planets less that about 10 Earth masses (not gas giants) with a predominantly helium atmosphere??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of asking this question would be:&#xA;Do any models show terrestrial planets retaining a primordial atmosphere or is expected that most all will be secondary? I am guessing that very little helium is &quot;outgassed&quot;, whereas hydrogen can be added to an atmosphere by photodissociation and other processes. I know that early star UV emissions, collisions, etc. will probably strip a &quot;close in&quot; planet's original atmospheres.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="2365" LastEditDate="2017-06-04T21:06:38.717" LastActivityDate="2017-06-08T18:44:31.887" Title="Helium in terrestrial planet atmospheres?" Tags="&lt;exoplanet&gt;&lt;planetary-atmosphere&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21212" PostTypeId="1" CreationDate="2017-06-04T17:12:47.820" Score="1" ViewCount="72" Body="&lt;p&gt;My telescope is blurry and I am not sure why I used a 254mm primary mirror f/5 with a secondary mirror at 62.5 and I also used a 32mm eye piece I know if I stand back 20cm from eyepiece I see a clear image but any closer and it gets blurry can someone tell me what I am missing&lt;/p&gt;&#xA;" OwnerUserId="17085" LastActivityDate="2017-06-04T17:12:47.820" Title="My homemade telescope" Tags="&lt;newtonian-telescope&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21214" PostTypeId="1" AcceptedAnswerId="21219" CreationDate="2017-06-05T09:24:46.270" Score="4" ViewCount="203" Body="&lt;p&gt;Can an intervening planet in orbit around a nearby star cause periodic occultation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please excuse the crude text graphics:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Tabby ...... PlanetP ......................................... Earth&#xA;                |&#xA;             StarX&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;StarX and Tabby's star are close to each other (maybe part of some local cluster, like the Centauri brothers) so the line of sight with our Sun/Earth has maintained over the recent handful of years.&lt;/li&gt;&#xA;&lt;li&gt;PlanetP orbits StarX with a period of around 750 Earth days.  Its ecliptic is at an angle (somewhat perpendicular) to the line of sight with Earth.&lt;/li&gt;&#xA;&lt;li&gt;PlanetP's orbit rotates (precession?) a bit around StarX, so the amount of occultation of Tabby's star by PlanetP varies from 15% to 22% to whatever.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17090" LastEditorUserId="2153" LastEditDate="2017-06-05T21:23:43.050" LastActivityDate="2018-05-18T15:50:10.427" Title="Can the explanation of Tabby's star KIC 8462852 be an intervening planet in orbit around another star?" Tags="&lt;star&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21215" PostTypeId="1" CreationDate="2017-06-05T13:20:57.767" Score="0" ViewCount="54" Body="&lt;p&gt;The redshift-distance relation is $cz=H_0d$. So if we have $z$ we can calculate $d$. But is there a way to calculate or covert or write this $d$ in terms of right ascension (RA) and declination (DEC) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="17094" LastEditorUserId="11143" LastEditDate="2017-06-05T13:30:44.133" LastActivityDate="2017-06-05T18:28:35.440" Title="relative distance and right ascention" Tags="&lt;coordinate&gt;&lt;redshift&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21217" PostTypeId="2" ParentId="21215" CreationDate="2017-06-05T18:28:35.440" Score="1" Body="&lt;p&gt;No. If you only know the distance to your object, that tells you absolutely nothing about where it is. That's like saying, I know the grocery store is 5 miles from my house, how can use this information to figure out exactly where it is? The answer is that you just can't, you need more information. – zephyr &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 3D space, you need three coordinates to locate an object. Ra, Dec are two (position on a sphere), z (or some other radial distance measure for nearby objects) is the third. Astronomy usually tries to make do with just Ra, Dec because that's easy to measure. – Alex&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(comments converted to answer under cc licence)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-06-05T18:28:35.440" CommentCount="0" CommunityOwnedDate="2017-06-05T18:28:35.440" ContentLicense="CC BY-SA 3.0" />
  <row Id="21218" PostTypeId="2" ParentId="21214" CreationDate="2017-06-05T18:56:23.520" Score="2" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First your scale is all wrong. If the scale of the diagram is such that the planet orbits Star X of a centimetre (on the diagram) then the distance between the stars would be 50 km or so. In other words, it would be an unimaginable coincidence for a planet to line up exactly. And since the stars are moving relative to each other (even if they are in a cluster) by the time the planet has orbited once, the planet would no longer be lined up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hypothesis doesn't explain much of the observed strangeness, the very large variation in brightness, the odd shape of the brightness graph. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, if the planet can orbit of star X, where is the star. We can't see a star in the right location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this hypothesis doesn't explain the observations, so would have to be rejected.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-06-05T18:56:23.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21219" PostTypeId="2" ParentId="21214" CreationDate="2017-06-05T21:16:40.280" Score="3" Body="&lt;p&gt;The answer is no, because the light curves simply do not match up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomers have detected planets before using &lt;a href=&quot;https://en.wikipedia.org/wiki/Methods_of_detecting_exoplanets#Transit_photometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;the transit method&lt;/a&gt;. Large planets cause dips in a star's light curve when they move between their parents star and Earth; if these dips are regular and strong enough, we can conclude that there is a planet-like body orbiting the star. It's not perfect (best used on conjunction with other methods), but if there &lt;em&gt;is&lt;/em&gt; a planet orbiting a star that is detectable like this, we should see a couple key features:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Periodicity, corresponding to the planet's orbital period.&lt;/li&gt;&#xA;&lt;li&gt;Relative regularity in the depths of the transits.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Transits aren't always easy to see, but sometimes you can rule them out as the cause of light curves. The light curves of KIC 8462852 are one of these cases. &lt;a href=&quot;https://arxiv.org/abs/1509.03622&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boyaijan et al. (2015)&lt;/a&gt; provided some of their light curves from the observations of the star. Here's one from Figure 1(b), showing a number of &quot;dips&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cjBbw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cjBbw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is neither regularity nor periodicity in the data, meaning that your proposed explanation is not possible. Looking at the dips in more detail (see the rest of the paper) confirms the conclusion.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2017-06-05T21:16:40.280" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21222" PostTypeId="1" AcceptedAnswerId="21223" CreationDate="2017-06-06T07:29:57.980" Score="7" ViewCount="1853" Body="&lt;p&gt;I recently learned about the terms &lt;em&gt;prograde&lt;/em&gt; and &lt;em&gt;retrograde&lt;/em&gt;. I've seen these terms used to describe the rotations and orbits of moons with respect to their planet, and planets with respect to their star. But I haven't been able to find what the relationship is between the rotation and orbit of the Sun with respect to the rotation of the Milky Way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, basically what I'm trying to do here is ask two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is the orbit of the Sun around the center of the Milky Way prograde or retrograde with respect to the rotation of the Milky Way?&lt;/li&gt;&#xA;&lt;li&gt;Is the axial rotation of the Sun prograde or retrograde with respect to the rotation of the Milky Way?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Regarding question #2, I know it might be considered a slightly weird question to ask, simply because the plane of the Solar System is significantly tilted relative to the plane of the Milky Way (by about 60°), and furthermore the Sun itself has its equatorial plane tilted by a few more degrees relative to the plane of the Solar System (about 7°). But if we mentally &quot;flatten out&quot; the Solar System such that the Sun's equatorial plane aligns with the plane of the Milky Way, then we can talk about whether its axial rotation is in the same direction as the rotation of the Milky Way, so I think this is a valid application of the concept.&lt;/p&gt;&#xA;" OwnerUserId="17098" LastActivityDate="2018-01-21T14:19:34.520" Title="Is the Sun prograde or retrograde with respect to the rotation of the Milky Way?" Tags="&lt;the-sun&gt;&lt;orbit&gt;&lt;milky-way&gt;&lt;rotation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21223" PostTypeId="2" ParentId="21222" CreationDate="2017-06-06T08:55:07.900" Score="11" Body="&lt;p&gt;Prograde. This is usually the case for stars in spiral galaxies outside the most central regions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A galaxy can be either &lt;em&gt;rotation-dominated&lt;/em&gt; or &lt;em&gt;dispersion-dominated&lt;/em&gt;, depending on whether its kinematics is dominated by ordered or random motion. Spiral galaxies such as the Milky Way belong to the first class, while ellipticals belong to the second. Irregular galaxies can be somewhere in between, with random motions but with an overall rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In spirals, stars and gas generally rotate in the same direction (although they can have more random motion in their bulges, i.e. in the central parts). The Sun follows the overall rotation of the Milky Way, except for roughly 20 km/s in the direction wrt. the overall rotation; this direction is known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_apex&quot; rel=&quot;noreferrer&quot;&gt;Solar apex&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, following a major merger event, the product can be two kinematically decoupled components. This is described well in &lt;a href=&quot;https://arxiv.org/abs/1403.1263v1&quot; rel=&quot;noreferrer&quot;&gt;Corsini (2014)&lt;/a&gt;. Such a merger is required to perturb the system sufficiently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On smaller scales, however, the turbulence of the gas clouds giving birth to stars are more random. Hence the rotation of a star may very well be in another direction than the overall disk rotation. As you say, the plane of the Solar System is tilted by 63º wrt. the plane of the Milky Way, in the direction that the Sun is traveling. This can be seen e.g. in this infrared IRAS image, where the angle between the bright Milky Way disk and the blue stripes — which is emission from dust in the plane of the Solar System — crosses at roughly 60º:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iwvR8.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iwvR8.jpg&quot; alt=&quot;IRAS&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;credit: &lt;a href=&quot;http://coolcosmos.ipac.caltech.edu/image_galleries/IRAS/allsky.html&quot; rel=&quot;noreferrer&quot;&gt;Caltech&lt;/a&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rotation of the Sun is also prograde wrt. the Milky Way (or else we would say that the tilt was 180º – 63º = 117º).&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2017-06-06T10:57:22.370" LastActivityDate="2017-06-06T10:57:22.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21224" PostTypeId="1" CreationDate="2017-06-06T14:54:56.263" Score="3" ViewCount="395" Body="&lt;p&gt;I'm looking for visualizations that show the movement of stars in barred galaxies but not having much luck. I've found many that simulate spiral galaxy arms in general, but none that include a bar or a nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to word my question in a way that is not dependent on my own assumptions, but some of my assumptions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the area that includes the bar is the nucleus.&lt;/li&gt;&#xA;&lt;li&gt;the nucleus and the bulge are not necessarily the same thing. (Bars within the nucleus is a result of gravity waves. Bulge is an artifact of how the galaxy was formed.)&lt;/li&gt;&#xA;&lt;li&gt;the edge boundary of the nucleus is a gravitational &quot;tipping point&quot;.&lt;/li&gt;&#xA;&lt;li&gt;stars are not crossing this boundary (ie: they either belong to the nucleus or to the arms, not both).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the movement of stars within the nucleus of a barred spiral galaxy&lt;/strong&gt;, and how do their galactic orbits differ from stars outside the nucleus?&lt;/p&gt;&#xA;" OwnerUserId="15795" LastEditorUserId="15795" LastEditDate="2017-06-07T17:35:36.577" LastActivityDate="2017-06-21T09:25:07.923" Title="What is the movement of stars within the nucleus of a barred spiral galaxy" Tags="&lt;galaxy-center&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21226" PostTypeId="2" ParentId="10698" CreationDate="2017-06-06T19:02:04.957" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;is there any objectively observable increase of brightness over the&#xA;  last few months?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. It's entirely your subjective perception that has changed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun is pretty close to a black-body radiator. That means the amount of energy we receive from it depends on two factors: its temperature, and its apparent size (the angular size of the solar disk as seen from Earth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The apparent size of the Sun has obviously not changed. Way too many things would go completely haywire if that happened - I'll leave that as an exercise to the reader.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its temperature has not changed either. This could be measured in various ways - but there's another way which is simpler: this being a black-body radiator, its temperature and its apparent color are correlated. The color of the Sun has not changed, obviously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TLDR: The Sun is still working as it always had.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2017-06-06T19:02:04.957" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21227" PostTypeId="2" ParentId="21197" CreationDate="2017-06-07T09:54:13.833" Score="2" Body="&lt;p&gt;It's possible to transmit an intelligible signal to a target roughly 4.3 light-years away using 0.1 milliwatts. The trick for that particular level of power is that both the sender and receiver have to be carefully positioned several hundred AU from a star, with those stars lined up between them, in order to take advantage of a gravitational lensing effect. If only one of the two radios can be positioned at a star's gravitational focus, then the necessary power rises, but can be under 40 watts, according to one of the diagrams reprinted at &lt;a href=&quot;https://www.centauri-dreams.org/?p=10123&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.centauri-dreams.org/?p=10123&lt;/a&gt; .&lt;/p&gt;&#xA;" OwnerUserId="17083" LastActivityDate="2017-06-07T09:54:13.833" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21229" PostTypeId="2" ParentId="21224" CreationDate="2017-06-07T14:35:00.617" Score="0" Body="&lt;p&gt;Think of what happens since the beginning. At first, at very large scale, you have a region of space a little more dense than what is around it, and it starts attracting more and more gas, forming a cloud. Gas falls in, in a spiral movement, giving an overall spin to the thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the meantime, stars form, most of them have a movement that follows the one of the initial gas cloud. However, gravity does its work and bring more and more stars to the core. They gather little by little in the &quot;galactic plane&quot; - I just learned this is due to &lt;a href=&quot;https://astronomy.stackexchange.com/q/1145/2153&quot;&gt;angular momentum conservation&lt;/a&gt; - could be similar thing that makes planets a little flatter at the poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand exactly how the arms and the bar form, but the thing is you have this globally revolving stars population, and more and more of them coming towards the galactic center under the effect of gravity ; &lt;a href=&quot;https://physics.stackexchange.com/questions/25706/density-of-stars-near-the-center-of-the-milky-way&quot;&gt;here&lt;/a&gt; is explained for our own galaxy, within 100 parsec from the Galactic center there is a stellar density about a 100 times higher as the one experienced in the vicinity of our sun)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;stars revolving far from the center will keep the overall rotating around the center&lt;/strong&gt; movement because they stay roughly unperturbed by each other - everything can change there if a merger with a bigger or similar-sized galaxy, and it can become an elliptical and erratically moving structure in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, a densely packed &lt;strong&gt;core&lt;/strong&gt; means there are more and more chances for the stars to do close fly-by and change there course erratically. This also happens in star clusters. This explain why in the center, &lt;strong&gt;there is no clear overall movement as the one you observe on the galactic arms&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11264" LastEditorUserId="11264" LastEditDate="2017-06-08T06:15:33.920" LastActivityDate="2017-06-08T06:15:33.920" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21230" PostTypeId="1" CreationDate="2017-06-07T15:37:36.453" Score="3" ViewCount="88" Body="&lt;p&gt;I'm looking for an introductory textbook on the Magellanic Clouds, something suited for a student about to obtain their Physics BS degree (ie: &lt;em&gt;no&lt;/em&gt;t a professional astronomer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware of &lt;a href=&quot;http://adsabs.harvard.edu/abs/1997macl.book.....W&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Magellanic Clouds, Westerlund (1997)&lt;/a&gt;, but that book is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) Not introductory&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) 20 years old&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any newer and ideally more beginner friendly textbooks out there?&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="9527" LastEditDate="2020-09-25T08:09:16.163" LastActivityDate="2020-09-25T08:09:16.163" Title="Introduction to the Magellanic Clouds?" Tags="&lt;galaxy&gt;&lt;resource-request&gt;&lt;magellanic-cloud&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21231" PostTypeId="1" AcceptedAnswerId="21232" CreationDate="2017-06-08T08:52:35.567" Score="2" ViewCount="346" Body="&lt;p&gt;Last night I was randomly surfing astronomy pages and came across this.I searched up Wikipedia and other sites,but it doesn't exactly give me enough information about their characteristics.&lt;/p&gt;&#xA;" OwnerUserId="17116" LastEditorUserId="616" LastEditDate="2021-03-04T02:34:55.910" LastActivityDate="2021-03-04T02:34:55.910" Title="What are Thorne-Żytkow objects?" Tags="&lt;neutron-star&gt;&lt;red-giant&gt;&lt;thorne-zytkow-object&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21232" PostTypeId="2" ParentId="21231" CreationDate="2017-06-08T09:52:48.047" Score="7" Body="&lt;p&gt;I'll try to add a bit of context to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Thorne%E2%80%93%C5%BBytkow_object&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt;, though the major references are all available there.  The article covers things like the potential formation mechanisms but I guess doesn't really put them into scientific perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider, for a moment, a red giant of one solar mass, far up the red giant branch. It has a deep convective envelope, at the bottom of which there is a shell of hydrogen fusing into helium, all surrounding an inert, highly degenerate helium core.  The core at this point is much like an isolated white dwarf.  So, back in the late 1970's, Kip Thorne and Anna Żytkow tried to model &lt;a href=&quot;http://adsabs.harvard.edu/abs/1977ApJ...212..832T&quot; rel=&quot;nofollow noreferrer&quot;&gt;(Thorne &amp;amp; Żytkow 1977)&lt;/a&gt; what would happen if you replaced the white dwarf with a neutron star.  Such hypothetical objects thus became known as &lt;em&gt;Thorne-Żytkow objects&lt;/em&gt; (TZOs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TZOs have mostly remained theoretical exotica, at least in part because it's very difficult to distinguish them from &quot;ordinary&quot; red supergiants. The main giveaway would be the presence of unusual elements in the atmosphere, produced by the much hotter nuclear reactions happening in the fusion shell around the neutron star core. There have been just a handful of papers published about them, almost all by theoretical groups, in the intervening 40 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only candidate that I think anyone takes seriously is &lt;a href=&quot;https://en.wikipedia.org/wiki/HV_2112&quot; rel=&quot;nofollow noreferrer&quot;&gt;HV 2112&lt;/a&gt;. As far as I know, &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014MNRAS.443L..94L&quot; rel=&quot;nofollow noreferrer&quot;&gt;Levesque et al. (2014)&lt;/a&gt; found the object serendipitously as part of a survey of red supergiants in the SMC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;PS, following comment:&lt;/em&gt;&#xA;The consensus view is that a single star couldn't form a TZO because if the core collapses into a neutron star, models predict that the envelope will be ejected.  That is, single star's only evolve into &quot;naked&quot; neutron stars.  Under this assumption, the only way to form a TZO is to have a neutron star collide with or accrete enough material from another star. This could happen if two wandering stars collide. Alternatively, it could happen if one star in a binary collapses into a neutron star.  In particular, it's known that the collapse process is probably not symmetric, so the neutron star gets a &quot;kick&quot; in some direction. If the neutron star is kicked into the companion, it will presumably undergo some viscous drag and settle into the centre.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastEditorUserId="1577" LastEditDate="2017-06-09T12:29:34.317" LastActivityDate="2017-06-09T12:29:34.317" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21233" PostTypeId="1" CreationDate="2017-06-08T15:13:42.520" Score="0" ViewCount="104" Body="&lt;p&gt;Are there physical phenomena, such as wave function collapse or black holes, which irrevocably destroy information about their prior states?&lt;/p&gt;&#xA;" OwnerUserId="17117" LastEditorUserId="10172" LastEditDate="2017-06-08T20:43:03.070" LastActivityDate="2017-06-09T12:54:18.507" Title="Are there any physical phenomena that destroy information?" Tags="&lt;black-hole&gt;&lt;cosmology&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21234" PostTypeId="1" CreationDate="2017-06-08T16:51:32.163" Score="10" ViewCount="236" Body="&lt;p&gt;As the fateful day draws closer, the United States will host a celestial event that comes once in three generations. I plan to witness this first hand at the exact center of greatest eclipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the planning of this, I have immersed myself in the work of Eddington back in 1919, confirming Einstein’s General Theory of Relativity (and its explanations for gravity) was right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally, could not just stand-by and just look at this event and not do something more. Knowing that failure, due to weather, equipment, and or other forces are stacked against me, I have to make an attempt because a chance like this will never present itself again in my lifetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal in coming here today is simple, I need some advice from someone who knows telescopes and cameras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equipment I plan on using is :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a 150 mm Celestron 6SE with an automatic GOTO mount, and an equatorial wedge. &lt;/li&gt;&#xA;&lt;li&gt;The camera will be a Sony A 6000, with an adapter.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;TThe field of view is large because of the 1500 mm focal of the telescope. I chose device because of a trade-off that has to be made, Eddington used a 100 mm refractor scope when conducting his experiment. With any shift in the apparent position of the stars behind the sun as Einstein predicted, the 150 mm telescope should, in theory, coupled with a camera that has a 24-megapixel sensor, have the necessary resolution to image this shift like Eddington's image. If any shift to the stars is to be captured, they will appear within close proximity surrounding the eclipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My problem is as follows, the telescope I will be using, will be the lens, and now the lens has only one f-stop. At f/10 and only having the ISO and shutter speed to work with, what exposure would I use ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing that a long exposure would only manage to capture the outer corona, and possibly engulfing and obscuring any stars that shift. Likewise, a shorter exposure would capture the inner corona and thus not image any stars at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do however plan to bracket 5 stops (shutter speeds) to better my odd of catching any shift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would anyone have any other useful tips trick to further better my odds in my endeavor? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks In advance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob Brooks &lt;/p&gt;&#xA;" OwnerUserId="17118" LastEditorUserId="10196" LastEditDate="2017-06-08T23:25:46.670" LastActivityDate="2017-06-08T23:25:46.670" Title="Planning to revisit Arthur Eddington's experiment from 1919" Tags="&lt;amateur-observing&gt;&lt;fundamental-astronomy&gt;&lt;photography&gt;" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21235" PostTypeId="2" ParentId="21211" CreationDate="2017-06-08T18:44:31.887" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Do you have any reputable estimates as to how much helium has been out&#xA;  gassed over the last 4 billion years&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reputable estimates - no.  Bad guesses, maybe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short answer:   only trace amounts.  It's impossible to have enough rare radioactive material to produce a real helium atmosphere.  Even if the planet retains all it's helium from radioactive decay it should be measurable in parts per million, not a real atmosphere.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of alpha particles that a radioactive element creates is pretty straight forward, just subtract the &lt;a href=&quot;http://particleadventure.org/hadrons.html&quot; rel=&quot;noreferrer&quot;&gt;Baryons&lt;/a&gt; in the final state from the initial state.   Beta decay changes the type of hadron, but not the number, so, from &lt;a href=&quot;https://en.wikipedia.org/wiki/Decay_chain#Uranium_series&quot; rel=&quot;noreferrer&quot;&gt;Uranium 238 into Lead 206&lt;/a&gt; - loss of 32 baryons, and all of them through Alpha decay - Alpha particles become helium atoms, so each U-238 decays into 8 helium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Decay_chain#Thorium_series&quot; rel=&quot;noreferrer&quot;&gt;Thorium 232 into Lead 208&lt;/a&gt; - 6 helium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Decay_chain#Actinium_series&quot; rel=&quot;noreferrer&quot;&gt;Uranium 235 into Lead 207&lt;/a&gt; - 7 helium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Potassium-40&quot; rel=&quot;noreferrer&quot;&gt;Potassium 40&lt;/a&gt;, the other common radioactive element only undergoes beta decay into CA40 (89.3%) and AR40 (10.7%), so it's irrelevant in the helium calculation.  There are smaller amounts of other elements that emit alpha particles, like &lt;a href=&quot;https://en.wikipedia.org/wiki/Radiogenic_nuclide#Radiogenic_nuclides_used_in_geology&quot; rel=&quot;noreferrer&quot;&gt;PT-190&lt;/a&gt;, but they're rare enough to be ignored.   The bigger unknown is how many elements with shorter half-life were common when the Earth was young but are no longer present.   I have no idea how to calculate that so that will be a bit of an unknown.  One possible approach would be to say that the forming planet would be sufficiently hot to lose much of it's helium, but that's a bit of a cheat.   I think, early on, predictions are rough due to unknown quantities of shorter-half life elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;U-238 is easiest to calculate with a half-life of about the age of the Earth, so about half of Earth's U-238 has completed its decay change and produced 8 helium per U-238.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thorium-232, with a half life of about 14 billion years, so only about 20% of the initial thorium-232 has decayed over the age of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and U-235, with a half-life of about 700 million years, means that almost 99% of it has decayed over the age of the Earth.  Current estimates say that about 0.72% of Uranium is U-235 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I touched on this above, but the big unknown for this scenario is in the range of 100 million year half-life.   That is, elements that are long-lasting enough to survive the Earth's formation but wouldn't remain in high concentrations today.   That's &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_radioactive_isotopes_by_half-life#1015_seconds&quot; rel=&quot;noreferrer&quot;&gt;in this range&lt;/a&gt;, bottom of the 10^12 seconds, top of the 10^15 seconds.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Curium-247 has a half-life of about 15 million years and it decays into lead-207 releasing 10 alpha particles in the process.   If planet formation happens relatively quickly, that could supply some helium to a planet, but I'm going to ignore it as having slightly too short a half-life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uranium-236 has a half life of about 23 million years and it decays into Thorium, emitting just 1 alpha particle.  Presumably much of it would become thorium by the time of planet formation, so I'm going to ignore it as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Plutonium-244&quot; rel=&quot;noreferrer&quot;&gt;Plutonium-244&lt;/a&gt;, mostly synthetic, but it is the rarest primordial element on Earth.   From the Wikipedia article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Accurate measurements, beginning in the early 1970s, have detected&#xA;  primordial plutonium-244, making it the second-shortest-lived&#xA;  primordial nuclide after 146Sm. The amount of 244Pu in the pre-Solar&#xA;  nebula (4.57×109 years ago) was estimated as 0.008 of amount of&#xA;  238U.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If estimates put it at a bit less than 1% as common as U-238, I'm going to ignore as well, as 1% won't make a big difference.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, &lt;a href=&quot;https://en.wikipedia.org/wiki/Isotopes_of_samarium#Samarium-146&quot; rel=&quot;noreferrer&quot;&gt;Samarium-146&lt;/a&gt; is worth considering with a half-life of about 100 million years (or 68 million depending on which source you use).   It decays into &lt;a href=&quot;https://en.wikipedia.org/wiki/Isotopes_of_neodymium&quot; rel=&quot;noreferrer&quot;&gt;Neodymium-142&lt;/a&gt; releasing a single alpha particle.   Samarium 147 and 148 also undergo alpha-decay but very slowly, half-life of 100 billion and 7 quadrillion respectively.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So these are the 4 primary helium &quot;outgasing&quot; isotopes relevant after planet formation.    U-238 (8 helium), U-235 (7), Th-232 (6) and SM-146 (1).    And as for how much, you have to ask how long the planet has been around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SM-146 decays fairly quickly, half life 100 million years.&#xA;U-235 - 700 million years&#xA;U-238 - 4.5 billion years&#xA;T-232 - 14 billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take a planet about the age of Earth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SM-146 - essentially all of it&#xA;U-235 - 99%&#xA;U-238 - 50%&#xA;T-232 - 20%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the trick is, how much of these elements is a terrestrial planet likely to have at formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we go by &lt;a href=&quot;http://www.livescience.com/15084-radioactive-decay-increases-earths-heat.html&quot; rel=&quot;noreferrer&quot;&gt;radioactive decay&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth%27s_internal_heat_budget#Radiogenic_heat&quot; rel=&quot;noreferrer&quot;&gt;radiogenic heat&lt;/a&gt; estimates, TH-232 and U-238 combine to generate 20 trillion watts (first article).  That' 6 x 10^20 joules per year or about 3.75 x 10^33 &lt;a href=&quot;http://www.convertunits.com/from/MeV/to/Joule&quot; rel=&quot;noreferrer&quot;&gt;mEV&lt;/a&gt; per year, by radioactive decay of those 2 elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Decay_chain#Thorium_series&quot; rel=&quot;noreferrer&quot;&gt;Thorium decay chain emits&lt;/a&gt; about 42.6 MeV per decay for 7 alpha particles.  (about 7 MeV per helium) and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Decay_chain#Uranium_series&quot; rel=&quot;noreferrer&quot;&gt;U-238 decay chain&lt;/a&gt;, about 51.7 MeV for 8 alpha particles.  (ballpark 6.5 MeV).   That includes Neutrinos though and that energy needs to be factored out.   I couldn't find precise numbers but about 10 MeV (or 20%) is lost into Neutrinos (technically anti-neutrinos) &lt;a href=&quot;http://www.nuclear-power.net/nuclear-power/fission/energy-release-from-fission/&quot; rel=&quot;noreferrer&quot;&gt;by this website&lt;/a&gt;.   So, very ballpark, lets say 5 MeV heat creates one alpha particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;current rates of decay of Thorium &amp;amp; U-238 should produce about 7.5 x 10^32 alpha particles, which corresponds to about 1.2-1.3 billion moles of Helium or about 5 million KG, between 5 and 6 tons of helium produced in the Earth's crust and mantle every year - based on radioactive measurements which are probably pretty accurate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thorium is currently about (ballpark) 4 times as abundant as U-238 but because it decays 3 times more slowly the helium production and heat generation in the Earth's core is similar (&lt;a href=&quot;https://en.wikipedia.org/wiki/Earth%27s_internal_heat_budget#/media/File:Evolution_of_Earth%27s_radiogenic_heat.svg&quot; rel=&quot;noreferrer&quot;&gt;this chart suggests&lt;/a&gt; thorium's heat output is higher, but they are close).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The helium production from Thorium has been pretty consistent, as 80% of the Thorium at the formation of the planet is still here.&#xA;The production from U-238 was about twice as high when the Earth formed.&#xA;The production from U-235 is currently negligible compared to the other two, but was much higher when the Earth was young.   working back from U-235 being about 0.7% of the total Uranium and it having gone through between 6 and 7 half-lifes since the formation of the Earth, (and accounting for U-238 being twice as abundant when the Earth formed), that means between 32 and 64 (lets say 50) x 0.7, so there was some 30-40% as much U-235 as U-238 when the Earth formed.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bad math again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thorium, 2.5-3 tons per year now, 25% more than that when Earth formed.  lets say 3 tons per year, average&lt;/p&gt;&#xA;&#xA;&lt;p&gt;U-238, 2.5 tons per year now, twice that when the earth formed.   lets ballpark average that out to 3.5-4 tons per year on average.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not going to average out U-235 because it starts out too high and decays too quickly but if there was about 30-40% as much U-235 as U-238 when the Earth formed I can just multiply U-238's numbers by 35% and by 7/8, working out to about 1 ton per year average.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm doing per year/on average just for quick bad math, so 3 + 3-4 + 1 = about 7.5 tons per year x 4.5 billion years - quick estimate for how much Helium was produced by those 3 elements - very bad estimate, 33 billion tons.   Most of this Helium is produced and would be trapped in the Earth's interior until plate tectonics allows it to escape.  The rate of escape is above my pay-grade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Samarium-146 is the hardest one to estimate.   It's essentially all decayed into Neodymium-142 but both Samarium (it has several stable isotopes) and Neodymium are abundant, and this chain produces a single helium.   It likely doesn't make a huge difference in the total.   Table of &lt;a href=&quot;https://en.wikipedia.org/wiki/Abundance_of_elements_in_Earth%27s_crust#cite_note-3&quot; rel=&quot;noreferrer&quot;&gt;elemental abundance in Earth's crust&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atmospherically speaking, 33 billion tons is nothing.   Even if we adjust to to 40 or 60 or 300 billion tons, that's still essentially too little to be an atmosphere.  Helium is currently &lt;a href=&quot;https://en.wikipedia.org/wiki/Helium&quot; rel=&quot;noreferrer&quot;&gt;about 5.2 ppm&lt;/a&gt; in the atmosphere and giving the Earth's atmosphere a mass of about 5,600 trillion tons, 5.2% by molecule, adjusted for the mass ratio of helium (4) to Nitrogen/Oxygen (29), there's about 4 billion tons of Helium in the Earth's atmosphere today.   If we were to say that 400 billion tons of helium was produced on Earth by radioactive decay (about 10 times my estimate), that's still a trace amount, just 100 times the current amount or 520 PPM, which is less than the Argon (&lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Eart&quot; rel=&quot;noreferrer&quot;&gt;currently 9,300 PPM&lt;/a&gt;).   Argon-40 which comes from the radioactive decay of &lt;a href=&quot;https://en.wikipedia.org/wiki/Isotopes_of_potassium&quot; rel=&quot;noreferrer&quot;&gt;Potassium-40&lt;/a&gt; (about 10% of the time), is more common because Potassium-40 is much more common than the Uranium, Thorium and Samarium-146 that decay alpha-particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a very long answer that basically says there's not enough Alpha-particle emitting elements that last long enough to survive planetary formation, that produce anything more than a trace amount of helium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth also retains some terrestrial helium, &lt;a href=&quot;https://en.wikipedia.org/wiki/Helium-3#Terrestrial_abundance&quot; rel=&quot;noreferrer&quot;&gt;about 7% of Earth's helium&lt;/a&gt; is terrestrial, 93% formed by radioactive decay.   Most of Earth's terrestrial helium is in it's mantle, very little in it's crust. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some interesting and loosely related reading.  &lt;a href=&quot;http://articles.latimes.com/2014/feb/19/science/la-sci-sn-yellowstone-helium-degassing-20140218&quot; rel=&quot;noreferrer&quot;&gt;Yellowstone is able to release more helium&lt;/a&gt; than the radioactive material produce because it's crust is unique, very old and it's trapped helium for over 2 billion years, only recently started to release it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As to your original question.  As I understand it, a planet needs to be very large to acquire significant amounts of helium.     By particle, the universe is about 92% hydrogen and 8% helium (or 75%-25% by mass).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.space.com/18385-jupiter-atmosphere.html&quot; rel=&quot;noreferrer&quot;&gt;Jupiter's outer atmosphere&lt;/a&gt; is slightly helium richer than that, about 10% helium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uranus' uppermost atmosphere &lt;a href=&quot;https://www.space.com/18708-uranus-atmosphere.html&quot; rel=&quot;noreferrer&quot;&gt;is about 15% helium&lt;/a&gt;, but only it's outer edge.   Uranus, unlike Jupiter and Saturn, isn't primarily a hydrogen-helium planet, it's more of a water/ice planet, sometimes called an ice giant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that planets need to be very massive to obtain significant amounts of the most common elements, hydrogen and helium, then, as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Helium_planet&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; implies, they need to get hot to lose their hydrogen, so helium atmosphere planets probably need to be quite large, perhaps more massive than Neptune.   It might be quite a it more difficult for a typical terrestrial planet to do that, unless if it was very massive and a terrestrial planet that massive might resemble a gas giant anyway due to a very thick atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a tough time thinking of any other way a helium atmosphere is likely to happen, unless you got the planet below 20 degrees K and all the other gases turned to liquid or ice, but I suspect that's not the answer you're looking for.  A planet probably needs to be quite large to retain significant amounts of helium I would think. outside of a very odd circumstance like a helium white dwarf colliding with another large object inside a stellar nursery, even then, helium gas doesn't clump like ices and rocky material, but maybe such a scenario could put enough helium into the planetary formation.  (bit of a crazy what-if), and I'm not sure I buy it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to reason this one out so, corrections are welcome and encouraged.   I'll try to clean up any grammar a little later.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-06-08T18:44:31.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21236" PostTypeId="1" CreationDate="2017-06-08T20:39:07.123" Score="2" ViewCount="969" Body="&lt;p&gt;Taking Hubble's law into account, what is the velocity of the most distant galaxies relative to Earth? Yes, I know that it is actually space that is expanding, (so, it is not their real velocity), but at what speed relative to Earth are they receding from us?&lt;/p&gt;&#xA;" OwnerUserId="17081" LastEditorUserId="2365" LastEditDate="2017-06-10T06:00:48.640" LastActivityDate="2017-06-20T17:37:19.393" Title="Velocity of most distant galaxies" Tags="&lt;expansion&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21237" PostTypeId="1" CreationDate="2017-06-08T23:19:16.643" Score="2" ViewCount="86" Body="&lt;p&gt;On Mars, we refer to a &quot;day&quot; as a &quot;sol&quot;. NASA has developed a whole vocabulary around this term, like &quot;yestersol&quot; and &quot;tosol&quot;, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do any other worlds have unique terms like this? How would one refer to a day on Venus, for example, or Mercury? Is &quot;sol&quot; used to avoid confusion with &quot;day&quot;, like on Mars, or is &quot;sol&quot; an exclusively Martian term and &quot;day&quot; should be used for any other world until a unique term is agreed upon?&lt;/p&gt;&#xA;" OwnerUserId="529" LastActivityDate="2017-06-09T06:06:51.560" Title="Which worlds have unique words for a &quot;day&quot;?" Tags="&lt;mars&gt;&lt;venus&gt;&lt;time&gt;&lt;standards&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21238" PostTypeId="1" CreationDate="2017-06-08T23:24:34.497" Score="6" ViewCount="216" Body="&lt;p&gt;For a planet, we can colloquially refer to its period of rotation as a &quot;day&quot; and its period of revolution around its parent star as a &quot;year.&quot; Some worlds have unique terms, such as Martian days being referred to as &quot;sols&quot;, but the principle is the same: one word for rotation around its axis, another for revolutions around the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an equivalent word for the period of a moon's revolution around a planet? On Earth it roughly lines up with a &quot;month,&quot; but multi-moon planets would have different periods for each moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I was in charge of timekeeping for a mission to Europa or Ganymede, what term would I use to refer to the period of time it takes for the moon to complete one orbit of Jupiter?&lt;/p&gt;&#xA;" OwnerUserId="529" LastActivityDate="2020-01-02T13:47:23.630" Title="Is there a timekeeping word for the orbit of a moon?" Tags="&lt;the-moon&gt;&lt;natural-satellites&gt;&lt;time&gt;&lt;standards&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21239" PostTypeId="2" ParentId="21238" CreationDate="2017-06-09T05:42:40.050" Score="4" Body="&lt;p&gt;You should use one of the precisely defined terms of specific measures of orbital period, which are described on &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_period&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Wikipedia page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might get away with just &quot;orbital period&quot; is you're not using it in a context that requires precision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For length of day there's &lt;a href=&quot;https://en.wikipedia.org/wiki/Synodic_day&quot; rel=&quot;nofollow noreferrer&quot;&gt;synodic day&lt;/a&gt; as well as sidereal period.  There's a discussion on the difference between these &lt;a href=&quot;http://cseligman.com/text/sky/rotationvsday.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-06-09T05:42:40.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21240" PostTypeId="2" ParentId="21237" CreationDate="2017-06-09T06:06:51.560" Score="4" Body="&lt;p&gt;Word are created in response to need. There are only two planets on which we have any significant presence on the surface: Earth and Mars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When running operations on Mars, the day/night cycle is significant, so a word was needed for the length of this cycle. Especially since the sol is so close to the day, there is greater potential for confusion. Other planets either have very long rotation periods or no solid surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On, for example, Mercury you can use &quot;day&quot; until there is a need for a new term.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-06-09T06:06:51.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21241" PostTypeId="2" ParentId="21233" CreationDate="2017-06-09T12:54:18.507" Score="2" Body="&lt;p&gt;This is one of the current questions of cosmology/physics.  So far all evidence points to the negative inside a black hole so far as quantum numbers go, but it's not clear what you're considering as &quot;prior state.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple counter-example is a camera. Once the photons are absorbed by the film or solid-state sensor, all phase information is lost.  Is that what you were looking for?   &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-06-09T12:54:18.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21242" PostTypeId="1" CreationDate="2017-06-09T13:36:08.370" Score="2" ViewCount="64" Body="&lt;p&gt;I am a layman in astronomy. Can someone please explain it to me whether the M-theory can be used to explain what is happening in the centre of the black hole where the singularity exists? Or is it still necessary to peek inside a black hole to gain that information? &lt;/p&gt;&#xA;" OwnerUserId="16495" LastActivityDate="2017-06-09T19:35:21.277" Title="The M-theory to understand black holes" Tags="&lt;fundamental-astronomy&gt;&lt;radio-astronomy&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2017-06-12T16:14:46.807" ContentLicense="CC BY-SA 3.0" />
  <row Id="21245" PostTypeId="2" ParentId="6336" CreationDate="2017-06-09T17:02:48.710" Score="2" Body="&lt;p&gt;There is a GCSE in astronomy offered by Edexcel: &lt;a href=&quot;https://qualifications.pearson.com/en/qualifications/edexcel-gcses/astronomy-2017.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://qualifications.pearson.com/en/qualifications/edexcel-gcses/astronomy-2017.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The assessment has two parts, an exam, and a series of two observation task (aided and unaided).&lt;/p&gt;&#xA;&lt;p&gt;The course is suitable for 14-16-year-olds.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2020-11-28T01:03:34.983" LastActivityDate="2020-11-28T01:03:34.983" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21246" PostTypeId="2" ParentId="21242" CreationDate="2017-06-09T19:35:21.277" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can someone please explain it to me whether the M-theory can be used&#xA;  to explain what is happening in the centre of the black hole where the&#xA;  singularity exists?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In Layman astronomy - yes, M-Theory (and probably a few string theories) do explain in detail what happens at the singularity, but it's a mathematical construct, not a testable hypothesis, so what it tells you might be wrong and little more than creative but irrelevant mathematics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking inside a black hole, well, that's not a good idea either.  Even if you were able to see what happened, you wouldn't be able to send a message back telling anyone, so that's also no help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum physics is all a model anyway, but it's a &lt;strong&gt;good&lt;/strong&gt; model because it provides accurate answers and it predicted the existence of particles that were later discovered.   Relativity is a good model too that stood up to testing.   Good models are useful and highly accurate and provide answers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;M-theory is a mathematical construct.  That's a different thing entirely.   It could still end up being right, but until it's testable, it's really just a mathematical exercise.   They really shouldn't call it a theory, it should be M-Hypothesis and String-Hypothesis.  They aren't real theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I may quote Shakespeare from Henry IV, just for fun:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;GLENDOWER&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I can call spirits from the vasty deep.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &#xA;  &lt;p&gt;HOTSPUR&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Why, so can I, or so can any man;&#xA;But will they come when you do call for them?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm personally much more comfortable with the &quot;&lt;em&gt;both accepted theories, relativity and quantum theory, break down at the singularity, so we don't know what happens&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you prefer that M-Theory says what happens, preferring that answer is entirely optional, but I think the majority of people in this field are more comfortable with the we don't know answer.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-06-09T19:35:21.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21247" PostTypeId="2" ParentId="21236" CreationDate="2017-06-09T19:54:17.643" Score="3" Body="&lt;p&gt;The speed relative to Earth is greater than the speed of light. This is consistent with Relativity as the space between us and the distant galaxies has stretched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\mathrm{Hubble\ constant}) \times (\mathrm{radius\ of\ the\ observable\ universe\ in\ Mpc})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which, as noted by Pela in a comment, is roughly 3.3 times the speed of light.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-06-09T19:54:17.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21249" PostTypeId="1" AcceptedAnswerId="21253" CreationDate="2017-06-10T14:17:21.297" Score="0" ViewCount="3359" Body="&lt;p&gt;Can it still be considered theory ?, because I can't find it's major or significant assumptions.&lt;/p&gt;&#xA;" OwnerUserId="17130" LastActivityDate="2017-06-10T20:00:08.790" Title="What are the significant assumption of the Inflation Theory?" Tags="&lt;cosmological-inflation&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2017-06-12T21:49:36.483" ContentLicense="CC BY-SA 3.0" />
  <row Id="21250" PostTypeId="1" AcceptedAnswerId="21254" CreationDate="2017-06-10T18:41:12.100" Score="4" ViewCount="70" Body="&lt;p&gt;In relation to possible asteroid mining on the moon, where goes asteroid mass after the impact? Will it be burried, scathered, or evaporated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iJ7Ss.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iJ7Ss.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/1ykhi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1ykhi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/WWbN4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WWbN4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Atrll.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Atrll.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ausew.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ausew.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17131" LastActivityDate="2017-06-10T20:12:35.140" Title="Where goes original asteroids mass after impact?" Tags="&lt;asteroids&gt;&lt;impact&gt;&lt;crater&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21251" PostTypeId="1" CreationDate="2017-06-10T19:12:05.883" Score="3" ViewCount="151" Body="&lt;p&gt;I had this question on a test and got it wrong but I can't understand the explanation.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/EkHnQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EkHnQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't get why we receive a Paschen series absorption. Could anyone explain?&lt;/p&gt;&#xA;" OwnerUserId="17134" LastEditorUserId="7472" LastEditDate="2017-07-19T10:39:16.620" LastActivityDate="2017-07-19T10:39:16.620" Title="Spectroscopy question about a Paschen series absorption" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21252" PostTypeId="1" AcceptedAnswerId="21257" CreationDate="2017-06-10T19:18:35.100" Score="9" ViewCount="1234" Body="&lt;p&gt;As far as I understand, in the events detected by LIGO, about 4% of the total mass of merging binary black holes was converted to gravitational waves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does this energy come from, i.e. what exactly gets converted into gravitational waves? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it simply the kinetic energy of the merging objects (velocities of these objects before merger are huge, up to 60% of c if I recall correctly), so does it mean emitting gravitational waves makes them orbit slower, but retain their original masses? Or do the compact objects really lose &quot;real&quot; mass, meaning they become lighter and in case of BHs their radius changes accordingly? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, let's assume two BHs, both with 50 solar masses, orbiting each other far enough (say 1 light year) so that GWs nor kinetic energy has no significance to these initial mass measurements. During the merge, they should radiate about 5 solar masses in GWs. Would the resulting black hole have mass of 95 or 100 solar masses?&lt;/p&gt;&#xA;" OwnerUserId="10813" LastEditorUserId="10813" LastEditDate="2017-06-10T19:36:49.330" LastActivityDate="2017-06-15T23:26:31.370" Title="From where does the energy for gravitational waves come from?" Tags="&lt;black-hole&gt;&lt;gravitational-waves&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21253" PostTypeId="2" ParentId="21249" CreationDate="2017-06-10T19:24:49.290" Score="2" Body="&lt;p&gt;The most significant assumption is that very shortly after the Big Bang, the universe expanded very rapidly for a short period of time. Much faster than normal expansion due to Big Bang. It was an exponential expansion and occurred around ${10}^{-32}$ seconds after Big Bang and lasted for tiny fraction of second. The universe continues to expand but not at that exponential rate. Yes it is still a theory and is accepted by many scientists.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="16926" LastEditDate="2017-06-10T20:00:08.790" LastActivityDate="2017-06-10T20:00:08.790" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21254" PostTypeId="2" ParentId="21250" CreationDate="2017-06-10T20:12:35.140" Score="2" Body="&lt;p&gt;All three.  The impact velocity will be huge and the kinetic energy of astroid extremely large. The asteroid will be pulverized into millions of fragments of asteroid. Many of these will be as small as dust particles and will be vaporized from the heat associated with collision. Hence some small fragments will evaporate will other larger fragments will be buried or blown away from impact site. The scatter field is an ellipse about the crater elongated along direction of motion of meteor.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastActivityDate="2017-06-10T20:12:35.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21255" PostTypeId="1" CreationDate="2017-06-11T12:34:14.010" Score="0" ViewCount="150" Body="&lt;p&gt;When were the Alt-az system and the RA-dec system first used?&lt;/p&gt;&#xA;" OwnerUserId="17139" LastEditorUserId="2365" LastEditDate="2017-06-12T06:05:54.917" LastActivityDate="2017-06-12T11:25:19.190" Title="When were the horizontal and equatorial celestial coordinate systems invented?" Tags="&lt;star&gt;&lt;observational-astronomy&gt;&lt;history&gt;&lt;navigation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21256" PostTypeId="2" ParentId="21255" CreationDate="2017-06-11T14:59:14.300" Score="0" Body="&lt;p&gt;Astrolabe, somehow can be taught as recent sky apps in which you can anticipate the exact position of stars, a very older version of course.But sextant is actually a tool for measuring angles (altitude of sun or any other star). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These definitions should not be considered so strict. Since ancient astronomers probably have used astrolabe as a tool for their research as well.&lt;/p&gt;&#xA;" OwnerUserId="7761" LastActivityDate="2017-06-11T14:59:14.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21257" PostTypeId="2" ParentId="21252" CreationDate="2017-06-11T19:00:50.610" Score="7" Body="&lt;p&gt;Radiating gravitational waves makes an inspralling binary orbit closer and faster. (Rob Jefferies)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The source of the energy for both increased kinetic energy, and the gravitational radiation is the same: gravitational potential energy. (PM 2Ring)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two black holes at a distance of 1 light year have a huge amount of potential energy, about 10^48 Joules of potential energy. As they spiral, a significant amount of that energy is radiated as gravitational waves&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is real mass lost. The mass of resulting black hole is smaller than the sum of the two merging black holes, though at no point does any black hole itself become smaller.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-06-11T19:00:50.610" CommentCount="3" CommunityOwnedDate="2017-06-11T20:43:51.520" ContentLicense="CC BY-SA 3.0" />
  <row Id="21258" PostTypeId="2" ParentId="18769" CreationDate="2017-06-11T19:49:20.110" Score="2" Body="&lt;p&gt;We don't know for sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we knew the answer, then we'd have a new standard model and then the reconnection speed would be nicely comparable to what's predicted. (zibadawa timmy)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the typical models of magnetic reconnection do not account for the extra resistivity that results from small-scale electromagnetic waves, which can be upwards of ~12-15 orders of magnitude larger than that caused by Coulomb collisions. (honeste vivere)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-06-11T19:49:20.110" CommentCount="1" CommunityOwnedDate="2017-06-11T19:49:20.110" ContentLicense="CC BY-SA 3.0" />
  <row Id="21260" PostTypeId="2" ParentId="20362" CreationDate="2017-06-12T03:52:39.537" Score="0" Body="&lt;p&gt;Yes, these celestial bodies in certain positions can produce this alignment, but the brightness of the moon is the dead give away, besides the fact that the sun couldn't possibly illuminate the side of the moon facing earth from 93 million miles behind it. At least not with this brightness. &lt;/p&gt;&#xA;" OwnerUserId="17146" LastActivityDate="2017-06-12T03:52:39.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21261" PostTypeId="1" AcceptedAnswerId="21266" CreationDate="2017-06-12T04:21:26.333" Score="6" ViewCount="2028" Body="&lt;p&gt;I recently learned (if I understood correctly) that the famous Ultra Deep Field photo(s) required something like a 10-day exposure on a very tiny piece of the sky to achieve. How can this occur when the telescope is constantly in motion zipping around earth at ~17,000 mph? Wouldn't the earth and/or possibly the moon get in the way every 90 minutes or so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose my question applies to just about any photo it takes since it would be out of alignment of the target in less than a second. &lt;/p&gt;&#xA;" OwnerUserId="10117" LastEditorUserId="10117" LastEditDate="2017-06-12T15:14:13.033" LastActivityDate="2017-06-12T15:14:13.033" Title="How does the Hubble telescope capture long-exposure photos?" Tags="&lt;photography&gt;&lt;hubble-telescope&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21262" PostTypeId="2" ParentId="21261" CreationDate="2017-06-12T05:22:42.137" Score="3" Body="&lt;p&gt;The HST orbits the Earth roughly in a plane. Also, the moon orbits the Earth roughly in a plane and the Earth orbits the sun in roughly a plane, and all three of these planes are within 30° or so of each other. So there's actually a rather large slice of sky (or two slices in opposite directions, if you prefer) where there are no regular interruptions to viewing. Furthermore, without knowing the exact design of the telescope, I'd nevertheless assume that it could through means of a shutter or other methods self-interrupt during times that a known interruption would obscure the intended shot. There's no particular reason for the telescope to take a single continuous exposure, after all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as your concerns about targets becoming &quot;out of alignment,&quot; keep in mind that the HST is designed to be very sensitive for imaging very distant objects (with low apparent brightness). This means one of two things, depending on which end of the distance spectrum you're considering:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Most targets for HST are very distant - the Deep Field galaxies are tens of billions of light-years away. The total displacement of the HST over even many days is so small as to be irrelevant at such distances. It does not cause a noticeable difference in the image alignment.&lt;/li&gt;&#xA;&lt;li&gt;Nearer objects imaged by HST generally have much higher apparent brightnesses. For example, as viewed from Earth even the brightest galaxies are much dimmer than Jupiter, which HST has also imaged. Greater brightness means a shorter exposure is needed for the same amount of light to be gathered by the sensor. Furthermore, nearby objects have well-known relative movements that can be corrected for if necessary.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4042" LastActivityDate="2017-06-12T05:22:42.137" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21265" PostTypeId="2" ParentId="21255" CreationDate="2017-06-12T11:25:19.190" Score="1" Body="&lt;p&gt;I remember answering a question on the origin of declination last year (&lt;a href=&quot;https://astronomy.stackexchange.com/questions/14275/why-is-declination-positive-in-the-northern-hemisphere/14278#14278&quot;&gt;Why is declination positive in the northern hemisphere?&lt;/a&gt;) which led me to discover &lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1942PASP...54...77W&amp;amp;db_key=AST&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt; on the origin of celestial co-ordinate systems. In it they seem to indicate that the ancient Greeks were the first to use Right Ascension to describe locations in the sky around 300 B.C. &#xA;Whereas Alt-Az co-ordinates (or variants of) have been around much longer because the horizon is an easy point of reference to start from. The first documented use comes from Babylonian times but the ancient Egyptians are also thought to have used a similar system as far back as 2000 B.C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference paper: &lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1942PASP...54...77W&amp;amp;db_key=AST&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1942PASP...54...77W&amp;amp;db_key=AST&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10437" LastActivityDate="2017-06-12T11:25:19.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21266" PostTypeId="2" ParentId="21261" CreationDate="2017-06-12T13:11:43.450" Score="10" Body="&lt;p&gt;The answer to this is that such images are not taken continuously. The HST did &lt;em&gt;not&lt;/em&gt; stare at one part of the sky for 10 continuous days, but rather it stared at one part of the sky for short periods over a long time which, when added up, amounted to 10 total days of observations. To quote &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble_Ultra-Deep_Field#Observations&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; on the subject of how the Hubble Ultra Deep Field (HUDF) image was taken:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The observations were done in two sessions, from September 23 to October 28, 2003, and December 4, 2003, to January 15, 2004. The total exposure time is just under 1 million seconds, from 400 orbits, with a typical exposure time of 1200 seconds. In total, 800 ACS exposures were taken over the course of 11.3 days, 2 every orbit, and NICMOS observed for 4.5 days. All the individual ACS exposures were processed and combined by Anton Koekemoer into a single set of scientifically useful images, each with a total exposure time ranging from 134,900 seconds to 347,100 seconds. To observe the whole sky to the same sensitivity, the HST would need to observe continuously for a million years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the above, ACS is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Advanced_Camera_for_Surveys&quot; rel=&quot;noreferrer&quot;&gt;Advanced Camera for Surveys&lt;/a&gt; and NICMOS is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Near_Infrared_Camera_and_Multi-Object_Spectrometer&quot; rel=&quot;noreferrer&quot;&gt;Near-Infrared Camera and Multi-Object Spectrometer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That, I think, is the basic answer to your question, but there's a few interesting and related points I'd like to point out. First, the HUDF image was actually comprised of many individual images rather than a single 10-day exposure image (as the quote indicates). This would necessarily have to be the case, even if Hubble had the ability to stare at the HUDF region of the sky for 10 days without interruption. Cameras work by taking in light. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Charge-coupled_device&quot; rel=&quot;noreferrer&quot;&gt;Charge-Coupled Device&lt;/a&gt; (CCD) which actually collects the light (and converts it to an electric charge) can only hold so much charge before it becomes saturated. At some point, you have to &quot;stop recording&quot; and read out the CCD which clears the accumulated charge. That way, you can resume again with a fresh slate. A 10-day continuous exposure would have easily saturated Hubble's CCD. So even if Hubble did spend 10 continuous days observing this patch of the sky, it would have been broken into many, many smaller exposures. This means, even if the Moon did get in the way during part of such a 10-day exposure, they could have just not made any observations during that time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other related fact I wanted to point out is that Hubble has something known as the &lt;a href=&quot;http://www.stsci.edu/hst/proposing/documents/primer/Ch_2_Systemoverview3.html#2042824&quot; rel=&quot;noreferrer&quot;&gt;Continuous Viewing Zone&lt;/a&gt; (CVZ). This is a region of the sky which is never obscured by the Earth, Moon, Sun, etc. as Hubble orbits. As the name implies, it allows for continuous viewing, without interruption. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble_Deep_Field&quot; rel=&quot;noreferrer&quot;&gt;Hubble Deep Field&lt;/a&gt; (HDF) (the precursor to the HUDF) was observed in this region. However, there are reasons why you wouldn't want to observe in this region such as it being polluted by &quot;Earthshine&quot;.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-06-12T13:11:43.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21267" PostTypeId="2" ParentId="21210" CreationDate="2017-06-12T21:55:23.153" Score="2" Body="&lt;p&gt;I have not studied this galaxy cluster specifically. However, I can speak about systems like galaxy clusters that have a large number of bodies that are in close enough proximity to have a non-negligible gravitational effect on each other. For any such n-body system, where n is large (a solar system, a galaxy, a galaxy cluster,...), it can be assumed that, 1. there is a net angular momentum of the system, and 2. it is reasonable to assume that this angular momentum is non-negligible. I state 1 and 2 based on the following (&lt;strong&gt;very&lt;/strong&gt; general model of solar system and galaxy cluster formation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such n-body systems form by the collapse of a very large amount of gas, dust, and other minerals. When fully extended, these clouds are cold, the the velocities of any given molecule or atom in them is comparatively slow. However, it is very improbable that the total angular momentum of the cloud of gas is zero, and more importantly, that it is close to zero. From this point, it is easy to reason that, so long as no angular impulse is applied to the system and no mass is added or removed from the system, the gas cloud can collapse or disperse in whatever way is consistent with the conservation of angular momentum, and the angular momentum of the system will not change. Thus, a galaxy cluster like Abel 1689 should have a non-negligible angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have skipped over the most important logical step in the above argument. I simply assumed again that the angular momentum of the gas cloud would not be even close to zero, effectively shifting the level of the explanation up one level. Unfortunately, the best answer to &quot;why do the gas clouds have non-negligible angular momentum?&quot; requires explaining where the gas clouds came from. The extent of my understanding here is that the gas clouds that formed galaxy clusters only existed during the very early universe. Once they collapsed into galaxies, there is no process that could have caused a galaxy cluster (or even a galaxy) to disperse so completely that it forms a new very low density gas cloud that can re-collapse. Thus, the gas that made the galaxy cluster must have had an angular momentum if the galaxy cluster system is to have angular momentum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know if the sum total of the angular momentum in the universe is said to be zero or some finite value. However, I do know that even if it is zero, localized regions of matter in a system that has zero angular momentum do not also need to have zero angular momentum. Thus, the Abell 1689 system could be one such local region with non-zero angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Continuation:&lt;/strong&gt; According to &lt;em&gt;The Cosmic Perspective: Seventh Edition&lt;/em&gt;, when considering galaxies that have not yet collided with another galaxy, the initial angular momentum of the protogalactic cloud is a large contributor in determining if a galaxy will be spiral or elliptical galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding elliptical galaxy clusters, TCP has this to say: &quot;Observations of galaxies in clusters support the idea that at least some elliptical galaxies result from collisions and subsequent mergers. Elliptical galaxies dominate the galaxy populations at the cores of dense clusters of galaxies, where collisions should be most frequent. This fact may mean that any spirals once present became ellipticals through collisions.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, TCP suggests in this paragraph that at least certain clusters contain dense cores of elliptical galaxies that resulted from &lt;em&gt;collisions and mergers&lt;/em&gt; of galaxies that had markedly different angular momenta, resulting in a new galaxy that has much less angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this doesn't really answer your question either. I have looked also for scholarly articles to find some consensus on what the typical angular momentum characteristics are, and found very little. Most discussions of galaxy cluster formation discuss filament formation and collapse. The one article of note I found was from Tom Crawford at the University Of Chicago: &#xA;&lt;a href=&quot;http://askanastronomer.org/bhc/2015/10/05/hierarchy-of-the-universe/&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;askanastromomer.org&quot;&gt;askanastronomer.org&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The most obvious and uncontroversial class of structures beyond the galaxy is the galaxy group or cluster. These structures consist of tens to thousands of individual galaxies that are bound by gravity to one other and orbit a common center. That center is not defined by some huge central object, as the Sun defines the center of our solar system, but instead by the center of all the mass in the cluster, including dark matter. In fact, the dark matter dominates the mass budget of galaxy clusters, outweighing the normal matter by about a five-to-one ratio.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This description suggests more than just &quot;random&quot; motions of gravitationally bounded objects (galaxies), but does not definitively say anything about the net angular momenta of these systems.&lt;/p&gt;&#xA;" OwnerUserId="14937" LastEditorUserId="14937" LastEditDate="2017-06-13T02:25:20.813" LastActivityDate="2017-06-13T02:25:20.813" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21268" PostTypeId="2" ParentId="21204" CreationDate="2017-06-12T21:57:20.333" Score="1" Body="&lt;p&gt;Stars are, to all intents, infinitely far away and hence present themselves as points - thus the small changes in the atmosphere's refractive index caused by different patches of density and temperature of air make them appear to twinkle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets, on the other hand, are relatively close by and (with the exception of the Moon and the Sun itself) whilst we cannot resolve any solar system body into a disk with the naked eye, they are still larger than (most) variations in the atmosphere, hence they don't appear to twinkle or the twinkling is much reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The evidence suggests you saw a planet, quite likely Jupiter or Saturn.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2017-06-12T21:57:20.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21269" PostTypeId="1" CreationDate="2017-06-13T06:07:12.907" Score="-2" ViewCount="365" Body="&lt;p&gt;I know that radio telescope can &lt;em&gt;'see'&lt;/em&gt;* further than an optical telescope but I would like to know how much further? (let's say the optical has a magnification of 50x and the radio has a dish of 90m in diameter).&lt;br&gt;&#xA;*When I say see I mean how far into space image wise.&lt;/p&gt;&#xA;" OwnerUserId="17010" LastEditorUserId="17010" LastEditDate="2017-06-14T06:01:02.040" LastActivityDate="2017-06-14T06:01:02.040" Title="how much further can a radio telescope than an optical telescope" Tags="&lt;radio-telescope&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21270" PostTypeId="2" ParentId="21210" CreationDate="2017-06-13T07:33:00.643" Score="0" Body="&lt;p&gt;Some galaxy clusters do have global rotation or line of sight velocity gradients, probably as a result of previous mergers. Evidence for this comes from an analysis of the positions and redshifts of the constituent galaxies (&lt;a href=&quot;https://arxiv.org/abs/astro-ph/0702184&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hwang &amp;amp; Lee 2007&lt;/a&gt;). In principle one could look for similar signatures in X-rays from the hot intracluster medium, but the technology is still lacking (&lt;a href=&quot;https://arxiv.org/abs/1305.5519&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bianconi et. al. 2013&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most recent survey I could find is that of&#xA;&lt;a href=&quot;https://arxiv.org/abs/1604.06256&quot; rel=&quot;nofollow noreferrer&quot;&gt;Manolopolou &amp;amp; Plionis (2017)&lt;/a&gt; using &#xA;SDSS DR10 data. Abell 1689 is not on their list. According to earlier studies (e.g. &lt;a href=&quot;https://academic.oup.com/mnrasl/article/366/1/L26/1050203/The-complex-velocity-distribution-of-galaxies-in&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lokas et al. 2006&lt;/a&gt;), the velocity structure of this cluster is complex, with multiple structures along the line of sight. The similarity of the cluster gravitational potential inferred from gravitational lensing and from X-ray emitting gas under the assumption of hydrostatic equilibrium suggests contributions from rotation or turbulence in the intracluster medium are small (&lt;a href=&quot;https://www.aanda.org/articles/aa/full_html/2015/02/aa23242-13/aa23242-13.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tchernin et all. 2015&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-06-13T07:33:00.643" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21271" PostTypeId="1" CreationDate="2017-06-13T08:38:54.210" Score="1" ViewCount="609" Body="&lt;p&gt;Of course, I know what the ecliptic is and what equator is. The problem is I still can not understand why we need two systems for measuring the same thing - are each of them useful only in specific occasions? For example, which is preferred in marine navigation? I am interested in this practical side of the question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you give an example which positions a star, using both coordinate systems?&lt;/p&gt;&#xA;" OwnerUserId="17139" LastEditorUserId="2365" LastEditDate="2017-06-13T15:27:13.307" LastActivityDate="2018-01-09T20:07:54.853" Title="What is the difference between equatorial and ecliptical coordinate system?" Tags="&lt;star&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21273" PostTypeId="2" ParentId="1812" CreationDate="2017-06-13T11:23:19.557" Score="-2" Body="&lt;p&gt;If Mars was placed near the Sun and Venus relocated to Mars' current orbit, neither one would be habitable. To make both Mars and Venus habitable(in Theory), a large lunar moon is required because it would cause a tidal crust movement that would help to ignite the planet's magnetic field. For Venus, Mercury would become its new moon as it could not only ignite a strong magnetic field, but it could correct the planet's rotation by making it rotate in a normal 24hr Earth rotation, and allow it CO2 atmosphere to cool down as it will be able to recycle it into the ground. As for Mars, Jupiter's moon Io would be ideal as it can increase the planet's surface pressure and magnetic field. Both Venus and Mars are capable of terraforming by themselves. All that's needed is a large lunar moon like Io and Mercury respectively.  &lt;/p&gt;&#xA;" OwnerUserId="17159" LastActivityDate="2017-06-13T11:23:19.557" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21274" PostTypeId="2" ParentId="21269" CreationDate="2017-06-13T14:34:53.563" Score="3" Body="&lt;p&gt;Radio telescopes cannot &quot;see further&quot; than optical telescopes but electromagnetic radiation in the range detected by radio telescopes may come from a further distance (i.e. from an earlier time) than detectable by optical telescopes and/or radio waves may penetrate things which are obscured at optical wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classic example of the first case is the cosmic microwave background (CMB) - actually the oldest detectable radiation in the universe (assuming our cosmology is even half-close to being right). This is the radiation that was emitted when the early expanding universe cooled just enough to allow hydgrogen atoms to form - the captured electrons gave off radiation and the universe's density was no longer great enough to ensure that radiation caused a balancing ionisation. That happened so long ago/so far away that it now looks like something very cold (about 3 degrees above absolute zero) and so is not detectable at visible wavelengths but is at microwave wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second case includes things like radiation from dust clouds in our galaxy - the dust blocks visual radiation but not radio waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the premiss of your question is wrong in that it is not a question of dish size and certainly not magnification (the correct parallel with an optical telescope is also dish - or rather lens or mirror - size incidentally.)&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2017-06-13T14:34:53.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21275" PostTypeId="1" CreationDate="2017-06-13T14:47:53.727" Score="1" ViewCount="63" Body="&lt;p&gt;If our universe is a 4-manifold (i.e. with every point in space-time already existing) does this imply that there is zero net energy (as net energy would do work and so change the universe)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, how do we account for the fact that our current experience of the universe is very far from net zero energy? Are there stores of negative energy out there? (Or, more likely, have I misunderstood this question?!)&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2017-06-13T14:47:53.727" Title="Does a 4-manifold universe imply zero net energy?" Tags="&lt;cosmology&gt;&lt;space-time&gt;&lt;special-relativity&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21276" PostTypeId="1" AcceptedAnswerId="21279" CreationDate="2017-06-13T15:34:48.997" Score="7" ViewCount="758" Body="&lt;p&gt;My course book on astronomy states the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Older stars seem have higher oxygen abundances than iron. Explanation is that back in the days when these older stars were being formed type II supernova's were common, while type Ia weren't. So later, when the type Ia was becoming more common, the - now younger - stars were formed with higher iron abundances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are type Ia supernova's better for iron enrichment than type II, and were these type II supernova's in any way better for higher oxygen abundances - or just less good at producing iron (and why)?&lt;/p&gt;&#xA;" OwnerUserId="17161" LastEditorUserId="17161" LastEditDate="2017-06-13T16:40:26.853" LastActivityDate="2017-06-13T21:27:42.660" Title="Why do type Ia supernovas produce more iron than type II" Tags="&lt;supernova&gt;&lt;stellar-evolution&gt;&lt;metallicity&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21277" PostTypeId="2" ParentId="21271" CreationDate="2017-06-13T15:55:12.680" Score="1" Body="&lt;p&gt;To locate an object you need a system of coordinates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to define a plane (to define the latitude) and a direction (from which to measure longitude). On Earth, the obvious plane is the Equator. There is no obvious direction from which to measure longitude (which in the past led to different definition of longitude in France and Britain)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no single, natural plane in the sky. Several are possible: You can project the equator into the sky. You can use the plane of the Earth's orbit around the sun. You could use the plane of the galactic disc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the Equator is simple, but rather Earth-centric. The Ecliptic system is more natural for solar system objects. The galactic system nice for deep space objects. The lack of a single natural plane in the sky means that several different coordinates systems can co-exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example Rigel, at RA 5h14'32'', Dec −08° 12′ 05''&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has a Lat-Long of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;-31.11°  76.09°   Ecliptic&lt;/li&gt;&#xA;&lt;li&gt;-4.40°    219.66°  Galactic&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-06-13T15:55:12.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21278" PostTypeId="1" CreationDate="2017-06-13T19:20:58.747" Score="0" ViewCount="233" Body="&lt;p&gt;Is it possible to establish one's longitude and latitude by observing the stars? Can you use observations of the stars to find the relative distance between two locations, which would be useful in map-making? What kind of crude tools would help (a sextant is too complex, but maybe a backstaff?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What difference would there be if you were doing this on a planet that was not the Earth, and had a different location in the galaxy, a different day and year length?&lt;/p&gt;&#xA;" OwnerUserId="17167" LastEditorUserId="2365" LastEditDate="2017-06-13T22:35:13.880" LastActivityDate="2017-06-15T17:17:27.997" Title="Celestial navigation from scratch" Tags="&lt;observational-astronomy&gt;&lt;coordinate&gt;&lt;history&gt;&lt;celestial-mechanics&gt;&lt;navigation&gt;" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="21279" PostTypeId="2" ParentId="21276" CreationDate="2017-06-13T21:27:42.660" Score="6" Body="&lt;p&gt;&lt;strong&gt;Context&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron has the highest nuclear binding energy of all the elements (not completely true, but sufficiently accurate in an astronomical context). So, fusion of light elements into iron or something lighter is an exothermic process - you gain energy doing it, allowing the star to function. This is what happens in the last stages of a type II supernova. The core of a massive star in its last moments of life is hot and dense enough to fuse silicon into iron. Just before the supernova explosion, there is an iron ball of about 1.4 solar masses at the centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The progenitor of a supernova type Ia is a binary system where a &quot;normal&quot; star loses mass to a compact stellar remnant (a white dwarf). Once the white dwarf Contexthas accreted enough mass to be above a limit of 1.4 solar massses, fusion starts again, completely disintegrating the compact object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explosion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A SN Ia completely destroys the white dwarf progenitor in a runaway fusion process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a SN II, the pressure on the central iron ball exceeds the degeneracy pressure exerted by the electrons in the iron atoms' electron shell. The Fermi principle in quantum mechanics states that no Fermion (such as an electron) may occupy the same quantum mechanical state as another. The pressure exerted here is so large that the electrons of the iron atoms can no longer obey it and are forced into the nucleus, where they react with the protons to form neutrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Iron abundance&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do SN Ia enrich their environment with more iron than SN II? This is not so much a matter of iron production, but about how much of that iron ends up in interstellar space where it can be part of a new generation of stars. In a SN Ia, the progenitor is completely destroyed, scattering all its constituent atoms into its host galaxy. A SN II forms a compact remnant, either a neutron star or a black hole. A lot of the later, heavier fusion products end up not being carried outward in the supernova explosion but become part of the compact remnant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that a lot of the heavy elements scattered from a supernova of the &quot;exploding massive star&quot; result from an abundance of neutrinos escaping the central explosion and reacting with the outer shell of lighter elements that is blasted away.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastActivityDate="2017-06-13T21:27:42.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21280" PostTypeId="1" CreationDate="2017-06-14T07:20:01.990" Score="3" ViewCount="2021" Body="&lt;p&gt;In order for the universe to be as it is, containing stars and galaxies, it is a requirement that the early universe was at a relatively low entropy state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why did the big bang produce this low entropy universe?&lt;/p&gt;&#xA;" OwnerUserId="13125" LastActivityDate="2018-12-13T02:04:27.763" Title="Why did the big bang produce a low entropy universe?" Tags="&lt;astrophysics&gt;&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21281" PostTypeId="2" ParentId="21251" CreationDate="2017-06-14T12:00:20.560" Score="4" Body="&lt;p&gt;The Paschen series absorption requires a population of H atoms in the $n=3$ level and a radiation field containing infrared photons $\leq 1875$ nm, that can boost them in to the $=4,5,6...$ levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If hydrogen gas is at 25,000K, you can use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Saha_ionization_equation&quot; rel=&quot;noreferrer&quot;&gt;Saha equation&lt;/a&gt; to work out the fractional ionisation. This depends on &lt;em&gt;both&lt;/em&gt; the temperature and density of the gas. For the temperature you have been given, the ratio of ionised to netutral hydrogen is $1.7\times 10^{25}/n_e$, where $n_e$ is the electron number density in m$^{-3}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The typical particle number densities of an interstellar gas cloud are much, much less (many orders of magnitude) than $10^{25}$ m$^{-3}$, so the gas is &lt;em&gt;totally ionised&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is the case then there can be no Paschen absorption and no Balmer absorption &lt;em&gt;due to the cloud&lt;/em&gt; since there are no atoms in the $n=2,3$ states. There may be small amounts of Paschen and Balmer emission &lt;em&gt;from the cloud&lt;/em&gt; via recombination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus what you basically see is the spectrum of the star and since you aren't told what kind of star it is, then it's impossible to rule out the presence of emission lines or absorption lines of any kind. It is thus a dreadful and unanswerable question.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-06-14T12:00:20.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21283" PostTypeId="2" ParentId="21197" CreationDate="2017-06-14T13:57:04.510" Score="2" Body="&lt;p&gt;A partial answer to your question emerged today on arXiv. &lt;a href=&quot;https://arxiv.org/abs/1706.03795&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hippke (2017)&lt;/a&gt; analyses this exact problem - dealing with power requirements for the transmitter, arrangements for detection, the information transfer rate, extinction etc. etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their conclusion is that there is no problem in principle with getting something to work at about 1 bit/s per Watt of power that the probe can obtain from photovoltaic power and &lt;em&gt;if&lt;/em&gt; a dedicated 39-m telescope is observing the probe using a coronagraphic imager. They also conclude that the best wavelength to work at is 400nm for a ground-based receiving telescope and 300 nm if a 10-m space based telescope is available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic transmission device is a $\sim 1$m diameter telescope, firing a coherent laser beam back towards Earth. The data rate scales as the area of the transmitting telescope. They do not discuss how such a thing could be built or sent to $\alpha$ Cen. A promised follow-up paper is to discuss the use of gravitational lensing to improve signal-to-noise.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-06-14T13:57:04.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21284" PostTypeId="1" CreationDate="2017-06-14T14:44:34.917" Score="2" ViewCount="710" Body="&lt;p&gt;I know larger stars can fuse heavier and heavier elements up to iron where it stops because fusing iron requires more energy than it releases, causing a collapse and supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does fusing iron in a stellar core use more energy than it releases?&lt;/p&gt;&#xA;" OwnerUserId="17172" LastEditorUserId="2365" LastEditDate="2017-06-14T16:52:04.690" LastActivityDate="2017-06-14T22:41:40.250" Title="Why does fusing iron in a stellar core use more energy than it releases" Tags="&lt;supernova&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ClosedDate="2017-06-15T14:09:13.490" ContentLicense="CC BY-SA 3.0" />
  <row Id="21285" PostTypeId="2" ParentId="21278" CreationDate="2017-06-14T16:47:59.443" Score="2" Body="&lt;p&gt;One's latitude is easy to establish. You just need to find the angle of elevation of the pole, the point about which the stars appear to revolve each night. A stick and a plumb line can do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One's Longitude is nearly impossible to find unless you have an accurate clock. To find Longitude you need to measure the position of the stars at a known absolute time (not a local solar time) If you have a working watch you are sorted. If not you have difficulty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are on another planet you may be lucky and have some way of keeping absolute time, such as another planet in the system with visible moons. By observing the motion of the moons you could tell the absolute time. Something like this can be done with Jupiter's moons on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being on another planet could change things: If the planet rotates, you can find latitude, but if, for example, the rotation of the planet was very slow, there could be very long days, and very long nights. One problem you may face is not having any way to keep records of observations, unless you have stack of paper and ink. Exploring and mapping a whole planet is not a task one person can achieve.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-06-14T16:47:59.443" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21286" PostTypeId="2" ParentId="21284" CreationDate="2017-06-14T17:11:56.980" Score="1" Body="&lt;p&gt;An atomic nucleus contains protons and neutrons. There is a balance of forces: the (residual) strong nuclear force attracts protons and neutrons together. The electro-magnetic force pushes them apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When nuclei fuse, energy is required to push them together, as you have to push against the electromagnetic repulsion. But when the nuclei come close enough, the nuclei are attracted by the strong force, and energy is released as the nuclei fall into each other. For small nuclei, the energy released in fusion is greater than the energy needed to overcome electromagnetic repulsion. For large nuclei (which have a larger positive charge) the reverse is true, more energy is required to push the nuclei together than is released.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron is at the balance point, at which the energy required to overcome electomagnetic repulsion just exceeds what is given out on fusion. So iron is the smallest nuclei to absorb energy when it engages in fusion. And so when substantial amounts of iron form in the core of a star, the core will no longer be able to heat up to maintain the balance with gravity, the core collapses and the star explodes. See &lt;a href=&quot;https://astronomy.stackexchange.com/questions/2231/why-is-iron-responsible-for-causing-a-supernova?rq=1&quot;&gt;Why is iron responsible for causing a supernova?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron is therefore one of the most stable nuclei (not the most stable: that is Nickle-62, but for various reasons this is not likely for form inside stars) Iron acts like the &quot;ash&quot; of nuclear fusion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more detailed answer would require understanding the quantum mechanics of how binding energy in nuclei works, which can explain why it is Iron that has close to maximum binding energy.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-06-14T19:58:32.780" LastActivityDate="2017-06-14T19:58:32.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21287" PostTypeId="2" ParentId="13596" CreationDate="2017-06-14T18:01:47.553" Score="1" Body="&lt;p&gt;For pulsars the relation is complex. The magnetic field tends to &quot;freeze in place&quot; after the star collapses, but I'm sure there are all kinds of perturbation mechanisms that may change its initial orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For blackholes (the Kerr-Newman metric, rotating and charged black hole) the relation is very simple: the rotation axis of the BH and the symmetry axis of its magnetic field always coincide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Kerr%E2%80%93Newman_metric#Some_aspects_of_the_solution&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Kerr%E2%80%93Newman_metric#Some_aspects_of_the_solution&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a simple and intuitive justification, although perhaps not rigorous. If the magnetic field was not aligned with rotation, the BH would appear to be &quot;blinking&quot;. But a BH cannot &quot;blink&quot;, nor can it otherwise send any signals to the external universe. Therefore, its magnetic field axis must coincide with its rotation axis.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2017-06-14T18:01:47.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21288" PostTypeId="2" ParentId="21284" CreationDate="2017-06-14T20:53:58.917" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does fusing iron in a stellar core use more energy than it releases?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It doesn't, at least not in the alpha ladder. The alpha ladder starts with the carbon-12 produced by the triple alpha process. A carbon-12 and an alpha particle (helium-4) combine to form oxygen-16, which in turn combines with an alpha particle to form neon-20, and so on, up to titanium-44, chromium-48, then iron-52, then nickel-56, and then zinc-60.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the production of zinc-60 that kills stars rather than the production of nickel-56. The reactions up to and including the production of nickel-56 are exothermic (i.e., they release heat). The production of zinc-60 is endothermic (it consumes heat, in the form of a gamma particle). The energy-producing reactions that kept the star from collapsing on itself end. Moreover, the temperatures needed to produce zinc-60 are so very high that photons can photodisintegrate zinc-60, recreating the nickel-56 nuclei and alpha particles that created those zinc-60 nuclei. The alpha ladder pretty much stops with nickel-56. (Other processes create elements beyond nickel.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nickel-56 ejected from supernovae is rather short-lived, decaying with a half life of 6 days into cobalt-56. This too is radioactive, decaying with a half life of 77 days into iron-56. The unique signatures of these two decays, and the slightly delayed transition from nickel-56 decay to cobalt-56 decay, are one of the key signs that a supernova has occurred.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2017-06-14T22:41:40.250" LastActivityDate="2017-06-14T22:41:40.250" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21289" PostTypeId="2" ParentId="21278" CreationDate="2017-06-15T00:30:58.833" Score="3" Body="&lt;p&gt;To expand a little on James's answer: The pole angle method doesn't care what time of year it is, the celestial pole isn't going anywhere. ;) However, if you're in the tropics, the altitude of the celestial pole is rather low, which can make accurate observation difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you're doing this on another planet, (or even in Earth's southern hemisphere) you may not have a convenient pole star, and doing this sort of thing without decent equipment isn't going to be particularly accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine longitude you need to have a good way of determining the time relative to a fixed reference, like the Greenwich meridian. The Earth makes one rotation on its axis (relative to the stars) in a (&lt;a href=&quot;https://en.wikipedia.org/wiki/Sidereal_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;sidereal&lt;/a&gt;) day, so one day is equivalent to 360° of longitude, one hour corresponds to 15°, and one minute of time corresponds to 15 minutes of longitude, which is 15 nautical miles at the equator, and around 11.5 nautical miles at 40° latitude. In other words, if your time calculations are off by one minute, your position calculations can be off by 10 to 15 miles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton described a method of determining the time from the moon, you can read about it in Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_distance_(navigation)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lunar distance (navigation)&lt;/a&gt; article. From the Earth, the Moon's angular diameter is roughly 0.5°, and it takes roughly an hour for it to travel through that distance relative to the stars. So if you want time measurements accurate to the minute you need &lt;em&gt;very&lt;/em&gt; good observations. However, that's the easy part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need &lt;em&gt;very&lt;/em&gt; accurate calculations that tell you where the Moon's supposed to be at that point in time. And that's not easy to do - the Moon's motion is tricky! I won't go into the details here, but Wikipedia has a good introduction to &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;lunar theory&lt;/a&gt;. If you're curious about the kind of formulas that get used, take a look at the links in &lt;a href=&quot;https://astronomy.stackexchange.com/q/20973/16685&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it wasn't just the difficulty of making sufficiently accurate shipboard observations that prevented Newton's method from being adopted, it was also the difficulty of preparing sufficiently accurate lunar tables. Actually, Newton's method of lunar distances &lt;em&gt;was&lt;/em&gt; used, principally from 1763 (when the necessary tables and method were first published) until about 1850, when it was superseded by the marine chronometer. However, as Wikipedia mentions, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The method saw usage all the way up to the beginning of the 20th&#xA;  century on smaller vessels that could not afford a chronometer or had&#xA;  to rely on the this technique for correction of the chronometer.&lt;br&gt;&#xA;  [...]&lt;br&gt;&#xA;  Captain Joshua Slocum, in making the first solo circumnavigation in&#xA;  1895–1898, somewhat anachronistically used the lunar method along with&#xA;  dead reckoning in his navigation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To quote Captain Slocum:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Even expert lunarians are considered as doing clever work when they&#xA;  average within eight miles of the truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course, on another planet you may have a better-behaved moon, or no moon at all. Or, as James mentioned, you may be very lucky and have a nearby large planet with fast-moving moons that are visible to the naked eye.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="16685" LastEditDate="2017-06-15T17:17:27.997" LastActivityDate="2017-06-15T17:17:27.997" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21290" PostTypeId="2" ParentId="21280" CreationDate="2017-06-15T13:15:17.677" Score="5" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Second_law_of_thermodynamics#Various_statements_of_the_law&quot; rel=&quot;noreferrer&quot;&gt;second law of thermodynamics&lt;/a&gt; states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the total entropy of an isolated system can only increase over time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If applied to our universe, one can conclude that the entropy of our universe must necessarily increase over time (or, improbably, stay constant), resulting in the final &lt;a href=&quot;https://en.wikipedia.org/wiki/Heat_death_of_the_universe&quot; rel=&quot;noreferrer&quot;&gt;heat-death of our universe&lt;/a&gt;. Just as Georges Lemaître saw that the Universe was expanding and surmised that it must have been smaller in the past (implying a Big Bang), we can also surmise that if the entropy of our Universe is increasing to a maximum entropy wherein the Universe experiences a heat death, it must necessarily have been lower in the past. This begs the question, why was it lower? How did the universe come to be in a state with such low entropy such that stars and galaxies formed. Why did it not spring into existence, already in the heat death stage?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to this is, I don't know. And I'm not sure anyone could know or give a solid answer, backed up by facts and observational evidence. At best, there are some guesses and hypotheses which I can go over a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, let me point out that the second law of thermodynamics may not in fact apply to our Universe as a whole. The problem with the laws of physics is that they often apply in localized cases, say when we're talking about a science lab or a galaxy. But when you try to apply them to the universe as a whole, there's just too many unknowns and usually the rules change. For example, is our universe an isolated system? Nobody can answer that because nobody can view the entire Universe. We're limited to only observing the &lt;a href=&quot;https://en.wikipedia.org/wiki/Observable_universe&quot; rel=&quot;noreferrer&quot;&gt;observable Universe&lt;/a&gt; so already we can't definitively say that the Universe adheres to one of the primary pre-conditions of the second law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not to mention, you have the difficulty that it may be impossible for all, absolute laws to apply to the Universe as a whole, when we already know some of them don't. The laws of physics all rely on each other. If you change one, you affect all the others. We know (or at least some believe) that the &lt;a href=&quot;http://www.preposterousuniverse.com/blog/2010/02/22/energy-is-not-conserved/&quot; rel=&quot;noreferrer&quot;&gt;law of conservation of energy does not necessarily hold for the Universe as a whole&lt;/a&gt;. At any local point in space and for any localized event, it does, but not when applied to the entire Universe. So already the laws of physics are messed up and that will have unknown consequences for other laws of physics when applied to the Universe as a whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy in our universe is not conserved because it is expanding. This expansion is constantly dumping new energy into our universe, effectively making our universe not &quot;isolated&quot; even if it may be based on the dictionary definition of the word. As such, it's entirely possible for the entropy to be driven down by this energy input via expansion. The initial inflationary epoch, which likely violated energy conservation globally, could have resulted in our low, initial entropy state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as I said, a lot of this is guessing and hypotheses. I don't think there's a good answer out there to this good question, and if there is, I'd really like to hear it.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-06-15T13:15:17.677" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21294" PostTypeId="2" ParentId="21252" CreationDate="2017-06-15T23:14:41.693" Score="1" Body="&lt;p&gt;As Rob correctly pointed out, the emission of gravitational waves reduces the orbital energy and result in an inspiral. This reduction in total energy also reduces the mass of the final BH, since $E=mc^2$. The bulk of the gravitational wave energy is emitted (and energy=mass lost) in the final chirp, when the separation approaches the Schwarzschild radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To quantify this, let's just make a simple energy budget calculation, starting from two equal-mass BHs of mass $M_\bullet$ orbiting each other at distance $d$ on a circular orbit. Then the orbital energy is&#xA;$$&#xA;  E_{\mathrm{orbit}} = -\frac{GM^2_\bullet}{2d} = -M_\bullet c^2 \frac{R_s}{4d} &#xA;$$&#xA;where $R_s=2GM/c^2$ the Schwarzschild radius of each BH and we have assumed that $d\gg R_s$ such that the orbit is Keplerian. The total initial energy is then given by the rest mass energies plus the orbital energy as&#xA;$$&#xA;  E_{\mathrm{total}} = M_\bullet c^2 \left[2-\frac{R_s}{4d}\right].&#xA;$$&#xA;After coalescence, a remnant of mass $M_{r}$ emerges. The energy deficit is the difference between the initial and final energies&#xA;\begin{equation}&#xA;  \delta E = M_\bullet c^2 \left[2-\frac{R_s}{4d}\right] - \frac{M_rc^2}{\sqrt{1-v^2/c^2}},&#xA;\end{equation}&#xA;where $v$ is the speed of the remnant w.r.t. to the centre of mass of the progenitors. This energy has been lost by gravitational wave radiation. If this corresponds to a certain amount $\mu$ of rest mass, then from&#xA;$&#xA;  \delta E = \mu c^2&#xA;$&#xA;we find &#xA;$$&#xA;  M_r = \sqrt{1-v^2/c^2}\left[2M_\bullet -\mu - M_\bullet\frac{R_s}{4d}\right].&#xA;$$&#xA;Now for $v=0$ and $R_s\ll d$, the mass deficit $\delta m\equiv 2M_\bullet-M_r$ is identical to $\mu$: the radiated energy corresponds to the mass deficit; the final hole has 95$M_\odot$ if $M_\bullet=50M_\odot$ and $\mu=5M_\odot$. In particular, the gravitational wave energy cannot be taken merely from the orbital energy as suggested by another answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass deficit is even larger than the radiated energy if the remnant has undergone a considerable velocity kick, such that $v\neq0$ (caused by asymmetric gravitational wave radiation).&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2017-06-15T23:26:31.370" LastActivityDate="2017-06-15T23:26:31.370" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21296" PostTypeId="1" AcceptedAnswerId="21302" CreationDate="2017-06-16T12:25:28.367" Score="2" ViewCount="459" Body="&lt;p&gt;Can someone explain this diagram?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The text is in Dutch, free translation: &quot;You are given a graph: a histogram of the 10 000 most apparent bright stars. Explain 1, 2 and 3.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ty2nB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ty2nB.png&quot; alt=&quot;A histogram of the 10 000 stars with highest apparent magnitude&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the number of counted stars of types O and M is small due to &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;O is bright but just very rare and,&lt;/li&gt;&#xA;&lt;li&gt;M is common but quite dim.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The puzzling bit in this diagram is the sudden dip from a local maximum F (why is there a maximum) to G (why is G a local minimum). The drop from K to M is explained above - that being that M class stars are just very dim (is this all there is to say here?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that if there wasn't a dip at 3, and the graph would just continue, that the local max at K can be explained as being there &lt;em&gt;just because there is a dip at G&lt;/em&gt;. If there wasn't a dip it wouldn't be a maximum and it'd just look like a bell-curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why is there a dip at G?&lt;/li&gt;&#xA;&lt;li&gt;Why does it rise so smooth from O to F? Is it just because B - F are more common?&lt;/li&gt;&#xA;&lt;li&gt;Why is there a maximum at F?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17161" LastEditorUserId="2531" LastEditDate="2017-09-14T16:33:56.523" LastActivityDate="2017-09-14T16:33:56.523" Title="Can someone explain this diagram showing the spectral type distribution of bright stars" Tags="&lt;stellar-evolution&gt;&lt;spectral-type&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21297" PostTypeId="2" ParentId="21296" CreationDate="2017-06-16T14:28:28.623" Score="3" Body="&lt;p&gt;I'd say there are fewer M stars on list because they are least bright. they may outnumber other star types but only nearest ones would make list of brightest stars. Type O stars are rare hence the small number. This info is available from a Wikipedia page on 'stellar classifications' and refer mostly to the mail sequence stars on the  Hertzsprung–Russell diagram. The upper peak around type K (your point 3) is maybe explained by the addition of Red Giants and Bright Giants, but this is just a guess really from the HR diagram.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="16926" LastEditDate="2017-06-16T15:44:19.563" LastActivityDate="2017-06-16T15:44:19.563" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21299" PostTypeId="1" CreationDate="2017-06-16T16:52:25.957" Score="1" ViewCount="185" Body="&lt;p&gt;I just need to confirm or refute something I'm believing since a very long time, maybe without real evidence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we distinguish stellar clusters by their total mechanical energy ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true that all &quot;open&quot; star clusters have a positive energy (i.e. more kinetic energy that potential energy), while all &lt;em&gt;globular clusters&lt;/em&gt; are &quot;closed&quot; systems with negative total energy (i.e less kinetic energy than total potential energy) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for example, the Peiades M45 are &quot;open&quot; : they have $E &amp;gt; 0$, while M4, M53 and M80 (globulars) have $E &amp;lt; 0$ ?  Is this true ?&lt;/p&gt;&#xA;" OwnerUserId="14736" LastActivityDate="2017-07-17T12:42:18.963" Title="Is it true that open clusters have $E &gt; 0$ and globular clusters have $E &lt; 0$?" Tags="&lt;star&gt;&lt;celestial-mechanics&gt;&lt;newtonian-gravity&gt;&lt;star-cluster&gt;&lt;globular-clusters&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21300" PostTypeId="2" ParentId="21299" CreationDate="2017-06-16T18:29:23.760" Score="1" Body="&lt;p&gt;Stars in an open star cluster are - according to wikipedia - &quot;loosly bound by mutual gravity&quot;. Whereas a globular cluster is more dense and will be &quot;less loosly&quot; bound - to use the same terminology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the stars in the open star cluster still are gravitationally bound, the total energy $E$ must be less than zero. This criterium that $E&amp;lt;0$ is &lt;em&gt;always&lt;/em&gt; correct for (gravitationally or otherwise) bound systems for exactly the reason you stated, the kinetic energy is 'less positive' than the potential is negative. Otherwise the particles (stars in this case) would break free from the gravitational interactions with the other particles - and that's the complete opposite of a bound system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the fate of most open star clusters is that it will fall apart. This can be due to &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;mass loss of the cluster or,&lt;/li&gt;&#xA;&lt;li&gt;internal processes like a slingshot due to close encounters between constituent stars.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From the get-go these open clusters are quite unstable, so $E$ will be close to zero, but still negative ofcourse because it's a bound system. But it is instable enough for the following to happen. Young, bright stars radiate a lot of energy, and this means that the radiation pressure on the surronding cloud from which these stars are born is high. The light from these young stars can blow large amounts of mass in the form of this gas and dust out of the system - which has an influence on the total energy of the system! Because the mass of the system drops, the total energy rises (read: becomes &lt;em&gt;less negative&lt;/em&gt;) - wich makes it easier for the constituent stars to reach escape velocity. If you don't like thinking in terms of energy, the escape velocity is $v_e = \sqrt{2GM/r}$, so $v_e \propto M $. If $M$ drops, so does the escape velocity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second bullet speaks for itself, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can deduce that $E$ &lt;em&gt;has to be negative&lt;/em&gt; from a conservation viewpoint. When the stars in the cluster are born from the same cloud, this cloud already is gravitationally bound, so $E&amp;lt;0$, and unless some energy is lost at some point, the cloud will stay bound. But like I said and as you said, most of these clusters don't stay bound for very long, and eventually will evaporate. &lt;/p&gt;&#xA;" OwnerUserId="17161" LastEditorUserId="17161" LastEditDate="2017-06-17T12:38:03.060" LastActivityDate="2017-06-17T12:38:03.060" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21301" PostTypeId="1" CreationDate="2017-06-16T19:17:55.900" Score="0" ViewCount="109" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xhpO6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xhpO6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I understand this 59 degree mentioned below, what does it mean? I could not get it. The photo is a diagram from an article dedicated to the invention of the ecliptic. The photo is from my tablet (the title could be seen)&lt;/p&gt;&#xA;" OwnerUserId="17139" LastActivityDate="2017-06-18T14:34:40.550" Title="Can someone explain me this diagram from an article?" Tags="&lt;orbit&gt;&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21302" PostTypeId="2" ParentId="21296" CreationDate="2017-06-16T22:16:06.153" Score="11" Body="&lt;p&gt;It's all to do with the relationships between mass, spectral-type and luminosity and the initial mass function of stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think your explanation of points 1 and 2 are completely correct. O and B stars are rarely born and short-lived; so even though they have enormous luminosities relatively few make it into a list of stars ordered by apparent brightness. A further important reason is that although they are immensely luminous, the more distant examples that might have made it into the list, cannot make it into the list because they are in the Galactic plane (they are young and still close to where they were born) and thus severely affected by extinction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M-dwarfs are indeed the most common type of star, but not common enough to overcome the fact that their luminosities are much, much lower than hotter stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dip in G stars is actually a peak in the K-stars caused by giants. Most low(ish) mass stars will evolve towards more or less the same point in the HR diagram forming the &lt;a href=&quot;https://en.wikipedia.org/wiki/Red_clump&quot; rel=&quot;noreferrer&quot;&gt;red clump&lt;/a&gt;. These stars are luminous, older and not concentrated in the galactic plane, can therefore be seen to great distances and produce a peak in the number counts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can see this using actual data rather than a cartoon, I producedthe HR diagram for the brightest 10,000 stars in the &lt;a href=&quot;http://cdsarc.u-strasbg.fr/viz-bin/Cat?I/311&quot; rel=&quot;noreferrer&quot;&gt;revised Hipparcos catalogue&lt;/a&gt; (about V&amp;lt;6.7). You can clearly see the red clump of giants. Now collapse this distribution onto the B-V axis and you can see there is a local minimum at $0.6&amp;lt;B-V&amp;lt;0.8$ corresponding to G-stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Absolute colour-magnitude diagram for the brightest 10,000 stars&lt;/em&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Etj5J.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Etj5J.png&quot; alt=&quot;Hipparcos HR diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The corresponding B-V distribution for these stars. G-stars have $0.6&amp;lt;B-V&amp;lt;0.8$&lt;/em&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Nkm1e.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nkm1e.png&quot; alt=&quot;Histogram of B-V&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-06-17T15:04:40.940" LastActivityDate="2017-06-17T15:04:40.940" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21304" PostTypeId="1" CreationDate="2017-06-17T04:41:58.217" Score="0" ViewCount="2471" Body="&lt;p&gt;Googling is giving me both Galileo Galilei and Nicolas Copernicus as the answer. I am sorry and will remove this question if it is an opinion based question.&lt;/p&gt;&#xA;" OwnerUserId="17192" LastActivityDate="2017-06-17T21:36:47.880" Title="Who is considered as the father of modern astronomy?" Tags="&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ClosedDate="2017-06-18T00:54:44.573" ContentLicense="CC BY-SA 3.0" />
  <row Id="21305" PostTypeId="2" ParentId="21026" CreationDate="2017-06-17T09:17:57.123" Score="1" Body="&lt;p&gt;Your hypothesis is correct: it is the &lt;em&gt;tidal interaction&lt;/em&gt; between neighbours which generates spin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of your proto halo (galaxy, stellar disc etc) as an ellipsoid (set by it's inertial tensor). Should the matter around it not be spherically distributed, it will apply a torque onto that ellipsoid and make it spin. The assumption in cosmology is that even though the primordial angular momentum you mention may be null, tidal interaction between neighbouring proto structures is sufficient to allow then to exchange (hence acquire individually) angular momentum, while preserving a zero sum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Reference on this subject is given by this &lt;a href=&quot;http://cdsads.u-strasbg.fr/cgi-bin/nph-iarticle_query?1984ApJ...286...38W&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This diagram may enlighten the issue: it represents the tidal torquing of the blue ellipsoid by the tides represented by the pink ellipsoid.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/nDyvV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nDyvV.png&quot; alt=&quot;Tidal torquing of the blue ellipsoid by the tides represented by the pink ellipsoid&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: the expression spin up is unfortunate it seems because it is in fact spin &lt;em&gt;conservation&lt;/em&gt; at the later stage of collapse of the proto halo which makes it rotate faster while preserving it spin (= total internal angular momentum).&lt;/p&gt;&#xA;" OwnerUserId="1227" LastEditorUserId="1227" LastEditDate="2017-06-17T09:27:24.983" LastActivityDate="2017-06-17T09:27:24.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21306" PostTypeId="1" AcceptedAnswerId="21308" CreationDate="2017-06-17T12:57:10.130" Score="0" ViewCount="88" Body="&lt;p&gt;If it's true that the deeper we look into space the further back in time the light is coming from then does that mean that just behind the cosmic background radiation the entire sky is filled with an unviewable image of the singularity that was the big bang and can that mean that we are sat at the centre of the Big Bang?&lt;/p&gt;&#xA;" OwnerUserId="17194" LastActivityDate="2017-06-17T17:17:33.283" Title="Centre of the singularity that is the big bang" Tags="&lt;cmb&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2017-06-18T16:42:32.513" ContentLicense="CC BY-SA 3.0" />
  <row Id="21307" PostTypeId="2" ParentId="21301" CreationDate="2017-06-17T12:58:28.737" Score="1" Body="&lt;p&gt;The Earth's equator is at an approximate 23.4 degree angle to the plain of its orbit around the Sun. That means that to an Earth-bound observer the Sun appears to move north and south of the celestial equator (which is merely the projection of the Earth's equator on to the sky) during the course of the year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact this is what gives us the seasons - so in the the Sun is at maximum northern declination (the equivalent of latitude when projected on to the sky) at the time of the Summer Solstice (around 21 June - hence summer in the northern hemisphere) and at maximum southern declination at the time of the winter solstice (around 21 December - hence northern winter). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this movement also means different rising and setting points - in the Northern hemisphere the Sun's rise is between north and ast in the summer and between south and east in the winter (unless, that is, you are within the Artic circle in which case it will not set/rise at all for at least some period in the summer and winter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a variety of online tools that show you this effect - this one: &lt;a href=&quot;https://www.timeanddate.com/sun/iraq/baghdad?month=12&amp;amp;year=2017&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.timeanddate.com/sun/iraq/baghdad?month=12&amp;amp;year=2017&lt;/a&gt; - shows that in Baghdad the sun rises - in 2017 - at 61 degrees East on 21 June and at 118 degrees East on 21 December - a difference of 58 degrees.&lt;/p&gt;&#xA;" OwnerUserId="823" LastEditorUserId="823" LastEditDate="2017-06-18T14:34:40.550" LastActivityDate="2017-06-18T14:34:40.550" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21308" PostTypeId="2" ParentId="21306" CreationDate="2017-06-17T17:17:33.283" Score="0" Body="&lt;p&gt;Since the Universe evolved from something like a singularity, where all of space was contracted to a point, yes, we are at the center of the Big Bang. So is every other point in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CMB represents the light from a universe-filling plasma that lasted for roughly 378,000 years. It is a feature of the &lt;em&gt;observable&lt;/em&gt; universe, rather than something that is actually there in the present universe; in particular, we can never reach it, as it's the same distance away from every point in the universe. As such, there's not much sense in talking about what's behind it, much like talking about what's behind a shadow.&lt;/p&gt;&#xA;" OwnerUserId="15299" LastActivityDate="2017-06-17T17:17:33.283" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21310" PostTypeId="2" ParentId="21304" CreationDate="2017-06-17T21:36:47.880" Score="1" Body="&lt;p&gt;I would say Copernicus. He did his work maybe 100 years earlier and formulated the heliocentric view of the universe. &lt;/p&gt;&#xA;" OwnerUserId="16926" LastActivityDate="2017-06-17T21:36:47.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21311" PostTypeId="1" CreationDate="2017-06-17T22:34:59.363" Score="3" ViewCount="99" Body="&lt;p&gt;I have been reading some cosmology and need to ask the experts at this point some specific points. I must admit I am a little confused. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few models of the universe and may be necessary at this point to see if I can sort out the main differences.  In particular James Hartle and Hawking's no boundary quantum state origin of the universe. I believe there may be some minor slight of hand imaginary time element to this to explain away a definite beginning point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this theory compete with the &quot;standard&quot; inflationary model by Allan Guth, the one with the inflaton field and infinite inflation complete with multiverse? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also Paul Steinhardt and Neil Turok who are not warming up on inflation, and each propose a cyclic model. Are these cyclic models the same? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do any of these models overlap the string theorists  &quot;big bangs&quot; from colliding large extra dimensions called &quot;Brains&quot;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are cosmologists these days heading in any definite direction with some of these complex and difficult for the layman to get a handle on?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I didn't mean to leave Penrose out of this , he worked with Hawking for some time when Hawking was working on his cosmological theories but I do not know which theory Penrose proposed apart from saying that his lectures appear to shy away from inflationary models proposed by Allan Guth.&lt;/p&gt;&#xA;" OwnerUserId="17197" LastActivityDate="2017-06-17T22:34:59.363" Title="What is the difference between Steinhardt's and Neil Turok's cyclic cosmology and Guth's inflationary model and Hartle, Hawking no boundry cosmology?" Tags="&lt;cosmology&gt;&lt;cosmological-inflation&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21313" PostTypeId="1" CreationDate="2017-06-18T10:20:09.980" Score="0" ViewCount="119" Body="&lt;p&gt;I'm trying to calculate RA and DEC with nutation correction. The direct calculation in MEEUS Page 151 can't be used for stars close to celestial poles and I thought I'd turn RA/DEC into ecliptical longitude and latitude and then add delta psi to the longitude and transform back to RA/DEC. Unfortunately my calculations are wrong somewhere and I can't figure out where I made a mistake.&#xA;You can find my spreadsheet here: &lt;a href=&quot;https://1drv.ms/f/s!AukrJ2RhCvoHh0w8MygI0qajeoXt&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://1drv.ms/f/s!AukrJ2RhCvoHh0w8MygI0qajeoXt&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17201" LastActivityDate="2017-06-18T13:07:32.707" Title="applying nutation to ecliptical coordinates (apparent position of star)" Tags="&lt;amateur-observing&gt;&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21314" PostTypeId="2" ParentId="21313" CreationDate="2017-06-18T13:07:32.707" Score="1" Body="&lt;p&gt;The standard methods of applying nutation retain equatorial components of position, and for near-polar objects they avoid the drawbacks of the abbreviated approximation by converting the equatorial components of the position into rectangular components and applying the nutation matrix for the date.  In principle this method need not involve approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The topic is explained in principle in chapter 3 of 'Explanatory &#xA;Supplement to the Astronomical Almanac' (ed. P K Seidelmann, 2nd ed, 1992/2006), where the full nutation matrix is specified on p.115 in terms of classical components of nutation in longitude and obliquity, dpsi and deps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculations to evaluate these components has undergone much development over the years, the latest version has a very long list of &gt;1000 trigonometrical components, but a selection of specifications and an approximation is given in the suite of algorithms and codes available at the website for the IAU SOFA software ('Standards of Fundamental Astronomy'), at  [&lt;a href=&quot;http://www.iausofa.org]&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.iausofa.org]&lt;/a&gt; , and of course any user can apply further shortening approximations to the series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The IAU suite has three versions of the calculation of the nutation components in longitude and obliquity, routines iau_NUT80 (older standard),  iau_NUT00A (full and very long post-2000 standard calculation), and iau_NUT00B (abridged version of the modern calculation).  There is also a routine iau_NUMAT for calculating the nutation matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(An intermediate approximation for the components of nutation, useful for applications where abridgment 2000B is too drastically approximated but 2000A is too long, is given in the US Naval Observatory's 'NOVAS' software suite (Naval Observatory Vector Astrometry software) [&lt;a href=&quot;http://aa.usno.navy.mil/software/novas/novas_info.php]&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://aa.usno.navy.mil/software/novas/novas_info.php]&lt;/a&gt;. The NOVAS suite also offers a more integrated approach to calculating astrometric problems , where IAU-SOFA provides individual short modules to be selected and assembled by the user.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, the question doesn't mention applying a precession correction as well, but in case that is wanted (which it very often is, in conjunction with nutation), the references already given above provide the needed resources.&lt;/p&gt;&#xA;" OwnerUserId="17202" LastActivityDate="2017-06-18T13:07:32.707" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21315" PostTypeId="2" ParentId="21301" CreationDate="2017-06-18T13:35:32.043" Score="1" Body="&lt;p&gt;This is almost too short to be answer. Using the &quot;rise&quot; formula at &lt;a href=&quot;https://astronomy.stackexchange.com/questions/14492/need-simple-equation-for-rise-transit-and-set-time/14508#14508&quot;&gt;Need Simple equation for Rise, Transit, and Set time&lt;/a&gt; and plugging in 32.54 degrees for Babylon's latitude, we get:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;When an object's declination is +23.5 degrees (the Sun at summer solstice), it rises 61.77 degrees east of north.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When an object's declination is -23.5 degrees (the Sun at winter solstice), it rises 118.23 degrees east of north, which is more conveniently expressed as 61.77 degrees east of south (note the expected symmetry here)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The azimuth degree difference between these positions is 56.46 degrees, close to the 59 degrees in the diagram.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The calculations for the Sun setting are very similar.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Why the difference?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If the Sun were a point and there were no refraction, the numbers above would be more accurate. Because the Sun is a disk and because of refraction, the sun actually rises when the geometric position of the center of its disk is 50 minutes below the horizon.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Minor: I'm using +-23.5 degrees as an approximation of the Sun's declination at the solstices: the actual number is a little different.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To be fair, even after applying some corrections, I couldn't quite get 59 degrees or even anything about 58.5 degrees, but I'm sure this is what they're talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TODO: I invite someone to create a graph showing how the &quot;59 degrees&quot; varies with latitude.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2017-06-18T13:35:32.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21316" PostTypeId="2" ParentId="2408" CreationDate="2017-06-18T16:47:58.750" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IfH0m.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IfH0m.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't really an answer, but more of a visualization (click for full size of 1024x768) based on the &#xA;formulas from &lt;a href=&quot;https://astronomy.stackexchange.com/questions/14492/need-simple-equation-for-rise-transit-and-set-time/14508#14508&quot;&gt;Need Simple equation for Rise, Transit, and Set time&lt;/a&gt; -- hope it helps &lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2017-06-18T16:47:58.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21318" PostTypeId="1" CreationDate="2017-06-18T21:16:51.140" Score="0" ViewCount="579" Body="&lt;p&gt;My question is seen in the title: What is the right ascension overhead at Greenwich at noon on 21 march&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a little bit confused since the right ascension is set by the vernal equinox and don't understand how this is related to Greenwich.&lt;/p&gt;&#xA;" OwnerUserId="15386" LastActivityDate="2020-06-20T03:03:10.397" Title="What is the right ascension overhead at Greenwich at noon on 21 march" Tags="&lt;right-ascension&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21320" PostTypeId="2" ParentId="21236" CreationDate="2017-06-18T23:17:53.677" Score="3" Body="&lt;p&gt;The most distant galaxy currently known is at a redshift $Z=11.1$ (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2016ApJ...819..129O&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oesch et al. 2016&lt;/a&gt;). According to &lt;a href=&quot;http://www.astro.ucla.edu/~wright/CosmoCalc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ned Wright's cosmology calculator&lt;/a&gt;, for a set of concordance cosmological parameters, this corresponds to a comoving distance of 9.88 Gpc (32 billion light years) and a recession velocity (now) from Hubble's law of &lt;strong&gt;2.28c&lt;/strong&gt; (i.e. 684,000 km/s). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course galaxies could exist at greater distances, although they do need time to form after the big bang. Theoretical simulations (e.g. &lt;a href=&quot;http://www.as.utexas.edu/~vbromm/araa.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bromm 2011&lt;/a&gt;) suggest star  forming galaxies might be present at redshifts of 20, only 200 million years after the big bang. Such galaxies, if they exist, would have recession velocities of 2.5c. &#xA;&lt;em&gt;This is the maximum possible recession velocity for any galaxy that is observable by us now (or in the near future)&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no possibility of &lt;em&gt;observable&lt;/em&gt; galaxies having recession velocities as high as 3.3c, since that would place them at redshifts of $&amp;gt;1000$ and they would need to have formed even before the epoch of hydrogen recombination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as Pela and Zephyr are pointing out, there are (likely) galaxies that exist now, &lt;em&gt;but that we cannot observe now and never will be able to measure&lt;/em&gt;, that are more distant than this and which are receding faster. We are reasonably sure that the universe probed by current observations of the cosmic microwave background is isotropic to comoving distances of about 14.1 Gpc (46 billion light years; &quot;the edge of the observable universe&quot;). Thus galaxies that are currently there, but which are just perturbations in the CMB as we see them now, will have recession velocities of 3.3c. Equally, there could be more distant galaxies than that in a much larger, or possibly infinite, universe and these could have much larger recession velocities still (that we also cannot observe or measure), and indeed they, like the galaxies we do observe, are possibly accelerating away from us.  &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-06-20T17:37:19.393" LastActivityDate="2017-06-20T17:37:19.393" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21322" PostTypeId="1" AcceptedAnswerId="21328" CreationDate="2017-06-19T05:55:05.480" Score="3" ViewCount="189" Body="&lt;p&gt;I came across several papers (e.g. &lt;a href=&quot;https://arxiv.org/abs/1112.0928v1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Looking for outflow and infall signatures in high mass star forming regions P. D. Klaassen, L. Testi and H. Beuther, 2011&lt;/a&gt;) talking about infall signatures in star formation. Can you please explain the physical meaning of it?&lt;/p&gt;&#xA;" OwnerUserId="11434" LastEditorDisplayName="user1569" LastEditDate="2017-06-20T11:41:51.057" LastActivityDate="2017-06-20T11:41:51.057" Title="What are infall signatures in stars?" Tags="&lt;astrophysics&gt;&lt;radio-astronomy&gt;&lt;observational-astronomy&gt;&lt;star-formation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21323" PostTypeId="2" ParentId="21318" CreationDate="2017-06-19T08:27:31.047" Score="1" Body="&lt;p&gt;At the vernal equinox, the sun crosses the celestial equator, and by definition, it has an RA of 0h (at the current epoch).  At mid-day on that day, the sun will be more or less due south, and as the line of zero RA forms a great circle, the RA &quot;overhead&quot; will also be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun won't be exactly zero RA except at the moment of the equinox, and also the length of the solar day varies throughout the year, which results in the time at which the sun is due south is not quite 12 noon by clock time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the above is independent of whether or not you are observing from Greenwich.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-06-19T08:27:31.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21324" PostTypeId="1" AcceptedAnswerId="21351" CreationDate="2017-06-19T09:19:00.817" Score="4" ViewCount="300" Body="&lt;p&gt;I was exploring Mars in Google maps and I found these dark spots, like tree shadows on the north pole. I don't have any idea what they might be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jIQ70.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jIQ70.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.com/maps/space/mars/@83.9730921,-126.6534232,1920m/data=!3m1!1e3&quot; rel=&quot;nofollow noreferrer&quot;&gt;View this in Google maps.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17209" LastEditorUserId="11143" LastEditDate="2017-06-19T14:21:30.270" LastActivityDate="2017-06-22T02:13:34.843" Title="What are these dark spots on Mars' north pole?" Tags="&lt;mars&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21325" PostTypeId="1" CreationDate="2017-06-19T10:55:52.590" Score="0" ViewCount="55" Body="&lt;p&gt;Could some stars be orbiting galaxies in outward spirals instead of perfectly round circulation? Patterns of matter movement relate strongly to the galaxy centre.&lt;a href=&quot;https://i.stack.imgur.com/l1om1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l1om1.png&quot; alt=&quot;Patterns of matter movement in galaxy associate to centre&quot;&gt;&lt;/a&gt; A special relationship between the galaxy and the central object suggest the centre may be an origin point in 'outflow'. The BH central object behaves strangely by instigating the release outwardly of massive amounts of matter. This may form stars that spiral out across the galaxy. The grand design of features in galaxies may related to general outward spiralling.&lt;a href=&quot;https://i.stack.imgur.com/ALKev.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ALKev.jpg&quot; alt=&quot;X-ray emission from the monster galaxy and surrounding hot (30-70 million degrees C) cluster gas. The bright central source is the supermassive black hole at the core of Perseus A itself. Low density regions are seen as dark bubbles or voids, believed to be generated by cyclic outbursts of activity from the central black hole. Image credit:  NASA/CXC/IoA/A.Fabian et al.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15083" LastActivityDate="2017-06-19T12:22:38.393" Title="Spiralling in Galaxy evolution" Tags="&lt;galaxy&gt;&lt;supermassive-black-hole&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21327" PostTypeId="2" ParentId="21325" CreationDate="2017-06-19T12:22:38.393" Score="-2" Body="&lt;p&gt;Basically, yes, it's possible.  Our moon is slowly moving away due to energy transfers from Earth related to tidal drag.  This is very slow spiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know whether a star can experience sufficient tidal drag relative to the center of its galaxy to cause a change in orbit - and there are a few billion other objects in the galaxy producing gravitational force in other directions, making the analysis kind of messy.&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-06-19T12:22:38.393" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21328" PostTypeId="2" ParentId="21322" CreationDate="2017-06-19T14:09:55.333" Score="3" Body="&lt;p&gt;The paper you linked is specifically trying to understand massive $(M\gtrsim 8\:\mathrm{M_\odot})$ star formation. It is still an open question as to exactly how very massive stars form. Do they form all at once in a giant accretion event (&quot;top-down&quot; formation) or do they form by having a bunch of low mass stars merge (&quot;bottom-up&quot; formation) or by some other process altogether? The authors of the paper state that it is particularly tricky to study high mass star formation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Evolution in low mass systems is often characterized by changes in the spectral energy distributions (SEDs). This is not as easily done in high mass systems (Molinari et al. 2008)&lt;sup&gt;1&lt;/sup&gt;, which is why studying the gas dynamics is very important.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Basically, they can't easily figure out what's going on by looking at the SED of the high mass star forming region, so instead, they're going to look at how the gas is flowing and moving during the formation. Klassen et al. break their observations into observing &quot;infall&quot; and &quot;outflow&quot; (via two different molecular transitions). What they mean by this is that one molecular transition (that of the $J=4-3$ transition in $\mathrm{HCO^+}$ and $\mathrm{H^{13}CO^+}$ molecules) will show evidence of gas flowing &lt;em&gt;into&lt;/em&gt; the core, i.e. infall, while another molecular transition (that of the $J=8-7$ transition in the $\mathrm{SiO}$ molecule) will show evidence of gas flowing &lt;em&gt;out of&lt;/em&gt; the core, i.e., outflow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During star formation, you tend to get both types of flows. Infall will occur because the gas is simply accreting onto the core due to gravity. Outflow will occur because the accreting matter and central core get very hot, causing an energy release outwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see precisely how they track infall flows in Section 3.2. They have a paragraph discussing precisely how they determine that the gas is falling into the core from the observations of the $\mathrm{HCO^+}$ and $\mathrm{H^{13}CO^+}$ molecules.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Double peaked or asymmetric H$\mathrm{HCO^+}$ line profiles in which the brightest emission is blueward of the source rest velocity can be interpreted as due to infall if the $\mathrm{H^{13}CO^+}$ is single peaked at the rest velocity. If the optically thin $\mathrm{H^{13}CO^+}$ also has a double peaked profile, the double peak in both lines is likely due to multiple components along the line of sight. Churchwell et al. (2010)&lt;sup&gt;2&lt;/sup&gt; describe the different mechanisms that could produce a double peak in an $\mathrm{HCO^+}$ line profile towards massive star forming regions, and conclude that self absorption best fits observed line profiles. The blue or red skew of the absorption profile can then be used to distinguish infall (blue) or outflow (red).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I won't go into detail to precisely explain the spectroscopic interpretation of the above quote, but suffice it to say, that is their method of detecting matter infalling towards the core of the high mass star.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;sup&gt;1&lt;/sup&gt; &lt;a href=&quot;http://cdsads.u-strasbg.fr/abs/2008A%26A...481..345M&quot; rel=&quot;nofollow noreferrer&quot;&gt;Molinari, S., Pezzuto, S., Cesaroni, R., et al. 2008, Astronomy and Astrophysics, 481, 345&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;sup&gt;2&lt;/sup&gt; &lt;a href=&quot;http://cdsads.u-strasbg.fr/abs/2010A%26A...513A...9C&quot; rel=&quot;nofollow noreferrer&quot;&gt;Churchwell,  E., Sievers,  A., &amp;amp; Thum, C. 2010, Astronomy and Astrophysics, 513, 9&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-06-19T14:09:55.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21329" PostTypeId="1" CreationDate="2017-06-19T14:49:05.823" Score="2" ViewCount="151" Body="&lt;p&gt;In this &lt;a href=&quot;https://worldbuilding.stackexchange.com/q/83817/1996&quot;&gt;worldbuilding question&lt;/a&gt;, the moon is gone:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;an extra-solar gas giant passed close by Earth in which the Moon was&#xA;  knocked from orbit to become a planet in an independent orbit, as well&#xA;  as sending Mercury crashing into its parent star.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So by introducing a rogue planetoid, the orbit of both the Moon and Mercury needs to be affected, with at least the former surviving the encounter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this possible? If so, what are the requirements (mass, velocity, proximity)?  Would it be possible for it to happen and affect the Earth minimally, so that it remains habital?&lt;/p&gt;&#xA;" OwnerUserId="17210" LastEditorUserId="17210" LastEditDate="2017-06-20T12:38:56.253" LastActivityDate="2017-06-23T04:11:28.833" Title="How to lose the moon?" Tags="&lt;orbit&gt;&lt;orbital-migration&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21332" PostTypeId="1" AcceptedAnswerId="21333" CreationDate="2017-06-19T19:57:10.977" Score="1" ViewCount="201" Body="&lt;p&gt;Did the apparent magnitude of the Sun change in the past 1000 years or so? If so, for what reasons?&lt;/p&gt;&#xA;" OwnerUserId="17171" LastActivityDate="2017-06-19T21:20:13.883" Title="How much has the apparent magnitude of the Sun changed?" Tags="&lt;the-sun&gt;&lt;apparent-magnitude&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21333" PostTypeId="2" ParentId="21332" CreationDate="2017-06-19T20:18:55.350" Score="2" Body="&lt;p&gt;No. The apparent magnitude of the sun is constant to within 0.2%  How long it has been that way is only speculation since measurements aren't available. 1000 years is a very short time period in the life cycle of a star. Its safe to say yes its been constant for 1,000 years. There is a small variation based on the 11 year sun spot cycle but this variation is reported at less than 0.1%. There is also a larger variation due to the difference in Earth-to-Sun distance caused by the earths elliptical orbit. This effect is about 7%. This is no due to the sun but rather due to the eccentricity of Earths orbit. There are also minor effects with a cycle time or much greater than 1,000 years. These also due to variations in Earths orbit.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="16926" LastEditDate="2017-06-19T21:20:13.883" LastActivityDate="2017-06-19T21:20:13.883" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21334" PostTypeId="1" CreationDate="2017-06-19T21:39:20.507" Score="5" ViewCount="513" Body="&lt;p&gt;How did scientists come to the conclusion that it is space that is expanding and not galaxies moving away from each other as in a giant explosion?&lt;/p&gt;&#xA;" OwnerUserId="17081" LastEditorUserId="1056" LastEditDate="2017-06-27T09:01:50.507" LastActivityDate="2017-06-27T09:01:50.507" Title="How did scientists come to the conclusion that it is space that is expanding?" Tags="&lt;expansion&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21335" PostTypeId="2" ParentId="21334" CreationDate="2017-06-19T22:24:40.627" Score="6" Body="&lt;p&gt;The simplest explanation relies on the statistical unlikelihood of us being at a special place and time in the cosmos.  Let's say, for a moment, that we see galaxies moving away from us because of some cosmic explosion that set them in motion.  In that case, we would expect to see the galaxies in one direction moving faster than the galaxies in another.  Looking towards the center of the explosion, we would see galaxies on the other side moving much faster than if we looked in the other direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, however, is not what we see.  What we observe is all galaxies moving away from us, with a speed proportional to their distance from us.  That means that the further away they are, the faster they are moving.  It doesn't matter what direction we look in; a galaxy 500 Mpc away in one direction will have a recessional velocity approximately the same as one 500 Mpc away in the other direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This forces us into one of two conclusions:  either we are at the exact center of the universe, which is statistically impossible, or space itself is expanding, so everything is moving away from everything else.  The animation below shows how this would work.  From any given point in space, it would look like everything is getting further away from you.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/t8pIu.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t8pIu.gif&quot; alt=&quot;An animation of the expansion of space.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explaining the recession of galaxies due to an explosion also raises many questions:  What caused the explosion?  Why would it start at one particular point?  What were things like before the explosion?  All of these are questions which do not have accepted answers.  Instead, the expansion of space ties directly into the idea of the Big Bang.  The expansion of space itself is what drove the Big Bang.  It is not that the universe is expanding into something.  Instead, space itself is expanding.  However, the topic of the Big Bang is a different subject, so you can look up information on that yourself.&lt;/p&gt;&#xA;" OwnerUserId="13426" LastActivityDate="2017-06-19T22:24:40.627" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21337" PostTypeId="1" AcceptedAnswerId="21389" CreationDate="2017-06-20T09:05:33.183" Score="0" ViewCount="400" Body="&lt;p&gt;I am doing some research for my fantasy novel, but I'd like to really understand how my planet system would cause life to grow if at all possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea would be, there is a large core planet that is orbiting a sun. Two smaller planets (or moons) are in an orbit around it on polar opposite sides of the core planet. Is this even possible? Would the necessary means to spark life occur?&lt;/p&gt;&#xA;" OwnerUserId="17221" LastEditorUserId="11143" LastEditDate="2017-06-20T13:21:30.497" LastActivityDate="2017-06-25T14:02:56.457" Title="Can smaller habitable planets (or moons) orbit a larger habitable planet?" Tags="&lt;orbit&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21338" PostTypeId="2" ParentId="21334" CreationDate="2017-06-20T12:53:15.700" Score="3" Body="&lt;p&gt;Phiteros' answer is spot on. We believe it is space that is expanding since the alternative has a near zero chance of being true. But I wanted to provide another, related piece of the puzzle $-$ &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmological_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Cosmological Principle&lt;/a&gt;. This is notion that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the spatial distribution of matter in the universe is homogeneous and isotropic when viewed on a large enough scale.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What this means is that no matter where you are in the Universe, be it on Earth or in a galaxy far far away, the universe will appear to look the same to you in every direction you look. If this principle is true, it implies something very important about our universe: &lt;em&gt;There is no &quot;special&quot; location in the Universe which distinguishes it from the rest of the Universe.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the universe was expanding from a single point due to an explosion, that point of expansion would be the center of the universe and thus hold a special significance. If you were at that point, you'd know it. Any observations at that point would clearly look different than if you were anywhere else in the Universe. If the Cosmological Principle is true, that point can't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone really spends the time to think about it, this isn't a great reason as to why we know space itself is expanding (Phiteros' answer is the great reason). What I'm basically saying is, I'm making up a principle that says there can't be a center of the universe and if that principle is true, the center of the universe can't exist. It's a little tautological, but there is good evidence that the Cosmological Principle is true. We've performed large scale observations of the distribution of galaxies and matter throughout our universe and the evidence shows that on large scales, the universe does indeed appear to be homogeneous and isotropic.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-06-20T12:53:15.700" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21339" PostTypeId="1" CreationDate="2017-06-20T12:59:20.197" Score="1" ViewCount="160" Body="&lt;p&gt;My question is similar to the one posted by Abtract, whom however never got a useful answer. I am trying to calculate from scratch the exact times of sunrise and sunset anywhere on earth taking into account the 'equation of time' and other time dependent variables. This requires computing the line of sight to the rim of the sun at any universal time instant. To do this one needs an invariant as possible coordinate systems, preferably centered at the sun. I have tried using the ecliptic plane and its normal as basis, but with only moderate success (up to about a minutes difference from published values). Now many texts define the ecliptic plane as the one which contains the yearly path of the earth around the sun or, vice versa, the apparent path of the sun around the earth.&#xA;However, unless stated as an &quot;average path&quot; over a certain (long - say 100 years) time frame, then the normal of this plane must wobble monthly due to the pertubations caused by the moon (ignoring the other planets) and the intersection between the earth's equatorial plane and the ecliptic also varies. So is the ecliptic actually the path of the earth-moon barycenter (in which case it would be more invariant) or is there another more stringent definition I can use to get accurate timings? So exactly what is the best polar coordinate system and what are it's commonly named coordinates and parameters.&lt;/p&gt;&#xA;" OwnerUserId="17223" LastActivityDate="2017-06-20T12:59:20.197" Title="Ecliptic plane, its precise definition useful for calculations" Tags="&lt;observational-astronomy&gt;&lt;orbital-elements&gt;&lt;declination&gt;&lt;inclination&gt;&lt;ascension&gt;" AnswerCount="0" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21340" PostTypeId="2" ParentId="21334" CreationDate="2017-06-20T13:05:19.250" Score="6" Body="&lt;p&gt;This excellent question continues to confuse laymen as well as professional astronomers. In the standard, general relativistic (GR), cosmological model, the Universe is described as expanding, with galaxies lying approximately still in space and being &amp;quot;dragged along&amp;quot;. Thus, every observer sees the other galaxies receding. The Universe started out in an &lt;span class=&quot;math-container&quot;&gt;$\sim$&lt;/span&gt;infinitely dense state — either finite or infinite in size — and has been expanding ever since, not &lt;em&gt;into&lt;/em&gt; something, but &amp;quot;by itself&amp;quot;.&lt;/p&gt;&#xA;&lt;strong&gt;Special relativistic model&lt;/strong&gt;&#xA;&lt;p&gt;A more intuitive, &amp;quot;explosion-like&amp;quot;, description of the recession of galaxies would in principle be possible: In this model, Big Bang happened at a specific point in an otherwise empty space. Particles traveled outwards from this point at all possible velocities less than &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; (&lt;em&gt;left&lt;/em&gt; panel in the figure below). The slow particles are still close to this special point; the faster ones are farther away, and the fastest ones — traveling at almost &lt;span class=&quot;math-container&quot;&gt;$v=c$&lt;/span&gt; — comprise a &amp;quot;surface&amp;quot; of the cloud of debris. In this model, the Universe is not isotropic. But to any observer who is not at the edge of this cloud of debris, it &lt;em&gt;would&lt;/em&gt; look isotropic, at least for a sufficiently small region of the cloud, and the observer would still see galaxies recede at velocities proportional to their distance from him/her. All it requires is that the observer's observable Universe is sufficiently small.&lt;/p&gt;&#xA;&lt;p&gt;Thus, the answer provided by Phiteros is unfortunately wrong, and is not a proof that the SR model is wrong.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0jlaV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0jlaV.png&quot; alt=&quot;expansion&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;strong&gt;Time dilation&lt;/strong&gt;&#xA;&lt;p&gt;The time dilation of various &amp;quot;standard clock&amp;quot; processes in distant galaxies (e.g. the time it takes for a supernova to fade away) has been proposed as a way to distinguish the GR model from the SR model. However, it turns out that the time dilation factor in the same in the two models, namely &lt;span class=&quot;math-container&quot;&gt;$1+z$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; is the redshift of the galaxy. So this doesn't work either. It &lt;em&gt;does&lt;/em&gt;, however, rule out models that don't predict any time dilation at all, such as the &amp;quot;tired light&amp;quot; hypothesis.&lt;/p&gt;&#xA;&lt;strong&gt;Supernova magnitudes&lt;/strong&gt;&#xA;&lt;p&gt;The best &amp;quot;proof&amp;quot; (in quotes because nothing is ever really proven in physics, only verified or falsified) of GR is given in &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0310808&quot; rel=&quot;nofollow noreferrer&quot;&gt;Davis &amp;amp; Lineweaver (2004)&lt;/a&gt;, I think. Because of the mechanism that makes supernovae type Ia explode, these objects are thought to reach a known brightness at their peak luminosity, and can thus be used as &amp;quot;standard candles&amp;quot;. Plotting their magnitudes (in the &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; band, taken from &lt;a href=&quot;https://arxiv.org/abs/astro-ph/9812133&quot; rel=&quot;nofollow noreferrer&quot;&gt;Perlmutter et al. 1999&lt;/a&gt;) &lt;span class=&quot;math-container&quot;&gt;$m_B$&lt;/span&gt; vs. their redshift, the result can be compared to the predictions of various cosmological models. In the standard GR model, the result depends on the &lt;em&gt;history&lt;/em&gt; of the expansion, which in turn depends on the densities of the various constituents of the Universe (baryons, dark matter, dark energy, radiation). This is the reason that SN1a can be used to constrain the densities of the components. In the special relativistic (SR) explosion model, however, the result only depends on the velocity of the emitter &lt;em&gt;at the time of emission&lt;/em&gt;, and the velocity of the observer &lt;em&gt;at the time of observation&lt;/em&gt;. Whatever happened in between doesn't matter (in this models nothing special happened in between&lt;span class=&quot;math-container&quot;&gt;$^\dagger$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;p&gt;The result is shown in the figure below for various GR models (&lt;em&gt;thin black lines&lt;/em&gt;) and for the SR model (&lt;em&gt;thick black line&lt;/em&gt;). The SR model is ruled out at &lt;span class=&quot;math-container&quot;&gt;$23\sigma$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JdrbK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JdrbK.png&quot; alt=&quot;Mz relations&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;Figure 5. from Davis &amp;amp; Lineweaver (2004).&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^\dagger$&lt;/span&gt;&lt;sub&gt;As the authors note, &amp;quot;1) SR could be manipulated to give an evolving Hubble’s constant, and 2) SR could be manipulated to give a non-trivial relationship between luminosity distance, &lt;span class=&quot;math-container&quot;&gt;$d_L$&lt;/span&gt;, and proper distance, &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt;. However, it is not clear how one would justify these ad hoc corrections&amp;quot;.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-06-21T13:02:06.377" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21341" PostTypeId="2" ParentId="21337" CreationDate="2017-06-20T14:24:23.713" Score="2" Body="&lt;p&gt;First off, let me point out that if you have a central planet which is orbited by other bodies, those other bodies must necessarily be considered moons and cannot be planets. This is due to the &lt;a href=&quot;https://en.wikipedia.org/wiki/IAU_definition_of_planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;official definition of a planet&lt;/a&gt; stating that it must orbit the central star. The objects orbiting your central planet are not primarily orbiting the central star and thus would be considered moons or satellites.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Orbital Dynamics&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;What you're proposing is that your central planet is orbited by two moons that are on opposite sides of the planet. This condition provides some strong constraints on your system as dictated by the laws of physics.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If we are to consider that the moons are orbiting around the central planet, then the planet must be fairly massive. If all three of your bodies are of a similar or comparable mass, you'd have a much different (and not necessarily stable) type of orbit than what you're describing (see &lt;a href=&quot;http://www.atlasoftheuniverse.com/orbits.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this link&lt;/a&gt; for example). The fact that you have two moons orbiting your planet, without your planet having a noticeable orbit about the moons, implies the planet has most of the mass of the system. This results in one of two scenarios, either &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;your central planet is very large (e.g, the size of Jupiter or Saturn) with moons the size of Mercury or &lt;a href=&quot;https://en.wikipedia.org/wiki/Ganymede_(moon)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ganymede&lt;/a&gt;, or&lt;/li&gt;&#xA;&lt;li&gt;your central planet is relatively small (e.g., the size of Earth or Venus) with moons the size of asteroids (e.g., &lt;a href=&quot;https://en.wikipedia.org/wiki/Moons_of_Mars&quot; rel=&quot;nofollow noreferrer&quot;&gt;Phobos and Deimos&lt;/a&gt;, the moons of Mars)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of course you could have any gradation between the two possibilities I just listed. The main point is, your planet must be much more massive than your moons in order to be the dominant gravitational player in its system.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You also state your moons are orbiting such that they're on opposite sides of the planet. Such an orbital configuration is known as a 1:1 &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_resonance&quot; rel=&quot;nofollow noreferrer&quot;&gt;resonance&lt;/a&gt;. It necessarily implies that your moons are orbiting at the exact same distance from the planet and take the same amount of time to complete a single orbit. While such a system is technically stable, I can't say with certainty that it is stable for a long period of time. More than likely, small perturbations will occur that will build up over time and cause your moons to leave their 1:1 orbital resonance. This will result in unpredictable outcomes, but my best guess is that the moons may start gravitationally interacting if they get close enough after no longer being on the same side of the planet. Eventually this may result in them exchanging energy and one getting kicked to a closer orbit while another getting kicked to a further orbit. But that's just a guess.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So short answer, yes your system is possible but it comes with some constraints and is likely not going to be stable for a long time period.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Habitability&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;You also ask about the habitability of such a system. This is a bit tricky and there's no single answer. There are so many conditions that affect habitability. To make matters worse, we only have a single example of a habitable planet. However, there are general conditions we believe must be true for a planet to be habitable (for humans at least).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big condition is that your system must be at a good temperature such that water can exist as a liquid. If your central planet is a Jupiter/Saturn analog, this is not going to happen since such planets have large, mostly hydrogen gaseous envelopes. There's not really any pools of water for humans. However, your moons could still host water. It really depends on how far out your planet is from the star. These types of planets form very far from the star, where water doesn't easily exist as a liquid so either your planet must migrate in to the inner stellar system early on (and stop at a respectable distance), or else some sort of tidal effect has to keep your moons much warmer than they would be other wise (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Enceladus&quot; rel=&quot;nofollow noreferrer&quot;&gt;Enceladus&lt;/a&gt;, for example). The former means your chosen moon orbits are likely to be less stable and the latter means you're going to need other moons to help provide the heating, again making your orbits less stable. If your central planet is Earth-like, it can certainly host water and be as Earth-like as you want, but the moons are unlikely to be big enough to be habitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various other conditions for habitability such as nice atmospheres or magnetospheric protection from the star but one could write a whole book on this subject. I'll leave off by pointing you to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/15074/are-there-any-common-characteristics-of-habitable-planets&quot;&gt;this question on habitability&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-06-20T14:24:23.713" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21342" PostTypeId="2" ParentId="21224" CreationDate="2017-06-21T09:25:07.923" Score="2" Body="&lt;p&gt;It's a little hard to know how to answer your question, partly because you're using &quot;nucleus&quot; in a confusing fashion. So let me try answering what &lt;em&gt;might&lt;/em&gt; be your general question, which is &quot;how do stars move in the region spanned by the bar?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&quot;Nucleus&quot; is generally used to refer to the inner few hundred parsecs -- or even, these days, just the inner ten parsecs or so. Since bars extend to anywhere from several hundred parsecs to as much as ten thousand parsecs in radius, they are usually well outside the &quot;nucleus&quot;. [I am deliberately ignoring so-called &quot;nuclear bars&quot; -- which are generally only a few hundred parsecs in size -- in order to keep things simple ;-)])&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars in a disk galaxy generally all orbit in the same direction and in the same plane. Some of the orbits are close to circular, while others are more elongated. These elongated orbits &lt;em&gt;precess&lt;/em&gt; -- that is, the axis of the apparent ellipse rotates with time, so the star traces out a rosette. In a barred galaxy, there are a lot of stars which are on elongated orbits which are &lt;em&gt;aligned&lt;/em&gt;, and which precess at the same rate. Thus, they maintain a common oval region that is dense with stars (the &quot;bar&quot;) which rotates as a whole at a more-or-less constant speed (even as the stars making up the bar are doing their own individual orbits at different speeds). The combined gravity of all these stars in an oval configuration is what keeps the whole thing synchronized and consistent, although there are some stars on more chaotic orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There &lt;em&gt;is&lt;/em&gt; a kind of magic radius called &quot;corotation&quot;. This is where an imaginary star in a circular orbit would go once round the center of the galaxy in the same time that the bar as a whole make one complete turn. Stars &lt;em&gt;inside&lt;/em&gt; corotation go once around &lt;em&gt;faster&lt;/em&gt; than the bar turns, while stars outside this radius would go once round slower than the bar. Stars orbiting inside the corotation radius can form part of the bar (and generally &lt;em&gt;have&lt;/em&gt; to be part of the bar if they're not on some very chaotic orbit). Stars outside corotation &lt;em&gt;cannot&lt;/em&gt; be part of the bar; their orbits will actually tend to be elongated perpendicular to the bar, though as you get further away from the bar, its gravitational influence becomes less and less important, so that stars further out can have orbits that are pretty much circular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is good evidence -- both theoretical and observational -- that bars cannot extend further out than the corotation radius, and that bars in fact tend to extend about 3/4 or 4/5 of the way to corotation (with the stars further out but still inside corotation probably having very chaotic orbits).&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2017-06-21T09:25:07.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21344" PostTypeId="1" AcceptedAnswerId="21345" CreationDate="2017-06-21T12:33:22.333" Score="8" ViewCount="2544" Body="&lt;p&gt;I would like to know the time it takes to go from sunset to dark based on GPS location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, ideally I would also like to have a formula that determines the percentage of sunlight at each minute past sunset. For example, after 15 minutes, there is only 50% sunlight left. &lt;/p&gt;&#xA;" OwnerUserId="17171" LastEditorUserId="17171" LastEditDate="2017-06-21T20:45:51.990" LastActivityDate="2017-06-21T20:45:51.990" Title="How can I calculate the time between sunset and actual darkness?" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;light&gt;&lt;sun-rays&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21345" PostTypeId="2" ParentId="21344" CreationDate="2017-06-21T13:35:33.857" Score="15" Body="&lt;p&gt;One may define three forms of &lt;a href=&quot;https://en.wikipedia.org/wiki/Twilight&quot; rel=&quot;noreferrer&quot;&gt;twilight&lt;/a&gt; on Earth.  Although the actual amount of light depends on weather, topography, and land cover, they are defined as:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Civil twilight: Solar angle &gt; -6°&lt;/li&gt;&#xA;&lt;li&gt;Nautical twilight: Solar angle &gt; -12°&lt;/li&gt;&#xA;&lt;li&gt;Astronomical twilight: Solar angle &gt; -18°&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Human eyes see logarithmically so % of sunlight left is not a very useful measure, unless you are interested in incoming shortwave radiation for solar panel or other energy calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question, you need to calculate at what time the Sun will be at those elevations below the horizon; this is calculated in a similar fashion to sunrise and sunset times.  However, on top of that you also need to consider atmospheric refraction.  See &lt;a href=&quot;https://en.wikipedia.org/wiki/Position_of_the_Sun&quot; rel=&quot;noreferrer&quot;&gt;position of the Sun&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Sunrise_equation&quot; rel=&quot;noreferrer&quot;&gt;sunrise equation&lt;/a&gt;.  The formulae are quite complicated, but you may be able to simplify them if you're only interested in the duration, not the absolute times.&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;The time after sunset at which those occur are a function of latitude and time of year.  On this &lt;em&gt;carpet plot&lt;/em&gt; from Wikipedia you can read the length of different forms of twilight at 70°N (north of the Arctic circle):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/a/a9/Carpet_Plot_of_Sun-Elevation_over_a_whole_Year_-_North_of_the_Polar_Circle.png/1024px-Carpet_Plot_of_Sun-Elevation_over_a_whole_Year_-_North_of_the_Polar_Circle.png&quot; alt=&quot;carpet plot, 70°N&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1462" LastEditorUserId="1462" LastEditDate="2017-06-21T13:49:38.783" LastActivityDate="2017-06-21T13:49:38.783" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21346" PostTypeId="2" ParentId="21207" CreationDate="2017-06-21T20:39:04.673" Score="2" Body="&lt;p&gt;As Rob Jeffries suggested, in the WFPC2 observations the progenitor star fell located on one of the wide-field (&quot;WF&quot;) chips of WFPC2. (I verified this by downloading one of the preview images from the Mukulski Archive.) This was purely by chance, since the WFPC2 observations were taken years before the discovery of the failed supernova. It's good luck that the region of the galaxy where the failed SN took place was imaged by WFPC2, and only slightly bad luck that it wasn't in a region imaged by the planetary-camera (&quot;PC&quot;) chip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WFPC2 is/was an array of four equal-sized CCD chips, each with 800 x 800 15-micron-wide pixels. However, the optics were set up so that 3 of the chips (the WF chips) were imaged using a focal ratio of F/12.9, while the fourth (the PC chip) was imaged separately with a focal ratio of F/28.3. In essence, a magnified optical beam was focused on the PC chip, giving a scale of about 3.0 milli-arcseconds per micron, or 0.0455 arc seconds per pixel. Meanwhile, the WF chips received a scale of 6.67 milli-arcseconds per microns, or 0.097 arc seconds per pixel. This undersamples the actual resolution of the telescope (which has a point-spread function with a full-width half-maximum of about 0.09 arc seconds in the I-band).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The UVIS (UV-visual) part of WFC3 has two 2k x 4k CCDs (with, I think, 15 x 15 micron pixels) and a focal ratio of F/31, which works out to 0.0395 arc seconds per pixel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in practice, WFC3 will always resolve things better than the WF chips of WFPC2 can, because its pixels can properly sample the telescope resolution and the WF pixels cannot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see the difference in the upper &quot;progenitor&quot; (WFPC2) images, compared with the lower &quot;2015&quot; (WFC3/UVIS) images from &lt;a href=&quot;http://arxiv.org/abs/1609.01283&quot; rel=&quot;nofollow noreferrer&quot;&gt;the actual paper&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ROukc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ROukc.jpg&quot; alt=&quot;From Figure 1 of the original paper&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the stronger visible pixelation in the WFPC2 images (top two panels), due to the fact that the WF pixels are larger in angular size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect any other apparent differences in the press-release images are probably due to the processing used to create the color image. In particular, the side-by-side (WFPC2 vs WFC3/UVIS = &quot;before and after&quot;) color image from the press release shows both images at the same spatial scale, as part of a single image file, which means that the same pixel density is present throughout the combined image. But since each &lt;em&gt;original&lt;/em&gt; WF pixel is about 2.5 times the (angular) size of a WFC3 pixel, the WFPC2 image had to be rebinned to match WFC3 pixel scale (1 WF pixel = 6.25 WFC3 pixels) before the combined (JPEG/PNG/TIFF) color image could be made. The interpolation used for this rebinning process may be what introduced the extra smoothing in the WFPC2 image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edited to add comment about rebinning used to generate color press-release image.]&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2017-06-26T07:39:36.690" LastActivityDate="2017-06-26T07:39:36.690" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21347" PostTypeId="1" CreationDate="2017-06-21T21:30:38.670" Score="1" ViewCount="327" Body="&lt;p&gt;chemistry tells us that when mixing hydrogen and oxygen the result is an explosion and water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my question is: does this happen on astronomical scale? for example, did they ever observe planet-size hydrogen cloud collide with planet-size oxygen cloud to produce a huge explosion and planet made primary of water?&lt;/p&gt;&#xA;" OwnerUserId="17237" LastActivityDate="2021-01-27T15:58:02.510" Title="Hydrogen and oxygen in space" Tags="&lt;planetary-formation&gt;&lt;hydrogen&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21348" PostTypeId="1" AcceptedAnswerId="21349" CreationDate="2017-06-21T22:50:54.420" Score="9" ViewCount="4212" Body="&lt;p&gt;It may be surprising to know that all four giants in our solar system have rings orbiting around them.  Yet only Saturn has a system of rings visible to the naked eye.  Why is that?  What's so different about Saturn in comparison to Jupiter, Uranus and Neptune that makes it the only planet to have a visible ring system?&lt;/p&gt;&#xA;" OwnerUserId="10085" LastActivityDate="2017-06-23T19:38:28.597" Title="Why Only Saturn Has Visible Rings" Tags="&lt;planetary-formation&gt;&lt;planetary-ring&gt;&lt;formation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21349" PostTypeId="2" ParentId="21348" CreationDate="2017-06-22T01:21:39.970" Score="9" Body="&lt;p&gt;By &quot;visible to the naked eye&quot;, I take it you mean &quot;visible from Earth with a small telescope&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn's rings are &lt;a href=&quot;https://en.wikipedia.org/wiki/Rings_of_Saturn#Physical_characteristics&quot; rel=&quot;noreferrer&quot;&gt;largely water ice&lt;/a&gt;, and so they reflect more sunlight back to us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter's rings, have &lt;a href=&quot;https://en.wikipedia.org/wiki/Rings_of_Jupiter#Discovery_and_structure&quot; rel=&quot;noreferrer&quot;&gt;lower proportions of ice&lt;/a&gt;, and lots of smaller dust particles that tend to scatter light forward rather than back to us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ring systems of Uranus and Neptune are made of really dark material, so also don't reflect much light in our direction.&lt;/p&gt;&#xA;" OwnerUserId="17238" LastEditorUserId="17238" LastEditDate="2017-06-22T01:30:06.147" LastActivityDate="2017-06-22T01:30:06.147" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21350" PostTypeId="2" ParentId="21347" CreationDate="2017-06-22T01:59:39.177" Score="5" Body="&lt;p&gt;Hydrogen and oxygen only react when there is sufficient energy. For instance, the autoignition temperature of hydrogen at 1 atmosphere is &lt;a href=&quot;https://en.wikipedia.org/wiki/Autoignition_temperature#Autoignition_point_of_selected_substances&quot; rel=&quot;noreferrer&quot;&gt;536 °C&lt;/a&gt;. This is why you can do that experiment with mixed hydrogen and oxygen in a balloon, that only explodes when you touch the balloon with a lit taper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space is cold. Molecular clouds have temperatures in the &lt;a href=&quot;http://coolcosmos.ipac.caltech.edu/cosmic_classroom/cosmic_reference/molecular_clouds.html&quot; rel=&quot;noreferrer&quot;&gt;tens of degrees Kelvin&lt;/a&gt;. In the rare (if not impossible) case that a molecular cloud of pure oxygen did exist, and it encountered a molecular cloud of hydrogen, there wouldn't be enough energy for an explosion.&lt;/p&gt;&#xA;" OwnerUserId="17238" LastEditorUserId="17238" LastEditDate="2017-06-22T04:26:46.253" LastActivityDate="2017-06-22T04:26:46.253" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="21351" PostTypeId="2" ParentId="21324" CreationDate="2017-06-22T02:13:34.843" Score="2" Body="&lt;p&gt;They look similar to the &quot;dark dune spots&quot; of the southern hemisphere, associated with accumulation of melting liquid at the bottom of dunes during warming seasons: &lt;a href=&quot;https://en.wikipedia.org/wiki/Geyser_(Mars)&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Geyser_(Mars)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17238" LastActivityDate="2017-06-22T02:13:34.843" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21352" PostTypeId="2" ParentId="21347" CreationDate="2017-06-22T05:29:06.950" Score="2" Body="&lt;p&gt;While an oxygen ball of gas isn't possible, an Oxygen rich atmosphere on a planet, through photosynthesis is possible.  Earth for example.   A hydrogen rich planet (Jupiter or Saturn) are also possible, as is a methane rich atmosphere (Titan).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two bodies like that were to collide, you probably would see an enormous flame by the oxygen and fuel combustion process.  That said, the gravitational impact would generate far more energy than any chemical combustion.  If you were to collide two planets with atmospheres that together could combust, I'm not sure it would make much difference than two planets combining with atmospheres that wouldn't combust.  The impact is what would stand out. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-06-22T05:29:06.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21356" PostTypeId="2" ParentId="7801" CreationDate="2017-06-23T02:46:54.967" Score="1" Body="&lt;p&gt;According to absorber theory of Wheeler, photon  is a derived concept. Photon is the action between two charges. The action is called &quot;action at a distance&quot; i.e. aaad. Hence photon must start from an emitter ended on an absorber. Even the photon is just emitted from an emitter, it knowns its ending absorber which perhaps is an eye of a man who will bore in hundreds years latter.&lt;/p&gt;&#xA;" OwnerUserId="17247" LastActivityDate="2017-06-23T02:46:54.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21357" PostTypeId="2" ParentId="21329" CreationDate="2017-06-23T04:11:28.833" Score="-1" Body="&lt;p&gt;The Earth and moon orbit around a barycentre that is &quot;about 3/4 the radius of Earth&quot; away from the Earth's centre: &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbit_of_the_Moon&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Orbit_of_the_Moon&lt;/a&gt;. Removing the moon would almost certainly affect the Earth's orbit around the sun due to the change in the centre of mass, and probably a resultant change in momentum of the Earth system (such as it was). That's without even considering the gravitational tug of the planetoid on the Earth, and that of the moon as it gets snatched away. I would think it spectacularly unlikely for the interaction to leave Earth travelling along the orbit of the centre of mass of the old Earth-moon system. But, I haven't actually modelled it, so I might be wrong there.&lt;/p&gt;&#xA;" OwnerUserId="17238" LastActivityDate="2017-06-23T04:11:28.833" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21358" PostTypeId="2" ParentId="13429" CreationDate="2017-06-23T04:19:48.673" Score="1" Body="&lt;p&gt;@tefalya   Somebody told me  &quot;EM is the integral of electron density squared over the emitting volume.  In collisional ionized plasmas, it gives the amount of material emitting at a given temperature.  In the low density regime, temperature, emission measure, and abundance together with an atomic code completely specify the observed spectral energy distribution of coronal plasma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The normalization parameter you get out of APEC, MEKAL, etc. relate to the emission measure.  If you read the description of the APEC model in XSPEC, you will see that the normalization = 10^-14* n_e^2 dV/(4 !pi distance^2).  The Emission Measure EM is n_e^2 dV, and has units of cm^-3. So the emission measure is an observable, and by assuming that the volume is not changing significantly during the flare, the changing emission measure is a proxy for the changing electron density.&quot;&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2017-06-23T04:19:48.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21362" PostTypeId="1" AcceptedAnswerId="21371" CreationDate="2017-06-23T12:47:46.970" Score="-1" ViewCount="98" Body="&lt;p&gt;I have really little knowledge of Astronomy in general. Last Monday, I tested my new camera by doing a Night-Sky timelapse. After a little post-processing, i noticed dozens of light streaks going in all directions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't tell if those are meteors, airplanes or satelites. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you help me out? This is the video: &lt;a href=&quot;https://www.dropbox.com/s/y4jfiua8blzanpe/test2.2_h264-420_1080p_25_MQ.mp4?dl=0&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.dropbox.com/s/y4jfiua8blzanpe/test2.2_h264-420_1080p_25_MQ.mp4?dl=0&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some information about the shots: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Location: Central Switzerland&lt;/li&gt;&#xA;&lt;li&gt;Camera direction: Straight south&lt;/li&gt;&#xA;&lt;li&gt;Camera inclination app. 30°&lt;/li&gt;&#xA;&lt;li&gt;Date: 19.6.-20.6.&lt;/li&gt;&#xA;&lt;li&gt;Time: app.22:57-02:50&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot guys, Marc&lt;/p&gt;&#xA;" OwnerUserId="17252" LastActivityDate="2017-06-24T11:44:44.593" Title="Night sky timelapse - SO MANY METEORS?" Tags="&lt;amateur-observing&gt;&lt;photography&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21364" PostTypeId="2" ParentId="21348" CreationDate="2017-06-23T19:38:28.597" Score="3" Body="&lt;p&gt;Building on what hartacus said, there's a couple other reasons involved as well:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uranus and Neptune are much further out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At closest approach to earth the distance to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Saturn is about 746 million miles (1.2 billion km) &lt;/li&gt;&#xA;&lt;li&gt;Uranus is about 1.7 billion miles (2.5 billion km) &lt;/li&gt;&#xA;&lt;li&gt;Neptune is about 2.7 billion miles (4.3 billion km)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So Uranus and Neptune are very far away... and they're also smaller, AND their rings are not as dense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an amateur telescope, it's very hard to see Uranus as anything other than a small, dim, fuzzy, blueish disk.  Neptune is just a blue-green dot.  The rings are so insubstantial, in comparison, that they just don't stand out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of Jupiter, the problem is simply that they're so insubstantial.  The Jovian rings are very thin when compared to Saturn's.  Two flyby missions, Pioneer 10 and Pioneer 11, flew by Jupiter without noticing them at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combined with the difference in composition and albedo, they're just not visible to smaller Earth-based telescopes.&lt;/p&gt;&#xA;" OwnerUserId="10546" LastActivityDate="2017-06-23T19:38:28.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21365" PostTypeId="1" AcceptedAnswerId="21385" CreationDate="2017-06-23T19:41:25.513" Score="4" ViewCount="172" Body="&lt;p&gt;Did we ever witness a &quot;meteor shower&quot; which was actually dust or debris of some kind from our own Moon?&lt;/p&gt;&#xA;" OwnerUserId="1538" LastActivityDate="2017-06-25T04:41:50.413" Title="Meteor shower from Moon's &quot;debris&quot;" Tags="&lt;the-moon&gt;&lt;meteor&gt;&lt;meteorite&gt;&lt;meteor-shower&gt;&lt;meteoroid&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21369" PostTypeId="1" CreationDate="2017-06-24T07:25:51.873" Score="0" ViewCount="111" Body="&lt;p&gt;I know that after a star undergoes the process of mass loss, depending on the mass of the core the stellar remnants gets converted into a white dwarf star, neutron star or a black hole. Hence, if the mass of the stellar remnant exceeds the Tolman–Oppenheimer–Volkoff limit, then it gets converted to a black hole. So is there currently any equation which helps us to calculate the amount of time taken by the leftover core to get converted to a black hole or the rate at which the mass is gravitationally collapsing?&lt;/p&gt;&#xA;" OwnerUserId="16984" LastEditorUserId="2365" LastEditDate="2017-06-25T20:32:33.513" LastActivityDate="2017-06-25T20:32:33.513" Title="Question about stellar remnants" Tags="&lt;astrophysics&gt;&lt;stellar-evolution&gt;&lt;quantum-mechanics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21371" PostTypeId="2" ParentId="21362" CreationDate="2017-06-24T11:44:44.593" Score="1" Body="&lt;p&gt;Some are probably satellites, others are probably planes. Meteors are highly improbable in this case. Reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are no notable meteor showers visible at your given location and time.&lt;/li&gt;&#xA;&lt;li&gt;Even if there were, as someone in comments pointed out, meteors tend to appear as if they all came from the same point in the sky. This is clearly not the case&lt;/li&gt;&#xA;&lt;li&gt;Most meteors occurrences are very brief. A meteor lasting more than a few seconds is very rare (some well-known showers are exceptions to this, but they are certainly not active at your given time). Hence, when compressing a period of four hours into twelve seconds, one second in real-time gets compressed to less than $0.001$ seconds. Most captured meteors wouldn't be appear for that long on your timelapse.&lt;/li&gt;&#xA;&lt;li&gt;Although the difference between airplanes and satellites is hard to spot on a timelapse (the main difference is that planes have blinking lights, whereas satellites do not. The frequency of the blinking is too fast to be visible on this timelapse), most airplanes don't disappear halfway through the sky. They would appear at the horizon and remain visible until they disappear behind the horizon again. The visibility (and luminosity) of satellites, however, depends on the orientation of the satellite, which usually causes a satellite to 'disappear' mid-flight. Some tracks on your timelapse do cover the entire width of the screen, and those are probably planes&lt;/li&gt;&#xA;&lt;li&gt;If all of these are satellites, then that would be the highest amount of satellites I've ever seen on a time span of four hours.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you want to further track their origin, there exists software for both airplanes and satellites that can tell you exactly which ones will be visible at a given location and time. A quick Google search will give you plenty of suggestions. On my smartphone, I use an app called heavens-above to be aware of satellites passing by on an observing night, but I don't think it can calculate back to the past.&lt;/p&gt;&#xA;" OwnerUserId="11824" LastActivityDate="2017-06-24T11:44:44.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21372" PostTypeId="1" AcceptedAnswerId="24506" CreationDate="2017-06-24T13:02:11.677" Score="3" ViewCount="167" Body="&lt;p&gt;According to this article &lt;a href=&quot;http://news.nationalgeographic.com/2016/02/160209-meteorite-death-india-probability-odds/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://news.nationalgeographic.com/2016/02/160209-meteorite-death-india-probability-odds/&lt;/a&gt; a science professor at Tulane University made a paper to calculate the probability of a person being hit by a meteor on Earth . And also an astronomer made a calculation (not mentioning the procedure) with a similar figure. My question is, since the moon has no atmosphere and more meteors reach the surface, is there any mathematical formula, paper, etc. to calculate how likely a spot in moon's surface it's likely to be hit by a meteor? &lt;/p&gt;&#xA;" OwnerUserId="11562" LastEditorUserId="11562" LastEditDate="2017-07-21T17:56:47.743" LastActivityDate="2018-01-12T17:41:58.113" Title="is there any mathematical formula, paper or anything that shows the probability of a meteor hitting a particular spot of moon's surface?" Tags="&lt;the-moon&gt;&lt;meteor&gt;&lt;impact-probability&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21373" PostTypeId="1" AcceptedAnswerId="21377" CreationDate="2017-06-24T16:33:09.227" Score="3" ViewCount="89" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GMOSk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GMOSk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was helping classify galaxies at &lt;a href=&quot;https://www.galaxyzoo.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Galaxy Zoo&lt;/a&gt; when I came across this picture of a galaxy. The dull red body looks huge compared to the galaxy. Is it because it is closer to the earth? It would be interesting to know what is going on here.&lt;/p&gt;&#xA;" OwnerUserId="17262" LastActivityDate="2017-06-27T13:39:23.940" Title="Is the bright red body a star?" Tags="&lt;galaxy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21374" PostTypeId="1" CreationDate="2017-06-24T16:37:25.453" Score="1" ViewCount="100" Body="&lt;p&gt;I know that the radius of the observable universe is about 46.6 billion light years and is expanding at every instant of time due to dark energy . So is it possible that we our entire universe could be inside a black hole and that the edge of our observable universe represents the event horizon which in this case is expanding faster than the speed of light due to matter being constantly accreted inside it . Where that black hole is a part of another universe and that universe of another black hole and the pattern goes on . Is this possible ?&lt;/p&gt;&#xA;" OwnerUserId="16984" LastActivityDate="2017-06-24T17:02:56.173" Title="Universe and black holes" Tags="&lt;black-hole&gt;&lt;cosmology&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21375" PostTypeId="2" ParentId="21374" CreationDate="2017-06-24T17:02:56.173" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I know that the radius of the observable universe is about 46.6 billion light years and is expanding at every instant of time due to dark energy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's some ambiguity about dark energy. It's said to be responsible for the accelerating expansion of the universe as opposed to the expansion. See the Wikipedia &lt;a href=&quot;https://en.wikipedia.org/w/index.php?title=Dark_energy&amp;amp;oldid=785660829#Inflationary_Dark_Energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;dark energy&lt;/a&gt; article where you can read that the term &quot;dark energy&quot; was coined by Michael Turner in 1998. This was in a &lt;a href=&quot;https://arxiv.org/abs/astro-ph/9808133&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper&lt;/a&gt; that referred to the type 1a supernovae observations made by the Supernova Cosmology Project and the High-Z Supernova Search Team. Their observations indicated that the expansion of the universe is accelerating, which was contrary to expectations and led to Nobel prizes in 2011.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So is it possible that we our entire universe could be inside a black hole &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. We aren't in a black hole. A black hole is a place where the vertical light beam can't get out because the &quot;coordinate&quot; speed of light at the event horizon is zero. As &lt;a href=&quot;http://news.bbc.co.uk/1/hi/sci/tech/4249192.stm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stephen Hawking said&lt;/a&gt;, the universe is &lt;em&gt;like&lt;/em&gt; a black hole in reverse, but it isn't the same as a black hole.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and that the edge of our observable universe represents the event horizon which in this case is expanding faster than the speed of light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The edge of our observable universe is a horizon of sorts, see &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0310808&quot; rel=&quot;nofollow noreferrer&quot;&gt;Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the Universe&lt;/a&gt; by Davis and Lineweaver. But it isn't the same sort of horizon as the black hole event horizon. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;due to matter being constantly accreted inside it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That isn't supposed to be the reason why the universe is expanding. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where that black hole is a part of another universe and that universe of another black hole and the pattern goes on. Is this possible ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reputable people have talked about black holes as &lt;a href=&quot;https://www.google.co.uk/#q=black+hole+baby+universe&amp;amp;spf=1498323539447&quot; rel=&quot;nofollow noreferrer&quot;&gt;baby universes&lt;/a&gt; but there's no supporting evidence whatsoever. My understanding of general relativity and cosmology is such that I think such hypotheses are flat out wrong. So I would say no, it is not possible.      &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2017-06-24T17:02:56.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21376" PostTypeId="2" ParentId="21369" CreationDate="2017-06-24T17:04:34.580" Score="2" Body="&lt;p&gt;It depends entirely on the circumstances of black hole formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remnant may be a stable neutron star, but if it accreted enough matter later on (by fallback in a supernova, or from a binary companion), then it may collapse. This could happen at any time after formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the core of the collapsing star forms a proto-neutron star above the TOV limit of stability, then no force can prevent black hole formation. It will happen on the cooling timescale of the proto-neutron star at the centre. Once it is &quot;cold&quot; then the collapse would occur on a free fall timescale - which is about a millisecond in these circumstances. The cooling timescale of a proto neutron is governed by how long the copiously generated neutrinos can be trapped in the core, which could be as long as(!) a few seconds&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-06-25T11:26:44.450" LastActivityDate="2017-06-25T11:26:44.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21377" PostTypeId="2" ParentId="21373" CreationDate="2017-06-24T17:11:07.633" Score="5" Body="&lt;p&gt;It is a bright star. You can see the &quot;diffraction spikes&quot; that are the signature of a point-like object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The star will be in our Galaxy, so is certainly closer than the galaxy in the image. However, this is not why it appears to be big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image appears big because the starlight will be spread by the telescope objects over a finite area - the point spread function. In addition, when a star is very bright, the central CCD pixels saturate and charge can spill into adjacent pixels. Even if that were not the case, there is enough light in the &quot;wings&quot; of the point spread function to produce a large image.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-06-24T17:11:07.633" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21378" PostTypeId="1" AcceptedAnswerId="21387" CreationDate="2017-06-24T19:31:44.877" Score="3" ViewCount="416" Body="&lt;p&gt;Germane (GeH&lt;sub&gt;4&lt;/sub&gt;) has been detected in the atmosphere Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its formation from the elements seems to be thermodynamically unfavorable because its enthalpy of formation is positive and (I think) it has lower entropy than the free elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what the heck is it doing there, considering free germanium would sink and free hydrogen would float?&lt;/p&gt;&#xA;" OwnerUserId="17264" LastActivityDate="2017-06-30T15:10:42.527" Title="What is germane doing in the atmosphere of Jupiter?" Tags="&lt;jupiter&gt;&lt;planetary-atmosphere&gt;&lt;astrochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21380" PostTypeId="1" CreationDate="2017-06-25T00:33:22.597" Score="0" ViewCount="87" Body="&lt;p&gt;I'm an undergrad in physics and astronomy deciding to look into applying for time on a research telescope, and I'm not sure how to proceed with locating/finding telescopes that I can apply for time on. Specifically, I'm interested in looking at the chemical composition of protostellar disk(s). Any applicable information would be nice.&lt;/p&gt;&#xA;" OwnerUserId="11031" LastEditorUserId="11031" LastEditDate="2017-06-25T02:39:06.740" LastActivityDate="2017-06-25T20:45:43.480" Title="How should I find a telescope for research time?" Tags="&lt;telescope&gt;&lt;astrophysics&gt;&lt;space-telescope&gt;&lt;protostar&gt;&lt;astrochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21381" PostTypeId="1" AcceptedAnswerId="21383" CreationDate="2017-06-25T00:52:06.480" Score="5" ViewCount="1274" Body="&lt;p&gt;It seems every once in a while there's a &lt;a href=&quot;https://www.morningticker.com/2017/06/scientists-shocked-by-massive-solar-system-discovery/&quot; rel=&quot;noreferrer&quot;&gt;claim&lt;/a&gt; of a new planet in the Solar System.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand how local planets can stay undiscovered for this long when astronomers are accomplishing what seem to be &lt;em&gt;far&lt;/em&gt; more difficult feats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, it's been a whole century since they had accurate enough measurements to tell that Mercury was precessing in a way that didn't jive with Newtonian mechanics.&lt;br&gt;&#xA;Now, if there are Earth-sized planets out there, shouldn't their presence have a pretty sizable effect on the orbits of bodies such as Neptune and Pluto? How can there be such large planets with no measurable effects until 2017 technology came along?&lt;/p&gt;&#xA;" OwnerUserId="7775" LastActivityDate="2017-06-25T21:54:25.430" Title="How do we not know what's in the Solar System?" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;solar-system&gt;" AnswerCount="4" CommentCount="1" ClosedDate="2017-06-26T01:20:20.040" ContentLicense="CC BY-SA 3.0" />
  <row Id="21382" PostTypeId="2" ParentId="21381" CreationDate="2017-06-25T02:37:49.377" Score="8" Body="&lt;p&gt;The key here is the distance. A hypothetical extra planet such as the &quot;Planet X&quot; I assume you're thinking of would be so far out that its effect on the orbits of other planets would be negligible at best. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/emtd2.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/emtd2.jpg&quot; alt=&quot;Orbit&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hypothetical orbit, shown above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plus, the Kuiper Belt and the Oort cloud are massive and full of other bodies. Tiny gravitational influences on planets we can see could also be due to a number of those. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hypothesis that there might be a planet is in part due to the orbits of certain smaller Kuiper belt objects though, which might have been disrupted while passing by a large body like &quot;Planet X&quot;. So far though those orbits aren't sufficient evidence to conclude that such a planet does exist.&lt;/p&gt;&#xA;" OwnerUserId="11031" LastActivityDate="2017-06-25T02:37:49.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21383" PostTypeId="2" ParentId="21381" CreationDate="2017-06-25T02:44:52.360" Score="13" Body="&lt;p&gt;There could be another planet in the solar system that has not been discovered if that planet had a highly elliptical orbit with a single orbit taking many hundreds of earth years to complete. That planet could be in the part of its orbit further from sun and could stay in that area for a few hundred years, longer than we've had large telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At such a distance it would receive very little sunlight and the chances of visually detecting it would be very small. And we wouldn't know where to look either, which just compounds the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other more distant plants have been discovered but their discovery was due to the effects on the star or stars it is orbiting. We observe a planet passing in front of a star which briefly lowers the apparent magnitude of its parent star. This is a good clue of a planet in orbit. This is impossible in our solar system because an unknown planet would have to come between us and the sun. We would easily see it if that were the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have observed some discrepancies in the orbits of some of the outer planets. This has led to the speculation of another planet. It is very difficult to determine the cause of these small discrepancies though. All bodies orbiting our sun are not only effected by the suns gravity but by the gravitational attraction of all other bodies (including all objects in the asteroid and Kuiper belts). This is a very intense calculation. This attraction constantly changes due to the motion of all the bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simulations have been developed to study this but must take a lot of computing time even with today's high speed computers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps explain the situation.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="10111" LastEditDate="2017-06-25T21:54:25.430" LastActivityDate="2017-06-25T21:54:25.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21384" PostTypeId="1" AcceptedAnswerId="21454" CreationDate="2017-06-25T04:01:32.903" Score="1" ViewCount="122" Body="&lt;p&gt;When planets are formed do only gas and ice giants accrete hydrogen and helium? Does this accretion have to be beyond the &quot;snow line&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, would all hydrogen or helium in the atmosphere of a super-earth come from outgassing (radioactive decay, etc)?&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2017-06-29T16:05:11.433" Title="If a rocky world is large enough (not a gas or ice giant) can it have helium from the primordial disk in its atmosphere?" Tags="&lt;planetary-atmosphere&gt;&lt;planetary-formation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21385" PostTypeId="2" ParentId="21365" CreationDate="2017-06-25T04:41:50.413" Score="5" Body="&lt;p&gt;There is a class of meteorites called &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_meteorite&quot; rel=&quot;noreferrer&quot;&gt;Lunar Meteorites&lt;/a&gt; and several of those have been found on earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it doesn't necessarily indicate whether we have actually witnessed a certain lunar meteor shower. Considering that nearly &lt;a href=&quot;http://www.astronomy.com/magazine/ask-astro/2014/07/space-debris&quot; rel=&quot;noreferrer&quot;&gt;100 metric tons of space debris falls on earth each day&lt;/a&gt;, one would imagine that it'll be colossally impractical to attempt mapping a meteorite to the moon, using a reference to a recorded meteor shower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it might be worthwhile to take a look at &lt;a href=&quot;http://meteorites.wustl.edu/lunar/moon_meteorites.htm&quot; rel=&quot;noreferrer&quot;&gt;this link&lt;/a&gt; which addresses rather interesting questions:&lt;/p&gt;&#xA;&#xA;&lt;h3&gt; How did lunar meteorites get here? &lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Any rock on the lunar surface that is accelerated by the impact of a meteoroid to &lt;b&gt;lunar escape velocity&lt;/b&gt; of $2.38 km / s$ or greater will leave the Moon's gravitational influence. Most rocks ejected from the Moon become captured by the gravitational field of either the Earth or the Sun and go into orbit around these bodies. Over a period of a few years to tens of thousands of years, those orbiting the Earth eventually fall to Earth, whereas those orbiting the Sun may also eventually strike the Earth in a few tens of millions of years after they were launched from the Moon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt; How do we know that they come from the Moon? &lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Chemical compositions, isotope ratios, mineralogy, and textures of the lunar meteorites are all similar to those of samples collected on the Moon during the Apollo missions. These meteorites are rich in a mineral called &lt;b&gt;anorthite&lt;/b&gt; and have high concentrations of aluminum and calcium. Because of some unique aspects about how the Moon formed, the lunar highlands are composed predominantly of anorthite. To the best of our knowledge, anorthite is much less common on asteroids, and on the surfaces of any other planets or planetary satellites.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt; &lt;i&gt; Quotes derived and paraphrased from the linked article. &lt;/i&gt; &lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An up to date &lt;a href=&quot;http://meteorites.wustl.edu/lunar/moon_meteorites_list_alumina.htm&quot; rel=&quot;noreferrer&quot;&gt;catalog&lt;/a&gt; of all lunar meteorites found so far indicates that the heaviest lunar meteorite is a rock (pair) named &lt;a href=&quot;http://meteorites.wustl.edu/lunar/stones/kalahari008.htm&quot; rel=&quot;noreferrer&quot;&gt;Kalahari 008 &amp;amp; 009&lt;/a&gt;, weighing ~$14kg$.&lt;/p&gt;&#xA;" OwnerUserId="10589" LastActivityDate="2017-06-25T04:41:50.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21386" PostTypeId="2" ParentId="21381" CreationDate="2017-06-25T08:46:30.233" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Now, if there are Earth-sized planets out there, shouldn't their&#xA;  presence have a pretty sizable effect on the orbits of bodies such&#xA;  as Neptune and Pluto? How can there be such large planets with no&#xA;  measurable effects until 2017 technology came along?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Short answer:  It's not the mass, it's the distance that really matters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longer answer:  I can't do all the math, but to cover some of the basics.  The &lt;a href=&quot;https://en.wikipedia.org/wiki/Discovery_of_Neptune#Irregularities_in_Uranus.27s_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Discovery of Neptune&lt;/a&gt; is the famous example based on unexplained orbital inconsistencies.  Uranus' orbit wasn't following Newton's laws and the simplest explanation was that there was another undiscovered planet.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you say an Earth sized planet, Earth is about 1/17th the mass of Neptune, so an Earth &quot;Mass&quot; planet in Neptune's orbit would have had 1/17th the effect on Uranus' orbit.  That might have been too small to notice.  I don't want to guarantee that, but it's entirely possible that 1/17th the effect would have gone unnoticed for some time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uranus and Neptune's closest pass is about 10 AU from each other.   For comparisons sake, Uranus and Pluto's closest pass is about 11 AU from each other.   Planet 9 is currently about 20 times the distance from the Sun as Neptune.  That puts it (ballpark) about 600 AU from Neptune currently and in this case, closest pass doesn't matter because it's effectively been in about the same part of the sky since Uranus has been observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the inverse square rule, at some 60 times the distance, that's 1/3,600 times the gravitational effect, and if we give it a mass of about 1/2 the mass of Neptune, that's about 1/7,200th the effect.   That's teeny-tiny.  In fact, it's not much greater than the gravitational effect that Pluto has on Uranus at closest pass  (fun sidebar — &lt;a href=&quot;https://www.quora.com/How-large-would-Neptune-appear-at-Plutos-closest-approach&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pluto gets closer to Uranus than it gets to Neptune&lt;/a&gt; - if this source is right).    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, it's even worse than that because it's not the gravitational pull but the tidal force that should be considered.   Any tug that the theoretical Planet 9 gives to Neptune, it gives a similar tug to the entire inner solar system.  If an object tugs on everything mostly equally, the observed effect is very small.   It's the variation in tug on the outer planets to the inner panets that should be considered, and that's essentially the tidal force, where the 3rd power rule comes into play.   That makes any gravitational perturbations a distant 9th planet would have on the inner planets, essentially negligable.  Lower than a number of large Kuiper belt objects and that would be basically undetectable even with today's instruments.  There's too much gravitational &quot;noise&quot; to detect something that small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, basically Pulchritude's answer — the vast distance is the problem.   Any orbital perturbation from Planet 9 on the outer planets, Uranus and Neptune for even a large object at that distance is smaller than the effect from Pluto.    It might have 5,000 times Pluto's mass, but at 60 times the distance, the tidal force at 60 times the distance is about 200,000 times less.  (It's about 20 Neptune distances from the sun, but orbital perturbations are most noticeable at closest pass, and Neptune and Uranus pass within about 10 AU from each other), so that's where the 60 times more distant comes into play.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Planet 9 moves closer to its perihelion it should become more detectable by orbital wobbles but even at closest pass it's too far for the perturbations to be significant.  And it's likely to be discovered (assuming it exists), long before it reaches perihelion.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="7142" LastEditDate="2017-06-25T20:58:20.233" LastActivityDate="2017-06-25T20:58:20.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21387" PostTypeId="2" ParentId="21378" CreationDate="2017-06-25T09:06:47.370" Score="4" Body="&lt;p&gt;How much specific elements sink inside a planetary caldron with variations of heat and thermals and chemistry, depends not just on their density but also the element's chemistry.   &lt;a href=&quot;https://earthscience.stackexchange.com/questions/4798/what-percent-of-the-earths-core-is-uranium&quot;&gt;This question&lt;/a&gt; discusses Uranium in Earth's core and the top answer suggests (I believe correctly) there's essentially no Uranium in Earth's core, it's largely in the crust because Uranium readily oxidizes, so it forms lighter elements that don't sink but rise with the silicates and basalt, which is (roughly) similar density.   Uranium floats in the Earth's mantle because it binds with Oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Free Oxygen (O2) should destroy any Germane, but there's likely so little free Oxygen in Jupiter's atmosphere that the Germane is able to be present (in a fraction of parts per billion - it's very rare in Jupiter, but it's detectable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, given Jupiter's internal heat, thermodynamically unfavorable doesn't matter much.   There's enough heat that chemistry tends to operate more in an equilibrium balance, and not always in the exothermic direction, like we usually see in most natural reactions at temperatures we see on Earth.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, and at risk of stating the obvious, Germane is one of those &quot;weird&quot; heavy gases, due to the atomic electrical symmetry, and because it's a gas, not a solid and given Jupiter's significant thermal currents, it mixes in Jupiter's atmosphere.   It likely freezes out of Jupiter's upper atmosphere when it gets too cold, but it's present in the lower atmosphere, present enough to be detected anyway.  (I think it's less than 1 ppb).    - That's a brief answer anyway.   I've not read any of the recent articles on this.  They're all pay articles.    I invite anyone who has read them to give a more detailed answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So what the heck is it doing there, considering free germanium would&#xA;  sink and free hydrogen would float?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I want to add that gases behave differently than solids or liquids.   In chemistry, sometimes things mix and form solutions (like water and alcohol), sometimes they don't and for layers (like oil and water).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gases are particularly good at staying mixed and not layering, especially where wind and updrafts are present.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;why aren't scads of other simple components mentioned in discussions&#xA;  of Jupiter's atmosphere, like methylamine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is &lt;a href=&quot;https://en.wikipedia.org/wiki/Methylamine&quot; rel=&quot;nofollow noreferrer&quot;&gt;Methylamine&lt;/a&gt; in Jupiter's atmosphere.   Saturn's too.   &lt;a href=&quot;http://adsabs.harvard.edu/full/1937ApJ....86..321W&quot; rel=&quot;nofollow noreferrer&quot;&gt;This research study&lt;/a&gt; mentions &quot;bands&quot; of it.    &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1029/GL004i005p00203/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;This one also discusses it&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;urea, and oxamic acid, to name just a few?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Urea&quot; rel=&quot;nofollow noreferrer&quot;&gt;Urea&lt;/a&gt; is a solid at room temperature.   It decomposes when melted so it's not stable as a gas.   It melts at 133-135 C.   It might never be stable on Jupiter as a gas since it's not even stable as a liquid above 135 C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Oxamic_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oxamic Acid&lt;/a&gt;, isn't even stable by itself, it's water soluble.    It's also Oxygen abundant, and free Oxygen is rare on Jupiter.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So those last 2 aren't good examples, neither is a stable as a gas at all, Germane is stable in Jupiter's atmosphere, though it's present in very small quantities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important point to consider with trace gases in planetary atmospheres is formation and half life.  Methane, for example isn't stable in Earth's atmosphere.   It reacts with Oxygen, though at a few hundred parts per billion, it's rare enough that it doesn't react a lot.   Methane is also constantly being produced in the stomachs of mammals (gas) but it also gets destroyed, reacting with oxygen whenever it gets near a flame or lightning, combining with Oxygen to form CO2 and Water.   It has a half life of about 10-15 years, but because it's constantly being created by incomplete digestion and decomposition, it's always present in our atmosphere - in about 1 part per million.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CO2 has a longer half life in our atmosphere, mostly it isn't destroyed, it needs to be rained out.   It's also constantly being resperated, destroyed in photosynthesis and re-created by respiration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point of all this is that any trace element in a planet's atmosphere, must have a source of creation and a half life of, destruction if you will.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A curious exception on Earth is Argon.   Argon is a byproduct of radioactive decay and as a noble gas, it doesn't react with anything, so it gets created and it just stays in our atmosphere.   It gets created slowly but because it's never destroyed, it's currently about 1% of our atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GeH4 follows the same basic guideline.  It gets created inside Jupiter and it's stable enough in Jupiter's atmosphere that it's equilibrium between creation and destruction leads to about 1 part per billion.    I don't know nearly enough to say what it's atmospheric half life is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you said, there are &quot;scads&quot; of elements in Jupiter's atmosphere.   There re countless more that are NOT in it's atmosphere.   For an element to be present in an atmosphere it needs some degree of stability and some form of creation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter is nearly 90% hydrogen.   It's not practical for it's hydrogen to float around the heavier elements.   Think of it as a sea of hydrogen, with other stuff in the hydrogen, some mixes and floats, some sinks, some chemically reacts, but the hydrogen is always present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I only meant to touch on some of the basics and not give so long an answer.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I mean literally, how did you figure it out? What did you google? I&#xA;  ask because I'm dealing with hundreds of compounds and I'm hoping you&#xA;  found a nice centralized location for information like that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We're getting a little outside the guidelines here, but briefly, I looked up the molecular structure of each of the compounds you suggested and they didn't look like gases to me.  They looked too complex and crooked shaped, so I was suspicious from the get go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the answer to your question is simply that I checked Wikipedia, which can be long (Urea) or very short (Oxalic acid), and not always 100% reliable.   Once there I did a word search for &quot;boiling point&quot; and &quot;Melting point&quot; when boiling point didn't show up.   For any molecule to be a gas it needs a boiling point, so that's the first quick and easy thing to search.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also look for &lt;a href=&quot;https://en.wikipedia.org/wiki/Phase_diagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;phase diagrams&lt;/a&gt;, which are great, but rare for anything but the most common molecules.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Oxalic acid, I was wrong, it is a solid.   Wikipedia only mentioned it in an aqueous solution.   &lt;a href=&quot;https://pubchem.ncbi.nlm.nih.gov/compound/oxamic_acid#section=Chemical-and-Physical-Properties&quot; rel=&quot;nofollow noreferrer&quot;&gt;This site gives it a melting point&lt;/a&gt; but no boiling point, which suggests to me, it doesn't make it as a gas, or, at least, not easily.    Maybe under very specialized circumstances, perhaps in a pressure tank filled with noble gases, but I'm just speculating.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Googling &quot;Melting point&quot; &quot;boiling point&quot;, next to the name of the element isn't briliant, but it can provide a quick answer.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Longer addition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gases tend to be smaller molecules, noble gases or when more massive, they need to be symmetric.  A chemist probably wouldn't put it that way, but that's the gist of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take line 2 of the periodic table, Lithium (number 3) to Florine (number 9) - ignoring noble gases for now. These atoms have 4 electron pair orbitals in their outer shell which want to form a tetrahedron.   CO2 has 2 double bonds - making it a straight line, or non-polar is the chemistry term.   I used the word symmetrical in my initial answer.  Non-polar is, I think, more correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That non-polarity or symmetry is why CO2 doesn't bond easily with other things, because it's a straight line, with equal charges on each end.   because it doesn't bond easily with other molecules, it's a gas at relatively low temperature.    CH4, while a tetrahedral, not a line, is also non-polar and has an equal charge on all sides, so it also remains a gas at quite cold temperatures.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water (H20) is different.   Oxygen shares 2 single bonds or pairs of electrons with Hydrogen, it fills it's other 2 outer electron pairs itself, so it's bent shape and it's polar.   This polarity gives H20 a side with a positive charge and a side with a negative charge.   That polarity helps it bond with itself.  That's why water stays wet or stays ice in much warmer temperature than similar mass molecules like CH4 and heavier molecules like CO2.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a gas, non-polarity helps.  A straight or tetrahedral, or flat triangle or six sided polyhedral (kind of a cube shape) and a few carbon chains are, as far as I know, the only options for non-polarity.    Very light, polar molecules can be gas molecules too.  All this varies somewhat with temperature and pressure, so there's no exact answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch4/gases1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Common gases at room temperature.&lt;/a&gt; - note, this is not a complete list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll resort them and add a few.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elemental gases&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H2 (hydrogen)       2.02&#xA;N2 (nitrogen)       28.01&lt;br&gt;&#xA;O2 (oxygen)         32.00&lt;br&gt;&#xA;F2 (fluorine)       38.00&lt;br&gt;&#xA;Cl2 (chlorine)      70.91&lt;br&gt;&#xA;BR2 (bromine) - at about 60 degrees C.&#xA;I2 (Iodine) - at about 180 degrees C.&#xA;AT2 (Astatine) - at about 337 degrees C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth noting that all of these, with the exception of Hydrogen and Nitrogen, bond more readily with other elements and are unlikely to be present in a planetary atmosphere.   (Photosynthesis creating O2 being an exception to this rule).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noble gases&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He (helium)     4.00&lt;br&gt;&#xA;Ne (neon)       20.18&lt;br&gt;&#xA;Ar (argon)      39.95&lt;br&gt;&#xA;Kr (krypton)        83.80&lt;br&gt;&#xA;Xe (xenon)      131.30&lt;br&gt;&#xA;Radon (radioactive, short half life, but it is a gas)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These can be present in a planet's atmosphere as they are mostly chemically inactive.  (Xenon is slightly reactive).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gas molecules (non-polar)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CH4 (methane)       16.04&lt;br&gt;&#xA;NH3 (ammonia)       17.03&#xA;C2H6 (ethane)       30.07&lt;br&gt;&#xA;PH3 (phosphine)         34.00&lt;br&gt;&#xA;CO2 (carbon dioxide)        44.01&lt;br&gt;&#xA;C3H8 (propane)      44.10&lt;br&gt;&#xA;C4H10 (butane)      58.12&lt;br&gt;&#xA;BF3 (boron trifluoride)         67.80&lt;br&gt;&#xA;SF6 (sulfur hexafluoride)       146.05  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'll say more about these shortly). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gas molecules, polar&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HCN (hydrogen cyanide)      27.03&lt;br&gt;&#xA;CO (carbon monoxide)        28.01&lt;br&gt;&#xA;NO (nitrogen oxide)         30.01&lt;br&gt;&#xA;H2S (hydrogen sulfide)      34.08&lt;br&gt;&#xA;HCl (hydrogen chloride)         36.46&lt;br&gt;&#xA;N2O (dinitrogen oxide)      44.01&lt;br&gt;&#xA;NO2 (nitrogen dioxide)      46.01&lt;br&gt;&#xA;O3 (ozone)      48.00&lt;br&gt;&#xA;SO2 (sulfur dioxide)        64.06&lt;br&gt;&#xA;CF2Cl2 (dichlorodifluoromethane)        120.91   (only a little polar)&#xA;H20 (a gas under right temperature and pressure)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are more polar molecules than non-polar, but outside of the somewhat odd CF2CL2, all polar gases are relatively light, SO2 being the most massive, molecular weight of 64.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mentioned Methylamine which is basically ammonia (NH3) where one of the hydrogens is replaced by a methyl (CH3) group.   NH2CH3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://pubchem.ncbi.nlm.nih.gov/compound/dimethylamine#section=Odor&quot; rel=&quot;nofollow noreferrer&quot;&gt;DiMethylamine&lt;/a&gt; (CH3)2NH is also a gas at about 7 degrees C and up (boiling point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Playing around with temperature and variations on the gas molecules (replacing H with CH3, replacing H with NH2, replacing H with OH, but remember, Oxygen tends to be spoken for, like a perfect 10 at a dance, so that's not a good one, unless there's life and a source of oxygen (photosynthesis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly the &quot;Ane&quot; series, more accurately called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Group_14_hydride&quot; rel=&quot;nofollow noreferrer&quot;&gt;Group 14 hydrides&lt;/a&gt;.   Group 14:   carbon, silicon, germanium, tin, and lead, and the hydrides, Methane, Silane, Germane, Stanane, Plumbane.   All of these are polar and all are gas molecules.   Most are very reactive with Oxygen.  Methane requires a flame, but the other 4 react with Oxygen quite easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as temperature goes up, you add new gases, but heat tends to destroy complex chemistry, so there's a bit of a trade-off.    There's no easy answer as to what could be a gas and what couldn't, but starting with the building blocks and swaping might be a place to start.   That doesn't always work though.     CO2 is non polar and a gas.   SO2, even though Silicon is in the carbon group, is polar and bent.  It's not a gas  (it's closer to sand) with a very high melting point.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rwpRg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rwpRg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the disappointing answer is, sometimes you can go down the column in the periodic table and find another and another gas simply by replacement and sometimes you can't.  In the case of SIO2, the bonds are very different than CO2, and the melting point is over 3,000 degrees.   Explained in detail in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/34591/why-is-sio2-solid-while-co2-is-a-gas&quot;&gt;this question here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hexafluoride#Binary_hexafluorides&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hexaflouride series&lt;/a&gt; is interesting.  4 of them are stable gases at Earth's temperature and generally non reactive enough to breath and sound like James Earl Jones when you talk, but they're not likely to be found in a planet's atmosphere cause they're vulnerable to photodisintigration and not likely to be reformed in significant numbers   Other molecues are more likely to form.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, ofcourse You could have a planet with a 3,000 degree surface temperature and all sorts of elemental gases, see cool &lt;a href=&quot;http://www.ptable.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;periodic table with temperature slide&lt;/a&gt;) and at 6,000 degrees, all the elements are basically gases, but temperature that high, destroys any complex chemistry so you don't get complex molecules.  Also, at temperatures that high, the exo/endo thermic direction no longer applicable.  Molecules tend to form back and forth in an equilibrium ratio.  That bit I remember from School.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the Germane in Jupiter question, it does raise some other questions, what about the other Group 14 hydrides?   What about Saturn?  It's possible that Saturn has too little metallic hydrogen to create much Germane.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick in general is both formation and stability.   If a gas has too short a half life (like if it's vulnerable to photodisintigration like a lot of carbon chains are), then it's not likely to last. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter also has storms and wind powerful enough to bring up some elements that aren't necessarily gas - like dust in a sense, up high into it's atmosphere, some carbon chains, sulfur and phosphorus and &lt;a href=&quot;https://en.wikipedia.org/wiki/Ammonium_hydrosulfide&quot; rel=&quot;nofollow noreferrer&quot;&gt;ammonium hydro-sulfide&lt;/a&gt;, (which, despite having a boiling point, it's more accurate to say it separates into Ammonia and Hydrogen Sulfide at 56.6 degrees C, in the cold upper clouds of Jupiter the two elements could combine, into something like a dust.   Technically it's a salt.   &lt;a href=&quot;https://www.universetoday.com/15152/color-of-jupiter/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apologies if I got carried away.   I love thinking about planetary atmospheres.   Can't wait till the J.W.S.T. gets a picture of some in other solar systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My kitten is trying to delete what I wrote, so I'll post it now - will tidy up later.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-06-30T15:10:42.527" LastActivityDate="2017-06-30T15:10:42.527" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="21388" PostTypeId="2" ParentId="21381" CreationDate="2017-06-25T11:06:21.573" Score="13" Body="&lt;p&gt;As the other answers have said, it's a matter of distance. Not just directly (large distance = small gravitational effect), but also because large distance = long orbital period. If Planet 9's orbit is 10,000 years long, it has moved across a very small portion of its orbit over the past 100 years, so the force vector of its gravitational attraction has moved across a very small angle. That alone makes the wobble much harder to detect than the wobble caused by e.g. Mercury with its 3-month orbital period. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's &lt;a href=&quot;https://xkcd.com/1633/&quot; rel=&quot;noreferrer&quot;&gt;a graph&lt;/a&gt; that illustrates which objects we can see with current technology, and which are below the limits of current technology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Lv60d.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lv60d.png&quot; alt=&quot;XKCD: observable planets&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of those &quot;far more difficult feats&quot; (like detecting planets that orbit other stars) are done by &quot;cheating&quot;: we (mostly) can't see those planets directly, but we can detect their presence by the effect it has on the light of its star. That's a technique we can't use for objects in the outer solar system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's another visualization, one that shows the rapid progress we've made in how much of the solar system we can detect. In 1980, we had identified about 8000 asteroids. By 2010, we'd found half a million. The red area in the graph above is shrinking steadily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=S_d-gs0WoUw&quot; rel=&quot;noreferrer&quot;&gt;Asteroid visualization&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="361" LastEditorUserId="361" LastEditDate="2017-06-25T14:15:26.080" LastActivityDate="2017-06-25T14:15:26.080" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21389" PostTypeId="2" ParentId="21337" CreationDate="2017-06-25T14:02:56.457" Score="0" Body="&lt;p&gt;A planet and moon both habitable seems feasible to me, given the planet's orbit is in the stars 'goldilocks' orbital region.  As an example, if an Earth/moon system were recreated except larger.  So the Earth moon system was created early in our solar system by a large Mars sized collision with early Earth.  We ended up with a rare larger than normal moon, compared to Earth, 1/6th the size.  Looking at the other planets, no other moon comes close to being this big compared to the planet.  Anyway, if instead of a collision creating two bodies the size of the Earth and moon you increased both masses by 2.5, you could end up with an Earth 2.5 times the size of ours, likely the edge of feasible, and a moon the size of Mars, also the edge of feasible.  So yes both could sustain life, imo, assuming other conditions are met.  (Water, organics, solar output, magnetic fields, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the 1:1 resonant moons, I'd have to agree with the previous comment, very unlikely to be stable.  Now if you want an advanced civilization to create an artificial one, that could work.  Or use a very large Jupiter to have a couple of habitable planets in orbit and maybe even habitable planets at L3 and L4 (Lagrange points), though I am unsure of their stability too I don't think it's be too big a stretch.  (Jupiter's Lagrange asteroids are called Trojans and appear stable but none are particularly big.)&lt;/p&gt;&#xA;" OwnerUserId="17272" LastActivityDate="2017-06-25T14:02:56.457" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21392" PostTypeId="2" ParentId="4753" CreationDate="2017-06-25T16:55:39.257" Score="5" Body="&lt;p&gt;A chemical element is defined by the number of protons it contains, this largely defines its chemical properties. Elements can, within certain bounds, have varying number of neutrons (elements with the same number of protons but a different number of neutrons are called isotopes). Number of neutrons can have a subtle effect on chemical properties and a more significant effect on stability ie rate of radioactive decay. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the big chemical differences which define an element are determined by the proton number and a given element will only have a handful of isotopes within a fairy narrow range. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, elements are classified by the periodic table which lists the elements in groups according to atomic number (number of protons). When the periodic table was first proposed there were a number of gaps between known elements (at this point the existence of protons was not known). These gaps have subsequently been filled so there is no space for new elements untill you get to high atomic numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The periodic table is full in terms of what could be considered reasonably stable elements. There is no fundamental reason why you can't propose  elements with ever increasing atomic numbers. However the trend so far is that with increasing atomic number elements become more and more unstable. They can be created in particle accelerators but only exist for a tiny amount of time and don't exist in nature in any way which you might consider being a 'real' material like iron or copper. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been various predictions of theoretical islands of stability but even then we are talking about very short lived elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in terms of the way that we tend to understand the term there are no new elements to discover, as all reasonably stable possibilities are accounted for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that there could well be entirely new &lt;strong&gt;materials&lt;/strong&gt; composed of known elements or indeed previously unknown states of matter.   &lt;/p&gt;&#xA;" OwnerUserId="14091" LastEditorUserId="14091" LastEditDate="2018-11-02T11:04:16.520" LastActivityDate="2018-11-02T11:04:16.520" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21393" PostTypeId="2" ParentId="21380" CreationDate="2017-06-25T20:45:43.480" Score="1" Body="&lt;p&gt;As an undergraduate, you are a member of a university. Your first step then should be to talk to your Professors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a guide to what might be expected, look at the &lt;a href=&quot;https://www.as.arizona.edu/telescope-proposals&quot; rel=&quot;nofollow noreferrer&quot;&gt;University of Arizona telescope proposal page&lt;/a&gt; They accept proposals from members of Arizona University. You have to have a clear aim and plan. This is why you need to talk to your professors first.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-06-25T20:45:43.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21394" PostTypeId="1" AcceptedAnswerId="21429" CreationDate="2017-06-25T21:54:25.053" Score="4" ViewCount="269" Body="&lt;p&gt;Can someone explain my why &quot;white dwarfs&quot; and &quot;red giants&quot; were so named? And maybe that's a silly question but why not in the opposite way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;@EDIT&lt;/strong&gt;&#xA;Is this somehow co-related with Stefan–Boltzmann law?&lt;/p&gt;&#xA;" OwnerUserId="17275" LastActivityDate="2017-06-27T22:29:13.903" Title="Origin of terms &quot;white dwarfs&quot; and &quot;red giants&quot;?" Tags="&lt;star&gt;&lt;white-dwarf&gt;&lt;red-giant&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="21395" PostTypeId="1" AcceptedAnswerId="21397" CreationDate="2017-06-26T11:29:57.860" Score="2" ViewCount="244" Body="&lt;p&gt;Refer to the below minutephysics video, in which they say that in order to reach the Sun from earth, we must first stop revolving around the Sun ie make our tangential velocity zero, and then head straight for the sun, otherwise it is nearly impossible to reach the sun because even a very huge push towards the sun will either send us in an elliptical or hyperbolic orbit:-  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=LHvR1fRTW8g&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=LHvR1fRTW8g&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If even a little bit of orbital speed tangentially, prevents something from falling into the sun, then I wonder how does the sun formed itself at the beginning? Because, most collapsing material probably would be having at least some tangential velocity with respect to the center.&lt;/p&gt;&#xA;" OwnerUserId="17277" LastEditorUserId="17277" LastEditDate="2017-06-28T07:41:34.887" LastActivityDate="2017-06-28T07:41:34.887" Title="How was the Sun formed?" Tags="&lt;star-formation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21396" PostTypeId="1" AcceptedAnswerId="21400" CreationDate="2017-06-26T11:38:22.010" Score="4" ViewCount="116" Body="&lt;p&gt;For the method of waiting for a planet to pass in front of a star. rather than just measuring the light that is blocked do we also have to take into consideration the light that may not be coming to us because of gravitational lensing caused by the planet? Is the mass of a planet large enough to cause gravitational lensing significant enough for us to see an effect?&lt;/p&gt;&#xA;" OwnerUserId="5432" LastEditorUserId="11264" LastEditDate="2017-06-26T13:16:47.950" LastActivityDate="2017-07-05T12:08:14.500" Title="Do we have to take gravitational lensing into consideration when measuring the size of exoplanets?" Tags="&lt;exoplanet&gt;&lt;gravitational-lensing&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21397" PostTypeId="2" ParentId="21395" CreationDate="2017-06-26T11:58:30.280" Score="2" Body="&lt;p&gt;Where you're wrong: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;even little bit of orbital speed tangentially prevents something from falling into the sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You need to have huge tangential velocity to escape the sun attraction! The closer you are, the higher the velocity must be for celestial body to avoid falling on the sun.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;While Venus is orbiting the Sun at 35.02 Km/s, the Earth is also orbiting the Sun in the same direction at 29.78 Km/s. [&lt;a href=&quot;https://physics.stackexchange.com/questions/54010/why-does-venus-transit-so-slowly&quot;&gt;Reference&lt;/a&gt;]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not easy to understand, because there is no clear analogy of how stars form at our level on earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interstellar gas clouds eventually collapse to form stars are very large, very massive, very cold, and very tenuous. The star is at the beginning a little portion of the cloud that is more dense than the rest - it results of local and large scale instabilities (&lt;a href=&quot;https://astronomy.stackexchange.com/questions/8468/is-there-an-element-of-chance-chaos-in-stellar-evolution&quot;&gt;some&lt;/a&gt; call it an element of chaos).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Little by little, the density at the center of the cloud increase - at the same time, it has to overcome opposite forces arising from more probable friction. If the gas is cool enough compared to the density, at some point you start having more and more matter, which results in more and more gravitational attraction for surrounding gas. &lt;/p&gt;&#xA;" OwnerUserId="11264" LastEditorUserId="11264" LastEditDate="2017-06-27T16:50:05.693" LastActivityDate="2017-06-27T16:50:05.693" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21398" PostTypeId="1" AcceptedAnswerId="21404" CreationDate="2017-06-26T14:08:51.600" Score="0" ViewCount="236" Body="&lt;p&gt;According to Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“The Oort cloud may have billions of objects with absolute magnitude&lt;/p&gt;&#xA;&#xA;&lt;p&gt;brighter than 11 (corresponding to approximately 20-kilometre diameter)”&lt;/p&gt;&#xA;" OwnerUserId="17277" LastActivityDate="2017-06-27T15:48:55.420" Title="If the Oort cloud has billions of objects larger than 20 km, then why do we have so few comets?" Tags="&lt;solar-system&gt;&lt;oort-cloud&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21399" PostTypeId="1" AcceptedAnswerId="21439" CreationDate="2017-06-26T14:17:00.850" Score="3" ViewCount="871" Body="&lt;p&gt;The motivation for this question comes from &lt;a href=&quot;https://space.stackexchange.com/q/21773/12102&quot;&gt;this question&lt;/a&gt;. A star is named in the occultation prediction shown below, from AsteroidOccultations.com's &lt;a href=&quot;http://www.asteroidoccultation.com/observations/NA/MU69Binzel.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;News &amp;amp; Announcements&lt;/a&gt; for 2014 MU69 on &lt;code&gt;2017-Jun-03 03:11&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that the star is in the Omega Centauri cluster, and I believe it is between +12 and +14 visual magnitude roughly. The name given in the prediction is &lt;code&gt;NH -08194514G&lt;/code&gt; The location shown in the prediction is &lt;code&gt;R.A: 19 03 34.493 DEC -20 34 39.34&lt;/code&gt; but &lt;a href=&quot;https://en.wikipedia.org/wiki/Omega_Centauri&quot; rel=&quot;nofollow noreferrer&quot;&gt;according to Wikipedia&lt;/a&gt; the coordinates of the cluster are roughly &lt;code&gt;R.A. 13h 27m, DEC. -47deg, 29m&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need help understanding why the two coordinates seem so different (am I misunderstanding the prediction?) and finding out how to look up this particular star’s visual magnitude by using its name. I’ve tried a few quick searches but realized that I was over my head and not likely to get the right information.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From the &lt;a href=&quot;https://space.stackexchange.com/q/21773/12102&quot;&gt;linked question&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People have fanned out over South America and South Africa with telescopes with cameras and GPS clocks and have attempted to &lt;em&gt;watch the shadow of a roughly magnitude +13 star cast by a rock in the KUIPER BELT pass over the Earth!&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Early Twitter reports reported clear skies and images of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Omega_Centauri&quot; rel=&quot;nofollow noreferrer&quot;&gt;Omega Centauri&lt;/a&gt; star cluster. There is a Twitter hash tag for these events, &lt;strong&gt;#mu69occ&lt;/strong&gt;. Here is a 500 millisecond exposure from a few days before the occultation. The short exposure time is necessary to resolve small scale debris from the main occultation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/krWZu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/krWZu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Omega Centari Star Cluster, 500 millisecond exposure. &lt;a href=&quot;https://twitter.com/Alex_Parker/status/870169900970639360&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tweeted by&lt;/a&gt; Alex Partker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is actually to look for shadows cast by &lt;em&gt;even smaller debris, orbiting around the rock in the Kuiper belt.&lt;/em&gt; This is for New Horizons mission planning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NSp4X.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NSp4X.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; From AsteroidOccultations.com's &lt;a href=&quot;http://www.asteroidoccultation.com/observations/NA/MU69Binzel.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;News &amp;amp; Announcements&lt;/a&gt; for 2014 MU69&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2017-07-10T13:14:52.100" Title="Where and how to look up the coordinates and magnitude of the star in this occultation prediction?" Tags="&lt;coordinate&gt;&lt;star-catalogues&gt;&lt;occultation&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21400" PostTypeId="2" ParentId="21396" CreationDate="2017-06-26T14:59:03.910" Score="6" Body="&lt;p&gt;No, there is no effect here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational lensing magnification works by increasing the observed surface &lt;em&gt;area&lt;/em&gt; of the lensed object while preserving the surface &lt;em&gt;brightness&lt;/em&gt;. An exoplanet angular diameter is much smaller than that of its host star (even if we can't resolve either). Hence, all the magnification happens across a uniformly lit background, for which the measured increased brightness is zero. In effect, you stretch the image of part of the star into equally bright other parts of the star, resulting in zero magnification - the Einstein radius of the lens is much smaller than the radius of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might get a negligible effect when the star moves across the edge of the stellar disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why does even that not matter?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The geometry is very disadvantageous. Setting aside the small mass of the planet (which is another factor), gravitational lensing works best when the distance between source and lens is about the same as the distance between lens and observer. This is very much not the case in the scenario we look at here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gravitational lensing has no effect on a background of uniform brightness.&lt;/li&gt;&#xA;&lt;li&gt;The geometry of the lensing system makes any leftover effects tiny.&lt;/li&gt;&#xA;&lt;li&gt;The mass of the lens is small to begin with.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Also, there will be other effects that will completely dominate any lensing going on.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit to emphasize the size argument&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xSSVg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xSSVg.png&quot; alt=&quot;A simulation of a gravitational lens. Red is a small source, green the resulting lensed image.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this image you can see a small source in red, and the resulting lensed image in green. Note that the increase in brightness is purely an effect of the size of the lensed image, not more flux per area. If we assume that the blue circle is the size of the star, we can see that any lensing happening while the planet transits its star will simply stretch parts of a bright uniform background into other parts of the same bright uniform background. Of course, the extent of the lensed image here is greatly exaggerated in order to be able to show any lensing features at all. With a planet in front of a star you would not even get multiple images.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastEditorUserId="13681" LastEditDate="2017-07-05T12:08:14.500" LastActivityDate="2017-07-05T12:08:14.500" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21401" PostTypeId="1" AcceptedAnswerId="39813" CreationDate="2017-06-26T15:41:52.043" Score="9" ViewCount="159" Body="&lt;p&gt;I originally wrote &quot;...farthest distance from Earth...&quot; but changed to &quot;observer&quot; in case the occultation was observed from a space telescope. It may not matter much but I didn't want to over-constrain or burden the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The occultation of a roughly 13th magnitude star by a roughly 40 km diameter object in the Kuiper Belt; 2014 MU69 &lt;a href=&quot;https://space.stackexchange.com/q/21773/12102&quot;&gt;has been successfully observed&lt;/a&gt;! I'm hoping to be able to &lt;a href=&quot;https://astronomy.stackexchange.com/q/21399/7982&quot;&gt;look up the star's name direct&lt;/a&gt;. On &lt;code&gt;2017-Jun-03 03:07 UTC&lt;/code&gt; it was about 43.294 AU from the Sun, and 42.435 AU from Earth, (using the &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;noreferrer&quot;&gt;JPL Horizons ephemeris&lt;/a&gt;.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if this is the farthest distance from which a solar system body would have been seen to occult a star? &lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-12-01T07:50:54.823" LastActivityDate="2020-11-13T05:28:31.810" Title="What is the farthest reported distance from observer from which a solar system body has occulted a star?" Tags="&lt;observational-astronomy&gt;&lt;kuiper-belt&gt;&lt;occultation&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21402" PostTypeId="1" CreationDate="2017-06-26T15:55:12.337" Score="2" ViewCount="180" Body="&lt;p&gt;I want to back-predict solar eclipses as part of a research project.&#xA;My main goal is to obtain a list of all solar eclipses that took place over a tight grid (let's say of 100km by 100km) for the years 5000BC to 4000BC (just as an example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally, I'd like results to follow, more or less, the format here of &lt;a href=&quot;https://eclipse.gsfc.nasa.gov/JSEX/JSEX-SA.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://eclipse.gsfc.nasa.gov/JSEX/JSEX-SA.html&lt;/a&gt;, but any library you can suggest will be fine.&#xA;According to the previous reference, &quot;The Besselian elements and values of ΔT used in Solar Eclipse Explorer are the same as those used by Five Millennium Canon of Solar Eclipses: -1999 to +3000. For the purposes of calculating eclipse circumstances from a given place, the growing uncertainty in the value of ΔT and the corresponding longitude become unacceptably large outside time period of -1499 to 3000 (1500 BCE to 3000 CE).&quot; and hence I could not use their tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I will simply count how many eclipses could be seen at the center of each square on my grid for the 1000-year period, I can tolerate large inaccuracies in the timing of eclipses. Moreover, as long as errors are random, I can also afford missing some eclipses or having them missplaced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any resource I could use to compute eclipses as I described?&lt;/p&gt;&#xA;" OwnerUserId="17282" LastActivityDate="2017-06-26T15:55:12.337" Title="Back-predicting solar eclipses" Tags="&lt;solar-eclipse&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21403" PostTypeId="1" AcceptedAnswerId="21408" CreationDate="2017-06-26T16:08:38.090" Score="2" ViewCount="779" Body="&lt;p&gt;I have encountered two seemingly contradictory theories on this(I don't know which one of them is correct, or where I am wrong):-&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;From Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Measurements of cosmic microwave background (CMB) anisotropies indicate that the universe is close to flat. For the shape of the universe to be flat, the mass/energy density of the universe must be equal to the critical density. The total amount of matter in the universe (including baryons and dark matter), as measured from the CMB spectrum, accounts for only about 30% of the critical density. This implies the existence of an additional form of energy to account for the remaining 70%.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The above statement, if correct, I suppose indicates that there is a critical density for which mass/energy has to account for. This means that &lt;strong&gt;if suppose there was no Dark Matter,&lt;/strong&gt; &lt;strong&gt;our calculations/assumptions regarding the amount of Dark Energy would have been much higher&lt;/strong&gt;(because then Dark Energy would have to account for a greater share of critical density).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The universe is accelerating in expanding. Dark Energy is supposed to cause this acceleration in expansion. While ordinary matter plus Dark Matter, is supposed to reduce this acceleration/expansion through gravity. So, according to this theory, &lt;strong&gt;if there would have been no Dark Matter&lt;/strong&gt;, then there would be less Dark Energy required to account for the observed acceleration in expansion, so &lt;strong&gt;our calculations/assumptions regarding the amount of Dark Energy would have been lower.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17277" LastEditorUserId="11143" LastEditDate="2017-06-27T15:20:23.407" LastActivityDate="2020-08-26T13:24:24.363" Title="Relationship between Dark Energy and Dark Matter" Tags="&lt;dark-matter&gt;&lt;dark-energy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21404" PostTypeId="2" ParentId="21398" CreationDate="2017-06-26T16:27:37.960" Score="7" Body="&lt;p&gt;Most Oort cloud objects stay in the Oort cloud, and we never see them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comets don't last long. A comet is made of a mixture of ices and dust. A comet that enters the inner solar system has one of a few outcomes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;it could pass out of the solar system, and either never be seen again, or enter such a long orbit that it isn't seen again for an extremely long time (like Hale Bopp)&lt;/li&gt;&#xA;&lt;li&gt;it could fall into the sun (or travel so close that it is completely broken up) (like comet ISON)&lt;/li&gt;&#xA;&lt;li&gt;it could get caught into a short period and over time have all its ices boiled away by the heat of the sun. (asteroid 2015 TB145. Halley's comet is in the process of losing its ice with each orbit it grows smaller)&lt;/li&gt;&#xA;&lt;li&gt;it could hit a planet. (Shoemaker-Levy 9)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In all cases, the comet is lost, destroyed or deactivated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As comets have a short lifespan, they are rare.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-06-27T15:48:55.420" LastActivityDate="2017-06-27T15:48:55.420" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21405" PostTypeId="1" CreationDate="2017-06-26T16:40:09.797" Score="0" ViewCount="54" Body="&lt;p&gt;On board of a hypothetical spaceship, I have a hypothetical clock consisting of 2 mirrors and a light beam is being reflected between the mirrors to measure the time. Now if we hypothetically exceed the speed of light; what will I see is happening to my clock on board of my ship? (a) Light can't exceed the speed of light so according to me the clock has freezed, (b) Light always passes the observer at the speed of light, so my clock is keeping perfectly time but the speed of light inside my clock has surpassed the speed of light? (c) The clock will run backwards?  I just can't wrap my mind around this problem.&lt;/p&gt;&#xA;" OwnerUserId="17081" LastActivityDate="2017-06-26T22:58:10.743" Title="An hypothetical timepiece on board of a spaceship. Please Help" Tags="&lt;light&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2017-06-27T15:26:22.027" ContentLicense="CC BY-SA 3.0" />
  <row Id="21406" PostTypeId="1" AcceptedAnswerId="21407" CreationDate="2017-06-26T17:11:28.813" Score="0" ViewCount="1213" Body="&lt;p&gt;As time goes by, explanations for unusual things that exist in fiction get disproven. We know radiation probably isn't going to create a Giant ant that destroys everything, we're pretty sure we can't use ray guns, and the idea that falling into a black hole would put you face to face with Ming the Merciless is laughable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comic nerds might have guessed from the title why I'm asking, but I'm curious if there could still be any reason to latch on to to explain why superman is powered by a yellow sun, but not a red one? Does the higher temperature of a yellow sun create a higher level of some concentration of exotic particles, or once you get to the point of [solar fusion] is the output pretty much exactly the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: This question is not about superman, or his powers, just if there is any (temperature aside) meaningful difference between a yellow and red sun (that could be conveniently used by DC to still justify superman's powers).&lt;/p&gt;&#xA;" OwnerUserId="4026" LastEditorUserId="4026" LastEditDate="2017-06-26T18:00:39.330" LastActivityDate="2017-06-27T05:33:05.377" Title="Other than relative temperatures, is there any difference between a red sun and a yellow sun?" Tags="&lt;the-sun&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21407" PostTypeId="2" ParentId="21406" CreationDate="2017-06-26T17:51:14.933" Score="3" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small red dwarfs may be fully convective. Larger stars like the sun have a region in which the energy is transferred by radiation. So in small red dwarves, the star becomes fully mixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Red Giants are quite different and are large and old stars, swollen by their powerful cores.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both red and yellow stars produce large amounts of neutrinos. Yellow stars produce more than red dwarfs because they are more powerful. Red Giants would produce more than yellow dwarfs, as they more powerful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing that could explain superman.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-06-27T05:33:05.377" LastActivityDate="2017-06-27T05:33:05.377" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21408" PostTypeId="2" ParentId="21403" CreationDate="2017-06-26T20:28:59.493" Score="6" Body="&lt;p&gt;First of all, if there were no dark matter (DM), you wouldn't ask this question, since structures — including galaxies, stars, planets, and you — wouldn't have had the time to form in the early Universe before it had expanded too much for gravitational collapse to occur. But let's use magic and make the galaxies anyway:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The CMB (specifically the &lt;em&gt;power spectrum&lt;/em&gt; of the CMB) shows that the total density $\rho_\mathrm{tot}$ of mass/energy in the Universe is extremely close to the critical density $\rho_\mathrm{c}$. That is,&#xA;$$&#xA;\frac{\rho_\mathrm{tot}}{\rho_\mathrm{c}} \equiv \Omega_\mathrm{tot} \simeq 1.&#xA;$$&#xA;The &quot;$\Omega$&quot; is a common way to express densities; as a ratio to the critical density. The CMB also gives some constraints on the &lt;em&gt;total&lt;/em&gt; amount of mass (DM + &quot;normal&quot; matter, i.e. baryons), but it is better at constraining the &lt;em&gt;ratio&lt;/em&gt; of baryons-to-DM. Together with the matter density $\Omega_\mathrm{M}=\Omega_\mathrm{b}+\Omega_\mathrm{DM}$ obtained from observations of supernovae and, in particular, &lt;a href=&quot;https://en.wikipedia.org/wiki/Baryon_acoustic_oscillations&quot; rel=&quot;nofollow noreferrer&quot;&gt;baryonic acoustic oscillations&lt;/a&gt;, we then obtain the amount of DM. This fraction is roughly $\Omega_\mathrm{DM}=0.26$ (&lt;a href=&quot;https://arxiv.org/abs/1502.01589&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck Collaboration et al. 2016&lt;/a&gt;).&#xA;If there were no DM, then $\Omega_\mathrm{tot}$ just wouldn't be $1$, but rather $\Omega_\mathrm{tot} - \Omega_\mathrm{DM} \simeq 0.74$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Similarly, if there were no DM, there would be less matter to counteract the expansion of the Universe. That means that we wouldn't observe the same relation between the brightness and the distance of supernovae, from which we infer the presence of dark energy (DE). Instead, the brightnesses would be somewhat &lt;em&gt;lower&lt;/em&gt;, because the supernovae would be farther away due to the faster expansion. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In other words, you are right that if there were no DM, &lt;em&gt;and&lt;/em&gt; if we observed the same thing as we do, then there would be a contradiction. But if there were no DM, we  &lt;em&gt;wouldn't&lt;/em&gt; see what we see. Therein lies the resolution.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2017-06-28T07:58:45.953" LastActivityDate="2017-06-28T07:58:45.953" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="21409" PostTypeId="2" ParentId="21405" CreationDate="2017-06-26T22:58:10.743" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;On board of a hypothetical spaceship, I have a hypothetical clock consisting of 2 mirrors and a light beam is being reflected between the mirrors to measure the time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The mirrors, clock and you are all in the same reference frame and your relative velocities are &lt;em&gt;zero&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed the whole shebang is traveling at relative to something else is irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the clock on board the ship with you will calmly pass the seconds just as you'd expect an object at rest relative to you to do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now if we hypothetically exceed the speed of light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can't do that.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;what will I see is happening to my clock on board of my ship ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Beyond the speed of light has no meaning I know of in this context.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I just can't wrap my mind around this problem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's no problem : you can't travel faster than light : keep repeating that mantra and all will be well. :-)&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-06-26T22:58:10.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21411" PostTypeId="1" AcceptedAnswerId="21413" CreationDate="2017-06-27T07:52:43.937" Score="4" ViewCount="2358" Body="&lt;p&gt;Anytime I hear about astronomy there is observatory in Chile involved. Sounds like there are the best conditions to watch space. Anyone care to explain what is the reason? &lt;/p&gt;&#xA;" OwnerUserId="17287" LastActivityDate="2017-06-27T21:19:19.507" Title="Why are all space observatories in Chile?" Tags="&lt;observational-astronomy&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21412" PostTypeId="2" ParentId="21411" CreationDate="2017-06-27T08:47:35.210" Score="8" Body="&lt;p&gt;I write this as a theorist who's never visited a modern observatory, so I will gladly defer to any true observers who do come along. But as far as I know, the best sites for astronomical observatories (certainly in optical or near-IR) combine a few common features. Above all, you want to be high and dry: above as much of that pesky atmosphere as possible. You don't want turbulent air distorting the image (related to what we usually call &quot;twinkling&quot;), nor do you want clouds to stop you from seeing the sky at all. Ideally, you also want to be reasonably close to the equator, so that you get access to different parts of the sky at different times of year. You also want to be far away from light pollution but the &quot;high and dry&quot; objective tends to bring that anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With just these few things in mind, the ideal locations for telescopes become quite clear. The summits of &lt;a href=&quot;https://en.wikipedia.org/wiki/Mauna_Loa_Observatory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mauna Loa&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Mauna_Kea_Observatories&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mauna Kea&lt;/a&gt; on the Island of Hawaii (i.e. Big Island) are certainly up there. The top US telescopes are usually up there. For the Southern Hemisphere, the Atacama Desert in Chile also combines these features. For that reason, the &lt;a href=&quot;https://en.wikipedia.org/wiki/European_Southern_Observatory&quot; rel=&quot;nofollow noreferrer&quot;&gt;European Southern Observatory&lt;/a&gt;, when it formed in the late 50s and early 60s, decided on the Chilean site, and they've been building many of their most advanced telescopes there ever since.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The news stories you hear are probably based on telescopes at either the &lt;a href=&quot;https://en.wikipedia.org/wiki/Paranal_Observatory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Paranal&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/La_Silla_Observatory&quot; rel=&quot;nofollow noreferrer&quot;&gt;La Silla Observatories&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt;: As pointed out in &lt;a href=&quot;https://astronomy.stackexchange.com/questions/21411/why-are-all-space-observatories-in-chile/21412?noredirect=1#comment36407_21412&quot;&gt;a comment below&lt;/a&gt;, there are other significant observatories in Chile.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastEditorUserId="1577" LastEditDate="2017-06-27T14:42:55.590" LastActivityDate="2017-06-27T14:42:55.590" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21413" PostTypeId="2" ParentId="21411" CreationDate="2017-06-27T11:27:54.823" Score="14" Body="&lt;p&gt;[disclaimer: I am also not a true observer, but I have visited observatories before, and I do amateur observations regularly]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definitely not all observatories are in Chile. Chile, however (the Atacama desert in particular) is an excellent location for an astronomical observatory. The Atacama desert has one of the darkest night skies on Earth, which is excellent for astronomical observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Warrick mentioned, the ideal observatory is located at the highest possible altitude, such that there is as little air as possible between the telescope and the stars. Something else to consider, though, is that if the telescope should be manned by humans permanently it cannot be located too high due both to the difficulty for humans to function at extremely high altitudes and the price tag of getting the telescope parts up there. The two observatories in the Atacame desert of Chile are both located at an altitude of 2.5 km, which is definitely not the highest point on earth (the &lt;a href=&quot;https://en.wikipedia.org/wiki/Mauna_Loa_Observatory&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Mauna Loa&quot;&gt;Mauna Loa&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Mauna_Kea_Observatories&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Mauna Kea&quot;&gt;Mauna Kea&lt;/a&gt; observatories mentioned by Warrick are both located at higher altitudes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another requirement is that the location should be as cold as possible. This is not to torture the poor observers operating the telescope, but because warm air tends to be more turbulent. More turbulence in the air means less precise observations. However, as temperature tends to decrease with increasing altitude, this is usually automatically satisfied. Having a stable temperature also helps. If the telescope heats up during the day, the temperature of the air inside will differ from that of the surrounding air at night. Since the refractive index of air depends (although very slightly) on temperature, this will cause further deviation of light rays from their ideal straight paths, leading to less accurate measurements. For this reason professional telescopes usually contain some kind of temperature regulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A combination of the above reasons is why observatories are in the locations they are. Theoretically, the best observatory location has been found to be &lt;a href=&quot;https://en.wikipedia.org/wiki/Ridge_A&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ridge A&lt;/a&gt; on Antarctica. The reason you always hear about Chile is not its location. There are other locations that are at least as well suited as the Atacama desert for astronomical observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most important reason you always hear about Chile is probably that it is home to the VLT, or Very Large Telescope, which is the most productive ground-based telescope in the optical region (according to &lt;a href=&quot;https://en.wikipedia.org/wiki/Very_Large_Telescope&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Wikipedia&quot;&gt;Wikipedia&lt;/a&gt;). It uses interferometry combining the observations from four individual telescopes. This makes the combined effective aperture of the VLT the biggest in the world. As both the most productive telescope and the telescope with the largest effective aperture it produces a lot of results, and those results are quite likely to be important enough to appear in popular science articles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However Chile is definitely not the only possible location for observatories. Apart from Mauna Kea and Mauna Loa, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Roque_de_los_Muchachos_Observatory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Observatory del Roque de los Muchachos&lt;/a&gt; on La Palma is the first example that pops up in my mind. It is home to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gran_Telescopio_Canarias&quot; rel=&quot;nofollow noreferrer&quot;&gt;GTC&lt;/a&gt;, the worlds largest single-mirrored optical telescope.&lt;/p&gt;&#xA;" OwnerUserId="11824" LastEditorUserId="11824" LastEditDate="2017-06-27T21:19:19.507" LastActivityDate="2017-06-27T21:19:19.507" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21414" PostTypeId="1" AcceptedAnswerId="21416" CreationDate="2017-06-27T11:53:58.700" Score="4" ViewCount="126" Body="&lt;p&gt;I just read the answers to the question &lt;a href=&quot;https://astronomy.stackexchange.com/questions/21411/why-are-all-space-observatories-in-chile&quot;&gt;&quot;Why are all space observatories in Chile?&quot;&lt;/a&gt;.  In addition to the reasons provided, one of the answers also mentions that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...if the telescope should be manned by humans permanently, it cannot&#xA;  be located to high due to the difficulty for humans to function at&#xA;  extremely high altitudes, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which leads to my question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there any unmanned large telescopes located on Earth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I realize that, even if large telescopes are unmanned, they probably need to be serviced by humans on a regular basis. But are humans needed permanently on site? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also realize that, even though there are humans monitoring the measurements from &lt;em&gt;space telescopes&lt;/em&gt;, space telescopes are unmanned in the sense that there are no people at the space telescopes manning them.&lt;/p&gt;&#xA;" OwnerUserId="5537" LastActivityDate="2017-06-27T15:44:53.793" Title="Are there any unmanned large telescopes located on Earth?" Tags="&lt;observatory&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21415" PostTypeId="1" AcceptedAnswerId="21420" CreationDate="2017-06-27T12:28:44.010" Score="1" ViewCount="131" Body="&lt;p&gt;Mass-Energy of the Universe:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5% ordinary matter,&#xA;27% dark matter,&#xA;68% dark energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about other energies such as thermal energy, potential energy, kinetic energy, chemical energy, and radiant energy?&lt;/p&gt;&#xA;" OwnerUserId="17277" LastActivityDate="2017-06-27T15:08:42.063" Title="Why aren't other kinds of energies considered in the mass-energy of the universe?" Tags="&lt;universe&gt;&lt;cosmology&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21416" PostTypeId="2" ParentId="21414" CreationDate="2017-06-27T12:46:37.637" Score="4" Body="&lt;p&gt;The largest &quot;robotic&quot; (i.e., unmanned) telescope I'm aware of is the 2.4-meter &lt;a href=&quot;http://www.ucolick.org/public/telescopes/apf.html&quot; rel=&quot;noreferrer&quot;&gt;Automated Planet Finder&lt;/a&gt;. Other large robotic telescopes include the 2.0-meter Liverpool Telescope and its copies (Faulkes Telescope North and Faulkes Telescope South). You can read about the automated control system of the Liverpool telescope &lt;a href=&quot;http://telescope.livjm.ac.uk/TelInst/Robotic/&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are all located at existing observatories (e.g., APF at Lick Observatory, Liverpool Telescope at Observatorio del Roque de los Muchachos, La Palma, Canary Islands, Spain), which means there are people on site for potential maintenance work, etc., though said staff are really devoted to working on other telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest thing to an isolated, automatic telescope in an uninhabitable setting might be the small (0.5-m) Antarctic Survey Telescope at &lt;a href=&quot;https://en.wikipedia.org/wiki/Dome_A&quot; rel=&quot;noreferrer&quot;&gt;Dome A&lt;/a&gt; in Antarctica, which I believe is serviced annually by Chinese expeditions. It can apparently be remotely controlled via Iridium satellite communications, though actually retrieving the full data sets is part of what the annual expeditions are for.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2017-06-27T12:46:37.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21417" PostTypeId="2" ParentId="21415" CreationDate="2017-06-27T12:57:07.660" Score="0" Body="&lt;p&gt;I'm gonna give this a shot, without going into QFT vocabulary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those energies you've mentioned are not fundamental forces of the universe.&#xA;Instead, they are all expressions of the four fundamentals gravitation, electromagnetism, weak and strong force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;weak force&lt;/strong&gt; is the weirdest one, it's effect is not to move things, but to change particle species. For this particles must meet, thus it doesn't act on distance.&#xA;It is also as the name says weak, because in any particle encounter a change of species is very improbable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;strong force&lt;/strong&gt; is not active on large distances, only very small ones, of about a femtometer ($10^{-15}$m, the typical size of an atomic nucleus). Thus this one drops out as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;electromagnetic force&lt;/strong&gt; is also fairly strong compared to gravity, but due to having equal amounts of positive and negative charges in the universe, this is shielded on large length scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then gravity remains, which cannot be shielded. Thus on cosmic length-scales gravity dominates.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;thermal energy, potential energy, kinetic energy, chemical energy, and radiant energy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thermal energy is just the kinetic energy of microscopically moving particles, that hit each other often enough so that their momenta are redistributed all the time and thus on average they stay in the same place.&lt;br&gt;&#xA;Potential energy is the exact energy of whatever fundamental force field you're in.&lt;br&gt;&#xA;Chemical energy is electromagnetism + quantum mechanics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radiation energy requires a bit more explanation:&#xA;However thanks to &lt;a href=&quot;https://en.wikipedia.org/wiki/General_relativity&quot; rel=&quot;nofollow noreferrer&quot;&gt;general relativity&lt;/a&gt; we understand better how gravity rules over the universe: It is actually the energy-momentum-density of the universe that governs cosmic expansion. This is I guess what you refer to as mass-energy. And 'radiant energy' is the energy-momentum of photons, which do contribute to the energy-momentum of the universe. &lt;a href=&quot;https://en.wikipedia.org/wiki/Physical_cosmology&quot; rel=&quot;nofollow noreferrer&quot;&gt;Physical cosmology&lt;/a&gt; takes this into account, and is even able to derive that there was a &lt;a href=&quot;https://en.wikipedia.org/wiki/Photon_epoch&quot; rel=&quot;nofollow noreferrer&quot;&gt;phase in the life of our universe&lt;/a&gt; where photons were dominating the energy-momentum.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2017-06-27T12:57:07.660" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21419" PostTypeId="2" ParentId="21373" CreationDate="2017-06-27T13:39:23.940" Score="-2" Body="&lt;p&gt;Is this a photo from the Digitized Sky Survey (DSS)? If so, I have also seen these, but they are not true stars most of the time. They  are mostly &quot;artifacts&quot; left over from stacking or imaging.&lt;/p&gt;&#xA;" OwnerUserId="17291" LastActivityDate="2017-06-27T13:39:23.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21420" PostTypeId="2" ParentId="21415" CreationDate="2017-06-27T15:08:42.063" Score="2" Body="&lt;p&gt;Various missions such as WMAP and the Planck Satellite have measured the mass-energy content of the universe. You tend to see images like the one below produced by these scientific ventures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9JF5s.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9JF5s.png&quot; alt=&quot;WMAP results&quot;&gt;&lt;/a&gt;&#xA;&lt;sup&gt;Credit: &lt;a href=&quot;https://map.gsfc.nasa.gov/news/&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA/WMAP Science Team&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This breaks the entire mass-energy content of the universe into three buckets: Dark Energy, Dark Matter, and (the mass of) atoms. But what about the rest of the universe. Surely there's other mass-energy out there such as the energy in the curved space-time metric, or kinetic energy of atoms, or radiation energy in photons, or the mass-energy of neutrinos! Have the astronomers and physicists forgotten about those things? Well the answer is no, they're just not telling you about it because it doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Equation for the universe&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The universe's content and evolution is defined by something known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Friedmann equation&lt;/a&gt; (technically there's a set of equations, but that's not relevant here). One important and useful form of this equation is given below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{H(a)^2}{H_0^2} = \Omega_{0,R}a^{-4} + (\Omega_{0,b} + \Omega_{0,c})a^{-3} + \Omega_{0,k}a^{-2} + \Omega_{0,\Lambda} +\ ...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation may seem complicated, but let me explain the parts. The left hand side more or less is the expansion rate of the universe in the past (or future) compared to the present day expansion rate. The right hand side is the sum of all the various contributions to the mass-energy of the universe. The important parameters here are the various &quot;energy density parameters&quot;, denoted by $\Omega$. I've included the parameters for radiation ($\Omega_{0,R}$), baryonic matter ($\Omega_{0,b}$), cold dark matter ($\Omega_{0,c}$), space-time curvature ($\Omega_{0,k}$), and dark energy ($\Omega_{0,\Lambda}$), but you could add more energy density parameters as well. The ultimate point I'm going to show is that there are only a few energy density parameters that are actually important or significant and the rest can be ignored completely.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Thermal/Kinetic Energy&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;You mention things like thermal and kinetic energies (which are effectively the same thing since thermal energy just comes from the total kinetic energy of a collection of particles). We can't and don't count these because there's no way to absolutely define the thermal or kinetic energy of something. To define kinetic energy of a given mass requirements specifying it's velocity, but that velocity must be defined with respect to a particular reference frame. Special relativity tells us that there is no absolute reference frame though, which means I could define your velocity with respect to me, or the Sun, or the center of the Galaxy and get a different kinetic energy for you with each definition.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Radiation Energy&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Radiation energy is just the energy in light. The universe is flooded with photons (a very large portion of which are CMB photons) and these photons have non-zero energy. But that energy is almost completely negligible in the current universe. From WMAP results, &lt;a href=&quot;https://arxiv.org/pdf/1401.1389.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;it is estimated&lt;/a&gt; that $\Omega_{0,R} = 5\times10^{-5}$ which equates to contributing about $0.005\%$ of the Universe's mass-energy budget. It would be a tiny, tiny sliver in the above graph. Interestingly, this wasn't always the case. If you run the universe backwards, you find that very early on, $\Omega_{R}$ was the dominant source of mass-energy in the Universe.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Neutrino Energy&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Similar to radiation energy, neutrino energy is negligibly small. In part, this is because the neutrino mass is so tiny. I don't have an estimate for $\Omega_{0,\nu}$ but I can say it's probably as small or smaller than $\Omega_{0,R}$ and equally insignificant.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Gravitational and Chemical Energy&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;These types of energies are what's known as potential energies. They're a weird sort of energy because they're not actually real. They're potential. It's the energy you &lt;em&gt;would get&lt;/em&gt; if you performed some action (e.g., moving a body through a field). If you hold a ball up high, you might say that ball has gravitational potential energy, but the ball doesn't actually have that energy, it just has the potential to gain energy when you drop it. We simply say it does have that energy to make balancing energy more tidy and simple. When you do drop the ball, it gains energy by falling through the gravitational field and taking energy from it. So really, your question boils down to talking about the energy inherent in background gravitational fields. The same is true for chemical energy. Chemical processes such as extreme oxidation, aka fire, release energy, but that energy comes from breaking atomic bonds and that breaking of bonds involves converting electric potential energy to something else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately you just want to add in energy from the quantum fields that permeate the universe, $\Omega_{QF}$. I can't estimate this number, nor have I ever even seen someone try, but I can almost assure you that this number would be extremely tiny, even more so that the previous considered energies. Again, we find it's negligible and would by the tiniest mote of a sliver on the above pie chart.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Space-Time Curvature Energy&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;There's energy in space-time itself. This is known from General Relativity. This space-time energy of the Universe is denoted by $\Omega_{k}$. The interesting thing about this is that it appears the universe is flat! This implies, to decent precision, that $\Omega_{0,k} = 0$. This energy is more than negligible, it is non-existent (to within observational precision).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Conclusion&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I'm sure you could think up a lot of different types of other energies you might want to include in this, but hopefully you see the point I'm driving at. They're all completely negligible. The arguably largest contribution comes from radiation and that's been shown to be $0.005\%$ of the Universe's mass-energy budget. What we see is that the Universe is dominated by Dark Energy, has a bit of Dark Matter, a small amount of atoms, and negligible amounts of the rest of everything else. And overtime, the Dark Energy portion of that pie chart will get bigger and bigger until we'll be able to throw out atoms and dark matter from the plot as well since they'll be just as insignificant as neutrinos and radiation are now.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-06-27T15:08:42.063" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21421" PostTypeId="2" ParentId="21414" CreationDate="2017-06-27T15:44:53.793" Score="3" Body="&lt;p&gt;In addition to @PeterErwin's answer, there is the &lt;a href=&quot;https://www.iiap.res.in/iao_telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Himalayan Chandra Telescope&lt;/a&gt; (2m), located near the Indian Astronomical Observatory at Hanle at 4500m altitude but operated remotely from near Bangalore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to leave the realm of optical astronomy, the &lt;a href=&quot;http://www.almaobservatory.org/en/about-alma/location&quot; rel=&quot;nofollow noreferrer&quot;&gt;ALMA telescope array&lt;/a&gt; is located at 5000m altitude in the Atacama desert in Chile and operated from a nearby support facility located at an altitude of 2900m. Access to the array itself is kept minimal due to the extreme altitude.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastActivityDate="2017-06-27T15:44:53.793" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21422" PostTypeId="1" CreationDate="2017-06-27T16:59:53.640" Score="42" ViewCount="7915" Body="&lt;p&gt;When a space gas gets pulled together a star is formed. On the other hand, when a massive star dies, it collapses to a black hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would think that the initial mass of the gas would be bigger than the star that had existed for billion years and lost mass in the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what stopped the space gas from forming a black hole in the first place? &lt;/p&gt;&#xA;" OwnerUserId="17293" LastEditorUserId="10031" LastEditDate="2017-06-29T07:27:00.190" LastActivityDate="2017-06-30T13:33:09.110" Title="Why does gas form a star instead of a black hole?" Tags="&lt;star&gt;&lt;black-hole&gt;&lt;gravity&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21423" PostTypeId="2" ParentId="21411" CreationDate="2017-06-27T17:05:11.153" Score="0" Body="&lt;p&gt;The earth's atmosphere absorbs or attenuates many frequencies of the spectrum.  While radio and visible light make it through pretty well, other regions are blocked and usually require (expensive) space-based telescopes for observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Infrared is strongly absorbed by water vapor.  It turns out that the Atacama is dry enough that you can do quite a bit of IR observations from the ground.  Combined with the altitude, almost no other place on earth could support such an instrument.  IR observations are important for objects at high redshift, including information on the oldest galaxies.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2017-06-27T17:05:11.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21424" PostTypeId="2" ParentId="21422" CreationDate="2017-06-27T17:32:13.667" Score="73" Body="&lt;p&gt;Essentially gas does, it just happens to form a star first. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass is not the only factor in creating a black hole. You also need for this mass to reach a high density. In the process of doing this, a star usually forms. The energy producing processes in the star interior produce a pressure that balances the gravitation attraction. This prevents a star from reaching a critical density needed for black hole formation. As these energy producing processes run out of useable fuel the star will eventually collapse onto itself creating a black hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can't just take a large volume of gas and create a black hole. Other physical processes occur. &lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="2365" LastEditDate="2017-06-27T18:10:06.953" LastActivityDate="2017-06-27T18:10:06.953" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="21425" PostTypeId="2" ParentId="13455" CreationDate="2017-06-27T17:33:28.330" Score="3" Body="&lt;p&gt;Considering we have sent orbiters &lt;A HREF=&quot;https://en.wikipedia.org/wiki/MESSENGER&quot; rel=&quot;nofollow noreferrer&quot;&gt;to&lt;/A&gt; &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Magellan_(spacecraft)&quot; rel=&quot;nofollow noreferrer&quot;&gt;both&lt;/A&gt;, I find undetected moons to be ludicrous. Even if they didn't search, the gravitational disturbances would have disturbed their mapping enough to tell there were moons if there were any.&lt;/p&gt;&#xA;" OwnerUserId="8056" LastActivityDate="2017-06-27T17:33:28.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21426" PostTypeId="1" AcceptedAnswerId="21428" CreationDate="2017-06-27T19:31:07.353" Score="2" ViewCount="200" Body="&lt;p&gt;We currently possess a &lt;a href=&quot;http://www.celestron.com/browse-shop/astronomy/telescopes/nexstar-slt/nexstar-130slt-computerized-telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Celestron NexStar 130 SLT&lt;/a&gt; computerized reflector telescope and have an opportunity to do backyard astronomy with my 6-year old daughter from time to time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She is quite interested in space and astronomy, knows a lot of objects in the Solar System, which have atmosphere and what don't, knows major planet satellites and thinks there is life on Titan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, I find it quite challenging to actually do telescope observations with her:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;it takes time to set things up, assemble the equipment and align the telescope. She is not patient enough and that sometimes &quot;kills&quot; all the fun and excitement&lt;/li&gt;&#xA;&lt;li&gt;it is not easy for her to get used to looking through an eyepiece&lt;/li&gt;&#xA;&lt;li&gt;it is easy to touch the telescope which leads to a shaking image&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What are the general guidelines to observe the night sky with the 6-year old? What can I do to make the process less tedious, more fun and engaging?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(we also have the 15x70 Binoculars)&lt;/p&gt;&#xA;" OwnerUserId="17296" LastActivityDate="2017-07-07T03:59:03.157" Title="Observing the night sky with a 6-year old" Tags="&lt;telescope&gt;&lt;observational-astronomy&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21427" PostTypeId="2" ParentId="21426" CreationDate="2017-06-27T20:26:28.440" Score="2" Body="&lt;p&gt;Regarding your first dot point: I would suggest trying to have the equipment set up before she joins you, if possible. If you have a family member who can look after her/keep her distracted while you do this, she won't have to wait while you set up the equipment. (I understand this may be hard, depending on your situation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your second and third dot points: keep working on it. Learning things like how to look through an eyepiece, and that touching the telescope leads to a shaky image, is a valuable part of the experience. She may learn it slower at her stage in life, but it's stuff that everyone has to learn when they start using a telescope, and it can be really good to start early. You may need to be a bit patient with her while she's learning, but she may also learn patience through the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sounds like she's very enthusiastic about space, which is great. Talk through what you're seeing, and be excited about it; look at some things she knows about, and some things she doesn't yet, in a spirit of investigation.&lt;/p&gt;&#xA;" OwnerUserId="17238" LastActivityDate="2017-06-27T20:26:28.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21428" PostTypeId="2" ParentId="21426" CreationDate="2017-06-27T20:39:57.093" Score="3" Body="&lt;p&gt;Your issues seem to be rather in line with what one would expect from a 6-year old getting in touch with observational astronomy. Actually, adults can sometimes experience these difficulties too when they first get in touch with it. I'm speaking out of experience of having shown objects of the night sky through a telescope to visitors. Some people just don't understand that when they touch the telescope, it moves. I would recommend the following guidelines to &quot;fix&quot; your proposed problems (expanding on hartacus' answer, which appeared while I was writing this):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Someone else could keep her busy while you set up the telescope. Since it is computerized, you won't waste time looking for objects and the only downtime will be due to the moving of the telescope. You can minimise this by planning your observation ahead of time to minimise the distance between two following objects.&lt;/li&gt;&#xA;&lt;li&gt;As for the eyepiece, just tell her how to use it. She'll learn it in time. Even adults who have never used a telescope before generally need to be told how to look through it. &lt;/li&gt;&#xA;&lt;li&gt;Most telescopes have some mechanism to lock the primary axes of the telescope mount. You can use this to make sure the telescope always points to the same spot. If you find that it is the tripod that is moving, you could try to find a way to lock that as well. Maybe surround its legs with loose bricks?&lt;/li&gt;&#xA;&lt;li&gt;To minimise the last two difficulties at the same time, the best position to look through a telescope is to have both hands behind your back, move one eye as close to the eyepiece as possible &lt;em&gt;without touching it&lt;/em&gt;, and close your other eye.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As a final remark on looking through the telescope, she might not be seeing the same thing as you when she looks through the telescope, because her eyes are different from you. I don't know whether this is applicable, but most telescopes have a way to adjust their focus to make the image as sharp as possible. People with near - or farsightedness generally require the focus of a telescope to be different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These, however, are difficulties you will definitely overcome in time. When these are fixed, the most important part of the experience will be in the objects you show her. As hartacus mentioned, show her things she can relate to (planets are great), and tell her about the things she is looking at.&lt;br&gt;&#xA;Apart from planets, star clusters can be impressive too. She'll surely be astonised if you tell her how many stars there are in one cluster. Some open clusters can even be thought of as 'micro-constellations' (more specifically, &lt;a href=&quot;https://en.wikipedia.org/wiki/NGC_457&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;).&lt;br&gt;&#xA;There are some nebulae / galaxies that are nice to watch, but I would restrain from them in the beginning because most will appear as blurry grey spots through an average telescope, which is not that impressive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One last thing: you mentioned a binoculars. You can already see a lot with binoculars, although most impressive objects would be too faint. The main issue with the binoculars is that you can't aim them for her (unless you have a mount for them, which exists for some). If you can teach her to aim them herself however, there are some spots in the night sky which have 'mini-constellations': sets of stars in which you can recognize images, but they are too small to see with the naked eye, and too big to see with a telescope.&lt;/p&gt;&#xA;" OwnerUserId="11824" LastActivityDate="2017-06-27T20:39:57.093" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21429" PostTypeId="2" ParentId="21394" CreationDate="2017-06-27T20:57:02.560" Score="1" Body="&lt;p&gt;The terminology is related to Planck's Radiation Law. We have (from Wikipedia):&#xA;$$&#xA;{\displaystyle B_{\lambda }(\lambda ,T)={\frac {2hc^{2}}{\lambda ^{5}}}{\frac {1}{e^{\frac {hc}{\lambda k_{\mathrm {B} }T}}-1}}} &#xA;$$&#xA;where $ k_B $ is the Boltzmann constant, h the Planck constant, and c the speed of light, $ \lambda $ is the wavelength, and T is the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A typical white dwarf is very hot, with $ T &amp;gt; 10^5 K $ even. Here let's assume it is about 50,000 K. A typical red giant is of lower temperature than the sun. Let's take it to be about 4,500 K (it will usually be a bit lower). Then we can plot the Spectral Irradiance of both the sources and compare them (I had an older Matlab code).&#xA;&lt;a href=&quot;https://i.stack.imgur.com/GxgKp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GxgKp.png&quot; alt=&quot;Plot for the White Dwarf (the y-axis is $ W m^{-2} sr^{-1} nm^{-1} $&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The y-axis is $ W m^{-2} sr^{-1} nm^{-1} $.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9AAgB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9AAgB.png&quot; alt=&quot;Plot for Red Giant&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the white dwarf we see that the peak occurs at a significantly lower wavelength than visible light (at about 0.58 nm). Most white dwarfs are hotter so the peak would shift further to the left. Thus, we expect the white dwarf to appear white with a bluish tinge as blue is still marginally higher than the other wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the red giant, the peak occurs at about 640 nm, which is near the red end of the spectrum. Red giants colder than this would have peaks shifted to the right. So they appear orange-red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @AtmosphericPrisonEscape already mentioned, dwarf and giant relate to the physical size as compared to the original star in the main sequence.&lt;/p&gt;&#xA;" OwnerUserId="11510" LastActivityDate="2017-06-27T20:57:02.560" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21430" PostTypeId="2" ParentId="21426" CreationDate="2017-06-27T21:18:54.547" Score="2" Body="&lt;p&gt;My experience is that children struggle with telescopes. I think it's partly monocular vision (it's difficult for children to close just one eye), partly concentration and keeping still. You could try to get her accustomed to using binoculars in the day time, but 15 x 70 will be much too big and heavy - 8x25 would be better, and cheap. &lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-06-27T21:18:54.547" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21431" PostTypeId="2" ParentId="20505" CreationDate="2017-06-27T21:54:07.307" Score="4" Body="&lt;p&gt;The corona varies dramatically in brightness from the inner portion near the sun to the outer portion a few solar radii away. The innermost portion is the brightest; at ISO 100 and f5.6, an exposure of ~1/125 second will capture it. So it's roughly the same brightness as the full moon. The outer corona will require much longer exposures of perhaps a half second, but note that during this time the sun will move noticeably with a long lens if you're not using a tracking mount, so some blurring will likely occur. Overall, practicing on the crescent moon will give you a pretty good idea of the dynamic range of the corona.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diamond ring is dangerous to look at directly, as is any unobscured part of the sun's disk; you should look away at first glimpse, or earlier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's lots more detail scattered around the web, which you've probably found by now. A good overview is available at &lt;a href=&quot;http://www.mreclipse.com/Totality2/TotalityCh12-3.html#SEExpoTab&quot; rel=&quot;noreferrer&quot;&gt;http://www.mreclipse.com/Totality2/TotalityCh12-3.html#SEExpoTab&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17300" LastActivityDate="2017-06-27T21:54:07.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21432" PostTypeId="2" ParentId="21394" CreationDate="2017-06-27T22:29:13.903" Score="1" Body="&lt;p&gt;The use of giant and dwarf in reference to stars of the highest and lowest luminosity is attested by 1914, said to have been suggested by Danish astronomer Ejnar Hertzsprung, (1873-1967); hence red dwarf (attested by 1922) etc. (&lt;a href=&quot;http://www.etymonline.com/index.php?allowed_in_frame=0&amp;amp;search=dwarf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.etymonline.com/index.php?allowed_in_frame=0&amp;amp;search=dwarf&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The terms &quot;giant&quot; and &quot;dwarf&quot; refer to the position on the Herzprung-Russel diagram.It is apparent that there are some stars that are much brighter (and, we deduce, larger) then the temperature of their surface would suggest (these are the giant stars) There are others that are much dimmer dispite being very hot - white dwarfs. More recently, all non-giant stars have become classified as dwarfs.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-06-27T22:29:13.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21433" PostTypeId="2" ParentId="21422" CreationDate="2017-06-27T22:31:31.147" Score="24" Body="&lt;p&gt;In simple terms: Because the gas nukes itself apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the gas (H or He) is put under extremely high pressure, which happens before a black hole is formed, the atoms start nuclear fusion, which releases a lot of energy. That continuous stream of energy is what makes the sun bright and is also what keeps the sun from collapsing into itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the fusion burned up/fused too much of one element, another element becomes the dominant element  for fusion, which leads to different states during the life of a star. And once the star runs out of fuel, gravity wins. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &lt;a href=&quot;https://astronomy.stackexchange.com/a/21434/16377&quot;&gt;Rob says&lt;/a&gt;, you'd need way more mass than that of a &quot;massive star&quot; to counteract the nuclear fusion energy.&lt;/p&gt;&#xA;" OwnerUserId="16377" LastEditorUserId="16377" LastEditDate="2017-06-29T07:00:03.317" LastActivityDate="2017-06-29T07:00:03.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21434" PostTypeId="2" ParentId="21422" CreationDate="2017-06-27T22:41:17.080" Score="20" Body="&lt;p&gt;In the present day universe, this does not happen for two reasons. First, the gas is unstable to fragmentation as it collapses. The reason for this is that the &lt;a href=&quot;https://en.wikipedia.org/wiki/Jeans_instability#Jeans_mass&quot; rel=&quot;noreferrer&quot;&gt;Jeans mass&lt;/a&gt;, the smallest mass that is likely to collapse, scales as $T^{3/2}/\rho^{1/2}$, where $T$ is the temperature and $\rho$ the density. &lt;em&gt;If&lt;/em&gt; the gas can cool as it collapses, then the temperature remains roughly constant, the Jeans mass falls and the cloud breaks up into smaller cores. These cores are usually much smaller than the (at least) several solar masses required to form a black hole (see below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, each of those cores eventually gets hotter in the middle. For masses above $0.075M_{\odot}$, the core becomes hot enough for nuclear fusion. This maintains the high temperature and pressure, which keeps gravity at bay until the fuel runs out. After this, quantum mechanics in the form of electron degeneracy or neutron degeneracy pressure or even the repulsive force between nucleons might support the star (as a white dwarf or neutron star), but not if it is more massive than $\sim 3M_{\odot}$. For lower mass balls of gas (brown dwarfs or planets) they skip the nuclear fusion and go straight to being supported by electron degeneracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the &lt;em&gt;early&lt;/em&gt; universe, what you suggest &lt;em&gt;might&lt;/em&gt; actually happen and this &lt;em&gt;might&lt;/em&gt; be how supermassive black holes and quasars exist only a few hundred million years after the big bang. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Primordial gas made of just hydrogen and helium atoms cannot cool very efficiently (it is the presence of heavier atoms, produced by previous generations of stars, in present day gas clouds that leads to efficient cooling). Primordial clouds are thus less susceptible to fragmentation because they heat up as they get more dense and the Jeans mass cannot become small. In such circumstances it could be that a large black hole ($10^4$ to $10^5$ solar masses) &lt;em&gt;can&lt;/em&gt; form directly from a collapsing gas cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://www.eurekalert.org/pub_releases/2017-03/uoh-nme031417.php&quot; rel=&quot;noreferrer&quot;&gt;this press release&lt;/a&gt; for an alternative summary of this idea and links to recent academic papers on the topic (e.g. &lt;a href=&quot;https://arxiv.org/abs/1504.04042&quot; rel=&quot;noreferrer&quot;&gt;Agarawal et al. 2015&lt;/a&gt;; &lt;a href=&quot;https://arxiv.org/abs/1703.03805&quot; rel=&quot;noreferrer&quot;&gt;Regan et al. 2017&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-06-29T08:05:21.767" LastActivityDate="2017-06-29T08:05:21.767" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21435" PostTypeId="2" ParentId="21422" CreationDate="2017-06-28T15:34:16.360" Score="5" Body="&lt;p&gt;Not a full answer, but more than a comment. Can't we just pile in more and more mass to overcome the problems the other answers presented? Surely at some point in the Universe, e.g., at the very beginning, conditions were favorable for forming &lt;em&gt;extremely&lt;/em&gt; massive stars such that the inward pressure could overcome any outward pressure from heat/fusion. Possibly then a black hole might form right away, despite the barriers preventing total collapse. After all, there's always barriers preventing total collapse, be they radiation pressure, degeneracy pressure, etc. It's just a matter of overcoming it with enough mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out someone already asked this question and their answer is (sometimes) no. &lt;a href=&quot;https://arxiv.org/abs/1304.4601&quot; rel=&quot;nofollow noreferrer&quot;&gt;Johnson et al. 2013&lt;/a&gt; discusses the idea of primordial super massive black holes being formed by super massive stars. By super massive, I mean $\sim50,000\:M_\odot$ (the largest, currently known star on record is $\sim200$-$300\:M_\odot$). Their idea is that you could have such a huge amount of mass pile up early in the universe that you might have a &quot;star&quot; by some technical sense of the word but it would almost immediately (compared to cosmological timescales) form a black hole. Their ultimate result was that they found that trying to overcome the radiation pressure barrier you see in normal star formation by piling on insane amounts of mass just causes things to explode in &quot;the biggest explosion in the Universe&quot;.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="15835" LastEditDate="2017-06-30T13:33:09.110" LastActivityDate="2017-06-30T13:33:09.110" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21436" PostTypeId="1" CreationDate="2017-06-28T16:49:29.723" Score="4" ViewCount="203" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/21422/why-does-gas-form-a-star-but-not-a-black-hole&quot;&gt;This&lt;/a&gt; question got me thinking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In another universe, with a different value for the gravitational constant G, how much bigger would G need to be for star formation to be prevented?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words in a hypothetical universe, the star would attempt to ignite, but the force of gravity would overwhelm the star's outward pressure and you would just get a black hole instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not looking for a precise answer here... just ROM... 10 times bigger? 10,000 times bigger?&lt;/p&gt;&#xA;" OwnerUserId="17319" LastActivityDate="2017-08-02T10:53:08.487" Title="How strong would gravity need to be to prevent star formation" Tags="&lt;black-hole&gt;&lt;gravity&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21437" PostTypeId="1" CreationDate="2017-06-28T18:32:19.853" Score="4" ViewCount="148" Body="&lt;p&gt;According to the article cited behind, and to a post here in Astronomy SE , there are galaxies moving away from Milky Way faster than light, even at speeds of 2.3c . According to this article &#xA;&lt;a href=&quot;http://curious.astro.cornell.edu/the-universe/cosmology-and-the-big-bang/104-the-universe/cosmology-and-the-big-bang/expansion-of-the-universe/1066-can-two-galaxies-move-away-from-each-other-faster-than-light-intermediate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Can two galaxies move away from each other faster than light&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The two galaxies we've been discussing are not travelling through&#xA;  space, it is the space between them that is expanding. Or put in&#xA;  another way, they are stationary and all the space around them is&#xA;  being stretched out. This is why it doesn't violate the theory of&#xA;  relativity, because it is not motion in the traditional sense.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So does this motion between galaxies &quot;in not a traditional sense&quot; count at all to calculate relativistic mass? &lt;/p&gt;&#xA;" OwnerUserId="11562" LastActivityDate="2018-07-19T21:51:08.667" Title="Does the milky way have relativistic mass against galaxies which are moving away from it at high speeds?" Tags="&lt;galaxy&gt;&lt;speed&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21438" PostTypeId="2" ParentId="21436" CreationDate="2017-06-28T18:38:38.333" Score="1" Body="&lt;p&gt;There is no such limit, at least not a meaningful one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the gravity would be more strong, smaller hydrogene clouds could evolve to stars, even Jupiter or Earth sized, or much smaller. As you increase the strength of the gravity, the required mass for star formation decreases, but stars will still exist.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastActivityDate="2017-06-28T18:38:38.333" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21439" PostTypeId="2" ParentId="21399" CreationDate="2017-06-28T19:30:19.857" Score="2" Body="&lt;p&gt;I believe this is a misinterpretation of this tweet: &lt;a href=&quot;https://twitter.com/Alex_Parker/status/870169900970639360&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://twitter.com/Alex_Parker/status/870169900970639360&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the reply chain, it starts with &lt;a href=&quot;https://twitter.com/Alex_Parker/status/870168057032921088&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://twitter.com/Alex_Parker/status/870168057032921088&lt;/a&gt; where Alex Parker states &quot;Also, last night while we were packing up I took a very quick image of the Carina Nebula with one of our #MU69occ cameras.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tweet in question states &quot;This is what Omega Centauri looks like after just 500 milliseconds of exposure with one of the #MU69occ cameras.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, Alex Parker was taking images of various clusters with the camera he plans to use for the MU69occ event. The image targets have nothing to do with the occultation itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also went to &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ssd.jpl.nasa.gov/horizons.cgi&lt;/a&gt; and took a geocentric view of &quot;2014MU69&quot; for the month of June and it confirms this object is nowhere near Centaurus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I emailed Amanda M. Zangari (azangari@boulder.swri.edu) and her response was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&lt;pre&gt;&#xA;Hi Barry,&#xA;&#xA;I do not know where the star numbers come from.&#xA;&#xA;Gaia IDs for each of the stars are here:&#xA;&#xA;http://www.boulder.swri.edu/MU69_occ/gaiainfo.html&#xA;&lt;/pre&gt;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2017-07-10T13:14:52.100" LastActivityDate="2017-07-10T13:14:52.100" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21440" PostTypeId="1" CreationDate="2017-06-28T19:46:08.180" Score="2" ViewCount="483" Body="&lt;p&gt;If we want to extend the lifespan of our habitable zone safely, would it be safer to institute a yearly 'tax' on our suns outer layer of hydrogen, or a yearly 'subsidy' to our suns outer layer of hydrogen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the feasibility of said tax/subsidy is beyond the scope of this question)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to ask this question:&#xA;Assuming the core of the sun remains untouched,&#xA;but its outer layer of hydrogen were reduced by half,&#xA;would the suns main sequence remain intact for a longer time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To avoid ending up with a white dwarf and its associated bad weather, what would be the ideal point to reverse the tax and re-subsidize?&lt;/p&gt;&#xA;" OwnerUserId="17325" LastActivityDate="2017-06-29T19:41:46.863" Title="Which would be safer: removing hydrogen or adding hydrogen to our sun?" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;habitable-zone&gt;&lt;white-dwarf&gt;&lt;main-sequence&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21441" PostTypeId="2" ParentId="21440" CreationDate="2017-06-28T20:00:51.440" Score="4" Body="&lt;p&gt;My kneejerk reaction is that your only option is to remove a chunk of mass from the outer part of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun will respond (on a Kelvin-Helmholtz timescale), by contracting and becoming less luminous because the core temperature is lower in a less massive star. This will extend its main sequence lifetime, because only the central parts of the star are involved in fuelling the nuclear burning. The core is separated from the well-mixed outer convective envelope by a stable radiative zone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the default solar evolution involves it getting slowly more luminous with time, it is possible that mass could be extracted at just the right rate to keep the Sun at constant luminosity. This sounds like an interesting calculation to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alternative of adding &quot;fuel&quot; in the form of hydrogen won't work. The Sun's luminosity will increase because of its increased mass and increased central temperature. However, because the new fuel cannot be mixed into the core (only the outer part is mixed by convection), then the main sequence lifetime would be reduced.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-06-29T11:27:07.693" LastActivityDate="2017-06-29T11:27:07.693" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21442" PostTypeId="2" ParentId="21121" CreationDate="2017-06-28T20:55:57.937" Score="1" Body="&lt;p&gt;The phased array feeds (PAFs) on ASKAP are indeed located at the focal plane. The ASKAP PAFs have 188 receivers, and from this digital beam-forming will create 36 independent beams on the sky. Some more information can be found from the &lt;a href=&quot;https://www.csiro.au/en/Research/Astronomy/ASKAP-and-the-Square-Kilometre-Array/PAFs&quot; rel=&quot;nofollow noreferrer&quot;&gt;CSIRO&lt;/a&gt;. Imaging is then done by scanning the dish (or letting the sky roll overhead). The scanning can then be done much faster than with a single receiver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its important not to confuse these independent beams with purpose of the array itself. The main reason for using an array is interferometry. We use an array for this purpose to have multiple baselines (the distance between the antennas). Having multiple baselines allows the telescope to 'see' structure various angular resolutions. It's true that you can use the array in a 'fly's eye' mode, and then every dish is providing 36 independent beams. In this mode even larger areas of the sky can be scanned quickly. This is useful for searching for transient events such as fast radio bursts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the phasing within the PAF needs to be correlated and the beams formed digitally, and then the signals from each dish need to correlated. A huge task indeed! The computing breakdown for ASKAP is given &lt;a href=&quot;http://www.atnf.csiro.au/projects/askap/computing.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. A new supercomputer (Pawsey) has been built in WA for the purpose of processing ASKAP data.&lt;/p&gt;&#xA;" OwnerUserId="17323" LastEditorUserId="17323" LastEditDate="2017-06-29T05:53:19.213" LastActivityDate="2017-06-29T05:53:19.213" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21443" PostTypeId="1" AcceptedAnswerId="21445" CreationDate="2017-06-28T21:53:17.250" Score="14" ViewCount="7405" Body="&lt;p&gt;It was always my belief that at the center of many galaxies, there are supermassive black holes. If this is the case, then we should not see a &quot;light&quot; coming out from the center since light get's sucked in black holes. Doing a quick search of galaxies on google image, I came across this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RzQKC.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RzQKC.jpg&quot; alt=&quot;Androme&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6EK2a.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6EK2a.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nwIgt.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nwIgt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are famous galaxies, and real, and so I was wondering what this very white-yellow glow is in the center. If it is a collection of stars, then why are there so many stars in the center of a galaxy? If there is a black hole there, then why is there light?&lt;/p&gt;&#xA;" OwnerUserId="17327" LastActivityDate="2020-01-12T22:52:44.367" Title="What is this bright &quot;glow&quot; in the center of galaxies?" Tags="&lt;star&gt;&lt;black-hole&gt;&lt;galaxy&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21444" PostTypeId="2" ParentId="21440" CreationDate="2017-06-28T23:30:33.840" Score="1" Body="&lt;p&gt;This has been considered before. The phrase you want is &lt;a href=&quot;https://en.wikipedia.org/wiki/Star_lifting&quot; rel=&quot;nofollow noreferrer&quot;&gt;star lifting&lt;/a&gt; (aka stellar husbandry), which is the (theoretical) practice of removing mass from a star to extend its lifetime. Since the rate of fusion reactions in the core of star increases faster than increasing math, smaller stars have much longer lifetimes, up to several trillion years for a 0.1 M☉ red dwarf. (A 1M☉ star with 90% of its mass extracted is probably not exactly the same as a 0.1 M☉ star, but should behave similarly.) Further, the extracted matter could be used to either replenish the existing star or create new stars, extending the luminosity even further. Smaller stars do have lower luminosity, but adjusting planetary orbits is expected to be a minor side project for any civilization capable of star lifting, if they would even bother with planets at that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At present we have only extremely sketchy ideas of how we would even go about learning how to learn to do star lifting, it does not seem to violate any laws of physics. Stars already lose small amounts of mass over their lifespan, so it seems plausible that megascale engineering could enhance that and collect the lost mass.&lt;/p&gt;&#xA;" OwnerUserId="14088" LastEditorUserId="14088" LastEditDate="2017-06-29T19:41:46.863" LastActivityDate="2017-06-29T19:41:46.863" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21445" PostTypeId="2" ParentId="21443" CreationDate="2017-06-29T02:28:28.667" Score="15" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If this is the case, then we should not see a &quot;light&quot; coming out from the center since light get's sucked in black holes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are overestimating the size and the capabilities of a supermassive black hole. Contrary to pop sci portrayals of black holes, black holes are not giant vacuum cleaners in space that suck up anything and everything close by. While the supermassive black hole at the center of the Milky Way is indeed very massive (about four million times the mass of our Sun), it isn't very large physically. It's less than a couple dozen solar diameters across. It also isn't that hungry, gobbling up perhaps the equivalent of four or so Earth masses over the course of a year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the central bulge of a spiral galaxy contains several million stars in a fairly small volume. That central bulge is what you are seeing in those images. The supermassive black holes near the centers of those bulges gobbles only a tiny, tiny fraction of the light emitted by those millions of stars.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-06-29T02:28:28.667" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21446" PostTypeId="1" CreationDate="2017-06-29T05:12:41.747" Score="-1" ViewCount="48" Body="&lt;p&gt;As far as I'm concerned, most star systems in our galaxy are traveling in fairly circular paths around the black hole at the center. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hypothetically, if a large gravitational force pulled star systems closer to the center of our galaxy would binary star systems result? If two stars of different masses were travelling in a centripetal motion around the center of our galaxy, and they were pulled towards the center (which &lt;em&gt;i think&lt;/em&gt; would result in two different accelerations completely) could a situation result where two stars collide or join into a system? What is the most probable event to happen in this situation besides two stars not approaching eachother?&lt;/p&gt;&#xA;" OwnerUserId="7189" LastActivityDate="2017-06-29T15:09:50.860" Title="Star systems forming when large gravitational force is applied" Tags="&lt;supermassive-black-hole&gt;&lt;star-systems&gt;&lt;stellar-dynamics&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21447" PostTypeId="1" AcceptedAnswerId="21451" CreationDate="2017-06-29T08:09:23.900" Score="0" ViewCount="867" Body="&lt;p&gt;I need to use the sunrise equation but one of the variables is the sun declination. On the other hand, I have the values for solar elevation angle that I need. Are they the same thing?&lt;/p&gt;&#xA;" OwnerUserId="17171" LastActivityDate="2017-06-29T10:49:33.070" Title="Is there a difference between the solar elevation angle and sun declination?" Tags="&lt;the-sun&gt;&lt;declination&gt;&lt;sun-rays&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21448" PostTypeId="2" ParentId="21422" CreationDate="2017-06-29T09:27:43.363" Score="-4" Body="&lt;p&gt;When that space gas gets pulled together to form a star, the original gas and now a star their mass is the same (unchanged), but the size reduced, due to gravity. When the star collapses to be a black hole, again their mass the same but size shrunk, due to gravity. &#xA;It is not a valid way to ask in this manner: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So what stopped the space gas from forming a black hole in the first place?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The space gas does eventually shrink to become a black hole [if the initial mass is good enough for it to exist as a star then collapsed as a black hole], only the observer taking section by section of the whole process to observe [depends on which section is observed]. It's simple, when replace the object for analogy to this: heating the ice. Ice melts to become water, then gas. Now the OP is like asking: why ice doesn't turn to gas when heated but water? It sounds logical but in fact it is the construct of the words mystified a simple analogy.   &lt;/p&gt;&#xA;" OwnerUserId="17330" LastEditorUserId="17330" LastEditDate="2017-06-29T14:22:13.843" LastActivityDate="2017-06-29T14:22:13.843" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21450" PostTypeId="1" CreationDate="2017-06-29T10:37:59.630" Score="0" ViewCount="4400" Body="&lt;p&gt;We don't have a photo of our Milky Way the whole, all the pictures are artists' interpretations showing a bulge, a central bar, and several spiral arms. The milky way is spinning. I would like to know: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Does it spin with the arms trailing or leading?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) I read if the direction of &lt;a href=&quot;http://www.sfaaa.com/index.php/read-our-answers/316-does-the-milky-way-rotate-and-or-revolve-clockwise-or-counterclockwise-by-james-goss&quot; rel=&quot;nofollow noreferrer&quot;&gt;spin clockwise or counterclockwise&lt;/a&gt; depends on whether the Milky Way is observed from above or below. I don't understand how the direction of spin (&lt;a href=&quot;http://curious.astro.cornell.edu/94-the-universe/the-milky-way/dynamics/482-does-the-milky-way-spin-counter-clockwise-if-so-do-all-spiral-galaxies-spin-in-this-direction-and-why-beginner&quot; rel=&quot;nofollow noreferrer&quot;&gt;clockwise &amp;amp; counterclockwise&lt;/a&gt;) depends on the point of view of the observer.&lt;/p&gt;&#xA;" OwnerUserId="17330" LastEditorUserId="2365" LastEditDate="2017-07-02T21:45:17.247" LastActivityDate="2018-10-31T22:10:22.310" Title="In which direction does the Milky Way rotate?" Tags="&lt;milky-way&gt;&lt;galactic-dynamics&gt;&lt;spiral-arms&gt;" AnswerCount="1" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="21451" PostTypeId="2" ParentId="21447" CreationDate="2017-06-29T10:49:33.070" Score="1" Body="&lt;p&gt;Elevation and Declination are from different co-ordinate systems, so solar elevation is given in Alt/Az co-ordinates and refers to the elevation above the local horizon. Sun declination is measured in RA/Dec co-ordinates (which is an equatorial co-ordinate system) and measures the Sun's inclination above or below the equator on the celestial sphere.&lt;/p&gt;&#xA;" OwnerUserId="10437" LastActivityDate="2017-06-29T10:49:33.070" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21452" PostTypeId="1" AcceptedAnswerId="21472" CreationDate="2017-06-29T11:57:04.177" Score="5" ViewCount="336" Body="&lt;p&gt;I'm curious for an estimate of how wide a field the Hubble space telescope has observed. Say, for example, how many times it has changed its pointing, and what fraction of that were revisiting the same field.&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2017-06-30T22:39:36.317" Title="How large part of the sky has Hubble observed?" Tags="&lt;observational-astronomy&gt;&lt;hubble-telescope&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21453" PostTypeId="2" ParentId="21446" CreationDate="2017-06-29T15:09:50.860" Score="1" Body="&lt;p&gt;Most binary stars are believed to have formed together as they condensed out of a common rotating dense cloud of gas and dust. A few more form when, coincidentally, two are ejected from a cluster at the same time at very low speeds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In high density stellar systems, close flybys tend to disrupt wide binary systems.  This happens because energy from the flyby can be exchanged with the binary and sometimes that increases the binary orbital energy from negative to positive values (escape regime). So, if you increase the density of the galaxy (by increasing the gravitational source at the center), it is likely to reduce the number of binary systems via close encounters.  However, it would take a tremendous increase in density from the mean galactic density for this to begin to be non-negligible.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2017-06-29T15:09:50.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21454" PostTypeId="2" ParentId="21384" CreationDate="2017-06-29T16:05:11.433" Score="2" Body="&lt;p&gt;The snow line is the inner radius from the star where the water condenses onto dust and larger particles as ice (ie snowballs).  The hydrogen and helium never freezes out.  Planets start forming by accreting these particles (with or without ice but definitely not with H or He), but when a growing planet reaches a sufficiently high mass, it starts to pull in H and He (and more importantly, can hold onto them as they have low atomic mass and thus tend to escape quickly unless there is enough gravity). Exactly what mass that occurs at depends on a lot of factors (temperature, pressure, etc), but it happens below Neptune's mass and in part of the super-earth mass class.  The upper range in mass of the super-earth are also called mini-Neptunes or gas dwarf planets and from their typically low densities it is clear that they are mostly composed of H and He.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, beyond the snow line it is more likely for a planet to grow to a large mass since more of the surroundings has condensed out of the gas into small particles, therefore accretion is more efficient.  So, it may be that mini-Neptunes are usually forming beyond the snow line and then some of them migrate closer to its star.  But, I don't see any reason to doubt that some of the many mini-Neptunes with H and He found by Kepler/K2 inside the snow-line were formed in situ.  I am not an expert in this field, and perhaps the experts would disagree.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2017-06-29T16:05:11.433" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21455" PostTypeId="1" CreationDate="2017-06-29T16:26:28.473" Score="1" ViewCount="138" Body="&lt;p&gt;In chapter 2 of &quot;The Revolutions of the Heavenly Spheres&quot; where Copernicus maintains that the Earth is spherical, the Edward Rosen translation of the book states that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the elevations of the poles have the same ratio everywhere to the portions of the earth that have been traversed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd like to know what Copernicus is referring to in the aforementioned piece of text and how it provides evidence for the spherical shape of the Earth.&lt;/p&gt;&#xA;" OwnerUserId="17321" LastEditorUserId="11143" LastEditDate="2017-06-29T16:31:39.643" LastActivityDate="2017-06-29T16:53:52.303" Title="How are the elevations of the poles proportionate to distances of earth that have been traversed according to Copernicus" Tags="&lt;earth&gt;&lt;history&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21456" PostTypeId="2" ParentId="21455" CreationDate="2017-06-29T16:53:52.303" Score="2" Body="&lt;p&gt;Here, I believe he is referring to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Horizontal_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;altitude&lt;/a&gt; of the celestial poles at different latitudes on the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/A41n3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/A41n3.jpg&quot; alt=&quot;The celestial poles&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Earth is indeed a sphere then the altitude of the celestial poles above the horizon should be the same as your latitude on the Earth. That is, as you travel along a line of latitude, the altitude of the pole should change accordingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/W4OA3.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W4OA3.gif&quot; alt=&quot;local coordinates&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Copernicus was noting that this is indeed what we observe. At the North pole the North celestial pole is directly overhead, and on the equator it is on the horizon. This observation, therefore, is direct evidence that the Earth is spherical (or at least spheroidal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see the evidence of this in star trail photographs. The stars will always appear to rotate about the celestial poles. This image shows the star trails above Ecuador, at a latitude of about 1$^{\circ}$. You can clearly see that the stars are rolling directly over head; with the North and South celestial poles located on the horizon to the right and left, respectively.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/6F9Of.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6F9Of.jpg&quot; alt=&quot;Star trails in Ecuador&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17323" LastActivityDate="2017-06-29T16:53:52.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21459" PostTypeId="1" AcceptedAnswerId="21461" CreationDate="2017-06-29T22:08:19.650" Score="-1" ViewCount="182" Body="&lt;p&gt;Why do special theory of relativty (with the core message of E=mc²) and quantum mechanics do not go together? Why is the special theory of relativity in the quantum model not valid?&#xA;Can anyone list some reasons or explain it on examples?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: The question is based on mixing up something I read. So the question is bullsh***. Sorry.&lt;/p&gt;&#xA;" OwnerUserId="17034" LastEditorUserId="17034" LastEditDate="2017-07-03T11:25:07.220" LastActivityDate="2017-07-03T11:25:07.220" Title="Contradiction of special theory of relativity and quantum mechanics" Tags="&lt;special-relativity&gt;&lt;quantum-mechanics&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21460" PostTypeId="2" ParentId="20115" CreationDate="2017-06-29T22:58:14.610" Score="6" Body="&lt;p&gt;This is from an article in the &lt;a href=&quot;https://theconversation.com/how-far-theyll-go-moana-shows-the-power-of-polynesian-celestial-navigation-72375&quot; rel=&quot;noreferrer&quot;&gt;Conversation&lt;/a&gt;, by Duane W. Hamacher, and Carla Bento Guedes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Voyagers measure the angles between stars and the horizon using their hands. The width of your pinkie finger at arm’s length is roughly one degree, or double the angular diameter of the Sun or Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hold your hand with the palm facing outward and thumb fully extended, touching the horizon. Each part of your hand is used to measure a particular altitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Bmtfo.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bmtfo.png&quot; alt=&quot;Finding North&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hand method used by Nainoa Thompson to find the altitude of the Polaris. Journal of the Polynesian Society&#xA;In Hawai'i, the “North Star”, Polaris, is Hokupa'a, meaning “fixed star”. It lies close to the north celestial pole. The altitude of Hokupa'a indicates your northerly latitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the film, we see Moana Waialiki using this technique to measure the altitude of a group of stars. Look closely and you can see that she’s measuring the stars in Orion’s Belt. The position of Moana’s hand indicates the star above her index finger has an altitude of 21º. Given that the movie takes place about 2,000 years ago near Samoa, the position of Orion indicates they are travelling exactly due East.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xkziE.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xkziE.jpg&quot; alt=&quot;Moana&amp;#39;s measurement&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moana measures altitude of Orion’s belt stars. Walt Disney Studios Motion Pictures&#xA;Later in the film, we see Moana navigating by following Maui’s fish hook. In the various Polynesian traditions, the hook was used to pull islands from the sea. It is represented by the constellation Scorpius, which rises at dusk in mid-May. This indicates southeasterly travel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DKEeZ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DKEeZ.png&quot; alt=&quot;Scorpius&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at Scorpius - Maui’s Hook - in the same orientation as shown in the film. Stellarium&#xA;However, the positions of the stars are not fixed in time. Over the 3,500 years that Polynesians have been exploring the Pacific, the stars have gradually shifted due to precession of the equinoxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the latitude of Samoa, the Southern Cross has lowered from 60º altitude in 1500 BCE to 41º today. Those navigating by the stars must gradually adjust their measurements as the positions of stars slowly shift over time.&quot;&lt;/p&gt;&#xA;" OwnerUserId="17323" LastActivityDate="2017-06-29T22:58:14.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21461" PostTypeId="2" ParentId="21459" CreationDate="2017-06-30T03:02:11.693" Score="9" Body="&lt;p&gt;There is no contradiction between special relativity and quantum mechanics. Quantum field theory fully merges special relativity and quantum mechanics to describe relativistic electrons and protons (quantum electrodynamics) and quarks (quantum chromodynamics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problems lie with merging general relativity and quantum mechanics.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-06-30T03:02:11.693" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21463" PostTypeId="1" AcceptedAnswerId="21466" CreationDate="2017-06-30T09:31:49.777" Score="2" ViewCount="942" Body="&lt;p&gt;Suppose Europa is found habitable and humans start living on it. Europa is close to Jupiter and during the night Jupiter would be reflecting light to Europa, just like our full moon reflects light to earth. Our moon is small compared to Jupiter. Would humans still need artificial light during Europas nights?&lt;/p&gt;&#xA;" OwnerUserId="14136" LastEditorUserId="2365" LastEditDate="2017-06-30T14:36:59.773" LastActivityDate="2017-07-15T16:29:40.753" Title="How much light would be reflected from Jupiter to Europa (in Europas night)?" Tags="&lt;light&gt;&lt;jupiter&gt;&lt;habitable-zone&gt;&lt;night-sky&gt;&lt;europa&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21466" PostTypeId="2" ParentId="21463" CreationDate="2017-06-30T13:38:43.603" Score="8" Body="&lt;p&gt;It's a pretty straight forward calculation of 3 factors.   Distance from the sun, apparent size and albedo.   I'm going to compare Jupiter to our full moon, since we're all familiar with that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Jupiter&quot; rel=&quot;noreferrer&quot;&gt;Jupiter averages&lt;/a&gt; between 4.95 and 5.45 AU from the sun.   That puts it, in comparison with our full moon (about 1.02 AU on average), that means individual square meters on Jupiter would receive between 1/23.5 and 1/28.5 the light that hits our moon.  I'm going to round and say between 1/24 and 1/28.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brightness is also important.  The Moon is very dark with an albedo of about 12% - about the color of worn or greyish asphalt.  Jupiter's bond albedo is about .34, so it would reflect nearly 3 times more light.   That means per square meter, Jupiter is up to about 1/8th to 1/10th as bright.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big difference is size.   From Earth, the Moon takes up about 1/2 of 1 degree of arc.   Jupiter's diameter is about 37 times the Moon's diameter.    It's distance to Europa is about 1.73 times Earth's distance to the Moon, so it's apparent size would be about 21 times the diameter (about 10 degrees of arc), the reflective area is closer to 450 times.   If you divide that by the ratio of 8 to 10 depending on how close or far Jupiter is to the Sun in it's orbit, A full Jupiter would be the brightness of roughly between 45 and 55 average full moons.   It wouldn't glow as brightly, though some of it's lighter bands would be almost half way there.   But the entire package would be 45 to 55 times as bright (ignoring any atmospheric reflection, which would make it appear even brighter). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's probably bright enough to read under, but not so bright that it would hurt your eyes if you stared at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This only answers your question for a full jupiter, but the ratio, 45 to 55 times brighter should be about right for all the phases.   And about 450 times larger than an average Moon.   The Moon's size varies fairly significantly from Earth, though we don't always notice.   Europa has a much more circular orbit around Jupiter, so it's size would stay about the same from the surface of Europa.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-06-30T13:46:33.453" LastActivityDate="2017-06-30T13:46:33.453" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21467" PostTypeId="1" CreationDate="2017-06-30T14:27:55.493" Score="1" ViewCount="184" Body="&lt;p&gt;Is it possible that most galaxies (and our galaxy) were formed from leftovers of supernova explosions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to have a black hole 4,000,000 x solar mass the initial mass of the star should have been even greater. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I was thinking, is it possible that &lt;strong&gt;after the big bang, gas formed much bigger stars 1,000,000,000,000,000,000,000x bigger than the sun&lt;/strong&gt;, which later exploded forming: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;black holes in the middle of future galaxies&lt;/li&gt;&#xA;&lt;li&gt;Gas, which later created smaller stars, planets, etc.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Am I trying to reinvent a bicycle here? &lt;/p&gt;&#xA;" OwnerUserId="17293" LastEditorUserId="2365" LastEditDate="2017-06-30T14:38:18.833" LastActivityDate="2017-06-30T14:45:09.440" Title="Was Milky Way formed after Sagittarius A Explosion?" Tags="&lt;black-hole&gt;&lt;galaxy&gt;&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21468" PostTypeId="2" ParentId="21467" CreationDate="2017-06-30T14:45:09.440" Score="3" Body="&lt;p&gt;This is ruled out because that's not what we see when we look back in the past. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this was the case at the beginning, we would see these huge stars, with probably different characteristics from the ones formed only recently. But we don't see anything like that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what we see when we look very far away - that's few hundreds of millions of years after the beginning - we see signatures of first generation stars forming, very big stars indeed, but nothing close to what you're talking about. &lt;/p&gt;&#xA;" OwnerUserId="11264" LastActivityDate="2017-06-30T14:45:09.440" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21470" PostTypeId="2" ParentId="21463" CreationDate="2017-06-30T17:23:50.737" Score="4" Body="&lt;p&gt;This is based purely on Stellarium's simulation, so may not be accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Europa, full Jupiter has a magnitude of almost -17 and an angular diameter of 12 degrees:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0JDqE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0JDqE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes it about 56 times brighter and 24 times larger (in diameter, not area-- it's about 600 times larger in area) than our full moon (magnitude -12.3), more than good enough for reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's still not as bright as the Sun on Europa, which clocks in at a magnitude of -23.06:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6uKpd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6uKpd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you accept the quarter moon (magnitude -10.0) provides sufficient reading light, two of Jupiter's other moons (Io and Ganymede; Callisto is very dark and has a low albedo) can do this as well:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fY0GQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fY0GQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TBGXs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TBGXs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it seems Europa's sky is fairly well lit, even at solar night. But don't get your hopes too high: I performed this simulation at the intersection of Europa's equator and prime meridian. An earlier simulation in the northern hemisphere showed that Jupiter doesn't rise at all there.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2017-07-15T16:29:40.753" LastActivityDate="2017-07-15T16:29:40.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21471" PostTypeId="2" ParentId="21452" CreationDate="2017-06-30T21:10:13.030" Score="2" Body="&lt;p&gt;I found another discussion on the same topic here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://forum.cosmoquest.org/archive/index.php/t-18214.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://forum.cosmoquest.org/archive/index.php/t-18214.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is pretty old, but at least some numbers. According to that I gues its far bellow 10% these days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also interesting to search &lt;a href=&quot;https://hla.stsci.edu/hlaview.html#&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble Legacy Archive&lt;/a&gt; for &quot;footprints&quot; - but this does not cover the whole sky at once, you need to search for an object and see if there are some footprints in its area on a flashplayer-powered map.&lt;/p&gt;&#xA;" OwnerUserId="17348" LastActivityDate="2017-06-30T21:10:13.030" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21472" PostTypeId="2" ParentId="21452" CreationDate="2017-06-30T22:39:36.317" Score="5" Body="&lt;p&gt;25 years of say 16 orbits/day is a (very over-estimated) maximum of 146,000 pointings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest instrument is the various incarnations of the optical imager with a FOV of 6 (WFPC2) to 11 (ACS) square arcminutes. Thus the absolute maximum fraction of the sky (40,000 square degrees) that can have been covered is 1 per cent. And this &lt;em&gt;is&lt;/em&gt; a big overestimate, since many parts of the sky have received many repeated pointings (e.g. the Hubble deep field).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-06-30T22:39:36.317" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21473" PostTypeId="1" CreationDate="2017-07-01T06:44:14.013" Score="1" ViewCount="136" Body="&lt;p&gt;Many time I saw a moving star in the night generally been 08:00 to 11:00 PM. It is not any satellite, because it move in all directions, although it moves in a straight line. Sometimes from north to south, and sometimes from east to west and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also not a comet, because its brightness is continuous, and it moves continuously in a direction. It is also not a plane ✈, because in the night a plane is travelling with three-four coloured light. And they generally do not move on the way of planes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are those moving stars? &lt;/p&gt;&#xA;" OwnerUserId="17349" LastEditorUserId="1891" LastEditDate="2018-07-26T10:08:56.200" LastActivityDate="2018-07-26T10:08:56.200" Title="What are these objects on the night sky generally between 8:00 and 11:00 PM?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2017-07-06T16:17:26.893" ContentLicense="CC BY-SA 4.0" />
  <row Id="21475" PostTypeId="1" AcceptedAnswerId="21476" CreationDate="2017-07-01T08:08:38.787" Score="0" ViewCount="746" Body="&lt;p&gt;Lets say the spaceship is traveling at .9c to Proxima Centauri. How long does this trip take from the observational standpoint of the passenger vs. the observer from Earth?&lt;/p&gt;&#xA;" OwnerUserId="17350" LastActivityDate="2019-05-20T14:02:26.653" Title="How long would a near-light-speed trip to proxima centauri take from the traveler's and the earth-observer's perspectives?" Tags="&lt;light&gt;&lt;general-relativity&gt;&lt;time-dilation&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21476" PostTypeId="2" ParentId="21475" CreationDate="2017-07-01T13:32:59.967" Score="2" Body="&lt;p&gt;Proxima Centauri is about 4.25 light years away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.25 / 0.9 = 4.72 years, for the observer on earth. See fourth grade math for info on the calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the person on the ship, he is undergoing time dilation of about 2.29, which results in 4.72 / 2.29 = 2.06 years. See for example &lt;a href=&quot;https://www.fourmilab.ch/cship/timedial.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.fourmilab.ch/cship/timedial.html&lt;/a&gt; for info the calculation of the dilation.&lt;/p&gt;&#xA;" OwnerUserId="10084" LastActivityDate="2017-07-01T13:32:59.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21477" PostTypeId="2" ParentId="16650" CreationDate="2017-07-01T22:09:24.823" Score="1" Body="&lt;p&gt;According to &lt;a href=&quot;https://phys.org/news/2017-03-probing-worlds-nasa-james-webb.html#nRlv&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt;, the TRAPPIST-1 system is supposed to be studied by JWST with the potential to detect atmospheres' composition in infrared band. So at last 40 ly (approx. 12 pc) should work. Somewhere else I read there is an assumption to study earth-like exoplanetary atmospheres at distance of 10-20 pc. But it might be better to see it from different angle: the real limit is due to the produced light energy/intensity, not the distance itself. For example, see an article on &lt;a href=&quot;https://www.cosmos.esa.int/web/jwst-nirspec/exoplanets&quot; rel=&quot;nofollow noreferrer&quot;&gt;exoplanet transit spectroscopy&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17348" LastActivityDate="2017-07-01T22:09:24.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21478" PostTypeId="1" AcceptedAnswerId="21479" CreationDate="2017-07-02T03:43:41.690" Score="1" ViewCount="1936" Body="&lt;p&gt;These are two diagrams of Cassegrain reflecting telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One uses a concave secondary mirror, and the other uses a convex one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any difference between the two?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, does one cost more? Does one produce a better image?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;===&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: These images came from:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://astro-canada.ca/les_telescopes-reflecting_telescopes-eng&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://astro-canada.ca/les_telescopes-reflecting_telescopes-eng&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.a-levelphysicstutor.com/optics-telescs.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.a-levelphysicstutor.com/optics-telescs.php&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;===&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/j0i4Y.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j0i4Y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/n80ol.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/n80ol.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15340" LastEditorUserId="15340" LastEditDate="2017-07-02T04:53:51.600" LastActivityDate="2017-07-05T06:05:40.513" Title="In a reflecting telescope (Cassegrain), is there a difference between using a concave or convex secondary mirror?" Tags="&lt;telescope&gt;&lt;telescope-lens&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21479" PostTypeId="2" ParentId="21478" CreationDate="2017-07-02T05:53:52.833" Score="1" Body="&lt;p&gt;There are many flavors of reflecting telescopes based on the original &lt;a href=&quot;https://en.wikipedia.org/wiki/Cassegrain_reflector&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cassegrain design&lt;/a&gt; which has been attributed by history to &lt;a href=&quot;https://en.wikipedia.org/wiki/Laurent_Cassegrain&quot; rel=&quot;nofollow noreferrer&quot;&gt;Laurent Cassegrain&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Cassegrain reflector is named after a published reflecting telescope design that appeared in the April 25, 1672 Journal des sçavans which has been attributed to Laurent Cassegrain. Similar designs &lt;strong&gt;using convex secondaries&lt;/strong&gt; have been found in the Bonaventura Cavalieri's 1632 writings describing burning mirrors and Marin Mersenne's 1636 writings describing telescope designs. James Gregory's 1662 attempts to create a reflecting telescope included a Cassegrain configuration, &lt;strong&gt;judging by a convex secondary mirror&lt;/strong&gt; found among his experiments. (from &lt;a href=&quot;https://en.wikipedia.org/wiki/Cassegrain_reflector&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;; emphasis added)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt;, a &lt;a href=&quot;https://en.wikipedia.org/wiki/Gregorian_telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Gregorian Telescope&lt;/strong&gt;&lt;/a&gt; does indeed use a concave secondary mirror, placed &lt;em&gt;after the primary focus&lt;/em&gt;. The diverging beam is then re-imaged behind the primary. According to Wikipedia, examples of Gregorian telescope designs in use include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The MeerKAT, the Green Bank Telescope, the Arecibo Observatory, and the Allen Telescope Array are all radio telescopes employing off-axis Gregorian optics.&lt;/li&gt;&#xA;&lt;li&gt;The Vatican Advanced Technology Telescope, the Magellan telescopes, and the Large Binocular Telescope use Gregorian optics.&lt;/li&gt;&#xA;&lt;li&gt;The Giant Magellan Telescope will also use Gregorian optics.&lt;/li&gt;&#xA;&lt;li&gt;The Solar Optical Telescope on the Hinode satellite.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Advantages of the Gregorian include the option of putting a field stop at the primary focus, helpful when viewing near bright, interfering sources of light, and providing an upright, non-inverted image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In either the convex/Cassegrain-type or concave/Gregorian type of system, the curved secondary mirror provides several important advantages, including:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Slowing the convergence of the rays, which allows for the diameter to be much smaller than the primary (so as to block less light and be much lighter weight) while still bringing the focus beyond the hole in the primary and out to where an eyepiece and/or other optics can be conveniently located.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The length of the telescope can be quite short while the effective focal length seen at the eyepiece is still quite long, allowing higher magnification for a given eyepiece.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The cartoon illustration shown in the &lt;em&gt;&lt;a href=&quot;http://astro-canada.ca/les_telescopes-reflecting_telescopes-eng&quot; rel=&quot;nofollow noreferrer&quot;&gt;Canada under the stars&lt;/a&gt;&lt;/em&gt; website shows a meniscus with it's concave side used as a reflector. I think this is just an error by the illustrator. It wouldn't make sense to show a Gregorian telescope in a paragraph about Laurent Cassegrain, and the primary focus does not fall in front of the secondary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tilted the drawing so that the optical axis is horizontal, and highlighted the incoming and reflected rays of light striking the top edge of the secondary. You can see that the reflected ray is more parallel to the axis, which would only happen if the mirror were convex. The very large distance between the lens of the eyepiece, and the actual location of the exit for one's eye confirms that some artistic license has been taken with the optics here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference, look at the &lt;em&gt;realistic&lt;/em&gt; drawing of a Cassegrain optical system from Wikipedia shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The drawing from the &quot;&lt;a href=&quot;http://www.a-levelphysicstutor.com/optics-telescs.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;A-level Physics Tutor&lt;/a&gt;&quot; website is also problematic, showing the primary focus &lt;em&gt;far in front&lt;/em&gt; of the primary mirror of an optical telescope, followed by an eyepiece lens with the same diameter as the secondary mirror. Again this is artistic license, but there's no excuse for it in a proper physics tutorial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are examples of optical telescopes where the secondary mirror is &lt;em&gt;so strongly convex&lt;/em&gt; that the light then becomes divergent, expanding toward the primary before being refocused (made convergent again) by a tertiary mirror, such as the (crazy/beautiful) &lt;a href=&quot;https://en.wikipedia.org/wiki/Large_Synoptic_Survey_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Large Synoptic Survey Telescope&lt;/a&gt;. See also &lt;a href=&quot;https://www.lsst.org/about/tel-site/optical_design&quot; rel=&quot;nofollow noreferrer&quot;&gt;LSST Optical Design&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TrPjW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TrPjW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8Exmc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8Exmc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Drawing of a Cassegrain reflecting telescope, from &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Cassegrain.en.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VYHyt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VYHyt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Illustration of the optical system of the Large Synoptic Survey Telescope, from &lt;a href=&quot;https://www.lsst.org/about/tel-site/optical_design&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; (rotated).&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-07-05T06:05:40.513" LastActivityDate="2017-07-05T06:05:40.513" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21480" PostTypeId="1" CreationDate="2017-07-02T06:27:03.893" Score="2" ViewCount="2149" Body="&lt;p&gt;Can electromagnetic waves change their properties in the presence of gravity? &lt;/p&gt;&#xA;" OwnerUserId="17349" LastEditorUserId="37191" LastEditDate="2021-02-12T21:06:48.453" LastActivityDate="2021-02-18T13:52:42.837" Title="Can electromagnetic waves be affected by gravity?" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;light&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2021-02-18T14:23:49.550" ContentLicense="CC BY-SA 3.0" />
  <row Id="21481" PostTypeId="1" AcceptedAnswerId="21492" CreationDate="2017-07-02T06:31:29.043" Score="0" ViewCount="390" Body="&lt;p&gt;I have two ephemerides for &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_elements&quot; rel=&quot;nofollow noreferrer&quot;&gt;orbital elements&lt;/a&gt;—that is, each contains $a$ (semi-major axis), $e$ (ecc.), $i$ (incl.), $\Omega$ (long. asc. node), $\omega$ (arg. of peri.), $\tau$ (time of peri.), $T$ (period), and a timestamp $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to compute a new ephemeris at a point somewhere between these two ephemerides.  For example, the timestamp ($t$) of the new ephemeris can be calculated by a linear interpolation of the start ($t_0$) and end ($t_1$) ephemerides' times as:$$&#xA;    t = \text{lerp}(t_0,t_1, x) = t_0 \cdot (1-x) + t_1 \cdot x&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I can't seem to get an interpolation for the other parameters that doesn't have discontinuities in it.  &lt;strong&gt;My question: how &lt;em&gt;should&lt;/em&gt; I be interpolating ephemerides?  Maybe I need to do this through some intermediate parameters, like $\varpi$, $\lambda$, or $\nu$?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;sub&gt;&lt;sup&gt;For example, if I take $\tau$ from the first ephemeris and lerp everything else, I get a strong discontinuity when mean anomaly is calculated (as $M:=n (t-\tau)=(2 \pi/T) (t - \tau)$).  I suspect this is because the mean anomaly calculation is only valid at the ephemeris, and as e.g. $T$ changes due to interpolation, $M$ changes erroneously.&lt;/sup&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;sub&gt;&lt;sup&gt;&lt;a href=&quot;https://pastebin.com/q934rL9K&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is an example for Mercury.  The first three ephemerides come from NASA HORIZONS and are correct.  Interpolating the first two ephemerides produces $M=328^\circ$.  However, two seconds later, the interpolation (which is now interpolating from the last two ephemerides) produces $M=294^\circ$.&lt;/sup&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2576" LastEditorUserId="2576" LastEditDate="2017-07-03T02:12:35.543" LastActivityDate="2017-07-03T03:25:48.700" Title="How Can I Interpolate Ephemerides?" Tags="&lt;orbital-elements&gt;&lt;ephemeris&gt;&lt;mathematics&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21483" PostTypeId="2" ParentId="21480" CreationDate="2017-07-02T13:21:02.860" Score="3" Body="&lt;p&gt;Yes. Einstein predicted one such effect known as gravitational lensing. This occurs when light from a distant star or galaxy passes near a galaxy or galaxy cluster on its way to our planet. The nearer galaxy actually bends the EM waves and causes a focusing effect enabling us here on earth to actually see the more distant galaxy. If you do a Google search on 'gravitational lensing' you'll get some very cool pictures of this effect and a more complete explanation of this.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="7472" LastEditDate="2021-02-18T13:52:42.837" LastActivityDate="2021-02-18T13:52:42.837" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21484" PostTypeId="1" AcceptedAnswerId="21486" CreationDate="2017-07-02T17:17:10.787" Score="1" ViewCount="70" Body="&lt;p&gt;I live on the periphery of a large light dome, looking towards the West. I get about 1/3 of the sky towards mountains that is darker. According to ClearDarkSky.com I live in an area with artificial to night sky brightness ratio of about 9.00, Sky Brightness mags / arcsec&lt;sup&gt;2&lt;/sup&gt; V Band of about 19.5, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Bortle_scale&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bortle Scale&lt;/a&gt; of about 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the Clear Sky Clock (also at ClearDarkSky.com) I see the following sky values:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cloud Cover&lt;/li&gt;&#xA;&lt;li&gt;Transparence&lt;/li&gt;&#xA;&lt;li&gt;Seeing&lt;/li&gt;&#xA;&lt;li&gt;Darkness&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which of these values is most important in a light polluted area, especially when looking west in my case to the darker sky portion?  I'm thinking it is transparency as low transparency may result in more scattering of light pollution and thus more pollution?  But perhaps moon phase beyond a certain portion matters too, especially if the moon is in the west.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a large Dobsonian telescope and wonder which of these is going to have the most beneficial effect when viewing near the city?&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="2365" LastEditDate="2017-07-02T18:08:08.810" LastActivityDate="2017-07-02T19:00:40.403" Title="In an area with lots of light pollution, which clear sky clock parameter is most important?" Tags="&lt;light-pollution&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21485" PostTypeId="2" ParentId="21473" CreationDate="2017-07-02T18:49:40.160" Score="2" Body="&lt;p&gt;Your objects sound like satellites to me. Why do you think they are not satellites?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where I live at 40 degrees north latitude, these are the paths that we see the satellites moving (just a rough estimate: not a scientific count)&#xA;• 90% are in a polar orbit, so they generally move south to north or north to south (different parts of the same orbit). These generally follow lines of right ascension.&#xA;• 9% are inclined 50 to 70 degrees to the equator, so they cross the sky from NW to SE or SW to NE (different parts of the same orbit)&#xA;• 1% go across the southern half of the sky from west to east .&#xA;•0% go from east to west because there are very few satellites in such orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Jan mentioned, what is your latitude? That can affect which objects you see and where you see them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how long are they visible? They ones you see with your unaided eyes will cross the sky in 4 to 7 minutes although the time they are visible can be less. You will see more in the eastern half of the sky after sunset than in the western half because of the phase angle.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-07-02T18:49:40.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21486" PostTypeId="2" ParentId="21484" CreationDate="2017-07-02T19:00:40.403" Score="3" Body="&lt;p&gt;I would agree that transparency is the more important item, but it does depend on what you are observing. Deep sky objects such as clusters, nebulae, and galaxies will benefit from the darker skies by the better transparency. The Sun, Moon, planets and double stars will benefit more from better seeing than better transparency (but my guess is that you are not using a large dob for the Moon and planets.)&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-07-02T19:00:40.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21487" PostTypeId="1" AcceptedAnswerId="21488" CreationDate="2017-07-02T19:16:28.643" Score="4" ViewCount="1098" Body="&lt;p&gt;I studied the &lt;a href=&quot;https://en.wikipedia.org/wiki/Farnese_Atlas&quot; rel=&quot;nofollow noreferrer&quot;&gt;Farnese Atlas&lt;/a&gt; and I wonder how did the Ancients discover the ecliptic, the equator, etc. with the Ptolemaic system namely without knowing that the Earth rotates on itself, especially for the Ecleptic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask the question because if in their model the Earth does not rotate, so how do they explain the two revolutions of the sun around the earth:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The first one being the daily cycle (sunrise and sunset)&lt;/li&gt;&#xA;&lt;li&gt;The second one being the tropical year&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With a model where the Earth does not rotate, I do not understand how they found out that the Sun make a revolution of a one year around the Earth (whence the ecliptic) with a second daily revolution. How could they conjugate these two Sun's cycles?&lt;/p&gt;&#xA;" OwnerUserId="17365" LastEditorUserId="17365" LastEditDate="2017-07-03T16:44:54.960" LastActivityDate="2019-05-03T04:49:26.150" Title="How did the Ancients understand the ecliptic using the Ptolemaic system" Tags="&lt;history&gt;&lt;ecliptic&gt;&lt;geocentrism&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21488" PostTypeId="2" ParentId="21487" CreationDate="2017-07-02T21:28:19.340" Score="6" Body="&lt;p&gt;The ecliptic was understood to be the path of the sun relative to the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ptolemy believed the sky rotated daily around the Earth. All the planets, including the Moon and the Sun, shared in this motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the planets had secondary motions, relative to the sky. And in particular relative to the stars that they thought were fixed to the sky. This motion was discovered by careful observations of the sky, and was very well known, long before Ptolemy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motion of the stars was very simple: therefore when Ptolemy was describing his system, he only concerned himself with the motion of the planets (including the moon and sun) relative to the stars. It was clear that the planets all approximately followed a great circle relative to the stars. This great circle is the ecliptic. The Sun's motion was the simplest. It was exactly on the great circle, and it's position relative to the stars moved almost evenly. The other planets (including the moon) had orbits that were less even, but by introducing a system of Eccentrics, Deferents, Epicycles and Equants, Ptolemy was able to account for nearly all their motions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The precession of the equinoxes was noticed because the Greeks used a calendar based on the heliacal rising (the date that stars are just visible before sunrise) of certain stars. And it was noticed that the heliacal rising was changing relative to the seasons and the equinox. We understand the reason for this to be the precession of the equinoxes.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-07-02T21:28:19.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21489" PostTypeId="2" ParentId="21487" CreationDate="2017-07-02T21:37:36.347" Score="0" Body="&lt;p&gt;They saw the stars, planets, Moon and the sun (the Celestial Sphere) rotate around the Earth each day.  The Sun however, moved about 1 degree each day across the Celestial Sphere, so by the end of a year it moved 360 degrees. They called one full cycle of the Sun moving across the Celestial Sphere, a year. The planets (the word meant wanderers) moved each at slightly different and variable rates. The Sun, planets and the Moon moved along on roughly the same great circle on the Celestial Sphere.  Since they traveled along the same path but at different rates, they would eclipse each other every so often. I guess that is why they called that great circle the ecliptic.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2017-07-02T21:43:23.020" LastActivityDate="2017-07-02T21:43:23.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21490" PostTypeId="2" ParentId="21450" CreationDate="2017-07-02T22:01:46.217" Score="6" Body="&lt;p&gt;The Milky Way has arms that form due to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/18611/why-do-galaxies-have-arms/18620&quot;&gt;density waves&lt;/a&gt;. Like the majority of spiral galaxies, the arms are trailing. Individual stars orbit in circles (roughly), neither towards or away from the centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider a common map of the Milky way (imagined from a point North of the Earth, Celestial North is not the same as Ecliptic North, which are both about 60 degrees off from &lt;a href=&quot;https://www.physicsforums.com/attachments/there-planes-angles_no-earth-09oct2016-jpg.107279/&quot; rel=&quot;noreferrer&quot;&gt;Galactic North&lt;/a&gt;)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/UkfQ2.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UkfQ2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;The stars in the galaxy would be moving in a clockwise fashion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to view from the other side, it would be as if you had made a mirror image of the galaxy, so the motion would be counterclockwise. If you look at the back of a clock, then from the wrong side, the hand would move counterclockwise. However the rotation of the galaxy is still with its arms trailing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some galaxies that rotate with arms leading: &lt;a href=&quot;http://hubblesite.org/news_release/news/2002-03&quot; rel=&quot;noreferrer&quot;&gt;NGC 4622&lt;/a&gt; is one example.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-07-03T05:43:22.810" LastActivityDate="2017-07-03T05:43:22.810" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21491" PostTypeId="2" ParentId="14276" CreationDate="2017-07-02T23:18:20.487" Score="2" Body="&lt;p&gt;If you are using SOFA, I'd recommend reading the IERS Convention 2010. Here is the link: &lt;a href=&quot;http://iers-conventions.obspm.fr/2010officialinfo.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;IERS Convention 2010&lt;/a&gt;. Chapter 5 is where you should read... in it, it describes the procedures as well which functions in the SOFA library you should use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The above link has moved. The new link is: &lt;a href=&quot;https://www.iers.org/IERS/EN/Publications/TechnicalNotes/tn36.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IERS Convention 2010&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14357" LastEditorUserId="14357" LastEditDate="2020-06-10T06:26:21.680" LastActivityDate="2020-06-10T06:26:21.680" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21492" PostTypeId="2" ParentId="21481" CreationDate="2017-07-03T03:25:48.700" Score="0" Body="&lt;p&gt;OK, I see what you did. You looked at the osculating elements for Mercury's orbit around the solar system barycenter, by using Horizons with these parameters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qsEE5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qsEE5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you point out in the attached pastebin, the values jump around a lot:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;$$SOE&#xA;2451544.500000000 = A.D. 2000-Jan-01 00:00:00.0000 TDB &#xA; EC= 1.970095254610514E-01 QR= 3.160869485749435E-01 IN= 7.013139646115477E+00&#xA; OM= 4.812103959293303E+01 W = 2.595697040970977E+01 Tp=  2451500.244608603884&#xA; N = 3.993474279540898E+00 MA= 1.767327672710030E+02 TA= 1.777641486117292E+02&#xA; A = 3.936372330648511E-01 AD= 4.711875175547588E-01 PR= 9.014706864254217E+01&#xA;2451575.500000000 = A.D. 2000-Feb-01 00:00:00.0000 TDB &#xA; EC= 2.099217560035118E-01 QR= 2.935320561427645E-01 IN= 7.027294624675849E+00&#xA; OM= 4.850357410323192E+01 W = 3.527448059792021E+01 Tp=  2451590.708947353531&#xA; N = 4.355289183393867E+00 MA= 2.937606361003802E+02 TA= 2.698751266828741E+02&#xA; A = 3.715227679957090E-01 AD= 4.495134798486537E-01 PR= 8.265811633648384E+01&#xA;2451604.500000000 = A.D. 2000-Mar-01 00:00:00.0000 TDB &#xA; EC= 2.241380367466583E-01 QR= 3.093951924304613E-01 IN= 6.978935239526586E+00&#xA; OM= 4.831291617128525E+01 W = 3.114644418385429E+01 Tp=  2451590.362934530713&#xA; N = 3.916530358018232E+00 MA= 5.536824608413040E+01 TA= 7.970140671677829E+01&#xA; A = 3.987760801330001E-01 AD= 4.881569678355390E-01 PR= 9.191809256960804E+01&#xA;$$EOE&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;so interpolation doesn't work well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem: Mercury does NOT orbit the solar system barycenter; it orbits the Sun (&lt;a href=&quot;https://astronomy.stackexchange.com/questions/18169/what-point-does-earth-actually-orbit/18304&quot;&gt;What point does Earth actually orbit?&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The barycenter is the center of mass of our solar system. If you're far enough away from our solar system, you can treat the solar system as a point mass at that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside our solar system, that doesn't work as well for most planets (massive Jupiter is an exception). Example: when Sun-Mercury-Jupiter form a straight line (and Jupiter's on the same side of the Sun as Mercury), Jupiter's gravitational on influence on Mercury is stronger than it is on the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you run Horizons using the Sun as the center body:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8VeLA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8VeLA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you'll get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;$$SOE&#xA;2451544.500000000 = A.D. 2000-Jan-01 00:00:00.0000 TDB &#xA; EC= 2.056302512089075E-01 QR= 3.074991199665784E-01 IN= 7.005014199657344E+00&#xA; OM= 4.833053756455964E+01 W = 2.912428058698772E+01 Tp=  2451502.287118767854&#xA; N = 4.092345945977128E+00 MA= 1.727497133778637E+02 TA= 1.751155303115542E+02&#xA; A = 3.870982252717257E-01 AD= 4.666973305768729E-01 PR= 8.796910250314700E+01&#xA;2451575.500000000 = A.D. 2000-Feb-01 00:00:00.0000 TDB &#xA; EC= 2.056312618551657E-01 QR= 3.074988256430905E-01 IN= 7.005012539288613E+00&#xA; OM= 4.833050698201237E+01 W = 2.912403060593708E+01 Tp=  2451590.256131388247&#xA; N = 4.092344011669980E+00 MA= 2.996128340775531E+02 TA= 2.767940962437320E+02&#xA; A = 3.870983472501978E-01 AD= 4.666978688573051E-01 PR= 8.796914408304917E+01&#xA;2451604.500000000 = A.D. 2000-Mar-01 00:00:00.0000 TDB &#xA; EC= 2.056306629761345E-01 QR= 3.074990706849075E-01 IN= 7.005011440208267E+00&#xA; OM= 4.833046370516735E+01 W = 2.912368651066544E+01 Tp=  2451590.256076942664&#xA; N = 4.092343747828102E+00 MA= 5.829102946757737E+01 TA= 8.083993808120411E+01&#xA; A = 3.870983638882191E-01 AD= 4.666976570915307E-01 PR= 8.796914975460210E+01&#xA;$$EOE&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Notice that most of the parameters change only infinitesimally, since Mercury's orbit around the Sun is fairly stable (for a counterexample, see the Moon's orbit around the Earth, which varies quite a bit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a note, I also thought it would be nice if the planets orbited the barycenter and did some work on it, but the numbers just don't work out: &lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/playground4.m&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/playground4.m&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2017-07-03T03:25:48.700" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21493" PostTypeId="1" AcceptedAnswerId="21494" CreationDate="2017-07-03T04:47:17.517" Score="0" ViewCount="317" Body="&lt;p&gt;I am not talking about an antigravity effect that would cancel gravity, rather if there is a repulsive force, like when you use two magnets and flip one of them, where it becomes obvious that one is pushing away the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope the question is a good fit for the site, feel free to migrate it otherwise.&lt;/p&gt;&#xA;" OwnerUserId="1630" LastActivityDate="2017-07-03T05:55:10.903" Title="Does the inverse of gravity exists?" Tags="&lt;gravity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21494" PostTypeId="2" ParentId="21493" CreationDate="2017-07-03T05:55:10.903" Score="3" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No such effect has been discovered. Gravity seems to be always attractive, never repulsive. Nor has a particle with negative mass ever been discovered. In particular, antimatter appears to have positive mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not theoretically impossible, but a particle with negative mass would be very odd, as the faster it moved, the less momentum and energy it would have.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-07-03T05:55:10.903" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21495" PostTypeId="2" ParentId="21480" CreationDate="2017-07-03T07:50:07.110" Score="5" Body="&lt;p&gt;&lt;strong&gt;Yes (and somewhat No apparently too)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;If you follow Einstein's theory of general relativity then gravity has no direct effect on electromagnetic waves itself.&lt;/p&gt;&#xA;&lt;p&gt;Gravity however affects space time itself warping it towards the gravity source. Thus any electromagnetic wave passing close enough to the gravity source will bend towards that source when observed from the outside.&lt;/p&gt;&#xA;&lt;p&gt;From the point of the wave however, it follows a straight path and is not affected by gravity at all.&lt;/p&gt;&#xA;&lt;p&gt;If you mix in quantum theory, it's getting even more weird, because suddenly parts that work well for general relativity no longer work within the quantum model. This is part of current research in physics to bring general relativity and quantum theory into alignment.&lt;/p&gt;&#xA;" OwnerUserId="14441" LastEditorUserId="7472" LastEditDate="2021-02-18T13:52:25.777" LastActivityDate="2021-02-18T13:52:25.777" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21497" PostTypeId="1" AcceptedAnswerId="21501" CreationDate="2017-07-03T11:42:10.953" Score="2" ViewCount="191" Body="&lt;p&gt;I read many times that black holes is created when center of a very massive star collapses in upon itself but what does it mean ? Does weight of center of massive star increase so much that itself can't bear weight or I can say space itself cant bear weight of such center of massive star that it collapse. I tried to read answer on google but want to know in layman language as I am not astro physics researcher or student. Thanks &lt;/p&gt;&#xA;" OwnerUserId="14136" LastEditorUserId="14136" LastEditDate="2017-07-03T15:56:15.430" LastActivityDate="2017-07-03T15:58:50.803" Title="What does it mean 'black holes is created when center of a very massive star collapses in upon itself'?" Tags="&lt;black-hole&gt;&lt;supernova&gt;&lt;supermassive-black-hole&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21500" PostTypeId="2" ParentId="21497" CreationDate="2017-07-03T15:50:29.797" Score="-1" Body="&lt;p&gt;A star is always a balance of gravity trying to collapse it inward and nuclear fusion trying to blow it up. Usually, the forces are in balance, as the gravity causes more nuclear fusion from hydrogen all the way to iron. However, iron takes an exponentially higher amount of energy to fuse, so gravity wins and crunches the star inward. This weight forms the black hole and also generates a final burst of nuclear fusion, causing the burst of light that people see, and also forming the elements beyond iron. Hope that this answer helps!&lt;/p&gt;&#xA;" OwnerUserId="10161" LastActivityDate="2017-07-03T15:50:29.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21501" PostTypeId="2" ParentId="21497" CreationDate="2017-07-03T15:58:50.803" Score="3" Body="&lt;p&gt;This Wikipedia page on &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_evolution&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellar Evolution&lt;/a&gt; is a good starting point with lots of links to more info.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A star is a huge sphere of mostly hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like any large mass it has a gravitational field and that field acts on star itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it would always want to collapse to as dense a sphere as possible under it's own gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now as the density increases, particularly at the center, the pressure also increases and so does the temperature.  When these are high enough the core of the star will be able to fuse hydrogen (and later other materials) into heavier elements.  Doing so releases huge amounts of energy and this is what creates all that light we see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the energy released also prevents the star from collapsing more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it can stay that way (more or less) for billions of years.  It is this energy produced by fusion at the core that balances the gravitational force (weight) of the rest of the star trying to crush the core.  So fusion that generates more energy than it requires to happen is essential to keep a star &quot;alive&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a long time, the hydrogen at the core runs out and the star starts to burn heavier elements at the core (helium, carbon, etc. ).  It does that until those run out and (roughly) we reach elements like Iron remaining.  At that point fusing those elements no longer generates enough energy to stop the star collapsing and it starts to collapse again, but very, very rapidly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exactly what happens next depends mostly on the mass of the star.  It can end up as a white dwarf, neutron star or black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the final collapse is a very rapid process and quite violent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the useful fuel at the core runs out, the stars dies a spectacular death one way or another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is a black hole (in simple terms) ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity is a result of the way mass (or even energy) can distort space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond a certain point this distortion becomes so great that everything on one side becomes trapped inside a &quot;surface&quot; called the &lt;em&gt;event horizon&lt;/em&gt;.  An event horizon is what a black hole actually is - no event horizon, no black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Within the black hole, space is distorted in a way that's basically impossible to describe without some mathematics.  The effect is that there is no direction you can move in and no speed you can travel at which will not lead you to the center of the black hole - the singularity.  Don't even try and visualize it, it just isn't a common sense idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This even works with light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now an event horizon is not a surface you can feel.  You can actually travel across an event horizon (but only inward, not outward !).  So this change in the way space-time is distorted is not something that is sudden or obvious to the person doing the falling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to our star collapsing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it collapses (without energy from fusion to stop that), it becomes denser and denser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if it is large enough it will collapse to a black hole, because the core will become so dense from the collapse that it will be enough to form an event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our Sun is not large enough to do that.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-07-03T15:58:50.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21503" PostTypeId="1" CreationDate="2017-07-03T17:37:15.783" Score="1" ViewCount="803" Body="&lt;p&gt;Given a date, I'd like to return the lunar phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jean Meeus's Astronomical Algorithms book and the many packages like PyEphem that make use its algorithms provide ways to calculate when a primary phase (New, First Quarter, Full, and Last Quarter).  The secondary phases (Waxing Crescent, First Quarter, Waxing Gibbous, Waning Gibbous, and Waning Crescent) can be easily calculated as falling in between the primary ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The percent illuminated the moon (or any other body between the observer and the Sun) can also be calculated for a given point in time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also calculate the subsolar point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What isn't clear is how long the primary phases should be considered to be in play.  How are labels generally determined for these phases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next new moon occurs 2017/7/23 09:45:34 UTC.  When should it no longer be considered new but waxing crescent?  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When it is 1.0% illuminated? &lt;/li&gt;&#xA;&lt;li&gt;When the subsolar point moves 5 degrees west of 180E?  10 degrees?&lt;/li&gt;&#xA;&lt;li&gt;At 2017/7/24 00:00:01 local time?&lt;/li&gt;&#xA;&lt;li&gt;At 2017/7/24 00:00:01 UTC?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm inclined to use the last.  The moon is &quot;new&quot; for the entire UTC day 0.0% illumination falls on.  The following day it is waxing crescent and remains so until the day the first quarter falls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone know of a standard here or rule of thumb?&lt;/p&gt;&#xA;" OwnerUserId="7619" LastActivityDate="2019-07-28T15:49:14.043" Title="calculating lunar phases" Tags="&lt;moon-phases&gt;&lt;ephemeris&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21504" PostTypeId="1" AcceptedAnswerId="21506" CreationDate="2017-07-03T19:53:29.573" Score="1" ViewCount="114" Body="&lt;p&gt;Discovering a solar system that cannot be seen by the naked eye on earth seems like an easy task with telescopes (e.g. the hubble telescope). I'm guessing, that our sun is the reference point for the bearing of the telescope at the point of discovery of a solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do you find this solar system again? F.e., after discovering a &quot;new&quot; solar system, a year passes by, and there is need to know in which &quot;direction&quot; this solar system is in relation to ours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit off-topic, but a reason as to why this question might be important: In science fiction there is often the possibility to travel faster than light, reaching other solar system quickly - or for the sake of this question lets assume, we would be able to travel there too in a timely manner. How would they/we know where to go, since the universe is ever expanding.&lt;/p&gt;&#xA;" OwnerUserId="17370" LastActivityDate="2017-07-03T20:11:44.600" Title="How are solar systems found again after being discovered?" Tags="&lt;telescope&gt;&lt;star-systems&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21505" PostTypeId="2" ParentId="21504" CreationDate="2017-07-03T20:08:19.447" Score="3" Body="&lt;p&gt;Newly found solar systems will be associated with stars. There is a well defined system of coordinates for stars (and other celestial objects). Stars, other than the sun, are so far away that their coordinates change only a very small amount from year to year, and so it is easy to locate them. &lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-07-03T20:08:19.447" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21506" PostTypeId="2" ParentId="21504" CreationDate="2017-07-03T20:11:44.600" Score="5" Body="&lt;p&gt;I think you're overestimating how quickly stars appear to move as viewed from Earth.  Once a telescope like &lt;a href=&quot;https://www.nasa.gov/mission_pages/kepler/main/index.html&quot; rel=&quot;noreferrer&quot;&gt;Kepler&lt;/a&gt; identifies a solar system around a star, we know where that star is.  Stars have coordinates associated with them that astronomers can use to know where to point their telescopes to see it.  For example, &lt;a href=&quot;https://en.wikipedia.org/wiki/Betelgeuse&quot; rel=&quot;noreferrer&quot;&gt;look at the star Betelgeuse&lt;/a&gt;.  Look in the info box on the right, the right ascension and declination are the coordinates to find it.  On the timescale of one Earth year, or even a few hundred Earth years, the stars are mostly fixed, and that coordinate to find them still works.  Sure they move &lt;em&gt;slightly&lt;/em&gt;, but not enough to lose track of it, or even enough to dramatically change how the constellations appear to us.  Even if one does somehow move dramatically, it would take decades to centuries, and astronomers would be tracking it and updating the coordinate information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about it this way.  The constellations used in the sky today are named after many ancient Greek and Roman mythological characters.  This isn't something we decided recently, they were named hundreds or thousands of years ago and they're still recognizable in the same form they were then.  If stars moved across our sky as quickly as you appear to think, those constellations would all have to be renamed.  You need timescales of tens of thousands of years to make the sky unrecognizable, and even then it would have more to do with &lt;a href=&quot;http://www.ancient-wisdom.com/precession.htm&quot; rel=&quot;noreferrer&quot;&gt;the precession of the equinoxes&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="14255" LastActivityDate="2017-07-03T20:11:44.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21507" PostTypeId="1" CreationDate="2017-07-03T20:23:16.083" Score="1" ViewCount="130" Body="&lt;p&gt;Exciting discoveries of weak gravitational lensing suggest the &lt;a href=&quot;https://doi.org/10.1093/mnras/stx517&quot; rel=&quot;nofollow noreferrer&quot;&gt;existence of galactic filaments made of dark matter that are the size of galactic superclusters&lt;/a&gt;. My questions involves the future of the galactic filaments while space indefinitely expands between the galaxy clusters within each galactic filament. Will the filaments indefinitely expand with the indefinite expansion of space or will the filaments divide because of an event horizon? For example, event horizon paradigms suggest that the expansion of space will eventually isolate each galaxy cluster from each other galaxy cluster. Perhaps we do not yet have enough information to calculate this because we know little about dark matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also posted this question at the &lt;a href=&quot;https://groups.google.com/d/topic/sci.astro.research/l5-hGFMaVCU/discussion&quot; rel=&quot;nofollow noreferrer&quot;&gt;moderated news group sci.astro.research&lt;/a&gt;. A moderator noted: &quot;I think your last statement is &lt;em&gt;definitely&lt;/em&gt; true: we don't know enough about dark matter to calculate this.&quot;&lt;/p&gt;&#xA;" OwnerUserId="14131" LastEditorUserId="11264" LastEditDate="2017-10-04T12:59:22.647" LastActivityDate="2017-10-04T12:59:22.647" Title="Will galactic filament indefinitely expand while its galaxy clusters indefinitely disperse?" Tags="&lt;dark-matter&gt;&lt;galaxy-cluster&gt;&lt;gravitational-lensing&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21508" PostTypeId="2" ParentId="21503" CreationDate="2017-07-03T21:42:07.680" Score="1" Body="&lt;p&gt;A new moon occurs 2017/7/23 09:45:34 UTC. Instantly after that it is waxing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A crescent won't be visible for a few days. It reaches first quarter about 7 days later. The moon will appear to be about half full for maybe a couple of days. Nevertheless, first quarter, full moon, third quarter and new moon are moments in time, not intervals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may, sometimes be convenient to refer to &quot;the day that new moon occurs&quot;, and in casual use people will say the moon is full while it appears as a circle to the naked eye (usually lasting several nights) These are casually defined for casual use. For precise terminiology say &quot;it is now 3 days 4 hours after New Moon&quot; or similar.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-07-03T21:57:14.350" LastActivityDate="2017-07-03T21:57:14.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21509" PostTypeId="1" AcceptedAnswerId="21510" CreationDate="2017-07-04T03:49:25.540" Score="0" ViewCount="438" Body="&lt;p&gt;Suppose I have a 2D data array with number of counts in each pixel (i.e., this is the image array). Suppose I have another 1-1 mapped same-shape data array that gives the 1-sigma Gaussian standard deviation in the number of counts in each pixel in the image (i.e., this is the error array). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I do circular aperture photometry, I just calculate -2.5*log10(summed counts within aperture) + magnitudeZeropoint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, for the magnitude uncertainty on my aperture magnitude, I read that I'm supposed to do a quadrature sum of the errors within my aperture, then compute the fractional flux uncertainty as the ratio of my quadrature-summed uncertainty divided by my measured counts from the image, and then magnitude error = 2.5*log10(1 + fractional flux uncertainty). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How come the uncertainties have to be quadrature summed ($\sqrt{\Sigma \sigma_{x,y}^2}$), instead of just added up like I do with the image pixel values? The quadrature sum results in a smaller errorbar but is that realistic? Also, I emphasize that my data arrays are in units of counts (i.e., ADU's), not electrons.&lt;/p&gt;&#xA;" OwnerUserId="607" LastEditorUserId="2365" LastEditDate="2018-07-27T22:33:50.607" LastActivityDate="2018-07-27T22:33:50.607" Title="Aperture photometry uncertainties" Tags="&lt;observational-astronomy&gt;&lt;photometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21510" PostTypeId="2" ParentId="21509" CreationDate="2017-07-04T12:47:55.263" Score="1" Body="&lt;p&gt;Because the uncertainties in the number of counts detected in each pixel are considered to be independent. That means some of them are higher than the &quot;true value&quot; (the count rate you would measure if you observed for an infinitely long time), whilst others are lower. To some extent these independent uncertainties will cancel, and the net result is that for normally distributed errors the correct protocol is what you have described.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps a good way of seeing this is that suppose you took 100 &lt;em&gt;independent&lt;/em&gt; measurements of the same thing, each with its own, roughly equal uncertainty. If I asked what the uncertainty in the average was, you wouldn't just add up all the errors and divide by 100 because that would give an uncertainty that was identical to that of an individual measurement. Instead you would do the quadrature sum of the uncertainties and divide by 100, which reduces the uncertainty in the average by a factor of $\sqrt{100}$.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-07-04T12:47:55.263" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21511" PostTypeId="1" AcceptedAnswerId="21513" CreationDate="2017-07-04T17:40:26.077" Score="0" ViewCount="839" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Qs5qb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qs5qb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this image an original picture which is not photoshoped? I see a huge number of Hubble Telescope images on Google and I'm curious whether they are real images or not. If not, where I can find the real images captured by the Telescope which are not photoshopped?&lt;/p&gt;&#xA;" OwnerUserId="16098" LastActivityDate="2019-06-26T06:35:50.987" Title="Is this image captured by the Hubble Telescope an original image?" Tags="&lt;hubble-telescope&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21512" PostTypeId="1" CreationDate="2017-07-04T19:52:45.620" Score="4" ViewCount="178" Body="&lt;p&gt;So for example: a Neutron star has a radius of 10km and 2 solar masses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A singularity can have a radius of 1m and 10 solar masses, with a 1km event horizon radius. Basically it will just be a more compact object than a neutron star. How come we are so sure that singularity is an infinitely small point instead? &lt;/p&gt;&#xA;" OwnerUserId="17293" LastActivityDate="2017-07-04T21:55:31.817" Title="How do we know that Singularity is not an object with reasonable radius?" Tags="&lt;black-hole&gt;&lt;neutron-star&gt;&lt;event-horizon&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21513" PostTypeId="2" ParentId="21511" CreationDate="2017-07-04T20:13:10.727" Score="5" Body="&lt;p&gt;All images provided from large telescopes are a rendering of the data for viewing pleasure. Usually sensors are sensitive in a wide range of wave length, even outside what humans can perceive and color is added by using filters. Hubble uses infrared filters and the so-called &lt;a href=&quot;http://www.astronomymark.com/hubble_palette.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble palette&lt;/a&gt; (false color).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you mean, do these pictures show what a human observer will see? No, they don't. They are a visualisation of the data collected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All things observed at low light conditions are black and white, as humans lose color vision at low light.&lt;/p&gt;&#xA;" OwnerUserId="15824" LastEditorUserId="28526" LastEditDate="2019-06-26T06:35:50.987" LastActivityDate="2019-06-26T06:35:50.987" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21515" PostTypeId="2" ParentId="21512" CreationDate="2017-07-04T21:52:19.003" Score="1" Body="&lt;p&gt;Within the event horizon gravity is so strong that no matter how great is the force you apply, you can't move away. For an object to remain as an object it would need to be more than infinitely strong (an obvious impossibility)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This assumes that general relativity is a good model. It is possible that general relativity is incorrect at these extreme conditions. It is possible that some kind of string theory prevents ultimate collapse, but this is beyond established theory.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-07-04T21:52:19.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21516" PostTypeId="2" ParentId="21512" CreationDate="2017-07-04T21:55:31.817" Score="6" Body="&lt;p&gt;We don't know, but general relativity implies that no force can stop the object from shrinking to infinitesimally small.  Quantum Mechanics is more uncertain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just want to point out that 10 solar masses would have a radius 29.53 km at the event horizon, not 1 KM.   A Neutron Star is pretty close to its event horizon already. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with &quot;some distinct size inside the event horizon&quot; is that things change when the acceleration is greater than the speed of light.   If the acceleration is strong enough that the gluons are drawn towards the center faster than they an move (speed of light), then it's entirely reasonable that the structure wouldn't hold.   Another way of saying this is that the gravity becomes stronger than the Neutron Degeneracy Pressure and there's no known force that would prevent continued shrinkage once the Neutron Degeneracy Pressure is overcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Pauli exclusion principal says that two neutrons can't occupy the same space, but this can be overcome somewhat by the weight and pressure approaching infinity.   The neutrons can get closer and closer as their momentum approaches infinity and that happens at the singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, both relativity and QM (the two best theories for dealing with this question) say that a ridiculously small object may exist at the singuarity, not meters, but maybe the size of an atomic nucleus or size of a quark or even, possibly zero size, but without a quantum field theory of gravity it's impossible to say with much confidence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other possibilities exist like space might get stretched into some very strange properties, where virtual particles become real but fall back towards the singularity, similar to hawking radiation but without the escape, or, the angular momentum could form into a ring or band, not a point but all that is just guesses.    The mathematics of the two main theories both say that a very tiny point is entirely reasonable and the argument for what you suggest, an object with a distinct radius of say 1 meter, is far less supported by current theories.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's my layman's answer anyway.   I can delete this if I got too much of the physics wrong, but I thought I'd give you a quick answer.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-07-04T21:55:31.817" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21520" PostTypeId="1" CreationDate="2017-07-05T05:40:47.947" Score="0" ViewCount="249" Body="&lt;p&gt;Putting aside how such a planet could be formed and just lets assume it exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would an gas giant &quot;look&quot; like if it just consisted of the gas compound earths atmosphere consists of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could we just watch through it?&#xA;Would we see a burning core and therefore see something like a dim star?&#xA;Or would there be still causes that would block vision, even if there were no other components included?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And would such a planet be able to hold it self together?&lt;/p&gt;&#xA;" OwnerUserId="10102" LastActivityDate="2017-07-05T17:43:54.643" Title="What would a gas giant consisting of only earths atmosphere look like?" Tags="&lt;gas-giants&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2017-07-07T12:20:23.217" ContentLicense="CC BY-SA 3.0" />
  <row Id="21521" PostTypeId="1" CreationDate="2017-07-05T06:21:18.460" Score="1" ViewCount="305" Body="&lt;p&gt;I read that gaseous planets such as Jupiter and Saturn, have a solid core deep inside. So, I wondered whats the difference between them and rocky planets apart from the fact that they have - what can supposedly be considered, a very thick atmosphere, and that their atmospheric pressure would probably be too high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any other difference which I am missing? I mean, if we go by definition, couldn't all their gaseous content, be considered as an atmosphere?&lt;/p&gt;&#xA;" OwnerUserId="17277" LastActivityDate="2019-11-05T18:51:16.853" Title="If gaseous planets have solid core, can they theoretically be considered as rocky planets?" Tags="&lt;planet&gt;&lt;solar-system&gt;&lt;planetary-atmosphere&gt;&lt;gas-giants&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21522" PostTypeId="2" ParentId="21521" CreationDate="2017-07-05T10:08:16.473" Score="2" Body="&lt;p&gt;As a general rule, a gas giant planet is large enough that it retains a lot of hydrogen and helium, such as Jupiter and Saturn.  Their chemical makeup is similar to the universe, nearly 90% hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ice giants, like Uranus and Neptune are made up of material more similar to comets, lots of ices like H20, CO2, CH4, NH3 are the primary 4 ices that form into bodies beyond the frost line, but neither Uranus or Neptune have much free hydrogen or helium, either because they're too small or because of where they formed, to close to the sun where it was too hot for them to retain or keep the lightest 2 gases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a general rule, in addition to your definition, one possible definition, of a gas giant is an abundance of hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we get a bit of better look at planets in other solar systems, some of the terminology might need to change.    Super-Earths with a lot of water would be similar to hot Neptunes.   Any very massive planet that is able to collect and retain massive amounts of hydrogen will look like a gas giant, whether it has a super-earth origin or not.    Perfect definitions aren't possible.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-07-05T10:08:16.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21523" PostTypeId="1" AcceptedAnswerId="21527" CreationDate="2017-07-05T12:45:26.600" Score="-8" ViewCount="669" Body="&lt;p&gt;Why the 'dark matter' is still the main research trend? It looks like a waste of time, money and resources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=MvNCWMD6so4&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=MvNCWMD6so4&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17385" LastActivityDate="2017-07-05T16:27:39.130" Title="Why &quot;Dark matter&quot; is still considered?" Tags="&lt;dark-matter&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" ClosedDate="2017-07-06T13:42:05.043" ContentLicense="CC BY-SA 3.0" />
  <row Id="21525" PostTypeId="2" ParentId="21523" CreationDate="2017-07-05T14:41:45.893" Score="0" Body="&lt;p&gt;Science advances through observation, hypothesis, experimentation/prediction, analysis of data, and comparing the hypothesis with the observed data.  If the data continues to support one theory over another, through peer reviewed publications, then it will become the currently accepted theory of how something works.  I happen to think there are issues with the theory of dark matter and find MOND, Modified Newtownian Dynamics, to be far more likely to be the reason we have 'missing matter' in the universe.  But it does not explain all missing matter and phenomenon and so as scientists we should wait until the data supports our 'theory' and not assume others are misguided and wasting time, money, etc.   &lt;/p&gt;&#xA;" OwnerUserId="17272" LastActivityDate="2017-07-05T14:41:45.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21526" PostTypeId="2" ParentId="21520" CreationDate="2017-07-05T14:58:10.267" Score="-3" Body="&lt;p&gt;Jupiter's atmosphere is mostly Hydrogen, lying over liquid Hydrogen, on top of metallic Hydrogen, and finally at the center a large rocky core.  As you remove the hydrogen atmosphere the liquid would turn to gas, metallic to liquid, and on and on until all the hydrogen were gone and you have a massive core left approximately 12 to 45 times the mass of Earth.  (Not much is known about the core.)  So if you had only the rocky core, cooled down from it's current temperature of 36,000K, and added a nitrogen/oxygen Earth like atmosphere, you could see the bottom in an Earth like way.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would not be a dim star, though it would glow very hot until it cooled, and also the planet would hold together.  &lt;/p&gt;&#xA;" OwnerUserId="17272" LastActivityDate="2017-07-05T14:58:10.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21527" PostTypeId="2" ParentId="21523" CreationDate="2017-07-05T16:27:39.130" Score="11" Body="&lt;p&gt;I'm going to answer this question, not because I believe you want a thoughtful, coherent answer, but rather for others who may come across this question and look at the propaganda you've posted. I don't believe anything I can say will change your mind, but hopefully I can inform other, unsuspecting viewers&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The Context&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The video you've posted is from thunderbolts.info. This is a pseudo-scientific organization that widely promotes well-debunked and/or non-scientific concepts such as creationism or the ideas of the &lt;a href=&quot;http://rationalwiki.org/wiki/Electric_Universe&quot; rel=&quot;noreferrer&quot;&gt;Electric Universe&lt;/a&gt;. This video specifically discusses the concept of Dark Matter and tries to make an argument that scientists have now finally disproven that dark matter exists. To back up this claim, they cite two news articles and have a voice over from an &quot;astronomer&quot; named Barry Setterfield. The disrepute of the video is pretty clear when, within the first minute, they state:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2016 may be remembered as the year that the hypothesis of Dark Matter was finally, officially falsified.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I certainly don't remember that occurring in 2016.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to next take a look at the two news articles this video is based one. They're both published on phys.org. One important point I want to mention here is that phys.org is &lt;em&gt;not&lt;/em&gt; the author of these articles. This website is merely a physics-related news curator which accepts and culls news articles from a wide variety of sources. In both cases, these articles were submitted by the respective institutions or groups of people performing the research and so likely had a heavy slant to play up the importance or interpretation of their own work.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;News Article 1&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The first news article is titled &lt;a href=&quot;https://phys.org/news/2016-09-spiral-irregular-galaxies-current-dark.html&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Acceleration relation found among spiral and irregular galaxies challenges current understanding of dark matter&lt;/em&gt;&lt;/a&gt;. It is a report on a paper published by &lt;a href=&quot;https://arxiv.org/abs/1609.05917&quot; rel=&quot;noreferrer&quot;&gt;McGaugh et al. 2016&lt;/a&gt; in the Physical Review Letters. The main author, Stacy McGaugh is a strong proponent of a hypothesis called &lt;a href=&quot;https://en.wikipedia.org/wiki/Modified_Newtonian_dynamics&quot; rel=&quot;noreferrer&quot;&gt;MOND&lt;/a&gt;, which is a fairly weak, yet persistent contender for Dark Matter. It purports to explain the same things Dark Matter does, but without introducing new physics or mysterious, unexplained particles. It simply modifies the physics we already know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper published by McGaugh details their measurement of the rotation curves of galaxies. Ironically, they find the exact same mass deficit one would be expected to find and would explain via Dark Matter. The focus of their paper however, is on a tight correlation between the expected centripetal acceleration due to baryonic matter and the observed centripetal acceleration. Note the important word here: &lt;em&gt;correlation&lt;/em&gt;. All they've done is shown the two concepts are related via an empirical equation. They haven't disproved Dark Matter, they haven't even found evidence it doesn't exist, and they haven't found &quot;new physics&quot;. Their paper, in my personal opinion, is weak at best. They make heavy-handed assumptions and don't often explain the details, relegated it to sentences like their assumptions &quot;only affects details, not the basic result&quot; (granted it is a letter, not a full blown paper).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, I see nothing ground breaking or even important in this article or research paper and it seems clear to me that the phys.org article itself was likely written by Stacy McGaugh or one of the other paper authors.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;News Article 2&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The second article cited is titled &lt;a href=&quot;https://phys.org/news/2016-12-unexpected-interaction-dark-ordinary-mini-spiral.html&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Unexpected interaction between dark matter and ordinary matter in mini-spiral galaxies&lt;/em&gt;&lt;/a&gt;. The article itself is pretty devoid of anything substantial. I wasn't even sure what research the authors had performed and it was clear they simply wrote up this article to play up their own work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading &lt;a href=&quot;https://arxiv.org/abs/1609.06903&quot; rel=&quot;noreferrer&quot;&gt;their paper&lt;/a&gt;, it seems their work is focused on studying dwarf, disk galaxies. They posit that baryonic matter plays a more important role in galaxy formation than others might have previously suspected but importantly, for this video, they appear to have no qualms with declaring Dark Matter as real. In fact they heavily use the concept of Dark Matter in their models and assumptions. What this video does is to cherry pick certain statements from this paper, misinterpret them, and claim stronger statements about Dark Matter that the original paper never stated. These claims are only backed up by explanations which simply restate the idea without actually providing a real explanation. This video citing this article seems like a clear cut case of the logical fallacy of &lt;a href=&quot;https://en.wikipedia.org/wiki/Argument_from_authority&quot; rel=&quot;noreferrer&quot;&gt;Appealing to Authority&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Barry Setterfield&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Let's discuss the main narrator of the video, &lt;a href=&quot;http://rationalwiki.org/wiki/Barry_Setterfield&quot; rel=&quot;noreferrer&quot;&gt;Barry Setterfield&lt;/a&gt;. He is awarded the dubious title of &quot;Astronomer&quot; in the video (again an appeal to authority) and spends most of the video explaining the historical background and evidence for Dark Matter. I won't even get into the fact that he missed out on a lot of the history, choosing to jump right to Vera Rubin and completely missing Fritz Zwicky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, this narrator is a known young-Earth creationist. He has no scientific degrees (having only been through a couple of years of undergraduate work before dropping out). He has published papers describing how the speed of light is decaying over time which have not been well received. I should note that none of his published papers (as far as I can tell) are published in well-respected, peer-reviewed journals. Most are simply self-published, or else published in junk journals which have since gone under. His only claim to being an &quot;astronomer&quot; is that he helps run the christian backed, &lt;a href=&quot;http://www.newhopechristian.net/students/new-hope-observatory&quot; rel=&quot;noreferrer&quot;&gt;New Hope Observatory&lt;/a&gt;. I don't say any of this to cast aspersions, merely to provide context so that things aren't misinterpreted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Throughout this video, he constantly makes unwarranted claims. For example, from the second news article he concludes that the authors indicated dark matter halos do not extend past the baryonic matter and thus&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dark Matter can only occur in those locations where ordinary matter can be found, not on its own or elsewhere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From here he immediately jumps to the much stronger statement&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This negates the whole idea of Dark Matter existing at all&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So he misinterprets a paper's statement and moves on to make a much stronger claim. He attempts to back this up by making the studies he's talking about sound comprehensive. He specifically implies his above claims are validated by calling out the &quot;2,696 data points&quot; in the paper from the first news article. A large number of data points does not make someone correct and it doesn't bridge the gap between his citations and conclusions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He finally concludes (with no real evidence) that Dark Matter cannot possibly exist based on his interpretation of only two very small studies. He then quickly casts aside MOND, the popular alternative to Dark Matter, in favor of an explanation from the Electric Universe &quot;theory&quot;. &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Electric Universe&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;As stated above, the concept of the Electric Universe is the idea that almost the entire universe is made of plasma and therefore almost all dynamics, including large-scale galaxy formation, are dominated by electric and magnetic forces. It is true to state that a large percentage of the baryonic matter in the universe is in a plasma form, but it is &lt;em&gt;not&lt;/em&gt; correct to claim that implies electric forces should be the governing force when it comes to large-scale dynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setterfield calls out the idea that electric forces are $10^{39}$ times stronger than gravitational forces and so should completely overpower any gravitational affects. &lt;a href=&quot;http://scienceworld.wolfram.com/physics/FundamentalForces.html&quot; rel=&quot;noreferrer&quot;&gt;He is correct about the strength of the electric force compared to the gravitational force&lt;/a&gt;, but completely neglects the crucial fact that electric charges come in both positive and negative forms. As it turns out, almost all of the universe is electrically neutral on large scales and so these powerful electric forces are all cancelled out by the conflicting charges. Because of this, electric forces only play a significant role over something known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Debye_length&quot; rel=&quot;noreferrer&quot;&gt;Debye length&lt;/a&gt;, a well-known concept you'd learn in a first semester plasma physics course. The Debye length for every type of plasma is far, far short of being long enough to govern large-scale dynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setterfield ends the video by citing the work of &lt;a href=&quot;https://en.wikipedia.org/wiki/Anthony_Peratt&quot; rel=&quot;noreferrer&quot;&gt;Anthony Peratt&lt;/a&gt; who claims to have run simulations which showed that electric forces were capable of forming galaxies and even resulted naturally in the flat rotation curves that are currently explained by dark matter.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Anthony Peratt&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;One thing that strikes me as odd in this whole quest to demonize Dark Matter is that this video is quick to hold up &quot;respected&quot; scientists as proof that their research and ideas are valid, while work by many other respected scientists is invalid. This two-faced approach to scientific authority is just a bit too convenient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, Anthony Peratt is a &quot;plasma cosmologist&quot;, i.e., a proponent of the electric universe theory. He published a collection of papers where he simulated two &quot;clouds&quot; of electrically charged plasma and showed that as they evolved, they appeared to form something that looked like a galaxy. There's numerous problems with this though.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;His plasma clouds were not galactic in size and could not have produced real galaxies. He even admits he limited his initial plasma cloud size to &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The fraction of the length correspond[ing] to a local region that is capable of interacting with an adjacent local region in a neighboring filament.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Anything larger wouldn't cut it in his simulations.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;His results only &quot;looked&quot; like galaxies in that they had spirals and similar shapes. They did not really exhibit the complex nature of the large variety of real galaxies.&lt;/li&gt;&#xA;&lt;li&gt;His spirals were a result of &quot;winding&quot; the matter as it spins. Scientists have long ago come upon the &lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/W/Winding+Problem&quot; rel=&quot;noreferrer&quot;&gt;&quot;winding problem&quot;&lt;/a&gt; and know that arms in a spiral galaxy are &lt;em&gt;not&lt;/em&gt; a result of winding matter but instead are &lt;a href=&quot;https://en.wikipedia.org/wiki/Density_wave_theory&quot; rel=&quot;noreferrer&quot;&gt;density waves&lt;/a&gt; at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lindblad_resonance&quot; rel=&quot;noreferrer&quot;&gt;Lindblad resonances&lt;/a&gt;. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Overall, using this research as evidence of galaxies forming in an Electric Universe, devoid of Dark Matter, is hardly convincing and certainly not enough to overturn the century of work concerning Dark Matter by thousands of scientists across the world.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Dark Matter&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The last point I want to make is about Dark Matter itself. Dark Matter is a hypothesis. Arguably, the first solid proposal for Dark Matter came from Fritz Zwicky in 1933. Zwicky found that the standard assumptions, physics, and observations did not all agree and that somewhere, something must be wrong. He chose to interpret that the problem lie in the amount of mass he was observing and that there was invisible mass he wasn't observing that was changing his results. Alternative theories such as MOND choose to claim the problem lies in our physics equations instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, Dark Matter has had fairly widespread support since it's inception and has been used to explain a variety of diverse problems over the years. It is used to explain the dynamics of galaxy clusters, the rotations of galaxies, observations in gravitational lensing, anisotropies in the CMB, and many more things. The ideas concerning Dark Matter have constantly been refined and improved. &lt;a href=&quot;https://en.wikipedia.org/wiki/Massive_compact_halo_object&quot; rel=&quot;noreferrer&quot;&gt;MACHOS&lt;/a&gt; have been all but thrown out; &lt;a href=&quot;https://en.wikipedia.org/wiki/Weakly_interacting_massive_particles&quot; rel=&quot;noreferrer&quot;&gt;WIMPS&lt;/a&gt; are still a possible explanation; we believe Dark Matter must be &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Cold_dark_matter&quot; rel=&quot;noreferrer&quot;&gt;cold&lt;/a&gt;&quot;. This is how science works. A problem was found, a solution was posed, and over the years that solution has been reworked and molded to better explain observations. Importantly Dark Matter helps explain areas and concepts beyond its original invocation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's not to say the idea of Dark Matter isn't without its faults or criticisms. For example, such problems include&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Cuspy_halo_problem&quot; rel=&quot;noreferrer&quot;&gt;the cuspy halo problem&lt;/a&gt;,&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Dwarf_galaxy_problem&quot; rel=&quot;noreferrer&quot;&gt;the missing satellite problem&lt;/a&gt;,&lt;/li&gt;&#xA;&lt;li&gt;or &lt;a href=&quot;http://www.preposterousuniverse.com/blog/2014/07/18/galaxies-that-are-too-big-to-fail-but-fail-anyway/&quot; rel=&quot;noreferrer&quot;&gt;the too-big-to-fail problem&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But, these problems aren't an invitation to throw out nearly 100 years worth of work in favor of a poorly supported and conceptually flawed &quot;theory&quot;. Instead, they should be motivation to further refine our hypotheses and what we think we know about Dark Matter. There may come a day when something better than Dark Matter comes along, when another hypothesis is capable of explaining everything Dark Matter can and more. I will gladly drop Dark Matter on that day. But it is not this day.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-07-05T16:27:39.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21528" PostTypeId="1" CreationDate="2017-07-05T17:05:31.223" Score="1" ViewCount="154" Body="&lt;p&gt;Cloudy weather would defeat the purpose so I am guessing that the western states would be more ideal due to greater chance of clear sky.  Isolation would also be nice.  At a camp site on or near a mountain top.  Also, are there efforts to set up mobile solar observatories in the totality path?&lt;/p&gt;&#xA;" OwnerUserId="16998" LastEditorUserId="7411" LastEditDate="2017-08-05T15:34:20.073" LastActivityDate="2017-08-05T15:34:20.073" Title="Where would be the absolute most logical place in USA to observe the total solar eclipse on August 21, 2017?" Tags="&lt;solar-eclipse&gt;&lt;observatory&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21530" PostTypeId="2" ParentId="21520" CreationDate="2017-07-05T17:43:54.643" Score="1" Body="&lt;p&gt;Jupiter's atmosphere, if you could see through it, the inside of the planet would begin to glow visibly red from heat perhaps a couple/few thousand miles below it's upper clouds.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/du0nH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/du0nH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://lasp.colorado.edu/education/outerplanets/giantplanets_interiors.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, some two thousand miles of atmosphere above that visible glow is almost certainly well past the point of transparency.   Add to that, mixing and cloud formation would make seeing through all that impossible - no matter what the atmosphere was made of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hot gas rises and as it rises, some gas condenses from gas into liquid droplets (clouds), but these clouds form in updrafts so even though they start falling back towards the planet, the thick atmosphere and the updraft movement maintains the clouds for some time.   There's different layers of clouds too.   Jupiter's upper most clouds are Ammonia, but below that there's water clouds (and perhaps others).   go deep enough you might get sulfur clouds (I'm not sure) or more dense hydrocarbon (gasoline like) clouds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An atmosphere that thick maintaining transparency is impossible.   Hydrogen and Helium are (to my knowledge) just as transparent as Oxygen and Nitrogen, maybe more, but the mixing and condensation gives Jupiter a constant cloud cover, different colors mostly represent different temperatures and depths from which that gas came from.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you gave Jupiter an Oxygen-Nitrogen atmosphere but maintained it's inner core, the inner hydrogen would rise and mix with the Oxygen-Nitrogen.   Chemistry would inevitably happen and the Oxygen would bind with the hydrogen and make water.   Some of the Nitrogen would probably form Ammonia and in the end, Jupiter might not look that different.   Initially it might have a bluish tint, but only temporarily.  It would likely always have clouds and mixing some heavier/darker gases from below.   You'd never see it's core, no matter what.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you had a planet with only one gas in it's atmosphere and entire makeup - avoiding any chemistry, say a planet made entirely of hydrogen or entirely of Argon or Oxygen - then you might get specific colors relating to the specific element.   But such planets are pure fantasy.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My best guess is that Jupiter's appearance wouldn't change much if you gave it an Earth-like atmosphere, but Jupiter is basically made of gas, so if you did that, it would just mix and it's inner hydrogen would rise to the top and Oxygen-nitrogen would sink.   It's the trace gases that give Jupiter it's color anyway, not it's hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Layman's answer.  :-)&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-07-05T17:43:54.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21531" PostTypeId="1" AcceptedAnswerId="21536" CreationDate="2017-07-05T18:44:57.723" Score="4" ViewCount="153" Body="&lt;p&gt;We have seen galaxies around 13 billion years old. Since these galaxies formed so early in the history of the known universe, are there observables about these galaxies that are apparently different from much younger galaxies. I suppose all current physics laws had condensed out by then but was just curious about the formation and 'aging' of such old galaxies.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="16926" LastEditDate="2017-07-05T20:17:40.997" LastActivityDate="2017-07-05T21:59:34.963" Title="What have we learned from the observation of most distant galaxies" Tags="&lt;galaxy&gt;&lt;deep-sky-observing&gt;&lt;age&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2017-07-09T12:36:42.203" ContentLicense="CC BY-SA 3.0" />
  <row Id="21532" PostTypeId="2" ParentId="21528" CreationDate="2017-07-05T18:53:09.000" Score="3" Body="&lt;p&gt;I assume that when considering where to view the Solar Eclipse from, you're looking for the place with the best observance of totality. This elicits two factors, the degree of totality and the time elapsed during totality. The best observed eclipse will be a total eclipse, i.e., 100% totality, and for the longest time possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A map of the path of 100% totality is shown below. In it, you can see that grey band that passes across the US. Anywhere within that band, you'll be able to see a 100% total solar eclipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4BlZA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4BlZA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you view &lt;a href=&quot;https://eclipse2017.nasa.gov/sites/default/files/interactive_map/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this interactive map from NASA&lt;/a&gt;, you can see two specific points along the totality band that are called out, one labeled GE and one labeled GD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The GE marker signifies the point of greatest eclipse which is &lt;a href=&quot;https://eclipse.gsfc.nasa.gov/SEhelp/SEglossary.html#greatest&quot; rel=&quot;nofollow noreferrer&quot;&gt;defined as&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The instant when the axis of the Moon's shadow cone passes closest to Earth's center. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The GD marker signifies the point of greatest duration, quite sensibly &lt;a href=&quot;https://eclipse.gsfc.nasa.gov/SEhelp/SEglossary.html#greatdur&quot; rel=&quot;nofollow noreferrer&quot;&gt;defined as&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The instant when the length of the total (or annular) phase reaches a maximum along the central path of a solar eclipse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, if you're looking for the point of 100% totality that lasts for the longest amount of time, you'll want to be at the point of the greatest duration. Geographically, this is pretty close to &lt;a href=&quot;http://www.eclipse2017.org/2017/communities/states/MO/Carbondale_129.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carbondale, IL&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2017-07-05T20:48:49.357" LastActivityDate="2017-07-05T20:48:49.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21536" PostTypeId="2" ParentId="21531" CreationDate="2017-07-05T21:54:11.260" Score="5" Body="&lt;p&gt;&lt;em&gt;Nearby&lt;/em&gt; galaxies are seen in their old age. Distant galaxies are, on average, the same age as the local ones. But due to the finite speed of light, the farther you look, the younger you observe them. That means that the most distant galaxies are observed in their infancy, and they look &lt;em&gt;much&lt;/em&gt; different from the old ones.&lt;/p&gt;&#xA;&lt;strong&gt;Galaxy evolution&lt;/strong&gt;&#xA;&lt;p&gt;This is one reason to observe galaxies at all distances: We get a statistical picture of the galaxies through different epochs, and hence probe their evolution. From such observations, we see e.g. how they grow with time, how their star formation increase in the early Universe until the Universe was roughly four billion years (Gyr) old and the decreased (&lt;a href=&quot;https://arxiv.org/abs/astro-ph/9607172&quot; rel=&quot;noreferrer&quot;&gt;Madau et al. 1996&lt;/a&gt;), and how the metals and dust built up over time.&lt;/p&gt;&#xA;&lt;strong&gt;Galaxy formation&lt;/strong&gt;&#xA;&lt;p&gt;Observing the most distant ones is not just a matter of breaking a previous record in distance, or time. When we look at the most distant (spectroscopically confirmed) galaxy to date, &lt;a href=&quot;https://en.wikipedia.org/wiki/GN-z11&quot; rel=&quot;noreferrer&quot;&gt;GN-z11&lt;/a&gt;, we see 13.4 Gyr back in time. What's the big deal if we find another one that is 13.5 Gyr back in time? Well, since the Universe at that time was only a few hundred million years (Myr) old, that would give us quite good constraints on the minimum timescale of galaxy formation. Can a galaxy form in only 300 Myr or 200 Myr, or does it need 400 Myr?&lt;/p&gt;&#xA;&lt;strong&gt;Physical properties&lt;/strong&gt;&#xA;&lt;p&gt;As I said above, the most distant galaxies look vastly different from the local ones. For instance, disks hadn't formed yet, both because it takes some time to settle into a disk, and because galactic mergers were more common at that time, which disturbed the galactic dynamics. The physical properties of the interstellar medium were also different, e.g. because it was more metal-poor, since metals are formed by stars over time. Dust, which is made from metals, was also sparse. But how long time does it take for stars to form metals, and dust? Previously it was thought that most dust was produced in AGB stars which die peacefully and hence have good conditions for making dust (high density and cool gas). But these stars typically live for billions of years, and observations of the most distant galaxies showed us that they contained dust as well. So people started wondering if supernovae, which are the violent deaths of massive and hence short-lived stars, could do it, even though it was thought that the gas was much too hot too form dust right after the explosion, and much too dilute after it had cooled down. But indeed it seems they can (e.g. &lt;a href=&quot;https://arxiv.org/abs/0903.0217&quot; rel=&quot;noreferrer&quot;&gt;Kozasa et al. 2009&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-07-05T21:59:34.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21537" PostTypeId="2" ParentId="13526" CreationDate="2017-07-06T00:04:02.640" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;General relativity is often explained as saying spacetime is curved by gravity, what does this mean? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This sounds dodgey. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity is a manifestation of spacetime curvature which is caused by, among other things, massive objects like stars, planets etc. In other words gravity is a consequence of spacetime curvature.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How could we perceive a curve in spacetime when there is no external &quot;straight&quot; reference frame for instance?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Like the answers above, this is achieved through experimental observations or measurements. For e.g. Eddington measured the bending of light around the limb of the Sun which would not have been possible where it not for curvature. &lt;/p&gt;&#xA;" OwnerUserId="16812" LastActivityDate="2017-07-06T00:04:02.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21538" PostTypeId="1" CreationDate="2017-07-06T01:03:37.450" Score="0" ViewCount="318" Body="&lt;p&gt;I have limited information to shift a spectrum (in the X-ray 0.5-10 keV bandpass) from redshift &lt;span class=&quot;math-container&quot;&gt;$z=2$&lt;/span&gt; to the rest frame. I have a plot of normalized (photon) counts s&lt;span class=&quot;math-container&quot;&gt;$^{−1}$&lt;/span&gt; keV&lt;span class=&quot;math-container&quot;&gt;$^{−1}$&lt;/span&gt; as a function of energy (keV). Is there a way to shift the spectrum I have to rest frame? Or do I really need luminosity distance, etc.?&lt;/p&gt;&#xA;" OwnerUserId="17391" LastEditorUserId="9527" LastEditDate="2021-01-15T14:09:38.590" LastActivityDate="2021-01-15T14:09:38.590" Title="How to shift an AGN X-ray spectrum to rest frame" Tags="&lt;redshift&gt;&lt;luminosity&gt;&lt;spectra&gt;&lt;x-ray&gt;&lt;active-galaxy&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21541" PostTypeId="1" AcceptedAnswerId="27027" CreationDate="2017-07-06T11:55:49.510" Score="2" ViewCount="100" Body="&lt;p&gt;I've recently been introduced to relativistic celestial Reference Systems (RS) such as the Geocentric Celestial Reference System (GCRS) and Barycentric Celestial Reference Systems (BCRS). So, for example we use GCRS to model near-Earth satellites or use BCRS to model probes far into our solar system. I understand the &lt;em&gt;need&lt;/em&gt; for such RS i.e. to be in compliance with the theory of relativity - no favoured coordinate systems, limitation of the speed of propagation of information etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what are some layman definitions of advantages of such systems to really send the message home? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I notice that when describing $n$-body dynamics in our solar system for example, Barycentric coordinate time is used instead of proper time. This blows my mind! &lt;/p&gt;&#xA;" OwnerUserId="16812" LastEditorUserId="16812" LastEditDate="2017-07-07T06:42:04.787" LastActivityDate="2018-07-20T23:43:35.980" Title="Reference frames in relativity?" Tags="&lt;solar-system&gt;&lt;general-relativity&gt;&lt;near-earth-object&gt;&lt;frame-of-reference&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21543" PostTypeId="1" CreationDate="2017-07-06T12:50:54.293" Score="2" ViewCount="289" Body="&lt;p&gt;I'd imagine ancient methods are fun exercises in geometry. how has the measurement has evolved over time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first method is credited to &lt;a href=&quot;https://en.wikipedia.org/wiki/Eratosthenes&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eratosthenes&lt;/a&gt;. He knew that during the summer solstice in Syenne the angle of the Sun overhead at noon was $0^\circ$. Following this he made the same measurement in Alexandria and measured the angle of the Sun overhead using shadows and calculated it to be $\approx 7^\circ$ which is approximately $1/50$ of a circle. Hence, both the circumference and the radius are easily calculated and quite accurate for such crude methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sU4q4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sU4q4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16812" LastEditorUserId="16812" LastEditDate="2017-07-07T02:12:51.943" LastActivityDate="2017-07-07T02:12:51.943" Title="How have we measured the radius of the Earth (old and new)?" Tags="&lt;solar-system&gt;&lt;earth&gt;&lt;distances&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21544" PostTypeId="2" ParentId="21543" CreationDate="2017-07-06T13:41:32.510" Score="2" Body="&lt;p&gt;This is a pretty broad question so I'll limit myself to describing one instance of how the radius of the Earth is measured. Specifically, I'll talk about how it's measured today, with current technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As technology has progressed, our ability to precisely measure facets of the Earth has drastically improved. At this point, we can measure the exact shape and size of the Earth to such precision that asking about the &quot;radius of the Earth&quot; is no longer meaningful. Primarily because we know the shape of the Earth to such precision that we know it is not a perfect sphere with a fixed radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current standard used to describe the exact shape of the Earth is the &lt;a href=&quot;https://en.wikipedia.org/wiki/World_Geodetic_System&quot; rel=&quot;nofollow noreferrer&quot;&gt;World Geodetic System&lt;/a&gt;, the latest edition of which is referred to as WGS84. In effect, this is a coordinate system centered on the Earth that precisely describes the shape of the Earth. This is done by using sensitive satellite measurements to fit the &lt;a href=&quot;https://en.wikipedia.org/wiki/Spherical_harmonics&quot; rel=&quot;nofollow noreferrer&quot;&gt;spherical harmonics&lt;/a&gt; to the Earth's geoid. The current, most updated version of WGS84 has more than 4.6 million harmonic terms with an accuracy in &quot;radius&quot; at any given point of 10 km. That is, you could pick any precise location on the Earth and use WGS84 to calculate the radius of the Earth at that exact point to an accuracy of 10 km. This radius would of course vary as you moved about the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really want to break this down into a simple number, the WGS model, at the crudest level, considers the Earth to be an oblate spheroid with an equitorial radius of $R_{eq} = 6\:378\:137\:\mathrm{m}$ and a polar radius of $R_{p} = 6\:356\:752.3\:\mathrm{m}$.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-07-06T13:41:32.510" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21545" PostTypeId="1" CreationDate="2017-07-06T14:28:49.377" Score="9" ViewCount="1947" Body="&lt;p&gt;Our moon and Saturn's moon, Titan, have this feature.  Because of this we only observe one side (hemispere) of our moon. Why is this?  What is the Newtonian or astrophysics that would explain this?  It seems that it would be some kind of equilibrium that is arrived at over time, but I don't understand it.&lt;/p&gt;&#xA;" OwnerUserId="16998" LastActivityDate="2017-07-06T16:15:15.757" Title="Why do certain moons have their rotational period equal to their orbital period?" Tags="&lt;the-moon&gt;&lt;astrophysics&gt;&lt;rotation&gt;&lt;orbital-mechanics&gt;&lt;titan&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21546" PostTypeId="2" ParentId="21545" CreationDate="2017-07-06T14:47:14.577" Score="3" Body="&lt;p&gt;The simple answer is: tidal forces, which are a secondary effect of gravity. In the same way that the Moon causes low and high tides of the oceans here on Earth, the Earth also has a similar effect on the Moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force is of the same origin, however far stronger due to the mass of the Earth. These tidal forces cause a torque on the rotation on the Moon and this is why it only shows the same face to the Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fun fact, Pluto and it's only moon Charon show only one face during their orbit due to the tidal force. Also the tidal forces are slowing down the rotation of the Earth too. &lt;/p&gt;&#xA;" OwnerUserId="16812" LastActivityDate="2017-07-06T14:47:14.577" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="21547" PostTypeId="1" CreationDate="2017-07-06T15:59:36.603" Score="6" ViewCount="11782" Body="&lt;p&gt;Cameras are great, they allow us to compress several seconds of light exposure into a single instant. This allows us to take amazing pictures of the night sky, like this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZqLFa.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZqLFa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course you can see we also have a great view of the landscape, we know that this is a trick from a camera and we're collecting a timed sample of light. This isn't something a human eye can see because we can't compress time into a single vision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious though -- people knew about the milky way long before we could do this cool trick with camera. That means they must have seen at least something of the concentration of stars and gases from earth -- So it makes sense that without light pollution this could be seen with the naked eye to some extent -- What does the milky way look like from earth to the human eye without light pollution?&lt;/p&gt;&#xA;" OwnerUserId="4026" LastActivityDate="2018-04-21T14:10:18.900" Title="What does the milky way look like from earth to the human eye without light pollution?" Tags="&lt;amateur-observing&gt;&lt;milky-way&gt;&lt;light-pollution&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21548" PostTypeId="2" ParentId="21545" CreationDate="2017-07-06T16:15:15.757" Score="14" Body="&lt;p&gt;The answer to this is certainly tidal forces, but that doesn't explain the exact mechanism for how tidal forces result in &lt;em&gt;tidal locking&lt;/em&gt;, i.e., an orbiting body showing the same face to the central body as it orbits due to the rotation rate and revolution rate being equal. I'll describe this mechanism using the Earth-Moon system so I can be specific, but it equally applies for any system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To start, tidal forces are a result of differential gravitational forces across the distributed mass of a body. The Moon is not a point mass, it has an extended size. The force of gravity on the Moon by the Earth is dependent upon distance (as is the force of gravity for anything). What this means is that on the Earth-facing side of the Moon, the force of gravity is stronger and as you progress through the Moon to the Earth-opposing side, the force of gravity gets weaker. This means the Earth-facing side of the Moon gets pulled stronger and closer to the Earth while the Earth-opposing side, while still getting pulled towards the Earth, doesn't get pulled as strongly or as close. Ultimately, this results in the Moon being &lt;em&gt;deformed&lt;/em&gt; such that it becomes slightly oblate and stretched out in the direction of the Earth. This flexing of the surface is known as tides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's assume the Moon is not currently tidally locked with the Earth, and in fact spins slightly faster than it orbits. The Earth is causing tides on the Moon and the Moon is spinning on its axis. The tides, caused by the tidal forces, want to remain aligned with the Earth-Moon line since that's the direction the tidal forces are applied. However, it takes time and a lot of energy to deform the Moon. Once the Moon is deformed, it's going to spin along and try to pull that tidal deformity along with it, effectively moving the tidal bulge ahead of the Earth-Moon line. The Earth is still applying the tidal force along that Earth-Moon line to try and pull the tidal bulge back in line. This constant force trying to pull the tidal bulge back (or ahead if the Moon is spinning too slow) allows for a transfer of momentum to slow down the Moon (or again, speed it up if it is too slow). The main point here is that tidal locking is a equilibrium state because if the Moon is spinning too slow or too fast, the Earth trying to pull the tidal bulge into the Earth-Moon line will change the rotation rate of the Moon until it becomes tidally locked. Once it is tidally locked, that tidal bulge will always be along the Earth-Moon line and this force will disappear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That covers about half of the answer though. When trying to tidal lock the Moon to the Earth, you have to consider two time periods. The first, discussed in the preceding paragraph, is the rotation time of the Moon around its axis. The other is the revolution time of the Moon around the Earth. Both must match. The previous paragraph described how the rotation time of the Moon around its axis could be affected, but there's also a means of affecting the revolution time of the Moon around the Earth. Fortunately, this is by a nearly identical mechanism as above. In effect, the Moon also causes tidal bulges on the Earth and since the Earth is spinning these tidal bulges won't be directly aligned with the Earth-Moon line. This unaligned tidal bulge on Earth acts to transfer energy to the Moon's orbital speed, causing it to speed up or slow down. Incidentally, through conservation of angular momentum, this necessarily causes the Moon to drift away from us at a small but persistent rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summation, tidal forces cause tidal locking, but it happens through intricate and slow forces over a long time period affecting both the Moon's orbital velocity and rotation rate until a equilibrium is found. That equilibrium is tidal locking.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-07-06T16:15:15.757" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21549" PostTypeId="2" ParentId="21547" CreationDate="2017-07-06T16:24:02.010" Score="5" Body="&lt;p&gt;It is pretty obvious from a dark sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cv81p.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cv81p.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.skyandtelescope.com/astronomy-resources/how-many-stars-night-sky-09172014/&quot; rel=&quot;noreferrer&quot;&gt;Sky and Telescope&lt;/a&gt; have a simulation (using stellarium) of the sky with a limiting magnitude of 6.5 (about the limit with excellent eyesight and a very dark sky). The Milky Way is very clear. Light and dark tracks are visible. It is clearly something that every person would know about and be able to see nightly. It is no surprise that it is mentioned in the myths of many ancient peoples.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-07-06T16:24:02.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21552" PostTypeId="1" CreationDate="2017-07-07T04:28:48.307" Score="0" ViewCount="63" Body="&lt;p&gt;If bacteria managed to arrive then thrive somehow (if not already) on a gas giant could it cause the atmosphere to expand or contract in volume?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would living matter potentially induce energy into the planets system to heat it up?&lt;/p&gt;&#xA;" OwnerUserId="825" LastActivityDate="2017-08-06T20:55:26.317" Title="What possible effects of life replicating on gas giant atmosphere?" Tags="&lt;astrobiology&gt;&lt;abiogenesis&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21553" PostTypeId="1" AcceptedAnswerId="21557" CreationDate="2017-07-07T07:32:48.503" Score="5" ViewCount="162" Body="&lt;p&gt;I assume such a system would be hard to detect due to a lack of radiation, at least if the vicinity of the black hole does not contain a lot of mass which falls into it over time, thus creating (a lot of) radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since black hole formation in a supernova explosion is a process which destroys any planetary system around its original star, any planetary system around a black hole must have formed &lt;em&gt;after&lt;/em&gt; black hole formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few objections come to mind off-the-cuff: Because black holes have a large mass there is a minimum distance for planetary orbits below which they would disintegrate due to tidal forces. This may interfere with the usual evolution of planets (they clear their orbits and kick out interfering bodies, for example), because their &quot;orbital space volume&quot; becomes too large for them to keep clean, and the orbital period becomes long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is anything known or is there any speculation about such systems?&lt;/p&gt;&#xA;" OwnerUserId="11190" LastEditorUserId="11190" LastEditDate="2017-07-07T13:00:01.333" LastActivityDate="2017-07-07T15:40:13.457" Title="Is there any theoretical or empirical research about planetary systems with a black hole in their center?" Tags="&lt;black-hole&gt;&lt;planetary-formation&gt;&lt;planetary-systems&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21554" PostTypeId="1" AcceptedAnswerId="21567" CreationDate="2017-07-07T09:48:23.407" Score="0" ViewCount="421" Body="&lt;p&gt;&lt;a href=&quot;https://eclipse.gsfc.nasa.gov/SKYCAL/SKYCAL.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This nice page&lt;/a&gt; from NASA has the possibility to create an astronomical events calendar for a certain period. I tried to do that for the year 2017 for Central Europe. What comes out puzzles me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On July 8th the Moon is at it's south Declination. On July 12th the Moon goes through the descending node. On July 22nd it finds itself at the north Declination.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I had always though the south Declination means the southernmost point of the Moon on its orbit. Furthermore I had always thought the descending node is the point the Moon passes by going &quot;down&quot; in its motion. How can it reach a southernmost point and later go even further downwards?&lt;/li&gt;&#xA;&lt;li&gt;I had been inclined to think that the north and south Declinations would be about half way between nodes (give or take due to slight changes during the motion). How can the Moon be in one of these south or north points and only 4 days later pass a node? And take again 9 more days to reach a south/north Declination again?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It would be nice if the answer could, besides clarifying these concepts (ascending and descending node, south and north Declination), point me to a reading where they as well as other Moon orbital points and parameters are clearly defined and explained.&lt;/p&gt;&#xA;" OwnerUserId="17405" LastEditorUserId="17405" LastEditDate="2017-07-07T21:28:32.493" LastActivityDate="2017-07-08T04:21:05.140" Title="Nodes and declinations of Moon" Tags="&lt;the-moon&gt;&lt;fundamental-astronomy&gt;&lt;declination&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21556" PostTypeId="1" AcceptedAnswerId="21562" CreationDate="2017-07-07T12:54:51.023" Score="5" ViewCount="642" Body="&lt;p&gt;How was the first exoplanet discovered? What method was used to detect it?&lt;/p&gt;&#xA;" OwnerUserId="17380" LastEditorUserId="11264" LastEditDate="2017-07-07T22:29:35.127" LastActivityDate="2017-07-07T22:29:35.127" Title="How was the first exoplanet detected?" Tags="&lt;exoplanet&gt;&lt;history&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21557" PostTypeId="2" ParentId="21553" CreationDate="2017-07-07T13:02:24.440" Score="11" Body="&lt;p&gt;Since planets are known to exist around neutron stars (e.g. see &lt;a href=&quot;http://adsabs.harvard.edu/abs/1992Natur.355..145W&quot; rel=&quot;noreferrer&quot;&gt;Wolcznan &amp;amp; Frail (1992)&lt;/a&gt;; and see the list of &lt;a href=&quot;https://en.wikipedia.org/wiki/Pulsar_planet#List_of_pulsar_planets&quot; rel=&quot;noreferrer&quot;&gt;&quot;pulsar planets&quot;&lt;/a&gt; - technically, these were the first exoplanets ever discovered), then it is difficult to see any fundamental reason why there should not be examples orbiting black holes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, since it is probably feasible to form black holes via &lt;a href=&quot;https://arxiv.org/abs/1609.01283&quot; rel=&quot;noreferrer&quot;&gt;direct collapse&lt;/a&gt;, without an associated supernova then maybe the hazards for planets are not so severe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Against this, must be set the fact that black hole progenitors are likely to be more massive than neutron star progenitors and the role of winds and radiation on the formation of planets is still not well established.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets orbiting neutron stars are/were found from a timing analysis of pulsar signals. The same phenomenenon is not present for black holes and direct imaging or observing a transit is not remotely feasible at present.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-07-07T15:40:13.457" LastActivityDate="2017-07-07T15:40:13.457" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21558" PostTypeId="2" ParentId="21556" CreationDate="2017-07-07T13:16:10.017" Score="5" Body="&lt;p&gt;The first exoplanets to be discovered orbit about neutron stars, and pulsars in particular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pulsars that do not have planets and that are not a part of a multiple star system exhibit an extremely regular frequency. This frequency is modulated by the orbital frequency in binary systems that contain a pulsar. A pulsar with planets will exhibit a similar (but reduced) frequency modulation because the pulsar and planets orbit one another. The timing techniques used to determine whether pulsars are members of a binary systems enabled the discovery of the first exoplanets, in 1992.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-07-07T13:16:10.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21559" PostTypeId="1" CreationDate="2017-07-07T15:08:11.027" Score="1" ViewCount="94" Body="&lt;p&gt;This question is a followup of &lt;a href=&quot;https://astronomy.stackexchange.com/questions/21553/is-there-any-theoretical-or-empirical-research-about-planetary-systems-with-a-bl?noredirect=1#comment36771_21553&quot;&gt;Is there any theoretical or empirical research about planetary systems with a black hole in their center?&lt;/a&gt;; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it is possible to detect planets around stars with the change of light they cause, I would think it should also be possible to detect them when they orbit around an active Stellar Mass BH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I (can) understand from the first answer, there is no known case of planets orbiting around Stellar Mass BH. What could be preventing us to observe it?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there too few active Stellar Mass BH close enough for us to have a chance the system's plan is well aligned with us? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do we only see Stellar Mass BH when they have a stellar companion feeding them, and this discards to possibility for close-by planets?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Or is it because the light we can see from them is too fast-changing for a wobble or transition to be observable with today telescopes?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11264" LastEditorUserId="1891" LastEditDate="2020-12-08T11:58:42.900" LastActivityDate="2020-12-08T11:58:42.900" Title="Could we see planets around active Stellar Mass Black Hole?" Tags="&lt;black-hole&gt;&lt;exoplanet&gt;&lt;star-systems&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21561" PostTypeId="1" AcceptedAnswerId="21608" CreationDate="2017-07-07T18:00:50.707" Score="1" ViewCount="421" Body="&lt;p&gt;I already have a small 60mm Tasco refractor (with 300mm focal lenght) and a homemade 40mm refractor (with 1300mm focal lenght) and I want to buy a new telescope to have a better experience (I don't have an EQ mount yet)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I face the current choice: for the same price, I can have a 70mm refractor (900mm fl) with a great EQ mount, or a 114mm reflector (900mm fl) with a less great EQ mount...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I compare those 2 two? They are not the same type, and I feel like comparing oranges with apples. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the technical difference between them, and I know the details about collimation. I just don't know how to rate them and chose one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In short: Is it better a big reflector or a small refractor?&lt;/strong&gt;&#xA;(If I want to extend the information to other choices I might encounter)&lt;/p&gt;&#xA;" OwnerUserId="17411" LastEditorUserId="11822" LastEditDate="2017-07-08T08:18:03.543" LastActivityDate="2017-07-12T10:30:16.960" Title="How to compare a reflector telescope to a refractor one?" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21562" PostTypeId="2" ParentId="21556" CreationDate="2017-07-07T18:27:07.763" Score="10" Body="&lt;p&gt;This question is a bit tricky because it's uncertain as to what counts as &quot;discovering&quot; an exoplanet. Today, we count an exoplanet as having been discovered if it can be detected either by two separate techniques or three separate transits (if found via the transit method). Some of the very very early hunts for exoplanets did indeed find them, but many of their initial results were dubious or questionable (and sometimes wrong) and weren't really confirmed until much later. So do you count a planet as being discovered when it was first potentially detected or do you not count it until its been completely confirmed?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;First Exoplanet Detection&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The very first time someone could claim they discovered an exoplanet was in 1988. The astronomers Bruce Campbell, G. A. H. Walker, and Stephenson Yang announced that they had discovered a planet around the star &lt;a href=&quot;https://en.wikipedia.org/wiki/Gamma_Cephei#Planetary_system&quot; rel=&quot;noreferrer&quot;&gt;Gamma Cephei&lt;/a&gt; using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Methods_of_detecting_exoplanets#Radial_velocity&quot; rel=&quot;noreferrer&quot;&gt;radial velocity method&lt;/a&gt;. However, their results were not that great since, at the time, the detection was at the threshold of their technological capabilities and their results weren't entirely believable. It wasn't until years later that this planet was actually confirmed as existing. Whether you count this as the first real exoplanet discovery or not depends on what you define as a discovery, given that the actual confirmation didn't come until years after the first detection.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;First Confirmed Exoplanet Detection&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;In 1992, Aleksander Wolszczan and Dale Frail managed to find and confirm a planet around a the pulsar &lt;a href=&quot;https://en.wikipedia.org/wiki/PSR_B1257%2B12&quot; rel=&quot;noreferrer&quot;&gt;PSR B1257+12&lt;/a&gt;, again with the radial velocity method. At this time, their measurements were good enough to be believed and so considered &quot;confirmed&quot;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;First Confirmed Exoplanet Detection Around a Main-Sequence Star&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This case is particularly called out because it was the first time an exoplanet was detected and confirmed to exist around a normal main-sequence star. This occurred in 1995 around the star &lt;a href=&quot;https://en.wikipedia.org/wiki/51_Pegasi&quot; rel=&quot;noreferrer&quot;&gt;51 Pegasi&lt;/a&gt;. This discovery was made by Michel Mayor and Didier Queloz. Many people would consider this the first &quot;true&quot; exoplanet discovery and will cite this case if you ask when the first exoplanet discovery was made.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-07-07T18:27:07.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21563" PostTypeId="2" ParentId="21559" CreationDate="2017-07-07T19:59:11.810" Score="2" Body="&lt;p&gt;Stellar Black Holes (SBHs) are really really hard to find. If there's a free-floating one in space with no companion, our chances of finding it are almost nil. However, observations have more or less ruled out the possibility that space is freely floating with huge quantities SBHs. This was one of the proposed sources of Dark Matter, namely &lt;a href=&quot;https://en.wikipedia.org/wiki/Massive_compact_halo_object&quot; rel=&quot;nofollow noreferrer&quot;&gt;MACHOs&lt;/a&gt;. We don't believe MACHOs exist at this point because lensing observations towards the galactic core suggest they don't exist in any significant quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_black_holes#Stellar_black_holes_and_candidates&quot; rel=&quot;nofollow noreferrer&quot;&gt;list of known and candidate SBHs&lt;/a&gt; contains less than 30 entries. The reason SBHs are hard to find is that, as stated above, they don't seem to be too common, and also that they can only really be observed indirectly through their interactions with a binary companion. Often, one observes the X-ray radiation coming from infalling matter around the black hole, taken from the binary companion, rather than observing the black hole itself.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there too few active Stellar Mass BH close enough for us to have a chance the system's plan is well aligned with us? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer to this is probably yes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do we only see Stellar Mass BH when they have a stellar companion feeding them, and this discards to possibility for close-by planets?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again yes, we can only &quot;see&quot; the SBHs when they have a companion feeding them. This doesn't count the three cases where we've found SBHs through gravitational wave observations though, but all of those were far outside our galaxy. Whether a planet could exist in such a violent binary system is up for debate. There's only one known case that I know of where a planet is orbiting a similar system and that is &lt;a href=&quot;https://en.wikipedia.org/wiki/PSR_B1620-26&quot; rel=&quot;nofollow noreferrer&quot;&gt;PSR B1620-26&lt;/a&gt;. This is a white dwarf, pulsar pair with a planet around it. This would be completely analagous to an X-ray binary system with a black hole if only the pulsar were a bit more massive. So certainly it seems possible for a planet to be around such a system, but probably not very likely.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Or is it because the light we can see from them is too fast-changing for a wobble or transition to be observable with today telescopes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This certainly plays a part too. Even if you find a binary system with an SBH that is active and producing X-rays, you're still going to have a hard time teasing out an exoplanet signal, especially if that exoplanet is small (i.e., smaller than Jupiter). The transit method would be difficult due to the natural variability and hence noise in the X-ray emissions and the radial velocity method may not be sensitive enough to small planets. Such a measurement would be really hard.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As it turns out though, astronomers are thinking about these things. There was recently a paper published almost on this very topic. Check out &lt;a href=&quot;https://arxiv.org/pdf/1703.05762.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Irma &amp;amp; Stefano (2017)&lt;/a&gt; where they discuss trying to find planets around X-ray binaries that may potentially house a SBH.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-07-07T19:59:11.810" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21564" PostTypeId="2" ParentId="21552" CreationDate="2017-07-07T20:05:33.780" Score="1" Body="&lt;p&gt;Yes, but probably not by a huge amount. There are two obvious ways this could happen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Direct change in atmospheric composition. If our microbe does something like eat methane and excrete longer-chain hydrocarbons, that's going to eventually change the atmosphere enough to have some effect. (If the hydrocarbons weren't already forming, they probably aren't stable under local conditions, so there will be some equilibrium reached, the way the Earth's atmosphere isn't 100% oxygen despite hundreds of millions of years of photosynthesis.)&#xA;Change in solar heat balance. The change in atmosphere will probably also change the amount of greenhouse effect (one way or the other), and second-order effects or even just enough microbial biomass can change the albedo (again, either way). On Earth, these effects can combine to change the temperature by tens of Kelvins, which would be a few percent different in atmospheric volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In neither case is the added life going to change the total mass of the atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="14088" LastActivityDate="2017-07-07T20:05:33.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21566" PostTypeId="1" CreationDate="2017-07-08T04:10:34.377" Score="1" ViewCount="485" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Younger_Dryas_impact_hypothesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Younger Dryas Impact Theory&lt;/a&gt; describes an asteroid impact in Northern America, possibly the &lt;a href=&quot;http://ottawa-rasc.ca/wiki/images/thumb/5/57/Odale_articles_hudsonbayarc_Arc.jpg/500px-Odale_articles_hudsonbayarc_Arc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;eastern part of Hudson Bay&lt;/a&gt;, which contributed to the end of the last ice age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a comet strikes a glacier, would it make a print on the land below? If specifically the comet that made this &lt;a href=&quot;http://ottawa-rasc.ca/wiki/images/thumb/5/57/Odale_articles_hudsonbayarc_Arc.jpg/500px-Odale_articles_hudsonbayarc_Arc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;impact crater in Hudson Bay,&lt;/a&gt; and a glacier that was maybe comparable to Antarctica so 2 km thick.&lt;/p&gt;&#xA;" OwnerUserId="17416" LastEditorUserId="2365" LastEditDate="2019-11-20T23:02:59.550" LastActivityDate="2019-11-20T23:02:59.550" Title="If a comet strikes a glacier, does it make a print on the land below?" Tags="&lt;earth&gt;&lt;asteroids&gt;&lt;impact&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21567" PostTypeId="2" ParentId="21554" CreationDate="2017-07-08T04:21:05.140" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lSNrg.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lSNrg.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image above (click for full size) shows the moon's path for much of July. The celestial equator is in purple and the ecliptic is in green. Note that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;On July 8th (meaning between July 8th at 0000 UTC and July 9th at 0000 UTC), the moon does indeed reach it's southernmost declination, but is still north of the ecliptic.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On July 12th, the moon, whose declination is increasing slower than that of the ecliptic, crosses the ecliptic from north to south, it's descending node.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On July 22nd, the moon is at its northernmost declination, but still south of the ecliptic.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On July 25th, the moon, moving south slower than the ecliptic, crosses the ecliptic from south to north, it's ascending node.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As the page you reference notes, the moon's northern and southernmost declination this month are +-19.4 degrees, about 4 degrees less than the tilt of the ecliptic. Thus, the moon changes declination slower than the ecliptic, at least for this month.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2017-07-08T04:21:05.140" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21568" PostTypeId="2" ParentId="21538" CreationDate="2017-07-08T13:51:06.193" Score="2" Body="&lt;p&gt;The frequency is &lt;a href=&quot;https://en.wikipedia.org/wiki/Redshift#Measurement.2C_characterization.2C_and_interpretation&quot; rel=&quot;nofollow noreferrer&quot;&gt;related&lt;/a&gt; to the redshift by $$\frac{\nu_{\rm obs}}{\nu_{\rm emit}} = \frac{1}{1+z}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another useful &lt;a href=&quot;https://casper.berkeley.edu/astrobaki/index.php/Specific_Intensity#Proof_that_Specific_Intensity_is_conserved_along_a_ray&quot; rel=&quot;nofollow noreferrer&quot;&gt;relation&lt;/a&gt; is the fact that, for any redshift,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{I_{\nu}}{\nu^3} = {\rm constant}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $I_{\nu}$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Specific_radiative_intensity&quot; rel=&quot;nofollow noreferrer&quot;&gt;specific radiative intensity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific count rate (counts per area per time per energy interval) is proportional to $I_{\nu}/\nu$. (Do you see why?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That should give you enough information to derive how the measured count rate changes with $z.$ Of course, to know the actual value, not just the function of $z$, you'd need to know the actual count rate measured, not just the normalized value.&lt;/p&gt;&#xA;" OwnerUserId="17410" LastActivityDate="2017-07-08T13:51:06.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21570" PostTypeId="2" ParentId="21566" CreationDate="2017-07-08T15:48:47.113" Score="3" Body="&lt;p&gt;While it's a clever suggestion, the answer is almost certainly that they are not related because the &lt;a href=&quot;https://en.wikipedia.org/wiki/Nastapoka_arc&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nastapoka Arc&lt;/a&gt; &quot;crater&quot; is too big.   It's 450 km in diameter.  By comparison, the &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_impact_craters_on_Earth#Largest_craters_.2810_Ma_or_more.29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chicxulub crater&lt;/a&gt; is 180 km.   Even if an impact through mile of ice could, somehow, spread the impact out wider, the proposed Younger Dryas Impact object was almost certainly too small to leave a 340 km crater.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also the problem that glacial ice carves the landscape and a neat circular crater is unlikely, even if the size had been a closer fit, an impact crater that was run over by a glacier should look disturbed by the glacier, not a neat circle visible on the surface.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What geological evidence might remain for a large impact on a glacier is an interesting question given that the ice would significantly buffer any impact and continue to reshape the landscape below it as the glacier moves and/or melts and possibly floods the region.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mile high ice glaciers are relatively rare events, both localized &lt;a href=&quot;https://en.wikipedia.org/wiki/Timeline_of_glaciation&quot; rel=&quot;nofollow noreferrer&quot;&gt;and periodic&lt;/a&gt;, and large meteor impacts are also rare events.   A combination of the two, large enough to leave clear evidence through the ice, would at best, be a hugely rare event and there's a good possibility that the right combination of impact and ice hasn't happened at all over the last 2.5 million years (northern Hemisphere) or on Antarctica (3% of the surface area of the earth) - last 20-30 million years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the safer bet is that there's no impact crater from the proposed Younger Dryas impact specifically because of the thickness of the glacier.   It remains a debated hypothesis, so there's some uncertainty, but Ice should significanly buffer all but the largest impacts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Younger Dryas, there should be other physical evidence such as debris from the impacting meteor or comet, but debris cast over a glacier that first re-freezes over and later melts can be spread very far and be very hard to identify.   Much of it could have washed into the ocean.   A somewhat related example, it took 105 years before &lt;a href=&quot;https://www.technologyreview.com/s/514511/first-tunguska-meteorite-fragments-discovered/&quot; rel=&quot;nofollow noreferrer&quot;&gt;debris from the Tunguska&lt;/a&gt; event was identified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(and this question should probably be moved to Earth Science).  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-07-08T15:48:47.113" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="21571" PostTypeId="2" ParentId="1216" CreationDate="2017-07-08T16:34:52.897" Score="3" Body="&lt;p&gt;Little pieces of Halley's comet hit the Earth every spring. It's called the Eta Aquarid meteor shower, but I think you are thinking of something bigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cosmicttusk.com/halleys-comet-abbott-at-agu-on-the-530s-event/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dallas Abbot's paper&lt;/a&gt; notes that at the 533-540 point on the &lt;a href=&quot;http://www.gisp2.sr.unh.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;GISP2&lt;/a&gt; ice core there are Sn-rich particles, Ni-rich particles and cosmic spherules. These she identifies as evidence of a minor cometary impact. The resolution of the GISP2 core is sufficient to place this impact in the late spring, consistent with the Eta Aquarid meteor shower, which is known to be produced by Halleys comet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dallas notes that an increase in cometary dust in the atmosphere could partly explain the &lt;a href=&quot;https://en.wikipedia.org/wiki/Extreme_weather_events_of_535%E2%80%93536&quot; rel=&quot;nofollow noreferrer&quot;&gt;extreme weather events of 535-536&lt;/a&gt;, in combination with a minor volcanic eruption. Although there are other, competing hypotheses, and this is not considered to be settled science.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-07-08T16:34:52.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21572" PostTypeId="1" CreationDate="2017-07-09T08:34:04.940" Score="6" ViewCount="2832" Body="&lt;p&gt;There are lots of galaxies, for example, the Milky Way and so on. These galaxies consist of lots of stars. I want to know how galaxies are classified. Is it only by their shape that is caused by gravity? Or are there some other criteria?&lt;/p&gt;&#xA;" OwnerUserId="17424" LastEditorUserId="1891" LastEditDate="2019-12-20T15:44:45.233" LastActivityDate="2019-12-20T15:44:45.233" Title="What are the criteria for the classification of galaxies?" Tags="&lt;galaxy&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21574" PostTypeId="1" CreationDate="2017-07-09T17:46:41.523" Score="3" ViewCount="364" Body="&lt;p&gt;For many years now I've been using Voyager sky simulator by Carinasoft. The projection selections are great, but Equirectangular doesn't seem to one one of them. Is there a program out there that does this? I cooked up an Excel sheet to do the coordinate conversions, and I can import these into Autocad if I have to, but it seems like there should be an off-the-shelf package to do this.      &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For my current project I need to be able to show a 360 degree overview of changes due to precession from 3000-2000-1000 BC at a glance.  It may be my familiarity with the typical simple projection published for surveyors in a standard Ephemeris before the internet and GPS all but eliminated the demand for such publications.   &lt;/p&gt;&#xA;" OwnerUserId="17427" LastActivityDate="2017-07-10T04:14:18.553" Title="360 degree equirectangular projection of all naked eye stars with precessional motion for simulating" Tags="&lt;star&gt;&lt;coordinate&gt;&lt;constellations&gt;&lt;ecliptic&gt;&lt;sky-survey&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21575" PostTypeId="2" ParentId="21572" CreationDate="2017-07-09T19:55:44.963" Score="3" Body="&lt;p&gt;A galaxy is just a very large collection of stars (and interstellar matter such as dark matter, gas and dust) held together by gravity. Galaxies are classified mainly as elliptical, spiral, and irregular. There is not a specific criterion other than the fact that the stars all all bound by their mutual gravitational attraction. It is speculated that most galaxies have a black hole at its center. This is not necessary however for a collection of stars to be considered as a galaxy.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="16926" LastEditDate="2017-07-11T14:23:49.927" LastActivityDate="2017-07-11T14:23:49.927" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="21576" PostTypeId="2" ParentId="21572" CreationDate="2017-07-09T20:34:24.953" Score="8" Body="&lt;p&gt;There are three main classes of galaxies: Irregulars, Ellipticals, and Spirals. Irregular galaxies, as their name suggests, do not fit into the &quot;normal&quot; classification scheme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how do we distinguish between elliptical and spiral galaxies?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Brighness profile&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radial brightness profile of an elliptical galaxy follows a &lt;a href=&quot;https://en.wikipedia.org/wiki/De_Vaucouleurs%27_law&quot; rel=&quot;noreferrer&quot;&gt;deVaucouleur law&lt;/a&gt; ($r^{1/4}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spiral galaxies have an exponential radial brightness profile, although their central regions (&quot;bulge&quot;) also follows a deVaucouleur law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Star formation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars are formed in the spiral arms of spiral galaxies (and can be formed in irregulars), while elliptical galaxies tend to only have old, and consequently low mass, stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Components&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as we can tell, all galaxies consist of a dark matter halo and stars. In addition, spiral galaxies also have clouds of dust and gas. If conditions are right, these can form new stars. (Some ellpticals have a very thin, very hot gas component as well, but there is a lot less of it than in a spiral galaxy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Kinematics&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spiral galaxies are rotationally supported, while elliptical galaxies are mainly pressure-supported (i.e. they act like an ideal gas, with stars as gas molecules). There are some rotational features present in ellipticals, but they tend to be minor compared to the overall random motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A graphical overview of the various galaxy types is usually shown in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble_sequence&quot; rel=&quot;noreferrer&quot;&gt;Hubble tuning fork&lt;/a&gt; diagram. Note that this does not indicate an evolutionary progression from one type to the next.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastActivityDate="2017-07-09T20:34:24.953" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21577" PostTypeId="2" ParentId="21574" CreationDate="2017-07-10T04:14:18.553" Score="2" Body="&lt;p&gt;The &quot;cylinder&quot; projection of Stellarium does this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/khM4e.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/khM4e.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2017-07-10T04:14:18.553" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21581" PostTypeId="1" CreationDate="2017-07-10T11:04:10.597" Score="1" ViewCount="65" Body="&lt;p&gt;What's the name of the phenomenon when you look at a bright object (e.g. the moon) and see some perpendicular &quot;beams&quot; going top-down and right-left (bare eye observation, also slightly visible on iPhone pic)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ibb.co/j2b4AF&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ibb.co/j2b4AF&lt;/a&gt;&#xA;see the link for a picture &lt;/p&gt;&#xA;" OwnerUserId="17434" LastEditorUserId="-1" LastEditDate="2017-07-10T20:18:52.493" LastActivityDate="2017-07-11T18:02:46.633" Title="Light distorsion/ilusion when looking at bright objects" Tags="&lt;amateur-observing&gt;&lt;light&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" ClosedDate="2017-07-23T17:40:04.067" ContentLicense="CC BY-SA 3.0" />
  <row Id="21583" PostTypeId="1" CreationDate="2017-07-10T13:45:38.217" Score="4" ViewCount="627" Body="&lt;p&gt;When a moon is in orbit around it's planet but is so close that it will become tidally locked does it experience a large 'wobble' in between the phase when it still has its own rotation and the final phase when it is always showing the same face to the planet?  I can only imagine that it would 'wobble' very widely, back and forth like a pendulum until finally settling into a locked position.  &lt;/p&gt;&#xA;" OwnerUserId="17272" LastActivityDate="2017-12-09T16:12:01.040" Title="Tidal locking of a moon" Tags="&lt;tidal-locking&gt;" AnswerCount="2" CommentCount="7" ClosedDate="2017-07-11T16:55:07.463" ContentLicense="CC BY-SA 3.0" />
  <row Id="21584" PostTypeId="1" AcceptedAnswerId="21602" CreationDate="2017-07-10T16:31:29.087" Score="5" ViewCount="189" Body="&lt;p&gt;A Comet is orbiting around the sun with period of $T= 5.5$ years. The comet has radius $R$. After 5.5 years (A period) we understand that its radius has decreased about $20cm$. How can we calculate the R. We know that comet is approximately a sphere and it is made of Ice. We have the $G = 6.67 * 10^{-11}$ SI_Unit, $M_{sun} = 1.99 * 10^{30} Kg$, $\sigma = 5.67 *10^{-8}$ SI_Unit (Stefan-Boltzmann constant) , $L_{sun}=3.85 * 10^{26} w$, $l_f = 333 KJ/Kg$ , $l_v = 2260 KJ/Kg$. It would be fine that if the solution doesn't need the numerical value of density of ice but if it can't be solved without density of ice, you are allowed to use it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can calculate the distance to sun ($a$) with formula:&#xA;$T^2 = 4 *\pi^2 a^3 /(GM_{Sun})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a= 4.66* 10^{11} m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also can calculate $F= L_{sun} / (4 * \pi * a^2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also approximated the power that the comet gives is: $P = F * Area$ (think ice don't reflect any light)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't know how to calculate the R and formulate how it decreases every moment (because it affects the area).&lt;/p&gt;&#xA;" OwnerUserId="16441" LastActivityDate="2017-07-11T20:08:21.873" Title="A question about calculating the radius of a comet" Tags="&lt;astrophysics&gt;&lt;comets&gt;&lt;radiation&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21585" PostTypeId="2" ParentId="21583" CreationDate="2017-07-10T19:05:47.433" Score="4" Body="&lt;p&gt;There's no back and forth motion on the way towards tidal locking.   There's kind of two ways to look this kind of &quot;settling&quot;, there's overshooting and correction, basically a back and forth, and there's a very gradual force that grows smaller, basically moving towards zero but not overshooting.    I realize that's just re-writing your question, but it helps to understand the 2nd option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll use the Earth as an example, because the Earth is in the gradual process of becoming tidally locked to the Moon.  It's the same process, but easier to visualize. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the Earth rotates faster than the Moon orbits, the tidal bulge moves ahead of the moon.   This is called the tidal bulge offset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XgtZe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XgtZe.jpg&quot; alt=&quot;http://pages.uoregon.edu/jimbrau/BrauImNew/Chap07/FG07_23.jpg&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pages.uoregon.edu/jimbrau/astr121-2005/Notes/Chapter7.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the size of the bulge and the angle of the offset (and the mass and distance of the Moon) that causes the familiar gravitational effects, the Earth's rotation slowing down and the Moon moving away from the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DbFCu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DbFCu.jpg&quot; alt=&quot;https://ase.tufts.edu/cosmos/pictures/Explore_figs_8/Chapter5/Fig5_28copy.jpg&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ase.tufts.edu/cosmos/print_images.asp?id=5&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current angle of offset for the Earth's tidal bulge to the Moon is about 3 degrees, often exaggerated in diagrams so it's easier to see.  (&lt;em&gt;I should find a source for that - I've not been able to find a good one)&lt;/em&gt;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the Moon's point of view (and if we assume a circular orbit), the tidal offset stays in the same place.  It's a factor of the relative rotation speed.   As Earth slows down, the angle of offset will move towards zero.    At tidal locking, the angle of offset becomes zero and there's no force to continue in the other direction.    In a simple mathematical model, tidal locking wouldn't ever happen completely, it would just gradually get closer and closer, like an object shot from the Earth at exactly escape velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, because objects aren't perfectly symmetrical and because orbits aren't perfectly circular, the &quot;take forever to get to zero&quot;, doesn't happen.  But it's still a gradual approach to zero and no overshooting and returning because the force on the rotation approaches zero as the angle of offset approaches zero and the force needed to change a planet's rotation is enormous and there's essentially zero force to make that happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With more eccentric orbits, like Mercury and it's 3:2 rotation to orbit resonance, that might be different.   Not sure how eccentricity might play into this.   That's more complicated.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-07-10T19:05:47.433" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21586" PostTypeId="1" CreationDate="2017-06-29T23:48:25.723" Score="3" ViewCount="734" Body="&lt;p&gt;This might be an astronomy.stackexchange.com question, but I think it is more in line with a camera obscura.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the last eclipse I didn't plan ahead and grabbed some lousy toy &quot;spy&quot; telescope we had (about 2 inches long), set it up on a tripod and pointed it at a white foam board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to plan ahead more this time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can I use to project a decent image of the sun onto a foam board?  What kind of lens do I need?  Would a magnifying lens work?  How about a close-up diopter filter?  How about a 50mm lens?  (I'd like to use something I already have, or is cheap -- and I don't want to use a pin hole.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first thought was that a magnifying glass only focuses the sun down to a pin point (and would burn the board), but I'm recalling now that past that focal point you can project an image.&lt;/p&gt;&#xA;" OwnerUserId="17439" OwnerDisplayName="rrauenza" LastActivityDate="2017-08-25T06:47:13.603" Title="Projecting an image of the sun / eclipse" Tags="&lt;optics&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21587" PostTypeId="2" ParentId="21586" CreationDate="2017-06-30T11:16:24.183" Score="1" Body="&lt;p&gt;I've done this using two lenses.  In my case they were a 300 mm and a 50 mm.  I built a jig to hold the lenses in-line with each other, to allow for small adjustments in the placement, and for a tripod mount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put the long lens, in my case the 300 mm, in front normally.  That will form a focused image behind it.  Put the short lens backwards behind that.  This lens is used in reverse.  It takes the small image created by the long lens and projects it out what is normally its front.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio of the front lens focal length to the back lens focal length is how much angular &quot;gain&quot; you get from this setup.  In my case, the cone of project of the sun was 6 times wider coming out of the 50 mm than going into the 300 mm.  This means you don't have to get as far away from the overall setup to get a decent size picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not where this jig is now.  I'll try to remember to take a picture of it and post it here within the next day or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing you'll find is that the earth rotates annoyingly fast when you are looking at a magnified picture of the sun.  Use a telescope mount if you can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another issue is blocking out as much ambient light as you can.  The projected image of the sun is reasonably bright, but so is daylight.  Some sort of baffle over the top and the front except for a hole for the front lens helps a lot.  People can then view the projected image from the sides.&lt;/p&gt;&#xA;" OwnerDisplayName="Olin Lathrop" LastActivityDate="2017-06-30T11:16:24.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21588" PostTypeId="2" ParentId="21586" CreationDate="2017-07-01T16:03:23.527" Score="1" Body="&lt;p&gt;Safety: make absolutely sure nobody looks onto the light path. The concentrated light/heat/UV can cause eye damage. One way to do this is to set things up in a box, which (a) shades the viewing screen, and (b) lets you make the viewing opening too small for kids (or thoughtless adults) get their heads in - or make sure that your screen is close enough that there's no room to squeeze a head in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of the prime focus image is determined by focal length. To get a decent size prime focus image needs a LONG focal length. That's why a telescope style projection setup as in olin's answer can give you a bigger image from shorter length lenses. (and is less likely to just ignite your viewing screen). However, remember that you're using focused raw sunlight, so there's a lot of heat involved. That means that for a real telescope, it's safest to use only an all-metal style refractor, where there's nothing to melt. You probably also want to avoid using expensive eyepieces - cheaper ones usually have fewer elements, and it's less of a financial hit if the heat damages something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're using a scope with a 90 degree star diagonal, that means that you're projecting things side on, which makes it easier to shade the screen.&#xA;Or if you're going straight through, you can add a card across the front with a hole for the lens to provide shade. &lt;/p&gt;&#xA;" OwnerUserId="17016" OwnerDisplayName="JerryTheC" LastActivityDate="2017-07-01T16:03:23.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21590" PostTypeId="1" AcceptedAnswerId="21592" CreationDate="2017-07-11T05:45:34.973" Score="5" ViewCount="320" Body="&lt;p&gt;Am not asking about planetary/meteorite surfaces and am restricting myself to the interstellar medium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt;:&#xA;by interstellar medium I mean, the following: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Interstellar medium (ISM) is the matter that exists in the space between the star systems in a galaxy. This matter includes gas in ionic, atomic, and molecular form, as well as dust and cosmic rays. It fills interstellar space and blends smoothly into the surrounding intergalactic space.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(copied verbatim from a google search).&lt;/p&gt;&#xA;" OwnerUserId="17445" LastEditorUserId="11264" LastEditDate="2017-07-12T16:55:02.787" LastActivityDate="2017-07-12T16:55:02.787" Title="Do liquids exist in the interstellar medium?" Tags="&lt;interstellar-medium&gt;&lt;interstellar&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21592" PostTypeId="2" ParentId="21590" CreationDate="2017-07-11T08:07:39.237" Score="2" Body="&lt;p&gt;&lt;strong&gt;Basically, No.&lt;/strong&gt; for liquid as we commonly know them on earth. If you trow a water bucket in space, it gets quickly vaporized. From &lt;a href=&quot;http://scienceblogs.com/startswithabang/2009/06/29/water-in-space-what-happens/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this interesting ref&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;having a pressure vacuum will cause the water to boil almost instantly&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is however an exception I've been thinking of. Since glass has an amorphous structure, it can also be regarded as a liquid... And if some advanced form of life or intelligence managed to send some into the interstellar medium, then &lt;strong&gt;yes there is&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also possible that some stellar explosion send away magma of some form with an amorphous structure similar to that of glass (&lt;a href=&quot;https://en.wikipedia.org/wiki/Obsidian&quot; rel=&quot;nofollow noreferrer&quot;&gt;e.g. obsidian&lt;/a&gt;) &lt;/p&gt;&#xA;" OwnerUserId="11264" LastEditorUserId="11264" LastEditDate="2017-07-11T14:58:03.293" LastActivityDate="2017-07-11T14:58:03.293" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21593" PostTypeId="2" ParentId="21583" CreationDate="2017-07-11T08:38:08.257" Score="-1" Body="&lt;p&gt;For an animated lesson, see Isaac Author's amazing space engineering anthology, including &lt;a href=&quot;https://youtu.be/lmc5XqChJpY&quot; rel=&quot;nofollow noreferrer&quot;&gt;TIDAL LOCKING EXPLAINED&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="825" LastEditorUserId="825" LastEditDate="2017-12-09T16:12:01.040" LastActivityDate="2017-12-09T16:12:01.040" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21595" PostTypeId="2" ParentId="21586" CreationDate="2017-07-11T12:50:21.693" Score="1" Body="&lt;p&gt;Like the two above, I'm building a solar projector.  There's actually some pretty well-laid out plans and instructions (and lens kits) available for this.  That said, it's basically a telescope with the eyepiece slightly refocused, and a projection screen some distance away.&lt;br&gt;&#xA;You can use a pre-built telescope, spotting scope, or binocular to build this.  However, I would NOT recommend using one you like, as explained in the cautions at the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out the following link for some info, and the second link is to a lens kit that costs $11 after shipping.  It's the one used in the instructions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://today.cofc.edu/2017/06/13/solar-eclipse-viewer-diy/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://today.cofc.edu/2017/06/13/solar-eclipse-viewer-diy/&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.surplusshed.com/pages/item/l14766.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.surplusshed.com/pages/item/l14766.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to pick out your own lenses, here's an equation to help you figure out how big the projected image will be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H = 2D\tan (\frac{S}{2} \times \frac{L_1}{L_2} \times \frac{\pi}{180})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H$ is the image diameter&#xA;$S$ is the angular size of the sun (0.53 degrees)&#xA;$L_1$ is the focal length of your first lens (called an objective)&#xA;$L_2$ is the focal length of your second lens (called an eye lens or eyepiece)&#xA;$D$ is the distance between your eyepiece lens and your screen (in whatever units you want H to be)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 1:  your calculator should be set to radians (not degrees).  Google uses radians by default.&#xA;Note 2: if you have a cheap telescope, binocular, or spotting scope where the eyepiece is built in, you can substitute $\frac{L_1}{L_2}$ for the magnification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, if you have a 20x setup (say 500mm objective and 25mm eye), and the distance to the screen is 1 foot, your image will be just over 2&quot; in diameter.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cautions:&#xA;NEVER look directly at the sun or worse, directly through magnifying optics at the sun.&#xA;Due to risk of permanent damage to your telescope, never point a quality telescope at the sun.  Optics can melt, and plastic parts WILL melt given time.&lt;/p&gt;&#xA;" OwnerUserId="10820" LastEditorUserId="-1" LastEditDate="2017-08-25T06:47:13.603" LastActivityDate="2017-08-25T06:47:13.603" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21596" PostTypeId="2" ParentId="21572" CreationDate="2017-07-11T14:04:29.093" Score="7" Body="&lt;p&gt;Alex answers nicely how galaxies can be classified according to their morphology. In astronomy, galaxies are detected using a variety of detection techniques. Especially in the high-redshift (i.e. distant) Universe, galaxies are not easily detected and are only visible using specific methods (although some galaxies show up with multiple techniques). These methods each probe different things, and galaxies belonging to one class will thus have other parameters than galaxies belong to other classes, although there will always be ome overlap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physical properties defining whether a galaxy may be selected by a given technique is hence not only morphology, but also stellar mass, star formation rate, dust mass, size, clumpiness, kinematics, luminosity, the presence of active galactic nuclei, and many others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accordingly, we ofted classify galaxies from the method and the selection criterion used (and preferably use a three-letter acronym to describe them):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of these are:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Lyman-break galaxies (LBGs)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The technique used to find these galaxies revolutionized the field in the mid-90'es (&lt;a href=&quot;https://arxiv.org/abs/astro-ph/9602024&quot; rel=&quot;noreferrer&quot;&gt;Steidel et al. 1996&lt;/a&gt;). The reason is that a large field of view can be investigated, allowing to detect many galaxies at the same time. The idea is to observe the same field in several different wavelength bands. If large amounts of neutral hydrogen is present, wavelengths shortward of the &quot;Lyman-break&quot; at 912 Å, or 91.2 nm, needed to ionize hydrogen are absorbed, effectively making the galaxy invisible in all bands shortward of this. And because the light is redshifted as it travel through the Universe toward us, galaxies at different redshifts will drop out of different band (the method is also called the &quot;drop-out technique&quot;). An example is seen here, where the galaxy spectrum (&lt;em&gt;black line&lt;/em&gt;) drops steeply so that flux enters the red $R$ band and the green $G$ band, but not the ultraviolet $U$ band:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vveYs.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vveYs.png&quot; alt=&quot;LBG&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Credit: &lt;a href=&quot;http://www.dark-cosmology.dk/~jfynbo/LBG.html&quot; rel=&quot;noreferrer&quot;&gt;J. Fynbo&lt;/a&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above figure, the break has been redshifted to somewhere between the $G$ band and the $U$ band, constraining its redshift to roughly $z = 3$-$4$. To further constrain the redshift, spectroscopic follow-up is needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since large amounts of neutral hydrogen is needed, this technique will tend to select massive, and hence rather evolved galaxies.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Lyman alpha emitters (LAEs)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;When an electron decays from the first excited state to the ground state, a s-called Lyman alpha (Ly$\alpha$) photon is emitted. This can happen either when a hydrogen atom is perturbed (in a collision) and excited, or if hydrogen is ionised and recombines. Both mechanisms are at play when galaxies are born, where gas accretes onto a central potential (causing collisions), and young, massive star ionize the surrounding gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxies found from their ability to emit Ly$\alpha$ are called LAEs. They can be either found either spectroscopically — where are strong emission line will be seen at $\lambda = 1216$ Å — or photometrically by observing the field in a broadband and a narrowband centered at $\lambda = 1216$ Å and looking for excess flux in the narrowband.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this techniques tends to probe young galaxies, they will often be relatively small, but with a high star formation rate. And because dust absorbs Ly$\alpha$ more easily than other wavelengths, LAEs tend to be rather dust-free.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Sub-millimeter galaxies (SMGs)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If a galaxy, on the other hand, is very dusty, it can be difficult to detect in the optical and, especially, in the ultraviolet. The reason is that dust has a strong preference to absorb light with shorter wavelengths. But the energy absorbed must go somewhere, and is thus emitted again, although at longer wavelengths, i.e. in the infrared and in the sub-mm region. Galaxies found this way are referred to as SMGs. Because it takes some time for the dust mass to build up, this technique tends to probe evolved, massive galaxies.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Damped Lyman alpha absorbers (DLAs)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The three techniques described above all have in common that they detect galaxies from their &lt;em&gt;emission&lt;/em&gt;. A complimentary technique is looking for absorption features in the spectrum of a bright background source, e.g. a quasar. Because the light is redshifted on its way, especially hydrogen but also metals such as iron and magnesium produce absorption lines at various places in the spectrum corresponding to the wavelength that the quasar light has been redshifted to at a given point in space. Diffuse hydrogen filaments make narrow absorption line known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lyman-alpha_forest&quot; rel=&quot;noreferrer&quot;&gt;Lyman $\alpha$ forest&lt;/a&gt;, and when a large pocket of gas is present — which indicates the presence of a galaxy — a broad (&quot;damped&quot;) absorption line is produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example is seen in this spectrum of the quasar Q2348-011 lying at $z=3.0$. An intervening galaxy at $z=2.6$ causes the broad absorption at $\lambda\sim4400$ Å.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HqXLe.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HqXLe.png&quot; alt=&quot;DLA&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Credit: &lt;a href=&quot;https://arxiv.org/abs/1012.3175&quot; rel=&quot;noreferrer&quot;&gt;Laursen (2010)&lt;/a&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only in few cases is the galaxy responsible for the absorption found. This is partly because the light from the quasar outshines everything in its (projected) vicinity, but possibly also because the huge hydrogen cloud is a galaxy in the making, that perhaps hasn't form many stars yet. And since the probabiliy for sightlines toward quasars of hitting a &lt;em&gt;small&lt;/em&gt; galaxy is larger than hitting a &lt;em&gt;large&lt;/em&gt; galaxy (due to the total cross section of small galaxies being larger), galacitc counterparts of DLAs should tend to be small. Thus, DLAs are thought to probe young galaxies in the process of forming.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Other types&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Other types includes &lt;em&gt;distant red galaxies&lt;/em&gt; (DRGs), &lt;em&gt;(ultra)luminous infrared galaxies&lt;/em&gt; (LIRGs and ULIRGs), and &lt;em&gt;gamma-ray burst host galaxies&lt;/em&gt; (GHGs). One of the big challenges of astronomy is to ascertain how the galaxies of the different group fit in some big picture. Is there for instance an evolutionary sequence from DLA→LAE→LBG→SMG→DRG (see e.g. &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0512384v1&quot; rel=&quot;noreferrer&quot;&gt;Gawiser 2005&lt;/a&gt;)?&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2017-07-11T14:04:29.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21598" PostTypeId="1" CreationDate="2017-07-11T15:12:41.773" Score="0" ViewCount="182" Body="&lt;p&gt;In spite of being a simple words game, that would make me feel less uncomfortable, as there could be a mysterious field still less mysterious than matter that passes through itself etc. Moreover, couldn't be that there exist some high order terms that play a role only when mass reach big values, i. e. those comparable to a galactic mass? How/why this cannot be the case in general relativity? &lt;/p&gt;&#xA;" OwnerUserId="17437" LastEditorUserId="2365" LastEditDate="2017-07-11T16:58:26.930" LastActivityDate="2017-07-11T21:18:28.730" Title="Wouldn't be &quot;dark matter&quot; better called &quot;massless gravity&quot;?" Tags="&lt;gravity&gt;&lt;cosmology&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" ClosedDate="2017-07-11T23:06:33.417" ContentLicense="CC BY-SA 3.0" />
  <row Id="21600" PostTypeId="1" CreationDate="2017-07-11T16:56:55.080" Score="1" ViewCount="289" Body="&lt;p&gt;We know that circumpolar stars are stars that don't rise or set. If we are in a location with $\phi = 35^o N$. What is the fraction of circumpolar stars to all stars visible in this location?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the declination of this stars is $\delta &amp;gt; 90 - 35 = 55$ But I don't know what is the degree of all stars visible in this location and is the fraction as simple as dividing this two angles or not?&lt;/p&gt;&#xA;" OwnerUserId="16441" LastActivityDate="2017-07-13T02:12:01.047" Title="Fraction of circumpolar stars to all stars visible in a location" Tags="&lt;star&gt;&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21601" PostTypeId="2" ParentId="21581" CreationDate="2017-07-11T18:02:46.633" Score="2" Body="&lt;p&gt;It's a bunch of different phenomena, somewhat related, but still distinct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Naked eye&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you mention naked eye observations - seeing a small amount of short spikes around bright objects is normal. The causes can be multiple: diffraction of light in/around your eyelashes, dirt/secretions on the cornea, just regular aberrations in the eye's optical system, or in extreme cases it might actually signal vision problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your iPhone picture&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears that you're taking the picture through some fabric or some other kind of rectangular network-shaped material. In this case, the spikes are due simply to reflection of light on the individual threads. If you removed all fabric and kept only one thread without moving it at all, the light would reflect on it exactly in the same place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Telescopes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many reflector telescopes show a pattern of cross-shaped spikes around stars. Those are due to diffraction. Those telescopes have a secondary mirror supported by an X-shaped element called the &quot;spider&quot;. Light is diffracting on the spider and creates a similar pattern of spikes around stars.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2017-07-11T18:02:46.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21602" PostTypeId="2" ParentId="21584" CreationDate="2017-07-11T18:36:39.903" Score="4" Body="&lt;p&gt;Here's my approach to solving this problem. You provide some of the initial steps, but I'll still go through them just for completeness.&lt;/p&gt;&#xA;&lt;h3&gt;Orbital Distance&lt;/h3&gt;&#xA;&lt;p&gt;We're told the period is &lt;span class=&quot;math-container&quot;&gt;$T=5.5\:\mathrm{years}$&lt;/span&gt;. This means we can immediately calculate the orbital distance (or more precisely, the semi-major axis, &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt;). Since we're talking about a comet orbiting the Sun, we can simply use:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T^2 = a^3$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is the orbital period in units of &lt;span class=&quot;math-container&quot;&gt;$\mathrm{years}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is the semi-major axis in units of &lt;span class=&quot;math-container&quot;&gt;$\mathrm{AU}$&lt;/span&gt;. I find that &lt;span class=&quot;math-container&quot;&gt;$a = 3.116\:\mathrm{AU} = 4.66\times 10^{11}\:\mathrm{m}$&lt;/span&gt;. Good so far.&lt;/p&gt;&#xA;&lt;h3&gt;Volume of Evaporated Shell&lt;/h3&gt;&#xA;&lt;p&gt;We're told that the radius of the comet decreases by &lt;span class=&quot;math-container&quot;&gt;$\Delta R = 20\:\mathrm{cm}$&lt;/span&gt;. We can assume the comet is perfectly spherical and calculate the volume of the evaporated shell which should depend on both &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt;, the radius of the comet, and &lt;span class=&quot;math-container&quot;&gt;$\Delta R$&lt;/span&gt;. This will be necessary since we need to know the total amount of ice that was evaporated. The volume of this shell is given by&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V = \frac{4}{3}\pi(R^3-(R-\Delta R)^3)$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$V = \frac{4}{3}\pi(R^3-R^3+3R^2\Delta R-3R\Delta R^2 + \Delta R^3)$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$V = \frac{4}{3}\pi(3R^2\Delta R-3R\Delta R^2)$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$V = 4\pi R^2\Delta R\left(1-\frac{\Delta R}{R}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Note that I've made a specific choice here. I've dropped the &lt;span class=&quot;math-container&quot;&gt;$\Delta R^3$&lt;/span&gt; term from the second line. The reason being that it is a third order term and &lt;span class=&quot;math-container&quot;&gt;$\Delta R^3 \ll R\Delta R^2$&lt;/span&gt;. You could argue that I could drop the second order term, &lt;span class=&quot;math-container&quot;&gt;$R\Delta R^2$&lt;/span&gt; since &lt;span class=&quot;math-container&quot;&gt;$R\Delta R^2 \ll R^2\Delta R$&lt;/span&gt;, but I'm choosing to keep this second order term around so we wind up with an &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; to solve for in the final answer.&lt;/p&gt;&#xA;&lt;h3&gt;Power Input&lt;/h3&gt;&#xA;&lt;p&gt;The next step is to find the total energy input per second to this comet, e.g., the power. You've basically already defined this part. The flux on this comet is defined as&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F = \frac{L_\odot}{4\pi a^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The power input is simply the flux times the area of the comet.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P_{\mathrm{in}} = FA = \frac{L_\odot}{4\pi a^2} \pi R^2 = \frac{1}{4}L_\odot \frac{R^2}{a^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Note that this assumes the comet is in a circular orbit and thus always at the orbital radius of &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt;. If the comet's orbit had any sort of eccentricity, then &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; would be a function of radius and you'd have a much harder time.&lt;/p&gt;&#xA;&lt;h3&gt;Evaporation Energy&lt;/h3&gt;&#xA;&lt;p&gt;Now we need to calculate the total energy necessary to evaporate the evaporated shell of volume &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; from above. In order to evaporate a solid ice, you have to go through four stages of heating. First, you raise the ice's temperature to the melting point. The energy input for this is defined by the specific heat capacity for ice, &lt;span class=&quot;math-container&quot;&gt;$c_{\mathrm{ice}}$&lt;/span&gt;. Then you add energy to convert the ice into water. The energy input for this is defined by the latent heat of fusion, &lt;span class=&quot;math-container&quot;&gt;$\ell_f$&lt;/span&gt;. Now you can raise the temperature of the water until it reaches the next stage. This is defined by the specific heat of water, &lt;span class=&quot;math-container&quot;&gt;$c_{\mathrm{water}}$&lt;/span&gt;. Finally, you add energy to convert the water to gas, defined by the latent heat of vaporization, &lt;span class=&quot;math-container&quot;&gt;$\ell_v$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;All of these can be put together into a single equation.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$E_{\mathrm{evap}} = c_{\mathrm{ice}}(m_{\mathrm{shell}}\Delta T_1) + \ell_fm_{\mathrm{shell}}+c_{\mathrm{water}}(m_{\mathrm{shell}}\Delta T_2) + \ell_vm_{\mathrm{shell}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$E_{\mathrm{evap}} = (c_{\mathrm{ice}}\Delta T_1 + \ell_f +c_{\mathrm{water}}\Delta T_2 + \ell_v)m_{\mathrm{shell}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Each one of the terms in this equation accounts for the energy input of one of the stages I described above. Note that &lt;span class=&quot;math-container&quot;&gt;$\Delta T_1$&lt;/span&gt; is the temperature change from the initial temperature to the melting point (&lt;span class=&quot;math-container&quot;&gt;$273.15\:\mathrm{K}$&lt;/span&gt;). A reasonable temperature change might be &lt;span class=&quot;math-container&quot;&gt;$73.15\:\mathrm{K}$&lt;/span&gt; (assuming an initial temperature of &lt;span class=&quot;math-container&quot;&gt;$200\:\mathrm{K}$&lt;/span&gt;), based on the &lt;a href=&quot;http://www.astronomy.com/news/2014/08/rosetta-takes-comets-temperature&quot; rel=&quot;nofollow noreferrer&quot;&gt;temperature of comet 67P as determined by Rosetta&lt;/a&gt;. The &lt;span class=&quot;math-container&quot;&gt;$\Delta T_2$&lt;/span&gt; is the temperature change from the melting point to the boiling point which is necessarily &lt;span class=&quot;math-container&quot;&gt;$100\:\mathrm{K}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;You can look in a table somewhere and find that &lt;span class=&quot;math-container&quot;&gt;$c_{\mathrm{ice}} = 2.108\:\mathrm{kJ\:kg^{-1}\:K^{-1}}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$c_{\mathrm{water}} = 4.187\:\mathrm{kJ\:kg^{-1}\:K^{-1}}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Lastly, you need to define the mass of the evaporated shell, &lt;span class=&quot;math-container&quot;&gt;$m_{\mathrm{shell}}$&lt;/span&gt;. This is simply the volume already determined, multiplied by the density of ice/water. Technically those densities will be different, but they're close enough that we can just use &lt;span class=&quot;math-container&quot;&gt;$\rho_{shell} = 1000\:\mathrm{kg\:m^3}$&lt;/span&gt;. So ultimately we have:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$E_{\mathrm{evap}} = (c_{\mathrm{ice}}\Delta T_1 + \ell_f +c_{\mathrm{water}}\Delta T_2 + \ell_v)V \rho_{\mathrm{shell}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;For simplicity's sake, I'm going to define&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\eta \equiv (c_{\mathrm{ice}}\Delta T_1 + \ell_f +c_{\mathrm{water}}\Delta T_2 + \ell_v)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;so that&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$E_{\mathrm{evap}} = 4\pi \eta \rho_{\mathrm{shell}} R^2\Delta R\left(1-\frac{\Delta R}{R}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;h3&gt;Putting it all together&lt;/h3&gt;&#xA;&lt;p&gt;We now know the total energy necessary to evaporate the comet's &lt;span class=&quot;math-container&quot;&gt;$20\:\mathrm{cm}$&lt;/span&gt; shell as well as the energy input per second. We know that it receives this energy input per second for one orbital period of &lt;span class=&quot;math-container&quot;&gt;$5.5\:\mathrm{years}$&lt;/span&gt; which means we can say:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T = \frac{E_{\mathrm{evap}}}{P_{\mathrm{in}}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T = \frac{4\pi \eta \rho_{\mathrm{shell}} R^2\Delta R\left(1-\frac{\Delta R}{R}\right)}{\frac{1}{4}L_\odot \frac{R^2}{a^2}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T = 8\pi \eta \rho_{\mathrm{shell}} \Delta R \frac{a^2}{L_\odot} \left(1-\frac{\Delta R}{R}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Again, for simplicity, I'll define&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\xi \equiv 8\pi \eta \rho_{\mathrm{shell}} \Delta R \frac{a^2}{L_\odot}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;so that&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T = \xi \left(1-\frac{\Delta R}{R}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;It should be pretty easy to see now that&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\boxed{R = \frac{\Delta R}{1-T/\xi}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The rest is just plugging everything in.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-07-11T20:08:21.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21603" PostTypeId="2" ParentId="21598" CreationDate="2017-07-11T21:18:28.730" Score="1" Body="&lt;p&gt;I don't think the title of your question and your question are really in sync here and I'm not sure I will correctly answer your question. However, I will give it  a try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give you a hint of the conclusion. My answer would be, &quot;NO&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several clues / reasons why I come to this answer.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Gravity is not necessarily a real thing. When you read the literature concerning general relativity about gravity you will learn that acceleration can be experienced as gravity and in the fabric of space/time gravity is &quot;merely&quot; a curvature in this fabric that. This means that gravity is highly debatable as something that could be seen as an independent force / object in the universe.&lt;/li&gt;&#xA;&lt;li&gt;When gravity exists as a force, whether it is an exchange of gravitons between two objects with a mass, or whether it is a curvature in space / time due to the mass of an object, the phenomenon &quot;gravity&quot; is always related to the mass of an object and doesn't stand on its own as an independent force of nature.&lt;/li&gt;&#xA;&lt;li&gt;Dark matter is not a &quot;mysterious&quot; thing that passes through itself. Dark matter is just the container concept of matter out there that doesn't interact a lot with other matter, but just leaves its marks in the universe by showing a gravitational force as interaction with other matter. Since we have only little understanding of gravity and we still have difficulty measuring gravity as a force in the universe, we have difficulty understanding and measuring dark matter directly. This doesn't mean it is not out there.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, to get back to your second part of the question, &quot;massless gravity&quot;, this is not a viable alternative. Gravity exists either by massive objects exchanging gravitons and therefore exercizing gravity upon each other, or a massive object &quot;denting&quot; the fabric of space / time and therefore creating a gravity field. In either case gravity cannot be seen as a farce that exists without the existence of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, to the first part of your question. Whether we can observe objects directly or indirectly, this doesn't make it more or less realistic. Dark matter is generally accepted to be there. We just haven't found a way to measure it. We can only see that the visible matter in the universe can't explain the gravitational pull on the visible matter and therefore we assume there is some other matter out there we haven't been able to &quot;see&quot;. Hence, dark matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, dark matter does interact through gravity and our current successes in measuring gravitational waves might also result in a better understanding of gravity and how to measure the effect of dark matter on the gravitational pull in the universe. So, to answer your question: &quot;No, we call dark matter because we can only observe it through its influence on other matter due to gravity. There is no such thing as gravity without mass.&quot;&lt;/p&gt;&#xA;" OwnerUserId="6599" LastActivityDate="2017-07-11T21:18:28.730" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21604" PostTypeId="2" ParentId="21600" CreationDate="2017-07-12T01:55:24.583" Score="1" Body="&lt;p&gt;The answer depends on how detailed you want to be. For example,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Do you want (a) an approximate answer by assuming the stars are uniformly distributed, or (b) to use the real distribution of stars?&lt;/li&gt;&#xA;&lt;li&gt;Do you want to know the answer based on (a) the number of stars visible throughout the entire year, or (b) just a given night, or (c) just an instant of time?&lt;/li&gt;&#xA;&lt;li&gt;The magnitude limit of your sky affects how many stars are visible. More stars are visible from a dark site in the country than in a city.&lt;/li&gt;&#xA;&lt;li&gt;The other factor that has a small effect on the answer is whether you want to include atmospheric extinction or not. Stars near the horizon appear to be fainter than their published magnitude due to the atmosphere. Some stars close to the southern horizon (in your example) are not visible because they do not rise high enough.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Combination 1(a) and 2(c) [uniform distribution, an instant of time] can be calculated based on the area of sky that is circumpolar versus the area for half of the sky. (As @zephyr indicated, you need to use steradians to calculate the area of the sky.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combination 1(a) and 2(a) [uniform distribution, entire year] is also easy to calculate from the area of the sky that is circumpolar, and the area of the entire sky minus the area of the sky that never rises (declination less than -55 for your example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other combination requires using a catalog of star positions to &quot;count&quot; how many stars are visible at a given instant of time. In other words, based on the day and time (or the local sidereal time),  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;calculate the altitude of each star.&lt;/li&gt;&#xA;&lt;li&gt;if above the horizon, calculate if it is bright enough to be visible given the sky conditions and (optionally) atmospheric extinction&lt;/li&gt;&#xA;&lt;li&gt;if visible, tally the star as circumpolar or not&lt;/li&gt;&#xA;&lt;li&gt;after counting all of the stars, perform the final calculation of the ratio.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Such a calculation would be easy to perform in a spreadsheet using a catalog such as the Yale Bright Star Catalog (which may not be complete down to the faintest stars visible). I did a similar calculation many, many years ago. See the May 1994 Sky &amp;amp; Telescope magazine, pages 86-88.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-07-12T01:55:24.583" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21606" PostTypeId="1" CreationDate="2017-07-12T05:03:56.737" Score="2" ViewCount="93" Body="&lt;p&gt;I've been reading about &lt;a href=&quot;https://en.wikipedia.org/wiki/SN_1054&quot; rel=&quot;nofollow noreferrer&quot;&gt;SN 1054&lt;/a&gt; and how most professional investigators believe that it is the Crab Nebula today. But the wiki article has this to say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The location of the guest star as &quot;to the south-east of Tianguan, perhaps several inches away&quot; has perplexed modern astronomers, because the Crab Nebula is not situated in the south-east, but to the north-west of Zeta Tauri.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, then I remembered the &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;precession of the equinoxes&lt;/a&gt;. I know precession does not change the fixed background of stars, but can change their relative location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we rewind the precession 1000 years, &lt;strong&gt;does Zeta Tauri cross the zero meridian or the celestial equator?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7449" LastActivityDate="2017-07-12T05:03:56.737" Title="Was the Crab Nebula to the southeast of Zeta Tauri 1,000 years ago?" Tags="&lt;observational-astronomy&gt;&lt;supernova&gt;&lt;history&gt;&lt;precession&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21608" PostTypeId="2" ParentId="21561" CreationDate="2017-07-12T10:30:16.960" Score="1" Body="&lt;p&gt;Small and effective refractors are easier to make than small and effective reflectors, but reflectors are cheaper in general. Unless you have some special reason to want to avoid a larger aperture (eg because you are engaged in solar observation) or a special reason why you want the optical qualities of a refractor, you almost certainly want a bigger aperture and so for a fixed price you are likely to be better off with the reflector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not having a EQ mount may be a bit of an issue though. &lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2017-07-12T10:30:16.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21611" PostTypeId="1" AcceptedAnswerId="21624" CreationDate="2017-07-12T17:30:18.530" Score="2" ViewCount="745" Body="&lt;p&gt;If I use an equatorial mount for my telescope to view the moon, is there a way to take in count the moon movement around the earth, or the movement is too small compared to earth rotation?&lt;/p&gt;&#xA;" OwnerUserId="17411" LastActivityDate="2017-07-13T16:37:47.793" Title="How much the moon movement can influence an equatorial mount setup?" Tags="&lt;the-moon&gt;&lt;telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21612" PostTypeId="1" CreationDate="2017-07-12T19:37:45.693" Score="3" ViewCount="92" Body="&lt;p&gt;Usually, metal-ions in terrestrial chemistry will be solvated - either by gas-phase solvent molecules or by liquid-phase solvent molecules. But the interstellar medium (ISM) is never in local thermodynamic equilibrium. Moreover, the solvation of any system is driven by the system's proclivity to attain thermodynamic stability. But given the lack of local thermodynamic equilibrium in the ISM, solvation should not at all be an important factor in the ISM. Is it right or wrong? Any reference from the Astrochemistry/Astrophysics community will also be highly appreciated. &lt;/p&gt;&#xA;" OwnerUserId="17445" LastEditorUserId="2365" LastEditDate="2018-08-01T15:32:03.447" LastActivityDate="2018-08-01T15:32:03.447" Title="For chemical reactions occurring in the interstellar medium, which involve metal-ions, will the metal ions be solvated or not?" Tags="&lt;interstellar-medium&gt;" AnswerCount="0" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21613" PostTypeId="2" ParentId="20010" CreationDate="2017-07-12T21:11:02.607" Score="2" Body="&lt;p&gt;The conceptual problem is that the half-light radius is the radius of a &lt;em&gt;circle&lt;/em&gt; enclosing half the total light, assuming that the object is circular with a radial intensity profile equal to the Sersic function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need to integrate over circular rings, each of which has an area of $2 \pi r \, dr$ and an intensity (or surface brightness) of $S(r)$:&#xA;$$L(&amp;lt; R) = \int_{0}^{R} 2 \pi r \, S(r) \, dr$$&#xA;where $S(r)$ is the Sersic function evaluated at radius $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, you're just doing the integral of a line of fixed width:&#xA;$$\int_{0}^{R} S(r) \, dr$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Python terms, the (pre-calculated) function you integrate should be&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I = 2 * np.pi * x * Sersic(x)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;not&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I = Sersic(x)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(Alex is also correct that you have almost no resolution near the value of r_eff that you are trying to find.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2017-07-12T21:11:02.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21615" PostTypeId="2" ParentId="21572" CreationDate="2017-07-12T21:42:50.623" Score="3" Body="&lt;p&gt;The primary, traditional classification for galaxies in the local universe is based on &quot;morphology&quot; -- in other words, on their optically visible shape; this goes back to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble_sequence&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble Sequence&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll list the main categories and the defining shape, and then some other characteristics which are not part of the main criteria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Elliptical Galaxies:&lt;/strong&gt; These are circular or elliptical in projected shape (ellipsoidal or triaxial in 3D shape), with no visible disk and very little gas or dust, and little or no evidence for young stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stars are almost all old, and tend to orbit in random directions. Very luminous/massive ellipticals tend to have centrally concentrated radial profiles in the stellar density (now usually described using &lt;a href=&quot;https://en.wikipedia.org/wiki/Sersic_profile&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sersic profiles&lt;/a&gt; with high values of the index $n$); faint, low-mass &quot;dwarf ellipticals&quot; have more exponential stellar profiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;S0 (or Lenticular) Galaxies:&lt;/strong&gt; These have a prominent disk of stars, but one which lacks visible spiral arms and has little or no gas or dust, and little or no evidence for young stars. The disk may, however, have one (or sometimes two) stellar bars, and sometimes rings as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stars are mostly old and almost all orbit in the same direction within the disk, but the orbits may be somewhat elliptical rather than circular. They almost always have a prominent &quot;bulge&quot; of stars dominating the middle of the galaxy; the bulge may be a very centrally concentrated part of the disk, the vertically thickened part of a stellar bar, or a round collection of old stars with mostly random orbits (somewhat like a small elliptical galaxy) -- or a combination of all three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Spiral Galaxies:&lt;/strong&gt; These have a prominent disk of stars, gas and dust; the disk has spiral arms in it (hence the name). The subclassifications within this category (e.g., Sa vs Sb vs Sc vs Sd) are based on a combination of three factors: the relative prominence of a central bulge (if any); how tightly or loosely wound the spiral arms appear to be; and the degree to which the spiral arms are smooth versus being broken up into fragments and stellar clusters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stars and gas almost all rotate in the same direction, with orbits that are relatively circular. They are almost always a mix of young and old stars, with new stars being formed in the disk. They &lt;em&gt;may&lt;/em&gt; have a bulge in the center, but some do not; the bulges may be as diverse and complicated as those in S0 galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Irregular Galaxies:&lt;/strong&gt; As the name suggests, these are more raggedy, lopsided, and generally &quot;shapeless&quot;. They are usually rich in gas, and are almost always lower in mass than the other types; they are, like spirals, often forming stars at the present time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of different kind of dwarf (= faint, low-mass) galaxies which may or may not fall neatly into the above categories. For example, &lt;strong&gt;dwarf spheroidal&lt;/strong&gt; galaxies are very faint and low-mass; in terms of structure, stellar orbits, and the absence of gas or current star formation, they resemble ellipticals, but are very diffuse rather than centrally concentrated. Recent and still somewhat mysterious discoveries include &quot;ultracompact dwarf&quot; (UCD) galaxies and &quot;ultradiffuse galaxies&quot;.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2017-07-12T21:42:50.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21617" PostTypeId="2" ParentId="21600" CreationDate="2017-07-13T02:12:01.047" Score="0" Body="&lt;p&gt;I cannot find my original calculations at the moment. I assume that the Sky &amp;amp; Telescope article is copyrighted, but I found a PDF of it on the internet at &lt;a href=&quot;https://archive.org/details/Sky_and_Telescope_1994-05-pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://archive.org/details/Sky_and_Telescope_1994-05-pdf&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a table from the article which shows the number of stars visible at various times of the night for an observer at 40 N latitude:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7XtL1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7XtL1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-07-13T02:12:01.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21621" PostTypeId="1" CreationDate="2017-07-13T13:43:36.607" Score="1" ViewCount="447" Body="&lt;p&gt;Today at about 2 am I have spotted a light very similar (perceived speed and apparent brightness, motion due East) to ISS. At very first I thought to have spotted the station but immediately I doubted about as I feel that the observation time is too far from sunrise. ISS is indeed ruled out as there is not email from the spot the station service in my inbox. Can satellites orbiting at higher orbits be see even in deep night?  I feel that answers is yes based on simple geometry. However what could reflect so much light? I think there are not objects with reflecting surface so big to be compared to those of ISS... &lt;/p&gt;&#xA;" OwnerUserId="17437" LastActivityDate="2017-07-13T20:06:42.023" Title="Moving light in night sky" Tags="&lt;identify-this-object&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21622" PostTypeId="2" ParentId="21547" CreationDate="2017-07-13T14:02:31.493" Score="2" Body="&lt;p&gt;I saw it once although I was at the perifery of a &quot;rather big&quot; industrial town, but it was very cold and clean. At first I was even scared of some chemiluminescent pollution! It was far from the spectacular display of photographs or from what one can see as reported in the other answers above but nevertheless breathtaking. Two arched stripes of white luminescence, more exotic than just &quot;milky&quot; as the name suggests, if I can say so. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems other answers post pics that should be impossible to attain without light accumulation.&#xA;I would say for a first naked eye observation to expect just a splitted luminous white trace. Impressive nevertheless! &lt;/p&gt;&#xA;" OwnerUserId="17437" LastEditorUserId="17437" LastEditDate="2018-04-21T14:10:18.900" LastActivityDate="2018-04-21T14:10:18.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21623" PostTypeId="1" CreationDate="2017-07-13T16:35:13.823" Score="3" ViewCount="117" Body="&lt;p&gt;The Friedmann Equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\ddot{a}}{a} = -\frac{4\pi G}{3}\left(\rho + \frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know, where does the $3p/c^2$ come from. I know it is something like density of energy or something like that but I don't know where it comes from.&lt;/p&gt;&#xA;" OwnerUserId="16441" LastEditorUserId="11143" LastEditDate="2017-07-14T14:35:15.470" LastActivityDate="2018-09-13T15:29:54.920" Title="Question about $3p/c^2$ in Friedmann equation" Tags="&lt;cosmology&gt;&lt;density&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21624" PostTypeId="2" ParentId="21611" CreationDate="2017-07-13T16:37:47.793" Score="3" Body="&lt;p&gt;For most purposes, the difference between the Moon's motion across the sky and star's motion (the sidereal rate) is negligible. The Moon appears to move about 1/2 degree per hour relative to the stars, so while the stars travel 15 degrees per hour east to west, the Moon travels approximately 14.5 degrees per hour. Unless your mount is aligned accurately and has an accurate motor, you probably will not notice this difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some telescope drives do have a &quot;lunar rate&quot; so that it will track the Moon more accurately. But these approximations do not take into account the fact that the Moon is also moving north or south in declination as it orbits the Earth; the lunar rate only compensates for its west to east motion. (Maybe a goto scope will track the Moon more accurately since it theoretically could be calculating the position of the Moon and pointing to a changing position.)&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-07-13T16:37:47.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21625" PostTypeId="2" ParentId="21621" CreationDate="2017-07-13T17:07:55.077" Score="1" Body="&lt;p&gt;If you are in the Northern Hemisphere, it is currently close to midsummer. The Sun is never very far below the horizon in the North, even in the middle of the night. There are plenty of satellites that are visible even late into the night. &lt;a href=&quot;http://www.heavens-above.com/AllSats.aspx?lat=44.4652&amp;amp;lng=11.3379&amp;amp;loc=Italy&amp;amp;alt=187&amp;amp;tz=CET&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heavens Above&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most are not nearly as bright as the ISS can be. However, against a dark sky a satellite may appear brighter. Lots of the objects that are visible then are actually rocket bodies. They often have a higher orbit than functioning satellites.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-07-13T20:06:42.023" LastActivityDate="2017-07-13T20:06:42.023" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21627" PostTypeId="1" AcceptedAnswerId="21629" CreationDate="2017-07-13T22:28:25.870" Score="4" ViewCount="234" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Intergalactic_star&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, a rogue star or intergalactic star is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;not gravitationally bound to a galaxy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But according to &lt;a href=&quot;https://en.wikipedia.org/wiki/Intergalactic_star#cite_note-12&quot; rel=&quot;nofollow noreferrer&quot;&gt;a citation on the same Wikipedia page&lt;/a&gt;, the first intergalactic stars were discovered in the Virgo cluster of galaxies. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;These stars form a &lt;strong&gt;massive group&lt;/em&gt;&lt;/strong&gt;  aproximately 300,000 light years away from the nearest galaxy. &lt;br&gt;&lt;sub&gt;&lt;em&gt;(Emphasis mine)&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is there any clear, stated definition which differentiates galaxies from &quot;massive groups of rogue stars&quot;?&lt;/p&gt;&#xA;" OwnerUserId="11562" LastEditorUserId="11562" LastEditDate="2017-12-13T17:09:20.803" LastActivityDate="2017-12-13T17:09:20.803" Title="What differentiates a &quot;group of rogue stars&quot; from a galaxy?" Tags="&lt;galaxy&gt;&lt;star-systems&gt;&lt;intergalactic-space&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21629" PostTypeId="2" ParentId="21627" CreationDate="2017-07-14T06:48:48.610" Score="6" Body="&lt;p&gt;It is just a poor choice of words. The comment refers to a discovery paper by &lt;a href=&quot;https://arxiv.org/abs/astro-ph/9801228&quot; rel=&quot;noreferrer&quot;&gt;Ferguson et al. (1998)&lt;/a&gt;, where they used a very deep image in the Virgo cluster to establish that there were an excess of distant stars compared with a control field outside of the cluster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This excess of about 630 stars were all in one tiny HST field of view, so form a &quot;group&quot; in some sense, but there is no suggestion that these stars are not found throughout the Virgo cluster. Indeed subsequent work &lt;em&gt;has&lt;/em&gt; found evidence for intracluster stars in other areas of the cluster (e.g. &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0201219&quot; rel=&quot;noreferrer&quot;&gt;Durrell et al. 2002&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-07-14T06:48:48.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21633" PostTypeId="1" CreationDate="2017-07-14T13:50:56.083" Score="3" ViewCount="376" Body="&lt;p&gt;Anyone know if there is a List of Moving stars on the northern Europe hemisphere I can search for? Like Barnard's &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish to image them [when possible] and compare their yearly movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="17489" LastActivityDate="2017-07-22T12:01:39.137" Title="List of Moving stars on the northern hemisphere? [Stellar kinematics]" Tags="&lt;stellar-dynamics&gt;&lt;kinematics&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21636" PostTypeId="2" ParentId="21547" CreationDate="2017-07-14T14:40:09.617" Score="2" Body="&lt;p&gt;It depends where you are standing, which altitude and longitude. The nearest the equator and the higher altitude, the clearer the stars are. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dd4ju.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dd4ju.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a &lt;a href=&quot;https://www.google.fr/search?biw=895&amp;amp;bih=530&amp;amp;tbm=isch&amp;amp;sa=1&amp;amp;q=difference%20in%20sky%20clarity%20by%20longitude%20cohen&amp;amp;oq=difference%20in%20sky%20clarity%20by%20longitude%20cohen&amp;amp;gs_l=img.3...2113.3295.0.4661.6.6.0.0.0.0.111.501.5j1.6.0....0...1.1.64.img..0.0.0.PccMvz8kaoY#imgrc=NYFHIrRrPqj5YM:&quot; rel=&quot;nofollow noreferrer&quot;&gt;bunch of pictures&lt;/a&gt; by a guy called Cohen who alledgedly calculates his exposure time. I can confirm that the countryside in France, the sky is looks like that when your eyes are used to the dark. At Lattitude zero degrees on the equator you can see at least twice as many stars as in France, on top of of an equatorial mountain like mount Bromo, you will perhaps see 5 times as many stars as in temperate climates, so the MilkyWay is 5 times more visible than in our best conditions, elsewhere.&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2018-04-20T03:21:33.163" LastActivityDate="2018-04-20T03:21:33.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21641" PostTypeId="1" AcceptedAnswerId="21647" CreationDate="2017-07-14T23:19:38.800" Score="3" ViewCount="1509" Body="&lt;p&gt;I understand that a star runs out of fuel slowly and gradually by fusing heavier and heavier particles (because of pressure by gravity) until the last heaviest particle is not able to fuse at the core of the star. This is where gravity takes over, and mass collapses (emitting gases out) to become a smaller dense star (with no energy left to give, just the gravity to absorb). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand is why medium stars collapse to become a supernova which emits light while big stars collapse to form black holes which absorb all light.&lt;/p&gt;&#xA;" OwnerUserId="13587" LastEditorDisplayName="user1569" LastEditDate="2017-07-15T11:26:49.357" LastActivityDate="2017-07-16T20:58:59.873" Title="Why do medium stars collapse to form supernovae while big stars collapse to form black holes?" Tags="&lt;black-hole&gt;&lt;supernova&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21642" PostTypeId="1" CreationDate="2017-07-14T23:33:44.413" Score="1" ViewCount="1105" Body="&lt;p&gt;If a torch is switched on and pointed towards the moon, would the light from the torch reach to the moon?&lt;/p&gt;&#xA;" OwnerUserId="17495" LastEditorUserId="16292" LastEditDate="2017-07-20T16:18:52.820" LastActivityDate="2017-07-20T19:23:28.807" Title="Does the light from a torch reach to the moon?" Tags="&lt;the-moon&gt;&lt;light&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21643" PostTypeId="2" ParentId="21641" CreationDate="2017-07-15T00:09:12.673" Score="4" Body="&lt;p&gt;Small stars do not collapse to become supernovae. Our Sun, which is a not quite a small star, will not become a supernova. It will eventually become a red giant, shrink a bit, become a red giant again, and eventually become a white dwarf. In its end life it will go through some upheavals which will be rather minor compared to a supernova. This is the fate for stars between 1/2 of a solar mass and 8 or so solar masses. Stars smaller than 1/2 of a solar mass die extremely slowly, never even becoming a red giant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only very large stars become supernova. The smaller ones leave neutron stars as remnants, the larger ones, black holes. A tiny fraction of the very largest stars are thought to bypass the supernova stage and directly collapse to become black holes. These &lt;a href=&quot;https://en.wikipedia.org/wiki/Failed_supernova&quot; rel=&quot;nofollow noreferrer&quot;&gt;failed supernovae&lt;/a&gt; form because those stars are so massive they collapse before they have consumed all of their fuel.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-07-15T00:09:12.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21644" PostTypeId="1" CreationDate="2017-07-15T01:14:27.633" Score="0" ViewCount="166" Body="&lt;p&gt;So it was around 6 AM and i was looking at the sky and there were not many visible stars because the sun was about to rise. There were two stars that were brighter than the others and i was looking at them for about 5 mins. They weren't moving and the space between them seemed very little. Suddenly, one of those stars started moving down in direction towards the horizon and the other remained still. The star continued to go down. In 1 or 2 minutes it went behind the horizon and stopped being visible, while the other was still at the same place like earlier. At first I thought that the moving object was a satellite, but then i realised that a satellite would orbit the earth with constant speed, and the object i saw remained still for atleast 5 mins. So what else can it be???&lt;/p&gt;&#xA;" OwnerUserId="17496" LastActivityDate="2017-07-15T13:34:08.120" Title="A strange star-like object" Tags="&lt;star&gt;&lt;orbit&gt;&lt;satellite&gt;&lt;identify-this-object&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2017-07-17T13:03:35.907" ContentLicense="CC BY-SA 3.0" />
  <row Id="21645" PostTypeId="2" ParentId="21642" CreationDate="2017-07-15T02:12:50.370" Score="1" Body="&lt;p&gt;Yes. If there is nothing opaque in the way to obstruct it, it will reach the moon. However, the intensity of light (from a point source like a torch) diminishes with the square of distance. So, by the time it reaches the moon, it will be extremely faint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further information on the inverse square law as applied to light from a point source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Inverse-square_law#Light_and_other_electromagnetic_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Inverse-square_law#Light_and_other_electromagnetic_radiation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17238" LastActivityDate="2017-07-15T02:12:50.370" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="21647" PostTypeId="2" ParentId="21641" CreationDate="2017-07-15T05:21:08.310" Score="2" Body="&lt;p&gt;There are 2 main kinds of supernova.  The kind you are referring to is a &lt;strong&gt;type 2&lt;/strong&gt; supernova.  There are also type 1a, type 1b, and type 1c supernova.  Type 1a supernova are not caused by the implosion of huge stars, but rather, a white dwarf accumulating ever greater amounts of mass from a companion star until the &lt;a href=&quot;https://en.wikipedia.org/wiki/Chandrasekhar_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chandrasekhar limit&lt;/a&gt; for mass is reached.  Then we get the type 1a supernova, which leaves little core remnant and is more or less constant sized explosion with standard luminosity.  This is the reason type 1a supernova are used as standard candles to gauge the distance of objects in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &lt;a href=&quot;https://astronomy.stackexchange.com/questions/21641/why-do-medium-stars-collapse-to-form-supernovae-while-big-stars-collapse-to-form/21643#21643&quot;&gt;David Hammen's answer&lt;/a&gt; has pointed out, type 2 supernova endings occur only for stars that are more than 8x our sun's mass. Neutron stars and black holes are remnants left post-supernova, after the outer layers and majority of star has has blown off.  The size of the remnant is  predictable to an extent pre-supernova and depends on the mass of the star.  However this is only an approximation and there are always exceptions as the physics is complex and depends on too many variables. Type 1 supernovas leave little of no remnant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cause of Type 2 Supernova:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what causes a dying star of said mass or greater to blow itself up to smitherines in such a way that more energy is released in the short span of the explosion that that of an entire galaxy for same time period.  This is in contrast to the much more docile, gradual, and protracted conclusion of stars whose mass is less than 8xSun;  planetary nebula after a red giant collapse and leaves white dwarf as remnant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Massive Star Pre-Supernova&lt;/strong&gt;(type 2) and implosion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The core of a massive star will accumulate iron and heavier elements which are not exo-thermically fusible.  Iron is the end of the exothermic fusion chain. Any fusion to heavier nuclei will be endothermic.  Endothermic fusion absorbs energy from the surrounding layer causing it to cool down and condense around the core further.  This added inward pressure is in addition to the inward pressure of gravity and causes a gradient where the core continues to collapse and get denser while exothermic fusion maintains expansive outward pressure above the gradient.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the density of the inner core reaches a point where the pressure overcomes &lt;a href=&quot;https://en.wikipedia.org/wiki/Electron_degeneracy_pressure&quot; rel=&quot;nofollow noreferrer&quot;&gt;electron degeneration pressure&lt;/a&gt;.  Then another gradient is formed where inside the gradient atoms no longer exists because the electrons of the atoms will join their nucleus. The protons in the nucleus combine with the electrons to become neutrons, and release neutrinos.  There will be no more individual atoms inside of this gradient.  The sudden collapse inside the electron degeneration pressure gradient of the core triggers the initial supernova implosion as said core collapses in a few seconds to a tiny point object of nuclear density (3×10&lt;sup&gt;17&lt;/sup&gt; kg/m&lt;sup&gt;3&lt;/sup&gt;) no bigger than &lt;a href=&quot;http://xtreme.as.arizona.edu/NeutronStars/index.php/neutron-star-radii/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Manhattan&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Supernova Explosion:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sudden in-rush of both the heavy element core outside the electron degeneration gradient, and the exotherimically fusible outer layers of the star cause immense heat and pressure in a few seconds period of time causing much of the lighter than iron elements to simultaneously fuse all at once.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;**Anti-Matter:**&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Also, the extremely dense and energetic neutrino flux would be rushing outward from the mentioned EDP core collapse.  When the neutrinos encounter head on the equally energetic atomic nuclei rushing  inwards the following occurs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a neutron absorbs a neutrino, antimatter is formed as neutrons become anti-protons and positrons.&#xA;When this Anti-Matter encounters normal matter, we have the most efficient and complete mass to energy conversion possible: &#xA;Matter-AntiMatter Annihilation  (MAMA).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The combination of the simultaneous fusion and the MAMA causes all hell to break loose in the greatest release of energy in the shortest time, 2nd only to the gamma ray burst, known in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remnant Post-Supernova:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is left behind after most of the star has blown off is a solid extremely dense mass of neutrons around the size of Manhattan.  Thus the term neutron star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is nothing denser than a neutron star. (See note below) It is just pure mass with no empty space left. To get a perspective what we are taking about here, our normal atom on earth is almost all empty space. If we take the simplest atom which is hydrogen, and the nucleus (one proton) was scaled up to the size of a pea sized marble in the middle of a football stadium, the orbital of the single electron would extend all the way at the outer edge of the stadium, the electron no bigger than a grain of sand; almost all empty space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now just imagine a sphere of solid mass of such marbles filling the stadium with no space left. That is how we can picture the density of a neutron star, and it is the density limit of the universe. A black hole has this same density, just more massive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A neutron star is so dense that one teaspoon (5 milliliters) of its material would have a mass over 5.5×10&lt;sup&gt;12&lt;/sup&gt; kg (that is 1100 tonnes per 1 nanolitre), about 900 times the mass of the Great Pyramid of Giza.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the density limit constant, a neutron star can only get larger, occupy a greater volume, as it gains more mass. The mass and surface area determine the escape velocity, the velocity needed for an object to escape the host's gravity from the surface. The immense gravity causes the smallest neutron star to have an escape velocity of 100,000 km/s, or one third of the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the mass of the neutron star surpasses the level at which the &lt;a href=&quot;https://en.wikipedia.org/wiki/Escape_velocity&quot; rel=&quot;nofollow noreferrer&quot;&gt;escape velocity&lt;/a&gt; for the neutron star exceeds the speed of light, then even light cannot escape its gravitation, thus becoming a &lt;strong&gt;black hole&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: There is theoretically a point where the gravity and density of a neutron star will reach a threshold called the neutron degeneracy pressure.  This was hypothesized by Tolman–Oppenheimer–Volkoff, and limit by same name.  This would result in greater density called baryonic density.  Then if we go denser, we can get quark density, and so on.  These are all speculative theories since the sub-nuclear interactions are not well understood, and we can't look inside a black hole to see the evidence.&lt;/p&gt;&#xA;" OwnerUserId="16998" LastEditorUserId="16998" LastEditDate="2017-07-16T20:58:59.873" LastActivityDate="2017-07-16T20:58:59.873" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21651" PostTypeId="1" CreationDate="2017-07-15T09:04:16.077" Score="2" ViewCount="1856" Body="&lt;p&gt;I'm very new to astronomy and wondering what I can see with these specs, I honestly have no clue what a lot of these specs mean, and need help thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Features:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maximum magnification: 675x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Large aperture offers clearly illuminated views&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stable construction provides solid support and control&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metal tripod with slow motion control rod&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strong performance in both high and low magnification&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Easy sky focusing with finderscope&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Specification&lt;/em&gt;&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aperture: 60mm (2.36inch)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Focal Length: 900mm.f/15&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finderscope:  5x24&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diagonal Prisms: 90degree&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slow Motion Control Rod For Easy Vertial Micro Adjustment&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Height: 128cm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standar: 0.965” Acceddoness Include&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eyepiece: SR4mm H12.5mm H20mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3X Barlow Lens, 1.5X Erector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Telescopic Aluminum Tripod&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it also cost $99, and also wondering if it's worth for the price&#xA;here's a link to the site&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.dshop.com.au/buy/astronomical-telescope-675x-magnification/F90060M?gclid=CjwKCAjw-qbLBRB7EiwAftBCI7h_7lH3E9goJG4Top0N-OHaraxB0cM8IiZgfygRkHD8DZ1ThARCdRoCzA8QAvD_BwE&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.dshop.com.au/buy/astronomical-telescope-675x-magnification/F90060M?gclid=CjwKCAjw-qbLBRB7EiwAftBCI7h_7lH3E9goJG4Top0N-OHaraxB0cM8IiZgfygRkHD8DZ1ThARCdRoCzA8QAvD_BwE&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17501" LastActivityDate="2018-08-21T13:09:00.837" Title="what can I see with this telescope?" Tags="&lt;telescope&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21652" PostTypeId="2" ParentId="21641" CreationDate="2017-07-15T09:33:29.513" Score="7" Body="&lt;p&gt;A supernova is caused by the sudden (less than 1 second) collapse of the core of a massive (initially more than 8 solar masses in total) star. Less massive stars do not undergo core collapse and do not produce core-collapse supernovae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing that determines whether a supernova remnant is a neutron star or black hole is largely the mass of the core at the onset of core collapse, but may also depend on how effectively the rest of the star is blown into space by the supernova explosion and how much material falls back onto the collapsed core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the collapsed core has a mass of less than about 2 solar masses (and perhaps up to 3, though there is no observational evidence for this), then the collapsed core, consisting mainly of neutrons, can be supported against its gravitational weight and stabilised as a &quot;neutron star&quot;. The support is provided by neutron degeneracy pressure and the extreme repulsion between neutrons when they are squeezed together, which is caused by the strong nuclear force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the collapsed core is more massive, or if more material from the exploding star falls back onto the collapsed core, then even this strong nuclear repulsion will be unable to support its weight. In fact, above about 3 solar masses, General Relativity dictates that no force can possibly support the weight of the compact remnant and it will collapse further to become a black hole. No light (or anything else) can emerge from &lt;em&gt;inside&lt;/em&gt; a black hole; so-called Hawking radiation will be negligible; so the only light received from such objects (after any supernova has faded), will be due to material being accreted into the black hole that is heated as it spirals towards it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A third possibility exists, which is direct collapse to a black hole. It is possible that for very massive stars (at least 25-30 solar masses), that the collapsing core is never stabilised even as a proto neutron star, that no bright supernova is produced and that much of the mass of the star directly collapses into a black hole. This is a very active research topic since the discovery of 20-30 solar mass black holes by the LIGO gravitational wave experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarise, I can do no better than quote from an excellent review paper by &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/307647/fulltext/&quot; rel=&quot;noreferrer&quot;&gt;Fryer (1999)&lt;/a&gt;:&#xA;&quot;Thus for core collapse models we can define three regimes of compact object formation: (1) low-mass, core-collapse stars drive strong explosions with little fallback and produce neutron stars; (2) moderate-mass stars produce explosions, but the fallback is sufficient to form black holes; and (3) high-mass stars are unable to launch shocks and collapse directly to black holes. The question for core-collapse theorists, then, is to determine the limits for these regimes.&quot;,&#xA;where the separation between these regimes is (1) $\geq 8$ up to perhaps 20 solar masses; (2) 20-40 solar masses and (3) more than about 30-40 solar masses, but these boundaries depend on metallicity, rotation and a correct understanding of supernova and neutrino physics.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-07-15T12:49:10.167" LastActivityDate="2017-07-15T12:49:10.167" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21653" PostTypeId="1" CreationDate="2017-07-15T09:50:16.893" Score="6" ViewCount="377" Body="&lt;p&gt;The formation of the Moon had many hypothesis over the years, most of them involving a Mars-sized body called Theia, which presumably collided with the Earth. But even within the Theia hypothesis, is not very clear the conditions of that collision. So, what is the most plausible hypothesis about Theia today? &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Theia fully disintegrates and the remnants debris form the Moon in a head-on collision(the Big Impact hypothesis)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Theia-Earth suffer a glazing collision and Theia manages to escape to space, but some of the remnants of the collision form the Moon (&lt;a href=&quot;https://arxiv.org/abs/1207.5224&quot; rel=&quot;noreferrer&quot;&gt;Reufer, 2012&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Other hypothesis, &lt;a href=&quot;https://www.nasa.gov/topics/solarsystem/features/moon_formation.html&quot; rel=&quot;noreferrer&quot;&gt;such a double Theia system&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;All the hypothesis have problems, but to me the most believable is 2, because it solves the majority of the problems of the other ones, although there's not much consensus on that one, the only who defends it is Reufer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, some research shows that only 2% of Theia actually contributed to the formation of the Moon (I don't remember the reference), so 1 doesn't seem too much feasible because according to that one Theia fully disintegrates.&lt;/p&gt;&#xA;" OwnerUserId="9949" LastEditorUserId="9949" LastEditDate="2017-07-17T18:22:24.507" LastActivityDate="2021-02-12T09:09:08.927" Title="The most feasible hypothesis about Theia" Tags="&lt;the-moon&gt;&lt;solar-system&gt;&lt;planetary-formation&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21654" PostTypeId="1" CreationDate="2017-07-15T10:46:18.377" Score="1" ViewCount="676" Body="&lt;p&gt;As I was searching the answer to 'What comes after Black Holes', I found these two threads talking about the same thing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/1746/do-black-holes-die&quot;&gt;Do black holes die?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/2082/can-a-black-hole-reach-a-limit-where-it-can-no-longer-attract-more-matter&quot;&gt;Can a black hole reach a limit where it can no longer attract more matter?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers here mention that Black holes do die but with time and they &quot;&lt;strong&gt;evaporate&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you mean by &lt;strong&gt;Evaporation of a Black Hole&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How do these black holes lose their mass?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13587" LastEditorUserId="13587" LastEditDate="2017-07-15T17:42:36.673" LastActivityDate="2017-07-15T22:17:43.430" Title="How do black holes lose their mass?" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2017-07-17T14:19:20.197" ContentLicense="CC BY-SA 3.0" />
  <row Id="21658" PostTypeId="2" ParentId="21651" CreationDate="2017-07-15T13:22:05.977" Score="4" Body="&lt;p&gt;You can see everything you want in the sky, but for new astronomers there are some simple objects that are nice to look at:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Moon (especially in 1/4 phase you will see craters with good shadows)&lt;/li&gt;&#xA;&lt;li&gt;Jupiter and its largest moons&lt;/li&gt;&#xA;&lt;li&gt;Andromeda galaxy (M31)&lt;/li&gt;&#xA;&lt;li&gt;The Orion Nebula&lt;/li&gt;&#xA;&lt;li&gt;Star Cluster Pleiades&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can see all of them without telescope, but once you use the scope it will make you a &quot;big fan&quot; of astronomy. Choose a dark night without fog or clouds and a place with minimum lights around. The best place is a high one in a desert. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Very important! Don't ever watch the Sun, without a special filter! It will damage the eyes!&lt;/p&gt;&#xA;" OwnerUserId="17507" LastEditorDisplayName="user1569" LastEditDate="2018-08-21T13:09:00.837" LastActivityDate="2018-08-21T13:09:00.837" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21659" PostTypeId="2" ParentId="21644" CreationDate="2017-07-15T13:34:08.120" Score="1" Body="&lt;p&gt;If it disappeared towards east or so, as I  think because the other &quot;star&quot; should be Venus, that it can be a satellite - or even more an airplane - as seen from &quot;its rear&quot;. Basically at first it went far along the line of your line of sight looking still, until finally you could see the curvature of its path. From crude geometric feeling it was most likely an airplane, a lower altitude make it easy the seeing &quot;from the rear&quot; (think about what you see observing the rear light of a car moving on an horizontal road towards a down hill. It looks more or less two lights in the dark then it disappears. Or a ship approaching horizon). About absence of blinking: if it indeed was an airplane mirroring the sunlight, not way to see the less intense &quot;navigation&quot; lights.&lt;/p&gt;&#xA;" OwnerUserId="17437" LastActivityDate="2017-07-15T13:34:08.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21660" PostTypeId="1" AcceptedAnswerId="21674" CreationDate="2017-07-15T13:58:27.407" Score="5" ViewCount="1472" Body="&lt;p&gt;Over on the &lt;a href=&quot;https://scifi.stackexchange.com&quot;&gt;SF &amp;amp; Fantasy Stack&lt;/a&gt;, I've asked a question about a curious watch that shows 15 intervals on its watch face instead of the usual 12 hours. That watch belongs to someone living on Ganymede, close to the end of the 21st century.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;a href=&quot;https://scifi.stackexchange.com/q/163847/19561&quot;&gt;Why does Jet's old watch show 15 hours?&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Does Ganymede have any day/night cycle, or any other cycle, that would explain this curious watch?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In analogy to earth, we should expect a 30-hour Ganymede-day. But Ganymede is in tidal lock, always showing its same side to Jupiter. Also, &lt;a href=&quot;https://astronomy.stackexchange.com/q/10047/16292&quot;&gt;calculations show&lt;/a&gt; that the light reflected by Jupiter pales compared to that of the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would assume that Ganymede's day/night cycle is sun based, possibly complicated by passing through Jupiter's shadow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What do day and night on Ganymede look like?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="16292" LastActivityDate="2017-07-22T15:28:11.567" Title="Does Ganymede have a day/night cycle? If so, what does it look like?" Tags="&lt;orbit&gt;&lt;jupiter&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21662" PostTypeId="2" ParentId="21651" CreationDate="2017-07-15T14:44:05.713" Score="3" Body="&lt;p&gt;As you say you know nothing about what a lot of the specs mean, it's hard to know where to start.  Here are a few points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objective diameter 60mm: this is the diameter of the lens at the front (objective).  The size of the objective determines how faint are the things that you can see.  Under &#xA;the right conditions (very dark and clear night sky)  60mm will show many great astronomical objects, but it is generally the smallest size you will see for astro telescopes.  This is because the objective size also determines how much detail can be seen (again bigger is better), but also this is heavily dependent on the quality of the lens, and in a $100 telescope you shouldn't expect too much.  This has a knock on effect on how much usable magnification you can get out of the scope.  A rule of thumb is that the maximum usable magnification is about 50 times the size of the objective (in inches) - about 120 times in this case - assuming excellent optical quality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eyepieces:  I suspect that only the H 20 and H 12.5 will be usable as the SR 4, giving 225 magnification) will just produce a blur (and anything in the night sky you look at will be very hard to keep track as the earth rotates).  The optical design gives a narrow field of vie, and it is unlikely that you would be able to see the whole of the moon in the same field of view, which you may find disappointing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tripod:  no detail is given, but at this price it is unlikely to be very stable (in other words wobble like a jelly).  &lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-07-15T14:44:05.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21664" PostTypeId="1" CreationDate="2017-07-15T16:17:50.710" Score="0" ViewCount="91" Body="&lt;p&gt;I was looking through sky yesterday and saw something red in sky (naked eye ) like nebula or something then tried using stellarium and mobile observatory apps to find out what it was, But couldnt find anything like this in them. I also tried to capture it on my phone but nothing visible except brightest stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just know it was in cygnus - vulpecula - sagitta or somewhere near it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As i wear glasses and dont have access to dark sky, it was my first naked eye view of some sort of deep sky object. Ill become so happy if i find what that red object was.&lt;/p&gt;&#xA;" OwnerUserId="17511" LastActivityDate="2017-07-16T00:04:15.993" Title="What was this red deep sky object?" Tags="&lt;naked-eye&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2017-07-18T06:46:20.120" ContentLicense="CC BY-SA 3.0" />
  <row Id="21665" PostTypeId="1" CreationDate="2017-07-15T16:38:42.790" Score="3" ViewCount="350" Body="&lt;p&gt;I'm trying to build a 3D model of the Sun, Earth and Moon in in Javascript (using Three.JS, which relies on x,y,z coordinates for all objects) in order to simulate the Aug. 21 eclipse. I've successfully figured this out using the Kepler method that &lt;a href=&quot;https://ssd.jpl.nasa.gov/txt/aprx_pos_planets.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA describes&lt;/a&gt;, but would like to use the HORIZON ephemeris data for greater accuracy, given how sensitive the timing of an eclipse can be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read up on celestial coordinates, but still can't quite figure out how to take the output of the HORIZON tool and convert RA, Dec, etc into 3D coordinates, the way the Kepler document outlines. Can anyone point me to an authoritative primer one how to get from celestial coordinates to geometric coordinates?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coordinates based on AU are just fine -- I've already figured out how to scale that in the simulator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="17512" LastActivityDate="2017-10-16T17:26:03.473" Title="How to visually chart HORIZON ephemeris" Tags="&lt;planet&gt;&lt;positional-astronomy&gt;&lt;ephemeris&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21666" PostTypeId="2" ParentId="21654" CreationDate="2017-07-15T17:35:07.443" Score="1" Body="&lt;p&gt;&quot;Black holes evaporate with time&quot; means they lose mass and energy with time. This phenomenon is termed as Hawking radiation (named after Prof. Stephen Hawking who gave the theory) or simply black hole evaporation. According to him when a particle or antiparticle pop into existence at the event horizon instead of annihilating each other as they are supposed to, one of the the two will fall in the black hole and other will escape. Some of the mass inside the black hole is converted into energy which later converts into particles ( at the event horizon in this case)&#xA;The black hole loses energy/mass in form of  particle ( mass and energy are equivalent) &lt;/p&gt;&#xA;" OwnerUserId="17513" LastEditorUserId="2365" LastEditDate="2017-07-15T22:17:43.430" LastActivityDate="2017-07-15T22:17:43.430" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21667" PostTypeId="1" CreationDate="2017-07-15T19:01:24.133" Score="-2" ViewCount="100" Body="&lt;p&gt;&lt;strong&gt;Pluto is fuzed to all the asteroids, comets, other minor planets, etc. Would it be big enough to hold an atmosphere without forming a tail?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All minor planets, asteroids, comets, etc that aren't orbiting bigger planets are included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything bigger than a minor planet, or any object orbiting something that exceeds minor planet size, isn't included in the fusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So our moon, for example, because it's orbiting Earth and Earth is too big to be included, would be exempt from the fusion and wouldn't have its mass and matter added to Pluto's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pluto keeps its orbit, after the fusion it just gets bigger with the comets asteroids and other minor planets made one with it. Anything that would stay frozen on Pluto's surface before will still be frozen, anything would melt or boil on Pluto's surface would still melt or boil, etc. There's just more stuff there. Nu Pluto is still just as cold post-fuze as it was before. it's just a much much bigger lump of still-cold rock.&lt;/p&gt;&#xA;" OwnerUserId="17514" LastEditorUserId="17514" LastEditDate="2017-07-19T09:03:33.100" LastActivityDate="2017-07-19T09:03:33.100" Title="Pluto is fuzed to all the asteroids, comets, other minor planets, etc. Would it be big enough to hold an atmosphere without forming a tail?" Tags="&lt;gravity&gt;&lt;asteroids&gt;&lt;comets&gt;&lt;planetary-atmosphere&gt;&lt;mass&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="0" ClosedDate="2017-07-18T06:47:07.710" ContentLicense="CC BY-SA 3.0" />
  <row Id="21668" PostTypeId="1" CreationDate="2017-07-15T23:51:33.377" Score="0" ViewCount="8767" Body="&lt;p&gt;It's Saturday 15th July 2017 and currently 00:44. I saw this strange object at around 00:35. I was looking out my window and spotted a bright orange light moving at a steady speed across the sky, it was definitely not a star, at first I thought it was maybe a plane and one of its engines was on fire but it was moving too fast to be a plane and it didn't appear to lose any speed or height as you might expect in that situation. Then I though it could be a landing plane since I live near an airport, again the same conclusion. I watched this bright orange light fly across the sky and occasionally dim as it went behind a cloud although it was still visible. It came out the first cloud and wntered the second, however when it came out of the second cloud it had changed into several dim red lights in a flat line, with the accasional bright flash of green to replace the red lights. It continued to fly over the airport and eventually got too far away to see anymore. I read up and though I could have seen a satellite but I've seen iridium flares before and they've been very different to this, this incident has left me confused and I'd love some insight as to what this might have been. Thanks :)&lt;/p&gt;&#xA;" OwnerUserId="17517" LastEditorUserId="17517" LastEditDate="2017-07-16T21:22:23.577" LastActivityDate="2017-07-16T21:22:23.577" Title="I saw an orange light in the sky which changed into flashing lights, what is this?" Tags="&lt;light&gt;&lt;identify-this-object&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2017-07-18T12:49:45.830" ContentLicense="CC BY-SA 3.0" />
  <row Id="21669" PostTypeId="2" ParentId="21664" CreationDate="2017-07-16T00:04:15.993" Score="1" Body="&lt;p&gt;I think there are no deep sky objects (or extended objects) that show any color to the naked eye. They are too faint to activate the color sensors, so they appear white. I think you saw something nonastronomical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will have a better idea the next time you see it; get a better estimate of its location so that you can find it on a star chart. Use binoculars if you have them.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-07-16T00:04:15.993" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21670" PostTypeId="2" ParentId="21668" CreationDate="2017-07-16T08:28:50.020" Score="2" Body="&lt;p&gt;Your description matches that of a fireball, which are typically large meteors.  As they enter the atmosphere they heat up and glow.  The colour will depend on their speed and composition.  As they get hotter, they often break up into several pieces, which follow the same trajectory.  Much the same effect arise when old satellites or space debris re-enters the atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-07-16T08:28:50.020" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21671" PostTypeId="2" ParentId="21667" CreationDate="2017-07-16T11:21:30.953" Score="4" Body="&lt;p&gt;It depends what kind of atmosphere.   Pluto is cold enough (even at it's hottest) that CO2, NH3 and CH4 are mostly solids.  (look up boiling temperature of each and compare to Pluto's temperature if you like).  There's some photo-evaporation, and perhaps some equilibrium for the heavier gases, but Pluto's primary atmosphere is Nitrogen, which has a lower boiling/freezing temperature.   (the atmosphere is too thin for it to exist as a liquid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/4/4a/Solar_system_escape_velocity_vs_surface_temperature.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Based on this chart&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_escape&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jeans escape&lt;/a&gt;, Pluto would need an escape velocity of about 2.5 km/s (roughly the mass of Io, or about 7 times it's current mass) to retain it's lighter gases like CH4 and NH3 as well as it's nitrogen.   (Nitrogen would require a slightly lower escape velocity).    Very roughly speaking, escape velocity can be tied to mass.   If it was closer to the sun and hotter, it would need to be more massive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is, of-course, a ballpark answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With no magnetic field, pluto would still lose some atmosphere to the solar wind, but that far from the sun, that loss would be slow.   It wouldn't lose much atmosphere to heat-escape (or Jeans-escape), like it does now.   That's not to say Pluto would have a significant atmosphere if it had 7 or 8 times it's current mass but it would likely have a fair bit more.   Pluto's problem is that it's cold enough that much of it's atmosphere would freeze.    If it could get a thick enough atmosphere to maintain a permanent greenhouse gas effect, then it might maintain a real atmosphere.   Again the &lt;strong&gt;minimum&lt;/strong&gt; mass required for that would be in the 7-8 times it's current mass range, but the atmosphere would also need to be sustained and not freeze on it's surface.   Significantly more mass and a higher escape velocity would make retaining an atmosphere much more likely.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The colder the outer edge of a planet's atmosphere is, the lighter the planet (or moon or dwarf-planet) can be and still retain an atmosphere but there's also the problem of too cold and the atmosphere will freeze into ice on the surface.   Titan is a good example of this.  It's far enough away from the sun that it can maintain an atmosphere, mostly formed by out-gassing from it's icy surface.   A few billion years ago Jupiter's 4 Galilean moons may have had atmospheres similar to Titan, lost over time due to heat from the sun and perhaps also from Jupiter's very active radiation belts.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-07-17T03:10:02.867" LastActivityDate="2017-07-17T03:10:02.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21674" PostTypeId="2" ParentId="21660" CreationDate="2017-07-17T07:24:43.933" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/21470/16292&quot;&gt;This answer&lt;/a&gt; to a question about the amount of light reflected by Jupiter on Europa, has screenshots from &lt;a href=&quot;http://www.stellarium.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellarium&lt;/a&gt;, one of which shows the info I needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TBGXs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TBGXs.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h3&gt;Ganymede&lt;/h3&gt;&#xA;  &#xA;  &lt;p&gt;Type: moon&lt;br&gt;&#xA;  (...)&lt;br&gt;&#xA;  Sidereal period: 7.16 days&lt;br&gt;&#xA;  Sidereal day: 171h42m33.4s&lt;br&gt;&#xA;  &lt;strong&gt;Mean solar day: 171h59m36.0s&lt;/strong&gt;&lt;br&gt;&#xA;  (...)&lt;br&gt;&#xA;  &lt;sub&gt;&lt;em&gt;(Emphasis mine)&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, &lt;a href=&quot;https://ops-alaska.com/time/gangale_jupiter/jupiter.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this proposal&lt;/a&gt; for a calendar system for Jupiter's Galilean moons shows that there is no likely subdivision in 15 intervals (which was the cause of my question here).&lt;/p&gt;&#xA;" OwnerUserId="16292" LastActivityDate="2017-07-17T07:24:43.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21676" PostTypeId="2" ParentId="21651" CreationDate="2017-07-17T14:38:16.563" Score="7" Body="&lt;p&gt;I know you didn't specifically ask this, but I feel it answers the spirit of your question. I'm just going to go through the list of features and specifications and explain what they mean.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Features&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Maximum magnification: 675x&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should hopefully be obvious, but this claims to provide images that are 675 times larger that they would appear to the naked eye. However, don't get too excited about this number. When it comes to telescopes, the magnification power is not the most important number. Telescopes for amateurs will try to play off this ignorance though and very often talk about how good that telescope's magnification to make it sound impressive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Large aperture offers clearly illuminated views&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aperture is the physical size of the collecting mirror/lens. Usually aperture size is specified as the diameter of the main mirror/lens and larger is better. A larger aperture means the telescope can collect more light and thus you can see fainter objects you couldn't normally see with your naked eye. The phrase &quot;clearly illuminated views&quot; doesn't really mean much though, its just marketing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stable construction provides solid support and control&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess they're just telling you the telescope won't wobble on it's stand. Which is good. It would be hard to look at a fixed object in the sky while your telescope is wobbling all around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Metal tripod with slow motion control rod&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure it matters that the tripod is metal or not. But they do say it has a &quot;slow motion control rod&quot;. Basically, these are rods you can turn to move the telescope very slightly rather than swinging it with your hand. This just provides the ability to finely move the telescope to help point it precisely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Strong performance in both high and low magnification&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretty obvious statement, albeit a vacuous one. What counts as &quot;strong performance&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Easy sky focusing with finderscope&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The telescope comes with a finderscope. You can see that as the small &quot;mini&quot; telescope attached to the main telescope. When trying to point your telescope at a specific star, it can be hard to know if you're pointing at the right one. The finder scope provides some basic ability to view a larger region of the sky and try to get your telescope generally pointed in the right direction. From there, you can look through the main telescope and use the &quot;slow motion control rod&quot; to get it pointed exactly right.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Specifications&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Aperture: 60mm (2.36inch)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the aperture I mentioned above. They're telling you the diameter of the main lens. Bigger is better here because it means the telescope can collect more light and see fainter objects. That's why professional astronomical telescopes are constantly getting bigger and bigger. I'll note that they're probably playing mental games by saying its 60 mm when 6 cm would have sufficed. More than likely, 60 sounds more impressive than 6 which is why they chose to use mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Focal Length: 900mm.f/15&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tells you two pieces of information. First, that the focal length is 900 mm. This telescope is a Refraction telescope. The light enters the telescope and immediately hits the main lens. The main lens is curved and bends the light, focusing it to a point. This focused light it sent to the eyepiece which will then direct it right to your eye for viewing. They're telling you that the telescope focuses the light to the eyepiece over a distance of 900 mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other quantity is the &lt;a href=&quot;https://en.wikipedia.org/wiki/F-number&quot; rel=&quot;noreferrer&quot;&gt;f-number&lt;/a&gt;. It is the ratio of the focal length to the aperture size. You can hopefully easily see that 900/60 = 15 and thus the f-number is written as f/15. The &quot;/&quot; here does not imply division, it's simply how the f-number is written.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be honest, neither of these two quantities will really matter much for you. I think they're just quoted because they're common &quot;optics&quot; numbers you'd see with higher end cameras and they want this telescope to sound like a serious product. The only thing about this that affects anything for you is the field of view, that is how much of the sky you can see at one time. This will only matter when looking at extended objects like the Moon. A longer focal length means a larger field of view. The f-number is virtually useless for astronomical applications. It only really applies in photography.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Finderscope: 5x24&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is again the finderscope I mentioned above. It's used to get your telescope pointed more or less in the right direction so you can then look through your telescope and do fine pointing. The number 5x24 signifies that it magnifies by 5x and has a 24 mm diameter aperture. This is going to be cheap and may not focus very well so don't expect anything amazing. You'll only use it to point the telescope though so if you're fine with that then so be it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Diagonal Prisms: 90degree&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just the prism at the end of your telescope, and before the eyepiece, that let's the eyepiece be 90 degrees turned from the main telescope shaft. This is useful because you can sit next to your telescope and view through the lens on the side rather than having you align your eye directly with the telescope's main barrel. It just provides easier and more comfortable viewing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Slow Motion Control Rod For Easy Vertial Micro Adjustment&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Already described this above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Height: 128cm&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just giving you the height of the telescope. Although it's unclear precisely what height this is. Certainly the height changes as you move the telescope around on the mount. Is this the maximum height? Is it the height when the telescope is level? Who knows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Standar: 0.965” Acceddoness Include&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I actually have no idea what this means. Clearly they misspelled Standard. Other than that, I don't know what an &quot;Acceddoness&quot; is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Eyepiece: SR4mm H12.5mm H20mm&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just providing you with descriptions of the three eye pieces that come with this telescope. The measurements are the eyepiece aperture sizes. A larger size generally means higher magnification and higher field of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3X Barlow Lens, 1.5X Erector&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two additional optical pieces provided. The first is a &quot;Barlow lens&quot;. This will sit in front of your other eyepieces and serves to provide an additional 3x magnification. The second is the Erector lens. It will provide an additional 1.5 magnification but it also will &quot;erect&quot; your image. The optics of your telescope will flip your image upside down. This is just how optics works (and it's the same reason you appear upside down in a spoon). In truth, when looking at astronomical objects, it doesn't really matter if you're looking at it upside down or not, but cheaper telescopes like to provide the illusion of having fancy gadgetry like erector lenses. There's not much of a reason to flip the image to be &quot;correct&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Telescopic Aluminum Tripod&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's just a basic metal tripod that lets you sit your telescope on a stable mount. The legs can adjust to different heights and are thus &quot;telescopic&quot; which has nothing to do with the fact that it is a tripod for a telescope.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2017-07-17T16:37:25.550" LastActivityDate="2017-07-17T16:37:25.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21677" PostTypeId="1" AcceptedAnswerId="21679" CreationDate="2017-07-17T19:35:53.060" Score="2" ViewCount="248" Body="&lt;p&gt;I have read of this process many times, but I don't think I know the term specifically for the moment when hydrogen fusion begins. What is this moment called?&lt;/p&gt;&#xA;" OwnerUserId="17532" LastEditorUserId="2153" LastEditDate="2017-07-17T21:32:53.973" LastActivityDate="2017-07-19T08:00:27.030" Title="Term for the moment when hydrogen fusion begins in a star" Tags="&lt;star-formation&gt;&lt;terminology&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21678" PostTypeId="2" ParentId="21651" CreationDate="2017-07-17T19:43:10.747" Score="3" Body="&lt;p&gt;Previous answers are good; I will add:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnification is objective focal-length/eyepiece focal length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 3x barlow lens gives an effective focal length of 2700mm; that divided by 4 mm focal length maximum power eyepiece gives the advertised magnification of 675x.  Usable maximum power for a 60mm aperture, though, is about 120x, so already the advertisement is lying to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard for amateur eyepieces is 1.25&quot; diameter; .965&quot; is used only for very cheaply made telescopes and eyepieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The eyepieces are indicated to be of Huygens and Symmetrical design, which are inexpensive designs with a limited field of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Judging by the specifications of the other components, the mounting is also likely to be very cheaply made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, this advertisement is telling you to &quot;STAY AWAY&quot;!  I used such a telescope when I was a boy and it was an exercise in frustration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look for a beginner telescope from a reputable company that is honest about its capabilities, with fewer but more useful (and standard) accessories.  You can find a 90mm reflector with a standard-sized Plossl eyepiece and a usable mount for about the same price, and it will result in a much more enjoyable experience.&lt;/p&gt;&#xA;" OwnerUserId="14434" LastActivityDate="2017-07-17T19:43:10.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21679" PostTypeId="2" ParentId="21677" CreationDate="2017-07-17T21:41:19.813" Score="4" Body="&lt;p&gt;&lt;strong&gt;Ignition&lt;/strong&gt; or more fully &lt;a href=&quot;http://www.schoolsobservatory.org.uk/learn/astro/stars/formation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellar ignition&lt;/a&gt; is the term you want. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-07-19T08:00:27.030" LastActivityDate="2017-07-19T08:00:27.030" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21683" PostTypeId="1" CreationDate="2017-07-18T09:13:09.860" Score="0" ViewCount="179" Body="&lt;p&gt;I'm wondering which telescope I should buy, by the way, I'm new to astronomy and if you know any others that are under $100, It'd be appreciated. You don't need to take the price difference into consideration when replying, thank you!&lt;/p&gt;&#xA;&lt;h1&gt;1 : &lt;a href=&quot;http://www.dshop.com.au/buy/astronomical-telescope-675x-magnification/F90060M?gclid=CjwKCAjw-qbLBRB7EiwAftBCI7h_7lH3E9goJG4Top0N-OHaraxB0cM8IiZgfygRkHD8DZ1ThARCdRoCzA8QAvD_BwE&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.dshop.com.au/buy/astronomical-telescope-675x-magnification/F90060M?gclid=CjwKCAjw-qbLBRB7EiwAftBCI7h_7lH3E9goJG4Top0N-OHaraxB0cM8IiZgfygRkHD8DZ1ThARCdRoCzA8QAvD_BwE&lt;/a&gt;&lt;/h1&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Features&lt;/strong&gt;&lt;/em&gt;:&lt;/p&gt;&#xA;&lt;p&gt;Maximum magnification: 675x&lt;/p&gt;&#xA;&lt;p&gt;Large aperture offers clearly illuminated views&lt;/p&gt;&#xA;&lt;p&gt;Stable construction provides solid support and control&lt;/p&gt;&#xA;&lt;p&gt;Metal tripod with slow motion control rod&lt;/p&gt;&#xA;&lt;p&gt;Strong performance in both high and low magnification&lt;/p&gt;&#xA;&lt;p&gt;Easy sky focusing with finderscope&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Specifications&lt;/strong&gt;&lt;/em&gt;:&lt;/p&gt;&#xA;&lt;p&gt;Aperture: 60mm (2.36inch)&lt;/p&gt;&#xA;&lt;p&gt;Focal Length: 900mm.f/15&lt;/p&gt;&#xA;&lt;p&gt;Finderscope:  5x24&lt;/p&gt;&#xA;&lt;p&gt;Diagonal Prisms: 90degree&lt;/p&gt;&#xA;&lt;p&gt;Slow Motion Control Rod For Easy Vertial Micro Adjustment&lt;/p&gt;&#xA;&lt;p&gt;Height: 128cm&lt;/p&gt;&#xA;&lt;p&gt;Eyepiece: SR4mm H12.5mm H20mm&lt;/p&gt;&#xA;&lt;p&gt;3X Barlow Lens, 1.5X Erector&lt;/p&gt;&#xA;&lt;p&gt;Telescopic Aluminum Tripod&lt;/p&gt;&#xA;&lt;h1&gt;2 : &lt;a href=&quot;https://www.bhphotovideo.com/c/product/617033-REG/Celestron_21035_Travel_Scope_2_7_70mm_Refractor.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.bhphotovideo.com/c/product/617033-REG/Celestron_21035_Travel_Scope_2_7_70mm_Refractor.html&lt;/a&gt;&lt;/h1&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Specifications&lt;/strong&gt;&lt;/em&gt;:&lt;/p&gt;&#xA;&lt;p&gt;70mm Achromatic Refractor&lt;/p&gt;&#xA;&lt;p&gt;400mm Focal Length, f/5.7 Focal Ratio&lt;/p&gt;&#xA;&lt;p&gt;Anti-Reflection Fully Coated Optics&lt;/p&gt;&#xA;&lt;p&gt;1.25&amp;quot; Rack-and-Pinion Focuser&lt;/p&gt;&#xA;&lt;p&gt;Manual Alt-Azimuth Mount with Lock&lt;/p&gt;&#xA;&lt;p&gt;10mm and 20mm Eyepieces, Erecting Prism&lt;/p&gt;&#xA;&lt;p&gt;5x24 Finderscope&lt;/p&gt;&#xA;&lt;p&gt;Aluminum Photo-Style Tripod&lt;/p&gt;&#xA;&lt;p&gt;Backpack Holds OTA, Tripod, Accessories&lt;/p&gt;&#xA;" OwnerUserId="17501" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-07-18T13:46:43.507" Title="Which telescope should I buy?" Tags="&lt;telescope&gt;&lt;space&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2017-07-23T17:39:21.647" ContentLicense="CC BY-SA 3.0" />
  <row Id="21684" PostTypeId="1" AcceptedAnswerId="21685" CreationDate="2017-07-18T11:44:44.883" Score="2" ViewCount="781" Body="&lt;p&gt;Roughly how long did it use to take for the early Earth 4.5~ billion years ago to complete one revolution around the Sun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I know the Earth was spinning much faster, so a day would only take 2-3 hours, so I'd like to have the answer in terms of current Earth days for better understanding - if possible.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I Google'd the heck out of this question and couldn't find an answer.)&lt;/p&gt;&#xA;" OwnerUserId="17542" LastEditorUserId="17542" LastEditDate="2017-07-18T14:20:50.667" LastActivityDate="2020-12-16T19:34:31.060" Title="How long was an early Earth year?" Tags="&lt;the-sun&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2017-07-23T17:40:16.897" ContentLicense="CC BY-SA 3.0" />
  <row Id="21685" PostTypeId="2" ParentId="21684" CreationDate="2017-07-18T12:35:52.173" Score="1" Body="&lt;p&gt;There are two components to take into account for this question&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the earth rotation rate &lt;strong&gt;on itself&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;the orbiting period &lt;strong&gt;around the sun&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;A day back then was shorter than today's earth day. The earth was rotating quicker in the past, but its rotating period around the sun didn't change much because of Kepler's law.&lt;/p&gt;&#xA;&lt;p&gt;I cannot say how it was 5 billion years ago (anyway, did the earth had something you could call a &lt;em&gt;surface&lt;/em&gt; to track the day from then?), but you could find data derived from fossils analysis for the last billion years, e.g. &lt;a href=&quot;https://spacemath.gsfc.nasa.gov/earth/6Page58.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; from Nasa's SpaceMath, on how long a day was back then - &lt;em&gt;I realize now the figures below assume the orbiting period of the earth around the sun (~8766 hours) stayed constant over this period of time. &lt;a href=&quot;https://astronomy.stackexchange.com/a/19857/11264&quot;&gt;This assumption is wrong as explained here&lt;/a&gt; - however stating it &amp;quot;has been close to its current value for the last 2-3 billion years&amp;quot;, so below figures are an approximation&lt;/em&gt;&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;Period&lt;/th&gt;&#xA;&lt;th&gt;Age(years)&lt;/th&gt;&#xA;&lt;th&gt;*Days/year&lt;/th&gt;&#xA;&lt;th&gt;*Hours/day&lt;/th&gt;&#xA;&lt;th&gt;ratio&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Current&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;td&gt;365&lt;/td&gt;&#xA;&lt;td&gt;24.0&lt;/td&gt;&#xA;&lt;td&gt;1.000&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Upper Cretaceous&lt;/td&gt;&#xA;&lt;td&gt;70   million&lt;/td&gt;&#xA;&lt;td&gt;370&lt;/td&gt;&#xA;&lt;td&gt;23.7&lt;/td&gt;&#xA;&lt;td&gt;0.988&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Upper Triassic&lt;/td&gt;&#xA;&lt;td&gt;220  million&lt;/td&gt;&#xA;&lt;td&gt;372&lt;/td&gt;&#xA;&lt;td&gt;23.5&lt;/td&gt;&#xA;&lt;td&gt;0.979&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Pennsylvanian&lt;/td&gt;&#xA;&lt;td&gt;290  million&lt;/td&gt;&#xA;&lt;td&gt;383&lt;/td&gt;&#xA;&lt;td&gt;22.9&lt;/td&gt;&#xA;&lt;td&gt;0.954&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Mississippian&lt;/td&gt;&#xA;&lt;td&gt;340  million&lt;/td&gt;&#xA;&lt;td&gt;398&lt;/td&gt;&#xA;&lt;td&gt;22.0&lt;/td&gt;&#xA;&lt;td&gt;0.917&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Upper Devonian&lt;/td&gt;&#xA;&lt;td&gt;380  million&lt;/td&gt;&#xA;&lt;td&gt;399&lt;/td&gt;&#xA;&lt;td&gt;22.0&lt;/td&gt;&#xA;&lt;td&gt;0.917&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Middle Devonian&lt;/td&gt;&#xA;&lt;td&gt;395  million&lt;/td&gt;&#xA;&lt;td&gt;405&lt;/td&gt;&#xA;&lt;td&gt;21.6&lt;/td&gt;&#xA;&lt;td&gt;0.900&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Lower Devonian&lt;/td&gt;&#xA;&lt;td&gt;410  million&lt;/td&gt;&#xA;&lt;td&gt;410&lt;/td&gt;&#xA;&lt;td&gt;21.4&lt;/td&gt;&#xA;&lt;td&gt;0.892&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Upper Silurian&lt;/td&gt;&#xA;&lt;td&gt;420  million&lt;/td&gt;&#xA;&lt;td&gt;400&lt;/td&gt;&#xA;&lt;td&gt;21.9&lt;/td&gt;&#xA;&lt;td&gt;0.913&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Middle Silurian&lt;/td&gt;&#xA;&lt;td&gt;430  million&lt;/td&gt;&#xA;&lt;td&gt;413&lt;/td&gt;&#xA;&lt;td&gt;21.2&lt;/td&gt;&#xA;&lt;td&gt;0.883&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Lower Silurian&lt;/td&gt;&#xA;&lt;td&gt;440  million&lt;/td&gt;&#xA;&lt;td&gt;421&lt;/td&gt;&#xA;&lt;td&gt;20.8&lt;/td&gt;&#xA;&lt;td&gt;0.867&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Upper Ordovician&lt;/td&gt;&#xA;&lt;td&gt;450  million&lt;/td&gt;&#xA;&lt;td&gt;414&lt;/td&gt;&#xA;&lt;td&gt;21.2&lt;/td&gt;&#xA;&lt;td&gt;0.883&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Middle Cambrian&lt;/td&gt;&#xA;&lt;td&gt;510  million&lt;/td&gt;&#xA;&lt;td&gt;424&lt;/td&gt;&#xA;&lt;td&gt;20.7&lt;/td&gt;&#xA;&lt;td&gt;0.863&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Ediacarin&lt;/td&gt;&#xA;&lt;td&gt;600  million&lt;/td&gt;&#xA;&lt;td&gt;417&lt;/td&gt;&#xA;&lt;td&gt;21.0&lt;/td&gt;&#xA;&lt;td&gt;0.875&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Cryogenian&lt;/td&gt;&#xA;&lt;td&gt;900  million&lt;/td&gt;&#xA;&lt;td&gt;486&lt;/td&gt;&#xA;&lt;td&gt;18.0&lt;/td&gt;&#xA;&lt;td&gt;0.750&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QPiPI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QPiPI.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;From which I would roughly interpolate: &lt;strong&gt;600 &lt;strike&gt;days&lt;/strike&gt; &lt;em&gt;earth self-rotation&lt;/em&gt; per &lt;strike&gt;year&lt;/strike&gt; &lt;em&gt;full rotation around the sun&lt;/em&gt; 2 billion years ago&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pJdlg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pJdlg.png&quot; alt=&quot;600 days per year 2 billion years ago&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Edit&lt;/h1&gt;&#xA;&lt;p&gt;We are saying here that earth was rotating quicker in the past:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Detailed  studies  of  fossil  shells,  and  the  banded  deposits  in  certain  sandstones,  reveal a much different length of day in past eras! These bands in sedimentation and  shell-growth  follow  the  lunar  month  and  have  individual  bands  representing  the  number  of  days  in  a  lunar  month.&lt;/p&gt;&#xA;&lt;p&gt;By  counting  the  number  of  bands,  geologists can work out the number of days in a year, and from this the number of hours  in  a  day  when  the  shell  was  grown,  or  the  deposits  put  down.  The  table  above shows the results of one of these studies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11264" LastEditorUserId="2365" LastEditDate="2020-12-16T19:34:31.060" LastActivityDate="2020-12-16T19:34:31.060" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="21687" PostTypeId="2" ParentId="21651" CreationDate="2017-07-18T13:27:42.623" Score="5" Body="&lt;p&gt;You've already gotten some good answers in technical sense. Let me chip in and give some feedback on the usage of my own telescope and how it might compare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cost: AUD 99.95 (roughly USD 80 at time of writing). This is lower than I would be expecting to pay for a reasonable scope of this spec - as antlersoft noted, this is likely because of poorer quality components&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnification: the 675x top number quoted is correctly calculated but does not take into account the practical (&quot;seeing&quot; or atmospheric disturbance) or theoretical limitation before you lose detail (check out &lt;a href=&quot;http://www.telescope.com/Telescope-Power-Magnification/p/99813.uts&quot; rel=&quot;noreferrer&quot;&gt;http://www.telescope.com/Telescope-Power-Magnification/p/99813.uts&lt;/a&gt; for some information on this). 120x is a reasonable maximum for a perfect night's seeing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with the magnification statistic is that they are often highlighted by manufacturers who don't know or care about the real limits. To put this into context, I have a reflecting telescope with a 200mm objective and this is theoretically good up to 400x - however, I've never been anywhere near there because of &quot;seeing&quot;. The best I've managed is 200x and this is quite rare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other problem with high magnification for this telescope is that it has a small objective lens and therefore limited light gathering capability so, as you increase magnification, the image becomes much dimmer very quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll certainly see the Moon and it will probably look quite good in this scope. I often find the Moon uncomfortable bright in my scope and use a mask to artificially restrict my objective to something similar to this scope. It works well at 35x so your 20mm eyepiece (45x) would probably work well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I normally use 110x for Jupiter, so the 10mm (90x) would likely work and I'd expect to see the banding on the planet and the four major moons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use the same for Saturn and the 10mm (90x) would probably get you to see the rings though I'm not sure how clearly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, all this is based on the assumption that the build quality is good - it's all a best case ... personally I would suggest going for a named telescope from a known brand. I see you've posted something in a separate question (&lt;a href=&quot;https://astronomy.stackexchange.com/questions/21683/which-telescope-should-i-buy&quot;&gt;Which telescope should I buy?&lt;/a&gt;) , so I'll post a suggestion there too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck.&lt;/p&gt;&#xA;" OwnerUserId="17502" LastActivityDate="2017-07-18T13:27:42.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21688" PostTypeId="2" ParentId="21683" CreationDate="2017-07-18T13:46:43.507" Score="2" Body="&lt;p&gt;Of the two, I would go with the Celestron - it's from a named and respected manufacturer with sensible specifications and relevant additional equipment. The eyepiece selection is thoughtful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you pay the price because the Celestron's focal length at 400mm is much more limited. The best magnification you'll be able to get is 40x without a Barlow (good for the Moon but rather anaemic for the planets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other scope promises much higher magnifications but I worry that the manufacturer's name isn't stated, that there seems to be a number of redundant extra items and about the overall build quality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I would consider the &lt;a href=&quot;https://www.bhphotovideo.com/c/product/485261-REG/Celestron_21061_AstroMaster_70_AZ_2_7_70mm_Refractor.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.bhphotovideo.com/c/product/485261-REG/Celestron_21061_AstroMaster_70_AZ_2_7_70mm_Refractor.html&lt;/a&gt; as an alternative, though it is a little more expensive.&lt;/p&gt;&#xA;" OwnerUserId="17502" LastActivityDate="2017-07-18T13:46:43.507" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21689" PostTypeId="1" AcceptedAnswerId="21692" CreationDate="2017-07-18T14:56:29.307" Score="3" ViewCount="236" Body="&lt;p&gt;In which case, for every planet that we detect, there should be significantly more that we can't detect because they don't transit in line with the Earth and, thus, are invisible.  Now, I suppose that we can reasonably assume that most planets in a given solar system orbit in a fairly well defined plane, so if we detect one planet then there are probably others in the same plane.  But, if we have detected no planets in a system, isn't it simply more likely that the planets are orbiting in a plane that doesn't bisect the Earth?  If that is the case, since there is such a tiny degree of variation that would allow a transit to be observed, shouldn't there be a huge number of planets that are not visible because the inclination of their orbital plane is something other than 0 (with respect to the Earth)?&lt;/p&gt;&#xA;" OwnerUserId="17510" LastActivityDate="2017-07-18T16:45:30.380" Title="Isn't it likely that most extra solar planets don't transit their star in line with the Earth?" Tags="&lt;planet&gt;&lt;exoplanet&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21691" PostTypeId="2" ParentId="21665" CreationDate="2017-07-18T15:58:38.387" Score="1" Body="&lt;p&gt;I just went through this by myself.  The principal thing you have to do is choose the vector format for your ephemeris, and then download the ephemeris as a plain text (ascii) table. This is way easier than using the Spice tools.&lt;/p&gt;&#xA;" OwnerUserId="17547" LastActivityDate="2017-07-18T15:58:38.387" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21692" PostTypeId="2" ParentId="21689" CreationDate="2017-07-18T16:45:30.380" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Isn't it likely that most extra solar planets don't transit their star in line with the Earth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes this is certainly true. The chances that a planet's orbital plane is directly along our line of sight is pretty small. And you are also correct to assert that it is more likely that most planets will orbit in the same plane. So if we don't see a planet around a star, it doesn't mean the planets don't exist. &lt;a href=&quot;https://en.wikipedia.org/wiki/Evidence_of_absence&quot; rel=&quot;noreferrer&quot;&gt;Absence of evidence is not evidence of absence&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the early days of exoplanet detection, we had only a handful of known planets to study. This hardly precludes itself to statistical analysis. However, as the years have progressed, the number of known and confirmed planets has piled up, allowing astronomers to start inferring information based on statistical arguments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can argue that if we've spotted &lt;em&gt;X&lt;/em&gt; number of planets after observing &lt;em&gt;Y&lt;/em&gt; number of stars under &lt;em&gt;Z&lt;/em&gt; conditions for &lt;em&gt;T&lt;/em&gt; time then with a little bit of statistical analysis and math, there should be &lt;em&gt;A&lt;/em&gt; number of planets in our galaxy, many of which we can't observe with current techniques. In fact, that is precisely what many astronomers have done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/1202.0903&quot; rel=&quot;noreferrer&quot;&gt;Cassan et al. (2012)&lt;/a&gt; looked at microlensing detections and were able to infer, based on known detections and statistical analysis, that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Planets are at least as numerous as the stars in the Milky Way.&lt;/li&gt;&#xA;&lt;li&gt;17% of stars host a Jupiter-mass planets ($0.3$-$10\:M_J$).&lt;/li&gt;&#xA;&lt;li&gt;52% of stars host &quot;Cool Neptunes&quot; ($10$-$30\:M_\oplus$).&lt;/li&gt;&#xA;&lt;li&gt;62% of stars host an Earth-mass planet ($5$-$10\:M_\oplus$).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0004-637X/764/1/105/pdf&quot; rel=&quot;noreferrer&quot;&gt;Swift et al. (2013)&lt;/a&gt; performed a similar type of analysis but focused mostly on M Dwarfs. They inferred that M Dwarf stars are expected to have, on average, $\sim2$ planets per star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are certainly many more such analyses and almost all of them come to the conclusion that &lt;strong&gt;there are hundreds of billions of planets in our galaxy&lt;/strong&gt;. This can be confidently stated based on only the ~3500 exoplanets found so far.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-07-18T16:45:30.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21693" PostTypeId="1" AcceptedAnswerId="22090" CreationDate="2017-07-18T17:00:35.623" Score="0" ViewCount="663" Body="&lt;p&gt;NASA's predictions for the August 2017 solar eclipse show that about half of Kansas City will see a total eclipse and the other half won't (they'll just see a &quot;near&quot; total eclipse, perhaps even Bailey's Beads?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought this would be a great way of testing how accurate NASA's eclipse path prediction is: have people in Kansas City report, with location, whether they saw a total eclipse or not (of course, many people in Kansas City will probably head into the total eclipse zone just to see it, but hopefully few will remain outside it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It then occurred to me that someone must've already done this for previous eclipses (I know it's done for lunar occultations of other stars).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, googling only tells me that NASA's 2017 path predictions are more accurate than ever before, but not actually HOW accurate they are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where can I find data about accuracy of eclipse predictions?&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2017-08-13T20:01:18.537" Title="How accurate are NASA's eclipse predictions?" Tags="&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="21696" PostTypeId="2" ParentId="21627" CreationDate="2017-07-18T21:12:14.477" Score="-4" Body="&lt;p&gt;According to theoretical work by Dr. Ernest Sternglass, there are expected to be many such groups of &quot;rogue stars&quot;.  In Sternglass's model [the &quot;electron-positron pair model of matter&quot;] there are expected to be many stars and clusters of stars which are not near any galaxy because of the process by which stars and star clusters form.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Briefly, the model says that each kind of astronomical object, such as a single star, or a star cluster, or a small galaxy, or a large galaxy, or a galaxy cluster, etc., is formed by a single massive explosion, (known as a &quot;quasar&quot;).  A not-very-energetic quasar might form a star cluster, while a very-energetic quasar might form a large galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the model says that a quasar is a galaxy or star cluster, or other astronomical object, in the process of being born.&lt;/p&gt;&#xA;" OwnerUserId="13290" LastActivityDate="2017-07-18T21:12:14.477" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21697" PostTypeId="2" ParentId="21677" CreationDate="2017-07-18T23:40:34.470" Score="3" Body="&lt;p&gt;It is known as the nuclear turn-on in the Hertzsprung-Russell diagram; the opposite of the nuclear turn-off when hydrogen is exhausted in the core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In actual fact there is no &quot;moment&quot; when the star begins to fuse hydrogen, it is a gradual process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the models of &lt;a href=&quot;http://www.astro.ulb.ac.be/~siess/pmwiki/pmwiki.php/StellarModels/Z02&quot; rel=&quot;nofollow noreferrer&quot;&gt;Siess et al. (2001)&lt;/a&gt; suggest that a solar mass pre main sequence star gets 0.01% of its luminosity from hydrogen fusion at 2 million years old,  1% after 16 million years, but it is not until it is 56 million years old that 99% of the luminosity comes from hydrogen fusion.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-07-18T23:51:50.077" LastActivityDate="2017-07-18T23:51:50.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21698" PostTypeId="2" ParentId="21677" CreationDate="2017-07-19T00:47:33.960" Score="2" Body="&lt;p&gt;The onset of hydrogen fusion for a star (i.e. when it enters the main sequence) is called the &lt;a href=&quot;https://en.mimi.hu/astronomy/zero-age_main_sequence.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;zero-age main sequence&lt;/a&gt; (ZAMS). Conversely, the terminal-age main sequence (TAMS) refers to the point a star stops fusing hydrogen (and by definition leaves the main sequence); the time between the ZAMS and TAMS is called the main sequence lifetime. On the Hurtzsprung-Russell diagram, several stars at their ZAMS are grouped into a line:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ckAzH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ckAzH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt; Picture from &lt;a href=&quot;http://cnx.org/contents/im49RUTc@4/Evolution-from-the-Main-Sequen&quot; rel=&quot;nofollow noreferrer&quot;&gt;Evolution from the Main Sequence to Red Giants&lt;/a&gt;, courtesy of Rice University under a &lt;a href=&quot;https://creativecommons.org/licenses/by/4.0/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Creative Commons Attribution License 4.0&lt;/a&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some astronomers consider the ZAMS to be the &quot;birth&quot; of a star, in the sense that the object is no longer a protostar and meets the criteria of a star (&lt;a href=&quot;https://astronomy.swin.edu.au/cosmos/Z/Zero+Age+Main+Sequence&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;). These astronomers thus consider a star's age to be the time between now and its ZAMS. However, this is simply a matter of terminology, as others consider the formation of the star to be the moment when its molecular cloud collapses.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2017-07-19T01:01:51.950" LastActivityDate="2017-07-19T01:01:51.950" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="21699" PostTypeId="1" CreationDate="2017-07-19T02:50:17.693" Score="1" ViewCount="93" Body="&lt;p&gt;I'm curious, if someone were to be floating out in space. Would they eventually be captured by the gravitational pull from another planet or sun. To be more specific the person isn't anywhere near said planet or sun they are just passing between them. My question, is there some sort of current in between planets or sun that can have the slightest effect on an object passing through?&lt;/p&gt;&#xA;" OwnerUserId="17550" LastActivityDate="2017-07-20T18:35:21.217" Title="What would happen if a person was out in space? would they get picked up by the pull of gravity from inbetween our planets, or sun?" Tags="&lt;planet&gt;&lt;gravity&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21702" PostTypeId="1" CreationDate="2017-07-19T01:00:12.380" Score="5" ViewCount="285" Body="&lt;p&gt;Most astronomers say it is likely to happen. Is this true? Are there any scientists that reject this?&lt;/p&gt;&#xA;" OwnerUserId="19265" OwnerDisplayName="Sue K Dccia" LastActivityDate="2017-07-19T13:15:48.373" Title="Do most astronomers think that Andromeda will collide with Milky Way?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21703" PostTypeId="2" ParentId="21702" CreationDate="2017-07-19T01:57:06.293" Score="8" Body="&lt;p&gt;Relatively high-resolution measurements were made with the Hubble Space Telescope over a 8 year period. From these measurements, astronomers determined that the collision was a certainty.  It is predicted to occur in about 4 billion years. I suppose there may be some astronomers who disagree with the measurements or calculations but I can find no mention of that. The results were reported in an article in &quot;Nature&quot; (&lt;a href=&quot;http://www.nature.com/news/andromeda-on-collision-course-with-the-milky-way-1.10765#/b1&quot; rel=&quot;noreferrer&quot;&gt;http://www.nature.com/news/andromeda-on-collision-course-with-the-milky-way-1.10765#/b1&lt;/a&gt;) and published in &quot;Astrophysical Journal&quot;  Both are well-respected publications. The &quot;Astrophysical Journal&quot; reference is below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sangmo Tony Sohn; Jay Anderson; Roeland van der Marel (Jul 1, 2012). &quot;The M31 velocity vector. I. Hubble Space Telescope proper-motion measurements&quot;. The Astrophysical Journal. doi:10.1088/0004-637X/753/1/7.&lt;/p&gt;&#xA;" OwnerUserId="16926" OwnerDisplayName="john" LastEditorUserId="16926" LastEditDate="2017-07-19T13:15:48.373" LastActivityDate="2017-07-19T13:15:48.373" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21706" PostTypeId="1" CreationDate="2017-07-19T11:35:09.910" Score="2" ViewCount="428" Body="&lt;p&gt;Since the moon is tidally locked to earth, what about center of mass and geometric center? How far from each other are they? Can a celestial body be tidally locked to an other one if its mass distribution is perfectly homogeneous (center of mass at the very same place of geometric center)&lt;/p&gt;&#xA;" OwnerUserId="16500" LastActivityDate="2017-08-18T15:25:01.930" Title="Moon, center of mass &amp; geometric center, tidal locking" Tags="&lt;the-moon&gt;&lt;tidal-forces&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21708" PostTypeId="2" ParentId="21706" CreationDate="2017-07-19T14:51:54.503" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Since the moon is tidally locked to earth, what about center of mass and geometric center? How far from each other are they?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They differ by about two kilometers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can a celestial body be tidally locked to an other one if its mass distribution is perfectly homogeneous (center of mass at the very same place of geometric center)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, but that condition is not what I would call &quot;perfectly homogenous&quot;. What's needed is a non-spherical mass distribution. For example, a uniform density ellipsoidal body would have its center of mass and center of figure of coinciding, but would still be subject to torques. It's the moment of inertia that is key rather than the offset between center of mass and center of figure. An object whose center of mass and center of figure do not coincide necessarily has a non-spherical moment of inertia tensor.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-07-19T14:51:54.503" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="21711" PostTypeId="1" CreationDate="2017-07-19T17:24:25.040" Score="2" ViewCount="1007" Body="&lt;p&gt;I was wondering if the sun's radiation affects the gravity when it passes through space? Like can radiation strengthen or weaken the force of gravity? &lt;/p&gt;&#xA;" OwnerUserId="17550" LastEditorUserId="7411" LastEditDate="2017-07-20T09:54:46.827" LastActivityDate="2017-07-21T20:54:50.773" Title="Can radiation affect gravity?" Tags="&lt;gravity&gt;&lt;radiation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21712" PostTypeId="2" ParentId="13716" CreationDate="2017-07-19T17:43:25.830" Score="11" Body="&lt;p&gt;A belated answer, but neither of the existing answers properly explain this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proper explanation is simple. In Newtonian mechanics, tidal influences make all objects in retrograde orbits and those objects in prograde orbits below the equivalent of geosynchronous radius spiral inward. Only objects orbiting prograde above the equivalent of geosynchronous radius spiral outward. Our Moon orbits at about 385000 km, well above geosynchronous radius of 42164 km. Phobos orbits Mars with a semi-major axis of 9377 km, well below the areosynchronous radius of 20400 km. While our Moon is spiraling outward from Earth, Phobos is spiraling inward toward Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mutually tidally locked objects spiral neither inward nor outward in Newtonian mechanics. General relativity modifies this dynamic a bit, making mutually tidally locked objects spiral inward. &lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-07-19T17:43:25.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21713" PostTypeId="2" ParentId="21711" CreationDate="2017-07-19T21:02:46.273" Score="6" Body="&lt;p&gt;Radiation has energy, and energy exerts gravity. You could say that it &quot;strengthens the force of gravity&quot; (it definitely doesn't &lt;em&gt;weaken&lt;/em&gt; it), but it would be more appropriate to say that it contributes to the total gravitational field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect it immensely small, however, and is in almost all circumstances completely insignificant compared to the gravity of other forms of mass/energy. But most of the radiation in the Universe is the so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow noreferrer&quot;&gt;cosmic microwave background&lt;/a&gt;, and because this radiation is gradually redshifted with time, it used to be more energetic, and hence exert more gravity. It is fairly easy to calculate that until the Universe was roughly 50,000 years old, radiation actually dominated the gravitational field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prologue:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Zephyr points out below, the term &quot;exerting gravity&quot; is not really the correct way of describing gravity. In general relativity, objects are assumed follow straight lines in space, unless a force (e.g. electromagnetic) acts upon it. What &lt;em&gt;we&lt;/em&gt; interpret as &quot;gravity&quot; is the deviation from what &lt;em&gt;we&lt;/em&gt; think is &quot;straight ahead&quot;, caused by a geometrical change in the space (and time) — and thus in the mathematical rules we use to calculate e.g. how parallel lines behave, how many degrees there are in a triangle, etc — around a massive object. The deformation of space is given by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Einstein_field_equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein field equations&lt;/a&gt;, in which the so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor&quot; rel=&quot;nofollow noreferrer&quot;&gt;stress–energy tensor&lt;/a&gt; appears. This is a mathematical object consisting of 16 numbers, of which &lt;em&gt;one&lt;/em&gt; number (the &quot;$T^{00}$'th component&quot;) holds the energy density of &lt;em&gt;everything&lt;/em&gt;, i.e. radiation, dark matter, dark energy, stars, planets, bicycles, and — in particular — your mom.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2017-07-21T20:54:50.773" LastActivityDate="2017-07-21T20:54:50.773" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21715" PostTypeId="1" AcceptedAnswerId="21732" CreationDate="2017-07-19T23:14:09.293" Score="4" ViewCount="325" Body="&lt;p&gt;I was watching a science fiction anime from the mid 70's. In it, there was a star orbiting a planet. Since this seemed impossible, and I had never heard of it, I looked on the internet to see if this could be possible in a very unlikely situation or in theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On &lt;a href=&quot;https://www.quora.com/Is-it-possible-for-a-star-to-orbit-a-planet-Has-it-ever-been-observed&quot; rel=&quot;nofollow noreferrer&quot;&gt;a similar question on &lt;em&gt;Quora&lt;/em&gt;&lt;/a&gt;, after many no, no, no, no answers from astronomers/astrophysicists, some of them with very detailed models and theories with different scenarios and equations, one answer read,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Absolutely.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If there were two stars of identical mass orbiting each other in equilibrium, then you could throw a planet of any size at the barycenter of the two stars. Then optically you'd have not just one, but two stars orbiting a planet. The planet would be equidistant from the two stars at all moments in time, and would experience identical gravitational pull in two diametrically opposite directions and hence never deviate from the barycenter. This system would not be in a stable equilibrium though, and a passing celestial object of any mass could disrupt this system. Hence, this is probably something that is unlikely to be observed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is this possible?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11562" LastEditorUserId="11562" LastEditDate="2017-12-13T17:08:26.693" LastActivityDate="2017-12-13T17:08:26.693" Title="Can a binary star optically &quot;orbit&quot; a planet?" Tags="&lt;star&gt;&lt;planet&gt;&lt;orbit&gt;&lt;binary-star&gt;&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="21722" PostTypeId="2" ParentId="21699" CreationDate="2017-07-20T11:23:22.760" Score="1" Body="&lt;p&gt;Any point in space feels the gravity of the whole &lt;em&gt;observable&lt;/em&gt; universe around it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are in the middle of nothing, then the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_field&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravitational field&lt;/a&gt; you measure can be incredibly low. But from small differences of mass and energy density throughout the (constantly moving) universe, it fluctuates, and drag you around in a direction or another, as long as you are mass or energy or spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are close to a massive object, on the other hand, then you are strongly bound to it by gravity, you will measure a strong field. Now if you have an equivalently massive object on the opposite of your position, gravity field is high, but the force you feel from is low in this place, because it annihilates (even with different mass bodies, this is what happens at &lt;a href=&quot;https://en.wikipedia.org/wiki/Lagrangian_point&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lagrangian point&lt;/a&gt; L1).&lt;/p&gt;&#xA;" OwnerUserId="11264" LastActivityDate="2017-07-20T11:23:22.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21724" PostTypeId="1" AcceptedAnswerId="21814" CreationDate="2017-07-20T03:23:11.793" Score="6" ViewCount="1347" Body="&lt;p&gt;Is there any scientific reason for the jovian planets to be in descending order of size or is it purely coincidental? In terms of radii it goes Jupiter, Saturn, Uranus and Neptune. &lt;/p&gt;&#xA;" OwnerUserId="16812" OwnerDisplayName="Rumplestillskin" LastEditorUserId="16292" LastEditDate="2017-07-22T21:10:39.307" LastActivityDate="2017-07-25T23:18:31.253" Title="What is the reason the Jovian planets are in descending order of size?" Tags="&lt;planet&gt;&lt;solar-system&gt;&lt;astrophysics&gt;&lt;gas-giants&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21729" PostTypeId="1" AcceptedAnswerId="21738" CreationDate="2017-07-20T17:12:31.270" Score="2" ViewCount="56" Body="&lt;p&gt;Let the distribution of galaxies be a number like $n$. we want to find the total number of galaxies within the radius $r$ from a specified point. in which shape of the universe (flat, open, closed) the number we get is higher? ($n,r$ are the same in this universes and only the area (volume) in each universe is different). &lt;/p&gt;&#xA;" OwnerUserId="16441" LastActivityDate="2017-07-21T00:36:25.297" Title="galaxies in a specific radius in different shapes of universe" Tags="&lt;cosmology&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21730" PostTypeId="1" AcceptedAnswerId="21731" CreationDate="2017-07-20T17:44:54.523" Score="3" ViewCount="2837" Body="&lt;p&gt;I have some lenses - a 10 cm and 7 cm convex lense and a 4 cm and 1 cm concave lenses. I am trying to make a telescope by combining one of convex and concave lenses, but I am struggeling to do so. The picture that is created using these two lenses doesn't show anything other than the thing that can be seen by eyes. Is it possible to make simple telescope using these lenses?&lt;/p&gt;&#xA;&lt;p&gt;PS: All of numbers are lense diameters.&lt;/p&gt;&#xA;" OwnerUserId="17424" LastEditorUserId="37191" LastEditDate="2021-05-28T09:40:31.407" LastActivityDate="2021-05-28T09:40:31.407" Title="Challenges when trying to build a telescope using two lenses" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;diy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21731" PostTypeId="2" ParentId="21730" CreationDate="2017-07-20T18:19:03.170" Score="4" Body="&lt;p&gt;Diameter is important, but if you're trying to get any magnification out of that system, you need to look at the focal length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a very distant light source, like the Sun, focused through a convex lens, until the image of the source is as small and as clear as possible, then the focal length is the distance between the lens and the image you're creating. Like when you burn paper with the lens, focal length is the lens-paper distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For concave lenses focal length is a little more tricky to define, but they do have a focal length anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If F is the focal length of the forward lens (the objective), and f is the focal length of the lens near your eye (the ocular, or the eyepiece), then the magnification of the instrument is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M = F / f&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need an objective with a long focal length, and an eyepiece with a short focal length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both the objective and the eyepiece could be convex actually. It's just that the image will be reversed. If the eyepiece is concave, the image will be straight-up, but the field of view of the telescope will be narrow (like looking through a peephole).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use both convex lenses, the distance between them need to be close to the sum of their focal lengths, F + f. Start there and adjust it slightly for best results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sWFZr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sWFZr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use a concave lens for the eyepiece, then the distance between lenses needs to be the difference of their focal lengths, F - f.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Both lenses need to be perpendicular on their common axis, to minimize aberrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It helps if they are mounted in a tube, to reduce glare, but it's not absolutely necessary. They could simply be attached to a long rod or something - but you'll get a lot of glare into the eyepiece that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do not necessarily try to get very high magnification. If the diameter of the objective is D, measured in mm, then the maximum usable magnification is 2D. So an objective 100 mm in diameter could never provide more than 200x magnification - I mean it could, but the image would be blurry. Even then, this is a theoretical limit, and in practice the image will get pretty bad long before you reach the limit, unless you use high quality lenses designed specifically for telescopes, etc. Also, at high magnification it's difficult to hold the scope steady, and the image is jumping all over the place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try and make a 5x ... 8x instrument first, see how that goes. A lot of binoculars are in this range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've built several refractor telescopes from lenses when I was in high school. It's totally doable. It just takes some patience to figure out how to solve all problems.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2017-07-20T18:19:03.170" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21732" PostTypeId="2" ParentId="21715" CreationDate="2017-07-20T18:28:26.197" Score="6" Body="&lt;p&gt;Planets don't orbit stars. Stars don't orbit planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever there are two bodies bound by gravity, they are both orbiting their common center of mass. For example, both the Earth and the Moon orbit their common center of mass - but that's pretty close to the center of the Earth actually, so it seems like the Moon orbits the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a star to seem to orbit a planet, that would mean the planet is much heavier than the star. As far as we know, that's impossible. Their common center of mass would be much closer to the star, so it would seem like the planet is orbiting the star, as usual.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you have two stars orbiting each other (actually, orbiting their common center of mass) very closely, then you could have a planet circling around both. If the planet is circling around too closely, its orbit would not be stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if the planet's orbit is far enough from the pair of stars, the orbit could be stable a very long time. The minimum distance it at least 2x ... 4x the distance between stars, ideally much bigger. We have discovered such planets. Kepler-47c is a gas giant in the circumbinary habitable zone of the Kepler-47 system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This wiki page has more details:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Habitability_of_binary_star_systems&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Habitability_of_binary_star_systems&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: If you have a closely-bound binary star, with a planet right in between them, in the barycenter of the two stars, that's not a stable system. Any slight perturbation would pull the planet out of there, and then the net force would be pulling it further out. It would eventually fall into one of the stars. It's not self-stabilizing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't even imagine any conceivable mechanism that would put the planet in there to begin with.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2017-07-20T19:13:57.963" LastActivityDate="2017-07-20T19:13:57.963" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21733" PostTypeId="2" ParentId="21699" CreationDate="2017-07-20T18:35:21.217" Score="2" Body="&lt;p&gt;There are no currents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It all depends on the initial speed of the object, and the distance to the sun and planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's moving pretty slowly initially, it will eventually get captured in a loop orbit around some planet or maybe around the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's moving pretty fast initially, it will described a trajectory through the system and eventually leave, never to return.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is no different from what happens to asteroids, comets, the planets themselves, etc. This is all the play of gravity. This is called orbital mechanics. It's part of what is popularly known as &quot;rocket science&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an app called Universe Sandbox. Install it on your computer and play with some solar system simulations. Setup a simulation with a central star, and just one body near it. Launch the body with various speeds. See what happens. The various scenarios unfolding from there are the answer to your question.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2017-07-20T18:35:21.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21734" PostTypeId="2" ParentId="21642" CreationDate="2017-07-20T19:23:28.807" Score="2" Body="&lt;p&gt;Visible light from a torch has a 50% (blue light) to 90% (red light) chance of making it through the Earth's atmosphere, depending on the amount of dust and aerosols in the atmosphere and how close the Moon is to the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light travels in straight lines. The Moon has an angular diameter of only 0.5 degrees as seen from the Earth. Most ordinary torch beams diverge by much more than this, so only a small fraction of the photons from a torch would pass through a circle of angular diameter 0.5 degrees. Hence only a small fraction of the emitted photons would hit the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you said that the torch was pointed &quot;towards the Moon&quot;. Actually, you pointed it to where the Moon was 1.5 seconds ago, and the torch light reaches the distance of the Moon a further 1.5 seconds later. Thus the Moon has travelled 3 seconds in its orbit. However, since the orbital period (360 degrees) is a month, then it only covers an angle of 1.5 arcseconds in 3 seconds, so this shift is not relevant for your experiment (though it is for those bouncing lasers off reflectors on the Moon, which proves that light does in fact reach the Moon).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-07-20T19:23:28.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21736" PostTypeId="1" AcceptedAnswerId="21737" CreationDate="2017-07-20T23:44:00.107" Score="2" ViewCount="51" Body="&lt;p&gt;Based on atmospheric parameters, size, or distance from the star, is there any way to simulate or predict the magnetosphere of an exoplanet? Are there programs (similar to climate models for atmospheres) that can simulate magnetospheres?&lt;/p&gt;&#xA;" OwnerUserId="17584" LastActivityDate="2017-07-21T00:13:48.037" Title="Predicting/simulating magnetospheres of exoplanets?" Tags="&lt;exoplanet&gt;&lt;magnetic-field&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21737" PostTypeId="2" ParentId="21736" CreationDate="2017-07-21T00:13:48.037" Score="4" Body="&lt;p&gt;The magnetosphere is usually not generated by the atmosphere, but by the interior ferromagnetic flows of a planet. I say usually not, because Venus is an exception for the terrestrial planets in our solar system.&lt;br&gt;&#xA;But in general rocky planets will exhibit a wide range of interior flow conditions, depending on their individual composition and state of cooling.&lt;br&gt;&#xA;Gaseous planets can be a bit easier in that case, if one accepts uncertainties about the properties of metallic hydrogen.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you then simply go an assume you know all those complicated properties, then sure, you can simulate something. But it's not a scientifically testable prediction only playing around with parameters, as long as it's not tied to a specific object.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2017-07-21T00:13:48.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21738" PostTypeId="2" ParentId="21729" CreationDate="2017-07-21T00:36:25.297" Score="2" Body="&lt;p&gt;It seems clear to me that this is a homework question. I believe what the question is after is to look at the equations for distances in the universe as a function of curvature and see how this curvature affects things. Read, for example, sources like &lt;a href=&quot;http://www.astro.ufl.edu/~guzman/ast7939/projects/project01.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt; or &lt;a href=&quot;http://www.astronomy.ohio-state.edu/~dhw/A873/notes3.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with most of things in cosmology, there's lots of equations and many ways to write them. A relevant equation would be the following angular diameter distance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D_M = (1+z)^{-1}&#xA;\begin{cases}&#xA;\frac{1}{\sqrt{\Omega_k}}\sinh\left(\sqrt{\Omega_k}D_c/D_H\right), &amp;amp; k = -1 \\&#xA;D_c, &amp;amp; k = 0 \\&#xA;\frac{1}{\sqrt{\left|\Omega_k\right|}}\sin\left(\sqrt{\left|\Omega_k\right|}D_c/D_H\right), &amp;amp; k = +1 \\&#xA;\end{cases}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, you can see the three forms it takes for a closed universe ($k=-1$), a flat universe ($k=0$), and an open universe ($k=+1$). It should be pretty clear that if you chose some volume with some radius calculated by $D_A$, you'd get different volume sizes and thus different amounts of galaxies in each volume for the three universe types.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By looking at the equations and playing around with them, you should be able to convince yourself which universe type results in the largest volume.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-07-21T00:36:25.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21739" PostTypeId="1" AcceptedAnswerId="21799" CreationDate="2017-07-21T02:38:09.797" Score="2" ViewCount="119" Body="&lt;p&gt;I'm going to Nebraska for the 2017 eclipse, and am weighing my options. I'm considering whether to pack my 40mm Coronado PST or to make a Baader AstroSolar filter for my 66mm William Optics apo. I've used the PST for transits and annular eclipses past, but have never seen a total solar eclipse before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I see the &lt;a href=&quot;https://en.wikipedia.org/wiki/Baily%27s_beads&quot; rel=&quot;nofollow noreferrer&quot;&gt;diamond ring effect&lt;/a&gt; (aka &lt;em&gt;Baily's beads&lt;/em&gt;) with my PST, or should I stick with white light?&lt;/p&gt;&#xA;" OwnerUserId="17585" LastEditorDisplayName="user1569" LastEditDate="2018-06-07T07:30:02.373" LastActivityDate="2018-06-07T07:30:02.373" Title="Is the diamond ring effect visible in a 40mm hydrogen alpha telescope?" Tags="&lt;amateur-observing&gt;&lt;observational-astronomy&gt;&lt;solar-eclipse&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21745" PostTypeId="1" CreationDate="2017-07-21T12:13:38.653" Score="1" ViewCount="95" Body="&lt;p&gt;According to this,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Intergalactic_star&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Intergalactic_star&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rogue stars are theorized to be the intergalactic space because of this,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The way these stars arise is still a mystery, but the most common&#xA;  theory is that the collision of two or more galaxies can toss some&#xA;  stars out into the vast empty regions of intergalactic space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could a planet be toss from a galaxy just like it's theorized rogue stars are?&lt;/p&gt;&#xA;" OwnerUserId="11562" LastActivityDate="2017-07-21T12:13:38.653" Title="Could a planet be toss from a galaxy just like it's theorized rogue stars are?" Tags="&lt;planet&gt;&lt;intergalactic-space&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21746" PostTypeId="1" AcceptedAnswerId="21747" CreationDate="2017-07-21T13:37:52.610" Score="-1" ViewCount="205" Body="&lt;p&gt;In Friedman equation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the case k &amp;lt; 0, with a Universe containing only matter (pressure = 0)&#xA;  so that $\rho = \rho_0/ a^3$ . What is the solution a(t) in a&#xA;  situation where the final term of the Friedmann equation dominates&#xA;  over the density term? How does the density of matter vary with time?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The question from an introduction to modern cosmology from Andrew Liddle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Friedmann Equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\dot {a}/a)^2 = 8\pi G \rho / 3 - k/a^2$$ &#xA;My problem is I don't even understand the question? what does the first question mean? the answer of the book is $a(t) \alpha t$. I am not even sure that it is the only answer. Can anyone say what the question wants and is the answer of the book, the only answer?&lt;/p&gt;&#xA;" OwnerUserId="16441" LastEditorUserId="16441" LastEditDate="2017-07-21T17:47:45.757" LastActivityDate="2017-07-21T19:46:41.500" Title="Solving Friedmann equation for specific conditions" Tags="&lt;cosmology&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21747" PostTypeId="2" ParentId="21746" CreationDate="2017-07-21T14:29:41.047" Score="4" Body="&lt;p&gt;It wants to know what the general solution of the equation is in the circumstances given. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G\rho}{3} - \frac{k}{a^2}$$&#xA;$$ \left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G \rho_0}{a^3} - \frac{k}{a^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the &quot;second term dominates over the density term&quot; (which it will when $a$ is large), then&#xA;$$ \dot{a} = -k$$&#xA;(but $k$ has a negative value)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $a(t) \propto t$.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-07-21T19:46:41.500" LastActivityDate="2017-07-21T19:46:41.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21749" PostTypeId="2" ParentId="21633" CreationDate="2017-07-22T04:34:38.073" Score="6" Body="&lt;p&gt;What you're looking for is stars with &quot;High Proper Motion&quot;.  A Google search of &quot;High proper motion stars&quot; will provide a number of such lists.  A quick look at Wikipedia gives this link:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Proper_motion#Stars_with_high_proper_motion&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Proper_motion#Stars_with_high_proper_motion&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can follow the links to individual stars if you wish to determine their latitudes.&lt;/p&gt;&#xA;" OwnerUserId="17600" LastActivityDate="2017-07-22T04:34:38.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21752" PostTypeId="2" ParentId="21633" CreationDate="2017-07-22T07:51:28.350" Score="4" Body="&lt;p&gt;&lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-fsam&quot; rel=&quot;nofollow noreferrer&quot;&gt;SIMBAD&lt;/a&gt; allows you to search for stars given various criteria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-sam?Criteria=dec%20%3E%20-20%20%26%20pm%20%3E1000%20%26%20Vmag%20%3C%204&amp;amp;submit=submit%20query&amp;amp;OutputMode=LIST&amp;amp;maxObject=10000&amp;amp;CriteriaFile=&quot; rel=&quot;nofollow noreferrer&quot;&gt;searched&lt;/a&gt; for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dec &gt; -20 (easily visible from most locations in the Northern hemisphere)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vmag &amp;lt; 4 (easy naked eye objects)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;pm &gt; 1000 (more than 1 arcsecond of proper motion per year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got 8 results (including a double star counted twice)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#|       identifier        |typ|  coord1 (ICRS,J2000/2000)   |Mag U |Mag B |Mag V |Mag R |Mag I |  spec. type   |#bib|#not&#xA;-|-------------------------|---|-----------------------------|------|------|------|------|------|---------------|----|----&#xA;1|* tau Cet                |PM*|01 44 04.08338 -15 56 14.9262| 4.43 | 4.22 | 3.50 | 2.88 | 2.41 |G8V            | 963|   1&#xA;2|* alf CMa                |** |06 45 08.91728 -16 42 58.0171|-1.51 |-1.46 |-1.46 |-1.46 |-1.43 |A1V+DA         |1281|   3&#xA;3|* alf Boo                |RG*|14 15 39.67207 +19 10 56.6730| 2.46 | 1.18 |-0.05 |-1.03 |-1.68 |K1.5IIIFe-0.5  |1947|   0&#xA;4|* alf CMa A              |SB*|06 45 08.917 -16 42 58.02    |     ~|     ~|-1.09 |     ~|     ~|A1V            |1192|   0&#xA;5|* eta Cas                |SB*|00 49 06.29070 +57 48 54.6758| 4.04 | 4.02 | 3.44 | 2.94 | 2.58 |F9V+M0-V       | 544|   0&#xA;6|* alf CMi                |SB*|07 39 18.11950 +05 13 29.9552| 0.82 | 0.79 | 0.37 |-0.05 |-0.28 |F5IV-V+DQZ     |1635|   1&#xA;7|* tet UMa                |SB*|09 32 51.43390 +51 40 38.2811| 3.67 | 3.64 | 3.18 | 2.74 | 2.47 |F7V            | 353|   1&#xA;8|* gam Ser                |V* |15 56 27.18269 +15 39 41.8206| 4.31 | 4.34 | 3.84 | 3.37 | 3.13 |F6IV           | 500|   0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This includes some &quot;famous&quot; stars: Tau Cet is a nearby star that is similar to the sun. Alf CMa is Sirus. Alf Boo is Arcturus. Alf CMi is Procyon. Tet UMa (Theta Ursa Major) is circumpolar in much of the Northern hemisphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One arcsecond is a very small change. You should simulate the motion in software to understand the accuracy needed to detect such a motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be a difficulty insomuch as these stars are so bright that they will outshine any nearby stars making their relative motion harder to measure. You may be better searching for &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-sam?Criteria=dec%20%3E%2020%20%26%20Vmag%20%3C12%20%26%20Vmag%20%3E6%20%26%20pm%20%3E4000&amp;amp;submit=submit%20query&amp;amp;OutputMode=LIST&amp;amp;maxObject=10000&amp;amp;CriteriaFile=&quot; rel=&quot;nofollow noreferrer&quot;&gt;dimmer, but faster moving&lt;/a&gt; stars.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-07-22T12:01:39.137" LastActivityDate="2017-07-22T12:01:39.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21753" PostTypeId="1" AcceptedAnswerId="21756" CreationDate="2017-07-22T08:51:00.493" Score="2" ViewCount="199" Body="&lt;p&gt;The Phys.org article &lt;a href=&quot;https://phys.org/news/2017-07-mystery-space-aliens.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;'Mystery' signal from space is solved. It's not aliens&lt;/a&gt; cites a summary of the conclusion &amp;quot;by &lt;a href=&quot;http://phl.upr.edu/about/abel_mendez&quot; rel=&quot;nofollow noreferrer&quot;&gt;Abel Mendez&lt;/a&gt;, director of the Planetary Habitability Laboratory at the University of Puerto Rico at Arecibo in a blog post Friday, revealing the true nature of the signals.&amp;quot;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;After further fueling speculation by summoning the world experts in the hunt for life elsewhere in the universe—The SETI Berkeley Research Center at the University of California—the team issued its conclusion.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;We are now confident about the source of the Weird! Signal,&amp;quot; Mendez wrote.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;The best explanation is that the signals are transmissions from one or more geostationary satellites.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;The signals only appeared around Ross 128 because it is located &amp;quot;close to the celestial equator where many geostationary satellites are placed,&amp;quot; Mendez added.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The article also goes on to say:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Astronomers detected strange signals that seemed to be coming from a dwarf star about 11 light-years away, but have now determiend that the signals are interference from a distant geostationary satellite&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Mendez is further quoted the Space.com article &lt;a href=&quot;https://www.space.com/37516-strange-radio-signals-red-dwarf-star.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Weird Radio Signals Detected from Nearby Red Dwarf Star&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Each of these hypotheses has its issues, he said. For example, solar flares of the type that could be responsible generally occur at lower frequencies. In addition, Mendez wrote, there aren't a lot of other objects in the Ross 128 field of view, &amp;quot;and we have never seen satellites emit bursts like that.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;and in the Space.com article &lt;a href=&quot;https://www.space.com/37579-weird-radio-signal-ross-128-star-satellite.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Not Aliens: Weird Radio Signal from Star Likely Has Duller Explanation&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;This explains why the signals were within the satellite’s frequencies and only appeared and persisted in Ross 128; the star is close to the celestial equator, where many geostationary satellites are placed,&amp;quot; Mendez added. &amp;quot;This fact, though, does not yet explain the strong dispersion-like features of the signals (diagonal lines in the figure); however, it is possible that multiple reflections caused these distortions, but we will need more time to explore this and other possibilities.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Wikipedia lists the star &lt;a href=&quot;https://en.wikipedia.org/wiki/Ross_128&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ross 128&lt;/a&gt; at at declination of about &lt;code&gt;+0° 48'&lt;/code&gt;, and the latitude of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Arecibo_Observatory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arecibo telescope&lt;/a&gt; to be about &lt;code&gt;+18° 21'&lt;/code&gt;. Satellites in &lt;em&gt;geostationary&lt;/em&gt; orbits are in a circular ring approximately coincident with the equator at a distance from the center of the Earth of only about 42,000 km. From the Northern hemisphere, geostationary satellites will appear &lt;em&gt;several degrees south in declination&lt;/em&gt;. (See this &lt;a href=&quot;https://space.stackexchange.com/a/12409/12102&quot;&gt;interesting answer&lt;/a&gt;.)&lt;/p&gt;&#xA;&lt;p&gt;For Arecibo, I calculate about &lt;code&gt;-3.2°&lt;/code&gt; declination.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;QUESTION:&lt;/strong&gt; How could a radio telescope have picked up satellite signals from a satellite at about &lt;code&gt;-3.2°&lt;/code&gt; declination while observing a star at &lt;code&gt;+0.8°&lt;/code&gt; declination? And why wasn't this immediately ruled out - surely radio telescopists know about satellites by now, especially the crowded ring of geostationary satellites! Aren't there standard procedures to rule these out before making announcements?&lt;/p&gt;&#xA;&lt;p&gt;Could the half-angle acceptance really have been 4 degrees? Since the &lt;s&gt;brown&lt;/s&gt; &lt;strong&gt;red&lt;/strong&gt; dwarf is extremely close to the solar system, could proper motion put it 4 degrees away from the value listed in Wikipedia?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JGG5x.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JGG5x.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lS0Xy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lS0Xy.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; &amp;quot;The signal that seemed to emanate from the red dwarf star Ross 128, as detected by the Arecibo Observatory in May 2017 (enclosed in the red frame).&amp;quot; Credit: PHL @ UPR Arecibo  From &lt;a href=&quot;https://space.stackexchange.com/a/12409/12102&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-07-24T20:41:36.043" Title="&quot;Mysterious radio signals&quot; - how could a geostationary satellite and Ross-128 line up with Arecibo?" Tags="&lt;radio-astronomy&gt;&lt;artificial-satellite&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21756" PostTypeId="2" ParentId="21753" CreationDate="2017-07-22T09:57:53.703" Score="2" Body="&lt;p&gt;The original team suggested three main possibilities: &quot;(1) unusual stellar activity, (2) emissions from other background objects, or (3) interference from satellite communications&quot;  Clearly radio astronomers know about satellites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They noted &quot;in the absence of solid information about the signal, most astronomers would think that [radio interference or instrumental failures] would probably be the most likely explanation.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The satellite hypothesis has some problems it &quot;does not yet explain the strong dispersion-like features of the signals (diagonal lines in the [weird] figure [in the question])&quot; But on a balance of probablities this is the most likely explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are satellites in the region of Ross 128:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rX0Bv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rX0Bv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Location of Geostationary satellites operating between 4-8 GHz in the same region of the sky as Ross 128 (yellow dot).&lt;br&gt;&#xA;Credit: Enriquez et al. (SETI Berkeley), &lt;a href=&quot;http://seti.berkeley.edu/ross128.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://seti.berkeley.edu/ross128.pdf&lt;/a&gt;&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your analysis of the location of a satellite, based on the geometry of satellite orbiting over the equator is not supported by this image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The observers did not claim to have received a SETI signal. Indeed if you look at their &lt;a href=&quot;http://phl.upr.edu/library/notes/ross128&quot; rel=&quot;nofollow noreferrer&quot;&gt;original report&lt;/a&gt; they note that &quot;aliens&quot; is at the bottom of their possible explanations. The signal looks different from typical satellite bursts. They hoped the signal was astronomical in nature so there may have been a little wishful thinking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the most likely explanation is that we are observing multiple reflections from geostationary satellites transmitting on the 4-8 GHz band.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All quotations from the &lt;a href=&quot;http://phl.upr.edu/press-releases/theweirdsignal?utm_content=buffer0a5d3&amp;amp;utm_medium=social&amp;amp;utm_source=twitter.com&amp;amp;utm_campaign=buffer&quot; rel=&quot;nofollow noreferrer&quot;&gt;press release&lt;/a&gt; from the planetary habitability observatory.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-07-24T20:41:36.043" LastActivityDate="2017-07-24T20:41:36.043" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21758" PostTypeId="2" ParentId="21660" CreationDate="2017-07-22T15:28:11.567" Score="3" Body="&lt;p&gt;Here's some pertinent facts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparent magnitudes: Sun from Ganymede     = -23.11&#xA;                     Jupiter from Ganymede = -16.02 (full)&#xA;                     Difference            = - 7.09&#xA;                     Brightness difference = 2.512 E7.09 = 686 times brighter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth comparison:    Sun from Earth        = -26.74&#xA;                     Moon from Earth       = -12.74 (mean full)&#xA;                     Full moon/ Jupiter from Ganymede difference = 20.5 times&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ganymede facts:      Orbital period = 1 Ganymede day: 7.16 days&#xA;                     Orbital inclinations: .2 deg to Jupiter equator&#xA;                                          2.2 deg to ecliptic&#xA;                     Orbital eccentricity: .0011 (circular)&#xA;                     Orbital speed (ave.): 10.88 km/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter facts:       Mean radius: about 70000 km&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summary:&#xA;If you are on the &quot;far side&quot; of Ganymede you won't see Jupiter (maybe a glow on the horizon from Ganymede's thin atmosphere). The sun will be 2.512 E3.63 = about 28 times dimmer than seen from Earth (which is still pretty darn bright). A day will last 7.16 Earth days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The near side will be far more interesting. The sun will go through the same cycle at the same brightness as on the far side. However, Jupiter will &quot;hover&quot; in the sky (maybe a little wobble/ libration) and when full will be about 686 times dimmer than the sun in the sky. Jupiter will go through phases over the course of the 7.16 day cycle and be less bright of course as it approaches eclipse. The eclipse (by rough calculations) will last about three and a half hours. The planet will have an angular diameter of about 7.5 degrees (compare with 0.5 for sun/moon from Earth) and be about 20.5 times brighter than a full moon when Ganymede is between the sun and Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fun side note is that the sky won't be totally dark during eclipse. Light from the sun will reflect off aerosols and hazes in Jupiter's upper atmosphere making Jupiter &quot;glow&quot; in Ganymede's sky.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2017-07-22T15:28:11.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21759" PostTypeId="1" AcceptedAnswerId="21762" CreationDate="2017-07-23T02:54:31.617" Score="2" ViewCount="98" Body="&lt;p&gt;So I live in a suburb in Victoria, Australia. Less than an hour away from the city and I guess there is a bit of light pollution because from my backyard I can probably only see about 15 - 20 stars (probably less), I'm wondering what these stars are, and what I'll see when I get this telescope: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.opticscentral.com.au/saxon-707az2-refractor-telescope.html?___SID=U#.WXQNMtN940r&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.opticscentral.com.au/saxon-707az2-refractor-telescope.html?___SID=U#.WXQNMtN940r&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is going to sound stupid but how do I know when there are planets in the sky that I can see? I don't think I've ever actually seen a planet other than the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="17501" LastActivityDate="2017-07-23T10:02:34.760" Title="How do I know what I'll be able to see?" Tags="&lt;star&gt;&lt;planet&gt;&lt;telescope&gt;&lt;light&gt;&lt;space&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21760" PostTypeId="1" AcceptedAnswerId="21787" CreationDate="2017-07-23T08:56:25.460" Score="2" ViewCount="222" Body="&lt;p&gt;This &lt;a href=&quot;https://www.youtube.com/watch?v=5GIjfkFucSs&quot; rel=&quot;nofollow noreferrer&quot;&gt;nice video&lt;/a&gt; musing about (among other things) the US Naval Observatory's $3,500 expedition to &lt;a href=&quot;https://en.wikipedia.org/wiki/Baker_City,_Oregon&quot; rel=&quot;nofollow noreferrer&quot;&gt;Baker City&lt;/a&gt; or at least &lt;a href=&quot;https://en.wikipedia.org/wiki/Baker_County,_Oregon&quot; rel=&quot;nofollow noreferrer&quot;&gt;Baker County&lt;/a&gt; Oregon to observe the &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_eclipse_of_March_29,_2006&quot; rel=&quot;nofollow noreferrer&quot;&gt;June 18, 1918 solar eclipse&lt;/a&gt;, there is a photograph of the expedition showing both normal-sized portable telescopes, and in the background a very long telescope (at least 6 meters, can't see the end) held up by a ladder, with an &lt;em&gt;even longer&lt;/em&gt; tent extending backward from the shack sitting at the bottom of the telescope along the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is this instrument? How does it function optically, and what kind of data does it produce?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; Screenshot from the &lt;a href=&quot;https://www.youtube.com/watch?v=5GIjfkFucSs&quot; rel=&quot;nofollow noreferrer&quot;&gt;video&lt;/a&gt;, photo credits available at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/meZA6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/meZA6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-07-23T09:06:12.163" LastActivityDate="2017-07-24T18:12:21.550" Title="What is the large astronomical instrument shown in this 1918 eclipse expedition photo?" Tags="&lt;telescope&gt;&lt;solar-eclipse&gt;&lt;instruments&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21761" PostTypeId="1" CreationDate="2017-07-23T09:12:43.587" Score="0" ViewCount="158" Body="&lt;p&gt;I would like to know how many known quasars there are in the Universe as well as x-ray binary black holes (not Neutron stars) but I don't seem to be able to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also want to know how many black holes have been found via gravitational wave detections but I think the answer to this question is 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you guys know any credible sources that list the objects in the sky via category and gives detailed info of how we found them (x-ray binary or gravitational waves or quasars).&lt;/p&gt;&#xA;" OwnerUserId="17618" LastActivityDate="2017-07-23T17:38:36.403" Title="Is there an official list of objects in the sky?" Tags="&lt;black-hole&gt;&lt;gravitational-waves&gt;&lt;quasars&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21762" PostTypeId="2" ParentId="21759" CreationDate="2017-07-23T10:02:34.760" Score="3" Body="&lt;p&gt;Firstly, if you're planet spotting, don't worry too much about light pollution. The planets are some of the brightest objects in the sky and some (especially Jupiter) can easily be observed even with a full Moon - the full Moon (along with the Sun!) is the biggest contributor to light pollution!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at the list of brightest stars ( &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_brightest_stars&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/List_of_brightest_stars&lt;/a&gt; ), which also contains estimates for the brightness of the Sun, Moon and major planets. There aren't typically any stars brighter from Earth than Venus, Jupiter, Mars and Mercury and precious few brighter than Saturn. I'm going to suggest you probably have seen many of the planets - but just didn't recognise them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;+1 for Stellarium ( &lt;a href=&quot;http://www.stellarium.org/en_GB/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.stellarium.org/en_GB/&lt;/a&gt; ). It's free, intuitive and very visual to use. You can put in your local viewing location and it gives you a view for any time of the night, future or past. At the time of writing (23 July 2017), Saturn and Jupiter should be looking good for the Southern Hemisphere. This rotates throughout the year, and Stellarium will help with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do a web search for &quot;the sky at night in the southern hemisphere&quot; and you'll find a number of examples of websites with highlights to look for when you get out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Couple of final suggestions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Do allow plenty of time to let your eyes adjust to the darkness - it takes me a good 20-30 minutes to get going. You'll see many more stars then&lt;/li&gt;&#xA;&lt;li&gt;Do be aware that local conditions (houses / trees etc) might restrict visibility of things near the horizon. In any case, you find items over-head are much clearer&lt;/li&gt;&#xA;&lt;li&gt;Refractors (yours included) come with an 'erecting prism'. Despite the name, the real benefit isn't that you see the image the right way up (that isn't really an issue in astronomy), the prism just allows you a more sensible and comfortable viewing position - you'll understand when you get going&lt;/li&gt;&#xA;&lt;li&gt;Don't restrict yourself to planets. When you get going, learn about the deep sky objects too. For instance, the southern hemisphere has access to the famous Magellenic Clouds which we cannot see from the northern hemisphere!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17502" LastActivityDate="2017-07-23T10:02:34.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21763" PostTypeId="2" ParentId="21761" CreationDate="2017-07-23T10:44:16.687" Score="0" Body="&lt;p&gt;There are actually numerous lists, categorised in all sorts of ways. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on what level you are approaching this, I'd suggest &lt;a href=&quot;https://en.wikipedia.org/wiki/Lists_of_astronomical_objects&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia's list&lt;/a&gt; may be your best entry point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want raw data, the &lt;a href=&quot;https://heasarc.gsfc.nasa.gov/W3Browse/all/ngc2000.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NGC 2000 list&lt;/a&gt; is the way to go.&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="43" LastEditDate="2017-07-23T17:38:36.403" LastActivityDate="2017-07-23T17:38:36.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21764" PostTypeId="1" AcceptedAnswerId="21786" CreationDate="2017-07-23T11:04:37.590" Score="1" ViewCount="979" Body="&lt;p&gt;I ran &lt;a href=&quot;https://www.aanda.org/articles/aa/abs/2013/09/aa21843-13/aa21843-13.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;VSOP2013&lt;/a&gt; in Fortran, and I now have the following position $x, y, z$ as well as velocity $v_x, v_y, v_z$ data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I convert the position to &lt;a href=&quot;https://en.wikipedia.org/wiki/Ecliptic_coordinate_system#Spherical_coordinates&quot; rel=&quot;nofollow noreferrer&quot;&gt;heliocentric ecliptic spherical coordinates&lt;/a&gt; $l, b, r$? &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;PLANETARY EPHEMERIS VSOP2013  EMB     &#xA;Ecliptic   Heliocentric Coordinates:  X,Y,Z (au)  X',Y',Z' (au/d)  - Dynamical Frame J2000&#xA;&#xA;Julian Day 2415021.0   &#xA;-0.2054467990    0.9615310525    0.0002141233   -0.0171063024   -0.0036576954   -0.0000011815&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="17578" LastEditorUserId="7982" LastEditDate="2017-07-24T19:18:22.330" LastActivityDate="2017-07-24T19:18:22.330" Title="How to convert VSOP's xyz coordinates to heliocentric spherical lbr coordinates?" Tags="&lt;the-sun&gt;&lt;algorithm&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21765" PostTypeId="2" ParentId="21761" CreationDate="2017-07-23T13:45:42.570" Score="1" Body="&lt;p&gt;Lists of astronomical objects are called &quot;catalogues&quot; and other answers have already mentioned the NGC catalogue of galaxies and nebulae. However, you asked about quasars and black hole x-ray binaries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For quasars, there are several catalogues. None claim to be complete, as new discoveries happen all the time. The &lt;a href=&quot;https://heasarc.gsfc.nasa.gov/w3browse/all/milliquas.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;MILLIQUAS&lt;/a&gt; catalogue is a compendium of 460,222 type-I QSOs and AGN. It also contains about a million &quot;Quasar candidates&quot;. (sources that may be quasars, 80% confidence)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Black hole binaries, the difficulty is proving that the object is indeed a black hole. This usually requires calculating the mass of the accreting object, which is &lt;a href=&quot;https://arxiv.org/abs/astro-ph/9802177&quot; rel=&quot;nofollow noreferrer&quot;&gt;not always easy&lt;/a&gt;. The &lt;a href=&quot;https://heasarc.gsfc.nasa.gov/w3browse/all/xray.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;X-ray catalogues&lt;/a&gt; are classified by the mass of the normal stellar component. &lt;a href=&quot;https://heasarc.gsfc.nasa.gov/W3Browse/all/hmxbcat.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The High Mass x-ray binary catalogue&lt;/a&gt; notes objects that are &quot;black hole candidates&quot;. It is in no way a definitive list, since no such list exists.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-07-23T13:45:42.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21767" PostTypeId="1" CreationDate="2017-07-23T19:25:46.880" Score="3" ViewCount="112" Body="&lt;p&gt;Hello fellow amateur and professional astronomers,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On July July 17, 2017, I was at an observing session in Avrig, Sibiu county, Romania (45.7288° N, 24.3784° E). At approx 11:30 pm I pointed my green laser pointer (class III, wavelength 532 nm +/-10, YL-Laser 303) towards Arcturus in order to help my colleague align the SW AZEQ 5 mount within the 3-star procedure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the laser beam was on, at approx 1 arcminute beneath and to the left from Arcturus, with my naked eye, I observed a point of light (white/blueish) lighting up to the point which I estimate to be close to the visual magnitude of Jupiter or slightly higher for appprox 5 seconds, and then the &quot;flare&quot; faded out to black. As there was no &quot;regular&quot; star-like remanant, I was left to conclude that the initial flare did not originate from a star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ruled out a reflection of the laser beam from a satelite (as too far for my laser's power), a plane or a bird (due to the white/blueish light I saw, its fixed position and apparent distance to my eye), as well the possibility of a meteorite burst (as it had no apparent trajectory, but it appeared as a fixed point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am so intrigued and curious as to what possible optics or physics could have been involved and explanations or ideas you may have to explain the phenomenon? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I searched online for others' possible similar observations without success. &lt;/p&gt;&#xA;" OwnerUserId="17624" LastEditorUserId="2365" LastEditDate="2017-07-23T21:00:30.397" LastActivityDate="2017-07-24T02:40:27.013" Title="5 sec flare near Arcturus" Tags="&lt;star&gt;&lt;flare&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21769" PostTypeId="2" ParentId="16450" CreationDate="2017-07-24T00:23:08.853" Score="0" Body="&lt;p&gt;I inputed James K's value into my game for the lighting for my earth lit moon level, and here is what it looks like:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/tlRpd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tlRpd.jpg&quot; alt=&quot;Moon base lit by Earth&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd say this is quite beautiful!  I realize it would probably be dark enough that we wouldn't see it quite that way, but the robot you control in Planetary Settlers has an enhanced camera to adjust to dark or bright environments.  Note that the land is dark grey, but is lit turquoise by the lighting.  In this view, you are standing in a sand garden, so the color of that land appears a bit different (starts off light yellow, lit by turquoise light).  A big thank you for all of your inputs.  If you are interested in learning more about the game I am making, I just made my website for it here: &lt;a href=&quot;https://jonathanlcopeland.wixsite.com/planetary&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://jonathanlcopeland.wixsite.com/planetary&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you all for your wonderful answers!&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2017-07-24T00:23:08.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21770" PostTypeId="1" AcceptedAnswerId="32975" CreationDate="2017-07-24T01:11:39.960" Score="3" ViewCount="92" Body="&lt;p&gt;If we were to locate a pulsar in a close elliptical orbit around a black hole, could this reveal significant information about the black hole (e.g. time dilation / structure)?  Could this even tell us the structure of a black hole (e.g. if it is a singularity, or a sphere formed around the event horizon, or something else)?&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2019-08-09T10:47:22.633" Title="Could a closely orbiting pulsar be used to study a black hole?" Tags="&lt;orbit&gt;&lt;black-hole&gt;&lt;pulsar&gt;&lt;time-dilation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21772" PostTypeId="2" ParentId="21767" CreationDate="2017-07-24T02:40:27.013" Score="3" Body="&lt;p&gt;My first thought is that you saw a flare from a satellite, meaning a favorable reflection off of a solar panel or something. I use the software &quot;Heavensat&quot;, and it shows the satellite Nextsat (Norad number 30774) passed very close to Arcturus at 23:40 pm from your location (45.7288° N, 24.3784° E). Note that I am assuming that your time is 3 hours ahead of UT (Universal Time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EiQfL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EiQfL.jpg&quot; alt=&quot;Path of Nextsat on July 17 from reference location&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know if Nextsat typically flares, although reflections from satellites are common. Nextsat is in a low altitude orbit, so its motion would normally be detectible. Perhaps the duration of the brightening was shorter than you estimate (so that the motion was not detectible), or it was so unexpected that you did not notice the motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second thought would be a head on meteor like you suggested.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-07-24T02:40:27.013" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21776" PostTypeId="2" ParentId="20466" CreationDate="2017-07-24T05:54:39.303" Score="4" Body="&lt;p&gt;Several excellent answers have already been given. But no one has talked about logarithmic perception. (&lt;a href=&quot;https://en.wikipedia.org/wiki/Weber%E2%80%93Fechner_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Weber%E2%80%93Fechner_law&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;p&gt;We perceive everything logarithmic-ally. For humans, the difference between &lt;span class=&quot;math-container&quot;&gt;$10 metres$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$100 metres$&lt;/span&gt; is the same as between &lt;span class=&quot;math-container&quot;&gt;$100 metres$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$1 km$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/S5zcj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S5zcj.png&quot; alt=&quot;Weber-Fechner Law&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;An illustration of the Weber–Fechner law. On each side, the lower square contains 10 more dots than the upper one. However the perception is different: On the left side, the difference between upper and lower square is clearly visible. On the right side, both squares look almost the same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hence it is much better to measure distance on astronomical scales in &lt;strong&gt;parsecs&lt;/strong&gt; than &lt;strong&gt;metres&lt;/strong&gt; because humans understand the difference between &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; parsecs better than they would do if the same data was presented in meters.&lt;/p&gt;&#xA;" OwnerUserId="15901" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-07-24T05:54:39.303" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21779" PostTypeId="1" AcceptedAnswerId="21784" CreationDate="2017-07-24T08:42:13.807" Score="0" ViewCount="55" Body="&lt;p&gt;What are the most popular packages used to create graphs for astronomy journal papers?&#xA;Looking at various journals, the styles of the graphs appear to be quite uniform but I can't identify the package(s) that create them.&lt;/p&gt;&#xA;" OwnerUserId="16414" LastEditorUserId="16414" LastEditDate="2017-07-24T09:17:19.363" LastActivityDate="2017-07-24T11:56:46.543" Title="Graphics packages for publishing" Tags="&lt;software&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="0" ClosedDate="2017-07-25T14:37:30.990" ContentLicense="CC BY-SA 3.0" />
  <row Id="21782" PostTypeId="1" CreationDate="2017-07-24T11:22:43.490" Score="3" ViewCount="655" Body="&lt;p&gt;I'm currently trying to learn about how adaptive optics correct the blurriness caused by atmospherical seeing effects. This is my current understanding of how adaptive optics works (trying to offer some context, I have a very limited understanding at the moment). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wavefronts from celestial bodies become distorted as they pass through the Earth's atmosphere due to variations in refraction indices as they pass through different density air, temperature variations and so on. This distortion as far as I know, causes the wavefront to look squiggly instead of being straight parallel lines. Once the light reaches the telescope, it is reflected off a tip tilt mirror, then reflected off a deformable mirror, then a beam splitter splits the light rays and one half is passed through a computer sampling device which detects distortions in the wavefront. This is done up to a 1000 times per second, and somehow the computer causes actuators fitted at the back of the deformable mirror to move such that the reflected light off the deformable mirror is now undistorted (i.e. the wavefronts are a straight line, not squiggly). Thus adaptive optics fixes the distortion issue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a few questions arising from my current understanding:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What exactly does the tip tilt mirror do to help fix the distortion problem? Wiki simply says that it fixes the problem of aberration but doesn't offer a specific explanation of how. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why are straight wavefronts ideal when viewing celestial bodies? As in, how does the squiggliness of the wavefronts actually make the images look blurry?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I believe I am asking too many different types of questions in one post, so I'll make a separate post for this last question, but for anyone reading this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I keep reading that increasing the size of the primary mirror will improve resolution. I have seen justification of this using Dawe's limit such as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;By considering the formula R= 2.1*10^5*wavelength/ D, where R is the minimum angular separation required by two point sources for them to be resolved, so as primary mirror diameter increases, D increases, so R decreases meaning that resolution of the telescope is improved.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't really see how they can apply the R=... formula to primary mirrors because I have only seen it derived in the context of light diffracting when it passes through a small slit, and not mirrors (which don't allow light to pass through, but rather reflects it). Can someone explain to me why we can apply this formula to mirrors as well? &lt;/p&gt;&#xA;" OwnerUserId="17639" LastEditorDisplayName="user1569" LastEditDate="2017-07-24T12:56:37.187" LastActivityDate="2018-06-03T17:26:37.090" Title="How do tip tilt mirrors correct distortion in adaptive optics?" Tags="&lt;astrophysics&gt;&lt;optics&gt;&lt;adaptive-optics&gt;&lt;mirror&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21783" PostTypeId="1" CreationDate="2017-07-24T11:38:13.033" Score="4" ViewCount="157" Body="&lt;p&gt;One of the commenters on my answer to &lt;a href=&quot;https://worldbuilding.stackexchange.com/a/86733/3460&quot;&gt;https://worldbuilding.stackexchange.com/a/86733/3460&lt;/a&gt; questions that the blast would even have a noticeable impact if that supernova was in Alpha Centauri. And since my answer is just based on my intuition, I would like to figure that out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So radiation put aside, what would the blast of &lt;a href=&quot;https://en.wikipedia.org/wiki/SN_1006&quot; rel=&quot;nofollow noreferrer&quot;&gt;SN 1006&lt;/a&gt; be like for earth when arriving, supposed it went supernova just 4 light-years away?&lt;/p&gt;&#xA;" OwnerUserId="10102" LastEditorUserId="10102" LastEditDate="2017-07-24T12:54:30.243" LastActivityDate="2017-07-24T19:32:16.147" Title="Supposed SN 1006 was only 4 lightyears away what would the blasts impact be like on earth?" Tags="&lt;supernova&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21784" PostTypeId="2" ParentId="21779" CreationDate="2017-07-24T11:56:46.543" Score="3" Body="&lt;p&gt;The most commonly used packages for generating graphics in astronomy are probably&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/IDL_(programming_language)&quot; rel=&quot;noreferrer&quot;&gt;IDL&lt;/a&gt; -- this has long been the most popular, in part because there is a lot of astronomical data reduction and analysis code written in IDL.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/SM_(plotting)&quot; rel=&quot;noreferrer&quot;&gt;SuperMongo&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Python package &lt;a href=&quot;https://matplotlib.org&quot; rel=&quot;noreferrer&quot;&gt;matplotlib&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The apparently uniform style you refer to is probably a historical combination of IDL and SuperMongo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matplotlib (and other Python packages that act as wrappers around it) is on a trajectory to displace the others; I strongly suspect it is already much more common than SuperMongo, and may be roughly as common as IDL by now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will sometimes see figures produced by the &lt;a href=&quot;https://en.wikipedia.org/wiki/PLplot&quot; rel=&quot;noreferrer&quot;&gt;PLplot library&lt;/a&gt; as output from programs written&#xA;in Fortran, C/C++, etc. Plots produced using R are also starting to appear.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2017-07-24T11:56:46.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21786" PostTypeId="2" ParentId="21764" CreationDate="2017-07-24T17:55:35.923" Score="1" Body="&lt;p&gt;Looking at the numbers and comparing to &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL Horizons&lt;/a&gt;, the best match is to the position of the Earth-Moon Barycenter in heliocentric coordinates. That's of course consistent with the data being labeled as:  &lt;code&gt;EMB, Ecliptic Heliocentric Coordinates&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can live with some small uncertainties due to the details of how coordinate transforms may evolve over time, &lt;em&gt;I believe&lt;/em&gt; you can just apply some simple trigonometry. You can confirm &lt;a href=&quot;https://en.wikipedia.org/wiki/Ecliptic_coordinate_system#Rectangular_coordinates&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r = \sqrt{x^2 + y^2 + z^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$l = tan^{-1}(y/x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b = sin^{-1}(z/r)$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QGGmq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QGGmq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Data in Question: for JD 2415021.0&#xA;position (AU):     [-0.2054467990,  0.9615310525,  0.0002141233]&#xA;velocity (AU/day): [-0.0171063024, -0.0036576954, -0.0000011815]&#xA;&#xA;JPL Horizions Earth-Moon barycenter J2000 Heliocentric for JD 2415021.0&#xA;position (AU):     [-0.2054465857,  0.9615310983,  0.00021393097] &#xA;velocity (AU/day): [-0.0171063032, -0.0036576916, -0.00000118253]&#xA;&#xA;JPL Horizions Earth J2000 Heliocentric for JD 2415021.0&#xA;position (AU):     [-0.2054522688,  0.9615603123,  0.00021303115] &#xA;velocity (AU/day): [-0.0171136285, -0.0036594066, -0.00000182603]&#xA;&#xA;JPL Horizions Earth J2000 Solar System barycenter for JD 2415021.0&#xA;position (AU):     [-0.2022720210,  0.9679285296,  0.00010939456]&#xA;velocity (AU/day): [-0.0171209826, -0.0036556277, -0.00000165111]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-07-24T18:03:29.427" LastActivityDate="2017-07-24T18:03:29.427" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21787" PostTypeId="2" ParentId="21760" CreationDate="2017-07-24T18:12:21.550" Score="3" Body="&lt;p&gt;This construct was a telescope. Citing &lt;a href=&quot;http://articles.adsabs.harvard.edu//full/1921PMcCO...2..165M/0000198.000.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Owing to the difficulty of erecting a double tower, most other astronomers follow the mechanically simpler plan of laying the camera tube horizontally and allowing the sunlight to be fed into it by means of a plane mirror driven by clockwork to counteract the westward motion of the sun.&lt;br&gt;&lt;sub&gt;pg. 255&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1919PA.....27....1H&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Popular Astronomy article&lt;/a&gt; also discusses the eclipse observation in detail. They describe the primary telescopes used for photography. The first was a 65-foot telescope (that measurement being the length, not the aperture).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This instrument was in charge of W. A. Conrad, who was assisted by W. L. Veale. The camera was horizontal and was directed toward the sunrise point. Light from the corona was reflected from a silvered glass mirror 10 inches in diameter attached to a coelostat controlled by clockwork. The lens was 7 1/2 inches in aperture and the plates were double coated special coronal, 17x20 inches, made by the Eastman Kodak Co. The diameter of the sun's image on the plates is about 7 1/4 inches. Five exposures were made of 5, 15, 35, 15, and 5 seconds duration respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They also had a 105-inch telescope which was also laid horizontally and had the light directed into it via a coelostat. While I can't be completely sure which telescope is pictured in that image, given the size, my guess would be the 65-foot telescope.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-07-24T18:12:21.550" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21789" PostTypeId="2" ParentId="21783" CreationDate="2017-07-24T19:32:16.147" Score="2" Body="&lt;p&gt;Last week, a paper was posted on arxiv (&lt;a href=&quot;https://arxiv.org/abs/1707.04253v1&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/1707.04253v1&lt;/a&gt;) which studies the &quot;The Resilience of Life to Astrophysical Events&quot;, in particular it deals with the temperature change of oceans (as boiling oceans would evaporate the water of a planet which seems to be bad for biological life forms...). For a assumed SN explosion of Proxima Centauri the temperature change of Earth oceans would only be 0.1 degree Celsius - not enough to extinct life by boiling the oceans.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TWqE7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TWqE7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(extract from Sloan, Batista and Loeb, 2017)&lt;/p&gt;&#xA;" OwnerUserId="950" LastActivityDate="2017-07-24T19:32:16.147" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21791" PostTypeId="1" CreationDate="2017-07-24T21:51:35.100" Score="6" ViewCount="257" Body="&lt;p&gt;I'm wondering how our galaxy is tilted with respect to the Andromeda galaxy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seen from the Andromeda galaxy, what would our galaxy look like?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it appear as a spiral, or would it be seen edge-on, and appear long and thin?&lt;/p&gt;&#xA;" OwnerUserId="13290" LastEditorUserId="1056" LastEditDate="2017-07-31T21:45:46.403" LastActivityDate="2017-07-31T21:45:46.403" Title="Would our galaxy appear as a spiral when seen from the Andromeda galaxy?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21792" PostTypeId="2" ParentId="21791" CreationDate="2017-07-24T21:59:47.127" Score="6" Body="&lt;p&gt;The &quot;Galactic coordinates&quot; of the Andromeda galaxy are l=121 degrees, b=-22 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means the Andromeda galaxy is seen by us at an angle of 22 degrees to the plane of the Milky Way galaxy. Thus if you were at the Andromeda galaxy looking at the Milky Way, it would be neither face-on or edge on, but a bit closer to the latter than the former. You would still be able to see that the Milky Way had a spiral structure, because the plane of the Milky Way is much thinner than its diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB. This simple argument works because the Andromeda galaxy is more than ten times further away than the diameter of the Milky Way galaxy.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-07-24T21:59:47.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21793" PostTypeId="1" AcceptedAnswerId="21797" CreationDate="2017-07-24T22:36:51.033" Score="1" ViewCount="135" Body="&lt;p&gt;In a story I'm working on, a huge, 20 by 14 kilometer, Ni-Fe asteroid attains a 1760 km. orbit around earth (why/how? Long story). I figure it has an H magnitude of ~12. Would this body be visible naked-eye from the surface during the day?&lt;/p&gt;&#xA;" OwnerUserId="15814" LastActivityDate="2017-07-25T02:30:20.093" Title="Asteroid visible during the day?" Tags="&lt;asteroids&gt;&lt;apparent-magnitude&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21795" PostTypeId="2" ParentId="11554" CreationDate="2017-07-24T23:38:59.183" Score="1" Body="&lt;p&gt;It's sort of like a pool and water: a big empty pool is space, and water is air; if you pour one drop of water in the pool, that's not going to fill it up. Instead, the water tries to fill the pool evenly. Hope this helps/helped!&lt;/p&gt;&#xA;" OwnerUserId="17648" LastEditorUserId="69" LastEditDate="2017-07-25T00:35:14.677" LastActivityDate="2017-07-25T00:35:14.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21796" PostTypeId="1" CreationDate="2017-07-25T02:21:34.847" Score="0" ViewCount="551" Body="&lt;p&gt;I heard something(long time ago) that Milky Way can be seen in South hemisphere only, is it true? If so, what is the reason?&lt;/p&gt;&#xA;" OwnerUserId="8049" LastActivityDate="2017-07-25T05:03:12.090" Title="Can Milky Way be seen in South hemisphere only?" Tags="&lt;observational-astronomy&gt;&lt;milky-way&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21797" PostTypeId="2" ParentId="21793" CreationDate="2017-07-25T02:30:20.093" Score="5" Body="&lt;p&gt;The key to answering your question is to know the albedo of the asteroid compared to the Moon. The Moon is rather dark (reflecting 7% to 12% of the light if I remember correctly, about as black as worn asphalt). I think, but do not know with certainty, that the asteroid would not be much darker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the dimensions that you give, the asteroid would be approximately the same apparent size as the moon: 0.45 x 0.65° compared to the Moon's 0.5°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So both have approximately the same albedo and approximately the same size. Therefore I say that the asteroid would be visible in the daylight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also this is a sci-fi story, correct? If you can get the asteroid into that orbit, then anything is possible. :-)&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-07-25T02:30:20.093" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21798" PostTypeId="2" ParentId="21796" CreationDate="2017-07-25T02:32:58.840" Score="5" Body="&lt;p&gt;That is not true. The band of the Milky Way can be seen from any latitude on the planet. It just requires the right time of year and a dark sky.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-07-25T02:32:58.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21799" PostTypeId="2" ParentId="21739" CreationDate="2017-07-25T02:49:58.760" Score="1" Body="&lt;p&gt;Since no one else has answered, I will provide my opinion. Let me mention that I have not seen a total eclipse since 1991, so I do not remember the diamond ring affect explicitly, and my notes do not mention it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the diamond ring can only be seen with the naked eye. It consists of the inner corona on one side of the moon and the last bit of photosphere on the opposite side of the moon. I believe that the inner corona is not visible in any type of filter. That is why I am speculating that the diamond ring is a naked eye phenomena only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WARNING The diamond ring occurs as the partial phase ends and totality begins, and then when totality ends and the partial phase begins. Of course you need to be extremely careful when looking at the partial eclipse with the naked eye. Essentially, you cannot look! It would be better to use a hand held filter to block the photosphere while looking at the corona without the filter.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-07-25T02:49:58.760" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21801" PostTypeId="2" ParentId="21796" CreationDate="2017-07-25T05:03:12.090" Score="0" Body="&lt;p&gt;The band of milky way is inclined by more than $60\deg$ to the celestial equator. Hence, it can be &lt;strong&gt;easily seen from the southern hemisphere.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, &lt;strong&gt;the milky way center lies in Sagittarius (a southern hemisphere constellation)&lt;/strong&gt;, which means, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The Milky Way would look better in the Southern Hemisphere.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15901" LastActivityDate="2017-07-25T05:03:12.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21804" PostTypeId="1" AcceptedAnswerId="21806" CreationDate="2017-07-25T09:08:16.313" Score="6" ViewCount="1336" Body="&lt;p&gt;A black dwarf is a hypothetical object that is the end result of the cooling of a white dwarf. None yet exist, because there hasn't been enough time in the age of the universe for them to cool down sufficiently. The oldest known white dwarfs are at ~3800 K and would glow orange. It will take many billions of years before any are cool enough (~750 K I believe) to not emit light visible to the naked eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what would a black dwarf look like? I'm assuming it is illuminated by some other source of light such as a star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question could be rephrased: What is the visible absorption (and emission) spectrum of electron degenerate matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This is my attempt at an answer&lt;/strong&gt;: Probably not black. A lot would depend on whether electron-degenerate matter behaves as metallic or dielectric, based on its band-structure. My guess would be metallic, considering the electrons are dissociated from their parent nuclei in electron-degenerate matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the surface gravity of a black dwarf would be so strong, I'd expect the surface to be very smooth and relatively free of Lambertian scattering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From these two features, I'd make a tentative prediction that a black dwarf would appear as a mirror-like sphere. &lt;/p&gt;&#xA;" OwnerUserId="17655" LastActivityDate="2018-12-28T17:42:14.370" Title="What does a black dwarf look like?" Tags="&lt;white-dwarf&gt;&lt;black-body&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21805" PostTypeId="1" AcceptedAnswerId="21865" CreationDate="2017-07-25T10:54:29.017" Score="2" ViewCount="130" Body="&lt;p&gt;I was going through the &lt;a href=&quot;https://www.eso.org/public/images/eso1242a/zoomable&quot; rel=&quot;nofollow noreferrer&quot;&gt;zoomable image of Milky way&lt;/a&gt; and after zooming to this particular location, we can see a reddish area on the left side of a probably large star. Here is a screenshot I took after zooming to that particular area:&#xA; &lt;a href=&quot;https://i.stack.imgur.com/GWYHM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GWYHM.jpg&quot; alt=&quot;image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is this gas cloud or some weird lens effect ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Location ( center of the blue rectangle )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OMvSz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OMvSz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17658" LastEditorUserId="17658" LastEditDate="2017-07-25T11:53:45.327" LastActivityDate="2021-01-01T18:54:56.700" Title="Is this gas cloud or lens effect?" Tags="&lt;nebula&gt;&lt;gas&gt;" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="21806" PostTypeId="2" ParentId="21804" CreationDate="2017-07-25T11:19:13.963" Score="5" Body="&lt;p&gt;The material at the surface of a white dwarf is not degenerate. &#xA;The &quot;visible&quot; surface is defined as where the optical depth exceeds some threshold and this will occur at a low enough density that even at a few hundred kelvin, the ratio of the Fermi energy to the thermal energy is too low for significant degeneracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, at these temperatures, the electrons attached to atomic hydrogen and helium would not be ionised and the material would mostly be in atomic and molecular states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of appearance a reasonable comparison would be with that of the very cool T- and Y-dwarfs; although some of the more complex chemistry that occurs in those atmospheres would not occur in very cool white dwarfs because the trace quantities of heavier elements would probably have sunk out of the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I note that there probably are white dwarfs cooler than 3000K. This is the upper limit to the temperature of an invisible companion of to the pulsar PSR 2222-0137. &lt;a href=&quot;https://arxiv.org/abs/1406.0488&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kaplan et al. (2014)&lt;/a&gt; say that this most likely a massive (&lt;span class=&quot;math-container&quot;&gt;$\sim 1 M_{\odot}$&lt;/span&gt;) old white dwarf that has cooled so much that they cannot detect it at visible wavelengths - so probably you would call this a &quot;black dwarf&quot;, though the term is not used in the scientific literature.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-12-28T17:42:14.370" LastActivityDate="2018-12-28T17:42:14.370" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21809" PostTypeId="1" AcceptedAnswerId="21899" CreationDate="2017-07-25T16:27:52.187" Score="3" ViewCount="957" Body="&lt;p&gt;&lt;a href=&quot;https://eclipse2017.nasa.gov/safety&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA's eclipse safety page&lt;/a&gt; says that eclipse glasses should:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Not&lt;/strong&gt; be used if they are older than three years, or have scratched or wrinkled lenses&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why shouldn't three year old glasses be used? Is it because the coating somehow degrades with time? Were standards changed three years ago such that the older glasses can no longer be recommended? Is there just a stronger concern for scratches or damage over the years? Will eclipse glasses (or other solar filters) purchased today be unsafe three years from now?&lt;/p&gt;&#xA;" OwnerUserId="13389" LastActivityDate="2017-08-01T17:28:07.140" Title="Why are three year old eclipse glasses not recommended?" Tags="&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21810" PostTypeId="1" AcceptedAnswerId="21811" CreationDate="2017-07-25T19:55:39.257" Score="3" ViewCount="84" Body="&lt;p&gt;We have OBAFGKM to denote the strength of hydrogen lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently we added S, N, C, and the W classes when we learned of new kinds of stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use T, L, and Y to denote brown dwarfs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The D classes refer to electron-degenerate stars (white dwarfs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've looked around quite a bit and I haven't seen anything regarding neutron stars and black holes. Does this mean there's no common symbol or abbreviation? What do they use in star catalogs? (I've never found a catalog with stars of all degeneracies in it.)&lt;/p&gt;&#xA;" OwnerUserId="17264" LastActivityDate="2017-07-25T20:04:15.043" Title="What abbreviations, letters, or symbols do we use to denote highly degenerate stars?" Tags="&lt;star&gt;&lt;classification&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21811" PostTypeId="2" ParentId="21810" CreationDate="2017-07-25T20:04:15.043" Score="4" Body="&lt;p&gt;Neutron stars and black holes do not have assigned spectral types since they do not have a measurable optical/IR spectrum - which is the basis for assigning a spectral type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many classifications for systems containing black holes and neutron stars. They are not related to the letter-based spectral types of &quot;normal&quot; stars or white dwarfs, but more to do with their X-ray, radio or inferred physical properties (e.g low- or high-mass X-ray binary systems).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-07-25T20:04:15.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21813" PostTypeId="1" CreationDate="2017-07-25T21:17:50.153" Score="2" ViewCount="224" Body="&lt;p&gt;Would it be possible for two planets orbiting around each other to share a moon that orbits both separately? Would the moon in question would do something akin to a large figure 8 as it orbited both or would this be possible in the first place?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what effects might we see on the planets because of this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Side note: This stems from a Worldbuilding question, but I figured I would ask the astronomy question on the Astronomy StackExchange)&lt;/p&gt;&#xA;" OwnerUserId="17257" LastActivityDate="2017-07-25T21:17:50.153" Title="Shared moon in a double planet system" Tags="&lt;orbit&gt;&lt;gravity&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21814" PostTypeId="2" ParentId="21724" CreationDate="2017-07-25T23:18:31.253" Score="2" Body="&lt;p&gt;Since the initial formation of the solar system and after the initial mass formation of each of the planets let's see what what we can derive from the laws of physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any mass that could further accumulate on any planet would have to come from 3 places:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In the nearby vicinity of the planet's orbit.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;From outside the solar system and extra Pluto Kuiper belt objects.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;From the Sun as solar wind.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We can surmise that in the first 200 million to half billion years in the life of the solar system, the planets would have grown in the most mass by (1) sweeping up the asteroids and debris in their orbit and vicinity.  In fact, Jupiter with its huge mass continues to pull away stray asteroids to a lesser extent to this day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The occasional encounters with objects coming from (2) outside the solar system are too rare and far in between to be of significance.  The cosmic rays from external to the solar system are also not significant for mass gain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That leaves (3) the solar wind.  Whether a planet gains or loses mass from the solar wind is a balance between the &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a)  Gravitational escape velocity of the planet, which is directly proportional to the mass of the planet&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b)  Average temperature of planet's surface and atmosphere&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c)  Average speed or kinetic energy of the particles from the solar wind&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d)  Intensity of the planet's magnetic field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(e)  Intensity of the solar wind, which is inversely proportional to the square of the distance from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The (e) factor was and still is much too strong for the inner 4 planets to gain mass from the solar wind.  Being that the escape velocity would be too low to keep the atmosphere with the ongoing assault of the solar wind as is the case with Mercury and our moon.  We can thank the earth's magneto sphere for saving our atmosphere from the solar wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now when we get to where Jupiter's orbit is, about 10AU, the intensity of the solar wind is one hundredth of what it is on the earth.  The solar wind that does reach Jupiter is decelerated and deflected towards the poles by Jupiter's very strong magnetic field .  And since the escape velocity is so high due to Jupiter's mass, most of the solar wind is absorbed into Jupiter resulting in mass gain from solar wind over time. Not even hydrogen can escape Jupiter's  gravitation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same net gain from solar wind applies to the next 3 gas giants Saturn, Uranus, and Neptune.  Since these are in order further away, the solar wind intensity correspondingly less.  All three still have a high gravitational escape velocity due to their size and the ones further out have a very low surface temperature so the gases don't have enough average energy to escape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we go further out from Jupiter the same dynamic of mass gain through captured solar wind plays out, albeit less and less due to the distance from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the solar wind for purposes of planetary  mass gain may seem insignificant now, we must understand that this was not the case earlier on.&#xA;And over 5 billion years much mass can be gained in this manner.&lt;/p&gt;&#xA;" OwnerUserId="16998" LastActivityDate="2017-07-25T23:18:31.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21822" PostTypeId="1" AcceptedAnswerId="21907" CreationDate="2017-07-26T17:02:22.817" Score="4" ViewCount="746" Body="&lt;p&gt;I purchased a 10 pack of glasses for viewing the upcoming eclipse. They show to be ISO 12312-2:2015 certified. How can this be tested and verified?&lt;/p&gt;&#xA;" OwnerUserId="17685" LastActivityDate="2017-08-20T06:31:14.740" Title="How to test direct sun viewing glasses" Tags="&lt;solar-eclipse&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21827" PostTypeId="1" CreationDate="2017-07-27T07:12:14.157" Score="1" ViewCount="553" Body="&lt;p&gt;I have Earth Moon Barycenter coordinate (J2000) generated from VSOP2013 in Astronomical Units, &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;JD=2415020.5   &#xA;X=-0.1968857640    Y=0.9633224602     Z=0.0002147057    AU &#xA;X'=-0.0171376156   Y'=-0.0035078892   Z'=-0.0000011481  AU/day&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How to convert it to Geocentric?&lt;/p&gt;&#xA;" OwnerUserId="17578" LastEditorUserId="2618" LastEditDate="2017-07-27T14:59:45.100" LastActivityDate="2017-07-27T14:59:45.100" Title="Converting Earth Moon Barycenter coordinate to Geocentric coordinate" Tags="&lt;the-sun&gt;&lt;the-moon&gt;&lt;coordinate&gt;&lt;ephemeris&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21832" PostTypeId="2" ParentId="21827" CreationDate="2017-07-27T14:58:50.690" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How to convert it to Geocentric?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You don't. That's the position and velocity of the Earth-Moon barycenter with respect to the center of the Sun. Converting that to geocentric (a) doesn't make sense, and (b) even if it did, can't be done without knowing the location of the Earth-Moon barycenter in geocentric coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position of the Earth-Moon barycenter plus the geocentric inertial position of the Moon expressed in the J2000 ecliptic frame plus the Earth-Moon mass ratio lets you compute the position of the Earth with respect to the Sun (or the position of the Sun with respect to the Earth). Add information about the positions of other planets with respect to the Sun and you can compute the positions of those other planets with respect to the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot do the above with VSOP2013 alone because VSOP2013 does not contain information about the position of the Moon with respect to the Earth. You'll need a separate lunar ephemeris for that.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-07-27T14:58:50.690" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21833" PostTypeId="1" CreationDate="2017-07-27T18:52:47.803" Score="9" ViewCount="1471" Body="&lt;p&gt;I've recently become interested in the idea of building my own small-scale radio telescope. A quick online search finds a few instructions on how to build this using a satellite dish. These suggest a dish of radius of around 1 metre, but the only objects that you seem to be able to detect are the Sun, the Earth, the Moon (possibly) and communications satellites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sounds like it would be become boring fairly quickly, and I would like to observe some deep-sky objects. Some example objects that I fancy observing (depending on how strong their signals are) are Betelgeuse/Sirius, the Crab Nebula and the Andromeda Galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N.B. I am quite satisfied with pointing the dish at an object and the signal strength increasing. I don't expect to get any kind of image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evidently, you cannot pick these up using a metre-wide dish, so what is the minimum size required to pick up these objects (noting that I can't exactly build Arecibo in my back garden)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case it matters, I live in a rural area, ~5 miles from the nearest town.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;I originally emailed Jodrell Bank this question but they ignored me :P&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1494" LastEditorUserId="1494" LastEditDate="2017-07-28T10:04:15.187" LastActivityDate="2017-12-18T11:37:58.967" Title="How big a dish do I need for radio astronomy?" Tags="&lt;amateur-observing&gt;&lt;radio-astronomy&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21834" PostTypeId="2" ParentId="21805" CreationDate="2017-07-27T19:04:18.430" Score="0" Body="&lt;p&gt;As the comments have said, this is simply a camera effect known as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Lens_flare&quot; rel=&quot;nofollow noreferrer&quot;&gt;lens flare&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;If you look around at the various bright stars, you can see this in multiple places, e.g.:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SBB2o.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SBB2o.jpg&quot; height=&quot;300&quot;/&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;https://snag.gy/SGyKUu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;snag.gy&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;and also:&lt;/p&gt;&#xA;&lt;img src=&quot;https://snag.gy/5DtdHW.jpg&quot; height=&quot;300&quot; /&gt;&#xA;" OwnerUserId="1494" LastEditorUserId="7472" LastEditDate="2021-01-01T18:54:56.700" LastActivityDate="2021-01-01T18:54:56.700" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21835" PostTypeId="1" CreationDate="2017-07-27T19:56:49.413" Score="0" ViewCount="195" Body="&lt;p&gt;Is a cold (3800K, or so) white dwarf still be considered/in degenerate state. Or, how can a gas be degenerate and cold?&lt;/p&gt;&#xA;" OwnerUserId="2757" LastActivityDate="2017-07-27T21:08:42.903" Title="What is a cold white dwarf?" Tags="&lt;white-dwarf&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21836" PostTypeId="2" ParentId="21835" CreationDate="2017-07-27T21:08:42.903" Score="4" Body="&lt;p&gt;I'm not entirely sure what your source of confusion is, so I'll just describe how temperature and degeneracy relate and hope that clears it up.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Fermions&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;All particles which are fermions are subject to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Electron_degeneracy_pressure&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pauli Exclusion Principle&lt;/a&gt;. This means no two fermions can occupy the same quantum state. They will resist doing so. What that really means, doesn't matter too much so for now, just accept it as an axiom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrons are fermions. A star, before it becomes a white dwarf, is filled with electrons just minding their own business. Then, as the star begins to form itself into a white dwarf, the crushing pressure of gravity squeezes all the electrons into a tight space. Gravity tries to push these electrons into the same quantum state, but the Pauli exclusion principle forbids this.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Fermi Energy&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: The following is a gross over-simplication, but it gets the idea across without being too terribly wrong. Don't take it as gospel truth.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can define a set of all the possible quantum states. For simplicity, I'm just going to call them $\mathrm{state}\ 1$, $\mathrm{state}\ 2$, ..., $\mathrm{state}\ 67$, ..., $\mathrm{state}\ 10^{10483}$, etc. (ordered by energy). From above, we know each state can be occupied by only one electron. In general, these electrons will occupy a variety of non-sequential states. Pre-white dwarf, we might order the electrons by state and say&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\mathrm{electron}\ 1$ is in $\mathrm{state}\ 43$,&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{electron}\ 2$ is in $\mathrm{state}\ 12084$,&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{electron}\ 3$ is in $\mathrm{state}\ 4.187\times10^{78}$&lt;/li&gt;&#xA;&lt;li&gt;etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As the electrons are squeezed in the white dwarf, they'll try to get as close to the same quantum state as possible without violating the Pauli exclusion principle. A fully degenerate star would have electrons in the following states:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\mathrm{electron}\ 1$ is in $\mathrm{state}\ 1$,&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{electron}\ 2$ is in $\mathrm{state}\ 2$,&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{electron}\ 3$ is in $\mathrm{state}\ 3$&lt;/li&gt;&#xA;&lt;li&gt;etc.&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{electron}\ n$ is in $\mathrm{state}\ n$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In this completely degenerate state, all electrons have been forced to occupy all the lowest energy states. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Fermi_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fermi energy&lt;/a&gt; is the energy of the last electron in the highest state. If all electrons happen to be squeezed such that they're all below the Fermi energy, then we consider the object degenerate.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Temperature and Degeneracy&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Gravity might try to squash all the electrons down into the lowest quantum states, but the ability of gravity to achieve this depends on the temperature. Electrons at a higher temperature have higher energies and it becomes harder for gravity to suppress them. With a certain set of assumptions&lt;sup&gt;1&lt;/sup&gt;, one can say that for a star of a given temperature, $T$, and a given density, $\rho$, it will be degenerate if&lt;sup&gt;2&lt;/sup&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{T}{\rho^{2/3}} &amp;lt; 1261\ \mathrm{K\ m^2\ kg^{-2/3}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this equation, you can hopefully see that being colder actually &lt;em&gt;helps&lt;/em&gt; a white dwarf be degenerate. The two ways to make a star (or anything really) become degenerate are to cool it down and increase the density. For white dwarf formation, the fact that it's degenerate is more a function of the high density than it is the cool temperature.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;sup&gt;1&lt;/sup&gt;Assumptions and preceding equations are in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0805304029&quot; rel=&quot;nofollow noreferrer&quot;&gt;An Introduction to Modern Astrophysics&lt;/a&gt;, 2nd ed., pg. 566.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;sup&gt;2&lt;/sup&gt;As a preemptive measure, this equation does not apply universally. You cannot apply it to yourself and declare yourself degenerate!&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-07-27T21:08:42.903" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21837" PostTypeId="1" CreationDate="2017-07-27T23:50:10.197" Score="2" ViewCount="238" Body="&lt;p&gt;Now that we know that the universe is expanding, how much does a linear lightyear of space expand currently in a year?&lt;/p&gt;&#xA;" OwnerUserId="17699" LastActivityDate="2018-03-01T09:16:44.743" Title="How much does a linear lightyear of space expand in a year?" Tags="&lt;expansion&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21838" PostTypeId="2" ParentId="21837" CreationDate="2017-07-27T23:50:10.197" Score="7" Body="&lt;p&gt;A recent estimation of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble constant is 71.9 (km/s)/Mpc&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means there is an expansion of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(71.9\ \mathrm{\frac{km}{s\ Mpc}}\right)\left(\mathrm{\frac{60\ s}{min}}\right)\left(\mathrm{\frac{60\ min}{hr}}\right)\left(\mathrm{\frac{24\ hr}{day}}\right)\left(\mathrm{\frac{365\ day}{yr}}\right) = 2,267,438,400\ \mathrm{\frac{km}{yr\ Mpc}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To apply this expansion over a light year:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{2,267,438,400\ \mathrm{(km/yr)/Mpc}}{3,261,564\ \mathrm{lyr/Mpc}} = \boxed{695.2\ \mathrm{\frac{km/yr}{lyr}}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is about $7.3483\times10^{-11}\ \%$ per year or $432\ \mathrm{miles/year}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A linear light year of space expands by 695.2 km/yr with a given Hubble constant of 71.9 (km/s)/Mpc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;A better measurement released on 2018-02-22 using parallax found the Hubble constant to be &lt;a href=&quot;http://imgsrc.hubblesite.org/hvi/uploads/science_paper/file_attachment/308/finalpass5.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;73.45 ± 1.66 (km/s)/Mpc.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17699" LastEditorUserId="17699" LastEditDate="2018-03-01T09:16:44.743" LastActivityDate="2018-03-01T09:16:44.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21839" PostTypeId="1" CreationDate="2017-07-28T00:33:57.697" Score="1" ViewCount="121" Body="&lt;p&gt;In chapter 4 of &quot;The Revolutions of the Heavenly Spheres&quot; where Copernicus maintains that the objects in our solar system have circular orbits or orbits compounded of several circles, the Edward Rosen translation of the book states this &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It stands to reason, therefore, that their uniform motions appear nonuniform to us.&#xA;  The cause may be either that their circles have poles different [from the earth's] or&#xA;  that the earth is not at the center of the circles on which they revolve.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To add context: Copernicus believes that features of orbit that are inconsistent to circular motion such as retrograde motion, apogee, perigee, fast/slow movement, etc... Don't actually occur in the orbits of the Sun, Moon, and the planets. Stating that the actual causes of the inconsistencies could either be that &quot;their circles have poles different [from the earth's.]&quot; Or that these objects don't revolve around the Earth.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm mainly concerned with what Copernicus is referring to when he states that the orbits of these objects &quot;have poles different [from the earth's,]&quot; and how (if true) it could produce these inconsistencies in orbits. &lt;/p&gt;&#xA;" OwnerUserId="17321" LastEditorUserId="11143" LastEditDate="2017-07-28T12:45:28.637" LastActivityDate="2017-07-28T12:45:28.637" Title="What is Copernicus referring to when he states that &quot;circles have poles different [from the earth's]&quot;" Tags="&lt;orbit&gt;&lt;history&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21841" PostTypeId="2" ParentId="21839" CreationDate="2017-07-28T04:06:04.607" Score="1" Body="&lt;p&gt;Quoting page 12 of &lt;a href=&quot;https://books.google.com/books?id=LH4tWpJzzCcC&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://books.google.com/books?id=LH4tWpJzzCcC&lt;/a&gt; (translator uncredited):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ASIQD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ASIQD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and looking at the second sentence, we have &quot;they do not revolve around the same poles [...] but follow the oblique ecliptic&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the use of poles later in the page is the same, he's pointing out the north ecliptic pole (which is in Draco) is not the same as the north celestial pole (which is in Ursa Minor and close to the star Polaris).&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2017-07-28T04:06:04.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21842" PostTypeId="2" ParentId="21839" CreationDate="2017-07-28T04:07:50.073" Score="3" Body="&lt;p&gt;Remember that Copernicus was espousing a heliocentric Solar System.  What he means by &quot;pole&quot; is the object's axis of revolution and its axis of rotation.  The Earth rotates around its poles: i.e. the North and South pole.  A line between those poles is our axis of rotation.  The axis of revolution would be the axis about which the Earth orbits.  This would correspond to an imaginary line at the Sun. &#xA; Copernicus is presenting the idea that the axis of rotation is not the axis of revolution.&lt;/p&gt;&#xA;" OwnerUserId="13426" LastActivityDate="2017-07-28T04:07:50.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21843" PostTypeId="1" AcceptedAnswerId="21844" CreationDate="2017-07-28T04:52:27.497" Score="22" ViewCount="4381" Body="&lt;p&gt;To the extent of my understanding, Olber’s paradox states that if the universe was static and homogeneous, we should see a star at every point in the night sky and therefore the night sky should be equally as bright as day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since the night sky is dark and non-uniform, it can be said that the universe is &lt;em&gt;not&lt;/em&gt; static and &lt;em&gt;not&lt;/em&gt; homogeneous. However, if this was already known, what exactly is the “paradox”? Why isn’t it called Olber’s Observation or something else?&lt;/p&gt;&#xA;" OwnerUserId="17701" LastEditorUserId="1265" LastEditDate="2017-07-28T09:15:32.373" LastActivityDate="2017-08-01T22:09:34.390" Title="What exactly is the &quot;paradox&quot; in Olber's Paradox?" Tags="&lt;star&gt;&lt;universe&gt;&lt;history&gt;&lt;night-sky&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21844" PostTypeId="2" ParentId="21843" CreationDate="2017-07-28T05:46:06.057" Score="27" Body="&lt;p&gt;Olber's Paradox was created at a time before the idea of a finite universe was accepted.  (&lt;a href=&quot;https://en.wikipedia.org/wiki/Olbers%27_paradox#History&quot; rel=&quot;noreferrer&quot;&gt;It was thought of in the 1600's&lt;/a&gt;).  In order to resolve Olber's Paradox, you have to introduce the idea that either the universe had a beginning or it is of finite size.  (Note: the solution does not require an expanding universe).  So, at the time, it was a paradox.  Pretty much all astronomers considered the universe to be static and infinite.  Therefore, the fact that their observations didn't fit with what they expected made it a paradox.&lt;/p&gt;&#xA;" OwnerUserId="13426" LastEditorUserId="13426" LastEditDate="2017-07-29T05:13:00.973" LastActivityDate="2017-07-29T05:13:00.973" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="21849" PostTypeId="1" CreationDate="2017-07-28T12:20:46.620" Score="9" ViewCount="2696" Body="&lt;p&gt;As per the universal law of attraction, any two bodies (having some mass) experience a force of 'attraction' which is proportionate to ...and ...inverse proportionate ....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then comes my question: Why it should be force should be of type 'attraction' only ?  Why it should not be repulsion / any other kind of force ?&lt;/p&gt;&#xA;" OwnerUserId="17710" LastEditorUserId="1494" LastEditDate="2017-07-28T15:29:12.947" LastActivityDate="2018-01-27T00:07:04.193" Title="Why is gravity only an attractive force?" Tags="&lt;gravity&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21852" PostTypeId="2" ParentId="21849" CreationDate="2017-07-28T13:36:51.470" Score="11" Body="&lt;h3&gt;Because mass is positive&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;To expand your quote concerning the gravitational force into an equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_G = -\frac{Gm_1m_2}{r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force of gravity, $F_G$ is proportional to the product of the masses and inversely proportional to the distance, $r$, squared. Let's break this down and see what might cause $F_G$ to be positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this equation, $r$ cannot be negative because it's a distance between two locations. Two locations cannot be a negative distance apart. And even if they somehow were, the squared would take care of that anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G$ is the universal constant and always positive. You might argue that it could possibly be negative, but that's not possible. $G$ actually doesn't really exist. It doesn't describe anything fundamental to the physics of the universe. $G$ is simply a bookkeeping constant that allows us to get the right answer for the force based on any choice of units for mass and distance. Technically, if one uses the &quot;correct&quot; units for mass and distance (e.g., the &lt;a href=&quot;https://en.wikipedia.org/wiki/Planck_units&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck units&lt;/a&gt;), then $G=1$ and effectively doesn't exist. Since $G$ is just a scaling factor that depends on the choice of units, it will only be a positive number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That leaves us with the masses. These are the only things which could possibly be negative. Of course, to get a positive, repulsive force, one mass would have to be positive and the other negative. But what exactly is a negative mass? Mass is the metric which describes &quot;how much&quot; of something there is. How can you have less than nothing of something?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Why can mass not be negative?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;If you want to look at this another way, you can show that if mass could be negative, you'd get nonsensical results! Assuming of course, all other aspects of physics were the same. Recall from &lt;a href=&quot;https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion#Newton.27s_second_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Newton's second law&lt;/a&gt; that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = ma$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say there are two blocks sitting on a table. One block has a mass $m_1&amp;gt;0$ which is positive and the other has a mass $m_2&amp;lt;0$ which is negative. Ignore all other forces on these two blocks for the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I go up to $m_1$ and I apply a force to push this mass forward. The acceleration that is induced is: $a = F/m_1$. &lt;strong&gt;Necessarily&lt;/strong&gt;, the direction in which $m_1$ moves is the same direction in which I'm pushing. That's all well and good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I go over to $m_2$ and I apply the same force, attempting to push it forward on the table. The acceleration induced on $m_2$ will be: $a = -F/|m_2|$. Note I made $m_2$ positive and pulled out the negative sign. You can see that if my force is forward, the direction the mass moves will be backwards! But here's the problem, my hand is in the way because it's trying to push to mass. As the mass tries to move backwards into my hand, it will be applying a force back on my hand, which by Newton's third law, necessarily mean's my hand is applying more force on the block, which then applies more force on my hand, ... and suddenly infinite forces are being applied or equivalently, these objects are infinitely accelerating. This is described by the concept of &lt;a href=&quot;https://en.wikipedia.org/wiki/Negative_mass#Runaway_motion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Runaway Motion&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this seems strange to you, that's because it is. If negative masses existed, we'd live in a very weird universe. Fortunately, we live in a universe where physics makes sense, mass is positive, and by extension gravity is always attractive.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2018-01-27T00:07:04.193" LastActivityDate="2018-01-27T00:07:04.193" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="21853" PostTypeId="1" CreationDate="2017-07-28T14:35:40.693" Score="1" ViewCount="649" Body="&lt;p&gt;One of the simplest versions of Newton's equations of motion for the Sun, Earth and Moon can be obtained by making the approximation that the three bodies are perfect spheres. In this approximation, the pair potential between body i and j is  $$\frac{-G m_i m_j}{ r_{ij}},$$ where $r_{ij}$ is the distance between the bodies: one can solve the equations numerically and get the orbits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This simple model does not include multiple, additional factors, such as &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the fact that the Earth is not perfectly spherical (equatorial bulge),&lt;/li&gt;&#xA;&lt;li&gt;Earth's tides,&lt;/li&gt;&#xA;&lt;li&gt;the influence of other celestial bodies,&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amongst all these factors, what is the dominant one as for predicting the orbit of the Earth? In other words, what is the leading factor that should be included in the simple model above in order make its predictions for the Earth's orbit closer to the observations? &lt;/p&gt;&#xA;" OwnerUserId="16581" LastEditorUserId="2365" LastEditDate="2017-07-28T21:27:43.823" LastActivityDate="2017-09-03T04:35:51.803" Title="Equations of motion for Earth and Moon" Tags="&lt;celestial-mechanics&gt;&lt;newtonian-gravity&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21855" PostTypeId="1" AcceptedAnswerId="22062" CreationDate="2017-07-29T03:26:55.747" Score="9" ViewCount="196" Body="&lt;p&gt;The recent Phys.org article &lt;a href=&quot;https://phys.org/news/2017-07-milky.html&quot; rel=&quot;noreferrer&quot;&gt;Milky Way's origins are not what they seem&lt;/a&gt; explains the recent MNRAS article Anglés-Alcázar et al. 2017; &lt;a href=&quot;https://academic.oup.com/mnras/article-abstract/470/4/4698/3871367/The-cosmic-baryon-cycle-and-galaxy-mass-assembly&quot; rel=&quot;noreferrer&quot;&gt;The cosmic baryon cycle and galaxy mass assembly in the FIRE simulations&lt;/a&gt; DOI:(&lt;a href=&quot;https://doi.org/10.1093/mnras/stx1517&quot; rel=&quot;noreferrer&quot;&gt;https://doi.org/10.1093/mnras/stx1517&lt;/a&gt;), also in &lt;a href=&quot;https://arxiv.org/abs/1610.08523&quot; rel=&quot;noreferrer&quot;&gt;ArXiv&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Using supercomputer simulations, the research team found a major and unexpected new mode for how galaxies, including our own Milky Way, acquired their matter: intergalactic transfer. The simulations show that supernova explosions eject copious amounts of gas from galaxies, which causes atoms to be transported from one galaxy to another via powerful galactic winds. Intergalactic transfer is a newly identified phenomenon, which simulations indicate will be critical for understanding how galaxies evolve.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is quite a sizable paper about quite a sizable simulation, and I'm having difficulty even getting started to read it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt; I wanted to know, for a Milky Way-like galaxy, was most of the influx from other galaxies early on, or is this something that is predicted to happen somewhat steadily throughout the galaxy's lifetime?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxy m12i in the simulation represents a Milky Way-mass galaxy, and so I show a screen shot of one of the figures for that galaxy below. Already I am stuck because I don't understand what $z$ represents. It seems to be related to time, rather than red-shift and $z=0$ has a special meaning, but I don't know if that's the present, or the beginning of the simulation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; Selected parts of Figure 3 from &lt;a href=&quot;https://arxiv.org/abs/1610.08523&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. The selected column is data for galaxy m12i, which is representative of a Milky Way-mass galaxy in the simulation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Figure 3. Top panels: contribution of different processes to the evolution of galaxies m11 (left), m12i (middle), and m13 (right) from early times down to z = 0. Total stellar mass (top), ISM gas mass (middle), and gas accretion rate onto the galaxy (bottom) are indicated by the thick gray lines for each galaxy. Lines of different colors show the contributions from fresh accretion (purple), NEP wind recycling (blue), intergalactic transfer of gas (green), and galaxy mergers (ISM gas: orange; stars: red). All quantities represent average values over a timescale of ∼ 200 Myr. Bottom panels: fraction of stellar mass, ISM gas mass, and gas accretion rate contributed by each process as a function of redshift. Fractions are represented in terms of ranges, so that the contribution of each component at any redshift is given by the vertical extent of the corresponding color at that redshift.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/be1AB.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/be1AB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JMHTE.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JMHTE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-07-29T06:26:07.907" LastActivityDate="2017-08-11T14:54:43.120" Title="Half of the Milky Way's baryonic mass is from other galaxies - when did this happen?" Tags="&lt;galactic-dynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21865" PostTypeId="2" ParentId="21805" CreationDate="2017-07-29T15:08:14.890" Score="1" Body="&lt;p&gt;This particular object I tracked down by comparing the visual with the infrared image of the sky. It was a little tricky because no very bright star appears at that location. But after making comparisons, the star in the original image is &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=HD%20155161&amp;amp;NbIdent=1&amp;amp;Radius=2&amp;amp;Radius.unit=arcmin&amp;amp;submit=submit%20id&quot; rel=&quot;nofollow noreferrer&quot;&gt;HD 155161&lt;/a&gt;, also known as V* AH Sco, indicating that it is a variable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, Simbad lists it as a red supergiant at coordinates [17 11 17.02114,  -32 19 30.7132], and visual magnitude of about 7.5 Red supergiants are often unstable and are semi-regular varible stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That explains why it is much brighter in the VISTA infrared image than in the optical. Red supergiants have cool outer layers that give off much of their radiation as infrared, not as visible light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the &lt;a href=&quot;http://aladin.u-strasbg.fr/AladinLite/?target=V*%20AH%20Sco&amp;amp;fov=0.52&amp;amp;survey=P%2F2MASS%2Fcolor&quot; rel=&quot;nofollow noreferrer&quot;&gt;star on Aladin Lite&lt;/a&gt;, even in the infrared, there is not a hint of nebulosity around the star. The conclusion is that this is an optical effect, lens flare or similar.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-07-29T15:20:00.407" LastActivityDate="2017-07-29T15:20:00.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21866" PostTypeId="1" CreationDate="2017-07-29T18:21:00.743" Score="1" ViewCount="978" Body="&lt;p&gt;I am creating program in C which is derived from Javascript Solar Radiation calculator based on a book &lt;em&gt;Guide to HTML, JavaScript and PHP: For Scientists and Engineers&lt;/em&gt; By David R. Brooks. The original program calculates Sidereal time like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  // Sidereal time (Theta0)&#xA;  double GMST = 280.46061837+360.98564736629*( i-&amp;gt;JulianDate-2451545.);&#xA;  // Sidereal_time = GMST + 0.000387933*T2 - T3/38710000&#xA;  double Sidereal_time = GMST+0.000387933*T*T&#xA;       -T*T*T/38710000.;&#xA;  // Performance optimalizatuion of code for Sidereal=fmod(Sidereal,360.) or Sidereal%360&#xA;  double S_save=Sidereal_time/360.;&#xA;  if (S_save &amp;lt; 0.) Sidereal_time = Sidereal_time-ceil(S_save)*360.;&#xA;  else Sidereal_time = Sidereal_time-floor(S_save)*360.;&#xA;  if (Sidereal_time &amp;lt; 0.) Sidereal_time+=360.;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where i-&gt;JulianDate is date/time specified by user which is converted to Julian date. As you can see this approach is based on 2451545.0 which is the equivalent of &lt;code&gt;January 1, 2000, at 12:00:00 UT&lt;/code&gt; and &lt;code&gt;36525.0&lt;/code&gt; is Julian Year. I am not sure why there is 36525 instead &lt;code&gt;365.25&lt;/code&gt; (days of Julian Year). Can you explain this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once more the same code without comments:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  double GMST = 280.46061837+360.98564736629*( i-&amp;gt;JulianDate-2451545.);&#xA;  double Sidereal_time = GMST+0.000387933*T*T-T*T*T/38710000.;&#xA;  double S_save=Sidereal_time/360.;&#xA;  if (S_save &amp;lt; 0.) &#xA;     Sidereal_time = Sidereal_time-ceil(S_save)*360.;&#xA;  else &#xA;     Sidereal_time = Sidereal_time-floor(S_save)*360.;&#xA;  if (Sidereal_time &amp;lt; 0.) &#xA;     Sidereal_time+=360.;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Another question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I enter date older than January 1, 2000, at 12:00:00 UT, e.g. &lt;strong&gt;1945&lt;/strong&gt;, will the sidereal time be correct? Or do I need to change the number &lt;code&gt;36525.0&lt;/code&gt; to extend the range?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet, before the Sidereal time is calculated in the program there is this:&#xA;&lt;code&gt;double T=( i-&amp;gt;JulianDate-2451545.0)/36525.0;&lt;/code&gt; Which is similar thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sidereal_time is used to calculate hour_angle which is used to calculate Sun's elevation and Solar Zenit Angle Z. Will this variables be affected when I type 1945 or older date in the program?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;I have added log to my program which printed values into file. I have tested the program for range 1990-2010. You can see that the SideReal time same as L0 (Geometric mean longitude) is negative number when the date is older than 2000/01/01 as defined in the program to calculate T. The difference of 10 years is cca 0.09999 of Julian century. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Date:1990 1 1 12:0:0&#xA;Julian Century: -0.099986&#xA;L0 - geometric mean longitude of the sun:: -3319.117706&#xA;GMST: -1318039.123563&#xA;Sidereal time: -1318039.123559&#xA;Julian Date: 2447893.000000&#xA;&#xA;Date:1991 1 1 12:0:0&#xA;Julian Century: -0.089993&#xA;L0 - geometric mean longitude of the sun:: -2959.356420&#xA;GMST: -1186279.362275&#xA;Sidereal time: -1186279.362271&#xA;Julian Date: 2448258.000000&#xA;&#xA;Date:1992 1 1 12:0:0&#xA;Julian Century: -0.080000&#xA;L0 - geometric mean longitude of the sun:: -2599.595134&#xA;GMST: -1054519.600986&#xA;Sidereal time: -1054519.600983&#xA;Julian Date: 2448623.000000&#xA;&#xA;Date:1993 1 1 12:0:0&#xA;Julian Century: -0.069979&#xA;L0 - geometric mean longitude of the sun:: -2238.848201&#xA;GMST: -922398.854050&#xA;Sidereal time: -922398.854048&#xA;Julian Date: 2448989.000000&#xA;&#xA;Date:1994 1 1 12:0:0&#xA;Julian Century: -0.059986&#xA;L0 - geometric mean longitude of the sun:: -1879.086915&#xA;GMST: -790639.092761&#xA;Sidereal time: -790639.092760&#xA;Julian Date: 2449354.000000&#xA;&#xA;Date:1995 1 1 12:0:0&#xA;Julian Century: -0.049993&#xA;L0 - geometric mean longitude of the sun:: -1519.325629&#xA;GMST: -658879.331472&#xA;Sidereal time: -658879.331472&#xA;Julian Date: 2449719.000000&#xA;&#xA;Date:1996 1 1 12:0:0&#xA;Julian Century: -0.040000&#xA;L0 - geometric mean longitude of the sun:: -1159.564343&#xA;GMST: -527119.570184&#xA;Sidereal time: -527119.570183&#xA;Julian Date: 2450084.000000&#xA;&#xA;Date:1997 1 1 12:0:0&#xA;Julian Century: -0.029979&#xA;L0 - geometric mean longitude of the sun:: -798.817409&#xA;GMST: -394998.823248&#xA;Sidereal time: -394998.823247&#xA;Julian Date: 2450450.000000&#xA;&#xA;Date:1998 1 1 12:0:0&#xA;Julian Century: -0.019986&#xA;L0 - geometric mean longitude of the sun:: -439.056123&#xA;GMST: -263239.061959&#xA;Sidereal time: -263239.061959&#xA;Julian Date: 2450815.000000&#xA;&#xA;Date:1999 1 1 12:0:0&#xA;Julian Century: -0.009993&#xA;L0 - geometric mean longitude of the sun:: -79.294836&#xA;GMST: -131479.300670&#xA;Sidereal time: -131479.300670&#xA;Julian Date: 2451180.000000&#xA;&#xA;Date:2000 1 1 12:0:0&#xA;Julian Century: 0.000000&#xA;L0 - geometric mean longitude of the sun:: 280.466450&#xA;GMST: 280.460618&#xA;Sidereal time: 280.460618&#xA;Julian Date: 2451545.000000&#xA;&#xA;Date:2001 1 1 12:0:0&#xA;Julian Century: 0.010021&#xA;L0 - geometric mean longitude of the sun:: 641.213384&#xA;GMST: 132401.207554&#xA;Sidereal time: 132401.207554&#xA;Julian Date: 2451911.000000&#xA;&#xA;Date:2002 1 1 12:0:0&#xA;Julian Century: 0.020014&#xA;L0 - geometric mean longitude of the sun:: 1000.974670&#xA;GMST: 264160.968843&#xA;Sidereal time: 264160.968843&#xA;Julian Date: 2452276.000000&#xA;&#xA;Date:2003 1 1 12:0:0&#xA;Julian Century: 0.030007&#xA;L0 - geometric mean longitude of the sun:: 1360.735957&#xA;GMST: 395920.730132&#xA;Sidereal time: 395920.730132&#xA;Julian Date: 2452641.000000&#xA;&#xA;Date:2004 1 1 12:0:0&#xA;Julian Century: 0.040000&#xA;L0 - geometric mean longitude of the sun:: 1720.497244&#xA;GMST: 527680.491421&#xA;Sidereal time: 527680.491421&#xA;Julian Date: 2453006.000000&#xA;&#xA;Date:2005 1 1 12:0:0&#xA;Julian Century: 0.050021&#xA;L0 - geometric mean longitude of the sun:: 2081.244178&#xA;GMST: 659801.238357&#xA;Sidereal time: 659801.238358&#xA;Julian Date: 2453372.000000&#xA;&#xA;Date:2006 1 1 12:0:0&#xA;Julian Century: 0.060014&#xA;L0 - geometric mean longitude of the sun:: 2441.005465&#xA;GMST: 791560.999645&#xA;Sidereal time: 791560.999647&#xA;Julian Date: 2453737.000000&#xA;&#xA;Date:2007 1 1 12:0:0&#xA;Julian Century: 0.070007&#xA;L0 - geometric mean longitude of the sun:: 2800.766751&#xA;GMST: 923320.760934&#xA;Sidereal time: 923320.760936&#xA;Julian Date: 2454102.000000&#xA;&#xA;Date:2008 1 1 12:0:0&#xA;Julian Century: 0.080000&#xA;L0 - geometric mean longitude of the sun:: 3160.528038&#xA;GMST: 1055080.522223&#xA;Sidereal time: 1055080.522225&#xA;Julian Date: 2454467.000000&#xA;&#xA;Date:2009 1 1 12:0:0&#xA;Julian Century: 0.090021&#xA;L0 - geometric mean longitude of the sun:: 3521.274973&#xA;GMST: 1187201.269159&#xA;Sidereal time: 1187201.269162&#xA;Julian Date: 2454833.000000&#xA;&#xA;Date:2010 1 1 12:0:0&#xA;Julian Century: 0.100014&#xA;L0 - geometric mean longitude of the sun:: 3881.036260&#xA;GMST: 1318961.030447&#xA;Sidereal time: 1318961.030451&#xA;Julian Date: 2455198.000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT2: fmod(GMST,24); fmod(Sidereal_time,24);&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Date:1990 1 1 12:0:0&#xA;Julian Century: -0.099986&#xA;L0 - geometric mean longitude of the sun:: -3319.117706&#xA;GMST: -7.123563&#xA;Sidereal time: -7.123559&#xA;Julian Date: 2447893.000000&#xA;&#xA;Date:1991 1 1 12:0:0&#xA;Julian Century: -0.089993&#xA;L0 - geometric mean longitude of the sun:: -2959.356420&#xA;GMST: -7.362275&#xA;Sidereal time: -7.362271&#xA;Julian Date: 2448258.000000&#xA;&#xA;Date:1992 1 1 12:0:0&#xA;Julian Century: -0.080000&#xA;L0 - geometric mean longitude of the sun:: -2599.595134&#xA;GMST: -7.600986&#xA;Sidereal time: -7.600983&#xA;Julian Date: 2448623.000000&#xA;&#xA;Date:1993 1 1 12:0:0&#xA;Julian Century: -0.069979&#xA;L0 - geometric mean longitude of the sun:: -2238.848201&#xA;GMST: -6.854050&#xA;Sidereal time: -6.854048&#xA;Julian Date: 2448989.000000&#xA;&#xA;Date:1994 1 1 12:0:0&#xA;Julian Century: -0.059986&#xA;L0 - geometric mean longitude of the sun:: -1879.086915&#xA;GMST: -7.092761&#xA;Sidereal time: -7.092760&#xA;Julian Date: 2449354.000000&#xA;&#xA;Date:1995 1 1 12:0:0&#xA;Julian Century: -0.049993&#xA;L0 - geometric mean longitude of the sun:: -1519.325629&#xA;GMST: -7.331472&#xA;Sidereal time: -7.331472&#xA;Julian Date: 2449719.000000&#xA;&#xA;Date:1996 1 1 12:0:0&#xA;Julian Century: -0.040000&#xA;L0 - geometric mean longitude of the sun:: -1159.564343&#xA;GMST: -7.570184&#xA;Sidereal time: -7.570183&#xA;Julian Date: 2450084.000000&#xA;&#xA;Date:1997 1 1 12:0:0&#xA;Julian Century: -0.029979&#xA;L0 - geometric mean longitude of the sun:: -798.817409&#xA;GMST: -6.823248&#xA;Sidereal time: -6.823247&#xA;Julian Date: 2450450.000000&#xA;&#xA;Date:1998 1 1 12:0:0&#xA;Julian Century: -0.019986&#xA;L0 - geometric mean longitude of the sun:: -439.056123&#xA;GMST: -7.061959&#xA;Sidereal time: -7.061959&#xA;Julian Date: 2450815.000000&#xA;&#xA;Date:1999 1 1 12:0:0&#xA;Julian Century: -0.009993&#xA;L0 - geometric mean longitude of the sun:: -79.294836&#xA;GMST: -7.300670&#xA;Sidereal time: -7.300670&#xA;Julian Date: 2451180.000000&#xA;&#xA;Date:2000 1 1 12:0:0&#xA;Julian Century: 0.000000&#xA;L0 - geometric mean longitude of the sun:: 280.466450&#xA;GMST: 16.460618&#xA;Sidereal time: 16.460618&#xA;Julian Date: 2451545.000000&#xA;&#xA;Date:2001 1 1 12:0:0&#xA;Julian Century: 0.010021&#xA;L0 - geometric mean longitude of the sun:: 641.213384&#xA;GMST: 17.207554&#xA;Sidereal time: 17.207554&#xA;Julian Date: 2451911.000000&#xA;&#xA;Date:2002 1 1 12:0:0&#xA;Julian Century: 0.020014&#xA;L0 - geometric mean longitude of the sun:: 1000.974670&#xA;GMST: 16.968843&#xA;Sidereal time: 16.968843&#xA;Julian Date: 2452276.000000&#xA;&#xA;Date:2003 1 1 12:0:0&#xA;Julian Century: 0.030007&#xA;L0 - geometric mean longitude of the sun:: 1360.735957&#xA;GMST: 16.730132&#xA;Sidereal time: 16.730132&#xA;Julian Date: 2452641.000000&#xA;&#xA;Date:2004 1 1 12:0:0&#xA;Julian Century: 0.040000&#xA;L0 - geometric mean longitude of the sun:: 1720.497244&#xA;GMST: 16.491421&#xA;Sidereal time: 16.491421&#xA;Julian Date: 2453006.000000&#xA;&#xA;Date:2005 1 1 12:0:0&#xA;Julian Century: 0.050021&#xA;L0 - geometric mean longitude of the sun:: 2081.244178&#xA;GMST: 17.238357&#xA;Sidereal time: 17.238358&#xA;Julian Date: 2453372.000000&#xA;&#xA;Date:2006 1 1 12:0:0&#xA;Julian Century: 0.060014&#xA;L0 - geometric mean longitude of the sun:: 2441.005465&#xA;GMST: 16.999645&#xA;Sidereal time: 16.999647&#xA;Julian Date: 2453737.000000&#xA;&#xA;Date:2007 1 1 12:0:0&#xA;Julian Century: 0.070007&#xA;L0 - geometric mean longitude of the sun:: 2800.766751&#xA;GMST: 16.760934&#xA;Sidereal time: 16.760936&#xA;Julian Date: 2454102.000000&#xA;&#xA;Date:2008 1 1 12:0:0&#xA;Julian Century: 0.080000&#xA;L0 - geometric mean longitude of the sun:: 3160.528038&#xA;GMST: 16.522223&#xA;Sidereal time: 16.522225&#xA;Julian Date: 2454467.000000&#xA;&#xA;Date:2009 1 1 12:0:0&#xA;Julian Century: 0.090021&#xA;L0 - geometric mean longitude of the sun:: 3521.274973&#xA;GMST: 17.269159&#xA;Sidereal time: 17.269162&#xA;Julian Date: 2454833.000000&#xA;&#xA;Date:2010 1 1 12:0:0&#xA;Julian Century: 0.100014&#xA;L0 - geometric mean longitude of the sun:: 3881.036260&#xA;GMST: 17.030447&#xA;Sidereal time: 17.030451&#xA;Julian Date: 2455198.000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="17733" LastEditorUserId="17733" LastEditDate="2017-07-30T15:38:32.400" LastActivityDate="2017-07-30T17:02:57.020" Title="Sidereal time calculation" Tags="&lt;time&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21867" PostTypeId="1" CreationDate="2017-07-29T18:35:11.640" Score="0" ViewCount="157" Body="&lt;p&gt;Someone gave me an Orion StarBlast 4.5 EQ, and with it everything but the counterweight shaft. Because of this, the telescope is hard to use. It flops around quite a bit making viewing of objects difficult. Where can I buy just that part? I looked it up and all I could find is an &quot;upgrade&quot; to the rod for $70.&lt;/p&gt;&#xA;" OwnerDisplayName="Mlg Sound Effects" LastEditorDisplayName="user14781" LastEditDate="2017-08-01T07:59:32.393" LastActivityDate="2017-08-01T07:59:32.393" Title="Counter weight rod" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21868" PostTypeId="2" ParentId="21866" CreationDate="2017-07-29T19:40:59.933" Score="1" Body="&lt;p&gt;I have often seen programs use a Julian century, so the 36625 is probably the number of sidereal days in a century, whereas 36525 is the number of calendar days in a century (&lt;a href=&quot;https://en.wikipedia.org/wiki/Julian_year_(astronomy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Julian_year_(astronomy)&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sidereal time generally runs uniformly forward or backward :-). I think the calculation will work prior to 2000. I am not going to type it in to test it, but since you have (or will), you can test it. The sidereal time should be close to the same value for the same time and day regardless of the year.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastEditorDisplayName="user21" LastEditDate="2017-07-30T17:02:57.020" LastActivityDate="2017-07-30T17:02:57.020" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21869" PostTypeId="1" CreationDate="2017-07-30T01:14:30.173" Score="5" ViewCount="242" Body="&lt;p&gt;I once heard of a strange phenomenon, but now cannot find anything about it on the Internet.  I might be looking for the wrong keywords or misremember some relevant aspect.  So maybe someone here can help me out and point me in the right direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The described phenomenon goes like this, as I remember it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the course of the scientific research, people found out more and more about the objects surrounding us.  Moon, Sun, planets, other stars, other galaxies, etc.  But each time (more or less) a little later (after a new discovery was getting an established fact) it showed that the closest object of this kind to Earth was rather an exception among its peers.  For instance, the closest other galaxy is blue-shifted.  Or that the closest Pulsar is somehow an exception among its peers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be that the rule was not about the &lt;em&gt;closest&lt;/em&gt; but about the &lt;em&gt;first found&lt;/em&gt; object of a kind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I remember Prof Harald Lesch (a German physicist lecturing on TV in a popular format called &quot;Alpha Centauri&quot;) mentioning this peculiar phenomenon and fleshing it out with some examples, but now I'm not able to find the clip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the point about closing this question, but maybe leave it open for some weeks (or months), just in case somebody drops by who has something to contribute.  After a grace period the non-reaction can be counted as a vote for &quot;isn't common knowledge&quot;.  Without such a grace period closing the question can only be based on &quot;I don't know such a thing&quot; (which should not be enough).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concrete questions (because that was asked for, so it seemed to be unclear to some people) are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is such a phenomenon known?&lt;/li&gt;&#xA;&lt;li&gt;Is there a source describing it?&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Nice to have:  Which of the &lt;a href=&quot;http://www.br.de/fernsehen/ard-alpha/sendungen/alpha-centauri/alpha-centauri-videothek-videos100.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Alpha Centauri lectures&lt;/a&gt; was probably the one introducing me to the topic?&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Is there a name to it?&lt;/li&gt;&#xA;&lt;li&gt;What are examples of such objects?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1808" LastEditorUserId="1808" LastEditDate="2017-08-02T13:53:05.420" LastActivityDate="2018-08-22T18:32:02.743" Title="Rule of Exception in All Nearest Objects" Tags="&lt;star&gt;&lt;planet&gt;&lt;galaxy&gt;&lt;pulsar&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21871" PostTypeId="1" AcceptedAnswerId="21872" CreationDate="2017-07-30T06:37:52.663" Score="11" ViewCount="673" Body="&lt;p&gt;This &lt;a href=&quot;https://astronomy.stackexchange.com/a/21787/7982&quot;&gt;great answer&lt;/a&gt; about the US Naval Observatory's $3,500 expedition to Baker City Oregon to observe the &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_eclipse_of_June_8,_1918&quot; rel=&quot;noreferrer&quot;&gt;June 8, 1918 total solar eclipse&lt;/a&gt; links to the &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1919PA.....27....1H&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;noreferrer&quot;&gt;January 1919 Popular Astronomy article&lt;/a&gt; about the expedition. The article notes on page 6 the following coincidental observation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An interesting occurrence was the appearance of the brilliant new star in Aquila. This was seen by Mr. Peters and Mr. Conrad on the night of June 8 and the discovery was communicated by telegram to various observatories in the country. It had however been discovered earlier on the same night by other astronomers in the east.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the story behind this bright new star in Aquila? What was it? Who discovered it first?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2017-07-30T07:27:10.987" Title="What was the &quot;brilliant new star in Aquila&quot; on June 8, 1918, just after the solar eclipse?" Tags="&lt;star&gt;&lt;history&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21872" PostTypeId="2" ParentId="21871" CreationDate="2017-07-30T07:27:10.987" Score="11" Body="&lt;p&gt;It was Nova Aquilae 1918.  Wikipedia has some detail. &lt;a href=&quot;https://en.wikipedia.org/wiki/V603_Aquilae&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/V603_Aquilae&lt;/a&gt;&#xA;Apparently it was discovered by Zygmunt Laskowski, an amateur astronomer.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-07-30T07:27:10.987" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21875" PostTypeId="1" CreationDate="2017-07-31T01:22:53.373" Score="4" ViewCount="156" Body="&lt;p&gt;I did some research and a likely candidate is hot Jupiter type migration where the orbit didn't stabilize and crashed into the parent star.&lt;/p&gt;&#xA;" OwnerUserId="17732" LastActivityDate="2018-06-14T09:18:26.743" Title="Any observed example of a planet hitting its parent star?" Tags="&lt;impact&gt;&lt;orbital-migration&gt;" AnswerCount="0" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21878" PostTypeId="2" ParentId="21853" CreationDate="2017-07-31T10:42:01.327" Score="1" Body="&lt;p&gt;I may have found a partial answer to my question in the Navigational Astronomy chapter of &lt;a href=&quot;https://msi.nga.mil/NGAPortal/MSI.portal?_nfpb=true&amp;amp;_pageLabel=msi_portal_page_62&amp;amp;pubCode=0002&quot; rel=&quot;nofollow noreferrer&quot;&gt;'American Practical Navigator' by N. Bowditch&lt;/a&gt; . On page 228 &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] nutation, an irregularity in the motion of the Earth due to the&#xA;  disturbing effect of other celestial bodies, principally the Moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and on page 230 &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The precession of the Earth’s axis is the result of gravitational&#xA;  forces exerted principally by the Sun and Moon on the Earth’s&#xA;  equatorial bulge. The spinning Earth responds to these forces in the&#xA;  manner of a gyroscope. Regression of the nodes introduces certain&#xA;  irregularities known as nutation in the precessional motion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the dominant effects as for predicting the orientation of the Earth's axis seem to be the Earth's ellipticity combined with the attraction of the Sun and Moon. &lt;/p&gt;&#xA;" OwnerUserId="16581" LastActivityDate="2017-07-31T10:42:01.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21880" PostTypeId="1" AcceptedAnswerId="21894" CreationDate="2017-07-31T13:35:26.063" Score="3" ViewCount="191" Body="&lt;p&gt;A friend sent me an email which appears to be an extract from &lt;a href=&quot;http://www.cambridge-news.co.uk/news/perseid-meteor-shower-august-2017-13408672&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Cambridge News article&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A meteor shower astronomers have labelled 'a once in a lifetime opportunity' will take over the sky in August. The upcoming Perseid meteor shower will be the brightest in human history so far, Physics Astronomy reported.&lt;/p&gt;&#xA;&lt;p&gt;Anyone in the Northern Hemisphere should be able to view the shower on August 12. It is predicted to be so bright that stargazers should be able to catch a glimpse &lt;em&gt;&lt;strong&gt;even during the day&lt;/strong&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Astronomers believe that a shower this visible won't grace the sky for another 96 years. The Perseid meteor shower, one of the brighter light shows of the year, occurs annually between July 17 and August 24. The shower tends to peak around August 9-13.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;However, this article sources its information from the website &lt;a href=&quot;http://www.physics-astronomy.com/2017/07/get-ready-brightest-meteor-shower-in.html#.WX8wI4jytPb&quot; rel=&quot;nofollow noreferrer&quot;&gt;Physics Astronomy&lt;/a&gt;, a blog-type website which posts various articles on astronomical topics. Since Physics Astronomy doesn't seem to be a particularly reputable source, I will take this information with a pinch of salt.&lt;/p&gt;&#xA;&lt;p&gt;So, is this article true? Will you be able to see the Perseids in the daytime with the naked eye this year?&lt;/p&gt;&#xA;" OwnerUserId="1494" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-08-02T14:10:32.927" Title="Fact Check: How bright will the Perseids be in 2017?" Tags="&lt;amateur-observing&gt;&lt;meteor-shower&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21883" PostTypeId="2" ParentId="21880" CreationDate="2017-07-31T21:01:44.237" Score="3" Body="&lt;p&gt;There is variation in meteor showers from year to year, but for most showers, the big difference is how bright the moon is. If the moon is up, then it's hard to see many meteors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Perseids of 2017 will be rather ruined by the moon. There will be a waning gibbous moon, rising shortly after sunset and staying up for the rest of the night. It will make the sky too bright to see many meteors&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2018, with a new moon on the 11th, will be much better.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-07-31T21:01:44.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21884" PostTypeId="1" AcceptedAnswerId="21887" CreationDate="2017-07-31T21:24:42.810" Score="5" ViewCount="1441" Body="&lt;p&gt;One of the &lt;a href=&quot;https://eclipse2017.nasa.gov/sites/default/files/publications/safe_glasses_flyer.pdf&quot; rel=&quot;noreferrer&quot;&gt;NASA approved&lt;/a&gt; eclipse glasses manufacturers has the following warning on one of their &lt;a href=&quot;https://www.rainbowsymphonystore.com/products/plastic-eclipse-glasses&quot; rel=&quot;noreferrer&quot;&gt;eclipse glasses&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do &lt;strong&gt;NOT&lt;/strong&gt; look through a camera, telescope or binoculars with ANY eclipse shades on!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can see how the enlarged image of the Sun through a telescope or binoculars might overwhelm the eclipse glasses' filters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how would looking at the Sun using a camera's viewfinder or LCD screen reduce the effectiveness of eclipse glasses (assuming you're not using any zoom function)?&lt;/p&gt;&#xA;" OwnerUserId="2773" LastActivityDate="2017-08-01T18:41:47.887" Title="Why shouldn't you look at the Sun through a camera using eclipse glasses?" Tags="&lt;the-sun&gt;&lt;photography&gt;&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21886" PostTypeId="2" ParentId="21867" CreationDate="2017-07-31T23:30:02.603" Score="1" Body="&lt;p&gt;I have that same telescope in my collection! It served me well and I hope it will do the same for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This counterweight is not the one which comes with the telescope, but it should function much the same$-$it is $30.00, and you can find it &lt;a href=&quot;http://www.highpointscientific.com/skywatcher-counterweight-kit-star-adventurer-s20540?utm_source=google&amp;amp;utm_medium=cse&amp;amp;utm_term=SKY-S20540&amp;amp;gclid=Cj0KCQjwqvvLBRDIARIsAMYuvBF12HwLruQVZtkUV6DOeU8aO0J1mvXcTIzXlJcCHW7rbSia3jy7GUIaAib1EALw_wcB&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clear skies!&lt;/p&gt;&#xA;" OwnerDisplayName="user14781" LastActivityDate="2017-07-31T23:30:02.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21887" PostTypeId="2" ParentId="21884" CreationDate="2017-07-31T23:45:45.490" Score="4" Body="&lt;p&gt;Electronic viewfinders (such as the ones using LCD or OLED screens) are excluded from this discussion. There's no danger of eye injury there, except perhaps danger of damaging the camera's sensor if exposure is very long and the lens aperture is big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the classic or optical viewfinders that are targeted by the warning.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I can see how the enlarged image of the Sun through a telescope or binoculars might overwhelm the eclipse glasses' filters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A few observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's really the larger aperture of the instrument that is the problem here. The pupil of your eye is only a few mm in diameter. The objective lens or mirror of a telescope could be anywhere between dozens of mm to hundreds of mm (or thousands for very large instruments). The ratio of areas between telescope aperture and eye pupil is even greater - it's the square of the ratio of diameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say your eye's pupil is 2 mm in diameter. Let's say you use a 50 mm aperture telescope (small refractor or binoculars). The diameter ratio is 25x. The area ratio is 625x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the light captured by the very large area of the instrument is funneled into your eye through your pupil. With the instrument, now you're getting 625x more energy from the Sun, compared to the naked eye view. It's already dangerous to look at the Sun with the naked eye - with the instrument it's 625x more dangerous. And this is with a very small refractor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a class of &quot;solar filters&quot; that are made to be mounted on the instrument's eyepiece, or after the eyepiece. THESE ARE VERY DANGEROUS! All the increased, focused energy of the Sun is now absorbed by the filter, which can warp, melt, crack, or burst into flames. A filter failure at this point is likely to injure the user. White light filtering for solar observations must always occur ahead of the instrument, not after it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As for why classic, optical viewfinders were included in that list: it was out of an abundance of caution. Most of those viewfinders do not capture more light than your eye does, but some do. It's better to be safe than sorry. You can't expect everyone to be able to tell whether their viewfinder is dangerous or not. So, in a warning addressed to the general population, just tell them to stay away from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you install a full-aperture solar filter (like the Baader solar film) ahead of the viewfinder, then it's safe to use - provided the filter is attached firmly to the camera and cannot be blown away by some random gust of wind.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2017-08-01T18:41:47.887" LastActivityDate="2017-08-01T18:41:47.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21888" PostTypeId="1" AcceptedAnswerId="21890" CreationDate="2017-08-01T02:59:10.843" Score="1" ViewCount="493" Body="&lt;p&gt;I don't know very much about telescopes; I used one a few times as a kid, but that's about it. I understand the general design &amp;amp; use of reflectors &amp;amp; refractors, but I don't know enough to decipher focal length, etc. So, I don't feel comfortable trying to DIY my own contraption, that won't be nearly as good, assuming it works at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking to buy a minor telescope (a few hundred dollars [US] or less) to view the American Total Solar eclipse on Aug. 21, 2017; and possibly reuse for the Mercury transit in 2019. I'm a decently well off computer programmer, so blowing $200 (&amp;amp; some gas to drive up-state) for a one off just to experience the Solar eclipse is worth it to me, even if I don't use it again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I spent a few hours wandering around the internet for something, and I found a few styles of &quot;Projection Screens/Scopes&quot; that are really cool, but I can't find anywhere to buy them. The first is an image that several websites used as their article's banner image, but none say what it is or where to get it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Phtcf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Phtcf.jpg&quot; alt=&quot;Solar Telescope Projection Screen&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found another one, where a guy hacked together a projector scope from an old 8mm film projector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DzWVN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DzWVN.jpg&quot; alt=&quot;8mm projector scope&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found many DIY tutorials to roll-your-own out of dixie cups &amp;amp; Pringles cans, but as I said before: I'd rather not waste my time fumbling with that, since I'd probably screw it up; I'd rather pay money to get one that's good quality &amp;amp; done right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've wanted to buy a big light bucket for years now, so I can start looking at nebulae &amp;amp; planets, but never got around to it; mostly because I'm not sure what to buy. I've been looking at the &quot;Orion SkyQuest XX14i Dobsonian&quot; for a while, but never pulled the trigger. Given that the eclipse is only 3 weeks away, I figured something smaller would be a faster choice in the short term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the &quot;Celestron EclipSmart 50 50mm f/7.2 Alt-Az Solar Telescope&quot; that comes with a backpack storage bag, for \$100. It has a built-in (i.e. non-removable) solar filter, &amp;amp; a collapsible tripod. But that seems pretty cheap; is \$100 for a telescope like that normal, or is that a poor quality trap for laymen like me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do yall have any advice on something to get for the solar eclipse? And is there anywhere to get a professionally made projector screen like that, so I can watch the multi-hour process from a chair rather than bending over for 2 hours? (p.s. I'm 6 foot 3 inches tall) Also, if possible, being able to record video of it on my Samsung Galaxy S6 would be cool, but not required if that's a big deal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Yurelle&lt;/p&gt;&#xA;" OwnerUserId="17757" LastActivityDate="2017-08-01T04:34:40.340" Title="Solar Telescope Projector Screen? [&amp; intro telescope for solar eclipse]" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;photography&gt;&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21890" PostTypeId="2" ParentId="21888" CreationDate="2017-08-01T04:34:40.340" Score="1" Body="&lt;p&gt;For your purposes, I'd really recommend just purchasing the EclipSmart. The telescope itself is of reasonable quality. If you have a camera tripod, bring it with you to view the eclipse$-$the stock tripod is so flimsy as to be useless. If you dont have one, buy one. Additionally, you should invest in different eyepieces. The telescope comes with a 20-mm EP, which will provide relatively low magnification. I'd recommend getting a 6-mm EP, and possible an intermediate one as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clear skies!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I would recommend against getting the XX14i as your first telescope. I think instead you should get a smaller EQ, in the 4-6-in bracket. Having to manually find your target will teach you much about the skies, and having to do so with a smaller aperture will humble you. I worked a 4&quot; Alt-Az for a decade as a kid before moving to a 4.5&quot; EQ for a year, and then finally an XT12i as a highschool sophomore. Having to find my targets in my relatively bright skies made the sights I see through my 12&quot; that much more incredible. I think allowing yourself the same experience will be a decision you won't regret. Don't fall for &lt;a href=&quot;http://simostronomy.blogspot.com/2009/01/aperture-fever.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;aperture fever&lt;/a&gt;!&lt;/p&gt;&#xA;" OwnerDisplayName="user14781" LastActivityDate="2017-08-01T04:34:40.340" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21894" PostTypeId="2" ParentId="21880" CreationDate="2017-08-01T08:55:55.390" Score="3" Body="&lt;p&gt;To answer the second part of the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Will you be able to see the Perseids in the daytime with the naked eye this year?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. Regular meteor showers (those caused by the Earth passing through the orbit of a comet) have meteors that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... are caused by particles ranging in size from about that of a small pebble down to a grain of sand, and generally weigh less than 1-2 grams.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(source: &lt;a href=&quot;http://www.amsmeteors.org/meteor-showers/meteor-faq/#3&quot; rel=&quot;nofollow noreferrer&quot;&gt;American Meteor Society website&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most meteors are about as bright as the brightest stars (magnitude -1) or fainter.  Meteor showers do not contain fireballs, which are caused by objects which are way larger (&gt;1kg).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h3&gt;3. Can you see fireballs in daylight, ...&lt;/h3&gt;&#xA;  &#xA;  &lt;p&gt;Yes, but the meteor must be brighter than about magnitude -6 to be noticed in a portion of the sky away from the sun, and must be even brighter when it occurs closer to the sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(source: &lt;a href=&quot;http://www.amsmeteors.org/fireballs/faqf/#3&quot; rel=&quot;nofollow noreferrer&quot;&gt;American Meteor Society website&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is predicted to be so bright that stargazers should be able to catch a glimpse even during the day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So that claim is bogus and you should treat it the same way as a $3 bill.&lt;/p&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2017-08-01T08:55:55.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21897" PostTypeId="1" CreationDate="2017-08-01T12:55:58.917" Score="2" ViewCount="196" Body="&lt;p&gt;Have we ever seen a pulsar directly from above or below it's spin axis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering if a pulsar seen from this angle would appear to be constantly spinning or if it would just appear like a normal star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HVHtz.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HVHtz.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2844" LastActivityDate="2017-08-01T21:34:03.753" Title="What does a pulsar look like if viewed from above?" Tags="&lt;pulsar&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21898" PostTypeId="2" ParentId="21436" CreationDate="2017-08-01T16:48:51.313" Score="4" Body="&lt;p&gt;By way of starting a discussion, because I don't clearly see how to answer this question correctly or all the factors that may be involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you increase $G$, then by the virial theorem, the central temperature of the ball of gas (that will contract quasti-statically, since radiation cannot escape unimpeded) will be proportional to $GM/R$. So for a fixed mass, the ignition temperature for nuclear fusion will be reached at a larger radius and stars would be physically larger and less dense.&#xA;$$T_c \simeq \frac{GMm_u}{10Rk_B},$$&#xA;assuming an ionised hydrogen gas and where $m_u$ is an atomic mass unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to avoid &quot;starhood&quot;, you need to compress the gas either to within its Schwarzschild radius or to a density where it could be supported by electron degeneracy pressure without reaching a high enough temperature to ignite hydrogen fusion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the first of these. $R_s = 2GM/c^2$, so the central temperature reached at $R=R_s$ would be $\simeq m_u c^2/20k_B = 5\times 10^{11}$K, is &lt;em&gt;independent&lt;/em&gt; of $G$ and is far higher than the $10^{7}$K required for H fusion (ignore deuterium). So direct collapse to a black hole cannot circumvent nuclear fusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second possibility is likely to be more constraining.&#xA;There is a standard treatment given in &lt;a href=&quot;http://adsabs.harvard.edu/abs/1983bhwd.book.....S&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shapiro &amp;amp; Teukolsky (1983)&lt;/a&gt; thaat determines the radius $R$ at which electron degeneracy becomes important in supporting a star of a given mass $M$. The idea is that you work out the mass and radius for which the phase space occupied by an electron approaches $h^3$. The result is (for ionised hydrogen):&#xA;$$ R/R_{\odot} \simeq 0.07 (M/M_{\odot})^{-1/3}$$&#xA;However, digging into the maths a bit, this can be written as a proportionality&#xA;$$ R \propto G^{-1} M^{-1/3}$$&#xA;and thus the central temperature at which electron degeneracy pressure becomes important will be&#xA;$$T_c \propto GM/R  \propto G^2 M^{4/3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus (for a fixed $G$) what happens is that a contracting ball of gas will reach a critical $T_c$ for fusion if the mass exceeds some critical threshold (above about $0.075M_{\odot}$ in our universe) then H fusion will begin before degeneracy sets in. If you increase $G$ then this threshold actually becomes &lt;em&gt;smaller&lt;/em&gt; and smaller balls of gas will become stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about the Chandrasekhar mass? In our universe the threshold mass for fusion is much lower than the Chandrasekhar mass ($5.8M_{\odot}$ for pure hydrogen), so there is no problem in supporting an object just below the fusion threshold with electron degeneracy pressure. Dig a little deeper and we find that the Chandrasekhar mass is in fact $M_{Ch} \propto G^{-3/2}$. This means we can now express the critical central temperature in terms of mass &lt;em&gt;as a fraction&lt;/em&gt; of the Chandrasekhar mass.&#xA;$$ T_c \propto G^2 \left(\frac{M}{M_{Ch}}\right)^{4/3} (G^{-3/2})^{4/3} \propto \left(\frac{M}{M_{Ch}}\right)^{4/3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the threshold between a star that burns hydrogen and an object (brown dwarf) that will be supported by degeneracy pressure is a fixed fraction of the Chandrasekhar mass and does not depend on $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My conclusion is therefore that although changing $G$ will do interesting things to stellar evolution (changing the rate at which it occurs for instance - see &lt;a href=&quot;https://astronomy.stackexchange.com/questions/17937/solar-luminosity-proportional-to-g7?rq=1&quot;&gt;here&lt;/a&gt; where I show that the luminosity scales as $G^7$ and therefore lifetimes scale as $G^{-7}$ !), I do not think that changing $G$ allows direct collapse to a black hole without the central temperature becoming high enough to initiate hydrogen fusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, a failed star is possible in our universe - it just needs to be less than $0.075M_{\odot}$ and collapse is halted by electron degeneracy pressure. In a universe with an increased $G$ this threshold becomes &lt;em&gt;smaller&lt;/em&gt; as $G^{-3/2}$.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-08-02T10:53:08.487" LastActivityDate="2017-08-02T10:53:08.487" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21899" PostTypeId="2" ParentId="21809" CreationDate="2017-08-01T17:28:07.140" Score="2" Body="&lt;p&gt;According to the American Astronomical Society (AAS), the warning that eclipse glasses should not be used if they are more than 3 years old is &lt;strong&gt;outdated&lt;/strong&gt; (provided that they are compliant with the ISO 12312-2 standard).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;https://eclipse.aas.org/eye-safety/iso-certification&quot; rel=&quot;nofollow noreferrer&quot;&gt;page&lt;/a&gt; on their web site says the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Note: If your eclipse glasses or viewers are compliant with the ISO 12312-2 safety standard, you may look at the uneclipsed or partially eclipsed Sun through them for as long as you wish. Furthermore, if the filters aren't scratched, punctured, or torn, &lt;strong&gt;you may reuse them indefinitely&lt;/strong&gt;. Some glasses/viewers are printed with warnings stating that you shouldn't look through them for more than 3 minutes at a time and that you should discard them if they are more than 3 years old. &lt;strong&gt;Such warnings are outdated and do not apply to eclipse viewers compliant with the ISO 12312-2 standard adopted in 2015.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2773" LastActivityDate="2017-08-01T17:28:07.140" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21900" PostTypeId="2" ParentId="21897" CreationDate="2017-08-01T21:34:03.753" Score="1" Body="&lt;p&gt;My understanding is that we &quot;see&quot; pulsars (also called &quot;supernova remnants&quot;) by the radio and/or other electromagnetic radiations which they emit, which come to us in regular pulses, like the light-beam on the conning-tower at an airport.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because the beam of radiation shoots out along the magnetic-axis, while the object spins around its spin-axis, which might be considerably different from the magnetic-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we are directly above, it seems that we might not receive any of the radiation, so we might not be able to &quot;see&quot; the object.&lt;/p&gt;&#xA;" OwnerUserId="13290" LastActivityDate="2017-08-01T21:34:03.753" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21901" PostTypeId="2" ParentId="21843" CreationDate="2017-08-01T22:09:34.390" Score="8" Body="&lt;p&gt;The question has been addressed, but for completeness I should like to remark that the most thorough and readily understandable discussion of Olber's Paradox is that of E. Harrison in his book &lt;em&gt;Cosmology, the Science of the Universe&lt;/em&gt; (CUP 2000).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also worth remarking that we now know that the sky at night is not in fact dark: the light from the cosmic microwave background (CMB) dominates all other sources of radiation, including the sum total of all the stars.  Most of the CMB radiation comes out in the microwave region of the electromagnetic spectrum.  This was discussed by M. Longair  and R.A. Sunyaev in the journal &lt;em&gt;Astrophysical Letters&lt;/em&gt; (vol 4 pp65-70 1969) with the following sketch&lt;a href=&quot;https://i.stack.imgur.com/S0aRt.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S0aRt.jpg&quot; alt=&quot;The intensity of the radiation received from the sky over all of the electromagnetic spectrum.  The dashed lines were tentative estimates when this was first drawn.  The flux of the 2.7 CMB is shown as a blackbody dominating all other wavelengths.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Updates on this were published in 1990 by M. Ressell and M. Turner (&lt;em&gt;Comments on Astrophysics&lt;/em&gt;, vol 14 p323) and by R. Henry in 1999 (&lt;em&gt;Astrophys. J.&lt;/em&gt; Vol 516 pp. L49-L52).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the sky is not dark at night.  The light of the night sky is due to the CMB and not due to stars.  The finite lifetime of stars, the finite age of the universe and the cosmic expansion have together reduced the contribution from starlight. &lt;/p&gt;&#xA;" OwnerUserId="5301" LastActivityDate="2017-08-01T22:09:34.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21907" PostTypeId="2" ParentId="21822" CreationDate="2017-08-02T07:19:44.597" Score="3" Body="&lt;p&gt;The American Astronomical Society (AAS) answers this question on their &lt;a href=&quot;https://eclipse.aas.org/eye-safety/iso-certification&quot; rel=&quot;nofollow noreferrer&quot;&gt;How to Tell If Your Eclipse Glasses or Handheld Solar Viewers Are Safe&lt;/a&gt; page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically they say that since anyone can claim ISO 12312-2 compliance even if they haven't been certified, you should make sure you obtained your eclipse glasses from a vendor whose ISO 12312-2 compliance has been verified by the AAS Solar Eclipse Task Force:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We used to say that you should look for evidence that they comply with the ISO 12312-2 international safety standard for filters for direct viewing of the Sun. But now the marketplace is being flooded by counterfeit eclipse glasses that are labeled as if they're ISO-compliant when in fact they are not. So now we suggest that you make sure you get (or got) your eclipse viewers from one of the suppliers listed on our &lt;a href=&quot;https://eclipse.aas.org/resources/solar-filters&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reputable Vendors of Solar Filters &amp;amp; Viewers&lt;/a&gt; page.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They say that checking eclipse glasses to see if they meet the ISO 12312-2 standard would require a spectrophotometer, which most people do not have access to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Unfortunately, you can't check whether a filter meets the ISO standard yourself — doing so requires a specialized and expensive piece of laboratory equipment called a spectrophotometer that shines intense UV, visible, and IR light through the filter and measures how much gets through at each wavelength.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So they have assembled a task force to put together a list of manufacturers and vendors of eclipse glasses that they have verified to be ISO 12312-2 certified (their list is longer than NASA's list of &lt;a href=&quot;https://eclipse2017.nasa.gov/safety&quot; rel=&quot;nofollow noreferrer&quot;&gt;5 manufacturers&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The AAS Solar Eclipse Task Force has been working diligently to compile a list of such vendors, now posted on our &lt;a href=&quot;https://eclipse.aas.org/resources/solar-filters&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reputable Vendors of Solar Filters &amp;amp; Viewers&lt;/a&gt; page. We've checked manufacturers' ISO paperwork to make sure it's complete and that it comes from a recognized, accredited testing facility, and we've personally examined manufacturers' products. We've asked manufacturers to identify their authorized resellers, and we've asked dealers to identify the source of the products they're selling. Only when everything checks out do we add a vendor to our listing. If we don't list a supplier, that doesn't mean their products are unsafe — only that we have no knowledge of them or that we haven't convinced ourselves they are safe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They also give advice on how to tell if your eclipse glasses are not safe to use:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can you tell if your solar viewer is not safe? The only thing you can see through a safe solar filter from a reputable vendor is the Sun itself. If you can see ordinary household lights through your eclipse glasses or handheld viewer, it’s no good. Safe solar filters produce a view of the Sun that is comfortably bright (like the full Moon), in focus, and surrounded by black sky. If you glance at the Sun through your solar filter and find it uncomfortably bright, out of focus, and surrounded by a murky haze, it’s no good.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2773" LastActivityDate="2017-08-02T07:19:44.597" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21913" PostTypeId="1" CreationDate="2017-08-02T11:04:45.287" Score="0" ViewCount="532" Body="&lt;p&gt;My definition of traditional name is that name that appears in the stellar catalogs made in classical antiquity by Greeks and Romans and which were later translated and compiled by the Arabs to be later used again by Europeans from the Renaissance. &lt;a href=&quot;https://en.wikipedia.org/wiki/Traditional_Chinese_star_names&quot; rel=&quot;nofollow noreferrer&quot;&gt;I do not look for traditional Chinese&lt;/a&gt; or Mayan names, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most stars visible without a telescope were given names in Antiquity. These Greek or Babylonian names were transmitted and modified during the Middle Ages by Arabian compilers and many of them are still popular, such as Sirius or Vega.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other names are longer and more &quot;exotic&quot;, like Betelgeuse. I am thinking, for example, of really long and exotic names like &lt;a href=&quot;https://en.wikipedia.org/wiki/Alpha_Librae&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zubenelgenubi&lt;/a&gt; (Alpha Librae) or &lt;a href=&quot;https://en.wikipedia.org/wiki/Beta_Librae&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zubeneschamali&lt;/a&gt; (β Librae). Are there any stars whose &quot;ancient&quot; or &quot;medieval&quot; name are longer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_Arabic_star_names&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is a list of arabian star names&lt;/a&gt;, later modified by europeans, but it may be incomplete, because greek catalogues included about 1.000 stars and this list only includes 200.&lt;/p&gt;&#xA;" OwnerUserId="11487" LastEditorUserId="11487" LastEditDate="2017-08-02T13:07:43.540" LastActivityDate="2017-08-02T13:07:43.540" Title="What is the star whose western traditional name is the longest?" Tags="&lt;star-catalogues&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21915" PostTypeId="2" ParentId="21913" CreationDate="2017-08-02T12:40:19.197" Score="2" Body="&lt;p&gt;According to Google, the longest name for any star is “Shurnarkabtishashutu”, the Arabic for “under the southern horn of the bull”. The star is known as Zeta Tauri. You could also scroll thru Wikipedia's list of star names for different countries.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastActivityDate="2017-08-02T12:40:19.197" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21916" PostTypeId="2" ParentId="21849" CreationDate="2017-08-02T13:26:50.343" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is gravity only an attractive force?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR&lt;/strong&gt;&lt;br&gt;&#xA;Because mass is always positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;strong&gt;There are different notions of mass, but they're equivalent.&lt;/strong&gt;&lt;br&gt;&#xA;There are two distinct notions of mass: gravitational and inertial. The masses in Newton's law of gravitation, $F = \frac{Gm_1m_2}{r^2}$, are gravitational masses. The mass in Newton's second law of motion, $F=ma$, is inertial mass. Gravitational and inertial mass are implicitly assumed to be the same in Newtonian mechanics. General relativity makes this assumption explicit in the equivalence principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But what if they're not equivalent?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike mathematics, where one can simply make an assumption and see where it leads, assumptions in physics need to be validated. This assumption has been tested with many kinds of materials, both on the ground and in space. Variations on the Cavendish experiment using different kinds of materials have been made. Within the limits of the rather lousy accuracy of the gravitational constant (one part per ten thousand, at best), every one of these is consistent with the null hypothesis (gravitational and inertial mass are the same) and inconsistent with the hypothesis that different materials have measurably different gravitational and inertial masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth's Moon, with its very different near-side and far-side, provides an even better mechanism for testing this equivalence. Rather than the one part per ten thousand (at best) accuracy available to Cavendish-style experiments, the Moon shows that gravitational and inertial mass for sodium and iron are equivalent to within about &lt;a href=&quot;https://arxiv.org/abs/1203.2150&quot; rel=&quot;nofollow noreferrer&quot;&gt;one part per ten trillion&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So much for ordinary matter, but what about antimatter?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That an ordinary matter particle and its antimatter equivalent have the same (positive) inertial mass has been tested over and over in particle colliders around the world. Whether the equivalence principle also applies to antimatter remains a somewhat open question. While there are many reasons to think that the equivalence principle applies to antimatter as well as normal matter, testing that this is the case is very hard. The best results to date are from the ALPHA experiment, which tests whether neutral antihydrogen (a antiproton and an positron) falls up or down. The results are that antihydrogen's gravitational mass lies somewhere between -65 and 120 times its inertial mass. This is not anywhere close to conclusive, but it does lean towards antimatter having a positive gravitational mass, consistent with the equivalence principle.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-08-02T13:26:50.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21917" PostTypeId="2" ParentId="21880" CreationDate="2017-08-02T14:10:32.927" Score="2" Body="&lt;p&gt;The Perseids will &lt;strong&gt;not&lt;/strong&gt; be the brightest in recorded human history. The fact-checking website &lt;a href=&quot;http://www.snopes.com/perseid-meteor-shower-largest/&quot; rel=&quot;nofollow noreferrer&quot;&gt;snopes.com&lt;/a&gt; has this to say about this false claim:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The reason the web site that originally made this claim cites zero sources is likely due to the fact that there is a complete lack of factual information to support it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The 2017 Perseid shower won't even be as &quot;bright&quot; as it was last year, let alone in recorded history. &lt;a href=&quot;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20170001495.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Per NASA&lt;/a&gt;, the peak hourly rate in 2017 will be about 80% of the peak 2016 rate. Making matters worse, the 2017 shower will be obscured by the Moon. The 2016 rate wasn't the best show in recorded history, either, or even in recent history. The latter honor (&quot;best show in recent history&quot;) goes to the 1993 Perseid shower, which occurred a year after the parent comet Swift-Tuttle made a close approach to the Earth. By recent, I mean quite recent. The showers centered around the year 1862 were probably even better than those around 1992.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meteor showers result from the Earth intersecting the debris trail left behind by a comet. This debris trail is distributed along the comet's orbit, but the distribution is not uniform. The peak concentration in general is close to the comet itself. The distance between the Earth and the comet when the Earth comes closest to the comet's orbit has been increasing since 1992, making the intensity of the Perseid shower decrease rather than increase. The comet will make a very close approach in 2126, resulting in some nice shows for our children's children's children.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-08-02T14:10:32.927" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21918" PostTypeId="1" AcceptedAnswerId="21976" CreationDate="2017-08-02T20:14:41.553" Score="3" ViewCount="205" Body="&lt;p&gt;I recently read that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Gamma-ray bursts are catastrophic events, related to the explosion of massive stars 50 times the size of our sun. If you ranked all the explosions in the universe based on their power, gamma-ray bursts would be right behind the Big Bang,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm asking this question in the tentative to compare the Big Bang to a gamma-ray burst. I keep in mind that a key difference between both is that Big Bang was &lt;strong&gt;Not an explosion&lt;/strong&gt;, but want to know similarities to point out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Near the formation of a black-hole they &lt;a href=&quot;https://www.sciencedaily.com/releases/2017/07/170726132110.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;tell in this same article&lt;/a&gt; that matter and magnetic field play a key role the gamma-ray bursts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know magnetic fields are induced by charge displacement. Is it possible there was a kind magnetic field before protons and electrons formed, or during the first expansion phase?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are pictures I modified from Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Chronology_of_the_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;chronology of the universe&lt;/a&gt; to illustrate:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/kAVVS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kAVVS.png&quot; alt=&quot;Do we have a way to know how magnetic fields were at that time?&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Do we have a way to know how magnetic fields were at that time?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Close-up (image borrowed from &lt;a href=&quot;http://donsnotes.com/science/astronomy/&quot; rel=&quot;nofollow noreferrer&quot;&gt;donsnotes&lt;/a&gt; ):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WkN3X.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WkN3X.png&quot; alt=&quot;Any way we could have information on magnetism at this point of universe creation?&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Any way we could have information on magnetism at this point of universe creation?&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In the Electroweak Era?&lt;/li&gt;&#xA;&lt;li&gt;During the Great Unification Theory (GUT) Era?&lt;/li&gt;&#xA;&lt;li&gt;And then in the Plank Era?&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11264" LastEditorUserId="11264" LastEditDate="2017-08-04T12:51:38.010" LastActivityDate="2017-08-05T17:21:31.343" Title="Is there a way to know how the magnetic field was at the very beginning of our universe?" Tags="&lt;big-bang-theory&gt;&lt;magnetic-field&gt;&lt;gamma-ray-bursts&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21919" PostTypeId="1" CreationDate="2017-08-02T20:49:11.567" Score="1" ViewCount="164" Body="&lt;p&gt;Saw &lt;a href=&quot;http://www.easier.com/137022-breaking-news-planet-9-found.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; on someone from Hampshire UK reporting a possible discovery of Planet Nine. I thought that it would be too dim for anything but the most powerful telescopes to see. Is there any legitimacy to this article? &lt;/p&gt;&#xA;" OwnerUserId="6870" LastActivityDate="2017-08-03T19:17:13.727" Title="Is there any legitimacy to this report of planet nine being discovered by an amateur?" Tags="&lt;amateur-observing&gt;&lt;9th-planet&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21920" PostTypeId="2" ParentId="21919" CreationDate="2017-08-02T21:13:18.737" Score="3" Body="&lt;p&gt;This has not been confirmed so whether it is legit is unknown at this time.  A recent report (&lt;a href=&quot;https://arxiv.org/abs/1706.05348&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/1706.05348&lt;/a&gt;) calls into question the original analysis that leads to the initial report of planet nine. The report that fueled initial interest is given below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(EVIDENCE FOR A DISTANT GIANT PLANET IN THE SOLAR SYSTEM&#xA;Konstantin Batygin and Michael E. Brown&#xA;Division of Geological and Planetary Sciences, California Institute of Technology, Pasadena, CA 91125, USA; kbatygin@gps.caltech.edu&#xA;Received 2015 November 13; accepted 2016 January 10; published 2016 January 20)&lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="16926" LastEditDate="2017-08-03T01:45:20.770" LastActivityDate="2017-08-03T01:45:20.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21923" PostTypeId="1" AcceptedAnswerId="21929" CreationDate="2017-08-02T22:10:24.083" Score="0" ViewCount="74" Body="&lt;p&gt;When I use Stellarium, I am unable to see Azha in London or Atlanta, Georgia.  However, it is possible to see Azha in Tokyo and Auckland.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not changing the time, only the location.  I presume that my time is converted to their time.  So for example, if my time is 11:30 (London) then when I select Tokyo, I should see what the sky is like at 11:30 Tokyo time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see Azha in Auckland too.  I would've thought the ability to see a star is based on N/S hemisphere not E/W hemisphere.  Is it right I should be able to see Azha in the Eastern Hemisphere but in the West or am I using Stellarium incorrectly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asking because of the Eta Eridanids. They might be an obscure or a disappointing meteor shower. The reason for such differing locations was to see where in the world it's possible to see them. &lt;/p&gt;&#xA;" OwnerUserId="5401" LastEditorUserId="2365" LastEditDate="2017-08-03T08:04:30.807" LastActivityDate="2017-08-03T19:58:48.637" Title="Seeing Azha (Eta Eridani) in Northern Hemisphere" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21924" PostTypeId="2" ParentId="21528" CreationDate="2017-08-02T22:38:41.527" Score="2" Body="&lt;p&gt;You are correct that the western states have a higher probability of clear skies.  NOAA analyzed historical data for August 21 and has a &lt;a href=&quot;https://www.ncei.noaa.gov/news/ready-set-eclipse&quot; rel=&quot;nofollow noreferrer&quot;&gt;web page&lt;/a&gt; where they show a map indicating the chances of cloudiness in various parts of the country:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dFkl7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dFkl7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On that map, a white dot indicates a low probability of cloudiness, and the darker a dot is, then the higher the probability of cloudiness in that area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the same web page, they have an interactive map that shows the path of totality, and it lets you click on each dot to see the actual percentage chance that the eclipse will be visible, along with a breakdown of probabilities of clear skies, a few clouds, scattered clouds, broken clouds, and an overcast sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on that interactive map, it seems that Ontario, Oregon has the highest probability that the eclipse will be viewable at 93.3%:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RfaN0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RfaN0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2773" LastActivityDate="2017-08-02T22:38:41.527" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21925" PostTypeId="1" AcceptedAnswerId="21934" CreationDate="2017-08-03T01:35:32.300" Score="2" ViewCount="92" Body="&lt;p&gt;From what I understand, the TRAPPIST 1 system was discovered using the transiting method looking at 20 ultra-cool small stars. I'm not sure how many have been observed since this study.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I wonder is, what would we expect to see if all 20 had planets? I'm not sure how many planetary systems in this and future similar studies we would expect to see edge on. What about radial velocity measurements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did we just get lucky with TRAPPIST 1?&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2017-08-03T10:59:31.773" Title="TRAPPIST 1 and planet statistics?" Tags="&lt;exoplanet&gt;&lt;trappist-1&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21929" PostTypeId="2" ParentId="21923" CreationDate="2017-08-03T08:02:17.743" Score="0" Body="&lt;p&gt;Eta Eridani has a declination of about -8 degrees, so it is visible from most of the Northern Hemisphere. Eridanus is a constellation that is next to Orion, and like Orion, it is easiest to see during winter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Eta Eridanids are a meteor shower that have a peak on about 11th of August. The radiant on that date is just south of Eta Eridani, at about -11 degrees. On 11th of August, Eta Eridani rises at about 3am in London, About an hour before twilight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tokyo, which is much further south, and doesn't use Daylight saving, sees Eta Eridani rise at about 1am (local time). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you change location in Stellarium the time doesn't change. So if you change from London at 3am in August (UTC+1), the time in Tokyo will be 11 am (UTC+9).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Eta Eridanids are a minor shower with a ZHR of about 6 meteors per hour They will be visible whenever the radiant is above the horizon, but from more northern locations it is only visible for a short time. It also clashes with the Perseids. One may only see a couple of Eta Eridanids during the viewing window, which is why it was only identified in 2005.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-08-03T19:58:48.637" LastActivityDate="2017-08-03T19:58:48.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21934" PostTypeId="2" ParentId="21925" CreationDate="2017-08-03T10:59:31.773" Score="6" Body="&lt;p&gt;The probability of observing a planet transit is approximately $(R_p + R_s)/a$, where $R_p$ and $R_s$ are the planet and stellar radius respectively and $a$ is the semi-major axis of the planet's orbit. This assumes that planetary orbits are circular and randomly oriented with respect to our line of sight to the star (for which there is little or no counter-evidence).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of Trappist-1, it has an estimated radius of $R_p=0.11R_{\odot}$, the planetary companions have radii and semi-major axes of (in units of Earth radii and au respectively)  (1.09,0.011), (1.06, 0.015), (0.77,0.021), (0.92,0.028), (1.05, 0.037), (1.13, 0.045), (0.75, 0.059).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using approximation above, then the transit probabilities are 0.051, 0.037, 0.026, 0.020, 0.015, 0.012, 0.009 respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now these probabilities cannot be treated as independent of one another, since it is quite likely (though this is still the subject of research) that planets will naturally tend to inhabit the same orbital plane (the Trappist-1 planetary orbit inclinations are within 0.2 degrees of each other - flatter than the solar system). Thus you are much more likely to see second and subsequent transits in a system with one transiting planet than you are in a star picked at random. On the other hand, a system with lots of planets that are &lt;em&gt;not quite&lt;/em&gt; in the same orbital plane gives an increased chance of seeing one of them transit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, you can see from the probabilities above that seeing one and only one transiting system of this type from 20 targets, even if all of them had similar planetary systems,  is not at all unexpected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radial velocity studies of such systems are difficult, but currently being attempted. Trappist-1 has an apparent J magnitude of 11.4, which is &lt;em&gt;very&lt;/em&gt; challenging for current telescopes and instrumentation to get high resolution spectroscopy.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-08-03T10:59:31.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21937" PostTypeId="1" CreationDate="2017-08-03T12:25:10.137" Score="1" ViewCount="50" Body="&lt;p&gt;I am curious to know whether Jupiter is a reason that we have life on earth as it protects earth from comets, without Jupiter – comets would be more likely to hit us?&lt;/p&gt;&#xA;" OwnerUserId="17807" LastActivityDate="2017-08-03T12:25:10.137" Title="Is it true that Jupiter protects Earth?" Tags="&lt;jupiter&gt;&lt;comets&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2017-08-04T05:04:34.643" ContentLicense="CC BY-SA 3.0" />
  <row Id="21939" PostTypeId="1" AcceptedAnswerId="21969" CreationDate="2017-08-03T19:09:21.030" Score="3" ViewCount="273" Body="&lt;p&gt;Io is the only body in the Solar System whose crust is dominated by sulfur and its compounds. Carbon and nitrogen are also very common in the universe but we don't see them or their compounds in Io's crust. Why not?&lt;/p&gt;&#xA;" OwnerUserId="17264" LastActivityDate="2017-08-05T08:52:14.323" Title="Io: Why so much sulfur?" Tags="&lt;jupiter&gt;&lt;natural-satellites&gt;&lt;geology&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21941" PostTypeId="1" CreationDate="2017-08-03T19:51:49.890" Score="1" ViewCount="13411" Body="&lt;p&gt;According to the Globe Earth Theory, the Sun is in the middle of the Solar System and the Earth is the 3rd planet from the Sun, making Mercury and Venus between Earth and the Sun. If this is true, Mercury and Venus should only be seen from the lit side of the planet, since it is always between the Sun and Earth. How can you see it at night?&lt;/p&gt;&#xA;" OwnerUserId="17813" LastActivityDate="2017-08-04T01:28:11.297" Title="How can we see Venus at night?" Tags="&lt;venus&gt;&lt;night-sky&gt;" AnswerCount="2" CommentCount="8" ClosedDate="2017-08-13T20:59:49.727" ContentLicense="CC BY-SA 3.0" />
  <row Id="21942" PostTypeId="1" AcceptedAnswerId="21945" CreationDate="2017-08-03T20:07:33.077" Score="1" ViewCount="105" Body="&lt;p&gt;I'm looking at Saturn through a modest telescope at 50x. What point in space, expressed as the distance from Saturn, let's say it's 750 million miles from Earth, would I need to view Saturn with the naked eye and see it at the same apparent size as viewed from Earth at 50x. How do I estimate this in general for other objects as well. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="17814" LastEditorUserId="11264" LastEditDate="2017-08-04T09:26:54.980" LastActivityDate="2017-08-04T09:26:54.980" Title="Telescope magnification compared to distance from an object" Tags="&lt;telescope&gt;&lt;observational-astronomy&gt;&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21943" PostTypeId="1" CreationDate="2017-08-02T23:44:09.950" Score="1" ViewCount="504" Body="&lt;p&gt;I am interested in learning the locations and velocities of the near earth asteroids at the time of the crucifixion of Jesus Christ.&#xA; I would like to back-calculate the paths of all known near earth asteroids to find out if any was between the earth and the sun at the time of Jesus' crucifixion on Good Friday.  The Horizons system of NASA JPL only permits back calculations to about 1400AD.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to see if there was a clustering of asteroids on Good Friday between the earth and sun.  If so they could be the remnants of a gigantic collision that would have been capable of making a dust cloud in interplanetary space that could have produced a shadow on the entire earth which would account for the observations reported by the gospels and the Chinese astronomers and St. Dionysius and others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words if any asteroids are calculated to have been between the earth and sun on Good Friday they may be left-over debris from the possible collision that produced the dust cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this.  Do you know of any available computer program like the JPL  Horizons program that will work to give asteroid positions back in the time period between 29AD and 35AD?  (I give a range of years because biblical scholars give a range of possible years for the Crucifixion of Jesus.  The determination of a debris asteroid would help historians know the timing of the Crucifixion better, also.)&#xA;Thank you for your attention and assistance.&#xA;Best Regards,&#xA;Dr. Francis Kelly&#xA;President, The Catholic Association of Scientists and Engineers&lt;/p&gt;&#xA;" OwnerUserId="17817" OwnerDisplayName="Casepres" LastEditorUserId="2365" LastEditDate="2018-12-07T17:52:24.747" LastActivityDate="2019-04-18T18:42:30.553" Title="Asteroid locations at the time of the crucifixion of Jesus" Tags="&lt;asteroids&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21944" PostTypeId="2" ParentId="21943" CreationDate="2017-08-03T01:52:35.247" Score="12" Body="&lt;p&gt;The computer program you are asking for is literally impossible:  Even ignoring the affect of the greater universe and transient visitors such as comets on hyperbolic orbits, the Solar system is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Chaos_theory&quot; rel=&quot;noreferrer&quot;&gt;chaotic&lt;/a&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Complex_systems#Complexity_and_chaos_theory&quot; rel=&quot;noreferrer&quot;&gt;complex system&lt;/a&gt;.  We have a pretty good handle on the motion of the bodies we are, and have been aware of, during the time modern astronomy has been studying the skies, i.e., over the past few hundred years.  (Remember that in the earlier years of telescopic astronomy, we were &lt;em&gt;unaware of&lt;/em&gt; most of the currently known members of the Solar system, including &lt;em&gt;all&lt;/em&gt; of the asteroids, and &lt;em&gt;every&lt;/em&gt; massive object beyond Saturn.)  As you deviate further from the present (either past or future), the chaotic nature of the system comes more into play, especially on smaller bodies such as asteroids.  As a result, stating orbital positions either far in the past or far in the future requires a larger margin of error as the time differential grows.  Eventually you get to a point where the error is so great that we cannot say reliably where an object was or will be.  Perhaps if someone had come up with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion&quot; rel=&quot;noreferrer&quot;&gt;laws of planetary motion&lt;/a&gt; sooner than &lt;a href=&quot;https://en.wikipedia.org/wiki/Johannes_Kepler&quot; rel=&quot;noreferrer&quot;&gt;Kepler&lt;/a&gt;, we would have more reliable data further into the past, and therefore more accurate orbital elements over a longer range than we do, but that didn't happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The asteroid belt (white) and Jupiter's trojan asteroids (green)&lt;/strong&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/f/f3/InnerSolarSystem-en.png&quot; alt=&quot;&quot;&gt;&lt;br&gt;&#xA;From &lt;a href=&quot;https://en.wikipedia.org/wiki/Asteroid#/media/File:InnerSolarSystem-en.png&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, linked from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Asteroid&quot; rel=&quot;noreferrer&quot;&gt;Asteroid&lt;/a&gt; page&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to the chaotic nature of the Solar system, there's also the fact that the vast majority of the asteroids revolve around the Sun &lt;em&gt;outside&lt;/em&gt; of the Earth's orbit, so the chances of any asteroids &lt;em&gt;we know of&lt;/em&gt; having &quot;a gigantic collision ... that could have produced a shadow on the entire Earth&quot; (i.e., between the Earth and the Sun) are between &lt;em&gt;slim&lt;/em&gt; and &lt;em&gt;none&lt;/em&gt;.  (Note also that the &lt;em&gt;amount of material&lt;/em&gt; present in said &quot;dust cloud&quot; would have to be &lt;em&gt;immense&lt;/em&gt; in order to have a &lt;em&gt;noticeable&lt;/em&gt; affect on the amount of sunlight falling on the Earth.)  The only way such a collision could have occurred would be if two objects traveling &lt;em&gt;through&lt;/em&gt; the Solar system happened to collide in such a manner that the &quot;dust cloud&quot; the collision produced &lt;em&gt;happened&lt;/em&gt; to fall between the Earth and Sun - and also left the Solar system without a trace afterwards.  This would have required the objects being in similar trajectories crossing the plane of the ecliptic at a high angle and speed - a &lt;em&gt;chaotic&lt;/em&gt; event if there ever was one!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, basically you're looking for a way to &lt;em&gt;prove&lt;/em&gt; that a chaotic event happened sufficiently far in the past that chaotic errors make our knowledge of most of the Solar system (asteroids and such) &lt;em&gt;and outside influnences&lt;/em&gt; questionable - no, that's not going to happen.  Math doesn't work that way.&lt;/p&gt;&#xA;" OwnerUserId="17422" OwnerDisplayName="FKEinternet" LastActivityDate="2017-08-03T05:55:56.563" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21945" PostTypeId="2" ParentId="21942" CreationDate="2017-08-03T20:39:27.350" Score="5" Body="&lt;p&gt;Just divide the actual distance by the magnification.  In your example that would give 750 million /50, which is 15 million miles.  (Actually saturn is currently over 800 million miles from earth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this works is that a magnification of M means the object is M times its apparent linear size, which requires being M times closer.  This is a slight approximation, but is fine for distant objects.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-08-03T20:39:27.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21946" PostTypeId="2" ParentId="21941" CreationDate="2017-08-04T00:24:47.620" Score="6" Body="&lt;p&gt;We can see Venus at night (dusk) and in the morning (predawn), just not all night or every night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need some geometry to understand this, and it boils down to tangents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've shamelessly borrowed this diagram from a paper to show the idea.  It's discussing Mercury, but it's the same idea for Venus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x9toF.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x9toF.png&quot; alt=&quot;Elongation of Mercury&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll simplify things by using the pretty good approximation that Earth and Venus have circular orbits.  We'll also assume Earth is a perfect sphere - another good approximation.  We're also assuming Earth's rotation is not at an angle to the plane of it's orbit (that's the worse approximation !).  Finally we're going to assume the orbits are not inclined (not a bad approximation, but not strictly true).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maths would get messy if we didn't do those things, so it turns out that the maths works out that the number of degrees that's the maximum the Earth can rotate through and still see Venus after the sun goes down is approximately :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta = 90^\circ-cos^{-1}\left(\frac {r_{venus}} {r_{earth}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\frac {r_{venus}} {r_{earth}} \approx 0.72$ and this means $\theta \approx 46^\circ$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think about $46^\circ$ of rotation out of a quarther circle that's slightly more than half, and a quarter circle's worth of rotation of the Earth is about 6 hours, so that angle represents our ability to see Venus for (at most) three hours after sunset and three hours before sunrise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also discussed &lt;a href=&quot;http://curious.astro.cornell.edu/our-solar-system/62-our-solar-system/planets-and-dwarf-planets/venus/250-why-can-t-we-see-venus-at-night-when-it-is-an-inner-planet-intermediate&quot; rel=&quot;noreferrer&quot;&gt;in this article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The technical term for these angles is &lt;a href=&quot;https://en.wikipedia.org/wiki/Elongation_(astronomy)&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;elongation&lt;/em&gt;&lt;/a&gt;.  The actual maximum elongation is slightly over $47^\circ$ so the simplifications didn't do much damage to our estimate !  As it happens this year (in January) we had an elongation of Venus that was almost the maximum.  It varies (due to the factors we simplified) between $45^\circ$ and and slightly over $47^\circ$.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-08-04T00:24:47.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21947" PostTypeId="2" ParentId="21941" CreationDate="2017-08-04T00:32:44.723" Score="10" Body="&lt;p&gt;Before we begin: how far a planet is seen from the Sun is called &lt;strong&gt;elongation&lt;/strong&gt;, and it's measured in degrees. 0° elongation means it's right on top of the Sun (or behind); 180° elongation means it's opposite to the Sun (it's highest in the sky at midnight, when the Sun is on the other side of the Earth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Elongation_(astronomy)&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Elongation_(astronomy)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the image below, where the Sun and the Earth are shown, along with an inferior planet (such as Mercury) and a superior planet (such as Mars). Elongation is the angle marked ε.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6Adp3.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6Adp3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can you see it at night?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Both Venus and Mercury are always seen close to the Sun. You can never see them at midnight, true. But you can see them during twilight, or, in Venus' case, shortly after nightfall (and then it sets quickly), or shortly before dawn in the morning (and then it's masked by the light of day). This is not a paradox or an impossibility. Here's why:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury in particular can only be seen during twilight, very close to sunset or sunrise. Its orbit is so close to the Sun in space (only 39% the size of Earth's orbit), it does appear very close to the Sun even in the sky as seen from Earth. You could not see it during the day, of course, because the sky is so bright. But during twilight you can sometimes see it peeking just to the side of the Sun, very briefly - then it either sets in the West, or is swallowed by the light of day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The greatest elongation that Mercury can achieve is 28°. Hold your arm straight out in front of you and make a fist; twice the size of your fist is nearly 28°. But even that's quite exceptional; most of the time Mercury stays closer to the Sun, closer than half of that angle (less than one fist).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I should make it very clear: it's very rare that you can actually see Mercury with your naked eyes. It pretty much has to be right at maximum elongation. I've observed its phases (just like the Moon, or Venus) in a small dobsonian telescope a couple times; at max elongation it's close to the first quarter phase, or last quarter, and it looks like a very tiny replica of the Moon seen at that same phase. Kind of like this (except in my small telescope no craters or mountains were visible, because the image was just too tiny):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HSSCc.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HSSCc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Planetary_phase&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Planetary_phase&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus' orbit is quite a bit wider, about 72% the size of Earth's own orbit. So it can swing ahead or behind the Sun quite widely. But that's all it can do - swing ahead or behind. It can never go &quot;all the way around&quot; the Earth. You can never see it at midnight, when the Sun is on the other side of Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus' maximum elongation is 47°. Stretch your arm out and make a fist; now stick your thumb and your pinky sideways as far as they go (like the &quot;hang loose&quot; gesture, but stick them out all the way). The span from tip to tip is about 25°; twice that span is the maximum elongation of Venus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2iXwI.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2iXwI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine Venus at maximum elongation - two &quot;hang loose&quot; hand spans. The Sun has already set in the West about 1 hour and 40 minutes ago, so this is during the night, it's dark already, and now the Sun is one &quot;hang loose&quot; hand span below horizon. But Venus is at maximum elongation, which means it's one handspan &lt;em&gt;above&lt;/em&gt; horizon. That's quite hight in the sky actually, you can see it pretty well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there you have it. You can actually see Venus during the night. Not for long, because it still can't wander off too far from the Sun. It must remain within two &quot;hang loose&quot; hand spans from the Sun, or less than that distance, at all times. So Venus itself will set in the West soon after you see it. It cannot hang around until midnight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus also shows phases, just like the Moon or Mercury. I've watched those too in my telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get more precise angle measurements if you can use an old-school marine sextant, like captains on ships back in the day. The fist and &quot;hang loose&quot; gesture techniques are just approximations. Or just get a protractor and tape two straws to it at the desired angle, and sight along the straws to estimate the angle in the sky.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The superior planets (Mars, Jupiter, etc) are not like that. Their orbits are larger than Earth's own orbit. So they can actually be seen, once in a while, high in the sky at midnight, when the Sun is on the opposite side of Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please refer to the diagram at the top of this reply, it's quite self-explanatory.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, as you can see, the standard solar system model in astronomy is in excellent agreement with what you can see with your own eyes. There is no disagreement between observation and the model - if there was, &lt;em&gt;the model would be fixed&lt;/em&gt;. That's how science works.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2017-08-04T01:28:11.297" LastActivityDate="2017-08-04T01:28:11.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21948" PostTypeId="1" AcceptedAnswerId="21956" CreationDate="2017-08-04T02:04:54.490" Score="10" ViewCount="226" Body="&lt;p&gt;I'm quite interested to know the history of the study of the Earth's departure from spherical symmetry. What were the first methods used to model the non-spherical geometry of the Earth? For example, there are geodesy satellite and missions used now to determine the gravity field of the Earth but that is &quot;recent&quot; technology. They obviously had other methods prior to this.&lt;/p&gt;&#xA;" OwnerUserId="16812" LastActivityDate="2019-02-26T18:34:35.927" Title="How did we first determine the Earth's departure from sphericity?" Tags="&lt;solar-system&gt;&lt;earth&gt;&lt;history&gt;&lt;space-geometry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21951" PostTypeId="1" AcceptedAnswerId="21959" CreationDate="2017-08-04T09:07:16.783" Score="1" ViewCount="350" Body="&lt;p&gt;This morning I was in the shower and it struck me that the word planet might be related to the fact that the PLANEts orbit the sun in a PLANE. Did this naming somewhat happen by accident or was it consciously chosen? If it is the former then I'm simply mindblown by the fact that I didn't see this relation before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I quickly looked up the etymology on the word planet, which comes from Greek and means wanderer/wandering. Nonetheless, could there still be a relation, both words might stem from the same origin...?&#xA;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="17823" LastEditorUserId="17823" LastEditDate="2017-08-08T22:47:35.300" LastActivityDate="2018-06-07T19:26:18.003" Title="Is there a relation with the noun planet and to be in a plane?" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;naming&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="21952" PostTypeId="1" CreationDate="2017-08-04T10:56:00.750" Score="-1" ViewCount="150" Body="&lt;p&gt;I could never understand how in Interstellar, they say our every hour spent on the planet is equal to 7 years on Earth and when he returns back to her daughter, the father is older than her daughter. How is this possible when both the planet over there and Earth and for the fact everything in the universe is moving with the same time reference point. The age of daughter and father should have been the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know but is this termed as Time Dilation ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this work ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please avoid any formulae in explanation.&lt;/p&gt;&#xA;" OwnerUserId="13587" LastEditorUserId="13587" LastEditDate="2017-08-04T11:10:34.723" LastActivityDate="2017-08-07T16:21:06.873" Title="How do I imagine time difference?" Tags="&lt;time-dilation&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21955" PostTypeId="2" ParentId="21943" CreationDate="2017-08-04T15:14:00.813" Score="5" Body="&lt;p&gt;Just for clarification, it sounds like you're asking, in addition to asteroid tracking, about a possible astronomical explanation to &lt;a href=&quot;https://en.wikipedia.org/wiki/Crucifixion_darkness#Naturalistic_explanations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Crucifixion Darkness&lt;/a&gt;.   That would require a collision in transit between the Earth and the Sun that would make a debris cloud large enough for 3 hours of shade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such an astronomical event would be very rare because asteroid on asteroid collisions are rare and this one would need to be precisely timed and placed.   Ideally you'd want the debris field to be moving in roughly the same direction as the earth relative to the sun, to maximize the shading period of time.  Such an event is extremely extremely unlikely but not impossible.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am interested in learning the locations and velocities of the near&#xA;  earth asteroids at the time of the crucifixion of Jesus Christ.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://www.nasa.gov/audience/foreducators/postsecondary/features/F_Near_Earth_Program.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NEO Program&lt;/a&gt; has greatly increased the tracking of &lt;a href=&quot;https://en.wikipedia.org/wiki/Near-Earth_object#Near-Earth_asteroids&quot; rel=&quot;nofollow noreferrer&quot;&gt;Near Earth Asteroids&lt;/a&gt;.  Current estimates put the number of 1-km or greater asteroids at 981 +/- 19.  Smaller asteroids, the number grow exponentially.   Smaller asteroids are harder to see.   According to this, &lt;a href=&quot;https://www.vox.com/a/asteroid-day&quot; rel=&quot;nofollow noreferrer&quot;&gt;16,000 have been identified&lt;/a&gt; and there are thousands more.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed out in the other answer, Near Earth Asteroids are subject to gravitational perturbations and that makes predicting  their location a couple thousand years ago quite difficult, especially if they make more than one close pass to Earth because orbital changes get amplified with each pass, the closer the pass, the greater the amplification of uncertainty.   Tracking thousands of asteroids back nearly 2,000 years is an enormous task.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;any was between the earth and the sun at the time of Jesus'&#xA;  crucifixion on Good Friday.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Asteroids are too small to block the sun in any meaningful way.   We only study near Earth Asteroids to be prepared for collisions.   Other than the occasional collision (and possible space mining), Near earth asteroids are astronomically boring objects that are too small to see.  They would have gone mostly unnoticed to anyone before telescopes because they wouldn't have been visible, even in transit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a clustering of asteroids on Good Friday between the earth and sun. If&#xA;  so they could be the remnants of a gigantic collision that would have&#xA;  been capable of making a dust cloud in interplanetary space that could&#xA;  have produced a shadow on the entire earth which would account for the&#xA;  observations reported by the gospels and the Chinese astronomers and&#xA;  St. Dionysius and others.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Asteroids don't &quot;cluster&quot;.  They're too few and too far apart.  A collision of two is possible but rare.   A cluster of many, enough to block the sun for 3 hours is impossible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asteroid collision and a dust cloud is an interesting proposal.   Unlikely but worth exploring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a question about &lt;a href=&quot;https://space.stackexchange.com/questions/1382/how-frequently-do-asteroids-collide-with-each-other&quot;&gt;asteroid collisions with other asteroids&lt;/a&gt; in space.   While it's believed that they happen, they're rare events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;inter-asteroid / asteroid-comet / inter-comet collisions are extremely&#xA;  rare. They are in fact so rare, that there is basically no statistics&#xA;  about it. It is hard or near impossible to put a number of that.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;However, there are attempts to predict collisions. Keeping in mind&#xA;  that orbital parameters of small and distant objects can only be&#xA;  determined with some uncertainty, the idea is to go through the&#xA;  database of known objects and predict when a pair of objects comes&#xA;  into a certain proximity. This gives candidates for collision events.&#xA;  They are rather interesting for researchers, who are studying the&#xA;  structure or chemical composition of such objects. Another however,&#xA;  the amount of collisions actually happening after such a prediction is&#xA;  still incredibly low.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, while we know that asteroid on asteroid collisions do happen they are so rare that none has ever been observed.   For such an event to block out the sun, you'd need a couple very large asteroids, and the collision would probably need to be quite close to the Earth and in ideal placement between the Earth and the Sun where the debris field moved in relation with the Sun and Earth so as to maintain a shadow.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the collision would need to be high speed enough to create a large spray of dust, so not a glancing blow, but a full on impact.  That's improbabilities on top of improbabilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.astro.umd.edu/~msk/science/trails/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Meteor and comet trails&lt;/a&gt; (not to be confused with comet tails) follow an elliptical orbit.   Those are easy to trace back to their origin because all the debris is lined up and orderly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0JVMf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0JVMf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Impact collisions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Collisions behave very differently and you'd need a sufficiently large high speed collision to create any kind of large dust cloud to create 3 hours of shade on Earth.  Tracking debris from a 2,000 year old collision is very hard without having that debris in hand - then, maybe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An impact crater is what the footprint of a collision looks like.   (Resemblance to the Death Star is purely coincidental)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iSaGJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iSaGJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what a larger collision looks like (no longer visible to the eyes but detectable by gravitational and crustal density anomalies.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZFwwC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZFwwC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what an even larger impact looks like, where you get enough debris blown off where it forms into a moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XtSmW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XtSmW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explanation (&lt;a href=&quot;https://plutopod.tumblr.com/page/21&quot; rel=&quot;nofollow noreferrer&quot;&gt;scroll down to giant impact&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's thought that &lt;a href=&quot;https://en.wikipedia.org/wiki/Binary_asteroid&quot; rel=&quot;nofollow noreferrer&quot;&gt;Binary asteroids&lt;/a&gt; sometimes called double asteroids or asteroids with a moon are relatively common, and may be the result of impacts, but there in lies one problem.   A 2,000 year old impact might just look like a binary asteroid.   It wouldn't be possible to discern the date of the impact just by telescope and orbital tracking, but statistically a 2,000 year old impact would be abnormally young.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the largest, highest speed collisions, the two objects would just blow their debris apart - and there would basically be nothing to detect 2,000 years later.   Debris from a collision would either coalesce back into an asteroid or binary asteroid by gravity or simply drift apart.   In either case, evidence of a collision wouldn't be obvious outside of some hands on analysis of glassy compounds formed by impact or other chemical studies, but we'd actually have to get the debris into a lab to study it that closely.  By telescope, there would no evidence of such a collision and debris cloud after 2,000 years.   Any debris from the cloud would look just like all the other space dust with nothing distinguishable about it at least by telescope.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other problem with finding a theoretical impact that you're looking for is that you can't limit it to near earth asteroids.   Any comet or longer period asteroid that crosses the Earth could have had such an impact and any evidence of that comet being impacted would have left with the comet as it flew past the outer planets long ago, perhaps changing it's trajectory so it's no longer an earth crosser.   The number of objects to look for grows enormously when you consider that one of the objects could have come from much further out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A third problem with your theory is that such an impact, close enough to cause visible shade would very likely shower the Earth with debris and that would have been noted in the history books as well as a meteor shower like none before or since.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A long or medium period comet being struck by an asteroid might be a better explanation as the icy body being impacted would create a dirty gas cloud of some size while reducing debris impact on Earth as much of the ices would melt into gases and disburse under the sun's rays.   That would decrease the showered by meteors problem, but make finding the impacted object even harder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your space impact idea is a clever attempt to explain crucifixion darkness, but astronomically speaking, hugely improbable.   And the answer to your question is that evidence, if such an event did happen, would be very hard to detect, and with current technology, probably impossible.  Good question though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to think about evidence is that, when a collision happens on a planet, the planet's surface is relatively static and the record is preserved.  Obviously there's some weathering but information can be deduced by impacts that are hundreds of millions of years old.   In space, in the solar system, everything moves.   Things fly apart from each other while orbiting the sun.   A two thousand year old collision in space, even a large one, might leave no detectable record after 2,000 years, or if one object survived it could be virtually anywhere in the solar-system 2,000 years later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-08-04T18:54:08.653" LastActivityDate="2017-08-04T18:54:08.653" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="21956" PostTypeId="2" ParentId="21948" CreationDate="2017-08-04T15:59:30.473" Score="7" Body="&lt;p&gt;Isaac Newton was the first to put it together and realized the Earth had an oblate shape in 1687.    &lt;a href=&quot;https://en.wikipedia.org/wiki/Pendulum#Early_observations&quot; rel=&quot;noreferrer&quot;&gt;See 1672 - Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1666 &lt;a href=&quot;https://en.wikipedia.org/wiki/Pendulum#1656:_The_pendulum_clock&quot; rel=&quot;noreferrer&quot;&gt;Robert Hooke reasoned&lt;/a&gt; that pendulums could measure variation in gravity, but this remained unobserved until 1671, when &lt;a href=&quot;https://en.wikipedia.org/wiki/Jean_Richer&quot; rel=&quot;noreferrer&quot;&gt;Jean Richter&lt;/a&gt; observed that pendulums kept different time near the equator (French Guiana compared to Paris).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Around that time, 1673, &lt;a href=&quot;https://en.wikipedia.org/wiki/Pendulum#1673:_Huygens.27_Horologium_Oscillatorium&quot; rel=&quot;noreferrer&quot;&gt;Christian Huygens observed&lt;/a&gt; that the wide swing of a pendulum would move in a very slight oval, so to make pendulum clocks more accurate, the swings were kept to a low angle and the swing kept perpenducular   That, however, has to do with the Earth's rotation, but not your question specifically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton was the first to put it all together that the observed gravitational fluctuation was due to the equatorial bulge and faster speed of rotation near the equator.    &lt;a href=&quot;https://www.scientificamerican.com/article/earth-is-not-round/&quot; rel=&quot;noreferrer&quot;&gt;Article here&lt;/a&gt; also credits him with being the first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, related:   &lt;a href=&quot;https://en.wikipedia.org/wiki/History_of_geodesy&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/History_of_geodesy&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-08-04T17:28:40.043" LastActivityDate="2017-08-04T17:28:40.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21957" PostTypeId="2" ParentId="11615" CreationDate="2017-08-04T18:12:15.307" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://in-the-sky.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://in-the-sky.org&lt;/a&gt; will help you spot what planets are visible from your location with direction and time. I could see Jupiter, Saturn using the naked eyes as bright spots. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;90x magnification should be just fine for Jupiter and Saturn - you'd see the disc of Jupiter clearly with the four Galilean satellites and Saturn's rings&lt;/p&gt;&#xA;" OwnerUserId="17832" LastEditorUserId="2365" LastEditDate="2019-08-11T11:18:50.193" LastActivityDate="2019-08-11T11:18:50.193" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21958" PostTypeId="1" CreationDate="2017-08-04T19:36:36.123" Score="0" ViewCount="177" Body="&lt;p&gt;In chapter 6 of &quot;The Revolutions of the Heavenly Spheres&quot; where Copernicus maintains that the heavens are immense compared to the size of the Earth, the Edward Rosen translation of the book states this &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The upshot of the argument, then, is the claim that the earth as a part of the&#xA;  celestial sphere shares in the same nature and movement so that, being close to the&#xA;  center, it has a slight motion. Therefore, being a body and not the center, it too will&#xA;  describe arcs like those of a celestial circle, though smaller, in the same time. The&#xA;  falsity of this contention is clearer than daylight. For it would always have to be&#xA;  noon in one place, and always midnight in another, so that the daily risings and&#xA;  settings could not take place, since the motion of the whole and the part would be&#xA;  one and inseparable.&quot;&lt;br&gt;&#xA;  To add context. Copernicus is debating the classical point which assumes that &quot;the center is at rest, and the nearer an object is to the center, the slower its motion.&quot; Copernicus in the previous paragraph states that this assertion could be translated into a sphere. Where the center are the poles, and the constellations around those poles (i.e. Ursa Minor, Draco, etc...) would be rotating slower as they are closer to those poles. Yet Copernicus refutes this by saying that if those constellations truly moved slower than the rest, they wouldn't be able to make a full 360° rotation in a year.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Copernicus then in the paragraph above, argues that the Earth too must orbit around the center in order to account for the &quot;slowness&quot; in the poles. However Copernicus further refutes this by stating that there would be no day/night cycle in this case.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To elaborate on my question asked in the title. Why would the Earth need to orbit in order to account for this &quot;slowness.&quot; And why would there be no day/night cycle even though the firmament would be rotating and not the Earth along with it.&lt;/p&gt;&#xA;" OwnerUserId="17321" LastEditorUserId="17321" LastEditDate="2017-08-07T03:01:16.833" LastActivityDate="2017-09-06T19:46:16.293" Title="Why would there be no day/night cycle even though the firmament would still be rotating, according to Copernicus" Tags="&lt;earth&gt;&lt;history&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21959" PostTypeId="2" ParentId="21951" CreationDate="2017-08-04T21:23:26.650" Score="13" Body="&lt;p&gt;There is no close connection. &lt;em&gt;Planet&lt;/em&gt; is from Greek &lt;em&gt;asteres planetai&lt;/em&gt; meaning wandering stars (via Latin and French). &lt;em&gt;Plane&lt;/em&gt; is from Latin &lt;em&gt;planus&lt;/em&gt; meaning &quot;level or flat&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that both words are derived from the proto-indo-European root *pele-, (flat) However the evidence linking *pele to &lt;em&gt;planetai&lt;/em&gt; is weak. Thus there is no more connection between planet and plane than between planet and field (field is also derived from *pele, via the sound change in German from p to f) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.etymonline.com/index.php?allowed_in_frame=0&amp;amp;search=planet&quot; rel=&quot;noreferrer&quot;&gt;http://www.etymonline.com/index.php?allowed_in_frame=0&amp;amp;search=planet&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.etymonline.com/index.php?allowed_in_frame=0&amp;amp;search=plane&quot; rel=&quot;noreferrer&quot;&gt;http://www.etymonline.com/index.php?allowed_in_frame=0&amp;amp;search=plane&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.etymonline.com/index.php?allowed_in_frame=0&amp;amp;search=plane&quot; rel=&quot;noreferrer&quot;&gt;http://www.etymonline.com/index.php?term=*pele-%20(2)&amp;amp;allowed_in_frame=0&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-08-05T05:38:21.317" LastActivityDate="2017-08-05T05:38:21.317" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21960" PostTypeId="1" AcceptedAnswerId="21961" CreationDate="2017-08-05T03:01:37.400" Score="2" ViewCount="63" Body="&lt;p&gt;I'm in the Minneapolis/St. Paul area and around 9:30 on August 4th 2017 I saw a bright light that got dimmer moving from the tree line to the east up and to my left fairly slowly, and steadily getting dimmer. There were no flashing lights like an airplane. Around 5-10 minutes later I saw another light exactly like the first in nearly the same place going the same direction, they both faded and disappeared. I grew up in the north and have spent many nights out in hay fields staring at the sky but sadly I have never really studied astronomy and my guess at brightness is based on Google images. My first guess would have been a satellite but seeing two on the same path makes me think otherwise. Thoughts?&lt;/p&gt;&#xA;" OwnerUserId="17834" LastActivityDate="2017-08-05T04:58:46.773" Title="Two lights roughly 5 minutes apart about as bright as Mars or maybe Saturn" Tags="&lt;amateur-observing&gt;&lt;meteor&gt;&lt;luminosity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21961" PostTypeId="2" ParentId="21960" CreationDate="2017-08-05T04:58:46.773" Score="3" Body="&lt;p&gt;According to CalSky (&lt;a href=&quot;https://www.calsky.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.calsky.com&lt;/a&gt;), there were two Iridium flares visible low in the south-east from Minneapolis at 21:18 (Iridium 60) and 22:28 (Iridium 55) on that evening. Both would have been about 15 degrees elevation and both in the right ballpark for brightness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iridium flares occur when the rising / setting Sun's rays bounce off the antennae or solar panels of one or other of the Iridium communications satellites. By their very nature they occur around dusk or dawn. They'll flare up, persist for a few seconds and fade away.&lt;/p&gt;&#xA;" OwnerUserId="17502" LastActivityDate="2017-08-05T04:58:46.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21962" PostTypeId="1" CreationDate="2017-08-05T06:10:34.710" Score="1" ViewCount="1735" Body="&lt;p&gt;The telescope I'm using is a National Geographic 76 Diameter, 700 focal length, Equatorial Mount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's hard to use the equatorial mount, as this is my first telescope it's hard to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything I see is upside down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how to polar align my telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how to make move the telescope around properly because of this EQ mount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be really appreciated&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I watched videos of people aligning their finderscope but i don't even know how to move mine let alone align it.&lt;/p&gt;&#xA;" OwnerUserId="17501" LastEditorUserId="17501" LastEditDate="2017-08-05T06:29:37.843" LastActivityDate="2019-09-12T11:22:15.353" Title="I need help using this telescope!" Tags="&lt;telescope&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21966" PostTypeId="2" ParentId="21918" CreationDate="2017-08-05T07:59:03.533" Score="-1" Body="&lt;p&gt;Magnetic fields play a role in gamma-ray bursts (GRB) because you have a star that is collapsing into a black hole the magnetic fields serve to funnel a small portion of material into two jets coming out of the poles. If the jet happens to be pointing in the direction of the Earth, we see a very bright light source and that's what we call a GRB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly none of this has any analog in the Big Bang.&lt;/p&gt;&#xA;" OwnerUserId="17836" LastActivityDate="2017-08-05T07:59:03.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21967" PostTypeId="2" ParentId="21962" CreationDate="2017-08-05T08:12:22.717" Score="0" Body="&lt;p&gt;Many telescopes make images look upside down, flipped left-to-right, or both (i.e. rotated by 180 degrees). This is just the nature of imaging systems and the occasional image erector or &lt;a href=&quot;https://en.wikipedia.org/wiki/Star_diagonal&quot; rel=&quot;nofollow noreferrer&quot;&gt;star diagonal&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are right that equatorial mounts are more difficult to use. To ease the frustration, I suggest you just stop worrying about the mount and just point the telescope manually. You can start with easier targets and gradually try to find more difficult ones:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The easiest target is the moon. Just place the telescope on the floor with no concern for alignment and aim it at the moon. Look for craters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) After the moon, try to find binary stars. The easier pair to see is Mizar and Alcor, which are a pair of stars in the Big Dipper. Again, just point the telescope manually. Then gradually learn about more binary stars (I really like Albireo). Use the free planetarium software &lt;a href=&quot;http://www.stellarium.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellarium&lt;/a&gt; to help you find them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) After binary stars, try to find the moons of Jupiter. This is more difficult because Jupiter moves in the sky, but it is also very rewarding. You will definitely need Stellarium to tell you when/where to find Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later in the future you can return to trying to setup the equatorial mount. The only advantage of setting up the mount correctly is that you can then have the telescope track the sky for you. It's a nice feature, but one that you can survive without.&lt;/p&gt;&#xA;" OwnerUserId="17836" LastEditorUserId="7982" LastEditDate="2019-09-12T11:22:15.353" LastActivityDate="2019-09-12T11:22:15.353" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="21968" PostTypeId="2" ParentId="21952" CreationDate="2017-08-05T08:38:08.903" Score="2" Body="&lt;p&gt;I don't like analogies. They are always wrong (if they weren't wrong, they wouldn't be analogies) but they often look like explanations. Instead of an analogy, I will give you a thought experiment. It has no math, but you still have to concentrate to follow it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Imagine that you are in a tower and you drop a cannon ball. The cannon ball speeds up because gravitational potential energy is converted into kinetic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Now instead imagine that you shoot a cannon ball straight up. The cannon ball slows down because it's consuming energy to climb Earth's gravitational well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Now imagine that instead of a cannon ball, you shoot a photon. The photon is climbing up the gravitational well too, so it is not unreasonable to imagine that it must also be losing kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here is the thing: a photon cannot slowdown. We know that from experiments. The speed of light is a constant of the universe. So if the photon cannot slow down, how can it lose energy? Well, to lose energy the photon needs to change wavelength. Longer wavelength means less energy. So as the photon climbs up the gravity well, the wavelength gets longer. This is known as gravitational redshift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to finish the thought experiment, imagine that you have a friend at the International Space Station, looking at your photon. Imagine also that on the ground you have a clock, that is synchronized with the photon. For every crest of the wave, your clock does one tick. You would imagine that if you see that your clock ticks always happen together with the crest of each wave, your friend at the space station must agree with that fact too. But here is the thing: He is looking at the photon when it has a longer wavelength. If he has to agree that your clock ticks coincided with the crest of each wave, then he has to see your clock ticks happening farther apart than what you saw. In other words, he will see your clock ticking more slowly. This is time dilation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To conclude: What I wrote here is not a proof, but it is a thought experiment that hopefully conveys that time dilation is reasonable.&lt;/p&gt;&#xA;" OwnerUserId="17836" LastActivityDate="2017-08-05T08:38:08.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21969" PostTypeId="2" ParentId="21939" CreationDate="2017-08-05T08:52:14.323" Score="3" Body="&lt;p&gt;Io is not dominated by sulfur. Io is mostly silicate rock and iron. Sulfur is a thin coating on the surface and sulfur dioxide makes up most of the atmosphere, but that's because sulfur dioxide is a volatile. In other words, it has a much lower melting and boiling point than silicate rock. So while the rock remains solid, the sulfur dioxide is gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon is not common on Io's crust for the same reason that it is not common on Earth's crust. Carbon may be common in the universe, but it binds to oxygen to make a gas. When the solar system was forming most of the carbon was in the form of CO or CO2 and would not easily have coalesced into a solid body. Silicon is much less common than carbon, but Earth and Io are made of silicon because silicon makes solid stuff at the relevant temperatures.&lt;/p&gt;&#xA;" OwnerUserId="17836" LastActivityDate="2017-08-05T08:52:14.323" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21970" PostTypeId="1" CreationDate="2017-08-05T10:10:22.557" Score="2" ViewCount="188" Body="&lt;p&gt;I have seen Jupiter and its 4 largest moons in a telescope, several times. The moons are always aligned in a line. This is fine. Very interesting. But.. in case of Saturn, not only that we can see its ring, moons are not aligned in a line. I was wondering why is it so and I came to a conclusion that Saturns axial tilt must be bigger than Jupiters. I have visited english wikipedia and found out that it is true - and that the difference is about 23°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not a physicist. I do not even study astrophysics, but I wonder why the difference is so big. I came to another conclusion that the rotation speed may play a huge role here. The faster an object is spinning, the smaller axial tilt. And according to wikipedia, there is a difference and it seems that it fits in my theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So. I did my research, but now I am not sure if that´s all and if it´s right. I mean, maybe there is another factor that I forgot to count-in. I know that, according to scientists, Jupiter formed before Saturn. Does it mean that the difference in formation answer my question to why the axial tilt &amp;amp; the rotation speed vary so much? Or is it because of the distance to Sun?&lt;/p&gt;&#xA;" OwnerUserId="17842" LastEditorUserId="2365" LastEditDate="2017-08-05T10:39:46.967" LastActivityDate="2017-08-05T10:39:46.967" Title="Why does the axial tilt &amp; the rotation speed of Saturn &amp; Jupiter vary so much?" Tags="&lt;planet&gt;&lt;astrophysics&gt;&lt;planetary-formation&gt;&lt;planetary-science&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21971" PostTypeId="1" CreationDate="2017-08-05T11:18:26.977" Score="0" ViewCount="722" Body="&lt;p&gt;Telescope :  National Geographic 76/700 EQ&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried aiming my telescope at the moon but I just couldn't do it, it was so hard to do, my finder scope isn't helping at all and I watched a bunch of videos saying to use the screws to align the finder scope but the screws on my finder scope don't do anything. Any suggestions?? I really need help tonight is cloudy and I only get like 5 minutes every hour to look at the moon&lt;/p&gt;&#xA;" OwnerUserId="17501" LastActivityDate="2017-08-05T14:35:46.263" Title="How do I align my finder scope National Geographic 76/700 EQ" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21974" PostTypeId="2" ParentId="21971" CreationDate="2017-08-05T14:35:46.263" Score="1" Body="&lt;p&gt;The moon is generally big enough to not need a finder scope. In general, The easiest way to aim a telescope directly at something is to look parallel to the telescope, right above the edge. If you can just see the object you are aiming at, it should be at least visible in the finder scope (if properly configured, see below). In the case of the moon, try to make just the top half visible when looking parallel to your telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are having trouble using a finder scope, it might not be properly aligned. A finder scope has two wires forming a cross. A properly aligned finder scope will have an object in the middle of the telescope's view if it is exactly on the cross of the wires. If this is not the case, the position of the wires can be adjusted using the screws. This is best done at daytime by aiming the telescope at a recognizable landmark (somewhat) nearby, such as a tower. &lt;/p&gt;&#xA;" OwnerUserId="11824" LastActivityDate="2017-08-05T14:35:46.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21975" PostTypeId="1" AcceptedAnswerId="21977" CreationDate="2017-08-05T17:19:19.983" Score="3" ViewCount="102" Body="&lt;p&gt;There are a lot of objects of different sizes in the Kuiper belt. Are objects in the Kuiper belt stable? Is there any chance in the future that one of the larger objects would accumulate enough critical mass to start the process of becoming a proper planet (i.e., clearing its neighboorhood)?&lt;/p&gt;&#xA;" OwnerUserId="17732" LastEditorUserId="2153" LastEditDate="2017-08-05T22:23:54.000" LastActivityDate="2017-08-05T22:23:54.000" Title="Kuiper belt planet formation" Tags="&lt;planet&gt;&lt;planetary-formation&gt;&lt;kuiper-belt&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21976" PostTypeId="2" ParentId="21918" CreationDate="2017-08-05T17:21:31.343" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a way to know how the magnetic field was at the very beginning of our universe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know of one. But then again I think there was &lt;em&gt;no&lt;/em&gt; magnetic field at all at the very beginning of the universe. I can't point to any material that proves this beyond a doubt. But I can say that a quiescent black hole &quot;has no hair&quot;, so it has no magnetic field, and the evolution of the universe is said to be something like pulling away from a black hole. See Hawking saying the universe is like a time reversed black hole &lt;a href=&quot;http://news.bbc.co.uk/1/hi/sci/tech/4249192.stm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I recently read that &quot;Gamma-ray bursts are catastrophic events, related to the explosion of massive stars 50 times the size of our sun. If you ranked all the explosions in the universe based on their power, gamma-ray bursts would be right behind the Big Bang&quot;. I'm asking this question in the tentative to compare the Big Bang to a gamma-ray burst. I keep in mind that a key difference between both is that Big Bang was &lt;strong&gt;Not an explosion&lt;/strong&gt;, but want to know similarities to point out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There aren't many similarities. A gamma-ray burst is where some of the infalling matter is directed by magnetic fields generated by infalling matter and converted into gamma rays. The Big Bang is when the space of the universe started expanding.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Near the formation of a black-hole they &lt;a href=&quot;https://www.sciencedaily.com/releases/2017/07/170726132110.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;tell in this same article&lt;/a&gt; that matter and magnetic field play a key role the gamma-ray bursts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;True enough. But as far as I know matter and magnetic fields played no role in the Big Bang. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I know magnetic fields are induced by charge displacement. Is it possible there was a kind magnetic field before protons and electrons formed, or during the first expansion phase?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's possible I suppose. A magnetic field is associated with &lt;a href=&quot;https://en.wikipedia.org/wiki/Curl_(mathematics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;curl&lt;/a&gt; aka &quot;rot&quot; which is short for rotor. The early universe could have had an overall magnetic field, or gravitomagnetic field, if it was somehow rotating. See Mark Hadley's &lt;a href=&quot;http://www2.warwick.ac.uk/newsandevents/pressreleases/galaxy_sized_twist/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Galaxy sized twist in time pulls violating particles back into line&lt;/a&gt;. Also see &lt;a href=&quot;http://ns.umich.edu/new/releases/8467-the-universe-may-have-been-born-spinning-according-to-new-findings-on-the-symmetry-of-the-cosmos&quot; rel=&quot;nofollow noreferrer&quot;&gt;the universe may have been born spinning&lt;/a&gt;. But I think it's rather speculative stuff myself.    &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do we have a way to know how magnetic fields were at that time?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't think so. The early universe was said to be &lt;a href=&quot;https://www.google.co.uk/search?q=early+universe+maelstrom+-lego+-marvel&amp;amp;oq=early+universe+maelstrom+-lego+-marvel&amp;amp;gs_l=psy-ab.3...9026.13219.0.13473.8.8.0.0.0.0.104.677.7j1.8.0....0...1.1.64.psy-ab..0.0.0.wni7IJfoGOU&quot; rel=&quot;nofollow noreferrer&quot;&gt;maelstrom&lt;/a&gt;. Hence I think information about the initial state got scrambled.   &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2017-08-05T17:21:31.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21977" PostTypeId="2" ParentId="21975" CreationDate="2017-08-05T22:15:33.927" Score="3" Body="&lt;p&gt;The Kuiper Belt is fairly large (with inner and outer radii $\sim30\text{ AU}$ and $\sim50\text{ AU}$), but it does not contain much mass. The total mass is likely somewhere between $0.01$ and $0.1$ Earth masses&lt;sup&gt;&lt;a href=&quot;http://iopscience.iop.org/article/10.1086/423697/pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;[1]&lt;/a&gt;, &lt;a href=&quot;http://adsabs.harvard.edu/abs/2001AJ....122.1051G&quot; rel=&quot;nofollow noreferrer&quot;&gt;[2]&lt;/a&gt;&lt;/sup&gt;, which is essentially a mass range from five times the mass of Pluto to the mass of Mars. Now, it was likely much more massive in the past - its origin and mass loss are tied to the evolution of the orbits of ice giants in the early Solar System (see in particular &lt;a href=&quot;https://en.wikipedia.org/wiki/Nice_model#Formation_of_the_Kuiper_belt&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Nice model&lt;/a&gt;) - but we're talking about the formation of planets in the future. At present, there simply isn't enough mass to go around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's talk about &lt;a href=&quot;https://en.wikipedia.org/wiki/Clearing_the_neighbourhood&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;clearing the neighborhood&quot;&lt;/a&gt;. The IAU's &lt;a href=&quot;https://www.iau.org/static/resolutions/Resolution_GA26-5-6.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;definition of a planet&lt;/a&gt; merely states, as one of its three criteria, that a planet must have&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;cleared the neighbourhood around its orbit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wikipedia gives several quantitative measures of whether or not an object has cleared its neighborhood. The quantity $\Pi$ (&lt;a href=&quot;https://arxiv.org/abs/1507.06300&quot; rel=&quot;nofollow noreferrer&quot;&gt;Margot (2015)&lt;/a&gt; is given by&#xA;$$\Pi(m,a)=\frac{m}{M^{5/2}a^{9/8}}k$$&#xA;where $m$ is the mass of the body (Earth masses), $M$ is the mass of the star (solar masses), $a$ is the radius of its semimajor axis (AU), and $k\simeq807$. A body is a planet if $\Pi&amp;gt;1$.  If the mass of this hypothetical planet, $m_H$, is $0.01$ Earth masses, then no orbit in the range $30\text{ AU}&amp;lt;a&amp;lt;40\text{ AU}$ gives me $\Pi(.01M_{\oplus},a)&amp;gt;1$. If $m_H$ is $0.1$ Earth masses, then almost all $a$ in the same range give me $\Pi(0.1M_{\oplus},a)&amp;gt;1$. That said, it is &lt;em&gt;extremely&lt;/em&gt; unlikely that a body could amass even a small fraction of the total mass in the Kuiper Belt - remember the inner and outer boundaries!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantity $\Lambda$ (&lt;a href=&quot;http://www.boulder.swri.edu/~hal/PDF/planet_def.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stern &amp;amp; Levison (2002)&lt;/a&gt;) is given by&#xA;$$\Lambda(m,a)=\frac{\mu^2}{a^{3/2}}\times\left[\text{Terms accounting for the properties of the body's orbit}\right]$$&#xA;where $\mu$ is the ratio of the mass of the body to the mass of the star. The author's list a value on the order of $\Lambda=3\times10^{-3}$ for Pluto; if we assume that the orbital terms are about the same for this body, we find $\Lambda(0.01M_{\oplus},a)\sim0.075$ and $\Lambda(0.1M_{\oplus},a)\sim7.5$, with the cutoff being $\Lambda=1$. Again, it is not possible for the body to accumulate the total mass of the Kuiper Belt, so we can safely assume that $\Lambda\ll1$.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2017-08-05T22:15:33.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21978" PostTypeId="1" AcceptedAnswerId="21979" CreationDate="2017-08-05T22:59:59.063" Score="1" ViewCount="1124" Body="&lt;p&gt;I am excited to see the 2017 eclipse in South Carolina and would like to understand where exactly the eclipse-path intersects I-95, so that I can plan a day-trip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have approximated at roughly Santee,SC, but would really like to overlay and pinpoint the eclipse-path intersection with I-95 in Google maps (&lt;a href=&quot;http://xjubier.free.fr/en/site_pages/SolarEclipsesGoogleEarth.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;blue line in .kml file&lt;/a&gt;).  Unfortunately, the KML file does not show I-95&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Hasn't someone already done this with the web version with Google maps?&lt;/li&gt;&#xA;&lt;li&gt;Is there a simple process to either embed I-95 in the Google Earth .kml or &lt;/li&gt;&#xA;&lt;li&gt;Command Google Earth to overlay the blue eclipse path over the internet connected map data?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I was able to plot Santee, SC (on I-95) on the .kml file:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VK7A3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VK7A3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any edits to sharpen or clarify questions are appreciated.  I pose this question to the astronomy community because I think that someone has already done this.  See you at the eclipse!&lt;/p&gt;&#xA;" OwnerUserId="17847" LastEditorUserId="17847" LastEditDate="2017-08-05T23:06:55.477" LastActivityDate="2017-08-07T15:45:15.143" Title="2017 Eclipse intersection with I-95" Tags="&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21979" PostTypeId="2" ParentId="21978" CreationDate="2017-08-05T23:06:24.077" Score="2" Body="&lt;p&gt;NASA has a very good interactive &lt;a href=&quot;https://eclipse2017.nasa.gov/sites/default/files/interactive_map/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;online map&lt;/a&gt; which also shows highways.  It looks like I-95 intersects with the path by a wildlife refuge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.imgur.com/ldLA9HL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/ldLA9HL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kVc5L.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kVc5L.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A note on planning your trip: there will most likely be a very large number of people who also want to go out to see the eclipse.  So be prepared for lots of traffic.  Also, depending on where you plan on going, you will probably have to leave very early in order to get a good spot.&lt;/p&gt;&#xA;" OwnerUserId="13426" LastEditorUserId="7982" LastEditDate="2017-08-07T15:45:15.143" LastActivityDate="2017-08-07T15:45:15.143" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21981" PostTypeId="1" CreationDate="2017-08-06T01:22:26.140" Score="0" ViewCount="137" Body="&lt;p&gt;I purchased a National Geographic 76/700 EQ telescope and there was a piece missing. I'm wondering if it's needed or if I should return it, if you could also tell me what it does that'd be great. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an image of what is missing :&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/sVvpc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sVvpcm.png&quot; alt=&quot;Missing part&quot; title=&quot;Click for full image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17501" LastEditorDisplayName="user17206" LastEditDate="2017-08-06T04:57:24.047" LastActivityDate="2017-08-06T16:41:53.713" Title="Missing piece in my telescope!" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21983" PostTypeId="1" CreationDate="2017-08-06T04:20:13.963" Score="1" ViewCount="179" Body="&lt;p&gt;So I'm a beginner at astronomy and I was trying to look at the moon but while trying to aim at it I saw a ring-like object in the sky. Does anyone know what this would be? all I could see was a ring with nothing in the middle of it. Was this a special type of star?&lt;/p&gt;&#xA;" OwnerUserId="17501" LastActivityDate="2017-08-06T19:13:16.523" Title="I saw something unusual in the sky" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2017-08-27T11:44:43.857" ContentLicense="CC BY-SA 3.0" />
  <row Id="21985" PostTypeId="1" CreationDate="2017-08-06T07:49:15.203" Score="2" ViewCount="104" Body="&lt;p&gt;I would like to see some GEO and HEO images of Earth. But I can't find a good source. There are a lot of LEO images of Earth but I can'f find any raw images from higher (preferably 30,000km+) altitudes. There should be at least 400 satellites at that height, according to Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any website with those kind of images?&lt;/p&gt;&#xA;" OwnerUserId="17852" LastActivityDate="2017-08-06T18:41:26.827" Title="GEO and HEO images of Earth" Tags="&lt;orbit&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21987" PostTypeId="1" CreationDate="2017-08-06T14:29:02.467" Score="2" ViewCount="156" Body="&lt;p&gt;I tried to find the habitable planets, thus I suppose to calculate the effective temperatures that incident on the planets from their host stars, I used &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;T_\mathrm{eff}=\frac{L}{4\pi\sigma_\mathrm{SB}R^2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I don't know how to calculate the effective temperatures in polar regions?&lt;/p&gt;&#xA;" OwnerUserId="17854" LastEditorUserId="2365" LastEditDate="2017-08-06T22:28:06.997" LastActivityDate="2018-01-16T15:40:32.347" Title="Luminosity in polar regions" Tags="&lt;habitable-zone&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21988" PostTypeId="2" ParentId="21981" CreationDate="2017-08-06T16:41:53.713" Score="1" Body="&lt;p&gt;It looks likely to be one of the bolts for adjusting the polar alignment of the scope. There are usually two opposed hand operated bolts for fine adjustment of the left/right pointing of the mount on the tripod (so it points north/south - you're expected to position the tripod so its roughly correct to start with), and another two opposed bolts (might only be one in some designs) for tilting the top part of the mount  up/down so that the rotation axis in the top part (parallel to the telescope tube in the photo) (which pivots around the silver bolt left and level with the top of the red circle) points at the celestial pole (how much tilt this needs varies with your latitude).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that bit is missing - (it looks most likely to be the one you screw in to tilt the top part of the mount upwards more) - you won't be able to set the scope up properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you check the instructions on how to set the mount up, you should be able to tell which bit it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The reason why it's important, is that if you set the mount up properly, then when something starts to drift out of view due to the earth's rotation, then you can keep it in view by only adjusting one of the slow motion controls).&lt;/p&gt;&#xA;" OwnerUserId="17016" LastActivityDate="2017-08-06T16:41:53.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21989" PostTypeId="2" ParentId="21985" CreationDate="2017-08-06T18:41:26.827" Score="2" Body="&lt;p&gt;You might want to look at noaa.gov (&lt;a href=&quot;https://www.nnvl.noaa.gov/view/globaldata.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.nnvl.noaa.gov/view/globaldata.html&lt;/a&gt; in particular)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and similar sites for weather for other countries. In other words, the satellites in geostationary orbits (GEO) that take images are mostly weather satellite. I do not have a breakdown by type, but I would guess that a vast majority of the satellites in GEO are communication satellites, and they have no reason to take images.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that satellites in high earth orbit are there for astronomical purposes (to &quot;get away&quot; from the Earth, such as Chandra) or communication (such as the Russian Molniya satellites), so they are not imaging the Earth. &lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-08-06T18:41:26.827" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21990" PostTypeId="2" ParentId="21983" CreationDate="2017-08-06T19:13:16.523" Score="2" Body="&lt;p&gt;If you were using a telescope with a central obstruction (like a newtonian reflector (including dobsonian mounted ones) or an SCT or MCT, then what you saw is most likely an out of focus bright object (such as a star).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When things are in focus, a star shows as a point (actually, at very high magnifications, a tiny disc with concentric light and dark rings around it - this is called an Airey disc - caused by diffraction effects - but usually the magnification is low enough that it looks like a point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When things are out of focus, you seen the shape of the telescopes aperture that gets bigger the more out of focus you are. For a refractor, with no central obstruction, this is a circle. For things with a central obstruction, you see a ring (donut) shape. This is what causes donut shaped out of focus highlights on pictures taken with mirror lenses. (and why normal out of focus highlights with normal lenses are round or close to it (the camera iris blades often produce a polygon shape when the lens is stopped down)).&lt;/p&gt;&#xA;" OwnerUserId="17016" LastActivityDate="2017-08-06T19:13:16.523" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21992" PostTypeId="2" ParentId="21958" CreationDate="2017-08-07T07:42:01.990" Score="1" Body="&lt;p&gt;Find a spherical object and imagine it's a celestial sphere with the world at the middle... consider the daily rotation of the sphere around as a spinning top, and a yearly tilt through the constellations which is the same as an unstable spinning top which is about to tip over. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, rotate it and dont spin the days, and only spin the sphere for the yearly seasons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How fast are all points on the sphere rotating as it tilts, same as a spinning top is not straight, and omitting the individual rotations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All points on the sphere are going at the same speed, the top of an orange describes a 1cm axial tilt, and the sides of the sphere also describe a 1cm axial tilt... put the orange sideways and tilt it again, the rind is at the same speed as the side whichever way the orange is viewer. How to combine the slower star speed of rotation in 24 hours with the equal star speed of rotation through the seasons? The previous theory didn't give two speeds and axes of rotation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the 7 heavens other than the stars can be described as an epicycle on a 2D, as a turntable and epicycle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the stars, the 2D epicycle is more complicated, because it has to account for two speeds of rotation, the constant one around the seasons and the one that is slower in the middle. The epicycle geometry breaks down in 3D as a sphere, and only a sphere can explain the two types of rotations in the stars, and the 2100 year one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Copernicus said: A sphere cannot move as an epicycle. Take an orange and try to move it as an epicycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dm6Ev.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dm6Ev.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2017-09-06T19:46:16.293" LastActivityDate="2017-09-06T19:46:16.293" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21995" PostTypeId="2" ParentId="20374" CreationDate="2017-08-07T15:18:41.777" Score="0" Body="&lt;p&gt;The universe began with a Big Bang, which supplied the universe with huge amounts of matter in the form of free quarks and electrons. After the universe had expanded for a bit, it cooled enough for simple atoms like hydrogen-1 and helium-4 . This time was known as 'the dark ages' because light could not easily penetrate the matter, and besides, there were not any stars to create light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the matter created at the beginning of the dark ages was not evenly distributed. The regions with slightly more matter ended up attracting, through gravity, more and more matter to the protostar. The matter eventually condensed to a critical density, where fusion begins. This seems to be the leading hypothesis as I searched through several sources, including &lt;a href=&quot;http://sdcc3.ucsd.edu/~ir118/SIO87W13/FirstStars.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Formation of the First Stars and Galaxies&lt;/a&gt; and &lt;a href=&quot;http://www.astro.yale.edu/larson/papers/SciAm04.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;The First Stars in the Universe - Yale Astronomy&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10161" LastEditorUserId="10161" LastEditDate="2017-08-08T00:44:43.953" LastActivityDate="2017-08-08T00:44:43.953" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21996" PostTypeId="2" ParentId="21952" CreationDate="2017-08-07T16:21:06.873" Score="0" Body="&lt;p&gt;It's a bit odd, but I hope this helps. It's like a coordinate system with time on the x axis and space on the y. For example, if you move at a speed of 5 mph, that can be split between the x and y axes. If you're going mostly in the x direction, you're not moving that fast in the y. With time, the idea is that if you are moving extremely fast through space (our y axis in this case), there's not as much left over to travel through time (the x axis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we have a fixed &quot;speed&quot; through space-time, the faster we move through space, the slower we move through time (It's like placing a ladder, the length stays the same, but depending on how we angle it, it can go very high, or just barely off the ground)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously not a perfect explanation, but it at least gets the right idea for what you're looking at.&lt;/p&gt;&#xA;" OwnerUserId="17870" LastActivityDate="2017-08-07T16:21:06.873" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21997" PostTypeId="2" ParentId="21849" CreationDate="2017-08-07T16:30:48.713" Score="0" Body="&lt;p&gt;Theoretically, gravity can be &quot;attractive&quot; in the sense that objects move towards you when pushed. This can occur from negative mass (doesn't seem to make sense, but theoretically possible). Peter Engels and others have written a paper about it &lt;a href=&quot;https://arxiv.org/pdf/1612.04055.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and it's an interesting idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is that by cooling the atoms to almost absolute zero, they create a Bose-Einstein condensate and act likes waves in the realm of quantum dynamics.&lt;/p&gt;&#xA;" OwnerUserId="17870" LastActivityDate="2017-08-07T16:30:48.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21998" PostTypeId="2" ParentId="20981" CreationDate="2017-08-07T19:35:16.387" Score="3" Body="&lt;p&gt;I make telescopes and telescope optics (mirrors). The question is actually pretty complex. I get it a lot, because most people believe magnification is the end-all-be-all of telescopes, but reality is a little different. Here's what I mean by that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are correct that the actual magnification of the instrument (how much bigger the image is, compared to the object) is given by the formula:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;M = F / f&#xA;&#xA;M = magnification&#xA;F = focal length of the objective&#xA;f = focal length of the ocular (eyepiece)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The formula seems to suggest that you could get near-infinite magnification from any telescope by just reducing f, the focal length of the ocular. And it's technically true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the question is - how much can I increase magnification before it's too much to be useful?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you zoom into an image with Photoshop, at some point you start seeing the pixels; any zooming past that point does not reveal more details. Telescopes do not have pixels, but the image the objective provides does not have an arbitrary amount of details - just like with Photoshop, arbitrarily increasing magnification with a telescope past some point will only give you a messy blur (not pixelated, just blurry).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how much is too much? It depends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had perfect optics, used in perfect conditions, then the maximum useful magnification would be:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Mu = 2 * D&#xA;&#xA;Mu = maximum useful magnification&#xA;D = diameter of the objective in mm (a.k.a. aperture)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So for an instrument with a 75 mm aperture, the maximum useful magnification is around 150x. If you increase magnification past this point, the image is bigger, but it also gets progressively more blurry, so no additional details are revealed. But those are ideal numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, optical elements (lenses, mirrors) are not perfect. Only the most expensive instruments have optics that are anywhere near perfection. Imperfections add to the blur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even with perfect optics, if the telescope is not perfectly collimated (the lenses and mirrors are not perfectly aligned), that reduces performance too. Most refractors are collimated from factory and need no adjustment, but some cheap refractors may not be perfectly collimated. Other instruments such as dobsonian reflectors need periodic tuning to redo their collimation. When the instrument is miscollimated, the maximum useful magnification is reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atmosphere is turbulent. This is called &quot;seeing&quot; in astronomy. When seeing is bad, extra blur is added to the image, and you have to dial down the magnification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, magnification is not the ultimate parameter of a telescope. When you buy a car you don't ask the dealer &quot;what's the fastest car you have?&quot; You adjust the speed of the car to the conditions: drive at one speed on the freeway, at a different speed in a residential neighborhood, and at yet another speed in the parking lot at the local grocery store. Same with telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some objects require a lot of magnification - things like Mars at opposition, or certain double stars, or small craters on the Moon, these all benefit from the highest magnification you could get. But objects such as Jupiter are better seen at more like a medium or medium-high magnification (Jupiter is low contrast so too much magnification washes out detail in most cases - but not always). And then there are nebulae and clusters which are better seen at low magnification. Horses for courses.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So what's the actual answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, you have the upper limit set in stone: 2 * D, or 150x for a 75 mm scope. But that's more like a theoretical bound. In practice the actual value will rarely be equal to that, usually will be a little below it, and sometimes much below it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then you have to choose your magnification based on the actual object you're observing: high, medium, or low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why for experienced astronomers &lt;em&gt;magnification is just a number&lt;/em&gt;. They've learned to adjust the instrument to the local conditions, to the object being observed, and even to the characteristics of the observer (younger folks tolerate very high magnification better than older ones - something to do with the narrow size of the beam of light exiting the scope at high magnification, then hitting floaters and other defects that accumulate with age).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why most people end up with a handful of eyepieces, and choose the best one depending on each case.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2019-01-26T00:54:51.263" LastActivityDate="2019-01-26T00:54:51.263" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21999" PostTypeId="2" ParentId="21849" CreationDate="2017-08-07T20:40:51.220" Score="1" Body="&lt;p&gt;Along the same lines with previous answers suggesting &quot;mass cannot be negative,&quot; I'd like to add an insight for why that might probably be the case. If Higgs field and particles' varying degrees of interaction with the field is what gives rise to what we call mass, then the theory suggests that photons don't have mass (and constitute the velocity limit through space) because they don't interact with the field at all. I don't think the framework allows for negative interaction with the field or an &quot;anti-Higgs&quot; field.&lt;/p&gt;&#xA;" OwnerUserId="17732" LastEditorUserId="17732" LastEditDate="2017-08-11T15:17:31.817" LastActivityDate="2017-08-11T15:17:31.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22004" PostTypeId="2" ParentId="20374" CreationDate="2017-08-08T06:54:35.307" Score="1" Body="&lt;p&gt;It is a question of where, but also to better understand, you must know when it happened. Because earls structure was different from today's.&lt;/p&gt;&#xA;&lt;p&gt;From wiki on Universe's chronology,&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:CMB_Timeline300_no_WMAP.jpg#/media/File:CMB_Timeline300_no_WMAP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/6/6f/CMB_Timeline300_no_WMAP.jpg/1200px-CMB_Timeline300_no_WMAP.jpg&quot; alt=&quot;CMB Timeline300 no WMAP.jpg&quot;&gt;&lt;/a&gt;&lt;br&gt;By NASA/WMAP Science Team - Original version: NASA; modified by &lt;a href=&quot;//commons.wikimedia.org/wiki/User:Kaldari&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;User:Kaldari&quot;&gt;Ryan Kaldari&lt;/a&gt;, Public Domain, &lt;a href=&quot;https://commons.wikimedia.org/w/index.php?curid=11885244&quot; rel=&quot;nofollow noreferrer&quot;&gt;Link&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;you will learn that first stars are thought to appear when the Universe is about 3% of its current age, i.e. 400 million years old. This is the time it took for first atoms to cool down sufficiently and gather together in the clouds in question.&lt;/p&gt;&#xA;&lt;p&gt;To see &lt;em&gt;&lt;strong&gt;where&lt;/strong&gt;&lt;/em&gt; first star formation happened, we have the possibility to look back in time.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;we can see the star being born, and for this we need to use radio frequencies, because infrared was the only thing coming trough the surrounding gas clouds, from that time is so red shifted...&lt;/li&gt;&#xA;&lt;li&gt;if we are very lucky, we can see these stars exploding - since they are so big, they will quickly come to an end and explode - but because of surrounding gas clouds it is regarded as a &lt;a href=&quot;https://www.sciencedaily.com/releases/2012/11/121101141222.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;rare event&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;there are simulation tentative &lt;a href=&quot;http://uk.businessinsider.com/mapping-the-expansion-of-the-early-universe-2016-3?r=US&amp;amp;IR=T&quot; rel=&quot;nofollow noreferrer&quot;&gt;like this one&lt;/a&gt; where we use currently know physic laws to calculate the evolution of matter since it first emitted light (&lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow noreferrer&quot;&gt;CMB&lt;/a&gt; - the bluish-greenish thing on above image); the very small differences in this primordial light we are able to see allow the simulation to explain densities differences and the apparition of &lt;em&gt;filaments&lt;/em&gt;, &lt;em&gt;planes&lt;/em&gt; and &lt;em&gt;clusters&lt;/em&gt; (nodes at the intersection of planes and filaments). These things can be artistic, as you can see here filaments and a &lt;strong&gt;Protogalaxy&lt;/strong&gt; simulated:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RO7iw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RO7iw.png&quot; alt=&quot;Filament and cluster simulated&quot; /&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Whbrv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Whbrv.png&quot; alt=&quot;Protogalaxy simulated&quot; /&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.skyandtelescope.com/astronomy-news/gigantic-protogalaxy-in-the-cosmic-web-0805201523/&quot; rel=&quot;nofollow noreferrer&quot;&gt;from Sky &amp;amp; Telescope&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Clusters where the denser places of the universe, and &lt;strong&gt;due to this highest matter density, is where first stars where created.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Here is an images where &lt;a href=&quot;http://www.skyandtelescope.com/astronomy-news/lighting-up-the-cosmic-web/&quot; rel=&quot;nofollow noreferrer&quot;&gt;it is thought&lt;/a&gt; a filament is actually seen, lightened by a catastrophic event it hosts:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cKITY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cKITY.png&quot; alt=&quot;indirect filament observation&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The Lyman-alpha blob (blue fuzz) surrounding the quasar UM 287 (white dot in center) extends 1.5 million light-years long, far too big to be contained within the quasar's host galaxy or the host galaxy's halo. The blue fuzz is likely to be part of a cosmic web filament.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;S. Cantalupo / UCSC&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I invite you to get a better idea of what we can see of the distribution of stars from our planet &lt;a href=&quot;http://www.atlasoftheuniverse.com/superc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, where you can zoom in and out and get an understanding of scales.&lt;/p&gt;&#xA;" OwnerUserId="11264" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-08-08T06:54:35.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22010" PostTypeId="1" AcceptedAnswerId="22019" CreationDate="2017-08-08T17:35:31.210" Score="5" ViewCount="712" Body="&lt;p&gt;Ice giants are currently not considered suitable for life due to temperature, pressure and composition.  &lt;a href=&quot;http://www.ncbi.nlm.nigh.gov/pmc/articles/PMC4156692/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This paper&lt;/a&gt; gives some parameters for life in regards to temperature, pressure and composition:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Temperature:&lt;/strong&gt; $-15\ ^\mathrm{o}C$ to $122\ ^\mathrm{o}C$&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Pressure Upper Limit:&lt;/strong&gt; $1,100\ \mathrm{atm}$, and &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Composition to Include:&lt;/strong&gt; liquid water, $C$, $N$, $S$, and $P$.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm trying to find out what the conditions of the hot dense ammonia/water mantle border with the hydrogen/helium atmosphere are, in regards to temperature and pressure. We know the mantles are under very high pressures, have liquid water (though in a likely different state due to pressures), C, N and S. What are the pressures and temperatures at this mantle border?  Specifically in reference to Uranus and Neptune.&lt;/p&gt;&#xA;" OwnerUserId="17272" LastEditorUserId="11143" LastEditDate="2017-08-08T19:55:44.863" LastActivityDate="2017-08-09T06:17:23.843" Title="Why are ice giants not considered suitable for life?" Tags="&lt;life&gt;&lt;astrobiology&gt;&lt;abiogenesis&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22012" PostTypeId="1" CreationDate="2017-08-09T02:37:45.020" Score="4" ViewCount="2800" Body="&lt;p&gt;I've travelled to farms at night, and away from civilation. I don't really know alot about astronomy, but my goal is to see something like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/JgdLV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JgdLV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars, lots of them, but that &quot;thick purple cluster of gas right in the center&quot;. I want to see that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that on a very very dark day, we would be able to see this, but sometimes I only see stars, just lots and lots of stars, something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mwCeq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mwCeq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;I want to see this &quot;thick cloud cluster, which I assume is the milky way&quot;. How do I see this? How on some days do I only see just stars and stars, when I really want to see this main beauty?&lt;/p&gt;&#xA;" OwnerUserId="17327" LastActivityDate="2017-08-10T21:13:35.753" Title="How do I see the milky way?" Tags="&lt;star&gt;&lt;milky-way&gt;&lt;photography&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22017" PostTypeId="1" AcceptedAnswerId="22021" CreationDate="2017-08-09T05:39:22.920" Score="7" ViewCount="9104" Body="&lt;p&gt;People smarter than me say the sun will be 99% covered during the 2017 solar eclipse where I live (Portland, OR metro). I am curious to know what it might look like (photo or drawing). Does this photo approximate 99% coverage? Or will there be slightly more sun? Will I still see some of the corona?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CTnkp.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CTnkp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10117" LastActivityDate="2017-08-18T19:46:27.317" Title="What does a solar eclipse at 99% totality look like?" Tags="&lt;solar-eclipse&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22019" PostTypeId="2" ParentId="22010" CreationDate="2017-08-09T06:17:23.843" Score="4" Body="&lt;p&gt;A basic approach would be to estimate the air-pressure at the top of the mantle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uranus' surface gravity is a bit less than Earths, but, somewhat counter-intuitively, it's atmosphere is light enough that it's gravity increases as you move inside the planet towards the top of the mantle.  (&lt;a href=&quot;https://physics.stackexchange.com/questions/18446/how-does-gravity-work-underground&quot;&gt;the same is true for Earth&lt;/a&gt;).   For ballpark estimate, lets say the gravity works out to be about the same as Earths, because I don't want to calculate, but it might be a bit more than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Mass of Uranus' atmosphere, by Wikipedia is about &lt;a href=&quot;https://en.wikipedia.org/wiki/Uranus#Internal_structure&quot; rel=&quot;nofollow noreferrer&quot;&gt;0.5 times the mass of Earth&lt;/a&gt;.  It's surface area is about 16 times Earth (4 times the Radius), but the surface area of the Mantle would be less, about 80% of the radius which works out about 10.2 times the surface area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you spread 0.5 Earth masses over 10.2 times Earth's surface area, you get a pressure between 50,000 and 60,000 atm.   Some 50 times the upper limit you provided in your question.   That's probably a limiting factor for most &quot;ice giants&quot;.   There's simply too much hydrogen and helium atmosphere to allow liquid water at 1,100 ATM.   Ice giants might create their own hydrogen by heat and pressure on Methane, creating diamond like structures and free hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat below all that atmosphere (about 20% of Uranus' radius) would be outside your criteria too.  I couldn't find any good estimates for Uranus' mantle's temperature.   There's too many unknowns about why Uranus is as cold as it is on it's surface, so estimates on it's mantle temperature seem pretty tentative and unavailable, but what little I know about how temperature rises inside a planet, I don't see any way the top of Uranus' mantle would come close to the temperature criteria you gave.  Not at a depth of 20% of Uranus' radius.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider &lt;a href=&quot;https://en.wikipedia.org/wiki/Uranus#/media/File:Tropospheric_profile_Uranus_new.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this chart from Uranus' wikipedia&lt;/a&gt; at 200,000 mbar Uranus temperature is approaching 340 K.    200,000 mbar is about 200 atm.   The mantle begins, by my ballpark estimate at about 50,000 atm.   The temperature wouldn't be close unless the planet had a very long time to cool down, perhaps tens of billions of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could run ballpark estimates on Neptune too, but I suspect the results would be similar.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an ice giant to meet your criteria, it would need to have comparatively very little hydrogen and helium compared to Uranus.  Perhaps if an ice giant was close enough to it's star that it had a hot enough surface temperature to lose much of it's hydrogen and helium through &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_escape#Thermal_escape_mechanisms&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jeans escape&lt;/a&gt;, and then, it had a long time to cool down, it might get the right criteria of liquid water at the proper range of temperature and pressure.  I don't see any other scenario where it would work. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-08-09T06:17:23.843" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22020" PostTypeId="2" ParentId="22017" CreationDate="2017-08-09T07:11:09.823" Score="2" Body="&lt;p&gt;The sun will still be very very bright and I advise you not to look at at directly.  If you were to, you would not see any corona, since it is very faint, relative to the sun (even 1%).  The sun will look a bit like your photo, but I'd say with a little more light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there will be a very noticeable darkening of the environment around you, to the extent that any local wildlife will likely start make preparations for nighttime.  It will also cool down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As totality approaches, gaps between shadows made by, for example leaves on trees, will progressively show the shape of the partially obscured sun - similarly with pinhole cameras.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-08-09T07:11:09.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22021" PostTypeId="2" ParentId="22017" CreationDate="2017-08-09T07:22:45.630" Score="6" Body="&lt;p&gt;At 99% the sun will be reduced to a small sliver. There are &lt;a href=&quot;https://www.timeanddate.com/eclipse/in/usa/portland-or&quot; rel=&quot;nofollow noreferrer&quot;&gt;simulations&lt;/a&gt; on the internet. It will become very notably darker. However, the sun is 400000 times brighter than the full moon, so 1% of the Sun is still 4000 times brighter than the Moon, and so no corona will be visible. You can still damage you eye looking at that sliver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the centre of totality is less than an hour's drive from Portland, if you have a car, and can get out early, you can see totality and the corona (weather permitting)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-08-09T17:02:38.247" LastActivityDate="2017-08-09T17:02:38.247" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22025" PostTypeId="1" AcceptedAnswerId="22027" CreationDate="2017-08-09T10:46:37.587" Score="-1" ViewCount="83" Body="&lt;p&gt;I made a simulation of total eclipse 2017 in Kansas city by StarryNight5.0 software and this software shows this eclipse will not be total in this city. Web sites says it is total. Where I was wrong with this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cbLlK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cbLlK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here is the &#xA;&lt;a href=&quot;http://olumcamp.ir/astronomy/SE_site/SolarEclipse2017Kansas.snf&quot; rel=&quot;nofollow noreferrer&quot;&gt;file&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15641" LastEditorUserId="15641" LastEditDate="2017-08-09T10:52:23.717" LastActivityDate="2017-08-09T12:54:09.163" Title="Why my simulation of eclipse 2017 is wrong!" Tags="&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2017-08-11T03:02:29.057" ContentLicense="CC BY-SA 3.0" />
  <row Id="22027" PostTypeId="2" ParentId="22025" CreationDate="2017-08-09T12:54:09.163" Score="3" Body="&lt;p&gt;If I have understood your attached file correctly, it shows your location as latitude 45.3, and longitude -80.4.  I don't know where that is, but it's not Kansas City.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-08-09T12:54:09.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22028" PostTypeId="2" ParentId="22012" CreationDate="2017-08-09T13:12:12.593" Score="0" Body="&lt;p&gt;You might want to try a very dark night instead of a &quot;dark day&quot;. :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;cluster of gas&quot; that you mention is from countless stars. From a dark site, it looks like clouds, but you soon realize that they are not moving. It is not as well defined as in the photo, but it is impressive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to needing a dark site far from lights, you need to look at the correct time of day and night. You want to look when the constellation of Sagittarius is at its highest. &lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-08-09T13:12:12.593" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22029" PostTypeId="2" ParentId="22012" CreationDate="2017-08-09T18:08:03.220" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;thick purple cluster of gas right in the center&quot;. I want to see that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First off, the human eye is a wonderful device, but it's no match for modern cameras. We can't see colors at such low levels of light. We're all colorblind in those conditions. Only cameras continue to see color when illumination is low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, when looking at objects such as nebulae or galaxies, including our own, all you see is a black-and-white image. Even with telescopes, the human eye still doesn't see colors in these objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is exactly one single exception - the Orion nebula, which is by far the brightest of all. Some observers, after a bit of practice, can begin to distinguish a few shades of color there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets, and some stars, are different. These objects are much more bright. The human eye has no trouble seeing color where color exists here. E.g. Jupiter is a slight shade of yellow, Betelgeuse is quite clearly red, etc.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;purple&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure the Milky Way is not purple. That's probably just a processing artifact. Go on Astrobin and search for &quot;milky way&quot; for a lot more examples - they tend to not look very purple usually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.astrobin.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.astrobin.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I want to see this &quot;thick cloud cluster, which I assume is the milky way&quot;. How do I see this? How on some days do I only see just stars and stars, when I really want to see this main beauty?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are two main obstacles to doing astronomy at the bottom of an ocean of air: atmospheric turbulence (a.k.a. &quot;seeing&quot;) and light pollution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing affects observing high resolution objects such as planets and double stars when seen in telescopes. It's caused by air eddies and vortices that muddy the image. It's of no interest here so we can ignore it for this discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light pollution affects observing faint objects such as galaxies (including our own), nebulae, and star clusters. It's caused by light from artificial sources (city lights, industrial lights, cars, etc) going up, being reflected on particles in the air, then going back down. It makes the sky seem to &quot;glow&quot;. It drowns out those objects which are not very bright - and galaxies tend to not be bright.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's only one way to avoid light pollution: go far away from human settlements and facilities. The desert and the open sea are not light polluted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google &quot;light pollution map&quot; for some websites that will help you choose a dark site. Here's one example out of many:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jshine.net/astronomy/dark_sky/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.jshine.net/astronomy/dark_sky/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently I've visited Death Valley. The surrounding area has essentially zero light pollution. Even driving 30 minutes out of Ridgecrest you're already under a very dark sky. I took the opportunity to show my kids how a dark sky really looks like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Milky Way is very clearly visible. Wherever you look, there are stars upon stars. Take a pair of binoculars, point them at random places in the sky, and even more stars are visible. There are star clusters that you could chance upon just by sweeping the sky with the binoculars, and they look like someone threw a handful of diamonds on black velvet. Little shooting stars are visible often. It's extraordinary.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It should be noted that the Milky Way is not visible during the winter months (December – February). It's best to try to watch it in June – August. Even later in the year (September – November) it's still visible in the West, just before it sets, a short time after nightfall. If you're an early riser, you can see it before daybreak in the spring (March – May).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to wait until the sky goes completely dark after sunset. That means 1.5 hours after sunset, or more. To determine the precise moment of sunset, go on Google or Wolfram Alpha and search for &quot;sunset today&quot;. Then add 90 minutes to the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will also need to time your visit to a night when the Moon is not visible. The Moon must be below horizon, otherwise it would cause some &quot;light pollution&quot; of its own. A few days before and after the new moon are the best time for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could go on Wolfram Alpha and search for &quot;next new moon&quot; or even &quot;new moon in september&quot; and it will give you the date. Several days plus or minus that date should be fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.wolframalpha.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TLDR:&lt;/strong&gt; A moonless night in the summer, in a desert or some other unpopulated area, 1.5 hours after sunset or later.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2017-08-10T20:27:27.607" LastActivityDate="2017-08-10T20:27:27.607" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22032" PostTypeId="1" AcceptedAnswerId="22056" CreationDate="2017-08-09T21:51:28.007" Score="8" ViewCount="3503" Body="&lt;p&gt;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Three-body_problem&quot; rel=&quot;noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Three-body_problem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have three or more celestial bodies rotating each other in a stable manner ever been observed, or it is only a theoretical problem? &lt;/p&gt;&#xA;" OwnerUserId="17914" LastEditorUserId="2365" LastEditDate="2017-08-10T12:29:19.443" LastActivityDate="2017-08-11T10:33:00.037" Title="Is any 3 body system known?" Tags="&lt;orbit&gt;&lt;stellar-dynamics&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22033" PostTypeId="2" ParentId="22032" CreationDate="2017-08-09T23:30:00.490" Score="3" Body="&lt;p&gt;Yes, there are triple star systems, bound by gravity, that seem stable. See a list here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Category:Triple_star_systems&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Category:Triple_star_systems&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also systems with 4 stars or more, although these are fewer, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could argue that star clusters and galaxies are the ultimate N-body problem, for very large values of N.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2017-08-09T23:30:00.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22039" PostTypeId="2" ParentId="22032" CreationDate="2017-08-10T12:28:49.437" Score="8" Body="&lt;p&gt;The three body problem is a theoretical problem in Newtonian mechanics. It is possible to solve, exactly, the two body problem: Both bodies move in conic sections, typically ellipses, relative to their common centre of mass. The general three body problem cannot be solved exactly. There are special cases that can be solved exactly, of greater interest are situations in which, although the equations cannot be solved exactly, the system is stable in the long term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several possible configurations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The &quot;Sun and planets&quot; configuration&lt;/strong&gt;: In this configuration one body is much more massive than the others, it is relatively fixed, and the second and third body orbit it. The second and third body are in orbits that remain sufficiently distant from each other that they remain in their orbits around the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The sun-planet-moon configuration&lt;/strong&gt; In this configuration, one body is much more massive than the other two. The second and third body are in a close orbit around each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The circumbinary configuration&lt;/strong&gt; in this configuration, two large bodies are in a very close orbit around each other, the third body is distant, and orbits much further out. This is the &quot;tattoine&quot; configuration, with two suns close to each other orbited by a distant planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No real system is a theoretical three body system, since there are always perturbations (from non-spherical objects, from tidal distortions or from the fluctating gravitational field of the rest of the universe) But there are lots of examples of systems with more than two bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solar system provides examples of all these configurations. The system of &quot;Sun-Earth-Moon&quot; is a three body system. The system of Sun+planets has at least 9 bodies, in orbit and stable for billions of years (although not proved stable, and known to be chaotic over large timescales). Circumbinary systems are rarer (as they are less stable) but [Kepler 16b] (&lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler-16b&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Kepler-16b&lt;/a&gt;) provides an example of one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is common to all three configuations is that they can be approximated by two body solutions. For example, the moon orbits the Earth in an elliptical orbit, that approximates a two body system, and the Earth-moon system orbits the sun in an elliptical orbit too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you won't find in nature is three bodies in a highly chaotic orbit which deviates greatly from an elliptical orbit, nor will you find orbits like the figure-of-8 orbit, as these orbits are not stable. In a fairly short amount of time either on body is ejected, or two bodies collide.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-08-10T12:28:49.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22041" PostTypeId="2" ParentId="22012" CreationDate="2017-08-10T21:13:35.753" Score="-1" Body="&lt;p&gt;Darkest skies I've seen were in the outback of Australia, perhaps 20km from the nearest (small) town.  Still couldn't see purple clouds though.  The naked eye view was more like your 2nd image.&lt;/p&gt;&#xA;" OwnerUserId="17930" LastActivityDate="2017-08-10T21:13:35.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22042" PostTypeId="1" AcceptedAnswerId="22060" CreationDate="2017-08-10T21:57:45.090" Score="2" ViewCount="833" Body="&lt;p&gt;I'm just getting into image processing and analysis using Python, and I'm looking for astronomical images (preferably raw, unprocessed images) to practice on. Can anyone suggest some places I can download public-access telescope images from?&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2017-08-11T14:26:41.800" Title="Practicing image processing and analysis" Tags="&lt;data-analysis&gt;&lt;photography&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22043" PostTypeId="2" ParentId="22042" CreationDate="2017-08-10T22:18:11.857" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://hubblesource.stsci.edu/sources/illustrations/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://hubblesource.stsci.edu/sources/illustrations/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.nasa.gov/multimedia/imagegallery/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.nasa.gov/multimedia/imagegallery/index.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are two sites I have downloaded images from. I had to right click the image to download but it worked.  Maybe more of a challenge for you are images of rain waves. I don't know where to find them though. I think all of images from above sites have been processed so they may not fit your needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These sites may help you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ned.ipac.caltech.edu/level5/March01/Andernach/Ander6.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ned.ipac.caltech.edu/level5/March01/Andernach/Ander6.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hubblesite.org/get_involved/hubble_image_processors/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://hubblesite.org/get_involved/hubble_image_processors/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last site is for amateur image processors and has access to raw image data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good Luck.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="10196" LastEditDate="2017-08-11T08:17:21.753" LastActivityDate="2017-08-11T08:17:21.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22044" PostTypeId="2" ParentId="10359" CreationDate="2017-08-10T22:22:44.767" Score="5" Body="&lt;p&gt;If the starshade was to stay in place over the ground telescope, it would have to be in a geosynchronous orbit, which puts it almost 36,000 km from Earth's surface. To cover an angular distance of 1 arcsecond (1/3600 of a degree), it would have a size of around 10km. Certainly not impossible, but technologically impractical at the moment. Also, the starshade would only shade a small region of the sky, so it wouldn't be useful from other observing sites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the space telescopes using a starshade in geosynchronous orbit; that would be impossible, at least for scopes orbiting Earth, like Hubble is. The starshade is orbiting Earth at a different distance, and therefore speed, than the telescopes. As far as I can see, it would be near impossible to keep them aligned for anything more than a few minutes. I haven't done any number-crunching on this, though.&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2017-08-10T22:22:44.767" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22045" PostTypeId="1" CreationDate="2017-08-11T02:38:31.173" Score="1" ViewCount="29" Body="&lt;p&gt;Everyone knows you need clear skies to see what the stars or the Sun is doing. Weather is not too particularly hard to determine. There are many resources out there that will tell us the weather in any city. However, they very often lack other atmospheric obstructions. Naturally, light pollution is one of the biggest concerns. But another major concern is erratic atmospheric changes due to things like forest fires. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for good resources that will tell me not only weather, but also light pollution and other obstructions such as smoke from forest fires.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There's a solar eclipse coming up and I'm driving 500 miles to see it.)&lt;/p&gt;&#xA;" OwnerUserId="17933" LastEditorUserId="7982" LastEditDate="2021-02-28T11:46:32.970" LastActivityDate="2021-02-28T11:46:32.970" Title="What are reliable resources for weather and all other atmospheric obstructions?" Tags="&lt;planetary-atmosphere&gt;&lt;atmospheric-effects&gt;&lt;light-pollution&gt;&lt;weather&gt;&lt;resource-request&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22046" PostTypeId="2" ParentId="22045" CreationDate="2017-08-11T02:44:49.577" Score="3" Body="&lt;p&gt;You might try &lt;a href=&quot;https://clearoutside.com/forecast/50.7/-3.52&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://clearoutside.com/forecast/50.7/-3.52&lt;/a&gt;  It gives weather and visibility among other things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have an app for iPhone and android. Advertises site was developed by astronomers.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastActivityDate="2017-08-11T02:44:49.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22047" PostTypeId="2" ParentId="22032" CreationDate="2017-08-11T03:25:37.343" Score="4" Body="&lt;p&gt;There are many systems with 3 and more bodies, but more or less universally they are highly hierarchical systems that behave approximately like a bunch of two-body systems. For example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The solar system has 4 giant planets, 4 rocky planets, dozens of moons, and thousands of asteroids and Kuiper belt objects. Most bodies approximately behave like they are just orbiting the Sun, and the moons approximately behave like they are just orbiting their host planet. Celestial mechanics is largely the study of the higher order corrections on top of that.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are triple-star systems, but they usually have one close binary, and a third, more distant star. The close binary behaves mostly like the two stars just orbit each other, and the distant third star approximately behaves like it just orbits the center of mass of the binary.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A galaxy has hundreds of billions of stars. But they are orbiting their common gravitational potential. In the very distant future most of the stars in the galaxy will be shed away, but that will take much longer than the present age of the universe.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These kinds of systems are often dynamically unstable, but if the stability timescale is longer than the age of the universe, it doesn't matter.&lt;/p&gt;&#xA;" OwnerUserId="17836" LastActivityDate="2017-08-11T03:25:37.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22049" PostTypeId="1" CreationDate="2017-08-11T06:45:42.423" Score="4" ViewCount="444" Body="&lt;p&gt;The classical planets (Mercury, Venus, the Sun, Mars, Jupiter and Saturn) were considered to be &quot;wandering stars&quot; - and the word &quot;planet&quot; means just that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia says &lt;a href=&quot;https://en.wikipedia.org/wiki/Star#Observation_history&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;By the 17th century, the idea of the stars being the same as the Sun was reaching a consensus among astronomers.&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the planets were considered stars in those times, the question is when did the astronomers realize that they are something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same question in put in another words: When did the scientists first hypothesize that the planets do not emit their own light, like stars, and they just reflect the star from the sun? And when was that proven for the first time?&lt;/p&gt;&#xA;" OwnerUserId="17932" LastActivityDate="2017-08-11T18:47:16.313" Title="When it was discovered that the classical planet are not stars?" Tags="&lt;planet&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22052" PostTypeId="2" ParentId="22049" CreationDate="2017-08-11T08:15:07.270" Score="4" Body="&lt;p&gt;The moon has been known to be lit by the sun since antiquity. Since the moon was known to be lit by the sun, it is not much of a leap to believe the other planets were too.  The Pythagoreans believed all the Planets (including the sun and fixed stars) were lit by the light of the &quot;central fire&quot;. So the notion that the planets reflected light is ancient. This notion is consistent with Ptolemy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for proof, at least since Galileo in about 1610.  Galileo observed the phases of Venus, which he interpreted as a spherical body being lit by the sun from various angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is notable that this does not appear to have been a &quot;wow&quot; moment. It confirmed what Galileo already believed about the orbit of Venus, he does not appear to be have been greatly surprised that the planets show as discs in a telescope. I surmise that the fact that planets reflect the light of the sun was generally believed by Muslim and Western astronomers since ancient time.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-08-11T08:15:07.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22053" PostTypeId="2" ParentId="22049" CreationDate="2017-08-11T08:17:55.710" Score="1" Body="&lt;p&gt;I don't know when it was first hypothesised that the planets were different from the stars.  Given that stars twinkle, and planets don't (at least not so much), it could well have been in classical times.  However the first proof would have come with the use of telescopes to look at the sky, so around about Galileo's time.  This would have given proof because the planets would resolve into disks, and the bands of Jupiter were noted almost immediately.  Stars however remained pointlike irrespective of magnification applied.  Also telescopes would have shown that that Mercury &amp;amp; Venus, and to a lesser extent Mars exhibit phases, like the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a further point as to when it was realised that the stars and the sun were the same type of object.  Again I don't know when this was first hypothesised, but fairly compelling evidence would have come with the advent of stellar spectrometry, which I think was was mid 19th Century.  This is because the stars would show similar patterns of absorption lines as the sun.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-08-11T08:17:55.710" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22056" PostTypeId="2" ParentId="22032" CreationDate="2017-08-11T10:33:00.037" Score="12" Body="&lt;p&gt;Answering in the spirit of the question, I think he's asking if there is ever a chaotic 3 body system that's long term stable, or, to put it another way, a 3 body system without a standard hierarchy where it's stable.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is no.   3 body systems without Heirarchy are never stable for very long.   They can certainly exist for a while, but they aren't long term stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What do we mean by Hierarchy&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun-Earth-Moon is an example of a long-term-stable 3 body system in heirarchy.  It will remain stable for billions of years.    The Sun is pretty much in the center.   The Earth-Moon system orbits the sun and the Moon also orbits the Earth.   In a sense, the Moon orbits both the Sun and the Earth.    The Moon can orbit the Earth because it's inside the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere#True_region_of_stability&quot; rel=&quot;noreferrer&quot;&gt;stable part of Earth's Hill Sphere&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hierarchies beyond 3 might exist, such as Sun, Planet, Moon, Moon's satellite but there are no natural satellites of Moons in our solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Multiple planets&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, multiple planets around a comparatively massive Central star can be long term stable, and just for clarity, lets call a system stable if it lasts a billion years.    Mercury, Venus, Earth, Mars, Juptier, Saturn, Uranus, Neptune are obviously stable and mathematical models say that they should remain stable for at least the next 3.5 billion years, after which, there's a chance &lt;a href=&quot;http://www.scholarpedia.org/article/Stability_of_the_solar_system#Marginal_stability_of_the_Solar_System.&quot; rel=&quot;noreferrer&quot;&gt;Mercury could become chaotic&lt;/a&gt; and fly out of it's orbit, flying into the sun, crashing into another planet or perhaps leaving the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any of these orbital systems, for any kind of long term prediction of where the planets will be in a billion years, requires n-body calculations which are very complex and inexact, but they can be calculated within a range of error, which makes it possible to say that a system is stable, for a couple billion years anyway.   You can't, for example, ignore the other planets and calculate where the Earth will be in 50 million years.  The planets affect Earth's orbit but they don't destabilize it.   This is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Perturbation_(astronomy)&quot; rel=&quot;noreferrer&quot;&gt;perturbation theory&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the final example, &lt;strong&gt;binary star system&lt;/strong&gt; with a planet orbiting the two stars.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://what-if.xkcd.com/150/&quot; rel=&quot;noreferrer&quot;&gt;This XKCD&lt;/a&gt; discusses the binary star system, and from same:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A planet orbiting two stars can't get too close to them or its orbit&#xA;  becomes unstable. If it gets too close, the irregular tugging from the&#xA;  gravity of the two stars as they orbit will eventually cause the&#xA;  planet to crash into one of them or get flung out of the system.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For a system with two similar-sized stars, this &quot;critical radius&quot; is&#xA;  around six times the distance between the two stars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What this means is that any 3 body system has to follow the rules if it's going to be stable.   You can have several planets orbiting a central star, and those planets will perturb each other, sometimes casting a planet out but you can also have stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can have Sun, Planet, Moon, but the Moon must be inside the stable region of the Hill Sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you can have one object orbiting two central objects, like Tatooine, or, like the 4 satellites that orbit Pluto-Charon.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But anything outside those rules is an unstable 3 body or n-body system and it may take some time, but it's ultimately unstable and unlikely to last anywhere close to the billion(s) of years or so that stable orbits tend to last.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a star with two planets who's orbits cross each other (note, Pluto and Neptune don't actually cross, they just appear to cross when drawn in 2 dimensions).  Their orbits don't come close to crossing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Horseshoe_orbit&quot; rel=&quot;noreferrer&quot;&gt;Horeshoe orbits&lt;/a&gt; are 3 body orbits that might last a fairly long time, but I don't believe they are truly long-term stable (if anyone can verify that, please do), but intuitively, I don't see how horeshoe orbits could last very long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have a 3 body system where there's no structured orbit and all 3 bodies kind of &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/f/f9/N-body_problem_%283%29.gif&quot; rel=&quot;noreferrer&quot;&gt;dance around each other&lt;/a&gt;.   That system isn't stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If properly balanced, a 3 body system outside the rules of stability might last for hundreds or thousands or orbits, but in celestial mechanical time that's not stable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically some 3 body orbits have been worked out, but they are stable in the same way that a pencil is stable if you balance it perfectly on it's point in a room with no wind.   While &lt;a href=&quot;http://www.sciencemag.org/news/2013/03/physicists-discover-whopping-13-new-solutions-three-body-problem&quot; rel=&quot;noreferrer&quot;&gt;this article&lt;/a&gt; says that they found 13 solutions, none of these solutions would actually work because space isn't perfectly smooth.  It's full of objects with gravity.   So, like the balanced pencil that gets knocked over by a slight gust of wind, a passing rogue planet would throw any of these &quot;solutions&quot; out of whack and they would progressively destabilize.  Where as a stable system (sun/earth/moon) can hande some perterbations and remain stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This diagram below is an example of a perfectly stable 3 body orbit in a computer program, but like the pencil, a small push on any one of the 3 objects would send it towards instability.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/siT7H.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/siT7H.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that's not too long.  I can clean up if needed.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-08-11T10:33:00.037" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22058" PostTypeId="2" ParentId="22049" CreationDate="2017-08-11T11:48:30.947" Score="5" Body="&lt;p&gt;On p.33 of the &quot;Cambridge Concise History of Astronomy&quot; it states that Hellenistic astronomers interpreted the variation in the brightness of the planets as a sign that they were closer or further away at various points of time. Unfortunately it doesn't have a deeper discussion here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But on p.45 it discusses Ptolemy's thinking - he thought Mercury was, at its closest point, as close as the Moon and that the fixed stars were at the same distance as Saturn at its furthest point - the latter point suggesting that while the planets were seen as different, they might not have been thought of as being &lt;em&gt;that different&lt;/em&gt; beyond the way they wandered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To back that up further there is, for instance, a copy of an image from the Nuremberg Chronicle of 1493 on p.77 - here the planets are merely shown as wandering stars.&lt;/p&gt;&#xA;" OwnerUserId="823" LastEditorUserId="823" LastEditDate="2017-08-11T18:47:16.313" LastActivityDate="2017-08-11T18:47:16.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22060" PostTypeId="2" ParentId="22042" CreationDate="2017-08-11T13:48:44.553" Score="3" Body="&lt;p&gt;I know you've already accepted an answer, but I feel that it doesn't provide a true representation of the type of image processing astronomers actually do. Actual image processing in astronomy is rarely done with JPG, PNG, or TIFF files. Instead, astronomers mostly work with FITS files and cameras/astronomical instruments actually produce FITS files when taking images. The advantage here is that FITS files can record much more information than just image information such as where/when the image was taken, multiple layers of the image, encoded WSC information, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As part the mission objective of various organizations such as the National Science Foundation and NASA, &lt;strong&gt;all&lt;/strong&gt; raw astronomical data taken in the United States by observatories funded by the United States are free and open to the public (after a limited time period in which the observing scientist gets exclusive rights to the data for their science objectives - usually several months to a year). This includes both ground (e.g. Keck, SDSS, Arecibo) and space observatories (e.g., Hubble or Spitzer). This is also true of many observatories funded by other countries as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A great place to start would be to head over to the &lt;a href=&quot;https://hla.stsci.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble Legacy Archive&lt;/a&gt;. &lt;a href=&quot;https://hla.stsci.edu/hlaview.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Enter the site&lt;/a&gt; and try searching for an object or coordinates. For example, you can search for M101 (otherwise known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Pinwheel_Galaxy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pinwheel Galaxy&lt;/a&gt;). You'll see a list of images come up and next to them, you can download the FITS file generated by Hubble when it took the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another great resource would be the &lt;a href=&quot;http://www.sdss.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sloan Digital Sky Survey&lt;/a&gt;. This is a systematic astronomical observation of as much of the sky as possible using a variety of instruments. You can practice looking at images and spectra. I should warn you that it will take a bit more effort to figure out where the data is and how to interpret it, but feel free to ask questions here when you run into trouble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general though, if you're looking for a particular type of image or data, find an observatory that records it and look for their publicly available data sets.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2017-08-11T14:26:41.800" LastActivityDate="2017-08-11T14:26:41.800" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22061" PostTypeId="1" CreationDate="2017-08-11T13:57:05.490" Score="5" ViewCount="77949" Body="&lt;p&gt;I keep hearing that you can go blind during a solar eclipse. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the sun actually any brighter during a solar eclipse? This is how the media seems to frame it, and they suggest you need special glasses to view the eclipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But after some reading, it seems that the warning is about just not staring at the sun in general, and there is nothing special about the eclipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the case?&lt;/p&gt;&#xA;" OwnerUserId="17944" LastEditorUserId="17944" LastEditDate="2017-08-17T16:58:38.680" LastActivityDate="2017-08-21T04:28:38.590" Title="Is the sun any brighter during a solar eclipse" Tags="&lt;solar-eclipse&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22062" PostTypeId="2" ParentId="21855" CreationDate="2017-08-11T14:54:43.120" Score="3" Body="&lt;p&gt;I think what you want to concentrate on is the bottom panel in your first figure, which shows the amount of accretion from different sources as a function of redshift (and thus time; see below). This shows that the contribution from &quot;intergalactic transfer&quot; (green cuve = material in winds outflowing from one galaxy that ends up in another) was almost zero until around $z \sim 3$ (age of universe $\sim 2$ Gyr), peaked at around $z \sim 2$ (age of universe $\sim 3.5$ Gyr) and then declined slowly from there, so that it's now about 10% of the peak rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;$z$&quot; &lt;em&gt;is&lt;/em&gt; the redshift, which is being used as a proxy for time: $z = 0$ corresponds to the present day (age of universe = 13.7 Gyr), $z = 1$ corresponds to an age of about 6 Gyr, $z = 2$ is about 3.3 Gyr, and so forth. You can use Ned Wright's &lt;a href=&quot;http://www.astro.ucla.edu/~wright/CosmoCalc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cosmology Calculator&lt;/a&gt; to convert individual redshifts to ages of universe (click on the &quot;Flat&quot; button to use a standard cosmology).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&quot;Gyr&quot; = gigayear = 1 billion = $10^{9}$ years, in case that's not clear.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2017-08-11T14:54:43.120" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22063" PostTypeId="2" ParentId="22061" CreationDate="2017-08-11T15:08:55.010" Score="3" Body="&lt;p&gt;No, the intrinsic brightness of the sun does not change. Part of the sun light is blocked from reaching earth however making the sun appear dimmer or less intense. This has a tendency to cause people to stare at the sun to see this effect. The sun is so bright even a portion of the light reaching your eyes is intense enough to damage them. So do not look at the sun during the eclipse, or any other time, without proper eye protection.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastActivityDate="2017-08-11T15:08:55.010" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22064" PostTypeId="2" ParentId="22061" CreationDate="2017-08-11T18:01:01.117" Score="6" Body="&lt;p&gt;The Sun is the same as ever. It's just that people tend to stare at it for long times during the eclipse. A quick glance doesn't do much damage, but prolonged staring could be bad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those who see the phase of totality have an additional risk. After the Sun goes completely dark, and it then finally shows up again, some people feel compelled to keep watching that tiny sliver of light without protection for a long time. Don't. When the actual surface of the Sun is visible again after totality is over, it's time to put the googles back on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do not use regular sunglasses. They typically don't block the kinds of radiation they should block to allow you to watch the event safely. There are several options here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Very dark welding glasses.&lt;/strong&gt; Ask a welder or go to a welding shop. They have these special glasses to protect themselves from the radiation from the welding arc, which is similar to the Sun's radiation. These glasses are numbered to show how dark they are. For viewing the Sun, use the darkest ones, labeled #14. A welding mask with a #14 glass is perfectly safe no matter how long you're looking at the Sun. #13 glass might be safe too for short durations, but it's not recommended. Anything lower than that is not safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Goggles made specifically for watching the Sun.&lt;/strong&gt; There are many kinds of them. As long as they pass the ISO 12312-2 international safety standard they're fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend the goggles made of Baader solar film. They show the Sun in its original color (not red, green, orange, etc). There are several kinds here, too; the Baader AstroSolar Silver/Gold Film passes the ISO standard, these are what I use. You can buy them online in various places and they're very cheap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More info:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://eclipse2017.nasa.gov/safety&quot; rel=&quot;noreferrer&quot;&gt;https://eclipse2017.nasa.gov/safety&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://eclipse.aas.org/resources/solar-filters&quot; rel=&quot;noreferrer&quot;&gt;https://eclipse.aas.org/resources/solar-filters&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2017-08-11T18:01:01.117" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22066" PostTypeId="1" CreationDate="2017-08-11T20:44:11.773" Score="1" ViewCount="86" Body="&lt;p&gt;Solar spots seams to be a good indicator of solar activity. Does anyone know the reason for the Maunder and Dalton minima?&lt;/p&gt;&#xA;" OwnerUserId="17947" LastActivityDate="2017-08-11T20:44:11.773" Title="Does anyone knows the origin of the Maunder Minimum and Dalton minimum?" Tags="&lt;fundamental-astronomy&gt;" AnswerCount="0" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22067" PostTypeId="1" AcceptedAnswerId="22068" CreationDate="2017-08-11T23:17:06.903" Score="2" ViewCount="201" Body="&lt;p&gt;If someone was far enough above the milky way to see the entire diameter, how much can he actually see with naked eye? I'm aware that on Earth, we only see stars within 1,000 light-years (roughly) by naked eyes. I found myself wondering if someone is more than 1,000 light years away from milky way then we won't be able to see it?&lt;/p&gt;&#xA;" OwnerUserId="17948" LastActivityDate="2017-08-12T16:02:21.550" Title="If someone was above the galaxy, how much of it can he see?" Tags="&lt;milky-way&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22068" PostTypeId="2" ParentId="22067" CreationDate="2017-08-11T23:51:58.340" Score="3" Body="&lt;p&gt;The Andromeda galaxy is not too different from ours, and yet you can see it with your naked eyes at 2.5 million light years distance. You do need to be under a very dark sky, far from cities, and it does look like a mere white smudge, but it is visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now compare that with the fact that the Milky Way itself is quite visible to the naked eye if you're far from cities...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/22012/how-do-i-see-the-milky-way&quot;&gt;How do I see the milky way?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...and basically interpolate somewhere in between &quot;little white smudge&quot; and &quot;river of stars filling up half the sky&quot; to get an idea of how our galaxy would look like from its vicinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would not be bright anyway, and you would have to adapt your eyes to complete darkness first. But you would definitely see its outline. Whether you would distinguish anything of the shape of its arms or not, is harder to say, and would probably depend a lot on the actual distance.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2017-08-11T23:51:58.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22069" PostTypeId="1" CreationDate="2017-08-12T04:43:46.707" Score="1" ViewCount="6070" Body="&lt;p&gt;So a few nights ago I walked outside on my front lawn and sat down on the tailgate of my truck. I was facing north. I was looking down and towards the house across the street. Out of no where a bright light caught my eye. I looked up and didn't see anything. A couple of seconds later another light appeared, then disappeared. I pin pointed it to the left of the left of 'Kochab' star in 'Ursa Minor.' I was locked onto the sky at that point... So I kept watching and it kept happening. It was irregular. Zero regularity; it kept doing this. Bright flashes of light and subtle growing of light for a 20 minute period moving towards the north east. Randomly doing this with 2 seconds, sometimes with 5-7 second intervals. Even the type of flashing was irregular. Sometimes flashing following by flashing and other times growing light following by growing light. I've never seen anything like this in my life. I know what the space station looks like, I've seen it countless times as well as space junk. They're constant lights. And we all know what planes look like in the night sky. Constant red lights with regular blinking of white light. It was traveling in a straight path with little to no noticeable curvature. In no way could it have been reflecting light. No doubt it was projecting its own light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything similar you've known or heard about? Please help explain this.&lt;/p&gt;&#xA;" OwnerUserId="17951" LastActivityDate="2017-08-12T12:39:27.610" Title="Please help explain this unknown light in the sky. Not ISS. Not a star or comet. Too slow" Tags="&lt;light&gt;&lt;space&gt;&lt;satellite&gt;&lt;identify-this-object&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2017-08-14T15:58:53.997" ContentLicense="CC BY-SA 3.0" />
  <row Id="22074" PostTypeId="2" ParentId="22069" CreationDate="2017-08-12T12:39:27.610" Score="4" Body="&lt;p&gt;That sounds like a high altitude nonfunctioning satellite to me. For example, a rocket body or upper stage that is tumbling. I have seen these types of satellites flash  many times before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pattern is irregular because of the spinning and changing angle to the Sun.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-08-12T12:39:27.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22075" PostTypeId="1" AcceptedAnswerId="22077" CreationDate="2017-08-12T12:59:52.297" Score="2" ViewCount="115" Body="&lt;p&gt;I want to do project on astronomical dataset. Regarding that project, I try SDSS who give public dataset but it is quite complicated for me. I gone through public dataset and downloaded some .FITS file one is ~3.2mb and other one is ~3 GB but I don't know how to deal with this file and generate images or other files from .FITS files. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me, this is my Master's project in computer science. &lt;/p&gt;&#xA;" OwnerUserId="17961" LastEditorUserId="13681" LastEditDate="2017-08-12T14:44:00.433" LastActivityDate="2017-08-15T14:56:33.843" Title="Stuck with SDSS data" Tags="&lt;astrophysics&gt;&lt;data-analysis&gt;&lt;software&gt;&lt;sky-survey&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ClosedDate="2017-08-20T21:07:04.670" ContentLicense="CC BY-SA 3.0" />
  <row Id="22076" PostTypeId="2" ParentId="22075" CreationDate="2017-08-12T14:09:10.407" Score="3" Body="&lt;p&gt;You should go to the site &lt;a href=&quot;https://fits.gsfc.nasa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://fits.gsfc.nasa.gov/&lt;/a&gt; where you can read about the FITS data format. This site also has utilities for examining and viewing the data. Most of your concerns about this data should be answered by the documentation on this web site.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastActivityDate="2017-08-12T14:09:10.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22077" PostTypeId="2" ParentId="22075" CreationDate="2017-08-12T14:42:44.427" Score="3" Body="&lt;p&gt;The header of a FITS file is ASCII, and points you to further information. Calling &lt;code&gt;head -n 1 example.fits&lt;/code&gt; directs you to &quot;'Astronomy and Astrophysics', volume 376, page 359; bibcode: 2001A&amp;amp;A...376..359H&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A software to view FITS images is &lt;a href=&quot;http://ds9.si.edu/site/Home.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ds9&lt;/a&gt; (yes, that makes it hard to google...). Alternatively, there's &lt;a href=&quot;http://archive.eso.org/cms/tools-documentation/skycat.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;skycat&lt;/a&gt;.&#xA;Both can be used to view and do basic analysis on FITS images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For FITS binary tables, it is probably best to write your own analysis software. For a quick look, &lt;a href=&quot;https://heasarc.gsfc.nasa.gov/ftools/fv/&quot; rel=&quot;nofollow noreferrer&quot;&gt;fv&lt;/a&gt; might do the job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Accessing the data yourself&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to write your own software for analysis, take a look at either &lt;a href=&quot;https://heasarc.gsfc.nasa.gov/fitsio/fitsio.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;cfitsio&lt;/a&gt;, a C library for reading and writing FITS files, or &lt;a href=&quot;http://www.stsci.edu/institute/software_hardware/pyfits&quot; rel=&quot;nofollow noreferrer&quot;&gt;pyfits&lt;/a&gt;, a python module which does the same and which seems to have merged with the &lt;a href=&quot;http://www.astropy.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;astropy&lt;/a&gt; software package. The cfitsio page in the link above has links to documentation on both the cfitsio library and the FITS standard in general.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastEditorUserId="13681" LastEditDate="2017-08-15T14:56:33.843" LastActivityDate="2017-08-15T14:56:33.843" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22078" PostTypeId="2" ParentId="22067" CreationDate="2017-08-12T16:02:21.550" Score="1" Body="&lt;p&gt;Just to add some bad mathematics to this problem, lets use Andromeda.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short answer.   The numbers agree what Florin Andrei said.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longer answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Andromeda_Galaxy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Andromeda has&lt;/a&gt; an apparent magnitude of 3.44 from about 2.54 million light years distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tricky part of the question is, how close would you be, above Andromeda's center to see the galaxy as a whole, but not so close that individual stars stand out.    The width of the center of galaxies is tricky and I had no success looking up specific numbers.  Galaxies have fuzzy boundaries anyway.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average width of the Milky way is given as about 1,000 light years, but it's thickest in the middle.  Ballpark estimates, I would guess the thickness of the Milky way to be maybe 3,000 - 6,000 light-years at the center (&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/a/a1/ESO_-_The_Milky_Way_panorama_%28by%29.jpg/825px-ESO_-_The_Milky_Way_panorama_%28by%29.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;see picture&lt;/a&gt;).   The thickness of Andromeda, a larger galaxy, perhaps 7,000-10,000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as for how close you could be where individual stars wouldn't stand out and you'd see the galaxy as a whole.  From Earth, the most distant star that can be seen by the naked eye is about 4,000 light years away.   &lt;a href=&quot;https://ww2.kqed.org/quest/2007/09/28/the-unaided-eye/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source.&lt;/a&gt;  Given that Andromeda would appear like a white smudge, similar to the Milky way does from the Southern hemisphere, individual stars would stand out much less over the white smudge, so an additional 1,000 light years or so from the fuzzy edge should be enough to make the galaxy appear as a galaxy without visible individual stars.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we give a distance of 8,000 light years from Andromeda, or, about 310-320 times closer than we are now.  That would make it about a 100,000 times brighter and give it a brightness of about about -9.1, which might be a tricky number to conceptualize.   It's significantly less bright than a full moon but many times brighter than Venus, but it would be spread out over a large area of the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you moved a bit further away, the spiral arms of Andromeda might be more visible, say 15,000 or 20,000 light years above it's center, but the brightness would drop to a -7.6 or -6.8 respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is pretty ballpark cause it's comparing brightness from our view at a significant angle to the view from directly above.  The brightness might be less obstructed from directly above so my estimates might be low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to give an idea of what a -9.1 over much of the sky would mean, it's probably easier to look at that in terms of number of zero magnitude stars.   See &lt;a href=&quot;http://www.skyandtelescope.com/astronomy-resources/astronomy-questions-answers/if-you-combine-the-magnitudes-of-all-visible-stars-how-bright-a-star-will-you-come-up-with/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The entire night sky (360 degrees, all directions) has a brightness of about 500-550 zero magnitude stars.   A -9.1 would be the equivalent of about 4,200 zero magnitude stars, so at about 8,000 light years, Andromeda would look like a not very bright white smudge over perhaps nearly half of the sky, with dimmer regions outside the center.  But the overall brightness would be several times brighter than our moonless night sky on a clear night, but still about 20-25 times dimmer than the full moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At greater distances, the brightness drops with the square of the distance, but from far enugh away to see the spiral arms, they would probably be very faint, but perhaps discernable.  Hard to say. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-08-12T16:02:21.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22081" PostTypeId="1" AcceptedAnswerId="22088" CreationDate="2017-08-13T05:57:12.153" Score="2" ViewCount="121" Body="&lt;p&gt;This &lt;a href=&quot;https://space.stackexchange.com/a/22601/12102&quot;&gt;nifty answer&lt;/a&gt; describes and provides links to the data of the recovery of near earth object 2012 TC4, and the answer also shows the first actual image of the body for this pass. At apparent magnitude +27 (at about 52 million km) it may be the faintest asteroid to have been imaged to date. It's worth visiting that answer first before reading on here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://space.stackexchange.com/questions/22596/#comment64062_22601&quot;&gt;This comment&lt;/a&gt; points out that 2012 TC4 appears on the &lt;a href=&quot;https://echo.jpl.nasa.gov/asteroids/goldstone_asteroid_schedule.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goldstone Asteroid Schedule&lt;/a&gt;, a list of upcoming planned observations of asteroids by antennas at the &lt;a href=&quot;https://deepspace.jpl.nasa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Deep Space Network&lt;/a&gt;. Here are the entries for 2012 TC4 in the &lt;code&gt;Upcoming Observations&lt;/code&gt; and &lt;code&gt;Past Observations&lt;/code&gt; tables:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Upcoming Goldstone Observations&#xA;DATES           TARGET    H     Notes:&#xA;2017 Oct  9-14  2012 TC4  26.7  Close approach = 7.9 Earth radii on Oct. 12. NHATS&#xA;&#xA;Past Goldstone Observations&#xA;DATES           TARGET          Notes:&#xA;2012 Oct        2012 TC4        Echoes not detected. Likely equipment issues.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I noticed at the top of the page there is a link to the &lt;a href=&quot;http://www.naic.edu/~pradar/&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Arecibo asteroid radar schedule&lt;/a&gt;, and in the fifth line of the list shown in the screen shot below, I see that for 2012 TC4 there will be what is labeled there as a &quot;WORLDWIDE CAMPAIGN!&quot; of observations. The all-caps and exclamation point are a direct quote from Arecibo's table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the extent of the worldwide campaign to observe 2012 TC4? How many facilities? Is it coordinated? Perhaps simultaneous reception of reflected signals by multiple sites? Will it be coordinated with other observations such as infrared or optical wavelengths?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plot below the Arecibo table is made from JPL Horizons data in the Earth-Moon barycenter frame. It shows +/- 24 hours from closest approach.&#xA;The hyperbola (red) is the calculated trajectory of 2012 TC4 with dots every 2 hours and the short, (roughly) circular arc is the Moon, as descried in &lt;a href=&quot;https://space.stackexchange.com/q/22596/12102&quot;&gt;this question&lt;/a&gt;. The closest approach to Earth's center is projected to be about 51,000 km. These are preliminary calculations based on the first observation and may be refined a bit as the time of closest approach nears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uKezm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uKezm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/is3GK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/is3GK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aUIXC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aUIXC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KZAVf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KZAVf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-08-13T08:30:47.737" LastActivityDate="2017-08-14T22:32:28.287" Title="What is the extent of the &quot;worldwide campaign&quot; to observe near-earth object 2012 TC4?" Tags="&lt;observational-astronomy&gt;&lt;radio-astronomy&gt;&lt;near-earth-object&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22082" PostTypeId="1" AcceptedAnswerId="22099" CreationDate="2017-08-13T08:30:19.413" Score="1" ViewCount="249" Body="&lt;p&gt;The eoPortal article for &lt;a href=&quot;https://eoportal.org/web/eoportal/satellite-missions/content/-/article/tess&quot; rel=&quot;nofollow noreferrer&quot;&gt;TESS (Transiting Exoplanet Survey Satellite)&lt;/a&gt; gives a fairly in-depth summary of the space telescope design and planned operation. Some further background information on its observation mode and orbit can be found in the video linked below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the caption (also shown below), I'm guessing that &quot;star noise&quot; is possibly related to shot noise; the counting statistics of the incoming photons or perhaps the photoelectrons thereby produced, and &quot;sky noise&quot; is the cumulative effect of all the other actual light present in the spatially binned area of a given star, including &lt;a href=&quot;https://en.wikipedia.org/wiki/Zodiacal_light&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zodiacal light&lt;/a&gt; within the solar system as well as all other sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, and even if not, I don't really understand the units of $\sigma$ as parts per million per square-root hours, and why sky noise would vary as $I^{-1}$ while star noise &lt;em&gt;expressed in these units&lt;/em&gt; varies as $I^{-1/2}$, where $I$ is the &lt;em&gt;intensity of the light&lt;/em&gt; from a given star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; Figure 12: Top: Expected 1σ photometric precision as a function of stellar apparent magnitude in the IC band. Contributions are from photon-counting noise from the target star and background (zodiacal light and unresolved stars), detector read noise (10 e-), and an assumed 60 ppm of incorrigible noise on hourly timescales. Bottom: The number of pixels in the photometric aperture that optimizes the signal-to-noise ratio (image credit: TESS Team)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Excellent background on TESS: &lt;a href=&quot;https://youtu.be/mpViVEO-ymc&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/mpViVEO-ymc&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9xz7b.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9xz7b.jpg&quot; alt=&quot;TESS photometric precision&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-12-15T15:10:28.030" LastActivityDate="2018-12-15T15:10:28.030" Title="What do &quot;star noise&quot; and &quot;sky noise&quot; refer to for TESS (Transiting Exoplanet Survey Satellite)?" Tags="&lt;observational-astronomy&gt;&lt;photometry&gt;&lt;mathematics&gt;&lt;tess&gt;&lt;zodiacal-light&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22083" PostTypeId="5" CreationDate="2017-08-13T08:35:05.260" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2017-08-13T08:35:05.260" LastActivityDate="2017-08-13T08:35:05.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22084" PostTypeId="4" CreationDate="2017-08-13T08:35:05.260" Score="0" Body="Questions about the Transiting Exoplanet Survey Satellite (TESS)" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-08-13T15:35:19.800" LastActivityDate="2017-08-13T15:35:19.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22088" PostTypeId="2" ParentId="22081" CreationDate="2017-08-13T12:18:55.970" Score="0" Body="&lt;p&gt;The JPL press release mentions that more than 12 observatories, labs and universities will be involved but doesn't mention which.&#xA;&lt;a href=&quot;https://www.jpl.nasa.gov/news/news.php?feature=6906&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.jpl.nasa.gov/news/news.php?feature=6906&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ii8O0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ii8O0.png&quot; alt=&quot;Paragraph from JPL press release about the tracking campaign&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The University of Maryland has a website on the 2017 apparition of 2012 TC4&#xA;&lt;a href=&quot;http://2012tc4.astro.umd.edu/index.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://2012tc4.astro.umd.edu/index.shtml&lt;/a&gt;&#xA;Here's a snapshot of the list of planned observations shown there&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Ue4lK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ue4lK.png&quot; alt=&quot;List of planned observations of 2012 TC4 in 2017&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17969" LastEditorUserId="-1" LastEditDate="2017-08-14T22:32:28.287" LastActivityDate="2017-08-14T22:32:28.287" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22090" PostTypeId="2" ParentId="21693" CreationDate="2017-08-13T19:47:55.213" Score="2" Body="&lt;p&gt;The path of totality, which eclipse maps show to be 70 miles wide, is actually narrower than that, by up to 1 mile, according to some experts.&lt;/p&gt;&#xA;&lt;p&gt;An &lt;a href=&quot;http://www.kansascity.com/news/local/article166394247.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; by the Kansas City Star discusses this:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Those maps, provided by NASA and others, show a crisply defined, 70-mile-wide path of totality where the moon will block 100 percent of the sun. But they are not as precise as they appear, at least on their edges.&lt;/p&gt;&#xA;&lt;p&gt;The southern edge of the path as shown on the maps could be off by as little as the length of a football field or as much as a half-mile, eclipse mapping experts say. Likewise for the northern edge, meaning the path of totality might be just 69 miles wide.&lt;/p&gt;&#xA;&lt;p&gt;“This is an issue. This is really an issue, but it’s not advertised. … Yeah, all the maps are wrong,” Mike Kentrianakis, who is the solar eclipse project manager for the American Astronomical Society and who routinely consults with NASA, told The Star.&lt;/p&gt;&#xA;&lt;p&gt;Xavier Jubier, a French engineer whose calculations have been used to create the interactive Google maps of the eclipse, confirmed to The Star by email that the actual path of the totality is slightly narrower than the 70 miles shown on current maps.&lt;/p&gt;&#xA;&lt;p&gt;Ernest Wright, who created maps and other multimedia presentations on the eclipse for NASA, said he thought the map might be narrower by about 100 meters, slightly longer than a football field. He also said it’s possible that Kentrianakis is correct in his estimation that the path is narrower by a half-mile or more.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The article explains the reason the path of totality is narrower than the eclipse maps indicate is because they use a 41-year old value for the radius of the Sun, which is now known to be too small a value:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Wright explained that eclipse maps are made based on what is known about the relative sizes and positions of the moon and the sun. “We have really good information about the orbit of the moon, the positions of the sun, the positions of the Earth. All of that is really well nailed down,” Wright said. “In order to get more accuracy, we need to take into account the mountains and valleys on the moon, and the elevations on the Earth. And we’re starting to do that, as well.”&lt;/p&gt;&#xA;&lt;p&gt;The size of the moon, in fact, has been measured to within a meter, and its position in the heavens has been measured to within a centimeter. “But the last sort of uncertainty might surprise you,” Wright said. “It’s the size of the sun.”&lt;/p&gt;&#xA;&lt;p&gt;Jubier said that the current maps are accurate using the the 696,000-kilometer radius and other standards agreed upon in 1976 at a meeting of the International Astronomical Union. “This is perfectly accurate but we know it does use a solar diameter that is not large enough. Why don’t we change the value(?)” Jubier wrote. “Well simply because the IAU (International Astronomical Union) has not yet approved a new value. This is part of the research we’re doing and for which we’re looking for funding.”&lt;/p&gt;&#xA;&lt;p&gt;He continued, “So technically speaking if the Sun is larger than the adopted IAU value, and we know it is, the eclipse path is necessarily narrower and our tools can simulate this, yet the standard maps for the public will still retain the currently adopted solar radius until a new value has been accepted.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2773" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-08-13T20:01:18.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22091" PostTypeId="1" AcceptedAnswerId="22104" CreationDate="2017-08-13T22:25:57.353" Score="2" ViewCount="281" Body="&lt;p&gt;By putting solar eclipse glasses directly between a remote control IR emitter and device receiver, could some unsafe glasses be detected? My guess is most fake glasses are just blocking visible light at best, while safe ones should block the IR signal as well. It would not guarantee they are 100% safe, but it might be able to detect bad ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know to check for the ISO logo, cert number, and buy from a reputable vendor recommenced by the American Astronomical Society (which I have). It would still be nice to double check pairs of glasses before giving them out to friends and family.&lt;/p&gt;&#xA;" OwnerUserId="17971" LastActivityDate="2017-08-14T22:04:49.087" Title="Can you test solar eclipse glasses with a remote control?" Tags="&lt;solar-eclipse&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22092" PostTypeId="1" AcceptedAnswerId="22103" CreationDate="2017-08-13T22:46:01.073" Score="2" ViewCount="190" Body="&lt;p&gt;Like many, we bought solar glasses to view the eclipse, and are now in a strange place where we can't tell if they're safe or not.  The big fuss seems to be about ISO 12312-2 compliance.  Apparently there are some glasses which are claiming to be certified and putting the logo in the glasses when they are, in fact, not up to spec.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the only &quot;solution&quot; that has been given is to buy from one of the &quot;reputable&quot; vendors &lt;a href=&quot;https://eclipse.aas.org/resources/solar-filters&quot; rel=&quot;nofollow noreferrer&quot;&gt;documented by the AAS&lt;/a&gt;.  Many smaller vendors are crying foul, saying the big guys are just trying to push out competition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is this: if I had a pair of glasses, what would be the &lt;em&gt;proper&lt;/em&gt; way of determining if they met ISO 12312-2 standards?  Is there supposed to be a way for a customer to trace their purchase back and verify that the company that made them at least has their paperwork in order? (A company that has their paperwork and is still counterfeiting anyway is a different challenge).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously the astronomy community is interested in ensuring their products are safe.  How were they supposed to do that in the ideal circumstances?  If I can't confirm their certification as a small-time buyer, how would a large-scale buyer of, say, 100,000 glasses or viewers determine that the product they are buying is genuine &lt;em&gt;before&lt;/em&gt; someone burns their eyeballs?  Surely when there are hundreds of thousands of dollars on the line, there would be a process to demonstrate compliance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I am specifically looking for how one is supposed to verify these glasses without relying on a whitelist provided by a single group such as the AAS (other than perhaps the ISO community itself).  Surely there is a way to verify claims of ISO compliance without relying on the good intentions of a professional society.  Otherwise, what is the point of having certifications in the first place?&lt;/p&gt;&#xA;" OwnerUserId="11087" LastEditorUserId="11087" LastEditDate="2017-08-14T01:08:37.713" LastActivityDate="2017-08-14T18:43:44.453" Title="How can I tell if my solar glasses are up to ISO standards?" Tags="&lt;solar-eclipse&gt;&lt;standards&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="22093" PostTypeId="2" ParentId="21962" CreationDate="2017-08-13T23:55:55.930" Score="5" Body="&lt;p&gt;Seeing things upside-down is normal with an astronomical telescope.  You have to take that into account when pointing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sounds like you have at least four other distinct problems here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) You don't know how to align your equatorial mount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For someone just starting out, you don't have to do this precisely, just enough so it doesn't get in your way too much.  First, adjust the tilt of the polar axis to match your latitude.  If you live at 40 degrees north, you want your polar axis to be tilted 40 degrees up from horizontal.  Then, turn the mount so the upward-tilted part of the polar axis points north.  Again, your first few times out with the telescope you won't need to get this precisely right.  You should find that if the polar axis is correctly aligned, it should point near (although not precisely at) the North Star.  (This assumes you live north of the equator; if you live south of the equator, you'll want the up-tilted part to point south).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) You don't know how to use an equatorial mount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is tricky to use a German equatorial mount (like the one in the picture) on objects near the meridian (the imaginary line connecting the north and south poles in your sky) since as you move the telescope around, it will need to swap places with your counterweight.  This just takes getting used to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) You don't know how to align your finderscope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is actually easiest to do this during the day.  Select the lowest power eyepiece you have (the one with the biggest lens will be lowest power-- don't use a barlow if they gave you one).  Point the telescope at a distant, but distinctive, terrestrial object on the horizon.  A tall tree sticking up, a church steeple, a smoke stack, something like that.  Do this just by sweeping the telescope along the horizon line while looking through the low-power eyepiece.  Adjust the focus on the eyepiece so distant objects appear sharp.  When you have your target centered in the eyepiece, leave the telescope there (the telescope should stay where you point it if it is balanced properly; if the telescope wants to move around, you might try locking the axes--make sure to unlock them again when you want to move the telescope).  Then look through the finder scope-- you may have to adjust the focus on the finderscope.  Be very gentle with the finderscope, you don't want to move the telescope around.  You should be able to see your target through the finderscope.  Loosening and tightening the screws on the finder mount will change the aim of the finder slightly.  You want to adjust the finder aim until the finder crosshairs are on your target.  Then doublecheck that the target is still centered in the eyepiece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) You don't know how to use your finderscope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the finderscope is properly aligned, you can use it to point the telescope.  Center the target you want to look at in the finder cross hairs.  It should appear somewhere in the view of the telescope through the low power eyepiece.  Again, it is easiest to practice this during the day on distant terrestrial objects on the horizon, before you start looking for the moon and planets at night.&lt;/p&gt;&#xA;" OwnerUserId="14434" LastActivityDate="2017-08-13T23:55:55.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22094" PostTypeId="2" ParentId="22091" CreationDate="2017-08-14T02:44:51.010" Score="1" Body="&lt;p&gt;I was considering a similar line of thinking.  However, there were a few hangups which I believe prevent yee-average-human from testing the glasses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many sites &lt;a href=&quot;http://www.skyandtelescope.com/observing/celestial-objects-to-watch/solar-filter-safety/&quot; rel=&quot;nofollow noreferrer&quot;&gt;suggest&lt;/a&gt; a transmittance of 0.0032 (welding shade 12) to be the minimum to be safe (and typically recommend 13 or 14, which has even lower transmittance).  That means you're only seeing 0.3% of the light get through the goggles.  Your test would need to have enough sensitivity to pick up that tiny amount of IR light, and to not only receive a signal but be able to meaningfully measure its amplitude. You would also need to ensure that UV light is blocked as well... and all the other IR bands (I am not confident as to how wide-band protection one actually needs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even then, you'd have to deal with the issue of having built your own contraption to test a device which must protect your eyes from a permanent damage that you literally cannot see.  I'd want to be quite the confident electrical engineer and systems test expert to make sure that the tests I do &lt;em&gt;actually&lt;/em&gt; test what needs to be tested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I have a &lt;a href=&quot;https://astronomy.stackexchange.com/questions/22092/how-can-i-tell-if-my-solar-glasses-are-up-to-iso-standards&quot;&gt;related question&lt;/a&gt; asking how one can track down the ISO certifications and see if your particular vendor has legitimate glasses or not.  I run into the same issue: how can you &lt;em&gt;trust&lt;/em&gt; that this device performs as indicated, because your eyesight depends on it.&lt;/p&gt;&#xA;" OwnerUserId="11087" LastActivityDate="2017-08-14T02:44:51.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22095" PostTypeId="1" AcceptedAnswerId="22096" CreationDate="2017-08-14T03:30:44.110" Score="5" ViewCount="413" Body="&lt;p&gt;Whenever I see the term dust for the particles of a nebula, I ask myself whether this is actually based on some reliable measure of particle size. Of course we can all agree it looks like dust from this distance. And we all know how such appearances often lead to assumptions which are incorrect and whose incorrectness we are often blind to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any reliable evidence by which this &quot;dust&quot; is known not to be pebbles, rocks, boulders etc.? Which in reality would seem vastly more likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How might we measure size from such distances?&lt;/p&gt;&#xA;" OwnerUserId="12000" LastActivityDate="2017-08-14T08:05:35.343" Title="How big is nebula dust?" Tags="&lt;nebula&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22096" PostTypeId="2" ParentId="22095" CreationDate="2017-08-14T08:05:35.343" Score="9" Body="&lt;p&gt;The size of cosmic dust grains is in general given not by some size, but by a size &lt;em&gt;distribution&lt;/em&gt;. The only direct measurements of such a distribution are made on dust collected on plates of satellites, which is of course a very local measurement. When we think that distributions look similar — though not exactly alike — in other locations of the Universe, so far away that we will never be able to go there, it is because we know how a given distribution and composition affects light traveling through an ensemble of dust grains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A particle has a given probability of absorbing a photon with a given wavelength, and in general this probability peaks around wavelengths of the order of the size of the particle (for small particles). Thus, if we know how the spectrum of some light source looks if there were no dust around (and we do for many sources, e.g. stars), then the difference between the known intrinsic spectrum and the observed spectrum can be &lt;em&gt;modeled&lt;/em&gt; assuming some distribution and composition. Often the composition needs not be assumed, but can be constrained from the &lt;em&gt;emission&lt;/em&gt; of the dust at infrared wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, the model that fits best the observations is a steep power law of the form $P(r)\propto r^{-a}$ with an index of roughly $a\sim3.5$; that is, the probability of finding a small grain is much larger than the probability of finding a large grain (more precisely, for $a=3.5$, grains of size $r=x$ is $10^{3.5}\simeq3\,000$ times more common than grains of size $r=10x$, and $10^7\!$ more common than grains of size $r=100x$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For very large grains, the probability of absorbing a photon becomes independent of the wavelength of the photon. Whereas the small grains as described above have a &quot;color preference&quot;, large grains are said to be &quot;gray&quot;. This is the case for boulders, rocks, pebbles, and even sand-grain-sized particles. Thus, if a cloud consisted of such particles, the spectrum of a background source would simply be diminished by a constant factor at all wavelengths. This is very rarely observed — rather the sources are diminished much more at the short wavelengths than at the long wavelengths, as expected if there are more small grains than large grains.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2017-08-14T08:05:35.343" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="22098" PostTypeId="2" ParentId="22091" CreationDate="2017-08-14T15:40:33.043" Score="0" Body="&lt;p&gt;No, because among other reasons it's UV, not IR which poses the major ocular damage risk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;THere's also the fact that you can't possibly measure a 30 or 40 dB attenuation factor in the method you're proposing.  Your source is very weak and your receiver is binary (it responds or it doesn't).&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-08-14T15:40:33.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22099" PostTypeId="2" ParentId="22082" CreationDate="2017-08-14T15:56:21.950" Score="2" Body="&lt;p&gt;The key formula for signal-to-noise calculations in photometry can be written as something like&#xA;$${\rm SNR} = \frac{S_{\rm star}}{\sqrt{S_{\rm star} + S_{\rm sky} + \sigma_R^{2}}},$$&#xA;where $S_{\rm star}$ is the photons counted from your target star, $S_{\rm sky}$ is the amount of unrelated photons in your photometry aperture (due to sky or other stars) and $\sigma_R$ is the readout noise in your detector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numerator is of course your (interesting) signal and the denominator is the noise. The reason the noise terms are added in the way they are is that each term inside the square root is a &lt;em&gt;variance&lt;/em&gt;. i.e. The variance of the detected signal from the star is equal to the number of detected photons (a property of Poissonian counting statistics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for more complexity we can express the signals as being photons per unit time (I'll use a small $s$ for these), such that $S_{\rm star} = s_{\rm star} t$, where $t$ is the integration time. &#xA;$${\rm SNR} = \frac{s_{\rm star} t}{\sqrt{s_{\rm star}t + s_{\rm sky}t + \sigma_R^{2}}},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the thing defined as $\sigma$ in your plots is the &lt;em&gt;inverse&lt;/em&gt; of the SNR, plotted on a logarithmic axis.&#xA;$$\sigma = (s_{\rm star }t)^{-1}\sqrt{s_{\rm star}t + s_{\rm sky}t + \sigma_R^{2}} $$&#xA;$$\log \sigma = -\log s_{\rm star} -\log t +0.5 \log(s_{\rm star}t + s_{\rm sky}t + \sigma_R^{2})\ \ \ (1)$$&#xA;This is being plotted against the $I$ &lt;em&gt;magnitude&lt;/em&gt;, which is proportional to $-2.5\log s_{\rm star}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we now look at each of these contributions individually, we see that the noise due to &quot;shot noise&quot; on the star should indeed be a straight line on your plot. From eqn (1), ignoring other sources of noise&#xA;$$ \log \sigma = -0.5\log s_{\rm star} -0.5\log t$$&#xA;Thus in units of parts per million and scaled for the square root of the exposure time, this will appear as a straight line in a log-log plot with gradient of $-0.5/-2.5=+0.2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now just look at the contribution due to the sky background, from eqn (1), ignoring other sources of noise&#xA;$$\log \sigma = -\log s_{\rm star} + 0.5 \log s_{\rm sky} -0.5 \log t$$&#xA;So the gradient here is -1 in log space (the sky term is just a constant offset in the log-log plot) and +0.4 in terms of (log) ppm/magnitude. Hence the gradient due to the sky noise term is twice as steep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, looking at the readout noise contribution&#xA;$$\log \sigma = -\log s_{\rm star} + \log \sigma_{R} -\log t$$&#xA;Therefore this has the same gradient as the sky contribution (as you can see in the plot), but the offset in the log-log plot should depend on the exposure time. This makes sense because the brightness at which readout noise will dominate &lt;em&gt;does&lt;/em&gt; depend on how long the exposure is. So there is some unspoken (at least in the caption you show) assumption about how long the exposure time is in order to place the contribution from readout noise on the same graph, although the gradient is understood.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-08-14T15:56:21.950" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22103" PostTypeId="2" ParentId="22092" CreationDate="2017-08-14T18:18:19.640" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;if I had a pair of glasses, what would be the proper way of determining if they met ISO 12312-2 standards?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The process involves a lot of lab equipment, calibrated sources of infrared, visible light, and ultraviolet, and calibrated detectors. Extremely unlikely you'll be able to whip something up in your garage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buy from vendors with good reputation; these are online stores that sell nothing but astronomy equipment, and they know what they're dealing with. If you're an astronomer, you know who those are. If you aren't, go on various astronomy forums and see which vendors people are recommending. Or see the AAS page for a list of vendors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amazon is now cracking down on bad eclipse glasses, but realistically speaking they are not an online store specializing in astronomy, so their product vetting may or may not be perfect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's getting a little late now, so better hurry up and get your eclipse glasses already.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can always dash to the local welding supplies store and get a #14 welding glass. It's completely safe and passes the ISO standard. Maybe get the whole mask to go with it, so your face won't get sunburned.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2017-08-14T18:43:44.453" LastActivityDate="2017-08-14T18:43:44.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22104" PostTypeId="2" ParentId="22091" CreationDate="2017-08-14T18:26:49.850" Score="1" Body="&lt;p&gt;Short answer: No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer: No. You're testing at a single wavelength. The Sun emits continuously at a variety of wavelengths from deep infrared to far ultraviolet. Testing with a single kind of radiation doesn't tell you much about the filter behavior at the other wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buy from the online vendors that specialize in selling astronomy equipment - the ones that all astronomers buy their gear from. They tend to know the stuff they're selling. The list of vendors on the AAS page is good.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2017-08-14T18:26:49.850" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22106" PostTypeId="1" AcceptedAnswerId="22107" CreationDate="2017-08-14T19:39:20.953" Score="2" ViewCount="144" Body="&lt;p&gt;I read an &lt;a href=&quot;https://m.phys.org/news/2017-08-stars-orbiting-supermassive-black-hole.html#&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; on a group of stars orbiting the supermassive black hole Sagittarius A. Did these stars get caught by the black hole's gravity like planets capture moons, or were they born from the remnants of the star that went supernova and turned into black hole?&lt;/p&gt;&#xA;" OwnerUserId="16136" LastEditorUserId="2365" LastEditDate="2017-08-15T21:25:36.493" LastActivityDate="2017-08-15T21:25:36.493" Title="Stars orbiting supermassive black holes" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22107" PostTypeId="2" ParentId="22106" CreationDate="2017-08-14T21:45:31.210" Score="5" Body="&lt;p&gt;They are captured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formation of supermassive black holes is rather uncertain. They may have formed from core collapse of very ancient stars  (and then grown) or they may have some other mechanism of formation in the early universe. However we do know that the first formed early on in the history of the universe. They are as old as the galaxy, or perhaps even older.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand the stars that orbit the black hole are of regular age. &#xA;They are not particularly old. Therefore the black hole formed long long before the stars that orbit it. And so those stars must have been captured by some mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Star formation doens't normally happen in supernova remnants. A supernova can provide the shock that starts stellar formation in molecular clouds.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="11143" LastEditDate="2017-08-15T13:58:32.100" LastActivityDate="2017-08-15T13:58:32.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22110" PostTypeId="1" CreationDate="2017-08-15T04:42:08.983" Score="0" ViewCount="164" Body="&lt;p&gt;Sounds kinda crazy and cheap but has anyone ever tried to take a flexible (plastic) mirror and attach a string to the back side then pull the string so it makes a concave surface? I realize it wouldn't be the most accurate device in the world, but if I wanted to build a telescope with a child I think it would be a cheap and fun way to make it. Thoughts? &lt;/p&gt;&#xA;" OwnerUserId="17982" LastActivityDate="2017-08-15T06:40:48.543" Title="Possible to make a cheap reflective objective lens with a mirror?" Tags="&lt;telescope&gt;&lt;telescope-lens&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22112" PostTypeId="2" ParentId="22110" CreationDate="2017-08-15T06:40:48.543" Score="1" Body="&lt;p&gt;That won't work because the shape will be too distorted to be useful. For astronomical purposes, a good quality mirror is needed since the objects you look at are dim and small. If you want a usable astronomical telescope then the optics need to be high quality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to illustrate the properties of telescopes, a simple refracting telescope can be built using the lens from (low power) reading glasses as the objective and a (high power) hand lens as the eyepiece. I've also tried with a shaving mirror, but found the distortion to be too great to form a useful image.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-08-15T06:40:48.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22118" PostTypeId="1" AcceptedAnswerId="22120" CreationDate="2017-08-15T12:12:21.503" Score="4" ViewCount="610" Body="&lt;p&gt;It looks as if it was polished by applying emery to its spinning equator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial guess was that it was simply an illusion caused by light direction, but the terrain map shows the same phenomena.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Google Moon photo and terrain maps (&lt;a href=&quot;https://www.google.com/moon/&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hQEBG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hQEBG.png&quot; alt=&quot;Google Moon photo map&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/fDWvt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fDWvt.png&quot; alt=&quot;Google Moon terrain map&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;North pole mosaic assembled from photos by NASA's Lunar Reconnaissance Orbiter (&lt;a href=&quot;https://www.nasa.gov/content/nasa-releases-first-interactive-mosaic-of-lunar-north-pole&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;&lt;a href=&quot;https://www.nasa.gov/sites/default/files/gigapan_small_str01.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://www.nasa.gov/sites/default/files/gigapan_small_str01.png&quot; alt=&quot;North pole mosaic from NASA&amp;#39;s Lunar Reconnaissance Orbiter&quot;&gt;&lt;/a&gt;&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;South pole shot by NASA's Clementine (&lt;a href=&quot;https://svs.gsfc.nasa.gov/3443&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;&lt;a href=&quot;https://i.stack.imgur.com/N4Mst.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/N4Mst.jpg&quot; alt=&quot;South pole shot by NASA&amp;#39;s Clementine&quot;&gt;&lt;/a&gt;&lt;/h3&gt;&#xA;" OwnerUserId="17994" LastEditorUserId="2365" LastEditDate="2017-08-15T14:37:17.477" LastActivityDate="2017-08-15T20:33:51.840" Title="Why is the Moon smooth near the equator and bumpy near the poles?" Tags="&lt;the-moon&gt;&lt;geology&gt;&lt;topographic-map&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22120" PostTypeId="2" ParentId="22118" CreationDate="2017-08-15T14:32:26.923" Score="9" Body="&lt;p&gt;There are a few factors that create this effect:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First in the visual light images, the view around the equator is seen back lit, with almost no shadow, the few around the pole is seen with low lighting, and low lighting brings out features of the terrain. The Polar mosaics have an unreal quality, since they are made of images all taken when the sun is locally at its maximum above the horizon, so the light appears to be coming from all directions at once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first mosaic and the topographic map have a projection that is &quot;conformal&quot; this means that circular craters appear circular on the map, but it does introduce very large distortions of scale this, combined with shading which has been added to the map, makes regular-sized craters near the poles appear huge. The shading scales with the map, so craters near the pole have much more dramatic shading than those near the equator, that is an artistic effect, not real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally on one face of the moon, that which faces us, and is in the middle of the pictures, there are large &quot;maria&quot;. These are relatively flat lava plains. There are no maria at the poles. Most are on the near side of the moon: see &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1850/why-are-most-lunar-maria-on-the-visible-side?rq=1&quot;&gt;Why are most lunar maria on the visible side?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is a combination of low angle of light at the poles, the way that google has projected and shaded their map and the distribution of lava plains on the moon. The poles are not more bumpy than the equatorial region in general. They are more bumpy that the lunar maria, but the apparent effect you describe is an illusion.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-08-15T20:33:51.840" LastActivityDate="2017-08-15T20:33:51.840" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22121" PostTypeId="1" AcceptedAnswerId="22127" CreationDate="2017-08-15T22:27:43.160" Score="13" ViewCount="233" Body="&lt;p&gt;Our Sun follows an &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_cycle&quot; rel=&quot;noreferrer&quot;&gt;11-year cycle&lt;/a&gt; of low-to-high activity. During high activity, there are much more sunspots and much higher chance of flares and other disruptive behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we observe the activity of other stars? I'm particularly interested in other G-types like the Sun. Do they also follow an 11-year cycle?&lt;/p&gt;&#xA;" OwnerUserId="7449" LastActivityDate="2017-08-16T07:08:12.850" Title="Do other G-type stars follow an 11-year cycle of activity?" Tags="&lt;star&gt;&lt;the-sun&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22122" PostTypeId="1" CreationDate="2017-08-16T00:34:19.693" Score="0" ViewCount="1736" Body="&lt;p&gt;I am aware that planets orbit in an elliptical fashion and that the sun occupies one of two focal points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us say that the left focal point is f1 and the right f2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the sun at the same focal point for each planet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not which ones are different?&lt;/p&gt;&#xA;" OwnerUserId="17997" LastEditorUserId="823" LastEditDate="2017-08-18T18:19:49.793" LastActivityDate="2017-12-24T11:44:34.837" Title="Which focus point does the sun occupy for each planet?" Tags="&lt;orbit&gt;&lt;solar-system&gt;" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22123" PostTypeId="2" ParentId="22122" CreationDate="2017-08-16T02:32:33.973" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am aware that planets orbit in an elliptical fashion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, or at least to a good approximation.  Because the planets affect each other's motion to a small extent (and also because of relativistic effects) the motion is not quite elliptical.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and that the sun occupies one of two focal points.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, because of the complexities of the gravitational interactions of all the bodies, the Sun does not occupy a single position, but has a complex motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to read about &lt;a href=&quot;https://en.wikipedia.org/wiki/Barycenter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Barycenter&lt;/a&gt; and have a look at &lt;a href=&quot;http://www.orbitsimulator.com/gravity/articles/ssbarycenter.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this explanation of the complex motions this results in&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in reality this all happens in more than two dimensions - the orbits are inclined at different angles as well.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let us say that the left focal point is f1 and the right f2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An ellipse does indeed have two foci, but the orbits of planets and the motion of the Sun are not quite ellipses anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even if we ignore these small deviations from an ellipse, each planet's orbit is in a different orientation and has different foci.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the sun at the same focal point for each planet?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So &quot;no&quot;.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-08-16T02:32:33.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22127" PostTypeId="2" ParentId="22121" CreationDate="2017-08-16T07:08:12.850" Score="12" Body="&lt;p&gt;The answer is yes. Early work showed that just over half of sun-like stars exhibit magnetic activity cycles similar to the Sun. These cycles can be monitored by looking at chromospheric activity, measured in the emission line cores of their Ca II H and K lines; by looking at spot modulation of their light curves; or for a few stars, looking at their coronal X-ray emission over time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subsequent work has found cycles in most stars where there is good data (&lt;a href=&quot;https://arxiv.org/abs/1604.06701&quot; rel=&quot;noreferrer&quot;&gt;Olah et al. 2016&lt;/a&gt;). The cycles vary in length from star-to-star, where they have been found. Some stars show multiple cycle periods (e.g.  &lt;a href=&quot;https://arxiv.org/abs/0904.1747&quot; rel=&quot;noreferrer&quot;&gt;Olah et al. 2009&lt;/a&gt;). There is obviously an observational bias towards finding shorter periods, but the Sun's cycle is towards the longer period end of what has been found. There seems to be a rough correlation between cycle period and rotational period - the cycle period is 100-300 times as long as the rotation period. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An excellent short review is given by &lt;a href=&quot;https://arxiv.org/abs/1511.06116&quot; rel=&quot;noreferrer&quot;&gt;Korhonen (2015)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-08-16T07:08:12.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22128" PostTypeId="2" ParentId="22122" CreationDate="2017-08-16T07:44:53.763" Score="3" Body="&lt;p&gt;Ignoring inclination and by simplifying the orbits into Kepler's ellipses, you still couldn't determine left and right because the relation would be relative to a 360 degree plane, like hands on a clock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this simplified approach, the Perihelion, the Sun, the center of the ellipse, the other Foci and the Aphelion are all in a straight line along the major axis.  The Sun and the Perihelion point in the same direction relative to the &quot;fixed&quot; stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/phgt0.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/phgt0.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, another way to ask your question is to ask where and at what angle do the planets perihelion happen and this is a little bit easier to look up because the perihelion is a real event, unlike the foci which a largely irrelevant mathematical representation.  Each planet passes through it's perihelion once every orbit.   It's not like clockwork, as there's variation from the other planets, but it's roughly in the same place, once every orbital year.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would think a table of planet's perihelion would be fairly easy to look up in a table or chart, but surprisingly, I didn't see any, but individual planets can be found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.timeanddate.com/astronomy/perihelion-aphelion-solstice.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth is at Perihelion&lt;/a&gt; usually in the first week of January.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury was at perihelion (&lt;a href=&quot;https://in-the-sky.org/news.php?id=20170323_11_100&quot; rel=&quot;nofollow noreferrer&quot;&gt;according to this website&lt;/a&gt;) on March 23, 2017 and about every 88 days after that  (about June 19 and Sept 15, 2017)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using NASA-&lt;a href=&quot;https://eyes.nasa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eyes on the Solar System&lt;/a&gt;, a ballpark estimate puts Earth and Mercury's Perihelion are some 30-40 degrees apart, and &lt;a href=&quot;http://astropixels.com/ephemeris/astrocal/astrocal2017gmt.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Venus will be at Perihelion&lt;/a&gt; on Oct 3, 2017, and it's perihelion direction relative to the Sun is even closer, between Earth's and Mercuries, but that pattern ends with Mars.  &lt;a href=&quot;http://earthsky.org/sky-archive/mars-at-perihelion-in-late-october&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mars' next Perihelion&lt;/a&gt; will be on September 16, 2018.   it appears to be nearly 180 degrees opposite Mercury.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20110221111450AAvF2au&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jupiter is in a different direction&lt;/a&gt; as well.   Directions can be anywhere on the 360 degree orbital plane (ignoring inclination).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth noting that over long periods of time (or not that long for Mercury), The planet's perihelion moves.   Earth's moves around a full circle in roughly 112,000 years.   This is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Apsidal_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;Apsidal precession&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also worth noting that if you do apply Kepler ellipses, the location of the focal point relative to the center of the ellipse is a product of eccentricity and distance.  The focal point and center of the ellipse are just mathematical points in empty space anyway.  They have no real use, but if you calculated about where each planet's focal point is relative to the sun, they'd not only be in different directions relative to the sun, but they'd all be different distances from the sun too and with difference inclination.   That would be a kind of fun, but entirely useless mathematical exercise if you wanted to do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to stress that none of this is astronomically useful.  It's not relevant to observation or location of planets because Kepler's laws, while a brilliant leap forward at the time and accurate enough to overcome the geocentric model, they are incomplete and so, the foci are a product of a roughly accurate but incomplete model.    &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-08-16T17:36:27.350" LastActivityDate="2017-08-16T17:36:27.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22129" PostTypeId="1" CreationDate="2017-08-16T08:48:25.343" Score="1" ViewCount="145" Body="&lt;p&gt;How can we find Dec and R.A of a star by having Dec and R.A of another star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example in the following picture , we have that for Errai (Dec: 77 , R.A 23H 40M). How can we find Dec and R.A of Gamma Delphinus. Note that we can't use Polaris for example because of cloud or something else. I want answer for when we have telescope and when we don't have telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qO8Ct.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qO8Ct.png&quot; alt=&quot;Picture from Starry Night&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16441" LastActivityDate="2017-08-18T06:40:52.440" Title="finding Dec and R.A of a star by having Dec and R.A of another star" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22130" PostTypeId="2" ParentId="22129" CreationDate="2017-08-16T09:06:10.277" Score="2" Body="&lt;p&gt;You can measure the position in the sky of any star; this is its altitude and azimuth. The star's position in the sky depends on its actual position in space, your position on Earth and the rotation of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measuring the position requires some kind of angle-measurer. Such as a quadrant or sextant. You also need knowledge of which direction is North. This can be determined, even if the pole star is not visible, from the location of other stars. A telescope can increase accuracy but the same measurements can be done either with or without a telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use your position, and the position of Gamma Cephei in the sky (by altitude and azimuth) to determine the local sidereal time. You can then use the position of Gam Del and the sidereal time to determine the ra and dec of Gam Del.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other questions discuss the conversion between alt-az to ra-dec &lt;a href=&quot;https://astronomy.stackexchange.com/questions/12677/software-to-convert-ra-and-dec-into-alt-and-az&quot;&gt;Software to convert RA and DEC into ALT and AZ&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-08-18T06:40:52.440" LastActivityDate="2017-08-18T06:40:52.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22134" PostTypeId="2" ParentId="22122" CreationDate="2017-08-16T11:45:17.633" Score="5" Body="&lt;p&gt;In addition to the other answers, that are right, a different doubt seem to arise from the wording of the question:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Let us say that the left focal point is f1 and the right f2.&lt;/p&gt;&#xA;&lt;p&gt;Is the sun at the same focal point for each planet?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If the question is if the Sun is in f1 or f2, the answer is that it doesn't matter. Ellipses are symmetrical, and therefore both foci are equivalent and any naming of them is just conventional. Just by saying that planets move in elliptical orbits and the Sun occupies one focus, Kepler's first law is unambiguously established.&lt;/p&gt;&#xA;&lt;p&gt;If you need to mention a particular focus, one is just the Sun position and the other one is often referred as the empty focus.&lt;/p&gt;&#xA;" OwnerUserId="14097" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-12-24T11:44:34.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22139" PostTypeId="1" CreationDate="2017-08-17T02:51:12.430" Score="2" ViewCount="62" Body="&lt;p&gt;... today is August the 17th? I just saw the moon (4:30 CEST) and I have the strong impression that the unlit (by the Sun) part of the Moon is anyway &quot;visible&quot; as a very very dark grey portion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will be impossible to find it against the dark sky if not for the companion bright slice. What I see basically fills the round, just is smaller as compared to the bright part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This latter effect it is explainable by the fact that bright lights appear to us bigger as compared to less luminous objects when put in direct comparison. This is matter of perception. As such my question is: I am also just &lt;em&gt;perceiving&lt;/em&gt; the rest of the moon because I know is there or indeed I &lt;em&gt;see&lt;/em&gt; something real? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, it is reasonable that Earth can illuminate the Moon, however I am not sure. Moreover, the &quot;Earth phase&quot; as seen by the Moon should be crucial here. If I am right, this might explain why sometimes I see the above phenomenon but more often I don't.&lt;/p&gt;&#xA;" OwnerUserId="17437" LastEditorUserId="17437" LastEditDate="2017-08-17T02:56:59.347" LastActivityDate="2017-08-17T02:56:59.347" Title="Does Earth illuminate the unlit part of the Moon as" Tags="&lt;the-moon&gt;&lt;moon-phases&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2017-08-18T04:00:46.857" ContentLicense="CC BY-SA 3.0" />
  <row Id="22140" PostTypeId="2" ParentId="7736" CreationDate="2017-08-17T03:51:34.323" Score="6" Body="&lt;p&gt;The Earth viewed from moon will appear larger, in proportion to how much larger the Earth's diameter is versus the moons diameter.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Earth diameter 7900mi&lt;/li&gt;&#xA;&lt;li&gt;Moon diameter 2100mi&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the Earth-view from the moon would appear 3.75 times as large as the Moon appears in the sky on earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked up the moon's typical angular diameter, it is 0.5 degree.  So the Earth's typical angular diam would be 1.9 degree.&lt;/p&gt;&#xA;" OwnerUserId="18023" LastEditorUserId="2365" LastEditDate="2017-08-17T06:15:47.283" LastActivityDate="2017-08-17T06:15:47.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22145" PostTypeId="1" CreationDate="2017-08-17T06:41:23.770" Score="1" ViewCount="91" Body="&lt;p&gt;Where I am (Vancouver, WA), the Sun will be 99%+ covered. Will I be able to see the &lt;a href=&quot;https://en.wikipedia.org/wiki/Baily%27s_beads&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Beads of Baily&quot;&lt;/a&gt; when the sun is 99%+ covered in my area ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Downtown Portland, OR is expected to be 99.4% (according to local news). I expect Vancouver, WA, a suburb just north of downtown Portland, to be over 99% as well (99.3-99.2%??)&lt;/p&gt;&#xA;" OwnerUserId="10117" LastEditorUserId="2531" LastEditDate="2017-08-18T07:04:34.097" LastActivityDate="2017-08-18T07:04:34.097" Title="Will I be able to see the &quot;Beads of Baily&quot; at 99%+ totality?" Tags="&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22154" PostTypeId="1" AcceptedAnswerId="22155" CreationDate="2017-08-17T16:20:32.860" Score="2" ViewCount="342" Body="&lt;p&gt;Can I see Uranus and Neptune with an iOptron telescope with a 600mm focal length. If so, can I expect to see them in colour, or simply as white objects?&lt;/p&gt;&#xA;" OwnerUserId="18040" LastEditorUserId="2365" LastEditDate="2017-08-17T19:13:47.183" LastActivityDate="2017-08-17T19:13:47.183" Title="Can I see Uranus and Neptune with a household telescope?" Tags="&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22155" PostTypeId="2" ParentId="22154" CreationDate="2017-08-17T17:38:05.417" Score="2" Body="&lt;p&gt;Yes, you can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use a Newtonian reflector, 200mm objective, 1000mm focal length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uranus is sometimes right on the fringe of naked eye but looks a lot clearer as a small but recognisable blueish disc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neptune, sadly, just looks like a star to me ... but you can find it.&lt;/p&gt;&#xA;" OwnerUserId="17502" LastActivityDate="2017-08-17T17:38:05.417" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22157" PostTypeId="2" ParentId="22145" CreationDate="2017-08-17T18:57:52.723" Score="2" Body="&lt;p&gt;No, at 99% a narrow sliver of sun will be visible, the width of that sliver will be about 1.5% of the diameter of the sun at its thickest point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Baily's beads are caused by the sun shining through valleys on the edge of the moon. Such a valley would have to be about 50km deep (or surrounded by hills 50km high).  You do get some beads at the thin edge of the sliver but these would be too bright to see against the sliver of sun (which would still be blinding bright)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get Bailey's beads you need totality, with the sun shining through valleys. Baily's which is why they are normally only visible for a few seconds before true totality, usually a couple of seconds after the predicted start of totality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see the worksheet I used to make these estimations at &lt;a href=&quot;https://ggbm.at/sZ3EYp7z&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ggbm.at/sZ3EYp7z&lt;/a&gt; however it is undocumented.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-08-17T19:03:03.553" LastActivityDate="2017-08-17T19:03:03.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22158" PostTypeId="1" CreationDate="2017-08-17T19:45:55.170" Score="1" ViewCount="55" Body="&lt;p&gt;I live in Champaign, Illinois (&lt;a href=&quot;https://www.google.com/maps/place/Champaign,+IL/@40.11461,-88.34715,12z/data=!3m1!4b1!4m5!3m4!1s0x880cd08dde97691b:0x30f57847b475bfcf!8m2!3d40.1164204!4d-88.2433829&quot; rel=&quot;nofollow noreferrer&quot;&gt;see map&lt;/a&gt;) and am planning to watch the August 21&lt;sup&gt;st&lt;/sup&gt; solar eclipse. This community is within the 95% totality range, but not within the path of totality. Will I be able to see the Baily's beads, and more importantly &lt;strong&gt;will I be able to see the corona&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="10161" LastActivityDate="2017-08-17T19:45:55.170" Title="Will you be able to see the corona in a 90%+ partial eclipse?" Tags="&lt;the-sun&gt;&lt;solar-eclipse&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2017-08-18T07:02:46.913" ContentLicense="CC BY-SA 3.0" />
  <row Id="22159" PostTypeId="1" CreationDate="2017-08-17T23:08:10.017" Score="1" ViewCount="168" Body="&lt;p&gt;I, like many others, am excited to see the eclipse. I want to view the eclipse through my telescope, however, I don't have any solar safety equipment for solar viewing and I cannot find any online that I can get before the eclipse. Could anyone lead me to a reputable site where I could purchase solar (preferably mylar) film before the eclipse?&lt;/p&gt;&#xA;" OwnerUserId="5533" LastActivityDate="2021-01-11T20:27:31.523" Title="Where can I still find solar film?" Tags="&lt;the-sun&gt;&lt;telescope&gt;&lt;amateur-observing&gt;&lt;solar-eclipse&gt;&lt;resource-request&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22160" PostTypeId="2" ParentId="22159" CreationDate="2017-08-18T00:27:28.180" Score="2" Body="&lt;p&gt;Amazon has solar film sheets if that's what you're looking for. It is sold by Amazon for &lt;a href=&quot;http://thousandoaksoptical.com/products/solar-filters/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thousand Oaks Optical&lt;/a&gt;. Amazon sometimes has next day or 2 days delivery.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="616" LastEditDate="2021-01-11T16:49:02.343" LastActivityDate="2021-01-11T16:49:02.343" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22161" PostTypeId="1" CreationDate="2017-08-18T02:58:49.203" Score="6" ViewCount="124" Body="&lt;p&gt;I've seen many pictures of the corona during a solar eclipse, and the structure and delicacy is amazing. Clearly, the appearance would change over time, depending on the fine details of the solar wind and the structure of the mass shed from the solar surface. I wonder, though: might the appearance, to the unaided eye, change during a single eclipse?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could actually be treated as two questions (all example durations and times taken from the August 2017 eclipse):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Might the corona's appearance visibly change during totality at a single location (a maximum of 2 minutes 40 seconds)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Might the corona's appearance visibly change between the first Earth-wide totality (16:48 UTC, when the Moon's shadow touches down in the Pacific) and the last Earth-wide totality (20:02 UTC, when the Moon's shadow last leaves the Atlantic)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Searching for solar wind velocities yields a broad range of values, but $400km/s$ seems to be a common value. With the solar diameter being about $1{,}400{,}000km$, that's about a solar diameter per hour, which suggests there will be no visible change at a single location, but there certainly could be changes when viewed across the entire 3+ hours of totality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, can we expect changes in the coronal appearance during this eclipse, either at a single location or across the entire eclipse path?&lt;/p&gt;&#xA;" OwnerUserId="9889" LastEditorUserId="9889" LastEditDate="2017-08-19T14:34:25.753" LastActivityDate="2017-08-19T14:34:25.753" Title="Might someone see changes to the corona during a solar eclipse?" Tags="&lt;the-sun&gt;&lt;solar-eclipse&gt;&lt;solar-wind&gt;&lt;coronal-mass-ejection&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22162" PostTypeId="2" ParentId="22161" CreationDate="2017-08-18T05:25:14.143" Score="2" Body="&lt;p&gt;Yes. There is a chance of one or more solar flares or CME's occurring during the eclipse and these would be noticeable during totality. There are actually two NASA jets equipped with telescopes and high-speed cameras to film the Corona and any such event. The plane will track the eclipse throughout its duration. &lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="16926" LastEditDate="2017-08-18T06:28:47.030" LastActivityDate="2017-08-18T06:28:47.030" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22166" PostTypeId="1" CreationDate="2017-08-18T13:01:19.277" Score="8" ViewCount="1248" Body="&lt;p&gt;I have a 6cm aperture newtonian refractor telescope through which I saw the Orion Nebula yesterday night. As all you can hope with a telescope of such a small aperture, I saw some faint cloudy structure, but when I looked away from the cloud through the telescope it appeared to be more distinct and dense. Why is it so?&lt;/p&gt;&#xA;" OwnerUserId="18018" LastEditorUserId="18018" LastEditDate="2017-08-19T11:26:41.610" LastActivityDate="2017-09-05T12:36:12.947" Title="Why does the cloud of Orion Nebula appear more distinct when I am not directly looking at it?" Tags="&lt;observational-astronomy&gt;&lt;orion&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22167" PostTypeId="2" ParentId="22166" CreationDate="2017-08-18T13:59:20.870" Score="17" Body="&lt;p&gt;This is called the &lt;strong&gt;Averted vision&lt;/strong&gt;. It involves not looking directly at the object, but looking a little off to the side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How does it works?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the retina is less sensitive at the center, and this is very remarkable on the very faint objects (like nebulas or galaxies)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a common technique used by astronomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More info &lt;a href=&quot;http://oneminuteastronomer.com/86/averted-vision/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How to make the most of Averted vision?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To maximise the effects of your Averted Vision, you first need to know which eye is your &lt;strong&gt;dominant eye&lt;/strong&gt;. To do so:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Point with your finger a distant object&lt;/li&gt;&#xA;&lt;li&gt;Then blink an eye, then the other eye.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;The eye in which your finger is perfectly aligned with the object is your dominant eye&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then, for the best Averted vision:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Look at the &lt;strong&gt;top-right&lt;/strong&gt; of the object if your dominant eye is the &lt;strong&gt;right&lt;/strong&gt; one&lt;/li&gt;&#xA;&lt;li&gt;Look at the &lt;strong&gt;top-left&lt;/strong&gt; of the object if your dominant eye is the &lt;strong&gt;left&lt;/strong&gt; one&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7085" LastEditorUserId="7085" LastEditDate="2017-09-05T12:36:12.947" LastActivityDate="2017-09-05T12:36:12.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22169" PostTypeId="1" AcceptedAnswerId="22323" CreationDate="2017-08-18T18:48:38.160" Score="1" ViewCount="512" Body="&lt;p&gt;Reviewing several simulations of the solar eclipse, it is apparent that the path of the moon relative to the sun curves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &lt;a href=&quot;https://www.timeanddate.com/eclipse/in/usa/kansas-city?iso=20170821&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.timeanddate.com/eclipse/in/usa/kansas-city?iso=20170821&lt;/a&gt; for example, the moon enters at 135 degrees and leaves at 217 degrees direction - that is almost a 90 degree right turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the path curved not straight?&lt;/p&gt;&#xA;" OwnerUserId="18057" LastActivityDate="2017-08-25T14:28:14.513" Title="Why does the path of the moon in an eclipse curve (relative to the sun)" Tags="&lt;solar-eclipse&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22170" PostTypeId="2" ParentId="22166" CreationDate="2017-08-18T19:41:48.427" Score="3" Body="&lt;p&gt;The effect you describe is called &quot;averted vision&quot;.   It works because the center of the retina is optimized for details.  We use it for reading and anything that needs sharp vision.   The other parts of the retina are more sensitive to light and and motion but not to detail&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Digital cameras are like this too.   Having many tiny pixels makes for a sharp image but the small pixels are less sensitive to low light and create more noise or grain in the image.  A digital camera is more sensitive to light if the pixels are physically larger but of course then you can have fewer of these in a same size sensor&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The eye in effect has smaller &quot;pixels&quot; that are closer together but less sensitive to light in the center of the visual field and larger more sensitive &quot;pixels&quot; in the off center areas.   So, by looking off to the side you can cause the image of the nebula to fall on the part of the retina that is more sensitive to light.&lt;/p&gt;&#xA;" OwnerUserId="18060" LastActivityDate="2017-08-18T19:41:48.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22171" PostTypeId="2" ParentId="22169" CreationDate="2017-08-18T19:46:27.153" Score="4" Body="&lt;p&gt;You have misunderstood the website. When the eclipse begins, in the morning, the sun will be in south-east, at a bearing of 135 degrees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At maximum eclipse (at about midday) the sun will be almost due south, at a bearing of 178 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the eclipse ends, in the afternoon, the sun will be in the south-west, at about 217 degrees bearing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun and the moon will move in the sky, due to the rotation of the Earth, this says nothing about the straightness of the path of the moon relative to the sun.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-08-18T19:46:27.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22173" PostTypeId="2" ParentId="22166" CreationDate="2017-08-18T20:54:27.147" Score="0" Body="&lt;p&gt;&quot;Averted vision&quot; is a nice term, and perhaps expertly technical as well. It's the same principle we are taught as shooters/small arms handlers to utilize for night-work when enhanced vision equipment is not available; we simply know it as a function of our peripheral vision, right up there with using the same visual technique to more easily detect motion - &lt;em&gt;especially&lt;/em&gt; in a darkened environment. The phenomenon is due to both the distribution and the disparity in numbers of rods versus cones within the human retina. In any event, it's pretty cool once you get used to using it to your advantage :-) &lt;/p&gt;&#xA;" OwnerUserId="18063" LastEditorUserId="18063" LastEditDate="2017-08-18T21:03:55.993" LastActivityDate="2017-08-18T21:03:55.993" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22175" PostTypeId="1" CreationDate="2017-08-19T03:18:25.263" Score="2" ViewCount="122" Body="&lt;p&gt;I was watching a summary video about the Banach-Tarski paradox and it led me to a question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I (a) assume the universe is infinite and (b) assume it is possible to receive a single photon of light from any star within it that has a line-of-site to the Earth and (c) assume that the photon has travelled an infinite distance, then such a photon, if enough information could be extracted from it, would prove the universe is infinitely large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today we have finite observational and recording capabilities.  I believe (not being a physicist or astronomer myself) that we can't prove the universe is larger than some size, X, because we cannot detect lower than a particular number of photons, Y, to ascertain the distance they travelled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, assuming my assumptions aren't assinine (they very well could be), and ignoring our present observational limitations, what is the smallest number of photons from which we could discern information about the distance those photons travelled, and what distance would that represent?&lt;/p&gt;&#xA;" OwnerUserId="18065" LastActivityDate="2017-08-19T07:11:41.087" Title="Based on the smallest number of photons that can deliver information about a star, what is the largest size of the universe?" Tags="&lt;universe&gt;&lt;observational-astronomy&gt;&lt;photons&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22179" PostTypeId="2" ParentId="22175" CreationDate="2017-08-19T07:11:41.087" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So, assuming my assumptions aren't assinine (they very well could be) ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your words, not mine. Your &quot;very well could be&quot; is the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Assume the universe is infinite&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe might well be infinitely large, but the observable universe is all that we can possibly see. By all signs, the universe is 13.8 billion years old. We can't see the stuff that is beyond the observable universe because not enough time has passed for light to have traveled that far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make matters worse, the early universe was very hot and dense. Photons traveled but a short distance before being absorbed. We can't see those very early photons because they no longer exist. This changed about 379,000 years after the universe formed, at which time the universe had sufficiently cooled so as to allow photons to roam free. The photons let loose by that clearing of the universe form the cosmic microwave background radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;em&gt;Assume it is possible to receive a single photon of light from any star within it that has a line-of-site to the Earth&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per the big bang theory, the first stars formed about 100 to 200 million years after the formation of the universe. Stars that formed at that time beyond the observable universe: We can't see them. Due to the continuing (and possibly accelerating; google &quot;dark energy&quot;) expansion of the universe, we will never see those stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a famous paradox related to your question: Why aren't we bathed night and day by light from an infinite number of stars if the universe is uniform and is infinitely old&gt; This is Olber's paradox. The solution is simple: The universe is not infinitely old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;em&gt;Assume that the photon has travelled an infinite distance, then such a photon, if enough information could be extracted from it, would prove the universe is infinitely large&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only information that can be extracted from a single photon is its frequency, and even that is a bit fuzzy due to quantum mechanics. A photon from a remote star is indistinguishable from a photon of the same frequency emitted a nearby lightbulb.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-08-19T07:11:41.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22186" PostTypeId="2" ParentId="21987" CreationDate="2017-08-19T14:00:47.133" Score="1" Body="&lt;p&gt;The formula you have written resembles the Stefan-Boltzmann law, and its correct form is $$T^4 = \frac{L}{4\pi R^2 \sigma_{SB}}$$.  It relates the luminosity of a blackbody of a given radius to its effective temperature. Earth as a whole can be assumed to be a black body at an effective temperature. So you can calculate the effective temperature of the &quot;earth&quot;. This is very well explained in the example section of this wiki page: &lt;a href=&quot;https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For calculating the effective temperature of the polar region, first assumption is that polar region is a black body. Then the question to ask is: what is the effective teperature of the polar region given that the it reaches equilibrium when the Sun light falls on it. To calculate this further, you would need to calculate the angle at which the sun light fall on the polar region you are interested in and also the distance travelled by the sun light in the atmosphere. These parameters will vary with respect to the time of the year, time of the day and the region you are interested in (North pole, South pole). Some calculated temperatures are given on the same wiki page. I hope that helps. &lt;/p&gt;&#xA;" OwnerUserId="10253" LastActivityDate="2017-08-19T14:00:47.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22189" PostTypeId="1" CreationDate="2017-08-19T15:36:12.720" Score="0" ViewCount="444" Body="&lt;p&gt;As I look for a spot for viewing the total eclipse, how much will cloud cover affect the viewing experience? Is there a threshold list like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under x % won't have a noticeable difference&#xA;Over x % you won't be able to see the stars&#xA;Over x % you won't be able to see the sun's atmosphere&#xA;Over x % you will only notice the darkening&lt;/p&gt;&#xA;" OwnerUserId="18081" LastActivityDate="2020-03-27T12:02:15.317" Title="How much does cloud cover affect viewing the total eclipse" Tags="&lt;solar-eclipse&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22190" PostTypeId="2" ParentId="22189" CreationDate="2017-08-19T16:17:16.987" Score="1" Body="&lt;p&gt;I don't know if there is a list but I'me sure your local weather surface will give you some idea. Cloud cover could have a big effect of course. But it will be the same as it is for any day or night. Total overcast will allow you to notice the darkness only. Just use your common sense on what cloud cover you deem too risky for travel. There is an internet site: &lt;a href=&quot;https://clearoutside.com/forecast&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://clearoutside.com/forecast&lt;/a&gt; which is supposedly developed by astronomers to help determine viewing conditions. Just input a latitude and longitude and it gives you info related to viewing conditions. Different days are on left after local forecast. Click the triangle in the lower right of the date and it gives you a detailed forecast for the location you entered. This includes predicted cloud cover by the hour.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="16926" LastEditDate="2017-08-19T17:03:24.153" LastActivityDate="2017-08-19T17:03:24.153" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22198" PostTypeId="1" AcceptedAnswerId="22199" CreationDate="2017-08-19T22:04:40.433" Score="2" ViewCount="161" Body="&lt;p&gt;I have attempted to research this topic due to curiosity, but I have yet to find a credible answer, the best I could find is &lt;a href=&quot;http://www.express.co.uk/news/science/764801/MILLIONS-black-holes-capable-of-swallowing-Earth-asteroids&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; article, which has no listed sources, and the author has also done similar &quot;click bait&quot; style headlines with little to no evidence.&lt;/p&gt;&#xA;" OwnerUserId="16418" LastActivityDate="2017-08-20T00:14:10.227" Title="How many wandering black holes are there in the milky way?" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22199" PostTypeId="2" ParentId="22198" CreationDate="2017-08-20T00:14:10.227" Score="3" Body="&lt;p&gt;Yes as StephanG said there are thought to be 100 million black holes in our galaxy composed of approximately 100 billion stars. Most of these are thought to be stellar mass black holes. Not the super massive black hole assumed to be at the center of our galaxy. Some references are:&lt;a href=&quot;http://hubblesite.org/explore_astronomy/black_holes/encyc_mod3_q7.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://hubblesite.org/explore_astronomy/black_holes/encyc_mod3_q7.html&lt;/a&gt; and &lt;a href=&quot;https://www.sciencenews.org/article/we-share-milky-way-100-million-black-holes&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.sciencenews.org/article/we-share-milky-way-100-million-black-holes&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastActivityDate="2017-08-20T00:14:10.227" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22200" PostTypeId="2" ParentId="22189" CreationDate="2017-08-20T05:14:40.953" Score="0" Body="&lt;p&gt;Well,I don't mean to answer a question with a question, but: on a cloudy day, can you see the sun through the clouds?&lt;/p&gt;&#xA;" OwnerUserId="16723" LastActivityDate="2017-08-20T05:14:40.953" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22201" PostTypeId="1" CreationDate="2017-08-20T05:37:17.660" Score="-6" ViewCount="210" Body="&lt;p&gt;Help needed, please. I need the input of the experts here. I have always been troubled by the formula ..F = G * m1 * m2 / r^2.. , which appears to be counter-intuitive. It should be ...F = G ( m1 + m2 ) / r^2.. I present here two examples to support my hypothesis.&lt;br&gt;&#xA;A )  Comparing 4 different objects falling to Earth at a height of 100 m.&#xA;r^2 = (6371 + .01)^2 km. = 40590915 ... G = 6.67*10^-11 ...m1 = 5.972*10^24 kg&lt;br&gt;&#xA;when m2 = 1 gm ,                     F = 9813338.722204&lt;br&gt;&#xA;when m2 = 1 kg ,                     F = 9813338.722204&lt;br&gt;&#xA;when m2 = 1 trillion kg( 12 zeroes), F = 9813338.722206&lt;br&gt;&#xA;when m2 = 1 quintillion  kg ( 18 ) , F = 9813340.365429&lt;br&gt;&#xA;     Here, heavier objects do fall faster, but impossible to observe or detect, even &#xA;with our most delicate instruments. Only when m2 is overwhelmingly massive, do we begin to notice the difference.&lt;br&gt;&#xA;B ) orbits of Earth and Moon around the Sun&lt;/p&gt;&#xA;&#xA;&lt;p&gt;m1 = mass of Sun = 1.989*10^30&lt;br&gt;&#xA;                                                                                                                                    m2A = mass of earth = 5.972*10^24 and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;m2B = moon = 7.3477*10^22....&#xA;average distance of Sun to Earth = 149.6 million km &#xA;Since Moon orbits earth, then average distance to Sun also the same.&#xA;r^2 =( 149.6 million + radius of Sun 595500 ) ^2 = 150295500 km.&lt;br&gt;&#xA;when m2 = moon , F = 5873.11734&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when m2 = Earth , F = 5873. 13476 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, both the Earth and Moon are falling towards the Sun at the practically same rate, and as such, the Sun's gravitational influence should not affect the orbit of Moon around the Earth. So we don't need to invoke the Hill Theory to explain why the moon is not pulled away from its orbit.&#xA;Please let me know where I've erred , and enlighten me , hopefully with a reply coherent to me. Thanks&lt;/p&gt;&#xA;" OwnerUserId="18088" LastActivityDate="2017-08-21T06:26:15.410" Title="Is there something wrong about the gravity formula?" Tags="&lt;gravity&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2017-08-21T15:25:08.153" ContentLicense="CC BY-SA 3.0" />
  <row Id="22202" PostTypeId="2" ParentId="21822" CreationDate="2017-08-20T06:31:14.740" Score="-1" Body="&lt;p&gt;I purchased 100 pair of glasses from Walmart online from AllStar Gadgets to use at school with students. It says they are certified. How can you be sure? &lt;/p&gt;&#xA;" OwnerUserId="18089" LastActivityDate="2017-08-20T06:31:14.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22205" PostTypeId="2" ParentId="10204" CreationDate="2017-08-20T14:36:21.307" Score="1" Body="&lt;p&gt;&lt;strong&gt;YES&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have summarized pretty much everything available on the subject. There are loads of people saying you cannot, but fortunately for us, the &lt;em&gt;&lt;a href=&quot;https://www.rasc.ca/tov/safety?page=0%2C6&quot; rel=&quot;nofollow noreferrer&quot;&gt;Royal Astronomical Society of Canada&lt;/a&gt;&lt;/em&gt;, that you quote, includes evidence to back up their claims. They are a &lt;a href=&quot;https://en.wikipedia.org/wiki/Royal_Astronomical_Society_of_Canada&quot; rel=&quot;nofollow noreferrer&quot;&gt;respected institution&lt;/a&gt; and the &lt;a href=&quot;https://uwaterloo.ca/optometry-vision-science/people-profiles/b-ralph-chou&quot; rel=&quot;nofollow noreferrer&quot;&gt;author&lt;/a&gt; of this article appears immensely qualified. The article itself does not include evidence that welding goggles can be stacked, but the inclusion of such for all the major points adds credibility to the entire article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does appear like the glasses are not exactly additive, this &lt;a href=&quot;https://www.reddit.com/r/solareclipse/comments/6ttztj/stacking_welding_glass/dlnkm4f/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reddit post&lt;/a&gt; goes through some calculations, but they are pretty close such that aiming for 1 tint higher will more than offset it and if anything is overkill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8KOnA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8KOnA.jpg&quot; alt=&quot;Transmittance curve of glass and polycarbonate SN14 welding filters.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18094" LastActivityDate="2017-08-20T14:36:21.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22208" PostTypeId="2" ParentId="22201" CreationDate="2017-08-20T18:37:44.873" Score="2" Body="&lt;p&gt;To enlarge on Rob's comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $m_2 = 1 \mathrm{g},\ F = 9813338.722204$ and therefore a one gram mass would accelerate at $a =F/m = 9813338.722204/0.001 = 9813338722.204 \mathrm{ms^{-2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $m_2 = 1 1 \mathrm{kg},\ F = 9813338.722204$ and therefore a one kilogram mass would accelerate at $a =F/m = 9813338.722204/1 = 9813338.722204 \mathrm{ms^{-2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, different masses would accelerate at very different rates. Smaller masses would fall much faster than larger masses. As Rob says, that might have been noticed. Instead we observe that larger masses accelerate at the same rate as smaller ones, because larger masses have a larger force of gravity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, both masses accelerate at the same rate: $9.81 ms^{-2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You seem to be confusing &quot;Force due to gravity (or weight)&quot; with &quot;acceleration due to gravity&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic problem with your idea is that it doesn't match even simple observations. Newton's model, however, is a very good fit to observations of gravity. Therefore your model is not just wrong, it very wrong and it is of no use.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-08-21T06:26:15.410" LastActivityDate="2017-08-21T06:26:15.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22209" PostTypeId="1" AcceptedAnswerId="22211" CreationDate="2017-08-20T22:52:31.523" Score="3" ViewCount="335" Body="&lt;p&gt;In &lt;a href=&quot;https://arstechnica.com/science/2017/08/heres-what-happens-to-your-retina-if-you-view-an-eclipse-without-protection/&quot; rel=&quot;nofollow noreferrer&quot;&gt;an article on Ars Technica&lt;/a&gt; about safely viewing a solar eclipse, they say this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;First, what not to do: Don’t view the eclipse with your naked eye or unfiltered telescopes, binoculars, sunglasses (yeah, even if they’re really dark), camera lenses, or other optics devices. Don’t use anything that focuses light, even if you’re wearing eclipse glasses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This may sound like a stupid question, but are eyeglasses (by which I mean, lenses to correct poor vision) safe to wear during an eclipse along with eclipse glasses? I don't know if the light focusing is strong enough to be concerned about.&lt;/p&gt;&#xA;" OwnerUserId="13850" LastActivityDate="2017-08-20T23:09:02.967" Title="Is it safe to view a solar eclipse with eyeglasses along with eclipse glasses?" Tags="&lt;solar-eclipse&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22210" PostTypeId="2" ParentId="22209" CreationDate="2017-08-20T23:05:49.850" Score="3" Body="&lt;p&gt;Yes, it is fine to wear eclipse glasses over your eyeglasses.  What the article in Ars Technica is talking about is looking through a telescope with eclipse glasses.  The telescope would focus the light, burning right through the eclipse glasses.  But wearing normal eyeglasses behind eclipse glasses is perfectly fine.  &lt;/p&gt;&#xA;" OwnerUserId="13426" LastActivityDate="2017-08-20T23:05:49.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22211" PostTypeId="2" ParentId="22209" CreationDate="2017-08-20T23:09:02.967" Score="4" Body="&lt;p&gt;Yes, it's safe :).  Your eyeglasses are just correcting your vision. The net effect is just to give the equivalent of normal vision - and in any case, you'd normally be holding the eclipse glasses in front of your eyeglasses, in the same way that front-of-scope solar filters work to reduce the incoming light to a safe level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you're longsighted (which uses converging (positive) lenses to correct your vision, the glasses are close enough to your eye that you're nowhere near the focal point - so the light doesn't get concentrated much (and probably gets unconcentrated again by your eye, anyway).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're shortsighted, like me, your glasses will be diverging (negative) lenses which spread the light out rather than concentrating it (and again, are close enough to your eye that's there's not enough distance for it to affect things much.&lt;/p&gt;&#xA;" OwnerUserId="17016" LastActivityDate="2017-08-20T23:09:02.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22217" PostTypeId="1" AcceptedAnswerId="22411" CreationDate="2017-08-21T08:52:23.993" Score="7" ViewCount="406" Body="&lt;p&gt;For atmospheric refraction of visible light, &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_refraction#Values&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia gives&lt;/a&gt; the order of 1 arc minute at 45° altitude above the horizon, and 5.3 arc minutes at 10°. This is caused by the dielectric polarizability of all of the bound electrons in all the atoms of the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the much lower HF frequencies of radio, the free electrons and ions will contribute, and some forms of radio communication have relied on refraction at large incident angles to deflect terrestrial signals back to the Earth at a distant ground station.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I expect that at the lower frequencies used in radio astronomy, corrections to the observed location of radio sources due to ionospheric refraction could be much larger than those at visible wavelengths, but I am not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How large does can this effect ever get? At what frequency? Are there ever corrections as large as 1 degree? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I started thinking about this after asking &lt;a href=&quot;https://space.stackexchange.com/q/22724/12102&quot;&gt;How many stations could one hear with an AM/FM radio in front of the ISS' cupola window?&lt;/a&gt; which includes the image below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; from the Radio Jove Project's exercise &lt;a href=&quot;https://radiojove.gsfc.nasa.gov/education/educ/radio/tran-rec/exerc/iono.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Effects of Earth's Upper Atmosphere on Radio Signals&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/r3nps.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/r3nps.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2020-09-21T03:23:14.577" LastActivityDate="2020-09-22T06:54:36.450" Title="How large does refraction become in radioastronomy?" Tags="&lt;radio-astronomy&gt;&lt;atmospheric-effects&gt;&lt;adaptive-optics&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22224" PostTypeId="1" CreationDate="2017-08-21T16:01:26.820" Score="2" ViewCount="116" Body="&lt;p&gt;I'm writing a report about the solar eclipse happening over north America and was wondering why looking at a solar eclipse is worse than looking at the sun regularly. In addition what alternate materials can be used to make eclipse glasses that are safe to look through. &lt;/p&gt;&#xA;" OwnerUserId="18122" LastActivityDate="2017-09-11T23:33:04.107" Title="2017 solar eclipse report" Tags="&lt;observational-astronomy&gt;&lt;solar-eclipse&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22225" PostTypeId="1" AcceptedAnswerId="22294" CreationDate="2017-08-21T17:54:42.843" Score="1" ViewCount="1731" Body="&lt;p&gt;I was given a 1995 vintage Tasco &quot;302003&quot; &lt;a href=&quot;https://en.wikipedia.org/wiki/Newtonian_telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Newtonian&lt;/a&gt; telescope recently and have had some exciting success with it.  I have managed to view the Moon in beautiful detail and also had my first glimpse of Jupiter (with &lt;s&gt;Enceladus&lt;/s&gt; Ganymede &amp;amp; Europa in the same shot!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was able (with very considerable difficulty) to photograph both Moon and Jupiter scenes by holding my camera up to the eyepiece.  Far from ideal but it has really fired my imagination about doing more astronomical photography.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a picture of my telescope and it's accessories.  Despite spending the last 15 years in the back room of a charity shop, it's in brand new condition and all the parts are present including the tripod and all fixings!...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Tvrjg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Tvrjg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;(Blue arrows indicate places where eyepiece lenses can be inserted).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My camera is a Nikon D40 (APS-C format, DX lens ring system).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to modify this telescope's eyepiece (a rack-and-pinion tube-length adjustable type) so that I can attach my camera in order to take photographs of astronomical subjects?  Photographs of the Moon would be my primary goal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am happy to make severe modifications to the telescope if necessary.  But I realise that certain measurements would be absolutely critical to making good photographs possible.  I think I can make an adaptor that goes between the eyepiece socket and the camera.  But I don't know all of the parameters I will have to know to design such an adaptor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problems I believe I need to solve...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I think that holding the camera's central axis perfectly concentric with the eyepiece's axis is critical.&lt;/li&gt;&#xA;&lt;li&gt;I need some sort of &lt;a href=&quot;https://en.wikipedia.org/wiki/Relay_lens&quot; rel=&quot;nofollow noreferrer&quot;&gt;Relay Lens&lt;/a&gt; between the secondary mirror and the camera body.&lt;/li&gt;&#xA;&lt;li&gt;Keeping the rack-and-pinion tube adjustment (actually a fairly sturdy mechanism) is important and would allow better focusing of the picture as the camera is not likely to have enough auto-focus range to cope with the 700mm F length.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I do have a small Milling machine and most of the basic machine tools.  I have a ready supply of aluminium stock and industrial plastic stock (RG1000) that I can build an adaptor from.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Is this project feasible?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I'm not looking for perfect photos, but rather something that is better and more convenient than holding my cellphone camera close to the eyepiece and holding my breath! :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My astronomy/optics knowledge is practically zero but I'm a fairly savvy mechanical engineer and can build things to a specification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for reading, I look forward to your responses.&lt;/p&gt;&#xA;" OwnerUserId="18091" LastEditorUserId="18091" LastEditDate="2017-09-02T17:53:55.763" LastActivityDate="2020-10-18T18:23:59.730" Title="Retrofitting a DSLR camera to a super-cheap Tasco reflector telescope" Tags="&lt;photography&gt;&lt;newtonian-telescope&gt;" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="22226" PostTypeId="1" AcceptedAnswerId="22229" CreationDate="2017-08-21T17:59:33.237" Score="1" ViewCount="215" Body="&lt;p&gt;I was wondering if the astronauts on the ISS are able to see the shadow from the moon on the earth at all during the eclipse.&lt;/p&gt;&#xA;" OwnerUserId="18128" LastEditorUserId="18128" LastEditDate="2017-08-21T18:24:22.107" LastActivityDate="2017-08-21T18:32:52.297" Title="Can Astronauts Aboard ISS See the Shadow of The Moon on The Earth During an Eclipse?" Tags="&lt;solar-eclipse&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22227" PostTypeId="2" ParentId="22226" CreationDate="2017-08-21T18:30:15.527" Score="0" Body="&lt;p&gt;Well, I did just a bit more googling and found &lt;a href=&quot;http://earthsky.org/space/iss-astronauts-aug-21-2017-total-solar-eclipse&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; which does indeed show images of a shadow on the earth.  Pretty cool.&lt;/p&gt;&#xA;" OwnerUserId="18128" LastActivityDate="2017-08-21T18:30:15.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22228" PostTypeId="1" AcceptedAnswerId="22272" CreationDate="2017-08-21T18:31:54.907" Score="2" ViewCount="227" Body="&lt;p&gt;I live in a place where we weren't on the path of totality but were in the 90% range. I was expecting it to be a lot darker than it ended up being. What does it mean to be in the 90% range on this map? I was expecting it to block out 90% of the visible light, but while it was noticeably dimmer it wasn't close to what I thought 10% of light would look like.&lt;a href=&quot;https://i.stack.imgur.com/B0lSn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B0lSn.jpg&quot; alt=&quot;Nasa solar eclipse map&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18130" LastActivityDate="2017-08-24T17:53:22.607" Title="Eclipse Visible Light Not Dark as Expected" Tags="&lt;solar-eclipse&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22229" PostTypeId="2" ParentId="22226" CreationDate="2017-08-21T18:32:52.297" Score="2" Body="&lt;p&gt;Yes. Here's an ISS photo &lt;a href=&quot;https://eclipse2017.nasa.gov/eclipse-shadow-iss&quot; rel=&quot;nofollow noreferrer&quot;&gt;from the 2002 eclipse over the Indian ocean&lt;/a&gt;.&#xA;This year, 2017, the &lt;a href=&quot;https://eclipse2017.nasa.gov/iss-observations&quot; rel=&quot;nofollow noreferrer&quot;&gt;ISS will see the moon's shadow on earth once&lt;/a&gt;, over southern Illinois, at 43.9% totality.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2017-08-21T18:32:52.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22230" PostTypeId="1" CreationDate="2017-08-21T19:04:10.947" Score="4" ViewCount="138" Body="&lt;p&gt;I live in NYC and just experienced the eclipse. What fascinated me the most was that despite the fact that the sun was about 70% blocked, the area was not noticeably dimmer. If the news wasn't saying there was a solar eclipse I wouldn't have realized it was happening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much of the sun needs to be blocked by the moon in order to noticeably dim an area?&lt;/p&gt;&#xA;" OwnerUserId="1549" LastActivityDate="2017-08-24T00:49:48.673" Title="How much of the sun needs to be blocked in order to see a noticeable drop in brightness" Tags="&lt;the-sun&gt;&lt;the-moon&gt;&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22231" PostTypeId="1" AcceptedAnswerId="22290" CreationDate="2017-08-21T19:07:23.570" Score="8" ViewCount="347" Body="&lt;p&gt;Let's say we could travel anywhere in the solar system, at what distance would it be safe for a human to stare at the sun (without eye protection) and not get eye/retina damage?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Not sure where this question is good for, here, or perhaps Biology.SE, so let me know.)&lt;/p&gt;&#xA;" OwnerUserId="10152" LastActivityDate="2017-08-23T21:01:36.667" Title="At what distance would it be safe for humans to stare at the sun?" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22232" PostTypeId="1" CreationDate="2017-08-21T19:43:25.870" Score="1" ViewCount="99" Body="&lt;p&gt;I live in Austin, TX where the eclipse on August 21, 2017 reached its maximum point with ~66% of the sun covered by the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why didn't it seem any darker than a normal time with 0% eclipse?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="18132" LastActivityDate="2017-08-22T04:45:33.597" Title="Why didn't it seem darker outside with 66% of the sun covered by the moon?" Tags="&lt;solar-eclipse&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ClosedDate="2017-08-29T12:00:40.360" ContentLicense="CC BY-SA 3.0" />
  <row Id="22233" PostTypeId="1" CreationDate="2017-08-21T20:01:02.097" Score="25" ViewCount="20074" Body="&lt;p&gt;People watch sunsets all the time.  You don't see people using special &quot;sunset glasses&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet with an eclipse, warnings are posted everywhere not to view it without special glasses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this?&lt;/p&gt;&#xA;" OwnerUserId="9582" LastEditorUserId="9582" LastEditDate="2017-08-22T15:49:19.500" LastActivityDate="2018-10-08T18:15:19.183" Title="Why is it okay to watch a sunset but not an eclipse?" Tags="&lt;the-sun&gt;&lt;solar-eclipse&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22234" PostTypeId="1" CreationDate="2017-08-21T20:03:53.410" Score="3" ViewCount="184" Body="&lt;p&gt;During a solar eclipse, some of the visible area of the sun is obscured by the moon.  The exact percentage varies with time and with the observer's location.  My question is, does the sky darken by this same percentage?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, suppose the sun is 80% covered.  Does sky light diminish by 80%?  Are objects illuminated by 20% of their normal amount?&lt;/p&gt;&#xA;" OwnerUserId="18133" LastActivityDate="2019-07-13T12:48:46.810" Title="Does the sky darken proportionally during a partial solar eclipse?" Tags="&lt;light&gt;&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22235" PostTypeId="1" AcceptedAnswerId="22236" CreationDate="2017-08-21T20:15:26.540" Score="4" ViewCount="155" Body="&lt;p&gt;The ecliptic (Earth's orbital plane) is inclined 23.4 degrees. This is the same as Earth's axial tilt. The Moon's orbit is inclined 5.145 degrees to the ecliptic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the maximum latitude where a solar eclipse produces an umbra &lt;strong&gt;should&lt;/strong&gt; be 28.545 N and 28.545 S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is clearly not the case. Wikipedia has a huge list of solar eclipses with charts. &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_eclipse_of_October_12,_1939&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's one&lt;/a&gt; that goes very near the south pole. How is this possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Actually I missed something. Since the Moon and Sun must be in the same place, and the Sun can never exceed 23.4 degrees declination, solar eclipses must be bound by 23.4 N/S. I should not have added the two angles. Thanks to user:berrycenter for pointing it out.&lt;/p&gt;&#xA;" OwnerUserId="7449" LastEditorUserId="7449" LastEditDate="2017-08-22T23:28:03.060" LastActivityDate="2017-08-23T19:22:31.857" Title="Why do some solar eclipses' umbra cross arctic/antarctic regions?" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;solar-eclipse&gt;&lt;ecliptic&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22236" PostTypeId="2" ParentId="22235" CreationDate="2017-08-21T20:15:26.540" Score="6" Body="&lt;p&gt;I wanted to ask this question even though I realized the correct geometry in the middle of typing it. So I will answer this myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, &lt;strong&gt;the sub-lunar point&lt;/strong&gt; never exceeds 28.545 N/S. If you're standing at the sub-lunar point, the Moon will be directly overhead. The Moon never wanders outside this latitude bracket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that does not mean the Moon's shadow can't wander outside it. Here's one way it could happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rjrKl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rjrKl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; this is my attempt to draw something approximately to scale. I think I got the Earth/Moon sizes about right, but the Moon-Earth distance should around 4x greater. And the solar rays are of course not parallel but I don't think it's possible to draw or perceive tiny angles for this scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The green line is the ecliptic. The Sun and Moon are supposed to be aligned along a horizontal line, but for some reason I drew the solar rays half and half beside the green line which is deceptive. I'll try to edit the pic when I can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In technical parlance, the Moon could be slightly above or slightly below its &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_node&quot; rel=&quot;nofollow noreferrer&quot;&gt;Node&lt;/a&gt;. If the Moon was &lt;strong&gt;exactly&lt;/strong&gt; at its node, then the umbra would absolutely be centered somewhere between 28.5 N/S (actually 23.4 N/S because the Sun never goes beyond that). But as you can see, the farther it gets from its node, the steeper the shadow impacts Earth and the farther towards a pole it gets. Too far from the node, and the shadow will not intercept Earth at all and we won't have an eclipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you draw a line from the center of the Moon to the center of the Earth, you can see that the sub-lunar point is definitely within 28.5 degrees N/S. But the shadow &lt;strong&gt;does not&lt;/strong&gt; necessarily follow that imaginary line. The shadow always follows the &lt;strong&gt;line from the Sun to the Moon.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could even imagine a very extreme case where the shadow falls on &quot;the other side&quot; of the pole. I think that may be the case &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_eclipse_of_August_12,_1942&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; since that eclipse looks really short.&lt;/p&gt;&#xA;" OwnerUserId="7449" LastEditorUserId="7449" LastEditDate="2017-08-23T19:22:31.857" LastActivityDate="2017-08-23T19:22:31.857" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22237" PostTypeId="2" ParentId="22233" CreationDate="2017-08-21T20:32:36.550" Score="26" Body="&lt;p&gt;During a sunset, the Sun is lower in the sky than during most of the day - &lt;em&gt;much&lt;/em&gt; lower. Therefore, light from the Sun travels through &lt;a href=&quot;https://earthscience.stackexchange.com/a/7663/1399&quot;&gt;about 120 miles of dense atmosphere&lt;/a&gt;, compared to the roughly 2 miles it travels through from straight up. Here's a rough sketch (not to scale) to demonstrate this. It is clear that $B&amp;gt;A$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wl1Z4.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wl1Z4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light scatters in the atmosphere; in fact, &lt;a href=&quot;https://en.wikipedia.org/wiki/Rayleigh_scattering&quot; rel=&quot;noreferrer&quot;&gt;one type of scattering is why the sky is blue&lt;/a&gt;. The longer travel distance means that &lt;a href=&quot;http://scienceline.ucsb.edu/getkey.php?key=3269&quot; rel=&quot;noreferrer&quot;&gt;there is much more scattering&lt;/a&gt; of ultraviolent light, which in turn means that the light you see is less intense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, you say. But doesn't the eclipse still block a lot of light? Well, unless there's 100% coverage - totality - there's still plenty of light coming from the uncovered part of the Sun, and that matters. &lt;a href=&quot;https://eclipse2017.nasa.gov/faq&quot; rel=&quot;noreferrer&quot;&gt;The uncovered part is as bright as it normally is&lt;/a&gt;, and looking at that part is just as dangerous with or without the eclipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's one more thing to consider, which is that people watching a sunset don't look at the Sun; they look at the clouds and sky &lt;em&gt;around&lt;/em&gt; the Sun. If you look directly at the Sun, your eyes will be damaged, no matter what's happening, an eclipse or a sunset.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2017-08-21T22:17:56.627" LastActivityDate="2017-08-21T22:17:56.627" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="22238" PostTypeId="1" AcceptedAnswerId="22242" CreationDate="2017-08-21T21:32:41.807" Score="5" ViewCount="291" Body="&lt;p&gt;I want to say first off, I am a complete beginner, starting out my interest in astronomy (so go easy!). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been able to see some nice sights on some nights, however, I had a strange experience the other night when viewing Saturn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had been viewing it the night before, using my highest magnification eyepiece and got some really nice views. It can be noted my telescope has a focal length of 1000mm and aperture of 100mm and I was using a 5mm eyepiece, which was really pushing the limits of the useable magnification, as far as I am aware.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So anyway, the next night had clear skies again, if anything maybe clearer. I was excited to see Saturn again, but this time once I had found it, I put in my 5mm eyepiece and just couldn't get it into the same focus I had the previous night. I googled it and somewhere said that although the skies can be good, the &lt;em&gt;seeing&lt;/em&gt; can be poor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is meant by this statement and how am I able to tell if there is going to be a night with poor &lt;em&gt;seeing&lt;/em&gt; in advance so I know which eyepieces I can use for that particular session? The app I use to check the skies only tells me if it is going to be clear so I don;t know what to be looking out for.&lt;/p&gt;&#xA;" OwnerUserId="17096" LastActivityDate="2017-08-22T15:36:22.093" Title="Bad seeing on a clear night?" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22239" PostTypeId="1" AcceptedAnswerId="22241" CreationDate="2017-08-21T23:01:40.443" Score="3" ViewCount="393" Body="&lt;p&gt;I've seen the solar noon formula described as &lt;code&gt;snoon = 720 – 4*longitude – eqtime&lt;/code&gt;. I'm curious if it is equally reliable within a relatively small margin of error (+-60 seconds) to simply get the midpoint between sunrise and sunset, given that data is already calculated and available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any occasion when solar noon wouldn't be equal time from sunrise and sunset, give or take the aforementioned margin of error?&lt;/p&gt;&#xA;" OwnerUserId="13380" LastActivityDate="2020-02-05T22:46:20.937" Title="Getting Solar Noon from Sunset and Sunrise" Tags="&lt;the-sun&gt;&lt;time&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22241" PostTypeId="2" ParentId="22239" CreationDate="2017-08-21T23:34:43.327" Score="3" Body="&lt;p&gt;The approach of using the midpoint of the sunrise and sunset times is reasonably accurate. I do not know what the accuracy is, but probably within 1 or 2 minutes. It would not be as accurate in polar regions where the sun is close to being visible all day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason it is not precise is that the sun is moving north (December through June) or south. This changes the time of sunset with respect to sunrise if the Sun were not moving north or south. Solar noon would be earlier or later, respectively, from the midpoint.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-08-21T23:34:43.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22242" PostTypeId="2" ParentId="22238" CreationDate="2017-08-22T00:17:07.253" Score="2" Body="&lt;p&gt;Seeing refers to turbulence in the atmosphere that distorts wavefronts passing through and results in the blurring of distant objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, there is no foolproof way of predicting what the seeing will be from night to night, though building up a database of local seeing and meteorological measurements might allow you to find some crude correlations using principal component analysis.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-08-22T00:17:07.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22244" PostTypeId="1" CreationDate="2017-08-22T02:57:58.400" Score="8" ViewCount="4249" Body="&lt;p&gt;Proxima Centauri is the closest star to the earth other than the sun yet it isn't visible to the naked eye, but stars that are further away are easily found in the night sky. Why is that?&lt;/p&gt;&#xA;" OwnerUserId="18139" LastActivityDate="2017-08-22T14:01:44.427" Title="Why are nearby stars like Proxima Centauri and Barnard's star not visible to the naked eye?" Tags="&lt;star-systems&gt;&lt;apparent-magnitude&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22245" PostTypeId="2" ParentId="22244" CreationDate="2017-08-22T04:30:05.023" Score="4" Body="&lt;p&gt;Stars vary a lot in both size and brightness. The nearby stars you mentioned are less bright than many stars that are much further away. We see the bright stars but not the less bright ones. Its like comparing the effectiveness of the old 2-battery flashlight to the 5-battery one. Both are flashlights but one is much brighter than the other.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastActivityDate="2017-08-22T04:30:05.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22248" PostTypeId="2" ParentId="22244" CreationDate="2017-08-22T05:16:48.630" Score="20" Body="&lt;p&gt;John's answer is correct.   For a few more details:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars brightness works out to roughly the 4th power of their relative mass.  This falls off a bit for very large stars, but for smaller stars it's in the ballpark.   (Red-giant phases not included.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Proxima_Centauri&quot; rel=&quot;noreferrer&quot;&gt;Proxima Centauri&lt;/a&gt; is about 12% the mass of our sun, and about 14% the sun's diameter.   At 12% its mass, its brightness, using the 4th power rule, is about 0.02% or 1/5,000th as bright.   That's total light.  Because it's smaller and colder, much of that light is in the infra-red range that we can't see.   In terms of visible light or visual magnitude, it's about 1/20,000th as bright as our sun.   That means, for it to be as bright as our sun from Earth in visible light, it would need to be about 1 million KM away.   If it was as far away as Pluto, some 40 AU on average, it wouldn't look very different than Venus at Venus' peak brightness, with a distinct red color.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Larger stars can have a brightness over 100,000 times brighter than our Sun.   So, because the mass of the star affects the brightness to about the 4th power, what it really comes down to is size matters.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distance matters too, but only to the square of the distance, so mass of the star matters more.   Here's an &lt;a href=&quot;https://www.space.com/21640-star-luminosity-and-magnitude.html&quot; rel=&quot;noreferrer&quot;&gt;article with a chart&lt;/a&gt; of the visually brightest stars in the sky.  Negative numbers and smaller numbers correspond to greater brightness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might go without saying, but virtually none of the stars that can be seen with the naked eye are red dwarf stars even though they're the most common type of star.  They're too dim for us to see at multiple light-years distance.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2017-08-22T14:01:44.427" LastActivityDate="2017-08-22T14:01:44.427" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22249" PostTypeId="2" ParentId="22233" CreationDate="2017-08-22T05:25:44.970" Score="6" Body="&lt;p&gt;There is also a second reason for this. The density of the atmosphere decreases as you go up in the atmosphere. At sunset, the sun's rays hit the atmosphere at an angle and refract through the atmosphere. The refraction is proportional to the incident angle, so it happens more at the horizon than during the day when the sun is high in the sky. The refraction also causes dispersion of light. This is part of the reason why the sun becomes red at sunset. The redness is also caused by the increasing scattering of blue light as the light passes through more atmosphere. The other colors, including ultraviolet radiation are systematically removed from the sun as it drops below the horizon. Because the light is refracted, the sun's image appears above the horizon even though the sun has actually already gone below the horizon. At this point, the damaging ultraviolet light can no longer reach your eyes. &lt;/p&gt;&#xA;" OwnerUserId="18143" LastActivityDate="2017-08-22T05:25:44.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22256" PostTypeId="1" CreationDate="2017-08-22T11:23:17.720" Score="1" ViewCount="199" Body="&lt;p&gt;Can one star be observed at different wave length show different redshift ? &lt;/p&gt;&#xA;" OwnerUserId="18159" LastActivityDate="2017-08-23T08:04:06.080" Title="Redshift observation at different wave length" Tags="&lt;redshift&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22257" PostTypeId="1" AcceptedAnswerId="22264" CreationDate="2017-08-22T12:20:18.593" Score="0" ViewCount="112" Body="&lt;p&gt;I am a middle school teacher from India(UTC+5:30). On August 21, 2017 morning at (17.385044, 78.486671) we observed an unusually early sunrise. It felt like its 7'O clock in the morning at 5:45 AM. And we had an unusually bright day. I'm sure it has to do something with the total solar eclipse but I don't exactly know what it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My students are persistently asking me the reason for this early sunrise but I don't have enough knowledge of astronomy to answer them. Can someone kindly answer why we observed this phenomenon?&lt;/p&gt;&#xA;" OwnerUserId="18112" LastActivityDate="2017-08-22T13:24:39.443" Title="On the day of total solar eclipse why was early sunrise observed in India?" Tags="&lt;the-sun&gt;&lt;solar-eclipse&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22259" PostTypeId="2" ParentId="22257" CreationDate="2017-08-22T12:35:36.130" Score="1" Body="&lt;p&gt;I doubt the eclipse had anything to do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take into consideration that a solar eclipse is just the shadow of the moon projected over the globe when the sun is in opposition to the earth taking the moon as the reference frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be a psychological effect given your student's expectations of observing some kind of astronomical event. It could also be that the meteorological conditions provided a particularly clear sky that day. Take also into account that this was a Monday. Probably on Mondays there is less pollution given that during the weekend there where less people commuting, which would mean clearer skies and therefore the feeling of brightness.&lt;/p&gt;&#xA;" OwnerUserId="18161" LastActivityDate="2017-08-22T12:35:36.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22260" PostTypeId="2" ParentId="22256" CreationDate="2017-08-22T12:43:06.660" Score="1" Body="&lt;p&gt;It's not impossible, but does require a bit if a special set-up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally, the light from a star is redshifted mainly due to its velocity toward/away from us (a small, but important, contribution comes from gravitational redshift, when the light climbs up the potential of the star). Because stars tend to rotate, light coming from the side of the star rotating &lt;em&gt;away&lt;/em&gt; from us is slightly more redshifted than the light coming from the side rotating &lt;em&gt;toward&lt;/em&gt; us. If a large &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_prominence&quot; rel=&quot;nofollow noreferrer&quot;&gt;prominence&lt;/a&gt; is present on one side, then because the prominence is hotter than the average surface temperature of the star and thus emits at shorter wavelengths, the combined spectrum will exhibit a larger redshift at short wavelengths.&#xA;Since the flux from the prominence alone is, after all, small compared to the total flux, the effect would be rather small, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more common mechanism that may cause a differential redshift is stellar outflows. If large amounts of gas are being expelled (e.g. in protostars or AGB stars), then atoms and molecules in the outflowing material will absorb some of the light from the star. These absorption lines are then blue shifted wrt. the &quot;systemic&quot; velocity (i.e. the velocity of the star wrt. us), again skewing the redshift as a function of wavelength. A spectral line may even be seen both in emission and absorption, known as a so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/P_Cygni#P_Cygni_profile&quot; rel=&quot;nofollow noreferrer&quot;&gt;P-Cygni profile&lt;/a&gt;. The figure below (from the Wiki article linked to) shows the same intrinsic wavelength (here H$\alpha$) both redshifted and blueshifted (by rotation and thermal motion), but the blueshifted light is absorbed by the outflowing atmosphere:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lCqUR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lCqUR.png&quot; alt=&quot;PCygni&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2017-08-22T12:43:06.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22261" PostTypeId="2" ParentId="22256" CreationDate="2017-08-22T12:50:02.377" Score="0" Body="&lt;p&gt;If I am not mistaken, when analyzing the spectrum of a star, in order to measure its redshift, we need to find several well known absorption lines (which occur at different wavelengths) shifted by the same amount: the redshift (actually we also need to characterize as well as possible the star too). It follows that you do not get different redshifts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when looking at a nearby object at different wavelengths, the sun for example, we may be looking at different parts of it. These may be moving toward or away from us. There, you would see a redshift for parts moving away from us, but this is just the doppler effect from a moving object, nothing to do with the universe expanding. &lt;/p&gt;&#xA;" OwnerUserId="18161" LastEditorUserId="18161" LastEditDate="2017-08-23T08:04:06.080" LastActivityDate="2017-08-23T08:04:06.080" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22262" PostTypeId="2" ParentId="12228" CreationDate="2017-08-22T13:13:23.567" Score="3" Body="&lt;p&gt;The most simple Lidov-Kozai model is that of a massless object ($m_1$ in your diagram) rotating a massive object ($m_0$), which is itself in an orbit with another massive object ($m_2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a hierarchical 3-body system ($m_2$ is assumed to always be far enough from $m_0$ and $m_1$). It is easier to look at as two 2-body orbits:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inner orbit - $m_0$ and $m_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Outer orbit - $m_2$ and $m_1$+$m_0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $m_1$ is massless, the outer orbit is not affected by it and is a simple Kepler orbit of 2 bodies ($m_0$ and $m_2$) with fixed parameters. Because of that, the outer orbit defines the coordinates system and lie in XY plane, with its angular momentum $\vec{L}_{out}=L_{out}\hat{z}$.&#xA;What we really have here is a motion of a test-particle ($m_1$ a.k.a the perturbed) around a massive object ($m_0$) in a binary system (with $m_2$). The inner orbit can be viewed as a Kepler orbit with a perturbation due to $m_2$ (a.k.a the perturber). Its parameters do change with time and are described by the Lidov-Kozai mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this model (which is probably what you're asking for):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question A&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The least massive object is the massless object $m_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question B&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What evolve are the parameters of the inner orbit ($m_0$ and $m_1$) - its eccentricity, inclination, angular momentum etc. How? in a periodic manner. The periodic change in the eccentricity literally means that the inner orbit becomes more circular, then more eccentric, then more circular again and again.&#xA;The eccentricity-inclination change is easier to see because of the following constant of motion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{1-e^2}\cos i = const. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It is not exactly $L_z$, but a scaled version of it $\frac{L_z}{\mu\sqrt{GMa_{in}}}$. $\mu,M$ being the reduced and total mass of the inner orbit respectively)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact of this being constant is not easy to see, but if given as a fact and the fact the $e$ is periodic, you can see that the inclination $i$ is periodic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question C&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When deriving this model one is &quot;averaging&quot; out (over one full period) the true-anomaly (the exact position of the mass inside the orbit) of $m_1$ around $m_0$ -&gt; meaning we refer to the inner orbit as an &quot;elliptic ring&quot;, and the same goes for the outer orbit. We also assume no energy exchange between both orbits (rings), so both inner/outer semi-major axes are also fixed (from the relation $E=-\frac{GM}{2a}$, where M is the total mass of the orbit)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally speaking (without any averaging) - this is still a chaotic 3-body problem and everything can happen - the inner orbit might just completely be destroyed by m1 being thrown out of the system, for an instance.&lt;/p&gt;&#xA;" OwnerUserId="18160" LastEditorUserId="18160" LastEditDate="2018-04-11T13:40:30.627" LastActivityDate="2018-04-11T13:40:30.627" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22263" PostTypeId="1" AcceptedAnswerId="22267" CreationDate="2017-08-22T13:14:30.673" Score="2" ViewCount="196" Body="&lt;p&gt;Recently I bought a couple of books on cosmology from a second-hand bookshop. One was published in 1993 and the other is from the 1970.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know, from a recent book, that the inflaton theory is probably newer than both books and that the Higgs Boson has been discovered, but that is all that I know has changed recently in the field of cosmology. Has anything else changed since then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which topics of the books should I understand are considered outdated?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The two books are &lt;em&gt;Modern Cosmology&lt;/em&gt; by Jagjit Singh (published 1961 and rereleased in 1970) and &lt;em&gt;Afterglow of Creation&lt;/em&gt; by Marcus Chown (published in 1993) &lt;/p&gt;&#xA;" OwnerUserId="1494" LastEditorUserId="1494" LastEditDate="2017-08-29T15:48:01.370" LastActivityDate="2017-08-29T15:48:01.370" Title="How wary should I be of old cosmology books?" Tags="&lt;cosmology&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22264" PostTypeId="2" ParentId="22257" CreationDate="2017-08-22T13:24:39.443" Score="2" Body="&lt;p&gt;If your sunrise had been earlier than expected, it would have been the same around the world, and would have been noticed very widely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a chart showing times of sunrise/set at your location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tquKy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tquKy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-08-22T13:24:39.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22265" PostTypeId="1" CreationDate="2017-08-22T14:23:31.217" Score="2" ViewCount="150" Body="&lt;p&gt;I was surprised to see a very measurable difference in temperature at my home during the recent solar eclipse.  The temperature was about 93 degrees fahrenheit but dropped to around 86 degrees.  It made me wonder how an unusual eclipse might affect a planetary body.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If there were a situation where someone had put an object into a stable synchronous orbit that would block a percentage of solar radiation, how big would it have to be for an appreciable effect?  An example would be from Futurama's &lt;em&gt;Crimes of the Hot&lt;/em&gt; episode Wernstom's giant parabolic mirror that reflected 40% of all sunlight.&lt;/li&gt;&#xA;&lt;li&gt;If an effect similar to a solar eclipse would be prolonged over a specific area, how would that effect the local atmosphere?  I would assume it would create a stable &quot;cold front&quot; in that area.  Is there any information on how that would affect other passing cold or warm fronts?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2199" LastActivityDate="2017-08-23T17:08:06.177" Title="Long term temperature effects of an eclipse" Tags="&lt;solar-eclipse&gt;&lt;atmospheric-effects&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22266" PostTypeId="2" ParentId="22238" CreationDate="2017-08-22T15:36:22.093" Score="1" Body="&lt;p&gt;To expand on Rob J's answer:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try for nights when the forecast is not only clear but no wind.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And almost invariably the best seeing is on the coldest winter nights, again in absence of wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other things that affect seeing include: altitude (top of a mountain is already above the densest part of the atmosphere), and making sure your telescope is set up long enough to be in thermal equilibrium with the local environment, so that there's no updrafts right next to the telescope.&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-08-22T15:36:22.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22267" PostTypeId="2" ParentId="22263" CreationDate="2017-08-22T15:57:35.893" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How wary should I be of old cosmology books?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's a sliding scale. Generally the older the book, the more the chance of something being wrong. But it depends what books you're talking about. Are they text books? Or are they popscience books? Some books, even newish books, have a habit of presenting speculation as established fact. There's been a lot of trouble about inflation and the multiverse recently. Peter Woit &lt;a href=&quot;http://www.math.columbia.edu/~woit/wordpress/?p=9134&quot; rel=&quot;nofollow noreferrer&quot;&gt;talks&lt;/a&gt; &lt;a href=&quot;http://www.math.columbia.edu/~woit/wordpress/?p=9289&quot; rel=&quot;nofollow noreferrer&quot;&gt;about it&lt;/a&gt; on his Not Even Wrong blog. He says &lt;em&gt;“the political campaign for the multiverse continues today with a piece by Amanda Gefter at Nautilus. It’s a full-throated salvo from the Linde-Guth side of the multiverse propaganda war they are now waging”&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I know, from a recent book, that the inflaton theory is probably newer than both books &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inflation started in circa 1980. Guth coined the phrase and generally gets the credit, but it arguably goes back to Erast Gliner in 1965 and his paper &lt;a href=&quot;https://drive.google.com/file/d/0BzWboAYQRwu2ZzU1WTFQYXliams/edit&quot; rel=&quot;nofollow noreferrer&quot;&gt;algebraic properties of the energy-momentum tensor and vacuum like states of matter&lt;/a&gt;. It didn’t get much attention.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and that the Higgs Boson has been discovered&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When inflation was first mooted Alan Guth talked about the Higgs field and GUTs, saying magnetic monopoles were &lt;em&gt;&quot;topologically stable knots in the Higgs field expectation value”&lt;/em&gt;. But that was old inflation, which was replaced by new inflation, which was replaced by eternal inflation. As far as I know the Higgs field is not thought to be linked to the inflaton field. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Has anything else changed since then?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, like pela said, before 1998 people thought the expanding universe was slowing down. There are other changes too. Some people used to say the universe was once the size of a grapefruit. Now since WMAP some people say the observable universe was once the size of a grapefruit, and the universe is infinite. For myself I cannot see how that fits with big bang cosmology, which I think is largely correct.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which topics of the books should I understand are considered outdated?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's hard to say. Dark matter, dark energy, cosmological constant, flatness of the universe, size of the universe, age of the universe. If you could name the books maybe I could look at the chapters.   &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2017-08-22T15:57:35.893" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22268" PostTypeId="1" AcceptedAnswerId="22270" CreationDate="2017-08-22T19:09:53.450" Score="3" ViewCount="154" Body="&lt;p&gt;From my understanding over the course of the seasons the moon and sun rise at varying locations. Though the day to day differences are no noticeable, to me at least. For instance, in the summer, the sun takes a longer path overhead as a result of the earths tilt towards the sun in the summer months -- in the winter, short paths. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the seemingly negligible difference in path lengths of the sun from day to day &lt;strong&gt;why then do solar eclipses only happen on a single day and not over the course of many days?&lt;/strong&gt; It seems that if the moon and sun paths during the day were to intersect they should do it multiple days in a row as their path lengths across the sky during the day change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume the answer to this lies particularly with the moons orbit around the earth and its shape/inclination. But I am far from my last astronomy courses and have lost all the jargon and knowledge needed to determine this.  &lt;/p&gt;&#xA;" OwnerUserId="18174" LastActivityDate="2017-08-23T07:50:57.563" Title="Why don't multiple eclipses happen over the course of a few days?" Tags="&lt;solar-eclipse&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22270" PostTypeId="2" ParentId="22268" CreationDate="2017-08-22T19:36:43.650" Score="6" Body="&lt;p&gt;The reason is because the moon moves much faster (in terms of angle per day) than the sun. In fact it moves on average 12.5 degrees per day, and all in the same direction.  Given that the sun is only about 0.5 degrees in diameter, it doesn't take long for the moon to overtake it, and once it's past  it doesn't go backwards.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-08-22T19:36:43.650" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22272" PostTypeId="2" ParentId="22228" CreationDate="2017-08-23T02:57:37.803" Score="5" Body="&lt;p&gt;As seen in this &lt;a href=&quot;https://astronomy.stackexchange.com/questions/22230/how-much-of-the-sun-needs-to-be-blocked-in-order-to-see-a-noticeable-drop-in-bri&quot;&gt;comment&lt;/a&gt;, normal daylight is around 100,000 lux, but 10,000 lux is still considered daylight. So, unless you're getting close to 99%, it will still be fairly light outside.&lt;/p&gt;&#xA;" OwnerUserId="18130" LastActivityDate="2017-08-23T02:57:37.803" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22274" PostTypeId="1" AcceptedAnswerId="22275" CreationDate="2017-08-23T05:18:52.860" Score="1" ViewCount="75" Body="&lt;p&gt;An annular eclipse is similar to total eclipse except our Moon &quot;apparently&quot; appears smaller than the Sun, my question is since in both cases there will be a glowing ring forming... so how to tell them apart?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastActivityDate="2017-08-23T06:40:59.593" Title="How to tell between annular and total eclipse (directly)?" Tags="&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22275" PostTypeId="2" ParentId="22274" CreationDate="2017-08-23T06:40:59.593" Score="3" Body="&lt;p&gt;There is a massive difference in the brightness of the &quot;ring&quot;. During a total eclipse, the photosphere of the sun is completely hidden and the corona comes into sight. The corona is much much less bright than the photosphere of the sun, so during a total eclipse, the sky becomes dark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During an annular eclipse the photosphere is never completely obscured. The bright 'surface' of the sun is still visible, so the sky never becomes completely dark, the sun remains dangerously bright, and the corona is not visible. An annular eclipse is like the partial phase of a total eclipse, because part of the sun is still visible.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-08-23T06:40:59.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22278" PostTypeId="2" ParentId="22268" CreationDate="2017-08-23T07:50:57.563" Score="4" Body="&lt;p&gt;Eclipses last for a couple to maybe a few hours, from a certain point of view.   The time that an eclipse lasts is most commonly discussed relative to a specific location.   For eclipse watchers, you park your car, get your glasses, and watch the show that lasts for a couple minutes for the total eclipse.  (Longer for the partial).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this map&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/llss1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/llss1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The eclipse hit Oregon at 10:17 AM Pacific time (1:17 PM Eastern time).   It passed over the east coast of South Carolina at 2:46 PM Eastern time, or 1 hour and 29 minutes later.   If you count the time the eclipse passed over parts of the Pacific and Atlantic oceans, it probably lasted over 2 hours and if you count partial eclipses, a fair bit longer than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty is, to see the eclipse for 2 hours, you'd need to be in a fast plane, following the path.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun is about 1/2 of 1 degree of arc in the sky, or 1/720th the full circle.  A Lunar &lt;a href=&quot;https://community.dur.ac.uk/john.lucey/users/lunar_sid_syn.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;synodic orbit&lt;/a&gt; takes 29.5 days, and 1/720th of 29.5 days is about 1 hour and that's partial to partial.  The total eclipse is a small fraction of that and lasts a few minutes max, depending on how close the Moon is to the Earth and it's relative size to the sun.   So for a total eclipse, it only takes a couple minutes for the Moon to pass through a complete one.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why eclipses last longer than that the 1 hour it takes the Moon to move 0.5 degrees relative to the sun is because the Earth is quite large.   At nearly 8,000 miles in diameter, it takes the a bit over 3 1/2 hours for the Moon's shadow or umbra, moving on average about 2,230 mph, to pass over nearly 8,000 miles and 4 1/2 hours time if you include the partial eclipse.   That's only if it hits near the equator.  If the eclipse only grazes the Earth near a pole it lasts much less time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 1 hour and 27 minutes is in part cause it takes the Moon about 1 hour to pass over the United States around Noon, and because the Earth rotates with the Moon, in the same direction, in a sense, the US is trying to keep up with the Moon's shadow, so that extends the time to about 1.5 hours to cross the US.   If the Earth rotated faster, say, roughly 8-10 hour days depending on your latitude, then it might be possible to observe hour long total eclipses because the Earth's rotation could roughly match the Moon's relative movement to the sun under optimal conditions, but a 24 hour rotation is too slow for that to happen.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the problem with timing these things, we're on a rotating sphere observing an orbiting object.  Eclipses vary in duration for those reasons.   For the longest eclipse to the stationary observer, look for an eclipse near the equator and near mid-day.  A larger, close moon might provide a total eclipse a little longer, but the closer the Moon is the faster it orbits so for a larger-closer moon, the partial eclipse period would be somewhat shorter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that wasn't unnecessarily complicated.   Dr. Chuck's answer is correct.  It doesn't take long for the Moon to cross the Sun's position in the sky.  About 1 hour edge to edge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://scienceblogs.com/startswithabang/2012/06/01/how-long-does-a-solar-eclipse-last/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article&lt;/a&gt; goes into even more detail on these points with some videos if interested.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-08-23T07:50:57.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22281" PostTypeId="2" ParentId="22234" CreationDate="2017-08-23T11:11:57.737" Score="1" Body="&lt;p&gt;Assuming the intensity of the sunlight is proportional to the area of the Sun which is showing, we can roughly calculate the intensity of sunlight, &lt;span class=&quot;math-container&quot;&gt;$I(d)$&lt;/span&gt; as:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$I(d) = 1 - \frac{\mathrm{Area}_{\mathrm{overlap}}}{\mathrm{Area}_{\mathrm{Sun}}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$\mathrm{Area}_{\mathrm{overlap}}$&lt;/span&gt; is the overlap between the Moon and the Sun (the green area).&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RIbRZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RIbRZ.jpg&quot; alt=&quot;&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://jwilson.coe.uga.edu/EMAT6680Su12/Carreras/EMAT6690/Essay2/circles%5B1%5D.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;uga.edu&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;Some calculation finds the overlap to be:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{Area}_{\mathrm{overlap}} = 2r^2\cos^{-1}\left(\frac{d}{2r}\right) - \frac{d}{2}\sqrt{4r^2 - d^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; is the distance between the centres of the Moon and the Sun. Similarly, the area of the Sun is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{Area}_{\mathrm{Sun}} = \pi r^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Putting this together, you get the graph:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4JhJV.jpg&quot; alt=&quot;Intensity changing with time&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Where the Y axis is brightness (ranging from 100% to 0%) and the X axis is &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; (ranging from &lt;span class=&quot;math-container&quot;&gt;$2r$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;p&gt;I hope this answers your question: the brightness does not change proportionally to coverage/time.&lt;/p&gt;&#xA;&lt;h3&gt;Note:&lt;/h3&gt;&#xA;&lt;p&gt;This assumes that &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; increases linearly with time (the moon doesn't speed up or slow down as it moves across the Sun).&lt;/p&gt;&#xA;" OwnerUserId="1494" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-07-13T12:48:46.810" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22282" PostTypeId="1" CreationDate="2017-08-23T11:18:24.977" Score="5" ViewCount="243" Body="&lt;p&gt;Is it possible that we mistake some other wavelength emission to be 21 cm because of redshift?&lt;/p&gt;&#xA;" OwnerUserId="11434" LastActivityDate="2017-08-23T12:51:20.380" Title="How to know that the 21cm Hydrogen line is the actual emission and not any other redshifted line?" Tags="&lt;astrophysics&gt;&lt;fundamental-astronomy&gt;&lt;radio-astronomy&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22283" PostTypeId="2" ParentId="22282" CreationDate="2017-08-23T12:51:20.380" Score="9" Body="&lt;p&gt;No, not really.  The first thing is that we know that  ${H}$ is far more abundant than other elements or simple molecules in the universe.  The next thing is that the 21 cm line comes from a relatively unusual hyperfine splitting, and there just aren't any other sources near that wavelength and in intensity levels that can easily be detected.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most cases, astronomers have other methods of estimating objects' speeds, so any measurement of 21cm apparent redshift that was inconsistent with those estimates would be an indication that the wrong line was being observed.&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-08-23T12:51:20.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22284" PostTypeId="1" CreationDate="2017-08-23T17:02:25.897" Score="1" ViewCount="152" Body="&lt;p&gt;So I have a CMD diagram and on the left, I have apparent magnitude and at the bottom, I have the color index. I would just estimate the magnitude of stars leaving the main sequence and continue from there, however, I was wondering if there's a way to do it more accurately?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would take stars between certain magnitudes and calculate the average magnitude, but again, I'm estimating from where to where the MSTO point is... &lt;/p&gt;&#xA;" OwnerUserId="18079" LastActivityDate="2017-08-23T19:33:36.523" Title="How do you determine the apparent magnitude of the stars leaving main sequence with CMD diagrams?" Tags="&lt;astrophysics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22285" PostTypeId="1" AcceptedAnswerId="22289" CreationDate="2017-08-23T17:03:06.537" Score="0" ViewCount="1202" Body="&lt;p&gt;On Equator, why do objects cast the shortest shadows at 11:32 and not 12:00? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, this happens in Costa Rica.&lt;/p&gt;&#xA;" OwnerUserId="2541" LastEditorUserId="823" LastEditDate="2017-08-25T10:51:54.673" LastActivityDate="2017-08-25T10:51:54.673" Title="On the Equator, why are the shortest shadows seen at 11:32, and not at Noon?" Tags="&lt;the-sun&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22286" PostTypeId="2" ParentId="22265" CreationDate="2017-08-23T17:08:06.177" Score="1" Body="&lt;p&gt;It was suggested that we could cool down Venus by putting a powder cloud in the Lagrange point between it and the Sun. This will cool Venus in a few months to the point where its atmosphere will start liquify, creating a global CO2 ocean. The surface of an ocean is much easier to colonize than creating flying cities.&lt;/p&gt;&#xA;" OwnerUserId="2541" LastActivityDate="2017-08-23T17:08:06.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22287" PostTypeId="2" ParentId="22285" CreationDate="2017-08-23T19:15:51.783" Score="6" Body="&lt;p&gt;The time the sun is actually directly overhead depends on where you are in your time zone. In the eastern part of the time zone, the sun will be overhead sooner, and in the western part it will be overhead later. Where you are, it happens to be directly overhead at 11:32. &lt;/p&gt;&#xA;" OwnerUserId="7752" LastActivityDate="2017-08-23T19:15:51.783" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22288" PostTypeId="2" ParentId="22284" CreationDate="2017-08-23T19:33:36.523" Score="1" Body="&lt;p&gt;There is no clearly-defined &quot;turn-off&quot; magnitude. Stars move from the zero age main sequence (roughly speaking, where core hydrogen burning begins to dominate the luminosity) to the terminal age main sequence (where the core hydrogen runs out) relatively slowly, increasing in luminosity as they do so. Their evolution then accelerates redward in the colour magnitude diagram. The turn-off is roughly the most luminous star for which this acceleration in evolution is yet to occur, but this is confused/blurred by observational uncertainties, binarity, differential reddening and spreads in age and rotation rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice what you do if you really want to do a proper job of estimating the age, with uncertainties, is you do a population synthesis (including binaries and observational uncertainties) for clusters of stars with a variety of ages. You then compare these with the observations using maximum likelihood or Bayesian techniques (e.g. &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0609764&quot; rel=&quot;nofollow noreferrer&quot;&gt;Naylor &amp;amp; Jeffries 2006&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-08-23T19:33:36.523" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22289" PostTypeId="2" ParentId="22285" CreationDate="2017-08-23T20:29:57.813" Score="9" Body="&lt;p&gt;If you look at the position of the sun at the same time each day, over the course of a year, you will find it traces a &quot;figure of 8&quot; pattern in the sky called an &lt;em&gt;&lt;a href=&quot;http://solar-center.stanford.edu/art/analemma.html&quot; rel=&quot;noreferrer&quot;&gt;analemma&lt;/a&gt;&lt;/em&gt;. This may seem an extremely obscure thing to have a name, but it is important for sundial makers, and comes from sundial cant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this means that the time that the sun will cross the meridian will vary from day to day. At the equator, the sun will travel directly overhead only on the equinoxes. In June, the sun will be in the North at midday, In December it will be in the South.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each country is in a time-zone. It is a matter of convenience that the time in the country is all the same, rather than each location using their local solar time. At your location in Costa Rica you have found that solar noon occurs at 11:32. The Costa Rican's could choose to put their clocks forward an hour, but if they did, then local noon would be at 12:32.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect of time-zones is most extreme in the far west of China, where Beijing time is in use. There, local noon occurs at about 15:00. China chooses to have a single time zone, in contrast to other large countries like the USA or Russia.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-08-23T20:29:57.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22290" PostTypeId="2" ParentId="22231" CreationDate="2017-08-23T21:01:36.667" Score="7" Body="&lt;p&gt;If the sun is visible as a disc, then it is certainly can cause damage. As you move further from the sun the disc appears smaller, but the surface brightness remains the same. The eye can resolve a disc of about 1 arcminute, and the sun is about 30 arcminutes when viewed from Earth. So you would need to be at least 30 times further from the Sun than the Earth is, which puts you almost exactly at Neptune.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From that distance, the sun would be a magnitude -19 star. If you stared at it, it could probably still damage your eyes, however, your eyes are constantly jumping (saccades) which might move the image of the sun around the retina enough to prevent damage. There is, of course, no atmosphere to protect you from UV, so let's hope that's not a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond Neptune the sun gets dimmer, as it can't appear to get any smaller, and there is probably little risk by the time you reach the main part of the Kuiper belt. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-08-23T21:01:36.667" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22291" PostTypeId="1" CreationDate="2017-08-23T22:28:49.230" Score="3" ViewCount="764" Body="&lt;p&gt;What is currently the closest &lt;em&gt;known&lt;/em&gt; heavenly body, brown dwarf, rogue planet or other to our solar system? At one time it was Alpha Centauri. &lt;/p&gt;&#xA;" OwnerUserId="18118" LastEditorUserId="7982" LastEditDate="2017-08-26T06:21:53.010" LastActivityDate="2017-08-26T06:21:53.010" Title="What is the closest heavenly body to our solar system" Tags="&lt;solar-system&gt;&lt;rogue-planet&gt;&lt;brown-dwarf&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22292" PostTypeId="2" ParentId="22291" CreationDate="2017-08-23T23:24:39.070" Score="8" Body="&lt;p&gt;The closest body we currently know to exist is &lt;a href=&quot;https://en.wikipedia.org/wiki/Proxima_Centauri&quot; rel=&quot;noreferrer&quot;&gt;Proxima Centauri&lt;/a&gt; which is part of the Alpha Centauri system at about 4.25 light years away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be rogue planets that are closer but we haven't been able to detect any. Objects down to brown dwarf size that are closer than Alpha Centauri have not been found by extensive surveys. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a list of close objects to the solar system: &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_nearest_stars_and_brown_dwarfs&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/List_of_nearest_stars_and_brown_dwarfs&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="17582" LastEditorUserId="2365" LastEditDate="2017-08-24T06:16:21.553" LastActivityDate="2017-08-24T06:16:21.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22294" PostTypeId="2" ParentId="22225" CreationDate="2017-08-23T23:43:47.180" Score="2" Body="&lt;p&gt;The problem you have with many newtonian reflectors - especially smaller ones - is that to use them for prime focus photography (where you're effectively using the telescope like a big telephoto lens) , you need to set things up so that the camera sensor ends up at the focal plane of the telescope. And in a DSLR, the sensor is some way inside the camera body (behind the shutter, which is behind the mirror.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a newtonian scope, the focal plane is usually somewhere inside the focuser tube, and many newts don't have enough focuser inwards travel to let you get the focal plane the 40mm or so beyond the end of the focuser that you need to get it to reach a DSLR sensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(With other types of scope that use a separate star diagonal, things are usually easier - you can leave the star diagonal off, which has the same effect as a big chunk of inward focus).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a newtonian, if you can't reach prime focus, there are three basic options - either take photos through the eyepiece (afocal photography) as you already tried,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or move the main mirror further up the tube (which moves the focal plane further out, giving you more in focus). It also means the light cone is fatter when it reaches the secondary mirror, so you may lose some light as the outer part of the light cone misses the secondary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or you can try using a barlow lens - that replaces the last bit of the light cone with a longer, narrower angle version (which increases the magnification and moves the focal plane further out). That MAY or MAY NOT let you reach infinity focus; it's not guarenteed, but may help. You can get barlows that have a T-Thread on the output end, so you can attach a T ring directly to the barlow (yours doesn't look like one of this type) and then hook the camera up that way. (as an added benefit, since you end up with a closed tube setup (the barlow lens closes off the access to the camera, it also keeps spiders and dust out of the camera :) ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a look at the bottom bit of the barlow (with the lens) you may find that you can unscrew it. You might be able to make a short top section with a T thread for the T ring at the back, and a suitable thread for the lens at the front; you'd probably need to make it the same diameter as the eyepiece barrel (to fit in the focuser) with a fatter bit at the back for the T thread and to stop it sliding all the way into the focuser.&#xA;I've used a commercial 1.25&quot; barrel T threaded barlow (somewhat shorter than yours) for lunar and solar (WITH A SUITABLE FRONT OF SCOPE FILTER - DON'T TRY LOOKING AT THE SUN WITHOUT ONE) imaging with an APS sensor camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You may need to check your scope - older ones often used a .965&quot; barrel rather than the current 1.25&quot; (or 2&quot;) barrel size used by most modern astro scopes.)&lt;/p&gt;&#xA;" OwnerUserId="17016" LastActivityDate="2017-08-23T23:43:47.180" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="22295" PostTypeId="2" ParentId="22228" CreationDate="2017-08-23T23:49:50.867" Score="5" Body="&lt;p&gt;The real reason is that the human eye is not a linear sensor. A 10x decrease in illumination does not produce a 10x reduction in your perception of light, but more like a 2x reduction at most.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, during the 90% phase of a partial eclipse, the level of illumination is massively reduced. You could check it with a device that could measure ambient light, and it would show you the huge decrease. But your eyes tell a different story because they're not built to operate on a linear scale.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2017-08-23T23:49:50.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22296" PostTypeId="2" ParentId="22230" CreationDate="2017-08-24T00:49:48.673" Score="1" Body="&lt;p&gt;Based on the CNN's reporting it appeared to get noticeably darker at near 95% coverage. That's just from watching the onset of darkness for locations experiencing totality. &lt;/p&gt;&#xA;" OwnerUserId="16926" LastActivityDate="2017-08-24T00:49:48.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22297" PostTypeId="1" AcceptedAnswerId="22300" CreationDate="2017-08-24T04:58:54.533" Score="1" ViewCount="185" Body="&lt;p&gt;I'm running the program found at &lt;a href=&quot;https://www.projectpluto.com/source.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.projectpluto.com/source.htm&lt;/a&gt; in order to find the Right ascensions and declinations of celestial bodies at various points in time. An odd thing I've noticed is that the declination is reported in hours, minutes, seconds as opposed to degrees, minutes, seconds and I'm not sure how to use that. Especially considering the hours seem to lie on [-24,24] which doesn't make sense for the [-90,90] range for declination measured in degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sun RA=10:11:50.1 Dec= 11:08:50.8&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/I7IFH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I7IFH.png&quot; alt=&quot;image showing right ascensions and declinations of various planets&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is sampled from Los Angeles, California at 8 PM PDT (3:00 UTC) so I would expect when I ran these numbers to calculate elevation and azimuth that the elevation would be a good bit under the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried scaling the hours by 90/24 so that it would be back in the [-90,90] range but when I tried running those numbers through &lt;a href=&quot;http://www.convertalot.com/celestial_horizon_co-ordinates_calculator.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this tool&lt;/a&gt;, it ended up reporting that the sun would actually be just above the horizon which is definitely wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9iudr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9iudr.png&quot; alt=&quot;wrong calculation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So with that said, how should I be interpreting this data to get the result I expect? I'm pretty new to this, so I may be lacking something incredibly fundamental.&lt;/p&gt;&#xA;" OwnerUserId="18198" LastEditorUserId="18198" LastEditDate="2017-08-24T05:05:38.970" LastActivityDate="2017-08-24T06:09:44.280" Title="Measuring declination in negative hours of arc?" Tags="&lt;positional-astronomy&gt;&lt;declination&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22300" PostTypeId="2" ParentId="22297" CreationDate="2017-08-24T06:09:44.280" Score="2" Body="&lt;p&gt;It's not hours, minutes, and seconds, its degrees, arcminutes and arcseconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 360 degrees in a circle, 60 arcminutes in a degree and 60 arcseconds in an arcminute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the planets lie close to the ecliptic, and the ecliptic tilted at 23 degrees relative to the equator, the declension of the planets rarely exceeds +-23 degrees, giving an appearance that their dec. is like a time.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-08-24T06:09:44.280" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22302" PostTypeId="1" AcceptedAnswerId="22310" CreationDate="2017-08-24T08:55:01.147" Score="2" ViewCount="831" Body="&lt;p&gt;Both the laser collimator and cheshire eyepiece are tools used in astronomical telescope collimation. What are the main advantages/disadvantages of each? Which one might be best suited to start with for telescope newbies?&lt;/p&gt;&#xA;" OwnerUserId="18205" LastEditorUserId="7982" LastEditDate="2017-08-26T06:08:04.793" LastActivityDate="2017-08-28T18:48:48.493" Title="Laser Colimator vs Cheshire eyepiece" Tags="&lt;telescope&gt;&lt;collimation&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ClosedDate="2017-08-26T11:55:43.597" ContentLicense="CC BY-SA 3.0" />
  <row Id="22304" PostTypeId="1" AcceptedAnswerId="22308" CreationDate="2017-08-24T12:14:57.983" Score="2" ViewCount="1142" Body="&lt;p&gt;I found two sources sort of answering this question, but conclusions I made of them contradict each other. &#xA;&lt;a href=&quot;https://www.reddit.com/r/askscience/comments/29jrvq/what_is_the_smallest_possible_distance_from_pluto/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.reddit.com/r/askscience/comments/29jrvq/what_is_the_smallest_possible_distance_from_pluto/&lt;/a&gt; - here is stated that Pluto and Neptune never come closer than two billions kilometres from each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet from picture in here (I am aware it might be not to scale) &lt;a href=&quot;https://astronomy.stackexchange.com/questions/11782/can-pluto-and-neptune-collide-anytime-in-future&quot;&gt;Can Pluto and Neptune collide anytime in future?&lt;/a&gt; it seems that Pluto and Neptune got a lot closer to each other shortly after 1999 than a half of Neptune orbit radius, judging by their speeds and positions in 1999.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how is it? How close can they get to each other and when is it going to happen next time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tried to use wikipedia, but while article about Pluto contains information at least about when it was in Aphelion, article about Neptune does not any such information (or I could not find it).&lt;/p&gt;&#xA;" OwnerUserId="18206" LastActivityDate="2017-08-26T09:36:06.957" Title="When and where are Pluto and Neptune closest to each other?" Tags="&lt;solar-system&gt;&lt;pluto&gt;&lt;neptune&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22305" PostTypeId="1" CreationDate="2017-08-24T13:09:14.527" Score="1" ViewCount="313" Body="&lt;p&gt;An exoplanet named after its host star receives the suffix letter &quot;b&quot; if it was the first planet discovered to orbit that star, or &quot;c&quot; if it was the second, or &quot;d&quot; if it was the third, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why this naming system starts from &quot;b&quot; and not from &quot;a&quot;? Is the letter &quot;a&quot; reserved for another object in the planetary system? If yes, for what is &quot;a&quot; reserved?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, here is a list of the major objects in the Gliese 667 triple-star system:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gliese 667A (a star)&lt;/li&gt;&#xA;&lt;li&gt;Gliese 667B (a star)&lt;/li&gt;&#xA;&lt;li&gt;Gliese 667C (another star)&lt;/li&gt;&#xA;&lt;li&gt;Gliese 667Cb (a planet orbiting star C)&lt;/li&gt;&#xA;&lt;li&gt;Gliese 667Cc (another planet orbiting star C)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="18209" LastEditorUserId="18209" LastEditDate="2017-08-24T13:31:34.203" LastActivityDate="2017-08-24T13:31:34.203" Title="Why does exoplanets' naming start from &quot;b&quot; and not from &quot;a&quot;?" Tags="&lt;exoplanet&gt;&lt;naming&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2017-08-26T06:31:41.740" ContentLicense="CC BY-SA 3.0" />
  <row Id="22306" PostTypeId="2" ParentId="22305" CreationDate="2017-08-24T13:22:29.897" Score="0" Body="&lt;p&gt;A  quick &lt;a href=&quot;https://en.wikipedia.org/wiki/Exoplanet_naming_convention&quot; rel=&quot;nofollow noreferrer&quot;&gt;lmwtfy&lt;/a&gt; reveals the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the WMC naming system, the brightest member of a star system&#xA;  receives the letter &quot;A&quot;. Distinct components not contained within &quot;A&quot;&#xA;  are labeled &quot;B&quot;, &quot;C&quot;, etc. Subcomponents are designated by one or more&#xA;  suffixes with the primary label, starting with lowercase letters for&#xA;  the second hierarchical level and then numbers for the third.[2] For&#xA;  example, if there is a triple star system in which two stars orbit&#xA;  each other closely with a third star in a more distant orbit, the two&#xA;  closely orbiting stars would be named Aa and Ab, whereas the distant&#xA;  star would be named B. For historical reasons, this standard is not&#xA;  always followed: for example Alpha Centauri A, B and C are not&#xA;  labelled Alpha Centauri Aa, Ab and B.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-08-24T13:22:29.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22307" PostTypeId="1" AcceptedAnswerId="22314" CreationDate="2017-08-24T13:38:04.260" Score="9" ViewCount="127" Body="&lt;p&gt;A long time ago I read a very interesting way of studying the sun's internal structure via sound waves. This new field of study, probability already known here, is called helioseismology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much can we learn from gas giants by &quot;listening&quot; to sound waves?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somewhat tangent to the question: Does this extend to all other kind of giants such as hot Jupiter, ice giants, etc...?&lt;/p&gt;&#xA;" OwnerUserId="11062" LastEditorDisplayName="user1569" LastEditDate="2017-08-24T16:25:15.580" LastActivityDate="2017-08-25T00:28:23.430" Title="Gas giants and seismology" Tags="&lt;astrophysics&gt;&lt;gas-giants&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22308" PostTypeId="2" ParentId="22304" CreationDate="2017-08-24T15:38:22.080" Score="3" Body="&lt;p&gt;Per page 13 &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1965AJ.....70...10C&amp;amp;defaultprint=YES&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1965AJ.....70...10C&amp;amp;defaultprint=YES&amp;amp;filetype=.pdf&lt;/a&gt; (it's Harvard, so relatively reliable):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the closest approach of Pluto to Neptune is locked in near aphelion&#xA;  and the minimum distance between the bodies is approximately 18 a.u.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which works out to roughly 1.7 billion miles or 2.7 billion kilometers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image from the other answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lxgCE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lxgCE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is deceptive if not outright incorrect (tineye shows 15 results, some of which seem reliable, but I couldn't find the originating source).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how much Pluto moves from 1979 to 1989, compared to how much Neptune moves in the same period. It seems that Pluto moved a lot more, but, since Pluto was about the same distance away as Neptune, it should be moving at nearly the same speed, not faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Stellarium shows, Neptune and Pluto are about 40 degrees apart in 1999 (as viewed from the Sun) significantly more than the diagram in the other question indicates (this might be an artifact of the planes being tilted 17 degrees and an unusual perspective view):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kWCAd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kWCAd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Pluto is invisible but at the center of the cross-- the bright object nearby is 13 Oph, not Pluto).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A perhaps more interesting question: what is the closest approach of the two orbits, even if the planets in question aren't occupying that given point in the orbit.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2017-08-25T02:20:23.577" LastActivityDate="2017-08-25T02:20:23.577" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22309" PostTypeId="2" ParentId="22228" CreationDate="2017-08-24T16:04:43.133" Score="3" Body="&lt;p&gt;To expand on Florin Andrei's answer, it's not just that the human eye is not a linear sensor, although that is definitely an important part of the reason. The eye itself adjusts to the ambient light level by changing pupil size, which will mask overall changes in brightness. In addition, the eye, like most senses, is primarily a &lt;em&gt;relative&lt;/em&gt; sensor. Colors are judged relative to neighboring colors, and brightness is judged relative to neighboring brightness. &quot;Neighboring&quot; applies to time as well as space: a sudden change in brightness from 100,000 lux to 10,000 lux would be quite noticeable, but when it happens over an hour it is imperceptible. The final change to totality is quicker, and also substantial enough that you would be able to detect it even if it were slow. When it's dark enough, the color mix is different, shadows disappear, and it's actually harder to see in an absolute way. &lt;/p&gt;&#xA;" OwnerUserId="7752" LastEditorUserId="7752" LastEditDate="2017-08-24T17:53:22.607" LastActivityDate="2017-08-24T17:53:22.607" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22310" PostTypeId="2" ParentId="22302" CreationDate="2017-08-24T20:17:36.783" Score="3" Body="&lt;p&gt;Why not both? (if you can afford them)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both will do the job, but each has its strengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cheshire eyepiece&lt;/strong&gt; (or, more properly, Cheshire/sight tube combo, which is the popular combination nowadays):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easier to use when the scope is grossly out of alignment. Like, when you put it together for the first time, or after it dropped on the ground and rolled downhill for a minute. But can you do that with a laser? Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easier to align the secondary mirror, and it tends to produce dead-on results for this purpose. But can you do that with a laser? Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not immune from precision issues. Most parts of the Cheshire are usually well aligned (because they are machined on a lathe), but the crosshairs on the sight tube may or may not be perfectly centered. But can lasers also suffer from alignment problems? Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recommended Cheshire / sight tubes: the ones made by Catseye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.catseyecollimation.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.catseyecollimation.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They make much more than just Cheshire eyepieces, and all their tools are very high quality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Laser collimator&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's faster and arguably easier to use for normal daily collimation before you begin your normal observing sessions. But could you use a Cheshire instead? Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A laser is only as good as its own alignment precision. A misaligned laser will not collimate your scope properly. But do Cheshire sight tube combos suffer from precision issues with their crosshairs? Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recommended laser collimators: the laser collimator and the tuBlug combo made by Howie Glatter. Unfortunately Howie has passed away recently, so either wait until someone picks up where he left off, or find a used laser/tuBlug combo somewhere on eBay or classified ads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.collimator.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.collimator.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Howie's lasers are essentially guaranteed to have zero collimation errors from factory. From anecdotes told by users, these devices tend to keep their perfect collimation even after being dropped to the ground repeatedly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tuBlug is not mandatory, but it makes primary mirror alignment much easier.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that you also need a precision center mark on the primary, for these methods to work well. Hopefully it's been placed there precisely from factory; otherwise there are ways to re-center it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;How to verify if the laser is centered:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Place it in the focuser. Do not tighten the screws too much, allow the laser to rotate in the focuser a little bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turn the laser on. Watch where the spot hits the primary. Now rotate the laser in the focuser. If the spot describes a circle as you rotate the laser, then the laser is misaligned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different method: Make a V-shaped trough from 2 pieces of wood or aluminum. Placed the through horizontally on a sturdy table. Put the laser on the trough. Turn it on, projecting the spot on a distant flat object, like a wall or fence across the street.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now rotate the laser in the trough and watch the spot. Does it stay put? Then the laser is fine. Does it move in a circle? Then the laser is misaligned - find the adjustments screws (if it has them) and adjust it until it's perfectly square.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2017-08-28T18:48:48.493" LastActivityDate="2017-08-28T18:48:48.493" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22311" PostTypeId="1" AcceptedAnswerId="22312" CreationDate="2017-08-24T20:24:58.173" Score="1" ViewCount="217" Body="&lt;p&gt;I have heard that when something passes the event-horizon of a black hole nothing can stop it from getting pulled in. Is that true for objects partially past the event-horizon? Would it be possible to attach a camera to a stick or rope, put it inside the event-horizon, take a picture, and pull it back up and look at the picture?&lt;/p&gt;&#xA;" OwnerUserId="18215" LastActivityDate="2017-08-24T20:58:23.563" Title="Is it possible to throw a camera attached to a rope within the event-horizon of a black hole and then pull it back in?" Tags="&lt;black-hole&gt;&lt;event-horizon&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22312" PostTypeId="2" ParentId="22311" CreationDate="2017-08-24T20:58:23.563" Score="7" Body="&lt;p&gt;According to the observer holding the rope, the camera would never make it past the event horizon. It would appear to slow down to a standstill just above the horizon (though would effectively disappear because of redshift). At some point the rope would snap, but the camera would still not appear to cross the horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to an observer with the camera, they would cross the horizon. But if they were lowered slowly, the rope would snap before they got there, since infinite force is needed to &quot;hover&quot; just above the event horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They would then find themselves inexorably falling inwards in the same way that you and I are inexorably drawn into the future and can't revisit the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That different observers in different frames have different versions of events is not unusual in relativity.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-08-24T20:58:23.563" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22314" PostTypeId="2" ParentId="22307" CreationDate="2017-08-25T00:28:23.430" Score="5" Body="&lt;p&gt;I'm certainly not an expert, but from what I understand, the type of observation done in helioseismology might not work with gas giants, hot jupiters, ice giants, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun is dense enough and it has enough gravity that, while its a plasma, it's relatively solid most of the way through.   Unlike Red Dwarf stars which are lighter and &lt;a href=&quot;https://en.wikipedia.org/wiki/Red_dwarf&quot; rel=&quot;noreferrer&quot;&gt;undergo convection&lt;/a&gt; (0.35 solar masses and under are modeled to be fully convective), the sun is layered and doesn't undergo convection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While oversimplified, the sun behaves like a drum in regards to it's internal soundwaves.  The sun is also very very loud, inside and on the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you played a drum in a vacuum, you couldn't hear it, but you could observe the motion of the drum's surface and work out what it would sound like by visual observation only.  Same with guitar strings.  The vibration is visible in wavelength and amplitude.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exaggerated, the pattern on a drum would look something like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Cun3h.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cun3h.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://plus.maths.org/content/what-makes-object-musical&quot; rel=&quot;noreferrer&quot;&gt;Source of picture&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's basically how the Sun's soundwaves are studied.   It's often compared to Sonar but I don't like that comparison because sonar works by sending a soundwave out and seeing what gets reflected back.  It's more like the visual observation of sound vibrations, or, like studying a drum by looking at it's vibrations instead of listening to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a picture of what those vibrations look like on the sun (again, greatly exaggerated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/C5bQU.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/C5bQU.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.cora.nwra.com/~werne/eos/text/solar_waves_main.html&quot; rel=&quot;noreferrer&quot;&gt;Source.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted in the article, a drum has one node.  The sun has many nodes, so it's more complicated.  It's the studying of those many nodes that gives insight into the inside of the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with doing that on a gas giant planet is, (and I'm partially guessing here), I don't believe a gas giant planet is solid enough to behave like a drum.   The convection (and gas giant planets are full of convection in their upper layers) would likely dilute any studyable patterns or nodes from it's interior, much like how wind reduces sound waves.   I don't believe this method would work on gas giant planets.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-08-25T00:28:23.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22315" PostTypeId="1" CreationDate="2017-08-25T02:21:38.023" Score="0" ViewCount="52" Body="&lt;p&gt;I know that virtual particles are constantly be created and annihilated even in a complete vacuum, I like to know if there exist such a mechanism to spawn an anti-particle black hole can be any size not necessary stellar? I think black hole lose mass by gradually throwing out one of the pair of virtual particles before they could cancel out each other, so would this also apply to an anti-particle black hole as well if it does exists?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastActivityDate="2017-08-25T02:21:38.023" Title="Do anti-particle black hole exist and can they evaporate?" Tags="&lt;black-hole&gt;&lt;hawking-radiation&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2017-08-31T10:32:51.903" ContentLicense="CC BY-SA 3.0" />
  <row Id="22318" PostTypeId="1" AcceptedAnswerId="22321" CreationDate="2017-08-25T06:52:09.953" Score="1" ViewCount="749" Body="&lt;p&gt;I own an orion laser collimator which I had not used in a very long time. I decided to collimate an old Celestron NextStar 114 and I realized that the laser did not produce a circle but an oval.&#xA;Apparently the telescope is roughly collimated using a collimation cap. &#xA;There seems to be a corrector less in the focuser which I don't dare to disassemble. Could this oval be produced by this corrector or perhaps the laser collimator needs collimation too or maybe there is a problem in the secondary mirror?&lt;/p&gt;&#xA;" OwnerUserId="18205" LastActivityDate="2017-08-25T13:08:47.857" Title="Laser Collimator producing oval not circle on the Primary" Tags="&lt;telescope&gt;&lt;collimation&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22320" PostTypeId="2" ParentId="22318" CreationDate="2017-08-25T09:50:56.140" Score="2" Body="&lt;p&gt;First of all, what do stars look like through the scope, particular just inside and outside focus?  If they are oval, then read on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not familiar with this telescope, but I suspect the most likely sources of the problem are&#xA;1. The corrector lens you mention (which is a relatively unusual thing to have) is distorted.  If you can't remove this, then you are stuck with it.&#xA;2. The adjustment screws on the secondary mirror may be too tight, thereby distorting the shape of the .  Try loosening each screw by a small amount, say 1/8th of a turn, and see if this improves things.  Make sure the telescope tube is horizontal when you do this in case the secondary slips off its mount.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-08-25T09:50:56.140" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22321" PostTypeId="2" ParentId="22318" CreationDate="2017-08-25T13:08:47.857" Score="2" Body="&lt;p&gt;This is a slightly tricky one. For a conventional Newtonian reflector, shining a laser collimator back over the secondary to the primary should present a small dot on the primary - the task then is to adjust the secondary to center the dot on the primary - there is hopefully a mark on the primary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a reasonable general primer: &lt;a href=&quot;https://starizona.com/acb/basics/using_collimating_newt.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://starizona.com/acb/basics/using_collimating_newt.aspx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HOWEVER, the 114 is special, having a Barlow lens embedded in the bottom of the focuser. This would have the effect of spreading the laser light back into the telescope, leading to a projected shape (a disk or oval) on the primary. It's not really possible to conclude whether the collimator is collimated from this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reading the manual ( &lt;a href=&quot;http://www.nexstarsite.com/Manuals.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.nexstarsite.com/Manuals.htm&lt;/a&gt; ), the manufacturer recommends a simple sight test: remove the eyepiece, and look down the focuser, adjusting the primary to center your eye (hence not worrying about the secondary alignment).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you should be able to remove the Barlow lens by racking the focus tube all the way out and unscrewing it (do remember which way around it is mounted ...), which would allow the use of the laser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplest solution would be to use a low tech Cheshire collimator, I think. These forum discussions might give useful background:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astronomyforum.net/celestron-nexstar-telescope-forum/94220-celestron-114eq-collimation-woes.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.astronomyforum.net/celestron-nexstar-telescope-forum/94220-celestron-114eq-collimation-woes.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.cloudynights.com/topic/117354-nexstar-slt-114-viewing-collimation/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.cloudynights.com/topic/117354-nexstar-slt-114-viewing-collimation/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17502" LastActivityDate="2017-08-25T13:08:47.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22323" PostTypeId="2" ParentId="22169" CreationDate="2017-08-25T14:28:14.513" Score="4" Body="&lt;p&gt;The apparent curvature of the moon's transit across the sun is primarily due to the change in perspective of the viewer as s/he is rotated to the east by the earth's normal daily-cycle rotation - approximately 15 degrees absolute per hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the beginning of the eclipse, the viewer is standing &quot;upright,&quot; and sees the moon touch the sun at approximately 1 O'clock.  Over the next three hours, as the eclipse progresses, the viewer is &quot;laid on her/his side&quot; relative to his/her initial orientation due to the rotation of the earth; as if tilting his/her neck to the east. Since most of us were looking (essentially) towards the south for this eclipse, it was as if we had tilted our head to the left, so the exit of the moon - instead of appearing at approximately 7 O'clock as expected for a linear transit, the exit point appeared to be at 9 O'clock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnitude of this effect is dependent upon several variables, including the viewer's latitude during observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect is also exaggerated by the fact that the sun transited through its peak azimuth between the beginning (rising towards its highest azimuth - approximately 63 degrees) and the end of the eclipse (setting away from its highest azimuth).  While the earth itself only rotated approximately 45 degrees absolute  during the 3 hour event, the apparent rotation of the moon's transit line across the sun was approximately 60 degrees (in the Kansas City simulation).&lt;/p&gt;&#xA;" OwnerUserId="18227" LastActivityDate="2017-08-25T14:28:14.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22324" PostTypeId="1" CreationDate="2017-08-25T16:30:15.950" Score="1" ViewCount="68" Body="&lt;p&gt;When was Mars 686 , Venus 224 and Mercury 87 days sidereal period was discovered , was it discovered by the ancients or after Copernicus  &lt;/p&gt;&#xA;" OwnerUserId="18229" LastActivityDate="2017-08-25T16:30:15.950" Title="Planets sidereal periods" Tags="&lt;venus&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22325" PostTypeId="1" CreationDate="2017-08-25T18:58:21.360" Score="1" ViewCount="71" Body="&lt;p&gt;I know there are a number of parameters that determine the occurrence of a solar or lunar eclipse. The number should be 18; 6 degrees of freedom for 3 rigid bodies. But can't figure out why eclipses are not repeated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we have a night with a full moon, the next night we again have a close-to-full moon, so it can be considered a distribution over several days. What prevents the same happening for eclipses? Is it possible to have two semi-eclipses in two consecutive days? If not, why does such a big difference occur over just a day?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="8051" LastEditorUserId="2773" LastEditDate="2017-08-25T21:54:11.370" LastActivityDate="2017-08-25T21:54:11.370" Title="What prevents consequent eclipses?" Tags="&lt;solar-eclipse&gt;&lt;lunar-eclipse&gt;&lt;space-geometry&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2017-08-26T20:13:32.820" ContentLicense="CC BY-SA 3.0" />
  <row Id="22326" PostTypeId="2" ParentId="22325" CreationDate="2017-08-25T19:29:37.087" Score="2" Body="&lt;p&gt;You have to remember that the alignment for an eclipse is a very rare situation. It's not as simple as the occurrence of a new moon. The planet and the moon have to align just so in relation to the sun for an eclipse to occur. Sorry, no numbers here but just trying to emphasize that several factors have to occur between 2 fast moving bodies.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastActivityDate="2017-08-25T19:29:37.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22327" PostTypeId="1" CreationDate="2017-08-25T22:37:41.677" Score="-1" ViewCount="4526" Body="&lt;p&gt;This evening while siting in the garden I saw an unusual group of light objects in the sky. On their own each of the 5 objects would look like a satellite and I would not give it a second thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mentioned objects clearly moved in the same direction across the sky and with a speed and light pattern of a satellite. The other unusual thing was that the objects were rapidly changing formation moving away and towards one-another with no discernible pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I would like to say as I Live in the London area I am quite accustomed to seeing planes in the night sky and do not think the object I saw was a plane or planes. The light emitted was white and constant and the formation of the objects was fairly scattered and changing very fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone please suggest what was it that I saw?&lt;/p&gt;&#xA;" OwnerUserId="18234" LastActivityDate="2017-08-26T12:49:40.187" Title="Group of satellite like objects moving through the sky in a formation" Tags="&lt;observational-astronomy&gt;&lt;satellite&gt;&lt;formation&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2017-08-31T10:33:44.560" ContentLicense="CC BY-SA 3.0" />
  <row Id="22328" PostTypeId="1" CreationDate="2017-08-25T22:52:50.977" Score="1" ViewCount="71" Body="&lt;p&gt;If you look at this &lt;a href=&quot;https://dept.astro.lsa.umich.edu/ugactivities/Labs/BrightStars/SFAStarChartsAll.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; they have days realting to right acension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did they calculate nov 20th == 0h, dec 6 == 1h?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was trying to calculate this in astropy, but I end up with nov 20th, 8pm == 22h.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from astropy import units as u&#xA;from astropy.time import Time&#xA;from astropy.coordinates import SkyCoord, EarthLocation, AltAz&#xA;&#xA;time_str =  &quot;2017-11-20 20:00&quot;&#xA;&#xA;#Greenville SC, eclipse just went through there.&#xA;LON =  -82.3940&#xA;LAT = 34.8526&#xA;loc = EarthLocation(lat=LAT*u.deg, lon=LON*u.deg, height=0*u.m)&#xA;&#xA;utcoffset = -4*u.hour  # Eastern Daylight Time&#xA;obstime = Time(time_str,scale='utc') - utcoffset&#xA;&#xA;cord = SkyCoord(alt=90*u.degree,az=0*u.degree,obstime=obstime,frame = 'altaz',location = loc)&#xA;print obstime,cord.fk5.ra.hour, cord.fk5.dec.deg&#xA;&#xA;&amp;gt;&amp;gt; 2017-11-21 00:00:00.000 22.505946538 34.7569396915&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="18233" LastActivityDate="2017-08-26T09:12:13.587" Title="How are the days relating to right ascension around a polar star chart decided?" Tags="&lt;amateur-observing&gt;&lt;python&gt;&lt;right-ascension&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22329" PostTypeId="1" AcceptedAnswerId="22343" CreationDate="2017-08-26T02:36:46.923" Score="5" ViewCount="1534" Body="&lt;p&gt;The equation which relates Terrestrial [Dynamical] Time (TT, or TDT) and International Atomic Time (TAI) is the following, according to the &lt;a href=&quot;http://tycho.usno.navy.mil/systime.html&quot; rel=&quot;noreferrer&quot;&gt;U.S. Naval Observatory&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$TDT = TAI + 32.184$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I have understood, both TT and TAI are continuous time systems (differently from UTC, which has leap-second gaps) and described by the same unit (the SI second), but counted from a different epoch (the origin T&lt;sub&gt;0&lt;/sub&gt; of one is 32.184 seconds after the other).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears that both time systems are essentially the same. So what is the actual difference between them? If the difference between them is the nature of how they are computed (e.g., TT is a theoretical timescale; and TAI is a statistical timescale based on several clocks), why the 32.184 shift?&lt;/p&gt;&#xA;" OwnerUserId="18209" LastActivityDate="2017-08-27T10:23:20.173" Title="What is the difference between Terrestrial Time (TT) and International Atomic Time (TAI)?" Tags="&lt;time&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22331" PostTypeId="1" CreationDate="2017-08-26T06:31:57.623" Score="5" ViewCount="512" Body="&lt;p&gt;I just read most of Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Giant-impact_hypothesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;article on the Giant Impact Hypothesis&lt;/a&gt;. Basically a large object impacts Earth and creates debris that soon coalesces into the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is something I did not find mentioned: &lt;strong&gt;How would the Moon get its circular orbit?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A debris field from any impact would at first follow a very elliptical orbit, with a perigee very close to the surface of the Earth and an apogee much farther away. In orbital mechanics, where you start is where you end up. In fact it doesn't even have to be where you start. Pick any point in the orbit and you will return there again. In other words, the original elliptical orbit doesn't change its shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...Unless there are pertubations or impacts that change your velocity. How exactly did that happen? Does the GIH just assume something else did that fortuitously?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also seen youtube video simulations like &lt;a href=&quot;https://www.youtube.com/watch?v=cn5vh7hZ2o0&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;. It's hard to tell if the camera angle is changing or if the orbital plane is unstable (how?), but you can clearly see the debris goes around and impacts the surface again and again. I.e., &lt;strong&gt;the orbit remains highly elliptical. I've never seen a simulation that shows the debris circularizing.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also want to note that this question is highly relavent not just to the Moon, but many other things like orbital capture as well. I've never understood orbital capture theories as they all seem to leave out how the initial elliptical orbit later becomes circular, e.g., Triton.&lt;/p&gt;&#xA;" OwnerUserId="7449" LastEditorUserId="7449" LastEditDate="2018-04-26T23:46:55.647" LastActivityDate="2018-04-26T23:46:55.647" Title="Does the Giant Impact Hypothesis explain how the Moon circularized its orbit?" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;orbital-migration&gt;&lt;eccentric-orbit&gt;&lt;theia&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22332" PostTypeId="2" ParentId="22304" CreationDate="2017-08-26T08:56:15.193" Score="2" Body="&lt;p&gt;Here is a quick supplemental answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stability_of_the_Solar_System#Neptune.E2.80.93Pluto_resonance&quot; rel=&quot;nofollow noreferrer&quot;&gt;Neptune-Pluto Resonance&lt;/a&gt; subsection of the Wikipedia article &lt;a href=&quot;https://en.wikipedia.org/wiki/Stability_of_the_Solar_System&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stability of the Solar System&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Neptune–Pluto system lies in a 3:2 &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_resonance&quot; rel=&quot;nofollow noreferrer&quot;&gt;orbital resonance&lt;/a&gt;. C.J. Cohen and E.C. Hubbard at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Naval_Surface_Warfare_Center_Dahlgren_Division&quot; rel=&quot;nofollow noreferrer&quot;&gt;Naval Surface Warfare Center Dahlgren Division&lt;/a&gt; discovered this in 1965. Although the resonance itself will remain stable in the short term, it becomes impossible to predict the position of Pluto with any degree of accuracy, as the uncertainty in the position grows by a factor e with each &lt;a href=&quot;https://en.wikipedia.org/wiki/Lyapunov_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lyapunov time&lt;/a&gt;, which for Pluto is 10–20 million years into the future. Thus, on the time scale of hundreds of millions of years Pluto's orbital phase becomes impossible to determine, even if Pluto's orbit appears to be perfectly stable on 10 MYR time scales (Ito and Tanikawa 2002, MNRAS).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The resonance they discovered means that for a moderate period of time (at least tens of millions of years) the relative motion between Neptune and Pluto will be repetitive, with $3\times T_{Neptune} \approx 2\times T_{Pluto}$. &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1965AJ.....70...10C&amp;amp;defaultprint=YES&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;The paper&lt;/a&gt; cited in the &lt;a href=&quot;https://astronomy.stackexchange.com/a/22308/7982&quot;&gt;other excellent answer&lt;/a&gt; is the report of this discovery by Cohen and Hubbard in 1965 mentioned here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure 5 of that paper shows the motion of Neptune and Pluto in a rotating frame. The frame rotates with the &lt;em&gt;average orbital rotation&lt;/em&gt; of Neptune, so over very long times you can see Neptune slowly rocks back and forth a bit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0JZNO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0JZNO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you like Python you can reproduce these fairly easy. The package &lt;a href=&quot;http://rhodesmill.org/skyfield/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Skyfield&lt;/a&gt; uses the same NASA JPL Ephemerides as the &lt;a href=&quot;https://i.stack.imgur.com/3vQXQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL Horizons&lt;/a&gt; site but is easier to use. You can see that 95% of this script is just making it look nicer and getting the position of the planets is just a few lines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the results for only 6,000 years, a much smaller period than shown in Cohen and Hubbard 1965, so Neptune only makes a small segment of its 25,000 cycle, near one end where it's &quot;moving&quot; slowly. The first set of plots are in inertial (non-rotating) J2000 ecliptic coordinates, and the second is rotating with the average orbital motion of Neptune, so that Neptune appears nearly fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first plot shows Neptune-Pluto separation versus calendar year. The minimum in this period seems to be around year -77, with a distance of 2.65 billion km or 17.73 AU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/htnRG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/htnRG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3vQXQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3vQXQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8i2BS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8i2BS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def Rpos(pos, angle):&#xA;    x, y, z = pos&#xA;    ca, sa  = np.cos(angle), np.sin(angle)&#xA;&#xA;    xr = x*ca - y*sa&#xA;    yr = y*ca + x*sa&#xA;&#xA;    return np.vstack((xr, yr, z))&#xA;&#xA;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;from mpl_toolkits.mplot3d import Axes3D&#xA;&#xA;from skyfield.api import Loader &#xA;&#xA;loader = Loader('~/Documents/Skydata', verbose = True)&#xA;ts     = loader.timescale()&#xA;&#xA;de405  = loader('de405.bsp')  # about 65 MB&#xA;de421  = loader('de421.bsp')  # about 17 MB&#xA;de422  = loader('de422.bsp')  # about 650 MB&#xA;&#xA;de = de422&#xA;&#xA;neptune = de['neptune barycenter']&#xA;pluto   = de['pluto barycenter']&#xA;&#xA;# years   = np.arange(1600, 2201)&#xA;years   = np.arange(-2999, 3001)&#xA;&#xA;time    = ts.utc(years, 1, 1) # January 1st of each calendar year&#xA;&#xA;npos = neptune.at(time).ecliptic_position().km&#xA;ppos = pluto.at(time).ecliptic_position().km&#xA;&#xA;npsep = np.sqrt(((npos-ppos)**2).sum(axis=0))&#xA;&#xA;if True:&#xA;&#xA;    plt.figure()&#xA;    plt.plot(years, npsep)&#xA;    plt.show()&#xA;&#xA;aukm = 149597870.700&#xA;print &quot;minimum separation (km): &quot;, npsep.min()&#xA;print &quot;minimum separation (AU): &quot;, npsep.min()/aukm&#xA;&#xA;if True:&#xA;&#xA;    fig = plt.figure()&#xA;&#xA;    ax1  = fig.add_subplot(1, 2, 1)&#xA;    for pos in [npos, ppos]:&#xA;        x, y, z = pos&#xA;        ax1.plot(x, y)&#xA;    ax1.plot([0], [0], 'or')&#xA;    ax1.set_xlim(-5E+09, 7E+09)&#xA;    ax1.set_ylim(-5E+09, 7E+09)&#xA;&#xA;    ax2  = fig.add_subplot(1, 2, 2, projection='3d')&#xA;    for pos in [npos, ppos]:&#xA;        x, y, z = pos&#xA;        ax2.plot(x, y, z, linewidth=1.0)&#xA;    ax2.plot([0], [0], [0], 'or')&#xA;    ax2.set_xlim(-5E+09, 7E+09)&#xA;    ax2.set_ylim(-5E+09, 7E+09)&#xA;    ax2.set_zlim(-6E+09, 6E+09)&#xA;&#xA;    ax2.view_init(elev=20., azim=-110)&#xA;&#xA;    plt.show()&#xA;&#xA;nargperi = 276.3 * (np.pi/180.) # Neptune argument of perihelion (radians)&#xA;nT       = 164.79 # Nepture orbital period (years)&#xA;&#xA;nangle   = np.arctan2(npos[1], npos[0])#  - nargperi&#xA;&#xA;nangle_mean = 2.*np.pi*np.mod(years/nT, 1.0)  # this is a bit sloppy&#xA;&#xA;nposr = Rpos(npos, -nangle_mean)&#xA;pposr = Rpos(ppos, -nangle_mean)&#xA;&#xA;if True:&#xA;&#xA;    fig = plt.figure()&#xA;&#xA;    ax1  = fig.add_subplot(1, 2, 1)&#xA;    for pos in [nposr, pposr]:&#xA;        x, y, z = pos&#xA;        ax1.plot(x, y)&#xA;    ax1.plot([0], [0], 'or')&#xA;    ax1.set_xlim(-7E+09, 7E+09)&#xA;    ax1.set_ylim(-7E+09, 7E+09)&#xA;&#xA;    ax2  = fig.add_subplot(1, 2, 2, projection='3d')&#xA;    for pos in [nposr, pposr]:&#xA;        x, y, z = pos&#xA;        ax2.plot(x, y, z, linewidth=1.0)&#xA;    ax2.plot([0], [0], [0], 'or')&#xA;    ax2.set_xlim(-7E+09, 7E+09)&#xA;    ax2.set_ylim(-7E+09, 7E+09)&#xA;    ax2.set_zlim(-7E+09, 7E+09)&#xA;&#xA;    ax2.view_init(elev=20., azim=-116)&#xA;&#xA;    plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-08-26T09:36:06.957" LastActivityDate="2017-08-26T09:36:06.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22333" PostTypeId="2" ParentId="22331" CreationDate="2017-08-26T09:01:54.233" Score="1" Body="&lt;p&gt;Well, I don't really think the GI Hypothesis can answer the orbital plan of the moon, I would say that the circular orbit of the moon today regarding the GIH is also due to other variables, like Jupiter, the Sun, etc... If the GIH is true, then the moon should still be on a quite elliptical orbit still, since the gravitational forces of other celestial bodies are not strong enough to transformer the orbit in the 4 million years since (Concerning my personal knowledge on gravity and the impact it has on long duration of time)&lt;/p&gt;&#xA;" OwnerUserId="18240" LastActivityDate="2017-08-26T09:01:54.233" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22334" PostTypeId="2" ParentId="22328" CreationDate="2017-08-26T09:12:13.587" Score="1" Body="&lt;p&gt;The date on the charts shows the meridian (north south line) at 8pm local time on that date.  Over a period of a month, the RA of the meridian will move 2hr (so that it gets back to where it started from after 12 months), and hence over 16 days, as you ask about, it will move 1.05 hours.  So you can either say that's close enough to 1 hr, or worry about the small discrepancy - that's assuming the 1hr you cite is actually 1.00hrs.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-08-26T09:12:13.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22338" PostTypeId="2" ParentId="22331" CreationDate="2017-08-26T11:34:31.657" Score="8" Body="&lt;p&gt;I want to point out that the Moon's orbit isn't circular now.   &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbit_of_the_Moon&quot; rel=&quot;noreferrer&quot;&gt;A 0.055 mean eccentricity&lt;/a&gt; isn't that circular.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, onto your question.   I think you're making a bad assumption on the &quot;must have started very elliptic.   Individual objects that are ejected from a planet need to follow their orbital path.   So any object flung from the earth would need likely either escape the Earth or fall back into it because you can't launch something, with a singular push, into a circular orbit.   All the individual bits of debris should have had highly eccentric orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the formation of the Moon was much more complicated than that.   First, it was enough matter to have it's own gravitational field, basically influencing itself, and the forming moon would be the combination of all those individual objects, so it wouldn't necessarily be eliptic at all.   The eccentricity of billions of objects could have (and likely did) cancel each other out to a large degree.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the impact was off-center, it set the Earth spinning and most of the debris moved in the same direction around the Earth.   But the coalescing mass of material that didn't fall back to the Earth could have been in a reasonably circular orbit at formation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We should also define what number constitutes &quot;highly&quot; elliptic.   The Moon is thought to have started out 3-5 Earth Radii distant.   Any material that passed closer than 3 radii would have been inside the Roche limit and had a hard time coalescing into the Moon.    &lt;a href=&quot;http://curious.astro.cornell.edu/about-us/37-our-solar-system/the-moon/the-moon-and-the-earth/31-how-close-was-the-moon-to-the-earth-when-it-formed-intermediate&quot; rel=&quot;noreferrer&quot;&gt;This article&lt;/a&gt; suggests that the debris would have had a hard time being ejected further than 5 Earth Radii, though I'm not sure how that conclusion was reached.  But you have a few things going on.  Debris colliding with other debris, and some (presumably quite a bit) of debris falling back to the Earth and, mentioned above, the debris having a gravitational effect on itself.   Ultimately momentum needs to be conserved and calculating the formation of the Moon requires a super-computer, but in general I think the initial orbit could have been relatively circular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Moon formed close to the Earth, which most models suggest it did and since it couldn't pass inside 3 Earth Radii and stay solid, those numbers put a limit on how eccentric the initial orbit probably was.   If we use the 5/3 Earth Radii as an estimate, Ra=5, Rp=3, then the eccentricity is 0.25, and that's an estimate for the upper limit of the initial eccentricity after formation.  It may well have been quite a bit less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CH3qB.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CH3qB.gif&quot; alt=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/imgmec/kep.gif&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for the sake of argument, lets give the newly formed moon an eccentricity of .25 or, maybe a little higher with a &lt;a href=&quot;http://www.lunarplanner.com/LunarPerigee/&quot; rel=&quot;noreferrer&quot;&gt;more distant apogee&lt;/a&gt;, but we have to keep the perigee at 3 Earth radii or greater.   Your question still stands, how did the Moon's initial eccentricity become somewhat circular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_circularization&quot; rel=&quot;noreferrer&quot;&gt;tidal circularization&lt;/a&gt;.   I can't find a good article on it, but for many 2 body systems, tides circularize orbits.   It's well known that the tides on Earth, caused by the Moon's gravity rotate ahead of the Moon and this creates a tug on the moon that pushes (pulls?) the Moon into a higher orbit, further away from the Earth.   This same secondary effect circularizes the Moon's orbit because the closer the Moon is too the Earth, the greater the push.  The math behind this gets very complicated and it's above my paygrade to put numbers behind it and the Wikipedia article I've linked is very lacking, so I invite anyone to provide more details if they can. But Tidal circularization is a commonly accepted theory and given the strength of the tides early on, it's likely any eccentricity in the Moon's orbit circularized relatively quickly, at least, astronomically speaking.   (Tens of millions of years is a long time to you and me, but not to a stable orbit).&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-08-26T11:34:31.657" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22340" PostTypeId="2" ParentId="22327" CreationDate="2017-08-26T12:49:40.187" Score="4" Body="&lt;p&gt;Definitely a UFO - Unidentified Flying Object, not necessarily aliens :-).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few possibilities I can think of (would be nice to know how fast the objects were going, and whether or not they disappeared), arranged by most plausible to least plausible:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Birds&lt;/strong&gt; If it was just after sunset, or just before sun rise, birds could have been flying up far above, where the sun was shining, though it was dark on the ground.  This would be moving fairly slowly through the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Shiny Balloon&lt;/strong&gt; It could have been a shiny helium balloon drifting by with the lights of the city reflecting off it as it moved, causing the apparent motion.  If this were the case, the &quot;dots&quot; should have stayed fairly close together, and it should have been fairly slow moving unless it was low and the wind was blowing fast.  It would seem the pattern would be randomized, and not necessarily a &quot;formation&quot; though, unless formation just meant they were travelling together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Meteor fragments&lt;/strong&gt; It could have been fragments of a meteor that broke up, but they would likely be travelling straight, and not flying in any recognizable formation (but would be going nearly same direction, likely spreading out with time, and burning up / disappearing).  This would have gone fairly fast through the sky and disappeared, or at least shrunk if they were large fragments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Aliens&lt;/strong&gt; At first, seemed the only plausible explanation, but there are others after thinking about it a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Satellites&lt;/strong&gt; would not change formation like that, but there is a possibility atmospheric distortions (like the &quot;waves of heat&quot; you can see coming off the road on a hot day) could have caused the effect with a single passing satellite.  However, you probably would have noticed the stars doing the same thing if that were the case.  I am also not aware of multiple satellites being grouped together like that, and they would not be able to change formation like that.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2017-08-26T12:49:40.187" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22341" PostTypeId="1" CreationDate="2017-08-26T18:43:11.213" Score="0" ViewCount="120" Body="&lt;p&gt;Suppose we have a variable star with a change in magnitude between maximum and minimum of 5 mag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much would you need to stop down a telescope (block some of the area) so that the star at maximum brightness appears to be the same brightness as when it is at minimum magnitude with no stop.&lt;/p&gt;&#xA;" OwnerUserId="18247" LastEditorUserId="2365" LastEditDate="2017-08-29T06:14:02.987" LastActivityDate="2017-08-29T06:21:48.180" Title="Stopping down a telescope to reduce brightness by 5 magnitudes" Tags="&lt;variable-star&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22343" PostTypeId="2" ParentId="22329" CreationDate="2017-08-26T22:55:29.870" Score="5" Body="&lt;p&gt;It's historical. TT is a successor to The International Astronomy Union's ephemeris time, and the offset between TT and TAI is a result of getting TT to match ET when they switched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Excerpts from &lt;a href=&quot;https://en.wikipedia.org/wiki/Terrestrial_Time&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Terrestrial_Time&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Terrestrial Time (TT) is a modern astronomical time standard defined by the International Astronomical Union, primarily for time-measurements of astronomical observations made from the surface of Earth.[1] For example, the Astronomical Almanac uses TT for its tables of positions (ephemerides) of the Sun, Moon and planets as seen from Earth. In this role, TT continues Terrestrial Dynamical Time (TDT or TD),[2] which in turn succeeded ephemeris time (ET). TT shares the original purpose for which ET was designed, to be free of the irregularities in the rotation of Earth.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The present definition of TT is a linear scaling of Geocentric Coordinate Time (TCG), which is the proper time of a notional observer who is infinitely far away (so not affected by gravitational time dilation) and at rest relative to Earth. TCG is used so far mainly for theoretical purposes in astronomy. From the point of view of an observer on Earth's surface the second of TCG passes in slightly less than the observer's SI second. The comparison of the observer's clock against TT depends on the observer's altitude: they will match on the geoid, and clocks at higher altitude tick slightly faster.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Time coordinates on the TT and TCG scales are conventionally specified using traditional means of specifying days, carried over from non-uniform time standards based on the rotation of Earth. Specifically, both Julian Dates and the Gregorian calendar are used. &lt;strong&gt;For continuity with their predecessor Ephemeris Time (ET), TT and TCG were set to match ET at around Julian Date 2443144.5 (1977-01-01T00Z). More precisely, it was defined that TT instant 1977-01-01T00:00:32.184 exactly and TCG instant 1977-01-01T00:00:32.184 exactly correspond to the International Atomic Time (TAI) instant 1977-01-01T00:00:00.000 exactly.&lt;/strong&gt; This is also the instant at which TAI introduced corrections for gravitational time dilation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Bold emphasis mine).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more on International Atomic time see &lt;a href=&quot;https://en.wikipedia.org/wiki/International_Atomic_Time&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/International_Atomic_Time&lt;/a&gt; . Basically, it's a weighted average of multiple atomic clocks - and the altitude effect mentioned in the middle paragraph above is why they added the gravitational time dilation corrections. From the Wikipedia link above:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;n the 1970s, it became clear that the clocks participating in TAI were ticking at different rates due to gravitational time dilation, and the combined TAI scale therefore corresponded to an average of the altitudes of the various clocks. Starting from Julian Date 2443144.5 (1 January 1977 00:00:00), corrections were applied to the output of all participating clocks, so that TAI would correspond to proper time at mean sea level (the geoid). Because the clocks had been on average well above sea level, this meant that TAI slowed down, by about one part in a trillion. The former uncorrected time scale continues to be published, under the name EAL (Echelle Atomique Libre, meaning Free Atomic Scale).[7]:215&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17016" LastEditorUserId="17016" LastEditDate="2017-08-27T10:23:20.173" LastActivityDate="2017-08-27T10:23:20.173" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22344" PostTypeId="1" AcceptedAnswerId="22350" CreationDate="2017-08-27T00:43:40.357" Score="2" ViewCount="362" Body="&lt;p&gt;While this question could fall under Space Exploration or Physics (I don't have the &quot;best&quot; reputation on THAT forum), I feel that it is very much related to how the universe itself functions, especially regarding Einstein. While looking to see if anyone has confirmed our universe having a 4th spatial dimension (or some non temporal or spatial dimension) due to the fact that space curves, for me curvature requires a higher dimension (paper bending or a stick bending, always +1 to its dimension), I was met with the explanation that this &quot;curvature&quot; was intrinsic and not extrinsic (requiring another dimension). That was fine and all until I realized that wormholes would be exclusively extrinsic, how could you make a wormhole if bending space is intrinsic? Obviously you can't. But Einstein  put a lot of time into something that apparently wouldn't work, so personally I am siding with Einstein here. But that still leaves the question; Are wormholes possible thus making our universe a 3+1+1 dimensional universe (I wouldn't group it up with the 3 because that would say that matter could move like that, it can't) or are wormholes impossible and so our universe remains 3+1 dimensional?&lt;/p&gt;&#xA;" OwnerUserId="16449" LastActivityDate="2017-08-27T20:36:14.803" Title="Are wormholes even possible?" Tags="&lt;space-time&gt;&lt;general-relativity&gt;&lt;wormhole&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22345" PostTypeId="1" AcceptedAnswerId="22348" CreationDate="2017-08-27T02:02:47.970" Score="7" ViewCount="525" Body="&lt;p&gt;I was fortunate enough to see totality during the eclipse of 2017 August 21. &#xA;It did leave me wondering about the 1919 Eddington experiment, though.  During totality, it only got dark enough to see a few of the brightest objects like Venus and Jupiter.  It was more like twilight than night that way.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how were they able to see and photograph stars in the 1919 eclipse?  Do skies get darker during some total eclipses than others?  Was there more light scattering in my case, from being in a populated area?  Or do the right optics let you photograph stars during twilight-like conditions?&lt;/p&gt;&#xA;" OwnerUserId="18250" LastActivityDate="2017-08-27T05:12:53.850" Title="Dark enough to see stars during total solar eclipse?" Tags="&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22348" PostTypeId="2" ParentId="22345" CreationDate="2017-08-27T04:58:28.080" Score="11" Body="&lt;p&gt;One thing to double check is if you can see stars easily at night from the same location. If there are city lights nearby, combined with humidity or dust you can have a bright sky due to terrestrial effects.&lt;/p&gt;&#xA;&lt;p&gt;Another thing to consider is the time needed for dark adaptation. If you walk from a brightly lit room out under a dark sky, you can't see many stars right away. There needs to be time for your eyes to adapt to the dark. If you were looking at the screen of a camera or phone or another light source, that would further slow down the adaptation process.&lt;/p&gt;&#xA;&lt;p&gt;The astrometric data for the 1919 eclipse test of General Relativity was done with a telescope, not the naked eye. Stars are unresolved, which means that no matter what magnification, they are still small points of light who's size (visual or on film) is determined by other effects, like diffraction, aberrations and focus as well as the resolution of the retina or the film. In short, they stay small. However, sky brightness is reduced by magnification squared. If you expand a patch of sky by 10X then each point is 100 times dimmer.&lt;/p&gt;&#xA;&lt;p&gt;Both of these are offset by the increase in light from the larger aperture of the telescope, &lt;em&gt;but its mostly the sky brightness that decreases with magnification&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The interesting stars were the ones closest to the Sun, because they would have the largest deflection, which was really small, only of the order of one second of arc. There the problem was the brightness of the solar corona, more than the sky. If you look closely at the photographic negative below, you can see pairs of faint horizontal lines. In the middle of each pair is the position of a star to be observed. Even with the telescope and extended photographic exposure, the images are faint.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Iory0.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Iory0.jpg&quot; alt=&quot;1919 eclipse test of General Relativity cropped&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above and below:&lt;/strong&gt; 1919 eclipse negative, from &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:1919_eclipse_negative.jpg&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. Original from: F. W. Dyson, A. S. Eddington, and C. Davidson, &amp;quot;A Determination of the Deflection of Light by the Sun's Gravitational Field, from Observations Made at the Total Eclipse of May 29, 1919&amp;quot; Philosophical Transactions of the Royal Society of London. Series A, Containing Papers of a Mathematical or Physical Character (1920): 291-333, on 332.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Negative of the 1919 solar eclipse taken from the report of Sir Arthur Eddington on the expedition to verify Einstein's prediction of the bending of light around the sun. For positive version, see &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:1919_eclipse_positive.jpg&quot; rel=&quot;noreferrer&quot;&gt;Image:1919 eclipse positive.jpg&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Original caption:&lt;/p&gt;&#xA;&lt;p&gt;In Plate 1 is given a half-tone reproduction of one of the negatives taken with the 4-inch lens at Sobral. This shows the position of the stars, and, as far as possible in a reproduction of this kind, the character of the images, as there has been no retouching. A number of photographic prints have been made and applications for these from astronomers, who wish to assure themselves of the quality of the photographs, will be considered as as far as possible acceded to.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WJHUC.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WJHUC.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; Screen shot from the March 2009 Physics Today article &lt;a href=&quot;http://old.philosophy.ox.ac.uk/__data/assets/pdf_file/0003/38685/kennefick_phystoday_09.pdf&quot; rel=&quot;noreferrer&quot;&gt;Testing relativity from the 1919 eclipse—a question of bias&lt;/a&gt;. The telescope on the right, in the square tube is the 4-inch instrument that provided the photographs used in the analysis. Heliostat mirrors driven by clocks were used to reflect light from the region near the Sun into the fixed telescope.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/S6GU7.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S6GU7.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-08-27T05:12:53.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22349" PostTypeId="1" AcceptedAnswerId="22362" CreationDate="2017-08-27T05:59:53.903" Score="0" ViewCount="279" Body="&lt;p&gt;It came to my mind one day that why does this universe even exist and why do we even exist and like we are enclosed in the so called atmosphere and a floating rock in space called the Earth, so is there any possibility that the universe is also enclosed in something and if it is, then what is outside that enclosure?&lt;/p&gt;&#xA;" OwnerUserId="18252" LastActivityDate="2017-08-28T06:50:42.567" Title="Why do we we even exist?" Tags="&lt;earth&gt;&lt;universe&gt;&lt;big-bang-theory&gt;&lt;fundamental-astronomy&gt;&lt;space&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" ClosedDate="2017-08-28T10:09:51.433" ContentLicense="CC BY-SA 3.0" />
  <row Id="22350" PostTypeId="2" ParentId="22344" CreationDate="2017-08-27T12:22:23.180" Score="4" Body="&lt;p&gt;The idea of wormholes is completely abstract.  We have no evidence they actually exist and they are a result of following the strict mathematics of general relativity to &lt;em&gt;and beyond&lt;/em&gt; it's intended limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you push a theory past it's &quot;design limits&quot; you can't expect the results to be valid.  Sometimes they are, sometimes they aren't.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;for me curvature requires a higher dimension (paper bending or a stick bending, always +1 to its dimension)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The problem here is that you're thinking in a common, everyday sense about something that fundamentally is &lt;em&gt;not&lt;/em&gt; understandable using common everyday sense.  One of the major problems people face learning even special relativity is that common sense ideas (like &lt;a href=&quot;https://en.wikipedia.org/wiki/Relativity_of_simultaneity&quot; rel=&quot;nofollow noreferrer&quot;&gt;simultaneity&lt;/a&gt;) go out the window, and this is even more difficult in general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand how space-time is distorted in such extreme ways requires an understanding of the mathematics.  The simple explanations that involve things like rubber-sheet analogies just don't give you the right idea at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A typical tool to explore space-time and how it distorts is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Penrose_diagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carter-Penrose diagram&lt;/a&gt;.  Another one is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kruskal%E2%80%93Szekeres_coordinates&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kruskal-Szekeres coordinates&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So really the problem here is you trying to use your common everyday sense in a situation that it does not work in.  The reason physicists do all that complex mathematics in the first place is because that &lt;em&gt;is&lt;/em&gt; what works.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-08-27T12:22:23.180" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22351" PostTypeId="1" AcceptedAnswerId="22353" CreationDate="2017-08-27T12:44:33.610" Score="0" ViewCount="113" Body="&lt;p&gt;Recently I have found the article &lt;a href=&quot;https://arxiv.org/abs/astro-ph/9511022&quot; rel=&quot;nofollow noreferrer&quot;&gt;Does There Really Exist the Problem of the Dark Matter in Spiral Galaxies?&lt;/a&gt; which states that dark matter does not exist because astronomers are wrong assuming stationarity of galaxy rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to them Doppler measurements give only one component of speed which is insufficiebt to restore full three dimensional vector. So they claim that stars in galaxies are on escape tragectories but reproduce correst rotational curves if we assume stationarity of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are evidences that this is not true, what about measurements of proper motions, not doppler shifts only?&lt;/p&gt;&#xA;" OwnerUserId="18255" LastEditorDisplayName="user1569" LastEditDate="2017-08-27T17:34:30.497" LastActivityDate="2017-08-28T11:42:15.673" Title="What are evidences of stationarity of rotation of galaxies" Tags="&lt;galaxy&gt;&lt;rotation&gt;&lt;dark-matter&gt;&lt;doppler-effect&gt;&lt;spiral-arms&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22353" PostTypeId="2" ParentId="22351" CreationDate="2017-08-27T19:19:50.797" Score="5" Body="&lt;p&gt;If stars were actually about to escape, they would do so on a so-called &lt;em&gt;dynamical timescale&lt;/em&gt;:&#xA;$$&#xA;t_\mathrm{dyn} \sim \sqrt{\frac{R^3}{8GM}},&#xA;$$&#xA;where $G$ is the gravitational constant, and $R$ and $M$ are the radius and the mass of the galaxy, respectively. For a Milky Way-sized galaxy, this is equates to roughly 30 million years$^\dagger$. But disk galaxies are observed over time spans of billions of years. If they had for some reason been gravitationally bound and then &quot;unleashed&quot;, they would have to be so at times synchronized such that we, today, in all directions, observe them in a state that has a very short time span compared their lives throughout the history of the Universe. That is, distant galaxies (which are young because we look far back in time) were unleashed early in their lives, and nearby galaxies (which we see in the &quot;old age&quot; because we look less far back in time) were unleashed recently. This not only seems unlikely, but would also put us in a special place in the Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that the paper you link to is over 20 years old, non-refereed and has no citations (and that the author in general has very few citations). Of course this by no means implies that he's wrong, but it does give a hint that it's a good idea to be cautious of the conclusions of the paper. Since that paper was written, much more independent evidence for dark matter has been obtained.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;Since the roughly constant rotation speed of disk galaxies with galaxy mass (a few 100s of km/s) implies that $R$ grows approximately linearly with $R$ (and not with $R^3$ as one might expect), smaller galaxies — which exist in much larger numbers than large galaxies — have even shorter dynamical timescales.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2017-08-28T11:42:15.673" LastActivityDate="2017-08-28T11:42:15.673" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22355" PostTypeId="1" AcceptedAnswerId="22375" CreationDate="2017-08-27T21:15:50.500" Score="3" ViewCount="212" Body="&lt;p&gt;I've read that when a body reaches about 13 times the mass of Jupiter that deuterium fusion starts and the body becomes a brown dwarf instead of a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a multiple question.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Are there any other factors besides mass and possibly metal abundance (composition) that would determine this &quot;turn on&quot; point ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the &quot;brown dwarf&quot; was detected indirectly and no spectroscopic information was available, could we assume that metalicity would be the same as the larger star ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Would a higher metallicity body &quot;turn on&quot; at a lower mass? (Would a difference between [Fe/H] = -0.19 and [Fe/H] = +0.07 make much of a difference in &quot;turn on mass&quot;?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;*I did miss the related question when I posted this one, but I do have a question here that wasn't answered. It was about how stellar metallicity effects the onset of fusion. I think it may be as simple as whether or not adding &quot;metals&quot; makes the core hotter or colder for a given mass. I also wanted to know if I could assume that both &quot;stars&quot; in a binary system should have the same metallicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Now that I've read Rob's excellent (as usual) answer, I would be curious as to how long deuterium fusion lasts in brown dwarfs from 13 to 80 Jupiter masses and what happens (in models) when it stops.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="2531" LastEditDate="2017-08-29T07:06:36.357" LastActivityDate="2017-08-29T07:06:36.357" Title="&quot;Turn on point&quot; for fusion?" Tags="&lt;brown-dwarf&gt;&lt;nucleosynthesis&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22356" PostTypeId="1" CreationDate="2017-08-27T23:17:52.270" Score="4" ViewCount="123" Body="&lt;p&gt;Hopefully this is not a silly question, but I was wondering why earth is not always in the light (and any other planet for that matter). I mean, we have billions of stars just in our galaxy that all give off light from all directions. That light has obviously reached is as we can see them shimmering in the night sky. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that in mind, what has happened to the light from the stars? Has it weakened somehow? (didn't think that was possible) Surely if we are being hit with light from all directions, it makes sense that we should always be illuminated. So what's going on here? &lt;/p&gt;&#xA;" OwnerUserId="17096" LastActivityDate="2017-08-28T05:40:14.470" Title="Why is it not always light?" Tags="&lt;light&gt;&lt;sun-rays&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2017-08-30T15:17:52.223" ContentLicense="CC BY-SA 3.0" />
  <row Id="22357" PostTypeId="2" ParentId="22356" CreationDate="2017-08-28T05:40:14.470" Score="2" Body="&lt;p&gt;Olber's Paradox (Heinrich Wilhelm Olbers, 1758–1840): If the Universe were infinite, or there were infinitely many stars, then no matter which direction you look, you'd see a star. Thus, the night sky would be as bright as the Sun. Therefore, the Universe is not infinite, and does not contain an infinite amount of stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the light from distant stars does 'get weaker' because of the inverse square law; this is the same reason lights look dimmer when they're further away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if gravitational redshift has anything to do with it, too. Anyone?&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2017-08-28T05:40:14.470" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22362" PostTypeId="2" ParentId="22349" CreationDate="2017-08-28T06:50:42.567" Score="-2" Body="&lt;p&gt;That's an interesting question, but as the comments say might be a better place to ask this question in the Philosophy SE site, but the question about the universe being englobed maybe by another much bigger 'space' is something we might never be able to prove right or wrong. This 'existential' question might refer to the holograph theory, the multiverse theory. The thing is, at the moment the only possible explanation of whats beyond the 'observable universe' boundary is nothing, neon, an infinite multi-dimensional plane of nothingness. &lt;/p&gt;&#xA;" OwnerUserId="18240" LastActivityDate="2017-08-28T06:50:42.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22363" PostTypeId="1" AcceptedAnswerId="22365" CreationDate="2017-08-28T07:06:08.223" Score="6" ViewCount="1711" Body="&lt;p&gt;I know mercury is tidal locked to our Sun like our Moon is to Earth, is it merely coincident or there is some sciences behind it? Some man-made communication satellites are also tidal locked to Earth! I think it is intentional so that there is minimum interruption for signal transmission, they used thrusters and gyro to help them to always point toward Earth.&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="907" LastEditDate="2017-08-28T14:38:51.550" LastActivityDate="2017-08-31T20:28:02.397" Title="Is it coincident that all planets with tight orbit are tidally locked to their parent body?" Tags="&lt;planet&gt;&lt;tidal-locking&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22365" PostTypeId="2" ParentId="22363" CreationDate="2017-08-28T09:50:32.157" Score="9" Body="&lt;p&gt;This is a physical result: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;The change in rotation rate necessary to tidally lock a body &lt;strong&gt;B&lt;/strong&gt; to a larger body &lt;strong&gt;A&lt;/strong&gt; is caused by the torque applied by &lt;strong&gt;A&lt;/strong&gt;'s gravity on bulges it has induced on &lt;strong&gt;B&lt;/strong&gt; by tidal forces.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stellar bodies get tidally locked with time. You even get &lt;a href=&quot;https://physics.stackexchange.com/questions/12541/tidal-lock-radius-in-habitable-zones&quot;&gt;graphs like there&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On how long it generally takes, from &lt;a href=&quot;http://en.wikipedia.org/wiki/Tidal_locking#Timescale&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t_{\text{lock}} \approx \frac{\omega a^6 I Q}{3 G m_p^2 k_2 R^5} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $Q$ and $k_2$ depend on the constituents and age of the satellite, which themselves evolve with time. Difficult to estimate. Moreover, you rarely know the age of your system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it explains that for long-time established system of relatively close bodies, tidal lock is generally the case. The more tide there is, the quicker it gets.&lt;/p&gt;&#xA;" OwnerUserId="11264" LastEditorUserId="11264" LastEditDate="2017-08-28T14:00:45.340" LastActivityDate="2017-08-28T14:00:45.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22369" PostTypeId="5" CreationDate="2017-08-28T12:49:51.440" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2017-08-28T12:49:51.440" LastActivityDate="2017-08-28T12:49:51.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22370" PostTypeId="4" CreationDate="2017-08-28T12:49:51.440" Score="0" Body="Early Planetary object based on the Giant Impact Hypothesis which collided with early earth about 4 Billion years ago." OwnerUserId="18240" LastEditorUserId="18240" LastEditDate="2017-08-29T05:48:39.473" LastActivityDate="2017-08-29T05:48:39.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22371" PostTypeId="1" AcceptedAnswerId="22372" CreationDate="2017-08-28T16:24:37.047" Score="3" ViewCount="356" Body="&lt;p&gt;The word we use to refer to what's inside the cosmic horizon is the 'universe', so what would you call the empty space outside of the horizon?&lt;/p&gt;&#xA;" OwnerUserId="1494" LastActivityDate="2017-08-30T20:13:48.360" Title="What's the name of outside the cosmic horizon?" Tags="&lt;universe&gt;&lt;naming&gt;&lt;cosmological-horizon&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22372" PostTypeId="2" ParentId="22371" CreationDate="2017-08-28T16:30:09.483" Score="8" Body="&lt;p&gt;The outside is also the universe, the inside is just the observable universe.&lt;/p&gt;&#xA;" OwnerUserId="17582" LastActivityDate="2017-08-28T16:30:09.483" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22373" PostTypeId="1" CreationDate="2017-08-28T17:40:28.663" Score="0" ViewCount="317" Body="&lt;p&gt;What is the condition for the number density ($n$) of a gas to be &lt;strong&gt;ultra-relativistic or non-relativistic&lt;/strong&gt; and &lt;strong&gt;degenerate or ideal&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found problems in this subject when I was reading about White-Dwarfs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A book said $n \gg n_{QUR}$ or $n \ll n_{QUR}$ is the condition for being degenerate or ideal (classical), respectively. And $n_{QUR} = 8 \pi (kT/hc)^3$. Is this condition right? I know where $n_{QUR}$ comes from but I don't know what it really shows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that book it is said that $n \ll (mc/h)^3$ is the condition for being ultra-relativistic for a degenerate gas and $kT\gg mc^2$ is the condition for being ultra-relativistic for an ideal gas. Where do these equations come from? What is $mc/h$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In another way, my question is mainly about this diagram and about how we can get the lines in this diagram. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/LlVaX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LlVaX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;sub&gt;&lt;strong&gt;Credit:&lt;/strong&gt; The Physics of Stars, A.C.Phillips&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$k$ Boltzmann constant&lt;/li&gt;&#xA;&lt;li&gt;$h$ Planck constant&lt;/li&gt;&#xA;&lt;li&gt;$c$ light speed&lt;/li&gt;&#xA;&lt;li&gt;$n$ is number &lt;em&gt;density&lt;/em&gt;, not the number of particles&lt;/li&gt;&#xA;&lt;li&gt;$n_{QUR}$ is the quantum concentration number for ultra-relativistic particles&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16441" LastEditorUserId="11143" LastEditDate="2017-08-28T18:24:23.907" LastActivityDate="2017-08-28T19:58:44.273" Title="What is the condition for the number density of a gas to be ultra-relativistic or non-relativistic and degenerate or ideal" Tags="&lt;astrophysics&gt;&lt;white-dwarf&gt;&lt;gas&gt;&lt;degenerate-matter&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22374" PostTypeId="2" ParentId="22373" CreationDate="2017-08-28T19:58:44.273" Score="1" Body="&lt;p&gt;This is fairly standard bookwork that I am not going to reproduce and you should understand that the lines on your diagram are &quot;fuzzy&quot; in the sense that they mark the loci where you can neither use one approximation or another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So first, the division between a classical gas and a quantum gas (note that all the gases in your plot are ideal - the term ideal refers to the particles being point-like and non-interacting, which is as true for the electrons in a white dwarf as those in the atmosphere of the Sun). The condition here is that for a classical gas, the density/temperature combination should be such that the phase-space (3d momentum $\times$ volume) occupied by each particle is much greater than $\hbar^3$. This ensures that there are more available quantum states than indistinguishable electrons to fill them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second criterion is to compare the momentum of particles with $mc$. Clearly if these are comparable then the particles are relativistic. In a classical gas, the momentum depends on the square root of temperature, as you quote. In a Fermi-Dirac quantum gas, the maximum momentum only depends on the density of particles (as $n^{1/3}$). Hence the differing criteria.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-08-28T19:58:44.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22375" PostTypeId="2" ParentId="22355" CreationDate="2017-08-28T20:28:36.093" Score="5" Body="&lt;p&gt;I don't have detailed model calculations to hand and I'm not sure they have been done for the small range of metallicity you mention. However, in handwaving terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Mass and composition are the most important variables. Helium abundance, in addition to metallicity, could play a role. Rotation could be a very second order effect. Very rapid rotation would lower core temperatures at the same mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) This is the general assumption if a star plus brown dwarf companion formed in a binary or even if they formed in the same cluster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Higher metallicity leads to larger opacity in the atmosphere and a larger brown dwarf. This is a small effect in fully convective objects like these, except perhaps when they are very young and have superadiabatic surface layer. However, as this is when D burning takes place, possibly there is a small effect - higher metallicity leads to a larger radius for the same mass and a lower core temperature. So the mass threshold would be slightly higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) The duration of D burning is mass dependent. The graph below, from &lt;a href=&quot;https://arxiv.org/abs/astro-ph/9705201&quot; rel=&quot;nofollow noreferrer&quot;&gt;Burrows et al. (1997)&lt;/a&gt;, shows luminosity vs time for brown dwarfs, planets and stars. The D burning phase is the early plateau in luminosity, lasting a few million years for things which are almost stars to maybe one or two hundred million years at 13 Jupiter masses. When D fusion ceases, then cooling recommences and the luminosity will continue to fade (forever), unless the object is massive enough to ignite hydrogen (i.e. a star: the difference between the green lines and blue lines on the plot).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OoRsc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OoRsc.png&quot; alt=&quot;Burrows 97 calculations&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-08-29T07:04:49.257" LastActivityDate="2017-08-29T07:04:49.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22376" PostTypeId="1" CreationDate="2017-08-28T21:41:09.223" Score="0" ViewCount="5462" Body="&lt;p&gt;I too saw a star like object moving across the sky from Bournemouth towards say Parkstone, it appeared to be a star moving for quite a long time until it disappeared from my view!&#xA;I have seen this the last two nights and so have my family, it's not the Space Station as I see this regularly as my Star Walk app, you can add in when you know what the object is! But unless I know the name of this object I can't track it on my app, does anyone know what it could Be?&#xA;It is not a plane as far too high like a star moving but fast across the sky? This was timed 22:10 on Sunday August 28th 2017&lt;/p&gt;&#xA;" OwnerUserId="18273" LastActivityDate="2017-08-30T12:15:27.293" Title="I too saw a star like object moving across the sky it appeared to be a star moving!" Tags="&lt;identify-this-object&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2017-08-30T11:59:15.387" ContentLicense="CC BY-SA 3.0" />
  <row Id="22378" PostTypeId="2" ParentId="22371" CreationDate="2017-08-28T23:28:21.000" Score="3" Body="&lt;p&gt;Since the universe is expanding, the things (galaxies, quasars, etc) that are currently in the observable region, on our side of the cosmic horizon, will someday be over the horizon. It stands to reason, then, that whatever is currently over the horizon is just the things that were once on our side, but long ago.&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2017-08-28T23:28:21.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22379" PostTypeId="1" AcceptedAnswerId="22384" CreationDate="2017-08-29T01:25:28.413" Score="1" ViewCount="195" Body="&lt;p&gt;I know that astronomers have detected supermassive black holes in the early universe. What was the age of the universe at which these black holes were detected?  In other words how long is it thought to take for these supermassive black holes to form and what are the mechanisms by which they formed, other than from supermassive stars?&lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="16926" LastEditDate="2017-08-29T03:32:29.097" LastActivityDate="2017-08-29T07:30:24.940" Title="Primordial supermassive black holes. How old are they?" Tags="&lt;star&gt;&lt;astrophysics&gt;&lt;supermassive-black-hole&gt;&lt;early-universe&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22383" PostTypeId="2" ParentId="22341" CreationDate="2017-08-29T06:21:48.180" Score="1" Body="&lt;p&gt;Treating as a school question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each change of 1 magnitude changes brightness by how much? So &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnitude_(astronomy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;by what factor will a change of 5 magnitudes give&lt;/a&gt;? (5 magnitudes is a very convenient number for this question)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will need to reduce the area of the telescope by the same factor.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-08-29T06:21:48.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22384" PostTypeId="2" ParentId="22379" CreationDate="2017-08-29T07:30:24.940" Score="2" Body="&lt;p&gt;The first supermassive black holes must have been formed earlier than about 750 million years after the big bang, since there is evidence for luminous quasars at redshifts of up to 7 and with likely black holes of a billion solar masses or more at their centres (e.g. &lt;a href=&quot;https://arxiv.org/abs/1310.7960&quot; rel=&quot;nofollow noreferrer&quot;&gt;Momjian et al. (2013)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The candidate mechanisms to produce such objects in a short period of time are hyper-Eddington accretion onto stellar black hole seeds; the direct collapse of large, primordial gas clouds into black holes; or the merger of stellar-sized black holes in dense clusters, followed by gas accretion. These possibilities are accessibly reviewed by &lt;a href=&quot;https://arxiv.org/abs/1703.03083&quot; rel=&quot;nofollow noreferrer&quot;&gt;Smith, Bromm &amp;amp; Loeb (2017)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-08-29T07:30:24.940" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22387" PostTypeId="2" ParentId="22331" CreationDate="2017-08-29T08:40:33.027" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the Giant Impact Hypothesis explain how the Moon circularized its orbit?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, it does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The giant impact hypothesis says that the Moon formed a handful of Earth radii from the center of the Earth. The Moon currently orbits about 60 Earth radii from the center of the Earth. This means that whether the Moon's initial orbit was highly eccentric or almost circular is irrelevant with regard to the Moon's current orbit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A debris field from any impact would at first follow a very elliptical orbit, with a perigee very close to the surface of the Earth and an apogee much farther away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is incorrect. A debris field will circularize itself rather quickly. Collisions put some particles on an escape trajectory, other particles on a collision trajectory, with what's left tending toward having nearly circular orbits.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-08-29T08:40:33.027" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22388" PostTypeId="1" AcceptedAnswerId="22399" CreationDate="2017-08-29T09:01:01.853" Score="6" ViewCount="1835" Body="&lt;p&gt;How to generate an uniform distribution of stars on the sky ? What I want is make a simulation of random points following uniform distribution on a part of the sky (making the assuption that we have the same amount of stars in any direction). The problem is how I do that because of the cos(dec) on the RA ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I did in python is this&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;RA = np.random.uniform(-180,180)&#xA;Dec = np.random.uniform(-90,90)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But it is totally false, since we will have more points on the poles (if I do many points, this will clearly not be uniform)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot in advance!&lt;/p&gt;&#xA;" OwnerUserId="18281" LastEditorUserId="18281" LastEditDate="2017-08-29T09:34:05.203" LastActivityDate="2017-08-29T16:38:53.570" Title="Generate an uniform distribution on the sky" Tags="&lt;positional-astronomy&gt;&lt;python&gt;&lt;mathematics&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22389" PostTypeId="2" ParentId="22376" CreationDate="2017-08-29T09:09:17.680" Score="1" Body="&lt;p&gt;This is very probably another artificial satellite, there are hundreds of them &lt;/p&gt;&#xA;&#xA;&lt;p&gt;references : &#xA;&lt;a href=&quot;http://www.pixalytics.com/sats-orbiting-earth-2016/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.pixalytics.com/sats-orbiting-earth-2016/&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.universetoday.com/42198/how-many-satellites-in-space/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.universetoday.com/42198/how-many-satellites-in-space/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18281" LastEditorUserId="18281" LastEditDate="2017-08-30T12:15:27.293" LastActivityDate="2017-08-30T12:15:27.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22391" PostTypeId="1" AcceptedAnswerId="22403" CreationDate="2017-08-29T09:31:31.667" Score="6" ViewCount="2213" Body="&lt;p&gt;While looking up at the stars I wondered, what are the odds that some of them are already dead. I did some research and found a very interesting article &lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2013/08/13/are_the_stars_you_see_in_the_sky_already_dead.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. It states that the odds are rather small, so it is not like about 50% of the stars we see are already dead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I still wonder if there is an example of a star we know to be already gone, while it is still visible on earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, the only way to find out is to watch the stars light. So if the light &quot;goes out&quot;, we know the star has been gone for quite some time. Assuming this to be fact, I would have to conclude that such an example does not (and cannot) exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true? Or is there in fact a star that is already dead that we know of, which can be seen in the sky?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: By &quot;seeing the star&quot; I am talking about a star that looks just like any other star in the night sky (not, for instance, a visible supernova).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: If there is such a star, how did we find out it is already gone?&lt;/p&gt;&#xA;" OwnerUserId="16218" LastEditorUserId="1891" LastEditDate="2019-04-21T01:02:42.003" LastActivityDate="2019-04-21T01:02:42.003" Title="Is there an example of a star that is already dead, but can still be seen on Earth?" Tags="&lt;star&gt;&lt;light&gt;&lt;supernova&gt;&lt;night-sky&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22392" PostTypeId="2" ParentId="22391" CreationDate="2017-08-29T10:00:18.873" Score="4" Body="&lt;p&gt;Maybe Betelgeuse in the Orion constellation. It seems it is nearly at the end of its life and it is at around 600 light years from us. (But well it can also be in something like  greater than 10000yrs of lifetime so it is difficult to say (still very low at astronomical scales)...)&lt;/p&gt;&#xA;" OwnerUserId="18281" LastActivityDate="2017-08-29T10:00:18.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22393" PostTypeId="2" ParentId="22388" CreationDate="2017-08-29T10:15:15.670" Score="1" Body="&lt;p&gt;This is really more computer science or mathematics than astronomy, but here's a function &lt;code&gt;random_point_on_unit_sphere()&lt;/code&gt;  that creates random points on a unit 3D sphere, which is then used by another function &lt;code&gt;random_point_on_sky()&lt;/code&gt; to convert to RA and dec values, using &lt;a href=&quot;http://www.astropy.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;astropy&lt;/a&gt;. Finally, the function &lt;code&gt;print_random_star_coords()&lt;/code&gt; print a number of RA,dec pairs.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;from astropy.coordinates import SkyCoord&#xA;from astropy import units as u&#xA;&#xA;def random_point_on_unit_sphere():&#xA;    while True:&#xA;        R   = np.random.rand(3) #Random point in box&#xA;        R   = 2*R - 1&#xA;        rsq = sum(R**2)&#xA;        if rsq &amp;lt; 1: break       #Use r only if |r|&amp;lt;1 in order not to favor corners of box&#xA;    return R / np.sqrt(rsq)     #Normalize to unit vector&#xA;&#xA;def random_point_on_sky():&#xA;    p     = random_point_on_unit_sphere()&#xA;    r     = np.linalg.norm(p)&#xA;    theta = 90 - (np.arccos(p[2] / r)    / np.pi * 180)            #theta and phi values in degrees&#xA;    phi   =       np.arctan(p[1] / p[0]) / np.pi * 180&#xA;    c     = SkyCoord(ra=phi, dec=theta, unit=(u.degree, u.degree)) #Create coordinate&#xA;    return c.ra.hms, c.dec.deg                                     #Many different formats are possible, e.g c.ra.hour for decimal hour values&#xA;&#xA;def print_random_star_coords(nstars):&#xA;    for n in range(nstars):&#xA;        RA,dec = random_point_on_sky()&#xA;        print(RA,dec)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2017-08-29T11:27:25.407" LastActivityDate="2017-08-29T11:27:25.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22395" PostTypeId="2" ParentId="22391" CreationDate="2017-08-29T10:24:41.223" Score="7" Body="&lt;p&gt;You are, I think, asking two very similar but linked questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Are there stars - in our galaxy, as stars in other galaxies are not individually visible - that are 'dead' but still alight to us. The Slate article you linked is rather good in this respect: i) the lifespan of a given star is much longer than the time taken for light to reach us so it is unlikely that this is the case and ii) most stars have an extended extinction compared to the time taken for light to reach us, with few going out in a supernova flash. Hence it is pretty unlikely that this is the case at the moment - but it will happen on occasion&lt;/li&gt;&#xA;&lt;li&gt;If there is a specific star which is 'dead' but still alight to us - could we know? Not really, as no useful information will be travelling to us faster than the speed of light. The best we might do is to observe a supernova and then say &quot;this event happened x years ago&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17502" LastActivityDate="2017-08-29T10:24:41.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22397" PostTypeId="2" ParentId="22388" CreationDate="2017-08-29T11:09:52.617" Score="6" Body="&lt;p&gt;Random points on the surface of a sphere can be generated by allowing the azimuthal angle $\phi$ to take a uniformly distributed random value between 0 and $2\pi$. To convert this to RA in degrees you multiply by $180/\pi$. To convert to hours, minutes and seconds you divide the $\phi$ in degrees by 15, which gives the hours, divide the remainder by 60 which gives the minutes and then divide the remainder of that by 60 to give the seconds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The declination angle $\delta$ is not uniformly distributed, because the area covered at an angle $\delta$ goes as $\cos \delta$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the potential confusion here between declination and the polar angle $\theta$, which is conventionally measured &lt;em&gt;down&lt;/em&gt; from the pole. The area of a strip of width $\Delta \theta$ at a given $\theta$ on a unit sphere is $$\Delta A = 2\pi  \sin \theta\ \Delta \theta = 2\pi  \cos (\pi/2 - \delta)\ \Delta \delta = 2\pi  \cos \delta\ \Delta \delta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To translate this into something we can use, we say that the probability of finding a star in a small range of declinations&#xA;$$ dP \propto \cos \delta\ d\delta,$$&#xA;so the cumulative probability of finding a value of $\delta$ between $-90^{\circ}$ and $\delta$ is&#xA;$$ P(\delta) = \frac{\int^{\delta}_{-90^{\circ}}\cos \delta^{\prime}\ d\delta^{\prime}}{\int^{+90^{\circ}}_{-90^{\circ}}\cos \delta^{\prime}\ d\delta^{\prime}},$$&#xA;where the denominator makes sure the proportionality is correctly normalised. From this, we obtain&#xA;$$P(\delta) = \frac{1}{2}(1 + \sin \delta) \tag*{(1)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The procedure for then obtaining a random declination is to assign a uniform random number to the cumulative probability $P$ between 0 and 1. Then from equation (1) we have&#xA;$$ \sin \delta = 2P -1$$&#xA;$$\delta = \sin^{-1}(2P-1),$$&#xA;where $P$ is a random number between 0 and 1.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-08-29T15:50:44.580" LastActivityDate="2017-08-29T15:50:44.580" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22398" PostTypeId="2" ParentId="22263" CreationDate="2017-08-29T11:19:53.180" Score="1" Body="&lt;p&gt;Any book on Cosmology published prior to 2000 does not contain the results or discussion thereof of the WMAP and Planck surveys of the cosmic microwave background. These constitute &lt;em&gt;the&lt;/em&gt; best constraints on cosmological theories, including the relative contributions of dark energy, dark matter and the flatness or otherwise of the universe, and the best supporting evidence for the big bang model and for inflationary ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 1994 book &lt;em&gt;may&lt;/em&gt; not even contain the Nobel prize-winning esults on the cosmic microwave background by COBE, which was the first to see in detail the ripples that the later missions used to perform precision cosmology.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-08-29T11:19:53.180" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22399" PostTypeId="2" ParentId="22388" CreationDate="2017-08-29T12:03:23.933" Score="4" Body="&lt;p&gt;Heres more python than you can shake a telescope at. I just used &lt;a href=&quot;https://i.stack.imgur.com/G1rdk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;@RobJeffries' algorithm&lt;/a&gt;. This is just a python script, the real answer to the question is &lt;a href=&quot;https://i.stack.imgur.com/G1rdk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;@RobJeffries' answer&lt;/a&gt; and I've just scripted it. The mathematics behind generating statistically uniform distributions is explained very nicely there as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Python is below the plots. You can see on an X-Y graph they thin out at the top and bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 3D plot is live, you can use your cursor and move the sphere around. Of course not HERE :-), but in your own python window if you run the script.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G1rdk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G1rdk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/avktz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/avktz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;&#xA;from mpl_toolkits.mplot3d import Axes3D&#xA;&#xA;halfpi, pi, twopi = [f*np.pi for f in [0.5, 1, 2]]&#xA;&#xA;degs, rads = 180/pi, pi/180&#xA;&#xA;# do radians first, then convert later&#xA;&#xA;nstars = 2000&#xA;&#xA;ran1, ran2 = np.random.random(2*nstars).reshape(2, -1)&#xA;&#xA;RA  = twopi * (ran1 - 0.5)&#xA;dec = np.arcsin(2.*(ran2-0.5)) # Hey Barry Carter!&#xA;&#xA;funcs = [np.cos, np.sin]&#xA;&#xA;cosRA,  sinRA  = [f(RA)  for f in funcs]&#xA;cosdec, sindec = [f(dec) for f in funcs]&#xA;&#xA;x = cosRA * cosdec&#xA;y = sinRA * cosdec&#xA;z = sindec&#xA;&#xA;decs = (np.arange(11)-5) * halfpi / 6.&#xA;RAs  = (np.arange(12)-5) * halfpi / 6.&#xA;&#xA;theta = np.linspace(0, twopi, 101)&#xA;&#xA;costh, sinth = [f(theta) for f in funcs]&#xA;zerth = np.zeros_like(theta)&#xA;&#xA;fig = plt.figure()&#xA;&#xA;ax  = fig.add_subplot(1, 1, 1, projection='3d')&#xA;&#xA;ax.plot(x, y, z, '.k')&#xA;&#xA;# lines of declination&#xA;xvals = costh * np.cos(decs)[:, None]&#xA;yvals = sinth * np.cos(decs)[:, None]&#xA;zvals = zerth + np.sin(decs)[:, None]&#xA;for x, y, z in zip(xvals, yvals, zvals):&#xA;    plt.plot(x, y, z, '-g', linewidth=0.8)&#xA;&#xA;# lines of Right Ascention&#xA;xvals = costh * np.cos(RAs)[:, None]&#xA;yvals = costh * np.sin(RAs)[:, None]&#xA;zvals = sinth + np.zeros_like(RAs)[:, None]&#xA;for x, y, z in zip(xvals, yvals, zvals):&#xA;    plt.plot(x, y, z, '-r', linewidth=0.8)&#xA;&#xA;ax.set_xlim(-1.1, 1.1)&#xA;ax.set_ylim(-1.1, 1.1)&#xA;ax.set_zlim(-1.1, 1.1)&#xA;&#xA;ax.view_init(elev=30, azim=15)&#xA;&#xA;plt.show()&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;    plt.plot(degs*RA, degs*dec, '.k')&#xA;    plt.xlim(-180, 180)&#xA;    plt.ylim(-90, 90)&#xA;    plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-08-29T16:38:53.570" LastActivityDate="2017-08-29T16:38:53.570" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22402" PostTypeId="2" ParentId="22217" CreationDate="2017-08-29T14:02:53.480" Score="2" Body="&lt;p&gt;I found some interesting information &lt;a href=&quot;http://www.arrl.org/files/file/Technology/pdf/119962.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;in this vulgarization paper&lt;/a&gt; by Ian Poole.&lt;/p&gt;&#xA;&lt;p&gt;A first point is electron density in ionosphere changes between day and night, so the resulting bend will be different:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/oWsj3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oWsj3.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;This &lt;a href=&quot;http://www.astrosurf.com/luxorion/qsl-mars-communication.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;very interesting site explains&lt;/a&gt; notably that there is a&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;cut-off frequency for the ionosphere beyond which it loses its capacity to reflect shortwaves. Depending the latitude, the season and the solar activity mainly, during the day this frequency is  around 3-10 MHz and goes down to about 2-6 MHz during the night&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The article includes an illustration of the transition from angular deflection to complete reflection as a function of angle (click for full size):&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/z4H7O.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z4H7Om.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;caption:&lt;/strong&gt; Space wave, ground wave, and ionospheric waves. Above a critical angle, waves escape in free space while waves emitted under a low incidence angle can reach very far countries. This is valid between approx. 1-500 MHz.&lt;/p&gt;&#xA;&lt;p&gt;But the best site I found on the subject &lt;a href=&quot;http://www.cv.nrao.edu/course/astr534/Introradastro.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;is that one.&lt;/strong&gt;&lt;/a&gt; It states that&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Ionospheric reflection (not absorption) prevents photons with wavelengths &amp;gt; 30 m (f&amp;lt; 10 MHz) to reach the ground&#xA;[...]&lt;/p&gt;&#xA;&lt;p&gt;Total internal reflection in the ionosphere at longer wavelengths makes the Earth look like a silvery ball from space, like the glass face of an underwater wristwatch viewed obliquely.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It goes on saying the atmosphere is not perfectly transparent at any radio frequency. And moreover it adds noise. It explains why the best sites for radio observation at higher frequencies are exceptionally high and dry.&lt;/p&gt;&#xA;" OwnerUserId="11264" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-09-03T03:50:35.763" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22403" PostTypeId="2" ParentId="22391" CreationDate="2017-08-29T18:02:08.933" Score="12" Body="&lt;p&gt;There is an accepted answer already, but there is a couple known cases of a star we know has gone supernova, and yet we can still see it.  &lt;a href=&quot;https://www.space.com/31418-hubbles-einstein-cross-supernova-strikes-back.html&quot; rel=&quot;noreferrer&quot;&gt;This source&lt;/a&gt; describes one such unique circumstance.  The star that exploded happened to be in a galaxy that was behind another massive one from our point of view.  The alignment was just right such that the light formed an &lt;a href=&quot;https://en.wikipedia.org/wiki/Einstein_Cross&quot; rel=&quot;noreferrer&quot;&gt;Einstein Cross&lt;/a&gt;.  The light from each point on the cross takes a different path to get here, and each of those paths are different lengths.  Thus, the different points on the cross show the star at 4 different times in its final years.  The Space.com article I linked was written in 2015. Scientists first noticed the supernova in 2014, and each image of the supernova arrived within a year of when it was first noticed.  If I'm understanding that article correctly, as of now (2017) all of the images of that supernova show the supernova aftermath, and not the star beforehand.  However, there was a period of time where we could see the star both pre and post-supernova at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You imply in your question that you are focusing on stars within our own galaxy.  I don't know of any such situations of Einstein Cross events closer to home that let us do this.&lt;/p&gt;&#xA;" OwnerUserId="14255" LastEditorUserId="14255" LastEditDate="2017-08-29T19:14:45.050" LastActivityDate="2017-08-29T19:14:45.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22404" PostTypeId="2" ParentId="22391" CreationDate="2017-08-29T18:14:59.760" Score="8" Body="&lt;p&gt;Most high-redshift galaxies — i.e. the &lt;em&gt;very&lt;/em&gt; distant ones, &lt;em&gt;billions&lt;/em&gt; of light years away — are detected due to the UV light that comes primarily from very hot stars (in particular, copious amounts of Lyman $\alpha$ photons at 121.6 nm are emitted from the gas surrounding the hot stars which is ionized and then recombines). Because these stars &quot;burn their fuel&quot; so fast, they typically live only for a few to a few tens of &lt;em&gt;millions&lt;/em&gt; of years. That means that those stars for sure are long gone.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2017-08-29T18:14:59.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22405" PostTypeId="1" CreationDate="2017-08-29T20:19:44.740" Score="2" ViewCount="238" Body="&lt;p&gt;The Saha Equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{N_{i+1}}{N_{i}} = \frac{g_{i+1}}{g_{i}}\frac{g_e}{N_e h^3}(2 \pi m_{e} kT)^{3/2} \exp(- \chi_i / kT)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\chi_i$ is ionization potential. My problem is about the sign ($+$ or $-$) of $\chi_i$. I know that its formula is something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\chi_i = (13.6\ \mathrm{eV}) Z^2 \left(\frac{1}{n_i^2} - \frac{1}{n_j^2}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and in this case $n_j$ is $\infty$. But I get confused about it's sign ($+$ or $-$). Can someone simply say what is the sign of the formula for ionization potential and the sign of the formula for the Energy we need to go from $n_i$ to $n_j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean: What is the complete formula for $E_m - E_n$ and also $\chi_{n}$? (with the right sign so they can be used in the first Saha Equation above)&lt;/p&gt;&#xA;" OwnerUserId="16441" LastEditorUserId="11143" LastEditDate="2017-08-30T13:22:29.320" LastActivityDate="2017-08-30T13:39:44.587" Title="A simple question about Saha Equation" Tags="&lt;astrophysics&gt;&lt;stellar-astrophysics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22406" PostTypeId="1" CreationDate="2017-08-29T21:17:09.173" Score="1" ViewCount="454" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/131/is-the-moon-moving-further-away-from-earth-and-closer-to-the-sun-why&quot;&gt;Here&lt;/a&gt; it has been discussed why the moon is moving away from earth at a rate of 1.5 inches per year. Is there any way we can estimate the value (e.g, 1.5 inches per year) mathematically? &lt;/p&gt;&#xA;" OwnerUserId="18291" LastActivityDate="2018-06-18T09:24:14.997" Title="Estimating the Rate at which Moon is Moving Away from the Earth" Tags="&lt;the-moon&gt;&lt;gravity&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22407" PostTypeId="1" CreationDate="2017-08-29T21:31:26.480" Score="0" ViewCount="137" Body="&lt;p&gt;I found an answer at &#xA;&lt;a href=&quot;https://physics.stackexchange.com/questions/26795/how-to-calculate-the-amount-of-night-time-during-a-flight&quot;&gt;https://physics.stackexchange.com/questions/26795/how-to-calculate-the-amount-of-night-time-during-a-flight&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I was stuck at the part:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cosθ_1=\sinφ_1\sinφ+\cosφ_1\cosφ\cos(λ−λ_1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cosθ_2=\sinφ_2\sinφ+\cosφ_2\cosφ\cos(λ−λ_2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which  $ (λ(t),φ(t))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be derived (after some tedious calculations).&lt;/p&gt;&#xA;" OwnerUserId="18292" LastEditorUserId="2365" LastEditDate="2017-08-29T22:22:59.487" LastActivityDate="2017-12-29T18:36:57.730" Title="calculate sunset in flight" Tags="&lt;astrophysics&gt;&lt;mathematics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22411" PostTypeId="2" ParentId="22217" CreationDate="2017-08-30T08:02:29.810" Score="4" Body="&lt;p&gt;The refractive deviations in position are very similar for both radio and optical astronomy, until you consider very low frequency radio waves (&lt;span class=&quot;math-container&quot;&gt;$&amp;lt;200$&lt;/span&gt; MHz) when the effect becomes rapidly larger.&lt;/p&gt;&#xA;&lt;p&gt;For plane parallel refraction an approximation for the deviation you are talking about is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\Delta \theta \simeq (n-1) \cot \theta,$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; is the observed elevation, &lt;span class=&quot;math-container&quot;&gt;$\Delta \theta$&lt;/span&gt; is the change in elevation from its true value due to refraction and &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is the refractive index averaged over airmass.&lt;/p&gt;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.gb.nrao.edu/%7Ermaddale/Education/AtmosphereWeatherForRadioAstronomy.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this source&lt;/a&gt; from the Green bank radio telescope, they use something like this, with an added model for how &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; varies with height, scaled by the atmospheric pressure. The largest value of &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; quoted is 1.00031 at ground level. This is basically the same as the refractive index of air at visible wavelengths and amounts to about 60 arcseconds at elevations of 45 degrees.&lt;/p&gt;&#xA;&lt;p&gt;So, to my surprise, the effects of refraction on radio telescope pointing are similar to those for optical telescopes. It simply turns out that the real part of the refractive index (that controls the phase velocity of light and hence refraction) is just as close to 1 for radiowaves as it is for visible light.&lt;/p&gt;&#xA;&lt;p&gt;Here is &lt;a href=&quot;https://www.fig.net/resources/proceedings/fig_proceedings/fig_2002/Js28/JS28_rueger.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;another source&lt;/a&gt; that gives some algorithms to calculate the effective (small) real refractive index for radio waves, with similar results.&lt;/p&gt;&#xA;&lt;p&gt;This source claims the calculations are valid for mm-wavelengths and upwards. Of course they cannot be valid as one approaches the ionospheric cut-off at around 40 MHz (wavelengths of metres), where the refractive index will depart sharply from unity and the deflections must get commensurately larger.&lt;/p&gt;&#xA;&lt;p&gt;I did manage to find something about positional refraction at these low frequencies. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Low-Frequency_Array_(LOFAR)&quot; rel=&quot;nofollow noreferrer&quot;&gt;LOFAR&lt;/a&gt; radio array can apparently do work down to frequencies as low as 10MHz, but the practical limit may be a little higher. Anyway, they do need to account for refraction in the ionosphere and I found &lt;a href=&quot;http://lwa.phys.unm.edu/chicago08/talks/anderson_software.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this presentation&lt;/a&gt;, which contains a section on refraction and in particular, the figure below.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4auFC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4auFC.png&quot; alt=&quot;Refraction at low frequencies&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Thus for low frequency radio astronomy (&amp;lt;200 MHz), refraction is certainly a bigger effect than in the optical. e.g. At elevations of 45 degrees, the refracted positions are shifted by about 0.1 and 0.4 &lt;em&gt;degrees&lt;/em&gt; at 50MHz and 30MHz respectively.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2020-09-22T06:54:36.450" LastActivityDate="2020-09-22T06:54:36.450" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="22414" PostTypeId="1" CreationDate="2017-08-30T09:24:50.983" Score="3" ViewCount="567" Body="&lt;p&gt;The three principal moments of inertia of Earth and Moon are reported, for example, in T. J. Ahrens, &lt;em&gt;Global Earth Physics - A handbook of Physical Constants&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found reports of the moment of inertia factor $\dfrac{C}{M R^2}$, where $C$ is the polar moment of inertia, $M$ the mass and $R$ the mean radius, for planets of the solar system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the three principal moments of inertia for the planets of the solar system known?&lt;/p&gt;&#xA;" OwnerUserId="16581" LastEditorUserId="15641" LastEditDate="2017-09-01T21:04:00.257" LastActivityDate="2017-11-01T21:56:15.770" Title="Moments of inertia of planets" Tags="&lt;planet&gt;&lt;orbital-mechanics&gt;" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="22418" PostTypeId="1" CreationDate="2017-08-30T13:00:34.920" Score="4" ViewCount="1224" Body="&lt;p&gt;How does the bar in a barred spiral galaxy form? What prevents it from being spirals all the way down like water spiralling down a plughole?&lt;/p&gt;&#xA;" OwnerUserId="1494" LastActivityDate="2017-09-05T20:16:24.897" Title="How is the bar formed in a Barred-Spiral galaxy?" Tags="&lt;galaxy&gt;&lt;formation&gt;&lt;spiral-arms&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22419" PostTypeId="2" ParentId="22406" CreationDate="2017-08-30T13:13:00.270" Score="2" Body="&lt;p&gt;This is such a complex problem that I think your chances of getting an accurate answer are almost nil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the equations are likely to be highly complex if you want to be accurate down to the tenth of an inch. For example, you'd need to model the exact shape of the Earth due to the Moon's tides, most likely accurate down to the meter or so. Scientists have done this &lt;em&gt;on average&lt;/em&gt;, but finding how these types of models minutely change on an hourly basis is near impossible. The data just doesn't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, filling in the variables you'll have in your equations is likely going to be impossible because many won't be known. For example, do you know the flexure/elasticity/tensile strength/etc. of the entire Earth's crust at every point on Earth over time? My guess is you don't. You'd need to know that though to do really accurate calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way we know the recession rate of the Moon is through measurements. Crude models might get you crude order of magnitude numbers, but to match the precision of observations would require models that are far more informed than we currently are.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-08-30T13:13:00.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22420" PostTypeId="2" ParentId="22405" CreationDate="2017-08-30T13:39:44.587" Score="1" Body="&lt;p&gt;Just to provide a &lt;a href=&quot;https://en.wikipedia.org/wiki/Ionization_energy#Electrostatic_explanation&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;, the Ionization potential for a &lt;strong&gt;hydrogen-like&lt;/strong&gt; atom is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\chi_n = \frac{(13.6\:\mathrm{eV})Z^2}{n^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the energy necessary to ionize an electron in a hydrogen-like atom that &quot;begins&quot; at the $n^\mathrm{th}$ energy level. This equation is positive because you need to &lt;em&gt;input&lt;/em&gt; energy to the electron to break it out of its potential well. Note that you may see a nearly identical equation for the energy of the electrons which will be negative. That implies the electrons themselves have negative energy because they're in the electric potential well and you need a positive amount of energy to ionize them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when you look at the Saha equation, you see a negative sign in the exponential term: $\exp(-\chi_i/kT)$. It was already established that $\chi_i$ is positive and so also are $k$ and $T$ simply by their nature. The negative sign comes from the derivation and meaning of this exponential term. Describing the full derivation would involve an entire lecture, but to point you along the right track, the exponential term is known as the &lt;em&gt;Boltzmann Factor&lt;/em&gt; and ultimately comes from the fact that particle energies are very often distributed according to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Boltzmann_distribution&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boltzmann distribution&lt;/a&gt;. That negative sign is simply an artifact of the Boltzmann distribution concerning how particles like to probabilistically arrange their energies. If that sign was a positive, the universe would just blow up so be careful with your signs!&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-08-30T13:39:44.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22423" PostTypeId="2" ParentId="1855" CreationDate="2017-08-30T15:52:39.317" Score="0" Body="&lt;p&gt;I'm not sure if this is relevant for you, but it was for me. This will calculate true elevation, but it will not be the same as what is observed due to atmospheric refraction. There is a Meeus solar calculator and that uses this to correct for that:&#xA;1.02 / (60 * tan((alpha + (10.3/(alpha + 5.11))) * pi/180))&lt;/p&gt;&#xA;" OwnerUserId="18039" LastActivityDate="2017-08-30T15:52:39.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22424" PostTypeId="2" ParentId="1895" CreationDate="2017-08-30T17:08:23.173" Score="1" Body="&lt;p&gt;user68441 is in error.  We are not part of the Sagittarius Dwarf Elliptical Galaxy.  Our velocity perpendicular to the galactic plane is only 4.95±0.09&#xA;km/s with respect to the Local Standard of rest.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each cycle takes about 70 million years with an amplitude of 100pc (Matese et al. 1995), it will be roughly 30 million years before we cross the plane again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the thickness of the disk is roughly 1000LY, the solar system will stay within the disk, unless perturbed by an interaction with another star.&lt;/p&gt;&#xA;" OwnerUserId="17340" LastActivityDate="2017-08-30T17:08:23.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22426" PostTypeId="2" ParentId="22371" CreationDate="2017-08-30T19:24:50.560" Score="-2" Body="&lt;p&gt;This relates to what I was saying about &lt;em&gt;the universe was once the size of a grapefruit&lt;/em&gt; morphing into &lt;em&gt;the observable universe was once the size of a grapefruit&lt;/em&gt;. Along with &lt;em&gt;the universe is infinite&lt;/em&gt;. IMHO there is no evidence for the latter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be pedantic, I guess the answer to your question is &quot;the non-observable universe&quot;. That’s because the word universe is &lt;a href=&quot;http://www.etymonline.com/index.php?term=universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;derived&lt;/a&gt; from “uni” as in unicycle and “verse” as in vice versa. It means &lt;em&gt;turned into one&lt;/em&gt;. It means &lt;em&gt;everything&lt;/em&gt;. Hence the word &quot;multiverse&quot; doesn't float my boat.  &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2017-08-30T19:24:50.560" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22429" PostTypeId="1" AcceptedAnswerId="30349" CreationDate="2017-08-31T04:17:00.973" Score="7" ViewCount="567" Body="&lt;p&gt;In Monterey, California, with a &amp;quot;half-moon&amp;quot; high in the sky at about 9 pm (2017-08-30), the moon had a distinct red outline on the outer edge of the NE quadrant; for awhile I saw it also on the outer edge of the SW quadrant.&lt;/p&gt;&#xA;&lt;p&gt;I observed this phenomena from two locations, about a mile distant from each other.&lt;/p&gt;&#xA;&lt;p&gt;I've never seen that before. What would cause it?&lt;/p&gt;&#xA;&lt;p&gt;I took a picture of it with my camera, but you can't tell anything &amp;quot;out of the ordinary&amp;quot; from the image.&lt;/p&gt;&#xA;&lt;h2&gt;UPDATE&lt;/h2&gt;&#xA;&lt;p&gt;It's a clunky picture (I've been watching &lt;em&gt;My Three Sons&lt;/em&gt; lately) - red not visible, fuzzy, and makes it like a full moon, but here it is, at any rate:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JXAiQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JXAiQ.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18308" LastEditorUserId="31410" LastEditDate="2021-05-27T13:54:54.937" LastActivityDate="2021-05-27T13:54:54.937" Title="Why did the moon have a distinct reddish edge tonight?" Tags="&lt;the-moon&gt;&lt;moon-phases&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="22435" PostTypeId="1" AcceptedAnswerId="22436" CreationDate="2017-08-31T13:20:53.203" Score="5" ViewCount="1506" Body="&lt;p&gt;Here's what Canis Minor looks like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://www.universetoday.com/wp-content/uploads/2008/10/cmsky_cortner_lab-571x432.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, all I see is two stars. Canis Minor is a constellation of the giant Orion's hunting dog. Two stars can give quite a meaning. Two stars can mean a stick, a telescope, &lt;em&gt;et cetera&lt;/em&gt;. But how could that be interpreted as a dog?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is a Greek constellation, are there any sources that explain why ancient Greeks considered those two stars to represent a dog?&lt;/p&gt;&#xA;" OwnerUserId="16636" LastActivityDate="2017-09-04T03:03:24.580" Title="Exactly how does Canis Minor look like a dog?" Tags="&lt;star&gt;&lt;constellations&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22436" PostTypeId="2" ParentId="22435" CreationDate="2017-08-31T13:24:03.077" Score="8" Body="&lt;p&gt;Quoting from wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Canis Minor was one of the original 48 constellations formulated by Ptolemy in his second-century Almagest, in which it was defined as a specific pattern (asterism) of stars; Ptolemy identified only two stars and hence no depiction was possible. The Ancient Greeks called the constellation &lt;em&gt;προκυων/Procyon&lt;/em&gt;, &quot;coming before the dog&quot;, transliterated into Latin as &lt;em&gt;Antecanis&lt;/em&gt;, &lt;em&gt;Praecanis&lt;/em&gt;, or variations thereof, by Cicero and others. Roman writers also appended the descriptors &lt;em&gt;parvus&lt;/em&gt;, &lt;em&gt;minor&lt;/em&gt; or &lt;em&gt;minusculus&lt;/em&gt; (&quot;small&quot; or &quot;lesser&quot;, for its faintness), &lt;em&gt;septentrionalis&lt;/em&gt; (&quot;northerly&quot;, for its position in relation to Canis Major), &lt;em&gt;primus&lt;/em&gt; (rising &quot;first&quot;) or &lt;em&gt;sinister&lt;/em&gt; (rising to the &quot;left&quot;) to its name Canis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It looks like the original (Greek) identification of the constellation by Ptolemy didn't attribute anything to it, let alone a dog. But over time, the constellation became known based on its proximity to Canis Major. Originally known as Procyon meaning &quot;coming before the dog&quot;, i.e., it rises before Canis Major. This relation to Canis Major seems to have led it to being considered as a representation of a dog as well. The Minor attribute appears to be due to the fact that it is fainter overall than Canis Major.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really though, you can't pick apart constellation patterns too much. Much of them don't look that much like their attributions and only evolved over thousands of years of stories and imagination. It's not like one guy sat down one day and said, what does this pattern of stars remind me of, and that was that.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2017-08-31T13:36:06.053" LastActivityDate="2017-08-31T13:36:06.053" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22438" PostTypeId="2" ParentId="22407" CreationDate="2017-08-31T16:35:23.957" Score="1" Body="&lt;p&gt;After choosing an arbitrary time t, you have the two equations you reference and two unknowns: $ λ$ and $φ$. The problem is that the equations cannot be solved directly, so an iterative approach is required. For example, &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;assume that $(λ−λ_1)=θ_1$. Note that this also lets you calculate $λ$&lt;/li&gt;&#xA;&lt;li&gt;solve for $φ$ from the first equation&lt;/li&gt;&#xA;&lt;li&gt;use that value of $φ$ in the second equation to solve for $λ$&lt;/li&gt;&#xA;&lt;li&gt;If the values of $λ$ from equation 1 and 2 are different, try another value for $λ$ in the first equation.&lt;/li&gt;&#xA;&lt;li&gt;repeat until the assumed value for $λ$ in equation 1 and the calculated value from equation 2 are equal.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It has been a long time since I did spherical trig, but I think the solution on the Physics SE is inefficient. It looks like they are calculating the sun's altitude at multiple point (multiple times of t) along the path, so the two equations two unknowns needs to be solved multiple times -- that is inefficient. It should be possible to determine the equation of the great circle through the departure and destination point. Essentially you need to calculate the longitude of where it cross the equator and the inclination. That probably needs to be solved iteratively, but it only needs to be solved once. Then for any time t, the position $λ$ and $φ$ can be calculated directly, and the sun's altitude can be calculated directly. &lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-08-31T16:35:23.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22440" PostTypeId="1" CreationDate="2017-08-31T17:28:48.023" Score="0" ViewCount="122" Body="&lt;p&gt;According to the theory of relativity. The speed of light is same for every observer. But how can we know that this applies to more than a local region? What if light travelled to us from a distant region where the speed of light was different or where the relativistic principle did not apply?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we design an experiment that refutes both these ideas?&lt;/p&gt;&#xA;" OwnerUserId="18170" LastEditorUserId="2365" LastEditDate="2017-08-31T22:08:45.850" LastActivityDate="2017-08-31T22:08:45.850" Title="is this really true that speed of light is same as we look from different parts of universe?" Tags="&lt;light&gt;&lt;special-relativity&gt;&lt;cosmological-principle&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22441" PostTypeId="2" ParentId="22363" CreationDate="2017-08-31T20:28:02.397" Score="0" Body="&lt;p&gt;Tidal forces (that is, the difference in gravitational pull from one side on an object compared to the other) can slow the body's rotation. For example, the near side of the Moon feels Earth's gravity stronger than the far side, because it's closer. This tidal friction robs the orbiting body of angular momentum, slowing its rotation. The long-term effect of this, is that the orbiting body will slow enough that one side will always face the planet. What's more, the tidal force will also dampen out any wobble in the orbiting body. The Moon does wobble a little as it orbits the Earth -- this is called libration -- so that in total we see a little more than 50% of the Moon's surface throughout the month. &lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2017-08-31T20:28:02.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22442" PostTypeId="2" ParentId="22440" CreationDate="2017-08-31T21:38:32.267" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;According to theory of relativity . Light speed is same for every observer . How can we prove it is right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By doing experiments. Pela mentioned the &lt;a href=&quot;https://simple.wikipedia.org/wiki/Michelson%E2%80%93Morley_experiment&quot; rel=&quot;nofollow noreferrer&quot;&gt;Michelson-Morley experiment&lt;/a&gt;. That's what got Einstein thinking. That's why he came up with the theory of relativity. However other experiments will prove that the speed of light is &lt;em&gt;not&lt;/em&gt; the same for every observer. It's the same for every &lt;em&gt;inertial&lt;/em&gt; observer, because of the wave nature of matter. But see &lt;a href=&quot;http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/speed_of_light.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Is The Speed of Light Everywhere the Same?&lt;/a&gt; on the PhysicsFAQ website. It's written by &lt;a href=&quot;http://www.math.ucr.edu/home/baez/physics/cv.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Don Koks&lt;/a&gt;, a physicist in Adelaide. He says this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;That the speed of light depends on position when measured by a non-inertial observer is a fact routinely used by laser gyroscopes that form the core of some inertial navigation systems.  These gyroscopes send light around a closed loop, and if the loop rotates, an observer riding on the loop will measure light to travel more slowly when it traverses the loop in one direction than when it traverses the loop in the opposite direction.  This is known as the Sagnac Effect.  The gyroscope does employ such an observer: it is the electronics that sits within the gyro.  This electronic observer detects the difference in those light speeds, and attributes that difference to the gyro's not being inertial: it is accelerating within some inertial frame.  That measurement of an acceleration allows the body's orientation to be calculated, which keeps it on track and in the right position as it flies&quot;.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your measurement of the speed of light is not totally constant, it depends on your acceleration. If you were in a room in an accelerating rocket, you would be what Koks called the &quot;global&quot; observer, and you would say that light at the ceiling travels faster than light down near the floor.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But how can we know that this applies to more than a local region? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By understanding general relativity. See the general relativity section of the PhysicsFAQ article: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Einstein talked about the speed of light changing in his new theory.  In the English translation of his 1920 book &quot;Relativity: the special and general theory&quot; he wrote: &quot;according to the general theory of relativity, the law of the constancy of the velocity [—Either Einstein or his translator obviously mean &quot;speed&quot; here, since velocity (a vector) is not in keeping with the rest of his sentence. People often say &quot;velocity&quot; when they clearly mean &quot;speed&quot;.] of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity [...] cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity [speed]  of propagation of light varies with position.&quot; This difference in speeds is precisely that referred to above by ceiling and floor observers.&quot;&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What if light travelled to us from a distance region where the speed of light was different or where the relativistic principle did not apply? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You wouldn't measure any difference, because you measure the speed of light here, not somewhere else. &lt;/p&gt;&#xA;" OwnerUserId="9772" LastEditorUserId="9772" LastEditDate="2017-08-31T21:45:42.100" LastActivityDate="2017-08-31T21:45:42.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22443" PostTypeId="5" CreationDate="2017-08-31T22:13:50.163" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2017-08-31T22:13:50.163" LastActivityDate="2017-08-31T22:13:50.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22444" PostTypeId="4" CreationDate="2017-08-31T22:13:50.163" Score="0" Body="Questions about streams of plasma released from the Sun or other stars (stellar winds)." OwnerDisplayName="user14291" LastEditorDisplayName="user14291" LastEditDate="2017-09-01T10:29:12.887" LastActivityDate="2017-09-01T10:29:12.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22445" PostTypeId="5" CreationDate="2017-08-31T22:16:35.293" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2017-08-31T22:16:35.293" LastActivityDate="2017-08-31T22:16:35.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22446" PostTypeId="4" CreationDate="2017-08-31T22:16:35.293" Score="0" Body="Questions about the techniques and practice of observing the night sky." OwnerDisplayName="user14291" LastEditorDisplayName="user14291" LastEditDate="2017-09-01T10:28:42.933" LastActivityDate="2017-09-01T10:28:42.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22448" PostTypeId="1" CreationDate="2017-09-01T00:48:36.133" Score="4" ViewCount="10678" Body="&lt;p&gt;Planets, including Mercury, Venus, Earth, and Mars are rocky planets and are small compared to gas giants. And all these planet are near to Sun. But mass is directly proportional to force. So why don't any planets lie between rocky planets. And if gas giants lie farther always they why does Pluto lie after all the gas giants?&lt;/p&gt;&#xA;" OwnerUserId="18170" LastEditorUserId="11143" LastEditDate="2017-09-01T15:16:38.247" LastActivityDate="2020-11-05T18:00:39.840" Title="Why are rocky and small planets nearer to the Sun whereas big, gas giants are farther?" Tags="&lt;planet&gt;&lt;planetary-formation&gt;&lt;pluto&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22449" PostTypeId="2" ParentId="22448" CreationDate="2017-09-01T00:53:19.053" Score="0" Body="&lt;p&gt;As far as our understanding of planetary formation goes, rocky planets tended to form closer to the Sun because the materials they're made of -- silicates and heavier gases -- 'fall' inwards towards the Sun. In the areas closer to the Sun, it's too warm for the lighter gases to condense there, and so the larger gas and ice giants tended to form further from the Sun. As for trans-Neptunian objects like Pluto...we're not exactly sure. Can anyone else shed some light on the current research on the formation of the Kuiper Belt and Oort Cloud?&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2017-09-01T00:53:19.053" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22451" PostTypeId="1" AcceptedAnswerId="22452" CreationDate="2017-09-01T06:14:24.683" Score="11" ViewCount="268" Body="&lt;p&gt;General Relativity says that gravity moves at the speed of &lt;em&gt;c&lt;/em&gt;.  How precise are the measurements?&lt;/p&gt;&#xA;" OwnerUserId="17699" LastActivityDate="2020-08-07T01:35:12.833" Title="How precise are the observational measurements for the speed of gravity?" Tags="&lt;gravity&gt;&lt;speed&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22452" PostTypeId="2" ParentId="22451" CreationDate="2017-09-01T06:14:24.683" Score="17" Body="&lt;p&gt;The accuracy is extremely good thanks to LIGO and a binary neutron star merger.  It's accurate to about 3 parts in 1 quadrillion!&lt;/p&gt;&#xA;&lt;p&gt;See the paper &lt;a href=&quot;https://arxiv.org/abs/1710.05834&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gravitational Waves and Gamma-rays from a Binary Neutron Star Merger: GW170817 and GRB 170817A&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The paper constrains &amp;quot;the difference between the speed of gravity and the speed of light to be between &lt;span class=&quot;math-container&quot;&gt;$−3×10^{-15}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$+7×10^{−16}$&lt;/span&gt; times the speed of light.&amp;quot;&lt;/p&gt;&#xA;&lt;h3&gt;Older response before the LIGO observation:&lt;/h3&gt;&#xA;&lt;p&gt;One of the more recent of gravity was first measured in 2003 using Jupiter and a quasar.  The measurements were for the purpose of proving a finite speed and weren't very precise.  &lt;a href=&quot;https://www.nrao.edu/pr/2003/gravity/&quot; rel=&quot;nofollow noreferrer&quot;&gt;See the paper for more&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;More recently in 2013 a Chinese group built a model using Earth's tides that helped them narrow it down.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;... [T]he speeds of gravity are from 0.93 to 1.05 times the speed of light with a relative error of about 5%. This provides first set of strong evidences to show that the speed of gravity is the same as the speed of light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This is so far the most accurate measurement I've seen.  &lt;a href=&quot;https://link.springer.com/content/pdf/10.1007%2Fs11434-012-5603-3.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;See the paper for more.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;In the near future, &lt;a href=&quot;https://arxiv.org/abs/1707.06101&quot; rel=&quot;nofollow noreferrer&quot;&gt;LIGO may be able to provide more accurate measurements&lt;/a&gt; by comparing the distance among detectors and the delay of observation.&lt;/p&gt;&#xA;&lt;p&gt;Also related, see the question &amp;quot;&lt;a href=&quot;https://physics.stackexchange.com/questions/146655/how-is-the-most-accurate-value-of-g-measured/&quot;&gt;How is the most accurate value of  G  measured?&lt;/a&gt;&amp;quot;&lt;/p&gt;&#xA;" OwnerUserId="17699" LastEditorUserId="17699" LastEditDate="2020-08-07T01:35:12.833" LastActivityDate="2020-08-07T01:35:12.833" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22453" PostTypeId="1" AcceptedAnswerId="22457" CreationDate="2017-09-01T07:20:59.733" Score="2" ViewCount="2723" Body="&lt;p&gt;Does the Sun have any atmosphere of its own? Is it just like the atmospheres that the planets have.&lt;/p&gt;&#xA;" OwnerUserId="5343" LastEditorUserId="7982" LastEditDate="2021-02-28T11:26:49.020" LastActivityDate="2021-02-28T11:26:49.020" Title="Does the Sun have any atmosphere?" Tags="&lt;the-sun&gt;&lt;space&gt;&lt;stellar-atmospheres&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22454" PostTypeId="2" ParentId="12134" CreationDate="2017-09-01T09:00:39.667" Score="2" Body="&lt;p&gt;A Pluto orbiter would have been possible, but then the entire mission would have to be designed around that. New Horizons was designed as a small fast probe that could reach Pluto relatively quickly (almost 10 years), but not be able to stop there. An orbiter would have to take a lot more fuel with it, and approach Pluto slower. Such a mission would therefore take a lot longer to reach Pluto, i.e. it could easily take double the time or more, and would be more expensive.&lt;/p&gt;&#xA;" OwnerUserId="18324" LastActivityDate="2017-09-01T09:00:39.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22455" PostTypeId="1" CreationDate="2017-09-01T09:21:05.960" Score="8" ViewCount="307" Body="&lt;p&gt;When initial searches for a Kuiper Belt object for New Horizons to fly to after passing Pluto did not find good targets, the Hubble telescope was used, and it resulted in the current targeted flyby for 2019. Initial searches used ground based telescopes. When the search risked running out of time without a good target, the Hubble was brought in to help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to what you read about the current generation of large earth based telescopes that use adaptive optics, those telescopes have a much larger angular resolution and light gathering area than the Hubble. So why was the Hubble better capable of finding a good target?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt; &lt;br&gt;&#xA;Part of the answer may be that the current generation of adaptive optics telescopes &lt;a href=&quot;https://astronomy.stackexchange.com/questions/14795/why-actually-arent-ground-based-observatories-using-adaptive-optics-for-visib/14825&quot;&gt;only do adaptive optics in infrared&lt;/a&gt;, at least if KBOs are best observed in visible light which I don't know but I've moved that to a &lt;a href=&quot;https://astronomy.stackexchange.com/questions/22466/what-wavelengths-are-used-to-search-for-distant-solar-system-objects-like-kbos-a&quot;&gt;separate question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding everyone suggesting that atmospheric absorbtion is to blame, how does that square with this: The 8.3m Subaru telescope (which was one of the telescopes used in the ground search) has a light collecting area of 53m2. The Hubble has a collecting area of 4.5m2. So atmospheric absorbtion would need to be 91.5% for them to collect the same amount of light. Sure atmospheric absorbtion is high for some infrared wavelengths, but surely not that high over all the relevant wavelengths. &lt;/p&gt;&#xA;" OwnerUserId="18324" LastEditorUserId="18324" LastEditDate="2017-09-01T21:34:36.157" LastActivityDate="2018-03-12T09:36:58.180" Title="Why was the (small) Hubble better able to find KBO targets for New Horizons than large adaptive optics ground telescopes?" Tags="&lt;telescope&gt;&lt;hubble-telescope&gt;&lt;adaptive-optics&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22456" PostTypeId="2" ParentId="22451" CreationDate="2017-09-01T12:51:54.653" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How precise are the observational measurements for the speed of gravity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not very. See the Wikipedia article on &lt;a href=&quot;https://en.wikipedia.org/wiki/Speed_of_gravity#Possible_experimental_measurements&quot; rel=&quot;nofollow noreferrer&quot;&gt;the speed of gravity&lt;/a&gt; which refers to dissent surrounding the claims made by Sergei Kopeikin and Edward Fomalont. The dissenters include Clifford M Will and Steve Carlip who are fairly well respected. The &lt;a href=&quot;https://arxiv.org/abs/1602.04188&quot; rel=&quot;nofollow noreferrer&quot;&gt;2016 paper&lt;/a&gt; about the LIGO detection said the measured speed was less than 1.7c. The &lt;a href=&quot;https://arxiv.org/abs/1707.06101&quot; rel=&quot;nofollow noreferrer&quot;&gt;more recent paper&lt;/a&gt; by Neil Cornish and others said it was between 0.55c and 1.4c. This is not at all precise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in seismology &lt;a href=&quot;https://en.wikipedia.org/wiki/Seismic_wave#Primary_waves&quot; rel=&quot;nofollow noreferrer&quot;&gt;p waves and s waves&lt;/a&gt; travel at different speeds. So IMHO it wouldn't be unreasonable to expect different types of waves in space to travel at different speeds. As to whether one should expect light waves and &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_wave#Effects_of_passing&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravitational waves&lt;/a&gt; to travel at exactly the same speed is a slightly different matter, complicated by the way a gravitational wave is thought to be a quadrupole wave.  &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2017-09-01T12:51:54.653" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22457" PostTypeId="2" ParentId="22453" CreationDate="2017-09-01T14:57:43.657" Score="12" Body="&lt;h2&gt;Yes the Sun has an atmosphere.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Disclaimer: I'm not sure if you meant this, but your question implies the Sun is a planet. It, of course, is a star and not a planet. Just wanted to make that clear.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;What is an atmosphere?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;When you ask if the Sun has an atmosphere, you're actually asking a tricky question. What do you mean by atmosphere? How do you define the boundary of the Sun, above which anything is considered an atmosphere? This is pretty easy for planets like Earth since they have a nice solid surface. But the Sun is a giant ball of plasma heated to thousands of degrees. There's no easy or clear division between the &quot;surface&quot; and &quot;atmosphere&quot;. Any discussion of the Sun's atmosphere involves defining what we mean by the surface of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Optical Depth&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;That being said, astronomers have come up with (arbitrary) ways to define the surface of the Sun. One common metric is to use &lt;a href=&quot;https://en.wikipedia.org/wiki/Optical_depth&quot; rel=&quot;noreferrer&quot;&gt;optical depth&lt;/a&gt;. Optical depth is a unitless number that defines one's ability to &quot;see&quot; through a gas (or plasma). An optical depth of 1 or higher means the gas is opaque and cannot be seen through. An optical depth less than one means the gas is transparent and can be seen through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when you have something the Sun or even fog, the optical depth varies with the distance you're looking into that object. I'll talk about fog since it's familiar, but the same idea applies to the Sun's atmosphere. Say you're standing in a forrest and its very foggy out. There's a tree 1 meter away from you that you can see. You could measure your optical depth, $\tau$, of the fog between you and tree and might find that $\tau = 0.15$. Since $\tau$ is less than one, that implies you can see the tree, but the value of $\tau$ also implies how well you can see it. If $\tau = 0$, there's nothing between you and the tree to impede your ability to see it. Let's say there's another tree that's 5 meters away. Now there's more fog between you and the tree and while you can still see it, it is harder to see it. The optical depth of the fog between you and the tree 5 meters away might be $\tau = 0.75$. It's still less than one, implying the tree is visible, but because there's more fog between you and the tree, the optical depth is higher. Finally, there may be a tree 10 meters away with so much fog between you and the tree that the optical depth is $\tau = 1.5$. You can't see this tree because there's too much fog in the way. Hopefully you now realize that anything which is at a distance where $\tau &amp;gt; 1$ is not visible to you. That effectively defines a &quot;surface&quot; around you precisely when $\tau = 1$. Anything beyond that point is not visible and anything closer is visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're talking about the Sun, you can look at the Sun, but you'll only see light which originates from a point where $\tau &amp;lt; 1$. There are countless photons bouncing around inside the Sun, but you can't see them because they're in an opaque part of the Sun. Astronomers use the optical depth as a metric for defining the &quot;surface&quot; of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Keep in mind the above description is highly simplified, almost to the point of being wrong. The optical depth is a useful metric for defining a surface, but it doesn't imply there's an exact radius for the surface or even that the surface is constant for every wavelength. There are a lot of other factors that make this much more complicated than I'm describing here. Hopefully you get the general idea though.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The Sun's atmosphere&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;For the Sun, the atmosphere would be anything above the surface. Nominally the surface is defined as the point where the $\tau = 2/3$ (despite what I said above, and for reasons I won't go into here). The atmosphere above this surface is complicated and difficult to study. The atmosphere, just above the surface, is violent, turbulent, filled with outbursts and magnetic fields, and &lt;em&gt;extremely&lt;/em&gt; hot. Below are some pictures of this region of the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wbbEF.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wbbEF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;sub&gt; Left: Image of the corona during a solar eclipse. Right: Image of the corona from &lt;a href=&quot;https://sohowww.nascom.nasa.gov/&quot; rel=&quot;noreferrer&quot;&gt;SOHO&lt;/a&gt;. An occulting mask has been placed over the Sun.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atmosphere of the Sun extends far beyond that though. In fact, Earth is currently moving through the Sun's atmosphere. It's very tenuous out near Earth, but still exists. The Sun's atmosphere hitting our planet is the reason for aurora. Beyond the lower portions, the atmosphere is generally referred to as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_wind&quot; rel=&quot;noreferrer&quot;&gt;Solar Wind&lt;/a&gt;. This solar wind actually extends far out, beyond Pluto even. Exactly how far is difficult to define, but estimates place our Sun's atmosphere extending out to about &lt;a href=&quot;https://en.wikipedia.org/wiki/Bow_shocks_in_astrophysics#Around_the_Sun&quot; rel=&quot;noreferrer&quot;&gt;$\sim230\:\mathrm{AU}$&lt;/a&gt;. At that point is the bow shock, where our Sun's atmosphere slams into the interstellar medium surrounding us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our own Sun's atmosphere is hard to study on a large scale since we're inside it, but we have been able to observe this bow shock around other stars, as shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/w5sPL.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/w5sPL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;sub&gt;LL Orionis bow shock in Orion nebula. The star's atmosphere collides with the nebula flow. Hubble, 1995&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-09-01T14:57:43.657" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22459" PostTypeId="1" CreationDate="2017-09-01T18:02:55.207" Score="3" ViewCount="79" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xQNDz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xQNDz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw this blazing thing in the sky from Chennai, India at around 21:00 IST. Is this by any chance the Florence asteroid ? If not, what is this ? I'm pretty sure that this cannot be a firecracker because it was huge with large flames and it actually lasted in sky for at least 30 seconds till it reached heights where commercial aircrafts fly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: This object was moving from the ground and upwards towards the atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="18326" LastActivityDate="2017-09-01T21:01:52.703" Title="Is this the Florence asteroid?" Tags="&lt;asteroids&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22460" PostTypeId="2" ParentId="22455" CreationDate="2017-09-01T18:56:03.027" Score="2" Body="&lt;p&gt;The magnitude of these Kuiper Belt Objects is incredibly small, to begin with. The atmosphere distorts stars normally and scatters light even on the clearest of nights. In addition to that, these closer objects can be found with infrared detection. The atmosphere absorbs infrared wavelengths extremely well, which makes space based observations a necessity. The Hubble telescope also detects ultraviolet, visible, and near infrared making that an ideal telescope for these small Kuiper Belt Objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uBMBi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uBMBi.jpg&quot; alt=&quot;Atmospheric absorption of various wavelengths&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To deal with the atmospheric absorption, ground based telescopes such as the Subaru telescope get built on mountains so that there is less atmosphere to look through and the chance of cloud cover. However, the problem with the search for KBO was that it needed to be done in a short period of time so that New Horizons could be directed to it with less fuel. Hubble is ideal for that because it can look at the objects in the right direction all day while ground based telescopes could only do that at night assuming it is a clear enough night to see those objects. Normally Hubble is reserved for the most exclusive scientific projects and searches because of how its data quality is amazing. New Horizons had already cost so much that it was worth using a bit of time searching for its next destination instead of just ground based telescopes. &lt;/p&gt;&#xA;" OwnerUserId="18327" LastEditorUserId="18327" LastEditDate="2017-09-06T20:46:52.873" LastActivityDate="2017-09-06T20:46:52.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22461" PostTypeId="2" ParentId="22455" CreationDate="2017-09-01T19:14:27.600" Score="4" Body="&lt;p&gt;Adaptive optics only mitigate the air turbulence that blurries the images - and even that is only a partial recovery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All other issues remain. Air absorbs various wavelengths. Air has a certain amount of glow from various sources (light pollution, etc) which masks faint objects. Etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no real substitute for a large telescope operating in vacuum.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2017-09-01T19:14:27.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22462" PostTypeId="1" CreationDate="2017-09-01T19:37:57.123" Score="3" ViewCount="227" Body="&lt;p&gt;My telescope's sidereal tracking is lousy: it just doesn't work. To overcome this, when taking long-exposure photographs through it, what's the best way to reduce star trailing?&lt;/p&gt;&#xA;" OwnerUserId="1494" LastActivityDate="2017-09-02T20:13:42.040" Title="How to reduce star trailing when taking long exposure photographs?" Tags="&lt;photography&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22463" PostTypeId="2" ParentId="22459" CreationDate="2017-09-01T19:48:29.100" Score="0" Body="&lt;p&gt;No, it's not Florence.  Florence is quite faint - 9th magnitude currently- so about 1/15th of the brightness of the faintest star visible to the unaided eye. If it were a long enough exposure to show Florence, there would likely be other stars visible in the picture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And although Florence is moving quite quickly, I doubt it's moving as fast as appears in your photo, but without knowing how long the exposure is, it's hard to be definitive.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, the brightness appears to vary, which suggests that is could be a satellite, or a meteor, or possibly an aeroplane.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastEditorUserId="11822" LastEditDate="2017-09-01T19:55:52.623" LastActivityDate="2017-09-01T19:55:52.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22464" PostTypeId="2" ParentId="22462" CreationDate="2017-09-01T20:28:59.787" Score="5" Body="&lt;p&gt;Sounds like you should get a new mount. :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being more serious, even expensive mounts have periodic tracking errors (due to manufacturing tolerance of gears and moving equipment) that need to be corrected. These are the basic ways to do the corrections:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Add a guide scope to the side of your scope, use an eyepiece with cross-hairs, and watch a star while you take a photo. When the scope drifts, you push buttons to move the scope to keep the star centered.&lt;/li&gt;&#xA;&lt;li&gt;Do the same thing with a guide scope but use a dedicated guide camera that provides the corrections to the tracking.&lt;/li&gt;&#xA;&lt;li&gt;There are also guide cameras that you can insert just in front of the camera. They use a small &quot;pick-off&quot; mirror to look at a star in the field (off axis guider). The guide camera provides the corrections to the tracking.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Of course, all of these methods require a mount that can be moved slowly and smoothly in both right-ascension and declination to keep the guide star centered. Also, the mount needs to be an equatorial mount and accurately aligned. (More expensive mounts can be alt-azimuth mounts and use a gadget to rotate the camera during the exposure.)&lt;/p&gt;&#xA;" OwnerUserId="17355" LastEditorUserId="17355" LastEditDate="2017-09-02T20:13:42.040" LastActivityDate="2017-09-02T20:13:42.040" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22465" PostTypeId="2" ParentId="22459" CreationDate="2017-09-01T21:01:52.703" Score="6" Body="&lt;p&gt;Congratulations, you have seen a fireball. An especially bright meteor. It was bright and fast. This is a rock, a very small asteroid, hitting our atmosphere and &quot;buring up&quot;. It is also possibly a satellite or space junk re-entering. Re-entering satellites look move slower than asteroids. You can see fireball videos on youtube, as well as re-entries, to compare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This event was also &lt;a href=&quot;http://fireballs.imo.net/imo_view/event/2017/3030&quot; rel=&quot;noreferrer&quot;&gt;witnessed from Puducherry&lt;/a&gt;, the details seem to match up quite well. The time is slightly out, but the location is spot on. You can add your report at &lt;a href=&quot;http://fireballs.imo.net/members/imo/report_intro&quot; rel=&quot;noreferrer&quot;&gt;http://fireballs.imo.net/members/imo/report_intro&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, its not Florence, which will not be entering our atmosphere, and so will not appear to be burning, it will be a single point of light. It will be moving much slower, and won't be visible with the naked eye.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-09-01T21:01:52.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22466" PostTypeId="1" CreationDate="2017-09-01T21:31:50.940" Score="2" ViewCount="55" Body="&lt;p&gt;What wavelengths of light are used/most suited to search for distant solar system objects like KBOs and Oort cloud members? I suppose they are brightest in reflected sunlight, so they are best searched for in visible and near infrared. Is that correct?&lt;/p&gt;&#xA;" OwnerUserId="18324" LastActivityDate="2018-07-20T23:42:03.310" Title="What wavelengths are used to search for distant solar system objects like KBOs and Oort cloud members?" Tags="&lt;oort-cloud&gt;&lt;wavelength&gt;&lt;kbo&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22467" PostTypeId="2" ParentId="22448" CreationDate="2017-09-02T02:51:38.110" Score="3" Body="&lt;p&gt;The planets all have differences based on where they are and their compositions. The outer gas giants are so large today because of how quick they were to accrete material. In a protoplanetary disk, once an object gets massive enough, it can attract others through gravity, making the process much quicker than granules sticking together through electromagnetic forces or random collisions. The reason the outer planets were able to get massive much quicker than the inner planets is that there is more material with larger circumferences and that material was able to stay a &lt;a href=&quot;https://lco.global/spacebook/planets-and-how-they-formed/&quot; rel=&quot;nofollow noreferrer&quot;&gt;solid rather than become gaseous&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a larger mass, the outer planets attracted material quicker and the process was self-perpetuating. In addition, these gas giants had the opportunity to collect more gas that the sun did not end up taking as they are farther out. Hot Jupiters can confuse scientist because of how close they are to their host star when they could not have formed there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Kuiper Belt and Oort cloud are also two interesting topics in planetary formation and are most likely relics of planetary formation. This would result from the material that was created around that area, but also material that was &lt;a href=&quot;https://solarsystem.nasa.gov/planets/kbos/indepth&quot; rel=&quot;nofollow noreferrer&quot;&gt;flung out by gas giants&lt;/a&gt; to larger orbits. The Oort Cloud could also have been formed from interactions of the sun with &lt;a href=&quot;https://science.nasa.gov/science-news/science-at-nasa/2010/23nov_aliencomets/&quot; rel=&quot;nofollow noreferrer&quot;&gt;neighboring stars&lt;/a&gt; but we do not have direct observation of this cloud yet. &lt;/p&gt;&#xA;" OwnerUserId="18327" LastActivityDate="2017-09-02T02:51:38.110" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22468" PostTypeId="2" ParentId="22462" CreationDate="2017-09-02T03:52:12.120" Score="2" Body="&lt;p&gt;This is not quite a full answer, but a useful addition to your image processing stack are deconvolution methods designed with astronomy in mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a free software tool for image processing called Iris which includes some useful deconvolution methods not available in typical image processing applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a &lt;a href=&quot;http://www.astrosurf.com/buil/iris/tutorial12/doc30_us.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;brief tutorial&lt;/a&gt; showing what's possible with these tool.  Note in particular the RL2 method, rather than the RL method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that these tool should not be used generally to correct long exposure star trails, but can correct small trails.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastEditorUserId="10196" LastEditDate="2017-09-02T13:12:52.660" LastActivityDate="2017-09-02T13:12:52.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22469" PostTypeId="1" AcceptedAnswerId="22470" CreationDate="2017-09-02T09:48:40.060" Score="6" ViewCount="6913" Body="&lt;p&gt;I was talking to my 9-year-old son about the Florence asteroid that is passing close to the earth. He wanted to view youtube videos about that asteroids, and we saw the starting of this one &lt;a href=&quot;https://www.youtube.com/watch?v=UzxMbkrNbcg&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=UzxMbkrNbcg&lt;/a&gt; and when I realized that the title video has 2036 as the year in which an asteroid crashes into the Earth. I explained that that was science fiction (like Back to the future) and only an animation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He asked me: &quot;If an asteroid or similar object was to collide with the Earth,.... how many years or months before the crash occurs would the scientific community know?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose that the advance time depends on technology, so assume the current level of technology.&lt;/p&gt;&#xA;" OwnerUserId="18328" LastEditorUserId="2365" LastEditDate="2017-09-02T11:48:07.740" LastActivityDate="2019-08-06T05:07:34.093" Title="If an asteroid was to collide with Earth, how much warning would we have?" Tags="&lt;earth&gt;&lt;asteroids&gt;&lt;crater&gt;&lt;future&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22470" PostTypeId="2" ParentId="22469" CreationDate="2017-09-02T12:11:10.157" Score="5" Body="&lt;p&gt;Asteroids come in all shapes and sizes, and generally the bigger they are the easier they are to detect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small asteroids, from a the size of sand grains (properly called meteoroids) to a few meters across hit the Earth everyday without causing harm and are undetectable until they hit (we call the trace as they hit the atmosphere a meteor) They cause no damage. For such objects we get 0 hours warning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somewhat larger objects can cause some local damage For example the Chelyabinsk asteroid caused damage from broken glass in Russia. Objects this large &lt;em&gt;may&lt;/em&gt; be detected a few hours to a few days. The Chelyabinsk asteroid wasn't, and hit with no warning. We have observed some small asteroids just prior to them hitting the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have found large numbers of objects whose orbit can bring them close to Earth. &lt;em&gt;Every object we have found has an orbit that &lt;strong&gt;misses Earth&lt;/em&gt;&lt;/strong&gt;.  With each new object that is located we can forecast its orbit for several centuries. However new objects are being found all the time. It is &lt;em&gt;possible&lt;/em&gt; that a object will be discovered that has the potential to cause damage to a wider area. If the object happened to be on an orbit that had it collide with Earth then there &lt;em&gt;could&lt;/em&gt; be less than a year's warning. That is very unlikely. It is slightly more likely that an object will be discovered to be on a collision course only after several orbits, in which case there would be years or decades warning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very large objects that could threaten the survival of humankind are rare. There are some such objects, (larger than Florence). We hope that we have now detected all very large near Earth asteroids, and discoverd them to be non-threatening (like Florence is not a current threat to Earth)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally a comet may be on a collision course with Earth. Comets that are spotted are usually detected a few months before they could threaten Earth. But comets are much rarer than asteroids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, between 0 hours and a century, with several decades being likely for an object that could do widespread damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The particular object referred to in the video is Apophis. It is roughly 400m asteroid that is &lt;strong&gt;not going to hit the Earth&lt;/strong&gt; for as long as we can forecast. It will, however, pass close by in 2029, and again in 2036. In 2005 the uncertainty in orbit was such that we could not rule out a collision in 2036. Further observations in 2006 and after allowed and a more accurate orbit to be calculated, and showed that it would certainly miss.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-09-02T14:11:10.687" LastActivityDate="2017-09-02T14:11:10.687" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="22471" PostTypeId="1" CreationDate="2017-09-02T12:54:12.417" Score="0" ViewCount="2362" Body="&lt;p&gt;Black hole is the densest .... in universe. We know that dimension of universe is 4D . So, what is the dimensions of black hole? Is string theory explains what is it's dimension?&lt;/p&gt;&#xA;" OwnerUserId="18170" LastActivityDate="2017-09-02T17:10:49.957" Title="what is the dimension of Black Hole?" Tags="&lt;black-hole&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2017-09-06T13:39:05.590" ContentLicense="CC BY-SA 3.0" />
  <row Id="22472" PostTypeId="2" ParentId="22471" CreationDate="2017-09-02T14:31:13.930" Score="1" Body="&lt;p&gt;The singularity itself is, by definition, zero dimensional (well...you could argue one dimensional as it exists over time, but it has zero spacelike dimensions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The event horizon is a two dimensional sphere, just like the surface of a beach ball - whether it can be said to have any thickness is unknown, there might be a thickness down at the Planck scale, but that's currently unknown. &lt;/p&gt;&#xA;" OwnerUserId="11884" LastActivityDate="2017-09-02T14:31:13.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22473" PostTypeId="2" ParentId="22471" CreationDate="2017-09-02T17:10:49.957" Score="0" Body="&lt;p&gt;I think, knowing that what dimension a black hole has depend on approach it with it's geometry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we discuss about a black hole and specially it's event horizon, we encounter to some geometry to study it. We should knowing the geometry of space-time and is space-times Flat and curved, if we use the geometry of Minkowski Space-time or other geometry. the Space-time structure is very important here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general relativity, it's unknown that what happens in GKP and with which manifold we must mathematically discuss about it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this question that what is the dimensions of black hole, is physically vague and mathematically also. &lt;/p&gt;&#xA;" OwnerUserId="15641" LastActivityDate="2017-09-02T17:10:49.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22474" PostTypeId="1" CreationDate="2017-09-02T20:54:04.777" Score="2" ViewCount="654" Body="&lt;p&gt;The sizes of the stars in the picture below have more to do with their brigtness than the actual size of the star.  I'm looking for a formula that relates the viewing size of a star with its brightness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bxPjR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bxPjR.jpg&quot; alt=&quot;The Central Stars of Orion&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I've worked on so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Wikipedia on &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_magnitude#History&quot; rel=&quot;nofollow noreferrer&quot;&gt;Apparent Magnitude, History&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The viewing brightness can be related to the apparent magnitude by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B=0.4^A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before I sit down and count pixels and try to do a curve fit to find the  the relationship, I'm guessing this work has been done before.  I wonder if anyone has any leads that would get me to a good equation that I can use to generate the view sizes of stars.&#xA;&lt;hr&gt;&#xA;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My main goal is to make a reasonable 'cartoon' of the various constellations for the astronomy classes I'm teaching.  My rough work so far looks like this.  (I'm fairly certain the vertical scale is off.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Pp3gO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pp3gO.png&quot; alt=&quot;This is a cartoon of some of the stars of Orion.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16774" LastEditorUserId="16774" LastEditDate="2017-09-03T04:11:20.003" LastActivityDate="2017-09-06T07:07:02.637" Title="How to Find the Viewing Size of a Star" Tags="&lt;star&gt;&lt;size&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22475" PostTypeId="2" ParentId="22414" CreationDate="2017-09-02T21:01:15.070" Score="2" Body="&lt;p&gt;The planets are very, very close to spherical (the Earth has a variaton of about 0.1% between polar and equatorial radii).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rotational inertial for a solid sphere is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I=\frac{2}{5}mr^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you give us details about why you might be interested in looking at the rotational inertia of planets in greater detail, there might be more motivation for putting effort into finding an answer for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just saying.&lt;/p&gt;&#xA;" OwnerUserId="16774" LastActivityDate="2017-09-02T21:01:15.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22476" PostTypeId="1" CreationDate="2017-09-02T22:11:46.843" Score="3" ViewCount="394" Body="&lt;p&gt;Does the Earth's orbit around the sun exhibit nodal precession, as does the moon's around the Earth?  If it does, that must mean that the ecliptic is in a periodic wobble with respect to the celestial sphere?&lt;/p&gt;&#xA;" OwnerUserId="18331" LastActivityDate="2017-10-06T08:16:59.133" Title="nodal precession: earth's orbit of sun" Tags="&lt;orbit&gt;&lt;precession&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22477" PostTypeId="1" CreationDate="2017-09-02T22:53:35.933" Score="0" ViewCount="1080" Body="&lt;p&gt;As I am aware, lunar eclipses can take place several times a year on Earth. However, I am curious whether it would be possible for this event to only occur once annually on a specific date (and preferably on the same face of the planet) if the moon's angular orbit around Earth were to be altered from 5°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it bluntly:&#xA;How would the components causing a lunar eclipse need to be altered for a singular 'blood moon' to be able to appear reliably every year on say the 25th of September? If it is at all possible.&#xA;Additionally, would these alterations cause any other significant effects between the Earth and Moon?&lt;/p&gt;&#xA;" OwnerUserId="18332" LastActivityDate="2017-09-03T07:19:56.083" Title="What would be the required conditions for a lunar eclipse to occur only once on a precise annual basis?" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;moon-phases&gt;&lt;lunar-eclipse&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22479" PostTypeId="2" ParentId="22477" CreationDate="2017-09-03T07:19:56.083" Score="3" Body="&lt;p&gt;A lunar eclipse occurs when a full moon occurs at the same time as a lunar node (when the moon crosses the ecliptic - the plane of the Earth's orbit around the sun)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time between full moons is 29.53... days (although it varies slightly due to the moon's orbit not being circular) This is the synodic month. The time between nodes is 27.21... days. This is the draconic month (because of the dragon that eats the sun!) It is shorter than the time between full moons because of the curved motion of the Earth around the sun, and because the gravity of the sun distorts and rotates the orbit of the moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now on Earth, 223 synodic months is almost the same as 242 draconic months. And so an eclipse will recur every 223 months, or 6585 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you want an eclipse to recur every year. For that you would need the length of the synodic month, and the draconic month to be an exact fraction of a year. If you change the size of the orbit of the moon you will vary both the length of the synodic and draconic month. If you change the size of the Earths orbit you can change the difference between a synodic, and draconic month, but you also change the length of the year. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you end up with a juggling game: You have two variables to change, the Earth-Moon distance and the Sun-Earth distance. And you get three values: Synodic, Draconic month and length of a year. You aim to get both of the months an exact factor of the year. It's probably possible to achieve this, but you may end up freezing or baking the Earth. You have also changed the calendar (so &quot;September&quot; may not exist)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you need to consider the perturbations of other planets and so with time the length of the month will vary, and your three-way co-incidence will break eventually.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-09-03T07:19:56.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22480" PostTypeId="1" AcceptedAnswerId="22496" CreationDate="2017-09-03T08:43:25.363" Score="3" ViewCount="118" Body="&lt;p&gt;Below is a small GIF made from a small subset of images contained in the original 36MB GIF found at &lt;a href=&quot;https://cneos.jpl.nasa.gov/images/news/florence.p5us.1Hz.s382.sep01.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://cneos.jpl.nasa.gov/images/news/florence.p5us.1Hz.s382.sep01.gif&lt;/a&gt; as I just found in the Sky and Telescope article &lt;a href=&quot;http://www.skyandtelescope.com/astronomy-news/asteroid-florence-has-two-moons/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Asteroid Florence Has Two Moons&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although it's tempting to interpret the image as an optical image, I belive this is pulsed-doppler data from the Goldstone Deep Space Network center. I am guessing that one axis is range (delay) and the other is doppler shift. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still I can't figure out the geometry that would produce an &quot;eclipse&quot; (in this case the moon passing behind the larger asteroid) with it disappearing so &quot;far&quot; from the asteroid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone help me out here? Is one axis delay (time) and the other doppler shift? One can start with &lt;a href=&quot;https://cneos.jpl.nasa.gov/news/news199.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this description&lt;/a&gt; but there is more to it than is discussed there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; &quot;A radar image shows asteroid 3122 Florence and tiny echoes from its two moons. &lt;a href=&quot;https://cneos.jpl.nasa.gov/images/news/florence.p5us.1Hz.s382.sep01.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is an animation&lt;/a&gt; that shows them more clearly. The direction of the radar illumination (and thus the direction toward Earth) is at the top.&quot; From &lt;a href=&quot;http://www.skyandtelescope.com/astronomy-news/asteroid-florence-has-two-moons/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. NASA / Jet Propulsion Laboratory. This is a small subset of the frames contained in the original 36 MB GIF, and the size has been decreased by a factor of 2 in order to fit in SE's 2 MB limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Pn291.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pn291.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-09-03T13:58:25.577" LastActivityDate="2017-09-04T00:18:14.350" Title="What is the physical geometry of this apparent &quot;eclipse&quot; of a tiny moon of Asteroid Florence?" Tags="&lt;observational-astronomy&gt;&lt;asteroids&gt;&lt;radar&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22481" PostTypeId="1" AcceptedAnswerId="22483" CreationDate="2017-09-03T09:19:49.480" Score="0" ViewCount="86" Body="&lt;p&gt;I read that Dragonfly44 is a galaxy mostly consists of dark matter, and it should contain a super massive black hole too. If there really is a dark SMBH, how does it lose mass? I think there is no way for dark matter to be converted into energy and vice versa, no?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastActivityDate="2017-09-03T12:30:24.003" Title="How does dark SMBH evaporate?" Tags="&lt;dark-matter&gt;&lt;supermassive-black-hole&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22482" PostTypeId="1" CreationDate="2017-09-03T09:52:48.133" Score="1" ViewCount="91" Body="&lt;p&gt;I was out camping with my family at Crater Lake in Oregon. And I was looking up at the stars for a little while when I noticed a star-like object blink, then disappear. I thought I might have been going crazy for a second because I was watching the sky for a while, but then I noticed it blink again but in a different spot, further down. I was watching for a bit and a few seconds later, it let off another bright blink for about a second in a different spot again, and then disappeared. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It did this for about a solid 5 minutes before completely vanishing and never blinking again in the middle of the sky. I realised this blinking was going in a downward zig-zag pattern, there was no way it was a plane, and no way it was a satellite. I showed my family and they saw it too when it was happening, so it wasn't just me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The object was almost exactly the same brightness as all the other stars but maybe a bit more bright. I have no idea what to think of this, and the object was as big as all the other stars too, I found it weird though that it lit up, then disappeared for a few seconds to just reappear in a different spot kinda far away over and over again.&lt;/p&gt;&#xA;" OwnerUserId="18337" LastEditorUserId="2365" LastEditDate="2017-09-03T12:09:30.650" LastActivityDate="2017-09-03T12:09:30.650" Title="Blinking Star-Like object in the sky? What could it be?" Tags="&lt;identify-this-object&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ClosedDate="2017-09-04T05:48:18.753" ContentLicense="CC BY-SA 3.0" />
  <row Id="22483" PostTypeId="2" ParentId="22481" CreationDate="2017-09-03T12:30:24.003" Score="2" Body="&lt;p&gt;The &quot;no hair&quot; theory of black holes states that a black hole formed from the collapse of normal matter would be exactly the same as a black hole formed from dark matter. You could even form a black hole from photons, if you could get enough of them close together at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What originally collapsed to form the black hole has no relevance to whether Hawking radiation is given off. A black hole formed of dark matter WIMPS would behave exactly the same as a black hole formed from normal matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark matter can be converted into &quot;energy&quot; (I'll assume that you mean light, ie photons), in the same way that normal matter can: for example by antimatter annihilation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual formation of black holes, however, requires them to be mostly made of normal matter, since the matter has to lose energy as it falls together, and to do this it must interact with itself (essentially this is friction). As dark matter does not seem to interact with itself or with normal matter, it would not form a major part of a black hole, even in a galaxy dominated by an excess of dark matter. The supermassive black hole may be exceptionally small in such a galaxy, as there is less suitable matter to fall into it.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-09-03T12:30:24.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22484" PostTypeId="1" AcceptedAnswerId="22504" CreationDate="2017-09-03T13:27:14.147" Score="0" ViewCount="109" Body="&lt;p&gt;For example: a system with a &lt;strong&gt;0.2&lt;/strong&gt; solar mass &lt;strong&gt;white dwarf&lt;/strong&gt; orbiting a &lt;strong&gt;0.25&lt;/strong&gt; solar mass &lt;strong&gt;red dwarf&lt;/strong&gt;. Does the red dwarf steals mass from the white dwarf or vice versa.&lt;/p&gt;&#xA;" OwnerUserId="18338" LastActivityDate="2017-09-04T17:49:47.073" Title="What if a white dwarf is less massive than her partner?" Tags="&lt;astrophysics&gt;&lt;binary-star&gt;&lt;star-systems&gt;&lt;white-dwarf&gt;&lt;red-dwarf&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22485" PostTypeId="1" CreationDate="2017-09-03T16:16:22.943" Score="0" ViewCount="98" Body="&lt;p&gt;One day when I was playing basketball I saw a circle around the bright Sun. It is a perfect circle with a radius of about 20-30 km. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this circle of light form? I have seen the same also around the moon. But this does not happen every day. Why ?&lt;/p&gt;&#xA;" OwnerUserId="18170" LastEditorUserId="2365" LastEditDate="2017-09-03T21:39:31.220" LastActivityDate="2017-09-03T21:39:31.220" Title="why there is a circle of light surrounding Sun and Moon?" Tags="&lt;light&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22486" PostTypeId="2" ParentId="22485" CreationDate="2017-09-03T17:13:42.980" Score="2" Body="&lt;p&gt;Most probable you have been experiencing a weather phenomena: Ice crystals in the sky reflect or bend the light to form a &lt;a href=&quot;https://en.m.wikipedia.org/wiki/22%C2%B0_halo&quot; rel=&quot;nofollow noreferrer&quot;&gt;halo&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, this depends on the temperatures and moisture content of the atmosphere and cannot be seen every day.&lt;/p&gt;&#xA;" OwnerUserId="15824" LastActivityDate="2017-09-03T17:13:42.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22487" PostTypeId="1" AcceptedAnswerId="22491" CreationDate="2017-09-03T17:35:55.630" Score="2" ViewCount="121" Body="&lt;p&gt;Galaxies commonly interact and collide with other galaxies. Is it possible that clusters of galaxies similarly interact and collide? Have people tried to study this phenomenon previously? I couldn't find anything specific in the literature.&lt;/p&gt;&#xA;" OwnerUserId="18339" LastActivityDate="2017-09-06T16:44:16.360" Title="Is it possible for galaxy clusters to interact?" Tags="&lt;galaxy&gt;&lt;galaxy-cluster&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22488" PostTypeId="1" CreationDate="2017-09-03T18:19:33.640" Score="6" ViewCount="555" Body="&lt;p&gt;I need help identifying a blob. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last night I took a picture of Polaris, and there is a weird blob in it that I can't identify.  Star charts and other pictures on the 'net do not show this blob.  I took three pictures (with slightly different positions) and the blob has the same shape and position (relative to Polaris) in each.  The blob does not exist in other pictures that I took (pointing in completely different directions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taken with a Canon t5i, 300mm lens, 3 minute exposure on a motorized equatorial mount.  Viewing conditions were terrible, with lots-o-moon and smoke from forest fires.  From Colorado.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polaris is the brightest star, slightly left and down from center. Yildun, the next star in the handle of the little dipper, is near the upper left corner. The blob in question is very near, and slightly up and to the left of Polaris.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple Lens Flare:  I don't think it's lens flare (LF).  With LF, you can draw a straight line through the photo where the flare, the bright object, and the center of the photo all line up. Also, the LF tends to match the shape of the bright object.  I don't think it's LF because the blob doesn't line up with the center of the photo and a bright object, and there are no bright objects that have the same shape as the blob (with that weird tail to it).  Also, I have three photos with the same blob. In each case, Polaris (the brightest point in the photo) moves slightly because the camera moved slightly but the blob moves exactly with Polaris and not somehow relative with the center of the frame.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under careful analysis, there &lt;em&gt;IS&lt;/em&gt; lens flare, but it is where you'd expect it and not where the blob is. If you draw a line though Polaris and the center of the frame, there is a flare on the opposite side of the center and the same distance away from the center as Polaris is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Schmutz in the camera:  I don't think it's dust/dirt in the camera/sensor.  If that were the case then I would have seen it in other photos from the night, and I didn't.  Also, the blob would have stayed in the same position in the frame and not moved with Polaris as it did in the 3 different photos that I have of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not-So-Simple Lens Flare:  Maybe.  Although I cant imagine how this would happen unless some lens elements were not radially-symmetric and/or elements were not lined up on a common axis.  But I don't know everything about camera design so maybe this is possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a new blob in the sky:  Possible.  Although I think the odds of &quot;user error&quot; are much higher.  Maybe someone else wants to take a new photo of Polaris?  (wishful thinking)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/N95pt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/N95pt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e8D4O.jpg&quot; alt=&quot;Polaris with weird blob[1]&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="18341" LastEditorUserId="7982" LastEditDate="2020-02-27T05:19:07.040" LastActivityDate="2020-02-27T05:19:07.040" Title="Weird Blob near Polaris?" Tags="&lt;photography&gt;&lt;identify-this-object&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22491" PostTypeId="2" ParentId="22487" CreationDate="2017-09-03T19:50:23.207" Score="7" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically, structure is expected to form first on small scales (stars and stellar clusters), and later on increasingly larger scales — galaxies, groups, and eventually galaxy clusters (see e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2006ccha.book.....L&quot; rel=&quot;noreferrer&quot;&gt;Longair 2006&lt;/a&gt;). This is confirmed, at least to some extent, observationally. For instance, galaxies have been detected out to a redshift of $z=11.2$ (400 million years after Big Bang; &lt;a href=&quot;https://arxiv.org/abs/1603.00461&quot; rel=&quot;noreferrer&quot;&gt;Oesch et al. 2016&lt;/a&gt;), while cluster have only been detected out to $z=2.5$ (2.6 &lt;em&gt;billion&lt;/em&gt; years after Big Bang; &lt;a href=&quot;https://arxiv.org/abs/1604.07404&quot; rel=&quot;noreferrer&quot;&gt;Wang et al. 2016&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, when clusters formed, the Universe had already expanded so much that interaction between them is rather rare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does happen, however. One of the most important examples is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Bullet_Cluster&quot; rel=&quot;noreferrer&quot;&gt;Bullet Cluster&lt;/a&gt;, which consists of two colliding galaxy clusters. The reason I mention this is that it really beautifully confirmed the existence of dark matter. The image below (from &lt;a href=&quot;https://apod.nasa.gov/apod/ap060824.html&quot; rel=&quot;noreferrer&quot;&gt;NASA's Astronomy Picture of the Day&lt;/a&gt;) shows the two clusters after the collision. The stars and galaxies are so far apart that collisions between galaxies are rare, and collisions between stars virtually never happen. Thus, they have just passed right through each other, as seen in the image. The gas between the galaxies, however, collides, slows down, and is separated from the galaxies. This heats the gas to millions of degrees, emitting X-rays (seen in red). The &lt;em&gt;blue&lt;/em&gt; stuff is a map of the mass distribution, made using &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_lens&quot; rel=&quot;noreferrer&quot;&gt;gravitational lensing&lt;/a&gt;. This mass is clearly separate from the gas, but coincides with the galaxies, and gives a much higher mass than the visible mass, roughly 5.5 times more, which is exactly what is found for the ratio between dark and normal matter using other methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6yJCJ.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6yJCJ.jpg&quot; alt=&quot;bullet&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2017-09-03T19:50:23.207" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22492" PostTypeId="2" ParentId="22488" CreationDate="2017-09-03T19:55:56.680" Score="6" Body="&lt;p&gt;That looks like a lens effect. Polaris is the brightest object and there is an internal reflection in the lens or the camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a nice photo under difficult conditions. Can you see NGC 188, an odd open cluster?&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-09-03T19:55:56.680" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22493" PostTypeId="1" CreationDate="2017-09-03T20:23:51.107" Score="3" ViewCount="140" Body="&lt;p&gt;What kind of computer do you need to do adaptive optics of big telescopes? Are we talking supercomputer clusters or can my desktop do it, or my smartphone? I suppose it depends on the size of the telescope and the wavelength that is targeted. I'm interested in both the kind of computers that are used and the actual number of FLOPs per second that is required.&lt;/p&gt;&#xA;" OwnerUserId="18324" LastActivityDate="2017-09-05T13:03:16.160" Title="How difficult is adaptive optics computationally?" Tags="&lt;adaptive-optics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22494" PostTypeId="1" CreationDate="2017-09-03T20:48:12.547" Score="0" ViewCount="587" Body="&lt;p&gt;I somewhere see that in a white dwarf, we have this equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho = 2 N_e * m_p / V = 2 n_e * m_p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $N_e$ is the number of particles (in this case e, electrons) and $n_e$ is the number density. and $m_p$ mass is mass of proton. My problem is with the 2 in this equation. why $2 * n_e$ and not just $n_e$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also about Electric Potential Energy between a Ion and electron, somewhere said:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ E = (Z e * 1e) * n_e^{1/3} / (4 \pi \epsilon_0) $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand the $n_e$ term. Is the above really a equation or is it an approximation? I think it can't be an equation  because $ne \neq 1/r^3$.&lt;/p&gt;&#xA;" OwnerUserId="16441" LastEditorUserId="2365" LastEditDate="2018-07-27T22:33:27.537" LastActivityDate="2018-07-27T22:33:27.537" Title="Question about number density of electrons" Tags="&lt;white-dwarf&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22495" PostTypeId="2" ParentId="22494" CreationDate="2017-09-03T21:44:46.210" Score="3" Body="&lt;p&gt;The 2 in your equation is the average number of mass units per electron. e.g. totally ionised carbon has 6 electrons and 12 mass units. In your formula, the mass of a proton is used as a mass unit (close enough in this case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your second equation $n_e^{-1/3}$ is an approximation for the inter-particle separation. This &lt;em&gt;is&lt;/em&gt; an approximation and there is a small numerical factor that would need to be introduced in an accurate treatment.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-09-03T21:44:46.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22496" PostTypeId="2" ParentId="22480" CreationDate="2017-09-04T00:18:14.350" Score="3" Body="&lt;p&gt;The vertical axis is signal delay, and therefore distance from the radar antenna; this is more or less true spatial information. As the caption indicates, delay is increasing from top to bottom in the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horizontal axis is frequency shift (due to the Doppler effect). To the left are signals with a lower frequency relative to what was sent from the radar antenna, i.e. reflections from points moving away from the antenna; to the right, signals with a higher frequency, coming from points moving towards it.¹ This axis is more difficult to interpret.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a single rigid body, which Florence presumably is, the only way it can have points moving in different directions at the same time is by rotating. On one edge of the body, points that have just come into view of the radar are moving quickly towards it. This relative movement slows down as the points approach the centre of the radar-visible disk; then, they move away increasingly fast until they disappear from view on the opposite edge. (The speed change isn’t linear, which means the body will be somewhat warped on the radar image.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only this effect allows the radar to distinguish between points that are at the same distance. If the asteroid wasn’t rotating at all, or if we happened to look exactly onto one of its poles, it would appear only as a thin vertical line in the image. For a rotating body, note that the radar cannot tell to which direction the rotation is: A mirror asteroid rotating the other way would look exactly the same to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the moons. They are not rigidly connected to Florence, and apparently not tidally locked to it either, so they will not follow its rotation. With that, the starting point “moving in different directions at the same time = rotation” which worked on the main asteroid breaks down. When one of the moons is moving towards the radar, that tells us exactly nothing about its relative position to Florence; the visualization is misleading in that respect. Therefore, it is absolutely possible for a moon to be straight behind Florence even when it appears to its side in this particular visualization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Emily Lakdawalla has a nice blog post on this topic, with animated drawings: &lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/2011/3248.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;How radio telescopes get “images” of asteroids&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;¹ After correcting for the movement of Florence – or the whole Florence-and-moons system – relative to Earth, of course.&lt;/p&gt;&#xA;" OwnerUserId="14006" LastActivityDate="2017-09-04T00:18:14.350" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22497" PostTypeId="2" ParentId="22435" CreationDate="2017-09-04T03:03:24.580" Score="2" Body="&lt;p&gt;While there's no hope for Canis Minor (unless you're willing to look at fainter stars, at which point you might be able to pull it off), here's H. A. Rey's vision of some of the nearby constellations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MNccj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MNccj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H. A. Rey sees Canis Minor as a dog's leash (maybe for one of those invisible pretend dogs?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, if you really want to give it a go, here are all the stars in Canis Minor that are magnitude 5.5 or brighter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TVPpB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TVPpB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2017-09-04T03:03:24.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22499" PostTypeId="1" CreationDate="2017-09-04T06:56:49.807" Score="1" ViewCount="41" Body="&lt;p&gt;If a planet passed near the moon and caused its orbit to decay down to earth- how long could it take and how would that effect the visible phases of the moon on earth?&lt;/p&gt;&#xA;" OwnerUserId="18351" LastActivityDate="2017-09-04T06:56:49.807" Title="If the moon's orbit decays- how would it effect the moon's phases?" Tags="&lt;orbit&gt;&lt;natural-satellites&gt;&lt;orbital-elements&gt;&lt;orbital-mechanics&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" ClosedDate="2017-09-06T13:39:27.313" ContentLicense="CC BY-SA 3.0" />
  <row Id="22500" PostTypeId="2" ParentId="22455" CreationDate="2017-09-04T07:48:28.503" Score="0" Body="&lt;p&gt;This [1] article suggests that one of the advantages that Hubble and other space telescopes have is that they can better image very faint objects because they don't have to deal with atmospheric glow. Adaptive optics don't help with that, and a larger collecting area also collects more background glow. Also the atmospheric glow is intenser in infrared than in visible light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other listed differences: Ground telescopes can't make as accurate brightness measurements due to atmospheric turbulence (AO apparently doesn't help with that); ground telescopes can have better angular resolution due to larger sizes; ground telescopes can use bigger, heavier, better spectrographs than are practical in spacecraft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://www.ucolick.org/~max/History_AO_Max.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Introduction to Adaptive Optics and its History, Claire Max, at Center for Adaptive Optics, U. Calif, 2001&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18324" LastEditorUserId="18324" LastEditDate="2018-03-12T09:36:58.180" LastActivityDate="2018-03-12T09:36:58.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22502" PostTypeId="2" ParentId="22493" CreationDate="2017-09-04T16:58:38.923" Score="2" Body="&lt;p&gt;If you're only trying to correct a small field, using a single guide star (or a single laser guide star), then the requirements aren't very arduous. The MACAO adaptive-optics module used on the Very Large Telescope at the European Southern Observatory dates from the early 2000s and is based around two 400 MHz PowerPC 604 chips (one for the &quot;supervisory computer&quot;, the other for the &quot;real time computer&quot;). You can read more about it &lt;a href=&quot;ftp://ftp.eso.org/pub/adoptics/macao_vlti_fdr/vlt-spe-eso-15600-2062.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; (PDF file).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For next-generation multi-conjugate adaptive optics, which tries to correct a larger field using multiple guide stars, the requirements are bit more stringent: my impression is that you need several GFLOPs in the real time computer (some of this may be handled by using specialized DSP chips instead of general CPUs), and additional processing power for external monitoring and analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, crudely speaking, your smartphone could probably handle single-guide-star AO (except it would need to be re-engineered a bit and running a real-time operating system, and you'd need an extra computer for external monitoring); for multi-guide-star AO, you'd end up needing the equivalent of a powerful (multi-CPU) server or (small) cluster.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2017-09-04T16:58:38.923" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22503" PostTypeId="1" AcceptedAnswerId="22505" CreationDate="2017-09-04T17:30:48.273" Score="8" ViewCount="763" Body="&lt;p&gt;There seems to be three things described in the recent Nature Astronomy paper &lt;a href=&quot;https://www.nature.com/articles/s41550-017-0224-z&quot; rel=&quot;noreferrer&quot;&gt;Millimetre-wave emission from an intermediate-mass black hole candidate in the Milky Way&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Molecular cloud CO–0.40–0.22 with an extremely broad velocity width, near the centre of our Milky Way galaxy.&lt;/li&gt;&#xA;&lt;li&gt;a compact object with a mass of about &lt;span class=&quot;math-container&quot;&gt;$10^5 M⊙$&lt;/span&gt; is lurking in this cloud.&lt;/li&gt;&#xA;&lt;li&gt;a point-like continuum source called CO–0.40–0.22*&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What are the objects CO–0.40–0.22 and CO–0.40–0.22* and what is the difference between the two. How did the originally discovered CO–0.40–0.22 get its name?&lt;/p&gt;&#xA;&lt;p&gt;The news of the paper is described in the Science item &lt;a href=&quot;http://www.sciencemag.org/news/2017/09/long-rumored-midsized-black-hole-may-be-hiding-out-milky-way&quot; rel=&quot;noreferrer&quot;&gt;Long-rumored midsized black hole may be hiding out in the Milky Way&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;There is a nice writeup in Gizmodo for lay scientists: &lt;a href=&quot;http://gizmodo.com/astronomers-think-they-discovered-the-missing-link-in-b-1798703443&quot; rel=&quot;noreferrer&quot;&gt;Astronomers Think They Discovered The Missing Link in Black Hole Evolution&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The paper is not paywalled and can be read here: &lt;a href=&quot;https://www.nature.com/articles/s41550-017-0224-z.pdf&quot; rel=&quot;noreferrer&quot;&gt;https://www.nature.com/articles/s41550-017-0224-z.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Abstract:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;It is widely accepted that black holes with masses greater than a million solar masses (M⊙) lurk at the centres of massive galaxies. The origins of such ‘supermassive’ black holes(SMBHs) remain unknown, although those of stellar-mass black holes are well understood. One possible scenario is that intermediate-mass black holes (IMBHs), which are formed by the runaway coalescence of stars in young compact star clusters merge at the centre of a galaxy to form a SMBH. Although many candidates for IMBHs have been proposed, none is accepted as definitive. Recently, we discovered a peculiar molecular cloud, CO–0.40–0.22, with an extremely broad velocity width, near the centre of our Milky Way galaxy. Based on the careful analysis of gas kinematics, we concluded that a compact object with a mass of about 105 M⊙ is lurking in this cloud. Here we report the detection of a point-like continuum source as well as a compact gas clump near the&#xA;centre of CO–0.40–0.22. This point-like continuum source (CO–0.40–0.22*) has a wide-band spectrum consistent with 1/500 of the Galactic SMBH (Sgr A*) in luminosity. Numerical simulations around a point-like massive object reproduce the kinematics of dense molecular gas well, which suggests that CO–0.40–0.22* is one of the most promising candidates for an intermediate-mass black hole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-09-04T18:38:43.200" Title="How did the object CO-0.40-0.22 get its name, and how is it distinct from CO-0.40-0.22*?" Tags="&lt;black-hole&gt;&lt;galaxy&gt;&lt;milky-way&gt;&lt;radio-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22504" PostTypeId="2" ParentId="22484" CreationDate="2017-09-04T17:49:47.073" Score="3" Body="&lt;p&gt;If they are close enough to exchange mass, the Red dwarf will always lose mass to the white dwarf.  That doesn't mean there won't be some exchange going back the other way, but the white dwarf will always win the exchange rate and it won't even be close.   It makes no difference whether it's heavier or lighter.   What matters primarily, is how tightly held it's outer mass is to each star and how strong the tidal forces are.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The white dwarf's mass is very tightly bound to it by gravity and being so small, it's not going to experience strong tidal variation.  The red dwarf's outer layers are far less gravitationally bound to it, and being much larger, it experiences greater tidal bulging.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much mass is exchanged would depend on how close they are, but around any star, and if they are close enough, the white dwarf will always steal mass from the main sequence star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A curious thing happens with main sequence stars.   As their mass goes up, their surface gravity goes down.   A star with 0.25 times the mass of our sun, &lt;a href=&quot;https://en.wikipedia.org/wiki/Red_dwarf#Description_and_characteristics&quot; rel=&quot;nofollow noreferrer&quot;&gt;extrapolating from this chart&lt;/a&gt; would have a radius about 32-33% of our Sun, and being 1/3rd the size, using the inverse square rule, it's surface gravity would be about 9/4ths or 2.25 times the gravity on the surface of our sun - roughly speaking.  There would be other factors like rotation rate and solar storms, and, as mentioned above, how close they are, but there's no way that a main sequence star of any size could steal more mass than it gives up from a densely packed White dwarf.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-09-04T17:49:47.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22505" PostTypeId="2" ParentId="22503" CreationDate="2017-09-04T18:09:02.663" Score="9" Body="&lt;p&gt;An earlier paper on the object (&lt;a href=&quot;https://arxiv.org/abs/1512.04661&quot; rel=&quot;noreferrer&quot;&gt;Oka et al. (2015)&lt;/a&gt;) explains that CO-0.40-0.22 is a &amp;quot;high-velocity compact cloud&amp;quot;. The first discovery of such an object was two decades ago (&lt;a href=&quot;https://arxiv.org/abs/astro-ph/9810434&quot; rel=&quot;noreferrer&quot;&gt;Oka et al. (1998)&lt;/a&gt;), when CO 0.02-0.02 was found. The naming convention used for that object was&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;&amp;quot;CO&amp;quot; + Galactic longitude + Galactic latitude&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;For instance, CO 0.02-0.02 has latitude &lt;span class=&quot;math-container&quot;&gt;$0.02^{\circ}$&lt;/span&gt; and longitude &lt;span class=&quot;math-container&quot;&gt;$-0.02^{\circ}$&lt;/span&gt;; CO-0.40-0.22 has latitude &lt;span class=&quot;math-container&quot;&gt;$-0.40^{\circ}$&lt;/span&gt; and longitude &lt;span class=&quot;math-container&quot;&gt;$-0.22^{\circ}$&lt;/span&gt;. Naming based on coordinates is &lt;a href=&quot;https://ned.ipac.caltech.edu/level5/March01/Andernach/Ander3.html&quot; rel=&quot;noreferrer&quot;&gt;used for many other radio sources&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The * in the name of the continuum point source is, I believe, in continuation of the tradition set by the discovery of &lt;a href=&quot;https://en.wikipedia.org/wiki/Sagittarius_A*&quot; rel=&quot;noreferrer&quot;&gt;Sagittarius A*&lt;/a&gt;, the point source in &lt;a href=&quot;https://en.wikipedia.org/wiki/Sagittarius_A&quot; rel=&quot;noreferrer&quot;&gt;Sagittarius A&lt;/a&gt;. It is often necessary to distinguish between the two. Quoting from &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0305074&quot; rel=&quot;noreferrer&quot;&gt;Goss et al. (2003)&lt;/a&gt;,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Eight years after the discovery, one of us (Brown) invented the name Sgr A* to distinguish the compact source from the other components in the galactic center and to emphasize the unique nature of this source. Brown (1982) proposed a model of Sgr A* consisting of twin precessing jets with a period of 2300 years. The model has not stood the test of time but the name immediately was accepted. As an example, the VLBI results discussed by Lo et al. (1985) uses the name Sgr A*; the review article by Lo (1987) also uses this nomenclature.&lt;/p&gt;&#xA;&lt;p&gt;Bob Brown provides the following rationale for the name: “ Scratching on a yellow pad one morning I tried a lot of possible names. When I began thinking of the radio source as the “exciting source” for the cluster of H II regions seen in the VLA maps, the name Sgr A* occurred to me by analogy brought to mind by my Phd dissertation, which is in atomic physics and where the nomenclature for excited state atoms is He*, or Fe* etc.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The difference between CO-0.40-0.22 and CO-0.40-0.22* is simple: The former is a high-velocity compact cloud, and the latter is a point radio source inside it, possibly an intermediate-mass black hole. The formation, evolution and interactions of the two are likely tied together, but, as with Sagittarius A and Sagittarius A*, they need to be distinguished.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-09-04T18:38:43.200" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22506" PostTypeId="1" CreationDate="2017-09-04T21:51:19.333" Score="1" ViewCount="136" Body="&lt;p&gt;I've been looking into neutrinos. One thing I'm puzzled about is how fast they travel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the Wikipedia &lt;a href=&quot;https://en.wikipedia.org/wiki/SN_1987A#Neutrino_emissions&quot; rel=&quot;nofollow noreferrer&quot;&gt;SN1987A&lt;/a&gt; article where you can read this: &lt;em&gt;“approximately two to three hours before the visible light from SN1987A reached Earth, a burst of neutrinos was observed at three separate neutrino observatories”&lt;/em&gt;. Some &lt;a href=&quot;https://www.google.co.uk/#q=sn1987a+7.7hours&quot; rel=&quot;nofollow noreferrer&quot;&gt;reports&lt;/a&gt; describe an initial burst of neutrinos which arrived 7.7 hours before the light. It's thought that the neutrinos were emitted before the light because of the way stars collapse, wherein photons take longer to get out. I don't have much of an issue with that. But what's interesting is that in 168,000 years travelling through space, the photons didn’t overtake the neutrinos. That suggests to me that neutrinos are massless, which ties in with what Fermi said in section VII of his &lt;a href=&quot;http://microboone-docdb.fnal.gov/cgi-bin/RetrieveFile?docid=953;filename=FermiBetaDecay1934.pdf;version=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;1934 paper&lt;/a&gt;. He plotted three curves and noted that the μ=0 curve matched the empirical curves. He concluded that &lt;em&gt;“the rest mass of the neutrino is either zero, or, in any case, very small in comparison to the mass of the electron”&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is that if neutrinos travel at exactly c they have no mass at all. And I can't find any evidence that they don't. I can find plenty of evidence for electrons travelling at speeds less than c. The neutrino is said to be a fermion like an electron. But I can't find any evidence of neutrinos travelling at speeds less than c. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody give me a reference to any cosmological evidence that neutrinos travel at less than c? Or any other hard scientific evidence? By evidence I don't mean an inference such as &lt;em&gt;&quot;neutrinos go missing so they must oscillate therefore they must have mass therefore they must travel at less than c&quot;&lt;/em&gt;. I mean evidence. &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2017-09-05T00:52:26.003" Title="Is there any cosmological evidence that neutrinos travel at a speed other than c?" Tags="&lt;supernova&gt;&lt;speed&gt;&lt;neutrinos&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22507" PostTypeId="2" ParentId="22506" CreationDate="2017-09-04T23:04:02.187" Score="5" Body="&lt;p&gt;A paper by &lt;a href=&quot;http://arxiv.org/abs/1603.07568&quot; rel=&quot;noreferrer&quot;&gt;Wie et al. (2016)&lt;/a&gt; uses the coincidence between gamma rays from Gamma ray bursts and individual 3-30 TeV neutrinos to estimate the difference in propagation speed between neutrinos and photons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They find&#xA;$$ \frac{|v-c|}{c} &amp;lt; 2.5 \times 10^{-18} $$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They say this is about 7 orders of magnitude more precise than any previously published limit. However, it does rely on the genuine coincidencce between the GRBs and these individual neutrino events, which does seem to be in question or even doubtful in some cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass limit this places on the neutrinos appears to be a rather uninteresting $&amp;lt;6.8$ keV, but it is a direct measurement of speed rather than an extrapolation of the speed from a mass limit and special relativity.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-09-04T23:04:02.187" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22508" PostTypeId="2" ParentId="22506" CreationDate="2017-09-05T00:52:26.003" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I can find plenty of evidence for electrons travelling at speeds less than c. The neutrino is said to be a fermion like an electron. But I can't find any evidence of neutrinos travelling at speeds less than c.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This seems to be the core of your doubt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd really suggest you read the answers to questions on this subject on Physics SE, but a few notes on this specific point.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A fermion can be massless (Weyl and Majorana types can be).&lt;/li&gt;&#xA;&lt;li&gt;Experiment suggests neutrinos may have a very small mass (order of a few eV).  This is often pointed to as evidence suggesting the standard model is not complete.&lt;/li&gt;&#xA;&lt;li&gt;There is no absolute agreement on the mass (or lack of) of neutrinos.  It's a good way to get particle physicists arguing. :-)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So your confusion is understandable : so is everyone else.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;By evidence I don't mean an inference such as &quot;neutrinos go missing so they must oscillate therefore they must have mass therefore they must travel at less than c&quot;. I mean evidence. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That &lt;em&gt;is&lt;/em&gt; evidence. :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the problems with what you're asking for is that it presumes that neutrinos don't oscillate.  If someone gets an experimental result apparently demonstrating they do, you should not ignore it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how familiar you are with the theory of &lt;a href=&quot;https://en.wikipedia.org/wiki/Neutrino_oscillation&quot; rel=&quot;noreferrer&quot;&gt;neutrino oscillation&lt;/a&gt; and why it's so important, but I've linked to Wikipedia's page on the subject as a starting point just in case.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[Regarding SN1987]  I don't have much of an issue with that. But what's interesting is that in 168,000 years travelling through space, the photons didn’t overtake the neutrinos.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This simply places an upper limit on the mass of neutrinos.  Wikipedia's article quotes this as 16 eV from that experimental data and I've no reason to argue with that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the failure to overtake doesn't say any more than this.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;That suggests to me that neutrinos are massless, which ties in with what Fermi said in section VII of his 1934 paper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To give Fermi his due, note this quote from that paper (well, the translation of it) :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Hence we conclude that the rest mass of the neutrino is either zero, or, in any case, very small in comparison to the mass of the electron.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well the current values place the upper limit at around 1/30,000 th of the electron mass.  That's &quot;very small&quot; in my book.  So Fermi had it nailed pretty good : good scientist, that lad. :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just a few items of importance show how much has changed :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;late 1940's : the development of QED&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.aps.org/programs/outreach/history/historicsites/savannah.cfm&quot; rel=&quot;noreferrer&quot;&gt;The experimental discovery of the neutrino only happened in 1956&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;In 1962 the muon neutrino was detected.&lt;/li&gt;&#xA;&lt;li&gt;The standard model was developed in the late 1970's&lt;/li&gt;&#xA;&lt;li&gt;In 1998 the first evidence for neutrino oscillations was found.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So you do need to factor in the new physical models for the neutrino into your thinking about them.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-09-05T00:52:26.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22509" PostTypeId="2" ParentId="22476" CreationDate="2017-09-05T00:58:58.350" Score="1" Body="&lt;p&gt;The ecliptic is in periodic wobble wrt the celestial sphere, but that's because of nutation, not precession. Precession (nodal precession) changes where the Sun is during the equinoxes and solstices. The Spring Equinox, for example, has moved from Aries (it's still called &quot;the first point of Aries&quot;) to near the Pisces-Aquarius border&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2017-10-06T08:16:59.133" LastActivityDate="2017-10-06T08:16:59.133" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22511" PostTypeId="1" AcceptedAnswerId="22512" CreationDate="2017-09-05T03:43:46.603" Score="6" ViewCount="801" Body="&lt;p&gt;I know an asteroid can have moons, but how about rings? Does it also get affected by the Roche limit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The asteroid Florence has two moons. &lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="2153" LastEditDate="2017-09-05T04:11:44.757" LastActivityDate="2017-12-12T12:59:09.640" Title="Can an asteroid have rings?" Tags="&lt;asteroids&gt;&lt;planetary-ring&gt;&lt;roche-limit&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22512" PostTypeId="2" ParentId="22511" CreationDate="2017-09-05T04:03:19.523" Score="7" Body="&lt;p&gt;The simple answer is yes, an asteroid can have rings. The current known example is &lt;a href=&quot;https://en.wikipedia.org/wiki/10199_Chariklo&quot; rel=&quot;noreferrer&quot;&gt;10199 Chariklo&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Rings_of_Chariklo&quot; rel=&quot;noreferrer&quot;&gt;whose rings&lt;/a&gt; were discovered in 2014 (see &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014Natur.508...72B&quot; rel=&quot;noreferrer&quot;&gt;Braga-Ribas et al. (2014)&lt;/a&gt;). It actually has two rings, at distances of 391 km and 405 km, 7 km and 3km in width. As the discovery paper notes, they may be the (young) remnants of a debris disk around Chariklo, and have some moderately-sized bodies embedded in them. The observations indicated that water ice is likely present. Another small body, &lt;a href=&quot;https://en.wikipedia.org/wiki/2060_Chiron&quot; rel=&quot;noreferrer&quot;&gt;2060 Chiron&lt;/a&gt;, may also have rings (see &lt;a href=&quot;https://arxiv.org/abs/1501.05911&quot; rel=&quot;noreferrer&quot;&gt;Ortiz et al. (2015)&lt;/a&gt;). Again, water ice is thought to be a major component of the putative rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both ring systems were hypothesized in part due to evidence from &lt;a href=&quot;http://occult.mit.edu/research/stellarOccultations.php&quot; rel=&quot;noreferrer&quot;&gt;stellar occultations&lt;/a&gt;, i.e. cases where the minor body passed in front of a star. Stellar occultations have proved useful in many cases involving small bodies, such as Kuiper Belt Objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Ortiz et al. noted, Chariklo and Chiron are the two most massive members of a class of minor planets called &lt;a href=&quot;https://en.wikipedia.org/wiki/Centaur_(minor_planet)&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;centaurs&lt;/em&gt;&lt;/a&gt;. They orbit (unstable) in between the giant planets of the Solar System. Insights into ring formation around centaurs could perhaps provide information about centaur formation and evolution. It seems likely that the rings form from either&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The impact of a smaller body on the surface of the centaur&lt;/li&gt;&#xA;&lt;li&gt;The breakup of a satellite orbiting the centaur (as it crosses the Roche limit, as you speculated)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, rings have &lt;em&gt;not&lt;/em&gt; been observed around Trans-Neptunian Objects in similar stellar occultation data, leading the authors to hypothesize that centaurs are a separate population. More observations could shed some light on this hypothesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A final interesting point I'd like to make is that &lt;a href=&quot;https://en.wikipedia.org/wiki/Minor-planet_moon&quot; rel=&quot;noreferrer&quot;&gt;many moons around minor planets&lt;/a&gt; have been found, but only one or two ring systems are known.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2017-09-05T04:03:19.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22514" PostTypeId="2" ParentId="22455" CreationDate="2017-09-05T09:16:55.830" Score="2" Body="&lt;p&gt;I think you have hit the nail on the head in your question. KBOs are seen in reflected sunlight and they are incredibly faint, since the amount of reflected light reaching the earth goes as the inverse fourth power of their distance from us (see my answer to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/254/why-cant-we-observe-the-oort-cloud-with-a-telescope&quot;&gt;this question&lt;/a&gt; on trying to view Oort cloud objects). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see such objects requires deep imaging observations with low levels of background contamination. That background is minimised by having images with an extremely small point spread function (PSF) - the kind of PSF that can only be achieved by space-based telescopes or ground-based telescopes using adapative optics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the solar spectrum of course is strongly peaked in the visible region and ground-based adaptive optics systems are not effective in this wavelength range (ground-based AO works in the near-infrared, but in addition to the KBOs being intrinsically fainter at these wavelengths there is also the issue of background noise contributed both by the Earth's atmosphere and the telescope itself). Therefore the Hubble Space Telescope is the instrument of choice.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-09-05T09:16:55.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22517" PostTypeId="2" ParentId="22493" CreationDate="2017-09-05T13:03:16.160" Score="2" Body="&lt;p&gt;If you look around, you should be able to find home-user AO systems which run, albeit at slightly lower bandwidth than you might like, on your PC (or an equivalent chip in the vendor's box).   It depends a lot on the number of actuators on your deformable mirror, as do most matrix operations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be interested in the offerings at &lt;a href=&quot;https://www.thorlabs.com/navigation.cfm?guide_id=2037&amp;amp;gclid=EAIaIQobChMIlbm-n4yO1gIVEZ7ACh3p9gKMEAMYASAAEgLbgPD_BwE&quot; rel=&quot;nofollow noreferrer&quot;&gt;ThorLabs&lt;/a&gt; , &lt;a href=&quot;http://diffractionlimited.com/products/cameras-accessories/adaptive-optics/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Diffraction Limited&lt;/a&gt; or possibly &lt;a href=&quot;http://www.stellarproducts.com/adaptive/AO5lit.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellar Products&lt;/a&gt;, although that last one looks like abandonware. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-09-05T13:03:16.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22518" PostTypeId="2" ParentId="22474" CreationDate="2017-09-05T13:34:46.120" Score="2" Body="&lt;p&gt;Based on the math in the Wikipedia article about the &lt;a href=&quot;https://en.wikipedia.org/wiki/Airy_disk#Mathematical_details&quot; rel=&quot;nofollow noreferrer&quot;&gt;Airy Disk&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intensity of the starlight as a function of angular distance $\theta$ from the center is reasonably approximated by a Gaussian $$ I(\theta) = I_0 \exp(-\theta^2/(2\sigma^2)) $$ where $\sigma$ is a measure for the width of the intensity profile, which for an ideal optical system and diffraction-limited images is $$ \sigma \approx 0.42 \lambda N $$ (where $\lambda$ is the wavelength and $N$ is the f-number) and which is independent of the peak intensity $I_0$.  For a non-ideal or not-diffraction-limited system, the Gaussian should still provide a reasonable approximation, but then $\sigma$ is (much) larger than for the ideal system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The apparent edge of a star in an image occurs where the intensity $I(\theta)$ drops below a certain minimum $I_\text{limit}$, which (from the first equation) corresponds to an angular distance $$ \theta_\text{edge} = \sigma\sqrt{2\log(I_0/I_\text{limit})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can rewrite this in terms of visual magnitude $V$.  For fixed $\sigma$, $I_0$ is a measure for the brightness of the star.  Brightness is related to magnitude $V$ through $$ V = -2.5 \log_{10}(I/I_\text{ref}) $$ (with $I_\text{ref}$ the reference intensity for the magnitude scale) so $$\log(I_0/I_\text{limit}) = 0.4 \log(10) (V_\text{limit} - V_0)$$ and&#xA;\begin{eqnarray} &#xA;   \theta_\text{edge} &amp;amp; = &amp;amp; \sigma \sqrt{0.8 \log(10) (V_\text{limit} - V_0)} \approx 1.36 \sigma \sqrt{V_\text{limit} - V_0}&#xA;\\ V_\text{limit} - V_0 &amp;amp; = &amp;amp; \frac{\theta_\text{edge}^2}{0.8 \log(10) \sigma^2} \approx 0.543 \left( \frac{\theta_\text{edge}}{\sigma} \right)^2&#xA;\end{eqnarray}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the difference between the limiting magnitude and the star's magnitude is proportional to the apparent angular area of the star in the image.&lt;/p&gt;&#xA;" OwnerUserId="9401" LastActivityDate="2017-09-05T13:34:46.120" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22519" PostTypeId="1" AcceptedAnswerId="22540" CreationDate="2017-09-05T14:11:54.843" Score="0" ViewCount="583" Body="&lt;p&gt;I do not understand completely the meaning of critical density when talking about atomic or molecular line emission in astrophysical gas clouds: I know that it is way more complex than this, but just taking the two levels approximation (i.e. an atom/molecule with two quantum levels, 1 and 2, where 2 is the upper one) the critical density would be&#xA;$n_c = A_{12}/\gamma_{12}$, where $A_{12}$ is the Einstein coefficient for spontaneous radiative emission (i.e. de-excitation) and $\gamma_{12}$ the collisional de-excitation probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that if the gas density is $n &amp;lt; n_c$ the radiative de-excitations dominate, so I would expect to see the photons; conversely if $n&amp;gt;n_c$ collisions dominate, so the particles will lose energy through collisions before having enough time to spontaneously decay and emit photons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This argument of mine, however, seems to be wrong, because everybody use $n_c$ exactly in the opposite way: I see the photons only if $n&amp;gt;n_c$. I would expect that for $n \ll n_c$ the particles are experiencing almost no collisions, so neither collisional excitation, and they do not decay because they are not excited in the first place; but I feel that something wrong has been written in the previous lines, so I would like to read someone's opinion about it before to make even more pie in the sky conjectures.&lt;/p&gt;&#xA;" OwnerUserId="7919" LastActivityDate="2017-09-06T16:29:04.133" Title="Meaning of critical density in radiative transfer" Tags="&lt;radiative-transfer&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22521" PostTypeId="1" CreationDate="2017-09-05T17:07:09.347" Score="1" ViewCount="134" Body="&lt;p&gt;I may be way off the mark here, but wouldn't it be possible to use kugelblitz (artificial black holes formed from a concentration of radiation rather than matter) to significantly delay heat death? The black hole would reconcentrate diffuse matter and turn it into usable energy. It may not be especially practical due to the limited range of the kugelblitz's gravitational field, but if nothing else, couldn't it be used to create energy by feeding it matter? With its smaller size, it would release the produced energy through Hawking radiation much faster than stellar or supermassive black holes.&lt;/p&gt;&#xA;" OwnerUserId="18370" LastActivityDate="2017-09-05T17:07:09.347" Title="Use of kugelblitz to delay heat death" Tags="&lt;black-hole&gt;&lt;entropy&gt;" AnswerCount="0" CommentCount="10" ClosedDate="2017-09-09T08:08:57.773" ContentLicense="CC BY-SA 3.0" />
  <row Id="22522" PostTypeId="1" CreationDate="2017-09-05T18:11:58.290" Score="4" ViewCount="206" Body="&lt;p&gt;Let's say that Peter has a set of superb telescopes and he's watching the sky every night, for 3-4 months. He may not know it just yet, but there is an asteroid heading towards Earth, on a crash course. He is too lazy to plot an orbit, because he doesn't believe he could die like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over the months the Earth travels quite a lot around the Sun, resulting in various amounts of parallax, depending on the distance. The location of the telescopes are also a factor, and so is the (rotation of the) axis of Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To simplify this and to get rid of Earth radius parallax, Peter is a researcher on the South Pole with those mighty telescopes.&lt;br /&gt;&#xA;Over the course of months, what kind of a trajectory would a doomsday asteroid make? As it gets bigger/brighter (which is also relative, because the asteroid itself can rotate)/closer, how would its position change, compared to the stars in the background?&lt;br /&gt;&#xA;Would the change slow down or accelerate? Would it be a small piece of a spiral (I don't mean complete spirals, just the fact that it wouldn't be a parabola or a circle)? What do you think Peter would see if he animated its movement during the course of months on a virtual sky sphere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The National Geographic article &lt;a href=&quot;https://news.nationalgeographic.com/2015/10/151023-astronomy-asteroid-encounter-Earth-stargazing-nasa/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Asteroid Called ‘Spooky’ Will Buzz Earth on Halloween&lt;/a&gt; has an animation of the trajectory of asteroid 2015 TB145 (nicknamed &lt;em&gt;Spooky&lt;/em&gt; because it looks like a skull).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;16 minutes apart visual timelapse:&lt;br /&gt;&#xA;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/e/ec/2015_TB145_discovery.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://upload.wikimedia.org/wikipedia/commons/e/ec/2015_TB145_discovery.gif&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18372" LastEditorDisplayName="user1569" LastEditDate="2017-10-26T12:41:00.310" LastActivityDate="2018-08-30T15:09:44.207" Title="What kind of a visual trajectory would an asteroid have if it were to hit Earth?" Tags="&lt;orbit&gt;&lt;earth&gt;&lt;observational-astronomy&gt;&lt;asteroids&gt;&lt;coordinate&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22524" PostTypeId="1" CreationDate="2017-09-05T18:51:25.180" Score="1" ViewCount="47" Body="&lt;p&gt;We know that energy density is given by this formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u = \frac{1}{V}\int_0^\infty\epsilon_pf(\epsilon_p) g_s V/h^3 4\pi p^2 dp$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\epsilon_p=pc$ for relativistic particles. also $f(\epsilon_p)= \frac{1}{\exp((\epsilon_p -\mu)/kT) \pm1}$ where minus is for bosons and plus is for fermions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we evaluate this for photons which have $\mu =0$ we get&#xA;$u = aT^4$ where $ a= 4 \sigma /c$ is radiation constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If electrons and positrons are in equilibrium with photons (so $\mu = 0$ and $g_s =2$) We will get $u=(7/8) a T^4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;equilibrium equation:&#xA;$\gamma + \gamma \iff e^+ + e^-$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now think we have electron+ positron + three types of neutrinos + three types of anti neutrinos which are at all in equilibrium with photons. $g_s=1$ for neutrinos and anti neutrinos. now how can we get an equation for energy density? what we must use for $g_s$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also not sure about $u = (7/8) aT^4$, Is it the energy density of electrons or positrons (so the total energy density is $2u$) or energy of both of them is in this equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: All of the particles are relativistic. These question conditions is for early universe.&lt;/p&gt;&#xA;" OwnerUserId="16441" LastEditorUserId="11143" LastEditDate="2017-09-05T19:42:41.497" LastActivityDate="2017-09-05T19:42:41.497" Title="Question about finding energy density of a fermionic-bosonic gas" Tags="&lt;astrophysics&gt;&lt;early-universe&gt;&lt;photons&gt;" AnswerCount="0" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22525" PostTypeId="2" ParentId="14833" CreationDate="2017-09-05T20:04:24.737" Score="1" Body="&lt;p&gt;The Universe might have expanded at changing rates. $\Omega_M$ can also influence the redshift. It is not precisely known whether the universe is really flat or open / closed. This can also influence the redshift calculations. (Note: It is also believed that the Hubble constant changed over the evolution of the Universe, decreasing or increasing). Someone else has mentioned before me, and I learned the same at my astronomy classes: massive stars could have formed 0,1Gyrs after TBB. They must have had extremely short lives though. In the first billion years quasars were very trendy, usually they have the great z values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And don't take this the hard way, but astronomers tend to work with ridiculously big margins of error. Also, some functions can be used in a specific value interval only. The farther the data is out of that interval (meaning it's too big/small), the more probable it is that it won't yield a realistic result. Extreme data like this should be taken with a grain of salt.&lt;/p&gt;&#xA;" OwnerUserId="18372" LastActivityDate="2017-09-05T20:04:24.737" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22526" PostTypeId="2" ParentId="22418" CreationDate="2017-09-05T20:16:24.897" Score="4" Body="&lt;p&gt;&quot;Galactic bars develop when stellar orbits in a spiral galaxy become unstable and deviate from a circular path. The tiny elongations&#xA;in the stars’ orbits grow and get locked into place, forming a bar. The bar becomes even more pronounced as it collects more and&#xA;more stars in elliptical orbits. Eventually, a high fraction of the stars in the galaxy’s inner region join the bar. This process has been&#xA;demonstrated repeatedly with computer-based simulations.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source:&#xA;&lt;a href=&quot;http://hubblesite.org/hubble_discoveries/science_year_in_review/pdf/2008/barred_spiral_galaxies_and_galactic_evolution.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://hubblesite.org/hubble_discoveries/science_year_in_review/pdf/2008/barred_spiral_galaxies_and_galactic_evolution.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18372" LastActivityDate="2017-09-05T20:16:24.897" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22530" PostTypeId="1" AcceptedAnswerId="22536" CreationDate="2017-09-06T06:37:55.823" Score="0" ViewCount="1362" Body="&lt;p&gt;I have been given an old telescope (Newtonian, 910mm) and I got a mount for my DSLR and I'm trying to take some pictures of various objects.  In particular, I'm trying to photograph Saturn, but I'm having some difficulty getting a decent image.  This is about the best I could get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RGo4b.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RGo4b.jpg&quot; alt=&quot;Saturn&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was taken with a 2x Barlow T-mounted to my DSLR via an extender, basically a metal tube about 2.5 inches long (without both of these my focuser doesn't have enough range unless I drop an eyepiece into the extender, then the Barlow isn't needed).  I shot on a one second delay to allow any vibration from the shutter opening to subside.  Focus was manual via the telescope itself, and was at the sharpest spot I could find.   I had been outside for some time and temperature difference between the air and the scope were minimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if there are any pointers or maybe an idea of what I might be doing wrong.  This above exposure was 1/10s, ISO 5000.  I thought maybe the ISO setting was too high, but the image below was at 1/15s, ISO 2000 and doesn't look any better:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jcvQ1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jcvQ1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There seems to be some chromatic effect, in the first image the upper-left side has a blue tint which works its way across the spectrum to red in the lower-left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scope was collimated using the star method as I don't have a tool, so it may not be perfect but it's pretty good.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing that I am really puzzled about is that through the lens the image did not look like either of these pictures.  Although a bit small, I could make out the variations in color and some of the texture of the planet and its rings.  The image was sharp and there were no chromatic aberations.  Aside from having the rough shape of Saturn, the pictures have almost no resemblance to what I saw right before I took the pictures.  (This is looking directly through the lens of the camera through the optical viewfinder, live preview also looks ok, but the screen resolution isn't as great.)&lt;/p&gt;&#xA;" OwnerUserId="2341" LastEditorUserId="2341" LastEditDate="2017-09-06T14:58:03.360" LastActivityDate="2017-09-06T14:58:03.360" Title="How to photograph Saturn" Tags="&lt;photography&gt;&lt;saturn&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22531" PostTypeId="2" ParentId="17863" CreationDate="2017-09-06T06:46:29.677" Score="0" Body="&lt;p&gt;You are free to pick your reference point any way you want.  You could pick a $\text{HJD}$ at which the star is brightest, or dimmest, or exactly average, or you could pick the $\text{HJD}$ of the first observation, or you could even set $T_0 = 0$.  If two people pick different $T_0$, then the phases they find differ by a constant value (&lt;a href=&quot;https://en.wikipedia.org/wiki/Modular_arithmetic&quot; rel=&quot;nofollow noreferrer&quot;&gt;modulo&lt;/a&gt; 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is convenient to set $T_0$ at or before the smallest $\text{HJD}$ in your observations, so that all $\text{HJD} - \text{T_0}$ are nonnegative, because the $\text{INT}$-like functions provided by calculators, computer languages, and software libraries don't all round negative numbers in the same direction.  Many round all numbers towards zero, so that $+3.4$ yields $+3$ (as we want) and $-3.4$ yields $-3$, but here we want $-3.4$ to yield $-4$, otherwise you get negative phases before $T_0$ and positive phases after $T_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your observations of brightness are evenly spaced in time, then you can use a &lt;a href=&quot;https://en.wikipedia.org/wiki/Fast_Fourier_transform&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fast Fourier Transform&lt;/a&gt; to find the approximate period and phase.&lt;/p&gt;&#xA;" OwnerUserId="9401" LastActivityDate="2017-09-06T06:46:29.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22532" PostTypeId="2" ParentId="22474" CreationDate="2017-09-06T07:07:02.637" Score="1" Body="&lt;p&gt;This is the problem faced by all star map makers and has been &quot;solved&quot; by each map-maker. The solution has some aesthetic considerations if you use purely area Sirius becomes huge ... So instead of using a formula measure up the symbol sizes used in your favorite star atlas for each magnitude range. &#xA;Check copy-right for your use.&lt;/p&gt;&#xA;" OwnerUserId="10761" LastActivityDate="2017-09-06T07:07:02.637" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22536" PostTypeId="2" ParentId="22530" CreationDate="2017-09-06T08:55:51.683" Score="4" Body="&lt;p&gt;Hmm - good question, Michael; and thanks for sharing your images with us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For me, it's about eliminating possible causes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Collimation - OK (the star test should be fine for your purposes)&lt;/li&gt;&#xA;&lt;li&gt;Focus - probably OK (suggest you try the same kit - scope, Barlow, DSLR - during the day and focus on a far distance object if possible to double-check)&lt;/li&gt;&#xA;&lt;li&gt;Chromatic Aberration - Newtonians are less prone certainly, but it can happen. The most common cause is atmospheric dispersion caused by looking at an object very low in the sky. Your profile doesn't say where you're observing but Saturn is very low in the Northern Hemisphere sky at the moment&#xA;Magnification - one problem for you is that your kit is probably not optimal for small objects. I have a 1000mm FL Newtonian and need to use a x3 Barlow and a specialist CMOS/CCD planetary camera for Saturn as it looks too small on my DSLR (which gets used for larger objects)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You said &quot;The thing that I am really puzzled about is that through the lens the image did not look like either of these pictures&quot; - please could you clarify what you meant by lens here; were you talking about using the scope, Barlow and a separate eyepiece, or were you using scope, Barlow and DSLR with live preview lens?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is much more common for people to image the planets using a video stacked approach. The idea behind this is that they are very bright (so exposure times can be very short) and the limiting factor is random blurring caused by atmospheric seeing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take a video file of say 1000 frames at 20fps then there is software available that can combine these into good shots (e.g. AutoStakkert, Registax). What DSLR are you using? For what it's worth, the human eye is very good at picking details and resolving out the seeing on bright objects, so my first photo attempts were dreadful compared to what I could see in the eyepiece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd recommend you try reading up on this approach for inspiration. Some (but not all) DSLRs do allow video recording modes, so it might be worth a try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn is definitely a tricky object to go for - at the time of writing Jupiter isn't in a good Northern Hemisphere space (I'd normally suggest trying it out as a somewhat easier target). Have you tried practising by imaging parts of the Moon - it's a much easier target to build confidence and practical experience? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please don't be disheartened ... it took me a couple of years to get semi-decent pictures of Saturn&lt;/p&gt;&#xA;" OwnerUserId="17502" LastActivityDate="2017-09-06T08:55:51.683" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22539" PostTypeId="2" ParentId="20505" CreationDate="2017-09-06T15:22:12.253" Score="1" Body="&lt;p&gt;This is not an answer, but an update on how the photography of the eclipse went.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ended up on ISO 400 and f/5.6 (limited by my 300mm lens) doing a series of nine 0.7 stop bracketing, ranging from 1/3200 to 1/80 second exposures, which turned out to be just about bang on for corona and prominences as you should be able to see on the attached images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BxFxA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BxFxA.jpg&quot; alt=&quot;1/80s&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Jkoa7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jkoa7.jpg&quot; alt=&quot;1/3200s&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16360" LastActivityDate="2017-09-06T15:22:12.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22540" PostTypeId="2" ParentId="22519" CreationDate="2017-09-06T16:29:04.133" Score="2" Body="&lt;p&gt;Your misunderstanding is in the second paragraph. If $n&amp;lt;n_c$ then there are insufficient collisions to excite the molecule into the upper level. Thus in equilibrium there are very few excited molecules and since the luminosity is proportional to the number of molecules in the excited level then little radiation is seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The critical density is defined as that (temperature-dependent) density at which the rate of collisional excitation to the upper level of a molecular (or atomic, I suppose) transition, equals the rate of radiative de-excitation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate of de-excitation is given by $n_2\ A_{21}$, where $n_2$ is the number density of the molecular species &lt;em&gt;in the excited state&lt;/em&gt; and $A_{21}$ is the relevant spontaneous emission coefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate of excitation is given by $\gamma_{12} n_1 n$,&#xA;where $\gamma_{12}$ is the collision probability (which depends on the speed of molecules in the gas and hence temperature), $n_1$ is the number density of the molecule of interest in the lower energy state and $n$ is the number density of &lt;em&gt;all molecules&lt;/em&gt; in the gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a rate of collisional de-excitation given by $\gamma_{21}\ n_2 n$. However there must be a relationship between $\gamma_{12}$ and $\gamma_{21}$ because for a gas in thermodynamic equilibrium, we know that &#xA;$$n_2/n_1 = \exp(-E_{21}/k_BT),$$ and so if collisions are in detailed balance then&#xA;$$\gamma_{21}\ n_1 \exp(-E_{21}/k_BT) n = \gamma_{12} n_1 n$$&#xA;$$ \gamma_{21} = \gamma_{12} \exp(E_{21}/k_BT)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, so now consider a gas where there are radiative losses from state 2 and allow it to come into equilibrium such that&#xA;$$\frac{dn_2}{dt} = 0$$&#xA;$$ -A_{21}n_2 - \gamma_{21}n_2 n + \gamma_{12} n_1 n = 0$$&#xA;$$ \frac{n_2}{n_1} = \exp(-E_{21}/k_BT)\ \frac{1}{1 + A_{21}/\gamma_{21}n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If now we say that $A_{21} \ll \gamma_{21}n$, then radiative de-excitation is (relatively) negligible compared with collisional de-excitation then the $n_2/n_1$ ratio approaches the level expected by the Boltzmann ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If on the other hand $A_{21} \gg \gamma_{21}n$, then the ratio of $n_2/n_1$ becomes much &lt;em&gt;lower&lt;/em&gt; and radiation removes excited molecules faster than collisions create them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we ask how much radiation we are going to see from this gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) If $A_{21} \ll \gamma_{21}n$, which we can restate as $n \gg n_c$, the &quot;critical density&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The luminosity from the molecules will be &#xA;$$L \propto n_2 A_{21} \simeq n_1 A_{21} \exp(-E_{21}/k_BT)$$&#xA;If $k_BT &amp;lt; E_{21}$ this means that $n_1 \gg n_2$ and the luminosity per molecule in the gas has a fixed value that is independent of density $n$. There is just a fixed amount of radiative losses per molecule of the species in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) If $A_{21} \gg \gamma_{21}n$, which we can restate as $n \ll n_c$, The luminosity from the molecules will be &#xA;$$L \propto n_2 A_{21} \simeq n_1 A_{21} \exp(-E_{21}/k_BT) \frac{\gamma_{21} n}{A_{21}}$$&#xA;and so this is &lt;em&gt;smaller&lt;/em&gt; than for the first case by a factor of $n/n_c$. So now there is less radiation per molecule of the interesting species because fewer of those molecules are in an excited state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB This is a different thing to the &quot;quenching&quot; density for &lt;em&gt;forbidden&lt;/em&gt; lines that I refer to in my comment.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-09-06T16:29:04.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22541" PostTypeId="2" ParentId="22487" CreationDate="2017-09-06T16:38:08.103" Score="0" Body="&lt;p&gt;Watch this video, I think it will give you an answer, sometimes it is better at once to see , then 10 times to read:&#xA;&lt;a href=&quot;https://youtu.be/rENyyRwxpHo&quot; rel=&quot;nofollow noreferrer&quot;&gt;Laniakea: Our home superclaster &lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17507" LastEditorUserId="17507" LastEditDate="2017-09-06T16:44:16.360" LastActivityDate="2017-09-06T16:44:16.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22542" PostTypeId="2" ParentId="22455" CreationDate="2017-09-06T17:54:25.353" Score="7" Body="&lt;p&gt;I suspect it's a combination of two things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Stable, guaranteed high-resolution imaging across the entire field of view, something not possible with ground-based adaptive optics;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Very low background in the optical for &lt;em&gt;HST&lt;/em&gt; (Hubble), versus a very high background for ground-based AO in the near-infrared.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Most adaptive optics systems are only able to correct in a small region (the &quot;isoplanatic patch&quot;) around a bright &quot;guide star&quot; (say, half an arc minute in radius at most); even with artificial laser guide stars, you still need a moderately bright guide star for so-called &quot;tip-tilt&quot; correction. This means you can only do adaptive optics searches in limited parts of the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;HST&lt;/em&gt;, on the other hand, delivers high-resolution imaging throughout its entire field of view (several arc minutes across), all the time, regardless of where it's pointed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make matters worse, the trajectory of New Horizons is close to the Galactic plane, so there are lots of &lt;em&gt;faint&lt;/em&gt; background stars. This makes it harder to pick out possible Kuiper Belt objects, making a very accurate and stable point spread function (such as that of &lt;em&gt;HST&lt;/em&gt;) even more important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These searches are best done in the optical, to minimize the sky background. The absence of atmospheric sky glow (mostly scattered light from the Sun and Moon) for &lt;em&gt;HST&lt;/em&gt; makes it easier to quickly detect faint sources like KBOs. The fact that adaptive optics systems, as you and Rob Jeffries note, work almost entirely in the near-infrared, where the atmospheric background is much higher, makes it even worse for them.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="88" LastEditDate="2018-01-15T16:06:04.613" LastActivityDate="2018-01-15T16:06:04.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22543" PostTypeId="1" AcceptedAnswerId="22544" CreationDate="2017-09-06T19:00:07.833" Score="1" ViewCount="164" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8oTgZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8oTgZ.jpg&quot; alt=&quot;partial eclipse&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was taking shots of the 2017 eclipse using my D7200 camera with a 1000 ND filter. As I was going through the pictures after downloading them to a pc, I noticed that there's what appears to be a tiny light source present off to the side of the sun. I have circled it in the picture below. It is also present in the pictures during totality, which I took without the ND filter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't seem like a camera or a lens artifact, since it showed up pretty consistently in different shot configurations, but I did not know that a 300 mm zoom with an 1000 ND filter could capture a light source besides the sun and the moon in full day light. If it is not an artifact, what could be the source of this light?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/96dYD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/96dYD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18388" LastEditorDisplayName="user1569" LastEditDate="2017-09-07T19:34:28.003" LastActivityDate="2017-09-07T19:34:28.003" Title="A real light source, or an artifact?" Tags="&lt;the-sun&gt;&lt;light&gt;&lt;solar-eclipse&gt;&lt;photography&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22544" PostTypeId="2" ParentId="22543" CreationDate="2017-09-06T19:38:39.493" Score="1" Body="&lt;p&gt;I agree with the comment by StephenG. This is a &lt;a href=&quot;https://www.howtogeek.com/194866/how-to-identify-stuck-pixels-and-remove-them-from-your-digital-photos/&quot; rel=&quot;nofollow noreferrer&quot;&gt;stuck pixel&lt;/a&gt;. As the link suggests, they are a very common flaw in digital cameras, and not an astronomical feature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this because it appears, on zooming in, to be a single pixel defect, most light sources even with the best focus, cover more than 1 pixel. The filter you are using is very powerful removing nearly all light. An object would have to be exceptionally bright to be seen through such a filter. If it was real then it would be so bright that when you removed the filter, that one spot would have totally overexposed your image. Since stuck pixels are pretty common, and you might not have noticed them in a picture taken at day, or of a star-field, it is a safe bet that that is the problem.  &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-09-06T19:38:39.493" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22545" PostTypeId="1" AcceptedAnswerId="22637" CreationDate="2017-09-06T21:22:37.243" Score="8" ViewCount="3064" Body="&lt;p&gt;My specific example for the question is the future collision of the galaxies Milky Way (our own galaxy) and Andromeda in a couple billion years. The star in question is obviously the sun in this case. I want to know the chances of a collision with another star and if it's significant or not. &lt;/p&gt;&#xA;" OwnerUserId="4160" LastActivityDate="2019-05-16T21:19:07.533" Title="What are the chances of a star colliding with another during a galactic collision?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;galaxy&gt;&lt;milky-way&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22546" PostTypeId="1" CreationDate="2017-09-07T00:07:57.433" Score="5" ViewCount="8675" Body="&lt;p&gt;I know we can't see the center.  But do we see any of the central bulge?  Are we just seeing nearby stars in the Orion Spur where we are?  In the Sagittarius Arm?  In the other direction, are we seeing any of the arms further out from ours?  Where are the dust clouds we see, in relation to these arms?&lt;/p&gt;&#xA;" OwnerUserId="18392" LastActivityDate="2017-10-25T14:44:00.400" Title="What specific parts of the Milky Way do we see when we look up?" Tags="&lt;milky-way&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22547" PostTypeId="2" ParentId="22546" CreationDate="2017-09-07T00:10:03.393" Score="0" Body="&lt;p&gt;The Universe Today has a great video on just this topic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.universetoday.com/115203/what-part-of-the-milky-way-can-we-see/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.universetoday.com/115203/what-part-of-the-milky-way-can-we-see/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT to clarify:&lt;br&gt;&#xA;If you look in the direction of the constellation Sagittarius, you're looking towards the center of the Milky Way Galaxy. You're correct that we can't see all the way to the center because of obscuring dust. Since we're looking out from inside one of the spiral arms, we can only see about 1500 light-years, which amounts to roughly 1.5 - 2% of the Galaxy's diameter.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/CzMXg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CzMXg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17535" LastEditorUserId="17535" LastEditDate="2017-09-07T01:39:53.943" LastActivityDate="2017-09-07T01:39:53.943" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22554" PostTypeId="1" CreationDate="2017-09-07T09:14:09.683" Score="3" ViewCount="72" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_meteorite&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; the first meteorite of Lunar origin wasn't identified until 1982. Today 306 Lunar meteorites have been identified.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why weren't they identified as Lunar already a decade earlier, when&#xA;the Apollo samples had been analyzed?&lt;/li&gt;&#xA;&lt;li&gt;How do Lunar meteorites differ, the Lunar surface consists of&#xA;meteoroid material anyway?&lt;/li&gt;&#xA;&lt;li&gt;Were there suggestions before 1982 that certain meteorites were of&#xA;Lunar origin, and did they turn out to be correct?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2017-09-07T09:14:09.683" Title="Why weren't Lunar meteorites identified until 1982?" Tags="&lt;the-moon&gt;&lt;history&gt;&lt;meteorite&gt;" AnswerCount="0" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22555" PostTypeId="1" AcceptedAnswerId="22631" CreationDate="2017-09-07T10:28:35.037" Score="3" ViewCount="359" Body="&lt;p&gt;I've found that the speed of expansion is about $70\frac{km}{s\cdot 10^{6}pc}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what is the acceleration of expansion in terms of $\frac{km}{s^{2}\cdot 10^{6}pc}$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know how fast this value of &quot;70&quot; is changing, ie what is $\mathrm{d}H_0/\mathrm{d}t$?&lt;/p&gt;&#xA;" OwnerUserId="18401" LastEditorUserId="2365" LastEditDate="2017-09-11T17:43:19.860" LastActivityDate="2017-09-11T21:33:05.400" Title="What is the acceleration for the expansion of the Universe?" Tags="&lt;cosmological-inflation&gt;" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="22557" PostTypeId="1" CreationDate="2017-09-07T11:56:34.063" Score="3" ViewCount="145" Body="&lt;p&gt;Could the expansion of universe be a side effect of gravity? I mean maybe gravity &quot;attracts things&quot; at a &quot;small&quot; distances and repells them at &quot;larger&quot; distances?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we have any theory/hypothesis/paper on this topic? Is it possible to prove or disprove?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPD.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another, more clear way to ask the same question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Could gravity have a repulsive term that dominates at cosmological&#xA;  distances and could it be such a term that causes (accelerated)&#xA;  expansion?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Thanks to @uhoh and @Richard, for helping to clarify question a bit.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="18401" LastEditorUserId="18401" LastEditDate="2017-09-10T10:37:00.567" LastActivityDate="2017-09-10T19:52:37.580" Title="Could the expansion of universe be a side effect of gravity?" Tags="&lt;cosmological-inflation&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22558" PostTypeId="1" AcceptedAnswerId="22564" CreationDate="2017-09-07T12:37:34.103" Score="2" ViewCount="883" Body="&lt;p&gt;Would it explode in a supernova or  will it form a new star?&lt;/p&gt;&#xA;" OwnerUserId="18338" LastActivityDate="2017-09-08T11:33:48.710" Title="What happens if a white dwarf collides with a star like our sun or a red dwarf?" Tags="&lt;star&gt;&lt;astrophysics&gt;&lt;white-dwarf&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22559" PostTypeId="1" AcceptedAnswerId="22562" CreationDate="2017-09-07T14:37:39.730" Score="3" ViewCount="377" Body="&lt;p&gt;In 1.5 million years, the star &lt;a href=&quot;https://en.wikipedia.org/wiki/Gliese_710&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gliese 710&lt;/a&gt; is expected to pass through the Oort cloud. Now obviously this will shower the inner solar system with comets, but the Wikipedia page doesn't say anything about the star's impact on the orbits of the planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the star not throw the Earth and other planets into highly elliptical orbits, destroying all life? Or, simply, do we not know because this would involve solving the 11-body problem?&lt;/p&gt;&#xA;" OwnerUserId="1494" LastActivityDate="2017-09-07T18:29:48.650" Title="What will be the impacts on the planets' orbits from Gliese 710?" Tags="&lt;star&gt;&lt;solar-system&gt;&lt;future&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22560" PostTypeId="1" AcceptedAnswerId="22563" CreationDate="2017-09-07T17:12:14.613" Score="5" ViewCount="348" Body="&lt;p&gt;I ask myself the reason of why the sun in not uniform on the edge of the moon at this moment of the eclipse (cf. picture) is not unifom is because of the mountains of the moon ? Or can it due to erruptions of sun ? If not is there another reason to explain that ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CHVFH.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CHVFH.jpg&quot; alt=&quot;Eclipse of sun&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18281" LastActivityDate="2019-01-30T19:41:41.570" Title="Mountains on the moon during solar eclipse?" Tags="&lt;the-sun&gt;&lt;the-moon&gt;&lt;solar-eclipse&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22561" PostTypeId="2" ParentId="22558" CreationDate="2017-09-07T17:50:29.653" Score="4" Body="&lt;p&gt;I'm far from an expert in this, but I have a reasonable guess at an answer.  The &lt;a href=&quot;https://en.wikipedia.org/wiki/White_dwarf#Binary_stars_and_novae&quot; rel=&quot;nofollow noreferrer&quot;&gt;White Dwarf page on Wikipedia&lt;/a&gt; describes what would probably happen.  It seems that the WD would consume mass from the companion star until it can either sustain fusion, or fuse so much so quickly that it explodes into a Type 1a supernova.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If a white dwarf is in a binary star system and is accreting matter&#xA;  from its companion, a variety of phenomena may occur, including novae&#xA;  and Type Ia supernovae. It may also be a super-soft x-ray source if it&#xA;  is able to take material from its companion fast enough to sustain&#xA;  fusion on its surface. A close binary system of two white dwarfs&#xA;  can radiate energy in the form of gravitational waves, causing their&#xA;  mutual orbit to steadily shrink until the stars merge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It does appear to depend on the mass of the companion, and whether they actually collide or merely pass close.  If they didn't form together, it is unlikely that they will be captured in each other's orbit later on, so the WD might just pass through the system, disrupting planets as it goes but leaving the other star unharmed.  There doesn't appear to be a one-size-fits-all answer here.&lt;/p&gt;&#xA;" OwnerUserId="14255" LastActivityDate="2017-09-07T17:50:29.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22562" PostTypeId="2" ParentId="22559" CreationDate="2017-09-07T18:29:48.650" Score="7" Body="&lt;p&gt;It'll have basically no effect on the planet's orbits, except perhaps on the theoretical planet 9 and even then, not very much.  As close as it's expected to get Gilese will till be too far away to affect the 8 planets much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gilese 710's mass is about 60% the mass of our Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen two estimates, of 13,000 Astronomical Units (AU) and 13,365 or about 77 light days or 0.21 light years at closest pass, and some estimates put a +/- margin of error at as much as 20% of that, but lets use 13,365 AU as an estimate.   For a star, that's exceptionally close but it's enormously distant for a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neptune, the most distant planet, averages about 30.1 AU from the Sun, so Gilese will be about 440 times further away from Neptune than Neptune is from the Sun.   That will give it a negligible gravitational effect of less than 1 part in 300,000 on Neptune's orbit.  That's, by crude calculation, a bit less than the effect Saturn has on Earth's orbit.  That's enough for a small perturbation, but you'd need some pretty high tech equipment to even detect it.   It's about 50 times less than the gravitational effect Neptune had on Uranus when it was noticed that Uranus' orbit was behaving oddly and that observation lead to the discovery of the planet Neptune.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, basically, for the 8 known planets, the outer planets will get small pushes, noticeable by very meticulous observation only and the inner planets, essentially no effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's effect on Planet 9 would be greater but still small.  The estimate for planet 9 at Aphelion is about 1,200 AU or 1/11th the closest pass of Gilese.  That's close enough for a noticable push.  Not enough for any major changes like pushing it's eccentricity to where it passes anywhere close to Neptune, or ejecting it from the solar system, but Planet 9 could be jostled somewhat, perhaps as much as a 1% change to it's orbital period and/or eccentricity, which for a planet's orbit, 1% is a large change over a single orbital period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bigger effect, as you mentioned, will be on more distant objects, several thousands to tens of thousands of AU from the Sun, or several times more distant than the theoretical Planet 9 at it's most distant.   Objects that distant have never been observed, but are theorized to be out there in vast numbers based on a handful of observed objects with very high eccentricity.  Oort cloud objects at those distances will get cast every which way, some cast out of the solar-system and a small percentage, pushed to higher elliptical orbits where they could pass through the inner solar system.   Our sun will do the same thing to Gilese 710, casting it's Oort cloud objects in every direction and each star will steal a number of orbiting objects from the other.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial effect, the inner solar system might pass through the outer edge of Gilese 710's outer oort cloud.  I don't think anyone knows how extensive it's Oort cloud is at 13,000 AU, but any initial increase in objects passing through the inner solar system will be from Gilese 710's Oort cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any of our Solar system's Oort cloud objects that Gilese 710 sends towards the inner solar system will take (roughly) between 50,000 and 400,000 years to reach the inner solar-system, so Gilese 710 will be perhaps still visible to the naked eye when the proposed rain of comets begins, but it will be long past being the brightest star in the sky if/when any objects it sends towards the inner solar system make it here.   The reason for this is that it moves much faster than any objects it's likely to send our way.   Gilese 710 will pass through the solar system at about 50,000 kph and any objects it sends towards the inner solar system, because they're so spread out and most are likely to just get a small push from the passing star, they will leisurely drift towards the inner solar system at roughly between 500 and 1,500 kph, requiring about 50,000 years for the first ones to begin to reach the inner solar system and become comets or perhaps crash into the Earth or Moon or other planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, fun fact.   At closest pass, Gilese 710 should be a hair brighter than Jupiter at Jupiter's brightest, nearly -2.8 apparent magnitude.  It will be the brightest star in the sky for ballpark 15,000 - 20,000 years, over that time-frame it will move across the fixed stars a good percentage of the 180 degrees of horizon.   In a human lifetime it might move a bit over 1 degree of arc at closest pass, so it won't be very noticeable that it's moving, but that's very rapid proper motion for a star.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Too long?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've excluded the mathematical formulas cause I don't do the that well and it would get a little ugly.  I can't do the math of a 3 body system, so this is an approximation, but I my approximation should be in the ballpark.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-09-07T18:29:48.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22563" PostTypeId="2" ParentId="22560" CreationDate="2017-09-07T19:13:14.353" Score="7" Body="&lt;p&gt;The beads of bright light can be seen thanks to the Moons rugged topography. They're called &lt;a href=&quot;https://en.wikipedia.org/wiki/Baily%27s_beads&quot; rel=&quot;noreferrer&quot;&gt;Baily's beads&lt;/a&gt; in honor of an English astronomer who explained the phenomenon, Francis Baily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the photograph you posted, you can also see &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_prominence&quot; rel=&quot;noreferrer&quot;&gt;Solar prominences&lt;/a&gt;. It's plasma ejected into the corona. Its temperature is lower than that of the coronal plasma and it emits more visible light.&lt;/p&gt;&#xA;" OwnerUserId="211" LastActivityDate="2017-09-07T19:13:14.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22564" PostTypeId="2" ParentId="22558" CreationDate="2017-09-07T21:42:33.020" Score="5" Body="&lt;p&gt;Fun question.   Without doubt it would be very violent and spectacular, but not much is known about &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_collision&quot; rel=&quot;nofollow noreferrer&quot;&gt;stellar collisions&lt;/a&gt; and only a few have ever been observed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most stellar collisions happen due to tight orbits where the stars spiral in towards each other or perhaps, 3 or more body chaotic star systems with unstable orbits that lead to an impact.   Space is so large that actual star on star collisions are exceptionally rare.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A spiral in towards each other is almost certainly more common, it also quite different in terms of outcome.  Much of the gravitational kinetic energy of the collision goes into angular momentum, so, while still spectacular, it's quite a bit less explosive than a direct head-on collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.cbsnews.com/news/gravity-waves-from-black-hole-merger-detected/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Black hole/black hole collisions&lt;/a&gt; have been observed by detection of gravity waves when two black holes orbit around each other and the orbit decays.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.smithsonianmag.com/science-nature/all-the-gold-in-the-universe-could-come-from-the-collisions-of-neutron-stars-13474145/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Neutron Star/Neutron star&lt;/a&gt; collisions may be responsible for the very heavy elements in the Universe and White dwarf-White dwarf collisions may lead to a supernova similar to a type 1a Supernova (&lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_collision#Type_Ia_supernova&quot; rel=&quot;nofollow noreferrer&quot;&gt;per Wikipedia source&lt;/a&gt;).   All these collisions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a tight orbit, a white dwarf will steal material from a near-by star and if the orbit is close enough, lead to a decaying orbit and collision, but that would happen over time.  If the white dwarf reaches the &lt;a href=&quot;https://en.wikipedia.org/wiki/Chandrasekhar_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chandrasekhar limit&lt;/a&gt; in that scenario, it will undergo type 1a supernova.  If there's not enough mass for that, it should eventually absorb the nearby star, leading to a new star, admittedly a strange one with very high metallicity and perhaps abnormally hot and short lived for it's mass, but it should basically be a star once the collision settles down, but probably a star that burns both hydrogen and helium, but wouldn't collect enough helium for a helium flash.   I don't see any other alternative with that kind of below Chandrasekhar mass spirally in collision than it being a very metallic star, hot for it's mass and with a shorter lifespan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a head on collision - that's very different and, I'm basically guessing, but you have a white dwarf star - about the size of Earth with a mass probably at least 30%-40% of the sun.   White dwarfs don't have a specific minimum size, but for a star to go white dwarf within the 13.6 billion years or so that stars have been around, that star would have needed enough mass to have completed lifecycle and in the process, it loses some mass, so perhaps, 30% of the mass of the sun might be the current lowest possible mass of a white dwarf, but as billions of years pass, that low end will grow smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we take a low-mass white dwarf at 30%-40% the mass of the sun, it's still a super-dense object and crashing that into a star at their combined escape velocity, not the more common decaying orbital spiral, that's a messy calculation.   I'd only be guessing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Escape velocities aren't additive, but still, white dwarf and small star would collide in this theoretical at an impressive velocity.    If it was a glancing blow, the white dwarf would almost certainly punch straight through, blowing a lot of matter off the star and perhaps affecting it's rotation rate.   It would be an impressive site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it was a direct hit, I don't know if the white dwarf would punch through the star or not.   It might still punch through.   The White dwarf would pull the surface of the star towards it as it got close and that would be the first contact between the two.   The Sun's escape velocity is about 618 km/s and the mutual gravitation the White dwarf would be traveling to the sun a bit above that, but the White dwarf's surface gravity is much greater, so the mater pulled off the sun into the white dwarf would be hitting the white dwarf at a few thousands of km/s.  (Small white dwarf, so lets say 3,000 km/s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's still far too slow for fusion to happen, so this is basically still just collision mathematics.   As the white dwarf plowed deeper into the star, towards the much hotter core, the heat and velocity of impact probably would increase fusion rates, but if it just passed right through the star that would be a pretty temporary and minimal effect.   The larger effect would be the disruption to the star after a very dense planet sized object punched it's way through it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the star was viscus enough to stop the white dwarf from making it all the way through, then the combined mass and dense center and added heat from impact should significantly increase the rate of fusion of the original star, even though some material would likely be blown off, the net result should be a heavier star that burns hotter and faster and if the new mass is comfortably above 1.4 solar masses, you likely have a super-nova in the making.   But as to whether or not a white dwarf would be to punch all the way through a star upon direct collision - I have no idea.  I guess that it would, but that's just a guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fun related question is, how long would it take the white dwarf to emerge from the other side, assuming it's a relatively center hit and assuming it passes through (which it might, but I'm not sure if it does).   I'll try to give that a look too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, just my 2 cents on this question.   Corrections welcome.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-09-08T11:33:48.710" LastActivityDate="2017-09-08T11:33:48.710" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22566" PostTypeId="1" AcceptedAnswerId="22572" CreationDate="2017-09-08T00:12:43.333" Score="3" ViewCount="79" Body="&lt;p&gt;Does wavelength play a role in cosmological redshift? Do we see certain wavelengths affected less by expansion or notice any delay in arrival of certain wavelengths?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How accurate are any observations?&lt;/p&gt;&#xA;" OwnerUserId="17699" LastEditorUserId="17699" LastEditDate="2017-09-08T00:42:31.940" LastActivityDate="2017-09-08T08:15:52.823" Title="Does wavelength affect redshift caused by the metric expansion of space?" Tags="&lt;cosmology&gt;&lt;expansion&gt;&lt;redshift&gt;&lt;cosmological-inflation&gt;&lt;wavelength&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22570" PostTypeId="2" ParentId="22555" CreationDate="2017-09-08T08:10:00.977" Score="0" Body="&lt;p&gt;You have 1pc=3.26ly=3.26*300000*365.25*24*3600 km = ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Parsec&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Parsec&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you could easily find H0 :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry I've read to quickly the question. The acceleration rate of the universe is given by second equation of Friedmann Lemaître.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\ddot{a}}{a} =  -\frac{4 \pi G}{3}\left(\rho+\frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\rho$ and p density of the fluids (matter, dark energy...) you consider in the cosmological (standard) model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Friedmann_equations&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Accelerating_expansion_of_the_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Accelerating_expansion_of_the_universe&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18281" LastEditorUserId="18281" LastEditDate="2017-09-08T14:19:15.397" LastActivityDate="2017-09-08T14:19:15.397" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22572" PostTypeId="2" ParentId="22566" CreationDate="2017-09-08T08:15:52.823" Score="5" Body="&lt;p&gt;Standard cosmological models predicts that the cosmological redshift and the speed of light are wavelength-independent. This result is confirmed observationally e.g. by &lt;a href=&quot;https://arxiv.org/abs/1605.09380&quot; rel=&quot;noreferrer&quot;&gt;Ferreras &amp;amp; Trujillo (2016)&lt;/a&gt;, who used $\sim500\,000$ &lt;a href=&quot;http://www.sdss.org&quot; rel=&quot;noreferrer&quot;&gt;SDSS&lt;/a&gt; galaxy spectra down to a precision of $\Delta z \sim 10^{-6}$ and $\Delta z \sim 10^{-5}$ for galaxies at $z&amp;lt;0.1$ and $z&amp;gt;0.1$, respectively.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2017-09-08T08:15:52.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22577" PostTypeId="1" CreationDate="2017-09-08T17:18:04.617" Score="5" ViewCount="229" Body="&lt;p&gt;I would like to know please whether or not I can look for exoplanets without a telescope.&lt;/p&gt;&#xA;" OwnerUserId="18419" LastEditorUserId="37191" LastEditDate="2021-05-18T22:36:27.727" LastActivityDate="2021-05-18T22:36:27.727" Title="Can I find exoplanets without a telescope?" Tags="&lt;observational-astronomy&gt;&lt;exoplanet&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22581" PostTypeId="2" ParentId="12095" CreationDate="2017-09-08T19:53:29.687" Score="2" Body="&lt;p&gt;Not quite the webservice, but there is a JavaScript library Orb2.js with examples including X-Y(-Z) in ecliptic coordinates. So maybe you can compute it directly, not relying on some external service:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.lizard-tail.com/isana/lab/orbjs/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.lizard-tail.com/isana/lab/orbjs/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I do not use it actively, it just a pointer to site to explore, not a definitive solution for you.&lt;/p&gt;&#xA;" OwnerUserId="17348" LastActivityDate="2017-09-08T19:53:29.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22582" PostTypeId="2" ParentId="2441" CreationDate="2017-09-09T01:12:30.610" Score="0" Body="&lt;p&gt;An observer falling into a black hole does not see himself fall into the singularity unimpeded. The black hole will always evaporate before infinity, therefore the infalling observer will fall to the center of an evaporated black hole and find nothing special other than universal heat death. &lt;/p&gt;&#xA;" OwnerUserId="18425" LastActivityDate="2017-09-09T01:12:30.610" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22583" PostTypeId="1" CreationDate="2017-09-09T03:18:22.947" Score="0" ViewCount="3918" Body="&lt;p&gt;We know neutron star is very densest in the universe and it weighs the thousand times of our Sun. And Black hole having strong gravitational pull not even light can escape From it . So , whose gravitational pull is greater ?&lt;/p&gt;&#xA;" OwnerUserId="18170" LastActivityDate="2017-09-15T23:44:35.573" Title="whose gravitational pull is more stronger , Neutron Star or Black hole?" Tags="&lt;star&gt;&lt;black-hole&gt;&lt;neutron-star&gt;" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22585" PostTypeId="2" ParentId="22546" CreationDate="2017-09-09T05:55:57.260" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But do we see any of the central bulge?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With the naked eye and a dark sky, the galactic center is brighter than other areas of the galactic plane.  Some of the light from that region does make it to earth to make that direction identifiable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the ESO night sky composite from Brunier and Tapissier (which is approximately what someone could see with the naked eye given dark skies and good viewing), you can see that while the galactic center is brighter, other directions along the plane seem fairly similar.  That makes it very difficult to say what fraction of scattered glow is coming from one arm or another.  It all just blends together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VtiqH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VtiqH.jpg&quot; alt=&quot;ESO0932 Milky Way Galaxy&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.eso.org/public/news/eso0932/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.eso.org/public/news/eso0932/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If all we can see are stars in our galaxy that are relatively close to us, then the Milky Way should look relatively uniform, right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not necessarily.  The visible part of the galaxy is relatively thin.  It doesn't have a fixed distance, but a commonly quoted one is 1000ly.  That means that even if light were magically cut off at 3000ly or so, it could appear non-uniform.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2017-09-09T05:55:57.260" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22586" PostTypeId="2" ParentId="22583" CreationDate="2017-09-09T06:20:06.773" Score="3" Body="&lt;p&gt;The black hole has greater pull.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you stand beside a massive object (planet, star, black hole or anything else with mass) you will feel a gravitational pull, and unless you do something about it you will start to accelerate towards the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the gravitational pull (or more precisely the gravitational acceleration) depends on &lt;em&gt;two&lt;/em&gt; things: How massive the object is, and &lt;em&gt;how far you are from its centre&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutron stars are all between about 1.2 and 3 times the mass of the sun, any larger and they become black holes. Objects that are believed to be black holes are know with masses of about 4 times the mass of the sun. A black hole can be of any mass, there is no known way for smaller black holes to form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you fix the distance (say 150 million km), then the gravitational pull depends only on the mass of the object. A neutron star can be at most about three times the mass of the sun, black holes are nearly all larger than that, so the gravitational pull of the black-hole is greater. But if you find a way to make a small black hole, then it would have lower mass and so less gravitational pull, at the same distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also look at surface gravity. Black holes don't have a proper surface, but you could use the event horizon instead. Here the answer is easy: The gravitational pull becomes infinite at the event horizon (from a certain point of view), Neutron stars have very strong, but not an infinitely strong pull so the gravitational pull of a black hole is greater than a neutron star.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-09-09T06:20:06.773" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22592" PostTypeId="1" CreationDate="2017-09-09T10:48:58.253" Score="1" ViewCount="1702" Body="&lt;p&gt;My friend posted a picture of the moon as if there was an eclipse because it was so red but there wasn't a scheduled eclipse at the time (it was last night, from the Philippines). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a little concerned because how else is that possible? It was in Manila so I guess there was a lot of smoke in the air but I don't know if that's the reason behind the moon's redness.&lt;/p&gt;&#xA;" OwnerUserId="18432" LastEditorUserId="2365" LastEditDate="2017-09-09T11:26:00.070" LastActivityDate="2017-09-09T11:33:43.397" Title="Can the moon turn red even without an eclipse?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22593" PostTypeId="2" ParentId="22592" CreationDate="2017-09-09T11:21:15.200" Score="2" Body="&lt;p&gt;Yes, pollution from (for example) wildfires can cause the moon to appear red. It is normally not noticed unless the moon is low in the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earthsky magazine is reporting that &lt;a href=&quot;http://earthsky.org/earth/red-moon-sun-wildfire-smoke-images-aug-sept-2017&quot; rel=&quot;nofollow noreferrer&quot;&gt;red sunsets and red moons are being reported across the USA.&lt;/a&gt; caused by wildfires. Even when the moon is quite high in the sky. I believe that fires from logging in the jungles of Borneo and the Philippines can have the same effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has also been answered on &lt;a href=&quot;https://earthscience.stackexchange.com/questions/12223/why-were-both-the-sun-and-the-moon-red-today&quot;&gt;https://earthscience.stackexchange.com/questions/12223/why-were-both-the-sun-and-the-moon-red-today&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-09-09T11:33:43.397" LastActivityDate="2017-09-09T11:33:43.397" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22594" PostTypeId="1" CreationDate="2017-09-09T12:31:06.623" Score="4" ViewCount="310" Body="&lt;p&gt;Which is the nearest &lt;em&gt;known or suspected&lt;/em&gt; black hole from our solar system and how far is it from us? Is there an easily-viewed catalog that &lt;em&gt;astronomers use&lt;/em&gt; to keep track of these?&lt;/p&gt;&#xA;" OwnerUserId="18434" LastEditorUserId="7982" LastEditDate="2017-09-10T07:22:47.953" LastActivityDate="2017-09-10T11:42:44.183" Title="How far is the nearest black hole from our solar system? Is there an official list?" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="22595" PostTypeId="2" ParentId="11997" CreationDate="2017-09-09T13:55:31.130" Score="1" Body="&lt;p&gt;The issue is the definition, as always. What is the Zodiac? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mean the zone where the sun stays in the sky during the year, a strip around the ecliptic,  with current astronomical constellations, the solar path crosses 13 constellations. If you include the moon, Venus and the planets, the answer is 21 or 25, depending on the time scale. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the definition is also dependent on the constellation patches and regions, so it is relative. Moreover, as the Universe moves, and stars move too, it could change in the past and it could change in the future.&#xA;Millions or billions of years in the future, supposing Earth and the Sun are still alive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference here &lt;a href=&quot;http://www.ips-planetarium.org/?page=a_mosley1999b&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.ips-planetarium.org/?page=a_mosley1999b&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2391" LastEditorUserId="2365" LastEditDate="2017-09-09T19:57:21.527" LastActivityDate="2017-09-09T19:57:21.527" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22597" PostTypeId="1" CreationDate="2017-09-09T17:04:56.380" Score="15" ViewCount="1253" Body="&lt;p&gt;Neptune is profoundly anomalous when it comes to moons among the gas giants.  Jupiter has its four massive moons, Saturn has a several large spherical bodies and Uranus, a world similar in size to Neptune has five or so major moons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet Neptune, the ice giant which sports one of the largest hill sphere radii in the solar system only has one major moon, of which is likely to be ripped to shreds in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find it peculiar because Uranus has five major moons yet Neptune only has one major moon.  What explains this disparity?  Did some disruption happen to earlier moons Neptune may have had?  What's known about this?&lt;/p&gt;&#xA;" OwnerUserId="18424" LastEditorUserId="18424" LastEditDate="2017-09-09T17:58:30.017" LastActivityDate="2017-09-10T05:03:52.420" Title="Why does Neptune only have one major moon?" Tags="&lt;neptune&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22598" PostTypeId="2" ParentId="22597" CreationDate="2017-09-09T18:56:54.690" Score="16" Body="&lt;p&gt;It seems possible that &lt;a href=&quot;https://en.wikipedia.org/wiki/Neptune&quot; rel=&quot;noreferrer&quot;&gt;Neptune&lt;/a&gt; once had a system of moons similar to other &lt;a href=&quot;https://en.wikipedia.org/wiki/Outer_planets&quot; rel=&quot;noreferrer&quot;&gt;outer planets&lt;/a&gt;. However &lt;a href=&quot;https://en.wikipedia.org/wiki/Triton_(moon)&quot; rel=&quot;noreferrer&quot;&gt;Triton&lt;/a&gt;, a &lt;a href=&quot;https://en.wikipedia.org/wiki/Kuiper_belt&quot; rel=&quot;noreferrer&quot;&gt;Kuiper belt&lt;/a&gt; object was captured by Neptune, and this disrupted the existing moons. It would have initially had an &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_eccentricity&quot; rel=&quot;noreferrer&quot;&gt;eccentric orbit&lt;/a&gt;, intersecting with the orbits of other moons, setting them into unstable orbits and ultimately either colliding with each other, hitting the planet or being ejected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minor moons formed from the debris of these collisions after Triton had entered a more stable orbit. So the disruption was the capture of Triton.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="13772" LastEditDate="2017-09-10T05:03:52.420" LastActivityDate="2017-09-10T05:03:52.420" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22599" PostTypeId="1" AcceptedAnswerId="22604" CreationDate="2017-09-10T00:12:39.327" Score="2" ViewCount="1328" Body="&lt;p&gt;I've made a home-made collimation cap and am trying to use the &lt;a href=&quot;http://garyseronik.com/a-beginners-guide-to-collimation/&quot; rel=&quot;nofollow noreferrer&quot;&gt;procedure&lt;/a&gt; linked to in &lt;a href=&quot;https://astronomy.stackexchange.com/a/12032/2341&quot;&gt;another answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I've run into two issues right off the bat.  In &lt;a href=&quot;http://50.87.248.98/~ellenroo/garyseronik/wp-content/uploads/manual/Labels.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this picture&lt;/a&gt; from the linked page, it shows a &quot;primary mirror donut&quot;. All I see is a black circle with nothing in between.  I'm using a piece of bright white paper that I printed the alignment patterns on to attach to the eyepiece and I thought this would be bright enough to see, but apparently not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second issue is that this method seems to rely on there being four &quot;spider vanes&quot; which are holding the secondary mirror in place.  However, my telescope only has one, making it seemingly impossible to line anything up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what it looks like through the cap:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Gw2IQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gw2IQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It &lt;em&gt;looks&lt;/em&gt; like the dark circle in the middle it fairly well aligned, but I can't tell for sure if it needs any adjustment.  And I'm not sure how to deal with the fact that there is nothing inside the circle to align with.  I have tried taking an LED flashlight and shining it on the top of the paper while i look through the pinhole and if I focus the LED circle on the edge of the paper at 90 degree intervals it appears to be aligned, but I don't know if this is sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, the halo around the outside seems to be the edge of part of the inside of the side tube, and it does not move when I adjust the collimation screws.&lt;/p&gt;&#xA;" OwnerUserId="2341" LastActivityDate="2017-09-10T09:45:56.787" Title="Collimating with a home-made collimation cap" Tags="&lt;newtonian-telescope&gt;&lt;collimation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22600" PostTypeId="1" AcceptedAnswerId="22601" CreationDate="2017-09-10T03:35:12.387" Score="4" ViewCount="539" Body="&lt;p&gt;I see this kind of projection of the celestial sphere into 2D frequently, usually with an image of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow noreferrer&quot;&gt;milky way galaxy&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/Milky_Way&quot; rel=&quot;nofollow noreferrer&quot;&gt;cosmic microwave background&lt;/a&gt;. This particular image shows the path of visible transits of the inner solar system planets across the sun as seen from locations outside of the solar system from the Motherboard article &lt;a href=&quot;https://motherboard.vice.com/en_us/article/gyy4eq/at-least-9-exoplanets-could-see-earth-with-present-day-human-technology&quot; rel=&quot;nofollow noreferrer&quot;&gt;At Least 9 Exoplanets Could See Earth With Present-Day Human Technology&lt;/a&gt; which links to: R. Wells, K. Poppenhaeger, C. A. Watson, R. Heller; &lt;em&gt;Transit Visibility Zones of the Solar System Planets&lt;/em&gt;, Monthly Notices of the Royal Astronomical Society, stx2077, &lt;a href=&quot;https://doi.org/10.1093/mnras/stx2077&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://doi.org/10.1093/mnras/stx2077&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the 2D elliptical projection of the Celestial Sphere called, and how can I make one? Is there a python-friendly package, but more importantly, is there a &lt;em&gt;mathematical expression&lt;/em&gt; for this transform?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iTtCu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iTtCu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-09-10T03:46:36.820" LastActivityDate="2017-09-10T03:47:56.233" Title="What is the 2D elliptical projection of the Celestial Sphere called, and how can I make one?" Tags="&lt;python&gt;&lt;mathematics&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22601" PostTypeId="2" ParentId="22600" CreationDate="2017-09-10T03:42:32.340" Score="4" Body="&lt;p&gt;I want to say that's a &lt;a href=&quot;https://en.wikipedia.org/wiki/Mollweide_projection&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mollweide projection&lt;/a&gt;. I know that they're pretty common in astronomy; many images of the CMB use them. I was actually working with one recently. Given latitude $\varphi$ and longitude $\lambda$, the $x$ and $y$ coordinates of an object are&#xA;$$x=R\frac{2\sqrt{2}}{\pi}(\lambda-\lambda_0)\cos\theta,\quad y=R\sqrt{2}\sin\theta$$&#xA;for projecting from a sphere of radius $R$. $\theta$ is the solution to&#xA;$$2\theta+\sin2\theta=\pi\sin\varphi$$&#xA;Similar images can be found &lt;a href=&quot;http://wise.ssl.berkeley.edu/sky_coverage.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; (labeled as Mollweide projections), which are of the same shape and proportions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TUupm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TUupm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matplotlib will let you do Mollweide projections; a nice tutorial can be found &lt;a href=&quot;http://balbuceosastropy.blogspot.com/2013/09/the-mollweide-projection.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;a href=&quot;https://matplotlib.org/basemap/users/moll.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Basemap&lt;/a&gt; is a possibility, although I haven't used it before. It has quite a few different built-in projections. There are some interesting features available in the different matplotlib implementations, and using the &lt;a href=&quot;https://pypi.python.org/pypi/pyephem/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;code&gt;ephem&lt;/code&gt;&lt;/a&gt; package, you can use astronomical data. That said, I've found that the transformations aren't that hard to implement by hand; I was able to get some basic projections done from scratch.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2017-09-10T03:47:56.233" LastActivityDate="2017-09-10T03:47:56.233" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22602" PostTypeId="1" AcceptedAnswerId="22996" CreationDate="2017-09-10T04:03:34.487" Score="6" ViewCount="263" Body="&lt;p&gt;The lengthy paper &lt;a href=&quot;https://arxiv.org/abs/1212.5225&quot; rel=&quot;noreferrer&quot;&gt;Nine-Year Wilkinson Microwave Anisotropy Probe (WMAP) Observations: Final Maps and Results&lt;/a&gt; mentions &quot;spinning dust&quot; about 51 times. It is a potential contributor to the measured spectrum and the details of its emission spectrum has more than one proposed model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;question:&lt;/strong&gt; Is there a simple way to understand the physics behind the emission spectrum of &lt;em&gt;spinning dust&lt;/em&gt; and how it would differ from &lt;em&gt;non-spinning dust?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rlg3M.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rlg3M.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2017-10-04T22:06:30.520" Title="What is the physics of the &quot;spinning dust&quot; contribution to Cosmic Microwave Background measurements?" Tags="&lt;cmb&gt;&lt;dust&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22603" PostTypeId="2" ParentId="22594" CreationDate="2017-09-10T08:26:40.590" Score="4" Body="&lt;p&gt;According to table 2 of &lt;a href=&quot;https://arxiv.org/abs/1510.08869&quot; rel=&quot;nofollow noreferrer&quot;&gt;Corral Santana (2015)&lt;/a&gt; (The BlackCAT catalogue of stellar mass black holes in X-ray transients - which other than gravitational waves, is how they are found), the nearest &lt;strong&gt;known&lt;/strong&gt; black hole is the primary of the V616 Mon system (otherwise known as A0620-003) at a distance of $1.06 \pm 0.10$ kpc ($3460 \pm 390$ ly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes can only be found at present when they are in close binary systems with other stars (or black holes). This is the tiny minority of an estimated Galactic population of 100 million black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nearest black hole is likely invisible to us at present. A statistical estimate can be made (see &lt;a href=&quot;https://astronomy.stackexchange.com/questions/4725/statistically-what-would-the-average-distance-of-the-closest-black-hole-be&quot;&gt;my answer here&lt;/a&gt;) that the nearest will be within 20 pc or so, depending on the parent population and detailed physics of supernovae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The online &lt;a href=&quot;http://www.astro.puc.cl/BlackCAT/&quot; rel=&quot;nofollow noreferrer&quot;&gt;BlackCAT catalogue&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="15936" LastEditDate="2017-09-10T11:42:44.183" LastActivityDate="2017-09-10T11:42:44.183" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22604" PostTypeId="2" ParentId="22599" CreationDate="2017-09-10T09:45:56.787" Score="3" Body="&lt;p&gt;The Gary Seronik website that you cite is a good one - it's not leading you astray!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a couple of important things to look at:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Look straight down the main tube of your telescope at the primary (largest) mirror. Is there a small paper ring or another mark dead in the centre of that mirror? If so, good. If not, read through this webpage &lt;a href=&quot;http://garyseronik.com/centre-dotting-your-scopes-primary-mirror/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://garyseronik.com/centre-dotting-your-scopes-primary-mirror/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;What are the lighting conditions when you're performing this? The photo you included is helpful but seems to imply that you might be pointing the scope at a bright light. This will wash out all contrast and make the procedure difficult. I collimate my scope just using ambient light during the daytime&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Don't worry about only having one support rather than four. The key will be to identify your centre dot (you can just see it on Seronik's webpage, first picture ) and to see how it is centred when you look through the collimation cap onto the secondary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe read this article: &lt;a href=&quot;http://www.skyandtelescope.com/astronomy-resources/how-to-align-your-newtonian-reflector-telescope/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.skyandtelescope.com/astronomy-resources/how-to-align-your-newtonian-reflector-telescope/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or search for Newtonian Collimation on Youtube?&lt;/p&gt;&#xA;" OwnerUserId="17502" LastActivityDate="2017-09-10T09:45:56.787" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22605" PostTypeId="2" ParentId="22557" CreationDate="2017-09-10T11:56:56.683" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Could the expansion of universe be a side effect of gravity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, everything is possible I suppose. And we do see plenty of material that talks about the expansion of the universe as a kind of &lt;a href=&quot;https://www.google.co.uk/search?source=hp&amp;amp;q=repulsive+gravity&amp;amp;oq=repulsive+gravity&amp;amp;gs_l=psy-ab.3..0l3j0i22i30k1.936.3574.0.4075.18.17.0.0.0.0.150.1992.1j16.17.0....0...1.1.64.psy-ab..1.17.1989.0..46j35i39k1j0i131k1j0i46k1j0i67k1.dPNy5qyeJQI&quot; rel=&quot;nofollow noreferrer&quot;&gt;repulsive gravity&lt;/a&gt;. But I would say no, the expansion isn't due to gravity, and it isn't some side effect of gravity. It's something different. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I mean maybe gravity &quot;attracts things&quot; at a &quot;small&quot; distances and repels them at &quot;larger&quot; distances?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It doesn't. Gravity attracts things at all distances. It follows the inverse square rule. Some people like Mordehai Milgrom say it doesn't quite follow the inverse square law, see &lt;a href=&quot;https://en.wikipedia.org/wiki/Modified_Newtonian_dynamics&quot; rel=&quot;nofollow noreferrer&quot;&gt;MOND&lt;/a&gt;. But he isn't saying it repels at large distance. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do we have any theory/hypothesis/paper on this topic? Is it possible to prove or disprove?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We have general relativity. In general relativity &lt;em&gt;gravity sucks&lt;/em&gt;. It doesn't blow. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Could gravity have a repulsive term that dominates at cosmological distances and could it be such a term that causes (accelerated) expansion?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Like I said, everything is possible. But I will say no, the universe is expanding because of dark energy, not because of gravity. I rather like Erwin Schrödinger’s cosmic pressure myself. See the 2014 &lt;a href=&quot;https://books.google.co.uk/books?id=lXGFAwAAQBAJ&amp;amp;pg=PA268&amp;amp;lpg=PA268&amp;amp;dq=Einstein+%22cosmic+pressure%22&amp;amp;source=bl&amp;amp;ots=uaef6D4cQs&amp;amp;sig=zDKcggRJ01ywTpiBTNPpdmdBYIc&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwjCgfDlmNnUAhVRa1AKHSG_AYgQ6AEIKzAB#v=onepage&amp;amp;q=Einstein%20%22cosmic%20pressure%22&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cambridge companion to Einstein&lt;/a&gt; by Michel Janssen and ‎Christoph Lehner: &lt;em&gt;“Schrödinger [1918] had pointed out another way of treating the cosmological constant: moving it from the left-hand side of Equation [7], where it represents a contribution to space-time curvature, to the right-hand side, where it represents a contribution to the energy-matter distribution. Then it would correspond physically to a kind of cosmic pressure”&lt;/em&gt;. Unfortunately Einstein didn’t like it. You can read more in Alex Harvey’s 2012 paper &lt;a href=&quot;https://arxiv.org/abs/1211.6338&quot; rel=&quot;nofollow noreferrer&quot;&gt;How Einstein Discovered Dark Energy&lt;/a&gt;. He gives a translation of Schrödinger’s paper and refers to &lt;a href=&quot;http://einsteinpapers.press.princeton.edu/vol7-trans/47&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein’s response&lt;/a&gt;. If you were to ask a modern cosmologist, I think the answer would be that Schrödinger was pretty close to the mark. See &lt;a href=&quot;https://books.google.co.uk/books?id=GUySYQaDM1cC&amp;amp;pg=PA252&amp;amp;lpg=PA252&amp;amp;dq=lambourne+cosmic+pressure&amp;amp;source=bl&amp;amp;ots=1SQg2svc2s&amp;amp;sig=Rc2w1PAWH_SpCXr1KGurqVCxOqo&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwjqtLWw6d7UAhUHIcAKHc_QDxMQ6AEIJzAB#v=onepage&amp;amp;q=lambourne%20cosmic%20pressure&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;page 252&lt;/a&gt; of Relativity, Gravitation and Cosmology by Robert J Lambourne: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/msU68.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/msU68.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2017-09-10T11:56:56.683" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22606" PostTypeId="2" ParentId="575" CreationDate="2017-09-10T13:53:06.543" Score="3" Body="&lt;p&gt;Also keep in mind that building a barn door tracker yourself is going to be exciting, you'll spend time calculating at what speed the motor should be turning, find the proper gears, cut the wood. It's a lot of DIY excitement that I enjoyed myself. However you must keep in mind that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Motorizing it is going to cost your min 100$&lt;/li&gt;&#xA;&lt;li&gt;Using a door hinge is practical, but also imprecise, you must align the north pole to less than a degree while your pieces of wood would be misaligned.&lt;/li&gt;&#xA;&lt;li&gt;You'll need to adjust precisely to motor speed to match earth rotation, this involves precise measurement of the distance between your hinge and the screw that drives your barn door tracker.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Build a mechanically tracker will be challenging, but as I wend though I can give you a couple of clues:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The hinge must be as wide as it can practically be, this will reduce the angle error of the hinge itself.&lt;/li&gt;&#xA;&lt;li&gt;The screw must be as far as possible so that your motor turns faster&lt;/li&gt;&#xA;&lt;li&gt;It would be helpful to have a mechanical part on the opposite side of the hinge to take the sideways efforts and reduce the angular error of the hinge, otherwise take a 5-6mm screw, something large that can be rigid. The more heavy duty the better your alignment will be.&#xA;Steppers are noisy, use at least 1/32th stepping driver or use a full sine wave 3 phases brushless motor to avoid vibrations and noise.&lt;/li&gt;&#xA;&lt;li&gt;Using a belt drive reduces the noise: most of the noise coming out of my stepper seems to come from the spinning axle. Noises and vibration are getting transferred to the gears and then to the board where they resonate. Using a driving belt dampens the vibrations from the stepper.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="18442" LastEditorUserId="18442" LastEditDate="2017-09-10T16:02:05.217" LastActivityDate="2017-09-10T16:02:05.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22607" PostTypeId="1" AcceptedAnswerId="22614" CreationDate="2017-09-10T14:34:47.587" Score="0" ViewCount="142" Body="&lt;p&gt;Does a nova, caused by runaway fusion in hydrogen accreted onto a white dwarf from a companion star, create a visible nebula?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a planet were in a circumbinary orbit around a binary pair that produced a nova, would it survive?&lt;/p&gt;&#xA;" OwnerUserId="18338" LastEditorUserId="2365" LastEditDate="2017-09-10T17:40:50.733" LastActivityDate="2017-09-10T19:26:36.337" Title="Do novas create a nebula? Could a planet survive if its star exploded in a nova?" Tags="&lt;nebula&gt;&lt;nova&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22608" PostTypeId="1" CreationDate="2017-09-10T14:55:06.270" Score="0" ViewCount="115" Body="&lt;p&gt;I am pretty new to astronomy and the project I am currently working on requires some knowledge of equatorial coordinate system.&#xA;As long as I have been researching this question, I understood that vernal equinox is the origin of this system. At this moment I am stuck with task of finding distance on celestial sphere between west(on horizon) and vernal equinox.&#xA;So, how do you count this distance?&lt;/p&gt;&#xA;" OwnerUserId="18443" LastActivityDate="2017-09-13T11:49:25.590" Title="Distance from W(West) to vernal equinox" Tags="&lt;right-ascension&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22609" PostTypeId="1" AcceptedAnswerId="22616" CreationDate="2017-09-10T15:53:07.167" Score="42" ViewCount="18785" Body="&lt;p&gt;A possible answer for this is that, &lt;strong&gt;light emitted from the galaxies travelled a billion miles all the way to earth, where the hubble space telescope picked up this light through its sensors, and was able to construct an image of the galaxy&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but if this is true, and galaxies are billions of miles away, shouldn't the light particles emitted from the galaxies be &lt;strong&gt;scattered&lt;/strong&gt; all over the place? after all they have been travelling from millions of years, and have probably collided with asteroids and other foreign objects. What were the chances that about 95% of the photons actually reached earth, giving us a very detailed image. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the andromeda galaxy which has a distance of 1.492 × 10^19 mi from earth. If light emitted from the galaxy travels in all directions, then how is it that we can still map out the entire galaxy, evident from the photo below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shouldn't like half of the galaxy be missing since photons could have hit other objects, and &quot;never have reached earth&quot;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DVW2m.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DVW2m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17327" LastActivityDate="2020-10-25T02:27:27.563" Title="How do we have photos of galaxies so far away?" Tags="&lt;galaxy&gt;&lt;light&gt;&lt;photography&gt;" AnswerCount="6" CommentCount="22" FavoriteCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="22611" PostTypeId="2" ParentId="575" CreationDate="2017-09-10T17:38:38.597" Score="3" Body="&lt;p&gt;There are several options.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do star trail photos with just a tripod. If you have dark skies, Point the camera near the pole star, do a long exposure, and get star trails as the sky appears to rotate around the pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're in a light polluted area, then you'll find your maximum exposure time is limited by the light pollution - expose for long enough and the light pollution background will saturate one or more colour channels - which could prevent you using a long enough exposure to get long trails.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also take pictures of the moon - since it's basically a gray rock in bright sunlight, you don't need a long exposure at all for this. But you do need a long focal length to make the moon big enough to fill a decent part of the frame, and at 150 mm it's likely to be disappointingly small. (You need around 1500mm or so to fill most of an APS sized sensor frame).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also do constellation photography with just a tripod. The trick here is to keep the exposure short enough that the rotation of the earth doesn't cause significant star trailing.  As a rough rule of thumb, divide 400 by the lens focal length to get the approximate exposure time in seconds (thus varies in practice depending on where you're aiming, and how picky you are about trailing, but should give you a starting point). Thus with a 50mm lens, you can get away with around 8 seconds, around 2-3 seconds with a 150mm lens, and around 33 seconds with a 12mm lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can improve results by stacking several exposures in software - try the free Deep Sky Stacker - which will reduce noise and usually give you a cleaner result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For longer exposures, you'll need some sort of tracking mount to compensate for the earth's rotation and prevent star trailing. You can either build a barn door tracker, as mentioned in other replies, which rotate a camera platform around a hinge whose axis points at the celestial pole, or look at a commercial solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several mounts designed for use with cameras, such as Vixen's polarie, Skywatcher's star adventurer mount (or similar from Vixen or Ioptron), or the Astrotrak system (or Kenko's Sky Memo (not sure if this is still available)). Note that with some of these you also need to budget for a couple of tripod heads - one to point the mount at the celestial pole, and one to point the camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you could look at a motorised german equatiorial mount - which would also let you use a telescope - but to get good long exposure images with a telescope usually requires a good mount, which range from expensive to very expensive - and may also require autoguiding (automatic correction using a second guide camera and software) to apply corrections for long exposures. Long exposure deep sky imaging is more accessible than it was, but is still expensive to do well (and easily capable of soaking up as much money as you want to throw at it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that popular go-to alt-az mounted are limited for long exposure imaging, as although goto alt-az mounts can track objects - which is fine for visual use - they suffer from an effect called field rotation which makes the field of view gradually rotate, causing star trailing after a while (around 30s or so - it depends on latitude and where you're pointing). German equatorial mounts, with one axis pointed at the celestial pole, don't suffer from this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For planetary imaging, the planets are bright enough that you can capture video with a webcam (or similar planetary astro camera) and telescope, and then process the video to stack the best frames. This is a different type of imaging to the long exposure deep sky stuff.&lt;/p&gt;&#xA;" OwnerUserId="17016" LastActivityDate="2017-09-10T17:38:38.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22612" PostTypeId="5" CreationDate="2017-09-10T17:43:08.520" Score="0" Body="" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-09-10T17:43:08.520" LastActivityDate="2017-09-10T17:43:08.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22613" PostTypeId="4" CreationDate="2017-09-10T17:43:08.520" Score="0" Body="For questions about stellar explosions on white dwarfs caused by runaway fusion reactions in an accreted hydrogen atmoshphere and visible as a &quot;new star&quot;. Distinct from a supernova. " OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-09-10T17:43:08.520" LastActivityDate="2017-09-10T17:43:08.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22614" PostTypeId="2" ParentId="22607" CreationDate="2017-09-10T19:26:36.337" Score="1" Body="&lt;p&gt;A nova may leave behind a &lt;a href=&quot;https://en.wikipedia.org/wiki/Nova_remnant&quot; rel=&quot;nofollow noreferrer&quot;&gt;nova remnant&lt;/a&gt;: a shell of gas, expanding at about 1000km/s. In recurrent novae, they may be lit by light echoes from subsequent eruptions. They are much less massive than planetary nebulae, and at much lower energy than supernova remnants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Novae displaying nebula shells or remnants include: GK Per, RR Pic, DQ Her, FH Ser, V476 Cyg, V1974 Cyg, HR Del and V1500 Cyg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets in such a system would be bathed in high energy gas, this would not be enough to vaporise the planet, but it would make life on such a world extremely difficult.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-09-10T19:26:36.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22615" PostTypeId="2" ParentId="22609" CreationDate="2017-09-10T19:31:09.377" Score="26" Body="&lt;p&gt;As Rob Jeffries says, the universe is mostly empty space. A photon can easily travel thousands of light years without interacting with anything. Most of the interaction would occur when photons entered the earth's atmosphere. The Hubble avoids this. These photos were most likely from combining several viewing sessions giving basically an extended time period for observing the galaxy.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="2531" LastEditDate="2017-09-10T21:17:33.587" LastActivityDate="2017-09-10T21:17:33.587" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22616" PostTypeId="2" ParentId="22609" CreationDate="2017-09-10T19:36:24.047" Score="53" Body="&lt;p&gt;There are two reasons that often — but not always — light from galaxies millions and even billions of lightyears away make it through the Universe and down to us:&lt;/p&gt;&#xA;&#xA;&lt;strong&gt;Particle &lt;em&gt;number&lt;/em&gt; and particle &lt;em&gt;size&lt;/em&gt;&lt;/strong&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;First, the intergalactic medium (IGM) is &lt;em&gt;extremely&lt;/em&gt; dilute. The number density of particles out there is of the order $n\sim10^{-7}\,\mathrm{cm}^{-3}$, or roughly 26 orders of magnitude lower that the air at sea level! That means that if you consider a tube from Andromeda to the Milky Way with cross-sectional area of $1\,\mathrm{cm}^{2}$, it will contain roughly one microgram of matter (thanks to Rob Jeffries for catching a factor $10^6$ error).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Second, even if a photon comes close to an atom, it will only be absorbed if its energy matches closely some transition in the atom. Since most of the atoms are ionized (and thus should be called plasma instead, but in astronomy the distinction if often not made), there are no electrons to absorb the photon. The photons are more likely to interact with the free electrons via Thomson scattering, but the Thomson cross section is immensely small $(\sim10^{-24}\,\mathrm{cm}^{2})$, so even if you consider the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;noreferrer&quot;&gt;CMB photons&lt;/a&gt; — which have traveled through the Universe almost since the Big Bang — only around 5% of them have interacted with electrons on their way.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In other words: The amount of transmitted light depends on two factors: 1) The amount of matter along the line of sight, and 2) that matter's ability to absorb the light. In the IGM, both are tremendously small. When the light enters the inter&lt;em&gt;stellar&lt;/em&gt; medium (ISM) inside our galaxy, it may encounter denser clouds with atoms that are able to absorb the light. But usually (although not always) &quot;dense&quot; is still very dilute compared to Earth's atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;strong&gt;Mathematical expression&lt;/strong&gt;&#xA;&#xA;&lt;p&gt;In general, if a beam of light traverses a region of particles, each with a cross section $\sigma$ (measured e.g. in cm$^2$), passing $N$ particles per area of the beam (measured e.g. in cm$^{-2}$), then the opacity of the medium is given by the &lt;em&gt;optical depth&lt;/em&gt; $\tau$, defined by&#xA;$$&#xA;\tau \equiv N \, \sigma.&#xA;$$&#xA;The transmitted fraction $f$ of photons is then&#xA;$$&#xA;f = e^{-\tau}.&#xA;$$&#xA;In general $\sigma$ depends on the wavelength, and thus part of the spectrum may pass unhindered, while another part may be completely absorbed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure below (from &lt;a href=&quot;http://spiff.rit.edu/classes/phys443/lectures/qso/qso.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;) shows the spectrum of a quasar lying at a distance of 22 billion lightyears, i.e. $10\,000$ times farther away than Andromeda. You see that there are several thin absorption lines (caused by intervening hydrogen clouds whose densities are a factor of 10-100 higher than the IGM), but still most of the light makes it down to us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WSHff.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WSHff.png&quot; alt=&quot;qso&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the light we see from this quasar was emitted so long ago, the Universe was considerably smaller at that time, and thus the density was larger. Nonetheless, only a small fraction is absorbed. The farther away the light is emitted, the longer ago it was, which means smaller Universe, and higher density, and thus the more light is absorbed. If you consider this quasar (from &lt;a href=&quot;http://classic.sdss.org/news/releases/20000413.qso.img2.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;) which lies 27 billion lightyears away, you see that much more light is absorbed in part of the spectrum. Still, however, much light make it through to us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fP9Zm.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fP9Zm.gif&quot; alt=&quot;qso2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that it is only the short wavelengths that are absorbed is quite interesting — but that's another story.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2017-09-12T06:25:05.640" LastActivityDate="2017-09-12T06:25:05.640" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22618" PostTypeId="1" CreationDate="2017-09-11T02:17:12.653" Score="0" ViewCount="82" Body="&lt;p&gt;This is a question regarding sound waves and space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that sound waves must travel through a medium &quot;air&quot;, enter our brains, and then we can decipher what exactly the sound is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in space, there is no air, and so does that mean that &lt;strong&gt;no sound waves are being produced&lt;/strong&gt;, or that &lt;strong&gt;there are waves, but we cannot hear them because they never reach us&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And how can we tell? We cannot &quot;hear&quot; space, so who says its because theres no air, rather then just because &quot;its not being produced in the first place?&quot;&lt;/p&gt;&#xA;" OwnerUserId="17327" LastActivityDate="2017-09-11T02:17:12.653" Title="Does space not make sound, or can we just not hear it?" Tags="&lt;space&gt;&lt;planetary-atmosphere&gt;&lt;wavelength&gt;&lt;vacuum&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2017-09-12T15:43:57.383" ContentLicense="CC BY-SA 3.0" />
  <row Id="22619" PostTypeId="2" ParentId="22609" CreationDate="2017-09-11T02:27:27.400" Score="14" Body="&lt;p&gt;Sorry if this logic seems a bit circular, but we can get unobscured pictures of galaxies because they are unobscured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As has been mentioned - space is really, really big and really, really empty. This is hard for us to contemplate, because there's so much stuff right next to us - but this is actually a really unusual condition. The next star to the Sun is over 4 light-years away, but we get almost all (99.9999999999...%) the light from it that heads in our direction - the same with light from further away - we get a huge number of photons sent to us from objects very far away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hubble also uses the simple camera techniques of lensing and long exposures to take images of distant objects - so more light is received to construct the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, the other part of this, it is almost impossible to take a picture of a galaxy (or star) that is behind another galaxy or dust cloud. For example, we can't easily see past the centre of our own galaxy, because there's a lot of dust and gas and stars in the way. The picture in your question, on the other, seems to be Andromeda, which is above the plane of the galaxy. Our galaxy is quite thin compared to its diameter, and we're a decent way out of the galactic centre, meaning there's a lot less stuff in the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there are &lt;a href=&quot;https://en.wikipedia.org/wiki/NGC_3628&quot; rel=&quot;nofollow noreferrer&quot;&gt;some galaxies&lt;/a&gt; we've taken images of which are obscured by dust:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/c5Ukh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c5Ukh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14176" LastEditorUserId="14176" LastEditDate="2017-09-12T22:10:44.853" LastActivityDate="2017-09-12T22:10:44.853" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22620" PostTypeId="2" ParentId="22609" CreationDate="2017-09-11T03:02:18.510" Score="23" Body="&lt;p&gt;There's a misconception in your question I don't think the other answers have addressed. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If light emitted from the galaxy travels in all directions, then how&#xA;  is it that we can still map out the entire galaxy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Light &lt;strong&gt;is&lt;/strong&gt; emitted from the galaxy in all directions. Only a tiny, tiny fraction of it is directed to Earth, and of that, an even tinier fraction is collected by any given telescope. But we can still see it, because galaxies are very, very bright. Andromeda contains about a trillion stars. &lt;/p&gt;&#xA;" OwnerUserId="18448" LastActivityDate="2017-09-11T03:02:18.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22628" PostTypeId="1" AcceptedAnswerId="22630" CreationDate="2017-09-11T13:55:44.103" Score="5" ViewCount="152" Body="&lt;p&gt;We knew the earth rotates, but the moon is tidally locked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To tell that another planet is rotating, you have to be able to distinguish its surface (I guess?). Who first did this?&lt;/p&gt;&#xA;" OwnerUserId="18467" LastActivityDate="2017-09-11T14:31:43.077" Title="Who first observed another planet rotating?" Tags="&lt;planet&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22630" PostTypeId="2" ParentId="22628" CreationDate="2017-09-11T14:31:43.077" Score="6" Body="&lt;p&gt;Mercury's surface was virtually unknown until Mariner 10 flew by and Venus is covered in clouds. Of the terrestrial planets, that leaves Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was likely &lt;a href=&quot;https://en.wikipedia.org/wiki/History_of_Mars_observation#cite_note-jbaa94-28&quot; rel=&quot;noreferrer&quot;&gt;Christiaan Huygens&lt;/a&gt; in November 1659 who first realised Mars was rotating when he tried to map it. His estimate of roughly 24 hours was fairly close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe Galileo's earlier observations on Jupiter were with a telescope not powerful enough to resolve smaller 'surface' features (like the Great Red Spot), which may have either proven that Jupiter rotates, or at least that something on the surface is moving across it.&lt;/p&gt;&#xA;" OwnerDisplayName="user10106" LastActivityDate="2017-09-11T14:31:43.077" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22631" PostTypeId="2" ParentId="22555" CreationDate="2017-09-11T14:41:43.757" Score="4" Body="&lt;p&gt;The figure below shows the evolution of the Hubble parameter $H$ from 10 billion years (Gyr) ago, to 10 Gyr into the future:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZexIv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZexIv.png&quot; alt=&quot;Hz&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the change in $H$ is modest nowadays, compared to the past. The &quot;acceleration&quot; of the expansion at any point in time is given by the tangent to the curve at that time. Today, the change is $dH/dt \simeq -1.2\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}\,\mathrm{Gyr}^{-1}$, or roughly $-10^{-17}\,\mathrm{km}\,\mathrm{s}^{-2}\,\mathrm{Mpc}^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, it will take roughly a billion years before the expansion rate has decreased by $1\,\mathrm{km}\,\mathrm{s}^{-1}$, and it asymptotically approaches some $57\,\mathrm{km}\,\mathrm{s}^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $H$ is defined as $(da/dt)\,/\,a$, where $a$ is the &lt;em&gt;scale factor&lt;/em&gt; (the &quot;size&quot; of the Universe), a constant $H$ implies that $a\propto e^{Ht}$; that is, the size of the Universe increases exponentially.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2017-09-11T20:24:30.357" LastActivityDate="2017-09-11T20:24:30.357" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22632" PostTypeId="2" ParentId="22555" CreationDate="2017-09-11T17:04:59.803" Score="0" Body="&lt;p&gt;The acceleration of the universe is written as $\dot{H}$, where $H$ is the Hubble Rate (today, the Hubble Rate is given the name the &lt;em&gt;Hubble Constant&lt;/em&gt;, $H_0$, and is given the value of 70 km/s/10&lt;sup&gt;6&lt;/sup&gt;pc). The &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann_equations#Equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Friedmann Equation&lt;/a&gt; can be simplified to the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{H} + H^2 = -\frac{4\pi G}{3}\left(\rho + \frac{3p}{c^2}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\rho$ is the density, and $p$ is the pressure. $G$ is the gravitational constant, and $c^2$ is the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to find the acceleration, simply substitute the relavent values into the equation below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{H_0} = -\frac{4\pi G}{3}\left(\rho_0 + \frac{3p_0}{c^2}\right) - H^2_0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;I will try to find the relavent values, but as the previous answer says, the value is roughly $-10^{-17}\,\mathrm{km}\,\mathrm{s}^{-2}\,\mathrm{Mpc}^{-1}$&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1494" LastEditorUserId="1494" LastEditDate="2017-09-11T21:33:05.400" LastActivityDate="2017-09-11T21:33:05.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22634" PostTypeId="1" CreationDate="2017-09-11T17:37:02.500" Score="-1" ViewCount="69" Body="&lt;p&gt;According to the Wikipedia article &lt;a href=&quot;https://en.wikipedia.org/wiki/Elliptical_galaxy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Elliptical galaxy&lt;/a&gt;, elliptical galaxies have a much lower concentration of gas between the stars than spiral galaxies. I know that until the gas is below a certain concentration, its atoms don't travel very far between collisions and it follows the normal gas laws and the gas actually flows into the stars when they pass by. I know that when the interstellar gas is at such a low concentration that atoms on average travel a much larger distance than the average diameter of a star in the elliptical galaxy it's in, the gas doesn't follow the normal gas laws and gas gets mainly absorbed into stars by happening to be headed towards one and colliding with it rather than by flowing towards one and the ratio of the rate of absorption of gas into stars to the concentration of interstellar gas is lower, and a star has about the same expected time before colliding with another star as an interstellar gas atom has before colliding with a star. I also know that when the interstellar medium is at the equilibrium concentration, there are 3 possible driving forces for the reduction of interstellar gas; star formation, flow of interstellar gas into stars, and random collisions of interstellar gas atoms with a star which can only occur at very low concentration; and 2 driving forces for the increase in interstellar gas, the slow escape of gas molecules from stars and collisions between stars because collisions occur at a high enough speed to produce an explosion instead of combining into a larger star. Which driving force for the reduction of interstellar gas is the largest? Also, which driving force for the increase in interstellar gas is larger? If it's star formation, that means the interstellar gas is at a high enough concentration to follow the normal gas laws but since the amount of interstallar gas is at equilibrium, the main driving force for the increase in interstellar gas is collisions between stars. If it's the slow escape of gas from stars, the main driving force for the decrease in interstellar gas could be the flow of gas into stars or the random collisions of gas molecules with stars but not star formation. If the main driving force for the reduction in interstallr gas concentration is collisions of gas molecules with stars, then a star has about the same expected time beofre a collision with another star as an intersteller gas atom has before colliding with another star so the expected time before a star collides with another star is less than the age of the galaxy and if it's flow of gas into stars, the main driving force for the increase in interstellar gas could be either of the two driving forces but if the driving force for the increase is collisions between stars, the expected time before a star collides with another star is less than the age of the galaxy, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if the interstellar gas concentration has not yet reached equilibrium, it could be that not enough time has gone by for the interstellar gas to not follow the normal gas laws and star formation occurs mainly by the gas coalescing into stars. It could also be that the interstellar gas started diving exponentially until it stopped following the normal gas laws then the rate at which it divides exponentially greatly reduced so it has not yet had time to reach the equilibrium concentration.&lt;/p&gt;&#xA;" OwnerUserId="13685" LastEditorUserId="13685" LastEditDate="2017-09-11T18:34:23.917" LastActivityDate="2017-09-11T18:34:23.917" Title="Is the expected time for a star in an elliptical galaxy to collide with another star less than the average age of elliptical galaxies?" Tags="&lt;galactic-dynamics&gt;&lt;interstellar-medium&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22635" PostTypeId="2" ParentId="22634" CreationDate="2017-09-11T18:18:41.690" Score="5" Body="&lt;p&gt;The collision timescale for a star in the solar neighborhood is&lt;sup&gt;1&lt;/sup&gt;&#xA;$$t_c\simeq5\times10^{10}\text{ Gyr}\left(\frac{R}{R_{\odot}}\right)^{-2}\left(\frac{v}{30\text{ km s}^{-1}}\right)^{-1}\left(\frac{n}{0.1\text{ pc}^{-3}}\right)^{-1}$$&#xA;where $R$ is the radius of a star, $v$ is its speed relative to the stars around it, and $n$ is the stellar number density. That number isn't in years; it's in &lt;em&gt;billions&lt;/em&gt; of years. You're looking at collision timescales on the order of $\sim10^{19}\text{ yrs}$, many orders of magnitude larger than the current age of the universe. Even changing $v$ and $n$ a bit can't really lower that significantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, there are lots of stars in a galaxy. Current estimates put 100 billion or so stars in the Milky Way, give or take. So yes, it's certainly likely for a galaxy to have two of its stars collide sometime before now (and I'm ignoring mergers in binary systems, as well as globular clusters, which also see many stellar collisions). But the collision timescale for any &lt;em&gt;single&lt;/em&gt; star is much, much greater than the current age of its parent galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can probably see why galaxies are usually modeled as &lt;a href=&quot;http://www.ast.cam.ac.uk/~vasily/Lectures/SDSG/sdsg_5_coll.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;collisionless systems&lt;/em&gt;&lt;/a&gt;. The collisionless version of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Boltzmann_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boltzmann equation&lt;/a&gt; is used, which simplifies computations quite a bit.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1 I'm referencing my textbook, &lt;em&gt;Foundations of Astrophysics&lt;/em&gt;, Ryden &amp;amp; Peterson. Eqn 22.17, page 516.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2017-09-11T18:27:50.117" LastActivityDate="2017-09-11T18:27:50.117" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22636" PostTypeId="2" ParentId="22555" CreationDate="2017-09-11T20:04:02.873" Score="1" Body="&lt;p&gt;Pela's answer gives the numerical value, I thought I would just explain the difference between $\dot{H}$, which is negative, and an acceleration, which is positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann_equations#Equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Friedmann acceleration equation&lt;/a&gt; is given by &#xA;$$\frac{\ddot{a}}{a} =  -\frac{4 \pi G}{3}\left(\rho+\frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3}, \tag*{(1)}$$&#xA;whilst the Hubble parameter is defined as&#xA;$$H = \frac{\dot{a}}{a} \tag*{(2)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In equation (1) the first term gets smaller with time since both matter density $\rho$ and pressure $p$ due to matter and  radiation become smaller. The second term due to the cosmological constant $\Lambda$ is positive and eventually dominates and we are in that regime &lt;em&gt;now&lt;/em&gt;. This is what is meant by an accelerating universe, where the second derivative of the scale factor $\ddot{a}$ is positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you have modified your question to ask what the time derivative of the Hubble parameter (not constant!) is. Differentiating equation (2) with respect to time, we have&#xA;$$\dot{H} = \frac{\ddot{a}}{a} - H^2$$&#xA;$$\dot{H} = -\frac{4 \pi G}{3}\left(\rho+\frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3} - H^2 \tag*{(3)}\, .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{H}$ is the gradient of the plot shown in Pela's answer and it is &lt;em&gt;negative&lt;/em&gt;. That is $H$ is getting smaller. As the universe gets bigger, $H$ reaches its asymptotic value when $\ddot{a}/a = H^2$. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-09-11T20:04:02.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22637" PostTypeId="2" ParentId="22545" CreationDate="2017-09-11T20:47:28.170" Score="9" Body="&lt;p&gt;Let's see what we get from some back-of-the-envelope estimates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine throwing one star (e.g., the Sun) at the other galaxy. How likely is it we'll hit a star in the other galaxy? Well, it's basically proportional to how big a target each star in the other galaxy is (its cross-sectional area) compared to the size of the whole galaxy, multiplied by the total number of stars in the target galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume it's the Milky Way-Andromeda scenario, so each galaxy has about 100 billion stars, and each star is roughly the same size as the Sun (some are much larger, most are smaller). The actual target area for an individual star is a circle with twice the star's radius (we're counting one star just grazing the other as a collision). Let's also assume the stars are more or less evenly distributed in a circular disk. Since &quot;100,000 light years&quot; is a common (and not completely crazy) estimate of the Milky Way's size, that's a circle of radius = 50,000 light years (about &lt;span class=&quot;math-container&quot;&gt;$10^{16}$&lt;/span&gt; meters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: 100 billion stars in the target galaxy, each with target radius &lt;span class=&quot;math-container&quot;&gt;$\sim 2 R_{\odot}$&lt;/span&gt;, gives us a total target area of &lt;span class=&quot;math-container&quot;&gt;$10^{11} \times \pi (2 R_{\odot})^{2} \approx 10^{30}$&lt;/span&gt; m&lt;span class=&quot;math-container&quot;&gt;$^{2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of the target galaxy is &lt;span class=&quot;math-container&quot;&gt;$\pi R_{gal}^{2} \approx 10^{42}$&lt;/span&gt; m&lt;span class=&quot;math-container&quot;&gt;$^{2}$&lt;/span&gt;. So the chance of our Sun hitting a star in the other galaxy is &lt;span class=&quot;math-container&quot;&gt;$\approx 10^{30} / 10^{42} = 10^{-12}$&lt;/span&gt; -- or about one in a trillion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The odds of &lt;em&gt;any&lt;/em&gt; star from our galaxy &lt;em&gt;not&lt;/em&gt; hitting a star in the other galaxy would be &lt;span class=&quot;math-container&quot;&gt;$(1 - 10^{-12})^{10^{11}} \approx 0.90$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there's only about a 10% chance of one (or more) of the galaxy's 100 billion&#xA;stars hitting a star in the other galaxy. And the chances of any &lt;em&gt;one particular star&lt;/em&gt; (like our Sun) hitting a star in the other galaxy is about one in a trillion.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="-1" LastEditDate="2019-05-16T21:19:07.533" LastActivityDate="2019-05-16T21:19:07.533" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22638" PostTypeId="1" AcceptedAnswerId="22643" CreationDate="2017-09-11T22:02:18.217" Score="5" ViewCount="909" Body="&lt;p&gt;First, I'd like to take the definition of a &quot;double-planet&quot; as two bodies orbiting each other where the center of gravity is not inside the larger body. Also, the system would have to fill other planet requirements (like emptying it's own orbital area around the star).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's say that the &quot;binary system&quot; is in an orbit close enough to a star to be tidally locked to the star. How would the tidal effects of the star's gravity effect the orbit of the planets around each other??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take an example of a &quot;double-planet&quot; (of $0.075$ and $0.030\:\mathrm{M_{Earth}}$ with a separation of $1.17 \times 10^6\:\mathrm{km}$ orbiting their CM in about $450\:\mathrm{days}$. They are $0.085\:\mathrm{AU}$ from a star of mass $0.35\:\mathrm{M_{Sun}}$ and thus orbit the star in $15.3\:\mathrm{days}$. The planetary rotation/co-orbital period is much longer than the orbital period around the star, so I would think that the planets would have characteristics of a lone TL planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering how such a system would evolve over time. Would someone on the surface of one of the planets experience experience night and day in 15.3 day cycles? Would the difference in tidal effects of the star on the different sized planets cause the orbit to &quot;precess&quot; or have some other effect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine designing a long term calendar for this system!!&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="11143" LastEditDate="2017-09-12T14:29:03.693" LastActivityDate="2017-09-19T07:44:06.730" Title="A &quot;tidally locked&quot; double planet?" Tags="&lt;orbit&gt;&lt;exoplanet&gt;&lt;tidal-locking&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22640" PostTypeId="2" ParentId="22224" CreationDate="2017-09-11T22:57:46.877" Score="2" Body="&lt;p&gt;Normally people don't focus their eyes on the sun; its always in the sky but noone usually has a reason to look at it so you never focus on it, and most know not to. During an eclipse thats exactly where you want to look and focus on, and focussing on it makes it even more dangerous (guessing due to time spent with same area of your retina receiving the focussed energy from it rather than moving around your view and unfocussed spot).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During totality (darkest part of 100% eclipse) your eyes start to adjust for the dark sky and open up, allowing even more light in; as sun comes out of totality it gets very very bright again quite quickly, and your eyes might not be ready for it - use protection as you see the diamond ring start to get bigger/brighter (or only look at the diamond ring on the way into totality for safest view, when its getting darker).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same effect as totality is almost had by using eclipse glasses or other dark viewers that cover your head/eyes, and switching from that view back to unprotected view of the sun could be equally as dangerous for the same reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm not an expert).&lt;/p&gt;&#xA;" OwnerUserId="10521" LastEditorUserId="10521" LastEditDate="2017-09-11T23:03:15.520" LastActivityDate="2017-09-11T23:03:15.520" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22641" PostTypeId="2" ParentId="22224" CreationDate="2017-09-11T23:27:58.570" Score="0" Body="&lt;p&gt;It's not inherently less safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's just that it's safe to look at the Sun during the total phase, but people keep staring at it after totality is over, when it's not safe anymore. That's where the problem is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the photosphere starts showing again, the safe period is over. You can look at the &quot;diamond ring&quot; for, like, a second or two - but that's it. Unfortunately, some people are lulled into a sense of security and keep looking at the Sun a long, long time after totality is over. They keep staring at it fixedly until it hurts and they feel compelled to avert their gaze - but it's quite possible that the damage is done already at that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the gradual nature of what happens after totality that traps people into a dangerous situation. Totality is over, you've seen the diamond ring for a second or two - that's it, show's over, avert your gaze, put the goggles on.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The safety of eclipse goggles cannot be confidently evaluated by amateurs. Are there ways to make eclipse goggles that are safe? Yes. Can anyone guarantee that those are safe with 100% confidence? No. So don't do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only goggles that are guaranteed to be safe are those that comply with the ISO 12312-2 international safety standard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://eclipse.aas.org/eye-safety/iso-certification&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://eclipse.aas.org/eye-safety/iso-certification&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I prefer goggles made from the Baader AstroSolar Silver/Gold Film, which complies with the ISO standard, and show the Sun in its original color - as opposed to many other goggles that color the Sun green, red, orange or other unrealistic hues.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2017-09-11T23:33:04.107" LastActivityDate="2017-09-11T23:33:04.107" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22642" PostTypeId="1" CreationDate="2017-09-12T02:43:31.310" Score="2" ViewCount="979" Body="&lt;p&gt;Is anyone aware of where one can obtain &lt;strong&gt;ephemeris data in Geocentric Celestial Reference Systems (GCRS) coordinates&lt;/strong&gt; for a geodesy satellite such as LAGEOS (It doesn't really matter which mission)? Or, alternatively, simulated data in &lt;strong&gt;GCRS coordinates&lt;/strong&gt;. Specifically, Cartesian position and velocity coordinates. I can't seem to find them anywhere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; My confusion has increased. I will try and rephrase my question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to obtain an ephemeris file and perform a transformation such that the ephemeris data can be expressed in GCRS coordinates? And if so, what is the process?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Purpose of question:&lt;/strong&gt;&#xA;What I am interested in is ephemeris data for a satellite (doesn't really matter which mission) that is subject to classical Newtonian acceleration plus relativistic corrections as recommended by the IERS such as Schwarzschild, Lense-Thirring and Geodetic accelerations.&lt;/p&gt;&#xA;" OwnerUserId="16812" LastEditorUserId="16812" LastEditDate="2017-09-20T07:58:47.600" LastActivityDate="2017-09-20T07:58:47.600" Title="where to obtain ephemeris data for geodesy satellites?" Tags="&lt;data-analysis&gt;&lt;satellite&gt;&lt;artificial-satellite&gt;&lt;near-earth-object&gt;&lt;ephemeris&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22643" PostTypeId="2" ParentId="22638" CreationDate="2017-09-12T04:05:43.380" Score="7" Body="&lt;p&gt;Your scenario isn't stable.   A simple way to explain this is to imagine that the planet's orbit each other at the same rate they orbit the star (your scenario has them orbiting even slower).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same rate of orbit, the synodic period essentially approaches infinity.  (see diagram of the Moon's synodic orbit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZRqZy.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZRqZy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When this happens, the inner planet is at L1 to the outer planet and the outer planet is at L2 to the inner planet.   That's ofcourse, not quite right as the different masses would have different Hill Spheres, and the mutual gravitation of two objects massive to have a barrycenter outside the more massive planet would combine into a slightly faster orbit, but it's close enough to demonstrate the instability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The L1 and L2 points are on the border of the Hill Sphere and well outside the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere#True_region_of_stability&quot; rel=&quot;noreferrer&quot;&gt;true region of stability&lt;/a&gt;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No system can be stable where the Moon (or binary planet system) orbits around each other slower than they orbit around the central star.   They need to be at least 50% to 67% closer in than that would allow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the Orbital Period or &quot;T&quot; squared = distance (&quot;a&quot; semi-major axis) cubed, &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6&quot; rel=&quot;noreferrer&quot;&gt;per Kepler's laws&lt;/a&gt;, the upper limit for a stable orbital period of a moon around a planet or a binary planet would be about 19%-35% the period of the planet's year.    Any longer the Moon or binary system would risk instability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be able to set up a synchronous system where the planets orbited each other 4 times for every orbit around the sun, where they'd end up in the same (basically eclipse) position every perihelion with the planet experiencing greater tides (smaller and/or more fluid planet) closer to the sun at perihelion.  I suspect a 4-1 orbital period ratio is about as close as you're likely to get for any length of time for a stable binary planet-star, tidal locked system and that would be odd, but I see no reason why it wouldn't be stable.   It would only be possible around a small star where the planets were fairly close so that all 3 could have significant tidal effects on each other.   Without strong tides you don't get tidal locking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also can't have tidal locking with both objects when they have different orbital periods.   Tidal locking means that the rotation period is equal to the (sidereal) orbital period.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-09-19T07:44:06.730" LastActivityDate="2017-09-19T07:44:06.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22645" PostTypeId="2" ParentId="22642" CreationDate="2017-09-12T05:00:25.187" Score="2" Body="&lt;p&gt;The &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL Horizons&lt;/a&gt; website gives ephemerides for a large number of Earth satellites as well as deep space spacecraft, planets, moons, asteroids, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you configure it as shown below, selecting the units you'd like, the full state vectors &lt;code&gt;(x, y, z, vx, vy, vz)&lt;/code&gt;, use geocentric coordinates (&lt;code&gt;@500&lt;/code&gt;) you will get the following output. You can choose various time step sizes (down to 1 minute) and time frames. You can choose direct download as csv text rather than HTML if you like - recommended for long files.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks to me that because of the special status and use of the LAGEOS satellites, the ephemeris only predicts a relatively short period into the future, unlike other spacecraft which have predictions for years sometimes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This should get you started!&lt;/em&gt; If you have more questions about this post as comments and I or someone else will try to help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BTVB3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BTVB3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the Table settings. I believe that the ICRF/J2000.0 with Ecliptic and Mean Equinox settings &lt;strong&gt;will give you something close to GCRS&lt;/strong&gt; but right now I am not sure how close. I'll keep reading and ask for help in comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bB6Ax.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bB6Ax.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a sample output from the above configuration. It scrolls left-right and up-down.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Revised: Sep 11, 2017                 LAGEOS-2                         -122195&#xA;&#xA;LAGEOS-2 (1992-070B, &quot;Laser Geodynamics Satellite&quot;, NASA/ASI)&#xA;60-cm diameter sphere, 426 corner cube reflectors (VIS), 4 germanium (IR)&#xA;Deployed: 1993-Oct-22 from Space Shuttle Columbia (STS 52)&#xA;&#xA;&#xA;*******************************************************************************&#xA;Ephemeris / WWW_USER Mon Sep 11 21:51:05 2017 Pasadena, USA      / Horizons    &#xA;*******************************************************************************&#xA;Target body name: LAGEOS-2 (spacecraft) (-122195) {source: LAGEOS-2}&#xA;Center body name: Earth (399)                     {source: DE431mx}&#xA;Center-site name: BODY CENTER&#xA;*******************************************************************************&#xA;Start time      : A.D. 2017-Jan-01 00:00:00.0000 TDB&#xA;Stop  time      : A.D. 2017-Jan-05 00:00:00.0000 TDB&#xA;Step-size       : 1440 minutes&#xA;*******************************************************************************&#xA;Center geodetic : 0.00000000,0.00000000,0.0000000 {E-lon(deg),Lat(deg),Alt(km)}&#xA;Center cylindric: 0.00000000,0.00000000,0.0000000 {E-lon(deg),Dxy(km),Dz(km)}&#xA;Center radii    : 6378.1 x 6378.1 x 6356.8 km     {Equator, meridian, pole}    &#xA;Output units    : KM-S                                                         &#xA;Output type     : GEOMETRIC cartesian states&#xA;Output format   : 2 (position and velocity)&#xA;Reference frame : ICRF/J2000.0                                                 &#xA;Coordinate systm: Ecliptic and Mean Equinox of Reference Epoch                 &#xA;*******************************************************************************&#xA;            JDTDB,            Calendar Date (TDB),                      X,                      Y,                      Z,                     VX,                     VY,                     VZ,&#xA;**************************************************************************************************************************************************************************************************&#xA;$$SOE&#xA;2457754.500000000, A.D. 2017-Jan-01 00:00:00.0000,  5.049112760756619E+03,  9.246781656953866E+03,  5.722633641721932E+03, -2.978199703429669E+00,  3.689975896939779E+00, -3.350502047340569E+00,&#xA;2457755.500000000, A.D. 2017-Jan-02 00:00:00.0000, -6.140907600397200E+03, -8.148707798015072E+03, -6.918983669661846E+03,  2.434461456961566E+00, -4.250750074318191E+00,  2.808377534161328E+00,&#xA;2457756.500000000, A.D. 2017-Jan-03 00:00:00.0000,  6.849580412560958E+03,  6.106324592083777E+03,  7.740446107133596E+03, -1.906840987637830E+00,  4.973834339923835E+00, -2.290764139470047E+00,&#xA;2457757.500000000, A.D. 2017-Jan-04 00:00:00.0000, -7.504132672898423E+03, -4.707277560275819E+03, -8.539479988453819E+03,  1.304309397636298E+00, -5.242849004192330E+00,  1.679103108590806E+00,&#xA;2457758.500000000, A.D. 2017-Jan-05 00:00:00.0000,  7.740851830184564E+03,  2.245705288312970E+03,  8.942780571589567E+03, -6.155844902136471E-01,  5.667380276723395E+00, -9.687665040214430E-01,&#xA;$$EOE&#xA;**************************************************************************************************************************************************************************************************&#xA;Coordinate system description:&#xA;&#xA;  Ecliptic and Mean Equinox of Reference Epoch&#xA;&#xA;    Reference epoch: J2000.0&#xA;    XY-plane: plane of the Earth's orbit at the reference epoch&#xA;              Note: obliquity of 84381.448 arcseconds wrt ICRF equator (IAU76)&#xA;    X-axis  : out along ascending node of instantaneous plane of the Earth's&#xA;              orbit and the Earth's mean equator at the reference epoch&#xA;    Z-axis  : perpendicular to the xy-plane in the directional (+ or -) sense&#xA;              of Earth's north pole at the reference epoch.&#xA;&#xA;  Symbol meaning:&#xA;&#xA;    JDTDB    Julian Day Number, Barycentric Dynamical Time&#xA;      X      X-component of position vector (km)                               &#xA;      Y      Y-component of position vector (km)                               &#xA;      Z      Z-component of position vector (km)                               &#xA;      VX     X-component of velocity vector (km/sec)                           &#xA;      VY     Y-component of velocity vector (km/sec)                           &#xA;      VZ     Z-component of velocity vector (km/sec)                           &#xA;&#xA;Geometric states/elements have no aberrations applied.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-09-12T05:34:31.247" LastActivityDate="2017-09-12T05:34:31.247" CommentCount="23" ContentLicense="CC BY-SA 3.0" />
  <row Id="22650" PostTypeId="2" ParentId="22609" CreationDate="2017-09-12T16:18:45.817" Score="1" Body="&lt;p&gt;There's been some good answers already, but I'd like to add my two-pennyworth:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do we have photos of galaxies so far away?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because there's nothing much between them and us that interferes with the light that reaches our cameras. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A possible answer for this is that, light emitted from the galaxies travelled a billion miles all the way to earth, where the Hubble space telescope picked up this light through its sensors, and was able to construct an image of the galaxy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's a billion miles to &lt;em&gt;Saturn&lt;/em&gt;. Well actually the distance varies with the orbits, but see &lt;a href=&quot;https://www.space.com/18477-how-far-away-is-saturn.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Space.com article&lt;/a&gt;: &lt;em&gt;&quot;At their most distant, when they lie on opposite sides of the sun from one another, they are just over a billion miles (1.7 billion km) apart&quot;&lt;/em&gt;. The &lt;a href=&quot;https://www.google.co.uk/search?q=andromeda+galaxy+how+far+miles&amp;amp;oq=andromeda+galaxy+how+far+miles&amp;amp;gs_l=psy-ab.3..0i22i30k1l2.82831.84293.0.84392.6.6.0.0.0.0.134.542.2j3.5.0....0...1.1.64.psy-ab..1.5.538...0.UhwIvdw4S-A&quot; rel=&quot;nofollow noreferrer&quot;&gt;Andromeda galaxy&lt;/a&gt; is circa fifteen billion &lt;em&gt;billion&lt;/em&gt; miles away. Or circa fifteen quintillion miles. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;but if this is true, and galaxies are billions of miles away, shouldn't the light particles emitted from the galaxies be scattered all over the place?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't forget that photons have an E=hf wave nature. And that even though they are scattered in the air, you can still see the Moon. Yes, there's a bit of light going astray in space. But not so much that the night-time sky is some blank foggy fug. You can see Saturn too. And the stars. And the galaxies, but they are &lt;a href=&quot;https://www.google.co.uk/search?q=andromeda+galaxies+very+dim&amp;amp;oq=andromeda+galaxies+very+dim&amp;amp;gs_l=psy-ab.3...7848.14077.0.14358.16.16.0.0.0.0.127.1150.13j3.16.0....0...1.1.64.psy-ab..1.0.0.3AYbIEWlZKk&quot; rel=&quot;nofollow noreferrer&quot;&gt;rather dim&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;after all they have been travelling from millions of years, and have probably collided with asteroids and other foreign objects. What were the chances that about 95% of the photons actually reached earth, giving us a very detailed image. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The chances are high. We have pictures of planets and things &lt;em&gt;because&lt;/em&gt; the chances are high. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider the Andromeda galaxy which has a distance of 1.492 × 10^19 mi from earth. If light emitted from the galaxy travels in all directions, then how is it that we can still map out the entire galaxy, evident from the photo below?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I was covered in lights, I would emit light in all directions, and you would see me because some of that light goes into your eye. The Andomeda galaxy is similar. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Shouldn't like half of the galaxy be missing since photons could have hit other objects, and &quot;never have reached earth&quot;? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. And if half the photons didn't reach Earth, you'd just see a dimmer galaxy, that's all. &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2017-09-12T16:18:45.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22651" PostTypeId="1" AcceptedAnswerId="22656" CreationDate="2017-09-12T17:51:45.460" Score="63" ViewCount="18199" Body="&lt;p&gt;Does our galaxy moves through space? Or does it stay in a single location? If it does move, what causes it to move?&lt;/p&gt;&#xA;" OwnerUserId="18490" LastEditorUserId="2365" LastEditDate="2018-07-21T22:39:43.003" LastActivityDate="2020-02-07T11:39:43.960" Title="Does the Milky Way move through space?" Tags="&lt;galaxy&gt;&lt;universe&gt;&lt;milky-way&gt;" AnswerCount="5" CommentCount="12" FavoriteCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="22652" PostTypeId="1" CreationDate="2017-09-12T18:00:10.953" Score="0" ViewCount="106" Body="&lt;p&gt;Can we, with arbitrary precision, calculate what's the place on earth that's closest to the sun on a given date? If so, what's the math like? Say, if I wanted to calculate this within 1m, 10cm or 1cm precision, what exactly should I take into account?&lt;/p&gt;&#xA;" OwnerUserId="11265" LastActivityDate="2017-09-13T15:19:56.943" Title="Can we, with arbitrary precision, calculate what's the place on earth that's closest to the sun on a given day?" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;distances&gt;&lt;time&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22653" PostTypeId="2" ParentId="22651" CreationDate="2017-09-12T18:19:41.650" Score="38" Body="&lt;p&gt;Galaxies move &lt;em&gt;through&lt;/em&gt; space with velocities of the order of a several 100 km per second; small velocities for small groups (~100 km/s; e.g &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0008201&quot; rel=&quot;noreferrer&quot;&gt;Carlberg et al. 2000&lt;/a&gt;) and large velocities for rich clusters (~1000 km/s; e.g &lt;a href=&quot;http://adsabs.harvard.edu/abs/1993ApJ...404%E2%80%A638G&quot; rel=&quot;noreferrer&quot;&gt;Girardi et al. 1993&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to this so-called &quot;peculiar velocity&quot;, galaxies also also carried away from each other due to the expansion of the Universe, at a velocity proportional to the distance from each other (the &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble%27s_law&quot; rel=&quot;noreferrer&quot;&gt;Hubble flow&lt;/a&gt;&quot;). But this is not a motion &lt;em&gt;through&lt;/em&gt; space; rather it is space itself that is expanding (and hence the velocity may exceed the speed of light for sufficiently large distances).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One may define a &quot;global reference frame&quot; with respect to which velocities are measured. Any reference is valid, but it makes sense to use the frame in which all galaxies are, on average, in rest (when the Hubble flow is subtracted)$^\dagger$. In this frame, the Local Group that the Milky Way is a part of moves with some $620\, \mathrm{km \, s}^{-1}$ (as noted in John Duffield's answer above), whereas the center of &lt;strong&gt;the Milky Way has a velocity$^\ddagger$ of&lt;/strong&gt; $\mathbf{565 \pm 5 \, km \, s^{-1}}$ (&lt;a href=&quot;https://arxiv.org/abs/1807.06205&quot; rel=&quot;noreferrer&quot;&gt;Planck Collaboration et al. 2018&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What causes this movement of galaxies? Galaxies that are not too far from each (i.e. closer), &quot;feel&quot; each others mutual gravitational forces. A galaxy in a group or cluster moves around in the common gravitational field, but for galaxies that are farther away, the Hubble flow carries them away from each other too fast for them to attract each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This movement can be traced back to the tiny quantum mechanical fluctuations in the primordial soup of particles during &lt;a href=&quot;https://en.wikipedia.org/wiki/Inflation_(cosmology)&quot; rel=&quot;noreferrer&quot;&gt;cosmic inflation&lt;/a&gt;, i.e. less than $\sim10^{-32}\,\mathrm{s}$ after the Big Bang. As time went by, ever-so-slight overdensities grew in amplitude, until they collapsed to form the structure we see in the Universe today. During this collapse, matter grew turbulent, whirling clumps around that eventually became the galaxies that orbit each other.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;Formally, one uses the frame in which the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;noreferrer&quot;&gt;cosmic microwave background&lt;/a&gt; is isotropic.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^\ddagger$&lt;sub&gt;Taking into account our motion around the Galactic center, our Sun (currently) moves through space at $369.82\pm0.11\,\mathrm{km\, s^{-1}}$.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-08-16T06:47:46.247" LastActivityDate="2018-08-16T06:47:46.247" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22654" PostTypeId="2" ParentId="22652" CreationDate="2017-09-12T18:19:59.867" Score="6" Body="&lt;p&gt;Calculation of the &lt;a href=&quot;https://rl.se/sub-solar-point&quot; rel=&quot;noreferrer&quot;&gt;subsolar point&lt;/a&gt; is relatively straightforward. The actual point moves at nearly 500m/s in a circle, as the Earth spins. The linked website does the calculation in a few lines of python using a free library, The value is probably correct to within a few 10s of kilometres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you are looking for a different order of accuracy:  If you (for some reason) need 1cm accuracy, you would need to take into account the actual, non-spherical shape of the Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth is not a sphere, it can be better modelled as an oblate ellipsoid. But for 1cm accuracy, you would need to consider all of the mountains and hills. And you would need a very accurate time, not just a date. That is not a practical calculation. I can think of no practical application.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-09-12T18:19:59.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22655" PostTypeId="2" ParentId="22609" CreationDate="2017-09-12T18:38:43.327" Score="0" Body="&lt;p&gt;Let me give some simple explanations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, no, no. 95% of the photons don’t reach Earth. Even if 5% of photons emitted (within a few seconds) just by one star, say, by our Sun had reached Earth, our planet would have been completely scorched! Now then, Andromeda has hundreds of billions of stars (or suns). Nothing of that reaches us, except for an infinitesimally small number. It’s mind-boggling how small the percentage of photons that reach us is! You can try to calculate that very roughly. It is very easy to calculate what percentage of photons emitted by the Sun reaches Earth. And the Sun is only 8 minutes away from Earth, while Andromeda is more than 2.5 million years away! So, actually, it’s not that difficult to imagine how many photons reach us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, why don't asteroids, planets or stars block everything? Andromeda is way too large to be blocked like that! It’s easier to block the view of the Pacific Ocean from space by placing a few specks of dust in between! The diameter of Andromeda is more than 200 million light years. Can we block it from view? Actually it can be blocked by something as large as a nebula close to our solar system. Such a nebula must be many light years in diameter; it must be dense enough; and not too far away. Thankfully nothing like that blocks this beautiful galaxy from our view. However it happens with some other galaxies and deep space objects. As to very distant nebulas, they won’t block Andromeda from our view because they will look way too small against the background of Andromeda which is much further away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is light not scattered?  Why should it be scattered that much to make Andromeda blurry? When the Moon is on the horizon, its light travels through many hundreds of miles of dense atmosphere almost parallel to the surface of Earth; yet, we can still train our telescopes on it and see the various features of the Moon. It would not be a very clean view but we would still see a lot. Now, in space light travels through an almost complete vacuum, especially empty is the void between galaxies. So, there’s no reason for light to be scattered too much. Photons and many other particles are stable enough and can travel much larger distances: billions of light years. Another way to look at it is to ask a question how much photons should deviate from their straight path so that Andromeda becomes blurry to us. Well, they have to go sideways a lot, and the diameter of Andromeda is too huge for that. That doesn’t seem logical, as photons travel in straight lines. Large objects, like stars and black holes will affect their path but the diameter of Andromeda is so huge that it is not an option, unless we artificially place trillions of black holes along the line between Andromeda and our solar system in an attempt to warp the image of Andromeda or to make these black holes gobble up all the light from the galaxy! So, when astronomers say most of the light reaches us, they mean that intergalactic space is almost complete vacuum, and the photons that go exactly in our direction are “free” to go. Yet, only an infinitesimally small number of them go exactly in our direction and it is still enough for nice photos. Why? That's why:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absolute magnitude (relative luminosity against that of an object $40$ times brighter than the Sun at a distance of $33$ light years away) of Andromeda is around $-21.5$. Our Sun is only around $5$. The higher the number the dimmer the object. An object with an absolute magnitute of $1$ would be $2.5^{5-1}=40$ times brighter than the Sun. The difference between Andromeda and our Sun is $-21.5-5=-26.5$. This means Andromeda is very roughly $2.5^{26.5}\approx 40,000,000,000$ times brighter than the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to how large it is in the night sky, well, lengthwise it is roughly six times the diameter of the moon but you can only see the bright central part. To see the whole extent you need a large aperture telescope and long exposure photography to gather more light and produce a better, more detailed image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope, this primitive explanation will be of some help. Andromeda is visible today if weather permits :)&lt;/p&gt;&#xA;" OwnerDisplayName="user18491" LastEditorDisplayName="user18491" LastEditDate="2017-09-18T06:41:05.167" LastActivityDate="2017-09-18T06:41:05.167" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22656" PostTypeId="2" ParentId="22651" CreationDate="2017-09-12T18:49:05.520" Score="79" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the Milky Way move through space?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes it does. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm very fascinated with space, although I don't have a degree or any formal education, I'm still very in love with everything about it and want to learn constantly. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Good man Mike. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One thing I ask myself is if our galaxy moves through space?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It does. When we look at the Cosmic Microwave Background Radiation we see a &quot;dipole anisotropy&quot; due to the motion of the Earth relative to it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MsvXR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Image courtesy of William H. Kinney's &lt;a href=&quot;https://ned.ipac.caltech.edu/level5/Sept02/Kinney/Kinney_contents.html/%22title%22&quot; rel=&quot;noreferrer&quot;&gt;Cosmology, inflation, and the physics of nothing&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background#CMBR_dipole_anisotropy&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; for more:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;From the CMB data it is seen that the Local Group (the galaxy group that includes the Milky Way galaxy) appears to be moving at 627±22 km/s relative to the reference frame of the CMB (also called the CMB rest frame, or the frame of reference in which there is no motion through the CMB) in the direction of galactic longitude l = 276°±3°, b = 30°±3°.[82][83] This motion results in an anisotropy of the data (CMB appearing slightly warmer in the direction of movement than in the opposite direction).[84]&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;627 km/s is quite fast. See &lt;a href=&quot;https://astrosociety.org/edu/publications/tnl/71/howfast.html#6&quot; rel=&quot;noreferrer&quot;&gt;this article&lt;/a&gt;, which says it's 1.3 million miles an hour. The speed of light is just under 300,000 km/s or 670 million miles per hour, so the Milky Way is moving through the Universe at circa 0.2% of the speed of light. Also see &lt;a href=&quot;https://physics.stackexchange.com/a/25929/76162&quot;&gt;the CMBR&lt;/a&gt; physics answer by ghoppe which talks about the CMBR reference frame, which is in effect the reference frame of the universe.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Or does it stay in a single location? If it does move, what causes it to move?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm afraid I don't know why it's moving. Perhaps it's because the Universe is full of things moving in fairly random directions. Like a gas.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hopefully the question makes sense, if not I can elaborate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It certainly makes sense to me! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 13/09/2017&lt;/strong&gt; : as Dave points out in the comments, there are other motions, including the motion of the solar system around the galaxy, which is circa 514,000 mph. (See the Wikipedia &lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_year&quot; rel=&quot;noreferrer&quot;&gt;Galactic Year&lt;/a&gt; article). And the motion of galaxies isn't neat and tidy either. &lt;/p&gt;&#xA;" OwnerUserId="9772" LastEditorUserId="9772" LastEditDate="2017-09-13T16:59:08.633" LastActivityDate="2017-09-13T16:59:08.633" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="22657" PostTypeId="1" AcceptedAnswerId="22658" CreationDate="2017-09-12T20:59:56.340" Score="-1" ViewCount="139" Body="&lt;p&gt;How did the humans take photos for the galaxies, black holes, stars, nebulas etc in the outer space?!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess they went out of the Earth planet and nearly out of the Milky Way galaxy by space-ships or probes and that is how they took photos for the stars, nebulas, blackholes, and galaxies in the outer space&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my guess is wrong then how did they do it?!&lt;/p&gt;&#xA;" OwnerUserId="13592" LastActivityDate="2017-09-12T21:34:55.800" Title="The photos of the outer space" Tags="&lt;black-hole&gt;&lt;galaxy&gt;&lt;space&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22658" PostTypeId="2" ParentId="22657" CreationDate="2017-09-12T21:34:55.800" Score="8" Body="&lt;p&gt;The same way I can take photos of my neighbourhood without leaving my house; light travels from the surrounding houses, through my windows, and I capture that light in a photograph. No need to leave my house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stars you can see at night would be like the houses in my street. The Milky Way Galaxy would be like the sky-glow I can see coming from the middle of the city I live in. I can see all of those things without leaving my house. On a dark night, I can even see sky-glow from the next town over, which is a bit like seeing a distant galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way, we don't need to leave Earth to get photos of things in outer space; light travels from those objects, through the Earth's atmosphere (which is like the windows in my house), and we capture that light in photographs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment (and for the near future), we don't have any spaceships which can travel outside the Galaxy -- which would take millions of years at the speed of our fastest ships yet -- and our first interplanetary spacecraft (the Voyager probes, launched in the 1970's (before I was born), have only just left the Solar System. It would take them tens of thousands of years even to reach the nearest star (but they're not even aimed in that direction).&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2017-09-12T21:34:55.800" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22659" PostTypeId="2" ParentId="22651" CreationDate="2017-09-12T21:34:56.060" Score="10" Body="&lt;p&gt;Simply put:&#xA;The important thing about the motion of our Milky Way in the Universe is simply gravitational attraction among galaxies and clusters of galaxies that causes movement. At a small scale we are attracted to the Andromeda galaxy and are on a “collision” course with it (Of course we won’t collide because galaxies are way too diffuse so to speak). The most important gravitational attractions that cause our Galaxy to move even faster is the Great Attractor (a large collection of galaxies) and the much more awe-inspiring Shapley Supercluster of galaxies. It looks like these are the major clusters of galaxies to which our Milky Way Galaxy is gravitationally attracted, and it causes our Galaxy to move at around 600 km/sec with reference to the microwave background radiation. This background radiation or CMB (cosmic microwave background) is the absolute reference against which to measure velocities in our Universe. The CMB is the remnant of and the proof of the Big Bang event. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as was mentioned before, the space expands in our universe producing the effect of galaxies (that are far away for each other) to look like they run away from each other very fast. It holds only for galaxies that are very far away in space. That’s because space expands like a rubber band with most distant points moving away from each other faster. In other words, galaxies that are extremely far away from each other, well, they run away from each other almost with the speed of light or even faster! No, they are not moving faster than the speed of light – it’s just that space in the universe expands very fast between two very distant points, even faster than the speed of light! It is just like the further apart the two points are on the elastic rubber band the faster they will move apart when we extend the band. The difference here is that our three-dimensional space acts like a rubber band if it is a three-dimensional space we live in after all (here's a lot to be discovered in the future, I guess)?!&lt;/p&gt;&#xA;" OwnerDisplayName="user18491" LastEditorUserId="1891" LastEditorDisplayName="user18491" LastEditDate="2020-02-07T11:39:43.960" LastActivityDate="2020-02-07T11:39:43.960" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22660" PostTypeId="1" AcceptedAnswerId="22668" CreationDate="2017-09-13T00:24:46.303" Score="8" ViewCount="1481" Body="&lt;p&gt;According to Wiki the age of the universe is 13 billion years old, and I was taught that background radiation made the universe uniform in all directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doesn't this define a sphere of space in the universe with Earth at the center? This sphere would have a radius of 13 billion light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We (on Earth) could not see anything outside this sphere. As light did not exist more than 13 billion years ago. So for us this defines some kind of edge for the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what happens when an alien in another galaxy 5 billion light years away from Earth also looks towards the stars. Don't they see a uniform background radiation and measure the same age of the universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That alien would also have a sphere of 13 billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have two spheres, one for Earth and one for our alien friend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two spheres would overlap each other by 5 billion years. Which means, if we look in the opposite direction from that alien's galaxy. We can see an additional 5 billion years further than they can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we measure the total width of the two overlapping spheres we get a distance of 18 billion light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I now grasp my hair in confusion. Is this possible that two visible spheres could overlap to create a distance longer than the oldest ray of light? And from there I fall into more questions. Can two people measure the same universe age from different spots in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why I think my idea is is wrong, but how could it be wrong and why?&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2019-08-01T06:25:41.200" Title="Is the age of the universe relative to an observer's location in that universe?" Tags="&lt;universe&gt;&lt;light&gt;&lt;general-relativity&gt;&lt;age&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22661" PostTypeId="2" ParentId="12068" CreationDate="2017-09-13T01:42:45.213" Score="1" Body="&lt;p&gt;We'd analyze the DNA and RNA sequences (assuming they had any) and the coding of amino acids, and compare them with the sequencing and coding on Earth. As far as we are evolutionary from Archaea and Bacteria, we're still a lot closer than an alien life form would be.&lt;/p&gt;&#xA;" OwnerUserId="18503" LastActivityDate="2017-09-13T01:42:45.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22663" PostTypeId="2" ParentId="22608" CreationDate="2017-09-13T02:18:06.557" Score="1" Body="&lt;p&gt;The right ascension of the point on the horizon and due west is LST - 6 hours, where LST is the Local Sidereal Time. The LST can be calculated from knowing the date, time, and longitude. The declination of the west point is 0 degrees. So the angle between that right ascension and 0 hours right ascension (the vernal equinox) is just the difference between the two.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastEditorUserId="17355" LastEditDate="2017-09-13T11:49:25.590" LastActivityDate="2017-09-13T11:49:25.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22668" PostTypeId="2" ParentId="22660" CreationDate="2017-09-13T06:23:09.827" Score="15" Body="&lt;p&gt;You are labouring under the misapprehension that how far we can see directly gives the age of the universe. Whilst it is true that the oldest light we can see was emitted some 13.7 billion years ago, the stuff that emitted that light is now roughly 46 billion light years away, thanks to expansion of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe itself probably extends significantly beyond this and &lt;em&gt;may&lt;/em&gt; be infinite. That does not conflict with the big bang model or relativity. If the universe is infinite then it has always been infinite. Places that are separated by more than the oldest light ray as you put it, are not in causal contact and, depending on the cosmological parameters, may never be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a homogeneous, isotropic universe (and don't forget we can only make assumptions about regions beyond our observable universe), all observers would agree on the age of the universe and that the universe was once very much smaller and all would see (on average) the same sort of universe in all directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To ensure that places that are outside of causal contact now are homogeneous requires that they &lt;em&gt;were&lt;/em&gt; in causal contact in the past. This is the nature of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Horizon_problem&quot; rel=&quot;noreferrer&quot;&gt;&quot;horizon problem&quot;&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took the picture below from the &quot;slideplayer&quot; website, I am unsure who the originator is, but it serves its purpose here and I think encapsulates your question. It appears in this diagram that A and B can never have &quot;communicated&quot; and can never have originated from the same place. The solution provided by inflation is to have a massive exponential expansion of space in the first fraction of a second. In short, the distances between points in space (or galaxies if you like, although there were no galaxies at the epoch of inflation) are suddenly increase by many, many orders of magnitude. This give the &lt;em&gt;appearance&lt;/em&gt; of faster-than-light motion, although the cosmic speed limit only applies to local measurements and not to the expansion of space itself. The end result is a universe that appears homogeneous well beyond the limits of a radius of the age of the universe in light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/41ENE.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/41ENE.jpg&quot; alt=&quot;Light horizon&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-09-14T08:58:35.830" LastActivityDate="2017-09-14T08:58:35.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22669" PostTypeId="2" ParentId="22608" CreationDate="2017-09-13T07:10:46.167" Score="0" Body="&lt;p&gt;Unless you want to develop the answer from first principles, then I suggest you use formulas already published.  One source is &quot;Astronomical Algorithms&quot;, by Jean Meeus.  &lt;a href=&quot;http://www.willbell.com/math/mc1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.willbell.com/math/mc1.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have a copy of this boo, so I can't guarantee that it contains what you want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no connection with the author or the publisher.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-09-13T07:10:46.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22671" PostTypeId="2" ParentId="22651" CreationDate="2017-09-13T10:37:24.167" Score="9" Body="&lt;p&gt;&lt;strong&gt;NO&lt;/strong&gt;, if don't want to consider it to be doing so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything in the universe only &quot;moves&quot; when considered from a different frame of reference. From the &lt;a href=&quot;https://en.wikipedia.org/wiki/Frame_of_reference&quot; rel=&quot;noreferrer&quot;&gt;observational frame of reference&lt;/a&gt; of the Milky Way galaxy, every &lt;em&gt;other&lt;/em&gt; object in the universe is moving, and the Milky Way is stationary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I walk to work, what I am actually doing is dragging the surface of the Earth, and everything on it, towards me with my feet, until my office arrives.&lt;/p&gt;&#xA;" OwnerUserId="32" LastActivityDate="2017-09-13T10:37:24.167" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="22673" PostTypeId="1" CreationDate="2017-09-13T12:21:18.450" Score="2" ViewCount="528" Body="&lt;p&gt;I'm working with data from radio surveys. My images are .fits files with BUNIT Jy/beam. I want to calculate the net flux from a continuous source (not point source). When I simply added all the intensity values and converted it into Jy using the beam parameters, my professor said that it is not enough. He said, &quot;You have to account for noise by taking 3 or 4-sigma level depending on the distribution of the noise&quot;. The distribution of noise (at places other than the source) is nearly Gaussian as I can see from a fits viewer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;take 3-sigma level&quot;, does he mean that I should fit a Gaussian to the noise, find sigma and then choose the pixels with intensity values greater than 3*sigma? If I do this, obviously I lose a lot of pixels. Also, if this is correct, should I subtract the mean of intensity of &quot;noise&quot; pixels from my source intensity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An RMS noise is given in the header, with a caution that it varies highly around some sources. Is this RMS noise value in anyway related to the sigma that I calculated earlier?&lt;/p&gt;&#xA;" OwnerUserId="18525" LastActivityDate="2018-06-24T13:19:10.500" Title="How to calculate the integrated source flux density of a continuous source?" Tags="&lt;radio-astronomy&gt;&lt;radio-telescope&gt;&lt;mathematics&gt;&lt;radio&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22676" PostTypeId="2" ParentId="22651" CreationDate="2017-09-13T13:24:56.243" Score="6" Body="&lt;p&gt;Monty Python does not seem very accurate in scientific issues, but this time it does it's job quite well in the Galaxy Song also mentioning the milky ways movement.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Just remember that you're standing on a planet that's evolving&#xA;And revolving at nine hundred miles an hour,&#xA;That's orbiting at nineteen miles a second, so it's reckoned,&#xA;A sun that is the source of all our power.&#xA;The sun and you and me and all the stars that we can see&#xA;Are moving at a million miles a day&#xA;In an outer spiral arm, at forty thousand miles an hour,&#xA;Of the galaxy we call the 'Milky Way'.&#xA;Our galaxy itself contains a hundred billion stars.&#xA;It's a hundred thousand light years side to side.&#xA;It bulges in the middle, sixteen thousand light years thick,&#xA;But out by us, it's just three thousand light years wide.&#xA;We're thirty thousand light years from galactic central point.&#xA;We go 'round every two hundred million years,&#xA;And our galaxy is only one of millions of billions&#xA;In this amazing and expanding universe.&#xA;&#xA;The universe itself keeps on expanding and expanding&#xA;In all of the directions it can whizz&#xA;As fast as it can go, at the speed of light, you know,&#xA;Twelve million miles a minute, and that's the fastest speed there is.&#xA;So remember, when you're feeling very small and insecure,&#xA;How amazingly unlikely is your birth,&#xA;And pray that there's intelligent life somewhere up in space,&#xA;'Cause there's bugger all down here on Earth.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And yes, these figures are accurate: (at the time of publish) &#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Galaxy_Song#Accuracy_of_astronomical_figures&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Galaxy_Song#Accuracy_of_astronomical_figures&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18530" LastActivityDate="2017-09-13T13:24:56.243" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22677" PostTypeId="1" CreationDate="2017-09-13T15:17:43.940" Score="6" ViewCount="2062" Body="&lt;p&gt;It seems everyone is on the same page about there being no &quot;absolute velocity&quot; due to the fact that everything is relativistic. However, this leaves me confused. This seems to be disconnected with the concept that the speed of light is in fact an unchanging constant. But to state that all speed is relative to an object is to essentially say, &quot;speed does not exist, but is only relative to an observer&quot;. But yet, the speed of light is a calculable constant that never changes. How then, can we say, there is no such thing as absolute speed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I throw a flashlight in space that is turned on 50mph, does the speed of the emitted light from the flashlight increase by 50 miles per hour in the direction I threw it? Of course not. We know this. But if this is truly the case, how then can there not be an absolute velocity which could be calculated by determining the speed of light in relevance to the object we're measuring (since the speed of light is constant)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help because this is bothering me.&lt;/p&gt;&#xA;" OwnerUserId="18534" LastEditorUserId="1378" LastEditDate="2017-09-13T18:30:29.400" LastActivityDate="2019-08-12T01:10:06.573" Title="Does Absolute Velocity Exist?" Tags="&lt;light&gt;&lt;space&gt;&lt;speed&gt;&lt;special-relativity&gt;&lt;velocity&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22678" PostTypeId="2" ParentId="22652" CreationDate="2017-09-13T15:19:56.943" Score="0" Body="&lt;p&gt;Word choice is important.  It's easy to calculate to arbitrary &lt;strong&gt;precision&lt;/strong&gt; but to arbitrary &lt;strong&gt;accuracy&lt;/strong&gt;, not so much.  Aside from the shape of the Earth, defining a surface of the sun is difficult as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ytPl9.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ytPl9.gif&quot; alt=&quot;precision vs. accuracy&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-09-13T15:19:56.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22679" PostTypeId="2" ParentId="22660" CreationDate="2017-09-13T15:53:04.813" Score="1" Body="&lt;p&gt;Well, the “first” hypothesis was that the universe is infinite and the number of space objects is infinite. Then, the Big Bang theory came along and we found out that our universe is roughly 13 billion years old and the radius of the observed universe is 13 billion light years, and we postulated that the Universe is probably connected: there’s no edge and if you could stop time and move in one direction you would probably arrive at the same point just like you would do on the surface of the Earth. Now we know that it was not a very precise idea: the radius of the observed universe turned out to be a bit larger: around 14 billion years, and it happens that the most distant galaxies move away from us faster than light, so this radius (14 billion years) is only the radius of the currently visible universe (we see how it looked 14 billion years ago when the universe was much smaller). Now this radius has reached 45 billion light years but we can only see the past (or 14 billion years). So wee see the past, not the present size. The fact that the galaxies that are very far away from each other move away from each other faster than light is attributed to the expansion of the space itself and doesn’t violate the General Theory of Relativity (we may say it makes it somewhat imprecise, just like relativity made Newton’s theory slightly imprecise). Our space is sometimes illustrated as an elastic rubber band on which the farther away two points are, the faster they will move apart, when you extend the rubber band. And our space is expanding in somewhat similar fashion. We don’t see the invisible part of the universe (the whole universe may be 150-600 billion light years in diameter -- this size will never be visible and it's much more than the distance of 45 billion light years to which the most distant visible galaxies must have moved by now but we still observe them as if they were 13-14 billion light years away). Also we don’t know if our universe is infinite or finite. You can imagine a finite one as a “connected universe” that is if we could stop time and go in one direction eventually arriving in our starting point. If, however, it is not the case and our universe is open and not connected, we may call it infinite. The both theories are fine. Unfortunately we don’t know which one is correct! If the Universe is not connected, there might be abnormalities like “edges”, I mean, there may be a point in Universe where galaxies end and there are no more galaxies further beyond or something much weirder than that. This is unfortunately beyond our visible Universe and we have no way of knowing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing is that your question revolves around the concept of space. Well, it’s roughly three-dimensional in real life, if we don’t consider very small (as in quantum mechanics) or cosmologically large distances. Yet, we know that our space might warp, so things are not that simple. Right now we don’t know much about the actual configuration of our metric space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I see, many folks have difficulty in imagining the connected Universe. Well, let’s simplify everything greatly. Let’s say our Universe is just like Earth, and let’s imagine Earth grows in size, so towns that are far away from each other move away from each other the fastest. Then you can take the length of the equator and call it the diameter (or just use it instead of the diameter) of our “test” universe. Half the distance of the equator will be the radius (or just use it instead of the radius) of our “test” universe, and it will be the same everywhere, and everything will be expanding as we initially postulated. It explains the paradox with aliens and two circles you mentioned. Note that towns are not growing in size because, as is the case with galaxies, we presume here that our towns are held by &quot;gravitation&quot; from expanding. Also note that we use the surface of the Earth in this example, which is two-dimensional, and the universe is three-dimensional (probably, or, let’s say, approximately to avoid imprecision). Hope, this primitive spherical illustration is of some help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an explanatory and helpful quote from the article &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Metric_expansion_of_space&quot; rel=&quot;nofollow noreferrer&quot;&gt;Metric Expansion of Space&lt;/a&gt;&lt;/em&gt; from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The universe could be infinite in extent or it could be finite; but the evidence that leads to the inflationary model of the early universe also implies that the &quot;total universe&quot; is much larger than the observable universe, and so any edges or exotic geometries or topologies would not be directly observable as light has not reached scales on which such aspects of the universe, if they exist, are still allowed. For all intents and purposes, it is safe to assume that the universe is infinite in spatial extent, without edge or strange connectedness.[15]&#xA;  Regardless of the overall shape of the universe, the question of what the universe is expanding into is one which does not require an answer according to the theories which describe the expansion; the way we define space in our universe in no way requires additional exterior space into which it can expand since an expansion of an infinite expanse can happen without changing the infinite extent of the expanse. All that is certain is that the manifold of space in which we live simply has the property that the distances between objects are getting larger as time goes on. This only implies the simple observational consequences associated with the metric expansion explored below. No &quot;outside&quot; or embedding in hyperspace is required for an expansion to occur. The visualizations often seen of the universe growing as a bubble into nothingness are misleading in that respect. There is no reason to believe there is anything &quot;outside&quot; of the expanding universe into which the universe expands.&#xA;  Even if the overall spatial extent is infinite and thus the universe cannot get any &quot;larger&quot;, we still say that space is expanding because, locally, the characteristic distance between objects is increasing. As an infinite space grows, it remains infinite&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may want to read the full article. I think it’s not bad and roughly in line with modern science. It’ll give you way more information than my simplified explanations. You may also want to get acquainted with the four-dimensional Euclidean geometry – I think it might make some things easier to grasp when talking about our metric space. So, I recommend the article &lt;em&gt;Metric Expansion of Space&lt;/em&gt; in Wikipedia as a good starter – I hope it’s not very challenging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Please note that my explanations are substantial simplifications. Various topologies of the Universe or metric space (various combinations in terms of finiteness, infinity, connectedness, and curvature) might be theoretically possible. Let me add one more advanced notion here: Wheeler suggested that the topology of space-time might be fluctuating (space-time foam), thus bringing quantum mechanics into here. Well, I guess, we still have a very limited knowledge when it comes to our actual metric space. &lt;/p&gt;&#xA;" OwnerDisplayName="user18491" LastEditorUserId="11143" LastEditDate="2017-09-14T14:55:50.213" LastActivityDate="2017-09-14T14:55:50.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22681" PostTypeId="1" AcceptedAnswerId="22682" CreationDate="2017-09-13T17:19:11.417" Score="9" ViewCount="432" Body="&lt;p&gt;We have encountered galaxies that act as lenses, magnifying the light coming from behind them. This is super fascinating and it makes me wonder what other optic tools we might have at our disposal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question of what we know about our own galaxy is obviously fascinating because it's kind of the the one we have the worst view at. There's so much noise around us it's hard to get a macroscopic view of the galaxy. It occurs to me that if we could stick a mirror out and look at our galaxy from the outside, maybe we have a better view. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, we don't have the technology to make a very useful mirror. And it wouldn't be very efficient. If anything, we would simply launch a probe out there and send back pictures and science data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then I got to thinking maybe there's a naturally occurring mirror we could leverage. We know there are entities that can bend light, are there any that bounce it? It would be incredibly advantageous if there was one that allowed us to see our own galaxy, but also it seems unlikely. But are there -any- objects out there that reflect light like a mirror allowing us to see around them or behind them or their &quot;backside&quot;?&lt;/p&gt;&#xA;" OwnerUserId="2239" LastActivityDate="2017-09-14T14:31:30.837" Title="Are there any mirrors in space?" Tags="&lt;observational-astronomy&gt;&lt;optics&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22682" PostTypeId="2" ParentId="22681" CreationDate="2017-09-13T18:12:59.520" Score="7" Body="&lt;p&gt;Very likely not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's a gravitational process like galaxy lensing, then a &quot;mirror&quot; would require a very large deviation. That's extremely unlikely with this process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With other processes, e.g. classic reflection, it seems even more unlikely. We don't know any natural processes that could craft mirrors at cosmic scale.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2017-09-13T18:12:59.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22683" PostTypeId="2" ParentId="22677" CreationDate="2017-09-13T18:30:12.077" Score="4" Body="&lt;p&gt;Chris, you are actually on the verge of understanding how special relativity works. You're very close. You only need to take one extra step.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;to state that all speed is relative to an object is to essentially say, &quot;speed does not exist, but is only relative to an observer&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is correct. Absolute space, as in newtonian physics, does not exist. Same for absolute time. Both are just relations between objects, and between events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as you hang on to the notions of absolute space and absolute time, relativity cannot be understood. If you let go of them, relativity actually seems natural.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Einstein simply noticed this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) All previous experiments (like Michelson-Morley) have shown that whenever you measure the speed of light, you always get the same result, no matter what.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) Maxwell's equations for electromagnetic fields (such as light) include the speed of light, and the value of that speed is frame-independent. Which means, they always show the speed of light to be the same, no matter how the observer is moving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both A and B tell the same thing: speed of light is always the same. But how can that be, when different observers are moving through absolute space at different speeds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is - absolute space is not a thing. Space is just a background for a special type of relation between objects, called &quot;distance&quot;. Time is just a background for a special type of relation between events, called &quot;duration&quot;. But neither space nor time are things, separate from objects and events. Only when you put together a whole lot of &quot;distance&quot; type of relations, space seems to appear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The really important thing is speed of light. It's a fundamental constant of nature, like the Planck energy, etc. This universe is built in such a way that speed of light and other constants like it remain the same no matter what.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But space and time are derivative notions. They are not primordial realities like the speed of light. So they depend on the observer. For me, the bundle of distance relations we call &quot;space&quot; looks a certain way. For you, the same bundle looks a bit differently. This is okay, because space is not absolute; it's derived from a lot of other things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you throw the flashlight, the speed of light from it is the same for everyone. But how can that be, since different observers are moving in different ways? Simple: they all see the bundles of distance relations differently; those differences are such that speed of light always remains the same.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;And that's special relativity in a nutshell.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2017-09-13T18:46:07.373" LastActivityDate="2017-09-13T18:46:07.373" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22684" PostTypeId="2" ParentId="22681" CreationDate="2017-09-13T18:39:56.170" Score="4" Body="&lt;p&gt;You could argue that a neutron star satisfies your final criterion (letting us see behind them). The intense &lt;a href=&quot;https://en.wikipedia.org/wiki/Neutron_star#Gravity_and_equation_of_state&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravitational field at a neutron star's surface&lt;/a&gt; serves as a gravitational lens - bending not just light from other stars, but emissions from the neutron star itself! &lt;a href=&quot;https://van.physics.illinois.edu/qa/listing.php?id=16591&quot; rel=&quot;nofollow noreferrer&quot;&gt;For normal stars, this effect is small&lt;/a&gt;, but it becomes more dramatic for compact objects like neutron stars.  Rob Jeffries wrote a nice answer explaining the geometry of the situation &lt;a href=&quot;https://physics.stackexchange.com/a/321954/56299&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sort of &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_lens&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravitational lensing&lt;/a&gt; is probably the only phenomenon that approaches what you're looking for. A &lt;a href=&quot;https://en.wikipedia.org/wiki/Reflection_nebula&quot; rel=&quot;nofollow noreferrer&quot;&gt;reflection nebula&lt;/a&gt; sort of acts like a mirror, reflecting the light from one or more nearby stars, but because of the scattering (and the somewhat low dust density, relative to an actual mirror), we can't use nebulae to actually see the stars. We can get a decent idea of what sort of light they're emitting (although the scattered light is still a bit different), but then again, we can already get that by just observing the stars normally.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2017-09-13T18:39:56.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22686" PostTypeId="2" ParentId="22677" CreationDate="2017-09-13T19:28:07.903" Score="-4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does Absolute Velocity Exist?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes it does. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It seems everyone is on the same page about there being no &quot;absolute velocity&quot; due to the fact that everything is relativistic. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a &quot;relativist&quot;, but I'm not on that page. Because of the CMBR. Google on &lt;a href=&quot;https://www.google.co.uk/search?source=hp&amp;amp;q=cmbr+reference+frame&amp;amp;oq=cmbr+reference+frame&amp;amp;gs_l=psy-ab.3..35i39k1j0i22i30k1.911.3857.0.4002.21.20.0.0.0.0.196.2321.0j16.16.0....0...1.1.64.psy-ab..5.16.2316.0..0j0i131k1j0i10k1.UzMTmrx0HjU&quot; rel=&quot;nofollow noreferrer&quot;&gt;CMBR reference frame&lt;/a&gt; and check out the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background#CMBR_dipole_anisotropy&quot; rel=&quot;nofollow noreferrer&quot;&gt;CMBR dipole anisotropy&lt;/a&gt;. It isn't an absolute frame in the strict sense of the word. But you can use it to gauge your speed through the universe. And the universe is as absolute as it gets. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, this leaves me confused. This seems to be disconnected with the concept that the speed of light is in fact an unchanging constant. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm afraid it isn't. Check out &lt;a href=&quot;http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/speed_of_light.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Is The Speed of Light Everywhere the Same?&lt;/a&gt; on the PhysicsFAQ website: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Does the speed of light change in air or water? Yes. Light is slowed down in transparent media such as air, water and glass.&quot;&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;These gyroscopes send light around a closed loop, and if the loop rotates, an observer riding on the loop will measure light to travel more slowly when it traverses the loop in one direction than when it traverses the loop in the opposite direction.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;In this passage, Einstein is not talking about a freely falling frame, but rather about a frame at rest relative to a source of gravity.  In such a frame, the not-quite-well-defined &quot;speed&quot; of light can differ from c, basically because of the effect of gravity (spacetime curvature) on clocks and rulers.&quot;&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is no. The speed of light is not everywhere the same. But you measure the local speed of light to be the same because you define our seconds and your metres using light. So when you use them to measure the speed of that light, you always get the same answer. See &#xA;&lt;a href=&quot;https://arxiv.org/abs/0705.4507&quot; rel=&quot;nofollow noreferrer&quot;&gt;Comments on &quot;Note on varying speed of light theories&quot;&lt;/a&gt; by Joao Magueijo and John Moffat. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But to state that all speed is relative to an object is to essentially say, &quot;speed does not exist, but is only relative to an observer&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Speed exists. It might be relative to something else, even if that something is the universe. But even though it's only relative, that's no reason to say it doesn't exist. A bullet is just a harmless lump of lead that you can weigh in your hand. But a speeding bullet will kill you dead. Think about how that sits with &lt;em&gt;&quot;speed does not exist, but is only relative to an observer&lt;/em&gt;&quot;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But yet, the speed of light is a calculable constant that never changes. How then, can we say, there is no such thing as absolute speed?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can't. We can say there's no such thing as an absolute reference frame, such that you won't find your experimental results in your sealed box are affected by the speed of that box. But like I said, the universe is as absolute as it gets, and you can measure your speed relative to the universe. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If I throw a flashlight in space that is turned on 50mph, does the speed of the emitted light from the flashlight increase by 50 miles per hour in the direction I threw it? Of course not. We know this. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Correct. Because light has an E=hf wave nature. The speed of a wave depends on the medium it's passing through, not on the speed of the emitter. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But if this is truly the case, how then can there not be an absolute velocity which could be calculated by determining the speed of light in relevance to the object we're measuring (since the speed of light is constant)? Please help because this is bothering me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's simpler than you think. Look at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Second&quot; rel=&quot;nofollow noreferrer&quot;&gt;definition of the second&lt;/a&gt; and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Metre#Speed_of_light_definition&quot; rel=&quot;nofollow noreferrer&quot;&gt;definition of the metre&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The SI definition of second is the duration of 9 192 631 770 periods of the radiation...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The metre is the length of the path travelled by light in vacuum during a time interval of 1/299792458 of a second.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You measure the local speed of light to be constant because of the way you define your seconds and your metres, and because of the &lt;a href=&quot;https://www.google.co.uk/search?source=hp&amp;amp;q=wave+nature+of+matter&amp;amp;oq=wave+nature+of+matter&amp;amp;gs_l=psy-ab.3..0j0i20k1j0i22i30k1l2.104.2688.0.2743.22.17.0.0.0.0.349.2578.0j8j3j2.13.0....0...1.1.64.psy-ab..9.13.2573.0..0i67k1j0i10k1.GhzUjjVXe8g&quot; rel=&quot;nofollow noreferrer&quot;&gt;wave nature of matter&lt;/a&gt;. Check out things like &lt;a href=&quot;https://en.wikipedia.org/wiki/Pair_production&quot; rel=&quot;nofollow noreferrer&quot;&gt;pair production&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Electron_diffraction&quot; rel=&quot;nofollow noreferrer&quot;&gt;electron diffraction&lt;/a&gt;. Ask yourself this: if you and your rods and clocks were made out of sound waves, such that you defined your metres and second using the motion of sound waves, how would you ever measure the local speed of sound to be anything other than 340 m/s?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 14/09/2017:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check out what I said above, see this in the &lt;a href=&quot;http://einsteinpapers.press.princeton.edu/vol7-trans/156?highlightText=%22spatially%20variable%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein digital papers&lt;/a&gt; dating from 1920:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;As a simple geometric consideration shows, the curvature of light rays occurs only in spaces &lt;strong&gt;where the speed of light is spatially variable&lt;/strong&gt;. From this it follows that the entire conceptual system of the theory of special relativity can claim rigorous validity only for those space-time domains where gravitational fields (under appropriately chosen coordinate systems) are absent.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed of light in vacuo varies. Also see &lt;a href=&quot;https://physics.stackexchange.com/a/25929/76162&quot;&gt;this PSE answer&lt;/a&gt; which quotes professor &lt;a href=&quot;http://www.astro.ubc.ca/people/scott/faq_basic.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Douglas Scott&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;However, the crucial assumption of Einstein's theory is not that there are no special frames, but that there are no special frames where the laws of physics are different. There clearly is a frame where the CMB is at rest, and so &lt;strong&gt;this is, in some sense, the rest frame of the Universe&lt;/strong&gt;. But for doing any physics experiment, any other frame is as good as this one. So the only difference is that in the CMB rest frame you measure no velocity with respect to the CMB photons, but that does not imply any fundamental difference in the laws of physics.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Special relativity pre-dated the CMBR. Also see this in the &lt;a href=&quot;http://einsteinpapers.press.princeton.edu/vol7-trans/197?highlightText=%22therefore%20there%20exists%20an%20ether%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein digital papers&lt;/a&gt;, again from 1920:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Recapitulating, we may say that according to the general theory of relativity space is endowed with physical qualities; in this sense, therefore, &lt;strong&gt;there exists an ether&lt;/strong&gt;&quot;.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Wikipedia and above for the &lt;a href=&quot;https://en.wikipedia.org/wiki/Second&quot; rel=&quot;nofollow noreferrer&quot;&gt;definition of the second&lt;/a&gt; and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Metre#Speed_of_light_definition&quot; rel=&quot;nofollow noreferrer&quot;&gt;definition of the metre&lt;/a&gt;. They are both defined using light. &lt;/p&gt;&#xA;" OwnerUserId="9772" LastEditorUserId="9772" LastEditDate="2017-09-14T12:32:49.970" LastActivityDate="2017-09-14T12:32:49.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22687" PostTypeId="2" ParentId="22681" CreationDate="2017-09-14T01:54:22.860" Score="1" Body="&lt;p&gt;I think if you shoot a laser beam tangent to just outside event horizon of  black hole then it might reflect light as mirror reflects incident ray hitting its surface at zero degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but again this is bending of light not reflecting it.&lt;/p&gt;&#xA;" OwnerUserId="14102" LastActivityDate="2017-09-14T01:54:22.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22694" PostTypeId="1" AcceptedAnswerId="22697" CreationDate="2017-09-14T14:16:06.500" Score="5" ViewCount="820" Body="&lt;p&gt;This is in a way related to this &lt;a href=&quot;https://astronomy.stackexchange.com/questions/16311/how-can-there-be-1-000-stellar-ancestors-before-our-sun/16313#16313&quot;&gt;question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question is: Our star is third generation star, which is explained by existing Barium. That Barium was created by other stars. Now, those stars must have been Supernova. When they exploded, it was either a neutron star or a black hole.&#xA;Where is that Neutron star or black hole?&lt;/p&gt;&#xA;" OwnerUserId="17151" LastEditorUserId="17151" LastEditDate="2017-09-15T06:21:34.523" LastActivityDate="2020-10-02T14:22:15.243" Title="Parent stars of our Sun - Where are its remains?" Tags="&lt;the-sun&gt;&lt;black-hole&gt;&lt;neutron-star&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22695" PostTypeId="2" ParentId="22681" CreationDate="2017-09-14T14:31:30.837" Score="0" Body="&lt;p&gt;IMHO the most likely way a cosmological mirror might come into existence is if some localized conditions led to a &quot;cloud&quot; of particles which cause &lt;a href=&quot;https://en.wikipedia.org/wiki/Nonlinear_optics#Optical_phase_conjugation&quot; rel=&quot;nofollow noreferrer&quot;&gt;phase conjugation&lt;/a&gt; .  Since this requires, in general, not only a highly ordered collection of particles/material but also an appropriate 'pump' (optical or otherwise),  I'd be both thrilled and astonished if we were to detect such a configuration.&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-09-14T14:31:30.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22696" PostTypeId="1" CreationDate="2017-08-29T11:54:50.547" Score="35" ViewCount="709" Body="&lt;p&gt;Suppose that you deposit an astronomer, armed with our current knowledge of orbital mechanics, on a dome on the far side of the Moon, so that the Earth is perpetually hidden from them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And, of course, assume that this person has no specific knowledge about the system they're in beyond what they can glean from observations. If you will, imagine that they learned all our modern orbital mechanics and related physics in alpha centauri, and then got teleported to our Moon.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it is reasonable to expect that this person should be able to deduce from observations of the sky that the body they are on is one half of a binary system, and they should be able to measure the orbital characteristics (semi-major axis, ellipticity, inclination) as well as the  position of the barycentre (much closer to the other body, corresponding to a much more massive partner). What observations are needed to deduce this? What level of observational accuracy is needed for those observations, and to what historical epoch does it correspond to? (I.e. would Tycho Brahe's kit have been sufficient? Would Galileo's? Would the ancient Greeks'? Or would this require a late-19th-century (or even later) observatory?)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(As pointed out in MartinV's answer, our astronomer might find it hard to distinguish between situations with an orbiting pair vs one single huge body. Thus, if convenient, you can assume that, via short ~100km forays from the dome, our astronomer is able to measure the lunar radius by measuring solar inclinations at different points with known distances between them, &lt;a href=&quot;https://en.wikipedia.org/wiki/Spherical_Earth#Hellenistic_astronomy&quot; rel=&quot;noreferrer&quot;&gt;à la Erathostenes&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="207" OwnerDisplayName="Emilio Pisanty" LastEditorUserId="207" LastEditDate="2017-09-15T15:21:28.450" LastActivityDate="2017-11-15T15:02:54.753" Title="If you lived on the far side of the Moon, how could you infer the existence of Earth?" Tags="&lt;the-moon&gt;&lt;observational-astronomy&gt;&lt;history&gt;&lt;orbital-mechanics&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22697" PostTypeId="2" ParentId="22694" CreationDate="2017-09-14T16:14:47.747" Score="11" Body="&lt;p&gt;We have no way of identifying where the Sun was born, what the surrounding environment was, or where the Sun's siblings are right now. This is easy to see from some bare numbers: the Sun's current orbital period around the Milky Way is some 250 million years, and it's been around for some 4.5 billion years, making for some 20 orbits around the galaxy since its birth. There's a lot of stellar-movement dynamics that can happen over those twenty orbits, and any siblings have long parted ways. Similarly, any stellar remnants that are close to us right now are close because we're passing by, not because we were born in their neighbourhood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, whatever the source of the heavy elements in the solar system, it's several billions of years too late to figure out any individual sources.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;That said, it is important to emphasize that there is no such unique source to begin with. As explained in the answer you linked to, there are plenty of such predecessors:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are grains of material trapped inside meteorites that consist of solids that were already present in the pre-solar material. These are important because these grains were thought to have formed in individual stellar events and their isotopic compositions can be studied. These tell us that the Sun formed from material that has been inside many different stars of different types.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Moreover, if you want to place those predecessors, you are fighting against the internal mixing of the interstellar gas on a galactic scale:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Mixing in the interstellar medium is reasonably effective. The material spewed out from supernovae and stellar winds 5-12 billion years ago has had plenty of time to mix throughout the Galaxy before the Sun's birth. Turbulence and shear instabilities should distribute material on galactic length scales in a billion years or less.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This means that the Sun was &lt;em&gt;not&lt;/em&gt; born out of the ashes of its neighbours. Instead, it was born out of the ashes of stars that might have died on the opposite side of the galaxy, several billion years before the Sun's birth, whose ashes then got thoroughly distributed by the mixing of the interstellar medium.&lt;/p&gt;&#xA;" OwnerUserId="207" LastActivityDate="2017-09-14T16:14:47.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22699" PostTypeId="1" AcceptedAnswerId="22700" CreationDate="2017-09-15T06:48:05.767" Score="4" ViewCount="352" Body="&lt;p&gt;Is there any information about the future of the neutron star?&#xA;What I am interested what happens to it as the time passes. Say a billion years. Would it still be a neutron star? Would it collapse to black hole? Would it simply dissipate?&lt;/p&gt;&#xA;" OwnerUserId="17151" LastActivityDate="2017-09-15T17:38:52.650" Title="Neutron star, aging proccess" Tags="&lt;neutron-star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22700" PostTypeId="2" ParentId="22699" CreationDate="2017-09-15T07:07:31.920" Score="8" Body="&lt;p&gt;A newly formed neutron star cools (via neutrino emission) to an effectively &quot;zero-temperature&quot; configuration in a matter of a few minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is meant by that is that it contracts to a radius where the separation between its neutrons is so small that neutron degeneracy pressure and the pressure associated with mutual repulsion between neutrons (from the strong nuclear force) becomes large enough to support the star, and since neither of these depend on temperature, the star can continue to cool off without any change in radius. In other words, the radius it has when it is at an interior temperature of $10^{10}$K (and a surface temperature of about $10^{8}$K), a few minutes after formation, and the radius it has for any cooler temperature are essentially the same. Hence neutron stars &lt;em&gt;do not&lt;/em&gt; contract, become denser and collapse into black holes as they cool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A neutron star cools for the first million years or so by various neutrino emission mechanisms in the interior. Thereafter, at a surface temperature of about 100,000 degrees, the dominant cooling is radiation from the surface, which then proceeds on timescales of only millions of years. The neutron star is then thought to get colder and colder (NB no old, cold neutron stars have ever been observed, but they are so small and so faint at low temperatures that they could never be found with current technology, but there should be about a billion of them out there in our Galaxy). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The future of the neutron star is therefore to be a fading cinder. The vast majority of the billion or so neutron stars thought to exist in our galaxy are probably at temperatures of around 1000-10,000 degrees, and probably maintained at that temperature by the accretion of &lt;em&gt;small&lt;/em&gt; amounts of gas from the interstellar medium. They may also be reheated by Ohmic dissipation of their magnetic fields or viscous dissipation of their rotational kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it is possible, that if a neutron star accreted enough material, then at some point in the future it could exceed the maximum possible mass for a neutron star and become a black hole. But most neutron stars appear to be born at masses well below this limit and it would likely take many galaxy lifetimes to accrete enough material for them to make that transition. Neutron stars born into binary systems may experience more rapid accretion. This accretion is observed (in X-rays) and could result in black hole formation on shorter timescales.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-09-15T17:38:52.650" LastActivityDate="2017-09-15T17:38:52.650" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="22702" PostTypeId="2" ParentId="22696" CreationDate="2017-09-15T15:05:19.893" Score="3" Body="&lt;p&gt;&lt;strong&gt;This is a really good question - and quite subtle.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earliest opportunity might be that inter-month changes in the stellar parallax of the Sun might lead the observer to conclude that either i) the Moon is a single, very large, rotating body or ii) it is part of a multi-body co-rotating system. However i) would seem to be inconsistent with a near and strongly curved horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not then, certainly when we develop a quantitative model of orbital mechanics involving mass and gravity&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I don't think stellar parallax would directly help us as it (in the modern day) just tells us that we are in orbit around the sun and little about the Earth-Moon system itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at how a Ptolemy equivalent on the Moon (call him Moon-Ptolemy) might see it. He would have no way of distinguishing the Earth-Moon system from his assumption that he is just sitting on a solid object at the center of creation. Of course he wouldn't see a &quot;moon&quot; in orbit around him, but he would see the Sun, stars and major planets. Stellar parallax (to him, the Sun &quot;moving through the Zodiac) would just tell him that Sun is rotating around his Moon, as are the planets. The existence of planetary epicycles would be a curiosity required to make his model work - but it does work and he has no notion of the Earth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moon-Galileo might (or might not) be able to develop the heliocentric model - he misses out on one key insight that Earth-Galileo had: that the Earth wasn't special because other planets also had moons. Moon-Galileo would find the orbital system of Jupiter interesting but not a key insight, so he might not develop the new model. Even so, someone else would.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, in a qualitative scientific world, there would still be nothing to help the Moon observer deduce the existence of the Earth behind the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect the truth would become unavoidable when orbital mechanics developed sufficiently to incorporate mass and gravity into calculation. It might have been around the time of Moon-Kepler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure I agree with the comments looking at observations of the planets - I don't see how they help distinguish between an Earth-Moon system as opposed to a simple, very large, rotating Moon body with no co-orbiter (which would be a natural assumption to make). Even the monthly changes in parallax caused by the rotation of the Moon around the Earth might be waved away by suggesting the simple rotation of a much larger Moon body - though our hero might certainly question the compatibility of this with the apparent curvature and distance to their Moon horizon.&lt;/p&gt;&#xA;" OwnerUserId="17502" LastActivityDate="2017-09-15T15:05:19.893" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22703" PostTypeId="1" AcceptedAnswerId="22705" CreationDate="2017-09-15T15:21:35.357" Score="2" ViewCount="622" Body="&lt;p&gt;I'm trying to determine what a star of comparable mass and luminosity to the Sun would look like from the surface of the Earth if it were 1 light year away - I chose Alpha Centauri since it is close to us, and similar to the Sun. Could a star like the Sun be that close without exerting significant influence over it, or would they have an orbital relationship? I'm trying to design a star system for a science fiction novel, and would like there to be a star close enough and bright enough to be visible during the day, and cast a shadow at night, but far enough away to provide no heat energy, and be free from the Sun's gravity.&lt;/p&gt;&#xA;" OwnerUserId="18582" LastActivityDate="2017-09-15T17:39:59.993" Title="How bright would Alpha Centauri appear (in apparent Magnitude) if it were 1 light year away?" Tags="&lt;apparent-magnitude&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22704" PostTypeId="2" ParentId="22696" CreationDate="2017-09-15T15:28:45.980" Score="7" Body="&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth_tide&quot; rel=&quot;noreferrer&quot;&gt;body tide&lt;/a&gt; seismometer on the lunar far side would pick up both the solar tide, and the &lt;a href=&quot;https://www.space.com/26246-lunar-tide-seen-from-space.html&quot; rel=&quot;noreferrer&quot;&gt;20 inch body distortion produced by the earth&lt;/a&gt;.&#xA;While &quot;tidally locked&quot;, the moon is not in a perfectly circular orbit, and also wobbles a bit; &lt;a href=&quot;https://en.wikipedia.org/wiki/Libration&quot; rel=&quot;noreferrer&quot;&gt;libration&lt;/a&gt;. Your seismometer should pick up both effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Watching the parallax of mars cycle every 28 days, as suggested in comments above, might be a simpler way to go.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2017-09-15T15:28:45.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22705" PostTypeId="2" ParentId="22703" CreationDate="2017-09-15T17:38:13.370" Score="2" Body="&lt;p&gt;The change in magnitude is given by:&#xA;$$m_2-m_1=-2.5\ \log_{10}(B)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where B is the change in brightness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alpha Centauri is about 4.4ly so inverse square says it would be 19.4x brighter at 1ly, so a magnitude differential of -3.2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its apparent magnitude would therefore go from -0.27 to -3.47. This is brighter than Sirius (-1.46) but not close to the Sun (-26.7) nor Venus. Venus doesn't cast a shadow, so neither would Alpha Centauri at 1ly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On gravity, Jupiter is roughly one thousandth the mass of Alpha Centauri but would be about 63,000x closer than the star at 1ly. It would therefore have a gravitational effect on the Sun of:&#xA;$$63\,000^2/1000=\mathrm{4\ million\ times}$$&#xA;that of Alpha Centauri at 1ly!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sounds like this works for your story.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Data brightness sources: &lt;a href=&quot;https://www.space.com/21640-star-luminosity-and-magnitude.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.space.com/21640-star-luminosity-and-magnitude.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass and distance sources: &lt;a href=&quot;https://en.wikipedia.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17502" LastEditorUserId="2365" LastEditDate="2017-09-15T17:39:59.993" LastActivityDate="2017-09-15T17:39:59.993" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22706" PostTypeId="1" AcceptedAnswerId="22725" CreationDate="2017-09-15T19:38:21.457" Score="0" ViewCount="175" Body="&lt;p&gt;I would like to know which part of the Milky Way is currently up. Preferably using astropy. The ingredients I have are the current time, and my current position on earth. How would I use these to get the minimal and maximal galactic longitude currently up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I know how to convert from Alt-Az to Galactic coordinates, but I don't know at which azimuth the galactic plane crosses the horizon at altitude=0.)&lt;/p&gt;&#xA;" OwnerUserId="18583" LastActivityDate="2017-09-17T17:09:54.717" Title="Get visible galactic longitudes in astropy" Tags="&lt;milky-way&gt;&lt;coordinate&gt;&lt;astropy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22707" PostTypeId="1" AcceptedAnswerId="22710" CreationDate="2017-09-15T21:41:20.977" Score="8" ViewCount="199" Body="&lt;p&gt;In other words, if a group of stars born within the same nebula, e.g. the Pleiades, are initially gravitationally bound - what causes them to finally get separated and proceed individually? Is it the influence of  &amp;quot;external&amp;quot; mass? Or, are they now and we're they always, even at birth, having the escape velocity? (If the latter is true then I think they should not be considered truly bound from the beginning...)&lt;/p&gt;&#xA;" OwnerUserId="17437" LastEditorUserId="2531" LastEditDate="2021-05-09T13:58:16.633" LastActivityDate="2021-05-09T13:58:16.633" Title="Why do stars born in a cluster finally disperse?" Tags="&lt;astrophysics&gt;&lt;stellar-dynamics&gt;&lt;star-cluster&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22708" PostTypeId="1" AcceptedAnswerId="22713" CreationDate="2017-09-15T22:54:11.800" Score="1" ViewCount="235" Body="&lt;p&gt;The seasonal temperature is ultimately due to the precession of the Earth around the axis. But what I'm curious about is...  is it due more to the side experiencing winter being farther from the sun or is it more due to the fact that the days are shorter and the nights are longer?&lt;/p&gt;&#xA;" OwnerUserId="7668" LastEditorDisplayName="user1569" LastEditDate="2017-09-16T15:34:09.433" LastActivityDate="2017-09-16T15:34:09.433" Title="What contributes the most to the seasonal temperature variation?" Tags="&lt;earth&gt;&lt;rotation&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22709" PostTypeId="2" ParentId="22583" CreationDate="2017-09-15T23:01:42.807" Score="0" Body="&lt;p&gt;Neutron Stars and Black Holes are the most dense things in the entire universe. Neutron stars come about when a star collapses and has a supernova. As this happens, the core is smashed with all of the weight of the star put into a single point. Because of the immense pressure, the protons in the core turn into neutrons. This in turn creates a super dense and super heavy dead core of the star it once was. Now even though neutron stars have some of the most immense densities in the entire universe, black holes have them beat. Black holes come when extremely large stars compress their cores so tightly the density of the point hits infinity which creates a black hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this answers your question.&lt;/p&gt;&#xA;" OwnerUserId="18584" LastActivityDate="2017-09-15T23:01:42.807" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22710" PostTypeId="2" ParentId="22707" CreationDate="2017-09-15T23:39:33.523" Score="12" Body="&lt;p&gt;The boundedness or otherwise of clusters remains to be established in most cases. The vast majority of clusters become unbound and disperse at a much younger age than the Pleiades. Or they may be born unbound as you suggest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stars in a cluster have a distribution of velocities and there will always be a tail of high speed stars that will be able to escape the gravitational potential in an &quot;evaporation process&quot;, even in a cluster which is globally bound (i.e. the sum of kinetic and gravitational potential energies is negative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that this isn't usually sufficient to explain the disappearance of clusters. A more effective disruptive mechanism are the Galactic tides. Stars that stray beyond a tidal radius can be stripped from the cluster. The reduction in the cluster's mass then reduces the tidal radius allowing more stars to escape in what eventually would become a runaway process. The tidal radius for a cluster near the Sun is about $5 M^{1/3}$ light years, where $M$ is the cluster mass expressed in units of the solar mass. For the Pleiades, $M \sim 1000 M_{\odot}$, so stars that stray beyond 50 light years can be lost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases, the evaporation process causes a cluster core collapse. The escaping stars take away kinetic energy and the remaining stars become more tightly bound. This is what has happened in most of the massive and long-lived globular clusters. Less massive open clusters like the Pleiades are much more susceptible to tidal dissipation.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-09-16T18:32:28.380" LastActivityDate="2017-09-16T18:32:28.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22711" PostTypeId="2" ParentId="22583" CreationDate="2017-09-15T23:44:35.573" Score="2" Body="&lt;p&gt;Gravitational pull is determined by only two things: mass and distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The greater the mass, the greater the pull. The shorter the distance, the greater the pull.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same distance (measured to the center), objects with the same mass will exert the same gravitational pull, no matter what they are. A black hole, a neutron star, a normal star, a giant marshmallow - they all pull the same at the same distance if their masses are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only difference is this: normal stars are quite fluffy and bloated, you can't get too close to their center because pretty soon you'll run into their surface, and that's where the formula for the pull changes. So the maximum pull from a normal star is not too big, simply because you can't get &quot;close enough&quot; to the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutron stars are much more dense. You can get a lot closer to their center before you hit the surface. So at the same mass, a neutron star lets you get closer so as to generate a greater pull.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes are the most dense of all. In theory you could get arbitrarily close to the center, because the BH is all in the center (sort of, not really, probably). So at the same mass, a BH could theoretically generate the greatest pull.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Black hole having strong gravitational pull not even light can escape From it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's complicated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/2240/what-is-a-singularity-what-is-at-the-center-of-a-black-hole-specifically-regar/2254#2254&quot;&gt;What is a singularity? What is at the center of a black hole? Specifically regarding space-time&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2017-09-15T23:44:35.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22712" PostTypeId="1" AcceptedAnswerId="22715" CreationDate="2017-09-16T00:53:54.637" Score="1" ViewCount="138" Body="&lt;p&gt;Would a short-period comet, with a 2-mile nucleus, be able to come within 200,000 miles of Earth without breaking up? &lt;/p&gt;&#xA;" OwnerUserId="18588" LastEditorUserId="2365" LastEditDate="2017-09-16T06:36:56.290" LastActivityDate="2017-09-16T18:36:36.950" Title="Would a comet break up during a close pass of the Earth" Tags="&lt;comets&gt;&lt;near-earth-object&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22713" PostTypeId="2" ParentId="22708" CreationDate="2017-09-16T01:29:22.847" Score="5" Body="&lt;p&gt;[&lt;img src=&quot;https://i.stack.imgur.com/rfNkc.png&quot; alt=&quot;enter image description here&quot;&gt; The earth is actually closer to the sun in the Northern hemisphere's winter. The seasonal temperature variation is predominantly due to the angle the earth makes with the sun. In the northern hemisphere in winter the angle is such that the earth is tilted with north pole away from the sun and the sunlight hitting the earth is spread over a much larger area than if it was pointed towards the sun. Due to this tilt the sun is also lower in the sky and has the effect shown in the 2nd figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/L4tKT.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L4tKT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="2365" LastEditDate="2017-09-16T06:39:19.100" LastActivityDate="2017-09-16T06:39:19.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22714" PostTypeId="2" ParentId="22707" CreationDate="2017-09-16T02:41:10.507" Score="1" Body="&lt;p&gt;I’d like to add that globular clusters are usually much larger than open/irregular ones. Often we are talking about many thousands or even millions of stars in a globular cluster. That’s a lot of mass and, therefore, most of stars in such clusters are well bound by gravitational force. Of course when a cluster passes near a large mass (e.g. a galactic center or a massive dust cloud) stars are gravitationally attracted and some may live the cluster, actually sometimes a lot may leave forming a “leaving” group in the form of a huge tail of stars. Such tails (tail-like collections) of stars may be very large. Simply put, it all comes down to the magnitude of gravitational pull exerted on them. Large pulls are called tidal shocks and tidal waves as they are complicated, and the kinematics is not as easy as one may think it is. Anyway, large globular clusters under the force of gravity (there’s a lot of force there) are very tight in the center and can even form massive black holes in the center. Stars that are far away from the core of the cluster are more likely to leave under the influence of external gravitational forces at some point in time as the clusters rotate around the center of the galaxy and things change over time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Open clusters are much smaller and are not so tightly bound by gravitation force. So, it is no surprise that stars in such clusters are much more susceptible to massive gravitational sources that can pull them out of their clusters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Escape velocities are more likely to be attained later on (somewhat more or less haphazard) or, to be more precise, during the lifespan of the cluster, rather than in the very beginning, especially for large clusters. Smaller stars (in terms of mass) are of course more likely to leave their clusters, which is basic physics and it needs no further comment in my opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The brightest globular cluster 47 Tucanae (too-kAH-nee) in the night sky very probably contains a huge black hole near the center which “sucks” in nearby objects and stars. Those won’t escape. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/890jL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/890jL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;So, it's all about gravitational forces in our Galaxy that causes stars to get separated from their clusters.&lt;/p&gt;&#xA;" OwnerDisplayName="user18491" LastActivityDate="2017-09-16T02:41:10.507" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22715" PostTypeId="2" ParentId="22712" CreationDate="2017-09-16T06:36:21.880" Score="2" Body="&lt;p&gt;It wouldn't break up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As my Granddad used to say &quot;Close is only good in horseshoes&quot;. If the comet doesn't impact the Earth will be able to pass by unchanged. 200000 miles is a long way, about the distance of the moon from Earth. Asteroids have come this close before (eg 2011 XC2). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comets can be broken up by close encounters with the sun, but the sun is rather hotter than the Earth [citation needed]. 200000 miles is also well outside the Roche limit for the Earth-comet system, and at the speed that comets travel relative to the Earth, tidal effects won't have enough time to act.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the comet will pass the Earth quite happily. It is a rather small comet, and even the comet's coma won't interact with the Earth. After closest approach it is possible that the Earth could pass through the comet's tail, which would cause a meteor storm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even at 100000 miles, there is not much effect on satellites nearly all of which orbit at 22000 miles or less (22000 is geostationary orbit) It is plausible that they could be damaged by meteoroids. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dust that comes of comets is all quite small. If you are looking for a &quot;near miss that still causes damage&quot; try this. The comet makes a close pass of Jupiter that causes it to split into two parts. The larger part misses Earth by 100000 miles but the smaller part (a 300m chunk that wasn't noticed until too late because story) &lt;a href=&quot;http://impact.ese.ic.ac.uk/cgi-bin/crater.cgi?dist=10&amp;amp;distanceUnits=1&amp;amp;diam=300&amp;amp;diameterUnits=1&amp;amp;pdens=1000&amp;amp;pdens_select=0&amp;amp;vel=50&amp;amp;velocityUnits=1&amp;amp;theta=70&amp;amp;wdepth=&amp;amp;wdepthUnits=1&amp;amp;tdens=2500&quot; rel=&quot;nofollow noreferrer&quot;&gt;hits Merchtem 10km outside Brussels&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-09-16T16:20:37.917" LastActivityDate="2017-09-16T16:20:37.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22716" PostTypeId="1" AcceptedAnswerId="22718" CreationDate="2017-09-16T06:51:01.510" Score="1" ViewCount="346" Body="&lt;p&gt;General relativity tells us space is expanding.Then why does the distance between planets and heavenly bodies increase?&lt;/p&gt;&#xA;" OwnerUserId="18592" LastActivityDate="2017-09-16T09:24:34.463" Title="If universe is expanding why is the distance between planets not increasing?" Tags="&lt;planet&gt;&lt;space&gt;&lt;general-relativity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22718" PostTypeId="2" ParentId="22716" CreationDate="2017-09-16T09:24:34.463" Score="1" Body="&lt;p&gt;Distances between planets and stars do not expand as long as they are bound together by gravity sufficiently. This is true even at the scale of an entire Galaxy.&lt;/p&gt;&#xA;" OwnerUserId="11264" LastActivityDate="2017-09-16T09:24:34.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22721" PostTypeId="1" AcceptedAnswerId="22727" CreationDate="2017-09-16T16:41:27.010" Score="9" ViewCount="6088" Body="&lt;p&gt;Let's assume that there is a perfectly spherical planet and there is a moon which is also perfectly spherical. Let's assume that there is no atmospheric drag and no other gravitational pull. If the moon is put in a perfectly circular orbit around the planet somehow, would the moon eventually &quot;fall&quot; towards the planet and form an elliptical orbit or would it continue to follow the perfectly circular orbit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: What i actually wanted to ask is that would the &lt;em&gt;gravity&lt;/em&gt; of the planet cause the moon to &quot;fall&quot; towards the planet or would the gravity allow the moon to continue its orbit without bending its path any further towards the planet. I know that no planet can be a true sphere or a cube due to the shapes of the particles.&lt;/p&gt;&#xA;" OwnerUserId="18598" LastEditorUserId="18598" LastEditDate="2017-09-16T21:54:18.403" LastActivityDate="2017-09-17T04:17:14.810" Title="Is a perfectly circular orbit possible?" Tags="&lt;orbit&gt;&lt;gravity&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22722" PostTypeId="2" ParentId="22712" CreationDate="2017-09-16T18:20:51.380" Score="2" Body="&lt;p&gt;No, that isn't close enough. The &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Roche_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Roche Limit&lt;/a&gt; for break up of a rigid body, caused by tidal forces is given by&#xA;$$d = 1.26 R_C \left(\frac{M_E}{M_C}\right)^{1/3},$$&#xA;where $M_E$ is the mass of the Earth, $M_C$ and $R_C$ are the mass and radius of the comet, and $d$ is how close the comet can come to the Earth before tidal forces disrupt it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a comet with a 1.6 km radius and assume it has a density of &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Comet&quot; rel=&quot;nofollow noreferrer&quot;&gt;0.6 g/cc&lt;/a&gt;, then $M_C \simeq 10^{13}$ kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting this into the Roche Limit formula, we get $d =  17,000$ km. So the comet would have to get within 10,000 km of the surface of the Earth before it broke up.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-09-16T18:36:36.950" LastActivityDate="2017-09-16T18:36:36.950" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22723" PostTypeId="1" CreationDate="2017-09-16T18:46:33.667" Score="2" ViewCount="296" Body="&lt;p&gt;&lt;a href=&quot;https://www.nasa.gov/sites/default/files/thumbnails/image/17-051.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;This is a picture (mosaic?) of Jupiter's south pole taken by Juno&lt;/a&gt; (also shown below). It has gotten a lot of press, but I haven't seen anything, even in the scientific paper that accompanied its release, that answers the first question that came to mind when I saw it : is this a view into the inside of Jupiter? I seem to recall seeing IR images of Jupiter where the poles appear quite hot, which would be consistent with a &quot;clear sky&quot; providing a view of deeper, hotter layers. But how deep can see in this picture?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MEfUb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MEfUb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I am aware that this is a visible light image, and that visible light does not penetrate the cloud layer of Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is whether part of the south pole of Jupiter is cloud-free, at least to a depth where atmospheric extinction limits our ability to see the clouds. This interpretation of the picture would assume that blue light penetrates deeper than red light into the atmosphere (clouds not included) of Jupiter, and that this is the reason that some of the clouds, presumed to be deeper, look blue, while some areas between them are essentially black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example of a picture of Jupiter in IR, which shows the poles as being relatively hot. The troposphere of Jupiter (or any planet) is generally hotter the deeper you go, so the poles appearing hotter is consistent with less cloud cover at the poles. [&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/2/2c/Thermal_emission_of_Jupiter.jpg&quot; alt=&quot;description]&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware that perhaps no-one, including the Juno team, knows the answer to my question. Some useful partial answers would include an estimate of the depth to which visible light could penetrate Jupiter's atmosphere in the absence of clouds, and whether blue light would penetrate deeper than red.&lt;/p&gt;&#xA;" OwnerUserId="18600" LastEditorUserId="7982" LastEditDate="2021-02-28T11:45:56.230" LastActivityDate="2021-02-28T11:45:56.230" Title="Are the darker blue/black areas in this picture what the atmosphere of Jupiter looks like without clouds?" Tags="&lt;jupiter&gt;&lt;planetary-atmosphere&gt;&lt;atmospheric-effects&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22724" PostTypeId="2" ParentId="22721" CreationDate="2017-09-16T18:50:03.100" Score="26" Body="&lt;p&gt;&quot;Perfectly&quot; is a funny word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perfect circles are a mathematical abstraction. Real objects are not &quot;perfect&quot;.  So supposing a &quot;perfectly spherical planet&quot; is to suppose something that does not and &lt;em&gt;could not&lt;/em&gt; exist. All real planets are made of atoms and anything made of little clumps of matter cannot be perfectly spherical.  Even if you built a planet that was as spherical as possible, it would be distorted by its rotation and the tides. So there are no perfectly spherical planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you say &quot;put in a perfectly circular orbit&quot;. This is like drawing a line that is exactly $\pi$cm long. Again you are supposing something that does not, and &lt;em&gt;could not&lt;/em&gt; exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we can do is consider a mathematical model of gravity. If you model the sun and the planet as &quot;particles&quot; (ie point masses) and you model gravity with Newton's law of universal gravity, and if you give the model the system with the exact amount of energy to give a perfect circle, then the system will remain in a perfect circle, it will never become elliptical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use general relativity to model gravity, then the release of gravitational radiation will mean that no circular orbits are possible, all orbits will spiral inwards, however it would not become elliptical. Something similar will happen with quantum models of gravity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your question can only be answered in the context of a mathematical model of gravity. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-09-16T22:17:48.390" LastActivityDate="2017-09-16T22:17:48.390" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22725" PostTypeId="2" ParentId="22706" CreationDate="2017-09-16T19:12:28.480" Score="0" Body="&lt;p&gt;The following works for me; it uses the rootfinder from scipy to find the zero.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from astropy.coordinates import EarthLocation, AltAz, Galactic&#xA;from astropy.time import Time&#xA;from scipy.optimize import newton&#xA;import astropy.units as u&#xA;&#xA;dwl = EarthLocation(lat=6.39688*u.deg, lon=52.6275*u.deg)&#xA;mytime = Time('2017-09-16T21:00:00')&#xA;&#xA;def altitude_of_longitude(longitude):&#xA;    ''' Give the altitude (in deg) of a galactic longitude (in deg) '''&#xA;    return Galactic(l=longitude*u.deg, b=0*u.deg).transform_to(&#xA;                    AltAz(location=dwl, obstime=mytime)).alt.value&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now use the rootfinder from scipy, with two different initial guesses, hoping that they end up at the two different solutions where the galactic plane crosses the horizon. This gives:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;newton(altitude_of_longitude, 0)&#xA;&amp;gt;&amp;gt;&amp;gt; 16.231753986033333&#xA;newton(altitude_of_longitude, 180)&#xA;&amp;gt;&amp;gt;&amp;gt; 196.23547201834657&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The only remaining part is to check which part is visible:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;altaz = Galactic(l=0*u.deg, b=0*u.deg).transform_to(AltAz(location=dwl, obstime=mytime))&#xA;altaz.alt&#xA;&amp;gt;&amp;gt;&amp;gt; −9°08′51.8623′′&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Apparently the longitude at 0° is under (negative altitude), so the visible part is between 16° and 196° galactic longitude.&lt;/p&gt;&#xA;" OwnerUserId="18583" LastEditorUserId="18583" LastEditDate="2017-09-17T17:09:54.717" LastActivityDate="2017-09-17T17:09:54.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22727" PostTypeId="2" ParentId="22721" CreationDate="2017-09-16T23:26:00.513" Score="8" Body="&lt;p&gt;Short answer:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes.  If you ignore the tidal effect and relativity and any change in mass (planets radiate light and lose atmosphere and add space dust and meteors all the time, so mass isn't constant), then in a two body system with no outside effects, the orbit would remain perfectly circular.   There would be no outside force to affect the circular orbit.    A circular orbit is impossible because nothing can be that exact, but on a computer simulation you could set it up and it would remain circular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your scenario to work you'd need to give both the planet and moon infinite hardness, so they wouldn't bend at all and fixed mass and space would need to be completely empty of anything else.  Needless to say that's impossible.  But only in Newtonian gravity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relativity creates a very very tiny decay in orbits, in your system of a planet/moon that would be close to negligible but there would be a very tiny spiraling inward.    The relativistic effect on an orbit was first noticed with Mercury's orbit around the sun (and Mercury isn't falling into the sun, it was noticed by other effects - but lets not get into that here).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, any loss in mass, gain in mass or orbital drag (because space is full of tiny particles, fast moving particles, photons and neutrinos, all of which cause a tiny but at least in simulation, calculatable drag), then the two body system would have an imperceptibly small spiral and not be a perfect circle.  You could say in a sense that it becomes elliptical but it would be more like a constant very tiny force where, once it was elliptical, it could wonder back to more circular.   Not all perturbations or drag on an orbit make that orbit more elliptical, it can work in either direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth noting that &quot;falling&quot; or decaying towards the planet wouldn't &quot;create&quot; an elliptical orbit.   A circle is an ellipse.   You asked specifically about 2 body systems, where, ignoring tides, falling in or out would be more of a slow spiral.   An ellipse isn't the result of a decaying or perturbed orbit.  An ellipse is the baseline orbit.  Perturbations and orbital decay happen on top of the ellipse (if that makes sense), they don't cause the ellipse. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a 3 or more body system you get orbital perturbations on the orbits.   Those often remain stable, they're just variations that mostly move back and forth.  See &lt;a href=&quot;https://en.wikipedia.org/wiki/Milankovitch_cycles#Orbital_shape_.28eccentricity.29&quot; rel=&quot;noreferrer&quot;&gt;Eccentricity variation&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Milankovitch_cycles#Apsidal_precession&quot; rel=&quot;noreferrer&quot;&gt;Apsidal Precession&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-09-17T02:24:13.337" LastActivityDate="2017-09-17T02:24:13.337" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22728" PostTypeId="1" CreationDate="2017-09-16T23:53:36.607" Score="0" ViewCount="116" Body="&lt;p&gt;It's 5.50 am now in Dhaka, Bangladesh.  I just saw a moving star around 5 am. It was a starry sky and that star was moving in a straight line. It was very fast, faster than regular planes (though I didn't hear any sound) and disappeared in the Eastern sky.  It was a little brighter and a little closer than the other stars. Can anyone tell, what was it? &lt;/p&gt;&#xA;" OwnerUserId="18606" LastActivityDate="2017-09-18T18:45:21.597" Title="UFO or something else?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22730" PostTypeId="1" CreationDate="2017-09-17T02:09:14.977" Score="2" ViewCount="235" Body="&lt;p&gt;The solar wind contributes a very small amount of matter to the gas giants.  As the sun approaches the red giant phase, the solar wind increases significantly.  10% of the sun's mass lost as solar wind over its lifetime, which is equal to 100 Jupiters in mass.  After the red giant phase the sun releases the outer, mostly hydrogen, outer layer in the form of the planetary nebula. The planetary nebula is about 50% of the mass of the sun, which is equal to 500 Jupiters in mass, the remaining 40% of the sun will be left as the white dwarf. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all this additional mass, most of which is hydrogen, is added to the gas giants how massive would they be and would any of them reach brown or red dwarf status?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smallest brown dwarf is 13 Jupiters in mass, which is able to fuse deuterium. Smallest red dwarf is at 75 Jupiters in mass, which can fuse hydrogen.  &lt;/p&gt;&#xA;" OwnerUserId="17272" LastEditorUserId="17272" LastEditDate="2017-09-19T14:03:02.667" LastActivityDate="2017-09-19T14:03:02.667" Title="Jupiter/Saturn's fate after Sun goes red giant and planetary nebula?" Tags="&lt;star&gt;&lt;jupiter&gt;&lt;red-giant&gt;&lt;red-dwarf&gt;" AnswerCount="0" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22731" PostTypeId="2" ParentId="22723" CreationDate="2017-09-17T03:56:29.063" Score="4" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/22723/are-the-darker-blue-black-areas-in-this-picture-what-the-atmosphere-of-jupiter-l#comment38964_22723&quot;&gt;userLTK's comment&lt;/a&gt; is correct, you cannot see very deep at all.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Information about the image&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;First off, there are three imagers on the Juno spacecraft: &lt;a href=&quot;https://en.wikipedia.org/wiki/JunoCam&quot; rel=&quot;nofollow noreferrer&quot;&gt;JunoCam&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/UVS_(Juno)&quot; rel=&quot;nofollow noreferrer&quot;&gt;UVS&lt;/a&gt;, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Jovian_Infrared_Auroral_Mapper&quot; rel=&quot;nofollow noreferrer&quot;&gt;JIRCAM&lt;/a&gt;. You can see a &lt;a href=&quot;https://en.wikipedia.org/wiki/Juno_(spacecraft)#Scientific_instruments&quot; rel=&quot;nofollow noreferrer&quot;&gt;full list of instruments here&lt;/a&gt;. JunoCam is described as a: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;visible-light camera/telescope of the Juno Jupiter orbiter, a NASA space probe launched to the planet Jupiter on 5 August 2011. It was built by Malin Space Science Systems. The telescope/camera has a field of view of 58 degrees with four filters (3 for visible light).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The image you're seeing was taken by JunoCam and is a visible image. In fact, this image is combined by images taken over multiple days and from multiple orbits to get different vantage points and show all areas lit up. It isn't however a mosaic $-$ the 58 degree field of view allows JunoCam to image the entirety of Jupiter in one image. Instead, this image is simply the carefully crafted combination of many different images of the south pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FWIW, UVS is an imager in the ultraviolet and JIRAM is an imager in the infrared.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;How deep are you seeing into Jupiter?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Not very deep at all. Because the image is in the visible spectrum, the depth into Jupiter before it becomes too opaque is going to be very, very short. You may not even be seeing down to Jupiter's &quot;surface&quot; where the atmospheric pressure equals 1 bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, JIRCAM can see down to about 5-7 bars if its lucky, enough to penetrate the very upper layers of the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Infrared images of the poles&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Obviously I can't be certain what you saw before concerning infrared images of Jupiter's poles, but my guess is they may have been something like the image below. This image was taken by JIRAM&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bTr4H.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bTr4H.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-09-17T03:56:29.063" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="22732" PostTypeId="2" ParentId="22721" CreationDate="2017-09-17T04:17:14.810" Score="2" Body="&lt;p&gt;No. Tidal friction will perturb your orbit out of spherical. Because your planet and moon aren't optimally shaped this will happen faster than if they were allowed to take on the liquid drop shape they would naturally have. Once having achieved the balanced shape and balanced orbit around the barycenter, your system is still not quite circular due to general relativistic effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the nature of the beast; circular orbits are inherently unstable and want to fall into precessing ellipses.&lt;/p&gt;&#xA;" OwnerUserId="8056" LastActivityDate="2017-09-17T04:17:14.810" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22733" PostTypeId="1" AcceptedAnswerId="22734" CreationDate="2017-09-17T07:35:34.063" Score="1" ViewCount="85" Body="&lt;p&gt;Is Corona Mass Ejection able to reach the outer planets such as Jupiter and Saturn? If so can it blows away the icy rings or the distance is simply beyond the reach of maybe like Earth it is protected by a magnetic field?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastActivityDate="2017-09-17T17:28:42.763" Title="Can CME destroy planetary rings?" Tags="&lt;gas-giants&gt;&lt;planetary-ring&gt;&lt;coronal-mass-ejection&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22734" PostTypeId="2" ParentId="22733" CreationDate="2017-09-17T09:00:37.880" Score="4" Body="&lt;p&gt;Yes it can reach both Saturn and Jupiter. But it's not powerful enough to remove their rings. Even Earth's radiation belts don't get blown away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Jupiter's magnetic field is $\approx 770 \mu T$ (at the surface of its equator) which is way more powerful than Earth's ($\approx 40 \mu T$ at the equator). Saturn's magnetic field on the other hand is a tad less powerful than that of the Earth ($\approx 20 \mu T$ at its equator). So Earth's magnetosphere shields it's particles from CMEs. Jupiter is further away and its magnetosphere is way stronger than Earth's. So it shields everything effectively. Saturn is even further away and although its magnetic field is comparable to that of the Earth, the distance from the Sun assures us that a CME wouldn't have a significant impact of any sort there.&lt;/p&gt;&#xA;" OwnerUserId="17425" LastEditorUserId="17425" LastEditDate="2017-09-17T17:28:42.763" LastActivityDate="2017-09-17T17:28:42.763" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22735" PostTypeId="1" AcceptedAnswerId="22737" CreationDate="2017-09-17T10:45:16.000" Score="2" ViewCount="116" Body="&lt;p&gt;What would people, standing on Earth, see, hear and smell, if a short period comet, with a two mile diameter nucleus, came by at 100,000 miles above us?&lt;/p&gt;&#xA;" OwnerUserId="18588" LastEditorUserId="37191" LastEditDate="2021-03-07T07:02:38.070" LastActivityDate="2021-03-07T07:02:38.070" Title="What would people on Earth see during a close comet pass?" Tags="&lt;observational-astronomy&gt;&lt;comets&gt;&lt;near-earth-object&gt;&lt;smell&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22737" PostTypeId="2" ParentId="22735" CreationDate="2017-09-17T11:57:18.350" Score="3" Body="&lt;p&gt;Hear and smell: Nothing. The comet and its coma are well outside the atmosphere, no sound or scent can travel through space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coma of the comet could be 30000km across, bigger than the moon, and the comet is half the distance to the moon, so the coma could appear about 20 times larger than the full moon. However, it would be very dim. Although its total brightness might be quite high, the surface brightness would be still very low. It is likely that only the central part of the coma would be visible, as a hazy blur, with no clearly defined edge. Some jets might be visible towards the nucleus. In a dark sky it would look quite dramatic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nucleus would be only about 4 arcseconds across, however it is close enough to be visible as a star-like body in the centre of the fuzzy nucleus. Telescopes would reveal the shape of the body, but telescopes on Earth would show few details. The comet would slowly track across the sky, clearly moving relative to the stars if you observe it for any length of time. The exact speed would depend on its orbit, but it wouldn't be moving like a meteor, the encounter would take a few hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tail, which might be visible in dark sky locations, would be pointing away from the sun, rather than behind the direction of travel.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-09-19T17:19:19.160" LastActivityDate="2017-09-19T17:19:19.160" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22739" PostTypeId="1" AcceptedAnswerId="22741" CreationDate="2017-09-17T21:41:45.970" Score="0" ViewCount="112" Body="&lt;p&gt;Would a direct hit by a short-period comet with a 3 mile diameter nucleus cause enough damage on Earth to threaten or destroy human life and the ability to produce crops?&lt;/p&gt;&#xA;" OwnerUserId="18588" LastActivityDate="2017-09-18T06:22:49.423" Title="Would a direct hit by a small comet be catastrophic?" Tags="&lt;comets&gt;&lt;impact&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22741" PostTypeId="2" ParentId="22739" CreationDate="2017-09-18T00:22:57.163" Score="3" Body="&lt;p&gt;Here, have some fun with this...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://impact.ese.ic.ac.uk/ImpactEffects/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://impact.ese.ic.ac.uk/ImpactEffects/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't go into long term effects, but the short term is pretty devastating for a significant portion of the planet.  'Nuclear winter' effects seem inevitable for those not atomised, scorched, crushed, or covered in ejecta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not good, nope.&lt;/p&gt;&#xA;" OwnerUserId="15741" LastActivityDate="2017-09-18T00:22:57.163" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22750" PostTypeId="1" AcceptedAnswerId="22751" CreationDate="2017-09-18T17:33:18.477" Score="6" ViewCount="2257" Body="&lt;p&gt;In 1931, &lt;a href=&quot;https://en.wikipedia.org/wiki/Subrahmanyan_Chandrasekhar&quot; rel=&quot;noreferrer&quot;&gt;Chandrasekhar&lt;/a&gt; was able to show that there is a certain critical mass (&lt;a href=&quot;https://en.wikipedia.org/wiki/Chandrasekhar_limit&quot; rel=&quot;noreferrer&quot;&gt;Chandrasekhar's limit&lt;/a&gt;) beyond which a white dwarf cannot exist, since the electronic fluid at that point cannot support its weight no matter how compressed it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The core of such a star will simply collapse inward. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The critical mass, Chandrasekhar showed, is 1.4 times that of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I first heard this, I was totally confused. Sun is a star too. The Sun will eventually enter the Red Giant Phase, and then it will eventually become a white-dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How and why did Chandrasekar express his limit relative to Sun?&lt;/p&gt;&#xA;" OwnerUserId="18633" LastEditorUserId="2365" LastEditDate="2017-09-18T18:44:00.033" LastActivityDate="2017-09-18T19:55:19.987" Title="Why is Chandrasekar Limit expressed relative to Sun?" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22751" PostTypeId="2" ParentId="22750" CreationDate="2017-09-18T17:43:57.473" Score="29" Body="&lt;p&gt;The mass of the sun is just a unit of convenience in a astronomy. The sun's mass and luminosity, in particular, are relatively easy for us to measure precisely, and when we're talking about stars, provide a convenient scale where the numbers won't be too &quot;astronomical&quot; (be too high a power of $10$ to picture easily). You could derive the Chandrasekhar limit in grams, kilograms, or slugs from the relevant physics, if you wanted. The relevant equation (&lt;A HREF=&quot;https://en.wikipedia.org/wiki/Chandrasekhar_limit&quot; rel=&quot;noreferrer&quot;&gt;from the Wikipedia article&lt;/A&gt;) is:&#xA;$$M_{\mathrm{limit}} = \frac{\omega_3^0 \sqrt{3\pi}}{2} \left(\frac{\hbar c}{G}\right)^{3/2} \frac{1}{(\mu_e m_\mathrm{H})^2},$$&#xA;where $\mu_e$ is the average molecular weight per electron (stellar composition dependent), $m_\mathrm{H}$ is the mass of hydrogen, and $\omega_3^0$ is a numerical constant that is approximately $2.018236\ldots$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted that the Chandrasekhar limit is a limit on the mass of the final white dwarf, not of the object that will produce the white dwarf.&lt;/p&gt;&#xA;" OwnerUserId="14244" LastEditorUserId="14244" LastEditDate="2017-09-18T19:55:19.987" LastActivityDate="2017-09-18T19:55:19.987" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="22752" PostTypeId="2" ParentId="22728" CreationDate="2017-09-18T18:45:21.597" Score="2" Body="&lt;p&gt;According to this (&lt;a href=&quot;https://spotthestation.nasa.gov/sightings/view.cfm?country=Bangladesh&amp;amp;region=None&amp;amp;city=Dhaka#.WcATdWdSz0o&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://spotthestation.nasa.gov/sightings/view.cfm?country=Bangladesh&amp;amp;region=None&amp;amp;city=Dhaka#.WcATdWdSz0o&lt;/a&gt;) the ISS was around at that time of morning. The sunrise was 5:45, so it could have been illuminated by the sun before it rose.&lt;/p&gt;&#xA;" OwnerUserId="15202" LastActivityDate="2017-09-18T18:45:21.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22753" PostTypeId="2" ParentId="8390" CreationDate="2017-09-18T21:02:43.883" Score="0" Body="&lt;p&gt;Note that how fast - and which direction - you need to move also depends on where you're pointing; it's not a one rate fits everything solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you're pointing at the celestial pole, you don't need to move the scope at all. While if you're pointing at the celestial equator, that needs the fastest tracking rate. What you end up doing is tracking along a circle that represents the declination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which means that in order to work out what the declination is, you're going to need angle encoders on both your alt and az axes.&lt;/p&gt;&#xA;" OwnerUserId="17016" LastActivityDate="2017-09-18T21:02:43.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22754" PostTypeId="1" AcceptedAnswerId="23160" CreationDate="2017-09-18T22:59:51.580" Score="3" ViewCount="197" Body="&lt;p&gt;I have read about and heard about the theory that most of the heavy elements we have here on Earth came from colliding neutron stars rather than by way of supernovae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how strong this theory is, but I wonder if a LIGO (and backup observational/ gamma ray) detection would help answer this question in any way.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2017-10-23T21:47:45.510" Title="Will the detection of colliding neutron stars by LIGO help answer the question of where heavy elements came form?" Tags="&lt;gravitational-waves&gt;&lt;elemental-abundances&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22756" PostTypeId="1" CreationDate="2017-09-19T08:07:34.043" Score="2" ViewCount="333" Body="&lt;p&gt;I really liked the clever way the author explained stuff in this book but I already understood most of it. Neil himself states in the book that if you are not in a hurry and want to know more about the topic the are plenty of bigger books you can look into. So my question is: what is sort of a next step after this book on this topic (astronomy, astrophysics)? &#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="18645" LastActivityDate="2017-09-19T13:10:25.913" Title="What should I read next after Neil deGrasse Tysons &quot;Astrophysics for people in a hurry&quot;?" Tags="&lt;astrophysics&gt;&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ClosedDate="2017-09-20T02:01:47.543" ContentLicense="CC BY-SA 3.0" />
  <row Id="22758" PostTypeId="5" CreationDate="2017-09-19T09:58:47.667" Score="0" Body="&lt;p&gt;Questions asking &lt;em&gt;&quot;What is a good book about X?&quot;&lt;/em&gt; or more generally, &lt;em&gt;&quot;What should I read to learn about X?&quot;&lt;/em&gt;, typically where there is not a single authoritative reference.  Also requests for multimedia, like video lectures.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastEditorUserId="10196" LastEditDate="2017-09-19T13:31:39.040" LastActivityDate="2017-09-19T13:31:39.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22759" PostTypeId="4" CreationDate="2017-09-19T09:58:47.667" Score="0" Body="Questions asking &quot;What is a good book about X?&quot; or more generally, &quot;What should I read to learn about X?&quot;, typically where there is not a single authoritative reference.  Also requests for multimedia, like video lectures." OwnerUserId="10196" LastEditorUserId="10196" LastEditDate="2017-09-19T13:31:35.683" LastActivityDate="2017-09-19T13:31:35.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22762" PostTypeId="1" CreationDate="2017-09-19T13:03:32.530" Score="1" ViewCount="366" Body="&lt;p&gt;So I am currently reading the book &lt;a href=&quot;http://phdcomics.com/noidea/&quot; rel=&quot;nofollow noreferrer&quot;&gt;We have no idea&lt;/a&gt; by Jorge Cham and Daniel Whiteson.&lt;br /&gt;&#xA;In one of their chapters they talk about how gravity is a force that only attracts and doesn't repel, most of the time. Dark energy is what makes up 68% of the universe and the only way we know it exists is from its rapid expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expansion isn't slowing down, however, it is speeding up. Gravity is a force that is said to be a force that only attracts and doesn't repel, unlike electromagnetism, the weak force and the strong force. Gravity should be slowing down that expansion, and I am wondering if anyone can explain this or is this concept out of our reach.&lt;/p&gt;&#xA;" OwnerUserId="18630" LastEditorDisplayName="user1569" LastEditDate="2017-10-26T12:41:13.137" LastActivityDate="2017-12-25T18:52:57.920" Title="Does gravity repel when dark energy is involved?" Tags="&lt;gravity&gt;&lt;expansion&gt;&lt;dark-energy&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22763" PostTypeId="2" ParentId="22756" CreationDate="2017-09-19T13:10:25.913" Score="1" Body="&lt;p&gt;I liked Black holes &amp;amp; time warps by Kip Thorne. &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/0393312763&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's the link.&lt;/a&gt; While being easy to understand, he does go into some details regarding black holes and the warping of time, which I think are interesting to the casual reader. If you have a science degree, you might not learn much new informations, but it's a good read.&lt;/p&gt;&#xA;" OwnerUserId="17425" LastActivityDate="2017-09-19T13:10:25.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22764" PostTypeId="1" AcceptedAnswerId="22809" CreationDate="2017-09-19T13:15:01.363" Score="10" ViewCount="1878" Body="&lt;p&gt;As far as I understand it, one can not hide spacecraft, because the heat from the engine would be very visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I tried searching for the farthest distance an actual spacecraft has been detected besides its radio signals that are intentionally sent to us. Unfortunately I could not find any data on this.&lt;/p&gt;&#xA;" OwnerUserId="18649" LastActivityDate="2019-01-12T23:56:00.813" Title="Whats the farthest distance a spacecraft has been detected (except by radio)" Tags="&lt;distances&gt;&lt;space-probe&gt;" AnswerCount="5" CommentCount="10" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22766" PostTypeId="1" AcceptedAnswerId="22770" CreationDate="2017-09-19T14:12:55.137" Score="3" ViewCount="8220" Body="&lt;p&gt;I know stars explode because of the fuel causing a fusion compound pushing it apart and the fuel runs out and it 'bounces' for lack of a better term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the fact that more massive stars are supposed to explode more quickly: why doesn't the Sun explode considering its massive size?&lt;/p&gt;&#xA;" OwnerDisplayName="user18650" LastEditorUserId="28132" LastEditDate="2019-06-01T01:42:57.263" LastActivityDate="2019-06-01T01:42:57.263" Title="Why doesn't the Sun explode?" Tags="&lt;the-sun&gt;&lt;explosion&gt;" AnswerCount="5" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22768" PostTypeId="2" ParentId="22764" CreationDate="2017-09-19T15:49:19.047" Score="5" Body="&lt;p&gt;When it was 5700km from Earth, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Rosetta_(spacecraft)#Deep_space_manoeuvres&quot; rel=&quot;noreferrer&quot; title=&quot;Rosetta&quot;&gt;Rosetta&lt;/a&gt; spacecraft was misidentified briefly as an asteroid as it flew by Earth for a gravity assist. I cannot find any more information on other spacecraft observed without radio from further away (I'm happy to be corrected!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If landers count; as mentioned in the comments, spacecraft orbiting Mars have observed their lander brethren on the surface.&lt;/p&gt;&#xA;" OwnerDisplayName="user10106" LastActivityDate="2017-09-19T15:49:19.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22769" PostTypeId="2" ParentId="22766" CreationDate="2017-09-19T16:10:28.680" Score="1" Body="&lt;p&gt;Our sun (I'm assuming you're from Earth or at least the solar system) is actually not all that big, &lt;a href=&quot;https://spaceplace.nasa.gov/sun-compare/en/&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;compared to other stars&quot;&gt;compared to other stars&lt;/a&gt;. The gravitational pull of the mass of the sun is kept in check by the fusion that this pull provides. Thus the Sun is at exactly the equilibrium of these two forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the Sun doesn't explode because its forces are balanced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also won't explode in the future because the mass of the Sun is not enough to trigger a supernova. It will more &quot;swell up and blow away&quot;.&lt;/p&gt;&#xA;" OwnerUserId="15202" LastEditorUserId="15202" LastEditDate="2017-09-19T16:15:36.640" LastActivityDate="2017-09-19T16:15:36.640" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22770" PostTypeId="2" ParentId="22766" CreationDate="2017-09-19T16:11:37.860" Score="3" Body="&lt;p&gt;Our sun is a particularly average sized star on the main sequence. It is not going to ever go &quot;supernova&quot; but instead will slowly swell and darken towards red, eventually swallowing Mercury and Venus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EjsYd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EjsYd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(from &lt;a href=&quot;http://www.oswego.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.oswego.edu/&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very boring in the grand scheme of things. Which is good for us :-)&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2017-09-19T16:11:37.860" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22771" PostTypeId="2" ParentId="22764" CreationDate="2017-09-19T16:26:36.540" Score="9" Body="&lt;p&gt;I saw the Cassini spacecraft visually through a 20 inch (500 mm) telescope when it flew by the Earth in August 1999. At the time of my sighting, it was just over 72000 miles (116000 km) away. The complete description is on the &lt;a href=&quot;http://www.satobs.org/seesat/Aug-1999/0217.html&quot; rel=&quot;noreferrer&quot;&gt;Seesat-L archives&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not have the information, but I think the Apollo spacecraft were seen visually at a greater distance.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-09-19T16:26:36.540" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22772" PostTypeId="2" ParentId="22766" CreationDate="2017-09-19T20:37:19.693" Score="2" Body="&lt;p&gt;Whether a star explodes or not is given by how quickly nuclear fusion happens inside it. If fusion takes place at a steady pace, the star does not explode. If lots of material fuse at once, the star explodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our Sun, like most other stars, just keeps fusing material at a slow and steady pace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It takes a different kind of star for the explosion to happen. Those stars that do explode are called supernovas, and there are a few different kinds of them. They could simply be much bigger than the Sun, and the conditions within these giant stars eventually trigger an explosion. Or they could be a white dwarf with a companion star nearby that keeps depositing material on the dwarf - when the pile is big enough, an explosion occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of that applies to our Sun.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2017-09-19T20:37:19.693" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22773" PostTypeId="2" ParentId="22754" CreationDate="2017-09-19T22:52:35.490" Score="1" Body="&lt;p&gt;Yes it will, in the sense that one of the major sources of uncertainty is how common these events are. The detection of (a population) of such events will give us a handle on that number, at least for the present-day milky way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, the one detection of GW170817 has moved us from an upper limit situation to getting the rate (in the local universe) to an order of magnitude.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-10-16T21:15:13.780" LastActivityDate="2017-10-16T21:15:13.780" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22776" PostTypeId="1" AcceptedAnswerId="22778" CreationDate="2017-09-20T06:47:02.403" Score="12" ViewCount="1358" Body="&lt;p&gt;Ignore pulsars and the like, just a neutron star in empty space – does it emit light and neutrinos?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="14006" LastEditDate="2017-09-20T08:52:02.820" LastActivityDate="2017-09-20T11:49:44.433" Title="Do neutron stars shine?" Tags="&lt;neutron-star&gt;&lt;neutrinos&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22778" PostTypeId="2" ParentId="22776" CreationDate="2017-09-20T11:41:07.973" Score="10" Body="&lt;p&gt;Neutron stars do indeed shine and some like &lt;a href=&quot;https://en.wikipedia.org/wiki/RX_J1856.5-3754&quot; rel=&quot;noreferrer&quot;&gt;this one&lt;/a&gt; have been observed in visible light. Like white dwarf stars, if they were scaled up to be the size of the Sun, they would outshine the Sun and most other stars in the galaxy. It is only their small size that make them hard to observe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for neutrinos, it seems they do too. This &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0012122&quot; rel=&quot;noreferrer&quot;&gt;link&lt;/a&gt; might help explain neutrino activity in / from neutron stars further.&lt;/p&gt;&#xA;" OwnerDisplayName="user10106" LastEditorDisplayName="user10106" LastEditDate="2017-09-20T11:49:44.433" LastActivityDate="2017-09-20T11:49:44.433" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22781" PostTypeId="2" ParentId="22764" CreationDate="2017-09-20T13:20:59.270" Score="-3" Body="&lt;p&gt;This is unfortunately a vague question.   If, for example, the spacecraft were reasonably large, you could detect it indirectly due to its gravitational pull on other (putative) nearby objects,  or even gravitational lensing.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far as heat-sensing or not,  the spacecraft would have to be very cold indeed to not radiate in the microwave range at levels inconsistent with the cosmic background. How sensitive a radiotelescope would be needed is a separate issue.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, if you had a &lt;a href=&quot;https://what-if.xkcd.com/141/&quot; rel=&quot;nofollow noreferrer&quot;&gt;really powerful lidar&lt;/a&gt; you'd easily &quot;ping&quot; the spacecraft -- perhaps vaporizing it in the process. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-09-20T13:20:59.270" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22783" PostTypeId="2" ParentId="22764" CreationDate="2017-09-20T15:41:38.107" Score="5" Body="&lt;p&gt;The &lt;a href=&quot;http://gbot.obspm.fr/&quot; rel=&quot;nofollow noreferrer&quot;&gt;GBOT&lt;/a&gt; (Ground Based Optical Tracking) is systematically looking at the Gaia Probe at L2 of the Earch-Sun System at about &lt;strong&gt;1.5 Mio KM&lt;/strong&gt; to locate it precisely (+/- 150 meters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has been tested on the &lt;a href=&quot;http://gbot.obspm.fr/gbotpipeline/spie9149-25.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;WMAP and Planck&lt;/a&gt; Space probes around L2 before that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe the spotting of Rosetta was even farther, but I can't comprehend the protocol: &lt;a href=&quot;http://www.minorplanetcenter.net/iau/mpec/K07/K07V69.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.minorplanetcenter.net/iau/mpec/K07/K07V69.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks &lt;a href=&quot;https://astronomy.stackexchange.com/users/7982/uhoh&quot;&gt;uhoh&lt;/a&gt; for figuring out the distance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the first reported observation at 2007-11-07 07:37 UTC the distance between Rosetta and Earth was about &lt;strong&gt;5.35 million km&lt;/strong&gt; according to &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL Horizons&lt;/a&gt;. And the flyby was on the 13th, with a closest approach of about 11,700 km from the center of the Earth (1.84 Earth radii) so it all checks out! &lt;a href=&quot;https://i.stack.imgur.com/4mIkr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Graph&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="18649" LastEditorUserId="18649" LastEditDate="2017-09-23T16:51:28.520" LastActivityDate="2017-09-23T16:51:28.520" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22785" PostTypeId="1" CreationDate="2017-09-21T04:12:56.673" Score="0" ViewCount="197" Body="&lt;p&gt;In September, Saturn is 9.9au away from Earth. If we use the parallax method, what should be the base line between the two people?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I know that $D = B / \theta$, but how do I find out theta? I'm pretty sure the answer is pretty obvious, but my mind really is not working right now, so any help is appreciated&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance. &lt;/p&gt;&#xA;" OwnerUserId="18672" LastEditorUserId="10196" LastEditDate="2017-09-21T07:08:55.027" LastActivityDate="2017-09-21T12:57:40.987" Title="Saturn observation : how to calculate requirements for Parallax method" Tags="&lt;saturn&gt;&lt;parallax&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22786" PostTypeId="1" CreationDate="2017-09-21T05:09:15.810" Score="4" ViewCount="1038" Body="&lt;p&gt;From several texts I have read, I have learned that right ascension is measured on a scale of 24 hours. I understand that Earth rotates almost exactly 360 degrees relative to distant objects, which is what RA is used for. The 24 hour system should only apply to the 361 degree rotation period (solar day). Why then do we split the sidereal day of the earth into 24 hours, when if it were split into 23 hours and 56 minutes we could know where an object would be x hours away after x hours?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More (possibly) relevant information:&#xA;I understand the Hour Angle HR is the measure of how far Aries is from the local meridian (RA of current meridian of longitude). This way you could calculate how many hours away the object is, but once again, this would not be exactly correct as 4 minutes has to be subtracted from the 24 hour scale, multiplying the whole time to something like 99.7% (14.44 degrees per hour), correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, this is really confusing me and I would appreciate some help. I did see a similar question but that had nothing to do with the sidereal day v. solar day issue I'm having. &lt;/p&gt;&#xA;" OwnerUserId="18673" LastActivityDate="2017-09-21T19:57:04.883" Title="Why is right ascension measured on a 24 hour scale rather than a 23 hours and 56 minutes scale?" Tags="&lt;solar-system&gt;&lt;earth&gt;&lt;coordinate&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22792" PostTypeId="2" ParentId="22786" CreationDate="2017-09-21T10:03:48.273" Score="2" Body="&lt;p&gt;It's 23 hours, 56 minutes, and 4.1 seconds, more or less. That's one sidereal day expressed in solar time. Solar time is neither a particularly convenient nor particularly useful time scale when looking at the stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, one could use sidereal time, which is what hour angle refers to. There are 24 sidereal hours in a sidereal day, 60 sidereal minutes in a sidereal hour, and 60 sidereal seconds in a sidereal minute.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-09-21T10:03:48.273" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22795" PostTypeId="1" AcceptedAnswerId="23226" CreationDate="2017-09-21T12:57:10.283" Score="0" ViewCount="925" Body="&lt;p&gt;I'm trying to calculate the state of the Earth at a given time using Keplerian orbital elements from &lt;a href=&quot;https://ssd.jpl.nasa.gov/?planet_pos&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;. It has the following instruction:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;5&quot;&gt;&#xA;  &lt;li&gt;Compute the coordinates, $\mathbf r_{ecl}$, in the J2000 ecliptic plane, with the x-axis aligned toward the equinox:&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So after this calculation we are supposed to know direction of &lt;em&gt;some&lt;/em&gt; equinox. But there exists the phenomenon of precession of equinoxes, so this direction shouldn't in general be fixed. I assume the equinox mentioned in that document is some &quot;mean&quot; equinox for the range of dates the data are valid for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how do I calculate the true equinox position/direction, given these data?&lt;/p&gt;&#xA;" OwnerUserId="735" LastActivityDate="2017-10-21T19:18:28.547" Title="How to calculate true equinox position?" Tags="&lt;earth&gt;&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22796" PostTypeId="2" ParentId="22785" CreationDate="2017-09-21T12:57:40.987" Score="1" Body="&lt;p&gt;Are you treating $\Theta$ as the only unknown, i.e.  you believe you know the baseline distance exactly?    If so, then  you compare your angular sighting of a distant star vs. Saturn with the same for the other observer.    The devil is in the error analysis.  Your two angles will be absurdly close to each other in magnitude, which means a very small error/uncertainty in value leads to a large error in the calculated distance $D$ to Saturn.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should find some first-year-level text on measurement &amp;amp; statistical error analysis to get a better feel for the difficulty here.&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-09-21T12:57:40.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22797" PostTypeId="2" ParentId="22786" CreationDate="2017-09-21T16:44:12.323" Score="3" Body="&lt;p&gt;The problem is not that right ascension goes from 0 to 24 hours (or 0 to any other number). The problem is that the Earth is revolving around the Sun, and our time system is based on the Sun. Because of the motion around the Sun, one solar day (24 hours 0 minutes 0 seconds) is different than one sidereal day (23 hours 56 minutes 4 seconds). &lt;strong&gt;So you need two clocks: one for the sun, and one for the stars&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example may help. Let's change right ascension to be from 0 to 100. At midnight (00:00) on day 1, let's define that 0 right ascension is on the meridian. At 23:56, the Earth has made one complete rotation relative to the stars, so 0 right ascension is back on the meridian. At 24:00, the Earth has made one complete rotation relative to the Sun (one day), but the right ascension has increased to 0.28 during the extra 4 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Day 2. At 23:52, the Earth has completed a second full rotation relative to the stars, so the right ascension is again 0. But after two complete days (24:00 on day 2), the right ascension is now 0.56.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-09-21T16:44:12.323" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22798" PostTypeId="2" ParentId="22786" CreationDate="2017-09-21T19:57:04.883" Score="5" Body="&lt;p&gt;Right ascension is a historical oddity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To specify a point in the sky you need a coordinate system, the one which we have come to use has it's origin at the Point of Aries on the Equator and the Ecliptic (a reasonable choice), It uses the Equator for one axis, and the meridian through the point of Aries for the other, again these are convenient and reasonable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coordinate system also requires a unit to be chosen for measuring angular distance. For measuring declension, we use degrees. The more mathematically pure may prefer radians, but degrees are a commonly used unit for measuring angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For measuring the angular distance around the equator we have come to use &quot;hours&quot;, by dividing the full turn into 24 parts. This is a historical oddity. It has a few conveniences: The sidereal day is only slightly shorter than 24 hours, so a star with RA of 1hr will be due south about 1 hour after a star with a RA of 0hr (not exactly but close enough for a rule of thumb). A telescope with an equatorial mount that is set to make a full turn in 24 hours will track stars well enough for a human observer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing the equator into (23+56/60) parts would be inconvenient since its not a whole number. For these reasons Flamsteed used Hours for his catalogue in 1712, and we have followed his tradition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the use of &quot;hours&quot; may be roughly related to the apparent motion of the stars, but it is really just a way of specifying an angle, and using 24 is easier than (23+56/60)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-09-21T19:57:04.883" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22802" PostTypeId="1" AcceptedAnswerId="22810" CreationDate="2017-09-22T06:05:08.370" Score="1" ViewCount="3185" Body="&lt;p&gt;I'm trying to do a project for a databases class where I have a user enter their current location and I tell them what constellations/planets/etc are visible in their area. I've seen this data compiled into images, PDFs, and website entries, but I can't seem to find a csv for it. I get that this data is bound to be massive or extremely complex, but I feel like there should be something out there. Any clues?&lt;/p&gt;&#xA;" OwnerUserId="18692" LastActivityDate="2017-09-24T18:44:15.077" Title="Where can I find a visible star dataset?" Tags="&lt;data-analysis&gt;&lt;constellations&gt;&lt;night-sky&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22803" PostTypeId="2" ParentId="22802" CreationDate="2017-09-22T06:23:33.537" Score="2" Body="&lt;p&gt;You can probably extract the data from &lt;A HREF=&quot;http://www.stellarium.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellarium&lt;/A&gt;. Also, does it need to be all sky, or not? What Wavelength? If you want an all sky visible catalogue, the United States Navy produced the &lt;A HREF=&quot;http://www.usno.navy.mil/USNO/astrometry/optical-IR-prod/nomad&quot; rel=&quot;nofollow noreferrer&quot;&gt;NOMAD&lt;/A&gt; catalog by combining &lt;A HREF=&quot;http://irsa.ipac.caltech.edu/Missions/2mass.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;2MASS&lt;/A&gt;, &lt;A HREF=&quot;http://www.usno.navy.mil/USNO/astrometry/optical-IR-prod/ucac&quot; rel=&quot;nofollow noreferrer&quot;&gt;UCAC&lt;/A&gt;, and the &lt;A HREF=&quot;http://www.usno.navy.mil/USNO/astrometry/optical-IR-prod/usno-b1.0&quot; rel=&quot;nofollow noreferrer&quot;&gt;USNO-B&lt;/A&gt; catalog. You could also use the &lt;A HREF=&quot;https://www.aavso.org/apass&quot; rel=&quot;nofollow noreferrer&quot;&gt;APASS&lt;/A&gt; all sky catalog, meant for helping astronomers with photometric calibration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't care about whether the wavelength is in the visible, CalTech's &lt;A HREF=&quot;http://irsa.ipac.caltech.edu/frontpage/&quot; rel=&quot;nofollow noreferrer&quot;&gt;IRSA&lt;/A&gt; has a bunch of infrared source catalogs (the smallest/shallowest one being the IRAS catalog). The deepest all sky infrared catalogs are the &lt;A HREF=&quot;http://irsa.ipac.caltech.edu/Missions/wise.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;WISE&lt;/A&gt; sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other good databases to dig in to include the &lt;A HREF=&quot;https://heasarc.gsfc.nasa.gov/W3Browse/all/hipparcos.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hipparcos&lt;/A&gt; set of parallaxes, or some part of the &lt;A HREF=&quot;http://gea.esac.esa.int/archive/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gaia catalog&lt;/A&gt;.&lt;/p&gt;&#xA;" OwnerUserId="14244" LastEditorUserId="14244" LastEditDate="2017-09-24T18:44:15.077" LastActivityDate="2017-09-24T18:44:15.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22807" PostTypeId="1" AcceptedAnswerId="23285" CreationDate="2017-09-22T13:24:41.967" Score="1" ViewCount="392" Body="&lt;p&gt;Would someone know of a good reference on the oldest known predictions of constellation positions?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How far ahead could accurate positions be predicted? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, could current constellation positions be accurately predicted between A.D. 69 to A.D. 86?&lt;/p&gt;&#xA;" OwnerUserId="18704" LastEditorUserId="11264" LastEditDate="2017-10-24T16:02:37.907" LastActivityDate="2017-10-26T11:28:49.137" Title="What is the oldest known prediction of constellation positions?" Tags="&lt;history&gt;&lt;constellations&gt;" AnswerCount="2" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="22809" PostTypeId="2" ParentId="22764" CreationDate="2017-09-22T16:06:57.690" Score="11" Body="&lt;p&gt;&lt;a href=&quot;https://www.nasa.gov/image-feature/goddard/2017/large-binocular-telescope-snags-first-glimpse-of-osiris-rex&quot; rel=&quot;noreferrer&quot;&gt;OSIRIS-Rex has been spotted&lt;/a&gt; on its approach to Earth, at a distance of approximately 7 million miles (12 million kilometers) away, with a brightness of approximately 25th magnitude. The Large Binocular Telescope was used for this observation, this has a pair of 8.4-meter mirrors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cv1Nc.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cv1Nc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="361" LastEditorUserId="361" LastEditDate="2017-09-23T19:51:33.023" LastActivityDate="2017-09-23T19:51:33.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22810" PostTypeId="2" ParentId="22802" CreationDate="2017-09-22T16:47:15.663" Score="2" Body="&lt;p&gt;If you are looking for stars that are visible to the unaided eye, the &lt;a href=&quot;http://tdc-www.harvard.edu/catalogs/bsc5.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Yale Bright Star Catalog&lt;/a&gt; is a good source. It includes 9100 stars to approximately magnitude 7 and is available in ASCII and binary formats -- perfect for your database project. Based on the date, time, latitude and longitude, you can calculate which stars are above the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As someone pointed out, the planets move day-to-day, week-to-week, or month-to-month through the constellations. (The Moon moves much faster than that, going through the complete zodiac in about 27.3 days.) You would need to calculate the position of the planets in order to determine if they are visible or not. That might be beyond the scope of your &quot;database&quot; project.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-09-22T16:47:15.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22813" PostTypeId="2" ParentId="22807" CreationDate="2017-09-23T03:14:57.860" Score="3" Body="&lt;p&gt;No, there is no possible way that humans from any known civilization in the first century AD would have been able to extrapolate current constellation positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prior to the work of &lt;a href=&quot;https://en.wikipedia.org/wiki/Johannes_Kepler&quot; rel=&quot;nofollow noreferrer&quot;&gt;Johannes Kepler&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Isaac_Newton&quot; rel=&quot;nofollow noreferrer&quot;&gt;Isaac Newton&lt;/a&gt;, humans did not have sufficient knowledge of the basic principles of celestial movement to have any kind of basis for making such calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, even if they had the basic principles, they would also have needed much more accurate and detailed information about the makeup of the Milky Way galaxy. The information currently available, that goes into the NASA models, is the product of a great many man-years invested in both observations and analysis of those observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, even if they had the basic principles and the necessary details with which to create a proper model, they lacked the computing power necessary to actually run the model - computation of that time being limited to pencil and paper (or perhaps an abacus).&lt;/p&gt;&#xA;" OwnerUserId="38" LastActivityDate="2017-09-23T03:14:57.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22820" PostTypeId="2" ParentId="10946" CreationDate="2017-09-23T08:13:15.123" Score="2" Body="&lt;p&gt;If we &quot;hold&quot; the Earth and &quot;move&quot; the Sun away, the Moon wouldn't stay with the Earth, but would follow the Sun. It is the only satellite in the Solar System that is attracted to the Sun stronger than to its own host planet:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;our Moon is unique among all the satellites of the planets, is so far as it is the only planetary satellite whose orbital radius exceeds the threshold value, which means it is the one satellite on which the Sun's gravitational acceleration exceeds the host planet's gravitational acceleration.  Consequently, it is the only moon in the solar system that is always falling toward the Sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mathpages.com/home/kmath405/kmath405.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Moon Always Veers Toward the Sun&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18718" LastActivityDate="2017-09-23T08:13:15.123" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22822" PostTypeId="1" CreationDate="2017-09-23T10:08:24.740" Score="6" ViewCount="515" Body="&lt;p&gt;Each vane holding the secondary mirror in front of the primary causes a diffraction spike. Couldn't the secondary mirror be held with only one vane?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If stability is the issue, couldn't it be substituted be steering the vane to hold the secondary in place with precision, similar to active optics, regardless of winds? And why does the James Webb Space Telescope use three vanes when it will operate in undisturbed microgravity and shadow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Maybe the star spikes are desired 'cause they make stars look nice?)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2019-07-16T06:25:37.863" Title="Why not use only one support vane for the secondary mirror, to avoid multiple diffraction spikes?" Tags="&lt;telescope&gt;&lt;space-telescope&gt;&lt;adaptive-optics&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22825" PostTypeId="2" ParentId="22822" CreationDate="2017-09-23T11:19:34.977" Score="3" Body="&lt;p&gt;I cannot answer about the JWST.&#xA;Regarding amateur telescopes, there is the option of using curved supports, which are intended to eliminate diffraction spikes and reduce overall diffraction.  &lt;a href=&quot;http://www.fpi-protostar.com/crvmnts.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.fpi-protostar.com/crvmnts.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Athough diffraction spikes are eliminated by curved supports, diffraction occurs symmetrically around bright objects.&lt;/p&gt;&#xA;" OwnerUserId="5355" LastEditorUserId="2365" LastEditDate="2017-09-23T21:00:55.517" LastActivityDate="2017-09-23T21:00:55.517" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22830" PostTypeId="1" CreationDate="2017-09-24T14:38:54.977" Score="1" ViewCount="611" Body="&lt;p&gt;What would the moon look like to someone at the South Pole?  To a first approximation, it's bisected by the horizon. But how far above and below does it get?&lt;/p&gt;&#xA;" OwnerUserId="7142" LastEditorUserId="2365" LastEditDate="2017-09-24T16:17:14.987" LastActivityDate="2017-09-24T16:17:14.987" Title="View of moon from South Pole?" Tags="&lt;orbit&gt;&lt;the-moon&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22832" PostTypeId="2" ParentId="22830" CreationDate="2017-09-24T16:16:16.407" Score="5" Body="&lt;p&gt;The moon orbits close to the plane of the ecliptic, which is tilted at 23.5 degrees to the equator, the moon is slightly off the ecliptic, and can be about 5 degrees above the ecliptic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, on the winter solstice (June 21) if full moon happens to be furthest from crossing the ecliptic, the moon would be 28.5 degrees above the horizon: roughly a third of the way between the horizon and the zenith. It can get the same distance below the horizon, and during summer the full moon would not be seen at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our moon is unique in being close to the plane of the ecliptic, and not in the plane of the equator, which suggests its formation was not like that of other moons in the solar system.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-09-24T16:16:16.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22834" PostTypeId="2" ParentId="22546" CreationDate="2017-09-24T21:07:59.963" Score="0" Body="&lt;p&gt;So, to sum up, the answer to my question is that on a clear night, in the right position, with the naked eye you can see the central bulge of the Milky Way galaxy, even though it's up to 26,000 light years from Earth.  You can't identify the particular spiral arm that you are seeing, in any direction, as such.  But you can locate, at least with a telescope if not the naked eye, objects that are in our Orion Spur, the Sagittarius Arm, and the Perseus Arm in the other direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=rcQiLAktC-M&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=rcQiLAktC-M&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18392" LastEditorUserId="18392" LastEditDate="2017-09-25T14:34:16.203" LastActivityDate="2017-09-25T14:34:16.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22835" PostTypeId="2" ParentId="22766" CreationDate="2017-09-24T21:13:47.120" Score="9" Body="&lt;p&gt;There are two things to discuss here: (a) why the Sun does not explode; and (b) why the Sun will not explode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) An explosion occurs when the timescale for the energy release by some process is much shorter than the timescale on which a system can adjust to damp the energy release process. In the present day Sun, nuclear fusion is a very slow process: on average it takes many billion years for a proton to fuse with another. This timescale is quite temperature dependent, so you might have thought the centre of the Sun might heat up quickly, leading to a runaway &quot;explosion&quot;. However, an increase in temperature leads to an increase in pressure that would expand the Sun, reducing the core density and temperature and decreasing the rate of nuclear fusion again. The timescale for the Sun to react in this way is just millions of years, so this acts like a thermostat that keeps the reactions under control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Stars more massive than the Sun burn through their hydrogen and other heavier fuels, and end up with an inert iron core from which no further energy can be extracted. The core subsequently collapses and the gravitational potential energy released by the core collapse powers a supernova explosion. The reason that the core does not quietly collapse into a black hole (at least in the supernovae we see!) is that the core &quot;bounces&quot; when &lt;em&gt;neutrons&lt;/em&gt; in the core are squeezed very close together. This provides a quantum mechanical effect called degeneracy pressure that resists a complete collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ironically, it is degeneracy pressure in the solar core that &lt;em&gt;prevents&lt;/em&gt; our Sun reaching the supernova stage. After burning through its hydrogen and helium the core of the Sun would consist of carbon and oxygen; but it will never become hot enough to commence fusion of these elements, because &lt;em&gt;electron&lt;/em&gt; degeneracy pressure will support it from contracting and getting any hotter. Only more massive stars will ever attain the core temperatures required to ensure progress towards a terminal iron core.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-09-24T22:38:23.997" LastActivityDate="2017-09-24T22:38:23.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22836" PostTypeId="1" AcceptedAnswerId="22849" CreationDate="2017-09-24T21:15:41.500" Score="1" ViewCount="258" Body="&lt;p&gt;I'm working with some NEOWISE images that seem to have incomplete header tags. I've been advised by the SAO (because I'm using DS9 to view them) that I have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;a partial PC matrix defined, yet it does not conform to the FITS WCS standard. With a combo of CDELT and PC keywords, the CDELT keywords hold the scaling factor, the PC keywords, the rotation matrix, which should be normalized. If you want the rotation matrix to also contain the scale factor, use CD keywords&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's one of the headers I need to fix:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;SIMPLE  =                    T / conforms to FITS standard                      &#xA;BITPIX  =                  -32 / array data type                                &#xA;NAXIS   =                    2 / number of array dimensions                     &#xA;NAXIS1  =                 1459                                                  &#xA;NAXIS2  =                  903                                                  &#xA;WCSAXES =                    2 / Number of coordinate axes                      &#xA;CRPIX1  =                729.5 / Reference pixel for axis 1                     &#xA;CRPIX2  =                451.5 / Reference pixel for axis 2                     &#xA;PC1_1   =     -0.0333333333333 / Coordinate transformation matrix element       &#xA;PC2_2   =      0.0333333333333 / Coordinate transformation matrix element       &#xA;CDELT1  =                   1. / [deg] Coordinate increment at reference point  &#xA;CDELT2  =                   1. / [deg] Coordinate increment at reference point  &#xA;CUNIT1  = 'deg     '           / Units of coordinate increment and value        &#xA;CUNIT2  = 'deg     '           / Units of coordinate increment and value        &#xA;CTYPE1  = 'GLON-AIT'           / Projection for axis1                           &#xA;CTYPE2  = 'GLAT-AIT'           / Projection for axis1                           &#xA;CRVAL1  =                   0. / [deg] Coordinate value at reference point      &#xA;CRVAL2  =                   0. / [deg] Coordinate value at reference point      &#xA;LONPOLE =                   0. / [deg] Native longitude of celestial pole       &#xA;LATPOLE =                  90. / [deg] Native latitude of celestial pole        &#xA;RADESYS = 'ICRS    '           / Equatorial coordinate system                   &#xA;BAND    = 'W1      '           / Wise Band                                      &#xA;END&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I've already managed to fix a couple of the invalid tags, but apart from these minor edits, this is unaltered from the file that my research supervisor gave me. The FITS standard &lt;a href=&quot;https://heasarc.gsfc.nasa.gov/docs/fcg/standard_dict.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; gives descriptions of each of the tags, but I'm not sure how to implement the SAO's advice. Can anyone help me fix the tags?&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2017-09-25T09:38:11.190" Title="Problems with FITS headers" Tags="&lt;data-analysis&gt;&lt;python&gt;&lt;fits-header&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22837" PostTypeId="2" ParentId="20333" CreationDate="2017-09-24T21:35:19.003" Score="1" Body="&lt;p&gt;There's also a software package, called &lt;a href=&quot;http://ds9.si.edu/site/Download.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;DS9&lt;/a&gt;, which I use for viewing both the image and the header. I edit the headers in Python using astropy.&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2017-09-24T21:35:19.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22838" PostTypeId="1" CreationDate="2017-09-24T21:45:55.930" Score="1" ViewCount="1248" Body="&lt;p&gt;I have been seeing all these &quot;hoaxes&quot; on the internet about Planet X and how it would destroy Earth, on September 23rd 2017. It didn't happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started to wonder if it's possible - with our current technology - to map the path of an asteroid with calculations of how an asteroid will pass near or collide with Earth. &lt;a href=&quot;https://en.wikipedia.org/wiki/(89959)_2002_NT7&quot; rel=&quot;nofollow noreferrer&quot;&gt;I've heard one is going to &quot;safely&quot; pass Earth on January 13th 2019.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18736" LastEditorUserId="2153" LastEditDate="2017-09-25T01:58:03.330" LastActivityDate="2017-09-25T03:26:56.367" Title="Can we actually predict an asteroid impact?" Tags="&lt;earth&gt;&lt;asteroids&gt;&lt;impact&gt;&lt;near-earth-object&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ClosedDate="2017-09-25T11:41:45.317" ContentLicense="CC BY-SA 3.0" />
  <row Id="22840" PostTypeId="2" ParentId="22838" CreationDate="2017-09-24T23:15:22.027" Score="1" Body="&lt;p&gt;&lt;strong&gt;Yes&lt;/strong&gt; it is possible to predict the path of an asteroid, and if it will strike the Earth. In the first instance the accuracy of the predicted path depends upon the quality and spacing of the positional. The better the position measurements (minimum 3) the better the prediction. If the position measurements are spread around the asteroids orbit, the predicted orbit is more accurate. &#xA;The second factor is more subtle. Because the solar system has many planets, there are many forces on an asteroid. These cannot all be put into an exact solution but need to be calculated step by step. The uncertainty in the predicted path accumulates the further into the future you push the predictions. When an asteroid makes a close pass to a planet a slight difference in the approach can make a large difference in the path after approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter being the most massive planet alters the orbits of a lot of asteroids and comets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are projects to find and measure the positions of asteroids that could strike Earth. Look up SpaceGaurd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is because of these uncertainties that NASA etc give a probability of an asteroid hitting Earth on it's next close approach.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comets are more difficult, because as gas/dust leave the comet it acts like a rocket engine and so moves the comet off its predictable path. The effect is small but noticeable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classically, the positions of asteroids is measured by taking an image and measuring where the asteroid is on the image relative to stars with accurately known positions. Sounds simple, but needs skill to do well, and the help of good software to take into account like distorts caused by the telescope/camera optics ... The calculations also involve knowing the location of the Earth and camera, relative to the stars.&lt;/p&gt;&#xA;" OwnerUserId="10761" LastActivityDate="2017-09-24T23:15:22.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22842" PostTypeId="2" ParentId="22838" CreationDate="2017-09-25T03:26:56.367" Score="2" Body="&lt;p&gt;There's a wide variety of answers for how much warning.   Most of what the &lt;a href=&quot;https://www.nasa.gov/audience/foreducators/postsecondary/features/F_Near_Earth_Program.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Near Earth Asteroid Program&lt;/a&gt; or focuses on is &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_Earth-crossing_minor_planets&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth Crossing Objects&lt;/a&gt; or &lt;a href=&quot;https://cneos.jpl.nasa.gov/about/neo_groups.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Near Earth Objects&lt;/a&gt;.   Those pose the greatest risk of impact and they are also, somewhat conveniently, the easiest to track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hardest to track is a high eccentricity long period orbit that approaches from the sun side of the planet.   &lt;a href=&quot;https://en.wikipedia.org/wiki/Chelyabinsk_meteor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chelyabinsk&lt;/a&gt; was a sun side asteroid (an asteroid that hits during the day generally approaches from the sun side.   Those are the hardest to see, and even though Chelyabinsk was too small to worry about (20 meters diameter estimate).  It's possible that it would have been observed by the luck of having a telescope pointed in the right direction had it approached from the night side of the planet.   Still, an object that size, it's unlikely that they'd have had more than a few hours warming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The larger the object, the easier it is to see and once an object is observed for some time and it's trajectory has been studied, it's path can be predicted years, even decades ahead of time.   The warning time drops sharply for an object not yet observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Potentially_hazardous_object&quot; rel=&quot;nofollow noreferrer&quot;&gt;Potentially Hazardous Objects&lt;/a&gt; are at least 100 meters across, sometimes 150 meters is used as a minimum.   25-50 times the reflective surface of Chelyabinsk, so an object that size could be seen a fair bit further ahead of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, Planet X doesn't exist and if it did exist, it would have cleared it's orbital path by now and it wouldn't be sending asteroids and/or comets towards Earth once every orbit and predicting an exact date (Sept 23rd) from an unobserved planet X. That's just silly.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is stuff out there in our solar system that we haven't seen.   Lots of stuff, maybe even a hundred or more unidentified dwarf planets, possibly another planet or two.  (&lt;a href=&quot;https://www.explainxkcd.com/wiki/index.php/1633:_Possible_Undiscovered_Planets&quot; rel=&quot;nofollow noreferrer&quot;&gt;Click here for cool diagram&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is, orbits tend to clear out over time and our solar-system is over 4 billion years old.   Most of the stuff that had an orbit that was going to crash into the Earth did so long ago and Planet Nine, assuming it exists, has cleared out it's orbit and it rarely sends stuff flying towards Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space is also so big that Earth is, by comparison, a very tiny target, so an asteroid coming from the outer reaches of the solar system and hitting Earth is like shooting a hole in one from the space shutting.   Stuff is far more likely to miss than hit the Earth.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to answer your question on an object coming from a distant orbit, sent our way by a theoretical planet, it's hard to say when we'd see it in that scenario if it's a brand new observation and it would also depend on whether it came towards the Earth on the sun side or the night side.  (Objects can hit the Earth from any direction), but sun side observations are the hardest to see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The good news is, high eccentricity impacts of objects large enough to be considered potentially hazardous are rare enough that you are unlikely to be affected by an Asteroid Impact in your lifetime.   It's somewhere around 140 on the list of things you should worry about.   And low eccentricity orbits that could hit Earth, or NEOs, those are being carefully tracked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other good news is that Planet X (if it even exists) doesn't threaten Earth.  No object in distant orbit around our sun would be a threat.  In fact, having a planet out there would actually send more asteroids away than it would cause to fly in towards the inner solar system.   They would reduce, not increase the number of impacts, similar to Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a large object passing through might post an asteroid/comet impact threat. &lt;a href=&quot;https://en.wikipedia.org/wiki/Gliese_710#Computing_and_details_of_the_closest_approach&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gilese 710&lt;/a&gt; could do that, but it won't pass close enough to have that effect for about 1.36 million years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't want to make light of this.  Asteroid impacts do happen and Earth may have been hit by a very destructive asteroid or comet about 12,900 years ago (the &lt;a href=&quot;https://en.wikipedia.org/wiki/Younger_Dryas_impact_hypothesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Younger Dryas impact hypothesis&lt;/a&gt;) and a global changing event like that might happen every, 50,000 or 100,000 years or so, which means it's 100%, absolutely worth looking out and seeing as much as possible, because Earth will get hit by a large object again.    But such impacts are very rare on a human lifetime scale. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not sure I ever answered your question.  How much warning for a large object, say 150-250 meters?   If the trajectory makes it tough to see, maybe only a few hours.   If it approaches from the night side, maybe a few days or a few weeks.   Longer if someone gets lucky and has a big telescope pointed in the right place, but I'm kinda guessing.    As our technology improves, the warning period will increase.  (I invite correction, especially on this last paragraph).  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-09-25T03:26:56.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22846" PostTypeId="2" ParentId="22836" CreationDate="2017-09-25T07:11:15.187" Score="1" Body="&lt;p&gt;Using Fits &lt;a href=&quot;http://heasarc.gsfc.nasa.gov/ftools/fv&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fits View (FV)&lt;/a&gt;View (FV), you can either modify the header by terminal commands, or using the GUI which makes you visualize all the keywords and values.&lt;/p&gt;&#xA;" OwnerUserId="1038" LastActivityDate="2017-09-25T07:11:15.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22849" PostTypeId="2" ParentId="22836" CreationDate="2017-09-25T09:38:11.190" Score="2" Body="&lt;p&gt;You are missing the PC1_2 and PC2_1 elements of your pixel transformation matrix. If your coordinate transformation is a rotation, you would use&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;PC1_1 = cos(t)&#xA;PC1_2 = sin(t)&#xA;PC2_1 = -sin(t)&#xA;PC2_2 = cos(t)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;for the PC matrix. The pixel scale (in degree/pixel, according to CUNITi) goes into CDELTi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed out, you can combine the PC and CDELT steps into a single CD matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://montage.ipac.caltech.edu/docs/headers.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://www.aanda.org/articles/aa/full/2002/45/aah3859/aah3859.right.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for more details.&lt;/p&gt;&#xA;" OwnerDisplayName="user18746" LastActivityDate="2017-09-25T09:38:11.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22854" PostTypeId="1" AcceptedAnswerId="23008" CreationDate="2017-09-25T12:53:36.360" Score="5" ViewCount="140" Body="&lt;p&gt;How do bodies like &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/2041-8213/aa876c/meta;jsessionid=75628DF4577F6DF4346E97FC333FBD72.c4.iopscience.cld.iop.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;WASP-12b&lt;/a&gt; lose mass to its central body? The process is never really explained in popular media, instead showing visualisations of a thin veil of gas being &quot;sucked&quot; in a rather straight line (and then into a ring around the star) rising straight from the surface of the planet. I highly doubt the mechanics which these illustration insinuate: Gravitation isn't working like a vacuum cleaner on a dusty rug. Either the rug goes with it, or the dust stays on the rug. The only mechanism that I can think of are particles from the upper atmosphere which happen to pick up enough thermodynamic energy so that they are faster than the escape velocity of the planet. But is this a process which effects a big enough mass transfer to be of significance? Do I overlook something? How is such a mass defect going on really?&lt;/p&gt;&#xA;" OwnerUserId="844" LastEditorUserId="844" LastEditDate="2017-10-04T13:17:32.257" LastActivityDate="2017-10-07T00:03:12.027" Title="How does a planet lose mass to its host star?" Tags="&lt;supernova&gt;&lt;binary-star&gt;&lt;white-dwarf&gt;&lt;hot-jupiter&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22855" PostTypeId="1" AcceptedAnswerId="22856" CreationDate="2017-09-25T13:39:30.040" Score="2" ViewCount="969" Body="&lt;p&gt;I have a question about the view of the Milky Way that we can see from earth with the naked eye.  I believe this to be light from the stars contained in one of the &quot;arms&quot; of the galaxy.  I had always assumed that, from our vantage point, I was looking outward through one of the arms.  This was based on the appearance of the apparent curvature over the sky above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it occurred to me that this could also be looking in towards the center of the galaxy.  And the fact that it appears to curve around us over the night sky is just an illusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, perhaps at different times in the year we are alternatively looking inward and then later looking outward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can any shed some, errr, light on this?&lt;/p&gt;&#xA;" OwnerUserId="18755" LastActivityDate="2019-09-19T12:34:53.970" Title="Milky Way, view outward or inward?" Tags="&lt;milky-way&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22856" PostTypeId="2" ParentId="22855" CreationDate="2017-09-25T17:17:26.497" Score="8" Body="&lt;p&gt;An image of the whole sky (360 degrees) centred on Sagittarius and rotated so the galaxy is in the middle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0wftz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0wftz.jpg&quot; alt=&quot;Milky way&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image from the &quot;2MASS infrared survey&quot; actually uses infra-red light, as it penetrates the interstellar dust better than visible light. It shows that any apparent curvature is an illusion. The milky way goes straight across the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also shows that when we look towards Sagittarius we look towards the galactic centre, and when we look away from it (towards Auriga) we are looking through to stars in the outer spiral arms. The night side of the Earth faces the galactic centre during July, and faces outwards in January. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also visible in the image are the two Magellanic clouds (small but nearby galaxies outside the milky way), the clear bulge around the centre of the galaxy and the dark bands of dust along the milky way. The central bulge is much brighter in infrared than the spiral arms. Using infrared makes these features more visible. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="14006" LastEditDate="2017-09-26T20:55:13.207" LastActivityDate="2017-09-26T20:55:13.207" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22857" PostTypeId="2" ParentId="11634" CreationDate="2017-09-25T17:57:16.783" Score="19" Body="&lt;p&gt;The League of Lost Causes wrote the definitive &lt;a href=&quot;http://www.leagueoflostcauses.com/blog/2013/08/astronomy-101.html&quot; rel=&quot;noreferrer&quot;&gt;How to identify that light in the sky?&lt;/a&gt; guide:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/t6k78.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t6k78.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sup&gt;Licensed under &lt;a href=&quot;https://creativecommons.org/licenses/by-nc-nd/3.0/&quot; rel=&quot;noreferrer&quot;&gt;Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user1569" LastActivityDate="2017-09-25T17:57:16.783" CommentCount="1" CommunityOwnedDate="2017-09-25T17:57:16.783" ContentLicense="CC BY-SA 3.0" />
  <row Id="22858" PostTypeId="1" CreationDate="2017-09-25T18:17:06.883" Score="2" ViewCount="82" Body="&lt;p&gt;A star emits energy through the fusion of hydrogen to helium in the core through the proton proton cycle or the carbon nitrogen cycle. Over time, hydrogen in the core is increasingly burned, which causes the fusion process to decrease over time. As a result, the star cools down and decreases the radiation pressure in the core. As a result, the core draws a little more together under the influence of its own weight, which increases the temperature. This temperature rise leads to a hydrogen fusion in a shell outside the core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how is the temperature increased as the radiation pressure is decreased? Is this increase caused by conduction or convection or....?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastActivityDate="2017-09-26T07:28:08.010" Title="What causes the raise of temperature in the outer layer of a red giant when the core gets contracted?" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22859" PostTypeId="1" CreationDate="2017-09-25T18:53:31.273" Score="9" ViewCount="298" Body="&lt;p&gt;Or was it always at about 5.14 degrees inclination or has the inclination changed over time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Orbit_of_the_Moon#/media/File:Earth-Moon.PNG&quot; rel=&quot;noreferrer&quot;&gt;See diagram&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;James K's answer to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/22830/view-of-moon-from-south-pole&quot;&gt;this question&lt;/a&gt; got me thinking about this, and I don't mean to call him out, but this part of his answer doesn't seem quite right when he wrote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Our moon is unique in being close to the plane of the ecliptic, and&#xA;  not in the plane of the equator, which suggests its formation was not&#xA;  like that of other moons in the solar system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For starters, I imagine the physics of the initial orbit of an impact moon would be somewhat complicated, but it seems likely, given an angled impact like the one that's believed to have formed our moon (and hitting at an angle is statistically more common than a direct impact anyway), &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in this scenario, the planet gets significant angular momentum and this angular momentum should dictate the planet's new equator and axial tilt relative to it's orbit.  I would think the moon should form roughly along that same equatorial plane, but I'm just guessing.   Perhaps it could form several degrees off - not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 2nd point, is, a rapidly rotating planet with a large equatorial bulge and a close moon, if the Moon formed off the equator, would it's orbit migrate over the equator where the gravitation was greatest and would that happen relatively quickly or not at all?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's basically the question.   Was the Moon always at roughly 6 degrees off the Earth's equator or has it only moved off an orbit over Earth's equator more recently, perhaps due to the gravitational effect of the Sun?    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or are there other factors.   Mars' axis is thought to have changed rather significantly due to Jupiter's gravitational effects perhaps 100,000 years agao and it's Moons orbit over Mars' equator which suggests that Mars' equatorial bulge dragged the Moons with it? - or is my thinking way off on that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thinking is that a planet's equatorial bulge would drive moons towards a 0 degree inclination around it's equatorial bulge and our Moon is different because of proximity to the sun which also has a strong gravitational effect.   The 5.14 degrees of inclination is a balance between the Earth's equator and the solar gravity.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2020-12-06T18:04:27.510" Title="When the Moon formed and early on, did it orbit over the Earth's equator?" Tags="&lt;the-moon&gt;&lt;orbital-mechanics&gt;&lt;orbital-migration&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22860" PostTypeId="1" CreationDate="2017-09-25T19:19:02.710" Score="0" ViewCount="234" Body="&lt;p&gt;I seen the other day the moon lit directly below or towards my horizon perfectly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question, is there a term for when a moon's crescent is aligned with the viewer's horizon? &lt;/p&gt;&#xA;" OwnerUserId="18757" LastActivityDate="2017-09-25T19:53:57.030" Title="Is there a term for a horizontal moon crescent" Tags="&lt;the-moon&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22861" PostTypeId="2" ParentId="22860" CreationDate="2017-09-25T19:53:57.030" Score="2" Body="&lt;p&gt;No formal astronomical term, but it is sometimes said to be &quot;on its back&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples from &quot;&lt;a href=&quot;https://books.google.co.uk/books?id=nR1EBgAAQBAJ&amp;amp;pg=PA62&amp;amp;lpg=PA62&amp;amp;dq=hang%20your%20hat%20on%20the%20moon%27s%20horns&amp;amp;source=bl&amp;amp;ots=7ii8Gs9yVv&amp;amp;sig=UpM23Wj0JdbrNaF_vXPgs4LCBHg&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwiIj-XyjMHWAhUoKsAKHQ_YAAkQ6AEIKDAA#v=onepage&amp;amp;q=hang%20your%20hat%20on%20the%20moon&amp;#39;s%20horns&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;weather lore&lt;/a&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When the moon lies on its back&lt;br&gt;&#xA;  Then the sou-west wind will crack.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When the moon is on its back,&lt;br&gt;&#xA;  She sucks the wet into her lap.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It's sure to be a dry moon if it lies on its back&lt;br&gt;&#xA;  so you can hang your hat on its horns. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-09-25T19:53:57.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22862" PostTypeId="1" CreationDate="2017-09-25T21:27:23.347" Score="2" ViewCount="52" Body="&lt;p&gt;Let's imagine a moon orbiting a planet at 75000km distance (assume circular orbit). Now let's say that the Hill Sphere of this planet only extends to 78,000km (small planet in the habitable zone of a small star).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with the moon/Earth situation, this moon is slowly drifting farther away from the planet. (ie Deimos and Mars).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What factors effect the time frame of this moon being lost so that it would be in orbit of the star and not the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know of a computer simulation showing this happen? Wondering what the trajectory and duration of the transition period would look like.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2017-09-25T21:27:23.347" Title="Is there a computer simulation that anyone knows of that describes the loss of a moon as it's orbit extends beyond the Hill Sphere of the planet?" Tags="&lt;orbit&gt;" AnswerCount="0" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22863" PostTypeId="2" ParentId="16734" CreationDate="2017-09-25T21:30:45.117" Score="0" Body="&lt;p&gt;Just to add to James' answer, we have amazing images of the stars very near to the BH at the galactic core. One star in particular, called S2, has had its complete &lt;a href=&quot;https://en.wikipedia.org/wiki/S2_(star)#/media/File:Orbit_of_S2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;orbit mapped&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2017-09-25T21:30:45.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22872" PostTypeId="2" ParentId="22858" CreationDate="2017-09-26T07:28:08.010" Score="3" Body="&lt;p&gt;The temperature increase of the core and shell is governed by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Virial_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;virial theorem&lt;/a&gt;. As contraction occurs, gravitational potential energy is released. According to the virial theorem, half of this is radiated (or conducted or convected) away and half goes into the internal energy of the contracting material, thus raising its temperature. You can find a simple proof &lt;a href=&quot;https://physics.stackexchange.com/questions/343230/why-the-he-core-gets-hotter-and-hotter-is-it-because-it-shrinks&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radiation pressure is unimportant, but continues to increase in any case. Only the surface of the star becomes cooler, not the interior.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-09-26T07:28:08.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22882" PostTypeId="1" AcceptedAnswerId="22894" CreationDate="2017-09-26T23:09:38.810" Score="3" ViewCount="221" Body="&lt;p&gt;My question is if I put you down somewhere in the Milky Way galaxy in any given period of time, is there a way for you to tell when and where you are based on stars or constellations? Or some known astronomical event like a supernova? How accurate would that be? &lt;/p&gt;&#xA;" OwnerUserId="18782" LastEditorUserId="11143" LastEditDate="2017-09-27T13:31:56.870" LastActivityDate="2017-09-27T16:15:54.750" Title="Knowing time any where and any when" Tags="&lt;galaxy&gt;&lt;space-time&gt;&lt;time&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22884" PostTypeId="1" CreationDate="2017-09-27T05:00:31.910" Score="1" ViewCount="267" Body="&lt;p&gt;I started using the software that comes with Jean Meeus book Astronomical Algorithms. I am now using the updated code from naughter(dot)com/aa.html. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am able to calculate the heliocentric longitude and latitude for an asteroid using the orbital elements. Below is for the asteroid Juno. I can't find any source code or instructions for converting the asteroid coordinates I calculated below to geocentric longitude and latitude. Can anyone recommend a method to find asteroid geocentric longitude and latitude? Thanks. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;double Longitude = 0.0; //West Greenwich  &#xA;double Latitude  = 51.4826; //North Greenwich  &#xA;double TZOffSet = 0.0;&#xA;&#xA;&#xA;//Asteroid Juno Elements;   Using epoch J2000&#xA;CAAEllipticalObjectElements elements;  &#xA;&#xA;//Calculation Julian Date  2017, 9, 25 &#xA;elements.JDEquinox = JD;   &#xA;&#xA;elements.T = (JD - 2451545.0) / 36525.0;&#xA;&#xA;//semi-major axis of the orbit&#xA;elements.a = 2.6700912;&#xA;&#xA;//eccentricity of the orbit&#xA;elements.e = 0.25498122;&#xA;&#xA;//inclination of the plane of the ecliptic&#xA;elements.i = 12.98211;&#xA;&#xA;//Longitude of ascending node&#xA;elements.omega = 169.91138;&#xA;&#xA;//longitude of perihelion&#xA;elements.w = 248.10807; &#xA;&#xA;//-------------------------------------------------------------------------&#xA;CAAEllipticalObjectDetails details = CAAElliptical::Calculate(JD, elements, true);&#xA;&#xA;1) Heliocentric Ecliptic Longitude  = 215.6670202&#xA;2) Heliocentric Ecliptic Latitude = 9.377841131&#xA;3) True Geocentric RA = 13.90415220&#xA;4) True Geocentric Declination = -3.802673727&#xA;5) True Geocentric Distance = 4.136531368&#xA;6) True Geocentric Light Time = 0.0238906126&#xA;7) Astrometric Geocentric RA = 13.90397025&#xA;8) Astrometric Geocentric Declination = -3.80217202&#xA;9) Astrometric Geocentric Distance = 4.13653445&#xA;10) Astrometric Geocentric Light Time = 0.023890630&#xA;11) Elongation = 26.74464206&#xA;12) Phase Angle  = 7.928218201&#xA;//-------------------------------------------------------------------------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="14787" LastActivityDate="2017-10-27T12:08:38.570" Title="Finding Asteroid Geocentric Longitude and Latitude" Tags="&lt;asteroids&gt;&lt;coordinate&gt;&lt;ephemeris&gt;&lt;algorithm&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22889" PostTypeId="2" ParentId="22882" CreationDate="2017-09-27T08:16:42.487" Score="2" Body="&lt;p&gt;Some of this might stray into WorldBuilding StackExchange territory, and 'when' will be tough to approximate as galactic rotation and stars going through different phases of their lives can make constellations and reference points obsolete with time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As StephenG mentioned, pulsars, owing to their (usually) unchanging and detectable rotation periods, can be used as reference points. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Pioneer_plaque&quot; rel=&quot;nofollow noreferrer&quot;&gt;plaques attached to Pioneer 10 and 11&lt;/a&gt; provide the solar system's location relative to nearby pulsars. Pulsar coordinates were also added to the Voyager Golden Album too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alternative would be to adopt relatively fixed locations on your 'local' celestial sphere and determine your position relative to them, and therefore where your destination is. This assumes you know where your destination is relative to your reference points. The galactic core's location, the direction of galactic rotation, the Magellanic Clouds or other extra-galactic sources could be your reference points. They would hardly be accurate but if you are attempting a Voyager-esque journey through the Milky Way, it's a start.&lt;/p&gt;&#xA;" OwnerDisplayName="user10106" LastActivityDate="2017-09-27T08:16:42.487" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22893" PostTypeId="2" ParentId="22884" CreationDate="2017-09-27T11:50:09.443" Score="2" Body="&lt;p&gt;This can be accomplished through the following steps :&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;compute the position of the Earth in the same heliocentric frame,&lt;/li&gt;&#xA;&lt;li&gt;express both positions (Earth and asteroid) in heliocentric rectangular coordinates,&lt;/li&gt;&#xA;&lt;li&gt;change the origin of the frame to get the location of the asteroid in geocentric rectangular coordinates,&lt;/li&gt;&#xA;&lt;li&gt;convert the position of the asteroid in geocentric latitude and longitude.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerDisplayName="user15104" LastActivityDate="2017-09-27T11:50:09.443" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22894" PostTypeId="2" ParentId="22882" CreationDate="2017-09-27T13:26:41.590" Score="4" Body="&lt;p&gt;This question is someone open-ended since you don't provide many constraints, so I'll impose the following two conditions of my own to help me give an answer:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I'll assume that after you're randomly placed in space and time, that you can make any observation one could make here on Earth - that is, you have whatever equipment is necessary to observe anything about the Universe we can currently observe.&lt;/li&gt;&#xA;&lt;li&gt;I'll assume that you have access to all current astronomical knowledge (in a digital library or some such thing) such as star catalogs, physics books, published papers, etc.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h3&gt;How do we know our location?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;In determining location, because we exist in three dimensional space, we have to specify position with three coordinates. The simplest coordinate system in my mind, in terms of what we can measure, are the three coordinates below:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;our radial distance from the center of the galaxy,&lt;/li&gt;&#xA;&lt;li&gt;our height above or below the galactic plane,&lt;/li&gt;&#xA;&lt;li&gt;and our azimuthal angle from some reference point.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;em&gt;Yes I know these coordinates aren't orthogonal, but they relate well to measurements one could make and simple coordinate transformations could make them orthogonal.&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first one is pretty easy. We can currently measure the distance to &lt;a href=&quot;https://en.wikipedia.org/wiki/Sagittarius_A*&quot; rel=&quot;noreferrer&quot;&gt;Sag A*, the black hole at the center of the galaxy&lt;/a&gt;. It's not particularly easy and measurements have fluctuated over the years with respect to our position on Earth (see e.g., &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0306220&quot; rel=&quot;noreferrer&quot;&gt;Eisenhauer et al. 2003&lt;/a&gt; or &lt;a href=&quot;https://arxiv.org/abs/1607.05726&quot; rel=&quot;noreferrer&quot;&gt;Boehle et al. 2016&lt;/a&gt;), but it is a measurement which could allow you to define #1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second coordinate could be achieved by simply mapping the local star field. A survey akin to &lt;a href=&quot;https://en.wikipedia.org/wiki/Sloan_Digital_Sky_Survey&quot; rel=&quot;noreferrer&quot;&gt;SDSS&lt;/a&gt; could accomplish this. This is precisely what was done in this &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1995AJ....110.2183H&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;noreferrer&quot;&gt;Humphreys &amp;amp; Larson 1995&lt;/a&gt;. It would take time and a lot of work, but you could make this measurement pretty easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far so good. Now we come to coordinate #3 and suddenly we hit a problem. We need to know our azimuthal position with respect to some reference point. That is, if you look down on the galaxy and consider our angle from some line, what would that line be and what would that angle be? Since the clear intent is to relate our new position to where the Earth is, the most logical reference line is the line connecting the Earth to the center of the galaxy. But how do we know where that line is with respect to our new position? I'm afraid I don't have a good answer here. Others suggested using pulsars but that isn't foolproof. First, you don't know when you're being deposited so it could be billions of years in the past or future when all current knowledge of pulsars is no longer relevant. Second, you might be on the other side of the milky way where we can't actually see anything currently and thus have no knowledge of pulsars over there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your best bet might be to map out the star distribution of as wide a swath of the galaxy as you can see to try and understand what the spiral arms look like near you. &lt;a href=&quot;https://en.wikipedia.org/wiki/Milky_Way#Spiral_arms&quot; rel=&quot;noreferrer&quot;&gt;We've already done this for our current position in the galaxy.&lt;/a&gt; Galactic arms move and change slowly, taking hundreds of millions if not billions of years to change significantly. If you can recognize the general structure of the arms, that would give you a decent reference to your azimuthal position.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;How do we know the time?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I'm going to break this down into two possible answers. The first assumes you've been placed in a time not too distant from our current time (think less than a million years) and you're still pretty near to our current position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case above, pulsars are the way to go. Pulsars pulse with extreme regularity, so that they're great time keepers. But if your goal is to know how much time has passed, you need to look at the time rate of change of the pulses. As it turns out, the pulsars don't pulse with perfect regularity. &lt;a href=&quot;https://www.sciencedaily.com/releases/2012/10/121008091556.htm&quot; rel=&quot;noreferrer&quot;&gt;The time between pulses decays very slowly over time due to the pulsars losing angular momentum very slowly&lt;/a&gt;. The general step to calculating a reasonably precise time is as follows.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Find a pulsar that is currently known and has current measurements of the pulse rate and decay in pulse rate. To make sure you've identified a given pulsar as the correct one, you'll have to map its relation to other stars via a wide survey (the star fields will have changed, but if you're not too far away from our time you can predict with reasonable accuracy where most stars will be or have been in the past).&lt;/li&gt;&#xA;&lt;li&gt;Measure the pulse rate (and decay rate if you wish, but that may take years or even decades).&lt;/li&gt;&#xA;&lt;li&gt;Compare your currently measured pulse rate to the present day measured rate and decay rate and do some calculations. You should be able to figure out your current time with respect to the time now fairly precisely. I can't say how precise, but I wouldn't be surprised if you could get within a decade or so of the current time (depending on how accurate your pulsar models are).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The only caveat to the above approach is that pulsars can &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Glitch_(astronomy)&quot; rel=&quot;noreferrer&quot;&gt;glitch&lt;/a&gt;&quot;. Normally the pulse rate decays smoothly, but every once in a while a pulsar might just glitch and quickly change pulse rates. It is unclear exactly why these glitches occur and they're unpredictable so you have to make the assumption that your pulsar hasn't glitched between now and whatever time you're in. If it has, your time may be off wildly. To mitigate this, you should of course use as many pulsars as you can find. If one pulsar gives a different number than the rest, you can assume it glitched and throw it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second case is more tricky. Now I'm going to assume you're in a time really distant from our own. Millions or possibly billions of years away from now. In that case, any hope in finding a precise time (or even a precise year) is wildly presumptuous. At best, you can figure out your general time within the universe by measuring the current age of the universe. We currently know the age to be $13.799\pm 0.021\ \mathrm{billion\ years}$ based on the &lt;a href=&quot;https://arxiv.org/abs/1502.01589&quot; rel=&quot;noreferrer&quot;&gt;latest and greatest measurements&lt;/a&gt;. You can see our uncertainty is down to 21 million years. If you make similar measurements of the Universe's age in your time, then the best you'll be able to do is define your time with respect ours with an accuracy of ~20 million years. But if you're a billion years in the future, that's a pretty good measurement in my opinion.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-09-27T13:26:41.590" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22896" PostTypeId="1" CreationDate="2017-09-27T15:44:42.243" Score="4" ViewCount="3561" Body="&lt;p&gt;As far as I know, Earth and Mars were relatively similar when the solar system formed. So most of the CO2 was used to form carbonic rocks (right?). I found that tectonic movements are the reason that the CO2 on Earth could be released again, which don't exist on Mars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how comes that the atmosphere has 95% CO2?&lt;/p&gt;&#xA;" OwnerUserId="18799" LastEditorUserId="2365" LastEditDate="2017-09-27T16:38:24.090" LastActivityDate="2017-09-27T19:18:11.380" Title="Where does the CO2 in Mars atmosphere come from?" Tags="&lt;planetary-atmosphere&gt;&lt;mars&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22897" PostTypeId="2" ParentId="22882" CreationDate="2017-09-27T16:15:54.750" Score="1" Body="&lt;p&gt;The location part of the problem is mostly dealt with in Zephyr's answer. I completely concur, but the last part - the azimuth - is indeed tricky, but you also have to define what you mean by position in azimuth. If you are in the recent past or future then some definition would be possible based on the directions of a network of radio-loud, high-redshift quasars. Indeed, accurate measurements of their redshifts could then be combined with a cosmological model to tell you at what exact cosmic epoch your measurements were made at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, a cosmological model could be combined with a precise measurement of the temperature of the cosmic microwave background to give you your cosmic age. This would work at any cosmic epoch (assuming we have got the cosmological model correct) and be far more precise than some alternatives you could think of, like measuring the current age of the Galaxy using the luminosity of the coolest white dwarfs or the ages of stars in the oldest globular clusters (maybe accurate to a billion years).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you are billions of years into the future or past then defining the azimuthal location in any well-defined way becomes impossible. The Galaxy rotates with a not precisely known speed that varies with radius. Distant quasars come and go (or more precisely, turn on and off); local galaxies move with respect to our Galaxy with not precisely known (tangential) motions. Perhaps you could make a definition with respect to the cosmic ray dipole alignment, but this would change as the Galaxy's motion changes due to the influence of other galaxies in the local group and supercluster.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-09-27T16:15:54.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22898" PostTypeId="1" AcceptedAnswerId="22902" CreationDate="2017-09-27T16:48:34.630" Score="-1" ViewCount="166" Body="&lt;p&gt;If GWs propagate at near the speed of light, how did Earth (and its constituent elements) arrive at its current location billions of years before the GW arrived here (or light from distant stars for that matter)?&lt;/p&gt;&#xA;" OwnerUserId="10117" LastEditorUserId="10117" LastEditDate="2017-09-27T16:55:14.753" LastActivityDate="2017-09-27T19:55:27.060" Title="How did Earth get here before gravitational waves?" Tags="&lt;gravitational-waves&gt;&lt;speed&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22899" PostTypeId="2" ParentId="22898" CreationDate="2017-09-27T18:10:26.380" Score="3" Body="&lt;p&gt;Earth didn't get here before GW's. GW's have been around since the early universe. We have just recently developed the technology to detect the large gravitational waves associated with very massive disruptions in the fabric of space-time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also know that our sun is not a 1st generation star, so much of the matter used to create our sun and planets came from a much earlier super-nova explosion.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="16926" LastEditDate="2017-09-27T18:19:25.580" LastActivityDate="2017-09-27T18:19:25.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22900" PostTypeId="2" ParentId="22896" CreationDate="2017-09-27T18:16:53.410" Score="5" Body="&lt;p&gt;The early atmospheres of Venus, Earth and Mars all had a lot of CO2, Carbon and oxygen are some of the most common elements in molecular clouds, and so it is no surprise that their most common and stable compound should be common in atmospheres.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the Earth, life happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The carbon in the atmosphere was fixed by autotrophs (particularly blue-green algae and the later plants) into the bodies of living things, and then into rocks formed from those bodies: Chalk, Limestone, Coal and Oil. This left an atmosphere dominated by Nitrogen, and later by the toxic waste products from those algae: oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars was too small to hold onto lighter gases like Nitrogen, and these escaped into space, there is just a trace of the heavier gases: CO2, but at a very low pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus retains a dense atmosphere dominated by CO2&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-09-27T18:16:53.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22901" PostTypeId="2" ParentId="22896" CreationDate="2017-09-27T19:18:11.380" Score="3" Body="&lt;p&gt;In addition to James K's answer, many gases exist in the disk of material that forms into planets.   What's a gas as opposed to a liquid or solid also depends on temperature and pressure.   The most abundant &quot;gases&quot; in our solar-system are hydrogen, helium, CO2, H2O, CH4, NH3, N2, O2, CO, Neon (and maybe some others I've overlooked), in something sort of close to that order.   Many of these &quot;gases&quot; are also ices that form up the majority of comets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During planetary formation, the ring of debris is too warm for hydrogen,  Helium and Neon to be anything but a gas, though some hydrogen is bound to heavier elements and abundant in planet formation.  Helium and Neon and the other Noble Gases don't bind well with other elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other abundant &quot;gases&quot;, H20, CO2, NH3, CH4, etc, can exist as ices past their respective &lt;a href=&quot;https://en.wikipedia.org/wiki/Frost_line_(astrophysics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;frost lines&lt;/a&gt;.   That's why those molecular compounds are much more common in the outer part of the solar system.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth and Mars formed inside the frost-line so they formed mostly out of rocky material, after which, much of the lighter elements, primarily gases and liquids, were acquired by comet impacts after formation, though some probably existed prior too formation.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the late heavy bombardment there were many large impacts and one consequence of very large comet or meteor impacts is a rise in temperature, so, even in the early solar-system when the sun wasn't as luminous as it is now, the planets, Earth and Mars spent some of the time quite hot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's two primary ways a planet can lose it's atmosphere, &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_escape&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jeans Escape&lt;/a&gt; and by the solar wind.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solar wind is made up of almost entirely charged particles, so if a planet has a magnetic field, the impact of those high speed particles is largely deflected, where as, if they impact the upper atmosphere, the planet can lose it's atmosphere like tiny billiard balls being knocked off one at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the method of those two ways a planet can lose it's atmosphere are different, it comes down to basically the same thing.   When gas molecules on the outer edge of a planet's atmosphere move faster than escape velocity they are likely to escape the planet into space and because lighter gas molecules move faster than heavier ones.   That's the &lt;a href=&quot;https://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_distribution&quot; rel=&quot;nofollow noreferrer&quot;&gt;Maxwell-Boltzmann law&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/Root-mean-square_speed&quot; rel=&quot;nofollow noreferrer&quot;&gt;Root-mean-square formula&lt;/a&gt;, planets are more likely to retain heavier gases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus and Mars have both lose most of their lighter gases, H20, CH4, NH3 but CO2 is heavy enough to have been retained by both those planets, though it's worth pointing out that we don't know how much CO2 Mars had millions and billions of years ago.   Mars may have lost much of it's CO2 over time as well.  It just lost it more slowly than the lighter gases.   Mars was able to retain some of it's water, however, in the form of ice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus, like Mars, lost nearly all of it's gaseous water.   We know that Venus used to have much more water because it's very high &lt;a href=&quot;http://www.nature.com/nature/journal/v363/n6428/abs/363428a0.html?foxtrotcallback=true&quot; rel=&quot;nofollow noreferrer&quot;&gt;D to H ratio&lt;/a&gt; wouldn't be possible unless it had lost a significant percentage of it's water, likely 99.9% or higher. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth is massive enough and Earth has a strong magnetic field, so Earth is able to retain it's lighter gases like CH4, NH3 and atmospheric H20, though Earth loses Hydrogen and Helium to space.    NH3 is kind of interesting because it dissolves in water quite readily, so as soon as the young Earth had oceans, those oceans were likely Ammonia/water with whatever else readily dissolved in the liquid such as Iron.   The early oceans are thought to have been brown colored from the dissolved Iron.  But Earth's young atmosphere was likely mostly CO2 and CH4 with most of the H20 in liquid form, (and at times, much of it was ice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As life and photosynthesis began releasing Oxygen, Oxygen readily bonded with the CH4 and dissolved Iron in the oceans, turning the oceans blue and in time, making the Atmosphere free of CH4 and more abundant with O2.   It's unclear (and perhaps unlikely?) if Mars and Venus ever underwent that Oxygenation period that Earth did.   Having abundant liquid oceans and having photosynthetic life made a significant difference to Earth's atmosphere.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mentioned plate tectonics and that plays a role too as does the chemistry that happens in the oceans and below the surface of a planet.    The main reason Mars is 95% CO2 now is because it's small and it doesn't have a good magnetic field, only small localized ones.    It lost most of it's atmosphere and nearly all of it's lighter gas molecules.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-09-27T19:18:11.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22902" PostTypeId="2" ParentId="22898" CreationDate="2017-09-27T19:55:27.060" Score="4" Body="&lt;p&gt;Let's look at &lt;a href=&quot;https://en.wikipedia.org/wiki/First_observation_of_gravitational_waves&quot; rel=&quot;nofollow noreferrer&quot;&gt;GW150914&lt;/a&gt;, the first directly detected gravitational wave event. &lt;a href=&quot;https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.116.061102&quot; rel=&quot;nofollow noreferrer&quot;&gt;The source was a binary black hole, with a calculated luminosity distance of $410^{+160}_{-180}\text{ Mpc}$&lt;/a&gt; - roughly 1.3 billion light-years, give or take. This means that the signal was produced about 1.3 billion years ago. The system itself is much older, of course; the progenitor stars formed &lt;a href=&quot;http://www.nature.com/nature/journal/v534/n7608/full/nature18322.html?foxtrotcallback=true&quot; rel=&quot;nofollow noreferrer&quot;&gt;about 10 billion years before the merger&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth didn't come onto the scene until roughly 4.5 billion years ago. At that point, the binary black holes had yet to merge together; the system was producing gravitational waves, but nothing as strong as the ones detected by LIGO. The gravitational waves we detected were produced 3 billion years &lt;em&gt;after&lt;/em&gt; the Earth was formed - and it took them 1.3 billion more to reach Earth!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is, the signal LIGO detected was created after Earth was formed, and created far away. They did not start out at the same time Earth did.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2017-09-27T19:55:27.060" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22903" PostTypeId="1" CreationDate="2017-09-27T23:04:23.867" Score="11" ViewCount="398" Body="&lt;p&gt;The recent &lt;a href=&quot;https://www.ligo.caltech.edu/news/ligo20170927&quot; rel=&quot;noreferrer&quot;&gt;4th discovery of gravitational waves&lt;/a&gt; where the black hole masses were 31 and 25 times the mass of the sun, and they released 3 solar masses of energy as gravitational waves, made me wonder what would it feel like to be orbiting this binary black hole system on a planet at 1AU when the merger happened? &lt;/p&gt;&#xA;" OwnerUserId="6870" LastActivityDate="2019-09-02T01:03:04.190" Title="What would it feel like to be orbiting near a merging black hole?" Tags="&lt;black-hole&gt;&lt;gravitational-waves&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22917" PostTypeId="1" AcceptedAnswerId="22934" CreationDate="2017-09-29T00:54:45.450" Score="2" ViewCount="160" Body="&lt;p&gt;I read that the stars in the center of globular clusters are incredibly close to one another and that these stars can be spaced 100 to 1000 times closer than the stars near the sun. Because these stars are so closely spaced, would their heliospheres interact with each other? There are binary stars whose coronas interact and create powerful x-rays. Would something like that happen with those central globular cluster stars, but on a much larger scale, what with the larger amount of stars involved?&lt;/p&gt;&#xA;" OwnerUserId="18831" LastActivityDate="2017-09-30T14:48:35.637" Title="Do the heliospheres of stars at the center of globular clusters interact?" Tags="&lt;globular-clusters&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22920" PostTypeId="1" AcceptedAnswerId="22926" CreationDate="2017-09-29T06:56:50.980" Score="4" ViewCount="157" Body="&lt;p&gt;Just curious I like to know if one fine day our Sun turns into a white dwarf at the later stage of its life, would the magnetic bubble, heliosphere gets thinner and eventually disappear? also at this moment is it changing in size?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastActivityDate="2017-09-29T14:18:23.990" Title="Would the heliosphere disappears if our sun becomes a white dwarf?" Tags="&lt;the-sun&gt;&lt;magnetic-field&gt;&lt;white-dwarf&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22921" PostTypeId="1" CreationDate="2017-09-29T08:05:12.043" Score="4" ViewCount="189" Body="&lt;p&gt;From what I can understand, &lt;em&gt;Cosmic Microwave Background Radiation&lt;/em&gt; (CMBR) is microwave now, but was initially visible light emitted long time ago.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The microwave is producing a wave that’s about 12 centimeters long. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now CMBR has right now this (time changing) distribution (&lt;a href=&quot;http://www.astro.ucla.edu/~wright/CMB.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;peak at 2mm&lt;/a&gt;):&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Cmbr.svg#/media/File:Cmbr.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/c/cd/Cmbr.svg/1200px-Cmbr.svg.png&quot; alt=&quot;Cmbr.svg&quot;&gt;&lt;/a&gt;&lt;br&gt;By &lt;a href=&quot;//commons.wikimedia.org/wiki/User_talk:Quantum_Doughnut&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;User talk:Quantum Doughnut&quot;&gt;Quantum Doughnut&lt;/a&gt; - Own work&#xA;&lt;a href=&quot;//commons.wikimedia.org/wiki/File:Lissajous_small.svg&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;File:Lissajous small.svg&quot;&gt;&lt;/a&gt;&#xA;This diagram was created with gnuplot., Public Domain, &lt;a href=&quot;https://commons.wikimedia.org/w/index.php?curid=12958270&quot; rel=&quot;nofollow noreferrer&quot;&gt;Link&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it looks to me that in some time, when its wavelength falls at around 12cm), some of its energy should be transferred to water molecules the way it is in a microwave oven. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is it the case? Will some of the CMBR energy actually transferred to water molecules? &lt;/li&gt;&#xA;&lt;li&gt;Will this energy be too faint to matter at our scale?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11264" LastEditorUserId="11264" LastEditDate="2017-09-29T09:51:30.807" LastActivityDate="2017-11-02T17:06:18.230" Title="Will universe's lakes get boiled by the CMBR?" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22926" PostTypeId="2" ParentId="22920" CreationDate="2017-09-29T14:18:23.990" Score="3" Body="&lt;p&gt;Yes the heliosphere will disappear. As you know, the heliosphere exists due to the solar wind, which consists of charged particles that follow the magnetic field lines starting from the Sun and extending far into the universe (of course at some point they return to the Sun, as the loops need to be closed, but it's so far away that we usually assume it to be at infinity). Now when the Sun goes Red Giant and then loses its outer shells (planetary nebula), it will become a white dwarf. Then there won't be any wind originating from its surface and therefore there won't be any &quot;magnetic bubble&quot; left.&#xA;The heliosphere stays roughly the same in size as long as the Sun remains in its Main Sequence stage, as a star remains relatively stable during this time of its life.&lt;/p&gt;&#xA;" OwnerUserId="17425" LastActivityDate="2017-09-29T14:18:23.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22929" PostTypeId="1" AcceptedAnswerId="22933" CreationDate="2017-09-29T23:50:33.577" Score="2" ViewCount="319" Body="&lt;p&gt;Would any moons/rings around either of the gas giants stay intact or get ripped apart by the tidal forces between the two giants? If any moons/rings are able to form around either of the giants, is there a limit to how large a ring system could be, or how far a moon could orbit?&lt;/p&gt;&#xA;" OwnerUserId="18831" LastActivityDate="2017-09-30T07:46:41.513" Title="Could binary gas giants have ring and moon systems?" Tags="&lt;natural-satellites&gt;&lt;gas-giants&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22930" PostTypeId="1" AcceptedAnswerId="22932" CreationDate="2017-09-30T00:35:09.407" Score="2" ViewCount="776" Body="&lt;p&gt;Let's say you had a binary planet system of two earth-sized worlds with about the same mass. Could a stable moon orbit around the barycenter of both planets? If so, is there a limit to the mass of such a moon? Would it have to orbit in a Lagrangian point?&lt;/p&gt;&#xA;" OwnerUserId="18831" LastActivityDate="2017-09-30T07:12:06.717" Title="Could a moon orbit two planets at once?" Tags="&lt;natural-satellites&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22931" PostTypeId="2" ParentId="22930" CreationDate="2017-09-30T06:37:46.880" Score="5" Body="&lt;p&gt;Yes, but.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Circumbinary configurations can be stable over long time periods, but in general they are a less stable configuration than &quot;two moons orbiting one planet&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To increase the likelihood that the configuration will be stable, the two planets should be in a tight orbit (which can be unstable itself) and the moon should be distant and small. To get the moon to be distant enough the whole system should be distant from the sun.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-09-30T06:37:46.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22932" PostTypeId="2" ParentId="22930" CreationDate="2017-09-30T07:12:06.717" Score="6" Body="&lt;p&gt;The Pluto-Charon system is similar to this.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QbXxv.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QbXxv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's unlikely that you'd have two planets of nearly the same mass because some variation is far more common, but a moon orbiting a binary planet system is certainly possible.   It's easier if it's not that close to the sun, for the same reason that it's difficult for Mercury and Venus to have moons.   A larger hill-sphere and less solar gravitation would make your system more stable in the long term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the Moon couldn't be in a Lagrange point in your system because, for starters, L1, 2 and 3 are never stable and the potentially stable Lagrange points, L4 and L5 require a large central planet and a significantly less massive object, (less than 1/26th the mass) orbiting the central planet.   The L4 and L5 points are to the orbiting object.  That doesn't work in a binary system of 2 objects of similar mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any working Lagrange point scenario you'd have two moons and one planet, not binary planets and one moon.    You could, in theory have two moons at Lagrange points to each other orbiting a binary planet system but that would be adding even more improbabilities, but theoretically possible and stable for some time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XP2TT.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XP2TT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-09-30T07:12:06.717" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22933" PostTypeId="2" ParentId="22929" CreationDate="2017-09-30T07:46:41.513" Score="3" Body="&lt;p&gt;It would depend how close the gas giants were too each other, but in theory it would be possible.   Similar to the binary star system (image below), provided you ignore the AUs and scale this down from stars to planets, the same principal applies to orbits around two gas giants orbiting each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PWDRw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PWDRw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.jpl.nasa.gov/spaceimages/details.php?id=PIA09227&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They would have relatively small hill spheres depending on how close they were and corresponding, even smaller &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere#True_region_of_stability&quot; rel=&quot;nofollow noreferrer&quot;&gt;true regions of stability.&lt;/a&gt; But planetary ring systems tend to be quite close to the planet anyway and larger ring systems need to be almost exclusively inside the Roche Limit.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The binary gas giant orbits would have to be outside each other's respective Roche Limits anyway and provided there was sufficient distance between then, like maybe 4-6 times the larger of the two Roche limits (ballpark guess), then it's possible to have both gas giant planets with their own ring systems.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's not possible is to have two orbiting disks around two bodies where the orbiting disks intersect each other, like they had in the &lt;a href=&quot;http://ncc-1031.com/news/battle-at-the-binary-stars-is-2nd-part-of-star-trek-discovery-premiere/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Startrek Discovery premier episode&lt;/a&gt;.   And that was forming star systems, not gas giants, but it's still impossible for any orbiting ring or debris disk to intersect the way they did it.   It was a cool looking image, but clearly impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pbfZ1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pbfZ1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also want to point out that a ring system around both planets like the top of the first image with the writing &quot;disk around both stars&quot;, that's not possible with a typical ring system.  Ring systems need to be in tight orbits around their planet, though faint/disburse ring systems like &lt;a href=&quot;https://www.jpl.nasa.gov/spaceimages/details.php?id=PIA03550&quot; rel=&quot;nofollow noreferrer&quot;&gt;Saturn's E ring&lt;/a&gt; can exist further out.   The further out a ring system, the more likely it would be disturbed by the binary planet. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-09-30T07:46:41.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22934" PostTypeId="2" ParentId="22917" CreationDate="2017-09-30T14:48:35.637" Score="2" Body="&lt;p&gt;Similar to what @Barrycenter said in his comment, There's a wide range of distances between 100 or 1,000 times closer on average and as close as binary stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alpha Centauri A and B get &lt;a href=&quot;https://en.wikipedia.org/wiki/Alpha_Centauri&quot; rel=&quot;nofollow noreferrer&quot;&gt;as close as 11.2 AU&lt;/a&gt;.   The Sun and Alpha Centauri are roughly 270,000 AU apart so at 1,000 times closer, the close stars in a globular closer would still be (ballpark), 250-300 AU from their nearest neighbors.  In a crowded ball of stars you're not likely to get binary systems that far apart, so most of the interaction that you're looking for would be in close binary systems.   Neighboring stars in a globular cluster would mostly be too far apart to have much interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also a world of difference between the corona and the heliosphere.  The Sun's corona extends about &lt;a href=&quot;https://www.nasa.gov/content/goddard/nasas-stereo-maps-much-larger-solar-atmosphere-than-previously-observed/&quot; rel=&quot;nofollow noreferrer&quot;&gt;12 solar radii&lt;/a&gt; or about .05 AU.   The heliosphere, while, it's one, not a sphere and two, probably has some variation in it's boundary, extends roughly 100 AU (&lt;a href=&quot;https://astroquizzical.com/astroquizzical/is-our-heliosphere-touching-the-astrosphere-of&quot; rel=&quot;nofollow noreferrer&quot;&gt;121 by this estimate&lt;/a&gt;), over 2,000 times further than the corona.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/garCu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/garCu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Binary systems like Alpha Centauri A &amp;amp; B probably have a shared heliosphere and/or some form of heliopause between them where their two solar winds meet.   It would be interesting to study that in more detail.  I'm not sure how much is known about the heliopause between two binary stars, especially given that they orbit each other.   I'm not sure what form it would take - perhaps that would make a good question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Binary stars need to be much closer than 11.2 AU to have corona to corona interaction.  Stars that close are probably quite rare but they do happen.   Here's one example of a binary star system so close &lt;a href=&quot;https://www.universetoday.com/132763/exploding-binary-stars-will-light-sky-2022/&quot; rel=&quot;nofollow noreferrer&quot;&gt;they're actually thought to be touching&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-09-30T14:48:35.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22935" PostTypeId="1" CreationDate="2017-09-30T19:59:35.807" Score="0" ViewCount="188" Body="&lt;p&gt;First premise: I have already made this question in the worldbuilding section, but since it is a very specific question many have recommended me to write in the astronomy section so here we are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second premise: I'm Italian so please excuse any mistake&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'm writing a story about a fast and cataclysmic climate change. Obviously there are some artistic licenses but I would like nothing in my story to be completely impossible, only (at most) very unlikely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The climate change that I would like to include in my story is very extreme, so I had to discard any terrestrial forcing to cause it (no change of ocean currents, supervolcanoes and methane eruptions) because they could not act fast enough and would not be so catastrophic (compared to what I imagine). So I focused on astronomical forcing. I excluded the impossibly forcing (for example, a magical and radical reduction in solar luminosity) and those that would be more catastrophic than the climate change itself (i.e. a planetary impact).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last possible (though very very unlikely) option I have left is a dramatic change in the Earth's orbit. I have some knowledge in physics and astronomy so I did a personal search to figure out what could cause such an orbital change.The most plausible option would be to pass a rogue planet (or star) into the solar system, but since I want to complicate my life, I would like the orbital change to happen without any notice, which is quite impossible if it is the passage of a planet or a star in the solar system to cause it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only plausible explanations I have found to fully satisfy my scenario are either the passage of a primordial black hole in the solar system (a black hole with Jupiter's mass would have a schwarzschild radius of 2.8 meters and, if it did not come in direct contact with some celestial body, it may take in the solar system without notice) or passing a high-density cluster of dark matter (assuming an self-interacting form of dark matter) near or inside the solar system. Both are immensely unlikely options and very speculative (primordial black holes are hypothetical objects and we still have no idea what the dark matter really is and how it behaves) but I do not think they are impossible in the beginning, or is it wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming I want to move the Earth 2x10^7 km (if I do not mistake at this distance the average temperature of the planet would be between 15 and 18 ° C lower than now) farther from the sun, could this change take place in less than a decade and the earth's orbit would remain stable or simply the planet would be expelled from the solar system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also wondered what effects this sudden and drastic lowering of the Earth's temperature would have on the weather (especially considering that the poles would cool down much faster than the equator) but I think it's a pretty OT question so I leave it as optional&lt;/p&gt;&#xA;" OwnerUserId="18848" LastActivityDate="2017-10-31T09:50:05.780" Title="The consequences and the mechanisms of a shift of the Earth away from the sun" Tags="&lt;orbit&gt;&lt;black-hole&gt;&lt;astrophysics&gt;&lt;planetary-atmosphere&gt;&lt;dark-matter&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22936" PostTypeId="1" CreationDate="2017-10-01T00:33:59.940" Score="-1" ViewCount="157" Body="&lt;p&gt;I would like to know please whether or not I can pick up alien radio signals from my backyard.&lt;/p&gt;&#xA;" OwnerUserId="18419" LastEditorUserId="2365" LastEditDate="2017-10-01T12:50:32.487" LastActivityDate="2017-10-01T12:57:57.617" Title="Can I pick up alien radio signals from my backyard?" Tags="&lt;seti&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2017-10-07T18:08:10.887" ContentLicense="CC BY-SA 3.0" />
  <row Id="22939" PostTypeId="2" ParentId="22935" CreationDate="2017-10-01T09:00:49.520" Score="3" Body="&lt;p&gt;It sounds more like world-building than astronomy to me, but since they punted it here, I'll give a crack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a few ways you could approach this and stay within good science but making change relatively quick is difficult.  This might be possible if the planet is in a situation like mars, with no moon and between two more massive planets, so it's more vulnerable to orbital perturbations.   Mars axial tilt is believed to have changed quite a bit and fairly quickly.   While the total solar energy is constant, a variation in axial tilt can lead to a variation in ice buildup where, generally speaking a lower tilt is better for ice buildup, as the higher tilt leads to warmer summers and that melts ice more easily.   Mars axial tilt &lt;a href=&quot;http://phoenix.lpl.arizona.edu/mars102.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;varies from 13 to 40 degrees&lt;/a&gt;, but that variation can take hundreds of thousands or millions of years, which is probably longer than you're looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nearly any orbital change, even a rapid one would probably take tens of thousands of years minimum.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geological changes are probably a little faster.  While it's still in the direction of theoretical, and I think you're right to avoid a large volcano or something like that because the effect from the volcanic ash would be greater than any long term effect from climate change.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;About 3 million years ago, the formation of the Isthmus of Panama is thought to have significantly changed ocean currents that cooled the Earth by a few degrees that lead to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Pleistocene&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pleistocene epoch&lt;/a&gt; and ice age cycle.   You'd need to get creative to figure out a way for a land bridge like that to make a radical change, but it's possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method, if the oceans are shallow on your planet, you could have a valley like the &lt;a href=&quot;https://phys.org/news/2009-12-mediterranean-sea-years.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mediterranean valley about 5.3-5.6 million years ago&lt;/a&gt;, with a natural land bridge keeping the ocean water out.   That would be a hot, dry, sandy valley.   Then when the land bridge gives way the ocean quickly flows in.   If your valley is big enough the oceans could drop several feet and if you have land or small islands near the poles, the falling ocean level could increase polar land which could trigger ice formation.   Glaciers don't form over water, only land.   This is a bit of a stretch cause you'd need two geographical oddities, very shallow oceans over the poles and a very large dry valley ready to be flooded, but it could turn a planet towards a major ice age.   Ice tends to keep growing once it starts.   It's a natural feedback mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the best method would be a passing rogue planet giving your planet a gravity assist.  You'd want a larger rogue planet and it would need to pass fairly close to your planet to tug your planet into a longer orbit.  You'd probably want a gas giant mass and as it passed, it would probably need to be visibly larger than our moon in the night sky.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mentioned 20 million km, or a 13% increase in distance and about a 22% drop in solar energy received.  You wouldn't need that much because of feedback mechanisms.   Ice ages can be triggered by as small as a 1% difference and for radical, rapid climate change, pushing your planet just 4-6% further away should be enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'd need to make your planet's year longer, using Kepler's 3rd law, you'd use the 3/2 power rule, so 4% further would make the planet's year 6% longer or an additional 22 days to a 365 day year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to how fast that would make the planet cold.  That's hard to say.   You'd probably notice longer colder winters and frosting pretty much immediately, but for ice buildup and enormous glaciers you're still probably looking at at least a few hundred years, but that's kind of a ballpark guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could do the same thing with a theoretical primordial black hole but they've kind of fallen out of fashion since it's believed they don't exist.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to get creative you could have the rogue planet pass in-between say your earth like planet and a mars like planet, pulling the Earth orbit more eccentric and longer and the mars planet shorter, possibly leading to the two planets having crossing or nearly crossing orbits - which could lead to all kinds of scenarios and predictions like a possible planetary collision in 6,000 years or something or a near pass and enormous tides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would also stir up the asteroids and comets in our system likely leading to an increase in impacts following the pass for the next 100 or more years.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rogue planet would create tides too if it appeared larger than the moon in the sky and maybe an earthquake or two, but wouldn't need to be a global disaster.    Rogue planets passing that close would be very rare, but not impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;rftwqa&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-10-01T09:00:49.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22940" PostTypeId="1" CreationDate="2017-10-01T11:12:57.070" Score="2" ViewCount="55" Body="&lt;p&gt;So I was looking at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;pp-chain&lt;/a&gt; reactions that take place inside stars in a bit more detail.&lt;br&gt;&#xA;I got confused about the massive differences in reaction efficiencies concerning two reactions $^{3}_{2}He$.&lt;br&gt;&#xA;In the lecture notes I've googled (the ones that I've understood best so far), the reaction&lt;br&gt;&#xA;$$^{3}_{2}He + ^{3}_{2}He \rightarrow p + p + ^{4}_{2}He$$&#xA;has an efficiency factor $S_1 \approx 6MeV-barn$ and being quoted as 'strong reaction'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the other reaction I'm confused about in contrast is &#xA;$$^{3}_{2}He + ^{4}_{2}He \rightarrow ^{7}_{4}Be + \gamma$$&#xA;with a way lower reaction probability of $S_2 \approx 4\cdot 10^{-4} Mev-barn$ and being quoted as 'electromagnetic interaction'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So my question&lt;/strong&gt;: How can we understand the huge discrepancy between $S_1$ and $S_2$? Particularly, when is a reaction a 'strong' interaction?&lt;br&gt;&#xA;Isn't the strong force necessarily always involved, else there wouldn't be a potential minimum in the nucleus to tunnel to...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The S-Factor is just a part of the velocity dependent reaction cross-section $\sigma(v)$ so that $\sigma(E) = \frac{exp(-\tau)}{E} S(E)$. Here $\tau$ is the tunneling probability from Quantum theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also appreciate any good textbook reference, as so far I'm working with google and random lecture notes I'm finding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;br&gt;&#xA;Thoughts on this:&lt;br&gt;&#xA;In both cases particles must tunnel towards each other first. Also we know that $^{4}_{2}He$ is a bound state for the nuclear force.&lt;br&gt;&#xA;So what might be happening is that in $^{3}_{2}He + ^{3}_{2}He$ the nuclei rearrange themselves into the energetically favoured bound state, kicking out the surplus protons.&lt;br&gt;&#xA;In the case of $^{3}_{2}He + ^{4}_{2}He$ the bound state already exists, and thus in most cases the reaction will be reversed. Only a few of those reactions (why?) allow the spontaneous formation of a higher-shell object like $ ^{7}_{4}Be$, thus the much lower reaction rate.&lt;br&gt;&#xA;If this story so far makes sense, then I still don't understand when and why  $^{7}_{4}Be$ is formed, I'd be grateful for explanation.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditDate="2017-10-02T10:46:35.313" LastActivityDate="2017-10-02T10:46:35.313" Title="pp-Chain reactions involving $^{3}_{2}He$ - differences in S-factor" Tags="&lt;stellar-evolution&gt;&lt;nucleosynthesis&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22941" PostTypeId="2" ParentId="22936" CreationDate="2017-10-01T12:57:57.617" Score="4" Body="&lt;p&gt;This is like asking &quot;can I spot polar bears from my back garden?&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is nothing that would stop a determined polar bear from coming into my garden, so it is not impossible. But unless you live in Churchill, its probably not going to happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get a radio telescope, and it would be able to receive a sufficiently powerful signal. But detecting alien signals is hard: so far we haven't detected any. Even with big detectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you detect aliens: Yes.  Will you detect aliens: Probably not.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-10-01T12:57:57.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22942" PostTypeId="1" AcceptedAnswerId="22957" CreationDate="2017-10-01T13:13:35.953" Score="1" ViewCount="511" Body="&lt;p&gt;Why do we always think of the universe as expanding? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely, from our perspective an expanding universe would look exactly the same as a universe in which everything is shrinking away from everything else? &lt;em&gt;&quot;Oh, but the red shift!&quot;&lt;/em&gt; I hear people exclaim. Well how about this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The speed of light is simply slowing down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hear me out. If the speed of light is a measuring stick against which all things are measured, and if the speed of light is slowing down, then it means everything shrinks and high frequency waves generated in the past are observed with a lower frequency today, which causes us to make observations that the universe is expanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is wrong with my alternative?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this, if the speed of light were experiencing, say, exponential decay, wouldn't we &lt;strong&gt;expect&lt;/strong&gt; to see the universe's 'expansion' accelerating? (which I thought was a relatively new and profound discovery)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Apologies if the question is silly, I've never made a formal study of physics and the universe, but this has always baffled me and I haven't been able to find an explanation&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="18853" LastEditorUserId="18853" LastEditDate="2017-10-01T13:21:34.557" LastActivityDate="2017-10-02T17:23:06.020" Title="Is an expanding universe equivalent to a universe in which everything is shrinking?" Tags="&lt;universe&gt;&lt;light&gt;&lt;expansion&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22943" PostTypeId="2" ParentId="13488" CreationDate="2017-10-01T16:57:56.733" Score="7" Body="&lt;p&gt;To add to the excellent answer by barrycarter, there are 2 planetarium-like codes, that I know of, that run on a mac and would make excellent tools for viewing certain astronomical objects. The codes are &lt;a href=&quot;http://stellarium.org/&quot; rel=&quot;noreferrer&quot;&gt;Stellarium&lt;/a&gt; and &lt;a href=&quot;https://celestia.space/&quot; rel=&quot;noreferrer&quot;&gt;Celestia&lt;/a&gt;. Both turn your computer into your own planetarium where you can search and view objects in space.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorDisplayName="user21" LastEditDate="2019-02-18T22:52:19.810" LastActivityDate="2019-02-18T22:52:19.810" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22945" PostTypeId="2" ParentId="21653" CreationDate="2017-10-01T22:34:50.753" Score="-2" Body="&lt;p&gt;As I recall, when Theia collided with Earth it created oceans, and the modern atmosphere, which led to life. As a side-effect, a ring was created around the Earth, with colliding rocks forming two moons. That's right; Earth had two moons 4 billion years ago. Eventually, they collided to form the modern moon.&lt;/p&gt;&#xA;" OwnerUserId="18855" LastActivityDate="2017-10-01T22:34:50.753" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22946" PostTypeId="1" CreationDate="2017-10-01T22:59:33.920" Score="1" ViewCount="199" Body="&lt;p&gt;The book “Foundations of Astrophysics” by Ryden and Peterson states that one measures azimuth in the horizontal coordinate system eastwards starting on the northernmost point of the horizon circle. Thus, east becomes 90 degrees, et cetera. Wikipedia and other sites back this up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it seems as if the position of the vernal equinox is always given relative to the southern part of an observer’s celestial meridian. During the vernal equinox, the sun culminates in the south, setting the siderial time to 0h0min. Knowing the difference between the local siderial time and a star’s right ascension gives you the current position of a star, relative to the southern half of the celestial meridian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This website describes this too: &lt;a href=&quot;http://www.polaris.iastate.edu/NorthStar/Unit4/unit4_sub2.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.polaris.iastate.edu/NorthStar/Unit4/unit4_sub2.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did I miss something? We so explicitly defined azimuth as being relative to the most northern point on the meridian. Now we’re doing all our measurments relative to the south. I’m not complaining: stars culminate in the south, so this makes it easier. But where did we switch? And when must I switch? Did I make a mistake somewhere?&lt;/p&gt;&#xA;" OwnerUserId="18857" LastActivityDate="2018-01-10T23:18:58.953" Title="How to deal with the celestial meridian?" Tags="&lt;coordinate&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22947" PostTypeId="2" ParentId="11037" CreationDate="2017-10-02T00:19:39.913" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;... Sun's reflection from southern Atlantic Ocean looks very charming. But does it really happen like that?&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Exactly as shown in your Screenshot?, no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a large reflective surface, like a calm ocean, you get a 'virtual image' where the Sun appears reflected from a point underwater; but the wavyness and imperfections make the reflection diffused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yJisn.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yJisn.gif&quot; alt=&quot;Sun reflecting off of the Earth&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.physicsclassroom.com/class/refln/Lesson-4/Reflection-and-Image-Formation-for-Convex-Mirrors&quot; rel=&quot;noreferrer&quot;&gt;http://www.physicsclassroom.com/class/refln/Lesson-4/Reflection-and-Image-Formation-for-Convex-Mirrors&lt;/a&gt; .&lt;/p&gt;&#xA;" OwnerUserId="18054" LastActivityDate="2017-10-02T00:19:39.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22949" PostTypeId="2" ParentId="22942" CreationDate="2017-10-02T05:08:49.847" Score="1" Body="&lt;p&gt;This hypothesis has far too many contradictions with our current physical models. I'll list a few:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How would you actually know if the galaxies are shrinking, or if they are moving further apart and the view is just zooming out to keep them in frame?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We would see how it agrees with our modern theories. Relativity has matched our measurements with extreme accuracy, so let's just analyze why your hypotheses would go against its basic principles:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Oh, but the red shift!&quot; I hear people exclaim.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;General relativity predicted that our universe was dynamic a decade before the first observations of redshift confirmed it. A static universe is a physical impossibility; anything, from the tiniest thermal fluctuations, would send it perpetually collapsing or expanding.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The speed of light is simply slowing down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This would violate the second postulate of special relativity, which states the speed of light is always $c$. A massless particle cannot change its speed.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2017-10-02T05:08:49.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22957" PostTypeId="2" ParentId="22942" CreationDate="2017-10-02T17:23:06.020" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Surely, from our perspective an expanding universe would look exactly the same as a universe in which everything is shrinking away from everything else?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is false. If everything was shrinking away from everything else, that would imply that the recessional rate of distance we observe due to this shrinkage would be independent of distances. Afterall, all galaxies are shrinking at the same rate so the rate they &quot;recede&quot; is only dependent upon their shrinkage rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we observer that the more distant an object is, the faster it recedes from us. This is the premise of Hubble's law and is only explained if we assume the universe is expanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, there are many objects in our visible universe which are receding from us faster than the speed of light. This is only possible because these objects are not moving through space, but rather space itself is expanding. In your scenario, these objects are shrinking in space (I presume) and there's no possible way they could shrink fast enough to appear to be receding faster than the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Oh, but the red shift!&quot; I hear people exclaim. Well how about this:&#xA;  &lt;br&gt; &lt;br&gt;The speed of light is simply slowing down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There have been people who have tried to suggest that the universal constants (e.g., the speed of light, the mass of an electron, etc.) are not actually constant and instead time-varying (albeit slowly). No such theory has ever been successful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again though, your idea suffers from a few problems, even if you assume it to be true. There are multiple kinds of redshift. Redshift from receding objects occurs because space itself is expanding and causing the wavelengths to expand as a result - otherwise known as cosmological redshift. In your scenario, the redshift would be caused by the (faster-than-light) recession itself (since the galaxies are physically moving away from us), also known as doppler redshift. You then complicate the problem by making the value of the speed of light change over time. This would cause light, at the moment of emission, to be redshifted both by the doppler effect and by the slowing of $c$. As the light travelled, the time-varying of $c$ would have no effect on the wavelength. In this scenario we would have long ago realized that moving objects get a double redshift, namely the doppler effect and the time-varying $c$ effect. We just don't see that.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the speed of light is a measuring stick against which all things are measured, and if the speed of light is slowing down, then it means everything shrinks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those two ideas don't relate at all. There's nothing suggesting the speed of light slowing down implies everything shrinks.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;high frequency waves generated in the past are observed with a lower frequency today&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As explained above, your proposal would cause the redshift to occur at the time of transmission, not en route. So there wouldn't be &quot;high frequency waves generated in the past&quot; They would have been emitted as a lower frequency. Which itself is troubling because how then do you explain the CMB changing frequency over time?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is wrong with my alternative?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Aside from the above mentioned points, a huge number of things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;All the physical constants of nature are tied together. You can't just change one and hope that the others don't notice. If the speed of light is constantly decreasing that has drastic implications throughout all of physics. For example, if the speed of light changes, suddenly the electromagnetic force strength changes, atoms aren't stable, and the universe explodes! What is your proposal for stopping that occurrence?&lt;/li&gt;&#xA;&lt;li&gt;Just how exactly are things shrinking? Are the atoms getting closing together? Are the atoms themselves shrinking? At some point you're going to hit a limit and your object becomes a black hole. Are you really predicting that eventually the universe will be nothing but black holes? My guess is that if you did such a calculation, you'd find the time it took to reach that point is about a second, but the Universe has been around for a few billion years.&lt;/li&gt;&#xA;&lt;li&gt;What caused this initial shrinkage? Motion requires energy input. Where did this energy come from? How did it apply itself so uniquely as to cause every individual thing to shrink into itself and only itself?&lt;/li&gt;&#xA;&lt;li&gt;How did the universe get to be so distributed? Your proposal basically implies the Big Bang didn't happen since you attempt to explain things the Big Bang theory already explains. If your explanation differs and you believe yours is the correct one, then the Big Bang Theory cannot be correct. If that is the case, then how did the universe get into its current state if not the Big Bang?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm sure I could go on and on, but hopefully you can see there are more than a few holes in your idea. No offense, but it takes years and years of studying to contribute to the forefronts of science. Ground-breaking ideas aren't going to be had by someone who read about a few concepts from Wikipedia or A Brief History of Time and decided they could come up with a better theory. That's the academic equivalent to reading about the rules of (American) football and deciding you can play in the NFL.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-10-02T17:23:06.020" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22959" PostTypeId="1" AcceptedAnswerId="22989" CreationDate="2017-10-02T19:49:27.047" Score="4" ViewCount="214" Body="&lt;p&gt;If two brown dwarfs got close enough to exchange gas with each other, could it trigger nuclear fusion within their cores? Is it possible for two brown dwarfs to orbit so closely and become a rocheworld in the first place?&lt;/p&gt;&#xA;" OwnerUserId="18831" LastActivityDate="2017-10-04T05:33:04.830" Title="If two binary brown dwarfs began to exchange gas with each other, would they both ignite into stars?" Tags="&lt;star-formation&gt;&lt;brown-dwarf&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22960" PostTypeId="1" AcceptedAnswerId="22964" CreationDate="2017-10-02T20:38:36.543" Score="4" ViewCount="709" Body="&lt;p&gt;There's a theory that the reason Mercury has such an enormous iron core is that it was once a much larger planet before it got impacted by an object, resulting in most of its mass getting blasted away. If this theory is true, how large was Mercury before the impact? Could it have been the size of Earth or Mars, or maybe even a super-earth?&lt;/p&gt;&#xA;" OwnerUserId="18831" LastActivityDate="2020-03-21T18:34:08.497" Title="How large was Mercury before it shrunk?" Tags="&lt;planet&gt;&lt;mercury&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22961" PostTypeId="2" ParentId="22959" CreationDate="2017-10-02T20:51:19.920" Score="7" Body="&lt;p&gt;It is possible for two brown dwarfs to form a contact binary. This would not, of itself, cause core temperatures in either to rise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the two brown dwarfs were to merge, the mass of the resulting body could be enough to generate sufficient heat and pressure in the core for nuclear reactions to start. If both brown dwarfs had a mass of about 60 Jupiters, the combined size of 120 Jupiters would certainly be big enough to form a red dwarf.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-10-02T20:51:19.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22962" PostTypeId="1" AcceptedAnswerId="23293" CreationDate="2017-10-02T22:01:54.133" Score="0" ViewCount="52" Body="&lt;p&gt;I've been studying images from the WISE and NEOWISE missions. WISE used infrared cameras to survey the entire sky. When its primary mission ended in 2010, it was repurposed to look for Near-Earth Objects (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2011ApJ...731...53M&quot; rel=&quot;nofollow noreferrer&quot;&gt;described here&lt;/a&gt;). What changes to the spacecraft and software were needed in order to change its mission profile? Does it use the same cameras and instruments now as it did when performing a whole-sky survey? My thinking is that a whole-sky survey requires different equipment than a survey to look for small, fast-moving (compared to the stars) asteroids.&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2017-10-25T19:07:20.327" Title="Why is NEOWISE used for galactic observation?" Tags="&lt;space-telescope&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22963" PostTypeId="1" AcceptedAnswerId="22966" CreationDate="2017-10-02T22:46:21.850" Score="6" ViewCount="2329" Body="&lt;p&gt;Are there any known binary systems in which one member is a black hole and the other is not a black hole (main sequence star, giant, neutron star, white dwarf, whatever)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Googling seems to turn up only discussion of binaries in which both components are black holes. Is there some astrophysical reason why the kind of system I'm talking about cannot form? I think I remember that gravitational wave observatories have talked about black hole-neutron star mergers, but are such systems purely hypothetical so far?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess Sagittarius A* is sort of close to being an example, if you count it and the stars orbiting it (or orbiting their common barycenter) as a multiple star system.&lt;/p&gt;&#xA;" OwnerUserId="15381" LastActivityDate="2017-10-03T20:36:33.813" Title="Binaries consisting of a black hole and a non-black hole?" Tags="&lt;black-hole&gt;&lt;binary-star&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22964" PostTypeId="2" ParentId="22960" CreationDate="2017-10-02T23:39:09.160" Score="5" Body="&lt;p&gt;If we look at the planet's cores and I'm going to ignore liquid vs solid and focus on size overall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars:  &lt;a href=&quot;https://en.wikipedia.org/wiki/Mars#Internal_structure&quot; rel=&quot;noreferrer&quot;&gt;Core estimated 1,794&lt;/a&gt; +/- 65 km radius.  The planet is 3,390 km radius.   About 53% of the planet's radius is its core.   Mars also has more sulfur in it's core and more Iron in it's mantle than Earth, suggesting that it probably didn't mix as well as Earth did, but I'm not sure that would significantly effect the size of it's core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus:  Core estimates have some uncertainty, but &lt;a href=&quot;https://www.universetoday.com/36161/core-of-venus/&quot; rel=&quot;noreferrer&quot;&gt;by this article&lt;/a&gt;, it's core is thought to be about 3,000 km, about 49.5% of it's 6,052 km radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Structure_of_the_Earth#Core&quot; rel=&quot;noreferrer&quot;&gt;Earth's core&lt;/a&gt; is about 3,400 km and it's radius 6,371 km, about 53.4% of it's total radius.    If we use the 49.5%-53.4% as a guideline, Mercury's 2,440 km radius (85% core, so it's core is about 2,074 KM), so a rough estimate using the other 3 planets as a guideline, 3,880 - 4,190 km radius.    That puts it 500-800 km larger than Mars in radius, roughly 2,000 km smaller than Venus in radius.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's obviously just a rough estimate, but assuming it was similar to the other 3 inner planets, it's probably in that range.   I should probably also account for compression.  The more massive the planet the greater the compression of it's core, but that would probably only vary a couple percentage points off the estimate and it wouldn't make a big difference.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-10-02T23:39:09.160" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22965" PostTypeId="2" ParentId="22963" CreationDate="2017-10-03T00:56:54.647" Score="6" Body="&lt;p&gt;In general those binaries will be hard to detect, but there is no reason why they wouldn't form.&lt;br&gt;&#xA;A whole class of objects, &lt;a href=&quot;https://en.wikipedia.org/wiki/X-ray_binary&quot; rel=&quot;noreferrer&quot;&gt;X-Ray binaries&lt;/a&gt; are in fact thought to be hosting a pair of BH and non-BH objects. The idea here is that the X-ray luminosity originates from mass accreted from the companion through Roche-Lobe-overflow.&lt;br&gt;&#xA;The mass from the companion or 'donor' falls down the gravitational well of the BH, rams into its accretion disc and creates &lt;a href=&quot;https://en.wikipedia.org/wiki/Bremsstrahlung&quot; rel=&quot;noreferrer&quot;&gt;Bremsstrahlung&lt;/a&gt; as it deaccelerates violently. This radiation falls into the wavelength of X-Rays and escapes the binary system, to be picked up by our telescopes later.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a small number of cases we've been successful in determining what the binary companion is, depicted in this famous graphic by Orosz:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4vg3R.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4vg3R.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cmi2.yale.edu/bh/week4/pages/page6.html#&quot; rel=&quot;noreferrer&quot;&gt;See also here for more information&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This graphic depicts the binary distance, approximate size of the accretion disc, and size of the companion star. The companion stars are very huge, because they reached their red giant stage, not because they have some other anomalous properties.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This stage is a normal ending of a star's life, just like what our sun will experience. Just in this case there's this black hole in the vicinity that has quite strong tides, which are able to strip away parts of the evolved star's outer shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This graphic is a few years old now, so I'm not sure what the recent state of affairs concerning the total number of those objects is. Also there are many more unconfirmed BH binary candidates amongst the X-ray binary population, and I don't even know anything about BH-exotic things pairings. But maybe someone else can comment on that.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2017-10-03T00:56:54.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22966" PostTypeId="2" ParentId="22963" CreationDate="2017-10-03T01:31:18.340" Score="14" Body="&lt;h2&gt;Black hole and main sequence star/giant star&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We can observe binary systems containing a black hole by looking for emissions from accretion disks which may form when matter is transferred from the companion star. &lt;a href=&quot;https://en.wikipedia.org/wiki/X-ray_binary&quot; rel=&quot;nofollow noreferrer&quot;&gt;X-ray binaries&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Microquasar&quot; rel=&quot;nofollow noreferrer&quot;&gt;microquasars&lt;/a&gt; are particularly notable types.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compact object doesn't have to be a black hole - neutron stars also sometimes show up - but in some cases, measurements indicate that the remnant is a black hole. In other cases, either the mass of the compact object is poorly constrained or could still be a massive neutron star, so for many systems, the nature of this object remains unknown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Cygnus_X-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cygnus X-1&lt;/a&gt;:&lt;/strong&gt; The companion star here is HDE 226868, a massive O-type star.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/V404_Cygni&quot; rel=&quot;nofollow noreferrer&quot;&gt;V404 Cygni&lt;/a&gt;:&lt;/strong&gt; The companion star is a K-type star (much less massive than HDE 226868); the system forms a &lt;em&gt;microquasar&lt;/em&gt;, with occasional high-energy x-ray emission.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Black hole and white dwarf&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It seems a logical assumption that black hole/white dwarf systems should be possible, and indeed they are. Mass loss through accretion by a companion object can change a star's evolutionary future (see the &lt;a href=&quot;https://en.wikipedia.org/wiki/Algol_paradox&quot; rel=&quot;nofollow noreferrer&quot;&gt;Algol paradox&lt;/a&gt; for a particularly weird case), but for many systems with a black hole, the &quot;normal&quot; star should still progress through the main sequence and post-main-sequence evolution, possibly becoming a white dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matter transfer can still happen at this stage, of course, so emission may continue, although the nature of the radiation could be changed. It's true that there aren't many known systems of this type, but there almost certainly more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;http://chandra.harvard.edu/press/17_releases/press_031317.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;X9&lt;/a&gt;:&lt;/strong&gt; The companion star is a white dwarf orbiting extremely close to the black hole. &lt;a href=&quot;https://arxiv.org/abs/1702.02167&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bahramian et al.&lt;/a&gt; are confident that the companion is a white dwarf, and the main object is likely a black hole.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Black hole and neutron star&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;These type of systems are possibly the most interesting from a gravitational point of view. They should produce gravitational waves, which could be detectable by LIGO.&lt;sup&gt;1&lt;/sup&gt; &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/2041-8205/832/2/L21&quot; rel=&quot;nofollow noreferrer&quot;&gt;LIGO has yet to observe such a system&lt;/a&gt;, which seems a bit puzzling, although something like the blue supergiant in &lt;a href=&quot;https://en.wikipedia.org/wiki/Cygnus_X-3&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Cygnus X-3&lt;/strong&gt;&lt;/a&gt; could evolve into a neutron star in the future. This is a case where more data and observations are needed.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; You've probably heard the rumors that LIGO detected a possible neutron star-neutron star merger, which would imply that its sensitivity to neutron star-binary systems could indeed be high enough for detections, as expected. It turned out that &lt;a href=&quot;https://www.ligo.caltech.edu/news/ligo20170927&quot; rel=&quot;nofollow noreferrer&quot;&gt;it was simply a binary black hole system&lt;/a&gt;&lt;sup&gt;&lt;a href=&quot;https://dcc.ligo.org/public/0145/P170814/010/GW170814.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;[1]&lt;/a&gt;&lt;/sup&gt; - not quite as exotic as some people were hoping for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;#x20;&amp;#x20;&amp;#x20;&amp;#x20;&amp;#x20;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="16401" LastEditDate="2017-10-03T20:36:33.813" LastActivityDate="2017-10-03T20:36:33.813" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22969" PostTypeId="2" ParentId="22963" CreationDate="2017-10-03T06:20:36.423" Score="6" Body="&lt;p&gt;Prior to the discovery of gravitational wave sources, the &lt;em&gt;only&lt;/em&gt; confirmed stellar-sized black holes, were those detected in binary systems with more &quot;normal&quot; stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These were most detected via X-rays emitted by hot material that is accreted from the &quot;normal&quot; companion. The two main categories are high-mass and low-mass X-ray binaries: in the former, the black hole accreted the wind from a massive stellar companion; in the latter, the mass is transferred by Roche lobe overflow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black hole status is confirmed by measuring a dynamical mass for the unseen companion that exceeds any plausible upper limit for a stable neutron star. Read the review by &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0306213&quot; rel=&quot;noreferrer&quot;&gt;McClintock (2003)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-10-03T06:20:36.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22977" PostTypeId="1" AcceptedAnswerId="22979" CreationDate="2017-10-03T14:00:22.553" Score="10" ViewCount="701" Body="&lt;p&gt;My understanding of why Enceladus is geologically active is that tidal forces from Saturn and - to a lesser extent - from the nearby larger moon Dione provide heat to the moon's interior, just like Jupiter does for Io. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should the other mid-to-large-sized moons Mimas, Tethys, Dione and Rhea not also exhibit similar activity from the same forces? Granted, Dione and Rhea are further from Saturn and thus less susceptible to tidal stresses, but what about Tethys and Mimas? What special attributes does Enceladus possess that has allowed it to be geologically active in modern times? Is it purely down to the fact that the interior of Enceladus is more ice than rock compared to the other moons, or are there more factors to consider?&lt;/p&gt;&#xA;" OwnerDisplayName="user10106" LastEditorUserId="11806" LastEditDate="2017-10-04T00:53:57.067" LastActivityDate="2017-10-04T00:53:57.067" Title="Why is Enceladus the only geologically active moon among its neighbours?" Tags="&lt;natural-satellites&gt;&lt;saturn&gt;&lt;enceladus&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22978" PostTypeId="1" AcceptedAnswerId="23662" CreationDate="2017-10-03T14:59:47.143" Score="3" ViewCount="1085" Body="&lt;p&gt;I'm trying to compute solar eclipses for a long period of time, see &lt;a href=&quot;https://astronomy.stackexchange.com/questions/21402/back-predicting-solar-eclipses&quot;&gt;Back-predicting solar eclipses&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially, I wanted at several points on Earth whether an eclipse was visible at some point in time.&#xA;I have since changed the approach and I will now compute whether an eclipse could be seen from the center of the Earth at a given moment. Later, after knowing roughly the time of the eclipse, I will iterate over a grid on Earth to check where it could be actually seen.&#xA;So far, my first question is: is it right to check for eclipses at the center of the Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, I wrote a small code in Python to test how good I am at predicting eclipses, checking against &lt;a href=&quot;https://eclipse.gsfc.nasa.gov/SEsearch/SEsearchmap.php?Ecl=01501207&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://eclipse.gsfc.nasa.gov/SEsearch/SEsearchmap.php?Ecl=01501207&lt;/a&gt;&#xA;for these coordinates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lat.: 27.4782° N&#xA;Long.: 109.9299° W&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#!/usr/bin/python&#xA;&#xA;from skyfield.api import load, Topos&#xA;from skyfield.positionlib import ICRF&#xA;&#xA;planets = load('Ephemerides/de422.bsp')&#xA;ts = load.timescale()&#xA;&#xA;earth = planets['earth']&#xA;sun = planets['sun']&#xA;moon = planets['moon']&#xA;&#xA;time = ts.utc(150,12,06,23,49,21.9)&#xA;place = earth + Topos('27.4782 N','109.9299 W')&#xA;&#xA;observe_moon = place.at(time).observe(moon).apparent().position.au&#xA;observe_sun = place.at(time).observe(sun).apparent().position.au&#xA;distance = ICRF(observe_moon).separation_from(ICRF(observe_sun))&#xA;distance = distance.degrees&#xA;&#xA;print distance&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The distance I obtain is above 10 degrees, while at the time provided, if I did things correctly, the eclipse was at its maximum at those coordinates. What is wrong in my code or my understanding?&lt;/p&gt;&#xA;" OwnerUserId="17282" LastActivityDate="2017-11-25T13:09:44.797" Title="Eclipses with SkyField" Tags="&lt;solar-eclipse&gt;&lt;python&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22979" PostTypeId="2" ParentId="22977" CreationDate="2017-10-03T16:11:29.737" Score="7" Body="&lt;p&gt;This puzzle is something called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Enceladus#Mimas.E2.80.93Enceladus_paradox&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Mimas paradox&lt;/em&gt;&lt;/a&gt;. The two bodies are nearly twins, but Mimas is substantially less active than Enceladus. There are currently several proposed solutions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://meetingorganizer.copernicus.org/EGU2015/EGU2015-10569.pdf&quot; rel=&quot;noreferrer&quot;&gt;Mimas cooled quickly&lt;/a&gt;, before it entered its resonance with Tethys (which contributes to tidal heating, just as Dione does for Enceladus). &lt;a href=&quot;https://en.wikipedia.org/wiki/Enceladus#Radioactive_heating&quot; rel=&quot;noreferrer&quot;&gt;A &quot;hot start&quot;&lt;/a&gt; from rapid radioactive decay (of &lt;a href=&quot;http://news.bbc.co.uk/2/hi/science/nature/6445323.stm&quot; rel=&quot;noreferrer&quot;&gt;aluminum-26 and iron-60&lt;/a&gt;) shortly after Enceladus' formation could have allowed its subsurface ocean to stay liquid.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.lpi.usra.edu/meetings/lpsc2006/pdf/2200.pdf&quot; rel=&quot;noreferrer&quot;&gt;Enceladus is rockier&lt;/a&gt; and therefore has a higher density. This would have led to quicker cooling, and would also mean that tidal dissipation could be higher for Enceladus than for Mimas. It does seem a bit paradoxical - one would imagine that more rock would mean less water ice and therefore a smaller subsurface ocean - but this isn't a problem.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It's worth noting that Tethys is thought to be made almost entirely of ice. If the rocky-Enceladus hypothesis is true, then that could be a major factor in the lack of geological activity on Tethys. However, Tethys likely once had a subsurface ocean. Dione currently does, and experiences tidal heating; however, its semi-major axis is large enough that the effects are not as strong as on Enceladus.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2017-10-03T16:11:29.737" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22981" PostTypeId="2" ParentId="22921" CreationDate="2017-10-03T16:33:49.030" Score="2" Body="&lt;p&gt;Not really: microwaves heat mainly by using a &lt;em&gt;lot&lt;/em&gt; of radiation, the particular wavelengths matters only for the ability to (somewhat) heat evenly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You probably know of the Sun, a thing in the sky which shines most brightly in the optical, that is in wavelengths much shorter than the ones in a microwave. Yet it is clearly very capable of heating you and everything around you far above the ambient temperature of space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the wavelength of radiation is not crucial for heating (it matters a bit depending on the material), it is the amount of incoming radiation that matters. If the CMB could start to boil water because it was redshifted from 2 mm to 12 cm, it would need to be at such high intensity that water would &lt;em&gt;already&lt;/em&gt; be boiling, because water absorbs more radiation of 2 mm wavelength than of 12 cm:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/mau4m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mau4m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18887" LastActivityDate="2017-10-03T16:33:49.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22983" PostTypeId="2" ParentId="22978" CreationDate="2017-10-03T17:24:43.253" Score="1" Body="&lt;p&gt;To answer your first question, you will rarely or never see a &lt;em&gt;total&lt;/em&gt; eclipse at the center of the Earth because the moon's shadow isn't long enough. You would, however, see partial annular eclipses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do see a partial annular eclipse at the center of the Earth, there will definitely be one (not necessarily total) somewhere on the surface of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The opposite, however, is not true. It's quite possible for locations on Earth to see a solar eclipse, but the shadow not to pass through the Earth's center, so a center-of-Earth (geocentric) observer wouldn't see even an annular or partial eclipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard debugging tip: try with an eclipse closer to today to see if it's the code, inaccuracy of Skyfield, or just that the umbra didn't hit the Earth's center.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2017-10-03T17:24:43.253" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22984" PostTypeId="2" ParentId="22946" CreationDate="2017-10-03T17:31:24.347" Score="1" Body="&lt;p&gt;I think you are overlooking these factors:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Iowa State University is at 42 degrees North latitude, so the Sun and celestial equator that they are showing in their figures are always to the south at the meridian. Naturally, observers in the southern hemisphere would see the sun and celestial equator to their north at the meridian.&lt;/li&gt;&#xA;&lt;li&gt;The lines of right ascension go from pole-to-pole.&lt;/li&gt;&#xA;&lt;li&gt;So the hour angle that the website is describing is the angle between the plane of the meridian and the plane of right ascension. This angle applies regardless of where you look in the sky: north, south, or overhead.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-10-03T17:31:24.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22985" PostTypeId="1" AcceptedAnswerId="22986" CreationDate="2017-10-03T21:42:24.127" Score="1" ViewCount="87" Body="&lt;p&gt;I am wondering why it was so hard to locate Luke in Star Was Episode 7, if you already have the small part of the map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have a sector of &lt;em&gt;our&lt;/em&gt; galaxy that roughly equals the one missing in this &lt;a href=&quot;http://ko.starwars.wikia.com/wiki/%EC%9D%80%ED%95%98%EA%B3%84?file=TFA_galaxy_map.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;image&lt;/a&gt;. (I found no image on the web of the part in question. In the movie version on netflix you can take good look at it at 52:56.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible and how hard would it be to identify a planet within that part of the map, given our knowledge of our galaxy and&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;we only know the small part and have no information about its size.&lt;/li&gt;&#xA;&lt;li&gt;there might be &quot;landmarks&quot; (planets, stars, nebulars, and the like) missing.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If it is not possible, what minimum requirements would have to be met (on the available equipment, the map, our knowledge, ..) to make it possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note: this is not a Star Wars question, that is just the motivation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About which galaxies do we have enough knowledge to be able to do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ko.starwars.wikia.com/wiki/%EC%9D%80%ED%95%98%EA%B3%84?file=TFA_galaxy_map.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ko.starwars.wikia.com/wiki/%EC%9D%80%ED%95%98%EA%B3%84?file=TFA_galaxy_map.jpg&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18892" LastActivityDate="2017-10-03T22:35:43.560" Title="Find a planet with incomplete information of a star chart" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;observable-universe&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2017-10-13T13:21:16.667" ContentLicense="CC BY-SA 3.0" />
  <row Id="22986" PostTypeId="2" ParentId="22985" CreationDate="2017-10-03T22:35:43.560" Score="2" Body="&lt;p&gt;I don't know if this will be closed as too general, but I think there's a fun question in here about galaxy mapping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick is to have a good map of the galaxy.   We can't see large sections of the Milky way, so we might have a better map of Andromeda than the Milky way overall from our perspective.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, there are landmarks.    &lt;a href=&quot;https://en.wikipedia.org/wiki/Pioneer_plaque&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pioneer&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Voyager_Golden_Record&quot; rel=&quot;nofollow noreferrer&quot;&gt;Voyager&lt;/a&gt; used Pulsars as landmarks in a diagram on their plaques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aXySa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aXySa.jpg&quot; alt=&quot;http://news.nationalgeographic.com/content/dam/news/2017/08/14/voyager-pulsar-map/pulsar-map-voyager.ngsversion.1502807521895.adapt.1900.1.jpg&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;a href=&quot;http://www.oarval.org/pulsars.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;by this estimate&lt;/a&gt;, at least 200,000 pulsars in the milky-way.  A map of 14 of them won't do someone on the other side of the galaxy if our map has some errors (&lt;a href=&quot;https://www.forbes.com/sites/startswithabang/2017/08/17/voyagers-cosmic-map-of-earths-location-is-hopelessly-wrong/#60c130dd69d5&quot; rel=&quot;nofollow noreferrer&quot;&gt;which Voyager's might&lt;/a&gt;) or if they don't have a good chart on the other side of the galaxy.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Pulsar#Maps&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pulsars are very precise&lt;/a&gt; and they can make  very landmarks if you have enough of them and properly spaced.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, stars move, so maps age and lose accuracy over time, but in your star-wars example, a human lifetime wouldn't change things much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other possible landmarks include X-ray binaries,  very large stars (less common than smaller ones), black holes, neutron stars, the age of stars could be used, contact binaries, nebulas (or is that nebulae), stellar nurseries.   Identifying an individual planet would be easy given good data and a complete map.   A typical modern computer could do it, but the trick is having a good map.   There's about 100 billion stars in the Milky way, so it comes down to how good the map is, but you're correct, if the small map is good, R2's larger galaxy map is irrelevant.   To be relevant, the smaller map should have been maybe 100-500 stars and R2's map nearly the entire galaxy, and there would be ways to make the map useless if it fell into enemy hands, by not including pulsars or star size, making it very very difficult to identify which part of the galaxy by just 100 stars, none of them distinct.   The principal of needing both maps is valid and could be done, but you're right that the size of the 2nd map was ridiculous, also, why would they need to track his journey like in the picture?.    Movies often ignore the finer points like that and just focus on what looks cool.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-10-03T22:35:43.560" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22989" PostTypeId="2" ParentId="22959" CreationDate="2017-10-04T05:33:04.830" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If two brown dwarfs got close enough to exchange gas with each other, could it trigger nuclear fusion within their cores?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Very roughly, a star can undergo fusion once it is approximately 7.5-8.3% the mass of the Sun, depending on the metallicity of the brown dwarfs. If one of the brown dwarfs is on the brink of fusion, the gases that it would accrete from its less massive counterpart would make it theoretically possible for the dwarf to be pushed over its threshold and become a star, similar to type Ia supernovae. Of course, this would only happen to one brown dwarf.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it possible for two brown dwarfs to orbit so closely and become a rocheworld in the first place?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Orbiting brown dwarf binaries are unable to exchange gas between one another. The only way that binary star systems are able to exchange gas is if one of the stars exceeds its Roche lobe, which is the region around a star in a binary system where orbiting material is bound to the star, and the other star attracts the first star's material. As brown dwarfs do not grow in radius, it is impossible for a brown dwarf's matter to exist outside of its Roche lobe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources:&lt;br&gt;&#xA;&lt;em&gt;Science&lt;/em&gt; (vol 313, p 936)&lt;br&gt;&#xA; &quot;Mass transfer and accretion in close binaries – A review&quot; by C.B. Boyle in &lt;em&gt;Vistas in Astronomy&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="18888" LastActivityDate="2017-10-04T05:33:04.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22996" PostTypeId="2" ParentId="22602" CreationDate="2017-10-04T11:52:07.937" Score="5" Body="&lt;p&gt;&quot;Spinning dust&quot; is a mechanism proposed to explain a particular feature in the foreground emission of &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow noreferrer&quot;&gt;CMB&lt;/a&gt;; a bump around $\nu\sim20\,\mathrm{GHz}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dust grains acquire charge through photoelectric emission and collisions with electrons and ions (&lt;a href=&quot;https://arxiv.org/abs/astro-ph/9802239&quot; rel=&quot;nofollow noreferrer&quot;&gt;Draine &amp;amp; Lazarian 1998&lt;/a&gt;). As zephyr comments, if the dust is a poor conductor, its charges will, in general, be unevenly distributed, causing small dust grains to exhibit an electric dipole moment $\mathbf{\mu}$. But the molecules making up the grains may themselves have a dipole moment, and even a perfect conductor&#xA;will, in general, have its charge centroid displaced from the mass centroid (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1975duun.book..155P&quot; rel=&quot;nofollow noreferrer&quot;&gt;Purcell 1975&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Collisions and radiation may cause the grains to start spinning, and in the presence of magnetic fields (which are very common in the interstellar medium), this spin in turn causes the particles to emit radiation with a power $P$, according to &lt;a href=&quot;https://en.wikipedia.org/wiki/Larmor_formula&quot; rel=&quot;nofollow noreferrer&quot;&gt;Larmor's formula&lt;/a&gt; which for a rotating dipole can be written as:&#xA;$$&#xA;P = \frac{2}{3} \frac{\mu_\perp^2 \omega^4}{c^3}.&#xA;$$&#xA;Here&#xA;$\mu_\perp$ is the component of $\mu$ perpendicular to the rotation axis,&#xA;$\omega$ is the angular frequency of the rotation, and&#xA;$c$ is the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The emitted radiation matches the rotation frequency which lies in the (tens of) GHz-region, corresponding to wavelengths in the microwave region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, the radiation from non-spinning dust will be thermal, thus lying in the infrared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper you link to (&lt;a href=&quot;https://arxiv.org/abs/1212.5225&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bennett et al. 2013&lt;/a&gt;) shows the difference in Fig. 22 (although thermal radiation peaks outside the observational range of WMAP): The spinning dust peaks around $\nu\sim20\,\mathrm{GHz}$, while the thermal radiation peaks around $\nu\sim2\,000\,\mathrm{GHz}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eHR9q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eHR9q.png&quot; alt=&quot;WMAP&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that spinning dust also emits thermal radiation, and in fact thermal fluctuations within the grains cause the charges to shift around quite rapidly, changing $\mu_\perp$ and $\omega$ (&lt;a href=&quot;https://arxiv.org/abs/1603.05277&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hoang et al. 2015&lt;/a&gt;), thus effective smearing out the spectral lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A symmetrical dust grain will arguably be hit by particles/photons, on average, as much from one side as from another. A mechanism to accelerate an asymmetrical grain's spin is proposed by &lt;a href=&quot;http://adsabs.harvard.edu/abs/1975duun.book..155P&quot; rel=&quot;nofollow noreferrer&quot;&gt;Purcell (1975)&lt;/a&gt;. Below is Fig. 1 from his paper. A particle that strikes in a concavity is more likely to interact with the grains twice. If the gas it colder than the grain, it will be heated and leave the grain with a larger velocity than it entered with, causing the grain in the figure to start spinning counter-clockwise; it the gas is hotter (which is more often the case), it will cause the grain to spin clockwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6bDyB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6bDyB.png&quot; alt=&quot;turbine&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2017-10-04T22:06:30.520" LastActivityDate="2017-10-04T22:06:30.520" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="23005" PostTypeId="2" ParentId="2533" CreationDate="2017-10-05T10:41:48.463" Score="0" Body="&lt;p&gt;Interesting. This process could impact the black hole formation in the first place. Consider a rotating star that dies and starts to shrink due to gravitational forces. As it shrinks, all its mass will be more and more compacted in a smaller radius. This will have two consequences: 1) the gravitational force atracting different parts of the body will will grow up with the inverse of squared radius and 2) its rotational speed will increase due to angular momentum conservation and the expanding force, due to the rotation, will grow up with the inverse of radius cubed. This means the expanding force will grow more quickly than the contracting one and, at least in Newtonian view, the expanding force will win. From this point of view, it looks like a rotating star would never form a black hole... &lt;/p&gt;&#xA;" OwnerUserId="18919" LastActivityDate="2017-10-05T10:41:48.463" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23007" PostTypeId="1" AcceptedAnswerId="23010" CreationDate="2017-10-05T21:24:40.257" Score="2" ViewCount="613" Body="&lt;p&gt;I'm on the most northern part of Germany (Schleswig-Holstein) it is 11.20pm right now and the moon is here as bright as I never saw it before I was just outside and it is as bright as if it was dim daylight. When the clouds moved away from the moon and I looked into the moon for a little moment it was so bright, that it actually glared me for a little moment. Is there anything specific going on that is just favored this event? Or is it something that simply happens sometimes and I just never noticed it before?&lt;/p&gt;&#xA;" OwnerUserId="10102" LastActivityDate="2017-10-08T06:48:08.520" Title="Why the moon is blinding me right now?" Tags="&lt;the-moon&gt;&lt;moon-phases&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23008" PostTypeId="2" ParentId="22854" CreationDate="2017-10-05T22:21:06.943" Score="4" Body="&lt;p&gt;In the case of WASP-12b, at least, the close proximity to the star has actually deformed the planet so much that it is overflowing its Roche-lobe, the area around a planet or star that . We can show this mathematically by finding the approximate Roche-lobe of the planet using:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{r_1}{A}=.46224\sqrt[3]{\frac{M_1}{M_1+M_2}}$  for $\frac{M_1}{M_2}&amp;lt;.8$ (which it is)&lt;br&gt;&#xA;where $A$ is the orbital separation and $r_1$ is the radius of the Roche-lobe around $M_1$.&lt;br&gt;&#xA;Solving for $r_1$ we get $97788\pm 2318$ miles. The planet's radius is $77759 \pm 3909$ miles. Taking the best possible scenario, there would be $\approx$ 13,000 miles between the planet's surface and its Roche-lobe &lt;em&gt;if&lt;/em&gt; the planet was a perfect sphere. It is unknown how much the star affects the planet, but it would not be a stretch to think that the major axis of the planet would be 13,000 miles longer than the minor axis, allowing for the planet's gas to be shoved out of the Roche-lobe and &quot;sucked&quot; into the star.  &lt;/p&gt;&#xA;" OwnerUserId="18888" LastEditorUserId="18888" LastEditDate="2017-10-07T00:03:12.027" LastActivityDate="2017-10-07T00:03:12.027" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23009" PostTypeId="2" ParentId="23007" CreationDate="2017-10-05T22:32:22.447" Score="4" Body="&lt;p&gt;Other than it being a Full Moon and the Moon is high in the sky, there is nothing unusual going on.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-10-05T22:32:22.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23010" PostTypeId="2" ParentId="23007" CreationDate="2017-10-05T23:23:14.793" Score="4" Body="&lt;p&gt;I can't give a definitive answer, but food for thought.   The Moon orbits on about a 5 degree inclination to the Earth's orbit around the sun.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6I4uS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6I4uS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the full Moon happens when the Moon and Sun are on opposite sides of the Earth, the Moon is most directly opposite the Sun near the Spring and fall equinox.   That makes the Harvest Moon (full moon closest to the fall equinox) and the full Moon closest to the spring equinox, more directly opposite the sun, so they reflect slightly more light to the Earth.   This effect basically just makes the Moon a bit rounder than other full moons.   If you've ever noticed, mid summer or mid winter, the full moon can appear fatter than it is tall.  That's not an illusion, it's due to the 5 degree inclination.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This effect on overall brightness is small, but your eyes might be noticing a rounder moon and seeing it as brighter.  It's theoretically possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://earthsky.org/astronomy-essentials/close-and-far-moons-in-2017&quot; rel=&quot;nofollow noreferrer&quot;&gt;This year's largest moon was in March&lt;/a&gt;, so today's full moon is actually smaller than average.  Astronomical measurements will always say that the closest full moon is the brightest one because it's larger in the sky and it reflects more total light towards the Earth, but apparent brightness to your eyes might have little to do with it's increasing or decreasing &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_diameter&quot; rel=&quot;nofollow noreferrer&quot;&gt;angular diameter&lt;/a&gt; and likely has more to do with how clear the night sky is.  Our eyes can fool us.  It's possible that the smaller moon appears brighter even though it reflects less light because it also scatters less light across the rest of the sky, similar to how a smaller flashlight might appear brighter if you stare into it, compared to a flashlight with a wider opening that gives off more total light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A clear night is probably the best explanation.  I'd pick that over smaller angular diameter or greater roundness.   On a clear night, the Full Moon can appear brighter because less light gets scattered around it, making it look bright against a sky that's slightly darker, but you don't notice the sky's variation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The full moon is also higher in the sky in winter, because it moves mostly opposite the sun.   Being higher in the sky mans there's less atmosphere between the Moon and you.   Directly overhead should be brighter than at an angle to a person viewing from Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;doing some looking I came across &lt;a href=&quot;https://www.space.com/11162-10-surprising-moon-facts-full-moons.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this website&lt;/a&gt;.  It claims:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A 95-percent illuminated moon appears half as bright as a full moon&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Believe it or not, the moon is half as bright as a full moon about 2.4&#xA;  days before and after a full moon.  Even though about 95 percent of&#xA;  the moon is illuminated at this time, and to most casual observers it&#xA;  might still look like a &quot;full&quot; moon, its brightness is roughly 0.7&#xA;  magnitudes less than at full phase, making it appear one-half as&#xA;  bright.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Perhaps the extra few degrees of fullness really does make a noticeable difference.   I'd assumed it wouldn't make a big difference but this article suggests it does.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-10-08T06:48:08.520" LastActivityDate="2017-10-08T06:48:08.520" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="23014" PostTypeId="1" AcceptedAnswerId="23018" CreationDate="2017-10-06T08:44:19.660" Score="1" ViewCount="122" Body="&lt;p&gt;I don't think this would be possible, but I'm curious, and as what ever query I googled I only found posts about black holes indeed existing, what black holes are, that even the Milky Way has a black hole in its center or that We can't know if dwarf galaxies have non-SMBH at their center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So sorry if it is a stupid question but I didn't want to go on with assuming something without knowing: &lt;strong&gt;So is it theoretically possible for a Galaxy to have a star at its gravitational center of mass?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10102" LastEditorUserId="11143" LastEditDate="2017-10-06T13:12:15.737" LastActivityDate="2017-10-06T13:18:00.257" Title="Could a dwarf galaxy host a star at the center instead of a SMBH?" Tags="&lt;galaxy&gt;&lt;galaxy-center&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23018" PostTypeId="2" ParentId="23014" CreationDate="2017-10-06T13:18:00.257" Score="8" Body="&lt;p&gt;A galaxy's center of gravity is not determined by the most massive object, but by all objects in the galaxy. Even supermassive black holes (SMBHs) do not dominate the gravitational field except very, very close to the center. By far, most of the stars in a galaxy couldn't care less about the SMBH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The region within which a BH dominates over that of the stars (the &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sphere_of_influence_(black_hole)&quot; rel=&quot;noreferrer&quot;&gt;sphere of influence&lt;/a&gt;&quot;$^\dagger$) is given by (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/1972ApJ...178..371P&quot; rel=&quot;noreferrer&quot;&gt;Peebles 1972&lt;/a&gt;)&#xA;$$&#xA;r = \frac{G M_\mathrm{BH}}{\sigma^2},&#xA;$$&#xA;where $G$ is the gravitational constant, $M_\mathrm{BH}$ is the mass of the black hole, and $\sigma$ is the velocity dispersion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Milky Way (MW), there's a SMBH (Sagittarius A*) of roughly $M_\mathrm{BH} \simeq 4\times10^6\,M_\odot$. In that region, the stellar velocity dispersion is roughly $50$–$100\,\mathrm{km}\,\mathrm{s}^{-1}$ (e.g. &lt;a href=&quot;https://arxiv.org/abs/1006.0064&quot; rel=&quot;noreferrer&quot;&gt;Genzel et al. 2010&lt;/a&gt;). Plugging in those values, you'll find that Sgr A* dominates the kinematic out to roughly 3 pc, or 10 lightyears, which is nothing compared to MW's radius of $\sim10^5$ lightyears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take the most massive conceivable stars (a hypothesized &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_population&quot; rel=&quot;noreferrer&quot;&gt;Pop III star&lt;/a&gt; of $M\sim10^3\,M_\odot$) in &lt;a href=&quot;https://en.wikipedia.org/wiki/Willman_1&quot; rel=&quot;noreferrer&quot;&gt;Willman 1&lt;/a&gt;, the smallest known dwarf galaxy — which has stellar velocity dispersion of the order $5$–$10\,\mathrm{km}\,\mathrm{s}^{-1}$ — you'll find that such a star will dominate the gravitational potential out to a distance of only $\lesssim0.1\,\mathrm{pc}$, again completely negligible compared to the galaxy's radius of $\sim25\,\mathrm{pc}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, although it's possible for a dwarf galaxy to host a massive star, it will just be a star like all the others, and will in no way define the galaxy's gravitational center.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;Not to be confused with the event horizon which is even smaller.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2017-10-06T13:18:00.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23021" PostTypeId="1" AcceptedAnswerId="23022" CreationDate="2017-10-06T15:46:09.543" Score="2" ViewCount="110" Body="&lt;p&gt;I was watching the YouTube video &lt;a href=&quot;https://www.youtube.com/watch?v=mQ08miQEa_k&quot; rel=&quot;nofollow noreferrer&quot;&gt;Perseid Meteor Shower - Mojave Desert, California&lt;/a&gt; which looks East before sunrise, presumably in mid-August of 2016.  I've made a GIF of a few seconds near the beginning where a fireball appears at about &lt;code&gt;00:03&lt;/code&gt; in the video, and a faint luminescent trail appears to drift and change shape. I've timed the two stars in Orion Bellatrix and Betelgeuse and the motion is about seven seconds of video to move about 30 minutes in R.A. (near declination 0), so based on that the three or four second visibility in the video works out to about 10-15 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this common? Am I interpreting what I'm seeing correctly? Is such luminosity visible by eye, or only using a camera with a good lens?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wvyuY.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wvyuY.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2017-10-06T18:48:59.827" Title="How long can afterglow last after a fireball?" Tags="&lt;planetary-atmosphere&gt;&lt;meteor&gt;&lt;atmospheric-effects&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23022" PostTypeId="2" ParentId="23021" CreationDate="2017-10-06T16:46:02.847" Score="2" Body="&lt;p&gt;This is an persistent ionization trail. As the meteoroid goes through the upper atmosphere at high speed (sometimes more than 45km/s) electrons are stripped from atoms. As the electrons recombine, the gas glows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Persistent trails are formed by brighter meteors, and can be visible to the naked eye. They may be visible for up to about 45 minutes, but that depends on the meteor. Generally brighter meteors have brighter trails. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The movement and streaking in the trail can reveal the presence of upper atmosphere winds. As radio waves bounce off the ionised trails, they can be used for long distance radio communictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Persistent trails were &lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2014/10/19/persistent_train_astrophoto_of_a_bright_meteor_and_vapor_trail.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;discussed&lt;/a&gt; by the &quot;Bad Astronomer&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fOhqx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fOhqx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.w7ftt.net/66leonid2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image taken during the 1966 Leonid Meteor storm&lt;/a&gt;, produced by a Mag -6 meteor&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-10-06T18:48:59.827" LastActivityDate="2017-10-06T18:48:59.827" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23027" PostTypeId="1" AcceptedAnswerId="23029" CreationDate="2017-10-07T09:19:46.087" Score="14" ViewCount="1973" Body="&lt;p&gt;The previous generation of stars famously are the origin of all the heavier elements (up until iron?) in the solar system. So a big portion of the solar system mass actually is made up of Carbon, Silicon, Iron and the like because of that. But in the center, and only in the center, there is a star with presumably almost no heavy elements inside. How can that be? Am I wrong about the actual mass concentrations or is there really an imbalance, i.e. is the element distribution really lighter towards the center of the solar system? I would presume that the previous generation of stars just ended in a more or less uniform cloud of debris, from which the solar system formed. But if so, why aren’t there star systems where the star has a very different composition and is kind of a spluttering, dirty fusion machine (metaphorically, I mean)?&lt;/p&gt;&#xA;" OwnerUserId="18954" LastActivityDate="2017-10-09T17:26:43.303" Title="How did the lighter elements end up in the center of the solar system? Solar System Formation" Tags="&lt;solar-system&gt;&lt;star-formation&gt;&lt;planetary-formation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23029" PostTypeId="2" ParentId="23027" CreationDate="2017-10-07T12:43:10.807" Score="24" Body="&lt;p&gt;The solar system contains very little of elements heavier than Helium - less than 2% by mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is reflected in the chemical abundances measured in the photosphere of the Sun. i.e. The Sun does contain heavier elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question is the wrong way around; it is not that the heavier elements have not sunk into the middle, it is that the vast majority of hydrogen and helium that was in the same place as the planets when they formed, did not end up as part of the planets. In fact, even this is only partially true. The mass of planetary material in the solar system is also dominated by the hydrogen and helium in the gas giants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the conundrum is only why the smaller planets don't have a similar composition to the Sun. The answer to that is temperature and gravity. A small, hot planet just doesn't have the gravity to retain fast moving hydrogen and helium atoms, unless they are trapped in some compound (like water!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the small planets close to the Sun are depleted of light elements.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-10-07T12:43:10.807" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23031" PostTypeId="1" AcceptedAnswerId="23033" CreationDate="2017-10-07T19:57:57.553" Score="5" ViewCount="251" Body="&lt;p&gt;Since the night two days ago, the moon here in Israel appears to be orange as it rises, and then gradually becomes white as it continues travelling across the sky.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This phenomenon comes in a great timing as I am starting to become interested in astronomy (I have &lt;strong&gt;no&lt;/strong&gt; special knowledge in astronomy, just to make it clear).&#xA;So I googled up and came across &lt;a href=&quot;https://wonderopolis.org/wonder/what-makes-the-moon-look-orange&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; explanation.&lt;br&gt;&#xA;I'll try to summarize: it says that it happens because of Earth's atmosphere, that tends to scatter short wave-lengths (such as blue and green) and let long wave-lengths (such as red and orange) pass through. When the moon rises, it's in the position where the light coming from it to our eyes goes through the thickest layer of atmosphere (as opposed to when it's high up in the night skies) and that gives the orange effect. &#xA;I totally get this explanation, but it raises a couple of questions for me, whose answers I couldn't find. Maybe you could help me out. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Why doesn't the orange moon phenomenon occur all the time?&lt;/strong&gt;&lt;br&gt;&#xA;It does not happen quite often. What's so special about the past three nights? I mean, the rising moon's light always has to go through a thick atmosphere layer, doesn't it?&lt;/li&gt;&#xA;&lt;li&gt;Maybe that's not entirely astronomic, but, to complete the big picture - &lt;strong&gt;why does our atmosphere scatter short wave-lengths?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;This explanation suggests that all light coming to us from the sun is (at least) slightly reddish. Is this measurable (if true)? Because to the naked eye, sunlight refrected through a prism shows no tendency towards red. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thank you for your time reading the question, I hope you could rescue me from my ignorance... &lt;/p&gt;&#xA;" OwnerUserId="17218" LastEditorUserId="17218" LastEditDate="2017-10-07T20:11:34.733" LastActivityDate="2017-10-08T10:07:05.163" Title="A [very basic?] question about light scattering and colors" Tags="&lt;the-moon&gt;&lt;light&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23032" PostTypeId="1" CreationDate="2017-10-07T20:16:56.800" Score="1" ViewCount="64" Body="&lt;p&gt;Yesterday I watched &quot;The Orville&quot; episode 4 and there was interesting moment where a kid and old man saw the space for the first time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Kid: What is this?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Capitan: Kid, that is space.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Old man: I used to imagine how it would be. It was nothing like this.&#xA;  How far does it extend?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Kelly: No one knows. Probably to infinity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In real life we also occasionally ask ourselves if space is finite. And I think it's not a relevant question. We tend to adjust everything to Earth standars where everything is limited and finite. It's very hard for us to accept that something can be infinite, that we even think universe is inside of something and has a material &quot;shell&quot; or a wall. But even if there was wall it should be infinite. So in reality our &quot;wall&quot; is probably empty space or hard vacuum or infinite universes that we will never see because they are infinitely far away, well maybe most of them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's that simple to accept that space is infinite, that we should not even ask that question, since it's so irrelevant and space can only be infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for my english and the question, since you probably know this and much more, I just wanted to write something.&lt;/p&gt;&#xA;" OwnerUserId="18958" LastActivityDate="2017-10-09T07:00:46.680" Title="Is it relevant to ask the limits of space?" Tags="&lt;space&gt;" AnswerCount="0" CommentCount="7" ClosedDate="2017-10-09T13:00:05.977" ContentLicense="CC BY-SA 3.0" />
  <row Id="23033" PostTypeId="2" ParentId="23031" CreationDate="2017-10-07T20:58:19.070" Score="2" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;There may be a particular effect. Smoke from wildfires or cities, fine dust in the atmosphere etc. can have the effect of making the moon appear redder. However it is most likely an example of observation bias.  You noted that the moon appeared orange, and on subsequent nights you were looking for this effect. In fact it happens each moon rise. You just didn't notice it before (or if you did, it didn't make enough of an impression on you.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are two forms of scattering that are applicable. Rayleigh scattering, and Mie scattering. There is also absorption of some colours. Rayleigh scattering occurs as the electromagnetic wave that is light causes the molecules of the air to act like little electic dipoles. The light then bounces of these electrically active molecules. Shorter wavelengths are affected more than longer ones.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The light emitted by the sun is very slightly yellow. It is slightly reddened by scattering through the atmosphere. However it still contains a mixture of all visible colours, and so a prism will still show a complete rainbow. The reddening is not apparent to the eye, but can be measured by instruments.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-10-08T06:14:17.207" LastActivityDate="2017-10-08T06:14:17.207" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23034" PostTypeId="1" CreationDate="2017-10-08T09:13:34.063" Score="1" ViewCount="1150" Body="&lt;p&gt;If we assume that there is nothing except the sun in the universe then how far the light can travel ??&lt;/p&gt;&#xA;" OwnerUserId="18964" LastActivityDate="2017-10-26T14:58:47.570" Title="Maximum distance light can travel?" Tags="&lt;observable-universe&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" ClosedDate="2017-12-31T12:13:43.257" ContentLicense="CC BY-SA 3.0" />
  <row Id="23035" PostTypeId="2" ParentId="23031" CreationDate="2017-10-08T10:07:05.163" Score="1" Body="&lt;p&gt;Besides the arguments in the previous answer as well as in the comments, consider the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) the rising moon is not always the same orange or red simply because this depends on the height of the sun, as it should be clear if scattering is taken into account (see point 2). This is the major reason / answer to your main question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) under approximation, the intensity of scattered depends inversely on the fourth power of the wavelength. This dependency is what does amplify differences between the sun light that reaches our eyes directly (or &quot;straight but reflected&quot; by the moon) and light that has been diffused first. An easy but correct explanation for the ten to - 4 relationship is given by Wikipedia under Rayleigh scattering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Do not neglect the response of our eyes. It is not linear in terms of intensity nor sensitivity is the same at different wavelengths. Whatever you look at, in atmosphere or out of it, will look different to a spectrometer. The maximum number of photons emitted by the Sun is in the region we perceive as green,for instance. This is true both in space or after passing the atmosphere. &lt;/p&gt;&#xA;" OwnerUserId="17437" LastActivityDate="2017-10-08T10:07:05.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23036" PostTypeId="1" CreationDate="2017-10-08T10:35:03.297" Score="1" ViewCount="54" Body="&lt;p&gt;I am new to astronomy and I was given the task of finding out if there are objects which can't be resolved by a telescope of resolution 5 arcseconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was given the values of RA and Dec of these objects in degrees. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   RA              DEC&#xA; 201.1999388    27.49294258&#xA; 201.2048319    27.49212959&#xA; 201.1984702    27.4978016&#xA; 201.2025157    27.51610266&#xA; 201.1997827    27.51743447 ...etc&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How can I find these in the simplest way? Is there any simple formula which can be used to achieve this which is computationally efficient as well?&lt;/p&gt;&#xA;" OwnerUserId="18932" LastActivityDate="2017-10-09T06:23:50.610" Title="How to find if a set of objects lies within a given resolution (5 arcsec) on a telescope?" Tags="&lt;fundamental-astronomy&gt;&lt;observational-astronomy&gt;&lt;coordinate&gt;&lt;photometry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23037" PostTypeId="1" CreationDate="2017-10-08T13:42:18.310" Score="-5" ViewCount="173" Body="&lt;p&gt;The USA claims to have traveled to the moon in 1969.&#xA;BUT...&#xA;The temperature on the moon at day is 100 C  and at night is -173 cold. How did the astronauts not die??&#xA;How did they survive?&lt;/p&gt;&#xA;" OwnerUserId="18965" LastEditorUserId="16926" LastEditDate="2017-10-08T14:29:59.770" LastActivityDate="2017-10-09T02:07:13.133" Title="Moon trip in 1969 and temperature" Tags="&lt;the-moon&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2017-10-13T13:18:35.230" ContentLicense="CC BY-SA 3.0" />
  <row Id="23038" PostTypeId="2" ParentId="23037" CreationDate="2017-10-08T14:26:09.553" Score="2" Body="&lt;p&gt;The astronauts wore temperature controlled and pressurized suits and helmets. &lt;/p&gt;&#xA;" OwnerUserId="16926" LastActivityDate="2017-10-08T14:26:09.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23039" PostTypeId="2" ParentId="23037" CreationDate="2017-10-09T02:07:13.133" Score="4" Body="&lt;p&gt;Excessive heat loss during lunar night was not a concern; the astronauts needed sunlight to perform their EVAs. Excessive heat gain during lunar day was a concern, at least initially. Because of these concerns, &lt;strong&gt;all&lt;/strong&gt; of the moon landings occurred shortly after lunar dawn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first few lunar landings only lasted a couple of days, with the spacecraft lifting off the Moon's surface well before lunar noon. None of the feared consequences of excessive heating arose. The astronauts wore good boots. This enabled longer stays on the surface of the Moon in later Apollo missions. Even so, the practice of landing shortly after lunar dawn continued. Those early morning landings provided excellent relief of the terrain during landing and it provided the opportunity for longer stays on the surface of the Moon.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-10-09T02:07:13.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23040" PostTypeId="1" CreationDate="2017-10-09T02:41:36.147" Score="4" ViewCount="62" Body="&lt;p&gt;If the universe is only 40 to 50 galactic revolutions old, how has it acquired its present state so &quot;quickly&quot; relative to this time scale?&lt;/p&gt;&#xA;" OwnerUserId="18968" LastActivityDate="2017-10-09T16:41:48.627" Title="Galactic revolutions relative to the age of the universe" Tags="&lt;galaxy&gt;&lt;cosmology&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2017-10-10T10:03:44.447" ContentLicense="CC BY-SA 3.0" />
  <row Id="23041" PostTypeId="1" CreationDate="2017-10-09T06:11:08.757" Score="-1" ViewCount="309" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/a9DgR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from the fact that nothing we know of, excluding pure energy can escape from black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They affect both space and time in a manner that nothing else can quite compare to, and usually have smaller black holes orbiting the supermassive black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from this, not much is known about these naturally occurring eaters of everything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, additional information tells us that blackholes both have a North and South Pole, and also drift through space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also to be noted, they have strong electromagnetic fields near the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now although unconfirmed, it's assumed that these magnetic fields are generated by an object orbiting the blackholes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now onto magnetic fields/waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite my hours upon hours of researching, I couldn't find any facts stating that electromagnetic fields/waves have any weight value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I did find out that it is possible to transmit information using electromagnetic fields/waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, until proven wrong I can safely theorize that black holes do generate electromagnetic fields, possibly even generated after the horizon and escaping. (However, I am uncertain, just as anybody else is.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, in theory, if you can figure out how to map objects with electromagnetic fields/waves, you could very well study the internal area of a black hole. Of course, the previous theory would also have to be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to investigate the inside of a black hole using electromagnetic waves generated after the horizon?&lt;/p&gt;&#xA;" OwnerUserId="18967" LastEditorUserId="2365" LastEditDate="2017-10-09T17:43:14.527" LastActivityDate="2017-10-09T20:28:28.207" Title="Can we peek into a black hole with magnetic waves?" Tags="&lt;black-hole&gt;&lt;space-time&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2017-11-06T20:58:05.833" ContentLicense="CC BY-SA 3.0" />
  <row Id="23043" PostTypeId="2" ParentId="23036" CreationDate="2017-10-09T06:23:50.610" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Great-circle_distance#Formulas&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; would be your best resource. Just note that the variables are expressed in geographical latitude and longitude, and your values are Right Ascension and Declination. Longitude and RA are the same, but &lt;em&gt;Declination&lt;/em&gt; is 0 at the equator and is positive toward the north pole, whereas &lt;em&gt;latitude&lt;/em&gt; is 0 at the north pole and is positive toward the south pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't forget to mind your units (it'll probably be in degrees since that's what your RA and Dec are). If you're doing this with a computer, you'll probably want to convert to radians first since most computer trig functions are in radians. Whatever program you use, just be sure to check what units it prefers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, be sure to convert your separation to arcseconds and you should be all set.&lt;/p&gt;&#xA;" OwnerUserId="18969" LastActivityDate="2017-10-09T06:23:50.610" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23044" PostTypeId="1" AcceptedAnswerId="23047" CreationDate="2017-10-09T06:39:12.690" Score="6" ViewCount="290" Body="&lt;p&gt;According to Hubble's law, the universe is expanding and this expansion is proportional to distance between the two objects and thus objects are moving away at a very fast rate,  then why is it predicted that after 4 billion years the Milky Way shall collide with Andromeda galaxy?&lt;/p&gt;&#xA;" OwnerUserId="18971" LastEditorUserId="7472" LastEditDate="2019-06-15T07:49:13.520" LastActivityDate="2019-06-15T07:49:13.520" Title="Collision with Andromeda Galaxy" Tags="&lt;cosmology&gt;&lt;milky-way&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23045" PostTypeId="2" ParentId="23040" CreationDate="2017-10-09T06:41:41.353" Score="5" Body="&lt;p&gt;Gravitationally bound dynamical systems evolve towards their equilibrium states on a dynamical timescale of $\sim (G \rho)^{-1/2}$, where $\rho$ is their average density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume the Galaxy started of with $10^{12}$ solar masses in a sphere of radius $10^5$ pc, then the dynamical timescale is 1 billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider an object at the &quot;edge&quot; of such a system, Newtonian gravity and a bit of rotational mechanics, we find that the fastest possible bound orbital period is also of order $(G\rho)^{-1/2}$. The Sun is not at the edge and the average density inside the solar orbit is much higher and so the orbital period can be a bit shorter than this (a factor of four).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus what we see is exactly what is expected for a $&amp;gt;10$ billion years old universe - a relatively settled dynamical system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could extend this argument to the timescale for major galaxy mergers. The distance between galaxies is of order ten times their size and thus the mean density of galaxy systems is 1000 times lower than for an individual galaxy. This &lt;em&gt;increases&lt;/em&gt; the relevant dynamical timescale to $1 \times 1000^{1/2} \sim 30$ billion years. That the Milky Way and Andromeda will collide some 17-18 billion years post big-bang is also therefore not an unexpected event.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-10-09T16:41:48.627" LastActivityDate="2017-10-09T16:41:48.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23047" PostTypeId="2" ParentId="23044" CreationDate="2017-10-09T08:40:49.327" Score="10" Body="&lt;p&gt;As you say, expansion is proportional to distance. And in fact, compared to the size of the (observable) Universe, Andromeda and our Milky Way are &lt;strong&gt;close together&lt;/strong&gt;. Indeed, farthest objects are &lt;em&gt;now&lt;/em&gt; more than 20 &lt;strong&gt;b&lt;/strong&gt;illion Ly from us, whereas Andromeda is only 2.5 &lt;strong&gt;m&lt;/strong&gt;illion Ly away. That is a &lt;strong&gt;1/10000&lt;/strong&gt; factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The space expansion between them is not enough to set them apart, because of their general motion. On small scale (until galaxy cluster scale), the motion is primarily ruled by gravity. &lt;a href=&quot;https://en.wikipedia.org/wiki/Galaxy_groups_and_clusters&quot; rel=&quot;noreferrer&quot;&gt;Galaxy cluster&lt;/a&gt; are in fact the largest gravitationally bound systems we know. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;They typically contain no more than 50 galaxies in a diameter of 1 to 2 megaparsecs&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since Andromeda is the closest galaxy, it is inside &quot;our&quot; galaxy cluster, and stays bound to ours by gravity.&lt;/p&gt;&#xA;" OwnerUserId="11264" LastActivityDate="2017-10-09T08:40:49.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23048" PostTypeId="1" AcceptedAnswerId="23143" CreationDate="2017-10-09T08:54:49.013" Score="5" ViewCount="337" Body="&lt;p&gt;I watched this simulation made by LIGO (Laser Interferometer Gravitational-Wave Observatory) where 2 black holes merge into one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=I_88S8DWbcU&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=I_88S8DWbcU&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was therefore thinking that theoretically nothing can escape a black hole given the features of the space and the black hole remain the same due to the strong gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I was thinking if 2 black holes stay together close enough, the &quot;event horizon&quot; would reshape as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We all know the gravity force is F=G(Mm/r^2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a point next to the event horizon but &quot;inside&quot; the black hole the gravity there will be so strong that the Escape velocity will be higher than C (speed of light).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we place another massive object (like another blackhole in the proximities of this point, but far enough, the gravity there will be counteracted and therefore this point will be able to escape its black hole, reshaping the event horizon and making it smaller. (We will have created a &quot;concavity&quot; in the event horizon)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a hypothetical scenario where we create thousands of micro blacholes orbiting the main blackhole, in a gravitational equlibrium. Would we be able to reshape and &quot;push&quot; the event horizon strong enough that objects inside could now use the chance and escape?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could we even pull the matter of the black hole and make it less dense along the time even transforming it into a start destroying the singularity?&lt;/p&gt;&#xA;" OwnerUserId="18972" LastActivityDate="2017-10-15T03:29:39.540" Title="Reshaping black holes to &quot;push&quot; the event horizon" Tags="&lt;black-hole&gt;&lt;event-horizon&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23052" PostTypeId="1" CreationDate="2017-10-09T13:25:38.350" Score="1" ViewCount="46" Body="&lt;p&gt;What is a singularity (black holes)?  What happens to time and space at that point? What does it leads to? &lt;/p&gt;&#xA;" OwnerUserId="18978" LastActivityDate="2017-10-09T13:25:38.350" Title="Singularity in Black holes?" Tags="&lt;black-hole&gt;&lt;singularity&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2017-10-09T16:11:20.490" ContentLicense="CC BY-SA 3.0" />
  <row Id="23054" PostTypeId="1" AcceptedAnswerId="23056" CreationDate="2017-10-09T16:42:35.000" Score="6" ViewCount="453" Body="&lt;p&gt;The total stellar mass of a galaxy cannot be measured directly. Astronomers can infer it from the galaxy spectrum by stellar population synthesis. However, since high-mass stars dominate the spectrum but contribute little to the total stellar mass, the inferred total stellar mass is sensitive to initial mass function (IMF) which is usually assumed a priori. One can also measure the total stellar mass by kinematics, but it is the total mass of stars, dark matter and gas enclosed that is measured in this way, and we still need some assumptions (say, on the dark matter profile) to obtain the total STELLAR mass. In a word, it's very difficult to measure the total stellar mass of a galaxy precisely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I wonder why people are still interested in this parameter. What can it tell us about the formation and evolution of galaxies? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that what can be measured relatively more accurately is the properties of high-mass stars which are also the main sources of stellar feedback and metal enrichment. Why not just focus on these stars? What makes low-mass stars (or to be more concrete, their mass) important to the galaxy as a whole? &lt;/p&gt;&#xA;" OwnerUserId="18981" LastEditorUserId="1056" LastEditDate="2017-10-17T00:54:36.713" LastActivityDate="2017-10-19T22:55:38.617" Title="What can the total stellar mass of galaxies tell us?" Tags="&lt;star&gt;&lt;galaxy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23055" PostTypeId="2" ParentId="23027" CreationDate="2017-10-09T17:26:43.303" Score="0" Body="&lt;p&gt;Toward the end of a stars lifecycle the star looses the ability to continue using hydrogen ect). To power itself, it begins to, (in a desperate attempt to live.) Create heavier elements such as iron. Now iron can't certainly not sustain a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such the iron destroys the star, thus marking the end of its life cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small amounts of iron probably exist in stars, (from the universes point of view.) But it's not enough to actually effect the star it's in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron is of course not the only things starsform that ultimately kill the star, I'm just using it as an example here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None the less stars are quite interesting, just as anything else in space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. This is just my vary basic understanding of space, and I'm learning a lot on this website.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a nice day/night.&lt;/p&gt;&#xA;" OwnerUserId="18967" LastActivityDate="2017-10-09T17:26:43.303" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23056" PostTypeId="2" ParentId="23054" CreationDate="2017-10-09T18:12:40.710" Score="5" Body="&lt;p&gt;Actually, the light of most galaxies is dominated by red giants, which are predominantly not very massive stars (more massive than the Sun sure, but less than a few solar masses). These are much more common than short-lived high mass stars, which in any case don't exist in many galaxies (e.g. old ellipticals).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In galaxies that not gas-rich - again most ellipticals, the stars form most of the baryonic mass and most of the stellar mass is in low-mass stars. Even in galaxies where there is lots of gas (the dust is a negligible mass component), there are ways of measuring this too. Estimating the total mass of baryonic versus dark matter is essential in understanding the formation and evolution of galaxies, since they behave differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars less massive than about 0.9 solar masses have had insufficient time to evolve. The baryonic mass in these stars is locked away and cannot be recycled into other stars. Knowing the fraction of mass in this situation is important in understanding the chemical evolution of galaxies.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-10-09T18:12:40.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23058" PostTypeId="1" AcceptedAnswerId="23062" CreationDate="2017-10-09T19:00:04.603" Score="15" ViewCount="4299" Body="&lt;p&gt;How does the ISM resist gravity? That's the only force acting on it, and all other particles seem to collect together to form stars. What makes the ISM so special among other particles?&lt;/p&gt;&#xA;" OwnerUserId="16849" LastActivityDate="2019-07-04T09:39:38.590" Title="Why isn't interstellar medium pulled in towards nearest massive object?" Tags="&lt;interstellar-medium&gt;&lt;interstellar&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23059" PostTypeId="2" ParentId="23041" CreationDate="2017-10-09T19:40:18.690" Score="1" Body="&lt;p&gt;You seem to believe that since electromagnetic radiation has no mass, it is not affected by black holes. Instead, think about it like this; on Earth, if you drop a 10 pound bowling ball and a 20 pound bowling ball, they will both hit the ground at the same time. Similarly, even though a photon is a massless particle, it is still affected by the gravity of the black hole and therefore cannot escape the event horizon. Instead, when black holes consume matter, the matter accelerates extremely quickly, producing electromagnetic radiation outside of the event horizon that we can detect with specialized telescopes. A popular example of this is the recent interaction between the gas cloud G2 and the supermassive black hole in the center of our galaxy, Sagittarius A*. Sag A* sucked in the gas and in the process released a large amount of x-ray and gamma radiation that could be easily detected by our telescopes. &lt;/p&gt;&#xA;" OwnerUserId="18888" LastActivityDate="2017-10-09T19:40:18.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23060" PostTypeId="2" ParentId="23041" CreationDate="2017-10-09T20:28:28.207" Score="7" Body="&lt;p&gt;I'll try to answer paragraph by paragraph, since there are some misconceptions here.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Aside from the fact that nothing we know of, excluding pure energy can escape from black holes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nothing can escape from beyond the event horizon, including &quot;pure energy&quot; (whatever that means)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They affect both space and time in a manner that nothing else can quite compare to, and usually have smaller black holes orbiting the supermassive black holes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Binary black holes do exist, but are quite rare. There is no reason to think that supermassive black holes are orbited by smaller black holes (although black holes are quite common, so it isn't impossible)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Aside from this, not much is known about these naturally occurring eaters of everything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We actually know quite a lot, although some is theoretical. Predictions of General relativity are well supported by observation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also to be noted, they have strong electromagnetic fields near the horizon. Now although unconfirmed, it's assumed that these magnetic fields are generated by an object orbiting the blackholes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Black holes are surrounded by magnetic fields, generated in the accretion disc of matter orbiting the blackhole.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Despite my hours upon hours of researching, I couldn't find any facts stating that electromagnetic fields/waves have any weight value.&#xA;  Although I did find out that it is possible to transmit information using electromagnetic fields/waves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Electromagnetic waves are &quot;Light&quot;. Short wavelength electromagnetic waves are ultraviolet, X-rays and gamma rays. Long wavelength electromagnetic waves are radio waves. They don't have mass, but are affected by gravity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;That said, until proven wrong I can safely theorize that black holes do generate electromagnetic fields, possibly even generated after the horizon and escaping. (However, I am uncertain, just as anybody else is.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Electromagnetic waves (ie light) cannot escape from a black hole. Nothing escapes from black hole. You can investigate the accretion disc by &quot;looking&quot; at it (ie detecting electromagnetic radiation). You can't investigate the inside of a black hole, since nothing escapes. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it possible to investigate the inside of a black hole using electromagnetic waves generated after the horizon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hence, No. This isn't possible.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-10-09T20:28:28.207" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23061" PostTypeId="2" ParentId="23058" CreationDate="2017-10-09T20:55:48.360" Score="12" Body="&lt;p&gt;First, consider that gravity is weak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nearest star system to the Sun is &lt;a href=&quot;https://en.wikipedia.org/wiki/Alpha_Centauri&quot; rel=&quot;noreferrer&quot;&gt;Alpha Centauri&lt;/a&gt;, at a distance of about 4 light-years. Consider the acceleration due to the Sun's gravity at half that distance:&#xA;$$a_S=\frac{GM_\odot}{r^2}\simeq3.7\times10^{-13}\text{ m/s}^2$$&#xA;&#xA;where $M_\odot$ is the mass of the Sun. That's an incredibly small acceleration, meaning that most massive bodies have a very small gravitational influence on the ISM. Particles are attracted to massive objects . . . but very, very weakly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, the ISM sometimes &lt;em&gt;does&lt;/em&gt; collapse. The number density varies in different parts, from $\sim10^{-3}$ particles per cubic centimeter to $\sim10^6$ particles per cubic centimeter (&lt;a href=&quot;https://astronomy.stackexchange.com/a/681/2153&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;https://physics.stackexchange.com/a/162259/56299&quot;&gt;2&lt;/a&gt;). Above about $10^4$ particles per cubic centimeter, though, you venture into the regime of &lt;a href=&quot;https://en.wikipedia.org/wiki/Molecular_cloud&quot; rel=&quot;noreferrer&quot;&gt;molecular clouds&lt;/a&gt;, which sometimes &lt;em&gt;do&lt;/em&gt; collapse to form stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a final note, gravity isn't the only force acting on the ISM. &lt;a href=&quot;http://www.scholarpedia.org/article/Galactic_magnetic_fields&quot; rel=&quot;noreferrer&quot;&gt;Galactic magnetic fields&lt;/a&gt;, for instance, can influence the ISM's dynamics in various scenarios, including preventing or enabling the collapse of molecular clouds (see &lt;a href=&quot;http://adsabs.harvard.edu/full/2005mpge.conf..147F&quot; rel=&quot;noreferrer&quot;&gt;Ferrier (2005)&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2017-10-10T00:29:45.477" LastActivityDate="2017-10-10T00:29:45.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23062" PostTypeId="2" ParentId="23058" CreationDate="2017-10-09T21:02:08.630" Score="31" Body="&lt;p&gt;It is not true that the particles in the interstellar medium (ISM) are only acted upon by gravity. For instance,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In many cases a significant part of the ISM is ionized, in which case it interacts with magnetic field which permeates the gas and may in some cases be quite strong.&lt;/li&gt;&#xA;&lt;li&gt;In the vicinity of massive and hence luminous stars, radiation pressure may exert a strong force on the ISM. They also emit copious amounts of cosmic rays (i.e. relativistic particles) that transfer momentum to the surrounding gas.&lt;/li&gt;&#xA;&lt;li&gt;Supernova explosions create hot bubbles that expand and sweep through the ISM, resulting in shock waves and galactic outflows. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In most cases, however, what may prevent a gas cloud from collapsing is simply its temperature.&#xA;Despite all the above processes, and despite gravity being the weakest force, gas clouds do sometimes collapse to form stars. The criterion for doing so is that the gas is dense enough, and that its internal pressure (or thermal energy) is weak enough. This is described by &lt;a href=&quot;https://en.wikipedia.org/wiki/Jeans_instability&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jeans instability&lt;/a&gt;, which formulates the criterion for a cloud of gas to collapse through equating pressure forces, or thermal energy, to gravity. One way to express this is the &lt;em&gt;Jeans mass&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$M_J$&lt;/span&gt; (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1902RSPTA.199....1J&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jeans 1902&lt;/a&gt;) which is the critical mass of a cloud where thermal energy is exactly balanced by gravitational forces:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;M_J = \rho \left( \frac{\pi k_B T}{4 \mu m_\mathrm{u} G \rho} \right)^{3/2} \\&#xA;\propto \frac{T^{3/2}}{\rho^{1/2}}.&#xA;$$&lt;/span&gt;&#xA;Here, &lt;span class=&quot;math-container&quot;&gt;$k_B$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$m_u$&lt;/span&gt; are Boltzmannn's constant, the gravitational constant, and the atomic mass unit, while &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; are the temperature, the mean molecular mass, and the density of the gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second line of the equation it is emphasized that &lt;span class=&quot;math-container&quot;&gt;$M_J$&lt;/span&gt; increases with temperature, and &lt;em&gt;de&lt;/em&gt;creases with density. In other words, if the gas is too hot, or too dilute, the total mass needed to collapse must be higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, gas will not collapse to form stars if the temperature is above some &lt;span class=&quot;math-container&quot;&gt;$10^4\,\mathrm{K}$&lt;/span&gt;. If the temperature is higher, the particles simply move too fast. Since various processes may easily heat the ISM to millions of degrees, the gas has to cool before it can collapse. On way to do this is by cooling radiation: Fast-moving atoms collide (either with each other or, more often, with electrons). Some of the kinetic energy of the atoms is spent exciting their electrons to higher levels. When the atoms de-excite, photons are emitted which can leave the system. The net result is that thermal energy is removed from the cloud, until at some point it has cooled enough to collapse.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2019-07-04T09:39:38.590" LastActivityDate="2019-07-04T09:39:38.590" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23066" PostTypeId="1" CreationDate="2017-10-10T03:03:52.647" Score="4" ViewCount="141" Body="&lt;p&gt;Suppose you traveled off Earth and back at relativistic speeds, so that when you come back much longer time elapsed on Earth than you experienced. Suppose you had an excellent telescope through which you could observe the details of the surface of the Earth down to human scale, all the time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What will you see?  Will you see people moving incredibly fast? Will you see people progressing from birth to old age in a matter of seconds?&lt;/p&gt;&#xA;" OwnerUserId="18987" LastActivityDate="2017-10-10T14:10:11.877" Title="If you looked at Earth while travelling at relativistic speeds, will you see sped-up activities?" Tags="&lt;time-dilation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23068" PostTypeId="2" ParentId="23066" CreationDate="2017-10-10T06:46:00.963" Score="5" Body="&lt;p&gt;It doesn't have to be a visible telescope, any electromagnetic radiation will act in the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll give you the answer, but if you want the details you need to look at a description of the resolution to the so-called  &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Twin_paradox&quot; rel=&quot;nofollow noreferrer&quot;&gt;twin paradox&lt;/a&gt;, for for example &lt;a href=&quot;https://www.cpp.edu/~ajm/materials/twinparadox.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that the spaceship travels  outwards and back to Earth at the same speed, with a  brief period of acceleration in between. From the point of view of the observer on the spaceship, they will receive signals at at &lt;em&gt;lower&lt;/em&gt; frequency on the outward half of the journey. The change in frequency to depends on how fast they go and is given by the relativistic Doppler effect formula.&#xA;$$f_{\rm out} = f \left(\frac{c - v}{c + v}\right)^{1/2} = \alpha f,$$&#xA;where $\alpha &amp;lt; 1$ and $f$ is the frequency of the light/signal on Earth. So the Earth would appear &lt;em&gt;slowed down&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the way back the observed frequency &#xA;$$f_{\rm in}= f \left(\frac{c + v}{c - v}\right)^{1/2} = \frac{f}{\alpha}$$ and the Earth would appear &lt;em&gt;speed up&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the observer on the spaceship they spend half the travel time in each of these phases. The average frequency they receive is&#xA;$$\bar{f} = \frac{(f_{\rm out}+f_{\rm in})}{2} = \frac{f}{2} \left(\alpha + \frac{1}{\alpha}\right),$$ which is always $&amp;gt;f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that &lt;em&gt;on average&lt;/em&gt; things on Earth appear speeded up, such that the Earth is older upon return, but what is seen from the spaceship is divided into distinct outbound and inbound phases.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-10-10T14:10:11.877" LastActivityDate="2017-10-10T14:10:11.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23073" PostTypeId="1" AcceptedAnswerId="23075" CreationDate="2017-10-10T18:02:20.003" Score="3" ViewCount="338" Body="&lt;p&gt;The cores of gas giants are kept under incredible pressure by the weight of the gasses of the giant. In a &lt;a href=&quot;https://en.wikipedia.org/wiki/Chthonian_planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;chthonian planet&lt;/a&gt;, said gases are stripped away, revealing the core of the gas giant, which should alleviate the pressure compressing the core of the gas giant. In a chthonian planet candidate, &lt;a href=&quot;http://exoplanet.eu/catalog/Kepler-52_c/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kepler 52-c&lt;/a&gt;, its density is over 600 times that of Jupiter. What is keeping the planet at such a high density?&lt;/p&gt;&#xA;" OwnerUserId="18888" LastEditorUserId="37191" LastEditDate="2020-12-28T22:33:08.407" LastActivityDate="2020-12-28T22:33:08.407" Title="What keeps Chthonian planets so dense?" Tags="&lt;exoplanet&gt;&lt;chthonian&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23075" PostTypeId="2" ParentId="23073" CreationDate="2017-10-10T22:43:36.970" Score="5" Body="&lt;p&gt;10 Jupiter masses at around 2 Earth radii?&lt;br&gt;&#xA;That for sure doesn't exist / would be quite the sensation to discover.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you look at data of any kind, one should pay attention to the measurment errors at least as much as to the actual value.&lt;br&gt;&#xA;A regular physics result (for example for a measurment of the gravitational acceleration $g$ of where you stand) looks like &#xA;$$g=(9.81 \pm 0.02) \frac{m}{s^2} $$&#xA;or if for any reason you have asymmetric errors&lt;br&gt;&#xA;$$g=(9.81^{+0.02}_{-0.01}) \frac{m}{s^2} $$&#xA;and errors always give an idea about how uncertain the method is with which the value was derived. &#xA;Now if you do take a look at the errors reported for the Mass quoted on the website you see that they are &#xA;$$M_{planet} = (10.41^{+0.0}_{-10.41}) $$&#xA;or so to say highly asymmetric, which should make one suspicious.&lt;br&gt;&#xA;A look then into the original publication makes it clear that this cited mass is in fact only an absolute upper limit.&lt;br&gt;&#xA;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2013MNRAS.428.1077S&quot; rel=&quot;noreferrer&quot;&gt;The authors of the paper&lt;/a&gt; were using two methods to estimate the masses of planets.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Looking for &lt;a href=&quot;https://en.wikipedia.org/wiki/Transit-timing_variation&quot; rel=&quot;noreferrer&quot;&gt;transit timing variations&lt;/a&gt; of known, and seen transiting systems. That means they had the system Kepler 52, with transiting planets K52b,c . K52b because it transits way more often than c has a well-determined period (Period with small errors!) and because of that any deviation in expected future transit time could be attributed to the \textbf{maximum masses} of K52c.&lt;/li&gt;&#xA;&lt;li&gt;The more massive and the more compact a system is, the quicker it will destabilize. This fact is often used in reverse, to take the system age and at given distances derive maximum masses below which the system must lie, or else it would have already flown apart.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Both methods can only give maximum masses and I'll just leave here fig. 5 from the original paper with the planet you're interested in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NjrDG.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NjrDG.png&quot; alt=&quot;(c) http://adsabs.harvard.edu/abs/2013MNRAS.428.1077S&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now remembering, that $1 M_J \approx 320 M_{\oplus}$, you see where your 10 Jupiter masses for K52c come from: That's the planets possible maximum mass for system stability. The TTV method gives already a constraint that is 100 times lower ($37.4 M_{\oplus} \approx 0.11 M_J$).&lt;br&gt;&#xA;Thus $37.4 M_{\oplus}$ is the planets true maximum mass.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is clearly an error on the side of exoplanet.eu, but then probably there are too many planets and papers to read for whoever puts those data in there.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summarizing&lt;/strong&gt;&lt;br&gt;&#xA;What we have here is only a maximum mass. Also the wrong one. To say what is now more probable, if $M_{K52c} = 37.4 M_{\oplus}$ or $M_{K52c} = 3.74 M_{\oplus}$ i'm not certain enough if I understand their anticorrelation method for the TTV signals.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2017-10-10T22:43:36.970" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23076" PostTypeId="1" CreationDate="2017-10-11T02:03:09.123" Score="1" ViewCount="174" Body="&lt;p&gt;I was looking to Scorpio the other day, and noticed that the last star in the tail (the telson's tip) -- known as Upsilon Scorpii, υ Scorpii, or Lesath -- seemed yet more red than the &quot;heart&quot; of the scorpion -- Antares or Alpha Scorpii. Was it that I was drunk, or some kind of sky pollution? (Actually, a friend confirmed it was red, and she seemed sober than me.) Or is it, indeed, red? I couldn't find any information on-line. I saw &lt;a href=&quot;http://www.usno.navy.mil/USNO/astrometry/optical-IR-prod/icas/fchpix&quot; rel=&quot;nofollow noreferrer&quot;&gt;this database&lt;/a&gt;, but don't know how to find a star by its name, or if they have information about the color.&lt;/p&gt;&#xA;" OwnerUserId="813" LastEditorUserId="2365" LastEditDate="2017-10-11T15:55:20.460" LastActivityDate="2017-10-11T15:55:20.460" Title="What's the color of Upsilon Scorpii?" Tags="&lt;star&gt;&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="23077" PostTypeId="1" CreationDate="2017-10-11T06:01:53.310" Score="-1" ViewCount="667" Body="&lt;p&gt;Maybe this is more appropriate for the Physics or Maths, but anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am struggling with applying Canonical units with formulas. I am struggling specifically with how they are used, am I to convert all the time? Use the whole numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example is a a simple velocity of a circular orbit: &lt;code&gt;v= SQRT Mu/r&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;r  =2DU (12,756.23KM)&#xA;Mu =1DU3/TU3 (3.9860KM3/s2)&#xA;&#xA;v = 4.5641KM/S or 0.707DU &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In order to convert it back to DU I had to use a percentage calculator (I'm bad at fractions). Is there a simpler way of converting back and forth?&lt;/p&gt;&#xA;" OwnerUserId="19014" LastEditorUserId="37191" LastEditorDisplayName="user1569" LastEditDate="2021-02-11T09:24:47.750" LastActivityDate="2021-02-11T09:24:47.750" Title="Astrodynamic Canonical Units" Tags="&lt;orbit&gt;&lt;astrophysics&gt;&lt;units&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2017-10-14T06:25:36.617" ContentLicense="CC BY-SA 3.0" />
  <row Id="23080" PostTypeId="2" ParentId="23076" CreationDate="2017-10-11T10:58:00.607" Score="2" Body="&lt;p&gt;I don't know what could cause &lt;a href=&quot;https://en.wikipedia.org/wiki/Upsilon_Scorpii&quot; rel=&quot;nofollow noreferrer&quot;&gt;Upsilon Scorpii&lt;/a&gt; to appear red as it is not noted as being a highly variable star. It's spectral type is B2, which implies it would be blue/white in colour. If you know the spectral type of a star, which many stellar databases will note, you can work out its colour from this.&lt;/p&gt;&#xA;" OwnerDisplayName="user10106" LastActivityDate="2017-10-11T10:58:00.607" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23085" PostTypeId="1" AcceptedAnswerId="23111" CreationDate="2017-10-11T16:44:37.593" Score="6" ViewCount="141" Body="&lt;p&gt;September 21, 2017 a bright fireball over Holland was reported by hundreds of people. Several all-sky cameras recorded it as well as a special radio beacon set up to capture radio signatures of meteors - the BRAMS beacon at Dourbes, Belgium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm puzzled by what I see in the radio spectrogram from one of the BRAMS stations at Ophain, Belgium :&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ZA02d.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZA02d.png&quot; alt=&quot;BRAMS radio spectrogram from Ophain&quot;&gt;&lt;/a&gt;&#xA;The orange blob in the center is the radio signature of this fireball.&#xA;The horizontal line is the beacon signal (0 Hz), time runs horizontally, frequency runs vertically, positive above and negative below the beacon line.&#xA;What is displayed are Doppler shifts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The track for this fireball has been computed since and relative to the radio receiver at Ophain and radio beacon at Dourbes it is receding from both.&#xA;However, the Doppler blob is above the beacon line which is only possible when the fireball recedes from Ophain but approaches the beacon, approaches both, or approaches Ophain but recedes from the beacon - NOT when it recedes from both!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way I can make sense out of this is to assume that the ionized&#xA;gas trail from the fireball reflecting the radiowaves was NOT receding from the receiver and transmitter beacon - only the fireball was?!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or are there any other explanations?&lt;/p&gt;&#xA;" OwnerUserId="19025" LastEditorUserId="19025" LastEditDate="2017-10-11T16:49:39.817" LastActivityDate="2017-10-12T20:25:43.820" Title="What explains this puzzling radio signature from a fireball?" Tags="&lt;meteor&gt;&lt;radio&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23086" PostTypeId="2" ParentId="23034" CreationDate="2017-10-11T16:49:46.737" Score="-1" Body="&lt;p&gt;Well, one reckons that light from the sun will travel to the end of our universe, if in fact it has an end.  One way to calculate the actual distance to the end of our universe is as follows:&#xA;(1)  assume that our universe is a very large &quot;black hole&quot; ----- understanding that a &quot;black hole&quot; is simply a volume of space from which light cannot escape.  (2)  use Schwartzschild's famous formula for the radius of a &quot;black hole&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;R  =  2GM/c.c&lt;/strong&gt;  where &quot;M&quot; is mass of universe.  If one use numeric value M = 1.58E58 grams (following Sternglass), then one calculates that sun-light will reach the end of our universe after travelling approximately 2.35E30 cm.&lt;/p&gt;&#xA;" OwnerUserId="13290" LastActivityDate="2017-10-11T16:49:46.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23087" PostTypeId="2" ParentId="23034" CreationDate="2017-10-11T22:26:28.350" Score="0" Body="&lt;p&gt;Using your scenario, and lets ignore dark energy expansion and red-shifting at greater distances,.  There would be no distance where the sun will disappear.  That might seem counter intuitive but it's true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Earth, we receive about 10^17 photons per square CM.  (&lt;a href=&quot;https://www.reddit.com/r/askscience/comments/2n6zo0/how_many_photons_per_second_would_hit_a_1_cm/&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;).  Using the inverse square rule, to receive 10 photons per second on a pair of human eyes, that distance would be 100 million AU - give or take, or about 1,580 light years.    We probably can't see just 10 photons, but in pitch black, with eyes adjusted to the dark, maybe we could.   That's about the distance where the sun would give of steady but very very dim light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further than that, what happens has more to do with the quantum nature of light than the brightness and distance.   As you double the distance you see 1/4th the frequency of photons.   At 5,000 light years you see 1 photon per second, basically a very dim blinking light, but just as bright as it is at 1,580 light years, just appearing less often.     Double the distance again and it flashes once every 4 seconds.    There's no distance where you stop seeing the sun, you just see it less and less often, but it never appears dimmer than 1 photon.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-10-11T22:26:28.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23089" PostTypeId="1" AcceptedAnswerId="23106" CreationDate="2017-10-12T03:06:11.860" Score="9" ViewCount="579" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/File:Asteroids_by_size_and_number.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;This diagram from wikipedia&lt;/a&gt; (shown below) graphs asteroid Diameter against Number of asteroids, for the Solar System.  (Graphic is courtesy of the Marco Colombo, DensityDesign Research Lab).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably the Diameter data for the larger asteroids is strongly supported by observations.  But for progressively smaller bodies direct observational data is presumably less available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pattern shows a strong, but not perfect, linear inverse correlation between log(Diameter ) and log(Number).  This seems quite reasonable for large bodies. It also sounds plausible that such a pattern continues into the realm of small asteroids and meteoroids.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what is the physical justification for extending the log:log pattern into the realm of small bodies which are too small to see from Earth?  Presumably the current estimate of the total mass of the Astroid Belt (4% of the mass of the Moon) is consistent with this distribution model.  But can we confidently rule out the possibility that the linear log:log pattern breaks down for small objects allowing, for example, a much larger number (and mass) of small bodies to be present in the Asteroid Belt, sufficient say for the total combined mass to equal 100% of our Moon's mass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DkY2O.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DkY2O.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="233" LastEditorUserId="233" LastEditDate="2017-10-17T02:10:17.173" LastActivityDate="2017-10-17T02:10:17.173" Title="How is the total mass of all the Solar System's small asteroids and meteoroids estimated?" Tags="&lt;asteroids&gt;&lt;mass&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23090" PostTypeId="1" CreationDate="2017-10-12T03:08:27.300" Score="27" ViewCount="6729" Body="&lt;p&gt;Equivalent formulations of this question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Would it make any sense to draw a map of the Sun?&lt;/li&gt;&#xA;&lt;li&gt;Is the Sun heterogeneous with respect to latitude and longitude? (I know that it is heterogeneous with respect to altitude/depth)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I know that the Sun does not rotate uniformly, so any map would line up differently as time passes.  However, this does not rule out a slice of the Sun at a given latitude having permanent geographical features, since this slice &lt;em&gt;does&lt;/em&gt; rotate uniformly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, even geographical features on Earth are not really permanent, they change on long time scales.  I don't expect the Sun to have any eternal features, I'm really just thinking about features that change on long time scales, or at least longer than a few months (e.g. not sunspots).&lt;/p&gt;&#xA;" OwnerUserId="19032" LastEditorUserId="7472" LastEditDate="2017-10-12T17:52:39.037" LastActivityDate="2017-10-12T18:25:51.127" Title="Does the Sun have permanent geographical features?" Tags="&lt;the-sun&gt;" AnswerCount="3" CommentCount="14" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23091" PostTypeId="2" ParentId="23090" CreationDate="2017-10-12T04:46:42.707" Score="31" Body="&lt;p&gt;As you suspect the Sun does not have any permanent surface features. Up till 1951 the longest actually observed sunspot group lasted 134 days see &lt;a href=&quot;http://adsabs.harvard.edu/full/1951ASPL....6..146P&quot; rel=&quot;noreferrer&quot;&gt;http://adsabs.harvard.edu/full/1951ASPL....6..146P&lt;/a&gt;. That duration gives you an idea how long sunspot groups last. There are associated with coronal holes which would last about as long. However, the structure (shape size) of the group and any coronal hole would be constantly changing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Temporary maps are made and there is a heliographic co-ordinate system, with a latitude and longitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sunspots show the rotation of the Sun, and so a north and south solar pole can be found. North is in the same side of the ecliptic as the Earth's north pole). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solar longitude is much more arbitary. The zero helographic meridian is defined as the plane (solar pole to solar pole) passing through where the solar equator crossed the plane of the ecliptic (earths orbital plane) at Greenwich mean noon on January 1, 1854. see &lt;a href=&quot;http://wso.stanford.edu/words/Coordinates.html&quot; rel=&quot;noreferrer&quot;&gt;http://wso.stanford.edu/words/Coordinates.html&lt;/a&gt;&#xA;Since then astronomers have just used a constant rotation rate based upon the suns equatorial rotation rate. These are called Carrington Solar Coordinates&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar system is used for the longitude of the gas giant planets like Jupiter and Saturn.&lt;/p&gt;&#xA;" OwnerUserId="10761" LastActivityDate="2017-10-12T04:46:42.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23093" PostTypeId="2" ParentId="14613" CreationDate="2017-10-12T05:29:47.140" Score="0" Body="&lt;p&gt;The Universe is infinite and always has been. Yes, it is expanding, but what does that mean when it exists everywhere? Yes, it can appear to have exploded from a point, but that &quot;point&quot; is a universe that is infinite in size and has no edge. There is no way to look at it or describe it from &quot;outside&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diagrams you often see of the history of the Big Bang show a poor concept of expansion and are just misleading. They imply an outside view is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the excellent answer to a similar question here:  &lt;a href=&quot;https://physics.stackexchange.com/questions/136860/did-the-big-bang-happen-at-a-point&quot;&gt;https://physics.stackexchange.com/questions/136860/did-the-big-bang-happen-at-a-point&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5526" LastEditorUserId="5526" LastEditDate="2017-10-14T01:52:53.523" LastActivityDate="2017-10-14T01:52:53.523" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23101" PostTypeId="2" ParentId="23058" CreationDate="2017-10-12T17:50:29.193" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How does the ISM resist gravity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It doesn't.&lt;/strong&gt; There are two distinct sources of gravity: internal and external. The internal or self-gravity of the ISM may in fact result in collapse and subsequent star formation, as explained in &lt;a href=&quot;https://astronomy.stackexchange.com/a/23062/633&quot;&gt;another answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The external gravitational attraction from any star or gas cloud on the ISM is too weak to be of relevance and can be neglected, as demonstrated in yet &lt;a href=&quot;https://astronomy.stackexchange.com/a/23061/633&quot;&gt;another answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the ISM is subjected to the combined gravitational pull from all stars, gas, and dark matter in the Galaxy, i.e.\ the &lt;strong&gt;gravity of the Galaxy&lt;/strong&gt; itself. In response to this pull, the ISM orbits the Galaxy on near-circular orbits, as do most of the stars (in a disc galaxy such as ours). Thus, the ISM is &lt;em&gt;not special&lt;/em&gt; in this respect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the ISM not fall into the inner Galaxy (where it is pulled to)? This is simply because it has too much angular momentum. The situation is exactly the same as for the Earth pulled to the Sun, yet orbiting (nearly) on a circle around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, note that magnetic forces and radiation pressure from nearby stars are much weaker than Galactic gravity and can be neglected when considering the Galactic orbits of the ISM.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2017-10-12T17:50:29.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23102" PostTypeId="2" ParentId="23077" CreationDate="2017-10-12T18:05:56.787" Score="2" Body="&lt;p&gt;It's not entirely clear what your problem is, but the best way is to stick to a certain unit system for your variables and use the correct physical relations. Then, if you want to express the variables in another unit, you must convert them. The standard (SI) units are inconvenient in astronomy, as they give some unwieldy numerical values, hindering intuitive understanding/interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in the Solar system or other planetary systems, a useful unit system is as follows.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;distances in AU (astronomical unit)&#xA;masses in Solar masses&#xA;time in years&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then Newton's constant of gravity is $G=4\pi^2$ and the velocity $V_c=\sqrt{GM/r}$ is in AU/yr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you can use another unit set (for example one for which $G=1$ at the expense of a strange mass or time unit), but you should not switch between different unit systems.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2017-10-12T18:05:56.787" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23103" PostTypeId="2" ParentId="23090" CreationDate="2017-10-12T18:24:01.503" Score="9" Body="&lt;p&gt;If you're not limited to the visible spectrum, the Sun has more-or-less permanent coronal holes at its poles which are nearly always present (albeit constantly changing in shape and size). These coronal holes are responsible for generating the &quot;fast&quot; solar wind. These change on fast enough time scales though that its not useful to make a map of them (or other coronal holes which may crop up).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is an image of one such coronal hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3NNDH.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3NNDH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-10-12T18:24:01.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23104" PostTypeId="2" ParentId="23090" CreationDate="2017-10-12T18:25:51.127" Score="2" Body="&lt;p&gt;The Sun has a north and south pole, and an equator. These can be considered geographic features.&lt;/p&gt;&#xA;" OwnerUserId="19050" LastActivityDate="2017-10-12T18:25:51.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23105" PostTypeId="1" CreationDate="2017-10-12T19:14:50.523" Score="3" ViewCount="794" Body="&lt;p&gt;So the story goes like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A long time ago, &lt;a href=&quot;https://en.wikipedia.org/wiki/Age_of_the_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;13.799±0.021 billion years&lt;/a&gt; to be exact, something happened. It was a &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Bang&quot; rel=&quot;nofollow noreferrer&quot;&gt;big bang, &lt;strike&gt;loud explosion&lt;/a&gt;&lt;/strike&gt; and universe came to existence. It grew and grew, and now we know that the observable universe is about &lt;a href=&quot;https://en.wikipedia.org/wiki/Observable_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;93 billion light years&lt;/a&gt; in diameter. So it means that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;objects that were once close ... are now up to around 45.7 billion&#xA;  light years away rather than up to 13.799 billion light years away ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hence my question is that everything is traveling faster than the speed of light?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is, then what does it mean in terms of &lt;a href=&quot;https://en.wikipedia.org/wiki/Special_relativity&quot; rel=&quot;nofollow noreferrer&quot;&gt;special relativity&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I did read:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/q/1948/698&quot;&gt;Is the universe expanding faster than speed of light?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/q/1860/698&quot;&gt;How fast is the universe expanding?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="698" LastEditorUserId="698" LastEditDate="2017-10-16T13:52:59.530" LastActivityDate="2017-10-16T14:04:47.810" Title="How can the universe be expanding faster than speed of light?" Tags="&lt;universe&gt;&lt;light&gt;&lt;big-bang-theory&gt;&lt;expansion&gt;&lt;observable-universe&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ClosedDate="2017-10-15T17:16:53.577" ContentLicense="CC BY-SA 3.0" />
  <row Id="23106" PostTypeId="2" ParentId="23089" CreationDate="2017-10-12T19:16:54.337" Score="8" Body="&lt;p&gt;Your question body is different from your question title and it seems you really want to ask what you did in the question body so I'll address that.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Short Answer:&lt;/strong&gt; The simple power law which applies for larger asteroids and comets actually doesn't extend that well to smaller bodies and shouldn't be trusted too much in that range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Long Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're right to be suspicious of using a simple power law to relate asteroid size to quantity at small sizes. Especially because there are known deviations from that power law at certain radii. As &lt;a href=&quot;https://en.wikipedia.org/wiki/Asteroid#Size_distribution&quot; rel=&quot;noreferrer&quot;&gt;wikipedia&lt;/a&gt; notes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The number of asteroids decreases markedly with size. Although this generally follows a power law, there are 'bumps' at 5 km and 100 km, where more asteroids than expected from a logarithmic distribution are found.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who's to say there aren't other bumps or deviations at radii that we can't observationally confirm. Our confidence in any power law's applicability for small asteroids and comets basically comes from modelling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This power law you're asking about is essentially something referred to as an &quot;Initial Mass Function&quot; (IMF). A &lt;em&gt;lot&lt;/em&gt; of work has been done to define an IMF for stars, relating the mass of stars to the number of stars at that mass. This stellar IMF follows a power law more or less, but strongly deviates for very low masses. At the surface, it would be reasonable to assume an IMF for asteroids and comets likewise deviates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way you answer this, is through models. See for example &lt;a href=&quot;http://www.boulder.swri.edu/~bottke/Reprints/Cuzzi_2010_Icarus_208_518_Toward_Initial_Mass_Functions_Asteroids_KBOs.pdf&quot; rel=&quot;noreferrer&quot;&gt;Cuzzi, Hogan, and Bottke (2010)&lt;/a&gt; (that link is to the formal journal paper, but for some reason some plots are obscured so you can also look at their &quot;unofficial&quot; &lt;a href=&quot;https://arxiv.org/abs/1004.0270&quot; rel=&quot;noreferrer&quot;&gt;arxiv version&lt;/a&gt;). In this paper, the authors attempt to construct a representative asteroid/comet population by modeling their formation via accretion of protoplanetary dust grains into asteroids, comets, &lt;a href=&quot;https://en.wikipedia.org/wiki/Kuiper_belt&quot; rel=&quot;noreferrer&quot;&gt;KBO&lt;/a&gt;, and planetesimals. They make certain assumptions about the initial gas disk around the Sun and bake in physics of how these dust grains formed into larger conglomerates. Their end goal is to produce an IMF that identifies how many asteroids/comets exist at a given size. They back up a lot of their modelling with data from observations so they stick to reality as much as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll let you peruse their results since they actually produce many different IMFs using a variety of different assumptions and starting conditions. However, I think one can summarize their results in a way which addresses your main question. As they state it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Asteroids were born big&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What they mean is that asteroids (and comets), rather than forming by successive growth via accretion of many small particles, seem to just clump together into larger bodies, completely bypassing the formation of many smaller meter-sized asteroids and comets (that's not to say there aren't meter-sized objects). You can actually see this in their various plots (fig. 4 for example). Much like the stellar IMF, the power law drastically changes once you get to small sizes and the power law which applied for larger sizes is no longer applicable. At which size the general power law no longer becomes applicable depends upon certain tunable parameters. Cuzzi et al. shows many different results with their tunable parameters set to many different possible values.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-10-12T19:16:54.337" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23107" PostTypeId="1" AcceptedAnswerId="23113" CreationDate="2017-10-12T20:00:52.947" Score="8" ViewCount="413" Body="&lt;p&gt;We've been talking about Cassini with my 6-year old daughter recently and I've mentioned that it picked up more speed by passing close to Venus (twice), Earth and Jupiter. But then, I've struggled to explain how did it pick up speed, why and how does it work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you recommend to explain/demonstrate the concept of &quot;gravity assist&quot; to a 6-year old?&lt;/p&gt;&#xA;" OwnerUserId="17296" LastActivityDate="2017-10-13T13:04:43.173" Title="Explaining gravity assist to a 6-year old" Tags="&lt;gravity&gt;&lt;space-travel&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23108" PostTypeId="2" ParentId="23105" CreationDate="2017-10-12T20:08:24.090" Score="4" Body="&lt;p&gt;You often hear the phrase: &lt;em&gt;Nothing can travel faster than the speed of light&lt;/em&gt;. Which is of course wrong. Public knowledge of physics is filled with somewhat true, but slightly wrong statements from physics. Something just seems to get garbled when taken from the real theories and converted into something the average bystander could understand. A more correct statement would be: &lt;em&gt;Information cannot travel through spacetime faster than the speed of light&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the two important parts this: &lt;em&gt;Information&lt;/em&gt; and &lt;em&gt;through spacetime&lt;/em&gt;. This allows for plenty of cases where something can actually travel faster than light. For example, a galaxy may recede from us at speeds greater than light (and many are) because those galaxies are not travelling &lt;em&gt;through&lt;/em&gt; spacetime. Rather spacetime itself is expanding and that can expand as fast as it pleases. Special relativity doesn't limit the expansion speed of spacetime itself, only information traveling through spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is the &lt;a href=&quot;https://en.wikipedia.org/wiki/EPR_paradox&quot; rel=&quot;nofollow noreferrer&quot;&gt;EPR paradox&lt;/a&gt; by which particles may be quantum entangled and somehow &quot;communicate&quot; over vast distances instantly to achieve the same quantum state at the same time. Clearly if the particles are actually communicating with each other, they'd have to do so at speeds faster than the speed of light. The key in this case is that no &lt;em&gt;information&lt;/em&gt; can be transferred between the two particles. That is, they can seemingly know about each others' state via FTL communication but this fact does not allow one to transmit information back and forth (e.g., sending ones and zeros to transmit a message). Since no information can be exchanged FTL this is not a violation of SR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;FWIW, I'm not sure if you're being facetious concerning your &quot;loud explosion&quot; comment, but the Big Bang most certainly wasn't an explosion.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="698" LastEditDate="2017-10-16T14:04:47.810" LastActivityDate="2017-10-16T14:04:47.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23109" PostTypeId="2" ParentId="23105" CreationDate="2017-10-12T20:12:31.837" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;It was a loud explosion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First, we have to clarify what the Big Bang was. The name is a misnomer, as it was neither loud nor an explosion. All we know is that the Universe is currently expanding — that is, space is literally being created between all matter. We have plenty of answers here explaining this more intuitively, such as &lt;a href=&quot;https://astronomy.stackexchange.com/questions/13999/how-does-the-concept-of-a-universe-with-no-center-work/14002#14002&quot;&gt;mine&lt;/a&gt; (which also answers your question about its implications on special relativity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, if we were to look further back in time, we'd expect there to be less and less space between matter. Our equations tell us that at 13.8 billion years in the past, the amount of space in the Universe should have been zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a problem. In general relativity, you can't have a metric with zero space. Thus, we know general relativity is incomplete. It cannot explain what happened at the moment of the Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[the] universe came to existence&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No such claim is made. All we know is that &lt;em&gt;something&lt;/em&gt; must have happened back then. Our current physical models do not accurately represent the extremely early Universe.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hence my question is that everything is traveling faster than the speed of light?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The expansion of the Universe is not happening at the same rate as it was back then. It is slow enough that everything within our Hubble sphere is receding at a lower velocity.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2017-10-12T20:12:31.837" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23110" PostTypeId="2" ParentId="23107" CreationDate="2017-10-12T20:20:58.397" Score="4" Body="&lt;p&gt;I tend to think the best explanations are the simple ones which explain the core concepts of the concept without getting too fancy or detailed. That being said, I'd just explain what happens during a gravity assist.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cassini flew by Jupiter and when it got close enough Jupiter's gravity began to pull Cassini towards it. This is exactly the same as what pulls things back to the ground on Earth. Because Cassini was being pulled towards Jupiter, it was basically &quot;falling&quot; onto Jupiter and was able to pick up some extra speed by falling. However, since Cassini was traveling sooo incredibly fast, it actually didn't fall onto Jupiter and was able to fly away instead with the extra speed it picked up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't say I've explained something like that to a 6 year old (and confirmed they understood it) but I think that is both factually accurate and simple enough for a young child to understand.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-10-12T20:20:58.397" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23111" PostTypeId="2" ParentId="23085" CreationDate="2017-10-12T20:25:43.820" Score="3" Body="&lt;p&gt;I checked with the people who run the BRAMS project and they confirmed my reasoning, and added another possible explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if the fireball moves away from the receiver AND transmitter, the ionization trail it leaves behind can expand and be blown by mesospheric winds towards the receiver and/or transmitter. It's this trail that mostly reflects the radio waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And theoretically it's still possible but unlikely that the observed Doppler signal was coming from another meteor trail unrelated to this particular fireball, but coinciding with the exact time frame the fireball appeared in.&lt;/p&gt;&#xA;" OwnerUserId="19025" LastActivityDate="2017-10-12T20:25:43.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23112" PostTypeId="2" ParentId="22946" CreationDate="2017-10-12T20:47:48.160" Score="0" Body="&lt;p&gt;The northern reference point is used in navigation.&#xA;The southern reference point is used in positional astronomy.&#xA;This is purely a convention, and as long as you know what reference point angles refer to, there should be no confusion.&lt;/p&gt;&#xA;" OwnerUserId="19025" LastActivityDate="2017-10-12T20:47:48.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23113" PostTypeId="2" ParentId="23107" CreationDate="2017-10-12T21:38:18.643" Score="8" Body="&lt;p&gt;I like the trowing a ball against a speeding train analogy.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lrXL9.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lrXL9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important factor is that Jupiter is moving, quite fast.  About 13 km/s.   The other important factor is that Jupiter is very massive, so it has a lot of gravity and it can draw objects towards it, causing their direction to bend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tricky concept for a 6 year old is that gravity is a zero-sum game.   The object that falls towards Jupiter, in this case it's a hyperbolic orbit but regular orbits work too, so as it falls towards it, it adds velocity then flying away from it, it loses the same amount of velocity, so the net velocity relative to Jupiter is zero.  No change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But because Jupiter is moving, Jupiter can tug the spacecraft along with it.  The velocity relative to Jupiter is unchanged but the velocity relative to the sun changes, Jupiter is like the train that the ball bounces off of.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The opposite can happen too.   Jupiter can be used to slow a spacecraft down. It depends on the angle of approach, like throwing a ball against a train while it moves away from you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also do this experiment below (with 2 balls not 3 to represent the gravity assist, though it's a cooler experiment with 3).  It's the same principal.   Bouncing off a moving object (Jupiter) is very different than bounding off a flat surface that isn't moving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=2UHS883_P60&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=2UHS883_P60&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that said, I'd still point out that this is a tricky concept and not to worry if they don't understand it.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-10-13T00:34:33.863" LastActivityDate="2017-10-13T00:34:33.863" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23114" PostTypeId="2" ParentId="19316" CreationDate="2017-10-13T00:14:54.513" Score="1" Body="&lt;p&gt;In essence read of the peak flux value and if a point source this will be the same as your integrated flux.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to elaborate on this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this question your source appears to be comparable to your beam. so I would treat as a point source. - note you might also want to consider that many people apply a &quot;match filter&quot; to their data if dealing with faint point source emission - this smooths the map - increasing the effective beam size within the map and also there may be some flux loss (see for example: &lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/bib_query?arXiv:1707.00990&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/cgi-bin/bib_query?arXiv:1707.00990&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will see this (mJy/beam to mJy) is also discussed here: &lt;a href=&quot;http://www.eaobservatory.org/jcmt/help/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.eaobservatory.org/jcmt/help/&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a real point source, a peak brightness value reported in units of mJy is the same as a peak brightness value reported in mJy/beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what happens if we have a map in mJy/beam and we want to obtain an integrated intensity value, a total flux value? We first sum up a number of pixels and now we want to get our units correct from mJy/beam to mJy…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total Flux = flux summed over a number of pixels/(number of pixels in a beam)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then your units are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[mJy*pixels/beam] / [pixels/beam] = [mJy].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we know that For a Gaussian:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beam Area =  2 × π × σ² [arcsec]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the σ of the Gaussian beam can be calculated from the JCMT FWHM values at 850 and 450 microns (reminder the beam components are provided in Dempsey’s 2013 paper).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FWHM = 2 σ √(2 ln 2)   [arcsec]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can use the FWHM to obtain σ to calculate the Beam Area and report the beam area in terms of pixels:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;number of pixels in a beam = Beam Area [arcsec] / (pixel length)²&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally I should add you will also want to do an aperture correction - see Dempsey et al 2013's calibration paper if you are using an aperture of a different size than 60&quot;.&lt;/p&gt;&#xA;" OwnerUserId="19057" LastActivityDate="2017-10-13T00:14:54.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23115" PostTypeId="1" CreationDate="2017-10-13T05:36:34.110" Score="2" ViewCount="101" Body="&lt;p&gt;The tilt of the Earth is the main factor that determines the seasons. It's also true that the eccentricity of Earth's orbit is very close to zero, implying that aphelion and perihelion distances are relatively close. If the eccentricity of Earth's orbit were close to but less than one, then would the distance between the Earth and the Sun become a dominant factor in determining the seasons?&lt;/p&gt;&#xA;" OwnerDisplayName="user15317" LastActivityDate="2017-11-12T22:56:57.603" Title="Is the eccentricity of a planets orbit relevant to finding whether the distance between said planet and its host star determines seasons?" Tags="&lt;orbit&gt;&lt;eccentric-orbit&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23118" PostTypeId="2" ParentId="23105" CreationDate="2017-10-13T11:59:53.973" Score="1" Body="&lt;p&gt;I do not fully understand how space and time do what they do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm sure that &quot;&lt;em&gt;universe expanding faster than speed of light&lt;/em&gt;&quot; doesn't mean that &lt;strong&gt;&lt;em&gt;locally, something&lt;/em&gt;&lt;/strong&gt; (i.e. any bit of information) is moving faster than light. It means to me that particles that already are too far away to exchange any information are &lt;em&gt;becoming increasingly separated by &quot;space&quot;&lt;/em&gt; at a rate that can be arbitrarily big.&lt;/p&gt;&#xA;" OwnerUserId="11264" LastActivityDate="2017-10-13T11:59:53.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23119" PostTypeId="2" ParentId="23107" CreationDate="2017-10-13T13:04:43.173" Score="1" Body="&lt;p&gt;I might modify Zephyr's answer just a hair -- and I have not tried this on a 6-year old, let alone the average adult, who is much more difficult to explain things to than any child is.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity is understood, so show  that gravity pulls both the satellite towards Jupiter and Jupiter towards the satellite.  But since  $ m_{Jupiter} &amp;gt;&amp;gt; m_{satellite} $, Jupiter speeds up a teensy bit while the satellite speeds up a whole lot in the opposite direction.  (Plus they manage to miss colliding :-)  )&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-10-13T13:04:43.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23122" PostTypeId="1" CreationDate="2017-10-13T15:31:11.873" Score="3" ViewCount="146" Body="&lt;p&gt;I have been looking at the website &lt;a href=&quot;https://thenightsky.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://thenightsky.com&lt;/a&gt; and wanted to create a star map my self, from a specific place on a specific date. Where would I be able to find this?&lt;/p&gt;&#xA;" OwnerUserId="19071" LastActivityDate="2020-10-18T20:33:54.090" Title="Star map, where?" Tags="&lt;space&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2020-10-19T22:42:51.403" ContentLicense="CC BY-SA 3.0" />
  <row Id="23124" PostTypeId="1" AcceptedAnswerId="23125" CreationDate="2017-10-13T17:05:17.867" Score="2" ViewCount="273" Body="&lt;p&gt;I was wondering about a planet who orbits two stars. Could it be that it moves as in the image?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qGacD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qGacD.png&quot; alt=&quot;Planet orbit with stars&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19073" LastEditorUserId="2365" LastEditDate="2017-10-13T18:01:48.190" LastActivityDate="2020-08-15T10:31:56.767" Title="Planet orbit circumbinary planet" Tags="&lt;planet&gt;&lt;orbit&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23125" PostTypeId="2" ParentId="23124" CreationDate="2017-10-13T18:02:08.290" Score="3" Body="&lt;p&gt;While such orbits can be proven to exist mathematically, they are unstable, and so no real planet has such an orbit. Such an orbit is analogous to balancing a pencil on its point. Mathematically there is a point of balance, but it is still impossible to do.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-10-13T18:02:08.290" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23128" PostTypeId="2" ParentId="23115" CreationDate="2017-10-13T22:33:34.873" Score="1" Body="&lt;p&gt;Yes it would. If the sun was much further away the season on earth would certainly be more winter-like. The reason distance is not the deciding factor for earth is the variation in distance is fairly small.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastActivityDate="2017-10-13T22:33:34.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23129" PostTypeId="2" ParentId="23122" CreationDate="2017-10-14T02:56:48.550" Score="4" Body="&lt;p&gt;As mentioned in the comments, download and install the free app Stellerium. You can input the location and date of interest and it will display the sky. The app is like having your own planetarium. There is another similar app named Celestia. It is also free. I would recommend Stellerium.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastActivityDate="2017-10-14T02:56:48.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23134" PostTypeId="1" CreationDate="2017-10-14T09:32:10.443" Score="2" ViewCount="30" Body="&lt;p&gt;The heliopause is where the Solar wind is counter balanced by the particle winds from other stars. This boundary seems to not be spherical and also to fluctuate in shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the shape of the heliopause be estimated by the pressure from surrounding stars and open clusters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the fluctuations be attributed to specific stars in the Solar neighborhood? Are there single stars both so active and so nearby that they dominate the fluctuations of the Heliopause? Did Voyager in effect observe the effects of a specific event on a specific star's surface when the heliopause boundary billowed back and forth over it? &lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2017-10-14T09:32:10.443" Title="Can the shape and fluctuations of the heliopause be attributed to specific nearby stars and clusters?" Tags="&lt;star&gt;&lt;interstellar-medium&gt;&lt;solar-wind&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23137" PostTypeId="1" CreationDate="2017-10-14T15:50:41.593" Score="1" ViewCount="115" Body="&lt;p&gt;As I understand it, the Sun will transition from its current phase as a 'yellow dwarf', to a red giant, and finally to a white dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even during these phases I understand there are transitional considerations, including the increases in luminosity nearing the end of the main phase, reduced temperatures, various ejections and other things being put into solar system, and even the formation of a nebula during the white dwarf phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At each stage, sub-stage, and during each transition of the Sun, is there a region the Earth could theoretically orbit and support life similarly to how it does today?&lt;/p&gt;&#xA;" OwnerDisplayName="user19087" LastEditorDisplayName="user19087" LastEditDate="2017-10-15T18:44:23.480" LastActivityDate="2017-10-15T18:44:23.480" Title="At each stage and during each transition of the Sun, is there a region the Earth could theoretically orbit and support life?" Tags="&lt;the-sun&gt;&lt;orbit&gt;&lt;earth&gt;" AnswerCount="0" CommentCount="15" FavoriteCount="0" ClosedDate="2017-10-20T10:59:54.173" ContentLicense="CC BY-SA 3.0" />
  <row Id="23138" PostTypeId="2" ParentId="19248" CreationDate="2017-10-14T19:23:33.487" Score="2" Body="&lt;p&gt;The current popular theory, that the Moon formed from accretion, would nearly eliminate any rotation of the Moon (relative to Earth). I say 'nearly', because objects striking the forming (accreting) Moon on the side further from Earth would have added slightly more rotational energy than objects striking the nearer side. This would give the Moon a slightly (negligible) retrograde rotation in its early history.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Moon has been rotationally locked to the Earth since its formation (current popular theory), no rotational energy has been lost due to tidal forces, so the term 'tidally locked' is misapplied to the Moon.&lt;/p&gt;&#xA;" OwnerUserId="19089" LastActivityDate="2017-10-14T19:23:33.487" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23139" PostTypeId="2" ParentId="17858" CreationDate="2017-10-14T20:14:17.197" Score="-1" Body="&lt;p&gt;No. Not with the limitations of our current technology. 99% of anything you've heard describing the state, makeup, nature of or appearance, of any exoplanet, is wishful thinking and conjecture. Our current technology is still rather limited. Today, we are able to come up with reasonable guesstimates of the sizes of exoplanets and their distances from their host stars, all with large degrees of uncertainty. That's it - size and distance - nothing else - at least for today (2017).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we look at a star and see that its light has regular periodic dips, we measure the length of, the strength of, and the period of those dips (that is all we can see). Then we do a lot of math and guess at the size (mass) of the exoplanet and its distance from its host star. Even the size (both mass and diameter) of these planet-hosting stars are mere guesses. Only 16 of all the visible stars have ever been directly imaged (resolved to a disc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, people do love to dream. Some scientists LOVE to embellish. Some scientists like to walk around like peacocks with their plumage spread in display, hoping folks won't see their behind behind their tail feathers.&#xA;(I got to tell you - I really enjoyed thinking up that last sentence.)&lt;/p&gt;&#xA;" OwnerUserId="19089" LastActivityDate="2017-10-14T20:14:17.197" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23140" PostTypeId="1" CreationDate="2017-10-14T21:54:04.270" Score="1" ViewCount="170" Body="&lt;p&gt;I am trying to get data from Gaia. Especially for Cepheids or RR Lyrae. On Gaia websites it say they have released some of the data however I cannot find it. I am looking for two columns with time and magnitude(light curve).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for help&lt;/p&gt;&#xA;" OwnerUserId="19090" LastActivityDate="2017-10-14T21:54:04.270" Title="Photometry data from Gaia" Tags="&lt;gaia&gt;&lt;light-curve&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23142" PostTypeId="1" AcceptedAnswerId="23145" CreationDate="2017-10-14T23:32:51.583" Score="2" ViewCount="126" Body="&lt;p&gt;If there were two earth-like planets in a binary system and one of them had landmasses, and the other one had oceans covering a significant amount of its surface, could you see the continents of the first planet reflected onto the oceans of the second one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, could this also work with oceanic moons?&lt;/p&gt;&#xA;" OwnerUserId="18831" LastActivityDate="2017-10-15T11:52:01.360" Title="Could a binary system of two planets with oceans reflect each other?" Tags="&lt;natural-satellites&gt;&lt;earth-like-planet&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23143" PostTypeId="2" ParentId="23048" CreationDate="2017-10-15T03:29:39.540" Score="3" Body="&lt;p&gt;The only places around one black hole where gravity would be strong enough to pull an object out from inside the horizon of another black hole are all inside the horizon of the first hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that &lt;em&gt;at minimum&lt;/em&gt; an object just inside the horizon of a black hole is traveling toward the center at the speed of light (as its speed would be measured from outside the hole, if we could see it) and getting faster. For a second black hole to catch up to the object to pull it out, it would itself have to be moving faster than the object - faster than light - which is obviously impossible but anyway would simply merge the two black holes, without rescuing anything. &lt;/p&gt;&#xA;" OwnerUserId="4042" LastActivityDate="2017-10-15T03:29:39.540" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23145" PostTypeId="2" ParentId="23142" CreationDate="2017-10-15T11:52:01.360" Score="5" Body="&lt;p&gt;The surface of the ocean isn't smooth, so there would not an image formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at how the Earth's oceans reflect the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/g7VxT.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g7VxT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(credit: NASA/ESA)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously the sun is reflected by the ocean, but the rough surface of the ocean scatters the light enough for no image to be formed. It would be the same if there were a planet being reflected: No detail would be visible.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-10-15T11:52:01.360" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23148" PostTypeId="1" CreationDate="2017-10-15T16:03:21.110" Score="-1" ViewCount="183" Body="&lt;p&gt;What are the consequences going to be if an asteroid or a celestial body collides with a space station rotating on Earth orbit?&lt;/p&gt;&#xA;" OwnerUserId="19099" LastEditorUserId="2365" LastEditDate="2017-10-15T17:11:40.693" LastActivityDate="2017-10-16T15:20:09.223" Title="What would be the result if an asteroid collides with a space station?" Tags="&lt;asteroids&gt;&lt;space&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2017-10-18T21:02:06.657" ContentLicense="CC BY-SA 3.0" />
  <row Id="23149" PostTypeId="2" ParentId="23148" CreationDate="2017-10-15T17:22:00.453" Score="1" Body="&lt;p&gt;It has a very, very low chance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the asteroid is larger as some centimeters, it will irrecoverable destruct the satellite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, mm sized space trash can damage a satellite already, although the tipical impact velocities are in the order of some $\frac{km}{s}$. As a comparision, cannon balls are flying typically with lesser speed as $1 \frac{km}{s} {}^*$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A collision with an asteroid on a solar orbit would have the impact speed in the order of some tens of $\frac{km}{s}^{**}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A direct collision with an asteroid larger as the satellite itself would have enough energy to vaporize the satellite. We, from the Earth, would see a space explosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*: Orbital velocity on low earth orbit is $7.8 \frac{km}{s}$, but most satellites are flying &lt;a href=&quot;https://en.wikipedia.org/wiki/Retrograde_and_prograde_motion&quot; rel=&quot;nofollow noreferrer&quot;&gt;prograde&lt;/a&gt; with a not very high &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_inclination&quot; rel=&quot;nofollow noreferrer&quot;&gt;inclination&lt;/a&gt;. Thus, the we can estimate their &quot;typical&quot; relative collision velocity in the order of some $\frac{km}{s}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;**: The configuration is similar in heliocentris orbits, but the velocities are higher with an order.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="1891" LastEditDate="2017-10-16T15:20:09.223" LastActivityDate="2017-10-16T15:20:09.223" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23150" PostTypeId="2" ParentId="23148" CreationDate="2017-10-15T19:15:51.087" Score="-1" Body="&lt;p&gt;There's an asteroid (2012 TC4) currently in the news, due to pass close to the earth, at around 42000 km (26000 miles), within the moons orbit and just above comsat altitude. It's believed to be in the 15-30m size range.&#xA;(see &lt;a href=&quot;http://www.bbc.co.uk/news/science-environment-41583704&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.bbc.co.uk/news/science-environment-41583704&lt;/a&gt; though the link may not stay around long).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparison (from the link above),&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When a 20m-wide asteroid exploded over Chelyabinsk in central Russia in 2013, it hit the atmosphere with energy estimated to be equivalent to 500,000 tonnes of TNT, causing a shockwave that damaged buildings and injured more than a thousand people.  [Source: BBC]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So if something that size hit a space station, you're not going to have much station left. Even a fairly small rock would do a lot of damage. Fortunately, the chance of being hit by something of significant size is very low, and a lot of the junk floating around (like the stuff that makes meteor showers) are dust particle sized lumps of cometary material that might damage but not destroy a station. &lt;/p&gt;&#xA;" OwnerUserId="17016" LastActivityDate="2017-10-15T19:15:51.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23151" PostTypeId="1" CreationDate="2017-10-16T02:11:39.880" Score="2" ViewCount="92" Body="&lt;p&gt;Some face-on spiral galaxies show barred structure in their central bulges, whereas some edge-on spiral galaxies have boxy bulges. I wonder if there is any relation between these two. In particular, do boxy bulges indicate barred spirals?&lt;/p&gt;&#xA;" OwnerUserId="19102" LastActivityDate="2017-10-16T14:39:28.413" Title="Do boxy bulges indicate barred spirals?" Tags="&lt;galaxy&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23154" PostTypeId="1" CreationDate="2017-10-16T12:32:08.560" Score="2" ViewCount="48" Body="&lt;p&gt;Why the most luminous stars (point sources of light) like the ones in the image below are rendered as crosses? If they are artifacts generated by some Hubble instrument, what causes them? Can them be corrected in some way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does the orientation of the crosses depend on?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bBin6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bBin6.jpg&quot; alt=&quot;Westerlund 1 supercluster&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;https://www.nasa.gov/image-feature/goddard/2017/hubble-hones-in-on-a-hypergiants-home&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="16793" LastActivityDate="2017-10-16T12:32:08.560" Title="Why stars are shown as crosses in some Hubble images?" Tags="&lt;hubble-telescope&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2017-10-16T17:37:34.750" ContentLicense="CC BY-SA 3.0" />
  <row Id="23156" PostTypeId="2" ParentId="23151" CreationDate="2017-10-16T14:39:28.413" Score="2" Body="&lt;p&gt;Yes, they do.&lt;/p&gt;&#xA;&lt;p&gt;Bars in massive galaxies undergo a vertical &amp;quot;buckling&amp;quot; phase about one or two billion years after they form; when this settles down (after about a billion years), the result is that the inner 1/3-2/3 of the bar is vertically thick, with the outer part of the bar still as flat as the rest of the galaxy disk.&lt;/p&gt;&#xA;&lt;p&gt;When seen side-on, the thick part of the bar looks like a peanut (double-lobed), often with a hint of an X-shape; when seen at an intermediate angle, it looks more rectangular and &amp;quot;boxy&amp;quot;. This is the situation &lt;a href=&quot;http://aasnova.org/2016/08/29/featured-image-the-milky-ways-x/&quot; rel=&quot;nofollow noreferrer&quot;&gt;for the Milky Way&lt;/a&gt;, in fact. (When seen with the bar end-on, the result looks very round, like a stereotypical spherical bulge.)&lt;/p&gt;&#xA;&lt;p&gt;You can also see this when the galaxy &lt;em&gt;isn't&lt;/em&gt; perfectly edge-on: some bars in galaxies inclined at 45 or 60 degrees (where 90 degrees = edge-on) have rectangular interiors with thinner, offset outer regions; this corresponds to the vertically thick/peanut-shaped inner part of the bar and the vertically thin outer part.&lt;/p&gt;&#xA;&lt;p&gt;We've even seen the signature in a couple of galaxies of a bar in the middle of forming the peanut-shaped bulge (the &lt;a href=&quot;http://aasnova.org/2016/07/20/do-galactic-bars-buckle-to-form-bulges/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&amp;quot;buckling phase&amp;quot;&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;p&gt;(In principle, you could get a boxy bulge from some kinds of minor mergers, where a small galaxy has merged with a larger galaxy. But if these exist in real galaxies, they're much rarer than the boxy bulges that are part of bars.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-10-16T14:39:28.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23157" PostTypeId="1" CreationDate="2017-10-16T15:44:24.550" Score="5" ViewCount="332" Body="&lt;p&gt;I am very used to working with seeing in arcseconds, but sometimes I come across the seeing in cm, e.g. &lt;a href=&quot;http://www.royac.iac.es/seeing.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked online for a way to convert from one to the other, but I can't find any. How would this be done?&lt;/p&gt;&#xA;" OwnerUserId="5286" LastEditorDisplayName="user1569" LastEditDate="2017-10-17T00:56:18.887" LastActivityDate="2017-10-17T00:56:18.887" Title="Seeing units, arcsec VS cm" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23159" PostTypeId="2" ParentId="23157" CreationDate="2017-10-16T16:38:33.923" Score="4" Body="&lt;p&gt;Seeing is typically measured as the FWHM of the seeing disk, but can also be expressed though the &lt;a href=&quot;https://en.wikipedia.org/wiki/Fried_parameter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fried parameter&lt;/a&gt; $r_0$, which measures the size, or strength, of the parcels of gas that cause the turbulence in the atmosphere. The relation between the two is (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/1995PASP..107..265V&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vernin &amp;amp; Munoz-Tunon 1995&lt;/a&gt;)&#xA;$$&#xA;\text{seeing} = 0.98\frac{\lambda}{r_0},&#xA;$$&#xA;where $\lambda$ is the wavelength of the light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r_0$ can by calculated by integrating the &quot;turbulence strength&quot; along the line of sight$^\dagger\!\!$, and  scales as $\lambda^{6/5}\!$. Because of this dependence, the &quot;usual&quot; seeing has only a weak ($\lambda^{-1/5}$) dependence on wavelength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_seeing&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Wikipedia article on seeing&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At visible wavelengths, $r_0$ varies from 20 cm at the best locations to 5 cm at typical sea-level sites&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$\dagger$&lt;sub&gt;Actually, $r_0$ refers to a distance corresponding to the line of sight towars zenith; observing at an angle $\zeta$ from zenith introduces a factor $\cos^{3/5}\!\zeta$.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2017-10-16T20:10:54.180" LastActivityDate="2017-10-16T20:10:54.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23160" PostTypeId="2" ParentId="22754" CreationDate="2017-10-16T16:59:11.923" Score="2" Body="&lt;p&gt;I just found out the answer to my question from a live press release on You Tube that has been covered by blogs like this: &lt;a href=&quot;http://news.nationalgeographic.com/2017/10/gravitational-waves-discovered-neutron-stars-pictures-science/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://news.nationalgeographic.com/2017/10/gravitational-waves-discovered-neutron-stars-pictures-science/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What LIGO does is tell you what is causing the gamma ray burst that can be studied different ways. The highly anticipated announcement I just watched explained that LIGO observed the neutron star merger (they're calling the event a kilonova). With the help of VIRGO, the area of the sky could be narrowed down better than previous LIGO detections. LIGO told us the masses of the bodies, the distance to the objects, how much mass was ejected and thus what the end product of the collision is expected to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have observed many gamma ray bursts and this one was not particularly bright, but LIGO told astronomers to check this one out and so it was being observed within a few hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electromagnetic spectrum of this event was collected and very broad lines were detected which are believed to be caused by heavy elements being ejected at near light speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By our knowledge of how often these &quot;kilonovae&quot; should happen and the volume of elements heavier than iron thought now to have been emitted (16 Earth masses was the number mentioned in the press release), it can be said that a majority of many of the heavy elements we see around us may have come from these events.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="7997" LastEditDate="2017-10-23T21:47:45.510" LastActivityDate="2017-10-23T21:47:45.510" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23161" PostTypeId="1" AcceptedAnswerId="23164" CreationDate="2017-10-16T17:31:12.293" Score="85" ViewCount="11504" Body="&lt;p&gt;I'm fairly certain people here will have heard about it, already, but apparently, two supernova leftovers clashed some 130 million years ago and some billion billion kilometres away ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I &lt;em&gt;haven't&lt;/em&gt; heard yet, however, is &lt;em&gt;why&lt;/em&gt; we should care.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean sure, it's an interesting phenomenon and measuring it can't have been easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now that we've heard it ... what changes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll admit it, I don't know particularly much about astronomy, but I'm curious:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the significance of having achieved this? Why does it matter whether or not we know?&lt;/p&gt;&#xA;" OwnerUserId="19110" LastEditorUserId="2531" LastEditDate="2017-10-17T05:48:51.023" LastActivityDate="2017-10-23T07:02:05.017" Title="Why is the discovery of merging neutron stars important?" Tags="&lt;gravitational-waves&gt;&lt;neutron-star&gt;" AnswerCount="2" CommentCount="10" FavoriteCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="23162" PostTypeId="1" AcceptedAnswerId="23166" CreationDate="2017-10-16T17:41:38.067" Score="29" ViewCount="2698" Body="&lt;p&gt;The NPR News item &lt;a href=&quot;http://www.npr.org/sections/thetwo-way/2017/10/16/557557544/astronomers-strike-gravitational-gold-in-colliding-neutron-stars&quot; rel=&quot;noreferrer&quot;&gt;Astronomers Strike Gravitational Gold In Colliding Neutron Stars&lt;/a&gt; mentions and quotes &amp;quot;&lt;a href=&quot;http://astro.berkeley.edu/faculty-profile/dan-kasen&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;Daniel Kasen&lt;/strong&gt;&lt;/a&gt;, a theoretical astrophysicist at the University of California, Berkeley:&amp;quot;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;He spent late nights watching the data come in and says the colliding stars spewed out a big cloud of debris.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;That debris is strange stuff. It's gold and platinum, but it's mixed in with what you'd call just regular radioactive waste, and there's this big radioactive waste cloud that just starts mushrooming out from the merger site,&amp;quot; Kasen says. &amp;quot;It starts out small, about the size of a small city, but it's moving so fast — a few tenths of the speed of light — that after a day it's a cloud the size of the solar system.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;According to his estimates, this neutron star collision produced around 200 Earth masses of pure gold, and maybe 500 Earth masses of platinum. &amp;quot;It's a ridiculously huge amount on human scales,&amp;quot; Kasen says. He personally has a platinum wedding ring and notes that &amp;quot;it's crazy to think that these things that seem very far out and kind of exotic actually impact the world and us in kind of intimate ways.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Has the merger of &lt;a href=&quot;https://en.wikipedia.org/wiki/Neutron_star&quot; rel=&quot;noreferrer&quot;&gt;neutron star&lt;/a&gt; binaries &lt;em&gt;been necessary to explain&lt;/em&gt; abundances of the heavy elements such as gold and platinum, or is this just an anecdotal item? How important are binary neutron stars for the abundances of heavy elements such as gold? Is there a particular or notable paper I can read on this?&lt;/p&gt;&#xA;&lt;p&gt;I have already read &lt;a href=&quot;https://astronomy.stackexchange.com/a/23160/7982&quot;&gt;this answer&lt;/a&gt; but I'm looking for a better explanation of the need for this kind of merger to explain abundances. I'm pretty sure there is nothing in any observed gamma ray events that shows spectral lines of gold or any identifiable heavy element (due to the incredible doppler broadening), so the connection must actually come from simulations.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-03-25T21:21:07.107" Title="Are binary neutron star mergers needed to explain the abundance of gold?" Tags="&lt;gravitational-waves&gt;&lt;neutron-star&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23163" PostTypeId="2" ParentId="23161" CreationDate="2017-10-16T18:06:04.730" Score="40" Body="&lt;p&gt;Because its awesome (&lt;a href=&quot;http://www.smbc-comics.com/index.php?id=2088#comic&quot; rel=&quot;noreferrer&quot;&gt;SMBC&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this guy called Copernicus suggested that the Earth orbits the Sun (not the other way round) - What changes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This guy Newton had a theory for how a mass responds to force, and how gravity works - So what?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another guy called Maxwell had this idea of how light could actually be waves of electromagnetic fields - does this matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A guy called Monet decided to paint some pictures of some waterlilies. Who cares?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last February, some guys from Denver carried a ball over a line more often than some guys from Carolina carried the ball over another line. So what?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worth finding things out because it means finding things out. It is worth understanding our world because it is there to be understood. Discovery is its own reward. It is not measured in by £ or $. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The observations of GW170817 show that heavy elements are created by the merger of neutron stars. Heavy elements like gold, platinum on Earth were likely created in a neutron star merger in The Milky Way, billions of years ago, that enriched the interstellar dust. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it doesn't matter to you that's fine. If Monet's Waterlilies or the Superbowl leave you cold, that's no problem either. But not everything of value is of use.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-10-16T18:06:04.730" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23164" PostTypeId="2" ParentId="23161" CreationDate="2017-10-16T18:44:49.387" Score="163" Body="&lt;p&gt;Reasons why this is important:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the &lt;em&gt;first&lt;/em&gt; simultaneous detection of a gravitational wave and electromagnetic signal, and the strongest GW signal yet in terms of signal to noise (&lt;a href=&quot;http://www.google.com/calendar/render&quot; rel=&quot;noreferrer&quot;&gt;Abbott et al. 2017a&lt;/a&gt;). It spectacularly corroborates the reality of the GW detection technology and analysis. The progenitor has been unambiguously located in a (relatively) nearby galaxy (&lt;a href=&quot;http://iopscience.iop.org/article/10.3847/2041-8213/aa9059/meta&quot; rel=&quot;noreferrer&quot;&gt;Soares-Santos et al. 2017&lt;/a&gt;), allowing a host of other telescopes to obtain detailed measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It shows that GWs travel at the speed of light, a further verification of Einstein's General Relativity (&lt;a href=&quot;http://iopscience.iop.org/article/10.3847/2041-8213/aa920c&quot; rel=&quot;noreferrer&quot;&gt;Abbott et al. 2017b&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It shows that most of the very heavy elements such as gold, platinum, osmium etc. are plausibly produced by merging neutron stars and constrains the rate of such mergers in the local universe (e.g. &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/2041-8213/aa905c/meta&quot; rel=&quot;noreferrer&quot;&gt;Chornock et al. 2017&lt;/a&gt;; &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/2041-8213/aa90b6/meta&quot; rel=&quot;noreferrer&quot;&gt;Tanvir et al. 2017&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It shows that short gamma ray bursts — some of the most energetic explosions in the universe — can be caused by neutron star mergers (e.g. &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/2041-8213/aa8f94/meta&quot; rel=&quot;noreferrer&quot;&gt;Savchenko et al. 2017&lt;/a&gt;; &lt;a href=&quot;https://mail.google.com/mail/?hl=en&amp;amp;shva=1#inbox&quot; rel=&quot;noreferrer&quot;&gt;Goldstein et al. 2017&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the closest detected short gamma ray burst (with a known distance). That the progenitor has also been characterised allows a closer investigation of the interesting physics underlying the ejection and jet mechanisms thought to be responsible for the gamma rays and later X-ray and radio emission (e.g. &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/2041-8213/aa9057/meta&quot; rel=&quot;noreferrer&quot;&gt;Margutti et al. 2017&lt;/a&gt;; &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/2041-8213/aa905d/meta&quot; rel=&quot;noreferrer&quot;&gt;Alexander et al. 2017&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It provides observational constraints on how matter behaves at extremely high densities, testing our understanding of fundamental physics to its limits — for example, the details of the gravitational wave signal moments before merger are diagnostic of the interior conditions of neutron stars at densities of $\sim 10^{18}$ kg/m$^3$ (&lt;a href=&quot;https://journals.aps.org/prd/abstract/10.1103/PhysRevD.81.123016&quot; rel=&quot;noreferrer&quot;&gt;Hinderer et al. 2010&lt;/a&gt;; &lt;a href=&quot;https://journals.aps.org/prd/abstract/10.1103/PhysRevD.82.024016&quot; rel=&quot;noreferrer&quot;&gt;Postnikov et al. 2010&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It provides an independent way of measuring the expansion of the universe. Merging binary gravitational wave sources are known as &quot;standard sirens&quot;, because the distance to the GW source pops straight out of the analysis and can be compared with the redshift of the identified host galaxy (&lt;a href=&quot;http://www.nature.com/nature/journal/vaap/ncurrent/full/nature24471.html&quot; rel=&quot;noreferrer&quot;&gt;Abbott et al. 2017c&lt;/a&gt;). The result agrees with measurements made using the cosmic microwave background and the distance-redshift relation calibrated by other means, verifying our estimation of distances, at least in the local universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, this event will turn out to be important because it was &lt;em&gt;lucky&lt;/em&gt;; in the sense that the source was detected well-inside the sensitivity horizon of LIGO (&lt;a href=&quot;https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.119.161101&quot; rel=&quot;noreferrer&quot;&gt;Abbott et al. 2017a&lt;/a&gt;). The detection itself, was not unexpected given the rates predicted based on studying the neutron star binary systems in our own Galaxy (e.g. &lt;a href=&quot;https://arxiv.org/abs/1308.4676&quot; rel=&quot;noreferrer&quot;&gt;Kim et al. 2015&lt;/a&gt;), but the fact that it was so close —&#xA; within the closest 5% of the sensitive survey volume where it could have been detected — is fortunate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, if someone thinks none of the above is interesting or important, then nothing I can write will convince them otherwise. The vast majority of people I speak to are curious and fascinated to find out about our cosmic origins and how the universe works.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-10-23T07:02:05.017" LastActivityDate="2017-10-23T07:02:05.017" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="23165" PostTypeId="1" CreationDate="2017-10-16T19:31:13.550" Score="4" ViewCount="181" Body="&lt;p&gt;Diamond Fuji is a phenomenon in which the (typically) setting sun is positioned directly atop Mt Fuji. Some relatively detailed predictions for viewing times at a nearby park are &lt;a href=&quot;http://www.vill.yamanakako.yamanashi.jp.e.kh.hp.transer.com/holiday_diamondfuji.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;available here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://d1fwj4bqr5098n.cloudfront.net/wp-content/uploads/sites/4/2016/12/13120736/Diamon-Fuji_1-1024x684.jpg&quot; alt=&quot;Diamond Fuji&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will be traveling to Tokyo in early December, and would like to observe Diamond Fuji to the extent possible (if at all). I can rely somewhat on tourist-friendly websites, but information is sparse and imprecise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, one should be able to compute the time and viewing locations of Diamond Fuji for any day of the year: the summit of Mt Fuji lies on the line connecting the viewer to the sun. What is the best way to work out this sort of calendar/map?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;As a non-expert I suspect &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1855/how-to-get-the-longitude-latitude-from-solar-zenith-azimuth&quot;&gt;this question&lt;/a&gt; is relevant but I am not certain.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="19113" LastActivityDate="2017-10-16T19:31:13.550" Title="Viewing &quot;Diamond Fuji&quot;" Tags="&lt;amateur-observing&gt;&lt;positional-astronomy&gt;" AnswerCount="0" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23166" PostTypeId="2" ParentId="23162" CreationDate="2017-10-16T20:05:09.697" Score="48" Body="&lt;p&gt;The creation of some very heavy neutron-rich elements, like gold and platinum, requires the rapid capture of neutrons. This will only occur in dense, explosive conditions where the density of free neutrons is large. For a long time, the competing theories and sites for the r-process have been inside core-collapse supernovae and during the merger of neutron stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that it has become increasingly difficult for supernovae to produce (in theoretical models) sufficient r-process elements to match both the quantity and detailed abundance ratios of r-process elements in the solar system (see for example &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/1742-6596/312/4/042008/meta;jsessionid=F07BC1557DE69F29EB2F59B976ED49C5.ip-10-40-2-120&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wanajo et al. 2011&lt;/a&gt;; &lt;a href=&quot;https://arxiv.org/abs/1207.2527&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arcones &amp;amp; Thielmann 2012&lt;/a&gt;). The conditions required, particularly a very neutron-rich environment in the neutrino-driven winds, are just not present without the fine tuning of parameters (see below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, the models invoking neutron star mergers are much more robust to theoretical uncertainties and successfully produce r-process elements. The question mark appears to be only over their frequency at various times in the evolution of a galaxy and exactly how much enriched material is ejected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The announcement of GW170817 makes this all more plausible. A neutron star merger has been seen. The behaviour of the optical and infrared emission after the event matches the expectations of merging neutron star models (e.g. &lt;a href=&quot;http://www.nature.com/nature/journal/vaap/ncurrent/full/nature24298.html?foxtrotcallback=true&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pian et al. 2017&lt;/a&gt;; &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/2041-8213/aa90b6/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tanvir et al. 2017&lt;/a&gt;). Of particular note is the developing  opacity and fading in the blue and  visible, with the spectrum &#xA;becoming dominated by the  infrared with broad spectral features. This is the expectation for an expanding cloud of material that is heavily polluted by the presence of lanthanides and other r-process elements (&lt;a href=&quot;http://iopscience.iop.org/article/10.3847/2041-8213/aa905c/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chornock et al. 2017&lt;/a&gt;). The reasonable agreement between the observations and models suggests that indeed a large quantity of r-process elements were produced in this explosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To go from there to the claim that the origin of gold is solved (as claimed in the press conference) is a step too far. The amount of r-process material produced has large uncertainties and is model-dependent.&#xA;The rate of mergers is only constrained to about an order of magnitude &lt;em&gt;in the local universe&lt;/em&gt; and is not measured/known in the early universe. What could be said is that this channel for r-process production has been directly observed and so must be taken into account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, r-process production by the supernova channel is not yet ruled out. Some simulations at least, that involve rotation and magnetic fields appear to be still &quot;in the game&quot; (e.g. &lt;a href=&quot;https://arxiv.org/abs/1501.06567&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nishimura et al. 2016&lt;/a&gt;). It could be that the presence of significant r-process material in very old metal-poor stars requires a supernova channel, since the merger of neutron stars takes some considerable time to occur (e.g. &lt;a href=&quot;https://arxiv.org/abs/1503.02954&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cescutti et al. 2015&lt;/a&gt;; &lt;a href=&quot;https://arxiv.org/abs/1610.02405&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cote et al. 2017&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall picture is still uncertain. A review by &lt;a href=&quot;http://arxiv.org/abs/1901.09044&quot; rel=&quot;nofollow noreferrer&quot;&gt;Siegel (2019)&lt;/a&gt; concludes that the best fit to the available evidence is that some rare type(s) of core collapse supernovae (known as &quot;collapsars&quot;) are still the best bet to explain the Milky Way r-process elements. The primary evidence for this is the presence of Europium (an r-process element) enhancements in some very old halo stars and the general trend of decreasing Eu/Fe with increasing Fe, suggesting a more alpha-element-like production site for the r-process - i.e. supernovae.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2019-03-25T21:21:07.107" LastActivityDate="2019-03-25T21:21:07.107" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23170" PostTypeId="1" AcceptedAnswerId="23175" CreationDate="2017-10-17T10:37:56.847" Score="4" ViewCount="342" Body="&lt;p&gt;Since now two neutron stars have been detected merging via gravitational waves, I was wondering what is the current detection threshold that the LIGO detectors can achieve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering that the first observed objects were two black holes with a combined mass of more than 60 solar masses and they now detected two neutron stars with a combined mass of only about 3 solar masses I was wondering what was the threshold that these detectors can actually detect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously there are much larger stars out there which orbit each other, but their size and distance from each other make gravitational waves too difficult to detect. So what masses and at what distances can we expect to be detected in the future?&lt;/p&gt;&#xA;" OwnerUserId="14441" LastActivityDate="2017-10-17T14:47:37.983" Title="What is the detection threshold of gravitational waves for LIGO?" Tags="&lt;gravitational-waves&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23171" PostTypeId="1" AcceptedAnswerId="23173" CreationDate="2017-10-17T13:20:20.980" Score="26" ViewCount="9248" Body="&lt;p&gt;I understood that everything was made of &lt;a href=&quot;https://en.wikipedia.org/wiki/Atom&quot; rel=&quot;noreferrer&quot;&gt;Atoms&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An atom is the smallest constituent unit of ordinary matter that has&#xA;  the properties of a chemical element. Every solid, liquid, gas, and&#xA;  plasma is composed of neutral or ionized atoms&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However with &lt;a href=&quot;https://en.wikipedia.org/wiki/Neutron_star&quot; rel=&quot;noreferrer&quot;&gt;Neutron Stars&lt;/a&gt;;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the basic models for these objects imply that neutron stars&#xA;  are composed almost entirely of neutrons, which are subatomic&#xA;  particles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does this mean neutrons can exist outside of an atom, and that a Neutron star isn’t comprised of an element that is recognisable on the periodic table?&lt;/p&gt;&#xA;" OwnerUserId="13157" LastEditorUserId="13157" LastEditDate="2017-10-17T16:01:04.073" LastActivityDate="2017-10-23T09:23:50.323" Title="Which atom is a Neutron star made of?" Tags="&lt;neutron-star&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23173" PostTypeId="2" ParentId="23171" CreationDate="2017-10-17T13:40:10.530" Score="55" Body="&lt;p&gt;Yes neutrons can exist outside the atom (or nucleus). In free space a neutron will &lt;a href=&quot;https://en.wikipedia.org/wiki/Beta_decay&quot; rel=&quot;noreferrer&quot;&gt;beta decay&lt;/a&gt; into a proton, and electron and an anti-neutrino on a timescale of 10 minutes. However, in the dense interiors of a neutron star, the electrons form a degenerate gas, with all possible energy levels filled up to something called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Fermi_energy&quot; rel=&quot;noreferrer&quot;&gt;Fermi energy&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the Fermi energy of the electrons exceeds the maximum energy of any possible beta-decay electron, then beta decay is blocked and free neutrons become stable. This is what happens inside a neutron star and you end up with mostly neutrons with a small fraction perhaps a few per cent electrons and protons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the outer parts of the neutron star, the protons and neutrons can still arrange themselves into nuclei (but not atoms), but these nuclei are extremely neutron-rich (they would not normally exist in nature) and are only stabilised against beta decay by the process I described above. The very outer envelope may consist of completely ionised iron-peak element nuclei and there may be an ultrathin (few cm) layer of recognisable ionised hydrogen, helium and carbon (e.g. &lt;a href=&quot;http://lanl.arxiv.org/pdf/0911.0672v1&quot; rel=&quot;noreferrer&quot;&gt;Wynn &amp;amp; Heinke 2009&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the density reaches about $3 \times 10^{16}$ kg/m$^3$ it becomes more favourable for the neutrons and protons to organise themselves into &quot;macro-nuclei&quot; - long strings and sheets of nuclear material, known colloquially as &lt;a href=&quot;https://en.wikipedia.org/wiki/Nuclear_pasta&quot; rel=&quot;noreferrer&quot;&gt;nuclear pasta&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At higher densities still, the pasta dissolves into a soup of mostly neutrons with about 1 per cent protons and electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diagram below (from &lt;a href=&quot;https://arxiv.org/abs/1109.3511&quot; rel=&quot;noreferrer&quot;&gt;Watanabe et al. 2012&lt;/a&gt;) shows roughly how these layers are arranged. It should be stressed that this is based on theoretical modelling, with the theory becoming less certain the further into the neutron star you go. Testing these ideas involves nuclear and particle experiments, observations of pulsars, of neutron star cooling, of X-ray bursts, mass and radius estimates in binary systems, pulsar glitches, etc., etc. None of the &lt;em&gt;details&lt;/em&gt; have been observationally confirmed beyond dispute, but the basic picture below fits what we know. In particular, the the crust and the n,p,e fluid regions are well understood in theory. The details of the nuclear pasta phases are still the subject of a lot of theoretical work, as are the details of superfluidity in the interior, and what happens in the very central regions (solid neutron core, extra hadronic phases, boson condensation, quark matter) is still theoretically difficult and observationally untested except perhaps to say that the softest equations of state have been ruled out by the existence of $2M_{\odot}$ neutron stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cvdzF.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cvdzF.png&quot; alt=&quot;Neutron star structure&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-10-23T09:23:50.323" LastActivityDate="2017-10-23T09:23:50.323" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="23174" PostTypeId="1" CreationDate="2017-10-17T14:09:24.617" Score="7" ViewCount="127" Body="&lt;p&gt;When the two neutron stars merged in the latest event, large amounts of gamma rays were emitted.  Given that their deorbitting is a result of gravitational wave production, where does the energy for the gamma rays come from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only sources I can think of are the rotational kinetic energies of the merging stars or the potential energy released by the final collapse into a single spherical mass.&lt;/p&gt;&#xA;" OwnerUserId="19141" LastActivityDate="2017-10-18T06:28:45.103" Title="Energy Source for Gamma Rays from Merging Neutron Stars" Tags="&lt;neutron-star&gt;&lt;gamma-ray-bursts&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23175" PostTypeId="2" ParentId="23170" CreationDate="2017-10-17T14:18:21.640" Score="5" Body="&lt;p&gt;I'm afraid this is not straightforward&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amplitude of the gravitational wave strain signal from a merging compact binary (neutrons star or black hole) is&#xA;$$h \sim 10^{-22} \left(\frac{M}{2.8M_{\odot}}\right)^{5/3}\left(\frac{0.01{\rm s}}{P}\right)^{2/3}\left(\frac{100 {\rm Mpc}}{d}\right),$$&#xA;where $M$ is the total mass of the system in solar masses, $P$ is the instantaneous orbital period in seconds and $d$ is the distance in 100s of Mpc. $h \sim 10^{-22}$ is a reasonable number for the sensitivity of LIGO to gravitational wave strain where it is most sensitive (at frequencies of 30-300 Hz).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can see that to increase the observability you can increase the mass, decrease the period or decrease the distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here are the complications. LIGO is only sensitive between about 30-300 Hz and the GW frequencies are &lt;em&gt;twice&lt;/em&gt; the orbital frequency. Thus you cannot shorten the period to something very small because it would fall outside the LIGO frequency range and you also cannot increase the mass to something too much bigger than the black holes that have been already seen because they merge before they can attain high enough orbital frequencies to be seen. (The frequency at merger is $\propto M^{-1}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A further complication is that the evolution of the signals is more rapid at lower masses. That is - the rate of change of frequency and amplitude increase rapidly with total mass. That is why the recent neutron star merger was detectable for 100s by LIGO, whereas the more massive black hole mergers could only be seen for about 1 second. But what this means is that you have fewer cycles of the black hole signal that can be &quot;added up&quot; to improve the signal to noise, which means that higher mass sources are &lt;em&gt;less&lt;/em&gt; detectable than a simple application of the formula I gave above would suggest. A further complication is that there is a geometric factor depending on how the source and detectors are orientated with respect to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, these are complications, but the formula can still be used as an approximation. So if we take the GW170817 signal, the total mass was about $2.8M_{\odot}$, the source was at 40 Mpc, so at frequencies of 200 Hz (corresponding to a period of 0.01 s) you might have expected a strain signal of about $3\times 10^{-22}$.  This gave a very readily detectable signal. The discovery paper (&lt;a href=&quot;https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.119.161101&quot; rel=&quot;noreferrer&quot;&gt;Abbot et al. 2017&lt;/a&gt;) says the &quot;horizon&quot; for detection was approximately 218 Mpc for LIGO-Livingston and 107 Mic for LIGO-Hanford. As the source was much closer than these numbers then it is unsurprising that the detection was strong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the formula above and a fixed orbital period of 0.01 s, we can see that the horizon distance will scale as $\sim M^{5/3}$. So a $10 M_{\odot} + 10 M_{\odot}$ black hole binary might be seen out to $218 \times (20/2.8)^{5/3} = 5.7$ Gpc (this will be an overestimate by a factor of a few because of the issue of the rapidity of the evolution towards merger that I discussed above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more through and technical discussion can be read &lt;a href=&quot;http://ligo.org/science/Publication-S6CBCRangeDoc/index.php&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;, although this is a couple of years out of date and LIGO's reach has been extended by about a factor of five since these calculations were done.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-10-17T14:18:21.640" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23176" PostTypeId="2" ParentId="23174" CreationDate="2017-10-17T14:24:25.110" Score="3" Body="&lt;p&gt;&lt;strong&gt;Not&lt;/strong&gt; Nuclear fusion;&lt;br&gt;&#xA;There is also a huge amount of nuclear fusion that &lt;strong&gt;creates heavy elements&lt;/strong&gt; from all the iron there is in neutron star: they in fact are not only made of neutrons. But as said in comment, building of heavier elements rather consumes energy than it generates some.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure scientists  has a clear agreement on how this works. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Gravitational collapse&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example I found &lt;a href=&quot;https://www.researchgate.net/publication/1779936_Red_Hole_Gamma-Ray_Bursts_A_New_Gravitational_Collapse_Paradigm_Explains_the_Peak_Energy_Distribution_and_Solves_the_GRB_Energy_Crisis&quot; rel=&quot;nofollow noreferrer&quot;&gt;this (rather old) abstract&lt;/a&gt; which states it could be a result of &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_collapse&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravitational collapse&lt;/a&gt;, without singularity (since here the result of the merger might not be a black-hole, or the lightest ever: &lt;a href=&quot;https://www.space.com/38471-gravitational-waves-neutron-star-crashes-discovery-explained.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;around 2.7 solar masses&lt;/a&gt;!)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;theoretical developments (string theory, quantum gravity, critical collapse), which suggest that complete gravitational collapse can occur without singularities or event horizons &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thermonuclear event again: fission&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.space.com/22231-gamma-ray-bursts-neutron-stars.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;From Space.com&lt;/a&gt;, they say Kilonova energy could be explained by rapid atomic decay of heavy elements&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The mergers of dense cosmic bodies that are thought to cause short gamma-ray bursts can also blast out neutron-rich gas that rapidly generates heavy elements such as gold and platinum, scientists say. These &quot;r-process&quot; elements can undergo radioactive decay and release an enormous amount of energy — 1,000 times or so that given off by stellar explosions such as novas. These powerful events are thus known as &quot;kilonovas&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11264" LastEditorUserId="11264" LastEditDate="2017-10-18T06:28:45.103" LastActivityDate="2017-10-18T06:28:45.103" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23178" PostTypeId="2" ParentId="23170" CreationDate="2017-10-17T14:47:37.983" Score="2" Body="&lt;p&gt;Figure 1 of &lt;a href=&quot;https://arxiv.org/abs/1704.04628&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt; shows the horizon distance (distance to which a circularly polarised overhead signal would be detected at SNR 8) for larger mass systems up to total mass of 1000 solar masses, assuming a search with compact binary coalescence templates. For higher masses the signal amplitude is generally larger, but they merge at lower frequencies so the signals are generally shorter-lived in the sensitive band of the detectors. As they're shorter they also, unfortunately, look a lot more like classes of instrumental &lt;a href=&quot;https://www.zooniverse.org/projects/zooniverse/gravity-spy/about/research&quot; rel=&quot;nofollow noreferrer&quot;&gt;glitches&lt;/a&gt;, so if they're not that strong (just above a threshold of roughly SNR 8) the background level can be large and lead to lower significance of any candidates.&lt;/p&gt;&#xA;" OwnerUserId="16795" LastActivityDate="2017-10-17T14:47:37.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23179" PostTypeId="1" AcceptedAnswerId="23189" CreationDate="2017-10-17T15:23:08.570" Score="6" ViewCount="812" Body="&lt;p&gt;So I was wondering, if &lt;a href=&quot;https://en.wikipedia.org/wiki/Drizzle_(image_processing)&quot; rel=&quot;nofollow noreferrer&quot;&gt;drizzle integration&lt;/a&gt; actually increases details in your image if the drizzled image results in higher resolution in terms of arcseconds / pixel than the telescope would allow (due to the uncertainty principle the telescope's resolving capability is dependent on its diameter, more commonly referred to as the diffraction limit &lt;a href=&quot;https://en.wikipedia.org/wiki/Diffraction-limited_system#Implications_for_digital_photography&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Airy_disk#Cameras&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt; )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my case I have around 0.52&quot; per px on my main camera and my GSO RC8&quot; allows for around 0.5&quot; resolution max due to the uncertainty principle. Drizzling an image would then give me 0.26&quot; per px on the drizzled image when increasing width and height of the image by factor 2, which is higher resolving power than what the telescope would actually be able to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from sharper and more defined stars when drizzle integrating, will drizzle integration actually increase my resolving capability?&lt;/p&gt;&#xA;" OwnerUserId="10271" LastEditorUserId="907" LastEditDate="2017-10-19T12:25:01.537" LastActivityDate="2017-10-20T17:00:40.927" Title="Can Drizzle Integration increase resolution beyond theoretical resolution of telescope?" Tags="&lt;photography&gt;&lt;angular-resolution&gt;&lt;rc-telescope&gt;&lt;drizzle&gt;&lt;post-processing&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23180" PostTypeId="5" CreationDate="2017-10-17T15:23:44.940" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2017-10-17T15:23:44.940" LastActivityDate="2017-10-17T15:23:44.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23181" PostTypeId="4" CreationDate="2017-10-17T15:23:44.940" Score="0" Body="Use this tag for questions that involve an RC (Ritchey–Chrétien) type of telescope" OwnerUserId="10271" LastEditorUserId="10271" LastEditDate="2017-10-18T21:31:47.390" LastActivityDate="2017-10-18T21:31:47.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23182" PostTypeId="5" CreationDate="2017-10-17T15:24:52.203" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2017-10-17T15:24:52.203" LastActivityDate="2017-10-17T15:24:52.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23183" PostTypeId="4" CreationDate="2017-10-17T15:24:52.203" Score="0" Body="Use this tag for questions related to post-processing of astronomical data / images" OwnerUserId="10271" LastEditorUserId="10271" LastEditDate="2017-10-17T16:31:00.787" LastActivityDate="2017-10-17T16:31:00.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23184" PostTypeId="1" AcceptedAnswerId="23186" CreationDate="2017-10-17T15:30:12.347" Score="7" ViewCount="4109" Body="&lt;p&gt;My question is straight forward.  Is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Kilonova&quot; rel=&quot;nofollow noreferrer&quot;&gt;kilonova&lt;/a&gt; bigger than a supernova?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does a kilonova differ from a nova in terms of...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;luminosity&lt;/li&gt;&#xA;&lt;li&gt;duration&lt;/li&gt;&#xA;&lt;li&gt;energy&lt;/li&gt;&#xA;&lt;li&gt;volume of space occupied by the ejecta&lt;/li&gt;&#xA;&lt;li&gt;speed of the ejecta&lt;/li&gt;&#xA;&lt;li&gt;other metrics?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="19146" LastEditorUserId="19146" LastEditDate="2017-10-18T14:15:56.133" LastActivityDate="2021-01-29T13:53:18.523" Title="Is a kilonova bigger than a supernova?" Tags="&lt;supernova&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23185" PostTypeId="2" ParentId="13813" CreationDate="2017-10-17T15:34:54.527" Score="3" Body="&lt;p&gt;As others have said, in General Relativity gravitational waves travel at the speed of light. However, with the observation of GW170817, a gravitational wave signal from the merger of two neutron stars, coincident with a gamma-ray burst (GRB170817A), we can now set &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/2041-8213/aa920c&quot; rel=&quot;nofollow noreferrer&quot;&gt;stringent limits&lt;/a&gt; on the difference between the speed of photons and gravitational waves. The gravitational wave signal recorded the merger time at approximately 1.7 seconds before the arrival of the gamma-rays, and the source was in a galaxy approximately 40 Mpc away, so (using 1.7 seconds as an upper limit on the time difference, and a lower limit on the distance of 26 Mpc from the uncertainty on the distance measurement) an upper limit on the difference in speeds is $\lesssim 10^{-15}$ (see &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/2041-8213/aa920c&quot; rel=&quot;nofollow noreferrer&quot;&gt;Abbott &lt;em&gt;et al&lt;/em&gt;, Ap. J. Lett. 848, 2, (2017)&lt;/a&gt; for details).&lt;/p&gt;&#xA;" OwnerUserId="16795" LastActivityDate="2017-10-17T15:34:54.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23186" PostTypeId="2" ParentId="23184" CreationDate="2017-10-17T16:26:59.917" Score="13" Body="&lt;p&gt;Although it's a little tricky to say what &quot;bigger&quot; means in this context, the answer is, in most senses, no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A supernova puts out about ten to a hundred times as much energy in the form of light, and hundred or more times as much matter is ejected. (A core-collapse supernova undoubtedly puts out much more energy in the form of neutrinos as well.) What matter &lt;em&gt;is&lt;/em&gt; ejected by a kilonova &lt;em&gt;does&lt;/em&gt; go out faster (30-60,000 km/s, versus about 10,000 km/s for supernova ejecta).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, a kilonova puts out much more energy in the form of gravitational waves, so they're bigger in that sense.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2017-10-18T14:05:34.917" LastActivityDate="2017-10-18T14:05:34.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23187" PostTypeId="1" AcceptedAnswerId="23201" CreationDate="2017-10-17T17:37:27.987" Score="16" ViewCount="494" Body="&lt;p&gt;I'm trying to read &lt;a href=&quot;https://dcc.ligo.org/public/0145/P1700294/006/ApJL-MMAP.pdf&quot; rel=&quot;noreferrer&quot;&gt;Multi-messenger Observations of a Binary Neutron Star Merger&lt;/a&gt; the &quot;cast of thousands&quot; OPEN Access ApJ letter 848:L12 (59pp), 2017 October 20 &lt;a href=&quot;https://doi.org/10.3847/2041-8213/aa91c9&quot; rel=&quot;noreferrer&quot;&gt;https://doi.org/10.3847/2041-8213/aa91c9&lt;/a&gt; and get a feeling for the sequence of events that took place as the gravitational waves and gamma ray burst reached Earth at about 12:41 UTC 2017-Aug-17.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There seems to be five instruments involved in the first detection and direction determination; &lt;a href=&quot;https://en.wikipedia.org/wiki/LIGO#Observatories&quot; rel=&quot;noreferrer&quot;&gt;LIGO-Hanford and LIGO-Livingston&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Virgo_interferometer&quot; rel=&quot;noreferrer&quot;&gt;VIRGO&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Fermi_Gamma-ray_Space_Telescope&quot; rel=&quot;noreferrer&quot;&gt;Fermi-GBM&lt;/a&gt;, and &lt;a href=&quot;https://en.wikipedia.org/wiki/INTEGRAL&quot; rel=&quot;noreferrer&quot;&gt;INTEGRAL&lt;/a&gt;. The first three are gravitational wave detectors and the last two are gamma-ray telescopes in Earth orbit. Figure 2 of the paper (part of which is shown below) provides a densely packed infographic of the early observations. In the top left one can see an inset that spans from twelve seconds before the merger during the gravitational wave (GW) ramp in frequency, to six seconds after, where the majority of the gamma ray burst (GRB) is detected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somehow the combination of the GW and GRB set in motion a sequence of events that triggered a world-wide observation campaign to look for the event in all the remaining electromagnetic spectrum from radio through visible and UV to X-rays. Neutrino data streams were checked as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; I'd like to ask about the sequence of events, the alerts, and the rapid automatic and manual analysis of the GW and GRB data which triggered the alerts. Which detector or combination first &quot;saw&quot; the event as some kind of flagged event? Did one trigger a rapid analysis of the other? Were these automated alerts triggering software to re-analyze, or SMS text messages to thousands of cell phones triggering everyone to sit down at their work stations?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; Figure 2 (partial) showing the timeline seconds before and hours and days after (logarithmic scale). The GW and GRB data were used (see Figure 1) to start the search for the rest of the electromagnetic search.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Fi1nU.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fi1nU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; Figure 1 showing the localizations made from different sets of GW and GRB detectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NynJj.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NynJj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-11-16T02:10:09.027" LastActivityDate="2018-11-16T02:10:09.027" Title="&quot;Who saw&quot; the binary neutron star merger first? What was the sequence of events? (GRB/GW170817)" Tags="&lt;gravitational-waves&gt;&lt;neutron-star&gt;&lt;gamma-ray-bursts&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="23189" PostTypeId="2" ParentId="23179" CreationDate="2017-10-17T18:56:52.220" Score="8" Body="&lt;p&gt;Drizzling can't actually do any better than the theoretical resolving power of the combination of your telescope and the atmosphere. What it &lt;em&gt;can&lt;/em&gt; do is at least partly compensate for having pixels that are too large to properly sample the resolution of your telescope + the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem&quot; rel=&quot;noreferrer&quot;&gt;Nyquist Sampling Theorem&lt;/a&gt; basically says that you should have at least 2 pixels per resolution element in order to properly sample the resolution. Or, conversely, if you want to achieve a resolution of X arcsec, you want pixels that are no more than X/2 arcsec in size. So if the theoretical combination of your telescope diameter and the atmosphere had a resolution of 0.5 arcsec (so we're talking about really good atmospheric conditions), you want pixels that are 0.25 arcsec or smaller; otherwise, the pixelization overwhelms the intrinsic resolution. But note that you don't really gain anything from going to even smaller pixels: you'll just have a nice blurry image without visible pixelization. Having pixels that are 0.1 arcsec in size won't make the image any sharper than pixels that are 0.2 arcsec in size, if the resolution of your system (telescope + atmosphere) is ~ 0.5 arcsec.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your hypothetical case, 0.52-arcsec pixels are good enough for resolutions down to about 1 arcsec. Ideal drizzling would allow you to effectively have pixels that were about 0.25 arcsec in size, which would accommodate a resolution of about 0.5 arcsec. But you can never do &lt;em&gt;better&lt;/em&gt; than the intrinsic resolution of your telescope + the atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7982" LastEditDate="2017-10-18T06:10:00.507" LastActivityDate="2017-10-18T06:10:00.507" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23190" PostTypeId="1" CreationDate="2017-10-17T19:45:46.163" Score="2" ViewCount="401" Body="&lt;p&gt;Is the gravity of the planets Jupiter and Saturn strong enough to alter the orbit of Mars, or do they just give a small wiggle after which Mars is following his former orbit?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastEditorUserId="18888" LastEditDate="2017-10-18T06:06:28.137" LastActivityDate="2017-10-19T00:18:20.453" Title="How much does the orbit of Mars deviate when it is in linear conjunction with Jupiter and Saturn?" Tags="&lt;planet&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2017-10-20T11:00:17.637" ContentLicense="CC BY-SA 3.0" />
  <row Id="23192" PostTypeId="2" ParentId="23171" CreationDate="2017-10-17T22:29:07.937" Score="20" Body="&lt;p&gt;A more historical/linguistic than physical answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Democritus proposed that matter could not be divided infinitely, but that at some point one would reach a smallest possible piece, which he called &lt;em&gt;atomos&lt;/em&gt;, for &quot;uncut&quot;. He was entirely correct as far as we know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before 1930 or so, we modern humans mistakenly applied his word &quot;atom&quot; to what we then saw as the smallest piece of a chemical element that maintained its properties, because no one had ever subdivided one, and chemical processes behaved as if these were small indivisible particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that kind of &quot;atom&quot; was split, we were stuck with the word. What Democritus called &quot;atom&quot; today we would call &quot;quark or lepton&quot;, things that we think are fundamental and not splittable. But we could be wrong there too, if future physicists split these things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the ordinary things in our lives--you, me, the Earth, your house, your food, etc., are made of &quot;atoms&quot; in the 1920 sense, interacting chemically, electrically, and gravitationally. Even the Sun is made of atoms, though in that case they're not quite unsplittable because they fuse and change into different kinds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certain rare and exotic things in the universe are not made of atoms in the 1920 sense, like neutron stars. Though they are made of small indivisible quarks and leptons; atoms in the Democritus sense.&lt;/p&gt;&#xA;" OwnerUserId="19160" LastActivityDate="2017-10-17T22:29:07.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23193" PostTypeId="2" ParentId="7778" CreationDate="2017-10-17T23:47:00.227" Score="2" Body="&lt;p&gt;On the basis of the recent detection of GW170817 and a host of other observational evidence, it seems that a neutron star merger is one way to get mass out from a neutron star - maybe some tenths of a solar mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also evidence that the ejected material from the collision is neutron-rich, at least initially, and then produces neutron-rich nuclei through the r-process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is impossible to have small lumps of stable neutron star matter. A high density is required to prevent the neutrons decaying (see &lt;a href=&quot;https://physics.stackexchange.com/questions/143166/what-is-the-theoretical-lower-mass-limit-for-a-gravitationally-stable-neutron-st&quot;&gt;https://physics.stackexchange.com/questions/143166/what-is-the-theoretical-lower-mass-limit-for-a-gravitationally-stable-neutron-st&lt;/a&gt; ) The (theoretical) minimum mass for a stable neutron star is of order 0.1-0.2 solar masses, though none have been seen in nature.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-10-17T23:47:00.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23194" PostTypeId="1" AcceptedAnswerId="23294" CreationDate="2017-10-18T03:22:28.210" Score="1" ViewCount="172" Body="&lt;p&gt;Could the black hole consume the radiation? If so, could the pulsar radiation absorbed by the black hole make it grow steadily more massive over the course of billions or trillions of years?&lt;/p&gt;&#xA;" OwnerUserId="18831" LastActivityDate="2017-10-25T19:20:25.847" Title="Could a pulsar in orbit around a black hole feed the black hole with its radiation?" Tags="&lt;black-hole&gt;&lt;pulsar&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23198" PostTypeId="1" CreationDate="2017-10-18T08:10:59.977" Score="6" ViewCount="117" Body="&lt;p&gt;The &lt;A HREF=&quot;https://astronomy.stackexchange.com/questions/23161/why-is-the-discovery-of-merging-neutron-stars-important&quot;&gt;recent&lt;/A&gt; detection of a binary neutron star inspiraling and colliding raises an interesting question in my mind. &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Type_Ia_supernova&quot; rel=&quot;noreferrer&quot;&gt;Type Ia supernovae&lt;/A&gt; are believed to be caused by white dwarf/regular star pairs and/or white dwarf pairs. Short gamma ray bursts are now known to be produced by neutron star/neutron star collisions. Black hole/black hole collisions haven't produced any verifiably detected electromagnetic radiation. What about some of the other mixed type collissions? Especially white dwarf/neutron star or neutron star/black hole?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that stellar binaries tend to have similar mass, so mixed type remnants will likely be rather rare. Hence, why I didn't ask about white dwarf/black hole binaries. Even so, I'm curious how the large difference in density would affect things. Especially if a neutron star would be tidally shredded into an accretion disk by a black hole before they can collide, making a relatively slow event, or will the process be more rapid and violent, akin to the events we've seen so far?&lt;/p&gt;&#xA;" OwnerUserId="14244" LastEditorUserId="14244" LastEditDate="2017-10-18T08:18:28.570" LastActivityDate="2017-10-18T08:18:28.570" Title="Do mixed type remnant collisions produce anything interesting?" Tags="&lt;supernova&gt;&lt;gravitational-waves&gt;&lt;gamma-ray-bursts&gt;&lt;stellar-remnants&gt;&lt;kilonova&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23201" PostTypeId="2" ParentId="23187" CreationDate="2017-10-18T11:27:13.927" Score="18" Body="&lt;p&gt;The initial Fermi trigger can be found &lt;a href=&quot;https://gcn.gsfc.nasa.gov/other/524666471.fermi&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;, and the following sequence of alerts that were sent out by the LIGO Scientific Collaboration/Virgo Collaboration (LVC) and various electromagnetic observers following-up the event can be found in the GCN circular archive &lt;a href=&quot;https://gcn.gsfc.nasa.gov/other/G298048.gcn3&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. This doesn't quite give the whole story of the time line of events, but is a good start and the &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/2041-8213/aa91c9/meta&quot; rel=&quot;noreferrer&quot;&gt;linked &quot;Multimessenger&quot; paper&lt;/a&gt; is fairly comprehensive in giving the complete story.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As shown in the figure, the signal arrived first in the gravitational wave detectors, with gamma-rays arriving at Fermi and INTEGRAL approximately 1.7 seconds after the observed merger time. However, the online analysis software on-board Fermi was quickest off the mark in detecting the burst and generating an automated trigger just 14 seconds after the signal arrived (I don't know any more of the internal details of Fermi triggering, how the alerts get sent out to people, or what subsequent manual interventions are required). Within a little over 7 minutes the online automated software searching for gravitational wave signals using compact binary coalescence templates (see, &lt;a href=&quot;https://arxiv.org/abs/1107.2665&quot; rel=&quot;noreferrer&quot;&gt;this paper&lt;/a&gt; and &lt;a href=&quot;https://arxiv.org/abs/1604.04324&quot; rel=&quot;noreferrer&quot;&gt;this paper&lt;/a&gt;) had produced a candidate using data from the LIGO Hanford detector alone (LIGO Livingston data had been automatically vetoed by the software due to the presence of a glitch [Figure 2 of &lt;a href=&quot;https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.119.161101&quot; rel=&quot;noreferrer&quot;&gt;this paper&lt;/a&gt;], and Virgo data hadn't yet propagated to the location where the analysis was run) - this automatically notified (via email or text) a number people in the LVC that something interesting had happened. Less than 10 seconds after the gravitational wave candidate had been recorded, an automated code called RAVEN (see, e.g., Section 4.1 of &lt;a href=&quot;https://arxiv.org/abs/1611.07947&quot; rel=&quot;noreferrer&quot;&gt;this paper&lt;/a&gt;) noted the temporal coincidence between the Fermi trigger and the gravitational wave candidate. After being notified of the gravitational wave trigger, various people within the LVC initiated a conference call and started manually looking at the data and saw the obvious chirp-like signal in time-frequency representations of the data. Roughly 33 minutes after the gravitational wave trigger, and 40 minutes after signal arrival, it was decided (people were involved) to issue an announcement (the first entry in the GCN list &lt;a href=&quot;https://gcn.gsfc.nasa.gov/other/G298048.gcn3&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;) that there was a joint Fermi gamma-ray burst trigger and gravitational wave trigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding signal arrival times at the various gravitational wave detectors: it arrived first at Virgo, followed by the LIGO Livingston detector, and finally at the LIGO Hanford detector.&lt;/p&gt;&#xA;" OwnerUserId="16795" LastEditorUserId="16795" LastEditDate="2017-10-24T08:05:41.223" LastActivityDate="2017-10-24T08:05:41.223" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23203" PostTypeId="1" CreationDate="2017-10-18T17:27:56.097" Score="1" ViewCount="545" Body="&lt;p&gt;If you keep adding electrons or protons to a Neutron star, what would happen? Are there decay processes that convert them to neutrons in the absence of the oppositely charged particle? &lt;/p&gt;&#xA;" OwnerUserId="17582" LastActivityDate="2019-04-07T18:07:17.720" Title="Can a Neutron Star become charged?" Tags="&lt;neutron-star&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23204" PostTypeId="1" AcceptedAnswerId="23243" CreationDate="2017-10-18T18:27:42.413" Score="2" ViewCount="446" Body="&lt;p&gt;My understanding of &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Type_Ia_supernova&quot; rel=&quot;nofollow noreferrer&quot;&gt;Type Ia supernovae&lt;/A&gt; is that they are expected, in most cases, to destroy the white dwarf(s) that went in to them, leaving behind no high density remnants (i.e. no white dwarf, neutron star, or black hole). Black hole/black hole collisions are expected to leave behind a black hole, of course, with &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Binary_black_hole#Observation&quot; rel=&quot;nofollow noreferrer&quot;&gt;less mass than the sum of the masses of the black holes that went in to the collision&lt;/A&gt;. Do we expect a kilonova to leave behind nothing but gas and radiation or some kind of stellar remnant? If it leaves behind a remnant, what class to we expect it to be (white dwarf, neutron star, or black hole) and what mass? The addition of mass seems redundant, but the dividing lines between the masses of these objects are based on upper limits on the mass of the less dense class (&lt;A HREF=&quot;https://en.wikipedia.org/wiki/Chandrasekhar_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chandrasekhar limit&lt;/A&gt; for white dwarfs, &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Tolman%E2%80%93Oppenheimer%E2%80%93Volkoff_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tolman–Oppenheimer–Volkoff limit&lt;/A&gt; for neutron stars), and don't actually apply as lower limits for the mass of the high density class. For example, neutron stars are sometimes referred to as &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Neutron_star#Giant_nucleus&quot; rel=&quot;nofollow noreferrer&quot;&gt;giant nuclei&lt;/A&gt;, which would put the lower limit of their mass at $1$ or $2$ atomic mass units, depending on whether the presence of a neutron and stability to radioactive decay are requirements. Yes, those times people are being poetic due to the difference in stabilization mechanisms (gravity vs nuclear forces), but the point remains that it may be possible for a neutron star to be theoretically stable at less than $1.4M_\odot$. The only lower limit to black hole mass I know of would be the lifetime limit from &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;black hole evaporation&lt;/A&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put another way, do we think the density at any point in either of the inspiraling neutron stars gets high enough to form an event horizon? If that happens, this seems like a plausible way for producing black holes that are smaller than those produce by &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Supernova#Core_collapse&quot; rel=&quot;nofollow noreferrer&quot;&gt;core collapse supernovae&lt;/A&gt;, kind of like how &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Critical_mass&quot; rel=&quot;nofollow noreferrer&quot;&gt;critical mass&lt;/A&gt; for atom bombs can be achieved by either bringing together enough fissile material or compressing the available material enough (e.g. the &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Fat_Man&quot; rel=&quot;nofollow noreferrer&quot;&gt;fat man&lt;/A&gt; detonation mechanism vs the &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Little_Boy&quot; rel=&quot;nofollow noreferrer&quot;&gt;little boy&lt;/A&gt; one).&lt;/p&gt;&#xA;" OwnerUserId="14244" LastActivityDate="2020-10-09T15:30:21.640" Title="Does a kilonova leave a high mass remnant?" Tags="&lt;black-hole&gt;&lt;supernova&gt;&lt;stellar-remnants&gt;&lt;kilonova&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23206" PostTypeId="2" ParentId="23179" CreationDate="2017-10-18T19:36:27.933" Score="6" Body="&lt;p&gt;Drizzling is just one technique in the field of &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Super-resolution_imaging&quot; rel=&quot;nofollow noreferrer&quot;&gt;super resolution imaging&lt;/A&gt;. Note that claims that you can't &quot;pass the resolution limit&quot; of a camera are wrong, in principle. In practice, it's really hard. What limits your ability to construct an image of what the camera is looking at are: the amount of random noise in the pixels, and your understanding of the camera's &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Point_spread_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;point spread function (PSF)&lt;/A&gt; (though some astronomers prefer to use the term &quot;point response function&quot; [PRF] if you include the effects of both the optics and the detector). In principle, if you invest enough time and money to nail down the PSF and increase the signal to noise ratio without limit (beware of bias), the object you're imaging isn't changing, and you have enough accuracy in measuring how you point the camera when dithering, then, in principle, you can push the resolution of the reconstructed image as far as you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, it's usually cheaper and more time efficient to just get a bigger camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a concrete mathematical example of how image reconstruction works, consider the sum of two offset &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Normal_distribution&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gaussians&lt;/A&gt;, with $\sigma=1$, $\mu_1=-0.1$, and $\mu_2=+0.1$. Here's a graph of such a curve overplotted with a single Gaussian with the same $\sigma$, scaled to have the same height at $x=0$. You can only see one line plotted, though I assure you there are two. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/IeHvf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IeHvf.jpg&quot; alt=&quot;Plot of sum of Gaussians vs. single close Gaussian&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see the difference between them, we have to plot the difference (plotting the residuals, in data analysis parlance). This next graph is the difference, zoomed in by a factor of 100 in the $y$-direction.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ujKLE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ujKLE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the theoretical difference is not zero. So, if what you're imaging isn't changing, you can model the background accurately enough, and you have a sufficiently accurate model of the camera's PSF, you could detect the difference between the high Gaussian and two slightly offset Gaussians scenarios if you invest enough time and energy into the project, in principle, even if your pixels are twice the full-width at half maximum (FWHM) of the Gaussian. $\mathrm{FWHM} = \sqrt{8\ln(2)}\sigma$ for Gaussians, so in this graph twice the FWHM is about $4.7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to further the goal of understanding the PSF is to intentionally make it easy to model (as opposed to 'as small as possible'), as is done in many forms of &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Aperture_masking_interferometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;non-redundant aperture mask imaging&lt;/A&gt;. I believe that the trade-off there is that you lose both field of view and the (obvious) loss of the light that gets blocked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, though, you're usually better off getting a bigger camera, or compensating for the atmosphere (e.g. &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Lucky_imaging&quot; rel=&quot;nofollow noreferrer&quot;&gt;lucky imaging&lt;/A&gt; and &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Adaptive_optics&quot; rel=&quot;nofollow noreferrer&quot;&gt;adaptive optics&lt;/A&gt;) if your imaging is seeing limited. Thus, the limits imposed by the &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nyquist sampling theorem&lt;/A&gt; are more practical, not theoretical. It's also prone to mislead you if you don't remember that pixels are not points, they're little buckets that collect light that falls into a finite area. Talking about it &quot;sampling&quot; some function, implicitly uses the &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Mean_value_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;mean value theorem&lt;/A&gt; to say that the amount of light collected divided by the pixel's area gives the value of the image's brightness there. What's often dropped is the caveat that it's the brightness &lt;em&gt;somewhere inside the pixel&lt;/em&gt;, not necessarily at the pixel's center. This is usually a very small effect, which is why it's often ignored, but when you're talking about working at the resolution limit and having pixel sizes comparable to that resolution, the distinction may become important. Roughly speaking, if the PSF is highly &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Curvature&quot; rel=&quot;nofollow noreferrer&quot;&gt;curved&lt;/A&gt; within the pixel, the offset from pixel center of the mean value can become significant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to think about what &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Drizzle_%28image_processing%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;drizzle&lt;/A&gt; gets you is that by moving the pixels between images, you move around the light buckets. In principle, with enough images at different offsets you can probably push the pixel density of the image to the limit of your ability to reconstruct the size of the offsets (usually the pointing of the camera/telescope, though I know of no reason why the sensor couldn't be moved around on some stepper motors attached to micrometers), or the sharpness of the divides between the pixels (the ability to gather light is neither uniform over a pixel, nor could the pixels fill the detector from edge to edge completely, these are just really good approximations most of the time).&lt;/p&gt;&#xA;" OwnerUserId="14244" LastEditorUserId="14244" LastEditDate="2017-10-20T17:00:40.927" LastActivityDate="2017-10-20T17:00:40.927" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="23207" PostTypeId="2" ParentId="23203" CreationDate="2017-10-18T21:21:03.340" Score="5" Body="&lt;p&gt;Charge is a conserved quantity. If you add a net charge to an object, then it becomes charged.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-10-18T21:21:03.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23208" PostTypeId="1" CreationDate="2017-10-18T22:35:33.887" Score="-3" ViewCount="191" Body="&lt;p&gt;if this happens in the future what will be the consecuences of this?.&#xA;Consecuences for the life or the climate conditions.&lt;/p&gt;&#xA;" OwnerUserId="19200" LastActivityDate="2017-11-12T08:16:03.330" Title="spin–orbit resonance, between moon and the earth" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23209" PostTypeId="2" ParentId="23190" CreationDate="2017-10-19T00:18:20.453" Score="3" Body="&lt;p&gt;To properly answer this question, you should apply the 3 body problem, which, gets very long and mathy.  It's technically not solved but it can be approximated with very good accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying basic Newtonian formulas can give you an estimate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun has about 1,048 times the mass of Jupiter, and if we line up Mars at it's Aphelion (1.67 AU) to Jupiter at it's Perihelion, (4.95 AU).  Mars at it's furthest and Jupiter at it's closest, Mars is still about twice as close to the Sun as it is to Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity, I'm going to call Mars at Aphelion an &quot;MAU&quot; - Mars Aphelion Unit, so Mars, furthest from the sun, 1 &quot;MAU&quot;, Jupiter closest to Mars, about 2 &quot;MAU&quot; and Jupiter closest to the Sun, 3 &quot;MAU&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in perfectly lined up conditions, Jupiter never has more than 1/4,000th the gravitational pull on Mars than the Sun does and we also have to factor in the tidal force.   It gets tricky here, cause all 3 objects exert a force on the other (or stretch space time which tells the objects where to go if you like the relativity version), but I'm going to stick with forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter tugs on Mars and the Sun at the same time so how far it pulls Mars from the sun is more of an an applied tidal force.  At closest point, Jupiter's tug on Mars from 2 &quot;MAU&quot; and it's tug on the Sun from 3 &quot;MAU&quot; means it tugs on Mars 2.25 times stronger.  Take the inverse, the effective gravitational pull - at closest, is the inverse of that, or 44%, so you can only apply 56% of Jupiter's gravitational pull towards Mars relative to the sun.  And that's at peak.  As Mars moves closer to the sun and/or further from Jupiter that number drops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect of Jupiter on Mars orbit relative to the sun is never more than about 1 part in 7200 of that of the sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Mars moved towards Jupiter in it's orbit it speeds up and effectively moves away from the Sun, and as it moves away from Jupiter, this effect is largely cancelled out over time.   In effect it's a slight stretching when moving towards Jupiter and squashing when moving away, of Mars' elliptical orbit every time around the sun.  These are sometimes called orbital perturbations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 part in 7,200 works out to about 0.14% and that's the peak Jupiter effect.    Rounding down, a .01% variation on Mars' orbit seems a plausible estimate and that might even be a little high.  Applying that to the diameter of Mars' orbit, that works out to a stretching/squashing of about 20,000 to 25,000 km.  Compared to Mars natural variation based on it's eccentricity of over 26 million miles, that's pretty insignificant.   This stretching and squashing is also largely balanced out as Jupiter circles around the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we apply the acceleration formula, I get a somewhat different number.   g = G*M/r^2.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational Constant:   6.67E-11&#xA;Mass of Jupiter 1.898E27 kg&#xA;r = distance between the two.   Closest pass is about 3.28 AU, or 4.91E11 meters&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The peak gravitational acceleration Mars feels from Jupiter is about 5.26E-7 m/s^2.   That sounds negligible, but Mars takes 187 days to orbit the sun and it spends about half that time closer to Jupiter than the Sun is to Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a (bad) estimate, because that's the maximum gravitational force at closest pass, if we estimate that over a Mars year, that works out to about 90 days of &quot;falling&quot; towards Jupiter and 90 days falling away.  Obviously the period is longer than 90 days but the force is less when Mars is further away so that's a guestimate.   Apply (d=1/2at^2) to 90 days or about 7.8 million seconds.  The distance using that formula comes to about 16,000 km.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, both estimates are pretty close, one of 16,000 the other of 20-25,000 km.   This also works well with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbit_of_Mars#Table_of_orbital_parameters&quot; rel=&quot;nofollow noreferrer&quot;&gt;4 or 5 significant figures&lt;/a&gt; usually given for Mars' obit.   4 figures on distance works out to nearest 100,000 km, which wouldn't be affected by Jupiter's stretching and squashing of Mars' orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now as for Saturn, at 30% the mass of Jupiter and roughly twice the distance from Mars and the Sun, you're looking at less than 8% of the gravitational force and with tidal effects figured in, maybe just 4%-6% the orbital perturbations.   Jupiter is the one that really matters.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbital stretching and squashing is what lead to the discovery of the planet Neptune, because astronomers noticed that Uranus orbit was effected in a similar way to how Jupiter affects Mars.  Observing those perturbations told them about where to look for Neptune and it was found not long after.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to point out that over tens of thousands of orbits, these perturbations can lead to changes in Mars' orbit, primarily it's eccentricity.   It's semi-major axis and period of the orbit remain mostly unchanged, but it's eccentricity operates on two cycles.  Mars' eccentricity varies on a &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbit_of_Mars#Changes_in_the_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;96,000 year and a two million year cycle&lt;/a&gt;.  Like Mars, &lt;a href=&quot;https://en.wikipedia.org/wiki/Milankovitch_cycles#Orbital_shape_.28eccentricity.29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth's eccentricity&lt;/a&gt; also has cycles driven by the other planets, 95,000, 125,000 and 413,000 years.   Mars' eccentricity is prone to greater variation, presumably (I think) because it gets closer to Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a rule, these eccentricity oscillations move back and forth unless two planets end up in resonance with each other.   When that happens the variations can continue to grow leading to significant movement of one or both of the planets over millions of years.  This is more likely to happen shortly after the solar-system forms.  Our solar system is thought to be long-term stable with no orbital resonances between planets now, and none expected for at least a couple billion years.   There's a small chance that Jupiter and Mercury will fall into resonance in a few billion years.   &lt;a href=&quot;http://www.scholarpedia.org/article/Stability_of_the_solar_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's a nice detailed article&lt;/a&gt; on orbital eccentricity variation and stability of the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a ballpark answer anyway.   Corrections to my math are welcome and if anyone can do the 3 body problem, feel free.  That would be much more accurate, but I think 16,000 - 25,000 km per orbit is in the ballpark.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-10-19T00:18:20.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23210" PostTypeId="1" AcceptedAnswerId="23218" CreationDate="2017-10-19T00:52:59.397" Score="4" ViewCount="160" Body="&lt;p&gt;I read that after LIGO and Virgo interferometers detected gravitational waves signal from a kilonova, many ground based telescopes quickly turns their mirrors towards the spot and &quot;sees&quot; gold and platinum ,etc other heavy elements over the next few days. Is it really possible to directly observe heavy elements being formed consider that the event occurs at a distance of around 130 millions light year away?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastActivityDate="2017-10-19T15:13:25.757" Title="Did the scientists really observed gold and platinum being created from kilonova?" Tags="&lt;neutron-star&gt;&lt;kilonova&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23212" PostTypeId="1" AcceptedAnswerId="23221" CreationDate="2017-10-19T01:51:03.373" Score="3" ViewCount="331" Body="&lt;p&gt;I am wondering since neutron star is smaller than the average main-sequence stars and it doesn't produce energy via fusion so why do we still call it a star?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastActivityDate="2017-10-19T12:17:52.567" Title="Why is neutron star a star?" Tags="&lt;neutron-star&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23213" PostTypeId="1" AcceptedAnswerId="23234" CreationDate="2017-10-19T02:00:15.727" Score="-1" ViewCount="639" Body="&lt;p&gt;Since there is no bacteria and oxygen in space, what will happen to a dead chicken's body if released into space?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming it's far enough for any nearby planet to collide with, will it rot? Freeze forever?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should we release the chicken into intergalactic space, will that change anything?&lt;/p&gt;&#xA;" OwnerUserId="19203" LastEditorUserId="19203" LastEditDate="2017-10-26T11:31:08.103" LastActivityDate="2017-10-26T11:31:08.103" Title="What happens to organic material in space" Tags="&lt;amateur-observing&gt;&lt;space&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2017-10-20T10:56:33.850" ContentLicense="CC BY-SA 3.0" />
  <row Id="23216" PostTypeId="1" AcceptedAnswerId="23220" CreationDate="2017-10-19T03:09:48.520" Score="5" ViewCount="347" Body="&lt;p&gt;In the news, of course, is the theory that much of the mass of the heaviest elements in Earth would have come from colliding neutron stars.  This graphic gives one such analysis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lTYGJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lTYGJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;sup&gt;Origin of the Elements in the Solar System by Jennifer Johnson is licensed under a &lt;a href=&quot;http://creativecommons.org/licenses/by-sa/4.0/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Creative Commons Attribution-ShareAlike 4.0 International License&lt;/a&gt;. (Larger version at &lt;a href=&quot;http://www.astronomy.ohio-state.edu/~jaj/nucleo/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.astronomy.ohio-state.edu/~jaj/nucleo/&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how accurate this is, but the URL looks good.  Also, this seems to be a new theory, and AFAIK this is the first actual evidence for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is correct, what are the implications for the quantity of heavy elements that would be found on planets outside our solar system?  In particular, I guess, the very heavy ones: gold, platinum, thorium, uranium...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've been watching the skies for signs of neutron star collisions for several years now (IIUC), and I guess we're starting to get an idea of the frequency of this event.  And I've seen estimates of how much gold etc. was produced by this particular event.  How many planets would one such collision supply with the quantity of heavy elements in Earth?&lt;br /&gt;&#xA;(I guess we only really know about gold in the Earth's crust, and to some extent in the mantle; are these elements concentrated in the core?  Any idea about heavy elements in other planets of our solar system?)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: given the frequency of the event per galaxy, the estimated number of stars with planets in a galaxy, etc., is it likely that most planets are deficient in heavy elements, as compared with Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Side comment: if so, then nuclear wars might be rare or impossible on those planets.)&lt;/p&gt;&#xA;" OwnerUserId="19205" LastEditorDisplayName="user1569" LastEditDate="2017-10-19T21:19:10.923" LastActivityDate="2017-10-19T21:19:10.923" Title="Neutron stars and distribution of heavy elements" Tags="&lt;star&gt;&lt;element&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23217" PostTypeId="1" AcceptedAnswerId="23228" CreationDate="2017-10-19T04:56:19.283" Score="-1" ViewCount="153" Body="&lt;p&gt;Velocity decides if 2 stars with different masses in a binary star formation will collide or not, but if 2 stars with same mass existed but with no velocity (i.e they are not orbiting each other, just staying at their places at some distance), would they collide ? I guess the answer is no, but just not sure of how and where do the gravitational wave friction come into the scenario.&lt;/p&gt;&#xA;" OwnerUserId="13587" LastActivityDate="2017-10-19T15:30:44.917" Title="Would 2 stars in binary star formation with same mass and no velocity collide into each other?" Tags="&lt;binary-star&gt;" AnswerCount="1" CommentCount="10" ClosedDate="2017-10-20T10:58:44.740" ContentLicense="CC BY-SA 3.0" />
  <row Id="23218" PostTypeId="2" ParentId="23210" CreationDate="2017-10-19T06:47:19.207" Score="8" Body="&lt;p&gt;No they didn't. The deduction that (very) heavy elements are being produced is somewhat indirect and involves creating a theoretical model of the expanding &quot;fireball&quot; and comparing that model against measurements of brightness vs time at a range of wavelengths and against spectra (mostly in the infrared) as the afterglow faded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally, when we view a stellar spectrum, one can see individual absorption lines that correspond to particular, identifiable elements. In the afterglow of this explosion that is not possible. It is moving so fast that the absorption features that were seen are smeared out by the Doppler effect &lt;em&gt;and&lt;/em&gt; the atomic physics required to predict the strength of absorption is too uncertain to make accurate identifications of particular elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is likely though is that the heavy elements, known as lanthanides are providing the large opacity in the &quot;photosphere&quot; of the fireball and are responsible for the rapid fading of the visible signal, the rather more lengthy decline in the infrared and the appearance of extremely broad absorption features that develop in the infrared. The heating of the ejecta by the radioactive decay of very neutron-rich nuclei is also a vital component of explaining the development of the afterglow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See for example &lt;a href=&quot;https://arxiv.org/abs/1710.05858&quot; rel=&quot;noreferrer&quot;&gt;Pian et al. (2017)&lt;/a&gt; and &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/2041-8213/aa905c/meta&quot; rel=&quot;noreferrer&quot;&gt;Chornock et al. (2017)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-10-19T08:45:47.453" LastActivityDate="2017-10-19T08:45:47.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23219" PostTypeId="1" CreationDate="2017-10-19T07:16:22.743" Score="1" ViewCount="295" Body="&lt;p&gt;If a matter distribution is spherically symmetric, the circular velocity is straightforwardly related to the mass via:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_{circ} = \sqrt{GM/r}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an axisymmetric potential, on the other hand hand, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_{circ}(R) = \sqrt{R\frac{d\Phi}{dR}}\neq \sqrt{GM/r}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\Phi$ the gravitational potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't quite understand is the following: the potential of the Milky Way is not spherical. How is the measured velocity then reliably related to a mass?&lt;/p&gt;&#xA;" OwnerUserId="6965" LastActivityDate="2017-10-20T08:39:02.510" Title="Circular velocity in axisymmetrici potential" Tags="&lt;rotation&gt;&lt;dark-matter&gt;&lt;mass&gt;&lt;velocity&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2017-10-20T11:09:26.813" ContentLicense="CC BY-SA 3.0" />
  <row Id="23220" PostTypeId="2" ParentId="23216" CreationDate="2017-10-19T07:23:15.063" Score="4" Body="&lt;p&gt;I may do some more detailed calculation, but I think there is a misconception that has provoked your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The content of (e.g) gold, with respect to other elements in the solar system can be readily measured in meteorites. The Earth is expected to contain a similar abundance, but is very depleted in hydrogen and helium and somewhat depleted of lighter elements like carbon and sodium. The surface abundance of gold in the Earth is then a bit lower, because it sinks during the Earth's formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We expect the relative abundances of chemical elements to be quite smoothly distributed in space and time. Material that is produced in supernovae and neutron star collisions is mixed quite quickly in the interstellar medium. Thus the material that makes up the solar system is contaminated by the excreta of as many as a billion stars that lived and died before it's birth. Even if a neutron star merger occurred right next to the molecular cloud that formed the Sun, that cloud may have contained a million solar masses of gas, that would have diluted the signature of the kilonova event. We expect this to be true for all stars born in our Galaxy, except those that were born right near the beginning, where individual &quot;pollution&quot; events might become apparent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate of these events in the local universe is only known observationally to an order of magnitude. In the past, it is also rather poorly known, but can be indirectly deduced from the rate of short Gamma ray bursts (of which we see only a small, uncertain fraction), if we assume that all sGRBs are merging neutron stars and that all merging neutron stars produce sGRBs. The rate in our Galaxy is extrapolated from the properties of the meagre number of identified double neutron stars and their orbital decay rates and is about 1 every 50,000 years (so about 200,000 over the lifetime of the Galaxy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These rates require that about 0.0001 to 0.001 solar masses of very heavy elements is produced in each of these explosions to reproduce the solar system abundances. In the recent event, the amount of heavy elements in the ejecta appear to be in that range.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-10-19T07:23:15.063" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23221" PostTypeId="2" ParentId="23212" CreationDate="2017-10-19T09:10:21.357" Score="5" Body="&lt;p&gt;It seems Walter Baade and Fritz Zwicky may have coined the term in 1934 (&lt;a href=&quot;https://en.wikipedia.org/wiki/Neutron_star#History_of_discoveries&quot; rel=&quot;noreferrer&quot;&gt;see here&lt;/a&gt;) and the name stuck around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've known some astronomers to label them 'stellar remnants' for the reason you gave; they are not stars in the sense they do not produce energy via nuclear fusion, and that they are not made entirely of neutrons as the name would suggest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They seem to be called stars for lack of another layman's term (from our point of view on Earth, not all neutron stars are pulsars), and that they still share some characteristics of stars: they emit their own light, they have masses greater than the Sun and can even have planets orbiting them.&lt;/p&gt;&#xA;" OwnerDisplayName="user10106" LastActivityDate="2017-10-19T09:10:21.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23222" PostTypeId="2" ParentId="23219" CreationDate="2017-10-19T09:55:47.313" Score="3" Body="&lt;p&gt;What you do is predict the orbital velocity for a more realistic hypothesised mass distribution. There is an easy analytic solution for a disc. More complicated distributions require a numerical solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you get far enough &lt;em&gt;outside&lt;/em&gt; an arbitrary mass distribution, then that solution will asymptotically look similar to $v_{\rm circ} = \sqrt{GM/r}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomers get into trouble when explaining the evidence for dark matter from galaxy rotation curves to the layperson. They often use the analogy of the solar system, where the orbital speed goes as $r^{-1/2}$. The trouble is that that a Galaxy is not like the solar system because the mass is &lt;em&gt;distributed&lt;/em&gt; and in general cannot be assumed to be concentrated at a point and Newton's shell theorem does not apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice what you can do is posit a more realistic distribution of matter based on star counts and the gas that can be observed. You can then predict (using a computer model) what the rotation velocity should be at any point if that is all the mass that exists. This model will converge to the simple $r^{-1/2}$ law once you get well outside the bulk of the mass in the model. The model can be compared to the observed velocities of gas, globular clusters, planetary nebulae and other velocity tracers - some of which do indeed exist well outside &lt;em&gt;the bulk&lt;/em&gt; of the visible galaxy matter. What you find is that they move much faster than predicted, indicating that considerably more unseen matter must exist inside their galactic orbits. You can then add that (dark) matter in various ways. Non-spherically symmetric and spherically symmetric dark matter profiles (of the same total mass) would produce different orbital velocities at different radii for precisely the reason you identify. By looking at tracers at different radii you can map out the distribution of unseen mass. Once happy with your model you can integrate the density of seen and unseen matter to estimate the total mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line is that the $\sqrt{GM/r}$ is just a &quot;back of the envelope&quot; demonstration of the effect. Real research uses more accurate models or works in a regime far outside the bulk of the visible matter where the &quot;back of the envelope&quot; approximation works.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-10-19T09:55:47.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23223" PostTypeId="2" ParentId="23212" CreationDate="2017-10-19T12:17:52.567" Score="2" Body="&lt;p&gt;According to wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A star is a luminous sphere of plasma held together by its own gravity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then it goes on to say that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Near the end of its life, a star can also contain degenerate matter &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(see also &lt;a href=&quot;https://en.wikipedia.org/wiki/Degenerate_matter&quot; rel=&quot;nofollow noreferrer&quot;&gt;nuetron degeneracy&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.dictionary.com/browse/star&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.dictionary.com/browse/star&lt;/a&gt; has:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Any of the heavenly bodies, except the moon, appearing as fixed luminous points in the sky at night.&lt;/li&gt;&#xA;  &lt;li&gt;Astronomy. any of the large, self-luminous, heavenly bodies, as the sun, Polaris, etc.&quot;.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nothing in those definitions explicitly requires fusion, though the mention of plasma seems to infer it. The quantum mechanical effects in a nuetron star are an analogue to an ideal gas in classical physics. The degeneracy pressure thereof prevents the star from collapsing into a black hole.&lt;/p&gt;&#xA;" OwnerUserId="19215" LastActivityDate="2017-10-19T12:17:52.567" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23224" PostTypeId="2" ParentId="2115" CreationDate="2017-10-19T15:06:28.497" Score="1" Body="&lt;p&gt;Think of a neutron as a proton + electron pushed into it. Neutron stars are a lot like that; any electrons that have met a proton will be very strongly forced into forming a neutron.  This happens because an object containing very many protons and electrons has collapsed below the electron degeneracy pressure limit.  Presumably, there weren't exactly the same number of protons and electrons before the collapse; and a fraction of those particles did not find their counterparts before the collapse stopped.   It's thus quite possible that a particular neutron star has a small positive charge as a whole, and it probably contains some protons and electrons (not counting neutrons). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, being composed of protons, neutrons, and electrons does not make matter an element.  Being composed of protons and neutrons does not make matter an atomic nucleus.  The forces that hold the neutron star together are unrelated to the forces that hold together an atomic nucleus, and its properties are very different, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A positively charged storm cloud might have many more protons than electrons and still not be called an &quot;element&quot;.&lt;/p&gt;&#xA;" OwnerUserId="11041" LastActivityDate="2017-10-19T15:06:28.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23225" PostTypeId="2" ParentId="23210" CreationDate="2017-10-19T15:13:25.757" Score="-1" Body="&lt;p&gt;Yes and no, as the previous answers suggest. It is confirmed that matels heavier than Fe are created in merging neutron stars. However, better models are needed to infer the creation of specific metals from spectra.&lt;/p&gt;&#xA;" OwnerUserId="19227" LastActivityDate="2017-10-19T15:13:25.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23226" PostTypeId="2" ParentId="22795" CreationDate="2017-10-19T15:16:04.540" Score="0" Body="&lt;p&gt;Caption of &lt;em&gt;Table 1&lt;/em&gt; in the OP's reference contains:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Keplerian elements and their rates, with respect to the mean ecliptic and equinox of J2000&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, &quot;the equinox&quot; mentioned in step 5 cited in the OP refers to the &lt;em&gt;mean equinox&lt;/em&gt; of J2000. Now the main problem is what the mean equinox actually is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.oxfordreference.com/oso/viewauthority/10.1093$002foi$002fauthority.20110803100145829;jsessionid=6ECE3B5BF4561B199EA3EA67453DC05C&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oxford Reference&lt;/a&gt;, &lt;em&gt;mean equinox&lt;/em&gt; is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The direction to the equinox at a particular epoch, with the effect of nutation subtracted. The mean equinox therefore moves smoothly across the sky due to precession alone, without short-term oscillations due to nutation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This means that the term &lt;em&gt;equinox&lt;/em&gt;, as used in astronomy, doesn't only mean the discrete events of vernal and autumnal equinoxes, but also the direction which the perpendicular$^\dagger$ to the Earth's axis points to at any moment of time. Thus we can conclude that the equinox of J2000 is the direction this perpendicular points to at the epoch J2000, not at 20 March 2000 or any other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to get the true orientation of the Earth at the &lt;em&gt;event of equinox&lt;/em&gt; we should use a precession+nutation model to advance the orientation at the Epoch to the date required.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;$^\dagger$ the one lying in the ecliptic plane, of course&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="735" LastEditDate="2017-10-21T19:18:28.547" LastActivityDate="2017-10-21T19:18:28.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23227" PostTypeId="1" CreationDate="2017-10-19T15:23:16.520" Score="1" ViewCount="147" Body="&lt;p&gt;If this is off topic, please change it to world-building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to make a low cost heliography telescope, which is not digital, and which can be used to generate an worldwide numerical fingerprint using sun spot positions as a mathematical code?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every city in the world would have the same heliograph. Nearly all the sunlit heliographs of the world would agree what the sunspot fingerprint code is. The sun generated code would then decide who's turn it is to record financial verification tasks, government accounts, and other corruption issues. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would other cosmic signals be more practical to graph and suited to that end?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the sun change too slowly to be useful for generate a daily random code? what about weekly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have any tips of how to design a simple analogue heliograph that converts sunspots into random numerical code, (i.e. to make a crypto currency called Sun-Peso/ to decide an olympic host country)? &lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="35237" LastEditDate="2020-09-20T01:03:50.087" LastActivityDate="2020-09-20T01:03:50.087" Title="Is it possible to generate random numbers from images of the sun?" Tags="&lt;the-sun&gt;&lt;telescope&gt;" AnswerCount="0" CommentCount="10" ClosedDate="2017-10-20T11:02:23.143" ContentLicense="CC BY-SA 4.0" />
  <row Id="23228" PostTypeId="2" ParentId="23217" CreationDate="2017-10-19T15:30:44.917" Score="4" Body="&lt;p&gt;It seems the crux of your question lies in a misunderstand of how gravity works. So I'll try to answer your question by correcting your perceptions about gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity is a force imparted by objects with mass on all other objects with mass. The range of a gravitational force is infinite which means that galaxies millions and billions of light years away are technically applying a gravitational force on you, its just so infinitesimally small as to be not noticeable. The general equation for the force of gravity between two objects of mass $M_1$ and $M_2$ at a separtion of $r$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_g = G\frac{M_1M_2}{r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $G$ is the gravitational constant equal to $6.67\times10^{-11}\ \mathrm{m^3\ kg^{-1}\ s^{-2}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had nothing in the universe except the Earth ($M_1 = 5.97\times10^{24}\ \mathrm{kg}$) and the Moon ($M_2 = 7.35\times10^{22}\ \mathrm{kg}$) at their current distance of $r = 3.8\times10^7\ \mathrm{m}$, you'd find that $F_g = 2\times10^{22}\ \mathrm{N}$ where $N$ a &lt;a href=&quot;https://en.wikipedia.org/wiki/Newton_(unit)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Newton&lt;/a&gt; - a unit of force equivalent to $\mathrm{kg\ m\ s^{-2}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this scenario you calculate the gravitational force experienced by each body. The important point here is that &lt;em&gt;both bodies experience this singular force&lt;/em&gt;. The Moon is pulled towards the Earth with a force equal to $2\times10^{22}\ \mathrm{N}$ and the Earth is likewise pulled towards the Moon by the same force equal to $2\times10^{22}\ \mathrm{N}$. They cannot and do not cancel each other out because each body is feeling one and only force of gravity. The Moon knows it experiences a gravitational force from the Earth and so it responds to that force by moving towards the Earth. It also happens to apply a gravitational force on the Earth to pull the Earth towards itself, but that force is the &lt;em&gt;same force&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in your scenario, two neutron stars which are just sitting next to each other in space would experience a singular gravitational force between them and pull themselves towards each other, eventually colliding. You can hopefully see the above argument works whether the two objects are the Moon and Earth, two neutron stars, or even Earth and you. Due to the mutual, single gravitational force within the system, the two bodies will be attracted to one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, everything I've just said is the &quot;Newtonian&quot; description of gravity. A more complete and accurate picture is achieved by considering this from a General Relativity standpoint.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-10-19T15:30:44.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23229" PostTypeId="1" CreationDate="2017-10-19T17:55:04.693" Score="4" ViewCount="1198" Body="&lt;p&gt;So the visible light from stars can be used to identify the elements in that star by looking at the spectral emission lines and comparing those lines to emission lines of various elements (and their isotopes) here on earth.  These well-known line wavelengths can also be used to determine the relative motion of that star to the earth by looking at the red (or blue) shift of wavelengths of those spectral lines.  It would then be reasonable to assume that our own sun emits those same well-defined spectral lines, which should show up at specific wavelengths (based on the elements in the sun).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, sunlight appears (mostly) white, and using a prism shows a rather consistent rainbow of color across the visible spectrum, rather than what one would expect to see (based on analysis of emission spectrum from other stars) of spikes at specific wavelengths.  So, what is causing the full-spectrum light from the sun to appear at the surface of the earth if what I'm assuming is specific spectral lines being generated from fusion on the sun itself?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure I'm missing something in my understanding, but I'm not sure what it is.  I understand (from this question: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/18032/how-deep-is-the-a-fraunhofer-line-in-the-solar-spectrum-is-it-from-the-suns&quot;&gt;How deep is the &amp;quot;A&amp;quot; Fraunhofer line in the solar spectrum? Is it from the Sun&amp;#39;s or Earth&amp;#39;s atmosphere?&lt;/a&gt;) that the earth's atmosphere absorbs specific wavelengths of light, but why would the sun's light be consistent across its visible spectrum if the typical star light is generated at specific spectrum wavelengths?&lt;/p&gt;&#xA;" OwnerUserId="19229" LastActivityDate="2017-10-20T06:10:52.747" Title="Solar visible light spectrum" Tags="&lt;the-sun&gt;&lt;spectroscopy&gt;&lt;sun-rays&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23230" PostTypeId="2" ParentId="23229" CreationDate="2017-10-19T18:55:40.357" Score="6" Body="&lt;p&gt;When we look at spectral lines in a star's spectrum, we're actually looking for absorption lines, not emission lines. A star's spectrum usually resembles that of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Black_body&quot; rel=&quot;nofollow noreferrer&quot;&gt;black body&lt;/a&gt;, continuous and smooth. However, there are elements in the star's atmosphere that absorb some of the emitted light; these create characteristic absorption lines in the spectrum we observe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a reason that a star's spectrum closely resembles that of a black body. H&lt;sup&gt;-&lt;/sup&gt; opacity means that some of the solar photosphere is optically thick for a wide range of frequencies. This in turn means that the emission no longer comes from just a few frequencies corresponding to a few elements. This spectrum appears continuous, and therefore like a black body.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2017-10-20T00:50:41.277" LastActivityDate="2017-10-20T00:50:41.277" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23231" PostTypeId="2" ParentId="23229" CreationDate="2017-10-19T19:18:50.073" Score="6" Body="&lt;p&gt;What you're missing is that the resolution of a prism isn't high enough to resolve the relatively narrow spectral lines. What's more, the light that gets generated by fusion reactions doesn't reach the surface of the sun for a very long time, and it get's scattered and split a large number of times along the way, removing any signature of fusion from their origins. The reason we can be confident that fusion is happening in there is because we can detect &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Solar_neutrino&quot; rel=&quot;noreferrer&quot;&gt;solar neutrinos&lt;/A&gt; that exit the sun almost completely unimpeded, and have the expected spectrum from the fusion process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more, have a look at &lt;A HREF=&quot;http://www.astro.cornell.edu/academics/courses/astro201/kirchhoff.htm&quot; rel=&quot;noreferrer&quot;&gt;Kirchhoff's laws of spectroscopy&lt;/A&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to the sun's spectrum. We have a few reasons to be confident in our ability to measure absorption lines of the sun independent of the atmosphere. First, the amount of atmosphere we look at the sun through depends on the angle of the sun in the sky (a concept called &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Air_mass_(astronomy)&quot; rel=&quot;noreferrer&quot;&gt;airmass&lt;/A&gt;). So, if we observe the sun at different angles, and keep track of how the absorption changes with airmass, we can extrapolate the spectrum back to zero airmass. Second, there is at least one element that was first discovered by it's absorption lines in the solar spectrum: helium (named for the Greek god of the sun, Helios), so we can be confident that the measurements we're making correspond to physical reality. Third, we have performed measurements using satellites that are above the atmosphere. &lt;/p&gt;&#xA;" OwnerUserId="14244" LastEditorUserId="14244" LastEditDate="2017-10-19T19:27:53.537" LastActivityDate="2017-10-19T19:27:53.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23232" PostTypeId="1" CreationDate="2017-10-19T22:09:08.083" Score="2" ViewCount="582" Body="&lt;p&gt;Stars give off light and are visible at nighttime but Planets do not give off light . So why are Planets visible at nighttime ?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastActivityDate="2017-10-20T13:44:25.267" Title="Planets visible at night" Tags="&lt;planet&gt;&lt;night-sky&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23234" PostTypeId="2" ParentId="23213" CreationDate="2017-10-19T23:45:29.960" Score="3" Body="&lt;p&gt;I think I understand the spirit of this question and I'll touch some of the basics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space is, obviously, mostly, but not entirely empty.   There's space dust and gas particles bouncing around in what we would call a near perfect vacuum on Earth, but there are still some particles out there, mostly hydrogen and helium, but a disburse scattering of other particles too.   In &quot;empty&quot; space, those collisions would be pretty close to negligible over the short term.   Bombarding an organic material with hydrogen (very very slowly) and the occasional helium or alpha particle would affect it gradually.   I have no idea the time scale though.   It might still look very much the same after 1,000 years.   Stuff would break down in space very very slowly.   Much slower than on Earth where we have wind, water and bacteria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if it was struck by a coronal mass ejection from the sun, the particles are disburse enough that it wouldn't make much visible difference, though charged hyrogen, alpha particles and electrons would make some chemical changes.   It might grow somewhat carcinagenic, after being struck by a CME but the effects wouldn't be visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The major effect would be from photons.    An object in Earth's orbit, facing the sun would get quite hot, like the surface of the Moon.   This heating is almost entirely from the Sun's light.  It's temperature would depend on it's color and how much light was absorbed vs reflected.  (look up albedo and black body temperature for more info) but a chicken in orbit around the Earth would cook, somewhat slowly, it might take a few hours or maybe the better part of a day, cause one side would be hot the other side cold.   The rate of rotation would matter too.  Ideally it would rotate very very slowly, so that the side facing the sun would have time to warm up.  One rotation every 2-3 hours might work.   If it rotates too fast it might never get enough heat to cook on the inside.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also not a good method of cooking cause the UV rays would make some chemical changes and probably make it carcinogenic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In space, there's no conduction or convection (the primary methods of heat exchange on Earth).   Any heating/cooking would come from light from a nearby star and enough light will cook a chicken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In deep space, far from any stars, it would freeze.   You know how water feels colder then air when they are the same temperature - that's because of convection.   Touching the water pulls heat out of you more effectively than air.   Similarly, a chicken will cook faster in 212 degree water than 212 degree air.    In space, there's no convection into empty space (if you ignore the unpleasant vacuum effect), it wouldn't feel as cold as very cold air, but your body would still lose heat by radiation and so it would freeze.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I pointed out above, any decay by cosmic rays, once frozen would be very very slow.   Perhaps millions of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the gist of it.   I think that addresses the spirit of your question.   These processes are well understood, so there would be little reason to study this in practice by releasing organic material into space.    A related concern is that astronauts in space were hit by cosmic rays and long term exposure wouldn't be healthy for a living person.   The space shuttle is largely shielded because it's inside Earth's magnetic field, but outside of that, radiation shielding becomes a problem that needs to be addressed.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-10-19T23:45:29.960" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23235" PostTypeId="2" ParentId="23229" CreationDate="2017-10-20T06:10:52.747" Score="5" Body="&lt;p&gt;Two questions are asked here. Sean Lake addresses the first - there is nothing unusual about sunlight; it's spectrum contains many dark absorption lines due to various chemical elements. You just need a reasonably high spectral resolution (high dispersion) in order to see them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The light from the Sun comes from the photosphere. We do not see what is going on in the centre with light. Light that is emitted at the centre of the Sun is re-absorbed almost immediately. It is when photons have a reasonable chance of escaping that defines where the visible surface is. In other words, the photosphere marks the point where (moving inwards) the opacity to light increases rapidly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This opacity is not the same at all wavelengths. Where there is a spectral line, due to an electron transition in an atom, it is higher. At these wavelengths, photons can only escape if they are emitted higher in the photosphere, at cooler temperatures. Such light is less intense than at other wavelengths, hence we see an &quot;absorption line&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Outside of absorption lines the photospheric opacity must be lower, but non-zero. The main opacity here is caused by bound-free transitions involving the H$^{-}$ ion. Electrons have a continuum of free states they can occupy outside of ions and atoms, so such processes occur over a continuous range of wavelengths. The reverse of these absorption processes (i.e. free-bound emission) is what provides a continuum emission spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun's spectrum is therefore due to material at a variety of temperatures (that is why it &lt;em&gt;isn't&lt;/em&gt; a blackbody). Photons in absorption lines (they are not black, just faint) are emitted at maybe 4500K, whilst photons in the continuum come from hotter regions at perhaps a bit more than 6000K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were not for the H$^{-}$ ion, the Sun would appear slightly smaller, hotter and with deeper absorption lines. But there would still be other bound-free or free-free processes occurring in the deeper, hotter layers of the Sun that would provide sources of continuous absorption and emission.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-10-20T06:10:52.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23236" PostTypeId="2" ParentId="23219" CreationDate="2017-10-20T08:39:02.510" Score="1" Body="&lt;p&gt;The measured rotational velocities can &lt;strong&gt;reliably but not very accurately&lt;/strong&gt; constrain the enclosed mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the deviation between $GM(r)/R$ and $R\partial\Phi/\partial R$ (note that this is a partial derivative at $z=0$) is typically small. Moreover, more careful modelling (decomposing the galaxy into disc and spheroidal components) usually reduces the resulting errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, even when doing so, the relation you're referring to only holds for objects on exactly circular orbits in an exactly axially symmetric gravitational potential. However, in reality even the gas does not follow exactly circular orbits, nor are galaxies exactly axially symmetric. Common deviations are spirals and bars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one includes all these effects into the modelling, the conclusions are not changed significantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When referring to the Milky Way, the rotation velocity of a circular orbit at the Solar position (known as the &quot;Local Standard of Rest&quot; LSR) has been measured relatively accurately to be $V_0=238\,$km$\,$s$^{-1}$ and the distance to the Galactic centre to be $R_0=8.27\,$kpc. From these numbers, the simple mass estimator gives&#xA;$$&#xA;M(&amp;lt;R_0) \approx R_0 V_0^2/G \approx 10^{11} \textsf{M}_\odot&#xA;$$&#xA;for the mass interior to the Solar orbit. This is larger than what one sees in terms of stars and gas, but only by about a factor $\sim2$. The measurement of rotation velocities in the outer Milky Way is notoriously difficult and the evidence for dark matter in our host galaxy is therefore based on other arguments (such as the velocities and orbits of satellite galaxies).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For external galaxies, however, one can peruse the gas velocities measured at large distances, when the observed flatness of the rotation curve (constant $V_0$ in above formulat) demands substantial amounts of matter, roughly $M\propto R$.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2017-10-20T08:39:02.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23237" PostTypeId="2" ParentId="23232" CreationDate="2017-10-20T10:01:56.507" Score="4" Body="&lt;p&gt;Planets and moons reflect sunlight, as &lt;a href=&quot;https://astronomy.stackexchange.com/questions/23232/planets-visible-at-night#comment39889_23232&quot;&gt;mentioned in this comment by AtmosphericPrisonEscape&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'll add a bit more info to make this post more informative :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That also means that just like the moon, &lt;a href=&quot;https://en.wikipedia.org/wiki/Planetary_phase&quot; rel=&quot;nofollow noreferrer&quot;&gt;the planets have phases&lt;/a&gt;, but they're just too subtle to see with the naked eye, and for any planet further from the Sun than the Earth is, we don't see all the phases.  They're full when the planet is furthest from us (on the other side of the Sun), so it'd be hard to tell the phase just from how bright the planet appears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But through a telescope, you can easily see the phases of Venus, and you can tell it's smaller when it's full (because it's farther away):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UJ1w2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UJ1w2.jpg&quot; alt=&quot;Phases of Venus&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;em&gt;Source: &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Phases_Venus.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikimedia Commons&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1462" LastEditorUserId="1462" LastEditDate="2017-10-20T10:27:26.457" LastActivityDate="2017-10-20T10:27:26.457" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23239" PostTypeId="1" AcceptedAnswerId="23242" CreationDate="2017-10-20T10:12:55.630" Score="1" ViewCount="145" Body="&lt;p&gt;Given a state of an ideal 3 body system (i.e., without external interference) in time $t$: the velocity $v_{i,t}$, mass $m_{i,t}$ and position $x_{i,t}$ for $i\in \{1,2,3\}$, using numerical method it is possible to determine any state in time $\hat{t}$, where $\hat{t}&amp;gt;t$. But is it possible to determine a unique state in time $\bar{t}$ where $\bar{t}&amp;lt;t$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In another word, is it sufficient to know a single state of the system for any given time to deduce the states of the system for all time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equivalently, will two different states of the 3-body system results in the same state (maybe at different time) in the future?&lt;/p&gt;&#xA;" OwnerUserId="19243" LastActivityDate="2017-10-20T15:08:27.190" Title="Is the three-body system &quot;unique&quot;?" Tags="&lt;orbit&gt;&lt;stellar-dynamics&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23240" PostTypeId="1" CreationDate="2017-10-20T11:45:58.393" Score="1" ViewCount="50" Body="&lt;p&gt;Given the latitude and longitude of a heliographic coordinate, I want to place the point on a CCD image and simulate its motion for a particular duration. Though I am able to track the motion of a point on the Sun, I am unable to place the point at the exact location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, (0,90) should ideally be placed at the right edge of the Sun, but with the current code, it falls outside the image. The issue seems to be in scaling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please suggest the parameters that I need to take into account while scaling.I have included the plate scale of the CCD, the ratio of the radius of the Sun to the Earth-Sun distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%% plate scale,pixel size = 0.6 arcseconds/pixel&#xA;p = 1/0.6&#xA;&#xA;%% Radius of the Sun/Earth-Sun distance&#xA;a = 0.0046491838&#xA;&#xA;%% scaling factor, converting it to degrees&#xA;s = a*p*3600&#xA;&#xA;%% Central pixels &#xA;xC = 2048;&#xA;yC = 2048;&#xA;&#xA;%% latitude and longitude&#xA;l = 0&#xA;b = -90&#xA;&#xA;x = round((b*s*cos((l*pi)/180))+xC)&#xA;y = round((l*s)+yC)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;--&gt; &lt;code&gt;x =-463&lt;/code&gt; &amp;amp; &lt;code&gt;y = 2048&lt;/code&gt; and hence the point is not getting plotted.&lt;/p&gt;&#xA;" OwnerUserId="18678" LastEditorUserId="18678" LastEditDate="2017-10-21T05:07:47.340" LastActivityDate="2017-10-21T05:07:47.340" Title="Tracking a heliographic coordinate on CCD" Tags="&lt;the-sun&gt;&lt;photography&gt;&lt;angular-resolution&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23241" PostTypeId="2" ParentId="23232" CreationDate="2017-10-20T13:44:25.267" Score="-1" Body="&lt;p&gt;This is incorrect.  All bodies give off light according to Planck's Law.  The amount of emitted light may be small compared with reflected light, or small compared to the sensitivity of your detector.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may seem like nitpicking, but failure to incorporate emittance when analyzing spectra will lead to sadness. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-10-20T13:44:25.267" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23242" PostTypeId="2" ParentId="23239" CreationDate="2017-10-20T15:08:27.190" Score="5" Body="&lt;p&gt;What you're really asking about is less to do with Astronomy and more to do with mathematics. You're basically asking if, given a system of differential equations, will a unique solution exist for all time? For an answer, you should check out the &lt;a href=&quot;https://arxiv.org/ftp/arxiv/papers/1605/1605.05317.pdf&quot; rel=&quot;noreferrer&quot;&gt;Existence and Uniqueness theorems of differential equations&lt;/a&gt;. You'd be better to ask questions like this on the Mathematics stack exchange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, to discuss the particular, astronomical case you've asked about, the answer is yes, you can run that system both forwards and backwards if you know some initial state. Newtonian mechanics is completely deterministic in that if you know all the equations of motion involved, as well as the entire state of the system at a given time, you can figure out the state of that system at any other time, both in the past and the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To speak to your particular 3-body orbiting problem though, I'll say that the system of equations is not solvable in closed form - that is, you can't write down an analytic solution to the equations like you could for the 2-body case. As &lt;a href=&quot;https://astronomy.stackexchange.com/questions/23239/is-the-three-body-system-unique#comment39913_23239&quot;&gt;userLTK states&lt;/a&gt;, you can write down approximate solutions in the &lt;a href=&quot;http://scienceworld.wolfram.com/physics/RestrictedThree-BodyProblem.html&quot; rel=&quot;noreferrer&quot;&gt;restricted 3-body problem&lt;/a&gt;, where one mass is significantly less than the other two and orbits under specific conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a solution at any time $\hat{t}$, you need to use numerical methods. Of course numerical methods are inherently flawed. Numerical errors build up the longer you simulate due to time steps which are not infinitesimal, from errors within the numerical algorithms used, and general floating point errors. In theory, if you had a computer with infinite precision and infinite computing power, you could solve a 3-body (or n-body) system perfectly, but we live in the real world where such things are impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove though, that you can figure out the state at any time in the past or present for a three body body, I've written a basic simulation in Python 3. It can run both forwards and backwards from a given start condition and start time. Essentially it puts three nearly identical masses in contrived starting positions and velocities. Below the code are plots of results.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;from numpy.linalg import norm&#xA;from matplotlib.pyplot import *&#xA;from time import time&#xA;&#xA;# Define physical constants&#xA;G = 6.67408E-11          # Gravitational Constant, m^3 kg^-1 s^-2&#xA;&#xA;# Define body 1 parameters&#xA;m1 = 2.2E30                # Mass, kg&#xA;x1 = np.array([0,1E11])  # Position, m&#xA;v1 = np.array([-3.5E4,0])  # Velocity, m/s&#xA;&#xA;# Define body 2 parameters&#xA;m2 = 1.9E30&#xA;x2 = np.array([1E11*np.cos(210*np.pi/180),&#xA;               1E11*np.sin(210*np.pi/180)])&#xA;v2 = np.array([3E4*np.cos(300*np.pi/180),&#xA;               3E4*np.sin(300*np.pi/180)])&#xA;&#xA;# Define body 3 parameters&#xA;m3 = 2E30&#xA;x3 = np.array([1E11*np.cos(330*np.pi/180),&#xA;               1E11*np.sin(330*np.pi/180)])&#xA;v3 = np.array([3E4*np.cos(60*np.pi/180),&#xA;               3E4*np.sin(60*np.pi/180)])&#xA;&#xA;# Define simulation parameters&#xA;n = 3                    # Number of bodies, unitless&#xA;t = 0                    # Simulation time, s&#xA;dt = 1E4                 # Simulation time step, s&#xA;tEnd = 1E8               # Simulation end time, s&#xA;m = np.array((m1,m2,m3)) # All masses&#xA;x = np.vstack((x1,x2,x3))# All positions&#xA;v = np.vstack((v1,v2,v3))# All velocities&#xA;xHist = [[list(x1)],[list(x2)],[list(x3)]]&#xA;vHist = [[list(v1)],[list(v2)],[list(v3)]]&#xA;&#xA;# Simulate until end time is reached&#xA;start = time()&#xA;while True:&#xA;    # Calculate acceleration&#xA;    a = []&#xA;    for i in range(n):&#xA;        a.append(0)&#xA;        for j in range(n):&#xA;            if i == j: continue&#xA;            a[-1] += - G * m[j] / norm(x[i]-x[j])**3 * (x[i] - x[j])&#xA;&#xA;    # Update velocities&#xA;    for i,vi,ai in zip(range(n),v,a):&#xA;        vi += ai * dt&#xA;        vHist[i].append(list(vi))&#xA;&#xA;    # Update positions&#xA;    for i,xi,vi in zip(range(n),x,v):&#xA;        xi += vi * dt&#xA;        xHist[i].append(list(xi))&#xA;&#xA;    # Update time and end simulation if past tEnd&#xA;    t += dt&#xA;    if dt &amp;gt; 0 and t &amp;gt; tEnd: break&#xA;    if dt &amp;lt; 0 and t &amp;lt; tEnd: break&#xA;&#xA;end = time()&#xA;print('Simulation finished in {:.4f} seconds.'.format(end-start))&#xA;&#xA;# Convert xHist and vHist to np arrays&#xA;xHist = np.array(xHist)&#xA;vHist = np.array(vHist)&#xA;&#xA;# Plot everything up&#xA;figure()&#xA;for i,c in enumerate(['or','ob','oc']):&#xA;    # Plot starting positions&#xA;    plot(xHist[i,0,0], xHist[i,0,1], c)&#xA;for i,c in enumerate(['-r','-b','-c']):&#xA;    # Plot path of star&#xA;    plot(xHist[i,:,0], xHist[i,:,1], c)&#xA;gca().set_aspect(1)&#xA;gca().set_xticks([])&#xA;gca().set_yticks([])&#xA;show(block = False)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note, the plots show the initial positions of the stars as the points and then trace out their paths over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Path of three masses for $t&amp;lt;t_0$&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/RvlD3.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RvlD3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;In this scenario, the three masses end up in the contrived scenario. I ran the simulation backwards by setting the timestep &lt;code&gt;dt&lt;/code&gt; and the end time &lt;code&gt;tEnd&lt;/code&gt; to be negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Path of three masses for $t_0&amp;lt;t$&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Tc5Mc.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Tc5Mc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;From here, the simulation is run forwards with positive &lt;code&gt;dt&lt;/code&gt; and &lt;code&gt;tEnd&lt;/code&gt;, starting from the same contrived scenario as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note how chaotic and unstable this system is. The entire system only remains a 3-body system for less than 6 years. Before that the three masses are separate and doing their own thing. They &quot;coincidentally&quot; meet (because I set it up so they should), orbit around each other for a little less than 6 years, and one gets ejected, resulting in the other two continuing to orbit one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Contrived Scenario with $m_1=m_2=m_3$, $|v_1|=|v_2|=|v_3|$, and all positions $120^o$ from each other&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/fYl6M.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fYl6M.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Just for fun, a &quot;stable&quot; 3-body problem with all stars of equal mass and orbits. This is really an unstable equilibrium orbit and any perturbations will screw it up and you'll see what you saw in the above two graphs. In fact, if you run this long enough, the numerical instabilities of my code will result in the orbits breaking down. This numerical instability, as I said above, is inherent in &lt;em&gt;any&lt;/em&gt; numerical solution and cannot be overcome, only minimized. I find that, using the numerical method I did, my system is resistant to numerical instabilities for about 7 years. If I want to run this any longer (and continue to be accurate), I need more robust numerical methods.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-10-20T15:08:27.190" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23243" PostTypeId="2" ParentId="23204" CreationDate="2017-10-20T15:08:59.293" Score="6" Body="&lt;p&gt;I think it is generally expected that the merger of two neutron stars will lead to the formation of a black hole. What is more uncertain is whether that black hole forms straight away, or an intermediate state of a hyper(/supra)massive neutron star forms (see, e.g. Sec IIC of &lt;a href=&quot;https://arxiv.org/abs/1307.5888&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hotokezaka &lt;em&gt;et al&lt;/em&gt;, PRD, 044026, 2013&lt;/a&gt; for a definition of hypermassive and supermassive neutron stars), which then collapses to a black hole after a short time. A few papers looking at the stability of such a hypermassive neutron stars formed from a binary merger are - &lt;a href=&quot;https://arxiv.org/abs/astro-ph/9910565&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0010493&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;, &lt;a href=&quot;https://arxiv.org/abs/1307.5888&quot; rel=&quot;nofollow noreferrer&quot;&gt;3&lt;/a&gt; and &lt;a href=&quot;https://arxiv.org/abs/1403.6327&quot; rel=&quot;nofollow noreferrer&quot;&gt;4&lt;/a&gt; (check out the references and citations to them for more info). In the first three reference I think the collapse to a black hole is expected within a second (see Table 2 in &lt;a href=&quot;https://arxiv.org/abs/1307.5888&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hotokezaka &lt;em&gt;et al&lt;/em&gt;, PRD, 044026, 2013&lt;/a&gt; for expected lifetimes before collapse times for various neutron star equations of state and masses), whilst in the final one a range of collapse times from fractions of a second up to several tens of thousands of seconds are found.&lt;/p&gt;&#xA;&lt;p&gt;A search for a post-merger remnant from the binary neutron star merger GW170817 is now available &lt;a href=&quot;https://arxiv.org/abs/1710.09320&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. In the introduction of this paper it describes in more detail some of the post-merger remnant possibilities that I mentioned above. The searches don't find any evidence for a signal, but given their sensitivities this isn't surprising (they'd require more than the total mass of the system to have been converted to gravitational waves to see anything!)&lt;/p&gt;&#xA;" OwnerUserId="16795" LastEditorUserId="35106" LastEditDate="2020-10-09T15:30:21.640" LastActivityDate="2020-10-09T15:30:21.640" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23245" PostTypeId="1" CreationDate="2017-10-20T20:07:19.073" Score="5" ViewCount="483" Body="&lt;p&gt;Does such a heavenly body exist called a black dwarf? It would be a Brown dwarf that has cooled to the point it no longer gives off heat or light.&lt;/p&gt;&#xA;" OwnerUserId="18118" LastActivityDate="2017-10-22T07:22:32.253" Title="Is there such a thing as a black dwarf" Tags="&lt;brown-dwarf&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23246" PostTypeId="2" ParentId="23245" CreationDate="2017-10-20T20:53:45.043" Score="10" Body="&lt;p&gt;A black dwarf is a &lt;em&gt;white&lt;/em&gt; dwarf that has cooled to the point that it is no longer emitting light. However it takes a long time for a white dwarf to cool down. The exact timescale is rather uncertain, but at least hundreds of billions of years, perhaps hundreds of trillions of years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The upshot is that that no black dwarfs exist in the universe today. The Universe is only about 14 billion years old, so there hasn't been time for any black dwarfs to form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper &lt;a href=&quot;http://xxx.lanl.gov/pdf/astro-ph/9701131v1&quot; rel=&quot;noreferrer&quot;&gt;A Dying Universe: The Long Term Fate and Evolution of Astrophysical Objects&lt;/a&gt; discusses brown dwarfs, black dwarfs, and other objects that may exist in an aging universe.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-10-20T20:53:45.043" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23247" PostTypeId="1" AcceptedAnswerId="23251" CreationDate="2017-10-21T01:57:38.483" Score="1" ViewCount="127" Body="&lt;p&gt;Cataclysms such as novas, kilonovas, supernova, magnetars, quasars, gamma ray bursts are rarely visible from Earth.   Under what conditions (such as distance, day vs. night, naked eye vs. level of optics) would such phenomena be visible from the Earth's surface, and what would they look like in terms of shape, initial duration, brightness, and so forth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, I have seen questions like &lt;a href=&quot;https://physics.stackexchange.com/questions/61872/how-long-does-a-supernova-last&quot;&gt;how long does a supernova last&lt;/a&gt; and I have read that they can be visible for some time afterwards as a nebula forms, but this doesn't tell me what the initial burst would look like.  For instance, would it be possible to see a supernova in a neighboring galaxy such as Andromeda?  If so, how long would it be visible with the naked eye?  Would it look like a flash, last a few seconds, several?  I assume with a supernova not in our own galaxy there would be no subsequent light from the expanding shell since it would be too far away, whereas in our own galaxy if it were close enough we would subsequently see it, but in that case would there be a short burst of light, following by darkness and then the re-emergence of light, or would it be a continuously visible phenomena?&lt;/p&gt;&#xA;" OwnerUserId="2341" LastEditorUserId="2341" LastEditDate="2017-10-21T04:07:32.170" LastActivityDate="2017-10-21T10:53:59.763" Title="Identifying cataclysmic stellar phenomena" Tags="&lt;amateur-observing&gt;&lt;supernova&gt;&lt;kilonova&gt;&lt;nova&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23248" PostTypeId="1" AcceptedAnswerId="23252" CreationDate="2017-10-21T03:59:59.700" Score="3" ViewCount="359" Body="&lt;p&gt;Perusing the topics here. I am convinced that this is the right place to get some answers. Although it does appear that my question should be relegated to the &quot;childrens table&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curious to understand more about the topic in archeoastronomy that is used to date ancient alignments utilizing calculations based on precession (which I very basically understand as an imaginary circle drawn in space by the earth's axis-line as the earth wobbles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example a &quot;temple&quot; in Peru has been dated as being aligned with the summer SOLSTICE(es)[edited] as of 6000 years ago by unnamed experts in the field. Other unnamed experts date it at 28000 years old - to which the first set of &quot;experts&quot; grudgingly concede is &quot;remotely&quot; possible - with an absolute qualification that there was no one here to build it that far back.(Statements of absolute certainty like this and, without proof make me itchy)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First part of question is: Why the huge discrepancy in the dating? Isn't the concept of precession a proven, quantifiable fact? (OK, that was two questions.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second part of question is: How do the first set of experts find themselves entrenched in their dating if they must also grudgingly agree that the second dating is remotely possible? (Both groups agree that neither the Inca nor any known predecessors actually built it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything which can help me to come to grips with this dating rationale would be great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not versed in astronomy whatsoever. I can barely find Sirius on a clear night. And, math is also a foreign concept as well. &#xA;What I DO understand is Ancient Civilizations. &#xA;So, I would like to understand this topic more so that I can discuss it with others in the field - without sounding like a complete idiot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit to clarify before more protests stack up] I acknowledge that I did not recall and therefore omitted the fact that non-solar stellar alignments AT THE TIME of summer solstice are the basis for this alignment-dating. At least that is the theory I am discussing. (There are others) One window frames the Sun at sunrise while another frames some star. The third window... I cannot locate a theory of what it was supposed to be for. I just wanted to understand the basis for DRIFT over time. And, the accepted answer does that very nicely. Thank you everyone&lt;/p&gt;&#xA;" OwnerUserId="19249" LastEditorUserId="37191" LastEditDate="2021-02-10T09:59:18.207" LastActivityDate="2021-02-10T09:59:18.207" Title="Relationship between precession and dating ancient alignments" Tags="&lt;precession&gt;&lt;alignment&gt;&lt;archaeoastronomy&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="23249" PostTypeId="2" ParentId="23208" CreationDate="2017-10-21T04:00:30.060" Score="2" Body="&lt;p&gt;It's not going to happen, but if it did, what it would mean is that the Earth would rotate very slowly, at the same rate the Moon orbits the Earth, currently 27.3 days, but for practical purposes, &lt;a href=&quot;https://community.dur.ac.uk/john.lucey/users/lunar_sid_syn.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;29.5 days relative to the sun&lt;/a&gt;.   Presumably even longer, because as the Earth slows, the Moon moves further away, so figure 30-something days for 1 synodic orbit.   I'll say 30 days, but it would probably be a bit longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that a complete day/night cycle would be 720 hours.  Days might be scorching hot and nights would get very cold and as a result, you'd probably see stronger winds, but the Coriolis effect would be measurably decreased, so prevailing winds might be less, or perhaps similar to Venus, where the upper atmosphere rotates in one direction very fast.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might see fewer hurricanes.  Snow would probably all melt during the day, so there would be less chance of permanent snow buildup outside of very cold regions that get very little solar energy during winter.  Plants would also need to adapt to 360 hours of sunlight and 360 hours of night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There would be no more tides and it's possible that the Moon's orbit and the slowing of the Earth's relatively rapid rotation assists the magnetic field so we might lose the Magnetic field too - which would be pretty harmful to life in the long run.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon would also stay in essentially the same place in the sky all the time.  That's a product of tidal locking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the gist of it anyway, but feel free to comment if I skipped anything.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-11-12T08:16:03.330" LastActivityDate="2017-11-12T08:16:03.330" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23251" PostTypeId="2" ParentId="23247" CreationDate="2017-10-21T06:30:02.977" Score="4" Body="&lt;p&gt;Any of these phenomena would be visible to the naked eye if they occured in the Milky Way, or a neighbouring galaxy (such as the Magellanic clouds) and was not obscured by dust. At the distance of stars, all of these would appear star-like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supernovae have been seen, most recently in 1987, when there was a supernova in the Large Magellanic cloud. It had a peak magnitude of 2.9, (about as bright as gamma ursa minor, the star at the bottom of the Little bear constellation, furthest from the polestar). It was visible, but not spectacularly bright. Supernovae that have occurred in the Milky way have been very bright, outshining all the other stars. A supernova in 1006 may have been brighter than Venus. If it is close enough a supernova could be seen during the day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these are highly energetic, and if you were close enough to see any detail (such as a nebula) with the naked eye, you would be toast. The exposive events (Supernovae, Kilonovae, Gamma ray bursts) would fade over time. Magnetars and Quasars are stable and would continue to shine. Remnant nebulae from Supernovae are visible in telescopes, the most well known is the Crab Nebula, M1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A supernova in Andromeda would likely be naked eye visible, but not impressively bright, though it depends on the acutal power of the explosion, and other factors, like dust in Andromeda.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supernovae are visible for months. Kilonovae fade faster. Gamma ray bursts may only last for a few seconds (though may be associated with a supernova that lasts longer). The visible element of a GRB might be visible, there was one that might have reached naked eye visibility, though there is no evidence that anyone actually saw it (it would only have been marginally visible in ideal conditions). Quasars are stable objects that don't fade over time, they are active supermassive black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of other, more local things that can produce a &quot;star&quot; that brightens then fades: Planes have headlights, and as they turn the appear like a star that brightens then fades over a few seconds. If they are distant enough, the other lights might not be visible.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-10-21T10:53:59.763" LastActivityDate="2017-10-21T10:53:59.763" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23252" PostTypeId="2" ParentId="23248" CreationDate="2017-10-21T07:25:33.457" Score="5" Body="&lt;p&gt;The answer is fairly simple.   It takes about 22,000 years for the Earth to perform 1 spin or move 360 degrees along it's axial precession.   The 22,000 year cycle means that the temple was lined up at 6,000 years ago and at 28,000 years ago, as well as 50,000 and 72,000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, if we build a statue that points directly at the North Star, In a couple thousand years it will visibly no longer be pointing at the North star, but in 22,000 years, after a full circle, it will be pointing at the North star again.   More specifically the North Star will apparently move in a circle that takes 22,000 years to complete, moving away from and back towards the celestial north pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any skepticism on the 28,000 year date wouldn't have anything to do with alignment as the alignment works just as well with either date, and it works for any period a full circle in the past, like 50,000 years ago.  The skepticism is likely related to doubt that it's that old.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar to the hour hand on a clock that takes 12 hours for the hour hand to return to where it started, the Earth takes about 22,000 years to return to the direction the North Pole is pointing.   That means, no time between 6,000 and 28,000 years ago will line up.  Those are the two most recent alignment periods.   It moves in a circle in 3D, more like the picture below than like a clock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oIioN.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oIioN.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an astronomical note, I want to point out that 22,000 is somewhat inaccurate.   The &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth's axial precession&lt;/a&gt; in and of itself is about 25,771 years, but that's only part of the movement.  There's also a larger, slower &lt;a href=&quot;https://en.wikipedia.org/wiki/Milankovitch_cycles#Apsidal_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;apsidal precession&lt;/a&gt; which adds to or subtracts from the 25,771 year wobble.   These together affect the alignment of the Earth to the mostly fixed stars, and that period varies, but it was about 22,000 years between 6,000 and 28,000 years ago.   It can take as little as 20,800 years or as many as 29,000.   Average is about 23,000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many Native American structures lined up quite well to astronomical alignments, so dating ancient structures by measuring when it lined up a reasonable approach.    As a related, somewhat similar example, the Mayan 2012 prophecy is also based on an astronomical alignment and they were, as I recall, a year or two late, but it was still a very good estimate given when they made it.   Understanding these alignments requires seeing things in 3 dimensions and that can be a little confusing to try to picture.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-10-22T12:49:09.197" LastActivityDate="2017-10-22T12:49:09.197" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23256" PostTypeId="2" ParentId="23245" CreationDate="2017-10-21T17:11:04.460" Score="3" Body="&lt;p&gt;A black dwarf might be defined as a cooled, compact object with a temperature and surface area that are small enough to make it invisible in the optical part of the spectrum, with most radiation emerging in the infrared. However, because there is always &lt;em&gt;some&lt;/em&gt; radiation in the short wavelength Wien tail of a spectrum this definition would be distance-dependent, so instead you could define a temperature below which you could argue that so little light emerges at visible wavelengths that it would be dark. Again, this is somewhat distance-dependent, but also depends on the size of the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, at astronomical distances, cool compact objects like old white dwarfs and neutron stars, or low-mass and older brown dwarfs are invisible at optical wavelengths when their temperatures fall significantly below 3000K. At 3000K, the V-K colour of such objects is $\sim 6$, meaning they are 25 times as bright in the near infrared. This increases to $\sim 100$ at a temperature of 2200K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of brown dwarfs (particularly the T-dwarfs) that can only be seen in the infrared. The coldest known white dwarf is probably older than 10 billion years, has a temperature below 3000K, and has not been detected at optical (or IR) wavelengths (&lt;a href=&quot;https://arxiv.org/abs/1406.0488&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kaplan et al. 2014&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are likely about a billion or so old, cold neutron stars in the Milky Way that cannot be seen via electromagnetic radiation at all unless they interact with something else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Black_dwarf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia definition&lt;/a&gt; of an object so cold it doesn't emit light is simply unworkable. All objects above absolute zero will emit electromagnetic radiation (light) and it is hard to see how an object could cool to below the cosmic radiation background.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-10-22T07:22:32.253" LastActivityDate="2017-10-22T07:22:32.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23257" PostTypeId="1" AcceptedAnswerId="23259" CreationDate="2017-10-21T19:02:06.263" Score="4" ViewCount="5350" Body="&lt;p&gt;I am currently reading &lt;em&gt;A Brief History of Time&lt;/em&gt; and while the book itself does not contain much mathematics in it, I want to learn more about the concepts Prof. Hawkings talks about more in-depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I am aware that I will never get the expansive knowledge that a university tuition in the field would give me, but I have already branched off into something else that does not allow me to pursue that field even if I really wanted to so I want to make do. (I also understand that any knowledge I do get will be basically useless to me, but I love space that much that, in spirit, the knowledge would feel useful to me.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, before I read books more technical, should I brush up on my math? I am aware that a more applicable subject like physics has more correlation with what I am interested in, but I don't want to read about physics if I have little to no knowledge what the hell each equation means. (Also, how I will learn is through online courses/websites/books. I know it is not the ideal way to go about it, but again, I will do what I can.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you believe I should somehow go back a few years and try my hardest at physics and follow the most tried path, I am all ears. It is just that I am now going to a technical college for IT and if I want to leave, I would have to spend a lot more time in education than most - meaning (as I live in the UK) will probably have to pay extra when I pass a certain age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this question is not exactly linked to astrophysics, but I do not know where else to ask it. If I had read &lt;em&gt;A Brief History of Time&lt;/em&gt; earlier in my life, I would not be in this sticky situation. Thanks in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ADDITION: I have always had a love for space and I was interested in physics, but my teacher was uninspiring and I was going through a tough period in my life. I am willing to work towards it now, even if my teacher is the worse thing to ever exist.)&lt;/p&gt;&#xA;" OwnerUserId="19266" LastEditorUserId="19266" LastEditDate="2017-10-21T19:18:28.550" LastActivityDate="2018-06-25T20:52:53.217" Title="If I want to learn about astrophysics, should I study up on mathematics?" Tags="&lt;astrophysics&gt;" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23258" PostTypeId="2" ParentId="23257" CreationDate="2017-10-21T23:16:27.967" Score="3" Body="&lt;p&gt;It would help quite a bit but not absolutely necessary. Don't get me wrong, astrophysics, like most types of physics, is highly mathematical. If you wanted to study it in depth you would need much math.  But astrophysics is a broad field and much can be learned, at least at a beginners level, with only a little calculus. This is my opinion of course. As StephenG said, knowing basic physics would be more helpful. I agree with that statement. &lt;/p&gt;&#xA;" OwnerUserId="16926" LastActivityDate="2017-10-21T23:16:27.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23259" PostTypeId="2" ParentId="23257" CreationDate="2017-10-22T01:42:53.920" Score="4" Body="&lt;p&gt;Yes, absolutely. At bare minimum, you'll need to study some geometry to understand how distances are calculated and how the shapes of orbits are related. Next, you'll want to study some algebra, so you can understand the likes of the inverse square law for how the intensity of light drops off as you move away from an object emitting light equally in all directions. Algebra also helps in understanding the magnitude system astronomers use to describe how bright objects are in the night sky. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After geometry and algebra comes trigonometry, especially spherical trigonometry, because of its use in dealing with how we describe the location of objects in the sky, and how to relate positions in one coordinate system to positions in another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if you want the really deep understanding, you'll want to go for integral and differential calculus. As my high school physics teacher used to say, the difference between an ordinary formula and one that uses calculus is like the difference between prose and poetry - the calculus encodes deeper meaning and is applicable in more situations than any single formula that can be derived from it.&lt;/p&gt;&#xA;" OwnerUserId="14244" LastActivityDate="2017-10-22T01:42:53.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23261" PostTypeId="1" AcceptedAnswerId="23274" CreationDate="2017-10-22T22:40:47.950" Score="3" ViewCount="242" Body="&lt;p&gt;I used to think that most terrestrial planet atmospheres were secondary (from out gassing). However, after reading this &lt;a href=&quot;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20140017487.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20140017487.pdf&lt;/a&gt; and noticing that the ratio of CO2 to N2 is far higher on Venus than that just expected from out gassing, I have done a 180.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking that Earth has a primarily secondary atmosphere because of the Moon creating event and smaller planets like Mars just aren't massive enough to retain original nitrogen, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this cast a shadow on the idea of early extreme stellar activity ripping off atmospheres of close in planets?&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2017-10-24T02:42:43.227" Title="Is it thought that the majority of Earth size or larger terrrestrial planets have retained most of their original atmospheres?" Tags="&lt;planet&gt;&lt;planetary-atmosphere&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23264" PostTypeId="5" CreationDate="2017-10-23T08:43:36.377" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2017-10-23T08:43:36.377" LastActivityDate="2017-10-23T08:43:36.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23265" PostTypeId="4" CreationDate="2017-10-23T08:43:36.377" Score="0" Body="Tag used for topics related to drizzle-integration (a super-sampling algorithm for stacking images)" OwnerUserId="10271" LastEditorUserId="10271" LastEditDate="2017-10-24T16:03:29.340" LastActivityDate="2017-10-24T16:03:29.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23266" PostTypeId="1" AcceptedAnswerId="23268" CreationDate="2017-10-23T08:48:34.743" Score="4" ViewCount="394" Body="&lt;p&gt;I know that light shifts from high frequency(like gama rays) to low frequency (like radio waves) when traveling throw a gravitational field, this is called red shifting.&#xA;My question is: Is there a limit to this, what happens to the light wave when it can no longer stretch, does it get destroyed ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, and sorry for the dumb question, just curios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;From my knowledge and the internet :) &lt;a href=&quot;https://en.wikipedia.org/wiki/Electromagnetic_spectrum&quot; rel=&quot;nofollow noreferrer&quot;&gt;Electromagnetic spectrum&lt;/a&gt; it states that light is an electromagnetic wavelength and the frequency which the wavelength travels up and down and the distance it's travel it's called spectrum and in that spectrum there is a frequency that is visible to us, but the spectrum is very large, my question is just for curiosity, I want to know if we know what happens when the light gets beyond the spectrum, can it? And I know that the expanding of space-time makes the light stretch from a high frequency spectrum to a low one, but what happens when you go to the limit of the spectrum ? can the wavelength be stretch to the point of &quot;breaking&quot; ? &lt;/p&gt;&#xA;" OwnerUserId="19291" LastEditorUserId="19291" LastEditDate="2017-10-23T14:19:44.650" LastActivityDate="2017-10-24T20:49:48.453" Title="Can space-time expansion destroy light?" Tags="&lt;redshift&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23268" PostTypeId="2" ParentId="23266" CreationDate="2017-10-23T11:25:51.883" Score="6" Body="&lt;p&gt;The stretching has no limit. Light can be redshifted to infrared, then to microwaves, to radio waves. There is no limit. This is because there is no medium that carries the waves to be stretched. There is no limit of the spectrum. It runs from arbitrarily long wavelengths, to arbitrarily short. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of thinking of waves being stretched you could use a model suggested by Feymann. Each photon is like a small clock with one hand. THe colour of the light is the time it takes for the hand to go round. The photons in visible light turn a thousand trillion times a second. Radio waves only a million times a second. There is no limit to how slowly the hand can turn. Red-shifting is the slowing of the hand. There is no need to think of something stretching, and nothing to &quot;break&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-10-23T13:14:36.210" LastActivityDate="2017-10-23T13:14:36.210" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23270" PostTypeId="1" CreationDate="2017-10-23T19:55:48.393" Score="2" ViewCount="53" Body="&lt;p&gt;I have an automated telescope that I have written some software for. I am currently using TPoint to compensate for some of the irregularities of my system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However with TPoint you apply an offset (based off of the given coefficients and the telescopes pointing position) to the target and that is where you point to acquire the target in your viewfinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know how to invert this process, such that given your known telescope point position, you evaluate your &quot;true&quot; pointing position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 6 parameters: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;IA   = Azimuth axis index error (encoder offset).&lt;/li&gt;&#xA;&lt;li&gt;IE   = Elevation axis index error (encoder offset).&lt;/li&gt;&#xA;&lt;li&gt;NPAE = Az/El axis non-perpendicularity.&lt;/li&gt;&#xA;&lt;li&gt;CA   = Elevation axis / Pointing axis non-perpendicularity.&lt;/li&gt;&#xA;&lt;li&gt;AW   = E-W misalignment of Azimuth axis.&lt;/li&gt;&#xA;&lt;li&gt;AN   = N-S misalignment of Azimuth axis.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And the offset equations come out to be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;azimuth_Correction = (-IA) + (-NPAE * Tan(boundedElevation)) + (-CA /&#xA;  Cos(boundedElevation)) + (-AN * Sin(azimuth) * Tan(boundedElevation))&#xA;  + (-AW * Cos(azimuth) * Tan(boundedElevation));&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;elevation_Correction = (IE) + (-AN * Cos(azimuth)) + (AW *&#xA;  Sin(azimuth)));&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where: boundedElevation is the &quot;capped&quot; value so that the elevation value gives the correct value for Tangent. This is:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;MAX_TANGENT_Q1 = Atan(10);&lt;/li&gt;&#xA;&lt;li&gt;MAX_TANGENT_Q2 = (Atan(-10) + PI); &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The TPoint can be found documentation &lt;a href=&quot;https://plone.unige.ch/EULER/t4-euler-telescope/rapports-de-mission/pointage/manuelTpoint.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;HERE&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16825" LastEditorUserId="16825" LastEditDate="2017-10-24T12:46:58.523" LastActivityDate="2017-10-24T12:46:58.523" Title="Any method to apply inverse TPoint model?" Tags="&lt;telescope&gt;&lt;observational-astronomy&gt;&lt;amateur-observing&gt;&lt;rotation&gt;&lt;coordinate&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23271" PostTypeId="1" AcceptedAnswerId="23281" CreationDate="2017-10-23T23:29:50.607" Score="8" ViewCount="654" Body="&lt;p&gt;This &lt;a href=&quot;https://astronomy.stackexchange.com/a/23237/7982&quot;&gt;great answer&lt;/a&gt; by @gerrit discussing planetary phases seen in visible light contains the image I've included below. According to the &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Phases_Venus.jpg&quot; rel=&quot;noreferrer&quot;&gt;Wikimedia Commons&lt;/a&gt; link these are &lt;a href=&quot;http://www.eso.org/public/about-eso/esoglance/&quot; rel=&quot;noreferrer&quot;&gt;ESO&lt;/a&gt; images from the Venus transit 2004 &lt;a href=&quot;http://www.eso.org/public/outreach/eduoff/vt-2004/&quot; rel=&quot;noreferrer&quot;&gt;VT-2004&lt;/a&gt; observing campaign. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ESO, or &lt;a href=&quot;https://en.wikipedia.org/wiki/European_Southern_Observatory&quot; rel=&quot;noreferrer&quot;&gt;European Southern Observatory&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Organisation Européenne pour des Recherches Astronomiques dans l'Hémisphere Austral) is a 16-nation intergovernmental research organization for &lt;strong&gt;ground-based astronomy&lt;/strong&gt;. Created in 1962, ESO has provided astronomers with state-of-the-art research facilities and access to the southern sky. (emphasis added)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These images are very high quality, and the last one, dated &lt;code&gt;8/6/04&lt;/code&gt; (08-June-2004) appears to be taken at/near inferior conjunction, when Venus is almost directly between the observer and the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; I'd like to understand with what instrument this image was taken, and how the technical challenges imaging Venus with such a small angular separation between it and the Sun &lt;em&gt;were overcome&lt;/em&gt;! I did a quick check and during the UTC day of June 8th, 2004, Venus moved from about 0.6 to 1.1 degrees from the Sun as viewed by Earth. How was Venus imaged so nicely? Was this really imaged from the ground through Earth's daytime atmosphere, or from space? In either case, how was a high quality telescope pointed so close to the Sun and shielded from stray light and protected from damage?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; Phases of Venus, from &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Phases_Venus.jpg&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UJ1w2.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UJ1w2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-10-24T00:06:33.157" LastActivityDate="2017-10-24T13:46:10.930" Title="How (the heck) was this photo of Venus at inferior conjunction (between us and the Sun) taken?" Tags="&lt;observational-astronomy&gt;&lt;photography&gt;&lt;venus&gt;&lt;planetary-transits&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23272" PostTypeId="1" CreationDate="2017-10-24T01:36:24.223" Score="3" ViewCount="57" Body="&lt;p&gt;I'd like to know if there's a repository out there of pictures taken everyday of the night sky. Not looking forward to anything fancy. Maybe a group of people out there have set up something out of their home telescopes or something.&lt;/p&gt;&#xA;" OwnerUserId="19301" LastActivityDate="2017-10-24T01:36:24.223" Title="Is there a Free repository of pictures taken everyday of the night sky?" Tags="&lt;telescope&gt;&lt;night-sky&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23274" PostTypeId="2" ParentId="23261" CreationDate="2017-10-24T02:42:43.227" Score="3" Body="&lt;p&gt;Too long for a comment, so . . . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important to remember that the field of primordial atmospheres has a good bit of uncertainty to it.  I've tried to read up on it as a layman and there's a good deal of contradicting theories and not much certainty.  Your first article addresses this uncertainty several times.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The reason I mention the ratio of CO2 to N2 is because I read&#xA;  &lt;a href=&quot;http://abyss.uoregon.edu/~js/ast121/lectures/lec14.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, for example, that out gassing produces the molecules in&#xA;  about a 4.6:1 ratio.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could you pull the quote to that 4.6:1 ratio.  I didn't see it in that article and that sounds much too accurate for an estimate in this field.   4.6 to 1 suggests about a 2% range of uncertainty.   I find that hard to believe.   Certainly a ratio with that kind of accuracy should cite a respected research study.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And noticing that the ratio of CO2 to N2 is far higher on Venus than&#xA;  that just expected from out gassing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What ratio is expected from outgassing?  Your &lt;a href=&quot;http://abyss.uoregon.edu/~js/ast121/lectures/lec14.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;2nd article&lt;/a&gt; mentions a neat and tidy percentages:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The composition of outgassing is similar for Venus, Earth and Mars and&#xA;  is composed of 58% H2O, 23% CO2, 13% SO2, 5% N2 and traces of noble&#xA;  gases (Ne, Ar, Kr). The latter evolution of this outgassing is driven&#xA;  primarily by the surface temperature and chemistry of the planet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I call BS, cause one, this doesn't site a source and some basic google-research on Earth volcanic activity suggests the numbers are far from that neat and tidy - see charts below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2OHde.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2OHde.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;http://earth.usc.edu/~stott/Catalina/intro.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TzfiM.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TzfiM.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.geo.mtu.edu/volcanoes/vc_web/background/S_chem.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 2nd source, above, doesn't even mention Nitrogen (N2), &lt;a href=&quot;https://en.wikipedia.org/wiki/Volcanic_gas#Composition&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia does&lt;/a&gt;, though they don't give percentages beyond 60% for water, 10-40% for CO2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth's volcanic gas today is probably not a good measuring stick because of 3 plus billion years of life (some nitrogen fixing, lots of long chain carbon molecules) and some 4 billion years of plate tectonics.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And to my knowledge, a sample of volcanic gas from Venus has never been studied.  In fact, there are no known active volcanoes on Venus at all.   &lt;a href=&quot;https://en.wikipedia.org/wiki/Volcanology_of_Venus&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.   Outgassing can perhaps be estimated by atmospheric content, but I'm skeptical on the accuracy of those numbers they gave.  I think your &lt;a href=&quot;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20140017487.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;first article&lt;/a&gt; from NASA far more reliable.   They quote a ton of sources and several times in the article, indicate legitimate uncertainty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I took away from the NASA article:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, when the solar system was very young, the t-tauri phase of the sun blew most of the atmosphere off the 4 inner planets (or 3 Earth-like if you want to ignore Mercury, which the first article does).  This is backed up in part by the lack of Xe-129 which should form as a gas based on the 15 million year half life of I-129.   If the atmospheres weren't blown off, we should see a certain percentage Xe-129 in the atmospheres of Earth/Mars/Venus.   The percentage is much too low, suggesting the blow-off did happen over the first, perhaps 100 million years, continuing well after the end of the T-tauri phase, when the sun was still very active with solar flares.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The giant impact that is thought to have created the Moon (about 30 million years after planet formation) may have blown off a chunk of Earth's atmosphere as well, but the Solar wind probably did too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the primordial atmosphere came from? Whether out gassing or comets (or a combination of both), I don't think that's a resolved question.  What seems true, based on the first article, is that Venus's water evaporated and it lost it's water by a combination of photo-chemistry and it's higher temperature and perhaps some jeans escape.   No water on it's surface may have slowed down any hope it had for plate tectonics and that reduced the rate that it outgassed.    (a general summary of the NASA article).  To get a better idea of Venus history, they'd like some rocky samples, but they don't have any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth maintained it's oceans, perhaps because it had more water to begin with, or a lower temperature or combination of the two.  The lunar impact may have sped up Earth's outgassing as it liquified the entire crust for thousands, perhaps as long as a couple million years.  It's unclear if Venus ever had it's crust liquified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But after Earth cooled after the giant impact, it had enough water for oceans and oceans made it possible to form carbonate rock and remove much of the CO2 from it's atmosphere.  Life assisted with the removal of CO2 and fixed some nitrogen too.  Ammonia also dissolves in water, so much of the nitrogen that came to Earth by comets could have been stored in the oceans, while ammonia that came to Venus by comet would have been vaporized as gas and light enough where some of it would inevitably be lost to jeans escape and subject to photo-chemical dissociation, and, a weaker magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth and Venus have similarish amounts of CO2, it's just that Earth has most of it's CO2 stored in carbonate rock.    Earth and Venus may have similar amounts of Nitrogen too, though Venus atmosphere has about 4 times as much as Earth's atmosphere, currently, that doesn't take into account Nitrogen in the oceans and fixed or bonded within the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think Venus &quot;extra&quot; CO2 is a mystery because it didn't have oceans long enough to pull the CO2 out of it's atmosphere.  We also don't know enough about the materials that formed Venus might have been different than those that formed Earth, and we don't know that much about the process that followed, so I don't think it's fair way to say it has too high a CO2 to N2 ratio.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's entirely possible that the giant impact did blow away a portion of Earth's atmosphere, leaving it somewhat atmosphere depleted compared to Venus, but your NASA article points out that there's some uncertainty on how much atmosphere the Earth had after the giant impact leading to uncertainty on how quickly the Earth cooled.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It could, I think, work in the other direction, where the enormous heat of impact and liquification of the crust may have enabled more atmosphere to outgas, giving Earth a thicker atmosphere early on but post impact, where, the Venus equivalent of that gas might still be trapped inside Venus.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a lot about this subject that remains unresolved.   This is just my takeaway from the articles.   Take this answer with a grain of salt.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-10-24T02:42:43.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23275" PostTypeId="1" CreationDate="2017-10-24T03:35:41.767" Score="22" ViewCount="2795" Body="&lt;p&gt;Self-explanatory, but I would like a comparison as well. Is the light enough to see by? How disrupted will the pitch darkness on the spot opposite of the 'solar pole' be?&lt;/p&gt;&#xA;" OwnerUserId="19304" LastActivityDate="2017-10-24T06:51:23.477" Title="What's the brightness of Alpha Centauri from Proxima Centauri?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;magnitude&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23277" PostTypeId="2" ParentId="23275" CreationDate="2017-10-24T06:51:23.477" Score="30" Body="&lt;p&gt;Not close to being able to read by.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proxima Centauri is about 13,000 AU from the two binary Centari stars.   Together they have about &lt;a href=&quot;https://en.wikipedia.org/wiki/Alpha_Centauri#Nature_and_components&quot; rel=&quot;noreferrer&quot;&gt;twice the luminosity&lt;/a&gt; of the sun but at 13,000 AU, that's roughly 2/169,000,000 the visible light that the Earth gets from the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The brightness variation of the full moon to the Sun is about 1 to 440,000, so, some rough math, the two stars would shine about 1/190th as bright as the full moon.   That would make those 2 stars by far the brightest stars in the Proxima Centauri sky, but far short of reading light.   Together, the two stars would be about 10 times as bright as Venus at Venus' peak.  (Venus is about 1/2000th as bright as the moon, &lt;a href=&quot;http://www.astronomynotes.com/starprop/s4.htm&quot; rel=&quot;noreferrer&quot;&gt;-4.4 apparent magnitude to -12.6&lt;/a&gt; for the moon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That would, I think, be bright enough to be visible during the day under an earth like sky a lot of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two stars would be quite close to each other too, maximum visible distance between them would be roughly 1/4 the diameter of the full moon, where they would still be visible separate stars when far apart, but they would pass close to each other too, perhaps appearing to touch to the naked eye.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two stars complete a full orbit around each other in 79.9 years, so the variation would be noticeable over a human lifetime.     &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-10-24T06:51:23.477" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23278" PostTypeId="1" AcceptedAnswerId="23289" CreationDate="2017-10-24T07:47:11.730" Score="3" ViewCount="203" Body="&lt;p&gt;After reading &lt;a href=&quot;https://astronomy.stackexchange.com/a/6388/7982&quot;&gt;this answer&lt;/a&gt; I wondered if there have been document comets with orbits on non-return, hyperbolic trajectories as they approached the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that they would not be gravitationally bound to the sun, so either they came from outside the solar system, or they managed to gain energy through a collision, or close pass with a large planet like Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quickest way to find the comet with the &quot;highest&quot; orbit is to ask for the one with the largest &lt;a href=&quot;https://en.wikipedia.org/wiki/Specific_orbital_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;specific orbital energy&lt;/a&gt;, the total energy (kinetic + potential) divided by the mass, (Joules/kg). For a bound orbit the value would be negative, for an unbound, non-returning orbit it would be positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What documented comet has/had the highest specific orbital energy? It should be an identified object with a known or measured orbit, not just a generic, nameless &quot;edge-of-the-&lt;a href=&quot;https://en.wikipedia.org/wiki/Oort_cloud&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oort Cloud&lt;/a&gt; placeholder.&quot; &lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-10-25T04:18:05.050" LastActivityDate="2017-10-25T16:03:48.300" Title="What documented comet has/had the highest specific orbital energy?" Tags="&lt;orbit&gt;&lt;comets&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23281" PostTypeId="2" ParentId="23271" CreationDate="2017-10-24T13:46:10.930" Score="8" Body="&lt;p&gt;I think you'll have to live with an incomplete answer. It appears this was taken by an amateur astronomer using a simple 4.5 inch telescope. You can find, on the ESO website, a collection of many images taken during the 2004 transit. &lt;a href=&quot;https://www.eso.org/public/outreach/eduoff/vt-2004/photos/vt-photos-page49.html&quot; rel=&quot;noreferrer&quot;&gt;On this page&lt;/a&gt;, is the image you are asking about (along with some other images by the same person). The credit for that image is given as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Phases of Venus&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Statis Kalyvas&lt;/em&gt;&#xA;  &lt;br&gt;4.5-inch Mizar telescope&#xA;  &lt;br&gt;Nikon Coolpix 750&#xA;  &lt;br&gt;February 27 - June 8, 2004&#xA;  &lt;br&gt;Thessaloniki, Greece&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm assuming (because I can't find any reference to state otherwise) that the name &quot;Mizar telescope&quot; is just the personal name Statis Kalyvas gave to his telescope. Although you can see that the equipment was nothing an amateur astronomer couldn't buy. A pretty typical telescope with a standard camera. Nothing &quot;professional&quot; or extremely high tech.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can probably guess at some other pieces of the puzzle though. From the image below, you can see that the 2004 transit began (from Thessaloniki's local time) at 7:13 AM and ended at 1:25 PM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4bGc7.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4bGc7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the observation period of the transit itself happened during the day, Statis definitely would have needed a &lt;a href=&quot;https://www.telescope.com/Accessories/Telescope-Eyepiece-Filters/Orion-Full-Aperture-Glass-Telescope-Solar-Filters/pc/-1/c/3/sc/48/e/4.uts&quot; rel=&quot;noreferrer&quot;&gt;solar filter&lt;/a&gt; for his telescope, which would have blocked upwards of 99.999% of the light from the Sun, allowing him to view the transit safely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the other images leading up to the transit, my guess is that he made his observations during in the evening when the Sun was near the horizon and not very bright (due to much of the light being scattered). Venus should've been higher in the sky than the Sun and still visible at that time. An eyepiece which gives a small field of view would've been useful here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can possibly guess that the image was taken near the horizon because a 4.5 inch telescope &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_resolution#Explanation&quot; rel=&quot;noreferrer&quot;&gt;should have a resolution&lt;/a&gt; of $1.21\ \mathrm{arcsec}$ at best whereas, near conjunction, Venus &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_diameter#Use_in_astronomy&quot; rel=&quot;noreferrer&quot;&gt;is almost an arcminute in size&lt;/a&gt;. Venus should be well resolved, but it looks a bit fuzzy. This is likely attributed to the high &lt;a href=&quot;https://en.wikipedia.org/wiki/Air_mass_(astronomy)&quot; rel=&quot;noreferrer&quot;&gt;airmass&lt;/a&gt; near the horizon. The drastic difference between the clarity of the image on 3/4 and 13/4 (just ten days later) leads me to believe the telescope was capable of better resolution and was simply impeded by the atmosphere near the horizon.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-10-24T13:46:10.930" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23283" PostTypeId="2" ParentId="23266" CreationDate="2017-10-24T20:49:48.453" Score="0" Body="&lt;p&gt;Another way to look at redshift is to consider the wavelength, which gets &quot;stretched&quot;. There's no limit to how long a wavelength can be - it could be millions of kilometres (or even light years!) - but there's a limit to the size of the antenna we can use to detect it. And redshifted photons have less energy, so extremely long wavelengths mean extremely low energy. &lt;/p&gt;&#xA;" OwnerUserId="14094" LastActivityDate="2017-10-24T20:49:48.453" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23285" PostTypeId="2" ParentId="22807" CreationDate="2017-10-25T07:23:01.953" Score="2" Body="&lt;p&gt;What you're asking for in the comments, on ancient predictions, that's a tricky subject.  You're basically talking about proper motion and Edmund Halley was the first person to observe it.   He recognized that 3 stars were about 1/2 a degree from where they were observed 1,850 years earlier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Proper_motion#History&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/2 of 1 degree in the night sky is larger than you might think.  It's about the diameter of the full moon, but 1,850 years is a long time.   That's why it went unnoticed until Halley took a closer look at the ancient star charts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's noted in the Wikipedia article that the ancient Greeks suspected proper motion happened but they were unable to observe it and nobody can make a decent prediction based on something that changes so slowly they are unable to observe it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With today's observation methods, it's just a matter of running a computer simulation to see what the constellations will look like 100 years from now (virtually unchanged) or 10,000 years from now (some of them would begin to be visibly skewed but likely still recognizable).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every star has it's own proper motion, so each constellation would undergo it's own rate of change.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-10-26T11:28:49.137" LastActivityDate="2017-10-26T11:28:49.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23289" PostTypeId="2" ParentId="23278" CreationDate="2017-10-25T14:42:35.207" Score="5" Body="&lt;p&gt;That is a very good timing for this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a comet with the name &lt;strong&gt;C/2017 U1 (PANSTARRS)&lt;/strong&gt;, that was just discovered a few days ago (on 18th October) which has an orbital &lt;strong&gt;eccentricity of &lt;code&gt;1.18&lt;/code&gt;&lt;/strong&gt;, which is by far highest eccentricity recorded so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The eccentricity describes how strongly &quot;elliptical&quot; an orbit is and refers to the variable &lt;code&gt;e&lt;/code&gt; in the formula of &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kepler's orbit&lt;/a&gt;. &lt;code&gt;1.18&lt;/code&gt; makes the orbit a &lt;strong&gt;hyperbolic&lt;/strong&gt; one, meaning it is not an actual orbit, but has higher velocity than the required escape velocity and is basically just passing by in our solar system. And it's the first recorded comet so far that has an hyperbolic orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This comet is the first candidate for an interstellar comet ever in recorded history. It is already pretty certain that it originated from interstellar space, but not official yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quote from &lt;a href=&quot;http://www.minorplanetcenter.net/mpec/K17/K17UI1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;minorplanetscenter.net&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Further observations of this object are very much desired.  Unless there&#xA;  are serious problems with much of the astrometry listed below, strongly&#xA;  hyperbolic orbits are the only viable solutions.  Although it is probably&#xA;  not too sensible to compute meaningful original and future barycentric orbits,&#xA;  given the very short arc of observations, the orbit below has e ~ 1.2 for&#xA;  both values.  If further observations confirm the unusual nature of this&#xA;  orbit, this object may be the first clear case of an interstellar comet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sources:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA; &lt;li&gt;&lt;a href=&quot;http://www.minorplanetcenter.net/mpec/K17/K17UI1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.minorplanetcenter.net/mpec/K17/K17UI1.html&lt;/a&gt;&lt;/li&gt;&#xA; &lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/C/2017_U1_(PANSTARRS)&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/C/2017_U1_(PANSTARRS)&lt;/a&gt; &lt;/li&gt;&#xA; &lt;li&gt;&lt;a href=&quot;https://imgur.com/t/science_and_tech/vM4oh&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://imgur.com/t/science_and_tech/vM4oh&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10271" LastEditorUserId="10271" LastEditDate="2017-10-25T16:03:48.300" LastActivityDate="2017-10-25T16:03:48.300" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="23290" PostTypeId="1" CreationDate="2017-10-25T15:22:45.300" Score="1" ViewCount="102" Body="&lt;p&gt;Which physical quantity is the real / definitive tracer of galaxy evolution? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me explain better: if you aimed at understanding as much as possible about the evolution of galaxies and you had to choose only &lt;em&gt;one&lt;/em&gt; quantity to perfectly follow/observe (directly/indirectly) across the cosmic time, which quantity you would chose? why? Is it possible? If not, why? What combination of quantities you would choose?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stellar Mass, SFR, Metallicity, Dark Matter Mass, Angular momentum...&lt;/p&gt;&#xA;" OwnerDisplayName="user19333" LastActivityDate="2017-10-26T16:02:16.383" Title="Which physical quantity is the real/definitive tracer of galaxy evolution?" Tags="&lt;galaxy&gt;&lt;universe&gt;&lt;observational-astronomy&gt;" AnswerCount="0" CommentCount="15" ClosedDate="2017-10-30T13:14:40.570" ContentLicense="CC BY-SA 3.0" />
  <row Id="23291" PostTypeId="1" CreationDate="2017-10-25T16:34:08.763" Score="3" ViewCount="80" Body="&lt;p&gt;While detectors for gravitational waves are interferometers themselves, the question here is about an increase in resolution resulting from two or more detectors far away from each other. &lt;/p&gt;&#xA;" OwnerUserId="19335" LastEditorUserId="18643" LastEditDate="2017-12-01T18:14:07.633" LastActivityDate="2017-12-01T18:14:07.633" Title="Can two detectors for gravitational waves be combined into a very long baseline interferometer?" Tags="&lt;gravitational-waves&gt;&lt;interferometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23292" PostTypeId="2" ParentId="23203" CreationDate="2017-10-25T18:49:41.113" Score="2" Body="&lt;p&gt;The pressures are so great in a neutron star that most of the electrons combine with protons and become neutrons.  A statistical equilibrium is set up in which neutrons, because they are unstable, decay into electrons and protons while some electron proton pairs form neutrons.  For typical neutron star temperature/pressures, at any one time about 1/7 of the nucleons are protons (with an equal number of electrons) and 6/7 are neutrons.   If there were more electrons than protons, then the excess of electrons would persist since they would not have enough partners to combine with.   Of course, this net charge would attract protons and other positive ions from nearby space and this would reduce the neutron star's charge.  So, if you added electrons, you would change the charge of the neutron star, but the net charge would decay over time as positive particles are pulled in from space.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2018-04-11T20:08:17.243" LastActivityDate="2018-04-11T20:08:17.243" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23293" PostTypeId="2" ParentId="22962" CreationDate="2017-10-25T19:07:20.327" Score="1" Body="&lt;p&gt;After WISE ran out of cryogens to cool its detectors it was re-purposed to look for asteroids.  This was possible because the detectors for 3.4 microns and  4.6 microns could work at higher temperatures than the other detectors and passive cooling gets the spacecraft to 74 K.  It had already finished an all sky survey so there were no new deep space targets for it to study and, with only two detectors, the science would not be as significant, but it has been great as a detector of moving objects like asteroids since, for this, just one band-pass is enough. &lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2017-10-25T19:07:20.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23294" PostTypeId="2" ParentId="23194" CreationDate="2017-10-25T19:20:25.847" Score="2" Body="&lt;p&gt;Having the pulsar emiting radiation without decaying, the black hole not decaying via Hawking radiation and also the orbits of both objects not decaying and in a particular dynamic, I would say yes, veeeery slowly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thing is the radiation from the pulsar comes from particles that fall in it (dust, etc) and are &quot;reoriented and shot&quot; (roughly speaking) as high-energy radiation near the poles because the magnetic field.&#xA;The black hole would attract even more particles than the pulsar so considering both objects in a space with same density of particles, the black hole by itself should suck in more matter/energy compared with matter/energy from the pulsar beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also consider orbital implications. A black hole (unless supermassive at center of galaxies) and pulsars have equivalent masses. Not the same, but not different in orders of magnitude if the black hole is a stellar remnant as the pulsar.&#xA;That means that the orbit will be, probably, more like a double-star system, with the black hole moving also around the center of masses (you can thing about it as Pluto-Charon orbit system).&#xA;Now. Consider how stars are formed (from a dust cloud, it starts to collapse, gain angular momentum, etc). Chances are that the orbits and the spins of both objects are in a very similar planes. Think that the beams of the pulsar are shot mostly from the poles (with some inclination though).&#xA;I think is almost impossible for a naturally formed stars that one becomes pulsar, the other black hole, and also they orbit in a way that the beams of the pulsar are going to feed the black hole. The beams would just scape from the system, forming ~90º relative to the line from the pulsar-black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last, about the &quot;feeding&quot;. Lets consider that all the above is possible, just very unlikely. We have a scenario where the orbit of a pulsar is ~90º compared with its spin, like Sun-Uranus. Then a fraction of the time it feeds the black hole (twice a &quot;year&quot;).&#xA;Black holes eat stars sometimes. The size of the black hole grows when it eats, and is one of the theories about the mass of supermassive black holes in the center of most galaxies (just eating... the other is direct formation via a sudden collapse of super-massive dust cloud even avoiding a star formation).&#xA;I try o point that this feeding via a pulsar beam would be super slow compared with direct feeding of surroinding matter like the supermassive black holes do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say to the black hole: Sir, eat the pulsar :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And maybe it would become true: considering very long periods of time, orbits decay. The black hole would maybe collide/absorb the pulsar (as recently detected via gravitational waves)&#xA;Finally, depending on the period of time we are talking about, we have to consider black hole evaporation throught &lt;a href=&quot;https://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hawking ratiation&lt;/a&gt; (it would make it smaller). &#xA;On the other hand, I considered pulsars powered by matter orbiting/falling in it. there &lt;a href=&quot;https://en.wikipedia.org/wiki/Pulsar#Categories&quot; rel=&quot;nofollow noreferrer&quot;&gt;other types&lt;/a&gt;, so a rotation-powered one, after trillions of years, could maybe decay in an almost neutron star.&#xA;I don't know about the numbers so maybe these effects are almost zero, not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PD. This is my first answer in StackExchange and I'm not a native english speaker, feedback appreciated ;)&lt;/p&gt;&#xA;" OwnerUserId="19337" LastActivityDate="2017-10-25T19:20:25.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23295" PostTypeId="2" ParentId="22762" CreationDate="2017-10-25T19:23:57.303" Score="0" Body="&lt;p&gt;We do not yet know the answer to this question.  We observe the universal expansion, which should be slowing down from mutual gravitational attraction, is not slowing down as much as it should and might not be slowing down at all about now.  One explanation for this is that the cosmological constant, an added term in Einstein's equations of General Relativity is non-zero.  This term allows for gravity to effectively repel by causing expansion in low density regions.  For the entire universe, the expansion resulted from an initial impulse (the Big Bang) and gravity has been working against this, trying to compress it.  With the cosmological constant included, as more and more regions of the universe reach sufficiently low densities, gravity reverses and enhances expansion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, some think the expansion is due to some particle or another force that provides a negative pressure in the universe.   &lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2017-10-25T19:23:57.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23296" PostTypeId="2" ParentId="22522" CreationDate="2017-10-25T19:38:42.670" Score="4" Body="&lt;p&gt;There is no simple answer to this question other than: you have to do the math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most often you would see the asteroid brightening more rapidly and moving across the sky faster as the days pass, but not necessarily.  It could arrive directly along our orbital path and just brighten but show no motion in the sky.  It could start out lit up fully (full phase) and then arrive with only partial lighting from the sun, so it does not necessarily brighten.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You generally would not know the phase, unless you calculate it, because you will not be able to resolve it until it gets really close.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorDisplayName="user1569" LastEditDate="2017-10-26T12:40:35.630" LastActivityDate="2017-10-26T12:40:35.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23301" PostTypeId="1" CreationDate="2017-10-26T12:34:32.667" Score="4" ViewCount="90" Body="&lt;p&gt;Is it possible for a planet's axis to precess or wobble in such a way that its season cycle is of lesser duration than its year? The gyroscopic precession will thus give seasons to one or the other hemisphere during a possibly hundreds-year-long revolution around its star. Earth's precession is very much longer than it's revolution period, but is it possible that these could be reversed and a planet still be habitable?&lt;/p&gt;&#xA;" OwnerUserId="19350" LastActivityDate="2017-10-27T16:16:40.517" Title="Season cycle &lt; Year?" Tags="&lt;planet&gt;&lt;precession&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23303" PostTypeId="2" ParentId="23034" CreationDate="2017-10-26T14:58:47.570" Score="0" Body="&lt;p&gt;It will travel until absorbed by another object.&#xA;Since you assumed no other objects except sun it does not loose any energy or  change direction. Will go for ever unless hit a hydrogen or helium or whatever.&lt;/p&gt;&#xA;" OwnerUserId="19351" LastActivityDate="2017-10-26T14:58:47.570" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23304" PostTypeId="2" ParentId="6297" CreationDate="2017-10-26T15:39:05.953" Score="0" Body="&lt;p&gt;In his story &lt;em&gt;Far Centaurus&lt;/em&gt;, A.E. Van Vogt coined the term &quot;bachelor sun&quot;, which might fit, but this is usually taken as a prediction of black holes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Far_Centaurus&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Far_Centaurus&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19350" LastActivityDate="2017-10-26T15:39:05.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23308" PostTypeId="1" CreationDate="2017-10-27T04:18:00.820" Score="0" ViewCount="174" Body="&lt;p&gt;I was staring up at the stars around 11:30 PM and noticed an object moving. It had no lights, and was not bright whatsoever. It was moving smoothly and in a straight line, except extremely fast, and dead silent. It was fairly large too, compared to how big stars look. Never would have noticed it had I not been looking at that particular region of the sky at the exact moment I did. It had to have crossed the sky in less than a minute. In fact, I feel like I've seen this same kind of thing once or twice before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is what did I see? The fact that it was extremely dim, extremely fast, and produced no sound really confuse me.&lt;/p&gt;&#xA;" OwnerUserId="19359" LastActivityDate="2017-10-29T22:58:32.593" Title="Confusing satellite-like object seen in night sky" Tags="&lt;satellite&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2017-10-30T13:19:40.670" ContentLicense="CC BY-SA 3.0" />
  <row Id="23311" PostTypeId="1" AcceptedAnswerId="23322" CreationDate="2017-10-27T07:42:24.407" Score="6" ViewCount="159" Body="&lt;p&gt;I once read that the solar wind is more or less blowing a &quot;bubble&quot; into the interstellar medium. However, the numbers I found while trying to verify this don't give me a clear picture yet:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Interplanetary_medium&quot; rel=&quot;noreferrer&quot;&gt;interplanetary medium&lt;/a&gt; seems to have an average density of 5 particles per cm³ (at least near earth).&lt;/li&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Interstellar_medium&quot; rel=&quot;noreferrer&quot;&gt;interstellar medium&lt;/a&gt; density varies widely, from 10⁻⁴ to 10⁶ particles per cm³.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What I did not find was the density of the interstellar medium directly around the solar system. I know that the solar wind is interacting with the interstellar medium, which is responsible for the &lt;a href=&quot;https://en.wikipedia.org/wiki/Heliosphere&quot; rel=&quot;noreferrer&quot;&gt;termination shock&lt;/a&gt;, but I'm wondering if the density difference is large enough that one can say that the sun is actually blowing a &quot;bubble&quot; into the interstellar medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is the density inside the heliosphere much less than directly outside of it?&lt;/p&gt;&#xA;" OwnerUserId="16093" LastActivityDate="2017-10-27T19:27:19.257" Title="Does the sun blow a bubble into the interstellar medium?" Tags="&lt;interstellar-medium&gt;&lt;density&gt;&lt;solar-wind&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23312" PostTypeId="2" ParentId="1564" CreationDate="2017-10-27T08:23:12.883" Score="5" Body="&lt;p&gt;&lt;strong&gt;On the numbers&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rogue planets are thought to be more abundant than stars in the milky-way and depending on where you make the cut-off, rogue dwarf-planets for example, are expected to enormously outnumber the number of stars.   &lt;a href=&quot;https://science.nasa.gov/science-news/science-at-nasa/2011/18may_orphanplanets&quot; rel=&quot;noreferrer&quot;&gt;older article quoting the MOA survey&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.pbs.org/newshour/science/billions-rogue-planets-wander-universe-without-home&quot; rel=&quot;noreferrer&quot;&gt;This article - OGLE survey&lt;/a&gt;, or, Portland's Optical Gravitational Lensing Experiment estimates:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Based on a statistical analysis of more than 2,600 microlensing&#xA;  events, drawn from six years of observations on about 50 million&#xA;  stars, the OGLE team estimates that there is perhaps one Jupiter-mass&#xA;  rogue planet for every four stars in the galaxy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not an experiment it's a survey, but I guess they couldn't resist the name OGLE, so they went with that, so, props on the name, but I digress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's still some disagreement between MOA and OGLE, though I think the consensus is that OGLE gives a slightly better estimate of 1 Jupiter-mass rogue for every 4 stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the nature solar-system formation, we expect that a number of planets get cast out during the early period and larger orbiting objects are good at casting out smaller ones when they pass to close, but it's almost impossible for smaller objects to cast out a larger one.    We expect that the number of smaller rogue planets should increase significantly, perhaps even exponentially.  But the OGLE survey had difficulty locating smaller planets, 1-10 Earth masses.   That may be because they are more difficult to see, or they might be less common than expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Binary star systems, for example, with two or more much more massive than Jupiter objects could cast out larger objects like Jupiter-mass planets fairly easily, so there are significant unknowns on the number of smaller than Jupiter rogue planets.  Estimates are high, observations are low.   &lt;a href=&quot;https://www.nasa.gov/wfirst&quot; rel=&quot;noreferrer&quot;&gt;NASA's WFIRST&lt;/a&gt; might be powerful enough to resolve this after it's launch in 2020.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://news.stanford.edu/news/2012/february/slac-nomad-planets-022312.html&quot; rel=&quot;noreferrer&quot;&gt;This article&lt;/a&gt; a study by the Kavli institute, suggests an aggressive estimate of 100,000 rogue-planets (or Nomad planets) for every star in the Milky-way, based on the 2011 MOA estimate of two Jupiter-Mass objects for every star and extrapolating based on how common Pluto sized objects are thought to be relative to Jupiter sized (about 50,000 to 1).    If we adjust for the 1 to 4 instead of 2 to 1, that estimate is shaved down to about 12,000 pluto or larger rogue planets per star.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatever the real numbers are, if we count all the way down to dwarf planet size, Rogue planets should significantly outnumber stars and an estimate of 3-4 magnitudes isn't unreasonable.   For planet sized, say Mercury to Neptune, better observations would help with this estimate a great deal, but a bad estimate of Mercury and up Rogue planets outnumbering stars 10 or 20 to 1 isn't crazy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm fussing on the numbers because it's relevant to how many pass through the solar system.   Using &lt;a href=&quot;https://astronomy.stackexchange.com/questions/3601/how-often-do-stars-pass-close-1ly-to-the-sun&quot;&gt;this question/answer&lt;/a&gt; as an approximation and because the escape velocity from a solar system is usually tiny compared to the escape velocity from a galaxy or the orbital speed around the galaxy, generally speaking, rogue planets should have star-like orbits around the Milky-way.    If a star passes within one light-year of our sun about once every million years, even if you set a relatively high estimate of 100,000 rogue planets (pluto size and up), that's still one every 10 years, and that's within 1 light year.   If you want to estimate within the Kuiper Belt - where we might expect to observe a pluto sized rogue planet passing through.   The target area of the Kuiper belt is about a million times smaller than the 1 lightyear radius, so a rogue planet passing through the Kuiper belt - by this math, once every 10 million years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why it's never been observed is that they are usually very far away and too small to be noticed.  Even with today's best equipment, it's nearly impossible to see a Pluto sized rogue planet passing through a lightyear or half a light year away and if they only pass through every 10 or 100 or more years, well, that's why we've never seen one.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's possible, in our lifetime, someone might see one or two rogue planets pass within 1 light year of us, or inside the Oort cloud just past 1 light year, but we'll almost certainly need an upgrade in telescopes first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;on capture&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a two body system, gravitational capture is impossible and a rogue planet to our solar system is very much like a two body system.   The problem is the rogue object accelerates towards the larger object (the star) and it adds velocity.   That added velocity means it flies away just as fast.    In a two body system, objects are either already in orbit or they fly past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a loose sense, all the stars in the galaxy create an N-body system where capture becomes possible, but the problem is that the stars are so far apart that any secondary body gravitational effects from other stars is close to negligible.  The chance of Alpha Centauri (for example) slowing down an object so our Sun can catch it, is close to zero.   If it happened at all, it would be a very distant and very tenuous capture.   That's not to say it never happens.  Over 4 billion years of our solar system, capture may have happened several times, but I wanted to point out why capture is difficult and likely rare.  Most Rogue planets that got close enough would likely just pass through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Capture becomes statistically more likely when a massive 3rd body is close.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a Pluto-sized rogue planet passed close enough to one of our 4 large planets, or, the theoretical planet X, then it's direction and velocity could be significantly changed, and the odds of capture way up.   Needless to say, those 4 planets and the theoretical &quot;Planet Nine&quot; make very tiny targets in the vastness of space, but that's one way capture could happen and over the 4 plus billion years the solar system's been around, capture probably has happened from time to time.   The trick is getting a close enough look at distant orbiting objects to see if their chemistry indicates they aren't from our solar-system.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to Rogue planet capture, when two stars pass close enough to each other, two stars exchanging Oort cloud objects probably happens too and might be the more common method of object exchange.  Oort cloud objects already have a relatively slow orbital speed around their star which makes striping them away from one star and being captured by another, somewhat more likely.   It would also depend on the relative velocity of the two stars also and how close they passed.   But this method of exo-object capture might be quite a bit more common than rogue object capture.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an object like Sedna is eventually identified as not being from our solar-system, it would be hard to know if it had been a rogue planet or a &quot;stolen&quot; object from the Oort or Kuiper belt from a close passing star from long ago.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-10-27T08:23:12.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23313" PostTypeId="1" AcceptedAnswerId="23316" CreationDate="2017-10-27T08:52:39.237" Score="4" ViewCount="112" Body="&lt;p&gt;I wonder what is the contribution of metals and dust locked into planets, with respect to the amount in the interstellar medium (ISM). In other words, when we measure the metallicity and the dust mass of a galaxy we are sensitive to metals and dust in the various phases of the ISM, and thus the observed value is a lower limit. Is the &quot;correction&quot; a second-order variation? &lt;/p&gt;&#xA;" OwnerDisplayName="user19333" LastActivityDate="2017-10-27T12:29:57.137" Title="Metals and dust locked into planets" Tags="&lt;galaxy&gt;&lt;observational-astronomy&gt;&lt;interstellar-medium&gt;&lt;dust&gt;&lt;metallicity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23314" PostTypeId="2" ParentId="1621" CreationDate="2017-10-27T09:43:48.050" Score="0" Body="&lt;p&gt;Transmutation of elements from one to another had been understood since 1901 as a result of natural radioactive decay, but when Rutherford projected alpha particles from alpha decay into air, he discovered this produced a new type of radiation which proved to be hydrogen nuclei (Rutherford named these protons). Further experimentation showed the protons to be coming from the nitrogen component of air, and the reaction was deduced to be a transmutation of nitrogen into oxygen in the reaction&#xA;14N + α → 17O + p &#xA;This was the first-discovered nuclear reaction. We could use this reaction to terraform Neptune to our liking.&lt;/p&gt;&#xA;" OwnerUserId="19365" LastActivityDate="2017-10-27T09:43:48.050" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23316" PostTypeId="2" ParentId="23313" CreationDate="2017-10-27T12:17:18.700" Score="6" Body="&lt;p&gt;The amount of metals locked up in planets is completely negligible compared to that of the interstellar medium (ISM).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, in a typical stellar system, the mass fraction of planets is less than 1%, and of this by far the most mass will be in gaseous planets, rather than rocky planets. For instance, in the Solar System, the mass fraction of all planets is 0.1%, while rocky planets comprises only a fraction of $6\times10^{-6}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, typically only a small fraction of the gas in a galaxy is locked up in stars. For small galaxies — which are the most abundant — as little as ~5% of the gas is in stars (e.g. &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0101458&quot; rel=&quot;noreferrer&quot;&gt;Schombert et al. 2001&lt;/a&gt;), although for large galaxies like the Milky Way, which are more efficient at converting gas into stars, the ISM-to-stellar gas fraction can be of order unity (e.g. &lt;a href=&quot;https://arxiv.org/abs/1509.07240&quot; rel=&quot;noreferrer&quot;&gt;Morokuma-Matsui &amp;amp; Baba 2015&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the fraction of metals locked up in planets is of the order $10^{-5}\text{–}10^{-7}\!$, depending somewhat on the type of galaxy considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the fraction of metals locked up in &lt;em&gt;dust&lt;/em&gt; is considerable, with typical values of roughly 1/3 of the metals in dust. Thus, if you want to know the metallicity of a galaxy by looking at emission or absorption lines from metals, you must either account for this factor (hoping that it's a universal factor which it may or may not be, see e.g. &lt;a href=&quot;https://arxiv.org/abs/1303.1141&quot; rel=&quot;noreferrer&quot;&gt;Zafar &amp;amp; Watson 2013&lt;/a&gt;), or measure metals that tend not to deplete to dust grains, e.g. zinc or sulfur.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2017-10-27T12:29:57.137" LastActivityDate="2017-10-27T12:29:57.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23318" PostTypeId="2" ParentId="23301" CreationDate="2017-10-27T13:35:20.673" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it possible for a planet's axis to precess or wobble in such a way that its season cycle is of lesser duration than its year?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For this, definitely yes, if you tweak some parameters. On the Wikipedia page for &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_precession#Equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Axial Precision&lt;/a&gt;, you'll find the set of equations and parameters that allow you to calculate the precessional rate for Earth's axis. Mainly, the Earth's precession is caused by a variety of factors but primarily due to the Moon and Sun (and thus related orbital/body parameters). If you look at those equations, and start tweaking values in them, you'll find you can force the axial precession period to be less than a &quot;year&quot;. I'll go through and construct just such a system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First though, I need to figure out the orbital parameters of the planet such that it has a 100-year long revolution period. This is acheived by using &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kepler's Third Law&lt;/a&gt;, relating orbital period, orbital distance, and mass of the Sun (and Earth, but that's negligible). The relevant form of the equation, for our purposes is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P^2 = \frac{a^3}{M_{\mathrm{Sun}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, $P$ is the orbital period in years, $a$ is the orbital distance in AU (1 AU is the distance between Earth and the Sun), and $M_{\mathrm{Sun}}$ is the mass of the Sun in units of solar masses. If you're dictating that $P=100\ \mathrm{years}$, then I find I can set $a = 10\ \mathrm{AU}$ and $M_{\mathrm{star}} = 0.1\ M_{\odot}$ and make this equation true. What that means is that this new planet orbits a star which is 10 times less massive than our own star (making it a &lt;a href=&quot;https://en.wikipedia.org/wiki/Red_dwarf&quot; rel=&quot;nofollow noreferrer&quot;&gt;red dwarf&lt;/a&gt;) and orbits 10 times farther than Earth. My choice of parameters is all serendipitous and you could certainly find other parameters that work as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, given those parameters above, we can try to calculate the precessional rate for this new planet. I'll assume this new planet has an orbiting moon, just like Earth does. The two important equations for calculating the Earth's precessional rate are to account for the contributions from the Sun and the Moon. The equations are given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{\mathrm{Sun}} = \frac{3}{2}\left(\frac{GM_{\mathrm{Sun}}}{a_{\mathrm{Earth}}(1-e_{\mathrm{Earth}})^{3/2}}\right)\left(\frac{(C-A)}{C}\frac{\cos \epsilon}{\omega}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{\mathrm{Moon}} = \frac{3}{2}\left(\frac{GM_{\mathrm{Moon}}(1-1.5\sin^2 i_{\mathrm{Moon}})}{a_{\mathrm{Moon}}(1-e_{\mathrm{Moon}})^{3/2}}\right)\left(\frac{(C-A)}{C}\frac{\cos \epsilon}{\omega}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In these equations, $G$ is the Gravitational Constant, $M$ is mass in kg, $a$ is the orbital distance in meters, $e$ is the orbital eccentricity, $i$ is the inclination of the Moon's orbital plane wrt to the ecliptic, $(C-A)/C = 0.003273763$ is a measurement of the Earth's shape, $\epsilon$ is the Earth's axis' angle, nominally $23.43928^\mathrm{o}$, and finally $\omega$ is the rotation rate in rad/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging in all the values for the Earth-Moon-Sun system (which you can find on the wiki page), you find that $P_{\mathrm{Sun}} + P_{\mathrm{Moon}} = 7.789\times 10^{-12}\ \mathrm{rad/s}$ and to convert to a time scale, you can multiply by  $2\times10^{-7}$. Doing so gives you the nominal Earth precessional period of about 26,000 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given all of that, we can start playing with parameters to get it such that the precessional period for our new system is less than 100 years. I already said that $a_{\mathrm{Earth}}$ is 10 times larger and that $M_{\mathrm{Sun}}$ is 10 times smaller so we could get the period of 100 years. I found that I could make the following additional changes.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The moon is now 5.8 times more massive than our Moon.&lt;/li&gt;&#xA;&lt;li&gt;The moon is now a quarter of the distance of our Moon.&lt;/li&gt;&#xA;&lt;li&gt;The planet is spinning twice as fast.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With those changes, I find that the precessional period falls down to about 50 years, or half an orbit. You can continue to play with the numbers to get a different result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important consideration here is whether or not such a system is likely to even form. I think, given the system I've created, it is not outside the realm of possibilities. To have such a large, almost binary planetary system be so far away from a small star is unlikely, but I don't think impossible. The real difficulty here is that you've set the orbital period to such a long timescale. If you were to relax that a bit to something much shorter (say 10 years) you'd get a more realistic (and likely habitable) system.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Earth's precession is very much longer than it's revolution period, but is it possible that these could be reversed and a planet still be habitable?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Certainly there's nothing to say a planet with a precessional period shorter than the revolution period is uninhabitable. What will make it uninhabitable is being too far from the star. The extreme orbital time you've provided necessitated having a planet that is likely too far from the star to be in the &quot;habitable zone&quot;. You just have to play with all the parameters and find something that fits all your needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to consider is that if a planet has such a large and close moon as I've constructed, it will likely cause the planet to slow its rotation rate over short (astronomical) timescales. The planets rotation rate has a significant affect on the precessional rate and slowing it down over short timescales may mean your system doesn't keep the precessional rate under the rotation period for too long.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="6841" LastEditDate="2017-10-27T16:16:40.517" LastActivityDate="2017-10-27T16:16:40.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23319" PostTypeId="1" CreationDate="2017-10-27T16:55:34.720" Score="3" ViewCount="236" Body="&lt;p&gt;The question &lt;a href=&quot;https://astronomy.stackexchange.com/q/23203/7982&quot;&gt;Can a Neutron Star become charged?&lt;/a&gt; and the associated comments and answers started me wondering; Are there measurements of, or experimental limits to the residual charge of the Sun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the Sun has a complex atmosphere and a net solar wind, I there may be some subtleties related to what sphere is used to establish the net charge Q inside, so rather than me define a radius it may be better to see what evidence is available. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there is an ionized component of the solar wind and protons are 2,000 times heavier than electrons, I'm thinking that an initially neutral Sun would loose electrons more rapidly than protons, until a strong enough static field was produced that would add the extra &quot;push&quot; for the protons to leave at the same rate, but that's an extremely simplistic model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm asking primarily for some kind of &lt;strong&gt;measurement or experimental data rather than purely rationalization or hand-waving&lt;/strong&gt;. Has there ever been an attempted measurement of the static residual charge of the Sun?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2020-12-25T22:00:14.550" LastActivityDate="2020-12-25T22:00:14.550" Title="What are the experimental limits to the residual charge of the Sun?" Tags="&lt;observational-astronomy&gt;&lt;the-sun&gt;&lt;solar-wind&gt;" AnswerCount="0" CommentCount="21" ContentLicense="CC BY-SA 4.0" />
  <row Id="23320" PostTypeId="1" AcceptedAnswerId="23323" CreationDate="2017-10-27T18:01:20.393" Score="8" ViewCount="102142" Body="&lt;p&gt;Last night, I was on my balcony at 1AM (PST) and I looked up and saw two stars near the horizon (I'd guess ~30 degrees above the horizon), and they were &quot;twinkling&quot; about twice as fast as other stars higher in the sky, and I could clearly see them changing from red to white to blue repeatedly. Other stars in the sky only appeared white to me, and didn't seem to &quot;twinkle&quot; as rapidly as these two stars did. The red and blue make me think of red-shift and blue-shift, but I don't know how I would see both from the same object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was I seeing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if it helps, but I am in the Los Angeles area, and I was looking &lt;del&gt;in a roughly north direction.&lt;/del&gt; almost exactly to the east, according to google maps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I tried taking a picture, but light pollution from the nearby street lights wouldn't permit me taking a decent picture. However, I noticed a group of three stars close together in nearly a perfect almost vertical line, and managed to find that in Stellarium. I think I found the two stars I am seeing: Procyon and Sirius&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/L6FFf.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L6FFf.jpg&quot; alt=&quot;What I found in Stellarium&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything about either of these stars that would make them show as red/blue?&lt;/p&gt;&#xA;" OwnerUserId="19370" LastEditorUserId="19370" LastEditDate="2017-10-28T08:13:18.270" LastActivityDate="2019-05-02T08:06:59.223" Title="What is this rapidly twinkling red, blue, and white star I saw?" Tags="&lt;star&gt;&lt;night-sky&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23321" PostTypeId="2" ParentId="23320" CreationDate="2017-10-27T18:08:36.097" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://neave.com/planetarium/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Using a star chart&lt;/a&gt; and based on your location, direction of observation, and time of observation, I'd guess you were seeing the star &lt;a href=&quot;https://en.wikipedia.org/wiki/Capella&quot; rel=&quot;nofollow noreferrer&quot;&gt;Capella&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Capella is quite a striking star because it's actually &lt;em&gt;four&lt;/em&gt; stars. There are two sets of binary pairs that are almost exactly along the same line of sight such that the light from the four stars together co-mingles and from Earth looks like a single star. Interestingly, the different stars have different colors, meaning that as you stare at it, it can appear white, red, or blue, just as you described. Note that the human eye cannot observe redshift and blueshift like you describe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the star twinkled more than those higher up is due to the atmosphere itself. The closer to the horizon a star is, the more it will twinkle due to the starlight passing through more atmosphere before reaching you.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-10-27T18:08:36.097" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23322" PostTypeId="2" ParentId="23311" CreationDate="2017-10-27T19:27:19.257" Score="3" Body="&lt;p&gt;At the moment, the Solar System is inside the &lt;a href=&quot;https://en.wikipedia.org/wiki/Local_Interstellar_Cloud&quot; rel=&quot;nofollow noreferrer&quot;&gt;Local Interstellar Cloud&lt;/a&gt; (which is itself within the &lt;a href=&quot;https://en.wikipedia.org/wiki/Local_Bubble&quot; rel=&quot;nofollow noreferrer&quot;&gt;Local Bubble&lt;/a&gt;). The LIC has &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/1742-6596/767/1/012016/pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;a neutral hydrogen number density of $\sim0.1\text{ cm}^{-3}$&lt;/a&gt;, which is a bit lower than the mean galactic neutral hydrogen number density. The Local Bubble itself is less dense than the LIC, &lt;a href=&quot;http://www-ssg.sr.unh.edu/ism/LISM.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;by about one or two orders of magnitude&lt;/a&gt;. On the whole, this region of space is less dense than the ISM, on average.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we take atomic hydrogen density to be a good benchmark, then the LIC and surrounding regions are in fact less dense than the Solar System on average.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2017-10-27T19:27:19.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23323" PostTypeId="2" ParentId="23320" CreationDate="2017-10-27T21:19:40.463" Score="10" Body="&lt;p&gt;It's most probably Sirius.  At this time of year (at 1 am local time) it's low in the sky in the East, so there is a lot of atmosphere in the way, and as Sirius is a bright bluish star, it will show all the colours described as it twinkles.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-10-27T21:19:40.463" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23330" PostTypeId="1" AcceptedAnswerId="23365" CreationDate="2017-10-28T16:32:25.990" Score="1" ViewCount="531" Body="&lt;p&gt;What is beyond the edge of the observable universe? Scientists say that the Big Bang is the cause of all creation in the known universe, but it would seem the bang must have happened in the center of the universe ?&lt;/p&gt;&#xA;" OwnerUserId="19385" LastEditorUserId="19385" LastEditDate="2017-11-08T05:35:23.077" LastActivityDate="2017-11-08T05:35:23.077" Title="What is beyond the observable universe?" Tags="&lt;universe&gt;&lt;big-bang-theory&gt;&lt;observable-universe&gt;" AnswerCount="1" CommentCount="8" ClosedDate="2017-11-02T20:02:27.310" ContentLicense="CC BY-SA 3.0" />
  <row Id="23333" PostTypeId="1" AcceptedAnswerId="23336" CreationDate="2017-10-28T19:48:03.000" Score="3" ViewCount="1110" Body="&lt;p&gt;It will be cold and dark in 10^(10^100) years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What else can be said of the state of things in the far distant future? Will everything have fallen into supermassive black holes and then radiated back into space so there is no matter left?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the universe revert to an exotic state as it was in primordial times, or will it slow and stay unchanged indefinitely? What kind of structure will it have?&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2017-10-29T02:26:16.373" LastActivityDate="2019-11-13T20:12:59.587" Title="What will the universe be like in a googolplex years time?" Tags="&lt;black-hole&gt;&lt;universe&gt;&lt;cosmology&gt;&lt;cosmological-inflation&gt;&lt;future&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23335" PostTypeId="1" CreationDate="2017-10-28T20:41:46.767" Score="1" ViewCount="124" Body="&lt;p&gt;When galaxies move away from us (caused by the expanding spacetime) their light seems to show a redshift. But what is really needed for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What time lap is for example needed to see a difference in frequency, and how does this precisely work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this also possible for gravitational waves, or are amplitudes the only solution to derive the redshift from?&lt;/p&gt;&#xA;" OwnerUserId="10445" LastActivityDate="2017-10-29T09:37:29.317" Title="How is the redshift of starlight precisely measured?" Tags="&lt;galaxy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23336" PostTypeId="2" ParentId="23333" CreationDate="2017-10-28T20:47:02.597" Score="6" Body="&lt;p&gt;That is nearly long enough to reach heat-death, which is estimated as about &lt;span class=&quot;math-container&quot;&gt;$10^{10^{120}}$&lt;/span&gt;. What that means is rather speculative, since it depends on various events that we have never observed, such as the spontaneous formation of black holes by quantum tunnelling. Such events are utterly rare, but are predicted to occur at very long timescales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any matter, will have at some time over a googolplex years quantum tunnelled into having enough mass to collapse as a black hole. So no matter will exist&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any black hole will have evaporated in Hawking radiation.  So no black holes exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Redshift will reduce the energy of photons to the point where their wavelengths are comparable to the scale of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing is happening in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://en.wikipedia.org/wiki/Timeline_of_the_far_future&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Timeline_of_the_far_future&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there is a: immovable object/unstoppable force paradox here, for in a universe that is spatially big enough, there are pockets of order among the chaos of heat death. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="30509" LastEditDate="2019-11-13T20:12:59.587" LastActivityDate="2019-11-13T20:12:59.587" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23337" PostTypeId="2" ParentId="23335" CreationDate="2017-10-28T21:55:11.103" Score="2" Body="&lt;p&gt;I don't really think I fully understand the first part of your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Redshift is measured by comparing the wavelengths of a redshifted pattern of absorption or emission lines with the wavelengths they would have in an object at rest. All lines are shifted by the same factor of $1+z$, where $z$ is the redshift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational waves cannot be used to obtain a redshift because the intrinsic wavelength of the waves is not known. An inspiralling binary emits waves with a frequency equal to twice the orbital frequency. This in turn depends on the mass, but the observed frequency is also redshifted. There is thus a &quot;degeneracy&quot; between the mass and redshift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can do with gravitational waves is estimate the physical distance to the source using the wave amplitude and how quickly the amplitude changes with time. If the host galaxy of the source can then be identified, then you can use the redshift of that galaxy to get an estimate of the Hubble parameter (e.g. see &lt;a href=&quot;http://www.nature.com/nature/journal/vaap/ncurrent/full/nature24471.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Abbott et al. 2017&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-10-29T09:37:29.317" LastActivityDate="2017-10-29T09:37:29.317" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23338" PostTypeId="2" ParentId="14185" CreationDate="2017-10-28T23:48:31.203" Score="-2" Body="&lt;p&gt;To create a black hole a body must occupy a radius that is smaller than its Schwarzschild limit. If a body has a mass greater than 3.2 solar masses then the theoretical behaviour of the neutron is that the force acting upon the neutrons due to the pressure of the collective mass of the star would cause the neutron to collapse. Then the way that a neutron star would form a black hole is by having a companion star which looses matter to the neutron star. If the mass of the neutron star is close to the mass of 3.2 solar masses then the star will theoretically collapse into a black hole when enough matter is accreted to exceed the critical limit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The caveat is that there is no force in the universe that can accelerate  energy faster than the speed of light. When neutron collapses it’s surface must accelerate at a rate that exceeds the speed of light, so this in fact is not allowed. An alternate theory is that the matter does not collapse into a black hole nut rather exists at an energy density that is just above its Schwartzschild limit. In fact recent observations made by Ligo of colliding black holes and by the recent observations of Sagittarius A*'s G2 gas cloud collision in fact fits much more closely with this model than these object being in fact true black holes. When the G2 gas cloud interacted with Sagittarius A* it did not create a large output of energy, but rather there was a just small release of energy consistent with a black star that exists at an energy state just above its true event horizon. These objects do not emit light and so are virtually indistinguishable from black holes, except when they interact with other objects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When LIGO first observed a gravitational wave there was a small gamma ray burst 0.04 seconds after the gravitational wave struck. There should be no Gamma Ray Bursts when black holes merge. There was a paper written in June of 2016 that used statistical models to suggest that the GRB did not occur. (Someone's job was at stake.) When you look at the method they used it virtually eliminates any possible observation of a GRB that would be associated with a dark star merger due to the weak GRB it would produce. Thus, no black hole merger could be observed in the future associated with a gravitational wave observation, unless it as a neutron star merger, which produces a much larger GRB. Recently LIGO announced just such an event where two neutron stars merged and this coincided with a GRB. When asked why this has not been observed before the officials at LIGO said it was because the merging of black holes do not produce GRBs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with creating black holes is that the matter at the event horizon, or simply the energy, must accelerate faster than the speed of light, which is not allowed by relativity. The comeback is that spacetime itself is collapsing faster than the speed of light. One must then ask how this could happen if the relativity does not allow super luminous speeds. To accelerate spacetime to super luminous speeds something must have dragged spacetime to this acceleration. But, that is not allowed. In fact what the observations are really showing is that it is most probable that black holes do not exist in nature. To back this up the search for extraterrestrial planets has found zero black holes orbiting other stars. Since 1968 only about 20 stellar mass black hole candidates have been found anywhere. If they exist where are they? Every observation made is consistent with black stars and only mathematical models are consistent with black holes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;WE must also consider that the present theory predicts that black holes will explode at the end of their lifetimes as they evaporate. No spikes of energy of this nature has been observed and yet there should some if the theory is correct. We must also consider that CERN should have produced black holes at the energy levels they have employed and no black holes have been created. This was not expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black stars paper.&#xA;&lt;a href=&quot;https://arxiv.org/pdf/1611.03853.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/pdf/1611.03853.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better solution to the runaway endpoint problem&#xA;&lt;a href=&quot;https://journals.aps.org/prd/abstract/10.1103/PhysRevD.77.044032&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://journals.aps.org/prd/abstract/10.1103/PhysRevD.77.044032&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statistical trick and logic used to eliminate future black star merger observations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://iopscience.iop.org/article/10.3847/2041-8205/827/2/L38/meta;jsessionid=5902D1B053AA93E11FDDF2A02A79BFBF.ip-10-40-2-120&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://iopscience.iop.org/article/10.3847/2041-8205/827/2/L38/meta;jsessionid=5902D1B053AA93E11FDDF2A02A79BFBF.ip-10-40-2-120&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The non-event of the Sagittarius A* G2 collision&#xA;&lt;a href=&quot;http://earthsky.org/space/milky-ways-black-hole-more-active&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://earthsky.org/space/milky-ways-black-hole-more-active&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19389" LastEditorUserId="19389" LastEditDate="2017-10-29T00:15:47.737" LastActivityDate="2017-10-29T00:15:47.737" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23345" PostTypeId="2" ParentId="23308" CreationDate="2017-10-29T16:32:05.323" Score="1" Body="&lt;p&gt;At first blush, this sounds like it was a large, low-Earth orbiting satellite, like the ISS for example. Depending upon the angle above the horizon the transit time could be very low s you described. Your position would determine whether the satellite was illuminated by the Sun or not at that local time.&lt;/p&gt;&#xA;" OwnerUserId="19399" LastActivityDate="2017-10-29T16:32:05.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23346" PostTypeId="1" CreationDate="2017-10-29T16:40:42.513" Score="5" ViewCount="3045" Body="&lt;p&gt;For a series of supernovae I have the observed magnitude and I want to convert these values to flux.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that I need to apply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$m - M_r = -2.5\log_{10} \frac{F}{F_r}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$M_r$&lt;/span&gt; is a 'reference magnitude' with corresponding known 'reference flux'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exactly should I use as my reference values?&lt;/p&gt;&#xA;" OwnerUserId="19400" LastEditorUserId="1891" LastEditDate="2019-11-26T19:20:11.220" LastActivityDate="2019-11-26T19:20:11.220" Title="How do you convert an observed magnitude to a flux?" Tags="&lt;cosmology&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2017-11-06T21:00:07.293" ContentLicense="CC BY-SA 4.0" />
  <row Id="23348" PostTypeId="1" CreationDate="2017-10-29T21:07:55.680" Score="1" ViewCount="298" Body="&lt;p&gt;Seeing the news about &lt;a href=&quot;https://en.wikipedia.org/wiki/A/2017_U1&quot; rel=&quot;nofollow noreferrer&quot;&gt;A/2017 U1&lt;/a&gt; made me wonder what would happen if a similar object, but with a mass close to Earth's, had a close call with Earth. Could it shift Earth's orbit enough to quickly and significantly change the conditions for life?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/835/19405&quot;&gt;This answer&lt;/a&gt; sort of gets it it, but it's not clear how much of an effect the change in orbit would have.&lt;/p&gt;&#xA;" OwnerUserId="19405" LastEditorDisplayName="user1569" LastEditDate="2017-10-31T16:57:12.963" LastActivityDate="2018-09-26T22:01:50.780" Title="How much could Earth's orbit shift without causing an ecological catastrophe?" Tags="&lt;orbit&gt;&lt;earth&gt;&lt;orbital-mechanics&gt;&lt;habitable-zone&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23351" PostTypeId="2" ParentId="23308" CreationDate="2017-10-29T22:58:32.593" Score="0" Body="&lt;p&gt;My guess, as @Tim said, is the ISS. I've seen it a few times, and it looks exactly as you've described. There are a few websites and mobile apps that can inform you of any satellites that are passing over your position; my go-to is &lt;a href=&quot;http://heavens-above.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;heavens-above.com&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2017-10-29T22:58:32.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23352" PostTypeId="2" ParentId="23348" CreationDate="2017-10-29T23:06:25.637" Score="1" Body="&lt;p&gt;Depending on how one defines &quot;quickly&quot; and &quot;significantly,&quot; I think a gravitational influence strong enough to do that would tear apart the Earth through tidal effects long before the change in orbit would make the planet unsuitable for life.&lt;/p&gt;&#xA;" OwnerUserId="7108" LastActivityDate="2017-10-29T23:06:25.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23353" PostTypeId="1" CreationDate="2017-10-30T01:15:22.850" Score="1" ViewCount="719" Body="&lt;p&gt;Assume that a star has a uniform mass density. Then how does the temperature&#xA;scale as a function of the distance to the center?&lt;br /&gt;&#xA;&lt;em&gt;Note: I am not looking for an answer that involves derivations or integrations.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here is what I did:&lt;br /&gt;&#xA;The star is in equilibrium so the pressure at distance &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; must equal the gravity force at distance &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt;. Pressure is given by &lt;span class=&quot;math-container&quot;&gt;$P=nKT$&lt;/span&gt;, and so &lt;span class=&quot;math-container&quot;&gt;$P\times Area=F_g$&lt;/span&gt;.&lt;br /&gt;&#xA;Then &lt;span class=&quot;math-container&quot;&gt;$nKT \times 4\pi r^2=F_G$&lt;/span&gt;. The mass of the part of the star contained within radius &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; from the center is &lt;span class=&quot;math-container&quot;&gt;$\frac{4}{3}\pi r^3\times m_sn$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$m_s$&lt;/span&gt; is the mass of each particle and &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is the number of particles per volume.&lt;br /&gt;&#xA;Then we can rewrite &lt;span class=&quot;math-container&quot;&gt;$nKT \times 4\pi r^2=m(\frac{G\frac{4}{3}\pi r^3\times m_snF}{r^2})$&lt;/span&gt; and simplify to &lt;span class=&quot;math-container&quot;&gt;$KTr=\frac{m}{3}Gm_s$&lt;/span&gt;. Isolating &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; we get that &lt;span class=&quot;math-container&quot;&gt;$T=\frac{mGm_s}{3Kr}$&lt;/span&gt;.&lt;br /&gt;&#xA;This seems to imply that the temperature decreases the further we are from the core, however I am not sure what &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; should stand for in &lt;span class=&quot;math-container&quot;&gt;$F_g=mg$&lt;/span&gt;, should &lt;span class=&quot;math-container&quot;&gt;$m=m_s$&lt;/span&gt;, i.e be the mass of a single particle? Or does &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; itself depend on the radius in which case the answer I got is invalid.&lt;br /&gt;&#xA;Can someone verify that the answer I got is right?&lt;/p&gt;&#xA;" OwnerUserId="19407" LastEditorUserId="37191" LastEditDate="2021-02-12T21:02:38.440" LastActivityDate="2021-02-12T21:02:38.440" Title="Assume that a star has a uniform mass density. Then how does the temperature scale as a function of the distance to the center?" Tags="&lt;star&gt;&lt;gravity&gt;&lt;astrophysics&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23354" PostTypeId="1" AcceptedAnswerId="23474" CreationDate="2017-10-30T01:31:43.043" Score="2" ViewCount="95" Body="&lt;p&gt;If I've taken an image of the Galactic bulge with the pixel units in MJ per steradian:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Y4BGt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y4BGt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I've looked down each column of the image and added up the values:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/OGZQh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OGZQh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;then do the units change? My thinking is that I'm taking something that is measured per steradian (which is a solid angle) and integrated it along a single axis, which would be in degrees. Would this make the units of my graph now MJ per radian? Or am I simply adding values in MJ per sr, so the units don't change?&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2017-11-10T22:52:44.657" Title="Units of integrated flux density (irradiance) along longitudes" Tags="&lt;fundamental-astronomy&gt;&lt;units&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23355" PostTypeId="1" AcceptedAnswerId="23357" CreationDate="2017-10-30T05:46:57.370" Score="0" ViewCount="73" Body="&lt;p&gt;Do we have a theory that can tell us how much change gravitational waves can make to space time. If it is infinite then I think any time we can see biggest change in space time and how much could it affect to humans?&lt;/p&gt;&#xA;" OwnerUserId="14136" LastEditorDisplayName="user1569" LastEditDate="2017-10-30T13:13:31.407" LastActivityDate="2017-10-30T13:13:31.407" Title="Maximum disturbance to space time of gravitational waves, and will this affect humans?" Tags="&lt;gravity&gt;&lt;space-time&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2017-10-30T13:20:08.267" ContentLicense="CC BY-SA 3.0" />
  <row Id="23357" PostTypeId="2" ParentId="23355" CreationDate="2017-10-30T06:22:45.007" Score="1" Body="&lt;p&gt;Gravitational waves cause a stretch and squeeze in space-time. It has two polarizations Plus and Cross polarization. This serves as an axis for the stretch and squeeze when a gravitational wave hits a body. However, the order of stretching and squeezing is actually really small. It changes space-time by an order of $10^{-21}$m. That's around the diameter of a proton. There are Numerical relativity simulations where you can see for yourself how much the change in space-time occurs for a certain pair of mass collisions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out the Waveforms for SEOBNR, Taylor etc. You might get a good idea. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;FYI : Question belongs in &lt;a href=&quot;https://physics.stackexchange.com/&quot;&gt;physics.stackexchange&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="18932" LastActivityDate="2017-10-30T06:22:45.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23362" PostTypeId="1" AcceptedAnswerId="23363" CreationDate="2017-10-30T12:35:51.693" Score="0" ViewCount="63" Body="&lt;p&gt;I wish to create a 3D visualisation of NEO's by date in relation to earth, but calculating their position at that time has got me stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this question I was able to fine Nasa's NEO object data:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/602/how-can-an-amateur-astronomer-verify-the-position-of-near-earth-objects&quot;&gt;Previous Question&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Listing Of NEO's:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://api.nasa.gov/neo/rest/v1/feed?start_date=2015-09-07&amp;amp;end_date=2015-09-08&amp;amp;api_key=DEMO_KEY&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nasa Date range NEO Data&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://api.nasa.gov/neo/rest/v1/neo/3726710?api_key=DEMO_KEY&quot; rel=&quot;nofollow noreferrer&quot;&gt;Detailed NEO Data&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the information in these JSON streams. Is it possible to calculate the position of the NEO in relation to earth? It doesn't seem to offer any positional information that I can see, although this is not my field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is there a better data set which will assist me in this?&lt;/p&gt;&#xA;" OwnerUserId="19417" LastActivityDate="2017-10-30T23:39:16.057" Title="Calculate NEO object position with nasa Near Earth Object data" Tags="&lt;positional-astronomy&gt;&lt;celestial-mechanics&gt;&lt;near-earth-object&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23363" PostTypeId="2" ParentId="23362" CreationDate="2017-10-30T19:05:42.073" Score="0" Body="&lt;p&gt;The file you had linked as 'Detailed NEO Data' lists all the orbital parameters needed to determine the position of the object as a function of time. It lists these parameters for the objects orbiting the earth. This should be just what you need. They are located at bottom of file. There are routines to convert the orbit in orbital parameters to cartesian coordinates. This may be easier to use. You can google on orbital parameters conversion to find these routines. I think I have a copy in Fortran but not many people use Fortran now.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="16926" LastEditDate="2017-10-30T23:39:16.057" LastActivityDate="2017-10-30T23:39:16.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23364" PostTypeId="2" ParentId="23353" CreationDate="2017-10-30T19:19:17.650" Score="2" Body="&lt;p&gt;I am not sure what you mean by looking for an answer without derivations? Of course you have to solve the maths to get a mathematical answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The handwaving answer is that the star has to be hotter in the middle, because at constant density the pressure depends on temperature and you need a negative pressure gradient to support the weight of the gas above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation of hydrostatic equilibrium is&#xA;$$\frac{dP}{dr} = -\rho g,$$&#xA;where $\rho$ is the mass density and $g$ is the gravity at radius $r$, which by Newton's shell theorem is due to the total mass interior to $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation is easily derived by considering the force equilibrium on a slab of gas in the star (your error above was in not recognising that it is the &lt;em&gt;difference&lt;/em&gt; in pressure across the slab that provides the upward force).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the density (and composition and ionisation state) are constant, then this equation can be written as&#xA;$$ \frac{\rho k_B }{m} \frac{dT}{dr} = -\rho \frac{4\pi G r^3 \rho}{3r^2},$$&#xA;where $m$ is the mass per particle and a perfect gas has been assumed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating with respect to radius and assuming $T(r=0)=T_0$, we obtain&#xA;$$T = T_0 - \frac{2\pi G\rho m}{3 k_B}r^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if the star has radius $R$, then this will be radius at $T=0$, since nothing could be supported beyond that. In which case&#xA;$$T = \frac{2\pi G\rho m}{3 k_B}(R^2-r^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or in terms of mass $M$&#xA;$$T = \frac{GM m}{2 k_B R^3}(R^2-r^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Constant density is of course a poor approximation. The above treatment gives a central temperature of 7 million Kelvin for the Sun, assuming $m\simeq 10^{-27}$ kg.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-10-30T23:59:59.687" LastActivityDate="2017-10-30T23:59:59.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23365" PostTypeId="2" ParentId="23330" CreationDate="2017-10-30T23:36:36.810" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is beyond the observable universe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's no way to know because we can't see it.  No information from the unobservable part of the universe can ever get to us.  If we can't see it, we can't know what's there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over time, the observable universe grows and we see more of the universe, though this is actually a pretty complicated due to the rate of expansion and I'd rather not open that can of worms, but for further reading, see &lt;a href=&quot;https://astronomy.stackexchange.com/questions/14336/are-there-any-galaxies-which-fell-out-of-sight-horizon-due-to-cosmic-expansion&quot;&gt;this question&lt;/a&gt; or &lt;a href=&quot;https://astronomy.stackexchange.com/questions/19909/will-we-start-seeing-galaxies-disappear-due-to-universe-expansion&quot;&gt;this one&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But avoiding the complexities of dark energy expansion, we have a pretty good idea that the unobservable universe is very much like what we can see.   This makes logical sense because it likely formed the same way at roughly the same time and out of the same stuff.   So, we expect it's just more galaxies and more space.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to this, if there was a point where the galaxies stopped being there at a theoretical 'edge', not far from the edge of the observable universe, we'd expect to see that lack of gravitation resulting in the galaxies moving away less fast and that's not observed.   The galaxies appear to be moving away from us equally fast in all directions, so we have good reason to believe that the universe is uniform out considerably beyond what we can observe.   It's at least several times larger than what we can see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we think of the big bang at happening at a point, which is often depicted in sciency-tv shows, then we shouldn't expect uniformity due to gravity and faster moving parts spreading out more, but what we see when we observe the universe is uniformity.   That means that if it did explode from a singular point, it must be so large that we can't see any variation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more popular approach is that the big bang didn't happen at a point but it happened everywhere.   Google Everywhere Stretch for some explanations of this.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth noting that not all shapes (or &lt;a href=&quot;http://dictionary.obspm.fr/index.php/index.php?showAll=1&amp;amp;formSearchTextfield=manifold&quot; rel=&quot;nofollow noreferrer&quot;&gt;manifolds&lt;/a&gt;) have centers.   The surface of the Earth has no center on the surface.   So even if we could see the entire universe, finding it's center might still not be possible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determining the center of the observable universe, however, is easy.   We can see in all directions just about equally so we're the center.  But we can't get a good grasp on the size and shape of the entire universe, or just universe is more correct.   Since we can't see all of it, finding the center is impossible.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The bang must have happened in the center of the universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that the big bang isn't well understood.  It's what physicists call a singularity which means, the math breaks down.     The point exploding outwards is often represented, but shouldn't be considered accurate.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematical approaches can come up with a variety of possibilities like hyperplanes intersecting or 4 or 5 dimensional spheres.   (Stephen Hawking had a 5 dimensional sphere argument where no big bang happened, it's just all a circle - that's in one of his books, I think blackholes and baby universes).   He was able to create a universe without a big bang.   That the universe is infinite and the big bang happened across infinite space is also a popular idea today.    The singular point that blew up into the universe isn't big right now, though that viewpoint still applies to the observable universe, but it's fallen out of favor for the entire universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My two cents anyway, as a layman.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-10-30T23:36:36.810" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23368" PostTypeId="1" AcceptedAnswerId="23369" CreationDate="2017-10-31T14:24:56.167" Score="5" ViewCount="581" Body="&lt;p&gt;Imagine I pointed a laser into the night sky. How likely is it that a particular photon will ever hit anything?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question bothered me for a while. I know very little about astronomy. When I asked my roommate he laughed and said that it must be very little, since the universe is mostly empty. I know it is true, but it seems to me that volume of the objects in the universe does not matter. Rather, it should be the area of objects when projected on a sphere that matters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On one hand, universe is pretty empty, so hitting something outside the Solar System might be hard. But on the other hand, there are a lot of possible targets. These targets, however, appear smaller when the distance increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which one of these, or other, forces will dominate? Would it make a difference if we ignored gravity?&lt;/p&gt;&#xA;" OwnerUserId="19427" LastActivityDate="2017-11-01T21:48:43.050" Title="What is a probability of a particle launched from earth hitting any object in the universe?" Tags="&lt;universe&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23369" PostTypeId="2" ParentId="23368" CreationDate="2017-10-31T15:17:45.140" Score="8" Body="&lt;p&gt;I can't give you an exact number, but the chances your photon will hit anything within the next several billion years is exceedingly small. Nay, infinitesimally small. Almost no chance that it will happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you're really asking, stated as a physics question is, &lt;em&gt;what is the mean free path of a photon in the present universe?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider an object (be it an electron, photon, or baseball) traveling through some medium (be it the Sun, space, or Earth's atmosphere), we can calculate how far, on average, it can travel before hitting something. This distance is known as the &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Mean_free_path&quot; rel=&quot;noreferrer&quot;&gt;mean free path&lt;/a&gt;&lt;/em&gt;. You want to know how far a photon can travel through our universe before it hits something, in other words, what is the photon's mean free path?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that a photon's mean free path in our universe is larger than the observable universe. What that means is your photon can travel from Earth to the edge of the observable universe and still not hit anything (on average).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To support this fact, I point you to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;noreferrer&quot;&gt;Cosmic Microwave Background&lt;/a&gt; (CMB) radiation. This is radiation which was produced more than 13 billion years ago in the early universe and has been traveling through the universe since, almost completely unimpeded. In fact, if you tally up all the photons in the universe, you'll find that &lt;a href=&quot;https://physics.stackexchange.com/questions/196366/does-the-number-density-of-photons-n-gamma-approx-108-mathrm-m-3-refer/196373#196373&quot;&gt;a very large fraction are CMB photons&lt;/a&gt;. This tells you that these photons, despite being produced 13 billion years ago and traveling through the universe since then, have never hit anything since and most are still traveling through space, unimpeded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are correct when you say that its the projected area of objects on our sky that matters, but you have to take into account the extremely minuscule projected area the matter in the rest of the universe has, primarily due to how far away it is.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2017-10-31T18:26:53.213" LastActivityDate="2017-10-31T18:26:53.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23370" PostTypeId="1" CreationDate="2017-10-31T18:31:15.817" Score="3" ViewCount="377" Body="&lt;p&gt;I am by no means an astronomy expert, having only taken 2 semesters in college. But I am an avid follower of news reports that look for signs of life on other planets. Today it was announced that 20 new planets &quot;with the potential for life&quot; have been found. Along with the usual signs (temperature and water) these planets were also cited because one of them has a year of 395 Earth days, a very similar year to our own. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm always bothered by these reports because they are all predicated on the premise that life on other planets will closely resemble our own. What if life on other planets doesn't need water? What if they are not carbon-based life forms? What if they can live in extremely hot (or extremely cold) temperatures?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It just seems like a massive mistake to (effectively) say, &quot;this planet doesn't look anything like ours, there must be no life there&quot; or to imply that only Earth qualities (or similar) can support life. Are there any fields of astronomy that look beyond the &quot;near-Earth&quot; life comparisons and look for life in other (drastically different) forms? &lt;/p&gt;&#xA;" OwnerUserId="19429" LastActivityDate="2017-11-01T10:36:59.730" Title="What if we are looking for the wrong signs of life on other planets?" Tags="&lt;planet&gt;&lt;life&gt;&lt;earth-like-planet&gt;&lt;origin-of-life&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23371" PostTypeId="2" ParentId="23353" CreationDate="2017-10-31T18:31:23.120" Score="-1" Body="&lt;p&gt;Based on info from &lt;a href=&quot;http://www.astro.caltech.edu/~george/ay20/Ay20-Lec7x.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;CalTech&lt;/a&gt;,  a constant density is unrealistic.  I wouldn't place much faith in any equations derived from such an assumption. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/50z5i.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/50z5i.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-10-31T18:31:23.120" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23373" PostTypeId="2" ParentId="23370" CreationDate="2017-11-01T06:12:00.417" Score="9" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;What if we are looking for the wrong signs of life on other planets?&lt;/p&gt;&#xA;&lt;p&gt;Are there any fields of astronomy that look beyond the &amp;quot;near-Earth&amp;quot;&#xA;life comparisons and look for life in other (drastically different)&#xA;forms?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I'm always bothered by these reports because they are all predicated&#xA;on the premise that life on other planets will closely resemble our&#xA;own.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Short answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;It doesn't matter what people think.   What matters is what we can observe.   Just because they will probably look for life on Earthlike planets first, doesn't mean they won't look elsewhere.  The answer will depend on what they find, not what anyone expects to find.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Long answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;I think there are a lot of good comments below your question, but I want to try to address a few points.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;On looking for life.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The scientists that look into space through our biggest telescopes aren't exactly &amp;quot;looking&amp;quot; for life on other planets because life is too hard to see.   Mostly, they look at stars.   Our best telescopes today &lt;a href=&quot;https://en.wikipedia.org/wiki/Methods_of_detecting_exoplanets&quot; rel=&quot;noreferrer&quot;&gt;can't see exoplanets at all&lt;/a&gt;.  Exoplanets are detected indirectly by variations in the stars they orbit.    If exoplanets are too far away most of the time to even be seen, looking for signs of life on them is obviously out of reach.   We're not actively looking for life because we can't.   We need much bigger telescopes for that.&lt;/p&gt;&#xA;&lt;p&gt;So, basically &amp;quot;20 Earthlike planets discovered&amp;quot; is the upper limit of our technology right now.  That's today's cutting edge discovery.   I understand that it might be boring.  I know that they're just adding more planets to the directory, but that's the limit that our best equipment can deliver right now.&lt;/p&gt;&#xA;&lt;p&gt;Almost nothing is known about all those planets, just the very basics, like how close they are to their star, and about how large they are and sometimes, about how dense they are giving us a limited idea of their composition and if they are hot, cold or perhaps in the Goldilocks zone.   If there's only 5 lines of data per exoplanet, the tag-line &amp;quot;Earth-like&amp;quot; often makes the best headline, not because NASA is focused on only Earthlike planets, but because there's not a lot of detail.&lt;/p&gt;&#xA;&lt;p&gt;It's kind of like &amp;quot;I see a bird, but it's far away and I don't have my binoculars.  It's flying south&amp;quot;, but I can't see what kind of bird, I can make out the size, to some extent, I can see that there's just one, but I can't make out the shape.   So, I write in my bird journal &amp;quot;One largeish bird flying south - too far away to see what kind&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;That's basically where we stand with exoplanet observation right now.&lt;/p&gt;&#xA;&lt;p&gt;This will change some when the &lt;a href=&quot;https://www.jwst.nasa.gov/&quot; rel=&quot;noreferrer&quot;&gt;James Webb Space Telescope&lt;/a&gt; is operating.   The JWST should be able to see star-light filtered through some exoplanet atmospheres, which will tell us a whole lot more than we currently know about some of those exoplanets.  Information such as how thick an atmosphere and what it's made of will tell us a lot, and give us some information on the climate of the planet.  JWST will be a HUGE (all caps for emphasis) step forward in the search for alien life, though it still won't observe life directly, but it will hopefully get a good read on some exoplanet's atmospheres, and from that, deductions can be made, such as abundant oxygen isn't expected without some form of photosynthesis.   Oxygen is too reactive to be likely to exist on it's own.&lt;/p&gt;&#xA;&lt;p&gt;A combination of CH4 and O2 could be especially interesting and one of the first things to look for, but they also have to be careful.  There's always the chance of more than one explanation for a planet's atmosphere.  We shouldn't expect any kind of guarantee of discovery, but JWST should be an enormous step forward in what we can observe about exoplanets.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;On reporting&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Scientists in this field, their job is essentially to analyze and report the data.   Their job is not, for the most part, to speculate on life but to simply report what can be verified (and today, relatively little other than &amp;quot;I found a planet&amp;quot; can be verified).&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Hypothetical_types_of_biochemistry&quot; rel=&quot;noreferrer&quot;&gt;Theoretical biology&lt;/a&gt; is great.  Don't get me wrong.  It's fun and it's real science, but the majority of telescope observation focuses on observe, analyze and report.  That might sound dry, but that's mostly how science works.  &amp;quot;20 Earthlike planets found&amp;quot; was impossible 10-15 years ago.  It really is cutting edge.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;On Theoretical Alien Life:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Scientists, today (&lt;a href=&quot;http://bigthink.com/videos/neil-degrasse-tyson-life-on-europa-jupiters-icy-moon&quot; rel=&quot;noreferrer&quot;&gt;Neil deGrass Tyson among them&lt;/a&gt;) who would love to get a closer look at some of the Moons of Jupiter and Saturn, because they are thought to hold the best chance of having life beyond Earth in our solar-system.   The desire to find alien life, if it's out there, is one of the highest held goals of NASA and Scientists in this field.   Nobody's ignoring or dismissing anything.  They will likely point JWST towards Earthlike planets first, but that's only natural.&lt;/p&gt;&#xA;&lt;p&gt;I'd like to also point a basic rule of astronomical observation.   &amp;quot;Weird&amp;quot; is actually good.   Weird leads to discoveries like dark matter and dark energy.  Like the Isaac Asimov &lt;a href=&quot;https://www.brainyquote.com/quotes/quotes/i/isaacasimo109758.html&quot; rel=&quot;noreferrer&quot;&gt;&amp;quot;That's funny&amp;quot;&lt;/a&gt; quote, finding things they don't expect is often the best find.   Scientists love that.   If weird atmospheres are found by JWST, it won't matter if they are on Earthlike planets or not.  Anything weird or unexpected will be of great interest and closely looked at.&lt;/p&gt;&#xA;&lt;p&gt;Like those theoretical Alien Mega-structure or &lt;a href=&quot;https://sciencealert.com/that-alien-megastructure-star-just-started-going-haywire-again-and-scientists-are-freaking-out&quot; rel=&quot;noreferrer&quot;&gt;Dyson sphere stars&lt;/a&gt; (which &lt;a href=&quot;https://www.space.com/32864-alien-megastructure-star-telescope-changes.html&quot; rel=&quot;noreferrer&quot;&gt;probably aren't Dyson spheres&lt;/a&gt;).  I'm highly skeptical that it's aliens, but I'm glad they're looking into it.  Anything unexpected is worth studying.&lt;/p&gt;&#xA;&lt;p&gt;When JWST starts to get better data on atmospheres, then you'll see more interesting headlines than &amp;quot;Earthlike&amp;quot;.  For now, and for the next year or so, &amp;quot;More Earthlike planets found&amp;quot;, is as far as our technology can take us.&lt;/p&gt;&#xA;&lt;p&gt;Finally, I want to point out that on theoretical biology, there are issues.  (The other answer covers this better than I did so I'll abbreviate), but there are reasons to suspect that life might need to be carbon and liquid water based, so planets thought to have liquid water should be the first places to look.&lt;/p&gt;&#xA;&lt;p&gt;And ultimately, it doesn't matter what people think is out there.   What matters is what JWST (and telescopes that follow) will discover.   What we know about the Universe is based on what has been observed, not on the various prejudices and assumptions people made.   Sometimes what is observed goes against the scientific belief of the day, and that's what's great about science.  Evidence trumps belief.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-11-01T10:24:44.543" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23374" PostTypeId="2" ParentId="23370" CreationDate="2017-11-01T06:43:16.317" Score="3" Body="&lt;ol&gt;&#xA;&lt;li&gt;We only have a sample of 1 - the Earth - for analysis or proposition of potential life-compatible conditions on other planets.&lt;/li&gt;&#xA;&lt;li&gt;Only carbon-based life can exist in this universe. Boron and Silicon are the next 2 elements for potential life, and neither have the ability to provide polymeric bonds required for complex biochemical assembly.&lt;/li&gt;&#xA;&lt;li&gt;Water is essential internally in chemical reaction, cellular construction and metabolism, vital elemental fluidity and other critical cellular operations. No life formation of longer-term survivability is possible without a water source.&lt;/li&gt;&#xA;&lt;li&gt;There are organisms that can live in environments that are extremely high temp (&gt; 150°C), highly acidic (&amp;lt; pH 1.0), cold temp (&amp;lt;0°C) high pressure (ocean bottom), and other stressful conditions, but there is no evidence that these extremophiles were the first organisms and all evaluations conclude they only survive within their specific environment, not transferrable to other environments.&lt;/li&gt;&#xA;&lt;li&gt;Carbon-based organisms have fixed limitations on their survivability based on the fossil record and the astrophysics, geology and biology knowledge that exists. These life characteristics place specific limitations and requirements on planets, stars, solar systems and galaxies on many levels and parameters. Elaborate and thorough examination of other potential life compatible extra-stellar bodies is necessary to determine the conditions that may allow life and the nature and specificity of conditions required for life on Earth reveal the levels and type of critical conditions that must be compatible. They are firm and extensive at many levels.&lt;/li&gt;&#xA;&lt;li&gt;Astrobiology, one of the fastest growing and scientifically lacking areas of science, examines the requirements of exoplanets to support life. The more the requirements are studied, the more the increase of understanding of the high specificity and wide range of life supportive conditions is revealed.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="19432" LastEditorUserId="2365" LastEditDate="2017-11-01T10:36:59.730" LastActivityDate="2017-11-01T10:36:59.730" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23376" PostTypeId="2" ParentId="23368" CreationDate="2017-11-01T21:48:43.050" Score="3" Body="&lt;h1&gt;&lt;a href=&quot;https://what-if.xkcd.com/109/&quot; rel=&quot;nofollow noreferrer&quot;&gt;xkcd worked this out!&lt;/a&gt;&lt;/h1&gt;&#xA;&lt;p&gt;It was part of a related question: what if you pointed an infinite power laser in random directions, how much damage would happen?  I'll try to summarize the parts of his analysis relevant to this question.&lt;/p&gt;&#xA;&lt;p&gt;If you choose a truly random direction, you'd have about a 50% chance of hitting the Earth. For the times you don't hit Earth, 89,999 times out of 90,000 your beam will pass out of the Milky Way without hitting anything. The objects you have the highest chance of hitting are the Sun or Moon, but the odds of hitting either of them is about 1 in 180,000. He goes on to show that your odds of hitting one of Jupiter's moons is about 1 in 1 trillion. Hitting a star is even harder, &amp;quot;even if you aim for the core [of the Milky Way].&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;He doesn't seem to have worked it out to the level of individual photons hitting individual hydrogen atoms, but given that space is mostly empty anyway the angular size method he uses should be a reasonable approximation.&lt;/p&gt;&#xA;" OwnerUserId="14255" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-11-01T21:48:43.050" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23378" PostTypeId="1" AcceptedAnswerId="23380" CreationDate="2017-11-02T10:01:36.620" Score="4" ViewCount="2239" Body="&lt;p&gt;I recently saw &lt;a href=&quot;https://i.imgur.com/EpuhHJa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;the below image&lt;/a&gt; circulating around twitter/facebook/reddit. It is titled &quot;Andromeda's actual size if it was brighter&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NIsYM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NIsYM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the &lt;a href=&quot;https://www.reddit.com/r/spaceporn/comments/1u0dxs/andromedas_actual_size_if_it_was_brighter/cedhp35/&quot; rel=&quot;nofollow noreferrer&quot;&gt;comments&lt;/a&gt; provides a link to &lt;a href=&quot;https://apod.nasa.gov/apod/ap130801.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;another comparison&lt;/a&gt; provided by APOD, and the following comment mentions that the first one I linked seems a bit off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the first image accurate in term of relative sizes? What is the relative angular size of Andromeda's galaxy and the Moon?&lt;/p&gt;&#xA;" OwnerUserId="7209" LastEditorUserId="7209" LastEditDate="2017-11-03T18:14:57.120" LastActivityDate="2017-11-03T18:14:57.120" Title="Is this photocomposition of Andromeda's galaxy and the Moon accurate?" Tags="&lt;the-moon&gt;&lt;galaxy&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23380" PostTypeId="2" ParentId="23378" CreationDate="2017-11-02T11:17:16.900" Score="8" Body="&lt;p&gt;The &lt;a href=&quot;http://astropixels.com/galaxies/M31-01.html&quot; rel=&quot;noreferrer&quot;&gt;Andromeda Galaxy's angular size&lt;/a&gt; is about 178x63 arc-minutes and the &lt;a href=&quot;https://lco.global/spacebook/using-angles-describe-positions-and-apparent-sizes-objects/&quot; rel=&quot;noreferrer&quot;&gt;Moon&lt;/a&gt; is about 31 arc-minutes wide. Relatively speaking, the Andromeda Galaxy's apparent size is nearly 6x2 Moons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So from the photo, their relative sizes in the sky appears to be very roughly accurate. The galaxy appears to have been superimposed from another image just to show what the galaxy might look like in the sky if it were brighter.&lt;/p&gt;&#xA;" OwnerDisplayName="user10106" LastEditorDisplayName="user10106" LastEditDate="2017-11-02T11:43:08.273" LastActivityDate="2017-11-02T11:43:08.273" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23382" PostTypeId="1" CreationDate="2017-11-02T15:00:05.640" Score="2" ViewCount="202" Body="&lt;p&gt;So, to add a bit of context, I’m doing the NaNoWriMo challenge, which consists of writing a novel of 50k words in one month. My whole idea is based around the following question: What would happen if the sun were to suddenly disappear?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make it short, the story is about a colony on a faraway planet, which is settled for few generations already. And one day or let’s say, in less than a month, the sun disappear. So far, so good. Except I need a valid reason as to why the sun disappear… I have found some ideas (dust from volcanic eruption, change of orbit, dying sun, etc…) but none of them looked good enough. (I’m open for suggestion if any.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m now investigating the possibility that the planet would gain a satellite big enough to hide the sun on the whole planet (like an eclipse but lasting forever). This &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6188/can-a-natural-satellite-exist-in-a-geostationary-orbit&quot;&gt;topic&lt;/a&gt; lead me to think that if the satellite is in geostationary orbit, it is possible. &lt;strike&gt;But there is still some questions I can’t get my head around.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How fast a new satellite can “appear”?&lt;/li&gt;&#xA;&lt;li&gt;Is it possible that a bigger but “lighter” (with less gravitational force) satellite can orbit around a smaller but “heavier” planet?&lt;/li&gt;&#xA;&lt;li&gt;This &lt;a href=&quot;https://astronomy.stackexchange.com/questions/22265/long-term-temperature-effects-of-an-eclipse&quot;&gt;topic&lt;/a&gt; reminded me that no sun means no heat. Can the geothermic activity of a planet be enough to prevent the surface from being impossible to live on?&lt;/strike&gt; (Off topic.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, basically, the question is, if it's possible to have a forever eclipse, how ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m sorry in advance if the questions are not clear enough, even though I understand the basics, I don’t necessarily have all the technical terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you !!&lt;/p&gt;&#xA;" OwnerUserId="19452" LastEditorUserId="19452" LastEditDate="2017-11-04T00:36:33.343" LastActivityDate="2017-11-04T00:36:33.343" Title="Is it possible that a satellite creates a forever eclipse?" Tags="&lt;the-sun&gt;&lt;planet&gt;&lt;solar-eclipse&gt;&lt;natural-satellites&gt;" AnswerCount="2" CommentCount="9" ClosedDate="2017-11-03T12:58:48.190" ContentLicense="CC BY-SA 3.0" />
  <row Id="23383" PostTypeId="2" ParentId="23382" CreationDate="2017-11-02T16:53:59.743" Score="6" Body="&lt;p&gt;The only potentially valid way to cause a long-lasting eclipse that I can think of is to have your moon orbit at the planet's L1 &lt;a href=&quot;https://en.wikipedia.org/wiki/Lagrangian_point&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lagrange point&lt;/a&gt; between the planet and the star (and the moon has to be the right size). If it had just the right orbit and was the right size, it could then cause a long-lasting eclipse. Note I'm not saying permanent because such a system wouldn't be &lt;em&gt;permanent&lt;/em&gt;. I'm not going to run through the math$^1$ to determine if such a system is stable or for how long. I'd guess it may be stable for a few orbits at least.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main problem is getting your moon there in the first place. One could conceivable say that by some freak chance a third fairly massive body came close to your planet/moon (either from outside the solar system or near the edge) and gravitationally perturbed your Moon's orbit such that it ended up in the L1 point. But the chances of such an event are so slim that it would be incredulous if it actually happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And just to add a technicality, once your moon moved to the L1 point, it would now be orbiting the Sun and wouldn't be a moon anymore according to the definition of a moon.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;$^1$ Out of curiosity, I ran through some of the math very loosely and found that if you put the Moon in Earth's L1 position, the Moon survives for about 3-5 months before perturbations knock it out of the L1 position. Of course our Moon isn't the right size to cause an eclipse at that distance, but this gives a &lt;em&gt;very&lt;/em&gt; rough estimate for how &quot;permanent&quot; the scenario I described above really is.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2017-11-03T13:26:15.700" LastActivityDate="2017-11-03T13:26:15.700" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23384" PostTypeId="2" ParentId="23382" CreationDate="2017-11-02T16:58:34.860" Score="-1" Body="&lt;p&gt;I'm afraid this hypothetical object blocking the sun would have to be artificial. If the object is in synchronous orbit, and if it is to perpetually eclipse the sun, then the planet would have to be tidally locked to the sun. If that's the case, then the object blocking the sun would have to be at the sun itself.&lt;br /&gt;&#xA;If you want to put an object in orbit around the sun that's large enough to block the sun's view from the planet, the object would have to have a delta-v lower than the amount required to sustain that orbit. Because of this, the object blocking the sun would require extra energy to sustain it's orbit in between the planet and the sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the three questions,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I guess it depends on the civilizations technological ability to detect it. Theoretically, nothing can truly be hidden in space. There should always be some method to detect anything, up until the inverse square law diminishes the information to nothing. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes. Mercury is smaller than Callisto, but has more mass than two Callisto's combined. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Depends on the kind of planet. If it is very volcanic, maybe? Anyhow, I wouldn't think it would last long. There's a lot of surface area on a planet to radiate heat out into space. It would get really cold, really fast. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Edit: I forgot about the Langrange points. The object could orbit on the point between the star and the planet, but I'd suspect the orbit would not be very stable for such a large object. At least, on a cosmic timescale. &lt;/p&gt;&#xA;" OwnerUserId="19454" LastEditorDisplayName="user1569" LastEditDate="2017-11-03T13:20:05.717" LastActivityDate="2017-11-03T13:20:05.717" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23385" PostTypeId="1" AcceptedAnswerId="23628" CreationDate="2017-11-02T20:48:16.810" Score="7" ViewCount="296" Body="&lt;p&gt;Pixel based detectors, particularly optical &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Charge-coupled_device&quot; rel=&quot;nofollow noreferrer&quot;&gt;CCDs&lt;/A&gt; like what was used in the &lt;A HREF=&quot;http://www.sdss.org/instruments/camera/&quot; rel=&quot;nofollow noreferrer&quot;&gt;SDSS camera&lt;/A&gt;, are ubiquitous in astronomy. Is there any dead area on the detectors? Not the obvious gaps between each individual sensor, but on the sensor itself. That is, does a typical detector have any gaps between the pixels from microscopic wires laid on the front, alternating doping regions in the silicon, or just areas where a photon can hit and the photo-electron is unlikely to be collected? What fraction of the area is dead?&lt;/p&gt;&#xA;" OwnerUserId="14244" LastEditorUserId="14244" LastEditDate="2017-11-11T02:06:57.410" LastActivityDate="2017-11-21T12:46:38.220" Title="What does the active area on a typical CCD based detector look like?" Tags="&lt;data-analysis&gt;&lt;detectors&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23390" PostTypeId="2" ParentId="23346" CreationDate="2017-11-03T16:24:20.803" Score="3" Body="&lt;p&gt;The reference magnitude and flux really depends on what filter (i.e. range of wavelengths) the observation was made at (e.g., see the &lt;a href=&quot;https://en.wikipedia.org/wiki/UBV_photometric_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;UBV system&lt;/a&gt;, or a more complete set of photometric passbands &lt;a href=&quot;https://en.wikipedia.org/wiki/Photometric_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;). If you know the filter, and it is in one of the common ultraviolet, optical, or near infrared filters, then &lt;a href=&quot;http://www.astronomy.ohio-state.edu/~martini/usefuldata.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this website&lt;/a&gt; and &lt;a href=&quot;https://archive.is/20121204144725/http://www.astro.utoronto.ca/~patton/astro/mags.html#conversions#conversions&quot; rel=&quot;nofollow noreferrer&quot;&gt;this website&lt;/a&gt; provide useful reference fluxes. Using the common &lt;a href=&quot;https://en.wikipedia.org/wiki/AB_magnitude&quot; rel=&quot;nofollow noreferrer&quot;&gt;AB&lt;/a&gt; magnitude system then a source with a measured magnitude of zero in any band by definition has a reference flux of 3631 Jy (where 1 &lt;a href=&quot;https://en.wikipedia.org/wiki/Jansky&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jansky&lt;/a&gt; = 10&lt;sup&gt;-26&lt;/sup&gt; W Hz&lt;sup&gt;-1&lt;/sup&gt; m&lt;sup&gt;-2&lt;/sup&gt;).&lt;/p&gt;&#xA;" OwnerUserId="16795" LastActivityDate="2017-11-03T16:24:20.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23392" PostTypeId="1" AcceptedAnswerId="23495" CreationDate="2017-11-03T23:08:50.140" Score="2" ViewCount="151" Body="&lt;p&gt;Not sure if this is the right StackExchange for this, but seems appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any off-the-shelf software that can display a star field (view of the night sky) as if seen from a particular location and looking in a particular direction?  Ideally, I'd like to be able to specify the location (Lat/Lon coordinates) and the direction in which the virtual camera is pointing, and get an image of the starfield.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ability to modify the parameters of the virtual camera would be of additional benefit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Free software would be best, but paid may be ok as well, depending on the price.&lt;/p&gt;&#xA;" OwnerUserId="19471" LastActivityDate="2017-11-12T11:45:38.433" Title="Software for simulating a night sky" Tags="&lt;star&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23393" PostTypeId="2" ParentId="23392" CreationDate="2017-11-03T23:39:27.647" Score="6" Body="&lt;p&gt;Yes. &lt;strong&gt;Stellarium&lt;/strong&gt; is a very nice package that meets your needs. I use it on my mac and it works great. There are versions for Windows and Linux as well. It is a free package. You can enter your location and see the night sky in any direction. It has zoom capability and you can rotate sky to look in any direction. Their website describes it as a planetarium for your computer. It has a database of over 600,000 stars. The website is &lt;a href=&quot;http://www.stellarium.org&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="16926" LastEditDate="2017-11-03T23:47:37.227" LastActivityDate="2017-11-03T23:47:37.227" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23396" PostTypeId="1" AcceptedAnswerId="23398" CreationDate="2017-11-04T13:38:46.490" Score="5" ViewCount="1197" Body="&lt;p&gt;If I get the numbers right, at about 17 Jupiter masses a mainly gaseous planet would ignite into some type of star. What about rocky planets? Is there a theoretical limit to their size, so that they can grow up to size x? (I'm imagining a &quot;liquid&quot; planet would congeal into rock after certain gravity value, and would somehow fit into the &quot;rocky class&quot; as well).&lt;/p&gt;&#xA;" OwnerUserId="1707" LastActivityDate="2019-09-26T11:43:35.097" Title="How big can a rocky planet get?" Tags="&lt;planet&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2017-11-04T19:38:24.520" ContentLicense="CC BY-SA 3.0" />
  <row Id="23398" PostTypeId="2" ParentId="23396" CreationDate="2017-11-04T17:45:29.813" Score="4" Body="&lt;p&gt;Giant planets have an abundance of H and He. Hydrogen burning occurs at relatively low temperatures. Rocky planets, on the other hand, don't have much hydrogen and even less helium. For them, the first relevant process would be C burning, which occurs at about 8 solar masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is valid only as a thought experiment. Rocky planets loose or don't accrete H/He mainly because of insufficient mass. If they were massive, they would consist of H/He and therefore &quot;ignite&quot; at 17 Jupiter masses.&lt;/p&gt;&#xA;" OwnerUserId="11461" LastEditorUserId="30487" LastEditDate="2019-09-26T11:43:35.097" LastActivityDate="2019-09-26T11:43:35.097" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23399" PostTypeId="1" CreationDate="2017-11-04T23:49:26.327" Score="1" ViewCount="458" Body="&lt;p&gt;If we got close enough to an elliptical galaxy would it look similar to a globular cluster?&lt;/p&gt;&#xA;" OwnerUserId="2757" LastEditorUserId="2365" LastEditDate="2017-11-05T08:50:35.640" LastActivityDate="2017-11-06T01:16:33.270" Title="Do elliptical galaxies look similar to globular clusters?" Tags="&lt;galaxy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23400" PostTypeId="2" ParentId="23399" CreationDate="2017-11-05T03:01:39.863" Score="0" Body="&lt;p&gt;It would look as shown below.&lt;a href=&quot;https://i.stack.imgur.com/vBY7j.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vBY7j.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/VibNy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VibNy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;And yes it would resemble a globular cluster except the stars move differently. Elliptical galaxies have an approximately ellipsoidal shape. Globular clusters are often associated with galaxies. The Milkey Way contains over 150 identified globular clusters. Shown below are globular clusters. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/WxDhn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WxDhn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;   &lt;a href=&quot;https://i.stack.imgur.com/oQqfi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oQqfi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="16926" LastEditDate="2017-11-05T13:19:23.063" LastActivityDate="2017-11-05T13:19:23.063" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23401" PostTypeId="1" CreationDate="2017-11-05T15:34:13.427" Score="1" ViewCount="1082" Body="&lt;p&gt;I have a dataset with events that are occurring over a period of 5 years. I am trying to determine whether these events are related to moon phase. I downloaded the &quot;fraction of moon illuminated&quot; for the time period, but am struggling to define the cutpoints for crescent, quarter, and gibbous. These are defined by the USNO as 0.25, 0.5, and 0.75 (&lt;a href=&quot;http://aa.usno.navy.mil/data/docs/MoonFraction.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://aa.usno.navy.mil/data/docs/MoonFraction.php&lt;/a&gt;) but I am seeking to divide the period into 8 bins of equal time. I think it is a sine function, where the period is 29.53059 days. If I divide this by 8, I get: 3.69132375 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me determine the fraction illuminated (y) that corresponds to these outpoints?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 days&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.69132375 days&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7.3826475 days&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11.07397125 days&lt;/p&gt;&#xA;&#xA;&lt;p&gt;14.765295  days&lt;/p&gt;&#xA;&#xA;&lt;p&gt;18.45661875  days&lt;/p&gt;&#xA;&#xA;&lt;p&gt;22.1479425  days&lt;/p&gt;&#xA;&#xA;&lt;p&gt;25.83926625  days&lt;/p&gt;&#xA;&#xA;&lt;p&gt;???&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="19472" LastActivityDate="2017-11-06T13:50:13.393" Title="Trying to categorize moon phase using fraction illuminated" Tags="&lt;the-moon&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23403" PostTypeId="2" ParentId="23401" CreationDate="2017-11-05T18:14:32.540" Score="0" Body="&lt;p&gt;The fraction of the moon illuminated does look like it could be defined by a sine/cosine function;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Period; 29.53059 days. Amplitude; 0.5. Y-Shift; 0.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Y = Acos(Bx)+C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Y = 0.5cos((2π/29.53059)(x-3.69132375/2))+0.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.69132375 is 0.96193976625 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;7.3826475 is 0.69134171618&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11.07397125 is 0.30865828381&lt;/p&gt;&#xA;&#xA;&lt;p&gt;14.765295 is 0.03806023374&lt;/p&gt;&#xA;&#xA;&lt;p&gt;18.45661875 is 0.03806023374&lt;/p&gt;&#xA;&#xA;&lt;p&gt;22.1479425 is 0.30865828381&lt;/p&gt;&#xA;&#xA;&lt;p&gt;25.83926625 is 0.69134171618&lt;/p&gt;&#xA;" OwnerUserId="19491" LastActivityDate="2017-11-05T18:14:32.540" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23405" PostTypeId="2" ParentId="23401" CreationDate="2017-11-05T22:13:07.693" Score="3" Body="&lt;p&gt;I used pyephem to calculate the fraction illuminated at each time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Program&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import ephem&#xA;m = ephem.Moon&#xA;d1 = ephem.Moon('2017') # d1 is the date of the first new moon in 2017&#xA;days = [3.69132375*i for i in range(1,8)]&#xA;for day in days:&#xA;    m.compute(d1+day)&#xA;    print(day, m.moon_phase)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Output&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3.69132375   0.1482372306829685&#xA;7.3826475    0.5259582805101828&#xA;11.07397125  0.8929835990614957&#xA;14.765295    0.9932503040700423&#xA;18.45661875  0.8004766309895268&#xA;22.1479425   0.4698580677294108&#xA;25.83926625  0.15043495343265506&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that this is dependent on the particular month. There is an asymmetry as the motion of the moon is not perfectly even, it moves in an elliptical orbit.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="11143" LastEditDate="2017-11-06T13:50:13.393" LastActivityDate="2017-11-06T13:50:13.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23406" PostTypeId="1" CreationDate="2017-11-05T23:26:08.497" Score="2" ViewCount="92" Body="&lt;p&gt;The title says it : I am looking for the word to use for a point, located on Earth surface (maybe an hypothetical perfectly spherical Earth), that has the same coordinates as a specific star, and therefore can be considered to be &quot;under&quot; the star in layman's terms.&lt;/p&gt;&#xA;" OwnerUserId="19497" LastActivityDate="2017-11-06T15:57:50.290" Title="Specific word for the point of the terrestrial surface &quot;under&quot; a star or other object in the sky" Tags="&lt;coordinate&gt;" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="23407" PostTypeId="2" ParentId="23399" CreationDate="2017-11-06T01:16:33.270" Score="1" Body="&lt;p&gt;Your question is not very well defined since you don't say what observational properties you wish to compare. From your emphasis on distance, I assume that you want to compare the visual appearance of elliptical galaxies with globular clusters of the &lt;em&gt;same angular extent&lt;/em&gt; on the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In which case ellipticals and globulars are easily distinguishable with the aid of a telescope (I assume you do not mean naked eye objects, since the nature of globular clusters is pretty difficult to make out with the naked eye, but if you do mean that then of course most smudges on the sky are indistinguishable with the naked eye). The reason for this is that true elliptical galaxies contain factors of $\sim 10^4$ more stars than even the most massive globular clusters and are about 100 times the physical diameter. Thus, if they had the same angular extent, an elliptical galaxy be $\sim 100$ times further away, would look much smoother and less grainy than a globular cluster, because it contains $10^4$ times more stars per unit area, but would have around the same surface brightness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there are such things as &lt;a href=&quot;https://en.wikipedia.org/wiki/Dwarf_spheroidal_galaxy&quot; rel=&quot;nofollow noreferrer&quot;&gt;dwarf spheroidal galaxies&lt;/a&gt;, but here the problem is the other way around. They have about the same number of stars as globular clusters, but have diameters a factor of 100 larger. Thus if they had the same angular extent they would look similar to a globular cluster in terms of &quot;graininess&quot;, but since they would be 100 times further away, they would be $10^4$ times fainter!&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-11-06T01:16:33.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23408" PostTypeId="1" CreationDate="2017-11-06T01:59:06.410" Score="4" ViewCount="122" Body="&lt;p&gt;Stars move relative to many frames of reference.  We can use the CBR for this question, I suppose, but it may not matter.  Their momentum p=mass x velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars lose mass as it is converted to energy. Percentage-wise it isn't much, but it is tons per second, so it isn't negligible.&#xA;What happens to the momentum of the star as it loses mass? Presumably it is conserved, so does the velocity of the star increase?  Or does the overall system carry off some momentum with the energy, leaving the stellar velocity constant?&lt;/p&gt;&#xA;" OwnerUserId="19498" LastEditorUserId="2365" LastEditDate="2017-11-06T23:09:18.420" LastActivityDate="2017-11-06T23:09:18.420" Title="What happens to the momentum of a star as its mass is converted to energy?" Tags="&lt;star&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2017-11-06T20:58:52.390" ContentLicense="CC BY-SA 3.0" />
  <row Id="23409" PostTypeId="2" ParentId="23408" CreationDate="2017-11-06T02:14:47.937" Score="2" Body="&lt;p&gt;Momentum is conserved. But it is the momentum of the system as a whole. The energy released by nuclear fusion escapes as a result of neutrinos and radiation. If these emissions are isotropic (the same in all directions), or at least with no hemispheric asymmetry, then NO net momentum is removed from the star, it is not accelerated and therefore its velocity with respect to any reference frame remains unchanged. It's momentum in that rest frame &lt;em&gt;is&lt;/em&gt; then reduced by the percentage of mass that it has lost. The lost momentum exists in the escaping radiation and neutrinos which will appear anisotropic in any frame of reference other than the rest frame of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice most radiation loss is fairly symmetric, if not spherically, then at least not in the sense that more radiation does not emerge from one hemisphere. I am not sure about neutrinos, but I see no reason for asymmetric emission and these are a fairly minor component in any case.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-11-06T02:14:47.937" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23411" PostTypeId="2" ParentId="23406" CreationDate="2017-11-06T10:34:09.457" Score="1" Body="&lt;p&gt;Positions are usually given relative to the observer, not relative to the object. So instead of saying &quot;I am under the star X&quot;, you would say &quot;star X is overhead&quot;. The word for an object directly overhead is zenith: &quot;the star X is at the zenith&quot;.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-11-06T10:34:09.457" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23413" PostTypeId="2" ParentId="23401" CreationDate="2017-11-06T11:04:59.427" Score="0" Body="&lt;p&gt;The Moon's phase is naturally divided into 4 segments, not 8.&#xA;0 to 50% illuminated: waxing crescent&#xA;50 to 100%; waxing gibbous&#xA;-100 to -50%: waning gibbous&#xA;-50 to 0%: waning crescent&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the negative sign is used to indicate waning phases. Likewise the 4 phases of the Moon are&#xA;0% New Moon&#xA;50% First Quarter&#xA;100% Full Moon&#xA;-50% Last Quarter (or Third Quarter)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I did not see the values 25, 0.5, and 0.75 on the page that you referenced.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-11-06T11:04:59.427" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23417" PostTypeId="5" CreationDate="2017-11-06T15:08:59.700" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2017-11-06T15:08:59.700" LastActivityDate="2017-11-06T15:08:59.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23418" PostTypeId="4" CreationDate="2017-11-06T15:08:59.700" Score="0" Body="This is a special tag that is designed to be used only by the system for questions that have had all of their other tags removed. Do not add this tag to existing questions, it is reserved for use by the system to identify posts with no valid tags." OwnerUserId="7472" LastEditorUserId="7472" LastEditDate="2017-11-06T20:56:29.807" LastActivityDate="2017-11-06T20:56:29.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23419" PostTypeId="2" ParentId="23406" CreationDate="2017-11-06T15:57:50.290" Score="1" Body="&lt;p&gt;The point on the earth where the sun is directly overhead is commonly called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Subsolar_point&quot; rel=&quot;nofollow noreferrer&quot;&gt;subsolar point&lt;/a&gt;.  I've also heard sublunar point which would be the point corresponding to the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not aware of any more general term for this point relating to any other astronomical objects.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2017-11-06T15:57:50.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23420" PostTypeId="1" AcceptedAnswerId="23421" CreationDate="2017-11-06T16:00:44.267" Score="2" ViewCount="301" Body="&lt;p&gt;I would like to know the name of this eye-shaped galaxy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.wired.com/2009/07/eyegalaxy/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Strange eye-shaped galaxy has black-hole iris, Betsy Mason, 23.07.2009, Wired.com&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jygsg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jygsg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19484" LastEditorDisplayName="user1569" LastEditDate="2017-11-06T20:55:44.850" LastActivityDate="2017-11-06T20:55:44.850" Title="Whats the name of this eye-shaped galaxy?" Tags="&lt;naming&gt;&lt;definition&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23421" PostTypeId="2" ParentId="23420" CreationDate="2017-11-06T17:34:05.717" Score="4" Body="&lt;p&gt;That is galaxy NGC 1097, a barred spiral galaxy in the constellation Fornax. It has been distorted by interactions with neighbouring galaxies and has a weakly active galactic nucleus, with jets coming from the nucleus. The jets contain stars from a galaxy that has recently (in astronomical terms) been cannibalized. The ring in the centre is a region of star formation, surrounding the nucleus of the galaxy at about 5000 light years.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-11-06T17:34:05.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23423" PostTypeId="1" CreationDate="2017-11-07T00:51:19.550" Score="5" ViewCount="210" Body="&lt;p&gt;If you were orbiting a star undergoing a core-collapse supernova, what would be the delay between the neutrino pulse from the collapse and the first visible effects on the surface? Basically, I'm trying to get a scientifically-plausible gauge of the warning time given by said neutrino pulse.&lt;/p&gt;&#xA;" OwnerUserId="19515" LastActivityDate="2017-11-07T07:27:15.707" Title="Delay between neutrino pulse and visible flash of supernova" Tags="&lt;astrophysics&gt;&lt;supernova&gt;&lt;neutrinos&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23424" PostTypeId="2" ParentId="21782" CreationDate="2017-11-07T01:59:28.570" Score="3" Body="&lt;p&gt;Not an expert, but offer one solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The adaptive optics correction is accomplished by a tip-tilt mirror and a deformable mirror. Usually the atmospherical wavefront distortion (or say, the phase screen) looks like this: &#xA;&lt;a href=&quot;https://i.stack.imgur.com/AbDD1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AbDD1.jpg&quot; alt=&quot;phase screen&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note its ramp-like shape. In other word, the 2nd and 3rd Zernike terms are large. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deformable mirror has limited correction range: it deals well with the higher-order Zernike terms, but not the 2nd and 3rd Zernike terms. This is why the tip-tilt mirror is here: it introduces a general ramp correction to suppress the 2nd and 3rd Zernike terms, then the afterwards wavefront contains only the higher order Zernike terms, and hence can be corrected well by the deformable mirror.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need a little Fourier optics knowledge to fully understand. Briefly speaking, The image $I$ (this is also called the point-spread-function, the linear system response of a Fourier optical system) captured on the sensor follows: &#xA;$$I(\frac{x}{\lambda f}) = |\mathcal{F}\{p(y)*e^{j (2\pi/\lambda) w(y)}\}|^2,$$&#xA;where $x$ and $y$ are the coordinates on the sensor and the aperture, $\lambda$ the wavelength, $f$ the lens focal length, $\mathcal{F}$ denotes Fourier transform, $p$ and $w$ are the aperture shape and the distortion wavefront, respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplity let $p(y) = 1$ for every $y$ (i.e. infinity aperture size). If $w(y) = 0$, then $I(x)$ is a perfect Dirac function. This simply says your star has no blurring. To the contrary, for example $w(x) = \alpha x^2$ for some $\alpha$, then it results in a defocused blurring point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same formula above also helps answering your additional question. Increasing the aperture size (the main mirror will finally be mapped to the aperture) is equivalent to, in the Fourier domain, decreasing the blurring size.&lt;/p&gt;&#xA;" OwnerUserId="16553" LastEditorUserId="16553" LastEditDate="2018-06-03T17:26:37.090" LastActivityDate="2018-06-03T17:26:37.090" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23425" PostTypeId="2" ParentId="23423" CreationDate="2017-11-07T02:29:19.250" Score="7" Body="&lt;p&gt;The neutrino pulse associated with the core-collapse supernova SN1987A in the Large Magellanic Cloud, were estimated to have arrived at the Earth about 3 hours before the rise of the supernova light curve (e.g. the review by &lt;a href=&quot;http://thucydides.sjca.edu/~beall/astrophysics/beall_2006_chjaa_neutrinos_sn.pdf&quot; rel=&quot;noreferrer&quot;&gt;Beall 2006&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If neutrinos are massless or have small masses (which they do), then this is your answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they have masses (and they do), then the answer &lt;em&gt;could&lt;/em&gt; depend on the neutrino mass and how far away you are from the supernova, since particles with mass will travel slower than the speed of light. Therefore the &quot;delayed&quot;, but faster, optical emission will catch-up with the neutrinos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The neutrino masses are probably of the order of &lt;a href=&quot;https://en.wikipedia.org/wiki/Neutrino#Mass&quot; rel=&quot;noreferrer&quot;&gt;0.1 eV&lt;/a&gt;, whilst the energies of neutrinos from core-collapse supernovae are around 30 MeV in energy. i.e. The neutrinos travel with a Lorentz factor $\gamma = (1-v^2/c^2)^{-1/2} \sim 3\times 10^{7}$. Thus&#xA;$$\frac{1}{1 - v^2/c^2} \sim 9\times 10^{14}$$&#xA;and so $v/c \sim 1 - 1/1.8\times 10^{15}$. That is, the neutrinos probably travel at about 1 part in $10^{15}$ slower than the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the distance of the Sun, this amounts to a delay of $5 \times 10^{-13}$ s, at the distance of the Andromeda galaxy about  0.006s, but for a galaxy at a distance of a Gpc, the delay would be about 100 s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So actually, unless you are detecting the supernova on the other side of the observable universe, it makes no odds and certainly makes no difference at the distance of the Large Magellanic Cloud. So I say your answer is about 3 hours.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-11-07T04:02:21.780" LastActivityDate="2017-11-07T04:02:21.780" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23428" PostTypeId="1" CreationDate="2017-11-07T05:30:37.820" Score="10" ViewCount="3295" Body="&lt;p&gt;Surly the moon is not projecting a direct beam of light which only lights up the small area of clouds around the moon. What phenomenon is happening that only allows us to see that center focus glow and not all the clouds glowing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IGGuS.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IGGuS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19519" LastEditorUserId="1891" LastEditDate="2017-11-09T06:44:50.433" LastActivityDate="2020-09-30T03:10:08.947" Title="Why do we only see the clouds directly infront of the moon glow from moon light?" Tags="&lt;the-moon&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23432" PostTypeId="1" AcceptedAnswerId="23434" CreationDate="2017-11-07T07:27:37.527" Score="3" ViewCount="566" Body="&lt;p&gt;I have heard many times that this planet is gaseous like jupiter this planet is like super earth like rocky planet etc. so is there any planet which is mix of gase and rock ?&lt;/p&gt;&#xA;" OwnerUserId="14136" LastEditorUserId="14136" LastEditDate="2017-11-14T09:01:07.133" LastActivityDate="2017-11-14T09:01:07.133" Title="Is there any planet out there which is half gaseous and half rocky ? is this possible?" Tags="&lt;planet&gt;&lt;exoplanet&gt;&lt;planetary-formation&gt;&lt;gas-giants&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23433" PostTypeId="1" AcceptedAnswerId="23435" CreationDate="2017-11-07T08:07:04.067" Score="10" ViewCount="912" Body="&lt;p&gt;I was reading about gigantic storms on giant gas planets, the great red spot on Jupiter and hexagon on Saturn to name a few, how about our Sun which is consist of plasma(hot gas which have some of their electrons stripped) why don't we see any big hurricane? &lt;/p&gt;&#xA;" OwnerUserId="6760" LastActivityDate="2017-11-08T03:59:37.950" Title="Why is there no hurricane on the Sun?" Tags="&lt;the-sun&gt;&lt;weather&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23434" PostTypeId="2" ParentId="23432" CreationDate="2017-11-07T09:04:37.037" Score="5" Body="&lt;p&gt;Probably.   Relatively little is known about exoplanets because they're very hard to get a good look at, but there's no reason why a rocky world couldn't accumulate enough ices and/or gas to also resemble a gas giant.   Now half rocky half gas giant (hydrogen/helium) might be rare.   Half rocky half &quot;ices&quot; is very possible and those have likely already been observed.  It helps to understand planet formation and elemental abundance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's likely a size limit to rocky planets with thin atmospheres because mass tends to accumulate and retain atmosphere and with hydrogen and helium being the most abundant gases in the universe, once a planet gets massive enough, it should collect and retain hydrogen and helium and begin to resemble Jupiter or Saturn, even if it started out as a rocky world.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rocky worlds, need elements like Magnesium, Silicon, and Iron, often bound with oxygen. &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth#Chemical_composition&quot; rel=&quot;noreferrer&quot;&gt;close to 90% of the mass of rocky worlds&lt;/a&gt; in our solar system comes from those 4 elements and 90% isn't a bad estimate for rocky exoplanets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on our &lt;a href=&quot;https://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements#Solar_system&quot; rel=&quot;noreferrer&quot;&gt;solar-system:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron makes up about 0.117%, Silicon 0.065% and and Magnesium 0.051%, for about .234% of the mass of the solarsystem.  Oxygen bound to those elements, using Earth's composition as an estimate adds about another 50%, up to about 0.35%.   Hydrogen and Helium make up about 97.6%, which leaves about 2% of the mass of the solar system in the form of ices and heavier gases, not hydrogen and helium.  Those are primarily water, ammonia, CO2 and Methane, with smaller amounts of other gases/ices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These numbers are rough but good enough for an estimate.   In our solarsystem there's about 6 times as much ice and heavier gases by mass than there is rocky material and there's close to 40 times as much hydrogen and helium as the other elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets need to be quite large to &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_escape&quot; rel=&quot;noreferrer&quot;&gt;hold hydrogen and helium&lt;/a&gt;.   Jupiter and Saturn are the only planets in our solar-system that are hydrogen and helium abundant.    Even the smaller gas giants, Uranus and Neptune have comparatively little hydrogen and helium.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there's probably no such thing as a small gas giant.   Gas giants need to be large or they don't exist at all.  (I should make a footnote that as a gas giant loses atmosphere, you can get a small gas giant, but that would be in a transition phase).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't like the term &quot;Ice giants&quot;, though it's often used to define Neptune and Pluto because they are primarily ices (water, methane, CO2, NH3), not primarily hydrogen/helium.     I don't like calling them &quot;ice giants&quot; when they can also be hot, so I prefer the term Neptune like planets or Neptunes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uranus (a Neptune type planet) is mostly made up of ices and heavier gases with an estimate of just &lt;a href=&quot;https://www.universetoday.com/18870/what-is-uranus-made-of/&quot; rel=&quot;noreferrer&quot;&gt;3%-10% hydrogen and helium&lt;/a&gt;.  Estimates of about 70%-90% of ice/not hydrogen/helium gas and the remaining, about 7%-20% more dense material.   That would be some other elements, perhaps some solidified carbon (diamonds), higher sulfur and lower Iron by percentage than Earth as well as some Earth like silicates. magnesium-oxides, Iron and Iron-oxides, though the high internal temperature might not leave much in the way of chemical bonds towards the center of the planet where temperatures reach 9,000 degrees C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neptune is similar to Uranus but more massive and more dense.  It's thought to have more water and perhaps a larger internal mantle, though like Uranus, the majority of it's mass is ices and heavy gases (not hydrogen/helium).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neptune and Uranus have hydrogen in the form of Methane, Ammonia and Water, but I'm counting that hydrogen as part of their &quot;heavy gases/ices&quot;.   They both have a comparatively low percentage of free hydrogen (3%-10%) making a clear distinction between Neptune like planets and gas giants like Jupiter and Saturn &lt;a href=&quot;https://www.space.com/18472-what-is-saturn-made-of.html&quot; rel=&quot;noreferrer&quot;&gt;with which the majority of their mass is hydrogen&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It helps to think of how planets form.   Smaller planets have too little gravity to retain hydrogen and helium gas and solar-systems are mostly too warm for hydrogen to freeze, so planets are made up of solid material for the most part, either rocky or ices that clump and stick together during formation.   When they have enough gravity, then they can begin to hold onto an atmosphere and further from the star makes this easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon Titan, as an example, is small to have an atmosphere, but it's quite far from the sun and largely made up of ices on it's surface, so as the ice on it's surface melts, it basically replenishes it's atmosphere.  Titan is still out-gassing, and it losing atmosphere, but it loses atmosphere slowly enough that it retains an atmosphere, but in a sense, it's in a transition, where it's atmosphere is outgassed.  When it runs low on surface ice that readily thaws, it should begin to resemble the icy moons of Jupiter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, back to your original question.   Many combinations are possible, some are not.   Baby Jupiters (the mass of Neptune) are probably unlikely, though a gas giant close to it's star that's lost a lot of it's atmosphere could resemble a baby Jupiter, but like Titan, that would be transitional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rocky Jupiters are unlikely because hydrogen outnumbers rocky material by over so much, over 200 to 1 in most of the milky-way.   Now, some solarsystems are likely more &quot;metalic&quot; than others, so that ratio will have some variation, but it's once there's enough mass for a gas giant to form, hydrogen is likely to be the abundant element and Helium #2 and any rocky core would be dwarfed by the hydrogen and helium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rocky Neptune however, no reason why not.   A planet with 8 earth masses of Earth like material and 8 Earth masses of ices (and 5%-10% hydrogen) would basically be a rocky Neptune.   It would likely be a little smaller and certainly denser, but not all that different in outward appearance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water worlds is a common term that might qualify as a &quot;rocky-neptune&quot;.   We don't know exactly what they are made of, but density estimates suggest a high percentage of water (and presumably CO2, CH4, NH3), which is the majority of Uranus and Neptune.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a chart of exoplanets of less than 20 Earth masses.   I suspect there's a considerable margin of error in these estimates, but it more or less agrees with the rocky-neptune argument.   Planets with 2 or 3 grams per cc would be in that range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See chart:  www.hpcf.upr.edu/~abel/phl/hec_plots/exoplanet_df.png&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://phl.upr.edu/library/notes?tmpl=%2Fsystem%2Fapp%2Ftemplates%2Fprint%2F&amp;amp;showPrintDialog=1&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Baby Neptunes of just 1 or 2 Earth masses might be possible too, but they'd probably need to be quite cold and far from their star or they'd be in danger of losing their atmosphere.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iceworlds like Pluto can get quite small, but Pluto has very little atmosphere.  Planets generally need to be fairly large to retain their atmosphere.   If they are too cold, that atmosphere freezes.   Titan, as I mentioned above, is in what could be called a slow transition where it's atmosphere comes from it's surface and it loses it slowly.   Ceres, based on it's density, is an icy moon like object too, though it's lost nearly all it's surface ice, but it probably has a lot of water/&quot;ices&quot; below it's surface.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-11-07T09:04:37.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23435" PostTypeId="2" ParentId="23433" CreationDate="2017-11-07T14:22:29.730" Score="12" Body="&lt;p&gt;We see sunspots, which are giant, planet-sized storms on the surface of the Sun. There are however many differences between sunspots, the Great Red Spot on Jupiter, and tropical cyclones (e.g., hurricanes) on the Earth. Tropical cyclones are low pressure systems fueled by evaporation of warm ocean water and sustained by the Earth's somewhat rapid rotation rate. The Great Red Spot is a high pressure system that is sustained by Jupiter's quite rapid rotation rate. Sunspots are low temperature systems fueled by the Sun's magnetic field and carried along by the Sun's rather low rotation rate.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-11-07T14:22:29.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23436" PostTypeId="2" ParentId="23432" CreationDate="2017-11-07T16:57:02.010" Score="3" Body="&lt;p&gt;I think @userLTK answer is pretty complete, but I would like to add one from a different perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomers, as all humans, tend to categorize things we see using words. This is necessary for fast communication but sometimes there are objects that are at the edge of one category and another (think in a brown dwarf).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reality is pretty analogic at big scales and we tend to cut it, &quot;digitalize&quot; it in order to understand what are we talking about, but almost always there will be examples in the limits from one categorize to the next, or even superposition, having a same object being in two categories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed, the amount of gas a planet has depends primarily on its mass. At the same time, there is one category of planets -by mass- missing in our solar system: super-earths (and probably mini-neptunes... again, this is about what granularity we want to use in definitions to a proper communication, and the bounds of each category. I won't get into that).&#xA;So if Earth has 1 earth mass and Neptune-Uranus about 17-15 earth masses, you can easily imagine a planet having the right mass in the middle of those for being half rock and half gas. Maybe 3, 5 or 8 Earth masses, no idea about the exact mass for the exact 50%. It also depends on its formation and evolution.&lt;/p&gt;&#xA;" OwnerUserId="19337" LastActivityDate="2017-11-07T16:57:02.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23437" PostTypeId="1" AcceptedAnswerId="23447" CreationDate="2017-11-07T17:06:07.253" Score="5" ViewCount="347" Body="&lt;p&gt;So every four years, we add an extra day. A leap year. This is because a year is actually $365_{1/4}$ days long. But what happens between leap years? After two years, it would get dark at 8:00 AM instead of 8:00 PM. This is obviously not the case. What's going on here?&lt;/p&gt;&#xA;" OwnerUserId="19532" LastActivityDate="2017-11-11T07:37:53.187" Title="Why isn't the day backwards between leap years?" Tags="&lt;earth&gt;&lt;time&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23438" PostTypeId="2" ParentId="23433" CreationDate="2017-11-07T20:30:15.920" Score="10" Body="&lt;p&gt;In addition to the answer above, spiraling storms like hurricanes or the great red spot, are quite orderly and require the right conditions and energy transfer.  The great red spot keeps relatively consistent latitude and it's been there for centuries, so it's obviously stable and ordered, though it may be shrinking.  The cause of the great red spot isn't known, but efficient heat transfer of Jupiter's vast internal heat, and the principal of hot internal gas rising and cool surface gas falling and Jupiter's very strong Coriolis effect likely helped create and maintain it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For hurricanes on Earth, a few specific things to happen.   There needs to be an energy source to sustain them, which is why they only form over warm oceans, mostly during the summer and fall seasons when oceans are warmest.   The rapid evaporation of warm ocean water feeds the hurricane and the condensation of that evaporated water vapor in the upper atmosphere, drives the low pressure system.   The spiral is the most efficient form of heat transfer and of light air rising/warm air falling.  The high speed surface winds increase the evaporation rate over the ocean, so once the spiral forms and stabilizes, it's self sustaining, until it drifts over colder water or land.  Hurricanes are orderly with very efficient heat transfer and ordered layers of rising and falling air. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z4XRB.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z4XRB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over 90% of tropical depressions don't become hurricanes.   Generally speaking, a perpendicular direction between the cool air above and the warm air below is required to get the spiraling wind started.  That's partly why the IPCC has previously predicted a possible decrease in hurricane formation, because the formative conditions need to be just right and a more turbulent upper atmosphere might decrease hurricane formation even though the warmer oceans works in the opposite direction.   All this was footnoted with some uncertainty and predicting changes in wind direction is tricky, so it shouldn't be held against the IPCC.  The point is, Hurricanes need the right balance.  They don't form easily, though once formed they tend to stabilize and grow, until they drift off the warm ocean water that feeds them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air is also quite light, and the heat energy transfer of water's phase change is significant enough to create the 100 plus mph winds in an orderly spiral.  On both Jupiter and Earth the right conditions are met for large, high wind speed, spiral storm formation.  Like Earth, Jupiter also has clouds and rain, both water and ammonia, which likely assist in it's heat transfer by phase change (though I'm nowhere near smart enough to say how much that contributes regarding Jupiter's red spot, on Earth the phase change of water is essential for hurricane formation.   Without abundant warm surface water - no hurricane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun, by comparison, is all plasma.  There's no phase change that efficiently increases the transition of heat and energy, though there are probably variations in ionization, but I'll get to that later.   The Sun's surface is also quite chaotic and it has magnetic storms, making the neat birth of a spiral storm by perpendicular wind gusts one above the other, less likely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnetic storms are twisted and I don't want to say that nothing spirals or twists on the surface of the sun, because that's not true.  But the magnetic storms on the surface of the sun aren't like the neat and tidy cone shaped spirals of hurricanes.  They reach well above the sun's atmosphere, not in the atmosphere, and the shape is different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Cl5nl.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cl5nl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the material that makes up the Sun's &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_transition_region&quot; rel=&quot;noreferrer&quot;&gt;transition region&lt;/a&gt; or &quot;atmosphere&quot; isn't good for hurricane formation.   To quote from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Below, most of the helium is not fully ionized, so that it radiates&#xA;  energy very effectively; above, it becomes fully ionized. This has a&#xA;  profound effect on the equilibrium temperature (see below).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Below, the material is opaque to the particular colors associated with&#xA;  spectral lines, so that most spectral lines formed below the&#xA;  transition region are absorption lines in infrared, visible light, and&#xA;  near ultraviolet, while most lines formed at or above the transition&#xA;  region are emission lines in the far ultraviolet (FUV) and X-rays.&#xA;  This makes radiative transfer of energy within the transition region&#xA;  very complicated.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Below, gas pressure and fluid dynamics usually dominate the motion and&#xA;  shape of structures; above, magnetic forces dominate the motion and&#xA;  shape of structures, giving rise to different simplifications of&#xA;  magnetohydrodynamics. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The transition region itself is not well studied in part because of&#xA;  the computational cost, . . .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hurricanes could theoretically form as a result of fluid dynamics, but the rapid rate that partially ionized helium radiates heat makes the formation of large circulating structures, which are basically engines of convection, impractical and unnecessary.   There's no need for efficient convection when the energy transfer is very efficient by radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun's atmosphere isn't like the atmosphere of the Earth and in the upper layers of Jupiter where the atmosphere is fairly effective and holding onto it's heat (We wouldn't have warm and cold fronts if it wasn't).  Those regions of warm and cold air that mostly maintain their temperature drive the convective process.   You need jets of warm and cold air to flow past each other in a hurricane.  The Sun's efficient radiation of partially ionized helium works against that principal.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also a relatively low Coriolis effect on the surface of the sun, which assists with the formation of hurricanes.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, the conditions are not at all right.   The Sun's turbulence, it's relatively low rotation rate, no phase-change to feed the system and it's partially ionized helium in it's lower &quot;atmosphere&quot;, all work against the formation of spiraling, cone shaped, high speed wind systems.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Brown dwarfs with much cooler surface temperatures, hurricanes might be entirely possible.   The math behind atmospheric convection mechanisms is complicated, so this is more of a general explanation but the Sun isn't a good candidate for hurricanes on many levels.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-11-07T22:10:13.887" LastActivityDate="2017-11-07T22:10:13.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23439" PostTypeId="2" ParentId="23437" CreationDate="2017-11-08T00:16:38.030" Score="5" Body="&lt;p&gt;Leap seconds keep the clock in sync with solar time of day. Leap years keep the calendar in sync with the seasons. The two (leap seconds and leap years) have nothing to do with one another. We have leap seconds because a day is now a tiny bit longer than 86400 seconds. We have leap years because a year (tropical year) is about 365.24219 days long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we didn't have leap years but did have leap seconds, solar noon and noon by the clock will continue to occur at more or less the same time (ignoring cyclical changes such as the equation of time and daylight saving time). The calendar would become out of sync, becoming almost a month off from the seasons in just a century. Adding an extra day every four years doesn't quite do the trick; this is why 2100, for example, will not be a leap year.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-11-08T00:16:38.030" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23445" PostTypeId="2" ParentId="12095" CreationDate="2017-11-08T16:25:41.803" Score="2" Body="&lt;p&gt;For orbital elements (and more), I can't think of a better source than the &lt;a href=&quot;http://minorplanetcenter.net/web_service&quot; rel=&quot;nofollow noreferrer&quot;&gt;web service&lt;/a&gt; offered by the Minor Planet Center. Not everything you are asking for comes out of the box, but you can compute them from orbital elements.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;MPC Web Service&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The MPC is providing a web service interface to users, allowing them to programmatically fetch minor planet properties data from the MPC's database. Data can be returned in either JSON or XML format. The default format is XML, and will be used unless JSON format is specified as a parameter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A &lt;a href=&quot;https://gitlab.com/tostt/mpc-rest-client&quot; rel=&quot;nofollow noreferrer&quot;&gt;Java interface&lt;/a&gt; to this service exists on GitLab.&lt;/p&gt;&#xA;" OwnerDisplayName="user15104" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-03-04T08:28:02.053" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23446" PostTypeId="2" ParentId="23428" CreationDate="2017-11-08T16:40:55.307" Score="8" Body="&lt;p&gt;This is an experiment that you can do at home:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Put a paper fixed in front of a light bulb&lt;/li&gt;&#xA;&lt;li&gt;Move your head around and see how the bright spot on the paper accompany your moves.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But, why is that? How is it that not all the paper is lighted the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When light travels through a translucent material some of it gets reflected back, some bounces inside of it and some goes through. As always the shortest path is the easiest, so the light coming straight &quot;bulb-paper-eye&quot; gets through easily (brightest spot) and as you go further from the center you are not seeing direct light anymore, but diffused light, so it gets dimmer and dimmer because to get to your eye it has to bounce in the right direction. That doesn't imply that the rest of the paper is not getting light, it is just not bouncing in your direction and the same happens with the clouds.&lt;/p&gt;&#xA;" OwnerUserId="19545" LastActivityDate="2017-11-08T16:40:55.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23447" PostTypeId="2" ParentId="23437" CreationDate="2017-11-08T17:39:32.990" Score="5" Body="&lt;p&gt;As a simplification, think of the length of the year as where the Earth is in its orbit. After 1 year of 365 days, the Earth has not returned to the same spot in its orbit. If we did not have leap years, then the date when seasons occur would change. Leap years keep the seasons in synch with the date. After 365+365+365+366 days (that is, 3 regular years and 1 leap year = 365.25*4), the Earth is at the same spot in its orbit. (This is approximately true because the year is not exactly 365.25 days long).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, when it gets dark is related to the length of the day, and that is independent of the length of the year. In order for it to get dark at 8 am, you would need to reset your clocks to 0 after 1 exact orbit (365.25 days).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for a little more detail. There is a slight difference between a year measured &quot;when the Earth returns to the same spot in its orbit&quot; as I described above and a year measured &quot;from one vernal equinox to the next&quot;. (The first definition is the sidereal year, and the second definition is the tropical year). The &lt;a href=&quot;http://astro.unl.edu/naap/motion3/sidereal_synodic.html&quot; rel=&quot;noreferrer&quot;&gt;&quot;Sidereal vs. Synodic Motions&quot;&lt;/a&gt; webpage explains why the sidereal year is about 20 minutes longer than the tropical year. Our calendar is based on the tropical year. So after 1 tropical year (365.242 mean solar days), the Earth still has not returned to the same spot in its orbit! Details, details, &lt;strong&gt;details&lt;/strong&gt;!&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-11-08T17:39:32.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23448" PostTypeId="2" ParentId="23428" CreationDate="2017-11-08T21:17:59.123" Score="5" Body="&lt;p&gt;The light is scattered by the particles in the clouds. These particles are relatively large, compared to the wavelength of light, and so the type of scattering is called &quot;Mie Scattering&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light that hits a cloud particle may be scattered in any direction, but it is much more likely to be scattered forwards than than to be scattered to the side. What this means is that around the moon the clouds appear lit up, to the side the clouds appear dark. The amount of scattering doesn't depend much on the colour of the light, so the scattered light appears the same colour as the moon (ie white)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hbW3Z.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hbW3Z.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an image from hyperphysics &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/atmos/blusky.html&quot; rel=&quot;noreferrer&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/hbase/atmos/blusky.html&lt;/a&gt; which illustrates how Mie scattering creates a white glare around the sun, or moon, and how Rayleigh scattering from air molecules causes the sky to appear blue.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-11-08T21:17:59.123" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23449" PostTypeId="1" AcceptedAnswerId="23451" CreationDate="2017-11-08T23:49:33.760" Score="3" ViewCount="2269" Body="&lt;p&gt;The closest star system after our sun is Alpha Centauri, a three star system with Proxima Centauri, a red dwarf, slightly closer to us than the other 2, around 4.3 light years. I tried &lt;a href=&quot;https://en.wikipedia.org/wiki/Alpha_Centauri&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; and &lt;a href=&quot;https://www.space.com/32560-alpha-centauri-what-we-know.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;other artic&lt;/a&gt;les that describe Alpha Centauri and there is plenty of information, but no where could a find an answer to said question.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My presumption is that if the distance is measurable, so should the variance in this distance.&lt;/p&gt;&#xA;" OwnerUserId="16998" LastActivityDate="2020-03-19T09:46:10.593" Title="Is the Alpha Centauri star system moving closer to us?" Tags="&lt;distances&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23450" PostTypeId="1" CreationDate="2017-11-08T23:56:56.833" Score="5" ViewCount="193" Body="&lt;p&gt;Due to the nature of my job, I have to spend log periods of time (4 - 6 months each), at different countries across Europe, Middle East and Far East. I have a 12 Inch Dobsonian at home, but I rarely spend 1 month a year, at home. Travelling with the Dob is out of question due to its weight and I don't always have access to low lying windows or a rooftop when abroad, making a dob pretty useless. I really want to have at one, at least 8&quot;, portable option while I'm travelling. Celestron NexStar 8 SE is looking good to me at the moment. It seems to have half the weight of an 8&quot; scope with equatorial mount and from what I've heard, it can be used table-top, without tripod. I can also carry the OTA in hand luggage, to avoid damage to optical components. Can someone confirm if&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;(i) Does 8SE work on tabletop without tripod?&lt;/li&gt;&#xA;&lt;li&gt;(ii) How durable is the mount? Can it survive transportation in checked baggage, if placed in its original packaging?&lt;/li&gt;&#xA;&lt;li&gt;(iii) Any other suggestions for a scope that is at or above 8&quot; and can be carried around easily when travelling internationally?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="19552" LastEditorUserId="1891" LastEditDate="2017-11-09T06:45:24.730" LastActivityDate="2017-11-09T21:47:31.590" Title="How to choose portable, yet powerful telescope" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23451" PostTypeId="2" ParentId="23449" CreationDate="2017-11-09T00:55:53.643" Score="6" Body="&lt;p&gt;Yes, it's moving closer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_nearest_stars_and_brown_dwarfs#/media/File:Near-stars-past-future-en.svg&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_nearest_stars_and_brown_dwarfs#Future_and_past&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In about 30,000 years Alpha Centauri will be 3 (and change) light years away, then it will start moving further away.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone wants to provide a more detailed explanation on stellar movement and charts, please feel free.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-11-09T00:55:53.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23454" PostTypeId="2" ParentId="7747" CreationDate="2017-11-09T06:02:47.337" Score="6" Body="&lt;p&gt;A way to do this is to first find the xyz in ECEF of the point on the ground (a transform for a spherical Earth is just the spherical to xyz coordinates transform from most physics text books where lat = 90-theta and lon = phi and r = radius of Earth).  Then, find the satellite slant range (which depends only on the elevation angle and the altitude of the satellite).  Then compute the unit vector toward the satellite in local ENU (east, north, up) coordinates (use Gram Schmidt to find local E and local N given U is the local point and the north pole is (0,0,1). Now, reuse your spherical transform to change az and el into local ENU.  That part is a little weird, but if you think about it, the local horizon elevation is just like the theta of a spherical system, and az is phi (U corresponds to elevation = 90deg).  Finally, compute the same unit vector in xyz (multiply each component by your ENU unit vectors and add the terms), and then &quot;walk along that vector&quot; by the slant path range.  You will now have the xyz of the satellite.  Finally, from the xyz of the satellite compute the lat and lon (and as a check, you should get back your same altitude).&lt;/p&gt;&#xA;" OwnerUserId="19556" LastActivityDate="2017-11-09T06:02:47.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23463" PostTypeId="2" ParentId="23450" CreationDate="2017-11-09T19:37:35.447" Score="1" Body="&lt;p&gt;I was reluctant to do an answer because I do not own a NexStar 8 SE, so I can't speak from direct experience. But this information may prove valuable, so here it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smaller NexStar scopes (like the 4) can be used tabletop-style. The scope is small enough so that the base is sitting just fine on a flat surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 8 might be a little too big for that - I'm not 100% sure. However, if that's the case, there's a simple remedy. Take a piece of plywood or plastic, bigger than the base, thick enough to be rigid, and improvise a way to attach it to the scope base (use bolts / nuts with the same thread as the ones on the tripod). The wider area should stabilize it, and the extra piece of material is flat and easy to carry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At your local home improvement store they surely carry plywood conveniently cut into round pieces of various diameters. Just get one of those, get the appropriate size bolts and washers, and that should be it. Maybe put 3 pads on the bottom of the plywood so it doesn't sit on the bolt heads. Get a spray can of shellac or urethane and give it a nice protective coating when you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The NexStar mount should withstand the challenges of travel pretty well. It's the optics I'm more worried about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As others have said, most tables are too wobbly for astronomy. Use a sturdy table, or some surface that doesn't move around, like the wide edge of a concrete balcony or something.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another option is a collapsible dobsonian. Not the pseudo-collapsible ones like the big SkyQuest series, but a true collapsible scope like the Sumerian dobs. A 16&quot; dob built this way folds down into the volume of a regular travel suitcase. A 10&quot; collapsible dob is like a small suitcase. These are built for the specific purpose of being taken on long trips away from home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sumerianoptics.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.sumerianoptics.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 8&quot; Alkaid Sumerian scope, at 800 mm focal length, is a pretty small dob already. The poles are probably pretty short. Should be easy to transport, and doesn't require much space when fully unfolded; you may actually have to place it on a stool or something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obsession Telescopes also makes dobsonians of all kinds. Their Classic series are bulky when packed, but the Ultra Compact series is basically a true collapsible dob which might be worth considering:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.obsessiontelescopes.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.obsessiontelescopes.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't need a table for these scopes, since they are sitting on the floor. Table wobbliness issues are eliminated. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are equatorial platforms made for all these scopes, so you could even use them for photography.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not own Sumerian or Obsession telescopes, but the feedback I hear through the grapevine is good. These are quality instruments that could be used as a general-purpose telescope even at home.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You could also take the DIY route and build a collapsible dob yourself, if you're so inclined. Many have done this, with great results.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2017-11-09T21:47:31.590" LastActivityDate="2017-11-09T21:47:31.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23467" PostTypeId="1" AcceptedAnswerId="23471" CreationDate="2017-11-10T09:20:32.613" Score="3" ViewCount="227" Body="&lt;p&gt;&lt;strong&gt;Explanation of potential problem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're not completely sure exactly how unlikely an event Abiogenesis was, given both the extreme unlikelihood of nucleotides randomly assembling into anything making sense and the vastness of primordial ocean and extent of time it existed. And still, despite their vastness, when involved with probabilistic chain of events we're quickly getting numbers by which planetary scale pales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may very well be that the event was unlikely enough to be nearly impossible not only on the planetary scale, but even on the galactic scale - or even on scale of the observable universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My proposed solution to this problem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless, with the &lt;em&gt;actual&lt;/em&gt; universe being &lt;em&gt;infinite&lt;/em&gt;, it would be bound to happen &lt;em&gt;somewhere&lt;/em&gt;. Still, occurrences of life could be separated by many times the width of observable universe, and none of them could ever contact another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...which would offer a possible explanation to the Fermi Paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does my proposed solution address the problem I outlined above, or is there a flaw in my reasoning/understanding?&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="6" LastEditDate="2017-11-10T14:21:38.427" LastActivityDate="2017-11-10T19:19:03.473" Title="Can an infinite universe explain the Fermi Paradox?" Tags="&lt;observable-universe&gt;&lt;fermi-paradox&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23468" PostTypeId="1" CreationDate="2017-11-10T09:56:25.333" Score="7" ViewCount="219" Body="&lt;p&gt;Since when are our observation capabilities good enough to discover objects similar to the first discovered interstellar object &lt;a href=&quot;https://en.wikipedia.org/wiki/1I/%CA%BBOumuamua&quot; rel=&quot;noreferrer&quot;&gt;1I/ʻOumuamua&lt;/a&gt; (October 2017) with high probability?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could we make rough estimate how frequent are passes of similar objects through the inner Solar System?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could be the probability that during this century we will be able to gain more observational knowledge about interstellar objects?&lt;/p&gt;&#xA;" OwnerUserId="188" LastEditorUserId="2365" LastEditDate="2018-08-01T15:31:33.563" LastActivityDate="2018-08-24T08:54:50.000" Title="Since when are we able to discover interstellar objects similar to 1I/ʻOumuamua?" Tags="&lt;asteroids&gt;&lt;oumuamua&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23470" PostTypeId="1" CreationDate="2017-11-10T17:21:20.440" Score="4" ViewCount="185" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GoPgW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GoPgW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was told that it was &quot;cedar&quot;, but that is an unclear term botanically.&lt;/p&gt;&#xA;" OwnerUserId="19336" LastEditorUserId="7982" LastEditDate="2019-07-12T15:12:46.440" LastActivityDate="2019-07-12T15:12:46.440" Title="What kind of reddish wood is it exactly that was historically used to line the inside of observatory cupolas (domes), in order to absorb moisture?" Tags="&lt;history&gt;&lt;observatory&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="23471" PostTypeId="2" ParentId="23467" CreationDate="2017-11-10T19:19:03.473" Score="5" Body="&lt;p&gt;The true explanation to the Fermi paradox is not known. There is a very long list of proposed explanations, some of which are listed here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Fermi_paradox#Hypothetical_explanations_for_the_paradox&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Fermi_paradox#Hypothetical_explanations_for_the_paradox&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you're saying is literally the first one on the list, also known as the Rare Earth hypothesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Rare_Earth_hypothesis&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Rare_Earth_hypothesis&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2017-11-10T19:19:03.473" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23472" PostTypeId="2" ParentId="23470" CreationDate="2017-11-10T20:42:34.847" Score="5" Body="&lt;p&gt;Probably western red cedar, &lt;em&gt;thuja plicata&lt;/em&gt;, especially if the observatory is in North America. Western red cedar is easily obtainable in North America. It has a light, red, wood that is particularly resistant to decay, and ideal for use in exposed positions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The soft red-brown timber has a tight, straight grain and few knots. It is valued for its distinct appearance, aroma, and its high natural resistance to decay, being extensively used for outdoor construction in the form of posts, decking, shingles, and siding.[28] It is commonly used for the framing and longwood in lightweight sail boats and kayaks. In larger boats it is often used in sandwich construction between two layers of epoxy resin and/or fibreglass or similar products. Due to its light weight—390 to 400 kg/m3 (24 to 25 lb/cu ft) dried—it is about 30% lighter than common boat building woods, such as mahogany. For its weight it is quite strong but can be brittle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thuja_plicata&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia: Thuja plicata&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be worth pointing out that while &quot;cedar&quot; refers to a group of &lt;a href=&quot;https://en.wikipedia.org/wiki/Cedar_wood&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cedar woods&lt;/a&gt;, trees of the proper &lt;a href=&quot;https://en.wikipedia.org/wiki/Cedrus&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cedrus&lt;/a&gt; genus are rare and localized:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cedrus (common English name cedar) is a genus of coniferous trees in the plant family Pinaceae. They are native to the mountains of the western Himalayas and the Mediterranean region, occurring at altitudes of 1,500–3,200 m in the Himalayas and 1,000–2,200 m in the Mediterranean.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Cedrus&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia: Cedrus&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13887" LastEditorUserId="7982" LastEditDate="2017-11-12T02:49:40.387" LastActivityDate="2017-11-12T02:49:40.387" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23473" PostTypeId="1" CreationDate="2017-11-10T22:05:34.160" Score="4" ViewCount="106" Body="&lt;p&gt;I would like to augment my GPS navigation with remote sensing of 'the sky.'  I am looking for any astronomical body that can be perceived during the day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe Pulsars emit something easily detected, seems IR is obvious..  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any input is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="19587" LastActivityDate="2018-02-09T02:08:50.910" Title="What EM spectrum can stars be most cheaply observed using Remote Sensing during the day?" Tags="&lt;amateur-observing&gt;&lt;radio&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23474" PostTypeId="2" ParentId="23354" CreationDate="2017-11-10T22:52:44.657" Score="1" Body="&lt;p&gt;You could say &quot;MJ in bins of x degrees in l&quot; where x is the degrees spanned by a pixel.  Or, you could divide the fluxes by the steradians in each column and keep it MJ/sr or &quot;Mean MJ/sr within longitudinal bins&quot;. Now the width of the bins is not important as long as they are small.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2017-11-10T22:52:44.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23476" PostTypeId="1" AcceptedAnswerId="23486" CreationDate="2017-11-11T01:27:49.797" Score="7" ViewCount="1326" Body="&lt;p&gt;&lt;strong&gt;Let's assume&lt;/strong&gt; that you had a black hole binary system and everything said here is possible. Their large masses would lead to a large emission of gravitational waves. The loss of orbital energy and angular momentum to this gravitational radiation should ultimately cause the two black holes to merge into a single black hole containing the combined mass of the merged black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this said, does a greater combined mass of the newly formed black hole mean that the radius of the &lt;strong&gt;new&lt;/strong&gt; event horizon is greater than the radius of the event horizons of the singular black holes? &#xA;I don't know if there is a formula connecting mass to radius, or if this is essentially a conceptually hypothetical concept, and the answer is simply that the larger the mass of the black hole, the larger the radius of its event horizon&lt;/p&gt;&#xA;" OwnerUserId="19589" LastEditorUserId="2365" LastEditDate="2017-11-11T14:44:43.297" LastActivityDate="2018-03-20T16:00:08.367" Title="The size of the radius of the event horizon of a black hole created by the merger of a black hole binary system" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;gravitational-waves&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23477" PostTypeId="1" AcceptedAnswerId="23482" CreationDate="2017-11-11T03:13:50.580" Score="5" ViewCount="517" Body="&lt;p&gt;When I was young (I won't tell you when) I saw one of the four &lt;a href=&quot;https://en.wikipedia.org/wiki/Galilean_moons&quot; rel=&quot;noreferrer&quot;&gt;Galilean moons&lt;/a&gt; of Jupiter suddenly disappear while watching through a small refractor. It wasn't a coincidence, I'd seen the prediction in some announcements in Sky and Telescope and it happened like clockwork. The orbital plane of the Galilean moons had rotated so that it included the Sun, and the moons were Eclipsing each other, or so I remember.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will this happen again? If so, when will they start, and where can I find predictions of individual eclipses?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2017-11-11T08:58:13.473" Title="When will the next series of mutual eclipses of Jupiter's moons begin?" Tags="&lt;jupiter&gt;&lt;natural-satellites&gt;&lt;eclipse-non-earth-related&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23482" PostTypeId="2" ParentId="23477" CreationDate="2017-11-11T08:02:58.107" Score="5" Body="&lt;p&gt;Mutual eclipses and occulations of Jupiter's moons occur during Jovian Equinox, twice in each orbit of Jupiter, and so roughly every 6 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last series of events was in 2015, so the next will start in 2021. There will be a series of events between January and August. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The site &lt;a href=&quot;http://lnfm1.sai.msu.ru/neb/nss/nsszph517he.htm&quot; rel=&quot;noreferrer&quot;&gt;http://lnfm1.sai.msu.ru/neb/nss/nsszph517he.htm&lt;/a&gt; can generate ephemerides for these events.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-11-11T08:02:58.107" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23483" PostTypeId="2" ParentId="23477" CreationDate="2017-11-11T08:58:13.473" Score="4" Body="&lt;p&gt;@JamesK &lt;a href=&quot;https://astronomy.stackexchange.com/a/23482/7982&quot;&gt;nailed it&lt;/a&gt;, I'll just add a little more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting from &lt;a href=&quot;http://lnfm1.sai.msu.ru/neb/nss/nsszph517he.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;the link&lt;/a&gt; in that answer, I've found the figures below on &lt;a href=&quot;http://lnfm1.sai.msu.ru/neb/nss/expl517.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;. It looks like this cycle will be difficult to watch in the first months of 2021, until Jupiter moves far enough away from the Sun as seen from Earth. Also, the southern declination will make it more difficult to watch from the highest northern latitudes. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Parameters of the Earth-Sun-Jupiter configuration &#xA;  during the mutual occultations and eclipses of the Galilean satellites in 2021&lt;/strong&gt;. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;The number of events per week (Monday to Sunday)&lt;/strong&gt;. &#xA;  Note: Only 30 % of the events can be seen on one observatory. &#xA;  Note: Events before 3 March 2021 are not observable (Twilight). &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z7fru.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z7fru.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Planetocentric planetoequatorial latitudes (deg) of the Earth and the Sun. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2D0SA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2D0SA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Geocentric angular distance (deg) between Jupiter and the Sun. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5gigf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5gigf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Declination of Jupiter (deg). &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PfUeS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PfUeS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The Sun-Earth phase angle on Jupiter (deg). &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ddrX5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ddrX5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2017-11-11T08:58:13.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23485" PostTypeId="2" ParentId="23476" CreationDate="2017-11-11T11:52:04.890" Score="0" Body="&lt;p&gt;Yes, there is a simple formula connecting mass with radius. It is called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schwarzschild radius&lt;/a&gt;. It is really well explained in Wikipedia, with examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quoting: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Schwarzschild radius is the radius of a sphere such that, if all&#xA;  the mass of an object were to be compressed within that sphere, the&#xA;  escape velocity from the surface of the sphere would equal the speed&#xA;  of light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Formula:&#xA;$$r_s = \frac{2MG}{c^2}$$&#xA;As you can see, with exception of radius and mass itself, all other are constants (c = speed of light, G = gravitational constant). So the double the mass, the double the radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore yes, after merging the size would be greater (I think this is your original question), althought not sure if it is the exact sum due to some kind of energy loss during the merge. As a side note, there is no need of having a binary system of black holes colliding in order to determine or change the size of a black hole, it can grow just by eating surrounding particles, gas, etc.&lt;/p&gt;&#xA;" OwnerUserId="19337" LastActivityDate="2017-11-11T11:52:04.890" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23486" PostTypeId="2" ParentId="23476" CreationDate="2017-11-11T14:06:55.637" Score="8" Body="&lt;p&gt;The &quot;radius&quot; (there is no physical surface) of the event horizon of a rotating black hole, depends on both it's mass $M$ &lt;em&gt;and&lt;/em&gt; angular momentum $J$, and is given by the equation&#xA;$$ r = \frac{GM}{c^2} +\sqrt{\left(\frac{GM}{c^2}\right)^2- \left(\frac{J}{Mc}\right)^2}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is thus difficult (for me anyway) to give a straightforward answer to your question. When two black holes merge, they will each have their own masses and angular momenta, plus there will be angular momentum in the orbit. The gravitational waves emitted during the merger can take away mass from the system as a whole (e.g. the final mass of the first observed black hole merger was three solar masses less than the summed masses of the merging components).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in general yes, for non-spinning black holes the event horizon grows as the total mass. But if the merger results in a black hole with maximal spin, where $J = GM^2/c$, then the final event horizon could be half the size given by the Schwarzschild radius ($2GM/c^2$), even though the total mass is larger than the black holes that contributed to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I think what is certainly true, is that it is not possible to add mass (even in the form of another black hole) to a given black hole and for the event horizon to get smaller, whatever the angular momentum contributed by that mass (section 4.2 of &quot;Black Holes&quot;, by Raine &amp;amp; Thomas, 2015, Imperial College Press).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-11-15T10:29:39.123" LastActivityDate="2017-11-15T10:29:39.123" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23488" PostTypeId="1" CreationDate="2017-11-12T02:15:24.647" Score="2" ViewCount="93" Body="&lt;p&gt;I think the question &lt;a href=&quot;https://physics.stackexchange.com/questions/193771/are-cosmic-rays-net-neutral-in-charge&quot;&gt;here&lt;/a&gt; is not answered directly. According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_ray#Composition&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Of primary cosmic rays, which originate outside of Earth's atmosphere, about 99% are the nuclei of well-known atoms (stripped of their electron shells), and about 1% are solitary electrons (similar to beta particles). Of the nuclei, about 90% are simple protons (i.e., hydrogen nuclei); 9% are alpha particles, identical to helium nuclei; and 1% are the nuclei of heavier elements, called HZE ions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Together, electrons and positrons need to be neutral in charge, right? &#xA;Only 1% are electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can they be neutral in charge?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="2365" LastEditDate="2018-07-27T22:33:11.093" LastActivityDate="2018-07-27T22:33:11.093" Title="Are electrons and positrons equal in cosmic rays?" Tags="&lt;cosmic-ray&gt;" AnswerCount="0" CommentCount="11" ClosedDate="2017-11-17T14:38:38.073" ContentLicense="CC BY-SA 3.0" />
  <row Id="23489" PostTypeId="1" CreationDate="2017-11-12T03:04:27.380" Score="-4" ViewCount="4970" Body="&lt;p&gt;The Earth is spinning at about a thousand miles per hour. Shouldn't we be thrown from its surface? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes I know someone would comment &quot;gravity&quot; but physically we should be thrown from Earths surface. Scientists say that gravity is holding us here but that seems to contradict physics. Maybe someone can better explain it to me.&lt;/p&gt;&#xA;" OwnerUserId="19600" LastEditorUserId="464" LastEditDate="2017-11-12T16:22:19.790" LastActivityDate="2020-10-30T13:42:04.420" Title="Why doesn't the Earth's rotation throw us off the surface?" Tags="&lt;gravity&gt;&lt;astrophysics&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23490" PostTypeId="2" ParentId="23489" CreationDate="2017-11-12T03:54:53.907" Score="4" Body="&lt;p&gt;It doesn't contradict physics, though it can be confusing.   I've read that Ptolemy rejected the rotating Earth hypothesis because he felt it should generate very high winds.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;flying off&quot; force, more correctly, the centrifugal force can be calculated and it's much smaller than gravity.  (I'm not good at formulas, so I'll skip the math for now).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HPQB5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HPQB5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to think about this is to consider Newton's first law, and an object in motion stays in motion at the same speed and in the same direction.   The Earth is large enough that standing on it's rapidly rotating surface feels like you're standing still because you're mostly moving in a straight line.  It takes 6 hours to turn 90 degrees.    Imagine being in a car and you feel the force to the side when the car turns.   Now, imagine that car is moving very fast, but it takes 6 hours to turn left.   The turn is slow enough that the &quot;lifting&quot; force in the case of the Earth or the sideways force in the theoretical car taking the 6 hour turn, is barely noticed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 1000 mph velocity at the equator isn't noticed at all because it's primarily in the linear direction with only a very slow turn, and being largely straight and constant at a constant velocity.  The change in direction is too small to be noticable.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to take Earth's rotation into account for long distances.  It affects Earth's prevailing winds and it needs to be taken into account with space-flight, missile launches and snipers, for very long distance shots, need to take the curvature and rotation of the Earth into account when shooting at far away targets.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's far to week to throw people off the Earth.   Earth would need to rotate some 17 times faster for that to happen. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-11-12T03:54:53.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23491" PostTypeId="2" ParentId="23489" CreationDate="2017-11-12T07:11:06.747" Score="16" Body="&lt;p&gt;OK, the reason we don't get flung off the surface of the Earth is that the rotational forces are not large enough to do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that Earth formed because material was pulled together by it's own gravity. If Earth rotated so quickly, that material would be thrown off and Earth, as we know it, would not have formed. That's a bit simplistic, but it's reasonable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maths ahead. :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force of gravity we experience is :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F_g = g_e M_o = \frac {GM_eM_o}{R_e^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the &quot;e&quot; subscript is for Earth and the &quot;o&quot; for the person.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the basic Newtonian equation for gravitational force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational force is always directly towards the center of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the equation for the force we experience due to rotation is a bit more complicated looking because we rotate about an axis (north-south) and the force from rotation varies with latitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F_c = M_o \frac {v^2} r$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$v = \frac {2\pi r} T$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$r = R_e \cos\theta$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is the period of rotation and &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; is the latitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That works out as :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F_c = M_o \frac {4\pi^2r}{T^2}=M_o\frac{4\pi^2R_e\cos\theta}{T^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the direction of that force is &lt;em&gt;perpendicular to the axis of rotation&lt;/em&gt;.  So we need to find out how much force we experience from it radially - away from the center and that is  :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F'_c = M_o \frac{4\pi^2R_e \cos^2\theta}{T^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For us to be flung off the Earth we'd need &lt;span class=&quot;math-container&quot;&gt;$F'_c&amp;gt;F_g$&lt;/span&gt;.  So is it ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest value for &lt;span class=&quot;math-container&quot;&gt;$F'_c$&lt;/span&gt; is at the equator (&lt;span class=&quot;math-container&quot;&gt;$\theta = 0$&lt;/span&gt;) and &lt;a href=&quot;https://www.google.com/search?q=4*pi%5E2*(radius%20of%20earth)*cos(0)%5E2%20%2F%20(1%20sidereal%20day)%5E2&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Google search for 4*pi^2*&amp;#40;radius of earth&amp;#41;*cos&amp;#40;0&amp;#41;^2 / &amp;#40;1 sidereal day&amp;#41;^2&quot;&gt;we&lt;/a&gt; &lt;a href=&quot;https://www.google.com/search?q=%28gravitational%20constant%29*%28mass%20of%20earth%29%20%2F%20%28radius%20of%20earth%29%5E2&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Google search for &amp;#40;gravitational constant&amp;#41;*&amp;#40;mass of earth&amp;#41; / &amp;#40;radius of earth&amp;#41;^2&quot;&gt;get&lt;/a&gt; (in SI units) :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{aligned}&#xA;F'_c &amp;amp;= M_o \times 0.034 \;{\rm m/s^2}&#xA;\\&#xA;F_g  &amp;amp;= M_o \times 9.8\phantom{00} \;{\rm m/s^2}&#xA;\end{aligned}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the rotational force is simply too small to overcome gravity.  We can't get thrown off the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how fast would Earth have to rotate that we would get thrown off the Earth ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That would need :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{4\pi^2R_e}{T^2} &amp;gt; g_e$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T^2 &amp;lt; \frac{4\pi^2R_e}{g_e}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T &amp;lt; 2\pi\sqrt{\frac{R_e}{g_e}} = 5066 \text{ seconds}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we'd need a day shorter that one and a half hours to be flung off the planet.  For the brief period we could cling on desperately, that would be dizzying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Earth rotated that fast it would not last long anyway - first the surface would be separated and then the inner layers would be flung off.  Earth would not have formed in the first place (at least not in anything like it's current form) because we need gravity to be strong enough to keep things together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that it is because of the rotational forces that Earth is not quite a sphere - it's a bit swollen at the equator than the poles (because the net force inward at the poles is larger while at the equator there is this small but appreciable force from the rotation).  This slight oblateness does change the numbers a tiny bit, but not much and not worth the effort of going into detail.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastEditorUserId="1891" LastEditDate="2019-04-30T20:21:59.403" LastActivityDate="2019-04-30T20:21:59.403" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23493" PostTypeId="1" AcceptedAnswerId="23496" CreationDate="2017-11-12T09:44:53.797" Score="0" ViewCount="1089" Body="&lt;p&gt;What is the minimum, maximum and average distance of the Venus and Jupiter? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only found 670,130,000 km and 670,198,000 km as an average distance (&lt;a href=&quot;http://www.planetsedu.com/distances-between-the-planets/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://theplanets.org/distances-between-planets/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="19604" LastEditorDisplayName="user1569" LastEditDate="2017-11-12T14:07:04.650" LastActivityDate="2017-12-03T09:39:36.583" Title="What's the distance between Venus and Jupiter?" Tags="&lt;distances&gt;&lt;jupiter&gt;&lt;venus&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23494" PostTypeId="1" AcceptedAnswerId="23590" CreationDate="2017-11-12T11:23:29.710" Score="4" ViewCount="325" Body="&lt;p&gt;I'm a Machine Learning Student and I'm working on Strong Lensing Image Classification. this is a hot topic nowadays but strangely after days of searching and communicating with different people I still can't find an accessible Dataset of these objects!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know any place I can find a simple(or not) Dataset containing images, labels and some extended features of Strong Gravitational Lensings?&lt;/p&gt;&#xA;" OwnerUserId="6718" LastEditorUserId="11264" LastEditDate="2017-11-12T17:31:42.537" LastActivityDate="2018-01-16T22:40:38.277" Title="Strong Gravitational Lensing Dataset" Tags="&lt;data-analysis&gt;&lt;gravitational-lensing&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23495" PostTypeId="2" ParentId="23392" CreationDate="2017-11-12T11:45:38.433" Score="0" Body="&lt;p&gt;I use cartes du ciel - free opensource software. You can construct your own catalogues and then use them, or import external data from other packages - I have used it in conjunction with the 'TopCat' academic package in scientific papers - it's that good! Also you can connect to your scope, set FOV for various eyepieces, etc.&lt;/p&gt;&#xA;" OwnerUserId="19605" LastActivityDate="2017-11-12T11:45:38.433" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23496" PostTypeId="2" ParentId="23493" CreationDate="2017-11-12T15:03:42.337" Score="4" Body="&lt;p&gt;The furthest distance Venus is from the Sun is 0.73 AU&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The furthest distance Jupiter is from the Sun is 5.46 AU&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest distance Jupiter is fro the Sun is 4.95 AU&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we assume the orbits are concentric circles (they're not !) then :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Minimum distance = 4.95-0.73 = 4.22 AU&lt;/li&gt;&#xA;&lt;li&gt;Maximum distance = 5.46+0.73 = 6.19 AU&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And a simple average of these gets you an average distance of about 5.2 AU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problems with that simple estimate are :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The orbits are not circles, but ellipses&lt;/li&gt;&#xA;&lt;li&gt;The orbits are at an inclination&lt;/li&gt;&#xA;&lt;li&gt;A more detailed calculation for circular concentric orbits produces a tricky double integral with no easy solution&lt;/li&gt;&#xA;&lt;li&gt;Over time (the only way to do a true average) the orbits would change from even &quot;simple&quot; ellipses.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A number of online sources quote the average distance as 4.48 AU.  This appears to be nothing more than the semi-major axis of Venus's orbit subtracted from the semi-major axis of Jupiter, which is probably an under-estimate of an average separation, as it neglects that they can be at opposite sides of the Sun sometimes and the same side at others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Minor update :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those interested in seeing the effect of a more detailed calculation I tried treating the orbits as concentric circles and doing the full integral on WolframAlpha.com .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The required integral in this case is (units are AU) :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L = \frac {5.2044}{4\pi^2} \int_0^{2\pi}\int_0^{2\pi} \left[ ( sin(v)-0.139 cos(u) )^2 + ( cos(v)-0.139 cos(v) )^2 \right]^{1/2}du\, dv$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $0.139$ is the ratio of Venus's orbital radius to Jupiter's and $5.2044$ AU is the mean radius of Jupiter's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the result of this tedious integral $L=5.2293$ AU, which is not exactly much of an improvement on the much simpler $5.2$ AU given earlier.  But it at least dispels then notion on some online pages that it could be $4.48$ AU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last (?) minor update :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above calculation just did a geometric average over the two orbit &lt;em&gt;shapes&lt;/em&gt;, but made no attempt to consider the motion.  Well it occurred to me to revise that and allow for two bodies in a circular motion in the same plane.  Guess what ?  &lt;em&gt;Same answer&lt;/em&gt; !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, almost.  In this case it's $5.2297$, which is an insignificant difference as we're already ignoring inclination and using circular orbits.  I'm inclined to put down the difference to a numerical calculation margin of error (the integrals do not have simple closed forms and would easily propagate errors, I think).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral boils to an average over the synodic period and is a simpler single integral which reduces to :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L = \sqrt{a^2+b^2} \frac 1 {2\pi} \int_0^{2\pi} \sqrt{1-2\frac {ab}{a^2+b^2} cos\theta}\, d\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do prefer this result myself as it's taking account of orbital motion and over time, rather than purely geometric.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastEditorUserId="10196" LastEditDate="2017-12-03T09:39:36.583" LastActivityDate="2017-12-03T09:39:36.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23499" PostTypeId="1" CreationDate="2017-11-12T16:05:43.930" Score="27" ViewCount="6992" Body="&lt;p&gt;My ten year old daughter asked me this question at breakfast and I initially said yes, but on second thought I was also puzzled.&lt;/p&gt;&#xA;" OwnerUserId="19609" LastActivityDate="2017-11-13T03:13:44.000" Title="Are there more stars in the universe than grains of sand in the Earth?" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23501" PostTypeId="2" ParentId="23499" CreationDate="2017-11-12T16:30:08.517" Score="34" Body="&lt;p&gt;A quick google gave me these (approximate) figures:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;7.5 x 10&lt;sup&gt;18&lt;/sup&gt; grains of sand in all the beaches and deserts of the world&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;7 x 10&lt;sup&gt;22&lt;/sup&gt; stars in the observable universe&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If these are reasonable estimates, then there are approximately nine thousand stars in the observable universe for each grain of sand on Earth. (By observable universe, I mean in all the galaxies that we can detect with our most powerful telescopes.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.npr.org/sections/krulwich/2012/09/17/161096233/which-is-greater-the-number-of-sand-grains-on-earth-or-stars-in-the-sky&quot; rel=&quot;noreferrer&quot;&gt;Robert Krulwich: Which Is Greater, The Number Of Sand Grains On Earth Or Stars In The Sky?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13887" LastEditorUserId="13887" LastEditDate="2017-11-12T16:46:19.570" LastActivityDate="2017-11-12T16:46:19.570" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="23502" PostTypeId="1" CreationDate="2017-11-12T16:35:12.300" Score="2" ViewCount="97" Body="&lt;p&gt;I'm working on an assignment, where I have to classify stars using the raw data of the telescope. I'm now trying to calibrate the pixel to armstrong ratio, and for that I need to identify the absorption lines present on the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been researching for two days now, and I haven't been able to find a 'trustworthy' source of information/confirmation on where to set the Balmer absorption lines for a star like the one in the image (I'm positive H alpha (the one at the far right), beta (on top) and gamma (i would say the one at the left of H beta) are quite prominent).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XI9jy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XI9jy.png&quot; alt=&quot;A spectral profile of 01_HIP105140_3s&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, I am trusting on &lt;a href=&quot;http://www.fieldtestedsystems.com/starspectra/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt; to do it. Is there a real source of info, be it a book, an institutional webpage or something, where I can confirm and that I could reference for an assignment? Am I wrong on the position of the Balmer Series?&lt;/p&gt;&#xA;" OwnerUserId="19610" LastActivityDate="2017-11-12T16:35:12.300" Title="Source for Identification of Balmer Hydrogen Absorption Lines" Tags="&lt;spectroscopy&gt;&lt;spectra&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23503" PostTypeId="1" CreationDate="2017-11-12T22:14:41.317" Score="0" ViewCount="205" Body="&lt;p&gt;I would like to know please how can we get to the exoplanet Proxima b. I heard it is the potentially habitable exoplanet closet to Earth at only 4.2 light years away. So exciting!&lt;/p&gt;&#xA;" OwnerUserId="18419" LastActivityDate="2017-11-23T17:13:46.967" Title="How can we ge to the exoplanet Proxima b?" Tags="&lt;universe&gt;&lt;space&gt;&lt;nasa&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2019-02-11T13:24:03.280" ContentLicense="CC BY-SA 3.0" />
  <row Id="23508" PostTypeId="1" CreationDate="2017-11-13T03:22:46.910" Score="-1" ViewCount="97" Body="&lt;p&gt;Sorry if this this doesn't really fit this forum, but WorldBuilding does not like questions like these. I cannot find the minimum density of hydrogen a Bussard ramjet (an interstellar ship propelled by hydrogen collected from the interstellar medium) needs to work, or the densities of these certain areas(galactic core, thicker arms, Large Magellanic Cloud, Small Magellanic Cloud, Magellanic Bridge, and the Magellanic stream, and intergalactic space). I am interested if this is a viable mode of transportation in different parts of the universe relatively near to us.&lt;/p&gt;&#xA;" OwnerUserId="19304" LastActivityDate="2017-11-13T20:00:58.003" Title="Would Bussard Ramjets work in other places in or near the galaxy?" Tags="&lt;milky-way&gt;&lt;interstellar-medium&gt;&lt;interstellar-travel&gt;&lt;intergalactic-space&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2017-11-13T20:17:57.307" ContentLicense="CC BY-SA 3.0" />
  <row Id="23514" PostTypeId="1" CreationDate="2017-11-13T13:48:05.927" Score="3" ViewCount="127" Body="&lt;p&gt;If our solar system some how captured a rogue planet, and it orbited the sun in the opposite direction to all the other planets,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what major affect would it have on the stability, would its drag cause all the other planets to slow down ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;could it be possible for it to be sustained at all ?&lt;/p&gt;&#xA;" OwnerUserId="18817" LastActivityDate="2017-11-13T17:48:46.623" Title="what would a captured planet do if it was in a retrograde orbit?" Tags="&lt;planet&gt;&lt;solar-system&gt;&lt;exoplanet&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23516" PostTypeId="1" CreationDate="2017-11-13T15:33:29.947" Score="5" ViewCount="1677" Body="&lt;p&gt;We take it for granted these days that the stars are unimaginably bigger than the planets and the moons. But when you look at the sky, it does not appear this way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon looks bigger and brighter than the stars. This leads me to wonder when astronomy first discovered the relative size of the stars. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Did people believe that the moon was bigger than the stars for a long time? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If so, where was this believed, and up until when? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What discoveries or observations led people to conclude that the stars are larger?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="19643" LastEditorUserId="11264" LastEditDate="2017-11-15T13:49:03.263" LastActivityDate="2017-11-15T13:49:03.263" Title="When did astronomy first discover that the stars are bigger than the moon?" Tags="&lt;star&gt;&lt;planet&gt;&lt;the-moon&gt;&lt;history&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23517" PostTypeId="2" ParentId="23514" CreationDate="2017-11-13T15:57:00.583" Score="0" Body="&lt;p&gt;Answering one question:  gravitational drag is primarily a function of where the two bodies are.  This means that any time one planet &quot;lags&quot; another the lead planet is slowed slightly by the gravitational pull.  A planet in retrograde orbit would simply reverse the portions of relative orbital position where gravity slows or accelerates each planet's speed.&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-11-13T15:57:00.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23519" PostTypeId="2" ParentId="23514" CreationDate="2017-11-13T17:48:46.623" Score="4" Body="&lt;p&gt;Totally dependent on the details.  A rogue planet could be any mass from Pluto to 10 Jupiter masses - quite different gravitationally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And exactly what it's trajectory was makes a huge difference.  If it came barreling into the inner system it could cause chaos, particularly a large body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would certainly have to pass thought the &lt;a href=&quot;https://en.wikipedia.org/wiki/Oort_cloud&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oort cloud&lt;/a&gt; and possibly the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kuiper_belt&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kuiper Belt&lt;/a&gt;.  Any appreciable mass doing would disturb many, many objects which have the potential to fall into the inner solar system (or not).  So that alone could be dangerous for us (or not).  It's a dice throw - quite random from our point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stable is relative.  If it disturbed orbits in such a way that in a hundred thousand years Earth would become uninhabitable that might be not too bad - a lot can happen to a species in that time scale, especially one armed with nukes. :-)  So this isn't a question with a single answer.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-11-13T17:48:46.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23520" PostTypeId="2" ParentId="23508" CreationDate="2017-11-13T20:00:58.003" Score="1" Body="&lt;p&gt;Yes it's theoretical, but on some levels, estimates can be calculated.  (probably better for space-ex than here), but it's a fun design to think about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic design, the ship generates a magnetic field and catches atoms in space as it moves.  I've read that the &quot;catch&quot; might be about a square mile in area, so you can calculate how much hydrogen and alpha particles and other particles (far less) and electrons (some).   As the ship picks up speed it's catch rate would increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's two primary problems.  The materials that it would catch in space, mostly charged particles.  Protons and Alpha Particles, they are not easy to fuse.  There are much better fusion materials available if you make your own deuterium for example, or lithium deuteride for easier storage, or 3He.   Fusing mostly protons and alpha particles is very hard and might never be practical.    You could in theory calculate the energy output - 4 protons into 1 helium yeilds 0.7% mass to energy ratio, which isn't bad, but won't get you close to the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other problem is that the magnetic catch that the ship generates, you have to take into account every particle it encounters and factor in the drag.   If each proton can deliver 0.7% of it's mass into energy, there becomes a velocity where that proton hits the ship where you lose about that much upon impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 3rd impracticality is that space is very empty and hydrogen is very light and quite abundant.   It doesn't make much sense to collect it in empty space when you can simply carry your own.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line is that it's probably not a useful design.  Could it work?   Sure.   It would work best riding a stream of particles behind you, like riding a jet from a distant solar flare or supernova.   You'd still need to solve the the hydrogen fusion problem and if you solve that, you could probably design better ways to fly than a ramjet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, yes, it could work.  Solar panels can drive a car and sails can move a battleship, but neither is particularly fast.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-11-13T20:00:58.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23522" PostTypeId="1" CreationDate="2017-11-13T23:20:46.553" Score="1" ViewCount="199" Body="&lt;p&gt;If we know that the reason a star burns without oxygen and fuel is that it converts hydrogen into helium and we have these two elements on earth then why can’t we make a star on our planet&lt;/p&gt;&#xA;" OwnerUserId="19651" LastActivityDate="2017-11-15T15:40:30.147" Title="Why Can’t we make a star" Tags="&lt;fundamental-astronomy&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2017-11-16T19:17:39.350" ContentLicense="CC BY-SA 3.0" />
  <row Id="23525" PostTypeId="2" ParentId="23516" CreationDate="2017-11-14T05:40:49.540" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;When did astronomy first discover that the stars are bigger than the&#xA;  moon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your choice of words &quot;discovery&quot; implies evidence, not just hypothesis.  The discovery that stars were like our sun wasn't possible until the distance to some nearby stars was measured and with that, a brightness comparison could be made between stars and our sun.   &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedrich_Bessel#Work&quot; rel=&quot;noreferrer&quot;&gt;That discovery was made in 1838&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first person with that theory goes back to 450 BC.   &lt;a href=&quot;http://curious.astro.cornell.edu/physics/49-our-solar-system/the-sun/general-questions/159-when-did-people-discover-that-the-sun-is-a-star-intermediate&quot; rel=&quot;noreferrer&quot;&gt;Short summary&lt;/a&gt; and &lt;a href=&quot;http://solar-center.stanford.edu/FAQ/Qsunasstar.html&quot; rel=&quot;noreferrer&quot;&gt;longer summary&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Anaxagoras, who lived in Athens, Greece, around 450 BC (about 2450&#xA;  years ago), thought that the Sun and stars were fiery stones, that the&#xA;  stars were too far away for their heat to be felt, and that the Sun&#xA;  was perhaps more than a few hundred miles in size. With that&#xA;  Anaxagoras was, as far as we know, the first one to suggest that the&#xA;  Sun is a star. His ideas were met with disapproval and he was finally&#xA;  imprisoned for impiety, because his ideas did not fit the prejudices&#xA;  of the time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, while he suggested that stars were like our sun, his size was way off and it's unclear how large he thought the Moon was, or even, if he had a good idea how large the Earth was.   &lt;a href=&quot;https://www.windows2universe.org/?page=/citizen_science/myw/w2u_eratosthenes_calc_earth_size.html&quot; rel=&quot;noreferrer&quot;&gt;Erastosthenes' estimate&lt;/a&gt; came later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astro.cornell.edu/academics/courses/astro201/aristarchus.htm&quot; rel=&quot;noreferrer&quot;&gt;Aristarchus of Samos&lt;/a&gt; came up with a much better, but still quite inaccurate, measure of the Sun and Moon's size and he recognized that the Moon was smaller than the Earth and the Sun much larger.  It's unclear what he thought about stars though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And much more recently:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Christiaan Huygens of Holland lived from 1629 to 1695. He determined&#xA;  the distance to the star Sirius, assuming that that star was as bright&#xA;  as the Sun and appeared faint only because of its great distance. He&#xA;  found that the distance to Sirius must be very great. &lt;strong&gt;At this time,&#xA;  then, the idea that the Sun is a star was considered seriously by&#xA;  scientists&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the idea that stars were like our sun was popular and perhaps even accepted, for over 100 years before the size and brightness of stars was first accurately measured.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Did people believe that the moon was bigger than the stars for a long&#xA;  time?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is hard to say.  Aristarchus of Samos ideas were less popular than Aristotle's  (who came a little earlier) and Aristarchus' theories weren't central to Ptolomy's model which became the model for over 1000 years.  I've often wondered what might have been different if Aristotle and Aristarchus could have exchanged ideas.  Aristotle might have learned something and re-written his model, but Aristarchus was born too late.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ptolomy, who's model was accepted by the Roman Catholic Church, &lt;a href=&quot;http://farside.ph.utexas.edu/Books/Syntaxis/Almagest/node3.html&quot; rel=&quot;noreferrer&quot;&gt;didn't pay much attention to sizes&lt;/a&gt; and, in fact, his model doesn't account for size well at all.  It's one of the ways that his model breaks down, but his basic model had the Moon closer to the Earth than the Sun, so presumably he recognized that the Sun was larger than the Moon and it's not unreasonable to guess that Ptolomy might have said the planets are smaller than the moon.  The stars, however, were for the most part, not addressed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Fyk7Q.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fyk7Q.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your question is, in particular, what Aristotle, Ptolemy up through scholars in the middle ages thought about the size of stars in comparison to the moon, then, I have to say that I don't know what was assumed about the size of the Moon and stars, though Giovanni Bruno was burned at the stake for suggesting stars were like our sun.  It's entirely possible most scholars thought stars were smaller than the Moon from the time periods between Aristotle to Kepler.  If someone knows more specifics, I encourage them to answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not long after Galileo and Kepler, it seems to have become more generally accepted that stars were large and very distant and the Moon was relatively small by comparison. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-11-14T06:08:34.900" LastActivityDate="2017-11-14T06:08:34.900" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23534" PostTypeId="2" ParentId="23522" CreationDate="2017-11-14T20:56:49.913" Score="2" Body="&lt;p&gt;The process by which stars convert hydrogen to helium isn't burning; it is the fusion of the nuclei of atoms. To get nuclei to fuse you need to confine them in a very small space and get them very hot. On Earth we can make things very hot, and we can put things under lots of pressure. The trouble is that we can't do both at the same time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual way we put things under pressure is squeezing them in some sort of vice. But if the thing we are squeezing is many millions of degrees, it will just vapourize our equipment. If we heat up some hydrogen, it expands and we don't get the confinement required for sustained fusion. (But this is how a thermonuclear bomb works)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a star is a huge ball of gas. The weight of the gas compresses the inner core of the star to an enormous pressure and traps the heat. A litre of the sun's core would weigh about 150kg at a pressure of about 250 billion atmospheres, and have a temperature of 15million degrees. This combination of temperature and pressure cannot be achieved on Earth. The hydrogen in the sun's core is confined by its own gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back on Earth we are trying to confine hydrogen with magnetic fields. (See the ITER Tokamak) With this system we will be able to start fusion, but the amount of energy we get out is less than the energy we put in. However solving this problem is merely a technical challenge. If we can work out how to use magnetic fields to confine hydrogen at very high temperatures we will be able to generate power from fusion.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-11-14T20:56:49.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23535" PostTypeId="1" CreationDate="2017-11-14T21:33:43.057" Score="0" ViewCount="62" Body="&lt;p&gt;I have found several right ascension and declination coordinates for the poles of bodies in the solar system, but there seems to be no comprehensive source that is easily uncovered with Google etc. containing any other than those of the planets and a handful of other bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most importantly i am missing them for the Moon, but i'd like them for all regular satellites and any other bodies if possible.&lt;/p&gt;&#xA;" OwnerUserId="19671" LastEditorUserId="2618" LastEditDate="2017-11-15T10:56:53.950" LastActivityDate="2017-11-15T11:30:45.440" Title="Is there any data source for the rotational axes of bodies in the solar system?" Tags="&lt;the-moon&gt;&lt;pole&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23538" PostTypeId="2" ParentId="23535" CreationDate="2017-11-15T08:02:50.693" Score="0" Body="&lt;p&gt;Wikipedia has a nice table located &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_tilt&quot; rel=&quot;nofollow noreferrer&quot;&gt;here,&lt;/a&gt; and &lt;a href=&quot;http://www.astronomynotes.com/tables/tablesb.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here.&lt;/a&gt; Maybe of further use is the list of references associated with these tables. There is also a nice illustration depicting the rotational axes of other planets shown &lt;a href=&quot;https://www.redshift-live.com/ext/en/magazine/articles/Redshift_Archive/17239-Rotational_Axes__total_disorder-1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here.&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="16926" LastEditDate="2017-11-15T08:17:02.730" LastActivityDate="2017-11-15T08:17:02.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23540" PostTypeId="2" ParentId="23535" CreationDate="2017-11-15T11:30:45.440" Score="2" Body="&lt;p&gt;Collecting and reporting the requested information is the job of the IAU Working Group on Cartographic Coordinates and Rotational Elements. They issue reports every few years; the most recent report was in 2009. The report is currently available at &lt;a href=&quot;http://aa.usno.navy.mil/publications/reports/Archinaletal2011a.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;usno.navy.mil&lt;/a&gt; and &lt;a href=&quot;https://astropedia.astrogeology.usgs.gov/alfresco/d/d/workspace/SpacesStore/28fd9e81-1964-44d6-a58b-fbbf61e64e15/WGCCRE2009reprint.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;astrogeology.usgs.gov&lt;/a&gt;, and elsewhere. Table 1 provides data for the Sun and planets (including Pluto), table 2 for satellites, and table 3 for other solar system bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question asks specifically about the Moon. The Moon's orientation is hideously complex. I won't even begin to replicate the expression in table 2 of the cited report. The report specifically addresses the Moon's orientation, admitting that the expression in table 2 is approximate. The best source is an ephemeris such as JPL's Development Ephemeris. These data are available in the form of SPICE kernels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Archinal, Brent Allen, et al. &quot;Report of the IAU Working Group on Cartographic Coordinates and Rotational Elements: 2009.&quot; &lt;em&gt;Celestial Mechanics and Dynamical Astronomy&lt;/em&gt; 109.2 (2011): 101-135. DOI 10.1007/s10569-010-9320-4&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2017-11-15T11:30:45.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23542" PostTypeId="2" ParentId="22696" CreationDate="2017-11-15T15:02:54.753" Score="2" Body="&lt;p&gt;An observer on the far side of the moon would have hard time to explain it stands on a single planet, because of the movement of the most noticeable thing in the sky: the sun!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, because of the eccentricity of the moon's orbit around earth, the day length, i.e. the sun's &quot;speed in the sky&quot;, depends on where you stand in your lunar orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And from observations it can do, e.g. other planets that are nearly perfectly round in the solar system (and for well known reasons), she should be forced to rule out the &quot;I'm standing on an elliptical single celestial body&quot; hypothesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot calculate the variation of day length on the far side of the moon in a reasonable amount of time, sorry about that.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another effect I'll try to illustrate with Wikipedia images: the sun trajectory's elevation in the sky would change  year after year (cycle: between 8 and 9 earthlings years), due to moon's &lt;a href=&quot;https://en.wikipedia.org/wiki/Apsidal_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;apsidal precession&lt;/a&gt; and its tilted orbit plan:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Moon_apsidal_precession.png#/media/File:Moon_apsidal_precession.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/5/53/Moon_apsidal_precession.png/1200px-Moon_apsidal_precession.png&quot; height=&quot;350&quot; alt=&quot;Moon apsidal precession.png&quot;&gt;&lt;/a&gt;&lt;br&gt;By &lt;a href=&quot;//commons.wikimedia.org/wiki/User:Rfassbind&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;User:Rfassbind&quot;&gt;Rfassbind&lt;/a&gt; - Own work., Public Domain, &lt;a href=&quot;https://commons.wikimedia.org/w/index.php?curid=42565168&quot; rel=&quot;nofollow noreferrer&quot;&gt;Link&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Lunar_perturbation.jpg#/media/File:Lunar_perturbation.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/6/61/Lunar_perturbation.jpg&quot; alt=&quot;Lunar perturbation.jpg&quot;&gt;&lt;/a&gt;&lt;br&gt;By Geologician, Homunculus 2 - from English Wikipedia, &lt;a href=&quot;http://creativecommons.org/licenses/by/3.0&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Creative Commons Attribution 3.0&quot;&gt;CC BY 3.0&lt;/a&gt;, &lt;a href=&quot;https://commons.wikimedia.org/w/index.php?curid=23379433&quot; rel=&quot;nofollow noreferrer&quot;&gt;Link&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11264" LastActivityDate="2017-11-15T15:02:54.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23543" PostTypeId="2" ParentId="23522" CreationDate="2017-11-15T15:40:30.147" Score="3" Body="&lt;p&gt;It's very easy to create a star: you just put an amount of hydrogen gas quite near to each other and gravity will do the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that the amount of gas you need for that is so large that even replacing the whole earth with hydrogen atoms is far from enough: you need far more hydrogen atoms for that and as such you can't do something on the earth when you need more space than the whole content of the earth :-)&lt;/p&gt;&#xA;" OwnerUserId="14381" LastActivityDate="2017-11-15T15:40:30.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23544" PostTypeId="1" AcceptedAnswerId="23545" CreationDate="2017-11-15T15:47:00.647" Score="3" ViewCount="57" Body="&lt;p&gt;Most of the articles online say that the peak will be in a couple of days and around 2:00 AM in the night. I am currently living in Dubai, UAE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I use Google Sky Maps, it shows both the Leonids Meteor Shower and Leo Constellation below the horizon, when using the search feature. It only comes above the horizon at around 6:00 AM, when it is 45 minutes since sun rise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I am not sure whether it'd be dark enough to see them. What is my best course of action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;This link: &lt;a href=&quot;https://www.timeanddate.com/astronomy/meteor-shower/leonids.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.timeanddate.com/astronomy/meteor-shower/leonids.html&lt;/a&gt; shows Leonids will appear at 65° Azimuth and 1.7° altitude. How'd I interpret that?&lt;/p&gt;&#xA;" OwnerUserId="19683" LastEditorUserId="19683" LastEditDate="2017-11-15T17:24:19.853" LastActivityDate="2017-11-15T19:16:17.623" Title="Will I be able to see the Leonid Meteor Shower during mid night?" Tags="&lt;meteor-shower&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23545" PostTypeId="2" ParentId="23544" CreationDate="2017-11-15T17:17:46.053" Score="2" Body="&lt;p&gt;Leo starts to rise at about midnight, local time, so you should be fine, weather permitting.  Note also that you may be able to see meteors even before then, as Leo is merely that part of the sky from which the meteors appear to originate, but you will likely be able to see them some distance away from Leo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know where the Google rise time comes from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT; the link you added shows Leo rising at about midnight, confirming that the google source is incorrect, or not being correctly interpreted.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastEditorUserId="11822" LastEditDate="2017-11-15T19:16:17.623" LastActivityDate="2017-11-15T19:16:17.623" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23549" PostTypeId="1" CreationDate="2017-11-16T03:28:43.950" Score="2" ViewCount="246" Body="&lt;p&gt;Could it be possible that dark matter is emitting EM radiations on a scale that is not known or perceivable by humans yet and undetectable by man-made instruments? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A type of radiation that our instruments were just not designed to pick up on? Being insensitive and not advanced enough? Thus we do not yet have all the pieces to solve the puzzle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; We need to remember that there are infrared and gamma rays on either end of the spectrum and we are aware that dark matter is definitely not emitting visible light, or any of the other type of EM waves. Considering that gamma rays were discovered roughly only a hundred years ago, isn't it possible that the spectrum could be further extended on either side and that we're still not aware of it?&lt;/p&gt;&#xA;" OwnerUserId="19116" LastEditorUserId="19116" LastEditDate="2017-11-23T15:20:48.410" LastActivityDate="2017-11-23T15:20:48.410" Title="Dark matter emitting EM radiations beyond our currently known spectrum?" Tags="&lt;dark-matter&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23554" PostTypeId="2" ParentId="23549" CreationDate="2017-11-16T06:54:36.703" Score="2" Body="&lt;p&gt;The namesake of dark matter is because it does not emit electromagnetic radiation. Humans have built detectors on all ranges on the electromagnetic spectrum and a single particle will likely not emit wavelengths that are very large. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, dark matter may produce light in collisions with other dark matter particles or normal matter. That is the premise of large scintillator detectors. &lt;/p&gt;&#xA;" OwnerUserId="18327" LastActivityDate="2017-11-16T06:54:36.703" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23555" PostTypeId="1" AcceptedAnswerId="23589" CreationDate="2017-11-16T06:59:01.243" Score="3" ViewCount="117" Body="&lt;p&gt;Reading that the &lt;a href=&quot;https://en.wikipedia.org/wiki/Leonids&quot; rel=&quot;nofollow noreferrer&quot;&gt;Leonids&lt;/a&gt; meteor shower is almost upon us, I clicked some links (e.g. &lt;a href=&quot;http://earthsky.org/astronomy-essentials/everything-you-need-to-know-leonid-meteor-shower&quot; rel=&quot;nofollow noreferrer&quot;&gt;EarthSky.org&lt;/a&gt; and &lt;a href=&quot;https://news.nationalgeographic.com/2017/11/how-to-see-leonid-meteor-shower-skywatching-space-science/&quot; rel=&quot;nofollow noreferrer&quot;&gt;NatGeo&lt;/a&gt;) to read more of the prediction; especially the part about when to best view them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had a &lt;a href=&quot;https://en.wikipedia.org/wiki/D%C3%A9j%C3%A0_vu&quot; rel=&quot;nofollow noreferrer&quot;&gt;Déjà vu&lt;/a&gt; when I saw &quot;early morning hours&quot; because it seemed to me that &quot;that's what they always say&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing that it's not a coincidence that the constellation for which the shower is named always seems to rise around or after midnight at the time of year when the shower occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'd like to ask if this is true, that they pretty much always are best viewed after midnight (except when issues of weather or the Moon interfere), and if there has ever been a shower where this wasn't the case — one that was best viewed &lt;em&gt;before midnight&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-11-16T07:25:55.520" LastActivityDate="2017-11-17T21:18:43.073" Title="Meteor shower predicted to be best viewed BEFORE midnight (for reasons other than the moon)?" Tags="&lt;meteor-shower&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23560" PostTypeId="2" ParentId="23555" CreationDate="2017-11-16T09:39:48.020" Score="3" Body="&lt;p&gt;Not that I know of, and it would somewhat surprise me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of meteors from a shower you will see depends on the time of night through a number of reasons (not considering the weather and the moon):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The distribution of meteoroids on Earth's orbit is not homogeneous. The highest potential occurs when Earth moves through the peak of the distribution. This is associated with a certain time on Earth.&lt;/li&gt;&#xA;&lt;li&gt;Meteors from a shower appear tot come from the same spot in the sky. This spot is called the &lt;em&gt;radiant&lt;/em&gt; and moves through the sky along with the stars. The higher the altitude (height) of the radiant, the more meteors you will be able see.&lt;/li&gt;&#xA;&lt;li&gt;Meteors enter Earth's atmosphere because Earth moves into the meteoroid cloud. This means that you will see more meteors when the spot on Earth where you are standing is facing the meteoroid cloud. This is always the direction in which the Earth is moving. If you are on the other side, the meteoroid cloud is 'behind' you and possible meteors will have to move faster than the Earth in order to enter the atmosphere. As pointed out by user Florin Andrei in the comments, this effect is also called the &quot;bugs on the windshield&quot; effect, because &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Earth is like a car moving fast, and meteors are like insects that get splattered. Of course you get more bugs being splattered on the piece of glass that's facing forward. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Depending on your location, 1. and 2. could combine to maximise their effect some time before midnight. However, the effect of 3. is quite big. Because of the way the Earth rotates, this will always happen early in the morning, before sunrise.&lt;/p&gt;&#xA;" OwnerUserId="11824" LastEditorUserId="11824" LastEditDate="2017-11-16T20:40:00.473" LastActivityDate="2017-11-16T20:40:00.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23563" PostTypeId="1" AcceptedAnswerId="23567" CreationDate="2017-11-16T13:03:37.163" Score="6" ViewCount="1088" Body="&lt;p&gt;Today, I thought to make a list of Astronomical objects, A to Z. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It goes as follows :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;A&lt;/strong&gt; - Asteroid&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;B&lt;/strong&gt; - Black Hole&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;C&lt;/strong&gt; - Comet&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;D&lt;/strong&gt; - Dwarf Planet&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;E&lt;/strong&gt; - Elliptical Galaxy&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;F&lt;/strong&gt; - Falling Star&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;G&lt;/strong&gt; - Globular Cluster&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;H&lt;/strong&gt; - Hypergiants (Stars)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;I&lt;/strong&gt; - Interstellar Medium&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;J&lt;/strong&gt; - Jets (Astrophysical)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;K&lt;/strong&gt; -  $\color{red} ?$&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;L&lt;/strong&gt; - Lenticular Galaxy&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;M&lt;/strong&gt; - Meteor&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;N&lt;/strong&gt; - Nebula&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;O&lt;/strong&gt; - Open Cluster&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;P&lt;/strong&gt; - Pulsar&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Q&lt;/strong&gt; - Quasar&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;R&lt;/strong&gt; - Rogue Planets&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;S&lt;/strong&gt; - Satellites&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;T&lt;/strong&gt; -   $\color{red} ?$&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;U&lt;/strong&gt; - Universe&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;V&lt;/strong&gt; - Variable Star&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;W&lt;/strong&gt; - White Dwarf&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;X&lt;/strong&gt; - X-ray Star&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Y&lt;/strong&gt; - Yellow dwarf star&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Z&lt;/strong&gt; -    $\color{red} ?$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But I couldn't think of something well-known from the letters &lt;strong&gt;K, T&lt;/strong&gt; and &lt;strong&gt;Z&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to avoid using name of something specific object for ex. &lt;strong&gt;K&lt;/strong&gt; - Kuiper Belt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please help me in completing this list?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="16477" LastEditorUserId="16477" LastEditDate="2017-11-16T19:54:04.030" LastActivityDate="2017-11-17T21:08:30.620" Title="Astronomical ABCD...!" Tags="&lt;fundamental-astronomy&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23564" PostTypeId="1" CreationDate="2017-11-16T13:17:15.373" Score="1" ViewCount="76" Body="&lt;p&gt;Something obligatory up-to-dated.&lt;/p&gt;&#xA;" OwnerUserId="19705" LastActivityDate="2017-11-16T13:17:15.373" Title="Does there exist any of open database or service with ALL discovered stars and their parameters?" Tags="&lt;star&gt;&lt;resource-request&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23566" PostTypeId="1" AcceptedAnswerId="23571" CreationDate="2017-11-16T14:33:22.057" Score="3" ViewCount="146" Body="&lt;p&gt;I remember seeing in a video an instrument that was used to measure a star's radius with nothing more than the light it emitted. &#xA;The device worked by opening or closing its mechanism in increments, while watching a display. The display started out showing some shape, maybe some lines or a circle, with blurriness. The closest you could get to making the shape blurry-less would determine the instrument's approximation of the star's radius. &#xA;My question is, what is the name of this instrument? &lt;/p&gt;&#xA;" OwnerUserId="19454" LastActivityDate="2017-11-16T19:53:00.490" Title="Device used to measure stellar radii" Tags="&lt;visualization&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23567" PostTypeId="2" ParentId="23563" CreationDate="2017-11-16T14:49:50.047" Score="7" Body="&lt;p&gt;Seems a little contrived, essentially you just have to dig deep enough to find a specific enough category that matches the letter. But here you go :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;K - KBO (Kuiper Belt &lt;em&gt;Object&lt;/em&gt;) , if you still don't like that you could use&lt;/p&gt;&#xA;&#xA;&lt;p&gt;K - Kreutz Sungrazer (A type of comet, &lt;a href=&quot;https://en.wikipedia.org/wiki/Kreutz_sungrazer&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Kreutz_sungrazer&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T - T-Tauri Star (A type of variable star, &lt;a href=&quot;https://en.wikipedia.org/wiki/T_Tauri_star&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/T_Tauri_star&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Z - Z Camelopardalis star (&lt;a href=&quot;https://en.wikipedia.org/wiki/Z_Camelopardalis_star&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Z_Camelopardalis_star&lt;/a&gt; redirects to Dwarf Nova, scroll down)&lt;/p&gt;&#xA;" OwnerUserId="19004" LastActivityDate="2017-11-16T14:49:50.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23568" PostTypeId="2" ParentId="23566" CreationDate="2017-11-16T16:48:46.353" Score="3" Body="&lt;p&gt;Measuring a star's radius directly (rather than inferring it through empirical relations to other observables) is extremely difficult. As &lt;a href=&quot;https://astronomy.stackexchange.com/questions/23566/device-used-to-measure-stellar-radii#comment40531_23566&quot;&gt;barrycenter points out,&lt;/a&gt; it can't be done for most stars because they're too far away. However, we have been successful in measuring the radius of a small selection of very close stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process is primarily done using infrared-wavelength interferometry. For example, this method has been used to &lt;a href=&quot;https://en.wikipedia.org/wiki/Betelgeuse#Recent_studies&quot; rel=&quot;nofollow noreferrer&quot;&gt;directly measure the radius of Betelguese&lt;/a&gt;, using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Infrared_Spatial_Interferometer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Infrared Spatial Interferometer (ISI)&lt;/a&gt;. I have no source for this, but I believe the USNO has performed a similar measurement for about 20 stars (I recall this from a seminar I saw years ago).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using interferometry is the only way I know that a stellar radii has directly been observed. I really don't have any guess as to what you saw.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-11-16T16:48:46.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23569" PostTypeId="2" ParentId="23563" CreationDate="2017-11-16T18:14:45.630" Score="3" Body="&lt;p&gt;for &lt;strong&gt;K&lt;/strong&gt; we have &lt;a href=&quot;https://en.wikipedia.org/wiki/Kilonova&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kilonova&lt;/a&gt; - a &quot;transient astronomical event that occurs in a compact binary system when two neutron stars or a neutron star and a black hole merge into each other&quot;. On 17 August 2017, a &lt;a href=&quot;https://en.wikipedia.org/wiki/GW170817&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravitational wave detection&lt;/a&gt; coinciding with a kilonova made a big news as the first such observation and an important milestone in science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My suggestions for &lt;strong&gt;T&lt;/strong&gt; are less good. &lt;a href=&quot;https://en.wikipedia.org/wiki/TAU_(spacecraft)&quot; rel=&quot;nofollow noreferrer&quot;&gt;TAU&lt;/a&gt; was a space probe proposed in 1987. Not exactly an astronomical body, though, and rather unknown. If you don't mind constellations, &lt;a href=&quot;https://en.wikipedia.org/wiki/Taurus_(constellation)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Taurus&lt;/a&gt; would be my go-to suggestion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for  &lt;strong&gt;Z&lt;/strong&gt;, my best suggestion would be &lt;a href=&quot;https://en.wikipedia.org/wiki/Alnitak&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zeta Orionis&lt;/a&gt; - a star system in the Orion's belt. Its primary star is the brightest Class O star in the night sky. Also known as Alnitak. &lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2017-11-16T18:14:45.630" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23570" PostTypeId="2" ParentId="23549" CreationDate="2017-11-16T18:48:26.373" Score="2" Body="&lt;p&gt;If it could produce EM radiation, that means it could interact via EM fields. That means there would be internal friction in the dark matter clouds, because friction ultimately boils down to electric forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the dark matter distributions we're observing seem to indicate that there is no friction between dark matter particles, whatever they are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ergo, dark matter cannot produce EM radiation at any frequency or wavelength.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2017-11-16T18:48:26.373" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23571" PostTypeId="2" ParentId="23566" CreationDate="2017-11-16T18:49:42.847" Score="5" Body="&lt;p&gt;I am not sure what you are referring to; possibly it is &lt;em&gt;intensity interferometry&lt;/em&gt; using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hanbury_Brown_and_Twiss_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hanbury Brown &amp;amp; Twiss&lt;/a&gt; method, but even this requires two telescopes. This kind of interferometry was used to estimate angular diameters for some tens of stars but is not generally used these days for this purpose. Instead, one constructs interferometers from separated telescopes using delay lines to compensate for path-length differences and maintain coherence between the different paths that the light takes. This then uses &lt;em&gt;amplitude interferometry&lt;/em&gt; to reconstruct the angular diameter of a star from the interference pattern seen in different telescope baseline configurations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radii (or to be more exact, the angular size) of many nearby stars and also of many giant stars at rather larger distances, have been measured using interferometric techniques. An accurate distance is then required to estimate an absolute radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relevant papers are by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2010ASPC..432....3B&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boyajian 2010&lt;/a&gt; and &lt;a href=&quot;https://arxiv.org/abs/1306.2974&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boyajian et al. (2013)&lt;/a&gt;. There are a number of such interferometric experiments, mostly working in the infrared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also direct radius measurements for stars that are occulted by the Moon and other solar system bodies - the time-dependent diffraction pattern can yield an angular size for the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Absolute radii are also provided by eclipsing binary systems (independent of whether they have a known distance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10182/how-do-astronomer-measures-the-size-of-any-celestial-objects&quot;&gt;How do astronomer measures the size of any celestial objects?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-11-16T19:53:00.490" LastActivityDate="2017-11-16T19:53:00.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23572" PostTypeId="2" ParentId="23563" CreationDate="2017-11-16T18:56:58.097" Score="3" Body="&lt;p&gt;Z - Zodiacal light&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T - why not T-dwarf ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;K - err K-dwarf or K-giant if you want variety.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think your F is a bit lame.  You could go for FU Ori object.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-11-16T18:56:58.097" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23573" PostTypeId="1" CreationDate="2017-11-16T21:27:59.370" Score="2" ViewCount="303" Body="&lt;p&gt;I am reading a book on the Coopernican Revolution where the author states that Tycho Brahe was able to make twenty times more accurate measurements than any of his predecessors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I remember having heard a lecture where the speaker said that Tycho did not significantly improve measurements since Ptolemy because no new equipment (telescope) had been invented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, which statement is correct? &lt;/p&gt;&#xA;" OwnerUserId="1293" LastActivityDate="2017-11-17T01:02:44.187" Title="Did Tycho Brahe improve measurements twentyfold?" Tags="&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23574" PostTypeId="2" ParentId="23573" CreationDate="2017-11-16T23:54:27.140" Score="2" Body="&lt;p&gt;I've read 10 fold, not 20 but I'm not sure which is more accurate.   He didn't use a telescope but that doesn't mean he didn't have new equiptment.   He had equipment specially built, some of it, 10 feet high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;https://www2.hao.ucar.edu/Education/FamousSolarPhysicists/tycho-brahes-observations-instruments&quot; rel=&quot;nofollow noreferrer&quot;&gt;article describes some of his equipment.&lt;/a&gt;  Nobody before him used equiptment this good.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He more accurately measured the placement and precise angles in the sky of many stars and planets with 10-20 times better accuracy than anyone before him.   He didn't use a telescope but that didn't matter.   He didn't look deeper into the solar-system or universe, but he mapped it more accurately.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was Brahe's accurate placement of Mars that helped Kepler work out his laws.   There was even a rumor that Kepler killed Brahe to get full time use of his observatory and his notes on Mars, but &lt;a href=&quot;http://www.historytoday.com/richard-cavendish/death-tycho-brahe&quot; rel=&quot;nofollow noreferrer&quot;&gt;exhumation provided no evidence of that&lt;/a&gt;.   Brahe was known for his voracious appetite and partying and is thought to have died from excessive drinking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The telescope didn't really help Kepler with his work either though he used one.  Kepler's skill was primarily in mathematics, which helped him work out the laws of orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galileo, who came about the same time as Kepler, but in a different part of Europe, built a better telescope, that was also a 10 fold leap forward.   The Telescope was instrumental in some of his key discoveries like Jupiter's moons and the phases of Venus.   I don't know if Kepler had access to Galileo's telescope.  It was considered a military secret.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-11-16T23:54:27.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23575" PostTypeId="2" ParentId="23573" CreationDate="2017-11-17T01:02:44.187" Score="5" Body="&lt;p&gt;Accuracy relative to other authors is hard to estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, here's a scholarly article that shows a modern determination of the actual accuracy of Brahe's measurements in absolute terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsbit.harvard.edu/full/1978JHA.....9...42W&quot; rel=&quot;noreferrer&quot;&gt;http://adsbit.harvard.edu/full/1978JHA.....9...42W&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conclusion is that it depends on the instrument and it depends on the actual measurement, but as a general rule he was able to obtain an accuracy better than 1 arc minute, often down to 0.5 arcmin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparison, the resolving power of the healthy, young, average human eye is around 1 arcmin. So Brahe was able to somehow make measurements with an accuracy better than the smallest details resolved by the human eye, even though there were no magnifying instruments available in his time. I believe this was achieved by averaging multiple measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, he was very near the limit of what was theoretically achievable before the invention of the telescope. He was certainly the best of his era by a large margin, and far better than distant predecessors such as Ptolemy. What mattered here was the methodology, and the unprecedented scale of the instruments (the armillary sphere, etc).&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2017-11-17T01:02:44.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23577" PostTypeId="1" CreationDate="2017-11-17T06:00:34.000" Score="2" ViewCount="572" Body="&lt;p&gt;Air mass relates to how much of the Earth's atmosphere the light from an observed object must pass through before being observed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At Zenith, and sea level, the value of &quot;air mass&quot; is set to 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light from objects lower to the horizon at an observing site must travel through more atmosphere, and hence the brightness of the object is somewhat diminished. In this situation the value of air mass increases above 1.000, presenting greater dimming for the observed object from increased atmospheric light scattering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As example, when making precise observations of the magnitude of an object in astronomy, there is a need to calculate the air mass for an object at the time of observation, determined in large part by its altitude angle from zenith, or from the geometric horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various formulas for calculating air mass. Precise calculations consider several factors and can be quite complicated.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Air_mass_(astronomy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;For more details see - Air Mass (Astronomy). Wikipedia.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question:&lt;/strong&gt;  Is there a simple formula to calculate the air mass at Zenith (only), for observing sites about sea level. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a calculation provides a comparison of the merit of an observing site at higher elevation over one at lower elevation in regard to the factor of air mass.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine for most elevations the factor is quite small. Naturally there are plenty of other factors to consider when it comes the the quality of an observing site - humidity, frequency of clear skies, human light sources and light pollution, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my understanding, the atmosphere in most calculations is considered to be 9 km in thickness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very simple calculation could be the site elevation subtracted from 9 km, divided by 9 km, providing a ratio value somewhat less than 1. However this calculation does not account for the changing density of the atmosphere with increasing elevation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a simple formula, what is it?&lt;/p&gt;&#xA;" OwnerUserId="13359" LastActivityDate="2018-01-17T12:46:29.220" Title="Calculating Air Mass at Zenith for observing sites above sea level?" Tags="&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23581" PostTypeId="1" AcceptedAnswerId="23582" CreationDate="2017-11-17T08:23:52.170" Score="4" ViewCount="62" Body="&lt;p&gt;Would it be true to assume that the mass of a galaxy is approximately the mass of all molecular clouds detected within it, let's say, through CO observations and then converted to H$_2$?&lt;/p&gt;&#xA;" OwnerUserId="17425" LastEditorUserId="17425" LastEditDate="2017-11-17T10:54:42.060" LastActivityDate="2017-11-17T10:54:42.060" Title="Relation between molecular cloud mass and galactic mass" Tags="&lt;mass&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23582" PostTypeId="2" ParentId="23581" CreationDate="2017-11-17T08:43:12.703" Score="8" Body="&lt;p&gt;The mass of an average galaxy appears to be totally dominated by dark matter, so your calculation would not give the galaxy mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if all you wanted was the baryonic (non dark matter) mass then what you suggest will be very much a lower limit. For example you can look at this paper by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2001ASPC..228..249C&quot; rel=&quot;noreferrer&quot;&gt;Chabrier (2001)&lt;/a&gt;, who estimates that gas forms less than half the baryonic mass in the Milky Way. This fraction will be even lower in elliptical galaxies that are largely gas-free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am no expert, but I also think that even the gas component is dominated by ionised and atomic gas where you wouldn't expect to see CO emission and which are better traced by recombination lines or 21cm radiation respectively. In other words, the molecular gas is a fraction of the gas, which is a fraction of the baryonic mass, which is a small fraction of the total mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an extensive review of the &quot;CO method&quot; by &lt;a href=&quot;https://arxiv.org/pdf/1301.3498.pdf&quot; rel=&quot;noreferrer&quot;&gt;Bolatto et al. (2013)&lt;/a&gt;. The conclusion appears to be that the molecular gas mass can be estimated to around 30%, but that the conversion factor can depend on the star forming environment and on the metallicty of the gas in poorly understood ways.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-11-17T10:37:40.887" LastActivityDate="2017-11-17T10:37:40.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23586" PostTypeId="2" ParentId="23577" CreationDate="2017-11-17T14:31:57.000" Score="1" Body="&lt;p&gt;I would use the wikipedia variable density atmosphere model&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Air_mass_(astronomy)#Variable-density_atmosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Air_mass_(astronomy)#Variable-density_atmosphere&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which gives airmass as a function of zenith distance for an exponentially decaying atmosphere of scale-height $H$. This formula is valid for all heights above sea level but the airmass given will be in terms of airmass at zenith from that height. To make your adjustment to an equivalent sea-level airmass then you would multiply by the ratio of the density at your observing altitude to the density at sea level.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-11-17T14:31:57.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23589" PostTypeId="2" ParentId="23555" CreationDate="2017-11-17T21:18:43.073" Score="2" Body="&lt;p&gt;The Draconids have a radiant that, from the the Northerly Latitudes (N America or Europe) is highest in the sky in the early evening of October 7th or 8th. This means that unlike most showers (which are best before dawn) the Draconids are most active after dusk. They are not visible to many Southern Hemisphere observers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are capable of producing short-spikes of activity, and there is a prediction that 2018 may generate a meteor storm, with over 1000 meteors per hour. It will coincide with a new moon, so could be rather special.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astronomy.com/observing/observe-the-solar-system/2010/04/meteors-and-meteor-showers&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomy.com&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://earthsky.org/astronomy-essentials/earthskys-meteor-shower-guide&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earthsky.org&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-11-17T21:18:43.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23590" PostTypeId="2" ParentId="23494" CreationDate="2017-11-17T22:35:53.067" Score="4" Body="&lt;p&gt;The most comprehensive catalogue of strong gravitational lenses that I am aware of is the &lt;a href=&quot;https://www.cfa.harvard.edu/castles/&quot; rel=&quot;nofollow noreferrer&quot;&gt;CASTLeS&lt;/a&gt; database with about 120 lens systems. Several of them have extended features, arcs, etc., while others are point sources. They provide the redshifts of lens and source where available, as well as time delays between multiple images and the velocity dispersion (a proxy for the mass) of the lens for a few.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would have to create a comparison dataset without strong lensing yourself, but that should not be difficult: Go to &lt;a href=&quot;https://archive.stsci.edu/prepds/udf/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble's Ultra Deep Field web page&lt;/a&gt;, download the images with the same filters as used in CASTLeS and cut out likely looking galaxies. The UDF contains mostly very high-redshift galaxies, so the chance of accidentally picking a strong-lens system is very slim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Disclaimer: I've only ever used ACS UDF images, not the NICMOS ones; if it's too hard, just skip the IR data.)&lt;/p&gt;&#xA;" OwnerUserId="13681" LastActivityDate="2017-11-17T22:35:53.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23591" PostTypeId="1" CreationDate="2017-11-17T23:43:59.190" Score="5" ViewCount="192" Body="&lt;p&gt;I am interested in studies and/or publications about planets that wander in space and are not orbiting around a star or its remnants. These dark and frozen worlds should've got ripped away from their respective system by a large passing body (another star? passing black hole) or any other astronomical event (which one?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curious, whether there exist statistical estimates on the number of such planets and the rate at which they are produced. I am also interested in reading about the statistical models that can produce those estimates and encode researchers' beliefs. If there are no such models, why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A provoking thought: According to Greek, such planets are really fitting their name πλάνητες ἀστέρες (planētes asteres, &quot;wandering stars&quot;) or simply πλανῆται (planētai, &quot;wanderers&quot;). It is frightening to imagine such world that drifts deeper and deeper into empty space. Forever.&lt;/p&gt;&#xA;" OwnerUserId="9761" LastEditorUserId="16477" LastEditDate="2017-11-20T16:46:27.847" LastActivityDate="2017-11-29T23:06:49.213" Title="Studies on planets without parent star" Tags="&lt;exoplanet&gt;&lt;rogue-planet&gt;&lt;resource-request&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23597" PostTypeId="2" ParentId="23577" CreationDate="2017-11-18T12:29:25.467" Score="1" Body="&lt;p&gt;A very crude approximation for &lt;em&gt;zenith&lt;/em&gt; air mass can be had using a simple &lt;a href=&quot;https://en.wikipedia.org/wiki/Scale_height&quot; rel=&quot;nofollow noreferrer&quot;&gt;scale height&lt;/a&gt; model and assume a uniform temperature and composition (which is not true) so you can treat density as proportional to pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case using $P(x) = P_0 exp(-x/H)$ where H is the scale height (very roughly 8,000 meters) and remembering that the integral of $exp(-x)$ is $-exp(-x)$ and that you've normalized sea level to 1.0, the zenith airmass can also be roughly approximated as $exp(-x/H)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So just for example, at 5,000 &lt;em&gt;feet&lt;/em&gt;, that's $exp(-(5000/3.3)/8000)$ or 0.83.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2017-11-18T12:29:25.467" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23607" PostTypeId="1" CreationDate="2017-11-19T19:00:01.500" Score="0" ViewCount="179" Body="&lt;p&gt;Suppose I want to build or buy a telescope for the following uses:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Astrophotography (mainly galaxies and nebulae).&lt;/li&gt;&#xA;&lt;li&gt;Transits of exoplanets and binary systems like Cygnus X-1.&lt;/li&gt;&#xA;&lt;li&gt;To search for BH+star systems.&lt;/li&gt;&#xA;&lt;li&gt;To measure the rotation curve of some spiral galaxies&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is that ALL feasible and/or cheap at amateur/semi-amateur level?&lt;/p&gt;&#xA;" OwnerUserId="2391" LastActivityDate="2020-01-10T01:07:31.520" Title="Telescope selection for concrete uses" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23615" PostTypeId="1" AcceptedAnswerId="23619" CreationDate="2017-11-20T14:15:28.067" Score="2" ViewCount="336" Body="&lt;p&gt;I know that meteors that We see as shooting stars travel tens of thousands of kilometers per hour and are tiny as a grain of sand, but after hitting the atmosphere and burn I suppose that its ashes would touch the ground at some point. How long does it take? Can it be measured or observed?&lt;/p&gt;&#xA;" OwnerUserId="19545" LastActivityDate="2017-11-21T03:45:34.940" Title="How much time does it take for a meteor (shooting star) to touch the ground after its combustion?" Tags="&lt;gravity&gt;&lt;planetary-atmosphere&gt;&lt;meteor&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2017-12-07T20:51:35.837" ContentLicense="CC BY-SA 3.0" />
  <row Id="23616" PostTypeId="2" ParentId="19027" CreationDate="2017-11-20T19:30:54.693" Score="1" Body="&lt;p&gt;The hierarchical picture of cosmology posits that the smallest structures formed first, early in the evolution of the Universe, and that more massive systems assemble from these initial structures.  The smallest structures therefore have the highest concentrations according to the picture you present, as the Universe had a higher mean density at the time. As the Universe evolves, the average density decreases, and the largest systems form from the smaller structures. Therefore, the higher the mass of the system, the smaller the concentration, whereas the smallest structures keep their initial concentration from their formation time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key here is that smaller structures are older than larger structures, therefore have a higher concentration.&lt;/p&gt;&#xA;" OwnerUserId="9826" LastActivityDate="2017-11-20T19:30:54.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23617" PostTypeId="2" ParentId="23607" CreationDate="2017-11-20T19:48:36.000" Score="3" Body="&lt;p&gt;(1) is very doable. The others don't seem doable, at least not easily. If you practice astrophotography for a while you'll have a better understanding of the difficulties involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's best (but not mandatory) if you begin doing plain visual astronomy, with something like an 8&quot; dobsonian, for a year or two, to get yourself familiarized with how it all works. Once that's accomplished, the transition to astrophotography should be easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For AP (astrophotography) you'll need: a mount, a camera, a telescope. The mount is by far the most important. The camera is very important. Oddly enough, the telescope is the least important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The accuracy, overall quality, and load carrying capacity of the mount will determine how well the whole system works. Just get the best mount you can afford.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Entry level mount examples: Celestron CGEM, Sky-Watcher EQ6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Midrange mounts: Losmandy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;High-end mounts: Software Bisque Paramount, Astro-Physics GTO&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Cameras: for solar system objects like planets or the Moon, you'll need cameras with smaller sensors that can take lots of frames quickly (planetary cameras). For galaxies and nebulae (DSO - deep space objects) you'll need cameras that perform well with long exposure (DSO cameras); the best DSO cameras have cooler elements installed on the sensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are OSC (one-shot color) cameras that don't require filters. There are monochrome cameras that require a filter box with R, G, B filters; you need to take separate exposures through each filter. OSC cameras are easier to use; monochrome cameras tend to provide higher performance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planetary camera examples: ZWO ASI 224 MC, ZWO ASI 178 MM&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DSO camera examples: ZWO ASI 1600 MM, ZWO ASI 071 MC-Cool&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Telescope:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For DSO it is very highly recommended to start with a relatively small refractor with short focal length. Something like the Orion ED80. Add a focal reducer / field flattener too. The long exposures require very high precision tracking from the mount, and a long focal length only creates problems. Keep it short until you learn how to make everything work. You could use a larger instrument like an SCT, but make sure your mount can carry that weight, and make sure you've fine tuned the process so the very long focal length of the SCT doesn't amplify the tracking errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For planetary imaging, the bigger the aperture the better. Use an SCT such as a Celestron C8 or bigger (up to C14 if the mount can carry it).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Capture software:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For DSO: Nebulosity (beginner), Sequence Generator Pro (advanced)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For planetary: FireCapture&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Image processing software:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For DSO: Deep Sky Stacker (beginner), PixInsight (advanced)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For planetary: AutoStakkert for stacking. Subsequent processing could be done with Registax, or Astra Image.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Guiding:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For DSO, you'll do very long exposures. The mount alone is not precise enough. You need to guide it. This is done by using a smaller telescope (guidescope) installed on the main telescope, using a small camera (guide camera).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The finderscope that's normally used for visual astronomy could be used as a guidescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a DSO camera and a planetary camera, you could use the planetary camera as a guide camera for DSO photography.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll have to connect the camera to a software that can control the mount and issue corrections to it. The most popular app is PHD2 a.k.a. Open PHD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When doing DSO imaging, you'll spend most time during your first year basically learning all the subtle aspects of guided exposures.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Learning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'The Deep-sky Imaging Primer' by Charles Bracken. A bit old but still very relevant.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2017-11-20T20:15:03.880" LastActivityDate="2017-11-20T20:15:03.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23618" PostTypeId="1" CreationDate="2017-11-20T23:25:34.023" Score="5" ViewCount="172" Body="&lt;p&gt;NASA recently discovered &lt;a href=&quot;https://www.nasa.gov/feature/solar-system-s-first-interstellar-visitor-dazzles-scientists&quot; rel=&quot;nofollow noreferrer&quot;&gt;'Oumuamua&lt;/a&gt;, an interstellar asteroid that appears to have a very elongated shape, unseen in solar system asteroids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mZzKF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mZzKF.jpg&quot; alt=&quot;oumuamua&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What physical mechanisms, in the progenitor stellar system or from interstellar transit, could lead to the formation of an asteroid with such a shape?&lt;/p&gt;&#xA;" OwnerUserId="941" LastEditorUserId="11264" LastEditDate="2017-11-27T15:04:54.167" LastActivityDate="2017-11-27T15:38:08.337" Title="Potential formation mechanisms of the interstellar asteroid 'Oumuamua" Tags="&lt;solar-system&gt;&lt;asteroids&gt;&lt;interstellar-medium&gt;&lt;interstellar&gt;&lt;oumuamua&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="23619" PostTypeId="2" ParentId="23615" CreationDate="2017-11-21T03:45:34.940" Score="3" Body="&lt;p&gt;I think the best model we could use for this is nuclear test fallout, whose dispersion was monitored quite extensively.  It's very, very dependent on weather patterns at altitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Castle_Bravo#Fallout&quot; rel=&quot;nofollow noreferrer&quot;&gt;Castle Bravo test&lt;/a&gt; mushroom cloud reached at altitude of about 40km and hence hit the Jet Stream.  In addition to this the wind speeds at lower altitudes were higher than ground speeds and this was not checked for, so the combination of factors spread the fallout downwind about 280 miles (about 430 km) in about 16 hrs.  This was one reason for the irradiation of many people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What these tests and other nuclear incidents show us is that material ejected into the high atmosphere can remain there a long time, even circling the globe &lt;a href=&quot;https://www.washingtonpost.com/national/as-fukushima-fallout-circles-the-globe-nuclear-sleuths-sift-it-for-clues/2011/03/30/AFfnQBHC_story.html?utm_term=.9674a018df89&quot; rel=&quot;nofollow noreferrer&quot;&gt;like traces from Fukushima&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory trace particles from a meteor could remain in the upper atmosphere for days, months, even years.  Most would be expected to fall within days, however.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2017-11-21T03:45:34.940" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23620" PostTypeId="1" CreationDate="2017-11-21T03:46:31.580" Score="10" ViewCount="467" Body="&lt;p&gt;News articles have mentioned that the recently discovered asteroid &lt;a href=&quot;https://www.nasa.gov/feature/solar-system-s-first-interstellar-visitor-dazzles-scientists&quot; rel=&quot;noreferrer&quot;&gt;'Oumuamua&lt;/a&gt; is theorized to be greatly elongated based on observations of brightness variation. However, this hypothesis seems to be based on the assumption that the object's albedo distribution is reasonably uniform.&#xA;What if it isn't? Could large periodic brightness variations also be explained by 'Oomuamua having two highly contrasting hemispheres like &lt;a href=&quot;https://en.wikipedia.org/wiki/Iapetus_(moon)&quot; rel=&quot;noreferrer&quot;&gt;Iapetus&lt;/a&gt;? If the astronomers involved have already eliminated this competing explanation, how would they have done so and with what additional observational data?&lt;/p&gt;&#xA;" OwnerUserId="19769" LastActivityDate="2017-11-21T12:26:11.533" Title="Could asteroid 'Oumuamua actually be round?" Tags="&lt;asteroids&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23626" PostTypeId="1" CreationDate="2017-11-21T11:53:58.033" Score="2" ViewCount="275" Body="&lt;p&gt;Where does the energy from a supernova come from? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that the iron core collapses into a large ball of neutrons - is that effectively a star-sized nuclear explosion? What is the reaction product called? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the effect go supercritical, instead of finding some equilibrium rate of burn? I've read discussion that the reaction energy does not appear to be enough to explain the energy seen in a supernova? How big is the gap?&lt;/p&gt;&#xA;" OwnerUserId="23008" OwnerDisplayName="James McLellan" LastEditorUserId="11264" LastEditDate="2017-11-22T07:29:17.523" LastActivityDate="2017-11-22T07:29:17.523" Title="Mechanics of Supernovae" Tags="&lt;supernova&gt;&lt;stellar-astrophysics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23627" PostTypeId="2" ParentId="23620" CreationDate="2017-11-21T12:26:11.533" Score="9" Body="&lt;p&gt;The hypothesis that the object has an extreme aspect ratio comes from both the max/min ratio of the light curve &lt;em&gt;and the shape&lt;/em&gt; of the light curve, along with some assumptions of the uniformity of the albedo as you mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you read the BBC article &lt;a href=&quot;http://www.bbc.com/news/science-environment-42053634&quot; rel=&quot;noreferrer&quot;&gt;Bizarre shape of interstellar asteroid&lt;/a&gt; and click from that page on the link that says &lt;strong&gt;which is published in Nature journal.&lt;/strong&gt; it will take you to an open access draft of the Nature Letter. In it you can see the shape of the light curve. I've chopped up some sections to show below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;dip&quot; is quite sharp. Remember magnitudes are a log scale, and so the dip would be even sharper on a linear scale. Since 5 magnitudes are a factor of ~100, a 2.5 magnitude dip is a factor of ~10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what you would get when a high-aspect ratio objects so that is viewed end-on. A black spot would darken the curve much more gradually, being viewed for a longer fraction of the cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not proof of the high-aspect ratio hypothesis, and the authors don't claim this. However, it's the best way to fit the curve with the fewest assumptions, as far as I can tell.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Screen shot of the on-line Accelerated Preview of the Nature Letter &lt;a href=&quot;https://www.nature.com/articles/nature25020&quot; rel=&quot;noreferrer&quot;&gt;A brief visit from a red and extremely elongated interstellar asteroid&lt;/a&gt; K. Meech et al. doi:10.1038/nature25020&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3pi00.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3pi00.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/07FbZ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/07FbZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jKTpy.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jKTpy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kPvAl.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kPvAl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2017-11-21T12:26:11.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23628" PostTypeId="2" ParentId="23385" CreationDate="2017-11-21T12:46:38.220" Score="3" Body="&lt;p&gt;An astronomical CCD is in principle sensitive across its entire active surface; there are no physical gaps between pixels. The pixels themselves are controlled and maintained by the underlying electronic circuitry, which sets up potential barriers between individual pixels. In practice, there are variations in the actual sensitivity within individual pixels: sensitivity is highest in the center and falls off towards the edges, although it never goes to zero anywhere within the pixel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.publish.csiro.au/AS/pdf/AS05013&quot; rel=&quot;nofollow noreferrer&quot;&gt;This paper (PDF)&lt;/a&gt; is a recent analysis of a (frontside-illuminated) CCD; Figure 13 shows the measured pixel response functions (how sensitive an individual pixel is across its surface), and Figures 14 and 15 show the photometric sensitivity map (what the sensitivity is for uniform illumination). In the latter case, you can see that the sensitivity never varies more than about 20% or so, depending on the wavelength of the light. Some of the variation corresponds directly to wiring within the CCD absorbing some of the light, but this has a very minor effect on the sensitivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The comparison made in Florin Andrei's answer to a consumer-camera-style CMOS sensor, as shown in the image, is misleading, both because CMOS pixels are physically somewhat more discrete in ways that CCD pixels are not, and because astronomical CCDs do not have per-pixel filters glued on top of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The argument that &quot;there has to be a narrow slice of silicon in between pixels to prevent short-circuits&quot; is not correct for CCDs; the electrical separation between pixels is maintained by the underlying electronics, not by any material barrier. In fact, you have to be able to &lt;em&gt;create&lt;/em&gt; temporary &quot;short-circuits&quot; between adjacent pixels in a column in order to transfer the accumulated electrons during readout. That's what the &quot;charger-transfer&quot; in &quot;charger-transfer device&quot; means.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2017-11-21T12:46:38.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23630" PostTypeId="2" ParentId="23626" CreationDate="2017-11-21T13:56:26.390" Score="6" Body="&lt;p&gt;The energy derives from gravitational potential energy. The core of a bit more than a solar mass collapses from the size of the Earth to a 10km radius. Some of the gravitational energy (a tiny percentage) released is transferred to the overlying envelope and blasts it into space. Further energisation takes place due to radioactive decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ball of neutrons is not a star-sized nuclear explosion   ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand what you mean by reaction product? The proto-neutron star consists mainly of neutrons(!) and neutron-rich nuclei.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The collapse occurs because electron degeneracy pressure is circumvented by neutronisation (inverse beta decay), which removes electrons from the gas. At that point, the collapse occurs on a freefall time of $&amp;lt;1$ second. There is no &quot;burn-rate&quot; as such; just a collapse and then a &quot;core bounce&quot; as the equation of state hardens because of neutron degeneracy pressure and the strong nuclear force between closely packed neutrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;observed&quot; energy (i.e. the electromagnetic output and ejecta kinetic energy)  is only of order 1% of the energy released from the gravitational collapse. Most of it emerges in the form of unseen neutrinos. So there is no energy gap and plenty of energy to power what is observed.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-11-21T14:47:10.357" LastActivityDate="2017-11-21T14:47:10.357" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23631" PostTypeId="1" CreationDate="2017-11-21T14:08:36.710" Score="2" ViewCount="556" Body="&lt;p&gt;I'm trying to convert some pixel coordinates I have into WCS coordinates, ideally into a WCS region for use in some further analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I've been able to load and parse a NuSTAR FITS file, do some analysis to make my selections and get the pixel coordinates for the image. For example, after my analysis I would end up having selected a rectangle of &lt;code&gt;x&lt;/code&gt; values from pixels 480 to 518, and &lt;code&gt;y&lt;/code&gt; values from 478 to 516.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that I tried to use the astropy WCS module to convert them, but it doesn't seem to find the required data in the header to actually do the conversion and just says that pixel 480 gives coordinate 480. That or I'm just doing something wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I looked through the FITS header myself, found that some keys (&lt;code&gt;TCRPX&lt;/code&gt;, &lt;code&gt;TCRVL&lt;/code&gt;, &lt;code&gt;TCDLT&lt;/code&gt;) give the reference pixel, reference pixel degree coordinate and pixel axis scale values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now I'm really stuck since what I need to get is a region file like those produced by ds9, in the format of:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# Region file format: DS9 version 4.1&#xA;global color=green dashlist=8 3 width=1 font=&quot;helvetica 10 normal roman&quot; select=1 highlite=1 dash=0 fixed=0 edit=1 move=1 delete=1 include=1 source=1&#xA;fk5&#xA;circle(19:58:21.593,+35:11:57.46,10&quot;)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To then run this through the &lt;code&gt;nuproducts&lt;/code&gt; tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TL;DR: Does anybody know how to convert the NuSTAR FITS physical pixel coordinates to FK5 coordinates?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: Wrote some &lt;a href=&quot;https://github.com/RobertRosca/MPhys-Project-Work/blob/master/FITS%20Coord%20Conversion.jl&quot; rel=&quot;nofollow noreferrer&quot;&gt;code to extract the coordinates&lt;/a&gt;. It converts the pixel to degree fine, but converting the FK5 $\alpha$ angle to sexagesimal doesn't work correctly for... some reason.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT2: Everything works now, &lt;a href=&quot;https://gist.github.com/RobertRosca/c91a124d29e26ada181b9bdea4d54169&quot; rel=&quot;nofollow noreferrer&quot;&gt;here's a notebook where I run through it all&lt;/a&gt; in case anybody runs into a similar problem. I go through extracting the right-ascension and declination in degrees from the pixel coordinates given by the FITS file, then converting these degrees to sexagesimal (h:m:s) angles. From my tests it seems to work quite well, results are a tad off  due to what I assume are some floating point arithmetic errors.&lt;/p&gt;&#xA;" OwnerUserId="19779" LastEditorUserId="6" LastEditDate="2019-06-20T21:08:26.007" LastActivityDate="2019-07-20T22:02:10.257" Title="Converting FITS (NuSTAR) coordinates to WCS" Tags="&lt;observational-astronomy&gt;&lt;coordinate&gt;&lt;python&gt;&lt;fits-header&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23634" PostTypeId="1" AcceptedAnswerId="27782" CreationDate="2017-11-21T20:40:30.450" Score="6" ViewCount="244" Body="&lt;p&gt;&quot;Oumuamua came from the direction of Vega, a bright star in the Lyra constellation. But even at 85,700 miles per hour, it took so long to reach our solar system that Vega wasn't in the same position 300,000 years ago.&quot;&#xA;&lt;a href=&quot;http://www.cnn.com/2017/11/20/world/first-interstellar-object-solar-system/index.html&quot; rel=&quot;noreferrer&quot;&gt;CNN&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is information on the motions of stars from &lt;a href=&quot;https://www.youtube.com/watch?v=Ag0qsSFJBAk&quot; rel=&quot;noreferrer&quot;&gt;ESA's Gaia mission (4 min. youtube with description)&lt;/a&gt;, so it should be possible to project the motion of Oumuamua back it time to see if it was near any stars. Has anyone done this?&lt;/p&gt;&#xA;" OwnerUserId="17743" LastActivityDate="2021-04-25T21:54:32.720" Title="Did interstellar asteroid Oumuamua pass near other stars?" Tags="&lt;star&gt;&lt;asteroids&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23644" PostTypeId="1" AcceptedAnswerId="23650" CreationDate="2017-11-22T11:27:07.780" Score="6" ViewCount="371" Body="&lt;p&gt;If Oumuamua wasn't an asteroid, could we detect that it is not an asteroid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disclaimer: I do NOT suggest that Oumuamua is in fact an alien spacecraft, a giant coffin, or a statue of Weird Al.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it apparently has a very unusual shape, if this is something else, with a different material (neither rock nor ice), could we detect that something is unusual?&lt;/p&gt;&#xA;" OwnerUserId="7085" LastEditorUserId="7085" LastEditDate="2017-11-22T13:44:57.980" LastActivityDate="2017-12-13T05:39:36.543" Title="Could we detect if Oumuamua asteroid wasn't an asteroid?" Tags="&lt;asteroids&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23646" PostTypeId="1" CreationDate="2017-11-22T15:47:10.750" Score="2" ViewCount="155" Body="&lt;p&gt;Do comets have a limited number of orbits?&lt;/p&gt;&#xA;" OwnerUserId="19795" LastEditorUserId="2365" LastEditDate="2017-11-23T06:23:49.797" LastActivityDate="2017-11-23T06:23:49.797" Title="Do comets have a limited number of orbits?" Tags="&lt;comets&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2017-12-06T18:13:01.580" ContentLicense="CC BY-SA 3.0" />
  <row Id="23647" PostTypeId="2" ParentId="23646" CreationDate="2017-11-22T19:33:19.603" Score="7" Body="&lt;p&gt;Everything in the universe has a limited number of orbits.   But the way to make sense of your question is to ask &lt;a href=&quot;https://astronomy.stackexchange.com/questions/18551/how-long-does-a-comet-live&quot;&gt;how long comets last&lt;/a&gt;.  Comets lose ice when they fly past the sun, so they get smaller each time.   How much smaller depends on how close they get to the sun and their composition, or how easily they &quot;melt&quot; or break apart.   (Ice doesn't exactly melt in space, it sublimates when it heats up)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sungrazing_comet&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sungrazers&lt;/a&gt; have a very limited number of orbits.   Halley's also is expected to have a limited lifespan of &lt;a href=&quot;https://www.thenakedscientists.com/articles/questions/how-long-does-it-take-comets-melt&quot; rel=&quot;nofollow noreferrer&quot;&gt;about 100 more orbits&lt;/a&gt; before it stops being a comet and basically becomes a rocky body.   Halley's passes closer to the sun than Venus at its perihelion, so it loses a lot of ice, each pass, which helps make it's tail impressive. (when Earth and Halley's are properly aligned that is).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question also depends on your definition of a comet.   Will Halley's still be a comet after it loses it's ice and it's tail? (sometimes called a &lt;a href=&quot;https://www.space.com/32748-weird-tailless-comet-oort-cloud-discovery.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Manx comet&lt;/a&gt;).  That's a matter of interpretation.   It will get smaller, but it will likely remain in orbit for some time after it loses it's ice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbiting objects that cross planetary orbits are also, on average, less stable, but I suspect that's another subject.   Halley's might be destined to crash into the Earth, or the Moon at some point.  It's orbit nearly intersects the Earth's orbit in two places.  But don't worry.  That's not going to happen anytime soon, if it happens at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DpRn3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DpRn3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-11-23T05:20:41.640" LastActivityDate="2017-11-23T05:20:41.640" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23650" PostTypeId="2" ParentId="23644" CreationDate="2017-11-23T04:21:00.557" Score="7" Body="&lt;p&gt;I think its a good question and has probably popped up in the back of many people's minds for milliseconds, both astronomers and lay people, though most of them would still be hesitant to admit it on an internet forum. And of course, our understanding of what is possible or even &amp;quot;normal&amp;quot; in the universe is vanishingly small.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;edit 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Scientific American: &lt;a href=&quot;https://www.scientificamerican.com/article/alien-probe-or-galactic-driftwood-seti-tunes-in-to-oumuamua/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Alien Probe or Galactic Driftwood? SETI Tunes In to 'Oumuamua&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;The Atlantic: &lt;a href=&quot;https://www.theatlantic.com/science/archive/2017/12/yuri-milner-oumuamua-interstellar-asteroid/547985/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomers to Check Mysterious Interstellar Object for Signs of Technology&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Breakthrough Initiative: &lt;a href=&quot;http://breakthroughinitiatives.org/news/14&quot; rel=&quot;nofollow noreferrer&quot;&gt;Breakthrough Listen to Observe Interstellar Object 'Oumuamua'&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;From the NPR news item &lt;a href=&quot;https://www.npr.org/sections/thetwo-way/2017/12/12/570167473/astronomers-want-to-know-does-this-interstellar-visitor-have-a-message-for-us&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomers Want To Know: Does This Interstellar Visitor Have A Message For Us?&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;Researchers working on long-distance space transportation have previously suggested that a cigar or needle shape is the most likely architecture for an interstellar spacecraft, since this would minimize friction and damage from interstellar gas and dust,&amp;quot; the group said in its &lt;a href=&quot;http://breakthroughinitiatives.org/news/14&quot; rel=&quot;nofollow noreferrer&quot;&gt;announcement&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;While a natural origin is more likely, there is currently no consensus on what that origin might have been, and Breakthrough Listen is well positioned to explore the possibility that 'Oumuamua could be an artifact.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;The possibility that this object is, in fact, an artificial object — that it is a spaceship, essentially — is a remote possibility,&amp;quot; Andrew Siemion, a member of the initiative and director of Berkeley's &lt;a href=&quot;http://seti.berkeley.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Search for Extraterrestrial Intelligence Research Center&lt;/a&gt;, told &lt;a href=&quot;https://www.washingtonpost.com/news/speaking-of-science/wp/2017/12/12/is-this-visitor-from-beyond-our-solar-system-an-asteroid-or-a-sign-of-life/?utm_term=.b4e5955fee99&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Washington Post on Monday&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;We don't want to be sensational in any way, and we are very realistic about the chances this is artificial,&amp;quot; Yuri Milner, the Russian billionaire financing the project, told the &lt;a href=&quot;https://www.scientificamerican.com/article/alien-probe-or-galactic-driftwood-seti-tunes-in-to-oumuamua/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Scientific American&lt;/a&gt;, &amp;quot;but because this is a unique situation we think mankind can afford 10 hours of observing time using the best equipment on the planet to check a low-probability hypothesis.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;And they'll be checking on that hypothesis by scanning the object for possible artificial transmitters through a radio telescope at West Virginia's Green Bank Observatory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;edit 1:&lt;/strong&gt; A few days after you've asked the question, a similar thought was uttered by the BBC correspondent &lt;a href=&quot;https://twitter.com/peaseroland?lang=en&quot; rel=&quot;nofollow noreferrer&quot;&gt;Roland Pease&lt;/a&gt; in the Science in Action episode &lt;a href=&quot;http://www.bbc.co.uk/programmes/w3csvrgt#play&quot; rel=&quot;nofollow noreferrer&quot;&gt;Interstellar Asteroid&lt;/a&gt;. Talking with astronomer &lt;a href=&quot;https://star.pst.qub.ac.uk/%7Eaf/Alan_Fitzsimmons/About_Me.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Alan Fitzsimmons&lt;/a&gt; of Queens University, Belfast (who's image of X was featured in a NASA &amp;quot;&lt;a href=&quot;https://apod.nasa.gov/apod/ap171103.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomy Picture of the Day&lt;/a&gt;&amp;quot;), Pease says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You say ‘cigar-shaped”, that’s to me “rocket-shaped”. Is there any way you would know that this is an interstellar rocket?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So it looks like you're in excellent company!&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The asteroid was discovered and has been observed mostly (if not completely) by it's reflection of radiation from the Sun. The Sun is roughly a blackbody emitter and most of the radiation is in the visible. Because of the dimness of the light even at maximum, there isn't much signal once dispersed spectroscopically.&lt;/p&gt;&#xA;&lt;p&gt;There is a spectrum published in ArXiv &lt;a href=&quot;https://arxiv.org/abs/1710.09977&quot; rel=&quot;nofollow noreferrer&quot;&gt;Palomar Optical Spectrum of Hyperbolic Near-Earth Object A/2017 U1&lt;/a&gt; on 2-Nov-2017 by Joseph R. Masiero though! Despite the low light level (V ~ +21) and expanded size due to poor seeing at low elevation resulting in a signal-to-noise ratio of order unity, the 5 meter Hale telescope (200 inch Mt. Palomar) was used to try to obtain some spectroscopic information. Ten 300-second runs (total of 50 minutes of collection) were used to produce the heavily binned spectrum below. I think those are 1-sigma error bars, so the spectrum doesn't say much more than &amp;quot;slightly red-ish&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/90WMd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/90WMd.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;See also Ye et al. &lt;a href=&quot;https://arxiv.org/abs/1711.02320&quot; rel=&quot;nofollow noreferrer&quot;&gt;1I/2017 U1 (`Oumuamua) is Hot: Imaging, Spectroscopy and Search of Meteor Activity&lt;/a&gt; for further background:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The moderate spectral gradient of ‘Oumuamua indicates that its surface is devoid of ultrared material that is common on outer solar system objects like KBOs. Various classes of KBOs are typically very red in color, with a&#xA;spectral gradient &lt;span class=&quot;math-container&quot;&gt;$\gtrsim (\text{100 nm})^{-1}$&lt;/span&gt;, likely due to the irradiation of organic material resulting from the bombardment of energetic particles (Brunetto et al. 2006). The less reddish color of ‘Oumuamua suggests that the object was either formed close to its original central star, or has lost its ultrared material due to close approach(es) to its original or other stars. It is difficult to say which scenario is more likely due to the chaotic nature of small body dynamics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;According to Sky and Telescope's &lt;a href=&quot;http://www.skyandtelescope.com/astronomy-news/update-on-interstellar-object-oumuamua/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Update on `Oumuamua, Our First Interstellar Object&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The object had already dimmed to 20th magnitude by the time the Pan-STARRS 1 telescope swept it up, so observers worldwide had to scramble to get major facilities to gather more observations before the one-of-a-kind interloper becomes impossibly faint.&lt;/p&gt;&#xA;&lt;p&gt;One lucky break came immediately after the MPC's initial announcement. Observer Joe Masiero (Jet Propulsion Laboratory) just happened be in the middle of a run with the 5-m Hale Telescope at Palomar Observatory, and he quickly obtained a spectrum that's reproduced below and explained in more detail here. There's no obvious absorption that would signify the presence of particular compounds or minerals.&lt;/p&gt;&#xA;&lt;p&gt;But the gentle slope, increasing toward the red (right) end, suggests that `Oumuamua is not a particularly good match to objects in the Kuiper Belt (which tend to be much redder). Instead, it's the kind of spectrum that a rocky surface would exhibit after being &amp;quot;weathered&amp;quot; by long-term exposure to space radiation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So I'd say that the answer to your question is &amp;quot;No&amp;quot;, there's very little besides its orbit and light curve and approximate color that has been or probably could have been detected. With more warning, and serious interest, it's possible an observation campaign could have been organized to get more aperture area and time to make a better spectrum, but since S/N only scales as the square root of those, a lot of other observations would have had to have been rescheduled to make it happen.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-12-13T05:39:36.543" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23653" PostTypeId="1" CreationDate="2017-11-23T12:33:30.900" Score="0" ViewCount="68" Body="&lt;p&gt;if the universe transitioned from a meta-stable vacuum to a stable vacuum we assume a universal death scenario. However since it cannot propagate faster than the speed of light, from any particular starting point, the universal expansion has crossed an intrinsic &quot;event horizon&quot; so how can this actually happen in any conceivable time scale?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also - cheeky extra question - could energy rich galaxies dislocate in some way from energy poor regions of spacetime&lt;/p&gt;&#xA;" OwnerUserId="19808" LastActivityDate="2017-11-24T00:49:29.540" Title="false to stable vacuum transition" Tags="&lt;astrophysics&gt;&lt;observable-universe&gt;&lt;galaxy-cluster&gt;&lt;fate-of-universe&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2017-12-06T20:59:28.107" ContentLicense="CC BY-SA 3.0" />
  <row Id="23654" PostTypeId="2" ParentId="23503" CreationDate="2017-11-23T17:13:46.967" Score="2" Body="&lt;p&gt;It's habitability is uncertain.   Red Dwarf stars that small could be unkind to their planets and strip them of most of their atmospheres.   We'll get a better look at the exoplanets around Proxima Centauri when the &lt;a href=&quot;https://www.jwst.nasa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;James Webb Space Telescope&lt;/a&gt; is up and running.  It's scheduled to launch in 2019 and it should be able to tell us something about exoplanet's atmospheres, which would go a long way towards estimating any habitability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually getting there, even with a small unmanned craft is a long ways away.   There's still significant technological difficulties with &lt;a href=&quot;https://en.wikipedia.org/wiki/Breakthrough_Starshot&quot; rel=&quot;nofollow noreferrer&quot;&gt;Breakthrough Starshot&lt;/a&gt;.  The lasers aren't set up yet for one.  For two, the aim has to be just about perfect, and three, we currently have no means of receiving a signal from a craft that far away.   We have to improve our communications tech.   Now, in theory, we know how to do it.   We could build giant satellite dishes on the Moon, for example, or orbiting in space, but that's very expensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another issue with starshot is that they have no means of slowing down, so they'd pass through the solar-systems they were sent to examine in a rapid fly-by, perhaps 10% of the speed of light, with only a few hours within the solar-system.   Fly-by missions are still useful, and given the difficulty of transmitting data an orbiting mission in another solar-system might not make practical sense, because such a mission could easily gather more information than it could send back.    The recent New Horizons / Pluto mission was a fly-by and it gathered a lot of information, &lt;a href=&quot;https://www.nasa.gov/feature/new-horizons-returns-last-bits-of-2015-flyby-data-to-earth&quot; rel=&quot;nofollow noreferrer&quot;&gt;that took over a year to send back to Earth&lt;/a&gt;.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rapid fly-bys shouldn't be dismissed though.   They can gather a lot of information that we have trouble seeing from Earth, even with newer and bigger telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More ambitious missions, and ships with their own fuel so they can break and remain in a distant solar-system are harder.   That requires much more equipment, much more mass and as a result, would accelerate more slowly.   Manned missions 4.2 light years away are so far out of reach that it makes little sense to talk about them outside of science fiction.   A manned mission to Mars is hard enough and that's over 100,000 times closer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starshot is many years away.   We might see Starshot launch in a decade or two - if it moves quickly.   Then it'll be another 2 decades or so for the ship to reach the Proxima system and another 4 years and change for any information to be returned.   35-45 years is probably on the ambitious estimate before we receive the first starshot data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the new information about exoplanets that we're going to get over the next few years will come from telescopes like James Webb and others that follow.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-11-23T17:13:46.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23655" PostTypeId="2" ParentId="23591" CreationDate="2017-11-23T17:33:22.697" Score="6" Body="&lt;p&gt;It would be more accurate to call them surveys than studies.   A study looks at an object in detail.  A survey counts and categorizes objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rogue planets are enormously difficult to see, and only a few very large ones have actually been observed directly.  For the largest rogue planets, it's unclear if they actually are ejected planets as opposed to failed stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.scientificamerican.com/article/wandering-in-the-void-billions-of-rogue-planets-without-a-home/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article discusses rogue planets&lt;/a&gt; in pretty good detail.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 2011 &lt;a href=&quot;https://www.nature.com/articles/nature10092.epdf?referrer_access_token=geHqH2LOeheWU0uQ5zmABtRgN0jAjWel9jnR3ZoTv0Nk9JKZpv6lE-Lv9RamE8LN_ITJRcpMAMnSPNrwy6k6bSv1eIUy6t5iaklA6KT6lNrV3u7Tq63z5DJrzdRm91bMCy-ICeVl6b6pe4ra9NjfPrf8lsyw8lYocA5V-YsUN0gBrM2qGsMIMFEgS-aWJ5ZD&amp;amp;tracking_referrer=www.scientificamerican.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;the MOA survey&lt;/a&gt; was published, estimating a high number of Jupiter sized Rogue planets, perhaps outnumbering stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More recently the &lt;a href=&quot;https://en.wikipedia.org/wiki/Optical_Gravitational_Lensing_Experiment&quot; rel=&quot;nofollow noreferrer&quot;&gt;OGLE Project&lt;/a&gt; estimated about 1/10th that many rogue Jupiters, which is more in line with statistical predictions.   Smaller planets are more likely to get ejected from their solar-systems than larger ones.    Rogue Juiters outnumbering stars seemed unlikely and my understanding is that most scientists are more in line with the OGLE estimate than MOA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both estimates rely on microlensing which amounts to very small and temporary changes when rogue planets drift between the Earth and a star.  There's a fairly large margin of error for both surveys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NASA's &lt;a href=&quot;https://wfirst.gsfc.nasa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;WFIRST telescope&lt;/a&gt; should answer a lot of questions on rogue planets.   It's not expected to be launched till the mid 2020s though.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-11-24T06:53:10.420" LastActivityDate="2017-11-24T06:53:10.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23656" PostTypeId="1" CreationDate="2017-11-23T18:06:31.353" Score="4" ViewCount="81" Body="&lt;p&gt;This morning at 8:00am UK time, give or take about 3 minutes, I saw a very bright light in the sky which seemed to get a little brighter, lasting for maybe 5 seconds after I first noticed it, then faded out to nothing for a few seconds, then came back for another few seconds and disappeared entirely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was at around the following co-ordinates, and I was looking towards the South:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lat: 51.980619 Long: -0.8443739614449441&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd say it was at something like 30 degrees inclination but that's a really rough guess. It was somewhere around a third to half way between the horizon and overhead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there are duplicate questions about this kind of thing BUT I believe this is different because:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I saw it during daylight. The sun was up, albeit behind clouds.&lt;/li&gt;&#xA;&lt;li&gt;The light was VERY bright, when I first saw it I thought it was Venus but then it obviously faded out. I think it was probably even brighter than Venus would have been.&lt;/li&gt;&#xA;&lt;li&gt;I have already ruled out Iridium flares by looking at Heavens Above. I have seen Iridium flares once or twice before and I think it looked similar.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Presumably there aren't too many objects in the sky that can get that bright. Are there any other objects which I could track apart from Iridium sats? Can anybody else find anything it might have been?&lt;/p&gt;&#xA;" OwnerUserId="19811" LastEditorUserId="2531" LastEditDate="2018-08-31T22:14:27.293" LastActivityDate="2018-08-31T22:14:27.293" Title="Bright, brief light in sky this morning (daytime) - satellite flare?" Tags="&lt;satellite&gt;&lt;flare&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2018-09-04T08:22:39.287" ContentLicense="CC BY-SA 4.0" />
  <row Id="23659" PostTypeId="1" CreationDate="2017-11-24T00:08:06.990" Score="3" ViewCount="179" Body="&lt;p&gt;Specifically in the milky way, if that changes anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like it should be in the disk, since stars there are richer in heavier elements, but I'm not sure.&lt;/p&gt;&#xA;" OwnerUserId="19815" LastActivityDate="2017-11-24T14:23:50.237" Title="Where are heavier elements than iron produced?" Tags="&lt;cosmology&gt;&lt;milky-way&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23660" PostTypeId="2" ParentId="23653" CreationDate="2017-11-24T00:49:29.540" Score="3" Body="&lt;p&gt;The standard picture of vacuum decay events is very much like nucleation in a phase transition: it starts somewhere, and then expands at a constant speed - lightspeed in this case. So the future lightcone of the nucleation event will be true vacuum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an accelerating expanding universe like our current $\Lambda$CDM cosmology future lightcones actually have a finite radius as measured by co-moving coordinates (e.g. from the perspective of matter moving passively with the expansion of the universe). Beyond a certain (current) distance there are galaxies that will never be reached by a light signal we send now - when it gets to the distance they currently have they will have moved much further away, and it never manages to close the gap. Same thing for vacuum decay: the front will always expand outward at lightspeed, but large parts of the observable universe will expand away so fast that it will never catch up with them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In about 100-200 billion years galactic clusters will become separated by exponentially growing gulfs and quickly lose contact (they stay the same size because they are gravitationally bound). They become separate &quot;island universes&quot;. Beyond this point vacuum decay would just be a local issue to one island universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extra question: if a galaxy has too much energy it will implode as a black hole - mass and energy are equivalent in relativity. Whether that counts as dislocation is unclear. But overall regions with more matter density tend to contract while less dense regions expand more, which is why we get the cosmic web of sheets and filiaments of dark matter plus galaxies, and expanding voids. Galaxies in the voids (they do exist) could be described as energy poor (even though their stars are doing just fine) and they would experience everything moving away from them not just because of universal expansion but also gravitational contraction towards remote masses.&lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2017-11-24T00:49:29.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23662" PostTypeId="2" ParentId="22978" CreationDate="2017-11-24T05:29:57.807" Score="2" Body="&lt;p&gt;I believe the differences here are caused by the different timescales that are being used by SkyField vs NASA. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that SkyField uses the proleptic Gregorian calendar for dates in the past. However, NASA used the Julian Calendar for dates before 1582, so for example, the eclipse on 0150-12-06 (Julian) falls on 0150-12-05 in Gregorian. Also, I would use &lt;code&gt;ts.ut1&lt;/code&gt; instead of &lt;code&gt;ts.utc&lt;/code&gt;, as ut1 is the timescale used by NASA. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the geocenter can be a efficient way to check for eclipses worldwide. When observing from the geocenter, the penumbral shadow cone touches the Earth when the following condition is true:&#xA;$$\mu ≤ arcsin(\frac{r_m + r_e}{d_m}) + arcsin(\frac{R_s - r_e}{D_s})$$&#xA;Where $\mu$ is the angular separation between the Sun and the Moon, $D_s$ is the distance of the Sun, $d_m$ the distance of the Moon, and the three constant radii are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sun: $R_s = 695700 km$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth: $r_e = 6378 km$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moon: $r_m = 1737.4 km$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This indicates that an eclipse, at least partial, is occuring somewhere on the Earth. The formula uses a spherical Earth as an approximation of its shape, so in some rare cases it can detect very near-misses by 20-25 km, but overall it is a good way to narrow down the search. Also, it is unnecessary to test this when the angular separation is outside the range 1.3° - 1.7°, since values below 1.3° always produce an eclipse, and values above 1.7° never produce one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also like to point out that uncertainties, mostly in Earth's rotation speed, make it very difficult to know where eclipses occured on the surface of the Earth for dates in the far past. The uncertainty becomes very large outside the range BC2000 - AD3000. This is the reason why NASA hasn't computed eclipse paths outside that range. According to research, the uncertainty could be as large as over 4 hours (several thousands of kilometers for eclipse paths) for dates before BC4000. See &lt;a href=&quot;https://eclipse.gsfc.nasa.gov/SEcat5/uncertainty.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; explanation on NASA's Eclipse Web Site.&lt;/p&gt;&#xA;" OwnerUserId="19814" LastEditorUserId="19814" LastEditDate="2017-11-25T13:09:44.797" LastActivityDate="2017-11-25T13:09:44.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23669" PostTypeId="2" ParentId="23659" CreationDate="2017-11-24T14:10:12.743" Score="8" Body="&lt;p&gt;The question of &quot;where&quot; the heavy elements are produced is slightly ambiguous. I gather from your mention of the Milky Way that you are interested in &quot;where in the Milky Way&quot; are the heavy elements produced, rather than what type of stars are they produced in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, to answer the former, you need to understand the latter. So I point you to my answer &lt;a href=&quot;https://physics.stackexchange.com/questions/7131/origin-of-elements-heavier-than-iron-fe?rq=1&quot;&gt;on Physics SE&lt;/a&gt; for a summary of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heavy elements can be divided in a number of ways. Most are produced by neutron capture in either the slow &lt;a href=&quot;https://en.wikipedia.org/wiki/S-process&quot; rel=&quot;noreferrer&quot;&gt;s-process&lt;/a&gt; or the more rapid &lt;a href=&quot;https://en.wikipedia.org/wiki/R-process&quot; rel=&quot;noreferrer&quot;&gt;r-process&lt;/a&gt;. Different elements are produced in different quantities by these two mechanisms and are created in different types of star. s-process elements are created inside massive stars before they become supernovae, but also in stars of more modest mass that never go through the supernova stage. The r-process is now thought to occur predominantly in neutron star mergers, but there is probably also a contribution from type II supernovae, particularly for the lighter r-process elements. A summary, which still has plenty of uncertainties, but probably gives a reasonable summary of current thing is shown below (diagram produced by &lt;a href=&quot;http://www.astronomy.ohio-state.edu/~jaj/nucleo/&quot; rel=&quot;noreferrer&quot;&gt;Jennifer Johnson&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Gy0lF.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gy0lF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question then amounts to where these stars were when they created these heavy elements towards the ends of their lives, and as a function of time since the Milky Way was born. The key facts here are that the progenitors of type II supernovae are massive stars with short lives. They are born in gas-rich regions and do not have time to travel far from this location before they explode. Thus most of the elements they produce are produced in the bulge and disc regions of the Milky Way. This is especially true for the bulge, which seems to have formed quite quickly at the start of the life of the Milky Way and there should (and indeed does seem to be) a radial gradient in metallicity increasing towards the disc plane and towards the Galactic centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Type Ia supernovae result from exploding white dwarfs can produce elemts at the iron-peak and somewhat beyond, mostly by &quot;standard&quot; fusion than by neutron capture. These events may have been produced by comparatively lower mass and long-lived progenitors, and may also have had long lives as white dwarfs before the supernova event. This means that type Ia supernova progenitors are spread more widely. They have had plenty of time to migrate from where they were actually born, so we expect their heavy element production to be more dispersed - still concentrated towards the disc, but much less so than the progenitors of type II supernovae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The progenitors of the asymptotic giant branch stars that can produce lots of s-process elements can also be quite old. The elements produced by these stars would be distributed in a similar way in space and time to those produced by type Ia supernovae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally we have merging neutron stars. These are quite problematic and not much is known about their progenitors or evolutionary paths and histories. They must originate from massive stars, with short lives that are born close to the Galactic plane or bulge. However, after both stars in the systems have exploded as supernovae and left behind neutron star remnants, the final neutron star merger can take place much later, due to a gradual inspiral triggered by the emission of gravitational waves. This gives the neutron star pair a long time (hundreds of millions or billions of years) to move about in the Galaxy before the merger takes place. In addition, the supernova explosions can give a &quot;kick&quot; to the neutron stars that can displace them well out of the Galactic plane with fast velocities. I expect therefore that the products of neutron star mergers could have been produced over a very large volume in our Galaxy. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-11-24T14:23:50.237" LastActivityDate="2017-11-24T14:23:50.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23671" PostTypeId="1" CreationDate="2017-11-25T00:49:42.587" Score="2" ViewCount="616" Body="&lt;p&gt;I understand the difference between the two to be that they both cross Earth's orbit, but Atens have an orbit smaller than Earth's, and Apollo have an orbit larger, but I don't understand why they aren't simply lumped together in one category. &lt;/p&gt;&#xA;" OwnerUserId="19831" LastEditorUserId="2365" LastEditDate="2017-11-27T06:28:09.183" LastActivityDate="2017-11-30T10:02:29.690" Title="Why is the difference between Apollo and Aten asteroids considered meaningful?" Tags="&lt;asteroids&gt;&lt;near-earth-object&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23674" PostTypeId="1" AcceptedAnswerId="23679" CreationDate="2017-11-25T06:38:18.393" Score="0" ViewCount="104" Body="&lt;p&gt;I recently learnt that some comets like &lt;a href=&quot;https://en.wikipedia.org/wiki/C/1980_E1_(Bowell)&quot; rel=&quot;nofollow noreferrer&quot;&gt;C/1980 E1 (Bowell)&lt;/a&gt; interacted with objects like gas giants in the solar system and gained enough energy to exit the solar system. But from Physics, I know that only if the total mechanical energy of a body is positive, can the body exit the solar system. An interaction with a gas giant should merely convert some of the potential energy of the body to kinetic energy and the total energy of the body should remain unchanged. If so, how can such bodies achieve positive mechanical energy to exit the solar system?&lt;/p&gt;&#xA;" OwnerUserId="16469" LastActivityDate="2017-11-25T08:20:37.570" Title="How do comets and other solar system bodies gain energy to exit the solar system?" Tags="&lt;astrophysics&gt;&lt;asteroids&gt;&lt;comets&gt;&lt;orbital-mechanics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23676" PostTypeId="1" AcceptedAnswerId="23717" CreationDate="2017-11-25T07:49:41.123" Score="1" ViewCount="10638" Body="&lt;p&gt;i have a beginner telescope of 50mm and want to see planets it is arefractor with f/12 ratio and two eyepieces of 5mm and 10mm and i want to see mars, Jupiter, Saturn and venus. my max zoom ix 120x and focal length is 600mm   &lt;/p&gt;&#xA;" OwnerUserId="19838" LastActivityDate="2018-04-28T18:40:00.597" Title="what magnification is required to see jupiter, saturn, venus and mars with a 50mm refractor scope with f/12 ratio and 600mm focal lenth" Tags="&lt;planet&gt;&lt;telescope&gt;&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23677" PostTypeId="1" AcceptedAnswerId="23715" CreationDate="2017-11-25T07:58:43.753" Score="2" ViewCount="111" Body="&lt;p&gt;I have started studying Extragalactic Magnetic Fields, and mainly focusing on &lt;a href=&quot;https://en.wikipedia.org/wiki/Biermann_battery&quot; rel=&quot;nofollow noreferrer&quot;&gt;Biermann Battery Theory&lt;/a&gt;. I have gathered much information on other Dynamo Theories, as well as about our current Observational and Theoretical status on Magnetic Field Genesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm looking for though, is a general &quot;picture&quot; on Extragalactic Magnetic Fields, as a whole. Faraday Rotation, AGN Jets, are basically the topics I have tracked, but I'm pretty sure there are a lot of others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm asking, if someone has studied in-depth extragalactic structures and magnetic field, possibly could guide me on some of the main and hot topics around them. This way I would have more knowledge on the topic which would be great for me!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean just Topic-Titles, would be great!&lt;/p&gt;&#xA;" OwnerUserId="19839" LastActivityDate="2017-11-28T09:28:39.573" Title="Extragalactic Magnetic Fields - General Information That I'm Missing" Tags="&lt;galaxy&gt;&lt;magnetic-field&gt;&lt;galaxy-center&gt;&lt;resource-request&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23679" PostTypeId="2" ParentId="23674" CreationDate="2017-11-25T08:20:37.570" Score="5" Body="&lt;p&gt;The mechanism by which the path, speed and energy of a celestial object or an artificial spacecraft is altered by massive planets and bodies is &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravity_assist&quot; rel=&quot;noreferrer&quot;&gt;Swing-by or gravity assist&lt;/a&gt;. Any encounter between a comet and a gas-giant like Jupiter is almost an elastic mechanical collision (though no physical contact occurs) and in any mechanical collision, energy exchange between the bodies undergoing collision takes place. In this case, when the comet C/1980 E1 passed closed to Jupiter, Jupiter passed on some of its mechanical energy to the comet, rendering its total mechanical energy positive and hence setting the comet free and no longer bounded to the solar system. It is the total mechanical energy of the Jupiter+C/1980 E1 that was conserved, not just that of the comet. But the small decrease in the mechanical energy of Jupiter would not have a significant impact on the massive planet.&lt;/p&gt;&#xA;" OwnerUserId="16469" LastActivityDate="2017-11-25T08:20:37.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23683" PostTypeId="1" CreationDate="2017-11-25T09:52:09.360" Score="2" ViewCount="74" Body="&lt;p&gt;There's no lack of photons, how well could a telescope resolve Sun spots and other surface features of the Sun? How would a high resolution Solar (space) telescope differ from for example Hubble which avoids the Sun by a 50° angle or so?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2017-12-01T00:43:37.650" Title="How well can the Sun's surface be resolved by an observatory?" Tags="&lt;the-sun&gt;&lt;observational-astronomy&gt;&lt;optics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23686" PostTypeId="2" ParentId="23676" CreationDate="2017-11-25T13:48:28.640" Score="3" Body="&lt;p&gt;Venus, Mars, Jupiter, and Saturn are visible with your eyes, so a telescope and magnification are not required to see them. Depending on the quality of the telescope, the magnification will show you some details.&#xA;Venus - you will be able to see the phase.&#xA;Mars - for several months every 26 months, you will be able to see some dark markings and a polar cap.&#xA;Jupiter - you will see the cloud bands and 4 of its moons.&#xA;Saturn - you will see its rings and moon Titan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enjoy the view!&lt;/p&gt;&#xA;" OwnerUserId="17355" LastEditorUserId="17355" LastEditDate="2017-11-26T18:33:54.223" LastActivityDate="2017-11-26T18:33:54.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23689" PostTypeId="1" CreationDate="2017-11-25T15:06:43.190" Score="0" ViewCount="216" Body="&lt;p&gt;At the center of massive galaxies there is a huge and massive black hole that radiates gamma rays. So at the center of the Milky Way there is also a huge and massive black hole and it also radiates gamma rays. Do these gamma rays reach our solar system? What is the age of black hole in the center of our galaxy? Does it still radiates gamma rays?&lt;/p&gt;&#xA;" OwnerUserId="19850" LastEditorUserId="88" LastEditDate="2017-11-28T14:48:59.550" LastActivityDate="2017-12-28T17:13:19.543" Title="Black holes at the center of galaxies" Tags="&lt;supermassive-black-hole&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23690" PostTypeId="2" ParentId="23689" CreationDate="2017-11-25T17:01:46.887" Score="1" Body="&lt;p&gt;Firstly, you probably mean that a supermassive black hole (SMBH) is in the centre of our &lt;em&gt;Galaxy&lt;/em&gt;, not our &lt;em&gt;solar system&lt;/em&gt;. That is indeed correct. The age of the black hole will be approximately the same as the Galactic age. It turns out that the oldest stars in our Galaxy are around 13 billion years old. This means of course that our Galaxy is at least that old. Hence it is safe to assume the central SMBH has the same age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very simplistically, we can say gamma rays are emitted from the accretion disk around the SMBH. This happens because of the great speed the particles are swirling around it and the friction that occurs between them. As you know the gamma rays are very energetic photons and have mean free paths that are huge. Therefore they can reach our solar system.&lt;/p&gt;&#xA;" OwnerUserId="17425" LastEditorUserId="2365" LastEditDate="2017-11-25T17:25:00.100" LastActivityDate="2017-11-25T17:25:00.100" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23691" PostTypeId="2" ParentId="23671" CreationDate="2017-11-25T17:54:43.377" Score="2" Body="&lt;p&gt;Asteroids can be classified in different ways. One way is on their spectral type, which relates their composition. Another way is to classify according to their orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most asteroids orbit between Mars and Jupiter. Others are classified according to the position of their orbits relative to the planets. So there are the asteroids with an orbit that is longer than Mars. Those with an orbit that is less than Mars, but more than Earth, even a few with an orbit that is less than the Earth. There is a natural gap between the asteroids with a period less than one year, and those with a period of more than a year, as the resonant effects of the Earth's gravity would clear out asteroids with a period that is very close to a year, or else trap them as Trojans. This gap makes a convenient point to divide the groups of asteroids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of these various classes the Earth-crossing asteroids are of particular interest. There are those with an orbit that takes less than one year  (Atens) and those that take more than one year (Apollos). There is no great difference, just a way of ordering the list of known asteroids.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-11-25T17:54:43.377" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23692" PostTypeId="1" AcceptedAnswerId="23694" CreationDate="2017-11-25T18:07:51.087" Score="6" ViewCount="2878" Body="&lt;p&gt;At the center of the Milky Way, we have a supermassive black hole. It's name is &quot;Sagittarius A*&quot;. I was wondering that why is it called so? I mean, why &quot;A*&quot; why not just &quot;A&quot;. Is there a naming system for super massive black holes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="16477" LastActivityDate="2017-11-30T17:27:21.373" Title="Why is Sagittarius A* called so?" Tags="&lt;supermassive-black-hole&gt;&lt;naming&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23693" PostTypeId="2" ParentId="22762" CreationDate="2017-11-25T18:10:24.530" Score="1" Body="&lt;p&gt;In addition to not knowing, there's other reasons to suspect gravity wouldn't reverse.   There's no known force that reverses over distance (OK, the strong force, kinda/maybe but over very short distances).  Also, the expansion of space explains things that we seem to observe or suspect occurred like galaxies moving away from us faster than the speed of light and the rapid expansion that's suspected to have happened very shortly after the big bang.   Gravity can't do either of those things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also the problem of how would gravity know that things are so far apart that they should start repelling, where nearby galaxies are still too close but more distant galaxies the repulsion kicks in?   It doesn't make a whole lot of sense that gravity would account for both attraction and expansion of space.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's not to say it's impossible.  We don't know how gravity works or what it is, so, maybe.  Never say never, until something is understood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@eshaya's answer, &quot;We don't know&quot; is probably the more straight forward answer to this question, but I wanted to point out some problems with repulsive gravity theory as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition (and I don't understand precisely why), but the gravitational waves discovery has &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_energy#Modified_gravity&quot; rel=&quot;nofollow noreferrer&quot;&gt;ruled out some theoretical gravity/dark energy proposals&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-11-25T18:10:24.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23694" PostTypeId="2" ParentId="23692" CreationDate="2017-11-25T18:26:57.977" Score="4" Body="&lt;p&gt;Check this &lt;a href=&quot;https://en.wikipedia.org/wiki/Sagittarius_A*#History&quot; rel=&quot;nofollow noreferrer&quot;&gt;wiki link&lt;/a&gt; out. &quot;The name Sgr A* was coined by Brown in a 1982 paper because the radio source was &quot;exciting&quot;, and excited states of atoms are denoted with asterisks.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no unified naming system for black holes. They are usually named after their host galaxy. Others are identified by the name of the survey in which they were observed. A few black holes are catalogued by their constellation and the order in which they were discovered&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The 1982 paper by Brown and Lo can be found &lt;a href=&quot;http://articles.adsabs.harvard.edu/full/1982ApJ...253..108B&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17425" LastEditorUserId="17425" LastEditDate="2017-11-27T18:11:18.110" LastActivityDate="2017-11-27T18:11:18.110" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23695" PostTypeId="2" ParentId="16403" CreationDate="2017-11-25T19:29:21.307" Score="0" Body="&lt;p&gt;No. In order for a star to sustain itself, fusion must take place to avoid collapse due to gravity. The earth is made from heavy elements (nickel, iron, etc) which are nearly impossible to fuse in stars. Therefore, due to this, the Earth cannot be a star due to the addition of more mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, if the additional mass was hydrogen, I'm conflicted to say what will happen, perhaps it'll be a low mass star that will shine for a limited time until it cannot sustain itself or fusion just might never happen. &lt;/p&gt;&#xA;" OwnerUserId="19851" LastEditorUserId="19851" LastEditDate="2017-11-25T19:54:46.557" LastActivityDate="2017-11-25T19:54:46.557" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23696" PostTypeId="1" AcceptedAnswerId="23697" CreationDate="2017-11-25T20:20:13.800" Score="3" ViewCount="908" Body="&lt;p&gt;If we see a nebula by looking through a powerful telescope, how can we tell whether we are looking at the remnant of a supernova or at a planetary nebula?&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="19799" LastActivityDate="2017-11-25T21:55:10.963" Title="How to visually tell the difference between a planetary nebula and a supernova remnant?" Tags="&lt;telescope&gt;&lt;observational-astronomy&gt;&lt;supernova&gt;&lt;nebula&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23697" PostTypeId="2" ParentId="23696" CreationDate="2017-11-25T21:39:39.463" Score="1" Body="&lt;p&gt;A Supernova Remnant contains a Black Hole or a Neutron Star while a Planetary Nebula contains a White Dwarf. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, Supernovae Remnants are likely to have great velocities, so Doppler is another plausible choice. - Not through telescope though.&lt;/p&gt;&#xA;" OwnerUserId="19839" LastEditorUserId="2365" LastEditDate="2017-11-25T21:55:10.963" LastActivityDate="2017-11-25T21:55:10.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23698" PostTypeId="1" CreationDate="2017-11-26T04:03:30.713" Score="3" ViewCount="190" Body="&lt;p&gt;I imagine a situation must happen sometimes where a telescope watching a very dark object may catch a bright flash of light or a reflection off of a satellite. If that's even a concern at all (maybe all such telescopes are already designed to exclude that possibility), what countermeasures against blinding someone's eyes or frying the digital photo sensors are there?&lt;/p&gt;&#xA;" OwnerUserId="363" LastActivityDate="2018-03-14T11:18:40.433" Title="How are telescopes protected from blinding light and satellite glares?" Tags="&lt;telescope&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23703" PostTypeId="2" ParentId="23698" CreationDate="2017-11-26T11:38:04.220" Score="6" Body="&lt;p&gt;Random bright objects can occasionally appear in astronomical images. Examples include asteroids and satellites. However, these objects are not moving at the sidereal rate across the sky, and thus unless the telescope is set to track them, they leave a trail across the image. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is unlikely to cause damage and CCD detectors are fairly robust to saturation of exposures in any case. However, it may ruin the image.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-11-26T11:38:04.220" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23704" PostTypeId="1" AcceptedAnswerId="23988" CreationDate="2017-11-26T21:20:56.977" Score="4" ViewCount="637" Body="&lt;p&gt;The star occupying 6th place on &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_nearest_stars_and_brown_dwarfs&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia's list of nearest stars and brown dwarfs&lt;/a&gt; can be called either &lt;strong&gt;Lalande 21185&lt;/strong&gt;, &lt;strong&gt;BD+36 2147&lt;/strong&gt;, &lt;strong&gt;Gliese 411&lt;/strong&gt; or &lt;strong&gt;HD 95735&lt;/strong&gt;. That seems to me like a fine, quadruple choice of equally bad names. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean, I know these come from positions held in various star catalogues and maybe aren't that bad compared to monstrosities like &lt;strong&gt;WISE J085510.83−071442.5&lt;/strong&gt;, but can't we keep those for stars like 50 light yrs away and invent something better for our &lt;em&gt;nearest neighbours&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This year we saw a &lt;a href=&quot;http://www.drewexmachina.com/tag/nearby-stars/&quot; rel=&quot;nofollow noreferrer&quot;&gt;batch of confirmations&lt;/a&gt; for planets orbiting in habitable zones of many of these stars. In not so distant future we may find ourselves discussing possibility of sending probes to some of these systems. Aren't better names long overdue?&lt;/p&gt;&#xA;" OwnerUserId="19829" LastEditorUserId="19829" LastEditDate="2017-12-15T11:28:35.313" LastActivityDate="2017-12-20T00:46:02.173" Title="Why nobody cares to give the nearest stars better names?" Tags="&lt;star&gt;&lt;naming&gt;&lt;star-catalogues&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23705" PostTypeId="2" ParentId="23704" CreationDate="2017-11-26T23:39:29.820" Score="9" Body="&lt;p&gt;There are a thousand or so stars within 50 light years, most are very dim red dwarfs, and most are unexceptional. If we invent &quot;names&quot; then we would have to learn the names of all the stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want to name all the stars that can be seen with a telescope, we would have to come up with a billion or so names.  The idea of naming stars rapidly becomes impractical - nobody is ever going to learn thousands of names which were chosen on whim. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what could be a &quot;better name&quot; than just making up a name based on one's heroes, friends, gods, or jokes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead we can make a list of stars, and &lt;em&gt;name them systematically&lt;/em&gt;. We could use the part of the sky that the star appears in, and some letter or  number. Or we could use the coordinates that the star appears at to label it. Or we could just give each star a number. This way we don't need to invent names, we just use our system. When we do this we get &quot;names&quot; like &quot;omicron ceti&quot; (using the letter &quot;omicron&quot; and the part of the sky &quot;ceti&quot;)&#xA;or &quot;HD 128621&quot; (star number 128621 in the list prepared by Henry Draper). or WISE 0410+1502 (a brown dwarf discovered by the Wide-field Infrared Survey Explorer, and located at 04hr10min, 15deg02min in the sky)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cataloguing is simply a better way of naming stars as there are so many of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a star is exceptional in some way, then there may be a reason to give it a name - Barnards Star (moves very fast), Sirius (very bright). But most stars are unexceptional, and there are so many of them. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-11-26T23:59:13.577" LastActivityDate="2017-11-26T23:59:13.577" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23707" PostTypeId="1" CreationDate="2017-11-27T03:00:04.077" Score="2" ViewCount="116" Body="&lt;p&gt;Does collision of two asteroids necessarily create only very small objects? (and how big is it &quot;very small&quot;?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am especially interested about collision of two asteroids with similar orbits (and thus relatively small difference of the speed vectors in the moment of collision).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can such a collision create big (say 1km, or maybe 100m) objects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it in this case likely that a created object would have something like a comet tail consisting of dust and small objects (probably gravitionally attracted to the resulting asteroid remnant)?&lt;/p&gt;&#xA;" OwnerUserId="19862" LastActivityDate="2017-11-27T16:50:18.293" Title="What size objects can be created from collision of two asteroids?" Tags="&lt;asteroids&gt;&lt;asteroid-belt&gt;" AnswerCount="2" CommentCount="0" ClosedDate="2017-12-07T20:51:53.267" ContentLicense="CC BY-SA 3.0" />
  <row Id="23712" PostTypeId="2" ParentId="23707" CreationDate="2017-11-27T11:12:20.447" Score="2" Body="&lt;p&gt;Since planets are forming from asteroid collisions, I would say those collisions won't necessarily create small objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine throwing a snowball against a window.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you throw it lightly, some snow will stick to the window and increase the mass of it (at least until it melts away) and the rest of the snowball will be scattered everywhere around but won't stick to the window.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you throw the snowball too hard, the window will break, resulting in even more tiny pieces. But still, some snow will stick to some pieces of glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's exactly what happens on asteroid impacts and depends on the speed, mass and structure of each asteroid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of those debris should reach from microscopic dust up to gigantic rocks with several kilometer wide diameter.&lt;/p&gt;&#xA;" OwnerUserId="19860" LastEditorUserId="6841" LastEditDate="2017-11-27T16:33:09.380" LastActivityDate="2017-11-27T16:33:09.380" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23713" PostTypeId="2" ParentId="23671" CreationDate="2017-11-27T14:32:02.440" Score="3" Body="&lt;p&gt;The first point to make is that categorization isn't an exact science.  It's done for convenience and practicality and sometimes, familiarity.  It's not going to be perfect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second point to make is that scientists like categorizing things.  Take the &lt;a href=&quot;https://en.wikipedia.org/wiki/Taxonomic_rank&quot; rel=&quot;nofollow noreferrer&quot;&gt;Taxonomic ranking&lt;/a&gt; of living creatures.  Everything from a paramecium  to an elephant has a species, a genus, a family, an order, a class, a phylum, a kingdom and a domain.   Categories are everywhere in science and they're useful.   &lt;a href=&quot;https://en.wikipedia.org/wiki/Taxonomic_rank#Significance_and_problems&quot; rel=&quot;nofollow noreferrer&quot;&gt;Taxonomy classifications aren't perfect&lt;/a&gt;, but classifications don't need to be perfect to be useful and provide distinction and quick and familiar identification.   The same is true for asteroid classification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both Atens and Apollos are already lumped into one category called &lt;a href=&quot;https://www.britannica.com/topic/Earth-crossing-asteroid&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth-crossing asteroids&lt;/a&gt;, so, when needed, they can be discussed together.   It's more common to separate them by the two types, but it's not required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth-crossing asteroids or Earth-Crossers are a sub-category of &lt;a href=&quot;https://en.wikipedia.org/wiki/Near-Earth_object&quot; rel=&quot;nofollow noreferrer&quot;&gt;Near Earth Objects&lt;/a&gt;, and Near Earth Objects or NEOs are of particular interest because every so often, an undesirably large object hits the Earth.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;NEOs can be divided into Atens, Apollos, Amors, Comets, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Atira_asteroid&quot; rel=&quot;nofollow noreferrer&quot;&gt;Atiras&lt;/a&gt;, sometimes called Internal Earth Objects or IEOs, but there are relatively few Atiras and they're sometimes included with the Atens, though technically, Interior Earth Objects aren't Atens if their aphelion is too far inside the Earth's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NEOs can be classified as &lt;a href=&quot;https://en.wikipedia.org/wiki/Potentially_hazardous_object&quot; rel=&quot;nofollow noreferrer&quot;&gt;potentially hazardous (PHOs)&lt;/a&gt; if they are over 140 meters across, per Wikipedia (NASA gives a different definition of PHOs below - so take that definition with a grain of salt).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the infrequent but &quot;could happen at any time&quot; impact possibility, a well funded effort is underway to catalog, track and calculate the trajectory of every NEO larger than 100 meters, called the &lt;a href=&quot;https://www.nasa.gov/audience/foreducators/postsecondary/features/F_Near_Earth_Program.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Near Earth Object Project&lt;/a&gt;.  Because so much attention is given to NEOs, the distinction and naming of Atens and Apollos makes perfect sense to me and doesn't seem unreasonable at all, but as pointed out above, there aren't any clear-cut rules when it comes to categorization.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's most relevant to the NEO project, in addition to size is &lt;a href=&quot;https://en.wikipedia.org/wiki/Minimum_orbit_intersection_distance&quot; rel=&quot;nofollow noreferrer&quot;&gt;Minimum Orbit intersection distance or MOID&lt;/a&gt;.   PHO's by the NASA article, as opposed to Wikipedia, are categorized by Earth-MOID of less than 0.05 AU and an absolute magnitude (which is related to surface area, which is related to size) of 22 or brighter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not all crossing asteroids (determined by aphelion and perihelion) have orbits that come close.   Because orbits take place in 3 dimensions, orbits can be crossing, but never get close.  Pluto, for example, is a Neptune-crosser but the orbits never get close.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I added MOID cause I thought it was interesting, but back to categorization.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other 7 planets have their own versions of Atens and Apollos, called inner and outer grazers.  See &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_Jupiter-crossing_minor_planets&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jupiter crossing objects&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 7 planets also potentially have libration asteroids, distinct from Crossers.   Libration asteroids are (captured, in horeshoe orbits, or Trojan).   The 4 inner planets have very few libration asteroids.   Jupiter has tens of thousands.  The outer 3 known planets have several known libration objects each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also resonance objects (Jupiter's Hildas, Neptunes Plutinos).   Objects can also be defined by outer orbits (Trans) or inner orbits (Cis), and by prograde or retrograde, which isn't the same that planets appear to do, but orbits relative to the sun can move opposite to the 7 planets.   In the inner solar-system, there are very few retrograde orbiting objects but out beyond the 8 known planets, retrograde is more common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dwarf-planet Pluto, for example, is a Plutino, which is a subset of a Neptune-resonant-object, and a subset of a Trans-Neptunian object, which is a subset of a Kuiper-Belt object (KBO).  I mention that only to point out how many orbital categories there are.   Scientists like categories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's other categories include Centaurs, Scattered Disk Objects, Oort Cloud Objects and, of-course, Asteroid belt objects and lets not forget comets, which are distinct from Asteroids, though the distinction gets a little fuzzy for dead comets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another categorization example, Jupiter has so many trojans, that it's two camps are separated into Greeks and Trojans, which is odd, because Jupiter's Trojan camp is a subset of Jupiter's Trojan asteroids.  That's a naming system that stuck, but which I think is less than ideal.  You can also call them L4 Trojans (Greek camp) or leading the planet's orbit and L5 Trojans (Trojan camp), following the planet's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall point is that there's many of categories of asteroids.  Atens and Apollos seem to me, fine and relevant, not arbitrary.   (want an arbitrary definition, how about when a moon-planet system becomes a duel planet system - that's pretty darn arbitrary).  Many definitions are arbitrary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(too long?)&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-11-27T14:32:02.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23714" PostTypeId="2" ParentId="23618" CreationDate="2017-11-27T15:38:08.337" Score="1" Body="&lt;p&gt;If you don't mind speculation, it seems to me that a massive impact - like the one that is thought to have formed our moon, but even larger, could eject shards like that.    That such a shard could escape it's solar-system and fly through space between solar-systems isn't unreasonable.   Asteroids (to my understanding) are somewhat loosely formed and an elongated shape like that is unlikely.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small moon or larger asteroid near a super-nova might get turned into an elongated shape like that by the blast as another speculative answer.   A 3rd possibility might be a ejection from a high velocity and somewhat violent debris disk, perhaps from a planet that fell inside the Roche limit of a white dwarf star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's still an object that's strange for two completely separate reasons.  It being the first ever object observed to be passing through our solar-system and it's thought to be very unusual shape (footnote - better images would be nice).   Still, the intersection of two unrelated firsts scores high in statistical improbability, even if the shape can be explained, it still leaves open the question of why two never before seen events coincided with the first ever observed exo-asteroid.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-11-27T15:38:08.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23715" PostTypeId="2" ParentId="23677" CreationDate="2017-11-27T16:25:39.527" Score="2" Body="&lt;p&gt;On this question, the &lt;a href=&quot;https://arxiv.org/pdf/astro-ph/0207240.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Widrow (2002)&lt;/a&gt; review is still pretty up-to-date. You can also have a look on the introduction of &lt;a href=&quot;https://www.imprs-astro.mpg.de/sites/default/files/beck.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Beck (2013)&lt;/a&gt;, a recent PhD thesis on the subject with a straightfoward introduction to the subject.&lt;/p&gt;&#xA;&lt;p&gt;The main questions remain the same for decades now:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is the origin of the magnetic seed fields?&lt;/li&gt;&#xA;&lt;li&gt;How to amplify magnetic field up to observed values in the local Universe?&lt;/li&gt;&#xA;&lt;li&gt;How to get large scale magnetic field, in particular if the magnetic seed fields are produced locally?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h1&gt;Origin of the magnetic seed fields&lt;/h1&gt;&#xA;&lt;p&gt;Where do we start? That is no easy question, with a diversity of answers. One of the most standard ones is the Biermann battery you mentioned (one significant problem with Biermann batteries is the timescale needed to amplify such small fields; see below) in the early Universe. Another one is that the magnetic field can be generated at small scales by first generations of AGN and supernovæ and then distributed at larger scales (dynamo can help but distribution at larger scale can still be an issue; see below). Lastly, you can have some exotic processes to generate seed fields in the first instants after the Big Bang, typically linked to inflation processes and such.&lt;/p&gt;&#xA;&lt;h1&gt;Magnetic field amplification&lt;/h1&gt;&#xA;&lt;p&gt;If there is an initial magnetic field — which is not so easy to demonstrate — there are few mechanisms that could amplify magnetic fields efficently enough to explain the observed magnetic field. The most standard one is the &lt;span class=&quot;math-container&quot;&gt;$\alpha-\omega$&lt;/span&gt; dynamo, combining a twist of the magnetic field lines in a differentially rotating disk with turbulent motions in the ISM that allows a regeneration of the magnetic fields (supernovæ explosions, stellar winds and other instabilities could do the work). Another possible mechanism is the magneto-rotational instability (MRI), that can easily generate a significant, large-scale field from almost any given seed field. There are other candidates such as Parker instability, or other flavors of dynamos.&lt;/p&gt;&#xA;&lt;p&gt;A central question then is of timescale: if these mechanisms can generate magnetic fields, the needed timescale to amplify a seed field to the current value may be far larger than the age of the Universe.&lt;/p&gt;&#xA;&lt;h1&gt;Distribution&lt;/h1&gt;&#xA;&lt;p&gt;Magnetic field is everywhere, not only in the ISM and at larger scale in galaxies, but also in galaxy  clusters, in the IGM and so on. If the origin of the magnetic seed fields is of cosmic origin (exotic processes in the early Universe), you can explain the observed distribution; if not, things get tricky again. Turbulence, galaxy collisions and mergers may be of help, but the question is still largely open.&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-11-28T09:28:39.573" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23716" PostTypeId="2" ParentId="23707" CreationDate="2017-11-27T16:50:18.293" Score="2" Body="&lt;p&gt;The speed of the collision changes everything.   The formation of comets and asteroids is likely to take place the &lt;a href=&quot;https://en.wikipedia.org/wiki/Protoplanetary_disk&quot; rel=&quot;nofollow noreferrer&quot;&gt;circumstellar disk&lt;/a&gt; of matter, perhaps even before the fist plant's formed.  As the disk flattens and elliptical orbits tend to lose eccentricity by collision.  Low velocity impacts between objects are more common.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Protoplanetary disks are made up of rocks and ices.  It's sometimes said that they are made of gas too but that's not very accurate because Gas is high velocity even in low temperature and it doesn't stick around very well.  Ice, however, tends to be sticky and sticky is good.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These swirling disks are also charged because much in space is charge and there's the solar magnetic field which may play a role in formation.   Once a certain size is reached, gravity takes over, but before then, the formation of objects like asteroids isn't well understood.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As larger planets form, however, the opposite tends to happen.  Large planets can fling material into more eccentric orbits and impact velocity tends to increase again as the disk forms into planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The collision of two asteroids in space is likely to result in two smaller objects and some scattered debris.   A head-on collision can turn both into tiny pieces depending on the velocity.   A more common glancing blow would probably still leave two objects intact with some lost material.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's likely that in the solar-system's circumstellar disk, 100 meter and 1 km objects were formed by low velocity collisions, assisted by electrical stickiness.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-11-27T16:50:18.293" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23717" PostTypeId="2" ParentId="23676" CreationDate="2017-11-27T21:21:29.397" Score="2" Body="&lt;p&gt;The 10 mm ocular should work fine. It will give you 60x of magnification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 5 mm ocular is probably a bit too much for this scope. You can definitely try it, but the expectation here is that the image will be a bit too washed out. With 50 mm of aperture you can't really expect to push magnification much above 100x (twice the aperture in mm), and even at around 100x it would work only if it's a high performance instrument. But, like I said, try and see what happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can see in 50 mm of aperture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will see the phases of Venus. It will look like the Moon, just very tiny. This is best done when Venus is at the greatest distance from the Sun, and then it will look like a disk cut in half.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter - you will definitely see the disk. Seeing the equatorial belts will be more challenging, but it should be doable; it may take a few attempts if it doesn't work on first try. Jupiter is low contrast and it's hard to see its features. Let the image float through the field of view, relax and just take it in; after a few minutes your brain adapts to the view and starts perceiving more details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter's moons are definitely visible, like 4 tiny stars that seem to follow the planet around. Sometimes some of them are hidden, so don't be surprised if you can't see them all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn's ring is clearly visible. You won't see details such as the Cassini division, but the ring itself will be shown across the planet's disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon is an excellent target for telescopes of all sizes, and there's a lot to explore there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars will probably not show a lot of details in 50 mm of aperture. Anyway, this planet is only worth observing for a few weeks around each opposition, which happens once every two years (plus a few weeks). At best you will resolve Mars as a tiny brick-red disk, but I doubt you'll see terrain features. But go ahead and give it a try; when the Hellas Basin is full of fog, or frosted over, it's very bright, so in a small aperture it might be perceived as a white speck of light on one side of the disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm actually kind of curious myself if this would work. I'll give it a try next time around, since I have a few 50 mm doublets to play with.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It's important to focus the instrument well before using it. E.g. point it at the Moon and adjust the focus until the image is as sharp as possible. You can also focus on stars - adjust focus until the image of the star is as small as it gets. You can also adjust focus on whatever object you're currently observing.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2017-11-27T21:21:29.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23718" PostTypeId="1" CreationDate="2017-11-28T00:52:25.237" Score="2" ViewCount="740" Body="&lt;p&gt;Referencing the following data for December 11th 2017.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.timeanddate.com/moon/@4799274?month=12&amp;amp;year=2017&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.timeanddate.com/moon/@4799274?month=12&amp;amp;year=2017&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a dark sky for stargazing does it matter if the moon is at 37.8% so long as it is below the horizon when I'm observing? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the moon will be set at 1:32PM on the 11th and not rise again until 1:59AM on the 12th I would assume the moon would not affect viewing between those times. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore is there a guideline as to how close to moonrise the moon will start to have an affect?&lt;/p&gt;&#xA;" OwnerUserId="19885" LastActivityDate="2017-12-31T00:54:32.753" Title="Does New Moon Affect Stargazing After MoonSet?" Tags="&lt;light-pollution&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23727" PostTypeId="1" CreationDate="2017-11-28T15:48:03.657" Score="1" ViewCount="237" Body="&lt;p&gt;Will we be able to see our galaxy if we travel far enough? When we leave earth far enough, we'll be able to see Earth from distant, is it same with the galaxy? if we travel far enough, will we see the shape of our galaxy with naked eye?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we have to travel really far to see our galaxy like a dot? or just far enough to realize the size and shape of our galaxy?&lt;/p&gt;&#xA;" OwnerUserId="19904" LastActivityDate="2017-11-29T15:00:40.950" Title="Will we be able to see our galaxy if we travel far enough?" Tags="&lt;galaxy&gt;&lt;light&gt;&lt;naked-eye&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23728" PostTypeId="2" ParentId="23727" CreationDate="2017-11-28T16:39:06.673" Score="5" Body="&lt;p&gt;In theory, yes; in practice probably not ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) It's perhaps helpful to take the Andromeda galaxy as a first working example. It is somewhat bigger than our own Milky Way galaxy but not so big as to be a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we look at Andromeda, it appears to occupies an area about 3 degrees by 1 degree. By comparison, the Moon appear about 0.5 degree across. Andromeda is a good target to see, a little faint but it is just about visible with the naked eye on a good night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Andromeda is about 2.5 million light years away so, even travelling at the speed of light, it would take 2.5 million years to get far away enough from the Milky Way so we could look back and see our galaxy like we see Andromeda today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;( In Star Trek, the writers seemed broadly to limit themselves to adventures within our galaxy for the same reason: within the realms of sci-fi, perhaps they could justify getting around the Milky Way within a reasonable period of time - not so to travel between the galaxies )&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii) To reduce the apparent size of our galaxy to a dot would require travelling at least 10x further or probably more - depending on how you are observing. 50 million light years, anyone?&lt;/p&gt;&#xA;" OwnerUserId="17502" LastEditorUserId="17502" LastEditDate="2017-11-28T16:49:41.613" LastActivityDate="2017-11-28T16:49:41.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23730" PostTypeId="1" AcceptedAnswerId="23747" CreationDate="2017-11-28T20:30:26.127" Score="-1" ViewCount="144" Body="&lt;p&gt;To produce artificial gravity on a space station, we simply spin it around some central point, and the acceleration makes objects &quot;fall&quot; outwards. If our space station is so far in deep space that it cannot detect any light from stars whatsoever, we can still tell if we are spinning or not based on the artificial gravity produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? What are we spinning relative to? If I were to take a stationary space station and spin everything else in the universe around it, would it produce the same effect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or to put it another way, suppose I took all of the matter in the Universe and made it into a gigantic disk. Would it be possible to spin the disk? What would it be spinning relative to?&lt;/p&gt;&#xA;" OwnerUserId="14089" LastActivityDate="2017-12-02T21:45:11.330" Title="What defines stationary vs spinning in space?" Tags="&lt;gravity&gt;&lt;space-time&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23733" PostTypeId="2" ParentId="23727" CreationDate="2017-11-29T02:01:10.723" Score="2" Body="&lt;p&gt;Yes you can but it depends partly on the direction you travel relative to the galactic plane and the strength of your telescope. As you might guess, if we moved away along the galactic plane from our current position, we may see our galaxy looking like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FSBrZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FSBrZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be very hard to actually discern the shape from this view. If you move from the present location perpendicular to the galactic plane, you may, after many millions of light years, get a view closer to this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hEZ7e.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hEZ7e.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much easier to tell the shape of a galaxy.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastActivityDate="2017-11-29T02:01:10.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23742" PostTypeId="1" AcceptedAnswerId="23746" CreationDate="2017-11-29T11:23:37.273" Score="-1" ViewCount="357" Body="&lt;p&gt;When DST is introduced does it mean that the sun is in zenith  one hour later (1pm winter local time) or one hour earlier (11am winter local time)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The location is New York.&lt;/p&gt;&#xA;" OwnerUserId="19938" LastEditorUserId="19938" LastEditDate="2017-11-29T14:33:43.533" LastActivityDate="2017-11-29T17:23:36.987" Title="At which winter hour is sun in zenith in DST?" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;time&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23745" PostTypeId="2" ParentId="23727" CreationDate="2017-11-29T15:00:40.950" Score="1" Body="&lt;p&gt;Theoretically it is completely feasible. In practice...just don't. In order to get there in a few thousand years, you would have to travel well beyond the speed of light (which is impossible if you didn't know). Once you got to a point where the entire galacy was within your vision, the entire galaxy would look distorted, since our galaxy is 100,000 light years accross, the light from the edges of the Galaxy would be thousands of years old, and would not match up with how the spiral should appear. If you were able to look at the earth  through the most powerful telescope ever created, you would see things as they were well before you left (that is assuming you did travel faster than the speed if light to get to the point where you currently are). &lt;/p&gt;&#xA;" OwnerUserId="19943" LastActivityDate="2017-11-29T15:00:40.950" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23746" PostTypeId="2" ParentId="23742" CreationDate="2017-11-29T17:23:36.987" Score="1" Body="&lt;p&gt;In general, daylight saving time (DST) is not used in the winter. In New York, DST is used in spring, summer, and fall. (Since the start date was changed from April to March, technically DST does begin before winter ends.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Sun is on the meridian at noon local time (and at its highest point), that would correspond to 1 pm DST.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-11-29T17:23:36.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23747" PostTypeId="2" ParentId="23730" CreationDate="2017-11-29T20:09:25.367" Score="2" Body="&lt;p&gt;Issac Newton described an experiment in which a bucket containing water is spun. As the water in the bucket starts to rotate, it becomes concave. The reason for this can be understood in terms of rotating frames of reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton supposed (as an &lt;em&gt;axiom&lt;/em&gt; of mechanics) that there are frames of reference (a system of locating particles in space relative to an origin, and in time) in which his 3 laws of motion are true, and momentum is conserved. Such a frame is called &quot;inertial&quot;. If one inertial frame exists then any frame with an origin that moves at constant velocity relative to the inertial frame is also inertial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I mentioned, the existence of inertial frames is an axiom of Mechanics. The truth of the axiom is verified by observation. It can't be proved. If a frame of reference is rotation with respect to an inertial frame, then the rotating frame is not inertial, and Newton's laws (in their simplest form don't hold). Instead of &quot;F=ma&quot; there are extra terms for the Coriolis and centrifugal force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a spinning space station, it is spinning relative to an inertial frame, and that is why there is a centrifugal force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Newton, the bucket experiment proves that there is a notion of &quot;absolute space&quot;. This interpretation is not accepted by all, and in particular, Ernst Mach rejected the idea of any absolute space. His ideas were influential on Einstein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read more about Newton's bucket experiment at &lt;a href=&quot;http://www-history.mcs.st-andrews.ac.uk/HistTopics/Newton_bucket.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www-history.mcs.st-andrews.ac.uk/HistTopics/Newton_bucket.html&lt;/a&gt; and you can read about Ernst Mach's ideas regarding absolute space at &lt;a href=&quot;https://en.wikipedia.org/wiki/Mach%27s_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Mach%27s_principle&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-11-29T20:09:25.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23751" PostTypeId="1" AcceptedAnswerId="23763" CreationDate="2017-11-30T07:03:36.813" Score="5" ViewCount="851" Body="&lt;p&gt;I was recently reading an &lt;a href=&quot;https://www.space.com/14719-spacekids-temperature-outer-space.html&quot; rel=&quot;noreferrer&quot;&gt;article&lt;/a&gt; on space.com about : &lt;strong&gt;What's the Temperature of Outer Space&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They said :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some parts of space are hot! Gas between stars, as well as the solar wind, both seem to be what we call &quot;empty space,&quot; yet they can be more than a thousand degrees, even millions of degrees.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But here is my question - &lt;strong&gt;&quot;How do we &lt;em&gt;define&lt;/em&gt; temperature in outer space?&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We perceive temperature as the average translational kinetic energy associated with the disordered motion of atoms and moleculules, so how can the temperature be so high at some places, if there isn't any (or very, very few) molecules out there in &lt;strong&gt;vacuum&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="16477" LastActivityDate="2017-12-05T15:07:40.340" Title="How do we define temperature in outer space?" Tags="&lt;space&gt;&lt;temperature&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23753" PostTypeId="1" CreationDate="2017-11-30T08:50:47.527" Score="2" ViewCount="582" Body="&lt;p&gt;I stay in Bangalore, India (12.9716° N, 77.5946° E). For my latitude, the Sun crosses zenith (the highest point in the sky) on 24th of April every Summer, which happens to be about 2 months prior to the June solstice (summer solstice to us here in the Northern Hemisphere).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Shouldn't the Sun be reaching its highest point on the longest day, i.e. on June 21st, instead of April 24th?&lt;br /&gt;&#xA;Doesn't the Sun take the longest arc across the sky on the day it passes through zenith?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference, the angle at which the Sun rises/sets in my location is about 13 degrees, which happens to be almost vertical to the horizon.&lt;/p&gt;&#xA;" OwnerUserId="19966" LastEditorDisplayName="user1569" LastEditDate="2017-11-30T16:32:46.090" LastActivityDate="2018-03-05T19:32:29.680" Title="Question regarding the path of sun crossing zenith before the summer solstice" Tags="&lt;the-sun&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23756" PostTypeId="2" ParentId="23671" CreationDate="2017-11-30T10:02:29.690" Score="1" Body="&lt;p&gt;There are four classes of asteroid in this Near-Earth Orbit taxonomy.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Atiras are those asteroids which have orbits interior to Earth's orbit. Their aphelion distance (farthest from the Sun) is less than 0.983 au.&lt;/li&gt;&#xA;&lt;li&gt;Atens are Earth-crossers with semimajor axes smaller than Earth's. Their semimajor axes are less than 1.0 au and their aphelion distance is greater than 0.983 au.&lt;/li&gt;&#xA;&lt;li&gt;Apollos are Earth-crossers with semimajor axes larger than Earth's. Their semimajor axes are greater than 1.0 au and their perihelion distance (nearest to the Sun) is less than 1.017 au.&lt;/li&gt;&#xA;&lt;li&gt;Amors have orbits exterior to Earth's orbit but interior to Mars'. Their semimajor axis is greater than 1.0 au and their perihelion distance is greater than 1.017 au, so they never cross Earth's orbit.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CYlwv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CYlwv.jpg&quot; alt=&quot;https://cneos.jpl.nasa.gov/about/neo_groups.html&quot;&gt;&lt;/a&gt;&#xA;(Image: NASA/JPL)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why Apollos and Atens are separate classes is that Apollos aphelion distance is not restricted. Apollos cross Earth's orbit but can come from far away - the Main Belt, for example - so they could have a quite large semimajor axis. &lt;strong&gt;Most of the time they are farther from the Sun than Earth is.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, Atens have a small orbit - smaller than Earth's but they cross Earth's orbit. So &lt;strong&gt;most of the time they are closer to the Sun than Earth is.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2017-11-30T10:02:29.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23761" PostTypeId="2" ParentId="23751" CreationDate="2017-11-30T13:51:53.700" Score="2" Body="&lt;p&gt;In the absense of any particles, we evaluate temperature by the radiation wavelengths seen. In &quot;empty space&quot; only the cosmic background radiation exists, so we mark it pretty cold.  see, for example, &lt;a href=&quot;https://www.uu.edu/dept/physics/scienceguys/2003July.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this note&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you look at a chunk of space containing even a few particles, then the temperature is defined based on the mean kinetic energy of the particles, or more precisely, the amount of entropy present. See &lt;a href=&quot;https://en.wikipedia.org/wiki/Temperature#Temperature_as_an_intensive_variable&quot; rel=&quot;nofollow noreferrer&quot;&gt;temperature as intensive variable&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-11-30T13:51:53.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23763" PostTypeId="2" ParentId="23751" CreationDate="2017-11-30T16:01:30.603" Score="4" Body="&lt;p&gt;First of all, the medium they are mentioning is far from empty even though it is much less dense than what we do know on Earth (see &lt;a href=&quot;https://astronomy.stackexchange.com/questions/679/how-does-interstellar-matter-density-vary/681#681&quot;&gt;this question&lt;/a&gt; for more details).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, the problem with temperature in astrophysics is that the medium you study is in general far from thermodynamic equilibrium (thermodynamic equilibrium means that there are no flows of enery or matter within the system and with the outside of the system).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most commonly, however, we can use another flavor of equilibrium to estimate temperature. This is the &lt;strong&gt;kinetic equilibrium&lt;/strong&gt;: since most collisions are elastic (meaning that the energy is conserved), particles velocities will follow a Maxwell-Boltzmann distribution. Depending on the ionization fraction (how much is your medium ionized?) this temperature will either the kinetic temperature of electrons or the kinetic temperature of hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, if you want to understand these high temperatures, you have to take into account the &lt;strong&gt;photoionization rate&lt;/strong&gt; and the &lt;strong&gt;photoelectric heating&lt;/strong&gt; (a process in which you eject electron from the matter of the studied medium due to UV absorption). The thermal state of the medium then depends on the equilibrium between the energy absorption by the matter and the re-emission of this energy in thermal radiation. In this framework, the energy that can be absorbed comes from the interstellar radiation field (ISRF). In the solar neighborhood, it is dominated by six different types of radiation: galactic synchrotron emission of relativistic electrons (electrons accelerated to tremendous speed in the galaxy), the cosmic microwave background, the infrared emission of the matter heated by the light emitted by stars, emission from ionized plasma, the light emitted by stars itself, and lastly X-rays from hot plasma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://arxiv.org/pdf/1412.5182.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Klessen &amp;amp; Glover (2016)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://arxiv.org/pdf/astro-ph/0106359.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ferrière (2001)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="88" LastActivityDate="2017-11-30T16:01:30.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23764" PostTypeId="2" ParentId="23692" CreationDate="2017-11-30T17:27:21.373" Score="4" Body="&lt;p&gt;The short answer is that, as George notes in his answer, there is no naming system for supermassive black holes. I wouldn't say they're really &quot;named&quot; after their host galaxies; more that when one wants to refer one, you usually do it by using the name of the galaxy (e.g., &quot;the SMBH in M31&quot; or &quot;M87's SMBH&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Sgr A*&quot; is actually the name of a &lt;em&gt;radio source&lt;/em&gt;, a subcomponent of the radio source &lt;a href=&quot;https://en.wikipedia.org/wiki/Sagittarius_A&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sgr A&lt;/a&gt; which happens to be associated with the Galaxy's SMBH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The very early system for naming radio sources -- before better radio telescopes found so many that the system became unworkable -- was basically &quot;name of constellation&quot; + &quot;decreasing brightness rank&quot;. So Sgr A was simply the brightest radio source in the constellation of Sagittarius. (In fact, the system was abandoned before most constellations even got to B.) Most such sources are not black holes; for example, Casiopeia A is a supernova remnant, as is Taurus A (the Crab Nebula).&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2017-11-30T17:27:21.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23765" PostTypeId="2" ParentId="23718" CreationDate="2017-12-01T00:12:18.617" Score="1" Body="&lt;p&gt;For most people the Moon does not significantly impact on stargazing, even when it is full, unless they are trying to view something in a direction close to the Moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For people doing astrophotography, the Moon has a bigger impact since the sky is noticeably brighter and results in fainter objects not being able to be photographed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for astronomical observatories, some operations are impossible at worst and severely restricted at best since the telescopes are usually larger than &quot;home&quot; varieties and cameras much more sensitive but most noticeably a few days either side of the full moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, depending on what you are actually doing, a ~38% Moon will not be a problem and if you are looking more than about 30° away from the Moon you will not even notice it.&lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2017-12-01T00:12:18.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23766" PostTypeId="2" ParentId="23704" CreationDate="2017-12-01T00:31:50.067" Score="1" Body="&lt;p&gt;Almost without exception, the stars that have proper names are those stars that are visible to the naked eye. The rest require an instrument such as a telescope to be seen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While a proper name may appeal to people reading about discoveries of things like exoplanets, using a catalogue designation makes it easier for the researchers. Most of the catalogue designations originate in a straightforward process of labelling new discoveries according to some method. For some, WISE J085510.83−071442.5 might be a monstrosity but for the people involved in WISE and subsequent studies it is a clear designation that helps with communication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not really any different than using &lt;em&gt;Sirius&lt;/em&gt; or &lt;em&gt;α Canis Majoris&lt;/em&gt;. Some people will use &lt;em&gt;α Canis Majoris&lt;/em&gt; to try to clearly communicate that they mean the brightest star in the constellation &lt;em&gt;Canis Major&lt;/em&gt;, and some people hearing that will complain that they don't simply call it &lt;em&gt;Sirius&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="18643" LastEditorUserId="18643" LastEditDate="2017-12-20T00:46:02.173" LastActivityDate="2017-12-20T00:46:02.173" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23767" PostTypeId="2" ParentId="23683" CreationDate="2017-12-01T00:43:37.650" Score="-1" Body="&lt;p&gt;Solar telescopes typically have filters to reduce the amount of light reaching their detector. Depending on their purpose they may use, for example, neutral density filters to reduce the amount of light to a level where they can resolve details without blocking any of the visible spectrum, hydrogen alpha filters to observe the Sun at a wavelength of 656 nanometres for observing features in the Sun's atmosphere such as solar prominences and the chromosphere, or blocking the solar disc to observe the Sun's corona, flares and other ejecta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hubble, on the other hand, is designed with a very different purpose in mind such as taking long-exposure images to obtain the deep- and ultra-deep field full spectrum images. It avoids the Sun as even a very short exposure to the Sun would damage its instruments.&lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2017-12-01T00:43:37.650" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23768" PostTypeId="2" ParentId="23631" CreationDate="2017-12-01T03:15:52.750" Score="1" Body="&lt;p&gt;You can upload your FITS files to &lt;a href=&quot;http://astrometry.net&quot; rel=&quot;nofollow noreferrer&quot;&gt;astrometry.net&lt;/a&gt; (or use an API) and get the coordinates. You can optionally get new FITS files back with the coordinates included in the metadata. From there you will be able to proceed with your processing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, instead of uploading everything, you can also install a copy of the astrometry.net package locally if you're running Linux.&lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2017-12-01T03:15:52.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23769" PostTypeId="1" CreationDate="2017-12-01T03:25:53.017" Score="1" ViewCount="72" Body="&lt;p&gt;I just watch an interesting video about &lt;a href=&quot;https://youtu.be/a7GuT83cc-4&quot; rel=&quot;nofollow noreferrer&quot;&gt;Moonphase watch&lt;/a&gt;. mmm... but why do I need to know about moonphase anyway, I just can't think of any practice for the moonphase indicator on the watch?&lt;/p&gt;&#xA;" OwnerUserId="19994" LastActivityDate="2017-12-01T03:25:53.017" Title="Why do I need to know about moonphase?" Tags="&lt;the-moon&gt;&lt;moon-phases&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23770" PostTypeId="2" ParentId="23291" CreationDate="2017-12-01T03:30:00.960" Score="3" Body="&lt;p&gt;They do use two or more detectors to triangulate the origin of a source. Most recently the Advanced Virgo detector in Italy was used in conjunction with the two LIGO detectors in the USA. This has been published in &lt;em&gt;Physical Review Letters&lt;/em&gt; (&lt;a href=&quot;https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.119.141101&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Phys. Rev. Lett. 119, 141101 – Published 6 October 2017&quot;&gt;Phys. Rev. Lett. 119, 141101 – Published 6 October 2017&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With two detectors the sky area for &quot;pinpointing&quot; the source is quite a large area, but with three detectors the area is reduced by a factor of about 20.&lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2017-12-01T03:30:00.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23773" PostTypeId="1" AcceptedAnswerId="23775" CreationDate="2017-12-01T11:57:55.560" Score="3" ViewCount="5481" Body="&lt;p&gt;Kepler's 3rd Law (regarding the relation between orbital period and the semi-major axis of an orbit) applies to all elliptical orbits. But as I understand it, the only reason it is safe to use the approximation that the semi-major axis of an orbit is approximately equal to the average orbital distance is because the planets of our solar system are generally not very eccentric. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/R3DI2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R3DI2.jpg&quot; alt=&quot;Picture to help explain words&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, the Earth is about 1 AU away from the Sun and has a very low orbital eccentricity, so one can say that the semi-major axis of the Earth's orbit is also approximately 1 AU in length. Since &lt;code&gt;P^2 (yr) = a^3 (AU)&lt;/code&gt;, one can deduce that the orbit of Earth should take about a year. But, what if the Earth had a higher eccentricity - If the orbital eccentricity of the Earth were higher, would it still be wise to approximate the orbital distance as the semi-major axis of an orbit?&lt;/p&gt;&#xA;" OwnerDisplayName="user15317" LastActivityDate="2018-05-02T15:26:03.247" Title="Can one approximate the semi-major axis of an orbit as the average orbital distance for eccentric orbits?" Tags="&lt;orbit&gt;&lt;orbital-elements&gt;&lt;orbital-mechanics&gt;&lt;kepler&gt;&lt;eccentric-orbit&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23775" PostTypeId="2" ParentId="23773" CreationDate="2017-12-01T12:58:08.313" Score="2" Body="&lt;p&gt;The gist of this, is that your assumption is incorrect.   It's the semi-major axis that defines the period, not the average distance.   Newton worked this out when he invented calculus and derived Kepler's laws.   (Look up Derivation of Kepler's laws for some explanations).   Here's the &lt;a href=&quot;https://en.wikipedia.org/wiki/Specific_relative_angular_momentum&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia version of the Math&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should add that Kepler's three laws, and Newton's derivation of them still don't perfectly define orbits.  They assume that only the larger mass matters, when in reality, the two objects tug on Each other.    The Sun doesn't just tug on the Earth but the Earth tugs on the Sun, which speeds up the orbit a tiny bit (the mass ratio of 330,000 to 1 makes this close to negligible, but it's there).  More accurate Newtonian orbital calculations between two massive objects require more advanced mathematics.  Kepler's laws only work if the central object is much more massive than the 2nd object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring the finer details and just looking at the three laws, Newton's derivation works out exactly to the semi-major axis.  The average distance doesn't enter into the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Average distance of an orbiting object to it's Sun or central object is problematic anyway because the planet moves more slowly when it's more distant, so there's a few ways to calculate average distance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;By time&lt;/strong&gt; the planet spends in each part of it's orbit (same as area as defined by Kepler's 3rd law)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, you can measure average distance &lt;strong&gt;by arc&lt;/strong&gt; or angle, taking smaller and smaller angles and running an average.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can take average distance &lt;strong&gt;by length&lt;/strong&gt; of the ellipse to the specific foci.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's three separate methods for measuring average distance.  Needless to say, this gets a little complicated.  The good news is, you don't need to do any calculation for average distance because the Semi-major axis is actually correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the semi-major axis equal the average distance, perhaps by arc or by length?   Not sure.   I'll try to work that out.  I know that it doesn't equal average by time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that make sense.  I'm not sure I explained that as well as I should.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On average distance.  The standard average distance with orbits is by time or area.  Kepler's 3rd law says equal areas over equal time, so it's really two ways of saying the same thing, equal time and equal surface area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TYz4M.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TYz4M.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think your question is interesting and I'll try to work out a relation between semi major axis and average distance.  Off hand, it's a little complicated, but it seems to me that ratio between average distance and semi major axis changes with eccentricity but, I'd rather work that out before I say for certain.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-12-01T13:33:42.897" LastActivityDate="2017-12-01T13:33:42.897" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23777" PostTypeId="1" CreationDate="2017-12-01T13:51:30.040" Score="5" ViewCount="900" Body="&lt;p&gt;Edit: just found &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6183/why-do-most-of-the-planets-rotate-counterclockwise-i-e-the-same-way-the-sun&quot;&gt;duplicate&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since orbital velocity is greater on small orbits, and smaller on large orbits, why don't planets spin retrograde around their own axis? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One portion of planet's mass is closer to its star, and should move prograde relative to center of mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of the planet's mass far from the star orbits higher and should move retrograde relative to center of mass, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which should create a resulting retrograde spin, with rotation speed depending on planet's diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's wrong with this? ( since fact is : most planets spin prograde )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4fk31.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4fk31.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lHaJu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lHaJu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this not to scale second illustration, planet is replaced by two masses linked by a beam. Kepler's law of equal surfaces swept per unit of time is used&#xA;to display the retrograde rotational force that should be applied relative to center of mass, which orbits black path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT : Ok I think I found where my mistake is(?) Differential gravitational pull (yellow arrows) counteracts retrograde rotational force (green arrow) Which results in no rotational force at all. Anyway I still don't get how conservation of momentum of everything orbiting the sun, also applies to spin direction of orbiting objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4r6K5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4r6K5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16500" LastEditorUserId="16500" LastEditDate="2017-12-03T15:23:56.727" LastActivityDate="2017-12-04T10:13:40.173" Title="Why most planets don't spin retrograde?" Tags="&lt;orbit&gt;&lt;rotation&gt;&lt;velocity&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ClosedDate="2017-12-06T19:20:11.910" ContentLicense="CC BY-SA 3.0" />
  <row Id="23780" PostTypeId="2" ParentId="23777" CreationDate="2017-12-01T15:57:17.397" Score="2" Body="&lt;p&gt;Basically, go back to the pre-planet particle clouds, and as explained in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.geek.com/news/geek-answers-why-do-all-planets-rotate-and-orbit-in-the-same-direction-1564529/&quot; rel=&quot;nofollow noreferrer&quot;&gt;geek.com&lt;/a&gt;, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To answer the question we have to go back to the beginning, to the&#xA;  formation of a solar system like ours. Before a star and its planets&#xA;  exist, there’s just a cloud of disorganized gas and small molecules.&#xA;  This is often called a molecular cloud or “stellar nursery.” The Eagle&#xA;  and Orion Nebulae are some of the more famous stellar nurseries we&#xA;  have been able to observe with the Hubble Space Telescope. These&#xA;  clouds are composed mostly of molecular hydrogen, which wouldn’t be&#xA;  able to congregate outside the dense molecular clouds. These clouds&#xA;  can be of any size — not just massive structures like the Orion&#xA;  Nebula. Over time the energy of the molecules in the cloud pushing&#xA;  outward can be overcome by slower molecules collapsing together&#xA;  farther in. As long as there is sufficient mass in the molecular&#xA;  cloud, it continues collapsing in toward the center until it reaches a&#xA;  high enough mass to fuse hydrogen and become a new star. The spin that&#xA;  we see quite clearly now is related to the process of the molecular&#xA;  cloud collapsing. The original cloud was very, very large and made up&#xA;  of many individual molecules and small clumps of matter. On that&#xA;  scale, there is some small amount of rotation within the cloud. It&#xA;  could be caused by the gravity of nearby stellar objects, local&#xA;  differences in mass as the cloud churns, or even the impact of a&#xA;  distant supernova. The point is, most molecular clouds have at least a&#xA;  little rotation.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;As the cloud collapses to form a star, it has what physicists call&#xA;  angular momentum. This is the movement an object has as it rotates&#xA;  around a central point. In a large system like a molecular cloud, each&#xA;  particle has some angular momentum, and it all adds together across a&#xA;  very wide area. That’s a lot of momentum, and it is conserved as the&#xA;  cloud continues to collapse. But how does that get us to objects that&#xA;  spin and orbit?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Imagine a figure skater spinning around with arms outstretched. That&#xA;  is a model of angular momentum just like a collapsing cloud of gas.&#xA;  When the arms are drawn inward, the rotational velocity goes up&#xA;  because the total angular momentum is conserved unless there is some&#xA;  external force acting on it. There are such forces acting on the&#xA;  figure skater, but less so on a collapsing molecular cloud. So if a&#xA;  molecular cloud was maybe a light year across, then collapsed down to&#xA;  be just a fraction of that, it would be a huge change in size. Just&#xA;  like the figure skater pulling her arms in, the velocity must increase&#xA;  to conserve angular momentum and therefore form a spinning&#xA;  protostellar disc. It is from this orbiting matter that all the&#xA;  planets form, and of course, they are also spinning and orbiting in&#xA;  the same direction because of the conservation of angular momentum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Quoting from wikipedia as to the mystery of  the misbehaving planets: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;All eight planets in the Solar System orbit the Sun in the direction&#xA;  that the Sun is rotating, which is counterclockwise when viewed from&#xA;  above the Sun's north pole. Six of the planets also rotate about their&#xA;  axis in this same direction. The exceptions—the planets with&#xA;  retrograde rotation—are Venus and Uranus. Venus's axial tilt is 177&#xA;  degrees, which means it is spinning almost exactly in the opposite&#xA;  direction to its orbit. Uranus has an axial tilt of 97.77 degrees, so&#xA;  its axis of rotation is approximately parallel with the plane of the&#xA;  Solar System. The reason for Uranus's unusual axial tilt is not known&#xA;  with certainty, but the usual speculation is that during the formation&#xA;  of the Solar System, an Earth-sized protoplanet collided with Uranus,&#xA;  causing the skewed orientation.[6]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It is unlikely that Venus was formed with its present slow retrograde&#xA;  rotation, which takes 243 days. Venus probably began with a fast&#xA;  prograde rotation with a period of several hours much like most of the&#xA;  planets in the solar system. Venus is close enough to the Sun to&#xA;  experience significant gravitational tidal dissipation, and also has a&#xA;  thick enough atmosphere to create thermally driven atmospheric tides&#xA;  that create a retrograde torque.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-12-01T15:57:17.397" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23783" PostTypeId="1" AcceptedAnswerId="23895" CreationDate="2017-12-01T20:32:21.203" Score="0" ViewCount="103" Body="&lt;p&gt;If I understand correctly, red dwarfs consume all of their hydrogen and will eventually shrink into helium white dwarfs, whereas the higher mass stars will consume a fair amount of their hydrogen fuel, not necessarily all of it. Correct me if I'm wrong.&lt;/p&gt;&#xA;" OwnerUserId="16048" LastActivityDate="2017-12-09T19:22:57.190" Title="Do lower mass stars consume more hydrogen than higher mass stars?" Tags="&lt;star&gt;&lt;red-dwarf&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23784" PostTypeId="1" AcceptedAnswerId="23801" CreationDate="2017-12-02T00:21:59.573" Score="0" ViewCount="268" Body="&lt;p&gt;I understand that as the universe continues to expand, eventually our supply of hydrogen will be exhausted and then it's &quot;lights out.&quot;  Is that really going to happen or is there something &quot;eternal&quot; about the universe?  It matters because if the universe is eternal, and since we are a part of the universe, we too are eternal as well in some sense. &lt;/p&gt;&#xA;" OwnerUserId="5306" LastActivityDate="2017-12-04T12:15:43.190" Title="Is the universe eternal?" Tags="&lt;cosmology&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23785" PostTypeId="1" AcceptedAnswerId="23797" CreationDate="2017-12-02T03:08:23.653" Score="2" ViewCount="71" Body="&lt;p&gt;Problem number 23 in chapter 1 of Purcell's &quot;Electricity and Magnetism&quot; Second edition page 37 shows that 90% of the energy stored in the electrostatic field of a spherical charge distribution of radius R is contained in a spherical region of 10R. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy by this description is diffuse and only partially resides within what we would call the physical bounds of the particle. Gravity has the same form as the electromagnetic force and I would expect would have this same character of distribution of energy (i.e. 90% of the energy of the gravitational mass contained within a radius of 1R is stored within a sphere of 10R).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true of gravitational energy and has this little detail been considered in the galaxy rotation curve explanations? I would think that something this elementary would be well accounted for in galaxy rotation descriptions. Much of the electromagnetic and gravitational energy seems diffused into the area surrounding mass and I think that galaxy rotation curves require more diffuse gravitational energy distribution don't they?&lt;/p&gt;&#xA;" OwnerUserId="20025" LastEditorUserId="2531" LastEditDate="2017-12-04T18:03:34.187" LastActivityDate="2017-12-04T18:07:06.130" Title="Is diffuse energy distribution considered in the galaxy rotation curve explanations?" Tags="&lt;astrophysics&gt;&lt;galactic-dynamics&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23787" PostTypeId="1" CreationDate="2017-12-02T05:50:06.743" Score="1" ViewCount="221" Body="&lt;p&gt;If Earth had rings, how long would it take for the rings to loose orbit? One of the answers in this question &lt;a href=&quot;https://physics.stackexchange.com/questions/281828/if-earth-had-rings&quot;&gt;https://physics.stackexchange.com/questions/281828/if-earth-had-rings&lt;/a&gt; said we did at one time is it possible? &lt;a href=&quot;https://i.stack.imgur.com/hSxVI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hSxVI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="1891" LastEditDate="2018-03-01T03:02:47.733" LastActivityDate="2018-03-01T03:02:47.733" Title="If Earth had rings, how long would it take for the rings to loose orbit?" Tags="&lt;orbit&gt;&lt;gravity&gt;&lt;earth&gt;&lt;planetary-ring&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2018-03-07T21:05:38.827" ContentLicense="CC BY-SA 3.0" />
  <row Id="23790" PostTypeId="2" ParentId="23773" CreationDate="2017-12-02T08:13:37.403" Score="4" Body="&lt;p&gt;The maths says that the semi-major axis is not a good measure of average distance for high eccentricity (elliptical) orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are basically two ways to measure this : (1) an average over the entire orbit on a purely geometric basis, and (2) the average over time.  These give quite different results - qualitatively different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Average distance on geometric basis.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a way this is the most obvious one.  The mathematics is :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_g = \frac a {2\pi} \int_0^{2\pi}r\,d\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we use for an ellipse :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r = \frac \rho {1+\epsilon\,cos\theta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is (eventually :-) ) :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_g = a\,\sqrt{1-\epsilon^2} \tag{1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Average over time.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a tricker calculation, but in fact it's the most &quot;human&quot; average.  We want :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_t = \frac 1 T \int_0^T r\,dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maths on this is tedious and not at all useful, but the result ( with a little help from Wolframalpha.com ) is :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_t = a\,\frac 2 {\pi} \frac{2+\epsilon^2}{(1-\epsilon^2)^{\frac 5 2}}tan^{-1}\left( \sqrt{\frac {1-\epsilon}{1+\epsilon}} \right) \tag{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now by any definition that's a messy equation, but the main point is to note the key difference between the time averaged expression (2) and the geometric one (1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The geometric expression gets &lt;em&gt;smaller&lt;/em&gt; as eccentricity gets larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time averaged one gets &lt;em&gt;larger&lt;/em&gt; as eccentricity gets larger (a slight simplification).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact for both $\frac R a$ starts off at $1.000$ but when $\epsilon\to 1$ then  $R_g\to 0$ and $R_t\to \infty$ !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The geometric expression doesn't allow for the orbital speed of the object.  As you increase eccentricity an orbits ends up with having a faster close approach, so less time close up to it's primary.  As it moves away from the primary, the orbital speed also decreases, so it spends much more time at large distances than at short distances.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastEditorUserId="10196" LastEditDate="2017-12-03T14:37:33.070" LastActivityDate="2017-12-03T14:37:33.070" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23793" PostTypeId="1" AcceptedAnswerId="23798" CreationDate="2017-12-02T16:32:12.100" Score="3" ViewCount="510" Body="&lt;p&gt;So from what I've read, Oumuamua travels at around &lt;code&gt;26km/s&lt;/code&gt; when in interstellar space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also remember reading a while back that Voyager was travelling at around &lt;code&gt;18km/s&lt;/code&gt; when leaving the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing as Voyager took around 35 years (I think) to exit the solar system, I'd guess that Oumuamua entered our solar system around 20–25 years or so ago?&lt;/p&gt;&#xA;" OwnerUserId="7845" LastActivityDate="2017-12-03T13:16:09.113" Title="When would Oumuamua have entered the solar system?" Tags="&lt;solar-system&gt;&lt;asteroids&gt;&lt;speed&gt;&lt;interstellar&gt;&lt;oumuamua&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23794" PostTypeId="1" AcceptedAnswerId="26203" CreationDate="2017-12-02T16:58:42.090" Score="8" ViewCount="278" Body="&lt;p&gt;&lt;a href=&quot;https://earthscience.stackexchange.com/questions/12801/is-there-any-new-water-formed-on-or-brought-to-earth#comment23432_12801&quot;&gt;This comment&lt;/a&gt; (below the question &lt;a href=&quot;https://earthscience.stackexchange.com/q/12801/6031&quot;&gt;Is there any new water formed on, or brought to, Earth?&lt;/a&gt;) links to the 2002 Article in the New Scientist &lt;a href=&quot;http://smallcomets.physics.uiowa.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Was all the Earth’s water brought here by comets?&lt;/a&gt; which states:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Evidence that there is a constant rain of small comets, each a loose snowball held together by a carbon shell, first came in the 1980s from satellites that looked down on Earth and observed “atmospheric holes”- dark spots 50 to 100 kilometres across- that were appearing in the dayglow of the atmosphere. The dayglow is ultraviolet light emitted when sunlight interacts with the oxygen in the upper atmosphere. Water is the only common substance that could create such holes, and each could be made by a small comet containing between 20 and 40 tonnes of water that broke up into a cloud of vapour as it hit the atmosphere. The case for a cometary origin for the atmospheric holes grew in the 1990s when the trail of a cosmic snowball was picked up as it vaporised over the Atlantic by the visible imaging cameras on polar satellites.&lt;/p&gt;&#xA;&lt;p&gt;From the number of atmospheric holes, &lt;strong&gt;it’s possible to calculate that the atmosphere is being hit by a 20 to 40-tonne snowball comet every 3 seconds&lt;/strong&gt;. According to Frank, this is enough to add an inch of water to the Earth’s surface every 20,000 years. Over the 4.5-billion-year lifetime of the Earth they could have provided all the water in the oceans. (emphasis added)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The article also links to this University of Iowa page &lt;a href=&quot;http://smallcomets.physics.uiowa.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;smallcomets.physics.uiowa.edu/&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Is this still believed to be happening? Have there been additional observations? Is the earth receiving water in the form of 20 to 40 ton comets ariving every three seconds?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What ever happened to the &amp;quot;Small Comet Theory&amp;quot; of water influx on present-day Earth? Has it been discounted, modified, verified?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-05-07T18:29:51.403" Title="What ever happened to the &quot;Small Comet Theory&quot; of water influx on present-day Earth?" Tags="&lt;comets&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23796" PostTypeId="2" ParentId="23730" CreationDate="2017-12-02T21:45:11.330" Score="1" Body="&lt;p&gt;Unlike linear motion, rotation is relative to absolute space.  However, near a large rotating body, like a black hole, frame dragging would have a relative effect.&lt;/p&gt;&#xA;" OwnerUserId="18939" LastActivityDate="2017-12-02T21:45:11.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23797" PostTypeId="2" ParentId="23785" CreationDate="2017-12-03T00:18:13.850" Score="3" Body="&lt;p&gt;You appear to be asking whether gravitational potential energy acts as a significant contributor to the total mass-energy of a galaxy - a question that is only relevant within the context of General Relativity, where indeed other forms of energy besides the rest-mass energy contribute to the gravitational effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that GR is &lt;em&gt;not&lt;/em&gt; required to interpret galaxy rotation curves because the influence of the other forms of mass-energy (or more formally, other contributors to the stress-energy tensor) are very small. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The contribution of gravitational potential energy can be assessed using the ratio $GM/rc^2$, which is approximately the ratio of gravitational potential energy to rest mass energy. For example, if $10^{12}$ solar masses exists within a 100 kpc radius (reasonable numbers for the Milky Way); then the ratio is $5\times 10^{-7}$. Of course, if there is no dark matter then this would be even smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should also add that this tiny correction, like the gravitational potential energy, is &lt;em&gt;negative&lt;/em&gt;. That is, it &lt;em&gt;reduces&lt;/em&gt; the gravitational influence of the mass/energy in a galaxy. The (positive) total kinetic energy of the Galactic components (also tiny) would also need to be added, but since bound systems have a total energy (kinetic plus gravitational potential) $&amp;lt;0$, there is still a net reduction in the gravitational mass.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2017-12-04T18:07:06.130" LastActivityDate="2017-12-04T18:07:06.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23798" PostTypeId="2" ParentId="23793" CreationDate="2017-12-03T00:38:47.660" Score="5" Body="&lt;p&gt;One problem with answering this question is that it's not precisely clear &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6519/where-does-the-solar-system-end&quot;&gt;where the solar-system ends&lt;/a&gt; or when (if) &lt;a href=&quot;http://blogs.discovermagazine.com/crux/2013/03/20/did-voyager-1-leave-the-solar-system-or-not/#.WiM8KTdOnIU&quot; rel=&quot;nofollow noreferrer&quot;&gt;Voyager I crossed that border&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we set the estimate at 19 billion km (&lt;a href=&quot;https://science.nasa.gov/science-news/science-at-nasa/2013/12sep_voyager1&quot; rel=&quot;nofollow noreferrer&quot;&gt;per article&lt;/a&gt;) or 127 AU. (about 4 times the distance between Neptune and the Sun), then an estimate is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some conversions first.  Voyager left earth in 1977 and the announcement was in 2013, so it's usually listed as 36 years (it might be 35 and some months) but that's not all that relevant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;18 km/s x 31.5 million seconds in a year (close enough) is 567 million km per year.   It should have taken Voyager about 33.5 years at that velocity to reach 19 billion km.   However, Voyager's velocity wasn't consistent.  As it flies from the sun it slows down and it had several gravity assists along the way which sped it up and changed it's direction.   Mathematical calculations get a little tricky, but here's a chart of Voyager's velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/V6qUC.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V6qUC.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: that's Voyager 2's velocity not Voyager 1.  I'm having trouble finding Voyager 1's velocity chart).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's it's direction, more of a fish-hook shaped path than a straight line, though it's been mostly straight since Saturn which is over 90% of it's journey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JsA2g.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JsA2g.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oumuamua's &lt;a href=&quot;https://en.wikipedia.org/wiki/%CA%BBOumuamua#Trajectory&quot; rel=&quot;nofollow noreferrer&quot;&gt;velocity at infinity&lt;/a&gt;, as it approached our solar-system was 26.33 km/s.   It speeds up as it approaches the sun and it will slow down as it leaves.   When it passed the 19 billion km &quot;border&quot;, you have to add the escape velocity from our sun at that distance, which is about 3.7 km/s, so it entered the solar-system traveling at (close enough) 31 km/s or 976 million km/year.  At that velocity it would take 19.5 years to reach the sun (granted it will miss the sun), and in theory another 19.5 years to leave the solar system.   But that estimate doesn't take into account that it speeds up as it gets close to the sun, so it takes less time.   17-18 years (ballpark) might be a good estimate.  That's 17 (or so) years traveling in, another 17 (or so) to travel back out and reach the made-up border.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike Voyager, nearly all of Oumuamua's journey is nearly straight, except for a quick curve around the sun.  That makes it's journey close to &quot;V&quot; shaped when viewed from a distance.  (I only mention that in relation to Voyager's more curvy fish-hook flight path, which was by design to pass by several planets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QMOON.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QMOON.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's probably someone here who can do the math (I'd prefer not to), and they can work out a more precise answer of when it entered the fuzzy 19 billion km border, when it will reach it's perihelion or closest to the sun, and when it will pass outside that border again.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-12-03T13:16:09.113" LastActivityDate="2017-12-03T13:16:09.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23800" PostTypeId="1" CreationDate="2017-12-03T07:49:34.133" Score="1" ViewCount="390" Body="&lt;p&gt;I am wondering how the Earth got its original angular velocity. What was the cause that made the Earth rotate about its own axis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about its motion around the Sun? What gave the Earth the push to keep rotating around the Sun?&lt;/p&gt;&#xA;" OwnerDisplayName="user19510" LastEditorUserId="88" LastEditDate="2017-12-03T09:54:31.113" LastActivityDate="2017-12-03T09:54:43.870" Title="Rotation of the Earth about its own axis" Tags="&lt;solar-system&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2017-12-07T20:51:04.787" ContentLicense="CC BY-SA 3.0" />
  <row Id="23801" PostTypeId="2" ParentId="23784" CreationDate="2017-12-03T08:39:20.933" Score="1" Body="&lt;p&gt;The metaphysical aspects of your question can't be answered. Our bodies aren't eternal, and astronomy can't answer questions about the soal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observations of the universe suggest that it will continue expanding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you note, eventually all the stars will run out of hydrogen, collapse to white dwarfs, which then slowly cool. What happens next is rather speculative. At this time there will be nowhere that is warm in the universe. Everywhere will cool to very close to absolute zero. Complex chemistry will no longer be possible, and there will be no life in the universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some stars will fall into black holes. Others will be ejected from the galaxy. If protons decay, then very very very slowly, the matter will break down, leaving only black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If black holes emit Hawking radiation, then they too will break down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this model, the spacetime of the universe is eternal, but nothing within it survives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia has a list of some of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Timeline_of_the_far_future&quot; rel=&quot;nofollow noreferrer&quot;&gt;events of the far future&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-12-03T19:53:25.337" LastActivityDate="2017-12-03T19:53:25.337" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23802" PostTypeId="2" ParentId="23793" CreationDate="2017-12-03T08:53:04.530" Score="3" Body="&lt;p&gt;I like the &lt;a href=&quot;https://astronomy.stackexchange.com/a/23798/7982&quot;&gt;&quot;science&quot; answer posted by userLTK&lt;/a&gt;. Here I'll cheat and use a projection from the current ephemeris in &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL Horizons&lt;/a&gt;. Set up like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fQuV9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fQuV9.png&quot; alt=&quot;JPL Horizons setup&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and using the script below to plot the results, I get the years 1995 and 2039 for the times that the comet passes 127 AU. That's about +/- 22 years from perihelion, &lt;strong&gt;nicely confirming userLTK's ballpark estimate&lt;/strong&gt;. Outstanding is the question of the direction of the comet relative to any possible &quot;tail&quot; of the heliopause opposite the direction of the Sun. I think there is no &lt;em&gt;hard, observational information about this&lt;/em&gt; currently, so sticking with an estimate of a radius seems acceptable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ror92.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ror92.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the current ephemeris' header:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;JPL/HORIZONS              1I/'Oumuamua (A/2017 U1)         2017-Dec-03 00:25:53&#xA;Rec #:838451 (+COV)   Soln.date: 2017-Nov-21_20:25:17      # obs: 115 (34 days)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Python script:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;&#xA;fname = 'Comet horizons_results.txt'&#xA;&#xA;with open(fname, 'r') as infile:&#xA;&#xA;    lines = infile.read().splitlines()&#xA;&#xA;    iSOE = [i for i, line in enumerate(lines) if &quot;$$SOE&quot; in line][0]&#xA;    iEOE = [i for i, line in enumerate(lines) if &quot;$$EOE&quot; in line][0]&#xA;&#xA;    lines = lines[iSOE+1:iEOE]&#xA;&#xA;    lines = [line.split(',') for line in lines]&#xA;    JD  = np.array([float(line[0]) for line in lines])&#xA;    pos = np.array([[float(item) for item in line[2:5]] for line in lines])&#xA;    vel = np.array([[float(item) for item in line[5:8]] for line in lines])&#xA;&#xA;r = np.sqrt((pos**2).sum(axis=1))&#xA;s = np.sqrt((vel**2).sum(axis=1))&#xA;&#xA;years = 2017 + (JD - 2457754.5)/365.2564  # roughly&#xA;&#xA;plt.figure()&#xA;plt.subplot(2, 1, 1)&#xA;plt.plot(years, r)&#xA;plt.plot(years, 127 + np.zeros_like(years), '-k')&#xA;plt.title(&quot;distance from solar system barycenter (AU)&quot;, fontsize=16)&#xA;plt.xlim(1980, 2060)&#xA;plt.ylim(0, None)&#xA;plt.subplot(2, 1, 2)&#xA;plt.plot(years, s)&#xA;plt.title(&quot;speed relative to solar system barycenter (AU/day)&quot;, fontsize=16)&#xA;plt.xlim(1980, 2060)&#xA;plt.ylim(0, None)&#xA;plt.suptitle(&quot;Asteroid 'Oumuamua (A/2017 U1) from JPL Horizons&quot;, fontsize=18)&#xA;plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-12-03T08:58:50.180" LastActivityDate="2017-12-03T08:58:50.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23804" PostTypeId="2" ParentId="23800" CreationDate="2017-12-03T09:54:43.870" Score="1" Body="&lt;h2&gt;Origin of Earth rotation&lt;/h2&gt;&#xA;&lt;p&gt;There are two main phenomena that explain Earth rotation. First one is linked to the history of the solar system: there is angular momentum in the interstellar medium (ISM), so the collapsing cloud that formed our solar system had an initial angular momentum (rotation of the galaxy and turbulence in the ISM are typically a good candidate for that; see for example &lt;a href=&quot;https://arxiv.org/pdf/astro-ph/0301093.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mac Low &amp;amp; Klessen 2004&lt;/a&gt;). By conservation of the latter, particles in the collapsing cloud accelerate (the norm of the angular momentum &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$mrv$&lt;/span&gt;, with &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; the mass of the system, &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; the radius and &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; the velocity; as the cloud contracts, &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; decreases and then &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; has to increase) and due to centrifugal force, a disk will tend to form. By conservation of the angular momentum in the protoplanetary disk, you will also tend to form bodies that rotates.&lt;/p&gt;&#xA;&lt;p&gt;The second phenomena is linked to the formation history of the Earth. Simply put, Earth formation history is due to collisions of planetesimals (let say they are big chunks of rock, formed by a collision history of smaller chunks of rock, formed by...); these planetesimals are orbiting the forming star in the protoplanetary disk, their rotation being linked to the disk formation history (see above). When planetesimals collide, two things can basically happen: either the collision is perfectly central (meaning that the momentum of one of the two colliding objects is pointed toward the center of mass of the second object) and there is no transfer of angular momentum, either the collision is lateral and there is a transfer of angular momentum. Depending on the configuration, you can either transfer momentum that will make the other body rotating on way or the other. That being said, you can then object that there should be a equirepartition of collisions pushing the proto-Earth to rotate in both directions. Which is true, but external planetesimals are rotating faster than internal ones. Therefore, from the proto-Earth perspective (meaning, in a physical sense, in its reference frame), internal planetesimals are seen as coming backward, from front, and external planetesimals are seen as coming forward, from behind. Long story short, there are both transfering momentum in the same &amp;quot;direction&amp;quot;, forcing the proto-Earth to rotate in the same direction as the rotation of the protoplanetary disk (see for example &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1965SvA.....9...96A&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Artem'ev &amp;amp; Radzievskii 1965&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;h2&gt;Rotation around the Sun&lt;/h2&gt;&#xA;&lt;p&gt;This one's easy: gravity.&lt;/p&gt;&#xA;&lt;p&gt;The Earth is constantly falling towards the Sun because of gravity, and the centripetal acceleration counteracts perfectly gravity, mainting the Earth on its orbit.&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-12-03T09:54:43.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23806" PostTypeId="1" CreationDate="2017-12-03T13:27:49.123" Score="-1" ViewCount="363" Body="&lt;p&gt;When a meteorite shower strikes a water body, what will be the pH of the water after the strike.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will it always be $&amp;lt;7$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will it always be $&amp;gt;7$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will it always be $=7$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer varies from case to case&lt;/p&gt;&#xA;" OwnerUserId="15901" LastActivityDate="2017-12-03T20:21:20.913" Title="pH of meteorite" Tags="&lt;meteor&gt;&lt;meteor-shower&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ClosedDate="2017-12-07T20:50:47.713" ContentLicense="CC BY-SA 3.0" />
  <row Id="23808" PostTypeId="1" CreationDate="2017-12-03T17:59:02.293" Score="0" ViewCount="609" Body="&lt;p&gt;Why are the orbits of the planets elliptical ? The circular orbit also satisfies all the conditions of gravitational force. Then why did nature choose the elliptical orbits ?&lt;/p&gt;&#xA;" OwnerDisplayName="user19510" LastEditorUserId="10196" LastEditDate="2017-12-03T18:22:21.663" LastActivityDate="2017-12-03T21:43:46.460" Title="Why are planetary orbits not all circular?" Tags="&lt;solar-system&gt;" AnswerCount="3" CommentCount="3" ClosedDate="2017-12-06T18:08:45.100" ContentLicense="CC BY-SA 3.0" />
  <row Id="23809" PostTypeId="2" ParentId="23808" CreationDate="2017-12-03T18:32:39.717" Score="0" Body="&lt;p&gt;While a circular orbit is possible and consistent with equations of gravitation, elliptical orbits are almost a necessity due to the surroundings of any orbit. If you take our solar system as an example, the center of mass will change as not all planets orbit at the same speed. The center of mass is found with the equation:&#xA;$$x_c=\frac{\sum\limits_{i=1}^{n}{m_{i}x_i}}{\sum\limits_{i=1}^{n}{m_{i}}}$$&#xA;As the planets mass distribution around the sun changes, the point which the planets orbit changes slightly, disrupting any circular orbits. All orbiting bodies will have some disruption, preventing orbits to always be perfectly circular. In addition, losing energy and angular momentum to other sources can cause bodies to lose their orbital shapes and degenerate into elliptical orbits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, please use this &lt;a href=&quot;https://physics.stackexchange.com/questions/69997/why-dont-planets-have-circular-orbits&quot;&gt;link&lt;/a&gt; also posted as a comment.&lt;/p&gt;&#xA;" OwnerUserId="18327" LastActivityDate="2017-12-03T18:32:39.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23810" PostTypeId="2" ParentId="23808" CreationDate="2017-12-03T20:09:22.587" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Planets don't have elliptical or circular orbits. The elliptical orbit is the solution to the two body Newtonian gravity problem. As their are more than two particles in the solar system, the orbits aren't ellipses or circles.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A circle is an ellipse with eccentricity = 0. It's special in some ways, but for orbits it's just another elliptical orbit. The universe doesn't care what the eccentricity of the planets orbits is (providing it's not too big) There is no special reason why the eccentricity should be exactly 0. The eccentricity is essentially a random value. If you choose a number at random between 0 and 0.1 the probability of choosing exactly zero is 0.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The real question is &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1979/why-are-most-planetary-orbits-nearly-circular?rq=1&quot;&gt;Why are most planetary orbits nearly circular&lt;/a&gt; Since orbits of any eccentricty are possible it is (initially) surprising that all orbits are so close to circluar.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-12-03T20:09:22.587" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23811" PostTypeId="2" ParentId="23806" CreationDate="2017-12-03T20:21:20.913" Score="2" Body="&lt;p&gt;Meteorites are rocks, and rocks are normally quite insoluble. When an insoluble body is mixed with water the pH of the water isn't changed.#&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meteorites contain quite a lot of various metallic oxides, which are bases. This includes magnesium oxide which can dissolve a little, and is Alkali. The energy released by a large impact could also cause some nitrogen oxides to form from the atmosphere (which are acidic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However these are small effects. The pH of the water is little changed. The seas are big, and any effect is soon diluted.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-12-03T20:21:20.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23812" PostTypeId="2" ParentId="23808" CreationDate="2017-12-03T21:43:46.460" Score="3" Body="&lt;p&gt;For a given amount of angular momentum, a circular orbit is the orbit with lowest energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some orbits are elliptical (actually, all orbits approximate to ellipses) with a non-zero eccentricity, because they have yet to reach their lowest energy state, or because they have been excited from their minimum energy state circular orbit.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-12-03T21:43:46.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23813" PostTypeId="2" ParentId="12384" CreationDate="2017-12-04T01:43:55.533" Score="-1" Body="&lt;p&gt;Galaxies are oriented in different directions, if we consider the orientation to be relative to their central axis of rotation. Then when we consider things like galaxies and their &quot;direction&quot; of spin, we tend to orient them so that &quot;up&quot; is the direction that, when we look at them from &quot;above&quot;, they rotate in an anticlockwise direction - the same as observing Earth's rotation looking down from the North Pole. We tend to use a &quot;right-hand rule&quot; for this rotation, so that when the direction of rotation matches the curl of the fingers on the right hand, the thumb points in the &quot;up&quot; direction. What this means is that every galaxy rotates the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the smaller scale what we see is that things in systems tend to rotate in the same direction - looking &quot;down from above&quot; on our Solar System we see that the planets all revolve in the same direction around the Sun (anticlockwise). The vast majority of smaller bodies also revolve in this same direction. The Sun rotates in this same direction. And the planets themselves tend to rotate in the same direction (exceptions being Venus and Uranus). This is understandable as the accretion disc would have been rotating during Solar System formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, for the most part at least, as far as we have been able to determine, everything spins the same way within its own environment.&lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2017-12-04T01:43:55.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23814" PostTypeId="2" ParentId="21172" CreationDate="2017-12-04T03:43:57.680" Score="2" Body="&lt;p&gt;The practical length of the time exposure will depend on your equipment. Cameras used in astrophotography are generally able to be used with exposure times ranging from milliseconds to hours. As your question stands, cooling the chip will have no effect on the length of the exposure time. What it can affect is the amount of noise in the image. Cooling the sensor generally reduces the amount of noise. Cooling beyond -20°C may have limited value, particularly if you are not using research-grade instruments. While you may further reduce the noise, you may still be at the mercy of the quality of the optics, drive mechanisms, etc. and you may have less noise in the image but the limiting factor is not your image noise but the system limit for drift and distortion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some professional instruments use liquid nitrogen cooling while others use thermoelectric cooling, depending on the manufacturer and the intended use. The instruments that use sub-zero cooling also usually have a high quality sensor with a pixel well depth that supports longer exposures, without bleeding effects from adjacent pixels. It all depends on the sensor, camera, and other parts of the system, so your mileage may vary.&lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2017-12-04T03:43:57.680" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23821" PostTypeId="2" ParentId="23777" CreationDate="2017-12-04T10:13:40.173" Score="2" Body="&lt;p&gt;That is an intersting question as this is far from being obvious. Actually, for quite some time it was believed that planets should indeed spin retrograde (we can at least track it back to Laplace in his &lt;em&gt;Exposition du Système du Monde&lt;/em&gt; published in 1976 where he developed his nebular hypothesis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to prove that planet should spin prograde? We'll have to go back to their formation history. Let's start with an ideal case: particles moving in eliptical orbits around a central object of mass $M$, with a proto-planet of mass $m_0$, and rotating in a eliptical orbit. Let's look at what happens at a given point at a distance $r$ from the central object. First, for a planetesimal in a tangential external orbit, then for a planetesimal in a tangential internal orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tangential external orbit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can write Newton's Second Law ${\bf F} = m {\bf a}$ and project it on the current trajectory axis (normal and tangent), knowing that the acceleration is always something like $v^2/R$, with $R$ the curvature radius (that is different from $r$, the distance to the central object). We get for the normal component:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{GM}{r^2} \cos \alpha = \frac{v_{ext}^2}{R_{ext}^2},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_{ext}$ being the curvature radius of the planetesimal in an external orbit. For the proto-planet:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{GM}{r^2} \cos \alpha = \frac{v^2}{R^2},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha$ having the same value for both object, since it denotes the angle between the direction toward the curvature center and the central object and the orbits being tangential at this specific point, where $r$ the distance to the central object is also the same. Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{v^2}{R^2} = \frac{v_{ext}^2}{R_{ext}^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $R &amp;lt; R_{ext}$, we have $v &amp;lt; v_{ext}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tangetial internal orbit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the same argument, we can show that $v &amp;gt; v_{int}$ (I let it as an exercice; extra bonus point if you do it with the energy conservation law).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Consequences&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you go in the reference frame of the protoplanet, you will see the internal planetesimal, rotating at a lower speed, coming &lt;em&gt;backward&lt;/em&gt;, towards the proto-planet, and the external planetesimal, rotating at a faster speed, coming &lt;em&gt;forward&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/opetb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/opetb.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $v'_{ext}$ and $v'_{int}$ are the velocity of the external and internal planetesimal in the reference fram of the proto-planet. That is why the rotation of the proto-planet tends to be in the same direction as the whole system.&lt;/p&gt;&#xA;" OwnerUserId="88" LastActivityDate="2017-12-04T10:13:40.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23822" PostTypeId="1" CreationDate="2017-12-04T10:38:29.687" Score="4" ViewCount="118" Body="&lt;p&gt;I've read through all the other related answers to &quot;flatness&quot; questions but I need a bit more clarification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that a triangle in a 2D universe would not equal 180 degrees in both a closed and open universe. If I am correct, not being flat during the initial conditions would mean exponential expansion or collapse as time went by.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the above, could we explain flatness as uniform density of the universe, meaning all of the universe is the same approximate &quot;consistency&quot; on a large scale (homogeneity).&lt;/p&gt;&#xA;" OwnerUserId="19040" LastActivityDate="2017-12-06T13:03:28.903" Title="Does Flat Universe mean even distribution?" Tags="&lt;expansion&gt;&lt;cosmological-inflation&gt;&lt;density&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23823" PostTypeId="2" ParentId="23784" CreationDate="2017-12-04T12:15:43.190" Score="0" Body="&lt;p&gt;The universe is eternal...in a sense. The universe is rapidly expanding, and with that, the amount of space in between everything. The point when astrophysicists consider the universe to have &quot;ended&quot; is when all stars have died and there components are too far apart to create new ones. When the planets of solar systems are too far from their stars to see their light. When our universe is ruled by black holes who swallow up what is left of our once beautiful universe. When (theoretically) the redshift effect would occur on such a vast scale that we would no longer be able to see the light from the stars in the visible universe( by that point we will all have been dead for billions of years). To the point where the universe can no longer harbor life. The universe will still exist in a technical sense, but it will be a cold and unrecognizable universe from the one we call home today.&lt;/p&gt;&#xA;" OwnerUserId="19943" LastActivityDate="2017-12-04T12:15:43.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23826" PostTypeId="2" ParentId="23822" CreationDate="2017-12-04T21:11:56.703" Score="4" Body="&lt;p&gt;No, being flat and being homogeneous is not equivalent$^\dagger\!\!\!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flatness refers to the geometry, which depends on the total energy density $\rho$; if it is above or below a certain critical threshold $\rho_\mathrm{cr}$, we call the Universe &quot;closed&quot; or &quot;open&quot;, respectively, while if $\rho$ is exactly equal to $\rho_\mathrm{cr}$, we call it &quot;flat&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Homogeneity refers, as you say, to matter being evenly distributed (on large scales). If the Universe is not only homogeneous, but also isotropic (i.e. looks the same in all directions), &lt;em&gt;then&lt;/em&gt; there are three different possible geometries, namely flat, open, or closed. But a homogeneous universe doesn't have to be isotropic, and both open and closed universes can have evenly distributed matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, &#xA;$$&#xA;\mathrm{homogeneity} \nRightarrow \mathrm{flatness}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas the local geometry depends on the local density$^\ddagger$, the &lt;em&gt;global&lt;/em&gt; geometry does not depend on how the matter is distributed. For instance, you could in principle have a universe with no upper limit of structure size. In &lt;em&gt;our&lt;/em&gt; Universe, we find observationally that the largest structures have sizes of roughly half a gigalightyear. Above this scale, it is homogeneous (although it could be inhomogeneous on scales larger than the particle horizon). But another universe might have structure on all scales and thus not be homogeneous, but still meet the criterion $\rho = \rho_\mathrm{cr}$ and thus be flat.&#xA;You could also, as discussed in &lt;a href=&quot;https://astronomy.stackexchange.com/a/13193/5264&quot;&gt;this answer about homogeneity and isotropy&lt;/a&gt;, imagine a universe originating at a central point away from which the density always decreases (i.e. is isotropic around this point), but which still has $\rho = \rho_\mathrm{cr}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus,&#xA;$$&#xA;\mathrm{flatness} \nRightarrow \mathrm{homogeneity}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that while your statement about the geometry determining the fate of the Universe would be correct if it contained only matter and radiation, it seems that $\sim70$% of the density making it flat has the annoying propety of accelerating the expansion of the Universe. Thus, whereas a flat, matter-dominated universe would expand asymptotically toward a finite size, &lt;em&gt;our&lt;/em&gt; Universe is dominated, it seems, by something dubbed &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;dark energy&lt;/a&gt; making it expand exponentially despite being flat.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;For a universe, at least. For a roadkill, it might be equivalent.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^\ddagger$&lt;sub&gt;For instance, space &quot;bends&quot; enough around a massive cluster of galaxies to make &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_lens&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravitational lenses&lt;/a&gt;.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2017-12-06T10:45:56.623" LastActivityDate="2017-12-06T10:45:56.623" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23827" PostTypeId="1" CreationDate="2017-12-04T22:00:18.227" Score="2" ViewCount="206" Body="&lt;p&gt;I'm having trouble understanding the basic model for how matter bends space. The classic illustration is the canvas and the ball but that model needs a reference frame where one can define a downward direction to work. For the life of me I can't figure out why matter just doesn't &quot;sit&quot; in space.&#xA;I'm sorry this question isn't better formulated, maybe someone can point me towards a video or something?&lt;/p&gt;&#xA;" OwnerUserId="20079" LastEditorUserId="17425" LastEditDate="2017-12-05T08:49:49.693" LastActivityDate="2017-12-05T17:47:06.397" Title="Canvas and bowling ball" Tags="&lt;space-time&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23828" PostTypeId="1" AcceptedAnswerId="23829" CreationDate="2017-12-04T23:18:59.390" Score="2" ViewCount="377" Body="&lt;p&gt;I overheard a discussion between friends:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If your spaceship was transported to a random location in our &lt;a href=&quot;https://en.wikipedia.org/wiki/Milky_Way&quot; rel=&quot;nofollow noreferrer&quot;&gt;galaxy&lt;/a&gt;, you could identify nearby stars by their signature, and then &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_triangulation&quot; rel=&quot;nofollow noreferrer&quot;&gt;triangulate&lt;/a&gt; your position. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another friend replied. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The signature of a star isn't sufficiently unique. You can't identify a star by &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_spectroscopy&quot; rel=&quot;nofollow noreferrer&quot;&gt;spectroscopy&lt;/a&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A third friend chimed in&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There would be enough unique &lt;a href=&quot;https://en.wikipedia.org/wiki/Pulsar&quot; rel=&quot;nofollow noreferrer&quot;&gt;pulsars&lt;/a&gt; to work out your position from. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I came away not convinced on the stellar spectroscopy question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;strong&gt;Can you identify a star's 'signature' from its spectroscopy?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2894" LastActivityDate="2018-02-27T08:22:06.933" Title="Can you identify a star's 'signature' from its spectroscopy?" Tags="&lt;star&gt;&lt;spectra&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23829" PostTypeId="2" ParentId="23828" CreationDate="2017-12-04T23:58:05.623" Score="5" Body="&lt;p&gt;Realistically, no - there are too many stars with the same spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pulsar idea is a good one, though. There is a reason our location relative to some bright pulsars was sent out on the Voyager plaques: we are confident that a civilisation finding one could then find our sun, and us, as the pulsar rates are unique enough.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2017-12-04T23:58:05.623" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23830" PostTypeId="2" ParentId="23827" CreationDate="2017-12-05T00:55:28.317" Score="0" Body="&lt;p&gt;I don't think this should be voted down.  It's a perfectly fine question to ask about trying to understand a concept. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one sense, the two dimensional surface stretching downward into a 3rd dimension can be entirely replicated, but impossible to draw with 3 dimensions stretching down into a 4th dimension.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can only see 3 dimensions so visualizing a 4th dimension is very difficult for most of us, but mathematically it's relatively simple to do.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of 3 coordinates (X,Y and Z), (a 3d &lt;a href=&quot;https://en.wikipedia.org/wiki/Cartesian_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cartesian coordinate system&lt;/a&gt;), you have 4 coordinates, (X,Y,Z and W).   Where a 3 dimensional &quot;rubber sheet&quot;, X,Y,Z stretches it into W space, similar to how the 2d rubber sheet X,Y stretches into into Z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9qdZB.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9qdZB.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example of 2 d space in 3 dimensions is the surface of a sphere.   The surface of the Earth can be measured by 2 coordinates, Longitude and Latitude, but it obviously stretches around 3 dimensions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Where does space stretch into&quot;, well, the rubber sheet analogy isn't perfect.  It's more accurate (I think) to think of the depressions in the rubber sheet as gravitational wells, where objects speed up as they approach and slow down as they fly away, they also change direction and in the case of light, which can't speed up or slow down, it blue shifts going in, red shifts coming out, and it's affected by time dilation.   Changing properties of space and gravitational force towards are perhaps more accurate ways to think about it than stretched space (in my non-expert opinion).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But does space stretch at all?  Yes, it does.  The radius becomes longer than it should around massive objects.   &lt;a href=&quot;http://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/general_relativity_massive/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article explains the effect with our sun&lt;/a&gt;.   It's small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Euclidean result tells us that for each mile we come closer to the&#xA;  sun, the circumference of the circle is diminished by 2π miles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dTE2p.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dTE2p.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because of the presence of the sun, space around the sun is not&#xA;  exactly Euclidean. According to general relativity, for each mile that&#xA;  we come closer to the sun, the circle does not lose 2π miles in&#xA;  circumference; it loses only (0.99999999)x2π miles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If someone here wants to explain this in more physical terms in their own words, please feel free.  I'm a little bit over my head, but the article above explains some of the misleading aspects of the rubber sheet analogy. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-12-05T06:04:44.863" LastActivityDate="2017-12-05T06:04:44.863" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23832" PostTypeId="2" ParentId="23828" CreationDate="2017-12-05T07:18:07.840" Score="3" Body="&lt;p&gt;The answer is no, but not only for the reason you suppose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each star probably does have a different chemical composition and hence a different spectrum. However, many stars have a very similar chemical composition, and those stars born in a cluster an almost identical composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore stellar spectra whilst having the potential to be a unique identifier,  are not currently measured with sufficient precision to make this possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, another reason why what you suggest won't work is that by and large, we can only obtain spectra for stars inside a modest radius from the Sun, especially if the stars are near the Galactic plane. One could work in the infrared, but the spectra of bright stars are generally less distinctive in the infrared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could go with a database of positions, absolute magnitudes &lt;em&gt;and&lt;/em&gt; spectral types and probably work out where you were if it were reasonably close to the Sun. Note this only works if you are transported quickly (in terms of stellar evolutionary lifetimes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better bet is to look at the positions of local group galaxies (e.g. the magellanic clouds, Andromeda), which will be in noticeably different positions. Also, globular clusters can play a role, because many of these do have distinctive chemical abundances and multiple populations and are distributed around the Galaxy and out of the Galactic plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pulsars idea doesn't work at all because: (a) The radiation from pulsars is beamed into a small angle, so they would not be visible in the same way from another Galactic location. (b) By the time you had got to another Galactic location (unless faster than light travel is postulated) some pulsars will have turned off, all will have significantly changed their pulses and periods, since the total lifetime of the pulsar phenomenon is of order a million years.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-02-27T08:22:06.933" LastActivityDate="2018-02-27T08:22:06.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23837" PostTypeId="1" AcceptedAnswerId="23843" CreationDate="2017-12-05T14:23:25.197" Score="4" ViewCount="864" Body="&lt;p&gt;Some new large telescopes in the near future, on Earth and in space, will use multiple mirrors. The individual mirror pieces will be hexagonal in shape. I wonder why this shape is prefered over the circle sector shape (like pie slices). I would think that corners and joints between mirrors cause practical problems. And a circle sector slice has half as many corners and two rather than three or six neighboring pieces, and shorter interfacing edges with neighboring pieces overall. So shy is the hexagonal shape prefered?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="1030" LastEditDate="2017-12-05T14:28:04.107" LastActivityDate="2017-12-05T22:28:59.857" Title="Why do telescopes use hexagonal mirror pieces instead of pie slice shaped ones?" Tags="&lt;observatory&gt;&lt;mirror&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23838" PostTypeId="2" ParentId="23751" CreationDate="2017-12-05T15:07:40.340" Score="0" Body="&lt;p&gt;You are correct in your definition of heat, in that it does need a medium. The &quot;heat&quot; in space, however, is radiant. It does need a medium in order to be called heat, but the energy is there. Scientists measure the radiant energy in space to see how hot it would be if it were in a medium. The space is not physically hot, but the energy there means that if you were to travel there, the energy would he transferred into you, and you would become hot.&lt;/p&gt;&#xA;" OwnerUserId="19943" LastActivityDate="2017-12-05T15:07:40.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23839" PostTypeId="2" ParentId="23827" CreationDate="2017-12-05T17:34:15.137" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm having trouble understanding the basic model for how matter bends space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No wonder, because it doesn't. It curves spacetime. But spacetime isn't the same thing as space. See &lt;a href=&quot;http://math.ucr.edu/home/baez/einstein/node2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Baez&lt;/a&gt; article: &lt;em&gt;&quot;Similarly, in general relativity gravity is not really a 'force', but just a manifestation of the curvature of spacetime. Note: not the curvature of space, but of spacetime. The distinction is crucial&quot;.&lt;/em&gt; Space isn't curved where a gravitational field is. Instead spacetime is. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The classic illustration is the canvas and the ball but that model needs a reference frame where one can define a downward direction to work.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's right, it uses gravity to &quot;explain&quot; gravity, so it isn't ideal. However there is a way to understand it, and IMHO it's simpler than you think. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For the life of me I can't figure out why matter just doesn't &quot;sit&quot; in space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It does just sit in space. And it causes gravity because it's a concentration of energy that &quot;conditions&quot; the surrounding space. It alters it, and this alteration diminishes with distance. Remember the stress-energy tensor, remember that stress is directional pressure, and think of matter as exerting an outward pressure in all directions in the surrounding space. Something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PfiWD.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PfiWD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;sub&gt;Earth image courtesy of NASA, lines drawn by me&lt;/sub&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then see the &lt;a href=&quot;http://einsteinpapers.press.princeton.edu/vol7-trans/192?highlightText=%22neither%20homogeneous%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein digital papers&lt;/a&gt;, where Einstein said a gravitational field is a place where &lt;em&gt;&quot;space is neither homogeneous nor isotropic&quot;&lt;/em&gt;. This inhomogeneity isn't linear. You can plot it, using say optical clocks dotted throughout an equatorial slice through the Earth and the surrounding space. You plot lower slower clocks as lower down in a 3D image, and higher faster clock rates higher up. When you've finished your plot looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eSrP8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eSrP8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;sub&gt;CCASA image by Johnstone, see Wikipedia&lt;/sub&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The upper image is a bit like looking at the lower image from underneath. But anyway, your plot is curved, and you made it using clocks in space, so we can say spacetime is curved. Note that whilst we started with a two-dimensional equatorial slice and ended up with a three-dimensional curved plot, there is no extra dimension in any real sense. All we're depicting is the &quot;coordinate&quot; speed of light. This might sound unfamiliar but again see the &lt;a href=&quot;http://einsteinpapers.press.princeton.edu/vol7-trans/156?&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein digital papers&lt;/a&gt;: &lt;em&gt;&quot;the curvature of light rays occurs only in spaces where the speed of light is spatially variable&quot;&lt;/em&gt;. Light doesn't curve because your plot of optical clock rates is curved. It doesn't curve because spacetime is curved. Einstein never actually said that. He said it curves because the speed of light varies. Hence the tilted light-cones on John Norton's course. Light curves wherever your plot isn't flat. Where the speed of light varies. It's actually rather like &lt;a href=&quot;https://fas.org/man/dod-101/navy/docs/es310/SNR_PROP/snr_prop.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;sonar&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/C69UT.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/C69UT.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Image courtesy of FAS and the US Navy, see course &lt;a href=&quot;https://fas.org/man/dod-101/navy/docs/es310/SNR_PROP/snr_prop.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;ES310 chapter 20&lt;/a&gt;&lt;/sub&gt;  &lt;/p&gt;&#xA;" OwnerUserId="9772" LastEditorUserId="9772" LastEditDate="2017-12-05T17:47:06.397" LastActivityDate="2017-12-05T17:47:06.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23840" PostTypeId="2" ParentId="23753" CreationDate="2017-12-05T17:53:20.517" Score="1" Body="&lt;p&gt;Question 1: Shouldn't the Sun be reaching its highest point on the longest day, i.e. on June 21st, instead of April 24th?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: No. The Sun (or any celestial object) is highest when its declination equals your latitude. That is when the object is at the zenith. If your latitude is between 23.4 S and 23.4 N, the Sun's declination equals your latitude on some day other than the longest day. (The Sun's declination varies between -23.4 and +23.4, approximately.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you live outside of that range of those latitudes, then the Sun is closest to the zenith on the longest day of the year, and therefore is highest in the sky on the longest day of the year.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-12-05T17:53:20.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23841" PostTypeId="2" ParentId="23837" CreationDate="2017-12-05T18:12:54.837" Score="0" Body="&lt;p&gt;For one thing,  hexagons tile the plane -- meaning you can build as large as you want with one shape.  Granted in practice this won't work because there are different curvatures, though again all hexagons at a given radius will have the same curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, the likelihood of damage at the tip of a &quot;pie-slice&quot; is great, the aspect ratio (length-to-width) sucks, and remember that minimizing the longest dimension is critical to controlling cost and fab time for any mirror section.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add most of the info in the initial comments as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, if you had pie slices it'd be just about impossible to 'refocus' the system, while local hexagons are relatively easy to deal with. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2017-12-05T18:12:54.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23842" PostTypeId="2" ParentId="23753" CreationDate="2017-12-05T18:18:45.373" Score="3" Body="&lt;p&gt;Question 2: Doesn't the Sun take the longest arc across the sky on the day it passes through zenith?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: There are two effects. (In the remaining description, I will be referring to an observer in the northern latitude.) As the Sun moves northward from 0 degrees declination, the radius of the circle that it travels is smaller. In other words, the total length of the 20 degree declination circle is smaller than the 10 degree declination circle which is smaller than the 0 degree declination circle (the celestial equator). But at the same time, the fraction of the declination circle that is visible above the horizon is greater. At some combination of declination, these two effects result in the maximum length of the declination circle above the horizon. If the declination is north or south of that optimized solution, the length of the arc is smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Skipping the calculations for now and going right to the solution, here is a graph showing the length of the arc for various latitudes and declinations. This graph plots the results relative to the length of the arc along the celestial equator which is 180 degrees regardless of your latitude (ignoring refraction and the Sun's diameter).&#xA;&lt;a href=&quot;https://i.stack.imgur.com/C1BpN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/C1BpN.png&quot; alt=&quot;Length of arc as Sun crosses the sky&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your latitude is between 0 and 23.4 degrees, the longest arc is before the Sun reaches the zenith. For example, an observer at 20 degrees latitude witnesses the longest arc when the Sun is at about 13 degrees declination which is 7 degrees before the Sun reaches the zenith. If you live north of 23.4 degrees, the Sun is never overhead. North of 30 degrees, the longest arc is when the Sun is highest in the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can go into more details and the math, but I do not know if I should do it here or maybe in a new question.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-12-05T18:18:45.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23843" PostTypeId="2" ParentId="23837" CreationDate="2017-12-05T19:37:15.977" Score="8" Body="&lt;p&gt;The ideal shape for the mirror is round. It's the easiest to make. It's the best-behaved while in use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hex tiles are already harder. The mirror is a revolution surface generated by a conic curve (circle, parabola, hyperbola, ellipse), which needs to be machined with a precision greater than 0.1 microns. That's &lt;strong&gt;extremely difficult&lt;/strong&gt; already with a round mirror, where the surface can be easily distorted near the edge in the process of manufacturing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the hex tiles, maintaining that precision near the corners is very hard - corners are like the edge but exponentially harder. Unless you take special precautions and unless you perform extra work, you're guaranteed to end up with surface errors near the corners (the surface is either too raised, or too low, or has a more complex distortion). Based on my experience making round mirrors, I would guess the surface near the corners would tend to be turned down (too low relative to the rest), which is a defect that's extremely hard to correct - one of the hardest, in fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a pie slice, all problems mentioned above increase greatly in magnitude. The blunt angle of the hex tile gives you some support in machining the surface, but the sharp angle of the pie slice gives you almost none. I can't even begin to imagine the difficulties one would encounter while making a high precision surface on a pie slice. The sharp end of the slice would be almost guaranteed to be worse quality than the rest.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What most people don't realize is that maintaining the precise shape of the mirror while in use is difficult too. You can't just lay it on a hard surface - the uneven support would distort the precise shape. Mirrors are supported on the back by a precise mechanism, with a number of support elements precisely sharing the weight in equal amounts (or, if not equal, then deduced from finite element analysis), and located in places carefully calculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a round or hex tile this is, if not easy, then at least a well-understood and manageable procedure. A pie slice would be very difficult to support in the same even manner near the sharp tip. You can't rely on the symmetry of the shape to provide the even load on the support points. The support force would have to be specially adjusted there. It would probably be doable, but harder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is only an issue with terrestrial telescopes, which operate under gravity. Space telescopes don't suffer from this issue.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Alternatives to hex tiles could also be square tiles, or triangle tiles. These are worse than hex tiles for all the reasons shown above - the corners have sharper angles which make everything more difficult. The blunt angles of the hex tiles alleviate these issues somewhat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TLDR:&lt;/strong&gt; Non-round mirrors are hard to make anyway, but hex tiles are less hard compared to the alternatives.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2017-12-05T22:28:59.857" LastActivityDate="2017-12-05T22:28:59.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23846" PostTypeId="1" AcceptedAnswerId="23848" CreationDate="2017-12-06T01:44:23.780" Score="0" ViewCount="95" Body="&lt;p&gt;According to 2nd law of thermodynamic,  in an isolated system entropy increases over tume. In an open system entropy might seems decreases but combined result of all related events results in overall entropy increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets say, a suffled deck of playing cards. If I sort the cards it seems entropy decreases, what's then the related  events that increases overall entropy. I am asking just to clear some of my confusions.&lt;/p&gt;&#xA;" OwnerUserId="15770" LastActivityDate="2017-12-06T04:46:06.743" Title="How entropy increases over time" Tags="&lt;space-time&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2017-12-06T08:56:25.100" ContentLicense="CC BY-SA 3.0" />
  <row Id="23848" PostTypeId="2" ParentId="23846" CreationDate="2017-12-06T04:46:06.743" Score="1" Body="&lt;p&gt;While more of a physics than astronomy question. Entropy as a whole increase in the universe due to statistical probability. By rearranging a deck of cards looking at the back of them, no matter what order, it is the same macrostate with different arrangements of the cards (microstates). A better example would be if you decreased entropy by picking up cards from the floor and put them together. In that situation, while in your small area the cards lost entropy, you expended energy to do that task. The energy you expend is enough to make the entropy of you and the cards remain constant and arguably increase. However, if you then take into account the entire universe, the rest of it is increasing in entropy at a large rate anyway. This means no matter what you do, entropy will increase overall due to other effects. &lt;/p&gt;&#xA;" OwnerUserId="18327" LastActivityDate="2017-12-06T04:46:06.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23855" PostTypeId="2" ParentId="23822" CreationDate="2017-12-06T13:03:28.903" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does a flat universe mean even distribution?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Strictly speaking no. But there's no actual evidence for any kind of &quot;higher dimensional curvature&quot;. See &lt;a href=&quot;https://arxiv.org/abs/1303.5086&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/1303.5086&lt;/a&gt; where the Planck mission reported on the Background geometry and topology of the Universe. They found no evidence for any kind of &quot;asteroids&quot; toroidal universe.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I understand that a triangle in a 2D universe would not equal 180 degrees in both a closed and open universe. If I am correct, not being flat during the initial conditions would mean exponential expansion or collapse as time went by.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's what people tend to say, but IMHO you shouldn't treat it as settled science. See &lt;a href=&quot;https://arxiv.org/abs/1302.1498&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;The Waters I am Entering No One yet Has Crossed&quot;: Alexander Friedman and the Origins of Modern Cosmology&lt;/a&gt;. It’s by Ari Belenkiy, who credits Friedmann with the expanding universe. He also refers to Friedmann's m-for-monotonic M1 universe which expands from a singularity and has an inflexion point, his M2 universe which expands from a non-zero radius, and his p-for-periodic P universe which expands and contracts. The thing is, that we don't actually know for sure that any of these options properly describe the universe. Particularly since we have good evidence that the universe was once very dense, but nevertheless expanded. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given the above, could we explain flatness as uniform density of the universe, meaning all of the universe is the same approximate &quot;consistency&quot; on a large scale (homogeneity).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would say this: given that we don't know that Friedmann cosmology is correct, we &lt;em&gt;might&lt;/em&gt; explain flatness as uniform density of the universe.&lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2017-12-06T13:03:28.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23856" PostTypeId="1" CreationDate="2017-12-06T13:14:31.393" Score="1" ViewCount="43" Body="&lt;p&gt;How is accurate spectral analysis of a planets atmosphere achieved, bearing in mind that its host star's resultant light emission is not a full spectrum?&lt;/p&gt;&#xA;" OwnerUserId="20155" LastEditorUserId="17425" LastEditDate="2017-12-06T18:05:16.963" LastActivityDate="2017-12-06T18:05:16.963" Title="Spectroscopy of Exoplanets" Tags="&lt;spectroscopy&gt;" AnswerCount="0" CommentCount="5" FavoriteCount="1" ClosedDate="2017-12-07T20:47:32.410" ContentLicense="CC BY-SA 3.0" />
  <row Id="23858" PostTypeId="1" AcceptedAnswerId="23860" CreationDate="2017-12-06T19:59:25.850" Score="2" ViewCount="148" Body="&lt;p&gt;It is said that inflation justifies that magnetic monopoles don't exist. Can anyone explain how inflation theory explains the non existence of magnetic monopoles?&lt;/p&gt;&#xA;" OwnerUserId="20162" LastEditorUserId="2153" LastEditDate="2017-12-06T20:53:50.780" LastActivityDate="2017-12-06T20:53:50.780" Title="How does inflation justify the nonexistence of magnetic monopoles?" Tags="&lt;cosmology&gt;&lt;big-bang-theory&gt;&lt;cosmological-inflation&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23860" PostTypeId="2" ParentId="23858" CreationDate="2017-12-06T20:51:59.170" Score="4" Body="&lt;p&gt;Let me start off by stating that scientists have not conclusively proven that magnetic monopoles do or do not exist. There has never had a confirmed detection of one so the jury is still out on if a magnetic monopole can or does exist at all. That being said, if monopoles do exist, no one has a good explanation for why we haven't detected one yet. There are hypotheses, but no solid proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get to your question now, I think it is an incorrect statement to say that inflation theory explains the non-existence of magnetic monopoles. What inflation theory does &lt;em&gt;potentially&lt;/em&gt; explain, is the extremely low density of magnetic monopoles in the universe. The idea here is that magnetic monopoles, for some unknown reason, are &lt;em&gt;extremely&lt;/em&gt; unlikely to form, even in the intensely energetic and hot crucible of the Big Bang. Some estimates say that for 1 magnetic monopole that formed, $10^{29}$ &quot;normal&quot; particles formed, making magnetic monopoles extremely sparse and rare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where inflation theory comes in is that it says the universe expanded exponentially fast. If the monopoles were already sparse before this extreme expansion, the expansion would have only served to cause the magnetic monopole density in the universe be practically zero. It is quite possible that, post-inflation, the magnetic monopole density in the universe is so low that only one exists within the observable universe. Good luck finding that single particle out of the $10^{80}$ particle haystack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my mind though, that doesn't really answer the question as to why we don't see magnetic monopoles. Inflation just makes them extremely sparse, but why did so few form to begin with (assuming they formed at all). As I said above, nobody knows. A potential answer (without any evidence) is that magnetic monopoles are extremely energetic particles (e.g., with energies greater than the LHC of 14 TeV). If they have such extreme energies, even the Big Bang wouldn't have formed many of them. But again, that's just a guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only time will tell what the real answer is.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-12-06T20:51:59.170" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23868" PostTypeId="2" ParentId="23773" CreationDate="2017-12-07T14:41:42.333" Score="2" Body="&lt;p&gt;This was intended to be a supplemental answer to &lt;a href=&quot;https://astronomy.stackexchange.com/a/23790/7982&quot;&gt;StephenG's answer&lt;/a&gt;. However, there appears to be a problem with the expression for time-averaged distance in that answer. I think it's great to seek a mathematical expression, but it should be confirmed numerically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did a quick numerical double check and verified those general trends, but there may still be a problem with one of the expressions there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a constant semi major axis of 1, the time-averaged distance rises from 1 at $\epsilon = 0$ to 1.5 at $\epsilon \rightarrow 1$, while for the $\theta$-averaged (&quot;geometrical&quot;) distance drops from 1 down to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think both of us should now add the &lt;em&gt;path-average&lt;/em&gt; for completeness by averaging over &lt;em&gt;ds&lt;/em&gt;. :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vJt0F.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vJt0F.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Python script:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def deriv(X, t):&#xA;    x, v = X.reshape(2, -1)&#xA;    acc = -x * ((x**2).sum())**-1.5&#xA;    return np.hstack((v, acc))&#xA;&#xA;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;from scipy.integrate import odeint as ODEint&#xA;&#xA;halfpi, pi, twopi = [f*np.pi for f in 0.5, 1, 2]&#xA;&#xA;time = np.linspace(0, twopi, 10001)[:-1]&#xA;&#xA;a = 1.0&#xA;eps = np.hstack((0, 0.2, 0.5, 0.7, 0.9, 0.99))&#xA;&#xA;orbits = []&#xA;for ep in eps:&#xA;    rperi = a * (1. - ep)&#xA;    vperi = np.sqrt(2./rperi - 1./a)&#xA;    X0 = np.array([rperi, 0, 0, vperi])&#xA;&#xA;    answer, info = ODEint(deriv, X0, time, atol = 1E-12, full_output=True)&#xA;    xy = answer.T[:2]&#xA;    orbits.append(xy)&#xA;&#xA;rs = [np.sqrt((xy**2).sum(axis=0)) for xy in orbits]&#xA;rmeans = [r.mean() for r in rs]&#xA;&#xA;plt.figure()&#xA;plt.subplot(3, 2, 1)&#xA;for x, y in orbits:&#xA;    plt.plot(x, y)&#xA;plt.ylim(-1, 1)&#xA;plt.plot([0], [0], 'ok')&#xA;plt.subplot(3, 2, 3)&#xA;for r in rs:&#xA;    plt.plot(time, r)&#xA;plt.subplot(3, 2, 5)&#xA;plt.plot(eps, rmeans)&#xA;plt.plot(eps, rmeans, 'ok')&#xA;plt.plot(eps, np.ones_like(eps), '--k')&#xA;plt.ylim(0, 1.6)&#xA;&#xA;&#xA;theta = np.linspace(0, twopi, 10001)[:-1]&#xA;rs = [a * (1-ep**2)/(1 + ep*np.cos(theta)) for ep in eps]&#xA;rmeans = [r.mean() for r in rs]&#xA;&#xA;plt.subplot(3, 2, 2)&#xA;for r in rs:&#xA;    x, y = [r*f(theta) for f in (np.cos, np.sin)]&#xA;    plt.plot(x, y)&#xA;plt.ylim(-1, 1)&#xA;plt.plot([0], [0], 'ok')&#xA;plt.subplot(3, 2, 4)&#xA;for r in rs:&#xA;    plt.plot(theta, r)&#xA;plt.subplot(3, 2, 6)&#xA;plt.plot(eps, rmeans)&#xA;plt.plot(eps, rmeans, 'ok')&#xA;plt.plot(eps, np.ones_like(eps), '--k')&#xA;plt.ylim(0, 1.6)&#xA;plt.suptitle(&quot;Time averages      Theta averages&quot;)&#xA;plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-12-13T05:05:12.613" LastActivityDate="2017-12-13T05:05:12.613" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23870" PostTypeId="1" CreationDate="2017-12-07T16:43:52.250" Score="2" ViewCount="152" Body="&lt;p&gt;Can some one tell me how you can roughly estimate the number of collisions when Milkyway and Andromeda merged with each other. I know it is somehow related to the mean free path of the star. But couldn't figure out how to do it exactly.&lt;/p&gt;&#xA;" OwnerUserId="20177" LastActivityDate="2017-12-07T16:43:52.250" Title="How to to compute the expected number of collisions between stars when Milkyway and Andromeda merged?" Tags="&lt;galaxy&gt;&lt;cosmology&gt;&lt;milky-way&gt;&lt;dwarf-galaxy&gt;" AnswerCount="0" CommentCount="6" FavoriteCount="1" ClosedDate="2017-12-08T18:19:20.677" ContentLicense="CC BY-SA 3.0" />
  <row Id="23872" PostTypeId="2" ParentId="891" CreationDate="2017-12-07T22:09:38.320" Score="1" Body="&lt;p&gt;The fastest hypervelocity stars are supposed to go about 900km/s 2million mph &#xA;&lt;a href=&quot;https://www.space.com/19748-hypervelocity-stars-milky-way.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.space.com/19748-hypervelocity-stars-milky-way.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quasars are an obvious example of matter with near light speed velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When accelerated matter in the beam approaches the speed of light, astrophysical jets become relativistic jets as they show effects from special relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Astrophysical_jet&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Astrophysical_jet&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/4LQZk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4LQZk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2017-12-07T22:15:37.533" LastActivityDate="2017-12-07T22:15:37.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23873" PostTypeId="1" CreationDate="2017-12-08T06:16:10.657" Score="2" ViewCount="311" Body="&lt;p&gt;The CBC News Technology and Science Q&amp;amp;A &lt;a href=&quot;http://www.cbc.ca/news/technology/espresso-habitable-planet-search-1.4437598&quot; rel=&quot;nofollow noreferrer&quot;&gt;This ESPRESSO machine doesn't make coffee but scans the skies for habitable planets&lt;/a&gt; includes a nice description of &lt;a href=&quot;https://www.eso.org/sci/facilities/develop/instruments/espresso.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ESPRESSO&lt;/a&gt; (Echelle SPectrograph for Rocky Exoplanet and Stable Spectroscopic Observations) and links to the YouTube video &lt;a href=&quot;https://i.stack.imgur.com/vugHl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Echelle SPectrograph for Rocky Exoplanet and Stable Spectroscopic Observations&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's something really amazing about the way that the spectrogram has been folded so nicely on to this image. Is this what it looks like on a 2D detector? (&lt;em&gt;sans colour&lt;/em&gt; of course) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I see what looks like eight distinct sections&lt;/strong&gt;, and a really nice spacing between adjacent lines otherwise. &lt;strong&gt;How is this accomplished?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click or open the image in a new tab for full size view (it's worth it; 1.8 MB, 1738x2530)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2UQx0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2UQx0l.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Frame grab from the &lt;a href=&quot;https://youtu.be/pPkVv3D8lS8&quot; rel=&quot;nofollow noreferrer&quot;&gt;video&lt;/a&gt; at about &lt;code&gt;00:55&lt;/code&gt;. Recorded with Python using &lt;a href=&quot;https://stackoverflow.com/q/43159565/3904031&quot;&gt;this script&lt;/a&gt;, then extracting frames using &lt;a href=&quot;https://blender.stackexchange.com/a/43531/5334&quot;&gt;this script in Blender&lt;/a&gt;. Click or open in new view for full size - it's worth it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/s1q2F.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/s1q2F.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; What an Echelle spectrogram &quot;normally&quot; looks like (again with added color). From the really interesting University of Virginia &lt;a href=&quot;http://faculty.virginia.edu/ASTR3130/lectures/spectroscopy/spectrographs.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ASTR 3130 (Majewski) Lecture Notes&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2020-11-28T04:48:46.317" LastActivityDate="2020-11-28T04:48:46.317" Title="How does the ESPRESSO Echelle spectrograph fold the spectrum so nicely?" Tags="&lt;observational-astronomy&gt;&lt;spectroscopy&gt;&lt;data-analysis&gt;&lt;instruments&gt;" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="23875" PostTypeId="1" AcceptedAnswerId="23879" CreationDate="2017-12-08T06:27:58.087" Score="2" ViewCount="59" Body="&lt;p&gt;The CBC News Technology and Science Q&amp;amp;A &lt;a href=&quot;http://www.cbc.ca/news/technology/espresso-habitable-planet-search-1.4437598&quot; rel=&quot;nofollow noreferrer&quot;&gt;This ESPRESSO machine doesn't make coffee but scans the skies for habitable planets&lt;/a&gt; includes a nice description of &lt;a href=&quot;https://www.eso.org/sci/facilities/develop/instruments/espresso.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ESPRESSO&lt;/a&gt; (Echelle SPectrograph for Rocky Exoplanet and Stable Spectroscopic Observations) and links to the YouTube video &lt;a href=&quot;https://i.stack.imgur.com/vugHl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Echelle SPectrograph for Rocky Exoplanet and Stable Spectroscopic Observations&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the spectrogram shown at about &lt;code&gt;00:55&lt;/code&gt; in the video I've noticed that each band is accompanied by a finely space array of points. I am assuming that they are some kind of calibration, but not sure how they are produced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the orders are changing, I can't really figure out if these are equally spaced in wavelength or in frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are these densely space calibration &quot;lines&quot; produced?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see the full spectrogram in &lt;a href=&quot;https://astronomy.stackexchange.com/q/23873/7982&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZWqeS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZWqeS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HW8Ef.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HW8Ef.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4IyW7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4IyW7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/g7z15.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g7z15.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Cropped subsections from a HD Frame grab from the &lt;a href=&quot;https://youtu.be/pPkVv3D8lS8&quot; rel=&quot;nofollow noreferrer&quot;&gt;video&lt;/a&gt; at about &lt;code&gt;00:55&lt;/code&gt;. Recorded with Python using &lt;a href=&quot;https://stackoverflow.com/q/43159565/3904031&quot;&gt;this script&lt;/a&gt;, then extracting frames using &lt;a href=&quot;https://blender.stackexchange.com/a/43531/5334&quot;&gt;this script in Blender&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2017-12-08T11:22:53.797" Title="How are the ESPRESSO Echelle spectrograph's calibration &quot;lines&quot; produced?" Tags="&lt;spectroscopy&gt;&lt;instruments&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23879" PostTypeId="2" ParentId="23875" CreationDate="2017-12-08T11:22:53.797" Score="1" Body="&lt;p&gt;You're right to think that this is a calibration technique. In use here are &lt;a href=&quot;https://en.wikipedia.org/wiki/Frequency_comb&quot; rel=&quot;nofollow noreferrer&quot;&gt;laser combs&lt;/a&gt; that use self-interference of short pulses that are reflected multiple times inside their own cavity to generate the spacing that you see.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spacing then is mainly limited by the inverse of the pulse length, and thus by current laser technology. Sharpness of individual lines is usually increased by putting a &lt;a href=&quot;https://en.wikipedia.org/wiki/Fabry%E2%80%93P%C3%A9rot_interferometer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fabry-Pérot&lt;/a&gt; interferometer in sequence after the laser. For additional precision and estimation of laser stability, one uses often some secondary lamp with theoretically or laboratorially well-known lines (Th-Ar for example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The technology demonstration for ESPRESSO was done with its precursor, &lt;a href=&quot;https://www.eso.org/public/announcements/ann15037/&quot; rel=&quot;nofollow noreferrer&quot;&gt;HARPS&lt;/a&gt;.&lt;br&gt;&#xA;Also A &lt;a href=&quot;http://walsworth.physics.harvard.edu/oldsite/Activities/Astronomy/astro_comb.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;site&lt;/a&gt; with more details, the lowest poster looks pretty good in explaining the basics, I find.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2017-12-08T11:22:53.797" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23881" PostTypeId="1" AcceptedAnswerId="23883" CreationDate="2017-12-08T11:46:10.277" Score="1" ViewCount="173" Body="&lt;p&gt;I've never seen this Hubble picture, is it an artist impression? It's fascinating because there are voids apparently cause by supernovae. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0CliP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0CliP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;sup&gt;From the video &lt;a href=&quot;https://youtu.be/41Y6xov0ppw?t=1708&quot; rel=&quot;nofollow noreferrer&quot;&gt;1 Hour of Epic Space Music: COSMOS - Volume 1&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deviant art mentioned in the information below the video is by Natasha Sharma, she has a different style than that. There are also real photos from hubble in the video, I can't tell if it is a genius space artist or a real photo.&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorDisplayName="user1569" LastEditDate="2017-12-11T13:37:06.690" LastActivityDate="2017-12-11T13:37:06.690" Title="Is this a real photo or an art work?" Tags="&lt;nebula&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23883" PostTypeId="2" ParentId="23881" CreationDate="2017-12-08T13:55:22.547" Score="3" Body="&lt;p&gt;It is &lt;a href=&quot;https://rhads.deviantart.com/art/From-Nothing-492031701&quot; rel=&quot;nofollow noreferrer&quot;&gt;'From Nothing'&lt;/a&gt;, made by &lt;a href=&quot;https://cosmicspark.deviantart.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;cosmicspark&lt;/a&gt;, a.k.a Tim Barton.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This picture was painted from nothing, in roughly 12 hours while I listened to a lot of interesting lectures on physics, including Dr. Lawrence Krauss' famous lecture &quot;A Universe From Nothing&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user1569" LastActivityDate="2017-12-08T13:55:22.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23884" PostTypeId="2" ParentId="2544" CreationDate="2017-12-08T14:55:18.633" Score="0" Body="&lt;p&gt;Q - How far away are approaching comets detected today? How long time before they would pass Earth orbit? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A - About 30AU's. That's the same distance as neptune.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Halley%27s_Comet&quot; rel=&quot;nofollow noreferrer&quot;&gt;Halley's comet&lt;/a&gt;, which is 70.56 km/s has been seen in 2003 at 28AU's away, near it's 35AU's aphelion, using three telescopes for multiple nights, Halley is 11km diameter and currently has apparent magnitude 28v.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a graph of apparent magnitude of a 15 kilometer comet called 2015_ER61 over 3 years from 2014 to 2017, which was detected by PANSTARRS (Panoramic Survey Telescope and Rapid Response System). It passed 0.1AU's away from us. PANSTARRS also saw the 50m interstellar asteroid in November 2017 as it was travelling away.&lt;a href=&quot;https://i.stack.imgur.com/m3gKc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/m3gKc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zY9VV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zY9VV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pan-STARRS currently consists of two 1.8 m Ritchey–Chrétien telescopes located at Haleakala in Hawaii. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exoplanets were rare when they were detected with apparatus similar to PANSTARRS, and then they launched Kepler to find more, and they have as of 1 December 2017, detected 3,710 planets in 2,780 systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same will happen in the future with comets, especially for interstellar comets, which is like having material from many light years away arrive in the solar system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficult task is approaching a 25KM/s comet at landing speeds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of ESA's GAIA photography work is thrown away! When it could be used to detect comets! It discards most of it's CMOS data, including all the 1000ds of comets that it would have detected had it been designed differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottleneck in GAIA is how much data it can transmit back to our planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a suggestion of a future mission which can possibly multiply our knowledge of comets and asteroids by 2 orders of magnitude, the suggested mission is called FAiRLiGHT based on our experience of processing large amounts of CMOS data for vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;100W of solar panels can supply a 10TB SSD type HDD in space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;10TB of SSD can keep a month's worth of space telescope observations on disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;50W is also used for a serial CPU processor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CMOS data is kept in a virtual 3D or Voxel array inside the 10TB memory, with simple memory indexing that represents Pixel X, Y , Time values. (it's cubic memory, representing space as a 2D plane with a time variant) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's mathematical simplicity. You keep super simple indexing for all the photos in the momory, so that the processor requires nearly zero instructions in order to read, write and recall memory very fast. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 50W processor can process about 10 million CMOS pixels every second searching for vector data on board the FAiRLiGHT space telescopes. To be precise it can process about 100 MFLOPs, floating point operations per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a comet/asteroid reconnaissance mission that is essentially keeping all the GAIA photos in large memory banks and running a processor on 10 million pixels per second searching for shooting-star type signals across multiple photos. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clearest vectors/comets are then sent back to earth, which results in a thousand fold compression of the original CMOS files.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They will be able to use Bi-focal and Tri-focal versions of GAIA telescope, which then triangulate exact positions for comets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vogue has recently been about algorythms to study peaks in stallar brightness which describe exoplanets, and if material from other stars can be detected more efficiently like exoplanets were, for example by using satellite based bifocal vector processing engines, vector processing will perhaps become a well studied field and comets and interstellar meteorites will be seen frequently and with precision. &lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2017-12-08T15:25:52.380" LastActivityDate="2017-12-08T15:25:52.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23885" PostTypeId="1" CreationDate="2017-12-08T21:04:01.427" Score="3" ViewCount="301" Body="&lt;p&gt;If my understanding is correct, the Cosmic Microwave Background always come from our event horizon. By event horizon I mean the edge of the observable universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the universe is finite does it mean that after some time (maybe in billions of years?) there will start being a hole in CMB. As our event horizon let us see CMB from regions further and further away, isn't there a time where it will reach the edge of the universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the case, based on the known number of atom and mass of the universe, is it possible to calculate when this will happen?&lt;/p&gt;&#xA;" OwnerUserId="20199" LastActivityDate="2017-12-11T22:18:30.253" Title="Would a future hole in the Cosmic Microwave Background show us the edge of the universe?" Tags="&lt;universe&gt;&lt;cmb&gt;&lt;event-horizon&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23886" PostTypeId="1" CreationDate="2017-12-08T21:41:43.067" Score="0" ViewCount="129" Body="&lt;p&gt;Is it really necessary to look for Earth-like habitable zone planets, when it is likely that the &quot;humans&quot; to reach that planet will be bio-engineered or entirely mechanical by the time we would be exploring &quot;in person?&quot;&lt;/p&gt;&#xA;" OwnerUserId="20200" LastEditorUserId="2365" LastEditDate="2017-12-08T21:44:54.443" LastActivityDate="2017-12-11T16:52:08.720" Title="Why bother looking for Earth-like planets?" Tags="&lt;habitable-zone&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23887" PostTypeId="2" ParentId="23886" CreationDate="2017-12-08T21:48:02.047" Score="4" Body="&lt;p&gt;The possibility of exploring exoplanets is so remote it is not a factor. We are interested in Earth-like planets as there is only one type of planet that we know can support life, and that is the Earth-like ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are interested in Earth-like planets as they may provide a habitat for alien life. The question of &quot;does extra-terrestial life exist&quot; is of intrinsic interest.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-12-08T21:48:02.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23888" PostTypeId="2" ParentId="23885" CreationDate="2017-12-08T23:17:27.703" Score="9" Body="&lt;p&gt;First, let me clear up a misunderstanding:&lt;/p&gt;&#xA;&lt;strong&gt;Particle horizon&lt;/strong&gt;&#xA;&lt;p&gt;The &amp;quot;edge&amp;quot; of the observable Universe is called the &lt;em&gt;particle horizon&lt;/em&gt;, and lies roughly 47 Gly (billion lightyears) away. It is always receding, both because the Universe expands and because light from increasingly large distances eventually reach us. In &lt;em&gt;comoving coordinates&lt;/em&gt; (the coordinate system that expands along with the Universe), it asymptotically reaches a finite size (some 63 Gly), but in physical coordinates it increases forever&lt;span class=&quot;math-container&quot;&gt;$^\dagger$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;strong&gt;Event horizon&lt;/strong&gt;&#xA;&lt;p&gt;The &lt;em&gt;event&lt;/em&gt; horizon is the boundary between the part of the Universe that contains the galaxies (and other stuff) from which light emitted today will be able to reach us, and the part where it won't. Currently, it is roughly 17 Gly away, but his horizon is always shrinking because the expansion of the Universe accelerates. In other words, stuff outside the event horizon is so far away that the expansion carries it away too fast for the photon amitted today to ever reach us.&lt;/p&gt;&#xA;&lt;p&gt;Thus, at Big Bang (BB), the event horizon was 63 comoving Gly away, because that was the maximum distance from which we would ever receive information.&lt;/p&gt;&#xA;&lt;strong&gt;The CMB will always lag behind the particle horizon&lt;/strong&gt;&#xA;&lt;p&gt;The farther away we look, the more we look back in time. &lt;em&gt;If&lt;/em&gt; there were galaxies a second after BB, and &lt;em&gt;if&lt;/em&gt; we had the technology to look that far, we would see them right after BB, and they would be 47 Gly away. However, galaxies weren't born until several hundred Myr (million years) after BB. And now to your question: The CMB wasn't emitted &lt;em&gt;at&lt;/em&gt; BB, so it doesn't origin from a region that is now 47 Gly away. Instead, it was emitted 380,000 yr after BB, and it was emitted from &lt;em&gt;all&lt;/em&gt; of the Universe, in &lt;em&gt;all&lt;/em&gt; directions.&lt;/p&gt;&#xA;&lt;p&gt;This emittance wasn't instantaneous, but almost (took something like 10,000 yr), so let's assume for now that it was. Let's call that time &lt;span class=&quot;math-container&quot;&gt;$t_\mathrm{CMB}$&lt;/span&gt;. In that case, because light travels at &lt;span class=&quot;math-container&quot;&gt;$c = 300\,000\,\mathrm{km}\,\mathrm{s}^{-1}$&lt;/span&gt;, if you were present one second after &lt;span class=&quot;math-container&quot;&gt;$t_\mathrm{CMB}$&lt;/span&gt;, you would see the CMB that came from all regions that were 300,000 km away. That is, the CMB would be the light from a thin shell of radius 300,000 km.&lt;/p&gt;&#xA;&lt;p&gt;Two seconds after BB, the CMB would be the light from a shell of radius 600,000 km, and so on. However, because the Universe is expanding, it is actually at little more than 600,000 km in radius.&#xA;And now, 13.8 Gyr after BB, and hence 13.7996 Gyr after the CMB was emitted, this shell is not almost 14 Gyr away, but in fact 45 Gly away which is almost, but quite, at the particle horizon.&lt;/p&gt;&#xA;&lt;p&gt;Because of the expansion, the distance to the CMB shell always increases. But so does the distance to the particle horizon, and because the CMB was emitted &lt;em&gt;after&lt;/em&gt; BB, the particle horizon will always &amp;quot;be ahead&amp;quot; of the CMB. As the CMB shell recedes, it becomes more and more redshifted, slowly fading into darkness but never disappearing. This is irrespective of whether the Universe is finite or infinite, except in the possible, but rather fantastic, event that the Universe is larger than the ~47 Gly we can see today, but smaller than the 63 comoving Gly that is the maximum size we will ever see.&lt;/p&gt;&#xA;&lt;strong&gt;Finite or infinite?&lt;/strong&gt;&#xA;&lt;p&gt;We don't know if the Universe is finite or infinite, but if the un-observable part of it looks like the observable Universe, then it is infinite. What we can say is that, at least in the observable Universe, the CMB was emitted &lt;em&gt;everywhere&lt;/em&gt; in space, and there is currently no reason to believe that it was different elsewhere (although it could bee, if it expanded differently in other parts). To the best of our knowledge, space is homogeneous, meaning that (on average) there are as many baryons, photons, dark matter particles, and bicycles in one place as there is in another. In other words, to the best of our knowledge, there is no edge. Even if it's finite, it is difficult to (and most, if not all, astronomers don't) imagine an edge; rather it is thought that it would &amp;quot;curve back on itself&amp;quot;, just as the surface of a ball is finite in area, but doesn't have an edge.&lt;/p&gt;&#xA;&lt;strong&gt;2D analogy&lt;/strong&gt;&#xA;&lt;p&gt;As an analogy, imagine living on a flat, inifinite Earth, populated evenly by people with loud voices. At some point in time, everbody yells &amp;quot;CMB&amp;quot;. First you'll hear your neighbors yell, after 1 sec you'll hear those living 340 meters away, after 1 minute those living 20 km away, after a year those 10 million km away, and so on and so on forever.&lt;/p&gt;&#xA;&lt;p&gt;Now imagine the same on a round Earth. After 16 hours, you would hear the folks living at your antipode, and if I understand correctly, this is when you expect to &amp;quot;hear a hole in the sound&amp;quot;. But now you would start hearing the ones that are &lt;em&gt;closer&lt;/em&gt;, but whose sounds have been traveling more than halfway around Earth. After 32 hours, you'd hear your own yell. And so on.&lt;/p&gt;&#xA;&lt;p&gt;But, our Earth analogy is also expanding. If it were expanding at a steady rate, we &lt;em&gt;would&lt;/em&gt; eventually hear our antipode friends. You might think that if it were expanding fast enough, their sound would never reach you, but as explained by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ant_on_a_rubber_rope&quot; rel=&quot;nofollow noreferrer&quot;&gt;ant-on-a-rubber-band puzzle&lt;/a&gt;, that's not the case).&lt;/p&gt;&#xA;&lt;p&gt;However, our fictional Earth is not only expanding, but in fact this expansion is accelerating. And in &lt;em&gt;that&lt;/em&gt; case, there &lt;em&gt;is&lt;/em&gt; a limit to how far away you can hear people.&lt;/p&gt;&#xA;&lt;p&gt;Back to the real Universe, there is a limit to from how distant regions you may receive CMB photons (or any other information).&#xA;Moreover, if the Universe were finite, and were not accelerating, you might think that you would eventually see the CMB that was emitted from our own neighborhood and had been traveling around the Universe.&#xA;However, it can be shown that, in this case, the Universe will reach a maximum size, start collapsing, and end in a Big Crunch &lt;em&gt;exactly&lt;/em&gt; at the time where you expect to see your antipode.&lt;/p&gt;&#xA;&lt;p&gt;Only in a finite Universe with a rather special mix of gravitating matter and accelerating dark energy might it be possible to delay the collapse or the eternal expansion to a point where we eventually look back at ourselves. But that's sort of fine-tuning.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^\dagger$&lt;/span&gt;&lt;sub&gt;Unless the Universe contains a yet unknown substance that is able to halt the acceleration. This is entirely possible, but in this answer I consider only a well-behaving Universe that acts according to current knowledge.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2017-12-11T22:18:30.253" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="23890" PostTypeId="2" ParentId="23886" CreationDate="2017-12-09T00:02:22.357" Score="0" Body="&lt;p&gt;Your question comes across as more of a complaint than a true question.  Whenever you ask, &quot;Is it really necessary...&quot; it is easy for someone to respond, &quot;Is &lt;em&gt;anything&lt;/em&gt; really necessary?&quot;  If people are interested in searching for other planets that may be habitable, why not just see what they find?  Most people would find it fascinating/interesting if we verified a &quot;second Earth&quot; were truly out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then make a wild assertion using &quot;humans&quot; in scare quotes, assuming that we'd be mechanical or entirely separate entities and therefore only non-human entities.  Would you feel the same way if such a planet were less than 5 light years away? 10? 15?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest that if in the next 100 years we have enough research to show that humans could walk around and breathe on a planet less than 10 light years away, that there won't be anything anyone can do to stop humans from beginning the development phase of ambitious projects to reach that world.&lt;/p&gt;&#xA;" OwnerUserId="20201" LastActivityDate="2017-12-09T00:02:22.357" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23892" PostTypeId="2" ParentId="23886" CreationDate="2017-12-09T11:33:37.903" Score="1" Body="&lt;p&gt;I recently joined a project on exoplanet habitability, and have worked with other physics research in the past. An underlying issue considered by people working in these areas is &quot;does it really matter?&quot; It's a good question, and hard question to answer even for scientists. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can expand your question to whether astronomy itself is worth studying. I would argue it is (one reason I joined the project). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exoplanets were discovered relatively recently, making interest in them sky rocket. Just like how astronomers study stars to find how our sun was formed and how it will live out its life, exoplanet research can reveal how our solar system was formed as well as how it might evolve in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to the search for life, simply by observing other systems with different sizes, and in different stages of their formation and evolution, we can better understand our own as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, research for the sake of knowledge is fun! :) But of course it sometimes leads to unexpected yet important discoveries.&lt;/p&gt;&#xA;" OwnerUserId="18327" LastEditorUserId="7982" LastEditDate="2017-12-11T16:52:08.720" LastActivityDate="2017-12-11T16:52:08.720" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23894" PostTypeId="2" ParentId="20298" CreationDate="2017-12-09T14:16:58.147" Score="1" Body="&lt;p&gt;The difference is due to two things. First, using geometry calculates the time from the center of the sun crossing the horizon in the east to the center of the sun crossing it in the west. But sunrise is when the edge of the sun's disk crosses the horizon and sunset when the other edge crosses it, so your day length is low by the time it takes the sun to &quot;move&quot; one solar diameter, or about 2 minutes (half a degree apparent diameter). At higher latitudes it would be a bit longer because the sun is &quot;moving&quot; at an angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other difference is the effect of refraction, which makes the sun appear higher than it really is when it's near the horizon; and because the angle at which the sun rises above the horizon or sinks below it increases with latitude, so does the length of time it takes the sun to overcome the effect of refraction, so the real length of the day at the equinox varies from about 15h 06m at the equator to over 15 1/2 hours inside the polar circles (in fact, at the poles the sun, or most of its disk, would be slightly above the horizon for the entire 24h day). This effect is more pronounced at the solstice, so the length of the day at the summer solstice can be half an hour or even more than an hour longer (again at higher latitudes) than the length of the night at the winter solstice.&lt;/p&gt;&#xA;" OwnerUserId="20210" LastActivityDate="2017-12-09T14:16:58.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23895" PostTypeId="2" ParentId="23783" CreationDate="2017-12-09T19:22:57.190" Score="1" Body="&lt;p&gt;Stars less than about 0.3 solar masses are fully convective, so hydrogen throughout the entire star can be transported to the core and consumed.  At higher masses, the core is radiative, so only hydrogen that starts in the core is consumed.&lt;/p&gt;&#xA;" OwnerUserId="10371" LastActivityDate="2017-12-09T19:22:57.190" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23897" PostTypeId="1" AcceptedAnswerId="23905" CreationDate="2017-12-10T00:56:30.970" Score="2" ViewCount="298" Body="&lt;p&gt;How close to a brown dwarf would an orbiting planet need to be in order to receive as much sunlight as Earth receives from the Sun?&lt;/p&gt;&#xA;" OwnerUserId="20214" LastActivityDate="2017-12-10T11:36:08.217" Title="How close to a brown dwarf would a planet need to be to receive as much light as Earth?" Tags="&lt;star&gt;&lt;light&gt;&lt;brown-dwarf&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23898" PostTypeId="2" ParentId="23897" CreationDate="2017-12-10T01:08:19.910" Score="-1" Body="&lt;p&gt;Brown dwarves don't have fusion in their core, thus they don't have their own light. They may have a little heat production from other processes (for example, contraction, or radioactive decay in their core). These aren't enough strong to heat even the dwarf significantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many brown dwarve were found orbiting a star closely, they may have surface temperature even in the order of some thousands K. But they get their temperature from their stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the Jupiter's temperature is only 40K more warm as it would be reasoned by the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of red dwarves, the situation is better. The distance depends on the star. You can very easily calculate it: check its &lt;a href=&quot;https://en.wikipedia.org/wiki/Absolute_magnitude&quot; rel=&quot;nofollow noreferrer&quot;&gt;absolute luminosity&lt;/a&gt;, so you can see its total power output, compared to the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radiation what a planet receives (per surface area) depends quadratically from its distance.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="1891" LastEditDate="2017-12-10T01:16:14.310" LastActivityDate="2017-12-10T01:16:14.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23899" PostTypeId="1" AcceptedAnswerId="23900" CreationDate="2017-12-10T04:08:20.747" Score="0" ViewCount="92" Body="&lt;p&gt;Why it is not even theoretically possible to travel in past but possible to travel in future ?&lt;/p&gt;&#xA;" OwnerUserId="20162" LastEditorUserId="20162" LastEditDate="2017-12-12T08:41:36.320" LastActivityDate="2017-12-12T08:41:36.320" Title="Time travel in past and future" Tags="&lt;space-time&gt;&lt;time&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2017-12-11T09:35:22.193" ContentLicense="CC BY-SA 3.0" />
  <row Id="23900" PostTypeId="2" ParentId="23899" CreationDate="2017-12-10T05:26:08.870" Score="5" Body="&lt;p&gt;The thing about time travel is it's hard to define. Technically we're constantly traveling into the future. &#xA;Now, if you went at near-light speed for a long time and then stopped and came back, you would have the &lt;em&gt;illusion&lt;/em&gt; of having time travelled into the future because time essentially went on without you. This is entirely physically possible, though not feasible of course.&#xA;But past time travel, what does that mean? Does it mean rewinding all time for everything? Or one entity aging backwards? Or time goes backwards while you remain the same age, like the reverse of the future travel?&#xA;In any case, unlike the future travel, there is no way, with our current model of physics, to do any of them. &#xA;And I must reiterate, future travel by the means I described is only an illusion. You didn't move though time, time just went on without you. Obviously, this won't happen the other way.&lt;/p&gt;&#xA;" OwnerUserId="18617" LastActivityDate="2017-12-10T05:26:08.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23902" PostTypeId="1" CreationDate="2017-12-10T05:44:16.863" Score="6" ViewCount="1507" Body="&lt;p&gt;So I've heard that all black holes that have been observed &lt;em&gt;rotate&lt;/em&gt; to some degree. But if it has zero radius, how is it even possible for the black hole itself to be spinning? Or is it just all the junk inside the event horizon that's rotating? &#xA;Maybe it spins with infinite angular velocity and a finite angular momentum? Please explain how this works.&lt;/p&gt;&#xA;" OwnerUserId="18617" LastActivityDate="2020-10-30T19:07:40.913" Title="How can a singularity rotate?" Tags="&lt;black-hole&gt;&lt;high-energy-astrophysics&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23903" PostTypeId="2" ParentId="23902" CreationDate="2017-12-10T07:48:44.220" Score="4" Body="&lt;p&gt;Don't think of the singularity as being an object made of matter. A black hole is a vacuum solution to the relativity equations. That means there is &lt;em&gt;nothing&lt;/em&gt;  inside the black hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A black hole doesn't contain matter, but it still has mass. The mass of the black hole can be observed in the curvature of space-time around the black hole. It also has angular momentum, this causes space-time around a spinning black hole to be dragged. This is weird. But this is what relativity predicts, and it's predictions are well supported by observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a black hole, there is no object in the middle, it is the space-time itself that is rotating.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-12-10T07:54:14.153" LastActivityDate="2017-12-10T07:54:14.153" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23904" PostTypeId="1" CreationDate="2017-12-10T11:19:02.010" Score="-1" ViewCount="59" Body="&lt;p&gt;I have my college project and I thought of finding the temperature of a star with the black-body curve of star but i need real data.&#xA;I searched but can't find the thing what I want. all I am getting is ready-made graphs of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need the datas in numbers and not just graphs.&lt;/p&gt;&#xA;" OwnerUserId="20217" LastActivityDate="2018-03-11T16:51:44.393" Title="Where can we find astronomical real world data (spectral radiance and wavelength particularly) of a star?" Tags="&lt;star&gt;&lt;black-body&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23905" PostTypeId="2" ParentId="23897" CreationDate="2017-12-10T11:36:08.217" Score="4" Body="&lt;p&gt;Brown dwarfs are born moderately hot and luminous and then they contract and cool. Thanks to electron degeneracy in their cores, they never become hot enough to ignite hydrogen (though there is a brief deuterium burning phase) and as a result their fate is to cool and fade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plot below (from &lt;a href=&quot;http://adsabs.harvard.edu/abs/1997ApJ...491..856B&quot; rel=&quot;nofollow noreferrer&quot;&gt;Burrows et al. 1997&lt;/a&gt;) shows how the luminosity behaves as a function of time since birth (age) for objects of various mass. The brown dwarfs are the green and red tracks (some would call the red tracks giant planets). The curves are labelled in Jupiter masses. The highest mass brown dwarf is 73 Jupiter masses, the next one down is 70$M_J$ and then the green curves count down in steps of 5$M_J$ to 15$M_J$, then the first red curve is at 13$M_J$ and the red curves count down in steps of 1$M_J$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The axes are logarithmic (to base 10). Thus when the y-axis says -2, it means the brown dwarf is one hundredth the luminosity of the Sun, -3 = one thousandth etc. The x-axis is logarithmic in units of billions of years. Most brown dwarfs observed in the Galaxy will be in the range 1-10 billion years (i.e. between 0 and 1 on this x-axis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flux received by a planet in orbit around a star/brown dwarf will be proportional to the its luminosity divided by orbital radius squared. From this we can write that the orbital radius where a similar flux to that of the Earth around the Sun will be received is&#xA;$$ R = \left( \frac{L_{BD}}{L_{\odot}} \right)^{1/2}\ {\rm au} $$&#xA;$$ \log\left(\frac{R}{1\ {\rm au}}\right) = 0.5 \log \left(\frac{L}{L_{\odot}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This you can get you answer immediately from the graph below. Choose the mass of your brown dwarf and its age. Find the log luminosity on the y-axis and halve this value. That is the the log of the orbital radius in astronomical units where a planet would receive the same amount of flux as the Earth from the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. Consider a 50$M_J$ brown dwarf that is a billion years old. Find the appropriate green curve and see that the y-axis value for x=0 (1 billion years) is -4.2. The log of the orbital radius (in au) at which a planet would receive the same amount of flux from the brown dwarf as does the Earth from the Sun would be -2.1. In linear units, this is $10^{-2.1} = 0.0079$ au, or 1.19 million km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/peSwg.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/peSwg.gif&quot; alt=&quot;Brown dwarf cooling tacks&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A further issue to consider is that the brown dwarf surface will be far cooler than the Sun and also changes (cools) with time. This means that the radiation received from the planet will push further and further into the infrared as the brown dwarfs gets older and hardly any of the received radiation will be in the visible part of the spectrum.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-12-10T11:36:08.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23910" PostTypeId="1" CreationDate="2017-12-11T01:25:20.107" Score="2" ViewCount="50" Body="&lt;p&gt;I know deuterium can be used as a good tracer to constrain the big bang nucleosynthesis. But can anyone tell me the in details how this is actually done?&lt;/p&gt;&#xA;" OwnerUserId="20177" LastActivityDate="2017-12-11T01:25:20.107" Title="How the observed deuterium abundance constrains big bang nucleosynthesis?" Tags="&lt;galaxy&gt;&lt;cosmology&gt;&lt;big-bang-theory&gt;&lt;observational-astronomy&gt;&lt;nucleosynthesis&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23912" PostTypeId="1" CreationDate="2017-12-11T06:08:02.463" Score="10" ViewCount="862" Body="&lt;p&gt;I am sorry this question is probably silly for professional astronomers, of which I am not one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I often hear at lectures that immediately after the Big Bang, the universe was small, say, the size of grapefruit or something like that. But because of inflation which stretched space at super-light (but finite), speed, the observable universe now, may not be the whole universe, and the whole universe may be infinite. If it is infinite, then it would seem that it had to be infinite in infancy as well, just very dense. In fact, it seems &quot;in the limit&quot;, it had to be infinite even at Big Bang.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was early universe small, or only dense but still infinite?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about at Big Bang, was it infinite?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When astronomers say early universe was small, do they simply mean &quot;the part of the universe which corresponds to our observable universe, was small&quot;?&lt;/p&gt;&#xA;" OwnerUserId="20226" LastEditorUserId="9301" LastEditDate="2017-12-12T15:50:40.787" LastActivityDate="2017-12-12T15:50:40.787" Title="Was the whole Universe close to Big Bang very small, or just very dense?" Tags="&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23917" PostTypeId="1" AcceptedAnswerId="23924" CreationDate="2017-12-11T12:08:13.153" Score="2" ViewCount="542" Body="&lt;p&gt;I am a complete newbie to astronomy and all the related.&lt;br&gt;&#xA;So if i say something stupid i am sorry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But i was wondering if it is possible to translate GPS coordinates like the following: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;lat: 41.420042122273024 &#xA;long: 2.1533203149999736 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;to its corresponding Galactic coordinates?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am grateful for all insights that you may offer me :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PS&lt;/strong&gt;: if this is the wrong exchange to ask this could you refer me to the right place to ask this question. &lt;/p&gt;&#xA;" OwnerUserId="20235" LastEditorUserId="7982" LastEditDate="2020-02-15T02:07:26.307" LastActivityDate="2020-02-15T02:07:26.307" Title="Can you translate a GPS coordinate to a Galactic coordinate?" Tags="&lt;galaxy&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23921" PostTypeId="2" ParentId="23902" CreationDate="2017-12-11T15:08:14.147" Score="6" Body="&lt;p&gt;There are multiple solutions to general relativity which allow for multiple different types of black holes. The &quot;normal&quot; black hole you see most people talk about, with a zero-volume, point singularity, is known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_metric#Singularities_and_black_holes&quot; rel=&quot;noreferrer&quot;&gt;Schwarzschild black hole&lt;/a&gt;. If the black hole is spinning, the Schwarzschild solution no longer applies and you're talking about a different type of black hole. This new type is referred to as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Rotating_black_hole&quot; rel=&quot;noreferrer&quot;&gt;Kerr black hole&lt;/a&gt; (other types include the &lt;a href=&quot;https://en.wikipedia.org/wiki/Reissner%E2%80%93Nordstr%C3%B6m_black_hole&quot; rel=&quot;noreferrer&quot;&gt;Reissner-Nordstr&amp;ouml;m&lt;/a&gt; and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kerr%E2%80%93Newman_metric&quot; rel=&quot;noreferrer&quot;&gt;Kerr-Newman&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a Kerr black hole, the singularity at the center is still zero-volume, but it is no longer a point. Instead it is a disk of zero height, often referred to as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Ring_singularity&quot; rel=&quot;noreferrer&quot;&gt;ring singularity&lt;/a&gt; or &quot;Ringularity&quot;. The angular momentum of the spinning black hole is then the angular momentum around the axis of rotation passing through this ring.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="1501" LastEditDate="2019-07-29T14:45:39.527" LastActivityDate="2019-07-29T14:45:39.527" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23922" PostTypeId="2" ParentId="23904" CreationDate="2017-12-11T15:12:43.987" Score="1" Body="&lt;p&gt;The best way to do this is to pick a particular astronomical instrument or science mission and go to their website. Almost all astronomy projects are required to release their data to the public and they almost always link to their data from the website for that mission/instrument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the Sloan Digital Sky Survey which produces spectra for a huge number of stars has access to their spectra &lt;a href=&quot;https://dr14.sdss.org/optical/spectrum/view?run2d=v5_10_0&amp;amp;plateid=4055&amp;amp;mjd=55359&amp;amp;fiberid=596&amp;amp;action=search&quot; rel=&quot;nofollow noreferrer&quot;&gt;right here&lt;/a&gt;. You can download the fits files directly and start playing around with the actual data yourself.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-12-11T15:12:43.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23923" PostTypeId="2" ParentId="23912" CreationDate="2017-12-11T15:25:34.300" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am sorry this question is probably silly for professional astronomers, of which I am not one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This question is by no means silly. Your question is a common one about cosmology (the study of where the universe came from, how it is evolving and what its fate will be). The media often butchers these concepts horribly, resulting in a lot of confusion (out of all scientific information, they seem to have the hardest time reporting cosmology accurately). Your inquisition is definitely a good thing.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When astronomers say early universe was small, do they simply mean &quot;the part of the universe which corresponds to our observable universe, was small&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, they're usually referring to the entire Universe. In my final paragraph, I explain what this implies for the observable universe.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If it is infinite, then it would seem that it had to be infinite in infancy as well, just very dense. In fact, it seems &quot;in the limit&quot;, it had to be infinite even at Big Bang.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're closer to the truth. When we talk about the expansion of the Universe, we're really saying that space is being created between all matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you mentioned, the Universe may be infinite. It is not like a ball, but rather like a flat grid, and its &quot;expansion&quot; just means that the distances between objects on the grid are getting larger. In essence, more space is being created between the objects. That's what we mean by expansion — that objects are moving away from each other, since more space is being created between them. Below is a gif I've made to demonstrate this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/amQeZ.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/amQeZ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more useful way to describe this is to say the grid is expanding — that space itself, as a coordinate system, is growing. As an analogy, imagine are walking your dog. Suddenly, the ground begins expanding between you. You and your dog will separated and continue receding away from each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the same thing is happening with our universe. The grid is in fact growing, and objects are being swept away with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, now that we've gotten the core concepts down, I'll introduce one more bit of terminology. The &quot;scale factor of the Universe&quot; refers to how much the Universe has expanded, compared to now. For example, if in a billion years the scale factor is 3, that means that every object in the Universe is 3 times farther from each other compared to now. If the scale factor 700 million years ago was 0.8, then everything was closer by a factor of 0.8 at that time. By definition, the scale factor is 1 right now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if the Universe is expanding now, we'd expect it to be smaller as we look further back in time — i.e. the scale factor would be less. General relativity predicts the scale factor to be zero at 13.8 billion years ago. This would mean that every object would be zero times its current distance from us — in other words, there would be no space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think a Universe without space is impossible, you're correct. We apparently have a contradiction. In GR, you can't have a spacetime with zero space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our modern physical theories work fine up a few fractions of a second after the moment of contradiction, and our observations do agree with the idea of an extremely dense early universe. However, our theories break down as we try to model the Universe at earlier and earlier times, until they no longer prove accurate, preventing us from explaining the most interesting moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why the moment of the Big Bang is one of the biggest mysteries in cosmology. Theories like quantum gravity have arisen to try to explain the conditions near the Big Bang, but none are sufficient as of now.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I often hear at lectures that immediately after the Big Bang, the universe was small, say, the size of grapefruit or something like that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Indeed, the problem stems from the ambiguity when one says &quot;universe&quot;. In this case, they're referring to the observable universe, which is actually spherical. The observable universe was indeed much smaller near the time of the Big Bang, compared to its radius now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because its radius actually depends on our Universe's scale factor*, which means that at the moment GR predicts the scale factor to be zero, it also predicts the size of the observable universe to be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously that can't be the case, since as we've explained above, it shouldn't be possible for the scale factor to be zero. However, we &lt;em&gt;can&lt;/em&gt; say with reasonable confidence that the observable universe was likely the size of a grapefruit at one point, if not smaller (although &quot;grapefruit&quot; seems an arbitrary choice for comparison. I can't actually find the paper that first uses this analogy, so what they originally meant is a bit unclear).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;*Measuring distances is actually a bit tricky in cosmology; in some cases, we want to talk about distances or motion of objects while neglecting the Universe's expansion. To save you the need to learn a lot of terminology, I'm right now taking into account the expansion of the Universe when talking about the observable universe's size. The observable universe also grows due to factors besides the Universe's expansion, i.e. light from further and further galaxies reaching us.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2017-12-11T15:37:11.747" LastActivityDate="2017-12-11T15:37:11.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23924" PostTypeId="2" ParentId="23917" CreationDate="2017-12-11T16:16:08.113" Score="5" Body="&lt;p&gt;Sure you can! &lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Galactic coordinates&lt;/a&gt; have the same origin as other J2000.0 systems; the solar system barycenter (center of mass). This is very close to the Sun, usually but not always inside the Sun, because the larger planets, especially Jupiter, pull it around a little bit. You can read a little more &lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/G/Galactic+Coordinate+System&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for example, and also read &lt;a href=&quot;https://astronomy.stackexchange.com/a/16340/7982&quot;&gt;@zephyr's excellent answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first you might ask why the origin of the galactic coordinates isn't the center of the galaxy. I'm pretty sure that the answer is that &lt;em&gt;we don't know where that is!&lt;/em&gt; We'd have to know the masses and locations of everything, and of course since most of the galaxy's mass is &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_matter&quot; rel=&quot;nofollow noreferrer&quot;&gt;dark matter&lt;/a&gt; we're not going to know where the center of mass is any time soon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the XY plane of the galactic coordinates has been chosen for now, based on an estimate of the Galaxy's apparent equator. Since it's a different plane than our solar system's plane, otherwise known as the ecliptic, the coordinates will be different even though the origin is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cW0Dv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cW0Dv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since galactic coordinates are centered near the Sun, the distance of our position from the origin will still be about 1 AU (150,000,000 km).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below I wrote a little script in Python using the easy-to-use &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_matter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Skyfield&lt;/a&gt; python package. At the moment that I've run the program , the coordinates are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;time (JD):   2458099.18846&#xA;time (UTC):  (2017, 12, 11, 16, 31, 23.049599826335907)&#xA;latitude (degs):    41.42&#xA;longitude (degs):   2.15&#xA;galactic (km):     [ -1.46347711e+08  -2.89156773e+06  -2.34254700e+07]&#xA;barycentric (km):  [  2.69276456e+07   1.33751808e+08   5.79665590e+07]&#xA;and just for fun...&#xA;galactic latlon (degs):  [-9.0922867698877123, 181.13191434418721]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here is the Python script:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;from skyfield.api import Loader, Topos&#xA;&#xA;load    = Loader('~/Documents/SkyData')&#xA;planets = load('de421.bsp')&#xA;earth, sun = planets['earth'], planets['sun']&#xA;&#xA;ts = load.timescale()&#xA;tnow    = ts.now()&#xA;tmonth = ts.utc(2017, 12, range(1,31))&#xA;topo = Topos(latitude_degrees=41.42, longitude_degrees=2.15)&#xA;&#xA;position = earth + topo&#xA;&#xA;print &quot;time (JD):  &quot;, tnow.tt&#xA;print &quot;time (UTC): &quot;, tnow.tt_calendar()&#xA;print &quot;latitude (degs):   &quot;, topo.latitude.degrees&#xA;print &quot;longitude (degs):  &quot;, topo.longitude.degrees&#xA;print &quot;galactic (km):    &quot;, position.at(tnow).galactic_position().km&#xA;print &quot;barycentric (km): &quot;, position.at(tnow).position.km&#xA;print &quot;and just for fun...&quot;&#xA;print &quot;galactic latlon (degs): &quot;, [x.degrees for x in position.at(tnow).galactic_latlon()[:2]]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-12-11T18:43:04.470" LastActivityDate="2017-12-11T18:43:04.470" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="23927" PostTypeId="1" AcceptedAnswerId="27981" CreationDate="2017-12-12T01:04:40.783" Score="6" ViewCount="199" Body="&lt;p&gt;As pointed out in &lt;a href=&quot;https://astronomy.stackexchange.com/a/16340/7982&quot;&gt;this answer&lt;/a&gt; &lt;em&gt;the axes&lt;/em&gt; of &lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_coordinate_system&quot; rel=&quot;noreferrer&quot;&gt;galactic coordinates&lt;/a&gt; are defined such that galactic longitude and latitude (0, 0) point toward the &lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_Center&quot; rel=&quot;noreferrer&quot;&gt;galactic center&lt;/a&gt;. But then the plane defined by galactic latitude equal to zero would be parallel to the galactic plane only if the Sun were &lt;em&gt;in the galactic plane as well&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that the case? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, by what procedure was the galactic plane actually defined, considering a substantial part of it is obscured by dust (at least in the visible) and most of the mass is &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_matter&quot; rel=&quot;noreferrer&quot;&gt;dark matter&lt;/a&gt; and therefore not observed at all (directly at least)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How was the galactic plane established? Primarily what kinds of observations were used, and how were they analyzed to help decide how this plane should be chosen?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-10-12T04:37:05.523" LastActivityDate="2018-10-12T06:23:51.900" Title="How was the galactic plane established?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23936" PostTypeId="1" AcceptedAnswerId="23939" CreationDate="2017-12-12T15:46:24.533" Score="4" ViewCount="123" Body="&lt;p&gt;Sorry if this is a dumb question, I could not find anything naming this in my astronomy books, and Google searches only seemed to show me pages about astrology, [sarc] because naturally, human relationships are influenced by celestial bodies. [/sarc]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking along the lines of a parent-child relationship or a symbiotic relationship, but I couldn't find anything that specifically names this kind of relationship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an official term for this relationship?&lt;/p&gt;&#xA;" OwnerUserId="20264" LastActivityDate="2017-12-13T02:22:04.203" Title="What is the official term for a relationship between a planet and its moons?" Tags="&lt;planet&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23937" PostTypeId="1" CreationDate="2017-12-12T20:36:56.810" Score="9" ViewCount="323" Body="&lt;p&gt;Two days ago I went to a mathematics conference where there was a paper presented on Jupiter's formation via the disk-instability model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that there are two different theories for the formation of the planets. One of them is the &lt;strong&gt;Core accretion&lt;/strong&gt; model and the other one is the &lt;strong&gt;disk-instability&lt;/strong&gt; model. I also tried to read this paper on planetary formation in which they said 161 planets were put on to test. 90% followed Core accretion model and rest of them followed the later model. Did not say anything explicit about Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter is too close to the Sun to follow the Disk-Instability model (I am not sure completely).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to know which model Jupiter follows for its formation.&lt;br&gt;&#xA;(I asked this question on physics stack-exchange too but unfortunately there were no answers)   &lt;/p&gt;&#xA;" OwnerUserId="20267" LastEditorUserId="2365" LastEditDate="2017-12-12T20:58:56.677" LastActivityDate="2017-12-27T06:53:43.867" Title="How was Jupiter formed" Tags="&lt;jupiter&gt;&lt;planetary-formation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23938" PostTypeId="1" CreationDate="2017-12-12T20:49:00.167" Score="5" ViewCount="313" Body="&lt;p&gt;My question is very similar to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/21481/how-can-i-interpolate-ephemerides&quot;&gt;this one&lt;/a&gt; although the answer is not what I was looking for.&#xA;Basically I'm working in a space game (KSP) and I'm making it multiplayer. The clients send a packet with their orbital parameters at a interval (30 miliseconds by default) but between those packets I want to interpolate and draw an orbit between those 2 packets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For normal vectors or speed I use linear interpolation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;private static float Lerp(float v0, float v1, float t)&#xA;{&#xA;    return (1 - t) * v0 + t * v1;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But It doesn't work well when I work with orbits&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;var inclination = Lerp(inclination, Target.inclination, lerpPercentage),&#xA;var eccentricity = Lerp(eccentricity, Target.eccentricity, lerpPercentage),&#xA;var semiMajorAxis = Lerp(semiMajorAxis, Target.semiMajorAxis, lerpPercentage),&#xA;var LAN = Lerp(LAN, Target.LAN, lerpPercentage),&#xA;var argumentOfPeriapsis = Lerp(argumentOfPeriapsis, Target.argumentOfPeriapsis, lerpPercentage),&#xA;var meanAnomalyAtEpoch = Lerp(meanAnomalyAtEpoch, Target.meanAnomalyAtEpoch, lerpPercentage),&#xA;var epoch = Lerp(epoch, Target.epoch, lerpPercentage),&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What function can I use to interpolate them?&lt;/p&gt;&#xA;" OwnerUserId="20266" LastActivityDate="2018-03-05T18:32:29.823" Title="How can I interpolate 2 orbits" Tags="&lt;orbit&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23939" PostTypeId="2" ParentId="23936" CreationDate="2017-12-13T02:22:04.203" Score="3" Body="&lt;p&gt;I don't believe there's an official term but there are a few familiar/common terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Parent Body/Satellite is one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Central Object/Orbiting object or Central Object/Satellite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Star/Planet or Planet/Moon imply one orbits the other.   For a non orbiting Moon or Planet &quot;Rogue&quot; is used before their name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A grey area comes into play when the two objects can be considered binary as opposed to one orbiting the other.   Some people define a binary orbit as one where the barycenter is outside the body of the more massive object, but I don't like that definition because by that definition, Jupiter and the Sun are a binary system and the Sun weighs 1100 Jupiters.   The Sun also has planets that don't meet that criteria, more distant than Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem with the Barycenter outside so it's binary argument is that when a star dies and goes white dwarf or even denser, then pretty much every planet now orbits a barycenter outside the star, but when it was a main sequence star, this wasn't the case.   The orbits didn't change, just the size of the central object.  That shouldn't change the definition of the orbital relationship.   I think the Barycenter argument, while convenient, is highly flawed for that reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pluto/Charon could be called a binary dwarf planet, but because Charon was likely created from a collision on the Body that's now Pluto, I think that's a poor definition and I think Charon should be called a moon or collision moon.   That said, because four tiny moons orbit both Pluto and Charon, calling them a binary-system has some merit too.  There's no neat and tidy way to make a precise dividing line, besides the barycenter, which is neat and tidy, but as I said above, inaccurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asteroids are sometimes found orbiting one another and I rarely hear that defined as a central asteroid with a smaller asteroid orbiting it, it's more often called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Binary_asteroid&quot; rel=&quot;nofollow noreferrer&quot;&gt;binary asteroid&lt;/a&gt;.  I've never heard it called a baby asteroid orbiting it's parent, but again, definitions fall into a grey area if one asteroid is quite a bit larger than the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://what-if.xkcd.com/150/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This XKCD comic&lt;/a&gt; addressed circumbinary orbits and &quot;the other kind&quot;, which doesn't have a name.   Not everything needs a name, and in this case, I don't there is an official one.  &quot;Central body&quot; &quot;Central object&quot; or &quot;Parent&quot; work just fine as familiar and common terms.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-12-13T02:22:04.203" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23940" PostTypeId="1" AcceptedAnswerId="23969" CreationDate="2017-12-13T02:51:11.027" Score="1" ViewCount="2132" Body="&lt;p&gt;Suppose an Earth-like planet orbits a neutron star, though specifically &lt;em&gt;not&lt;/em&gt; a pulsar. Is there any possible orbit, around any possible neutron star, in which the planet could support human life?&lt;/p&gt;&#xA;" OwnerUserId="20214" LastActivityDate="2017-12-14T19:22:27.883" Title="Could life survive on a planet orbiting a neutron star?" Tags="&lt;neutron-star&gt;&lt;life&gt;&lt;habitable-zone&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2017-12-14T21:09:31.953" ContentLicense="CC BY-SA 3.0" />
  <row Id="23942" PostTypeId="2" ParentId="15063" CreationDate="2017-12-13T05:40:35.263" Score="-1" Body="&lt;p&gt;Helmet mounted binoculars is best, but if all you had was your binoculars kneeling is better then standing and prone position is best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/V4Rmf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V4Rmf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ntX6d.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ntX6d.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20246" LastEditorUserId="20246" LastEditDate="2017-12-13T05:58:07.923" LastActivityDate="2017-12-13T05:58:07.923" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23953" PostTypeId="1" CreationDate="2017-12-14T03:25:51.490" Score="1" ViewCount="75" Body="&lt;p&gt;Could someone please tell me if this ratio is important in determining star formation rates? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio is : $$\frac{L_B}{L_{H\alpha}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, could someone please explain the implication of the ratio to me or direct me to a source? &lt;/p&gt;&#xA;" OwnerUserId="20284" LastEditorUserId="2365" LastEditDate="2017-12-14T04:44:42.930" LastActivityDate="2017-12-14T04:44:42.930" Title="Star formation rate question" Tags="&lt;galaxy&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23955" PostTypeId="1" AcceptedAnswerId="23959" CreationDate="2017-12-14T06:22:51.137" Score="8" ViewCount="514" Body="&lt;p&gt;Has the Milky Way ever collided with another galaxy? Or will the collision with Andromeda be our first?&lt;/p&gt;&#xA;" OwnerUserId="16136" LastEditorUserId="13426" LastEditDate="2017-12-14T06:44:31.630" LastActivityDate="2017-12-17T13:05:33.870" Title="Has the Milky Way collided with another galaxy?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23957" PostTypeId="1" AcceptedAnswerId="24391" CreationDate="2017-12-14T07:41:24.320" Score="8" ViewCount="417" Body="&lt;p&gt;So, I know the Moon experiences tidal acceleration from the Earth. And, from what I've read, if not for the fact that the sun would boil away the oceans and engulf both of them first, about 50 billion years from now the Moon would be geosynchronous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was curious whether tidal acceleration could ever cause a satellite to escape, and short answer seems to be, yes, it's possible. Is that accurate? And if so, which factors add to it?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A faster rotating planet also results in tides farther ahead, and has more energy to lose before being totally locked. Very closely related, a farther satellite orbits more slowly, resulting in tides farther ahead, and requires less energy to escape. This one seems obvious: the faster the planet is still rotating while the satellite is farther away, the more likely it escapes.&lt;/li&gt;&#xA;&lt;li&gt;A more fluid planet would experience stronger tides, which would again slow it down faster, but definitely accelerate the satellite faster. A more viscous fluid than water would experience weaker tides, but I think they'd be farther ahead. I think more fluid is definitely more effective, but it seems less obvious than the first point.&lt;/li&gt;&#xA;&lt;li&gt;A larger planet would have its orbit slowed less by tidal acceleration, but have a stronger pull to escape. A larger satellite would cause stronger tides and more tidal force, but would require more force to accelerate, and would slow the planet down more quickly. I really can't tell which has the stronger effect...&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So that's question 1: &#xA;Is it really possible for tidal acceleration to throw a satellite into space before the planet is tidally locked, and if so, does either objects' size impact it, or just fluidity and relative periods?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, I wondered about tidal deceleration, and crashing into the planet. It seems obvious that any satellite orbiting retrograde would never stop decelerating, so all would eventually get torn apart and crash onto their planet. For non-retrogade satellites, an article I read implied that all of them will have the same fate... Which implies that the planet's rotation will always speed up more slowly than the satellite's falling orbit. Is that true? If not, then it seems there must be cases where the planet's rotation catches up, and they again end up tidally locked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that's question 2:&#xA;Will all non-retrogade tidally decelerating satellites eventually crash into their planet, or is it possible for the planet to catch up given the right starting conditions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Haven't really found any more information on this. Is there someone familiar with relevant equations, who knows the answer?&lt;/p&gt;&#xA;" OwnerUserId="20292" LastEditorUserId="20292" LastEditDate="2018-01-03T18:48:36.423" LastActivityDate="2018-01-08T11:23:03.020" Title="Eventual outcome of tidal acceleration and deceleration" Tags="&lt;natural-satellites&gt;&lt;orbital-mechanics&gt;&lt;tidal-forces&gt;&lt;tidal-locking&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23958" PostTypeId="2" ParentId="13105" CreationDate="2017-12-14T08:35:22.977" Score="0" Body="&lt;p&gt;Well, I'm not going to pretend I am right, or what I say is how it is, but I have been the person to ask those same questions before, and heres what I know now. First, ditch the whole 'nothing' component of the big bang. Its easy to think of the big bang as being an event in time, but it isn't, its the t=0 on your time line. Before that would be negative time, and its not that there are not theories about negative time, but for observation's sake we'll leave it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At t=0 all the matter today was present then, there was just no distance between each point, and this is regardless of your position in the universe. It wasn't a distance from some point that was 0, rather all points, pick any two points and their distance measured to be 0. In regard to what the universe is expanding &quot;in to&quot;, if our universe was truly at a point of infinite density, it wouldn't be expanding into anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If its hard to come to grips with space expanding, and the confusion about where this empty space is coming from, then look at it in a different way. How long does it take to go from A to B, if its 0 seconds then distance is 0. But if its greater than 0, you have distance. And if its infinity, you have infinite distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all honesty there are still questions that need to be answered, and while there are many good models of the universe out there, they seem to hold an incomplete picture, evident by the still on going debate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to conclude, there was no before, and there was no nothing either. Matter in the universe has apparently remained constant, and there doesn't seem to be any indication that this would be broken as it has remained that way ever since. &lt;em&gt;I would ignore the time stuff I said for now&lt;/em&gt;, its better for comprehending this rather than the reality of it. Hope this helped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this isn't the best answer in the universe, hopefully its a step for you to find the answer you are looking for.&lt;/p&gt;&#xA;" OwnerUserId="16449" LastEditorUserId="16449" LastEditDate="2018-02-28T00:44:24.157" LastActivityDate="2018-02-28T00:44:24.157" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23959" PostTypeId="2" ParentId="23955" CreationDate="2017-12-14T10:08:37.383" Score="8" Body="&lt;p&gt;According to our current understanding of &lt;a href=&quot;https://en.wikipedia.org/wiki/Physical_cosmology&quot; rel=&quot;noreferrer&quot;&gt;physical cosmology&lt;/a&gt; (as contrasted with cosmologies of various cultural references) the Milky Way must necessarily have had collisions in a process called &lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/H/Hierarchical+Clustering&quot; rel=&quot;noreferrer&quot;&gt;hierarchical clustering&lt;/a&gt;.&lt;br&gt;&#xA;How we're certain of this is the following:   &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;No matter what exactly happened at the big bang, we know the state of the universe at roughly the age of ~380.000 years, where the universe becomes transparent (more or less instantaneous, but let's omit that discussion). At this moment, matter emits light that teaches us the distribution of matter in the young universe.&#xA;This is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;noreferrer&quot;&gt;Cosmic Microwave Background Radiation&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;In the CMB we see that matter is remarkably homogeneous, apart from very small fluctuations, that help us build the cosmological standard model. This means at the time of the CMB emission, the universe had nearly the same matter density everywhere. Which is the contrary of clustered matter blobs (galaxies) that we see today.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CQuFb.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CQuFb.png&quot; alt=&quot;CMB fluctuations&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;em&gt;Fig. 1: CMB temperature fluctuations as seen by WMAP.&lt;br&gt;&#xA;Those variations can be translated into density fluctuations, given a physical model. To our current understanding the form the seeds of galaxy growth.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Now we take those small fluctuations, add dark matter and evolve a model of the universe over time, &lt;a href=&quot;https://en.wikipedia.org/wiki/Millennium_Run&quot; rel=&quot;noreferrer&quot;&gt;in large-scale compuational facilities&lt;/a&gt;. Then we find that galaxies are a natural outcome of the small CMB fluctuations that become gravitationally unstable. Those unstable blobs of matter amplify the initial fluctuations, until whole pockets of matter collapse into dense, rotating objects.&lt;/li&gt;&#xA;&lt;li&gt;Those objects would be dwarf galaxies, and those collide and merge and thus grow into larger, mature galaxies. This is the process of hierarchical clustering, I was referring to above. This process can be simplified to &quot;small things collapse first&quot;, as opposed to &quot;large things collapse first&quot;. In the latter case the Milky Way would not have had collisions before.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summarizing&lt;/strong&gt;&lt;br&gt;&#xA;Yes, our galaxy had collisions before, and they were necessary to grow to it's current size. Scientists are searching for the relics of those collisions in a process called &lt;a href=&quot;http://www.astro.lu.se/~greg/milkyway.html&quot; rel=&quot;noreferrer&quot;&gt;galactic archaeology&lt;/a&gt; (the link goes to a personal website, which I find neatly made).&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2017-12-14T10:08:37.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23963" PostTypeId="1" CreationDate="2017-12-14T14:09:53.310" Score="3" ViewCount="1170" Body="&lt;p&gt;I'm trying to find a suitable CCD and telescope combination. So, I want to make sure that the image circle of the telescope circumscribes the CCD. But I'm unable to find anything that tells me how to calculate the same. &lt;/p&gt;&#xA;" OwnerUserId="20296" LastActivityDate="2017-12-14T23:42:34.043" Title="How do we calculate the image circle of a telescope at prime focus?" Tags="&lt;telescope&gt;&lt;photography&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23966" PostTypeId="1" CreationDate="2017-12-14T15:56:13.073" Score="1" ViewCount="155" Body="&lt;p&gt;What should I improve my telescope kit to see more details of Moon?&#xA;I have the gskyer telescope of 60mm aperture and 350mm focal length.&#xA;my eyepieces are 10mm,25mm. &#xA;and I have x3 Barlow.&#xA;my best viewing of the Moon when I combine the Barlow with 10mm eyepiece.&#xA;I want to see deep details of the moon, by the way, this helps me to zoom in more for the planets like Mars, Jupiter, and Saturn.&#xA;any help?&lt;/p&gt;&#xA;" OwnerUserId="20298" LastActivityDate="2018-01-15T21:27:10.500" Title="What should I improve my telescope kit to see more details of Moon?" Tags="&lt;the-moon&gt;&lt;telescope&gt;&lt;jupiter&gt;&lt;saturn&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23968" PostTypeId="1" CreationDate="2017-12-14T19:00:02.847" Score="1" ViewCount="117" Body="&lt;p&gt;what is the difference between 4mm and 24mm and 30mm eyepieces?&#xA;could I buy an eyepiece from another brand?&#xA;I have the Gskyer telescope.could I buy for Celestron eyepieces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="20298" LastActivityDate="2017-12-14T20:13:00.563" Title="what does it mean of different eyepieces?" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23969" PostTypeId="2" ParentId="23940" CreationDate="2017-12-14T19:22:27.883" Score="2" Body="&lt;p&gt;It is possible that there could be planets where terrestrial-type life could survive orbiting a neutron star, but they are likely extremely rare and transient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The habitable zone where liquid water can persist around a star has a range $\approx 0.7\sqrt{L_*}$ to $1.4\sqrt{L_*}$ AU, where $L_*$ is the luminosity of the star measured in solar luminosities. The exact limits are debatable and depends somewhat on the planet's properties, but this is close enough for this answer. A typical neutron star has a low luminosity, so the habitable zone will be very close and narrow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astro.umd.edu/~miller/nstar.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Neutron stars cool down very rapidly at first&lt;/a&gt;. A young, million-Kelvin neutron star has luminosity around 0.18 solar luminosities, but a 10 million year old neutron star is down to 0.000026, and after a billion years it will be down to a millionth of the sun. This rapid cooling means that the life zone moves inwards from around 0.3-0.6 AU for the young star to 0.003-0.007 for the 10 million year star to 0.0007-0.0014 AU for the old star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that a planet orbiting inside the zone at some time will become too cold before too long unless it is very close to the star since the zone drift slows down over time. It is hence unlikely that life would naturally be able to emerge on a planet like this, since planets that are in the habitable zone early only get a few million years before they freeze. The close planets that get a long period with decent temperature also have a period of extremely hot temperature before that, when most likely all volatiles are boiled away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Given that such a system has experienced a supernova in its past, volatiles will have been reduced even further).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, I don't see a fundamental reason a close planet might not be terraformed with volatiles added from elsewhere. If we consider a planet orbiting a $L_*=10^{-6}L_\odot$ neutron star $10^{-3}$ AU away (that is, 149,597 km from the star) it could maintain water. It would have a period of 16 minutes. There would be some relativistic corrections to the orbit, but they are minor (we are $10^5$ Schwarzschild-radii away).  It would be tidally locked (since tidal locking happens on a timescale $\propto a^{6}$), but currently there are a fair number of atmosphere models suggesting such worlds could remain stable/ Living on the shady side would also prevent the damaging UV and X-rays, which would likely strip away the atmosphere over time. The main problem is that there would not be any geodynamic dynamo to power a protective magnetic field; this is a planet that really needs it.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worth noting that there is a very different system around a neutron star that in principle (but likely not in practice) may be habitable, and that would be far away from an accretion disk. &lt;a href=&quot;https://arxiv.org/pdf/astro-ph/0205175.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;A quiescent accretion disk may have a luminosity about 1/3 of the sun&lt;/a&gt;, but when going active this increases by a factor of 1,000-10,000. So here one could imagine a planet orbiting at distance, getting powered by the accretion disk light. This will likely be too unstable to actually work over long periods and have the same hard radiation problem as the close one, but on paper (or in a suitable work of fiction) it is not impossible.&lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2017-12-14T19:22:27.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23971" PostTypeId="2" ParentId="23968" CreationDate="2017-12-14T20:13:00.563" Score="1" Body="&lt;p&gt;The 4, 24 and 30mm eyepieces give different magnification (equal to the the focal length of the telescope in mm, divided by the number for the eyepiece).&#xA;You can use eyepieces from other brands, but make sure they have the same size barrel (the bit that slides into the telescope).  I don't know the Gskyer brand, but the eyepiece size will be either .965 inch or 1.25 inch.  Provided you get the same size, any other brand will work.  It's probably best to get familiar with what is provided by Gskyer before buying anything else.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will likely find that using lower magnification at first gives the best outcome. &lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-12-14T20:13:00.563" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23972" PostTypeId="1" AcceptedAnswerId="23975" CreationDate="2017-12-14T21:34:39.397" Score="3" ViewCount="98" Body="&lt;p&gt;A 2km comet at 40AU's can dim a sun sized star at 100 light years by 3%, and a 10km comet can cause an eclipse (is that right?) So would 2,3,10 km objects be easy to measure via star dimming if we had enough telescopic data from space? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we measure lines of stars that are eclipsed/dimmed in sequence by transiting comets? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given 1000 frames of a small patch of space, where all the images are recorded into a 3D image stack (a 3D array, a voxel) X,Y and Time as dimensions, then an Intel processor can compare 20 million voxels per second of that collection of images to search for vector information of aligned patterns of star dimming, i.e. searching for vectors of comets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that process is repeated for many patches of space, perhaps we can detect many comets, and I don't know the real maths involved, if a satellite of that design would realistically only find 1 comet, or 5000?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's for a theoretical comet apprehension system which uses 2/3 identical satellites  each comprised 100Tb of SSD and a 20Mflops processors to search for vector information in the images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some kind of fairy idea going around in my head, because of Oumuamua and because I work on voxels and I know that image collections from space can be stacked into 3D arrays and be searched for vector patterns at 2million voxels per second on a 200W PC, an idea that I am confusedly astrally-projecting as a basis to find comets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit- A quantitative answer would be cool, say if the voids between stars on the milky way arm are 8 times the width of the comet, the comet would on average hit 2 stars every 8x8 frames.&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2017-12-16T17:38:15.433" LastActivityDate="2017-12-16T17:38:15.433" Title="Can we measure comets in a similar way to exoplanets?" Tags="&lt;telescope&gt;&lt;exoplanet&gt;&lt;comets&gt;&lt;apparent-magnitude&gt;&lt;oort-cloud&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23973" PostTypeId="2" ParentId="23966" CreationDate="2017-12-14T23:26:45.010" Score="1" Body="&lt;p&gt;The theoretical resolution of a telescope - and the maximum useful magnification that you can use - increases with aperture (That's the diameter of the main mirror or lens, and not the f-ratio (which is what photographers usually mean by aperture)). As a rough guide, the typical everyday maximum magnification is around the same as the aperture in mm (and is achieved with an eyepiece of focal length equal to the scope's f ratio).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given good seeing conditions (steady atmosphere, which can be rare) you can push the magnification to around twice that (with half the eyepiece focal length, or by using a barlow) - which is what manufacturers usually quote as the maximum magnification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As magnification goes up, the brightness of the view goes down, since you're spreading the same amount of light over a larger area. (Increasing aperture increases the light gathering area; doubling the aperture catches four times as much light, so increasing the aperture allows brighter views at the same magnification. This is important for fainter objects, but the moon is bright enough that you don't usually have problems even with small scopes)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the best way to see more detail is to switch to a good quality, larger aperture scope.&lt;/p&gt;&#xA;" OwnerUserId="17016" LastEditorUserId="17016" LastEditDate="2018-01-15T21:27:10.500" LastActivityDate="2018-01-15T21:27:10.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23974" PostTypeId="2" ParentId="23963" CreationDate="2017-12-14T23:42:34.043" Score="2" Body="&lt;p&gt;Image Circle will vary with scope design. Just about all scopes will cover the field of a 1.25&quot; eyepiece barrel, and scopes with 2&quot; focusers will usually cover the field of a 2&quot; barrel (There are some exceptions where the field is bigger than 1.25&quot; but doesn't cover the full 2&quot;). Basically, you need to check the spec for the scope - and in any case the size of the focuser usally sets an upper limit. (With smaller SCTs and Maks, the size of the baffle tube can also limit the image circle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most small refractors for astrophotography will cover an APS sized DSLR sensor.&#xA;If you're looking for something to cover a full frame DSLR sensor, the choice gets more restricted - quite a few of the field flatteners/reducers for smaller refractors are only designed for APS coverage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to go for an unusually large sensor, then your choice is even more restricted - something like Takahashi's FSQ106ED astrograph has an 88mm diameter image circle in native mode, large enough for some medium format sized sensors, or 44mm with the reducer or extender - but there's a high price tag to go with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you're looking at large format astonomical CCDs, they're usually very expensive too. Most of the more affordable astro cameras have smaller sensors, so are less demanding on image circle size.&lt;/p&gt;&#xA;" OwnerUserId="17016" LastActivityDate="2017-12-14T23:42:34.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23975" PostTypeId="2" ParentId="23972" CreationDate="2017-12-15T01:46:42.070" Score="4" Body="&lt;p&gt;The thing about detecting exoplanets by eclipses is that the eclipses are &lt;em&gt;repeatable&lt;/em&gt;.  You have the same star, whose light curve dims in the same way with each eclipse, which gives you information about the planet's orbital period and its apparent size relative to the star.  The Kepler mission observed many fluctuations in stellar brightness that were &lt;em&gt;not&lt;/em&gt; associated with exoplanets, but instead with changes in the instrument, fluctuations in the actual brightness of the star,  probably some other explanations that I can't think of off the top of my head, and actual mysteries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, if you can only observe it once, then you can't claim to explain it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Occultations of distant stars by planets or comets don't give you a light curve: the star goes from &quot;on&quot; to &quot;off&quot;. So a single occultation doesn't give you any information about the size, speed, or direction of the object doing the occulting.  And the occultations are sufficiently rare that I think you'd be extremely unlikely to find a &quot;line&quot; of them by chance --- though that would make a fun data-mining project for an astronomy student.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you &lt;em&gt;can&lt;/em&gt; do, if you have an object that you already know about, is to make many observations of the &lt;em&gt;same&lt;/em&gt; occulatation.  That's the mission of the &lt;a href=&quot;https://occultations.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;International Occultation Timing Association&lt;/a&gt;.  If you click through to their web page you'll see this image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IeTqg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IeTqg.jpg&quot; alt=&quot;Occultations by a binary asteroid&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each line here shows the observed brightness of a particular star observed from a particular location at the same time.  You can see clearly that some observers saw the star disappear very briefly, some for longer, and some (who might have been predicted to be in the very center of the path) saw no occultation at all.  Assembling all these individual observations, properly synchronized, shows clearly the shadows of the &lt;em&gt;two&lt;/em&gt; parts of a binary asteroid, each of order 50 km in diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that 90 Antiope is a main-belt asteroid.  The angular diameter decreases linearly as the distance increases, so an object the same physical size orbiting out by Neptune (perhaps fifteen times farther) would need fifteen times the density of observing locations to create an image with the same detail.&lt;/p&gt;&#xA;" OwnerUserId="1498" LastActivityDate="2017-12-15T01:46:42.070" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23976" PostTypeId="1" AcceptedAnswerId="24017" CreationDate="2017-12-15T03:30:50.297" Score="2" ViewCount="159" Body="&lt;p&gt;Red dwarfs have lifespans of trillions of years and are also the most abundant stars, making up 70% of the stars in the Milky Way. No red dwarf star has ever died and none will die for a very, very long time. Because it take so long for them to die, they're not releasing their hydrogen back into the interstellar medium to make new stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As more massive stars live and die over the course of years, those stars release their hydrogen into the interstellar medium, which can form new stars, 70% of which will be red dwarfs. Does this mean that, billions of years into the future, all of the matter will get locked within red dwarfs until they begin dying out trillions of years later?&lt;/p&gt;&#xA;" OwnerUserId="18831" LastActivityDate="2017-12-19T09:47:53.090" Title="Will all of the gas in the universe be converted into red dwarf stars?" Tags="&lt;star-formation&gt;&lt;red-dwarf&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23977" PostTypeId="5" CreationDate="2017-12-15T03:31:34.767" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2017-12-15T03:31:34.767" LastActivityDate="2017-12-15T03:31:34.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23978" PostTypeId="4" CreationDate="2017-12-15T03:31:34.767" Score="0" Body="Questions about red dwarfs, the smallest and most abundant stars in the universe." OwnerUserId="18831" LastEditorUserId="18831" LastEditDate="2017-12-15T16:01:10.853" LastActivityDate="2017-12-15T16:01:10.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23979" PostTypeId="2" ParentId="23972" CreationDate="2017-12-15T04:30:36.770" Score="1" Body="&lt;p&gt;You should consider how fast a comet at 40 AU would move past a star.   The visible arc of a sun sized star at 100 light years is about the same as a 9 km object at 40 AU.  Objects at 40 AU orbit at about 4.7 km/s, so that would be a 2 second eclipse, if perfectly lined up, and more often than not it would be partially lined up, for even less time.   A 2 second shift in brightness might be too short to get a good measurement on.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related to your question, rogue planets are being surveyed by shading and/or by movement of the background star by gravitational lensing, but those partial shading or lensing require much larger, more massive objects and from our perspective, moving much slower across the sky.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Optical_Gravitational_Lensing_Experiment&quot; rel=&quot;nofollow noreferrer&quot;&gt;See Optical Gravitational Lensing Experiment or OGLE survey&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 2 seconds of shading that a 40 AU object would give you in your scenario (or, for example, a 20 seconds of shading from a 4,000 AU object) would be more difficult to accurately identify.   Mercury, for example, takes 8 hours to travel a distance equivalent to the diameter of the sun.   8 hours of temporary eclipse offers much more easily track-able data, especially given that it repeats on regular intervals every orbit.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-12-15T09:32:53.177" LastActivityDate="2017-12-15T09:32:53.177" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23980" PostTypeId="1" CreationDate="2017-12-15T05:32:16.410" Score="2" ViewCount="155" Body="&lt;p&gt;&lt;strong&gt;update 2:&lt;/strong&gt; A&amp;amp;A: &lt;a href=&quot;https://www.aanda.org/component/article?access=doi&amp;amp;doi=10.1051/0004-6361/202039767&quot; rel=&quot;nofollow noreferrer&quot;&gt;Six transiting planets and a chain of Laplace resonances in TOI-178&lt;/a&gt; (and &lt;a href=&quot;https://arxiv.org/abs/2101.09260&quot; rel=&quot;nofollow noreferrer&quot;&gt;arXiv&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gizmodo: &lt;a href=&quot;https://gizmodo.com/enigmatic-star-system-has-5-planets-locked-in-perfect-h-1846124935&quot; rel=&quot;nofollow noreferrer&quot;&gt;Enigmatic Star System Has 5 Planets Locked in Perfect Harmony&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Bad Astronomy &lt;a href=&quot;https://www.syfy.com/syfywire/a-six-planet-system-dances-in-time-to-the-tune-of-gravity&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Six-Planet System Dances in Time to the Tune of Gravity&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;div class=&quot;youtube-embed&quot;&gt;&lt;div&gt;&#xD;&#xA;                &lt;iframe width=&quot;640px&quot; height=&quot;395px&quot; src=&quot;https://www.youtube.com/embed/-WevvRG9ysY?start=0&quot;&gt;&lt;/iframe&gt;&#xD;&#xA;            &lt;/div&gt;&lt;/div&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;update 1:&lt;/strong&gt; The BBC News article &lt;a href=&quot;http://www.bbc.com/news/science-environment-42644531&quot; rel=&quot;nofollow noreferrer&quot;&gt;Citizen science bags five-planet haul&lt;/a&gt; announces the system K2-138, also described as a &amp;quot;resonant chain&amp;quot; of planets. However, see the actual paper in &lt;a href=&quot;https://arxiv.org/abs/1801.03874&quot; rel=&quot;nofollow noreferrer&quot;&gt;ArXiv&lt;/a&gt; and &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/1538-3881/aa9be0&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ast. J.&lt;/a&gt; where the title calls it only a &amp;quot;near-resonant chain&amp;quot; instead; &lt;em&gt;The K2-138 System: A Near-Resonant Chain of Five Sub-Neptune Planets Discovered by Citizen Scientists&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;The Phys.org news item &lt;a href=&quot;https://phys.org/news/2017-12-discovery-planet-reveals-distant-solar.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Discovery of new planet reveals distant solar system to rival our own&lt;/a&gt; outlines the recent announcement of results using AI to help search &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler_(spacecraft)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kepler&lt;/a&gt; photometric (transit-method) data for exoplanets. Near the end is the paragraph:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Kepler-90i wasn't the only jewel this neural network sifted out. In the Kepler-80 system, they found a sixth planet. This one, the Earth-size Kepler-80g, and four of its neighboring planets form what is called a &amp;quot;resonant chain,&amp;quot; where the planets are locked by their mutual gravity in a rhythmic orbital dance. The result is an extremely stable system, similar to the seven planets in the TRAPPIST-1 system, so precisely balanced that the length of Kepler-80g's year could be predicted with mathematics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;My question is about these &amp;quot;resonant-chains&amp;quot; of planets. If I understand correctly this would be a group of planets where due to mutual perturbative effects they are in orbits who's periods are in mutual rational number ratios, e.g. 3:2 or 7:9.&lt;/p&gt;&#xA;&lt;p&gt;Is it known yet how long-lived (time domain) or narrow (frequency) domain these resonances are, or are likely to be? In other words, are they thought to be really &amp;quot;locked&amp;quot; into these fixed ratios for say tens or even hundreds of millions of orbits, or are the orbits just really close, but with occasional, irregular &amp;quot;slippage&amp;quot; events?&lt;/p&gt;&#xA;&lt;p&gt;Since the span of data and its precision is limited, I would expect that one can not conclude from the data alone, and it's likely some modeling has been done on resonant-chains of planets to improve the understanding of the phenomenon, so I'm asking to understand what's known and believed, rather than a precise answer about a given system. Since the problem is mostly orbital mechanics it lends itself more easily to simulation than most problems.&lt;/p&gt;&#xA;&lt;p&gt;As background, Wikipedia &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_resonance&quot; rel=&quot;nofollow noreferrer&quot;&gt;addresses&lt;/a&gt; &amp;quot;locked&amp;quot; orbital resonances:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Under some circumstances, a resonant system &lt;strong&gt;can be stable and self-correcting&lt;/strong&gt;, so that the bodies remain in resonance. Examples are the 1:2:4 resonance of Jupiter's moons Ganymede, Europa and Io, and the 2:3 resonance between Pluto and Neptune. (emphasis added)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2021-04-21T01:05:06.137" LastActivityDate="2021-04-21T01:05:06.137" Title="Just how &quot;locked&quot; are resonant-chains of exoplanets thought to be? (e.g. K2-138 and TOI-178)" Tags="&lt;exoplanet&gt;&lt;orbital-mechanics&gt;&lt;kepler&gt;&lt;neural-networks&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="23987" PostTypeId="1" AcceptedAnswerId="24003" CreationDate="2017-12-15T13:58:05.913" Score="7" ViewCount="88" Body="&lt;p&gt;I've been listening to an episode of the &lt;a href=&quot;https://www.nasa.gov/ames/nisv-podcast-Kimberly-Ennico-Smith-Michael-Person&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;NASA in Silicon Valley&quot; podcast about the SOFIA mission&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the techniques they use at SOFIA is stellar occultations to study different objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My impression was that the &lt;em&gt;occultations by Pluto happen much more often than by Triton&lt;/em&gt;. Is this true and why is that?&lt;/p&gt;&#xA;" OwnerUserId="17296" LastEditorUserId="37191" LastEditorDisplayName="user21" LastEditDate="2021-02-12T10:17:50.377" LastActivityDate="2021-02-12T10:17:50.377" Title="Occultations of Triton" Tags="&lt;observational-astronomy&gt;&lt;telescope&gt;&lt;occultation&gt;&lt;triton&gt;&lt;sofia&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23988" PostTypeId="2" ParentId="23704" CreationDate="2017-12-15T15:43:45.350" Score="0" Body="&lt;p&gt;Thanks to &lt;a href=&quot;https://astronomy.stackexchange.com/users/21/barrycarter&quot;&gt;barrycarter&lt;/a&gt; I now know that &lt;a href=&quot;https://www.iau.org/public/themes/naming_stars/&quot; rel=&quot;nofollow noreferrer&quot;&gt;International Astronomical Union is very much aware of the need for better names&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Alphanumeric designations are useful for astronomers to officially&#xA;  identify the stars they study, but in many instances, for cases of&#xA;  bright stars, and stars of historical, cultural, or &lt;strong&gt;astrophysical&#xA;  interest&lt;/strong&gt;, it can be more convenient to refer to them by a memorable&#xA;  name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They even organized in 2014 exactly what I thought should be done, just not for the nearest star systems but for &lt;em&gt;305 well-characterized exoplanets discovered prior to 31 December 2008&lt;/em&gt; and their host stars. The campaign was called &lt;a href=&quot;http://nameexoworlds.iau.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;NameExoWorlds&lt;/a&gt; and produced such splendid results like &lt;a href=&quot;https://en.wikipedia.org/wiki/PSR_B1257%2B12&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/55_Cancri&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Mu_Arae&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems we're in neeed for a new campaign - maybe this time called NameStellarNeighbourhood. Especially as we now know that &lt;a href=&quot;https://www.universetoday.com/99309/nearly-all-sun-like-stars-have-planetary-systems/&quot; rel=&quot;nofollow noreferrer&quot;&gt;nearly all sun-like stars have planetary systems&lt;/a&gt;, so being a close neighbour to the Solar System rather than having a planet strikes me as a much better distinction meriting a proper name.&lt;/p&gt;&#xA;" OwnerUserId="19829" LastEditorUserId="19829" LastEditDate="2017-12-16T14:19:59.653" LastActivityDate="2017-12-16T14:19:59.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23990" PostTypeId="1" CreationDate="2017-12-15T21:45:41.370" Score="4" ViewCount="82" Body="&lt;p&gt;By travelling to specific places, &lt;a href=&quot;http://www.asteroidoccultation.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;astronomers can see three thousand star comet transition events every year&lt;/a&gt; at m15 for the major (50km+) comets and asteroids averaging about 20 seconds, perhaps that's 25k transitions if they listed 2 seconds occultations by the same 50km comets, perhaps that's not an exhaustive survey, it may be 5/10/20k?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some easy and fun maths:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Asteroid#/media/File:Asteroids_by_size_and_number.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;There are about 500 &gt;50km comets, and 100k &gt;5km&lt;/a&gt; which would transition visibly at 40AU's, so that's 600,000 total eclipses of &#xA;&#xA;&lt;p&gt;There are 1000,000 &gt;1km comets that cause eclipses of average m20 stars, and there are 100 times more m20 stars than the m15 ones, that's 60mn star dimming events at m20.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, We can perhaps divide that number by 100 (difficult figure to know) for a static view, but you would have to run very fast with your telescope to see 60 million star dimming positions every year, Can we keep the figure at 60mn or is it logical to change that to 0.6?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must have misjudged it by 1-2 orders of magnitude, multiple times less for velocity or something. Can we surmise if it is 60k or 60million at m15 and m20? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a comet of 11km at 40AU's travels through the disk of the galaxy, it can hide stars for at least 0.5 seconds. If the stars are N times distanced than the comet's diameter from a perspective, the comet would hide two stars every NxN frames, so for N = 5, a different star would dim every 25 frames. What is the real maths to say what the figure N is at the disk of the galaxy for something like Halley's Comet at m15 and m20? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;this page sais: &lt;a href=&quot;http://www.asteroidoccultation.com/asteroid_help.htm#hgm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.asteroidoccultation.com/asteroid_help.htm#hgm&lt;/a&gt;&#xA;&lt;em&gt;In reality there are probably hundreds of asteroid occultations every night.  However many cannot be observed for practical reasons (e.g. star too dim, event duration too short, etc.).&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2017-12-16T17:21:16.320" LastActivityDate="2017-12-16T17:21:16.320" Title="How often do comet occultations of stars happen?" Tags="&lt;star&gt;&lt;asteroids&gt;&lt;comets&gt;&lt;meteorite&gt;&lt;occultation&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23992" PostTypeId="1" CreationDate="2017-12-16T02:10:33.840" Score="6" ViewCount="378" Body="&lt;p&gt;At some time in the future it is said our galaxy will collide with Andromeda. Yet at the same time the universe is expanding at a ever-accelerating pace. Using the old raisin-in-the-baking-loaf-of-bread analogy, the raisins grow farther apart and don’t “collide” with each other during the baking process. It seems to be the same way with galaxies and the expanding universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are these two seemingly conflicting postulates reconciled? What am I missing here?&lt;/p&gt;&#xA;" OwnerUserId="10117" LastActivityDate="2017-12-20T23:12:29.040" Title="How is the Milky Way on a collision course with Andromeda when the universe is expanding?" Tags="&lt;milky-way&gt;&lt;expansion&gt;" AnswerCount="3" CommentCount="0" ClosedDate="2017-12-23T15:44:28.373" ContentLicense="CC BY-SA 3.0" />
  <row Id="23994" PostTypeId="2" ParentId="23992" CreationDate="2017-12-16T03:42:44.247" Score="5" Body="&lt;p&gt;I'm not an expert, but I looked into this recently when toying with the idea of a scale model of the observable universe. To the best of my understanding, the following is approximately correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TLDR: The expansion happens on the very large scale of the universe. At the scale of the observable universe the Andromeda galaxy is a next-door neighbour, and at the local scale gravity is strong enough to override any effect of expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Milky Way is 100,000 light years across, 2.5 million light years brings us to the Andromeda galaxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxies cluster in galaxy groups, and 10 million light years span about 50 galaxies and brings us to the edge of what's called the Local Group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxy groups cluster into superclusters. We're part of the &lt;a href=&quot;https://www.space.com/27016-galaxy-supercluster-laniakea-milky-way-home.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Laniakea Supercluster&lt;/a&gt;, which spans about 100,000 galaxies and 500 million light years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any appreciable expansion (redshift 0.1) you need at least a billion light years. It's something that happens in the voids between superclusters, and doesn't become really pronounced until you're halfway across the observable universe (redshift 1, 7-8 billion light years as measured in light travel time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the &quot;raisins in a loaf&quot; analogy, I think it's not too wrong if you think of the raisins as superclusters rather than individual galaxies. Note that although the dough expands in a baking loaf, individual raisins do not, because there are other forces at work that hold them together.&lt;/p&gt;&#xA;" OwnerUserId="9682" LastEditorUserId="9682" LastEditDate="2017-12-16T15:26:23.323" LastActivityDate="2017-12-16T15:26:23.323" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23996" PostTypeId="1" CreationDate="2017-12-16T07:16:15.357" Score="2" ViewCount="114" Body="&lt;p&gt;How small an area on Earth is it possible to identify with any precision when any celestial object (star, planet, comet, nova) (visible with the naked eye alone) is at its zenith directly overhead?&#xA;--Would it be possible to identify an area as small as a town, or village, a square mile, or an acre, or even a building? &#xA;--How big an earthly location would be required to narrow it down if such a celestial object is directly above?&#xA;  Thanks.&#xA;PS:  I've tried to narrow the question down to get a simple to understand answer.  I'm not very technical, nor an astronomer.&lt;/p&gt;&#xA;" OwnerUserId="20324" LastEditorUserId="2365" LastEditDate="2017-12-16T12:03:19.517" LastActivityDate="2018-01-16T02:14:35.273" Title="How small a location is possible to identify when any visible star is at its zenith directly above?" Tags="&lt;amateur-observing&gt;&lt;fundamental-astronomy&gt;&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23997" PostTypeId="1" CreationDate="2017-12-16T10:06:31.720" Score="1" ViewCount="828" Body="&lt;p&gt;how to connect two eyepieces together with the x3 Barlow lens&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 10mm,25mm, and x3 Barlow. Is there any way to connect them together to archive the maximum magnification of 105mm?&lt;/p&gt;&#xA;" OwnerUserId="20298" LastActivityDate="2018-03-13T23:32:08.233" Title="how to connect two eyepieces together with the x3 Barlow lens" Tags="&lt;telescope&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24000" PostTypeId="2" ParentId="23997" CreationDate="2017-12-16T11:47:04.073" Score="1" Body="&lt;p&gt;You will get the maximum magnification by using the 10mm eyepiece with the barlow lens.  You cannot connect 2 eyepieces in the way you suggest, but even if you could add the 10 and the 25 to get a 35mm, that would give lower magnification, as I explained in a previous answer.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-12-16T11:47:04.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24002" PostTypeId="1" CreationDate="2017-12-16T18:27:30.433" Score="4" ViewCount="138" Body="&lt;p&gt;I'm reading about the history and making of ESO's Very Large Telescope and I've found &lt;a href=&quot;https://www.eso.org/sci/publications/messenger/archive/no.50-dec87/messenger-no50-30-32.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; that says the scientific community had a choice between three suggestions: a one-piece 16m-telescope, four 8m-telescopes or 16 4m-telescopes. They chose the second solution (four 8m-telescopes), because&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;there was a clear trend that some form of limited array was the right direction for the VLT.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't understand what &lt;em&gt;limited array&lt;/em&gt; means. Is it just a name they picked up from radio interferometry, or is it some special method of interferometry being used at the VLT? The article says, that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the study group [for the VLT] concentrated its efforts on the scientific advantages one could draw from the limited array concept and particularly on interferometry.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could anyone clear this up for me?&lt;/p&gt;&#xA;" OwnerUserId="18219" LastEditorUserId="1891" LastEditDate="2017-12-18T10:55:58.627" LastActivityDate="2017-12-18T10:55:58.627" Title="What is a &quot;limited array&quot;?" Tags="&lt;history&gt;&lt;interferometry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24003" PostTypeId="2" ParentId="23987" CreationDate="2017-12-16T18:42:50.307" Score="4" Body="&lt;p&gt;Pluto is actually smaller in diameter than Triton, and is also farther away, meaning that Triton covers roughly 1.4x (according to WA) the angle that Pluto does, making occultation that much more probable apriori, ignoring their actual orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to the above, New Horizons recently observed Pluto's atmosphere with far more detail than what we can do from here on Earth, so it makes sense to measure a different target.&lt;/p&gt;&#xA;" OwnerUserId="11222" LastEditorUserId="11222" LastEditDate="2017-12-17T06:27:46.287" LastActivityDate="2017-12-17T06:27:46.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24004" PostTypeId="1" CreationDate="2017-12-16T18:55:29.363" Score="0" ViewCount="51" Body="&lt;p&gt;Guessing, the solar system would be taken out of our galaxy of any reason, would there be any difference to the live on earth or even to the mechanics in within our solar system?&lt;/p&gt;&#xA;" OwnerUserId="20334" LastActivityDate="2017-12-16T19:30:38.450" Title="Would there be any difference for our solarsystem to be not in the Milky Way?" Tags="&lt;galaxy&gt;&lt;solar-system&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2017-12-18T12:16:55.473" ContentLicense="CC BY-SA 3.0" />
  <row Id="24006" PostTypeId="2" ParentId="24004" CreationDate="2017-12-16T19:30:38.450" Score="0" Body="&lt;p&gt;No it wouldn't make a difference unless you moved it near another star or body. If it was moved such that something was close enough to exert significant gravitational attraction it would have an effect on our solar system.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastActivityDate="2017-12-16T19:30:38.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24007" PostTypeId="1" AcceptedAnswerId="24008" CreationDate="2017-12-16T21:43:50.757" Score="2" ViewCount="153" Body="&lt;p&gt;The nearby stars seem homogenously spaced out enough to give a general &quot;stellar density.&quot;&lt;br&gt;&#xA;Does there exist some kind of mathematical expression that can determine the number of stars n light-years from the sun?&lt;br&gt;&#xA;Or—&lt;br&gt;&#xA;Is there a rough value for this supposed stellar density idea? (Ex: 0.3 stars per cubic light-year.)&lt;/p&gt;&#xA;" OwnerUserId="19454" LastActivityDate="2017-12-16T22:07:48.667" Title="Finding the number of stars n light-years from the sun" Tags="&lt;distances&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24008" PostTypeId="2" ParentId="24007" CreationDate="2017-12-16T22:07:48.667" Score="4" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_density&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia, quoting Erik Gregersen&lt;/a&gt;. &lt;em&gt;The Milky Way and beyond.&lt;/em&gt; gives a local value of 0.004 stars per cubic light year, or 0.14 stars per cubic parsec. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So within &lt;em&gt;n&lt;/em&gt; light years you would expect $0.004\times 4\pi n^3/3\approx 0.016n^3$ stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This predicts the nearest star to be about 4 light-years away, consistent with observation.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-12-16T22:07:48.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24009" PostTypeId="2" ParentId="23996" CreationDate="2017-12-16T22:32:55.130" Score="1" Body="&lt;p&gt;Lets assume your observer has a good watch, so she knows the exact UTC time. But lets also assume that she has no special kit, and has no special training. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;She might be able to estimate the location of &quot;directly up&quot; to an accuracy of about one hand width (at arms length) which corresponds to roughly 5 degrees.  To achieve better than this, she would need some tools for sighting the position of stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With an error of +- 5 degrees, the error on the Earth's surface of any calcuation based on her observation is also 5 degrees of the Earth's surface. That is roughly a circle of radius 550km (350 miles).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such an observation could prove you were in Mexico and not Canada. But it couldn't distinguish Conneticut and Washington DC, much less a square mile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get 1 mile accuracy you would need measurements at least 350 times better, This is better than one-sixtieth of a degree. The most skilled observers did achieve this level of accuracy with naked eye equipment, but not from a single observation. It could be done with a telescope designed for measuring the position of stars, provided the operator was sufficiently highly skilled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your observer doesn't have a good watch, she is subject to the problem of Longitude: It is relatively easy to find Latitude just from astronomical observations. It is much harder to find the longitude.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-12-16T22:32:55.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24010" PostTypeId="1" CreationDate="2017-12-16T22:35:04.237" Score="4" ViewCount="1464" Body="&lt;p&gt;Images from SOHO, SDO and other sun observatories are often coloured differently for different wavelengths or temperatures:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/sXxLa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sXxLa.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/WUxuv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WUxuv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.nasa.gov/mission_pages/sunearth/science/Sun-Wavelength-Chart.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.nasa.gov/mission_pages/sunearth/science/Sun-Wavelength-Chart.html&lt;/a&gt; has statements like &quot;This light is emitted from the upper transition region and the chromosphere and are typically colorized in red.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is the convention of which temperature or wavelength gets which colour established?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it simply so that scientists can immediately tell what temperature or wavelength a picture shows, or are the colours optimized to show particular features?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And why is there a white ring or circle in the sunshade in some SOHO imagery, and why different sizes compared to the disc in the two photos above?&lt;/p&gt;&#xA;" OwnerUserId="10776" LastEditorDisplayName="user1569" LastEditDate="2018-09-30T16:54:00.940" LastActivityDate="2020-05-05T21:16:20.337" Title="What's the rationale behind the false colours in solar observation photographs?" Tags="&lt;the-sun&gt;&lt;telescope&gt;&lt;photography&gt;&lt;soho&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24011" PostTypeId="1" CreationDate="2017-12-17T00:16:25.547" Score="2" ViewCount="49" Body="&lt;p&gt;What differentiates the distance modulus equation to find the difference between two apparent magnitudes? For example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ m_1 = -2.5\cdot log_{10}\left(\frac{F_1}{F_2}\right) + m_2$$&#xA;leads to the following, in the U band:&#xA;$$\implies m_U = -2.5\cdot log_{10}\left(F_U\right) + C_U$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the relationship involving taking the sum of the fraction of the flux that a star outputs at each wavelength rather than the ratio between the fluxes of two stars, here in the UV band for example: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ m_U = -2.5\cdot log_{10} \left(\displaystyle\int_0^\infty F_{\lambda}S_U \,\, d\lambda\right) + C_U$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*$S(\lambda)$ = the fraction of the star's flux detected at wavelength lambda.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Succinctly, what is the difference between the above two equations? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, how would this be used in a real-world situation to find the apparent magnitude of an object? My deduction is that apparent magnitude isn't as simple as taking the flux in a certain filter to find the apparent magnitude, and that each individual wavelength must be accounted for. However, with that notion, why can't we just remove the sensitivity function and change the bounds to those wavelengths within the U band? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other option would be that this is just more accurate than taking the flux within the U band (referring to the above example), because you would take smaller intervals, possibly over 1 $\buildrel _\circ \over {\mathrm{A}}$ intervals since infinitesimal ones would be impossible.&lt;/p&gt;&#xA;" OwnerUserId="18673" LastEditorUserId="18673" LastEditDate="2017-12-20T20:26:30.903" LastActivityDate="2017-12-20T20:26:30.903" Title="Finding individual band magnitudes: Why and how is the integration of individual flux values for each wavelength used?" Tags="&lt;telescope&gt;&lt;light&gt;&lt;observational-astronomy&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24012" PostTypeId="1" AcceptedAnswerId="24029" CreationDate="2017-12-17T04:49:47.493" Score="2" ViewCount="233" Body="&lt;p&gt;I am trying to make an accurate illustration of the sky from when Ptolemy was observing the planets in Greece. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to find out when Mars was in retrograde during those years. I have read both SE questions and answer about how to calculate it mathematically but both of those are a little over my head. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the silly astrology sites have the retrograde dates listed back the 1930s, I want to know what were the dates Mars was in retrograde around 129 BC. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/249493/mathematically-calculate-if-a-planet-is-in-retrograde&quot;&gt;https://physics.stackexchange.com/questions/249493/mathematically-calculate-if-a-planet-is-in-retrograde&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/288023/mathematically-calculate-if-a-planet-is-in-apparent-retrograde-motion&quot;&gt;https://physics.stackexchange.com/questions/288023/mathematically-calculate-if-a-planet-is-in-apparent-retrograde-motion&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1582" LastActivityDate="2017-12-18T05:35:48.293" Title="How to calculate when Mars was in retrograde around 129 BCE?" Tags="&lt;mars&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24013" PostTypeId="1" CreationDate="2017-12-17T04:51:03.367" Score="2" ViewCount="117" Body="&lt;p&gt;If we graph the variability of a small bright star which is lined up with the asteroid belt, is it less constant than stars that are far away from the asteroid belt? To what degree does variability change relative to distance away from the most dense areas of the solar system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I added the arrows to illustrate the question.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/NaXKE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NaXKE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2017-12-17T16:06:49.877" LastActivityDate="2017-12-18T04:13:38.537" Title="Are stars beyond the asteroid belt less constant?" Tags="&lt;star&gt;&lt;asteroids&gt;&lt;apparent-magnitude&gt;&lt;variable-star&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="24014" PostTypeId="1" CreationDate="2017-12-17T06:19:59.980" Score="3" ViewCount="236" Body="&lt;p&gt;I have trouble interpreting the following excerpt: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;“Nor could it [the Earth] be cylindrical, with the curved surface in the east-west direction, and the flat sides towards the poles of the universe, which some might suppose more plausible. This is clear from the following: for those living on the curved surface none of the stars would be ever-visible, but either all stars would rise and set for all observers, or the same stars, for an equal [celestial] distance from each of the poles, would always be invisible for all observers.” &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I understand that if the Sun rotated around the curved side of the cylinder, no one living on the curved side would see any ever-visible stars because their view of the celestial views is blocked. However Ptolemy states that “all stars would rise and set for all observers,” which I believe would be wrong because the two poles and the curved surface would be oriented in different directions on a cylinder. He also states that the other possibility is that “the same stars, for an equal [celestial] distance from each of the poles, would always be invisible for all observers.” However, I don’t understand how his explanation in the excerpt directly above would prove that “the same stars… would always be invisible for all observers.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here’s the link to my source (page 41, chapter 5:) &lt;a href=&quot;https://isidore.co/calibre/get/pdf/Ptolemy%26%2339%3Bs%20Almagest%20-%20Ptolemy%2C%20Claudius%20%26amp%3B%20Toomer%2C%20G.%20J__5114.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://isidore.co/calibre/get/pdf/Ptolemy%26%2339%3Bs%20Almagest%20-%20Ptolemy%2C%20Claudius%20%26amp%3B%20Toomer%2C%20G.%20J__5114.pdf&lt;/a&gt;    &lt;/p&gt;&#xA;" OwnerUserId="17321" LastEditorUserId="17321" LastEditDate="2017-12-17T06:45:00.377" LastActivityDate="2017-12-17T07:59:15.710" Title="How did Ptolemy prove in the following excerpt that the Earth is not cylindrical?" Tags="&lt;earth&gt;&lt;history&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24015" PostTypeId="2" ParentId="24014" CreationDate="2017-12-17T07:59:15.710" Score="6" Body="&lt;p&gt;If the Earth were a cylinder, and stationary in the centre of the universe, then the apparent daily motion of the stars must be due to the stars' actual motion in the heavens. The stars he thought to be on a sphere that encircles the Earth, and there is a pole to this sphere, at the Northern pole there is a bright star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Earth were a cylinder, and the stars moved in circles around the axis of the cylinder; the axis of the cylinder pointing directly at the poles of the celestial sphere; then for any observer on the curved surface the pole star would be on the Northern horizon. The Southern pole would be on the southern horizon an all the stars would rise and set. That is clearly not the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ptolemy then considers the possibility of a cylindrical Earth with the axis of the cylinder not pointing towards the pole of the sphere. Then some stars would rise and set and others would remain always below the horizon. Ptolemy (in Egypt) would see the Pole Star at about 30 degrees above the horizon. But he knew that the position of the pole star was not the same in the Northern part of the Empire. There were stars that were sometimes visible in Alexandria that remained always below the Horizon in Rome. In Rome the Pole Star was 40 degrees above the Horizon. This is not consistent with a cylindrical Earth.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-12-17T07:59:15.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24016" PostTypeId="2" ParentId="23955" CreationDate="2017-12-17T13:05:33.870" Score="2" Body="&lt;p&gt;Yes. We can see &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_stellar_streams&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;stellar streams&quot; surround the Milky Way&lt;/a&gt;, long streams of stars from globular clusters and dwarf galaxies that have been disrupted by the galaxy and are now slowly merging with it. &lt;a href=&quot;https://watermark.silverchair.com/278-3-727.pdf?token=AQECAHi208BE49Ooan9kkhW_Ercy7Dm3ZL_9Cf3qfKAc485ysgAAAa4wggGqBgkqhkiG9w0BBwagggGbMIIBlwIBADCCAZAGCSqGSIb3DQEHATAeBglghkgBZQMEAS4wEQQMozKcFAfHUiX6JKIQAgEQgIIBYfvZ4WXrQAC-N9abIpy2SojCXFhSK9IYxIEbjUmiMB8XOazt7Ilb6a-46MUSTfLE7zm7rBJZgWv7ruleigTLoAS4znI2hVALDI5agqn6fsUvCtqy4h8wRTnEz1Y-EwwaRYBcrfj11VU_eervIUIYTDTYrseNHHxWZzs3y9Wg1zz-LwOOqsDUrizo-iznCc8z3-qX4ydrYosINCP8I1Me-dQGmplpc4k-pfZYZBXVHupbf-qVGYlLVyGPKW0qJDi3AV9rXq8d6ZDxU303LKoX9Rc9UdTddXf-0Z_syKxYFDEswrhG_XBsCMopEncKK2uJ4T7_RYc9alRYJL4WInEpyaH4E0qLfSBtFWry8pEnpGm1O-j35OhAMjnCLTe0L-Q5qJylVibVAASFiXYPzzj7GzcGG6MVdk4dq__dL0wv888YzgIwCA1GDROOnWEXLZ6AdvuDZDqKuqWcQO8NT2uPRKLf&quot; rel=&quot;nofollow noreferrer&quot;&gt;There is an upper limit&lt;/a&gt; (based on the amount of blue stars in the halo) of about 60 past such mergers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://iopscience.iop.org/article/10.1086/422734/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;It does not look like the Milky Way formed from a big merger&lt;/a&gt;. Had it done so it would have a much thicker and fuzzier disk (lots of &quot;heating&quot; by having stars in random orbits). The Andromeda collision will turn the galaxies into an elliptical galaxy, which has even less structure. &lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2017-12-17T13:05:33.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24017" PostTypeId="2" ParentId="23976" CreationDate="2017-12-17T13:15:25.907" Score="5" Body="&lt;p&gt;No. The reason is that gas recycling only recycles only about 40-50% of the gas in a sun-like star, leaving the rest as a white dwarf that slowly cools off. Heavier stars are even less effective in converting their gas into a condensed core. So the process will tend to produce a fair fraction of heavier white dwarfs now in the early stages of the stelliferous era, followed by a slow emergence of lighter white dwarfs from red dwarfs in the rest of the era.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/astro-ph/9701131&quot; rel=&quot;noreferrer&quot;&gt;Adams and Laughlin&lt;/a&gt; estimate that the final mass fractions will be brown dwarfs (never fused) 9.7%, white dwarfs 88%, and 2.4% neutron stars and black holes. In the post-stelliferous era some of the brown dwarfs will merge and make a few extra red dwarfs . &lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2017-12-17T13:15:25.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24018" PostTypeId="1" CreationDate="2017-12-17T16:48:10.207" Score="2" ViewCount="190" Body="&lt;p&gt;I have trouble interpreting the following excerpt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;“Against the first of these three positions militate the following arguments. If we imagined [the earth] removed towards the zenith or the nadir of some observer, then, if he were at sphaera recta, he would never experience equinox, since the horizon would always divide the heavens into two unequal parts, one above and one below the earth…”   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For context, when Ptolemy is speaking about “the first of these three positions…” he’s speaking about the position that holds that the Earth is equidistant from both poles of the celestial sphere, but is not on the axis spanning between both poles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He states that he’s arguing against the first proposition, and suggests that if an observer were at sphaera recta (which I’ve taken to mean that the observer is at the equator with direction/horizon being perpendicular to the equator,) then he would never experience equinox because his “horizon would always divide the heavens into two unequal parts, one above and one below the earth…”  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that in this case, while the Earth is displaced from the center of the celestial sphere, the Sun is still orbiting the true center. However, if it was equinox, and the Sun was orbiting on the same plane as the true center. Wouldn’t this observer still experience equinox because the Earth is still on the same plane as the true center?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here’s the link to my source (page 41, chapter 6:) &lt;a href=&quot;https://isidore.co/calibre/get/pdf/Ptolemy%26%2339%3Bs%20Almagest%20-%20Ptolemy%2C%20Claudius%20%26amp%3B%20Toomer%2C%20G.%20J__5114.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://isidore.co/calibre/get/pdf/Ptolemy%26%2339%3Bs%20Almagest%20-%20Ptolemy%2C%20Claudius%20%26amp%3B%20Toomer%2C%20G.%20J__5114.pdf&lt;/a&gt;   &lt;/p&gt;&#xA;" OwnerUserId="17321" LastEditorUserId="17321" LastEditDate="2017-12-18T16:54:13.990" LastActivityDate="2017-12-18T16:54:13.990" Title="How does Ptolemy prove in the following excerpt that the Earth could not be outside of the axis spanning between the poles of the celestial sphere" Tags="&lt;earth&gt;&lt;history&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24019" PostTypeId="1" CreationDate="2017-12-17T18:10:54.390" Score="1" ViewCount="839" Body="&lt;p&gt;Lets say we had the technology to create a mini-Dyson structure, like a shutter style curtain between the Sun and Venus, blocking the sunlight in a controllable manner to cool down it's temperature. &#xA;Could it be the first step to reverse the greenhouse effect and make Venus more &quot;terraformable&quot;? Or at least more explorable like Mars? Would there be any unintended consequences?&#xA;Would this be useful on other hot extra-solar earth-like planets?&lt;/p&gt;&#xA;" OwnerUserId="20343" LastActivityDate="2017-12-18T06:23:47.570" Title="Would blocking sunlight on Venus help make it habitable?" Tags="&lt;venus&gt;&lt;earth-like-planet&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2017-12-18T09:34:58.243" ContentLicense="CC BY-SA 3.0" />
  <row Id="24020" PostTypeId="1" AcceptedAnswerId="24021" CreationDate="2017-12-17T20:06:38.563" Score="3" ViewCount="173" Body="&lt;p&gt;A hypothetical situation which will help me with a real-life problem. &#xA;Several stationary objects are placed at great distances from our Sun -- say 0.1 light year, 0.5 ly, 1.0 ly. Assuming no other influences; &#xA;What velocity would each have when it reached the Sun?&#xA;How long would each take to reach the Sun?&#xA;Assuming the object-Sun path was perpendicular to the ecliptic &amp;amp; the Sun had its actual motion, would the objects hit the Sun, or (roughly) by how much would they miss it?&lt;/p&gt;&#xA;" OwnerUserId="20344" LastEditorUserId="37191" LastEditDate="2021-02-12T20:30:49.047" LastActivityDate="2021-02-12T20:30:49.047" Title="Gravitational attraction of Sun on distant object" Tags="&lt;the-sun&gt;&lt;gravity&gt;&lt;newtonian-gravity&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24021" PostTypeId="2" ParentId="24020" CreationDate="2017-12-17T21:22:49.410" Score="8" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Equations_for_a_falling_body#Examples&quot; rel=&quot;nofollow noreferrer&quot;&gt;full equation&lt;/a&gt; for the time for an object to drop is &#xA;&lt;span class=&quot;math-container&quot;&gt;$$t =  \frac{ \arccos \Big( \sqrt{ \frac{x}{r} }\Big)  + \sqrt{ \frac{x}{r} \ ( 1 - \frac{x}{r} ) } }{ \sqrt{ 2 \mu }  }  \, r^{3/2},$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; is the radius of the sun, &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; is distance of the object, and &lt;span class=&quot;math-container&quot;&gt;$\mu=GM=1.327\times10^{20}$&lt;/span&gt;. (in SI units, so you will need to convert your distance to metres, and get a time in seconds.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &lt;span class=&quot;math-container&quot;&gt;$x&amp;lt;&amp;lt;r$&lt;/span&gt; you can assume &lt;span class=&quot;math-container&quot;&gt;$\frac{x}{r}=0$&lt;/span&gt; so the formula simplifies to &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\pi r^{3/2}}{2\sqrt{2\mu}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an object released at 0.1,0.5 or 1 light year that is 3000000 years, thirty million years or a hundred million years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the speed on impact, you can assume that the body will reach very close to the escape velocity of the sun, which is about 620 km/s. It matters little whether the body is released from 0.1, 0.5 or 1.0 lightyears, because the body will pick up speed only when it approaches the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last part of you question is unclear. If the body is at rest relative to the sun, and the various gravitational affects of other stars are ignored, then it will fall towards the sun. If on the otherhand the body is at rest relative to the centre of the galaxy, then it is moving extremely fast relative to the sun, much faster than the sun's escape velocity. It won't fall anywhere near to the sun, instead the sun will deflect it slightly as it falls towards the centre of the galaxy.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="7982" LastEditDate="2019-07-13T01:10:36.267" LastActivityDate="2019-07-13T01:10:36.267" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24022" PostTypeId="2" ParentId="24019" CreationDate="2017-12-17T23:09:23.130" Score="1" Body="&lt;p&gt;The fundamental problem is the heat capacity of the planet. If you keep it out of sunlight long enough it will eventually cool to the temperature of the background ratiation, bit it will take a &lt;em&gt;long&lt;/em&gt; time. If we just treat it as a blackbody surrounded by a 0 K universe &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;the cooling time will roughly be&lt;/a&gt; $t_{cool}\approx 2.5\cdot 10^{11}$ seconds if we want to reach 300 K. That is about 7900 years. Wait for about 29,000 years and you can start scooping up frozen carbon dioxide. And this assumes good heat conduction and that the interior is not hotter - I think there will be bad geological issues when trying to change the crust temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another issue is of course the total lack of sunlight. Martyn Fogg describes a neat solution in his book &lt;em&gt;Terraforming&lt;/em&gt;: orbit a reflector in a polar orbit, angled at 45 degrees so it reflects sunlight to the planet. This would for the right orbit give a daily cycle despite Venus rotation issues and the shade shield. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether this is a sensible start for terraforming or just evidence that the technology does not help very much depends on your level of optimism and timescales. For exploration you may want a method that does not transform your exploration target as much.&lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2017-12-17T23:09:23.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24023" PostTypeId="1" AcceptedAnswerId="24024" CreationDate="2017-12-18T01:47:23.820" Score="32" ViewCount="5368" Body="&lt;p&gt;Would the temperatures during such a collision be able to ignite nuclear fusion, bringing the dead star back to life? If so, would it only be able to fuse for a short time before running out of fuel, or would it turn into a full-fledged star?&lt;/p&gt;&#xA;" OwnerUserId="18831" LastActivityDate="2019-02-13T02:20:14.863" Title="If two white dwarfs collided, would they become a star?" Tags="&lt;star-formation&gt;&lt;white-dwarf&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24024" PostTypeId="2" ParentId="24023" CreationDate="2017-12-18T02:16:02.940" Score="38" Body="&lt;p&gt;The answer to your question is both yes and no, depending on the circumstances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two white dwarfs colliding would likely yield a &lt;a href=&quot;https://en.wikipedia.org/wiki/Type_Ia_supernova#Double_degenerate_progenitors&quot; rel=&quot;nofollow noreferrer&quot;&gt;Type Ia supernova&lt;/a&gt;, assuming the combined mass exceeded the &lt;a href=&quot;https://en.wikipedia.org/wiki/Chandrasekhar_Limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chandrasekhar limit&lt;/a&gt; (&lt;span class=&quot;math-container&quot;&gt;$\sim1.4$&lt;/span&gt; solar masses). The unstable object resulting from the collision could not be supported by electron degeneracy pressure; when the temperature rises rapidly due to the collision, there is nothing to stop it (compare this to thermal pressure in a &quot;normal&quot; star, which can make the star expand or contract to compensate for changes in temperature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rise in temperature triggers fusion, which then increases the temperature, which triggers more fusion . . . and so on and so on, in a &lt;em&gt;very&lt;/em&gt; short time span. This is the same process, more or less, as in an accretion-powered type Ia supernova. The resulting explosion destroys the object, ejecting matter into space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with normal Type Ia supernovae, this sort of collision would likely happen in a binary system, with two white dwarfs in close orbits losing energy to gravitational waves and spiraling in (the odds of two unrelated white dwarfs colliding is very, very low). I'm not sure how luminous this event would be in the gravitational wave spectrum; likely fainter than a neutron star-neutron star inspiral but still strong. The speeds of the components could be quite fast, meaning that a lot of energy would be released in the collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things are slightly more complicated if the combined mass is less than the Chandrasekhar limit. An example system that is predicted to undergo a collision is &lt;a href=&quot;https://en.wikipedia.org/wiki/SDSS_J0106-1000&quot; rel=&quot;nofollow noreferrer&quot;&gt;SDSS J010657.39-100003.3&lt;/a&gt;. The total mass of the two white dwarfs is &lt;span class=&quot;math-container&quot;&gt;$\sim0.6$&lt;/span&gt; solar masses, certainly below the limit. In roughly 37 million years (&lt;a href=&quot;https://academic.oup.com/mnrasl/article/413/1/L101/1747379&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kilic et al. (2011)&lt;/a&gt;), they will collide and form a &lt;a href=&quot;https://en.wikipedia.org/wiki/Subdwarf&quot; rel=&quot;nofollow noreferrer&quot;&gt;subdwarf&lt;/a&gt; - fusing helium, interestingly enough (compare this to the carbon/oxygen fusion in Type Ia supernova, which in those circumstances is unstable). Similar scenarios could play out in other systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sort of a star would not be a &quot;normal&quot; star in the sense that it wouldn't truly be on the main sequence - just a bit below it. It would be an O- or B- type subdwarf, less luminous than main sequence stars of the same spectral type. I saw a couple of articles on these objects claiming that they would be stars just like the Sun - it's not true! Given that white dwarfs are highly depleted in hydrogen - remember, they are the degenerate cores of old stars - there can be virtually no hydrogen burning, even in a shell. Core helium burning is the only option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a final note, one explanation for the formation of &lt;a href=&quot;https://en.wikipedia.org/wiki/R_Coronae_Borealis_variable&quot; rel=&quot;nofollow noreferrer&quot;&gt;R Coronae Borealis variables&lt;/a&gt; is the collision of two white dwarfs. I don't think we have solid evidence yet for preferring this over other models (e.g. something involving hydrogen loss from the outermost shell), but if it &lt;em&gt;is&lt;/em&gt; the correct pathway, then the answer to your question could be a definitive yes. Note that this would require intermediate-mass white dwarf progenitors.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2019-02-13T02:20:14.863" LastActivityDate="2019-02-13T02:20:14.863" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="24025" PostTypeId="2" ParentId="24013" CreationDate="2017-12-18T03:01:36.550" Score="3" Body="&lt;p&gt;No, stars that are seen through the asteroid belt are not any more variable than stars in any other part of the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All stars vary in brightness on measurable timescales, although some have been measured to have such small variations that they are recorded as &quot;standard&quot; stars and for most purposes can be considered to have a constant brightness and thus be used as reference for other brightness measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are referring to, as noted, are occultations. There are many hundreds of thousands of known asteroids in our Solar System, and billions of stars in our galaxy, and so there could be hundreds of predicted occultation events every day, and there are many people interested in occultation events. For one thing, we can calculate the distance and speed of an asteroid quite accurately but we can't directly photograph it to determine its size. But when an asteroid passes in front of a star, we can measure the duration of the occultation and set precise limits on its size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for all the asteroids in the Main Belt, the space between them is huge and for any one star the chances of an occultation are quite (vanishingly) small, so the stars seen through the asteroid belt are not any more variable than anywhere else in the sky.&lt;/p&gt;&#xA;" OwnerUserId="18643" LastEditorUserId="18643" LastEditDate="2017-12-18T04:13:38.537" LastActivityDate="2017-12-18T04:13:38.537" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24026" PostTypeId="2" ParentId="24020" CreationDate="2017-12-18T04:36:10.067" Score="3" Body="&lt;p&gt;James K gave a good answer to this, but I just want to add that if the Sun was unmoving relative to the center of the Milky-way, it would fall towards the center along with your object.   The sun orbits within the Milky way with tangential velocity about 230 km/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's rather pointless, but we can compare mass and distance and find the sweet spot where the gravitational pulls are equal on your object to the sun and towards the center of the Milky-way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the center of the Milky-way, where, the center is everything that draws your object towards it, not just the big black hole at the center, precise numbers are impossible, but a rough &lt;a href=&quot;https://www.universetoday.com/22790/mass-of-the-milky-way/&quot; rel=&quot;nofollow noreferrer&quot;&gt;estimate of 200 billion solar masses&lt;/a&gt; can be used.  That might be low, but close enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun is roughly 26,000 light years from the center.   Using the inverse square rule, the distance would be the square root of 200 billion, or about 450,000, and divide the distance (26,000 light years) with that number and you get about 1/17th of 1 light year.    That's the sweet spot where the two gravitational tugs would be equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This demonstrates the uselessness of gravitational strength at astronomical distances.   Take the Moon for example.   It's under greater gravitational tug from the Sun than the Earth, but it still orbits the Earth (or, you could say it orbits both), but stronger gravity doesn't govern the hill sphere.   It's fun for calculating &quot;how long would it take to hit the surface if I dropped it from . . . 1 light year&quot;, but beyond the occasional fun calculation, there's no real practical use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 3 bodies the equation gets much more exotic, especially if one is orbiting the other and you add a 3rd and you want to know which way it would fall.  The truth is, it would fall towards both, slowly, but if it got close enough to the sun it could get a gravitational kick, sending it further away for a time.   Total energy is always conserved, but calculating the movement of that 1 smaller object with both the sun and a uniform massive center is a more complicated.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2017-12-19T02:44:28.023" LastActivityDate="2017-12-19T02:44:28.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24029" PostTypeId="2" ParentId="24012" CreationDate="2017-12-18T05:16:55.580" Score="3" Body="&lt;p&gt;Mars' retrograde motion happens in the summer of 129 BC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Jet_Propulsion_Laboratory_Development_Ephemeris&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL Ephemerides&lt;/a&gt; are an excellent source of solar system predictions, and are widely used. DE422 is about 600 MB and covers your period of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since your profile shows that you can program a computer, I've used the Python package &lt;a href=&quot;http://rhodesmill.org/skyfield/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Skyfield&lt;/a&gt; to download it, do the interpolation and calculate the apparent positions of Mars and the Sun from Alexandria in 129 BC. If I understand correctly, this would be year -128 in UTC, since there is &lt;a href=&quot;https://en.wikipedia.org/wiki/Year_zero&quot; rel=&quot;nofollow noreferrer&quot;&gt;no year zero&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also get similar data online using the &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL Horizons&lt;/a&gt; web interface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll leave the problem of checking when the Sun was below the horizon but Mars was above the horizon in Alexandria on each day to you, but I've included the altitude methods in the script.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars is red, Sun is yellow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mkWiF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mkWiF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;from skyfield.api import Loader, Topos&#xA;&#xA;load = Loader('~/Documents/SkyData')  # avoids multiple copies of large files&#xA;ts   = load.timescale()&#xA;&#xA;data    = load('de422.bsp')&#xA;earth   = data['earth']&#xA;mars    = data['mars']&#xA;sun     = data['sun']&#xA;&#xA;ts      = load.timescale()&#xA;&#xA;Alexandria  = earth + Topos(latitude_degrees  = 31.2,&#xA;                            longitude_degrees = 29.9,&#xA;                            elevation_m       = 5.0)&#xA;&#xA;# NOTE! The year 129 BC is year -128 in UTC I believe&#xA;# SEE   https://en.wikipedia.org/wiki/Year_zero&#xA;&#xA;# Check when Mars is &quot;up&quot; and Sun is &quot;down&quot; on a given day&#xA;&#xA;hours    = np.arange(0, 24, 0.1)&#xA;one_day  = ts.utc(-128, 7, 1, hours)  # July 1st&#xA;&#xA;mars_obs = Alexandria.at(one_day).observe(mars)&#xA;sun_obs  = Alexandria.at(one_day).observe(sun)&#xA;&#xA;mars_alt, mars_az, mars_dist = mars_obs.apparent().altaz()&#xA;sun_alt,  sun_az,  sun_dist  = sun_obs.apparent().altaz()&#xA;&#xA;# Check RA and DEC in 129 BC&#xA;&#xA;days     = np.arange(1, 366)&#xA;one_year = ts.utc(-128, 1, days)&#xA;&#xA;mars_obs = Alexandria.at(one_year).observe(mars)&#xA;sun_obs  = Alexandria.at(one_year).observe(sun)&#xA;&#xA;mars_RA, mars_Dec, mars_dist = mars_obs.apparent().radec()&#xA;sun_RA,  sun_Dec,  sun_dist  = sun_obs.apparent().radec()&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;&#xA;    plt.subplot(3, 1, 1)&#xA;&#xA;    marsalt, sunalt = mars_alt.degrees, sun_alt.degrees&#xA;&#xA;    plt.plot(hours, marsalt, '-r')&#xA;    plt.plot(hours, sunalt,  '-y')&#xA;    plt.plot(hours, np.zeros_like(hours), '-k')&#xA;    plt.title('altitude of Mars and Sun from Alexandria, July 1, 129 BC')&#xA;    plt.xlabel('hours (UTC)')&#xA;    plt.ylabel('altitude (degs)')&#xA;    plt.ylim(-10, None)&#xA;    plt.xlim(0, 24)&#xA;&#xA;    marsra, marsdec = mars_RA._degrees, mars_Dec._degrees&#xA;    sunra,  sundec  = sun_RA._degrees,  sun_Dec._degrees&#xA;&#xA;    # get rid of ugly lines&#xA;    dsunra, dmarsra = sunra[1:] - sunra[:-1], marsra[1:] - marsra[:-1]&#xA;    sunra, marsra = sunra[:-1], marsra[:-1]&#xA;    sunra[dsunra&amp;lt;-100]   = np.nan&#xA;    marsra[dmarsra&amp;lt;-100] = np.nan&#xA;&#xA;    plt.subplot(3, 1, 2)&#xA;    plt.plot(days, mars_RA._degrees, '-r')&#xA;    plt.plot(days, sun_RA._degrees,  '-y')&#xA;    plt.title('RA of Mars and Sun from Alexandria, year of 129 BC')&#xA;    plt.xlabel('days')&#xA;    plt.ylabel('RA (degs)')&#xA;    plt.xlim(0, 365)&#xA;&#xA;    plt.subplot(3, 1, 3)&#xA;    plt.plot(mars_RA._degrees, mars_Dec.degrees, '-r')&#xA;    plt.plot(sun_RA._degrees,  sun_Dec.degrees,  '-y')&#xA;    plt.plot([0, 360],  [0, 0], '-k')&#xA;    plt.title('RA vs Dec of Mars and Sun from Alexandria, year of 129 BC')&#xA;    plt.xlabel('RA (degs)')&#xA;    plt.ylabel('Dec (degs)')&#xA;    plt.xlim(0, 360)&#xA;&#xA;    plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-12-18T05:35:48.293" LastActivityDate="2017-12-18T05:35:48.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24031" PostTypeId="2" ParentId="24019" CreationDate="2017-12-18T05:48:05.377" Score="1" Body="&lt;p&gt;The Venus, like Earth, has 3 major heat sources or losses:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Radiative heating from inside&lt;/li&gt;&#xA;&lt;li&gt;Solar heating&lt;/li&gt;&#xA;&lt;li&gt;Loss of heat due to thermal radiation to the space&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(1) is about the same as of the Earth - the planets were created from the same &lt;a href=&quot;https://en.wikipedia.org/wiki/Protoplanetary_disk&quot; rel=&quot;nofollow noreferrer&quot;&gt;protoplanetary disk&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) is significantly stronger, about the double of the Earth's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surprisingly, also (3) is the same as the Earth - the upper cloud layer what we can see (= which is enough high to be not covered by anything above) has about the same temperature as the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the Venusian surface, there is very hot ($\approx 450 {}^\circ C$), but it is the result of the greenhouse effect and not of (1). The internal radiative heating is small, &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth%27s_internal_heat_budget#Global_internal_heat_flow&quot; rel=&quot;nofollow noreferrer&quot;&gt;around $70-100 \frac{mW}{m^2}$ on the Earth&lt;/a&gt; and there is no reason to think that it would be on the Venus significantly different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On an average night, the surface temperature loses 5-15 ${}^\circ C$ due to (3). It depends mainly on how cloudy is the sky. This happens only on the surface - some meters below the surface, there is no daily temperature fluctuation any more. The soil is a very good heat insulator, if it is enough thick, which is well visible also on that how small is (1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if some space macroengineering project could cover the Venus with a large sunshade, then the Venusian temperature would decrease to a new equilibric state between (1), (2) and (3). This could be significantly smaller, as its current temperature, depending mainly on that, how big percent of the Venusian solar radiation is covered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time scale of the transition to the new equilibric temperature balance would happen in weeks, at most in months. It wouldn't have any effect to the deeper layers of the Venusian soil, thus &lt;em&gt;minig on the Venus&lt;/em&gt; would be still problematic, but covering it with an enough large foil, could reach even a mean Earth-like temperature, too.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There is another major problems. The largest is that the Venus practically doesn't rotate (more exactly, its rotation period is around a half year), thus an Earth-like mean temperature would mean a very hot sun-side and a very cold dark one. However, its tight athmosphere works as a strong heat balancer - on the current Venus, practically there is no difference between the day and night sides. Although a 90 atm CO2 + 4 atm N2 athmosphere isn't very humanfriendly, we would probably need to compansate it with spacesuits filled with 94 atm He + 0.2 atm O2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solar soil could be best positioned on the L1 Lagrange-point between the Venus and the Sun:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hFrl7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hFrl7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;This picture show the Earth-Sun Lagrange-points, but the Sun-Venusian Lagrange points are very similar&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun-Venus L1 is around 2million km away from the Venus, and it is unstable for point-like mass points. However, in the case of a not point-like body (a planetary-sized foil isn't point-like), or by manipulating the &lt;a href=&quot;https://en.wikipedia.org/wiki/Radiation_pressure&quot; rel=&quot;nofollow noreferrer&quot;&gt;radiation pressure&lt;/a&gt; of the Sun, could be made statically stable, or could be stabilized dynamically.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="1891" LastEditDate="2017-12-18T06:23:47.570" LastActivityDate="2017-12-18T06:23:47.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24038" PostTypeId="1" AcceptedAnswerId="24442" CreationDate="2017-12-18T10:32:25.083" Score="4" ViewCount="177" Body="&lt;p&gt;In radio astronomy, many common objects are refered to by their designation from the 3rd Cambridge Catalogue of Radio Sources, e.g. &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=3C%20353&quot; rel=&quot;nofollow noreferrer&quot;&gt;3C 353&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For hobbyist purposes, I would like to create a catalogue for Cartes du Ciel that shows the location of the 3C objects. However, the catalogue exists in three versions, 3C from 1959 being the original, and &lt;a href=&quot;http://simbad.harvard.edu/simbad/sim-ref?querymethod=bib&amp;amp;simbo=on&amp;amp;submit=submit%20bibcode&amp;amp;bibcode=1959MmRAS..68...37E&quot; rel=&quot;nofollow noreferrer&quot;&gt;3CR&lt;/a&gt; (1962) and &lt;a href=&quot;http://3crr.extragalactic.info/&quot; rel=&quot;nofollow noreferrer&quot;&gt;3CRR&lt;/a&gt; (1983) updating the original. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mrao.cam.ac.uk/facilities/surveys/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Later surveys&lt;/a&gt; at different frequency bands used similar designations, e.g. 4C, 5C and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the databases linked above, however, it seems that both only list the updates to the original catalogue (e.g. 3C 353 is listed in neither of them). Is there a digitized, comprehensive catalogue of all 3C objects? Or, even better, is there an archive where I could combine entries from several C surveys based on, say, 21cm flux?&lt;/p&gt;&#xA;" OwnerUserId="20361" LastActivityDate="2019-05-24T01:15:09.013" Title="Is there a complete edition of the Cambridge Catalogue of Radio Sources (esp. 3C) available digitally?" Tags="&lt;radio-astronomy&gt;&lt;star-catalogues&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24040" PostTypeId="1" CreationDate="2017-12-18T11:09:51.730" Score="4" ViewCount="215" Body="&lt;p&gt;I'm looking at this IAU WGCCRE Report: &lt;a href=&quot;http://aa.usno.navy.mil/publications/reports/Archinaletal2011a.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://aa.usno.navy.mil/publications/reports/Archinaletal2011a.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It gives expressions for Earth orientation relative to the International Celestial Reference Frame in terms of right ascension and declination of the north pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numbers say that the declination of the Earth rotation axis starts from 90 degrees and decreases by 0.557 degrees every century:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;δ0 = 90.00 − 0.557T&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which means that in about 16000 years (and 16000 years ago) Earth rotation axis will lay in ecliptic plane. It doesn't make any sense to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because this model should be applied on short time scales only? I didn't find any information on the time range for which the model is adequate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: my question is not about how Earth actually rotates, but about the model and how it should be used.&lt;/p&gt;&#xA;" OwnerUserId="20363" LastEditorUserId="20363" LastEditDate="2017-12-18T13:02:00.873" LastActivityDate="2017-12-19T09:52:14.383" Title="IAU rotation model" Tags="&lt;rotation&gt;&lt;iau&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24041" PostTypeId="2" ParentId="21833" CreationDate="2017-12-18T11:37:58.967" Score="12" Body="&lt;p&gt;I am a member of &lt;a href=&quot;https://astropeiler.de&quot; rel=&quot;noreferrer&quot;&gt;Astropeiler Stockert e.V.&lt;/a&gt;, and we are fortunate enough to be able to approach this problem coming from the &quot;large side&quot; :-) We have a 25m, 10m and 3m telescopes as well as an interferometer made from two 1m satellite dishes available. All these dishes can be used to do interesting things, but you'll need to match the instrument to your target (and, in a hobbyist setting, often also the target to your instrument).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, you should think about what frequency band you want to work on. 21cm (1420 MHz) is the classical hydrogen line, which lends itself well for mapping neutral hydrogen in the spiral arms of our galaxy. You can expect comparatively strong signals in a quiet frequency band there. Lower frequencies (and large bandwidths) are interesting for pulsars, higher frequencies give access to more interesting phenomena but will require a lot of work on the high frequency side. So my recommendation would be to start out with 21cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, what to look for? In general, the following areas of observation are easily accessible for amateurs:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;continuum emission of the Milky Way, perhaps up to creating maps&lt;/li&gt;&#xA;&lt;li&gt;spectral measurements of selected, &quot;bright&quot; (as in &quot;intense radio sources&quot;) objects&lt;/li&gt;&#xA;&lt;li&gt;Pulsars&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let's look at them in detail:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Continuum measurements&lt;/em&gt; are easily done using dishes of any size, but I'd start from 3m upwards to get more interesting results. Calibration of the backend receiver is not trivial, and you should plan a bit of time for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Spectral measurements&lt;/em&gt; also benefit from larger collecting areas, but you'll also need to set up a suitable backend. They are, however, possible from 3m upwards if you are happy with a bit of mapping of intra-galactic velocities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Pulsars&lt;/em&gt; require a lot of collecting area and bandwidth, so they are mostly the domain of large instruments. We are currently &lt;a href=&quot;https://www.eucara.nl/wp/wp-content/uploads/2016/04/05-Pulsare_V2.pdf&quot; rel=&quot;noreferrer&quot;&gt;observing&lt;/a&gt; several dozens of the brightest pulsars on our 25m dish. The 10m might be sufficient for some of the very brightest. And while there is a &lt;a href=&quot;https://www.eucara.nl/wp/wp-content/uploads/2016/04/03-AmateurPulsaDetect.pdf&quot; rel=&quot;noreferrer&quot;&gt;report of pulsar measurements using a 3m dish and an RTLSDR frontend&lt;/a&gt;, this is a feat that requires experience and dedication. So, I'd recommend this only for 8m and above dishes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Interferometry&lt;/em&gt; with 2x1m dishes (20 GHz) is rather interesting, but very involved in terms of analysis. Two receiving systems help mitigate local fluctuations, and with that setup, we can observe sources down to 2 Jy (with a looong integration time). This setup will give you access to one or two handful of interesting targets like M1, W51 or Cyg A. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I'd like to recommend the &lt;a href=&quot;https://www.eucara.nl/wp/&quot; rel=&quot;noreferrer&quot;&gt;EUCARA&lt;/a&gt; conference &lt;a href=&quot;https://astropeiler.de/european-conference-amateur-radio-astronomy-eucara&quot; rel=&quot;noreferrer&quot;&gt;series&lt;/a&gt; (European Conference on Amateur Radio Astronomy) and the &lt;a href=&quot;http://www.radio-astronomy.org/&quot; rel=&quot;noreferrer&quot;&gt;SARA&lt;/a&gt; group as great starting points. They have conference presentations available online that show what other amateurs are doing. &lt;/p&gt;&#xA;" OwnerUserId="20361" LastActivityDate="2017-12-18T11:37:58.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24043" PostTypeId="2" ParentId="24040" CreationDate="2017-12-18T12:52:48.037" Score="1" Body="&lt;p&gt;The values &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha_0 = 0.00 - 0.641 T$$&#xA;$$\delta_0 = 90.00 - 0.557 T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;provide a &lt;em&gt;first-order estimate&lt;/em&gt; of the movement of the right ascension ($\alpha_0$) and declination ($\delta_0$) of the direction of the Earth's north pole for short periods of time after the epoch, expressed to three digits of precision only. $T$ represents the time after epoch expressed in units of Julian centuries (36525 days). The centuries unit was chosen to make the values of order unity, &lt;em&gt;not because&lt;/em&gt; the values are meant to be used to extrapolate for centuries. You should be thinking of using this for a period of years, &lt;em&gt;until the next update is published&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because the motion sets off in that direction, but will move in approximately a circle about the pole of the ecliptic, and in about 26,000 years return to roughly the same direction. Watch &lt;a href=&quot;https://www.youtube.com/watch?v=qlVgEoZDjok&quot; rel=&quot;nofollow noreferrer&quot;&gt;Steven Senders' snappy video&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This movie was created with Blender and is used in the Spitz Fulldome Curriculum for the SciDome planetariums around the world. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Precession is the wobble of the Earth which makes the poles shift position over ~26,000 years as well as the position of the Vernal and Autumnal equinoxes in the sky. This movie illustrates the movement. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Hopefully this movie gives a little more meaning to the Dawning of the Age of Aquarius. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://www.spitzinc.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.spitzinc.com/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Earth's axis &lt;a href=&quot;https://en.wikipedia.org/wiki/Nutation&quot; rel=&quot;nofollow noreferrer&quot;&gt;precesses&lt;/a&gt; around the ecliptic pole due to gravitational interaction between the Earth's tilted equatorial bulge and (mostly) the Sun. There is also a faster and much smaller amplitude &lt;a href=&quot;https://en.wikipedia.org/wiki/Precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;nutation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; Earth's Rotation, Precession and Nutation. From &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Praezession.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CC8UU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CC8UU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; Screenshots from Steven Senders' video &lt;a href=&quot;https://www.youtube.com/watch?v=qlVgEoZDjok&quot; rel=&quot;nofollow noreferrer&quot;&gt;Precession of the earth&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8XS9R.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8XS9R.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EVRS7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EVRS7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GzNYR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GzNYR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-12-19T09:52:14.383" LastActivityDate="2017-12-19T09:52:14.383" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="24044" PostTypeId="2" ParentId="21104" CreationDate="2017-12-18T13:27:30.600" Score="2" Body="&lt;p&gt;To understand the basics of algorithms like that, I find it's often best to look at what amateur practicioners are writing (as opposed to diving directly into scientific papers). The description of the &lt;a href=&quot;http://neutronstar.joataman.net/technical/epoch_folding.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;folding algorithm&lt;/a&gt; at joataman.net is thorough and pragmatic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A great in-depth book on the topic is Lorimer/Kramer: &lt;a href=&quot;http://www.jb.man.ac.uk/pulsarhandbook/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Handbook of Pulsar Astronomy&lt;/a&gt;, Oxford University Press.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the actual work, I would suggest the &lt;a href=&quot;http://sigproc.sourceforge.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;sigproc&lt;/a&gt; package by Duncan Lorimer. Its home is at Sourceforge, but there are forks all over the place because the original software is a bit quirky when run on modern Linuxes.&lt;/p&gt;&#xA;" OwnerUserId="20361" LastActivityDate="2017-12-18T13:27:30.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24045" PostTypeId="2" ParentId="16237" CreationDate="2017-12-18T13:49:04.233" Score="2" Body="&lt;p&gt;I am not a professional astronomer, so take this answer with a grain of salt, but just from visiting a few facilities, I know some.&lt;/p&gt;&#xA;&lt;p&gt;For single dish applications, the 100m telescope in Effelsberg, Germany, uses a &lt;a href=&quot;http://www3.mpifr-bonn.mpg.de/div/electronic/content/receivers/21cm/report.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;7-beam receiver&lt;/a&gt; -- fewer than the Parkes array you mention, and I dont't know of any single dish setup with a larger number of beams.&lt;/p&gt;&#xA;&lt;p&gt;Regarding arrays, the Westerbork Synthesis Array in the Netherlands uses the &lt;a href=&quot;https://www.astron.nl/astronomy-group/apertif/apertif&quot; rel=&quot;nofollow noreferrer&quot;&gt;APERTIF&lt;/a&gt; arrays in most of its individual 25m dishes. With 121 elements per array, this seems to be on the higher-end side of granularity.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;update:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;From astron.nl/dailyimage for 31-01-2017 &lt;a href=&quot;https://www.rug.nl/fse/news/first-image-with-apertif_-a-new-life-for-the-westerbork-radio-telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;First image with Apertif: a new life for the Westerbork radio telescope&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The first images made with the upgraded telescope that demonstrate this new 'wide-angle' capability is shown here. The first image shows the dwarf galaxy Leo T. The image is colour-coded and shows the gas (in blue) in this galaxy together with many distant radio galaxies in the background shown in orange. For comparison, the field of view of the previous Westerbork system and the size of the full moon are also indicated.&lt;/p&gt;&#xA;&lt;p&gt;To make this new capability possible, ASTRON developed and built the hardware in-house. 121 small receivers are used in each telescope whose signals are combined electronically to produce the large field of view.&lt;/p&gt;&#xA;&lt;p&gt;The upgraded telescope will also be used to search for and study new variable sources in the radio sky. With the new Apertif receivers, observations of large parts of the sky can be done much faster, and projects that used to be impossible as they would take tens of years can now be done in a much shorter time. Westerbork is therefore poised to make many new discoveries in the radio sky.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mYrBX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mYrBX.jpg&quot; alt=&quot;Apertif first image&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20361" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-04-27T06:39:40.933" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24046" PostTypeId="2" ParentId="14525" CreationDate="2017-12-18T14:07:12.163" Score="2" Body="&lt;p&gt;There are two parts to answering this, both of which have already been touched in previous answers. I happen to be a member of a volunteer organisation that brought an old telescope back into operation, and looking through old documents, we could pretty well reconstruct what the discipline's pioneers did when facing this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mechanics&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to start out building &lt;em&gt;good mechanics&lt;/em&gt;: very solid foundations and precise drive systems so that there is no &quot;wobble&quot; in any dimension, no slack between individual drive gear teeth and all movements are continuous and smooth. This was discussed in Hobbes' answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Measuring equipment&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, you need &lt;em&gt;precise angular measuring systems&lt;/em&gt; for the position of your instrument. You also need to understand how your telescope and receiver mount move and bend when being tilted towards the horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pointing&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But those two combined will only give you a machine-centric view of the direction of the telescope. As discussed in Markus Roellig's answer, you will then start out with creating a pointing model for your telescope; measuring the position of known so-called calibration sources and comparing the measured positions with the positions from literature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The beginning&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how did the authors of radio source catalogues come up with the positions of the reference sources? Well, &quot;our&quot; telescope started out with a small cabin with an optical telescope mounted within the steel structure of the dish. Initially, it was pointed at local structures on the ground that were equipped with RF transmitters to perform an alignment of the optical telescope and the radio telescope's beam axis. Later, the setup could be used to identify optical counterparts to bright radio sources in the sky. From these beginnings, you could iteratively build more comprehensive maps and catalogues.&lt;/p&gt;&#xA;" OwnerUserId="20361" LastActivityDate="2017-12-18T14:07:12.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24048" PostTypeId="1" CreationDate="2017-12-18T16:00:14.760" Score="24" ViewCount="1209" Body="&lt;p&gt;&lt;strong&gt;I'm an artist (and science enthusiast) and I've been trying to find a comprehensive resource that would help me clearly identify likely sky colors (as perceived by human vision) for exoplanets that have atmospheres with a similar chemical composition to Earth.&lt;/strong&gt; It's trickier than I expected. I've cobbled together several resources, and built what I hope is a decently accurate chart of apparent sun color along with sky color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE: I have revised this post and added a new, improved sky chart (below). Science references for sky chart: &lt;a href=&quot;http://www.xenology.info/Xeno/5.4.2.htm&quot; rel=&quot;noreferrer&quot;&gt;Ref A&lt;/a&gt;, &lt;a href=&quot;http://homepages.wmich.edu/~korista/colors_of_the_sky-Bohren_Fraser.pdf&quot; rel=&quot;noreferrer&quot;&gt;Ref B&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/slxc3.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/slxc3.png&quot; alt=&quot;New Sky Chart&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How accurate is my chart? Is this a fair representation of sky/sun on alien worlds with heavily Nitrogen/Oxygen atmospheres? In what ways could I improve it?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chart is not meant to account for things like dust, the look of the sky at sunrise/sunset, or other atmospheric effects. This is meant to be a boilerplate for the baseline look of the sky during the day. However if you'd like to comment on how the sky may change during things like sunset, or with effects like volcanism, I'm game!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I'm more interested in relative color, than getting absolute color 100% right. The image was built in a vector program in RGB mode, so assuming your screen is calibrated in the standard way, we are probably seeing almost the same thing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the key ideas I've gleamed from my reading that I'm using to build this. I consider most of them tentatively held, and am very open to input:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Earth-like atmospheres would tend to be light to dark blue at the zenith, due to the scattering of low-wavelength blue light. If the planet's sun was very hot, the sky would look a deeper blue, while cooler stars would give the sky a lighter blue to almost white look. When the sun gets to 3000k and below, the sky starts to take on an orange/brown tinge.&lt;/li&gt;&#xA;&lt;li&gt;Like on earth, the horizon is the lightest in color and the zenith the deepest.&lt;/li&gt;&#xA;&lt;li&gt;Denser atmospheres would appear brighter (more washed out) and the primary color in the spectrum more &quot;pure&quot; (I'm unsure what the term &quot;pure&quot; means exactly when it comes to optical perception...would it look whiter?). Likewise, thinner atmospheres would be less bright than earth's and the colors more &quot;pure.&quot;&lt;/li&gt;&#xA;&lt;li&gt;With increasing pressure the sky color at the zenith becomes increasingly yellow. In my image this means that an earth-like sky at 10x earth pressure would appear blueish/green near the zenith.&lt;/li&gt;&#xA;&lt;li&gt;At lower temperatures I'm assuming the sun would appear tinged by the color listed under &quot;star temperature.&quot; Otherwise you'll probably only see the star's color when it's near the horizon. &lt;/li&gt;&#xA;&lt;li&gt;I'm assuming when you get down into K and M class suns, the surroundings on the planet would take on a progressively redder tinge due to the decreasing prevalence of blue wavelength light.&lt;/li&gt;&#xA;&lt;li&gt;I suspect the gradient of color from horizon to zenith will be steeper/gentler in some atmospheres. I've guessed that the gradient would be more apparent on a high G world (on the right).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;References: &#xA;&lt;a href=&quot;http://homepages.wmich.edu/~korista/colors_of_the_sky-Bohren_Fraser.pdf&quot; rel=&quot;noreferrer&quot;&gt;Reference #1&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.orionsarm.com/page/321&quot; rel=&quot;noreferrer&quot;&gt;Reference #2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Original Chart:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nIfgG.png&quot; height=&quot;300&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="20372" LastEditorUserId="20372" LastEditDate="2017-12-21T02:00:37.217" LastActivityDate="2021-05-15T18:22:47.287" Title="Terrestrial Exoplanet Skies – I've Built a Visual Sky Chart. Is it Accurate?" Tags="&lt;exoplanet&gt;&lt;planetary-atmosphere&gt;&lt;terrestrial-planets&gt;&lt;weather&gt;&lt;atmospheric-effects&gt;" AnswerCount="3" CommentCount="17" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24050" PostTypeId="1" CreationDate="2017-12-19T01:36:54.277" Score="2" ViewCount="50" Body="&lt;p&gt;A while ago I saw a report on TV, where an astronomer described a device that was used for detecting movement of stellar objects (mainly asteroids, I think) before the age of computers. In order to detect such a movement, two photos of the region of interest were made, through a telescope, with a certain delay between these photos. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Afterwards, the photos have been placed into that device. The device had some sort of &quot;peephole&quot; through which one could see one of the photos, and some sort of lever/switch to quickly toggle between the photos - either by flipping them, or by tilting a mirror to show either of the photos. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objects that had moved between the photos (relative to the other, &quot;fixed&quot; objects) could thus quickly and easily be identified with the naked eye, because they appeared as dots jumping forth and back when toggling between the photos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know the name of this device, or a more detailed technical description?&lt;/p&gt;&#xA;" OwnerUserId="20379" LastActivityDate="2017-12-19T01:36:54.277" Title="What is the name of a device for detecting moving objects?" Tags="&lt;telescope&gt;&lt;observational-astronomy&gt;&lt;movement&gt;" AnswerCount="0" CommentCount="7" ClosedDate="2017-12-28T23:17:09.747" ContentLicense="CC BY-SA 3.0" />
  <row Id="24051" PostTypeId="1" CreationDate="2017-12-19T02:24:54.517" Score="3" ViewCount="219" Body="&lt;p&gt;I understand that the hour angle and declination can be abbreviated as (HA Decl), and rectascension and declination as (RA Decl). The hour angle of a star is of course zero hours at the meridian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun however does have solar time 12 hours at the meridian, and the &quot;solar&quot; hour angle would better be 12 hours. Therefore I look for an abbreviation for use in computer programming for the pair (solar time, declination) or (sun hour angle, declination), and I can't make up them. I didn't anywhere find them naturally, say in Wikipedia or through Google.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I considered (SOL Decl), (SHA Decl), (ST Decl) or the german (SZ Decl). I don't want to use (HA Decl) for the sun, because HA is zero at the meridian, but as already said for the sun its hour angle should be 12 at the meridian and therefore for clarity and for programming purposes, another designation is needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My angles are local, there is no need for a compulsary prefix L- (local) or G- (Greenwich).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that in english usage is &quot;apparent&quot; sun hour angle, but in german it would be &quot;true&quot; sun hour angle, therefore the word element &quot;apparent&quot; should be absent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is not about the formulas proper for hour angle or solar time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="20380" LastEditorUserId="1891" LastEditDate="2017-12-20T10:41:25.743" LastActivityDate="2017-12-20T10:53:56.950" Title="What is the abbreviation for solar time, hour angle of the sun?" Tags="&lt;the-sun&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24053" PostTypeId="2" ParentId="23976" CreationDate="2017-12-19T09:47:53.090" Score="2" Body="&lt;p&gt;Not necessarily. In order for a gas to collapse it needs to reach &lt;em&gt;Jeans instability&lt;/em&gt;, which is determined by parameters on the radius of the nebulae and on its mass. It is a strongly local condition, and it doesn't necessarily determine which kinds of stars can come out of that unstable nebula, that could be determined &lt;em&gt;after&lt;/em&gt; the already collapsing nebula fragments in different fractions of matter, each one of those could continue collapsing (if they are Jeans-unstable). If the fragments &lt;em&gt;do&lt;/em&gt; collapse, the star that will be formed from that gas can be of basically any type of star, from red dwarf to blue luminous main sequence star, all depending on that fragment's mass. Obviously, this is true for basically every protostar in any given galaxy. Hence, will all new stars in a remote time from now be red dwarfs? &lt;strong&gt;&lt;em&gt;not necessarily, it heavily depends on the forming nebulae&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="19871" LastActivityDate="2017-12-19T09:47:53.090" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24058" PostTypeId="2" ParentId="24010" CreationDate="2017-12-19T11:24:10.003" Score="6" Body="&lt;p&gt;Some of the images shown in your question are &lt;strong&gt;not&lt;/strong&gt; &lt;em&gt;colored&lt;/em&gt; or colorized, it is not a false color; it's the real color. The calculations that are made to produce the false colors that are used on colorized image datasets are shown in the &lt;a href=&quot;http://www.heliodocs.com/php/xdoc_print.php?file=&amp;#36;SSW/sdo/aia/idl/pubrel/aia_lct.pro&quot; rel=&quot;nofollow noreferrer&quot;&gt;aia_lct.pro&lt;/a&gt; file, where the exact RGB values are provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where a false color is chosen it is done so (for SDO and some other images) using the AIA Standard Color Tables (which are generated in the aia_lct.pro file). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Information on the &quot;&lt;a href=&quot;http://helio.cfa.harvard.edu/trace/SSXG/ynsu/Ji/SDO_Data_Analysis_guide.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;VSO's Guide to SDO Data Analysis&lt;/a&gt;&quot; (&lt;strong&gt;.PDF&lt;/strong&gt;), on page 46, describes use of the subroutines in the &lt;a href=&quot;http://www.heliodocs.com/php/xdoc_print.php?file=$SSW/sdo/aia/idl/calibration/aia_colors.pro&quot; rel=&quot;nofollow noreferrer&quot;&gt;sdo/aia/idl/calibration/aia_colors.pro&lt;/a&gt; source code file from &lt;a href=&quot;http://www.heliodocs.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.heliodocs.com/&lt;/a&gt; or view &quot;&lt;a href=&quot;http://www.lmsal.com/solarsoft/sswdoc/solarsoft/ssw_install_howto.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Installing Solarsoft&lt;/a&gt;&quot; for ready to run versions and more info.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Section 6.6.1 &quot;&lt;a href=&quot;https://www.lmsal.com/sdodocs/doc/dcur/SDOD0060.zip/zip/entry/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Using AIA Standard Scaling and Colors&lt;/a&gt;&quot; for additional information about the aia_colors.pro function. The &lt;a href=&quot;https://www.lmsal.com/sdodocs/doc/dcur/SDOD0060.zip/zip/entry/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Introduction&lt;/a&gt; of that Guide contains additional information and links about the SDO mission.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is the &lt;a href=&quot;https://sdo.gsfc.nasa.gov/data/aiahmi/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Description of the colors NASA uses&lt;/a&gt; for different wavelengths (using the AIA Standard Color Tables):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;view-source:&lt;a href=&quot;https://sdo.gsfc.nasa.gov/data/aiahmi/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://sdo.gsfc.nasa.gov/data/aiahmi/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt;h4&amp;gt;Telescopes&amp;lt;/h4&amp;gt;&#xA;&amp;lt;select name=&quot;telescope&quot; id=&quot;telescope&quot; class=&quot;form-control&quot;&amp;gt;&#xA;    &amp;lt;option value=&quot;aia_0171&quot;&amp;gt;AIA 171 (gold)&amp;lt;/option&amp;gt;&#xA;    &amp;lt;option value=&quot;aia_0193&quot;&amp;gt;AIA 193 (bronze)&amp;lt;/option&amp;gt;&#xA;    &amp;lt;option value=&quot;aia_0304&quot;&amp;gt;AIA 304 (red)&amp;lt;/option&amp;gt;&#xA;    &amp;lt;option value=&quot;aia_0211&quot;&amp;gt;AIA 211 (purple)&amp;lt;/option&amp;gt;&#xA;    &amp;lt;option value=&quot;aia_0131&quot;&amp;gt;AIA 131 (teal)&amp;lt;/option&amp;gt;&#xA;    &amp;lt;option value=&quot;aia_0335&quot;&amp;gt;AIA 335 (blue)&amp;lt;/option&amp;gt;&#xA;    &amp;lt;option value=&quot;aia_0094&quot;&amp;gt;AIA 094 (green)&amp;lt;/option&amp;gt;&#xA;    &amp;lt;option value=&quot;aia_1600&quot;&amp;gt;AIA 1600 (yellow/green)&amp;lt;/option&amp;gt;&#xA;    &amp;lt;option value=&quot;aia_1700&quot;&amp;gt;AIA 1700 (pink)&amp;lt;/option&amp;gt;&#xA;    &amp;lt;option value=&quot;hmib&quot;&amp;gt;HMI Magnetogram (gray)&amp;lt;/option&amp;gt;&#xA;    &amp;lt;option value=&quot;hmibc&quot;&amp;gt;HMI Colorized Magnetogram (colorized)&amp;lt;/option&amp;gt;&#xA;    &amp;lt;option value=&quot;hmii&quot;&amp;gt;HMI Intensitygram (gray)&amp;lt;/option&amp;gt;&#xA;    &amp;lt;option value=&quot;hmiic&quot;&amp;gt;HMI Intensitygram (orange)&amp;lt;/option&amp;gt;&#xA;    &amp;lt;option value=&quot;hmiihi&quot;&amp;gt;HMI Intensitygram High Cadence (orange)&amp;lt;/option&amp;gt;&#xA;    &amp;lt;option value=&quot;hmiif&quot;&amp;gt;HMI Intensitygram Flat (orange)&amp;lt;/option&amp;gt;&#xA;    &amp;lt;option value=&quot;hmid&quot;&amp;gt;HMI Dopplergram (gray)&amp;lt;/option&amp;gt;&#xA;    &amp;lt;option value=&quot;COMP211193171&quot;&amp;gt;AIA Composite 211, 193, 171&amp;lt;/option&amp;gt;&#xA;    &amp;lt;option value=&quot;COMP304211171&quot;&amp;gt;AIA Composite 304, 211, 171&amp;lt;/option&amp;gt;&#xA;    &amp;lt;option value=&quot;COMP094335193&quot;&amp;gt;AIA Composite 094, 335, 193&amp;lt;/option&amp;gt;&#xA;    &amp;lt;option value=&quot;COMPHMI171&quot;&amp;gt;Composite AIA 171, HMI Magnetogram&amp;lt;/option&amp;gt;&#xA;&amp;lt;/select&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;https://sdo.gsfc.nasa.gov/data/&quot; rel=&quot;nofollow noreferrer&quot;&gt;webpage showing the images in your question&lt;/a&gt; in various sizes, including a 48 hour movie for each.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The full Electromagnetic Spectrum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3djch.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3djch.jpg&quot; alt=&quot;Full Spectrum - from Radio Waves to Gamma Rays&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Visible Spectrum, including &lt;a href=&quot;https://en.wikipedia.org/wiki/Fraunhofer_lines&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fraunhofer Line Letters and Wavelength Numbers&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MAwNz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MAwNz.png&quot; alt=&quot;Fraunhofer Lines&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Fraunhofer lines are typical spectral absorption lines. Absorption lines are dark lines, narrow regions of decreased intensity, that are the result of photons being absorbed as light passes from the source to the detector. In the Sun, Fraunhofer lines are a result of gas in the photosphere, the outer region of the sun. The photosphere gas is colder than the inner regions and absorbs light emitted from those regions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the wavelength's frequency for each Element of interest:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IDAEa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IDAEa.png&quot; alt=&quot;Fraunhofer Lines and Wavelengths&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Each Filter accepts a narrow band which has a specific use and specific properties. Filters are more expensive when the bandwidth is very narrow, single frequency Filters are unnecessary and unaffordable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different frequency ranges of the spectrum are used to resolve different aspects of the object being observed, similarly different bandwidths are used to further enhance those features.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Listing the bandwidths for every filter and the expected results would be exhaustive, one website that gives such info for each of their filters is Daystar where their &lt;a href=&quot;http://www.daystarfilters.com/WavelengthStudy.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;filter comparison page&lt;/a&gt; indeed confirms that the Sun can be viewed in black and white using a monochrome sensor. It's when you want to combine multiple channels that use of color becomes necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example of the differences between bandwidths for the &lt;a href=&quot;https://en.wikipedia.org/wiki/H-alpha&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hydrogen Alpha&lt;/a&gt; Filter (a deep-red visible spectral line in the Balmer series with a wavelength of 656.28 nm):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Filters with 0.8 ångström bandwidths will reveal prominences in high contrast&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Filters with 0.7 ångström bandwidths will reveal prominences in high contrast and occasionally, surface texture. Prominences are generally larger with .7Å filters than in narrower bandwidths&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Filters with 0.6 ångström bandwidths will reveal some surface contrast as well as prominences.  A .6Å filter can be a good compromise for those having a difficult decision&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Filters with 0.5 ångström bandwidths will reveal better surface contrast as well as great prominences&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Filters with 0.4 ångström bandwidths will reveal excellent surface contrast as well as fine chromosphere detail&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Filters with 0.3 ångström bandwidths will reveal superior surface contrast above any other filter. Views are generally somewhat dimmer.  Expect &quot;Pencil-Thin&quot; details on surface and prominence features&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is bandwidth that is used to derive contrast, not coloring.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The SDO (Solar Dynamics Observatory) is a sun-pointing semi-autonomous spacecraft that allows nearly continuous observations of the Sun. &lt;strong&gt;Unlike the human eye&lt;/strong&gt; which uses rods and cones, the SDO uses 3 instruments: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The AIA (Atmospheric Imaging Assembly) images the solar atmosphere in multiple wavelengths to link changes in the surface to interior changes. Data includes images of the Sun in 10 wavelengths every 10 seconds. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The EVE (Extreme Ultraviolet Variability Experiment) measures the Sun's extreme ultraviolet irradiance with improved spectral resolution, &quot;temporal cadence&quot;, accuracy, and precision over preceding measurements made by TIMED SEE, SOHO, and SORCE XPS. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The HMI (Helioseismic and Magnetic Imager) studies solar variability and characterizes the Sun's interior and the various components of magnetic activity. HMI produces data to determine the interior sources and mechanisms of solar variability and how the physical processes inside the Sun are related to surface magnetic field and activity. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;a href=&quot;https://www.jpl.nasa.gov/spaceimages/details.php?id=PIA00256&quot; rel=&quot;nofollow noreferrer&quot;&gt;Magellan craft used Synthetic Aperture Radar&lt;/a&gt; uses radio waves and not visible light to &lt;em&gt;image&lt;/em&gt; map objects. Magellan synthetic aperture radar mosaics from the first cycle of Magellan mapping are mapped onto a rectangular latitude-longitude grid to create this image. Data gaps are filled with Pioneer Venus Orbiter altimetric data, or a constant mid-range value. Simulated color is used to enhance small-scale structure. The simulated hues are based on color images recorded by the Soviet Venera 13 and 14 spacecraft. The image was produced by the Solar System Visualization project and the Magellan science team at the JPL Multimission Image Processing Laboratory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly SDO data outside the visible spectrum is colorized to allow invisible portions of the electromagnetic spectrum (and Magnetometer recordings) to be visualized by the human eye in a readily digestible image rather than serving the general public a large dataset of numbers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The visible spectrum is the portion of the electromagnetic spectrum that is visible to the human eye. Electromagnetic radiation in this range of wavelengths is called visible light or simply light. A typical human eye will respond to wavelengths from about &lt;a href=&quot;https://en.wikipedia.org/wiki/Visible_spectrum&quot; rel=&quot;nofollow noreferrer&quot;&gt;390 to 700 nm&lt;/a&gt;. In terms of frequency, this corresponds to a band in the vicinity of 430–770 THz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spectrum does not, however, contain all the colors that the human eyes and brain can distinguish. Unsaturated colors such as pink, or purple variations such as magenta, are absent, for example, because they can be made only by a mix of multiple wavelengths. Colors containing only one wavelength are also called pure colors or spectral colors.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Q: How is the convention of which temperature or wavelength gets which colour established?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Is it simply so that scientists can immediately tell what temperature or wavelength a picture shows, or are the colours optimized to show particular features?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The colors are not optimized to show particular features or provide contrast. The aia_lct.pro file is where the exact RGB values are calculated, see top of this answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Q: And why is there a white ring or circle in the sunshade in some SOHO imagery, and why different sizes compared to the disc in the two photos above?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The coronagraph blocks out the Sun with an occulting disk to reveal the much fainter structures in the Sun's atmosphere. Sometimes &lt;a href=&quot;https://sdo.gsfc.nasa.gov/gallery/main/item/26&quot; rel=&quot;nofollow noreferrer&quot;&gt;multiple images are combined to remove the disk&lt;/a&gt; and provide a unoccluded image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Large Angle and Spectrometric Coronagraph (&lt;a href=&quot;https://en.wikipedia.org/wiki/Large_Angle_and_Spectrometric_Coronagraph&quot; rel=&quot;nofollow noreferrer&quot;&gt;LASCO&lt;/a&gt;) in the Solar and Heliospheric Observatory (&lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_and_Heliospheric_Observatory&quot; rel=&quot;nofollow noreferrer&quot;&gt;SOHO&lt;/a&gt;) doesn't use the latest technology, sometimes there are &lt;a href=&quot;https://en.wikipedia.org/wiki/Large_Angle_and_Spectrometric_Coronagraph#Disturbances&quot; rel=&quot;nofollow noreferrer&quot;&gt;disturbances&lt;/a&gt; in the images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two kinds of disturbances that repeatedly occur:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Blackouts and Whiteouts, in broken lines, circle-like shapes, or over the whole picture. They are caused by the electronics box. There has never been a firmware update, since it was judged as too sensitive changing the flight-software.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Black and white pixels, occurring in patterns, without pattern or alone. Those &quot;missing blocks&quot; are telemetry dropouts, caused by radio interference or a disturbance in the data transfer to Goddard Space Flight Center.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;[The section below might benefit from some more editing if the OP requests it.]&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;AIA 193&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This channel highlights the outer atmosphere of the Sun - called the corona - as well as hot flare plasma. Hot active regions, solar flares, and coronal mass ejections will appear bright here. The dark areas - called coronal holes - are places where very little radiation is emitted, yet are the main source of solar wind particles.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Where: Corona and hot flare plasma&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Wavelength: 193 angstroms (0.0000000193 m) = Extreme Ultraviolet&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Primary ions seen: 11 times ionized iron (Fe XII)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Characteristic temperature: 1.25 million K (2.25 million F)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;AIA 304&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This channel is especially good at showing areas where cooler dense plumes of plasma (filaments and prominences) are located above the visible surface of the Sun. Many of these features either can't be seen or appear as dark lines in the other channels. The bright areas show places where the plasma has a high density.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Where: Upper chromosphere and lower transition region&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Wavelength: 304 angstroms (0.0000000304 m) = Extreme Ultraviolet&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Primary ions seen: singly ionized helium (He II)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Characteristic temperature: 50,000 K (90,000 F)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;AIA 171&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This channel is especially good at showing coronal loops - the arcs extending off of the Sun where plasma moves along magnetic field lines. The brightest spots seen here are locations where the magnetic field near the surface is exceptionally strong.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Where: Quiet corona and upper transition region&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Wavelength: 171 angstroms (0.0000000171 m) = Extreme Ultraviolet&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Primary ions seen: 8 times ionized iron (Fe IX)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Characteristic temperature: 1 million K (1.8 million F)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;AIA 211&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This channel (as well as AIA 335) highlights the active region of the outer atmosphere of the Sun - the corona. Active regions, solar flares, and coronal mass ejections will appear bright here. The dark areas - called coronal holes - are places where very little radiation is emitted, yet are the main source of solar wind particles.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Where: Active regions of the corona&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Wavelength: 211 angstroms (0.0000000211 m) = Extreme Ultraviolet&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Primary ions seen: 13 times ionized iron (Fe XIV)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Characteristic temperature: 2 million K (3.6 million F)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;AIA 131&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This channel (as well as AIA 094) is designed to study solar flares. It measures extremely hot temperatures around 10 million K (18 million F), as well as cool plasmas around 400,000 K (720,000 F). It can take images every 2 seconds (instead of 10) in a reduced field of view in order to look at flares in more detail.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Where: Flaring regions of the corona&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Wavelength: 131 angstroms (0.0000000131 m) = Extreme Ultraviolet&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Primary ions seen: 20 and 7 times ionized iron (Fe VIII, Fe XXI)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Characteristic temperatures: 10 million K (18 million F)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;AIA 335&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This channel (as well as AIA 211) highlights the active region of the outer atmosphere of the Sun - the corona. Active regions, solar flares, and coronal mass ejections will appear bright here. The dark areas - or coronal holes - are places where very little radiation is emitted, yet are the main source of solar wind particles.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Where: Active regions of the corona&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Wavelength: 335 angstroms (0.0000000335 m) = Extreme Ultraviolet&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Primary ions seen: 15 times ionized iron (Fe XVI)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Characteristic temperature: 2.8 million K (5 million F)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;AIA 094&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This channel (as well as AIA 131) is designed to study solar flares. It measures extremely hot temperatures around 6 million Kelvin (10.8 million F). It can take images every 2 seconds (instead of 10) in a reduced field of view in order to look at flares in more detail.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Where: Flaring regions of the corona&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Wavelength: 94 angstroms (0.0000000094 m) = Extreme Ultraviolet/soft X-rays&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Primary ions seen: 17 times ionized iron (Fe XVIII)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Characteristic temperature: 6 million K (10.8 million F)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;AIA 1600&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This channel (as well as AIA 1700) often shows a web-like pattern of bright areas that highlight places where bundles of magnetic fields lines are concentrated. However, small areas with a lot of field lines will appear black, usually near sunspots and active regions.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Where: Transition region and upper photosphere&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Wavelength: 1600 angstroms (0.00000016 m) = Far Ultraviolet&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Primary ions seen: thrice ionized carbon (C IV) and Continuum&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Characteristic temperatures: 6,000 K (11,000 F), and 100,000 K (180,000 F)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;AIA 1700&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This channel (as well as AIA 1600) often shows a web-like pattern of bright areas that highlight places where bundles of magnetic fields lines are concentrated. However, small areas with a lot of field lines will appear black, usually near sunspots and active regions.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Where: Temperature minimum and photosphere&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Wavelength: 1700 angstroms (0.00000017 m) = Far Ultraviolet&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Primary ions seen: Continuum&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Characteristic temperature: 6,000 K (11,000 F)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;AIA Composite 211, 193, 171&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This image combines three images with different, but very similar, temperatures. The colors are assigned differently than in the single images. Here AIA 211 is red, AIA 193 is green, and AIA 171 is blue. Each highlights a different part of the corona.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;AIA Composite 304, 211, 171&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This image combines three images with quite different temperatures. The colors are assigned differently than in the single images. Here AIA 304 is red (showing the chromosphere), AIA 211 is green (corona), and AIA 171 is dark blue (corona).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;AIA Composite 094, 335, 193&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This image combines three images with different temperatures. Each image is assigned a color, and they are not the same used in the single images. Here AIA 094 is red, AIA 335 is green, and AIA 193 is blue. Each highlights a different part of the corona.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[Webpage had no Info Popup for: AIA 171 &amp;amp; HMIB]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;HMI Magnetogram&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This image comes from HMI, another instrument on SDO. It shows the magnetic field directions near the surface of the Sun. White and black areas indicate opposite magnetic polarities, with white showing north (outward) polarity and black showing south (inward) polarity.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Where: Photosphere&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Wavelength: 6173 angstroms (0.0000006173 m) = Visible (orange)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Primary ions seen: Neutral iron (Fe I)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Characteristic temperature: 6,000 K (11,000 F)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;HMI Colorized Magnetogram&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This image comes from HMI, another instrument on SDO. It shows the magnetic field directions near the surface of the Sun. White and black areas indicate opposite magnetic polarities, with white showing north (outward) polarity and black showing south (inward) polarity.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Where: Photosphere&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Wavelength: 6173 angstroms (0.0000006173 m) = Visible (orange)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Primary ions seen: Neutral iron (Fe I)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Characteristic temperature: 6,000 K (11,000 F)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[Webpage had no Info Popup for: HMI Intensitygram - colored]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[Webpage had no Info Popup for: HMI Intensitygram - Flattened]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[Webpage had no Info Popup for: HMI Intensitygram]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[Webpage had no Info Popup for: HMI Dopplergram]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[Webpage had no Info Popup for: HMI Intensitygram - High Cadence]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;EVE/ESP and EVE/MEGS-P&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This plot shows time series of 5 strong EUV emission lines, showing how bright the flare is emitting at these various emissions. Also shown is a Dark value, which is a detector that is blocked from seeing the Sun, which shows energetic particles from the Sun that can penetrate the EVE instrument and cause false counts. This Dark diode will increase during solar storms.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Where: Solar Irradiance (full Sun)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="2365" LastEditDate="2020-05-05T21:16:20.337" LastActivityDate="2020-05-05T21:16:20.337" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24060" PostTypeId="1" AcceptedAnswerId="24061" CreationDate="2017-12-19T13:16:37.373" Score="3" ViewCount="95" Body="&lt;p&gt;Is there a range of redhsift dz, such that if I look at z+dz from a massive object at redshift z, I will see its progenitors?&#xA;(in a LCDM scenario of halo assembling, with smaller halos evolving towards larger ones across the cosmic time)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a detailed explanation of this (maybe naif) question.&lt;/p&gt;&#xA;" OwnerDisplayName="user19333" LastActivityDate="2017-12-19T13:59:43.920" Title="If I look beyond (about same region, higher redshift) a galaxy, will I see its progenitors?" Tags="&lt;fundamental-astronomy&gt;&lt;observational-astronomy&gt;&lt;observable-universe&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24061" PostTypeId="2" ParentId="24060" CreationDate="2017-12-19T13:59:43.920" Score="2" Body="&lt;p&gt;I think I understand what you mean and the answer is: specifically no; but generally yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look out into space, we can divide up what we can see into redshift slices of arbitrary thickness. The light we see from galaxies in each slice has been travelling for a certain amount of time, so we are seeing that &quot;piece of space&quot; at a particular cosmic epoch corresponding to its redshift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look past that galaxy to a higher redshift slice, then what we are seeing there is a &lt;em&gt;different&lt;/em&gt; patch of space at a younger cosmic epoch. i.e You cannot see the same galaxy as it was at a different time. We see each and every galaxy only once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;em&gt;in general&lt;/em&gt; you will be seeing younger galaxies at higher redshift. In an LCDM model then smaller galaxies build up into larger ones, so there will be a higher density of galaxies at higher redshifts (even after accounting for cosmic expansion), but the galaxies they will produce by merging etc. are &lt;em&gt;not yet&lt;/em&gt; visible to us. It is possible that we will see them in the future (depending on the exact values of the cosmological parameters and what redshift we are talking about) as that distant patch of space recedes from us and becomes older (in cosmic terms).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-12-19T13:59:43.920" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24063" PostTypeId="1" CreationDate="2017-12-19T14:39:36.237" Score="1" ViewCount="85" Body="&lt;p&gt;Imagine a tidally locked planet orbiting a red dwarf, where habitability is not situated close to terminator zone, but on a &quot;small&quot; cap normal to incident starlight (zenith) with permanent average +15 Celsius over the year, variations through year depending on orbit's eccentricity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there was in our solar system one (maybe not spherical) tidally locked body orbiting some close to circular orbit around the sun, how far from the sun would it be to meet those +15C average zenith cap temperature? Does it make habitable zone larger than it is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yKlBU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yKlBU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16500" LastEditorUserId="16500" LastEditDate="2017-12-19T16:32:08.170" LastActivityDate="2017-12-19T16:32:08.170" Title="Does habitable zone take tidally locked bodies into account?" Tags="&lt;solar-system&gt;&lt;orbital-elements&gt;&lt;habitable-zone&gt;&lt;tidal-locking&gt;&lt;red-dwarf&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24064" PostTypeId="1" AcceptedAnswerId="24065" CreationDate="2017-12-19T15:41:02.040" Score="43" ViewCount="18014" Body="&lt;p&gt;As stated in several sources, it's supposed that in every galaxy there is a black hole in the middle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, &lt;em&gt;why do these black holes in the middle of galaxies not suck up all the surrounding matter in the galaxy?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2752" LastEditorUserId="1804" LastEditDate="2017-12-24T08:34:26.287" LastActivityDate="2018-01-08T18:56:51.650" Title="Why do Black Holes in the middle of galaxies not suck up the whole galaxy?" Tags="&lt;black-hole&gt;&lt;galaxy&gt;" AnswerCount="11" CommentCount="8" FavoriteCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="24065" PostTypeId="2" ParentId="24064" CreationDate="2017-12-19T19:01:18.163" Score="85" Body="&lt;p&gt;You shouldn't think of black holes as &quot;sucking things in&quot;. Black holes interact with matter through gravity, just the same as any other object. Think of our Solar System. All the planets orbit around the sun because it has a lot of mass. Since the planets have some lateral motion (they're not moving directly towards or away from the sun), they circle around it. This is known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_momentum#Conservation_of_angular_momentum&quot; rel=&quot;noreferrer&quot;&gt;conservation of angular momentum&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When talking about gravity, all that matters is the mass of the objects involved. It doesn't really matter what kind of object it is*. If you were to replace the sun with a black hole that had the same mass as our sun, the planets would continue on the exact same orbits as before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the black holes at the centers of most spiral galaxies do accumulate mass. Some of these black holes have &lt;a href=&quot;https://en.wikipedia.org/wiki/Accretion_disk&quot; rel=&quot;noreferrer&quot;&gt;accretion disks&lt;/a&gt; around them. These are swirling disks of gas and dust that is slowly falling into the black hole. These gas and dust particles lose their angular momentum through interactions with gas and dust nearby and by radiating energy as heat. Some of these black holes have very large accretion disks, and can generate huge amounts of electromagnetic radiation. These are known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Accretion_disk&quot; rel=&quot;noreferrer&quot;&gt;active galactic nuclei&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, long story short, black holes don't &quot;suck&quot;. They just interact with things gravitationally. Stars, gas, and other matter in the galaxy has angular momentum, so it stays in orbit around the center of the galaxy. It doesn't just fall straight in. This is the same reason the Earth orbits around the Sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Disclaimer: When you talk about things like tidal forces, you do need to take into account the size of the objects. But for orbital mechanics, we don't need to worry about it because the distances between the objects are generally much larger than the objects themselves.&lt;/p&gt;&#xA;" OwnerUserId="13426" LastEditorUserId="13426" LastEditDate="2017-12-20T00:45:24.557" LastActivityDate="2017-12-20T00:45:24.557" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="24066" PostTypeId="1" AcceptedAnswerId="24072" CreationDate="2017-12-20T02:12:58.630" Score="7" ViewCount="725" Body="&lt;p&gt;When I read some papers about MHD dynamo theory, there is always a $\tau$ that means convection overturning time. What does &lt;em&gt;convection overturn&lt;/em&gt; mean?&lt;/p&gt;&#xA;" OwnerUserId="20401" LastEditorDisplayName="user1569" LastEditDate="2017-12-20T12:13:20.033" LastActivityDate="2017-12-20T12:13:20.033" Title="What is convection overturn?" Tags="&lt;the-sun&gt;&lt;stellar-dynamics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24069" PostTypeId="2" ParentId="24051" CreationDate="2017-12-20T04:53:48.973" Score="1" Body="&lt;p&gt;Use &lt;em&gt;SHA&lt;/em&gt;, &lt;em&gt;solarHourAngle&lt;/em&gt; or some variant thereof, as James K said, if your programming language allows it. Longer variable names can be more cumbersome to type and have more possibility of mis-typing the name, but can save a lot of time by providing clarity of purpose when reviewing code at a later date.&lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2017-12-20T04:53:48.973" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24072" PostTypeId="2" ParentId="24066" CreationDate="2017-12-20T07:28:58.227" Score="6" Body="&lt;p&gt;The convective overturn time is the typical timescale for a convective cell to rise in a gas. Imagine a &quot;lava lamp&quot; - it's the time for one of the blobs to rise from its lowest to highest point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am most familiar with its use in stars, where convective energy transport is modelled using a mixing length. This posits that the typical height travelled by an adiabatic convection cell is some multiple $\alpha$ of the pressure scale height $H_p$ (the typical length on which the internal pressure changes significantly). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The convective overturn (or turnover) time is then $\alpha H_p/v$, where $v$ is the velocity at which convective cells rise.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-12-20T07:28:58.227" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24077" PostTypeId="2" ParentId="23992" CreationDate="2017-12-20T13:06:26.743" Score="4" Body="&lt;p&gt;In general, space in the Universe is expanding. However, this expansion can be locally retarded by high enough densities of mass-energy. On very large scales, the mean density is low enough that you can treat the whole universe as expanding. But if you look on smaller scales, you can find underdense regions (voids) where the expansion is actually &lt;em&gt;faster&lt;/em&gt; than average, and overdense regions where the expansion is slower or has even ceased. Inside of clusters and galaxy groups (and of course inside galaxies themselves), the density is high enough that space is no longer expanding, and so it's perfectly possible for galaxies to move through local space towards each other and collide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in your raisin-loaf model, the trick is that regions right around each raisin -- and in local regions containing several raisins close to each other -- &lt;em&gt;aren't&lt;/em&gt; expanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that j-g-faustus' argument is somewhat incorrect: if expansion only occurs in voids between superclusters, then Edwin Hubble would never have found it by looking at galaxies within the nearest 30 million light years. Space inside individual clusters and groups is not expanding, but space within superclusters &lt;em&gt;is&lt;/em&gt; (just not as fast as space within voids).&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2017-12-20T13:06:26.743" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24079" PostTypeId="1" AcceptedAnswerId="24084" CreationDate="2017-12-20T19:13:35.797" Score="2" ViewCount="112" Body="&lt;p&gt;Recently Hubble saw several asteroids traveling erratically through a deep field perspective. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What region of the sky is this, and how near were these asteroids to the Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The photo is composed of multiple shots and the asteroids have copies because of Hubble's movement through frames as the photo assemblage shifted.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Was Hubble viewing towards the plane of the solar system and the asteroid belt?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How did they explain that there were five in the same narrow frame? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomynow.com/2017/11/03/hubble-sees-nearby-asteroids-photobombing-distant-galaxies/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://astronomynow.com/2017/11/03/hubble-sees-nearby-asteroids-photobombing-distant-galaxies/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KCe2G.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KCe2G.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="7982" LastEditDate="2017-12-21T00:37:10.757" LastActivityDate="2017-12-21T00:39:58.833" Title="How close to Earth were the asteroids moving through this Hubble composite image?" Tags="&lt;asteroids&gt;&lt;hubble-telescope&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24080" PostTypeId="2" ParentId="24064" CreationDate="2017-12-20T20:53:30.697" Score="3" Body="&lt;p&gt;You must also factor in the dark matter which is interacting gravitationally with all of the &quot;hot matter&quot; that can be seen in the galactic disk. Dark matter was  discovered by carefully mapping the orbits of objects in galaxies and finding that the matter that could be seen could not account for the observed orbital motion. One of the mysteries of dark matter is that it is not pulled into the black hole the way hot matter is. The dark matter has the practical effect of balancing out some of the gravitational pull of the supermassive black hole at the galaxy's center. &lt;/p&gt;&#xA;" OwnerUserId="20428" LastActivityDate="2017-12-20T20:53:30.697" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24082" PostTypeId="2" ParentId="23992" CreationDate="2017-12-20T23:12:29.040" Score="2" Body="&lt;p&gt;Just as molecules in the bread have random motion due to heat, galaxies have random motion due to variations in the initial state of the universe. For galaxies that are close by, those variations are smaller than the Hubble expansion, and the Hubble expansion is simply a &lt;em&gt;general trend&lt;/em&gt;; object that are farther away from us tend to, &lt;em&gt;on average&lt;/em&gt;, be moving away from us at higher speed, but individual objects can still be moving towards us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://w.astro.berkeley.edu/~mwhite/darkmatter/hubble.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this site&lt;/a&gt;, the Hubble constant is &quot;around 70 kilometers per second for every megaparsec in distance&quot;. According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Andromeda_Galaxy&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;, Andromeda is &quot;780 kiloparsecs&quot; away. Multiplying those together, we get that Andromeda's movement due to the Hubble expansion should be about 5 km/sec. Wikipedia also says that &quot;The Andromeda Galaxy is approaching the Milky Way at about 110 kilometres per second&quot;. That means that Andromeda's motion due to Hubble expansion is about 4% of its total movement. Andromeda has so much velocity relative to Earth that the Hubble expansion simply gets drowned out.&lt;/p&gt;&#xA;" OwnerUserId="19010" LastActivityDate="2017-12-20T23:12:29.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24083" PostTypeId="1" CreationDate="2017-12-21T00:17:41.800" Score="1" ViewCount="140" Body="&lt;p&gt;given that everyone wonders whether there is a chance that there is something to oumuamua that is not natural, what about tumbling (and/or aiming?)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is there a natural gravitational or other force that would naturally stabilize non-orbiting asteroids in space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would guess that there is a chance that we could detect tumbling. &lt;/p&gt;&#xA;" OwnerUserId="20432" LastEditorUserId="88" LastEditDate="2017-12-21T09:25:40.600" LastActivityDate="2017-12-21T09:25:40.600" Title="Is Oumuamua tumbling?" Tags="&lt;oumuamua&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24084" PostTypeId="2" ParentId="24079" CreationDate="2017-12-21T00:18:23.793" Score="3" Body="&lt;p&gt;I visited the linked article in Astronomy Now and found the URL of the image contained the string &lt;code&gt;STSCI-H-p1733a&lt;/code&gt;. STScI stands for Space Telescope Science Institute. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Searching that string leads to &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://hubblesite.org/images/news/release/2017-33&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://hubblesite.org/images/news/release/2017-33&lt;/a&gt; (four images)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://hubblesite.org/news_release/news/2017-33&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://hubblesite.org/news_release/news/2017-33&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://hubblesite.org/image/4080/gallery&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://hubblesite.org/image/4080/gallery&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;the last two of which identify the location as &lt;a href=&quot;https://en.wikipedia.org/wiki/Abell_370&quot; rel=&quot;nofollow noreferrer&quot;&gt;Abell 370&lt;/a&gt;, at about RA: 2h 40m, Dec: -1.6°. At about RA = 0h the ecliptic and equator coincide, but by 2h 40m the ecliptic has a declination of about +16°, so this photo is looking about 18° below the ecliptic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first paragraph of the caption contained in both of those links says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Like rude relatives who jump in front your vacation snapshots of landscapes, some of our solar system's asteroids have photobombed deep images of the universe taken by NASA's Hubble Space Telescope. These asteroids reside, on average, only about 160 million miles from Earth — right around the corner in astronomical terms. Yet they've horned their way into this picture of thousands of galaxies scattered across space and time at inconceivably farther distances.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the distance is about 160 million km, then they are suggesting these asteroids are in the inner part of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Asteroid_belt&quot; rel=&quot;nofollow noreferrer&quot;&gt;main asteroid belt&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth is moving at about 30 km/sec around the Sun, and the Hubble Telescope is &quot;only&quot; moving 7.6 km/sec with respect to the Earth, in roughly the same direction. So much and probably most of the motion is due to the Earth's velocity. I'm suggesting this because if the distance to the asteroids is really roughly 160 million kilometers, then they would be near opposition to the Sun and the Earth's motion would be roughly perpendicular to the viewing direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;em&gt;a lot of asteroids!&lt;/em&gt; According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Asteroid_belt&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One hundred asteroids had been located by mid-1868, and in 1891 the introduction of astrophotography by Max Wolf accelerated the rate of discovery still further. A total of 1,000 asteroids had been found by 1921, 10,000 by 1981, and 100,000 by 2000. Modern asteroid survey systems now use automated means to locate new minor planets in ever-increasing quantities.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Asteroids are discovered constantly, and sometimes this happens in images that were not intended to be searches for asteroids, as in this case. While most most asteroids orbit close to the plane of the ecliptic, some of their orbits do stray substantially above and below it, as in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the date of the News Release is 02-Nov-2017, I don't know the time frame of all of the exposures that contributed to this final composite image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a screen shot of the Planetarium Mode of Dominic Ford's &lt;a href=&quot;https://in-the-sky.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;in-the-sky.org&lt;/a&gt;. The marker shows the position of Abell 370 just below the equator, and the Ecliptic is the thicker line above it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4Lzh1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4Lzh1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-12-21T00:39:58.833" LastActivityDate="2017-12-21T00:39:58.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24085" PostTypeId="2" ParentId="24083" CreationDate="2017-12-21T00:38:37.210" Score="5" Body="&lt;p&gt;Assuming that this is a serious question... yes, Oumuamua will be tumbling, just like every other small body. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That it comes from outside the Solar System does not change that it will have three axes of rotation because it is small and thus has not accreted enough matter that it would stabilise into a single primary axis. (This doesn't preclude one of its axes being a primary spin axis.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And just like every other small body, if we collect enough data in the form of reflected sunlight from its surface, we can construct shape and rotation models and estimate its size and composition.&lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2017-12-21T00:38:37.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24086" PostTypeId="1" AcceptedAnswerId="24087" CreationDate="2017-12-21T00:49:51.377" Score="2" ViewCount="62" Body="&lt;p&gt;A perfectly spherical asteroid on a straight line trajectory with no rotational movement at all, enters the gravitational field of the sun (from elsewhere far far away). When it exits on its hyperbolic path, does it still point in the same direction it entered, or does it rotate to face the direction of travel?&lt;/p&gt;&#xA;" OwnerUserId="20433" LastEditorUserId="2365" LastEditDate="2017-12-21T01:27:00.543" LastActivityDate="2017-12-21T01:27:00.543" Title="Change In Direction of Asteroid Approaching Sun" Tags="&lt;the-sun&gt;&lt;asteroids&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24087" PostTypeId="2" ParentId="24086" CreationDate="2017-12-21T01:26:23.173" Score="1" Body="&lt;p&gt;No.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A body in orbit will maintain its angular momentum, it will maintain its rotation rate as it is deflected by the sun. If it is not rotating then the deflection won't change the direction it is facing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There may be some minor effects that could cause a little rotation but it won't &quot;steer&quot; to face forward.)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-12-21T01:26:23.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24088" PostTypeId="1" CreationDate="2017-12-21T03:45:05.523" Score="5" ViewCount="93" Body="&lt;p&gt;The Phys.org article &lt;a href=&quot;https://phys.org/news/2017-12-cosmic-filament-probes-galaxy-giant.html&quot; rel=&quot;noreferrer&quot;&gt;Cosmic filament probes our galaxy's giant black hole&lt;/a&gt; links to &lt;a href=&quot;https://arxiv.org/abs/1711.04190&quot; rel=&quot;noreferrer&quot;&gt;A Nonthermal Radio Filament Connected to the Galactic Black Hole?&lt;/a&gt; which discusses (among other things) the importance of achieving high dynamic range to study these filaments in the proximity of the strong emission associated with Sgr A*. It mentions the fluctuation of radio intensity from Sgr A* on time scales as short as five minutes (as well as to instrumental time-varying effects that also require attention and calibration).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper refers to &lt;a href=&quot;https://www.nature.com/articles/354046a0&quot; rel=&quot;noreferrer&quot;&gt;Zhao et al. 2016&lt;/a&gt;, which in turn links to &lt;a href=&quot;https://www.nature.com/articles/354046a0&quot; rel=&quot;noreferrer&quot;&gt;Zhao et al. 1999&lt;/a&gt;. But so far I can't track down either &lt;em&gt;the nature&lt;/em&gt; of the short term variations seen in the radio brightness of Sgr A* nor &lt;em&gt;the mechanism&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the nature of this variation (amplitude, power spectrum, any periodicity?) and what are the proposed mechanisms?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2021-03-30T18:28:14.437" Title="Nature and mechanism of the short-term variability in radio strength of Sgr A*?" Tags="&lt;radio-astronomy&gt;&lt;sagittarius-a&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24094" PostTypeId="1" AcceptedAnswerId="24100" CreationDate="2017-12-21T11:41:37.030" Score="2" ViewCount="345" Body="&lt;p&gt;If I'm given the latitude, I can easily find the declination by using&#xA;declination = 90 - latitude.&#xA;I can also find the angle of dip from the height but should I add or subtract it from the declination I found?&#xA;Also I have to consider atmospheric refraction but I don't know how.&lt;/p&gt;&#xA;" OwnerUserId="16506" LastEditorUserId="16506" LastEditDate="2017-12-24T12:06:49.603" LastActivityDate="2017-12-24T12:06:49.603" Title="What should be the declination of a star to be marginally circumpolar given the latitude &amp; height from ground?" Tags="&lt;observational-astronomy&gt;&lt;positional-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24098" PostTypeId="2" ParentId="24064" CreationDate="2017-12-21T16:44:16.387" Score="-3" Body="&lt;p&gt;its all about ENTROPY which is proportional to the surface area of the event horizon of a black hole(see below for a heuristic quantum  argument due to Moffat/Wang as to why this is so).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a Schwarzschild  solution gives a radius of 2Gm for the event horizon with m the black hole mass and G Newton's constant. Adding mass to a black hole thus increases its entropy. Given an isolated system of finite total energy, it has a finite maximum entropy which acts as an attractor for the dynamics of the system,placing a limit on the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J von Neumann defines a quantum version of entropy as follows: Let f be a normal state of a local algebra of observables O(D) acting on the Hilbert space H. Then we can write this f as a convex sum of pure states. For a system of finite energy this sum is finite since H is then finite dimensional&#xA;Von Neumann’s non-commutative equivalent of a partition is the density operator   ie the weighted sum of the projections  onto the minimal vector spaces corresponding to these pure states&#xA;    Then we have the well-known equivalence;&lt;br&gt;&#xA;For such a normal state f, the von Neumann entropy is defined as  the entropy of the weights . We interpret it as an (inverse) measure of the amount of information that the quantum system in a given state will yield through measurement. The larger the entropy of the quantum system, the less information can be extracted. &#xA;&lt;strong&gt;The von Neumann Entropy of a Black Hole&lt;/strong&gt;&#xA;The measurement process cannot be performed by an external observer to elements within the interior, beyond the event horizon. We thus partition the event horizon of the black hole with elements each of area   k squared, where k is the Planck length and assume the Planck area  corresponds classically to the minimal projection   of the pure vector state  . Let N be the total finite number of partitions. By the ‘no hair’ hypothesis there is no preferred location on the event horizon, so that each partition element must have the same weighting. The von Neumann entropy of this partition is thus proportional to S the surface area of the black hole.&lt;/p&gt;&#xA;" OwnerUserId="20449" LastEditorUserId="20449" LastEditDate="2017-12-22T14:06:48.020" LastActivityDate="2017-12-22T14:06:48.020" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24099" PostTypeId="2" ParentId="24064" CreationDate="2017-12-21T17:09:49.977" Score="2" Body="&lt;p&gt;For galaxies with large black holes, the surrounding matter is in &lt;em&gt;orbit&lt;/em&gt; around the black hole(s), the same way that the moon orbits the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is a direct analogy to &quot;&lt;em&gt;Why doesn't the Moon fall to the ground?&lt;/em&gt;&quot; or &quot;&lt;em&gt;Why don't the planets fall into the sun?&lt;/em&gt;&quot;.  The black hole is more massive than the Sun, but its effects are of the same type.&lt;/p&gt;&#xA;" OwnerUserId="14019" LastActivityDate="2017-12-21T17:09:49.977" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24100" PostTypeId="2" ParentId="24094" CreationDate="2017-12-21T17:46:01.693" Score="4" Body="&lt;p&gt;Just to be clear, the &quot;height from sea level&quot;  (mentioned in the title) does not change what star is circumpolar. Whether the ground is at 0 units above sea level or 1500 units above sea level does not make a difference. I think what you intended to write is the height above the ground makes a difference (&quot;angle of dip from the height&quot; mentioned in the question). At 0 units above the ground, a circumpolar star has a declination of D1. At 1500 units above the ground, a circumpolar star has a declination that is farther from the pole, or D1-angle. (I assume that you are in the northern hemisphere.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refraction raises the apparent altitude of an object, so you can see a circumpolar star that is farther from than the pole than is possible without refraction. Thus, a circumpolar star has declination = D1-angle-Refraction. The adopted value for refraction of an object on the horizon is 34 arcminutes (Ref 1: The Astronomical Almanac, 2001, page A12). The actual amount of refraction depends on the atmospheric pressure and temperature. An approximate formula for refraction (applicable for observed altitudes below 15 degrees) is R=P(0.1594 + 0.0196a + 0.00002a^2)/[(273+T)(1 + 0.505a + 0.0845a^2)] where T is the temperature (degrees C), P is the barometric pressure (millibars), and a is the altitude (degrees) (Ref 1: page B62)&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2017-12-21T17:46:01.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24101" PostTypeId="2" ParentId="24064" CreationDate="2017-12-21T19:00:03.677" Score="2" Body="&lt;p&gt;One quick answer for your question would be event horizon or Schwarzschild radius. Anything which is fairly close to this radius/horizon will eventually be sucked up by the black hole.  &lt;/p&gt;&#xA;" OwnerUserId="20267" LastActivityDate="2017-12-21T19:00:03.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24102" PostTypeId="2" ParentId="24064" CreationDate="2017-12-21T19:53:37.543" Score="11" Body="&lt;p&gt;I once heard of a Japanese cartoon/movie/show where space pirates threatened to compress the planet Jupiter into a black hole and thus destroy half the Milky Way galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sounds like an interesting idea, but... even if you could compress Jupiter into a black hole, its mass would remain the same, meaning that Jupiter (now a black hole) would still continue to move around our sun in its same orbit, and Jupiter's moons would still continue to orbit Jupiter as they did before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many people think that once a star collapses into a black hole, its &quot;sucking power&quot; (its gravitational force) increases.  This is simply not the case.  Believe it or not, many stars are &lt;em&gt;less&lt;/em&gt; massive &lt;em&gt;after&lt;/em&gt; they turn into a black hole than &lt;em&gt;before&lt;/em&gt;, when they were shining stars.  This is because, at the end of their lives, some stars shed a significant portion of their outer layer into space right before they collapse into a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read that if you compressed the Earth to the size of a cherry, its density would be so great that it would turn into a black hole.  Assuming that were true and it was actually done, the black hole of Earth would still continue orbiting the sun once every year, and Earth's moon would continue to orbit the Earth about once every 29.5 days.  (Now, the spin of the new black-hole-Earth about its axis would probably be different, but the time it would take to orbit the sun would not change.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surprisingly, once the Earth got compressed into a cherry-sized black hole, less space debris would fall into it than before (when the Earth was the size of... well, Earth).  This is because the newly-formed black-hole-Earth would take up much less space (volume) and asteroids and comets would be more likely to miss the cherry-sized (or slightly-larger-than-cherry-sized) volume that, if not missed, would cause the debris to be sucked into the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the debris missed the black-hole-Earth by even a kilometer (which might seem like a large distance to us, but is very miniscule in astronomical terms), it would be flung off in a different direction, possibly never to return.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, basically, a common misconception people have regarding black holes is that nothing has more gravity than a black hole, and that stars that form into black holes all of a sudden have increased gravity and therefore get more &quot;sucking power.&quot;  This is simply not true.  Black holes still have the same mass as before (sometimes less, depending on how they're formed), and how much &quot;sucking power&quot; they have is still dependent on how much mass they're made up of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it may be true that the most massive stars in the universe are indeed black holes (if you'd even call them &lt;em&gt;stars&lt;/em&gt; at that point), there exist many stars that are more massive (and therefore have more &quot;sucking power&quot;) than many black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the fact that our galaxy's center probably contains a super-massive black hole doesn't mean that the black hole would suck up any more matter than if it were the same amount of mass which happened to not be in black-hole-form.&lt;/p&gt;&#xA;" OwnerUserId="20452" LastEditorUserId="20453" LastEditDate="2017-12-22T08:54:27.920" LastActivityDate="2017-12-22T08:54:27.920" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24103" PostTypeId="1" AcceptedAnswerId="24106" CreationDate="2017-12-21T20:26:24.600" Score="4" ViewCount="171" Body="&lt;p&gt;I seem to recall reading something that said if the density of the universe was off by as little as one atom per cubic meter, the universe, as we know it, would not exist; one less and there wouldn't have been enough gravity to form galaxies and stars, one more and there would have been too much matter. (I don't know (or remember) what would have happened in the case of too much matter.)&lt;br&gt;&lt;br&gt;If I Google the title of this question, all I get is stuff about the expansion of the universe, nothing about whether or not it would exist. I've tried other variations but can't seem to find anything.&lt;br&gt;&lt;br&gt;Did I grossly misinterpret this? If not, does anyone know of any articles that can confirm or refute this? &lt;/p&gt;&#xA;" OwnerUserId="7108" LastActivityDate="2017-12-21T21:44:36.353" Title="Critical density of the universe (no, not that one)?" Tags="&lt;cosmology&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24106" PostTypeId="2" ParentId="24103" CreationDate="2017-12-21T21:44:36.353" Score="6" Body="&lt;p&gt;You're basically asking about the &lt;a href=&quot;https://en.wikipedia.org/wiki/Flatness_problem&quot; rel=&quot;noreferrer&quot;&gt;Flatness Problem&lt;/a&gt;. As quoted from &lt;a href=&quot;https://en.wikipedia.org/wiki/Flatness_problem#Implication&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the initial density of the universe could take any value, it would seem extremely surprising to find it so 'finely tuned' to the critical value $\rho_c$. Indeed, a very small departure of $\Omega$ from 1 in the early universe would have been magnified during billions of years of expansion to create a current density very far from critical. In the case of an overdensity ($\rho &amp;gt; \rho_c$) this would lead to a universe so dense it would cease expanding and collapse into a Big Crunch (an opposite to the Big Bang in which all matter and energy falls back into an extremely dense state) in a few years or less; in the case of an underdensity ($\rho &amp;lt; \rho_c$) it would expand so quickly and become so sparse it would soon seem essentially empty, and gravity would not be strong enough by comparison to cause matter to collapse and form galaxies. In either case the universe would contain no complex structures such as galaxies, stars, planets and any form of life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is more or less what &lt;a href=&quot;https://astronomy.stackexchange.com/questions/24103/critical-density-of-the-universe-no-not-that-one#comment41367_24103&quot;&gt;Phiteros says in his comment above&lt;/a&gt;. As he states, the distinction here is not that the universe wouldn't exist at all, as you state, but rather it wouldn't exist as we know it, with galaxies, stars, planets, or life of any form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain a bit more, the basic idea is that the universe will have started out with a more-or-less homogeneous and isotropic distribution of matter with some density $\rho$ (this is a pretty strong assumption in its own right, and worth addressing in a completely different answer). Note that the overall curvature of the universe is directly tied to this density so you may see some sources talk about this in terms of curvature instead of density. The main point, as described in the quote above is that the evolution of the universe, i.e., the ability for galaxies, stars, planets, life, etc. to form, is dictated by a &lt;em&gt;critical density&lt;/em&gt;, $\rho_c$. If $\rho = \rho_c$ (or as close to equal as to be insignificantly different), our universe can form as it is now. The flatness problem comes into play because the Big Bang Theory implies that if the universe did not start out with (almost) exactly $\rho = \rho_c$, it would have veered far off course from our current universe and resulted in either a &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Crunch&quot; rel=&quot;noreferrer&quot;&gt;Big Crunch&lt;/a&gt; or a &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Rip&quot; rel=&quot;noreferrer&quot;&gt;Big Rip&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This idea poses a really big problem for the Big Bang Theory because it requires such a precise setup for our universe, where any betting person would've said there's no way the universe could've started out with $\rho=\rho_c$. The main resolution to this problem is by introducing inflation into the model. The main idea here is that the Universe may or may not have started out with $\rho = \rho_c$, but so long as it was close enough, the hyper inflation at the beginning of the universe would have forced the difference to be so small as to be effectively zero, negating the effects described in the quote above. Another potential, and not necessary mutually exclusive answer is found in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Anthropic_principle&quot; rel=&quot;noreferrer&quot;&gt;Anthropic Principle&lt;/a&gt;, which I won't delve into here.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2017-12-21T21:44:36.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24107" PostTypeId="1" AcceptedAnswerId="24109" CreationDate="2017-12-22T01:04:03.813" Score="2" ViewCount="280" Body="&lt;p&gt;Is there a physical limit to how far we can go?  I am thinking there could be based on the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: There are parts of the universe expanding faster than the speed of light that we will never be able to get to (without faster than light travel)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: There are a finite number of galaxies between here and the distance where 1 takes place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3: If we travel even at the speed of light, it seems we would at some point pass that finite number of galaxies, and wouled not be able to reach the next one!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean there is a limit to how far we can go, and furthermore, we will experience the “big rip”?&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2017-12-22T05:03:05.570" Title="Is there a physical limit to how far we can go?" Tags="&lt;distances&gt;&lt;observable-universe&gt;&lt;fate-of-universe&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24108" PostTypeId="2" ParentId="24064" CreationDate="2017-12-22T01:15:17.757" Score="2" Body="&lt;p&gt;The simple answer is that everything else in the galaxy is going sideways fast enough to escape being sucked in. Instead, the force of the suck (if you like) causes the stars' paths to be pulled into a circle around the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This phenomenon is &quot;orbit&quot;. As other answers pointed out, it's the same reason the Earth does not fall into the Sun, or the Moon fall onto the Earth, and why the International Space Station is hurtling along at around 17,150 miles per hour. They're all going sideways, the force of some large object is turning that sideways motion into circular motion, and if they weren't going fast enough then they'd curve (&quot;fall&quot;) towards that large object and crash into it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's like if you whirl a bucket on the end of a string. The bucket is going sideways, but the string is pulling it towards you. The bucket doesn't fly off away from you because of the force from the string, and so it curves in a circle. The force from the string happens to not be enough to collapse the bucket inwards and hit you.&lt;/p&gt;&#xA;" OwnerUserId="20461" LastActivityDate="2017-12-22T01:15:17.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24109" PostTypeId="2" ParentId="24107" CreationDate="2017-12-22T05:03:05.570" Score="9" Body="&lt;p&gt;Your 3 points are spot on.   There is a finite number of galaxies we could theoretically reach for the reasons you say.  The further away the distant galaxy the greater the expansion of space between us and that galaxy and beyond a certain distance, galaxies can't be reached, even by the speed of light.   We can see galaxies that we couldn't possibly travel to.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does this mean there is a limit to how far we can go, and furthermore,&#xA;  we will experience the “big rip”?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No.  It only means that some galaxies that we can see, we could never travel to.   There's no limit to how FAR we can go but there's still a finite number of galaxies that we could reach with close to the speed of light travel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.forbes.com/sites/startswithabang/2016/05/12/the-limits-of-how-far-humanity-can-go-in-the-universe/#3645287f4ae5&quot; rel=&quot;noreferrer&quot;&gt;This article&lt;/a&gt;, using the current size estimate for the observable universe of 46 billion light years in radius, says that 14.5 billion light years distance is the limit that we could theoretically reach, about 3% of the observable universe or about 3 billion galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VXSq0.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VXSq0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the big rip.   Nobody knows if that's an accurate prediction.  It's a possible outcome, but we don't know enough about dark energy to say if the big rip will happen or not.   Nobody knows.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And it's a pretty psalm, but I don't see the prediction of dark energy in it.   Only that it says the Universe will age, but life is eternal - which is a nice thought.&lt;/em&gt;  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-12-22T05:03:05.570" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24113" PostTypeId="1" AcceptedAnswerId="24117" CreationDate="2017-12-22T10:25:05.667" Score="2" ViewCount="70" Body="&lt;p&gt;I’m asking this question in reference to a scenario I saw posted a couple of days ago, where it’s stated that if you would shrink earth to the size of a cherry, it would become a black hole.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is it also correct to assume that changing the size of earth while maintaining the mass, will not affect objects which are further away than the event horizon?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m asking this in reference to the same scenario as above. It is also said that changing the size of the earth, so that it becomes a black hole, will have no effect on the orbit of objects around the earth - e.g. the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m trying to understand this, with an image of how much the earth will bend spacetime. Both as a earth-sized earth and as a cherry-sized earth.&lt;/p&gt;&#xA;" OwnerUserId="20472" LastActivityDate="2017-12-22T22:47:23.803" Title="Is it correct to assume that the relationship between an objects mass and size, is what decides if it’s a black hole or not?" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;earth&gt;&lt;space-time&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24115" PostTypeId="2" ParentId="24064" CreationDate="2017-12-22T12:31:55.290" Score="2" Body="&lt;p&gt;This is a common misconception about black holes: that they somehow 'suck up' everything around them or pull things into them.  In reality you could replace the Sun now with a black hole of the same mass and not notice any immediate difference.  It's not like it would suddenly start hoovering in the planets around it, that's just not how it works.&lt;/p&gt;&#xA;" OwnerUserId="20478" LastActivityDate="2017-12-22T12:31:55.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24117" PostTypeId="2" ParentId="24113" CreationDate="2017-12-22T13:52:42.820" Score="3" Body="&lt;p&gt;Yes. A black hole is in some ways a very simple object. It has mass, charge, angular momentum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These three values determine the size of the event horizon. If the Earth's mass were compressed to less than the size of a cherry it would be inside its event horizon and so would be a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational field of the Earth out at the moon would be (almost) unchanged. The moon would orbit an Earth-mass black hole in the same way that it orbits the Earth now.  The small differences are due to irregularities in the shape of the Earth, which wouldn't exist if the Earth were collapsed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above the Earth's surface the curvature of space would be (almost) the same if the Earth were to be collapsed. If a rocket were to use its thrusters to hold itself stationary 6400km above the black hole, the inhabitants would experience 1g of gravity, just as we do standing on the surface, 6400km from the centre of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you go far enough below the Earth's surface, gravity weakens, because there is mass above you.  If you collapsed the Earth, then gravity (and curvature of spacetime) would increase all the way to the black hole.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-12-22T22:47:23.803" LastActivityDate="2017-12-22T22:47:23.803" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24119" PostTypeId="1" AcceptedAnswerId="24120" CreationDate="2017-12-22T20:00:10.107" Score="3" ViewCount="47" Body="&lt;p&gt;I made it my geeky aim to see which of two flats that we are considering moving into is sunnier. The first flat faces S and the second faces NNW; both are at the same height above ground and have similar terrain (buildings/trees) around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conventional wisdom would suggest the S-facing one is (much) sunnier, but I am not sure if this is really so, since I don't think it is the case that the Sun, when it's above the horizon, spends most of its time in the Southern-hemispace/hemifield. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, I'd like to get a more quantitative estimate of how much direct sunlight each flat gets per day (averaged across seasons). How can I use tools available on the Internet (probably GoogleMaps-based) to find this out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My only solution to date was to generate Sun trajectories (elliptics) for each flat, using the SunCalc.net website; these suggest the NNW-facing flat hardly ever sees the sun, except around the summer solstice. Is this a reliable estimate, or can anyone suggest another way to solve this riddle?&lt;/p&gt;&#xA;" OwnerUserId="20486" LastActivityDate="2017-12-22T20:40:11.173" Title="Computing the average hours of sunshine at given location&amp;orientation" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24120" PostTypeId="2" ParentId="24119" CreationDate="2017-12-22T20:40:11.173" Score="2" Body="&lt;p&gt;Between September and March (ie winter, assuming a Northern Hemisphere location) The sun rises in the South East, is Due south at midday and sets in the South west. Your South facing window would get sun for as long as the sun is up. The North West facing window would only get a glimpse of the sun around sunset (and it would then be low and weak)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Between March and September (Northern hemisphere summer) The sun rises in the North East, it is in the south between roughly 6am and 6pm, then sets in the North West. Your Southern window would get 12 hours of sun each day. The North West window would get some sun in the evening, and this sun would be low (weak but shining into the room)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall the South facing window would get a lot more sun than the North West window. The reason is that the sun is always above a point on the Earth that is between 23.5 degrees North or South of the equator. For anyone in the Northern Hemisphere, the sun is usually South of them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quantatative estimate of this could be done, but the actual values you got would depend on the Latitude and precise aspect of the two windows. However the result from SunCalc looks to be robust. The North West facing window will only get a little sun on summer evenings, and little or nothing during the winter.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-12-22T20:40:11.173" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24121" PostTypeId="1" AcceptedAnswerId="24137" CreationDate="2017-12-22T20:56:29.340" Score="2" ViewCount="341" Body="&lt;p&gt;Is the exact duration of the shortest day of a year (at the winter solstice in the northern hemisphere) always the same, for given latitude and longitude, or does it fluctuate? If it does, what is the range?&lt;/p&gt;&#xA;" OwnerUserId="20487" LastEditorUserId="20487" LastEditDate="2017-12-22T21:27:02.937" LastActivityDate="2017-12-23T21:55:45.083" Title="Is the shortest day duration constant?" Tags="&lt;orbit&gt;&lt;earth&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24122" PostTypeId="2" ParentId="24121" CreationDate="2017-12-22T21:37:05.287" Score="2" Body="&lt;p&gt;It is not exactly the same each year. The exact moment of winter solstice is not at the same calendar date (because of leap days for example). In 2017 the solstice was at Dec 21 16:22, and in 2016 it was Dec 21 10:33. See also &lt;a href=&quot;https://astronomy.stackexchange.com/questions/12940/22nd-is-shortest-day-in-some-places-but-the-21st-is-shortest-in-other-places-c?rq=1&quot;&gt;22nd is shortest day in some places, but the 21st is shortest in other places. Can this be true?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is a little variation in the length of the shortest day, but the length of the day changes slowly at this time of year, so the range of variation is only a few seconds.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-12-22T21:48:04.690" LastActivityDate="2017-12-22T21:48:04.690" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24123" PostTypeId="2" ParentId="24121" CreationDate="2017-12-22T22:33:07.597" Score="3" Body="&lt;p&gt;By looking at the link provided by James K, I was able to write a Python 3 script that calculates the duration of the longest night for a predefined location (Boston, MA) for 160 years. The script uses the &lt;em&gt;PyEphem&lt;/em&gt; module. The longest night becomes shorter by ~72 msec per year, and its length also oscillates with the period of ~19 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HoAtr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HoAtr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the script that I used:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import ephem&#xA;import matplotlib.pyplot as plt&#xA;import matplotlib&#xA;import numpy as np&#xA;&#xA;location = ephem.Observer()&#xA;location.lat, location.lon = '42.3601', '-71.0589'&#xA;&#xA;years = range(1900, 2060)&#xA;longest_nights = []&#xA;for year in years:&#xA;    location.date = ephem.next_winter_solstice('{}-12-12'.format(year))&#xA;    sunset = location.next_setting(ephem.Sun())&#xA;    sunrise = location.next_rising(ephem.Sun())&#xA;    if sunrise &amp;lt; sunset: # Look at the night before!&#xA;        location.date -= 1       &#xA;        sunset = location.next_setting(ephem.Sun())&#xA;    longest_nights.append(sunrise - sunset)&#xA;&#xA;longest_nights = np.array(longest_nights)&#xA;longest_nights = (longest_nights - longest_nights.mean()) * 24 * 3600 # In seconds&#xA;&#xA;# Find the slope&#xA;xp = np.poly1d(np.polyfit(years, longest_nights, 1))&#xA;np_years = np.array(years).reshape(len(years), 1)&#xA;&#xA;# Plot&#xA;matplotlib.style.use(&quot;ggplot&quot;)&#xA;plt.plot(years, longest_nights, &quot;-o&quot;)&#xA;plt.ylabel(&quot;Deviation from the mean, sec&quot;)&#xA;plt.xlabel(&quot;Year&quot;)&#xA;plt.title(&quot;Duration of the longest night in a year in Boston&quot;)&#xA;plt.plot(years, xp(years), &quot;-&quot;)&#xA;plt.arrow(2017, plt.ylim()[0], 0, plt.ylim()[1] - plt.ylim()[0], color='k')&#xA;#plt.savefig(&quot;solstice.png&quot;)&#xA;plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="20487" LastEditorUserId="20487" LastEditDate="2017-12-23T21:55:45.083" LastActivityDate="2017-12-23T21:55:45.083" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24124" PostTypeId="1" AcceptedAnswerId="24132" CreationDate="2017-12-23T06:32:59.940" Score="2" ViewCount="161" Body="&lt;p&gt;This question could be strictly linked to my another &lt;a href=&quot;https://astronomy.stackexchange.com/questions/24064/why-do-black-holes-inside-in-the-middle-of-the-galaxies-not-suck-up-the-whole-ga&quot;&gt;question related to black holes.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It’s known that the  horizon event is a place where dust,star components are ready to be incorporated to the black hole*.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Actually, my question can be divided in few points:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Might be reasonable to believe that ,if it’s true that in many spiral galaxies there is a black hole in the middle, could these objects have so a powerful gravitational force to consider them the creators of the galaxies?&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(*) &lt;em&gt;when the black holes are sucking up dust from the horizon event, are they increasing their mass?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;If so, it’s reasonable to believe that their gravitational force will increase in a way to attract farther objects?&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;em&gt;Could they eject the dust or anything else if they are “eating” so much?&lt;/em&gt;&lt;/strong&gt; &#xA;     &lt;em&gt;This can be in contradiction with above point.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2752" LastEditorUserId="2752" LastEditDate="2017-12-23T09:48:15.547" LastActivityDate="2017-12-23T12:06:53.400" Title="Behaviour of black holes" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;galaxy-center&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24128" PostTypeId="2" ParentId="18120" CreationDate="2017-12-23T10:17:04.987" Score="0" Body="&lt;p&gt;Science is based on guesses, paraphrasing Feynman.&#xA;We guess that something works in a certain way.&#xA;A good guess explains the existing data and makes predictions which can be tested&#xA;The best guess is the good guess which is simplest i.e. minimises the number of additional assumptions.&#xA;So the assumption by Newton that gravity works for planets in the same way as it does for throwing stones while walking along a beach was, in essence, just a guess. &lt;/p&gt;&#xA;" OwnerUserId="20449" LastActivityDate="2017-12-23T10:17:04.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24131" PostTypeId="2" ParentId="3628" CreationDate="2017-12-23T11:09:57.000" Score="2" Body="&lt;p&gt;This is a question that is fundamental to our understanding of how order can emerge from disorder. So its worth considering the ways that this can happen:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Local entropy reduction by random fluctuation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is an attractor for the dynamics (point, cycle or strange)  giving rise to self-organisation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The system is dissipative and open, local order is sustained by energy crossing the system boundary (eg your local library/information store is kept ordered through continual energy input).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Clearly 2. of the list is the reason for accretion disks to form stable rings. Then random collisions of the bits does the rest. If the bits are small particles you get Saturn, if they are large you get rocky planets.&lt;/p&gt;&#xA;" OwnerUserId="20449" LastEditorUserId="1891" LastEditDate="2017-12-30T22:06:18.657" LastActivityDate="2017-12-30T22:06:18.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24132" PostTypeId="2" ParentId="24124" CreationDate="2017-12-23T12:06:53.400" Score="6" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;No, it seems that the formation of a supermassive black hole is a consequence of galaxy formation, not a cause of it. The mass of the black hole is a very small proportion of the mass of the matter and dark matter in the galaxy. The details are not yet worked out but it is the distribution of dark matter that forms the galaxy, and then galaxy mergers contribute to the growth of a large black hole in the centre.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As indicated in the answer to your previous question, &quot;sucked&quot; is entirely the wrong metaphor for how black holes work. Black holes have lots of mass in a small space, so they &quot;pull&quot; gravitationally.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Now gravity gets weird when it is very strong, as it affects the time. From the perspective of a person outside the black hole, a clock on a body falling towards the horizon will slow down, and will appear to stop as the body reaches the black hole, This means that we will never see something fall into the black hole.  However the mass of the body that falls towards the black hole will increase the effective mass of the black hole, and so increase the radius of the event horizon. &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For objects further from the black hole, the acceleration towards the black hole is proportional to its mass. So it is not the case that an object will &quot;start to feel the pull&quot;, but instead the pull will increase.  Note that mass that is orbiting close to the black hole will act almost like part of the black hole from the point of view of someone further out. This is not a special property of black holes, just ordinary gravity.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Black holes form an accretion disk around themselves, a disk of matter orbiting the black hole at very high speeds and getting very very hot. They also produce high energy jets of particles moving close to the speed of light from their poles. These are produced from the accretion disk, not from the black hole itself. Objects that fall towards the black hole don't come back.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-12-23T12:06:53.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24137" PostTypeId="2" ParentId="24121" CreationDate="2017-12-23T15:56:29.407" Score="1" Body="&lt;p&gt;It appears to be quite constant when viewed on a coarse enough scale:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rVXap.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rVXap.png&quot; alt=&quot;Lengths of the winter solstice night from 1900 to 2100, at 60, 40, and 20 degrees north latitude. The three curves look rather flat, with over 18 hours of darkness at 60 degrees, over 14.5 hours at 40 degrees, and a bit over 13 hours at 20 degrees.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zooming in on just one of those, it fluctuates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ogxDy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ogxDy.png&quot; alt=&quot;Lengths of the winter solstice night from 1900 to 2100, at 60 degrees north latitude. This curve exhibits a downward trend, plus an oscillation with a period of 18.6 years, plus shorter-term variations.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variations at lower latitudes are similar but smaller in magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IWQEF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IWQEF.png&quot; alt=&quot;Deviations of the lengths of the winter solstice night from the mean from 1900 to 2100, at 60, 40, and 20 degrees north latitude.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation is reversed south of the equator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7bXtg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7bXtg.png&quot; alt=&quot;Deviations of the lengths of the winter solstice night from the mean from 1900 to 2100, at 60, and 30 degrees north and south latitude. The length of the night on the December solstice has been increasing rather than decreasing south of the equator.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Python script used to create the above (python3, not python2):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import ephem&#xA;import matplotlib&#xA;import matplotlib.pyplot&#xA;&#xA;start_year = 1900&#xA;end_year = 2100&#xA;&#xA;&#xA;class SolsticePlot :&#xA;    DayLength, DayLengthDev, NightLength, NightLengthDev = range(4)&#xA;&#xA;    _figure_titles = (&#xA;        'solstice_day_length',&#xA;        'solstice_day_length_dev',&#xA;        'solstice_night_length',&#xA;        'solstice_night_length_dev')&#xA;&#xA;    _plot_titles = (&#xA;        &quot;December solstice day length&quot;,&#xA;        &quot;December solstice day length deviation from mean&quot;,&#xA;        &quot;December solstice night length&quot;,&#xA;        &quot;December solstice night length deviation from mean&quot;)&#xA;&#xA;    _ylabels = (&#xA;        &quot;Day length (hours)&quot;,&#xA;        &quot;Day length deviation (seconds)&quot;,&#xA;        &quot;Night length (hours)&quot;,&#xA;        &quot;Night length deviation (seconds)&quot;)&#xA;&#xA;    def __init__ (self, ptype, show, save) :&#xA;        self.option = ptype&#xA;        self.show = show&#xA;        self.save = save&#xA;&#xA;    def figure_title (self) :&#xA;        return self._figure_titles[self.option]&#xA;&#xA;    def plot_title (self) :&#xA;        return self._plot_titles[self.option]&#xA;&#xA;    def ylabel (self) :&#xA;        return self._ylabels[self.option]&#xA;&#xA;    def use_day_length(self) :&#xA;        return self.option == self.DayLength or \&#xA;            self.option == self.DayLengthDev&#xA;&#xA;    def use_deviations(self) :&#xA;        return self.option == self.DayLengthDev or \&#xA;            self.option == self.NightLengthDev&#xA;&#xA;&#xA;    def _next_interval(self, obs, sun) :&#xA;        obs.date = obs.next_transit(sun) # Noon&#xA;        sun.compute (obs)&#xA;        if self.use_day_length() :&#xA;            return obs.next_setting(sun) - obs.previous_rising(sun)&#xA;        else :&#xA;            return obs.next_rising(sun) - obs.next_setting(sun)&#xA;&#xA;&#xA;    def _extremal_length(self, year, lat, lon) :&#xA;&#xA;        obs = ephem.Observer()&#xA;        obs.lat = str(lat)&#xA;        obs.lon = str(lon)&#xA;&#xA;        sun = ephem.Sun()&#xA;&#xA;        # Get day or night lengths in 6 day window about the solstice.&#xA;        solstice = ephem.next_solstice(42357+(year-2015)*365.25)&#xA;        ndays = 6&#xA;        obs.date = solstice-ndays/2&#xA;        sun.compute (obs)&#xA;        lengths = [24.0*self._next_interval(obs, sun) for _ in range(ndays)]&#xA;&#xA;        # Find the shortest or longest, as appropriate.&#xA;        if (lat &amp;gt;= 0.0) == self.use_day_length() :&#xA;            index = min(range(ndays), key=lengths.__getitem__)&#xA;        else :&#xA;            index = max(range(ndays), key=lengths.__getitem__)&#xA;        if (index == 0) or (index == ndays-1) :&#xA;            raise RuntimeError(&quot;Can't find shortest day&quot;)&#xA;&#xA;        return lengths[index]&#xA;&#xA;&#xA;    def ydata (self, years, lat, lon) :&#xA;        extrema = [self._extremal_length(y, lat, lon) for y in years]&#xA;        if self.use_deviations() :&#xA;            avg = sum(extrema) / len(extrema)&#xA;            return [3600*(x-avg) for x in extrema]&#xA;        else :&#xA;            return extrema&#xA;&#xA;&#xA;    def plot(self, years, latitudes, lon) :&#xA;        matplotlib.pyplot.figure(self.figure_title())&#xA;        for ilat,lat in enumerate(latitudes) :&#xA;            color = &quot;C{:d}&quot;.format(ilat+1)&#xA;            label = &quot;{:d}\N{DEGREE SIGN} N&quot;.format(lat)&#xA;            matplotlib.pyplot.plot(years, self.ydata(years, lat, lon),&#xA;                color=color, linestyle='solid', marker='+', label=label)&#xA;        matplotlib.pyplot.title(self.plot_title())&#xA;        matplotlib.pyplot.ylabel(self.ylabel())&#xA;        matplotlib.pyplot.xlabel('Year')&#xA;        matplotlib.pyplot.legend()&#xA;        if self.show :&#xA;            matplotlib.pyplot.show()&#xA;        if self.save :&#xA;            matplotlib.pyplot.savefig(&#xA;                self.figure_title()+'_'+tag+'.png', dpi=200)&#xA;        matplotlib.pyplot.close()&#xA;&#xA;&#xA;if __name__ == &quot;__main__&quot; :&#xA;    years =  list(range(start_year,end_year+1))&#xA;    show = False&#xA;    save = True&#xA;    matplotlib.style.use('seaborn-colorblind')&#xA;    lon = 0&#xA;    for tag, latitudes in [&#xA;        ('m60', [-60]),&#xA;        ('60',[60]),&#xA;        ('20',[20]),&#xA;        ('60_40_20', [60, 40, 20]),&#xA;        ('pm60_30',[60,30, -30, -60])] :&#xA;        for ptype in range(4) :&#xA;            SolsticePlot(ptype, show, save).plot(years, latitudes, lon)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Over the very long term (a hundred thousand of years or more), the Earth's rotation rate is getting slower. This is a true secular trend that will eventually result in increases in the lengths of both the day and the night on the day of the year with the least daylight of the year, regardless of other effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This very slow secular increase is counteracted or exacerbated by cyclical effects. One is the Earth's axial tilt. This varies cyclically, but with a period of 41000 years. Over the short term (hundreds of years is short in this context), the effects of changing axial tilt are much stronger than are the effects of the secular increase in the length of day. The Earth is currently in a phase where the axial tilt is decreasing. This makes the length of daylight increase on the &quot;shortest day of the year&quot;, and hence makes the length of that longest night decrease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A much shorter term cyclical effect results from the location of the Moon with respect to the Earth and Sun. The Earth and Moon together orbit the Sun. This means that when the Moon is above the plane of this orbit, the Earth will be slightly below, and vice versa. This subtly changes the lighting of the Earth. Where this occurs varies as the lunar node precesses over an 18.6 year period.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2017-12-23T17:04:39.070" LastActivityDate="2017-12-23T17:04:39.070" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24138" PostTypeId="2" ParentId="16450" CreationDate="2017-12-23T16:15:52.240" Score="-1" Body="&lt;p&gt;The reason the earth is blue is Rayleigh scattering -- same reason the sky is blue.  Think of it as a big, faintly glowing blue light bulb.&lt;/p&gt;&#xA;" OwnerUserId="20509" LastActivityDate="2017-12-23T16:15:52.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24139" PostTypeId="1" AcceptedAnswerId="24140" CreationDate="2017-12-23T18:41:51.170" Score="6" ViewCount="600" Body="&lt;p&gt;How is dark matter different from ordinary (visible) matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Does dark matter consists of invisible astronomical objects like white dwarfs, black holes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Has dark matter not been directly observed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a better explanation for dark matter. I could not understand completely the difference between dark matter and ordinary matter.&lt;/p&gt;&#xA;" OwnerUserId="20510" LastEditorUserId="2365" LastEditDate="2018-02-04T14:26:05.850" LastActivityDate="2018-02-04T14:26:05.850" Title="How is dark matter different from ordinary (visible) matter?" Tags="&lt;dark-matter&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24140" PostTypeId="2" ParentId="24139" CreationDate="2017-12-23T18:48:47.853" Score="6" Body="&lt;p&gt;Nobody knows it for sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is known:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;it interacts gravitationally&lt;/li&gt;&#xA;&lt;li&gt;it doesn't interact electromagnetically&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&quot;Dark&quot; doesn't mean &quot;black&quot; here, it means here &quot;invisible&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many calculations and theories, the most popular one says it may be from LSPs (least supersymmetric particles).&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="1891" LastEditDate="2017-12-23T21:54:44.257" LastActivityDate="2017-12-23T21:54:44.257" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24141" PostTypeId="1" CreationDate="2017-12-23T21:28:41.177" Score="3" ViewCount="1199" Body="&lt;p&gt;Assume you are given a star map that doesn't have the name of the stars on it. You know the latitude, longitude of the observer and the local time (year, month, day, local time). How could I proceed to find the stars, planets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MZw1j.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MZw1j.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes: This requirement is given by a problem with the photo shown above. You cannot go outside, nor use other resources. &#xA;The stellar map represents the sky of a city ($44^\circ 26'$ N, $26^\circ 06'$ E, local time 00:47, date 04.03.2016).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Mark on the map the cardinal points&lt;/li&gt;&#xA;&lt;li&gt;Identify on the map the planets from the solar system&lt;/li&gt;&#xA;&lt;li&gt;Identify Polaris&lt;/li&gt;&#xA;&lt;li&gt;Identy the circumpolarity zone. Identify at least 2 constellations outside the zone&lt;/li&gt;&#xA;&lt;li&gt;Find the galactic equator and the ecliptic&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="20265" LastEditorUserId="7982" LastEditDate="2017-12-26T17:31:44.830" LastActivityDate="2017-12-27T00:29:09.683" Title="Star map - finding stars" Tags="&lt;star&gt;&lt;night-sky&gt;" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24142" PostTypeId="2" ParentId="24139" CreationDate="2017-12-23T21:55:54.697" Score="9" Body="&lt;p&gt;There is certainly one and possibly two types of dark matter to solve two dark matter problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first dark matter problem is that most gravitating matter is dark. Most of the inferred mass that appears to be responsible for the observed gravitational dynamics of large scale structures like galaxies and clusters of galaxies cannot be seen (i.e. is dark). That is, we can't detect it through visible light or any other electromagnetic waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is possible that you could make up this mass out of very low mass stars,   cold white dwarfs, planets, black holes, lost golf balls or indeed any other form of cold, non-luminous matter. Such objects (bar the golf balls perhaps) certainly do exist, are known as &lt;em&gt;baryonic&lt;/em&gt; dark matter (because they are made of &quot;normal&quot; protons and neutrons - baryons), but there is unlikely to be enough of this stuff to explain the observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second problem is that most of the dark matter must be &lt;em&gt;non-baryonic&lt;/em&gt;. This is deduced from the lack of sufficient plausible baryonic dark matter to explain the dynamics of galaxies and clusters, but also from estimates of the primordial abundances of helium, deuterium and lithium. Big bang nucleosynthesis depends on the density of baryonic matter compared to the universe as a whole and it looks like baryonic matter can only be responsible for a fifth of the dark matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the bulk of the dark matter is supposed to be some non-baryonic stuff, probably in the form of particles that do not interact via the electromagnetic interaction and therefore do not emit or absorb light.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-12-23T21:55:54.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24143" PostTypeId="1" AcceptedAnswerId="24144" CreationDate="2017-12-23T22:15:50.150" Score="4" ViewCount="77" Body="&lt;p&gt;By using only the electromagnetic signature of a star, could a star be distinguished with reliable accuracy from any other star?&lt;br&gt;&#xA;To elaborate a little, say we have a collection of about 200,000 stars. By using EM spectrum alone (and any means of analysis, such as spectral analysis), and by tagging each star with its corresponding EM signature, could there be similarities too close to fail a distinction between any two or more stars?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, of course, I know that such a thing is possible. I'm not asking whether it is possible. I just want to know the odds of such a lack of distinction, considering all the factors of stellar evolution. It's either in the ballpark or not.&lt;/p&gt;&#xA;" OwnerUserId="19454" LastEditorUserId="2365" LastEditDate="2017-12-23T22:45:24.780" LastActivityDate="2018-12-20T08:52:43.107" Title="Stellar electromagnetic signatures" Tags="&lt;spectroscopy&gt;&lt;stellar-evolution&gt;&lt;stellar-structure&gt;&lt;spectral-type&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24144" PostTypeId="2" ParentId="24143" CreationDate="2017-12-23T23:29:18.593" Score="3" Body="&lt;p&gt;The spectrum of a star is almost certainly a unique fingerprint. Even though stars are born in clusters, formed from the reasonably chemically homogeneous environment of a giant molecular cloud, there are likely to be &lt;em&gt;small&lt;/em&gt; differences in their local environment. Furthermore their formation environments, and the stars themselves at later times, can be polluted by external events (e.g. nearby supernovae). Even if their chemical abundances were identical, their spectra would not be because of differences in their mass, rotation and hence their temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But&lt;/strong&gt;, the differences between stars may be smaller than the measurement precision, rendering their spectra effectively indistinguishable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give any sort of definitive answer requires some definition of what types of stars are being observed, their temperature, gravity and chemical abundance dispersion, at what spectral resolution and with what signal to noise ratio in what waveband.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment I would say this is &lt;em&gt;impossible&lt;/em&gt; for any star in our Galaxy. The differences in the spectra between one star and its most similar &quot;twin&quot; (the word might be appropriate here, as they were likely born in the same stellar nursery) are &lt;em&gt;far&lt;/em&gt; smaller than we can distinguish with current data and techniques.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-12-19T07:17:18.983" LastActivityDate="2018-12-19T07:17:18.983" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24145" PostTypeId="2" ParentId="24141" CreationDate="2017-12-24T00:54:54.230" Score="4" Body="&lt;p&gt;NOTE: Information on question and comments has been changing a lot, so please beware that some parts of the answer may be related to previous versions of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Short answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a question to check if you know the sky. Once you can identify constellations and planets on the sky, the answer on the map is quite obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Long answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer this question you need to be familiar with the northern hemisphere sky. If you don't know it, you would need a properly labelled map to answer the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suggested steps:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Assuming that the map just represents the visible portion of sky and the horizon is printed in it, first step would be to look for Polaris. The point of the horizon which is closest to Polaris is the North. The South is in the opposite edge of the map. To locate East and West, place the map with South lower and then East is at left and West is at right - just opposite as they would be in a map of Earth, but this is a map of Sky.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I think it would be easier to identify constellations before planets. The easier constellations are likely to be the circumpolar ones (those that we can always see in the northern hemisphere).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To find the planets, you should find the Ecliptic by looking at zodiacal constellations and then check for points that don't belong to those constellations. Of course, to do that you need to know the constellations. A helpful piece of information is that at 00:47 local time, the Sun, Mercury and Venus can't be seen. If the map shows the planets visible with naked eye, Mars, Jupiter and Saturn may be there in addition to the Moon.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: addition about 1st step (finding north)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your map is drawn inside of a circle. That circle represents the horizon, and Polaris is not in the centre of the map. Therefore, there is a point in the horizon that is the closest point to Polaris. That point is the north.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this just comes from the fact that Polaris is always in the north. If you look at Polaris in the sky, you are facing north - just as if you look at the Sun at noon, you are facing south.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To better understand this, I suggest taking a sky map and a compass, going outside and trying to find Polaris tonight in the actual sky - tonight or the next available starry night.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: finding Polaris&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find a constellation or a given star, you need to know the constellations or compare your map of the actual sky with a labelled star map. The question is not very different to (for example) &quot;find Ireland in a blank world map&quot;: either you know the world and where Ireland is, or you have a labelled map to compare. Star maps aren't that different from geographical maps. In Earth maps you know where you are by looking at geographical features, and in sky you know where you are by looking at stars positions. To find Polaris, look for Ursa Minor; finding Ursa Major or Casiopea may help, too.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addition after the map was included in the question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;North is in the right side of the map. Polaris is the dot halfway from the center and the right side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest clue to find Polaris is the conspicuous Ursa Major in the center of the map. As others have suggested, two stars of the Ursa Major point to Polaris. Furthermore, some stars from Ursa Minor are easy to spot over Polaris.&lt;/p&gt;&#xA;" OwnerUserId="14097" LastEditorUserId="14097" LastEditDate="2017-12-26T20:20:07.357" LastActivityDate="2017-12-26T20:20:07.357" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="24146" PostTypeId="1" AcceptedAnswerId="24152" CreationDate="2017-12-24T06:05:54.040" Score="2" ViewCount="52" Body="&lt;p&gt;Let me know if this is more appropriate for worldbuilding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, I'd like to know how tidal effects would behave in these circumstances (if it's even stable):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An icy moon orbiting a gas giant, with the gas giant orbiting a red dwarf. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a detail that may manipulate the answer:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I want the moon (if possible) to live in the red dwarf's habitable zone. It's okay if it periodically leaves the habitable zone, so long as it returns. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I just need to know &lt;strong&gt;where tidal locking occurs&lt;/strong&gt; (if at all), and &lt;strong&gt;how the moon receives most of it's heat&lt;/strong&gt; (through internal heating like tidal forces or though external heating via the red dwarf).&lt;/p&gt;&#xA;" OwnerUserId="20515" LastActivityDate="2017-12-24T12:49:58.760" Title="How would tidal effects behave in this setup: An icy moon orbiting a gas giant, as the gas giant orbits a red dwarf?" Tags="&lt;orbit&gt;&lt;gas-giants&gt;&lt;red-dwarf&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2017-12-26T23:08:01.413" ContentLicense="CC BY-SA 3.0" />
  <row Id="24147" PostTypeId="1" AcceptedAnswerId="24149" CreationDate="2017-12-24T06:55:17.390" Score="1" ViewCount="650" Body="&lt;p&gt;In artistic renditions of space scenes, there were stunning depictions of gasses or galaxies with bold red and blue colors immediately adjacent to one another, in contact as water is with oil in a container. I wonder, are there known arrangements of gas like that, with both ends of the color spectrum adjacent to one another? If so, what are some examples, what is known about them, and what causes the two colors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One reason I'd call it into question is that I'd imagine the properties of boldly blue gasses are different than those of red gasses. Maybe that assumption is wrong because the property of color isn't necessarily associated with other properties. Another assumption would be that gasses with very different properties wouldn't find themselves collecting adjacent to one another in space. That assumption could also be wrong - I'm not an astronomer and only have basic understanding of physics and chemistry.&lt;/p&gt;&#xA;" OwnerUserId="15917" LastEditorUserId="88" LastEditDate="2017-12-24T08:35:39.120" LastActivityDate="2017-12-24T10:58:40.170" Title="Are there clouds/clusters of gas in space that have red and blue color adjescent to one another?" Tags="&lt;light&gt;&lt;visualization&gt;&lt;spectra&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24149" PostTypeId="2" ParentId="24147" CreationDate="2017-12-24T09:59:15.410" Score="2" Body="&lt;p&gt;Images that you see of nebulae show more intense colour than you could ever see with your eyes. Most nebulae are too dim to for our colour vision to be very active when looking at them. Cameras are designed to try and mimic human sight, to produce an image with colours that is close to what we would see. Telescopes are doing science and the colours are there to enhance features that we couldn't otherwise see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A telescope like Hubble will image a region at several wavelengths. Some of these wavelengths might be in the infrared or ultraviolet, beyond human sight. Then a colour is assigned to each wavelength, and the images are mixed. (See this description of &lt;a href=&quot;http://bf-astro.com/hubblep.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;how to achieve this with photoshop&lt;/a&gt;) This brings out details that would be lost in natural colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there are two main reasons that dust and gas in a nebula might shine. One is that it is reflecting the light of nearby stars, the other is that it has been energised and is emitting its own light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen gas will emit light with a red colour (in the right conditions), and some stars shine in blue-white light. If an emission nebula is next to a reflection nebula then you will get blue-on-red. A good example of this is the Trifid Nebula in Sagittarius. As in &lt;a href=&quot;http://www.astronomersdoitinthedark.com/index.php?c=151&amp;amp;p=450&quot; rel=&quot;nofollow noreferrer&quot;&gt;this photo&lt;/a&gt; by &lt;a href=&quot;http://www.astronomersdoitinthedark.com/index.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Scott Rosen&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nMfY3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nMfY3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The lower, reddish part of the nebula contains the emission nebula — hydrogen gas which is emitting red hydrogen light since it is being heated by the hot young stars within the nebula. The upper blue portion of the Trifid is a reflection nebula — it consists of dust particles that are reflecting the light from the hot stars contained within the emission nebula. Because the hot stars are mostly blue, we see the reflection nebula as having a blue hue. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2017-12-24T10:58:40.170" LastActivityDate="2017-12-24T10:58:40.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24150" PostTypeId="2" ParentId="24141" CreationDate="2017-12-24T11:50:06.857" Score="1" Body="&lt;p&gt;I think the previous hints were already quite extensive, so I'll comment on the planets.&lt;br&gt;&#xA;You should know that Venus and Mercury can't be visible at midnight (which is not a random time for those planets, think why!).&lt;br&gt;&#xA;The total number of possibly visible planets in the whole solar system is 5. If you're taking an Olympiad on the topic then I'd assume you had an introductory class into astronomy at least, then it should be clear why it's 5.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have been given the information that there are 3 planets visible. So from 5-2=3 you know which planets that must be. Then the only thing you have to do is to order them according to magnitude to assign them their Names.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2017-12-24T11:50:06.857" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24152" PostTypeId="2" ParentId="24146" CreationDate="2017-12-24T12:49:58.760" Score="0" Body="&lt;p&gt;I'm not sure exactly what your question is, but Tidal locking is fairly common for moons.   If the Gas giant was in the habitable zone of the red dwarf star, then it would need to be fairly close and as a result the Moon would need to be fairly close to the Gas giant - it should be tidally locked to the gas giant, not the star.  (A moon being tidally locked to a star is probably impossible because the planet will always have the greater tidal effect on the moon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your scenario, you have a moon, tidally locked to a gas giant planet, which orbits a star.   Tides don't exist in a perfectly circular tidal lock because the pressure is constant, but there's no such thing as a perfectly circular orbit, so there will always be some tidal squeezing as the moon moves around the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real question is how eccentric the Moon's orbit is around the gas giant planet, because it's eccentricity will determine it's tidal squeezing and any libration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as periodic habitable zones, that would depend on how eccentric the orbit of the gas giant was around the star.  You could have it moving in and out of the habitable zone, but it would presumably be a short-period orbit, so it would be a matter of days outside, then days inside, perhaps a week or two outside and a week or two inside, unless you went highly eccentric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these are general points.  There's lots of variables.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no one answer for where the Moon would receive most of it's heat - you'd need to work out the specifics or orbit, distance and solar temperature and then give the Moon a specific orbit and then you could work out an approximation of where more heat comes from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this feels like worldbuilding to me.   It's too hypothetical for astronomy.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-12-24T12:49:58.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24156" PostTypeId="2" ParentId="1145" CreationDate="2017-12-24T14:40:19.187" Score="6" Body="&lt;p&gt;The mass of the galaxy (mostly in the form of dark matter) is in a roughly spherical blob. So if you look at mass, the galaxy isn't a disc, it is a spheroid.  But Dark Matter is invisible, and what we can see (stars, gas etc) is in a disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that Dark matter and the normal matter behave differently is that when gas flows there is &quot;friction&quot; (Dark matter doesn't interact with itself or normal matter). This causes the gas to heat up, and that heat energy is then released (as infra-red, light and so on) This means that over time the gas in the galaxy will tend to fall to a lower level.  However the gas also has angular momentum (it is rotating), and angular momentum must be conserved (it can't be radiated away like energy). So the gas will try to fall into a low energy configuration that can maintain angular momentum. The shape that achieves this is a disc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any gas clouds that are not orbiting in the plane of the disc will hit it,  and over time they will be pulled into the same disc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gas clouds produce stars, and so most stars will also be in the plane of the disc. Very old clusters of stars in globular clusters however can be found in a spherical pattern around the disc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So galaxies form disc shapes because the gas that makes stars falls into a disc shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, not all galaxies are discs. When disc-shaped galaxies collide, this can disturb the orbits of the stars, and you get a galaxy which is &quot;blob&quot; shaped, these are called elliptical galaxies, and are very common. Small galaxies also often don't have a disc structure either. These are called irregular galaxies. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-12-24T14:40:19.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24158" PostTypeId="1" AcceptedAnswerId="24159" CreationDate="2017-12-24T23:59:40.637" Score="2" ViewCount="178" Body="&lt;p&gt;I have been researching this on google, and I think there is a process. What is this process called? I have forgotten.&lt;/p&gt;&#xA;" OwnerUserId="16048" LastEditorUserId="16048" LastEditDate="2017-12-25T00:12:23.087" LastActivityDate="2017-12-25T03:11:15.520" Title="Is there any phenomenon where a supernova doesn't anything leave behind?" Tags="&lt;supernova&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24159" PostTypeId="2" ParentId="24158" CreationDate="2017-12-25T03:11:15.520" Score="4" Body="&lt;p&gt;You are probably thinking of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Pair-instability_supernova&quot; rel=&quot;nofollow noreferrer&quot;&gt;pair-instability supernova&lt;/a&gt;. Such a supernova releases more energy than the star's gravitational binding energy, completely disrupting the star and dispersing the contents.  Meaning that with nothing else in the universe, over infinite time every particle from the star will get infinitely far away from any/every other particle.  The mass, energy, etc. of the star doesn't vanish from the universe, though; it just gets spread throughout it.  But there will be no dense remnant left behind: no dwarf, no neutron star, no black hole; over long time scales there won't even be a discernible nebula remaining (though note the star may have shed off significant amounts of material prior to this event, which would interact with the ejecta and may result in some sort of permanent structure suggesting a star used to be there).&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2017-12-25T03:11:15.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24160" PostTypeId="1" AcceptedAnswerId="24168" CreationDate="2017-12-25T03:13:28.637" Score="2" ViewCount="1312" Body="&lt;p&gt;I was skimming through Neil deGrasse Tyson's book &lt;em&gt;Astrophysics for People in a Hurry&lt;/em&gt; when I noticed he says something curious on page 26:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;By now, one second of time has passed [since the big bang]. The universe has grown a few light years across.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could the universe have expanded to a few light years in diameter in just one second if nothing can travel than light? By definition of a light &lt;strong&gt;year&lt;/strong&gt;, wouldn't take a 1+ years for the universe to get this big?&lt;/p&gt;&#xA;" OwnerUserId="20525" LastActivityDate="2017-12-25T11:31:39.700" Title="How could the universe be a few light years across just 1 second after the big bang?" Tags="&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24161" PostTypeId="2" ParentId="24064" CreationDate="2017-12-25T04:08:17.477" Score="4" Body="&lt;p&gt;Gravity follows the law of inverse square.  To put it simply if you double the distance from a gravity source you quarter is effect.  So if you double the distance you are from the earth you feel 1/4g. It's important to note that as the distance increases it will never be 0, it will always be some non-zero value no matter the distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at galactic distances the force of gravity the central black hole has very little effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This only explains part of it.  The other part is conservation of angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force of gravity and the angular momentum is what is responsible for orbits.  In orbital mechanics you raise your orbit by adding speed, not altitude.  Your adding angular momentum which raises your orbit.  To lower your orbit you reduce you speed which reduces your angular momentum and your altitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for things to &quot;fall&quot; into a black hole they must be travailing at a speed where their orbit intersects the event horizon. This is rarely the case or those &quot;things&quot; wouldn't truly be in orbit to begin with. So the very fact that all the &quot;stuff&quot; that makes up the galaxy orbits the central black hole means that it cant just fall into it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These 3 things are always in balance in a stable orbit, force of gravity, speed and altitude ( or distance from the gravity source ).  If you change one of them the other 2 must also changes.  If you decrease speed your altitude goes down, and the gravity increases.  If you increase the gravity the speed must also increase or the altitude will decrease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you see things can't just fall into the black hole.  That said it is my view that eventually everything in the galaxy will fall into the central black hole, however this will take many billions of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this is overly simplifying things, and I am by no means an expert on this stuff.  But it's something I can picture in my mind, the balance between momentum and gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v&lt;/p&gt;&#xA;" OwnerUserId="20526" LastEditorUserId="20526" LastEditDate="2018-01-08T18:56:51.650" LastActivityDate="2018-01-08T18:56:51.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24162" PostTypeId="1" CreationDate="2017-12-25T05:40:30.910" Score="2" ViewCount="70" Body="&lt;p&gt;&lt;em&gt;(For the sake of this question, let's assume that all other factors are malleable.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially creating a permanent lunar eclipse, could a moon orbit around a planet so slowly that it is always opposite the sun from its planet?&lt;/p&gt;&#xA;" OwnerUserId="17257" LastActivityDate="2017-12-25T06:59:26.417" Title="Is it possible for a moon to only be out at night?" Tags="&lt;orbit&gt;&lt;natural-satellites&gt;&lt;lunar-eclipse&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24163" PostTypeId="1" AcceptedAnswerId="41902" CreationDate="2017-12-25T05:49:07.630" Score="7" ViewCount="201" Body="&lt;p&gt;&lt;strong&gt;edit:&lt;/strong&gt; The short form of the question is &lt;em&gt;Why are there TWO telescopes??&lt;/em&gt; A longer version follows. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Large_Binocular_Telescope&quot; rel=&quot;noreferrer&quot;&gt;Large Binocular Telescope&lt;/a&gt; (and &lt;a href=&quot;http://www.lbto.org/leech-hr8799.html&quot; rel=&quot;noreferrer&quot;&gt;LBT blog as source of images below&lt;/a&gt;) has two 8.4 meter diameter mirrors, allowing for the possibility of interference in one dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware that a single baseline, two-element &lt;a href=&quot;https://en.wikipedia.org/wiki/Michelson_stellar_interferometer&quot; rel=&quot;noreferrer&quot;&gt;stellar interferometer&lt;/a&gt; can be used to measure the separation between a pair of stars by looking at the Fourier transform of the intensity vs time of the interference. A single frequency peak is produced by one star, but if there are two close together, there will be &quot;beating&quot; in the time series that shows up as two peaks in the FT. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm wondering if the LBT was designed and built with a plan to do more than that. &lt;em&gt;Just for example&lt;/em&gt;, did plans include an ability to use interferometry to help reconstruct 2D images, or enhance the resolution beyond that which is already provided by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Large_Binocular_Telescope#Adaptive_optics&quot; rel=&quot;noreferrer&quot;&gt;adaptive optics&lt;/a&gt; used in each half separately? Were there other plans or justifications?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possibly helpful; &lt;a href=&quot;https://arxiv.org/abs/1608.00037&quot; rel=&quot;noreferrer&quot;&gt;Current Status of the Facility Instrumentation Suite at The Large Binocular Telescope Observatory&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Open image in new window for full size:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rxwF9.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rxwF9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-04-14T03:45:37.310" LastActivityDate="2021-03-11T14:27:38.997" Title="What are the raisons d'être for the Large Binocular Telescope &quot;binocularity&quot;?" Tags="&lt;interferometry&gt;&lt;post-processing&gt;&lt;large-binocular-telescope&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="24164" PostTypeId="2" ParentId="24162" CreationDate="2017-12-25T06:43:59.010" Score="4" Body="&lt;p&gt;If the Earth were alone in space and the hypothetical moon had negligible mass, you can calculate the distance at which the period would be one year using:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a = \left( \frac{GM_E T^2} {4 \pi^2} \right)^{1/3} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $GM_E$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Standard_gravitational_parameter&quot; rel=&quot;nofollow noreferrer&quot;&gt;standard gravitational parameter&lt;/a&gt; of the Earth (about 3.9860E+14 m^3/s^2) and $T$ is the period (about 365.2564 x 24 x 3600 sec). You should get about 2.2 million kilometers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that because the Earth is orbiting around the Sun, the added gravitational effect results in a location a bit closer, about 1.5 million kilometers from the Earth, where a low mass object would &lt;em&gt;tend to&lt;/em&gt; remain at the far side from the Sun of the Earth for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This location is one of the five &lt;a href=&quot;https://en.wikipedia.org/wiki/Lagrangian_point&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lagrangian points&lt;/a&gt;, and has the designator Sun-Earth L2. There is a similar point L1 &lt;em&gt;between&lt;/em&gt; the Earth an the Sun as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is not a stable orbit, even in a perfect &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler%27s_three-body_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;circular restricted three-body problem&lt;/a&gt; (CR3BP or CRTBP) scenario, the least little displacement would cause it to eventually wander away into it's own, somewhat chaotic orbit around the Sun within a few years. The reality of the more elliptical orbits of Solar System bodies plus gravitational perturbations would only reinforce this instability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So find a way to give the hypothetical moon a propulsion system to do &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_station-keeping&quot; rel=&quot;nofollow noreferrer&quot;&gt;station keeping&lt;/a&gt; maneuvers to keep it there, and you've got &lt;em&gt;almost&lt;/em&gt; what you are looking for. The problem is that this point is in the same neighborhood as the point where the Umbra of the Earth's shadow disappears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Earth's orbit is elliptical, the point where the umbra terminates moves a bit, and since there is no &lt;em&gt;exact definition of a Lagrange point's location&lt;/em&gt; for a messy, realistic Solar System, the question of whether a satellite in the vicinity of L2 would be in partial or total eclipse is debatable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you already have the propulsion to do station keeping, you could possibly move it in to the umbra and maintain it there as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sWd0G.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sWd0G.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-12-25T06:59:26.417" LastActivityDate="2017-12-25T06:59:26.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24168" PostTypeId="2" ParentId="24160" CreationDate="2017-12-25T11:31:39.700" Score="3" Body="&lt;p&gt;Your question hinges on the misconception that nothing can travel faster than light. No &lt;em&gt;physical&lt;/em&gt; object can move faster than light, but plenty of non-physical things can. For example, take the straight line $y = mx + c$. If one varies $c$, the x-intercept of the line also changes. It's possible that this x-intercept moves faster than light for a suitably small $m$. The caveat is that this intercept is a mathematical point, and it cannot be used to transfer information. Relativity isn't violated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the universe's expansion, the physical objects in the universe aren't moving faster than light, but they can appear to do so because space is expanding. Let's take Alice and Bob, separated by one meter, and let the space between them expand at some rate. There's no speed-of-light limit on how fast this rate can be. If space expands by a billion meters per second, then one second later, Alice and Bob would be separated by a billion and one meters, but they wouldn't have traveled faster than the speed of light since from their own point of view, they're both stationary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way, our best guess of the current universe is that the expansion is speeding up. A very far away object might eventually appear to recede from us faster than the speed of light. Relativity remains unviolated, although it means that we'll eventually be causally disconnected from these faraway objects (that is, we cannot affect or even observe them, and vice versa). Something of this sort - exponentially-increasing expansion - happened in the very early universe.&lt;/p&gt;&#xA;" OwnerUserId="20348" LastActivityDate="2017-12-25T11:31:39.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24171" PostTypeId="2" ParentId="24141" CreationDate="2017-12-25T14:15:49.917" Score="2" Body="&lt;p&gt;Interesting question :).  To find polaris, find Ursa Major, then use the pointer stars (the two in the bowl of the big dipper furthest away from the &quot;handle&quot; part) and use those to find Ursa Minor/Polaris. [Edit - since you're at around 45 degrees latitude, Polaris is going to be somewhere on a circle centred on the middle of the map with radius going about  halfway to the horizon, which gives you a cross check].&#xA;That gives you two constellations. The W shape of Casseopeia is also circumpolar - roughly the same distance opposite polaris from the big dipper star where the &quot;bowl&quot; joins the &quot;handle&quot;. That's three. Depending on the time of year, you should be able to find at least two more easily recognisable constellations - Orion and Leo are easy to remember, as are Lyra and Delphinus and Sagitta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find north, draw a line through Polaris from the center of the map. Where that hits the horizon is North. Going the opposite way is South. Rotate the map so North is at the top and West is to the right, East to the left (remember you're looking up at the Sky, so they're reversed from a normal looking-down-at-the-ground map). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For circumpolar stars, it depends on the map projection used. If it's using a planisphere projection, then you can draw a circle centered on the polaris and just touching the horizon (But in that case, I don't think the horizon is going to be a circle unless there's something special about 45 degrees).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the celestial equator, you want a circle 90 degrees out centred on Polaris (again, assuming a planisphere projection). That goes through the northernmost star in Orions belt, if orion is visible. Otherwise, at 45 degrees latitude, it's two thirds of the way from Polaris to the horizon on your north-south line. (North to Polaris is 45 degrees, Polaris to south is 135 degrees (45 to zenith then 45 to equator then 45 to horizon). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the planets (Mars, Jupiter, and Saturn) either remember roughly where jupiter and saturn are (since the date is close to the present), or order by magnitude. (Mercury and Venus are ruled out since as inner planets they're closer to the sun, and near midnight they're on the the other side of the earth and not visible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the map isn't using a planisphere style projection (where circles of equal declination map to circles centred on polaris on the map) then trying to draw the celestial equator and circumpolar stars needs a much better memory for how the declination lines fit with the constellations, since they won't be circular on the map&lt;/p&gt;&#xA;" OwnerUserId="17016" LastEditorUserId="17016" LastEditDate="2017-12-27T00:29:09.683" LastActivityDate="2017-12-27T00:29:09.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24172" PostTypeId="2" ParentId="24064" CreationDate="2017-12-25T16:32:58.660" Score="3" Body="&lt;p&gt;Well, I'm not a physics student, but I think people usually nurture the misconception of &quot;sucking power&quot; of a black hole for a reason. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us consider Newton's equation for gravity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = {Gm_im_j\over r_{ij}^2}$&#xA;for two bodies i and j, and $r_{ij}$ is the distance between the centre of mass of two bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the Sun today suddenly decided to become a black hole without shedding weight, it won't affect Earth's orbit, because even if Sun's volume has changed, $r_{ij}$ remains constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why Black Holes &quot;suck&quot; is because, since they occupy extremely less volume compared to planets and stars, you can get the $r_{ij}$ component to be really really small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Correct me if I'm wrong.&lt;/p&gt;&#xA;" OwnerUserId="20537" LastActivityDate="2017-12-25T16:32:58.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24173" PostTypeId="1" AcceptedAnswerId="24176" CreationDate="2017-12-25T16:34:31.613" Score="3" ViewCount="453" Body="&lt;p&gt;Are there examples of single stars that are a part of multiple constellations? What about stars that are part of one Western constellation and one non-Western? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="20538" LastActivityDate="2017-12-25T17:27:37.113" Title="Stars in multiple constellations" Tags="&lt;constellations&gt;&lt;night-sky&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24175" PostTypeId="2" ParentId="24064" CreationDate="2017-12-25T16:39:36.760" Score="2" Body="&lt;p&gt;Be patient, eventually it will unless the galaxy's rate of expansion exceeds the black hole's gravitational growth as it consumes the matter around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that scenario the galaxy will eventually diffuse, with its matter continuing to travel away from the black hole until it encounters another galaxy, at which point it stands a good chance of eventually getting sucked into that's galaxy's black hole. Nothing survives forever.. :-) &lt;/p&gt;&#xA;" OwnerUserId="20539" LastActivityDate="2017-12-25T16:39:36.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24176" PostTypeId="2" ParentId="24173" CreationDate="2017-12-25T17:22:14.553" Score="5" Body="&lt;p&gt;In traditional constellation shapes / outlines, yes. But the International Astronomical Union (IAU) standardised on a set of 88 constellations in 1922, and from 1924 to 1930 formalised the constellation boundaries, splitting the sky into separarate areas (that cover the complete sky) so there's no ambiguity now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, before the constellations were standardised in 1930, the same star (on the boundary of Auriga and Taurus) was known both as Gamma Aurigae and Beta Tauri.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the traditional constellations of Ophiuchus (The serpent bearer)  and Serpens (the snake) (Where Serpens ran across Ophiucus) was handled by splitting them into Ophiucus and Serpens Caput (snakes head) and Serpens Cauda (snakes tail). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;see &lt;a href=&quot;http://www.ianridpath.com/boundaries.htm&quot; rel=&quot;noreferrer&quot;&gt;http://www.ianridpath.com/boundaries.htm&lt;/a&gt; for more details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different cultures have different traditional constellations. If you have a look at the free Stellarium program (see &lt;a href=&quot;http://stellarium.org/&quot; rel=&quot;noreferrer&quot;&gt;http://stellarium.org/&lt;/a&gt;), that has a set of overlays showing constellations from a number of different cultures. (I don't remember if that's part of the standard installation, or an optional extra you can download).&lt;/p&gt;&#xA;" OwnerUserId="17016" LastEditorUserId="17016" LastEditDate="2017-12-25T17:27:37.113" LastActivityDate="2017-12-25T17:27:37.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24182" PostTypeId="2" ParentId="12327" CreationDate="2017-12-26T08:19:39.653" Score="3" Body="&lt;p&gt;A supplemental answer to &lt;a href=&quot;https://astronomy.stackexchange.com/a/19511/7982&quot;&gt;probably someone's answer&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Why this can indeed be called interferometry:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Once one thinks in terms of physical optics (e.g. $\text{exp}(j(\omega t - \mathbf{k} \cdot \mathbf{r} ))$ ) instead of ray optics, imaging is always an interference problem, and the math behind correlating signals from an array of radio telescopes to produce an image is &lt;em&gt;not so much different than&lt;/em&gt; the math behind calculating an intensity pattern at the focal plane of an optical telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the &lt;em&gt;Aperture Mask Interferometry&lt;/em&gt; section on page 5 of the &lt;a href=&quot;https://jwst.stsci.edu/files/live/sites/jwst/files/home/instrumentation/technical%20documents/jwst-pocket-guide.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;JWST Pocket Guide&lt;/a&gt; (see below) you can see that they have no problem using terms like &quot;interferometry&quot; and &quot;non-redundant baselines&quot; as one would when laying out a sparse pattern of dish antennas in an array. See for example a Google Maps image of the &lt;a href=&quot;https://en.wikipedia.org/wiki/MeerKAT&quot; rel=&quot;nofollow noreferrer&quot;&gt;Meerkat Array&lt;/a&gt; Core below, which has &lt;a href=&quot;https://astronomy.stackexchange.com/q/16621/7982&quot;&gt;already produced images&lt;/a&gt; with only 16 sites occupied with active dishes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the mask shown in the question is a &lt;em&gt;pupil mask&lt;/em&gt;, the pupil plane is conjugate to the telescope's aperture. So it's very similar to having seven large hexagonal holes in front of the JWST, or just using seven of the telescopes primary mirror's hexagonal mirror elements (with additional masking, these hexagons to not cover the full size of the individual elements, as shown in the illustration).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important distinction though is that these seven small apertures are of the same order in size as their separation, while in a large radio telescope array, the spacing between receivers is usually somewhere between &lt;em&gt;much larger&lt;/em&gt; and &lt;em&gt;much much larger&lt;/em&gt; than the aperture of individual elements. So the analogy breaks down at this point.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;What Aperture Masking Interferometry is for:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The purpose of using the mask is to enhance the system's resolution by narrowing the central peak of the point spread function compared to what you would get from the full aperture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By loosing contrast in the full field due to diffraction artifacts, as well as loosing ~90% of the transmitted intensity, you can &lt;em&gt;enhance contrast&lt;/em&gt; near the center of the field by narrowing the central diffractive peak, in this case in the range of 70 to 400 milli-arc seconds, according to the JAM Team's (JWST Aperture Masking Team) slides linked below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the first link:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Light admitted by 7 apertures in an otherwise opaque pupil mask interfere to produce an interferogram on the detector. This interfogram has a sharper core than is provided by normal &quot;direct&quot; imaging. The advantage is significant: while the ability to separate closely spaced objects with normal imaging is given by the familiar Rayleigh criterion (separation  δθ=1.22λ/Dδθ=1.22λ/D , where λ is the wavelength of light and D is the diameter of the telescope),  interferometry can resolve objects as close as δθ=0.5λ/Dδθ=0.5λ/D (the Michelson criterion).  AMI allows planetary or stellar companions that are up to ~9 magnitudes fainter than their host star and separated by 70–400 mas to be detected and characterized.  AMI  can also be used to reconstruct high-resolution maps of extended sources, such as active galactic nuclei.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can read more about it in: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;JWST's &lt;a href=&quot;https://jwst-docs.stsci.edu/display/JTI/NIRISS+Aperture+Masking+Interferometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;NIRISS Aperture Masking Interferometry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;ArXiv &lt;a href=&quot;https://arxiv.org/abs/1406.6882&quot; rel=&quot;nofollow noreferrer&quot;&gt;NIRISS aperture masking interferometry: an overview of science opportunities&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;The &lt;em&gt;JAM Team's:&lt;/em&gt; &lt;a href=&quot;https://jwst.nasa.gov/resources/Sivaramakrishnan2012_08_08_UMichAnnArbor.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Non-Redundant Aperture Masking Interferometry (AMI) and Segment Phasing with JWST-NIRISS&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OpPbi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OpPbi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Below:&lt;/strong&gt; MeerKAT &lt;em&gt;array core&lt;/em&gt; from Google Maps at (30.7136109S, 21.4399576E).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/l9OXm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l9OXm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2017-12-26T08:40:03.643" LastActivityDate="2017-12-26T08:40:03.643" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24183" PostTypeId="1" CreationDate="2017-12-26T09:12:22.847" Score="11" ViewCount="436" Body="&lt;p&gt;I understand that we can count pulses in time, but how do we know that the magnetosphere is rotating along with the surface of a neutron star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we measure the rotation of a neutron star in any other way like red-blue shifts? It should be pretty difficult due to the small radius and big distance but is it possible? Is it already done?&lt;/p&gt;&#xA;" OwnerUserId="15282" LastEditorUserId="1056" LastEditDate="2018-01-02T09:25:28.823" LastActivityDate="2018-01-02T09:25:28.823" Title="How do we know how fast the surface of a neutron star is rotating?" Tags="&lt;neutron-star&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24184" PostTypeId="1" CreationDate="2017-12-26T09:41:31.047" Score="1" ViewCount="306" Body="&lt;p&gt;If divide the distance to the sun by its diameter, I get the number &lt;strong&gt;107.517&lt;/strong&gt;&lt;br&gt;&#xA;Likewise, if divide the distance to the moon by its diameter, I get the number &lt;strong&gt;110.650&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both numbers are quite similar which means both objects look roughly the same size in the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'll do the same calculation for the Andromeda galaxy.&lt;br&gt;&#xA;Distance: 2,537,000 light years&lt;br&gt;&#xA;Diameter: 220,000 light years&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quotient: &lt;strong&gt;11.531&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get a number that's roughly 1/10th that of the moon.&lt;br&gt;&#xA;This means that Andromeda should look almost 10 times bigger than the moon or the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this calculation correct?&lt;br&gt;&#xA;Does the Andromeda galaxy really look 10 times bigger than the moon in the night sky?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it looks so big, why is it so difficult to see with the naked eye?&lt;/p&gt;&#xA;" OwnerUserId="20545" LastActivityDate="2017-12-26T09:41:31.047" Title="How big is the Andromeda galaxy in the night sky?" Tags="&lt;size&gt;&lt;angular-diameter&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2017-12-26T13:28:53.747" ContentLicense="CC BY-SA 3.0" />
  <row Id="24185" PostTypeId="2" ParentId="23937" CreationDate="2017-12-26T10:01:24.790" Score="1" Body="&lt;p&gt;Maxwell's Smith Prize Essay on the formation of the Rings of Saturn is relevant. Using structural stability as his criterion, he deduced that the 'rings' could only consist of a single almost infinite mass of very small particulates each orbiting in accord with Newtonian dynamics--the correct result as verified by NASA/ Cassini.&#xA; Applying the same approach to a forming  solar system this analysis shows that rings of particulates are a structurally stable attractor for the dynamics, provided the central star is much larger than the mean particulate size.&#xA;I am now  well outside my comfort zone so will leave further inferences to others :).&#xA;&lt;a href=&quot;http://www.clerkmaxwellfoundation.org/Saturn-s_Rings.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Maxwell's essay on rings of saturn&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20449" LastEditorUserId="20449" LastEditDate="2017-12-27T06:53:43.867" LastActivityDate="2017-12-27T06:53:43.867" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24190" PostTypeId="1" AcceptedAnswerId="24191" CreationDate="2017-12-26T14:57:56.090" Score="2" ViewCount="253" Body="&lt;p&gt;I'm a little confused. Could you help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is;&#xA;If we increase the radius of a telescope twice, how much does the light gathering power change? What about resolution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it increase 2^2?&#xA;For second part i know As the diameter of the telescope's objective increases, the resolving power increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="20510" LastActivityDate="2017-12-26T15:29:40.713" Title="Telescope light gathering power and resolution" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24191" PostTypeId="2" ParentId="24190" CreationDate="2017-12-26T15:29:40.713" Score="4" Body="&lt;p&gt;The collecting area of a telescope is roughly proportional to the square of its radius (for a circular aperture). I say roughly because you also have to factor in the small fraction of light blocked by the secondary if we are talking about reflecting telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle the angular resolution decreases as the reciprocal of the radius (that is, the smallest resolvable angle gets smaller). In practice, this may not be the case for ground-based telescopes without adaptive optics capabilities, where the smallest resolvable angle could be determined (for large telescopes at least) by turbulence in the atmosphere (a.k.a. the seeing).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2017-12-26T15:29:40.713" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24192" PostTypeId="2" ParentId="20058" CreationDate="2017-12-26T20:22:44.323" Score="2" Body="&lt;p&gt;Micrometer-sized dust condenses in red-giant atmospheres, then interacts with gas around young stars to grow to mm, then overcomes the &lt;a href=&quot;https://arxiv.org/pdf/1001.0488.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;bouncing barrier&lt;/a&gt; to reach cm to m sized &quot;pebbles.&quot;  (Answer derived from comments by A.P.E.; sadly, theirs links are dead.  &lt;strong&gt;Edit:&lt;/strong&gt; A.P.E. reposted an answer that I could accept.)&lt;/p&gt;&#xA;" OwnerUserId="15835" LastEditorUserId="15835" LastEditDate="2017-12-27T19:34:31.763" LastActivityDate="2017-12-27T19:34:31.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24193" PostTypeId="2" ParentId="24139" CreationDate="2017-12-26T20:59:59.047" Score="-1" Body="&lt;p&gt;Read &lt;strong&gt;Before the Big Bang&lt;/strong&gt; (1997) by Ernest Sternglass.  The short answer is that a big piece of &quot;dark matter&quot; is the source of a quasar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;non-standard&quot; and therefore controversial theory presented in this book says that most &quot;dark matter&quot; might be&lt;br&gt;&#xA;“fragments of the original primeval atom [of Georges Lemaitre’s model --- please google it if you need to]  ejected to large distances in the explosive ‘mini-Bang’ that had to accompany the formation of every cosmological structure [ie, every moon, planet, star, galaxy,  etc.] in a Lemaitre-type model … the existence of quasars and active cores of galaxies over a wide range of distances indicated that there were apparently delayed mini-Bangs in which new galaxies were created, as Maarten Schmidt had conjectured, together with vast amounts of dust and gas ejected into space” [p.211, &lt;strong&gt;BTBB&lt;/strong&gt;].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plus:  “All these results strongly suggested that some of the original fragments … from the Lemaitre [“primeval”] atom had managed to survive in the massive centers of large galaxies for very long periods … these massive electron[-positron] pair fragments were apparently ejected long after the Big Bang, as the Russian astrophysicist Novokov and the Israeli physicist Ne’eman had in fact suggested independently in the mid-1960s … their nuclei would be so massive that they would be invisible black holes, yet they could account for a dominant fraction of the total mass of the universe even today” [p.212, &lt;strong&gt;BTBB&lt;/strong&gt;].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Sternglass’s model, these fragments of the primeval atom are “seeds of galaxies and stars”, and are spread throughout our universe, ever since the Big Bang.  He says that a seed remains dormant for millions or billions of years,lurking in space, and then, after a long “count-down” process, during which the system divides in half, again + again + again,  it suddenly explodes, violently.  He calls this explosion a “mini-Bang”, and says that the “gamma-ray bursters” —(also called “quasars”)— which astronomers have observed since the 1970s, are in fact the “delayed mini-Bangs” which his model predicts.&lt;/p&gt;&#xA;" OwnerUserId="13290" LastEditorUserId="13290" LastEditDate="2017-12-26T21:13:01.917" LastActivityDate="2017-12-26T21:13:01.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24194" PostTypeId="2" ParentId="20058" CreationDate="2017-12-26T23:26:21.987" Score="7" Body="&lt;p&gt;In general, solids are made from from atoms that originate naturally in stellar nucleosynthesis,&lt;br&gt;&#xA;see for example the composition of our sun's atmosphere from an older paper of&lt;br&gt;&#xA;Asplund et al (2005):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/fDPlL.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fDPlL.gif&quot; alt=&quot;Solar element composition&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you see for example, that silicon (Si), an important rock-forming element, has an abundance relative to hydrogen of $10^{-3}$, or for every 1000 hydrogen atoms there is one silicon atom. Also we find lots of oxygen, carbon, magnesium, iron,... all known rock-forming elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above work has been performed using spectroscopy, and in a similar way we can detect solids in space mainly through three means: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Spectroscopy (again): Small particles of $\mu m$-sizes exist in vast numbers in the spectrae of red giants (I'll comment on the formation later), white Dwarves, in cold molecular clouds and even in far away Quasars. Those numbers are so huge that we can detect prominent features like the 10 $\mu m$-feature of the $Si-O$ bond that we find in many earthly minerals. See also the following example of a clear detection of silicate minerals in the envelope of a young star (the broad trough at 10 $\mu m$) (c)Uni Hawaii:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/PM8zW.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PM8zW.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As mentioned, when solids grow, their numbers diminish. Thus big solids around the sizes of mm become invisible in spectrae, but sometimes can be detected through scattered light, as they undergo &lt;a href=&quot;https://en.wikipedia.org/wiki/Mie_scattering&quot; rel=&quot;noreferrer&quot;&gt;Mie-scattering&lt;/a&gt; with visible or infrared light when being very near to a stellar source.&lt;br&gt;&#xA;Then one can take pictures like the one from Benisty et al. 2015 of the young stellar system MWC758, where we see dust spirals possibly caused by the existence of young planets:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/UAG5i.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UAG5i.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Reflection of solar light: As solids grow to cm, meter, kilometer-sizes they become invisible for the above methods. Only at sizes of tens to hundreds of kilometers and only in our solar system can we then detect them again through the reflection of solar light.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now that we have established that solids exist in space, let us review the theoretical perspective of growth mechanisms:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;dust of up to $\mu m$ condenses out in the relatively cool atmospheres of red giants (AGB stars), where the ionized environment helps to grow solids through charge separation. Details e.e. &lt;a href=&quot;https://homepage.univie.ac.at/thomas.posch/doc/Habil.pdf&quot; rel=&quot;noreferrer&quot;&gt;in here&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;this dust is blown away in the strong AGB star winds into the interstellar medium, where it helps to create new star systems (see also the second image I linked).&lt;/li&gt;&#xA;&lt;li&gt;All successive growth to larger sizes has to happen in the interaction with gas, thus it usually happens in the relatively dense environment around young stars&lt;/li&gt;&#xA;&lt;li&gt;The dust has now to overcome the &lt;a href=&quot;https://astrobites.org/2016/01/27/the-bouncing-barrier-of-silicates-and-ices/&quot; rel=&quot;noreferrer&quot;&gt;bouncing barrier&lt;/a&gt; in order to grow into cm to meter-sized particles that we call 'pebbles'. How particles overcome this barrier is unclear in general, but the article I've linked mentions a few solutions.&lt;/li&gt;&#xA;&lt;li&gt;Pebbles are defined by their property that they interact with the orbiting gas in protoplanetary discs through friction. This interaction allows them to trigger the &lt;a href=&quot;https://en.wikipedia.org/wiki/Streaming_instability&quot; rel=&quot;noreferrer&quot;&gt;streaming instability&lt;/a&gt;, a very efficient mechanism that is able to compress clumps of pebbles directly into km-sized solids.&lt;/li&gt;&#xA;&lt;li&gt;Now small comets grow through direct collisions, up to planetary sizes. Only in this stage geological mechanisms can play a role, like the igneous separation of minerals that you've mentioned.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That's it for now. I think with this description it is clear that the whole problem is not chicken-and-egg, but if you think otherwise then please comment.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2017-12-26T23:26:21.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24195" PostTypeId="1" CreationDate="2017-12-26T23:35:53.570" Score="2" ViewCount="725" Body="&lt;p&gt;I know the latitude and longitude of a city ($41^\circ 87'$ N, $87^\circ 62'$ W), the declination and right ascension of a star($16.51^\circ$, $68.98^\circ$) and the local time, date (27.12.2017, 1:20 AM). Could you please show me the exact calculations for hour angle, altitude?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will use the formula $LST=100.46+0.98d+15UT$, where d is the number of days since 1.1.2000 at 00:00 and UT is the universal time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 1:20 AM, $d=6570.055$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;UT is $UT=7:20$ or in hours $UT=7.3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$LST=6648.0639$ or $LST=168.0639$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$HA=LST-RA=168.0639-16.51=151.5539$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is my calculation?&lt;/p&gt;&#xA;" OwnerUserId="20265" LastEditorUserId="20265" LastEditDate="2017-12-30T00:07:13.273" LastActivityDate="2017-12-30T18:29:38.687" Title="Finding hour angle, altitude" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24196" PostTypeId="1" AcceptedAnswerId="24204" CreationDate="2017-12-27T01:42:53.457" Score="13" ViewCount="5367" Body="&lt;p&gt;As far as I understand it black holes radiate &lt;a href=&quot;https://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;noreferrer&quot;&gt;Hawking radiation&lt;/a&gt; and probably &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_wave&quot; rel=&quot;noreferrer&quot;&gt;gravity waves&lt;/a&gt;, which over time causes them to lose mass and eventually evaporate after almost unfathomable amounts of time.  I've also read that black holes eventually have difficulty gaining size after a certain point because gravitational forces working in a black hole's accreditation disk can eventually begin ejecting in-falling matter before it can be absorbed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I know black holes can become quite massive, that leads me to wonder if Black holes have a maximum possible size where they will begin to immediately radiate any additional mass that's added to it.  Or can a black hole effectively expand forever as long as the supply of matter is greater than the effects of radiation and ejection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there effectively a size limit to black holes?&lt;/p&gt;&#xA;" OwnerUserId="600" LastEditorUserId="600" LastEditDate="2018-07-04T18:23:27.020" LastActivityDate="2018-07-09T19:25:51.753" Title="Is there a maximum size for a black hole?" Tags="&lt;black-hole&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24202" PostTypeId="1" AcceptedAnswerId="24213" CreationDate="2017-12-27T07:38:38.040" Score="11" ViewCount="1496" Body="&lt;p&gt;Outside of a specially crafts environment, could liquid water exist on Mars? Not some time in the distant past, but within the last few years. Mars' atmosphere is too thin for water to exist, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps if the water was mixed with salts or other minerals, its sublimation point might be altered enough to exist? Mars can get warm during its summer days, up to about 20 degrees Celsius at the equator. Perhaps, in certain areas and time periods, we may get a kind of liquid water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the speculative nature of this question, but after NASA talked about the possibility of liquid water, it got me wondering if it was possible.&lt;/p&gt;&#xA;" OwnerUserId="20565" LastActivityDate="2017-12-31T07:51:57.747" Title="Can Liquid Water exist on Mars?" Tags="&lt;mars&gt;&lt;water&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24203" PostTypeId="2" ParentId="24202" CreationDate="2017-12-27T08:20:59.547" Score="0" Body="&lt;p&gt;It's not just the temperature that causes problems for liquid water, but also the atmosoheric pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Earth, water boils at ~96°C because the atmosoheric pressure is around 1 bar. If you lower the pressure, the boiling point for water is loser as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Mars, the atmospheric pressures range is from 30 Pa  on Olympus mon (highest mountain) to 1.155 Pa in the depths of hellas planitia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at this chart for &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Vapour_pressure_of_water&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vapor Pressure&lt;/a&gt; you will see, that 1.155 Pa (0.001155 kPa) is not enough pressure for liquid water when the ice on Mars is melting. It would turn instantly into gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition you could look at the &lt;a href=&quot;https://www.spscientific.com/freeze-drying-lyophilization-basics/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Triple Point&lt;/a&gt; of water and the chart at this site, and you see when water will be a solid, liquid or gas.&lt;/p&gt;&#xA;" OwnerUserId="19860" LastActivityDate="2017-12-27T08:20:59.547" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24204" PostTypeId="2" ParentId="24196" CreationDate="2017-12-27T09:11:07.980" Score="12" Body="&lt;p&gt;As David Hammen commented, the power emitted through Hawking radiation is proportional to $M^{-2}$. Thus the evaporation timescale for a black hole is proportional to $M^3$. This means that a more massive black hole is much more stable against evaporation than a lower mass black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other issue you mention is the limited rate that you can &quot;feed&quot; a black hole. There is inevitably a feedback; as gas is compressed towards the event horizon it gets hot and emits radiation. The pressure of this radiation can eventually balance the inward gravitational infall. For spherically symmetric accretion this leads to the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Eddington_luminosity&quot; rel=&quot;noreferrer&quot;&gt;Eddington limit&lt;/a&gt;, which sets the maximum spherical accretion rate, where $\dot{M}_{\rm max}\propto M$. That is, the maximum accretion rate is proportional to the black hole mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If accretion proceeds at the Eddington limit then the black hole mass grows exponentially with time and with a characteristic doubling timescale of around 50 million years (independent of the original mass - see &lt;a href=&quot;https://physics.stackexchange.com/questions/167250/is-there-a-limit-as-to-how-fast-a-black-hole-can-grow&quot;&gt;this Physics SE page&lt;/a&gt; for some mathematical details).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If black holes were limited to this accretion rate (though there is some evidence from the presence of very luminous quasars at high redshift that they may exceed it), then the maximum black hole mass will depend on the age of the universe and the size of the initial &quot;seed&quot; black holes. If we assume an initial mass of 100 solar masses, a doubling timescale of 50 million years and that the seed black holes formed 400 million years after the big bang (all plausible, but contestable), then there have been 266 doubling timescales since and the black hole could have grown by a factor of $10^{80}$ !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly there are no black holes with anywhere near this mass in the observable universe - the largest appear to be of order $10^{10}$ solar masses. Their growth is limited by their food supply. Supermassive black holes are found at the centres of galaxies. There is a poorly understood relationship between the black hole mass and the mass of the bulge of the galaxy it is in. The ratio peaks at about 1 percent for the most massive bulges (e.g. see &lt;a href=&quot;https://arxiv.org/pdf/0908.2028.pdf&quot; rel=&quot;noreferrer&quot;&gt;Hu 2009&lt;/a&gt;; &lt;a href=&quot;https://arxiv.org/pdf/1211.2816.pdf&quot; rel=&quot;noreferrer&quot;&gt;McConnell &amp;amp; Ma 2013&lt;/a&gt;). Since the most massive massive elliptical galaxies are of order $10^{12}$ solar masses, &lt;em&gt;this&lt;/em&gt; appears to set the maximum mass of a black hole in the present day universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The future is speculation. If the cosmic expansion rate continues to accelerate, then galaxy mergers will become increasingly uncommon and the opportunities for further black hole growth will be limited.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="7411" LastEditDate="2018-07-09T19:25:51.753" LastActivityDate="2018-07-09T19:25:51.753" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24205" PostTypeId="2" ParentId="24202" CreationDate="2017-12-27T09:30:01.513" Score="5" Body="&lt;p&gt;Had you asked a couple of months ago the answer would be &quot;probably yes&quot;, now it is more &quot;maybe&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pure water could not exist in the Martian Environment. The pressure is too low, so the water would turn to water vapour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, if you produce hyper-concentrated brine it can remain in liquid form for a while in some martian environments. Solutions of perchlorates have been shown to keep liquids from freezing even when conditions are as cold as minus 70 Celsius. T&lt;a href=&quot;https://www.nasa.gov/press-release/nasa-confirms-evidence-that-liquid-water-flows-on-today-s-mars&quot; rel=&quot;nofollow noreferrer&quot;&gt;his was thought to be the cause of recurring slope lineae&lt;/a&gt; (ie lines that repeatedly appear on slopes around the equator). The actual model of how water (or brine) could get there was challenging. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.nasa.gov/feature/jpl/recurring-martian-streaks-flowing-sand-not-water&quot; rel=&quot;nofollow noreferrer&quot;&gt;Further modelling&lt;/a&gt; this year suggests that the features appear to be &quot;granular flows&quot; ie sand, not water. They don't exist on shallow slopes. Water will flow downhill on any slope, but sand needs a steep slope to flow. Since the lines only appear on steeper slopes, they look more like sand flows not water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, water is involved, as hydrated salts have been detected on slopes with the lines. There is no clear mechanism for causing sand to start to slide, on some slopes and not others.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="19898" LastEditDate="2017-12-27T14:40:42.463" LastActivityDate="2017-12-27T14:40:42.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24208" PostTypeId="2" ParentId="24196" CreationDate="2017-12-27T13:21:39.480" Score="10" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a maximum size for a black hole?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As far as I understand it black holes radiate Hawking radiation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's what people say, but we have no actual evidence for Hawking radiation.  However even if Hawking was 100% correct, as Rob said, Hawking radiation has less and less effect as the black hole gets bigger and bigger. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and probably gravity waves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A black hole on its own won't emit any. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;which over time causes them to lose mass and eventually evaporate after almost unfathomable amounts of time. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's no scientific evidence that a black hole will disappear. However there is scientific evidence that black holes exist. There's definitely something very small and very massive &lt;a href=&quot;https://en.wikipedia.org/wiki/Sagittarius_A*#Central_black_hole&quot; rel=&quot;noreferrer&quot;&gt;at the center of our galaxy&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uy5hk.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uy5hk.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I've also read that black holes eventually have difficulty gaining size after a certain point because gravitational forces working in a black holes accreditation disk can eventually begin ejecting in-falling matter before it can be absorbed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, black holes are said to &quot;choke&quot; if they try to eat to much at once. See the physicsworld article &lt;a href=&quot;http://physicsworld.com/cws/article/news/2010/jan/08/supermassive-black-hole-struggles-to-swallow-milky-way&quot; rel=&quot;noreferrer&quot;&gt;Supermassive black hole struggles to swallow Milky Way&lt;/a&gt;. There are other issues to do with gamma ray bursters meaning black holes are messy eaters, but they still &quot;eat&quot;, as it were.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;While I know black holes can become quite massive, that leads me to wonder if Black holes have a maximum possible size where they will begin to immediately radiate any additional mass that's added to it. Or can a black hole effectively expand forever as long as the supply of matter is greater than the effects of radiation and ejection?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's the latter. Imagine you're near a supermassive black hole. It's so massive that any Hawking radiation is negligible. It has no accretion disk because it's eaten everything or blown it away. What happens next? &lt;i&gt;You fall in.&lt;/I&gt; So that black hole gets bigger.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there effectively a size limit to black holes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. If there was, and the black hole in the above scenario had reached it, you wouldn't fall in.   &lt;/p&gt;&#xA;" OwnerUserId="9772" LastEditorUserId="600" LastEditDate="2017-12-28T08:55:01.550" LastActivityDate="2017-12-28T08:55:01.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24210" PostTypeId="2" ParentId="13359" CreationDate="2017-12-27T23:49:57.220" Score="1" Body="&lt;p&gt;you may want to try RECONS &lt;a href=&quot;http://www.recons.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.recons.org/&lt;/a&gt; they have an animation of 10 year movement of Proxima Cen.&#xA;&lt;a href=&quot;http://www.recons.org/proxima.movie.large.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.recons.org/proxima.movie.large.gif&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20575" LastActivityDate="2017-12-27T23:49:57.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24211" PostTypeId="1" AcceptedAnswerId="24221" CreationDate="2017-12-28T04:22:42.577" Score="6" ViewCount="2031" Body="&lt;p&gt;If you were planing a maned mission to Titan and wanted to find a landing site that would get the most visible light for the longest amount of time, your first logical choice would be the dark side of the moon since it faces the sun for a longer duration than the face would. (&quot;face&quot; and &quot;dark side&quot; corresponding to the colloquial definitions they have when referring to our moon) Also, unlike the on the face, you wouldn't have to worry about Saturn eclipsing the sun. This would put you in a range of brightness roughly equivalent of twilight on Earth due to the distance of the sun and the thick atmosphere of Titan. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how bright would Saturn be when reflected on the face of Titan? Would it be bright enough to see anything if you wanted to go there, or would you need flashlights and night vision goggles?&lt;/p&gt;&#xA;" OwnerUserId="20578" LastActivityDate="2017-12-28T10:41:16.710" Title="How bright would it be on the &quot;face&quot; of Titan" Tags="&lt;light&gt;&lt;saturn&gt;&lt;luminosity&gt;&lt;extra-terrestrial&gt;&lt;titan&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24212" PostTypeId="1" AcceptedAnswerId="24270" CreationDate="2017-12-28T04:34:43.370" Score="2" ViewCount="200" Body="&lt;p&gt;I am on the west coast of the U.S. and for several weeks I have seen two large, twinkling objects in the eastern sky at night with the naked eye. In between them there are 3 smaller ones that are stacked vertically and very close to one another. All 5 rise in the sky as a group as the evening progresses. Could this be Orion? Thank you.  &lt;/p&gt;&#xA;" OwnerUserId="16474" LastEditorUserId="16474" LastEditDate="2017-12-31T06:05:26.147" LastActivityDate="2017-12-31T10:09:55.837" Title="Am I seeing the constellation Orion?" Tags="&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24213" PostTypeId="2" ParentId="24202" CreationDate="2017-12-28T06:04:25.773" Score="9" Body="&lt;p&gt;Liquid water can in principle exist at many locations on present-day Mars, but there are a few interesting twists to the story.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;At low elevation, the atmospheric pressure is high enough. The triple point pressure of H2O is 611 Pascal, which corresponds to about mid-elevation. At low elevations, such as the northern lowlands, the atmospheric pressure will exceed the triple point pressure, so the liquid phase of H2O becomes possible.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There will be substantial &quot;sublimation cooling&quot; or evaporative cooling. When the partial pressure of H2O is almost as high as the total pressure, vapor rapidly escapes into the atmosphere, which takes a lot of energy. If you put an ice cube at the equator of Mars, it will not melt, because the sublimation cooling is larger than the incoming solar energy, even at noon.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;However, if that ice cube is covered with a thin layer of dust, or contains dust that will accumulate on top as the ice sublimates, vapor loss will be reduced. In this circumstance, ice can melt to form liquid water. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;That leaves the question of how the ice cube would have gotten there initially. I call it the &quot;source problem&quot;. Places on Mars that get hot have already lost all their ice, and places on Mars that have ice do not get warm enough. So, it is difficult to get liquid water on Mars, but it is possible in principle.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;With salts, which lower the melting point, everything becomes much easier, although supplies of ice/frost and salt are needed. When water with salt evaporates, the salt concentration increases, so the solution is driven toward the eutectic point, where the melting point suppression is the strongest.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="14680" LastActivityDate="2017-12-28T06:04:25.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24221" PostTypeId="2" ParentId="24211" CreationDate="2017-12-28T10:41:16.710" Score="4" Body="&lt;p&gt;It's not hard to calculated brightness to distance, but that doesn't take into account cloud cover, which is important for Titan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just looking at distance first, Titan (based on Saturn's distance) averages 9.6 AU from the sun, or about 1/91st as bright as the sun appears from Earth, assuming equal atmospheres.  1/91st isn't a bright sunny day, but it's probably bright enough to read by.    The sun has an &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_magnitude#Table_of_notable_celestial_objects&quot; rel=&quot;nofollow noreferrer&quot;&gt;apparent magnitude&lt;/a&gt; of about -26.7 from Earth.   91 times less bright works out to -21.8 from Titan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn is a little more tricky because from Earth, we often see Saturn's rings but from Titan, you mostly don't.   Titan orbits almost entirely in line with Saturn's rings (&lt;a href=&quot;https://en.wikipedia.org/wiki/Titan_(moon)&quot; rel=&quot;nofollow noreferrer&quot;&gt;just 0.34854° inclination&lt;/a&gt;), which would make the rings appear like a thin band.   Visible but probably not very reflective.    You could also probably see ring shade and ring shine reflecting off the surface of Saturn from Titan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How bright would Saturn be?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Earth, Saturn's brightness varies from (mostly) &lt;a href=&quot;https://en.wikipedia.org/wiki/Saturn#Observation&quot; rel=&quot;nofollow noreferrer&quot;&gt;between 0 and 1 apparent magnitude&lt;/a&gt;.  That's a range of 2.5 depending on distance and angle, which is kind of a lot.    Titan would probably receive less variation due to no ring reflection (see Saturn as viewed from Earth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2uHQr.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2uHQr.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There would still be variation based on the angle of the sun reflecting off Saturn's surface.   Perhaps an estimate of between 0.2 and 1 is in the ballpark (as a guess) at Earth distance but Titan's viewpoint, almost entirely above the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using an an average distance between Earth and Titan of about 9.6 AU, and Titan's distance from Saturn, about 1/122 of 1 AU.  Titan is some 1170 times closer to Saturn than the Earth on average, which makes Saturn about 1.38 million times brighter from Titan than from the Earth.   which works out to between 15.3 and 15.4 apparent magnitude units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using my rough estimate, Saturn, from Titan would be in the range of -15.2 to -14.3 apparent magnitude, which would be some 400-800 times less bright than the Sun from Titan (it's not possible for a planet to outshine a star from the point of view of the moon), but several times brighter than the full moon from Earth.   You could probably see, well enough on Titan under a &quot;full Saturn&quot; - being some 5-10 times as bright as the full moon (peak -12.9/average -12.6).  It would also be much larger, so it wouldn't appear bright in the sky, but it would provide enough light to see well enough, but maybe not enough light for reading.   Enough to see though.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Permanent Clouds&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numbers above, however, doesn't take into account &lt;a href=&quot;https://en.wikipedia.org/wiki/Climate_of_Titan#Clouds&quot; rel=&quot;nofollow noreferrer&quot;&gt;Titan's cloud cover&lt;/a&gt; and it has nearly constant clouds.   Here is an &lt;a href=&quot;https://www.space.com/27644-titan-moon-sunny-seas-photo.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;article with a picture of a rare situation of sunlight reflecting&lt;/a&gt; off the surface of titan.   Mostly it's under a permanent haze.   I don't know how effective that haze is in blocking sunlight, maybe 90%, maybe even 99%, but it would certainly block a fair percentage, so, while you could probably see under sunlight, I think reading is no longer possible under those clouds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an interesting article that says Titan is brightest &lt;a href=&quot;https://www.space.com/36609-twilight-outshines-daylight-saturn-moon-titan.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;at twilight, not daylight&lt;/a&gt;, because it's atmosphere is very effective at scattering light.   It's atmosphere is thicker and much higher than Earths, and also more massive overall.     It's likely for any real work, Astronauts would probably want a permanent artificial light source and not rely much on light from the tiny sun though all those clouds.    But they might be able to see well enough to move around and explore without flashlights a little bit, but flashlights would help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Night vision goggles would do nothing.   They see heat or longer wavelength IR light outside the visible wavelengths.   They could spot heat sources, like other astronauts, or a rip in a space suit or hut - so they'd have uses seeing where the heat is escaping for example, but they wouldn't help you see the on surface of Titan.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Titan takes nearly 16 days to orbit Saturn anyway, and nobody would want to put up with 8 days of darkness.  Artificial light would be essential which, I would think, makes choosing a location based on sunlight not a key factor.  Perhaps someplace where there's less methane rain would be a priority, or less chance of solid ground becoming liquid under a warm spell like a nice &quot;dry&quot; water ice Island.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2017-12-28T10:41:16.710" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24223" PostTypeId="2" ParentId="16510" CreationDate="2017-12-28T11:45:13.227" Score="1" Body="&lt;p&gt;I am having the same issue too. Been considering using &lt;strong&gt;Python with Numpy, SciPy, and Astropy&lt;/strong&gt;, supplement that with &lt;strong&gt;GNU Octave&lt;/strong&gt; and &lt;strong&gt;PyRAF&lt;/strong&gt;. I heard some younger students would opt for the proprietary MATLAB in lieu of GNU Octave. My supervisor belongs to the older generation so I will have to &lt;em&gt;also&lt;/em&gt; learn IRAF to effectively communicate with him. Some calibration scientist recommended me to use the &lt;strong&gt;Herschel Interactive Processing Environment (HIPE)&lt;/strong&gt;, which has been developed for the ESA Herschel pipelines (like PACS and SPIRE) in mind and uses Jython (a hybrid of Python and Java). But I am learning to use that too, before either I become proficient in Python or developed some software with a proper GUI of my own... &lt;/p&gt;&#xA;" OwnerUserId="20598" LastActivityDate="2017-12-28T11:45:13.227" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24225" PostTypeId="2" ParentId="20058" CreationDate="2017-12-28T12:54:29.697" Score="1" Body="&lt;p&gt;The initial seeding of rocks is thought to be a rapid and energetic event resulting in the formation of chondrules which are millimeter-sized spherules that form as molten (or partially molten) droplets in space before coalescing to their parent asteroids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recent evidence suggests that the solar system comes from the hottest stars devoid of hydrogen at the surface and the resulting wolf-rayet nebula, which would give similar aluminium and iron isotopes found nearby. That's cool if we want to know where the energy came from to &lt;a href=&quot;https://en.wikipedia.org/wiki/Chondrule#Formation&quot; rel=&quot;nofollow noreferrer&quot;&gt;condense nebula into chondrules&lt;/a&gt;. Magnetic flares, nebula lightning and sunstorms may give the energy to seed the first rocks of a nebula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Accretion_(astrophysics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Wikipeda page of cosmic accretion theory is very mysterious.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="2153" LastEditDate="2017-12-30T21:03:51.980" LastActivityDate="2017-12-30T21:03:51.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24227" PostTypeId="2" ParentId="23689" CreationDate="2017-12-28T17:13:19.543" Score="2" Body="&lt;p&gt;Black hole in the centre of our galaxy is not active (which means it is not feeding much matter) as compared to black holes in the centre of other galaxies. So it does not emit that much gamma rays as other supermassive black holes does but still emits some amount of gamma rays. Along with gamma rays cosmic rays are also emitted which are ward off by radiation pressure of our sun's heliosphere. &lt;/p&gt;&#xA;" OwnerUserId="20162" LastActivityDate="2017-12-28T17:13:19.543" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24228" PostTypeId="1" CreationDate="2017-12-28T19:56:18.680" Score="5" ViewCount="98" Body="&lt;p&gt;The European Southern Observatory webpage has &lt;a href=&quot;http://www.eso.org/sci/facilities/paranal/decommissioned/isaac/tools/lib.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;a web page&lt;/a&gt; that has tabular spectrogram data from A. J. Pickles, University of Hawaii.  There are over 130 .dat files there.  Each one represents a spectrogram from a star.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;My question is at the bottom.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;First though, I will excerpt some lines from a few of the .dat files.  You can read the tabular data by unzipping the &quot;.dat.gz&quot; files and opening the resulting &quot;.dat&quot; in a plain text editor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ukwg0v.dat&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Description of File Name:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;UK&lt;/strong&gt; = They all start with UK&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;W&lt;/strong&gt; = Weak metalicity (can be: W for weak, S for strong, or nothing for normal)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;G&lt;/strong&gt; = Spectral class G (can be: O, B, A, F, G, K, M)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;0&lt;/strong&gt; = Spectral subtype 0 &lt;em&gt;(can be: 0-9)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;V&lt;/strong&gt; = Roman numeral 5 Yerkes luminosity class &lt;em&gt;(can be: Ia, Ib, II, III, IV, V, VI, VII)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Excerpt of File Contents:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#iRMS=9.222171502e-05 0&#xA;#    lk  ukf_wg0v  uks_wg0v        fh        fse&#xA;#&#xA; 1150.0  0.004218  0.000000  0.004218  0.000000&#xA; 1155.0  0.002700  0.000000  0.002700  0.000000&#xA; 1160.0  0.001559  0.000000  0.001559  0.000000&#xA;        ... [4,765 lines truncated] ...&#xA;24990.0  0.016113  0.000000  0.000000  0.016113&#xA;24995.0  0.016087  0.000000  0.000000  0.016087&#xA;25000.0  0.000000  0.000000  0.000000  0.000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/i3NO5.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i3NO5.gif&quot; alt=&quot;Plot of ukwg0v.dat&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ukg0i.dat&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#iRMS=1.052067091e-05 0&#xA;#    lk   ukf_g0i   uks_g0i        fh       fse         fk&#xA;#&#xA; 1150.0  0.000000  0.000000  0.000000  0.000000  0.000000&#xA; 1155.0  0.000000  0.000000  0.000000  0.000000  0.000000&#xA; 1160.0  0.000000  0.000000  0.000000  0.000000  0.000000&#xA;        ... [4,765 lines truncated] ...&#xA;24990.0  0.000000  0.000000  0.000000  0.000000  0.000000&#xA;24995.0  0.000000  0.000000  0.000000  0.000000  0.000000&#xA;25000.0  0.000000  0.000000  0.000000  0.000000  0.000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/il5yT.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/il5yT.gif&quot; alt=&quot;Plot of ukg0i.dat&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ukm3iii.dat&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#lRMS=0.06043180451 iRMS=0.04044797271 0&#xA;#    lk ukf_m3iii uks_m3iii        fh        fl        fd        fm        fsv&#xA;#&#xA; 1150.0  0.000000  0.000000  0.000000  0.000000  0.000000  0.000000  0.000000&#xA; 1155.0  0.000000  0.000000  0.000000  0.000000  0.000000  0.000000  0.000000&#xA; 1160.0  0.000000  0.000000  0.000000  0.000000  0.000000  0.000000  0.000000&#xA;        ... [4,765 lines truncated] ...&#xA;24990.0  0.302861  0.000000  0.000000  0.000000  0.000000  0.323665  0.000000&#xA;24995.0  0.301125  0.000000  0.000000  0.000000  0.000000  0.321810  0.000000&#xA;25000.0  0.300519  0.000000  0.000000  0.000000  0.000000  0.321163  0.000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/X2bTR.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X2bTR.gif&quot; alt=&quot;Plot of ukm3iii.dat&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;uko5v.dat&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#lRMS=0.0004083506647 iRMS=0.0004083506938 0&#xA;#    lk   ukf_o5v   uks_o5v        fh       fse         fl&#xA;#&#xA; 1150.0 12.354247 10.980931 12.354247  0.000000  0.000000&#xA; 1155.0  9.928268  8.291242  9.928268  0.000000  0.000000&#xA; 1160.0  8.410753  7.608442  8.410753  0.000000  0.000000&#xA;        ... [4,765 lines truncated] ...&#xA;24990.0  0.000000  0.000620  0.000000  0.000000  0.000000&#xA;24995.0  0.000000  0.001047  0.000000  0.000000  0.000000&#xA;25000.0  0.000000  0.001561  0.000000  0.000000  0.000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SlBxX.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SlBxX.gif&quot; alt=&quot;Plot of uko5v.dat&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://www.eso.org/sci/facilities/paranal/decommissioned/isaac/tools/lib/hilib.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;the accompanying paper&lt;/a&gt;, it helps to decode the columns:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;1st column:&lt;/strong&gt;  The wavelength in angstroms.  The rows start at 1150 Å and increment every 5 Å up to 25000 Å.  This can be seen across the &lt;strong&gt;X axis&lt;/strong&gt; of the plots.  &lt;em&gt;(I don't know why whoever made these plots decided to start them at 8000 Å, but the actual data begins at 1150 Å.)&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;2nd column:&lt;/strong&gt;  The corresponding flux at that wavelength.  Fluxes are in Fλ (F_lambda) units.  This is the &lt;strong&gt;Y axis&lt;/strong&gt; of the plots.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;3rd column:&lt;/strong&gt;  &quot;The standard deviation of the optical combination....  The normalized source contributions are averages where multiple components from a source exist.  The first line summarizes RMS errors of these individual averages.  The standard deviation column is generally zero in wavelength regions where only one source is present, but is sometimes replaced by the standard deviation of the average forming that single component.&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Subsequent columns:&lt;/strong&gt;  The source contributions.  Sometimes there are multiple spectra surveys that took measurements at a particular wavelength.  The average (arithmetic mean) of these sources appear in column 2.  The standard deviation of these sources appear in column 3.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I thought that the Y axis of a spectrogram was supposed to represent intensity of radiation, but in this data they refer to it as flux and are measuring it in terms of &quot;Fλ units&quot;.  I am very dumb with regard to spectroscopy.  Could somebody please explain to me how flux compares to intensity, and what exactly Fλ units are?&lt;/p&gt;&#xA;" OwnerUserId="20604" LastEditorUserId="20604" LastEditDate="2017-12-29T16:01:21.850" LastActivityDate="2017-12-29T16:01:21.850" Title="Need help understanding stellar spectroscopy data from ESO" Tags="&lt;spectroscopy&gt;&lt;metallicity&gt;&lt;spectrometry&gt;&lt;spectral-type&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24239" PostTypeId="1" CreationDate="2017-12-29T14:19:33.540" Score="4" ViewCount="395" Body="&lt;p&gt;I'm not sure if this is more a question for a 3D graphics forum, but I'm trying here first in case the answer has more to do with astronomy than 3D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am creating a model of the Earth and Moon in a piece of 3D software. To avoid the enormous scale of using real-world sizes, I am using a 1000:1 scale ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the dimensions I'm using:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Earth radius&lt;/strong&gt; = 6,371 m&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Moon radius&lt;/strong&gt; = 1,737 m&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Distance from Earth to&#xA;Moon&lt;/strong&gt; = 384,400 m&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since I have scaled everything equally, I would assume that the proportions of things would look accurate, but the moon looks very small indeed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a quick render from a camera placed just above the surface of the Earth, looking out at the moon:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cD7iT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cD7iT.jpg&quot; alt=&quot;Earth and Moon&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The camera's focal length is 24mm, which I understand to be relatively similar to the human eye. I was expecting the moon to look similar in size to how it looks when you look up at it in real life but, as you can see, it's a tiny little dot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there something obvious I'm missing here, or some explanation for this discrepancy?&lt;/p&gt;&#xA;" OwnerUserId="15788" LastActivityDate="2018-11-22T11:06:48.907" Title="Moon looks too small in 3D model" Tags="&lt;the-moon&gt;&lt;earth&gt;" AnswerCount="3" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="24240" PostTypeId="2" ParentId="13810" CreationDate="2017-12-29T15:15:04.247" Score="-1" Body="&lt;p&gt;If you have some spectral image in the form of a FITS-format file, like you have in the original image to the right of the P-V diagram: I think you will need to look up some reference on Fourier transform. This Fourier transform will take you from the RA-Dec image to the P-V diagram you want to generate. Also, you might need to read about the cross correlation method, as detailed in Tonry and Davis's 1979 paper: &lt;a href=&quot;http://adsabs.harvard.edu/abs/1979AJ.....84.1511T&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/1979AJ.....84.1511T&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method is called fitting which involves emission lines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The IRAF package you will need to use is called &quot;rvsao&quot;. There are some references on some webpages you can follow: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://tdc-www.harvard.edu/iraf/rvsao/papers/rvsao.adass97.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://tdc-www.harvard.edu/iraf/rvsao/papers/rvsao.adass97.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://arxiv.org/abs/astro-ph/9803252&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/astro-ph/9803252&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To generate a P-V digram you will need to map your radial velocity to a 1D-grid, say one depending on the radial position, the mapping should be one-to-one. Otherwise you can also generate a cube, depending what kind of original FITS data files you have from an observation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am only providing the details regarding the technical aspect, and am on purpose on explaining anything regarding &quot;distribution&quot;... Every relationship in physics which depends on space (and maybe time also) is some sort of a distribution. &lt;/p&gt;&#xA;" OwnerUserId="20598" LastEditorUserId="20598" LastEditDate="2017-12-31T12:49:11.620" LastActivityDate="2017-12-31T12:49:11.620" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24241" PostTypeId="1" CreationDate="2017-12-29T16:43:41.763" Score="4" ViewCount="98" Body="&lt;p&gt;The other night, I observed what I believe must have been a satellite, however it was at 1:30am so I was not sure that this would be possible. Being in the Southern Hemisphere we are just past Summer Solstice, so theoretically it could have been reflecting the sun's light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There were no flashing lights, just a star-like point of light. As it slowly traveled in a NE direction across the sky, it began to gradually fade as typical of a satellite passing into Earth's shadow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on my observation, I'd like to get an understanding of the approximate minimum altitude of the object that I observed, allowing for it to be out of the shadow of the Earth sunlight. I'm not a mathematician, but using Space Engine with the correct time &amp;amp; location, I had to position the camera to some 7000km in altitude before the sun was visible over the horizon. The object was N to NE from where I observed it, whist the midnight sun would have been shining away at the South Pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Date/Time: 27-Dec-2017, ~01:30 local time (GMT + 8)&#xA;Location of observation: Perth, Western Australia (~32S, 116E)&#xA;Position in sky : ~ 80 to 60 degrees elevation, moving north due northeast&#xA;Duration of observation : ~2 to 3 minutes&lt;/p&gt;&#xA;" OwnerUserId="20623" LastActivityDate="2017-12-30T17:34:40.290" Title="Altitude of Midnight Satellite?" Tags="&lt;observational-astronomy&gt;&lt;satellite&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24242" PostTypeId="2" ParentId="24183" CreationDate="2017-12-29T17:15:44.520" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How do we know how fast surface of neutron star is rotating?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I suppose we don't know the answer with total confidence, but I'd say we think we know with good confidence. Because we have robust evidence along with a model that matches that evidence.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I understand that we can count pulses in time, but how do we know that magnetosphere is rotating along with surface of neutron star?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I suppose we don't actually know for sure. But I can't think of any process whereby a stationary spherical body can emit regular pulses of radiation. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can we measure rotation of a neutron star in any other way like red-blue shifts? It should be pretty difficult due to small radius and big distance but is it possible? Is it already done?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, and no. At least as far as I know. But most of what I know came from &lt;a href=&quot;https://www.amazon.co.uk/Frozen-Star-George-Greenstein/dp/0881910112&quot; rel=&quot;nofollow noreferrer&quot;&gt;Frozen Star&lt;/a&gt; by George Greenstein. The first half of the book is all about neutron stars. It dates from 1984 and it could be a little out of date. But as far as I know there's not much wrong with it, and it's a good read regardless. Note that you can buy a second hand copy fairly cheaply.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zIRS0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zIRS0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2017-12-29T17:15:44.520" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24243" PostTypeId="1" CreationDate="2017-12-29T20:12:33.610" Score="3" ViewCount="1780" Body="&lt;p&gt;See additional information below added after the answer to the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the moon is between the earth and the sun, the gravitational force on the moon from the sun is greater than the gravitational force from the earth, hence the moon should continue its path towards the sun instead of staying in its orbit around the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the moon is at 90 degrees to the earth relative to the sun, the moon is moving almost directly towards the sun and then the gravitational force on the moon from the sun is roughly double the gravitational force from the earth, hence there is nothing to stop the moon's trajectory towards the sun and it should continue towards the sun away from the earth. What makes the moon &quot;turn&quot; back towards the earth and away from the sun at the point where it is right between the earth and the sun when at this point right when it turns, the gravitational force from the sun is roughly double the gravitational force from the earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware of the movement of the sun and also of the fact that the moon is moving on a different plane to that of the earth and I don't think that can explain my question above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your time, I would like to understand this situation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Additional information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer does not address the question, let me explain. Although it is true that the moon orbits the sun as described in the answer, this does not explain why the sun does not pull the moon out of its orbit around the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an object to change direction a net force has to be applied to it according to Newtons laws. a=F/m applied to the moon from the point in its orbit where the moon gets closer to the sun than the earth is to the sun, shows a force vector on the moon towards the sun applying twice the force as another force vector on the moon towards the earth. This results in a combined net force vector on the moon pointing away from the earth and this net force vector will progressively point further away from the earth and more towards the sun as the moon moves closer to the sun and therefore should pull the moon out of its orbit around the earth and pull it towards the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newtons laws must be true on the three separate objects; the sun, the earth and the moon in the example above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a Master of Science in Engineering and know my physics. I have some friends who are convinced they can prove that the earth is actually flat and that a search on &quot;flat earth&quot; will prove it, and they have asked me the question in this thread to prove to me that the heliocentric model is not true. I took up the challenge, however when looking at this simple example I have not been able to find an answer that is in accordance with normal physical laws. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope somebody can come up with an explanation that is in accordance with physical laws.&lt;/p&gt;&#xA;" OwnerUserId="20624" LastEditorUserId="1891" LastEditDate="2017-12-31T12:10:58.657" LastActivityDate="2017-12-31T12:10:58.657" Title="Why does the moon not leave its orbit around the earth due to the gravitational force from the sun?" Tags="&lt;the-sun&gt;&lt;orbit&gt;&lt;the-moon&gt;" AnswerCount="1" CommentCount="20" ClosedDate="2017-12-30T16:31:05.180" ContentLicense="CC BY-SA 3.0" />
  <row Id="24244" PostTypeId="1" AcceptedAnswerId="24246" CreationDate="2017-12-29T20:58:06.590" Score="2" ViewCount="59" Body="&lt;p&gt;Could a sun emit a large piece of matter that escapes it's gravity, and might eventually reach another solar system? And what would be the effect if a planet sized object reached our solar system and enters orbit around our sun?&lt;/p&gt;&#xA;" OwnerUserId="19197" LastActivityDate="2017-12-29T22:15:07.133" Title="Interstellar bodies" Tags="&lt;asteroids&gt;&lt;interstellar&gt;&lt;hypothetical&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24245" PostTypeId="1" AcceptedAnswerId="24247" CreationDate="2017-12-29T22:14:54.577" Score="1" ViewCount="38" Body="&lt;p&gt;I am planning some astrophotography. And comparing two times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On one night I plan a midnight shoot when a waning crescent moon rises at 1am. This is 1 week ahead of the new moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other night would be on a new moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the darkness of the sky be expected to be any different? I ask because a lot of articles I read about stargazing and astrophotography say the phases of the moon are important, but never mention the setting and rising of the moon as an important factor.&lt;/p&gt;&#xA;" OwnerUserId="17944" LastActivityDate="2017-12-29T22:20:11.290" Title="Does a set moon sky equal a new moon sky?" Tags="&lt;star&gt;&lt;the-moon&gt;&lt;photography&gt;&lt;moon-phases&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24246" PostTypeId="2" ParentId="24244" CreationDate="2017-12-29T22:15:07.133" Score="2" Body="&lt;p&gt;The sun couldn't. The sun gives off light and a solar wind. It doesn't emit lumps of solid matter. It is possible for a small body, such as a comet to get a slingshot around a planet and be ejected.  In the early solar system, when the planets orbits were unstable it is possible that larger bodies were ejected from the solar system. It is thought that rogue planets exist, far from any stars. Some larger than Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a rogue planet were to enter our solar system it would fall towards the sun, pass the sun and leave again. It is possible that it could interact with one of the other planets and enter into a long elliptical orbit. This would be very bad. A planet-sized object in an unstable orbit is not the sort of thing you want in your solar system. We would just have to hope that it never got close enough to Earth to affect it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rogue planet could significantly distort Earth's orbit, and if it collided with the Earth, enough energy would be released to melt the entire surface. Fortunately, rogue planets are rare enough that we would not expect one to enter the solar system for a very very long time.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-12-29T22:15:07.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24247" PostTypeId="2" ParentId="24245" CreationDate="2017-12-29T22:20:11.290" Score="2" Body="&lt;p&gt;After the moon has set, the sky is dark. There is a little &quot;lunar twilight&quot;, but this is pretty insignificant. Once the moon has set you don't need to worry about the moon. The sky will be dark at midnight.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2017-12-29T22:20:11.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24249" PostTypeId="2" ParentId="24243" CreationDate="2017-12-30T01:11:16.450" Score="4" Body="&lt;ul&gt;&#xA;&lt;li&gt;The speed of the Earth (and Moon) around the Sun is $\approx 29.8 \frac{km}{s}$.&lt;/li&gt;&#xA;&lt;li&gt;The speed of the Moon around the Earth is $\approx 1.05 \frac{km}{s}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The result is that the Moon has a nearly perfect circular orbit around the Sun, the effect of the Earth is only a perturbation in it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nKadC.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nKadC.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;In fact, the &quot;waves&quot; are much smaller (400.000km Moon orbital radius to  150.000.000km Earth orbital radius)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is mainly because $R_{Earth-Moon} \ll R_{Earth-Sun}$, and $M_{Moon} \ll M_{Earth} \ll M_{Sun}$. In general, 3-body gravitationally bound systems tend to be unstable, i.e. one of the components tends to fly away with time. But this system seems stable, because the ratio of the masses are big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same can't be said from the whole Solar System, (it is a notoriously unsolved problem) although also this seems stable in human time scales.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="1891" LastEditDate="2017-12-30T01:25:43.453" LastActivityDate="2017-12-30T01:25:43.453" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24251" PostTypeId="1" CreationDate="2017-12-30T04:32:26.920" Score="1" ViewCount="109" Body="&lt;p&gt;Years ago it was mentioned to me the New Yorker magazine had a Young Moon “contest”; who could find the youngest moon; the thinnest crescent after a new moon. I knew enough to know that I would have to know exactly when the new moon occurred and more importantly where. The Thin Crescent wouldn’t be too hard as it would follow closely the sun. Still, not knowing where the new moon happened meant that I could see anything from nothing to a one day old moon. I knew the moon traversed about 13 degrees of arc a day in so doing about 7% Of its disc was illuminated so my chance of seeing a very thin crescent depended on the new moon happening to the east of me by some 6 to 12 hours, as the “record” was something in the realm of 9hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question arose in me “Where on the globe would an observer see any of the quarterly phases exactly over their meridian?” This question led to almanacs that used hours, minutes, seconds and hundredths of seconds of measurements, all starting from the celestial spheres Greenwich or zero longitude, the Vernal Equinox, that semi-fixed(a 26,000 year cycle) starting place for celestial longitude. My question is, how can I find and use that “exact” beginning zero “longitude?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am being hobbled here by being forced to use the very nomenclature I don’t understand. I mean I understand it, it’s simple enough, I just don’t know where to start. How do I find right ascension 18h23m52.34s on the celestial sphere? Is there a bright star at the vernal equinox that is used to fix that spot in the sky?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I begin to think I’m communicating with a machine that can’t extrapolate...no matter, as long as I get an education. Ok then here’s a question: how can I find right ascention(ra)18h 23m 52s standing out in the desert some night? I don’t need declination as the object is large and I’ll know it when I see it. &lt;/p&gt;&#xA;" OwnerUserId="20625" LastEditorUserId="20625" LastEditDate="2017-12-31T17:30:46.437" LastActivityDate="2018-01-01T01:25:48.587" Title="Finding a location geographically of a celestial object, in this case the moon at one of its quarters" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2018-01-08T04:36:14.713" ContentLicense="CC BY-SA 3.0" />
  <row Id="24261" PostTypeId="2" ParentId="24239" CreationDate="2017-12-30T16:27:05.200" Score="2" Body="&lt;p&gt;This will become a little off-topic, but the topic itself is, too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The camera's focal length is 24mm, which I understand to be relatively similar to the human eye&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Focal length is only one of two parameters determining the field of view. In fact, for an object at given distance, it determines the size of its projection on a screen. For a simple, thin lens, sizes and distances (from the lens axis / lens) are given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\text{object size}}{\text{object distance}}=\tan\alpha=\frac{\text{projection size}}{\text{projection distance}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For objects far, far away, one can approximate $\text{projection distance}\approx\text{focal length}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the projection of the moon by a 24mm lens is just 0.22mm in diameter!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The second important parameter is the size of the image generating area.&lt;/strong&gt;&#xA;An expensive DSLR camera has an image sensor of 36x24mm, the cheaper ones (APS-C size sensor) 22x15mm, a mobile phone may be 5x4mm or smaller. Only on the phone sensor, the moon starts to become more than a small dot compared to the entire image, and due to all that megapixels, the image also has some details when zooming in.&lt;br&gt;&#xA;Also the human eye has a high resolution, so you can see the details, but the brain also plays some tricks to let the moon appear larger than it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can assume that the virtual image sensor of a 3D software also is in the order of 36mm, see next paragraph. But the moon is not only tiny on the image, since a FullHD image has just 2 megapixels, there aren't even any details visible when zooming in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 24mm lens on a 36x22mm sensor gives a horizontal field of view of 73°, which is about the value of the human eye, as stated in the comments. But this value does not only determine the field of view, it also determines how &quot;3D&quot; the image is. A photo taken from larger distance with high focal length appears flat compared to low distance and low focal length. And images taken at very low focal distance feel like taken by a mouse. Now, those guys using 55&quot; TVs as monitor on their desk would like to reduce the focal distance, because, well, they are like mice in front of that screen. So, this is the reason why many games allow to adjust this parameter.&lt;/p&gt;&#xA;" OwnerUserId="20639" LastActivityDate="2017-12-30T16:27:05.200" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24262" PostTypeId="2" ParentId="24241" CreationDate="2017-12-30T17:26:08.653" Score="2" Body="&lt;p&gt;I'll venture a guess that the 7000 kilometers (your software's camera position) was with respect to &lt;em&gt;the center&lt;/em&gt; of the Earth. Using 6371 km as a value for the Earth's radius, that would make the altitude only 629 km, which it turns out is quite a popular altitude for satellites in LEO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the simplifying case of midnight, one can draw a simple diagram. I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h = R_{\textrm{Earth}} \left(\frac{1}{\cos(35°)} - 1\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is about 1400 km at midnight, and the height will be lower before and after.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll wait to hear about the camera &quot;height&quot; and try to find another tool for you to use. Do you ever use Python programming language?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Red dots are Perth at local noon and midnight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yWO38.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yWO38.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="2365" LastEditDate="2017-12-30T17:34:40.290" LastActivityDate="2017-12-30T17:34:40.290" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24264" PostTypeId="2" ParentId="24195" CreationDate="2017-12-30T18:29:38.687" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eDAeC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eDAeC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should really be a comment, but... you can usually check your calculations using a program like Stellarium (screenshot above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the latitude/longitude and right ascension/declination you give, it appears you are viewing Aldebaran from Chicago, IL at 7:20am UTC on 2017-12-27 (since Chicago is on Central Time and not observing Daylight Saving Time, which is 6h behind UTC).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see from the above, the hour angle is about 3h16m19s (or 49.16 degrees), the azimuth (which is what I think you actually wanted instead of hour angle) is about 254 degrees, and the altitude is about 41 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process for converting right ascension and declination to azimuth and elevation isn't difficult, and I considered posting a general answer, but I wasn't sure what format to use. Most programming languages have an astronomy library (which will be more accurate), and there are plenty of sites that already walk through how we get the formula, not to mention my &lt;a href=&quot;https://astronomy.stackexchange.com/a/14508/21&quot;&gt;https://astronomy.stackexchange.com/a/14508/21&lt;/a&gt; which gives sourced formulas.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2017-12-30T18:29:38.687" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24265" PostTypeId="2" ParentId="24239" CreationDate="2017-12-30T21:12:25.363" Score="4" Body="&lt;p&gt;I decided to answer because the role of perception is not yet taken into account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way we see is not only dictated by the optics of our eyes as cameras. It depends on how the brain elaborate or interprete the signals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well known but still surprising that a finger at arm length can accommodate something like seven to ten moons (if not ten to thirteen, I can't look right now). Still, it quite possible that asking a person (not aware of that) something like  &quot;how many moon can you cover with a finger?&quot; we get an answer such as &quot;may be half&quot; or &quot; one? two? Tell me...&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically to Moon:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Moon_illusion&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Moon_illusion&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photographing the Moon with a smartphone or a point and shoot camera we get things like this picture (3.5 mm, aperture 2.6).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon looks much smaller in the photograph than as seen in reality. The result does not look too dissimilar from the pic posted by OP.&#xA;(Let me add that my photo was cropped a bit, so the native size of the moon is even smaller).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this is to say that whatever is aimed at giving a feeling of realism should consider perception of reality rather then size of reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is done in Stellarium and many others sky viewing apps where a &quot;realistic&quot; option is used to alter the angular size of the moon paradoxically away from reality. I.e. the moon is arbitrarily scaled up to fit our common perception. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is what the OP is dealing with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS I am pretty sure that my pic has been uploaded. For some reason it does not appear to be the case. If necessary let me know. By the way it was a house with sky and a small than commonly perceived Moon ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per &lt;a href=&quot;https://astronomy.stackexchange.com/questions/24239/#comment41614_24265&quot;&gt;this comment&lt;/a&gt; here is a cell phone image with a horizontal FOV of about 60°. However in this pic the moon is not that small. I suspect the brain still enlarges it as it is down at horizon and trees are close :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4KFTg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4KFTg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17437" LastEditorUserId="7982" LastEditDate="2017-12-31T17:22:19.310" LastActivityDate="2017-12-31T17:22:19.310" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24267" PostTypeId="2" ParentId="24212" CreationDate="2017-12-31T06:52:16.303" Score="4" Body="&lt;p&gt;Using the calsky.org website, this is a chart showing a view of the eastern night sky from a point near San Francisco, CA at 7pm this evening.  (The green tick mark on the horizon is due east).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orion is located between the moon and the horizon, and will be rising quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pjNt2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pjNt2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2017-12-31T06:52:16.303" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24270" PostTypeId="2" ParentId="24212" CreationDate="2017-12-31T10:09:55.837" Score="2" Body="&lt;p&gt;Now that you have extensively modified your question (in particular the stars now rising in the east), then yes this is Orion.  The upper bright star is Betelgeuse - you should be able to see that it is red - and the lower bright star is Rigel, which is bluish white.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 3 stars in the middle form Orion's belt.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2017-12-31T10:09:55.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24273" PostTypeId="1" CreationDate="2017-12-31T16:13:03.617" Score="1" ViewCount="337" Body="&lt;p&gt;If we take midnight of 1/1/2017 in any country and mark in space where the center of the Earth is located, at what date, in the same country, has the center completed a full orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am almost certain that has to be 1/1/2018, but I am not so sure about the hour. &lt;/p&gt;&#xA;" OwnerUserId="20652" LastEditorUserId="2365" LastEditDate="2018-01-01T18:58:41.910" LastActivityDate="2018-01-02T19:18:10.257" Title="How long does it take the center of the Earth to complete a full orbit?" Tags="&lt;earth&gt;&lt;planetary-transits&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24274" PostTypeId="1" CreationDate="2017-12-31T16:43:17.267" Score="8" ViewCount="142" Body="&lt;p&gt;Recently I've encountered an article in our local astronomical popularization magazine. It is about a well known task of &lt;a href=&quot;https://en.wikipedia.org/wiki/Sagittarius_A*&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sgr A*&lt;/a&gt;-black hole's mass estimation. The article is written as a step by step guide for young researchers. For example, the orbital parameters of the S2 star should be found using a drawing on a sheet of paper.&lt;/p&gt;&#xA;&lt;p&gt;That is why I believed there is no need for &lt;em&gt;very precise&lt;/em&gt; calculations&#xA;for this task.&lt;/p&gt;&#xA;&lt;p&gt;However, in addition to the &lt;span class=&quot;math-container&quot;&gt;$ x $&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$ y $&lt;/span&gt; coordinates of the S2 star, I was also given the corresponding uncertainty values &lt;span class=&quot;math-container&quot;&gt;$ \Delta x $&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$ \Delta y $&lt;/span&gt; (19  coordinate measurements for the time period from 1992 to 2003):&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{array}{|c|c|c|c|}&#xA;\hline&#xA;time    &amp;amp; x      &amp;amp;   \Delta x    &amp;amp; y      &amp;amp; \Delta y \\ \hline&#xA;1992.226 &amp;amp; 0.104 &amp;amp; 0.003 &amp;amp; -0.166 &amp;amp; 0.004 \\&#xA;1994.321 &amp;amp; 0.097  &amp;amp; 0.003 &amp;amp; -0.189 &amp;amp; 0.004 \\&#xA;...     &amp;amp; ..     &amp;amp; ..    &amp;amp; ..     &amp;amp;       \\&#xA;2003.353 &amp;amp; 0.077  &amp;amp; 0.002 &amp;amp; -0.030 &amp;amp; 0.002 \\&#xA;2003.454 &amp;amp; 0.081  &amp;amp; 0.002 &amp;amp; -0.036 &amp;amp; 0.002\\&#xA;\hline&#xA;\end{array}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Using &lt;a href=&quot;http://cseweb.ucsd.edu/%7Emdailey/Face-Coord/ellipse-specific-fitting.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;direct least squares method&lt;/a&gt; for S2 orbital fitting by an ellipse I've found parameters&#xA;&lt;span class=&quot;math-container&quot;&gt;$a,\ b,\ c,\ d,\ e,\ f$&lt;/span&gt; of the conic equation: &lt;span class=&quot;math-container&quot;&gt;$$ax^2+bxy+cy^2+ dx+ey+f=0$$&lt;/span&gt;  I've&#xA;also decided to simulate the &lt;strong&gt;error propagation&lt;/strong&gt; in a Monte Carlo fashion to find&#xA;the uncertainty in the estimates of &lt;span class=&quot;math-container&quot;&gt;$a,\ b,\ c,\ d,\ e,\ f$&lt;/span&gt;. And I succeeded in that.&lt;/p&gt;&#xA;&lt;p&gt;But, here is &lt;strong&gt;the question&lt;/strong&gt;: is there any &lt;strong&gt;other&lt;/strong&gt; appropriate,&#xA;&amp;quot;analytical&amp;quot; way (not a Monte Carlo method) to find the uncertainty in&#xA;the estimates of &lt;span class=&quot;math-container&quot;&gt;$a,\ b,\ c,\ d,\ e,\ f$&lt;/span&gt; which may be used in usual astronomers' practice?&lt;/p&gt;&#xA;" OwnerUserId="19494" LastEditorUserId="37191" LastEditDate="2021-03-18T13:00:31.020" LastActivityDate="2021-04-17T14:53:10.077" Title="Error propagation methods for orbit parameters" Tags="&lt;orbit&gt;&lt;mathematics&gt;&lt;sagittarius-a&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24275" PostTypeId="1" CreationDate="2017-12-31T23:05:42.570" Score="4" ViewCount="858" Body="&lt;p&gt;Where can I find free/open data for the observed (not calculated/theoretical) distance between the Earth &amp;amp; Moon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to see the observed distance of the Moon, hour by hour.&lt;/p&gt;&#xA;" OwnerUserId="20642" LastEditorUserId="20642" LastEditDate="2018-01-09T00:37:38.067" LastActivityDate="2019-03-24T01:00:31.153" Title="Observed lunar distance data?" Tags="&lt;the-moon&gt;&lt;earth&gt;&lt;distances&gt;&lt;data-analysis&gt;&lt;geocentrism&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24276" PostTypeId="1" AcceptedAnswerId="24278" CreationDate="2018-01-01T00:55:37.170" Score="4" ViewCount="284" Body="&lt;p&gt;I am enjoying Orion high in the sky this winter as always. I wondered if I could travel beyond it, say 2730 light years* and turn around, would I see Orion in reverse? If so, what of the neighboring constellations. I know brightnesses would be different but would I recognize the shape?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Happy new year to all.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I chose this number because Alnilam is 1360 light years away so I doubled the distance. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://www.space.com/3380-constellations.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.space.com/3380-constellations.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2018-01-11T20:27:39.567" Title="If we could travel far enough would we be able to see a constelation in reverse?" Tags="&lt;constellations&gt;" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24277" PostTypeId="2" ParentId="24251" CreationDate="2018-01-01T01:25:48.587" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How do I find right ascension 18h23m52.34s on the celestial sphere?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You use a star chart that shows the lines of right ascension and declination. Based on the stars that the line passes by, you can estimate the line in the real sky. It is similar to looking at a map of the Earth and using latitude and longitude to locate a city. (By the way, 0 hours right ascension, 0 degrees declination is the location of the vernal equinox on the celestial sphere. The bright star at that location is the Sun at the moment of the vernal equinox ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps your question should be &quot;How do I locate the young crescent Moon?&quot;. The answer to that question is that it would be better to calculate the azimuth of the Sun at sunset, and the altitude and azimuth of the Moon at 10 minute intervals afterwards. Based on knowing where the Sun was located on the horizon at sunset and the difference between the Moon and Sun, you can estimate where the Moon should be in the sky. There is no need to know where right ascension and declination are located among the stars (other than calculating the position of the Sun and Moon); the sky will be bright, so it is unlikely that there will be any stars visible that will help you locate the line of right ascension. &lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2018-01-01T01:25:48.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24278" PostTypeId="2" ParentId="24276" CreationDate="2018-01-01T02:45:39.110" Score="13" Body="&lt;p&gt;If the stars are all similar distance from us, &lt;a href=&quot;https://en.wikipedia.org/wiki/Pleiades#Distance&quot; rel=&quot;noreferrer&quot;&gt;like the Pleiades&lt;/a&gt;, then yes, you could find a place from where the stars would appear in a similar pattern and brightness, but the pattern would be a mirror image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't do it when the stars are significantly different distances because they wouldn't look the same.   Most constellations are a range of distances, which wouldn't look like a mirror image from any distance.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take Canis Major for example.   Sirius is the brightest star in the sky but in large part because it's only 8 light years away.   The 2nd brightest star in that constellation, &lt;a href=&quot;https://en.wikipedia.org/wiki/Epsilon_Canis_Majoris&quot; rel=&quot;noreferrer&quot;&gt;Epsilon Canis Majoris&lt;/a&gt; is about 430 light years away.   At that distance, Sirius might not even be visible to the human eye.  It certainly wouldn't be constellation worthy. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-01-01T02:45:39.110" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24279" PostTypeId="2" ParentId="24276" CreationDate="2018-01-01T03:50:40.160" Score="-1" Body="&lt;p&gt;Pack orion in a cuboid of custom dimensions &amp;amp; make sure maximum number of stars touching the faces of cuboid from inside.&#xA;Make a normal (perpendicular line from the face) reaching earth and then extend the line to other side of the cuboid such that cuboids centre is the centre for the line.&#xA;Be careful going to another end ;)&lt;/p&gt;&#xA;" OwnerUserId="20655" LastActivityDate="2018-01-01T03:50:40.160" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24280" PostTypeId="2" ParentId="24275" CreationDate="2018-01-01T04:32:53.720" Score="4" Body="&lt;p&gt;Found laser ranging data here: &lt;a href=&quot;http://www.geoazur.fr/astrogeo/?href=observations/donnees/lune/brutes&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.geoazur.fr/astrogeo/?href=observations/donnees/lune/brutes&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you can search for data for an arbitrary time period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data is what they call &quot;MINI&quot; format, which is hard to read, its basically a long string of numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a sample line:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5120160113152419452625024340653926601301910034002705017 087323+04325 5320a0702&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Luckily there is a specification for this format here: &lt;a href=&quot;http://www.geoazur.fr/astrogeo/observations/donnees/lune/mini-format.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.geoazur.fr/astrogeo/observations/donnees/lune/mini-format.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spec says the time-of-flight for the laser is charcters 24-37 from each line, measured in .1 picoseconds. So for the above line, the laser's round-trip flight time is 24340653926601 (.1 ps).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data does not contain the distance, so to calculate the distance from the flight time, I do the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide 24340653926601/2 to get 1-way flight time in .1 ps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the result 1.2170327e+13*.1 to get ps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the result 1.2170327e+12*1.0e-12 to get seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the result 1.2170327*299792458(the speed of light) to get the distance in meters: 364857224.599&lt;/p&gt;&#xA;" OwnerUserId="20642" LastEditorUserId="20642" LastEditDate="2018-01-07T06:49:34.703" LastActivityDate="2018-01-07T06:49:34.703" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24286" PostTypeId="2" ParentId="24273" CreationDate="2018-01-01T15:06:25.193" Score="3" Body="&lt;p&gt;It's not exactly one year because of the 28/29 February thing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact period is 365 days 6 hours 9 minutes 9 seconds [Random House Encyclopedia. New York: Random House: 174.]&lt;/p&gt;&#xA;" OwnerUserId="19839" LastActivityDate="2018-01-01T15:06:25.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24287" PostTypeId="1" CreationDate="2018-01-02T02:05:15.807" Score="4" ViewCount="26143" Body="&lt;p&gt;We purchased my young daughter a beginner telescope for Christmas and she's super excited to look at the large planets. Our telescope has a 70mm aperture and two eye pieces (10mm and 20mm) How much, if any, detail can we expect to see on Jupiter and Saturn? I don't want her going in expecting to see Saturn's rings and come out disappointed! We are very new to this and I want her to have the best experience possible with the equipment we have. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="20673" LastActivityDate="2019-02-01T22:08:14.357" Title="What can we expect to see with a telescope with a 70mm aperture and a 10mm eyepiece?" Tags="&lt;telescope&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="24288" PostTypeId="2" ParentId="12468" CreationDate="2018-01-02T04:04:30.470" Score="3" Body="&lt;p&gt;The answer that you are looking for rests on two main pillars. Are you limited in your ability to resolve objects because of the atmosphere like most ground-based telescopes, or because of the fundamentals of the wave nature of light (as per space telescopes)? I am assuming the latter in your question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the biggest rule-of-thumb equations when it comes to astronomy is the equation that you are looking for in your answer. Astronomers define angular resolution with regard to the &lt;em&gt;Rayleigh criterion&lt;/em&gt;: two point sources are regarded as just resolved when the principal diffraction maximum of one image coincides with the first minimum of the other (Wikipedia). See Airy Disk functions and the Wikipedia article on Angular Resolution for a better idea as to what this means. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The accurate, but still approximate, equation for the resolution of an object $\theta$ (radians), for a given wavelength of light $\lambda$, over some telescope's diameter $D$, is given below so long as the wavelength of light and the diameter is measured as the same length unit (i.e meters, feet, angstroms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \theta = \frac{1.22 \lambda}{D} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, the atmosphere is a pain, so, adaptive optics in most famous telescopes try and correct to this limit. To exceed this limit, major breakthroughs would need to happen in the wave nature of photons themselves. If your &quot;technology far beyond&quot; includes this, then, I don't think your question can be answered as you would hope aside from the standard &quot;who knows&quot;. Currently, only space-based telescopes operate at this limit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the most part, this is really all you need. It does not matter what object you are observing, this still always applies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited: Included unit specifications. &lt;/p&gt;&#xA;" OwnerUserId="20675" LastEditorUserId="20675" LastEditDate="2018-01-03T03:15:45.837" LastActivityDate="2018-01-03T03:15:45.837" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24295" PostTypeId="2" ParentId="24239" CreationDate="2018-01-02T08:24:10.187" Score="2" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;tl;dr&lt;/em&gt;-&lt;/strong&gt;  Your moon's dimensions look pretty decent, but the lighting effect is causing your moon to be mostly dark and dim, causing it to appear even smaller.  You could probably fix this pretty easily by adjusting the lighting model for the moon object.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To test your result, I tried this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I checked your numbers; they match Wikipedia's.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I made the scene you described in the question in a ray-tracer that I've created and haven't released (so we're definitely using different software).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Compared the result to the picture provided in &lt;a href=&quot;https://astronomy.stackexchange.com/a/24265/15986&quot;&gt;@Alchimista's answer&lt;/a&gt;, using the 60-degree camera angle they'd estimated for that picture and matched its dimensions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Here's the image of the moon I got (didn't put in the Earth; the stars in the background are randomly generated):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ebev9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ebev9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here're the moons cropped out from the ray-tracer and &lt;a href=&quot;https://astronomy.stackexchange.com/a/24265/15986&quot;&gt;@Alchimista's answer&lt;/a&gt;'s image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OYUj4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OYUj4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here's an overlay, cropping out the parts of the moon in the photo that didn't fit into the ray-tracer's moon's frame:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bevWA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bevWA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looks like a good fit!  There're a few small sources of error:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Ray-tracer didn't account for any atmospheric effects.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Aimed the camera directly at the moon, whereas the picture had the moon a bit off-center.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ray-tracer used exactly 60-degrees for the horizontal view angle, but that was only an estimate for the camera's.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Used numbers given in question for moon's distance/size.  Presumably the moon wasn't at exactly that estimated average distance in the picture.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The ray-tracer image and the picture used slightly different pixel heights/widths, since I eye-ball'd it.  They were pretty close (overlayed the windows), but probably not exact.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Just to avoid confusion, I didn't actually use the shared photo file, but rather a screenshot of the photo.  The relevant point there is that the pixel dimensions of the ray-traced image may not match the photo's pixel dimensions for this reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, I overlaid your moon with a red-framed version of mine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7zDtt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7zDtt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our moons look like they might be of about the same dimension, except they're drawn differently.  I drew mine as a 100% white light; any ray that hit it returned a perfectly white pixel.  Yours looks mostly dark, with a &lt;a href=&quot;https://en.wikipedia.org/wiki/Specular_highlight&quot; rel=&quot;nofollow noreferrer&quot;&gt;specular highlight&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a specular highlight isn't a good model for the moon; at least, not one as sharp as used in that scene.  As can be seen in the overlay with the photo's moon, the real-life moon has a more reflective effect with a brighter region (though that'll vary with the phase of the moon, e.g. a full moon vs. the crescent-moon that appears in the photo).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, the moon's probably going to look small without the typical scenery around it, though the choice of lighting effect makes it look yet smaller.&lt;/p&gt;&#xA;" OwnerUserId="15986" LastEditorUserId="15986" LastEditDate="2018-01-02T08:53:41.843" LastActivityDate="2018-01-02T08:53:41.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24296" PostTypeId="1" CreationDate="2018-01-02T09:39:42.790" Score="1" ViewCount="54" Body="&lt;p&gt;What size of the grating on GAIA RVS spectroscope ?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;How many lines on mm, or what defraction angle for 850 nm wavelenth light ?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I find some more details here: &lt;a href=&quot;http://sci.esa.int/gaia/44092-gaia-rvs-grating/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://sci.esa.int/gaia/44092-gaia-rvs-grating/&lt;/a&gt; and &lt;a href=&quot;http://sci.esa.int/gaia/44093-state-of-the-art-grating-for-gaia/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://sci.esa.int/gaia/44093-state-of-the-art-grating-for-gaia/&lt;/a&gt;&#xA;The GAIA RVS grating has dimensions of 20.5 cm × 15.5 cm and 302.11 grooves per mm.&lt;/p&gt;&#xA;" OwnerUserId="20687" LastEditorUserId="20687" LastEditDate="2018-01-07T20:08:27.027" LastActivityDate="2018-01-07T20:08:27.027" Title="What a grating on GAIA RVS spectroscope?" Tags="&lt;gaia&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24304" PostTypeId="1" AcceptedAnswerId="24305" CreationDate="2018-01-02T16:03:24.573" Score="3" ViewCount="898" Body="&lt;p&gt;I would like to know how the length of sunrise/sunset (and also time between civil/nautical/astronomical twilight) varies with latitude and with day of year. Ideally what I am looking for is an expression that takes latitude and day-of-year as arguments and outputs the time taken for the sun to rise or set at that latitude on that day.&lt;/p&gt;&#xA;" OwnerUserId="20697" LastActivityDate="2018-01-06T05:03:51.540" Title="Expression for length of sunrise/sunset as function of latitude and day-of-year" Tags="&lt;the-sun&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24305" PostTypeId="2" ParentId="24304" CreationDate="2018-01-02T17:14:46.567" Score="3" Body="&lt;p&gt;The equations necessary for a correct answer are somewhat lengthy, for a web interface that accepts the location and prints a calendar try: &lt;a href=&quot;https://sunrise-sunset.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://sunrise-sunset.org/&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a lengthy explanation of the mathematics see Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Equation_of_time#Calculating_the_equation_of_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;Calculating the equation of time&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Daytime&quot; rel=&quot;nofollow noreferrer&quot;&gt;Daytime&lt;/a&gt; webpages for information about the &lt;strong&gt;duration&lt;/strong&gt; of rising and setting, along with information about the differences in &lt;a href=&quot;https://en.wikipedia.org/wiki/Daytime#Variations_in_solar_noon&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar noon&lt;/a&gt; and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Twilight&quot; rel=&quot;nofollow noreferrer&quot;&gt;three types of twilight&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1X8Dy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1X8Dy.png&quot; alt=&quot;Twilight Types&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a &lt;a href=&quot;http://www.stargazing.net/kepler/sunrise.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;QBasic implementation by Keith Burnett&lt;/a&gt; for twilight times:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;'********************************************************************&#xA;'   Sun rise/set /twilight&#xA;'   From Explanatory Supplement&#xA;'   definitions and functions&#xA;DEFDBL A-Z&#xA;pi = 4 * ATN(1)&#xA;tpi = 2 * pi&#xA;twopi = tpi&#xA;degs = 180 / pi&#xA;rads = pi / 180&#xA;'&#xA;'   the function below returns the true integer part,&#xA;'   even for negative numbers&#xA;'&#xA;DEF FNipart (x) = SGN(x) * INT(ABS(x))&#xA;'&#xA;'   FNday only works between 1901 to 2099 - see Meeus chapter 7&#xA;'&#xA;DEF FNday (y, m, d, h) = 367 * y - 7 * (y + (m + 9) \ 12) \ 4 + 275 * m \ 9 + d - 730531.5 + h / 24&#xA;'&#xA;'   define some arc cos and arc sin functions&#xA;'&#xA;DEF FNacos (x)&#xA;    s = SQR(1 - x * x)&#xA;    FNacos = ATN(s / x)&#xA;END DEF&#xA;DEF FNasin (x)&#xA;    c = SQR(1 - x * x)&#xA;    FNasin = ATN(x / c)&#xA;END DEF&#xA;'&#xA;'   the function below returns an angle in the range&#xA;'   0 to two pi&#xA;'&#xA;DEF FNrange (x)&#xA;    b = x / tpi&#xA;    a = tpi * (b - FNipart(b))&#xA;    IF a &amp;lt; 0 THEN a = tpi + a&#xA;    FNrange = a&#xA;END DEF&#xA;'&#xA;'       the function below returns the time in 24 hour&#xA;'   notation - hhmm - given a decimal hours figure.&#xA;DEF FNdegmin$ (d)&#xA;c = ABS(d)&#xA;a = INT(c)&#xA;b = 60 * (c - a)&#xA;d$ = STR$(INT(a * 100 + b + .5))&#xA;d$ = RIGHT$(d$, LEN(d$) - 1)&#xA;WHILE LEN(d$) &amp;lt; 4&#xA;    d$ = &quot;0&quot; + d$&#xA;WEND&#xA;FNdegmin$ = d$&#xA;END DEF&#xA;CLS&#xA;'&#xA;'    get the date and geographical position from user&#xA;'&#xA;INPUT &quot;     lat : &quot;, glat&#xA;INPUT &quot;    long : &quot;, glong&#xA;INPUT &quot;    zone : &quot;, zone&#xA;INPUT &quot;rise/set : &quot;, riset&#xA;INPUT &quot; Sun alt : &quot;, altitude&#xA;INPUT &quot;   year  : &quot;, y&#xA;INPUT &quot;   month : &quot;, m&#xA;INPUT &quot;   day   : &quot;, d&#xA;day = FNday(y, m, d, 0)&#xA;PRINT &quot;  day no : &quot;; day&#xA;utold = pi&#xA;utnew = 0&#xA;sinalt = SIN(altitude * rads) 'go for the sunrise/sunset altitude first&#xA;sinphi = SIN(glat * rads)&#xA;cosphi = COS(glat * rads)&#xA;glong = glong * rads&#xA;DO WHILE ABS(utold - utnew) &amp;gt; .001&#xA;  utold = utnew&#xA;  days = day + utold / tpi&#xA;  t = days / 36525&#xA;  '     get arguments of Sun's orbit&#xA;  L = FNrange(4.8949504201433# + 628.331969753199# * t)&#xA;  G = FNrange(6.2400408# + 628.3019501# * t)&#xA;  ec = .033423 * SIN(G) + .00034907# * SIN(2 * G)&#xA;  lambda = L + ec&#xA;  E = -1 * ec + .0430398# * SIN(2 * lambda) - .00092502# * SIN(4 * lambda)&#xA;  obl = .409093 - .0002269# * t&#xA;  delta = FNasin(SIN(obl) * SIN(lambda))&#xA;  GHA = utold - pi + E&#xA;  cosc = (sinalt - sinphi * SIN(delta)) / (cosphi * COS(delta))&#xA;  SELECT CASE cosc&#xA;  CASE cosc &amp;gt; 1&#xA;    correction = 0&#xA;  CASE cosc &amp;lt; -1&#xA;    correction = pi&#xA;  CASE ELSE&#xA;    correction = FNacos(cosc)&#xA;  END SELECT&#xA;  utnew = FNrange(utold - (GHA + glong + riset * correction))&#xA;LOOP&#xA;PRINT &quot;    UT   : &quot;; FNdegmin$(utnew * degs / 15)&#xA;PRINT &quot;   zone  : &quot;; FNdegmin$(utnew * degs / 15 + zone)&#xA;END&#xA;'******************************************************************&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is an implementation by &lt;a href=&quot;https://astronomy.stackexchange.com/a/13053/18054&quot;&gt;Barry Carter&lt;/a&gt; of rising/setting &lt;strong&gt;duration&lt;/strong&gt; (see &lt;a href=&quot;https://github.com/barrycarter/bcapps/tree/master/ASTRO&quot; rel=&quot;nofollow noreferrer&quot;&gt;his GitHub&lt;/a&gt; for associated files):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;// http://astronomy.stackexchange.com/questions/12824/how-long-does-a-sunrise-or-sunset-take&#xA;&#xA;#include &amp;lt;stdio.h&amp;gt;&#xA;#include &amp;lt;stdlib.h&amp;gt;&#xA;#include &amp;lt;string.h&amp;gt;&#xA;#include &quot;SpiceUsr.h&quot;&#xA;#include &quot;SpiceZfc.h&quot;&#xA;#include &quot;/home/barrycarter/BCGIT/ASTRO/bclib.h&quot;&#xA;&#xA;// since the sun travels no more than 362 degrees per day, and has an&#xA;// angular diameter of no less than 31 minutes, the fastest possible&#xA;// sunrise is 31 minutes/362 degrees or 2 minutes; thus, 120 below in&#xA;// bc_between below is valid&#xA;&#xA;int main(int argc, char **argv) {&#xA;&#xA;  SpiceDouble radii[3];&#xA;  SpiceInt n;&#xA;  char direction[10];&#xA;&#xA;  furnsh_c(&quot;/home/barrycarter/BCGIT/ASTRO/standard.tm&quot;);&#xA;&#xA;  // the year 2015&#xA;  double stime = 1419984000, etime = 1451692800;&#xA;&#xA;  // test&#xA;  // double stime =  1388559600, etime = 1419984000;&#xA;&#xA;  // lat/lon from argv&#xA;  double lat = atof(argv[1]), lon = atof(argv[2]);&#xA;&#xA;  // the sun's radii&#xA;  bodvrd_c(&quot;SUN&quot;, &quot;RADII&quot;, 3, &amp;amp;n, radii);&#xA;&#xA;&#xA;  double *results = bc_between(9, lat*rpd_c(), lon*rpd_c(), 0., stime, etime,&#xA;                   &quot;Sun&quot;, -34/60.*rpd_c(), 120., radii[0]);&#xA;&#xA;  // NOTE: can NOT ignore result 0 and 1, but must check if they are borders&#xA;  for (int i=0; i&amp;lt;1000; i++) {&#xA;&#xA;    double rstart = results[2*i], rend = results[2*i+1];&#xA;&#xA;    // if we start seeing 0s, we are out of true answers&#xA;    // TODO: this may break near 1970&#xA;    if (abs(rstart) &amp;lt; .001) {break;}&#xA;&#xA;    // if the end result is too close to etime, result is inaccurate&#xA;    // same if start result is too close to stime&#xA;    if (abs(rend-etime)&amp;lt;1 || abs(rstart-stime)&amp;lt;1) {continue;}&#xA;&#xA;    // TODO: this really inefficient (and not even always correct?)&#xA;    if (bc_sky_elev(6, lat, lon, 0., rstart, &quot;Sun&quot;, 0.) &amp;gt;&#xA;    bc_sky_elev(6, lat, lon, 0., rend, &quot;Sun&quot;, 0.)) {&#xA;      strcpy(direction,&quot;SET&quot;);&#xA;    } else {&#xA;      strcpy(direction,&quot;RISE&quot;);&#xA;    }&#xA;&#xA;    // the &quot;day&quot; and length of sunrise/sunset&#xA;    printf(&quot;%f %f %s %f %f %f\n&quot;, lat, lon, direction, rstart, rend, rend-rstart);&#xA;  }&#xA;  return 0;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For an explanation see &lt;a href=&quot;https://en.wikipedia.org/wiki/Sunrise_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia's Sunrise Equation&lt;/a&gt; webpage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the algorithm for rising/setting &lt;strong&gt;times&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Sunrise/Sunset Algorithm&#xA;&#xA;Source:&#xA;    Almanac for Computers, 1990&#xA;    published by Nautical Almanac Office&#xA;    United States Naval Observatory&#xA;    Washington, DC 20392&#xA;&#xA;Inputs:&#xA;    day, month, year:      date of sunrise/sunset&#xA;    latitude, longitude:   location for sunrise/sunset&#xA;    zenith:                Sun's zenith for sunrise/sunset&#xA;      offical      = 90 degrees 50'&#xA;      civil        = 96 degrees&#xA;      nautical     = 102 degrees&#xA;      astronomical = 108 degrees&#xA;&#xA;    NOTE: longitude is positive for East and negative for West&#xA;        NOTE: the algorithm assumes the use of a calculator with the&#xA;        trig functions in &quot;degree&quot; (rather than &quot;radian&quot;) mode. Most&#xA;        programming languages assume radian arguments, requiring back&#xA;        and forth convertions. The factor is 180/pi. So, for instance,&#xA;        the equation RA = atan(0.91764 * tan(L)) would be coded as RA&#xA;        = (180/pi)*atan(0.91764 * tan((pi/180)*L)) to give a degree&#xA;        answer with a degree input for L.&#xA;&#xA;&#xA;1. first calculate the day of the year&#xA;&#xA;    N1 = floor(275 * month / 9)&#xA;    N2 = floor((month + 9) / 12)&#xA;    N3 = (1 + floor((year - 4 * floor(year / 4) + 2) / 3))&#xA;    N = N1 - (N2 * N3) + day - 30&#xA;&#xA;2. convert the longitude to hour value and calculate an approximate time&#xA;&#xA;    lngHour = longitude / 15&#xA;&#xA;    if rising time is desired:&#xA;      t = N + ((6 - lngHour) / 24)&#xA;    if setting time is desired:&#xA;      t = N + ((18 - lngHour) / 24)&#xA;&#xA;3. calculate the Sun's mean anomaly&#xA;&#xA;    M = (0.9856 * t) - 3.289&#xA;&#xA;4. calculate the Sun's true longitude&#xA;&#xA;    L = M + (1.916 * sin(M)) + (0.020 * sin(2 * M)) + 282.634&#xA;    NOTE: L potentially needs to be adjusted into the range [0,360) by adding/subtracting 360&#xA;&#xA;5a. calculate the Sun's right ascension&#xA;&#xA;    RA = atan(0.91764 * tan(L))&#xA;    NOTE: RA potentially needs to be adjusted into the range [0,360) by adding/subtracting 360&#xA;&#xA;5b. right ascension value needs to be in the same quadrant as L&#xA;&#xA;    Lquadrant  = (floor( L/90)) * 90&#xA;    RAquadrant = (floor(RA/90)) * 90&#xA;    RA = RA + (Lquadrant - RAquadrant)&#xA;&#xA;5c. right ascension value needs to be converted into hours&#xA;&#xA;    RA = RA / 15&#xA;&#xA;6. calculate the Sun's declination&#xA;&#xA;    sinDec = 0.39782 * sin(L)&#xA;    cosDec = cos(asin(sinDec))&#xA;&#xA;7a. calculate the Sun's local hour angle&#xA;&#xA;    cosH = (cos(zenith) - (sinDec * sin(latitude))) / (cosDec * cos(latitude))&#xA;&#xA;    if (cosH &amp;gt;  1) &#xA;      the sun never rises on this location (on the specified date)&#xA;    if (cosH &amp;lt; -1)&#xA;      the sun never sets on this location (on the specified date)&#xA;&#xA;7b. finish calculating H and convert into hours&#xA;&#xA;    if if rising time is desired:&#xA;      H = 360 - acos(cosH)&#xA;    if setting time is desired:&#xA;      H = acos(cosH)&#xA;&#xA;    H = H / 15&#xA;&#xA;8. calculate local mean time of rising/setting&#xA;&#xA;    T = H + RA - (0.06571 * t) - 6.622&#xA;&#xA;9. adjust back to UTC&#xA;&#xA;    UT = T - lngHour&#xA;    NOTE: UT potentially needs to be adjusted into the range [0,24) by adding/subtracting 24&#xA;&#xA;10. convert UT value to local time zone of latitude/longitude&#xA;&#xA;    localT = UT + localOffset&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here is an implementation of rising/setting &lt;strong&gt;times&lt;/strong&gt; in Basic:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;10 '         Sunrise-Sunset&#xA;20 GOSUB 300&#xA;30 INPUT &quot;Lat, Long (deg)&quot;;B5,L5&#xA;40 INPUT &quot;Time zone (hrs)&quot;;H&#xA;50 L5=L5/360: Z0=H/24&#xA;60 GOSUB 1170: T=(J-2451545)+F&#xA;70 TT=T/36525+1: ' TT = centuries&#xA;80 '               from 1900.0&#xA;90 GOSUB 410: T=T+Z0&#xA;100 '&#xA;110 '       Get Sun's Position&#xA;120 GOSUB 910: A(1)=A5: D(1)=D5&#xA;130 T=T+1&#xA;140 GOSUB 910: A(2)=A5: D(2)=D5&#xA;150 IF A(2)&amp;lt;A(1) THEN A(2)=A(2)+P2&#xA;160 Z1=DR*90.833: ' Zenith dist.&#xA;170 S=SIN(B5*DR): C=COS(B5*DR)&#xA;180 Z=COS(Z1): M8=0: W8=0: PRINT&#xA;190 A0=A(1): D0=D(1)&#xA;200 DA=A(2)-A(1): DD=D(2)-D(1)&#xA;210 FOR C0=0 TO 23&#xA;220 P=(C0+1)/24&#xA;230 A2=A(1)+P*DA: D2=D(1)+P*DD&#xA;240 GOSUB 490&#xA;250 A0=A2: D0=D2: V0=V2&#xA;260 NEXT&#xA;270 GOSUB 820: '  Special msg?&#xA;280 END&#xA;290 '&#xA;300 '        Constants&#xA;310 DIM A(2),D(2)&#xA;320 P1=3.14159265: P2=2*P1&#xA;330 DR=P1/180: K1=15*DR*1.0027379&#xA;340 S$=&quot;Sunset at  &quot;&#xA;350 R$=&quot;Sunrise at &quot;&#xA;360 M1$=&quot;No sunrise this date&quot;&#xA;370 M2$=&quot;No sunset this date&quot;&#xA;380 M3$=&quot;Sun down all day&quot;&#xA;390 M4$=&quot;Sun up all day&quot;&#xA;400 RETURN&#xA;410 '     LST at 0h zone time&#xA;420 T0=T/36525&#xA;430 S=24110.5+8640184.813*T0&#xA;440 S=S+86636.6*Z0+86400*L5&#xA;450 S=S/86400: S=S-INT(S)&#xA;460 T0=S*360*DR&#xA;470 RETURN&#xA;480 '&#xA;490 '  Test an hour for an event&#xA;500 L0=T0+C0*K1: L2=L0+K1&#xA;510 H0=L0-A0: H2=L2-A2&#xA;520 H1=(H2+H0)/2: '  Hour angle,&#xA;530 D1=(D2+D0)/2: '  declination,&#xA;540 '                at half hour&#xA;550 IF C0&amp;gt;0 THEN 570&#xA;560 V0=S*SIN(D0)+C*COS(D0)*COS(H0)-Z&#xA;570 V2=S*SIN(D2)+C*COS(D2)*COS(H2)-Z&#xA;580 IF SGN(V0)=SGN(V2) THEN 800 &#xA;590 V1=S*SIN(D1)+C*COS(D1)*COS(H1)-Z&#xA;600 A=2*V2-4*V1+2*V0: B=4*V1-3*V0-V2&#xA;610 D=B*B-4*A*V0: IF D&amp;lt;0 THEN 800&#xA;620 D=SQR(D)&#xA;630 IF V0&amp;lt;0 AND V2&amp;gt;0 THEN PRINT R$;&#xA;640 IF V0&amp;lt;0 AND V2&amp;gt;0 THEN M8=1&#xA;650 IF V0&amp;gt;0 AND V2&amp;lt;0 THEN PRINT S$;&#xA;660 IF V0&amp;gt;0 AND V2&amp;lt;0 THEN W8=1&#xA;670 E=(-B+D)/(2*A)&#xA;680 IF E&amp;gt;1 OR E&amp;lt;0 THEN E=(-B-D)/(2*A)&#xA;690 T3=C0+E+1/120: ' Round off&#xA;700 H3=INT(T3): M3=INT((T3-H3)*60)&#xA;710 PRINT USING &quot;##:##&quot;;H3;M3;&#xA;720 H7=H0+E*(H2-H0)&#xA;730 N7=-COS(D1)*SIN(H7)&#xA;740 D7=C*SIN(D1)-S*COS(D1)*COS(H7)&#xA;750 AZ=ATN(N7/D7)/DR&#xA;760 IF D7&amp;lt;0 THEN AZ=AZ+180&#xA;770 IF AZ&amp;lt;0 THEN AZ=AZ+360&#xA;780 IF AZ&amp;gt;360 THEN AZ=AZ-360&#xA;790 PRINT USING &quot;,  azimuth ###.#&quot;;AZ&#xA;800 RETURN&#xA;810 '&#xA;820 '   Special-message routine&#xA;830 IF M8=0 AND W8=0 THEN 870&#xA;840 IF M8=0 THEN PRINT M1$&#xA;850 IF W8=0 THEN PRINT M2$&#xA;860 GOTO 890&#xA;870 IF V2&amp;lt;0 THEN PRINT M3$&#xA;880 IF V2&amp;gt;0 THEN PRINT M4$&#xA;890 RETURN&#xA;900 '&#xA;910 '   Fundamental arguments&#xA;920 '     (Van Flandern &amp;amp;&#xA;930 '     Pulkkinen, 1979)&#xA;940 L=.779072+.00273790931*T&#xA;950 G=.993126+.0027377785*T&#xA;960 L=L-INT(L): G=G-INT(G)&#xA;970 L=L*P2: G=G*P2&#xA;980 V=.39785*SIN(L)&#xA;990 V=V-.01000*SIN(L-G)&#xA;1000 V=V+.00333*SIN(L+G)&#xA;1010 V=V-.00021*TT*SIN(L)&#xA;1020 U=1-.03349*COS(G)&#xA;1030 U=U-.00014*COS(2*L)&#xA;1040 U=U+.00008*COS(L)&#xA;1050 W=-.00010-.04129*SIN(2*L)&#xA;1060 W=W+.03211*SIN(G)&#xA;1070 W=W+.00104*SIN(2*L-G)&#xA;1080 W=W-.00035*SIN(2*L+G)&#xA;1090 W=W-.00008*TT*SIN(G)&#xA;1100 '&#xA;1110 '    Compute Sun's RA and Dec&#xA;1120 S=W/SQR(U-V*V)&#xA;1130 A5=L+ATN(S/SQR(1-S*S))&#xA;1140 S=V/SQR(U):D5=ATN(S/SQR(1-S*S))&#xA;1150 R5=1.00021*SQR(U)&#xA;1160 RETURN&#xA;1165 '&#xA;1170 '     Calendar --&amp;gt; JD&#xA;1180 INPUT &quot;Year, Month, Day&quot;;Y,M,D&#xA;1190 G=1: IF Y&amp;lt;1583 THEN G=0&#xA;1200 D1=INT(D): F=D-D1-.5&#xA;1210 J=-INT(7*(INT((M+9)/12)+Y)/4)&#xA;1220 IF G=0 THEN 1260&#xA;1230 S=SGN(M-9): A=ABS(M-9)&#xA;1240 J3=INT(Y+S*INT(A/7))&#xA;1250 J3=-INT((INT(J3/100)+1)*3/4)&#xA;1260 J=J+INT(275*M/9)+D1+G*J3&#xA;1270 J=J+1721027+2*G+367*Y&#xA;1280 IF F&amp;gt;=0 THEN 1300&#xA;1290 F=F+1: J=J-1&#xA;1300 RETURN&#xA;1310 '&#xA;1320 '   This program by Roger W. Sinnott calculates the times of sunrise&#xA;1330 '   and sunset on any date, accurate to the minute within several&#xA;1340 '   centuries of the present.  It correctly describes what happens in the &#xA;1350 '   arctic and antarctic regions, where the Sun may not rise or set on&#xA;1360 '   a given date.  Enter north latitudes positive, west longitudes&#xA;1370 '   negative.  For the time zone, enter the number of hours west of&#xA;1380 '   Greenwich (e.g., 5 for EST, 4 for EDT).  The calculation is&#xA;1390 '   discussed in Sky &amp;amp; Telescope for August 1994, page 84.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="18054" LastEditDate="2018-01-06T05:03:51.540" LastActivityDate="2018-01-06T05:03:51.540" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24306" PostTypeId="1" CreationDate="2018-01-02T19:11:51.427" Score="3" ViewCount="1951" Body="&lt;p&gt;I was very confused when I watched &lt;a href=&quot;https://youtu.be/IJhgZBn-LHg?t=138&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video from 2:18 till 2:45&lt;/a&gt;. And it seems very clear that if the Earth is really rotating from west to east, shouldn't be the Sun rising from west and setting in east? We all know certainly that the sun rises from east, but how come Earth is rotating in the way as shown in the linked video? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I somehow feel very stupid for asking this question...&lt;/p&gt;&#xA;" OwnerUserId="20698" LastActivityDate="2018-04-03T16:13:38.357" Title="If the Earth is rotating counter clockwise, how does the sun rise from east?" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;astrophysics&gt;&lt;rotation&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24307" PostTypeId="2" ParentId="24273" CreationDate="2018-01-02T19:18:10.257" Score="2" Body="&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Year#Sidereal,_tropical,_and_anomalistic_years&quot; rel=&quot;nofollow noreferrer&quot;&gt;sidereal year&lt;/a&gt; is 365.256363004 days (365 d 6 h 9 min 9.76 s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the well known Web &lt;a href=&quot;https://www.timeanddate.com/calendar/&quot; rel=&quot;nofollow noreferrer&quot;&gt;calendars&lt;/a&gt; and &lt;a href=&quot;https://www.timeanddate.com/date/dateadd.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;date calculators&lt;/a&gt; is at: &lt;a href=&quot;https://www.timeanddate.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.timeanddate.com/&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the result for your specific question 1/1/2017+1year = &lt;a href=&quot;https://www.timeanddate.com/date/dateadded.html?m1=1&amp;amp;d1=1&amp;amp;y1=2017&amp;amp;type=add&amp;amp;ay=1&amp;amp;am=&amp;amp;aw=&amp;amp;ad=&amp;amp;h1=00&amp;amp;i1=00&amp;amp;s1=00&amp;amp;ah=&amp;amp;ai=&amp;amp;as=&amp;amp;rec=&quot; rel=&quot;nofollow noreferrer&quot;&gt;1/1/2018&lt;/a&gt;. Adding exactly one year (on a calendar) gives you one year later, everything is the same except the 'year number'.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastActivityDate="2018-01-02T19:18:10.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24308" PostTypeId="2" ParentId="24306" CreationDate="2018-01-02T19:23:54.587" Score="1" Body="&lt;p&gt;Try standing in a room, facing a particular wall. Spread your arms out to your sides, and spin slowly with your right arm leading, so that you’re turning towards your right.  This is like the Earth rotating from west to east. Now notice that everything around you seems to be “rotating” in the other direction, from your right hand side to your left. This is the same as the Sun (and everything else in the sky) rising in the east and setting in the west. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a similar effect to being in a car, and overtaking a slower moving vehicle. You’re travelling forwards, but the other vehicle appears to be going backwards from your perspective. &lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2018-01-02T19:23:54.587" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24310" PostTypeId="2" ParentId="24287" CreationDate="2018-01-02T20:38:23.373" Score="0" Body="&lt;p&gt;As of right now, Saturn is not visible yet, it's too close to the Sun. You'll have to wait a few months until it becomes visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter is currently rising around 3 am, so it's only visible early in the morning before sunrise. Again, you'll have to wait until later this year for it to rise at a convenient hour in the evening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://heavens-above.com/PlanetSummary.aspx?lat=0&amp;amp;lng=0&amp;amp;loc=Unspecified&amp;amp;alt=0&amp;amp;tz=UCT&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://heavens-above.com/PlanetSummary.aspx?lat=0&amp;amp;lng=0&amp;amp;loc=Unspecified&amp;amp;alt=0&amp;amp;tz=UCT&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until then you can definitely observe the Moon. Your scope is powerful enough to show craters and mountains on the lunar surface. This is best done not when the Moon is full, but when there's some (or even a lot) of shadow on the disk. The terrain is much more easily visible near the edge between light and darkness on the lunar disk. When the Moon is completely full, the image is flat and lacks the contrast that makes the mountains visible &quot;in 3D&quot;. Any amount of shadow, from 1% to 99% makes that stuff much more visible near the transition line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, let me try and answer your question.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The aperture per se is more than enough to see some detail on Jupiter and Saturn. The question is the overall quality of the instrument, and the precision of its collimation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;50 mm of aperture is enough to see the rings of Saturn. I see them in my 50 mm finderscope, when paired with a strong eyepiece. The overall figure is very tiny, but you can definitely see the shape of the disk crossed diagonally by what looks like a small sharp line - the rings. I assume in your scope it will look bigger (I don't know what's the focal length of the primary lens).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Titan, Saturn's biggest satellite, should also be visible not too far from the planet once in a while, but it's easily confused with a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter is definitely seen as a disk. Whether or not you can see the equatorial belts depends on a number of factors - but again it should be doable in that aperture assuming everything else is shipshape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter's moons should be visible almost no matter what. You'll see them as anywhere between 1 and 4 little &quot;stars&quot; that are hanging around not too far from the planet, all in the same plane more or less. The configuration visibly changes within hours, and is very clearly different on successive nights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You (or your child) could do drawings of what's visible in the eyepiece on two successive nights and this way show how Jupiter's moons are quickly revolving around the planet. It could be a fun activity for kids if done right.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Make sure the scope is stable. Handheld operation is impossible in the field of astronomy, the image would simply bounce around too much. The scope must be installed on some kind of mount or tripod. Even then some tripods are a little shaky, so be careful with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your scope doesn't have a mount or a tripod, you can improvise something from as simple materials as a couple cardboard boxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://aquillam.wordpress.com/2010/01/04/a-cheap-tripod-alternative/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://aquillam.wordpress.com/2010/01/04/a-cheap-tripod-alternative/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything is better than handheld operation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If your scope has instructions for how to perform collimation, do it. If it's a refractor, it's unlikely it will require collimation, so then please disregard this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure the scope is focused well. Point it at a medium-bright star and tweak the eyepiece position until the star looks as small as possible. Or point it at the Moon and make the image look as sharp as possible. People with less than perfect vision may require a somewhat different ideal focus (unless they are wearing their glasses, in which case they are the same as 20/20 vision folks).&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="17437" LastEditDate="2018-01-03T19:12:53.837" LastActivityDate="2018-01-03T19:12:53.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24311" PostTypeId="1" CreationDate="2018-01-02T22:57:43.497" Score="3" ViewCount="71" Body="&lt;p&gt;I'm attempting to create a set of [reasonably] astronomically accurate planispheres from the perspective of observers positioned at the north or south poles of various nearby celestial objects. So basically looking out with a 180˚ FOV of each celestial hemisphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Stellarium and some information I found on this site, I've been able to piece together some decent enough reference images that I should be able to use for this project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: Earth's northern celestial hemisphere&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o2XgI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o2XgI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been able to make them for objects as far out as Pluto. But... I want to go farther. Let's say Alpha Centauri A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know Stellarium doesn't allow you to mount yourself to an object outside of the greater Solar System. This is unfortunate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I discovered Celestia, which I thought would be the answer. Using Celestia one can fly out to distant stars and wow, what a view!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Y7IyZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y7IyZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that Celestia can't bump the FOV past ~124˚. Manually forcing it to 180˚ causes the program to glitch out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here I am. I just can't seem to figure out a way to do this. I've been stuck on this for some time now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of any software or other tools that may be useful for this project? Celestia has shown me that it's possible. I feel like I'm really close.&lt;/p&gt;&#xA;" OwnerUserId="20701" LastActivityDate="2018-01-02T22:57:43.497" Title="Creating planispheres from the perspective of nearby objects" Tags="&lt;observational-astronomy&gt;&lt;night-sky&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24312" PostTypeId="2" ParentId="58" CreationDate="2018-01-02T23:59:27.890" Score="6" Body="&lt;h1&gt;At least 7.&lt;/h1&gt;&#xA;&lt;p&gt;We currently know of two star systems with 7 stars: &lt;a href=&quot;https://en.wikipedia.org/wiki/Nu_Scorpii&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nu Scorpii&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/AR_Cassiopeiae&quot; rel=&quot;nofollow noreferrer&quot;&gt;AR Cassiopeiae&lt;/a&gt;. The two have different structures, both of which are complicated but appear to be stable on stellar timescales.&lt;/p&gt;&#xA;&lt;h3&gt;&lt;a href=&quot;http://www.ctio.noao.edu/%7Eatokovin/stars/stars.php?cat=HIP&amp;amp;number=79374&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nu Scorpii&lt;/a&gt;&lt;/h3&gt;&#xA;&lt;p&gt;This system has two components, Nu Scorpii AB and Nu Scorpii CD.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Nu Scorpii A is a triple star system itself, with a binary system in the center. The inner two stars (Aa and Ab) cannot be resolved, in part because the fainter of the two is quite dim. The third star, Nu Scorpii Ac, is also faint. Nu Scorpii B is a single star orbiting Nu Scorpii A.&lt;/li&gt;&#xA;&lt;li&gt;Nu Scorpii C is also an individual star and is orbited by Nu Scorpii D, which is likely, like Nu Scorpii Aa and Ab, a binary system that cannot be resolved.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Nu Scorpii AB and Nu Scorpii CD have a long orbital period (compared to the days-long period of Nu Scorpii Aa and Ab), likely in the hundreds of thousands of years.&lt;/p&gt;&#xA;&lt;h3&gt;&lt;a href=&quot;http://www.ctio.noao.edu/%7Eatokovin/stars/stars.php?cat=HIP&amp;amp;number=115990&quot; rel=&quot;nofollow noreferrer&quot;&gt;AR Cassiopeiae&lt;/a&gt;&lt;/h3&gt;&#xA;&lt;p&gt;This system has a slightly different arrangement. It consists of a central triple-star system (AR Cassiopeiae AB) orbited by two binary systems (AR Cassiopeiae CD and FG).&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;AR Cassiopeiae A is an eclipsing binary with a period on the order of days, just like with Nu Scorpii A. AR Cassiopeiae B orbits these two at a large distance.&lt;/li&gt;&#xA;&lt;li&gt;AR Cassiopeiae C and D are similarly sized stars, although they are faint and hard to distinguish.&lt;/li&gt;&#xA;&lt;li&gt;AR Cassiopeiae F and G form a system similar to AR Cassiopeiae CD: two stars of similar masses, orbiting the primary triple-star system at the same distance (to within an order of magnitude).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;As with Nu Scorpii CD, AR Cassiopeia CD and FG orbit AR Cassiopeiae AB on a timescale of hundreds of thousands of years.&lt;/p&gt;&#xA;&lt;p&gt;The key thing to note about these high-multiplicity multiple-star systems is that their components are often binary systems themselves. In other words, you don't have a case where there's one or two central stars, with the rest independently orbiting them, or where you have a bunch of stars independently orbiting the same point. It's much stabler to create a few tight binaries and then set them in motion - much like a normal triple-star system, except each component is actually two stars.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;So, why haven't we found star systems with more than 7 stars in them? There are a few possible reasons:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We can't resolve certain components that we think are only single stars. We only know that Nu Scorpii Aab is a binary system by looking at its spectral lines and seeing how they shift over time as the components move in their orbits around each other. It's possible that we can't even detect these spectroscopic binaries in some systems.&lt;/li&gt;&#xA;&lt;li&gt;The initial protostellar cloud may not be massive enough for these systems to form easily. To form a system of 7 stars, unless one or more components were captured by the others (which seems unlikely, given that the arrangements are delicate), you need either one very, very massive cloud (if the &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1988ComAp..12..169B&amp;amp;defaultprint=YES&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;fission hypothesis&lt;/a&gt; somehow holds for the entirety of the system - which is probably not the case) or several massive clouds collapsing around the same time as each other, which also may not be plausible.&lt;/li&gt;&#xA;&lt;li&gt;It is likely that such systems are unstable to either complete collapse (i.e. ejecting members from the system) or losing members to passing stars. Given the high separations necessary for the components to orbit stably, it's not far-fetched that a passing star could perturb them enough to leave the system.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-01-03T15:40:31.283" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24329" PostTypeId="1" CreationDate="2018-01-03T16:18:13.040" Score="4" ViewCount="167" Body="&lt;p&gt;Is there any scientific explanation that causes polar lights (aurora borealis/australis) to have their unique shapes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Rayed Arcs&lt;/li&gt;&#xA;&lt;li&gt;Dual Bands&lt;/li&gt;&#xA;&lt;li&gt;Corona&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(Images on NASA Pdf)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This site (by NASA) proposes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The different shapes of auroras are a mystery that scientists are still trying to &#xA;  unravel. The shape seems to depend on where in the magnetosphere the electrons &#xA;  originate, what causes them to gain their energy, and why they dive into the &#xA;  atmosphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Subsection: Why do auroras come in different colors and shapes?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am specifically asking if there are any scientific observations (formulas or even simulations) that could describe such patterns (or even similar observations from fluid or gas dynamics). Furthermore, if it forms a rippled ring for instance, there should be a wave-like function, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Afterall, they could also be in a straight line..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://pwg.gsfc.nasa.gov/polar/EPO/auroral_poster/aurora_all.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aurora - NASA Pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20725" LastEditorUserId="6" LastEditDate="2019-01-29T14:43:15.610" LastActivityDate="2019-01-29T14:43:15.610" Title="What is the science behind the polar light shapes" Tags="&lt;light&gt;&lt;aurora&gt;&lt;corona&gt;" AnswerCount="0" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24331" PostTypeId="1" AcceptedAnswerId="24332" CreationDate="2018-01-03T18:16:37.800" Score="1" ViewCount="196" Body="&lt;p&gt;Everywhere I can hear, that if we press anything to the right size, we create black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we create neutron star by pressing, for example, our planet? And what if we press something bigger, like Jupiter?&lt;/p&gt;&#xA;" OwnerUserId="20727" LastActivityDate="2018-01-03T20:44:47.010" Title="Can we create neutron star by pressing earth? Or other objects?" Tags="&lt;black-hole&gt;&lt;neutron-star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24332" PostTypeId="2" ParentId="24331" CreationDate="2018-01-03T18:34:43.873" Score="7" Body="&lt;p&gt;It depends on what you mean by &quot;can&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to make the Earth into a black hole, you would have to compress it down to about the size of a marble.  That is impossible.  That is not just impossible with the technology of today, that is impossible with anything less than magic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we did make an Earth-mass black hole, it would be stable. (Except to Hawking radiation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you could somehow compress the Earth into a slightly smaller space, you would need to overcome electron-degeneracy and at some point before making your black hole you would form neutron matter. This is impossible. Nothing can do this, not now and not ever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An Earth-mass piece of neutron matter, however, would not be stable. Its own gravity is not sufficient to keep it from expanding, so it would very very rapidly explode. The same with Jupiter. A body needs to be larger than about 15% of the mass of the sun to have enough gravity as a neutron star to be stable. Jupiter has about 0.1% of the sun's mass&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can't create a small neutron star by compressing the Earth, unless we use magic, and if we stop squeezing it would explode.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-01-03T20:44:47.010" LastActivityDate="2018-01-03T20:44:47.010" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24333" PostTypeId="1" AcceptedAnswerId="24359" CreationDate="2018-01-03T21:59:16.940" Score="1" ViewCount="427" Body="&lt;p&gt;I'm a little bit confused about the setting circle of my EQ5 mount. Basically, I'm trying to find M42 nebula using Sirius star as reference. So, considere that I did the following steps:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A) Polar aligned for south pole&#xA;B) Pointed the telescope to Sirius star&#xA;C) Adjusted the RA wheel to 6h45m and DEC wheel to -16º42'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, I want to view the M42 nebula basing only in the RA and DEC wheels (without using the finderscope or whatever). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, I did:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;D) Changed the coordinates in the EQ5 mount to match RA 5h35m and DEC -5º23&#xA;E) Spent 5 minutes performing the step above (D)&#xA;F) Looked in the eyepiece&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So my questions are: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1)&lt;/strong&gt; After doing all these steps above, will the M42 nebula be 5 minutes away from the center of my eyepiece? So instead of using RA 5h35m I will need to use RA 5h40m?(??)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2)&lt;/strong&gt; If &lt;em&gt;yes&lt;/em&gt;, will I need to perform all the above steps every time that I want to see a new object? I mean, as fast as I can, to minimize the RA error...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3)&lt;/strong&gt; And if the answer is &lt;em&gt;yes&lt;/em&gt; again to Question (2), how useful is the setting circle under these conditions for you?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will appreciate any kind of help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[]s&lt;/p&gt;&#xA;" OwnerUserId="20730" LastActivityDate="2018-01-04T17:33:00.807" Title="Using setting circle for Right Ascension" Tags="&lt;newtonian-telescope&gt;&lt;declination&gt;&lt;right-ascension&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24335" PostTypeId="1" AcceptedAnswerId="24360" CreationDate="2018-01-04T03:19:34.123" Score="3" ViewCount="493" Body="&lt;p&gt;I have obtained the hindcast position of Mercury relative to the Sun for the epoch 2000-Jan-01 12:00:00 using two methods: PlanEph ver 4.2 (Dos-based) and JPL Horizons DE431 (online).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two methods give positions which are over 2000 Km apart. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Details are presented in the following figure:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NcsIJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NcsIJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I presume that such a large discrepancy is due to some error of specification or understanding on my part.  I wonder if anyone can suggest or help me find out what that error might be?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(EDIT)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;APPENDIX 1 JPL HORIZONS QUERY SETTINGS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Ephemeris Type:VECTORS    &#xA;Target Body: Mercury(199)    &#xA;Coordinate Origin: Sun (body center) [500@10]    &#xA;Time Span: Start=2000-01-01 11:59, Stop=2000-01-02 12:03,Step=1m    &#xA;Table Settings: output units=KM-S; CSVformat=YES    &#xA;Display/Output:download/save(plain text file).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;APPENDIX 2 Full listing of the JPL Horizons Output File&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(CSV format)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;*******************************************************************************&#xA; Revised: Jul 31, 2013              Mercury                             199 / 1&#xA;&#xA; GEOPHYSICAL DATA (updated 2008-Feb-07):&#xA;  Mean radius (km)      =  2440(+-1)     Density (g cm^-3)     =  5.427&#xA;  Mass (10^23 kg )      =     3.302      Flattening, f         =  &#xA;  Volume (x10^10 km^3)  =     6.085      Semi-major axis       = &#xA;  Sidereal rot. period  =    58.6462 d   Rot. Rate (x10^5 s)   =  0.124001&#xA;  Mean solar day        =   175.9421 d   Polar gravity ms^-2   =&#xA;  Mom. of Inertia       =     0.33       Equ. gravity  ms^-2   =  3.701     &#xA;  Core radius (km)      = ~1600          Potential Love # k2   =&#xA;&#xA;  GM (km^3 s^-2)        = 22032.09       Equatorial Radius, Re =    2440 km&#xA;  GM 1-sigma (km^3 s^-2)=   +-0.91       Mass ratio (sun/plnt) = 6023600&#xA;&#xA;  Atmos. pressure (bar) =                Max. angular diam.    = 11.0&quot;   &#xA;  Mean Temperature (K)  =                Visual mag. V(1,0)    = -0.42 &#xA;  Geometric albedo      =   0.106        Obliquity to orbit[1] = 2.11' +/- 0.1'  &#xA;  Sidereal orb. per.    =   0.2408467 y  Mean Orbit vel.  km/s = 47.362 &#xA;  Sidereal orb. per.    =  87.969257  d  Escape vel. km/s      =  4.435&#xA;  Hill's sphere rad. Rp =  94.4          Planetary Solar Const = 9936.9 (Wm^2)&#xA;&#xA;[1] Margot et al., Science 316, 2007&#xA;*******************************************************************************&#xA;&#xA;&#xA;&#xA;&#xA;*******************************************************************************&#xA;Ephemeris / WWW_USER Thu Jan  4 09:47:08 2018 Pasadena, USA      / Horizons    &#xA;*******************************************************************************&#xA;Target body name: Mercury (199)                   {source: DE431mx}&#xA;Center body name: Sun (10)                        {source: DE431mx}&#xA;Center-site name: BODY CENTER&#xA;*******************************************************************************&#xA;Start time      : A.D. 2000-Jan-01 11:59:00.0000 TDB&#xA;Stop  time      : A.D. 2000-Jan-01 12:03:00.0000 TDB&#xA;Step-size       : 1 minutes&#xA;*******************************************************************************&#xA;Center geodetic : 0.00000000,0.00000000,0.0000000 {E-lon(deg),Lat(deg),Alt(km)}&#xA;Center cylindric: 0.00000000,0.00000000,0.0000000 {E-lon(deg),Dxy(km),Dz(km)}&#xA;Center radii    : 696000.0 x 696000.0 x 696000.0 k{Equator, meridian, pole}    &#xA;Output units    : KM-S                                                         &#xA;Output type     : GEOMETRIC cartesian states&#xA;Output format   : 3 (position, velocity, LT, range, range-rate)&#xA;Reference frame : ICRF/J2000.0                                                 &#xA;Coordinate systm: Ecliptic and Mean Equinox of Reference Epoch                 &#xA;*******************************************************************************&#xA;&#xA;                JDTDB,            Calendar Date (TDB),                      X,                      Y,                      Z,                     VX,                     VY,                     VZ,                     LT,                     RG,                     RR,&#xA;    **************************************************************************************************************************************************************************************************************************************************************************&#xA;    $$SOE&#xA;    2451544.999305556, A.D. 2000-Jan-01 11:59:00.0000, -1.946394607859077E+07, -6.691260531388976E+07, -3.679595954330999E+06,  3.699453582609515E+01, -1.116598386942620E+01, -4.307714204076275E+00,  2.327713739463555E+02,  6.978310234741506E+07,  6.152788768051691E-01,&#xA;    2451545.000000000, A.D. 2000-Jan-01 12:00:00.0000, -1.946172639275932E+07, -6.691327522588462E+07, -3.679854414596554E+06,  3.699499188030234E+01, -1.116441595690670E+01, -4.307627980092747E+00,  2.327714970537363E+02,  6.978313925407939E+07,  6.149432668355465E-01,&#xA;    2451545.000694444, A.D. 2000-Jan-01 12:01:00.0000, -1.945950667956619E+07, -6.691394504380433E+07, -3.680112869688489E+06,  3.699544788177393E+01, -1.116284803117785E+01, -4.307541750189781E+00,  2.327716200939488E+02,  6.978317614060709E+07,  6.146076565335312E-01,&#xA;    2451545.001388889, A.D. 2000-Jan-01 12:02:00.0000, -1.945728693901455E+07, -6.691461476764810E+07, -3.680371319606457E+06,  3.699590383050987E+01, -1.116128009224054E+01, -4.307455514367438E+00,  2.327717430669925E+02,  6.978321300699817E+07,  6.142720458993068E-01,&#xA;    2451545.002083333, A.D. 2000-Jan-01 12:03:00.0000, -1.945506717110758E+07, -6.691528439741514E+07, -3.680629764350098E+06,  3.699635972651007E+01, -1.115971214009566E+01, -4.307369272625786E+00,  2.327718659728678E+02,  6.978324985325259E+07,  6.139364349330478E-01,&#xA;    $$EOE&#xA;    **************************************************************************************************************************************************************************************************************************************************************************&#xA;&#xA;Coordinate system description:&#xA;&#xA;  Ecliptic and Mean Equinox of Reference Epoch&#xA;&#xA;    Reference epoch: J2000.0&#xA;    XY-plane: plane of the Earth's orbit at the reference epoch&#xA;              Note: obliquity of 84381.448 arcseconds wrt ICRF equator (IAU76)&#xA;    X-axis  : out along ascending node of instantaneous plane of the Earth's&#xA;              orbit and the Earth's mean equator at the reference epoch&#xA;    Z-axis  : perpendicular to the xy-plane in the directional (+ or -) sense&#xA;              of Earth's north pole at the reference epoch.&#xA;&#xA;  Symbol meaning:&#xA;&#xA;    JDTDB    Julian Day Number, Barycentric Dynamical Time&#xA;      X      X-component of position vector (km)                               &#xA;      Y      Y-component of position vector (km)                               &#xA;      Z      Z-component of position vector (km)                               &#xA;      VX     X-component of velocity vector (km/sec)                           &#xA;      VY     Y-component of velocity vector (km/sec)                           &#xA;      VZ     Z-component of velocity vector (km/sec)                           &#xA;      LT     One-way down-leg Newtonian light-time (sec)                       &#xA;      RG     Range; distance from coordinate center (km)                       &#xA;      RR     Range-rate; radial velocity wrt coord. center (km/sec)            &#xA;&#xA;    Geometric states/elements have no aberrations applied.&#xA;&#xA;     Computations by ...&#xA;         Solar System Dynamics Group, Horizons On-Line Ephemeris System&#xA;         4800 Oak Grove Drive, Jet Propulsion Laboratory&#xA;         Pasadena, CA  91109   USA&#xA;         Information: http://ssd.jpl.nasa.gov/&#xA;         Connect    : telnet://ssd.jpl.nasa.gov:6775  (via browser)&#xA;                      http://ssd.jpl.nasa.gov/?horizons&#xA;                      telnet ssd.jpl.nasa.gov 6775    (via command-line)&#xA;         Author     : Jon.D.Giorgini@jpl.nasa.gov&#xA;    *******************************************************************************&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="233" LastEditorUserId="233" LastEditDate="2018-01-04T17:58:09.410" LastActivityDate="2018-01-04T18:21:00.150" Title="Difference in Ephemerides Position:- PlanEph 4.2 (VSOP87) vs JPL Horizons (DE431)" Tags="&lt;ephemeris&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24339" PostTypeId="1" AcceptedAnswerId="24356" CreationDate="2018-01-04T06:43:31.117" Score="6" ViewCount="942" Body="&lt;p&gt;Are there some stars between those two galaxies? In photos is only black. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if so, how many we could find there?&lt;/p&gt;&#xA;" OwnerUserId="20727" LastActivityDate="2018-01-04T15:27:14.507" Title="How many stars are between Milky way and Andromeda?" Tags="&lt;star&gt;&lt;galaxy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24351" PostTypeId="1" CreationDate="2018-01-04T12:06:38.233" Score="0" ViewCount="39" Body="&lt;p&gt;Are there any star system(planet system) where a star revolves around a star? Is it theoretically possible?&lt;/p&gt;&#xA;" OwnerUserId="20753" LastEditorUserId="11264" LastEditDate="2018-01-04T13:36:01.247" LastActivityDate="2018-01-04T13:36:01.247" Title="Star revolving a star" Tags="&lt;exoplanet&gt;&lt;star-systems&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2018-01-04T13:35:34.737" ContentLicense="CC BY-SA 3.0" />
  <row Id="24352" PostTypeId="2" ParentId="24351" CreationDate="2018-01-04T13:07:42.273" Score="1" Body="&lt;p&gt;What you are asking is about the existence of &lt;a href=&quot;https://simple.wikipedia.org/wiki/Binary_star&quot; rel=&quot;nofollow noreferrer&quot;&gt;binary star&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many binary stars in the universe. They are in fact more common  than single stars. Here are &lt;a href=&quot;https://www.quora.com/Why-are-binary-stars-more-common-than-single-stars&quot; rel=&quot;nofollow noreferrer&quot;&gt;interesting explanations&lt;/a&gt; for the latest.&lt;/p&gt;&#xA;" OwnerUserId="11264" LastActivityDate="2018-01-04T13:07:42.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24353" PostTypeId="1" CreationDate="2018-01-04T13:42:41.060" Score="5" ViewCount="101" Body="&lt;p&gt;Items reaching a Black Hole's Event Horizon seem to be frozen in time from the point of view of an outside observer. On the other hand, Black Holes collide and merge, in &lt;strong&gt;finite&lt;/strong&gt; steps, producing gravitational waves observable to outside observers. How is it possible these two facts not to contradict each other?&lt;/p&gt;&#xA;" OwnerUserId="13748" LastActivityDate="2018-01-04T21:26:44.407" Title="Items at the Event Horizon vs. colliding Black Holes" Tags="&lt;black-hole&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24356" PostTypeId="2" ParentId="24339" CreationDate="2018-01-04T15:27:14.507" Score="9" Body="&lt;p&gt;The stars between galaxies are called &lt;a href=&quot;https://en.wikipedia.org/wiki/Intergalactic_star&quot; rel=&quot;noreferrer&quot;&gt;&quot;&lt;em&gt;intergalactic stars&lt;/em&gt;&quot;&lt;/a&gt;, and one study (2012) claims to have identified about 675 of them between the &lt;em&gt;Milky Way&lt;/em&gt; and &lt;em&gt;Andromeda&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now, Vanderbilt astronomers report in the May issue of the &lt;a href=&quot;http://iopscience.iop.org/0004-637X&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Astronomical Journal&lt;/em&gt;&lt;/a&gt; that they have identified a group of more than 675 stars on the outskirts of the Milky Way that they argue are hypervelocity stars that have been ejected from the galactic core. They selected these stars based on their location in intergalactic space between the Milky Way and the nearby Andromeda galaxy and by their peculiar red coloration.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;-&lt;a href=&quot;https://news.vanderbilt.edu/2012/04/30/rogue-stars-intergalactic-space/&quot; rel=&quot;noreferrer&quot;&gt;&quot;Rogue stars ejected from the galaxy found in intergalactic space&quot;&lt;/a&gt;, Research News @Vanderbilt (2012-04-30)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15986" LastActivityDate="2018-01-04T15:27:14.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24357" PostTypeId="2" ParentId="13943" CreationDate="2018-01-04T15:53:45.860" Score="1" Body="&lt;p&gt;I think in the spirit of the question, the answer is yes. In an atmosphere like hydrogen or methane that can react with oxygen, oxygen is a resource that allows combustion to occur and therefore represents an energy reserve. &#xA;For example, here's a demonstration of an &lt;a href=&quot;https://www.youtube.com/watch?v=Ir1BddDKj8Q&quot; rel=&quot;nofollow noreferrer&quot;&gt;oxygen flame in a hydrogen atmosphere.&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="20756" LastActivityDate="2018-01-04T15:53:45.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24358" PostTypeId="2" ParentId="14129" CreationDate="2018-01-04T16:29:36.707" Score="1" Body="&lt;p&gt;There are two theories of how super massive black holes formed.&#xA;One is that they started as several stellar massive black holes that fell into the center of galaxies (which were forming around them) and came together creating a super massive black hole we see now.&#xA;Another theory is that they started as &quot;Dark stars&quot;, which are theoretical stars held together by dark matter, which can be much larger than a normal star. When the dark star collapses upon its self, it became a super massive black hole.&lt;/p&gt;&#xA;" OwnerUserId="20518" LastActivityDate="2018-01-04T16:29:36.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24359" PostTypeId="2" ParentId="24333" CreationDate="2018-01-04T17:33:00.807" Score="2" Body="&lt;p&gt;I think I mis-read your question before. Let me rephrase it to make sure I understand. In step D (moving the scope from Sirius to M42), that process took 5 minutes to complete. Or maybe it took 30 seconds to complete, but you then took a 4.5 minute coffee break :-). So during that 5 minutes of time since you calibrated the setting circle, M42 has drifted another 5 minutes to the west.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Yes, the object will be 5 minutes west of the eyepiece. Since your setting circle does not update during that time, you need to position the scope 5m to the west, or to 5h 30m right ascension (not 5h 40m R.A.).&lt;/li&gt;&#xA;&lt;li&gt;Yes. Before moving to the next object, you should reset the right ascension setting circle to the coordinate of the current object. Then move to the next object. Then compensate for the time it took to move the scope. Of course, switch to a low power eyepiece just before starting the move to increase the chance that the next object will be in the field of view.&lt;/li&gt;&#xA;&lt;li&gt;Mechanical setting circles are useful for small jumps. I have used them on my scope to locate Venus near the Sun. I have also used the circles on the scope at my club's observatory to locate planets in the day. (That is slightly different that your setup since the club's scope is permanently mounted, professionally made, and has really large circles -- all of which help to reduce any errors.) If I remember correctly, the &quot;rule of thumb&quot; is that mechanical circles should be about the same diameter as the optics in order to be &quot;accurate&quot;. A 200 mm reflector (the diameter of the mirror = 200 mm) should have circles that are about 200 mm in diameter. But anything should definitely get you close to the target.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2018-01-04T17:33:00.807" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24360" PostTypeId="2" ParentId="24335" CreationDate="2018-01-04T17:34:11.697" Score="2" Body="&lt;p&gt;The key difference between the two (VSOP87 from PlanEph versus DE431 from the JPL HORIZONS web site) is the time reference. PlanEph uses UT (Universal Time) while JPL uses TDB (Barycentric Dynamic Time; the acronym is from the French equivalent). On 1 Jan 2000, TDB was ahead of UT by 64.183927 seconds. This means that 12:01:04.183927 TDB on 1 Jan 2000 is the same as 12:00 UT on 1 Jan 2000. The values from JPL HORIZONS at this time are an X,Y,Z position of $X = -19459352\ \text{km}$, $Y=-66913992\ \text{km}$, $Z = -3680131\ \text{km}$ - a mere 30.5 kilometer root sum square difference compared to the VSOP87 vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;For completeness, here is my request to and response from the HORIZONS web site:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The request:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Current Settings&#xA;&#xA;Ephemeris Type : VECTORS&#xA;Target Body : Mercury [199]&#xA;Coordinate Origin : Sun (body center) [500@10]&#xA;Time Span : Start=2000-01-01 12:01:04.183927, Stop=2000-01-01 12:01:05.183927, Step=1 m&#xA;Table Settings : output units=KM-S; delta-T (TDB-UT)=YES&#xA;Display/Output : default (formatted HTML)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The response, sans verbiage:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;$$SOE&#xA;2451545.000742869 = A.D. 2000-Jan-01 12:01:04.1839 TDB [del_T=     64.183927 s]&#xA; X =-1.945935189324640E+07 Y =-6.691399174811687E+07 Z =-3.680130892116148E+06&#xA; VX= 3.699547967775486E+01 VY=-1.116273869590535E+01 VZ=-4.307535736975350E+00&#xA; LT= 2.327716286712980E+02 RG= 6.978317871203169E+07 RR= 6.145842537044049E-01&#xA;$$EOE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2018-01-04T18:21:00.150" LastActivityDate="2018-01-04T18:21:00.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24361" PostTypeId="2" ParentId="24353" CreationDate="2018-01-04T21:26:44.407" Score="1" Body="&lt;p&gt;I don't know if this question is a good fit for the Stackexchange format, because it's a complex topic that would require a lot of discussion. But I'll try to provide an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you're getting the wrong ideas from the image of the event horizon being frozen in time for the remote observer. While this is true, it also means the remote observer never actually gets any information from the event horizon proper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're a remote observer, and you see any change that happens related to a black hole, by definition that change is not happening at the event horizon or inside it. It happens outside that region. The event horizon might as well not even exist from this p.o.v.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The merger of the two black holes is what the remote observer sees. But a hypothetical local observer, inside the event horizon, would have a different perception.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the local observer, the whole history of anything outside the black hole takes place in an instant. The whole evolution of the universe, including the black hole colliding with another black hole, and the resulting object evaporating via Hawking radiation - all this stuff happens in an instant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You really need to think of it this way: the remote and the local observer experience two completely separate realities, where the flow of time in one is completely disconnected from the flow of time in the other.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2018-01-04T21:26:44.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24362" PostTypeId="1" CreationDate="2018-01-04T21:40:40.120" Score="-1" ViewCount="1450" Body="&lt;p&gt;I was watching the latest episode of Isaac Arthur and he was talking about a stellar laser or stellaser which was coined by a scientist named Steve Nixon. I can find no information about stellasers on the internet. What are they and what can they do?&lt;/p&gt;&#xA;" OwnerUserId="20760" LastActivityDate="2018-01-04T23:23:49.380" Title="What is a stellar laser or stellaser?" Tags="&lt;astrophysics&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2018-01-05T03:13:17.850" ContentLicense="CC BY-SA 3.0" />
  <row Id="24363" PostTypeId="1" CreationDate="2018-01-04T23:05:56.477" Score="6" ViewCount="1339" Body="&lt;p&gt;Recently got a Celestron 130 SLT (first &quot;real&quot; scope) and have a question. I have the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;130mm F/5 newtonian, 650mm focal length scope&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9 and 25mm factory EP's (more on that later)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Svbony 2x barlow&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very, very dark skies (100 miles from any cities, out in the desert, light pollution maps show &quot;zero light pollution&quot; on the maps anywhere I view from once I'm 5 minutes from home). I don't have expensive glass, I don't know what I'm doing... but at least I have this going for me :-) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4500 elevation with low mountains on either side, 10-25 miles away  (don't know that it matters) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;20/20 uncorrected vision (if that makes a difference)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curious what I should expect to be able to see with my scope, and at what magnification I should see them at? I plan to buy 3 EP's, and want to know the lowest and highest I should buy to see what I &quot;should&quot; be seeing. I don't want to pursue things that I won't be able to make out clearly anyways, or buy the wrong EP's for seeing those things. I see anywhere from 25 to 60 X your scope size, which is a wide range of mags. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have beautifully-dark skies, so I'm not sure if the EP has to do with light pollution also, or solely &quot;seeing&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will I be able to make out cassini's division? Will I be able to complete the messier list? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have nothing in-person to compare my scope to, so I'm relying on random internet information, which tells me my max magnification is anywhere from 150 to 250, and varying features and objects that I will be able to see. Any input will be greatly appreciated. I plan to start the lunar observing program soon (LOVE the moon, mostly because of my spaceflight interest and occupation), and have only looked at the moon so far, along with learning my constellations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(PS, whatever EP's I buy, I am willing to spend up to 80 a piece, on 3 total new ones (or 2, the 25mm seems ok)&lt;/p&gt;&#xA;" OwnerUserId="20761" LastEditorUserId="20761" LastEditDate="2018-01-04T23:45:41.060" LastActivityDate="2019-03-31T12:30:27.020" Title="What should I expect to see with my telescope with what EP's?" Tags="&lt;telescope&gt;&lt;space&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24364" PostTypeId="1" CreationDate="2018-01-04T23:23:22.270" Score="2" ViewCount="62" Body="&lt;p&gt;Say I have 2 records, in horizontal coordinates of the azimuth ($A_z,A_z^{'}$) and altitude ($\alpha,\alpha^{'}$) of the moon at the same time in two different places that are $\theta$ degrees apart (taking the angle from the centre of the earth and assuming the earth is a perfect sphere). Is it possible to calculate the parallax angle of the moon between these two points with only the values above? if so how and if not what other values would I need?&lt;/p&gt;&#xA;" OwnerUserId="20762" LastActivityDate="2018-01-04T23:23:22.270" Title="calculating paralax angle from observations" Tags="&lt;the-moon&gt;&lt;coordinate&gt;&lt;parallax&gt;" AnswerCount="0" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24365" PostTypeId="2" ParentId="24362" CreationDate="2018-01-04T23:23:49.380" Score="2" Body="&lt;p&gt;This &quot;laser star&quot; is a hypotethical object in an Asimov short story, which has a special star-like object with very uncommon properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is entirely Asimov's invention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the reality, no such thing was found until now. Not even any similar.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastActivityDate="2018-01-04T23:23:49.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24366" PostTypeId="1" AcceptedAnswerId="27026" CreationDate="2018-01-05T01:25:33.917" Score="6" ViewCount="110" Body="&lt;p&gt;I understand how in olden days the Sun-Planet distances were estimated using:-&lt;/p&gt;&#xA;&lt;p&gt;(i) measures of planet orbital periods (&lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;) from analysis of observations over several centuries;&lt;/p&gt;&#xA;&lt;p&gt;(ii) &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kepler's 3rd Law&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$(T^2 = k.a^3)$&lt;/span&gt; applied to the planet orbital periods to determine the relative lengths of the Sun-Planet semi-major axis distances (&lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt;) expressed in terms of the AU (astronomical unit = distance from centre of Earth to centre of Sun);&lt;/p&gt;&#xA;&lt;p&gt;(iii) particular Earth-Planet distances along Solar radials were measured using &lt;a href=&quot;https://en.wikipedia.org/wiki/Parallax&quot; rel=&quot;nofollow noreferrer&quot;&gt;Parallax&lt;/a&gt; techniques such as that by &lt;a href=&quot;https://en.wikipedia.org/wiki/Giovanni_Domenico_Cassini&quot; rel=&quot;nofollow noreferrer&quot;&gt;Casinni&lt;/a&gt; for Mars and &lt;a href=&quot;http://spiff.rit.edu/classes/phys445/gettys/venus_ex/venus_ex.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Transits of Venus&lt;/a&gt;;&lt;/p&gt;&#xA;&lt;p&gt;(iv) From one or more Earth-Planet radial distances the length of the AU could then be determined by simple algebra.&lt;/p&gt;&#xA;&lt;p&gt;(v) Other Sun-Planet distances could then be determined once the AU length was found.&lt;/p&gt;&#xA;&lt;p&gt;Modern ephemerides e.g. &lt;a href=&quot;https://en.wikipedia.org/wiki/Jet_Propulsion_Laboratory_Development_Ephemeris&quot; rel=&quot;nofollow noreferrer&quot;&gt;those issued by JPL&lt;/a&gt; use various techniques:-&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The orbits of the inner planets are known to subkilometer accuracy through fitting radio tracking measurements of spacecraft in orbit about them. Very long baseline interferometry measurements of spacecraft at Mars allow the orientation of the ephemeris to be tied to the International Celestial Reference Frame with an accuracy of 0′′.0002. This orientation is the limiting error source for the orbits of the terrestrial planets, and corresponds to orbit uncertainties of a few hundred meters.&lt;/p&gt;&#xA;&lt;p&gt;The orbits of Jupiter and Saturn are determined to accuracies of tens of kilometers as a result of fitting spacecraft tracking data.&lt;/p&gt;&#xA;&lt;p&gt;The orbits of Uranus, Neptune, and Pluto are determined primarily from astrometric observations, for which measurement uncertainties due to the Earth’s atmosphere, combined with star catalog uncertainties, limit position accuracies to several thousand kilometers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;from &lt;a href=&quot;https://www.google.co.uk/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=1&amp;amp;ved=0ahUKEwia5I_X0r_YAhWrJ8AKHbQrBXgQFggnMAA&amp;amp;url=https%3A%2F%2Fipnpr.jpl.nasa.gov%2Fprogress_report%2F42-196%2F196C.pdf&amp;amp;usg=AOvVaw2NWBqCO071OnZXuy-SkY_V&quot; rel=&quot;nofollow noreferrer&quot;&gt;Folkner et al 2014&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;I understand how spaceraft telemetry has been used to obtain highly accurate distances between Earth and the terrestrial planets.  But I am not clear about how highly-accurate distances can be determined between those planets and the centre of the Sun.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Do such (Sun-planet-distance) determinations still basically rely on (the more accurate Newtonian version of) Kepler's 3rd Law relating planet orbital period and Sun-planet semi-major axis distance?&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{T^2}{a^3} = \frac{4 \pi^2}{G(M+m)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;What other methods or assumptions are used/involved?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;EDIT - Afterthoughts&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;After further thought prompted by feedback from  u/atmosphericprisonescape&#xA;my question boils down to  &amp;quot;How are the positions (over time) of the Sun's Centre tied-in (with sub-kilometre accuracy) to the 4D space-time array of triangulated, high-accuracy, telemetry-derived, Terrestrial planet position determinations.&lt;/p&gt;&#xA;&lt;p&gt;I guess that Kepler's 3rd Law is not invoked as such. Rather, high-accuracy (sub-kilometre) determinations of the position of the Solar centre presumably require (in addition to the high-accuracy planet positions) some specific deterministic &amp;quot;motivation model&amp;quot;; i.e. a model of motion-determining factors.&lt;/p&gt;&#xA;&lt;p&gt;These factors would include Newtonian Inertia, Newtonian Gravitational forces and Non-Newtonian factors leading to additional orbital angular velocity (cf Non-Newtonian Perihelion Precession and General Relativity as described at &lt;a href=&quot;https://en.wikipedia.org/wiki/Apsidal_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt; wikipedia/Apsidal_precession&lt;/a&gt; ).&lt;/p&gt;&#xA;&lt;p&gt;The &amp;quot;motivation model&amp;quot; would constrain the relative positions of the Sun and planets via their involment as motion generators, reactors and enactors.&lt;/p&gt;&#xA;" OwnerUserId="233" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-07-20T21:08:09.147" Title="Determination of High-Accuracy Distances of Terrestrial Planets from the Sun" Tags="&lt;astrometry&gt;&lt;ephemeris&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24367" PostTypeId="2" ParentId="24363" CreationDate="2018-01-05T02:41:38.990" Score="6" Body="&lt;p&gt;A simple rule for the maximum magnification is: 2x the aperture in mm. In your case, that makes it 260x. But that's more like a guideline, since it's affected by so many factors. Speaking in general, trying to exceed it almost never provides any benefits, with some rare exceptions (high quality instrument, perfectly collimated, under a very stable atmosphere, observing close double stars - is one possible exception that comes to mind).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So eyepieces shorter than 2.5 mm (after dividing it by the barlow magnification, if any) would likely not be beneficial for you. Combined with your 2x barlow, that means eyepieces no shorter than 5 mm (+/- 1 mm, depending on some factors). In practice, such short eyepieces are pretty rare anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a longest usable focal length for eyepieces, which is the shortest usable focal length times 12, more or less, or about 30 mm in your case (plus or minus some change). Any longer than that and the exit pupil (the diameter of the ray of light exiting the scope) would be greater than the pupil in your eye, leading to loss of light and diminished performance. That's the theory. In practice, there's just not much benefit going much longer than that, as you will find out if you try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So stick to the 30 ... 2.5 mm interval, give or take 10 ... 20%. These limits are not set in stone, but you will rediscover them on your own, given enough time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A minimum of 3 eyepieces, and a maximum of 6 or so, should cover most use cases. But a lot of people just end up accumulating eyepieces over the years, or keep trading them.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Look into eyepieces with a wider apparent field of view. A typical Plossl gives you about 50 degrees AFoV. But the view is more interesting, and the scope is easier to use, in a wider AFoV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example and a place to start, I've used the Explore Scientific 82 degree series, which provide 82 degrees of AFoV. Large objects like the Moon or some nebulae or clusters look spectacular in such a wide eyepiece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not mandatory that you strive for a bigger AFoV, but it can be worth it. Also, lenses such as the ES82 series could be used even with high end telescopes, if you ever decide to upgrade.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;130 mm of aperture is definitely enough for a lot of fun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter will show off its equatorial belts. On a good night you will begin seeing swirls in between the belts - I did that with 150 mm of aperture when seeing was good, and you're within a rounding error from it. You will also see the 4 big moons of Jupiter, but only as little dots. Use medium-high magnification, don't push it too high, because Jupiter is low contrast and it washes out easily at high mag - just try and see what works best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give it time and allow your eye to adapt. I believe your scope can track, so enable tracking, relax, breathe calmly, and just take the image in over time. In a couple minutes, your brain/eye system will begin to adapt, and you'll start to see more details. Allowing some time to pass is especially important with low contrast objects like Jupiter, where details are so close to the eye's performance limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn will show you the Cassini division in its rings. You will also begin to see some equatorial belts on the planet, but just barely. High magnification is useful, but not necessarily the highest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury and Venus will show phases like the Moon, but no details beyond that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars can be observed for a few weeks every 2 years, at oppositions; there's a big opposition this year (2018) in the summer. It requires the best seeing (the most stable atmosphere), excellent collimation, and the shortest eyepiece (highest magnification) your telescope will take. Whichever polar cap is pointing at you should be visible, as a tiny white dot sitting on the edge. I have seen the Hellas Basin as a large, bright, white spot when it was full of frost or fog (not sure which), but it melts away quickly. Syrtis Major is probably easiest to see, as a dark, almost black line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nakedeyeplanets.com/mars-oppositions.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.nakedeyeplanets.com/mars-oppositions.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When seeing (air turbulence) is bad, Mars just doesn't work; it looks like an orange pill with no details. Either take a road trip to whichever part of the country has better seeing at that time, or use lots of patience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uranus and Nepture can definitely be seen in 130 mm. At high magnification, you will see them as tiny disks, clearly (but just barely) bigger than any star nearby. Uranus is a pretty blue-green shade, Neptune is a deep blue, almost indigo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon will provide endless entertainment, because there's so much to see. Zoom into a feature using high magnification, or zoom out and observe the whole thing at low magnification - it's fun either way.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the planets and the Moon, what's important is the so-called seeing, which is the inverse of turbulence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_seeing&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Astronomical_seeing&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When seeing is good (turbulence is low) then you can push the magnification all the way up to the limit. When seeing is bad, you have to reduce magnification, or else the image is soft and foggy. Mars only works well with the best seeing; it is possible that seeing might be consistently bad for the whole duration of an opposition - or it could be consistently good. It's weather basically, so it does whatever it wants to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the Clear Sky Charts site to get a seeing forecast:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cleardarksky.com/csk/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.cleardarksky.com/csk/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On that site, select a location closest to your home. Use the Seeing line in the chart for an estimate. Keep in mind this is weather forecast, so it's not always 100% accurate, but usually it works well enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When seeing is good, observe planets. When seeing is bad, observe DSOs (deep space objects, like galaxies, nebulae, star clusters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light pollution does not matter at all for the planets or for the Moon. You could observe these objects from the biggest city and they look about the same.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another important thing for planets is collimation. Read your scope's manual and practice what they preach. A scope out of collimation may severely underperform. It could mean the difference between seeing and not seeing the Cassini division. See here for some examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astrophoto.fr/collim.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.astrophoto.fr/collim.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a bewildering variety of tools and techniques for collimation. Start with the simple stuff (the manual) since it's better than nothing. Then keep reading. Here's a good page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.skyandtelescope.com/astronomy-resources/how-to-align-your-newtonian-reflector-telescope/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.skyandtelescope.com/astronomy-resources/how-to-align-your-newtonian-reflector-telescope/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or see here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://garyseronik.com/?s=collimation&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://garyseronik.com/?s=collimation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A laser collimator should in theory be one of the easiest and most precise tools, but beware of lasers that are not centered. Plug the laser in the focuser, and turn it (spin it) in place. Watch the laser spot on the primary mirror. If the spot moves in a circle as you keep turning the laser, it's not centered. Some lasers have adjustment screws for centering, some don't. High quality lasers like the Howie Glatter series are very precisely centered from the workshop and don't require adjustments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.collimator.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.collimator.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.catseyecollimation.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.catseyecollimation.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do own more complex collimators like the CATSEYE system, but in 99% of cases I just use a Howie Glatter laser with a 2&quot; TuBlug to do a quick check-up before each session. Any laser is fine as long as it's precision-centered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Collimation is something you'll keep learning for a long time, so don't worry, do what you can at first, and slowly refine your technique over time. But keep your scope collimated. Once you're comfortable with the technique, it only takes 5 minutes.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Air currents in your instrument can also affect the planets and the Moon. These are due to your scope being at a different temperature from the air (because it was stored indoors). The effect is like observing under bad seeing conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-1/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-1/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-2/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-2/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cure is easy: take the scope outside 1 hour before you begin your session, and let it &quot;breathe&quot;. A scope at thermal equilibrium with the air has no internal convection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your scope is probably not big enough to require a fan on the primary mirror, for forced convection, but you can definitely experiment with it if you like. At your aperture size this is low priority. Fix everything else first (collimation).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;DSOs (deep space objects) should be visible in your aperture, especially given the dark sky you're enjoying. More aperture is better, but the level of light pollution is also important. You are very lucky to have such a great location for DSOs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretty much the whole Messier catalog, over 100 objects, should look great in your scope under a dark sky. Among the easiest objects, my favorites are: M31 (the Andromeda galaxy), M13 (the Hercules Globular Cluster), M27 (the Dumbbell Nebula), M57 (the Ring Nebula), M42 (the Orion Nebula).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since you live under an ideal dark sky, you could just plug a low magnification eyepiece in it and slowly browse the sky. When you see something interesting, stop and switch to somewhat higher magnification. Chances are you'll stumble upon many DSOs this way. I do this with binoculars (low magnification) when I go to Death Valley, and it's amazing what you can discover randomly this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different DSOs require different magnifications, from low to high. You'll just have to experiment and see what works best in each case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing, collimation, thermal equilibrium don't mean a whole lot for DSOs, because they are fuzzy and low detail to begin with. But a very dark sky helps a lot.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You could also observe the Sun. Get a white light filter, the kind that fits on the top of your telescope, covering the whole aperture. Filters made from the Baader Solar Film are best, they are very good performers and tend to be cheap. Either buy the Baader foil and make the filter yourself, or purchase a complete filter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.astrozap.com/scripts/prodList.asp?idCategory=31&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.astrozap.com/scripts/prodList.asp?idCategory=31&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astrosolar.com/en/information/how-to/how-to-make-an-inexpensive-filter-cell/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://astrosolar.com/en/information/how-to/how-to-make-an-inexpensive-filter-cell/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just make sure the size of the filter fits the diameter of the OTA (the main tube) of your telescope - NOT the aperture size, but the external diameter of the OTA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will allow you to see sunspots. You can also observe the transit of Mercury over the solar disk, which happens once every several years. I think the next one is in 2019. You can also observe solar eclipses this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need to use the finderscope when observing the Sun, then make a small filter for it, too. I made some cute little solar filters for my finderscopes from Baader film and white plastic tubes used for sprinkler systems. Cheap, but effective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astrosolar.com/en/information/how-to/how-to-make-your-own-objective-solar-filter-for-your-camera-or-telescope/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://astrosolar.com/en/information/how-to/how-to-make-your-own-objective-solar-filter-for-your-camera-or-telescope/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, never point the scope at the Sun without the full aperture filter. And make sure the filter doesn't fall off. Or else all sorts of mayhem may ensue, up to and including someone losing an eye, or the house burns down, etc. Not kidding. Unfiltered, the solar energy coming out of the scope is like a plasma torch.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Buy a book called Turn Left At Orion. It will get you started with how to find and observe DSOs, complete with instructions on which time of the year is best for each object, how much magnification (this is a bit subjective), etc. It's a very good book, I highly recommend it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/0521153972&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amazon.com/Turn-Left-Orion-Hundreds-Telescope/dp/0521153972/?tag=amz-track-20&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would go with either a sturdy hardcover edition, or the spiral bound one - something that works well out in the field, not some snowflake paperback.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A planisphere is a good thing to have. You only use it 5 or 10 minutes each night, but it will quickly get you oriented in the sky when you do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/1928771017&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amazon.com/Guide-Stars-Ken-Graun/dp/1928771017/?tag=amz-track-20&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When reading the book or the planisphere, use a red light at the minimum intensity that allows you to read. This will preserve your night vision. There are many red flashlights out there made for astronomers, that allow you to dial down the intensity. This is only needed for DSOs. Night vision does not matter for the planets or the Moon.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's probably beneficial to get a small sky atlas too. Let's say you read in the news that a supernova has been discovered in a nearby galaxy, and it can be seen by amateurs. You could point the scope in its general direction, but you'll need a detailed local map of the region. The one point of light that is seen in the sky but is not seen on the map is the supernova. It also helps when observing other transients such as comets and asteroids (again, they stand out because they are not on the map). Finally, with some smaller DSOs it makes it easier to identify the object from the stellar background (and they should be marked on the map because they are permanent objects).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/1931559317&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amazon.com/Sky-Telescopes-Pocket-Atlas/dp/1931559317/?tag=amz-track-20&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also use various digital maps on a portable device, but I recommend that you master the paper maps first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you own a handful of eyepieces, a barlow, etc, you should keep them together in a carrying box. Many astronomers use a pluck foam case, and they just shape the foam to hold each and every component. I've never used the dividers that come with the case, since optical components are small, lightweight, rather fragile, and should be surrounded only by foam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.harborfreight.com/18-in-x-6-in-x-13-in-black-aluminum-case-62271.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.harborfreight.com/18-in-x-6-in-x-13-in-black-aluminum-case-62271.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another item I've found useful is a tiny folding table. It doesn't have to be huge, just big enough to hold a book, the sky atlas, etc. It's better than making a pile of stuff around you on the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cheap, collapsible barstool or something like it would help with long observing sessions. Definitely beats hunching over the scope for hours. Just make sure it's the right height. Big dobsonian owners tend to make or buy so-called observing chairs, which are adjustable, but those are probably overkill for you since your eyepiece doesn't move around that much. Your scope has a geometry similar to my SCT, which doesn't really require a true observing chair, just anything to sit on that's the correct height. Now if you had a big dob then that would be different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pretty common improvisation is an empty box or frame that's 3 different sizes in all directions (length, width, height). That allows some height adjustment, and it's pretty cheap to procure or make. Between sessions you could use it to store the telescope or something.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2018-01-06T01:58:21.773" LastActivityDate="2018-01-06T01:58:21.773" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24380" PostTypeId="1" AcceptedAnswerId="24381" CreationDate="2018-01-05T17:47:11.230" Score="1" ViewCount="701" Body="&lt;p&gt;I bought the Powerseeker 50AZ, which has D=50mm; F=600mm F/12 which comes with 3 eyepieces: 20mm, 12.5mm and 4mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can tell it's not possible to colimate this telescope. As suggested &lt;a href=&quot;https://astronomy.stackexchange.com/questions/8788/how-much-magnification-is-needed-to-see-planets-of-solar-system&quot;&gt;here&lt;/a&gt; this is a very important part to see sharp images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to observe jupiter and mars with it and jupiter I could see a dim small blurry image and mars a very tiny dot with absolutely no way of seeing details on any of those in what seemed the best focus I could get from them.&#xA;The only thing I can see details is the moon really. Also I have observed in what seemed very clear sky conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is:&#xA;How much can I expect to see from planets with this telescope? &#xA;With this current setup should I be seeing jupiter and mars in more details? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE:&#xA;Equipment link&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B0000UMLYI&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amazon.com/Celestron-PowerSeeker-50-Refractor-Telescope/dp/B0000UMLYI/ref=sr_1_1?ie=UTF8&amp;amp;qid=1515177254&amp;amp;sr=8-1&amp;amp;keywords=powerseeker%2B50az%2Btelescope&amp;amp;th=1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20507" LastEditorUserId="20507" LastEditDate="2018-01-05T18:36:02.643" LastActivityDate="2018-01-05T20:40:56.017" Title="Planetary observations: what to expect from the Powerseeker 50AZ?" Tags="&lt;telescope&gt;&lt;jupiter&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24381" PostTypeId="2" ParentId="24380" CreationDate="2018-01-05T18:03:40.153" Score="3" Body="&lt;p&gt;Currently, Mars is very far from the Earth, so it appears to be very small. I doubt that any amateur telescope will show any details on Mars in this condition. Perhaps when it comes to opposition and is close to Earth, you might be able to see some faint markings with your scope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter should show a sharp disk and nearly pinpoint moons with the 12.5 mm eyepiece (48 power). You should see two prominent bands on Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 4 mm eyepiece is probably poor quality, and since it provides 150 power, is too much power for a 50 mm scope. In ideal conditions, the general rule is 100 power for a 50 mm scope (2 times the scope size in mm). The clarity of the sky is not what makes a sharp image. There is a condition named &quot;seeing&quot; that indicates how sharp the viewing is. On some nights, the seeing is poor, so the images will be soft (not sharp) and perhaps shimmering, as if you are viewing the object through a stream of water. (Essentially, you are, except it is miles or kilometers of air!)&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2018-01-05T18:03:40.153" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24382" PostTypeId="2" ParentId="13943" CreationDate="2018-01-05T19:32:38.627" Score="3" Body="&lt;p&gt;Yes, it would work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between &quot;fuel&quot; and &quot;oxidizer&quot; is rather made for convenience, but it's not a fundamental distinction at the macroscopic level. The only differences are at the microscopic level, related to what the electrons do during the reaction. All such processes are redox reactions (reduction-oxidation), where one participant is the reducer, and the other is the oxidizer. That simply means the reducing agent loses electrons, while the oxidizing agent gains electrons; once the electron balance is final, the two components are bound together and the reaction is complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oxidizer doesn't even have to be oxygen, it can be many elements or compounds hungry for electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line is - you can burn either gas in an atmosphere made of the other. The gases don't care, as long as they react with each other. There are videos online where you can see chemists burning oxygen in an atmosphere of hydrogen, or burning hydrogen in an atmosphere of chlorine, etc. All these are redox reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to what's the minimum percentage of methane in the atmosphere that could sustain an oxygen flame, I am unsure. But it must be similar to the situation on Earth, where a low-enough oxygen contents in the air will eventually extinguish any flame. I would guess that 20% methane would work just fine for burning oxygen, but that's just a guess, so take it for what it's worth.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As others have pointed out, a methane atmosphere would indicate a very different chemistry at the planetary level, which may or may not be able to sustain life in general. It would definitely not sustain life as we know it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the Goldilocks zone for a methane atmosphere is different from the same zone for an oxygen atmosphere - it's further away from the star, because methane retains infrared better than oxygen (methane is a &quot;greenhouse gas&quot;).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=Ir1BddDKj8Q&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fire and Flame 19 - Oxygen Flame in Hydrogen&lt;/a&gt; (Royal Society of Chemistry)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=4rZ1Cy7Yx4E&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fire and Flame 21 - Oxygen Balloon in Hydrogen&lt;/a&gt; (Royal Society of Chemistry)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=nqVdgAYP6CM&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fire and Flame 22 - The Planet Jupiter&lt;/a&gt; (Royal Society of Chemistry)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=vtWp45Eewtw&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fluorine - Periodic Table of Videos&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=d49TzVF1gmY&quot; rel=&quot;nofollow noreferrer&quot;&gt;Exploding Hydrogen Bubbles&lt;/a&gt; (Periodic Table of Videos)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2018-02-05T19:42:38.677" LastActivityDate="2018-02-05T19:42:38.677" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24383" PostTypeId="2" ParentId="24380" CreationDate="2018-01-05T20:40:56.017" Score="2" Body="&lt;p&gt;Also see the answer by @JohnHoltz - as he said, the 4mm ocular is way too much magnification for this scope. Anything below 6mm is probably useless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't worry about collimation. Most refractors, if built properly, don't require maintenance such as collimation. I will edit my answer you've linked above to make that clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars is only observable for a few weeks around opposition, which occurs every 2 years approximately. The next opposition is later this year (2018) in the summer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nakedeyeplanets.com/mars-oppositions.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.nakedeyeplanets.com/mars-oppositions.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Realistically, even in a perfect instrument you may not see a lot of detail on Mars in 50 mm of aperture. If the optics are shipshape, and seeing (air turbulence) is not too bad, you may just barely see Syrtis Major (when it's facing us) like a dark smudge on the disk. If Hellas Basin is full of frost, you may see it as a bright spot. But that should be it. I have not tried to look at Mars in such a small aperture before, but I'll try it this year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to see the equatorial belts on Jupiter. I can see them in my 50 mm finderscopes.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2018-01-05T20:40:56.017" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24385" PostTypeId="1" CreationDate="2018-01-05T22:33:38.157" Score="2" ViewCount="79" Body="&lt;p&gt;I recently saw a 3D-plot of proper motions of galaxies within the Laniakea super-cluster (&lt;a href=&quot;https://sketchfab.com/models/754cb85550fb42588175f8a215718521&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://sketchfab.com/models/754cb85550fb42588175f8a215718521&lt;/a&gt;). Is a similar 3D map of the vector field of angular momentum possible? Would it be rather random, or would close galaxies within a cluster have strongly correlated angular momenta? Can we even estimate the axis of rotation from visual / spectroscopic data?&lt;/p&gt;&#xA;" OwnerUserId="20779" LastActivityDate="2018-01-05T22:33:38.157" Title="What is known about the angular momentum distribution within galaxy clusters?" Tags="&lt;astrophysics&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24386" PostTypeId="1" CreationDate="2018-01-06T06:45:28.573" Score="6" ViewCount="2244" Body="&lt;p&gt;Recently a galaxy A1689B11 was &lt;a href=&quot;http://www.sci-news.com/astronomy/most-ancient-spiral-galaxy-05401.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;discovered&lt;/a&gt;, having an age of 11 billion years. How that age was determined? News sources reported that some cutting edge technology with gravitational lensing was used. Can anyone explain the exact procedure?&lt;/p&gt;&#xA;" OwnerUserId="20162" LastEditorUserId="1891" LastEditDate="2021-01-13T22:14:11.110" LastActivityDate="2021-05-18T12:27:15.127" Title="How to determine the age of an old galaxy?" Tags="&lt;galaxy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24391" PostTypeId="2" ParentId="23957" CreationDate="2018-01-06T08:41:21.330" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So that's question 1: Is it really possible for tidal acceleration to&#xA;  throw a satellite into space before the planet is tidally locked, and&#xA;  if so, does either objects' size impact it, or just fluidity and&#xA;  relative periods?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;During the locking process, the angular momentum of one is fed into the other until the periods match.  The final destination depends on the initial momentum of both objects and the moment of inertia.  As an example, if you make the moon less massive, or the earth more massive or with a faster spin, then the final configuration of the system would have a larger orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could certainly create a system where the final theoretical configuration has a distance to the satellite that is beyond the primary's Hill sphere.  Such a system would be expected to eventually separate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fluidity might change the rate at which the final destination is reached, but not its configuration.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Will all non-retrogade tidally decelerating satellites eventually&#xA;  crash into their planet, or is it possible for the planet to catch up&#xA;  given the right starting conditions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each object is transferring the same amount of angular momentum.  But the response of each object depends (at least initially) on the relative magnitudes of their moments of inertia.  For a system where the satellite is (relatively) small, or where the difference in rotation rates is large, impact is inevitable.  The primary cannot speed up fast enough to keep up with the increasing speed of the descending partner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when the difference in periods is small, and the primary is small enough/far enough, then the objects would lock before impact.  &lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2018-01-06T08:41:21.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24395" PostTypeId="2" ParentId="24386" CreationDate="2018-01-06T11:54:22.493" Score="7" Body="&lt;p&gt;When the article speaks of the &quot;age&quot;, what it means is that we are seeing that galaxy as it was a long time ago. We are seeing it as a very young galaxy. The galaxy is very far away, so light has taken a long time to reach us from the galaxy. The light from the galaxy is 11 billion years old. We measure the age of the galaxy by finding its distance from us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the distance we use the Hubble Law, which states that the further away a galaxy is, the faster it moves away from us, due to the expansion of the universe. And we can measure the speed that the galaxy is moving away from us because when an object moves very fast, the light from that object becomes shifted towards longer wavelengths (red-shifted). By measuring light that is known to have a fixed wavelength when not red shifted, we can find the speed of the galaxy accurately, and then use the Hubble Law to find the distance, and hence the age of the galaxy. (&lt;a href=&quot;https://arxiv.org/abs/1710.11130&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A1689B11 has a redshift z=2.54 which corresponds to a light travel time of &lt;a href=&quot;http://www.astro.ucla.edu/~wright/CosmoCalc.html&quot; rel=&quot;noreferrer&quot;&gt;11.1 billion years&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational lensing makes the galaxy brighter, so it can be observed at these immense distances, it isn't directly used for finding the distance, but without gravitational lensing, the galaxy wouldn't be visible.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-01-06T16:19:41.803" LastActivityDate="2018-01-06T16:19:41.803" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24398" PostTypeId="2" ParentId="11082" CreationDate="2018-01-06T14:00:13.687" Score="9" Body="&lt;p&gt;No, the Main Sequence is more like a &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Main_sequence#Formation_and_evolution&quot; rel=&quot;noreferrer&quot;&gt;starting line&lt;/a&gt;&lt;/strong&gt;. Most stars spend a long time on one point of it (&lt;a href=&quot;https://en.wikipedia.org/wiki/Sun#Main_sequence&quot; rel=&quot;noreferrer&quot;&gt;10 billion years&lt;/a&gt; for the sun) while they fuse hydrogen into helium. Then they wander off it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Stellar_evolutionary_tracks-en-with-text.svg&quot; rel=&quot;noreferrer&quot;&gt;this diagram&lt;/a&gt; the black line is the Main Sequence. The colored lines show temporal sequences. This progress through time of an individual star is called its &lt;a href=&quot;https://en.wikipedia.org/wiki/Main_sequence#Evolutionary_tracks&quot; rel=&quot;noreferrer&quot;&gt;evolutionary track&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/155Fy.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/155Fy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numbers along the black line are solar masses (1=the sun). This diagram evolved through a sequence of images by Wikimedia users &lt;a href=&quot;https://commons.wikimedia.org/wiki/User:Rursus&quot; rel=&quot;noreferrer&quot;&gt;Rursus&lt;/a&gt;, &lt;a href=&quot;https://commons.wikimedia.org/wiki/User:G.A.S&quot; rel=&quot;noreferrer&quot;&gt;G.A.S&lt;/a&gt;, and Jesusmaiz.&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="71" LastEditDate="2018-01-08T08:17:47.550" LastActivityDate="2018-01-08T08:17:47.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24400" PostTypeId="1" CreationDate="2018-01-06T14:34:17.647" Score="4" ViewCount="211" Body="&lt;p&gt;For example, what would be some good references to read first? What sort of glass slab should I use? What size; diameter and thickness. Im a first-time builder trying to build one&lt;/p&gt;&#xA;" OwnerUserId="20802" LastEditorUserId="7982" LastEditDate="2018-01-29T08:24:20.213" LastActivityDate="2018-01-29T08:24:20.213" Title="Need advice on mirror-making for a home-made amateur reflector" Tags="&lt;mirror&gt;&lt;telescope-making&gt;&lt;mirror-making&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ClosedDate="2018-01-11T01:18:44.977" ContentLicense="CC BY-SA 3.0" />
  <row Id="24401" PostTypeId="2" ParentId="23957" CreationDate="2018-01-06T15:39:03.257" Score="4" Body="&lt;p&gt;You missed the the most important factor, which is, how close the sun is and as a result of the distance to the sun and relative mass, the Hill sphere and the approximate distance where &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere#True_region_of_stability&quot; rel=&quot;nofollow noreferrer&quot;&gt;the stable orbital distance ends&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angular momentum to rate the moon gets pushed away from the planet is an interesting combination of formulas, and could probably be calculated based on the factors you mentioned, density, mass, size, shape and mass of the liquid oceans on the surface, rate of rotation, etc.  There's probably a not all that complex approximation that could be worked out for difference scenarios to show how far a moon would migrate before the system tidally locks and how long it would take. (some less interactive systems might take trillions or quadrillions of years or longer to reach a relatively final position, compared to the approximately 50 billion or so timetable for the Earth-Moon system to tidally lock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we look at the Sun-Earth-Moon as an example, the most important on whether the Earth loses it's moon is how far the stable part of the Hill Sphere extends, which, if the masses of all 3 objects are what they are now, the primary factor is how close the Earth is to the Sun.   If the Earth was as close to the Sun as Venus, the Moon probably would escape, but it would take some time.   If the Earth was as close to the sun as Mercury (on average), it would have lost the Moon long ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, I think I have a simple inroad to your other questions, at least a good place to start.   Angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider a two body orbital system then the tidal effect on the orbit is fairly straight forward.  Considering systems where the bulge rotates ahead of the moon the moon is slowly pushed away from the planet and the planet's rotation gradually decreases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this two body system the moon never escapes because in a two body system, orbits remain stable over an enormous distance, the question essentially becomes, how far does the moon move away and how long does it take until the system is tidally locked where it, at least mathematically, reaches it's end state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 3rd body in the picture (the Sun) has an effect in a couple ways.   The relation between the Sun and Planet's relative mass and distance determines the approximate boundary of the stable region of the Hill Sphere, beyond which the moon likely escapes.   The Sun also tugs on the Moon, creating perturbations in it's orbit and making it non-curcular, where, in comparison, the two body system with the moon being pushed away moved towards becoming circular.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Red-dwarfs, where planets are close to the star, the planet becoming tidally locked to the red-dwarf star becomes much more likely (though some partial locking like Mercury's 3/2 orbit to rotation ratio may happen occasionally), but in most cases, with a red dwarf star and planets relatively close proximity, the planet becomes tidally locked to the star and effectively rotating slower relative to the moon, leading to a decaying orbit for moons towards their planets in red-dwarf systems.   The stable orbital region in such systems would also be much smaller.   The sun is a significant factor in the motion of moons to or from planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But getting back to the 2 body system and estimating the tidal effect, knowing that the moon will never escape from a closed two body system, the question then works out to how far the Moon can move from the planet and how long it will take until mutual tidal locking is reached.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How far the Moon can travel from the planet can be estimated by the total angular momentum of the system, so you're correct that the initial rotation speed is a factor, so is the relative mass between the Planet and Moon.  A more massive moon creates a larger tidal bulge on the planet, which results n the Moon moving away from the planet faster than a less massive moon.  At the same time, the more massive moon requires more angular momentum as it moves away from the planet, so it moves less far.   A less massive moon moves further but takes much much longer to get there and the tidal lock also takes longer to happen.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A more fluid planet would experience stronger tides, which would again&#xA;  slow it down faster, but definitely accelerate the satellite faster. A&#xA;  more viscous fluid than water would experience weaker tides, but I&#xA;  think they'd be farther ahead. I think more fluid is definitely more&#xA;  effective, but it seems less obvious than the first point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not correct to say a more fluid planet would experience stronger tides.  Saying that it would experience higher tides than if it was solid would be more accurate.  The tides the planet experiences depend on the mass and distance of the Moon and the size of the planet, so the Moon being equal, the tides would be the same whether the planet is liquid or solid.   Fluids are less rigid, so they are better at making tides, but there are tides on land too. On Earth they are called &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth_tide&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth-tides&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The higher tide does effectively push the Moon better and slow the planet more quickly and liquid responds to tides more efficiently, so planets with a liquid surface do move their moons more efficiently and faster than planets without a liquid surface.  The shape of the oceans matter too.   The wider Pacific has bigger tides than the narrower Atlantic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VCTqn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VCTqn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://marinebio.org/oceans/currents-tides/index.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't believe the viscosity of the liquid matters.   Liquids flow, solids don't and father ahead isn't correct either.   Planets generally rotate as a unit, though there are small discrepancies between the rate of rotation between the core and mantle, the entire planet rotates and the tidal bulge rotates with it.  Liquids are more efficient and bulging, but not either way the bulge moves ahead with the planet's rotation at equal angles ahead of the moon.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A larger planet would have its orbit slowed less by tidal&#xA;  acceleration, but have a stronger pull to escape. A larger satellite&#xA;  would cause stronger tides and more tidal force, but would require&#xA;  more force to accelerate, and would slow the planet down more quickly.&#xA;  I really can't tell which has the stronger effect...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here you have to be careful because it's a little more complicated.  The tidal force a planet experiences from it's moon has to do with the mass of the moon, the distance of the moon, but also the mass of the planet and the radius of the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://web.njit.edu/~gary/320/assets/tide_fig1.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://web.njit.edu/~gary/320/assets/tide_fig1.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://web.njit.edu/~gary/320/Lecture12.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the 2nd part of the formula is pretty tricky, $[(1 + R/d)^{-2} - (1 - R/d)^{-2}]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but if you work out the math for fairly distant moons, the difference in the tidal force from the far side of the planet to the near side is about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta F = \frac{-4GMmR}{d^3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So looking at the Earth-Moon system, ignoring the sun for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you make Earth the same size but twice the mass, the tidal force doubles but Earth's surface gravity would double as well which would give the ocean water twice the weight with equal mass, the force on the Moon should be similar, but a heavier Earth would take longer to slow down, so the effect would be that the Moon would move further, not less far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you keep the density the same and make the Earth larger, say 8 times the mass, so twice the diameter.  The tidal force the Moon exerts on our new super-earth would be 16 times greater but the gravity on the surface of the Earth just twice as much.   What that means is that by making the planet larger, the Moon actually moves away faster.   The larger (but equal density) planet should slow down at roughly an equal rate but as the Moon moves away that rate would slow down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the math on tides gets confusing for me.  The inverse tidal force that moves moons away from planets drops off significantly with distance.  The tidal force drops off with the cube of the distance.  The force from the tidal bulge may drop off even faster than that because the angle decreases with distance, (I would guess it drops off by the 4th power), so, as the Moon's distance from the Earth doubles, the speed it moves away drops by (roughly), 16 times.   (((I think))).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you make the Earth 8 times as massive, and twice the radius, the tidal force increases by 16, but if the Moon moves twice as far, the tidal force drops by 16, which is kind of neat that it cancels out.   The distance to gravity ratio, assuming the Moon doesn't change and the planet's density doesn't change, is 1 to 1, which is what I'd expect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's probably longer than it should be and it probably needs a lot of clean-up, but I need to call it a day, so I'll look at this tomorrow.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="88" LastEditDate="2018-01-08T11:23:03.020" LastActivityDate="2018-01-08T11:23:03.020" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24403" PostTypeId="2" ParentId="24400" CreationDate="2018-01-06T19:05:13.497" Score="4" Body="&lt;p&gt;I second &lt;a href=&quot;https://astronomy.stackexchange.com/questions/24400/home-made-amateur-reflector#comment41779_24400&quot;&gt;the comment&lt;/a&gt; about Texereau -- originally written in French; it's the best, bar none. You won't need anything else. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember the old adage that is quicker to make a 4&quot; mirror &lt;em&gt;and then&lt;/em&gt; an 8&quot; mirror than it is to make an 8&quot; mirror first. Don't bother with borosilicate. You should be able to get mirror kits and all the abrasives that you need online. Join a club if you can find one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;How to Make a Telescope, 2nd edition&lt;/em&gt; by Jean Texereau: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.willbell.com/tm/tm3.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.willbell.com/tm/tm3.htm&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B00EKYQCQS&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amazon.com/Telescope-second-Texereau-published-Atlantic/dp/B00EKYQCQS&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13887" LastEditorUserId="7982" LastEditDate="2018-01-07T15:18:45.940" LastActivityDate="2018-01-07T15:18:45.940" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24409" PostTypeId="1" CreationDate="2018-01-07T16:13:24.123" Score="5" ViewCount="180" Body="&lt;p&gt;I know we have some indication of &amp;quot;what's in the water&amp;quot; from the plumes of Enceladus, but would other places be the same. For instance, could the deep liquid water of Titan contain dissolved methane gas?&lt;/p&gt;&#xA;&lt;p&gt;I'm mainly contemplating what possible &amp;quot;solutions&amp;quot; would be beneficial to some form of microbiology.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="7472" LastEditDate="2020-11-03T15:16:43.290" LastActivityDate="2021-05-06T15:02:00.177" Title="What would theoretically be dissolved in the interior oceans of icy moons and planets?" Tags="&lt;planet&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24410" PostTypeId="1" AcceptedAnswerId="24412" CreationDate="2018-01-07T17:41:48.230" Score="14" ViewCount="777" Body="&lt;p&gt;So I just bought a globe for my extremely inquisitive soon to be 4-year old son. He knows the very basics of the solar system and knows that we live on planet Earth and also knows the concept of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While i was explaining to him what the illuminated equator line is he asked me How do people stand below this line? (Look at the globe from a 4 year old's eyes and you will wonder the same)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I explained to him that earth is not as small of a sphere as this globe depicts and almost everywhere it will be relatively flat but then he said there are so many places on this globe at the bottom if you live there earth will be above you and gravity will hold you upside down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I explain it to him?&lt;/p&gt;&#xA;" OwnerUserId="20820" LastActivityDate="2018-12-25T13:50:43.823" Title="Explain to a 4-year old why people below the equator are not tilted or suspended upside down" Tags="&lt;gravity&gt;&lt;earth&gt;" AnswerCount="8" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24411" PostTypeId="2" ParentId="24410" CreationDate="2018-01-07T18:04:01.160" Score="0" Body="&lt;p&gt;I would show him a magnet, and the magnet can hold something upside-down.    The pull is towards the Earth, just as the pull is towards the magnet.   Direction (up down) doesn't really matter only where the Earth is or where the magnet is, relative to the person or paperclip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SO the person on the &quot;bottom&quot; of the Earth is like the paperclip being held by the magnet above it.   He's being pulled by gravity towards the center of the Earth just like everyone else on the Earth.   Earth is always down - no matter where you are on the surface.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-01-07T18:04:01.160" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24412" PostTypeId="2" ParentId="24410" CreationDate="2018-01-07T18:24:57.037" Score="12" Body="&lt;p&gt;Explain that &quot;down&quot; is more or less toward the center of the Earth, and that direction changes with location. (For a four year old, skip the &quot;more or less&quot; part. It's just toward the center of the Earth, period.) The direction of &quot;down&quot; doesn't change much on trips across town, but it does change with larger trips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have taken your son on a long trip (just a few hundred miles or kilometers will do), show your son those two locations on the globe. The direction of &quot;down&quot; at that remote location and the direction where you live point in slightly different directions for a trip of a few hundred miles, vastly different directions for a longer trip.  Ask him if he felt tilted sideways at that remote location.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2018-01-07T18:24:57.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24413" PostTypeId="2" ParentId="24410" CreationDate="2018-01-07T18:51:29.863" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;While i was explaining to him what the illuminated equator line is he asked me How do people stand below this line? (Look at the globe from a 4 year old's eyes and you will wonder the same)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The issue your son sees is that, in his mind, gravity exists around &lt;em&gt;that&lt;/em&gt; globe and is always &lt;em&gt;down&lt;/em&gt;, because that's what's it's like where he's standing and he expects it (correctly !) to be the same for the model globe.  He has no conception of a space without gravity and cannot make the connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to explain that the Earth is in space, where the gravity is effectively zero and it's the Earth itself that pulls objects on and near it's surface to itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Earth there is is no other gravity &quot;down&quot;, unlike the globe he's being shown.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2018-01-07T18:51:29.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24414" PostTypeId="2" ParentId="24410" CreationDate="2018-01-07T21:37:14.420" Score="0" Body="&lt;p&gt;Show him water in outer space. The water doesn't fall down onto the floor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show him astronauts flying in outer space, they have no fall up and fall down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tell him that we would fall off if there was a bigger planet under the earth, (like a pea on a melon) otherwise we fall towards the mass of the planet. the planet also falls towards it's own center, that's why it's round. &lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2018-01-08T17:24:52.403" LastActivityDate="2018-01-08T17:24:52.403" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24415" PostTypeId="2" ParentId="5900" CreationDate="2018-01-08T02:07:38.520" Score="0" Body="&lt;p&gt;@questionhang, the moon's celestial north pole is always 1.543 degrees from the ecliptic north pole (Earth's orbital pole), which, according to &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_pole&quot; rel=&quot;nofollow noreferrer&quot;&gt;en.wikipedia.org/wiki/Orbital_pole&lt;/a&gt; was at RA 18h 0m 0.0s, D +66° 33′ 38.55″ at J2000.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an error or about 1.5 degrees if small enough for you, then use the ecliptic north pole.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, the problem is that the moon's north pole precesses through that whole circle every 18.6 years, so it changes fairly rapidly.  If one knew accurately where it was on a given date, one could to calculate where it should be now, but I don't find those numbers either.  For my purposes, I am ok using the ecliptic north pole.&lt;/p&gt;&#xA;" OwnerUserId="20822" LastActivityDate="2018-01-08T02:07:38.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24426" PostTypeId="1" AcceptedAnswerId="24429" CreationDate="2018-01-08T19:21:59.087" Score="0" ViewCount="899" Body="&lt;p&gt;I am working on a story. It would be helpful if someone could provide the distance in light years between star Epsilon Eridani and star Tau Ceti.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RunoA.png&quot; alt=&quot;star graph&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="19874" LastEditorDisplayName="user1569" LastEditDate="2018-01-09T11:06:06.783" LastActivityDate="2018-01-09T11:06:06.783" Title="Distance between stars Epsilon Eridani and Tau Ceti?" Tags="&lt;star&gt;&lt;distances&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24427" PostTypeId="2" ParentId="24400" CreationDate="2018-01-08T19:58:37.853" Score="4" Body="&lt;p&gt;Here's what I've learned from making telescope mirrors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with a mirror that's of a reasonable size and curvature. It is definitely possible to begin with a 12&quot; f/5 mirror, but the problem is you have to handle not just the higher volume of work, but also the learning process in a field that's completely new to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Begin with making a 6&quot; f/8 mirror. It's small enough, and the curvature is shallow enough, that you will be free to focus on the procedure itself. Complete that instrument, and play with it for a while. Later on just double the size, or go even bigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of thickness, just make sure the glass thickness / diameter ratio is 1 / 6. That's pretty thick, but it will prevent it from being elastically deformed while sitting in a simple mirror cell. A little thinner than that is fine usually, but not a whole lot thinner; with thin mirrors you could simply use a mirror cell with more support points (6 or 9 instead of 3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that you don't have to use borosilicate glass. Unless you have a fully automated instrument that does astrometry, plain glass will work about the same, you'll only have to adjust focus a bit more often. But borosilicate is less liable to crack during certain stages of the manufacture process when the mirror suffers a thermal shock (when applying warm pitch to create the polishing tool). Also, borosilicate is cheap enough, so why not use it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a place where you could buy supplies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://firsthanddiscovery.com/astronomy/telescope-making.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://firsthanddiscovery.com/astronomy/telescope-making.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's probably best to buy a whole kit instead of going the piecemeal route.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind this will be a long term project. You will not finish it in one week-end. You will keep working on it for months (unless you make it a full time occupation). However, because you can keep working on it, you could obtain good results. The first mirror I ever made was parabolized to a precision of lambda/25, which is far above the average quality of commercial mirrors. It's simply a function of being able to work on the project until you're satisfied with the results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OTOH, if it's taking too long, just accept the current results and move on. A lambda/6 mirror will work just fine, and if the scope is small it's probably not worth obsessing too much over quality.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Books&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mirror making&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'How To Make A Telescope' by Jean Texereau has been considered the foundational text of this field for a long time. Definitely read it, but keep in mind it's a little old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.willbell.com/tm/tm3.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.willbell.com/tm/tm3.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'Build Your Own Telescope' by Richard Berry is a little closer to the present day. His choice of materials and procedures may seem a bit more familiar to you. I don't see it as an either/or choice - read both Texereau and Berry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.willbell.com/tm/tm2.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.willbell.com/tm/tm2.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another excellent source of info for mirror making is the Stellafane website. It's an even more modern approach than both of those books, and it's basically the playbook they use at Stellafane in Vermont to help dozens of newbies make their own mirrors every year (with supervision). Last time I checked they were missing the chapters on sphericization and parabolization, but you can get that stuff from Texereau and Berry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://stellafane.org/tm/atm/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://stellafane.org/tm/atm/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I highly recommend you read all sources indicated above. Texereau is a classic, a bit old but full of wise advice. Berry is more modern, but it's his own method, it's what worked for him. Stellafane is a very practical approach that's still in use nowadays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, your own method will end up being a blend of all of the above, probably tracking the Stellafane process fairly closely, but with bits and pieces from the other authors too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Telescope making&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both Texereau and Berry have some recommendations, but they're very old. Read them for informative purposes but don't follow them. Instead, use Stellafane's guide:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://stellafane.org/tm/dob/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://stellafane.org/tm/dob/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is perfect for a 6&quot; to 8&quot; small dobsonian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later on, if you want to build a larger dobsonian, read the books by David Kriege and Albert Highe:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.willbell.com/tm/dobtel.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.willbell.com/tm/dobtel.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.willbell.com/tm/PortableNewtonianTelescopes.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.willbell.com/tm/PortableNewtonianTelescopes.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Kriege book is a bit older stuff, but he's very experienced (he's the owner of Obsession Telescopes). Highe is an engineer turned amateur telescope maker, and his book goes very deeply into the theory of engineering of such projects; you don't have to understand every single part of it, but it's very informative to just read it and see how he's doing it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Random useful links:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Stellafane mirror and telescope calculator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://stellafane.org/tm/atm/general/scope-calc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://stellafane.org/tm/atm/general/scope-calc.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mel Bartels' diagonal calculator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.bbastrodesigns.com/diagonal.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.bbastrodesigns.com/diagonal.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2018-01-08T20:14:29.360" LastActivityDate="2018-01-08T20:14:29.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24428" PostTypeId="2" ParentId="24410" CreationDate="2018-01-08T20:40:03.493" Score="1" Body="&lt;p&gt;I like the answer by D. H. However re-reading the question and the comment by C. W. it seems possible that yours boy got it right in a way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, the people below equator are tilted or upside down with respect to us. This still can be explained as suggested, of course we know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the two situations are different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It comes to me to realize that a sphere and Newton law are not enough to explain why somebody in Australia is not walking on its head. So the fact that &lt;em&gt;we sense up and down&lt;/em&gt; plays a crucial role. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately it is a mix of physiological psychological factors, I guess. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should make sure what your son in asking, in other words. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say that gravity is at center of Earth. The latter being spherical, gravity is &lt;em&gt;under our feet&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17437" LastEditorUserId="17437" LastEditDate="2018-12-25T13:50:43.823" LastActivityDate="2018-12-25T13:50:43.823" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24429" PostTypeId="2" ParentId="24426" CreationDate="2018-01-08T20:46:22.987" Score="2" Body="&lt;p&gt;As answered by &lt;a href=&quot;http://www.wolframalpha.com/input/?i=distance%20between%20tau%20ceti%20and%20epsilon%20eridani&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wolframalpha&lt;/a&gt;, (as linked by chirlu) The distance between Tau Ceti and Epsilon Eridani is 5.45 light years.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-01-08T20:46:22.987" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24430" PostTypeId="2" ParentId="18397" CreationDate="2018-01-08T23:00:21.407" Score="1" Body="&lt;p&gt;Just to beat this to death, &lt;a href=&quot;http://m.esa.int/Our_Activities/Space_Science/Gaia/Frequently_Asked_Questions_about_Gaia&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://m.esa.int/Our_Activities/Space_Science/Gaia/Frequently_Asked_Questions_about_Gaia&lt;/a&gt; notes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;What will be included in the first public data release?&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Gaia Data Release 1 includes the positions and G magnitudes – a broad,&#xA;  visible light passband spanning 330 nm to 1050 nm – for more than one&#xA;  billion stars using observations taken between 25 July 2014 and 16&#xA;  September 2015.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In addition, for a subset of data – about 2 million stars in common&#xA;  between the Tycho-2 Catalogue and Gaia – there will be a&#xA;  five-parameter astrometric solution, giving the positions, parallaxes,&#xA;  and proper motions for those objects. This is referred to as the&#xA;  Tycho-Gaia Astrometric Solution (TGAS).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://m.esa.int/Our_Activities/Space_Science/Gaia/Parallax&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://m.esa.int/Our_Activities/Space_Science/Gaia/Parallax&lt;/a&gt; notes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;By measuring the amount of displacement – the parallax angle – and&#xA;  knowing the distance between the Sun and the Earth, astronomers can&#xA;  determine the star’s distance using simple trigonometry.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Even for the nearest stars the amount of apparent movement is tiny:&#xA;  less than one arcsecond. Yet Gaia will measure the positions of one&#xA;  billion stars to microarcsecond accuracies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming &quot;will measure the positions&quot; means &quot;will measure the parallaxes&quot;, this appears to be referring to the final version of GAIA which will be published in the early 2020s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make things easier, &lt;a href=&quot;http://cdn.gea.esac.esa.int/Gaia/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://cdn.gea.esac.esa.int/Gaia/&lt;/a&gt; has a &quot;tgas_source&quot; subdirectory which just includes the 2,057,066 stars for which there is additional data (so you don't have to download the entire much larger GAIA archive).&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2018-01-08T23:00:21.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24431" PostTypeId="1" AcceptedAnswerId="24507" CreationDate="2018-01-08T23:10:24.633" Score="2" ViewCount="461" Body="&lt;p&gt;If I observe the moon pass my celestial meridian at a specific north polar clock 'time' will I observe every subsequent passage at ( 'current time' + 13.2* X 'days passed' mod 360* ) ? Does the moon pass celestial meridian at exactly the same time if observed from every point on this longitude?&lt;/p&gt;&#xA;" OwnerUserId="20197" LastEditorUserId="20197" LastEditDate="2018-01-08T23:53:23.210" LastActivityDate="2018-01-13T03:12:36.820" Title="Will time that moon crosses meridian always be periodic?" Tags="&lt;the-moon&gt;&lt;observational-astronomy&gt;&lt;positional-astronomy&gt;&lt;navigation&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24433" PostTypeId="1" AcceptedAnswerId="24579" CreationDate="2018-01-09T04:27:29.083" Score="1" ViewCount="68" Body="&lt;p&gt;I'm trying to create a widget that shows the 'time' on different planets. It will show how far through the day/night cycle (as a percentage) a point on the planet is. It has been easy to scale down the cycle from 24 hr to eg ~10hr day of Jupiter however I am struggling to work out how I can 'set' the time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Mars this has been straight forward, using the comparison of MTC and UTC of when the Curiosity rover landed but I can't find similar data for other planets to anchor time from earth to the other planets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a data base for this kind of information?&lt;/p&gt;&#xA;" OwnerUserId="20843" LastActivityDate="2018-01-23T10:24:21.353" Title="How can I calculate how far through the day the prime meridian is of different planets" Tags="&lt;rotation&gt;&lt;time&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24442" PostTypeId="2" ParentId="24038" CreationDate="2018-01-09T09:23:09.637" Score="3" Body="&lt;p&gt;You can use &lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR-3?-source=VIII/1A&amp;amp;-out.max=500&amp;amp;-out.form=ascii&quot; rel=&quot;nofollow noreferrer&quot;&gt;VizieR&lt;/a&gt; to retrieve the whole catalogue (3C and revised 3C). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;VizieR&lt;/strong&gt; is an online astronomical catalogue that allows you to access a quite comprehensive list of catalogues (16,929, &lt;a href=&quot;http://vizier.u-strasbg.fr/vizier//index.gml&quot; rel=&quot;nofollow noreferrer&quot;&gt;as of today&lt;/a&gt;); you can easily query and extract all kind of data from this online tool.&lt;/p&gt;&#xA;" OwnerUserId="88" LastActivityDate="2018-01-09T09:23:09.637" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24449" PostTypeId="1" AcceptedAnswerId="24496" CreationDate="2018-01-09T14:48:27.320" Score="3" ViewCount="664" Body="&lt;p&gt;I came upon this question where the situation is I have radial velocity measurements of a star with two circular orbiting exoplanets and the question is how to find out if the inclination angles of the planets are close to 90°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understood this question could be resolved using a transit method. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't see how inclination angle can be guessed through radial velocity measurements when they are close to 90°. What difference does it make on radial velocities ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&#xA;I think that the misunderstanding mentioned in the answers was really relevant, indeed in the scope of this problem an inclination of 90° the three points (Star, Exoplanet, observer) are aligned. Which means that a transit can be observed. &lt;/p&gt;&#xA;" OwnerUserId="20862" LastEditorUserId="20862" LastEditDate="2018-05-16T18:00:22.647" LastActivityDate="2018-05-16T18:00:22.647" Title="Guess inclination angle from radial velocity measurements" Tags="&lt;exoplanet&gt;&lt;planetary-transits&gt;&lt;inclination&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24450" PostTypeId="2" ParentId="24449" CreationDate="2018-01-09T15:53:36.073" Score="2" Body="&lt;p&gt;Radial velocity is a measurement in the change of speed of the star &lt;em&gt;towards &lt;strong&gt;us&lt;/em&gt;&lt;/strong&gt;, due to the planet's rotation around it. Speed variation change is detected thanks to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Doppler_spectroscopy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Doppler spectroscopy&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the planets rotation plan is orthogonal with our direction of view, then we cannot see the radial velocity change, because the star speed &lt;em&gt;towards&lt;/em&gt; us is constant. This would mean the planets you are talking about in your assignment were discovered using another method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a picture to illustrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Orbit3.gif#/media/File:Orbit3.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/5/59/Orbit3.gif&quot; alt=&quot;Orbit3.gif&quot;&gt;&lt;/a&gt;&lt;br&gt;By User:Zhatt - Own work, Public Domain, &lt;a href=&quot;https://commons.wikimedia.org/w/index.php?curid=280060&quot; rel=&quot;nofollow noreferrer&quot;&gt;Link&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above, the star speed changes in the plan that is 90° with our own, so there is no radial velocity variation we can detect with Doppler spectroscopy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit:&lt;/em&gt;&lt;br&gt;&#xA;In your case, you are not telling if inclination is 90° to &lt;em&gt;the line-of-sight&lt;/em&gt; or to the line &lt;em&gt;perpendicular to the line-of-sight&lt;/em&gt;. If it is &lt;strong&gt;90° to the line &lt;em&gt;perpendicular to the line-of-sight&lt;/em&gt;&lt;/strong&gt; (which is the usual in astronomy), then speed changes are maximum.&lt;/p&gt;&#xA;" OwnerUserId="11264" LastEditorUserId="11264" LastEditDate="2018-01-16T07:44:15.830" LastActivityDate="2018-01-16T07:44:15.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24451" PostTypeId="1" AcceptedAnswerId="24453" CreationDate="2018-01-09T17:47:16.377" Score="3" ViewCount="214" Body="&lt;p&gt;I was reading &lt;a href=&quot;https://www.vox.com/science-and-health/2018/1/8/16822272/black-hole-looks-like-what/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; posted on /r/space about photographing black holes, and they showed a picture of what's believed to be two black holes formed from two galaxies colliding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't realize the collision of galaxies could result in black holes, but it got me wondering. If you lived in one of those galaxies, what would happen to you? Would your planet be destroyed in the blink of an eye at some point? If so, when?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the sky gradually fill with more and more stars? If your galaxy became a black hole, how would you experience that? All at once or very gradually?&lt;/p&gt;&#xA;" OwnerUserId="13535" LastActivityDate="2018-01-10T20:27:15.113" Title="If one galaxy were to collide with another, what would it look like to an immortal inhabitant of one of their planets?" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24452" PostTypeId="2" ParentId="24410" CreationDate="2018-01-09T18:07:56.483" Score="1" Body="&lt;p&gt;I'd use a visual here - just tape some Lego people to a basketball, then explain that down is to the center of the basketball, and it only looks to him that there is a &quot;universal down&quot; because he too is standing on a much larger basketball. For someone that young, and for people of any age, a demonstration like that will make more sense than any explanation.&lt;/p&gt;&#xA;" OwnerUserId="14255" LastEditorUserId="14255" LastEditDate="2018-01-09T19:01:07.347" LastActivityDate="2018-01-09T19:01:07.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24453" PostTypeId="2" ParentId="24451" CreationDate="2018-01-09T19:13:09.453" Score="5" Body="&lt;p&gt;In the vast, vast majority of cases, not much happens. Galaxies are a whole lot of empty space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Sun was a soccer ball in a field in California, the nearest star of similar size, Alpha Centauri, would be another soccer ball all the way across the continent, in Greenland. This is a scaled view, where everything is proportional. Think of all that empty space in between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the galaxies would merge, or keep swirling around each other, but for the most part stars and their systems would just continue as they were. It is extremely unlikely that any given star would collide. That's true for most of the volume of the galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the center, however, things are a little different. Most galaxies already have very massive black holes at the center. These may collide and merge, or orbit each other a very long time, or continue separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formation of new black holes as a result of galactic collisions must be a very rare event. Even then, the entire galaxy would not &quot;become a black hole&quot;; rather, a small fraction of it would be added to an existing black hole or, in very rare cases, produce a new black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For most observers in such galaxies, there would not be dramatic changes, even over very long time intervals. This is because, again, galaxies are mostly empty space, and because the stars you see with the naked eye are very close to you, within a few dozen to a few hundred light years, which is a small portion of the whole galaxy.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: As shown in the comments, it's more accurate to think of Alpha Centauri as not a single soccer ball, but instead a larger soccer ball and a volleyball separated by a few hundred meters - it's a double star. Then Proxima Centauri would be a pingpong ball hovering 100 km away from Alpha.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2018-01-10T20:27:15.113" LastActivityDate="2018-01-10T20:27:15.113" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24454" PostTypeId="1" AcceptedAnswerId="24455" CreationDate="2018-01-09T19:20:27.900" Score="1" ViewCount="59" Body="&lt;p&gt;Let's say I am in Brasilia (latitude = 15.78°S). This is obviously south of the equator but north of the tropic of capricorn. So on the summer solstice I know that the Sun's declination is +23.26°. Does this mean that the sun will appear directly north at solar noon? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="20863" LastEditorUserId="20863" LastEditDate="2018-01-09T19:29:42.133" LastActivityDate="2018-01-09T19:36:08.043" Title="Sun's position on summer solstice" Tags="&lt;the-sun&gt;&lt;declination&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24455" PostTypeId="2" ParentId="24454" CreationDate="2018-01-09T19:36:08.043" Score="2" Body="&lt;p&gt;The Sun at solar noon (meaning that it is on the meridian) will be in one of these directions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Due south if the declination of the Sun is less that your latitude. (For Brasilia, that occurs from Nov 6 to Feb 5, approximately)&lt;/li&gt;&#xA;&lt;li&gt;Directly overhead (at the zenith) if the declination of the Sun is the same as your latitude. (For Brasilia, that occurs Nov 5 and Feb 6, approximately)&lt;/li&gt;&#xA;&lt;li&gt;Due north if the declination of the Sun is more than your latitude. (For Brasilia, that occurs from Feb 7 to Nov 4, approximately)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2018-01-09T19:36:08.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24456" PostTypeId="2" ParentId="24275" CreationDate="2018-01-10T00:45:38.930" Score="1" Body="&lt;p&gt;Here's a rough check on that, offered as a supplementary answer. Using the Python package &lt;a href=&quot;http://rhodesmill.org/skyfield/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Skyfield&lt;/a&gt; one can calculate the distance to the center of the Moon. Right now I don't know how to calculate the distance to the specific location of the Apollo 15 reflectors on the Moon, but the distance from the observatory to the &lt;em&gt;closest point&lt;/em&gt; on the Moon is about 200 km shorter than the distance determined from laser pulses as described in the other answer. This seems about right considering the radius of the Moon is about 1767 km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Output: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;altitude:  37.6454136245&#xA;azimuth:   193.116013331&#xA;distance (to center of Moon):  366418.551453&#xA;distance to closest point on moon:  364652.0&#xA;compare to:  364857 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Python script&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;&#xA;from skyfield.api import Loader, Topos &#xA;&#xA;load = Loader('~/Documents/fishing/SkyData')&#xA;data = load('de421.bsp')&#xA;ts   = load.timescale()&#xA;&#xA;planets = load('de421.bsp')&#xA;earth   = planets['earth']&#xA;moon    = planets['moon']&#xA;Grasse  = earth + Topos('43.753698 N', '6.922998 E', elevation_m = 372.)&#xA;&#xA;time    = ts.utc(2016, 1, 13, 15, 24, 19.4526250)&#xA;&#xA;alt, az, dist  = Grasse.at(time).observe(moon).apparent().altaz()&#xA;&#xA;print &quot;altitude: &quot;, alt.degrees&#xA;print &quot;azimuth:  &quot;, az.degrees&#xA;print &quot;distance (to center of Moon): &quot;, dist.km&#xA;&#xA;print &quot;distance to closest point on moon: &quot;, round(dist.km, 0) - 1767.&#xA;print &quot;compare to: &quot;, 364857&#xA;&#xA;&quot;&quot;&quot;&#xA;5&#xA;Format&#xA;&#xA;1&#xA;Color&#xA;&#xA;20160113&#xA;AAAAMMJJ&#xA;&#xA;1524194526250&#xA;HHMMSSsssssss&#xA;&#xA;24340653926601&#xA;2sssssssssssss  times 0.1 ps&#xA;&#xA;3&#xA;Reflector code  (3 = Apollo 15)&#xA;&#xA;01910&#xA;Station Code  (01910 = Grasse)&#xA;&#xA;034&#xA;Number of Echoes&#xA;&#xA;002705&#xA;Uncertainty  (0.1 ps)&#xA;&#xA;017&#xA;S/N ratio (0.1 ps)&#xA;&quot;&quot;&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; Moon landing sites, from &lt;a href=&quot;https://www.bobthealien.co.uk/moon/landingsites.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bob the Alien&lt;/a&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uaxbI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uaxbI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-01-10T00:56:32.513" LastActivityDate="2018-01-10T00:56:32.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24459" PostTypeId="2" ParentId="24410" CreationDate="2018-01-10T02:51:27.507" Score="4" Body="&lt;p&gt;The Earth is a very, very big ball just like the one you play with. It's so big you can't even tell you're standing on a ball. The bigger and heavier things are, the more they pull on other things. The Earth pulls everybody standing on it toward its middle so everybody thinks that they are right side up.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2018-01-10T02:51:27.507" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24460" PostTypeId="1" CreationDate="2018-01-10T03:28:38.617" Score="2" ViewCount="80" Body="&lt;p&gt;Whew! That title was a mouth-full, but accurate I hope. What I'm trying to figure out is a model for calculating the approximate temperature variation for a generally earth-like planet (similar atmosphere, large oceans, etc.) under two concurrent conditions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;High Eccentricity: 0.1, 0.2, 0.3 etc.  &lt;/li&gt;&#xA;&lt;li&gt;Long Diurnal Periods: 72 hours - 960 hours (the later could reflect a planet in a 2:1 orbital resonance around a M Star). &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Assume Obliquity = 0 so as to simplify things)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know conditions like the mass of the atmosphere and the size of the oceans play a key role in the retention and redistribution of heat. However I've been unable to locate any research papers or explicit models that would help. It's relatively straightforward to calculate real temperature (effective temperature modified for greenhouse effect) for both periapsis and apoapsis, however I've had no luck in finding resources for figuring out how much of this temperature variation would be mitigated by retention and redistribution. Same with long nights and long days. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a specific example would help, let's consider a planet with the following characteristics:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;30-day orbit&lt;/li&gt;&#xA;&lt;li&gt;15-day (720 hour) rotation (2:1 orbital resonance)&lt;/li&gt;&#xA;&lt;li&gt;30-day solar day (sunrise to sunrise)&lt;/li&gt;&#xA;&lt;li&gt;0 obliquity&lt;/li&gt;&#xA;&lt;li&gt;0.25 eccentricity&lt;/li&gt;&#xA;&lt;li&gt;Earth-like in other respects&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Calculations or even simply links to applicable research papers would be immensely helpful. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="20372" LastActivityDate="2018-01-10T03:28:38.617" Title="Predicting Temperature Variation on Planets with High Eccentricity or Long Diurnal Periods" Tags="&lt;exoplanet&gt;&lt;temperature&gt;&lt;earth-like-planet&gt;&lt;planetary-science&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24462" PostTypeId="2" ParentId="24410" CreationDate="2018-01-10T04:01:24.290" Score="4" Body="&lt;p&gt;Sounds like you have a budding astrophysicist on your hands!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Son, there's something called &lt;em&gt;gravity&lt;/em&gt; that kind-of pulls everything to the center of Earth. It's like a magnet, but everywhere you go on Earth it's there. It's in the mountains, in the ocean, even at daycare.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now present the magnet/paperclip example someone else suggested as a supplement to this explanation. Take the paperclip and let it attract to the magnet from the top. Now flip it over and explain that Earth's gravity works in a similar way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No need to get technical. I'm sure your future astrophysicist will discern the semantics of this concept at a later date.&lt;/p&gt;&#xA;" OwnerUserId="20869" LastActivityDate="2018-01-10T04:01:24.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24463" PostTypeId="1" CreationDate="2018-01-10T06:07:16.497" Score="1" ViewCount="37" Body="&lt;p&gt;Will the below official launch viewing areas be open for a night launch (Atlas V at 7:40 p.m. on 18 Jan 2018)? If not, what are the other options?&#xA;THE LC-39 OBSERVATION GANTRY&#xA;APOLLO/SATURN V CENTER&#xA;NASA CAUSEWAY&#xA;MAIN VISITOR COMPLEX&lt;/p&gt;&#xA;" OwnerUserId="20873" LastActivityDate="2018-01-10T06:07:16.497" Title="What are the best places near Kennedy Space Center to watch a night launch?" Tags="&lt;nasa&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2018-01-10T20:24:01.297" ContentLicense="CC BY-SA 3.0" />
  <row Id="24476" PostTypeId="1" CreationDate="2018-01-10T16:47:08.417" Score="3" ViewCount="52" Body="&lt;p&gt;This question might seem a bit off-topic, but I guess there are a lot of people here that know about optics, telescopes, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to simulate solar thermal systems that focus solar radiation, e.g. &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_power_tower&quot; rel=&quot;nofollow noreferrer&quot;&gt;heliostats&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar-powered_Stirling_engine&quot; rel=&quot;nofollow noreferrer&quot;&gt;dish Stirling&lt;/a&gt;. Usually they are simulated using raytracing or Gaussian cone optics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, I would like to simulate the optical system and get the resulting image of the sun on some surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end I would like to calculate the &quot;flux distribution&quot; (~brightness) of the reflected radiation on some sort of receiver. The receiver could be a plate with heat pipes beneath or a Stirling motor. I would like to create images like &lt;a href=&quot;http://solarenergyengineering.asmedigitalcollection.asme.org/data/journals/jseedo/929668/sol_136_03_031013_f002.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://solarenergyengineering.asmedigitalcollection.asme.org/data/journals/jseedo/929668/sol_136_03_031013_f002.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, I read that Zernike polynomials can only be used to simulate optical systems with circular apertures. This is not the case for my application. Currently I am evaluating the optical system using Monte Carlo raytracing (GPU based) , which yields the very high accuracy that is needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have not done any work on this so far, but for very large mirror arrays, e.g. 25000 heliostats, the raytracer takes quite a while to converge. And calculation time is a very critical point when in comes to optimizing the positions of the mirrors to avoid the shading and blocking amongst them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why we are looking for faster ways to evaluate the optical performance of these systems while still getting highly accurate results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you think it is possible to simulate them using Zernike polynomials? The simulation needs to consider surface errors of the mirrors, astigmatism and non-ideal focal lengths.&#xA;Can one use Zernike polynomials only for arbitraty optical systems?&lt;/p&gt;&#xA;" OwnerUserId="20895" LastEditorUserId="20895" LastEditDate="2018-01-11T07:21:11.207" LastActivityDate="2018-01-11T09:00:27.477" Title="Is it possible to model a solar thermal concentrator using Zernike polynomials?" Tags="&lt;telescope&gt;&lt;optics&gt;&lt;mirror&gt;" AnswerCount="0" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="24477" PostTypeId="1" AcceptedAnswerId="24480" CreationDate="2018-01-10T17:26:56.110" Score="9" ViewCount="3313" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UZEZi.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UZEZi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you'll notice, space time starts to deform near the black hole, but then at considerable distance from the black hole' surface, images in the background become ultra-distorted, but then as you look even closer to the black hole, the distortion starts to look normal again almost like nothing is happening except for a minor fish-eye lens effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If space-time distorts in a monotonic way as you approach the surface of a black hole, how does this ring of ultra-distortion exist? Why doesn't the image look even more distorted at the surface than at that ring?&lt;/p&gt;&#xA;" OwnerUserId="20897" LastEditorUserId="1891" LastEditDate="2018-01-11T19:30:38.897" LastActivityDate="2019-05-30T19:58:39.133" Title="What's that ring that only now appears in black hole illustrations?" Tags="&lt;black-hole&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24478" PostTypeId="1" AcceptedAnswerId="25529" CreationDate="2018-01-10T18:32:02.107" Score="2" ViewCount="1872" Body="&lt;p&gt;I have always seen answers to this point to measurements of how &quot;flat&quot; local geometry is, but my problem is, thats space not spacetime. I would say its safe to say that the geometry of space in our universe is indeed euclidean, but space isn't over time. The further I look, space should be less, the farther I go, space should be more. To me this seems as if spacetime is hyperbolic along the temporal axis, and that every &quot;instance&quot; cross-section is euclidean. So that ultimately is the question, is space&lt;strong&gt;time&lt;/strong&gt; hyperbolic?&lt;/p&gt;&#xA;" OwnerUserId="16449" LastActivityDate="2018-03-17T15:49:59.393" Title="Is spacetime hyperbolic?" Tags="&lt;space-time&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24479" PostTypeId="2" ParentId="24276" CreationDate="2018-01-10T18:47:40.627" Score="1" Body="&lt;p&gt;It depends on if the stars in the constellation are lined up, on a plane or on a sphere. If they line up as a plane, traveling the distance you are from the plane, to the other side, it should look reversed. However, if its lined up on a sphere, it may look distorted, how the inside of a sphere looks to the outside. And if the distances vary enough, it might look somewhere in between or distorted in another way. Typically the distances vary greatly. Heres a nice picture of the kinds of distortion (note that stars almost never line up in any way, distance-wise, in constellations, so this is just mainly for the general idea): &lt;a href=&quot;https://www.physics.wisc.edu/museum/Exhibits/Optics_Color/3%20Mirrors/3Mirrors_1i.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Link&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16449" LastActivityDate="2018-01-10T18:47:40.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24480" PostTypeId="2" ParentId="24477" CreationDate="2018-01-10T19:36:28.243" Score="12" Body="&lt;p&gt;This is due to photons actually going in circles around the black hole multiple times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Far away from the BH, the image is only slightly distorted. These are photons that deviated only a little bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you get closer, you're seeing photons that deviated more and more. At some point, you'll see photons that came from behind you, went around the BH, and hit your eye. Eventually, close enough to the BH, you'll see photons that did a full 360 degree turn. The BH acts like a fisheye camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking closer and closer to the disk, you get photons that did 1.5 turns, 2 turns, 3 turns and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you see is concentric ring images, each one of them containing the full picture of everything around the BH but compressed in a ring. There's essentially an infinity of concentric images, but you only see clearly the first few of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://imgur.com/VZnkFxP&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://imgur.com/VZnkFxP&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted that the black disk in the middle is not the event horizon. It's simply a region in space where no light comes from that could reach your eye. The actual event horizon is about 2x smaller. You could think of it as the image of the event horizon &quot;magnified&quot; by the spacetime distortion, but the analogy is not perfect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A black hole is, in many ways, like a very strange &quot;lens&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="7411" LastEditDate="2019-01-29T12:06:11.320" LastActivityDate="2019-01-29T12:06:11.320" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="24481" PostTypeId="1" AcceptedAnswerId="24482" CreationDate="2018-01-11T01:18:01.190" Score="1" ViewCount="58" Body="&lt;p&gt;The BBC News item &lt;a href=&quot;http://www.bbc.com/news/science-environment-42614401&quot; rel=&quot;nofollow noreferrer&quot;&gt;Light shed on mystery space radio pulses&lt;/a&gt; is interesting enough in its own right. But I noticed that it is reporting on news from the &lt;a href=&quot;https://aas.org/meetings/aas231&quot; rel=&quot;nofollow noreferrer&quot;&gt;231st meeting of the American Astronomical Society&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These kinds of meetings are usually annual; for example the &lt;a href=&quot;https://www.aps.org/meetings/meeting.cfm?name=APSAM19&quot; rel=&quot;nofollow noreferrer&quot;&gt;120th Annual meeting of the American Physical Society&lt;/a&gt; will be in 2019.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Were there enough astronomers in the US in 1787 to have an American Astronomical Society and it's first meeting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it turns out that the answer is that meetings were more frequent than annual, why would that be? &lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-01-11T01:50:52.103" LastActivityDate="2018-01-11T01:59:41.487" Title="Is the American Astronomical Society 230 years old?" Tags="&lt;history&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24482" PostTypeId="2" ParentId="24481" CreationDate="2018-01-11T01:59:41.487" Score="5" Body="&lt;p&gt;The first meeting of the AAS was held in September 1899 in Williams Bay, Wisconsin. The meetings were held once a year initially. The frequency varied somewhat for the next 60 or so years. The meetings appear to have settled down to twice a year about 40 years ago, one in January and another in late spring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://aas.org/meetings/past-aas-meetings&quot; rel=&quot;noreferrer&quot;&gt;https://aas.org/meetings/past-aas-meetings&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2018-01-11T01:59:41.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24494" PostTypeId="2" ParentId="6580" CreationDate="2018-01-11T19:53:07.820" Score="1" Body="&lt;p&gt;@PeterU - I strongly recommend that you read several introductory books to cosmology and modern physics. Some of the difficulties you're encountering are stemming from the fact that you're looking at popularized science, hear some of the analogies they use to make the public understand difficult scientific ideas, but you're taking those analogies at face value, not realizing they're just metaphors. This is not to accuse you of anything - these are complex matters and it takes time to figure them out. I know it took me many years, and I recognize some of my bewildered musings from the past in the questions you're asking here. I think you're doing fine, just don't get stuck in analogies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few clarifications:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When scientists say the universe is &quot;flat&quot;, they don't mean it looks like a pancake. They just mean its geometry is euclidean, that's all. The word &quot;flat&quot; is perhaps poorly chosen (it makes sense if you know a lot of geometry, but not everyone does). There are non-euclidean geometries that mathematicians are studying, but our universe does not seem to be built based on them. It is built based on an euclidean metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Non-Euclidean_geometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Non-Euclidean_geometry&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe is infinite and it's the same in all directions. There is no privileged up, down, left, right, back, or forth. They're all the same. And it goes on forever and ever.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The analogy with the ant walking on a balloon is just an analogy. It's hard to visualize a 3D space in its entirety. Instead, you could visualize a 2D space (the surface of the balloon) and try to figure out what happens then. But never forget it's just an analogy, nothing more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The insect cannot fly off the balloon, because then the analogy breaks. The purpose of the analogy is to show what a 2D observer sees in a particular 2D universe, and then try to extend that knowledge back to our 3D universe. But if your insects are flying around willy-nilly, then the toy is broken and it doesn't do what it's supposed to be doing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Never forget the difference between models (balloons) and reality (the universe).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a set number of Stars that formed and our known Universe only&#xA;  extends to a certain distance. Beyond that there can be other&#xA;  Universes in an infinite Space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not exactly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This universe we live in is infinite, as far as we can tell. It just goes on forever and ever. What you call &quot;space&quot; is this universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there is something called the &quot;observable universe&quot;. That's the chunk of universe we can see. That's finite. We cannot see all the way to infinity, because light doesn't go that fast. However, this is just a little chunk in an infinite whole.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the Hubble telescope has picked up the edge of the Universe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, it has not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modern instruments are getting closer to the theoretical limits of the observable universe. In other words, we are getting so good at astronomy, now we can see almost as far away as theoretically possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That doesn't mean we're seeing &quot;the edge of the universe&quot;. It only means we're seeing as far as the laws of nature allow us to see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When I say observable Universe I am referring to everything that was formed from the &quot;Big Bang&quot; , so that is finite but the void or Space that it is expanding into is Infinite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's not what the term means. &quot;Observable universe&quot; simply means the universe chunk that we can see, that's all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You imagine that the Big Bang has occurred at a single point and went out from there. That is not the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an infinite universe and &lt;strong&gt;the Big Bang has / is / will occur everywhere&lt;/strong&gt;, in every single point in that entire infinity. All points of the universe where the Big Bang has occurred have undergone a lot of expansion, and are still doing it now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are thinking about the universe in terms of newtonian physics, where space and time are linear and unchangeable. That is not the case. It's impossible to understand modern theories about the universe in that context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, the Big Bang was not a single point that popped. Get rid of that idea. Instead, do this: pick any point in this infinite universe. Now go back in time. If you go back far enough, you will reach a moment when things in that point were super-dense and super-hot. Going forward in time from that moment, things have expanded and cooled down. &lt;strong&gt;This is true for every single point in this infinite universe!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not a simple expansion out from a single point. The Big Bang has not occurred in a privileged place in the past. Instead, the Big Bang is a property that every point of the universe had, way back in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe is not only infinite - every single little chunk of it is expanding out from a very dense and very hot past, towards a very sparse and cool future. No chunk is privileged. They are all the same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the Multi Universe ? If this Universe were infinite then how and where would the other Universes be then ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's an analogy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that this universe is 3D. But imagine for a moment that it's 2D instead. Imagine it's like an infinite sheet of paper, stretching out forever, but only in 2 dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, a hand's width above that infinite sheet of paper place another infinite sheet. And a little higher another one. And so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these sheets of paper are infinite, yet they can coexist just fine. The trick is - they all exist in a 3D space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, you could have an infinity of 3D universes all coexisting in a higher dimensional multiverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WARNING: This is not a &quot;theory&quot;. It is just a hypothesis. We have no proof that the multiverse exists. Some scientists have hypothesized that this is the case because it would help explain certain features of this universe. But so far there is no proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is normal and it's how science works. This is not idle dreaming. Scientists encounter some strange phenomena. They put forth a hypothesis. Other scientists look for proof. If proof is not found, the hypothesis is scrapped. If proof is found, then it becomes a proper theory. The multiverse idea is currently in that stage before proof; we don't know yet how things will turn out. That is all.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2018-01-11T20:54:22.360" LastActivityDate="2018-01-11T20:54:22.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24495" PostTypeId="2" ParentId="24276" CreationDate="2018-01-11T20:27:39.567" Score="6" Body="&lt;p&gt;As others have said, Orion (and other constellations) only looks the way it does from a specific angle. The stars are distributed at different distances from Earth, so from a different position, including directly &quot;behind&quot; it, it would look quite different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a (terrestrial) example of something that looks different depending on the angle that it's viewed from (sidewalk &quot;trompe l'oeil&quot; art is popular in some cities). There's only one position where it has the intended effect; even from the same angle, being closer or further would break the illusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0Gxjt.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0Gxjt.jpg&quot; alt=&quot;Sidewalk trompe l&amp;#39;oeil art of a snail&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18499" LastActivityDate="2018-01-11T20:27:39.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24496" PostTypeId="2" ParentId="24449" CreationDate="2018-01-11T23:59:42.573" Score="2" Body="&lt;p&gt;It is possible that you have not understood the definition of inclination angle. This is the angle between our line of sight and the orbital axis of the star-exoplanet pair. Maximum amplitude radial velocity variations are seen when the inclination angle is 90 degrees. No radial velocity variations are seen when it is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to your question is just that if the inclination angle is close to 90 degrees you will see transits. If not, then you won't. Nothing definite can be ascertained from the radial velocity measurements alone. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-01-14T15:56:01.173" LastActivityDate="2018-01-14T15:56:01.173" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24503" PostTypeId="2" ParentId="24477" CreationDate="2018-01-12T16:05:32.097" Score="7" Body="&lt;p&gt;I know you've already accepted an answer, but I want to provide a more technical answer that gets more into to the physics involved. If you're really curious to read more, and have a propensity for math, I'd suggest you take a look at the incomparable &lt;a href=&quot;https://arxiv.org/pdf/astro-ph/9606001v1.pdf&quot; rel=&quot;noreferrer&quot;&gt;Lectures on Gravitational Lensing by Narayan and Bartelmann&lt;/a&gt;. This will be the basis of much of my answer and, in fact, is the basis of many later treatises on gravitational lensing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, I want to say that Florin's description of photons circling the black hole and producing multiple rings is a correct idea and really does occur, but I don't believe is pertinent to your question because those rings aren't visible in the image you posted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, what you're seeing is an effect in gravitational lensing whereby images of objects are distorted, magnified, and duplicated. What occurs, in simplified terms, is that near the black hole, the local space-time metric is warped (in other words the black hole exerts gravity) due to the massive singularity in the black hole. This warping of space-time causes the path of light, which would otherwise travel along a straight line, to be bent. It is possible, with some basic assumptions, to calculate precisely how the image of a background light source is distorted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main and well-defined ring that you see is known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Einstein_ring#cite_note-21&quot; rel=&quot;noreferrer&quot;&gt;Einstein Ring&lt;/a&gt;. You can see this ring really well in the (simulated) image below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/q1zwm.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q1zwm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the simple case of a single, non-rotating black hole, the physics is simple enough to actually do the calculations outright (albeit with some simplifying assumptions, e.g., the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_lensing_formalism#Thin_lens_approximation&quot; rel=&quot;noreferrer&quot;&gt;thin lens approximation&lt;/a&gt;). As described in the Lectures linked above:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Any source is imaged twice by a point mass lens.  The two images are on either side of the source, with one image inside the Einstein ring and the other outside.  As the source moves away from the lens, one of the images approaches the lens and becomes very faint, while the other image approaches closer and closer to the true position of the source and tends toward a magnification of unity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So you can see that you get duplicate images of any background object, which can clearly be seen in the image above. At about 7 o'clock in the image you can see two stars (one reddish, one blueish) which are outside the Einstein ring, and then a second image at about 1 o'clock which is inside the Einstein ring. The Einstein ring itself is a special case where the objects exactly on the ring are directly behind the black hole (from the observer's perspective). In this special case, you don't get two images anymore, but rather you get a ring of light. As objects approach this ring (in other words, as they approach being directly behind the black hole), their image gets distorted - the scientific jargon in the lensing community is they get &lt;em&gt;sheared&lt;/em&gt; - and they slowly transform from a double image into the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why you see the distortion you described. As an object behind the black hole approaches your line of sight, it appears as two images, one far outside the Einstein ring and one (very small) image close to the event horizon. Then, as the object approaches your line of site, the images of it approach the Einstein ring from both sides, where it gets brighter and more distorted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea mentioned by Florin about photons circling is true, and in fact you actually see multiple Einstein rings, but the other rings are very close to the black hole and in general you would not observe them. You can see these other Einstein rings as a slight glow around the black hole in the image above.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2018-01-12T16:05:32.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24504" PostTypeId="1" CreationDate="2018-01-12T16:59:33.020" Score="1" ViewCount="51" Body="&lt;p&gt;This is from a translation of Ptolemy's Almagest&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence it necessarily follows that the straight line draw n from the centre of the earth (which is the centre of the ecliptic) through the centre of the moon- to a point on the ecliptic, which determines the true position ([as it does] for all bodies).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand what is means when is refers to &quot;a line drawn from the centre of the earth through the centre of the moon to a point on the ecliptic. If the moon is on the ecliptic why must this line go through it. If it is not on the ecliptic how can a straight line go through its centre to a point on the ecliptic?&lt;/p&gt;&#xA;" OwnerUserId="20839" LastEditorUserId="2365" LastEditDate="2018-01-12T20:05:08.437" LastActivityDate="2018-01-12T20:05:08.437" Title="What does Ptolemy mean by &quot;a line drawn from the centre of the earth through the centre of the moon to a point on the ecliptic&quot;" Tags="&lt;the-moon&gt;&lt;coordinate&gt;&lt;ecliptic&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24506" PostTypeId="2" ParentId="21372" CreationDate="2018-01-12T17:29:24.770" Score="3" Body="&lt;p&gt;There is &lt;a href=&quot;https://www.forbes.com/sites/quora/2016/12/29/how-often-do-meteoroids-hit-the-moon/#444947a6f2be&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; source that claims 2800 kg of meteor material lands on the moon everyday.  They break down that amount to determine an average area per strike. There have been other studies like &lt;a href=&quot;http://www.dailymail.co.uk/sciencetech/article-3835015/A-facelift-Moon-81-000-years.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt; and &lt;a href=&quot;http://www.skyandtelescope.com/astronomy-news/space-rocks-hit-moon-often-1310201623/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this report.&lt;/a&gt;  I sure can't vouch for the accuracy of this first report but it's from a reputable (NASA) source.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="16926" LastEditDate="2018-01-12T17:41:58.113" LastActivityDate="2018-01-12T17:41:58.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24507" PostTypeId="2" ParentId="24431" CreationDate="2018-01-13T03:12:36.820" Score="3" Body="&lt;p&gt;Your formula is a crude approximation. It's ok if you just want a rough idea of the number of orbits the Moon makes over a long period of time, but it's not useful for calculating the daily motion of the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon's orbit is fairly eccentric (about 0.0549), so its daily speed varies quite a bit. The Babylonians discovered that the mean daily motion is approximately 13° 10' 35&quot;, but that it can range from 11° 6' 35&quot; to 15° 14' 35&quot;. The Moon's motion is rather complex, the Wikipedia article on &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;lunar theory&lt;/a&gt; is a good introduction, with a rundown of the history, and quite a few relevant formulae. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The fact that the Moon's motion is so hard to compute accurately (compared to the motions of the Sun and the visible planets) is a mixed blessing. On the one hand, it would have been great for navigators if it had been easier to prepare Moon position tables. On the other hand, it made it rather obvious that early models of the celestial motions were incorrect. No matter what they tried, the early cosmologists weren't able to produce a model of the Moon's motions that was consistent with observation. Eg, in the best lunar theory of Ptolemy's model of the cosmos, which was good enough at estimating the Moon's position to use for eclipse prediction, the apparent size of the Moon ought to change by around 100% every month, which it obviously doesn't do. :) So if it weren't for the Moon's tricky orbit we might still be using Ptolemy's geocentric model of the cosmos today, and the whole scientific revolution may never have happened.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2018-01-13T03:12:36.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24508" PostTypeId="1" AcceptedAnswerId="24524" CreationDate="2018-01-13T03:36:11.767" Score="2" ViewCount="114" Body="&lt;p&gt;To my understanding, white holes are hypothetical objects that are time-reversed black holes; black holes gravitationally &quot;suck in&quot; matter whereas white holes &quot;spit&quot; matter back out. As such, I think it's a nice idea that the Big Bang could be the &quot;spitting of matter&quot; via a white hole (assuming such an object exists). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This got me wondering how one could experimentally test for such an object. Since the expansion rate of the universe is increasing, could one possible explanation be that a white hole is continually &quot;spitting out&quot; matter such that an increase of flux of in-falling matter/space could be observed as an increasing rate of expansion of out-pouring matter/space? While I don't know of a way to detect this, I was wondering whether the following idea was fruitful or not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that a supermassive black hole is believed to be at the center of every galaxy, given that most galaxies (sans Andromeda) are receding away at a speed proportional to their distance relative to us observers, and given that the spatial distribution of matter in our universe is isotropic and homogeneous, could one build a simulation of such a concept? To emulate isotropy and homogeneity, one could use a random number generator corresponding to a uniform distribution to place &quot;galaxies&quot; along a grid; accounting for gravitational forces, one can simulate the time-evolution of these galaxies and compare rates of in-falling matter to find a correlation between the rate of in-falling matter and the expansion rate of the universe (likely as a function of time-elapsed, and age and size per black hole). While the simulation does not definitely prove anything, providing a correlation could be interesting in my opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a stupid idea, or is there a potential to build upon this idea? If it is not fruitful, I am curious as to why.&lt;/p&gt;&#xA;" OwnerDisplayName="user15317" LastActivityDate="2018-01-14T00:00:14.490" Title="Is it fruitful to search for a correlation between rates of in-falling matter in black holes and the expansion rate of our universe?" Tags="&lt;black-hole&gt;&lt;big-bang-theory&gt;&lt;expansion&gt;&lt;white-hole&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2018-01-29T21:10:16.103" ContentLicense="CC BY-SA 3.0" />
  <row Id="24510" PostTypeId="1" CreationDate="2018-01-13T05:46:53.327" Score="5" ViewCount="1119" Body="&lt;p&gt;&quot;Researchers have largely put to rest a long debate on the underlying mechanism that has caused periodic ice ages on Earth for the past 2.5 million years – they are ultimately linked to slight shifts in solar radiation caused by predictable changes in Earth's rotation and axis.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we know that the amount of solar irradiation has increased since the ice age. Do current changes in the orbit path still cause the average temperature to still be increasing until another such event causes another orbit change? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://oregonstate.edu/ua/ncs/archives/2009/aug/long-debate-ended-over-cause-demise-ice-ages-%E2%80%93-may-also-help-predict-future&quot; rel=&quot;noreferrer&quot;&gt;This appears to be the article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="20942" LastEditorUserId="6841" LastEditDate="2018-01-13T13:02:31.873" LastActivityDate="2018-01-14T15:04:17.263" Title="Do changes in the orbit of the Earth contribute to global warming" Tags="&lt;orbit&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24511" PostTypeId="1" CreationDate="2018-01-13T05:51:25.557" Score="3" ViewCount="81" Body="&lt;p&gt;All of the planets in the TRAPPIST-1 system are all orbiting very closely and are somewhat large. Would it be possible for at least one of the planets to have a stable moon in orbit around it? Or would the gravitational interactions between the seven planets and the TRAPPIST system's star make it too unstable for any moons? Would it be possible for any planets orbiting red dwarfs closely to have moons?&lt;/p&gt;&#xA;" OwnerUserId="18831" LastActivityDate="2018-01-13T05:51:25.557" Title="Is it possible for any of the planets in the TRAPPIST-1 system to have moons?" Tags="&lt;trappist-1&gt;&lt;red-dwarf&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24512" PostTypeId="2" ParentId="24510" CreationDate="2018-01-13T06:35:18.280" Score="7" Body="&lt;p&gt;They found in 1976 that the axial tilt and precession changes matched the geologically known ice age time periods. (see this pic for precise details)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graphs say the world will get a tiny bit warmer for a few hundred years and then colder, if it weren't for CO2 being at the highest level for many million years:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ozJR6.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ozJR6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Figure 1:  Orbital Parameters:  Eccentricity, Precession and Obliquity- click for larger image&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ECd5n.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ECd5n.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is it measured? &lt;a href=&quot;https://www.geog.ucsb.edu/ideas/Insolation.html&quot; rel=&quot;noreferrer&quot;&gt;It's mostly straight mathematics,&lt;/a&gt; say if the earth is 10% nearer to the sun it casts a larger shadow into outer space and collects more solar radiation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This figure is developed from A.L.Berger, 1978, Long Term Variations of Daily Insolation and Quaternary Climatic Changes, Journal of the Atmospheric Sciences, volume 35 (12), 2362-2367.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UgtcF.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UgtcF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;M.F.Loutre and A.Berger, 2000, Future Climate Changes: Are we entering an exceptionally long interglacial?, Climatic Change 46, 61-90&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effects of galactic cosmic rays on the atmosphere (via cloud nucleation) and those due to shifts in the solar spectrum towards the ultraviolet (UV) range, at times of high solar activity, are largely unknown. The latter may produce changes in tropospheric circulation via changes in static stability resulting from the interaction of the increased UV radiation with stratospheric ozone. More research to investigate the effects of solar behaviour on climate is needed before the magnitude of solar effects on climate can be stated with certainty.&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2018-01-13T09:52:05.590" LastActivityDate="2018-01-13T09:52:05.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24517" PostTypeId="2" ParentId="24510" CreationDate="2018-01-13T14:47:06.233" Score="4" Body="&lt;p&gt;The other answer is very good, but just to touch on a few details.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do we know that the amount of solar irradiation has increased&#xA;  since the ice age.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The total amount of solar radiation hitting Earth on a given year doesn't change very much.  The more significant changes, on average, has to do with the angle that the solar radiation hits, or, more specifically the angle the Earth is tilted and when the Earth is closest or furthest from the Sun.  Earth is currently closest to the Sun in January and Furthest in June.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;During summer, the sun is higher in the sky and stays out a longer percentage of the day and in winter, the period of daylight is shorter and the sun is lower.  That drives the seasons, as I'm sure you knew.  The tilt favors one hemisphere or the other, so when it's winter in the Northern Hemisphere, it's Summer in the Southern.    The total solar energy doesn't change, but the angle that the Earth tilts towards or away from the sun matters a lot.  It's the difference between winter and summer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summer and winter still happen during ice ages, but the variation of the angle of the sun increases with an axial tilt increase (up to 24.5 degrees) and decreases when the tilt decreases (22.5 degrees) - see obliquity in the chart below.   That creates a larger fluctuation between the seasons.   22.5, milder seasons, 24.5, more extreme seasons.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WsLaF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WsLaF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This hemisphere that gets more heat matters because the Northern Hemisphere has a lot more land and the Southern, a lot more ocean.   When more sunlight hits land, land wars up.  When more sunlight hits ocean-water oceans heat more slowly, but you also get evaporation and circulation.   Oceans are very effective heat sinks, they warm slowly during the summer and cool slowly during the winter.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, it's the way the Earth is tilted and when it's tilted that drives ice ages.   Glaciers also grow and melt slowly, so it's a gradual period of transition.  The Earth's response to these orbital changes takes some time.  a couple thousand years or more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a general rule, ice ages begin and grow when Northern Hemisphere summers are colder.  Colder summers allows fallen snow to stay longer, and where the snow survives the entire summer, that's when it can begin to accumulate and grow and you get glacial expansion.   When Northern Hemisphere summers are hotter, the reverse happens and glaciers begin to retreat.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How cold the winters are doesn't matter much, because snow can build up at 2 degrees below freezing or 50 degrees below freezing, so it's not the colder winters, but the colder summers that drive ice ages and likewise, it's warmer summers that end ice ages.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because there's nowhere for ice to grow in the SOuthern Hemisphere, besides ANtarctica where it's effectively permanent, the Southern Hemisphere variation doesn't matter.  It's pretty much only the Northern Hemisphere Summer that determines if glaciers will grow or shrink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You get some glacial buildup in some locations in the Southern Hemisphere like Mt. Kilimanjaro and some glaciers in South America, but nothing large enough to make an ice age.  Ice ages, currently are a Northern Hemisphere event.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Northern Hemisphere Summer variation is largely driven by two Milankovich cycles, Axial Tilt and Axial Precession/Apsidal Precession, which effect the same thing.    Tilt and Precession can add up or cancel each other out.  The 3rd cycle, Eccentricity is the only cycle that actually does change the total energy the Earth receives and it can enhance or reduce the other two.  As the other answer explains, it's just math, and how these 3 cycles add up, though both precession and Eccentricity aren't neat cycles, there's some stretching and squashing of the wavelengths, only Axial Tilt operates like clockwork.   The effect is a somewhat messy, almost chaotic looking rise and fall of Northern Hemisphere Summer TSI.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to get back to your question, the total amount of solar radiation hasn't increased much, if at all, since the end of the last ice age and that's not the key factor anyway.   It's the Northern Hemisphere summer radiation that matters most and that's actually been decreasing for the last 11,000 years since the last ice age, not increasing.  Glaciers take a long time to melt and there are various feed back mechanisms also in place, so temperature doesn't neatly follow the variation.   There's usually thousand to a few thousand years lag-time, and smaller variations might have little to no effect at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to think about it Think of a see-saw that requires a kick to move.  Cold wants to stay cold and Warm wants to stay warm.   Like the kid on the bottom of the sea-saw leans backwards and so angular momentum keeps him on the ground and he needs to kick upwards for the seasaw to move. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a shift to occur and glaciers to reverse, the solar variation needs to be enough to trigger a change over and above any feedback mechanisms in place.   That's why ice ages don't neatly follow the variation.   It's also why CO2 lags behind temperature, but that's another discussion.    &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do current changes in the orbit path still cause the average&#xA;  temperature to still be increasing until another such event causes&#xA;  another orbit change?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Currently Earth is in a relatively flat period in Northern Hemisphere Summer TSI.  Over the last 10,000 years or so, the Orbital changes have actually been cooling the Earth and beginning in the next couple thousand there will be a slight warming trend.   Orbital changes are not currently warming the Earth, they should, very slightly, have a cooling effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see a nice display of the warming followed by cooling of the Earth &lt;a href=&quot;https://xkcd.com/1732/&quot; rel=&quot;nofollow noreferrer&quot;&gt;in this chart here&lt;/a&gt;.  While it's got some cartoonist comments, it's actually a very good and science supported chart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And borrowing the chart from the other answer, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UgtcF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UgtcF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that the peak Northern Hemisphere summer solar radiation occurred around 11,000 years ago, but the temperature didn't peak until around 9,000 years ago.   As I mentioned above, this is because the Earth is slow to respond to the somewhat small orbital variations and because glaciers take a long time to melt.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current level of peak glacial melt and peak sea level wasn't reached until a bit less than 8,000 years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gy25b.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gy25b.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the last 11,000 years the Northern Hemisphere Summer TSI has been decreasing.   More recently and for the next couple thousand years there's been a leveling off and after that, in two thousand years or so, there will be a turn towards warming again, lasting 15,000 years or so.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No major orbital effects are anticipated for several tens of thousands of years, also noted in the other answer that we're in the middle of an unusually long between ice ages period, and any chance we might see a mini-ice ages have been thwarted by our greenhouse gas output.  These charts on orbital variation and Northern Hemisphere summer TSI are effectively moot with 400 PPM CO2.  They really only apply with pre-industrial levels.   400 PPM CO2 is probably enough to prevent any chance of an ice age.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/05ADi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/05ADi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The low points on the chart indicate a cooling push and the higher points a warming.   We're not do for a somewhat significant orbital cooling trend for about 60,000 years.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-01-13T14:47:06.233" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24518" PostTypeId="1" AcceptedAnswerId="24519" CreationDate="2018-01-13T16:20:18.080" Score="4" ViewCount="1113" Body="&lt;p&gt;&lt;a href=&quot;http://exoplanets.eu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://exoplanets.eu/&lt;/a&gt; displays R.A. on screen in hh:mm:ss format but when you export the table, it gives the R.A. as a decimal.  Likewise for Declination.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I turn for example 24.35417 (WASP-18) into 01:37:25.0 ?  And likewise for Declination -45.67778 to -45:40:40?  I'm looking for a method, not a website that will do it for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either a way to convert between formats or how do I export in non-decimal format?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="5401" LastActivityDate="2018-01-14T00:52:30.187" Title="Convert a Decimal into RA or Dec" Tags="&lt;star&gt;&lt;declination&gt;&lt;right-ascension&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24519" PostTypeId="2" ParentId="24518" CreationDate="2018-01-13T17:23:55.500" Score="4" Body="&lt;p&gt;DecRA is the decimal right ascension&#xA;RAh, RAm, RAs are the hms form&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$${\rm DecRA} = {\rm RAh}\times 15.0 + {\rm RAm}/4.0 + {\rm RAs}/240.0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ {\rm RAh} = {\rm INT}({\rm DecRA}/15.0) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\rm RAm} = {\rm INT}(({\rm DecRA}-{\rm RAh}\times 15.0)\times 4.0)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\rm RAs} = ({\rm DecRA}-{\rm Rah}\times 15.0 - {\rm RAm}/4.0)\times 240.0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where INT is the operation that &lt;em&gt;truncates&lt;/em&gt; to an integer&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;e.g. DecRA=24.35417&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\rm RAh} = {\rm INT}(24.35417/15.0) = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\rm RAm} = {\rm INT}((24.35417 - 1\times 15.0)\times 4.0) = 37$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\rm RAs} = (24.35417-1\times 15.0 - 37/4.0)\times 240.0 = 25.00$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DecDE is the decimal declination&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DEd, DEm, DEs are the dms form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;posneg is -1.0 for a position below the celestial equator (negative declination) and +1.0 for above the equator&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$${\rm DecDE} = {\rm DEd} + {\rm posneg}\times {\rm DEm}/60.0 + {\rm posneg}\times {\rm DEs}/3600.0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\rm DEd} = {\rm INT}({\rm DecDE})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\rm DEm} = {\rm INT}(({\rm DecDE} - {\rm posneg}\times {\rm DEd})\times 60.0\times {\rm posneg})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\rm DEs} = ({\rm DecDE} - {\rm posneg}\times {\rm DEd} - {\rm posneg}\times {\rm DEm}/60.0)\times 3600.0\times {\rm posneg}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;e.g. DecDE = -45.67778   posneg=-1.0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\rm DEd} = {\rm INT}(-45.67778) = -45$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\rm DEm} = {\rm INT}((-45.67778 - (-1.0)\times 45)\times 60.0\times (-1.0)) = 40$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\rm DEs} = (-45.67778 - (-1.0)\times 45 - (-1.0)\times 40/60.0)\times 3600\times (-1.0) = 40.0$$&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-01-14T00:52:30.187" LastActivityDate="2018-01-14T00:52:30.187" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24520" PostTypeId="1" CreationDate="2018-01-13T20:33:27.417" Score="1" ViewCount="53" Body="&lt;p&gt;I was wondering whether it has proven whether the universe is expanding and if so, what is its ratio, one more question that whether it has an effect for the centrifugal forces, i.e gravity and the weight of the planets and far- away stars. Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="20933" LastActivityDate="2018-01-13T20:33:27.417" Title="Modern Astrophysics" Tags="&lt;star&gt;&lt;astrophysics&gt;&lt;milky-way&gt;" AnswerCount="0" CommentCount="7" ClosedDate="2018-01-14T01:33:24.603" ContentLicense="CC BY-SA 3.0" />
  <row Id="24521" PostTypeId="1" AcceptedAnswerId="24523" CreationDate="2018-01-13T21:49:53.563" Score="2" ViewCount="403" Body="&lt;p&gt;When I was a kid I had several beautiful encyclopedias. In one of them there was a description of moon orbiting on of the gas giants that had huge ripples all along its surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It said in the text that it's plausible to think that the moon was torn apart and then got stuck back together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you remind me which one is it? I would like to read a little more about it.&lt;/p&gt;&#xA;" OwnerUserId="1135" LastActivityDate="2018-01-13T22:37:26.953" Title="Which moon has a scar as if it was torn in half and put back together?" Tags="&lt;natural-satellites&gt;&lt;gas-giants&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24522" PostTypeId="2" ParentId="24521" CreationDate="2018-01-13T22:10:31.080" Score="1" Body="&lt;p&gt;You are probably thinking of Miranda, a moon of Uranus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is currently thought that the varied terrain (including the 10km high cliff called Verona Rupes) is more likely due to past geological activity driven by tidal heating, although a massive impact has not been completely discounted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The impact hypothesis was first suggested after the Voyager fly-by in 1986.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-01-13T22:10:31.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24523" PostTypeId="2" ParentId="24521" CreationDate="2018-01-13T22:16:52.827" Score="2" Body="&lt;p&gt;I believe the moon in question is &lt;a href=&quot;https://en.wikipedia.org/wiki/Miranda_(moon)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Miranda&lt;/a&gt;, as James K suggested. It is hypothesized that it suffered a collision with another large body sometimes in the past. There are several large features that suggest this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Three large grooves, each &gt; 200 km long, cover parts of the surface (the &quot;scar&quot; you talked about). These are the &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_geological_features_on_Miranda#Coronae&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;coronae&lt;/em&gt;&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Observations indicate that some parts of the surface appear less than 100 million years old.&lt;/li&gt;&#xA;&lt;li&gt;Verona Rupes, a 10 km-high cliff, is the largest in the Solar System. Other, smaller, &lt;a href=&quot;https://en.wikipedia.org/wiki/Rupes&quot; rel=&quot;nofollow noreferrer&quot;&gt;rupes&lt;/a&gt; exist elsewhere on Miranda and across the Solar System, but not quite as high as this one.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Alternate explanations include heavy crustal activity in the past. Tidal effects from Uranus may also have played a role (either at different times than the putative impact, or instead of it entirely). Gas giants strongly interact with their moons through tidal forces; cases include Io's volcanism, driven by Jupiter, and the covered oceans of Europa and Enceladus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://voyager2.dvc.edu/faculty/kcastle/Planetary%20Features.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is a partial map of Miranda&lt;/a&gt;, displaying the coronae and several other features:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mqOHY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mqOHY.jpg&quot; alt=&quot;Partial map of Miranda&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2018-01-13T22:37:26.953" LastActivityDate="2018-01-13T22:37:26.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24524" PostTypeId="2" ParentId="24508" CreationDate="2018-01-14T00:00:14.490" Score="4" Body="&lt;p&gt;As a layman, I can tell you that there are some issues with your proposal.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest one is that White holes probably don't exist.  They're purely theoretical and, even worse than that, they're mostly not even well defined theoretical objects.  For a physical hypothesis to work, every element of the theory needs to be precisely defined.  White holes aren't well defined.  (now, let me add that, perhaps White holes are well defined by &lt;a href=&quot;https://en.wikipedia.org/wiki/White_hole&quot; rel=&quot;noreferrer&quot;&gt;Einstein's field equations&lt;/a&gt;, I'm not sure, but the specifics go above my paygrade).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And running time in reverse, a black hole really doesn't look anything like a big bang, and, granted, we don't know what the/a big bang looked like, but we know what a black hole swallowing matter looks like.   It absorbs matter by accretion disk and it shoots out jets along the accretion disks axis.   Running time backwards you can see streams of matter and the accretion disk reforming into stars and dust that flies away from the black hole but that gravitational consumption in reverse That looks NOTHING like the big bang is thought to have looked.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moments after the big bang are believed to have been a very hot, dense and rapidly expanding quark/photon/fundamental particle soup kind of thing.   Nothing in today's universe, backwards or forwards in time, looks anything close to that, so running time backwards doesn't really help.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The very early universe and the ratio of hydrogen to helium and the presence and near perfect uniformity of the background radiation all imply a mostly uniform, very hot, very dense, and only the slightest bit lumpy rapidly expanding origin, that only works under modern physical laws if space expands faster than the speed of light.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can re-create those conditions, but only for the tinyest fractions of a second, and with only a handful of particles, in particle accelerators. You might think that inside a black hole, similar conditions exist, but that's not entirely true.  Inside a black hole, gravity is dominant.   Due to faster than light expansion and uniformity, it's not believed that gravity was a dominant force in the moments after the big bang.  It was a force, but not a dominant force.   The space of space-time, for example, is very different inside a black hole than in the moments after the big bang. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A final issue with your question is whether the big bang was an explosion, or, more specifically that it originated from a source-point-origin at all, (as it's often presented), or whether it was an expansion that took place everywhere (everywhere stretch, so to speak).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A theoretical white hole ejecting matter would be a source point and a source point is subject to the inverse square rule and we'd expect to see a variation in density and evidence of a direction towards the origin.   We don't see that.  The observable universe appears remarkably homogeneous, which is only possible, from a source-point, if the size is enormously large.   The rate that distant galaxies move away from us is also very difficult to explain without some kind of dark energy expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variations in the image are exaggerated.  The cosmic background radiation is very homogeneous, implying we were inside the expansion, not part of some material that was ejected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ty5de.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ty5de.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.newscientist.com/article/dn8862-best-ever-map-of-the-early-universe-revealed/&quot; rel=&quot;noreferrer&quot;&gt;Source.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that you didn't say explosion specifically, but asked about a source object ejecting material, but both have a similar effect.  Ejected material spreads out from it's source.   The universe doesn't appear to have been formed in that way.   &lt;a href=&quot;https://profmattstrassler.com/articles-and-posts/relativity-space-astronomy-and-cosmology/history-of-the-universe/big-bang-expansion-not-explosion/&quot; rel=&quot;noreferrer&quot;&gt;Nice summary of the expansion theory here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's my layman's understanding anyway and I'm not sure I should have gone there.  Origin of the universe questions are kind of out there.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-01-14T00:00:14.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24528" PostTypeId="2" ParentId="24478" CreationDate="2018-01-14T14:42:53.613" Score="0" Body="&lt;p&gt;As OP still didn't react to the posted answers, I'll try my shot at this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, I'm not sure what you mean by hyperbolic.&#xA;A simple Hyperbola $f(x,y) \in R^2$ can be defined by a relation  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 - y^2 = a^2$$ with some constant $a^2$. &lt;a href=&quot;https://www.wolframalpha.com/input/?i=plot%20x%5E2-y%5E2%20%3D%201&quot; rel=&quot;nofollow noreferrer&quot;&gt;One can plot this.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the simple Minkowski spacetime metrics from special relativity states that there is a quantity $ds^2$ which stays invariant if an object is observed from different points in spacetime. This invariant quantity can be locally computed from&lt;br&gt;&#xA;$$ ds^2 = c^2 dt^2 - dx^2$$&#xA;which is a hyperbolic relation and also results from general relativity (the more general variant of special relativity) if one imposes a flat universe.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now Planck-observations show our universe to be approximately flat, expect for the ripples here and there. By ripples I mean sources of gravity, near which the relation between $dt$ and $dx$ becomes more complicated, and in particular non-hyperbolic.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you want to make any statement about 'spacetime', you can only do that about the 'spacetime of our current idea of the universe on a global scale'.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2018-01-14T14:42:53.613" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24529" PostTypeId="2" ParentId="24510" CreationDate="2018-01-14T15:04:17.263" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How do we know that the amount of solar irradiation has increased since the ice age.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Minor point: The Earth has been in an ice age for the last 2.6 million years, and still is in that ice age. You are asking instead about solar irradiation since the &lt;a href=&quot;https://en.wikipedia.org/wiki/Last_glacial_period&quot; rel=&quot;nofollow noreferrer&quot;&gt;most recent glacial period&lt;/a&gt;. This distinction is not just semantic. The Earth has been hot during much of the last 2.5 billion years. During those hothouse Earth periods, palm trees have grown even in polar regions (but obviously not a couple of billion years ago), with snow and ice only occurring at extremely high elevations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Major point: Solar irradiation at the top of the atmosphere has not increased since the end of the last glacial period. What has changed is the amount of solar irradiation at the Earth's surface at high northern latitudes during summer. The northern and southern hemispheres currently have very different distributions of land mass. The far north, with the exception of the Arctic Ocean, is mostly land, while the far south, with the exception of Antarctica, is mostly ocean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Climatologists use 65° north latitude as the bellwether that indicates how long term climate behaves. Glacial periods currently start (where &quot;currently&quot; means the last few million years) when summertime temperatures at those high northern latitudes remain mostly below freezing. Those very cold summertime temperatures means winter snow doesn't melt during the summer. It instead builds up over the years, and over the course of millennia, it covers the land and moves southward. Glacial periods end when summertime temperatures at high northern latitudes rise well above freezing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As explained in the other answers, the Milankovich cycles rank very high amongst the factors that drive those 65° N summertime temperatures. Other factors include&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The distribution of land over the Earth's surface. Ice ages have only occurred when there was/is a good amount of land near the poles. During the last 540 million years, periods of the Earth's history with no high latitude land masses have inevitably had hothouse as opposed to icehouse conditions.&lt;/li&gt;&#xA;&lt;li&gt;The amount of CO&lt;sub&gt;2&lt;/sub&gt; in the atmosphere. Ridiculously high CO&lt;sub&gt;2&lt;/sub&gt; levels have kept the Earth warm, even when the Milankovich cycles and land distribution would otherwise have favored ice formation.&lt;/li&gt;&#xA;&lt;li&gt;Hysteresis effects. Glaciations during the current ice age have changed from a 40000 year cycle to a 100000 year cycle, which many attribute to hysteresis effects.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2018-01-14T15:04:17.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24530" PostTypeId="1" AcceptedAnswerId="24548" CreationDate="2018-01-14T17:06:07.437" Score="5" ViewCount="1720" Body="&lt;p&gt;I'm prototyping a real time solar system / milky way discovery app.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you please help me on:&lt;br&gt;&#xA;-What is the best &lt;strong&gt;orbital data public API&lt;/strong&gt; I can use?&lt;br&gt;&#xA;-What is the &lt;strong&gt;standard numerical way to define orbit&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for something like &lt;a href=&quot;https://ssd.jpl.nasa.gov/sbdb.cgi#top&quot; rel=&quot;noreferrer&quot;&gt;JPL Small-Body Database Browser&lt;/a&gt;, but for every celestial objcets, and / or missions.&lt;/p&gt;&#xA;" OwnerUserId="20961" LastEditorUserId="20961" LastEditDate="2018-01-14T21:04:18.097" LastActivityDate="2018-01-16T01:40:56.470" Title="Solar system data API?" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;milky-way&gt;&lt;near-earth-object&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24531" PostTypeId="1" CreationDate="2018-01-14T17:25:56.653" Score="2" ViewCount="311" Body="&lt;p&gt;I want to write a program which, when given a date, a time, latitude and longitude, returns a list of the planets, Messier objects and NCC catalogue objects what are visible in the sky at that time and place?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any APIs which can do this or would I have to calculate the positions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Preferably, this should be accessible using JavaScript&lt;/p&gt;&#xA;" OwnerUserId="1494" LastEditorUserId="37191" LastEditDate="2021-04-28T09:10:48.723" LastActivityDate="2021-04-28T09:10:48.723" Title="How do you find the positions of celestial objects in the sky?" Tags="&lt;observational-astronomy&gt;&lt;astrometry&gt;&lt;javascript&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24532" PostTypeId="2" ParentId="11082" CreationDate="2018-01-14T18:58:30.977" Score="0" Body="&lt;p&gt;&lt;strong&gt;Short Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is no. The main sequence is a sequence in mass (and not a sequence in time).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most massive stars are located at the the top left (since they are the brightest and hottest/bluest). The lowest mass stars are are located at the bottom left ( since they dimmer and cooler/redder). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following the main sequence from the top left to the bottom right is thus a sequence from high to low mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A little bit more background&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The astronomers Hertzsprung and Russel were among the first to note that the brightness and colors of stars are not just random, but that the large majority of stars show a narrow relation between brightness and color.  The brightest stars are typically bluer (= hotter) and the dimmer stars are typically redder (= cooler).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When plotting the properties of stars in a diagram that shows the brightness on the vertical axis and the color (or temperature) on the horizontal axis, it turns out that the large majority of stars lie in a fairly narrow strip in this diagram. We call this stip the main sequence, simply because most stars lie on it. ( There are exceptions, for example red giants and white dwarfs do not lie on this sequence, but these are more rare). We now call this diagram the Hertzsprung-Russel diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most stars lie on this sequence because they spend about 90% of their life time there, without changing much. The sun is also one of the many stars on the main sequence. All stars on the main sequence are powered by nuclear fusion of hydrogen in their hot centers. This is such an efficient source of fuel for a star, that it lasts for 90% of its life. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computer models helped astronomers to understand how stars move through the Hertzsprung-Russel diagram when they get older.  When stars run out of hydrogen fuel in their centers they start changing and they leave the main sequence. This is when they can grow to become red giants. These changes relatively fast. This is why we do not see many stars away from the main sequence.  The tracks of how stars move through the diagram as they grow older are called evolutionary tracks.  These evolutionary tracks can be thought of as a sequences of time.  &lt;/p&gt;&#xA;" OwnerUserId="20962" LastActivityDate="2018-01-14T18:58:30.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24533" PostTypeId="1" CreationDate="2018-01-14T21:38:31.697" Score="3" ViewCount="137" Body="&lt;p&gt;If sometime in the distant future, the Universe stopped expanding, would this mean that the Universe was no longer infinite?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastEditorUserId="2365" LastEditDate="2018-01-14T21:54:46.203" LastActivityDate="2018-01-14T21:54:46.203" Title="Universe becoming finite" Tags="&lt;universe&gt;&lt;infinite&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24534" PostTypeId="2" ParentId="24533" CreationDate="2018-01-14T21:54:25.523" Score="3" Body="&lt;p&gt;If the density of the universe is high enough then the mass of the universe would be enough to eventually stop expansion and cause the universe to contract (the critical density is about $10^{-26}\, \mathrm{kg\, m^{-3}}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the case (it's probably not, the universe is probably flat, with dark energy causing expansion to speed up) then the universe is finite in time and always has been. The universe doesn't stop being &quot;infinite&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-01-14T21:54:25.523" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24537" PostTypeId="1" CreationDate="2018-01-15T09:35:54.380" Score="4" ViewCount="628" Body="&lt;p&gt;I've heard that $\mathrm{H\beta}$ and $\mathrm{H\delta}$ lines in galaxy spectra might be either emission lines or absorption lines, is that correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://en.wikipedia.org/wiki/Balmer_series#Role_in_astronomy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, Balmer lines are usually emission lines in the spectra of spiral and irregular galaxies probably due to the presence of type A stars. If ellipticals lack hot (O, B and A type?) stars, and are gas poor(?), then is it the remaining stars doing the absorbing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, are the CaII-H and -K doublet ever emission lines?&lt;/p&gt;&#xA;" OwnerUserId="20275" LastEditorUserId="11143" LastEditDate="2018-01-16T15:51:44.840" LastActivityDate="2018-01-16T15:51:44.840" Title="In galaxy optical spectra, why are Hbeta and Hdelta sometimes absorption lines?" Tags="&lt;galaxy&gt;&lt;spectra&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24540" PostTypeId="1" AcceptedAnswerId="24545" CreationDate="2018-01-15T13:32:03.520" Score="2" ViewCount="313" Body="&lt;p&gt;What is the standard coordination system used to describe the location at a celestial object approximately due east and at an angle of 10 degrees above the horizon?&lt;/p&gt;&#xA;" OwnerUserId="20976" LastActivityDate="2018-01-15T17:39:47.163" Title="Celestial Coordinates" Tags="&lt;coordinate&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24541" PostTypeId="2" ParentId="24540" CreationDate="2018-01-15T14:22:03.507" Score="3" Body="&lt;p&gt;There isn't one, because there cannot be one: which celestial object you see towards the east at 10 degrees above the horizon depends on:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Your position on Earth&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The time of day&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The time of year&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Because of Earth's rotation and motion around the sun, the positions of celestial objects change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among themselves, stars keep (for our purpose here) their relative position, and astronomical coordinate systems make use of that fact. The celestial sphere on which we project the stars rotates roughly around the North Star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Angular distance from the North Star (or more correctly, from the celestial equator) is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Declination&quot; rel=&quot;nofollow noreferrer&quot;&gt;declination&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you know from Earth's coordinate system, the position of the Null Meridian is arbitrary. On Earth, it's in London, in the sky, it's in the constellation pisces. This is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Right_ascension&quot; rel=&quot;nofollow noreferrer&quot;&gt;right ascension&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More information: &lt;a href=&quot;https://en.wikipedia.org/wiki/Equatorial_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that, depending on the application, other coordinate systems may be used in astronomy (such as galactic or heliocentric).&lt;/p&gt;&#xA;" OwnerUserId="13681" LastEditorUserId="13681" LastEditDate="2018-01-15T15:28:04.207" LastActivityDate="2018-01-15T15:28:04.207" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24543" PostTypeId="2" ParentId="24537" CreationDate="2018-01-15T15:28:36.120" Score="2" Body="&lt;p&gt;In the absence of emission lines from ionized gas, the (optical) spectrum of a galaxy will be the sum of the spectra of individual stars. These spectra will have absorption lines for hydrogen (H-alpha, H-beta, H-gamma, H-delta, etc.); the absorption lines are strongest for A stars and weaker for both earlier (O, B) and later (F, G, K, M) stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If ionized gas &lt;em&gt;is&lt;/em&gt; present (as is typical for spiral and irregular galaxies), then there will be hydrogen lines in emission. (This is normally due to O and B stars, which ionize gas with their strong UV emission; you've misread what the Wikipedia article actually says about A stars.) The emission is strongest for H-alpha, weaker for H-beta, and weaker still for the subsequent Balmer lines. This means that emission from H-alpha and (usually) H-beta overwhelms the stellar H-alpha and H-beta absorption, but H-gamma, H-delta, etc. can often still be seen in absorption because the emission from those lines is too weak to &quot;fill in&quot; the absorption lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elliptical galaxies usually have only F or G and later stars present; they usually lack both the gas &lt;em&gt;and&lt;/em&gt; the UV-emitting O and B stars necessary to ionize the gas, so you see only the stellar absorption lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Ca II H and K lines &lt;em&gt;can&lt;/em&gt; be seen in emission from individual M stars, due to emission from their chromospheres being strong enough to overwhelm the absorption in the photospheric spectrum (e.g., &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0602293&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/astro-ph/0602293&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2018-01-15T15:28:36.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24545" PostTypeId="2" ParentId="24540" CreationDate="2018-01-15T17:39:47.163" Score="4" Body="&lt;p&gt;The coordinate system for an object in the sky measured relative to the horizon is known as altitude-azimuth. The altitude is the angle above the horizon (10 degrees in your case), and the azimuth is the angle around the horizon, usually measured from North. (Due east would be 90 degrees azimuth in your example.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://en.wikipedia.org/wiki/Horizontal_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Horizontal Coordinate System on Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2018-01-15T17:39:47.163" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24548" PostTypeId="2" ParentId="24530" CreationDate="2018-01-16T01:40:56.470" Score="1" Body="&lt;p&gt;Here is a short answer. @barrycarter's helpful &lt;a href=&quot;https://astronomy.stackexchange.com/questions/24530/solar-system-data-api/24548#comment41933_24530&quot;&gt;comment&lt;/a&gt; links to a &lt;a href=&quot;https://astronomy.stackexchange.com/a/13489/7982&quot;&gt;much larger&lt;/a&gt; list of resources. Also, you may find answers below [this question] helpful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;a href=&quot;https://naif.jpl.nasa.gov/naif/toolkit.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Spice&lt;/a&gt; kernels for many (but not all) exploratory spacecraft as well, you can view them through the &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;Horizons&lt;/a&gt; interface or find them via ftp. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For most objects in Earth orbit, &lt;a href=&quot;https://en.wikipedia.org/wiki/Two-line_element_set&quot; rel=&quot;nofollow noreferrer&quot;&gt;Two Line Elements&lt;/a&gt; (TLEs) from &lt;a href=&quot;https://celestrak.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Celestrak&lt;/a&gt; or &lt;a href=&quot;https://www.space-track.org/auth/login&quot; rel=&quot;nofollow noreferrer&quot;&gt;Space-Trakck&lt;/a&gt; are useful. See &lt;a href=&quot;https://space.stackexchange.com/a/24445/12102&quot;&gt;this answer&lt;/a&gt; as well.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-01-16T01:40:56.470" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24549" PostTypeId="2" ParentId="23996" CreationDate="2018-01-16T01:58:32.043" Score="1" Body="&lt;p&gt;@JamesK's &lt;a href=&quot;https://astronomy.stackexchange.com/a/24009/7982&quot;&gt;excellent answer&lt;/a&gt; is probably what you are looking for. However, to address the question of how small a location is possible, the answer is highly dependent on how you define &quot;zenith&quot; or &quot;up&quot; in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you have three main choices:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The geometric vector pointing away from the center of mass of the Earth and passing through you. &lt;/li&gt;&#xA;&lt;li&gt;The vector pointing opposite the direction that local gravity is pulling you (the direction a &lt;a href=&quot;https://en.wikipedia.org/wiki/Plumb_bob&quot; rel=&quot;nofollow noreferrer&quot;&gt;plumb bob&lt;/a&gt; would indicate)&lt;/li&gt;&#xA;&lt;li&gt;The local perpendicular to a &lt;a href=&quot;https://en.wikipedia.org/wiki/Reference_ellipsoid&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference ellipsoid&lt;/a&gt; such as &lt;a href=&quot;https://en.wikipedia.org/wiki/World_Geodetic_System#A_new_World_Geodetic_System:_WGS_84&quot; rel=&quot;nofollow noreferrer&quot;&gt;WGS 84&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Wikipedia says that &lt;a href=&quot;https://en.wikipedia.org/wiki/Zenith&quot; rel=&quot;nofollow noreferrer&quot;&gt;zenith&lt;/a&gt; is the second one (opposite local gravitational &quot;down&quot;) which means that the answer would depend on how accurately you could measure this. If you had a plumb bob that could define your local zenith to say 0.01 degree (0.000175 radian) then using 6371 km as an average radius of the Earth, you could only define the location to about 1 kilometer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you used a &lt;em&gt;geometric&lt;/em&gt; or &lt;em&gt;geodetic&lt;/em&gt; definition, and were calculating the zenith mathematically, then you could obtain a location to an accuracy of centimeters, again, mathematically. But this may or may not have any practical use.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; from &lt;a href=&quot;https://en.wikipedia.org/wiki/File:GRAVIMETRIC_DATUM_ORIENTATION.SVG&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o9YNP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o9YNP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-01-16T02:14:35.273" LastActivityDate="2018-01-16T02:14:35.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24554" PostTypeId="1" CreationDate="2018-01-16T07:45:58.473" Score="1" ViewCount="39" Body="&lt;p&gt;What will happen to us?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter, for example, take away comets from their path, so they will not hit earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What else can/cannot happen? Have they any purpose for us?&lt;/p&gt;&#xA;" OwnerUserId="20727" LastActivityDate="2018-01-16T07:45:58.473" Title="What if Jupiter and other gas planets suddenly disappear?" Tags="&lt;gas-giants&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2018-01-16T09:25:32.623" ContentLicense="CC BY-SA 3.0" />
  <row Id="24556" PostTypeId="1" CreationDate="2018-01-16T08:15:13.230" Score="2" ViewCount="61" Body="&lt;p&gt;Also what is the accepted or most common way to typeset the wavelengths of a doublet? e.g. lambdalambda-3727,3729 with or without a hyphen? and without the Angstrom symbol?&lt;/p&gt;&#xA;" OwnerUserId="20275" LastActivityDate="2018-01-16T08:15:13.230" Title="Who first used two lambdas to typeset the wavelengths of a spectra line doublet?" Tags="&lt;spectra&gt;&lt;notation&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24560" PostTypeId="1" CreationDate="2018-01-16T15:54:45.250" Score="1" ViewCount="33" Body="&lt;p&gt;What is the difference between a black hole and a super massive black hole?&lt;/p&gt;&#xA;" OwnerUserId="21002" LastActivityDate="2018-01-16T15:54:45.250" Title="Black holes and super massive black holes" Tags="&lt;black-hole&gt;&lt;supermassive-black-hole&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2018-01-16T22:26:59.913" ContentLicense="CC BY-SA 3.0" />
  <row Id="24561" PostTypeId="1" AcceptedAnswerId="24564" CreationDate="2018-01-16T19:43:24.333" Score="2" ViewCount="231" Body="&lt;p&gt;Does the path of a rocket or anything going straight up look curved because the mass of the earth is curving space-time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume the earth, the rocket and all the celestial bodies whose gravitational field reaches the earth to be massless then will the rocket follow truly a straight line, because there would be nothing curving the space-time?&lt;/p&gt;&#xA;" OwnerUserId="21007" LastEditorUserId="2365" LastEditDate="2018-01-16T22:12:28.820" LastActivityDate="2018-01-16T22:26:24.990" Title="do rockets follow a straight path but is looks curved because of space time?" Tags="&lt;space-time&gt;&lt;general-relativity&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24563" PostTypeId="1" AcceptedAnswerId="24585" CreationDate="2018-01-16T21:48:27.840" Score="2" ViewCount="103" Body="&lt;p&gt;I am trying to get a list of all of Mercury's Perihelia and Aphelia over a period of multiple decades. For each event I need X,Y,X and Vx,Vy,Vz data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing the time of the first event it is easy to obtain approximate times of the subsequent events using the known orbital period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have managed to acquire values using the manual web interface to JPL Horizons.  Working on one event at a time I can get data for an epoch within one minute of each event.  But I want (a) timings to the nearest second and (b) to obtain a whole series of apsides events in one go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The JPL Horizons email interface seems to be the way to go. Using this I have been able to obtain data for a particular time range and step interval. But, as with the web interface, it is not possible (in one go) to cover the whole multi-decade period at the desired resolution (1s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now I want to use the TLIST facility to specify a list of individual event times (times at 1s intervals within a range from several seconds before and after the approximate time of each apsides event).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The documentation - see &lt;a href=&quot;https://ssd.jpl.nasa.gov/?horizons#email&quot; rel=&quot;nofollow noreferrer&quot;&gt;how to get Horizons email documentation mailed to you&lt;/a&gt; - includes this section on TLIST:-&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;!TLIST allows users to specify up to 10000 discrete times&#xA;!individually. If TLIST is used, START_TIME, STOP_TIME and&#xA;!STEP_SIZE are ignored. TLIST is a list of either &quot;Julian&#xA;!Day Numbers&quot; (JD) or &quot;Modified Julian Day Numbers&quot;&#xA;!(MJD; days since Nov 17, 1858 (JD 2400000.5)).&#xA;&#xA;!In either case, the input is one or more real numbers.&#xA;!The program puts TLIST in ascending chronological order&#xA;!and determines whether it is JD or MJD based on smallest&#xA;!day number.  Example usage:&#xA;!&#xA;!   TLIST= '2451454.8937833' '2451455.11256351'&#xA;!          '2451456.45576728'&#xA;!          '2451459.65778822'&#xA;!          '2451458.97618331'&#xA;!          '2451459.72787832'&#xA;!          '2451462.25656722'&#xA;!&#xA;!   Note that there can be more than one value on the line&#xA;!   OR one per line. Line length should be less than 80&#xA;!   characters. List is space or comma delimited. &quot;TLIST=&quot;&#xA;!   is used only once so it does not supercede any previous&#xA;!   assignment in the command file.&#xA;&#xA;&#xA; TLIST = 200*' '&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But I am having a problem in formatting the TLIST input data in a way that the program will accept.  Every attempt so far has returned an error message of the form &quot;BATVAR: cannot read TLIST value: --F4F5E805D2040CE6C2&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have emailed the system manager but have had no response so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a copy of the email text which I have been sending (format = To:   horizons@ssd.jpl.nasa.gov Subject: JOB ****** ).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be grateful if anyone can suggest what needs to be done with the data format for the TLIST command.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;!$$SOF&#xA;EMAIL_ADDR = '************@gmail.com'&#xA;!...Target= Mercury&#xA;COMMAND= '199'&#xA;!...Reference Site = Centre of Sun&#xA;CENTER= '500@10'&#xA;MAKE_EPHEM= 'YES'&#xA;TABLE_TYPE= 'VECTORS'&#xA;OUT_UNITS= 'KM-S'&#xA;REF_PLANE= 'ECLIPTIC'&#xA;REF_SYSTEM= 'J2000'&#xA;VECT_CORR= 'NONE'&#xA;VEC_LABELS= 'YES'&#xA;VEC_DELTA_T= 'YES'&#xA;CSV_FORMAT= 'YES'&#xA;OBJ_DATA= 'YES'&#xA;VEC_TABLE= '2'&#xA;!&#xA;! The following works OK (but I want to use TLIST)&#xA;!&#xA;!START_TIME= 'JD 2451590.2526841'&#xA;!STOP_TIME= 'JD 2451590.2596841'&#xA;!STEP_SIZE= '1 m'&#xA;!&#xA;!&#xA;! What I desire (but it fails):-&#xA;!&#xA;!TLIST=&#xA;!'2451546.2714931'&#xA;!'2451546.2715046'&#xA;!'2451546.2715162'&#xA;!'2451546.2715278'&#xA;!'2451546.2715393'&#xA;!'2451546.2715509'&#xA;!'2451546.2715625'&#xA;!&#xA;! Test (it failed)&#xA;!&#xA;   TLIST= '2451454.8937833,2451455.11256351,'&#xA;          '2451456.45576728,'&#xA;          '2451459.65778822,'&#xA;          '2451458.97618331,'&#xA;          '2451459.72787832,'&#xA;          '2451462.25656722'&#xA;!&#xA;!&#xA;! The Following Fail:-&#xA;!&#xA;!   TLIST= 'JD 2451454.8937833' 'JD 2451455.11256351'&#xA;!          'JD 2451456.45576728'&#xA;!          'JD 2451459.65778822'&#xA;!          'JD 2451458.97618331'&#xA;!          'JD 2451459.72787832'&#xA;!          'JD 2451462.25656722'&#xA;!&#xA;!   TLIST= '2451454.8937833' '2451455.11256351'&#xA;!          '2451456.45576728'&#xA;!          '2451459.65778822'&#xA;!          '2451458.97618331'&#xA;!          '2451459.72787832'&#xA;!          '2451462.25656722'&#xA;!&#xA;!TLIST= '2451546.27149310' '2451546.27149310'&#xA;! '2451546.27150460'&#xA;!&#xA;!TLIST= '2451546.2714931' '2451546.2714931'&#xA;! '2451546.2715046'&#xA;!&#xA;!TLIST= '&#xA;! 2451546.2714931&#xA;! 2451546.2715046&#xA;! 2451546.2715162'&#xA;!&#xA;!TLIST= '&#xA;! 2451546.2714931,&#xA;! 2451546.2715046,&#xA;! 2451546.2715162'&#xA;!&#xA;!TLIST= ''2451546.2714931', '2451546.2715046', '2451546.2715162''&#xA;!&#xA;!TLIST= '2451546.2714931', '2451546.2715046', '2451546.2715162'&#xA;!&#xA;!TLIST= '2451546.2714931' '2451546.2715046' '2451546.2715162'&#xA;!&#xA;!TLIST= '2451546.2714931 2451546.2715046 2451546.2715162'&#xA;!&#xA;!TLIST='2451546.2714931,2451546.2715046,2451546.2715162'&#xA;!&#xA;!TLIST=200*''&#xA;!&#xA;!TLIST= '2451454.8937833' '2451455.11256351'&#xA;!&#xA;!TLIST= '2451546.2'&#xA;!&#xA;!TLIST='2451546.2714931''2451546.2715046''2451546.2715162'!'2451546.2715278',&#xA;!'2451546.2715393',&#xA;!'2451546.2715509',&#xA;!'2451546.2715625'&#xA;!&#xA;!$$EOF&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="233" LastEditorUserId="233" LastEditDate="2018-01-19T00:45:42.303" LastActivityDate="2018-01-19T00:53:30.637" Title="JPL Horizons - how to format TLIST input times for Email query?" Tags="&lt;orbit&gt;&lt;ephemeris&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24564" PostTypeId="2" ParentId="24561" CreationDate="2018-01-16T22:26:24.990" Score="1" Body="&lt;p&gt;This is partly space exploration and partly physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rockets do not normally go straight up. They start off going up, to get into thinner air, then they steer to turn themselves eventually to a horizontal path, until they are going fast enough to stay in orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general relativity, an object that is free to move in the Earth's gravitational field will appear to accelerate towards the ground as it is following a path (called a geodesic) in curved spacetime. This is not specific to rockets; a cricket ball will follow a curved path when thrown due to the curvature of spacetime around the Earth. And if one is far  from any gravitational field, then a cricket ball will travel in a straight line. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To directly answer your question: The path of a rocket going straight up will not look curved, it will look like it is going straight up.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-01-16T22:26:24.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24566" PostTypeId="1" CreationDate="2018-01-17T02:50:49.127" Score="4" ViewCount="286" Body="&lt;p&gt;Question in the title. Let's say that I have a lot of garbage- mined-out asteroids, spent nuclear fuel, satellite debris, and just general trash. Also safe ways to get these into Earth orbit. Instead of sending this into the Sun, Jupiter, or out of the Solar System's plane, what if I took some random quasi-satellite, put it a bit behind the Moon on its orbital path, and started piling garbage on it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How large could it get before bad things start happen to Earth? For example, tides drowning the coasts. I'm not going to consider collisions, because of (insert megacorp here) Moon Tethers™.&lt;/p&gt;&#xA;" OwnerUserId="19304" LastActivityDate="2018-05-05T12:14:28.723" Title="How large could a second moon of Earth be?" Tags="&lt;the-moon&gt;&lt;earth&gt;&lt;satellite&gt;" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="24571" PostTypeId="2" ParentId="21104" CreationDate="2018-01-17T15:56:54.683" Score="2" Body="&lt;p&gt;The problem you are trying to solve is one generally known as spectral estimation. In addition to the practical references provided by jstarek, I suggest you also read some fundamentals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book by S.M.Kay &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/0130151599&quot; rel=&quot;nofollow noreferrer&quot;&gt;Modern Spectral Estimation&lt;/a&gt;, although almost 20 years old, is totally up to date and is a very well explained (easy to read) introduction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will see that the folding algorithm is just a flavour of the Welch periodogram, and that your cross-correlation approach fits as an MLE estimation scheme (nowadays also known as signal subspace techniques).&lt;/p&gt;&#xA;" OwnerUserId="21024" LastActivityDate="2018-01-17T15:56:54.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24572" PostTypeId="2" ParentId="13591" CreationDate="2018-01-17T16:04:46.343" Score="2" Body="&lt;p&gt;You may take a look at the lates parametrization &lt;a href=&quot;https://naif.jpl.nasa.gov/pub/naif/generic_kernels/pck/pck00010.tpc&quot; rel=&quot;nofollow noreferrer&quot;&gt;file&lt;/a&gt; by JPL-NAIF for the precession, nutation and pole orientation of the largest known bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although, for the large time scales you are asking, I expect you will need to propagate the data and make your own wild guess, or dig into appropiate literature about solar system physics.&lt;/p&gt;&#xA;" OwnerUserId="21024" LastActivityDate="2018-01-17T16:04:46.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24573" PostTypeId="1" CreationDate="2018-01-17T16:43:52.967" Score="3" ViewCount="28" Body="&lt;p&gt;I have 2 questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Could you please give me arxiv numbers of papers dealing with the Bousso bound for the particle horizon (known Universe), from masters level upwards, with also some papers a bit more philosophical or vulgarised?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It has been said that due to Bousso bound (or Bekenstein's, I don't remember), QFT or any other field theory should have a limit in the micro dimensions, therefore be discrete instead of continuous: do you know some arxiv papers with the same levels and approaches than in (1)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="21022" LastEditorUserId="2365" LastEditDate="2018-01-17T18:15:02.403" LastActivityDate="2018-01-17T18:15:02.403" Title="Bousso bound within particle horizon, discrete Q.field theory" Tags="&lt;observable-universe&gt;&lt;entropy&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24576" PostTypeId="1" AcceptedAnswerId="25609" CreationDate="2018-01-18T10:10:31.860" Score="6" ViewCount="2633" Body="&lt;p&gt;I asked this on the ‘movies’ stackexchange site but they suggested I ask here, it’s a purely hypothetical query based on a BBC drama show called ‘Hard Sun’, feel free to kill the question if it’s inappropriate here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the TV show’s series finale an expected ‘extinction level event’ happens and appears to be a black hole close to our sun, it already has developed a spiral of hot gas around it drawn from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly this is fiction but I’m keen to understand just how badly-considered this is. Firstly the show is set in London and everyone seems amazed when the image below rises from the dusk yet clearly it's been in existence for the time taken to create the spiral (hours at least?) so nations east of the UK would have seen it for that time and reported on it globally surely?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My real question is more around is there any potential scenario where a black hole of that size could develop suddenly and get to that size quickly enough for that image to be a surprise? My understanding, very much as a scientific-layman, is that a black hole of that size would have to have the mass of at least one of our suns to get to that size (probably more) - is that a valid assumption? If so where has the mass come from? Also wouldn't the sudden existence of something else with at least the same mass as our sun suddenly appearing at the centre of our solar system have enormous gravitational impact on everything in the system, including Earth? We'd have massive volcanic/tectonic impact wouldn't we?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So again sorry if this is inappropriate, I was just incredulous at how badly-considered the scientific aspects of this fiction were and wanted to check that my understandings were correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Screengrab:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/3yMDh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3yMDh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21028" LastEditorUserId="21028" LastEditDate="2018-01-19T11:06:05.170" LastActivityDate="2020-01-09T19:40:07.903" Title="Theoretical sudden appearance of a black hole in BBC drama show *spoiler warning*" Tags="&lt;black-hole&gt;&lt;gravity&gt;" AnswerCount="6" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="24578" PostTypeId="2" ParentId="24576" CreationDate="2018-01-18T11:37:00.597" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Firstly the show is set in London and everyone seems amazed when the image below rises from the dusk yet clearly it's been in existence for the time taken to create the spiral (hours at least?) so nations east of the UK would have seen it for that time and reported on it globally surely?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And we've kind of a large number of space based satellites that do nothing but monitor the Sun and all the people monitoring them would certainly spot this minor event.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My real question is more around is there any potential scenario where a black hole of that size could develop suddenly and get to that size quickly enough for that image to be a surprise ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;About the only way I can think of is someone creating a &lt;a href=&quot;https://en.wikipedia.org/wiki/Kugelblitz_%28astrophysics%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kugelblitz&lt;/a&gt;, although it's pretty far fetched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe a very freak collision of two high mass objects head-on, but that would create far worse problems - essentially an absolutely huge explosion and we would not be looking at the result quietly afterward.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My understanding, very much as a scientific-layman, is that a black hole of that size would have to have the mass of at least one of our suns to get to that size (probably more) - is that a valid assumption ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's a minimum size for the formation of a black hole by gravitational collapse of a star and it's larger than our Sun's mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is, however, no theoretical minimum for the size of a black hole if you can make one some other way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If so where has the mass come from ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ask the writers. :-)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also wouldn't the sudden existence of something else with at least the same mass as our sun suddenly appearing at the centre of our solar system have enormous gravitational impact on everything in the system, including Earth? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes that's a major issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any black hole large enough to start siphoning off the Sun's outer layers would need a significant mass and it would have a significant impact on our orbit - and do so more or less immediately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our orbital speed is enough to maintain our almost circular orbit around the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With something as massive as a small star added close to the Sun, our orbital speed would no longer be enough for a circular orbit and we'd be in an elliptical orbit about the Sun-BH pair.  It's more complicated than than that, but that's the broad strokes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that ellipse would be with our current distance as an aphelion, whereas we'd have a smaller perihelion distance.  So we'd get warmer.  And it would not take much to raise temperatures here uncomfortably.  Very roughly if we were to have a closest approach of 0.72 AU (Venus's current orbit) we'd find our temperatures rise by perhaps 24 degrees Celsius, which would be enough to make the planet unlivable for us - ( maybe 60 C at the equator and enough to melt significant parts of the polar ice and permafrost ).  And as this change would happen every six months as we orbited moving from close in to far out, we'd be permanently shifting climate.  And that ignores greenhouse effects - we could end up with a runaway greenhouse effect that raised temperatures to lethal levels everywhere.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We'd have massive volcanic/tectonic impact wouldn't we?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's a hard one to call.  This would depend on how far apart they Sun and the BH were and how massive the BH was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we'd have lots to worry about.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2018-01-18T11:37:00.597" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24579" PostTypeId="2" ParentId="24433" CreationDate="2018-01-18T11:56:21.743" Score="1" Body="&lt;p&gt;NASA's Navigation Ancillary Facility (NAIF) publishes planetary constants kernels (PCK) which are basically text files containing pole orientations for the largest known bodies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PCKs includes the parametric orientation of their prime meridian in agremeent with IAU standards (meaning datum is J2000.0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://naif.jpl.nasa.gov/pub/naif/generic_kernels/pck/pck00010.tpc&quot; rel=&quot;nofollow noreferrer&quot;&gt;latest PCK&lt;/a&gt; dates from 2011 and is actively used by most operational interplanetary flight projects across the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the &lt;a href=&quot;https://naif.jpl.nasa.gov/naif/toolkit.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NAIF SPICE library&lt;/a&gt; (available in Fortran, C, Matlab and IDL) you can load this kernel and read the orientation of the prime meridian for your body and date of interest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time offset between bodies can be reconstructed as angular offset between meridians. But you may find some time-related functions within SPICE which can make your task easier. &lt;/p&gt;&#xA;" OwnerUserId="21024" LastEditorUserId="21024" LastEditDate="2018-01-23T10:24:21.353" LastActivityDate="2018-01-23T10:24:21.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24581" PostTypeId="2" ParentId="8431" CreationDate="2018-01-18T12:52:11.617" Score="4" Body="&lt;p&gt;I find the manner in which the question is posed rather opinion-oriented. Nevertheless, I try to give a few objective add-ons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both missions have been great technical successes, having very different mission requirements, spacecraft designs and concepts of operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When looking at the scientific return, objectively Rosetta has blown it out (count it in number of Science/Nature/etc. publications and count the amount of new geophysical models for pristine solar system bodies. The comet has told us so much, that few weeks ago NASA has approved budget for supporting a &lt;a href=&quot;https://www.nasa.gov/press-release/nasa-invests-in-concept-development-for-missions-to-comet-saturn-moon-titan&quot; rel=&quot;nofollow noreferrer&quot;&gt;mission concept&lt;/a&gt; to revisit the body. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rosetta was, at the very origin of the project, conceived as an ESA-NASA cooperation, where JPL will build the lander. However, politics aside, Congress cut the flow and ESA had to turn to DLR to build a lander with limited budget, limited mass and limited volume. Therefore, Philae's platform had no steering (active AOCS) in contrast to Hayabusa, which was a fully-equipped platform. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The landing sequence for Philae was pre-planned. Philae didn't control its path, it was the comets inhomogeneous gravity at ~1m/s for several km. Propagate your uncertainties... Still, the landing was pin-point, damn the rebound on that hard rock. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hayabusa performed an inertial hovering followed by a touch-and-go sequence. The operation was rehearsed twice before. The sample was taken with the equivalent of a vacum cleaner on a dusty body with such low density you could put your arm through and come out on the opposite hemisphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One important requirement in sample collection is to prevent contamination of the sample by the system and the environment. Specially during re-entry. For the small sample of Itokawian dust that was recovered, isolation conditions could not be proven. So most likely some material from spacecraft outgassing went in there. This could not be proven one way or the other. Nevetheless, technically speaking, this operation: sample capture, retrieval and re-entry was a breakthrough. And in fact, OSIRIS-REX is just going to try and repeat that on asteroid Bennu with a load of lessons learnt. I personally look forward to the science return and wish for a lot more data coming down this time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you ask, why such a big boom for Rosetta, the key is, the few hours of experiments on the surface helped give context to large set of extended science observations by the orbiter. Something that had never been done before on a small body (be it a comet or an asteroid), actually only on Mars. This got the whole community very excited, for a good reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an engineer, I understand the voices of &quot;this wasn't a real landing&quot;, &quot;this was luck&quot;, &quot;it was a failure&quot; etc. But you know, JPL redefined the landing gear of their next comet sample return proposal when Philae's harpoons didn't work. This is what erros are made for, to make it better the next chance. And this involves all agencies and all countries collaborating instead of competing to win a front page. It also involves the general public understanding the importance of each mission out there. Some prove technology, some bring a science storm. Let's leave politics to politians ...&lt;/p&gt;&#xA;" OwnerUserId="21024" LastActivityDate="2018-01-18T12:52:11.617" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24583" PostTypeId="1" CreationDate="2018-01-18T18:26:14.763" Score="4" ViewCount="102" Body="&lt;p&gt;AFAIK, objects in the universe thought to be black holes show evidence of &lt;em&gt;radiatively inefficient accretion flow&lt;/em&gt;. But is this the case with Quasars? If not, is there a possibility that the core of a quasar is an object similar to a neutron star.&lt;/p&gt;&#xA;" OwnerUserId="21044" LastActivityDate="2018-01-20T13:03:27.480" Title="Is it possible that quasars are not powered by black holes but some other kind of compact, massive object?" Tags="&lt;supermassive-black-hole&gt;&lt;luminosity&gt;&lt;quasars&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24585" PostTypeId="2" ParentId="24563" CreationDate="2018-01-19T00:37:31.800" Score="2" Body="&lt;p&gt;By trial and error I found that the following email text (including intermediate comment lines) is acceptable to HORIZONS:-&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;!$$SOF&#xA;EMAIL_ADDR = 'xxxxxx@gmail.com'&#xA;!...Target= Mercury&#xA;COMMAND= '199'&#xA;!...Reference Site = Centre of Sun&#xA;CENTER= '500@10'&#xA;MAKE_EPHEM= 'YES'&#xA;TABLE_TYPE= 'VECTORS'&#xA;OUT_UNITS= 'KM-S'&#xA;REF_PLANE= 'ECLIPTIC'&#xA;REF_SYSTEM= 'J2000'&#xA;VECT_CORR= 'NONE'&#xA;VEC_LABELS= 'YES'&#xA;VEC_DELTA_T= 'YES'&#xA;CSV_FORMAT= 'YES'&#xA;OBJ_DATA= 'YES'&#xA;VEC_TABLE= '3'&#xA;TLIST= '2451546.27149305' '2451546.27150462'&#xA; '2451546.2715162'&#xA; '2451546.27152777'&#xA; '2451546.27153934'&#xA;!&#xA;! The following works OK but I want to use TLIST&#xA;!&#xA;!START_TIME= 'JD 2451590.9526841'&#xA;!STOP_TIME= 'JD 2451591.0596841'&#xA;!STEP_SIZE= '10 m'&#xA;!&#xA;!$$EOF&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However, I can't say that I have found the reason why all the previous submissions failed to execute.  For example if I move the final 7-line comment block (leaving the !$$EOF line in place at the end) and put it in front of the TLIST command line ... then the job fails.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless I have a format for using TLIST which works OK for me now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Useful Links&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi?CGISESSID=4f071eb036a53cdde6734789ff0cfa6e#top&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL HORIZONS web interface&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ssd.jpl.nasa.gov/?horizons#email&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ssd.jpl.nasa.gov/?horizons#email&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ssd.jpl.nasa.gov/dat/Configuring_Mail_Clients_to_Send_Plain_ASCII_Text.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ssd.jpl.nasa.gov/dat/Configuring_Mail_Clients_to_Send_Plain_ASCII_Text.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="233" LastEditorUserId="233" LastEditDate="2018-01-19T00:53:30.637" LastActivityDate="2018-01-19T00:53:30.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24586" PostTypeId="1" CreationDate="2018-01-19T03:00:32.407" Score="3" ViewCount="497" Body="&lt;p&gt;What is earth tipped toward? The simple answer is Polaris.  The axis (one end of it) points at Polaris, so that's what it's tipped to point at.  But when we saw &quot;toward&quot; what do we mean? When it is tipped &quot;toward&quot; one thing, what is it being tipped &quot;away from&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If earth were not tipped then the equator would lie in the same plane as earth's orbit.  Earth being tipped 23 degrees means it is tipped 23 degrees &quot;away from&quot; alignment with its orbit.  So a 0 degree tip would be &quot;not tipped&quot; and a 23 degree tip is tipped 23 degrees in some direction.  What direction is it tipped in?  Another way to ask this would be, what would it be tipped at if it was tipped a full 90 degrees in the general direction in which it is currently tipped 23 degrees?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a coordinate system with two axes in the plane of earth's orbit and one axis perpendicular to that plane.  Earth's orbit stays in the plane, and the angle that earth is tipped points to, let's say, negative X.  So the line through the north pole and south pole lies in the XY plane and the earth's orbit lies in the XZ plane.  I want to look at stars and be able to see those axes, so I want to know what stars they roughly point at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cleaner way to ask this might be, what would be the north star if earth wasn't tipped? What would be the south star? What if it was tipped 90 degrees? That gives four of the six stars I'm looking for. I'm not sure how to phrase the question for the other two.  I want to call these stars &quot;untipped north&quot;, &quot;untipped south&quot;, &quot;winter&quot;, &quot;summer&quot;, &quot;spring&quot;, and &quot;fall&quot; because the direction of the stars would be the same as the direction earth is from the sun during the extremes of those seasons (northern hemisphere).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plane of earth's orbit is aligned with a celestial (infinitely distant) great circle called the ecliptic.  It inversely happens to also be the path through the sky that the sun takes over the course of a year.  There is a latitude/longitude coordinate system where the equator is the ecliptic called &quot;ecliptic coordinates&quot;.  So the question is actually:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using ecliptic latitude and longitude, what 6 stars are at:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;0, 0&lt;/li&gt;&#xA;&lt;li&gt;0, 90&lt;/li&gt;&#xA;&lt;li&gt;0, 180&lt;/li&gt;&#xA;&lt;li&gt;0, -90&lt;/li&gt;&#xA;&lt;li&gt;90, 0&lt;/li&gt;&#xA;&lt;li&gt;-90, 0&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="21050" LastEditorUserId="21050" LastEditDate="2018-01-19T16:15:36.750" LastActivityDate="2018-01-24T00:56:03.870" Title="If the earth was not tilted, what would be the north star?" Tags="&lt;positional-astronomy&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24590" PostTypeId="1" AcceptedAnswerId="24591" CreationDate="2018-01-19T08:19:35.130" Score="138" ViewCount="23130" Body="&lt;p&gt;&lt;a href=&quot;https://xkcd.com/1944/&quot; rel=&quot;noreferrer&quot;&gt;XKCD 1944&lt;/a&gt; claims that there is &quot;more gold in the sun than water in the oceans&quot;. Is this really true? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1xFzK.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1xFzK.png&quot; alt=&quot;The end of the rainbow&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21057" LastEditorDisplayName="user1569" LastEditDate="2018-01-19T11:43:11.563" LastActivityDate="2018-11-07T03:36:30.600" Title="How much gold is there in our sun?" Tags="&lt;the-sun&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="24591" PostTypeId="2" ParentId="24590" CreationDate="2018-01-19T08:38:05.380" Score="119" Body="&lt;p&gt;The mass of the sun is 1.989 × 10&lt;sup&gt;30&lt;/sup&gt; kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.periodictable.com/Properties/A/SolarAbundance.html&quot; rel=&quot;noreferrer&quot;&gt;Abundance in the Sun of the elements&lt;/a&gt; gives a percentage 1 × 10&lt;sup&gt;-7&lt;/sup&gt; % &lt;a href=&quot;http://www.periodictable.com/Elements/079/data.html&quot; rel=&quot;noreferrer&quot;&gt;for gold&lt;/a&gt; &lt;sup&gt;*&lt;/sup&gt;, so that leaves you with a mass of 1.989 × 10&lt;sup&gt;21&lt;/sup&gt; kg of gold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://science.howstuffworks.com/environmental/earth/geophysics/question157.htm&quot; rel=&quot;noreferrer&quot;&gt;HowStuffWorks states&lt;/a&gt; that there is 1.26 × 10&lt;sup&gt;21&lt;/sup&gt; kg water on Earth, of which 98% is in the oceans, i.e. 1.235 × 10&lt;sup&gt;21&lt;/sup&gt; kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would mean the XKCD statement is true: there is 1.6 times as much gold in the sun as there is water in the oceans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;* They cite WolframAlpha as their source. Executing &lt;a href=&quot;https://www.wolframalpha.com/input/?i=SolarAbundance+%22Gold%22&quot; rel=&quot;noreferrer&quot;&gt;SolarAbundance &quot;Gold&quot;&lt;/a&gt; there confirms this (mass) percentage.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user1569" LastEditorDisplayName="user1569" LastEditDate="2018-01-19T13:45:39.640" LastActivityDate="2018-01-19T13:45:39.640" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="24596" PostTypeId="2" ParentId="24590" CreationDate="2018-01-19T12:42:05.953" Score="26" Body="&lt;p&gt;&lt;a href=&quot;http://www.knowledgedoor.com/2/elements_handbook/element_abundances_in_the_sun.html&quot; rel=&quot;noreferrer&quot;&gt;&quot;Element Abundances in the Sun - The Elements Handbook&quot;, &lt;em&gt;KnowledgeDoor&lt;/em&gt;&lt;/a&gt; claims that the base 10 log of the number of atoms of gold in the Sun for every $10^{12}$ atoms of hydrogen is $1.01 \pm 0.15$. If I'm reading their references correctly, that's from &lt;strong&gt;Abundances of the Elements: Meteoritic and Solar&lt;/strong&gt;, Anders, Edward, and Nicolas Grevesse, &lt;em&gt;Geochimica et Cosmochimica Acta&lt;/em&gt;, volume 53, number 1, 1989, pp. 197–214, doi:10.1016/0016-7037(89)90286-X&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atomic mass of gold is $197$ times the atomic mass of hydrogen (more precise figures are available, but irrelevant given the accuracy of the atomic proportions). So $2020$ kg of gold for every $10^{12}$ kg of hydrogen, meaning that ignoring all other elements and running with $1.99 \times 10^{30}$ kg for the mass of the Sun, it contains $4 \times 10^{21}$ kg of gold. Taking other elements into account - helium is actually significant - reduces that value to $3 \times 10^{21}$ kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is about twice as much as the mass of the ocean, which corresponds to 2 standard deviations ($\log_{10} 2 \approx 0.3$ vs the standard deviation of $0.15$ in the log 10 value of the abundance).&lt;/p&gt;&#xA;" OwnerUserId="21067" LastEditorUserId="600" LastEditDate="2018-01-21T18:19:26.057" LastActivityDate="2018-01-21T18:19:26.057" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24598" PostTypeId="1" AcceptedAnswerId="24636" CreationDate="2018-01-19T13:26:05.137" Score="3" ViewCount="2134" Body="&lt;p&gt;Every Azimuth equation I have been able to locate so far depends on inputting a time, and I do not want to have to brute force the solution by simply trying all of the times. (See below for standard equations.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My use case doesn't require great precision, within a few degrees is fine. So, ideally I'm looking for equation(s) that take only latitude (and possibly longitude?) as input and output max and/or min azimuth angles for that location.  Also, I'm only going to care about years within a century of now so I don't believe that the year to year variation will matter given my precision requirements, but if I'm wrong I'm happy to provide the current year as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Standard Solar Equations&lt;/strong&gt; Based on &lt;a href=&quot;http://www.esrl.noaa.gov/gmd/grad/solcalc/calcdetails.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NOAA's Solar Calculator&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Note: The sunrise and sunset results are theoretically accurate to within a minute&#xA; for locations between +/- 72° latitude, and within 10 minutes outside of those latitudes.  &#xA; However, due to variations in atmospheric composition, temperature, pressure and conditions, &#xA; observed values may vary from calculations.&#xA;...&#xA;Please note that calculations in the spreadsheets are only valid for dates between 1901 and 2099, &#xA; due to an approximation used in the Julian Day calculation.&quot; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;$\Xi = $Latitude (+ to N)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Phi = $Longitude (+ to E)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega = $Time Zone (+ to E)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d = $Date&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tau = $Time (hrs past local midnight)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\upsilon = $Julian Day = $d + 2415018.5 + \tau - \frac{\omega}{24}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma = $Julian Century = $\frac{\upsilon - 2451545}{36525}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho = $Geom Mean Long Sun (deg) = $\left(280.46646 + \sigma&#xA;\left(36000.76983 + 0.0003032\sigma\right)\right) \bmod 360$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\xi = $Geom Mean Anom Sun (deg) = $357.52911 + \sigma\left(35999.05029 - 0.0001537\sigma\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu = $Eccent Earth Orbit = $0.016708634 - \sigma\left(0.000042037 + 0.0000001267\sigma\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda = $Sun Eq of Ctr = $\sin\left(\xi^{rad}\right)\left(1.914602 - \sigma\left(0.004817+0.000014\sigma\right)\right) + \sin\left(2\xi^{rad}\right)\left(0.019993 - 0.000101\sigma\right) + 0.000289\sin\left(3\xi^{rad}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\kappa = $Sun True Long (deg) = $\rho + \lambda$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\iota = $Sun True Anom (deg) = $\xi + \lambda$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta = $Sun Rad Vector (AUs) = $\frac{1.000001018\left(1 - \mu^2\right)}{1 + \mu\cos\left(\iota^{rad}\right)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\eta = $Sun App Long (deg) = $\kappa - 0.00569 - 0.00478\sin\left(\left(125.04-1934.136\sigma\right)^{rad}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\zeta = $Mean Obliq Ecliptic (deg) = $23 + \frac{26 + \frac{21.448 - \sigma\left(46.815+\sigma\left(0.00059 - \sigma0.001813\right)\right)}{60}}{60}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\epsilon = $Obliq Corr (deg) = $\zeta + 0.00256\cos\left(\left(125.04-1934.136\sigma\right)^{rad}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta = $Sun Declin (deg) = $\left(\arcsin\left(\sin\left(\epsilon^{rad}\right)\sin\left(\eta^{rad}\right)\right)\right)^o$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = $var y = $\tan\left(\left(\frac{\epsilon}{2}\right)^{rad}\right)^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Gamma = $Eq of Time (minutes) = $4\left(y\sin\left(2\rho^{rad}\right)-2\mu\sin\left(\xi^{rad}\right)+4\mu y\sin\left(\xi^{rad}\right)\cos\left(2\rho^{rad}\right)-0.5y^2\sin\left(4\rho^{rad}\right)-1.25\mu^2\sin\left(2\xi^{rad}\right)\right)^o$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\gamma = $True Solar Time (min) = $\left(1440\tau + \Gamma + 4\Phi - 60\omega\right) \bmod 1440$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\beta = $Hour Angle (deg) = $if\left(\frac{\gamma}{4}&amp;lt;0\right) \{\\&#xA;\frac{\gamma}{4}+180\\&#xA;\} else \{\\&#xA;\frac{\gamma}{4}-180\\&#xA;\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Omega = $Solar Zenith Angle (deg) = $\left(\arccos\left(\sin\left(\Xi^{rad}\right)\sin\left(\delta^{rad}\right)+\cos\left(\Xi^{rad}\right)\cos\left(\delta^{rad}\right)\cos\left(\beta^{rad}\right)\right)\right)^o$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha = $Solar Azimuth Angle (deg cw from N) = &#xA;$if\left(\beta&amp;gt;0\right) \{\\&#xA;\left(\arccos\left(\frac{\sin\left(\Xi^{rad}\right)\cos\left(\Omega^{rad}\right)-\sin\left(\delta^{rad}\right)}{\cos\left(\Xi^{rad}\right)\sin\left(\Omega^{rad}\right)}\right)^o+180\right) \bmod 360\\&#xA;\} else \{\\&#xA;\left(540-\arccos\left(\frac{\sin\left(\Xi^{rad}\right)\cos\left(\Omega^{rad}\right)-\sin\left(\delta^{rad}\right)}{\cos\left(\Xi^{rad}\right)\sin\left(\Omega^{rad}\right)}\right)^o\right) \bmod 360\\&#xA;\}$&lt;/p&gt;&#xA;" OwnerUserId="21070" LastEditorUserId="21070" LastEditDate="2018-01-20T03:05:16.683" LastActivityDate="2018-01-23T18:06:59.810" Title="How to calculate the maximum and minimum solar azimuth at a given location?" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;azimuth&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24600" PostTypeId="1" AcceptedAnswerId="24601" CreationDate="2018-01-19T15:14:10.213" Score="5" ViewCount="530" Body="&lt;p&gt;This question proved much more difficult to formulate than I initially imagined. It concerns how far out in the Milky Way that we actually see with different telescopes. I know the Milky Way disk is about 100k light years wide, its halo a couple of times that wide, and most of the naked eye stars are within 1000 light years. I also know that when many stars are close together they are more visible, which means we can easily see for example the double cluster in Perseus at about 7000 light years with our naked eyes. But if you take a large telescope and peer into the distance, how easily do you see through the star fields until the point where you have no more foreground stars in the field of view, and all that is there are the numerous faint galaxies lurking in the back. For example, in the Hubble Deep field you see a few stars. Are we looking at the last few stars at the edge of the Milky Way, or are there more hidden in the field that we can't see because they are too far away/too faint?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that there are dust clouds in the Milky Way obscuring the view, so if you look towards the galactic center you would definitely not see all the way to the other side. But lets assume we're looking perpendicular to the galactic plane. Lets also say that you're using a 16&quot; telescope, allowing you to see maybe down to the 15th magnitude. Do you see the stars at the edge of the galaxy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of approaching the question... From what I've read it seems we estimate the Milky Way to contain about 250 billion stars. I also read that the Gaia satellite will map about 1 billion stars. Does that mean our best telescopes are only able to discern 1/250 of the stars in the Milky Way? Of course all stars on the other side the galaxy disk are hidden from us, so they surely account for a large portion of the invisible stars. But looking straight up from the galactic plane, can we see the last galactic stars before the empty intergalactic &quot;void&quot;?&lt;/p&gt;&#xA;" OwnerUserId="21073" LastEditorUserId="16061" LastEditDate="2019-08-08T11:49:37.730" LastActivityDate="2019-08-08T11:49:37.730" Title="Can we see the stars at the edge of the Milkyway?" Tags="&lt;star&gt;&lt;galaxy&gt;&lt;distances&gt;&lt;milky-way&gt;&lt;magnitude&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24601" PostTypeId="2" ParentId="24600" CreationDate="2018-01-19T15:29:38.923" Score="9" Body="&lt;p&gt;Take a look at the HR-diagram. You have forgotten to take into account that stars come in vastly different luminosities.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, we can barely see all red dwarves out to a distance of 10pc, while we know that they vastly outnumber the more luminous stars.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The luminous red giants however we'll be able to see on side of the edge of the disc, but even Gaia will not be able to look through to the other edge. It's limit magnitude is 20, while extinction through dust in the direction of the galactic center is often cited to be up to 100th magnitude.&lt;br&gt;&#xA;Hubble deep field galaxies for comparison are around 30th mag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is &lt;a href=&quot;https://en.wikipedia.org/wiki/Baade%27s_Window&quot; rel=&quot;noreferrer&quot;&gt;Baade's window&lt;/a&gt;, a region just above the central galactic plane that has low dust content on our line-of-sight there. We'll be able to spy quite some stars there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're interested in the Gaia mission, you can find information &lt;a href=&quot;http://www.astro.lu.se/~david/Gaia1.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2018-01-19T15:29:38.923" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24602" PostTypeId="2" ParentId="24586" CreationDate="2018-01-19T16:11:12.913" Score="5" Body="&lt;p&gt;It turns out that 0, 0 on ecliptic coordinates is not the &quot;winter&quot; direction, nor the &quot;summer&quot; direction.  And it turns out to be a good thing.  Zero longitude is defined using the &quot;spring&quot; direction, and this means that converting these points to another coordinate system is really easy.  The earth's axis is tipped 23 degrees, but imagine the axis &lt;em&gt;on which it is tipped&lt;/em&gt;.  Thats the spring/fall axis.  This means that if we tip the whole coordinate system 23 degrees, two points stay unaffected (the spring/fall points).  The tipped coordinate system is called equatorial coordinates and its more useful so everything uses it.  The point 0, 0 is the same in both coordinate systems, as is the antipode of 0, 0 which is 0, 180.  So you can get the spring and fall stars right away by browsing something like &lt;a href=&quot;http://www.sky-map.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.sky-map.org/&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the longitude coordinates are in hours (0 to 23) instead of degrees (0 to 360), and they are shown in the order longitude, then latitude.  Also note that stars you can see without a telescope have an apparent magnitude of 6 or less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zero longitude is the direction the sun is in when it is spring, so the 0, 0 star is actually the fall star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the other four stars: because the difference between the two coordinate systems is a single rotation around the spring-fall axis, the winter/summer points actually keep their longitude the same during the rotation.  And because the amount of rotation is 23 degrees:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;0, 90 becomes 23, 90&lt;/li&gt;&#xA;&lt;li&gt;and 0, -90 becomes -23, -90&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You can check this here: &lt;a href=&quot;https://lambda.gsfc.nasa.gov/toolbox/tb_coordconv.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://lambda.gsfc.nasa.gov/toolbox/tb_coordconv.cfm&lt;/a&gt; Note that equatorial coordinates are called &quot;celestial&quot; on this page.  Note that you have to enter a year because the tip of the earth's axis slowly changes.  Just enter the current year in both epoch boxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should make sense that the untipped north star would have an equatorial latitude of 90 minus 23 degrees, and the untipped south star would have an equatorial latitude of -90 plus 23 degrees.  Using the same tool to do the conversion:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;90, 0 becomes 66, -90&lt;/li&gt;&#xA;&lt;li&gt;-90, 0 becomes -66, 90&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you go back to sky map and poke around in those locations, you find:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The untipped north star is in Draco (18h 66d)&lt;/li&gt;&#xA;&lt;li&gt;The untipped south star is in Dorado (6h -66d)&lt;/li&gt;&#xA;&lt;li&gt;The fall star is in Pisces (0h 0d)&lt;/li&gt;&#xA;&lt;li&gt;The spring star is in Virgo (12h 0d)&lt;/li&gt;&#xA;&lt;li&gt;The winter/summer stars are in Gemini (6h -23d) and Sagittarius (18h 23d)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that the latter four stars are in zodiac constellations - this is because those constellations lie on the ecliptic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For winter/summer which is which?&#xA;Well the sun comes up in the east, so earth rotates counter clockwise, so it travels around the sun counter clockwise.  We know that after the line from the sun to the earth points toward fall, it points toward winter.  So whichever is left of Pisces in the sky would be the winter one:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The winter star would be in Gemini&lt;/li&gt;&#xA;&lt;li&gt;The summer star would be in Sagittarius&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that the four zodiac constellations we got are evenly spaced throughout the year, and that the months associated with them are 6 months off from the seasons we mapped them to.  This is because the zodiac months are assigned based on where the sun is in the sky.  You never see a zodiac's constellation at night during the month associated with it.&lt;/p&gt;&#xA;" OwnerUserId="21050" LastEditorUserId="21050" LastEditDate="2018-01-24T00:56:03.870" LastActivityDate="2018-01-24T00:56:03.870" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24603" PostTypeId="1" AcceptedAnswerId="24605" CreationDate="2018-01-19T18:14:57.567" Score="4" ViewCount="114" Body="&lt;p&gt;I recently came across the book intro to cosmology by j.v. Narlikar, where it mentions about quarks and other star and inadvertently uses mass of sun as reference. I want to know how me measure mass of a star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess from the parallax method we could calculate distance?&lt;/p&gt;&#xA;" OwnerUserId="20263" LastEditorUserId="1891" LastEditDate="2018-01-20T00:32:40.167" LastActivityDate="2018-01-20T00:32:40.167" Title="How do we measure the mass of a star?" Tags="&lt;star&gt;&lt;cosmology&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24605" PostTypeId="2" ParentId="24603" CreationDate="2018-01-19T20:18:41.527" Score="6" Body="&lt;p&gt;In general one can only measure the mass of a star if it is orbiting another body or is orbited by another body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In essence, the method exploits Kepler's third law, though the exact technique depends on what observations are available, on what timescale and with what resolution (spectral and angular) is available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other techniques can be calibrated using such objects as exemplars - for instance there are reasonably well-calibrated mass-luminosity or mass-color relationships.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-01-19T20:18:41.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24614" PostTypeId="2" ParentId="24583" CreationDate="2018-01-20T13:03:27.480" Score="5" Body="&lt;p&gt;The solution to your question is surprisingly simple, I think:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quasar that puts out energy around Eddington luminosity or higher, must accrete at a certain rate, corresponding to the energy output. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When consulting &lt;a href=&quot;http://abyss.uoregon.edu/~js/ast123/lectures/lec12.html&quot; rel=&quot;noreferrer&quot;&gt;a lecture&lt;/a&gt; and a &lt;a href=&quot;https://arxiv.org/abs/1510.03887&quot; rel=&quot;noreferrer&quot;&gt;random paper&lt;/a&gt; from the archive on this topic, it is evident that those accretion rates correspond to 1-10 solar masses per year or higher. A neutron star accreting 1-10 solar masses in one year will become a very massive black hole quickly enough.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2018-01-20T13:03:27.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24617" PostTypeId="1" AcceptedAnswerId="24618" CreationDate="2018-01-20T19:40:23.243" Score="5" ViewCount="1439" Body="&lt;p&gt;If you could put a supernova in the Hertzsprung–Russell diagram. Where would it be? Would it be near the white dwarfs or near the super giants?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that a supernova isn't a star but rather the explosion of a star, but if you technically could. Where would it be located?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="21118" LastEditorUserId="21118" LastEditDate="2018-01-20T19:45:24.830" LastActivityDate="2018-01-20T22:05:22.133" Title="Supernova in Hertzsprung–Russell diagram" Tags="&lt;supernova&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24618" PostTypeId="2" ParentId="24617" CreationDate="2018-01-20T22:05:22.133" Score="6" Body="&lt;p&gt;It would be located way off the diagram to the top and left  i.e. moderately hot and very, very luminous. Of course it then changes with time as the supernova fades and cools.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This paper by &lt;a href=&quot;https://arxiv.org/abs/1707.07695&quot; rel=&quot;noreferrer&quot;&gt;Faran et al. (2017)&lt;/a&gt; shows plots of luminosity and temperature versus time for type II supernovae. They start at 15,000K and luminosities of about a billion suns and then fade to about 5000K and by an order of magnitude in luminosity after 50 days.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-01-20T22:05:22.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24620" PostTypeId="1" CreationDate="2018-01-21T10:42:50.947" Score="2" ViewCount="238" Body="&lt;p&gt;As observed from the earth, the sun appears to move in an approximate circular orbit. For the motion of another planet like mercury as observed from the earth, would this be true? If not, why? What about other planets of the solar system?&lt;/p&gt;&#xA;" OwnerUserId="21133" LastActivityDate="2018-01-22T00:59:33.917" Title="Orbit of Mercury" Tags="&lt;orbit&gt;&lt;orbital-mechanics&gt;&lt;mercury&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24621" PostTypeId="1" AcceptedAnswerId="24973" CreationDate="2018-01-21T11:37:19.213" Score="3" ViewCount="280" Body="&lt;p&gt;In the episode of Cosmos dedicated to cometary objects and other things, the Oort cloud was graphically rendered as a spherical distribution of dots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This surprised me as I thought that it should be somehow a disk or a torus, given the solar system's rotational momentum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the Oort ensemble thought to be spherical? Is it deduced from the orbits of the various comets? I think so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why it does not spread into a more disk like object? &lt;/p&gt;&#xA;" OwnerUserId="17437" LastEditorUserId="17437" LastEditDate="2018-02-09T15:20:28.863" LastActivityDate="2018-02-09T15:20:28.863" Title="Oort cloud shape" Tags="&lt;solar-system&gt;&lt;oort-cloud&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24622" PostTypeId="2" ParentId="21222" CreationDate="2018-01-21T14:19:34.520" Score="-2" Body="&lt;p&gt;I think the answer is retrograde. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Earth orbiting satellites, prograde means that when viewed from above the North Pole the satellite moves counterclockwise. However, when viewed from above the north Galactic Pole the solar system orbits clockwise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these two norths consistent? Yes, even though the plane of the solar system is tilted by about 69 degrees from the plane of the Milky Way, it still &quot;points north&quot;, as you suggest in your question. &lt;/p&gt;&#xA;" OwnerUserId="21136" LastActivityDate="2018-01-21T14:19:34.520" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24624" PostTypeId="1" CreationDate="2018-01-21T16:43:24.463" Score="3" ViewCount="480" Body="&lt;p&gt;I am a college astronomy student and I have to find the velocity of Pluto at aphelion and perihelion. I have found the distance of perihelion to be $1.96\times10^9$ km. Aphelion has a distance of $3.27\times 10^9$ km. &#xA;It is given that Pluto's &quot;minimum orbital velocity is 3.27 km/sec&quot;. Which I think means the velocity of Pluto at aphelion is 3.27 km/sec. But then how do I find the velocity at perihelion? &lt;/p&gt;&#xA;" OwnerUserId="21138" LastEditorUserId="18643" LastEditDate="2018-08-24T14:26:52.047" LastActivityDate="2018-08-24T14:26:52.047" Title="How do I determine the velocity of Pluto at aphelion and perihelion?" Tags="&lt;pluto&gt;&lt;velocity&gt;" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="24625" PostTypeId="1" CreationDate="2018-01-21T19:20:55.320" Score="2" ViewCount="124" Body="&lt;p&gt;I'm trying to simulate the lunar eclipse on Jan. 31, similar to &lt;a href=&quot;http://time.com/4882923/total-solar-eclipse-map-places-view/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this simulation&lt;/a&gt; I made of last year's solar eclipse, using Three.js. I have the minute-by-minute coordinates for the Sun and the Moon in the J2000 frame -- where Earth is always at [0,0,0] -- and need to calculate the degree of occultation at each moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm treating the Sun as a point, since that's how 3D lighting works in Three.js, and came up with what I think is a clever solution to account for the penumbra: I calculated what size the Earth would have to be to create the penumbra from a point instead of a large spherical light source, like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ADFdJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ADFdJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This obviously isn't to scale, but the inner circle would be the actual earth, the black lines are the penumbral rays from the Sun, and the blue lines originate from the center of the sun but reach the plane of the moon (the little gray guy) at exactly the same point.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To test whether this actually works, I need to calculate the occultation of the moon and my imaginary Penumbral Earth (which is 8154.15km in radius), and compare those figures to &lt;a href=&quot;https://eclipse.gsfc.nasa.gov/LEplot/LEplot2001/LE2018Jan31T.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fred Espenak's calculations&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found &lt;a href=&quot;https://stackoverflow.com/questions/5188561/signed-angle-between-two-3d-vectors-with-same-origin-within-the-same-plane&quot;&gt;this StackOverflow post&lt;/a&gt; on how to calculate the angle between two vectors in 3D space. What I can't figure out his how to determine the apparent width of that angle and whether it is greater than the sum of the apparent radii of the Earth and Moon. I thought the Law of Cosines would be sufficient, but it accounts for the distance between the Earth and Moon as well. Here's a to-scale simulation showing the path of the moon over 24-hours on Jan. 31, with the Sun way off-screen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MBXTX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MBXTX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One idea I had was to measure the vector from the sun to the outer edge of the fake earth, and the vector from the sun to the inner edge of the moon, but cannot figure out how to determine the points on the sphere where these vectors would be tangent. (I &lt;a href=&quot;https://math.stackexchange.com/questions/2615021/calculating-the-tangent-line-from-a-point-to-the-surface-of-a-sphere?noredirect=1#comment5400837_2615021&quot;&gt;posed this question&lt;/a&gt; to Math StackExchange, but am hoping there's an easier way.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While SPICE (via the Python wrapper spiceypy) was perfect for getting the locations of the bodies, it's occultation function doesn't allow for an arbitrary location of the observer, which eventually I will need to account for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an easy--or difficult--equation for occultation? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="17512" LastActivityDate="2018-01-21T19:20:55.320" Title="How do I measure occultation for the upcoming lunar eclipse?" Tags="&lt;light&gt;&lt;lunar-eclipse&gt;&lt;occultation&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24628" PostTypeId="1" CreationDate="2018-01-21T21:42:13.100" Score="2" ViewCount="78" Body="&lt;p&gt;And why is it moving? Doesn't make very much sense to me, a moving antenna. Or is it kind of radar that is checking  for dangerous objects? These are screen shots from this &lt;a href=&quot;https://www.youtube.com/watch?v=RtU_mdL2vBM&quot; rel=&quot;nofollow noreferrer&quot;&gt;YouTube video&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SDkhb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SDkhb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uJc4F.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uJc4F.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20334" LastEditorUserId="7982" LastEditDate="2018-01-23T12:12:13.350" LastActivityDate="2018-01-23T12:12:13.350" Title="What is this antenna on the ISS?" Tags="&lt;iss&gt;" AnswerCount="1" CommentCount="8" ClosedDate="2018-01-24T02:26:42.053" ContentLicense="CC BY-SA 3.0" />
  <row Id="24629" PostTypeId="1" AcceptedAnswerId="24633" CreationDate="2018-01-21T21:50:26.590" Score="4" ViewCount="296" Body="&lt;p&gt;If a photon is moving past a black-hole, what is the minimum size of the black-hole that will capture the photon and prevent the photon from going onto orbit around the mass? Do black-holes exist that have a stream of photons orbiting them?   &lt;/p&gt;&#xA;" OwnerUserId="21145" LastEditorUserId="21145" LastEditDate="2018-01-21T22:24:53.170" LastActivityDate="2019-12-07T09:15:35.440" Title="minimum size for a black hole to capture a photon" Tags="&lt;black-hole&gt;&lt;light&gt;&lt;photons&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24631" PostTypeId="1" AcceptedAnswerId="24634" CreationDate="2018-01-21T22:06:06.313" Score="2" ViewCount="291" Body="&lt;p&gt;Is the Earth losing or gaining distance from the sun overall? Could asteroids over time add weight to the Earth and push it farther from the sun more than having a degrading orbit around the sun? Does asteroids and space dust increase the mass the the outer planets over sustaining or pulling the Earth from the Sun?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastActivityDate="2018-01-21T23:14:11.817" Title="Is Earth losing orbit?" Tags="&lt;orbit&gt;&lt;earth&gt;&lt;asteroids&gt;&lt;mass&gt;&lt;dust&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2018-01-23T20:10:14.273" ContentLicense="CC BY-SA 3.0" />
  <row Id="24632" PostTypeId="1" CreationDate="2018-01-21T22:39:22.470" Score="3" ViewCount="652" Body="&lt;p&gt;I have been working with the sunrise and sunset formula, using the equations from &lt;a href=&quot;https://en.wikipedia.org/wiki/Sunrise_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;. The problem is that, in my usage thereof, it seems that it only works for sea level; if I start with coordinates that correspond to any point above sea level (or below, for that matter), it will still output the sunrise and sunset for my coordinates at sea level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, today (21/01/2018), in Denver, CO (~1mi above sea level), sunrise was at 07:15:28. Yet my equations keep yielding 07:15:41, which is exactly on-target, were Denver at sea level. How do I account for the fact that higher altitudes experience sunrise earlier than lower altitudes of the same coordinates?&lt;/p&gt;&#xA;" OwnerUserId="17533" LastActivityDate="2018-01-24T18:23:46.807" Title="How do I adjust the sunrise equation to account for elevation?" Tags="&lt;the-sun&gt;&lt;observational-astronomy&gt;&lt;time&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24633" PostTypeId="2" ParentId="24629" CreationDate="2018-01-21T22:54:43.853" Score="5" Body="&lt;p&gt;There isn't really a minimum size. All black holes have an event horizon from which nothing, not even light can escape. The (Schwarzschild) radius of this event horizon is 3km for a black hole of a solar mass and scales linearly with mass. There is a region just outside the event horizon called the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Photon_sphere&quot; rel=&quot;noreferrer&quot;&gt;photon sphere&lt;/a&gt;, within which light can briefly orbit the black hole, but not in a stable fashion. It will either fly off in some other direction or spiral into the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can check out &lt;a href=&quot;http://physics.stackexchange.com/questions/52315/what-is-the-capture-cross-section-of-a-black-hole-region-for-ultra-relativistic&quot;&gt;this Physics SE&lt;/a&gt; question to see a derivation of the impact parameter (the effective radius at which light will be absorbed) and you will see that it is a small numerical factor ($\sqrt{27/4}$ for a non-rotating black hole) times the Schwarzschild radius (the event horizon of a non-rotating black hole), which is indeed proportional to the black hole mass.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-01-21T23:00:55.333" LastActivityDate="2018-01-21T23:00:55.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24634" PostTypeId="2" ParentId="24631" CreationDate="2018-01-21T23:14:11.817" Score="4" Body="&lt;p&gt;What determines the Earth's orbital separation from the Sun is the total mass of the Sun and Earth, as described by Kepler's third law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbital radius of the Earth is &lt;em&gt;increasing&lt;/em&gt; because the Sun is gradually losing mass both by emitting radiation and through the solar wind at a rate of around 5 million tonnes per second (e.g. &lt;a href=&quot;https://arxiv.org/abs/0801.3807&quot; rel=&quot;nofollow noreferrer&quot;&gt;Noerdlinger 2008&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any gain of mass through the accretion of dust or asteroids (by the Earth or Sun) is negligible by comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This mass loss rate is only sufficient to widen the Earth's orbit by about 1.5 cm per year.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-01-21T23:14:11.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24635" PostTypeId="2" ParentId="24620" CreationDate="2018-01-22T00:59:33.917" Score="1" Body="&lt;p&gt;The appearance of circular orbits and the reason why many of the ancient astronomers came to the conclusion that the Sun and stars orbit the Earth is because of the the Earth's rotation.  The rotation, somewhat obviously, operates on a daily (close to 24 hour) cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;fixed&quot; stars appear to orbit the Earth in what's closest to a perfect circle because the Earth's axis doesn't change relative to the Stars (&lt;a href=&quot;https://en.wikipedia.org/wiki/Milankovitch_cycles#Axial_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;it changes but very gradually&lt;/a&gt;), and compared to the stars distance, their movement and Earth's comparatively tiny orbital movement of 2 AU from one side of the sun to the other, makes the stars appear fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stars appear to complete one orbit or circle around the Earth every sidereal day, or &lt;a href=&quot;https://en.wikipedia.org/wiki/Sidereal_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;about 23 hours 56 minutes and 4 seconds&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UqsSO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UqsSO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun appears to orbit the Earth in an imperfect circle, the circle, as with the stars, is driven by the Earth's rotation but the variation to the circle, which shows itself in how high the sun is in the sky and how long a day is, has to do with Earth's axial tilt, where the Earth is in it's yearly orbit and on your latitude on the Earth.   If you lived on the North Pole the sun would appear to move in a circle during the summer, but it would disappear in winter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/c238q.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c238q.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand your question right, you're asking what those circles would look like if you lived on the surface of Mercury, or Venus or Neptune.  (You're not asking what those planets would look like to us, if that's your question google &quot;wandering stars&quot;, but I think your question is what the orbital circles would look like from different planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with the Earth, the 3 primary factors to consider are length of day, length of year and axial tilt and in Mercury's or Pluto's case, eccentricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury for example has essentially no axial tilt, but it has a 3/2 spin/orbit ratio which, effectively means that a day on Mercury lasts 2 years because it only spins 50% more rapidly than it orbits.   Mercury also has a significantly elliptical orbit where, at perihelion, the sun and stars appear to stop in the sky, and this stopping would appear to last a couple earth days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.messenger-education.org/images/archive/around_sun_img.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;enter link description here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because mercury orbits the Sun rather quickly it takes 87 days for the sun to return to the same place in the sky relative to the stars, but because Mercury effectively has 87.6 days worth of daylight and 87.6 days worth of night, the apparent return to where you started would appear every 2 years, every 175.2 days.   Still, on Mercury you'd have the impression of everything orbiting the sun in circles, though you might be close enough and have the right angle to make out the phases of Venus, which we couldn't see on Earth until Galileo's telescope.  The big weirdness from Mercury would be that the stars would appear to stop in the sky once every 87 days, but they'd also move very slowly so this might go unnoticed without some detailed measurement.   Also, on Mercury, you'd observe the sun growing visibly larger or smaller depending on where Mercury was in it's orbit and how close or far from the sun it was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus - the only thing remarkable about Venus is that its day is longer than it's year, so the circles would move in the reverse direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uranus is interesting because it's tilted nearly 90 degrees.  It's seasonal variation of the sun would be far greater than Earth's, but a year on Uranus is 84 Earth years, so the apparent solar movement would take a long time from our perspective.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DilLT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DilLT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, from Uranus and Neptune the orbital distance might be sufficient enough that &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_parallax&quot; rel=&quot;nofollow noreferrer&quot;&gt;parallax could be observed&lt;/a&gt; though the effect would still be small, but possibly observable, you might see, from the proper angle, some slight wobble of near-by stars like Alpha Centauri or Sirius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the gist.  Whether you call the Sun's apparent motion &quot;approximately circular&quot; is a matter of opinion I suppose, but I wouldn't call it circular.  For the closest to circular, you'd want a planet with close to 0 or close to 180 axial tilt (Mercury, Venus, Jupiter), and I don't particularly like the Mercury example cause the Sun stops in the sky once a year.  I suppose you could call that a circular movement with a pit-stop in the great race-track of the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There used to be a great simulated video of what the sun passing over Mercury would look like but I can't find it now.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-01-22T00:59:33.917" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24636" PostTypeId="2" ParentId="24598" CreationDate="2018-01-22T02:25:57.130" Score="1" Body="&lt;p&gt;The azimuth angle from due south to the point where an object rises or sets is a function of the latitude (lat) and declination (decl), as follows:&#xA;cos(angle) = -sin(decl)/cos(lat) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This ignores refraction and the radius of the object, so it will introduce some inaccuracy for the Sun. (The refraction and radius of the Sun amount to 50 arcminutes. The change in azimuth when the Sun rises 50 arcminutes is the error in using this formula. Unless you are at a latitude near a pole, the equation should be accurate to a fraction of a degree.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(edit Jan 23). The maximum declination of the Sun to use in the my formula above is equal to either the mean obliquity of the ecliptic (ζ in your equations) or the corrected obliquity (ϵ) depending on the desired accuracy. The minimum declination of the Sun is equal to the negative value of the obliquity. Thus, you can calculate the maximum and minimum azimuth of the rising and setting sun without knowing the date or time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obliquity is approximately 23.5 degrees, and if you use a latitude of 40 for example, my formula indicates the cosine of the rise point is -sin(23.5)/cos(40) which yields a rise point of 121.4 degrees east of south, or an azimuth of 58.6 degrees (180-121.4=58.6). Sunset is 121.4 degrees west of south, or an azimuth of 301.4 (180+121.4=301.4).&lt;/p&gt;&#xA;" OwnerUserId="17355" LastEditorUserId="17355" LastEditDate="2018-01-23T18:06:59.810" LastActivityDate="2018-01-23T18:06:59.810" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24637" PostTypeId="2" ParentId="24632" CreationDate="2018-01-22T02:40:08.780" Score="3" Body="&lt;p&gt;Higher elevations by themselves do not change the time of sunrise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horizon at sea level is 90 degrees from the zenith. The horizon at Denver is still an angle of 90 deg from the zenith if you assume that the ground is &quot;flat&quot;. In other words, even though you are 1 mile high in Denver, you cannot see the horizon at 0 mile elevation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are on a hill or mountain and looking down so that the distant horizon is more than 90 degrees from the zenith, then sunrise would occur early. If you are in a valley and the sun rises behind a hill or mountain, the sun rises later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need to know the elevation of the observer and horizon (or obstacles). From that you can calculate the altitude of the sky in that direction. From that you can calculate the time of rising.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(edit Jan 24) For points that are close to each other, simple trigonometry can be used.&#xA;$$\tan(\theta)=y/x$$ where y is the difference in elevation between points 1 and 2, and x is the distance. Sunrise occurs when the altitude of the Sun is $\theta-sun's\;radius-refraction$.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/5RrFj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5RrFj.png&quot; alt=&quot;Angle of obstruction on horizon&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more distant points with latitude (lat) and longitude (long), solving for the sides and angles of a triangles gives an approximate solution. (I am sure there are more precise formulas that account for the Earth's shape.) From the figure below,&#xA;$$a=radius\;of\;Earth+Elevation1$$&#xA;$$b=radius\;of\;Earth+Elevation2$$&#xA;$$\cos(\gamma)=\sin(lat1)\sin(lat2)+\cos(lat1)\cos(lat2)\cos(long1-long2)$$&#xA;$$c=\sqrt{a^2+b^2-2ab\cos(\gamma)}$$&#xA;$$\cos(\alpha)=\frac{b^2+c^2-a^2}{2bc}$$&#xA;$$\theta=\alpha-90$$&#xA;and sunrise occurs when the altitude of the sun is $\theta-sun's\;radius-refraction$. Naturally, you need to check all locations along the line of sight between the observer at point 1 and the obstruction at point 2 to find the highest obstruction.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/cmKDL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cmKDL.png&quot; alt=&quot;Obstruction on horizon for distant points. E-F is the theoretical horizon for flat terrain.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17355" LastEditorUserId="17355" LastEditDate="2018-01-24T18:23:46.807" LastActivityDate="2018-01-24T18:23:46.807" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24639" PostTypeId="1" AcceptedAnswerId="24644" CreationDate="2018-01-22T06:11:22.617" Score="1" ViewCount="398" Body="&lt;p&gt;If we went to deep space by a spacecraft and released water into space, what would happen? Would the water freeze to ice or would it remain as a liquid?&lt;/p&gt;&#xA;" OwnerUserId="11792" LastEditorUserId="9301" LastEditDate="2018-01-22T09:06:57.533" LastActivityDate="2018-01-22T09:06:57.533" Title="What would happen if we released water into deep space?" Tags="&lt;space&gt;&lt;water&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24642" PostTypeId="1" AcceptedAnswerId="24646" CreationDate="2018-01-22T07:42:36.997" Score="20" ViewCount="6993" Body="&lt;p&gt;If I had a directional photon-emitting source and placed it inside a black hole pointing upward and out towards the visible universe, I assume the photons traveling at the speed of light would slow and reverse direction back into the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if I took the same source and placed it outside of the black hole pointing inward towards the black hole center, can I assume that an emitted photon would travel towards the center faster than the speed of light it is already traveling at?&lt;/p&gt;&#xA;" OwnerUserId="21145" LastEditorUserId="21145" LastEditDate="2018-01-22T22:12:39.930" LastActivityDate="2018-02-13T08:07:25.697" Title="Speed of light in a black hole" Tags="&lt;black-hole&gt;&lt;photons&gt;" AnswerCount="3" CommentCount="15" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24644" PostTypeId="2" ParentId="24639" CreationDate="2018-01-22T08:26:30.313" Score="2" Body="&lt;p&gt;Water can't exist in a vacuum.  Water in space (atmosphere, no gravity) tends to form into a ball, but take away the atmosphere and it would rapidly fly apart.  It actually wouldn't look like boiling.  You get boiling on Earth because water is held in a pot by gravity.  In empty space the water would rapidly and somewhat explosively disburse &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://scienceblogs.com/startswithabang/2009/06/29/water-in-space-what-happens/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article&lt;/a&gt; says it would look something like what happens when you throw hot water into very very cold air and it rapidly turns to snow.  That's probably a pretty good visual image if you imagine no gravity and everything would just continue to fly apart, not slow by air resistance or fall to the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In cold space the water would freeze, but some of the liquid would also effectively also boil away, even in very cold temperature because liquid isn't stable in a vacuum.  As I said, some of it would boil, but it wouldn't look like boiling, It would fly apart faster than it would freeze but you'd also see it freeze fairly rapidly.   If the water was enclosed, freezing would take some time, but in a vacuum and spreading out, it would freeze into snow and ice crystals rapidly.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A percentage of the water would become gas and that would help carry some heat away from the remaining water, speeding up the freezing process.  That gas would fly away at high velocity and not be visible.   If you collected and weighed the ice-crystals after the water was tossed into space you'd probably find that a percentage of the water escaped as gas.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get an idea of water turning into ice by heat transfer and escape from forming into gas molecules in a video that show what happens to water in a vacuum chamber.   A vacuum chamber can't replicate the rapid spreading out due zero gravity but it demonstrates the water turning to ice, even at room temperature.   &lt;a href=&quot;https://www.youtube.com/watch?v=UG7nsZkVZc0&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's one video&lt;/a&gt; of that.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-01-22T08:26:30.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24646" PostTypeId="2" ParentId="24642" CreationDate="2018-01-22T10:24:21.857" Score="31" Body="&lt;p&gt;It doesn't work like that. An observer at the light source (and indeed any observers anywhere else) will always see light travelling (in vacuum) at the speed of light locally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a major problem with your thought experiment. It is &lt;em&gt;not possible&lt;/em&gt; for you to have a stationary light source within the event horizon of a black hole. It, and everything else in its vicinity, &lt;em&gt;must&lt;/em&gt; be moving inwards. This is as inexorable and unavoidable as is the passage of time for an observer outside the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, the best &quot;visual&quot; way of thinking about the situation inside the event horizon is to imagine your photons of light like salmon trying to swimming upstream, whilst you are on a boat flowing with the stream and releasing the salmon into the water. You will always see the salmon swimming at some speed with respect to your boat. Unfortunately if the stream flows fast enough then the salmon will make no progress and you will both be swept over a waterfall (the singularity) a little further downstream.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, your common-sense fails with the situation of firing light towards a black hole. Light is always measured to have a speed of $c$ &lt;em&gt;locally&lt;/em&gt;. It is following through with the consequences of this principle that leads to all the weird behaviour that black holes exhibit.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-01-22T22:34:56.990" LastActivityDate="2018-01-22T22:34:56.990" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="24647" PostTypeId="2" ParentId="24642" CreationDate="2018-01-22T10:25:43.177" Score="9" Body="&lt;p&gt;You cannot exceed the speed of light &quot;locally&quot;. But you can -see- imagine&lt;sup&gt;*&lt;/sup&gt; &lt;strong&gt;&lt;em&gt;distances increase quicker than the speed of light&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If by traveling you mean &quot;moving compared to local space time&quot;, then light cannot &lt;strong&gt;&lt;em&gt;travel&lt;/em&gt;&lt;/strong&gt; quicker than the speed of light. In your example, the distance increases faster than the speed of light, because &lt;strong&gt;&lt;em&gt;spacetime is dragged along&lt;/em&gt;&lt;/strong&gt; inside the black hole by its gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;*You in fact cannot &quot;see&quot; it, because you would need some information to be transferred somehow to do the same! This is not possible, because of this damn &lt;em&gt;c&lt;/em&gt; limitation.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="11264" LastEditorUserId="332" LastEditDate="2018-01-22T23:17:25.943" LastActivityDate="2018-01-22T23:17:25.943" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24648" PostTypeId="1" AcceptedAnswerId="24652" CreationDate="2018-01-22T11:34:59.733" Score="4" ViewCount="119" Body="&lt;p&gt;I am working on a little project where I want to simulate the solar system to teach myself working with python and OOP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to calculate the orbits &quot;real&quot;, from tangential velocities and gravitational accelerations. For this, in order to not only get circular orbits, but the real, eccentric orbits, I need kind of a snapshot of the solar system with positions and tangential velocities at those positions of the planets (and their moons, but I highly doubt that exists). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I think something like the velocities at the Apoapsis should be enough to derive a reasonably right orbit from it, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't even know if something like that exists, or where to get it from. Does anybody happen to know something of this?&lt;/p&gt;&#xA;" OwnerUserId="21166" LastEditorUserId="21166" LastEditDate="2018-01-22T11:45:43.127" LastActivityDate="2018-01-22T20:41:19.797" Title="snapshot of solar system" Tags="&lt;orbital-mechanics&gt;&lt;eccentric-orbit&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="24652" PostTypeId="2" ParentId="24648" CreationDate="2018-01-22T16:14:27.550" Score="2" Body="&lt;p&gt;You can get a position and velocity snapshot easily using the Python package &lt;a href=&quot;http://rhodesmill.org/skyfield/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Skyfield&lt;/a&gt;. It downloads a JPL &lt;a href=&quot;https://en.wikipedia.org/wiki/Jet_Propulsion_Laboratory_Development_Ephemeris&quot; rel=&quot;nofollow noreferrer&quot;&gt;Development Ephemeris&lt;/a&gt; and then interpolates the positions for whatever time you choose for your snapshot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also use &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL Horizons&lt;/a&gt; to produce one or thousands of positions of solar system bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://space.stackexchange.com/a/23409/12102&quot;&gt;This answer&lt;/a&gt; describes how to set up Horizons to save positions to your computer, and includes a script that can read the format of the saved tables. It also happens to be a simulator that does what you are talking about - statrs with a given position and propagates the orbits. &lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-01-22T20:41:19.797" LastActivityDate="2018-01-22T20:41:19.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24653" PostTypeId="1" AcceptedAnswerId="24659" CreationDate="2018-01-22T17:21:42.683" Score="0" ViewCount="83" Body="&lt;p&gt;It is said that planets are continuously moving away from the sun. At one time Mars was just the right distance to support life at least temperature wise? &#xA;Could Venus be far enough from the sun eventually to cool down? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related:&#xA;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/11958/are-planets-moving-away-from-the-sun&quot;&gt;Are planets moving away from the sun?&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/24631/is-earth-losing-orbit/24634#24634&quot;&gt;Is Earth losing orbit?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastActivityDate="2018-01-23T14:44:06.290" Title="Distance from the sun can equal a hospitable planet?" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;distances&gt;&lt;habitable-zone&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24654" PostTypeId="1" CreationDate="2018-01-22T18:48:58.050" Score="3" ViewCount="32" Body="&lt;p&gt;Pluto and Charon orbit about a common barycenter that isn't inside of either planet. Let's assume that this relationship was scaled up to the point where it could be detected by a search for extrasolar planets. Could any of the methods we currently use tell us the difference between one massive body vs two bodies of nearly identical mass?&lt;/p&gt;&#xA;" OwnerUserId="182" LastEditorUserId="1203" LastEditDate="2018-01-22T23:14:16.623" LastActivityDate="2018-01-22T23:14:16.623" Title="Extra-solar identification of binary planets" Tags="&lt;exoplanet&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24655" PostTypeId="2" ParentId="24598" CreationDate="2018-01-22T19:47:22.770" Score="2" Body="&lt;p&gt;Thanks to &lt;a href=&quot;https://astronomy.stackexchange.com/users/21113/ralf-kleberhoff&quot;&gt;Ralf Kleberhoff&lt;/a&gt; for Pointing me in the right direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 1: Find Summer Solstice Date&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the accuracy needs of this question simply finding the hemisphere and picking the most common date will work. (June 21 in the Northern Hemisphere and December 22 for the Southern Hemisphere). If greater accuracy is required. &lt;a href=&quot;https://astronomy.stackexchange.com/questions/13008/are-there-accurate-equinox-and-solstice-predictions-for-the-distant-past&quot;&gt;There&lt;/a&gt; &lt;a href=&quot;https://stellafane.org/misc/equinox.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;are&lt;/a&gt; &lt;a href=&quot;https://farside.ph.utexas.edu/Books/Syntaxis/Almagest/node36.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;many&lt;/a&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Summer_solstice&quot; rel=&quot;nofollow noreferrer&quot;&gt;resources&lt;/a&gt; &lt;a href=&quot;https://www.esrl.noaa.gov/gmd/grad/solcalc/&quot; rel=&quot;nofollow noreferrer&quot;&gt;available&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2: Find Sunrise and Sunset Times on Solstice Date&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is straightforward. However, for completeness. (Note: &lt;a href=&quot;https://en.wikipedia.org/wiki/Twilight&quot; rel=&quot;nofollow noreferrer&quot;&gt;90.833 is the solar elevation angle for sunrise and sunset.&lt;/a&gt;. However, 90 is sufficient for this accuracy and simplifies the equations.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ha = \arccos\left(\frac{\cos(90^o)}{\cos(\Xi^{rad})\cos(\delta^{rad})} - \tan(\Xi^{rad})\tan(\delta^{rad})\right)^o =  \arccos\left(-\tan(\Xi^{rad})\tan(\delta^{rad})\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$solarnoon = \frac{720 - 4\Xi - \Gamma}{1440}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$sunrisetime = solarnoon - \frac{4ha}{1440}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$sunsettime = solarnoon + \frac{4ha}{1440}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 3: Solve Azimuth Equation for Sunrise and Sunset Times&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply plug the $sunrisetime$ and $sunsettime$ into the equations provided in the question.&lt;/p&gt;&#xA;" OwnerUserId="21070" LastActivityDate="2018-01-22T19:47:22.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24656" PostTypeId="2" ParentId="24654" CreationDate="2018-01-22T22:58:09.057" Score="1" Body="&lt;p&gt;Yes indeed there could, especially in the case of a transiting planet system. Whilst the radial velocity curve would be nice and smooth, reflecting the position of the barycentre of the double planet, the transit light curve would show structure associated with the two planets obscuring the disc of the star. Essentially the transit shape would vary depending on the orbital phase of the double planet and on occasion the transit would occur in two steps as first one planet then the other &#xA;passed in front of the stellar disc. No such objects have been found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People are already looking for this in the context of searching for &quot;exo-moons&quot; - for example,&#xA;the &lt;a href=&quot;http://hek.astro.columbia.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Hunt for Exomoons with Kepler&quot;&lt;/a&gt; programme. Finding a double planet with equal sized components would be comparatively easy compared with the tenuous signals expected from exomoons.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-01-22T22:58:09.057" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24658" PostTypeId="1" CreationDate="2018-01-23T03:32:17.987" Score="1" ViewCount="25" Body="&lt;p&gt;After I back from Minimart, it's about 6pm though, I saw the sky, yeah a beauty evening sky, but I sense a something weird from the sky, It look like a star at first it moving in low speed, I Though that was a Kite or plane, but it cant be plane cuz it moved up then turned right then turned left and get back.&#xA;it size and it light same as a normal Stars.&#xA;It's move like as Kite, but it so bright like star.&#xA;after a few minutes, it's stoped move, but then its Moved Fastest Like A shooting stars, and yes it has a tail like Comet,that moved in Light speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;after that I got back to home,&#xA;I asked to my friend on FB,&#xA;I said that I saw a moving star,&#xA;I didnt say it when,&#xA;but my friend asked me back,&#xA;&quot;was that do along 6pm?&quot;&#xA;&quot;whoa, how can be You know that?&quot;&#xA;and then my friend said if that was an UFO&#xA;he ever saw that many times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(sorry for bad English)&lt;/p&gt;&#xA;" OwnerUserId="21189" LastEditorUserId="21189" LastEditDate="2018-01-23T03:46:49.963" LastActivityDate="2018-01-23T03:46:49.963" Title="A moving star at 6pm (Indonesia Time)" Tags="&lt;star&gt;&lt;identify-this-object&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2018-01-23T15:02:51.697" ContentLicense="CC BY-SA 3.0" />
  <row Id="24659" PostTypeId="2" ParentId="24653" CreationDate="2018-01-23T04:22:30.813" Score="5" Body="&lt;p&gt;It's actually the opposite.  Venus, Mars and I'll include the Earth as well, are likely getting more solar energy per square meter, not less, over time, even as they slowly move away from the sun.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.5 cm per year, from the Rob Jeffries' answer to the &lt;a href=&quot;https://astronomy.stackexchange.com/questions/24631/is-earth-losing-orbit/24634#24634&quot;&gt;2nd question you linked&lt;/a&gt; is too slow to make much difference.  That's 1 km every 66,000 years and about 70,000 km over the age of the solar-system at that rate.   Now, planets may have moved away at different rates, especially when the solar-system was young and orbits were more chaotic and when Jupiter was thought to be doing it's dance in and back out, so it's not precisely known how close Mars was to the sun, say, 3.7 to 4.1 billion years ago, but we do know that the Sun was a little smaller and not as hot at that time, perhaps about 25-30% less luminous overall, though it was also more active with solar storms.   But Mars probably got less solar energy 4.1 billion years ago than it gets now, and it will get more solar energy a billion or two billion years in the future, even though it will probably be a little further from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary factor is how much of an atmosphere it had.   Planets can't have liquid water oceans without a reasonably thick atmosphere and Mars is believed to have &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Mars#History&quot; rel=&quot;nofollow noreferrer&quot;&gt;had an atmosphere&lt;/a&gt; long ago.   How close it was to the sun matters too, but the thickness and composition of the atmosphere matters more.  Heat from the planets interior also matters and all the planets had more internal heat after formation, in the 3.7 to 4.1 billion years ago range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its also not 100% certain that Mars ever had relatively permanent liquid oceans and what we'd consider moderate temperatures.   Mars' geology suggests it had floods, but flooding could have been caused by local or temporary heating from volcanic activity or large meteor impacts.   It might also have been driven by a a more extreme axis leading to greater seasonal variation with glacial melt every summer (687 days in a year so summer lasted longer), and perhaps re-freezing every winter.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Mars_ocean_hypothesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mars Ocean Hypothesis&lt;/a&gt; is a good hypothesis, don't get me wrong, but it's worth pointing out that its not a certainty.  It's possible that the oceans were ice covered on top but liquid underneath from heat coming from inside the planet.   Nobody knows for certain how temperate Mars was 3.7 to 4.1 billion years ago.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-01-23T14:44:06.290" LastActivityDate="2018-01-23T14:44:06.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24662" PostTypeId="1" CreationDate="2018-01-23T07:12:26.887" Score="1" ViewCount="93" Body="&lt;p&gt;When we would be able to tell with surety that universe is finite or infinite? How can we exactly determine the density of the universe so, that we compare it with critical density, in order to determine its geometry? Does the infinite Universe entails infinite black holes or infinite stars?&lt;/p&gt;&#xA;" OwnerUserId="20162" LastEditorUserId="1891" LastEditDate="2018-01-25T06:07:22.000" LastActivityDate="2018-01-25T06:07:22.000" Title="Measuring the size and the geometry of Universe" Tags="&lt;black-hole&gt;&lt;cosmology&gt;&lt;infinite&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2018-01-24T03:03:51.017" ContentLicense="CC BY-SA 3.0" />
  <row Id="24664" PostTypeId="1" AcceptedAnswerId="24692" CreationDate="2018-01-23T09:03:45.100" Score="3" ViewCount="171" Body="&lt;p&gt;I've seen quite a few &lt;a href=&quot;https://amazingsky.net/2018/01/22/the-beauty-of-the-milky-way/&quot; rel=&quot;nofollow noreferrer&quot;&gt;time-laps films&lt;/a&gt; where the Milky Way is seen as bright and filling the sky, as some fantastic glowing cloud full of glittering stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if it is possible to see this (or something a little similar) through a telescope using my eyes (not a camera)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that you would want as little magnification as possible to see the sky like this, and it would be all about gathering as much light as possible? &lt;a href=&quot;https://astronomy.stackexchange.com/a/2198/21201&quot;&gt;This answer&lt;/a&gt; describes this scenario as &quot;Low Brightness&quot; (and low resolution) objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a &lt;a href=&quot;https://amazingsky.net/2018/01/22/the-beauty-of-the-milky-way/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sky Watcher Infinity 76P telescope&lt;/a&gt;. At the moment, it has a 30x erecting lens that tends to wobble in its helical mount, and doesn't give a good image. I've read claims that with a non erecting lens and a bit of care, this telescope can give good images.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Are non magnifying lenses available for the Sky Watcher? Or would I be better off taking a different approach completely, such as buying binoculars?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I'd like to be able to share this with my children, and they find the existing Sky Watcher lens hard to use – as well as its wobble, I think this is because it has a narrow &quot;exit pupil&quot; (?), so that you need to get your eye in just the right place. Can I get a low magnification lens that is more forgiving for the user?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can I expect to be able to see something from my backyard in a small UK town? Some of the shots in that film I link to seem to show clear bright sky pictures when there are other sources of light on the ground, or even the setting sun.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="21201" LastActivityDate="2018-01-24T21:13:12.867" Title="Can my Sky Watcher Infinity 76P &quot;make the sky bright&quot; &amp; do I want it to?" Tags="&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24667" PostTypeId="2" ParentId="24628" CreationDate="2018-01-23T12:10:24.243" Score="1" Body="&lt;p&gt;That is &lt;a href=&quot;http://winds.jpl.nasa.gov/missions/RapidScat/&quot; rel=&quot;nofollow noreferrer&quot;&gt;RapidScat&lt;/a&gt;, described in &lt;a href=&quot;https://space.stackexchange.com/a/8742/12102&quot;&gt;this answer&lt;/a&gt; in the Space Exploration SE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cdkOW.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cdkOW.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-01-23T12:10:24.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24670" PostTypeId="1" AcceptedAnswerId="26341" CreationDate="2018-01-23T17:59:32.663" Score="2" ViewCount="266" Body="&lt;p&gt;I am doing a homework on globular clusters and could not find trustworthy data on the average diameter of a globular cluster. Most clusters I included had a diameter of 120-150 ly.&#xA;But is &lt;strong&gt;that&lt;/strong&gt; the average diameter of such globular clusters?&lt;/p&gt;&#xA;" OwnerUserId="21212" LastActivityDate="2018-06-19T17:55:54.227" Title="Globular Cluster diameter" Tags="&lt;globular-clusters&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24671" PostTypeId="1" AcceptedAnswerId="24691" CreationDate="2018-01-23T20:17:56.650" Score="-1" ViewCount="85" Body="&lt;p&gt;The sun is losing mass(1.5cm per year) by emitting radiation and the orbits of the planets are widening because of the weakening gravity of the sun.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are most of the stars shedding mass via radiation also decreasing the total mass of the universe contribute to the increasing redshift of the universe? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from the loss in mass/gravity. How much force is coming from the radiation pushing on itself from inside the universe? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does all that lost mass from radiation go from the universe?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-01-24T22:54:43.687" LastActivityDate="2018-01-24T22:54:43.687" Title="Can the shedding of radiation of stars contribute to the red-shift of the universe?" Tags="&lt;star&gt;&lt;solar-system&gt;&lt;universe&gt;&lt;redshift&gt;&lt;solar-wind&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24672" PostTypeId="2" ParentId="2441" CreationDate="2018-01-23T21:07:22.637" Score="4" Body="&lt;p&gt;Thought provoking cosmologists!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm uber late to this discussion as I see it has been ongoing for literally years and don't know if there is still anyone monitoring this thread, but here' goes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I studied astrophysics at UC Berkeley in the late 80's so perhaps my info is a little dated, upfront apologies if so. I spent &lt;em&gt;a lot&lt;/em&gt; of time thinking about this problem for the past 30 years and have postulated a couple ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, these conjectures are based on the presumptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;time stops at the event horizon&lt;/li&gt;&#xA;&lt;li&gt;an in-falling observer into the EH looking backward would watch the universe rapidly age to heat death&lt;/li&gt;&#xA;&lt;li&gt;non-charged, non-rotating, solar-mass black hole&lt;/li&gt;&#xA;&lt;li&gt;a star of 2-3 solar masses is sufficient to overcome neutron degeneracy pressure and form a black hole (call it 2 for the discussion)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If true, then conjecture:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;start with star of say 3 solar masses&lt;/li&gt;&#xA;&lt;li&gt;we must consider an even horizon's existence and parameters from its &quot;birth&quot;&lt;/li&gt;&#xA;&lt;li&gt;minimum Schwarzschild radius is only 12 miles (2 solar masses)&lt;/li&gt;&#xA;&lt;li&gt;original main sequence star radius appx 100,000Km (100M+ Km for red giant)&lt;/li&gt;&#xA;&lt;li&gt;observer is orbiting star initially&lt;/li&gt;&#xA;&lt;li&gt;star burns through last percentages of helium and cascade collapses directly to black hole&lt;/li&gt;&#xA;&lt;li&gt;as star collapses, some amount of matter contracts to within 12 miles of star centroid (call it 2 solar masses)&lt;/li&gt;&#xA;&lt;li&gt;Event horizon now mathematically is formed and time STOPS for all matter at that radius&lt;/li&gt;&#xA;&lt;li&gt;Matter &lt;em&gt;outside&lt;/em&gt; that radius continues to fall in since time has not yet stopped &lt;em&gt;creating a compression sphere&lt;/em&gt; around the EH&lt;/li&gt;&#xA;&lt;li&gt;Matter already INSIDE the EH continues to fall. (it has momentum that must be conserved).  OR, does time STOP WITHIN the EH as well throughout the entire Schwarzschild sphere causing all that matter to FREEZE in position (reletive to the outside observer?  (unknown)  (perhaps time REVERSES?!)&lt;/li&gt;&#xA;&lt;li&gt;The outside observer would watch the matter at the EH STOP falling and radiating&lt;/li&gt;&#xA;&lt;li&gt;The in-falling matter looking back at the universe would now watch the universe rapidly age, perhaps even to its death?&lt;/li&gt;&#xA;&lt;li&gt;If so, this means that all in-falling EH matter, after the EH has formed, is trapped at the EH until the black hole evaporates.&lt;/li&gt;&#xA;&lt;li&gt;Which also leads to a huge COMPRESSION of the in-falling matter in successively faster time frames falling in from behind.&lt;/li&gt;&#xA;&lt;li&gt;In the given example, this is an entire solar mass of matter all rapidly compressing and increasing in pressure. (At the EH time has stopped so no interaction is going on from the perspective of our orbiting solar observer, but successively less time dilated layers further away from the EH compress into the equivalent of a whole new star burning its fuel in fempto-seconds to many seconds.  I.E.  SUPERNOVA)&lt;/li&gt;&#xA;&lt;li&gt;AND there is the gravitational attraction imbalance that has now been caused between the solar mass of matter at the EH and the 2 solar masses inside the SW&lt;/li&gt;&#xA;&lt;li&gt;In fact, all blackholes formed by stellar collapse should begin life as a 12 mile wide schwarzschild radius of 2 solar masses.&lt;/li&gt;&#xA;&lt;li&gt;Growth of blackhole sizes of this type (excluding primordial black holes) should ONLY be due to EH matter accretion or merging of black hole's EH's.&lt;/li&gt;&#xA;&lt;li&gt;No matter should ever fall into (or through) the EH in our lifetimes, or even the lifetime of the universe as long as we maintain that an observer falling into a blackhole sees the universe rapidly age behind him/her (the future)&lt;/li&gt;&#xA;&lt;li&gt;Therefore, all detection of radiation from black holes is due to interactions of matter very near the EH&lt;/li&gt;&#xA;&lt;li&gt;Which begs the question, does GRAVITY transcend the EH?&lt;/li&gt;&#xA;&lt;li&gt;If not, a black hole should lose &quot;2&quot; solar masses at is creation (can only test if we can measure the mass before an after its creation in perhaps a visible binary pair supernova&lt;/li&gt;&#xA;&lt;li&gt;But if gravity DOES transcend the EH as accepted, then the gravity of the solar masses at the EH should be exerting an opposing force on the matter inside, DECELERATING the collapse inside the EH!&lt;/li&gt;&#xA;&lt;li&gt;Also, there should be a &quot;ringing&quot; effect &quot;heard&quot; by our new gravity wave detectors as the matter inside doesn't just collapse infinitely into a singularity but &quot;bounces&quot; and reverberates from gravity and different layers of time dilation&lt;/li&gt;&#xA;&lt;li&gt;As a though, this COULD even result in a &quot;torus&quot; of sorts with dimensionalities being swapped or reversed (time/distance), rather than a singularity&lt;/li&gt;&#xA;&lt;li&gt;Add to this the planck density with these opposing internal forces and we possibly end up with some bizarre space-time topologies. &lt;/li&gt;&#xA;&lt;li&gt;PURE speculation: The environment INSIDE the black hole starts to look a whole lot like our own universe's history from the big bang (white hole?) if you simply change time's arrow.  (universe isn't expanding as we perceive but compressing into a torus with different distances from our viewpoint having different levels of time dilation)&lt;/li&gt;&#xA;&lt;li&gt;As an undergrad I wrote a paper that our universe is the inside of a black hole and I have seen many theories gravitating (sorry) to this solution over the past 30 years&lt;/li&gt;&#xA;&lt;li&gt;Including the most recent ideas that our universe is a compressed (3D) hologram on a 4 dimensional &quot;sphere&quot; that represents an event horizon equal to the entropy of our entire known universe.  Elegant.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Apologies for the terribly long winded comments here.  I'm sure the idea has more holes than swiss cheese.  Which is what the universe starts to look like with all these little pocket universes forming that we can't interact with!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question and theanswer that could take use to the next level of understanding of these concepts is this:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can an event horizon change shape?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the matter is time-dilation-locked to the event horizon, it cannot move (relative to the EH). If in-falling matter can witness the end of the universe, or even just a very long time, then the matter is time-dilation-locked by definition.  If it is NOT TD-locked, an in-falling observer SHOULD NOT BE ABLE TO SEE THE UNIVERSE RAPIDLY AGE BEHIND THEM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then for if the EH can change shape, either:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;matter needs to move with the EH  (acceleration? momentum? free energy?)&lt;/li&gt;&#xA;&lt;li&gt;OR the EH, being a mathematical definition, can move irrespective of the location of the matter, thereby changing the amount of time-dilation on the matter, slowing/speeding/stopping matter outside the EH, or UNKNOWN effects if it was already inside the EH.  (presumably an EH will always INCREASE in size, but what about shape?)&lt;/li&gt;&#xA;&lt;li&gt;back to shape:  Can an EH be ellipsoid?  Pancake?  If it can change from sphere to pancake, does this not mean that matter that was already inside the EH near the spherical radius has now EMERGED FROM THE BLACK HOLE as it were if that radius suddenly shrinks? (unless, again, it is somehow dragged with the EH)&lt;/li&gt;&#xA;&lt;li&gt;if this is the case, would not black hole mergers allow for matter from within the EH to escape there by constituting a whole new form of black hole emission &lt;em&gt;other than Hawking radiation&lt;/em&gt;? How would we detect this? How would we know?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I think the answer lies squarely with LIGO and more powerful versions of this instrument to be brought online in the future. Observing changes, arrival times, spectrum comparisons, and eventually direction, of gravity waves and their associated gamma ray bursts from merging black holes will help us pin down exactly what happens when event horizons collide!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for taking the time to review these ideas!&lt;/p&gt;&#xA;" OwnerUserId="21213" LastEditorUserId="1891" LastEditDate="2018-01-24T13:21:47.977" LastActivityDate="2018-01-24T13:21:47.977" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24673" PostTypeId="2" ParentId="24662" CreationDate="2018-01-23T21:29:15.893" Score="2" Body="&lt;p&gt;The relative recent &lt;a href=&quot;https://en.wikipedia.org/wiki/WMAP&quot; rel=&quot;nofollow noreferrer&quot;&gt;WMAP&lt;/a&gt; result shows, the Universe seems infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the &quot;is the Universe infinite&quot; question has a hidden problem: it doesn't matter, how far away can we see, we can't be sure, if anything out of it is finite or not. :-) Thus, it is about a definitely unknowable thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The WMAP measured essentially the curvature of the Universe and it found a zero curvature (i.e. it seems to be planar). If there is a minimal curvature, below a measurement precision, and we have a cyclic, spherical geometry, then its radius must be &lt;a href=&quot;https://physics.stackexchange.com/q/88786/32426&quot;&gt;at least 300billion light years&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastActivityDate="2018-01-23T21:29:15.893" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24674" PostTypeId="1" AcceptedAnswerId="24676" CreationDate="2018-01-24T00:01:20.040" Score="3" ViewCount="916" Body="&lt;p&gt;My task is to calculate current declination and right ascension for some artificial Earth satellite. For example ISS. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using latest TLE data for extracting keplerian orbital elements and then calculate current cartesian orbital state vectors in ECI frame. &#xA;From the position vector I directly calculate declination and right ascension, using cartesian to spherical coordinates conversion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right ascension gets calculated correctly but I am having problems getting the same result for declination as the one displayed by satellite trackers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speaking of satellite trackers, I don't understand how is their declination value even correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Declination is an angle measured north or south of the celestial equator.&#xA;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/9/98/Ra_and_dec_on_celestial_sphere.png/800px-Ra_and_dec_on_celestial_sphere.png&quot; alt=&quot;declination visually&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Celestial equator of course is on the same plane as the equator of Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then how is it possible that, when a satellite (ISS in this example) is almost directly above Earth's equator, satellite trackers calculate declination to be around -29° ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/CKCBMGv.png&quot; alt=&quot;tracker 1&quot;&gt;&#xA;&lt;img src=&quot;https://i.imgur.com/i4sOxe8.png&quot; alt=&quot;tracker 2&quot;&gt;&#xA;&lt;img src=&quot;https://i.imgur.com/B9kGttj.png&quot; alt=&quot;tracker 3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shouldn't the declination angle be zero? &#xA;My calculations for this &quot;ISS above equator&quot; example did give declination to be close to 0°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who is doing the calculations wrong and why?&lt;/p&gt;&#xA;" OwnerUserId="21215" LastActivityDate="2018-01-24T01:26:12.827" Title="Declination calculation" Tags="&lt;orbit&gt;&lt;satellite&gt;&lt;declination&gt;&lt;right-ascension&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24675" PostTypeId="1" AcceptedAnswerId="24808" CreationDate="2018-01-24T00:10:59.053" Score="6" ViewCount="189" Body="&lt;p&gt;I was reading about how Saturn has two moons, Janus and Epimetheus, that &lt;a href=&quot;https://www.wikiwand.com/en/Horseshoe_orbit&quot; rel=&quot;noreferrer&quot;&gt;swap orbits&lt;/a&gt; once every four years. Could something like this happen on a much larger scale, but with stars instead of a planet and moons? Let's say you had a large, massive star with two red dwarfs orbiting around it. Could those two red dwarf stars fall into a horseshoe orbit, regularly swapping orbits every few years or so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.wikiwand.com/en/Co-orbital_configuration&quot; rel=&quot;noreferrer&quot;&gt;more about horseshoe orbits&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18831" LastActivityDate="2018-01-31T21:19:42.867" Title="Is it possible for two stars to be in a horseshoe orbit around a much larger star?" Tags="&lt;star&gt;&lt;orbit&gt;&lt;binary-star&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24676" PostTypeId="2" ParentId="24674" CreationDate="2018-01-24T01:26:12.827" Score="2" Body="&lt;p&gt;You are forgetting parallax. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something which is very distant and lies on the celestial equator will have a declination of 0°, but if it is nearby then its declination will only be 0° to an observer located along Earth's equator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From your location in Zagreb, something like the ISS, even if it is crossing Earth's equator at a longitude of ~16° will appear to be in the southern sky, and your declination coordinate will likely be south of 0°.&lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2018-01-24T01:26:12.827" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24677" PostTypeId="1" CreationDate="2018-01-24T01:35:23.347" Score="16" ViewCount="4372" Body="&lt;p&gt;I wonder why the Earth-Moon distance is not the same at each perigee/apogee. Isn't the Moon's orbit a fixed ellipse with Earth at one of the foci? If it is so, shouldn't the distance at perigee/apogee be a fixed value?&lt;/p&gt;&#xA;" OwnerUserId="21216" LastEditorUserId="332" LastEditDate="2018-01-25T06:08:22.190" LastActivityDate="2018-02-22T22:33:29.553" Title="Why is the Earth-Moon distance not the same at each perigee/apogee?" Tags="&lt;the-moon&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24678" PostTypeId="1" CreationDate="2018-01-24T02:16:09.003" Score="3" ViewCount="66" Body="&lt;p&gt;My question is regarding the direction/orientation of earth. Is it a constant with reference to universal parameters?&lt;/p&gt;&#xA;" OwnerUserId="21217" LastActivityDate="2018-01-25T20:20:41.190" Title="How to fix a Direction in the context of universe?" Tags="&lt;universe&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2018-01-26T15:35:23.347" ContentLicense="CC BY-SA 3.0" />
  <row Id="24679" PostTypeId="2" ParentId="24678" CreationDate="2018-01-24T03:05:23.763" Score="1" Body="&lt;p&gt;I had a nice long answer all mapped out, then I came across a similar (identical?) question...&#xA;refer answer here:&#xA;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/19592/how-does-the-earths-plane-of-orbit-ecliptic-vary-over-time&quot;&gt;How does the earth&amp;#39;s plane of orbit (ecliptic) vary over time?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically it is changing about 1.4 degrees per 10000 years about now, but that will change chaotically over longer periods.&lt;/p&gt;&#xA;" OwnerUserId="21218" LastActivityDate="2018-01-24T03:05:23.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24680" PostTypeId="2" ParentId="24677" CreationDate="2018-01-24T06:17:13.963" Score="19" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Isn't the Moon's orbit a fixed ellipse with Earth at one of the foci?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, it's not. This isn't even true for the planets' orbits about the Sun. Each planet perturbs the orbits of the other planets, making Kepler's ellipses approximately correct rather than exact. The Moon's orbit is strongly perturbed by the Sun in a number of ways. The Moon's orbit deviates from being a fixed ellipse in a number of ways. One result of these solar perturbations (and to a much lesser extent, perturbations from Venus and Jupiter, and to an even lesser extent, from the other planets) is that the Moon's orbit precesses in a number of ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One such precession is the apsidal precession. The line from the Earth to the point at which the Moon reaches perigee does not point to a fixed position in space. It instead precesses with a period of about 8.85 years. This is what results in the so-called supermoons, which occur when the Moon's orbit is close to perigee when the Moon is full.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another such precession is the nodal precession. The line of nodes (where the Moon crosses from above to below the ecliptic, and vice versa), also precesses, but with a period of about 18.6 years. We only get eclipses when the Moon is very close to a node at a syzygy (either a full Moon, resulting in a lunar eclipse, or a new Moon, resulting in a solar eclipse).&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="21229" LastEditDate="2018-01-24T11:56:31.780" LastActivityDate="2018-01-24T11:56:31.780" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24681" PostTypeId="2" ParentId="24677" CreationDate="2018-01-24T06:50:12.477" Score="10" Body="&lt;p&gt;If the Moon and Earth were far away from any other gravitational bodies, then the orbit would be not only very consistent but also very close to circular as well.  Orbits like the Earth-Moon, where the mutual tidal force is strong and the inner body's rotational energy is transferred to the smaller body's orbital energy, those orbits tend to circularize over time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mathematics behind 3-body gravitation is pretty intense, and above my pay-grade, but I can explain with a visual.   The easiest way to picture this is with tidal forces.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We think of tidal forces as only affecting a solid body like waves on the Earth or the permanent tidal bulge on the moon, but all tidal forces are is a variation in the gravitational pull over different distances and because the Earth and Moon are bound to each other by gravity, that means the solar tidal force can be applied to the Earth-Moon system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fFGEQ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fFGEQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational pull from the Sun is stronger on the side of the planet closer to the sun and weakest on the opposite side.  This also happens relative to the Earth and Moon when one or the other is closer to the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gPG94.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gPG94.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the Earth/Moon orbit are in full moon or new moon, then the tidal force exerted by the sun is stronger on the closer body, weaker on the further body and the orbit effectively stretches in the direction of the arrows on the image above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the Earth-Moon orbit are in last quarter or first quarter, the tidal force exerted by the sun is in the perpendicular direction inwards, and the orbit effectively squashes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, the forces also have effects at the quarter points as well and everywhere in-between.  When the Moon is at waning crescent or waxing gibbous the Sun exerts more force on the closer object and less force on the further away object not resulting in a change in shape so much, but the force effectively accelerates the objects with respect to each other making them move slightly faster. The opposite happens at waning gibbous and waxing crescent: the Sun is effectively slowing down the relative speed between the Earth and the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, the Sun is constantly pulling or pushing the moon relative to the Earth, so there's a continuous stretching and squashing and accelerating and decelerating of the Moon's orbit around the Earth (or around the barycenter for you purists).   You might think this could shake the Moon loose from the Earth, and it would, if the Moon was about 30%-50% further away than it is now.   It's this tidal pulling and stretching that defines the vague border that's the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere#True_region_of_stability&quot; rel=&quot;noreferrer&quot;&gt;stable region of the Hill sphere&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This solar tidal effect is cyclical, operating every time the Moon completes a full moon cycle, which is a synodic orbit of about 29.5 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PkCCC.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PkCCC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon's &quot;Kepler orbit&quot; is a sidereal orbit of about 27.3 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What does this look like?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BmNWx.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BmNWx.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall effect, (noted in the other answer), is an unusually high &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_precession#Apsidal_precession&quot; rel=&quot;noreferrer&quot;&gt;lunar apsidal precession&lt;/a&gt; of just 8.85 years, or just over 118 sidereal (or Kepler) orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means is that the Moon's apogee and perigee shift about 3 degrees for each lunar orbit.  The Moon can't settle into a consistent orbit because of the solar gravitation acting upon it, and the tidal force on the Earth-Moon system is significant.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth, for comparison, has an &lt;a href=&quot;https://en.wikipedia.org/wiki/Apsidal_precession&quot; rel=&quot;noreferrer&quot;&gt;apsidal precession&lt;/a&gt;, mostly driven by Jupiter and Saturn, of about 112,000 years, or 112,000 orbits.   That's about a thousand times less angular change per orbit.   As a sidebar, the objects inside  the orbit, Venus for example, don't have much effect on Earth's orbit.  It's the outer planets that primarily drive apsidal precession.   Neptune, for example, has no outer-planets to speak of, and if planet 9 is found, it would be too far away, so Neptune's orbit is nearly circular.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="21229" LastEditDate="2018-01-24T13:21:00.300" LastActivityDate="2018-01-24T13:21:00.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24691" PostTypeId="2" ParentId="24671" CreationDate="2018-01-24T16:21:46.410" Score="1" Body="&lt;p&gt;The intuition in the question seems to be that if stars get lighter and this makes orbits widen, maybe the same effect explains why the universe &quot;widens&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars are converting some of their mass into light, but this is a small fraction: &lt;a href=&quot;https://physics.stackexchange.com/questions/367611/what-is-the-theoretical-efficiency-of-fusion&quot;&gt;the efficiency of fusion is about 0.5-1% of mass into energy&lt;/a&gt;. And many stars will not burn all the fusible matter in any case. Brighter stars also tend to lose mass through stellar winds, reaching rates up to $10%{-6} M_\odot$ per year or above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starlight tends to escape galaxies, while the stellar winds mostly feed the interstellar medium. Most mass hence remains in the galaxies and stellar activities cannot explain their accelerating expansion. The small amount of escaping starlight and very hot gas does make galaxies a bit less strongly bound in clusters, but it is small compared to estimated dark matter halo masses and (maybe) infall of intergalactic gas. &lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2018-01-24T16:21:46.410" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24692" PostTypeId="2" ParentId="24664" CreationDate="2018-01-24T16:40:04.750" Score="7" Body="&lt;p&gt;To make the images in the time-lapse (and similar Milky Way photos), the photographer uses time exposures with wide angle lenses, may stitch together multiple images from longer lenses, and uses extensive post-processing to balance contrast and include foreground or bright objects in the scene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest you can get to the same thing visually is to go someplace with very dark skies (middle of the ocean, or a wilderness with no nearby towns) and use the naked eye or wide-angle binoculars.  In dark skies, the Milky Way overhead is obvious and impressive, but it still won't match the color or vividness of the photos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your telescope isn't really useful for looking at the Milky Way as a whole, but it can probably give a satisfying view of the many star clusters that dot the Milky Way--the Pleiades and the Beehive are naked eye objects and so shouldn't be too hard to find with your telescope.  When these clusters are &quot;resolved&quot; into their individual stars, they are impacted less by the light-pollution you probably have in your backyard location.  I can't tell if the 76p eyepiece is one of the standard sizes that can be replaced (1.25&quot;, 0.965&quot; OD).  If it is, an inexpensive eyepiece of about 25mm focal length might improve your views of astronomical objects.&lt;/p&gt;&#xA;" OwnerUserId="14434" LastActivityDate="2018-01-24T16:40:04.750" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24693" PostTypeId="1" AcceptedAnswerId="24694" CreationDate="2018-01-24T17:10:12.740" Score="4" ViewCount="22150" Body="&lt;p&gt;I've been hearing about Halley's comet for a while now, and I'm just curious as to when and where to see it, and for how long. :)&lt;/p&gt;&#xA;" OwnerUserId="21239" OwnerDisplayName="Imtherealsanic" LastEditorUserId="1462" LastEditDate="2018-01-24T20:19:49.887" LastActivityDate="2019-10-20T20:57:47.987" Title="when are you able to see Halley's comet, and for how long?" Tags="&lt;comets&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24694" PostTypeId="2" ParentId="24693" CreationDate="2018-01-24T20:37:45.590" Score="13" Body="&lt;p&gt;Halley's comet orbits the sun and its orbit lasts 75 years. The orbit is a long elliptical orbit. For most of those 75 years it is a cold black dot, and frozen solid in the outer solar system. But a short period during that orbit it gets close to the sun. The ice and gas begin to boil of its surface and gets blown back by the solar wind of the sun. It then starts to look like a comet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last time Halley's comet was close to the sun was in 1986, and so the next time will be in 2061. You won't be able to see Halley's comet until then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3H7ne.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3H7ne.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;(image from &lt;a href=&quot;https://www.uwgb.edu/dutchs/PLANETS/Comets.HTM&quot; rel=&quot;noreferrer&quot;&gt;University of Wisconsin&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember 1986, and the comet was visible but hard to see. It was a very dim fuzzy dot, that was just about visible in my binoculars. This is because, unfortunately, the Earth was a long way from the comet. In 2061 Earth will be even further from the comet, however Earth will be better positioned when Halley is closest to the sun, so it should make a good show, and be &lt;a href=&quot;https://docs.google.com/viewer?url=http://www.abramsplanetarium.org/Programs/CaseForCometHalley2061.pdf&quot; rel=&quot;noreferrer&quot;&gt;clearly visible to the naked eye&lt;/a&gt;. Comet predictions are hard as much depends on exactly how much water vapour boils off, and this is hard to predict. Comets can have sudden changes in activity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, Halley's comet isn't the only comet. There are others that can be as visible as Halley will be in 2061. There are also long period comets that are undiscovered until they come close to the sun. These can be very bright. Comet Hale-Bopp was a &quot;great&quot; comet in 1997 and was visible for many months. Another great comet could come along any time but we can't predict when.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-01-25T17:05:03.307" LastActivityDate="2018-01-25T17:05:03.307" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24695" PostTypeId="2" ParentId="24664" CreationDate="2018-01-24T21:13:12.867" Score="4" Body="&lt;p&gt;The answer provided by @antlersoft is very much on point. Just adding to it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The SW76 is a fun little scope mainly concerned with ease of use. It's a lot simpler to handle than refractors of similar aperture, and it's very robust. If there ever was a child-friendly scope (in a good way), this would be it - but adults can have fun with it too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There might be several different models available? Not sure. I'm seeing reviews saying that it actually comes with two eyepieces, a 15x and a 30x. The 15x would provide a wider view, closer to binoculars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, it uses a proprietary format for the eyepieces, so you can't just buy some commonly available eyepiece and use it. This is my main criticism for an otherwise well-designed instrument. If you're missing the 15x lens, maybe you could find it cheap on eBay?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am the author of the answer you're quoting. Let me rephrase it. DSOs (deep sky objects) or &quot;low brightness&quot; objects are very sensitive to light pollution. If you live in the city, no matter what telescope you use, DSOs will not look that great. They just drown in the ambient light, reflected back down to Earth by the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will see most of the Messier objects (the ones that begin with the letter M, followed by a number) with your instrument, they just don't look super-impressive. E.g. the Orion Nebula (M42) is visible now in the winter months and you'll see it as a little smudge or &quot;cloud&quot; - just google &quot;how to find the orion nebula&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will never see the sky full of stars unless you go far away from the city. In my trips to the Death Valley I've carried binoculars with me, and it's amazing - while in the desert, point the lens in any direction and it's like a river of stars. But a very dark sky is mandatory to see it like this. This might be a little tricky to find in the Old World. There are many light pollution maps online that could help you find good spots. But as a rule of thumb - far away from people and industrial facilities is good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could find the 15x lens, your scope would be more or less equivalent to a pair of powerful binoculars, a 15x76 effectively, very similar to what I have (15x70). The 30x would work also, but the field of view would be 2x smaller (you're zooming in with the stronger lens).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Re: the usability issues&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what to say about the wobble. Hopefully nothing is broken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you're experiencing is not the narrow exit pupil. At 30x with a 76mm aperture the exit pupil is 2.5mm, which is plenty. Things get difficult below 0.5mm. It's probably a combination of small eye relief and narrow apparent field of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eye relief is the optimal distance between the lens and your eye. High end eyepieces have generous amounts of it, e.g. 20mm. Cheaper eyepieces tend to vary, some are fine, others pull the eye relief as close as only a few mm. That's not very comfortable to use, and it's tricky to find the sweet spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The apparent field of view is the size of the &quot;hole&quot; you're looking through. Not the lens diameter, which has nothing to do with it, but the apparent diameter of the image you see in the lens, from dark edge to dark edge. Some lenses are like looking through a keyhole. Other lenses are like looking through a big porthole on an ocean liner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two together determine how comfortable an eyepiece is to use. The good news is, once you learn the trick, you can find the sweet spot every time afterwards.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Your eyes will never compete with a digital camera sensor. That's just a fact of life. There are spectacular views to be seen in a telescope with your eyes, just understand that a camera can simply sit there and collect light patiently for minutes or even hours, and it will outperform your human &quot;hardware&quot; eventually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a matter of taste, but for me that still doesn't eliminate the thrill of seeing things with my own eyes in a telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your scope should be just fine for watching the Moon. It will show you craters and mountains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn, when it's up in the evening sky again in a few months, will have its ring visible in your scope. It will be pretty small, but should be there in the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moons of Jupiter, again later this year, will be visible in your scope like anywhere between 1 and 4 little &quot;stars&quot; that keep following the planet around. The configuration changes from one day to the next, demonstrating that these are not stars, but objects bound to the planet.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2018-01-24T21:13:12.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24696" PostTypeId="1" AcceptedAnswerId="24699" CreationDate="2018-01-24T21:54:11.093" Score="4" ViewCount="308" Body="&lt;p&gt;Let suppose you are at +3 degrees latitude (3 degrees north of Equator ... for example at Cali in Colombia). At what periods of the year will the sun be at your vertical (at your zenith) at noon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that at Sun is at equator zenith on 21st March and 21st September (spring and fall equinox). So at +3 degree north of equator will it be before or after 21st march? And before or after 21st September?&lt;/p&gt;&#xA;" OwnerUserId="21241" LastActivityDate="2018-01-25T01:04:33.790" Title="Sun Zenith angle and location on earth" Tags="&lt;the-sun&gt;&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24697" PostTypeId="1" AcceptedAnswerId="24717" CreationDate="2018-01-24T23:16:19.247" Score="-3" ViewCount="157" Body="&lt;p&gt;Is there a observable void outside our universe that can give a clue on how far the next on is? How many universes can surround ours? What does a universes gravity well look like?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-01-26T05:18:38.990" LastActivityDate="2018-01-26T20:16:40.750" Title="Is there any way to tell which way is the next universe?" Tags="&lt;universe&gt;&lt;redshift&gt;&lt;multiverse&gt;" AnswerCount="2" CommentCount="5" ClosedDate="2018-01-25T16:48:18.283" ContentLicense="CC BY-SA 3.0" />
  <row Id="24698" PostTypeId="1" AcceptedAnswerId="24714" CreationDate="2018-01-25T00:22:31.270" Score="2" ViewCount="2588" Body="&lt;p&gt;I am studying ways to make an anti-gravity machine (!!). I believe it can be done, or theoretically be done using gravity itself as a &quot;force&quot; that pushes outwards instead of inwards.&#xA;A similar question was asked &lt;a href=&quot;https://astronomy.stackexchange.com/questions/14280/why-are-planets-round&quot;&gt;here&lt;/a&gt;, which speaks about why objects in the universe are round, which as stated, is due to Gravity.&#xA;I understand Gravity is the strongest, most consistent &quot;force&quot; in the universe and to traverse it we have to push against it - jets, propulsion, etc.&#xA;Never the less, I am perplexed why gravity's forces make objects round - eventually or over time. Why can't it turn a planet into a cube for example? Why can't Gravity make objects flat, and perhaps round at the perimeter?&#xA;If I can answer this question, I am sure I can find a way to &quot;bend&quot; Gravity for my invention.&lt;/p&gt;&#xA;" OwnerUserId="21187" LastActivityDate="2019-11-15T17:57:13.470" Title="Why does gravity make everything round?" Tags="&lt;gravity&gt;" AnswerCount="4" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24699" PostTypeId="2" ParentId="24696" CreationDate="2018-01-25T01:04:33.790" Score="2" Body="&lt;p&gt;The Sun will be at your zenith a few days after 21st March, after it crosses the equator heading northwards* going towards summer, and a few days before 21st September, before it crosses the equator heading southwards* after summer has ended. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;*  from an Earth-centric viewpoint. We know, of course, that the Earth is rotating about its axis and revolving about the Sun, and that Earth's axial tilt is what causes the seasons and the apparent motion of the Sun across the sky both through the day and through the seasons.&lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2018-01-25T01:04:33.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24700" PostTypeId="2" ParentId="24698" CreationDate="2018-01-25T02:08:45.627" Score="10" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am studying ways to make an anti-gravity machine (!!).  I believe it can be done, or theoretically be done using gravity itself as a &quot;force&quot; that pushes outwards instead of inwards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It doesn't work that way.  No one has found a repulsive version of gravity and there is no evidence this is even possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in science, we go by the evidence.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A similar question was asked here, which speaks about why objects in the universe are round, which as stated, is due to Gravity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which the question's answer explains by the &lt;em&gt;attractive&lt;/em&gt; nature of gravity.  No evidence (that word again) of repulsion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I understand Gravity is the strongest, most consistent &quot;force&quot; in the universe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You understand wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is considered the &lt;em&gt;weakest&lt;/em&gt; of the fundamental forces.  This &lt;a href=&quot;https://physics.stackexchange.com/questions/4243/what-does-it-mean-to-say-gravity-is-the-weakest-of-the-forces#4255&quot;&gt;Physics SE answer explains why&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears to be strong only because on average, over a large scale, matter is generally neutrally charged, whereas mass (and energy) just gets bigger the more space you have (a little simplistic, but that's the general idea).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for &quot;consistent&quot;, well all the forces behave consistently or we would not be able to model them with mathematics.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and to traverse it we have to push against it - jets, propulsion, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More &lt;em&gt;evidence&lt;/em&gt; for no repulsion !&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Never the less, I am perplexed why gravity's forces make objects round - eventually or over time. Why can't it turn a planet into a cube for example? Why can't Gravity make objects flat, and perhaps round at the perimeter ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because these are unstable configurations.  Even if they could form from natural processes (which is unlikely as I'll explain) the resulting forces and stresses (for large objects) would be too great to prevent collapse into a more stable shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now smaller objects can form less rounded shapes - small asteroids can be all sorts of shapes.  Only their relatively small mass (and hence small gravity) allows them to resist compression into &quot;nicer&quot; round shapes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But larger objects have lots of gravity simply because the have lots of mass.  The materials they are made of will not withstand compressive forces on them and they will form roughly round shapes, because that's the most resistive configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why more-or-less spherical ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because these larger objects tend to form in a way that builds up material equally all over the surface, so there is some tendency to be spherically symmetrical just from the way material accretes as it grows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However once they get really big (planets, larger moons and so on) the forces on the materials will be so great that they will as a general rule, behave a little like fluids.  And fluids will end up in that nice spherical shape as they lack the rigidity to resist compression.  Again a little simplistic, but this gets really complicated quite quickly.  They key here is to stop thinking of e.g. rocky planets as rigid objects.  Maybe on a timescale of seconds they are, but on astronomical timescales they're not so rigid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for larger objects, more-or-less spherical is inevitable simply because they're large.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If I can answer this question, I am sure I can find a way to &quot;bend&quot; Gravity for my invention.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No anti-gravity to be seen here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really, learn the basics of physics if your interested and hopefully you'll find that not only is it interesting, but it can explain things without the need to invent other explanations.  Usually. :-)&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2018-01-25T02:08:45.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24701" PostTypeId="2" ParentId="24697" CreationDate="2018-01-25T03:00:59.263" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which direction is the next closest universe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which direction to the nearest unicorn?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How many universes can surround ours at a time?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We don't know the shape of our universe.  Whether it's infinite or finite.  Whether it's bound or unbound or whether it's part of a theoretical 4 dimensional multiverse or if 3 dimensions are some kind of natural limit.    Without knowing how far our universe extends, or what (if anything) lies beyond it, it's impossible to answer.    You may as well ask how many balls fit into a box, but you don't know the size of the balls or the box?   See the problem?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do universes redshift from each other?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In order to redshift, something needs to be both visible and have an established frequency, like the hydrogen signature of stars.   Since we've never seen another universe, and we don't even know how to look for one, or if they exist, and without knowing what they're made of.   Asking if they redshift can't be answered for many reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are alternative universe hypotheses out there, like the &lt;a href=&quot;https://en.wikipedia.org/wiki/Multiverse&quot; rel=&quot;nofollow noreferrer&quot;&gt;multiverse hypothesis&lt;/a&gt; and Neil deGrasse Tyson sometimes talks about &lt;a href=&quot;https://uncommondescent.com/intelligent-design/neil-degrasse-tyson-argues-for-multiverse/&quot; rel=&quot;nofollow noreferrer&quot;&gt;other (theoretical) universes&lt;/a&gt;, but it's all very hand-wavy and there's nothing measurable and without measurement or detection, there's no direction, distance to or number of.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe you should start with an easier one.   Point in the direction of the 4th dimension.  :-)&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-01-25T15:11:42.977" LastActivityDate="2018-01-25T15:11:42.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24707" PostTypeId="1" CreationDate="2018-01-25T07:43:20.093" Score="4" ViewCount="122" Body="&lt;p&gt;At the edge of black hole, matter and antimatter is created, while the matter escape, antimatter annihilate inside the black hole thus decreasing it mass. But due to immense gravitation pull, why the newly generated matter also get pulled into the black hole, since it will have very less velocity to escape ?&lt;/p&gt;&#xA;" OwnerUserId="20162" LastActivityDate="2018-01-25T16:30:56.903" Title="Hawking Radiation" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2018-01-29T19:50:28.897" ContentLicense="CC BY-SA 3.0" />
  <row Id="24709" PostTypeId="1" AcceptedAnswerId="25373" CreationDate="2018-01-25T09:40:12.120" Score="5" ViewCount="590" Body="&lt;p&gt;I'm trying to find/create a star map between the declinations of -45 to +60, and need it to be a rectangle. Is there any software/website that would let me create this? Is there a name for this kind of map/projection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I've attached a diagram to try and make myself clearer) &#xA;&lt;a href=&quot;https://i.stack.imgur.com/stJsC.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/stJsC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help :)  &lt;/p&gt;&#xA;" OwnerUserId="21255" LastActivityDate="2018-03-05T05:30:19.850" Title="Creating a rectangular star map" Tags="&lt;star&gt;&lt;fundamental-astronomy&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24711" PostTypeId="1" CreationDate="2018-01-25T11:09:30.347" Score="12" ViewCount="14806" Body="&lt;p&gt;I have heard, that on the northern hemisphere, you can estimate a direction of south if you imagine a line going through the tips of the border between illuminated and shadowed part of the Moon, and where this line crosses the horizon, that is estimated direction of the south (and vice versa from the southern hemisphere).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2ZKa8.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2ZKa8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This inspired me to think about how much information could one get simply by observing the Moon. I think you can estimate the latitude of your location. I live in the northern hemisphere and never have been anywhere else, so it's all just an imagination. That's why &lt;strong&gt;I need somebody to confirm or correct my conclusions&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the angle of the tilt of the Moon will be different as seen from different latitudes. Specificaly near the poles you would see the line between illuminated and shadowed part of the Moon very verticaly. At the equator horizontaly. And somewhere in the middle between the equator and the pole tilted under some angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would also see the Moon much lower on the horizon when standing near the poles and going near the zenith when standing on equator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also think you can use observation of the phase changes througout a month to find out whether you are at the southern or the northern hemisphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5pQFd.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5pQFd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is &lt;strong&gt;what I imagine&lt;/strong&gt; the Moon looks like. I have searched the internet for relevant information, but nobody seems to discuss the whole thing in detail. There is only one picture somewhat depicting the tilt of the Moon as seen from 3 different latitudes of the Earth and that's it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&#xA;A software for simulation of the skies (Stellarium) have been recommended to me by @barrycarter and after few hours playing with it, I can clonclude, that the graphics I provided above are accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting observation made while seeing the Moon during daytime. In the real life at the surface of the Earth this observation would not be possible, but the software allows it. Since the Sun is no longer positioned &quot;under horizon&quot;, it illuminates the Moon from different angle, than depicted in the grapics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made few screeenshots from view from the France (northern hemisphere) of this situation to illustrate what I mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GfF1z.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GfF1z.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to see, that at nighttime, the moon looks exactly as in the graphics, but as soon as the Sun rises, the illumination direction changes.&lt;/p&gt;&#xA;" OwnerUserId="21257" LastEditorUserId="21257" LastEditDate="2018-01-26T11:07:14.010" LastActivityDate="2020-10-16T22:34:10.773" Title="How does the Moon look like from different latitudes of the Earth?" Tags="&lt;the-moon&gt;&lt;moon-phases&gt;&lt;navigation&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24714" PostTypeId="2" ParentId="24698" CreationDate="2018-01-25T14:45:05.883" Score="4" Body="&lt;p&gt;The reason gravity creates round objects is because the equation describing the force of gravity is the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_G = -\frac{Gm_1m_2}{r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $G$ is the gravitational constant, $m_1$ and $m_2$ are two point masses which are gravitationally attracting one another, and $r$ is the distance between (the center of mass of) the two masses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think about if for a minute, it should be abundantly clear that gravity causes objects to be round because &lt;strong&gt;gravity is only dependent on distance and not on any angle or orientation.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means if I take a planet and throw additional mass onto it, gravity will cause that mass to be attracted to the center of mass of that planet. There's no other way in which gravity can act upon it. If everything is equally pulled towards the center, the only possible object you can form is a sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there are objects like asteroids which aren't round, but that's because the structural integrity of the rocks is actually stronger than gravity and gravity, while still trying to make the asteroid round, just can't overcome the strength in the rocks.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2018-01-26T14:03:32.337" LastActivityDate="2018-01-26T14:03:32.337" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="24715" PostTypeId="1" AcceptedAnswerId="24720" CreationDate="2018-01-25T14:45:58.990" Score="2" ViewCount="307" Body="&lt;p&gt;So I have been given this question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A typical Type Ia Supernova emits $10^{44}\ \mathrm{J}$. All nuclear weapons&#xA;  tested by humans have released a total of about $2\times10^{6}\ \mathrm{TJ}$ of energy. Assume the Supernova energy is released at the distance of the Sun. Determine how close to your retina the nuclear weapons energy would have to be released to make both appear to have the same apparent bolometric magnitude. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The issue I'm have is normally apparent magnitudes I would have to have joules per second where this is in just total joules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is it that with bolometric magnitude there is an automatic assumption that when total power is mention you assume joules a second?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please advise?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would just like to point out I am not looking for a direct answer to this question, I am just confused with why total energy has been used. I have read posts on wiki which seems to confuse me even more. I have just used the above question as an example from a book.&lt;/p&gt;&#xA;" OwnerUserId="21209" LastEditorUserId="11143" LastEditDate="2018-01-25T15:49:12.117" LastActivityDate="2018-01-26T16:41:11.647" Title="total energy and apparent bolometric magnitude" Tags="&lt;apparent-magnitude&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24716" PostTypeId="2" ParentId="24698" CreationDate="2018-01-25T14:48:27.423" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Never the less, I am perplexed why gravity's forces make objects round&#xA;  - eventually or over time. Why can't it turn a planet into a cube for example?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Think of an Earth-size cube as a planet with 8 HUGE mountains on it.  In fact, these mountains would be so huge that the bottom of the mountains couldn't support the top of the mountains.  As they collapse, the mountains would flow toward the centers of the cube faces.  The end result would be almost a sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might make it easier to visualize if you consider a cubical planet made entirely of water.  Clearly the tops of the &quot;water mountains&quot; would flow down into the &quot;valleys&quot;.&lt;/p&gt;&#xA;" OwnerUserId="20370" LastActivityDate="2018-01-25T14:48:27.423" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24717" PostTypeId="2" ParentId="24697" CreationDate="2018-01-25T16:14:05.763" Score="3" Body="&lt;p&gt;There is no way to know, if any other &quot;different Universe&quot; exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a terminologic problem with it. Universe means, on latinic, something what &quot;contains everything&quot;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If there is no way for the different &quot;Universes&quot; to interact, then how could we know that it exists?&lt;/li&gt;&#xA;&lt;li&gt;If there is a way to interact. For example, there is another spacetime where somehow we can pass through (or at least can see light or gravitation from it), then how could we call it as a &quot;different Universe&quot;? Then the Universe would be our current spacetime, and that other, together.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But the problem is a hypothetical one, as no alternate Universe were found until now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can play with different Universes in some theoretical constructions. For example, the spacetime of a rotating black hole looks like as if it would be some gateway to somewhere, out of our Minkowski-Universe (or to a different point of our current one). Google for Penrose-diagrams and for the topological extensions of the General Relativity, if you would like to play with it more.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="1891" LastEditDate="2018-01-26T20:16:40.750" LastActivityDate="2018-01-26T20:16:40.750" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24718" PostTypeId="1" AcceptedAnswerId="24721" CreationDate="2018-01-25T16:14:48.023" Score="3" ViewCount="165" Body="&lt;p&gt;I just read the &lt;a href=&quot;https://www.scientificamerican.com/article/a-new-recipe-for-hunting-alien-life/&quot; rel=&quot;nofollow noreferrer&quot;&gt;popular press&lt;/a&gt; version of a &lt;a href=&quot;http://advances.sciencemag.org/content/4/1/eaao5747&quot; rel=&quot;nofollow noreferrer&quot;&gt;new paper&lt;/a&gt;, and now I have a question about Titan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some quotes from the pop article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A rocky, ocean-bearing planet with more than 0.1 percent methane in its atmosphere should be considered a potentially inhabited planet, the researchers say. And if the atmospheric methane reaches levels of 1 percent or more? In that case, “potentially” doesn’t cut it—such a world would “likely” be home to alien life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thus, methane and carbon dioxide together, unaccompanied by carbon monoxide, on a rocky, ocean-bearing world would best be interpreted as an airtight sign of anoxic life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, some information about Titan from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Titan_(moon)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page about Titan&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Titan's atmosphere&#xA;    Stratosphere:&#xA;        98.4% nitrogen&#xA;        1.4% methane&#xA;        0.2% hydrogen&#xA;    Lower troposphere:&#xA;        95.0% nitrogen&#xA;        4.9% methane&#xA;        -OR, depending on who you ask-&#xA;        97% nitrogen&#xA;        2.7% methane&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The density of Titan is consistent with a body that is about 60% rock and 40% water.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;There are trace amounts of other hydrocarbons, such as [...] carbon monoxide [...].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I understand correctly, 'trace amounts' of CO is in line with the requirements stated in the Catling paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is Titan covered in microbes, or what?  Since the article didn't mention Titan, I presume I missed something.  What was it?&lt;/p&gt;&#xA;" OwnerUserId="19146" LastActivityDate="2018-01-25T20:05:24.717" Title="Did this paper just argue that there is surely life on Titan?" Tags="&lt;astrobiology&gt;&lt;titan&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24719" PostTypeId="2" ParentId="24707" CreationDate="2018-01-25T16:30:56.903" Score="3" Body="&lt;p&gt;For one thing, the mechanism behind Hawking Radiation is, as yet, unknown. Hawking, in his original formulation, never proposed a means by which this radiation was produced. The concept of particle-antiparticle pair production near the event horizon came as an after thought and there's no guarantee that is the way in which Hawking Radiation is actually produced. There are other potential explanations out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To address your more specific question though, I'd point out that your logic is a bit faulty. Under the assumption that Hawking Radiation can actually occur we necessarily know that energy is getting away from Black Holes which means that energy/matter must necessarily be able to achieve the escape velocity $-$ otherwise it wouldn't be escaping! So to propose the matter has less than the escape velocity is a non-starter since it's a fallacious premise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, the more direct answer to your question would be to explain how and why this Hawking Radiation can achieve escape velocity. Keep in mind first that the radiation which is escaping is doing so from &lt;em&gt;outside&lt;/em&gt; the event horizon in which you can escape with a velocity less than or equal to $c$. Primarily what is escaping as Hawking Radiation are photons which of course travel at $c$ and, being outside the event horizon, are perfectly capable of escaping the black hole.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2018-01-25T16:30:56.903" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24720" PostTypeId="2" ParentId="24715" CreationDate="2018-01-25T17:10:00.423" Score="0" Body="&lt;p&gt;If you take a look at how the magnitude system is derived, it usually starts with an equation similar to the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$100^{(m_1 - m_2)/5} = \frac{F_2}{F_1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ is the apparent magnitude and $F$ is the flux. The subscripts denote two different objects. This equation derives directly from the fact that, by definition, a magnitude 1 star is 100 times brighter than a magnitude 6 star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you look at the equation above, you'll see the right hand side is a ratio of fluxes, and necessarily it is unitless. What this actually means is that you can derive the magnitude system with any measure of &quot;brightness&quot; be it a flux which has units of $\mathrm{J\ s^{-1}\ m^{-2}}$, a luminosity which has units of $\mathrm{J\ s^{-1}}$ or power which also has units of $\mathrm{J\ s^{-1}}$. The only constraint is that you have to consider the same &quot;type of brightness&quot; (for want of a better term) for both sources so your ratio remains unitless. That is, you can't use flux for one star and power for another. They both have to be flux or they both have to be power. In your case, you're given the total energy output, which you can assume happens over the same time scale since they're both just explosions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be honest though, your problem doesn't really rely on that knowledge. In fact, your problem becomes exceedingly simple because it says to assume that $m_1=m_2$ which means the ratio of the flux from both sources is unity by the equation above. You just have to say $F_1 = F_2$, relate flux to luminosity and distance, and solve for your unknown.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2018-01-25T23:58:24.193" LastActivityDate="2018-01-25T23:58:24.193" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24721" PostTypeId="2" ParentId="24718" CreationDate="2018-01-25T20:05:24.717" Score="2" Body="&lt;p&gt;The underlying assumption in the paper is that the planet has water in liquid form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper suggests that finding both CO2 and CH4 in the atmosphere of a rocky planet in the &quot;habitable zone&quot; (in which water can exist in liquid form) would be hard to explain, without life. Thus one doesn't need to detect atmospheric oxygen. If you can detect both CO2 and CH4, this should be taken as evidence of anoxic life (probably in the form of microbes) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Titan has CH4, but it doens't have much CO2 and there is no liquid water.&#xA;Titan is not in the habitable zone. On Titan, H2O is a forms part of the &quot;rocks&quot;, and the oceans are made of hydrocarbons. Titan is not a good location to look for life as it doesn't have free liquid water. We have a photo from Titan's surface and it isn't covered in Microbes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uQxQy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uQxQy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-01-25T20:05:24.717" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24722" PostTypeId="2" ParentId="24678" CreationDate="2018-01-25T20:20:41.190" Score="1" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;How to fix a direction in the context of universe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Simple (or not so simple): Use the remote stars, or even better, use the extremely remote quasars. The former is what astronomers have done for centuries, culminating in the &lt;a href=&quot;http://adsabs.harvard.edu/full/1986HiA.....7...85K&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hipparcos Celestial Reference Frame&lt;/a&gt;. The latter is what is currently used to define the &lt;a href=&quot;https://www.iers.org/IERS/EN/DataProducts/ICRF/ICRF/icrf.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;International Celestial Reference Frame&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Note very well: The ICRF is not universal. There is no such thing as a universal reference frame thanks to the non-Euclidean nature of space and time. The ICRF does however quite work nicely across the expanse of the solar system.&lt;/p&gt;&#xA;&lt;br&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;My question is regarding the direction/orientation of earth. Is it a constant with reference to universal parameters?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;No, it is not. The Earth's axis of rotation precesses thanks to gravitational torques by the Moon and the Sun on the Earth's equatorial bulge.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-01-25T20:20:41.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24723" PostTypeId="2" ParentId="24698" CreationDate="2018-01-25T21:02:00.153" Score="6" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;I understand Gravity is the strongest, most consistent &amp;quot;force&amp;quot; in the universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;As noted in other answers, you understand incorrectly. Gravitation is the weakest of the four fundamental forces. Gravitation is such a minuscule force at the atomic level that it is hard to distinguish whether gravitation acts as predicted on antimatter (theoretically the same as how gravitation acts on matter).&lt;/p&gt;&#xA;&lt;p&gt;Gravitation emerges as a powerful force at the level of stars and planets because the two strongest fundamental forces, the strong and weak interactions, are extremely short range, and because the electromagnetic force is limited by the fact that macro scale objects tend to be electrically neutral. Gravitation remains in play because it has infinite extent and because mass, at least theoretically, is never negative.&lt;/p&gt;&#xA;&lt;p&gt;This, by the way, argues against your premise of creating antigravity. As far as we known, all objects (including antimatter) have a positive mass, and hence there is no such thing as antigravity.&lt;/p&gt;&#xA;&lt;br&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Why does gravity make everything round?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Gravity does not make everything round. Small objects tend to have a fractal shape. Somewhat larger objects, where gravitation plays some role, tend to look like misshapen potatoes. It's only with sufficiently large objects, where gravitation plays a dominant role, that things begin to appear round.&lt;/p&gt;&#xA;&lt;p&gt;An object needs to be a few hundred kilometers in diameter before self-gravitation can overcome the electromagnetic ability of smaller clumps of matter to maintain whatever non-spherical shape they happen to have. The size at which gravitation begins to make things &amp;quot;roundish&amp;quot; is called the potato radius.&lt;/p&gt;&#xA;&lt;p&gt;The potato radius depends on an object's composition. Rock has a much higher yield strength than does ice. Icy objects that formed in the outer solar system tend to be &amp;quot;roundish&amp;quot; when their diameter exceed 400 kilometers, but potato shaped when they are less than this size. For stony objects, the relevant size between lumpy potato and more or less roundishness is about twice that of an icy object.&lt;/p&gt;&#xA;&lt;br&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Why can't it turn a planet into a cube for example?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Because the potential energy due to self gravitation (and due to rotation in the case of a rotating object) at the surface of a cube is not uniform. This non-uniform energy distribution means the object has excess energy. This excess energy is minuscule for a cube the size of a die but is huge for a cube the size of a planet.&lt;/p&gt;&#xA;&lt;p&gt;The mantles of terrestrial planets are solid, more or less. (It is highly erroneous to think of the Earth's mantle as molten. It isn't.) That &amp;quot;more or less&amp;quot; is important. A better description is that the mantles of the terrestrial planets are elastoplastic, or perhaps even viscoelastoplastic. They deform elastically in response to short-term stresses and strains, plastically in response to longer term stresses and strains, and can even flow over even longer periods of time. This means that a cube the size of a planet will eventually deform into something close to the entropically-favored spherical or oblate spheroid shape.&lt;/p&gt;&#xA;&lt;p&gt;Gravitation and non-conservative interactions can shape very large and very diffuse objects into a shape that is markedly non-spherical. An interstellar gas cloud, for example, eventually collapses into a disk shaped object with a massive protostar in the center. Given the right conditions, an interstellar gas cloud initially collapses toward a spherical shape, but pancakes once enough material concentrates at the center of the cloud. Here volume is not anything close to a conserved quantity. Angular momentum is however close to a conserved quantity. This, coupled with collisions and gravitation, is what make such gas clouds collapse into circumstellar disks.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-01-27T16:23:18.433" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24726" PostTypeId="1" CreationDate="2018-01-26T01:15:06.390" Score="2" ViewCount="229" Body="&lt;p&gt;I am curious to see if I have calculated this correctly: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A Type I delta-Cepheid star with a one month (30day) period is put at the distance of the star Vega.&lt;br /&gt;&#xA;  a)  Determine the apparent magnitude this star would have.&lt;br /&gt;&#xA;  b)  Determine how many times more luminous the delta-Cepheid star is compared to Vega.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My workings:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Absolute Magnitudes given for Type I delta-Cepheid star:&#xA;$$M=-2.8log\left(P\right)-1.43$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the information given $P=30$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M=-2.8log(30)-1.43=-5.58$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Absolute magnitude is given by &#xA;$$m-M=5log\left(d\right)-5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using information for calculate value of $M$ and that vega = $8Pc$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m=5log\left(8\right)-5+5.58=5.10$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b)&#xA;Apparent magnitude is given by &#xA;$$m_1-m_2=-2.51log(\frac{F_2}{F_1})$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;as the two start are at the same distance then I will be left with just luminosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so adapting the above equation and subbing in the required values&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0-5.10=-2.51log\left(\frac{L_c}{Lv}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearrange&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$10^{\frac{5.10}{2.51}}=\frac{L_c}{L_v}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$107.61=\frac{L_c}{L_v}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so therefore &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L_c=108L_v$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where subscript c is delta-Cepheid  and subscript v is vega&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is this calculation correct&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="21209" LastEditorDisplayName="user1569" LastEditDate="2018-01-26T09:40:24.000" LastActivityDate="2021-03-07T17:06:35.513" Title="Type I delta-Cepheid star, calculation of apparent maginute and luminosity" Tags="&lt;distances&gt;&lt;cepheids&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24727" PostTypeId="1" AcceptedAnswerId="24728" CreationDate="2018-01-26T01:40:10.273" Score="0" ViewCount="78" Body="&lt;p&gt;Tonight between 4:30 am &amp;amp; 5 am, we were star gazing on a skyscraper. We saw hazy illuminated triangle patch. We took out star app but didn't find any comet passing through that region &amp;amp; I was sure it wasn't Andromeda Galaxy. I'm not sure what I watched. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note we thought it could be light of aeroplane, but we were sure it wasn't as it didn't move quickly. The second thought was some form of satellite. I think there isn't any satellite with that much light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please can anyone help me identify hazy triangle patch in the sky. I am attaching star map of my sky. The region I saw the hazy patch is marked with red circle on the map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm based out of Mumbai. I don't want my curiosity to die without knowing what I saw, I'm happy to be wrong if it isn't any stellar object. Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nK4Wk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nK4Wk.jpg&quot; alt=&quot;Sky Map from my area&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21269" LastActivityDate="2018-01-27T06:06:29.577" Title="Hazy patch with huge comet tail" Tags="&lt;star-gazing&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24728" PostTypeId="2" ParentId="24727" CreationDate="2018-01-26T06:36:25.463" Score="2" Body="&lt;p&gt;As you see from the app, there are no hazy objects there, and in the light polluted Mumbai skies, deep sky objects would not be easily visible. There are no bright naked eye comets visible at the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Satellites appear like moving stars, not hazy. The most likely hazy object would be a high cloud or part of a contrail, perhaps lit by the light of the city, or by the first light of dawn.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-01-27T06:06:29.577" LastActivityDate="2018-01-27T06:06:29.577" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24730" PostTypeId="2" ParentId="24576" CreationDate="2018-01-26T11:02:26.647" Score="0" Body="&lt;p&gt;I've thought a lot about the series finale, and it's confusing to say the least.  It is possible that the black hole could be some sort of micro singularity, theorised to be possible but never found.  That way it's mass isn't sufficient to throw all the planets into chaos the moment it comes near the solar system.  That being said, they had a timer on the end of the world since the first episode, so the question begs, how did they know it was coming?  Personally the most logical answer is that it's artificial in nature, perhaps a human space station experiment gone very, very wrong.  That would explain the timer, perhaps they knew all along that it was destined from that point to consume the sun.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring that, it isn't beyond science possibility that a naturally occurring micro black hole could be travelling at a slow enough speed, relative to the speed of our own solar system, to be captured within the gravity well of the star.  It's accretion disk would also likely slow down its speed as it sucks up matter from the star and increases its mass.  Needless to say, it's all extremely unlikely, but not impossible, that we could snag our own terrifying sun-eater by sheer chance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It really doesn't look good for the people of Earth in this series.  As the mass of the black hole increases, and the two orbit each other, it's going to play hell on the organisation of the planets within the solar system, likely shifting the orbit of Earth to a position far less pleasant for life here.  Truly, an awesome &quot;end of days&quot;.  Can't wait to see how it all pans out!&lt;/p&gt;&#xA;" OwnerUserId="21276" LastActivityDate="2018-01-26T11:02:26.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24731" PostTypeId="1" AcceptedAnswerId="24735" CreationDate="2018-01-26T07:37:19.573" Score="10" ViewCount="765" Body="&lt;p&gt;If Polaris is Earth's North Star as it is directly pointed at by our North pole, what would be the North Star of the Sun?&lt;/p&gt;&#xA;" OwnerUserId="21307" OwnerDisplayName="zander" LastEditorUserId="2365" LastEditDate="2018-01-26T17:26:11.277" LastActivityDate="2018-01-26T21:51:33.673" Title="What star is directly north of the suns north pole?" Tags="&lt;the-sun&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24734" PostTypeId="2" ParentId="24715" CreationDate="2018-01-26T14:50:01.537" Score="2" Body="&lt;p&gt;You have been given a terrible question and your confusion is totally justified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnitude system is a measure of the brightness of objects which is a measure of the energy received per unit time interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without knowing a characteristic timescale over which the energy is released in each case, then it is not answerable. Note that the timescale for energy release in the case of supernovae and bombs are different by many orders of magnitude - a type Ia supernova has a bolometric light curve (e.g. see Fig.6 of &lt;a href=&quot;https://arxiv.org/abs/1402.6842&quot; rel=&quot;nofollow noreferrer&quot;&gt;Scalzo et al. 2014&lt;/a&gt;) that lasts 2-3 weeks, peaking at around $10^{36}$ Watts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this, we can compare with the Sun's bolometric luminosity of $3.8\times 10^{26}$ Watts and apparent bolometric magnitude of -26.8, to estimate that the peak apparent bolometric magnitude of a type Ia supernova at the distance of the Sun would be -50.3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot do the same thing with nuclear explosions. They last a fraction of a second (in terms of their light output). Or are you meant to assume that you take the average flux over the 70 years that mankind has had nuclear weapons?? The question is a nonsense unless you are given some sort of timescale over which this explosive energy is released. About the only sane things you can do are (a) use ~70 years (actually, most of the energy was released during the 1950s and early 1960s) (b) use the duration of a single nuclear explosion or (c) (and I &lt;em&gt;imagine&lt;/em&gt; what was intended) use a similar duration as for a typical type Ia supernovae so that the question should have begun - &quot;Imagine that X and Y amounts of total energy are emitted over a similar timescale...&quot; &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-01-26T16:41:11.647" LastActivityDate="2018-01-26T16:41:11.647" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24735" PostTypeId="2" ParentId="24731" CreationDate="2018-01-26T17:44:42.900" Score="8" Body="&lt;p&gt;The North Pole of the sun points towards the constellation Draco. There is no bright star close to the North Pole of the Sun. The closest fairly bright star is Delta Draco, &quot;Altais&quot;. A magnitude 3 star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other marks on the image are the &quot;North Pole of the planets' orbits&quot; The Earth's rotation is inclined by 23 degrees from its orbit, so the Earth's rotational North Pole is not the same as the pole of its orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hDEdL.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hDEdL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-01-26T18:01:49.523" LastActivityDate="2018-01-26T18:01:49.523" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24736" PostTypeId="1" AcceptedAnswerId="24761" CreationDate="2018-01-26T20:38:12.253" Score="5" ViewCount="1700" Body="&lt;p&gt;I am learning co-moving units and I am interested to see how the conversion works. If Hubble constant is assumed to be 0.7, then I want to convert &lt;span class=&quot;math-container&quot;&gt;$1gm/cm^3$&lt;/span&gt; into comoving unit. If I want to express length in co-moving kilo-parsec per hubble constant (&lt;span class=&quot;math-container&quot;&gt;${\rm ckpc}/{\rm h}$&lt;/span&gt;), instead of cm, then how does the conversion happen?&lt;/p&gt;&#xA;" OwnerUserId="21282" LastEditorUserId="37191" LastEditDate="2021-02-11T09:22:25.683" LastActivityDate="2021-03-31T11:00:56.623" Title="How to convert density from physical unit to comoving unit?" Tags="&lt;cosmology&gt;&lt;coordinate&gt;&lt;units&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24737" PostTypeId="2" ParentId="24731" CreationDate="2018-01-26T21:51:33.673" Score="3" Body="&lt;p&gt;According to the 2001 Astronomical Almanac (page C3), the mean position of the pole of the Solar equator is at right ascension 286.13 degrees (19.07 hours), declination 63.87 degrees. As James indicated, this puts it close to the star delta ($\delta$) Draconis.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2018-01-26T21:51:33.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24738" PostTypeId="1" AcceptedAnswerId="24740" CreationDate="2018-01-27T02:12:52.240" Score="5" ViewCount="152" Body="&lt;p&gt;There are thousands of satellites and spent rocket parts in low earth orbit (LEO). Mostly near the lines of dawn and dusk they can be in sunlight while observatories below are still viewing the night sky, as are the observatories like Hubble (also in LEO).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are moving roughly 7 km/sec and as seen from Earth that motion will be roughly a half-degree per second, depending on distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the number of observatories imaging the sky at any moment and the number of objects in LEO, streaks across images must be happening quite regularly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While that would strongly impact a one-hour exposure on emulsion, these days solid state imaging is used and these are probably regularly buffered to provide the dynamic range necessary to capture dim objects with very bright images in the same field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While a fraction of a second streak by a satellite will not normally completely obliterate a long exposure, it is still a problem and it seems would have to be managed in an accepted and systematic way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How do observational astronomers manage streaks and other artifacts from objects in LEO? Since these things are (usually) carefully tracked and their trajectories predictable, do observatories, or at least observers, ever plan ahead for these events, or schedule to avoid them? Or is it just handled automatically in post-processing?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For subtext, see astronomer Alex Parker's &lt;a href=&quot;https://twitter.com/Alex_Parker/status/956629225695514625&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tweet&lt;/a&gt; (image shown below) and then astronomer Benjamin Pope's more recent &lt;a href=&quot;https://twitter.com/fringetracker/status/956970885486993409&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tweet&lt;/a&gt; about the Humanity Star, as well as his response &quot;Oh god why would you do this to us astronomers&quot;? response &lt;a href=&quot;https://twitter.com/RocketLab/status/956247341991739392&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also see: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The Washington Post: &lt;a href=&quot;https://www.washingtonpost.com/news/speaking-of-science/wp/2018/01/26/company-shoots-shiny-orb-into-orbit-astronomers-irked-over-space-graffiti/?utm_term=.008265c5398d&quot; rel=&quot;nofollow noreferrer&quot;&gt;Company shoots shiny orb into orbit and angers astronomers over ‘space graffiti.’&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;New York Times: &lt;a href=&quot;https://www.nytimes.com/2018/01/28/science/rocket-lab-humanity-star.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Is This Shiny Satellite Sky Art or ‘Space Graffiti’?&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Scientific American: &lt;a href=&quot;https://blogs.scientificamerican.com/life-unbounded/twinkle-twinkle-satellite-vermin/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Twinkle, Twinkle, Satellite Vermin&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Within Stack Exchange:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://space.stackexchange.com/a/21582/12102&quot;&gt;How will several thousand Low Earth Orbit satellites affect the night sky?&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://math.stackexchange.com/a/2620121/284619&quot;&gt;What is the name of the polyhedral shape of the Humanity Star?&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://space.stackexchange.com/q/24616/12102&quot;&gt;Does Humanity Star have non-reflective triangular panels? If so, what are their characteristics?&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://space.stackexchange.com/q/22476/12102&quot;&gt;Two week mark; has Mayak (Маяк) been spotted yet? Reflector deployed? Astronomy “ruined”?&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From Alex Parker's &lt;a href=&quot;https://twitter.com/Alex_Parker/status/956629225695514625&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tweet&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2G1vr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2G1vr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-01-29T10:39:47.307" LastActivityDate="2018-01-29T10:39:47.307" Title="How do observational astronomers manage streaks and other artifacts from objects in Earth Orbit?" Tags="&lt;observational-astronomy&gt;&lt;artificial-satellite&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24740" PostTypeId="2" ParentId="24738" CreationDate="2018-01-27T09:28:03.050" Score="6" Body="&lt;p&gt;Simple - as you say, in order to get lots of dynamic range in deep astronomical images, you generally need to split your exposure time down into a number of sub-exposures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you combine these images, you don't just add them up, you &quot;median stack&quot; them (take the median on a pixel by pixel basis) or perform a slightly more sophisticated flux-weighted averaging that kicks out the outliers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process gets rid of any moving objects in the final image as well as cleaning up all the cosmic rays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with this reflector that has been launched is that it will be the equivalent of a very bright star. It will totally saturate any single image it appears in and might actually be bright enough to damage, or at least leave a persistent image, in some infrared detectors.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-01-27T14:30:01.440" LastActivityDate="2018-01-27T14:30:01.440" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24746" PostTypeId="1" AcceptedAnswerId="24749" CreationDate="2018-01-27T16:52:24.900" Score="10" ViewCount="244" Body="&lt;p&gt;I've been using my Meade 90 ETX for a few years trying to become a better astronomer and am familiar enough with the basics to navigate my way around. I was wondering in what way an amateur astronomer such as myself can contribute to the scientific community. I would like to try and do more than just stargaze and take pictures.&lt;/p&gt;&#xA;" OwnerUserId="21237" LastActivityDate="2018-01-28T19:39:34.490" Title="How can I contribute to the scientific community using my telescope?" Tags="&lt;amateur-observing&gt;&lt;fundamental-astronomy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24747" PostTypeId="1" CreationDate="2018-01-27T18:29:21.033" Score="11" ViewCount="242" Body="&lt;p&gt;Is there any reputable published source on expected elemental abundances at the end of the era of stellar fusion? I am here interested in the contents of galaxies; much of intergalactic gas will be lost and remain primordial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/astro-ph/9701131&quot; rel=&quot;noreferrer&quot;&gt;Adams &amp;amp; Laughlin&lt;/a&gt; mention estimates of Timms that the final chemical composition will be $X\sim 0.2, Y \sim 0.6, Z \sim 0.2$ but does not analyse the composition in much detail beyond the chemical evolution of white dwarfs (which mostly happens in the later degenerate era). &lt;a href=&quot;https://arxiv.org/pdf/astro-ph/0406095v2.pdf&quot; rel=&quot;noreferrer&quot;&gt;Fukugita and Peebles&lt;/a&gt; also make use of a very approximate estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems entirely doable to run an IMF model, add some intergalactic gas infall assumpptions, add some recycling assumptions for different mass classes, generate a composition and iterate until a limiting value. But it also seems to be a lot of work that I strongly suspect somebody has already done somewhere far better than I could do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another, very crude, approach might be to argue that since we know the cosmic element abundances today after 13.6 billion years of stellar activity we can rescale them to a limiting value: $A(t)=A(\infty)(1-e^{-t/\tau})$ where $\tau$ is the timeconstant of stellar elemental conversion. That would give the desired $A(\infty)=A(t_\mathrm{now})/(1-e^{-t_\mathrm{now}/\tau})$ (at least for metals; hydrogen and helium would be depleted). This assumes that the rate of conversion does not change (highly doubtful, since it will be more red dwarf conversion of hydrogen to just helium in the future than in the past where many more heavy stars formed due to the peak of SFR). If we use the estimate that $Z$ will increase by a factor of 10 from the current $Z\approx 0.02$ we get $\tau=1.3097\cdot 10^{11}$ years which seems reasonable, and of course just $A(\infty)=10A(t_\mathrm{now})$. That would for example make 10% of the mass oxygen and 4.6% carbon. But it would be underestimating the r-elements produced in neutron star collisions (since these were more rare in the past) and cosmic ray spallation elements.&lt;/p&gt;&#xA;" OwnerUserId="19816" LastEditorUserId="19816" LastEditDate="2018-01-28T16:02:54.173" LastActivityDate="2018-01-28T16:02:54.173" Title="What is the limiting abundances of elements at the end of the stelliferous era?" Tags="&lt;metallicity&gt;&lt;astrochemistry&gt;&lt;future&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24749" PostTypeId="2" ParentId="24746" CreationDate="2018-01-28T08:33:04.513" Score="9" Body="&lt;p&gt;Amateurs do useful scientific work by being many-handed and widely spread.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many Supernovae are discovered by amateurs. You need a set-up that can image one galaxy after another and look for any &quot;new stars&quot; that appear in them. As you are looking for the appearance of a Mag-14 star you need a big mirror; 90mm might not be enough. However look to &lt;a href=&quot;http://www.bosssupernova.com/&quot; rel=&quot;noreferrer&quot;&gt;Backyard observatory supernova search&lt;/a&gt; for more information. With the right equipment, you might find 1 SN every 5000 galaxies imaged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also contribute by observing asteroid occultations. These occur when an asteroid passes in front of a star, blocking out its light. By timing the occultation you can get an idea of the size of the asteroid in one cross-section. If multiple people observe the same occultation from different locations you can get a picture of the shape of the asteroid. See the &lt;a href=&quot;http://www.asteroidoccultation.com/&quot; rel=&quot;noreferrer&quot;&gt;Asteroid occultation site&lt;/a&gt; which links to FAQs and a page for submitting reports.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-01-28T08:33:04.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24750" PostTypeId="1" CreationDate="2018-01-28T09:05:51.773" Score="1" ViewCount="48" Body="&lt;p&gt;When a star runs out of  hydrogen, it either goes through supernova or forms white dwarf and planetary nebula. So there should not be much hydrogen left to form a new star in the nebula. So how new stars are formed from these nebulas?&lt;/p&gt;&#xA;" OwnerUserId="20352" LastActivityDate="2018-01-28T10:59:21.067" Title="Doubt about star formation" Tags="&lt;supernova&gt;&lt;star-formation&gt;&lt;stellar-evolution&gt;&lt;nebula&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2018-01-28T13:51:04.067" ContentLicense="CC BY-SA 3.0" />
  <row Id="24753" PostTypeId="2" ParentId="24746" CreationDate="2018-01-28T10:44:20.980" Score="6" Body="&lt;p&gt;You can also help the study of variable stars and in particular their light curves - You do not require a big scope for this. See for example the &lt;a href=&quot;https://www.aavso.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;American Association of Variable Star Observers&lt;/a&gt;, or the &lt;a href=&quot;http://www.britastro.org/vss/&quot; rel=&quot;nofollow noreferrer&quot;&gt;British Astronomical Association Variable Star Section&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="11822" LastEditorDisplayName="user1569" LastEditDate="2018-01-28T19:39:34.490" LastActivityDate="2018-01-28T19:39:34.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24759" PostTypeId="2" ParentId="18791" CreationDate="2018-01-29T08:37:14.350" Score="-2" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sKAxi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sKAxi.png&quot; alt=&quot;WATER VAPOR&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/n6iRI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/n6iRI.png&quot; alt=&quot;TEMPERATURE&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://suzaku.eorc.jaxa.jp/GLI2/adeos/Earth_View/eng/adeos02e.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://suzaku.eorc.jaxa.jp/GLI2/adeos/Earth_View/eng/adeos02e.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can measure the error or water vapor by looking at an edge of something over a distance.  This could be possible by looking at the edge of a planet or the edge of something like a satellite.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The International Space Station travels in orbit around Earth at a speed of roughly 17,150 miles per hour (that's about 5 miles per second!). This means that the Space Station orbits Earth (and sees a sunrise) once every 92 minutes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its possible to find the ISS using the FREE SkyView on your phone and find the space station and perhaps study the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its interesting to think about humidity measurements that we see from a weather website and compare these to what you actually see.  It maybe that the ideal place for a telescope is just above the tree line?  (for people who don't like numbers) However, there are some really nice urban telescopes and hopefully in the future we will have more telescopes in nice cities even if not all the nights are clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its kinda fun to read an Indian paper on the topic... &#xA;&lt;a href=&quot;https://journals.ametsoc.org/doi/pdf/10.1175/1520-0450%281990%29029%3C0665%3AAPOTWV%3E2.0.CO%3B2&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://journals.ametsoc.org/doi/pdf/10.1175/1520-0450%281990%29029%3C0665%3AAPOTWV%3E2.0.CO%3B2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the Indian and Japanese papers both seem to show that the water vapor disappears at about 10-12 KM)?  However, some of the biggest telescopes in the world are being placed only about ~3000 meters up.  Its sometimes nice to place telescopes in urban areas where we can get to it.  Especially when the vapor doesn't really disappear until 10 KM which is way too high up anyway?!?!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: &lt;a href=&quot;https://www.eso.org/public/teles-instr/elt/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.eso.org/public/teles-instr/elt/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(also remember that temperature is not a linear function with altitude... it goes down then up then down then up then down... i think... just look at the graph...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why water vapor is more &quot;linear&quot; is still one of those cosmic mysteries... I liked the x^2 equation the other person posted.&lt;/p&gt;&#xA;" OwnerUserId="20984" LastEditorUserId="20984" LastEditDate="2018-01-29T08:44:06.193" LastActivityDate="2018-01-29T08:44:06.193" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24760" PostTypeId="2" ParentId="22488" CreationDate="2018-01-29T08:45:56.503" Score="1" Body="&lt;p&gt;Looking around in the central area, I found a 2nd pair of objects that do not look like stars. The upper one is a line but much longer than the trails of the other stars, and the lower one is also a fuzzy blob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If neither of these correspond to real objects either, then you have some kind of contamination on your sensor perhaps. Your comment says &lt;a href=&quot;https://astronomy.stackexchange.com/questions/22488/weird-blob-near-polaris/24760#comment38466_22492&quot;&gt;...The blob moves relative to the stars...&lt;/a&gt; in other exposures. If so, perhaps it does not move with respect to pixel coordinates?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, the short linear feature could possibly be a satellite in a very high orbit and so will appear to move quite slowly relative to the star field that your mount is tracking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9ufPY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9ufPY.jpg&quot; alt=&quot;mystery blob crop 1&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/Ri9g2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ri9g2.jpg&quot; alt=&quot;mystery blob crop 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iUu1V.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iUu1V.jpg&quot; alt=&quot;mystery blob crop&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-01-29T10:22:59.570" LastActivityDate="2018-01-29T10:22:59.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24761" PostTypeId="2" ParentId="24736" CreationDate="2018-01-29T11:27:22.957" Score="5" Body="&lt;p&gt;You're asking two different, but somewhat related questions: One has to do with a practical way of describing the expansion of the Universe; the other has to do with a way of dealing with our ignorance of the exact expansion rate of the Universe.&lt;/p&gt;&#xA;&lt;p&gt;I will answer your specific question in the last paragraph.&lt;/p&gt;&#xA;&lt;h3&gt;Comoving and physical coordinates&lt;/h3&gt;&#xA;&lt;p&gt;A comoving coordinate system is one in which particles that don't move &lt;em&gt;through&lt;/em&gt; space have fixed coordinates, even is that space is expanding (or contracting, or otherwise warping).&#xA;In contrast, the particles' &lt;em&gt;physical&lt;/em&gt; coordinates are the actual distances you would measure between them if you froze space and started laying out measuring rods.&lt;/p&gt;&#xA;&lt;h3&gt;Expansion, scale factor, and redshift&lt;/h3&gt;&#xA;&lt;p&gt;Cosmological distances are often measured in megaparsec (Mpc), i.e. one million parsec, or roughly 3.3 million lightyears.&#xA;&lt;em&gt;Our&lt;/em&gt; Universe expands, and we describe this expansion not by its size (because we don't know how large it is, or even if its finite), but by a &lt;em&gt;scale factor&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt;, which is &lt;em&gt;defined&lt;/em&gt; to be equal to &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; today. Thus, in the past, when everything was half the distance from each other compared to today, we had &lt;span class=&quot;math-container&quot;&gt;$a=0.5$&lt;/span&gt;. Because light moving through an expanding space is redshifted, light from a certain time when &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; had a certain value, is redshifted (&amp;quot;stretched&amp;quot;) by a certain factor. It turns out that if we observe some light to have been redshifted by a particular value of &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;, it must have been emitted at a time when the scale factor was &lt;span class=&quot;math-container&quot;&gt;$a = 1/(1+z)$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h3&gt;Definition of comoving coordinates&lt;/h3&gt;&#xA;&lt;p&gt;Comoving coordinates are defined to coincide with the physical coordinates &lt;em&gt;today&lt;/em&gt;. That is, if the physical distance to galaxy &amp;quot;CLASH 2882&amp;quot; is 3.4 Gpc, so is the comoving distance, always was, and always will be (save for a very small movement through space, but this can be neglected). The light we receive from CLASH 2882 is redshifted to twice its emitted wavelength, so &lt;span class=&quot;math-container&quot;&gt;$z=1$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$a=1/(1+1)=0.5$&lt;/span&gt;. At earlier times, when &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; was, say, &lt;span class=&quot;math-container&quot;&gt;$0.75$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$0.5$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$0.1$&lt;/span&gt;, the physical distance to CLASH 2882 was 2.5 Gpc, 1.7 Gpc, and 340 Mpc, but its comoving distance was always 3.3 Gpc.&lt;/p&gt;&#xA;&lt;p&gt;In other word, a comoving meter is only equal to a real, physical meter today, but was smaller in the past and will be larger in the future. To be explicit, we sometimes prepend a unit with a &amp;quot;c&amp;quot; or a &amp;quot;p&amp;quot;, like this:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;1\,\mathrm{pMpc} \equiv \frac{1\,\mathrm{cMpc}}{1+z}.&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;h3&gt;The use of comoving coordinates&lt;/h3&gt;&#xA;&lt;p&gt;Comoving coordinates are practical for instance when comparing properties of something that may or may not evolve with time, but also change with the size of space, e.g. densities. One example is the number density of galaxies. If galaxies had just been around forever, without evolving, then their number density would decrease with time proportionally to &lt;span class=&quot;math-container&quot;&gt;$1/a^3$&lt;/span&gt;. To study how galaxies form, evolve, merge, etc. it is helpful to factor this out, i.e. to use comoving coordinates. If the number density in comoving coordinates change, it must be due to some &amp;quot;real&amp;quot;, astrophysical effect, not just because of cosmological expansion.&lt;/p&gt;&#xA;&lt;p&gt;Another example is the density of neutral hydrogen, in &lt;span class=&quot;math-container&quot;&gt;$\mathrm{g}\,\mathrm{cm}^{-1}$&lt;/span&gt;. In physical units, if nothing happened this density would decrease as &lt;span class=&quot;math-container&quot;&gt;$1/a^3$&lt;/span&gt;, whereas in comoving units it would be constant. Nevertheless, when &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; was about &lt;span class=&quot;math-container&quot;&gt;$0.1$&lt;/span&gt; the global, comoving density dropped several orders of magnitudes, evident of a physically significant process referred to as the Epoch of Reionization.&lt;/p&gt;&#xA;&lt;h3&gt;Converting between physical and comoving units&lt;/h3&gt;&#xA;&lt;p&gt;So, if you want to convert between physical and comoving units, just remember that the comoving coordinates factor out the expansion of the Universe, and you have, for distances,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;d_\mathrm{physical} = a d_\mathrm{comoving} = \frac{d_\mathrm{comoving}}{1+z},&#xA;$$&lt;/span&gt;&#xA;and consequently, for densities,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;n_\mathrm{physical} = a^{-3} n_\mathrm{comoving} = n_\mathrm{comoving}(1+z)^3,&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Little &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt;&lt;/h1&gt;&#xA;&lt;p&gt;The little &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; commonly seen in length units (and units derived hereof) is a way of factoring out the exact magnitude of the expansion rate &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt; of the Universe. It is defined by&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;H_0 = 100h\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}.&#xA;$$&lt;/span&gt;&#xA;That is, if &lt;span class=&quot;math-container&quot;&gt;$H_0=70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$h=0.7$&lt;/span&gt;, and if &lt;span class=&quot;math-container&quot;&gt;$H_0=67.81\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$h=0.6781$&lt;/span&gt;. The use of &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; is something of a reminiscence from the times when we only knew that &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt; was of the order of &lt;span class=&quot;math-container&quot;&gt;$50$&lt;/span&gt;–&lt;span class=&quot;math-container&quot;&gt;$100\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$&lt;/span&gt;, and factoring out the exact value allowed people assuming different values to more easily compare their results. It may be argued — and has in fact been (&lt;a href=&quot;https://arxiv.org/abs/1308.4150&quot; rel=&quot;nofollow noreferrer&quot;&gt;Croton 2013&lt;/a&gt;) — that today, where &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt; is known rather precisely to be around &lt;span class=&quot;math-container&quot;&gt;$70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; only acts to confuse. It does, however, seem to prevail, especially in numerical, cosmological simulations.&lt;/p&gt;&#xA;&lt;h3&gt;&lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; notation&lt;/h3&gt;&#xA;&lt;p&gt;The way you express your question, namely that you&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;want to express length in co-moving kilo-parsec per hubble constant&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;shows, in my opinion, a common misinterpretation of the use of &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt;: The factor &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; is &lt;em&gt;not&lt;/em&gt; a part of the unit of some number; rather it should just be thought of as a number that is multiplied on the quantity. So, although the result is the same, I think it's better to write, say,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;d = 2300h^{-1}\,\mathrm{Mpc}&#xA;$$&lt;/span&gt;&#xA;than&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;d = 2300\,\mathrm{Mpc}/h.&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;h3&gt;&lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; in quantities&lt;/h3&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; appears whenever a derived quantity depends on &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt;. Because some quantities can be derived in different ways, you may actually encounter the same quantity multiplied by different powers of &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt;, depending on how it was derived. For instance, deriving a galaxy's mass from its luminosity introduces two factors of &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt;, so if its &amp;quot;true&amp;quot; (stellar) mass was, say, &lt;span class=&quot;math-container&quot;&gt;$M_\star=10^{10}\,M_\odot$&lt;/span&gt;,&#xA;then if you had assumed &lt;span class=&quot;math-container&quot;&gt;$h=0.7$&lt;/span&gt; you'd write&#xA;&lt;span class=&quot;math-container&quot;&gt;$M_\star=4.9\times10^9h^{-2}\,M_\odot$&lt;/span&gt;.&#xA;But if the same galaxy's mass was derived from dynamics, you'd only have one factor of &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt;, so you'd write&#xA;&lt;span class=&quot;math-container&quot;&gt;$M_\star=7\times10^9h^{-1}\,M_\odot$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h3&gt;Converting to &amp;quot;&lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt;-units&amp;quot;&lt;/h3&gt;&#xA;&lt;p&gt;In general, to convert a number to &amp;quot;&lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt;-units&amp;quot; (bearing in mind that &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; is not a part of the unit), you need to know how it was calculated. Derived distances, for example, usually scale inversely with &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt; (to the power of one), so it makes sense to measure in &lt;span class=&quot;math-container&quot;&gt;$h^{-1}\,\mathrm{Mpc}$&lt;/span&gt;, that is to divide the quantity by &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt;. Thus, assuming &lt;span class=&quot;math-container&quot;&gt;$67.8\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$&lt;/span&gt; we may write the distance to CLASH 2882 as&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{array}{rcl}&#xA;d &amp;amp; = &amp;amp; 3.3\,\mathrm{Gpc}\\&#xA;  &amp;amp; = &amp;amp; 2.3h^{-1}\,\mathrm{Gpc}.&#xA;\end{array}&#xA;$$&lt;/span&gt;&#xA;On the other hand, since derived times scale proportionally to &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt;, they may be measured, e.g., in &lt;span class=&quot;math-container&quot;&gt;$h\,\mathrm{years}$&lt;/span&gt; and thus, in velocities these two &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; factors cancel, such that velocities are just measured in, say, &lt;span class=&quot;math-container&quot;&gt;$\mathrm{km}\,\mathrm{s}^{-1}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h1&gt;The answer to your question&lt;/h1&gt;&#xA;&lt;p&gt;We are now ready to answer your specific question of converting from cm to comoving kpc with &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; factored out.&lt;/p&gt;&#xA;&lt;p&gt;You need to know two things: The assumed value of &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt;, and the redshift of the object you're considering. For the sake of this calculation, let's assume that &lt;span class=&quot;math-container&quot;&gt;$h=0.7$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$z=0.1$&lt;/span&gt;, and that we want to write the diameter of a galaxy whose physical diameter is, say, &lt;span class=&quot;math-container&quot;&gt;$10^{23}\,\mathrm{cm}$&lt;/span&gt; (a Milky Way-sized galaxy). Then&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{array}{rcl}&#xA;R_\mathrm{phys.} &amp;amp; = &amp;amp; 10^{23}\,\mathrm{cm}\\&#xA;                 &amp;amp; = &amp;amp; \frac{3\times10^{-22}\,\mathrm{kpc}}{\mathrm{cm}} 10^{23}\,\mathrm{cm}\\&#xA;                 &amp;amp; = &amp;amp; 32.4\,\mathrm{kpc}\\&#xA;                 &amp;amp; = &amp;amp; 32.4 \times h \, \times \, h^{-1}\,\mathrm{kpc}\\&#xA;                 &amp;amp; = &amp;amp; 22.7 h^{-1}\,\mathrm{kpc}.&#xA;\end{array}&#xA;$$&lt;/span&gt;&#xA;and&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{array}{rcl}&#xA;R_\mathrm{com.} &amp;amp; = &amp;amp; (1+z) \, R_\mathrm{phys.}\\&#xA;                &amp;amp; = &amp;amp; (1+0.1) \, 22.7\, h^{-1}\,\mathrm{kpc}\\&#xA;                &amp;amp; = &amp;amp; 25 h^{-1}\,\mathrm{kpc}.&#xA;\end{array}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Similarly, a density of &lt;span class=&quot;math-container&quot;&gt;$n_\mathrm{phys.}=1\,\mathrm{g}\,\mathrm{cm}^{-3}$&lt;/span&gt; could be expressed as&lt;span class=&quot;math-container&quot;&gt;$^\dagger$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$n_\mathrm{com.}=3.9 h^3\,\mathrm{g}\,\mathrm{cm}^{-3}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^\dagger$&lt;/span&gt;&lt;sub&gt; Note however, that if instead of particles per cubic centimeter you were to measure Solar masses per cubic Mpc, then depending on how you had measured the mass, there could be up to two factors of &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; canceling some of the &lt;span class=&quot;math-container&quot;&gt;$h^{-1}$&lt;/span&gt; in this expression.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2021-03-31T11:00:56.623" LastActivityDate="2021-03-31T11:00:56.623" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24766" PostTypeId="1" CreationDate="2018-01-29T17:27:14.517" Score="2" ViewCount="37" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Indian_Space_Research_Organisation&quot; rel=&quot;nofollow noreferrer&quot;&gt;ISRO&lt;/a&gt;'s next mission to the Sun is &lt;a href=&quot;https://en.wikipedia.org/wiki/Aditya-L1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aditya-L1&lt;/a&gt;, (also &lt;a href=&quot;https://www.isro.gov.in/aditya-l1-first-indian-mission-to-study-sun&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;), which will be placed at the Sun-Earth Lagrangian point L-1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why it is not placed near L-4 or L-5, since these are stable equilibrium point whereas L-1 is unstable equilibrium? What is the advantage to placing Sun-observing satellites like Aditya near the L-1 point?&lt;/p&gt;&#xA;" OwnerUserId="20162" LastEditorUserId="7982" LastEditDate="2018-01-29T22:05:26.373" LastActivityDate="2018-01-29T22:05:26.373" Title="Aaditya L1 mission ISRO" Tags="&lt;the-sun&gt;&lt;orbit&gt;&lt;orbital-mechanics&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24767" PostTypeId="2" ParentId="24711" CreationDate="2018-01-29T18:31:17.090" Score="6" Body="&lt;p&gt;Here are some answers to your various comments and questions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;you can estimate a direction of south&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;True, but it would be a very rough approximation at best. In order for it to be accurate, the bright limb of the Moon would need to face either due east or west on the celestial sphere. (In astronomy terminology, the position angle of the bright limb would need to be 90 or 270 degrees.) In reality, the bright limb of the Moon faces the Sun, which can be north or south of due east/west relative to the Moon. Therefore, the cusps are not north/south, so the cusps can point to the &quot;left or right&quot; of due south on the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second problem is using two points just 1/2 degree apart (the two cusps) and extending it to where it meets the horizon (along a great circle, too!). Depending on your location and the Moon's declination, that distance is most likely to be much larger than 1/2 degree. At 40 degrees N latitude, the moon can be approximately 27 to 73 degrees from the southern horizon. (These numbers are when the Moon is on the meridian.) The accuracy of extending such a short line over such a long distance of is going to be very low.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I think you can estimate the latitude of your location&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Technically, you can use any astronomical object to determine your latitude. Depending on how you do that, the Moon is probably the worse object to use. In order to estimate your latitude, you need to know the declination of the object (as well as other data or measurements). Because the Moon orbits the Earth, the declination changes &quot;quickly&quot;. The Sun or stars are better because their declination moves more slowly or are fixed, respectively. Polaris is the easiest object to use if you are in the northern hemisphere. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I also think you can use observation of the phase changes throughout a month to find out whether you are at the southern or the northern hemisphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I agree.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An interesting observation made while seeing the Moon during daytime. In the real life at the surface of the Earth this observation would not be possible, but the software allows it. Since the Sun is no longer positioned &quot;under horizon&quot;, it illuminates the Moon from different angle, than depicted in the graphics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Moon is visible during the day, so I do not understand the second sentence. For the rest of the comment, I suggest that you choose a day when the Moon is closer to the Sun and run the simulations. For example, choose a day when the Moon is within 3 or 4 days of New Moon, and watch the Sun and Moon as they cross the sky. The bright limb of the Moon always faces toward the Sun (along a great circle), so the angle of illumination on the Moon changes throughout the day -- just like Stellarium shows. (The Moon's image for Oct 2017 at 8 hours looks a little off to me, but it may just be the difficulty of estimating the great circle direction from the image. Displaying the lines of right ascension and declination may help.)&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2018-01-29T18:31:17.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24769" PostTypeId="1" AcceptedAnswerId="24777" CreationDate="2018-01-29T23:25:16.257" Score="4" ViewCount="124" Body="&lt;p&gt;If you could measure Earth's gravity with infinite precision, and did so both with Earth and its full population, and Earth with no people, would there be a direct difference in gravity? Is this even a valid way to think of gravity?&lt;/p&gt;&#xA;" OwnerUserId="21267" LastEditorUserId="14094" LastEditDate="2018-05-04T07:27:49.673" LastActivityDate="2018-05-04T07:27:49.673" Title="Do unattached objects contribute to a planet's gravity?" Tags="&lt;planet&gt;&lt;gravity&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24770" PostTypeId="2" ParentId="24769" CreationDate="2018-01-29T23:54:19.417" Score="7" Body="&lt;p&gt;Yes, it would change the attractive force by &lt;a href=&quot;https://www.wolframalpha.com/input/?i=70kg*6000000000%2Fearth%27s%20mass&quot; rel=&quot;nofollow noreferrer&quot;&gt;10^-14&lt;/a&gt; if there was a world without people. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the distance, you can measure gravity of people and buildings, in fact they used gravity to measure gaps in the pyramids of Giza once. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today gravity surveys can detect caves and reservoirs of 2m to 50m at 2m to 150m depth, so in the future, we will be able to discover many caves and fossils with nanotechnology gravimeters.&#xA; &lt;a href=&quot;http://physicsworld.com/cws/article/news/2016/mar/30/new-gravimeter-on-a-chip-is-tiny-yet-extremely-sensitive&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://physicsworld.com/cws/article/news/2016/mar/30/new-gravimeter-on-a-chip-is-tiny-yet-extremely-sensitive&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2018-01-30T12:43:18.817" LastActivityDate="2018-01-30T12:43:18.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24771" PostTypeId="1" AcceptedAnswerId="24772" CreationDate="2018-01-30T00:13:17.937" Score="10" ViewCount="2341" Body="&lt;p&gt;If it’s going to orbit around the earth, what kind of orbit would it be (what distance)?&#xA;Are there benefits to sending a telescope further away from the earth if it’s just to photograph galaxies and not objects that are in our solar system?&lt;/p&gt;&#xA;" OwnerUserId="21326" LastActivityDate="2018-01-31T11:22:22.863" Title="Is the James Webb telescope going to orbit around the earth?" Tags="&lt;observational-astronomy&gt;&lt;space-telescope&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24772" PostTypeId="2" ParentId="24771" CreationDate="2018-01-30T02:03:40.853" Score="20" Body="&lt;p&gt;Your first question - is JWST going to orbit Earth - is a little complicated. It will follow &lt;a href=&quot;https://jwst.nasa.gov/orbit.html&quot; rel=&quot;noreferrer&quot;&gt;a mission profile&lt;/a&gt; that will send it to &lt;a href=&quot;https://en.wikipedia.org/wiki/Lagrangian_point#L2&quot; rel=&quot;noreferrer&quot;&gt;the Sun-Earth $L_2$ Lagrangian point&lt;/a&gt;. It will take the telescope about three months to achieve its orbit in $L_2$. Now, $L_2$ is unstable, and so some &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_station-keeping&quot; rel=&quot;noreferrer&quot;&gt;station-keeping&lt;/a&gt; - essentially, course corrections by thrusters - will be needed. It won't be orbiting Earth, but orbiting the Sun &lt;em&gt;with&lt;/em&gt; Earth, in a nearby location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're not quite right that JWST will only be observing objects outside the Solar System. Some of the most exciting targets - &lt;a href=&quot;https://jwst.nasa.gov/galaxies.html&quot; rel=&quot;noreferrer&quot;&gt;young, evolving galaxies&lt;/a&gt; and &lt;a href=&quot;https://jwst.nasa.gov/origins.html&quot; rel=&quot;noreferrer&quot;&gt;exoplanetary atmospheres&lt;/a&gt; - lie far beyond. However, the telescope's capabilities for atmospheric analysis (thank you, &lt;a href=&quot;https://en.wikipedia.org/wiki/NIRSpec&quot; rel=&quot;noreferrer&quot;&gt;NIRSpec&lt;/a&gt;!), useful for determining the composition of exoplanetary atmospheres, will also be used for observations of the atmospheres of Mars and the giant planets, as well as compositional data for other bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it's not really helpful to send JWST elsewhere in the Solar System, given that most of its observations will be of objects outside the Solar System. Okay, maybe you could get higher-quality data on Mars if you sent it to Mars orbit, but that would be quite costly, and not effective for a spacecraft designed to be used by many scientists for a number of purposes. Keeping it at $L_2$ simplifies the orbital trajectories immensely, although it's still much too far away to repair, unlike Hubble. You also lessen the communication time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, you might say, why not just launch JWST into Earth orbit? We have scientific instruments there; why do we go to the trouble of putting &lt;a href=&quot;https://en.wikipedia.org/wiki/Lagrangian_point#Spacecraft_at_Sun%E2%80%93Earth_L2&quot; rel=&quot;noreferrer&quot;&gt;so many things at $L_2$&lt;/a&gt;? It turns out that &lt;a href=&quot;https://space.stackexchange.com/q/12353/5963&quot;&gt;debris high above Low Earth Orbit (LEO) would be too dangerous&lt;/a&gt; for the sensitive optics on the telescope. The best instruments are often also the most delicate, and JWST would be damaged if it was truly orbiting Earth, at LEO or elsewhere.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2018-01-30T18:03:14.300" LastActivityDate="2018-01-30T18:03:14.300" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="24773" PostTypeId="2" ParentId="24771" CreationDate="2018-01-30T02:04:11.737" Score="7" Body="&lt;p&gt;The James Webb Telescope will not be orbiting around the Earth, but the Sun, at a distance of 1.5 million kilometers or 1 million miles from the Earth. A benefit to sending it further away from the Earth is that there's less of the interference of light pollution from the Earth. The JWST's mirror is 21 feet wide, though, so its sensitivity to this will be miniscule, and its abilities unhindered. There are facts about it &lt;a href=&quot;https://jwst.nasa.gov/orbit.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://www.spaceanswers.com/astronomy/ten-reasons-why-nasas-james-webb-space-telescope-will-kick-some-cosmic-butt/&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="21328" LastActivityDate="2018-01-30T02:04:11.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24777" PostTypeId="2" ParentId="24769" CreationDate="2018-01-30T07:14:38.393" Score="2" Body="&lt;p&gt;I think the question has more to do with when a person walks around the surface, drives their car, or takes an elevator, or when an a cat jumps off a tree, or a bird flies through the sky, does that change the Earth's gravity, and the answer, as noted in the other answer, is yes, but very slightly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth's mass and angular momentum is constant (if we ignore atmospheric loss, radiation out and radiation in and meteors/space dust in).   What's not consistent is Earth's shape and variation in density, and Earth's shape and variation in density, does, very subtly, change it's gravity in the sense that Earth's gravity is lumpy, not smooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How big an effect Humans have on the planet's gravity by walking or driving around or building stuff is close to negligible. The largest effects on Earth's gravitational lumpiness are tectonic, not man made.   Parts of Japan, for example, &lt;a href=&quot;http://www.oregonlive.com/today/index.ssf/2011/05/japanese_earthquake_caused_plunge_in_elevation_under_country.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;sank nearly 3 feet&lt;/a&gt; during the big Earthquake in 2011.   The largest man made effect is the &lt;a href=&quot;https://www.smithsonianmag.com/smart-news/enough-ice-has-melted-antarctica-alter-earths-gravity-180952910/&quot; rel=&quot;nofollow noreferrer&quot;&gt;melting of glaciers&lt;/a&gt; driven by climate change.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-01-30T07:14:38.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24781" PostTypeId="2" ParentId="24771" CreationDate="2018-01-30T14:21:36.060" Score="5" Body="&lt;p&gt;It doesn't orbit us - it orbits the Sun, just like we do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;bizarrely and amazingly, it &quot;stays with us&quot; ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it keeps position outside us - it takes JWT 365 days to go around, just as it takes us 365 days to go around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sZ5Jc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sZ5Jc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this achieved?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have an incredibly long piece of string attached to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We (the Earth) spin around just like on a playground ride. We hold on to the string, and the JWT goes around &quot;outside&quot; us, and stays in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nah, no string but bizarrely gravity works exactly, precisely, like a string would - if you are at exactly the right distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various interesting spacecraft hang out at this &quot;L2&quot; region.  I have included an accurate drawing of our astounding GAIA spacecraft in the diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The JWT is kind of the &quot;strongman&quot; of our space telescopes. The Tom Brady -quarterback type in high school.  You know ... tallest, strongest, goes on to marry Giselle, etc. GAIA is kind of the &quot;class smartass and superbrain&quot; type. (Beyond all belief, GAIA is ... &lt;em&gt;mapping the milky way&lt;/em&gt;. No, really.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this was invented by the French (this guy called Lagrange - the &quot;L&quot; in &quot;L2&quot;), so today naturally the French just assume that JWT, etc, are all, basically, French things - after all, they thought of it, someone else just did some welding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual &quot;detailed&quot; manner in which GAIA for example flies is a Lissajous orbit.&#xA;(Yup - another French guy!) Coincidentally Lissajous sounds a bit like &quot;lasso&quot;, and on nice diagrams such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Lissajous_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; you can see it looks like a lasso. Giddyup, space telescopes!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NASA, err, France's blokes who worked this out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o0iFU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o0iFU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/CbcMh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CbcMh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Bonus factoid - you might be wondering &lt;em&gt;why&lt;/em&gt; the effort to put it in this particular place.  This was explained very nicely by @A.Leistra in a comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a careful look at the diagram and put yourself in the place of the yellow JWT...  Try it at different points around the diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now - from any where on the circle, look &lt;strong&gt;towards&lt;/strong&gt; the Earth and the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that ......... &lt;strong&gt;&lt;em&gt;they are in the same place&lt;/em&gt;&lt;/strong&gt; from your point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun &amp;amp; Earth are total troublemakers to space telescopes.  Because they are always in &lt;strong&gt;exactly the same direction&lt;/strong&gt;, the JWT can keep its back to them at all times, always.  The JWT has one big shield, which will always shield both the Sun &amp;amp; Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How clever are the French?&lt;/p&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="13071" LastEditDate="2018-01-31T11:22:22.863" LastActivityDate="2018-01-31T11:22:22.863" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24784" PostTypeId="1" AcceptedAnswerId="24785" CreationDate="2018-01-30T14:36:12.223" Score="3" ViewCount="124" Body="&lt;p&gt;First of all sorry if this question was already asked, but I made a small search here and on google and I couldn't find any answer, so.. Here I am.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was reading &lt;a href=&quot;https://jwst.nasa.gov/orbit.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA's webpage about the Webb telescope&lt;/a&gt;, and I noticed that they are saying that one of the sides will be very hot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought that one of the reasons to put the telescope on the L2 point was to avoid &quot;seeing&quot; the Sun, but apparently the L2 point is too far to have Earth fully cover the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So.. What is the maximum distance from Earth you have to be in order to have our planet block the Sun? Of course neglecting the effect of the atmosphere, which creates a sort of &quot;halo&quot; even if the Sun is fully covered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover what will the telescope see towards the sun? Just the star with a very tiny dot in the middle, or a very big black dot (optionally with the moon moving around it), or... What?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="16259" LastActivityDate="2018-02-05T18:55:00.703" Title="Maximum distance for earth eclipse" Tags="&lt;orbital-mechanics&gt;&lt;space-telescope&gt;&lt;eclipse-non-earth-related&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24785" PostTypeId="2" ParentId="24784" CreationDate="2018-01-30T15:51:26.803" Score="3" Body="&lt;p&gt;Since the Earth is approximately 109 times smaller than the Sun (regarding radius), and the distance between them is about 150 million km, the eclipse will be visible up to about 1.39 million km from the Earth. This distance will vary (1.36 - 1.41 mln km) because the distance Earth-Sun varies during the year. This is about 3.6 times more than the Moon-Earth distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this position, the apparent size of the eclipsed Sun will be only 1.08% smaller than the one on the Earth. The Moon has the radius about 1/4 of Earth's, and it will be visible as such on average. However&#xA;when in front of the Earth, it will be apparently almost 2 times bigger (regarding radius) than when behind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could orbit Earth at such distance, because Lagrangian points L1 and L2 are a bit farther (1.5 mln km). This means also that at L2 we could not see full Solar eclipse, but only annular with magnitude 0.93.&lt;/p&gt;&#xA;" OwnerUserId="21355" LastEditorUserId="21355" LastEditDate="2018-02-05T18:55:00.703" LastActivityDate="2018-02-05T18:55:00.703" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24786" PostTypeId="2" ParentId="24709" CreationDate="2018-01-30T16:29:46.737" Score="0" Body="&lt;p&gt;You'd be converting some images from spherical coordinate system to cartesian coordinate systems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find a texture mapping environment which can do the maths for you. I'd use Unity3D. You are converting spherical coordinate graphs to XY graphs. I think polar is XY and spherical does distance from origin so XYZ, X is the lattitude, Y the longitude. &lt;a href=&quot;https://www.google.fr/search?q=program%20to%20convert%20spherical%20textures&amp;amp;oq=program%20to%20convert%20spherical%20textures&amp;amp;aqs=chrome..69i57j69i64.9216j0j9&amp;amp;sourceid=chrome&amp;amp;ie=UTF-8&quot; rel=&quot;nofollow noreferrer&quot;&gt;perhaps check this&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations are 1 line of code. I learnt it myself having had 2/10 at my school maths exams. it's a copy/paste kind of equation written in every programming language. It's vaguely rated to orthographic projection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can say you are doing the inverse of spherical texture mapping. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can search for resources to convert the map of the earth to and from a sphere and a square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can check cubemap.skybox projection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.fr/search?q=spherical+texture+mapping&amp;amp;num=100&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ved=0ahUKEwjH1MqnjYDZAhUHJuwKHTuhCUwQ_AUICigB&amp;amp;biw=1034&amp;amp;bih=546&quot; rel=&quot;nofollow noreferrer&quot;&gt;spherical texture mapping&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will probably find many many programs that can do it, mathsworks, unity3d, blender, mathcad, texture mapping apps for games designers and cartographers, and they'll have different memory management, image conversion from bitmap/rgb32 and so forth and you'd find which is ok for your images. &lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2018-01-30T16:34:59.577" LastActivityDate="2018-01-30T16:34:59.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24788" PostTypeId="1" CreationDate="2018-01-31T01:33:36.057" Score="1" ViewCount="131" Body="&lt;p&gt;I'm new to astronomy and I have a Celestron 50Z but I'm thinking of buying a better telescope. &#xA;I see that there are many options out there but I'm not really sure as to which aspects I should consider that make a good telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone point me to the right direction? If you ask how much I'm willing to pay, I'd say up to US $300.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions not just on telescopes but as I said, things I should be considering which make a good telescope worth buying.&lt;/p&gt;&#xA;" OwnerUserId="20507" LastEditorUserId="7472" LastEditDate="2018-01-31T18:11:48.707" LastActivityDate="2018-01-31T18:11:48.707" Title="Thinking of buying a new telescope. What aspects should be considered?" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="8" ClosedDate="2018-01-31T15:49:58.373" ContentLicense="CC BY-SA 3.0" />
  <row Id="24789" PostTypeId="1" AcceptedAnswerId="24803" CreationDate="2018-01-31T04:00:08.573" Score="2" ViewCount="855" Body="&lt;p&gt;I know cosmic mean density for baryons is around $(2\ \mathrm{to}\ 5)\times10^{-31}\ \mathrm{g/cm^3}$, assuming that Hubble's Constant is $100\ \mathrm{km/s/Mpc}$. But what is the current cosmic mean density including all particles in the universe (dark matter etc.)?&lt;/p&gt;&#xA;" OwnerUserId="21282" LastEditorUserId="11143" LastEditDate="2018-01-31T14:38:27.510" LastActivityDate="2018-01-31T15:40:25.297" Title="What is the cosmic mean density?" Tags="&lt;cosmology&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24791" PostTypeId="2" ParentId="13954" CreationDate="2018-01-31T06:29:16.677" Score="0" Body="&lt;p&gt;I think you should try. Sattelite dishes are made to collect very weak singals. And dish need to be aligned in millimeter range. And they are factory made with accuracy.&#xA;start with 2ft parapolic offset dish which is much flat.&#xA;i will try this also. In place of lnb you can fit mirror and it will reflect image to center of  dish surface. Make a hole in center of dish surface and fit the eye peice lense to see.&lt;/p&gt;&#xA;" OwnerUserId="21370" LastEditorUserId="21370" LastEditDate="2018-01-31T07:59:21.650" LastActivityDate="2018-01-31T07:59:21.650" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24792" PostTypeId="2" ParentId="24788" CreationDate="2018-01-31T06:34:56.647" Score="3" Body="&lt;p&gt;There are many aspects that should be considered, but a few very important ones are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;New or secondhand?&lt;/strong&gt; With your budget, buying a new telescope will only get you so far. But if you try to look into secondhand products, you might be able to find something more advanced—well-kept telescopes last years. Of course, you must be careful with whom you are buying it from, and make sure to try it before buying as well as asking about the last time the mirror was cleaned, what eyepieces are included, etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Dobson or not?&lt;/strong&gt; There are many upsides to having a Dobson-type telescope: you get a much larger mirror for the same price, it's easy to set up and maintain... However, it does take more room than other telescopes, and you will have to do more maneuvers to aim and track things with it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On the same vein, &lt;strong&gt;does size matter?&lt;/strong&gt; If you are looking for the most compact and lightweight telescope possible? If yes, that will have an impact on the kind of telescope you should buy: maybe you will want a Maksutov instead of a Newton, maybe you will want a tripod mount because it can fold and takes the least room.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;What do you want to see?&lt;/strong&gt; As StephenG said earlier, it is important to know your expectations as for what you want to observe with your telescope. If you live in an urban environment and/or are unable to go to spots free of light pollution, that severely limits your possibilities to (mainly) planets and the Moon. To that effect, you might be more interested in a telescope with a bit more focal length. If you are indeed able to go to spots free of light pollution, then you probably want to maximize the diameter of your mirror, because if you get a mirror that is 20% larger, you will get 44% more light.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some other questions that may influence your choice are: is this a long-term investment or are you looking into making an upgrade in the next few years? Do you want to try astrophotography? How much time are you willing to take to set up your telescope before observations? Do you enjoy using a sky atlas to find the exact object you are looking for? Etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give you an example, I was a similar situation a few years ago. I had a budget of $400, and I wanted a telescope for visual observation only. Because I had access to good observation spots and size was not an issue, buying a Dobson was an easy choice. I joined an astronomy club and a member was going to upgrade his own Dobson (&lt;a href=&quot;https://www.astroshop.eu/gso-dobson-telescope-n-250-1250-dob/p,6680&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.astroshop.eu/gso-dobson-telescope-n-250-1250-dob/p,6680&lt;/a&gt;) to a bigger one, so I was able to test it and buy it secondhand knowing exactly what I was getting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should also have in mind the fact that a telescope will often require a few accessories (eyepieces, filters,...) that you might need from day one, so do look at what is included with the telescope you buy, as good eyepieces are things you can keep with your next telescopes, and bad accessories can affect the quality of your observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomicably,&lt;/p&gt;&#xA;" OwnerUserId="21371" LastActivityDate="2018-01-31T06:34:56.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24801" PostTypeId="1" CreationDate="2018-01-31T14:30:53.307" Score="2" ViewCount="112" Body="&lt;p&gt;Is it normal that super blue blood moon is much smaller than before the eclipse? If so, why? I'm quite positive it's not cloudy here. On my TV, they broadcast a live super blue blood moon from the 4th floor, and it looks big and red. You think they use a telescope?&lt;/p&gt;&#xA;" OwnerUserId="21351" LastEditorUserId="7472" LastEditDate="2018-01-31T15:11:11.673" LastActivityDate="2018-03-03T06:21:25.267" Title="Apparent size of the super blue blood moon" Tags="&lt;lunar-eclipse&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24802" PostTypeId="2" ParentId="24801" CreationDate="2018-01-31T14:48:58.387" Score="1" Body="&lt;p&gt;Yes, they probably use a telescope if it's &lt;em&gt;much&lt;/em&gt; bigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 'super' refers to the fact that the moon is closer to the earth than normal during a full moon (it's near to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Apsis&quot; rel=&quot;nofollow noreferrer&quot;&gt;perigee&lt;/a&gt; in its orbit), but the human eye cannot see the difference between the moon size an hour before and an hour after the perigee.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there's a phenomenon called &lt;a href=&quot;https://en.wikipedia.org/wiki/Moon_illusion&quot; rel=&quot;nofollow noreferrer&quot;&gt;moon illusion&lt;/a&gt; which makes the moon appear larger near the horizon, which might or might not be the case in your situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, the fact that it's a blue moon has nothing to with it; that's just the moment it takes place on the calendar.&lt;/p&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2018-01-31T14:48:58.387" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24803" PostTypeId="2" ParentId="24789" CreationDate="2018-01-31T15:40:25.297" Score="3" Body="&lt;p&gt;The simplest and shortest answer to this question is that since our universe is flat (to within our error bars), its density must be that of the critical density - that's what it means to be flat! In that case the overall density of the universe, accounting for matter, dark matter, dark energy, etc. is (assuming $H_0=100\ \mathrm{km/s/Mpc}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho_c = 1.8788\times 10^{-26}\ \mathrm{kg/m^3} = 1.8788\times 10^{-29}\ \mathrm{g/cm^3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to be more precise about it and actually use measurements of our universe, you can go about calculating it by using the density parameters. By definition, the density parameter is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Omega \equiv \frac{\rho}{\rho_c}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\rho_c = 1.8788\times 10^{-26}h\ \mathrm{kg/m^3}$ (notice that little $h$ now included). Numerous missions such as WMAP and Planck have measured these density parameters for the variety of mass-energies in our universe and you can look them up and put them all together to get the total mean density of our universe. Overall, you should find that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho = (\Omega_m + \Omega_c + \Omega_\Lambda + ...)\rho_c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the various $\Omega$s are defined for the various mass-energies in our universe. You can look up values for the density parameters from recent results such as from &lt;a href=&quot;https://lambda.gsfc.nasa.gov/product/map/current/params/lcdm_wmap9.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;WMAP&lt;/a&gt; or &lt;a href=&quot;https://arxiv.org/pdf/1502.01589.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck&lt;/a&gt;. Just decide on your chosen model and pull out the $\Omega$ values $-$ or more properly the $\Omega h^2$ values if you want to use your own value of $H_0$. It's not too difficult to plug everything in. You'll pretty much always find though, that the sum of the density parameters very nearly equals one and to an order of magnitude you can say $\rho=\rho_c$.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2018-01-31T15:40:25.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24804" PostTypeId="1" AcceptedAnswerId="24809" CreationDate="2018-01-31T16:26:04.153" Score="4" ViewCount="337" Body="&lt;p&gt;As the sun's rays pass through the Earth's atmosphere only the red light gets through. Is this the result of specific gases in the Earth's atmosphere filtering the red light?  Does all of the nitrogen and oxygen affect the sunlight shining through it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say the Earth (or just about any earth-like planet) has an atmosphere made mostly of green chlorine. When the sun shines through the chlorine atmosphere and hits the moon during a lunar eclipse, would the moon appear green? Would different gases result in the moon becoming different colors?&lt;/p&gt;&#xA;" OwnerUserId="18831" LastEditorUserId="7982" LastEditDate="2021-02-28T11:45:28.240" LastActivityDate="2021-02-28T11:45:28.240" Title="Do the gases in the Earth's atmosphere affect the color of a lunar eclipse?" Tags="&lt;lunar-eclipse&gt;&lt;atmospheric-effects&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24806" PostTypeId="1" AcceptedAnswerId="24827" CreationDate="2018-01-31T19:14:13.163" Score="3" ViewCount="200" Body="&lt;p&gt;According to the  Friedmann equations big bang happened everywhere, but if we account for inflation theory, &lt;a href=&quot;https://www.youtube.com/watch?v=OElAsQCwKTc&quot; rel=&quot;nofollow noreferrer&quot;&gt;big bang did not happened everywhere&lt;/a&gt;. How ? Anyone please explain. Please see the video for reference.&lt;/p&gt;&#xA;" OwnerUserId="20162" LastActivityDate="2018-02-02T08:50:02.357" Title="BIg Bang Happened everywhere" Tags="&lt;big-bang-theory&gt;&lt;cosmological-inflation&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24808" PostTypeId="2" ParentId="24675" CreationDate="2018-01-31T21:19:42.867" Score="4" Body="&lt;p&gt;Yes, it is possible, however such orbits are likely to be unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example in the paper&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ćuk, Matija, Douglas P. Hamilton, and Matthew J. Holman. &lt;em&gt;Long-term stability of horseshoe orbits.&lt;/em&gt; Monthly Notices of the Royal Astronomical Society 426.4 (2012): 3051-3056. &lt;a href=&quot;https://doi.org/10.1111/j.1365-2966.2012.21964.x&quot; rel=&quot;nofollow noreferrer&quot;&gt;doi:10.1111/j.1365-2966.2012.21964.x&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;it was found that if the mass of second body becomes larger than 1/1200 of the first body, horseshoe orbits become unstable. (Third body was assumed to have negligible mass).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonzero mass of the third body was investigated recently in a thesis:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Balaji, Bhaskaran. &lt;em&gt;Mass ratios in stellar triple systems that admit horseshoe orbits&lt;/em&gt;. Diss. Massachusetts Institute of Technology, 2016. &lt;a href=&quot;https://dspace.mit.edu/handle/1721.1/105626&quot; rel=&quot;nofollow noreferrer&quot;&gt;abstract&lt;/a&gt;, &lt;a href=&quot;https://dspace.mit.edu/bitstream/handle/1721.1/105626/963177537-MIT.pdf?sequence=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;pdf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and it was found that in addition to constraint of &lt;em&gt;Ćuk et al.&lt;/em&gt; nonzero mass of the third body  further decreases stability of such systems, so that the lifetime of horseshoe orbit $\tau \propto m_3^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For stars only coupling of more massive giants with lighter  red dwarfs would fulfill  first constraint and then the third star would never have 'negligible' mass, one could conclude that horseshoe orbits when all three bodies are stars are always unstable. And for larger masses of smaller stars the lifetime of such orbits would be too short to expect such orbits in our vicinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it would probably be possible to have stable horseshoe orbits of two stars (say a blue giant and red dwarf) and a planet.&lt;/p&gt;&#xA;" OwnerUserId="21391" LastActivityDate="2018-01-31T21:19:42.867" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24809" PostTypeId="2" ParentId="24804" CreationDate="2018-01-31T22:11:41.933" Score="5" Body="&lt;p&gt;From the way you have formulated your question, it seems to me that you are aware of the light diffusive process named Rayleigh Scattering and why it explains the various tints and colours of the sky depending on the Sun position. Other readers can refer to &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Rayleigh_scattering&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Rayleigh_scattering&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are used to our atmosphere and as such often discuss the RS merely on geometrical bases, as I shortly did just above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the intensity of RS does not only depends on the inverse of the fourth power of wavelength, but also on the size of the scattering particles and specifically on the square of the molecular polarisability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, changing the nature and composition  of the atmosphere does effect the intensity of RS. An example is SO2, which tends to intensify the yellow tint of sunsets in polluted area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should answer the first part of your questions. Let us move to the second part, specifically using the example atmosphere that came to your mind, chlorine as a green gas, i.e. molecular chlorine Cl2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the gas is not colourless. &lt;em&gt;Different respect to nitrogen and  oxygen, chlorine does absorb in the visible part of the electromagnetic spectrum&lt;/em&gt;.&#xA;It absorbs the colours opposite to green in the colours wheel, to keep things simple as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an inherently coloured atmosphere the relative importance of scattering will be less or at least variable depending on parameters such as atmosphere thickness, density etc. Let me ignore RS in such a toxic atmosphere :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is important now is that when you point your eyes up to the sky, the green colour you see is not that one whose propagation is hindered, but rather the one that can propagate (given that we are ignoring RS). In other words, the light will be always richer of green colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make the final answer simple let us look at the moon when the only light that hit it is that coming &lt;em&gt;throughout&lt;/em&gt; the planet atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With Earth, the Moon is bloody as for RS has removed the blue violet part of the solar spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hypothetical change on Earth atmosphere composition will also have a minor impact, as we have seen that the polarisability of the molecules counts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About Chloro, the planet having chlorine atmosphere. During the eclipse,  its moon will have a greenish tint, or likely appear green. In this case the moon is lit by sunlight depleted of the wavelengths absorbed by chlorine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To resume, the colour of the moon is always that less affected by the planet atmosphere but the mechanisms and the final result differ. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the atmosphere is intrinsically colourless, the colour of the moon is somehow &quot;complementary or so&quot; respect to that of the daily planet skydome. When the atmosphere absorbs, the moon and the sky dome will tend to the same colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I did assume that sunlight can pass the intrinsically coloured atmosphere.  Again depending on various parameters, this is of course not given. Most likely we won't see the Moon. We shall really care of our atmosphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the intrinsic properties of the moon do not change. But this is even more trivial. &lt;/p&gt;&#xA;" OwnerUserId="17437" LastEditorUserId="17437" LastEditDate="2019-02-28T14:36:03.540" LastActivityDate="2019-02-28T14:36:03.540" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24810" PostTypeId="1" AcceptedAnswerId="24812" CreationDate="2018-01-31T22:58:00.787" Score="75" ViewCount="13832" Body="&lt;p&gt;Wikipedia says the &lt;a href=&quot;https://en.wikipedia.org/wiki/Chicxulub_impactor&quot; rel=&quot;noreferrer&quot;&gt;Chicxulub impactor&lt;/a&gt; is thought to have been a 10-15 km diameter object. Would it have been visible to a (human&lt;sup&gt;*&lt;/sup&gt;) naked eye before impact? And if so, would it have appeared like a star that grew brighter and brighter each night?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;* I know, there were no humans at the time.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="14302" LastEditorUserId="37191" LastEditorDisplayName="user1569" LastEditDate="2021-03-18T09:14:43.840" LastActivityDate="2021-05-03T18:08:54.020" Title="Could the dinosaurs have seen the asteroid that killed them?" Tags="&lt;asteroids&gt;&lt;history&gt;&lt;comets&gt;&lt;naked-eye&gt;&lt;chicxulub&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="24811" PostTypeId="1" CreationDate="2018-01-31T23:30:04.053" Score="3" ViewCount="51" Body="&lt;p&gt;The amount of light collected for a period of time is proportional to the area of the collecting aperture. We use our eyes -pupils have diameters of approximately 5 mm-, binoculars and telescopes. How much more light can we collect with a 50mm pair of binoculars compared with our eyes? And with an 8-inch telescope? If we could have access to a 10-meter class professional telescope, how large would the gain be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;50^2/5^2= 100 times more light&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8in = 203.2 mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;203.2^2/5^2= 1652 times more light&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10000^2/5^2= 4000000 times more light&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did I solve this problem correctly?&lt;/p&gt;&#xA;" OwnerUserId="21392" LastActivityDate="2018-02-01T05:17:41.133" Title="Comparing Light Collection Between Pupil and Telescope?" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24812" PostTypeId="2" ParentId="24810" CreationDate="2018-01-31T23:54:49.133" Score="92" Body="&lt;p&gt;The answer is yes; for a few nights prior to the impact (assuming they had eyes with a similar sensitivity to our own and could look up!). It could also be much longer than this if the object was a cometary body.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Details:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Impacting solar system objects would have relative closing speeds from around &lt;a href=&quot;https://physics.stackexchange.com/questions/232541/why-is-meteor-speed-what-it-appears-to-be&quot;&gt;11 to 72 km/s&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;We could take the optimal case that the asteroid approaches whilst fully lit by the Sun (which probably precludes the minimum and maximum speed in the range quoted above) and then scale from another similar body - say the asteroid &lt;a href=&quot;https://en.m.wikipedia.org/wiki/4_Vesta&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vesta&lt;/a&gt;. This has a diameter of around &lt;span class=&quot;math-container&quot;&gt;$a=520$&lt;/span&gt; km, gets as close as &lt;span class=&quot;math-container&quot;&gt;$d=1.14$&lt;/span&gt; au from the Earth and has a maximum brightness of about &lt;span class=&quot;math-container&quot;&gt;$m=5.2$&lt;/span&gt; apparent magnitude (and is hence just visible to the naked eye) and an observed flux &lt;span class=&quot;math-container&quot;&gt;$f = f_0 10^{-0.4m}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$f_0$&lt;/span&gt; is a zeropoint for the magnitude scale.&lt;/p&gt;&#xA;&lt;p&gt;Thus the flux &lt;span class=&quot;math-container&quot;&gt;$f_n$&lt;/span&gt; received by a near-Earth asteroid of diameter &lt;span class=&quot;math-container&quot;&gt;$a_n$&lt;/span&gt;, at a distance &lt;span class=&quot;math-container&quot;&gt;$d_n$&lt;/span&gt; from Earth (in au) and with the same reflectivity would be&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ f_n = f\left(\frac{a_n}{a}\right)^2 \left(\frac{1+d}{1+d_n}\right)^2&#xA;\left(\frac{d}{d_n}\right)^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The magnitude of the dinosaur killer would then be&#xA;&lt;span class=&quot;math-container&quot;&gt;$$m_n = m -2.5\log (f/f_n)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;To be an at all conspicuous naked eye object, &lt;span class=&quot;math-container&quot;&gt;$f_n \geq f$&lt;/span&gt;. If we assume the dinosaur-killer had &lt;span class=&quot;math-container&quot;&gt;$a_n=10$&lt;/span&gt; km, then&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ d_n^2(1+d_n)^2 \leq 0.0022$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;An approximate solution is obtained by assuming &lt;span class=&quot;math-container&quot;&gt;$d_n \ll 1$&lt;/span&gt; and thus we find&#xA;&lt;span class=&quot;math-container&quot;&gt;$ d_n \leq 0.047$&lt;/span&gt; au or 7 million km.&lt;/p&gt;&#xA;&lt;p&gt;Moving at say 30 km/s, then it gets closer by 2.6 million km per day, thus hitting the Earth about 3 days after becoming a naked eye object. Obviously this would be longer for a slower approach speed or for a larger or more reflective asteroid. But shorter for a smaller, faster asteroid or if the asteroid approached from a direction not fully illuminated by the Sun or had a smaller albedo than Vesta.&lt;/p&gt;&#xA;&lt;p&gt;Another possibility is that the object is of a cometary origin with an icy composition. If that were so then it could be much brighter as a result of sublimation, outgassing and having a bright cometary nucleus and tail. he answer would still be yes, but the visibility period could be weeks (comets are rather unpredictably bright).&lt;/p&gt;&#xA;&lt;p&gt;It thus seems to me that there &lt;em&gt;is&lt;/em&gt; a plausible range of parameters and trajectories where a dinosaur-killing asteroid could be observed and then observed to grow brighter over a &lt;em&gt;few&lt;/em&gt; nights, but probably not much longer than that unless it was a comet.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2021-05-03T18:08:54.020" LastActivityDate="2021-05-03T18:08:54.020" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="24813" PostTypeId="1" AcceptedAnswerId="24839" CreationDate="2018-01-31T16:14:16.140" Score="2" ViewCount="186" Body="&lt;p&gt;I want to build a reflecting telescope in my home. I've attached a drawing of my plan that contains some explanations of what I understand so far, and highlights the parts of the design for which I have questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What type of mirror should i use, parabolic or spherical, and what are the factors considered when making that decision? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the glass used to make the mirror affect the &lt;em&gt;incoming&lt;/em&gt; light rays? What about the secondary mirror in the middle of the tube?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there other things to consider that would help for higher magnification? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some telescopes have a glass window at the front where light enters the tube, why is that? Would I need to add it here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kaWVb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kaWVb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21504" OwnerDisplayName="Sri vishnu Bharat" LastEditorUserId="7982" LastEditDate="2018-02-02T00:00:49.200" LastActivityDate="2018-02-02T00:00:49.200" Title="Telescope and its construction" Tags="&lt;optics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24814" PostTypeId="1" CreationDate="2018-02-01T03:41:35.917" Score="0" ViewCount="41" Body="&lt;p&gt;please correct me if I am mistaken, I was in the middle of a thought process. We take a unit area dA of sun,, which is emitting light directly to our line of sight. But there are also numerous other unit areas throughout sun's surface whose light does not reach our eyes directly. Then why do we still see the sun with same intensity?&lt;/p&gt;&#xA;" OwnerUserId="21282" LastActivityDate="2018-02-01T04:56:56.453" Title="Relation between intensity of Sun and the light that is emitted from surface of sun" Tags="&lt;radiative-transfer&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2018-02-01T13:59:53.913" ContentLicense="CC BY-SA 3.0" />
  <row Id="24815" PostTypeId="2" ParentId="24814" CreationDate="2018-02-01T04:56:56.453" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;We take a unit area dA of sun,, which is emitting light directly to our line of sight&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are, I think, imagining the light as if it was directed out from a point, with directions of photons outward from the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What actually happens is that the light is emitted from every point &lt;em&gt;on the surface&lt;/em&gt; and what's more every point on the surface can send light in any and all directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the Sun is a &lt;em&gt;diffuse&lt;/em&gt; source of light, not a point source of light.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But there are also numerous other unit areas throughout sun's surface whose light does not reach our eyes directly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So because of what I have described - light being emitted in all possible directions from all over the Sun's surface - we do see light from all parts of the surface.  And that light &lt;em&gt;does&lt;/em&gt; reach our eyes directly.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2018-02-01T04:56:56.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24816" PostTypeId="2" ParentId="24811" CreationDate="2018-02-01T05:17:41.133" Score="1" Body="&lt;p&gt;You have the idea right! The amount of light collected is indeed proportional to the area of the collecting surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you want to take it a bit further, try to think about the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Binoculars have one lens for each eye, does this mean you get twice the amount of light in the end?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Reflecting telescopes almost always have a secondary mirror which results in part of the light being masked and not collected by the primary mirror. Now imagine you have an 8-inch telescope with a 2-inch secondary mirror, can you calculate again the light collection ratio compared to the naked eye?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="21371" LastActivityDate="2018-02-01T05:17:41.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24818" PostTypeId="2" ParentId="24801" CreationDate="2018-02-01T05:42:12.687" Score="1" Body="&lt;p&gt;To complement what Glorfindel said, the term &quot;super Moon&quot; is quite hyperbolic, since the apparent diameter of a super Moon is only about 6.5% larger than that of a regular full Moon, so you would barely notice the difference anyway!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the super Moon you saw on TV, they might just have used a telephoto lens, as explained in this thread: &lt;a href=&quot;https://photo.stackexchange.com/questions/34517/how-do-people-shoot-very-large-moon-views&quot;&gt;Large Moon photos using telephoto lens compression&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21371" LastActivityDate="2018-02-01T05:42:12.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24820" PostTypeId="2" ParentId="24806" CreationDate="2018-02-01T09:20:15.823" Score="2" Body="&lt;p&gt;You seem to be misunderstanding what inflation does.   Lets say the entire universe is two stars and those two stars are gravitationally attracted to each other, but space is also expanding between them so the stars move apart from each other.   Both stars happened inside the big bang, and as they move apart, the space between them that inflation creates happened inside the big bang as well.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inflation doesn't happen beyond the big bang, it happens inside where the big bang happened.   It could also be said that the big bang is inflation, at least, that's what it is as far back as we can tell what it is.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I hope my layman's explanation isn't too far off).   The Minute Physics &lt;a href=&quot;https://www.youtube.com/watch?v=q3MWRvLndzs&amp;amp;list=PL6hbWOlJVc66AnUcGJn7NsAdZrTa45d_y&quot; rel=&quot;nofollow noreferrer&quot;&gt;Everywhere Stretch video&lt;/a&gt; explains it pretty well.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-02-01T09:20:15.823" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24822" PostTypeId="1" CreationDate="2018-02-01T11:59:09.653" Score="0" ViewCount="19332" Body="&lt;p&gt;So a couple days ago, my husband and I spotted a moving object in the sky just like a star. I know it wasn't a plane or a drone, and it was way up in the sky just a star. It just moved smoothly across the skies and than disappeared. I love to observe the sky so that was priceless for me and my husband.&lt;/p&gt;&#xA;" OwnerUserId="21412" LastEditorUserId="6" LastEditDate="2018-02-01T13:05:44.283" LastActivityDate="2018-02-01T13:05:44.283" Title="Spotted a moving star-like object" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2018-02-01T16:56:04.850" ContentLicense="CC BY-SA 3.0" />
  <row Id="24823" PostTypeId="2" ParentId="24822" CreationDate="2018-02-01T13:01:11.367" Score="2" Body="&lt;p&gt;Based on your description, that might be the &lt;a href=&quot;https://en.wikipedia.org/wiki/International_Space_Station#Sightings_from_Earth&quot; rel=&quot;nofollow noreferrer&quot;&gt;International Space Station&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The ISS is visible to the naked eye as a slow-moving, bright white dot because of reflected sunlight, and can be seen in the hours after sunset and before sunrise, when the station remains sunlit but the ground and sky are dark.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So if what you saw was not long after sunset, and if the object was quite bright and moving slowly across the sky, that might be the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to check for yourself, you can input your location on &lt;a href=&quot;http://www.heavens-above.com/PassSummary.aspx?satid=25544&quot; rel=&quot;nofollow noreferrer&quot;&gt;this webpage&lt;/a&gt; and change the search period to see if the ISS passed above your heads on that day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other possibilities include other satellites (not very bright but also moving smoothly in the sky), or—more rarely—particularly large shooting stars.&lt;/p&gt;&#xA;" OwnerUserId="21371" LastActivityDate="2018-02-01T13:01:11.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24825" PostTypeId="2" ParentId="24810" CreationDate="2018-02-01T13:11:47.180" Score="11" Body="&lt;p&gt;A carbonaceous condrite has the same reflectivity as the moon at around 7-13%.&lt;/p&gt;&#xA;&lt;p&gt;If there was ice, if the tail was 10 times smaller than Hale-Bopp, it would have auspiciously covered half of the sky.  it could have made an incredible display in the 1-2 days preceding the collision, because it was as close to the sun as Hale-Bopp, the brightest astronomical apparition in history, although it was 100 times closer to the earth in the final days than Hale-Bopp. Hale-Bopp's trail was 1.5 million km long, so the KT asteroid would have needed a tail 100 times smaller, to appear as bright as Hale Bopp in the final 2 days! &lt;a href=&quot;https://www.google.com/amp/s/www.space.com/amp/20354-dinosaur-extinction-caused-by-comet.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.google.com/amp/s/www.space.com/amp/20354-dinosaur-extinction-caused-by-comet.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{\text{moon distance}}{\text{moon diameter}} = 107\text{ moon diameters}$&lt;/span&gt; away, so you can cover the moon with a thumb at arms length ( ~1cm/107cm) and the impactor would be nearly thumb sized at:  &lt;span class=&quot;math-container&quot;&gt;$\text{comet diameter} \cdot107 = 1500\text{km}$&lt;/span&gt; away.&lt;/p&gt;&#xA;&lt;p&gt;If you were 1000km south of Chixculub, it would stay fairly constantly thumb sized for 10-20 seconds as it arrived from the South-East at 60 degrees, and in that time it would travel from 1500km above to 1000km on the horizon. The graph suggests it had a rotation period of about 2 to 10 seconds, so it would have revolved noticeably.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qRal3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qRal3.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;If you were 100km away from the impact, it would grow from thumb size to hand sized for 10-20 seconds. Then it would hit the atmosphere at 50km and the sky would ignite about 5 seconds prior to landfall.&lt;/p&gt;&#xA;&lt;p&gt;For most of the dinosaurs far away, it would be like holding a pea or a matchstick at arms length.&lt;/p&gt;&#xA;&lt;p&gt;Perhaps it would be brighter than Venus from 75000 km away; that's about 125 minutes prior to the impact.&lt;/p&gt;&#xA;&lt;p&gt;It would have been like a fire fly landing slowly on a tennis court if the dino was on the other side of the world.&lt;/p&gt;&#xA;&lt;p&gt;Mammals were nocturnal previous to the impact which is why mammals have developed whiskers, awesome hearing and only bichromatic photoreceptors of G.B, and mammal skin burns from UV. Only primates have R.G.B.&#xA;Dinosaurs had R.G.B because birds have the same gene to encode red photoreceptors as tortoises who can also see red, so scientists think that dino's had vision as good as birds, and have better vision than mammals, especially used for detecting movement and for flying.&lt;/p&gt;&#xA;&lt;p&gt;EDIT: On wiki, the estimated diameter of the comet has been changed to 10-80km. Hale-Bopp comet was 20-40km, and it was visible for months as one of the brightest comets in human history, with a similar aphelion of 0.914AU. That implies that the dinosaur comet may have been visible for many days prior to impact, because it would have been 5 times closer than Hale-Bopp at the end. It's possible that it was very bright in the sky for a week prior to impact.&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="31410" LastEditDate="2020-11-19T23:40:07.713" LastActivityDate="2020-11-19T23:40:07.713" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24827" PostTypeId="2" ParentId="24806" CreationDate="2018-02-01T13:54:32.470" Score="10" Body="&lt;p&gt;Okay, I think I know what Max Tegmark is talking about in the video. He is referring to the fact that, when you observe the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;noreferrer&quot;&gt;cosmic microwave background radiation&lt;/a&gt; (CMB) — i.e. the radiation that was &quot;released&quot; when the Universe had expanded and cooled sufficiently to allow protons and electron to combine into neutral hydrogen without immediately being ionized again — you find that it has &lt;em&gt;almost&lt;/em&gt; (to within $1$ in $\sim10^5$) the same wavelength everywhere, no matter in which direction you look, meaning that the temperature of the Universe was almost the same when it was emitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a universe like ours, but without inflation, this is puzzling. The reason is that, using the observed expansion rate and densities of the universe, you can calculate that at early times such a universe would simply expand so quickly that it would be impossible to reach &lt;em&gt;thermodynamical equilibrium&lt;/em&gt; on such large scales as the Universe we observe today. In fact, when we look at the sky, any two points separated by more than roughly 1 degree would never have been in &lt;em&gt;causal contact&lt;/em&gt;, i.e. been able to exchange information. Thus, there would be no way for one of these points to know e.g. what is the temperature of the other points, not to mention the rest of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This puzzle is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Horizon_problem&quot; rel=&quot;noreferrer&quot;&gt;the horizon problem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YMJLb.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YMJLb.jpg&quot; alt=&quot;CMB&quot;&gt;&lt;/a&gt;&#xA;&lt;sub&gt;Map of the CMB sky. The color coding shows the temperature of the radiation, and &lt;em&gt;red&lt;/em&gt; is roughly $18\,\mu\mathrm{K}$ hotter than &lt;em&gt;blue&lt;/em&gt;.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;em&gt;Credit: ESA/Planck Collaboration (plus my own annotations)&lt;/em&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What Tegmark points at is that although we're told that &quot;Big Bang happened everywhere&quot;, this doesn't make sense, because how can one part of this &quot;everywhere&quot; know about the other parts of &quot;everywhere&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One solution, obviously, is pure luck. After all, the regions on the sky are a little bit different. But this is a very unsatisfying solution, known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Fine-tuning&quot; rel=&quot;noreferrer&quot;&gt;fine-tuning&lt;/a&gt;, which we loathe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enter &lt;a href=&quot;https://en.wikipedia.org/wiki/Inflation_(cosmology)&quot; rel=&quot;noreferrer&quot;&gt;inflation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you assume the our Universe was much, much denser in the beginning, but then went through a period of &lt;em&gt;extreme&lt;/em&gt; (!) expansion, then in the very first tiny fraction of a fraction of a second, before this epoch which started when the Universe was $\sim10^{-36}$ seconds old, everything &lt;em&gt;was&lt;/em&gt; so close to each other that a region much, much larger than the part of the Universe that we can see (the &quot;observable Universe&quot;) could exchange information and reach thermodynamical equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that many people believe in this scenario, I think, is that it simultaneously solved two other puzzles, known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Flatness_problem&quot; rel=&quot;noreferrer&quot;&gt;flatness problem&lt;/a&gt; and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Bang#Magnetic_monopoles&quot; rel=&quot;noreferrer&quot;&gt;magnetic monopole problem&lt;/a&gt; (and that, so far, no better scenario has been proposed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to recap, Tegmark is &lt;em&gt;not&lt;/em&gt; saying that Big Bang didn't happen everywhere, but rather that, before Alan Guth proposed the inflation scenario in 1979, but after we first observed the CMB, this didn't make sense.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-02-02T08:50:02.357" LastActivityDate="2018-02-02T08:50:02.357" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24830" PostTypeId="1" CreationDate="2018-02-01T15:12:04.397" Score="8" ViewCount="403" Body="&lt;p&gt;Assume the Earth somehow stopped spinning once every ~24 hours, and instead started rotating at ~365 days per rotation, so it effectively would end up being tidal locked with the Sun.  In this scenario, the same side of the Earth would always be facing the Sun, and the opposite side of the Earth would always be in darkness.  Disregard all sorts of cataclysmic events that would transpire if something like this happened.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it would be safe to assume that life as we know it would quickly cease to exist except on the very edges of the day/night, but what would the approximate temperature stabilize at both on the point on the Earth directly facing the Sun, and a point on the surface of the Earth exactly away from the Sun?&lt;/p&gt;&#xA;" OwnerUserId="19229" LastActivityDate="2018-08-20T14:01:02.423" Title="Temperature of Earth if tidal locked with the Sun?" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24832" PostTypeId="1" AcceptedAnswerId="24833" CreationDate="2018-02-01T16:32:58.140" Score="2" ViewCount="153" Body="&lt;p&gt;I am going to be in Death Valley around August 8th. I was hoping to take a milky way shot with my camera when there. I would be shooting around 10pm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using google maps to scout out popular viewpoints in Death Valley for potential locations for the shot, as I will be getting terrestrial objects in the frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I have read , I believe the milky way will be visible most of the night, and best viewable in the earlier hours of the night. In addition it will be in the south sky. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So will it be precisely due south? Or will it be slightly south west or south east? If I could pinpoint the location, it would be easier to scout out a spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="17944" LastActivityDate="2018-02-01T19:55:33.473" Title="Location of the milky way in August in Death Valley" Tags="&lt;amateur-observing&gt;&lt;milky-way&gt;&lt;photography&gt;&lt;star-gazing&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24833" PostTypeId="2" ParentId="24832" CreationDate="2018-02-01T16:53:38.357" Score="3" Body="&lt;p&gt;The Milky Way will be in a straight line that goes from SSW, (a bearing of about 200 degrees) almost directly overhead and to the NE. The brightest part of the Milky way will be low in the sky in the SSW.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In future, you can check the visibility and direction of any astronomical object with planetarium software, such as the (free) &lt;a href=&quot;https://stellarium.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellarium&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-02-01T16:53:38.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24834" PostTypeId="1" CreationDate="2018-02-01T17:21:37.420" Score="2" ViewCount="155" Body="&lt;p&gt;I've just adapted a webcam for my telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm new to astronomy so I'm not sure if the view I get is right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I attached the webcam to my &lt;a href=&quot;https://en.wikipedia.org/wiki/Barlow_lens&quot; rel=&quot;nofollow noreferrer&quot;&gt;Barlow lens&lt;/a&gt;, the view on my laptop becomes a tiny circle of light, like a tunneling effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this normal? How could I make the view I see on screen lager?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what I'm currently seeing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UDx0w.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UDx0w.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21421" LastEditorUserId="7982" LastEditDate="2018-02-01T23:46:56.073" LastActivityDate="2018-02-01T23:46:56.073" Title="Webcam set up small circle tunnel vision type view" Tags="&lt;telescope&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24835" PostTypeId="1" CreationDate="2018-02-01T19:08:40.603" Score="2" ViewCount="58" Body="&lt;p&gt;Earth (and all of its 'neighbors') is a spheroid. Why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth could've been a cube, a tetrahedron (pyramid) or really &lt;em&gt;any&lt;/em&gt; shape! But why a sphere?&lt;/p&gt;&#xA;" OwnerUserId="21239" LastEditorDisplayName="user1569" LastEditDate="2018-02-02T08:42:50.140" LastActivityDate="2018-02-02T08:42:50.140" Title="Why is Earth a sphere?" Tags="&lt;earth&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2018-02-02T11:50:54.280" ContentLicense="CC BY-SA 3.0" />
  <row Id="24836" PostTypeId="2" ParentId="24832" CreationDate="2018-02-01T19:45:35.997" Score="1" Body="&lt;p&gt;In early August after the beginning of total darkness the Milky Way will be visible overhead. Most of it will stretch NE-SW, crossing &lt;strong&gt;the whole sky&lt;/strong&gt;. All this will change depending on the time of the night, of course. But it's impossible to miss, and it arches overhead - so as long as you can see the sky, you'll be fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any place in the Death Valley proper will be okay. Since there are few artificial lights, the sky is very dark everywhere. Even close to popular spots like Stovepipe Wells or Furnace Creek you will be fine - just drive 15 minutes away from the lights and you'll be in complete darkness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would only consider more mundane concerns such as ease of access, comfort, and safety. From an astronomical perspective it's all the same, as long as the overall horizon is low enough for your shot. I suggest scouting out the area during the day. Where the valley floor is pretty low, the horizon rises quite high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you start out from Ridgecrest, driving towards the Valley, as soon as you're close to the industrial facility NE of the town (not too close - a little before it, or a little past it) the sky is already very dark. That's maybe a 20 minutes drive from the town. Just pull over, and look up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definitely bring binoculars. This is not the usual city sky. Even in a small instrument, like regular binoculars, it's full of stars. You'll find little clusters and other interesting formations everywhere. Just make sure your elbows are resting steadily on the car's roof, and slowly scan the sky with the binocs. It looks amazing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This is probably obvious, but bears emphasizing for anyone reading this who are not familiar with the area - bring water, lots of it, and make sure your car's AC is working properly. You don't want to be stuck somewhere, without water, in 125 F / 51 C weather.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2018-02-01T19:55:33.473" LastActivityDate="2018-02-01T19:55:33.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24837" PostTypeId="1" CreationDate="2018-02-01T19:50:49.813" Score="-1" ViewCount="65" Body="&lt;p&gt;From the rooftop, certain lights look roughly as bright as Arcturus. If these lights emit a power of 10 W and are 5 km away. What does that tell you about Arcturus?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you assume all stars are equally luminous*, how far would Arcturus be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you measured the parallax to Arcturus, 90 mas**, how luminous would Arcturus be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am struggling with solving this problem. I realize that I might need the formula: $F = L/(4\pi D^2)$ for flux vs. luminosity and distance, but I do not know how to solve it. Please walk me through the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*The luminosity of the Sun is 3.86e26 W&lt;/p&gt;&#xA;&#xA;&lt;p&gt;** 1 Parsec = 3.08e16 m&lt;/p&gt;&#xA;" OwnerUserId="21392" LastEditorUserId="2365" LastEditDate="2018-02-01T20:38:14.530" LastActivityDate="2018-03-05T12:32:30.407" Title="Parallax and luminosity question about Arcturus, compared to terrestrial light?" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2018-03-06T05:27:23.530" ContentLicense="CC BY-SA 3.0" />
  <row Id="24838" PostTypeId="1" CreationDate="2018-02-01T19:57:35.507" Score="2" ViewCount="209" Body="&lt;p&gt;What is the inclination of Mercury's orbit w.r.t. the Sun's equatorial plane? &#xA;Additionally, what is the orientation of Mercury's orbit (long axis) w.r.t. the Milky Way?&lt;/p&gt;&#xA;" OwnerUserId="21423" LastActivityDate="2018-02-02T16:37:10.027" Title="What is the inclination of Mercury's orbit w.r.t. the Sun's equatorial plane?" Tags="&lt;the-sun&gt;&lt;mercury&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24839" PostTypeId="2" ParentId="24813" CreationDate="2018-02-01T20:12:18.890" Score="1" Body="&lt;p&gt;This is a very broad topic. You can literally write entire books on it. I'll stick a short answer here for future reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have two main options: refractor or reflector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Refracting_telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Refracting_telescope&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Reflecting_telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Reflecting_telescope&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refractor is easier. You'll need a relatively large lens with a long focal length - this will be your primary optics, at the top of your telescope. The length of your scope will be dictated by the focal length of the primary lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can measure the focal length by going outside when the Sun is up and trying to focus the Sun to a tiny pinpoint image - the distance then between lens and image is the focal length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an ocular or eyepiece you could use a smaller lens with a short focal length. Much better, you could use an eyepiece from an old microscope or binoculars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total distance between the objective (primary lens) and ocular (eyepiece) is the sum of their focal lengths. If the primary lens has focal length F1, and the eyepiece has focal length F2, then when looking at infinity the distance between them is F1 + F2. This will help you calculate the total length of the tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the ocular (eyepiece) is a divergent lens (the kind of lens that makes things seem smaller) then the total distance is F1 - F2. By the way, it's hard to measure the focal length of a divergent lens, because it doesn't make an image you can project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnification of the instrument will be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M = F1 / F2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do not worry too much about magnification. Make something that works first. More magnification is not always better. In the beginning, aim for something in the 10x ... 50x range. Refine it later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure the tube is rigid. A plastic pipe like the ones used in water installations should work well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure all lenses are centered and aligned with precision. You don't want your lenses to be leaning in any direction. Their main axis should coincide with the axis of the scope. This is important for performance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you get too many color fringes (little rainbows) around objects, try and reduce the diameter of the primary lens by placing a paper ring in front of it - this makes the image a bit darker but should diminish the pesky rainbows somewhat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mount the eyepiece on a smaller tube that's sliding into the main tube. Cut the main tube a little shorter than F1 + F2. This will allow you to move the eyepiece back and forth until the image is in good focus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you find a shop that makes prescription eyeglasses (for people with eye or vision problems), they have so-called &quot;blank&quot; lenses that they use to cut the glasses from. See if you can buy one of those for your primary optics. A focal length of about 1 meter should be fine (opticians call it a 1 dioptre lens, which you can find in pretty weak prescription glasses).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes you can pull a good lens out of a broken projector or something, but make sure the diameter is big enough (50 mm or bigger should be fine) and the focal length is big enough (I would say anywhere between 0.5 and 2 meters - that's between 2 and 0.5 dioptres - but keep it close to 1 m if possible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first telescope I made, many years ago, was very much like this. A blank lens from a prescription glasses maker, a plastic pipe, an eyepiece from a toy microscope. It worked reasonably well, I could see with it craters and mountains on the Moon.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Making a reflector is much harder, but the performance of the instrument is higher. Read the Stellafane guide, it's very comprehensive:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://stellafane.org/tm/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://stellafane.org/tm/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's described in there is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Dobsonian_telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;dobsonian telescope&lt;/a&gt;. That's a &lt;a href=&quot;https://en.wikipedia.org/wiki/Newtonian_telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;newtonian reflector&lt;/a&gt; on a simple alt-azimuth mount.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2018-02-01T20:29:44.017" LastActivityDate="2018-02-01T20:29:44.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24841" PostTypeId="1" CreationDate="2018-02-01T22:48:11.630" Score="2" ViewCount="305" Body="&lt;p&gt;So I know that the universe is flat in a 4D point of view, but what about a 3D point of view? The universe does not SEEM flat from our view, so how can it really be so? Please answer more simple, im not an astronomy pro.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if this is a dumb question, I don't have a degree in astrophysics or anything, i'm just a kid who is really interested in the universe, so i'm sorry (in advance) if this is a simply answered bad question.&lt;/p&gt;&#xA;" OwnerUserId="20518" LastActivityDate="2018-02-02T01:06:11.613" Title="If the Universe is Flat, why does it look 3D from our View?" Tags="&lt;universe&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24842" PostTypeId="1" CreationDate="2018-02-01T22:55:31.477" Score="10" ViewCount="140" Body="&lt;p&gt;One can ask whether two objects are spinning in the same direction or not. For example, the Sun, Earth, Luna, most of our solar planets (except Venus and Uranus), and the planets’ satellites all spin in approximately the same direction. Likewise, the orbits of these bodies are in the same direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the rotational plane of our solar system is tilted about 60 degrees relative to our galaxy’s. If you project one rotation onto the plane of the other, I have the impression from maps that they may have opposite orientations — i.e. if we face north to south, our solar system’s rotation is counterclockwise, and our galaxy’s rotation is clockwise. Correct? What about the rotations at other scales of the larger systems we are in, such as our local star cluster, our local galactic cluster, and so forth? Thank you.&lt;/p&gt;&#xA;" OwnerUserId="21425" LastActivityDate="2018-06-29T04:32:32.713" Title="To what extent do the spin of smaller celestial bodies reflect the spin of the larger system’s of which they are a part?" Tags="&lt;galaxy-cluster&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24843" PostTypeId="1" AcceptedAnswerId="24844" CreationDate="2018-02-01T23:20:56.673" Score="1" ViewCount="1021" Body="&lt;p&gt;I've read that Uranus has a lower surface gravity due to its lower density than Earth. However, does this mean Uranus has a weaker pull on its moons than Earth does for the same reason?&lt;/p&gt;&#xA;" OwnerUserId="16048" LastActivityDate="2018-02-03T11:55:40.587" Title="Does Uranus have a stronger gravitational pull on its moons than Earth does?" Tags="&lt;gravity&gt;&lt;uranus&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24844" PostTypeId="2" ParentId="24843" CreationDate="2018-02-01T23:52:06.310" Score="7" Body="&lt;p&gt;The &quot;strength of gravity&quot; (ie gravitational acceleration) is determined by the mass of the planet and the distance between the planet and the moon. The formula is &#xA;$$GM/r^2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Earth (mass= 1 Earth, distance to moon=390000km) the acceleration is $0.003 ms^{-2}$&lt;br&gt;&#xA;For Uranus and Titania  (mass =14.5 Earth, distance=435000km) the acceleration is $0.03 ms^{-2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, the gravitational pull of Uranus on its moon Titania is 10 times greater than the gravitational pull of Earth on her moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same formula applies to surface gravity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Earth, (mass = 1, radius = 6370km) the acceleration due to gravity on the surface is $9.8 ms^{-2}$&lt;br&gt;&#xA;For Uranus (mass =14.5 Earths, radius = 25400km) the acceleration due to gravity is $9.0ms^{-2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface gravity on Uranus is less that that of Earth, due mostly to the relatively low density of Uranus meaning that you are much further from the centre of the planet when you are at the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(These value vary due to the neither the Earth nor Uranus being perfectly spherical, and the effective gravity is also lower due to centrifugal effects)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-02-03T11:55:40.587" LastActivityDate="2018-02-03T11:55:40.587" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24845" PostTypeId="2" ParentId="24841" CreationDate="2018-02-02T01:06:11.613" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So I know that the universe is flat in a 4D point of view, but what about a 3D point of view?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You seem to be misunderstanding the term &quot;flatness&quot;. In this context, it does not mean the universe looks like a piece of paper in any sense. It instead refers to whether space is warped.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;i'm just a kid who is really interested in the universe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll try to simplify as much as possible. I'm assuming you've learned that triangles have 180°. For the mathematics you'll be dealing with in middle school, high school and even most of college, you'll only have to deal with that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more accurate statement is that triangles have 180° when our geometry is not warped. When triangles have 180°, we say we are dealing with Euclidean geometry (named after Euclid, an ancient Greek mathematician). There are in fact cases in higher mathematics when this does not hold, which we call Non-Euclidean geometries. Just as a side note: Euclidean geometry was developed thousands of years ago. On the other hand, Non-Euclidean geometries only began to develop in the 1800s, so they're pretty recent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, Euclidean geometry deals with flat spaces. For example, if you draw a triangle on a piece of paper, it will indeed have 180°. However, Non-Euclidean geometry deals with curved (warped) spaces. For example, if you draw a triangle on the surface of a ball (which is not flat), the triangle will actually have more than 180°. If you instead draw the triangle on a potato chip or saddle-shaped surface (also not flat), it will have less than 180°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So with regards to the Universe: let's say we draw a triangle connecting three stars in outer space (we can actually use any object or simply points, but I'm using stars as an example). Will that triangle be less than, equal to or more than 180°?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to our measurements, it will actually be equal to 180°; the Universe is indeed flat. This has major implications for physics and astronomy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;i'm sorry (in advance) if this is a simply answered bad question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Future advice: don't be afraid to ask questions. Yours wasn't bad and is pretty common.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2018-02-02T01:06:11.613" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24846" PostTypeId="1" AcceptedAnswerId="25475" CreationDate="2018-02-02T04:58:58.720" Score="1" ViewCount="1123" Body="&lt;p&gt;I saw some article on the web about making an eyepiece of a telescope with the lens from a CD-ROM. Will this really work? I have an objective lens of 100 cm focal length. What will be the magnification if I could make a refractive telescope? Can I see the planets and moon with that? Can anyone suggest the arrangement of the lenses?&lt;/p&gt;&#xA;" OwnerUserId="16995" LastEditorUserId="7472" LastEditDate="2018-02-02T20:12:10.733" LastActivityDate="2018-03-13T23:06:08.463" Title="Can I build a telescope with a lens from a CD-ROM as eyepiece?" Tags="&lt;telescope&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24848" PostTypeId="2" ParentId="24846" CreationDate="2018-02-02T05:44:32.183" Score="3" Body="&lt;p&gt;This article gives a construction guide for how to make an eyepiece using a CD ROM player lens: &lt;a href=&quot;http://www.instructables.com/id/How-to-make-a-eyepiece-for-telescope/&quot; rel=&quot;nofollow noreferrer&quot;&gt;How to make an eyepiece for a telescope&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to what the focal length of your  eyepiece will be, you can refer to &lt;a href=&quot;http://leung.uwaterloo.ca/MNS/102/Lectures%202007/Lect%20Mat/physics%20of%20cd%20pe930103.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;, in particular Figure 1 from which I derived the following relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f=0.73r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f$ is the focal length of your lens, and $r$ its radius (this is a rough approximation, of course). So given the fact that a typical CD drive lens is about 5mm in radius, its focal length will be only a few millimeters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, assuming the focal length of your telescope is 1000mm, the magnification will be a few hundred times. What you will be able to see with that depends on the width of your objective lens and how well you will have built your eyepiece, but you should probably be able to see the Moon with your custom eyepiece. As for the planets, it will be hard to get something good given how much you are magnifying—not impossible, but be aware that magnifying several hundred times requires a good weather, a good telescope and a good eyepiece.&lt;/p&gt;&#xA;" OwnerUserId="21371" LastActivityDate="2018-02-02T05:44:32.183" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24849" PostTypeId="2" ParentId="24838" CreationDate="2018-02-02T06:14:33.690" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the inclination of Mercury's orbit w.r.t. the Sun's equatorial plane?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_inclination#Orbits&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;, we can see that it is 3.38°.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Additionally, what is the orientation of Mercury's orbit (long axis) w.r.t. the Milky Way?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;This is a bit harder to estimate precisely, but we can refer to &lt;a href=&quot;http://coolcosmos.ipac.caltech.edu/image_galleries/IRAS/allsky.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this view from the IRAS satellite&lt;/a&gt; to get a rough figure of 63° between the ecliptic and the plane of the Milky Way (for more information on how to calculate that, check out &lt;a href=&quot;https://en.wikipedia.org/wiki/Aitoff_projection&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aitoff projection&lt;/a&gt;). Now just add/subtract Mercury's 7.01° of inclination to ecliptic and you have your result.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Because the angles of 63 and 7 degrees may not be along the same axis, there are ways to be a bit more precise: you can take Mercury's coordinates in the ecliptic coordinates system, and transform them into &lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;galactic coordinates&lt;/a&gt;. If you do that with only 3 different points on Mercury's orbit, you can find the angle made by the plane defined by those 3 points and the horizontal galactic plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use &lt;a href=&quot;https://omniweb.gsfc.nasa.gov/cgi/models/planet.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;this website&lt;/a&gt; to get your 3 sets of ecliptic coordinates for Mercury, and &lt;a href=&quot;https://ned.ipac.caltech.edu/forms/calculator.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;that website&lt;/a&gt; to transform them into galactic coordinates. You can transform those into regular Cartesian coordinates since the website that gave you the ecliptic coordinates also gives you the distance from Mercury to the Sun! Once you get there, it's easy to get the angle made by that plane and the horizontal: good luck!&lt;/p&gt;&#xA;" OwnerUserId="21371" LastEditorUserId="21371" LastEditDate="2018-02-02T16:37:10.027" LastActivityDate="2018-02-02T16:37:10.027" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24858" PostTypeId="1" AcceptedAnswerId="24860" CreationDate="2018-02-02T15:12:44.527" Score="1" ViewCount="438" Body="&lt;p&gt;The distant an interstellar object is, the more time is required for light to travel, which is known, measured as light years. So are the farthest stars and gas clusters in the Milky Way seen as how they looked millions of years ago &amp;amp; not how as they look today? &lt;/p&gt;&#xA;" OwnerUserId="20219" LastEditorUserId="20219" LastEditDate="2018-02-02T17:51:26.363" LastActivityDate="2018-05-02T11:30:13.927" Title="Is the Milky Way seen as how it looked millions of years ago?" Tags="&lt;galaxy&gt;&lt;distances&gt;&lt;milky-way&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24859" PostTypeId="1" AcceptedAnswerId="24863" CreationDate="2018-02-02T15:22:52.037" Score="10" ViewCount="6429" Body="&lt;p&gt;I am trying to understand how to calculate local sidereal time and have found the following formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{LST} = 100.46 + 0.985647 \cdot d + \text{long} + 15 \cdot \text{UT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $d$ is the number of days from J2000, including the fraction of a day&lt;br&gt;&#xA;UT is the universal time in decimal hours&lt;br&gt;&#xA;long is your longitude in decimal degrees, East positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They don't explain what the two constants are (&lt;code&gt;100.46&lt;/code&gt; and &lt;code&gt;0.985647&lt;/code&gt;), could anyone please explain what those constants are and how they were calculated in the first place?&lt;/p&gt;&#xA;" OwnerUserId="21446" LastEditorUserId="88" LastEditDate="2018-02-03T12:27:27.213" LastActivityDate="2018-02-03T12:27:27.213" Title="Local Sidereal Time" Tags="&lt;coordinate&gt;&lt;time&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24860" PostTypeId="2" ParentId="24858" CreationDate="2018-02-02T15:49:31.460" Score="5" Body="&lt;p&gt;Most of the matter in the Milky Way is contained within a radius of 100,000 light-years from the galactic center. This encompasses the disk and most of the halo, including halo stars and globular clusters. It also includes quite a few (though not all) of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Satellite_galaxies_of_the_Milky_Way&quot; rel=&quot;noreferrer&quot;&gt;Milky Way's satellite galaxies&lt;/a&gt;. This is rather large - the Sun is only about 25,000 light-years from the galactic center - but not truly inter-galactic scales. &lt;a href=&quot;https://en.wikipedia.org/wiki/Andromeda_Galaxy&quot; rel=&quot;noreferrer&quot;&gt;Andromeda, the nearest large spiral galaxy&lt;/a&gt;, is 2.5 million light-years away, and so we see it as it was 2.5 million years ago. However, we see most of the objects in the Milky Way as they were no more than 100-200,000 years ago, usually much closer to the present day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's true that &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_largest_galaxies&quot; rel=&quot;noreferrer&quot;&gt;some galaxies are a couple million light-years across&lt;/a&gt;, but they are quite rare. In fact, &lt;a href=&quot;https://en.wikipedia.org/wiki/Malin_1&quot; rel=&quot;noreferrer&quot;&gt;the largest known spiral galaxy, Malin 1&lt;/a&gt;, is a mere 650,000 light-years in diameter. I should make it clear, though, that most of these galaxies are large because they have extremely large &lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_halo&quot; rel=&quot;noreferrer&quot;&gt;halos&lt;/a&gt;, consisting of gas, pockets of stars, and dark matter. Most matter is not contained in the halo. Additionally, it is not always clear where a galaxy ends and where the intergalactic medium begins.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2018-02-02T15:49:31.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24861" PostTypeId="2" ParentId="24858" CreationDate="2018-02-02T16:34:18.437" Score="1" Body="&lt;p&gt;We can't see the stars on the other side of the Milky way, but (taking some numbers from HDE's answer), If we're 25,000 light years from the center and we're looking at the far side of the milky-way, to stars at the edge of the disk, about 75,000 light years from us, we see those star systems as they were 75,000 years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;75,000 years is very little time compared to the 230 million years that it takes our sun to orbit the milky way, and if the outer stars orbit slightly faster (lets say 225 million years), that's just 1/3000th of an orbit, so the stars wouldn't have moved much.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would change, if we could see it, is that stars close to each other would dance around.   75,000 years is enough time for stars to move relative to their closes neighbors.   &lt;a href=&quot;https://astronomy.stackexchange.com/questions/13359/are-there-good-animations-of-the-movement-of-proximate-stars-over-long-time-peri&quot;&gt;More info here&lt;/a&gt; and chart below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ynCxN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ynCxN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't have telescopes good enough to make out details like that on the other side of the galaxy, but that's an example of something we see that would have changed in 75,000 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a somewhat related sidebar,  Ole Rømer used this understanding to &lt;a href=&quot;https://en.wikipedia.org/wiki/R%C3%B8mer%27s_determination_of_the_speed_of_light&quot; rel=&quot;nofollow noreferrer&quot;&gt;calculate the speed of light&lt;/a&gt; by watching Jupiter and measuring when it's moon Io passed behind it viewed from Earth.  That only works for repeating orbits though.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-02-02T16:34:18.437" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24863" PostTypeId="2" ParentId="24859" CreationDate="2018-02-02T19:14:20.153" Score="10" Body="&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;LST = 100.46 + 0.985647 * d + long + 15*UT&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &#xA;  &lt;p&gt;They don't explain what the two constants are (100.46 and 0.985647), could anyone explain what those constants are and how they were calculated in the first place please?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are three constants there, 100.46, 0.985647, and 15.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of 100.46 degrees is the value needed to make the expression yield the correct value for GMST at 0&lt;sup&gt;h&lt;/sup&gt; UT on 1 January 2000. The value of 0.985647 degrees per day is the number of degrees the Earth rotates in one mean solar day, sans a multiple of 360. The value of 15 degrees per hour is the number of degrees the Earth rotates with respect to the mean fictitious Sun every hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding 0.985647: There is one extra sidereal day in a solar year than there are solar days. There are 365.2422 solar days in a year, so the Earth rotates $360*366.2422/365.2422=360.985647332$ degrees per solar day with respect to the stars. That first 360 is irrelevant (the result needs to be taken mod 360 in the end), resulting in the factor of 0.985647 (0.985647332 rounded to six significant digits).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding 15: Note that this is the number of degrees the Earth rotates per hour with respect to the Sun. Multiplying this by $366.2422/365.2422$ yields 15.04106864, the number of degrees the Earth rotates per hour with respect to the stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to achieve the same result is to fold that extra 0.04106864 degrees per hour into the number of days since noon on 1 January 2000.  Not surprisingly, 0.04106864*24 = 0.985647. This means that the $d$ in the approximate formula in the question must include the fractional days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to take care with this approximate formula. It is approximately true for the 200 year period centered around midnight on 1 January 2000, and you need to make sure that the $d$ is the number of days from noon on 1 January 2000, including fractional days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;strong&gt;Addendum&lt;/strong&gt;: Showing this is the same as the &lt;em&gt;Astronomical Almanac&lt;/em&gt; expression, sans a quadratic term&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Astronomical Almanac gives an expression for approximate mean sidereal time, in hours:&#xA;$$\mathit{GMST} = 6.697374558 + 0.06570982441908 D_0 + 1.00273790935 H + 0.000026 T^2$$&#xA;Where $\mathit{GMST}$ is the mean sidereal time in hours, $H$ is the universal time at the time in question, $D_0$ is the Julian date on the previous midnight of the time in question less 2451545.0, $D$ is the Julian date at the time in question (including fractional days) less 2451545.0, and $T$ is $D/36525$. The relationship between $D_0$, $D$, and $H$ is quite simple: $D_0 = D - H/24$. Substituting this in the above and omitting the quadratic term yields&#xA;$$\begin{aligned}&#xA;\mathit{GMST} &amp;amp;= 6.697374558 + 0.06570982441908 (D-H/24) + 1.00273790935 H&#xA;\\&#xA;&amp;amp;= 6.697374558 + 0.06570982441908D + H&#xA;\end{aligned}$$&#xA;(Strictly speaking, 1.00273790935-0.06570982441908/24 = 0.9999999999992 rather than 1.0, but that's just because that 1.00273790935 should be 1.0027379093508).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying by 15 yields the GMST in degrees:&#xA;$$\mathit{GMST}_{\text{deg}} = 100.4606184 + 0.9856473662862 D + 15 H$$&#xA;This is the expression in the question, sans the longitude and plus some extra digits.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2018-02-02T20:11:05.543" LastActivityDate="2018-02-02T20:11:05.543" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24864" PostTypeId="5" CreationDate="2018-02-02T20:33:08.320" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2018-02-02T20:33:08.320" LastActivityDate="2018-02-02T20:33:08.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24865" PostTypeId="4" CreationDate="2018-02-02T20:33:08.320" Score="0" Body="For questions about transits of planets in front of the Sun, as seen from another, more distant planet; for example, the transits of Mercury and Venus which are visible from the Earth." OwnerUserId="7472" LastEditorUserId="7472" LastEditDate="2018-02-03T22:53:41.753" LastActivityDate="2018-02-03T22:53:41.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24866" PostTypeId="2" ParentId="24843" CreationDate="2018-02-02T20:56:52.397" Score="3" Body="&lt;p&gt;Provided that you are the same distance from the centre and you are on or above the surface then the heavier (more mass) planet will have a greater gravitational attraction.  It is still possible that the heavier planet has a lower attraction at the surface since it may be bigger.  If you can find the mass and radius then you can use the formula that James gives to calculate the attraction at the surface.&lt;/p&gt;&#xA;" OwnerUserId="21104" LastActivityDate="2018-02-02T20:56:52.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24867" PostTypeId="1" AcceptedAnswerId="24868" CreationDate="2018-02-02T23:00:48.303" Score="2" ViewCount="65" Body="&lt;p&gt;I missed the sight yesterday (1 feb 2018) because I live right in the region where the sight was not visible. So when will the next Super Blue Blood Moon occur?&lt;/p&gt;&#xA;" OwnerUserId="21452" LastActivityDate="2018-02-03T04:24:14.113" Title="When will the next 'Super Blue Blood Moon' occur?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24868" PostTypeId="2" ParentId="24867" CreationDate="2018-02-03T04:24:14.113" Score="3" Body="&lt;p&gt;If you are looking for the perfect 'Super Blue Blood Moon', then, according to &lt;a href=&quot;https://www.space.com/39557-next-blood-moon-eclipses-2018-2019.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; (assuming you live in the US):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The next Super Blue Blood Moon won't happen until &lt;strong&gt;Jan. 31, 2037&lt;/strong&gt;, and it will be visible in North America, Europe, Asia, Australia and parts of Africa.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, if a Super Blood Moon is enough for you, you can see the next one on &lt;strong&gt;Jan. 21, 2019&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="21371" LastActivityDate="2018-02-03T04:24:14.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24871" PostTypeId="1" AcceptedAnswerId="24886" CreationDate="2018-02-03T07:58:04.050" Score="1" ViewCount="88" Body="&lt;p&gt;I recently watched Brian Greene documentaries on parallel universes and am trying to wrap my brain around multiple quantum universes that form at every event and Einstein's relativity that in space-time, past present and future coexist so we can observe the future of a far away galaxy from Earth by moving. &#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=MO_Q_f1WgQI&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=MO_Q_f1WgQI&lt;/a&gt; - Around 7th minute - explains how an alien moving towards earth has a now slice cutting in Earth's future&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is which quantum universe will be seen because future events in that galaxy would create multiple such universes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is regarding many worlds interpretation of quantum mechanics according to which multiple alternate futures are possible &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Many-worlds_interpretation&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Many-worlds_interpretation&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question has been answered by James K below.&lt;/p&gt;&#xA;" OwnerUserId="21460" LastEditorUserId="21460" LastEditDate="2018-02-05T18:46:09.577" LastActivityDate="2018-02-05T18:46:09.577" Title="Which quantum reality will be seen in a far away galaxy" Tags="&lt;space-time&gt;&lt;multiverse&gt;&lt;quantum-mechanics&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2018-02-05T08:52:05.353" ContentLicense="CC BY-SA 3.0" />
  <row Id="24878" PostTypeId="2" ParentId="24837" CreationDate="2018-02-03T12:18:18.663" Score="1" Body="&lt;p&gt;It tells you that Arcturus is a long way away and very bright. It was not realised just how far away and how bright stars were until the 17th century, and accurate calculations didn't come until the 18th. There are still many objects for which we only have a rough idea of of how distant and how bright they are. Even well known stars like Betelgeuse have not had their distance measured with great accuracy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a formula for Flux, (which you understand to be a formula for &quot;apparent brightness&quot;) and you know that distant light bulbs can appear as bright as stars so you can make an equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L_{\mathrm{bulb}}/(4\pi D_{\mathrm{bulb}}) = L_{\mathrm{star}}/(4\pi D_{\mathrm{star}})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know the L and D for the bulb, and in the first question you assume a value of the L for Arcturus, and deduce a distance. Substitute known and assumed values and solve for $D_{\mathrm{star}}$, taking care to get the units consistent (change everything to SI units: for this exercise). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second question you use parallax to calculate $D_{\mathrm{star}}$ (distance in parsecs = 1000/(parallax in mas)). You substitute this value to get the value for the Luminosity.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-02-03T12:18:18.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24881" PostTypeId="1" AcceptedAnswerId="24885" CreationDate="2018-02-03T15:30:35.173" Score="1" ViewCount="1061" Body="&lt;p&gt;I've been using a Meade ETX 90 for a few years now. Since it doesn't have the new computerized mount (with goto), I am looking into buying one to swap out. Where might I find this and is there a more technical term than just &quot;computerized mount&quot; that is used?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: This is an old scope that did not originally come with the ports that the autostar controller now goes into (on the newer models).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: I ended up getting a newer Meade ETX and plan to replace my current mount with the computerized one. Also, if you check out cloudynights.com you might be able to find one. I have seen one on there before.&lt;/p&gt;&#xA;" OwnerUserId="21237" LastEditorUserId="21237" LastEditDate="2018-04-23T01:11:52.040" LastActivityDate="2018-04-23T01:11:52.040" Title="Replacing Meade ETX with computerized mount" Tags="&lt;telescope&gt;&lt;refractor-telescope&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24883" PostTypeId="2" ParentId="23938" CreationDate="2018-02-03T17:37:06.757" Score="1" Body="&lt;p&gt;Several options are possible:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't interpolate at all. Just assume the planet stays fixed until the next update. The planet Earth moves $10^{-7}$ degrees in 30milliseconds. In the vastness of space that is a rounding error. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given two objects, their position and velocity and acceleration, and the values of their Keplerian elements at times t0 and t1, you want to estimate their position, velocity and acceleration at any time in between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use simple linear interpolation of their position. This has the advantage of being very quick, but it assumes a constant velocity. Over short time-spans this is a reasonable assumption. For most orbiting bodies this approximation should be accurate enough for differences in time of many minutes, as the changes in velocity are relatively small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can assume constant acceleration and interpolate using the SUVAT equation. This does very well if the objects don't move too far. For example it can model the flight of a ball in the Earth's gravitational field very well, giving a parabolic shape. This should give a fairly good approximation of the position over hours, or even days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can interpolate the orbital elements, linearly, and solve Kepler's equation to get the position of the body at any given time. For most bodies the Orbital elements change only slowly with time (except for the Anomaly) That is the shape of the orbit stays nearly the same, only the position on the orbit changes. Approximating the orbit of a planet by interpolation of its elements should give a good approximation for several hundred years of orbital time. However the process of solving the equation is slow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally you can numerically integrate the n-body problem for the solar system. Depending on the time step you choose this can be accurate for an indefinite amount of time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So these can be summarised as:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Constant position&lt;/li&gt;&#xA;&lt;li&gt;Constant velocity&lt;/li&gt;&#xA;&lt;li&gt;Constant acceleration&lt;/li&gt;&#xA;&lt;li&gt;Variable acceleration following an inverse square law.&lt;/li&gt;&#xA;&lt;li&gt;Variable acceleration n-body graviational field.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For a game engine, speed is good. If the gaps in time are on the order of milliseconds (of real time) then simply not interpolating should be undetectable.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-02-03T17:37:06.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24885" PostTypeId="2" ParentId="24881" CreationDate="2018-02-03T19:58:45.540" Score="1" Body="&lt;p&gt;Meade make the 497 auto star hand controller, which plugs in where your existing hand controller goes. This turns your mount into a fully computerised one. Alternatively you should be able to use a laptop to control the mount via the aux ports. It’s all in the manual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your telescope did not originally come with the auto star hand controller, here is a quote from &lt;a href=&quot;http://weasner.com/etx/faq.html#upgrade&quot; rel=&quot;nofollow noreferrer&quot;&gt;weasner.com&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Q. How do I upgrade a non-AutoStar ETX to an AutoStar capable ETX?&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Short answer: you can't. Meade does not offer an upgrade for the&#xA;  original model ETX (also known as the ETX-90RA or ETX Astro M). But&#xA;  you can mount the OTA on a different mount (see the Telescope Tech&#xA;  Tips page) or (if still available) get a JMI computer system (see the&#xA;  &quot;NGC-microMAX/NGC-MAX&quot; on the Accessory Reviews - Miscellaneous page).&#xA;  Or you could sell it and by a new AT model.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11822" LastEditorUserId="21237" LastEditDate="2018-02-03T22:16:17.550" LastActivityDate="2018-02-03T22:16:17.550" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24886" PostTypeId="2" ParentId="24871" CreationDate="2018-02-03T22:31:58.840" Score="4" Body="&lt;p&gt;So, the presenter explains that two observers who are not moving relative to one another will agree on the meaning of &quot;now&quot;.  However if one observer is moving, then they will no longer agree, and what one considers to be now will be in the past or future of the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He observes that a sufficiently distant observer, in another galaxy 100 million light years away that is moving at &quot;bicycle&quot; speed relative to us would have a 200 year discrepancy in their understanding of &quot;now&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this alien would have absolutely no way of observing our universe now. They could not see us in our &quot;now&quot;, nor in their &quot;now&quot; of 200 years ago, because they are 100 million light years away. They can only observe how our galaxy was 100 million years ago.  They can certainly not observe our galaxy in the future, even if they travel towards us at speeds approaching the speed of light.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the explanation of different &quot;now&quot; is about agreement on clocks, not about actual observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think then the program errs by treating Special Relativity as a full description of nature, rather than as a model. Relativity is a classical theory, in that in Relativity, if you know the initial state, you can predict the future state. There is no uncertainty principle in general relativity. It is correct to say that in Relativity the future is as real as the past. It is unclear if the same is true in Quantum Mechanics, it depends on how you interpret the maths. Whether the future exists or not cannot be answered by theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since different interpretations of QM are based on the same maths, there is no way of testing which interpretation of quantum mechanics is &quot;correct&quot; or even whether calling an interpretation &quot;correct&quot; has any scientific meaning. In the &quot;many worlds&quot; interpretation, the notion of a single space-time which you can take slices of is lost completely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I might note that I'm not a great fan of the explanations in that series, which seem to me to be confusing on a number of levels, and are more aimed at creating &quot;wow&quot; than being clear. For example, the mixed metaphor of &quot;slices of bread and aliens riding bikes&quot; seems to have confused you into thinking that there are aliens who are already seeing how our planet will be 200 years from now. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-02-03T22:31:58.840" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24887" PostTypeId="1" CreationDate="2018-02-03T23:08:58.647" Score="3" ViewCount="143" Body="&lt;p&gt;I've hypothesized for many years in my head that dark matter are the gravitational after effects of objects that have exploded. My understanding of physics is that when an object explodes, its center of mass remains in the same location. If that's the case, it seems that the gravity of that center of mass could remain. What observations or theories are there that counteract this hypothesis?&lt;/p&gt;&#xA;" OwnerUserId="21486" LastEditorUserId="1891" LastEditDate="2018-02-04T14:21:06.130" LastActivityDate="2018-11-03T19:56:45.450" Title="Dark matter virtual objects hypothesis" Tags="&lt;gravity&gt;&lt;dark-matter&gt;" AnswerCount="2" CommentCount="6" ClosedDate="2018-02-15T19:37:59.747" ContentLicense="CC BY-SA 3.0" />
  <row Id="24889" PostTypeId="1" AcceptedAnswerId="24892" CreationDate="2018-02-04T02:57:57.643" Score="1" ViewCount="79" Body="&lt;p&gt;In order to obtain a very high resolution spectrum to resolve closely spaced lines, or to measure doppler shits, spectrographs are operated with very high dispersion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But photographic plates and CCDs usually have rectangular formats with aspect ratios not very far from square. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Echelle spectrographs can put a highly dispersed spectrum into this format, but how? This &lt;a href=&quot;https://en.wikipedia.org/wiki/Echelle_grating&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; explains to some extent, but a concise discussion here would also be helpful.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2020-11-24T01:02:20.210" LastActivityDate="2020-11-24T01:02:20.210" Title="How does an Echelle spectrograph put such a long, high resolution spectrum into a square-ish format?" Tags="&lt;observational-astronomy&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24890" PostTypeId="2" ParentId="24287" CreationDate="2018-02-04T04:10:39.540" Score="2" Body="&lt;p&gt;I would buy a good 2x barrow lens to go with that. She might see Saturn's rings as a haze. Good for looking at the moon. Maybe Mars. A good tripod is a must.  I would suggest a camera mount. A light weight camera.  Secure the front of the telescope with 3 rubber bands to the tripod. To reduce vibration. 2 seconds pause on the camera. For vibration to end. A telescope that size is a fair place to start on Photos. A 3 second exposure. For stars planets &amp;amp; such. She can get some good photo's. Also other photo's. 70mm is a good daytime telescope size. Make sure she never looks towards the sun with it. For night work you need a 90mm or above. Not into astronomy. But use a 90mm Vexin for photo work. Ships in the bay. Objects up in the mountains that can not be got close to in Asia. A camera makes it so much more fun. You can build a camera mount not hard to do. &lt;/p&gt;&#xA;" OwnerUserId="20215" LastActivityDate="2018-02-04T04:10:39.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24892" PostTypeId="2" ParentId="24889" CreationDate="2018-02-04T09:28:02.600" Score="2" Body="&lt;p&gt;Echelle spectrometers operate in the higher spectral orders.  (CRIES+ uses orders 10-60 for IR light, I think.  Not sure about ESPRESSO, but you can count them...)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These high orders are highly overlapping, so, for example, if the input light is dispersed by the Echelle grating horizontally into a spectrum, then the light at a given point along the spectrum is has components contributed from several different orders.  But, since each contribution is a different wavelength from each order, the orders can be spread out &lt;em&gt;vertically&lt;/em&gt; using a second grating, the &quot;cross disperser&quot;, conveniently stacking them up for the camera.  You can see these components in &lt;a href=&quot;https://www.eso.org/sci/facilities/develop/instruments/espresso.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this diagram of the ESPRESSO spectrograph&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TL;DR:  Each horizontal line you see is a portion of a single spectral order, and the orders are spread vertically by a &lt;em&gt;cross disperser&lt;/em&gt; grating, oriented roughly perpendicularly to the first (Echelle) grating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found &lt;a href=&quot;http://slideplayer.com/slide/4178747/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this presentation with more details &amp;amp; the equations.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21481" LastActivityDate="2018-02-04T09:28:02.600" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24893" PostTypeId="2" ParentId="24887" CreationDate="2018-02-04T09:52:37.883" Score="1" Body="&lt;h3&gt;Newton's shell theorem explicitly contradicts this hypothesis&lt;/h3&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Shell_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Newton's shell theorem&lt;/a&gt; would seem to most directly comment on this hypothesis.&lt;/p&gt;&#xA;&lt;p&gt;Historically, this theorem was instrumental in providing motivation to support &lt;a href=&quot;https://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Newton's theory of gravity&lt;/a&gt;, which at that time was a speculative model that physicists around Newton's time were considering following &lt;a href=&quot;https://en.wikipedia.org/wiki/Hooke%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hooke's model for springs&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;A central premise of the shell theorem is that shells cancel out their own internal gravity.  For example, a perfectly spherical planet would be expected to have zero gravity as its center.  Also, Earth's gravity gets weaker as the center is approached.&lt;/p&gt;&#xA;&lt;p&gt;So, say that a star explodes.  Then according to Newton's law of gravity, the exploding shell should have zero gravitational pull within it.&lt;/p&gt;&#xA;&lt;h3&gt;This hypothesis would likely be modified Newtonian dynamics&lt;/h3&gt;&#xA;&lt;p&gt;For more information on alternative theories of gravity, Wikipedia's article on &lt;a href=&quot;https://en.wikipedia.org/wiki/Modified_models_of_gravity&quot; rel=&quot;nofollow noreferrer&quot;&gt;&amp;quot;&lt;em&gt;Alternatives to general relativity&lt;/em&gt;&amp;quot;&lt;/a&gt; may be helpful.&lt;/p&gt;&#xA;&lt;p&gt;In particular, this hypothesis would likely be described as &lt;a href=&quot;https://en.wikipedia.org/wiki/Modified_Newtonian_dynamics&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;modified Newtonian dynamics&lt;/em&gt; (MOND)&lt;/a&gt; since it explicitly contradicts Newtonian mechanics.&lt;/p&gt;&#xA;" OwnerUserId="15986" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-02-04T10:13:34.207" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24897" PostTypeId="1" CreationDate="2018-02-04T15:06:11.243" Score="1" ViewCount="62" Body="&lt;p&gt;Say an asteroid identical in all ways to the Chicxulub impactor (size, mass, materiel, orbital type, etc) was about to hit us.  (Again!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As of exactly early 2018,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;given our known instruments and indeed our known watch-out-for-asteroid-collisions systems,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in fact,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;when would it be a reasonable certainty, that we would know about it coming?  (Only days before hand?  Weeks - months - years - decades?)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Are there any gotchya factors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Such as ... &lt;em&gt;we'd likely miss it if it came from 'above'? 'from the direction of the sun'??&lt;/em&gt; etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any perhaps current limiting use cases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Something like ... &lt;em&gt;&quot;anything over K km we'd know for sure Y years ahead of time, smaller than that no chance...&quot;&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess, some folks on this list are familiar on a day to day basis with our &quot;known instruments and indeed our known watch-out-for-asteroid-collisions systems&quot; ...  but to the layman it's an unknown!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And technology is very topical, of course.  I guess the answer is changing in a Moore's Law -ish way from year to year.  (For example, would the JWST affect this calculation, or has nothing to do with the issue?  What about other systems or single instruments coming online soon?)&lt;/p&gt;&#xA;" OwnerUserId="13071" LastActivityDate="2018-02-04T15:06:11.243" Title="Detecting the Chicxulub impactor, right now" Tags="&lt;solar-system&gt;&lt;telescope&gt;&lt;asteroids&gt;&lt;sky-survey&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2018-02-09T10:32:49.820" ContentLicense="CC BY-SA 3.0" />
  <row Id="24898" PostTypeId="2" ParentId="17846" CreationDate="2018-02-04T15:26:06.953" Score="3" Body="&lt;p&gt;Here are some things you might find useful: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Aperture&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aperture is the most important factor that goes into a telescope purchase, NOT magnification! Doubling the diameter of the mirror (i.e. going from a 4&quot; telescope to an 8&quot;) results in a light collecting area four times greater. Planets and the moon will appear more detailed and be more enjoyable to observe. More faint objects such as nebulae and galaxies will be visible. Brighter deep sky objects will show more detail, but no color! I upgraded from an old, dusty 4.5&quot; to a 10&quot; with a new high quality eyepiece and was totally blown away. I was able to see the satellite galaxies of M31, the Andromeda galaxy, rich detail in the gas clouds of M42, the Orion nebula, and 40 other Messier objects during 5 nights of observing in December. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So hopefully I have gotten my point across: aperture is your number one priority! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mounts&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A telescope that is easy to use and setup will get used more often. A classic &lt;strong&gt;Newtonian&lt;/strong&gt; mount meets both criterion above. The telescope is often quite light and can be kept on its tripod in storage. One drawback with lower end Newtonian's is the inevitable cheap tripod that will come with the telescope. Cheap tripods are not very stable and often feel flimsy. You will find many options for small aperture, cheap Newtonian telescopes on Amazon, but remember, your goal is to maximize aperture! These are most often the telescopes that end up living most of their lives in a closet (I know from experience!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pros:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Lightweight, easy to move from storage to observing location&lt;/li&gt;&#xA;&lt;li&gt;Can be stored on tripod for easy of use&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cons:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Tripod will be cheap and flimsy and lower price ranges&lt;/li&gt;&#xA;&lt;li&gt;Included eyepieces will be junk&lt;/li&gt;&#xA;&lt;li&gt;Low aperture/price ratio&lt;/li&gt;&#xA;&lt;li&gt;No option of using 2&quot; eyepieces (I will cover these below!)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you want to maximize aperture while keeping costs low, the humble &lt;strong&gt;Dobsonian&lt;/strong&gt; is the way to go. The optical housing can be left on the base just like a Newtonian and moved from storage to your observing location. I will go ahead and tell you that the base+telescope is heavy. Taking this up and down the stairs can be a pain, but the telescope can be removed from the base for transportation. &lt;em&gt;A small stool will make observing much, much more enjoyable.&lt;/em&gt; I am 6'2&quot; and have a hard time using my Dobsonian without a one. A big factor when observing is comfort: hunching over for hours will cut your observing sessions short. This is quite a small thing, but it will make quite a big difference in terms of enjoying your telescope. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pros:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Best aperture/cost ratio&lt;/li&gt;&#xA;&lt;li&gt;Most accept 2&quot; wide angle eyepices&lt;/li&gt;&#xA;&lt;li&gt;Easiest in terms of operating &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cons:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Big and bulky&lt;/li&gt;&#xA;&lt;li&gt;Mount sits directly on the ground, so observing area must be relatively flat&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Final verdict&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's clear from your post you are interested in amateur astronomy, so I don't think you're going to buy a telescope and let it collect dust in the garage. My recommendation below falls into your budget:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B001DDW9V6&quot; rel=&quot;nofollow noreferrer&quot;&gt;Orion 8&quot; Dobsonian&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B005LHMLIQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;Meade Wide Angle Eyepiece&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This telescope will include a junky 1.25&quot; eyepiece in the box. The eyepiece I included is truly an order of magnitude better than the included one. Meade manufactures some of the best high quality telescope optics that are still affordable. This eyepiece has a greater field of view (it fits roughly 3 full moons from side-to-side in the field of view, whereas the included eyepiece snugly fits 2!) which simply enables you to see more of what you're looking at. The quality of the glass in the eyepiece is also far greater, so the image appears sharper and subtle details can be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last thing you want to do when observing light from a deep sky objects that is 10s to millions of light years away is lose it in a cheap eyepiece. While the 2&quot; eyepiece certainly isn't necessary, I highly recommend you go ahead and get it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps! &lt;/p&gt;&#xA;" OwnerUserId="20080" LastActivityDate="2018-02-04T15:26:06.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24900" PostTypeId="1" AcceptedAnswerId="24915" CreationDate="2018-02-04T17:35:01.273" Score="4" ViewCount="143" Body="&lt;p&gt;The new paper &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/2041-8213/aaa5fb/meta&quot; rel=&quot;noreferrer&quot;&gt;Probing Planets in Extragalactic Galaxies Using Quasar Microlensing&lt;/a&gt; (also in &lt;a href=&quot;https://arxiv.org/abs/1802.00049&quot; rel=&quot;noreferrer&quot;&gt;ArXiv&lt;/a&gt;) is too mind-boggling for me to understand exactly what is happening. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some very basic understanding of gravitational microlensing from other Q&amp;amp;A here (and &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_microlensing&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;) but I can't figure out what is happening in this paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are thousands of planets seen in another galaxy, which do not orbit stars? There is discussion of &quot;emission close to the event horizon of the supermassive black hole of the background quasar&quot;... there are too many unusual things going on at once for me to fathom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible for someone to sort out the basics of what's happening? The Quasar is in front, and the galaxy containing these planets is behind? What emission?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-02-06T08:15:07.267" Title="Need help understanding the recent observation of extra-galactic planets!" Tags="&lt;exoplanet&gt;&lt;gravitational-lensing&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24901" PostTypeId="1" AcceptedAnswerId="24903" CreationDate="2018-02-04T18:51:47.383" Score="0" ViewCount="77" Body="&lt;p&gt;If we orbit near a black hole to travel to future and come back to Earth, which of the quantum realities will be observed in the future Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is regarding many worlds interpretation of quantum mechanics according to which multiple alternate futures are possible &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Many-worlds_interpretation&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Many-worlds_interpretation&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="21460" LastEditorUserId="2365" LastEditDate="2018-02-05T19:15:09.510" LastActivityDate="2018-02-05T19:15:09.510" Title="Which quantum reality will be seen in a future Earth" Tags="&lt;space-time&gt;&lt;multiverse&gt;&lt;quantum-mechanics&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" ClosedDate="2018-02-05T08:26:57.517" ContentLicense="CC BY-SA 3.0" />
  <row Id="24902" PostTypeId="1" AcceptedAnswerId="24906" CreationDate="2018-02-05T00:36:05.510" Score="3" ViewCount="139" Body="&lt;p&gt;&lt;a href=&quot;http://www.independent.co.uk/news/science/planets-far-beyond-galaxy-discovered-first-time-astrophysicists-nasa-university-oklahoma-a8193531.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Recently&lt;/a&gt;, planets outside our galaxy have been discovered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would be the convention to naming these planets? As far as I know, it seems that there is no standard for naming exoplanets even in our galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if I'm publishing a scientific paper and I wish to name some of these recently discovered planets, what would be some guildelines on doing so?&lt;/p&gt;&#xA;" OwnerUserId="9662" LastActivityDate="2018-02-07T06:45:48.750" Title="How to name planets outside our galaxy?" Tags="&lt;exoplanet&gt;&lt;naming&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24903" PostTypeId="2" ParentId="24901" CreationDate="2018-02-05T03:21:17.340" Score="2" Body="&lt;p&gt;All of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you sit on Earth then (in the many worlds interpretation) each quantum mechanical observation creates multiple universes, and there is a copy of &quot;you&quot; in each one. So if I were to ask, which universe will I be in tomorrow, the answer is &quot;all of them&quot; (except the ones in which I die)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I go into space, float around and come back I will come back in all of the universes. Because I have been moving (relative to Earth) there will be a slight clock discrepancy in all the universes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I go into space, orbit a black hole, then come back. I will come back in all the universes (except for the ones that I die in). There will be a large discrepancy in clock time but that is irrelevant. I've not done any movie time-travel. It's just that my watch doesn't match the clocks on Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Copenhagen interpretation you come back in the one universe (with a certain probability that is close to 1). As the mathematics is identical, there is no experiment that can distinguish between different interpretations of QM.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-02-05T03:21:17.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24906" PostTypeId="2" ParentId="24902" CreationDate="2018-02-05T09:41:58.923" Score="9" Body="&lt;p&gt;&lt;a href=&quot;http://iopscience.iop.org/article/10.3847/2041-8213/aaa5fb/meta&quot; rel=&quot;noreferrer&quot;&gt;The original paper&lt;/a&gt; ( &lt;a href=&quot;https://arxiv.org/abs/1802.00049&quot; rel=&quot;noreferrer&quot;&gt;and a version on arXiv&lt;/a&gt; ) which this relates to does not seem to identify any actual planets.  It's clear from the abstract that it is simply suggesting there is evidence that what they describe as unbound planets seem to exist as an inference of the spectral data they have gathered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are &lt;em&gt;not&lt;/em&gt; identifying any individual planets at all.  They have not actually discovered any planets and there is nothing to name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the abstract :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Previously, planets have been detected only in the Milky Way galaxy. Here, we show that quasar microlensing provides a means to probe extragalactic planets in the lens galaxy, by studying the microlensing properties of emission close to the event horizon of the supermassive black hole of the background quasar, using the current generation telescopes. We show that a population of unbound planets between stars with masses ranging from Moon to Jupiter masses is needed to explain the frequent Fe Kα line energy shifts observed in the gravitationally lensed quasar RXJ 1131–1231 at a lens redshift of z = 0.295 or 3.8 billion lt-yr away. We constrain the planet mass-fraction to be larger than 0.0001 of the halo mass, which is equivalent to 2000 objects ranging from Moon to Jupiter mass per main-sequence star.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What they're saying is that, if you accept the hypothesis of their theory and data, there must exist a minimum of about 2000 objects of planet mass per star.  There's no way to identify any individual object and no means to verify their claims that I know of.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2018-02-05T09:41:58.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24910" PostTypeId="1" CreationDate="2018-02-05T16:36:59.017" Score="3" ViewCount="315" Body="&lt;p&gt;What would happen to Jupiter if a brown dwarf star collided with it, and vice versa what would happen to the brown dwarf star?&lt;/p&gt;&#xA;" OwnerUserId="21514" LastEditorUserId="2365" LastEditDate="2018-02-05T19:17:58.690" LastActivityDate="2018-02-06T20:00:06.657" Title="Jupiter vs brown dwarf star" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;planet&gt;&lt;jupiter&gt;&lt;brown-dwarf&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24912" PostTypeId="2" ParentId="24887" CreationDate="2018-02-05T16:51:38.817" Score="2" Body="&lt;p&gt;There are (at least) a couple of problems with your hypothesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first problem is that dark matter is just that - dark, in other words it emits no light that we can detect. The exploded remnants of stars are not dark at all. Indeed the winds and superwinds associated with exploding stars are often extremely hot and luminous in the UV and X-ray region. Even when these winds have slowed and cooled, they will then form warm dust (because they are enriched in heavy elements) that is then detectable in the infrared part of the spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second problem is that we know that the majority of dark matter is not just non-luminous &quot;normal&quot; matter. Estimates of the cosmic abundances of helium, deuterium and lithium produced in the big bang are sensitive to the overall density of this &quot;normal&quot; matter. These estimates tell us that only about 20% of the gravitating matter that appears to be required to explain the dynamics of the universe, the dynamics of galaxy clusters and the dynamics of galaxies, can be in the form of &quot;normal&quot; matter. But the atoms and particles released from an exploding star are just part of this &quot;normal&quot; matter budget - they have already been accounted for and so contribute nothing to solving this second dark matter problem.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-11-03T19:56:45.450" LastActivityDate="2018-11-03T19:56:45.450" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24915" PostTypeId="2" ParentId="24900" CreationDate="2018-02-05T21:34:29.403" Score="6" Body="&lt;p&gt;Instead of reading half-comprehensible newspaper articles, I found it more helpful to go to the &lt;a href=&quot;https://arxiv.org/abs/1802.00049&quot; rel=&quot;nofollow noreferrer&quot;&gt;original source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There, the authors explain their method very clearly:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There is a background galaxy, the Quasar galaxy. This object, being a Quasar, has an actively accreting SMBH. The SMBH, being an actively accreting one, will emit strongly in X-Rays. Those X-Rays, being emitted from the rotating accretion disc, deep inside the SMBH's gravitational well, will suffer from strong red- and blue-shifts.&lt;br&gt;&#xA;The paper works with only one line here, the Iron $\rm K\alpha$ line, observed by Chandra in 2009, 2012 and 2017.&lt;br&gt;&#xA;The red- and blue-shifted light of the SMBH, the 'image' will then travel through space and time.  &lt;/li&gt;&#xA;&lt;li&gt;Eventually this light will be lensed by a foreground galaxy, and in lucky cases the foreground galaxy can create Einstein rings, or at least multiple paths so that &lt;strong&gt;the same&lt;/strong&gt; quasar image crosses the lensed galaxy at multiple positions through the foreground galaxy.  &lt;/li&gt;&#xA;&lt;li&gt;The object of interest now, (RXJ1131-1231) consists of a quasar at $z=0.658$ and a foreground galaxy at $z=0.295$ that lenses the background quasar four times, giving four images in X-rays that can be spectrally resolved, to measure the distributions of red- and blueshifted $\rm K\alpha$-lines in each pixel.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And where the 'micro' in microlensing comes in is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Light from these Quasar images crosses different locations of the foreground galaxy and is further lensed by nearby stars in the region in the lens galaxy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So what is important here is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;No individual stars or planets are actually resolved&lt;/li&gt;&#xA;&lt;li&gt;The amount of microlenses must therefore be inferred by spectroscopy (that's why all the talk about red and blueshift).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now comes physics into play. The authors argue that they find anomalously high red- and blue-shifted $\rm K\alpha$-line distributions, that must be explained by planetary-mass objects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand how, let's consider fig. 1 from the paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fijul.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fijul.png&quot; alt=&quot;(c) Dai &amp;amp; Guerras 2018, arxiv:1802.00049v1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows the redshift distribution (The factor $G = \frac{E_{observed} }{ E_{rest-frame} }$) around the background Quasar accretion disc. Also shown is an example trace of a foreground object that could lens the emitted $K \alpha$ light from the disc.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;** The rest of the problem can now be understood in a very visual way: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the passing foreground object can actually lens the light in a way that would be seen in the spectrum is now determined by the ratio of the SMBH Gravitational radius and the foreground's object lens radius. This works as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational radius is just the Schwarzschild radius $r_g = GM/c^2$ and sets the scale of the above image. The lens radius $r_l$ is a more involved function of cosmological distance and lens mass. One can imagine it however, being a circle traveling on the green trajectory in fig 1. that will lens all the background light that it hits into one observational data point.&lt;br&gt;&#xA;The latter point must be emphasized, as the situation in fig 1. will be vastly under-resolved in the actual observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $r_l \gg r_g$ the lens will lens red- as well as blue-shifted $\rm K\alpha$-light into the image pixel and no net energyshift will be detected. If $r_l \le r_g$, then the lensed light will probably be very red or blue-shifted, according to the G-values it hits on it's trajectory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;** We're nearly done: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the lensing configuration of RXJ1131-1231 and their cosmological distances conspire to make the case $r_g = r_l$ correspond to Jupiter mass lens masses. This means objects of Jupiter masses and below will be able to contribute to a measurable Fe $\rm K\alpha$ redshift, but only if they are not buried in the larger lensing radius of a star (which would have $r_l \le r_g$)!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, as we already said, the authors only can spectroscopic information about the lensed images. So they try to reproduce the redshift-distribution of the spectrae of their images with a star-only distribution, which fails.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subsequent tests they make, indicate that they have to inject a large number of objects with $r_g=r_l$ into their test-lensing distribution to reproduce their data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summarizing:&#xA;&lt;strong&gt;This is a purely statistical argument for microlensing by a certain number of planetary-mass bodies in the X-Ray spectrae of 3 of their 4 macrolensed images. It's not a discovery and as another &lt;a href=&quot;https://astronomy.stackexchange.com/questions/24902/how-to-name-planets-outside-our-galaxy&quot;&gt;Q&amp;amp;A&lt;/a&gt; pointed out, there are no planets to name or point to.&lt;br&gt;&#xA;It is even not the only explanation for the X-Ray data seen, as there are other competing ones. &#xA;But the presented planetary explanation is the only one that can reach sufficient redshifts seen in the data.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(P.S. I hope I didn't make it more complicated than the original paper, I've actually tried to summarize it...)&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditDate="2018-02-06T08:15:07.267" LastActivityDate="2018-02-06T08:15:07.267" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24920" PostTypeId="1" AcceptedAnswerId="24935" CreationDate="2018-02-06T02:52:10.503" Score="4" ViewCount="510" Body="&lt;p&gt;I was wondering what are the green objects that you see scattered in the Hubble Ultra Deep Field image (Warning possibly large/slow download : &lt;a href=&quot;http://imgsrc.hubblesite.org/hu/db/images/hs-2004-07-a-full_jpg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://imgsrc.hubblesite.org/hu/db/images/hs-2004-07-a-full_jpg.jpg&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A crop of the image reduced in size :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1BPLV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1BPLV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21520" LastEditorUserId="10196" LastEditDate="2018-02-06T08:40:51.983" LastActivityDate="2018-02-06T15:49:34.603" Title="Green objects in Hubble Ultra Deep Field" Tags="&lt;hubble-telescope&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24921" PostTypeId="1" AcceptedAnswerId="24983" CreationDate="2018-02-06T04:57:00.363" Score="2" ViewCount="125" Body="&lt;p&gt;The size of Earth? Or bigger? I'd think they'd have to be pretty big to attract all that gas when the planets were being formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt; I'm writing a short story and the setting consists of late eighties/early nineties type of technology and knowledge of astronomy, so a rough estimate as an answer is fine&lt;/p&gt;&#xA;" OwnerUserId="21521" LastEditorUserId="21521" LastEditDate="2018-02-09T17:35:12.817" LastActivityDate="2018-02-09T17:35:12.817" Title="How big are the rocky/icy cores of Jovian planets?" Tags="&lt;planet&gt;&lt;gas-giants&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24922" PostTypeId="1" AcceptedAnswerId="24924" CreationDate="2018-02-06T05:56:03.627" Score="8" ViewCount="4992" Body="&lt;p&gt;My friend asked me &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Sun is also a star then why doesn't it twinkle like other stars&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I brought this question here so that I have (and have more) infomation on this topic.&lt;/p&gt;&#xA;" OwnerUserId="21522" LastActivityDate="2018-02-06T18:38:02.747" Title="Why do stars twinkle but the Sun doesn't (I'm asking this because the Sun is also a star)" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;solar-system&gt;&lt;light&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2018-02-06T16:37:59.023" ContentLicense="CC BY-SA 3.0" />
  <row Id="24924" PostTypeId="2" ParentId="24922" CreationDate="2018-02-06T06:48:31.077" Score="23" Body="&lt;p&gt;Stars twinkle because they are effectively a point of light. This point of light can be distorted and magnified by movement of patches of varying density in the atmosphere. These act as lenses causing the twinkle effect.&#xA;If you get to space, the stars don't twinkle. Stars appear as a point only because they are so very very distant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets twinkle less because they appear as little discs of light. If one point is being made brighter by the atmosphere perhaps another is being made less bright. By being a disc, the twinkling is smoothed out by the average amount of light making it to your eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun is a very large disc because it is closer than the other stars. The sun is so large in the sky that the distortion of the light is not very noticeable. However when the sun is setting and its light has to travel through a lot more air to reach the ground, you will notice a shimmering around the edge. This is due to the same kind of atmospheric distortion.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-02-06T18:38:02.747" LastActivityDate="2018-02-06T18:38:02.747" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="24935" PostTypeId="2" ParentId="24920" CreationDate="2018-02-06T15:49:34.603" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;what are the green objects that you see scattered in the Hubble Ultra Deep Field image&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Hubble Ultra Deep Field (HUDF) has approximately 10,000 objects in it. &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/427896/meta&quot; rel=&quot;noreferrer&quot;&gt;Between 25 and 50 of them are identified as extremely faint stars.&lt;/a&gt; That means the remaining ~10,000 objects are galaxies. To answer your question literally then, &lt;strong&gt;the green objects are galaxies&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I believe what you probably really mean to ask is, why are these galaxies green? To answer that, there are actually a number of points to address.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You seem to be under the belief that the concept of redshifting causes objects to appear red. While this is certainly a true statement, it is not exclusively true. Redshifting simply makes wavelengths longer (and thus shifted red-ward, towards the red end of the spectrum). It is certainly possible that an object which was once blue could be redshifted into the red and thus appear more red than it naturally is. However, an object that emitted most strongly in the UV could redshift into the green and thus appear more green than it naturally is. Some such effect may be going on in this image, but in actuality, I think there's other stuff going on.&lt;/li&gt;&#xA;&lt;li&gt;You have to account for image processing. It's very difficult to take a &quot;real color&quot; image of a galaxy. Very often (as was the case with the HUDF), images are taken in a few different color bands and combined together with some sort of image processing program. Its up to the discretion of the image processor to balance the colors correctly, get the right saturation, hues, luminosities, etc. There are actually a variety of different forms of the HUDF that include different bands and different forms of image processing where the images that appear green in your image don't appear green in those. An example is shown below of a rendition of the HUDF, taken directly from the same NASA/STScI Hubble website you found your image on. The coloration is significantly different than yours and many of the greenish galaxies from your image appear more blue.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UXMxd.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UXMxd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;I'm not saying this is the case with the galaxies in the HUDF, but some galaxies actually are green! A few examples include the &lt;a href=&quot;https://en.wikipedia.org/wiki/Pea_galaxy&quot; rel=&quot;noreferrer&quot;&gt;Pea Galaxy&lt;/a&gt; or the &lt;a href=&quot;https://en.wikipedia.org/wiki/Green_bean_galaxy&quot; rel=&quot;noreferrer&quot;&gt;Green Bean Galaxy&lt;/a&gt;. When you plot up galaxies based on color (note that this is the astronomical definition of color which doesn't necessarily have a 1-1 correspondence with visual color) you get a &lt;a href=&quot;https://en.wikipedia.org/wiki/Galaxy_color%E2%80%93magnitude_diagram&quot; rel=&quot;noreferrer&quot;&gt;Color-Magnitude diagram&lt;/a&gt; (CMD) wherein most galaxies are clumped into a red portion or a blue portion. However, there are a small group of galaxies in what is known as the Green Valley which do appear green for one reason or another. What actually causes their greenness, I think, is beyond the scope of this question and would be better suited for a new question.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2018-02-06T15:49:34.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24936" PostTypeId="1" CreationDate="2018-02-06T16:53:48.993" Score="2" ViewCount="1356" Body="&lt;p&gt;When calculating local sidereal time, &lt;code&gt;100.461&lt;/code&gt; &quot;is the value needed to yield the correct value of GMST&quot; - &lt;a href=&quot;https://astronomy.stackexchange.com/users/2618&quot;&gt;David Hammen&lt;/a&gt; (left as comment on &lt;a href=&quot;https://astronomy.stackexchange.com/a/24862/21446&quot;&gt;my previous question&lt;/a&gt;). How is this value calculated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If (hypothetically) I wanted to use January 1st 12:00 (noon) 2010 as my epoch, how would I calculate this number?&lt;/p&gt;&#xA;" OwnerUserId="21446" LastEditorUserId="21446" LastEditDate="2018-02-06T17:10:33.563" LastActivityDate="2018-02-07T13:50:21.197" Title="Calculating Local Sidereal Time" Tags="&lt;coordinate&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2018-02-09T10:32:30.810" ContentLicense="CC BY-SA 3.0" />
  <row Id="24938" PostTypeId="2" ParentId="24910" CreationDate="2018-02-06T20:00:06.657" Score="3" Body="&lt;p&gt;Brown dwarfs are heavier than Jupiter (jupiter has mass 0.000949 solar masses), although about the same size. Their compositions are fairly similar, with brown dwarfs &lt;a href=&quot;https://arxiv.org/abs/1008.5150&quot; rel=&quot;nofollow noreferrer&quot;&gt;depleted in deuterium if they are heavy enough&lt;/a&gt; (it is burned over a few million years after formation). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/astro-ph/9701131&quot; rel=&quot;nofollow noreferrer&quot;&gt;Adams &amp;amp; Laughlin&lt;/a&gt; modelled the collisions of brown dwarfs, mainly to estimate whether two dwarfs could merge to make a star heavier than 0.075 solar masses that would begin hydrogen fusion properly. They found that this could happen, and not too much mass appears lost as they splatter into each other if it is a somewhat glancing collision. Obviously, a near glancing hit might just scatter some of their atmospheres but not stop them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter would likely behave the same. The result would be a heavier brown dwarf. It would be heated significantly by the energy of the merger, which in turn would make it expand to a larger size until it cooled down. Jupiter would also contribute deuterium so if the dwarf was heavy enough it would further heat up for a while. Given Jupiter's minimal mass compared to the 0.075 limit it is not likely it would be enough to turn the new object into a red dwarf.&lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2018-02-06T20:00:06.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24940" PostTypeId="1" AcceptedAnswerId="24944" CreationDate="2018-02-06T23:39:40.927" Score="4" ViewCount="257" Body="&lt;p&gt;Sunspots rise vertically? Do they float on the surface like oil? Do they flow up or down vertically from the Sun's core like a lava lamp? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XPfPy.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XPfPy.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JYp2D.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JYp2D.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-02-07T20:07:17.110" LastActivityDate="2018-02-07T20:07:17.110" Title="How do sunspots move seen from the side?" Tags="&lt;star&gt;&lt;the-sun&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24941" PostTypeId="1" AcceptedAnswerId="24949" CreationDate="2018-02-07T01:38:24.443" Score="2" ViewCount="227" Body="&lt;p&gt;What hold the electrons in orbit? Is it gravity or the atomic attraction between the protons and neutrons? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked this question on Physics SE I hope I improved on the quality enough to do better here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2dLA5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2dLA5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oSHIG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oSHIG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the light we see from a neutron star from the neutrons or the electrons? &lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="2365" LastEditDate="2018-07-27T22:32:53.297" LastActivityDate="2018-07-27T22:32:53.297" Title="What is the light that we see from a neutron star come from?" Tags="&lt;gravity&gt;&lt;light&gt;&lt;neutron-star&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24942" PostTypeId="1" AcceptedAnswerId="24948" CreationDate="2018-02-07T05:56:18.820" Score="4" ViewCount="1069" Body="&lt;p&gt;How would the length of day and night change with changes in latitude on a planet with no axial tilt?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the poles of the planet be in constant sunlight?&lt;/p&gt;&#xA;" OwnerUserId="21555" LastActivityDate="2018-02-07T10:05:14.977" Title="How would latitude affect day/night lengths on a planet with no axial tilt?" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24944" PostTypeId="2" ParentId="24940" CreationDate="2018-02-07T06:28:21.167" Score="6" Body="&lt;p&gt;The sunspots are actually depressions and low thermal pressure zones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incredible magnetic fields are piercing the surface of the sun and the magnetic force pacifies the movement and the convection at the zone of the sunspot. that's why the surface becomes darker clouds near those magnetic fields. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elsewhere, the sun's surface has convection wells called solar granules a bit like a lava lamp, which are about 1000km diameter. You can check images of &quot;&lt;a href=&quot;https://www.google.fr/search?biw=776&amp;amp;bih=434&amp;amp;tbm=isch&amp;amp;sa=1&amp;amp;ei=iuZ6WomsHoS4kwXp1LmICA&amp;amp;q=solar%20convection&amp;amp;oq=solar%20convection&amp;amp;gs_l=psy-ab.3...0.0.0.6217.0.0.0.0.0.0.0.0..0.0....0...1c..64.psy-ab..0.0.0....0.ESxVznN7Qog&quot; rel=&quot;noreferrer&quot;&gt;solar convection&lt;/a&gt;&quot; to study maps of convection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pCV49.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pCV49.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a sunspot the movement of colder plasma flows downwards, and there is a hot spot trapped under the sunpot where the thermal energy escapes radially away from the sunspot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VR8nS.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VR8nS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a video attempting to illustrate the view under the sun:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=zU9O8SmCHFQ&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=zU9O8SmCHFQ&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a virtual simulation of them. The versions are wildly different: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=W5dldC5dPMk&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=W5dldC5dPMk&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=UN_9HMVXjY0&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=UN_9HMVXjY0&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=ARIJx0geaOE&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=ARIJx0geaOE&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here is a speeded up video of some sunspots:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/BejRxZAa66E&quot; rel=&quot;noreferrer&quot;&gt;https://youtu.be/BejRxZAa66E&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/l1VpMRMg4P4&quot; rel=&quot;noreferrer&quot;&gt;https://youtu.be/l1VpMRMg4P4&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/T5MnZUINzjY&quot; rel=&quot;noreferrer&quot;&gt;https://youtu.be/T5MnZUINzjY&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2018-02-07T12:38:08.020" LastActivityDate="2018-02-07T12:38:08.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24946" PostTypeId="2" ParentId="24902" CreationDate="2018-02-07T06:45:48.750" Score="0" Body="&lt;p&gt;It's a little bit like naming people off the shadows they cast walking past a light-source.   It's hard to tell which shadow belongs to what person.   A guy who walks by twice might get two names, two people who walk by at the same time might get one name between them.    The lack of specific identification is the difference between direct observation and a survey.   It's not even all that easy to tell if a person is tall or just closer to the light source - though some rough estimates can be made, specifics are difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't even name the planets in our own galaxy that are observed in microlensing events.    We could name the events, I suppose (and maybe they do).   But there's no way to identify and track individual planets from microlensing events.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-02-07T06:45:48.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24948" PostTypeId="2" ParentId="24942" CreationDate="2018-02-07T09:38:48.437" Score="5" Body="&lt;p&gt;The length of the day would be the same on (almost) every latitude, and just a bit longer than the night. The reason for that is that the planet's star (let's call it 'Sun') is a disk, not a point. So even if the middle of the Sun is below the horizon, part of it might still be above the horizon and it's technically 'day'. For the same reason, the length of an Earth day at &lt;a href=&quot;https://en.wikipedia.org/wiki/Equinox&quot; rel=&quot;noreferrer&quot;&gt;Equinox&lt;/a&gt; is slightly more than 12 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the poles, and a small region around them (depending on the apparent size of the Sun), it will always be day. At the North Pole of the planet, you will always see exactly the northern half of the Sun; the southern half will always be under the horizon. It will still take the Sun exactly the same amount of time to go 'around' the sky 360 degrees as it will on the equator or any other place on the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Equinox#Day_arcs_of_the_Sun&quot; rel=&quot;noreferrer&quot;&gt;diagrams in the Wikipedia article&lt;/a&gt; are a good visualization how &lt;em&gt;every&lt;/em&gt; day on such a planet would look like (click to enlarge):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZbdF6.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZbdF6m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/uBqsa.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uBqsam.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/qJovK.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qJovKm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7472" LastEditorUserId="7472" LastEditDate="2018-02-07T10:05:14.977" LastActivityDate="2018-02-07T10:05:14.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24949" PostTypeId="2" ParentId="24941" CreationDate="2018-02-07T09:45:34.973" Score="6" Body="&lt;p&gt;The visible &quot;surface&quot; of the neutron star contains no free neutrons. It consists of completely ionised atomic nuclei and free electrons. The exact composition is uncertain - either at or just below the surface will consist of iron-peak elements, becoming increasingly neutron-rich with depth, however the outer few cm (which is where any electromagnetic radiation arises) could consist of ionised hydrogen and helium that has been accreted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutron stars also have magnetospheres and these are also a source of radiation. Charged particles are accelerated away from the neutron star magnetic poles, spiral in the magnetic fields and emit (mainly the electrons, which undergo more extreme acceleration) synchrotron and curvature radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The high density &quot;neutron fluid&quot;, containing mostly free neutrons, with a small proportion of protons and electrons, is not reached until about 1-2 km below the neutron star surface.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-02-07T09:45:34.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24952" PostTypeId="2" ParentId="24936" CreationDate="2018-02-07T13:50:21.197" Score="1" Body="&lt;p&gt;The USNO page refers to &lt;a href=&quot;http://aa.usno.navy.mil/publications/docs/Circular_163.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;USNO Circular 163&lt;/a&gt; which in turn refers to a paper by &lt;a href=&quot;http://cdsads.u-strasbg.fr/cgi-bin/nph-data_query?bibcode=1982A%26A...105..359A&amp;amp;db_key=AST&amp;amp;link_type=ARTICLE&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aoki et al.&lt;/a&gt; and &lt;a href=&quot;http://cdsads.u-strasbg.fr/cgi-bin/nph-data_query?bibcode=1981CeMec..23...33M&amp;amp;db_key=AST&amp;amp;link_type=ARTICLE&quot; rel=&quot;nofollow noreferrer&quot;&gt;one by Moyer&lt;/a&gt;. These papers explain how the expressions created by Simon Newcomb at the end of the 19th century were updated to work with the new system of constants introduced in 1984. If you want to change the epoch, you could redo these calculations with the epoch of your choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further changes to celestial mechanics calculations have occurred since then, but apparently the USNO has decided that for the accuracy stated in the &lt;a href=&quot;http://aa.usno.navy.mil/faq/docs/GAST.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;USNO web page&lt;/a&gt;, 0.1 second, it is not necessary to overhaul the calculation to account for all the changes since 1984.&lt;/p&gt;&#xA;" OwnerUserId="7385" LastActivityDate="2018-02-07T13:50:21.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24953" PostTypeId="2" ParentId="24858" CreationDate="2018-02-07T14:25:12.403" Score="1" Body="&lt;p&gt;Yes, every object we see is how it was when the light reflected off it's surface or left it's surface(in case of self-luminous objects).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even the sun we see is 8 light minutes old. Proxima Centauri the nearest star to our Solar System appears as it was 4 years prior, as it is 4 light years away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Andromeda Galaxy appears to us as it was 2.5 million years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone is present in the Andromeda and looking at Earth, then they would see how Earth was 2.5 million years ago. Theoretically they could see ancient humans roam the Earth.&lt;/p&gt;&#xA;" OwnerUserId="21569" LastEditorUserId="21569" LastEditDate="2018-05-02T11:30:13.927" LastActivityDate="2018-05-02T11:30:13.927" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24955" PostTypeId="1" CreationDate="2018-02-07T18:25:46.560" Score="1" ViewCount="376" Body="&lt;p&gt;Yesterday, I was just star gazing and I noticed some star-like object moving with a high speed somewhere near the Orion's Belt and exhibiting all the seven colors i.e VIBGYOR and after a few moments, it disappeared. Any plausible explanations for this observation?  &lt;/p&gt;&#xA;" OwnerUserId="21572" LastEditorUserId="21572" LastEditDate="2018-02-09T17:44:45.613" LastActivityDate="2018-02-09T22:58:31.387" Title="Unidentified color changing object in sky" Tags="&lt;astrophysics&gt;&lt;identify-this-object&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2018-02-10T20:29:25.437" ContentLicense="CC BY-SA 3.0" />
  <row Id="24956" PostTypeId="1" CreationDate="2018-02-07T19:17:00.180" Score="1" ViewCount="52" Body="&lt;h2&gt;Topic&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I've read several articles online that state that &lt;strong&gt;our galaxy will collide with a neighboring galaxy in the future&lt;/strong&gt;, but there is &lt;strong&gt;no evidence&lt;/strong&gt; about it happening. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Question(s)&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;After reading a few articles on the topic, I started to wonder:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;When will it happen?&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Which galaxy will collide with our own?&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;&lt;strong&gt;Is it even going to happen?&lt;/strong&gt;&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(your post should answer all three)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a scary but oddly interesting theory, and a little enlightenment would be much appreciated! :)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Not a duplicate--&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I'm asking if the possibility that &lt;em&gt;&lt;strong&gt;any&lt;/em&gt;&lt;/strong&gt; galaxy will collide with our own, not just one.&lt;/p&gt;&#xA;" OwnerUserId="21239" LastEditorUserId="21239" LastEditDate="2018-02-07T23:43:47.593" LastActivityDate="2018-02-07T23:43:47.593" Title="Is the Milky Way Galaxy predicted to collide with a neighboring galaxy?" Tags="&lt;galactic-dynamics&gt;" AnswerCount="0" CommentCount="5" ClosedDate="2018-02-08T04:00:12.730" ContentLicense="CC BY-SA 3.0" />
  <row Id="24957" PostTypeId="1" CreationDate="2018-02-07T20:26:41.067" Score="3" ViewCount="145" Body="&lt;p&gt;I know it sounds crazy, but is it possible to &#xA;connect a reflector telescope and a refractor telescopes (like monocular for bird watching) together to build a better telescope?&lt;/p&gt;&#xA;" OwnerUserId="21573" LastEditorUserId="37191" LastEditDate="2021-05-19T15:18:37.077" LastActivityDate="2021-05-19T23:00:21.350" Title="It is possible to combine a reflector and a refractor telescope together?" Tags="&lt;observational-astronomy&gt;&lt;telescope&gt;&lt;refractor-telescope&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24958" PostTypeId="2" ParentId="24566" CreationDate="2018-02-08T01:14:56.030" Score="1" Body="&lt;p&gt;Your question doesn't work because it's based on a false premise - that you can have a stable 2nd moon in orbit. You can't. If you accept the fact that the 2nd moon won't be stable, the size it becomes dangerous is the size that you wouldn't want to fall to Earth - if it's garbage and loosely held together, 100-200 feet in diameter would probably be safe enough and make a big flash and a lot of noise if it hits the Earth, but not cause too much damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space garbage disposal is pretty easy - because there's so much empty space. In densely packed regions like low Earth orbit with lots of satellites, you'd want to avoid dumping garbage, but in the vastness of the Earth's Hill sphere, letting garbage just float in empty space wouldn't be all that dangerous, unless there was a lot of it. If there's enough manufacturing that you can't let garbage float, or if there's toxic minerals like lead or plutonium that you don't want to let fall to Earth, then creating a garbage dump in space makes sense, but it'd be way smarter to just use the Moon as a garbage dump than to build a 2nd one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Earth-Moon system could hold an additional moon in a stable orbit, then your question on how massive could be answered, but there's no practical reason and no benefit towards creating a garbage moon, it would have an unstable orbit and require periodic re-adjusting or it would be dangerous. Just make a garbage dump on the moon.  It's way smarter.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Would the L4 and L5 Lagrange points be unstable?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing a bit more research, it seems that Earth-Moon L4 and L5 are stable. NASA did some research into it. Perhaps that makes my answer not correct to this question.  An L5 garbage moon would be possible. - See &lt;a href=&quot;https://space.stackexchange.com/questions/8658/which-lagrange-earth-moon-point-is-better-for-a-space-settlement&quot;&gt;@Andy256 answer here.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted in the comments, a moon-sized object is a lot of material. The Moon is 20 times the mass of the asteroid belt. The number of comets and asteroids that would need to be captured, mined and discarded into the L5 orbit to create a 2nd moon is huge. We're not likely to create anything but a tiny moon out of captured objects because the Moon is comparatively enormous. I rather like the idea of using L5 as a place to capture, place and mine asteroids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's still a size limit though. If it grows to a few times bigger than the Moon, it's no longer the Moon's trojan, the Moon becomes its trojan and there's a mass limit for a stable trojan orbit, perhaps about 2 lunar masses, after which, the Moon's orbit would become unstable - and could be a world-ending scenario.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(old update below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;That's a tough call. The Earth-Moon system is far from ideal for stable L4/L5 Lagrange points because the Moon's orbit wobbles due to solar tugging. The mass ratio of 81 to 1 is large enough, with about 26 to 1 being the minimum required for stable L4/L5 saddle points, but the eccentricity and &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_precession#Apsidal_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;lunar apsidal precession&lt;/a&gt; could be problematic for long term stability.&lt;br&gt;&#xA;I'm not smart enough to say for certain whether they would be stable or not.  Perhaps another question, or it also fits here if someone else wants to answer.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="14094" LastEditDate="2018-05-05T12:14:28.723" LastActivityDate="2018-05-05T12:14:28.723" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24971" PostTypeId="2" ParentId="2137" CreationDate="2018-02-08T20:10:49.777" Score="1" Body="&lt;p&gt;There is another question being asked here, perhaps. We use the terms Extraterrestrial for that which is not from Earth, and Extrasolar for planets that do not orbit our star, Sol.  The term Universe is humanity's current term for everything as we understand or know it. It can be argued that it encompasses everything, even if it is not knowable to us. This is all in terms of human use of language to describe phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term 'Extrauniversal' could be used for something that is outside of 'Universe.' Of course, we, as humans, in our current experience have defined the Universe as 'everything,' in simple terms, but there is validity to a term that would incorporate something outside of everything.&lt;/p&gt;&#xA;" OwnerUserId="21596" LastActivityDate="2018-02-08T20:10:49.777" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24972" PostTypeId="2" ParentId="2193" CreationDate="2018-02-08T21:12:55.957" Score="-1" Body="&lt;p&gt;I have had all three &quot;styles&quot; of scopes. The &quot;Best&quot; scope is the one you get out and go outside with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The 4.5&quot; (114mm) reflector on a very cheap equatorial mount was my first scope. Not very good at all. It worked and I did see brightest planets, stars, Andromeda, Pleiades, and of course the Moon. No it wasn't a great scope, but I used and abused it. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The next was a cheap 90mm refractor on a cheap equatorial mount. Same viewing but I think just a tad better as far as clarity, but it had the exact same mount, and just as unsteady. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Next was the 8&quot; Dob, which I love, and still have, and the last scope I'll probably ever buy is my Stellarvue Access SV102. The mount I purchased separate, which is the Vixen Optics Advanced Polaris-M Motorized EQ Mount, with Tripod. I sold the 4.5&quot; (114) Reflector. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My experience has been that I stayed around long enough, went to enough star parties and evens, checked out all the astronomers in the area, went to McDonalds Observatory on Star Party Nights, and eventually got around to putting together a scope package that I use every night....well almost. There are pro's and con's to all scopes, but if I was a beginner again, I'd take the Dobsonian hands down for a starter scope. I still use it frequently. Just my opinion...&lt;/p&gt;&#xA;" OwnerUserId="21597" LastEditorUserId="88" LastEditDate="2018-02-09T10:18:17.310" LastActivityDate="2018-02-09T10:18:17.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24973" PostTypeId="2" ParentId="24621" CreationDate="2018-02-09T00:32:15.707" Score="1" Body="&lt;p&gt;I didn't really want to answer this one because there may be some new information that I'm not aware of and, to my understanding, the shape, mass and content of the Oort cloud is a subject of some ongoing study, so I invite correction.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formation of a solar-system is pretty complicated and there may still be significant unknowns on the process and with the Oort cloud, uncertainty on the precise shape, mass, density distribution and origin, whether it formed with the solar-system or whether much of it is captured.   I wanted to begin with the unknowns.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can't see the Oort cloud so we can't measure it directly.   Estimates can be made by observing very long period comets that fly into the inner solar system and by extrapolating their orbital periods, an estimate of the contents of the Oort cloud can be made.    One problem is, that estimate is based on very eccentric orbits only, because those are the only ones we see that enter our telescope's range.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one of my favorite &lt;a href=&quot;https://www.youtube.com/watch?v=tmNXKqeUtJM&quot; rel=&quot;nofollow noreferrer&quot;&gt;minute physics videos&lt;/a&gt; for it's simplicity and the fun diagrams they use.  The gist of the simple answer is that a cloud or nebulous mass of debris and dust, has a fixed angular momentum and a fixed orbital plane, so as the cloud of debris gets a push (usually from a not too far away supernova), and the debris begins to condense into a proto-solar-system, it maintains the orbital plane and angular momentum and as it condenses, it rotates very rapidly.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rapid rotation isn't all that relevant to your question, but that's why not all the matter can fall into the star, because there's usually too much angular momentum.  The same happens with gas giant planets, which is why Jupiter, for example, has 4 large formation moons (where as the Earth - somewhat smaller region where it formed, no formation moons, but it has an impact moon).   I'm making this more wordy than it should be, but the point is, stuff orbits the sun because of angular momentum.    The sum of this angular momentum has an orbital plane, and each individual object has an inclination to that orbital plane.    When planets absorb the matter in their orbital regions, the up and down directions, or inclinations of the objects in orbit tends to cancel out, but the net angular momentum and orbital plane remain constant.   (This is mostly, but not entirely true - for example, Planet 9, if it's eventually discovered, might explain why the 8 inner planets, on average don't line up with the sun's rotational plane.  Planet 9 may have taken some of the inclination in one direction with it as it was thrown far outside the solar-system leaving the inner 8 planets with an inclination in the other direction relative to the sun.   When (if) planet 9 is discovered, then we can check if it balances out our solar-systems tilted inclination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's the collisions that help the planets line up along the orbital plane of the solar system, because the ups and downs mostly cancel out.   If there are no collisions, then there's no cancelling out of the ups and downs and the objects in orbit remain in a kind of nebulous blob assortment - which over time, is probably best represented as a sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's not the entire answer, however. Take the asteroid belt for example.   There was (I think) probably not sufficient collisions to make the asteroid belt flat, and Ceres is probably not a failed planet, because it probably came later.  Based on Ceres' density it probably came from further out in the solar-system.  It may have started out as a moon (perhaps kicked out of Neptune's orbit by the miss-behaving Triton) or a dwarf-planet originally in the kuiperbelt.  It's density is too low to have formed by collisions in the asteroid belt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the Asteroid belt is flat (er mostly) likely due to gravitational shepherding from Jupiter and perhaps, by a very strong magnetic field from our young sun, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proper answer of our estimated shape of the oort cloud would be based on a survey of all the long-period (or highly eccentric) comets that we've observed and I'm not that interested in doing the research on that, but I would guess that there's enough variety of orbital inclination to support the generally circular shape - because I don't think the circular shape would be used so often if it didn't reflect the observation of long period comets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Kuiper belt, for example, (and I couldn't find a really specific answer to this), but it appears to be somewhat flat, kind of donut or torus shaped (Pluto has a higher than usual inclination).   The proper term, if you want to get technical, is inclination distribution.  No distribution = flat.  Full or high distribution = sphere.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Kuiper belt's relative/somewhat flatness may be driven by mostly Neptune sheparding rather than collisions (again I'm not precisely sure).   In fact it was the regularity of some inclinations of the most eccentric kuiper belt objects (just passing through in other words) that lead to the planet-9 theory in the first place.  If there's large planets around, they assist in the shepherding and flattening of smaller objects in their orbital viscinity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the Galaxy, to my knowledge, there's insufficient collisions to explain the flatness of the Milky-way (it's basically pizza shaped, with maybe a small ping pong ball or large marble in the middle).   It's my understanding that the Galaxy was flattened by a very strong magnetic field more than by collision (somebody correct me if I'm wrong).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the extent of my knowledge at least.   I invite someone smarter than me to answer this as well.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-02-09T01:13:21.203" LastActivityDate="2018-02-09T01:13:21.203" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24977" PostTypeId="1" CreationDate="2018-02-09T09:10:13.710" Score="4" ViewCount="323" Body="&lt;p&gt;Can the solar wind affect the orbital elements of a nearby asteroid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know if solar activity has an effect on the asteroid's orbit or gravitational stability and or can cause a gravitational disturbance to a meteoroid near the earth. &lt;/p&gt;&#xA;" OwnerUserId="21609" LastEditorUserId="2365" LastEditDate="2018-02-09T17:01:33.343" LastActivityDate="2018-02-10T15:05:23.493" Title="Solar Wind and Asteroid orbital behavior" Tags="&lt;asteroids&gt;&lt;solar-wind&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24981" PostTypeId="1" AcceptedAnswerId="24991" CreationDate="2018-02-09T14:42:53.817" Score="2" ViewCount="164" Body="&lt;p&gt;Space Exploration Stack Exchange were not very fond of this question and closed it, so I think it might fit better here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary: Assuming no foreknowledge of its existence, can Elon Musk's Tesla Roadster be detected and categorised from Earth?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Russell%27s_teapot&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bertrand Russell&lt;/a&gt; argued the following, an argument that is now known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Russell%27s_teapot&quot; rel=&quot;nofollow noreferrer&quot;&gt;Russell's Teapot&lt;/a&gt; (emphasis mine): &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Many orthodox people speak as though it were the business of sceptics to disprove received dogmas rather than of dogmatists to prove them. This is, of course, a mistake. If I were to suggest that &lt;strong&gt;between the Earth and Mars there is a china teapot revolving about the sun in an elliptical orbit&lt;/strong&gt;, nobody would be able to disprove my assertion provided I were careful to add that the teapot is too small to be revealed even by our most powerful telescopes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well now we have a &lt;a href=&quot;https://en.wikipedia.org/wiki/Elon_Musk%27s_Tesla_Roadster&quot; rel=&quot;nofollow noreferrer&quot;&gt;midnight cherry red automobile&lt;/a&gt;, between Earth and Mars, in an elliptical orbit around the Sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4b5r5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4b5r5.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume that this civilisation we have falls. The knowledge of this automobile is lost. Then in some centuries/millennia later humanity has managed to work its way back to a level of technology comparable to what we have today. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Could this automobile then be detected by &quot;our most powerful telescopes&quot;?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess that we could detect it as an orbiting object; it is large enough to be found. Can you confirm or refute this guess? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if the automobile can be found as an orbiting object, &lt;strong&gt;could it then be determined that this is not an ordinary space rock, but that there is something very odd about it?&lt;/strong&gt; That is to say: is it &lt;strong&gt;possible&lt;/strong&gt; to do &lt;strong&gt;any&lt;/strong&gt; kind of characterisation of the automobile that makes it stand out as an unusual space object. There is some uncertainty about the exact state of the automobile after such a long time, but there is no question that chemically and optically, this will be no space rock, even in several thousand years. The question is if we can detect that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Digging some more I found about the (re)discovery of &lt;a href=&quot;https://en.wikipedia.org/wiki/J002E3&quot; rel=&quot;nofollow noreferrer&quot;&gt;J002E3&lt;/a&gt;, which turned out to (probably) be the Apollo 12 Saturn-IVB upper stage. I quote from &lt;a href=&quot;http://adsabs.harvard.edu/abs/2003DPS....35.3602J&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper (emphasis mine)&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Additional &lt;strong&gt;spectral observations&lt;/strong&gt; were completed in May 2003 at the Air Force Maui Optical Supercomputing (AMOS) site. Through the modeling of common spacecraft materials, the observations of &lt;strong&gt;J002E3 show a strong correlation of absorption features to a combination of human-made materials including white paint, black paint, and aluminum&lt;/strong&gt;. Absorption features in the near IR show a strong correlation with paint containing a titanium-oxide semiconductor. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, in short: &lt;strong&gt;could&lt;/strong&gt; we do the same with this automobile and find that this is not a plain old space rock? How close would it have to be for this? &lt;/p&gt;&#xA;" OwnerUserId="11474" LastEditorUserId="11474" LastEditDate="2018-02-09T15:10:02.730" LastActivityDate="2018-02-10T08:35:08.790" Title="Musk's Roadster is now Russell's Teapot?" Tags="&lt;artificial-satellite&gt;&lt;classification&gt;&lt;detectors&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="24983" PostTypeId="2" ParentId="24921" CreationDate="2018-02-09T17:32:40.603" Score="2" Body="&lt;p&gt;10-30 Earth masses is an estimate from &lt;a href=&quot;http://adsabs.harvard.edu/abs/1982P&amp;amp;SS...30..755S&quot; rel=&quot;nofollow noreferrer&quot;&gt;this abstract&lt;/a&gt;, but &lt;a href=&quot;https://en.wikipedia.org/wiki/Jupiter#Internal_structure&quot; rel=&quot;nofollow noreferrer&quot;&gt;12-45&lt;/a&gt; is a more accurate answer.&lt;/p&gt;&#xA;" OwnerUserId="21521" LastActivityDate="2018-02-09T17:32:40.603" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24984" PostTypeId="2" ParentId="24977" CreationDate="2018-02-09T20:28:07.793" Score="3" Body="&lt;h2&gt;Sort of, via &lt;a href=&quot;https://en.wikipedia.org/wiki/Radiation_pressure#Solar_radiation_pressure&quot; rel=&quot;nofollow noreferrer&quot;&gt;radiation pressure&lt;/a&gt; and heating.&lt;/h2&gt;&#xA;&lt;p&gt;The Sun emits photons, which carry with them energy and momentum. Any flux of photons applies pressure to an object it hits; this is the basic principle behind &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_sail&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar sails&lt;/a&gt;. In general, the larger the object, the more force it feels, since pressure is force per unit area. There are two main ways light from the Sun can significantly affect minor planets, such as asteroids:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Yarkovsky_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Yarkovsky effect&lt;/a&gt;:&lt;/strong&gt; For a rotating body, there is a delay between when a portion of it receives heat and when the heat is re-radiated. The spinning object then has an offset between the angles of absorption and emission, which means that the radiation applies a net force. This can cause slow but steady changes; over 12 years, the asteroid &lt;a href=&quot;https://en.wikipedia.org/wiki/6489_Golevka&quot; rel=&quot;nofollow noreferrer&quot;&gt;6489 Golevka&lt;/a&gt; deviated from its predicted orbital path for 15 kilometers.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Yarkovsky%E2%80%93O%27Keefe%E2%80%93Radzievskii%E2%80%93Paddack_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Yarkovsky–O'Keefe–Radzievskii–Paddack effect&lt;/a&gt;:&lt;/strong&gt; The YORP effect occurs when irregularities in the surface of an object causes incoming light to be scattered in different directions. This changes the rotation direction and rate of the body. This has been observed in asteroids &lt;a href=&quot;https://en.wikipedia.org/wiki/54509_YORP&quot; rel=&quot;nofollow noreferrer&quot;&gt;54509 YORP&lt;/a&gt; (a rotational change of 250 degrees over four years) and &lt;a href=&quot;https://en.wikipedia.org/wiki/1862_Apollo&quot; rel=&quot;nofollow noreferrer&quot;&gt;1862 Apollo&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-02-09T20:28:07.793" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24986" PostTypeId="2" ParentId="24955" CreationDate="2018-02-09T22:58:31.387" Score="-1" Body="&lt;h1&gt;Probably a satellite flare&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;It sounds like a &lt;a href=&quot;https://en.wikipedia.org/wiki/Satellite_flare&quot; rel=&quot;nofollow noreferrer&quot;&gt;satellite flare&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coming, showing up clearly for a few moments, moving, then going away, are characteristic of a satellite flare, like Iridium flashes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check &lt;a href=&quot;http://www.heavens-above.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heavens Above&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_satellite_pass_predictors&quot; rel=&quot;nofollow noreferrer&quot;&gt;any other similar site&lt;/a&gt; to see if any predicted flares match with your location and time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The changing colours can be explained with &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Twinkling&quot; rel=&quot;nofollow noreferrer&quot;&gt;Twinkling&lt;/a&gt;&quot;. &lt;/p&gt;&#xA;" OwnerUserId="11474" LastActivityDate="2018-02-09T22:58:31.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24991" PostTypeId="2" ParentId="24981" CreationDate="2018-02-10T08:35:08.790" Score="3" Body="&lt;p&gt;Small objects in space are hard to detect and characterise. Compare the tesla with another small solar system body: &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi?find_body=1&amp;amp;body_group=sb&amp;amp;sstr=3729281&quot; rel=&quot;nofollow noreferrer&quot;&gt;2015TC25&lt;/a&gt;, It is currently about 0.3AU (3/10 of the distance from Earth to the sun) and it has a magnitude of about 30. That is perhaps detectable if you know exactly where to look. It would not be &quot;discoverable&quot; at this distance, with current technology. We know about because it made a close pass of Earth in 2015, and it was discovered during that pass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Tesla could be discovered if its orbit takes it close to the Earth. Observations of its spectrum would indicate it is not a regular asteroid, The cherry red paint would mark it as being utterly different from natural asteroids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How close depends on what technology you have. By comparision with 2015 TC25, it might be observable from 0.3AU, if you have enough time and money to spend.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-02-10T08:35:08.790" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24992" PostTypeId="1" CreationDate="2018-02-10T09:54:42.933" Score="1" ViewCount="228" Body="&lt;p&gt;I'm trying to understand orbital inclination by looking at the planetary orbit inclination table on &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_inclination&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Orbital_inclination&lt;/a&gt; (there are orbital inclinations for terrestrial and gas giant planets listed there).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By inclinations to ecliptic or Sun's equator, it looks like the Earth is the most (or least) inclined from all the listed objects having the smallest (or biggest) inclination value. However, the inclination to invariable plane seems to be bigger for some planets than for the Earth, and smaller for others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first, I thought that displaying the absolute value (dropping the minus sign) in case of a negative angle is the case. But if those three columns were just three different reference planes, the relative difference between planet inclinations should at least be the same in all cases. But it's not the case as, for example, Mercury's and Venus' orbital planes have 3.62° between themselves as per the inclination to ecliptic, but they have 0.48° as per the inclination to the Sun's equator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how should one read those numbers?&lt;/p&gt;&#xA;" OwnerUserId="21622" LastActivityDate="2018-02-10T11:28:48.547" Title="Is Earth the most inclined or not?" Tags="&lt;inclination&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24995" PostTypeId="2" ParentId="24992" CreationDate="2018-02-10T11:28:48.547" Score="1" Body="&lt;p&gt;The confusing thing is that inclination is not the only thing defining an orbital plane; it also matters where the ascending node is. &lt;em&gt;Only&lt;/em&gt; if all ascending nodes would be in the same direction, you could do the trick with the absolute values you mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't draw a picture right now, but imagine you have a table before you now, and use it as a reference plane (the Sun's equator). Put a sheet of paper on the table, and lift the right end a bit so that it inclines 7.155° w.r.t. the table. This is Earth's orbit, and it looks a bit like the picture in the Wikipedia article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take another sheet of paper, and lift the far end a bit so that it inclines 7.155° w.r.t. the table. This is supposed to be the orbital plane of a hypothetical planet with the same inclination to the Sun's equator, but it is &lt;em&gt;also&lt;/em&gt; inclined to the ecliptic (i.e. Earth's orbital plane).&lt;/p&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2018-02-10T11:28:48.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24997" PostTypeId="2" ParentId="2137" CreationDate="2018-02-10T11:48:37.130" Score="1" Body="&lt;p&gt;Universe according to Wikipedia is :-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Universe is all of space and time (spacetime) and its contents,[12] which includes planets, moons, stars, galaxies, the contents of intergalactic space and all matter and energy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, if we go by this meaning Universe is the space and time that was created as a result of the Big Bang that we currently live in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the theory of Big Bang is in fact true, this means that we (i. e. the Universe) is inside something and it is finite and if there is one Universe there can be many and all this is contained in something of extremely huge proportions which can be called as Multiverse. Now we don't know if this Multiverse is finite or infinite. If it is finite and this goes on and on and on each one will still be called as Multiverse atleast at this point of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, according to Wikipedia Multiverse is defined as :-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The multiverse (or meta-universe) is a hypothetical set of various possible universes including the universe which we live in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is more sensible to call everything that is created by the Big Bang as Universe because there may be other Universes which have completely different laws of Physics in which we won't even be able to exist.&lt;/p&gt;&#xA;" OwnerUserId="21569" LastActivityDate="2018-02-10T11:48:37.130" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24999" PostTypeId="2" ParentId="24977" CreationDate="2018-02-10T15:05:23.493" Score="0" Body="&lt;p&gt;I have lifted the text below from my &lt;em&gt;currently unanswered question&lt;/em&gt;, (to which I've just added a bounty) &lt;a href=&quot;https://space.stackexchange.com/q/18474/12102&quot;&gt;Did Rosetta improve on models of non-gravitational effects on comet 67P's orbit?&lt;/a&gt; in another SE site. Based on the current unclearness of the distinction between comets and asteroids as discussed in @zephyr's &lt;a href=&quot;https://astronomy.stackexchange.com/a/19973/7982&quot;&gt;colorful answer&lt;/a&gt; I'm going to leave it as is. (&lt;a href=&quot;https://astronomy.stackexchange.com/a/19958/7982&quot;&gt;Is this object an asteroid or comet, and how can it produce so many tails?&lt;/a&gt; and its answer are also worth a look.) Objects &quot;formerly known as&quot; asteroids will not outgas as much as those known as comets, but the math and background may be helpful as an additional answer here. Of course @HDE's &lt;a href=&quot;https://astronomy.stackexchange.com/a/24984/7982&quot;&gt;answer&lt;/a&gt; is nice and concise and to the point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Yarkovsky_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Yarkovsky&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Poynting%E2%80%93Robertson_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Poynting-Robertson&lt;/a&gt; effects are mentioned below as well. It's clear these are included in the modeling of at least some asteroids, and comets.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The calculation of the orbits of comets can be more difficult than those of most asteroids for a number of reasons. Some comets have such highly eccentric orbits that aphelion is too far for the comet to be observed continuously, or the period is so long that only one pass has been observed and a period can not be calculated, or it passes so close to the sun that it's orbit is highly modified. However comet &lt;a href=&quot;https://en.wikipedia.org/wiki/67P/Churyumov%E2%80%93Gerasimenko&quot; rel=&quot;nofollow noreferrer&quot;&gt;67P/Churyumov–Gerasimenko&lt;/a&gt; currently has a period of only about 6.4 years a perihelion/aphelion of 1.2 AU and 5.7 AU respectively. While it's orbit is within a so called &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_System#Inner_Solar_System&quot; rel=&quot;nofollow noreferrer&quot;&gt;frost line&lt;/a&gt;&quot; it remains further from the sun than the Earth's orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;https://space.stackexchange.com/a/18438/12102&quot;&gt;this answer&lt;/a&gt; I plot some data from a recent &lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA JPL Horizons&lt;/a&gt; ephemeris for comet 67P. The current default solution &lt;code&gt;soln ref.= JPL#K084/25, data arc: 1995-07-03 to 2016-05-30&lt;/code&gt; appears to use Marsden coefficients to model non-gravitational forces on the comet. &lt;a href=&quot;https://en.wikipedia.org/wiki/Brian_G._Marsden&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brian . Marsden&lt;/a&gt; was a British astronomer who contributed greatly to the field of cometary orbits. (See also &lt;a href=&quot;http://www.nytimes.com/1998/03/13/us/man-in-the-news-a-cheery-herald-of-fear-brian-geoffrey-marsden.html?_r=0&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.columbia.edu/cu/21stC/issue-3.2/reichhardt.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.) While exact modeling of non-gravitational forces on comets would be extremely complex, he introduced a simple empirical parameterization that provides a framework to discuss the magnitude and potential effects of these forces on the orbits of comets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the following convention: $\hat{\mathbf{e}}_R, \ \hat{\mathbf{e}}_T, \ \hat{\mathbf{e}}_N$ are unit vectors at the location of the comet in the radial, transverse, and normal directions where $\hat{\mathbf{e}}_R$ points away from the sun, $\hat{\mathbf{e}}_N$ is the direction of the angular momentum vector (perpendicular to the orbit plane) and $\hat{\mathbf{e}}_T$ is perpendicular to the first two and approximately in the direction of motion, non-gravitational accelerations &lt;a href=&quot;http://adsabs.harvard.edu/full/1983A%26A...126..307F&quot; rel=&quot;nofollow noreferrer&quot;&gt;can be parameterized&lt;/a&gt; using the empirical equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{a}_{NG} = (&#xA;A_1\hat{\mathbf{e}}_R \ +  \&#xA;A_2\hat{\mathbf{e}}_T \ +  \ &#xA;A_3\hat{\mathbf{e}}_N) \ g(r),    $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(r)= 0.111262\left(\frac{r}{2.808}\right)^{-2.15} \left(1+\left(\frac{r}{2.808}\right)^{5.093}\right)^{-4.6142}, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the acceleration coeficients $A_1,A_2,A_3$ commonly have units of $AU / day^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've reproduced these here to illustrate the basic idea. There are further considerations including a delay term and effects of rotation. However with this parameterization it is possible to discuss and at least get a handle on non-gravitational effects without a detailed physical model. These effects might the &lt;a href=&quot;https://en.wikipedia.org/wiki/Yarkovsky_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Yarkovsky&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Poynting%E2%80%93Robertson_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Poynting-Robertson&lt;/a&gt; effects, and of course recoil from material energetically ejected from the comet, especially as it approaches the sun and is heated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parameters $A_1,A_2,A_3$ in model can be used to express effects from physical models of comets, but they can also be used as fitting parameters to improve orbital solutions for comets based on observational data.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wtjMr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wtjMr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; linear and semi-log plots of $g(r)$ between 1.2 and 5.7 AU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/P1HAZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P1HAZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Example of the non-gravitational parameters used in the most recent &lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL Horizons&lt;/a&gt; ephemeris for comet 67P. The coefficients have units of $AU/day^2$. For a comparison, the gravitational acceleration at a distance of 1.2 $AU$ is about 0.0041 $m/s^2$ or about 2.1E-04 $AU/day^2$. This suggests that the non-gravitational forces used here have a parts-per-million effect per orbit which will become substantial over a large number of orbits.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-02-10T15:05:23.493" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25000" PostTypeId="1" CreationDate="2018-02-10T15:07:46.810" Score="1" ViewCount="50" Body="&lt;p&gt;The night sky from many parts of the Earth, away from any city lights and pollution, is full of starts, mostly the stars of the Milky Way Galaxy is what fills the night sky densely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many photos of outer space capturing the planets of the solar system there is always the black background, not even one single star to be found (I searched for photos of Saturn, Jupiter and even Earth, didn't find any peck of light around it which may be a star)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, why this happens? I've told myself regularly that there can be three reasons for that&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The exposure of the camera is too low to capture the stars that might luckily be in the background of a planet's photo.&lt;/li&gt;&#xA;&lt;li&gt;The stars are too spread out each other that in most of the photos there might not even be a star in the dark space behind the planet. &lt;/li&gt;&#xA;&lt;li&gt;The telescopes taking the photos aren't powerful enough to take the photos of the stars clearly (very much related to the 1st point, but instead of the exposure of the camera I believe it is the telescope that's not capable or maybe made for that)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I have tried looking for answers but didn't find much, maybe if the photo was taken from an angle from where the Milky way galaxy is just right behind the object i.e planet here, we may have seen some twinkling in the background? Or am I wrong? &lt;/p&gt;&#xA;" OwnerUserId="21629" LastActivityDate="2018-02-10T15:07:46.810" Title="Why stars are not visible in photos of Planets from space" Tags="&lt;solar-system&gt;&lt;photography&gt;&lt;intergalactic-space&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2018-02-10T15:29:28.097" ContentLicense="CC BY-SA 3.0" />
  <row Id="25001" PostTypeId="1" CreationDate="2018-02-10T21:16:07.180" Score="4" ViewCount="283" Body="&lt;p&gt;For example, when would someone with a telescope be able to see Starman and his Roadster (when will the Tesla roadster's elliptical orbit cross ours again, and how would that be calculated?) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For two circular orbits, I understand this is given by $$\frac{1}{S} = \frac{1}{T_2} - \frac{1}{T_1} $$ .&lt;/p&gt;&#xA;" OwnerUserId="18673" LastActivityDate="2018-10-31T13:13:21.457" Title="How would one calculate the synodic period of the Earth and an elliptical orbit?" Tags="&lt;orbit&gt;&lt;orbital-migration&gt;&lt;eccentric-orbit&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25002" PostTypeId="2" ParentId="25001" CreationDate="2018-02-11T03:48:28.690" Score="1" Body="&lt;p&gt;&lt;strong&gt;tl;dr:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The synodic period would be the same. For this object it will be about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\left( \frac{1}{365.25} - \frac{1}{557} \right)^{-1} \approx 1061 \ days.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can confirm that by reading this new ArXiv preprint about Roadster: &lt;a href=&quot;https://arxiv.org/abs/1802.04718&quot; rel=&quot;nofollow noreferrer&quot;&gt;The random walk of cars and their collision probabilities with planets&lt;/a&gt;. In the 2nd paragraph of Section 3 it says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The bodies reach the same orbital longitude on their synodic timescale of ∼ 2.8 yrs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's a slight difference, since the Roadster's orbit continues to be refined, and the paper is about evolution over long timescales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I think what you really want to know are &lt;em&gt;the times of closest approach&lt;/em&gt; when (at least) one orbit is significantly eccentric, and &lt;em&gt;those  will not be periodic&lt;/em&gt; or come at regular intervals.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Well the Roadster is quite far away now and moving farther rapidly. See &lt;a href=&quot;https://space.stackexchange.com/a/25121/12102&quot;&gt;this answer&lt;/a&gt; for some details and some nice images from medium-sized tracking telescopes. It's already dimmer than +22 magnitude as seen from Earth, and as far as I know there won't be any really close approaches to Earth for a long time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fQXGo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fQXGo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to answer your question, the period is about 557 days. You can see this for example if you go to JPL's &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;Horizons&lt;/a&gt; and look at the osculating elements. If you look a few weeks in the future when the Roadster is sufficiently far away from Earth not to be perturbed, the period around the Sun flattens out to 557 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TYEzA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TYEzA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CViZK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CViZK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above x2:&lt;/strong&gt; Top: Distance data from JPL's &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;Horizons&lt;/a&gt; plotted using Python &lt;a href=&quot;https://space.stackexchange.com/a/25079/12102&quot;&gt;from here&lt;/a&gt;. Bottom: Estimated apparent magnitude using distances from Horizons plus math shown in the text of &lt;a href=&quot;https://space.stackexchange.com/a/25121/12102&quot;&gt;this answer&lt;/a&gt;. Your milage may vary, but probably within +/-2 magnitudes of &quot;official&quot; predictions when they come out.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A really cool GIF can be seen in the Space.com article &lt;a href=&quot;https://www.space.com/39647-spacex-tesla-roadster-spotted-in-space.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Observatory Spots Elon Musk's Tesla Roadster Zooming Through Space (Video)&lt;/a&gt;. It is over 7 MB so I can't add it here. However, here are ten frames from somewhere in the middle. Still, you should go see the whole thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GIF:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vyp79.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vyp79.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;DEMIOS image below, from &lt;a href=&quot;http://www.elecnor-deimos.com/deimos-sky-survey-spacex-starman-tesla/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; as also &lt;a href=&quot;https://twitter.com/planet4589/status/961700958030966784&quot; rel=&quot;nofollow noreferrer&quot;&gt;tweeted&lt;/a&gt; by Jonathan McDowell. The little dot near the center moving to the right and up is Roadster in reflected sunlight, probably mostly from the white FH 2nd stage still attached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GIF:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fjvSZ.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fjvSZ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-10-31T13:13:21.457" LastActivityDate="2018-10-31T13:13:21.457" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25007" PostTypeId="1" AcceptedAnswerId="26672" CreationDate="2018-02-11T14:00:59.587" Score="4" ViewCount="181" Body="&lt;p&gt;There are star survey telescopes in space, and on Earth. I don't really know how to begin to know where they will all be looking for the next few months. On the off-chance that someone does, or can suggest how to find out in a systematic way, I thought I would ask here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These is JPL's &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi#top&quot; rel=&quot;noreferrer&quot;&gt;Horizons&lt;/a&gt; predictions for the Roadster spacecraft in its heliocentric orbit. They are based on both the last state vector readings from the spacecraft itself plus several optical sightings my medium sized telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the car is fairly small, it's still attached to the white, 2nd stage rocket body, so it continues to reflect a bit of light. You can read more and see some astronomical photos in &lt;a href=&quot;https://astronomy.stackexchange.com/a/25002/7982&quot;&gt;this answer&lt;/a&gt; and read &lt;a href=&quot;https://space.stackexchange.com/a/25079/12102&quot;&gt;this answer&lt;/a&gt; for info.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UXmpC.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UXmpC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-06-18T08:16:21.690" Title="Might the SpaceX Roadster intercept any ongoing surveys?" Tags="&lt;observational-astronomy&gt;&lt;deep-sky-observing&gt;&lt;sky-survey&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25010" PostTypeId="1" AcceptedAnswerId="25011" CreationDate="2018-02-11T16:17:04.320" Score="3" ViewCount="115" Body="&lt;p&gt;Is there a formula for describing a planet's position along an eccentric orbit over time? I am particularly interested in the elliptical case (0 &amp;lt; e &amp;lt; 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the planet has a higher velocity closer to the perihelion. Ideally I would like to be able to calculate a planet's position on it's orbital path in relation to how far through it's orbital period it is, if that makes sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, is there software (that doesn't require a university's budget) to work this out?&lt;/p&gt;&#xA;" OwnerUserId="21555" LastActivityDate="2018-02-11T17:01:28.340" Title="How to mathematically describe a planet's eccentric orbit?" Tags="&lt;orbit&gt;&lt;eccentric-orbit&gt;&lt;mathematics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25011" PostTypeId="2" ParentId="25010" CreationDate="2018-02-11T16:47:26.557" Score="9" Body="&lt;p&gt;The formula is Kepler's equation, but to understand it you need to know three values:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M$ is the &quot;Mean Anomaly&quot;. It increases linearly from 0 to 360 over the period of one orbit, measured from periapse to periapse. So if a planet has a period of 100 days, then the mean anomaly at day 0 is 0, at day 50 it 180degrees, at day 25 it is 90 degrees. This is only the position of the body that is in a circular orbit. It is the first stage of calculating the position of the body in an elliptical orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E$ is the &quot;Eccentric Anomaly&quot;. This is also not the actual location of the planet. It is defined by $\cos\,E = x/a$ where $x$ is the x-coordinate of the of the planet, measured &lt;em&gt;from the centre of the ellipse&lt;/em&gt; not the focus, and $a$ is the semi-major axis of the ellipse&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T$ is the &quot;True anomaly&quot;, the angle between the actual position of the planet and the line through the major axis of the ellipse, measure at the focus of the ellipse (ie at the sun). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are related by the Kepler equation. $$M = E- e\sin E,$$where $e$ is the eccentricity of the ellipse. You can calculate M directly, then solve this equation (numerically) to find E. Providing the eccentricty isn't very close to one, Newton's method is effective at finding a solution to this equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the true anomaly is given as &#xA;$$&#xA;T   = 2 \, \mathop{\mathrm{arg}}\left(\sqrt{1-e} \, \cos\frac{E}{2} , \sqrt{1+e}\sin\frac{E}{2}\right) &#xA;$$&#xA;(where $\mathrm{arg}(x,y)$ is available as &lt;code&gt;atan2(y,x)&lt;/code&gt; in many programming languages)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to use the Eccentric anomaly to calculate the coordinates of the planet directly using $\cos\,E=x/a$and $\sin\,E=y/b$ where a and b are the semi-major and -minor axes of the ellipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The word &quot;anomaly&quot; doesn't mean &quot;false result&quot;, but &quot;angular deviation (from periapse)&quot; and has been in use since Kepler's time.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-02-11T17:01:28.340" LastActivityDate="2018-02-11T17:01:28.340" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25012" PostTypeId="1" AcceptedAnswerId="25032" CreationDate="2018-02-11T17:11:07.517" Score="0" ViewCount="124" Body="&lt;p&gt;&lt;sup&gt;[I am sorry if this isn't the best place for this - it was either here or maybe on the &lt;a href=&quot;https://worldbuilding.stackexchange.com&quot;&gt;Worldbuilding SE&lt;/a&gt;.]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently re-read one of my favourite childhood sci-fi books (Chocky by John Wyndham), and in it a young boy is asked by an alien being, where in the Universe our planet is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It occurred to me that this is almost impossible to describe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays, with Wikipedia and the Internet, we can get some of the way by stating that we're in a galaxy that is approximately 10 billion years old, light takes approximately 100000 years to cross it, and which is part of a local group, itself part of a cluster, and that part of a super-cluster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately the alien might then ask a follow-up clarification question such as &lt;code&gt;what is a year&lt;/code&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we go to basics - in base 60 (as opposed to binary or base 10), our basic unit of time is called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Second&quot; rel=&quot;nofollow noreferrer&quot;&gt;second&lt;/a&gt;, and there are 60 seconds in a &lt;a href=&quot;https://en.wikipedia.org/wiki/minute&quot; rel=&quot;nofollow noreferrer&quot;&gt;minute&lt;/a&gt;, 60 minutes in an &lt;a href=&quot;https://en.wikipedia.org/wiki/hour&quot; rel=&quot;nofollow noreferrer&quot;&gt;hour&lt;/a&gt; and 24 hours in a &lt;a href=&quot;https://en.wikipedia.org/wiki/day&quot; rel=&quot;nofollow noreferrer&quot;&gt;day&lt;/a&gt; - which is the time our planet takes to rotate once completely. Our planet orbits a local star (that we call the sun) and each complete orbit takes approximately one &lt;a href=&quot;https://en.wikipedia.org/wiki/year&quot; rel=&quot;nofollow noreferrer&quot;&gt;year&lt;/a&gt;, and light takes around 8 minutes to get from our local star to our planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately our alien might then ask, &lt;em&gt;but how long is a second?&lt;/em&gt; - so we then get into atomic clocks, caesium 133 and phases of radiation, followed by the next inevitable question: what is caesium. So we have to get down to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Periodic_table&quot; rel=&quot;nofollow noreferrer&quot;&gt;Periodic Table of the Elements&lt;/a&gt; and decribe an atom, atomic weight, and how to define caesium in a way that a far off advance civilization could relate to - not to mention perhaps a tangent about Prime Numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if our alien was lucky enough to be in contact with someone like Professor Brian Cox, or a physyicist with astronomical leanings, how would such a description and conversation go?&lt;/p&gt;&#xA;" OwnerUserId="21644" LastEditorUserId="1891" LastEditDate="2018-02-11T18:09:33.307" LastActivityDate="2018-02-12T23:17:17.107" Title="Thought experiment about our location in the Universe" Tags="&lt;astrophysics&gt;&lt;distances&gt;&lt;stellar-astrophysics&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2018-02-13T07:29:43.817" ContentLicense="CC BY-SA 3.0" />
  <row Id="25013" PostTypeId="1" CreationDate="2018-02-11T17:51:54.620" Score="2" ViewCount="226" Body="&lt;p&gt;I wanted to point at the L4 and L5 of Sun-Jupiter to find Trojan asteroids but I don't know how to calculate the RA and Dec of these.&#xA;Do you know of any software that does it for you or how to calculate them?&lt;/p&gt;&#xA;" OwnerDisplayName="Mary" LastEditorUserId="2365" LastEditDate="2018-02-12T09:32:51.713" LastActivityDate="2018-02-12T20:21:29.340" Title="Does anybody know how to get Sun-Jupiter Lagrangian points L4 and L5 RA and Dec?" Tags="&lt;asteroids&gt;&lt;coordinate&gt;&lt;orbital-mechanics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25014" PostTypeId="1" CreationDate="2018-02-11T18:48:38.043" Score="2" ViewCount="519" Body="&lt;p&gt;as a follow-up to my previous &lt;a href=&quot;https://astronomy.stackexchange.com/questions/25012/thought-experiment-about-our-location-in-the-universe&quot;&gt;question&lt;/a&gt; I found out that the &lt;a href=&quot;https://podolakvandysolarsystem.wordpress.com/2016/04/11/voyager-golden-record-reading-the-instructions/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Voyager Golden record&lt;/a&gt; uses a star map of pulsars to precisely show where our solar system is...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible/feasible to base a navigation system on pulsars in order to replace GPS?&lt;/p&gt;&#xA;" OwnerUserId="21644" LastActivityDate="2018-02-11T23:47:50.653" Title="Could pulsar navigation replace GPS" Tags="&lt;pulsar&gt;&lt;navigation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25015" PostTypeId="2" ParentId="25014" CreationDate="2018-02-11T23:40:38.017" Score="2" Body="&lt;p&gt;Pulsar, or X-ray navigation is being actively tested right now. Currently an X-ray telescope mounted on the International Space Station completed a test where a small navigation computer was able to solve for it's own position and orbit in space relying only timing information from X-rays produced by a group of pulsars.&lt;/p&gt;&#xA;&lt;p&gt;I've written more about it in the question &lt;a href=&quot;https://space.stackexchange.com/q/24448/12102&quot;&gt;Is NICER/SEXTANT the first civilian “spacecraft” to determine it's own position in space without GPS or uplinked data?&lt;/a&gt; Spoiler alert, hold your cursor over the box:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;https://space.stackexchange.com/a/24450/12102&quot;&gt;Yes, it is the first&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You can also read about an investigation for X-ray communication in &lt;a href=&quot;https://space.stackexchange.com/q/19882/12102&quot;&gt;How will NavCube (actually) be important for the XCOM testing and demonstration?&lt;/a&gt; I believe this will be a significantly longer term technology development path.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.nasa.gov/content/about-nicer&quot; rel=&quot;nofollow noreferrer&quot;&gt;NICER&lt;/a&gt;, NASA’s Neutron star Interior Composition Explorer, will of course also be used for pure Astronomy research on neutron stars.&lt;/p&gt;&#xA;&lt;p&gt;Will this replace GPS? Not really, No.&lt;/p&gt;&#xA;&lt;p&gt;GPS serves the Earth's surface, and near Earth orbit (NEO) very nicely and efficiently. However, once you get a little farther, to MEO and GEO, it gets much more difficult to receive several simultaneous GPS signals because the satellites concentrate most of their power directly on to the Earth's disk, with only some side-lobes &amp;quot;leaking&amp;quot; past the terminator and into space.&lt;/p&gt;&#xA;&lt;p&gt;See the questions and of course their answers:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://space.stackexchange.com/q/14739/12102&quot;&gt;Will Glonass, Galileo, or BeiDou-2 satellites provide better cis-lunar navigation than GPS?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://space.stackexchange.com/q/14737/12102&quot;&gt;Has GPS been used beyond GEO? (duplicate)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://space.stackexchange.com/q/565/12102&quot;&gt;How far up have satellites used a GNSS for positioning, and how does the precision degrade with altitude?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;So for spacecraft in farther Earth's orbits it may be possible to augment or replace GPS or GNSS in general, and for deep space, there is currently no way to autonomously navigate today, so it will be the first.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/V64w2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V64w2.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; NICER X-ray timing telescope/concentrator array, from &lt;a href=&quot;https://www.nasa.gov/content/about-nicer&quot; rel=&quot;nofollow noreferrer&quot;&gt;About NICER&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-02-11T23:47:50.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25016" PostTypeId="1" AcceptedAnswerId="25026" CreationDate="2018-02-12T04:38:28.003" Score="0" ViewCount="1273" Body="&lt;p&gt;What is the farthest orbit that can be achieved around the Earth? How fast is that?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Not the question below:&#xA;Feel free to answer in a comment&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask because of this question: &lt;a href=&quot;https://worldbuilding.stackexchange.com/questions/104593/can-you-get-a-orbital-mass-driver-to-work-this-way&quot;&gt;https://worldbuilding.stackexchange.com/questions/104593/can-you-get-a-orbital-mass-driver-to-work-this-way&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an orbit exist where 2 objects going in opposite directions can intersect repeatedly with minor course corrections? &lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="2365" LastEditDate="2018-02-12T12:00:03.777" LastActivityDate="2018-02-12T12:00:03.777" Title="Farthest Orbit around the Earth?" Tags="&lt;orbit&gt;&lt;artificial-satellite&gt;&lt;space-travel&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2018-02-12T20:10:56.100" ContentLicense="CC BY-SA 3.0" />
  <row Id="25024" PostTypeId="2" ParentId="25013" CreationDate="2018-02-12T10:00:11.837" Score="4" Body="&lt;p&gt;The L4 and L5 points are unstable orbits. The Trojans move in perturbed (ie slightly non-keplerian) orbits that move them around the L4 and L5 points. At any time you will find Trojans in a pair of lobes between about 30 and 90 degrees ahead of Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find individual asteroids using &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA Horizons ephemeris&lt;/a&gt; &#xA;For example 624 Hektor, the largest Trojan (and placed in the Greek Camp around L4) will be found at RA 20 47 54.08, Dec -28 04 20.1 on Feb 10 2018. And 884 Piramus (in the Trojan Camp, L5) will be at RA 10 43 51.39 Dec +01 04 11.3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Jupiter is between them at RA 15 18 55.32 Dec -17 07 47.1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither are easy to find, with a Magnitude of 15 or 16, due to their greater distance than main belt asteroids, and very dark colour. The image below shows the distribution of Trojans around Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/om9i5.png&quot; alt=&quot;http://www.minorplanets.org/ExpeditionLaSagra/img/Jupiter_Trojans.png&quot;&gt;  &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-02-12T10:00:11.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25026" PostTypeId="2" ParentId="25016" CreationDate="2018-02-12T11:19:59.733" Score="3" Body="&lt;p&gt;The farthest Orbital Point from Earth is regulated by the fact that at what point the gravitational pull of the Sun overcomes the gravitational pull of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a post for this on &lt;a href=&quot;https://www.quora.com/What-is-the-farthest-possible-orbit-of-Earth&quot; rel=&quot;nofollow noreferrer&quot;&gt;Quora&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was the most up voted answer by user Paul Olaru on Quora:-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Approximately 1.5 million kilometers. It is limited by the point at which the influence of the sun's gravity becomes sufficiently stronger than Earth's that any object at that distance would become unbound from Earth &amp;amp; end up in independent orbit around then sun.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;That limit applies in general to any system of two bodies, and is called the Hill sphere. It's not actually perfectly spherical, but close enough for most practical purposes, and lies between the L1 &amp;amp; L2 Lagrange points. For circular orbits, the radius of the Hill sphere for any planet or moon can be calculated from&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;r = a*(m/(3M))^(1/3)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Where r is the radius of the Hill sphere, m is the mass of the planet or moon, M is the mass of the sun a planet orbits, or the planet that a moon orbits, and a is the distance from the sun (or planet) to its planet (or moon).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="21569" LastEditorUserId="21569" LastEditDate="2018-02-12T11:25:02.080" LastActivityDate="2018-02-12T11:25:02.080" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25027" PostTypeId="2" ParentId="25013" CreationDate="2018-02-12T11:34:01.467" Score="2" Body="&lt;p&gt;In order to find the rather large extent of space where Jupiter's Trojan asteroids can remain, one would use a map of the Celestial sphere, and look at the long stretch of points that happen to be +/- sixty degrees of wherever Jupiter happens to be at the moment. I show a screen shot from in-the-sky.org's &lt;a href=&quot;https://in-the-sky.org/staratlas.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Star Atlas&lt;/a&gt; page. The purple dot in the center next to the constellation name Libra is Jupiter, but because of some glitch it is not showing right now. Holding a cursor over the dot shows its name. The yellow diagonal line is the Ecliptic, and the +/- 60 degree regions will be at either end of the box. It's really really huge, there are no specific coordinates, and the only way to know an extremely faint dot would be an asteroid rather than a star would be an extremely detailed star map, or a series of exposures over hours or days showing the motion relative to the other objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; You can use any similar mapping program or website. You've asked about December, the programs can usually be walked backwards or forwards in time to whichever date you are interested in. The figure below is for mid-February, when I posted the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TP8YZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TP8YZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also enjoy Scott Manleys YouTube Videos:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://youtu.be/yt1qPCiOq-8&quot; rel=&quot;nofollow noreferrer&quot;&gt;Asteroids In Resonance With Jupiter&lt;/a&gt; (First 3:2 resonances, then Trojans are shown separately)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://youtu.be/xJsUDcSc6hE&quot; rel=&quot;nofollow noreferrer&quot;&gt;Asteroid Discovery - 1980-2012 - UHDTV&lt;/a&gt; (several versions on this YouTube channel)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Nmk4k.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nmk4k.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-02-12T20:21:29.340" LastActivityDate="2018-02-12T20:21:29.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25028" PostTypeId="1" CreationDate="2018-02-12T12:09:00.860" Score="3" ViewCount="115" Body="&lt;p&gt;I know that we cannot measure cosmological distances with parallax, but this is more a philosophical question. Am I correct in assuming that if we could measure distances using parallax (like stellar parlax used for nearby stars) the distance we would measure would correspond to the &quot;angular diameter distance&quot; ? &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_diameter_distance&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Angular_diameter_distance&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21660" LastEditorUserId="2365" LastEditDate="2018-02-15T22:03:40.633" LastActivityDate="2018-02-15T22:03:40.633" Title="Distance parallax and cosmological distance in universe in expansion" Tags="&lt;cosmology&gt;&lt;distances&gt;&lt;parallax&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25030" PostTypeId="1" CreationDate="2018-02-12T22:58:42.627" Score="5" ViewCount="419" Body="&lt;p&gt;What if we just can’t see far enough and the redshift actually shows the galaxies speeding up towards a gigantic black hole to where everything is converging, repeating a cycle that took place countless times “before” the big bang?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complete text in this link: &lt;a href=&quot;http://www.therealebook.com/Astronomy/Redshift_ContractingUniverse.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.therealebook.com/Astronomy/Redshift_ContractingUniverse.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in a contracting Universe, there is a region around the observer where all the galaxies are moving away; the galaxies that are moving towards the observer could be too far to be seen. If we can demonstrate this, we could challenge the conclusion that the Universe is expanding, based on the redshift effect observed in the “nearby” galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FqWZ2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FqWZ2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let’s start with a simpler representation: 1 dimension only (1D). Assume that all the galaxies move along a single line. We could represent an expanding and a contracting Universe as per the images below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lTZgL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lTZgL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Situation 1, we have the currently accepted model, where galaxies move away from each other, at increasing speed, which creates the redshift observed by Hubble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Situation 2, we assume that in the contracting Universe the galaxies have a speed that is inversely proportional to the distance from the center (as in a sink flow where the particles accelerate towards the drain), galaxies 1 and 3 will display a red shift effect, because they are moving away from each other, while galaxies 4, 5 and 6 will appear with blueshift effect. However, in order to see the blue shifted galaxies it is necessary to see beyond the center of the Universe, which would be not possible either because of the distance or because of the gigantic black hole in the center traps light from galaxies 4, 5 and 6 preventing it from reaching us at galaxy 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, assuming that from galaxy 2 we cannot see beyond galaxy 1, it is impossible to distinguish between situation 1 and 2.&#xA;With this we have demonstrated that for a 1D Universe it is impossible to determine if it is expanding or contracting, simple because we cannot see far enough. What about a 2D space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider two galaxies (1 and 2); consider that they form a angle beta, relative to the center of the universe (assuming that there is one); Consider that galaxy 1 is at the distance r from the center; consider that galaxy 2 is at a fraction of r (say xr) from the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can calculate the distance between galaxy 1 and 2 as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4VjO2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4VjO2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the equation above we can calculate alfa and from &lt;a href=&quot;https://i.stack.imgur.com/F241q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F241q.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; we can calculate d.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also suppose that the speed along the radius can be calculated with the equation v = c / r, where c is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore we can calculate the position of the two galaxies after a period of time t, and measure the distance d between them to verify if d increases (redshift) of decreases (blueshift).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see that no matter what the fraction x is, there is always a minimum value for the angle beta, such that any galaxy within this angle will be moving away from the reference galaxy. Now assume that the universe is so big that we cannot see galaxies beyond beta. In this way, all that we can see is redshifted galaxies!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that even for a contracting universe, depending on how far we can see, it is possible to detect only redshifted galaxies. Does not this disprove the theory of expanding universe based on redshift?&lt;/p&gt;&#xA;" OwnerUserId="21664" LastEditorUserId="9772" LastEditDate="2018-07-17T12:28:46.187" LastActivityDate="2020-10-09T23:35:03.047" Title="Could a contracting Universe create the redshift effect observed by Hubble?" Tags="&lt;redshift&gt;&lt;fate-of-universe&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25032" PostTypeId="2" ParentId="25012" CreationDate="2018-02-12T23:17:17.107" Score="2" Body="&lt;p&gt;It all depends on shared knowledge. What you are getting at is known in philosophy as &lt;a href=&quot;https://en.wikipedia.org/wiki/Indeterminacy_of_translation&quot; rel=&quot;nofollow noreferrer&quot;&gt;indeterminacy of translation&lt;/a&gt;. But note that the alien is already talking to the human: there has to be a language present, and languages do contain a lot of assumptions about the world (tenses imply that there is time, and the question itself implies space, for example). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, consider &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/0/0f/Earth%27s_Location_in_the_Universe_SMALLER_%28JPEG%29.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;a sequence of larger and larger scale maps of the neighbourhood&lt;/a&gt;, running from one showing the solar system to one showing the cosmic web. That seems to settle the &quot;where&quot; question using pattern matching fairly well...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...except that there is the obvious question of &lt;em&gt;what&lt;/em&gt; the maps show. In this particular example it is stellar baryonic matter, about 0.002 of the total mass/energy budget. If the alien is made of dark matter this will not make much sense except that it would on the larger scales recognise the topology of the web of halo structures it was familiar with. If it knew about baryonic matter it would infer that this was a map of baryonic stuff. Similarly a gaseous alien might think we have left out the 90% of baryonic matter that is intergalactic plasma. But the point is, the right kind of map or explanation would latch onto something in the knowledge of the alien and allow them to start inferring more meaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is not just sending a single message but a conversation there is a lot of room for testing and exploration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;So you are telling me where you are using a map showing 'pulsars'. Are they [untranslatable1] or [untranslatable2]?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What is the difference between [untranslatable1] and [untranslatable2]?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;[untranslatable1] is spatially localised but extends across time,&#xA; [untranslatable2] is temporally localised but spread out in space.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Then [untranslatable1] sounds more reasonable. We know them as compact stars that rotate fast and send radio waves.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Are 'stars' also spatially localised and extend across time?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It won't be easy, and it is quite possible that our map doesn't correspond to a known region (if the aliens are communicating in some weird non-local way; had they communicated in a local way they would have known the answer to the questions already). But pattern matching and interactive bootstrapping from the shared language and knowledge-base would be a way of getting around the indeterminacy of translation.&lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2018-02-12T23:17:17.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25033" PostTypeId="1" CreationDate="2018-02-12T23:18:08.347" Score="0" ViewCount="470" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RvU6V.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RvU6V.jpg&quot; alt=&quot;Earth and Moon&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have wondered, what'll happen when we destroy our planet or it becomes uninhabitable for human life? Is there another option than the complete extinction of the human race? Can anyone give me a logical response?&lt;/p&gt;&#xA;" OwnerUserId="21665" LastEditorUserId="18497" LastEditDate="2018-02-13T08:06:19.217" LastActivityDate="2018-02-13T12:46:56.173" Title="Is there another habitable planet in our galaxy?" Tags="&lt;earth-like-planet&gt;" AnswerCount="3" CommentCount="3" ClosedDate="2018-02-14T12:08:35.870" ContentLicense="CC BY-SA 3.0" />
  <row Id="25034" PostTypeId="1" AcceptedAnswerId="25040" CreationDate="2018-02-12T23:19:24.513" Score="1" ViewCount="74" Body="&lt;p&gt;Is a orbit possible where 2 objects going in opposite directions can intersect repeatedly with minor course corrections?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ra0BS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ra0BS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastActivityDate="2018-02-13T04:24:16.570" Title="Binary Orbit around the Earth?" Tags="&lt;orbit&gt;&lt;earth&gt;&lt;artificial-satellite&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25035" PostTypeId="2" ParentId="25033" CreationDate="2018-02-13T02:49:20.133" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;what'll happen when we destroy our planet or it becomes none-habitual for human life?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Very pessimistic idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We lack the technology to tell if a planet orbiting another star could be lived on by humans.  We barely have the technology to detect some planets orbiting a small number of other stars (compared to the number of stars in the galaxy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term for these planets is &lt;a href=&quot;https://en.wikipedia.org/wiki/Exoplanet&quot; rel=&quot;noreferrer&quot;&gt;exoplanet&lt;/a&gt; and Wikipedia says we have detected about 3750 to date orbiting about 2800 stars.  As there are somewhere between 100 and 400 &lt;em&gt;billion&lt;/em&gt; stars in the galaxy, we only know a tiny fraction of what's out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a very limited basis for listing some planets as possible candidates for supporting some form of life, but, with all respect to the serious scientists doing this work, this is nowhere near knowing if they really can support life at all, let alone humans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your question's answer is : we don't know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We cannot travel fast or far enough at present to reach another star in a reasonable time frame and we also could not do it with enough people.  We do not currently have any reasonable expectation that we will be able to do that in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even if we knew of a suitable planet, it would be unlikely we could ever travel there to e.g. colonize it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we're stuck here so keep recycling. :-)&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2018-02-13T02:49:20.133" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25036" PostTypeId="2" ParentId="25030" CreationDate="2018-02-13T03:20:32.557" Score="1" Body="&lt;p&gt;For a contracting universe, space would contract as a ratio, and hence even if we could only see a small distance, we would still see blue-shifted galaxies due to the fact that the galaxies further from the hypothetical centre of the universe (see &lt;a href=&quot;https://astronomy.stackexchange.com/questions/874&quot;&gt;here&lt;/a&gt; for why I say hypothetical) would be moving faster than those closer, and hence any two points in space must be moving towards each other. This holds true even in a finite universe, all points appear to contract towards the point of reference, as can be seen in the diagram below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/g7xKM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g7xKM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;The distance between Galaxies 1 and 3 and Galaxy 2 is clearly less after contraction than before.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2018-02-13T03:20:32.557" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25037" PostTypeId="2" ParentId="25033" CreationDate="2018-02-13T03:50:10.350" Score="2" Body="&lt;p&gt;I would like to add some points to StephanG's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are just considering another habitable planet in our galaxy why look anywhere else other than in our own Solar System?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Mars is half the size of Earth and also there are many proofs saying that there was &lt;a href=&quot;https://www.nasa.gov/press-release/nasa-confirms-evidence-that-liquid-water-flows-on-today-s-mars&quot; rel=&quot;nofollow noreferrer&quot;&gt;flowing water&lt;/a&gt; &lt;a href=&quot;https://www.nasa.gov/feature/jpl/recurring-martian-streaks-flowing-sand-not-water&quot; rel=&quot;nofollow noreferrer&quot;&gt;on Mars&lt;/a&gt; with which we can say that terraforming Mars is quite possible. There is even this &lt;a href=&quot;https://en.wikipedia.org/wiki/Terraforming_of_Mars&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; article on this.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then there is Titan which is Saturn's moon. When the Sun is nearing it's ending and becomes bigger and turns into s Supergiant, there is a possibility that we could colonize Titan. Here is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Life_on_Titan&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; article on this.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then there are Icy Moons i.e. &lt;a href=&quot;https://en.wikipedia.org/wiki/Enceladus&quot; rel=&quot;nofollow noreferrer&quot;&gt;Enceladus&lt;/a&gt; of Saturn and &lt;a href=&quot;https://en.wikipedia.org/wiki/Europa_(moon)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Europa&lt;/a&gt; of Jupiter that have maybe liquid water beneath their surface.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="21569" LastActivityDate="2018-02-13T03:50:10.350" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25039" PostTypeId="2" ParentId="25033" CreationDate="2018-02-13T04:10:20.770" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Is there another habitable planet in our galaxy?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What constitutes habitable?   Oceans, lakes and rivers, soil to grow plants, oxygen to breath, a magnetic field to help maintain the atmosphere, plate tectonics, an ozone layer to block UV rays, not too much or too little gravity, or too much or too little air pressure.   &lt;a href=&quot;https://www.smithsonianmag.com/smart-news/mars-surface-may-be-toxic-bacteria-180963966/&quot; rel=&quot;nofollow noreferrer&quot;&gt;No poison in abundance&lt;/a&gt; Oxygenated, not &lt;a href=&quot;https://en.wikipedia.org/wiki/Anoxic_waters&quot; rel=&quot;nofollow noreferrer&quot;&gt;Anoxic bodies of water&lt;/a&gt; or too many hydrogen sulfide producing bacteria, and lets add, just for fun, no 350 kph winds or 10 inch hale stones on a daily basis or dust storms that could choke a &lt;a href=&quot;http://starwars.wikia.com/wiki/Bantha&quot; rel=&quot;nofollow noreferrer&quot;&gt;bantha&lt;/a&gt;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On, and one more thing, if a planet meets all those criteria, especially the breathable oxygen atmosphere, there's probably life there already, and so there's the &quot;They were here first&quot; problem and life isn't always friendly, we might be toxic to them or they might be toxic or hostile towards us.   That's a problem with potentially habitable planets with Oxygen atmospheres.   If there's Oxygen, there's probably life and if there's life, that raises questions, not just moral ones, but safety questions as well.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a bit of a catch 22 in the habitable world question.   If there's oxygen there's probably life and if there's life, we need to be very careful before we set foot there.   The odds of finding a water/oxygen atmosphere world with no life is pretty remote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to answer your question - habitable planets?   Nobody knows.  There's no way of answering this question with what we know today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there a &lt;em&gt;potentially&lt;/em&gt; habitable planet in our galaxy?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one, almost certainly yes, though we don't have good enough telescopes to say yes for certain but there's enough planets out there that a few of them are expected to be &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Earth-like_planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth-like&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth-like, meaning, A rocky planet with plate tectonics, an atmosphere we could work with, oceans, continents and a magnetic field.   Such a planet could in theory, be terraformed and become Earth like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While terraforming is theoretical and there are significant unknowns and it would take some time, there's probably no shortage of Earth-like planets with roughly the right range of solar energy, gravity, atmosphere and liquid water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;With technology we could live in lots of places.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars, Titan, the Moon, Mercury, Venus (in acid-resistant dirigibles), Ceres, Icy bodies like Pluto or other Kuiperbelt Objects.   Those are all livable, with proper technology.   Colonies would likely be small at first, and it's perhaps not what I'd consider a fun place to live, but drilling a hole into the ice inside Pluto, for example, with an energy source, perhaps fusion, or if closer to the sun, solar power.   Settlements would be possible and not even that far beyond our technology today.   It'd be expensive to get all the necessary equipment out to the whichever location you choose (Lots of people have their eyes on Mars - but other locations would work too).  But such settlements aren't too far out of our current ability and we'll likely begin to see them perhaps in the next 50 years or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Traveling 20 or 30 light-years is hard.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want another planet, not just a small colony living on Mars or Ceres or Titan or Pluto (Ceres would be fun, you could probably jump like 100 feet in the air, or lack of air I should say) . . .but I digress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you want a real planet, that is another Earth, the hard part isn't finding one, the hard part is getting there.    I'm not going to say anything more about that other than inter-stellar space travel is really really really hard.   Gonzo, crazy difficult.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Outside of tiny settlements, Earth is it, for the foreseeable future.   We may be able to identify other planets in the not too distant future, but getting there is an entirely different ballgame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;What'll happen when we destroy our planet or it becomes none-habitual&lt;/strong&gt;&#xA;  &lt;strong&gt;for human life?&lt;/strong&gt; Is there another option then the complete wipe out of&#xA;  the Human race? Can anyone give me a logical response&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other scientists have pointed this out as a reason why funding for astronomy because, in time, it could play a role in saving our species.   &lt;a href=&quot;http://www.independent.co.uk/news/science/stephen-hawking-space-travel-will-save-mankind-and-we-should-colonise-other-planets-10058811.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stephen Hawking, for example&lt;/a&gt;.   So you're not the first to think about this. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;I believe that the long term future of the human race must be space&#xA;  and that it represents an important life insurance for our future&#xA;  survival, as it could prevent the disappearance of humanity by&#xA;  colonising other planets.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The good news is, actually wiping out our species is hard.   A large meteor impact like the one that killed the dinosaurs is a rare event - once every 100 million years or so for one that big and now there's a good chance we'd see it coming.   Mankind killing events are statistically rare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bad news - finding a 2nd Earth like planet and &lt;em&gt;getting there&lt;/em&gt; is very hard.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd advise not worrying cause worrying about the destruction of mankind strikes me as unproductive, but you can petition for more money to NASA or Space-X or other space exploration organization of your choosing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(too long?)&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-02-13T12:46:56.173" LastActivityDate="2018-02-13T12:46:56.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25040" PostTypeId="2" ParentId="25034" CreationDate="2018-02-13T04:24:16.570" Score="1" Body="&lt;p&gt;I'm not sure I see the point to this question, but if one of the orbiting objects is hollow and you allow for course correction, then yes, one object passing through the other is possible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see what force would keep the object with a hole in the middle properly positioned so it was in effect perpendicular to the movement of the other objects.   There's also problems with the orbits lining up just right rather than missing and giving each other a gravitational push, and with gravitational perturbations from the sun and the moon.   A system like that would be a lot easier and require less adjustment if the Sun and Moon weren't there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'd probably want both orbits to be over the equator to reduce fluctuations, but eliminating fluctuations would be impossible.  Every orbit would probably require some adjustment, and while it's theoretically possible, I'm not sure I see why anyone would want too - apart from perhaps it would look cool.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-02-13T04:24:16.570" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25041" PostTypeId="1" CreationDate="2018-02-13T05:15:05.150" Score="8" ViewCount="880" Body="&lt;p&gt;I understand polishing the mirror with a successively finer abrasive can create a curvature, but how is it possible to obtain a uniform curve that forms a parabola while polishing by hand?&lt;/p&gt;&#xA;" OwnerUserId="21669" LastEditorUserId="21237" LastEditDate="2018-02-13T15:12:28.877" LastActivityDate="2019-09-12T11:15:20.513" Title="How is it ensured that a parabolic shape forms while polishing a telescope mirror?" Tags="&lt;amateur-observing&gt;&lt;observational-astronomy&gt;&lt;telescope-lens&gt;&lt;telescope-making&gt;&lt;mirror-making&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25046" PostTypeId="2" ParentId="25030" CreationDate="2018-02-13T07:48:04.503" Score="8" Body="&lt;p&gt;Aside from being, I suspect, totally incompatible with GR, the model fails to match observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe appears to be homogeneous and isotropic. In the model you suggest, Hubble's law would be different depending on which direction you were looking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, looking away from the black hole, galaxy redshifts would increase to some asymptotic value, whereas looking towards the black hole they would increase to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the argument is that the central black hole is so far away that we cannot measure this asymmetry (i.e. well beyond the observable universe, given the homogeneity of the cosmic microwave background - but then does the model have an explanation for the CMB at all?), then the model also needs to explain why galaxies would move in this particular way, since they would not &quot;feel&quot; the influence of the black hole.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-02-13T07:48:04.503" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25048" PostTypeId="2" ParentId="24642" CreationDate="2018-02-13T08:07:25.697" Score="1" Body="&lt;p&gt;The speed of light will remain constant. Though the way it is perceived near a black hole changes with where and how it is perceived, it will remain constant. The speed of light does not increase or decrease just because it is near a black hole.&lt;/p&gt;&#xA;" OwnerUserId="21678" LastActivityDate="2018-02-13T08:07:25.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25050" PostTypeId="2" ParentId="1001" CreationDate="2018-02-13T10:23:12.843" Score="7" Body="&lt;p&gt;I was just thinking about that and here is my layman's explanation. Imagine you're tracing two dots on a crumpled piece of paper, the dots are moving, but as they are moving, so is the paper getting ‘uncrumpled’, the actual distance between the dots will be more than the sum of distances they have travelled. &lt;/p&gt;&#xA;" OwnerUserId="21681" LastEditorUserId="16685" LastEditDate="2018-02-14T01:55:26.773" LastActivityDate="2018-02-14T01:55:26.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25054" PostTypeId="1" CreationDate="2018-02-13T18:35:02.733" Score="23" ViewCount="7577" Body="&lt;p&gt;Io gets its volcanic heat from being flexed by Jupiter's gravity, and this seems to go on forever.  That sounds like a magic energy source, since Jovian gravity is perpetual and Io seems to remain in orbit perpetually.  Have we discovered magic?&lt;/p&gt;&#xA;" OwnerUserId="21690" LastEditorUserId="21237" LastEditDate="2018-02-15T18:15:01.357" LastActivityDate="2019-07-21T09:36:02.703" Title="Is Io a magic energy machine?" Tags="&lt;orbital-mechanics&gt;&lt;io&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25055" PostTypeId="2" ParentId="25054" CreationDate="2018-02-13T19:49:50.357" Score="41" Body="&lt;p&gt;The energy driving Io's volcanism is derived from its orbital energy and (to a lesser extent by the orbital energy of the other moons and the energy of Jupiter's rotation).  Similar mechanisms power the cryo-volcanism on Enceladus and the other icy moons.  The available orbital energy reservoir is huge compared to that required to drive the volcanic activity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does imply that the gravitational interactions on all the moons in the system are changing their orbits and rotations, albeit slowly.  Given unlimited time, the orbit of Io might decay to a point where it falls within Jupiter's Roche limit and is torn apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: the answer to &lt;a href=&quot;https://astronomy.stackexchange.com/q/25061/7982&quot;&gt;How can tidal heating lower Io's orbit&lt;/a&gt; describes this more precisely, and points out that tidal forces are transferring energy from Jupiter's rotation to the orbits of its moons, so Io's orbit in the long term gets larger (to the point of it leaving Jupiter orbit) rather than decaying.)&lt;/p&gt;&#xA;" OwnerUserId="14434" LastEditorUserId="14434" LastEditDate="2018-02-14T16:52:03.160" LastActivityDate="2018-02-14T16:52:03.160" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="25057" PostTypeId="2" ParentId="25041" CreationDate="2018-02-13T20:45:14.143" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How is it ensured that a parabolic shape forms while polishing a telescope mirror?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While a spherical shape is fairly easy to abtain, a parabolic shape definitely isn't!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a combination of special polishing strokes, and feedback via control and testing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://stellafane.org/tm/atm/mirror-refs/strokes.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://stellafane.org/tm/atm/mirror-refs/strokes.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normal polishing strokes (center over center, 1/3 diameter amplitude) tend to make a sphere. When all goes well, the surface of the mirror tends to become spherical, or very close to it. This is not different from grinding, only more precise. When two surfaces are rubbing against each other in a ball-and-socket type of joint, any irregularities will be removed and both surfaces will &quot;want&quot; to become spherical, or close to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When making telescope mirrors, the so-called &quot;normal stroke&quot; is what accomplishes that in the best way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dKj1j.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dKj1j.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here's the essential detail: no matter what other curve you're trying to obtain (parabola, hyperbola, ellipse), for the purpose of making telescope mirrors all these curves are very close to a circle, within a fraction of a micron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a typical process has an initial goal of making a sphere (cross-section is a circle). This is not mandatory, but it's easier to first make a sphere, and then apply &quot;corrections&quot; to it. You can definitely make a parabola directly, but the process is more complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For telescope mirrors, the parabola that's needed is basically a circle that's deeper in the center, and more shallow towards the edge. So, if you start with a spherical mirror (circular section) and use polishing strokes that tend to deepen the center in a gradual way, you can get pretty close to a parabola. This is known as a &quot;parabolizing stroke&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JAzZK.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JAzZK.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that as you apply that stroke, the center keeps deepening. At some point the correction will become deeper than a parabola and the mirror will become hyperbolic. That's sometimes what you actually need (if you make a hyperbolic mirror for a Ritchey-Chretien telescope). Or if you stop too early, the mirror is elliptic, which is used in other designs. The parabola is the thin edge between the elliptic and hyperbolic domains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You start with a mirror where the cross-section is a circle (spherical mirror). You begin to apply the W stroke. The circle immediately becomes a very shallow ellipse, which keeps deepening. At some point, for a very short time, that ellipse becomes a parabola. As you keep doing the W stroke, the parabola then changes quickly and now you have a hyperbola which just becomes more and more deep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key here is testing. You apply the parabolizing stroke for a short while, then stop and test the mirror. There are many testing techniques; the Couder mask is simple enough and very popular; or you could use an interferometer if you have (or can build) one. Is it still elliptic? Keep parabolizing. Has it become hyperbolic? Reset to a sphere and try again. Or adjust the stroke if either the edge or the center are being corrected too fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply parabolizing strokes. Test. Modify technique if necessary. Repeat. That's all there is to it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; I am using jargon in the text above. &quot;Deeper&quot; really means &quot;more strongly curved&quot;. &quot;Shallower&quot; really means &quot;less strongly curved&quot;. The difference between circle and parabola is that the parabola is more strongly curved in the center, while the curvature at the edge is less strong. The circle has constant curvature everywhere, by definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't necessarily mean the center of the mirror is physically more depressed; while doing any strokes, the whole surface keeps being polished down all the time. But the relative curvatures of center vs. edge may change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deeper vs shallower are terms that reflect the physical process, whether the tool applies more or less pressure on a certain area of the mirror (so polishing goes faster or slower there). But the geometric attribute that the process actually changes, that we are interested in, is the radius of curvature. Where the tool applies more pressure, the radius of the curvature decreases rapidly (the surface becomes more strongly curved).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deeper and shallower are practical terms related to process. More or less strongly curved are theoretical terms related to geometry. When you make mirrors, after a while these terms become linked in your mind.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It helps if you're familiar with the conic sections:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Conic_section&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Conic_section&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dqfIN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dqfIN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tRMfS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tRMfS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;red = circle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;orange = ellipse&lt;/p&gt;&#xA;&#xA;&lt;p&gt;green = parabola&lt;/p&gt;&#xA;&#xA;&lt;p&gt;blue = hyperbola&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="7982" LastEditDate="2019-09-12T11:15:20.513" LastActivityDate="2019-09-12T11:15:20.513" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25059" PostTypeId="2" ParentId="25054" CreationDate="2018-02-13T22:51:07.360" Score="23" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;And this seems to go on forever.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not strictly true.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Io seems to remain in orbit perpetually. Have we discovered magic?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No and no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Io's orbital speed is $v = 17.33 \mathrm{km/s}$. Its mass is $m = 8.9319 \cdot 10^{22} \mathrm{kg}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula for the kinetic energy is: $E = m \cdot v^2 / 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging the values for Io into it, its total kinetic energy is approximately $7 \cdot 10^{40}$ Joule according to &lt;a href=&quot;http://www.wolframalpha.com/input/?i=17330+*+(8.9319+*+10%5E22)%5E2+%2F+2&quot; rel=&quot;noreferrer&quot;&gt;Wolfram Alpha&lt;/a&gt;. That's a very, VERY large amount of energy. Io can afford to spend a small fraction of it to keep itself warm for a very long time without significant apparent changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But yeah, on a human time scale it does look like magic.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="7472" LastEditDate="2018-02-14T11:41:27.807" LastActivityDate="2018-02-14T11:41:27.807" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25060" PostTypeId="1" AcceptedAnswerId="25071" CreationDate="2018-02-14T03:40:58.857" Score="2" ViewCount="62" Body="&lt;p&gt;In the video linked in the Space.com article &lt;a href=&quot;https://www.space.com/39680-wfirst-space-telescope-cancellation-scientist-reactions.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;What Would It Mean for Astronomers If the WFIRST Space Telescope Is Killed?&lt;/a&gt;, (available in YouTube as &lt;a href=&quot;https://youtu.be/LbJpVHMV1m4&quot; rel=&quot;nofollow noreferrer&quot;&gt;WFIRST: The Best of Both Worlds&lt;/a&gt;) after &lt;code&gt;02:00&lt;/code&gt; Caltech cosmologist &lt;a href=&quot;http://web.ipac.caltech.edu/staff/wang/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Yun Wang&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;WFIRST will allow us to potentially make ground-breaking discoveries, finding out what dark energy is. So this will tell us if dark energy is an unknown form of energy, or if its a modification of general relativity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without going too deep in to GR, how might that happen? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What kinds of &lt;a href=&quot;https://www.wired.com/2017/01/case-dark-matter/&quot; rel=&quot;nofollow noreferrer&quot;&gt;WFIRST&lt;/a&gt; observations  and interpretations would be used to potentially address the nature of dark energy? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on comments, I'll reclassify the following as an afterthought:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this in any way related to news about a recent observational non-exclusion of &lt;a href=&quot;https://en.wikipedia.org/wiki/Erik_Verlinde&quot; rel=&quot;nofollow noreferrer&quot;&gt;Erik Verlinde&lt;/a&gt;'s theories (the guy who's &quot;&lt;a href=&quot;https://www.wired.com/2017/01/case-dark-matter/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Trying to kill dark matter&lt;/a&gt;&quot;), discussed in &lt;a href=&quot;https://www.universiteitleiden.nl/en/staffmembers/margot-brouwer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Margot Brouwer&lt;/a&gt;'s 2016 video &lt;a href=&quot;https://www.youtube.com/watch?v=c_a2jWYk8A0&quot; rel=&quot;nofollow noreferrer&quot;&gt;Verlindes new theory first tested with weak gravitational lensing&lt;/a&gt; about the &lt;a href=&quot;https://www.universiteitleiden.nl/en/staffmembers/margot-brouwer&quot; rel=&quot;nofollow noreferrer&quot;&gt;VLT Survey Telescope&lt;/a&gt; &lt;a href=&quot;http://kids.strw.leidenuniv.nl/index.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;kilo-degree survey&lt;/a&gt; results (see also Cosmos Magazine's &lt;a href=&quot;https://cosmosmagazine.com/physics/new-theory-of-dark-gravity-passes-first-test-but-einstein-s-still-on-top&quot; rel=&quot;nofollow noreferrer&quot;&gt;New theory of ‘dark gravity’ passes first test, but Einstein’s still on top&lt;/a&gt;, or am I mixing apples and oranges?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-02-14T06:16:16.057" LastActivityDate="2018-02-14T08:23:34.950" Title="How might WFIRST help us better understand dark energy?" Tags="&lt;cosmology&gt;&lt;dark-energy&gt;&lt;space-telescope&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25061" PostTypeId="1" AcceptedAnswerId="25083" CreationDate="2018-02-14T05:07:31.503" Score="13" ViewCount="809" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/25055/7982&quot;&gt;This answer&lt;/a&gt; to the question &lt;a href=&quot;https://astronomy.stackexchange.com/q/25054/7982&quot;&gt;Is Io a magic energy machine?&lt;/a&gt; suggests that the energy from the internal heating of Io due to &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_heating_of_Io&quot; rel=&quot;noreferrer&quot;&gt;tidal &quot;squishing&quot;&lt;/a&gt; as it moves cyclically closer and farther from Jupiter in its elliptical orbit will come from the energy of Io's orbit. A lower energy orbit is necessarily smaller, and that actually means that the velocity will be larger. (When you want to raise a satellite's orbit to a higher altitude, you actually use thrust &lt;em&gt;in the direction&lt;/em&gt; of motion to slow it down.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that tidal forces are a little complicated (cf. &lt;a href=&quot;https://astronomy.stackexchange.com/q/420/7982&quot;&gt;Why is the Moon receding from the Earth due to tides? Is this typical for other moons?&lt;/a&gt;), is it &lt;em&gt;a priori&lt;/em&gt; certain that the heating will lower Io's orbit, causing it to &lt;em&gt;speed up?&lt;/em&gt; (Consider that Earth's Moon's receding is due in part to Earth's liquid ocean, and Jupiter is a gas giant.) Is it just the &lt;a href=&quot;https://space.stackexchange.com/q/25329/12102&quot;&gt;perijove&lt;/a&gt; that will decrease, or the semi-major axis? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can a (seemingly, naively, on average) radial force cause a tangential acceleration? &lt;a href=&quot;https://en.wikipedia.org/wiki/Io_(moon)#Orbit_and_rotation&quot; rel=&quot;noreferrer&quot;&gt;Io&lt;/a&gt; is &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_locking&quot; rel=&quot;noreferrer&quot;&gt;tidally locked&lt;/a&gt; to Jupiter so its rotation around its own axis is synchronous to its rotation around Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit:&lt;/strong&gt; fwiw if the gravitational interactions between Io and Jupiter's other moons causes the problem to be too complex to answer easily, I'm more interested in the basic dynamics of tidal heating and effects on the orbit of a single moon, rather than specifically Io's situation.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="16323" LastEditDate="2018-02-14T08:12:27.063" LastActivityDate="2019-03-27T22:13:30.523" Title="How can tidal heating lower Io's orbit?" Tags="&lt;jupiter&gt;&lt;natural-satellites&gt;&lt;orbital-mechanics&gt;&lt;io&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25064" PostTypeId="1" CreationDate="2018-02-14T05:38:36.977" Score="2" ViewCount="92" Body="&lt;p&gt;Always I wonder about that if some day alien wants to talk to us, then how can we talk to them? They didn't know any of the language on the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What will be the basic building blocks for it?&lt;/p&gt;&#xA;" OwnerUserId="21700" LastEditorUserId="1891" LastEditDate="2018-02-18T19:01:40.367" LastActivityDate="2018-02-18T19:01:40.367" Title="What happens if one day aliens want to talk to us. How can we talk to them?" Tags="&lt;universe&gt;&lt;astrophysics&gt;&lt;fundamental-astronomy&gt;&lt;mathematics&gt;" AnswerCount="2" CommentCount="9" ClosedDate="2018-02-14T14:52:29.530" ContentLicense="CC BY-SA 3.0" />
  <row Id="25065" PostTypeId="1" AcceptedAnswerId="25089" CreationDate="2018-02-14T05:52:43.017" Score="1" ViewCount="147" Body="&lt;p&gt;I have a lab where given the mass of a star being $1.23$ times the math of our sun, some planets wobble of central start is $273$ days, and the velocity of 79 m/s. I found that the mass of the planet is $5.3\times 10^{27}$ kg and is approximately $.82$ AU away from its star. Here is a screen shot of my excel file:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bJXa9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bJXa9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I am asked would we expect this planet to be hotter or colder than earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this question is rather silly since we have no idea the composition of the planet or other related factors to make any conclusion. From what I found is based on the mass and radius it is probably a water world, and since it is pretty close to its star I would say that it is probably hotter than earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways, any suggestions or thoughts on this matter would be appreciated. &lt;/p&gt;&#xA;" OwnerUserId="18360" LastActivityDate="2018-12-17T13:37:17.943" Title="Would you expect this planet to be hotter or colder than earth?" Tags="&lt;planet&gt;&lt;distances&gt;&lt;doppler-effect&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25066" PostTypeId="1" AcceptedAnswerId="25118" CreationDate="2018-02-14T05:55:26.343" Score="1" ViewCount="188" Body="&lt;p&gt;In the video linked in the Space.com article &lt;a href=&quot;https://www.space.com/39680-wfirst-space-telescope-cancellation-scientist-reactions.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;What Would It Mean for Astronomers If the WFIRST Space Telescope Is Killed?&lt;/a&gt;, (available in YouTube as &lt;a href=&quot;https://youtu.be/LbJpVHMV1m4&quot; rel=&quot;nofollow noreferrer&quot;&gt;WFIRST: The Best of Both Worlds&lt;/a&gt;) after &lt;code&gt;02:00&lt;/code&gt; NASA astrophysicist &lt;a href=&quot;https://en.wikipedia.org/wiki/Neil_Gehrels&quot; rel=&quot;nofollow noreferrer&quot;&gt;Neil Gehrels&lt;/a&gt; says of &lt;a href=&quot;https://en.wikipedia.org/wiki/Wide_Field_Infrared_Survey_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;WFIRST&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It has the same image precision and power as the Hubble Space Telescope, but with one-hundred times the area of sky that is used.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Looking at the crude drawing on this &lt;a href=&quot;https://wfirst.gsfc.nasa.gov/observatory.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA page&lt;/a&gt; shows only a Cassegrain-like geometry, but there must be additional goodies to obtain a similar resolution with a 10x wider field. Does it have three, four, or five mirrors like the &lt;a href=&quot;https://en.wikipedia.org/wiki/Extremely_Large_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;E-ELT&lt;/a&gt; or the &lt;a href=&quot;https://en.wikipedia.org/wiki/Large_Synoptic_Survey_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;LSST&lt;/a&gt; (cf. &lt;a href=&quot;https://astronomy.stackexchange.com/q/15005/7982&quot;&gt;What is a quaternary mirror and why does the E-ELT need one?&lt;/a&gt; and answers therein) or corrector lenses? What is the magic sauce that makes this possible, and is there at least a schematic diagram of the optics somewhere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QRMn1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QRMn1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; From &lt;a href=&quot;https://wfirst.gsfc.nasa.gov/observatory.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-02-14T06:11:21.900" LastActivityDate="2018-02-18T23:42:38.207" Title="What allows WFIRST to have similar resolution to Hubble over a 100x larger solid angle?" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25068" PostTypeId="2" ParentId="25065" CreationDate="2018-02-14T08:07:26.180" Score="3" Body="&lt;p&gt;There are lots of unknowns however, you have a star that is more massive than the sun. More massive stars (on the main sequence) are brighter than smaller ones. So the star is brighter than the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also have a planet that is closer to the star than the Earth is to the sun. Both these effects would suggest that the planet would be warmer than the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed rough estimates suggest that this planet could have gone the way of Venus, and suffered a run-away greenhouse effect. The planet is considerably more massive than Earth. It would likely have a thick atmosphere that could contain lots of CO2.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-02-14T17:07:02.353" LastActivityDate="2018-02-14T17:07:02.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25069" PostTypeId="5" CreationDate="2018-02-14T08:21:29.703" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2018-02-14T08:21:29.703" LastActivityDate="2018-02-14T08:21:29.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25070" PostTypeId="4" CreationDate="2018-02-14T08:21:29.703" Score="0" Body="For questions regarding Io, one of the four moons of Jupiter discovered by Galileo." OwnerUserId="7472" LastEditorUserId="7472" LastEditDate="2018-02-14T11:40:51.190" LastActivityDate="2018-02-14T11:40:51.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25071" PostTypeId="2" ParentId="25060" CreationDate="2018-02-14T08:23:34.950" Score="1" Body="&lt;p&gt;The WFIRST's study of dark energy is pretty straight forward.   It's no different than the previous ways of studying dark energy.   It's not studied directly, cause we can't see it or re-create it, it's studied by the effect it has on the relative motion of distant galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WFIRST will get much more accurate readings on the relative velocity of galaxies both relative to each other in their neighborhood and relative to us, this will get a significantly more accurate estimate for dark energy than was previously possible and questions can be answered such as, is it speeding up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://wfirst.gsfc.nasa.gov/dark_energy.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA's WFIRST Web page&lt;/a&gt; discusses this in more detail&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The High Latitude Spectroscopic Survey&lt;/strong&gt; will measure accurate distances&#xA;  and positions of a very large number of galaxies. By measuring the&#xA;  changes in the distribution of galaxies, the evolution of dark energy&#xA;  over time can be determined. The High Latitude Survey will measure the&#xA;  growth of large structure of the universe, testing theory of&#xA;  Einstein's General Relativity.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Type Ia Supernovae (SNe) Survey&lt;/strong&gt; uses type Ia SNe as &quot;standard candles&quot;&#xA;  to measure absolute distances. Patches of the sky are monitored to&#xA;  discover new supernovae and measure their light curves and spectra.&#xA;  Measuring the distance to and redshift of the SNe provides another&#xA;  means of measuring the evolution of dark energy over time, providing a&#xA;  cross-check with the high latitude surveys.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;High Latitude Imaging Survey&lt;/strong&gt; will measure the shapes and distances of&#xA;  a very large number of galaxies and galaxy clusters. The shapes of&#xA;  very distant galaxies are distorted by the bending of light as it&#xA;  passes more nearby mass concentrations. These distortions are measured&#xA;  and used to infer the three-dimensional mass distribution in the&#xA;  Universe. This survey will determine both the evolution of dark energy&#xA;  over time as well as provide another independent measurement of the&#xA;  growth of large structure of the universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.syfy.com/syfywire/nasa%E2%80%99s-upcoming-wfirst-telescope-has-legit-superpowers&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dr. Yun Wang's comments&lt;/a&gt; appear to be written for a Science-TV show and/or &lt;a href=&quot;https://www.youtube.com/watch?v=LbJpVHMV1m4&quot; rel=&quot;nofollow noreferrer&quot;&gt;youtube video&lt;/a&gt;, and video like those are designed for a general audience so I wouldn't analyze that particular sentences meaning too closely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for Erik Verlinde's . . . ideas.   I'm not a fan of his proposals, but the answer is mostly the same.   WFIRST will get more accurate readings and better readings should help get more specific answers.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-02-14T08:23:34.950" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25073" PostTypeId="1" AcceptedAnswerId="25079" CreationDate="2018-02-14T08:42:15.350" Score="5" ViewCount="143" Body="&lt;p&gt;At what extent could be sure that no heavier elements than helium could be produced, if we know that matter in universe today is about 25% composed on helium, why could be so sure that the kind of reaction He-He-He diden't took place? &lt;/p&gt;&#xA;" OwnerUserId="21612" LastActivityDate="2018-02-14T10:12:41.913" Title="Why Helium could be produced in first minutes of big bang but not heavier elements" Tags="&lt;big-bang-theory&gt;&lt;matter&gt;&lt;helium&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25075" PostTypeId="2" ParentId="25064" CreationDate="2018-02-14T08:49:27.283" Score="2" Body="&lt;p&gt;Some basics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether we listen for or send signals into space, something to keep in mind is that space is full of EM radiation of all wavelengths, from very long wavelength background radiation, to a range of microwave, infrared, visible and UV light from stars to gamma-rays from accretion disks and collapsing stars.   Any signals we send and perhaps any signals we could receive would be like raindrops in an ocean.   There's an obvious range to listen called &lt;a href=&quot;http://www.setileague.org/askdr/hydrogen.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;the hydrogen line&lt;/a&gt;, which is the emptiest region of EM radiation in space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even so, across light years, signals would still be faint, so the first signal that's sent needs to be an easily recognizable signal and math is very good for that.   Pi in binary or a series of repeating prime numbers wouldn't happen naturally, so signals like that are a great way of saying &quot;We are here&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the first signals of prime numbers or other math are exchanged, and we see what direction the signal comes from and we know where to listen and where to send, then the communication can begin, presumably using more traditional language.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-02-14T08:49:27.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25077" PostTypeId="2" ParentId="25054" CreationDate="2018-02-14T10:02:02.317" Score="16" Body="&lt;p&gt;You don't have to look that far, actually. The tides here on Earth -- ebb and flow -- are &quot;powered&quot; by our Moon's gravity. It's not expressed as heat and volcanic activity, but the mechanisms are the same as on Io.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That energy (which can be, and is, turned into electricity by &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_power&quot; rel=&quot;noreferrer&quot;&gt;tidal power stations&lt;/a&gt;) is not &quot;free&quot;, though.  Earth used to rotate faster (making days shorter than today's 24 hours), and the moon used to be closer (it's still drifting away at ~4 cm per year)...&lt;/p&gt;&#xA;" OwnerUserId="6879" LastEditorUserId="6879" LastEditDate="2018-02-14T20:55:25.623" LastActivityDate="2018-02-14T20:55:25.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25078" PostTypeId="2" ParentId="1051" CreationDate="2018-02-14T10:12:23.050" Score="1" Body="&lt;p&gt;I think the other answer is correct on migration, but there's a flaw in the way this question is asked, which should be addressed.  It's also worth looking at the formation of Jupiter as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the rules of planet formation is that angular momentum remains largely constant.   Granted, some angular momentum gets transferred into heat, and some gets lost for any material that escapes from the system and a tiny amount is lost in the escape of thermal radiation (that's a &lt;a href=&quot;https://www.sciencealert.com/the-sun-is-slowing-down-and-scientists-think-they-finally-know-why&quot; rel=&quot;nofollow noreferrer&quot;&gt;bigger factor with stars&lt;/a&gt; than planets).  But these tiny variations aside, we can generally say that angular momentum of the contained material is conserved and not all of that falls into the planet.  Some remains in orbit around the planet, as moons, a ring or a dust cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars &lt;a href=&quot;https://en.wikipedia.org/wiki/Poynting%E2%80%93Robertson_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;empty out their closer-in orbital regions&lt;/a&gt; fairly quickly.  With planets, that happens much more gradually, so Jupiter likely retained an orbiting nebulous sphere of ice, dust and smaller debris for some time, even after it's moons began forming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Moons_of_Jupiter#Origin_and_evolution&quot; rel=&quot;nofollow noreferrer&quot;&gt;standard model for Jupiter's moons&lt;/a&gt; is that it may have gone through a few generations of moon formation, forming within the orbiting cloud of debris and in time, falling into the planet, while new moons formed and over time, the orbiting disk and gas thinned out.   Based on this model, Io is believed to be part of the latest generation of Jupiter's moons formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Wikipedia link above:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Simulations suggest that, while the disk had a relatively high mass at&#xA;  any given moment, over time a substantial fraction (several tenths of&#xA;  a percent) of the mass of Jupiter captured from the solar nebula was&#xA;  passed through it. However, only 2% the proto-disk mass of Jupiter is&#xA;  required to explain the existing satellites.&lt;a href=&quot;https://en.wikipedia.org/wiki/Moons_of_Jupiter#Origin_and_evolution&quot; rel=&quot;nofollow noreferrer&quot;&gt;3&lt;/a&gt; Thus there may have&#xA;  been several generations of Galilean-mass satellites in Jupiter's&#xA;  early history. Each generation of moons might have spiraled into&#xA;  Jupiter, because of drag from the disk, with new moons then forming&#xA;  from the new debris captured from the solar nebula.&lt;a href=&quot;https://en.wikipedia.org/wiki/Moons_of_Jupiter#Origin_and_evolution&quot; rel=&quot;nofollow noreferrer&quot;&gt;3&lt;/a&gt; By the time the&#xA;  present (possibly fifth) generation formed, the disk had thinned so&#xA;  that it no longer greatly interfered with the moons' orbits.&lt;a href=&quot;http://which%20can%20be%20simplified%20to%20the%20Roche%20limit:%20%20%7B%5Cdisplaystyle%20d=R%5Cleft(2%5C,%7B%5Cfrac%20%7B%5Crho%20_%7BM%7D%7D%7B%5Crho%20_%7Bm%7D%7D%7D%5Cright)%5E%7B%5Cfrac%20%7B1%7D%7B3%7D%7D%5Capprox%201.26R%5Cleft(%7B%5Cfrac%20%7B%5Crho%20_%7BM%7D%7D%7B%5Crho%20_%7Bm%7D%7D%7D%5Cright)%5E%7B%5Cfrac%20%7B1%7D%7B3%7D%7D%7D%20d=R%5Cleft(2%5C,%7B%5Cfrac%20%20%7B%5Crho%20_%7BM%7D%7D%7B%5Crho%20_%7Bm%7D%7D%7D%5Cright)%5E%7B%7B%7B%5Cfrac%20%20%7B1%7D%7B3%7D%7D%7D%7D%5Capprox%201.26R%5Cleft(%7B%5Cfrac%20%20%7B%5Crho%20_%7BM%7D%7D%7B%5Crho%20_%7Bm%7D%7D%7D%5Cright)%5E%7B%7B%7B%5Cfrac%20%20%7B1%7D%7B3%7D%7D%7D%7D.&quot; rel=&quot;nofollow noreferrer&quot;&gt;4&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Jupiter's fast rotation and tidal forces would suggest that it's moons should move away from it similar to our Moon moving away from Earth, but a cloud of orbiting debris tends to slow moon's orbits and cause them to fall into the planet.  Jupiter's powerful magnetic field and rapidly moving charged particles may also have an effect, the combination is too hard for me to say whether Io is moving inwards our outwards, there's too many moving parts and even an estimate of how those forces combine is above my pay-grade.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;But I digress, though I did want to point out that Io is not believed to have formed with Jupiter but formed later.   The question asks how orbiting debris can overcome tidal forces between Jupiter and other larger moons like Ganymede and Callisto. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cloud of orbiting debris in a disk around a planet can coalesce into a moon provided it's outside the &lt;a href=&quot;https://en.wikipedia.org/wiki/Roche_limit#Fluid_satellites&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;fluid&lt;/strong&gt; Roche limit&lt;/a&gt;.  A solid moon begins to break apart usually closer to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Roche_limit#Rigid-satellite_calculation&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;rigid&lt;/strong&gt; Roche limit&lt;/a&gt;, closer to the planet, due to some structural integrity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For moon formation, all that's required is that there be sufficient debris density and that the debris is outside the fluid Roche limit. It doesn't matter that the density of the orbiting ring is low, what matters is that once the coalescing starts, that the proto-moon is outside the Roche limit, &lt;a href=&quot;http://which%20can%20be%20simplified%20to%20the%20Roche%20limit:%20%20%7B%5Cdisplaystyle%20d=R%5Cleft(2%5C,%7B%5Cfrac%20%7B%5Crho%20_%7BM%7D%7D%7B%5Crho%20_%7Bm%7D%7D%7D%5Cright)%5E%7B%5Cfrac%20%7B1%7D%7B3%7D%7D%5Capprox%201.26R%5Cleft(%7B%5Cfrac%20%7B%5Crho%20_%7BM%7D%7D%7B%5Crho%20_%7Bm%7D%7D%7D%5Cright)%5E%7B%5Cfrac%20%7B1%7D%7B3%7D%7D%7D%20d=R%5Cleft(2%5C,%7B%5Cfrac%20%20%7B%5Crho%20_%7BM%7D%7D%7B%5Crho%20_%7Bm%7D%7D%7D%5Cright)%5E%7B%7B%7B%5Cfrac%20%20%7B1%7D%7B3%7D%7D%7D%7D%5Capprox%201.26R%5Cleft(%7B%5Cfrac%20%20%7B%5Crho%20_%7BM%7D%7D%7B%5Crho%20_%7Bm%7D%7D%7D%5Cright)%5E%7B%7B%7B%5Cfrac%20%20%7B1%7D%7B3%7D%7D%7D%7D.&quot; rel=&quot;nofollow noreferrer&quot;&gt;it's the Moon's density, not it's size that determines that Roche limit&lt;/a&gt; relative to the planet it orbits.   A moon information might initialy have a lower density, due to being less compact, so it may have a corresponding Roche limit that's more distant from the planet, but the variation is the cube root of the density, so the roche limit boundary wouldn't be all that much further out at the start of formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proto-moon doesn't need to add the ring-debris all at once, it only needs to be able to hold onto what it gets very close to, and that's a product of being outside the fluid Roche limit.   In time, the Moon clears out the region where it orbits, and as pointed out in the other answer, migration likely plays a role in moon formation, but migration isn't the reason Moons form, that's a product of sufficient density of the orbiting disk and gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(hope that makes sense, I'm not sure I explained the last part as well as I could have).&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-02-14T10:12:23.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25079" PostTypeId="2" ParentId="25073" CreationDate="2018-02-14T10:12:41.913" Score="3" Body="&lt;p&gt;The triple alpha process requires very high densities, very high (but not too high) temperatures, and a good amount of time. Large stars have all three. In particular, they have millions of years during which the very rare triple alpha process can proceed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperatures in the very early universe were extremely high, too high to create anything more complex than protium. By the time the universe had cooled enough so that it could produce deuterium (and thence helium), its density and temperature had already dropped too low for the triple alpha process to proceed. All of this happened in a few hundred seconds. Compare that with the millions of years during which large stars burn helium.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2018-02-14T10:12:41.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25081" PostTypeId="2" ParentId="25054" CreationDate="2018-02-14T10:26:26.880" Score="3" Body="&lt;p&gt;A worldbuilding answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of &quot;Magic Energy&quot; (I hate the word magic, lets say free energy) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say we have a colony far from the Sun.  Jupiter works, or, an outer planet, maybe even the theoretical Planet 9, where solar-energy panels would be pretty pointless, Uranium fuel might be scarce and 3He takes time to collect for fusion, so they need a power source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory, a devise could be built using material and mass asteroids and carbon fiber cables, and some kind of structure where there was a spherical outside that stretches and squishes in an orbit, ideally, like Io, between the planet and it's large moon where some eccentricity and variation would be maintained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stretching back and forth and the mass of the outer shell of this object could pull pistons and make electricity.   The amount of electricity would depend on the mass and size and flexibility of the spherical construct and the energy would come from a nearly limitless source - the planet's rotational energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether such a thing is practical is another question.  I have no idea if it would be practical at all, but it's theoretically possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'd also need a large moon, otherwise the orbit would circularize and energy would diminish.  But a Planet-Large Moon system, maybe it could work.  The downside is that so much energy would go into building the thing in the first place, I'm not sure it would be worth it.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-02-14T21:59:34.277" LastActivityDate="2018-02-14T21:59:34.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25083" PostTypeId="2" ParentId="25061" CreationDate="2018-02-14T10:36:12.360" Score="11" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How can tidal heating lower Io's orbit?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It doesn't, at least not to first order. The first order effect is that tidal heating acts to circularize Io's orbit. Counter to that, orbital resonances with Europa and Ganymede act to make Io's orbit more elliptical. This leads to a nice hysteresis loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose Io is in a fairly circular orbit. This results in reduced tidal stresses, thereby making Io cool down. A cooler and hence more rigid Io is less susceptible to tidal deformations than is a warmer and hence more plastic Io. Given two bodies in the same orbit, one warm and plastic, the other cool and rigid, the warmer body will suffer more tidal deformations than will the cooler one. This is captured by the object's $k_2$ Love number. The inevitable lag in response means that the response will not be symmetric about periapsis/apoapsis for an elliptical orbit, and the greater the plasticity, the greater the greater asymmetry. This is captured by the object's tidal quality factor $Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This cooling of Io as its orbit becomes close to circular enables the ever-present resonance effects to now dominate over the circularizing effects. Io's orbit slowly becomes more elliptical. That elliptical orbit increases tidal stresses on the cold, rigid Io, eventually making it start to warm up and become more plastic. The circularization effects grow as the orbit grows more elliptical and as Io's interior becomes more flexible and more plastic. Eventually the circularization effects dominate over the orbital resonance effect, making Io's orbit become more circular -- until the cycle repeats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rinse and repeat, at least as long as that three-way orbital resonance between Io, Europa, and Ganymede holds up. How long that three-way tidal resonance has existed and how long it will last is, as far as I know, unknown.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2018-02-14T13:49:34.597" LastActivityDate="2018-02-14T13:49:34.597" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="25086" PostTypeId="2" ParentId="25028" CreationDate="2018-02-14T11:34:55.203" Score="1" Body="&lt;p&gt;Yes, exactly. The angular diameter distance is defined such that it is compatible with &quot;regular&quot; trigonometric rules, just as the luminosity distance is compatible with regular relations between flux, distance, and luminosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only for at flat and non-expanding universe would these distances coincide with each other, and with the &quot;true&quot; distance, which is what you would measure if you lay out meter sticks (after freezing the Universe in time, so it doesn't expand more before you're done).&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2018-02-14T11:34:55.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25088" PostTypeId="2" ParentId="25054" CreationDate="2018-02-14T11:55:47.300" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is Io a magic energy machine?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, it's not. The source of Io's energy is Jupiter's rotation. Think of the our Moon. It is slowly spiraling outward away from the Earth. The Moon is not a magic energy machine. It steals angular momentum and orbital energy from the Earth's rotation. While the transfer of angular momentum from the Earth to the Moon is very close to even, the transfer of energy is not. The Earth-Moon system is dissipative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same goes for Io and its volcanos. Io is very slowly stealing angular momentum and energy from Jupiter's rotation, and in the process, vents  some of that stolen energy into space.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2018-02-14T11:55:47.300" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25089" PostTypeId="2" ParentId="25065" CreationDate="2018-02-14T12:23:43.487" Score="4" Body="&lt;p&gt;Yes, it is absolutely silly to predict an atmospheric/surface temperature without knowing it's composition or density. And we don't understand the formation process of atmospheres well enough to exclude any type of atmospheres around different stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is however one more or less objective quantity that you can calculate: &lt;a href=&quot;https://en.wikipedia.org/wiki/Planetary_equilibrium_temperature&quot; rel=&quot;nofollow noreferrer&quot;&gt;The equilibrium temperature&lt;/a&gt;. That is the temperature that the planet would have without an atmosphere.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is essentially given by the planetary distance (which you have) and the stellar irradiation (which you have still to find), and the planetary albedo (a free parameter between 0 and 1).&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="6" LastEditDate="2018-12-17T13:37:17.943" LastActivityDate="2018-12-17T13:37:17.943" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25091" PostTypeId="2" ParentId="25064" CreationDate="2018-02-14T14:50:07.233" Score="1" Body="&lt;p&gt;The general plan is to start very simply and build up - would you recognise the following?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.  .. ... .... .....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numbers, right? You'd probably want to save typing out lots of dots in the long run, so you'd want to convert it to a &quot;base-something&quot; system, e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0001 0010 0011 0100 0101&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conjunction with the original, it should be obvious that this is binary (the Aricebo message, if I remember correctly, used a form of base 10).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can start describing sequences of numbers that you think the aliens should understand - primes, Pi, or as Aricebo did, the atomic numbers of H, O, C, N...the elements that primarily form life on Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as the aliens have a concept of numbers, and a vaguely similar idea of mathematical logic to us, you can build up to most mathematical ideas in one way or another. Integrals, for example, could be done by sending a sequence corresponding to y=x^2, then defining two points in particular, and providing the resultant integration result. A civilization capable of picking up our signals &lt;em&gt;should&lt;/em&gt; (I make no assumptions about aliens!) recognise the concept.&lt;/p&gt;&#xA;" OwnerUserId="11884" LastActivityDate="2018-02-14T14:50:07.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25092" PostTypeId="1" AcceptedAnswerId="25097" CreationDate="2018-02-14T19:23:08.470" Score="2" ViewCount="75" Body="&lt;p&gt;I have a question: assume two identical stars, but one is 10 times farther than the other from the earth. If the nearest star has a relative magnitude of 5, what is the relative magnitude of the other?&lt;/p&gt;&#xA;" OwnerUserId="21241" LastEditorUserId="7472" LastEditDate="2018-02-16T06:37:53.503" LastActivityDate="2018-02-16T06:37:53.503" Title="Stars magnitude" Tags="&lt;magnitude&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25093" PostTypeId="1" AcceptedAnswerId="25096" CreationDate="2018-02-14T21:14:05.513" Score="1" ViewCount="155" Body="&lt;h1&gt;Description&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Lots of people have thought to of seen UFOs &lt;em&gt;(Unidentified Flying Objects)&lt;/em&gt;, in fact, &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_reported_UFO_sightings&quot; rel=&quot;nofollow noreferrer&quot;&gt;more than 83,000&lt;/a&gt; have been reported &lt;strong&gt;this year alone.&lt;/strong&gt; &lt;em&gt;(and it's only February!)&lt;/em&gt; But most of the time it's just a shooting star or an airplane. But some of the sightings don't seem to be any of the things a UFO is commonly mistook for, but there isn't any evidence supporting if it is really a UFO. I might not be alone on this theory, but I believe the sightings that potentially could be UFOs are messages from the cosmos.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There may not be much to back up my theory, but I just wish to learn some background knowledge:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Are UFOs &lt;em&gt;(and I guess all the UFO sightings reported, too)&lt;/em&gt; messages from another planet?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I also believe that if they are messages, they could possibly be responses to &lt;a href=&quot;https://voyager.jpl.nasa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA's Voyager spacecrafts&lt;/a&gt;, for their purposes were to inform any forms of life out in space what earth is like with the 'Sounds of Earth' music disk attached to it.&lt;/p&gt;&#xA;" OwnerUserId="21239" LastActivityDate="2018-02-14T21:21:39.753" Title="Is there a chance that UFOs could be messages from other planets?" Tags="&lt;exoplanet&gt;&lt;space-probe&gt;&lt;identify-this-object&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2018-02-15T18:30:21.113" ContentLicense="CC BY-SA 3.0" />
  <row Id="25094" PostTypeId="5" CreationDate="2018-02-14T21:19:50.810" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2018-02-14T21:19:50.810" LastActivityDate="2018-02-14T21:19:50.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25095" PostTypeId="4" CreationDate="2018-02-14T21:19:50.810" Score="0" Body="Questions asking about Unidentified Flying Objects (UFOs), which may be astronomical phenomena" OwnerUserId="21239" LastEditorUserId="6" LastEditDate="2018-02-15T18:14:53.110" LastActivityDate="2018-02-15T18:14:53.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25096" PostTypeId="2" ParentId="25093" CreationDate="2018-02-14T21:21:39.753" Score="10" Body="&lt;p&gt;&lt;strong&gt;No.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of sending a message is to have someone else read it. It defies logic to send a message over light years of space and then have it do anything but broadcast its presence very clearly to anyone who might possibly want to investigate it. If the goal of these UFOs was to tell us something, they wouldn't be such elusive specimens that only appear to specific people at specific times and places, and then disappear without a trace. If UFOs are in fact meant to communicate with humanity, they are indeed rather poorly designed, because they haven't accomplished that goal very well at all!&lt;/p&gt;&#xA;" OwnerUserId="10488" LastActivityDate="2018-02-14T21:21:39.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25097" PostTypeId="2" ParentId="25092" CreationDate="2018-02-14T22:47:48.360" Score="6" Body="&lt;p&gt;As @RobJeffries &lt;a href=&quot;https://astronomy.stackexchange.com/questions/25092/stars-magnitude#comment42948_25092&quot;&gt;mentions&lt;/a&gt;, you can calculate the difference from the equations which define magnitude and flux. But this particular case is even covered in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_magnitude#Standard_reference_values&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article on apparent magnitude&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For objects within the Milky Way with a given absolute magnitude, 5 is added to the apparent magnitude for every tenfold increase in the distance to the object.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So if the nearest star has magnitude 5, the farthest one has magnitude 5 + 5 = 10.&lt;/p&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2018-02-14T22:47:48.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25098" PostTypeId="1" CreationDate="2018-02-14T23:10:48.480" Score="1" ViewCount="31" Body="&lt;p&gt;Does the expansion of space only occur in the space between celestial objects or is it effecting the dimensional volume of the objects themselves? In simple terms is my ruler expanding due to the expansion of the universe?&lt;/p&gt;&#xA;" OwnerUserId="21145" LastEditorUserId="21145" LastEditDate="2018-02-14T23:31:30.907" LastActivityDate="2018-02-14T23:31:30.907" Title="Does the expansion of space imply an increase in the dimensional volume of objects?" Tags="&lt;expansion&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="2" ClosedDate="2018-02-15T18:15:55.850" ContentLicense="CC BY-SA 3.0" />
  <row Id="25101" PostTypeId="1" AcceptedAnswerId="25104" CreationDate="2018-02-15T06:20:19.643" Score="1" ViewCount="434" Body="&lt;p&gt;I just derived the pressure and temperature profile for a linear density model of a star. If I compare the profiles of pressure and temperature for a constant density model of stars, I see that the linear density model central pressure is higher, but the central temperature is lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the physical reason behind this?&lt;/p&gt;&#xA;" OwnerUserId="21282" LastEditorUserId="9301" LastEditDate="2018-02-15T18:09:52.867" LastActivityDate="2018-02-15T18:09:52.867" Title="Central temperature and pressure for constant density vs linear density model of stars" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25104" PostTypeId="2" ParentId="25101" CreationDate="2018-02-15T07:48:04.480" Score="1" Body="&lt;p&gt;Because the total mass of the star has to be conserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider a linear density model, for a star of fixed mass and radius, &#xA;you have to make the central density higher than the average density in a constant density model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, because the mass of each shell goes as $\rho r^2$, there needs to more shells with a. density higher than average. This means the average density of a mass shell is higher than in the constant density model and more of the mass is at smaller radii.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrostatic equilibrium demands that the pressure gradient is proportional to $-\rho M(r)/r^2$, where $M(r)$ is the mass inside radius $r$. At any radius, this will be larger on average in the linear density model because more of the mass is at smaller radii.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the pressure gradient is higher, then the central pressure will be higher. The value of the central temperature then just comes from gas laws.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-02-15T07:48:04.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25106" PostTypeId="1" CreationDate="2018-02-15T08:08:34.893" Score="0" ViewCount="343" Body="&lt;p&gt;It's certain that human life on Earth will not go forever. Therefore there will be one last generation of humans on Earth. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How will life look like for last few generations approaching the end?&lt;/li&gt;&#xA;&lt;li&gt;How will life end for the very last people, will they all die at once as if struck?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Assume the technology won't be much more advanced than today, no colonized planets, and we'd stay biologically very similar to now. Also assume that no meteor will struck us and end everything but that extinction will happen due to aging of the Earth and Sun.&lt;/p&gt;&#xA;" OwnerUserId="692" LastActivityDate="2018-02-15T09:25:15.790" Title="What will the last day of humanity look like?" Tags="&lt;earth&gt;" AnswerCount="1" CommentCount="7" ClosedDate="2018-02-15T14:24:05.113" ContentLicense="CC BY-SA 3.0" />
  <row Id="25108" PostTypeId="2" ParentId="25106" CreationDate="2018-02-15T09:25:15.790" Score="3" Body="&lt;p&gt;The last humans in this scenario likely die of starvation and/or drought during a natural short-term climate fluctuation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic issue is that as the sun heats up there will be a gradual reduction of carbon dioxide in the atmosphere (due to plant life but also silicate weathering, which speeds up in warm environments). There may also be a reduction in volcanic outgassing, although I think the timeframe is too slow. This is bad for plants, since they will get less CO$_2$ to photosynthesise from, and it will reduce the climate stabilising feedback so temperatures will start going up as the sun increases in luminosity. Eventually we get to a runaway greenhouse stage where water vapour alone triggers overheating, but this is possibly long after the biosphere is mostly dead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most complete model of this I have seen is &lt;a href=&quot;https://hal.archives-ouvertes.fr/file/index/docid/297542/filename/bg-3-85-2006.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt; by Franck, Bounama, &amp;amp; von Bloh. They estimate complex multicellular life to go extinct in 0.8–1.2 Gyr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this will happen long before the sun becomes even a subgiant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, if this model is correct we should expect that slowly but surely the future humans would be forced to migrate to cool regions and would be especially limited by the dwindling biomass available. But beside this they will be subject to normal random climate fluctuations that sometimes create decade-long droughts or other misfortunes, and it is likely one of those that actually gets the last individuals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the scenario assumed makes some &lt;em&gt;very&lt;/em&gt; strong assumptions. In order to remain recognisably humans over millions of years these humans would have to engage in deliberate genetic control, otherwise genetic drift will make them mutate into new species. It assumes that nobody does anything about the long-term trend. And so on. &lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2018-02-15T09:25:15.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25111" PostTypeId="2" ParentId="6188" CreationDate="2018-02-15T13:03:03.383" Score="1" Body="&lt;p&gt;Geostationary orbit requires:&#xA;* A precise distance between the bodies, giving a one-day orbital period.&#xA;* An equatorial orbit, so that the satellite is always above the same latitude (if not, it is called a geosynchronous orbit)&#xA;* A circular orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just getting one of these parameters exactly right by chance alone is extremely unlikely. If we found a satellite where all three are spot on, we would probably have to start considering the possibility that it was put there by some alien civilization.&lt;/p&gt;&#xA;" OwnerUserId="21752" LastActivityDate="2018-02-15T13:03:03.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25114" PostTypeId="1" CreationDate="2018-02-15T18:11:36.487" Score="3" ViewCount="377" Body="&lt;p&gt;I am reading about massive stars and their final consequences. what I understood is massive stars after throwing out their outer envelop with masses &gt; 1.4 solar Mass turn into neutron stars. But I also know that Massive stars can collapse to form a type II supernova. I need to understand how we determine which stars go through which final fate. And what initial mass they have to start their life with for both fates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ma be completely wrong with my understandings. Your clarifications will help a lot. &lt;/p&gt;&#xA;" OwnerUserId="21282" LastActivityDate="2019-09-05T13:10:08.697" Title="Core collapse supernova vs Neutron Star" Tags="&lt;star&gt;&lt;supernova&gt;&lt;neutron-star&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25115" PostTypeId="1" CreationDate="2018-02-15T22:07:08.507" Score="5" ViewCount="223" Body="&lt;p&gt;I read about &lt;a href=&quot;https://phys.org/news/2014-06-gravitational-fields-black-holes-eddy.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;natural gravitational eddies&lt;/a&gt; that travel in a wave that black holes have. They also have a strong magnetic field. Does those eddies follow magnetic field lines of the rotating black hole?  Can a galaxies have their own gravitational eddies like a black hole does?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wslp8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wslp8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-12-18T00:27:46.587" LastActivityDate="2018-12-18T00:27:46.587" Title="Gravitational eddies across the galaxy?" Tags="&lt;galaxy&gt;&lt;gravity&gt;&lt;space-time&gt;&lt;supermassive-black-hole&gt;&lt;time-dilation&gt;" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25116" PostTypeId="1" AcceptedAnswerId="25123" CreationDate="2018-02-16T03:09:27.750" Score="1" ViewCount="206" Body="&lt;p&gt;I was asked this question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How much weaker is the gravitational pull of the Sun on Neptune at 30 AU, than it is on Earth at 1 AU?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How would I even begin to determine this?&lt;/p&gt;&#xA;" OwnerUserId="21762" LastActivityDate="2018-02-22T09:18:39.997" Title="How would you relate the respective solar gravitational pulls of two planets?" Tags="&lt;planet&gt;&lt;gravity&gt;&lt;planetary-science&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25117" PostTypeId="1" AcceptedAnswerId="25151" CreationDate="2018-02-16T05:57:51.977" Score="5" ViewCount="475" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vTU2C.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vTU2C.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; (see small red lines near center of image) Tesla Roadster. 18 Feb. 2018, 11:28 UT. The image above comes from the average of ten, 300-seconds exposures, remotely taken with the 16”-f/3.75 robotic unit (“Pearl”) of Tenagra Observatories in Arizona. The Tesla Roadster is marked with two red lines. At the imaging time, the car was about 3.5 millions of km from the Earth, that is 9 lunar distances. The object was at mag. 21.0. Image by Gianluca Masi (The Virtual Telescope Project) and Michael Schwartz (Tenagra Observatories, Ltd). From &lt;a href=&quot;https://www.virtualtelescope.eu/2018/02/19/elon-musks-tesla-roadster-imaged-18-feb-2018/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Virtual Telescope Project&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; Zoomed in and cropped showing Roadster's image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VjuI2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VjuI2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;https://astronomy.stackexchange.com/a/16305/7982&quot;&gt;excellent answer&lt;/a&gt; from @MikeG shows exactly how to find a list of observations for a minor planet. But I'd like to ask about a list of observations of an artificial object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently the NASA JPL's &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;Horizons&lt;/a&gt; site is also providing predicted trajectory of an unusual spacecraft which is referred to as Roadster. It is the cherry-red Tesla Roadster attached to a roughly 3.6 x 8 meter white rocket body in heliocentric orbit, and it has been tracked by many different observatories because of its unique circumstances. (some GIFs in &lt;a href=&quot;https://astronomy.stackexchange.com/a/25002/7982&quot;&gt;this answer&lt;/a&gt;) Roadster's 4.6 minute rotational period was also recently &lt;a href=&quot;https://twitter.com/jotajotahermes/status/962545252446932993&quot; rel=&quot;nofollow noreferrer&quot;&gt;reported&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently Horizons is reporting solution #7. They have been updating their solutions regularly as new observations are made and reported because as Roadster recedes from Earth it is rapidly dimming and will soon pass magnitude +20, so each new solution helps for the reacquisition of the object as it dims.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This trajectory is based on JPL solution #7, a fit to 269 ground-based optical astrometric measurements spanning 2018 Feb 8.2 to 13.4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Would these observations still be catalogued in an astronomical database of observations? Since it's not a natural minor planet, it seems the &lt;a href=&quot;https://www.minorplanetcenter.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Minor Planet Center&lt;/a&gt; would not be hosting this information, but it seems it should still be made available to astronomers for identification purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But where?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/60dit.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/60dit.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The header of the Horizons output includes this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;*******************************************************************************&#xA;Revised: Feb 13, 2018          Tesla Roadster (spacecraft)             -143205&#xA;          (solution #7)&#xA;&#xA; Tesla Roadster (AKA: Starman, 2018-017A)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;TRAJECTORY:&#xA;  This trajectory is based on JPL solution #7, a fit to 269 ground-based &#xA;  optical astrometric measurements spanning 2018 Feb 8.2 to 13.4.&#xA;&#xA;  Trajectory name                       Start (TDB)          Stop (TDB)&#xA;  --------------------------------   -----------------   -----------------&#xA;  tesla_s7                           2018-Feb-07 03:00   2048-Jan-01 00:00&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-02-27T08:55:10.617" LastActivityDate="2018-02-27T08:55:10.617" Title="Where might astrometric observations of the Roadster spacecraft be catalogued?" Tags="&lt;observational-astronomy&gt;&lt;asteroids&gt;&lt;astrometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25118" PostTypeId="2" ParentId="25066" CreationDate="2018-02-16T06:24:55.500" Score="2" Body="&lt;p&gt;The ultimate resolution of a telescope depends upon its diameter. The larger the telescope aperture, the finer the resolution.&#xA;Hubble and WFIRST are both the same diameter 2.4m so they both have the same theoretical resolution.&lt;/p&gt;&#xA;&lt;p&gt;However the image quality on any telescope degrades the further you are away from the optical axis. In WFIRST there is a 3rd mirror that corrects for these inherent distortions for a much wider field of view.&lt;/p&gt;&#xA;&lt;p&gt;The last factor is the improvement in detector technology. The individual pixels are smaller, smaller gaps, and there chips are much larger. Also WFIRST has mosaic of these  CCD chips to cover the larger corrected field of view.&lt;/p&gt;&#xA;&lt;p&gt;The paper &lt;a href=&quot;https://wfirst.gsfc.nasa.gov/science/sdt_public/spie/929305.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Optical Design of the WFIRST-AFTA Wide-Field Instrument&lt;/a&gt; (Pasquale et al. 2014, Proc. SPIE-OSA Vol. 9293, 929305) discusses the design of the optical path (cycle #4). In addition to a small figure adjustment to the primary, the figures of the secondary and a tertiary mirror are optimized to provide high resolution over a much wider field than the HST. This design also provides a pupil at which filters can be located as well as an intermediate focus (1:1 with the final focus) as shown below.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The basic optical design is decidedly simple for fabrication, integration and test purposes. This approach begins with the base optical design; all three powered mirrors are optically co-axial and simple conics. We considered more complicated designs, including the use of tilted and de-centered mirrors and Zernike and/or anamorphic surface figures.[2,3] TMA designs are widely preferred for high A*Omega telescopes such as JWST and ATLAST.&lt;/p&gt;&#xA;&lt;p&gt;The primary mirror, referred to here as Telescope mirror 1 (T1), is a fast f/1.2 primary with a linear central obscuration of 31%. It is a light-weighted mirror using a hollow honeycomb core. T1 and Telescope mirror 2 (T2) together form an ~f/8, intermediate focus. This focus is uncorrected, with a large caustic image and strongly curved image surface. You will note the off-axis field bias (seen in Figure 1); this is typical in all TMA systems.&lt;/p&gt;&#xA;&lt;p&gt;Within the instrument, Mirror 3 (M3) is almost a 1:1 magnification relay of the intermediate image to the focal plane. Working in concert with T1 &amp;amp; T2, these 3 powered mirrors form the large corrected field of the TMA. A pupil is formed between M3 and the focal plane. With a diameter of approximately 100mm, the pupil allows for the insertion of bandpass filters and spectral dispersion elements via an element wheel. Finally, in order to package the system into the volume constraints, two fold mirrors were used.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OtNaT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OtNaT.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/na1dj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/na1dj.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10761" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-02-18T23:42:38.207" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25119" PostTypeId="1" CreationDate="2018-02-16T07:03:45.450" Score="4" ViewCount="352" Body="&lt;p&gt;Question in the title. Currently, Mars is a small orange dot in the sky. But what if green, blue, and some white were added? I wouldn't expect it to be too different- a large portion of it will remain pretty much the way it was, but with more snow (Tharsis bulge is REALLY high up).&lt;/p&gt;&#xA;" OwnerUserId="19304" LastActivityDate="2018-02-16T21:59:42.423" Title="What would a terraformed Mars look like from Earth?" Tags="&lt;solar-system&gt;&lt;light&gt;&lt;mars&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25122" PostTypeId="2" ParentId="25119" CreationDate="2018-02-16T08:33:45.657" Score="10" Body="&lt;p&gt;To terraform Mars, broadly speaking would make it more like an analogy of Earth. As you say, it might be a little colder, but we would expect a terraformed Mars to have oceans, snow and maybe some imported plant life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would Mars look from Earth? A good starting point would be to see it the opposite way around. &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomy_on_Mars#Astronomical_phenomena&quot; rel=&quot;noreferrer&quot;&gt;Here is how Earth looks from Mars&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like Earth from a distance, a terraformed Mars would likely appear as a pale blue or white dot. If you want more snow or ice than water, it would surely lean towards white rather than a deep blue.&lt;/p&gt;&#xA;" OwnerDisplayName="user10106" LastEditorDisplayName="user10106" LastEditDate="2018-02-16T09:05:39.423" LastActivityDate="2018-02-16T09:05:39.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25123" PostTypeId="2" ParentId="25116" CreationDate="2018-02-16T09:17:02.730" Score="3" Body="&lt;p&gt;For the most part, you can assume &lt;a href=&quot;https://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Newton's law of universal gravitation&lt;/a&gt;, (see link for more information). Basically, this law can be stated as the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the gravitational force between two celestial objects (or really, any two objects with mass), the force is given by the equation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F_G = G\frac{m_1 m_2}{r^2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the mass of the two objects, $m_1$, $m_2$, and the distance between them, $r$. The constant of proportionally given by $G$, the gravitational constant (not to be confused with $g$, the acceleration of an object with respect to the conditions on the surface on the Earth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the specific problem(s) that you are given, you can say that, using the information from Wikipedia, for the mass of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Sun&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sun&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth&lt;/a&gt;, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Neptune&quot; rel=&quot;nofollow noreferrer&quot;&gt;Neptune&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ m_\text{Sun} = 1.988 \times 10^{30} \text{ kg} $$&#xA;$$ m_\text{Earth} = 5.972 \times 10^{24} \text{ kg} $$&#xA;$$ m_\text{Neptune} = 1.024 \times 10^{26} \text{ kg} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question gives us a preliminary distance. The $r$ for the distance between the Sun and the Earth is given as $1 \text{ AU}$, the $r$ for the distance between the Sun and Neptune given as $30 \text{ AU}$. Note that $1 \text{ AU} = 1.496 \times 10^{11} \text{ m}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, for the gravitational pull (really another way of saying the gravitational force) between the Sun and Neptune, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{N \leftrightarrow S} = \left( 6.674×10^{-11} \text{ N m$^2$ kg$^{-2}$} \right) \frac{\left( 1.024 \times 10^{26} \text{ kg} \right) \left( 1.988 \times 10^{30} \text{ kg} \right)}{\left( 5.983 \times 10^{12} \text{ m} \right)^2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, for the gravitational pull between the Sun and the Earth, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{E \leftrightarrow S} = \left( 6.674×10^{-11} \text{ N m$^2$ kg$^{-2}$} \right) \frac{\left( 5.972 \times 10^{24} \text{ kg} \right) \left( 1.988 \times 10^{30} \text{ kg} \right)}{\left(1.496 \times 10^{11} \text{ m}\right)^2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for these quantities (note that the answer that one might get may be numerically different depending on the sources of the planatary constants),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{N \leftrightarrow S} = 3.796 \times 10^{20} \text{ N}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{E \leftrightarrow S} = 3.541 \times 10^{22} \text{ N}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit of extra information, the gravitational force felt by the Earth or Neptune from the Sun is exactly the same as the force that the Earth or Neptune acts upon the Sun, as per Newton's third law of motion. But, by now, I suspect that the problem is answered to where taking the ratio of the two is a trivial manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suggested by uhoh:&#xA;The force exerted by the Sun on Neptune (and vice versa) is about 93.3 times weaker than that of the force exerted by the Sun on the Earth(and vice versa).&lt;/p&gt;&#xA;" OwnerUserId="20675" LastEditorUserId="20675" LastEditDate="2018-02-22T09:18:39.997" LastActivityDate="2018-02-22T09:18:39.997" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25124" PostTypeId="5" CreationDate="2018-02-16T09:46:01.717" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2018-02-16T09:46:01.717" LastActivityDate="2018-02-16T09:46:01.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25125" PostTypeId="4" CreationDate="2018-02-16T09:46:01.717" Score="0" Body="For questions about the various phases on the Moon as observed from the Earth, also known as waxing and waning." OwnerUserId="7472" LastEditorUserId="7472" LastEditDate="2018-02-21T20:37:53.417" LastActivityDate="2018-02-21T20:37:53.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25126" PostTypeId="1" CreationDate="2018-02-16T11:15:42.077" Score="5" ViewCount="2924" Body="&lt;p&gt;How can I calculate the (non-bolometric) luminosity $L$ of a galaxy (or a star for that matter) over a given band from its AB apparent magnitude $m_{AB}$ over that band and its luminosity distance $d_L$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, consider the g-band which typically has a $\lambda_{eff} = 467 \text{ nm}$ and a $\Delta \lambda = 100 \text{ nm}$. Given this galaxy has an apparent AB magnitude of $m_g = 22.5$ and luminosity distance of $1991 \text{ Mpc}$ (i.e. $z = 0.355$ if you are curious), what is its luminosity? I know I shold use the following equation, but I don't know what value to pick for $M_\odot$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L/L_\odot = 10^{0.4(M_\odot - M)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried 5.12 from &lt;a href=&quot;http://www.astronomy.ohio-state.edu/~martini/usefuldata.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;a website&lt;/a&gt; which is the value of $M_\odot$ in g-band but it gives me a different answer compared to when I calculate luminosity using flux:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L = 4 \pi d_{L}^2 f_\nu \Delta \nu$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f_\nu$ is flux density and $\Delta \nu$ is the frequency width of the band (this is an approximation to the integration over frequency, assuming bands are step functions). So, how can I find luminosity using absolute magnitude?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please provide your references.&lt;/p&gt;&#xA;" OwnerUserId="20270" LastEditorUserId="20270" LastEditDate="2018-02-21T01:08:34.407" LastActivityDate="2018-02-25T09:04:43.653" Title="How to calculate luminosity in g-band from absolute AB magnitude and luminosity distance?" Tags="&lt;luminosity&gt;&lt;instruments&gt;&lt;absolute-magnitude&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25130" PostTypeId="2" ParentId="18387" CreationDate="2018-02-16T16:09:11.680" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://3753%20Cruithne&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cruithne&lt;/a&gt; and &lt;a href=&quot;http://news.bbc.co.uk/2/hi/science/nature/2251386.stm&quot; rel=&quot;nofollow noreferrer&quot;&gt;J002E3&lt;/a&gt; were detected using ground based astrophotography.&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/J002E3&quot; rel=&quot;nofollow noreferrer&quot;&gt;J002E3&lt;/a&gt; is thought to be &quot;the S-IVB third stage of the Apollo 12 Saturn V rocket ... It is thought that J002E3 left Earth orbit in June 2003, and that it may return to orbit the Earth in the mid-2040s.&lt;a href=&quot;http://3753%20Cruithne&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;&quot;&#xA;Ground telescope also found &lt;a href=&quot;https://en.wikipedia.org/wiki/6Q0B44E&quot; rel=&quot;nofollow noreferrer&quot;&gt;6Q0B44E&lt;/a&gt; which orbits beyond the moon.&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/2002_AA29&quot; rel=&quot;nofollow noreferrer&quot;&gt;2002 AA29&lt;/a&gt; was found via automatic sky survey.&#xA;Several of these objects are in &lt;a href=&quot;https://en.wikipedia.org/wiki/Horseshoe_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;horseshoe orbits&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless, the discovery of these objects usually comes via conventional astrophotography.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2018-02-16T16:09:11.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25131" PostTypeId="2" ParentId="25119" CreationDate="2018-02-16T16:52:00.897" Score="5" Body="&lt;p&gt;Mars currently has albedo 0.250, 81.7% of Earth's somewhat brighter 0.3. A terraformed Mars would presumably have a similar albedo. That would make it slightly brighter and bluer, about 0.2 ($2.5\log_{10}(0.3/0.25)$) magnitudes brighter. But this is not going to change the overall magnitude ($\approx -3$ at most) much. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various aspects of the terraforming assumptions may nudge this up (more ice, extra light from solar mirrors) or down (less clouds? deliberate ice darkening?)&lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2018-02-16T16:52:00.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25133" PostTypeId="1" AcceptedAnswerId="28785" CreationDate="2018-02-16T20:53:57.340" Score="0" ViewCount="209" Body="&lt;p&gt;Ship 1 is out side the black hole. Ship 2 is nearing a black hole. To ship 1 it would appear that ship 2 would be moving slower to the point it would seem like ship 2 stopped moving the closer to the black hole ship 2 gets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine that 2 naked black hole(an infinity dense point with out a surrounding black hole) merging would look like a blur from the speed of them, but and because of the time dilatation is this gif picture below a good rendering of how the merger would appear to ship 1? Then how fast does a black hole move when merging??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dxMmz.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dxMmz.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-02-18T00:12:20.663" LastActivityDate="2018-12-18T14:40:51.267" Title="What would a black hole merger look like?" Tags="&lt;black-hole&gt;&lt;speed&gt;&lt;time-dilation&gt;&lt;visualization&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25134" PostTypeId="2" ParentId="25119" CreationDate="2018-02-16T21:59:42.423" Score="0" Body="&lt;p&gt;I pulled these pictures from Google. I hope it is enough to answer with pictures?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1lwta.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1lwta.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://william-black.deviantart.com/art/Terraforming-Mars-392449850&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://william-black.deviantart.com/art/Terraforming-Mars-392449850&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Terraforming_of_Mars&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Terraforming_of_Mars&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/oAMjB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oAMjB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastActivityDate="2018-02-16T21:59:42.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25135" PostTypeId="1" CreationDate="2018-02-16T22:37:24.203" Score="5" ViewCount="309" Body="&lt;p&gt;I teach statistics and study Astronomy.  I'd like to create an activity for my statistics students using star data.  They could enter the data into Excel and create an HR diagram.  We are currently studying Correlation &amp;amp; Regression.  I wanted to show them that a scatter-plot can be VERY interesting.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where can I get stellar data that is already in a format that my students can use excel to create a scatter-plot?  Unbeknownst to them, they will create an HR diagram.  &lt;/p&gt;&#xA;" OwnerUserId="21778" LastEditorUserId="31410" LastEditDate="2021-05-27T13:55:46.293" LastActivityDate="2021-05-27T13:55:46.293" Title="Creating an HR diagram from tabulated stellar data" Tags="&lt;star&gt;&lt;hr-diagram&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25137" PostTypeId="1" AcceptedAnswerId="25146" CreationDate="2018-02-17T04:35:35.103" Score="2" ViewCount="80" Body="&lt;p&gt;If we have a stellar density model $\rho(r) = \rho_c(1-r/R)$, where the star is composed of ions behaving as perfect gas and electron with non-relativistic degeneracy. The central pressure is due to both gas and degeneracy pressure. My final goal is to find central temperature of this model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the linear density model, I calculated the central pressure only for gas to be $P_c = \frac{5GM^2}{4 \pi R^4}$. And central temperature for only gas is $T_c = \frac{5G \mu M_{H} M}{12KR}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the Pressure for non-relativistic degeneracy $P_{nrd} = C \rho^{5/3}$. I was wondering to find central pressure for degeneracy, may I use the Lane-Emden equations for polytrops and simply find central pressure and temperature from there? Then add the two temperatures to find the maximum central temperature for a gas+degenerate star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Lane-Emden model, they also consider density dependence on radial distance, but they express the density function as $\rho = \rho_c \theta^n$. Is it equivalent to $\rho(r) = \rho_c(1-r/R)$ for gas model?&lt;/p&gt;&#xA;" OwnerUserId="21282" LastActivityDate="2018-02-17T08:11:34.953" Title="Central temperature of a stellar model with combination of gas and degeneracy" Tags="&lt;stellar-astrophysics&gt;&lt;degenerate-matter&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25138" PostTypeId="1" CreationDate="2018-02-17T05:04:08.997" Score="1" ViewCount="185" Body="&lt;p&gt;I've stumbled across several articles about how &lt;a href=&quot;https://www.nature.com/articles/nphys3733&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Venusian polar high atmosphere is really cold&lt;/a&gt;. What these articles appear to fail to specify is what the surface temperatures of the poles are. I am interested in knowing the polar surface temperatures. I would think that convection would lower the polar surface temperature to lower amounts, but I don't know a lot about Venus's poles to say anything about this.&lt;/p&gt;&#xA;" OwnerUserId="19304" LastEditorUserId="19304" LastEditDate="2018-02-17T06:05:04.367" LastActivityDate="2018-02-17T17:29:26.473" Title="Venusian polar surface temperatures" Tags="&lt;solar-system&gt;&lt;planetary-atmosphere&gt;&lt;temperature&gt;&lt;venus&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25140" PostTypeId="2" ParentId="7863" CreationDate="2018-02-17T06:46:42.857" Score="1" Body="&lt;p&gt;What scientists says, is that mass didn't exist in the first time it was pure energy, and inflation happened at very high speeds (more than 50 times the speed of light) so even when particles and mass appeared (less than 1 in billion energy was converted to mass, matter and anti matter following: E = mc^2) there was a very high expansion velocity, such that Hydrogen and Helium formed in the rates (75% H,  25% He, and very little amount of Li) but no heavier elements says the scientists, densities rapidly and uniformly decreased (in terms of minutes due to expansion speeds), on the other hand a Blackhole requires very high mass densities to form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it happened that initial conditions were very different from a big star/supernova and lead to different results.&lt;/p&gt;&#xA;" OwnerUserId="21612" LastEditorUserId="21612" LastEditDate="2018-02-17T07:16:42.643" LastActivityDate="2018-02-17T07:16:42.643" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25146" PostTypeId="2" ParentId="25137" CreationDate="2018-02-17T08:11:34.953" Score="2" Body="&lt;p&gt;If you have a gas of non-degenerate ions and degenerate electrons, then by definition you have an upper limit to the temperature, since if it were higher the electrons would not be degenerate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below this temperature, the electron degeneracy pressure will always greatly exceed any perfect gas pressure from the ions. Since degeneracy pressure does not depend on temperature, then the central temperature could be &lt;em&gt;anything&lt;/em&gt; lower than the temperature below which the electrons can be assumed degenerate. For example, the mass, radius and central density of white dwarfs of a given mass will be very similar, although their central temperatures will change by more than an order of magnitude as they cool over billions of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wish to make the small corrections, either for partial degeneracy (the pressure is then slightly temperature dependent) or to include the ion pressure, then the only way to proceed is add the pressures and then solve the equations of stellar structure numerically.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-02-17T08:11:34.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25150" PostTypeId="2" ParentId="25138" CreationDate="2018-02-17T17:29:26.473" Score="1" Body="&lt;p&gt;Venus's surface temperature is hot everywhere at about 460⁰C. There may some small differences due to latitude, but not nearly as pronounced as on Earth or Mars. The supercritial CO2 that forms the lower atmosphere is a good conductor of heat, keeping surface temperatures very stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article you refer to notes that models of the Venusian atmosphere predict somewhat higher temperatures at high altitude and latitude than can be inferred from measurements made during aerobraking. This tells us that the models of the atmosphere need to be tweaked. It doesn't mean that there is a region of cool temperatures at the poles.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-02-17T17:29:26.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25151" PostTypeId="2" ParentId="25117" CreationDate="2018-02-17T18:09:29.957" Score="7" Body="&lt;p&gt;The Minor Planet Center inconspicuously puts this under &quot;Distant Artificial Satellites Observation.&quot;&#xA;&lt;a href=&quot;https://www.minorplanetcenter.net/iau/DASO/DASO.html&quot; rel=&quot;noreferrer&quot;&gt;DASO Circulars&lt;/a&gt; 561-566 have observations of the roadster 2018-017A.&#xA;It's also on the short list of artificial objects for which MPC can &lt;a href=&quot;https://www.minorplanetcenter.net/iau/artsats/artsats.html&quot; rel=&quot;noreferrer&quot;&gt;generate ephemerides&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-02-22T04:48:43.063" LastActivityDate="2018-02-22T04:48:43.063" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25152" PostTypeId="1" CreationDate="2018-02-17T20:05:45.507" Score="3" ViewCount="293" Body="&lt;p&gt;What effect does the size of a black hole have on the size or density of the photon sphere and it's proximity to the horizon? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: A person is in space looking at 2 black holes, one of them big and the other small. Could the bigger one appear smaller because of the way space bends around it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, can a large enough black hole be completely hidden behind the gravitational lensing? Could a dense photon sphere around a black hole look like a spherical distortion to distort the horizon or hide the black hole completely?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/llGpg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/llGpg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2019-02-18T17:09:20.767" LastActivityDate="2019-02-18T17:09:20.767" Title="How does the size of a naked black hole effect its photon sphere?" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;mass&gt;&lt;photons&gt;&lt;light-curve&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="25153" PostTypeId="2" ParentId="24677" CreationDate="2018-02-17T23:11:37.653" Score="5" Body="&lt;p&gt;The Moon's successive apogee/perigee distances from the Earth do indeed undergo changes: these changes are almost cyclical, and they have a main period close to 205.89 days (nearly 7 synodic months).  A main contributory factor to the changes in perigee distances is the periodic solar perturbation known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Evection&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;evection&lt;/em&gt;&lt;/a&gt;. Then, in decreasing order of maximum size, a second contribution is due to the perturbation known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Variation_(astronomy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;variation&lt;/em&gt;&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest of this answer summarises explanations how the evection (along with the variation) affects the perigee distances: also offered is a numerical example of extreme lunar-perigee data from the &lt;a href=&quot;https://babel.hathitrust.org/cgi/pt?id=uc1.31822039017108;view=1up;seq=5&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomical Almanac ('AA') for 2011&lt;/a&gt;: these data indicate how the combination of the two effects can account for nearly all of the observed range in lunar perigee distances. The natures and sizes of the two effects also indicate features by which the Moon's real orbit differs (considerably) from a simple Keplerian fixed ellipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The evection:&lt;/strong&gt; Older textbooks used to discuss the way in which the evection gives rise to changes in apogee/perigee distances -- for example &lt;a href=&quot;https://archive.org/details/anelementarytre02godfgoog&quot; rel=&quot;nofollow noreferrer&quot;&gt;H Godfray (1859), Elementary Treatise on the Lunar Theory&lt;/a&gt;. Godfray's explanation proceeds by showing the practical equivalence between two forms in which the moon's longitude and radius vector &amp;amp;c. can be expressed: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) A first form is the usual modern form of trigonometrical series representation, which effectively supposes a constant (mean) eccentricity for the Moon's orbit. Among many other terms is a main separate term for the evection, in a form first given by Euler, with an argument nowadays often expressed as $(2D-l)$ , where $D$ is the Moon's mean elongation from the Sun and $l$ is the Moon's mean anomaly, i.e. Moon's mean elongation from the current position of its mean perigee. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) The second form is an older representation of the Moon's motions, which supposes a cyclically variable eccentricity, and thus also a cyclically variable perigee distance, greatest equation, &amp;amp;c. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Godfray's book gives the explanations fairly fully for the effects on longitude and equation of center (at p.66, art.70 along with preceding derivations), and then a much briefer summary of the analogous demonstration of the effects on radius vector (at pp.76-77, art.85). (In a little detail: what is shown is that the lowest-order elliptical term and the evection term can be trigonometrically combined and rearranged, to give as their equivalent an approximation to a variable ellipse, in which the eccentricity cyclically fluctuates and the angular orientation of the apogee/perigee cyclically librates as well as showing its well-known mean overall rate of rotation. A corresponding modern trigonometrical development shows essentially the same relation between the two forms for the longitude series, going as far as third-order -- &lt;a href=&quot;https://books.google.com/books/about/Between_Theory_and_Observations.html?id=Sx41OOG1zdsC&quot; rel=&quot;nofollow noreferrer&quot;&gt;S A Wepster (2010)&lt;/a&gt;, at pp.100-104 in his historical and mathematical study of Tobias Mayer's 18th-century lunar theory and tables.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Independently of this older type of explanation, details in appendix A below show, with reference to modern data, how the main term of the evection reinforces the main elliptic term when the Sun is in line with the Moon's line of apses, and opposes it when the Sun is at 90° to that line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The variation:&lt;/strong&gt; Next in size after the evection, the variation (in radius vector) brings the Moon closer to the Earth at new and full moon, and takes it farther away at the lunar quarters. This effect was shown to result from the solar perturbing force by &lt;a href=&quot;https://books.google.com/books?id=Tm0FAAAAQAAJ&amp;amp;pg=PA239&quot; rel=&quot;nofollow noreferrer&quot;&gt;Newton, in Principia Book 1 Prop.66 corollaries 2-5&lt;/a&gt;, and in &lt;a href=&quot;https://books.google.co.uk/books?id=6EqxPav3vIsC&amp;amp;pg=PA262&quot; rel=&quot;nofollow noreferrer&quot;&gt;Book 3 Props.26-29&lt;/a&gt;; results later refined by numerous authors, notably &lt;a href=&quot;http://adsabs.harvard.edu/abs/1895AJ.....15..137H&quot; rel=&quot;nofollow noreferrer&quot;&gt;G W Hill, see especially his data for inverse radius vector e.g. at p.143 in (1895) Astron J 15, 137-143&lt;/a&gt;. (In Hill's paper, $\tau$ (tau) means the same as $D$ above.) The instantaneous amount of the variation depends on lunar phase, and so it also contributes to changes in perigee distance, because the mean period between perigees (~27.55 days) is about two days shorter than the mean period between new moons (~29.53 days), hence successive perigees occur at different phases in the lunation and are differently affected by the variation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Numerical example:&lt;/strong&gt;  Appendix A below cites recently-refined &lt;a href=&quot;http://adsabs.harvard.edu/abs/1988A%26A...190..342C&quot; rel=&quot;nofollow noreferrer&quot;&gt;modern values (Paris Observatory)&lt;/a&gt; for the amplitude of trigonometric terms affecting the Moon's radius vector. The main term of the evection is close to 3699 km in amplitude, and the main term of the variation is close to 2956 km.  Ignoring many smaller periodic effects, one may expect from what has already been mentioned, that when a new or full moon occurs at perigee (implying also that the Sun is in the line of apses), the main evection and variation terms both act to reduce the perigee distance, by about the sum of the two amplitudes, i.e. by about 6655 km. When on the other hand a perigee occurs at one of the lunar quarters (implying also that the Sun is at 90° to the line of apses), both the two terms have the opposite effect, i.e. to increase the perigee distance by about 6655 km. Thus the main terms of the evection and variation, when they both reach maximum in the same sense, can be expected to make for a range of about 13310 km in lunar perigee distances. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This trigonometric-based expectation can be compared with data from almost any recent Astronomical Almanac ('AA'). (In recent years, the lunar distance data in AA come from a numerically-integrated ephemeris, &lt;a href=&quot;https://en.wikipedia.org/wiki/Jet_Propulsion_Laboratory_Development_Ephemeris&quot; rel=&quot;nofollow noreferrer&quot;&gt;version DE405 for years 2003-2014&lt;/a&gt;, see &lt;a href=&quot;https://babel.hathitrust.org/cgi/pt?id=uc1.31822039017108;view=1up;seq=5&quot; rel=&quot;nofollow noreferrer&quot;&gt;AA for 2011&lt;/a&gt;, page L4. The integrations were fitted to modern lunar laser-ranging data, independently of classical trigonometric analysis.) The AA for 2011 (at hand while writing this answer) tabulates lunar distances daily at 0h TT (using units of earth-equatorial-radius, 6378.14 km), and provides the following example-data (see esp. pages D1, D8, D14). (i) A smallest tabulated local-minimum moon-distance for the year occurred on March 20 (0h) at 55.912 earth-radii, close to a perigee at March 19 19h and a full moon at March 19 18h 10m; and (ii) a largest tabulated local-minimum moon-distance for the year occurred on July 8 (0h) at 57.951, close to a perigee at July 7 14h, and to a lunar first-quarter at July 8 6h 29m.  At the dates for which the distances were tabulated, the phases and configurations were close but not exact, the moon was a very few degrees off exact perigee and also a little off exact syzygy or quadrature.  Neglecting this inexactness, one may reckon, for the reasons mentioned above and shown in the Appendix, that on both dates the evection and variation act in the same sense and rather close to their maxima; both of them reduced the perigee distance at date (i), and both increased it at date (ii). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By difference between data (i) and (ii) from AA 2011, the range of the tabulated local-minimum (near-) perigee distances was 2.039 earth-radii, equivalent to about 13000 km. This differs less than 2.5% from the combined peak-to-peak range (13310 km) of the main terms of evection and variation.  The calculation and comparison is of course rather rough, both by the inexactness of the configurations, and also because many smaller trigonometric terms are ignored. Nevertheless, it is close, and helps to indicate how the evection together with the variation can account for nearly all of the range in lunar perigee distances seen in a year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Appendix:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shown here are (A) how the effects mentioned above are also quantitatively inherent in the most recent analytical accounts of the lunar motions; and (B) how some (now historical) accounts have attempted to outline separately the gravitational causes of the evection -- a somewhat more awkward enterprise, involving approximations and engagement with older historical forms for expressing the motions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A: Quantitative description of the varying lunar perigee distances&lt;/strong&gt; is given here in terms of modern analytical expressions for the Moon's orbital longitude and radius vector. The following data are rounded from &lt;a href=&quot;http://adsabs.harvard.edu/abs/1988A%26A...190..342C&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;ELP 2000-85 - A semi-analytical lunar ephemeris adequate for historical times&quot;, by Michelle Chapront-Touzé and Jean Chapront (1988) Astronomy &amp;amp; Astrophysics 190, 342-352&lt;/a&gt;, especially at page 351: this represents one of several versions of the authors' 'ELP' (Ephémérides Lunaires Parisiennes), see also &lt;a href=&quot;ftp://ftp.imcce.fr/pub/ephem/moon/elp82b/README&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt; at one of the Paris Observatory websites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three largest trigonometric terms describing the time-varying differences between the Moon's true and mean radius vector, and its true and mean orbital longitude, are known respectively as the largest of the elliptical terms, and the main terms of the evection and variation. They are close to -- &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a)     $ -20905.355 \cos (l) -3699.111 \cos (2D-l)  -2955.968 \cos (2D) $ &#xA;(for the true radius vector (in km), relative to the mean distance of 385000.529 km), and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b)     $  +22639.586&quot; \sin (l) +4586.438&quot; \sin (2D-l)  +2369.914&quot; \sin (2D) $&#xA;(for the difference true minus mean orbital longitude, in arc&quot;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D$ and $l$ have the meanings already mentioned. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest elliptical term (left-hand term in (a) and (b)) can be considered as the largest (lowest-order) term in a trigonometrical series with arguments in multiples of $l$ only. These sub-series can be excerpted from the long series of terms, in many arguments, given on page 351 of the cited 1988 paper, thus: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c)  $ -20905.355 \cos (l) -569.925 \cos (2l) -23.210 \cos (3l) ... $ for radius vector, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d)  $ +22639.586&quot; \sin (l) +769.026&quot; \sin (2l) +36.124&quot; \sin (3l) ... $ for orbital longitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are approximately close to series for the equation of the center (in radius vector or orbital longitude) that could be developed for an exact Keplerian elliptical orbit with constant ('mean') eccentricity about 0.0549 (compare for example the forms given in &lt;a href=&quot;https://books.google.com/books/about/Methods_of_celestial_mechanics.html?id=eKgNAQAAIAAJ&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brouwer and Clemence (1961) Methods of Celestial Mechanics&lt;/a&gt;, pages 76-77, equations 73 and 75).  Together, series (c) and (d) express approximately a mean ellipse that the Moon could follow in the absence of perturbations.  Under this hypothetical condition, the lunar perigee distances for such a mean ellipse would of course always be the same, about 363502 km according to the three initial periodic terms excerpted here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then each second term in the three-term excerpts (a) and (b) above is the principal term responsible for the evection.  To see the effect of the evection terms, the argument $(2D-l)$ can be considered as effectively $(l -(2l-2D))$ , which differs from $l$, the argument of the elliptical inequalities, by a slowly-changing quantity $(2l-2D)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The period of $l$ ('anomalistic month') is about 27.55 days, but the period of $(2l-2D)$ is about 205.89 days (it is the mean interval between passages of the Sun past the Moon's line of apses, of which one direction points to apogee, the other to perigee).  (The mean interval between passages of the Sun past the Moon's mean apogee is double the foregoing, about 411.78 days, just under 14 mean synodic months.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two configuration-cases can usefully be pointed out:  (i) When the quantity $(2l-2D)$ is zero (which happens once in each 7-month cycle, when the sun's position is conjoined/opposed to the direction of the moon's apogee/perigee) then it can be seen from the series-excerpts above that the quoted evection term in each series reinforces the effect of the main elliptic term. (ii) In the other case, at the opposite extreme, when $(2l-2D)$ is 180° (which happens when the sun's position is 90° from the direction of the moon's apogee/perigee) it can be seen that the evection term in each series opposes the main elliptic term. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is like a 'beat' effect between two oscillations. On account of this, the maximum excursions from the mean, both in radius vector and in orbital longitude, are not the same in each cycle of $l$ : the local maxima fluctuate in amount, with a period of ~205.89 days, just under 7 mean synodic months.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expressions above thus show how the perigee distance of the moon varies, on account of the main evection term, over a range of about +/- 3699 km. The perigee distance is closer to the Earth in configuration-case (i), when the Sun conjoins/opposes the direction of the Moon's apogee/perigee; at this point the principal evection term(s) reinforce the elliptic terms), and the excursions in longitude are larger too. Then the perigee distance is larger in the second case, when the Sun is 90° away from the line of apses; at this point the evection term(s) and main elliptic terms are opposed, and here the excursions in longitude are smaller too.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In sum, the effects of the evection terms on perigee distance and on orbital longitude are approximately similar to effects that would arise from an increased orbital eccentricity in the first case, and from a reduced eccentricity in the second.  The results are modified by the variation according to the phase of the lunation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The (simpler) effect of the main term of the variation on radius vector has already been mentioned: the Moon is brought closer by about 2956km at new and full moon, and farther off by the same amount at the quarters. The exact perigee distances are also affected by other and generally smaller periodic terms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(These effects, when considered together, also show how full moons at about the closest possible perigee distances, and hence with largest apparent diameter, tend to occur at intervals of about 14 synodic months: these are the effects sometimes called 'super moons' which cause peaks of media interest.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;B:  Accounting gravitationally&lt;/strong&gt; for these selected features of the Moon's perturbations is somewhat awkward. From the mid-18th century to the early 20th, analytical solution-techniques typically treated at least the main known perturbing forces on the Moon as a whole, to give approximate series solutions for the lunar motions. Such methods generate masses of trigonometric terms, and leave it practically impossible to see which (if any) particular parts of the perturbing forces are responsible for the evection effects. Neither do modern numerical techniques show any easily-separable parts of the perturbation effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been at least two attempts to show, mainly geometrically and qualitatively, how the effects of the evection can arise gravitationally. For this purpose the evection is taken to be represented by fluctuations in orbital eccentricity, an equivalence discussed above and in the Godfray reference already cited. The more recent of the two expositions was given by &lt;a href=&quot;https://archive.org/details/cu31924072676756&quot; rel=&quot;nofollow noreferrer&quot;&gt;F R Moulton's (1914) Introduction to Celestial Mechanics&lt;/a&gt; (at chapter 9, esp. from p.321-360).  The original exposition was given by Newton in &lt;a href=&quot;https://books.google.co.uk/books?id=Tm0FAAAAQAAJ&amp;amp;pg=PA234&quot; rel=&quot;nofollow noreferrer&quot;&gt;Book 1 of the Principia, Proposition 66&lt;/a&gt;, especially corollary 9 (pp.243-5 in 1729 English translation from Latin). The explanations depend on examining the way in which the perturbing force alters the net power-law for the attraction of the Earth on the Moon, and does so differently in different parts of the Moon's orbit, making the inverse power a little more than 2 in some parts of the orbit and a little less in other parts. Beyond that it would take too much space to describe those explanations here, the originals are available in online archives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also worth noting that &#xA;(1) Absence of solar perturbing force would not render the moon's orbit circular or nearly so: the eccentricity is a free parameter corresponding to an arbitrary constant in the integration of the two-body problem: for example &lt;a href=&quot;https://books.google.co.uk/books/about/Fundamentals_of_Astrodynamics.html?id=UtJK8cetqGkC&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bate, Mueller, White (1971) Fundamentals of Astrodynamics&lt;/a&gt; at pages 19-21 give a notably transparent demonstration of this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) The solar force perturbing the Moon in its motion around the Earth is sometimes described as if represented by the absolute attraction of the Sun on the Moon: but it is really represented by the (vector) difference between the Sun's attraction on the Moon and the Sun's attraction on the Earth (Newton, Principia, &lt;a href=&quot;https://books.google.co.uk/books?id=Tm0FAAAAQAAJ&amp;amp;pg=PA21&quot; rel=&quot;nofollow noreferrer&quot;&gt;Corollaries 1, 2 and 6 to the laws of motion&lt;/a&gt; and Book 3, &lt;a href=&quot;https://books.google.co.uk/books?id=6EqxPav3vIsC&amp;amp;pg=PA262&quot; rel=&quot;nofollow noreferrer&quot;&gt;Proposition 25&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) The rotation (precession) of the line of apses in itself does not change the perigee distances, it alters the angular places of the perigee and the times when the moon reaches perigee. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) The orbit of the Moon is quite far from a Keplerian ellipse or any ellipse, it combines features of a variational orbit (nearly elliptical but with the Earth near the center not at a focus) and an ellipse of varying eccentricity and fluctuating line of apses. Newton already in an unpublished paper expressed an approximate recognition that the real orbit of the Moon is not exactly an eccentric Keplerian ellipse, nor exactly a central ellipse due to the variation, but &quot;an oval of another kind&quot; (see D T Whiteside (ed.) (1973), The Mathematical papers of Isaac Newton, Volume VI: 1684-1691, Cambridge University Press, at &lt;a href=&quot;https://books.google.com/books?id=lIZ0v23iqRgC&amp;amp;pg=PA533&quot; rel=&quot;nofollow noreferrer&quot;&gt;page 533&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17202" LastEditorUserId="17202" LastEditDate="2018-02-22T22:33:29.553" LastActivityDate="2018-02-22T22:33:29.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25154" PostTypeId="2" ParentId="7863" CreationDate="2018-02-18T01:22:11.470" Score="1" Body="&lt;p&gt;Who said it didn't produce a big black hole?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may very well live inside a big black hole. If you apply the hypothesized mass of our Universe to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schwarzschild radius&lt;/a&gt; equation, the resulting radius will not be too far (in order of magnitude) to the observable radius of the visible Universe. Indeed, what we call &quot;Big Bang&quot; may simply be the formation of our &quot;black-hole-universe&quot; from a previous star in another Universe (thus the theory of a &quot;Multiverse&quot;). That explains at least why our Universe is &lt;em&gt;finite&lt;/em&gt;, but light or matter apparently can't escape it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was first proposed at least 45 years ago (&lt;a href=&quot;https://www.nature.com/articles/240298a0&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;), I don't know why isn't it more popular, since it's so fascinating. (If you live in a poor country -- I believe science should be universal, and not just for the rich -- I suggest using &lt;a href=&quot;http://www.sci-hub.tw/&quot; rel=&quot;nofollow noreferrer&quot;&gt;sci-hub&lt;/a&gt;, like &lt;a href=&quot;http://sci-hub.tw/http://www.nature.com/nature/journal/v240/n5379/pdf/240298a0.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers to &lt;a href=&quot;https://astronomy.stackexchange.com/q/1460/813&quot;&gt;this question&lt;/a&gt; explain the idea with more detail.&lt;/p&gt;&#xA;" OwnerUserId="813" LastActivityDate="2018-02-18T01:22:11.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25155" PostTypeId="2" ParentId="18120" CreationDate="2018-02-18T02:54:25.073" Score="-1" Body="&lt;p&gt;i am by no means a scientist, thus nor an astrophysicist. i have a background in electrical engineering and a curiosity for cosmology. i ended up here essentially because i'm looking for answers to the question asked above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it seems to me that the following information is revelant to the question : a fairly recent (2017.09.20) article published on the NASA website mentions a study which reveals that the two methods used to compute Hubble's constant (one is based on observations of type 1a supernovae, the other on the CMB) disagree (although the Standard Model of Cosmology predicts their agreement) :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;« A recent study using the first method yielded an 8% greater expansion rate than the second method’s result. » - &lt;a href=&quot;https://science.nasa.gov/science-news/news-articles/hubbles-contentious-constant-news&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://science.nasa.gov/science-news/news-articles/hubbles-contentious-constant-news&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the article doesn't mention a clear explanation for this discrepency. for instance, perhaps there are holes in one or both computation methods. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;if i understand correctly : since it is believed that the CMB informs us about the early universe, but this is not so for type 1a supernovae, then another possible explanation is that both measurements are valid, and the discrepency means that something has changed over time. for instance, the article asks the question « Or are dark energy’s or dark matter’s properties changing over time? ». given the importance of Hubble's constant, perhaps this points to the fact that physics have changed over time.&lt;/p&gt;&#xA;" OwnerUserId="21795" LastActivityDate="2018-02-18T02:54:25.073" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25156" PostTypeId="1" CreationDate="2018-02-18T03:10:03.897" Score="2" ViewCount="1118" Body="&lt;p&gt;From my understanding of the word gas giant, it is a planet composed of entirely a gaseous atmosphere, and so planets Jupiter and onwards fall in this category. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, what would stop a spacecraft from just going right through any of these planets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For earth, a spacecraft would firstly rapidly disintegrate approaching the surface of earth, and if it happens to still have part of itself, it will collide with the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For gas giants like Uranus, &lt;strong&gt;if a spacecraft is able to counter the gravitational forces and slow down, could it actually &quot;go through&quot; Uranus&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="17327" LastActivityDate="2018-02-19T23:37:37.507" Title="Would a spacecraft just go &quot;through&quot; a gas giant?" Tags="&lt;planet&gt;&lt;jupiter&gt;&lt;gas-giants&gt;&lt;gas&gt;&lt;uranus&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25158" PostTypeId="1" CreationDate="2018-02-18T05:53:56.870" Score="1" ViewCount="464" Body="&lt;p&gt;if I am not using very odd programming language for Astrophysicists, here is my code for calculating path of sun. I am following &lt;a href=&quot;http://www.esrl.noaa.gov/gmd/grad/solcalc/solareqns.PDF&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; reference&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to calculate the local time as a function of zenith (e.g. sunset, sunrise, golden hour etc)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import java.time.LocalTime;&#xA;import java.util.Calendar;&#xA;import static java.lang.Math.acos;&#xA;import static java.lang.Math.cos;&#xA;import static java.lang.Math.sin;&#xA;import static java.lang.Math.tan;&#xA;&#xA;&#xA;/**&#xA; * Created by rudra on 17/02/18.&#xA; */&#xA;&#xA;public class PathOfSun {&#xA;&#xA;    static public double getTime() {&#xA;        LocalTime loacltime = LocalTime.now();//.getHour();&#xA;        double tzone = 5.5; //TODO: placeholder, will be calculated&#xA;        //    int localhour = time.getHour();&#xA;        int hour = loacltime.getHour();&#xA;        int min = loacltime.getMinute();&#xA;        int sec = loacltime.getSecond();&#xA;        double lat  =  28.6;//latlang.Lat;&#xA;        double lang =  77.2;//latlang.Lang;&#xA;        Calendar calendar = Calendar.getInstance();&#xA;        int dayofyear = calendar.get(Calendar.DAY_OF_YEAR);&#xA;        int daysthisyear = calendar.getActualMaximum(Calendar.DAY_OF_YEAR);&#xA;        double pi = Math.PI;&#xA;&#xA;        double gamma = (2 * pi / daysthisyear) * (dayofyear - 1 + (hour - 12) / 24);&#xA;        double eqtime = 229.18 * (0.000075 + 0.001868 * cos(gamma) - 0.032077 * sin(gamma) -&#xA;                0.014615 * cos(2 * gamma) - 0.040849 * sin(2 *gamma));&#xA;        double decl = 0.006918 - 0.399912 * cos( (gamma)) + 0.070257 * sin( (gamma)) -&#xA;                0.006758 * cos(2 *  (gamma)) + 0.000907 * sin(2 *  (gamma)) -&#xA;                0.002697 * cos(3 *  (gamma)) + 0.00148 * sin(3 *  (gamma));&#xA;        double toffset = 4 * lang*60 + 60*(tzone) + eqtime;&#xA;        double tst = 60 * hour + min + sec / 60.0 + toffset;&#xA;        double ha = (tst/(4*60)) - 180;&#xA;&#xA;        // Zenith&#xA;        double phi = acos(sin(Math.toRadians(lat)) * sin(decl) +&#xA;                      cos(Math.toRadians(lat)) * cos((decl)) * cos(Math.toRadians(ha*60)));&#xA;        // Azimuth&#xA;        double theta = pi - (acos((sin(Math.toRadians(lat)) * cos(phi) - sin(decl)) / (cos(Math.toRadians(lat)) * sin(phi))));&#xA;&#xA;        // Calculating the hourAngle as in second page&#xA;        // At sunrise&#xA;        double ha2 = acos(cos(Math.toRadians(90.833))/cos( Math.toRadians(lat))*cos((decl))-&#xA;                tan(Math.toRadians(lat))*tan((decl)));&#xA;&#xA;        double stime = (720 - 4*(lang + Math.toDegrees(ha)) - eqtime)/60.0;&#xA;&#xA;        System.out.println(&quot;DayOfYear&quot;+ dayofyear+ &quot;  DaysThisYear&quot;+daysthisyear);&#xA;        System.out.println(&quot;lat &quot;+lat+&quot;  lang &quot;+lang+&quot; ltime &quot;+ loacltime);&#xA;        System.out.println(&quot;eqtime &quot;+eqtime+&quot;  ha &quot;+ha+&quot; , &quot;+ Math.toRadians(ha)+ &quot;  ha2 &quot; + ha2+&#xA;                &quot;\ndecl &quot;+Math.toDegrees(decl)+&quot; azimuth &quot;+ Math.toDegrees(theta)+&quot; Zenith &quot;+Math.toDegrees(phi));&#xA;        System.out.println(&quot;Time of Sun &quot;+ stime);&#xA;&#xA;        return stime;&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The problem is its giving me more or less OK value for equationoftime and declination, but azimuth etc are grossly incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ANy idea where I am going wrong? I am following &lt;a href=&quot;https://www.esrl.noaa.gov/gmd/grad/solcalc/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; page for reference.&lt;/p&gt;&#xA;" OwnerUserId="21798" LastActivityDate="2018-02-19T22:11:52.997" Title="Path of sun calculation using java" Tags="&lt;the-sun&gt;&lt;azimuth&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25159" PostTypeId="1" CreationDate="2018-02-18T09:11:54.320" Score="2" ViewCount="57" Body="&lt;p&gt;I live in the Pacific North West (Washington State, US). I have watched the skies for years, looking for constellations that I recognize and keeping track of them. One of them is Orion. As I have watched Orion I noticed that it travels East to West as I look to the south. It traverses the line that is the Milkey Way... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is. If the stars that make up the constellation of Orion move East to West from my position, looking up and South. Why does the Milky Way seem to stay in the same position? How is it that the stars that make up Orion move but the stars that make up the milky way appear to be fixed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance for any input.&lt;/p&gt;&#xA;" OwnerUserId="21799" LastActivityDate="2018-02-18T19:07:14.453" Title="Movement of the stars relative to my position" Tags="&lt;milky-way&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25162" PostTypeId="2" ParentId="25159" CreationDate="2018-02-18T17:09:11.400" Score="2" Body="&lt;p&gt;The East to West motion of the stars of Orion is due to the rotation of the Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Milky Way also moves across the sky during the night for exactly the same reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stars that form Orion do not transverse the milky way. They remain in exactly the same position relative to the Milky Way. You are simply mistaken in thinking that the Milky way remains in the same position. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-02-18T17:09:11.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25163" PostTypeId="2" ParentId="25156" CreationDate="2018-02-18T19:01:13.583" Score="4" Body="&lt;p&gt;In short, No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side detail:&#xA;Uranus and Neptune consist likely of 20% gas and 80% rock, coming from simple density considerations. They have large inner cores with masses around $\rm 12-14 \; m_{earth} $, and something like $\rm 2-3 \; m_{earth} $ of gas on top of them.&lt;br&gt;&#xA;Jupiter and Saturn are true gas giants. They both have $\rm 5-20 \; m_{earth} $ of solids in them  and the rest of the $\rm 320 \; m_{earth} $ (=Jupiter) and $\rm 95 \; m_{earth} $ (=Saturn) are gas.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However even if Jupiter and Saturn were 100% gas, you couldn't fly through them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reasons for this is&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Enormous density, pressure and temperature. A gas mass of several tens of Earth masses will collapse under its own gravity to enormous central densities, temperatures and thus static pressures.&lt;br&gt;&#xA;The densities that this gas will be compressed to, can have densities comparable and exceeding that of water, which is a good rule-of-thumb for the center of Jupiter (typical calculations show gas densities around $\rm 20 g/cm^3$, &lt;a href=&quot;http://www.mpia.de/~mordasini/Site2c.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;).&lt;br&gt;&#xA;The static pressure will crush your spacecraft, the high density will near instantly slow you down to zero due to strong friction, and the high temperatures will just melt or evaporate your spacecraft.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Just because something is gaseous, doesn't mean you can soar through it, as you would on a calm day with a glider on a light breeze.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other factors that complicate even getting there:  &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;Re-entry heat. The cold upper atmospheres of the gas giants increase the mach numbers $v / c_{sound}$ at re-entry by up to a factor of two. As &lt;a href=&quot;https://space.stackexchange.com/questions/24902/what-aspects-of-reentry-heating-scale-as-the-8th-power/25334#25334&quot;&gt;recently discussed on space exploration&lt;/a&gt; the re-entry heating that a spacecraft experiences scales as the 8th power of the re-entry speed or more. So a spacecraft would experience around $2^8 = 256$ times more heat load. The engineering challenge to survive this already would be considerable.&lt;br&gt;&#xA;I bring this up, because as @uhoh pointed out, it is unclear whether you want to just slice the atmosphere or fly through the $r=0$ coordinate of the planet.    &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Core erosion. Funny enough, the solid core is probably not even there anymore. Recent work has shown, that under the enormous pressures given at the center of a gas giant a phenomenon called &lt;a href=&quot;https://arxiv.org/abs/1608.02685&quot; rel=&quot;nofollow noreferrer&quot;&gt;'core erosion'&lt;/a&gt; will take place. In this, typical rock-composing elements prefer to exist in a gaseous phase rather than in a solid phase at those high pressure, thus the solid core will dissolve into the surrounding gas.&lt;br&gt;&#xA;This will further increase friction when trying to fly through the gas.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditDate="2018-02-19T23:37:37.507" LastActivityDate="2018-02-19T23:37:37.507" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25164" PostTypeId="2" ParentId="25159" CreationDate="2018-02-18T19:07:14.453" Score="3" Body="&lt;p&gt;The Milky Way consists of numerous faint stars near our galactic equator.&#xA;Many of these are visible in binoculars.&#xA;As &lt;a href=&quot;https://www.google.com/search?q=milky+way+time+lapse&amp;amp;tbm=vid&quot; rel=&quot;nofollow noreferrer&quot;&gt;time lapse videos&lt;/a&gt; show, they move across the sky along with the bright stars.&#xA;The light and dark patches in those digital images are easier to see visually near the galactic center in Sagittarius than in the opposite direction near Orion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to its oblique angle, the galactic equator appears to go through a special cycle as everything revolves around the celestial pole.&#xA;From mid-northern terrestrial latitudes, it runs overhead NW-SE when Orion is highest, low in the north when Virgo and the Big Dipper are highest, and overhead NE-SW when Cygnus is highest.&#xA;We see no more than half of this cycle on any given night, which may explain the illusion you perceived.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-02-18T19:07:14.453" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25165" PostTypeId="1" CreationDate="2018-02-18T22:35:11.890" Score="6" ViewCount="438" Body="&lt;p&gt;I'm trying to work out how big solar panels have to be for a Saturn probe, but can't find any data for solar energy at Saturn.&lt;/p&gt;&#xA;" OwnerUserId="21804" LastEditorUserId="2365" LastEditDate="2018-02-18T23:07:08.317" LastActivityDate="2020-12-16T19:48:31.813" Title="How can I calculate solar intensity at Saturn" Tags="&lt;solar-system&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25166" PostTypeId="1" AcceptedAnswerId="25244" CreationDate="2018-02-19T01:20:08.437" Score="4" ViewCount="338" Body="&lt;p&gt;Reading another question, I came across the argument that inconsistencies in Neptune's and Uranus' orbits lead to the discovery of Pluto.  &lt;a href=&quot;http://pluto.jhuapl.edu/Participate/learn/What-We-Know.php?link=Discovery-of-Pluto&quot; rel=&quot;nofollow noreferrer&quot;&gt;This brief write-up&lt;/a&gt; mentions Percival Lowell and William H. Pickering by name as two people who looked for the theoretical 9th planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, were inconsistencies in Neptune's and Uranus' orbits detectable by early 20th century methods or were these scientists just chasing calculation or estimation errors with dreams of finding the next planet.  My (guess) is that Pluto is much too small, and too far away and just one of many similar mass kuiperbelt objects, so it's unlikely that it was really discovered by measured inconsistency in Uranus/Neptune orbits method, (despite the fact that &lt;a href=&quot;http://coolcosmos.ipac.caltech.edu/ask/154-Who-discovered-Pluto-&quot; rel=&quot;nofollow noreferrer&quot;&gt;there are articles that say that's precisely how Pluto was discovered&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a nutshell, that's the question.  Was the wobble that Pluto's causes on Neptune's orbit observable by 1910-1930 observation methods.  And related, was the wobble that the entire kuiperbelt (some 20 to 50 times the mass of little pluto, but much more spread out), was that wobble detectible by 1910-1930 methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hard math isn't necessary to me for an answer, as I find the math hard to follow sometimes, so feel free to answer with or without mathematical calculations.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-02-23T12:37:45.083" Title="How much are Neptune's and Uranus' orbits perturbed by Pluto and other KBOs?" Tags="&lt;orbit&gt;&lt;orbital-mechanics&gt;&lt;pluto&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25169" PostTypeId="1" AcceptedAnswerId="25178" CreationDate="2018-02-19T03:15:11.997" Score="2" ViewCount="445" Body="&lt;p&gt;Not the same question but is similar to other marked duplicate. &#xA;How much time dilation  does the center of a galaxy can exist and sustain human life from our point of view? What would a day equal to each others point of view looking at each other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not asking about moving celestial bodies moving from each other but the difference in the gravity wells from the center of our galaxy being generally denser then the edge of a galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JMO2E.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JMO2E.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is another way of asking if I wasn't clear above the line. Not to invalidate any answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point 1 being near the center of the galaxy and point 2 at the the most edge the galaxy. How much would the differences in the galaxies gravity well point 1 being deeper in the center compared to the point 2 the edge effect time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zNUYN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zNUYN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-02-23T17:31:53.460" LastActivityDate="2018-02-24T15:39:07.413" Title="How does space time differ within a galaxy?" Tags="&lt;galaxy&gt;&lt;space-time&gt;&lt;redshift&gt;&lt;life&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25170" PostTypeId="1" CreationDate="2018-02-19T05:19:14.730" Score="3" ViewCount="158" Body="&lt;p&gt;How does tip-tilt mirrors correct such issue? What is the physical origin of the error? &lt;/p&gt;&#xA;" OwnerUserId="21807" LastEditorUserId="7982" LastEditDate="2021-02-28T11:45:04.210" LastActivityDate="2021-02-28T11:45:04.210" Title="What is tilt anisoplanatism for Laser-guide-star usage in adaptive optic systems?" Tags="&lt;telescope&gt;&lt;atmospheric-effects&gt;&lt;adaptive-optics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25173" PostTypeId="2" ParentId="8254" CreationDate="2018-02-19T05:53:28.313" Score="2" Body="&lt;p&gt;The current best theory to describe the gravitation is the General Relativity. In it, gravity is not a force, but the change of the curvature of the spacetime as the result of the matter (energy) density in it. Essentially, it has 2 equation systems:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;how the matter (energy) density affects the curvature of the spacetime&lt;/li&gt;&#xA;&lt;li&gt;how things are moving in this curved spacetime&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This can result anything, even a repelling gravitational force. For example, parallel laser beams going into opposite direction repel eachother. The problem is that we would need 40-60 orders higher energy densities to have a measurable effect, so it remains only a theoretical calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &quot;common&quot; situations, i.e. if we have planets,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;being significantly smaller as a black hole&lt;/li&gt;&#xA;&lt;li&gt;and going with a speed significantly smaller as the speed of light,&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;the Newtonian gravity is a very good and very simple approximation. In it, gravity is always attracting and so is it.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="1891" LastEditDate="2018-02-19T07:23:41.867" LastActivityDate="2018-02-19T07:23:41.867" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25177" PostTypeId="1" AcceptedAnswerId="25181" CreationDate="2018-02-19T07:15:02.523" Score="0" ViewCount="118" Body="&lt;p&gt;According to the big Bang Theory, the universe have started from a point and it describes how the universe expanded from a very high-density and high-temperature state.The &lt;strong&gt;initial singularity&lt;/strong&gt; was a singularity of infinite density thought to have contained all of the mass and space-time of the Universe. If it was so dense,  why didn't it engulf the other parallel universes just like a supermassive black hole? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are no other universes,  why the existing black holes didn't suck the entire universe yet? &lt;/p&gt;&#xA;" OwnerUserId="20823" LastActivityDate="2018-02-19T09:07:06.460" Title="Singularity at the time of origins" Tags="&lt;big-bang-theory&gt;&lt;supermassive-black-hole&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25178" PostTypeId="2" ParentId="25169" CreationDate="2018-02-19T07:36:37.817" Score="3" Body="&lt;p&gt;The gravitational redshift factor and also the factor by which time dilation occurs in a gravitational field is approximately $GM/Rc^2$. Here, $R$ would be the radius from which light was emitted and $M$ is the amount of mass &lt;em&gt;within&lt;/em&gt; radius $R$ (making the crude assumption of spherical symmetry).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, for an external observer,  light emitted from a radius of 10 kpc, might have $M \sim 10^{11} M_{\odot}$ in the galaxy shown. The redshift/time dilation factor is only $5\times 10^{-7}$ and is basically negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could perhaps gets an order of magnitude bigger as one approaches the denser central bulge, but is still very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only appreciable redshift and time dilation effects will come in the immediate surroundings of any supermassive black hole. By immediate, I mean within about a hundred Schwarzschild radii, where $R_s = 2GM/c^2$. In our Galaxy that would mean within a billion &lt;em&gt;kilometres&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-02-19T07:36:37.817" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25181" PostTypeId="2" ParentId="25177" CreationDate="2018-02-19T09:07:06.460" Score="4" Body="&lt;p&gt;First thing to mention is all our limited knowledge (best guesses) from back than only goes back to the fraction of time just after the big bang not the creation moment itself. Moreover you have to remember that spacetime itself was also created at that moment as well as all the stuff in it. This means that using the term singularity as we understand from the core of a blackhole, which is to say an infinitely dense and infinitely small point/puncture &lt;strong&gt;IN&lt;/strong&gt; the fabric of space doesn't really marry up. The universe no matter how tightly packed conditions were at this early stage had no space to warp/puncture into.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe has no centre, it began everywhere, simultaneously and instantly with all matter distributed uniformly and in equal density. Without a central point all matter was pulling evenly on its self in all directions so wouldn't coalesce at a single point and collapse down as it does in a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as to your mention of supermassive black holes sucking in parallel universes &quot;they don't&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps&lt;/p&gt;&#xA;" OwnerUserId="21819" LastActivityDate="2018-02-19T09:07:06.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25186" PostTypeId="2" ParentId="25170" CreationDate="2018-02-19T16:26:24.387" Score="4" Body="&lt;p&gt;Isoplanatism  commonly refers to a region of angles over which a ground-based telescope observes effectively the same atmospheric turbulence (e.g. an &quot;isoplanatic patch&quot;), such that a laser guide star provides effective correction of atmospheric seeing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;an&lt;/em&gt;-isoplanatism refers to a lack of isoplanatism, or a way in which  the science target and the reference laser guide star differ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single laser guide star does not provide knowledge of tip or tilt errors since the up-going beam experiences tilts in addition the down-going signal; these two sets of tilts are entangled and in the case where the transmitter and receiver are identical the laser guide star appears stationary. See a detailed overview and history of this problem the introduction of Ragazzoni &#xA; 1996, &lt;a href=&quot;http://adsabs.harvard.edu/full/1996ApJ...465L..73R&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/full/1996ApJ...465L..73R&lt;/a&gt;. This is commonly solved by observing a &quot;natural guide star&quot; position, since its light only passes through the atmosphere once, and using that signal to maintain the position of the target star by changing the angle a small (&quot;tip-tilt&quot;) mirror. &lt;/p&gt;&#xA;" OwnerUserId="457" LastActivityDate="2018-02-19T16:26:24.387" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25187" PostTypeId="1" AcceptedAnswerId="25192" CreationDate="2018-02-19T17:33:36.913" Score="3" ViewCount="695" Body="&lt;p&gt;Sun's photosphere is about 19 times as hot as Earth's surface. &#xA;(Source: &lt;a href=&quot;http://solar-center.stanford.edu/vitalstats.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://solar-center.stanford.edu/vitalstats.html&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But Solar surface gravity is about 28 times terrestrial surface gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, shouldn't the density of gases at the photosphere of the sun be comparable to that on the surface of the Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the density of Sun's photosphere a million times lesser? &#xA;(Source: &lt;a href=&quot;http://solar-center.stanford.edu/vitalstats.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://solar-center.stanford.edu/vitalstats.html&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, what I am asking has been aptly worded in userLTK's answer, &quot;how can the photosphere be so light and almost vacuum like under such high gravity (28 earth gravity)&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the answers have suggested that this may be because the Sun's photosphere is ionised. This doesn't seem to be accurate. The Hydrogen ionisation at optical depth = 1 is only 0.04%.&#xA;(Source: &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1968SoPh....3....5G&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1968SoPh....3....5G&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="17185" LastEditorUserId="17185" LastEditDate="2018-02-20T12:32:11.290" LastActivityDate="2018-09-29T17:26:39.813" Title="Why is sun's photosphere a million times less dense than air at the surface of the Earth?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;planet&gt;&lt;solar-system&gt;&lt;earth&gt;" AnswerCount="5" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="25188" PostTypeId="1" AcceptedAnswerId="25215" CreationDate="2018-02-19T18:53:14.957" Score="2" ViewCount="184" Body="&lt;p&gt;&lt;em&gt;I'm very new at Astronomy, and my knowledge is sparse. I've tried to be conscientious about my Wikipedia research but there's going to be a lot of things I don't know. Thanks for your patience.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;BACKGROUND&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I had read that hot Jupiters can cause superflares due to magnetic recombination. But checking Wikipedia I came across the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Not all planetary transits can be detected by Kepler, since the planetary &gt; orbit may be out of the line of sight to Earth. However, the hot Jupiters &gt; orbit so close to the primary that the chance of a transit is about 10%. &gt; If superflares were caused by close planets the 279 flare stars &#xA;  discovered should have about 28 transiting companions; none of them &#xA;  actually showed evidence of transits, effectively excluding this &#xA;  explanation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;- &lt;a href=&quot;https://en.wikipedia.org/wiki/Superflare#Hot_Jupiters_as_an_explanation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;QUESTION&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Given this, and among modern astronomers in general, is the theory that hot Jupiters can cause superflares on the outs?&lt;/strong&gt; Or is the article just saying that hot Jupiters can't explain these particular stars, while the model remains valid elsewhere? Do we have any solid evidence of a gas giant causing solar flares, super or otherwise?&lt;/p&gt;&#xA;" OwnerUserId="21826" LastActivityDate="2018-02-23T03:09:49.053" Title="Can hot Jupiters cause solar flares?" Tags="&lt;star&gt;&lt;gas-giants&gt;&lt;solar-flare&gt;&lt;hot-jupiter&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25189" PostTypeId="2" ParentId="25187" CreationDate="2018-02-19T21:24:42.967" Score="1" Body="&lt;p&gt;Where the photosphere lies is not only a function of density $\rho$, but also of opacities $\kappa$. One will find it at a particular wavelength wherever the integral optical depth integral $\tau = \int \rho \; \kappa \; dz$ along the line of sight $z$ becomes one.&lt;br&gt;&#xA;As the opacities are a strong function of the composition of the atmosphere, it is insufficient to just take gravity and temperature and try to predict the location of the photosphere.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2018-02-19T21:24:42.967" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25190" PostTypeId="2" ParentId="25187" CreationDate="2018-02-19T22:09:38.240" Score="1" Body="&lt;p&gt;The Sun's photosphere is roughly a 400 km thick layer below which almost photons do not escape and above which almost all photons with an outward direction do escape the Sun. Even a fairly diffuse plasma (compared to Earth surface pressure) does a fairly good job of absorbing thermal photons. A somewhat diffuse plasma (e.g., the bottom of the photosphere) does an exceptionally good job of absorbing thermal photons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The somewhat high temperatures (4400 kelvins or more) in the Sun's photosphere means that a good portion of the gas (mostly hydrogen and helium, plus some trace elements) is ionized. The much lower temperatures (~300 kelvins) in Earth's troposphere means that essentially none of the gas (mostly molecular nitrogen and oxygen, plus some trace compounds) is ionized. The very different temperatures and very different makeup mean that the Sun's photosphere and the Earth's troposphere are incomparable.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2018-02-19T22:09:38.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25191" PostTypeId="2" ParentId="25158" CreationDate="2018-02-19T22:11:52.997" Score="2" Body="&lt;p&gt;The &lt;a href=&quot;https://www.esrl.noaa.gov/gmd/grad/solcalc/solareqns.PDF&quot; rel=&quot;nofollow noreferrer&quot;&gt;NOAA formula sheet&lt;/a&gt;&#xA;expresses some quantities in radians, some in degrees, and some in minutes of time.&#xA;It does say which is which, but it's easy to miss. A few minor code changes can help expose bugs:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Store angles in radians. Trigonometric expressions free of &lt;code&gt;Math.toRadians&lt;/code&gt; calls are easier to verify against a reference document.&#xA;Put that document's URL in a comment so it's easy to find.&lt;/li&gt;&#xA;&lt;li&gt;If you need to store something in other units,&#xA;&lt;a href=&quot;https://www.joelonsoftware.com/2005/05/11/making-wrong-code-look-wrong/&quot; rel=&quot;nofollow noreferrer&quot;&gt;make it obvious&lt;/a&gt;&#xA;with a naming convention such as &lt;code&gt;long_deg&lt;/code&gt; for longitude in degrees.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The other quirks are the $\cos(\phi)$ and $\cos(180 - \theta)$&#xA;on the left hand sides of those equations.&#xA;$\arccos$ has a restricted range of $[0, \pi]$,&#xA;which is OK for zenith angle but leads to 180-degree errors in azimuth&#xA;unless you code east/west cases or derive new formulas where you can use &lt;code&gt;Math.atan2&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solar &lt;a href=&quot;https://en.wikipedia.org/wiki/Hour_angle&quot; rel=&quot;nofollow noreferrer&quot;&gt;hour angle&lt;/a&gt; should equal 0 at noon and increase by 1 degree every 4 minutes.&#xA;Sunset hour angle should be about 90 degrees near an equinox, smaller in winter, larger in summer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sunrise/sunset formulas produce times in minutes past midnight UTC, so you'll need to apply a time zone offset to get local times.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-02-19T22:11:52.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25192" PostTypeId="2" ParentId="25187" CreationDate="2018-02-19T23:20:04.230" Score="3" Body="&lt;p&gt;In addition to the two answers above, I'll add that, first, your estimate of a million times seems wrong.   These numbers are rough approximations of a fluid, not fixed volume, and I'd take them with a grain of salt too, but Wikipedia gives the &lt;a href=&quot;https://en.wikipedia.org/wiki/Photosphere#Sun&quot; rel=&quot;nofollow noreferrer&quot;&gt;density of the photosphere&lt;/a&gt; as about 2×10−4 kg/m^3.   That's about 1/6,000 (not a millionth) of Earth's atmospheric density at the surface (1.2 kg/m^3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not clear if that density figure is an average for the entire photosphere or closer to the surface where we can measure, but if we measure &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Earth#Principal_layers&quot; rel=&quot;nofollow noreferrer&quot;&gt;the entire Earth's atmosphere&lt;/a&gt;, the density of our atmosphere drops significantly (how much depends on where you determine the top of the atmosphere is), but that's a problem, there's no absolute boundary, so comparing density is a futile exercise, but the ratio drops to far less than 6,000 to 1, when you take the density of Earth's entire atmosphere.  You could also compare the photosphere to Earth's Mesosphere and the photosphere probably becomes more dense, but no matter how you compare, it's always going to be apples to oranges and rather pointless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://astronomyonline.org/SolarSystem/SunDetails.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;photosphere is about 500 km thick&lt;/a&gt;.  I don't trust the numbers enough to calculate the pressure at the bottom of the photosphere with any accuracy.   If you take a 500,000 meter column at the above density, .0002 kg/m^3, that's 100 kg or 220 lbs per column, worked out to square inches (PSI), .14 psi - 1/100th the pressure on the surface of Earth - but these numbers are terrible and prone to high inaccuracy.  I just put this out there to show that the pressure and density at the bottom of the photosphere is still low, but not as low as 1 part in 6000 of Earth's surface.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The gist of your question&lt;/strong&gt; is, how can the photosphere be so light and almost vacuum like over 500 km thickness under such high gravity (28 earth gravity), and that's a fair question.   The answer, as others have pointed out is due to the high temperature and content that's mostly ionized hydrogen in a plasma state.   Plasma is a different state of matter than gas and it tends to be much more spread out.   The outward pressure of photons may be a key factor too (I'm not 100% sure on that point).&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-02-19T23:20:04.230" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25193" PostTypeId="1" AcceptedAnswerId="25199" CreationDate="2018-02-19T23:49:40.677" Score="2" ViewCount="86" Body="&lt;p&gt;News has come up lately about the impending approach of asteroid 2017 VR12 and how it might be visible to some backyard telescopes. However I've read conflicting or vague reports on what kinds of telescopes will be powerful enough to spot it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got a modest but still decent little Celestron NexStar 4 SE, which is a 4&quot; Schmidt-Cassegrain. Can anyone say for sure whether that has the power to see the asteroid? Currently I just have the stock eyepiece. Would I need to get a better one to help out?&lt;/p&gt;&#xA;" OwnerUserId="7519" LastActivityDate="2018-02-20T16:46:00.023" Title="Home telescope for viewing 2017 VR12" Tags="&lt;amateur-observing&gt;&lt;asteroids&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25199" PostTypeId="2" ParentId="25193" CreationDate="2018-02-20T06:48:51.410" Score="5" Body="&lt;p&gt;&lt;a href=&quot;https://www.minorplanetcenter.net/db_search/show_object?object_id=K17V12R&quot; rel=&quot;nofollow noreferrer&quot;&gt;2017 VR12&lt;/a&gt;&#xA;is expected to reach magnitude 11.8 on the night of March 6-7.&#xA;This is near the limit of a 4-inch scope; actual visibility depends on your observing site, your eyes, and the magnification.&#xA;You can get a more definite answer by trying to see stars of similar magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other hard part is to acquire the target.&#xA;It will look like a star, so you need a finder chart showing stars at least as faint as the asteroid.&#xA;If you haven't observed one before, try finding a&#xA;&lt;a href=&quot;https://theskylive.com/asteroids-and-dwarf-planets&quot; rel=&quot;nofollow noreferrer&quot;&gt;named asteroid&lt;/a&gt;&#xA;around magnitude 10 first.&#xA;In this close approach, 2017 VR12 will move 0.9 degree per hour north-south,&#xA;fast enough that you need either quarter-hour marks plotted on the finder chart or a continuously updated display.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://stellarium.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellarium&lt;/a&gt;&#xA;works pretty well for me; I downloaded some of the extra star catalogs, and I import the latest available orbital elements for the asteroid of interest.&#xA;If you wish to use your scope's go-to capability,&#xA;&lt;a href=&quot;https://astroadventures.wordpress.com/2014/06/08/how-to-find-a-comet-with-a-goto-mount/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this guide&lt;/a&gt;&#xA;may help, but you still have to figure out which &quot;star&quot; in the eyepiece field is the asteroid.&#xA;&lt;a href=&quot;http://earthsky.org/space/asteroid-2017-vr12-closest-march-6-7-2018&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article&lt;/a&gt;&#xA;identifies a few brighter stars you can try to watch it pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do recommend getting a ~10mm eyepiece in addition to the 25mm one you have.&#xA;It might help you see stars half a magnitude fainter, but the main benefit is higher magnification for planets.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-02-20T16:46:00.023" LastActivityDate="2018-02-20T16:46:00.023" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25205" PostTypeId="1" AcceptedAnswerId="25211" CreationDate="2018-02-20T11:15:50.613" Score="4" ViewCount="720" Body="&lt;p&gt;As far as I've known, sea tides are higher on a Full moon day. And that is due to the &quot;gravitational pull&quot; of the moon.&lt;br&gt;&#xA;But we know that moon is visible because of reflection of sun's rays. So, even on a new moon's day, the moon is still there, but it's not visible.&lt;br&gt;&#xA;So, why doesn't it affect the gravity on earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="21845" LastActivityDate="2018-02-20T15:27:05.650" Title="Why are tides high only on a full moon day?" Tags="&lt;the-moon&gt;&lt;tidal-forces&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25208" PostTypeId="1" AcceptedAnswerId="25210" CreationDate="2018-02-20T12:07:37.050" Score="28" ViewCount="3594" Body="&lt;p&gt;How did &lt;a href=&quot;https://en.wikipedia.org/wiki/Ole_R%C3%B8mer&quot; rel=&quot;noreferrer&quot;&gt;Ole Christensen Rømer&lt;/a&gt;, who in 1676, shortly after Galileo's death, was said to have measured the speed of light as 220,000 km/s by timing the orbits of Io around Jupiter? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My only guess is the moon Io orbiting Jupiter is simply a clock. It pulses away with an orbital period of 42 hours and provides to the earth, 21 hours of light each orbit; this does not change. Now, as viewed from &quot;above&quot; we go counterclockwise around the sun, and as we move towards Jupiter, the duration of light from Io would be less than 21 hours, whereas if we are moving away from Jupiter, the duration of light from Io would be more than 21 hours. Suppose we measure the time from Io's first appearance from behind Jupiter to its reappearance and call that its observed orbital frequency $f$ as opposed to its actual orbital frequency ${f_o}$ which is 1/42 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He must have figured out the Doppler effect and this equation...&#xA;$$f = \left( {1 + \frac{{\Delta V}}{C}} \right){f_o}&#xA;% MathType!MTEF!2!1!+-&#xA;% faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9&#xA;% wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8sr&#xA;% ps0lbbf9q8WrFfeuY-ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0-&#xA;% yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9Ff0dc9Gqpi0d&#xA;% meaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGMbGaeyypa0Zaae&#xA;% WaaeaacaaIXaGaey4kaSYaaSaaaeaacqqHuoarcaWGwbaabaGaam4q&#xA;% aaaaaiaawIcacaGLPaaacaWGMbWaaSbaaSqaaiaad+gaaeqaaaaa!3BBB!&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hashed through the numbers myself and I got an answer of about 10 minutes more or less from 42 hours depending on which way the earth is heading with reference to Jupiter.&lt;/p&gt;&#xA;" OwnerUserId="5306" LastEditorUserId="5306" LastEditDate="2018-02-21T23:39:24.243" LastActivityDate="2018-02-21T23:39:24.243" Title="How did Ole Christensen Romer measure the speed of light?" Tags="&lt;light&gt;&lt;history&gt;&lt;speed&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25210" PostTypeId="2" ParentId="25208" CreationDate="2018-02-20T14:54:19.177" Score="33" Body="&lt;p&gt;Ole Rømer did not measure a change in the frequency of light. He measured an apparent change in the orbital period of Io, one of Jupiter's moons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbit of Io can be measured very accurately by observing when it enters or leaves the shadow of Jupiter. When the Earth is moving away from Jupiter, Rømer noted that the orbit of Io appeared to be very slightly longer than when the Earth was moving parallel to Jupiter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the orbital period of Io is $p$ seconds. If you are not moving and you observe an eclipse at time $t$, the next eclipse will be at time $t+p$. However, if you are moving away from Io, the next eclipse will be seen at $t+p+x$, where $x$ is the time that light takes to travel from your position at time $t$ to your position $p$ seconds later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The delay over a single orbit was too small for him to measure (about 30 seconds). But the delay was cumulative, and over about 30 orbits of Io, the orbit was delayed by about a quarter of an hour. He noted:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[This] appears to be due to light taking some time to reach us from the satellite; light seems to take about ten to eleven minutes [to cross] a distance equal to the half-diameter of the terrestrial orbit. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course, Rømer didn't know the actual size of the half-diameter of the terrestrial orbit (the astronomical unit) However putting the modern value of about 150 million km this gives a value of the speed of light at about 220000km/s, which is not so far from the modern value of about 300000km/s&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-02-20T22:21:59.040" LastActivityDate="2018-02-20T22:21:59.040" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25211" PostTypeId="2" ParentId="25205" CreationDate="2018-02-20T15:06:07.297" Score="6" Body="&lt;p&gt;There are two main gravitational causes of tides: the Moon, and to a lesser extent the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the moon is full &lt;em&gt;or the moon is new&lt;/em&gt;, the Earth, Moon and Sun are roughly aligned, and the Lunar tide combines with the Solar tide to give a &quot;Spring tide&quot; that has a larger range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the moon is at first quarter, or third quarter (ie a half moon), the solar tide and the lunar tide are acting against each other, and you get a &quot;neap tide&quot; with a lower range than a spring tide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of the tide on a particular day at a particular location is strongly affected by the local shape of the coast and sea floor, Some places only have one tide each day, or tides of different size, due to these effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the name &quot;Spring tide&quot; is not from the season. You get Spring tides every New and every Full moon. It comes from an old English word meaning &quot;to bulge&quot; The tide &quot;bulges&quot; during New and Full moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you are not correct to say that tides are higher on the day of Full moon. The New Moon's gravity does create tides, and tides at New moon are as large as the tides at Full moon, and larger than the tides at half moon.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-02-20T15:27:05.650" LastActivityDate="2018-02-20T15:27:05.650" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25213" PostTypeId="2" ParentId="25187" CreationDate="2018-02-21T00:33:36.820" Score="1" Body="&lt;p&gt;The density of the solar atmosphere decreases with radius. The position of the solar photosphere is defined by where the optical depth (measured inwards) reaches unity. There is no direct connection between pressure/density and gravity; only between the pressure &lt;em&gt;gradient&lt;/em&gt; and gravity.&#xA;$$\frac{dP}{dr}= -\rho g$$&#xA;A larger gravity merely increases the pressure and density &lt;em&gt;gradient&lt;/em&gt; over what it would be in a lower gravitational field at the same density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The photosphere occurs at the density and temperature it does, because that is where the optical depth reaches unity. The species that dominates the opacity at these temperatures is the H$^{-}$ ion (hydrogen with an extra electron).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth's atmosphere is largely neutral and even at higher densities has a much lower opacity at visible wavelengths.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-02-21T00:43:31.893" LastActivityDate="2018-02-21T00:43:31.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25214" PostTypeId="2" ParentId="25187" CreationDate="2018-02-21T04:23:25.283" Score="4" Body="&lt;p&gt;I think there is a misconception underlying your question. You write&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;But Solar surface gravity is about 28 times terrestrial surface gravity.&lt;/p&gt;&#xA;&lt;p&gt;So, shouldn't the density of gases at the photosphere of the sun be comparable to that on the surface of the Earth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It does sound natural that a stronger gravity would imply a thicker atmosphere, but no matter how strong the gravity is, there is a point where the atmosphere fades away almost to nothing. At the height of the ISS, earth's gravity is around 80% of the gravity at sea level. But yet the ISS is orbiting in the vacuum of space. Similarly, above Jupiter's atmosphere, the gravity is still more than twice that of earth, but there is still vacuum. With the sun, it just so happens that the part of the atmosphere that produces the light we see is close enough to the (imprecisely defined) edge of space that its density has gone down almost to nothing.&lt;/p&gt;&#xA;&lt;p&gt;Certainly, if it were cooler, it would be denser. But some day the sun will be a white dwarf, about the size of the earth but with not that much less mass than it has now. Its gravity will be hugely strong, but there will still be a top of the atmosphere were it is less than a millionth of earth's atmosphere. As it cools, that atmosphere will cool so that it is no warmer than earth's atmosphere, but there will still be a point where the atmosphere is that thin. And even above that height, gravity will be much higher than even at the current surface of the sun.&lt;/p&gt;&#xA;" OwnerUserId="7752" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-02-21T04:23:25.283" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25215" PostTypeId="2" ParentId="25188" CreationDate="2018-02-21T04:44:07.267" Score="2" Body="&lt;p&gt;Just going off the &lt;a href=&quot;https://en.wikipedia.org/wiki/Superflare&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; you posted, it says the hot jupiter superflare theory was abandoned. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The flares were initially explained by postulating giant planets in&#xA;  very close orbits, such that the magnetic fields of the star and&#xA;  planet were linked. The orbit of the planet would warp the field lines&#xA;  until the instability released magnetic field energy as a flare.&#xA;  However, no such planet has showed up as a Kepler transit and this&#xA;  theory has been abandoned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-02-21T04:44:07.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25224" PostTypeId="2" ParentId="10946" CreationDate="2018-02-21T12:47:03.843" Score="0" Body="&lt;p&gt;Now, if the Moon needs to escape the Earth and go for Sun, it needs more speed to do so. It cannot escape Earth until it's speed is enough for escaping. It needs more velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbit of the Moon &lt;em&gt;around the Sun&lt;/em&gt; is essentially a circle with a radius of 150million km. Its orbit &lt;em&gt;around the Earth&lt;/em&gt; has only a 400 000 km radius, thus the effect of the Earth is only a minor perturbation of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking from the Sun, the Moon has a circular orbit around it, just like Earth, and their effect to eachother is nearly negligible.&lt;/p&gt;&#xA;" OwnerUserId="21875" LastEditorUserId="1891" LastEditDate="2018-02-23T12:29:12.000" LastActivityDate="2018-02-23T12:29:12.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25227" PostTypeId="1" AcceptedAnswerId="27029" CreationDate="2018-02-21T16:22:36.510" Score="4" ViewCount="171" Body="&lt;p&gt;I have read that &quot;earth rotation angle&quot; is the modern replacement for &quot;Greenwich sidereal time&quot;. Is it actually the case that modern professional astronomers are using earth rotation angle? Either way, what is the reason for using the one they use?&lt;/p&gt;&#xA;" OwnerUserId="7385" LastActivityDate="2018-07-20T23:03:35.963" Title="Do today's professional astronomers use earth rotation angle or sidereal time?" Tags="&lt;time&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25228" PostTypeId="1" AcceptedAnswerId="25326" CreationDate="2018-02-21T17:16:53.513" Score="1" ViewCount="93" Body="&lt;p&gt;I recently upgraded from a Meade ETX-RA to and ETX-EC and bought a used Audiostar controller. My EC came with a Meade autofocuser and one of the old (I think it’s 494) controllers that has the focus controls and the four lights on it. Is it possible to just plug the autofocuser AUX cord into the Audiostar and focus from that? I read the manual for the autofocuser and it seems like I should be able to. Just not sure which controls I am supposed to use.&lt;/p&gt;&#xA;" OwnerUserId="21237" LastActivityDate="2018-05-01T19:14:39.183" Title="Focusing with Meade Audiostar" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25229" PostTypeId="1" AcceptedAnswerId="25867" CreationDate="2018-02-21T23:07:46.103" Score="8" ViewCount="269" Body="&lt;p&gt;The new paper in Nature &lt;a href=&quot;https://www.nature.com/articles/nature25151&quot; rel=&quot;nofollow noreferrer&quot;&gt;A surge of light at the birth of a supernova&lt;/a&gt; (doi:10.1038/nature25151) describes the fortuitous capture of a supernova &quot;Shock Breakout&quot; - the earliest rise in brightness of a supernova only about 4 hours old. An &lt;em&gt;open access epdf&lt;/em&gt; is linked in the Washington Post article if you &lt;em&gt;click from inside it:&lt;/em&gt; &lt;a href=&quot;https://www.washingtonpost.com/news/speaking-of-science/wp/2018/02/21/a-self-taught-astronomer-spotted-something-no-scientist-had-ever-seen/&quot; rel=&quot;nofollow noreferrer&quot;&gt;A self-taught astronomer spotted something no scientist had ever seen&lt;/a&gt; which is a good read itself.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;His discovery, &lt;a href=&quot;http://nature.com/articles/doi:10.1038/nature25151&quot; rel=&quot;nofollow noreferrer&quot;&gt;reported Wednesday in the journal Nature&lt;/a&gt;, is a landmark for astronomy. Buso’s images are the first to capture the brief “shock breakout” phase of a supernova, when a wave of energy rolls from a star’s core to its exterior just before the star explodes. Computer models had suggested the existence of this phase, but no one had witnessed it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is this the earliest that such a phenomenon has been &quot;seen&quot; by a person, or does this observation also hold the record after including all of the automated searches for supernovae by robotic telescopes as well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rBMl7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rBMl7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; &quot;Astronomer Victor Buso poses in front of the telescope with which he witnessed the birth of the supernova 2016gkg. (Victor Buso)&quot; From &lt;a href=&quot;https://www.washingtonpost.com/news/speaking-of-science/wp/2018/02/21/a-self-taught-astronomer-spotted-something-no-scientist-had-ever-seen/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. Click/open for full size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H3fPU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H3fPUm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-12-28T01:45:50.467" LastActivityDate="2019-12-28T01:45:50.467" Title="Does this amateur observation hold a record even including robotic searches for supernovae?" Tags="&lt;observational-astronomy&gt;&lt;amateur-observing&gt;&lt;supernova&gt;&lt;serendipity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25234" PostTypeId="2" ParentId="25126" CreationDate="2018-02-22T11:25:24.393" Score="3" Body="&lt;p&gt;Here is my attempt to reconcile your calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the AB g-band apparent magnitude is 22.5, then the flux density in the g-band is given by &#xA;$$f_{\nu} = 10^{(-48.6-22.5)/2.5} =  3.63 \times 10^{-29}\ {\rm erg\ cm}^{-2} {\rm s}^{-1} {\rm Hz}^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the distance is 1991 Mpc, then the absolute g magnitude is&#xA;$$ M_g = m_g - 5\log d + 5 = -19.0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absolute AB magnitude of the Sun &lt;em&gt;in the g-band&lt;/em&gt; (from your source) is 5.12.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter tells us that the ratio of the &quot;&lt;em&gt;g-band&lt;/em&gt; luminosites&quot; of the galaxy and the Sun are given by&#xA;$$\frac{L_g}{L_{\odot,g}} = 10^{0.4(M_{{\odot},g} - M_g)} = 4.45\times 10^{9}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One cannot say more than this, in particular one cannot calculate the luminosity of the galaxy, without knowing more about its spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Also note that the equation above cannot be used to find the ratio of flux in one band to bolometric flux, as I think you are trying to do&lt;/strong&gt;. To see this, consider that the absolute V-band magnitude and bolometric magnitude of the Sun are almost the same. This does &lt;em&gt;not&lt;/em&gt; mean that all the flux from the Sun emerges in the V band!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your second method requires a figure for $\Delta \nu$, but you haven't said what you have used. The g-filter has a width of around $\Delta \lambda =100$ nm. Using $\lambda \nu = c$ we can say&#xA;$$ \Delta \nu  = |c\Delta \lambda/\lambda^2| = 1.28\times 10^{14}\ {\rm Hz}$$&#xA;$$ L_g = 4\pi d^2 f_{\nu} \Delta \nu = 2.2 \times 10^{42} {\rm erg/s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two figures would agree precisely if the &quot;g-band luminosity&quot; of the Sun were $4.9 \times 10^{32}$ erg/s, or in other words if 13% of the solar luminosity emerged in the g-band as defined above. This does not sound unreasonable, but requires a detailed integration of the solar spectrum over the actual g-band filter profile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Additional:&lt;/strong&gt; Just to check the numbers above, we can estimate the flux density of the Sun (in wavelength units, above the atmosphere) to be about 1.7 W m$^{-2}$ nm$^{-1}$ at 467 nm (just search on 'solar spectrum' for many examples) at a distance of $1.5 \times 10^{11}$ m. Putting these numbers together, I calculate a &quot;g-band luminosity&quot; for the Sun of &#xA;$$ L_{\odot,g}=4\pi d^2 f_{\lambda}\Delta \lambda = 4.8\times 10^{25} \rm W$$&#xA;which is indeed 13% of the solar bolometric luminosity.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-02-25T09:04:43.653" LastActivityDate="2018-02-25T09:04:43.653" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25236" PostTypeId="1" CreationDate="2018-02-22T14:13:52.437" Score="2" ViewCount="71" Body="&lt;p&gt;I just recently found out about the GRACE Nasa mission which basically mapped out the gravitational anomalies of earth as well as serving extensively in the study of earths geology and the movement of its distribution of water. While I continued reading about it I came across GRAIL which did a similar job but over our moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Link here to learn about GRACE: &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravity_Recovery_and_Climate_Experiment?scrlybrkr=5922e11d&quot; rel=&quot;nofollow noreferrer&quot;&gt;GRACE&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here for GRAIL: &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravity_Recovery_and_Interior_Laboratory&quot; rel=&quot;nofollow noreferrer&quot;&gt;GRAIL&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What i'm now wondering is are there space agencies or organizations considering missions to any of the other planets in our solar system and attempting to measure their gravitational anomalies in a similar manner hopefully giving vital information on their inner structure or formation.  &lt;/p&gt;&#xA;" OwnerUserId="15365" LastActivityDate="2018-02-24T17:51:24.120" Title="Are there currently any planned or proposed missions to measure the gravitational fields of other planets in our solar system?" Tags="&lt;gravity&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25237" PostTypeId="2" ParentId="25236" CreationDate="2018-02-22T18:27:57.983" Score="5" Body="&lt;p&gt;There are no such missions in an advanced state of planning, however we do have a good idea of the gravity field around Mars, gleaned from observations of anomalies in the orbits of the various orbital probes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These data were gathered into a &lt;a href=&quot;https://www.nasa.gov/feature/goddard/2016/mars-gravity-map&quot; rel=&quot;nofollow noreferrer&quot;&gt;map of the gravity field of Mars&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to that NASA article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The map was derived using Doppler and range tracking data collected by &lt;a href=&quot;http://deepspace.jpl.nasa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA's Deep Space Network&lt;/a&gt; from three NASA spacecraft in orbit around Mars: &lt;a href=&quot;http://marsprogram.jpl.nasa.gov/mgs/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mars Global Surveyor&lt;/a&gt; (MGS), &lt;a href=&quot;http://mars.nasa.gov/odyssey/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mars Odyssey&lt;/a&gt; (ODY), and the &lt;a href=&quot;https://www.nasa.gov/mission_pages/MRO/main/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mars Reconnaissance Orbiter&lt;/a&gt; (MRO). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The research is published as Antonio Genova et al. 2016, &lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/S0019103516001287?via%3Dihub&quot; rel=&quot;nofollow noreferrer&quot;&gt;Seasonal and static gravity field of Mars from MGS, Mars Odyssey and MRO radio science&lt;/a&gt;, Icarus, vol. 272, pp. 228-245; doi: 10.1016/j.icarus.2016.02.050&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jmzl1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jmzl1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cassini%E2%80%93Huygens&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cassini Spacecraft&lt;/a&gt; has made detailed gravitational measurements of Saturn and and along with imaging, internal structure details of its moon Enceladus. Also the &lt;a href=&quot;https://en.wikipedia.org/wiki/Juno_(spacecraft)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Juno Spacecraft&lt;/a&gt; has made detailed measurements of Jupiter's internal structure via maps of it's gravity field as well. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="7982" LastEditDate="2018-02-24T01:53:05.723" LastActivityDate="2018-02-24T01:53:05.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25238" PostTypeId="2" ParentId="10946" CreationDate="2018-02-22T20:00:19.970" Score="-2" Body="&lt;p&gt;Newton law: &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F=G*(m1*m2)/d² is the gravitationnal force between 2 things of mass m1 and m2, separated by a distance d. G is the gravitational constant (I don't remember the value).&lt;br /&gt;&#xA;--&gt; F_earth/moon=F_moon/earth=G*(m_moon*m_earth)/d²&lt;br /&gt;&#xA;Same thing for F_sun/moon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll notice that F_earth/moon is greater than the other force, so the Moon is more attracted by Earth than Sun.&lt;/p&gt;&#xA;" OwnerUserId="21894" LastEditorUserId="21894" LastEditDate="2018-02-23T21:09:55.800" LastActivityDate="2018-02-23T21:09:55.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25239" PostTypeId="1" AcceptedAnswerId="25243" CreationDate="2018-02-22T23:14:02.867" Score="-2" ViewCount="719" Body="&lt;p&gt;Does the gravity wells differ enough between galaxies to have different speeds in space time? How much slower in space time would the biggest galaxy have compared to the Milky Way? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xCO9C.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xCO9C.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-02-23T03:09:18.133" LastActivityDate="2018-02-23T20:47:47.700" Title="How does space time differ between galaxies?" Tags="&lt;gravity&gt;&lt;galaxy&gt;&lt;astrophysics&gt;&lt;space-time&gt;&lt;mass&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2018-02-23T22:02:59.533" ContentLicense="CC BY-SA 3.0" />
  <row Id="25240" PostTypeId="1" AcceptedAnswerId="25251" CreationDate="2018-02-23T02:19:48.223" Score="6" ViewCount="298" Body="&lt;p&gt;About 11 hours after the detection of GW170817 by Ligo/Virgo and the quickly succeeding detected GRB170817, a transient afterglow in the optical-nIR frequency appeared which showed many signs of &lt;em&gt;r&lt;/em&gt;-process nucleosynthesis. This transient was given the identification AT 2017 gfo.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;em&gt;SSS17a&lt;/em&gt;, now with the IAU identification of &lt;em&gt;AT 2017 gfo&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My questions are simple. What do the acronyms &quot;AT&quot; and &quot;gfo&quot; mean? And for that matter, what does the acronymn &quot;SSS&quot; stand for? I tried to do a quick search through &lt;a href=&quot;https://www.iau.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;the IAU website&lt;/a&gt; but all I could find were general instructions on how to come up with your own astronomical identifier. I suspect &quot;AT&quot; means &lt;a href=&quot;https://en.wikipedia.org/wiki/Transient_astronomical_event&quot; rel=&quot;nofollow noreferrer&quot;&gt;astronomical transient&lt;/a&gt;, but I can not confirm. &lt;/p&gt;&#xA;" OwnerUserId="16424" LastEditorUserId="16424" LastEditDate="2018-02-23T13:01:28.457" LastActivityDate="2018-02-23T14:46:09.403" Title="What do the acronyms in &quot;AT 2017 gfo&quot; mean?" Tags="&lt;astrophysics&gt;&lt;gravitational-waves&gt;&lt;naming&gt;&lt;gamma-ray-bursts&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25241" PostTypeId="2" ParentId="25188" CreationDate="2018-02-23T03:09:49.053" Score="2" Body="&lt;p&gt;Sun is more than one thousand times the mass of Jupiter, so it seems unlikely that even a very close approach by a &quot;hot Jupiter&quot; to its host star would cause a flare, and especially not a &lt;a href=&quot;https://science.nasa.gov/science-news/science-at-nasa/2012/28jun_alienflare&quot; rel=&quot;nofollow noreferrer&quot;&gt;superflare&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="21897" LastActivityDate="2018-02-23T03:09:49.053" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25242" PostTypeId="2" ParentId="25165" CreationDate="2018-02-23T03:53:37.480" Score="6" Body="&lt;p&gt;Saturn is about 10.1 AU from Sun, so by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Inverse-square_law#Light_and_other_electromagnetic_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;inverse-square law&lt;/a&gt;, its solar illumination is a little less than one-hundredth that of Earth, which is also called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_constant&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar constant&lt;/a&gt; with a value of 1362 W/m².&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$I = I_0 \frac{1\ \mathrm{AU}^2}{R^2} \approx 1362\ \mathrm{W\ m^{-2}}\frac{1^2}{9.583^2}\approx14.83\ \mathrm{W/m^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Saturn's relatively large eccentricity of 0.0565 puts it anywhere between 9.041 and 10.124 AU from the Sun, &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; varies from 16.66 and 13.29 W/m^2 during its 29.5 year orbital period.&lt;/p&gt;&#xA;" OwnerUserId="21897" LastEditorDisplayName="user8021" LastEditDate="2019-09-24T14:41:15.007" LastActivityDate="2019-09-24T14:41:15.007" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25243" PostTypeId="2" ParentId="25239" CreationDate="2018-02-23T04:22:44.770" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the gravity wells differ enough between galaxies to have different speeds in space time ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are using these terms in a very confused way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational field of an object depends on it's mass distribution.  It is normally described as a potential field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A gravitational field does not a have a speed that characterizes it, not do they have a speed through space-time (the field always propagates at the speed of light, regardless of source).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How much slower in space time would the biggest galaxy have compared to the Milky Way? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Keeping in mind that size has nothing to do with speed, note that are normally two aspects to the speed of an object &lt;em&gt;relative to something else&lt;/em&gt; (important to remember that speeds are always relative to some other object that &quot;defines&quot; what is at rest and that this choice is arbitrary - that's the theory of relativity for you).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxies can have a local speed due to e.g. motion within the local group like &lt;a href=&quot;https://en.wikipedia.org/wiki/Local_Group&quot; rel=&quot;nofollow noreferrer&quot;&gt;our own local group of galaxies&lt;/a&gt;. This aspect of motion is really just governed to &quot;normal&quot; gravitational effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a much larger scale they can have an apparent motion due to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Metric_expansion_of_space&quot; rel=&quot;nofollow noreferrer&quot;&gt;expansion of the universe&lt;/a&gt;.  This part of motion is due to a large scale effect only explained by using general relativity and which only has an observable effect over cosmological distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational field of an object has no effect on it's relative motion due to the expansion of space-time, whereas it's gravitational field does affect the relative motion of objects local to it and hence their relative motion to that object.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2018-02-23T04:22:44.770" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25244" PostTypeId="2" ParentId="25166" CreationDate="2018-02-23T04:45:09.003" Score="4" Body="&lt;p&gt;Since (1) the mass of the Pluto/Charon system is very small compared with Neptune, and (2) their orbits are in 2:3 resonance, so the perturbation of Neptune's orbit would have been undetectable. University of Rochester has a good &lt;a href=&quot;http://www.pas.rochester.edu/~blackman/ast104/perturbations.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;explanation&lt;/a&gt; of this. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Accidental Discovery of Pluto&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Later supposed perturbations of the orbits of Uranus and Neptune suggested&#xA;  the presence of yet another planet beyond the orbit of Neptune.&#xA;  Eventually, in 1930, a new planet Pluto was discovered, but we now&#xA;  know that the calculations in this case were also in error because of&#xA;  an incorrect assumption about the mass of the new planet. It is now&#xA;  believed that the supposed deviations in the orbits of Neptune and&#xA;  Uranus were errors in measurement because the actual properties of&#xA;  Pluto would not have accounted for the supposed perturbations. Thus,&#xA;  the discovery of Pluto was a kind of accident&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="21897" LastEditorUserId="6841" LastEditDate="2018-02-23T12:37:45.083" LastActivityDate="2018-02-23T12:37:45.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25247" PostTypeId="1" AcceptedAnswerId="25248" CreationDate="2018-02-23T06:18:41.610" Score="1" ViewCount="56" Body="&lt;p&gt;I was looking up the radius of Proxima Centauri and got a funny result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google says it's about 100,000 km, while WolframAlpha says its about 200,000 km.  Are the error bars that big, or does someone have some bad numbers?&lt;/p&gt;&#xA;" OwnerUserId="16774" LastActivityDate="2018-02-23T08:42:29.130" Title="Inconsistent online radius value for Proxima Centauri" Tags="&lt;size&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25248" PostTypeId="2" ParentId="25247" CreationDate="2018-02-23T08:42:29.130" Score="3" Body="&lt;p&gt;Wikipedia quotes $0.1542\pm 0.0045$ solar radii which is $107276\,km$ with an error of about $3130\, km$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They also give a reference to a 2017 paper available on Arxiv &lt;a href=&quot;https://arxiv.org/abs/1611.03495&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Proxima's orbit around Alpha Centauri&lt;/em&gt;, P. Kervella, F. Thévenin, C. Lovis&lt;/a&gt; that gives this value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wolfram's estimate is too far off to be an experimental error, IMO, and I think someone has just inadvertently used the diameter (twice the radius) rather than the radius.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2018-02-23T08:42:29.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25251" PostTypeId="2" ParentId="25240" CreationDate="2018-02-23T14:01:20.610" Score="5" Body="&lt;p&gt;You are correct. The AT is short for Astronomical Transient. The follows the year and an arbitary sequence of letters to uniquely identify the event, which is assigned by the Transient Name Server&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the official IAU designation for Supernova Candidates, however true supernovae will be re-designated &quot;SN YYYYX&quot;. The neutron star merger was not a supernova, so retains its AT designation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is described at (&lt;a href=&quot;http://cdsweb.u-strasbg.fr/cgi-bin/Dic?/557556&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;) with more details at &lt;a href=&quot;https://wis-tns.weizmann.ac.il/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Transient Name Server (TNS)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-02-23T14:46:09.403" LastActivityDate="2018-02-23T14:46:09.403" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25254" PostTypeId="1" AcceptedAnswerId="25259" CreationDate="2018-02-23T19:31:30.210" Score="2" ViewCount="71" Body="&lt;p&gt;I constructed a telescope and fixed my mobile's camera to the eye lens. I then viewed the sky via the mobile and I captured this image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PyhGj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PyhGj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the magnified version of it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UnkLS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UnkLS.jpg&quot; alt=&quot;Magnified&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if this actually the spectrum of a star because I feel like maybe the human oil secreted by my hands may have got stuck on the lens and that must've dispersed the light from some source. I'd like to know all of your views. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="21572" LastEditorUserId="7982" LastEditDate="2018-02-24T02:29:05.040" LastActivityDate="2018-03-01T07:38:42.097" Title="Spectrum of star" Tags="&lt;light&gt;&lt;astrophysics&gt;&lt;star-systems&gt;&lt;spectra&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25255" PostTypeId="2" ParentId="12384" CreationDate="2018-02-23T19:40:53.423" Score="1" Body="&lt;p&gt;Angular momentum of the entire universe remains constant. Assuming that the big bang was the only  thing present initially, it wouldn't  posses any angular momentum. Now, if there are unequal number of planets, galaxies, stars,etc. in the observable universe spinning in both directions, there'd be some net angular  momentum which would violate the conservation of angular momentum. Hence, in my opinion, there are equal number of celestial bodies spinning in both directions. &lt;/p&gt;&#xA;" OwnerUserId="21572" LastEditorUserId="21572" LastEditDate="2018-02-23T20:03:41.603" LastActivityDate="2018-02-23T20:03:41.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25256" PostTypeId="1" AcceptedAnswerId="25280" CreationDate="2018-02-23T20:13:24.370" Score="-1" ViewCount="91" Body="&lt;p&gt;While I was star gazing yesterday in the night time I observed a celestial body( probably a star ) flaunting various colors and was residing near the moon from the angle I saw it. It appeared to be stationary. Any plausible explanations for this phenomenon? Thanks. &lt;/p&gt;&#xA;" OwnerUserId="21572" LastActivityDate="2018-02-25T16:37:26.037" Title="Color changing celestial body near the moon" Tags="&lt;star&gt;&lt;star-systems&gt;&lt;star-gazing&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2018-03-07T21:07:03.193" ContentLicense="CC BY-SA 3.0" />
  <row Id="25259" PostTypeId="2" ParentId="25254" CreationDate="2018-02-24T02:06:18.527" Score="3" Body="&lt;p&gt;The other &lt;a href=&quot;https://astronomy.stackexchange.com/a/25257/7982&quot;&gt;answer&lt;/a&gt; (since deleted) is not right. The blob is a dozen pixels long, but monotonic in color change. It has nothing to do with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Bayer_filter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bayer filter&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's almost certainly chromatic aberration from some combination of the telescope and eyepiece optics and the camera optics. The short focal length and therefore tiny &lt;a href=&quot;https://en.wikipedia.org/wiki/Entrance_pupil&quot; rel=&quot;nofollow noreferrer&quot;&gt;entrance pupil&lt;/a&gt; size of a cell phone's camera only intercepts a small fraction of the bundle of rays from the eyepiece's &lt;a href=&quot;https://en.wikipedia.org/wiki/Exit_pupil&quot; rel=&quot;nofollow noreferrer&quot;&gt;exit pupil&lt;/a&gt; meant to fill a &lt;em&gt;human pupil&lt;/em&gt;. If it were offset from the center, it would sample rays near the edge of the path, subjected most strongly to chromatic aberration. So even if the star appeared in the middle of the recorded image, this offset could result in a first-order dispersion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the three color &quot;blobs&quot; (caused by the shape of the transmission spectra of the three color filter types) the RED and GREEN blobs overlap, but the blue blob is offset much farther from the other two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is consistent with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Dispersion_(optics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;dispersion&lt;/a&gt; characteristics of glasses and plastic, where the index of refraction changes more and more rapidly toward the blue end of the spectrum, as the energy of the photons get closer to a real absorption edge somewhere in the UV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dmx6K.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dmx6K.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/plh8E.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/plh8E.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-03-01T07:38:42.097" LastActivityDate="2018-03-01T07:38:42.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25260" PostTypeId="1" AcceptedAnswerId="25526" CreationDate="2018-02-24T03:03:04.130" Score="3" ViewCount="2623" Body="&lt;p&gt;If you create a electromagnetic field under a surface could you then create artificial gravity effect by wearing magnetized clothing and shoes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or could you bombard objects that you want to experience gravitational forces with artificial gravitons or graviolis? Or inject or drink a ferrofluid (non-toxic) like that used in rocket fuel? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A ferrofluid (portmanteau of ferromagnetic and fluid) is a liquid that&#xA;  becomes strongly magnetized in the presence of a magnetic field. A&#xA;  grinding process for ferrofluid was invented in 1963 by NASA's Steve&#xA;  Papell as a liquid rocket fuel that could be drawn toward a pump inlet&#xA;  in a weightless environment by applying a magnetic field.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Ferrofluid&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Ferrofluid&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I mean is create the forces of attraction and then make the objects you want to attract attractive, obviously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is as opposed to using other methods such as acceleration or centrifugal or centripetal force (or in addition to it). &lt;/p&gt;&#xA;" OwnerUserId="13092" LastEditorUserId="13092" LastEditDate="2018-02-24T03:10:41.670" LastActivityDate="2020-09-26T10:13:54.020" Title="Can you simulate artificial gravity by using an electromagnetic field?" Tags="&lt;gravity&gt;" AnswerCount="3" CommentCount="5" ClosedDate="2020-09-27T09:15:03.493" ContentLicense="CC BY-SA 3.0" />
  <row Id="25261" PostTypeId="1" AcceptedAnswerId="25264" CreationDate="2018-02-24T03:42:23.900" Score="1" ViewCount="62" Body="&lt;p&gt;The Phys.org article &lt;a href=&quot;https://phys.org/news/2018-02-astronomers-s0-star-ready-big.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomers discover S0-2 star is single and ready for big Einstein test&lt;/a&gt; describes the results of a careful analysis of radial velocity measurements of the star S0-2, which orbits the supermassive black hole at the center of our (Milky Way) galaxy with a period of only 16 years. The paper itself (Devin S. Chu et al, 2018) is available for open access at &lt;a href=&quot;https://arxiv.org/abs/1709.04890&quot; rel=&quot;nofollow noreferrer&quot;&gt;ArXiv&lt;/a&gt;, and also &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/1538-4357/aaa3eb/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;ApJ&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only mention of the word wavelength is in Section 2.2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The radial velocity of S0-2 is measured from both data sets by fitting a Gaussian to the Brγ absorption line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What exactly is&lt;/em&gt; a Brγ (&quot;Br&quot; + gamma) absorption line?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; &quot;The orbit of S0-2 (light blue) located near the Milky Way's supermassive black hole will be used to test Einstein's Theory of General Relativity and generate potentially new gravitational models. Credit: W. M. Keck Observatory (Click for full size)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9Iv1I.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9Iv1Im.jpg&quot; alt=&quot;stars orbiting supermassive black hole center of Milky Way&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-02-24T23:36:28.960" Title="What spectral feature was used to confirm that S0-2 is &quot;single and available&quot;? What is Brγ?" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25264" PostTypeId="2" ParentId="25261" CreationDate="2018-02-24T07:40:40.253" Score="4" Body="&lt;p&gt;Brγ is a line in the emission spectrum of Hydrogen, or single-electron (and therefore) hydrogen-like &lt;a href=&quot;https://en.wikipedia.org/wiki/Rydberg_atom&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rydberg&lt;/a&gt; atoms in general. In particular, it is the emission due to an electron moving from the 7th to the 4th energy level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Br is short for &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrogen_spectral_series#Brackett_series_(n%E2%80%B2_=_4)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brackett&lt;/a&gt;, after &lt;a href=&quot;https://en.wikipedia.org/wiki/Frederick_Sumner_Brackett&quot; rel=&quot;nofollow noreferrer&quot;&gt;Frederik Brackett&lt;/a&gt;, and gamma because &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\alpha: \ \ 5 \rightarrow 4,$ &lt;/li&gt;&#xA;&lt;li&gt;$\beta:  \ \ 6 \rightarrow 4,$ &lt;/li&gt;&#xA;&lt;li&gt;$\gamma: \ \ 7 \rightarrow 4,$ &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;...and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brγ has a wavelength of 2166nm, as you would get from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Rydberg_formula&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rydberg formula&lt;/a&gt;, which is well in the infra-red part of the spectrum. It is probably chosen as it has low absorption by the dust towards the centre of the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Screenshot from &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrogen_spectral_series#Brackett_series_(n%E2%80%B2_=_4)&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vDj5l.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vDj5l.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="7982" LastEditDate="2018-02-24T23:36:28.960" LastActivityDate="2018-02-24T23:36:28.960" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25265" PostTypeId="1" AcceptedAnswerId="25267" CreationDate="2018-02-24T12:52:19.997" Score="0" ViewCount="312" Body="&lt;p&gt;I come from a physics background where I don't believe something unless the evidence is shown to me, however I have a friend who believes Aliens have visited earth , which he believes based on conspiracy theories.&lt;br&gt;&#xA;My question is: is there any solid scientific reasoning as to why the conspiracy theories aren't true. As for him I need to give him evidence as to why they don't exist , and not the other way around .&lt;/p&gt;&#xA;" OwnerUserId="21919" LastActivityDate="2018-02-26T09:51:35.090" Title="How to prove to my friend that aliens have never visited earth" Tags="&lt;extra-terrestrial&gt;" AnswerCount="2" CommentCount="9" ClosedDate="2018-02-26T15:36:12.580" ContentLicense="CC BY-SA 3.0" />
  <row Id="25267" PostTypeId="2" ParentId="25265" CreationDate="2018-02-24T14:21:05.583" Score="2" Body="&lt;p&gt;If your friend believes in these conspiracy theories he's not likely to listen to scientific reasoning, but I'll give it a shot. Obviously there's no direct evidence, but there are convincing reasons to rule out this possibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, what would it take for life to travel to Earth? Well, the &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_potentially_habitable_exoplanets&quot; rel=&quot;nofollow noreferrer&quot;&gt;closest &lt;em&gt;potentially&lt;/em&gt; habitable planet&lt;/a&gt; is 4.2 light years away from us. That means it would take light, the fastest thing in the entire universe, over 4 years to travel that distance. That's an incomprehensibly vast distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, though, that the chance of such intelligent life on Proxima b is quite slim, since it faces stellar wind pressures over 2,000 times what we experience on Earth. Thus, it likely has had its atmosphere blown away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, so how about the next potentially habitable planets? The next closest one, Luyten b, is a more likely candidate for life; it seems to have a similar size and potentially similar temperature to Earth. However, it's three times farther than Proxima b. So for starter's, we'd be making two assumptions: that life exists on Luyten b, and that it is capable of traveling across an unfathomable distance like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Luyten b doesn't satisfy both of those qualities, then that's not good. The subsequent exoplanets on that list have major problems that make habitability extremely unlikely, and the only good candidate shows up at around 39 light years from us — another 3 times farther.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if we assume that any of these planets can support life capable of traveling to Earth, we raise the second question: why would they bother? There would be no certainty that they'd find anything on Earth. There'd be no certainty that they'd survive on Earth. Unless they have some method of magically confirming life, that entire trip seems pointless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the lack of definitive evidence makes this possibility negligible.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastEditorUserId="9301" LastEditDate="2018-02-24T20:00:54.627" LastActivityDate="2018-02-24T20:00:54.627" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="25269" PostTypeId="2" ParentId="25169" CreationDate="2018-02-24T15:39:07.413" Score="0" Body="&lt;p&gt;Quite simply,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in both cases the effect is absolutely tiny.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is your basic confusion, on both this and the other similar question:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;very heavy objects cause time dilation&lt;/li&gt;&#xA;&lt;li&gt;a galaxy is just &lt;em&gt;totally badass heavy&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;you naturally assumed then that galaxies cause time dilation&lt;/li&gt;&#xA;&lt;li&gt;Very surprisingly, this is simply &lt;em&gt;not correct&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The reason? Galaxies are very big. That is to say very thin.  They have incredibly low density.  Because of this there is simply no dilation at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Well, only a minuscule amount - but that is true of simply, say, standing on Earth.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the story!&lt;/p&gt;&#xA;" OwnerUserId="13071" LastActivityDate="2018-02-24T15:39:07.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25272" PostTypeId="2" ParentId="11733" CreationDate="2018-02-24T17:48:46.240" Score="1" Body="&lt;p&gt;Earth does have rings!  Not icy ones like Saturn, but ones made up mostly of solid masses a few meters across. A relatively sharp ring is centred on geostationary orbit. A rather more diffuse one (although probably with higher total mass) reaches up to altitudes of a few hundred km from the surface. The second one is constantly losing mass back to the planet due to friction from the atmosphere, but is replenished by a complex biological process.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-02-24T17:48:46.240" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25273" PostTypeId="2" ParentId="25236" CreationDate="2018-02-24T17:51:24.120" Score="2" Body="&lt;p&gt;We get this information (if not quite as accurately as from dedicated missions like GRACE, whenever we have a probe in orbit around a planet. By tracking it carefully we can get information about the variation in the gravitational field. &lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-02-24T17:51:24.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25275" PostTypeId="1" AcceptedAnswerId="25276" CreationDate="2018-02-25T02:49:38.930" Score="5" ViewCount="165" Body="&lt;p&gt;From the earth, and from the sun. Only interested in active, operational ones.&lt;/p&gt;&#xA;" OwnerUserId="21930" LastActivityDate="2018-02-25T05:32:03.357" Title="Which space telescope is the most distant?" Tags="&lt;telescope&gt;&lt;space&gt;&lt;space-telescope&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25276" PostTypeId="2" ParentId="25275" CreationDate="2018-02-25T03:49:21.177" Score="10" Body="&lt;p&gt;Currently New Horizions is temporarily hibernating; it's last activity was two months ago. So I'm going to post a supplementary answer here because it is &amp;quot;operational&amp;quot; in the sense that it still works and will be used again, even though it is not &amp;quot;active&amp;quot; at the moment.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The most recent and farthest-from-earth telescopic observations that I know of are from the &lt;a href=&quot;https://en.wikipedia.org/wiki/New_Horizons&quot; rel=&quot;noreferrer&quot;&gt;New Horizons&lt;/a&gt; spacecraft and it's &lt;a href=&quot;https://en.wikipedia.org/wiki/New_Horizons#Long-Range_Reconnaissance_Imager_(LORRI)&quot; rel=&quot;noreferrer&quot;&gt;Long-Range Reconnaissance Imager&lt;/a&gt; (LORRI), a 208 mm (8.2 inch) diameter Ritchey–Chretien telescope equipped with silicon-carbide optics and a cooled silicon CCD imager with a resolution of about 1 arcsecond:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The Long-Range Reconnaissance Imager (LORRI) is a long-focal-length imager designed for high resolution and responsivity at visible wavelengths. The instrument is equipped with a 1024×1024 pixel by 12-bits-per-pixel monochromatic CCD imager giving a resolution of 5 μrad (~1 arcsec)] The CCD is chilled far below freezing by a passive radiator on the antisolar face of the spacecraft. This temperature differential requires insulation, and isolation from the rest of the structure. The 208.3 mm (8.20 in) aperture &lt;a href=&quot;https://en.wikipedia.org/wiki/Ritchey-Chr%C3%A9tien_telescope&quot; rel=&quot;noreferrer&quot;&gt;Ritchey–Chretien&lt;/a&gt; mirrors and metering structure are made of silicon carbide, to boost stiffness, reduce weight, and prevent warping at low temperatures. The optical elements sit in a composite light shield, and mount with titanium and fiberglass for thermal isolation. Overall mass is 8.6 kg (19 lb), with the optical tube assembly (OTA) weighing about 5.6 kg (12 lb), for one of the largest silicon-carbide telescopes flown at the time (now surpassed by Herschel).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In December 2017 LORRI captured the images below of two Kuiper belt objects. From the February 9, 2018 NASA News item &lt;a href=&quot;https://www.nasa.gov/feature/new-horizons-captures-record-breaking-images-in-the-kuiper-belt&quot; rel=&quot;noreferrer&quot;&gt;New Horizons Captures Record-Breaking Images in the Kuiper Belt&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;NASA’s New Horizons spacecraft recently turned its telescopic camera toward a field of stars, snapped an image – and made history.&lt;/p&gt;&#xA;&lt;p&gt;The routine calibration frame of the “Wishing Well” galactic open star cluster, made by the Long Range Reconnaissance Imager (LORRI) on Dec. 5, was taken when New Horizons was 3.79 billion miles (6.12 billion kilometers, or 40.9 astronomical units) from Earth – making it, for a time, the farthest image ever made from Earth.&lt;/p&gt;&#xA;&lt;p&gt;New Horizons was even farther from home than NASA’s Voyager 1 when it captured the famous “Pale Blue Dot” image of Earth. That picture was part of a composite of 60 images looking back at the solar system, on Feb. 14, 1990, when Voyager was 3.75 billion miles (6.06 billion kilometers, or about 40.5 astronomical units [AU]) from Earth. Voyager 1’s cameras were turned off shortly after that portrait, leaving its distance record unchallenged for more than 27 years.&lt;/p&gt;&#xA;&lt;p&gt;LORRI broke its own record just two hours later with images of Kuiper Belt objects 2012 HZ84 and 2012 HE85 – further demonstrating how nothing stands still when you’re covering more than 700,000 miles (1.1 million kilometers) of space each day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Here is a 25-Feb-2018 screen shot of a map of the Solar System with New Horizon's current position plotted. You can read more about New Horizons at:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://pluto.jhuapl.edu/Mission/Where-is-New-Horizons/index.php&quot; rel=&quot;noreferrer&quot;&gt;Where is New Horizons&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.nasa.gov/mission_pages/newhorizons/main/index.html&quot; rel=&quot;noreferrer&quot;&gt;NASA's New Horizons main page&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9sMOm.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9sMOm.jpg&quot; alt=&quot;New Horizons position Feb. 2018&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Instrumentation on New Horizons is extensively described/explained in Spaceflight 101's &lt;a href=&quot;http://spaceflight101.com/newhorizons/instrument-overview/&quot; rel=&quot;noreferrer&quot;&gt;Instrument Overview&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The LORRI telescope is described for example in the ArXiv preprint &lt;a href=&quot;https://arxiv.org/abs/0709.4278&quot; rel=&quot;noreferrer&quot;&gt;Long-Range Reconnaissance Imager on New Horizons&lt;/a&gt; (click &amp;quot;PDF&amp;quot;). It has a spectral range of about 350 to 850 nm. The limiting magnitude for high precision positional measurements is about +11.5 for a short exposure, but with some binning and a roughly 10 second exposure that increases to about +17:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;For optical navigation, LORRI is required to be able to image a star of visual magnitude V=11.5 at SNR&amp;gt;7 in a single 100 ms exposure, with full width at half maximum (FWHM) &amp;gt;1 pixel. It is not desirable for too great a fraction of the energy from a point source to be imaged onto a single pixel, because stellar images become too undersampled. LORRI has a 4×4 pixel binning mode, for which its limiting magnitude requirement is V&amp;gt;17 in a single exposure of 9.9 s. This 4×4 pixel-binning mode will be used to search for the target KBO and to perform optical navigation on approach.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PKujo.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PKujo.png&quot; alt=&quot;New Horizon's Long Range Imager LORRI&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; Image and caption from &lt;a href=&quot;https://www.nasa.gov/feature/new-horizons-captures-record-breaking-images-in-the-kuiper-belt&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;With its Long Range Reconnaissance Imager (LORRI), New Horizons has observed several Kuiper Belt objects (KBOs) and dwarf planets at unique phase angles, as well as Centaurs at extremely high phase angles to search for forward-scattering rings or dust. These December 2017 false-color images of KBOs 2012 HZ84 (left) and 2012 HE85 are, for now, the farthest from Earth ever captured by a spacecraft. They're also the closest-ever images of Kuiper Belt objects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/L41HN.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L41HN.jpg&quot; alt=&quot;KBO objects imaged by New Horizons&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-02-25T05:32:03.357" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="25278" PostTypeId="5" CreationDate="2018-02-25T09:58:23.170" Score="0" Body="&lt;p&gt;Most questions about the International Space Station belong on our sister site &lt;a href=&quot;https://space.stackexchange.com/&quot;&gt;Space Exploration Stack Exchange&lt;/a&gt;. Please check if &lt;a href=&quot;https://space.stackexchange.com/questions/tagged/iss&quot;&gt;your question hasn't been asked already&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7472" LastEditorUserId="7472" LastEditDate="2018-02-26T23:59:07.497" LastActivityDate="2018-02-26T23:59:07.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25279" PostTypeId="4" CreationDate="2018-02-25T09:58:23.170" Score="0" Body="Most questions about the International Space Station belong on Space Exploration Stack Exchange." OwnerUserId="7472" LastEditorUserId="7472" LastEditDate="2018-02-26T23:58:52.433" LastActivityDate="2018-02-26T23:58:52.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25280" PostTypeId="2" ParentId="25256" CreationDate="2018-02-25T16:37:26.037" Score="5" Body="&lt;p&gt;Judging by the time of your post, I would probably say the star you saw was Aldebaran as it was very close to the moon on that night. It's the brightest star in the constellation Taurus so is easily seen close to the moon, where the moonlight washes out the dimmer stars around it. It is normally seen red/orange to the naked eye but atmospheric interference can cause it to 'twinkle' and the influence of the moon's light could possibly make it appear to look slightly different colours depending on its height in the sky. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This to me seems to be the most logical explanation and also seems that this was the most probable star you were seeing that night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be sure in future, get a star atlas or even download some apps on your phone to tell you the positions of stars in the sky. It will make stargazing more fun aswell!&lt;/p&gt;&#xA;" OwnerUserId="17096" LastActivityDate="2018-02-25T16:37:26.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25281" PostTypeId="1" CreationDate="2018-02-25T19:13:08.703" Score="2" ViewCount="84" Body="&lt;p&gt;The Milky Way has a bright center we can't see directly because of space dust. Are there's regions of the galaxy free of that dust that could receive all that light and radiation, right? Would it be possible to exist an 'ecosphere' around the galaxy's center where planets receive as much energy from it as the Earth receive from the Sun? Would it be possible to rogue planets orbiting the galaxy's center at that distance to sustain life?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or it's the case that interstelar dust always blocks most energy? Even in a 'polar orbit' around the galaxy's center?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, how many of them could exist? More than planets orbiting stars? &lt;/p&gt;&#xA;" OwnerUserId="21938" LastActivityDate="2018-02-25T19:13:08.703" Title="Life in sufficiently hot rogue planets" Tags="&lt;life&gt;&lt;rogue-planet&gt;" AnswerCount="0" CommentCount="8" ClosedDate="2018-02-26T15:25:12.493" ContentLicense="CC BY-SA 3.0" />
  <row Id="25282" PostTypeId="2" ParentId="25265" CreationDate="2018-02-25T22:24:14.433" Score="4" Body="&lt;p&gt;I don't think you can prove that aliens have never visited Earth. It's theoretically provable if one had a time machine, but without one, we can't really prove that aliens didn't visit Earth (say) 2 billion years ago. The physical evidence would be long gone (especially with plate tectonics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can say is that the burden of proof is on the person making the assertion. The null hypothesis is that aliens have never visited Earth. If your friend asserts that aliens have visited then the onus is on him to provide proof. If you can refute all the proofs he brings up, you can argue the null hypothesis remains the one to believe.&lt;/p&gt;&#xA;" OwnerUserId="20348" LastEditorUserId="20348" LastEditDate="2018-02-26T09:51:35.090" LastActivityDate="2018-02-26T09:51:35.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25283" PostTypeId="1" AcceptedAnswerId="25284" CreationDate="2018-02-24T22:47:16.533" Score="6" ViewCount="133" Body="&lt;p&gt;Could an asteroid or a rogue planet have enough magnetism and/or gravity to spin the Earth near the speed in which they are passing by Earth?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/If4oI.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/If4oI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" OwnerDisplayName="Muze" LastEditorUserId="53" LastEditDate="2018-02-26T01:09:08.997" LastActivityDate="2018-02-26T01:09:08.997" Title="Can Earth be spun like a magnet?" Tags="&lt;gravity&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25284" PostTypeId="2" ParentId="25283" CreationDate="2018-02-25T02:12:00.370" Score="9" Body="&lt;p&gt;Yes, a little, but not enough to matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnets can spin other magnets.   Think of magnets as a bar magnet with a north and a south, and you can think of them as linear not spherical if that helps with the imagery, because in a sense, they are linear with a distinct North and South magnetic pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iqm3C.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iqm3C.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth's magnetic field, for example has a distinct longitude and latitude that moves over time, but it's fixed at any given time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your question to work, it's easier to imagine a very powerful magnetic object flying past, and to think of the Earth's internal magnet as a compass arrow that wants to line up.   Magnets want to line up with magnetic field lines and a force would be created by a passing strong magnetic field that could orient the Earth's magnetic field to align with it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might think that this would turn the Earth, but it wouldn't.  Earth's magnetic field isn't &quot;the Earth&quot; it happens in the Earth's core, and it may happen mostly in the region where Earth's inner and outer core meet and the outer core is liquid (dense and viscous but still not solid).  The force would be applied to the magnetism inside the Earth, perhaps giving the Earth's inner core a slight tug and twirl and perhaps re-orienting the flowing charged liquid iron just outside of the core, but this force wouldn't turn the Earth as a whole, it would generate a force on the core and the &lt;a href=&quot;http://www.columbia.edu/cu/record/archives/vol22/vol22_iss1/Core_Spin.html&quot; rel=&quot;noreferrer&quot;&gt;core doesn't spin precisely with the Earth's surface&lt;/a&gt; anyway so any movement would likely not show up on the surface, for some time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that a percentage of earth's magnetism is carried by the liquid outer core, then the effect on rotation would be even less, more of a re-orientation than a direct angular momentum tug.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also difficult to generate a magnetic field strong enough to move something as massive as the Earth's core very much anyway.   Something to keep in mind as well is that &lt;a href=&quot;https://physics.stackexchange.com/questions/293310/why-does-magnetic-force-vary-in-proportion-to-the-cube-of-the-distance-instead-o&quot;&gt;the magnetic field drops by the cube of the distance&lt;/a&gt;, not the square, so you'd need a very powerful magnet, very close, to have any effect at all.   Anything large enough to have that strong a natural magnetic field, the gravity would have a much bigger effect than any interaction between the magnetism.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to answer your question, a tiny bit of angular momentum would be transferred when two magnetic fields fly past each other in space, so there would be some variation in rotation, but it would be tiny.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnetic fields as strong as Earth's are rare, you'd probably need a gas or ice giant planet or large rogue planet, and the interaction between passing magnetic fields is very weak compared to the gravitational interaction.  &lt;a href=&quot;https://www.kjmagnetics.com/blog.asp?p=earth-science&quot; rel=&quot;noreferrer&quot;&gt;Compared to it's mass, Earth is a very very weak magnet&lt;/a&gt;, so using magnetism to spin it would be ineffective and slow, but but in theory, if we wanted to set up a current and generate a magnetic field around the Earth, such a project could in theory spin the Earth's inner core like a induced magnetic field spins a magnet in a generator, but that would also depend on how permanent the magnetism is inside the Earth and it might not be fixed, but instead, a product of flowing magnetic metals.   Permanent magnets are easier to rotate than charged flowing liquid . . . but I digress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be an interesting calculation if somebody wanted to do the math behind this, but even without doing the math I can tell you that the effect would be teeny-tiny.&lt;/p&gt;&#xA;" OwnerUserId="6841" OwnerDisplayName="userLTK" LastActivityDate="2018-02-25T12:16:46.400" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25289" PostTypeId="1" AcceptedAnswerId="25295" CreationDate="2018-02-26T12:19:55.680" Score="1" ViewCount="103" Body="&lt;p&gt;What do the lines in this NASA graphic represent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does Solar cycle 22 have a peak of nearly 300 (assuming that peak represents the maximum observations at the same time) and &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_cycle_22&quot; rel=&quot;nofollow noreferrer&quot;&gt;this link&lt;/a&gt; says the peak is 212)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rhbly.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rhbly.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21949" LastEditorUserId="21237" LastEditDate="2018-02-26T13:58:07.143" LastActivityDate="2018-02-26T19:32:35.343" Title="What are the lines in this Solar Cycle graph?" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25291" PostTypeId="1" AcceptedAnswerId="25302" CreationDate="2018-02-26T15:50:11.213" Score="2" ViewCount="102" Body="&lt;p&gt;I have seen countless times the adjective &quot;compact&quot; in astronomy, e.g. compact stars, compact binaries, etc. I would like to know, what exactly is this supposed to mean? From &lt;a href=&quot;http://physics.open.ac.uk/~uckolb/research/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this source&lt;/a&gt;, I am led to believe that &quot;compact&quot; is meant to denote objects of the greatest matter densities in the universe, i.e. white dwarfs, neutron stars, and black holes. Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="16424" LastActivityDate="2020-01-29T03:55:49.420" Title="What does the word &quot;compact&quot; mean in astronomy/astrophysics?" Tags="&lt;astrophysics&gt;&lt;gravitational-waves&gt;&lt;naming&gt;&lt;gamma-ray-bursts&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25292" PostTypeId="1" AcceptedAnswerId="25296" CreationDate="2018-02-26T17:14:23.737" Score="6" ViewCount="347" Body="&lt;p&gt;Bear with me. I want to explain this to my 6 and half year old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, all things being equal, we could reach out to the Sun with a suitable implement, a gigantic space knife, could we cut a piece off the Sun? Take a slice from it? Do we end up with a body of Hydrogen and Helium, or is it impossible to extract anything from it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the Sun is a massive ball of plasma, the sheer mass compressing the gas through gravity until fusion is achieved. What I have been asked is, effectively, what happens if you try to take a piece of that plasma? Of our sun? &lt;/p&gt;&#xA;" OwnerUserId="21955" LastEditorDisplayName="user1569" LastEditDate="2018-02-27T08:40:30.903" LastActivityDate="2018-02-27T15:52:59.153" Title="Can I slice off a piece of the Sun?" Tags="&lt;the-sun&gt;&lt;space&gt;&lt;hydrogen&gt;&lt;helium&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25293" PostTypeId="2" ParentId="25292" CreationDate="2018-02-26T18:04:38.490" Score="1" Body="&lt;p&gt;We do not have, so far, have the technology to gather the Suns resources, besides Solar Panels here on Earth. A Dyson's Sphere is a theorized technology that could be able to harness the Suns energy. I am sure NASA is working on a solution, but at 5600 Deg. Celsius / 10,000 Fahrenheit, not even titanium can withstand a Close Encounter with our closest star, much less grabbing a piece of it. The &lt;a href=&quot;https://www.nasa.gov/content/goddard/parker-solar-probe&quot; rel=&quot;nofollow noreferrer&quot;&gt;Parker Solar Probe&lt;/a&gt; is a planned mission to the Sun, in the Summer of 2018. The probe is supposed to approach the Sun and orbit 3.67 million miles from the surface. We will have to see if it lasts very long. We will have to see!&lt;/p&gt;&#xA;" OwnerUserId="21598" LastEditorUserId="43" LastEditDate="2018-02-27T08:28:38.777" LastActivityDate="2018-02-27T08:28:38.777" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25295" PostTypeId="2" ParentId="25289" CreationDate="2018-02-26T19:04:07.797" Score="4" Body="&lt;p&gt;In the above plot by Hathaway,&#xA;the jagged line peaking at 284.5 in June 1989&#xA;is the monthly international sunspot number.&#xA;The smooth curve starting and ending a few years later is a&#xA;&lt;a href=&quot;https://solarscience.msfc.nasa.gov/predict.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;prediction&lt;/a&gt;&#xA;based on recent observations and statistics of past cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The peak of 212.5 in November 1989 is for a&#xA;&lt;a href=&quot;http://sidc.be/silso/node/52&quot; rel=&quot;nofollow noreferrer&quot;&gt;13-month smoothing&lt;/a&gt;&#xA;of the monthly number.&#xA;The Sunspot Index and Long-term Solar Observations (SILSO)&#xA;project at the Royal Observatory of Belgium&#xA;&lt;a href=&quot;http://sidc.be/silso/datafiles&quot; rel=&quot;nofollow noreferrer&quot;&gt;archives&lt;/a&gt;&#xA;these data and provides&#xA;&lt;a href=&quot;http://sidc.be/silso/monthlyssnplot&quot; rel=&quot;nofollow noreferrer&quot;&gt;this plot&lt;/a&gt;&#xA;of the monthly and smoothed sunspot numbers together.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Gi6sf.png&quot; alt=&quot;SILSO plot of monthly and 13-month smoothed sunspot numbers&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-02-26T19:32:35.343" LastActivityDate="2018-02-26T19:32:35.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25296" PostTypeId="2" ParentId="25292" CreationDate="2018-02-26T19:50:30.010" Score="7" Body="&lt;p&gt;The idea of removing material from the sun for either fuel or to lengthen its lifespan is called &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Star_lifting&quot; rel=&quot;noreferrer&quot;&gt;star lifting&lt;/a&gt;&quot;. Various hypothetical methods have been suggested, including heating parts of the atmosphere or squeezing it using magnetic fields. My own tentative suggestion is to combine these with chaos control to manipulate turbulence into making coronal mass ejections in desired directions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is not just that the sun is hot (5777 K at the surface), but that the stuff that is easy to scoop up with a huge magnetic field is a fairly sparse plasma - you don't get that much material despite the vast scales. Of course, you may want to compress the plasma into a smaller and more convenient ball, but now the ideal gas law tells you that it will heat up proportionally - to get a cool hydrogen ball you will need to bleed off all that heat slowly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is why I think Dyson spheres are probably the easiest way to start. &lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2018-02-26T19:50:30.010" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25297" PostTypeId="1" AcceptedAnswerId="25299" CreationDate="2018-02-26T20:44:40.840" Score="2" ViewCount="735" Body="&lt;p&gt;Let say Mercury, Venus, the Moon were all in alignment between the Earth and Sun. They are all closest to each other in respective elliptical orbit.  On the other side of the Sun you have the rest of the planets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would their gravity combine to have any gravitational effects on the Earth?&#xA;Or are they all too far away?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="7472" LastEditDate="2018-02-27T00:41:12.453" LastActivityDate="2019-08-27T21:30:59.947" Title="Do other planets affect the Earth?" Tags="&lt;the-sun&gt;&lt;planet&gt;&lt;gravity&gt;&lt;solar-system&gt;&lt;orbital-mechanics&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25299" PostTypeId="2" ParentId="25297" CreationDate="2018-02-26T21:43:19.780" Score="4" Body="&lt;p&gt;&lt;strong&gt;TL;DR:&lt;/strong&gt; No. Well, yes, but they are negligible.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Full answer: a conjunction like this 'nearly' happened on May 5th, 2000. NASA published &lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/alignment.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;an article&lt;/a&gt; about this event, stating:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;For example, the combined gravitational effect of all the planets together is much less than the effect of the Sun or the Moon on the Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;While unusual, such alignments have happened in the past without any consequences. The planets are simply too far away to have an effect on anything here on Earth - except our imaginations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;For hard numbers, see &lt;a href=&quot;http://www.tkcs-collins.com/truman/conjunct/conjunct.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article by Truman Collins&lt;/a&gt; which is linked there:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;h3&gt;Conclusion&lt;/h3&gt;&#xA;&lt;p&gt;What I discovered was that the tidal forces exerted on Earth from the five planets on the other side of the Sun are vastly smaller than those exerted by the Moon or the Sun. In fact, with the planets in the position they will be in on the big day, the largest tidal force from one of them will be from Jupiter, and it is about one five hundred thousandth the size of the tidal force exerted by the Moon on an average day!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7472" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-02-27T09:23:52.010" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25300" PostTypeId="2" ParentId="25297" CreationDate="2018-02-26T21:50:47.270" Score="4" Body="&lt;p&gt;Whether they are in a line is irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All masses have a gravitational effect. The gravitational effect of planets causes perturbation from an elliptical orbit. The gravitational pull of Jupiter (and to a lesser extent Saturn and the other planets) causes the Earth's orbit to change from nearly circular to slightly elliptical then back to circular. The planets also cause the angle that the elliptical orbit makes to rotate around the sun. Again this mostly caused by Jupiter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The eccentricity varies with roughly 100000 year period, and the angle varies over 112000 years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These effects are well understood and can be observed and measured. However alignments are inconsequential. Alignments of the planets have no effect on the Earth. They do not cause earthquakes. They do not cause volcanos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon does have significant effects, as it is the primary cause of tides.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-02-26T21:50:47.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25301" PostTypeId="1" CreationDate="2018-02-26T22:21:00.423" Score="4" ViewCount="223" Body="&lt;p&gt;What elements are required to fuse into phosphorus in a star. Can someone give me a step by step process please e.g &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;4 hydrogen nuclei fuse into one helium nuclei and then 3 helium nuclei&#xA;  fuse into carbon inside a red giant etc.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1H x 4 = 1HE, 1HE x 3 = 1C etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="21959" LastEditorUserId="2531" LastEditDate="2018-02-27T12:45:12.800" LastActivityDate="2018-02-27T12:45:12.800" Title="How is Phosphorus made in in stars?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;supernova&gt;&lt;red-giant&gt;&lt;astrochemistry&gt;" AnswerCount="0" CommentCount="6" ClosedDate="2018-03-06T05:30:12.623" ContentLicense="CC BY-SA 3.0" />
  <row Id="25302" PostTypeId="2" ParentId="25291" CreationDate="2018-02-27T00:09:27.980" Score="2" Body="&lt;p&gt;Compact stars are stars of great density; stellar remnants in which the fermion constituents have become degenerate. i.e. white dwarfs and neutron stars (though sometimes black holes would be included).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A compact binary is a binary star system containing at least one compact object.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-02-27T00:09:27.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25304" PostTypeId="2" ParentId="6312" CreationDate="2018-02-27T04:08:11.973" Score="4" Body="&lt;p&gt;I am by no means an authority on this area but prompted by user /u/called2voyage I will refer the 1994 Astronomical Journal paper by Shahid-Saless (Colorado) and Yeomans (JPL) &lt;a href=&quot;https://www.google.co.uk/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=6&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0ahUKEwic2LfchcXZAhXL-qQKHeJxAusQFghZMAU&amp;amp;url=http%3A%2F%2Fadsabs.harvard.edu%2Ffull%2F1994AJ....107.1885S&amp;amp;usg=AOvVaw1sOMUCKcBLnUaI427q9N7S&quot; rel=&quot;nofollow noreferrer&quot;&gt;Relativistic Effects on the Motion of Asteroids and Comets&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To paraphrase their abstract: They study the &lt;strong&gt;predicted&lt;/strong&gt; effects arising from relativistic perturbations on the motions of asteroids and comets and show that for a number of such objects, inclusion of relativistic contributions in the equations of motion give rise to significant improvements in the orbital solutions.  They go on to argue that ignoring such corrections to the equations of motion gives incorrect solutions.  They point out how the use of masses derived from relativistic ephemerides together with purely Newtonian equations of motion results in an inconsistent hybridized non-Newtonian, non-relativistic model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With respect to the posted question:-&#xA;&lt;strong&gt;(a) are the existing observational data on any Asteroids or Comets sufficient (in terms of accuracy and extent of observations) to determine/estimate rates of non-Newtonian perihelion advance?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...the authors short-list 15 asteroids with the largest &lt;strong&gt;predicted relativistic&lt;/strong&gt; perihelion precession rates (from a long-list of 156 commonly-studied objects).  For example Icarus has the greatest rate (0.101 arcseconds/year) and had (in 1993) been observed over 43 years thereby producing a predicted cumulative relativistic precession over this period of 4.34 arcseconds. Observations referred to are a mix of optical and radar (no spacecraft telemetry) and are those used to produce the ephemerides model from which orbital elements are extracted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors note that, assuming a (then) current observational accuracy of a few tenths of a second of arc, the relativistic contribution to precession should be detectable for Icarus and other asteroids near the top of the (ranked by precession-rate) list. As with the planets the majority of asteroid perihelion precession will be caused by the perturbing influences of other planets, but the authors do not present calculated values for these contributions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(b) are any such determinations publically available?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...the authors selected six asteroids whose motions are significantly affected by general relativistic effects. They then calculated a set of orbital elements for each orbit using JPL's development ephemeris DE200 (with Earth and Moon perturbations treated separately).  They did this in two ways, firstly with Newtonian equations of motion alone and secondly with the addition of non-Newtonian General Relativisitic equations of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors indicate that the residuals (differences between observed and predicted positions at various times) are improved (i.e. reduced) by application of non-Newtonian corrections to the equations of motion. For example the RMS residual for Icarus is improved by 30%. Thus it is technically possible to calculate differences in the orbital behavior (such as perihelion advance) predicted by (i) the pure-Newtonian model and  (ii) the Newtonian plus non-Newtonian model.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the paper does not actually present explicit accounts of total observed perihelion precession or a breakdown into Newtonian and non-Newtonian components.  In any case, due to complexity, such a breakdown could only apply over the period of observations which is limited (maximum 61 years in 1993) compared to the timescale (a few hundreds of years) over which (predominantly Newtonian) planetary-induced perturbations might be expected to average out.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it would be possible (assuming there are no major perturbing encounters with planets or the asteroid belt) to use the provided asteroid orbital element data in a numerical orbit simulator program to roughly model the perihelion precession rate of an asteroid orbiting the Sun in the presence of the other planets using Newtonian equations of motion (i) without and (ii) with relativistic modifications calculated from $\frac{F}{m} = \frac{GM}{r^2} \left(1+\frac{3V_t^2}{c^2}\right) = \frac{GM}{r^2} \left(1+\frac{3GM.SLR}{c^2.r^2}\right)$ (see &lt;a href=&quot;https://astronomy.stackexchange.com/questions/7900/can-general-relativity-indicate-phase-dependent-variations-in-planetary-orbital?rq=1&quot;&gt;Walter-Goldstein-Schwartzchild Formulation&lt;/a&gt; ).  That way the amount of asteroid precession resulting from each of the two different sources (Newtonian, non-Newtonian) could be modelled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: As /u/UhOh/ commented, an alternative, possibly superior, equation for the extra, relativistic acceleration is given in the Shahid-Sales &amp;amp; Yeomans paper cited above as Eqtn 3.11, on page 1886.&lt;/p&gt;&#xA;" OwnerUserId="233" LastEditorUserId="233" LastEditDate="2018-03-22T03:05:38.003" LastActivityDate="2018-03-22T03:05:38.003" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="25306" PostTypeId="1" AcceptedAnswerId="25318" CreationDate="2018-02-27T05:51:02.237" Score="4" ViewCount="431" Body="&lt;p&gt;Let's say I have the (RA,Dec) for two positions on the sky. I want to measure the position angle between these two positions. In other words, if I had the Cartesian coordinate vectors for these two positions (say [x,y,z]) then I think the position angle PA would simply be the dot product of the two vectors, so basically cos(PA). However, RA and Dec are spherical coordinates (corresponding to $\phi$ and $\theta$ respectively), not Cartesian coordinates. How do I get cos(PA) between the two positions on the sky if I only have their (RA,Dec) instead of [x,y,z]? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Position_angle&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Position_angle&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Position angle, usually abbreviated PA, is [a] convention for measuring angles on the sky in astronomy. The International Astronomical Union defines it as the angle measured relative to the north celestial pole (NCP), turning positive into the direction of the right ascension. In the standard (non-flipped) images this is a counterclockwise measure relative to the axis into the direction of positive declination.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="607" LastEditorUserId="13071" LastEditDate="2018-03-03T22:19:24.270" LastActivityDate="2019-08-01T10:09:24.720" Title="Measuring misalignment between two positions on sky" Tags="&lt;coordinate&gt;&lt;declination&gt;&lt;astrometry&gt;&lt;right-ascension&gt;&lt;alignment&gt;" AnswerCount="3" CommentCount="17" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25309" PostTypeId="1" AcceptedAnswerId="25310" CreationDate="2018-02-27T08:55:33.287" Score="0" ViewCount="41" Body="&lt;p&gt;Mike G's &lt;a href=&quot;https://astronomy.stackexchange.com/a/25151/7982&quot;&gt;great answer&lt;/a&gt; clued me in on how to find the collection of astrometric observations of spacecraft which is maintained by the Minor Planet Center. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've noticed that there are nonstandard format lines embedded within the observations. I can read the Date, RA, Dec, magnitude and band for the standard lines, but for the nonstandard lines with the &lt;code&gt;300&lt;/code&gt; in columns 56-65 (marked &lt;code&gt;xxx&lt;/code&gt;) I don't understand what this means and so far I haven't found a key that explains it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can just ignore them, but I'd really rather know what they are in case they are useful! or important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here: &lt;a href=&quot;https://www.minorplanetcenter.net/iau/DASO/Obs/DASO_000559_Obs.txt&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.minorplanetcenter.net/iau/DASO/Obs/DASO_000559_Obs.txt&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;00000000011111111112222222222333333333344444444445555555555666666666677777777778&#xA;12345678901234567890123456789012345678901234567890123456789012345678901234567890&#xA;               DateDateDateDateRARARARARARADecDecDecDecxxxxxxxxxxmm.mmb&#xA;2015-005B    tC2017 09 22.18235 01 55 07.45 +12 36 17.4          13.7 RqD0559Y00&#xA;2015-005B    tC2017 09 22.18335 01 55 20.26 +12 34 01.7          13.8 RqD0559Y00&#xA;2016-038C    tC2017 09 22.16232 01 16 49.19 +13 00 29.0          11.9 RqD0559Y00&#xA;2016-038C    tC2017 09 22.16336 01 17 04.66 +12 58 35.4          11.9 RqD0559Y00&#xA;2016-038C    tC2017 09 22.16433 01 17 19.05 +12 56 49.9          11.7 RqD0559Y00&#xA;2016-055A     V2017 09 21.35632 00 24 08.35 -01 15 45.6          17.0 VqD0559247&#xA;2016-055A     v2017 09 21.35632 1 008.0567   +50.9060     300           D0559247&#xA;2016-055A     V2017 09 21.35702 00 24 08.04 -01 16 01.2          17.1 VqD0559247&#xA;2016-055A     v2017 09 21.35702 1 008.0567   +50.9060     300           D0559247&#xA;2016-055A     V2017 09 21.48556 00 23 00.22 -02 01 37.3          16.9 VqD0559247&#xA;2016-055A     v2017 09 21.48556 1 008.0567   +50.9060     300           D0559247&#xA;2016-055A     V2017 09 21.48625 00 22 59.83 -02 01 53.0          17.4 VqD0559247&#xA;2016-055A     v2017 09 21.48625 1 008.0567   +50.9060     300           D0559247&#xA;2016-055A     V2017 09 21.48660 00 22 59.53 -02 02 02.0          17.0 VqD0559247&#xA;2016-055A     v2017 09 21.48660 1 008.0567   +50.9060     300           D0559247&#xA;2016-055A     C2017 09 21.88530 00 23 55.81 -01 22 12.6          17.1 VqD0559595&#xA;2016-055A     C2017 09 21.88701 00 23 54.89 -01 22 45.3          16.6 VqD0559595&#xA;2016-055A     C2017 09 21.88872 00 23 53.96 -01 23 18.6          16.9 VqD0559595&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;See also: &lt;a href=&quot;https://www.minorplanetcenter.net/iau/DASO/Obs/DASO_000562_Obs.txt&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.minorplanetcenter.net/iau/DASO/Obs/DASO_000562_Obs.txt&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and also: &lt;a href=&quot;https://www.minorplanetcenter.net/iau/DASO/Obs/DASO_000564_Obs.txt&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.minorplanetcenter.net/iau/DASO/Obs/DASO_000564_Obs.txt&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-02-27T09:38:45.367" LastActivityDate="2018-02-27T09:42:11.453" Title="What are these &quot;nonstandard format&quot; lines in Distant Artificial Satellites Observation circulars?" Tags="&lt;observational-astronomy&gt;&lt;asteroids&gt;&lt;astrometry&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25310" PostTypeId="2" ParentId="25309" CreationDate="2018-02-27T09:42:11.453" Score="2" Body="&lt;p&gt;Those are data from &quot;&lt;a href=&quot;https://www.minorplanetcenter.net/iau/info/RovingObs.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;roving observers&lt;/a&gt;&quot;, essentially any ground-based observation that isn't from an observatory blessed with a three character observatory code. The observatory code is columns 78 to 80 of the entries. Note that the observatory code for those roving observer entries is always 247&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The longitude, latitude, and elevation of the known observatories are known. This isn't the case with those &quot;roving observers&quot;, which is why they take two lines. The first line of a &quot;roving observer&quot; entry contains a &quot;V&quot; in column 15 and &quot;247&quot; as the observatory code. The second line of a &quot;roving observer&quot; entry contains a &quot;v&quot; in column 15. The bulk of the second line contains the longitude, latitude, and altitude of the observer.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2018-02-27T09:42:11.453" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25312" PostTypeId="2" ParentId="25292" CreationDate="2018-02-27T15:52:59.153" Score="1" Body="&lt;p&gt;A &quot;space knife&quot; is a non-starter: You can't cut a plasma with a knife because it behaves like a gas, not like a solid. You can't put anything into the sun or go anywhere near it because it is hot enough to vaporise &lt;em&gt;anything&lt;/em&gt;. Any concept that involves &quot;flying over the surface and scooping up hydrogen&quot; is not going to work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting close to the sun is hard. You are orbiting the sun at 30km/s If you want to get close to the sun, you will need to slow down a lot that requires a lot of fuel. Then to get away from the sun you will need to speed up again. This takes a lot more fuel, far more than you could hope to gather from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to collect some hydrogen without going anywhere near the sun.&#xA;You can collect hydrogen and helium from the solar wind. The sun blows plasma off its surface and this could be collected using magnetic fields.  This &quot;wind&quot; is not very dense: about 3-6 atoms per cm³. So you need to scoop a very large region for a long time to get a reasonable amount of gas. This gas is hot (36000 degrees) charged (its a plasma) and diffuse.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-02-27T15:52:59.153" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25315" PostTypeId="5" CreationDate="2018-02-27T21:46:52.247" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2018-02-27T21:46:52.247" LastActivityDate="2018-02-27T21:46:52.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25316" PostTypeId="4" CreationDate="2018-02-27T21:46:52.247" Score="0" Body="Questions relating to things that may occur in the future." OwnerUserId="21239" LastEditorUserId="21239" LastEditDate="2018-03-08T17:17:51.327" LastActivityDate="2018-03-08T17:17:51.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25318" PostTypeId="2" ParentId="25306" CreationDate="2018-02-27T23:22:38.093" Score="7" Body="&lt;p&gt;Assuming you mean the angle between the meridian line through A and the great circle that goes through points A and B, then it goes something like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define vectors from the origin to A and B assuming they lie on a unit sphere, such that $x_A= \cos \delta_A \cos \phi_A$, $ y_A = \cos \delta_A \sin \phi_A$ and $z_A= \sin \delta_A$, and similar for B. Here $\phi$ refers to right Ascension and $\delta$ is declination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The great circles in question define planes that pass through the origin. A normal to the plane defined by OAB is given by the vector product $\vec{n_1}=\vec{A}\times \vec{B}$. Similarly a great circle passing through the O, A and the NCP $(0, 0, 1)$ has a normal of $\vec{n_2}= \vec{A}\times (0,0,1) = (y_A, -x_A, 0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle you are looking for &lt;em&gt;is&lt;/em&gt; the angle between these two normal vectors, which can be found from scalar product in the usual way.&#xA;$$\cos \theta = \frac{\vec{n_1}\cdot \vec{n_2}}{|n_1||n_2|}=\frac{\cos \phi_{A}(y_{A}z_{B} - y_{B}z_A) + \sin \phi_{A}(x_Az_B - x_Bz_A)}{|\vec{A}\times \vec{B}|}$$&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-02-28T08:33:27.067" LastActivityDate="2018-02-28T08:33:27.067" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25320" PostTypeId="1" CreationDate="2018-02-28T06:55:11.327" Score="1" ViewCount="666" Body="&lt;p&gt;My question isn't why it doesn't have a circular orbit, and why an elliptical, my question is why do they move at all? What caused the debris in space to apply such a force on the planets that it has been circling around the Sun billions of years. &lt;/p&gt;&#xA;" OwnerUserId="21629" LastActivityDate="2018-02-28T16:17:16.953" Title="Why are the planets moving?" Tags="&lt;the-sun&gt;&lt;planet&gt;&lt;solar-system&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2018-03-03T22:20:39.320" ContentLicense="CC BY-SA 3.0" />
  <row Id="25322" PostTypeId="1" AcceptedAnswerId="25334" CreationDate="2018-02-28T09:38:46.503" Score="0" ViewCount="83" Body="&lt;p&gt;I'm playing with the idea to buy a telescope for years now and finally decided to do so. There are two which I consider: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://www.amazon.de/dp/B0038LX8XE/_encoding=UTF8?coliid=I3M0AUZT71KUYR&amp;amp;colid=31JR7TDSJ3GID&amp;amp;psc=0&quot; rel=&quot;nofollow noreferrer&quot;&gt;Celestron NexStar 127 SLT Mak&lt;/a&gt; and the &lt;a href=&quot;https://www.amazon.de/dp/B001MUP49E/_encoding=UTF8?coliid=I133NZEGXPG8Q7&amp;amp;colid=31JR7TDSJ3GID&amp;amp;psc=0&quot; rel=&quot;nofollow noreferrer&quot;&gt;Celestron NexStar 130 SLT&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already know the difference between a Refractor and Maksutov-Cassegrain and for which observations they're suited for best (and also that the Mak doesn't need collimating).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First question: Which would be the better choice for a beginner and eventually astrophotography?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second question: Is it possible to buy just the tube later (so I can reuse the motors and the tripod)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third question: As I've read, the tripod is pretty wobbly, so is it possible to replace the tripod later on (or is there a norm for the mounts)?&lt;/p&gt;&#xA;" OwnerUserId="21976" LastEditorUserId="7472" LastEditDate="2018-02-28T10:17:37.963" LastActivityDate="2018-03-01T07:35:37.773" Title="Which telescope to choose and some general questions" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2018-03-03T22:21:51.030" ContentLicense="CC BY-SA 3.0" />
  <row Id="25324" PostTypeId="2" ParentId="25320" CreationDate="2018-02-28T11:29:00.603" Score="5" Body="&lt;p&gt;This is a &lt;em&gt;simple explanation&lt;/em&gt; to give a basic understanding as to why the solar system rotates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's caused by gravity and velocity. If you put two objects with mass in space a distance apart and &lt;em&gt;at rest&lt;/em&gt;, they will accelerate directly toward each other, until &lt;em&gt;kerpow!&lt;/em&gt; they hit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, if they have some small, random relative velocity to begin with that's not directly toward or away from each other, then no amount of acceleration toward each other will zero that out. It means they will pass each other at some finite distance and keep going. If the random velocity is low enough, this dance will repeat and they will move in elliptical orbits around their center of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This kind of motion can be expresses in terms of its &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_momentum&quot; rel=&quot;noreferrer&quot;&gt;angular momentum&lt;/a&gt; and in this example (as in most) it is going to be a conserved quantity, so it's always going to be there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when things in the universe like solar systems or galaxies initially form, they have some random, residual, collective motion, and this angular momentum in the center of mass frame of whatever blob is coalescing doesn't go away. Objects in the solar system keep rotating because there's no other option, at least in the short term. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there was a large amount of gas in the solar system, that might slow things down a bit, and there is a teeny tiny loss of energy due to gravitational waves, although for planets this is so extremely small it can be ignored. But basically the &quot;wheels&quot; of the celestial sphere are well lubricated and solar system bodies will continue in their orbits for millions and perhaps billions of years because there's nothing we know about right now to stop them.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-02-28T11:29:00.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25326" PostTypeId="2" ParentId="25228" CreationDate="2018-02-28T15:01:02.667" Score="1" Body="&lt;p&gt;It turns out that if you hold the mode key for about five seconds and have the 1244 focuser plugged into the mount then a screen will pop up that allows you to use the arrow keys on the top of the remote to adjust the focus. Once you are done focusing, you simply press mode to exit.&lt;/p&gt;&#xA;" OwnerUserId="21237" LastEditorUserId="21237" LastEditDate="2018-05-01T19:14:39.183" LastActivityDate="2018-05-01T19:14:39.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25327" PostTypeId="2" ParentId="25320" CreationDate="2018-02-28T16:17:16.953" Score="2" Body="&lt;p&gt;It's a simple matter of statistics.  All the particles that did &lt;strong&gt;not&lt;/strong&gt; move in a near-elliptical orbit in the first place either drifted away or fell into the sun.  What you see is what's left. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to why orbits self-maintain, see uhoh's discussion of, basically, Keplerian dynamics. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-02-28T16:17:16.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25328" PostTypeId="1" CreationDate="2018-02-28T16:47:05.080" Score="3" ViewCount="891" Body="&lt;p&gt;If i know the radius r of a star in cm, how can I convert it to arcseconds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I have a star with r = 3.18e13 cm, and distance to the star d = 220 parsecs, what is the relation to convert the radius from cm to arcseconds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="10338" LastEditorUserId="10338" LastEditDate="2018-03-01T10:27:49.767" LastActivityDate="2018-03-04T23:30:06.157" Title="Finding the radius of a star in arcsec" Tags="&lt;star&gt;&lt;fundamental-astronomy&gt;&lt;size&gt;&lt;units&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25330" PostTypeId="2" ParentId="25306" CreationDate="2018-03-01T03:17:37.227" Score="4" Body="&lt;p&gt;The position angle P of a body ($\alpha_1, \delta_1$) with respect to another body ($\alpha_2, \delta_2$) can be calculated from&#xA;$$tan(P)={sin(\Delta\alpha)\over cos(\delta_2)tan(\delta_1)-sin(\delta_2)cos(\Delta\alpha)}$$&#xA;where $\Delta\alpha = \alpha_1-\alpha_2$. If the denominator is negative, the position angle lies in the range of 90 to 270 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: Jean Meeus, Astronomical Algorithms, Second Edition, &lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2018-03-01T03:17:37.227" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25334" PostTypeId="2" ParentId="25322" CreationDate="2018-03-01T07:35:37.773" Score="1" Body="&lt;p&gt;Macs also need to be collimated, but much more rarely than Newtonians.&#xA; They hold their alignment better (for more time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For astrophotography you also have to think about the field of view each scope will give with the camera a that you will use and the fact that their mounts are not equatorial (less important if you are into planetary photography).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that the Mac has a Vixen dovetail, which means you can later use it on other compatible head mounts or put some other scope with the same dovetail type to the SLT mount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be cheaper to wait more time and buy from the beginning the &quot;right&quot; equipment for your task, than buy something cheaper and then replace its parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers!&lt;/p&gt;&#xA;" OwnerUserId="5355" LastActivityDate="2018-03-01T07:35:37.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25335" PostTypeId="1" AcceptedAnswerId="25336" CreationDate="2018-03-01T09:37:21.327" Score="2" ViewCount="93" Body="&lt;p&gt;I've recently been analysing &lt;a href=&quot;http://exoplanet.eu/catalog/all_fields/&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Exoplanet Database&lt;/a&gt;, and have been looking at the 'molecules' field (molecules detected on the planet or in its atmosphere). They all make sense apart from &quot;O I&quot;. It's not iodine oxide (which would be I&lt;sub&gt;2&lt;/sub&gt;O), so I'm at a loss as to what it could be, especially since I'm very new to this field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is very much appreciated!&lt;/p&gt;&#xA;" OwnerUserId="21982" LastActivityDate="2018-03-04T20:07:05.107" Title="What does &quot;O I&quot; mean in the context of detected molecules?" Tags="&lt;exoplanet&gt;&lt;planetary-atmosphere&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25336" PostTypeId="2" ParentId="25335" CreationDate="2018-03-01T10:17:14.563" Score="3" Body="&lt;p&gt;This is the standard astronomer notation for neutral oxygen. Where chemists would write&#xA;$\mathrm{O}$ (or $\mathrm{O}^0$),&#xA;$\mathrm{O}^{+}$,&#xA;$\mathrm{O}^{++}$, and&#xA;$\mathrm{O}^{+++}$ (or $\mathrm{O}^{3+}$),&#xA;for neutral, once-, twice-, and thrice-ionized oxygen, respectively, astronomers almost exclusively write&#xA;$\mathrm{O}\,\mathrm{I}$,&#xA;$\mathrm{O}\,\mathrm{II}$,&#xA;$\mathrm{O}\,\mathrm{III}$,&#xA;$\mathrm{O}\,\mathrm{IV}$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is of course the case for other elements. For instance, you may have an $\mathrm{H}\,\mathrm{II}$ region (i.e. ionized hydrogen) surrounding hot stars, in a larger $\mathrm{H}\,\mathrm{I}$ (i.e. neutral) envelope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, we start at 1, where chemists start at 0.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-03-04T20:07:05.107" LastActivityDate="2018-03-04T20:07:05.107" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25337" PostTypeId="1" AcceptedAnswerId="25344" CreationDate="2018-03-01T10:53:01.917" Score="6" ViewCount="331" Body="&lt;p&gt;The recent  paper in Nature &lt;a href=&quot;https://www.nature.com/articles/nature25792&quot; rel=&quot;nofollow noreferrer&quot;&gt;An absorption profile centred at 78 megahertz in the sky-averaged spectrum&lt;/a&gt; seems (to me at least) to be an incredibly simple and elegant experiment with far-reaching implications.&lt;/p&gt;&#xA;&lt;p&gt;While the system has been implemented very carefully, it is essentially a spectrally resolved radiometer at radio frequencies below 200 MHz, measuring the sky-averaged power spectrum. And while many of &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_cosmic_microwave_background_experiments&quot; rel=&quot;nofollow noreferrer&quot;&gt;radiometers in space and on Earth&lt;/a&gt; measuring the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cosmic Microwave Background&lt;/a&gt; are fairly broad band, measure in the 10's to 100's of GHz range, and intended for very precise absolute intensity measurements of a smooth &lt;a href=&quot;https://en.wikipedia.org/wiki/Planck%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Plank distribution&lt;/a&gt;, the system described here measures between only 50 and 200 &lt;em&gt;MHz&lt;/em&gt;, and with a spectral resolution of only 6.1 kHz!&lt;/p&gt;&#xA;&lt;p&gt;The small set of carefully designed and optimized, antenna(s) and receiver(s) are simple and straightforward; a broadband dipole-type antenna above a large ground plane in a radio-quiet zone, some low noise electronics and what non-radio astronomers might call a very large bandwidth &lt;a href=&quot;https://en.wikipedia.org/wiki/Software-defined_radio&quot; rel=&quot;nofollow noreferrer&quot;&gt;Software-defined Radio&lt;/a&gt; or SDRs.&lt;/p&gt;&#xA;&lt;p&gt;And yet this small, simple receiver, sitting quietly on the ground in Australia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Outback&quot; rel=&quot;nofollow noreferrer&quot;&gt;Outback&lt;/a&gt; has produced observations are amazing! They've detected direct evidence of the first stars producing the first starlight in our early universe only about 200 million years after the Big Bang!&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Abstract:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;After stars formed in the early Universe, their ultraviolet light is expected, eventually, to have penetrated the primordial hydrogen gas and altered the excitation state of its 21-centimetre hyperfine line. This alteration would cause the gas to absorb photons from the cosmic microwave background, producing a spectral distortion that should be observable today at radio frequencies of less than 200 megahertz1. Here we report the detection of a flattened absorption profile in the sky-averaged radio spectrum, which is centred at a frequency of 78 megahertz and has a best-fitting full-width at half-maximum of 19 megahertz and an amplitude of 0.5 kelvin. The profile is largely consistent with expectations for the 21-centimetre signal induced by early stars; however, the best-fitting amplitude of the profile is more than a factor of two greater than the largest predictions2. This discrepancy suggests that either the primordial gas was much colder than expected or the background radiation temperature was hotter than expected. Astrophysical phenomena (such as radiation from stars and stellar remnants) are unlikely to account for this discrepancy; of the proposed extensions to the standard model of cosmology and particle physics, only cooling of the gas as a result of interactions between dark matter and baryons seems to explain the observed amplitude3. The low-frequency edge of the observed profile indicates that stars existed and had produced a background of Lyman-α photons by 180 million years after the Big Bang. The high-frequency edge indicates that the gas was heated to above the radiation temperature less than 100 million years later.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;I don't see anything about this data that strikes me as unmeasurable with even say 1970's technology if done carefully. The signal is huge, a distinct, broad (about 19 MHz centered at about 87 MHz), several percent (almost 0.6 K) deep &lt;em&gt;dip&lt;/em&gt; in an otherwise Rayleigh-Jeans regime blackbody distribution. In &lt;em&gt;20:20 hindsight&lt;/em&gt;, it seems to me this could have been detected a long time ago.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Am I missing something, or is this discovery remarkable not only for it's far reaching implication, but for its simplicity, elegance, and the fact that it could have been discovered quite a long time ago with '60's or '70's technology?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GAxE6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GAxE6.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Figure 1 from the &lt;a href=&quot;https://www.nature.com/articles/nature25792&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; Figures 2 and then 1 from the paper's &lt;a href=&quot;https://www.nature.com/articles/nature25792#extended-data&quot; rel=&quot;nofollow noreferrer&quot;&gt;Extended data&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KQQZC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KQQZC.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/e1HqW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/e1HqW.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-03-02T01:36:33.817" Title="Early universe's first starlight (indirectly) detected for the first time, could it have been detectable with 1970's technology?" Tags="&lt;cosmology&gt;&lt;radio-astronomy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25338" PostTypeId="2" ParentId="25328" CreationDate="2018-03-01T11:16:08.010" Score="4" Body="&lt;p&gt;Using basic circular maths: where $d$ is the distance of the star from the observer in AU, and $r$ is the star's radius in AU, and $a$ is the angle encompassed by the radius of the star in degrees:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;r = \frac{a}{360}2\pi{d}&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now rearrange it to make $a$ the subject:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;a = \frac{180r}{\pi{d}}&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get $a$ in arcseconds, you need to multiply the result by 3600 (because there are 3600 arcseconds in a degree):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;a = \frac{648000r}{\pi{d}}&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, convert your numbers from cm and parsecs to astronomical units:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r$ = 3.18e13cm = 2.126 AU&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d$ = 220pc = 4.538e7 AU&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put them into the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;a = \frac{648000\times2.126}{{4.538\times10^7}\pi}&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;a = 0.00966327′′&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Mike has pointed out, the final equation I've ended with can be further simplified if you use different units, to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;a = \frac{r}{d}&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $r$ is in AU and $d$ is in parsecs, giving an answer $a$ in arcseconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because the equation $a = \frac{648000r}{\pi{d}}$ has $\frac{648000}{\pi}$ in it - which is the definition of a parsec in AU.  Therefore, multiplying $d$ by this number to convert it from AU to parsecs, we get $a = \frac{648000r}{648000/\pi\times\pi{d}}$. Cancel out the pi on the bottom, and cancel the ${648000}$ on the top and bottom, and you're left with $a = \frac{r}{d}$ where $r$ is in au and $d$ is in pc.&lt;/p&gt;&#xA;" OwnerUserId="21982" LastEditorUserId="21982" LastEditDate="2018-03-01T16:59:04.737" LastActivityDate="2018-03-01T16:59:04.737" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25339" PostTypeId="1" AcceptedAnswerId="25748" CreationDate="2018-03-01T12:41:13.960" Score="7" ViewCount="517" Body="&lt;p&gt;In &lt;a href=&quot;https://astronomy.stackexchange.com/q/25337/7982&quot;&gt;this question&lt;/a&gt; I discuss the recent (open access) paper in Nature &lt;a href=&quot;https://www.nature.com/articles/nature25792&quot; rel=&quot;noreferrer&quot;&gt;An absorption profile centred at 78 megahertz in the sky-averaged spectrum&lt;/a&gt; at length. The abstract begins:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;After stars formed in the early Universe, their ultraviolet light is expected, eventually, to have penetrated the primordial hydrogen gas and altered the excitation state of its 21-centimetre hyperfine line. This alteration would cause the gas to absorb photons from the cosmic microwave background, producing a spectral distortion that should be observable today at radio frequencies of less than 200 megahertz(1).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where reference 1 is Jonathan Pritchard and Abraham Loeb (2012) &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0034-4885/75/8/086901/meta&quot; rel=&quot;noreferrer&quot;&gt;21 cm cosmology in the 21st century&lt;/a&gt;. From there I've found Leonid Chuzhoy and Zheng Zheng (2007) &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/522491/meta&quot; rel=&quot;noreferrer&quot;&gt;Radiative Transfer Effect on Ultraviolet Pumping of the 21 cm Line in the High-Redshift Universe&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand some basics about the hyperfine transition in hydrogen, and that the &quot;spin temperature&quot; of a gas in space can differ from the temperature of other partitions if it's being pumped, but these papers are more than a bit hard to read.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a simple way to explain the basics behind how the exposure of hydrogen to the UV light produced in early stars would cause the &quot;blip&quot; in the (now red-shifted) 21 cm part of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;noreferrer&quot;&gt;Cosmic Microwave Background&lt;/a&gt; radiation spectrum we see today?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-04-05T13:44:27.093" Title="How did UV from the earliest stars 'alter the state of the 21 cm line' such that it shows up in CMB today?" Tags="&lt;cosmology&gt;&lt;radio-astronomy&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25341" PostTypeId="1" CreationDate="2018-03-01T15:49:55.357" Score="4" ViewCount="648" Body="&lt;p&gt;Question in the title. Would any constellations be identifiable?&lt;/p&gt;&#xA;" OwnerUserId="19304" LastActivityDate="2018-03-01T16:37:25.047" Title="What did the night sky look like 100,000 years ago?" Tags="&lt;night-sky&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25342" PostTypeId="2" ParentId="25341" CreationDate="2018-03-01T16:37:25.047" Score="4" Body="&lt;p&gt;(Here I assume by &quot;constellation&quot; you mean &quot;pattern of stars&quot;, rather than &quot;region of the sky&quot;. To be a pedant one should use &quot;asterism&quot; for a pattern of stars.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some asterisms will be recognisable, but there can be some significant proper motion over that period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some stars will move more than others. Generally, closer stars will appear to have moved further than more distant stars. So the stars that make up the asterism of &quot;the plough&quot; (which are fairly close) will have moved more, but will still be in the same general region of the sky. On the other hand the bright stars of Orion are more distant. In 100000 years ago the general appearance of Orion would already have formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Videos showing &#xA;The past and future of the Plough &lt;a href=&quot;https://www.youtube.com/watch?v=txJH8RlIoXQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=txJH8RlIoXQ&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The future of Orion &lt;a href=&quot;https://www.youtube.com/watch?v=5KFQD6Bpnm0&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=5KFQD6Bpnm0&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-03-01T16:37:25.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25343" PostTypeId="1" CreationDate="2018-03-01T21:11:58.070" Score="6" ViewCount="666" Body="&lt;p&gt;This morning I was on the bus and as I looked out of the window I saw that the moon was larger than usual and red. I live in eastern Norway and it was about -20ºF / 30ºC outside. There's no forest fires, and I live in a small village. There's no eclipse scheduled either. The moon went behind the mountain line before I got to school ( bus ride happened  7:03 - 7:20 ), and nobody there had seen it. Can you explain this?&lt;/p&gt;&#xA;" OwnerUserId="21987" LastActivityDate="2018-10-22T18:00:36.603" Title="Are red moons possible without eclipses or smoke from fires?" Tags="&lt;the-moon&gt;&lt;lunar-eclipse&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25344" PostTypeId="2" ParentId="25337" CreationDate="2018-03-01T22:14:05.613" Score="5" Body="&lt;p&gt;I could not find a freely accessible version of this article, so I am mostly guessing here.&#xA;From this article &lt;a href=&quot;https://arxiv.org/pdf/1206.0267.pdf&quot; rel=&quot;noreferrer&quot;&gt;https://arxiv.org/pdf/1206.0267.pdf&lt;/a&gt; (especially sections 4 and 5), the main challenge seems to be distinguishing between the different components of the signal:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(1) the 21 cm signal coming from the high redshift Universe,&#xA;  (2) galactic and extra-galactic foregrounds,&#xA;  (3) ionospheric influences,&#xA;  (4) telescope response effects,&#xA;  (5) Radio frequency interference and&#xA;  (6) thermal noise&#xA;  Distinguishing between all these components means understanding each of them equally well and probably requires recent knowledge (in cosmology, radio receptors, etc)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In any case, the idea of using the 21-cm line to probe the early universe seems quite old:&#xA;&lt;a href=&quot;https://academic.oup.com/mnras/article/188/4/791/999622&quot; rel=&quot;noreferrer&quot;&gt;https://academic.oup.com/mnras/article/188/4/791/999622&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21988" LastActivityDate="2018-03-01T22:14:05.613" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25345" PostTypeId="1" CreationDate="2018-03-01T23:29:11.077" Score="3" ViewCount="4205" Body="&lt;p&gt;According to &lt;a href=&quot;http://earthsky.org/tonight/earth-comes-closest-to-sun-every-year-in-early-january&quot; rel=&quot;nofollow noreferrer&quot;&gt;earthsky.org&lt;/a&gt; the Earth gets &lt;strong&gt;3 million miles&lt;/strong&gt; closer to the Sun than its usual distance of about &lt;strong&gt;93 million miles&lt;/strong&gt; on average.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o8eL6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o8eL6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how close would the earth have to be to the sun for there to be problems with our survival?&lt;/p&gt;&#xA;" OwnerUserId="17327" LastEditorUserId="11143" LastEditDate="2018-03-02T16:19:26.723" LastActivityDate="2018-03-02T18:17:16.007" Title="How close would the sun have to get to Earth for there to be severe consequences?" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;distances&gt;&lt;life&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25346" PostTypeId="2" ParentId="25328" CreationDate="2018-03-01T23:34:33.403" Score="1" Body="&lt;pre&gt;&lt;code&gt;2 * arctan(r / d)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Use Wolfram Alpha, it takes free-form units like cm, parsecs, etc and usually does the right thing. And you can specify the unit for the answer you're looking for (e.g. arcseconds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=2+*+arctan(3.18e13+cm+%2F+220+parsecs)+in+arcseconds&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.wolframalpha.com/input/?i=2+*+arctan(3.18e13+cm+%2F+220+parsecs)+in+arcseconds&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the answer is 0.019 arc seconds&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2018-03-04T23:30:06.157" LastActivityDate="2018-03-04T23:30:06.157" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25348" PostTypeId="2" ParentId="25345" CreationDate="2018-03-02T07:55:14.947" Score="3" Body="&lt;p&gt;I'm going to use Gm (1 gigameter = $1\times10^9$m) and degrees Celsius for this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By getting closer to the Sun, probably the most major problem would be the increase in temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A while ago I wrote a program that calculates the &lt;a href=&quot;https://en.wikipedia.org/wiki/Effective_temperature#Surface_temperature_of_a_planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;effective surface temperature of a planet&lt;/a&gt;. Having dug it back out, I've played around with the values a bit. It's important to note that the &lt;em&gt;effective temperature&lt;/em&gt; is not the same as the &lt;em&gt;actual&lt;/em&gt; surface temperature, since it doesn't account for Earth's atmosphere. If Earth had no atmosphere, it would be the correct actual temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At our current distance of (avg.) 149.6Gm, the effective temperature is 257K, or -16°C (some websites cite other values ±2°C). &lt;a href=&quot;https://earthobservatory.nasa.gov/Features/GlobalWarming/page2.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA cites&lt;/a&gt; an &lt;em&gt;actual&lt;/em&gt; average temperature of 15°C. Assuming a linear relationship between effective temperature and actual temperature, we can assume that $T_{actual} = T_{eff}+31 \pm 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what would happen if we moved, say, 10Gm closer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, the effective surface temp is now -7°C, and I'm guessing the actual temperature would be about 24°C. This is an increase of 9°C. We would probably still be able to survive, but with difficulties. For starters, the &lt;a href=&quot;https://ourchangingclimate.wordpress.com/2011/01/19/sea-level-versus-temperature/&quot; rel=&quot;nofollow noreferrer&quot;&gt;sea level would rise&lt;/a&gt; by over 40m, leading the world to &lt;a href=&quot;http://flood.firetree.net/?ll=50.7813,18.8644&amp;amp;zoom=4&amp;amp;m=50&quot; rel=&quot;nofollow noreferrer&quot;&gt;look (at the best) like this&lt;/a&gt;. Not an easy situation, but nevertheless survivable.&#xA;Of course, there would be other effects like an increase in extreme weather, the mass extinction of species, and probably a whole load of things that are hard to foresee, but I'm not going to try to predict them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what about a move towards the Sun of 20Gm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effective temperature is now ~3°C, and the surface temperature is about 34°C. Things start to get bad now. &lt;a href=&quot;http://flood.firetree.net/?ll=50.7813,18.8644&amp;amp;zoom=4&amp;amp;m=60&quot; rel=&quot;nofollow noreferrer&quot;&gt;All of the ice caps have melted&lt;/a&gt;. Once fertile areas are now barren deserts. Surviving is hard, but possible, although famine is now a major problem across much of the world (especially around the equator). The Gulf Stream may have stopped, oddly enough cooling down some of western Europe and all of Britain. It's not looking good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about 50Gm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effective temperature is 42°C; the actual one is around 73°C. Where the Sun's habitable zone begins and ends is disputed, but now, at 0.65AU away from the sun, it's quite likely that we're not in it. It's very hard to predict what happens now. Humans would likely have to stay underground to remain alive, and food would be a major issue. The ecosystem would be pretty much destroyed, having not had a chance to adapt to the new temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For fun, if we moved 100Gm closer, we'd be at about 205°C. Ouch. We're closer than Mercury now, and look how that planet's coping. &lt;a href=&quot;https://en.wikipedia.org/wiki/Mercury_(planet)#Surface_conditions_and_exosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mercury ranges from&lt;/a&gt; -173°C to 427°C depending on various factors. Survival without massive life support is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any closer, and things just get worse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this is a suitable answer for your question!&lt;/p&gt;&#xA;" OwnerUserId="21982" LastEditorUserId="21982" LastEditDate="2018-03-02T09:25:05.963" LastActivityDate="2018-03-02T09:25:05.963" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25349" PostTypeId="2" ParentId="25345" CreationDate="2018-03-02T09:13:13.290" Score="0" Body="&lt;p&gt;The Earth passes within 91.4 million miles for just a brief time, then it moves back further out and 6 months later it's 94.5.  That 3.4% and change closer works out to 7% more solar energy, but that's just for closest and furthest points.   Over the closest and furthest month, the variation is smaller and (perhaps obviously), it averages out over the entire year.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's two ways to address to this question - &lt;a href=&quot;https://en.wikipedia.org/wiki/Milankovitch_cycles#Orbital_shape_(eccentricity)&quot; rel=&quot;nofollow noreferrer&quot;&gt;how eccentric the Earth's orbit can get&lt;/a&gt; which would make the &lt;a href=&quot;https://en.wikipedia.org/wiki/Perihelion_and_aphelion&quot; rel=&quot;nofollow noreferrer&quot;&gt;perihelion closer and aphelion further&lt;/a&gt;, or, the 2nd way, how much closer can you make the &lt;a href=&quot;https://en.wikipedia.org/wiki/Semi-major_and_semi-minor_axes#Orbital_period&quot; rel=&quot;nofollow noreferrer&quot;&gt;semi-major axis, which defines orbital period&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth's present eccentricity is 0.017 and decreasing (Wikipedia article above).  For simple approximation and low eccentricity orbits, the eccentricity 0.017 translates into (doubling it) a 0.034 (or 3.4%) variation perihelion to aphelion, which works out to a (1.034^2), about a 7% variation in solar energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth's peak eccentricity of 0.0679, (&lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c2&quot; rel=&quot;nofollow noreferrer&quot;&gt;this time I'll do the math&lt;/a&gt;), closest point (1-0.0679) = 0.9321 and farthest (1+0.0679)= 1.0679.   The ratio 1.0679/0.9321 = about 14.5% which works into a 31.3% variation in solar energy closest to farthest.     14.5% and 31% energy variation may sound like a lot, but consider that the winter/summer solar energy variation in latitudes away from the equator can be considerably over 100%.  So, 31% isn't world ending, in fact, the Earth handles those 31% days without too much trouble, though they can trigger or end ice ages depending on how it lines up with other Milankovich cycles.  We should also remember that it's 31% hottest to coldest, it's about 14.5% hottest to average, and again, just for one day then that number begins to drop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;14.5% size change isn't as much as it sounds either, that's about the variation between the largest full moon or &lt;a href=&quot;https://www.timeanddate.com/astronomy/moon/super-full-moon.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;supermoon and the smallest (micromoon)&lt;/a&gt;, though the average full moon variation is between perigee and apogee 11% and 12% in diameter.  How often do you look at a full moon and say &quot;that's bigger than the full moon 4 months ago&quot;.  That's not to say nobody notices, but many of us wouldn't recognize a supermoon without being told when to look for it, though it's fairly obvious if we could see them side to side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/N2aps.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/N2aps.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what effect does Earth's periodic 0.0679 eccentricity have?  Apart from triggering or perhaps ending an ice age, not that much..   You could push Earth's eccentricity pretty far without much danger to Earth, though past a certain point, crashing into Venus might become a concern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets have some fun and push Earth's eccentricity up to 0.15 (that's about half way between Mars' and Mercury's eccentricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f4az2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f4az2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perihelion 0.85% of the semi-major axis, Aphelion 1.15%.   At perihelion the solar intensity would be (1/.85)^2 = 38% stronger than normal.  That might cause scorching heat in some places where the Perihelion lined up with summer, and you might get some bitter cold weather when the Aphelion lines up with winter (note, this doesn't always happen, perihelion slowly cycles around the calendar every 26,000 years or so).  But when it lines up you might see some seasonal-weather extreemes, but Earth could (I think) survive even a 0.15 eccentricity.   The views of Venus at Perihelion would be impressive too, Venus would stay in the sky longer and be larger and brighter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Push Earth's eccentricity to about .26 and Earth and Venus would get uncomfortabaly/scary close and might crash into each other.   Fun to think about, but a potential planet killer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But apart from creating potentially greater variation in seasons, some very wild weather and perhaps triggering an ice age (perhaps the biggest ice age since snow-ball Earth) or accelerating climate change (it all depends on how the 3 Milankovich cycles line up with each other), You could push Earth's orbit surprisingly close to the Sun, half way to Venus or even closer at perihelion, without ending life on the planet.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you adjust Earth's semi-major axis, which makes the orbit faster or slower, and changes the length of the year - with that, there's much less variation.    @JThistle covered this.   If you push Earth's semi-major axis just a few percentage points closer to the sun, Earth would heat up and it would get bad quickly.  In fact, I think his estimates are conservative.  There are already some pretty good estimates out there on this, just google &quot;what will the Earth be like in 500 million years or a billion years&quot; for a few articles.  &lt;a href=&quot;https://www.sciencealert.com/here-s-a-mind-melting-snapshot-of-what-earth-will-look-like-in-1-billion-years&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's one&lt;/a&gt;.   The sun is growing slowly larger and more luminous, about 1% every 100 million years.  That will be a significant problem as soon as 500-600 million years from now, perhaps less.  Likewise, just a 2.5%-3% push closer to the sun would have a similar effect.  You can't reduce Earth's semi-major axis by much without causing serious problems.   Just 1% might be enough to melt Greenland over time, raise sea levels and prevent future ice ages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly if you push Earth just 1-2% further away, Earth could enter permanent ice ages (fortunately we can fix that by burning oil and coal), but sans man made greenhouse gas, just 1%-2% further away would trigger perhaps a permanent ice age for the next several million years and perhaps, dangerously low CO2 levels, at least until Antarctica drifts north enough for it's ice to melt.    It would be nice to be told to burn oil for the &quot;good of the planet&quot; though.  :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all very ballpark, but the Earth is, surprisingly, at a very good distance from the sun and just a few % one way or the other could be very bad, though in 15-25 million years when Antarctica is brushing up against South America and is no longer covered in ice, when that happens, Earth will have greater wiggle room away from the sun, but not closer to the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precise mathematical answers to this question are obviously impossible.  I'm just giving approximations.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-03-02T14:22:18.853" LastActivityDate="2018-03-02T14:22:18.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25351" PostTypeId="2" ParentId="25345" CreationDate="2018-03-02T18:17:16.007" Score="0" Body="&lt;p&gt;Without any significant change in orbital distance, there have been a variety of long-term (millions of years) tropic or ice ages, and transitions from one  to another led to rather a lot of extinctions.&lt;br&gt;&#xA;In the same vein,  a nice big solar eruption could wreak havoc independent of orbital distance changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I&quot;m not arguing with the other answers; just pointing out that time scale matters. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-03-02T18:17:16.007" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25352" PostTypeId="1" CreationDate="2018-03-03T02:40:11.467" Score="1" ViewCount="183" Body="&lt;p&gt;I'm worldbuilding and I'm trying to construct a standing stone calendar that'll accurately denote where to look for my planet's moon at its four major phases from 53.8 degrees South. So, I need the h&lt;sub&gt;0&lt;/sub&gt;, or hour angle at moonrise and moonset, for my moon at each of these four points in its orbit from this point on the surface of my planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, I can't even find any equations for how to find these numbers for Earth's moon, let alone another planet's moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, here goes...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My planet, Jasmi, has an axial tilt of 14.92 degrees and an orbital period about its star of 640.04 Earth days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jasča, my moon, has a semimajor axis of 373168 km and an eccentricity of about 0.01.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its orbit is 5 degrees inclined off the planet Jasmi's equatorial plane, toward the ecliptic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The longitude of the ascending node is 358.11 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jasča's argument of periapsis is 9.23 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since this is not supposed to be Earth, I could have measured the longitude of the ascending node from any old direction, but I'm already at my wits' end with trying to calculate everything properly and I want the night sky to be at least somewhat familiar to readers, so I just plopped our own Earth's celestial sphere onto Jasmi and offset its ecliptic to be 8.58 degrees closer to the celestial equator to account for the difference in planetary obliquity. So we're still setting longitude of the ascending node in degrees east of the First Point of Aries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any guidance y'all could offer would be very very much appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;lt;3 R&lt;/p&gt;&#xA;" OwnerUserId="21997" LastEditorUserId="21997" LastEditDate="2018-03-04T23:37:08.670" LastActivityDate="2018-03-04T23:37:08.670" Title="How do I calculate the hour angle of a planet's moon from a point on that planet's surface?" Tags="&lt;amateur-observing&gt;&lt;observational-astronomy&gt;&lt;natural-satellites&gt;&lt;orbital-elements&gt;&lt;moon-phases&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25353" PostTypeId="1" AcceptedAnswerId="25363" CreationDate="2018-03-03T03:28:08.083" Score="2" ViewCount="63" Body="&lt;p&gt;I witnessed a light fall from the sky. It was visibly about as bright as Mars, so brighter than meteorites I've seen, was just a point of light with no tail, and traveled much much faster than an aircraft. The horizon was blocked by a house, so I do not have any frame of reference for how far away it hit the ground (it could as well have been 3 miles or 10,000 miles away), but I do know the coordinates for where I was, and the direction is was relative to my position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some database I can use to search space events to learn what it was that I saw?&lt;/p&gt;&#xA;" OwnerUserId="21998" LastEditorUserId="2365" LastEditDate="2018-03-03T22:26:20.623" LastActivityDate="2018-03-04T09:07:16.667" Title="Is there a database for identifying the falling light I saw?" Tags="&lt;meteor&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25355" PostTypeId="1" CreationDate="2018-03-03T14:36:51.983" Score="0" ViewCount="1177" Body="&lt;p&gt;In the book &quot;The Crime of Claudius Ptolemy&quot; by Robert Newton, on page 81, it reads as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When we measure the time of a solstice, we measure the time when the&#xA;  meridian elevation angle of the sun has an extreme value, a maximum in&#xA;  the summer, and a minimum in the winter.... When we measure the time&#xA;  of an equinox, on the other hand, we measure the time when the&#xA;  meridian elevation takes on a specified value.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What does he mean by a &quot;specified value&quot;. What specified value?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how would the ancients have determined the time of the maximum, or the specified value? Time devices in ancient Greece, like water clocks, I suppose would have significant errors. Would they just start the water clock when the sun rose, then marked its level when the &quot;specified value&quot; occurred? That method would seem to have a lot of error.&lt;/p&gt;&#xA;" OwnerUserId="2800" LastActivityDate="2018-03-03T17:11:06.997" Title="How was the time of equinox measured in ancient times?" Tags="&lt;history&gt;&lt;equinox&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25356" PostTypeId="2" ParentId="25355" CreationDate="2018-03-03T16:49:45.153" Score="2" Body="&lt;p&gt;The &quot;specified value&quot; would be the latitude of the observer, and would be the average of the maximum and minimum values observed at the solstice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The altitude of the sun above the horizon is measured when it is due South. This is what &quot;meridian elevation&quot; means. So no clock is required. However this method can at best determine the date of the equinox. Precise observations over several years could be combined to obtain an time of equinox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position of the sun above the horizon can be simply determined by measuring the length of a shadow. However Ptolemy had access to astrolabes and quadrants for more accurate measurement of the sun's position.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-03-03T17:11:06.997" LastActivityDate="2018-03-03T17:11:06.997" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25357" PostTypeId="1" CreationDate="2018-03-03T17:05:11.760" Score="-1" ViewCount="143" Body="&lt;p&gt;I know that spectroscopy can be used to analyze the light of stars to determine the elements causing their emissions in the visible spectrum, but this requires knowing the emission spectra of the elements themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a standard source for this data?&lt;/p&gt;&#xA;" OwnerUserId="2800" LastActivityDate="2018-03-04T01:04:13.993" Title="Where can I find a database of elemental spectra?" Tags="&lt;spectroscopy&gt;&lt;spectra&gt;&lt;spectrometry&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2018-03-07T21:08:12.497" ContentLicense="CC BY-SA 3.0" />
  <row Id="25358" PostTypeId="2" ParentId="18259" CreationDate="2018-03-03T19:23:47.600" Score="1" Body="&lt;p&gt;The site to which you are referring is Sarmisegetusa in old Dacia, which was a well known royal citadel of the Illyrians in Roman times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sarmisegetusa was a fortress, not an observatory. The reconstructed posts there were used to hold up ceilings and walls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sarmisegetusa backs up against face of the mountain, so half of the horizon cannot even be seen from the site. That fact alone excludes the possibility that anyone was doing any astronomy there.&lt;/p&gt;&#xA;" OwnerUserId="2800" LastActivityDate="2018-03-03T19:23:47.600" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25362" PostTypeId="1" AcceptedAnswerId="25408" CreationDate="2018-03-03T22:31:05.993" Score="4" ViewCount="85" Body="&lt;p&gt;I have a working implementation of &lt;a href=&quot;http://www.stjarnhimlen.se/comp/tutorial.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schlyter's method&lt;/a&gt; for calculating the Sun's position.   I would like to know how Schlyter got the 3.24587E-5_deg in the following calculation for Mercury.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N =  48.3313_deg + 3.24587E-5_deg   * d    (Long of asc. node)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I could not find this number in JPL Horizons.   I'm guessing it's the difference between the N at two epochs divided by number of days between the two epochs, but I'm not sure.&lt;/p&gt;&#xA;" OwnerUserId="22010" LastActivityDate="2018-03-06T14:27:52.423" Title="Where to get rate of change for calculating ephemeris from JPL Horizons" Tags="&lt;ephemeris&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25363" PostTypeId="2" ParentId="25353" CreationDate="2018-03-03T22:45:45.273" Score="1" Body="&lt;p&gt;This could be a satellite, such as the ISS, which appears as a bright, fast moving star. You can look for times when the ISS is visible on websites like &lt;a href=&quot;http://www.heavens-above.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.heavens-above.com/&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be a meteor (not a meteorite, that is the stone, not the streak in the sky). Meteors travel much faster than satellites. On any given night you will see several meteors each hour. A meteor that is &quot;as bright as Mars&quot; (which varies in brightness, and can reach -3 but is more likely to be between -1 and 0) is a less common event, but not exceptionally rare. Reports of Fireballs (those with magnitude -4 or brighter) are collected by various bodies, such as the &lt;a href=&quot;https://ukmeteornetwork.co.uk/fireball-report/&quot; rel=&quot;nofollow noreferrer&quot;&gt;uk meteor network&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-03-04T09:07:16.667" LastActivityDate="2018-03-04T09:07:16.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25364" PostTypeId="2" ParentId="25357" CreationDate="2018-03-04T01:04:13.993" Score="2" Body="&lt;p&gt;The chemical composition of stellar atmospheres is usually obtained from &lt;em&gt;absorption&lt;/em&gt; lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analysis ingredients are a good spectrum, a model of the temperature and density structure of the stellar atmosphere and a catalogue of atomic and molecular transitions. It is these latter things that I think you seek.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such &quot;linelists&quot; are found in many places; consisting of line wavelengths, excitation potentials and oscillator strengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One very useful compilation/database is the &lt;a href=&quot;http://vald.astro.uu.se&quot; rel=&quot;nofollow noreferrer&quot;&gt;VALD&lt;/a&gt; atomic line database. Other catalogues exist for molecular transitions -e.g. &lt;a href=&quot;https://academic.oup.com/mnras/article/425/1/21/997702&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tennyson &amp;amp; Yurchenko 2012&lt;/a&gt;. However, these lists are usually just the starting point. Most spectroscopists tune their lists by a careful comparison with the spectrum of the Sun, or other standard stars.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-03-04T01:04:13.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25365" PostTypeId="1" CreationDate="2018-03-04T04:53:47.240" Score="-2" ViewCount="227" Body="&lt;p&gt;I would like to know please if there is any habitable exoplanet close to Earth.&lt;/p&gt;&#xA;" OwnerDisplayName="user22012" LastEditorUserId="7472" LastEditDate="2018-03-04T22:23:10.813" LastActivityDate="2018-03-09T01:49:25.110" Title="Is there any habitable exoplanet close to Earth?" Tags="&lt;exoplanet&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25369" PostTypeId="1" CreationDate="2018-03-04T14:48:04.903" Score="4" ViewCount="1207" Body="&lt;p&gt;The Milky Way galaxy is disc-shaped and earth is in its plane:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/JAIgG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JAIgG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; (&lt;a href=&quot;http://knowledgeofunknown.blogspot.ca/2011/04/milkyway-galaxy-dimensions.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the following image the milky way appears to be arcing across the sky:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/CauK8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CauK8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; (&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Milky_Way_Arch.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looks like an out of plane view of the milky way... How is it possible? Why an arch and not a thick straight line?&lt;/p&gt;&#xA;" OwnerUserId="6610" LastActivityDate="2020-03-15T07:47:47.937" Title="Why the Milky Way appears to be arching across the sky?" Tags="&lt;milky-way&gt;&lt;night-sky&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25370" PostTypeId="2" ParentId="25369" CreationDate="2018-03-04T15:07:35.407" Score="12" Body="&lt;p&gt;The Earth is a sphere (or is nearly a sphere). So to make a map of the whole Earth you would need to project it onto a flat surface. When you do this you create distortions. For example, on many maps, the straight line between New York and Japan looks like a long curve. It is not actually a curve, but when you stretch the surface of Earth to make it flat, some straight lines become curved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sky is also a sphere. If you ignore distance and just think of the direction of each point in the sky it would be a huge sphere that surrounds you. When we take a picture of the sky we must project that sphere onto a flat surface. That causes distortions, and it means that what is a straight line on the sphere of the sky is distorted by the projection into a curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you choose a projection in which the milky way goes straight across the middle, you can see that the milky way goes straight across the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The appearance of an &quot;arch&quot; is a distortion caused by stretching the image to fit on your flat screen. It is not real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oTKE2.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oTKE2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-03-04T15:07:35.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25371" PostTypeId="1" CreationDate="2018-03-05T03:07:32.400" Score="2" ViewCount="98" Body="&lt;p&gt;It is common knowledge that when we look at the stars we look back in time. And most stars are thousands of lightyears away from each other. The information that photons cary about the position of a stellar object can be really deceiving about their true position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do astronomers take lightspeed into account when they map stars ? If so, how do they figure out the true position of a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With all that in mind do galaxies really look like what we see in our photographs? If a galaxy has a diameter of bilions of lightyears, what we see could be a really skewed and warped image of the real galaxy. Can we computationaly reverse the warping ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&#xA;Image 1 and 2 explain what I mean by &quot;Warping&quot;. Stars closer to the core move faster and in combination with their distance from the viewer, information by light should give the impresion of a more &quot;spiraly&quot; galaxy as it whould take less time to reach the observer. Image 2 whould be the &quot;true&quot; shape of the galaxy if the effect could be reversed (I applied a counter clock-wise warp distortion)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An other factor of false position are gravitational lences. Although gravity from black holes do give the maximum distorsion, gravity from other stars could divert light reaching us. The diversion is negligent but can get big for astronomical distances (Image 3). Am I right ? If I am, do Astronomers even bother to make simulations to map the known sky as it whould be in real time ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/e58Ea.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/e58Ea.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22022" LastEditorUserId="22022" LastEditDate="2018-03-05T21:04:43.877" LastActivityDate="2018-03-05T21:04:43.877" Title="Information speed and our view of the cosmos" Tags="&lt;galaxy&gt;&lt;special-relativity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25372" PostTypeId="1" CreationDate="2018-03-05T03:19:41.683" Score="6" ViewCount="84" Body="&lt;p&gt;I know that the mean interval between total eclipses in a single place is about 375 years, which is not surprising given the small band of area of the totality. However, the penumbra of the eclipse partially shadows a much larger area than the totality:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DsFvB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DsFvB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;em&gt;Image Credit: NASA Scientific Visualization Studio&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were to take a wider area, such as the band of 90% coverage, how many years would there be between eclipses at a single place on average?&lt;/p&gt;&#xA;" OwnerUserId="2800" LastEditorUserId="616" LastEditDate="2018-03-05T16:22:14.213" LastActivityDate="2018-03-05T16:22:14.213" Title="How often do partial solar eclipses of 90% of totality occur in same place?" Tags="&lt;solar-eclipse&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25373" PostTypeId="2" ParentId="24709" CreationDate="2018-03-05T05:30:19.850" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Equirectangular_projection&quot; rel=&quot;nofollow noreferrer&quot;&gt;Equirectangular&lt;/a&gt;&#xA;or equidistant cylindrical projection&#xA;is the simplest way to meet these requirements.&#xA;Plate Carrée is the case where distortion is minimal at the equator.&#xA;Dominic Ford's &lt;a href=&quot;https://in-the-sky.org/data/constellations_map.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Map of the Constellations&lt;/a&gt; uses it,&#xA;as do the &lt;a href=&quot;https://svs.gsfc.nasa.gov/3895&quot; rel=&quot;nofollow noreferrer&quot;&gt;Deep Star Maps&lt;/a&gt;&#xA;by Ernie Wright of the NASA Scientific Visualization Studio.&#xA;In free software,&#xA;&lt;a href=&quot;https://sites.google.com/site/stellariumuserguide/sky-and-viewing-options#TOC-Projection&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellarium&lt;/a&gt; calls it &quot;Cylinder,&quot; and&#xA;&lt;a href=&quot;https://www.ap-i.net/skychart/en/documentation/chart_coordinates#projection&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cartes du Ciel&lt;/a&gt; calls it &quot;Cartesian.&quot;&#xA;If you plot your own, you could scale the horizontal dimension by&#xA;$\cos 30^\circ \approx 0.866$ to minimize distortion at&#xA;$\pm30^\circ$ declination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you are not plotting the polar regions,&#xA;other &lt;a href=&quot;http://www.progonos.com/furuti/MapProj/Normal/ProjCyl/projCyl.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;cylindrical projections&lt;/a&gt; are possible.&#xA;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Mercator_projection&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mercator&lt;/a&gt;&#xA;projection is available in both Stellarium and Cartes du Ciel.&#xA;Stellarium also offers the&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Miller_cylindrical_projection&quot; rel=&quot;nofollow noreferrer&quot;&gt;Miller cylindrical&lt;/a&gt; projection.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-03-05T05:30:19.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25374" PostTypeId="2" ParentId="1451" CreationDate="2018-03-05T07:20:20.363" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can the gravity well be infinitely deep?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only if the spin parameter of a rotating black hole is maximal so that the rotation speed of the horizon would be the speed of light $\rm (a=M)$ the integral of the $g_{\rm r r}$ component blows up, and the isometric embedding of the radial plane goes infinite at the horizon (see &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?db_key=AST&amp;amp;bibcode=1972ApJ...178..347B&amp;amp;letter=.&amp;amp;classic=YES&amp;amp;defaultprint=YES&amp;amp;whole_paper=YES&amp;amp;page=347&amp;amp;epage=347&amp;amp;send=Send%20PDF&amp;amp;filetype=.pdf#page=7&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bardeen 1972, Fig. 2&lt;/a&gt;). Here is Flamm's paraboloid for all spin parameters, as long as the spin is subextremal the distances are finite (when $\rm a=0$ the exterior Schwarzschild solution is recovered):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://org.yukterez.net/flamms.paraboloid.for.the.kerr.metric.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iSRUK.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5431" LastEditorUserId="5431" LastEditDate="2018-03-05T07:26:09.067" LastActivityDate="2018-03-05T07:26:09.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25375" PostTypeId="1" AcceptedAnswerId="25378" CreationDate="2018-03-05T07:37:57.100" Score="11" ViewCount="2680" Body="&lt;p&gt;I realize the Earth and the Moon both orbit around their shared center of mass, and that in the case of the Earth and Moon this center is &quot;inside&quot; the Earth. However, I'm looking for an approach to determine the distance the Earth travels in one &quot;orbit around the moon&quot; (as small as it may be).&lt;/p&gt;&#xA;" OwnerUserId="22027" LastActivityDate="2018-03-05T10:19:14.887" Title="What is the distance of the Earth's orbit around the Moon?" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25377" PostTypeId="2" ParentId="25371" CreationDate="2018-03-05T08:23:06.120" Score="4" Body="&lt;p&gt;The effect is small, but not negligible. It is not accounted for in astronomical catalogues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's work it out. We can start with the visible stars. Most of these are &lt;em&gt;closer&lt;/em&gt; than 1000 light years; let's say 100 light years. The typical velocity of these stars with respect to the Sun is tens of km per second; let's take an extreme example of 100 km/s in a direction at right angles to the star-Sun line. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;100 years at 100 km/s is 300 billion km. This sounds huge, but it is a tiny fraction of the size of the Galaxy and small compared with the distances between stars; but by how much does it change the apparent position in the sky? The change in angle (in radians) is given by the tangential distance moved in 100 years, divided by the distance to the star. This is the ratio of the speed of the star to the speed of light and is actually &lt;em&gt;independent of distance&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the example given, the change in position is $3.3\times 10^{-4}$ radians or 69 arcseconds. This is an enormous change in astronomical terms (not an angle you could easily resolve with the naked eye), but it is &lt;em&gt;not&lt;/em&gt; accounted for in astronomical catalogues, since we are interested in where to look at stars &lt;em&gt;now&lt;/em&gt; and neither do we know the accurate tangential velocity of everything in the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What &lt;em&gt;is&lt;/em&gt; accounted for in telescope pointing software is &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Stellar_aberration_(derivation_from_Lorentz_transformation)&quot; rel=&quot;nofollow noreferrer&quot;&gt;relativistic aberration&lt;/a&gt;, which is a change in the apparent position of a star due to changes in the relative motion as the Earth orbits the Sun. These corrections can be as large as 22 arcseconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for galaxies - you have your dimension estimates wrong. A large galaxy is 100 thousand light years across, with stellar speeds limited to a few hundred km/s. I don't understand what you mean by &quot;warping&quot;. Individual stars will have moved whilst the light travels towards us, others will have taken their place, stars will have died and others born, whole galaxies will have disintegrated or merged. All this is known. We can only observe what we can see &lt;em&gt;now&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A dynamical timescale is typically tens of millions of years in a Galaxy and far longer than it takes light to travel across the galaxy. To all intents and purposes, when we look at a distant galaxy we are seeing everything at roughly the same epoch in that galaxy's history.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-03-05T08:29:33.917" LastActivityDate="2018-03-05T08:29:33.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25378" PostTypeId="2" ParentId="25375" CreationDate="2018-03-05T10:19:14.887" Score="16" Body="&lt;p&gt;The mass of the Earth is about 81 times that of the Moon and the distance between their centres of mass is typically 384 400 km, so the centre of mass will be about 1/82 of this distance from the centre of mass of the Earth, which is about 4700km. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbits are not too far from circular, so the Earth travels about $2 \pi$ times that each month, so a little under 30000km.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-03-05T10:19:14.887" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25380" PostTypeId="1" CreationDate="2018-03-05T13:34:59.993" Score="3" ViewCount="143" Body="&lt;p&gt;I'm aware for over 100 years the reason has been known how impact craters are almost always circular, but my question really is has there been any recent insights in the area? What other topologies are physically possible? &lt;/p&gt;&#xA;" OwnerUserId="22032" LastActivityDate="2021-02-23T22:16:01.290" Title="Are all impact craters circular?" Tags="&lt;planet&gt;&lt;formation&gt;&lt;surface&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25382" PostTypeId="2" ParentId="25365" CreationDate="2018-03-05T16:14:46.610" Score="0" Body="&lt;p&gt;There are some possible planets in the Trappist-1 solar system.  Particularly planet 1e looks to be in the habitable zone but we won't know until the Webb telescope is launched next year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Close to Earth is a relative term, Trappist 1 is about 40 light years away.&lt;/p&gt;&#xA;" OwnerUserId="22042" LastActivityDate="2018-03-05T16:14:46.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25383" PostTypeId="1" AcceptedAnswerId="25574" CreationDate="2018-03-05T16:19:07.957" Score="2" ViewCount="114" Body="&lt;p&gt;I know that the line strength ratio tells us how hot the electron plasma in a nebula is, and also give information about the electron densities in the nebula. But how do you compute the line ratios? Say we consider two emission lines (both forbidden) decaying to the ground state. If I have the Einstein A coefficients and the statistical weights, how do I compute the line ratios? Any explanation (preferably with examples) will be highly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="22041" LastEditorUserId="616" LastEditDate="2018-03-16T09:28:38.380" LastActivityDate="2018-03-21T17:08:26.727" Title="Determining line ratios in planetary nebula" Tags="&lt;spectroscopy&gt;&lt;nebula&gt;&lt;radiative-transfer&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25384" PostTypeId="1" AcceptedAnswerId="25386" CreationDate="2018-03-05T16:27:09.260" Score="6" ViewCount="1410" Body="&lt;p&gt;I just realized the Earth is not a stationary object with the moon orbiting around it.  As shown in &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/5/59/Orbit3.gif&quot; rel=&quot;noreferrer&quot;&gt;this minimalist animation&lt;/a&gt; from Wikipedia, the Earth actually orbits a common barycenter with the moon.  If this was a carnival ride, people sitting in the moon and on the far side of the earth would certainly feel the centrifugal force acting on them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PiKVD.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PiKVD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is a very large system in reality, and the Earth also has quite a strong gravitational force felt on the surface anyway.  So when it comes to accurately measuring gravity or doing experiments that rely on gravity, how big of a factor is this centrifugal force?  Does the apearant force of gravity fluctuate from 9.9 m/s^2 to 9.7 or is it more along the lines of 9.800001 to 9.799999 (assuming the average is exactly 9.8, which is a simplification).  Or is there something I'm missing that means the force is non-existent?&lt;/p&gt;&#xA;" OwnerUserId="21077" LastActivityDate="2019-04-18T02:47:58.257" Title="Does the Earth orbiting around the Earth-Moon barycenter cause a measurable centrifugal force?" Tags="&lt;orbit&gt;&lt;tidal-forces&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25385" PostTypeId="1" AcceptedAnswerId="25388" CreationDate="2018-03-05T18:08:26.887" Score="0" ViewCount="249" Body="&lt;p&gt;In april 2002 the inner planets had aligned and this event was captured in this beautiful photo:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XjwTn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XjwTn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, according to &lt;a href=&quot;http://www.fourmilab.ch/cgi-bin/Solar&quot; rel=&quot;nofollow noreferrer&quot;&gt;this website simulations&lt;/a&gt; for the date, Jupiter was pretty much away. Is that possible? Can Jupiter saw as a straight line even be so far from the other planets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gJhJ7.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gJhJ7.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21949" LastActivityDate="2018-03-06T10:19:14.183" Title="2002 Planetary alignement" Tags="&lt;planet&gt;&lt;alignment&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25386" PostTypeId="2" ParentId="25384" CreationDate="2018-03-05T18:08:38.077" Score="6" Body="&lt;p&gt;That the Earth and Moon orbit about their center of mass from the perspective of an inertial frame of reference is a bit irrelevant. One thing that is quite relevant is that gravitational force is undetectable by a local measuring device. For example, people standing still on the surface of the Earth do not feel gravity. They instead feel the normal force pushing them up, away from the center of the Earth. The gravitational force on astronauts in the International Space Station is about 90% of what they experience on the Earth's surface, but they feel none of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another relevant factor is that the Earth as a whole, along with objects on the surface of the Earth, accelerate toward the Moon (and the Sun, and Jupiter, and Venus, and ...) gravitationally. The gravitational acceleration of those surface-bound objects toward those other bodies is not exactly the same as is that of the Earth as a whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between these accelerations results in a force that can be measured. This is the tidal acceleration. An extremely sensitive scale will show that you weigh slightly more when the Moon is on the horizon than when it is directly overhead. For a 61 kg person, this difference in weight between the Moon being on the horizon vs directly overhead is about 10&lt;sup&gt;-4&lt;/sup&gt; newtons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compared to the ~600 newton weight of that 61 kg person, this is a very small effect. This very small effect, along with an even smaller effect from the Sun (roughly half), are however responsible for the tides in the oceans.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2018-03-05T18:08:38.077" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25387" PostTypeId="2" ParentId="25380" CreationDate="2018-03-05T19:01:46.877" Score="1" Body="&lt;p&gt;Impact craters are not always circular. They vary based on the angle of incidence.&lt;/p&gt;&#xA;&lt;p&gt;One example is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Wetumpka_crater&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wetumpka crater&lt;/a&gt;. There is also research on &lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/S0032063314001093&quot; rel=&quot;nofollow noreferrer&quot;&gt;Asymmetric craters on Vesta: Impact on sloping surfaces&lt;/a&gt; which states e.g.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The global occurrence of these crater classes compared with a slope map clearly shows that these asymmetric crater types exclusively form on slopes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="37191" LastEditDate="2021-02-23T22:16:01.290" LastActivityDate="2021-02-23T22:16:01.290" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25388" PostTypeId="2" ParentId="25385" CreationDate="2018-03-05T19:43:22.670" Score="2" Body="&lt;p&gt;Below is an image of the solar system from that date. You can imagine drawing lines from the Earth to the various planets. You will see that Saturn, Mars, Venus and Mercury are quite closely aligned. And Jupiter is about 25 degrees out of that alignment. This matches with the photo that shows Jupiter further from the other planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UbtIY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UbtIY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that my image, &lt;a href=&quot;https://space.jpl.nasa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;derived from Nasa&lt;/a&gt; shows the orbits of the planets are not equally spaced. Your image does not represent reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The photo shows four planets quite close, and Jupiter a little way from them, just as the model of the solar system suggests. Of course the planets are in a straight line in the sky. They always are in a straight line, because they all are on the ecliptic. The alignment was significant because the planets appeared close together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the planets move around the sun in (almost) the same plane. The Earth also moves in this plane. The plane that the planets all move it is called the &quot;plane of the ecliptic&quot;. Imagine a circle drawn on a piece of paper. If you look at the circle from above you see a circle. If you look at it from an angle you see an ellipse, and if you look at it from the side you only see a line. The Earth is in the same plane as the other planets, so all the planets appear to be moving in a line. This means that the positions of the planets in the sky will &quot;always&quot; be in a line, even when there is no alignment. The plane of the ecliptic looks like a line, because Earth is on the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However when there is an alignment with the Earth (as there was in 2002) the planets will appear to be not only on the line of the ecliptic but close together in the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Of course the planets are not actually close together, and there are no gravitational or other malevolent influence of planets being in a line&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-03-06T10:19:14.183" LastActivityDate="2018-03-06T10:19:14.183" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25389" PostTypeId="1" CreationDate="2018-03-05T22:33:00.060" Score="2" ViewCount="727" Body="&lt;p&gt;Let's assume there is a right rectangular prism floating about in space, what would the gravity be like across its 6 flat surfaces? What gravity would you experience if you set foot on the rectangular prism's surface? Would it be the same as on Earth? Or would it be different the further along the surface you walk? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume that this right rectangular prism has a constant density and has a mass equal to Earth's. At any point on the surface of this prism, would gravity vary? That is, would it be an extreme variation than compared to one traveling across the surface of Earth? How extreme would this differential be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a rectangular prism with specific dimensions in mind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sN2mO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sN2mO.png&quot; alt=&quot;Rectangular Prism with dimensions&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22047" LastEditorDisplayName="user1569" LastEditDate="2018-03-06T20:11:54.107" LastActivityDate="2018-03-06T20:11:54.107" Title="Gravity on a right rectangular prism?" Tags="&lt;gravity&gt;&lt;astrophysics&gt;&lt;newtonian-gravity&gt;" AnswerCount="0" CommentCount="8" ClosedDate="2018-03-07T21:06:48.843" ContentLicense="CC BY-SA 3.0" />
  <row Id="25390" PostTypeId="1" CreationDate="2018-03-05T23:01:10.680" Score="2" ViewCount="84" Body="&lt;p&gt;I am trying to build a very low cost refractor telescope. It's for a project, so I need to be able to provide &quot;data&quot; about the telescope. But it seems I can't calculate anything (magnification, to be more specific) without knowing the focal length of my telescope. However, because I built it myself, I do not know what my focal length is. How can I find that out, or how can I calculate the magnification?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On another note, what is the best combination for the best magnification? I am not looking for anything great, just the best magnification possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can have a 50mm lens with a fl of 50cm, 30cm, and 20cm and a 38mm lens with each of the same focal lengths. Which is the best pair?&lt;/p&gt;&#xA;" OwnerUserId="22048" LastEditorUserId="1891" LastEditDate="2018-03-06T08:35:07.063" LastActivityDate="2018-03-06T08:35:07.063" Title="Is it possible to calculate the focal length of a refactor I built?" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25392" PostTypeId="2" ParentId="25380" CreationDate="2018-03-05T23:20:03.010" Score="5" Body="&lt;p&gt;There are elongated craters from grazing impacts, and these can sometimes get rather odd shapes like &lt;a href=&quot;https://en.wikipedia.org/wiki/Schiller_(crater)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schiller&lt;/a&gt; on the Moon and &lt;a href=&quot;https://en.wikipedia.org/wiki/Orcus_Patera&quot; rel=&quot;nofollow noreferrer&quot;&gt;Orcus Patera&lt;/a&gt; on Mars (the later is a bit contested, it might have formed from something other than an impact).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Double impact craters is another possibility. &lt;a href=&quot;https://hirise.lpl.arizona.edu/ESP_020894_1395&quot; rel=&quot;nofollow noreferrer&quot;&gt;Some craters occur simultaneously&lt;/a&gt; (&lt;a href=&quot;http://www.nbcnews.com/id/41836713/ns/technology_and_science-space/t/mars-meteorite-craters-make-mine-double/&quot; rel=&quot;nofollow noreferrer&quot;&gt;more direct picture&lt;/a&gt;). Splitting impactors is likely the one way of getting truly complex topologies. Most of the time they tend to follow each other, producing &lt;a href=&quot;https://en.wikipedia.org/wiki/Crater_chain&quot; rel=&quot;nofollow noreferrer&quot;&gt;crater chains&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If terrain shifts craters may be deformed, and depending on details in the rock they may also get odd shapes. &lt;a href=&quot;https://en.wikipedia.org/wiki/Meteor_Crater&quot; rel=&quot;nofollow noreferrer&quot;&gt;Meteor Crater&lt;/a&gt; in Arizona is somewhat squarish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Craters cannot generate that complex patterns since they are just explosions out from a center and impactors are usually single objects or random clouds. In principle one could arrange asteroids along a trajectory to spell out letters upon impact, but it is an exceedingly unlikely configuration so most impacts will only manage 'O'.&lt;/p&gt;&#xA;" OwnerUserId="19816" LastEditorUserId="19816" LastEditDate="2018-03-06T09:55:06.270" LastActivityDate="2018-03-06T09:55:06.270" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25396" PostTypeId="2" ParentId="25390" CreationDate="2018-03-06T01:20:26.487" Score="2" Body="&lt;p&gt;The focal length of a telescope is the focal length of the lens at the front of the scope. In your case it's either 50, 30, or 20 cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To magnify the image formed by one of those objective lenses, you need an eyepiece. None of the lenses you listed will work as an eyepiece unless you want a magnification of 2.5 (50/20).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eye pieces generally have a focal length of 25 to 30 mm or less. To get a quality image, you should buy a used eyepiece.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastEditorUserId="17355" LastEditDate="2018-03-06T04:16:35.923" LastActivityDate="2018-03-06T04:16:35.923" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25397" PostTypeId="1" AcceptedAnswerId="25401" CreationDate="2018-03-06T04:05:08.277" Score="4" ViewCount="370" Body="&lt;p&gt;Why do gravitational waves emitted by supermassive black holes binaries have low-frequency range (~10$^{-9}$ - 10$^{-2}$ Hz) than those emitted by stellar mass compact binaries (~10$^{-4}$  - 10$^2$ Hz)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below, a NASA graphic showing the wavelength ranges of various phenomenon found &lt;a href=&quot;https://imagine.gsfc.nasa.gov/features/satellites/archive/images/LISA_grav_wave_sepctrum_full.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. Click for full size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xOKBC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xOKBCm.jpg&quot; alt=&quot;NASA wavelength of various phenomenon&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22053" LastEditorUserId="7982" LastEditDate="2018-03-06T07:48:13.457" LastActivityDate="2018-03-06T08:04:24.080" Title="Why is frequency different for gravitational waves coming from SMBHs vs small compact binaries?" Tags="&lt;black-hole&gt;&lt;supermassive-black-hole&gt;&lt;gravitational-waves&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25398" PostTypeId="2" ParentId="25397" CreationDate="2018-03-06T05:40:45.267" Score="2" Body="&lt;p&gt;In general, the frequency of gravitational waves emitted by a binary system is $f=2f_{\text{orb}}$, where $f_{\text{orb}}$ is the orbital frequency of the system. In the late evolutionary phases during which gravitational waves from supermassive black hole binaries are detectable, the orbital periods are still comparatively large, measured in the months or years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Near coalescence, we still have an upper frequency bound, because &lt;a href=&quot;https://www.astro.umd.edu/~miller/teaching/astr498/lecture25.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;the maximum frequency $f_{\text{max}}$ is proportional to $\sqrt{G\rho}$&lt;/a&gt;, where $\rho$ is some sort of density for the system. In the case of a black hole, $\rho\propto M/R^3$, and since the &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schwarzschild radius&lt;/a&gt; $R_s\propto M$, we find that $f_{\text{max}}\propto\sqrt{M/M^3}=M^{-1}$. Therefore, the more massive the black hole, the lower the maximum orbital frequency. For supermassive black holes, this turns out to be in the nanohertz regime, give or take an order of magnitude.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2018-03-06T05:40:45.267" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25401" PostTypeId="2" ParentId="25397" CreationDate="2018-03-06T07:47:16.837" Score="3" Body="&lt;p&gt;The reason for the upper limit is that gravitational force scales as the product of masses over separation squared and that the closest approach of black holes is limited by their event horizons. The lower limit is because the amplitude of the waves gets smaller and becomes undetectable at lower frequencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more detail:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The frequency of gravitational waves is twice the orbital frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kepler's third law tells us that the orbital period $P$ is related to the total mass $M$ and orbital separation $a$ as&#xA;$$ P^2 \propto M^{-1} a^3 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbital frequency $\omega =2\pi/P$ so&#xA;$$\omega \propto M^{1/2} a^{-3/2} \tag{(1)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The frequency of gravitational waves therefore depends on both the mass &lt;em&gt;and&lt;/em&gt; separation of the binary system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the &lt;em&gt;maximum&lt;/em&gt; frequency occurs at closest approach, which for black holes is when their event horizons come into contact. The Schwarzschild radius is $2GfM/c^2$, where $f$ is the fraction of the system mass of each black hole. Thus the closest separation $a_{\rm min}$ is the sum of the Schwarzschild radii and is simply $2GM/c^2$, since $f_1+f_2=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $a_{\rm min}$ for $a$ in equation (1) we see that the maximum frequency&#xA;$\omega^{\rm max} \propto M^{1/2}a_{\rm min}^{-3/2} \propto M^{-1}$&#xA;Thus the maximum frequency gets bigger as the mass gets smaller, and as supermassive black holes are a million to a billion times more massive than stellar black holes, the maximum frequencies are correspondingly smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lower limits to frequency are not really limits at all. The frequencies can be anything smaller than the maximum. However the amplitude of the gravitational waves depends on the acceleration of the masses, which in turn decreases as they become more widely separated. This effectively gives a window between where the waves become detectable and the maximum frequency.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-03-06T08:04:24.080" LastActivityDate="2018-03-06T08:04:24.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25402" PostTypeId="1" CreationDate="2018-03-06T09:20:55.657" Score="1" ViewCount="97" Body="&lt;p&gt;As a body of a suitable size and mass moves within the Roche limit of a planet, it will disintegrate. This will form a debris field that will eventually form a planetary ring. Is there a name given to this debris field before it forms a ring? Something like a Roche train or Roche field seem appropriate but I can't find any reference to such&lt;/p&gt;&#xA;" OwnerUserId="22032" LastActivityDate="2018-03-06T12:37:29.957" Title="What is a debris field called before it becomes a planetary ring?" Tags="&lt;planetary-ring&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25403" PostTypeId="2" ParentId="25402" CreationDate="2018-03-06T11:05:51.323" Score="1" Body="&lt;p&gt;I don't think such an intermediate state exists for very long. Once you shred the moon, it's shredded and the particles orbit on locally Keplerian orbits.&lt;br&gt;&#xA;This would look like Saturns rings, which are also well inside the planet's Roche limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intermediate state you propose to name would have to form either&lt;br&gt;&#xA;a) a steaty state of in some dynamic variable or&lt;br&gt;&#xA;b) have a sufficiently long lifetime (which would be your job to prove)&lt;br&gt;&#xA;in order to earn its own name.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: As OP has asked for material on ring spreading, I can point to &lt;a href=&quot;http://adsabs.harvard.edu/abs/2012Sci...338.1196C&quot; rel=&quot;nofollow noreferrer&quot;&gt;A. Crida's work&lt;/a&gt; who had a bit of a controversial idea about the origin of the gas giant moons. However his calculations involve also classical calculations of collisional disc spreading.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditDate="2018-03-06T12:37:29.957" LastActivityDate="2018-03-06T12:37:29.957" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25404" PostTypeId="1" CreationDate="2018-03-06T13:48:45.060" Score="0" ViewCount="49" Body="&lt;p&gt;Today I saw something that passed like a jet. But am sure it's not so because jet will leave a long track of its passage. It moved in a straight line with sparkling track and what I saw was observed even from  200kms away from my destination. What could it be?&lt;img src=&quot;https://i.stack.imgur.com/XN2is.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="22062" LastEditorUserId="2365" LastEditDate="2018-03-06T14:25:38.383" LastActivityDate="2018-03-07T15:32:04.303" Title="It moved in a straight line with sparkling track.What could it be?" Tags="&lt;movement&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2018-03-07T20:52:53.327" ContentLicense="CC BY-SA 3.0" />
  <row Id="25405" PostTypeId="1" AcceptedAnswerId="25406" CreationDate="2018-03-05T15:49:00.290" Score="6" ViewCount="701" Body="&lt;p&gt;I don't mean necessarily a granular chart of all of the different rings at different distances, but is it days? Weeks?&lt;/p&gt;&#xA;" OwnerUserId="22065" OwnerDisplayName="Chris B. Behrens" LastActivityDate="2018-03-06T14:10:00.323" Title="What are the periods of Saturn's rings?" Tags="&lt;orbit&gt;&lt;saturn&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25406" PostTypeId="2" ParentId="25405" CreationDate="2018-03-05T16:44:20.443" Score="11" Body="&lt;p&gt;Hours to days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbital period is proportional to the 3/2 power of the orbital radius, and the orbital period of the moon &lt;a href=&quot;https://en.wikipedia.org/wiki/Methone_(moon)&quot; rel=&quot;noreferrer&quot;&gt;Methone&lt;/a&gt; (and thus of the particles in its associated ring arc) is very close to 24 hours. Working from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Rings_of_Saturn#Major_subdivisions&quot; rel=&quot;noreferrer&quot;&gt;orbital radii given in this table&lt;/a&gt;, I get: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;D ring: ~5 hours&lt;/li&gt;&#xA;&lt;li&gt;C ring: ~6-7 hours&lt;/li&gt;&#xA;&lt;li&gt;B ring: ~8-10 hours&lt;/li&gt;&#xA;&lt;li&gt;A ring: ~12-14 hours&lt;/li&gt;&#xA;&lt;li&gt;F ring: ~14.5 hours&lt;/li&gt;&#xA;&lt;li&gt;Janus/Epimetheus ring: ~16-17 hours&lt;/li&gt;&#xA;&lt;li&gt;G ring: ~19-20 hours&lt;/li&gt;&#xA;&lt;li&gt;Methone &amp;amp; Anthe arcs: ~24 hours&lt;/li&gt;&#xA;&lt;li&gt;Pallene ring: ~27 hours&lt;/li&gt;&#xA;&lt;li&gt;E ring: ~21 hours to ~90 hours&lt;/li&gt;&#xA;&lt;li&gt;Phoebe ring: 90-500 days&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2573" OwnerDisplayName="Russell Borogove" LastActivityDate="2018-03-05T16:44:20.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25407" PostTypeId="2" ParentId="25405" CreationDate="2018-03-06T03:30:42.803" Score="4" Body="&lt;p&gt;This is a supplemental answer to @RussellBorogove's &lt;a href=&quot;https://space.stackexchange.com/a/25856/12102&quot;&gt;excellent and intuitive answer&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Vis-viva_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;vis-viva equation&lt;/a&gt; one of the few equations I can actually remember:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v^2 = GM \left( \frac{2}{r} - \frac{1}{a} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can set $r=a$ for a circular orbit, giving $v=\sqrt{GM/a}$ where the product of the gravitational constant times Saturn's mass $GM$ is called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Standard_gravitational_parameter&quot; rel=&quot;nofollow noreferrer&quot;&gt;standard gravitational parameter&lt;/a&gt; and is about 3.793E+07 km^3/s^2. (beware of meters vs kilometers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The circumference is $C = 2 \pi a$, so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T = \frac{C}{v} = 2 \pi a \sqrt{a/GM} = \frac{2 \pi a^{3/2}}{\sqrt{GM}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which shows the 3/2 power scaling mentioned in @RB's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Methone Ring with a distance of 194,230 km, this gives 24.3 hours, same as the other answer of course.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One caveat is that for more accurate calculations, it's important to remember that for strongly oblate planets, the radial variation of the gravity field does not vary strictly as $1/r^2$ but will deviate significantly at shorter distances, and so the period &lt;em&gt;will not scale precisely as $a^{3/2}.$&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a spherically symmetric body, no matter what the radial profile you can put all of the mass at the center point instead of integrating throughout the sphere, as long as you are outside of it. This is called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Shell_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shell Theorem&lt;/a&gt; first proven by Sir Isaac Newton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the oblate figure is only cylindrically symmetric, not spherically symmetric, so you actually do have to integrate the force from every point within the planet. Most of this will already be captured in the $J_2$ term, and so I've just asked &lt;a href=&quot;https://space.stackexchange.com/q/25868/12102&quot;&gt;Equation for orbital period around oblate bodies, based on J2?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" OwnerDisplayName="uhoh" LastActivityDate="2018-03-06T10:28:23.650" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25408" PostTypeId="2" ParentId="25362" CreationDate="2018-03-06T14:27:52.423" Score="3" Body="&lt;p&gt;My email to Paul Schlyter and his reply, posted with permission:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On &lt;a href=&quot;http://www.stjarnhimlen.se/comp/tutorial.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.stjarnhimlen.se/comp/tutorial.html&lt;/a&gt; you note that&#xA;  Mercury's ascending node is given as:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;N =  48.3313_deg + 3.24587E-5_deg   * d&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;According to &lt;a href=&quot;https://ssd.jpl.nasa.gov/txt/p_elem_t1.txt&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ssd.jpl.nasa.gov/txt/p_elem_t1.txt&lt;/a&gt; Mercury's&#xA;  ascending node precesses in the negative direction.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;As &lt;a href=&quot;https://astronomy.stackexchange.com/questions/25362&quot;&gt;Where to get rate of change for calculating ephemeris from JPL Horizons&lt;/a&gt; notes, NASA's&#xA;  value matches the negative of your value almost exactly.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Are you using a different sign convention here, or is this an error?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reply:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It's neither - instead I use a somewhat different reference system.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I use the &quot;epoch of the day&quot; instead of the fixed epoch of, say,&#xA;  J2000.0. So add the rate of precession to the negative rate of change&#xA;  you found at NASA's site, and you should get my rate of change.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Since I use the &quot;epoch of the day&quot;, the positions I get are suitable&#xA;  to compute e.g. the rise, transit, and set times for the planets. Of&#xA;  course they are also good for computing the positions of several&#xA;  planets relative to one another. But if you need the positions to plot&#xA;  on a star map drawn for some fixed epoch, you need to apply a&#xA;  correction for precession to these positions.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Yes, it can be confusing to distinguish those different astronomical&#xA;  coordinate systems from one another. But that's a consequence of&#xA;  living on the planet Earth which orbits the Sun, rotates, and wobbles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2018-03-06T14:27:52.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25409" PostTypeId="1" CreationDate="2018-03-06T15:18:38.450" Score="1" ViewCount="311" Body="&lt;p&gt;I wanted to ask this, because while I was working on two fictional planets back on the Worldbuilding SE, one of them being habitable, which orbit the 61 Cygni A star of the 61 Cygni binary star system, I realised something: 61 Cygni A is a BY Draconis variable star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/61_Cygni&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is the Wikipedia article link for more information. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is: can a BY Draconis variable K-class star like 61 Cygni A permit the development of complex life, unlike its partner, 61 Cygni B, is a flare star that can barely get life past the primordial stage, if any planets are there? Or does it have some problems in that regard? If so, what are those problems that a habitable world in that star system could have, and other star systems of the BY Draconis variable star category? &lt;/p&gt;&#xA;" OwnerUserId="19719" LastEditorUserId="19719" LastEditDate="2018-03-07T16:20:44.223" LastActivityDate="2019-09-03T19:41:30.957" Title="Can BY Draconis variable stars sustain any form of complex life?" Tags="&lt;exoplanet&gt;&lt;astrophysics&gt;&lt;binary-star&gt;&lt;habitable-zone&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25410" PostTypeId="2" ParentId="25409" CreationDate="2018-03-06T16:13:33.940" Score="2" Body="&lt;p&gt;Life could easily form underground, where flares don't reach.  Never underestimate the power of 10^(googol) possible locations in the universe for living things to come into being.   No reason to limit to organic, or CO2 / O2 connected, or photon-dependent, life forms. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-03-06T16:13:33.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25411" PostTypeId="2" ParentId="25409" CreationDate="2018-03-06T17:02:00.637" Score="2" Body="&lt;p&gt;Suppose you have a star with luminosity varying between $L_{min}$ and $L_{max}$. Then the inner limit of the life zone will be in the interval $[c_1\sqrt{L_{min}},c_1\sqrt{L_{max}}]$ where $c_1\approx 0.7$ AU (or whatever value you think makes sense for the inner limit). The outer limit will be in $[c_2\sqrt{L_{min}},c_2\sqrt{L_{max}}]$ with $c_2\approx 1.4$ AU. So if $c_1\sqrt{L_{max}}&amp;lt;c_2\sqrt{L_{min}}$ there will be a region that is always in the life zone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.alcyone.de/SIT/mainstars/SIT000409.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;61 Cygni A has magnitude varying between 5.19 and 5.27&lt;/a&gt;, $L_{min}=0.666807$ and $L_{max}=0.717794$. That gives a constantly habitable zone between 0.61 and 1.14 AU. Narrower than the Sun's, but still there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A randomly changing luminosity might make climate somewhat unstable, but if it changes fast enough it might not couple very strongly. Now, the flares and other effects of being a variable star with big starspots may prove bad for volatile retention and life, but it is not clear how big that problem is. &lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2018-03-06T17:02:00.637" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25414" PostTypeId="2" ParentId="10597" CreationDate="2018-03-06T21:35:39.143" Score="2" Body="&lt;p&gt;The one far away is Ganymede, the brightest is Europa but not much of a way for the others (from any angle) I did find a diagram that lines up pretty well with your picture though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8qbbq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8qbbq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22070" LastActivityDate="2018-03-06T21:35:39.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25415" PostTypeId="1" CreationDate="2018-03-06T23:06:14.937" Score="5" ViewCount="165" Body="&lt;p&gt;Jupiter is posited, in some models, to have a rocky core about the size of a few Earths.  If we could see through the thick gaseous atmosphere, what topographical features might we find on the rocky core?  I know that no probe has ever sent data from Jupiter's (hypothesized) solid surface, and that orbiters probably can't penetrate the clouds to produce a topographic map, but do we at least have any educated guesses as to what sort of surface features are and are not likely to exist?  For example, given Jupiter's massive gravity, would the surface be extremely smooth?  Or would it not be out of the question to observe things like impact craters or mountain ranges?&lt;/p&gt;&#xA;" OwnerUserId="22072" LastEditorUserId="22072" LastEditDate="2018-03-06T23:13:56.823" LastActivityDate="2018-03-07T17:19:58.483" Title="What might the surface of Jupiter's rocky core look like?" Tags="&lt;jupiter&gt;&lt;surface&gt;&lt;topographic-map&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2018-03-07T21:07:18.593" ContentLicense="CC BY-SA 3.0" />
  <row Id="25416" PostTypeId="1" AcceptedAnswerId="26971" CreationDate="2018-03-06T23:24:41.413" Score="10" ViewCount="1560" Body="&lt;p&gt;So the standard theory of the solar nebula is that in the region of the gas planets, ice and rock could condense to form planetesimals, which could then accrete hydrogen and helium to form the gas giants. The giant planets are all mostly hydrogen and helium, but Uranus and Neptune have relatively large amounts of hydrogen compounds like methane (that's what gives them their color).&lt;/p&gt;&#xA;&lt;p&gt;My question is: why did that happen? How did Uranus and Neptune get their methane? My impression is that all the gas giants were far enough out for methane to condense into ice, so how did Uranus and Neptune end up preferentially with methane?&lt;/p&gt;&#xA;" OwnerUserId="2010" LastEditorUserId="7411" LastEditDate="2021-04-18T15:55:47.790" LastActivityDate="2021-04-18T15:55:47.790" Title="Why do Uranus and Neptune have more methane than Jupiter and Saturn?" Tags="&lt;solar-system&gt;&lt;gas-giants&gt;&lt;neptune&gt;&lt;uranus&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25417" PostTypeId="2" ParentId="25415" CreationDate="2018-03-06T23:29:20.777" Score="2" Body="&lt;p&gt;An answer to your question is not known at this time.  From the &lt;a href=&quot;https://en.wikipedia.org/wiki/Jupiter#Internal_structure&quot; rel=&quot;nofollow noreferrer&quot;&gt;Internal Structure&lt;/a&gt; section of Jupiter's Wikipedia page,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Jupiter is &lt;strong&gt;thought&lt;/strong&gt; to consist of a dense core with a mixture of elements, a surrounding layer of liquid metallic hydrogen with some helium, and an outer layer predominantly of molecular hydrogen. Beyond this basic outline, there is &lt;strong&gt;still considerable uncertainty&lt;/strong&gt;. The core is &lt;strong&gt;often described as rocky&lt;/strong&gt;, but &lt;strong&gt;its detailed composition is unknown&lt;/strong&gt;, as are the properties of materials at the temperatures and pressures of those depths (see below). In 1997, the existence of the core was suggested by gravitational measurements, indicating a mass of from 12 to 45 times that of Earth, or roughly 4%–14% of the total mass of Jupiter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Emphasis mine)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea that the core is &quot;rocky&quot; or composed of metallic hydrogen is still unknown. Determining more about this is one of the objectives of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Juno_(spacecraft)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Juno mission&lt;/a&gt; currently in orbit. We may obtain better models of Jupiter's internal structure in the coming years, but at this point we just don't know.&lt;/p&gt;&#xA;" OwnerUserId="14255" LastActivityDate="2018-03-06T23:29:20.777" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25418" PostTypeId="1" AcceptedAnswerId="25521" CreationDate="2018-03-07T09:50:36.767" Score="2" ViewCount="1553" Body="&lt;p&gt;There are some articles in wikipedia with the list of known (or candidates) black holes &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_black_holes&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_nearest_black_holes&quot; rel=&quot;nofollow noreferrer&quot;&gt;list of nearest black holes&lt;/a&gt;. Also there is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Milky_Way#/media/File:Artist%27s_impression_of_the_Milky_Way_(updated_-_annotated).jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;map of Milky Way&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is where can I get approximate position of the known black holes in our Galaxy (Milky Way) as &lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Galactic coordinate system&lt;/a&gt; or maybe some other galaxy coordinates? &lt;/p&gt;&#xA;" OwnerUserId="15004" LastActivityDate="2018-03-16T14:27:16.023" Title="Coordinates of known black holes in our Galaxy (Milky Way)" Tags="&lt;black-hole&gt;&lt;galaxy&gt;&lt;milky-way&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25422" PostTypeId="2" ParentId="25404" CreationDate="2018-03-07T15:32:04.303" Score="2" Body="&lt;p&gt;It is most likely a jet contrail. Not all contrails are persistent, and a high altitude jet can be seen over a great distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fireball may leave a trail, but if it does, then it is likely to persist for as long or longer than a jet contrail. A fireball appears to move much faster than a jet, crossing the sky in a few seconds. It also appears to be very bright.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image and description both point to this being a jet.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-03-07T15:32:04.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25423" PostTypeId="2" ParentId="25415" CreationDate="2018-03-07T16:32:32.320" Score="4" Body="&lt;p&gt;There's not going to be any surface features.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, let's assume you could look through a thick soup of metallic hydrogen. This material has densities starting at that of water, going to the densities of solid rock and even above, while still remaining liquid.&lt;br&gt;&#xA;On a sidenote: No space probe will ever be able to look into that mess. Only gravitational measurments will help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In recent years, &lt;a href=&quot;http://adsabs.harvard.edu/abs/2012PhRvL.108k1101W&quot; rel=&quot;nofollow noreferrer&quot;&gt;it has been shown&lt;/a&gt; that under such conditions typical minerals are thermodynamically not stable, and thus will dissolve into the hydrogen-helium mix.&lt;br&gt;&#xA;Thus, the core will not have any well-defined boundary, but it's going to look more like the following sketch (emphasis on sketch!):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SFSlQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SFSlQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which I took from &lt;a href=&quot;http://militzer.berkeley.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this author's website&lt;/a&gt;, which has also a lot of follow-up in-depth research material linked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In short&lt;/strong&gt;: Essentially any topographical features can be ruled out. That thing down there will be a liquefied, high-pressure mess.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is of course correct to say that Jupiter might not even have a core, but to build a planet with 1 Jupiter mass and one Jupiter radius one usually requires a certain amount of high-Z material in the core, which is why the core-idea is very attractive.&lt;br&gt;&#xA;Also, going around on conferences this year, it seems that &lt;a href=&quot;https://www.missionjuno.swri.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;JUNO&lt;/a&gt; does find a 5-25 Earth masses core, but there is no publication on this yet.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditDate="2018-03-07T17:19:58.483" LastActivityDate="2018-03-07T17:19:58.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25425" PostTypeId="1" CreationDate="2018-03-07T18:53:27.153" Score="3" ViewCount="106" Body="&lt;p&gt;I have purchased a &lt;a href=&quot;https://radiojove.gsfc.nasa.gov/software/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Radio Jove&lt;/a&gt; kit and set it up. I have also received some signals. I am using &lt;a href=&quot;http://www.radiosky.com/skypipeishere.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Radio Skypipe II&lt;/a&gt; software as discussed in this Radio Jove &lt;a href=&quot;https://radiojove.gsfc.nasa.gov/data_analysis/data_tutorial.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;tutorial&lt;/a&gt; (see also &lt;a href=&quot;http://www.reeve.com/Documents/Articles%20Papers/Reeve_RSPSoundcardSetup.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;) with a sound card to record and analyze the output of the receiver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what I get is like an image on the screen, which can be zoomed, and I'm not sure what I'm looking at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What further process should I follow to make some analysis of Jovian radio signals at 20.1 MHz beyond just looking at the image (&lt;a href=&quot;https://en.wikipedia.org/wiki/Waterfall_plot&quot; rel=&quot;nofollow noreferrer&quot;&gt;spectrogram? waterfall plot?&lt;/a&gt;) that the software produces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example of the Radio Skypipe II output from their &lt;a href=&quot;http://www.radiosky.com/skypipeishere.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Nbcn4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nbcn4.jpg&quot; alt=&quot;Radio Skypipe II example output&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22088" LastEditorUserId="7982" LastEditDate="2018-03-08T05:47:24.167" LastActivityDate="2018-03-08T05:47:24.167" Title="How to analyse Jovian signal received at 20.1 MHz using dual dipole antenna?" Tags="&lt;amateur-observing&gt;&lt;radio-astronomy&gt;" AnswerCount="0" CommentCount="7" ClosedDate="2018-03-08T17:20:39.093" ContentLicense="CC BY-SA 3.0" />
  <row Id="25426" PostTypeId="1" CreationDate="2018-03-07T22:26:31.467" Score="2" ViewCount="597" Body="&lt;p&gt;With binoculars or telescopes you'd be able to see to other countries if the earth were flat. And looking out at the ocean we can see where the earth curves and things like the top of ships, not the whole ship. Unless those who think the earth is flat believe they are looking at the end of the world when looking out at the ocean? How do they arrive at this position?&lt;/p&gt;&#xA;" OwnerUserId="22089" LastEditorUserId="6" LastEditDate="2018-03-08T02:57:15.567" LastActivityDate="2018-03-09T17:41:17.060" Title="Is the flat earth cosmology scientifically justifiable today?" Tags="&lt;earth&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ClosedDate="2018-03-08T11:41:42.130" ContentLicense="CC BY-SA 3.0" />
  <row Id="25428" PostTypeId="2" ParentId="25426" CreationDate="2018-03-07T23:21:18.453" Score="18" Body="&lt;p&gt;I realize this queston might be closed, and perhaps this should've been a comment, but it got too long:&lt;/p&gt;&#xA;&lt;p&gt;I was in a radio show last spring with a flat-earther. I really tried to understand him, rather than just ridicule him. We went to the beach with his 87x magnifying camera, to watch the coastline 19 km away. If my model of the geometry of Earth were right, we would be able to see the buildings on the other side of the belt, but not the shore. If his model were correct, I &lt;em&gt;thought&lt;/em&gt; we would be able to see the shore, but it turns out they have an ad hoc explanation that has to do with the vanishing point on a flat Earth lying &lt;em&gt;below&lt;/em&gt; the ground, thus making sufficiently distant objects disappear.&lt;/p&gt;&#xA;&lt;p&gt;So, apparently this is their explanation.&lt;/p&gt;&#xA;&lt;p&gt;When after five minutes I still didn't understand this, he told me it was no wonder that I didn't understand; it took him two  years to understand.&lt;/p&gt;&#xA;&lt;p&gt;Well, big surprise, we couldn't see the shore. We could in fact see a bit more of the buildings than we should be able to if there were no atmosphere. I assigned this to refraction (which can actually move the horizon several tens of kilometers away, or even ~100 km over water in the springtime when the air is getting warm but the water is still cold), but since he didn't believe in refraction, that didn't convince him.&lt;/p&gt;&#xA;&lt;p&gt;We then talked a bit about what mechanism might produce a flat Earth. The conversation sort of ended when I realized he didn't believe in gravity either…&lt;/p&gt;&#xA;&lt;h3&gt;Is a flat Earth scientifically justifiable today?&lt;/h3&gt;&#xA;&lt;p&gt;The question title has now been edited to a more easily answerable one. The answer is — unsurprisingly to most people on this site, I suppose — a big, resounding &amp;quot;&lt;strong&gt;NO&lt;/strong&gt;&amp;quot;! The flat Earth theory is &lt;em&gt;not&lt;/em&gt; a theory, in the scientific sense of the word. It's not even a hypothesis. You could call it a model, and in many cases, it's a fine model, namely on scales much smaller than the horizon distance. You may assume a flat Earth if you want to calculate the area of the parking lot. This is analogous to Newtonian dynamics being a fine model for small velocities and gravitational fields. It works fine if you want to calculate the force of an apple falling on Newton's head. But everybody knows it become less exact, the larger and faster the apple.&lt;/p&gt;&#xA;&lt;p&gt;I suspect that the use of the word &amp;quot;theory&amp;quot; in everyday language influences how people think about science. &amp;quot;It's just a theory&amp;quot;, they say, without realizing the enormous amount of work, experiments, verifications, and falsifications that go into building a scientific theory.&lt;/p&gt;&#xA;&lt;p&gt;Flat-Earthers in general don't realize, I think, that in science you can't just assume that something works in another way, without this assumption having severe consequences for many other theories.&lt;/p&gt;&#xA;&lt;h3&gt;Why not?&lt;/h3&gt;&#xA;&lt;p&gt;The list of reasons that a flat Earth is incompatible with a range of well-established theories is long. Here's a subset:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gravity&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gravity explains why a large mass becomes round. FE'ers have no explanation of why it could be flat.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;Electromagnetism&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Sun and Moon are circulating 5000 km over Earth, held up by electromagnetic forces. This is completely impossible within the frames of the accepted theory of electromagnetism, but they don't propose a new theory of EM.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;Astronomical distances&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Standard trigonometry reveals that in their model, the Moon would look different from different places of Earth. It does, but much, much less than if it were 5000 km away (because it is in fact 400,000 km away)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;Meteorology&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The guy I talked to had no explanation for why cyclones rotate clockwise and counter-clockwise in the Southern and Northen hemispheres, respectively. The Flat Earth Society &lt;a href=&quot;https://wiki.tfes.org/The_Coriolis_Effect&quot; rel=&quot;noreferrer&quot;&gt;explains&lt;/a&gt; this by &amp;quot;&lt;em&gt;The Wind Currents are put into gradual motion by the attraction of the Northern and Southern Celestial Systems, which are grinding against each other as gears at the equator line&lt;/em&gt;&amp;quot;. I… I don't know what to say about this…&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;Constellations&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Southern Cross can be seen from Australia and South America. I'm in Oslo at the moment, which according to their world map is right in the middle between Australia and South America, yet I can't see the Southern Cross.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;Satellites&#xA;&lt;ul&gt;&#xA;&lt;li&gt;FE'ers generally don't believe in Moon landings and space stations, but many satellites can be seen even with a relatively small telescope.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;The disappearance of ships and other objects below the horizon.&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Simple trigonometry shows that on a flat Earth, you should be able to see as far as the airmass allows you. On a round Earth, this is complicated by refraction, enabling us to see farther than if there were no atmosphere. But still, you can easily see how the Sun or Moon sets or rises partly below the horizon.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;Lots of other inconsistencies&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Stellar parallaxes.&lt;/li&gt;&#xA;&lt;li&gt;The different weight of the same object at equator and the poles.&lt;/li&gt;&#xA;&lt;li&gt;Eclipses (in particular the times at which they occur).&lt;/li&gt;&#xA;&lt;li&gt;The movement of Jupiter's moons (which can be seen with a regular binocular).&lt;/li&gt;&#xA;&lt;li&gt;Tides.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The problem is that there is no way to discuss these things with FE'ers. If you give an argument that they can't disprove, they will either claim that you're just wrong, or that &amp;quot;you think so because that's how you were schooled&amp;quot;, or that you're a part of the conspiracy.&lt;/p&gt;&#xA;&lt;p&gt;In particular the last point is important. Since the FE model, were it right, would imply that most of all physics would be wrong, that means that either&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;all physicists in the world are stupid, which makes it pretty amazing that we've made planes fly, computers work, bridges hold up traffic, compasses show us the way, as well as predicted stuff like Venus transits and the weather tomorrow. Or&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;all physicists are a part of a conspiracy, which would be equally amazing how none of the millions of physicists through history has confessed. Where's my bribe?!&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;h3&gt;Why am I not just ignoring them?&lt;/h3&gt;&#xA;&lt;p&gt;You may say that by writing this, I am just feeding the trolls, and giving people the impression that flat-Earthers are a problem for science. But in my opinion, the attitude that they hold toward science is much larger than just a few crackpots claiming what is obviously false. It's a symptom of something sad in our society, namely that &amp;quot;science is just &lt;em&gt;one&lt;/em&gt; opinion, and &lt;em&gt;my&lt;/em&gt; opinion is just a valid&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;I acknowledge that scientists can tend to be cocky and should try to be more open to understanding why people don't believe us. But what makes a few people believe in a flat Earth is the same thing that&#xA;makes people deny that human activities affect the climate,&#xA;or makes them believe that vaccines are evil,&#xA;and makes some politicians get away with using a term like &amp;quot;alternative facts&amp;quot;.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-03-09T17:41:17.060" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="25429" PostTypeId="2" ParentId="7900" CreationDate="2018-03-08T01:14:40.910" Score="1" Body="&lt;p&gt;An alternative expression for the extra, (supra-Newtonian, relativistic) acceleration (brought to my attention by user /u/uhoh) is presented by Shahid-Saless (Colorado) and Yeomans (JPL) in their 1994 Astronomical Journal paper: &lt;a href=&quot;https://www.google.co.uk/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=6&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0ahUKEwic2LfchcXZAhXL-qQKHeJxAusQFghZMAU&amp;amp;url=http%3A%2F%2Fadsabs.harvard.edu%2Ffull%2F1994AJ....107.1885S&amp;amp;usg=AOvVaw1sOMUCKcBLnUaI427q9N7S&quot; rel=&quot;nofollow noreferrer&quot;&gt;Relativistic Effects on the Motion of Asteroids and Comets&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their equation 3.11 for the Newtonian + Relativistic acceleration of a single target body orbiting the Sun is as follows:-&#xA;$$&#xA;\frac{d^2 \textbf{r}}{c^2dt^2} &#xA;= \frac{-\mu}{r^3}\textbf{r} &#xA;+ \frac{\mu}{r^3} \left[ \left(4\frac{\mu}{r}-\frac{v^2}{c^2}\right)\textbf{r} &#xA;+ 4 \frac{(\textbf{r}\textbf{.v})\textbf{v}}{c^2}\right]&#xA;$$&#xA;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{r}$  is the instantaneous position vector of the target body relative to the Sun,   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{v}$  is the instantaneous velocity vector of the target body relative to the Sun,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c$  is the speed of light,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu = GM/c^2$ is the Schwartzschild Gravitational Radius of the Sun, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G$ is the universal gravitational constant, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M$ is the (post-Newtonian) mass of the Sun, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{-\mu}{r^3} $ the first term on the RHS, is the Newtonian radial acceleration with the negative sign indicating acceleration towards the source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors present a derivation of the equation (which is beyond my expertise). They also use it to derive an expression for $\delta \omega$ the amount (in radians) of rotation of the line of apsides per complete ($2\pi$ radians) orbital revolution:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\delta\omega = \frac{6\pi\mu}{a(1-e^2)} &#xA;\equiv \frac{6\pi GM}{c^2 a(1-e^2)} &#xA;\equiv \frac{24\pi^3 a^2}{T^2c^2 (1-e^2)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a$ is semi-major axis of orbit, $e$ is eccentricity of orbit, and $T$ is the orbit period.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right-most version is identical to the &lt;a href=&quot;http://www.gsjournal.net/old/eeuro/vankov.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein 1915&lt;/a&gt; equation presented in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is noteworthy that the Shahid-Saless &amp;amp; Yeomans equation 3.11 indicates that when $\textbf{v}$ is not perpendicular to $\textbf{r}$ a part of the non-Newtonian acceleration will be directed in the direction transverse to the radial direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that caveats apply when moving between a General Relativistic model of Space-Time and a Euclidean-Galilean model -see Stan Liou's answer and the Shahid-Saless &amp;amp; Yeomans paper itself.&lt;/p&gt;&#xA;" OwnerUserId="233" LastActivityDate="2018-03-08T01:14:40.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25431" PostTypeId="1" AcceptedAnswerId="25435" CreationDate="2018-03-08T17:32:03.390" Score="1" ViewCount="76" Body="&lt;p&gt;In short: what mass is appropriate for absolute magnitude $M_V \sim -7$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some context: I'm processing the observational data (supernova light curve) and curve fitting gives me about $0.4M_\bigodot$ for ejecta mass. However, possible progenitor candidate has $M_V \sim -7$, which, I suppose, points to quite a large star. My background in astronomy is not so good (I was mostly doing theoretical physics, namely nonlinear optics and GR before I got to supernovae), so I'm doubting correctness of either my estimates or progenitor candidate parameters.&lt;/p&gt;&#xA;" OwnerUserId="22098" LastEditorUserId="2365" LastEditDate="2018-03-09T10:09:31.767" LastActivityDate="2018-03-09T10:09:31.767" Title="Mass-magnitude relation of evolved stars" Tags="&lt;supernova&gt;&lt;mass&gt;&lt;absolute-magnitude&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25433" PostTypeId="2" ParentId="25365" CreationDate="2018-03-09T01:49:25.110" Score="2" Body="&lt;p&gt;Habitability is related to size or mass (i.e. likely to have a rocky composition and maintain surface water), and distance from the host star (the “habitable zone” – not too hot, not too cold – which depends on stellar temperature and stellar flux). The Planetary Habitability Laboratory (University of Puerto Rico at Arecibo) publishes a &lt;a href=&quot;http://phl.upr.edu/projects/habitable-exoplanets-catalog&quot; rel=&quot;nofollow noreferrer&quot;&gt;catalog of potentially habitable exoplanets&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The PHL catalogue is divided into two lists. The first list covers the “conservative” habitable zone, where the combination of stellar temperature and flux is optimal (equivalent to an orbit between Earth and Mars around a Sun-like star). Exoplanets are also given an ESI (Earth Similarity Index) rating, a measure of how similar the planet is to the stellar flux, mass, and radius of Earth (Earth = 1.0).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As at March 2018 there are only 10 confirmed exoplanets within the conservative habitable zone, the closest (in order) being:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Proxima_Centauri_b&quot; rel=&quot;nofollow noreferrer&quot;&gt;Proxima Centauri&#xA;b&lt;/a&gt; (4.2 ly = 40&#xA;trillion km), ESI 0.85&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Gliese_667_Cc&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gliese 667 C&#xA;c&lt;/a&gt; (23.6 ly), ESI 0.84&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/TRAPPIST-1e&quot; rel=&quot;nofollow noreferrer&quot;&gt;TRAPPIST-1 e&lt;/a&gt; (39.6 ly),&#xA;ESI 0.85&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/TRAPPIST-1f&quot; rel=&quot;nofollow noreferrer&quot;&gt;TRAPPIST-1 f&lt;/a&gt;&#xA;(39.6 ly), ESI 0.68&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/TRAPPIST-1g&quot; rel=&quot;nofollow noreferrer&quot;&gt;TRAPPIST-1&#xA;g&lt;/a&gt; (39.6 ly), ESI 0.58&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/LHS_1140_b&quot; rel=&quot;nofollow noreferrer&quot;&gt;LHS 1140 b&lt;/a&gt; (40.7 ly), ESI&#xA;0.68&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;After these 6, the next nearest in the list is &lt;strong&gt;Kepler-186 f&lt;/strong&gt; at a distance of over 561 ly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that while Proxima b might be “in the zone”, it’s unlikely to be habitable. At only 7 million km from its host star (compare with Mercury, 46 million km from the Sun at perihelion), it’s likely to be tidally locked, restricting any habitable area to a thin border wedge between the two extreme regions. It also receives 400 times more X-ray radiation than the Earth; and the stellar wind pressure is 2000 times that of Earth, sufficient to blow away any atmosphere. Not an ideal candidate to look for life!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Kapteyn_b&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kapteyn b&lt;/a&gt; (12.8 ly) is in PHL’s conservative list and at 12.8 ly would be the 4th closest habitable planet, but PHL lists it as “unconfirmed”, as recent research has cast doubt on its existence: Robertson, Mahadevan &amp;amp; Roy (2015) concluded that &quot;Kapteyn b is not a planet in the Habitable Zone, but an artifact of stellar activity.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PHL’s second list covers exoplanets that are a bit less likely to be habitable, either because they are less able to maintain surface liquid water due to their composition (based on size/mass), or because they’re in a less hospitable orbit (the “optimistic habitable zone”). There are 33 confirmed exoplanets in this list, the closest being:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Ross_128_b&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ross 128 b&lt;/a&gt; (11.0 ly), ESI 0.86&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Luyten%27s_Star#Planetary_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gliese 273 b&lt;/a&gt; (12.4 ly), ESI 0.86&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/LHS_1723&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gliese 3323 b&lt;/a&gt; (17.4 ly), ESI 0.89&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Wolf_1061c&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wolf 1061 c&lt;/a&gt; (14 ly), ESI 0.82&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Gliese_832&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gliese 832 c&lt;/a&gt; (16 ly), ESI 0.81&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/TRAPPIST-1d&quot; rel=&quot;nofollow noreferrer&quot;&gt;TRAPPIST-1 d&lt;/a&gt; (39.6 ly), ESI 0.90&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14094" LastActivityDate="2018-03-09T01:49:25.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25434" PostTypeId="1" CreationDate="2018-03-09T01:53:53.597" Score="2" ViewCount="160" Body="&lt;p&gt;I have begun reducing images I collected with a telescope and CCD detector. I read in the the &quot;Handbook of CCD Astronomy&quot; by Howell that I need to trim overscan regions (as well as possible amplifier turn on/off) regions before I can compute the gain of the detector. However, it is not thoroughly clear where this boundary is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we treat the CCD detector as a matrix of pixels, then I know that the overscan region is along the edge columns. To visualize this, I computed the average and standard deviation along each column of pixels to view the variation in intensity from column to column. Below I inserted the left side of two amplifier matrices on my CCD detector: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/g5A4T.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g5A4T.png&quot; alt=&quot;Column-wise average and deviation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blue is the column-wise average and the green is the column-wise standard deviation. The y-axis is the (arbitrarily scaled) intensity and the x-axis is the pixel distance from the edge of the CCD array. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is essentially: &lt;strong&gt;how much of the blue and green curve should I cut-off?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: additionally, it should be noted that I am using the method outlined in &quot;CCD Data...&quot; by Massey &amp;amp; Jacoby (1992) &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1992ASPC...23..240M&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; to compute the gain. &lt;/p&gt;&#xA;" OwnerUserId="16149" LastEditorUserId="16149" LastEditDate="2018-03-12T21:12:54.650" LastActivityDate="2018-03-14T11:41:33.117" Title="How to trim CCD overscan and amplifier turn on/off regions" Tags="&lt;observational-astronomy&gt;&lt;data-analysis&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25435" PostTypeId="2" ParentId="25431" CreationDate="2018-03-09T09:57:56.960" Score="2" Body="&lt;p&gt;A star with absolute magnitude of -7 is big and powerful, It would be among the most powerful stars in the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The star would be near the end of its life, so comparing with other aging stars:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aldebaran is slightly more massive than the sun. It is a red giant (or orange giant according to some sources) It has an absolute magnitude of about -0.6. Aldebaran is too small to undergo a supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Betelgeuse is much bigger. Estimates of its mass put it at about 11 solar masses, but it could be bigger. It has a magnitude of about -6 (it is variable and there is some uncertainty on its distance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eta Carinae is one of the most powerful stars in this part of the galaxy. It is also variable, but has an absolute magnitude of -8—-9, and mass of 100-200 suns&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AG Carinae is a LBV with a mass of 50 suns and a minimum absolute magnitude of about -8, but becoming much brighter during outbursts. (as bright as -11)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So an evolved star with a magnitude of -7 could be supposed to be between the LBV stars and Betelgeuse, with a mass of around 15-20 suns. Stars do get very unstable this late in life, and there could be a great deal of uncertainty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Others here know more about supernovae mechanics, and may be able to explain any discrepancy between the supposed stellar mass, and the mass of the ejecta.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-03-09T09:57:56.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25438" PostTypeId="1" AcceptedAnswerId="25440" CreationDate="2018-03-09T13:20:37.677" Score="-1" ViewCount="125" Body="&lt;p&gt;As they say, &quot;In cosmos near earth but beyond atmosphere there are too much dangerous and powerful radiation frm quasars or gammy burst .,etc&quot;. So if there is so much cosmic radiations being emitted, why can't we extract energy from them?&lt;/p&gt;&#xA;" OwnerUserId="22112" LastEditorUserId="22113" LastEditDate="2018-03-09T20:10:53.443" LastActivityDate="2018-03-09T20:10:53.443" Title="Can we extract power from quasars?" Tags="&lt;quasars&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ClosedDate="2018-03-09T17:19:04.380" ContentLicense="CC BY-SA 3.0" />
  <row Id="25440" PostTypeId="2" ParentId="25438" CreationDate="2018-03-09T14:07:02.880" Score="6" Body="&lt;p&gt;We could, but there isn't enough energy to make it worthwhile. These types of radiation are referred to as &quot;highly energetic&quot; because each individual particle (or very short burst) has comparatively much energy, but averaged over time, there is far more energy around in the form of sunlight.  The highly energetic radiation is more dangerous because even a single particle can damage human cells, and over the years it would take to go to Mars (for instance) there would be a considerable risk of cancer.  Even then, most of that radiation comes from the Sun.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a loose analogy, consider the difference between being constantly blown by a steady wind (OK for energy generation, not dangerous) versus being hit by a bullet once a year (useless for energy generation, but deadly).&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-03-09T14:07:02.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25441" PostTypeId="2" ParentId="25438" CreationDate="2018-03-09T15:10:57.527" Score="3" Body="&lt;p&gt;Quasars are &lt;em&gt;not near the Earth&lt;/em&gt;. The nearest one is 600 million light years away. Most are much much further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are so very far away that they appear &lt;em&gt;very dim&lt;/em&gt;. The brightest quasar is only visible with a telescope, where it appears like a dim star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total amount of energy from a quasar per square metre on Earth is minute. It is not a sensible source of energy. Quasars are just too far away.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-03-09T15:10:57.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25443" PostTypeId="1" AcceptedAnswerId="25474" CreationDate="2018-03-09T16:51:49.270" Score="3" ViewCount="175" Body="&lt;p&gt;My understanding is that there is a gravitationally stable region between the Sun and Mercury where asteroids could - in theory - orbit for long periods without being perturbed. However these hypothetical vulcanoids are seemingly non-existent; it being likely that their orbits decay through other means besides the influence of a planet, such as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Yarkovsky_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Yarkovsky effect&lt;/a&gt;. We also have trojan asteroids, which share the orbits of planets such as Jupiter but spend their time in the leading and trailing Lagrange Points, keeping their orbits stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a Hot Jupiter such as &lt;a href=&quot;https://en.wikipedia.org/wiki/51_Pegasi_b&quot; rel=&quot;nofollow noreferrer&quot;&gt;51 Pegasi b&lt;/a&gt; were to host trojan asteroids, would their proximity to the planet allow their orbits to remain stable for longer, or would their orbits decay regardless? If a Hot Jupiter migrated inwards towards its star, is it more likely to have swept away any possible trojans?&lt;/p&gt;&#xA;" OwnerDisplayName="user10106" LastEditorDisplayName="user10106" LastEditDate="2018-03-13T14:56:31.873" LastActivityDate="2018-03-13T19:43:14.420" Title="Could Hot Jupiter systems sustain Trojans?" Tags="&lt;exoplanet&gt;&lt;trojan-asteroids&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25444" PostTypeId="1" AcceptedAnswerId="25451" CreationDate="2018-03-09T20:02:36.703" Score="3" ViewCount="205" Body="&lt;p&gt;Older pages like &lt;a href=&quot;http://astro.physics.uiowa.edu/~srs/2952_EXW/Locus19_EXW.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this &quot;Exploration of the Solar System&quot; course page&lt;/a&gt; describe the transition as being a few hundred kilometers down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More recent findings seem to put the boundary deeper. See&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The result was a surprise for the Juno science team because it indicated that the weather layer of Jupiter was more massive, extending much deeper than previously expected. The Jovian weather layer, from its very top to a depth of 1,900 miles (3,000 kilometers), contains about one percent of Jupiter’s mass (about 3 Earth masses).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.nasa.gov/feature/jpl/nasa-juno-findings-jupiter-s-jet-streams-are-unearthly&quot; rel=&quot;nofollow noreferrer&quot;&gt;- nasa.gov, March 7, 2018: &quot;NASA Juno Findings - Jupiter’s Jet-Streams Are Unearthly&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Below whirling jet streams, 3,000 kilometres deep, lies a dense, rotating core of liquid hydrogen and helium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.abc.net.au/news/science/2018-03-08/juno-jupiter-turbulent-atmosphere-jet-stream-liquid-core/9517628&quot; rel=&quot;nofollow noreferrer&quot;&gt;- abc.net.au, March 7, 2018: &quot;Jupiter: Juno uncovers deep jet streams, dense liquid core — and cyclone mystery&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the updated understanding of the mass of gas above it, what would the pressure and temperature be at that level?&lt;/p&gt;&#xA;" OwnerUserId="22118" LastActivityDate="2018-03-13T14:38:47.530" Title="What would the pressure and temperature of gas be, right above Jupiter's gas/liquid boundary?" Tags="&lt;jupiter&gt;&lt;planetary-atmosphere&gt;&lt;temperature&gt;&lt;gas&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25447" PostTypeId="1" AcceptedAnswerId="25448" CreationDate="2018-03-10T10:05:40.500" Score="3" ViewCount="154" Body="&lt;p&gt;After an initial collimation, I took my scope (150mm/750mm) out for a star test. While I could see concentric circles when the bright star is at the centre of the video, when the star is on the side of the view, I get an image typical of an uncollimated scope. Please refer the images below. The left image is how it appears when the bright star is at the centre. The right image is how it appears when the star is on the upper portion of the view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/83lNL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/83lNL.png&quot; alt=&quot;The left image is how it appears when the bright star is at the centre. The right image is how it appears when the star is on the upper portion of the view.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does my scope require further collimation, or is it fine the way it is now?&lt;/p&gt;&#xA;" OwnerUserId="22126" LastActivityDate="2018-03-13T17:15:30.707" Title="Newtonian reflector collimation question" Tags="&lt;telescope&gt;&lt;newtonian-telescope&gt;&lt;collimation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25448" PostTypeId="2" ParentId="25447" CreationDate="2018-03-10T14:31:42.533" Score="3" Body="&lt;p&gt;Your scope is f/5. There will always be some &lt;a href=&quot;https://en.wikipedia.org/wiki/Coma_(optics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;coma aberration&lt;/a&gt;.&#xA;Always, the image quality is worse as you get closer to the edge of the field of view (FOV).&#xA;Personally, given that the fact that you will want to observe your targets at the center of the FOV, this collimation would be acceptable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Happy photon hunting!&lt;/p&gt;&#xA;" OwnerUserId="5355" LastEditorUserId="616" LastEditDate="2018-03-13T17:15:30.707" LastActivityDate="2018-03-13T17:15:30.707" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25449" PostTypeId="1" AcceptedAnswerId="25450" CreationDate="2018-03-10T16:08:39.437" Score="1" ViewCount="267" Body="&lt;p&gt;Is that possible with some strange Quantum Effect could give rise to a completely new kind of a star? Gravastar(Gravitational Vacuum Star) an object hypothesized in astrophysics as an alternative to the black hole theory.The notion of gravastars builds on Einstein's theory of general relativity and imposes a universal &quot;smallest size&quot; that is known to exist according to well-accepted quantum theory. This size is known as the Planck length and is derived using the speed of light, Planck's constant and the gravitational constant. Quantum theory says that any scale smaller than the Planck length is unobservable and meaningless to physics and physicists. This limit can be imposed on the wavelength of a beam of light so as to obtain a limit of blue shift that the light can undergo. However, in the presence of an event horizon, the blue shift diverges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the original formulation by Mazur and Mottola, gravastars contain a central region featuring a p=-ρ false vacuum or &quot;dark energy&quot;, a thin shell of p=ρ perfect fluid, and a true vacuum p=ρ=0 exterior. The dark energy like the behavior of the inner region prevents collapse to a singularity and the presence of the thin shell prevents the formation of an event horizon, avoiding the infinite blue shift. The inner region has thermodynamically no entropy and may be thought of as a gravitational Bose-Einstein condensate. Severe red-shifting of photons as they climb out of the gravity well would make the fluid shell also seem very cold, almost absolute zero.&lt;/p&gt;&#xA;" OwnerUserId="22068" LastActivityDate="2018-03-10T19:27:45.490" Title="Gravastar Star With no Horizon with ultracompact and very density possible?" Tags="&lt;black-hole&gt;&lt;astrophysics&gt;&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25450" PostTypeId="2" ParentId="25449" CreationDate="2018-03-10T19:27:45.490" Score="4" Body="&lt;p&gt;Of course unknown &quot;strange Quantum Effects&quot; could do anything, but &lt;a href=&quot;https://phys.org/news/2016-10-ligo-black-holes-gravastars.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this letter&lt;/a&gt; reports calculations which suggest that at least one of the LIGO gravitational wave events is consistent with black holes but not with gravastars. That doesn't rule out the possibility of gravastars &lt;em&gt;and&lt;/em&gt; black holes existing, but it speaks quite strongly against the idea that all the phenomena attributed to black holes are actually due to gravastars.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-03-10T19:27:45.490" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25451" PostTypeId="2" ParentId="25444" CreationDate="2018-03-10T23:59:18.483" Score="5" Body="&lt;p&gt;Technically there isn't really a gas-liquid boundary because temperatures are well above the critical point of hydrogen &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrogen&quot; rel=&quot;nofollow noreferrer&quot;&gt;(33K and about 18bar)&lt;/a&gt;. It's a supercritical fluid. There are important changes at various depths though including the bottom of the circulating winds, the transition to metallic hydrogen and (it now appears) an increasing density of dissolved/suspended heavier elements making up a somewhat diffuse core. There is lots of &lt;a href=&quot;https://en.wikipedia.org/wiki/Jupiter#Internal_structure&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;pre-Juno&quot; information&lt;/a&gt; easily accessible, including temperature and pressure at the transition to metallic hydrogen (10000K and 20GPa). &lt;a href=&quot;https://www.nature.com/articles/d41586-018-02612-y&quot; rel=&quot;nofollow noreferrer&quot;&gt;This source&lt;/a&gt; gives a pressure of 100kBar at the bottom of the circulating winds layer (about 10 GPa), but I can't find a temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just found this very interesting diagram:&lt;a href=&quot;https://i.stack.imgur.com/mRvXj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mRvXj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; from &lt;a href=&quot;http://burro.case.edu/Academics/Astr221/SolarSys/OuterGiants/int.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;these lecture notes&lt;/a&gt; which shows how the expected temperatures and pressures inside all the giant planets relate the various phases of hydrogen. Notice how all of them except Jupiter do cross the short curved black line at the left, so that they do have a gas-liquid boundary at pressures of a few atmospheres and temperatures of a few tens of Kelvin&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="21928" LastEditDate="2018-03-13T14:38:47.530" LastActivityDate="2018-03-13T14:38:47.530" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25452" PostTypeId="1" CreationDate="2018-03-11T05:04:46.667" Score="2" ViewCount="391" Body="&lt;p&gt;I know that Venus is visible during sunset and sunrise, but is it also visible in deep dark night? &lt;/p&gt;&#xA;" OwnerUserId="22135" LastEditorUserId="2365" LastEditDate="2018-03-11T08:41:13.467" LastActivityDate="2018-03-12T04:10:47.347" Title="Is Venus visible throughout the night?" Tags="&lt;planet&gt;&lt;venus&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25453" PostTypeId="2" ParentId="25452" CreationDate="2018-03-11T08:40:00.603" Score="2" Body="&lt;p&gt;The short answer is no, but it might be visible for longer than you expect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Mercury or Venus is furthest from the sun they are said to be at &quot;Elongation&quot;. During favourable elongation that occurs at the end of 2018, Venus rises (in my location) at about 4am, four hours before the sun, and Venus will be in the sky when most people would say it is &quot;dark&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The previous Evening elongation in August 2018 is less favourable from the Northern Hemisphere, but from the southern hemisphere Venus won't set until the sky is dark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This contrasts with Mercury, which is only visible in the twilight as it is much closer to the sun. The other planets can be see at midnight when they are near opposition.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-03-11T08:40:00.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25456" PostTypeId="2" ParentId="25452" CreationDate="2018-03-12T04:10:47.347" Score="4" Body="&lt;p&gt;James K has provided a short answer &quot;no&quot;. The longer answer is that because Mercury and Venus are in closer orbits to the Sun than Earth's, they will always be in roughly the same area of the sky as the Sun, whereas the other planets aren't so restricted. Since something is generally only visible if it's above the horizon, it follows that if the Sun is below the horizon, Mercury and Venus will either be close to or below the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The very furthest that Venus gets from the Sun (its &quot;elongation&quot;) is 47.8°. If Venus is at its peak elongation around the time of the summer solstice, its angle above the horizon around sunrise (or sunset, depending on which side of the Sun Venus is on at the time) will be maximised. Add to this that since Venus is very bright at this elongation, it's actually visible even when the Sun is low above the horizon; and it remains visible for about three hours after sunset (or, if it's on the Sun's other side, it rises above the horizon about three hours before sunrise). In winter, on the other hand, the ecliptic is substantially lower and Venus's elevation at sunrise/sunset is therefore much less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Sun is more than 47.8° below the horizon, then Venus must also be below the horizon. You can never see Mercury or Venus in &quot;deep dark night&quot; (e.g. around midnight), whereas Mars, Jupiter and Saturn are readily visible (and quite bright!) if they happen to be above the horizon at midnight. [NB I'm excluding the situation of summer at extreme latitudes, since that isn't &quot;deep dark night&quot;].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Elongation_(astronomy)#/media/File:Positional_astronomy.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;This diagram on Wikipedia&lt;/a&gt; provides a simple representation of elongation and angular separation, and shows the difference in how the &quot;inner&quot; and &quot;outer&quot; planets can appear in the sky.&lt;/p&gt;&#xA;" OwnerUserId="14094" LastActivityDate="2018-03-12T04:10:47.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25458" PostTypeId="1" AcceptedAnswerId="25460" CreationDate="2018-03-12T05:41:09.960" Score="1" ViewCount="866" Body="&lt;p&gt;The Moon is inclined to the ecliptic 5.15 degrees, and the equator of the Earth tilts 23.44 degrees. Since the Moon's ascending node to the ecliptic rotates, the Moon's inclination to the Earth's equator varies between 18.3 ( = 23.44 - 5.15) to 28.6 ( = 23.44 + 5.15) degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Does only our moon have this property, or are there some other moons of other planets with a similar property?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From Wikipedia it seems many moons of other planets have a fixed inclination to the equator of their planet, so our Moon seems a special case. But is our moon's inclination property the only case?&lt;/p&gt;&#xA;" OwnerUserId="22142" LastEditorUserId="14094" LastEditDate="2018-03-12T11:50:11.757" LastActivityDate="2018-03-12T12:01:28.537" Title="Moon inclination" Tags="&lt;the-moon&gt;&lt;ecliptic&gt;&lt;inclination&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25460" PostTypeId="2" ParentId="25458" CreationDate="2018-03-12T12:01:28.537" Score="1" Body="&lt;p&gt;Our moon is special. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other planets have relatively much smaller moons. The other rocky planets either have very small moons (Phobos and Deimos are probably captured asteroids) or no moons at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The giant planets have a central system of moons that orbit prograde around their equator, and a wider system of satellites that orbit in many different orbits, and are not aligned to the planet or to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our Moon is the only moon that orbits roughly in the plane of the ecliptic, and the only one which shows the particular orbital dynamics in which the inclination varies as the ascending node rotates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This suggests that the formation of the moon was special, not part of normal planetary formation. The suggestion is that the moon was formed in a giant impact.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-03-12T12:01:28.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25462" PostTypeId="1" CreationDate="2018-03-12T13:33:56.847" Score="1" ViewCount="112" Body="&lt;p&gt;Need Help. I have a problem with my 50 mm objective with 500mm focal length refractor telescope. I can see moon clearly. I have 18 mm and 9 mm eyepieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see Jupiter and Venus as a bright point when focused at 30x or 60x and I see white disk when out of focus. While I have successfully resolved double stars with it and I saw Pleiades star cluster clearly with it, I cant see the moons of Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I buy 3x Barlow to see the bigger image? is low magnification is my problem? &#xA;Or there is another solution to this problem.?&lt;/p&gt;&#xA;" OwnerUserId="21370" LastEditorUserId="2365" LastEditDate="2018-03-12T14:34:46.913" LastActivityDate="2018-03-12T14:43:16.387" Title="planets are point like" Tags="&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25463" PostTypeId="2" ParentId="25462" CreationDate="2018-03-12T14:31:59.287" Score="0" Body="&lt;p&gt;Take a look at &lt;a href=&quot;http://www.deepskywatch.com/Articles/what-can-i-see-through-telescope.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.deepskywatch.com/Articles/what-can-i-see-through-telescope.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the example picture of Jupiter, shown to simulate its size (on a typical computer screen) when viewed at a magnification of 180x. If you use 60x magnification, your view will be a third of the size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planetary discs are really small!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you should see something that isn't point-like with some banding. And you certainly should be able to see the moons (which do look like a line of stars) and are visible even in a small pair of binoculars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fix could depend on what is wrong, which could be any of a range of problems, from &quot;silly mistakes&quot; (such as pointing the telescope a bright star instead of a planet) or lens that aren't what they claim to be. Atmospheric problems (like trying to view through a window), or technical problems (if a telescope isn't properly collimated). Your best bet might be to find local experts, a local astronomy club or similar.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-03-12T14:31:59.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25464" PostTypeId="2" ParentId="25462" CreationDate="2018-03-12T14:43:16.387" Score="3" Body="&lt;p&gt;You should definitely be able to see the 4 main moons of Jupiter. You won’t need additional magnification - they can be seen through 10 x binoculars. It’s possible that when you looked the moons were close to or behind Jupiter, so worth trying again. Or it may be that James is right and you were looking at a star. &#xA;As for the bands on Jupiter, these are quite subtle but at 60x magnification, 2 dark bands should be visible. Don’t expect to see the Great Red Spot, at least initially. &lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2018-03-12T14:43:16.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25465" PostTypeId="1" AcceptedAnswerId="25469" CreationDate="2018-03-12T15:19:09.880" Score="3" ViewCount="98" Body="&lt;p&gt;Mercury's surface was not mapped up-close until &lt;a href=&quot;https://en.wikipedia.org/wiki/Mariner_10&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mariner 10&lt;/a&gt; passed by in 1974. This was followed by &lt;a href=&quot;https://en.wikipedia.org/wiki/MESSENGER&quot; rel=&quot;nofollow noreferrer&quot;&gt;MESSENGER&lt;/a&gt; in 2008. I have seen several attempts of mapping Mars that came before space exploration, such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Martian_canal&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schiaparelli's maps from 1877&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regards to Mercury, were there any documented attempts to map its surface with ground-based observations prior to unmanned space exploration? If so, however unlikely, did any observed features from Earth match up with Mariner or MESSENGER close-ups?&lt;/p&gt;&#xA;" OwnerDisplayName="user10106" LastActivityDate="2018-03-12T19:41:40.627" Title="Mercury mapping attempts before Mariner 10?" Tags="&lt;history&gt;&lt;mercury&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25466" PostTypeId="1" AcceptedAnswerId="25499" CreationDate="2018-03-12T16:12:53.220" Score="5" ViewCount="527" Body="&lt;p&gt;In 2008, the European Southern Observatory reported, upon the end of a 16-year study described here &lt;a href=&quot;http://www.eso.org/public/usa/news/eso0846/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.eso.org/public/usa/news/eso0846/&lt;/a&gt;, that the 28 stars nearest the very large black hole in Sagittarius A included 22 in varied orbits compared to a &quot;swarm of bees&quot;, and six (all slightly farther from the BH) in a disc of very fast orbits. As the stars were &quot;too young&quot; to have migrated far, and as the tidal forces in black holes had apparently been felt to leave star formation in that region unlikely, further research was planned. I'm asking whether there might have been any results, given the torsion-based cosmology which pairs black holes with white ones and is most concisely described at arXiv.org &gt; &lt;a href=&quot;https://arxiv.org/abs/1510.08834&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/1510.08834&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spin of smaller celestial bodies may affect the spin of the larger systems of which they're a part to an extremely great extent, according to a version of inflationary cosmology developed by physicist Nikodem Poplawski since 2009. The simplest statement of his cosmology is at &lt;a href=&quot;https://arxiv.org/pdf/1007.0587.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/pdf/1007.0587.pdf&lt;/a&gt;, but, in spite of its title as an &quot;alternative&quot; to cosmological inflation, it actually describes a version of that theory, as pointed out in the Wikipedia article &quot;Inflation&quot;: Poplawski's version uses an extension of Einstein's General Relativity that's usually referred to as the Einstein-Cartan-Sciama-Kibble (ECSK) theory of gravity, and was developed, partly in consultation between Einstein and the French mathematician Cartan, to incorporate particulate spin (the rotation of subatomic particles) into relativity, since GR had been formulated several years before such spin was discovered. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Partly because ECSK made an assumption that fermions (loosely describable as &quot;matter particles&quot;) have a spatial extent (tiny, but about 37 times the Planck length) that's incompatible with a renormalization technique useful in quantum mechanics (which generally requires all subatomic particles to be pointlike), and perhaps partly because its application to cosmology allows a universe that's eternal to the past and to the future rather than one that's eternal only to the future, the ECSK theory has not been widely utilized in cosmology until recently, when physicists exasperated with the failure to discover any specialized subatomic particle that would cause inflation began turning to it. As you can see on p.7 in the piece I've linked you to, Poplawski's torsion-based cosmology may be falsified if there's no evidence for any prevalent direction of rotation, whereas inflationary cosmology that's based on a field of hypothesized &quot;inflaton&quot; particles cannot be falsified at all, leading to many recent objections to it by theoreticians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd also like to know (if possible) whether the direction of the orbits of the stars concerned is the same as the direction of rotation of the super-massive BH itself, assuming that the latter was not inferred from the former: This would help me to assess the plausibility of this BH cosmology, as one of its leading proponents (Nikodem Poplawski) feels the arrow of time to derive from the direction of matter's passage across the event horizons of BHs, and has also felt the torsion-based theory to be falsifiable by the lack of a prevalent direction of rotation in the universe. (These are details to me, but might appear in whatever conversations or observations might contain the answer to my question.)&lt;/p&gt;&#xA;" OwnerUserId="22115" LastEditorUserId="22115" LastEditDate="2019-08-23T15:48:35.553" LastActivityDate="2019-08-23T15:48:35.553" Title="Has a “white hole” theory been advanced to explain anomalous star formation in Sagittarius A?" Tags="&lt;black-hole&gt;&lt;cosmology&gt;&lt;milky-way&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="25467" PostTypeId="1" AcceptedAnswerId="25468" CreationDate="2018-03-12T17:24:51.830" Score="19" ViewCount="3735" Body="&lt;p&gt;How far is the sun from the edge of the galaxy in the following directions?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Moving along the shortest path parallel to the galactic plane (outward radially)&lt;/li&gt;&#xA;&lt;li&gt;Moving along the longest path parallel to the galactic plane (radially presumably passing through the center)&lt;/li&gt;&#xA;&lt;li&gt;Moving along the shortest path perpendicular to the galactic plane (&quot;up&quot;)&lt;/li&gt;&#xA;&lt;li&gt;Moving along the longest path perpendicular to the galactic plane (&quot;down&quot;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I realize these distances may be somewhat fuzzy (edge of galaxy is probably hard to define exactly, galaxy isn't exactly an ellipse, computing distances like this are hard even if the boundaries weren't fuzzy), so ranges and uncertainties are fine. I don't expect precision like computing whether certain points on the edge might be closer than just going straight out from the center and the like.&lt;/p&gt;&#xA;" OwnerUserId="11204" LastEditorUserId="11204" LastEditDate="2018-03-13T14:40:34.000" LastActivityDate="2018-03-13T14:40:34.000" Title="How far do we have to go to leave the galaxy?" Tags="&lt;the-sun&gt;&lt;milky-way&gt;&lt;positional-astronomy&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25468" PostTypeId="2" ParentId="25467" CreationDate="2018-03-12T18:03:49.493" Score="24" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/822/how-far-is-the-earth-sun-above-below-the-galactic-plane-and-is-it-heading-towar&quot;&gt;The sun is within a few parsec (15-25 pc) of the galactic plane&lt;/a&gt;, slightly above. The thin disk of the milky way (containing ~85% of the stars and gas) has a density going roughly like $\rho_0 \exp(-|z|/300 pc)$, while the thick disk (older stars, a few percent) has a scale height of 1000 pc instead. So if we want to move up (galactic north) above 90% of the star density we would have to move ~680 parsec, and going down 720 parsec. At this point at least the view would be splendid, but there would still be a fair number of thick disk and halo stars above and below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The disk has a scale length of 2.5–4.5 kiloparsec (kpc) and the sun is about 8 kpc away from the core, so we are already outside nearly 90% of the stars (not even counting the bulge). The radius is commonly given as 15.5-27.5 kpc, so we are 7.5-19.5 kpc from the nearest edge and 15.5-27.5 kpc from the far edge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we count the dark matter halo the distances get even fuzzier. The virial radius (&lt;a href=&quot;https://en.wikipedia.org/wiki/Virial_theorem#In_astrophysics&quot; rel=&quot;noreferrer&quot;&gt;a rough measure of where most mass is inside most of the time&lt;/a&gt;) is &lt;a href=&quot;http://adsabs.harvard.edu/abs/2006MNRAS.369.1688D&quot; rel=&quot;noreferrer&quot;&gt;200 kpc&lt;/a&gt;, so we are basically 200 kpc away from that distance in most directions.&lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2018-03-12T18:03:49.493" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="25469" PostTypeId="2" ParentId="25465" CreationDate="2018-03-12T19:41:40.627" Score="1" Body="&lt;p&gt;It's very hard to see anything on the surface of Mercury. It is always seen against the glare of the sun. It is always low in the sky, so we see it through a lot of atmospheres, the seeing is worse than it would be for an object higher in the sky. With CCD imaging technology you can see some details, by taking very short exposures and combining them&lt;a href=&quot;https://www.bu.edu/news/2000/05/26/new-ground-based-photos-of-mercurys-unseen-surface-obtained-by-bu-astronomers/&quot; rel=&quot;nofollow noreferrer&quot;&gt;(source)&lt;/a&gt;. You can't do that with conventional photographic plates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lowell produced a map at the start of the century showing supposed &lt;a href=&quot;https://planetarymapping.wordpress.com/2016/02/02/lowells-map-of-mecury-1902/&quot; rel=&quot;nofollow noreferrer&quot;&gt;linear markings that he thought were the result of cooling.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best that could be done in the middle of the twentieth century, direct observation &lt;a href=&quot;https://planetarymapping.wordpress.com/category/body/mercury/&quot; rel=&quot;nofollow noreferrer&quot;&gt;images from the Pic du Midi telescope&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/k4Ld2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k4Ld2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It wasn't until radar-based observations in the 60s that the rotational period was established. There were some mapping attempts at mapping in the 70s &lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/0019103572900231&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.sciencedirect.com/science/article/pii/0019103572900231&lt;/a&gt; (behind paywall)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be hard to identify any features on these drawings with know regions or features on the Mariner images.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-03-12T19:41:40.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25470" PostTypeId="1" CreationDate="2018-03-13T04:15:52.290" Score="4" ViewCount="488" Body="&lt;p&gt;Given that we've discovered hundreds of &lt;a href=&quot;https://en.wikipedia.org/wiki/Intergalactic_star&quot; rel=&quot;noreferrer&quot;&gt;intergalactic stars&lt;/a&gt; (IGs), and that most stars &lt;a href=&quot;https://www.space.com/24894-exoplanets-habitable-zone-red-dwarfs.html&quot; rel=&quot;noreferrer&quot;&gt;have planets&lt;/a&gt;, what do we know about the likelihood of IGs retaining their planets after being ejected from their galaxy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related to that, what are the chances of &lt;a href=&quot;https://en.wikipedia.org/wiki/Rogue_planet&quot; rel=&quot;noreferrer&quot;&gt;rogue planets&lt;/a&gt; being intergalactic as well? If there's a planet floating out in intergalactic space, is it more likely to be orbiting a star, or on it's own?&lt;/p&gt;&#xA;" OwnerUserId="6870" LastActivityDate="2018-03-13T15:20:01.637" Title="Is it likely that intergalactic stars would still retain their planets?" Tags="&lt;star&gt;&lt;planet&gt;&lt;rogue-planet&gt;&lt;intergalactic-space&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25472" PostTypeId="2" ParentId="25470" CreationDate="2018-03-13T15:20:01.637" Score="6" Body="&lt;p&gt;While near encounters with supermassive black holes are bad for the stability of solar systems [citation needed?], many intergalactic stars become intergalactic without a close encounter. They are &lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_tide&quot; rel=&quot;noreferrer&quot;&gt;part of streams and tidal tails that emerge when galaxies or globular clusters get tidally disrupted by another galaxy&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, even &lt;a href=&quot;https://arxiv.org/abs/1201.1446&quot; rel=&quot;noreferrer&quot;&gt;while encounters with black holes are usually disruptive simulations suggest that some planets may remain around the star&lt;/a&gt;. This is most likely for planets real close to the star, like hot jupiters. &lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2018-03-13T15:20:01.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25473" PostTypeId="2" ParentId="18844" CreationDate="2018-03-13T18:36:55.540" Score="7" Body="&lt;p&gt;Hill sphere is named after John William Hill (1812–1879) and its simple logic follows from the presence of three bodies (let's assume Sun is the largest mass with Earth as the secondary mass and a satellite of negligible mass orbiting the Earth as the third mass), where the radius of the Hill sphere will be the largest radius at which a satellite could orbit the secondary mass (Earth in this case). If its orbit exceeds the Hills radius, then it will fall to the gravitational influence of the first body (sun) and hence will no longer be a satellite of the secondary body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could write Newton's equations using the idea that the satellite has the same angular velocity as the secondary object. This is that, the angular velocity of the Earth around the sun equals to the angular velocity of the satellite around the sun. A demonstration about the derivation is given in the following link as well as that of the Roche limit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jgiesen.de/astro/stars/roche.htm&quot; rel=&quot;noreferrer&quot;&gt;http://www.jgiesen.de/astro/stars/roche.htm&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="22174" LastActivityDate="2018-03-13T18:36:55.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25474" PostTypeId="2" ParentId="25443" CreationDate="2018-03-13T19:43:14.420" Score="4" Body="&lt;p&gt;&lt;strong&gt;Yes&lt;/strong&gt;, hot Jupiters can have stable Trojans. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A great analysis of this very problem is given by &lt;a href=&quot;https://arxiv.org/pdf/1605.02507.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kislyakova et al. (2015)&lt;/a&gt;. They suggest that super-hot outgassing Trojans may be an explanation for the ultraviolet anomalies seen in the WASP-12 and HD 189733 systems. As part of their analysis, they explore the photogravitational stability of Trojans of various size, and while they show that meter-sized bodies can be ejected on timescales of around a million years, Io-sized objects can be stable over the lifetime of the system. It may also be worth noting that although they found many stable tadpole orbits, they did not find any stable horse-shoe orbits (although they do not show that such an orbit cannot exist).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the actual formation and capture of such a Trojan, &lt;a href=&quot;https://www.aanda.org/articles/aa/pdf/2007/07/aa6582-06.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Beauge' et al. (2007)&lt;/a&gt; modeled the inward migration of a massive planet, with its semi-major axis decaying over a time scale of $10^5 - 10^7~\mathrm{yr}$. Their results suggest that a Trojan satellite with a tadpole orbit close to $L_4$ will be maintained during the migration. &lt;/p&gt;&#xA;" OwnerDisplayName="user14781" LastActivityDate="2018-03-13T19:43:14.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25475" PostTypeId="2" ParentId="24846" CreationDate="2018-03-13T23:06:08.463" Score="2" Body="&lt;p&gt;Eyepieces are carefully designed. They try to achieve a sharp image with good contrast over a wide field of view. Even the simplest eyepieces of type Huygens consist of two lenses that match up. Good eyepieces can have 7 or more lenses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end every eyepiece is a fancy magnifying glass, so your CD ROM lens will work. It will just not make for a good image. It'll be much worse than the 10/25mm Plössl eyepieces that so many sell for cheap or sometimes even give away. They sell them for cheap because even though they are much better than any single lens, they are still on the lower end of the quality spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem is: the CD ROM lens has a tiny focal length which will result in very high magnification and a very dark image. A rule of thumb says: twice your mirror diameter in mm is about the useful upper limit for magnification. If you have an 8&quot; reflector, that would be 200mm x 2 = 400x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless an interesting experiment. You can try to put a cardboard ring in front of your telescope to give it a longer focal ratio. It might help a little while costing a lot of light and resolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see 5 planets without any magnification. To see details, I'd say the fun starts at around 150-200x even though they appear as a disc at much lower magnification.&lt;/p&gt;&#xA;" OwnerUserId="14066" LastActivityDate="2018-03-13T23:06:08.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25476" PostTypeId="2" ParentId="23698" CreationDate="2018-03-13T23:23:04.543" Score="2" Body="&lt;p&gt;Unless you happen to watch a nuclear explosion or a gamma ray burst or anything like that (very, very unlikely), no ordinary object in the night sky reflects or emits enough light to damage your eyes or fry your chip. Partly because they are very far away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing in the sky that can and will do that is the sun. So if you are watching at night, you are perfectly safe. So is your camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can happen is that watching the moon or Jupiter can wreck your eye's ability to view faint or dim objects. Eyes need up to 30 minutes (give or take) to adapt to solid darkness. Brighter objects will partly or completely reset that ability so you have to wait a while until your eyes are truly sensitive again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine this: The moon is the brightest thing in the night sky. You can watch it all night long. If you use a bigger telescope, a gray filter might be nice, but it's not bright enough to heat your eyeballs. Numbers on the internet range between 1/400000 and 1/1000000 as bright as the sun.&lt;/p&gt;&#xA;" OwnerUserId="14066" LastEditorUserId="14066" LastEditDate="2018-03-14T03:08:29.127" LastActivityDate="2018-03-14T03:08:29.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25477" PostTypeId="2" ParentId="23997" CreationDate="2018-03-13T23:32:08.233" Score="1" Body="&lt;p&gt;A Barlow lens doesn't multiply the focal length of your eyepiece, it multiplies the focal length of your telescope. Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's use a telescope of 1000mm focal length. 8&quot; f/5 Newtonian, for example. And you have a 10mm and 25mm Eyepiece and a 3x barlow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the barlow turns these 1000mm into 1000 x 3 = 3000mm focal length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnification is always: &lt;strong&gt;FL telescope / FL eyepiece&lt;/strong&gt;. That means, with the barlow, you have 3000 / 10 = 300x magnification or 3000 / 25 = 120 x magnification. Without barlow it's 100x (10mm) and 40x (25mm)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An eyepiece is a magnification lense. You can put two eyepieces on top of each other. You just won't see anything useful with that. Lens systems are carefully designed to result in an eyepiece that works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eyepieces with a bigger focal length give smaller magnification.&lt;/p&gt;&#xA;" OwnerUserId="14066" LastActivityDate="2018-03-13T23:32:08.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25478" PostTypeId="1" AcceptedAnswerId="25484" CreationDate="2018-03-14T03:11:48.187" Score="3" ViewCount="225" Body="&lt;p&gt;In response to today being &lt;a href=&quot;https://en.wikipedia.org/wiki/Pi_Day&quot; rel=&quot;nofollow noreferrer&quot;&gt;pi day&lt;/a&gt; (also see &lt;a href=&quot;http://www.piday.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.piday.org/&lt;/a&gt;), NASA has published it's fifth annual &lt;a href=&quot;https://www.jpl.nasa.gov/edu/teach/activity/pi-in-the-sky-5/&quot; rel=&quot;nofollow noreferrer&quot;&gt;pi in the sky day 5&lt;/a&gt; activity set (see also &lt;a href=&quot;https://www.jpl.nasa.gov/edu/news/2018/3/9/pi-goes-the-distance-at-nasa/&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA goes the distance&lt;/a&gt; and &lt;a href=&quot;https://www.jpl.nasa.gov/edu/events/2018/3/9/celebrate-pi-day-with-nasa/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Celebrate Pi Day with NASA&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This year's NASA activity includes the problem set &lt;a href=&quot;https://www.jpl.nasa.gov/edu/teach/activity/helium-heist-a-pi-in-the-sky-math-challenge/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Helium Heist: A 'Pi in the Sky' Math Challenge&lt;/a&gt;. Below is a cropped screen shot of the illustration from this &lt;a href=&quot;https://www.jpl.nasa.gov/edu/pdfs/piday2018_jupiter_handout.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;PDF&lt;/a&gt; version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part of the text that I found most striking is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It has been hypothesized that the helium was depleted out of the upper atmosphere and transported deeper inside the planet. The extreme pressure inside Jupiter condenses helium into droplets that form inside a liquid metallic hydrogen layer below. Because the helium is denser than the surrounding hydrogen, the helium droplets fall like rain through the liquid metallic hydrogen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Roughly what are the pressure and temperature thought to be at this depth in Jupiter? Is this regime at all accessible experimentally on Earth to confirm that these two fluids are immiscible, or is this behavior purely theoretical?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; cropped screen shot from &lt;a href=&quot;https://www.jpl.nasa.gov/edu/pdfs/piday2018_jupiter_handout.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; (NASA). Click for full size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dzoik.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dzoikm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-03-14T03:17:26.963" LastActivityDate="2018-03-14T10:28:36.900" Title="Helium rain inside metallic hydrogen within Jupiter, conditions and experiments?" Tags="&lt;jupiter&gt;&lt;hydrogen&gt;&lt;helium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25479" PostTypeId="1" AcceptedAnswerId="25482" CreationDate="2018-03-14T04:01:03.630" Score="4" ViewCount="358" Body="&lt;p&gt;Why only type -Ia supernovae are used as standard candle to calculate age of universe and why not type II, type Ib, type 1c ?&lt;/p&gt;&#xA;" OwnerUserId="20162" LastActivityDate="2018-03-14T06:25:02.610" Title="Use of type-I a supernovae as standard candle" Tags="&lt;universe&gt;&lt;age&gt;&lt;hubble-constant&gt;&lt;standards&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25482" PostTypeId="2" ParentId="25479" CreationDate="2018-03-14T06:24:33.247" Score="5" Body="&lt;p&gt;A type 1a supernova forms when a white dwarf grows through accretion to a certain size, at which it becomes unstable. This means that the precursor object is always a white dwarf of mass 1.39 solar masses. As the precursor object is always of the same type and the same size, the supernova is thought to be the same brightness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, type II supernovae form from the collapse of giant stars. The precursor object can have a range of masses from about 10 Solar masses, to a maximum of over 100. The brightness of the supernova depends on the mass of the precursor (and other factors such as the metallicity) and so type II supernovae are not all equally bright.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other types also have precursors which vary in size or other factors, only type 1a supernovae are thought to always have a white dwarf of mass 1.39 as a standard precursor, so only type 1a are suitable as a standard candle.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-03-14T06:24:33.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25484" PostTypeId="2" ParentId="25478" CreationDate="2018-03-14T10:28:36.900" Score="5" Body="&lt;p&gt;Different sources give significantly different values for the conditions under which hydrogen shifts from molecular to metallic. See the discussion in the comments associated with &lt;a href=&quot;https://astronomy.stackexchange.com/questions/25444/what-would-the-pressure-and-temperature-of-gas-be-right-above-jupiters-gas-liq/25451?noredirect=1#comment43745_25451&quot;&gt;this question&lt;/a&gt;. Pressures from about 20 to 300 GPa and temperatures very broadly around 10 000 K seem to be the most common.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experiments can only just reach these pressures, at lower temperatures, for tiny volumes of material, and the question of whether anyone has made metallic hydrogen at all remains controversial. Certainly no one can make it repeatably. (&lt;a href=&quot;https://en.wikipedia.org/wiki/Metallic_hydrogen&quot; rel=&quot;noreferrer&quot;&gt;wikipedia&lt;/a&gt; has a reasonable summary of the history) or make enough of it and keep it for long enough to check to what extent, if any, helium is soluble in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, analysing from theory whether it will be soluble shouldn't be that hard, since both hydrogen and helium are relatively simple atoms. &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0305-4608/9/5/007/meta&quot; rel=&quot;noreferrer&quot;&gt;An early paper&lt;/a&gt; concludes that the hydrogen would be at or close to the limit of how much helium could be dissolved. I'm sure there is more accurate recent work, but I couldn't find it quickly.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-03-14T10:28:36.900" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25485" PostTypeId="2" ParentId="23698" CreationDate="2018-03-14T11:18:40.433" Score="1" Body="&lt;p&gt;While I agree with all the other answers here that no satellite flare has any chance of being bright enough to damage eyes or instruments there are concerns about them wrecking observations. &lt;a href=&quot;https://en.wikipedia.org/wiki/Satellite_flare&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; reports that Iridium flares can reach magnitude -8 or brighter (Venus can be -4, the full moon is -12) and that could certainly be bright enough that scattered light from it would drown out some dim star or galaxy that you were trying to observe.  If the flare can be predicted then you can always shut off the detectors for a short period, but some flares come from out of control satellites that are tumbling unpredictably. &lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-03-14T11:18:40.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25486" PostTypeId="2" ParentId="20340" CreationDate="2018-03-14T15:12:18.323" Score="2" Body="&lt;p&gt;A black hole could be any size. It could be the size of a planetary system.&#xA;Given a specific distribution of matter, I think you don't necessarily need a singularity to be present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But given the fact that matter inside the black hole cannot escape it, it will eventually all move closer and closer and end up creating that singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if we imagine a rigid body black hole with structures to keep it from imploding, internal entropy would eventually make those structures collapse and you'd still get a singularity pretty fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's all based on my understanding on black holes, I am not a scientist. I'd love if someone could tell me if I understand correctly.&lt;/p&gt;&#xA;" OwnerUserId="22191" LastActivityDate="2018-03-14T15:12:18.323" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25487" PostTypeId="1" CreationDate="2018-03-14T15:19:08.630" Score="2" ViewCount="495" Body="&lt;p&gt;I'm working with some data and I have some values named &lt;code&gt;fluxmag0&lt;/code&gt; and &lt;code&gt;fluxmag0err&lt;/code&gt; that should correspond to the zero point values and error in flux units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The values are around $10^{13}$ for &lt;code&gt;fluxmag0&lt;/code&gt; and $10^9$ for &lt;code&gt;fluxmag0err&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to have the zero point error in magnitude units. If I take $2.5\log_{10}(\mathrm{fluxmag0err})$ the values are huge so I suppose it's not the way to compute this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I convert these values in terms of magnitudes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot in advance for your help &lt;/p&gt;&#xA;" OwnerUserId="22192" LastEditorUserId="2365" LastEditDate="2018-03-14T15:43:48.113" LastActivityDate="2018-03-14T15:57:56.610" Title="How to get the zeropoint magnitude?" Tags="&lt;apparent-magnitude&gt;&lt;magnitude&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25488" PostTypeId="2" ParentId="25487" CreationDate="2018-03-14T15:57:56.610" Score="2" Body="&lt;p&gt;Magnitude is defined relative to a standard flux of a magnitude 0 object $F_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M = -2.5\log(F/F_0)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if the error is $F_E$, then $F =F_0\pm F_E$, so the magnitude M is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ -2.5\log((F_0+F_E)/F_0) &amp;lt; M &amp;lt;-2.5\log((F_0-F_E)/F_0)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The errors here are tiny in comparison to the measured flux, only one part in 10000, so the range of magnitudes is similarly tiny +-0.0001&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-03-14T15:57:56.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25489" PostTypeId="1" AcceptedAnswerId="25490" CreationDate="2018-03-14T20:25:12.927" Score="1" ViewCount="66" Body="&lt;p&gt;There is a new moon on Saturday, and the forecast in my area of Winnipeg, Manitoba, Canada calls for clear skies on that night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to take a widefield astrophotograph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea what areas of the milky way will be visible, and where they will be located.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some steps I could take to determine this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there different dense (photogenic) portions of the milkyway visible at different times of the year?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there certain constellations associated with the dense portions of the milky way that I should be looking for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to further develop my knowledge of astronomy, for the purpose of widefield milkyway shots.&lt;/p&gt;&#xA;" OwnerUserId="17944" LastEditorDisplayName="user10106" LastEditDate="2018-03-15T12:00:16.670" LastActivityDate="2018-03-15T12:00:16.670" Title="How could I determine what areas of the Milky Way will be visible?" Tags="&lt;milky-way&gt;&lt;photography&gt;&lt;positional-astronomy&gt;&lt;constellations&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25490" PostTypeId="2" ParentId="25489" CreationDate="2018-03-14T21:05:04.587" Score="3" Body="&lt;p&gt;Have you tried to use Stellarium? &#xA;You can insert the date of the observation and see how the sky will look like from your location.&#xA; There is also setting to make the Milky Way be seen more intense.&lt;/p&gt;&#xA;" OwnerUserId="5355" LastActivityDate="2018-03-14T21:05:04.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25491" PostTypeId="1" AcceptedAnswerId="25494" CreationDate="2018-03-14T22:48:50.600" Score="5" ViewCount="452" Body="&lt;p&gt;I'm having a debate about how bright Martian nights would be, in terms of both the sky itself and ground illumination. &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10220/mars-night-sky-how-many-stars-visible&quot;&gt;This&lt;/a&gt; answer seems to indicate that the stars would be only dimly visible, and wouldn't provide enough light to see by. However, the question remains of how much illumination the moons would provide. I've seen it said that Deimos would appear about as bright as Venus does from Earth, and that Phobos would be about twenty times brighter, but it's not clear to me how that compares to e.g. our own moon's illumination. So how bright would it look? Would it be enough to see by?&lt;/p&gt;&#xA;" OwnerUserId="22197" LastActivityDate="2018-03-16T04:13:32.523" Title="Mars night sky -- Phobos and Deimos?" Tags="&lt;mars&gt;&lt;night-sky&gt;&lt;naked-eye&gt;&lt;phobos&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25492" PostTypeId="1" CreationDate="2018-03-14T23:16:49.283" Score="1" ViewCount="19" Body="&lt;p&gt;When information is encoded in a boundary of a bulk, following the holographic principle, it would raise a universe in the boundary with physics determined by the information encoded on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Can this information (once encoded in the boundary) be modified or changed? Can it be modified without changing nothing in the bulk?&lt;/p&gt;&#xA;" OwnerUserId="22190" LastActivityDate="2018-03-14T23:16:49.283" Title="How can the information encoded in a boundary be changed in holographic principle?" Tags="&lt;holographic-principle&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25494" PostTypeId="2" ParentId="25491" CreationDate="2018-03-15T04:49:24.023" Score="6" Body="&lt;p&gt;Taking Mars' average distance from the sun of 1.52 AU, the sun would be 43% as bright.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phobos has an &lt;a href=&quot;https://en.wikipedia.org/wiki/Phobos_(moon)&quot; rel=&quot;nofollow noreferrer&quot;&gt;albedo of about 0.071&lt;/a&gt; which is pretty dark.  Darker than the Moon with an average albedo of about 0.12, so it reflects about 59% as well as the Moon does.   That gives it a brightness to area of just over 25% the brightness of the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phobos is an odd shape, about 27x22x18 km.   Averaging that out to 22, then Phobos is 1/158 the diameter of the Moon, but it's closer to Mars' surface.  about 5,986 km, that's about 63.5 times closer than the Moon is to the surface of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working out the numbers, the size of phobos in the Martian sky would be on average 63.5/158 or 40% the diameter, or 16% the size of Earth's Moon from Earth.    Multiply that by 25% the brightness to area and that's just 4% the brightness of a full moon.  That would be peak brightness  (and if you want to fuss about Mars' Perihelion and say 5% the brightness of a full moon, fair enough, but lets not get ahead of ourselves).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brightness of our Moon varies depending how good a reflector it's surface is, and how much of the surface is in shadow.  The Moon is particularly cratered so &lt;a href=&quot;http://www.chicagotribune.com/news/weather/ct-wea-0404-asktom-20150403-column.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;a half moon is significantly less than half as bright as a full moon&lt;/a&gt;.  Phobos is so close to Mars that from the surface of Phobos, Mars takes up about 40 degrees of arc in the sky.  That casts a huge shadow, so there's no such thing, from Mars, as a full Phobos.  As Phobos approaches full, it enters Mars' shadow and it's brightness drops significantly  (whether it turns red like a blood moon or just vanishes - I'm not sure, perhaps that's best for another question).   But Phobos never becomes full, it gets a bit over 3/4s, then it enters Mars' shadow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Enough to see&quot; is a tough question, because sight depends on not just the light source but the reflectivity of your surroundings and there's some variation in how well individuals can see in the dark and, obviously, on how clear the night sky is on Mars on that particular night.  &lt;em&gt;I think 3%-4% of the brightness of a 3/4 moon would be pushing the limit though and probably not bright enough to see for the most people&lt;/em&gt;.    &lt;strong&gt;BUT&lt;/strong&gt; reading the comments, and reading the very cool Venus shadow article in the questions comments, my assumption might be wrong.   Limited sight might/(should?) be possible with some recognition of shadows and shapes.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-03-16T04:13:32.523" LastActivityDate="2018-03-16T04:13:32.523" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25497" PostTypeId="2" ParentId="20340" CreationDate="2018-03-15T09:17:37.443" Score="3" Body="&lt;p&gt;The answer specifically about the question of a neutron star disappearing inside an event horizon but remaining in some sort of equilibrium is no. At least, it is no according to General Relativity, which is the only respectable game in town at present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is no for two reasons. Firstly, in GR, the pressure that supports a star is also a source of gravity (or space-time curvature). The increasing pressure required to support a neutron star of increasing $M/R$ ultimately becomes self-defeating, irrespective of what mechanism or particle provides that pressure. This limit is reached at about 1.2 to 1.4 Schwarzschild radii (depending on how the pressure and density of the material are related), and stable objects cannot be smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the mathematics of GR clearly shows that inside the event horizon an object cannot be stationary and that its radial coordinate must decrease and a singularity (or a breakdown in GR as $r \rightarrow 0$ if you prefer) will be formed on a timescale of $\sim r_s/c$, where $r_s$ is the Schwarzschild radius. This is as inevitable as the increase of time is outside the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The details might be slightly different for a spinning (Kerr) BH. The formation of a singularity is still expected, but an isolated Kerr BH may form a &lt;a href=&quot;https://en.wikipedia.org/wiki/Ring_singularity&quot; rel=&quot;nofollow noreferrer&quot;&gt;ring-like singularity&lt;/a&gt;. This does not alter the impossibility of having a stable/static object inside the event horizon (in GR) and a &quot;singularity&quot; is expected to form.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-03-19T15:57:11.230" LastActivityDate="2018-03-19T15:57:11.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25499" PostTypeId="2" ParentId="25466" CreationDate="2018-03-15T11:10:12.430" Score="2" Body="&lt;p&gt;I'm going to say that the answer to the main question is &quot;no&quot;, for two reasons.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The Introduction to &lt;a href=&quot;http://adsabs.harvard.edu/abs/2018ApJ...853L..24N&quot; rel=&quot;nofollow noreferrer&quot;&gt;this recent paper by Naoz et al.&lt;/a&gt; mentions a variety of different suggested models and explanations, none of which is based on anything like alternate cosmological models. (The paper itself argues that the apparently problematic distribution of stellar orbits may be an artifact of failing to account for the presence of binary stars, so that the system isn't really anomalous at all.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It's not at all clear to me how an alternate cosmological model like the one you link to would be even remotely relevant to the details of star formation in the Galactic Center. In other words, I wouldn't expect anyone to be trying to explain star formation near Sgr A* using alternate cosmologies or extensions to GR (the star formation took place mostly well outside the region where GR becomes noticeably different from Newtonian gravity, so modifications to GR would be even less relevant).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As for the second question: we simply don't know the direction or amount of the central supermassive black hole's spin (yet).&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2018-03-15T11:10:12.430" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25500" PostTypeId="1" AcceptedAnswerId="25502" CreationDate="2018-03-15T12:14:24.233" Score="2" ViewCount="998" Body="&lt;p&gt;How does the &lt;a href=&quot;https://en.wikipedia.org/wiki/International_Terrestrial_Reference_System&quot; rel=&quot;nofollow noreferrer&quot;&gt;International Terrestrial Reference frame&lt;/a&gt; related to the &lt;a href=&quot;https://en.wikipedia.org/wiki/ECEF&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth-centered, Earth-fixed frame&lt;/a&gt;, as both of them rotate with the Earth?&lt;/p&gt;&#xA;" OwnerUserId="22207" LastEditorDisplayName="user1569" LastEditDate="2018-03-15T12:45:30.467" LastActivityDate="2018-03-15T16:15:18.423" Title="How does the ITRS frame related to ECEF frame?" Tags="&lt;frame-of-reference&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25501" PostTypeId="1" CreationDate="2018-03-15T12:41:01.833" Score="2" ViewCount="742" Body="&lt;p&gt;As I understand virtual particles pairs, e.g. electron and positron spontaneously pop into existence for tiny period of time, then they annihilate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When virtual electron and virtual positron annihilate we should see two photons emitted to opposite directions, right? If so, shouldn't we detect photons appearing from vacuum for every such pair?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, when a pair of virtual particles pops near event horizon of a black hole one particle may be captured by the black hole, while the other escapes it. Both particles have mass and energy. Thus caught particle brings the black hole its mass. Then how the black hole &lt;em&gt;loses&lt;/em&gt; mass through Hawking radiation? Shouldn't it gain mass instead?&lt;/p&gt;&#xA;" OwnerUserId="22203" LastEditorUserId="7472" LastEditDate="2018-03-20T20:03:24.793" LastActivityDate="2018-03-20T20:03:24.793" Title="Does Hawking radiation in fact bring mass into the universe?" Tags="&lt;black-hole&gt;&lt;hawking-radiation&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25502" PostTypeId="2" ParentId="25500" CreationDate="2018-03-15T15:45:07.157" Score="4" Body="&lt;p&gt;The simple explanation:&#xA;The International Terrestrial Reference (ITR) Frame &lt;em&gt;is&lt;/em&gt; an Earth-centered, Earth-fixed (ECEF) frame, but with the axes of the ITR frame precisely specified taking into consideration the movements of the Earth's crust due to plate tectonics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://confluence.qps.nl/qinsy/en/international-terrestrial-reference-frame-2014-itrf2014-29856813.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This website&lt;/a&gt; may be useful.&lt;/p&gt;&#xA;" OwnerUserId="5537" LastEditorUserId="5537" LastEditDate="2018-03-15T16:15:18.423" LastActivityDate="2018-03-15T16:15:18.423" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25503" PostTypeId="1" AcceptedAnswerId="25505" CreationDate="2018-03-15T16:38:10.963" Score="30" ViewCount="13453" Body="&lt;p&gt;Early this morning going out on the balcony, I looked up on a star chart app to verify it was Jupiter I was seeing.  Then I noticed the alignment of Mars, Saturn, and Pluto on the app.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Never being able to identify it before, I stared at where Pluto should be and I'm pretty sure I saw it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My only question is - since it's said that planets shine and stars twinkle, it did seem that Pluto was flicking a bit.  Is this normal?  Something to do with the relatively low luminosity and greater length of space?&lt;/p&gt;&#xA;" OwnerUserId="22212" LastActivityDate="2018-03-19T05:10:28.523" Title="In my mid-forties, I believe I have seen Pluto for the first time with the naked eye" Tags="&lt;pluto&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25504" PostTypeId="1" AcceptedAnswerId="25512" CreationDate="2018-03-15T16:42:34.193" Score="1" ViewCount="1463" Body="&lt;p&gt;How to perform the conversion of (r,v) vector from ECEF to ECI (J2000) coordinate frame?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What I tried&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I multiplied the matrix ([cosA -sinA 0] [sinA cosA 0] [0 0 1]) by the state vector, where A is Greenwitch mean sidereal time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation shows that the Z axis doesn't change. However, I have the ephemeris of a real satellite in both coordinate frames, and there Z axis differs! &#xA;Where is the issue?&lt;/p&gt;&#xA;" OwnerUserId="22207" LastEditorUserId="22207" LastEditDate="2018-03-16T13:53:13.003" LastActivityDate="2018-03-16T13:53:13.003" Title="ECEF to ECI conversion" Tags="&lt;coordinate&gt;&lt;frame-of-reference&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25505" PostTypeId="2" ParentId="25503" CreationDate="2018-03-15T18:06:47.107" Score="99" Body="&lt;p&gt;Pluto is something like magnitude 14. The limit for the human vision is somewhere between magnitude 6 (widely accepted) and 8-ish (highly trained observers with perfect vision in ideal conditions using special techniques - and it's a bit controversial anyway).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's zero chance that was Pluto. It was definitely a fixed star.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2018-03-15T18:06:47.107" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25506" PostTypeId="2" ParentId="25501" CreationDate="2018-03-15T18:08:27.533" Score="2" Body="&lt;p&gt;The black hole loses mass in order to generate the Hawking radiation. The energy carried by the radiation is exactly proportional to the mass lost by the black hole according to Einstein's formula E = mc^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No new mass is brought into the Universe this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Keep in mind the explanation below is not rigorous. As Stephen Hawking himself has said, this is more like a mental model, good enough to visualize things.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Virtual particles are called virtual because they don't really &quot;exist&quot; in the common sense. They have mass and they have energy. Due to Heisenberg's uncertainty principle, it's okay if particles with a total energy E almost &quot;exist&quot; for a very short while, as long as they return to nothingness afterwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The production / destruction of virtual particles could happen all the time, anywhere, not just at the edge of a black hole. This is fine, as long as each particle only seems to &quot;exist&quot; a short time. Conservation of energy is not violated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, at the edge of a black hole, something different happens. You have a pair of virtual particles popping up into existence. Soon after, one of the particles is captured by the black hole, while the other particle moves away from it. Since the captured particle is virtual, it doesn't really have mass and/or energy. No new mass is added to the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the two particles cannot re-annihilate each other. So you have one lone particle moving away from the black hole. It's not a virtual particle anymore, it's a real particle. But it has energy and perhaps it has mass also. Where does the energy come from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy required to make the new particle real comes from the gravitational field of the black hole. The black hole is losing mass in order to power the creation of real particles moving away from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, don't get stuck in details. This is not what really happens. This is more like a metaphor.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2018-03-15T19:10:34.563" LastActivityDate="2018-03-15T19:10:34.563" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25507" PostTypeId="1" CreationDate="2018-03-15T18:17:17.473" Score="18" ViewCount="639" Body="&lt;p&gt;In a recent &lt;a href=&quot;https://arxiv.org/pdf/1803.04716.pdf&quot; rel=&quot;noreferrer&quot;&gt;paper&lt;/a&gt; (Cosmic clocks: A Tight Radius - Velocity Relationship for&#xA;HI-Selected Galaxies by Meurer, et al.), it was noted in the conclusion that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[This] implies a constant orbital time of ∼1 Gyr at [the outermost] radius [of disc galaxies].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given what we knew about disc galaxies and dark matter up to this point, is this an unexpected conclusion? My sense is that it is strange that all disc galaxies have a constant rotation regardless of their size, but my intuition may be wrong. I know that dark matter speeds up the rotation of the outer parts of disc galaxies, but isn't it strange that they converge on this value?&lt;/p&gt;&#xA;" OwnerUserId="6" LastActivityDate="2019-12-08T05:07:57.390" Title="Is it &quot;weird&quot; that all disc galaxies rotate once every billion years?" Tags="&lt;galaxy&gt;&lt;astrophysics&gt;&lt;dark-matter&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25508" PostTypeId="1" AcceptedAnswerId="25511" CreationDate="2018-03-15T21:32:19.327" Score="2" ViewCount="119" Body="&lt;p&gt;I am looking for a official term that describes all stars that are not neutron stars or white dwarfs.  &quot;Still burning stuff stars&quot; or &quot;ordinary stars&quot; are all I have been able to come up with.  Is there an official term for stars that are still evolving and haven't &quot;died&quot; yet?&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2018-03-16T14:57:29.937" Title="Is there a term to describe all stars that are not neutron stars or white dwarfs?" Tags="&lt;star&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25509" PostTypeId="2" ParentId="25508" CreationDate="2018-03-15T21:41:58.597" Score="6" Body="&lt;p&gt;Every so often someone will note that a &quot;star&quot; is a ball of plasma that is supported by nuclear reactions at its core, and so &lt;em&gt;&lt;a href=&quot;https://xkcd.com/1475/&quot; rel=&quot;noreferrer&quot;&gt;technically&lt;/a&gt;&lt;/em&gt; neutron stars and white dwarfs are stellar remnants, and &lt;em&gt;not stars&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The technical term is therefore &quot;star&quot;. And if you think that could be ambiguous then just spell it out: &quot;Stars with supported by fusion reactions&quot;/&quot;Stars, excluding stellar remnants&quot;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-03-15T21:41:58.597" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25511" PostTypeId="2" ParentId="25508" CreationDate="2018-03-16T00:00:06.677" Score="5" Body="&lt;p&gt;It is tempting to think that non-degenerate stars covers it, since even the structure of the lowest mass main sequence &lt;em&gt;stars&lt;/em&gt; are not significantly affected by fermion degeneracy. Unfortunately the cores of more advanced evolutionary stages can become degenerate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would go with nuclear-burning stars, although this in turn excludes pre main sequence stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some ways it is easier to say a star that is not a compact stellar remnant.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-03-16T00:00:06.677" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25512" PostTypeId="2" ParentId="25504" CreationDate="2018-03-16T02:27:56.230" Score="1" Body="&lt;p&gt;A reference frame that has its Z axis fixed with respect to the Earth's rotation axis is not inertial. The Earth's rotation axis undergoes precession and nutation, and on a shorter term, undergoes polar motion as well.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2018-03-16T02:27:56.230" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25513" PostTypeId="2" ParentId="25503" CreationDate="2018-03-16T03:29:55.770" Score="16" Body="&lt;p&gt;As Florin correctly stated, it can't have been Pluto. You have probably looked at it and you have even gotten its light in your eyes. That little itty bitty shine just has no chance to make your retina do anything (edit: Interesting link in the comments. Might be that people actually can sense single photons. Doesn't help at all to see Pluto though). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://stellarium.org&quot; rel=&quot;noreferrer&quot;&gt;Stellarium&lt;/a&gt; is a nice tool in order to check which star you might have confused with Pluto. This is what it has to offer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UmpQ5.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UmpQ5.jpg&quot; alt=&quot;Stellariun screenshot showing Saturn, Mars and Pluto&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red crosshair is where Pluto should be and Stellarium doesn't bother to color a single pixel because Pluto is about 14.28 mag. 5 mag difference mean 100 times dimmer, so Pluto is at least 10000 times less bright than many of the stars that you see in this image, let alone the two planets (Mars is about 26 times closer than Pluto atm.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that you have seen &lt;a href=&quot;https://en.wikipedia.org/wiki/Pi_Sagittarii&quot; rel=&quot;noreferrer&quot;&gt;Pi Sagittarii&lt;/a&gt; (HIP 94141) which would be 2.85 mag and lines up nicely with the planets. Unless I'm wrong, it's roughly 37000 times brighter than Pluto&lt;/p&gt;&#xA;" OwnerUserId="14066" LastEditorUserId="14066" LastEditDate="2018-03-19T05:10:28.523" LastActivityDate="2018-03-19T05:10:28.523" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25516" PostTypeId="1" CreationDate="2018-03-16T06:33:18.547" Score="4" ViewCount="144" Body="&lt;p&gt;Today I read about the &lt;a href=&quot;http://memory-alpha.wikia.com/wiki/Romulan_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Romulan home star system&lt;/a&gt; and it looks like it might be impossible for such a star system to exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Memory Alpha describes Remus:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Remus was tidally locked, with one hemisphere always facing its primary while the other always faced away from it. The civilization of the Remans was located on the night side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Remus is farther from their star than Romulus. Romulus rotates many times during a single orbit of their star while Remus is tidally locked and rotates once per orbit of their star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a planet orbits even deeper in the gravity well of their star than Remus it should be even more likely to be tidally locked than plants further out.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could tidal interactions with neighboring planets and/or its moons prevent a planet from becoming tidally locked while being unable to prevent planets further out from being tidally locked? &lt;/p&gt;&#xA;" OwnerUserId="22226" LastEditorDisplayName="user1569" LastEditDate="2018-03-18T14:29:34.687" LastActivityDate="2018-03-18T14:29:34.687" Title="Can a solar system exist where the second planet rotates fast, and the third planet is tidally locked to their star?" Tags="&lt;orbital-mechanics&gt;&lt;tidal-locking&gt;&lt;orbital-resonance&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25519" PostTypeId="2" ParentId="25516" CreationDate="2018-03-16T13:03:30.287" Score="1" Body="&lt;p&gt;While it's true that planets closer to their star, being under the influence of a much stronger gravitational field than outer planets, are more likely to become tidally locked, that's not a hard and fast law.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no a priori reason that a more distant planet couldn't have been formed, e.g., as the collision of two solid but highly dissimilar (in density) objects.  The gross asymmetry in circumferential density would enhance tidal lock.  But I fear this is veering into WorldBuilding.SE more than astronomy.&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-03-16T13:03:30.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25520" PostTypeId="2" ParentId="25516" CreationDate="2018-03-16T13:34:39.837" Score="5" Body="&lt;p&gt;The tidal locking timescale depends on several factors: $$\tau_{lock} \approx \frac{0.4 \omega_0 a^5 m Q}{3 G M^2 k_2 r^3}$$ such as the initial spin rate $\omega_0$, the semimajor axis $a$, the mass $m$, the solar mass $M$, the radius $r$ and various dissipation parameters $Q$ and $k_2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two planets that merely differ in $a$ will have the inner one lock much earlier than the outer one. But give them slightly different $\omega_0$, $m$ and especially $r$ and the inner might dissipate more slowly than the outer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the possibility of spin-orbit resonances. Mercury is in  3:2 resonance, and tidal forces might lock planets into 2:1 or 5:2 resonances. That way the inner planet may be in a resonant lock while the outer is tidally locked 1:1. In fact, Venus is turning very slowly (and retrogradely) so it is not too far off, although the causes of the retrograde spin are unclear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, yes, you can have tidally locked planet orbiting outside non-locked planets.  &lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2018-03-16T13:34:39.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25521" PostTypeId="2" ParentId="25418" CreationDate="2018-03-16T14:27:16.023" Score="2" Body="&lt;p&gt;If I go from the &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_nearest_black_holes&quot; rel=&quot;nofollow noreferrer&quot;&gt;list of black holes&lt;/a&gt; to the page of a particular object like &lt;a href=&quot;https://en.wikipedia.org/wiki/A0620-00&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; there is a link to SIMBAD database like &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=1A%200620-00&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;. This SIMBAD database has galactic coordinates in this format&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gal coord. (ep=J2000) : 209.9563 -06.5399 [ 100 12 0 ]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where 209.9563 is a Galactic longitude, which changes from 0 to 360 degrees. -06.5399 is Galactic latitude, which is always around zero (+/- 10 degrees), because our galaxy is almost flat. Good explanation for galactic coordinates is on &lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;this wikipedia page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put the object on 2D map like &lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_coordinate_system#/media/File:Artist%27s_impression_of_the_Milky_Way_(updated_-_annotated).jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; only Galactic longitude and distance to the object is needed. For example, this is the map of all black holes which I could find in wikipedia in our Galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XCzpC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XCzpC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15004" LastActivityDate="2018-03-16T14:27:16.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25522" PostTypeId="1" AcceptedAnswerId="25525" CreationDate="2018-03-16T14:36:33.167" Score="2" ViewCount="103" Body="&lt;p&gt;I was wondering if the &quot;Esperance&quot; rock that was found on Mars by the Opportunity rover was formed due erosion or weathering? I have been reading a lot online but I did not seem to find anything. Does anyone know a type of rock on Mars that was not formed by erosion?&lt;/p&gt;&#xA;" OwnerUserId="22232" LastEditorUserId="22232" LastEditDate="2018-03-16T15:12:35.350" LastActivityDate="2018-03-16T22:31:34.250" Title="&quot;Esperance&quot; rock on Mars formed by erosion or weathering? or is there any rock on Mars not formed by erosion?" Tags="&lt;mars&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25523" PostTypeId="2" ParentId="25508" CreationDate="2018-03-16T14:57:29.937" Score="1" Body="&lt;p&gt;My advice would be to not seek a definition that is &quot;official,&quot; because official definitions will often not serve the application you have in mind.  Just say what you want to say, tailored to that application.  We'd like definitions to be instructions for sorting everything that is from everything that isn't, but in practice they actually can't work that way.  For example, what is the official definition of a &quot;computer&quot;, that works in any context?  At one level, the definition of a star should include whatever objects people who regard themselves as &quot;stellar astronomers&quot; study, because it is the commonality among all such stars that unites them.  But most &quot;official&quot; definitions of a &quot;star&quot; already exclude many of the objects that &quot;stellar astronomers&quot; study!  (The same holds for &quot;planets.&quot;)  So although official definitions do exist, they are often not what you want in your application.  That's why I say, just define what you mean yourself, and don't even try to be &quot;official.&quot;  For example, if your interest is in stellar evolution, and you are not interested in endpoints because they are not still evolving, then say stars that are still evolving.  If your interest is stars that obey an ideal gas law, you might favor a term like &quot;ordinarily gaseous stars,&quot; or some such thing.  Tailoring to the interest gives a better definition than something &quot;official.&quot;&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-03-16T14:57:29.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25524" PostTypeId="1" CreationDate="2018-03-16T16:36:40.470" Score="2" ViewCount="260" Body="&lt;p&gt;I have read that Euclidean space is considered as flat space and Minkowski space is flat space-time. So when we say our observable universe is flat are we saying that the space is flat or space-time is flat? I know the Planck data has proved the space to be flat but is it only space or space-time? Since in general relativity matter curves space-time and not just space.&lt;/p&gt;&#xA;" OwnerUserId="9277" LastActivityDate="2018-05-18T16:11:26.223" Title="Is the shape of observable universe and shape of space same?" Tags="&lt;space-time&gt;&lt;space&gt;&lt;observable-universe&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25525" PostTypeId="2" ParentId="25522" CreationDate="2018-03-16T22:31:34.250" Score="1" Body="&lt;p&gt;Rocks on Mars, just like rocks on Earth, can be igneous (formed from the cooling magma or lava) Sedimentary (formed from sediments that were laid down by water or wind) or Metamorphic (sedimentary rocks that have been changed by exposure to heat and pressure) (there may be other categories, such as meteorites)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After they have formed, rocks are subject to erosion and weathering. These processes don't form the rock, but can shape it or move it. The process of weathering produces sand, silt and clay, that can later form new rocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Esperance rock has a significant amount of &quot;clay&quot; (a sedimentary rock). The clay in Esperance was formed by the weathering of other rocks to very fine particles, these then gathered together in wet environment to form a clay. When the water evaporated, the clay hardened into a rock. The significance of Esperance is that the water under which it formed must have been chemically neutral, not strongly acidic.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-03-16T22:31:34.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25526" PostTypeId="2" ParentId="25260" CreationDate="2018-03-16T22:56:51.437" Score="8" Body="&lt;p&gt;There are two aspects to this question: first, creating a field that can exert a force and second, having a body that is coupled to that field and &quot;feels&quot; the acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Electric field:&lt;/strong&gt; You probably want a uniform field of acceleration, like on earth, where the same amount of gravity is present everywhere in our daily life (close to the surface of earth).  This can be achieved by putting sources into floor and ceiling of your ... space station.  With various setups this could create a homogeneous electric field (like a &lt;a href=&quot;https://en.wikipedia.org/wiki/Capacitor#Parallel-plate_model&quot; rel=&quot;noreferrer&quot; title=&quot;parallel plate capacitor&quot;&gt;parallel plate capacitor&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, use the field by putting charged bodies into it.  I could imagine a space suit with an eletrically charged layer, insulated on both sides.  I don't have an idea for a suit in the magnetic field though, because there are no magnetic charges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quick calculation: Assume a spherical human with a radius $r$ of half a meter.  The suit is a charged conducting sphere and charging it with $V_\text{suit}=1\mathrm{MV}$, which is a lot, even for &lt;a href=&quot;https://en.wikipedia.org/wiki/Marx_generator&quot; rel=&quot;noreferrer&quot; title=&quot;Marx generators&quot;&gt;Marx generators&lt;/a&gt;, then the charge stored in the suit is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q = V_\text{suit} 4\pi \epsilon_0 r = 6\cdot10^{-5}\,\mathrm{C}\,.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't look like a lot in the exponent, but Coulomb is an insanely large unit.  Now charging up the capacitor formed by floor and ceiling to some voltage $V_\text{deck}$, the total force on the suit is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = E \cdot Q = \frac{V_\text{deck}}{d}\, Q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the acceleration of the body should be one g:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = m a = m g$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a mass of lets say $m=70\,\mathrm{kg}$ this requires a voltage of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_\text{deck} = \frac{m g d}{Q} = 2\cdot10^7\,\mathrm{V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck with that :-)  The typical dielectric strength of air is $3\mathrm{MV}/\mathrm{m}$ so you would instantly get an electrical arc between floor and ceiling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Magnetic field:&lt;/strong&gt;  Magnetic fields behave a bit differently and the best option is probably a linearly varying magnetic field $\mathbf{B}$ from floor to ceiling (use &lt;a href=&quot;https://en.wikipedia.org/wiki/Helmholtz_coil&quot; rel=&quot;noreferrer&quot; title=&quot;Helmholtz coils&quot;&gt;Helmholtz coils&lt;/a&gt;).  If you manage to build a suit with some magnetic dipole moment $\mathbf{m}$ (for example using a permanent magnet, or running a current through the suit), the force is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{F} = \nabla (\mathbf{m} \cdot \mathbf{B})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dipole moment of a normal neodymium magnet is given to be about $1\,\mathrm{Am}^2$ &lt;a href=&quot;https://www.researchgate.net/publication/276915106_Physics_Education_Determination_of_Magnetic_Dipole_Moment_of_Permanent_Disc_Magnet_with_Two_Different_Methods&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.  For the same $m=70\,\mathrm{kg}$ body the magnetic field has to have a gradient of 7 Tesla per centimeter ...  or a difference of whopping 1400 Tesla from floor to ceiling (give or take one magnitude for not scaling the magnet to human weight).  As a comparison, the magnets at LHC generate fields of about 10 Tesla.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line: Earth is big, therefore gravity at earth's surface is strong, because of that it's hard to reproduce it using electric or magnetic fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Graviton:&lt;/strong&gt; The graviton is still a hypothetical particle.  Even if we confirm its existence, its behaviour should be (for the large part) very similar to gravity as we know it.  &quot;Bombarding&quot; something with gravitons is then equivalent to having a large mass nearby.  Understanding the theory will most probably not enable us to build graviton guns that work analogous to lamps producing photons.  Speaking of large masses, how about lining the floor with neutron star matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ferrofluid:&lt;/strong&gt; A paramagnetic material can be used to couple to the magnetic field, but I believe you need active stabilisation (circuits in the floor and ceiling).  There actually seems to be a theorem about that: &lt;a href=&quot;https://en.wikipedia.org/wiki/Earnshaw%27s_theorem&quot; rel=&quot;noreferrer&quot;&gt;Earnshaw's theorem&lt;/a&gt;.  There are other implications to drinking ferrofluids, besides the chemical toxicity, because it might accumulate in some spots of your body and more of the artificial gravity will attack at that spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess, we are still better off with gravity-like acceleration by rotation or propulsion ...&lt;/p&gt;&#xA;" OwnerUserId="22239" LastEditorUserId="22239" LastEditDate="2018-03-16T23:46:10.930" LastActivityDate="2018-03-16T23:46:10.930" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25529" PostTypeId="2" ParentId="24478" CreationDate="2018-03-17T15:49:59.393" Score="2" Body="&lt;p&gt;There are three kinds of different &lt;em&gt;hyperbolic&lt;/em&gt; in this question and the previous answer by AtmosphericPrisonEscape: the behavior of Lorentz transformations, the spatial hypersurfaces in cosmology (which seem to be flat), and the hyperbolic appearance of spacetime when tracing world lines in a universe with accelerated expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TL;DR: Spacetime is expanding exponentially and if you choose an unconventional hypersurface, that hypersurface is hyperbolic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lorentz transformations:&lt;/strong&gt; Lorentz covariance is the mathematical expression of Einstein's principle of the constancy of the speed of light. Changing the frame of reference of an observer is done by a Lorentz transformation which preserves the light cone and is closely related to the concept of causality.  &lt;a href=&quot;https://math.stackexchange.com/a/666294&quot;&gt;The answer to this question&lt;/a&gt; explains this for the Minkowski space, where a Lorentz transform is a &lt;em&gt;hyperbolic rotation.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this flat (Minkowskian) spacetime:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ ds^2 = c\, dt^2 - dx^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can use a Lorentz transform to switch to different coordinates $dt$ and $dx$ while the line element $ds$ is invariant, analogous to the hyperbola&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 1 = a^2 - b^2. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This kind of &lt;em&gt;hyperbolic&lt;/em&gt; is not the point of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cosmology:&lt;/strong&gt; If we assume large scale homogeneity and isotropy (Copernican principle), we can choose isochronous coordinates.  There you have a synchronous time $t$ and spatial hypersurfaces for given $t$ (&quot;time slices&quot;).  Freely falling observers are comoving, i.e. have fixed spatial coordinates $\mathbf x$.  Their world lines are perpendicular to the spatial hypersurfaces and their proper time matches the synchronous time $\tau = t$.  Galaxy clusters are such freely falling observers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take these assumptions and allow time dependence of spatial coordinates on time, you end up with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann%E2%80%93Lema%C3%AEtre%E2%80%93Robertson%E2%80%93Walker_metric&quot; rel=&quot;nofollow noreferrer&quot;&gt;FLRW metric&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ ds^2 = c\, dt^2 - a(t)^2 d\Sigma^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $d\Sigma^2$ describes the spatial hypersurface.  Now additionally to this being a hyperbolic equation, the hypersurfaces $d\Sigma^2$ can have curvature, especially constant positive (closed sphere), zero (Euclidean), or constant negative (open hyperboloid) curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A flat universe is, for example, given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ d\Sigma^2 = dr^2 + r^2 d\Omega,\\&#xA;   \text{where}\quad d\Omega = d\theta^2 + \sin(\theta)^2\, d\phi^2, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the metric of Euclidean 3-space in spherical coordinates.  Here the only relative acceleration between freely falling observers comes from the scale factor $a(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hyperbolic spacetime could be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ d\Sigma^2 = dr^2 + \sinh(r)^2 d\Omega $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Coordinate_systems_for_the_hyperbolic_plane#Polar_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;hyperbolic plane in spherical coordinates&lt;/a&gt;.  In such a spacetime, freely falling observers do not only measure a relative acceleration due to the scale factor, but an additional contribution due to the curvature of space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Expansion:&lt;/strong&gt; In the FLRW metric you see the scale factor $a(t)$.  The Friedmann equations describe the evolution of the scale factor.  It does not influence the qualitative shape of the spatial hypersurfaces, but in an expanding space their curvature becomes more and more flat (just like the surface of a sphere locally becomes more and more flat as the sphere expands).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A flat $d\Sigma$ and a constant $a$ gives you no acceleration between freely falling observers whatsoever.  An $a$ with constant change manifests as all observers flying apart, but no relative acceleration.  Finally, an $a$ with constant acceleration makes everything fly apart ever faster (currently observed).  The Friedmann equations tell us, that the behaviour of $a$ changed during the lifetime of the universe as different effects where dominating (radiation, matter, cosmological constant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently we observe an accelerating universe, so $a$ is exponentially increasing (just at the right end of the illustration).  If you look at the world line of two Galaxies, their physical distance increases exponentially.  Therefore the circumference of a chunk of space increases exponentially, so the hypersurface spanned by a line of freely falling observers is actually hyperbolic (white grid in the illustration).  However, I struggle to put that into an equation that clearly shows the hyperbolicity, so don't trust that too much, yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cCCU4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cCCU4.jpg&quot; alt=&quot;Timeline of expanding universe&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22239" LastActivityDate="2018-03-17T15:49:59.393" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25530" PostTypeId="1" AcceptedAnswerId="25531" CreationDate="2018-03-17T15:59:31.580" Score="1" ViewCount="947" Body="&lt;p&gt;As the questions says, I have the celestial coordinates of a star(ra, dec). My trigonometry is not so great so I haven't managed to figure out a solution to this. I want to find a place (lat, long) and a time when the star will be right above the computed point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. star coordinates: 7h 23m 23.97s -13° 25' 2.64&quot;, where would be the point(lat, long) and time where the star would be right above?&lt;/p&gt;&#xA;" OwnerUserId="22250" LastActivityDate="2018-03-18T07:52:06.023" Title="Given a star's ra, dec how can I compute one pair of lat, long and time where the star would be at zenith?" Tags="&lt;star&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25531" PostTypeId="2" ParentId="25530" CreationDate="2018-03-17T17:26:20.027" Score="4" Body="&lt;p&gt;Latitude is the easiest; it's the same as the star's declination: -13° 25' 2.64&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longitude and time depend on each other. A star with right ascension 7h 23m 23.97s will be at the zenith point at 7h 23m 23.97s local &lt;a href=&quot;https://en.wikipedia.org/wiki/Sidereal_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;sidereal time&lt;/a&gt; (by definition). You need to convert sidereal time to local time (at a given day), e.g. with the script described &lt;a href=&quot;https://astronomy.stackexchange.com/q/20296/7472&quot;&gt;here&lt;/a&gt;. On September 23rd (or whatever day the Fall equinox falls, thanks @MikeG) on the Greenwich meridian, this will happen at 7:23am UTC.&lt;/p&gt;&#xA;" OwnerUserId="7472" LastEditorUserId="7472" LastEditDate="2018-03-18T07:52:06.023" LastActivityDate="2018-03-18T07:52:06.023" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25532" PostTypeId="2" ParentId="25524" CreationDate="2018-03-18T00:06:11.343" Score="0" Body="&lt;p&gt;Space time is curved by gravity and there's gravity everywhere in the observable universe and perhaps everywhere in the (entire) universe or just &quot;universe&quot; is the proper term.   So, pretty much everywhere, there should be some local space-time curvature.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an analogy, the surface of the Earth is lumpy.  It's got mountains and valleys and high and low regions everywhere on it's surface.  No area of the Earth is completely smooth and perfectly zero sea-level, but despite that lumpiness, the Earth is round (an oblate spheroid but round is close enough).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth isn't a perfect example, but the Earth's surface is a representation of a two dimensional surface wrapped around a 3 dimensional spheroid and Earth's curvature is measurable from it's surface.   One method of doing this is to draw a triangle and measure the sum of the angles and the variation from 180 degrees relative to the size of the triangle gives you what you need to calculate the curvature.  Using a different and more practical method, Eratosthenes measured &lt;a href=&quot;https://en.wikipedia.org/wiki/Eratosthenes#Measurement_of_the_Earth&amp;#39;s_circumference&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth's curvature and size over 2000 years ago&lt;/a&gt;.  Eratosthenes's two measurements were taken about 500 miles apart was far enough to reduce any variation in elevation to a small error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth is about a thousand times as far across as the tallest mountain or deepest valley and it's got some gravitational variation, but it's still generally spheroid, both in gravitation and in shape.  The Universe is incalculably large compared to the local ripples and curves caused by gravity and the Universe, as far as we can tell, appears quite uniform, so the local pockets of high gravity and local curvature or quite irrelevant to the overall shape or potential curvature of the Universe, just as the mountains and valleys on Earth are irrelevant to it still being an oblate spheroid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Universe may not be flat, it just appears to be flat to our best measurements.   It might be curved, but so large, modern measuring techniques aren't able to measure the curvature.   This raises questions as to what a curved universe actually means (distinct from curved space-time that's easily measurable).   That can make some fun reading, but it's all theoretical.  Nobody knows the true shape or curvature of the universe though the universe being infinite and flat appears to be the more accepted idea today, but it's still an unanswerable question. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-03-19T08:58:02.303" LastActivityDate="2018-03-19T08:58:02.303" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="25533" PostTypeId="1" AcceptedAnswerId="25534" CreationDate="2018-03-18T13:13:39.870" Score="4" ViewCount="181" Body="&lt;p&gt;I was looking at a wikipedia article (&lt;a href=&quot;https://en.wikipedia.org/wiki/Pulsar#Formation,_mechanism,_turn_off&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Pulsar#Formation,_mechanism,_turn_off&lt;/a&gt;)&#xA;and I came across an interesting aspect that said pulsars turn off from rotation. What I don't understand is why emission of EM waves would slow a pulsar down?&lt;/p&gt;&#xA;" OwnerUserId="16048" LastActivityDate="2018-03-18T15:31:17.590" Title="Why do pulsars turn &quot;off&quot; from rotation?" Tags="&lt;pulsar&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25534" PostTypeId="2" ParentId="25533" CreationDate="2018-03-18T13:54:43.493" Score="7" Body="&lt;p&gt;The first thing you need to recall is that electromagnetic waves do carry momentum as well as energy.  This shows up in effects like &lt;a href=&quot;https://en.wikipedia.org/wiki/Radiation_pressure&quot; rel=&quot;noreferrer&quot;&gt;light pressure&lt;/a&gt;. Specifically a photon of wavelength $\lambda$ carries momentum $h/\lambda$. In and of itself, that doesn't answer your question though, since you are asking about the &lt;em&gt;rotation&lt;/em&gt; of the pulsar, and changed to its linear momentum don't affect that. However, there are two ways this can happen. First imagine a rotating sphere which is emitting radiation from its equator. Consider a point on that equator. The radiation it is emitting &quot;forwards&quot; (in the direction towards which it is rotating) will be blue shifted (as seen by a stationary distant observer) while the radiation emitted &quot;backwards&quot; will be correspondingly red-shifted. Since the blue-shifted radiation has shorter wavelength, it has more momentum, the net effect is a force on the source which tends to slow the rotation. (interestingly, the same effect is used at much lower energy levels to trap ultra-cold atoms in what is called &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Optical_molasses&quot; rel=&quot;noreferrer&quot;&gt;optical molasses&lt;/a&gt;&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second effect is a bit more technical. The radiation which is emitted towards the poles of the pulsar has a property called &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Circular_polarization&quot; rel=&quot;noreferrer&quot;&gt;circular polarization&lt;/a&gt;&quot; and a beam of circularly polarized light carries away &lt;a href=&quot;https://en.wikipedia.org/wiki/Spin_angular_momentum_of_light&quot; rel=&quot;noreferrer&quot;&gt;angular momentum&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="21928" LastEditDate="2018-03-18T15:31:17.590" LastActivityDate="2018-03-18T15:31:17.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25535" PostTypeId="1" AcceptedAnswerId="25538" CreationDate="2018-03-18T16:47:39.790" Score="2" ViewCount="365" Body="&lt;p&gt;Let's assume there is a small natural satellite oribiting a massive planet. The satellite rotates synchronously (revolution period = orbital period) on a circular orbit, both bodies are homogenous and rigid. The question is, how to find the Roche limit with regard to centrifugal forces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With use of &lt;a href=&quot;https://en.wikipedia.org/wiki/Roche_limit#Derivation_of_the_formula&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; I understand that the Roche limit is basically the orbit where difference of gravitational forces between the planet and two points in the satellite body (its centre and a surface facing to the planet) cancels the gravitational pull of the surface of the satellite to its centre. Wikipedia also states that more accurate is to include also the centrifugal force on the surface of the satellite due to its rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is not included the difference of the centrifugal forces acting on satellite centre and surface facing to the planet, caused by satellite's orbital movement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other forces I need to take into account, for example orbit of the whole sytem around a star?&lt;/p&gt;&#xA;" OwnerUserId="17348" LastActivityDate="2018-03-19T12:54:36.220" Title="Roche limit - calculation with centrifugal forces" Tags="&lt;gravity&gt;&lt;roche-limit&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25536" PostTypeId="2" ParentId="25535" CreationDate="2018-03-18T20:39:00.130" Score="3" Body="&lt;p&gt;The roche limit is a calculation of when pieces of a satellite will be pulled off the satellite. Any force that affects the whole satellite equally can be ignored, as this will affect the motion of the satellite as a whole but not (directly) whether pieces are pulled off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider an object which maintains a constant orientation, but rotates about another body, the paths each part of the object take differ only by a translation in space, and so each part will experience the same centrifugal force. So, &lt;strong&gt;the only rotation that needs to be considered is the rotation of the body around its own axis.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the special case of a tidally locked satellite in a circular orbit, the satellite's rotation around the planet, and the satellite's rotation around its own axis, are the same rotation, just differing in what axis we are considering the rotation to be around. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the wikipedia article's calculations, the centrifugal force for the rigid body case is considered around the satellite's own axis, but for the fluid body case it is considered as a rotation about the primary. This is an equally valid calculation for the special case of a tidally locked satellite in a circular orbit.&lt;/p&gt;&#xA;" OwnerUserId="22259" LastEditorUserId="22259" LastEditDate="2018-03-18T21:12:27.907" LastActivityDate="2018-03-18T21:12:27.907" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25538" PostTypeId="2" ParentId="25535" CreationDate="2018-03-18T21:13:15.553" Score="2" Body="&lt;p&gt;When a satellite is in a circular orbit around a planet you may consider its motion in a rotating frame of reference; the frame rotates at the same rate as the satellite orbits. In this frame the satellite doesn't move. But because this is a rotating frame the formula $F=ma$ is doesn't work, you need to add extra terms for the centrifugal force and the corilis force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the rotating frame there are two forces acting on the satellite: gravity and the centrifugal force, and since the satellite isn't moving (in this frame) these two forces must equal each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the satellite is modelled as a particle, all is well. But if the satellite is a rigid body the part of the satellite that is closer to the planet will experience more gravity than centrifugal force. The part of the satellite that is furthest from the planet will have the opposite: the centrifugal force will be greater than gravity. The result is that part of the satellite is pulled towards the planet, and part of the satellite is pushed away. This is called the &quot;tidal force&quot;, as it is the primary reason for tides in the oceans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The satellite has its own gravity, and normally that is stronger than the tidal force. But if the satellite is sufficiently close to the planet, the tidal force may become greater than the self-gravitation of the satellite. The point at which this occurs is the Roche limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic calculation of the Roche limit is based on finding when the tidal force (caused by the different gravitational and centrifugal forces over the satellite) exceeds the gravitational force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more subtle calculation can take into account other factors: The tidal force can distort the satellite, the satellite may have significant rotation, there may be significant strength in the materials that form the satellite. These factors can cause a satellite to break up earlier or later than a simple calculation suggests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, The &quot;centrifugal forces acting on satellite centre and surface facing to the planet, caused by satellite's orbital movement?&quot; are the tidal forces. So these are already accounted for in the simple calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big uncertainty in calculating the break-up of a satellite is the strength of the chemical bonds that hold the satellite together. Most artificial satellites orbit will inside their Roche limit, but they don't break up because they are held together by strong metallic bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a satellite is rigid, then it probably has some tensile strength and will hold together even if tidal forces are tending to break it apart. If a satellite is a &quot;rubble pile&quot; and doesn't have any significant strength, the assumption of it being &quot;rigid&quot; must be questioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These factors introduce uncertainties that are much greater than any other forces, such as solar tides.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-03-19T12:54:36.220" LastActivityDate="2018-03-19T12:54:36.220" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25540" PostTypeId="1" CreationDate="2018-03-19T04:29:19.903" Score="4" ViewCount="146" Body="&lt;p&gt;I read the following claim in an article (Lisa Grossman, &quot;Origin of Cosmic Oddities Proposed,&quot; &lt;em&gt;Science News&lt;/em&gt;, February 17, 2018, 8).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... IceCube has already traced an especially high-energy neutrino to a single active black hole ... IceCube [Neutrino Observatory in Antarctica] could eventually trace neutrinos back to [active black holes in] galaxy clusters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Aren't galaxy clusters at least millions of light-years from Earth?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Such neutrinos took millions of years to get to Earth.&lt;/li&gt;&#xA;&lt;li&gt;Other sources could have been in the path of the neutrinos in such a distant past.&lt;/li&gt;&#xA;&lt;li&gt;Are neutrinos affected by gravitational lenses as light can be?&lt;/li&gt;&#xA;&lt;li&gt;Is such tracing possible because neutrinos react so weakly with matter?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7981" LastActivityDate="2018-03-19T22:31:33.603" Title="How Does IceCube Trace a Neutrino to its Source?" Tags="&lt;neutrinos&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25541" PostTypeId="1" AcceptedAnswerId="25543" CreationDate="2018-03-19T04:53:31.767" Score="0" ViewCount="111" Body="&lt;p&gt;Is the generally accepted belief in the science community, that if one sits in a spaceship and move away from Earth where r (distance between earth and my ship) is continuously increasing, can one keep going forever? &lt;/p&gt;&#xA;" OwnerUserId="22265" LastEditorUserId="2365" LastEditDate="2018-03-19T12:56:42.070" LastActivityDate="2018-03-19T12:56:42.070" Title="what happens when we keep going radially away from earth?" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25543" PostTypeId="2" ParentId="25541" CreationDate="2018-03-19T09:26:17.040" Score="4" Body="&lt;p&gt;Yes. There are no edges of the universe that have been observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard cosmological models favoured by observations (open or flat universes) allow you to move outwards indefinitely. Closed universes are like the surface of a sphere, and you would eventually return to Earth unless the expansion was too fast; the same is true for the models with nontrivial topologies - but these have no evidence at present.&lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2018-03-19T09:26:17.040" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25544" PostTypeId="2" ParentId="25540" CreationDate="2018-03-19T10:00:37.747" Score="1" Body="&lt;p&gt;Neutrinos have nearly zero rest mass and travel nearly as &lt;a href=&quot;https://physics.stackexchange.com/questions/267035/where-are-all-the-slow-neutrinos&quot;&gt;fast as light&lt;/a&gt;.  That means that observations of neutrino flashes match up with observations of photons from the same source up to a &lt;a href=&quot;https://astronomy.stackexchange.com/a/23425/22239&quot;&gt;small delay&lt;/a&gt;.  So if you locate the origin of the neutrino (as angular position on sky) you can simply look it up in existing imagery or look there with a telescope, and the light you see basically traced the same path through time and space as the neutrino.  An active black hole as mentioned in the article should be visible in standard deep sky observations. (I couldn't find the article, so I'm relying on your quote here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If another source &quot;moves in the way&quot; just as the neutrino passes it, you will also see both sources at the same time with your telescope here on earth.  Then you either identify them, eg. by energy signature, or you cannot see the more distant one, or you simply cannot decide for one of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like all other objects that are freely falling through space and time, neutrinos follow geodesics (generalization of straight lines)  through spacetime and are affected by gravitational lenses just as photons are.  In the Newtonian picture, all objects, independent from their mass, are accelerated the same in a gravitational field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a way, it is true, that tracing the sources of neutrinos is easier because they are weakly interacting.  If they were interacting more strongly, you would see fewer of them and they might have been deflected, by scattering for example.  As a disadvantage, you get less secondary observations and less information from the source and the line of sight.  The electromagnetic waves we measure on earth contain riches of data in their spectrum (structure and composition of source, hydrogen and molecular clouds along the line-of-sight, ...).&lt;/p&gt;&#xA;" OwnerUserId="22239" LastEditorUserId="22239" LastEditDate="2018-03-19T22:31:33.603" LastActivityDate="2018-03-19T22:31:33.603" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25545" PostTypeId="2" ParentId="25540" CreationDate="2018-03-19T10:15:30.970" Score="3" Body="&lt;p&gt;You ask a number of questions. I'll try an answer them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, galaxy clusters (except the one we are part of) are many millions of light years away and so the neutrinos from them (which travel at a speed only a tiny fraction below that of light) have taken millions of years to get here. However, the light from those clusters has also taken millions of years to get here, so if we look in a certain direction and see a galaxy, and we detect a high energy neutrino coming from the same direction, then the galaxy was there when the neutrino was emitted (or passed through it). It may have moved since, but that would affect neither the light nor the neutrino.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutrinos are affected by gravity, and so the neutrino images of distant objects can be distorted in the same way as the optical images when there is a massive object (like another galaxy cluster) between it and us. However, if that were happening, we would see the distortion of the optical image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also possible (though extraordinarily unlikely) that a single neutrino might have been deflected by passing close to a very massive object  such as a black hole which is too small to make much of a difference to the optical image of the cluster. The chance of a neutrino from somewhere else being deflected by something we haven't detected, just enough to appear to come from a galaxy we can see, though, is absolutely amazingly minute. A bit like a bullet ricocheting off a falling meteorite in such a way that it seems to have come from that man over there holding a gun, even though it actually came from somewhere else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, on the question of detection. It is true that neutrinos interact weakly with matter, but these very high energy neutrinos interact less weakly than others. &lt;a href=&quot;https://arxiv.org/pdf/1305.7513.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This paper&lt;/a&gt; and in particular its figure 1 suggest that neutrinos of the energy reported ($2\times 10^{15}$ eV according to the article you refer to) would interact about 10 million times more strongly than the ones from the Sun or from nuclear reactors. Even so, most neutrinos will go straight through ice cube without interacting with it at all, despite the fact that it consists of a whole cubic kilometer of ice. Of the few that do interact, the interaction will usually &quot;kick loose&quot; a charged particle (often a muon) from the ice, which will be moving at close to the speed of light (in vacuum) itself, and in the same direction that the neutrino was moving. In particular, it will be moving faster than the speed of light in ice. When a charged particle moves faster than the speed of light in whatever material it's passing through it produces a &quot;shock wave&quot; of light, called Cerenkov radiation. Detectors in ice cube pick up that radiation and track the movement of its source, thereby getting a good idea of the direction from which the original neutrino came.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="21928" LastEditDate="2018-03-19T19:14:09.677" LastActivityDate="2018-03-19T19:14:09.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25547" PostTypeId="1" CreationDate="2018-03-19T13:43:19.187" Score="-3" ViewCount="143" Body="&lt;p&gt;Assuming  Prof. Poplawski's &quot;&lt;a href=&quot;https://doi.org/10.3847/0004-637X/832/2/96&quot; rel=&quot;nofollow noreferrer&quot;&gt;universe in a black hole&lt;/a&gt;&quot; ( and at &lt;a href=&quot;https://arxiv.org/pdf/1410.3881.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arxiv here&lt;/a&gt; ) concept is true--that our universe is the result of a nonsingular &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Bounce&quot; rel=&quot;nofollow noreferrer&quot;&gt;Big Bounce&lt;/a&gt; from a supermassive black hole (SMBH) under &lt;a href=&quot;https://en.wikipedia.org/wiki/Einstein%E2%80%93Cartan_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein-Cartan gravitation&lt;/a&gt;--then it seems possible to calculate the volume of this ur-object using the following method:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Assume all black holes have the same mass density.&lt;/li&gt;&#xA;&lt;li&gt;Assume there is some median black holes size.&lt;/li&gt;&#xA;&lt;li&gt;Assume every galaxy in our universe contains a Super Massive Black Hole (SMBH) at its center.&lt;/li&gt;&#xA;&lt;li&gt;This means there are 2 trillion such objects (the same number as the number of galaxies).&lt;/li&gt;&#xA;&lt;li&gt;Two trillion times the size (not mass) of the median SMBH gives us the volume of the ur-SMBH which can then expanded to include everything in our universe.&lt;/li&gt;&#xA;&lt;li&gt;Packing that volume into a single object gives us (hopefull) the size of the Ur-SMBH.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What size is that?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/37STZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/37STZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22274" LastEditorUserId="10196" LastEditDate="2018-03-19T21:30:12.770" LastActivityDate="2018-03-19T22:28:49.430" Title="Estimates VOLUME of the ur-Supermassive Black Hole of this universe" Tags="&lt;early-universe&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25548" PostTypeId="1" CreationDate="2018-03-19T18:47:36.073" Score="4" ViewCount="1575" Body="&lt;p&gt;Do we really know that the origin of gold was in a supernova explosion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am talking about the gold we find here on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So then how are we sure that Gold actually came from these explosions.  Did they go back some million years ago and figure out that Gold was made in supernova.&lt;/p&gt;&#xA;" OwnerUserId="22278" LastEditorUserId="7472" LastEditDate="2018-03-20T20:03:13.743" LastActivityDate="2018-03-20T20:03:13.743" Title="Does Gold really come from Supernovas?" Tags="&lt;supernova&gt;&lt;metal&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ClosedDate="2018-03-20T20:05:05.477" ContentLicense="CC BY-SA 3.0" />
  <row Id="25549" PostTypeId="1" AcceptedAnswerId="25551" CreationDate="2018-03-19T19:02:45.683" Score="2" ViewCount="101" Body="&lt;p&gt;Stephen Hawking proposed that a black hole emits radiation as it evaporates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes don't let any radiation escape their event horizon because even particles with the speed of light get trapped by their magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that mean when black holes emit radiation, they just re-absorb it?&lt;/p&gt;&#xA;" OwnerUserId="22022" LastEditorUserId="7472" LastEditDate="2018-03-19T23:56:32.337" LastActivityDate="2018-03-19T23:56:32.337" Title="Can we detect a black hole's radiation?" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25550" PostTypeId="2" ParentId="25548" CreationDate="2018-03-19T19:03:41.117" Score="9" Body="&lt;p&gt;We don't need to see things to understand them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is now generally understood that gold is formed in neutron star collisions.  The evidence for this partly theoretical: Models of neutron star collisions suggest they would produce large amounts of heavy elements, and regular supernovae would not produce gold in the right amounts to explain the concentrations found in the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, we have observational evidence from the collision of two neutron stars, detected by gravitational waves, which allowed for scientists to analyse the light coming from this collision in detail. We can't see gold, but we can see radiation that indicates the presence of radioactive isotopes that will decay to form gold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gold would float in the interstellar space for a while until it forms part of the dust around a newly forming star. That dust gathers together to form planets. Most of the actual gold we find came to the Earth carried in by meteorites, as the original gold will have fallen to the Earth's core when the Earth was molten.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have a good model for explaining how gold can be formed in neutron star collisions and evidence for this actually happening. That is enough for this to be the generally accepted theory.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-03-19T19:03:41.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25551" PostTypeId="2" ParentId="25549" CreationDate="2018-03-19T19:30:25.540" Score="2" Body="&lt;p&gt;The black hole itself does not emit the radiation, but rather the space just around it. Because of quantum effects, small particles and anti-particles will spontaneously create everywhere (and later annihilate themselves in a collision). If this happens just outside the event horizon, one particle will inevitably fall into it, &lt;em&gt;never to be seen again&lt;/em&gt;. The other particle is now 'free' and this is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;radiation Hawking predicted&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can we detect a black hole's radiation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, this radiation is way too small to be detected by itself, even if the black hole would be at our doorstep. However, the radiation makes the black hole less massive and smaller and finally it will vanish, producing a gamma ray flash which we hope &lt;em&gt;is&lt;/em&gt; detectable, even from a very large distance.&lt;br&gt;&#xA;As @JamesK notes in the comments, this evaporation will take extremely long for the 'common' black holes caused by supernovae and those found in galaxy centers, but smaller &lt;a href=&quot;https://en.wikipedia.org/wiki/Primordial_black_hole&quot; rel=&quot;nofollow noreferrer&quot;&gt;primordial black holes&lt;/a&gt;, formed in the very beginning of the universe, could be evaporating now and that's why we have a &lt;a href=&quot;https://en.wikipedia.org/wiki/Fermi_Gamma-ray_Space_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;telescope&lt;/a&gt; looking for those gamma ray flashes.&lt;/p&gt;&#xA;" OwnerUserId="7472" LastEditorUserId="7472" LastEditDate="2018-03-19T19:48:15.763" LastActivityDate="2018-03-19T19:48:15.763" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25552" PostTypeId="1" AcceptedAnswerId="25553" CreationDate="2018-03-19T19:58:20.893" Score="4" ViewCount="173" Body="&lt;p&gt;Could rays of light be ever trapped in a constant orbit around a black hole if approached from a certain angle ? Like light hitting a glass or liquid with critical angle.&lt;/p&gt;&#xA;" OwnerUserId="22022" LastEditorUserId="16477" LastEditDate="2018-03-23T23:19:29.463" LastActivityDate="2018-03-23T23:19:29.463" Title="Can light be trapped in orbit?" Tags="&lt;black-hole&gt;&lt;light&gt;&lt;photons&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25553" PostTypeId="2" ParentId="25552" CreationDate="2018-03-19T21:40:54.460" Score="9" Body="&lt;p&gt;In principle yes, but in practice no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Photon_sphere&quot; rel=&quot;noreferrer&quot;&gt;photon sphere&lt;/a&gt;, gravity is exactly so strong that a photon on a tangential trajectory would stay in orbit. For a non-rotating black hole of mass $M$, the radius of the photon sphere is $3/2$ times the radius of the &lt;em&gt;event horizon&lt;/em&gt; — i.e. the &quot;surface&quot; of the black hole — which itself is given by $r_s \equiv 2GM/c^2$, where $G$ and $c$ are the gravitational constant and the speed of light, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the orbits are unstable; any perturbation will cause a photon on the photon sphere to either escape or plunge into the black hole.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2018-03-19T21:40:54.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25554" PostTypeId="2" ParentId="25547" CreationDate="2018-03-19T22:28:49.430" Score="1" Body="&lt;p&gt;Some problems :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The energy total of the universe is predominantly made up of Dark Matter and Dark Energy, with only 5% for Baryonic matter (&quot;normal&quot; matter).  But as the proposed idea seems to completely change the model for the &quot;dark&quot; matter and energy values, it's hard to relate to that model.  We either need Dark Matter or some radical changes in physics to explain e.g. stellar motions in galaxies and I don't think the &quot;ur-SMBH&quot; is going to do that.  So we still need all that mass and energy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The SMBs at the center of galaxies are a small proportion of the normal matter in a galaxy.  For the Milky Way, our SMB mass is about $4\times 10^6 M_0$ whereas the Milky Way mass is about $10^{12}M_0$ which makes the SMB almost insignificant in calculating the mass total for your black hole.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Note you can't ignore energy.  GR tells us that energy contributes to gravitational fields, so it's not just about mass.  Every photon is significant, every neutrino, and that (again) means we need to include dark matter and dark energy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So your estimate of the ur-SMBH mass (and volume) would be omitting the most significant contributions to mass and energy in the Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Limiting the remainder to the idea of how you calculate the size of combined black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest black hole model is a Schwarzchild black hole : non rotating and alone in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radius of such a black hole is :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R = \frac {2GM}{c^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the volume and mass are &lt;strong&gt;not&lt;/strong&gt; proportional.  Instead we get :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V = \frac {4\pi}{3} \left(\frac {2GM}{c^2}\right)^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when we add two black hole together (and ignoring mass loss due to gravitational waves which I gather from &lt;a href=&quot;https://physics.stackexchange.com/a/238458/104964&quot;&gt;this answer on Physics SE&lt;/a&gt; is between about 5% and 11%), we find that :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M_{total} = M_1 + M_2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_{total} = \frac {4\pi}{3} \left(\frac {2G(M_1+M_2)}{c^2}\right)^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_{total} = \frac {4\pi}{3} \left( R_1+R_2 \right)^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_{total} = \frac {4\pi}{3} \left( R_1^3+R_2^3+3R_1^2R_2+3R_1R_2^2 \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_{total} = V_1 + V_2 + 4\pi \left( R_1^2R_2+R_1R_2^2 \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_{total} = V_1 + V_2 + 3V_1^{\frac 2 3}V_2^{\frac 1 3} + 3V_1^{\frac 1 3}V_2^{\frac 2 3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you were to merge two identical black holes of equal volume $V_0$ together you'd get a black hole of volume $8V_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if you combined $N$ black holes of equal size together you'd get :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_{total} = N^3V_0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This of course is a very naive calculation, but it does show why you can't just add the volumes together as you are suggesting.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2018-03-19T22:28:49.430" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25558" PostTypeId="2" ParentId="21566" CreationDate="2018-03-20T09:15:49.673" Score="3" Body="&lt;p&gt;&quot;If a comet strikes a glacier, does it make a print on the land below?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would depend upon the depth of glacial ice, the nature of &#xA;soil beneath it, and size-composition-velocity of the comet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being made of 'dirty ice' a comet's density is much less that an &#xA;asteroid's so even though their velocity is much grater as a rule &#xA;they would leave only a small cater at their target  and likely no &#xA;permanent mark where they hit deep ice.&#xA;However, prominent lateral imprints were formed by a relatively &#xA;recent earth impact as debris within the ice was ejected at&#xA;extreme velocity by explosive vaporization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asteroids striking deep glacial ice are an entirely different matter&#xA;Their fireball vaporized a tunnel though deep glacial ice by radiant &#xA;heat and their cold core penetrated poorly consolidated sediment for &#xA;many miles in some cases before explosive vaporization occurred to &#xA;create a wide variety of crater configurations in the 24 'recent'&#xA;events I've identified since '98.&lt;/p&gt;&#xA;" OwnerUserId="22293" LastActivityDate="2018-03-20T09:15:49.673" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25559" PostTypeId="2" ParentId="25548" CreationDate="2018-03-20T11:43:10.877" Score="5" Body="&lt;p&gt;Do we really know that the origin of gold was in a supernova explosion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up until the last few years that &lt;em&gt;was&lt;/em&gt; the generally accepted explanation, but things have changed somewhat due to two developments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About half the elements heavier than iron (including gold) must be produced via rapid neutron capture - the so-called r-process. This is not under debate, but the main astrophysical site of the r-process is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) It has become increasingly difficult to produce sufficient amounts of some of the heavier elements (including gold) in theoretical models of supernova explosions. The problem is that the exploding material is just not neutron-rich enough, unless very specific conditions to do with rapid rotation, magnetic fields and low overall metallicity are met.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) The simultaneous observation in gravitational waves (by aLIGO) and electromagnetic waves of &lt;a href=&quot;https://www.ligo.caltech.edu/page/press-release-gw170817&quot; rel=&quot;noreferrer&quot;&gt;GW170817&lt;/a&gt; has revealed that merging neutron stars are possibly a more effective site of the r-process element production. The GW detection clearly provides evidence that this event was produced by merging neutron stars and detailed infrared spectroscopy of the light from the accompanying &quot;kilonova&quot; explosion has revealed the presence of heavy lanthanide elements in the ejecta, which are a product of the r-process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there are &lt;em&gt;three&lt;/em&gt; difficulties in leaping to the conclusion that all the gold is produced by merging neutron stars. It may well be the dominant source but firstly, it remains to be established what the &lt;em&gt;rate&lt;/em&gt; of these events are in the local universe, let along what the rate of these events were before the Sun and solar system were born. There are order of magnitude uncertainties here, although one can make another inference/extrapolation to assume that merging neutron stars are responsible for short gamma ray bursts and use the rate of these to estimate the cosmic history of these events. Secondly it is uncertain just how much r-process element production takes place. This cannot be very well established from the observations of just one event and there are considerable theoretical uncertainties. Thirdly there are observations of very old halo stars that have atmospheres that are enriched with r-process elements. It seems unlikely (though perhaps not impossible) that there would have been time before the birth of these stars to produce this enrichment via the merging of neutron stars, which is a process that can take hundreds of millions or even billions of years. This there may be some proportion of r-process elements that &lt;em&gt;are&lt;/em&gt; produced in certain kinds of core-collapse supernovae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some more details and references can be found in my answer to this &lt;a href=&quot;https://astronomy.stackexchange.com/questions/23162/are-binary-neutron-star-mergers-needed-to-explain-the-abundance-of-gold?noredirect=1&amp;amp;lq=1&quot;&gt;closely related question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But maybe your question is broader than this. How can we be sure that heavy elements are produced inside stars? The answer to that lies in a long process of theory, followed by observation, followed by refinements to theory, but the basic answer is that we can see examples of very old stars born right at the beginning of our galaxy. These stars contain almost no heavy elements at all (less than a ten thousandth of what we infer to be in the Sun). We can also measure the presence of (some) heavy elements in the ejecta from high mass stars and we can also see fresh heavy elements appearing at the surfaces of giant stars that has been mixed up from the centres of the stars where they are being produced in nuclear reactions.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-03-20T11:49:12.030" LastActivityDate="2018-03-20T11:49:12.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25562" PostTypeId="5" CreationDate="2018-03-20T20:00:17.300" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2018-03-20T20:00:17.300" LastActivityDate="2018-03-20T20:00:17.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25563" PostTypeId="4" CreationDate="2018-03-20T20:00:17.300" Score="0" Body="For questions about Hawking radiation, emitted by black holes and causing them to slowly evaporate. Use together with the [black-hole] tag." OwnerUserId="7472" LastEditorUserId="7472" LastEditDate="2018-03-20T20:03:03.950" LastActivityDate="2018-03-20T20:03:03.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25564" PostTypeId="1" AcceptedAnswerId="25565" CreationDate="2018-03-20T20:31:06.410" Score="1" ViewCount="293" Body="&lt;p&gt;I have understood that merging black holes emit considerable amount of their mass as gravitational waves (up to 5-10%). Based on what I have understood from gravitational waves, the closer black holes are to merging, intensity (both frequency and amplitude) of these waves become stronger. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am thus assuming much of the total GW output happens during the last &quot;orbit&quot; of the inspiral, and that the origin of these waves is located within &quot;the final orbit&quot;, i.e. really close to event horizons of both black holes, my question is essentially, is it possible for such GW to collapse into a black hole themselves (something like a kugelblitz I believe), since there's just a lot of energy in GWs (say a few solar masses) in such small volume of space (say tens of kms cubed)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this might depend on the size of merging black holes, as both intensity of GWs as well as size of event horizons depend on the mass. I have a gut feeling that smaller black holes might be closer to this limit but maybe both the intensity and event horizon size depend linearly on the mass, and this does not matter at all.&lt;/p&gt;&#xA;" OwnerUserId="10813" LastActivityDate="2018-03-20T20:50:20.443" Title="Could gravitational waves near merging black holes collapse to a black hole themselves?" Tags="&lt;black-hole&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25565" PostTypeId="2" ParentId="25564" CreationDate="2018-03-20T20:50:20.443" Score="6" Body="&lt;p&gt;You might like to look at some of the &lt;a href=&quot;https://www.youtube.com/user/SXSCollaboration&quot; rel=&quot;noreferrer&quot;&gt;simulations from the SXS collaboration&lt;/a&gt;. Essentially the black holes and the gravitational waves are aspects of the same thing -- the curvature of space. In particular, the way black holes seem to merge in these simulations is that the two black holes come very close and space around them becomes more and more curved (and in more complex ways). Some of that curvature spreads out in the form of gravitational waves, becoming less intense as it gets further away, but some of it, very close to the two black holes reaches the &quot;critical&quot; level of curvature at which an event horizon appears (extending the event horizons of the two black holes) which then &quot;rings down&quot; becoming spherical and form the merged black hole. &lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-03-20T20:50:20.443" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25566" PostTypeId="2" ParentId="18559" CreationDate="2018-03-20T21:17:12.000" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://phys.org/news/2017-09-gravitational-stars-supernovae.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article&lt;/a&gt; basically seems to answer the question. They quote from an earlier study:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Although no CCSNe have currently been detected by gravitational-wave detectors, previous studies indicate that an advanced detector network may be sensitive to these sources out to the Large Magellanic Cloud (LMC). A CCSN would be an ideal multi-messenger source for aLIGO and AdV, as neutrino and electromagnetic counterparts to the signal would be expected. The gravitational waves are emitted from deep inside the core of CCSNe, which may allow astrophysical parameters, such as the equation of state (EOS), to be measured from the reconstruction of the gravitational-wave signal.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since we know from SN1987A that neutrinos from a supernova can be detected at that range, that seems to be a &quot;yes&quot;. The biggest uncertainty seems to be how much gravitational wave energy would be emitted by the supernova, and at what frequencies, which depends on a relatively detailed understanding of exactly how the matter moves around in the explosion, one simulation of which is illustrated in the (rather awesome) video in the article. &lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-03-20T21:17:12.000" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25567" PostTypeId="1" AcceptedAnswerId="25570" CreationDate="2018-03-20T22:23:39.307" Score="6" ViewCount="1006" Body="&lt;p&gt;In an attempting to answer the question “why aren’t changes in sunset and sunrise times symmetrical  around the solstices ?” I’ve got stuck on being able to explain the effect of the obliquity of the ecliptic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the general idea of the equation of time and how the difference in clock time and solar time answer my original question, but I’m struggling to grasp how the tilt of the Earth contributes to the equation of time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m trying to visualise the effect, and assuming it is to do with the point at which the sun is highest in the sky not being in the middle of sunset and sunrise (except at solsitces and equinoxes?). So, is this correct? Is there any good way of trying to visualise it ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m a bit confused by the wikipedia entry&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Equation_of_time#Obliquity_of_the_ecliptic&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Equation_of_time#Obliquity_of_the_ecliptic&lt;/a&gt; saying&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;“If this effect operated alone, then days would be up to 24 hours and 20.3 seconds long (measured solar noon to solar noon) near the&#xA;solstices, and as much as 20.3 seconds shorter than 24 hours near&#xA;the equinoxes” &lt;/li&gt;&#xA;&lt;li&gt;“the inclination of the ecliptic results in the contribution of a sine wave variation with an amplitude of 9.87 minutes and a period&#xA;of a half year to the equation of time. The zero points of this sine&#xA;wave are reached at the equinoxes and solstices”.&#xA;Is point 1 correct ?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Also found these links helpful to some extent, but haven’t fully explained it for me.&#xA;&lt;a href=&quot;https://web.archive.org/web/20150910174438/http://www.rmg.co.uk/explore/astronomy-and-time/time-facts/the-equation-of-time&quot; rel=&quot;noreferrer&quot;&gt;https://web.archive.org/web/20150910174438/http://www.rmg.co.uk/explore/astronomy-and-time/time-facts/the-equation-of-time&lt;/a&gt;&#xA;&lt;a href=&quot;http://totaleclipse.eu/Astronomy/EOT.html&quot; rel=&quot;noreferrer&quot;&gt;http://totaleclipse.eu/Astronomy/EOT.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22300" LastActivityDate="2018-03-23T19:34:18.097" Title="Effect of the obliquity of the ecliptic / tilt of the Earth on the equation of time" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;obliquity&gt;&lt;time&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25570" PostTypeId="2" ParentId="25567" CreationDate="2018-03-21T06:16:03.200" Score="5" Body="&lt;p&gt;As the &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar time&lt;/a&gt;&#xA;Wikipedia page illustrates, we can split a solar day into a constant 23h56m to complete a rotation relative to the stars&#xA;(a &lt;a href=&quot;https://en.wikipedia.org/wiki/Sidereal_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;sidereal day&lt;/a&gt;)&#xA;and a variable ~4m to compensate for one day's orbital motion around the Sun.&#xA;The Sun's apparent motion around the equator determines how much more than 360$^\circ$ the Earth must rotate from one solar noon to the next.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the Earth orbits the Sun in the ecliptic plane, which is oblique to the equator.&#xA;If that orbit were circular, the Sun's&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Ecliptic_coordinate_system#Spherical_coordinates&quot; rel=&quot;nofollow noreferrer&quot;&gt;ecliptic longitude&lt;/a&gt;&#xA;would change at a constant rate (see Note below).&#xA;One degree along the equator spans exactly 4m00s of&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Equatorial_coordinate_system#Spherical_coordinates&quot; rel=&quot;nofollow noreferrer&quot;&gt;right ascension&lt;/a&gt;;&#xA;one degree along the ecliptic, measured around the equator, may span ~8% more or less than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At an equinox, the ecliptic crosses the equator at a 23.4$^\circ$ angle,&#xA;and one degree of ecliptic longitude spans 3m40s of right ascension.&#xA;The Sun appears to move slower than average around the equator,&#xA;so the apparent solar day is slightly shorter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a solstice, the ecliptic is tangent to a $\pm$23.4$^\circ$ declination parallel,&#xA;and one degree of ecliptic longitude spans 4m22s of right ascension.&#xA;The Sun appears to move faster than average around the equator,&#xA;so the apparent solar day is slightly longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iAZGX.png&quot; alt=&quot;equatorial and ecliptic grids at March equinox&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/OaLZJ.png&quot; alt=&quot;equatorial and ecliptic grids at June solstice&quot;&gt;&#xA;&lt;em&gt;Stellarium images around the March equinox and the June solstice,&#xA;showing 30 degrees of ecliptic longitude and 2 hours of right ascension for comparison.&#xA;Equatorial grid is blue, ecliptic grid is orange, ecliptic is yellow.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation of time is the cumulative sum of differences between mean and apparent solar day length.&#xA;Like a mathematical function and its derivative, zero crossings of one should roughly correspond to maxima or minima of the other.&#xA;Differences of a few seconds per day, several weeks in a row, add up to a few minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The earliest/latest sunrise and latest/earliest sunset of the year do occur several days on either side of the summer/winter solstice.&#xA;However, on any given day, sunrise, noon, and sunset are all behind or ahead of mean solar time by approximately the same amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: the Earth's orbit is not quite circular.&#xA;Following &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Second_law_of_Kepler&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kepler's second law&lt;/a&gt;,&#xA;the Sun's ecliptic longitude changes ~3% slower at aphelion and faster at perihelion,&#xA;making the apparent solar day slightly shorter or longer respectively.&#xA;This &lt;a href=&quot;https://en.wikipedia.org/wiki/Equation_of_time#Eccentricity_of_the_Earth&amp;#39;s_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia entry&lt;/a&gt;&#xA;discusses it in more detail.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-03-23T19:34:18.097" LastActivityDate="2018-03-23T19:34:18.097" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25571" PostTypeId="1" CreationDate="2018-03-21T12:49:26.407" Score="0" ViewCount="95" Body="&lt;p&gt;What do people mean by the &quot;beam&quot; profile/model pertaining to&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Wilkinson_Microwave_Anisotropy_Probe&quot; rel=&quot;nofollow noreferrer&quot;&gt;WMAP&lt;/a&gt;?&#xA;Search results are all rather packed with jargon. Is  there a pictorial /layman explanation of it and how does it affect the data? &lt;/p&gt;&#xA;" OwnerUserId="22315" LastEditorUserId="616" LastEditDate="2018-03-21T21:26:14.640" LastActivityDate="2018-03-25T19:31:54.937" Title="WMAP beam profile" Tags="&lt;cmb&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25572" PostTypeId="1" CreationDate="2018-03-21T14:54:52.363" Score="1" ViewCount="1707" Body="&lt;p&gt;What simple mathematical function would reasonably well simulate sunlight intensity (disregarding weather) from dawn till dusk in relative numbers, 0-100 %. Are there significant differences (apart from maximum power and daylight duration) between temperate winter, summer or tropical days? The goal is more natural LED lighting for captive wild animals.&lt;/p&gt;&#xA;" OwnerUserId="22316" LastActivityDate="2018-04-06T19:26:14.817" Title="Sunlight intensity during a day" Tags="&lt;the-sun&gt;&lt;light&gt;" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="25574" PostTypeId="2" ParentId="25383" CreationDate="2018-03-21T17:08:26.727" Score="0" Body="&lt;p&gt;You need a model for the excitation of the line.  But since you say you have two forbidden lines, you can try assuming the density is well below the &quot;critical density&quot; (which is the density at which point the upper levels of the transitions get quenched by collisions instead of waiting as long as it takes for the &quot;forbidden&quot; radiative transition to occur).  Then if you are talking about two transitions from the same upper level, the line ratio is simply the ratio of the Einstein A values for the two transitions (note that the Einstein A value normally already includes the statistical weights of the lower levels).  If you are talking about two different upper levels, then you need a model for how those upper levels get populated, and the line ratios will be the ratios of the upper level populations times the Einstein A values.  It's not always easy to know how the upper levels get populated, so that's why it's hard to give you an answer to your question, it is very context dependent.  But the main simplification of using forbidden lines is that you don't have to worry about the photons getting reabsorbed before they get out.  (A detail is that line ratios are often given in terms of energy flux ratios, rather than photon ratios, so you'd need to multiply the Einstein A values by the photon energies if the transitions are at very different energies.)&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-03-21T17:08:26.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25577" PostTypeId="2" ParentId="25571" CreationDate="2018-03-21T21:49:52.317" Score="0" Body="&lt;p&gt;The &quot;beam profile&quot; is what a point source would look like when observed by the instrument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general the observed spatial distribution or profile of any source in a WMAP map will be the intrinsic profile or distribution of the source convolved with the beam profile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In layman's terms it is the amount of smoothing or blurring introduced into the WMAP pictures by limitations of the spatial resolution of the instrument.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-03-21T21:49:52.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25578" PostTypeId="1" CreationDate="2018-03-22T00:16:35.773" Score="-1" ViewCount="647" Body="&lt;p&gt;I am writing a one page essay front and back about the H-R diagram, I am supposed to describe what is on it and how we use them. But I don't know the difference between luminosity and absolute magnitude.&lt;/p&gt;&#xA;" OwnerUserId="22323" LastEditorUserId="11143" LastEditDate="2018-03-22T11:38:28.147" LastActivityDate="2018-03-22T11:46:01.813" Title="What is the difference between luminosity and absolute magnitude?" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25579" PostTypeId="2" ParentId="24576" CreationDate="2018-03-22T04:41:14.593" Score="1" Body="&lt;p&gt;Well...  maybe it's not a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If some kind of magnetic anomaly in the Sun was pushing plasma out in one direction, the object wouldn't have to be more massive than the sun to catch the spray.  The black spot could even be Mercury - any mass would capture an outbound plasma jet if the jet reached high enough.  This either pairs well with the sun being much too big or doubles down on the farce, depending on how much you know about stellar mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, that doesn't explain why it's black.  Maybe it's a small black hole.  I suppose if it was a small black hole, lighter than the sun, the Earth could have something like a conventional orbit.  But even in the absence of a plasma stream, I'm pretty sure the gravitational lens effect would mess about in visible ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or maybe flat earth theory is true in the show, and what we're really seeing is some kind of damage in the projection dome that encircles the earth!  It's not a black hole, it's the mother of all dead pixels.&lt;/p&gt;&#xA;" OwnerUserId="22325" LastActivityDate="2018-03-22T04:41:14.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25582" PostTypeId="1" AcceptedAnswerId="25621" CreationDate="2018-03-22T07:03:55.800" Score="2" ViewCount="164" Body="&lt;p&gt;The Gizmodo article &lt;a href=&quot;https://gizmodo.com/a-visiting-star-jostled-our-solar-system-70-000-years-a-1823954398&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Visiting Star Jostled Our Solar System 70,000 Years Ago&lt;/a&gt; refers to the the three year old ApJ. paper &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/2041-8205/800/1/L17/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Closest Known Flyby of a Star to the Solar System&lt;/a&gt; (also &lt;a href=&quot;https://arxiv.org/abs/1502.04655&quot; rel=&quot;nofollow noreferrer&quot;&gt;arXiv&lt;/a&gt;) which describes &lt;a href=&quot;https://en.wikipedia.org/wiki/Scholz%27s_star&quot; rel=&quot;nofollow noreferrer&quot;&gt;Scholz’s Star&lt;/a&gt; and it's pass close to our solar system about 70,000 years ago.&lt;/p&gt;&#xA;&lt;p&gt;The Gizmodo article then discusses the newly published paper in MNRAS &lt;a href=&quot;https://academic.oup.com/mnrasl/article-abstract/476/1/L1/4840245?redirectedFrom=fulltext&quot; rel=&quot;nofollow noreferrer&quot;&gt;Where the Solar system meets the solar neighbourhood: patterns in the distribution of radiants of observed hyperbolic minor bodies&lt;/a&gt;. However, while I can access the earlier paper, the new result is paywalled and I'm not able to access it (at least not soon).&lt;/p&gt;&#xA;&lt;p&gt;The second half of the abstract in particular is confusing me.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Here, we perform a systematic numerical exploration of the past orbital evolution of known hyperbolic minor bodies using a full N-body approach and statistical analyses to study their radiants. Our results confirm the theoretical expectations that strong anisotropies are present in the data. We also identify a statistically significant overdensity of high-speed radiants towards the constellation of Gemini that could be due to the closest and most recent known fly-by of a star to the Solar system, that of the so-called Scholz's star. In addition to and besides 1I/2017 U1 (‘Oumuamua), we single out eight candidate interstellar comets based on their radiants’ velocities.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I've heard of a radiant of a meteor shower - it's an apparent position of the source of meteors referenced to the position of the Earth at the time of the shower. But I've= never heard the term applied to the &lt;em&gt;orbit&lt;/em&gt; of a minor body unrelated to the Earth's position. What is the &amp;quot;radiant&amp;quot; of a minor body with a hyperbolic orbit in a simulation? Could it be the asymptotic incoming direction of the hyperbola?&lt;/p&gt;&#xA;&lt;p&gt;But my main question is simply how &amp;quot;statistically significant&amp;quot; is this result in reference to the enhancement in the direction of Gemini? I can't see the data. Is it just a tiny fluctuation, or a fairly obvious and robust feature?&lt;/p&gt;&#xA;&lt;p&gt;A screen shot of the data would be most welcomed!&lt;/p&gt;&#xA;&lt;p&gt;Possibly helpful: &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_hyperbolic_comets&quot; rel=&quot;nofollow noreferrer&quot;&gt;List of hyperbolic comets&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2020-07-25T06:08:44.770" LastActivityDate="2020-07-25T06:08:44.770" Title="How significant are the recent results that Scholz’s Star has perturbed several observed hyperbolic objects?" Tags="&lt;asteroids&gt;&lt;comets&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25583" PostTypeId="2" ParentId="25578" CreationDate="2018-03-22T11:46:01.813" Score="2" Body="&lt;p&gt;Luminosity is the total energy output of a star per unit time. For example, our own Sun has a luminosity of $3.828\times 10^{26}\ \mathrm{Joules/second}$. Our Sun outputs more energy in a single second than &lt;a href=&quot;https://en.wikipedia.org/wiki/Electric_energy_consumption&quot; rel=&quot;nofollow noreferrer&quot;&gt;the entire world consumes in a year&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absolute magnitude is, in some ways, analogous to luminosity in that it is a measure of the energy output of a star. The definition of absolute magnitude is that it is the magnitude a star would have if it were 10 parsecs from an observer. Obviously understanding this definition requires understanding the &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_magnitude&quot; rel=&quot;nofollow noreferrer&quot;&gt;magnitude system&lt;/a&gt;, which is outside the scope of the question and I will not be explaining it here.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2018-03-22T11:46:01.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25584" PostTypeId="1" AcceptedAnswerId="25588" CreationDate="2018-03-22T18:09:17.470" Score="2" ViewCount="204" Body="&lt;p&gt;Sometime in the summer months during the years from 1993 to 1998, my friend and I saw what looked like a gamma-ray burst (or what looked like an artist's impression of it). It looked like a star that got very bright (within a few seconds) and we saw what looked like a line shooting out from the sides of it that spread across our field of vision. All this happened in less than 10 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't confirm when it was any more exactly than that, unfortunately, as we were walking near my friends house at night in the summer, as we did hundreds of times in those years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia, the first naked-eye brightness gamma ray burst happened in 2008, so I imagine that's not it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What &lt;em&gt;could&lt;/em&gt; it have been?&lt;/p&gt;&#xA;" OwnerUserId="22340" LastActivityDate="2018-03-23T17:20:05.180" Title="Bright light in the sky, in the '90s, what could it have been?" Tags="&lt;amateur-observing&gt;&lt;light&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25585" PostTypeId="1" CreationDate="2018-03-22T18:24:25.887" Score="1" ViewCount="67" Body="&lt;p&gt;According to me following reason of mass accumulation of earth has been completely overlooked so far.&#xA;The sunrays that fall on the oceans travel deep into water and finally get converted into mass which eventually results in &quot;mass-accretion of the earth&quot;. So earth is continuously getting heavier due to the mass being added to it due to this reason year by year. &#xA;Therefore, should this also not be one of the reasons why the orbit-time of the earth (time taken by it to take one round of the sun) is bound to change (increase or decrease) with time? &lt;/p&gt;&#xA;" OwnerUserId="13871" LastEditorUserId="13871" LastEditDate="2018-03-22T18:30:19.460" LastActivityDate="2018-03-22T22:01:01.237" Title="Mass accumulation rate and orbit-time of earth" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25588" PostTypeId="2" ParentId="25584" CreationDate="2018-03-22T20:29:53.013" Score="3" Body="&lt;p&gt;Gamma ray bursts are very rarely bright enough in visible light to be seen with the naked eye. This is why they weren't detected until there was a network of satellites looking for atom bombs. You didn't see a GRB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If what you saw was &lt;em&gt;Bright&lt;/em&gt;, and &lt;em&gt;Fast&lt;/em&gt; it was likely to be a bright meteor, sometimes called a fireball. These appear as a very bright, fast-moving &quot;shooting star&quot;. They are rare, but not so rare that it would be possible to identify a particular event from the information that you give.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reflection from a satellite can look like a star that brightens and then fades. The Iridium satellites are well known for this though other satellites can also produce &quot;flares&quot;. Iridium flares have been visible since the late 90s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other mundane possibilities, but these would seem to be the most likely. Remember memory is a funny thing. After 20+ years thing that you &quot;remember&quot; might be different.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-03-23T17:20:05.180" LastActivityDate="2018-03-23T17:20:05.180" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25590" PostTypeId="2" ParentId="25585" CreationDate="2018-03-22T21:35:37.383" Score="7" Body="&lt;p&gt;OK. Let's do the numbers. The Earth presents an circular &quot;profile&quot; of radius about 6 million meters to the Sun so it's cross-sectional area is about $10^{14} m^2$. The solar irradiance at the top of the atmosphere is about 1400 $W/m^2$ so the total power received from the sun is about $10^{17}W$, by coincidence, more-or-less equal to $c^2$ in SI units. So the mass equivalent of the sunlight received by Earth is about $1 kg/s$ or about $3 \times 10^7 kg/year$. This is about one part in $10^{17}$ of the mass of the Earth per year, so even if were all absorbed and none of it re-radiated (which would be thermodynamically impossible) the increase in mass is so small that the orbital change would be unmeasurable. In fact as @James K has already pointed out in a comment, more or less the same amount of energy (and therefore mass) is radiated away as heat radiation, so even this tiny change does not really occur.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-03-22T21:35:37.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25592" PostTypeId="2" ParentId="25585" CreationDate="2018-03-22T22:01:01.237" Score="5" Body="&lt;p&gt;Steve Linton is of course correct, but I just wanted to point out a very basic misconception in your post. The orbital radius and period of the Earth hardly depend on the mass of the Earth at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The semi-major axis of the Earth's orbits depends (via Kepler's third law and conservation of angular momentum) on the &lt;em&gt;sum&lt;/em&gt; of the mass of the Sun and the mass of the Earth (the former of course being many orders of magnitude larger).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun is losing mass continually via radiation and via the solar wind. This is lost at a rate of $-9.3\times 10^{-14}$ solar masses per year ( &lt;a href=&quot;https://arxiv.org/abs/0801.3807&quot; rel=&quot;noreferrer&quot;&gt;Noerdlinger 2008&lt;/a&gt;). This results in the growth of the Earth's orbit by about 1.4 cm per year. Any mass accreted (or indeed lost by for instance the loss of atmosphere into space) has a &lt;em&gt;totally negligible&lt;/em&gt; effect on the Earth's orbit compared to this.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-03-22T22:01:01.237" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25593" PostTypeId="2" ParentId="25584" CreationDate="2018-03-22T22:14:34.350" Score="1" Body="&lt;p&gt;If what you saw occurred within an hour or so of dawn or dusk, then it sounds like an &lt;a href=&quot;https://www.wikihow.com/Find-an-Iridium-Flare&quot; rel=&quot;nofollow noreferrer&quot;&gt;Iridium flare&lt;/a&gt;. This is reflected sunlight off the large, reflective panel-like antennas of one of the Iridium communication satellites. Some of these can be particularly impressive, becoming much brighter than Venus over a few seconds and then smoothly fading away again on a similar timescale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The link I posted above gives you instructions for how to observe one again if you wish to confirm that this is indeed what you saw, though your opportunity to do so may &lt;a href=&quot;http://www.skyatnightmagazine.com/news/iridium-flares-cease-end-2018&quot; rel=&quot;nofollow noreferrer&quot;&gt;cease by the end of this year&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture below appeared in the &quot;Sky at Night&quot; magazine and was taken by Nikki Young.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, these satellites first appeared in the sky in the late 90s (1997), so perhaps you saw glints/flares from something else before then, but these would not have been as bright.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eM7cm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eM7cm.jpg&quot; alt=&quot;An iridium flare&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-03-22T22:22:08.347" LastActivityDate="2018-03-22T22:22:08.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25594" PostTypeId="2" ParentId="700" CreationDate="2018-03-23T05:49:25.870" Score="-2" Body="&lt;p&gt;It is said that neutron stars bend space/time so strongly that parts of the back are visible from the front! Of course a neutron star is essentially one very &lt;em&gt;very&lt;/em&gt; large ball of neutrons with all the light elements on the surface. Some scientists now believe that simple neutron star collisions do not generate all heavy elements but the existence of elements heavier than iron is due to black hole-neutron star collisions. If so then they don't have an event horizon despite their enormous gravity because the matter is too spread out, whereas for a true black hole its all concentrated in one place.&#xA;In fact it is believed that the escape velocity for a typical neutron star is around 1/3 to 1/2 the speed of light, still a large number and incidentally life &lt;em&gt;may&lt;/em&gt; be possible on a planet orbiting a neutron star with radiation tolerance sufficient even in a bacterium like Deinococcus radiodurans as long as the planet's orbit kept it well away from the jets.&#xA;A variant of this concept is when a neutron star hits a red supergiant briefly igniting helium fusion if the whole thing doesen't blow up first.&lt;br&gt;&#xA;&lt;a href=&quot;https://arstechnica.com/science/2014/06/red-supergiant-replaced-its-core-with-a-neutron-star/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arstechnica.com/science/2014/06/red-supergiant-replaced-its-core-with-a-neutron-star/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22350" LastEditorUserId="22350" LastEditDate="2018-03-25T07:53:21.547" LastActivityDate="2018-03-25T07:53:21.547" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25595" PostTypeId="1" AcceptedAnswerId="25598" CreationDate="2018-03-23T06:05:40.680" Score="-1" ViewCount="107" Body="&lt;p&gt;When we see a star from the earth that is 5 light years away, we are essentially looking at light emitted from the star 5 years ago. What if you were to instantly reach near the star, either by teleportation or Alcubierre drive, are you traveling back in time 5 years?&lt;/p&gt;&#xA;" OwnerUserId="22351" LastActivityDate="2018-03-23T13:26:04.073" Title="What if one teleportes to a star that is some light years away?" Tags="&lt;time-dilation&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2018-03-23T15:48:21.240" ContentLicense="CC BY-SA 3.0" />
  <row Id="25598" PostTypeId="2" ParentId="25595" CreationDate="2018-03-23T10:39:05.947" Score="0" Body="&lt;p&gt;Well not sure teleporting requires an answer.  There is not such a physics. However assuming the impossible I do not think we really need relativity SR in a technical sense. If I teleport instantaneously myself I would appear on the star as it will be seen from earth in 5 years. As such, in the earth calendar, I disappear for 5 y and &quot;in a way&quot; I move in the earth future as for an earth observer it will take 5 y to see me again. But he/she will see me as I was during my absence, ie his past. As far the star is concerned, I simply materialize there. The point is that from there,  I will see my last 5 y on earth as a &lt;em&gt;recorded&lt;/em&gt; movie. The same is for a star resident watching the movie with me. He see me before leaving so I wasn't able to give to him any sensible information. Even a pre alert of my arrival wouldn't alter anything on the succession of events.  This does not qualify as moving or moving information through time, especially not backwards.&#xA;At the end, the fictional teleporting seems to lead to less mind blowing results than a technological difficult but in principle possible travel at a relativistic speed. There is even no room for time dilation and it seems to me that the famous twins can use teleporting and re-meet with no even (apparent) paradox.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit.  Please note that the scenario can be made fully realistic (well almost, but because of &lt;em&gt;merely technological&lt;/em&gt; limits) by replacing&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;earth and star by widely separated islands; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;sound as the only way to communicate, such as big detonation while being away and whispering while sitting on the same rock; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a really supersonic vehicle for moving between the two islands, so that the time of the person travel can be taken to be negligible as compared to the time at sound speed. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Nothing can alter a cause effect succession of events even considering a third observer passing by or through the islands even on board a similar vehicle. &lt;/p&gt;&#xA;" OwnerUserId="17437" LastEditorUserId="17437" LastEditDate="2018-03-23T13:26:04.073" LastActivityDate="2018-03-23T13:26:04.073" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25600" PostTypeId="1" CreationDate="2018-03-23T17:01:39.907" Score="6" ViewCount="296" Body="&lt;p&gt;I know that the Earth-Sun Lagrange L1, L2, and L3 points are not considered stable over longer periods, especially when compared to L4 and L5... But, with the moon orbiting the Earth in the general path of the ecliptic, it would seem that the first time the moon got in between the Earth and the object at the L1 or L2 points, it would be enough to perturb the object a good bit, and the next lunar month would be even worse, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that if the moon was taken out of the equation, the L1 and L2 points would be much more stable. And, as a result of this line of thinking, I would think that the L3 point would be much more stable than L1 and L2, since the moon would be so much farther away and unimportant, so its effect could be ignored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I wrong in thinking this? Is the moon's influence not large enough to be a major factor? And, just to clarify, I am talking specifically about L1 and L2 with respect to a large 3rd body, like the moon, instead of some idealized 2 body only system.&lt;/p&gt;&#xA;" OwnerUserId="22362" LastActivityDate="2018-03-30T07:52:30.683" Title="How can Earth-Sun Lagrange points L1 &amp; L2 even be semi stable considering the moon?" Tags="&lt;the-moon&gt;&lt;earth&gt;&lt;orbital-mechanics&gt;&lt;celestial-mechanics&gt;&lt;lagrange-point&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25602" PostTypeId="1" CreationDate="2018-03-23T22:03:00.260" Score="1" ViewCount="97" Body="&lt;p&gt;I am having some issues with understanding what my adviser means by &quot;galactic roll,&quot; but I have not been able to find anything online regarding this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My project involves generating an array of az/el coordinates, converting them to ra/dec, then converting them to galactic lon/lat. I then use these coordinates to scan a Healpy map (all in Python).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She did mention starting off by setting horizontal roll = 0, and then converting horizontal coordinates to galactic coordinates. The galactic roll should then take on nonzero values, which is important to the data I am analyzing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read about Euler angles and quaternions and I think I'm on the right path. She provided some code for me to reference, and that uses quaternions and Euler angles. Does galactic roll refer to one of the Euler angles (perhaps $\theta$)? Another reference code she provided mentions parallactic angle; could that be it? Any help or reference material would be appreciated. Thank you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And before you mention this, yes I will ask her. I just want to give this an honest attempt first.)&lt;/p&gt;&#xA;" OwnerUserId="22366" LastActivityDate="2018-03-23T22:03:00.260" Title="Unsure how to obtain galactic roll?" Tags="&lt;coordinate&gt;&lt;python&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25603" PostTypeId="1" AcceptedAnswerId="25614" CreationDate="2018-03-23T23:27:17.480" Score="1" ViewCount="255" Body="&lt;p&gt;Will it suck the entire universe in? What will the black hole look like to us, assuming we do not immediately get sucked in?&lt;/p&gt;&#xA;" OwnerUserId="15166" LastEditorUserId="15166" LastEditDate="2018-03-23T23:48:08.680" LastActivityDate="2018-03-26T09:58:01.260" Title="If a black hole has a mass of a universe what will be the volume of it?" Tags="&lt;black-hole&gt;&lt;supermassive-black-hole&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25607" PostTypeId="2" ParentId="25603" CreationDate="2018-03-24T05:06:58.130" Score="0" Body="&lt;p&gt;Before I answer this, it's important to correct a few assumptions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) we can sit outside a universe-as-black-hole. This is impossible: since the universe includes everything that exists, then by definition we must be within it, so we can't look at it from the &quot;outside&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) a black hole &quot;sucks&quot; matter into it. It doesn't, any more than a large star &quot;sucks&quot; matter. If the Sun somehow collapsed and became a black hole (it can't, this is just a thought experiment), all the planets would continue to orbit pretty much as usual, since the Sun's mass wouldn't have changed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to your core question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If a black hole has a mass of a universe what will be the volume of&#xA;  it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schwarzschild radius&lt;/a&gt; is the radius defining the event horizon of a Schwarzschild black hole. If we take the mass of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Observable_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;observable universe&lt;/a&gt; as roughly $10^{53}$ kg, then using the formula $$R=\frac{2GM}{c^2}$$ the Schwarzschild radius of this mass is 15.7 billion light years [NB: by comparison, the comoving distance to the edge of the observable universe is about 46.6 billion light years]. The volume is then easily calculated as 1.6 x $10^{31}$ cubic light years or roughly $10^{79}$ $m^3$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparison, this is less than 4% of the volume of the observable universe.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia's &quot;quick facts&quot; on the observable universe give the mass as $10^{53}$ kg, but &lt;a href=&quot;https://en.wikipedia.org/wiki/Observable_universe#Mass_of_ordinary_matter&quot; rel=&quot;nofollow noreferrer&quot;&gt;the body of the article&lt;/a&gt; contains the following qualification:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The mass of the observable Universe is often quoted as $10^{50}$&#xA;  tonnes or $10^{53}$ kg. In this context, mass refers to ordinary&#xA;  matter and includes the interstellar medium (ISM) and the&#xA;  intergalactic medium (IGM). However, it excludes dark matter and dark&#xA;  energy. This quoted value for the mass of ordinary matter in the&#xA;  Universe can be estimated based on critical density. The calculations&#xA;  are for the observable universe only as the volume of the whole is&#xA;  unknown and may be infinite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My calculations are based on the mass of ordinary matter in the observable universe, representing 4.9% of the total &quot;mass/energy&quot; derived from the observed critical density and volume. Rob's answer includes dark matter (26.8% of total mass/energy) and dark energy (68.3% of total mass/energy). Both answers are thought experiments, since it's not possible to have a black hole with the mass of the universe &lt;em&gt;within our universe&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a &lt;a href=&quot;https://astronomy.stackexchange.com/questions/25603/if-a-black-hole-has-a-mass-of-a-universe-what-will-be-the-volume-of-it#comment44017_25603&quot;&gt;comment on the main question&lt;/a&gt;, userTLK makes an additional valid point that &quot;the escape velocity at [a black hole's] event horizon is &lt;em&gt;c&lt;/em&gt;. Dark energy might make that impossible. The vast distance and stretching of space and red-shifting of distant objects might make universe sized black holes impossible.&quot; &lt;/p&gt;&#xA;" OwnerUserId="14094" LastEditorUserId="14094" LastEditDate="2018-03-25T00:17:21.127" LastActivityDate="2018-03-25T00:17:21.127" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25608" PostTypeId="1" CreationDate="2018-03-24T06:32:00.713" Score="-4" ViewCount="796" Body="&lt;p&gt;How to calculate declination of the Sun if its right ascension is known?&lt;br&gt;&#xA;I am BSc (NS) student.&lt;/p&gt;&#xA;" OwnerUserId="22372" LastEditorUserId="7472" LastEditDate="2018-03-25T22:11:26.177" LastActivityDate="2018-03-25T22:11:26.177" Title="How to calculate declination of the Sun if its right ascension is known?" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2018-03-30T12:20:56.503" ContentLicense="CC BY-SA 3.0" />
  <row Id="25609" PostTypeId="2" ParentId="24576" CreationDate="2018-03-24T07:11:25.647" Score="3" Body="&lt;p&gt;I'm not sure this needs another answer, but I got hooked on the show, so why not.   (spoilers ahead).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Black Hole&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We don't know that it's a black hole, even with the image.  It kind of looks like a black hole drawing matter from the star, but there's a few problems with the image, and I'll get to that later.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple times in the show Elaine Renko (Agyness Deyn) says &quot;I don't know what it is&quot;.   She's presumably seen more of the data than anyone alive outside of the government insiders, so the show is careful not to commit to a specific scenario and the image only gives a little bit of information, so it could be a number of things.   Because they never said what it is, it's harder to criticize the science. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sudden appearance&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sudden appearances happen.  That's not to say something suddenly appears out of thin air, er, space, or quantum improbability or some kind of other dimension nonsense, but &quot;it suddenly appeared&quot; is actually somewhat reasonable because space is big and dark and something going unnoticed until it's right on top of us, is in the realm of possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Chelyabinsk_meteor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chelyabinsk meteor of 2013&lt;/a&gt; is an example of a destructive object just &quot;appearing&quot; in the sky because, as a sun-side meteor it was harder to see and it was smaller than the meteors that NASA tracks, so nobody saw it coming until it was a bright flash in the sky.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why NASA but nobody else.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;NASA has the best telescopes, though there are some very good ones around the globe and numerous amateur astronomers, so that particular aspect may not be probable but it's theoretically possible.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Firstly the show is set in London and everyone seems amazed when the&#xA;  image below rises from the dusk yet clearly it's been in existence for&#xA;  the time taken to create the spiral (hours at least?) so nations east&#xA;  of the UK would have seen it for that time and reported on it globally&#xA;  surely?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't like the spiral.  It's pretty but I don't like it.  (I'll get to that shortly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the suddenness of the visual at the London sunrise, consider what we see.   Two cops are chasing their suspect.   Other cops are in the garage with their boss who was shot a bit before.   Some paramedics are there, taking care of the shot person.   The cops chasing grace aren't going to stop and check their phones, so it's entirely possible that images began flooding the internets minutes before and the people in the show weren't aware.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How long was it visible?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're correct, if it's visible in London at sunrise, it's visible east of London earlier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically everywhere on Earth where you see the sun, you're seeing the same side of the sun.  The sun is in different parts of the sky but we all see the same side of the sun, so, unless there's clouds blocking the view, everyone east of London, still should have seen what they saw.   That raises the question, how long was the object visibly affecting the sun because it couldn't have been long or word would have spread.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The object appears in that image to be about one solar radii distance from the Sun's surface, assuming the angle is somewhat straight on, not viewed sideways, then we can estimate it's distance from the sun to be about 1 solar radii or 700,000 km.   The black dot is also quite small, maybe planet size, certainly smaller than Jupiter (which is 1/10th the diameter of the Sun), it appears larger than Earth (1/100th the diameter of the sun) and that would make it too large to be a black hole or neutron star, but perhaps a white dwarf, super earth, mini gas giant planet or planetary core fragment.  There's a fair bit of wiggle room in there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solar gravity, basically the escape velocity in reverse would have added about 430 km/s to whatever initial velocity relative to the sun that the object had as it approached the solar-system.  Most near-by exo-solar system objects have a relative velocity to the sun in the 1s or 10s of km/s, at least for those objects that orbit in the same direction through the milkyway, but hyper accelerated objects can reach into the 100s of kms per second relative to the Sun.    If that was a hyper-accelerated object approaching from the opposite side of the sun than the Earth, it's possible that it would be both difficult to see, and that it approached the solar-system very quickly.   The slower something moves, the more time there is to see it and it's gravitational influence, but a hyper-velocity, behind the sun approach might work, either impacting or making a close pass by the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that object is close to the sun, which it appears to be, even at hyper velocity, to the naked eye, it would appear largely still.   If we give this object a relative velocity to the sun of lets say 700-1000 km/s.  It would still take, from our perspective, 700-1,000 seconds (12-16 minutes), to travel one solar radii or 1/4 of 1 degree across the sky.   That, in theory, means that it might have only been visible for 12-16 minutes, or less if it was traveling faster.    That's a short enough time-frame that the cops could have chased their suspect unaware of the images that began flooding the internet.   A 12-16 minutes actually works and is theoretically possible, if hugely improbable.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My real question is more around is there any potential scenario where&#xA;  a black hole of that size could develop suddenly and get to that size&#xA;  quickly enough for that image to be a surprise? My understanding, very&#xA;  much as a scientific-layman, is that a black hole of that size would&#xA;  have to have the mass of at least one of our suns to get to that size&#xA;  (probably more) - is that a valid assumption?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're right.   Black holes have a minimum mass of about 2 and change solar masses and that much mass simply wouldn't go unnoticed, even with it's tiny size, it's tidal effects on the orbiting planets would be noticed even by armature astronomers  &quot;Hey, Neptune isn't where it's supposed to be&quot; - that kind of thing would get noticed.  Also, as it passed close to the sun it wouldn't just draw a spiral off the sun but it would make the sun distinctly egg shaped.   That much mass wouldn't work, but we don't know the objects mass.   We don't know if it's a black hole or what it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes don't suddenly develop either.  They're made in the cores of very large stars when they go supernova.   So your thinking is correct there too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where has the mass come from? Also wouldn't the sudden existence of&#xA;  something else with at least the same mass as our sun suddenly&#xA;  appearing at the centre of our solar system have enormous&#xA;  gravitational impact on everything in the system, including Earth?&#xA;  We'd have massive volcanic/tectonic impact wouldn't we?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The harm from a stellar mass object passing through the solar-system would be that it would change the planet's orbits and changing the Earth's orbit could be bad, pushing us outside our Goldilocks zone, but a 1 or 2 solar mass object would need to get pretty close to the Earth to cause the kind of tidal effects you're talking about.   That object appears close to the sun, and we don't know it's mass so the kind of tidal effects you're talking about aren't a problem.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Spiral&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't like the spiral because it looks like the object is pulling gas away from the Sun, but it's not changed the shape of the sun.  The spiral also appears well developed which would take some time.   I don't want to say it's impossible but the spiral around the object, while pretty, seems the least scientific part of the event to me.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Apologies for the long answer, but I wanted to point out that an unnoticed, very fast effect is theoretically possible, just hugely improbable.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-03-25T11:58:43.087" LastActivityDate="2018-03-25T11:58:43.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25611" PostTypeId="2" ParentId="25608" CreationDate="2018-03-24T09:40:28.547" Score="2" Body="&lt;p&gt;As discussed &lt;a href=&quot;https://en.wikipedia.org/wiki/Position_of_the_Sun#Overview&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, the path of the Sun &lt;em&gt;resembles&lt;/em&gt; a sine wave (see the image below), with a (semi-)amplitude equal to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_tilt#Earth&quot; rel=&quot;nofollow noreferrer&quot;&gt;axial tilt of the Earth&lt;/a&gt;, usually denoted $\varepsilon$, about 23.4 degrees. A simple formula, which might be accurate for your purposes, is the following: if $a$ denotes the right ascension (in hours), the declination $\delta$ is given by the following formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta = \varepsilon \sin(2\pi\frac{a}{24h})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual formula is more complicated. From the &lt;a href=&quot;https://en.wikipedia.org/wiki/Position_of_the_Sun#Equatorial_coordinates&quot; rel=&quot;nofollow noreferrer&quot;&gt;formulas linked to by @barrycarter&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha = \arctan(\cos\varepsilon \tan\lambda)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta = \arcsin(\sin\varepsilon \sin\lambda)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan\alpha = \cos\varepsilon \tan\lambda$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\tan\alpha}{\cos\varepsilon} = \tan\lambda$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda = \arctan\frac{\tan\alpha}{\cos\varepsilon}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging this into the declination formula, and using the fact that $\sin(\arctan(x)) = \frac{x}{\pm\sqrt{x^2+1}}$, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta = \arcsin(\sin\varepsilon \frac{\tan\alpha}{\pm\sqrt{\tan^2\alpha+\cos^2\varepsilon}})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(here the right ascension $\alpha$ is measured in radians or degrees, not hours).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3HptH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3HptH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(picture taken from &lt;a href=&quot;https://en.wikipedia.org/wiki/Ecliptic#/media/File:Constellations_ecliptic_equirectangular_plot.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, originally by Cmglee, Timwi)&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="7472" LastEditorUserId="7472" LastEditDate="2018-03-24T14:18:13.237" LastActivityDate="2018-03-24T14:18:13.237" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25614" PostTypeId="2" ParentId="25603" CreationDate="2018-03-24T20:44:45.830" Score="1" Body="&lt;p&gt;According to measurements of the cosmic microwave background the universe is geometrically flat - which means that the mass/energy density of the universe is close to the &quot;critical value&quot; of $\sim 10^{-26}$ kg/m$^{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radius of the observable universe is 46.6 billion light years, so the mass/energy contained within it is equivalent to $3.6\times 10^{54}$ kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Schwarzschild radius of a black hole is $2GM/c^2$. If the mass/energy of the universe is spherically symmetric then its Schwarzschild radius is 560 billion light years and thus larger than the observable universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note though that the Schwarzschild solution in General Relativity is &lt;em&gt;static&lt;/em&gt;. The universe is definitely not static.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-03-26T09:58:01.260" LastActivityDate="2018-03-26T09:58:01.260" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25616" PostTypeId="1" CreationDate="2018-03-25T00:37:38.150" Score="1" ViewCount="859" Body="&lt;p&gt;How to convert Ecliptic co-ordinates to Galactic? The wikipedia article on &lt;a href=&quot;https://en.wikipedia.org/wiki/Celestial_coordinate_system#Converting_coordinates&quot; rel=&quot;nofollow noreferrer&quot;&gt;Celestial Coordinate systems contains various conversions&lt;/a&gt;, but this one is currently absent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more specific example: I would like to convert Proper motion in ecliptic latitude and longitude from the ATNF catalog into proper motion in galactic coordinates. Using the following notation from &lt;a href=&quot;http://www.atnf.csiro.au/research/pulsar/psrcat/psrcat_help.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;the catalog documentation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;PMElong:     Proper motion in the ecliptic longitude direction (mas/yr) &#xA;PMElat:      Proper motion in ecliptic latitude (mas/yr) &#xA;...&#xA;ZZ:          Distance from the Galactic plane, based on Dist&#xA;XX:          X-Distance in X-Y-Z Galactic coordinate system (kpc)&#xA;YY:          Y-Distance in X-Y-Z Galactic coordinate system (kpc)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How would I calculate the following (my own addition) from &lt;code&gt;PMElong&lt;/code&gt; &amp;amp; &lt;code&gt;PMElat&lt;/code&gt;?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;PMZZ:        Proper motion perpendicular to Galactic plane&#xA;PMXX:        X-direction proper motion in X-Y-Z Galactic coordinate system (kpc/yr)&#xA;PMYY:        Y-direction proper motion in X-Y-Z Galactic coordinate system (kpc/yr)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="22382" LastActivityDate="2018-03-26T21:36:20.017" Title="Convert Ecliptic to Galactic Coordinates" Tags="&lt;coordinate&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25618" PostTypeId="1" AcceptedAnswerId="25619" CreationDate="2018-03-25T05:48:51.517" Score="0" ViewCount="71" Body="&lt;p&gt;All the nearby galaxies we see come in visible range, many are visible from ground based telescopes. And the farthest galaxy that hubble discovered recently, 13.4 billion light years away is in infra red.&#xA;Was this Galaxy ever in visible part of spectrum? &#xA;How long will it take galaxies like andromeda or whirlpool to go infrared so that they won't be visible to us if we are not using in infraRed lens?&#xA;What's the specific limit where the light makes transition from visible to infrared part of spectrum? &lt;/p&gt;&#xA;" OwnerUserId="22385" LastActivityDate="2018-03-25T08:02:01.227" Title="What's that limit, where we start to see infrared galaxies?" Tags="&lt;galaxy&gt;&lt;amateur-observing&gt;&lt;hubble-telescope&gt;&lt;infrared&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25619" PostTypeId="2" ParentId="25618" CreationDate="2018-03-25T08:02:01.227" Score="1" Body="&lt;p&gt;Galaxies give off a range of frequencies of light: Visible, but also infrared, microwave, and radiowaves. at longer wavelengths, ultraviolet, Xrays and gamma rays at short wavelengths. But stars give off quite a lot of their radiation in visible light, so nearby galaxies are quite bright in visible light and rather less bright in Ultraviolet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very distant galaxies are moving away from us due to the general expansion of the galaxy. This means that the light is stretched. Visible light gets stretched to infra-red, ultraviolet gets stretched to visible light. The distant galaxy is very very far away, and so very dim. It is a little bit brighter in infrared, although there is also a little visible light (and other wavelengths too) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxies never go completely invisible in visible light, even if they are red-shifted, because there is always ultraviolet light to red-shift to visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific limit is often taken to be 700nm as the boundary between red and infrared. Although this is fuzzy because some people can see a little further than others. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-03-25T08:02:01.227" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25621" PostTypeId="2" ParentId="25582" CreationDate="2018-03-25T11:46:27.953" Score="1" Body="&lt;p&gt;Not an answer, but just on radiants:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radiant means angle of approach.   It's applied to meteor showers because the Earth intersects the dust trail that causes the meteor shower at a specific place in orbit and dust trail has a specific angle of intersection relative to Earth and Earth's orbit.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2Pn13.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2Pn13.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://briankoberlein.com/2015/08/10/why-do-we-have-meteor-showers/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source of picture.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the &quot;radiant&quot; of a minor body with a hyperbolic orbit in a&#xA;  simulation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Radiant can't be applied to elliptical orbits though extremely elliptical orbits kind of have a radiant, but it can be applied to objects approaching the solar system.  A hyperbolic orbit generally implies an object that came from outside the solar-system though it could also be a gravity assisted object from inside the solar system that was accelerated from an elliptical to a hyperbolic orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an &lt;a href=&quot;https://www.imo.net/a2017-u1-first-interstellar-asteroid-ever-detected/&quot; rel=&quot;nofollow noreferrer&quot;&gt;article that mentions radiant in association with an exo-solar object&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;astronomers looked for the radiant it seemed to originate from.&#xA;  Calculations determined that A/2017 U1 came from a radiant located&#xA;  close to (R.A. = 18h 41m 25s ; Dec=+34° 15′) coordinates, which is&#xA;  near the actual position of Vega&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;exo-solar objects approach the solar system in a mostly straight line until they get close enough for the sun to bend their path.   Where they came from before their direction was curved by the sun's gravity is their radiant. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-03-25T14:10:06.023" LastActivityDate="2018-03-25T14:10:06.023" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25622" PostTypeId="1" CreationDate="2018-03-25T13:34:06.927" Score="1" ViewCount="123" Body="&lt;p&gt;I read that a blackhole has an entropy that is given by the area of its event horizon (measured in Planck units). Does that mean that this information is actually physically stored there? If yes, how can that be since there may be nothing there at all (no electrons for example that could carry this information in terms of spin up or spin down)? Is it perhaps stored in the vacuum somehow ? &lt;/p&gt;&#xA;" OwnerUserId="22388" LastEditorUserId="907" LastEditDate="2018-03-26T14:48:57.620" LastActivityDate="2018-03-26T16:25:25.687" Title="How is the entropy of a blackhole stored at its horizon?" Tags="&lt;black-hole&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25623" PostTypeId="2" ParentId="25571" CreationDate="2018-03-25T19:31:54.937" Score="2" Body="&lt;p&gt;The beam profile of a radio telescope characterizes its sensitivity to signals as a function of angle off axis.&#xA;This is usually divided into a main beam in the target direction and sidelobes in other directions.&#xA;As this figure from&#xA;&lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0067-0049/180/2/246&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hill et al. 2009&lt;/a&gt;&#xA;shows, the WMAP main beams and near sidelobes were rather complex:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CSWIb.png&quot; alt=&quot;WMAP main beams and near sidelobes&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image credit: WMAP Science Team. Each square covers 10x10 degrees.&#xA;Top row: beam maps based on multiple observations of Jupiter.&#xA;Middle row: beam models fitted to the observed maps.&#xA;Bottom row: residual differences between top and middle.&#xA;Left and right columns: dishes A and B, which point in different directions.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without further processing, the image of a point source resembles the beam profile.&#xA;To reconstruct a clear image, astronomers can&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Deconvolution&quot; rel=&quot;nofollow noreferrer&quot;&gt;deconvolve&lt;/a&gt;&#xA;the observed data with the beam profile.&#xA;Lest this operation amplify noise from the observed beam map,&#xA;they use a smooth mathematical model of it instead.&#xA;The WMAP team developed their beam model using physical optics software called DADRA (Diffraction Analysis of a Dual Reflector Antenna).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The far sidelobes were especially important for WMAP.&#xA;The sky has several radio sources, e.g. our galactic center,&#xA;much stronger than the desired background signal.&#xA;To minimize foreground contamination,&#xA;the WMAP team made an all-sky sidelobe map&#xA;based on multiple observations of the Moon.&#xA;Then they made a mask of the strong sources,&#xA;convolved it with the sidelobe map,&#xA;and subtracted the result from the observed signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For details of how and why they did all that,&#xA;see the &quot;Ancillary Data&quot; chapter of the&#xA;&lt;a href=&quot;https://lambda.gsfc.nasa.gov/product/map/dr5/pub_papers/nineyear/supplement/WMAP_supplement.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;WMAP 9-year Explanatory Supplement&lt;/a&gt;&#xA;and the&#xA;&lt;a href=&quot;https://lambda.gsfc.nasa.gov/product/map/dr5/map_bibliography.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;other publications&lt;/a&gt;&#xA;cited there.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-03-25T19:31:54.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25624" PostTypeId="2" ParentId="25616" CreationDate="2018-03-25T22:02:37.970" Score="2" Body="&lt;p&gt;Probably it's easiest to use some package for this, for example, if you use Python, you can use astropy:&#xA;&lt;a href=&quot;http://docs.astropy.org/en/stable/generated/examples/coordinates/plot_galactocentric-frame.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://docs.astropy.org/en/stable/generated/examples/coordinates/plot_galactocentric-frame.html&lt;/a&gt;.&#xA;Or if you are interested, here is a description on how to do it by hand, including the propagation the uncertainties:&#xA;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1987AJ.....93..864J&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/1987AJ.....93..864J&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you also need radial (line-of-sight) velocities, but I don't know if that is in the ATNF catalogue...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;About the radial velocities: we only see the pulsars move across the sky and know the distance, so we get two velocity components perpendicular to our line of sight, a $v_\alpha$ and a $v_\delta$, from the proper motion in Right Ascension ($v_\alpha=d\cdot \mu_\alpha$) and the propermotion in Declination ($v_\delta=d\cdot\mu_\delta$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably, it is useful for you to convert these proper motions from $\mu_\alpha, \mu_\delta$ to galactic longitude ($l$) and latitude ($b$) $\mu_l,\mu_b$, so you can get the velocities in cartesian coordinates: $v_y=d\cdot\mu_l$ and $v_z=d\cdot\mu_b$. This way, you have at least the two velocity components perpendicular to our LOS. Oh, and you maybe also want to subtract the rotation of the sun around the MW center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are ways to estimate the radial velocity sometimes: if the pulsar lives in a binary system, a spectrum of the companion maybe. Or a very crude way would be maybe something like &lt;a href=&quot;http://adsabs.harvard.edu/full/1977ApJ...216..842H&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/full/1977ApJ...216..842H&lt;/a&gt;... But I don't know much about it. &lt;/p&gt;&#xA;" OwnerUserId="22391" LastEditorUserId="22391" LastEditDate="2018-03-26T21:36:20.017" LastActivityDate="2018-03-26T21:36:20.017" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25625" PostTypeId="1" AcceptedAnswerId="25629" CreationDate="2018-03-26T02:37:12.170" Score="0" ViewCount="121" Body="&lt;p&gt;If I take a shot longer than 20 seconds, the stars get a tail. I have a Celestron Advanced GT mount with starshoot autoguide scope. I use a cannon can on a 9&quot; Celestron scope. I can't get a picture without having tails on my stars. It only looks good if I keep the pictures under 20 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did the polar alignment. The mount has a polar alignment scope threw the center of it. It's also balanced, the mount has a brids eye level. I use the new PhD software with the starshoot autoguide.. There is no slop in the mount, the gears have almost no backlash. &lt;/p&gt;&#xA;" OwnerUserId="22406" OwnerDisplayName="Jamie Shattler" LastEditorUserId="10196" LastEditDate="2018-03-27T03:03:12.303" LastActivityDate="2018-03-27T03:03:12.303" Title="Need help with telescope mount; stars have tails" Tags="&lt;telescope&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25626" PostTypeId="2" ParentId="25625" CreationDate="2018-03-26T04:01:38.357" Score="-2" Body="&lt;p&gt;you need a motorized equatorial mount, which tracks the stars in such a manner as to cancel the rotation of the earth and eliminate the tails on your photos. These are commercially available for many common types of telescope mounts.&lt;/p&gt;&#xA;" OwnerDisplayName="niels nielsen" LastActivityDate="2018-03-26T04:01:38.357" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25628" PostTypeId="2" ParentId="25622" CreationDate="2018-03-26T09:28:20.447" Score="3" Body="&lt;p&gt;Ultimately this is a highly theoretical question, and while appropriate here you may find perusing the &lt;a href=&quot;http://physics.stackexchange.com&quot;&gt;Physics&lt;/a&gt; site helpful to get deeper answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here's the basic idea for how I think about this.  It's based on asking &quot;where else would the information be?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine the universe is a 2-dimensional sheet.  A piece of paper, say.  And that's all there is for things inside this universe: no desk it's sitting on, no air around it, etc.  Whatever inhabitants of this universe wish to talk about must be contained within this 2-dimensional sheet of paper.  Now imagine there is a hole in this paper.  What can in-universe inhabitants possibly use to learn about this hole and whatever properties it might have?  What to us, as outside observers is &quot;the hole&quot; is inaccessible to them: it's not part of the paper (it's an absence of paper).  But the border of that hole is accessible to them.  So whatever information this hole conveys to this universe must be accessible through that border, and the border only: anything else either doesn't exist, or is some distance away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hole in this paper is our analogue for the black hole in our universe.  &quot;The inside&quot; of a black hole doesn't make sense to an external observer, because the universe&amp;mdash;the sheet of paper&amp;mdash; ends at the border of the hole.  There literally &lt;em&gt;is no inside&lt;/em&gt; to speak of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analogy gets complicated a bit by the fact that if you are not an external observer, but are infalling, then you can in fact pass through what the external observer calls the border. There's the whole &lt;a href=&quot;https://en.wikipedia.org/wiki/Black_hole_information_paradox&quot; rel=&quot;nofollow noreferrer&quot;&gt;information paradox&lt;/a&gt; as well when we bring in quantum mechanics.  But this is the basic gist of things: the black hole is a topological/geometric defect (as general relativity is a theory about the geometry and topology of space), and the only &quot;part&quot; of it which is contained in our universe is its border, the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual physics at and near the event horizon is still a matter of significant debate and research.  This little exercise does not explain why the information of the &lt;a href=&quot;https://en.wikipedia.org/wiki/No-hair_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;No-hair theorem&lt;/a&gt; is accessible, but nothing else is, for example.  It just offers some intuition about why whatever information is accessible must, in some sense, be stored &quot;on&quot; the event horizon.  As far as external observers are concerned, that's the only place it ever could be!&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2018-03-26T09:28:20.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25629" PostTypeId="2" ParentId="25625" CreationDate="2018-03-26T10:06:19.693" Score="3" Body="&lt;p&gt;If you can't get any longer than 20 seconds, chances are it's your polar alignment. The mount you have is a motorised GoTo EQ mount (I believe), but if these aren't aligned properly, then they can still have trouble. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, depending on your software, try a 2 or 3 star align to get a better accuracy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also check the scope is nicely balanced with the camera on the end, and that your mount can take the weight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you try all this and you are still getting trailing, it may be worth going to a professional to get the mount looked at. It may be faulty.&lt;/p&gt;&#xA;" OwnerUserId="17096" LastEditorUserId="17096" LastEditDate="2018-03-26T15:12:15.000" LastActivityDate="2018-03-26T15:12:15.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25630" PostTypeId="1" AcceptedAnswerId="25634" CreationDate="2018-03-26T13:38:20.877" Score="5" ViewCount="887" Body="&lt;p&gt;I know there are stars much larger than the sun and that we've discovered &quot;small&quot; black holes, so, let's say that the &lt;a href=&quot;https://en.wikipedia.org/wiki/Pistol_Star&quot; rel=&quot;noreferrer&quot;&gt;Pistol Star&lt;/a&gt;, which is at least 80x the mass of the sun, and the smallest known black hole, &lt;a href=&quot;https://en.wikipedia.org/wiki/XTE_J1650-500&quot; rel=&quot;noreferrer&quot;&gt;XTE J1650-500&lt;/a&gt;, around 4x the mass of the sun, collide. Would the pistol star &quot;eat&quot; the black hole, or is there some property that a black hole has that is immune to the star's power? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would think that the star would turn the black hole inside out, and like explode or something, but there's also the possibility that once a black hole rips space time the way it does maybe that can't be easily undone. I'm assuming the matter within the black hole isn't going to magically pop back into existence either as it's been broken down into it's most basic parts.&lt;/p&gt;&#xA;" OwnerUserId="22400" LastEditorUserId="2365" LastEditDate="2018-03-26T20:02:42.107" LastActivityDate="2018-03-26T20:02:42.107" Title="Can a star eat a black hole?" Tags="&lt;star&gt;&lt;black-hole&gt;&lt;gravity&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25631" PostTypeId="2" ParentId="25600" CreationDate="2018-03-26T14:58:03.520" Score="0" Body="&lt;p&gt;Problem solved (well, partially).  A project at &lt;a href=&quot;http://ccar.colorado.edu/ASEN5050/projects/projects_2016/Patel_Krishan/#conc&quot; rel=&quot;nofollow noreferrer&quot;&gt;colorado.edu&lt;/a&gt; goes into horrific detail, and end with the conclusion,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Lagrange points are a feature of the three-body problem being the&#xA;  equilibrium points from the CRTBP equations of motion. The collinear&#xA;  points are considered unstable and any perturbations for the exact&#xA;  location will cause exponentially departure. As such, fourth-body&#xA;  perturbations should be considered. We should have particular care&#xA;  because the Sun-Earth-Moon system is one such case. An initial look&#xA;  into the effect the Moon has on the Earth-Sun L1 and L2 points shows&#xA;  little concern. The mass ratio changes as little as 1% and even in the&#xA;  original case the Sun completely out-masses the other bodies. The&#xA;  positions of L1 and L2 change by a few thousand kilometers which is&#xA;  small in comparison to being approximately 1.5 million km away from&#xA;  Earth. Looking at the individual components of acceleration did not&#xA;  help. The Moon is about 2% of the Earth’s acceleration at its best&#xA;  (closest) and 3/4% at its worst. Compiling all the results I conclude&#xA;  that the Moon has about 1% of the effect that the Earth does on these&#xA;  Lagrange points. This seems small however over time this can add up in&#xA;  station keeping costs. Additionally, sensitive equipment such as the&#xA;  James Webb Space Telescope require steady orbits and knowing the&#xA;  perturbation will help keep it steady. In summary, the Moon does not&#xA;  have a large effect but a time aspect must be considered. Further work&#xA;  should include the four-body problem to completely understand changes&#xA;  to these equilibrium points.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-03-26T14:58:03.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25632" PostTypeId="1" CreationDate="2018-03-26T15:50:20.993" Score="2" ViewCount="76" Body="&lt;p&gt;I'm really confused with this one. Almost all documentation says you must start with your telescope pointing to the Celestial Pole and for the lucky ones who live in the north you can use Polaris.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I live in the Southern hemisphere (Chile), so besides no Polaris for me, the real question is: Should I start with the telescope pointing to the South or North Celestial Pole?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have had a month of frustration trying to align.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked into other answers here and other sites, but no one explicitly says where to point it and how to find it.&lt;/p&gt;&#xA;" OwnerUserId="22401" LastEditorUserId="2365" LastEditDate="2018-03-26T20:00:45.683" LastActivityDate="2018-03-26T20:00:45.683" Title="South or North Celestial Pole" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25633" PostTypeId="2" ParentId="25622" CreationDate="2018-03-26T16:25:25.687" Score="0" Body="&lt;p&gt;Mark Van Raamsdonk in his Lecture on Gravity and Entanglement (&lt;a href=&quot;https://arxiv.org/abs/1609.00026&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/1609.00026&lt;/a&gt;) provides the following answer to the question: He states that the statistical interpretation of black hole entropy remained mysterious for decades since there did not exists a framework to understand the microstates of black holes. However, according to him, our understanding of the microstates of a black holes has dramatically changed with the discovery of AdS/CFT (Anti-de-Sitter/Conformal Field Theory) correspondence [1] put forward by Juan Maldacena. The basic idea here is that certain non-gravitational quantum systems (CFTs), defined on fixed spacetimes on a sphere, are equivalent to quantum gravitational theories whose states correspond to different spacetimes with specific asymptotic behavior.Each state as well as observable in the non-gravitational system corresponds to a state and observable in the dual gravitational theory and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this context, he explains, that a Black hole in Anti-de-Sitter space is identified with a high energy state in the corresponding CFT. Such a theory has a discrete spectrum |E_i&gt; and a the thermal state corresponds to the usual canonical ensemble. Hence the entropy of the black hole is identified with the counting of microstates of this CFT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Anti-deSitter is a space with constant negative curvature&lt;/p&gt;&#xA;" OwnerUserId="22388" LastActivityDate="2018-03-26T16:25:25.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25634" PostTypeId="2" ParentId="25630" CreationDate="2018-03-26T18:15:06.053" Score="5" Body="&lt;p&gt;You ask if &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;there some property that a black hole has that is immune to the star's&#xA;  power?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The short answer is &quot;yes&quot; -- the property of being a black hole. Assuming the collision is at a reasonably low relative velocity, the black hole would consume most of the mass of the star, so that eventually you would be left with a much bigger black hole and a certain amount of matter and radiation heading away from the site where the collision took place. The details of how long this would take, how much matter and energy would escape and so on will vary depending on the exact structure of the star and the details of the collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially you can't disrupt a black hole much by throwing matter into it (if the matter you throw in is VERY dense, like another black hole or a neutron star there will be some temporary disturbance to the event horizon, but it will settle down in a fraction of a second), all it does is absorb the matter and get bigger. Even if the matter is super-hot dense fusing plasma from the centre of giant star. In the course of falling in, the matter gets distorted and heated (much hotter than the interior of even a large star) and sometimes whirled around the black hole, and this can appear as a huge explosion, but in the middle of it, the black hole is sitting there absorbing matter, largely unperturbed. &lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="21928" LastEditDate="2018-03-26T19:59:29.083" LastActivityDate="2018-03-26T19:59:29.083" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25635" PostTypeId="2" ParentId="25632" CreationDate="2018-03-26T18:24:48.547" Score="3" Body="&lt;p&gt;If your telescope has an equatorial mount, its polar axis should point toward the south celestial pole (SCP).&#xA;The bad news is that all the stars within 5 degrees of the SCP are magnitude 5 or fainter.&#xA;The good news is that visual observing does not require precise polar alignment.&#xA;Even if you are off by 5 degrees, the mount's drive should keep most targets in view at least 10 times as long as if the drive were not running.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't have to aim the telescope's optical tube at the pole, only the polar axis of its mount.&#xA;Find your geographic latitude, set the polar axis to the same angle above the horizontal, and point it south.&#xA;The &lt;a href=&quot;http://skymaps.com/downloads.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;skymaps.com Evening Sky Map&lt;/a&gt;,&#xA;Southern Edition, can help you get your bearings.&#xA;The SCP makes a right triangle with Crux and Canopus,&#xA;both of which are fairly high in the southern evening sky in March, April, and May.&#xA;If you really need a precise alignment, this&#xA;&lt;a href=&quot;https://www.assa.org.au/resources/equipment-reviews/polar-alignment/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Australian article&lt;/a&gt;&#xA;has a finder chart for&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sigma_Octantis&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\sigma$ Oct&lt;/a&gt;&#xA;and describes the drift method.&#xA;This &lt;a href=&quot;http://astrofotografiasur.blogspot.com/p/alineacion-polar-1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;article in Spanish&lt;/a&gt; may also be useful.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-03-26T19:00:40.567" LastActivityDate="2018-03-26T19:00:40.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25636" PostTypeId="1" CreationDate="2018-03-26T18:45:51.770" Score="1" ViewCount="496" Body="&lt;p&gt;What is the name of brightest star in the East 2am.  What does a satellite look like?&lt;/p&gt;&#xA;" OwnerUserId="22402" LastActivityDate="2018-03-26T18:45:51.770" Title="Brightest star in the Eastern sky 2017 in am" Tags="&lt;amateur-observing&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2018-03-26T20:50:04.523" ContentLicense="CC BY-SA 3.0" />
  <row Id="25637" PostTypeId="2" ParentId="10341" CreationDate="2018-03-26T19:44:04.893" Score="-5" Body="&lt;p&gt;No, there isn't a practical value. How much was wasted on the Hubble Telescope, then how much more was wasted correcting the lenses to actually get a clear picture? How much was wasted on a Mars Lander that crashed upon arrival?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If people want to pay for astronomy out of their pocket, fine. Don't dig into mine just so I can hear about how some galaxy was doing 10 billion years ago. If anything spend the money on oceanography- an area that is of much greater concern to those of us living here and not in the clouds.&lt;/p&gt;&#xA;" OwnerUserId="22403" LastActivityDate="2018-03-26T19:44:04.893" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25641" PostTypeId="1" CreationDate="2018-03-27T09:55:05.950" Score="2" ViewCount="465" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble_Ultra-Deep_Field&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble Ultra-Deep Field&lt;/a&gt; is an image showing about 10.000 galaxies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vkJQY.jpg&quot; alt=&quot;Hubble Ultra-Deep Field&quot;&gt;&lt;a href=&quot;https://i.stack.imgur.com/vkJQY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;&#xA;&lt;em&gt;Image Credit: NASA&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider Victor Buso's spectacular discovery of an early supernova in &lt;a href=&quot;https://en.wikipedia.org/wiki/NGC_613&quot; rel=&quot;nofollow noreferrer&quot;&gt;NGC 613&lt;/a&gt; (named SN 2016gkg):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IwlUJ.jpg&quot; alt=&quot;NGC 613 with Supernova 2016gkg&quot;&gt;&lt;a href=&quot;https://i.stack.imgur.com/IwlUJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;4&lt;/a&gt;&#xA;&lt;em&gt;Image Credit: Carnegie Institution for Science/Las Campanas Observatory/UC Santa Cruz&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This got me wondering about how many supernovae the Hubble Ultra-Deep Field may contain &lt;em&gt;without us knowing&lt;/em&gt; because we would need a light curve or reference image to identify one with certainty. Some supernovae also would be only a pixel or less and impossible to identify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to &lt;em&gt;calculate&lt;/em&gt; a meaningful lower bound?&lt;/p&gt;&#xA;" OwnerUserId="6593" LastActivityDate="2018-03-30T23:31:25.700" Title="How many Supernovae have been Imaged in the Hubble Ultra-Deep Field?" Tags="&lt;supernova&gt;&lt;hubble-telescope&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25642" PostTypeId="1" CreationDate="2018-03-27T05:32:39.850" Score="0" ViewCount="332" Body="&lt;p&gt;Black holes from my knowledge are created from the collapse of a star’s own gravity crushing itself, Now my question is well two questions, is it really infinitely dense to where space-time can be distorted to the point where it just stops? if so, this would mean objects caught in the event horizon would freeze and time would be stopped how is the black hole itself not effected by it’s own mass wouldn’t it just cease to exist in the first place?&lt;/p&gt;&#xA;" OwnerUserId="22413" OwnerDisplayName="arnelmujkic21" LastEditorUserId="6593" LastEditDate="2018-03-27T14:55:59.693" LastActivityDate="2018-03-28T10:53:22.457" Title="Do black holes really exist?" Tags="&lt;black-hole&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25645" PostTypeId="2" ParentId="25642" CreationDate="2018-03-27T15:13:15.823" Score="3" Body="&lt;p&gt;The literal answer to your question is &quot;yes&quot;. We can see &lt;a href=&quot;https://en.wikipedia.org/wiki/Black_hole#Observational_evidence&quot; rel=&quot;nofollow noreferrer&quot;&gt;clear astronomical evidence&lt;/a&gt; of black holes of a variety of sizes in this and other galaxies. It is possible of course that all that evidence is actually the result of some unknown physical process(es) and not black holes, but it would be a remarkable coincidence that it looks so much like we would expect black holes to look.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'll take your question as asking for a way to understand what black holes are. The best answer to that question is basically just equations -- &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_metric&quot; rel=&quot;nofollow noreferrer&quot;&gt;solutions&lt;/a&gt; to the governing equations of general relativity that have event horizons, but I'm going to assume that is not what you want here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we are left looking for a better intuitive explanation or understanding, bearing in mind that this is at best inexact, and at worst misleading. If you have a lot of matter in a small region of space, and you look at it from the outside you see that the light coming from the surface of the region is red-shifted and (which is basically the same thing) that clocks on the surface seem to run more slowly.  As you pack the matter in more tightly this effect becomes more pronounced. But because the clocks are running more slowly, whatever process is packing the matter in also runs more slowly. Eventually, as seen from the outside the clocks on the surface approach a complete standstill, and you will receive a final few extremely red-shifted photons from the surface and then nothing. What you have is now a black hole. From your point of view it is &quot;frozen&quot; and all you can see of it from the outside is the gravitational field surrounding it (and maybe an electric field as well). No information can ever leave it (I'm ignoring Hawking radiation and quantum mechanics in this answer, which is reasonable for big black holes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relativity being relativity, things can look quite different from another point of view. If you are falling freely into the black hole (and it is big enough that tides don't rip you apart) everything seems fine, as in this &lt;a href=&quot;https://www.youtube.com/watch?v=JcHneuh6DKo&quot; rel=&quot;nofollow noreferrer&quot;&gt;simulation&lt;/a&gt; Light from the rest of the universe is blue-shifted and increasingly distorted by gravity, but locally everything is OK. However, no matter what you do, within a finite time (a few days for the largest black holes we know about) you will reach a singularity -- a place where the equations of general relativity break down, and some other physics must presumably apply. We don't know what happens there, but we do know that no information about it can come back out of the black hole, so we can't send in any kind of probe to tell us about it.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-03-27T15:13:15.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25647" PostTypeId="1" AcceptedAnswerId="25648" CreationDate="2018-03-27T18:49:22.837" Score="1" ViewCount="772" Body="&lt;p&gt;I have a Meade DS-2080AT-LNT. It's an inexpensive scope that I've had for several years. It has a couple of eyepieces with it, but nothing that will let you get a really good look at planets. I doubt I can see any Messier objects as more than a blip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's more than good enough to take up and leave at my family's cabin in the far North and I think all of the kids that visit will enjoy it. But to get more out of the scope I was thinking about getting a Barlow lens. I see lenses of 2x and even 5x for sale. I don't know enough about telescopes to know what &quot;x&quot; of Barlow I should get for this scope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specs are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;80mm (3.1&quot;) Aperture: Plenty of brightness to reveal planets, clusters,&#xA;nebulas and more.&lt;/li&gt;&#xA;&lt;li&gt;Sturdy Fork Mount: Lightweight and solid aluminum mount and&#xA;tripod for stable views.&lt;/li&gt;&#xA;&lt;li&gt;494 AutoStar® Controller: Automatically locates more than 1400 objects&#xA;and points the telescope toward them for you at the push of a button.&lt;/li&gt;&#xA;&lt;li&gt;Series 4000 Super Plössl 1.25&quot; Eyepieces: (26mm and 9.7mm) Enjoy low&#xA;and high power viewing with crisp, wide fields of view.&lt;/li&gt;&#xA;&lt;li&gt;Altazimuth Set-up: Easy-to-use mount moves up/down, left/right.&lt;/li&gt;&#xA;&lt;li&gt;SmartFinder™/Red Dot Viewfinder™: Makes stars and other objects&#xA;easy to find. Electronic level sensor, north sensor, and precision&#xA;internal clock help get your scope aligned with the heavens quickly.&lt;/li&gt;&#xA;&lt;li&gt;AutoStar Suite™ DVD: Amazing planetarium software and instructional&#xA;video will help you learn about the night sky and how to use your&#xA;telescope. Print out star charts. Plan observing sessions. Displays over&#xA;10,000 night sky objects. Operates on any Windows®-based PC&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I appreciate any help!&lt;/p&gt;&#xA;" OwnerUserId="22414" LastActivityDate="2018-03-28T23:11:30.423" Title="How can I find out the maximum magnification of my telescope?" Tags="&lt;telescope&gt;&lt;telescope-lens&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25648" PostTypeId="2" ParentId="25647" CreationDate="2018-03-27T19:50:28.540" Score="5" Body="&lt;p&gt;The magnification is given by (focal length of telescope)/(focal length of eyepiece).  The spec I saw said the telescope focal length is 800mm, and the maximum magnification is 160x, so to get that would need a 5mm eyepiece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Barlow lenses have the effect of increasing the telescope focal length: the 2x barlow would therefore give an effective focal length of 1600mm, and using the 9.7 mm eyepiece would give magnification of 160 (approx).  However be aware that the occasions when 160 magnification gives really clear views will be quite rare, as it is at the limit of what an 80mm scope can achieve.  The rule of thumb for maximum useful magnification for a good scope is 50x the diameter of the objective lens (in inches).&lt;/p&gt;&#xA;" OwnerUserId="11822" LastEditorUserId="11822" LastEditDate="2018-03-28T23:11:30.423" LastActivityDate="2018-03-28T23:11:30.423" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25649" PostTypeId="1" AcceptedAnswerId="25665" CreationDate="2018-03-27T21:07:25.910" Score="11" ViewCount="1216" Body="&lt;p&gt;Supposedly when the sun becomes a red giant, it will grow big enough to swallow Earth.&lt;/p&gt;&#xA;&lt;p&gt;However, this requires the radius of the sun to expand by a factor of approximately 215×, meaning its volume would have to expand 10,000,000×.&lt;/p&gt;&#xA;&lt;p&gt;Maybe it's just me, but something about this doesn't feel intuitive :-) especially given that the sun isn't gaining much mass. Exactly what about fusing elements heavier than hydrogen implies that the reactants and/or products take up 10,000,000× as much volume? Nobody ever explains this when explaining the stellar lifetime, and I don't understand why it should be the case. (In fact, I would have expected fusion to result in a decrease in volume, given that nuclei are combining...)&lt;/p&gt;&#xA;&lt;h3&gt;Edit:&lt;/h3&gt;&#xA;&lt;p&gt;It seems there are 2 types of red giants, some that occur during hydrogen fusion, some helium.&lt;br /&gt;&#xA;If the answer is different for these two types, I'd at least like to know the answer for the helium type (though of course I'd appreciate one that goes further and addresses both).&lt;/p&gt;&#xA;" OwnerUserId="7775" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-04-15T10:42:51.240" Title="How can a red giant grow so big?" Tags="&lt;star&gt;&lt;red-giant&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25652" PostTypeId="2" ParentId="25649" CreationDate="2018-03-28T07:18:37.660" Score="6" Body="&lt;p&gt;There's &lt;a href=&quot;https://www.universetoday.com/122807/why-do-red-giants-expand/&quot; rel=&quot;noreferrer&quot;&gt;a nice description here&lt;/a&gt;. Remember that a star is made of gas (well plasma if you want to be picky), so it doesn't have a fixed volume. Once fusion starts a star will expand until it reaches a size where it can balance the amount of energy being produced by fusion with the amount radiated away from the surface. If it's too small, it will heat up, causing expansion which (depending on which parts of the star are expanding) will reduce the energy produced and increase the amount radiated away. A more detailed understanding of this requires keeping track of how temperature and density vary with depth in the star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a red giant, the energy is being produced not in the core, but in a spherical shell surrounding the core (because the core has more-or-less run out of fuel). This is actually a larger volume, so more energy is being produced. The star expands until it can radiate all that energy away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found &lt;a href=&quot;http://ads.harvard.edu/books/1989fsa..book/AbookC05.pdf&quot; rel=&quot;noreferrer&quot;&gt;a description with only moderate amounts of maths&lt;/a&gt; especially around page 132. So one thing is that you have a helium core, with hydrogen fusion going on just outside it. That means there is less mass &quot;above&quot; the fusion layer, so fusion is actually happening at lower pressures than when it was happening in the core. That requires higher temperatures in that layer, and when you do the equations, a much higher total energy output. That flow of energy, whether as radiation or convection reaches the outer layers of the star, and initially heats them up, causing them to expand (since the stars gravity is more or less unchanged and so cannot pull them down any harder). In expanding, they cool, which means they trap radiation more (colder gas is less transparent) and radiate away less, and they are heated again and expand again. This continues until a balance point is found (or not, for stars much bigger than the sun which can blow off much of their mass this way) and when you do the numbers it turns out that this balance requires a very large star.  Perhaps one way of thinking of it is that a star much less massive than the sun just slowly goes out. A star much more massive than the sub explodes. The Sun is poised between the two, so it &quot;nearly explodes&quot; but stops when it's outer layers have become very large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One additional remark is that the density in the upper parts of a red giant is pretty low -- by our standards it's a decent vacuum, contaminated with red-hot gas. However because the star is so big, it is still opaque, so we count that as part of the star.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="21928" LastEditDate="2018-03-28T12:32:57.087" LastActivityDate="2018-03-28T12:32:57.087" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25653" PostTypeId="2" ParentId="25649" CreationDate="2018-03-28T09:58:22.253" Score="5" Body="&lt;p&gt;The size of star in equilibrium is a balance of forces, the pressure produced by the hot plasma, heated by the nuclear reactions in the core, balanced by gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fusion rates are &lt;em&gt;strongly&lt;/em&gt; affected by temperature. Increase the temperature a little and you get a &lt;em&gt;lot&lt;/em&gt; more energy coming out. As the core runs out of hydrogen, it begins to collapse and heat up, forming an inert degenerate helium core surrounded by a shell of rapidly burning hydrogen. At this new equilibrium &lt;em&gt;a lot&lt;/em&gt; more energy is released. This positive feedback means that what seems like a small change (core to shell burning) has a massive effect on the energy output of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now when a star evolves, it puts out a lot more energy per second. Aldebaran produces 500 times more energy per second than the sun, despite being only a little larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this does cause the star to increase in size, but as the star gets larger, the outer layers are further from the centre of gravity, and so the force of gravity on them is reduced, in accordance with the inverse square law. With less gravity the growth in the size is amplified. So a large increase in power becomes a massive increase in size. This is why the growth in size is so much larger than simple intuition about the expansion of hot gas would predict.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the final stages of a star's evolution, the size of the star grows without limit, as the star's gravity is not sufficient to keep its outer layers bound to the star, and it becomes a planetary nebula.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-03-28T13:46:50.043" LastActivityDate="2018-03-28T13:46:50.043" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="25654" PostTypeId="2" ParentId="25642" CreationDate="2018-03-28T10:53:22.457" Score="-8" Body="&lt;p&gt;The honest answer is: Nobody knows for certain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They exist only in theories. Physicists regularly construct elaborate mathematical equations, and the Black Hole is something they essentially created to fill the gaps in certain theories, or to artificially 'complete' them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, for many of their theories to be correct, Black Holes MUST exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, though, we have no idea. Something that's infinitely dense and small destroys all logic, physics and observable reality.&lt;/p&gt;&#xA;" OwnerUserId="22083" LastActivityDate="2018-03-28T10:53:22.457" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25655" PostTypeId="1" CreationDate="2018-03-28T11:47:06.227" Score="1" ViewCount="45" Body="&lt;p&gt;I've asked a &lt;a href=&quot;https://astronomy.stackexchange.com/q/25007/7982&quot;&gt;question&lt;/a&gt; about active sky surveys that might intercept a 2% of 4π patch of sky in the next six months and potentially record this dim (+22 to +26 magnitude) object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've added a bounty with about one more day on the grace period, the only thing helpful was a comment that suggested I ask for a &quot;database of surveys&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any pages that list at least major active surveys? That would be a start at least.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-01-29T10:26:54.793" LastActivityDate="2019-01-29T10:26:54.793" Title="Is there any way to find out what sky surveys are *currently active?*" Tags="&lt;sky-survey&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25656" PostTypeId="2" ParentId="10341" CreationDate="2018-03-28T12:52:40.033" Score="1" Body="&lt;p&gt;Coming late to the party: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't want to jump into the game of weighing one scientific discipline against another - this is plain stupid. Saying that Romanistic is less valuable to mankind than Geoscience is nonsense. We can argue what a valuable science is, but the result of that discussion is purely subjective. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about philosophy - the origin of all scientific research? Zero immediate economic return. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pure fundamental research does not have any other goal except curiosity. Einsteins theory of relativity or quantum theory were considered without any practical application for decades! Now you own a gazilion devices at your home that contain laser LEDs and carry around GPS devices in your pocket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i just wanted to share a few pointers that you can find easily via Google:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.stfc.ac.uk/files/what-have-the-astronomers-ever-done-for-us-roger-davies&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;What have the Astronomers ever done for us&quot;&lt;/a&gt;  So if you ever used a calendar, a GPS to navigate your car, were in need of MRT or own a CCD-Camera the practical use of astronomy should be obvious to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.iau.org/public/themes/astronomy_in_everyday_life/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomy in Everyday Life&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://arxiv.org/pdf/1311.0508&quot; rel=&quot;nofollow noreferrer&quot;&gt;Why is Astronomy Important? - arXiv&lt;/a&gt; An article by Dr. Robert Aitken&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.nap.edu/read/9839/chapter/6#138&quot; rel=&quot;nofollow noreferrer&quot;&gt;Benefits to the Nation from Astronomy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomy is maybe the most basic scientific discipline. From its start mankind was looking up to the stars and wondering about the universe. &amp;lt;\rant&gt; &lt;/p&gt;&#xA;" OwnerUserId="1988" LastActivityDate="2018-03-28T12:52:40.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25658" PostTypeId="1" AcceptedAnswerId="25661" CreationDate="2018-03-28T19:36:39.753" Score="0" ViewCount="131" Body="&lt;p&gt;Okay, say we have a tidally locked planet that has one side facing its star. On the side that always faces the star, my question is will that side facing the star always be increasing in temperature? If the side is always facing the star, it doesn't have time to cool. &lt;/p&gt;&#xA;" OwnerUserId="16048" LastEditorUserId="2365" LastEditDate="2018-03-28T19:59:51.480" LastActivityDate="2018-03-28T20:04:44.517" Title="Would a tidally locked planet be always increasing in temperature?" Tags="&lt;tidal-locking&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25660" PostTypeId="2" ParentId="25658" CreationDate="2018-03-28T20:03:55.067" Score="4" Body="&lt;p&gt;You're asking if the temperature of the side facing the star will increase without bound. Almost anytime you're talking about a real, physical system, &quot;without bound&quot; just doesn't happen. Think about it - if the temperature was always increasing, eventually the planet will be hotter than the star its orbiting. At that point, why would it continue to get hotter? At some point, the system will settle into an equilibrium where the incoming radiation is balanced by the rate of cooling, at which point the planet won't get any hotter.&lt;/p&gt;&#xA;" OwnerUserId="10488" LastActivityDate="2018-03-28T20:03:55.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25661" PostTypeId="2" ParentId="25658" CreationDate="2018-03-28T20:04:44.517" Score="6" Body="&lt;p&gt;Hot objects radiate heat. The hotter something is, the faster it radiates heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The side of the planet that faces the sun would heat up until the rate at which it is absorbing heat from the star is equal to the rate at which it is radiating heat. Once it reaches this point it will be stable, and neither heat up more or cool down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a simple analogy, consider standing next to a heater or a fire. You will heat up, but you won't heat up forever. You will reach thermal equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-03-28T20:04:44.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25662" PostTypeId="2" ParentId="25655" CreationDate="2018-03-28T20:09:12.020" Score="1" Body="&lt;p&gt;If you really just want a list, &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_survey#List_of_sky_surveys&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; has a pretty good page on astronomical surveys. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some comprehensive information, including some tabulated values (such as sky coverage for each survey), try &lt;a href=&quot;https://arxiv.org/ftp/arxiv/papers/1203/1203.5111.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper from S. G. Djorgovski et al.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user14781" LastActivityDate="2018-03-28T20:09:12.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25663" PostTypeId="2" ParentId="25641" CreationDate="2018-03-28T20:18:33.340" Score="1" Body="&lt;p&gt;Supernovae hunters report finding about &lt;a href=&quot;http://www.bosssupernova.com/settingupasearch.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;1 supernova per 5000 images.&lt;/a&gt; Hunters will, naturally, focus on larger galaxies, as these are more likely to contain a supernova, whereas the deep field survey contains galaxies of all sizes, so perhaps about 1 supernova per 10000 galaxies imaged would be a reasonable estimate. Thus you might expect there to be about 1 supernova visible in this image, but as you note, if there is one, it may be too dim to be visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supernovae are all one pixel or less (they are so small and distant that they appear as point-like sources of light), they only appear to be bigger due to atmospheric blurring or imperfections in optics and/or electronics.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-03-28T20:18:33.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25664" PostTypeId="2" ParentId="25649" CreationDate="2018-03-28T20:54:59.553" Score="1" Body="&lt;p&gt;The intuitive way to think about it is to understand that there are multiple changes that, in essence, amplify each other.   Amplification in astronomy isn't all that uncommon.    It explains why gravity can make massive objects so small, because as the massive object gets smaller, the gravitation and weight of the object grows exponentially.  In a sense, the opposite happens with a Red giant.  The gravity at the surface grows low enough that the star kind of enters a run-away expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expansion of the star late in it's life is exponential.  That's why it can expand so much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sun was to double in size but it's mass was to remain unchanged.  In this hypothetical, the new Sun's surface gravity is divided by 4.   It's escape velocity is divided by the square root of 2, so the outer layer has much less weight, but the escape velocity still binds it to the star.    Everything being equal, expanding the sun should cause it to cool, but using the root mean square rule for thermal velocity, if the temperature is divided by 2, the velocity of the hydrogen and helium molecules is divided by the square root of 2.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this theoretical, they hydrogen atoms on the surface are moving slightly slower, but with 1/4 the gravity, they're more free and they can move further from the star based on their thermal velocity.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we keep expanding the sun, there comes a point where the outer hydrogen becomes incredibly loosely bound.   At red-giant size, say, 1 AU in radius or 215 current solar radii, the gravity is some 46,000 times lower and the hydrogen on the surface only experiences 0.006 m/s^2 gravitational acceleration, but those same hydrogen molecules at red giant temperature (some 3,000 degrees K), are moving about 5.5 km/s.  They can fly away from the surface for over a million km based on their thermal energy alone, compared to about 100 km at the surface of the sun currently (based on just under 8 km/s).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases, the outer layer of hydrogen and helium are in equilibrium, it's just that the gravity and red-giant size is so so so much lower that with the red-giant, equilibrium is this very loosely bound very disburse hot gas.   But that's only part of the reason. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider what else happens as the sun grows older.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3MBfK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3MBfK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.quora.com/How-can-you-describe-the-four-layers-of-the-sun&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The core, where the fusion happens is a comparatively small region in the center.   Wrapped around the core is the radiative zone and the conductive zone.  which help keep the heat from fusion trapped inside the sun.   As a result, over time, the inside of the sun grows hotter and as it grows hotter the core grows larger and it encompasses more and more of the radiative zone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we think of the radiative zone as a kind of blanket that traps heat inside the sun, as the core grows larger and more massive, the radiative zone is both stretched and it loses mass to the core, so it becomes thinner in two ways.   If the size of the core is doubled, the photons from the core have to travel through 1/4th as many molecules.   As the sun gets old enough and most of the fusion happens on the outer edge of the core, there's significantly less of a blanket to keep the heat trapped in.  It's not so much that more energy is being created, it's that that energy has an easier path to the outer region of the sun.    So you have an amplification effect, as the sun grows larger, the surface gravity drops by the square of the radius and the internal heat has less material to pass through to reach the outer layers, creating an outward pressure as well as delivering more heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The internal core collapse can play a role too.  Even as the inner core runs out of hydrogen to fuse and it begins to collapse, the act of collapsing generates significant heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure that's clear, but that's my attempt to explain what happens intuitively.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-03-28T21:00:30.010" LastActivityDate="2018-03-28T21:00:30.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25665" PostTypeId="2" ParentId="25649" CreationDate="2018-03-29T00:11:01.233" Score="9" Body="&lt;p&gt;In my mind, none of these explanations really cover the actual reason that red giants expand.  Indeed, this subject seems like an area where people just make up anything that sounds plausible, but it's often quite wrong (Fraser Cain mentions both light pressure and a higher volume in the fusing shell, but light pressure plays no role at all, and the volume of the shell isn't much different than the core, which is way &lt;em&gt;smaller&lt;/em&gt; than the Sun's core). So let's get the story straight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of the descriptions do include some of the key elements, including the fact that you have a shell of hydrogen fusion going on atop an inert degenerate helium core.  But the key reason for the expansion is that the way this shell self-regulates its fusion rate is quite a bit different from the way the Sun's core is self-regulating its fusion now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the Sun's core self-regulates its fusion rate to match the rate that energy (in the form of light) is diffusing out through the mass of the Sun.  The way it does this is essentially the same in all main-sequence stars: they regulate their core temperature, which is why the core temperature of more luminous main-sequence stars is a little higher.  But this is not at all the way the shell fusing in a red giant self-regulates its fusion rate-- it cannot regulate its temperature, because the temperature is handed to it by the gravity of the degenerate core its sits upon.  (This sets the temperature via the virial theorem, that is the key way the degenerate core affects the shell-- it sets its temperature.)  Since the shell doesn't regulate its own temperature, the temperature tends to be quite high, especially as the core gains mass (this is why the luminosity rises with time).  Fusion is very sensitive to temperature, so being stuck with a very high temperature makes the fusion rate go &lt;em&gt;berserk&lt;/em&gt;.  The rest of the star cannot support this spectacular fusion rate, so something else happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The star puffs out, and in so doing, we find the way the shell regulates its own fusion rate: it &lt;em&gt;lifts weight off the shell&lt;/em&gt;.  That reduces the pressure in the shell, which compensates for the high temperature and brings the fusion rate down to what the rest of the star can manage (which is set by the rate light can diffuse through the shell).  So there's the real reason-- the star must find a way to lift weight off its crazy-high-temperature shell to keep the fusion rate from going nuts, but the upshot is that the fusion rate is still pretty high, and gets higher as the mass of the core rises, forcing the shell temperature up ever more and forcing the star to puff out even more.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2018-03-29T00:46:44.123" LastActivityDate="2018-03-29T00:46:44.123" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25666" PostTypeId="1" CreationDate="2018-03-29T04:09:00.583" Score="0" ViewCount="377" Body="&lt;p&gt;How can I tell in each month, when here will be no moon shining or seen at night?&lt;/p&gt;&#xA;" OwnerUserId="22434" LastEditorUserId="1891" LastEditDate="2018-03-31T20:24:53.437" LastActivityDate="2018-04-04T19:11:31.467" Title="A moonless night" Tags="&lt;moon-phases&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25667" PostTypeId="1" CreationDate="2018-03-28T08:46:02.343" Score="21" ViewCount="412" Body="&lt;p&gt;I understand that radiation on planets near the galaxy’s center make life on these planets nearly impossible, and that one cannot truly “see” a black hole. However, if you could stand on the surface of a planet that orbits a star close to or within the galactic core, could you theoretically look up at the sky and see an absence of light/stars indicating the location of the central supermassive black hole?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be too far away to see, obstructed by debris, or too small to notice?&lt;/p&gt;&#xA;" OwnerDisplayName="BeccaLew" LastActivityDate="2018-03-29T05:16:34.507" Title="In a planetary system close to the galactic core, would it be possible to see the supermassive black hole?" Tags="&lt;space&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25668" PostTypeId="2" ParentId="25667" CreationDate="2018-03-28T09:15:48.097" Score="9" Body="&lt;p&gt;You couldn't see it as a black patch in the sky, because it's far too small. It's only 17 times the radius of our sun, which of course you can't see as a disc even from the outer reaches of our own solar system. What you &lt;em&gt;could&lt;/em&gt; easily see is the much larger area of light and other radiation from matter falling into it.&lt;/p&gt;&#xA;" OwnerUserId="9235" OwnerDisplayName="Mike Scott" LastActivityDate="2018-03-28T09:15:48.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25669" PostTypeId="2" ParentId="25667" CreationDate="2018-03-28T10:09:46.157" Score="30" Body="&lt;p&gt;If the black hole is active, meaning it is still capturing matter from its surrounding, it will have a large accretion disk all around, which is the only way to dissipate angular momentum for the matter falling into it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result of this dissipation all the matter will warm up and emit radiation. This disc will be fairly large and thus clearly visible as a bright object in the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;https://it.wikipedia.org/wiki/Buco_nero&quot; rel=&quot;noreferrer&quot;&gt;page&lt;/a&gt; shows an image captured by Hubble of such a disk:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/tzwG6.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tzwG6.jpg&quot; alt=&quot;disk around a black hole&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strictly speaking then, one cannot see directly the black hole, as its view will be covered by the bright emission of the disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the presence of the disk will allow to observe the black hole thanks to its gravitational effect.&lt;/p&gt;&#xA;" OwnerDisplayName="L.Dutch" LastActivityDate="2018-03-28T12:53:30.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25670" PostTypeId="2" ParentId="25667" CreationDate="2018-03-28T11:35:16.633" Score="8" Body="&lt;p&gt;There's a thing called gravitational lensing, which means that light coming from behind the black hole would bend towards it, and since the galactic core has a lot of stars, it might be that instead of a black spot in the sky, you'd see a great accumulation of light in and around the black hole's position. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how literal &quot;lensing&quot; is, so I don't know whether there is a focal point based on gravity and the energy of the light and whether it would matter where the planet is from this focal point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cfhtlens.org/public/what-gravitational-lensing&quot; rel=&quot;noreferrer&quot;&gt;http://www.cfhtlens.org/public/what-gravitational-lensing&lt;/a&gt; (google will provide more links if you search for it)&lt;/p&gt;&#xA;" OwnerUserId="19630" OwnerDisplayName="Real Subtle" LastActivityDate="2018-03-28T11:35:16.633" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25671" PostTypeId="2" ParentId="25667" CreationDate="2018-03-28T13:35:51.040" Score="4" Body="&lt;p&gt;&lt;strong&gt;Rainbow cloud.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A source of radiation from black holes is stuff spiralling into the black hole, heating up as it &quot;fell&quot; and released its gravitational potential energy.  Again this is black body radiation but this time the regular kind: the hotter the emitters are is the shorter the wavelength.  This radiation comes from next to the black hole, not out of the hole itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/24958/how-can-a-black-hole-emit-x-rays&quot;&gt;https://physics.stackexchange.com/questions/24958/how-can-a-black-hole-emit-x-rays&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The X-rays come from hot gas orbiting around the black hole in an&#xA;  accretion disk. As the gas orbits, magnetic stresses cause it to lose&#xA;  energy and angular momentum, thus spiralling slowly in towards the&#xA;  black hole. The orbital energy is transformed into thermal energy,&#xA;  heating up the gas to millions of degrees, so it then emits blackbody&#xA;  radiation in the X-ray band.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Once the gas gets closer than a few times the horizon radius, it&#xA;  plunges into the black hole, so while some X-rays can still escape&#xA;  just before the horizon, most are emitted a fair bit outside.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Telescopes to detect black holes look for the most energetic rays, which are emitted from the hottest areas of gas nearest the hole.  We cannot stand on a planet and look up and see xrays.  But consider: if there is very hot gas, next to it there is less hot gas, and next to that gas that is less less hot.  The cooler a blackbody is, the longer wavelength the emitted radiation is.  Somewhere in that progressively cooler cloud is gas which emits radiation in the visible wavelength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I here assert that the gradual change in temperature of this cloud as it is progressively farther from the hottest innermost gas should produce a gradual change in the frequencies emitted. The first visible light would be in the far violet, nearest the hole.  This will grade through blues and greens farther away and then to red at the farthest coolest part of the cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This prediction should be true not just for black holes, but for any cloud of gas heated from within.  Now let me look... here we go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/htObN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/htObN.jpg&quot; alt=&quot;rainbow nebula&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.space.com/12051-bright-nebula-photo-supergiant-star-betelgeuse.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.space.com/12051-bright-nebula-photo-supergiant-star-betelgeuse.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black hole rainbow cloud will be more symmetric than this one.  The star is spewing this stuff out willy-nilly but the hole is sucking gas in, so it will be a symmetric spiral.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="16293" OwnerDisplayName="Willk" LastActivityDate="2018-03-28T14:39:22.107" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25672" PostTypeId="2" ParentId="25666" CreationDate="2018-03-29T05:54:07.537" Score="3" Body="&lt;p&gt;A new moon (when the moon is not seen at night) happens regularly, every 29.5 days (roughly). This is roughly one day less than a month. So you can forecast the next few new moons by &quot;add a month, minus a day&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next New moon will be on 16 April. We would then expect new moons on:&#xA;15 May, 14 June, 13 July, 12 August and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A website like &lt;a href=&quot;https://www.timeanddate.com/moon/phases/uk/london&quot; rel=&quot;nofollow noreferrer&quot;&gt;timeanddate.com&lt;/a&gt; will do the accurate calculations for you (adjusting for your local time and so on.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-03-29T05:54:07.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25673" PostTypeId="1" CreationDate="2018-03-29T07:02:33.433" Score="1" ViewCount="499" Body="&lt;p&gt;What does the angle between the plane of the earth's orbit and the ecliptic (now about 23.5 degrees) vary between, and over what time scale?&lt;/p&gt;&#xA;" OwnerUserId="22438" LastActivityDate="2018-03-29T07:19:53.840" Title="What does the angle between plane of earth's orbit and ecliptic vary between?" Tags="&lt;orbit&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25675" PostTypeId="2" ParentId="25673" CreationDate="2018-03-29T07:19:53.840" Score="3" Body="&lt;p&gt;By definition, it is zero; the ecliptic &lt;em&gt;is&lt;/em&gt; the plane of earth's orbit (or rather, the inverse: the path the Sun seems to follow along the sky).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you're probably looking for variations in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_tilt#Earth&quot; rel=&quot;nofollow noreferrer&quot;&gt;axial tilt of the Earth&lt;/a&gt;. There's a lot of formulas in the article, and on the (astronomically speaking) short timescale of a few million years, it oscillates between 22 and 24.5 degrees with a period of about 41,000 years. See the graph below, the red dot is the year 2000:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KpyU6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KpyU6l.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(picture by Tfr000 on Wikipedia, data from Laskar (1986))&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2018-03-29T07:19:53.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25677" PostTypeId="1" AcceptedAnswerId="25679" CreationDate="2018-03-29T11:32:54.983" Score="0" ViewCount="330" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Problem: Suppose an observer's latitude is 45 S and a star's RA/DEC is&#xA;3h15min/41S. On what date will the sun set with the star.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I have been stuck on this problem for a while.&lt;/p&gt;&#xA;&lt;p&gt;I found out that the time of setting of a star is found using the formula:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\alpha+\arccos(-\tan(\delta).\tan(\phi))$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$\implies3\frac{1}{4}^h+\arccos(-\tan(-41)\tan(-45)) = \alpha_{sun}+\arccos(-\tan(\delta_{sun})\tan(-45))$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\implies 47.8 = \alpha_{sun}+arccos(tan(\delta_{sun}))$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I am not able to solve this any further.&lt;/p&gt;&#xA;&lt;p&gt;Is there something wrong in my approach? Something I am missing?&lt;/p&gt;&#xA;&lt;p&gt;If not how do I solve the above equation further?&lt;/p&gt;&#xA;&lt;p&gt;Thanks in anticipation!&lt;/p&gt;&#xA;" OwnerUserId="22114" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2020-01-23T13:52:33.487" Title="When does a star rise with the sun?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;declination&gt;&lt;right-ascension&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25678" PostTypeId="1" AcceptedAnswerId="25681" CreationDate="2018-03-29T15:09:57.173" Score="1" ViewCount="337" Body="&lt;p&gt;I am trying to relate Galactic Latitude (b) and Longitude (l) (spherical coordinates) to galactic cartesian x,y,z coordinates in the ATNF pulsar catalog. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.atnf.csiro.au/people/pulsar/psrcat/proc_form.php?version=1.58&amp;amp;Name=Name&amp;amp;GL=GL&amp;amp;GB=GB&amp;amp;ZZ=ZZ&amp;amp;XX=XX&amp;amp;YY=YY&amp;amp;startUserDefined=true&amp;amp;c1_val=&amp;amp;c2_val=&amp;amp;c3_val=&amp;amp;c4_val=&amp;amp;sort_attr=jname&amp;amp;sort_order=asc&amp;amp;condition=&amp;amp;pulsar_names=&amp;amp;ephemeris=short&amp;amp;coords_unit=raj%2Fdecj&amp;amp;radius=&amp;amp;coords_1=&amp;amp;coords_2=&amp;amp;style=Long%20with%20last%20digit%20error&amp;amp;no_value=*&amp;amp;fsize=3&amp;amp;x_axis=&amp;amp;x_scale=linear&amp;amp;y_axis=&amp;amp;y_scale=linear&amp;amp;state=query&amp;amp;table_bottom.x=43&amp;amp;table_bottom.y=24&quot; rel=&quot;nofollow noreferrer&quot;&gt;This query&lt;/a&gt; shows some sample data with &lt;code&gt;G_l&lt;/code&gt;, &lt;code&gt;G_b&lt;/code&gt;, &lt;code&gt;XX&lt;/code&gt;, &lt;code&gt;YY&lt;/code&gt;, and &lt;code&gt;ZZ&lt;/code&gt;; excerpt:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;------------------------------------------------------------------&#xA;#     NAME                     Gl     Gb     ZZ     XX     YY&#xA;                               (deg)  (deg)  (kpc)  (kpc)  (kpc)&#xA;------------------------------------------------------------------&#xA;1     J0002+6216    cwp+17     117.33 -0.07  -0.00  0.00   8.50&#xA;2     J0006+1834    cnt96      108.17 -42.98 -0.59  0.60   8.70&#xA;3     J0007+7303    aaa+09c    119.66 10.46  0.25   1.20   9.18&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The description of the variables from the &lt;a href=&quot;http://www.atnf.csiro.au/research/pulsar/psrcat/psrcat_help.html?type=normal&amp;amp;highlight=gl#gl&quot; rel=&quot;nofollow noreferrer&quot;&gt;ATNF documentation&lt;/a&gt; are as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;GL:          Galactic longitude (degrees) &#xA;GB:          Galactic latitude (degrees) &#xA;[...]&#xA;ZZ:          Distance from the Galactic plane, based on Dist&#xA;XX:          X-Distance in X-Y-Z Galactic coordinate system (kpc)&#xA;YY:          Y-Distance in X-Y-Z Galactic coordinate system (kpc)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;My understanding is that these variables should be related by the following equations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;tan(G_L) = YY/XX&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;tan(G_b) = XX/ZZ&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, when I test this assumption my calculated values are very different:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pYo4Q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pYo4Q.png&quot; alt=&quot;test-spreadsheet&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried exploring the possibility that the x,y,z coordinate system may be oriented differently than I expect, but I can find no orientation that yields similar results for &lt;code&gt;G_l&lt;/code&gt; or &lt;code&gt;G_b&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bxDtO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bxDtO.png&quot; alt=&quot;trig-exaustive-search&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where could I have gone wrong? I feel like I am losing my mind not being able to convert these with simple trig.&lt;/p&gt;&#xA;" OwnerUserId="22382" LastActivityDate="2018-03-29T19:18:35.893" Title="Why are the spherical and cartesian galactic coordinates in the ATNF Pulsar Catalogue different?" Tags="&lt;coordinate&gt;&lt;data-analysis&gt;&lt;pulsar&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25679" PostTypeId="2" ParentId="25677" CreationDate="2018-03-29T15:51:07.633" Score="2" Body="&lt;p&gt;The formula you gave is to find the hour angle of the star while setting, not the setting time itself. Suppose the hour angle of star is $HA\star$, and $RA=\alpha$,then the Local Sidereel Time is given by $LST=HA\star+\alpha$. At the time of setting, $HA\star=10^h1^m30.2^s$. Thus, $LST=13^h16^m30.2^s$at the time of setting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $RA\odot=6^h$ on $\text{June} 21^{st}$ and $HA_\odot =\arccos(\tan(23.43^{\circ}))=4^h17^m17^s$ at sunset. Thus the star sets with the sun after $\text{ June} 21^{st}$. Since $RA_\odot$ increases approximately by $2^h$ every month, we can assume that the star sets with the sen near $\text{July }21^{st}$. Let it set after $n$ days. At $\text{July }21^{st}$, $\delta_\odot = 20.09^{\circ}$ which gives us $HA_\odot=4^h34^m11^s$, and $RA_{\odot}=8^h9^m44^s$. The $HA_\odot$ at sunset is approximately the same for nearby dates. Since $\delta_\odot$ is also near the maximum, we can approximate $\Delta RA_\odot= \omega_\odot n$. Thus,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$13^h16^m30.2^s=8^h9^m44^s + \omega_\odot n +  4^h34^m11^s$$&#xA;$$ \omega_\odot n =0.5433^h$$&#xA;$$n=8$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At  $\text{July } 29^{th}$, $\delta_\odot=18.23^{\circ}$, $HA_\odot= 4^h43^m6^s$, $RA_\odot=8^h42^m12^s$ which gives us $LST=13^h25^m18^s$, which is a bit more than expected. Trial and error for few days before gives us that on $\text{July } 26^{th}$, the sun sets at LST $13^h15^m9^s$ which would be the most appropriate answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the sun sets along with the star at $\text{July } 26^{th}$&lt;/p&gt;&#xA;" OwnerUserId="16744" LastEditorUserId="16744" LastEditDate="2018-03-29T19:04:27.713" LastActivityDate="2018-03-29T19:04:27.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25681" PostTypeId="2" ParentId="25678" CreationDate="2018-03-29T18:47:17.953" Score="2" Body="&lt;p&gt;The ATNF Pulsar Catalogue's galactic longitude and latitude are heliocentric,&#xA;but the origin of their rectangular coordinates is near the center of our galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.atnf.csiro.au/people/pulsar/psrcat/psrcat_help.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;catalogue documentation&lt;/a&gt;,&#xA;section 6. Distances, says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Galactocentric coordinate system (XX, YY, ZZ) is right-handed&#xA;  with the Sun at (0.0, 8.5 kpc, 0.0)&#xA;  and the ZZ axis directed toward the north Galactic pole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From a heliocentric point of view&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_coordinate_system#Rectangular_coordinates&quot; rel=&quot;nofollow noreferrer&quot;&gt;(U, V, W)&lt;/a&gt;&#xA;= (8.5 - YY, XX, ZZ).&#xA;Then $\tan{l} = V / U$ and $\tan{b} = W / \sqrt{U^2 + V^2}$&#xA;as you'd expect.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-03-29T19:18:35.893" LastActivityDate="2018-03-29T19:18:35.893" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25683" PostTypeId="1" CreationDate="2018-03-30T01:06:24.740" Score="0" ViewCount="101" Body="&lt;p&gt;If you had a hypervelocity star travelling through space incredibly quickly, would the hypervelocity star's Oort cloud remain intact? Would the high speed of a hypervelocity star shed the comets and dwarf planets in its Oort cloud due to the star's high speeds, or would the Oort cloud get stretched into a teardrop shape? Could the Oort cloud remain unaffected?&lt;/p&gt;&#xA;" OwnerUserId="18831" LastActivityDate="2018-03-30T10:55:01.747" Title="Do hypervelocity stars have Oort clouds?" Tags="&lt;star&gt;&lt;oort-cloud&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25687" PostTypeId="1" CreationDate="2018-03-30T07:04:57.300" Score="2" ViewCount="84" Body="&lt;p&gt;Would the tidal forces of multiple moons interacting upon a planet speed up or slow down the day length of the planet? I was thinking that maybe the planet's day length would be slowed down due to all of the gravitational tugging, but on the other hand I was thinking that maybe all of those tidal forces could perhaps slingshot the planet ahead and make it spin at a faster rate. Would the day length even change at all? If the answer is yes, could all of those moons even affect the planet's precession?&lt;/p&gt;&#xA;" OwnerUserId="18831" LastActivityDate="2018-03-30T10:36:19.133" Title="Do the number of moons a planet has affect the speed of the planet's rotation?" Tags="&lt;planet&gt;&lt;rotation&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25688" PostTypeId="2" ParentId="25600" CreationDate="2018-03-30T07:52:30.683" Score="3" Body="&lt;p&gt;The L1, L2, L3 points are &lt;strong&gt;not stable.&lt;/strong&gt; (full stop) Small deviations grow exponentially, even in the perfectly circular restricted three-body problem. In reality, we have a non-circular many-body problem, when (1) these points are not strictly well-defined and (2) deviations from the simple case have to be accounted for to obtain the exact trajectories (and to keep any artificial satellites near these points).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The L1 &amp;amp; L2 points are attractive for space missions exactly because they are unstable, since this prevents the accumulation of debris (natural and artificial in origin) and hence reduces the probability of collisions. &lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2018-03-30T07:52:30.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25689" PostTypeId="1" CreationDate="2018-03-30T07:58:35.513" Score="1" ViewCount="297" Body="&lt;p&gt;I read about ring singularity on Wikipedia and didn't quite understand the flow of liquids inside a rotating body causing a ring shaped singularity with zero thickness. Moreover is it specifically ring shaped and not torus shaped in any case?&lt;/p&gt;&#xA;" OwnerUserId="22385" LastActivityDate="2018-03-30T09:27:05.790" Title="Why does kerr black holes have a ring shaped singularity rather than point?" Tags="&lt;black-hole&gt;&lt;astrophysics&gt;&lt;singularity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25690" PostTypeId="2" ParentId="25689" CreationDate="2018-03-30T09:27:05.790" Score="3" Body="&lt;p&gt;Firstly do not confuse the event horizon with the singularity. &lt;a href=&quot;https://en.wikipedia.org/wiki/Kerr_metric&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; gives the formulae for the Kerr metric. There are a number of places where this formula appears to break down because you appear to be dividing by zero, essentially whenever $\Sigma = 0$ or $\Delta = 0$, in the notation of that page,  corresponding to the ergosphere and the event horizon respectively. There are two event horizons and two ergospheres but they all have the topology of spheres, one inside another (with some touching at the poles). However those are not singularities in general because you can choose other coordinates for space and time where the division by zero goes away. However you can't do this at points on the equator of the inner ergosphere, which makes those points a ring singularity. It is a ring with zero thickness (at least until quantum effects come in) .There is &lt;a href=&quot;https://en.wikipedia.org/wiki/Kerr_metric#/media/File:Kerr-surfaces.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;a helpful picture&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flow of liquids doesn't really enter the picture here. One can think of the properties of this metric as a &quot;flow&quot; of space-time, but that's an analogy at best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There seems to be considerable doubt about what really happens inside a rotating black hole, or even, given that no information can escape from the black hole, about whether this is even a meaningful question.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-03-30T09:27:05.790" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25691" PostTypeId="1" AcceptedAnswerId="25692" CreationDate="2018-03-30T09:41:01.463" Score="2" ViewCount="271" Body="&lt;p&gt;When the asteroid hit the earth 65 million years ago, did the earth's gravity pull change? By how much?&lt;/p&gt;&#xA;" OwnerUserId="22459" LastActivityDate="2018-04-26T09:23:14.347" Title="When the asteroid hit the earth 65 mya did the earth's gravity pull change? By how much?" Tags="&lt;gravity&gt;&lt;earth&gt;&lt;meteorite&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25692" PostTypeId="2" ParentId="25691" CreationDate="2018-03-30T10:02:47.000" Score="6" Body="&lt;p&gt;The gravity of Earth did not change substantially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strength of gravity on Earth is about 9.8, but varies between 9.76 and 9.83 ms-2(due to Earth not being a perfect sphere)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be approximated from the mass of the Earth using $$g=GM_e/r_e^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $M_e = 5.97219\times 10^{24}kg$ and $r_e= 6371000m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the asteroid was about $10^{16}kg$ That changes the mass from $5.97219\times 10^{24}$ to $5.97219001\times 10^{24}$. It changes the gravity from 9.8 to 9.800000001&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words. The change in gravity was utterly insignificant. You probably get a greater change in gravity by walking downstairs.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-03-30T10:02:47.000" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25695" PostTypeId="2" ParentId="25687" CreationDate="2018-03-30T10:36:19.133" Score="3" Body="&lt;p&gt;Any moon exerts a tidal force on its primary, tending to cause the primary to bulge outward along the axis passing the moon and to narrow slightly in the other two directions. This effect is greater if the moon in question is larger and/or closer to the primary, but a solid primary may largely resist it, so that it changes shape little if at all. In the case of the Earth and its Moon, the solid Earth moves relatively little, while the water of the oceans is moved quite a bit. As a result, water has to flow from one place to another over the solid crust. That flow causes friction, which causes the &quot;bulge&quot; of water to lag behind the Moon and pull back on it by gravity. This process gradually slows the rotation of the Earth and drives the Moon further away.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process acts, in general to bring the rotation of the planet and the revolution of the Moon to the same rate, so that the planet keeps one face to the Moon. Usually this means slowing the planet's rotation, but a few moons are so close to the planet that it would be the opposite. This process will be less pronounced on a planet which  has a deeper liquid layer, since there will be less friction, or on a planet which has a stiff solid structure, since there will be less movement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of moons is not directly relevant. To a reasonable approximation they all work independently. &lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-03-30T10:36:19.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25696" PostTypeId="1" AcceptedAnswerId="25704" CreationDate="2018-03-30T10:40:28.723" Score="1" ViewCount="1059" Body="&lt;p&gt;I'm using a star database (the &lt;a href=&quot;http://www.astronexus.com/hyg&quot; rel=&quot;nofollow noreferrer&quot;&gt;HYG database&lt;/a&gt;) which has xyz positions and velocities for the stars. This has +X towards the vernal equinox, +Z towards north celestial pole, +Y towards Right Ascension 6 hours and it's based on J2000 and centered on the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to convert these to a Cartesian Galactic coordinate system centered on the sun, with +X pointing to its center and +Z angled so that it shoots out of the galactic plane on the north side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should just be a simple rotation, but after a lot of searching I can't seem to find the rotation to use, nor some way of converting it through a third system.&#xA;Does anyone know how to obtain the correct rotation matrix here? (or some other method of rotation) &lt;/p&gt;&#xA;" OwnerUserId="22460" LastEditorUserId="616" LastEditDate="2018-03-30T19:35:00.470" LastActivityDate="2018-04-29T20:05:14.000" Title="How to convert from Cartesian J2000 coordinates to Cartesian Galactic coordinates?" Tags="&lt;galaxy&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25697" PostTypeId="2" ParentId="25683" CreationDate="2018-03-30T10:49:36.163" Score="2" Body="&lt;p&gt;There is no problem with orbiting a star that moves very fast. Remember that space is mostly empty, so there is nothing like air resistance or friction that would tend to deform the orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars that become hypervelocity stars likely tend to lose their Oort clouds. They gain their velocity by passing very close to a massive body and getting a velocity increment, for example by having a secondary star or planet ripped away by tidal forces. That already doesn't bode well for the stability of any Oort cloud. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/1201.1446&quot; rel=&quot;nofollow noreferrer&quot;&gt;Simulations show that planets tend to be ejected from hypervelocity systems unless they are very tightly bound&lt;/a&gt;, since they have lower mass and respond more strongly to the velocity changes. This suggests that the lightly bound and small bodies in the Oort cloud will be scattered - likely at even higher velocities. &lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2018-03-30T10:49:36.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25698" PostTypeId="2" ParentId="25683" CreationDate="2018-03-30T10:55:01.747" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_kinematics#Hypervelocity_stars&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hypervelocity stars&lt;/a&gt; are moving at up to 1000 $km/s$ relative to the general movement of stars in their part of the galaxy (which is typically 100 $km/s$ relative to one another or the galactic centre). The Oort cloud of a star is a collection of rocky and icy bodies orbiting it up to 1 light year or so away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those bodies are too dense and massive to be deflected much by passing through the interstellar gas at these kinds of speeds and they are so widely spread that they could pass through the Oort cloud of another star without much risk of collisions, There are two things that might affect them, though:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Gravitational interactions with stars they pass could strip some Oort cloud bodies, or send then into the system as comets, or whatever&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Whatever caused this star to become a hypervelocity star in the first place (eg a close encounter with a supermassive black hole) might well have disrupted the Oort cloud, possibly fatally.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-03-30T10:55:01.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25699" PostTypeId="1" CreationDate="2018-03-30T11:46:30.753" Score="-4" ViewCount="91" Body="&lt;p&gt;Why is it the some massive stars go supernovae and than either form a black hole, pulsar or a magnetar. All stars essentially go through the same process when they’re alive and burning through their fuel. It almost looks as if stars create everything in the universe one way or another, is it too far-fetched to think that black holes (theortical), pulsars and magnetars to name a few, could be completely different objects? Like have we actually witnessed the creation of any of these objects??&lt;/p&gt;&#xA;" OwnerUserId="22413" LastEditorUserId="22413" LastEditDate="2018-03-30T11:56:29.133" LastActivityDate="2018-03-30T12:03:56.973" Title="creations after supernovas" Tags="&lt;supernova&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" ClosedDate="2018-04-04T12:48:01.013" ContentLicense="CC BY-SA 3.0" />
  <row Id="25700" PostTypeId="2" ParentId="25699" CreationDate="2018-03-30T12:03:56.973" Score="2" Body="&lt;p&gt;It's a question of pressure. When the hydrogen fuel in the centre of a star runs out, the helium &quot;ash&quot; starts to collapse under its own gravity, since there is no longer a source of energy within it to &quot;hold it up&quot;. For a small star, that's that. You end up with a small ball of highly compressed helium sitting their slowly cooling down.  For slightly bigger stars the process is a bit &quot;messier&quot; involving the helium fusing further to carbon and oxygen and other processes that may &quot;blow off&quot; the outer layers of the star. Nevertheless the outcome is basically the same, except that now its a ball of carbon and oxygen slowly cooling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out though, that there is a limit to how much pressure any matter made up of nuclei and electrons in the usual way can support, and for stars more than a few times bigger than the Sun, that limit is reached. So, as fusion goes on, and the core of the star becomes larger and more and more compressed, it eventually collapses under its own gravity, driving the electrons into the nuclei where the combine with protons to form neutrons and becoming a neutron star. This happens rather suddenly and the resulting shock has a rather dramatic effect on the rest of the star, which is what we call a type II supernova. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ball of neutrons is a neutron star, and may become a pulsar or a magnetar for a while before it calms down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even then, there is a limit to how much pressure a ball of neutrons can sustain, and if the star was very big the core may collapse further. If that happens it will become so dense that it will form a black hole and no further information will be able to escape it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course all of this is, in some sense theory. We can see some evidence of all of these types of object, and we can see supernovae happening far away, but it's a bit hard to do experiments on this scale, and the objects and events are a bit rare to quickly build up a certain picture by observing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also worth saying that there are some black holes that may never have formed from stars -- the really big ones at the centre of galaxies. They may just be the result of local density fluctuations in the early universe, which have subsequently grown. &lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-03-30T12:03:56.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25701" PostTypeId="1" CreationDate="2018-03-30T15:37:32.463" Score="5" ViewCount="639" Body="&lt;p&gt;I'm an undergrad interested in astophysics and this question struck my mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the &lt;a href=&quot;https://en.wikipedia.org/wiki/CNO_cycle&quot; rel=&quot;nofollow noreferrer&quot;&gt;CNO cycle&lt;/a&gt; requires carbon to start, and carbon is produced by thermonuclear fusion in stars, is it fair to think that the hotter stars in the first generation didn't have a CNO cycle?&lt;br /&gt;&#xA;Or should I instead think that the first generation of stars had an initial metalicity and therefore the CNO cycle was possible?&lt;/p&gt;&#xA;" OwnerUserId="22465" LastEditorUserId="616" LastEditDate="2018-03-30T19:33:10.380" LastActivityDate="2018-03-31T02:47:18.577" Title="The CNO cycle in first generation of stars" Tags="&lt;astrophysics&gt;&lt;cosmology&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25702" PostTypeId="2" ParentId="25701" CreationDate="2018-03-30T17:10:54.770" Score="6" Body="&lt;p&gt;First generation stars didn't do the CNO cycle initally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is estimated, that after the Big Bang three quarters of the matter is hydogren, one quarter helium, and trace amounts of heavier elements.  This is then also the composition of (as of now unobserved) first generation stars, having zero metallicity.  Even though those stars probably were very massive, they could not run the CNO cycle initially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luckily, the CNO cycle is not the only nuclear fusion process in stars.  In the proton-proton chain reaction, hydrogen is fused into helium.  The star is in equilibrium until the hydrogen in the core is &quot;used up&quot;.  Then it may collapse and heat until the &lt;a href=&quot;https://en.wikipedia.org/wiki/Triple-alpha_process&quot; rel=&quot;noreferrer&quot;&gt;triple-alpha process&lt;/a&gt; begins to produce carbon.  From there on there are more and more processes producing many different elements.  Due to convection and shell-burning, the CNO cycle might have been running in those first generation stars at later stages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, the most massive of the first generation (population III) stars may have ended up as black holes, gobbling up all of the produced metals (elements heavier than helium).  Many others may have exploded in &lt;a href=&quot;https://en.wikipedia.org/wiki/Pair-instability_supernova&quot; rel=&quot;noreferrer&quot;&gt;pair-instability supernovae&lt;/a&gt; ejecting all material into interstellar space, leaving no compact remnant behind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://iopscience.iop.org/article/10.1086/338487/pdf&quot; rel=&quot;noreferrer&quot;&gt;A Heger and S. E. Woosley: The Nucleosynthesis Signature of Population III&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22239" LastActivityDate="2018-03-30T17:10:54.770" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25703" PostTypeId="2" ParentId="669" CreationDate="2018-03-30T19:05:26.857" Score="7" Body="&lt;p&gt;The universe doesn't expand away from any centre perse. All the distances are expanding uniformly throughout the universe. This causes such an effect that for each individual observer, it looks as if the entire universe moves away from them. It can be demonstrated using this figure (from google):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ywFMw.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ywFMw.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A$ represents the universe at one moment, $B$ represents the universe at a later time. You can notice (barely) that $B$ is scaled up by a small amount. This represents the expansion of the universe. Now, suppose you put $B$ over $A$ as shown in $C$, then it looks like the universe expanded away from $X$. But if you place them as shown in $D$ then it looks like the entire universe is expanding from another point! This is all due to the homogenous expansion of the universe.&lt;/p&gt;&#xA;" OwnerUserId="16744" LastActivityDate="2018-03-30T19:05:26.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25704" PostTypeId="2" ParentId="25696" CreationDate="2018-03-30T19:24:09.227" Score="1" Body="&lt;p&gt;To transform an equatorial $(x, y, z)$&#xA;into a galactic $(x_G, y_G, z_G)$, the&#xA;&lt;a href=&quot;https://gaia.esac.esa.int/documentation//GDR1/Data_processing/chap_cu3ast/sec_cu3ast_intro.html#SS7&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gaia data release 1 documentation, section 3.1.7&lt;/a&gt;, uses&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{bmatrix} x_G \\ y_G \\ z_G\end{bmatrix}&#xA;= \mathbf{A}'_G&#xA;\begin{bmatrix} x \\ y \\ z \end{bmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathbf{A}'_G = \begin{bmatrix}&#xA;{−0.054876} ~ {−0.873437} ~ {−0.483835} \\&#xA;{+0.494109} ~ {−0.444830} ~ {+0.746982} \\&#xA;{−0.867666} ~ {−0.198076} ~ {+0.455984}&#xA;\end{bmatrix}&#xA;$$&#xA;is composed of three coordinate frame rotations&#xA;$$&#xA;\mathbf{A}'_G = \mathbf{R}_Z(-l_\Omega)&#xA;~ \mathbf{R}_X(90^\circ - \delta_G)&#xA;~ \mathbf{R}_Z(\alpha_G + 90^\circ)&#xA;$$&#xA;with the north galactic pole at equatorial coordinates&#xA;$(\alpha_G = 192.859^\circ, \delta_G = +27.128^\circ)$&#xA;and the galactic equator crossing the celestial equator at galactic longitude&#xA;$l_\Omega = 32.932^\circ$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equivalent to the transformation defined in the&#xA;&lt;a href=&quot;https://www.cosmos.esa.int/web/hipparcos/catalogues&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hipparcos catalogue&lt;/a&gt;, volume 1, section 1.5.3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1989A&amp;amp;A...218..325M&quot; rel=&quot;nofollow noreferrer&quot;&gt;Murray 1989&lt;/a&gt; and&#xA;&lt;a href=&quot;https://www.aanda.org/articles/aa/full_html/2011/02/aa14961-10/aa14961-10.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Liu 2011&lt;/a&gt;&#xA;give similar matrices whose elements differ after 6 decimal places, which matters only if you require sub-arcsecond precision.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-03-30T21:35:58.000" LastActivityDate="2018-03-30T21:35:58.000" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25705" PostTypeId="2" ParentId="25641" CreationDate="2018-03-30T23:21:35.787" Score="2" Body="&lt;p&gt;Supernovae probably occur about once per century in our own (quite large) galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were to assume similar star formation rates and a similar stellar birth-mass distribution in other galaxies and that the galaxies in the HST-UDF were similar to our own, then we could also assume that the supernova rate was similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The UDF was taken over about 4 months, or 1/300 of a century. So one might expect a supernova to occur in about 1/300 galaxies - so about 300 of the 10,000 galaxies observed will have hosted a supernova during the period that the image was built up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, even if this figure is correct (see below), many of these may have been missed because the galaxies were too far away or the supernovae were obscured by dust, or because they brightened and faded between observations (there was a 6 week gap in the observations). On the other hand, the UDF was imaged later by HST with other instruments, leading to further opportunities to identify supernova candidates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculated number could be awry in many ways. Most galaxies are smaller than the Milky Way, so one might expect fewer supernovae. On the other hand, most of these galaxies are far away and the star formation rate was likely higher in the past, leading to more supernovae. On the other hand, the supernova rate will be higher in some types of galaxies (Starbursts), but lower in others (ellipticals). Without a very detailed demography of the types, metallicity, masses and ages of the galaxies, and a detailed understanding of how the supernova rate varies with these parameters, your question is difficult to answer.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-03-30T23:31:25.700" LastActivityDate="2018-03-30T23:31:25.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25706" PostTypeId="2" ParentId="25701" CreationDate="2018-03-31T02:47:18.577" Score="4" Body="&lt;p&gt;I guess by what you mean &quot;first generation&quot; is the Pop III (aka zero-metallicity) stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These stars composed of H and He only (might have very small and negligible about of heavier elements) since they formed at the very early universe. Their evolution starts with fusion of H to get more He, and burning He to get more heavier elements. This is similar to what happens in the Sun. Therefore, carbon can be produced, and CNO cycle can occur at some point of the life time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also supported by the idea the idea that Pop III might be very massive and maintain its mass without significant loss by the wind mechanism (i.e., mass loss rate is higher with metallicity due to line emission of heavy elements). It has been shown by simulations that these massive stars can end their lives by thermonuclear explosion (i.e., similar to SNe Type Ia), instead of core-collapse one.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-03-31T02:47:18.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25708" PostTypeId="1" CreationDate="2018-03-31T10:20:42.563" Score="1" ViewCount="172" Body="&lt;p&gt;Laniakea and other super clusters are moving towards shapely super cluster .&#xA;Even the Great Attractor is moving towards it. Isn't it all about the Big Crunch?&lt;/p&gt;&#xA;" OwnerUserId="22477" LastActivityDate="2018-03-31T10:20:42.563" Title="Great attractor is doing the big crunch?" Tags="&lt;gravity&gt;&lt;galaxy-cluster&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2018-04-01T09:30:16.770" ContentLicense="CC BY-SA 3.0" />
  <row Id="25710" PostTypeId="1" CreationDate="2018-03-31T20:35:16.093" Score="0" ViewCount="63" Body="&lt;p&gt;Can I point to somewhere in the sky where the Big Bang happened? A pathway so to speak on where we (and our progenitor elements) came from?&lt;/p&gt;&#xA;" OwnerUserId="10117" LastActivityDate="2018-03-31T21:07:20.873" Title="Location of Big Bang" Tags="&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2018-04-01T02:02:07.360" ContentLicense="CC BY-SA 3.0" />
  <row Id="25711" PostTypeId="2" ParentId="25710" CreationDate="2018-03-31T21:07:20.873" Score="0" Body="&lt;p&gt;There is no such unique point, because a key element of the Big Bang model is called the &quot;cosmological principle,&quot; which means that every point in the universe is taken as essentially equivalent to any other.  The idea there is that as the universe ages, it changes, but it changes in the same ways everywhere so there is nothing going on that distinguishes one region from another.  This is a simplifying assumption that is supported by the fact that we see no boundary to the universe, and we see no evidence that anywhere else is different from here, the differences we see appear to be due only to the fact that we are seeing farthing into the past as we look farther away.  This assumption means you could pick any point you want, point to it, and say &quot;the Big Bang started there,&quot; and you would be no more right nor wrong than if you pointed anywhere else.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-03-31T21:07:20.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25712" PostTypeId="2" ParentId="23320" CreationDate="2018-04-01T03:38:28.897" Score="-2" Body="&lt;p&gt;They are not stars I’m from Prince Edward Island Canada those red blue and white looking disco balls in the sky are something the police or military are using. Look on U tube thier are people all over the world seeing the same thing . If u watch those for long enough they will drop every 3 hours approximately. And in most cases especially mine thier is heavy drone activity when these are present. I have really good footage of a lot of this stuff as I bought good camera equipment to gather evidence. If u have these near your house it’s not a good thing as your privacy is being affected from the law or military.&lt;/p&gt;&#xA;" OwnerUserId="22483" LastActivityDate="2018-04-01T03:38:28.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25715" PostTypeId="2" ParentId="23320" CreationDate="2018-04-01T09:49:17.373" Score="7" Body="&lt;p&gt;As you have already identified the objects you were seeing, I'll explain the effect you were seeing.  In that situation there are three things to consider: atmospheric chromatic dispersion, seeing and human color perception.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light entering the atmosphere is refracted, because of the changing speed of light in air compared to the vacuum of space.  The amount of refraction depends on the wavelength, causing &lt;em&gt;optical dispersion.&lt;/em&gt;  This effect is strongest for objects appearing near the horizon.  Basically the different color components of the star's image appear at slightly different elevations in the sky &lt;a href=&quot;https://cseligman.com/text/sky/atmosphericdispersion.htm&quot; rel=&quot;noreferrer&quot;&gt;(nice in-depth explanation)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turbulence in the atmosphere causes the &quot;twinkling&quot; or &lt;em&gt;seeing.&lt;/em&gt;  Moving pockets of hotter and colder air act like lenses that are projecting the star's light into varying directions, so a varying amount of light reaches your eye.  Together with the dispersion this produces a colorful twinkling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cone_cell&quot; rel=&quot;noreferrer&quot;&gt;color perception in the human eye&lt;/a&gt; doesn't work in faint light, this colorful twinkling is observed for the brightest stars only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The doppler shift is not the reason for the red and blue colors.  At a typical velocity dispersion of 30km/s of our surrounding stars, the doppler shift changes the wavelength by an imperceptible 10&lt;sup&gt;-4&lt;/sup&gt; fraction.  Also this would have to change extremely rapidly, which simply does not occur for our average night-sky stars or &lt;a href=&quot;https://www.youtube.com/watch?v=T8OAmcUnm4g&amp;amp;feature=youtu.be&amp;amp;t=4m32s&quot; rel=&quot;noreferrer&quot;&gt;most other objects,&lt;/a&gt; for that matter :-)&lt;/p&gt;&#xA;" OwnerUserId="22239" LastActivityDate="2018-04-01T09:49:17.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25716" PostTypeId="1" AcceptedAnswerId="25718" CreationDate="2018-04-01T12:22:41.403" Score="3" ViewCount="299" Body="&lt;p&gt;I would like to know how to calculate the time/date of the ISS passing in front of the moon (between the observer and the moon) or the sun, or to calculate the position in which the observer have to be to see the ISS passing in front of the moon/sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Th goal is to make a photograph of the ISS in the moon/sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible with tools such as Stellarium?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AOQ8V.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AOQ8V.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7085" LastEditorUserId="7085" LastEditDate="2018-04-04T09:14:21.307" LastActivityDate="2018-04-04T09:14:21.307" Title="How to calculate the position to be to see ISS passing in front of the moon or the sun?" Tags="&lt;the-sun&gt;&lt;the-moon&gt;&lt;photography&gt;&lt;iss&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25717" PostTypeId="1" CreationDate="2018-04-01T19:16:30.467" Score="3" ViewCount="414" Body="&lt;p&gt;Why Helium-burning stars are found in a horizontal branch on the H-R diagram? Why do they have roughly the same luminosity?&lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2018-04-02T15:49:23.917" Title="Why Helium-burning Stars are found in a Horizontal Branch?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25718" PostTypeId="2" ParentId="25716" CreationDate="2018-04-02T01:08:08.780" Score="3" Body="&lt;p&gt;The &lt;a href=&quot;https://transit-finder.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ISS Transit Finder&lt;/a&gt; offers such predictions up to 30 days ahead.&#xA;If a query yields no transits at the specified location,&#xA;the &quot;Show on Map&quot; button for a close pass shows where a transit would be visible.&#xA;As the &lt;a href=&quot;https://transit-finder.com/help&quot; rel=&quot;nofollow noreferrer&quot;&gt;documentation&lt;/a&gt; notes,&#xA;predictions beyond 10 days are rather uncertain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recent versions of Stellarium can show ISS's apparent position at a given place and time,&#xA;offering an alternative prediction for any given event.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-04-02T01:08:08.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25719" PostTypeId="1" CreationDate="2018-04-02T02:38:07.980" Score="23" ViewCount="7347" Body="&lt;p&gt;Consider it doesn't collide with any other objects. Would it be preserved perfectly in the vacuum or would its surface be damaged by anything like UV rays, radiation, gas, space dust, etc?&lt;/p&gt;&#xA;" OwnerUserId="22493" LastActivityDate="2018-04-03T15:25:42.547" Title="What would happen to a polished marble statue left in space for a million years?" Tags="&lt;universe&gt;&lt;astrophysics&gt;&lt;space&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25721" PostTypeId="2" ParentId="25719" CreationDate="2018-04-02T06:05:18.350" Score="2" Body="&lt;p&gt;Cosmic rays may impinge pressure on the statue, which will deteriorate its surface. Various electromagnetic rays (X-ray, Gamma rays and Infrared) can interact with the chemical elements of the statue. &lt;/p&gt;&#xA;" OwnerUserId="20162" LastActivityDate="2018-04-02T06:05:18.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25722" PostTypeId="2" ParentId="25719" CreationDate="2018-04-02T07:48:02.517" Score="38" Body="&lt;p&gt;There are three main space weathering processes that will affect the surface of the marble.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Cosmic rays, high energy particle from the sun and beyond, will hit the surface. This can change the chemistry of the surface. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Solar wind particles, hydrogen and helium, can become implanted in the surface&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Micrometeoroids will impact the surface, causing small craters, melting, and the inclusion of other elements such as iron.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These processes will tend to change the surface, developing a patina on a timescale of a hundred thousand years. The surface will darken (though as marble is not a typical rock in asteroids, there isn't any direct evidence of what happens with marble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Marble is largely CaCO3, and this is in a equilibrium with CaO and CO2. At  standard temperatures and even the very low partial pressure of CO2 in the atmosphere, this equilibium favours CaCO3. &lt;a href=&quot;https://www.researchgate.net/post/Does_anybody_know_how_to_reduce_decomposition_temperature_of_Calcium_Carbonate&quot; rel=&quot;noreferrer&quot;&gt;In our atmosphere one needs a temperature of 550⁰C to decompose Calcite&lt;/a&gt;. However in space there is no CO2, and so the Calcite would very slowly decompose to CaO. &lt;a href=&quot;http://meteorites.wustl.edu/metcomp/alca.htm&quot; rel=&quot;noreferrer&quot;&gt;Calcium in meteorites is mostly in the form of CaO.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-04-02T16:36:25.060" LastActivityDate="2018-04-02T16:36:25.060" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25723" PostTypeId="2" ParentId="25717" CreationDate="2018-04-02T13:08:29.337" Score="2" Body="&lt;p&gt;This is explained in &lt;a href=&quot;https://en.wikipedia.org/wiki/Horizontal_branch&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Wikipedia article&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Stars on the horizontal branch all have very similar core masses, following the helium flash. This means that they have very similar luminosities, and on a Hertzsprung–Russell diagram plotted by visual magnitude the branch is horizontal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To expand a little. In stars of a certain mass range, helium builds up in the core until it reaches a specific mass, at which point the &quot;Helium flash&quot; occurs and burning of helium to carbon and oxygen starts throughout the core. When things settle down, helium burning is going on in the core, which is more or less the same size, independently of the original mass of the star. Since this is the main power source of these stars, they all have about the same luminosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variation across the branch comes from how much remaining gas there is outside the helium-burning shell -- more gas means a larger cooler star radiating the same total amount of energy&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="21928" LastEditDate="2018-04-02T15:49:23.917" LastActivityDate="2018-04-02T15:49:23.917" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25728" PostTypeId="2" ParentId="24306" CreationDate="2018-04-02T21:15:38.190" Score="6" Body="&lt;p&gt;Try this non-Gedanken-but-real experiment:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hold an apple in your hand so the Sun (or a light) shines on it. Nerds may use a proper globe.&lt;/li&gt;&#xA;&lt;li&gt;Along the apple equator draw arrows to the right with an E for East: E&amp;rarr;E&amp;rarr;E&amp;rarr;E, .... Nerds be sure to use a non-permanent marker.&lt;/li&gt;&#xA;&lt;li&gt;Spin it counterclockwise (when looked at from above, i.e. the North pole). You're doing it right when all the Es move in the direction of the arrows. West to East.&lt;/li&gt;&#xA;&lt;li&gt;The Sun rises on all points on the apple that go from dark to bright; identify that line. It should be half a circle.&lt;/li&gt;&#xA;&lt;li&gt;Imagine a near indestructible humanoid killer machine from the future standing on that line (called &lt;em&gt;terminator&lt;/em&gt;, the line, that is). What direction does the machine look when it glances at the Sun the moment he (the Sun, that is) rises?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6593" LastEditorUserId="6593" LastEditDate="2018-04-03T16:13:38.357" LastActivityDate="2018-04-03T16:13:38.357" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25730" PostTypeId="1" CreationDate="2018-04-02T23:05:46.050" Score="4" ViewCount="227" Body="&lt;p&gt;I am trying to generate skymaps for a set of &lt;a href=&quot;http://www.mpe.mpg.de/617954/Fermi-GBM&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fermi GBM&lt;/a&gt; triggers, containing the probability that a trigger originated in a given part of the sky.  For each trigger, I have access to the RA, dec, error radius, and reliability values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://gcn.gsfc.nasa.gov/gcn/fermi_gbm_subthresh_archive.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This website&lt;/a&gt; is a perfect example of what I am trying to do.  There is a table with with RA, Dec, error, and reliability values, which were used to generate maps &lt;a href=&quot;https://gcn.gsfc.nasa.gov/notices_gbm_sub/gbm_subthresh_544041142.080000_map.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;(sample)&lt;/a&gt;.  My question is, how was this done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding of the problem is that I must first generate fits files based on the data, and then use either a python library or external program to render the skymap from the file.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have already looked at both &lt;a href=&quot;http://healpy.readthedocs.io/en/latest/tutorial.html#creating-and-manipulating-maps&quot; rel=&quot;nofollow noreferrer&quot;&gt;healpy&lt;/a&gt; and &lt;a href=&quot;http://docs.astropy.org/en/stable/io/fits/&quot; rel=&quot;nofollow noreferrer&quot;&gt;astropy&lt;/a&gt;, which provide a means to create and manipulate fits files.  Creating a blank fits file is easy, but I am at a loss as to how to populate the files I generate with relevant probabilities based on the data I have access to. Any help, even just a pointer in the right direction, would be greatly appreciated. Thank you for your time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit- description of the error radius parameter: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This parameter is the uncertainty in the position, in degrees. The GBM errors are not symmetric, and the given value is the average of the error ellipse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is what results in the asymmetrical blob shapes.  &lt;/p&gt;&#xA;" OwnerUserId="22510" LastEditorUserId="22510" LastEditDate="2018-04-03T20:21:16.340" LastActivityDate="2018-04-03T20:21:16.340" Title="Generate probability skymap" Tags="&lt;python&gt;&lt;gamma-ray-bursts&gt;&lt;fits-header&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25732" PostTypeId="1" CreationDate="2018-04-03T03:49:13.203" Score="3" ViewCount="144" Body="&lt;p&gt;I'd like to determine the distance at which a telescope with a certain set of parameters (sensitivity, collector area) can detect a blackbody with certain other parameters (temperature, emitting area).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let us suppose we have an infrared telescope with a sensitivity of '1Jy/sr' and a collector area of 10m^2 at a distance D from a 300m^2 perfectly black body at a temperature of 30 Kelvin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I determine D? What additional information do I need to perform the calculation?&lt;/p&gt;&#xA;" OwnerUserId="22518" LastActivityDate="2018-04-04T13:49:50.850" Title="Can a telescope detect a blackbody?" Tags="&lt;telescope&gt;&lt;infrared&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25733" PostTypeId="1" CreationDate="2018-04-03T03:50:52.397" Score="0" ViewCount="85" Body="&lt;p&gt;I'm quite lost on the definitions of words like omniverse, megaverse, multiverse, meta-universe, universe, and cosmos. I used to think the Universe and the Cosmos were essentially the same things, but now I'm reading the latter is an orderly system (not sure what's meant by that).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody help me understand what they each represent? In 1980, Carl Sagan said &quot;The cosmos is all that is, or ever was, or ever will be.&quot; This sounds like the Cosmos would be the greatest of all and all else would fall into that. Would it be accurate, for example, to say Cosmos &gt; Multiverse &gt; Universe? What about the other things I mentioned earlier?&lt;/p&gt;&#xA;" OwnerUserId="22519" LastEditorUserId="37191" LastEditDate="2021-02-01T21:49:30.570" LastActivityDate="2021-02-01T21:49:30.570" Title="Is the Cosmos a super set of the Universe and any hypothetical multiverse(s)?" Tags="&lt;universe&gt;&lt;multiverse&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2018-04-03T21:06:45.557" ContentLicense="CC BY-SA 3.0" />
  <row Id="25734" PostTypeId="2" ParentId="25733" CreationDate="2018-04-03T04:42:49.383" Score="4" Body="&lt;p&gt;&quot;Cosmos&quot; is the English cognate of the ancient Greek word which refers to the entirety of existence. It is equivalent to the Latin-derived word &quot;universe;&quot; the only difference between the two is the language of origin.&lt;/p&gt;&#xA;" OwnerUserId="4042" LastActivityDate="2018-04-03T04:42:49.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25736" PostTypeId="1" AcceptedAnswerId="25742" CreationDate="2018-04-03T09:29:30.007" Score="20" ViewCount="2662" Body="&lt;p&gt;This &lt;a href=&quot;https://space.stackexchange.com/a/26423/12102&quot;&gt;answer&lt;/a&gt; links to one of Scott Manley's excellent asteroid videos &lt;a href=&quot;https://www.youtube.com/watch?v=BKKg4lZ_o-Y&quot; rel=&quot;noreferrer&quot;&gt;Asteroid Discovery - 1970-2015 - 8K resolution&lt;/a&gt;. The animation highlights the positions of the meteors at the moment of their discovery, and by watching one can see the technology improve and notice patterns as instruments are pointed in different directions to avoid the light from the Sun and (at least sometimes) the Moon. (there is music, adjust volume accordingly)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are often fan-shaped patterns showing the directions that more sensitive telescopes with modest fields of view are pointed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I noticed that only during the year 2010, roughly between asteroid numbers &lt;code&gt;500,000&lt;/code&gt; and &lt;code&gt;520,000&lt;/code&gt; there are radial striations at certain distances from the Sun. I don't see this happen at any other time during the video.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this just a rendering artifact, or is it real? If real, what would cause the periodic radial modulation of sensitivity, and only in 2010?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;note 1:&lt;/strong&gt; YouTube allows for playback rates between 25% and 200%, and variable video resolutions. I found 25% and 1080p optimal for my current internet connection and screen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;note 2:&lt;/strong&gt; For those with GIFs disabled, one image is a GIF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;note 3:&lt;/strong&gt; 2nd image contains several cropped screenshots highlighting the &quot;periodic radial modulation of sensitivity&quot; in asteroid detection during 2010, for clarification purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0dd9S.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0dd9S.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/u9T70.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/u9T70.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uUfnW.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uUfnW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-04-03T10:55:05.813" LastActivityDate="2018-04-04T11:53:28.587" Title="Why did the distribution of asteroids discovered in 2010 have a radial modulation?" Tags="&lt;asteroids&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25737" PostTypeId="2" ParentId="25736" CreationDate="2018-04-03T10:34:28.203" Score="9" Body="&lt;p&gt;&lt;strong&gt;Answer based on a misunderstanding of the question, left here because it contains some useful background on WISE.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;pie-shaped&lt;/em&gt; patterns starting in 2010 are &lt;a href=&quot;https://www.youtube.com/watch?v=S_d-gs0WoUw&quot; rel=&quot;nofollow noreferrer&quot;&gt;results of the WISE mission&lt;/a&gt; (see the video description). The &lt;em&gt;radial&lt;/em&gt; pattern within those pie shapes is not explained by my answer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;NASA's Wide-field Infrared Survey Explorer (&lt;a href=&quot;https://www.space.com/33659-wise-space-telescope.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;WISE&lt;/a&gt;) is a space telescope launched in 2009 to map the entire sky in infrared wavelengths. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;WISE imaged the entire sky twice before running out of coolant in 2010. It then did a brief mission called NEOWISE, to look at near-Earth objects (NEOs) such as asteroids and comets, for four months before being placed into hibernation in February 2011. Less than three years later, in December 2013, the telescope was revived to continue its NEOWISE mission. That work continues today.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Looking at WISE: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CGcSO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CGcSO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can see the telescope is perpendicular to the solar panels, so it'll tend to look at objects perpendicular to the Earth-Sun line, which causes the bright bands you see in the video. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.space.com/33659-wise-space-telescope.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.space.com/33659-wise-space-telescope.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="361" LastEditorUserId="361" LastEditDate="2018-04-04T11:53:28.587" LastActivityDate="2018-04-04T11:53:28.587" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25738" PostTypeId="2" ParentId="25732" CreationDate="2018-04-03T13:42:48.277" Score="1" Body="&lt;p&gt;So the peak spectral radiance  (This from &lt;a href=&quot;https://en.wikipedia.org/wiki/Black-body_radiation#Planck&amp;#39;s_law_of_black-body_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;) is given by $$\displaystyle \nu _{\max }=T\times 1.04\times 10^{11}\ \mathrm {Hz} /\mathrm {K} $$ This works out to about $3\times 10^{12} Hz$. The and the spectral radiance will be given by Planck's Law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\displaystyle B_{\nu }(T)={\frac {2h\nu ^{3}}{c^{2}}}{\frac {1}{e^{\frac {h\nu }{kT}}-1}},}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;giving about $3\times 10^{-15} W/m^2/Hz$ so total power emitted by the $300 m^2$ black body is about $10^{-12} W/Hz$. That is spread over a spherical surface of radius $D$ for an intensity of $10^{-13}/D^2 $ of which you are capturing $10^{-12}/D^2 W/Hz$  or about $10^{14}/D^2 Jy$. So you are fine if $D&amp;lt; 10^7 m$ more or less.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="21928" LastEditDate="2018-04-04T13:49:50.850" LastActivityDate="2018-04-04T13:49:50.850" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25739" PostTypeId="2" ParentId="25719" CreationDate="2018-04-03T14:41:37.307" Score="0" Body="&lt;p&gt;Piggy backing on James K's answer above, there is a fourth process depending on the proximity to any star, which is thermal stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the statue rotates with respect to any &quot;near&quot; star, thermal stress will lead to surface weathering over time: &lt;a href=&quot;https://en.wikipedia.org/wiki/Weathering#Thermal_stress&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Weathering#Thermal_stress&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thermal stress weathering (sometimes called insolation weathering)[2] results from the expansion and contraction of rock, caused by temperature changes. For example, heating of rocks by sunlight or fires can cause expansion of their constituent minerals. As some minerals expand more than others, temperature changes set up differential stresses that eventually cause the rock to crack apart. Because the outer surface of a rock is often warmer or colder than the more protected inner portions, some rocks may weather by exfoliation – the peeling away of outer layers. This process may be sharply accelerated if ice forms in the surface cracks. When water freezes, it expands with a force of about 1465 Mg/m^2, disintegrating huge rock masses and dislodging mineral grains from smaller fragments.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="22525" LastEditorDisplayName="user1569" LastEditDate="2018-04-03T15:25:42.547" LastActivityDate="2018-04-03T15:25:42.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25740" PostTypeId="2" ParentId="25732" CreationDate="2018-04-03T16:52:47.410" Score="2" Body="&lt;p&gt;Let's assume a uniform laminar emitter, oriented so that it is at right angles to the line to the detector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific intensity emitted by a blackbody, at the surface of the blackbody is&#xA;$$B_{\nu} = \frac{2h\nu^3}{c^2} \frac{1}{\exp[h\nu/kT] -1}\ {\rm Wm}^{-2}{\rm Hz}^{-1},$$&#xA;where $B_{\nu}$ is the &quot;Planck function&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the definition of &lt;a href=&quot;https://en.wikipedia.org/wiki/Specific_radiative_intensity&quot; rel=&quot;nofollow noreferrer&quot;&gt;specific intensity&lt;/a&gt; we can calculate the flux (energy per unit time, per unit frequency) received at the detector as&#xA;$$ f_{\nu} \simeq B_{\nu}A \Delta \Omega \cos \theta\,$$&#xA;where $\Delta \Omega$ is the solid angle of the detector subtended at the laminar surface, $A$ is the area of the laminar surface, and $\theta$ is the angle with respect to a normal from the surface and here, $\cos \theta \simeq 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solid angle subtended by the detector at the surface is $\Delta \Omega = a/D^2$, where $a$ is the area of the detector, but we can also define a solid angle $\Delta \omega = A/D^2$, which is the solid angle of the surface at the detector. Thus&#xA;$$\frac{f_{\nu}}{\Delta \omega} =  B_{\nu}a  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the detector has a fixed sensitivity per unit solid angle then unless the object is unresolved, then it does not matter how far away it is; the value of $D$ does not come into it. It is a well-known fact in astronomy that resolved, extended objects have an observed &lt;a href=&quot;https://en.wikipedia.org/wiki/Surface_brightness&quot; rel=&quot;nofollow noreferrer&quot;&gt;surface brightness&lt;/a&gt; (flux density per unit solid angle) that is &lt;em&gt;independent of distance&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-04-04T13:49:03.160" LastActivityDate="2018-04-04T13:49:03.160" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25742" PostTypeId="2" ParentId="25736" CreationDate="2018-04-03T18:27:53.063" Score="17" Body="&lt;p&gt;I'm pretty sure that the radial pattern found in the data is a result of &lt;em&gt;WISE&lt;/em&gt;'s approximately 90 minute sampling cadence (dictated by the satellite's orbit), astrometric precision (about 0.2 arcseconds in the stacked images around launch, see &lt;A HREF=&quot;https://arxiv.org/abs/1008.0031&quot; rel=&quot;noreferrer&quot;&gt;Wright et al. 2010&lt;/A&gt;), and the number of free parameters in fitting the asteroid orbits based on that data. See, in the actual images the asteroids appear as points of light that shift appreciably between frames. IIRC, they expected 7 to 12 observations per asteroid. So you have 10-ish observations spanning 15 hours or so to fix the asteroid's orbital parameters around the sun. As you can imagine, there will be more parameters than can be perfectly fit with a single pass in this data set alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a guess: it's related to quantization in uncertainty estimates and how that feeds forward into the orbit fitting algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know the details behind the striping, but I'd bet it's related to the numerical precision used in the early processing of the data. They've either refined the orbits since using observations from passes separated by ~6 months, or modified how they're handling the numerical precision in their astrometric measurements since. More likely the former, but I'm sure if you asked Amy Mainzer (PI of NEOWISE, and head of the asteroid hunting part of the mission), Roc Cutri (head of the database creation and data processing part of the team), or any of the people on Mainzer's team, they could tell you more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relevant background: I was Ned Wright's grad student (original PI of &lt;em&gt;WISE&lt;/em&gt;), and he had me design and test an asteroid hunting algorithm in the run-up to launch (we ultimately didn't use it - it scaled like $N^2$, IIRC, and existing efforts in the literature scaled like $\log(N)$ or $N\log(N)$). I ended up working with the extragalctic part of the team, hence my uncertainty on the precise details of what the solar system team did. I think I asked them about the striping during a presentation, but I don't recall the answer, so I'm pretty sure the answer was mundane and unavoidable. &lt;/p&gt;&#xA;" OwnerUserId="14244" LastEditorUserId="14244" LastEditDate="2018-04-03T18:41:58.860" LastActivityDate="2018-04-03T18:41:58.860" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25743" PostTypeId="1" CreationDate="2018-04-03T19:46:09.743" Score="1" ViewCount="70" Body="&lt;p&gt;I am in Northampton England and every night in the SW sky is a very bright, white light that twinkles. It visibly appears to be moving towards the west but appears pretty much the same start place every night. Could anyone possibly tell me what it might be please?&lt;/p&gt;&#xA;" OwnerUserId="22529" LastActivityDate="2018-04-03T20:12:14.497" Title="What am I looking at?" Tags="&lt;star&gt;&lt;planet&gt;&lt;satellite&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25744" PostTypeId="2" ParentId="25743" CreationDate="2018-04-03T20:12:14.497" Score="1" Body="&lt;p&gt;Is it moving straight to the west (i.e. it keeps the same distance from the horizon) or does it 'move with the stars'?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you checked an online planetarium like &lt;a href=&quot;https://theskylive.com/planetarium#ra|3.402367700043592|dec|1.963283012898064|fov|62.59999999999999&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Sky Live&lt;/a&gt;? Based on the current information, it could just be &lt;a href=&quot;https://en.wikipedia.org/wiki/Sirius&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sirius&lt;/a&gt;, the brightest star in the night sky, or even Venus which is currently in her Evenstar period (but planets generally don't twinkle).&lt;/p&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2018-04-03T20:12:14.497" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25745" PostTypeId="2" ParentId="25736" CreationDate="2018-04-03T20:57:38.190" Score="6" Body="&lt;p&gt;The effect is already visible in the first video released in 2010:&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=S_d-gs0WoUw&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=S_d-gs0WoUw&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;For what it's worth, I tried to download the newest astorb.dat and plot it, but couldn't see the effect there. So it might very well be that the 2010 asteroids were based on preliminary WISE data and weren't all that accurate, and the later videos have not updated the old animations.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iiZLX.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iiZLX.png&quot; alt=&quot;Plot of asteroid locations&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;And indeed, from the comments of the 2010 video:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;odysseus9672:&lt;/strong&gt; @szyzyg I spoke with Prof Wright, the project PI, and he explained that the striping is due to the Minor Planet Center using an approximate fitting technique to the WISE data. The reasoning for this, I think (we've left Ned approved commentary here), is that the WISE data has a relatively short time baseline (~24 to 48 hrs), so the errors on the orbital fits will be quite large anyway, so there's no point in refining it much past what the error bars will support.﻿&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Scott Manley&lt;/strong&gt;: @odysseus9672 Glad to hear an explanation, this pretty much confirms my suspicions and makes me ask how many of the objects are likely to be lost again in the future. Most of the objects discovered in the leading edge scan in January will have passed through opposition with the Earth and most orbits haven't been refined. The second pass of the WISE survey will hopefully start getting longer arcs on some of these, but WISE will run out of coolant before it can observe every asteroid twice.﻿&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14071" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-04-03T20:57:38.190" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25746" PostTypeId="2" ParentId="25339" CreationDate="2018-04-03T22:01:02.133" Score="7" Body="&lt;p&gt;The Intergalactic medium at the relevant redshift is made of neutral hydrogen. What we can measure is the brightness temperature (the temperature that the IGM would have if it emitted as a blackbody) relative to the CMB. This quantity depends crucially on the following expression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{T_S - T_{CMB}}{T_S}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $T_S$ is called spin temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spin temperature is just a measure of the ratio between the number of hydrogen atoms in the first excited hyperfine state (spin parallel) and ground state (spin antiparallel).&#xA;It turns out that the spin temperature can be modified in three ways: one of them is Lyman resonant scattering (the other two are collisional coupling and scattering of CMB photons). &#xA;Using atomic physics it's not difficult to see that $T_S$ is a weighted mean of the temperature of the CMB and gas temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UV photons can change the spin temperature because a hydrogen atom in the lowest state, n=1 (if you are familiar with chemistry, a s state) with antiparallel spins, can absorb an UV photon and jump in a p (n=2) state, and then fall again in a s state but with parallel spins.&#xA;This mechanism is known as Wouthuysen–Field coupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before the emission takes place, the gas is at the same temperature of the CMB and both are equal to spin temperature, so the relative brightness temperature is zero.&#xA;When Lyman alpha emission begins the spin temperature decreases resulting in the observed absorption peak. &#xA;At some point the peak stops because the first stars emit X-rays and the gas becomes hotter than the background radiation.&lt;/p&gt;&#xA;" OwnerUserId="22531" LastActivityDate="2018-04-03T22:01:02.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25747" PostTypeId="1" CreationDate="2018-04-04T04:10:15.397" Score="4" ViewCount="857" Body="&lt;p&gt;I don't know if this is the correct term, but how would one calculate the time it takes for 3 planets to lie on the same line? &lt;/p&gt;&#xA;" OwnerUserId="18673" LastActivityDate="2018-04-04T14:17:33.950" Title="How would one calculate the synodic period between 3 planets?" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;solar-system&gt;&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25748" PostTypeId="2" ParentId="25339" CreationDate="2018-04-04T09:06:04.147" Score="4" Body="&lt;p&gt;Guiseppe Rossi's answer is excellent; I just want to add &lt;em&gt;why&lt;/em&gt; the mentioned UV radiation modifies the spectrum of the background radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(one word took the other, and it became a rather long comment.)&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The hyperfine level&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Neutral hydrogen in its ground state can be in two different configurations; either the proton and the electron may have &lt;em&gt;parallel&lt;/em&gt; spins ($\uparrow\uparrow$), or they may have &lt;em&gt;antiparallel&lt;/em&gt; spins ($\uparrow\downarrow$). When the spins are parallel, the atom has a slightly higher energy than when they're antiparallel. The atoms &quot;wants&quot; to make a spin flip to the lower energy configuration$^\dagger$, and will eventually do so, but since the line is forbidden, the lifetime of the parallel state is of the order $10^7\,\mathrm{yr}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relative population of the states is given by the Boltzmann distribution&#xA;$$&#xA;\begin{array}{rcl}&#xA;\frac{n_1}{n_0} &amp;amp; = &amp;amp; \frac{g_1}{g_0} \, e^{-\Delta E \, / \, k_\mathrm{B} T_S} \\&#xA;                &amp;amp; = &amp;amp; 3 \, e^{-0.068\,\mathrm{K} \, / \, T_S}\label{a}\tag{1},&#xA;\end{array}&#xA;$$&#xA;where subscripts 1 and 0 denote the $\uparrow\uparrow$ and $\uparrow\downarrow$ states, respectively, $n$ is the density, $g$ is the statistical weights (with $g_0,g_1 = 1,3$), $\Delta E = 5.9\times10^{-6}\,\mathrm{eV}$ is the energy difference of the states, $k_\mathrm{B}$ is the Boltzmann constant, and $T_S$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Excitation_temperature&quot; rel=&quot;nofollow noreferrer&quot;&gt;spin temperature&lt;/a&gt;, which I think is better thought of as &quot;a number that describes the relative populations&quot; than an actual temperature.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Departure from equilibrium&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;In thermal equilibrium, the spin temperature is equal to the &quot;real&quot;, kinetic temperature. Just after &lt;a href=&quot;https://en.wikipedia.org/wiki/Decoupling_(cosmology)&quot; rel=&quot;nofollow noreferrer&quot;&gt;decoupling&lt;/a&gt; of the radiation from matter at a redshift of $z\simeq1100$, the gas and the photons share the same energy, and since $T\gg 1$, we have that $n_1/n_0 \simeq 3$. But when the first stars begin to shine, they produce massive amounts of hard UV radiation which ionizes their surrounding medium. The ionized gas quickly recombines (in the beginning, at least), with $\sim2/3$ of the recombinations resulting in the emission of a Lyman $\alpha$ photon, i.e. a photon with an energy corresponding to the energy difference between the first excited state (one of the three $2P$ states) and the ground state (the $1S$ state) of the hydrogen atom (10.2 eV).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Ly$\alpha$ photons scatter multiple times on the neutral hydrogen. Each scattering excites an atom from $1S\rightarrow 2P$, which subsequently de-excites and emits an Ly$\alpha$ photon in another direction. But since the energy difference between the $2P$ and the $1S$ state is a million times larger than between the hyperfine states, there is equal chance of ending in the $\uparrow\uparrow$ and the $\uparrow\downarrow$ state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, $n_1/n_0$ is no longer $\simeq 3$, but is driven toward $\sim 1$. This is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Wouthuysen%E2%80%93Field_coupling&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wouthuysen–Field effect&lt;/a&gt; that Guiseppe Rossi mentions; from eq. \ref{a}, you see that this corresponds to a much smaller spin temperature, and thus the factor that Guiseppe Rossi mentions becomes negative. The full equation describing the brightness (or, equivalently, the flux received) as a function of redshift can be written (e.g. &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0311514&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zaldarriaga et al. 2004&lt;/a&gt;)&#xA;$$&#xA;T(z) = 23\,\mathrm{mK} \,&#xA;       \frac{T_S - T_\mathrm{CMB}}{T_S} \,&#xA;       (1+\delta)&#xA;       x_\mathrm{HI}(z)&#xA;       \frac{\Omega_\mathrm{b}h^2}{0.02}&#xA;       \left( \frac{0.15}{\Omega_\mathrm{m}h^2} \, \frac{1+z}{10} \right)^{1/2}\label{b}\tag{2}&#xA;$$&#xA;and when the $(T_S - T_\mathrm{CMB})\,/\,T_S$ factor is negative, you will get an absorption line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In eq. \ref{b},&#xA;$\delta$,&#xA;$x_\mathrm{HI}(z)$,&#xA;$\Omega_\mathrm{b}$,&#xA;$\Omega_\mathrm{m}$, and&#xA;$h$,&#xA;are&#xA;the local overdensity,&#xA;the neutral fraction of hydrogen,&#xA;the baryon and matter density parameter, and&#xA;the dimensionless (reduced) Hubble constant,&#xA;respectively, but this is of less importance.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the observed absorption line (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2018Natur.555...67B&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bowman et al. 2018&lt;/a&gt;) starts to drop around an observed frequency of $\nu_\mathrm{obs} = 65\text{–}70\,\mathrm{MHz}$, and since the rest frequency of the hyperfine line is $\nu_\mathrm{rest} = 1420\,\mathrm{MHz}$, this means that the first stars appeared around a redshift of $z = \nu_\mathrm{rest}/\nu_\mathrm{obs} - 1 \simeq 20$, corresponding to an age of the Universe of $\sim 180\,\mathrm{Myr}$ (i.e. million years — the largest absorption is reached at $z\simeq 17$, or $t\simeq 200\,\mathrm{Myr}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the big question is, according to eq. \ref{b} the dip should be of the order a few tens of mK, but is in fact roughly 0.5 K, i.e. an order of magnitude larger. One possible mechanism that could produce this effect is coupling of the gas with dark matter, something which is not usually considered possible but could happen if the dark matter particle has a very small charge (&lt;a href=&quot;https://arxiv.org/abs/1803.03091&quot; rel=&quot;nofollow noreferrer&quot;&gt;Barkana et al. 2018&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Time evolution of the 21 cm signal&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The figure below (from a great review by &lt;a href=&quot;https://arxiv.org/abs/1109.6012&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pritchard &amp;amp; Loeb 2012&lt;/a&gt;) shows how the 21 cm signal evolves with time. The dip discussed in this answer is the orange and red part.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;a href=&quot;https://i.stack.imgur.com/TVRZs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TVRZs.jpg&quot; alt=&quot;21cm&quot;&gt;&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;An analogy would be two magnets aligned parallel to each other with north in the same direction, preferring to flip around, but note Ken G's comment below; the transition doesn't necessarily involve a spin flip, and the analogy is not to be taken literally, since parallel magnets are alike, whereas parallel electrons/protons have opposite charges.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-04-05T13:44:27.093" LastActivityDate="2018-04-05T13:44:27.093" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25749" PostTypeId="1" CreationDate="2018-04-04T13:44:26.643" Score="3" ViewCount="269" Body="&lt;p&gt;I just made an observation that is very puzzling.  The observation concerns total solar eclipses and their spacing in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the saros cycle dictates recurrence of same saros eclipses roughly every 18 yrs 11 days apart.  However, the two American eclipses (Aug 21, 2017 and Apr 8, 2024) are spaced by 2422 days.  These are saros 145 and 139 respectively - seemingly unrelated eclipses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another pair of eclipses that are upcoming (Aug 2, 2027 and Mar 20, 2034), and they are also spaced exactly 2422 days apart.  These are saros 136 and 130 respectively.  Again, they are unrelated eclipses from each other - OK expected.  The only thing that these two pairs of eclipses have in common is that their paths CROSS each other 2422 days apart. So, here is my question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are these two pairs of totally unrelated eclipses &quot;dancing&quot; with each other with exactly the same spacing of 2422 days?  What mathematical phenomena could account for this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your time.&lt;/p&gt;&#xA;" OwnerUserId="22541" LastEditorUserId="616" LastEditDate="2018-04-04T13:57:18.437" LastActivityDate="2018-04-07T01:42:54.633" Title="Spacing (2422 days) between solar eclipses - unrelated saros numbers" Tags="&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25751" PostTypeId="2" ParentId="25747" CreationDate="2018-04-04T14:17:33.950" Score="4" Body="&lt;p&gt;Normally three planets would not line up along a simple line like that.  Planets have orbits with different inclinations, so at best they'd be in the same &lt;em&gt;plane&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working that out requires a detailed calculation of their motions.  The simplest approximation is to actually assume everything has the same inclination and treat the orbits as circles (which they're not precisely).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this simple model, a planet's angular position $\theta$ at any time $t$ is given by :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta(t) = \theta_0+t\frac {2\pi} T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $T$ is the orbital period and $\theta_0$ is the initial angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you want is the value of $t$ satisfying :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta_1(t) = \theta_2(t) + 2m\pi = \theta_3(t)+2n\pi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the $n$ and $m$ values are integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now doing this for two planets is easy, we get :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t_{12}=\left(\frac {\theta_{10}-\theta_{20}}{2\pi} - m\right)\frac{T_1T_2}{T_2-T_1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t_{13}=\left(\frac {\theta_{10}-\theta_{30}}{2\pi} - n\right)\frac{T_1T_3}{T_3-T_1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember we have to work out those $m$ and $n$ values to get a result !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the problem is that we need to get integer values and all the other values are real numbers.  This means that there may be no (exact) solution that produces integer values and hence no one time when they all line up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in the real world their are some other issues :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The orbital inclinations mean we need to work in three dimensions&lt;/li&gt;&#xA;&lt;li&gt;The orbits will be ellipses, not circles, and if we want even more accuracy they're not even convenient shapes like that (look up &lt;a href=&quot;https://en.wikipedia.org/wiki/Three-body_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;the three body problem&lt;/a&gt; to get an idea why).&lt;/li&gt;&#xA;&lt;li&gt;The speed of light is finite.  So what &quot;lined up&quot; means is not as simple as it seems.  There might be only one observer who would say they were lined up and every other observer would consider them not lined up.&lt;/li&gt;&#xA;&lt;li&gt;Planets have non-zero sizes, so we can allow them to be lined up within a range of values, not a simple set of numbers, but a whole set of ranges of values.&lt;/li&gt;&#xA;&lt;li&gt;Again with time in general relativity, whose time are using as a reference.  That's a really complicated one in general relativity (more than it might seem), but fortunately to a good approximation we can ignore relativity a lot of the time or use small corrective factors.&lt;/li&gt;&#xA;&lt;li&gt;All our measurements for the parameters of the orbit will have a finite accuracy &lt;em&gt;and&lt;/em&gt; some uncertainty range.  So, particularly over very long periods of time, our calculations can become inaccurate.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2018-04-04T14:17:33.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25752" PostTypeId="1" AcceptedAnswerId="25834" CreationDate="2018-04-04T16:32:18.053" Score="2" ViewCount="144" Body="&lt;p&gt;If I had a time travel machine that could drop me into any point in space and time, could I accurately predict earth’s “position”?  I know position is an improper term.  Would it be possible to know exactly where to land so that you are at the same spot on the surface of Earth as when you “departed”?&lt;/p&gt;&#xA;" OwnerUserId="10234" LastActivityDate="2018-04-10T15:27:13.490" Title="How well is Earth’s motion through the universe quantified?" Tags="&lt;space-time&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25753" PostTypeId="1" CreationDate="2018-04-04T17:03:29.950" Score="1" ViewCount="48" Body="&lt;p&gt;I don't know where to ask this question and I have found this site. If this is not the right place to ask it, please tell me and I will delete it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a computer engineer for 20 years but my true vocation is astronomy and astrophysics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking to start next autumn astrophysics degree at the university but I don't know if I need to do a transition year because I don't remember a lot of things of mathematics and physics that I have studied 20 years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also interested in studying mathematics at the university also, so I've thought that I can start first studying mathematics and then physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you think?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask it because physics has a lot mathematics and I don't know if it is better to study first mathematics and then physics.&lt;/p&gt;&#xA;" OwnerUserId="7566" LastEditorUserId="7566" LastEditDate="2018-04-04T17:33:57.193" LastActivityDate="2018-04-04T17:33:57.193" Title="Study astrophysics" Tags="&lt;astrophysics&gt;&lt;mathematics&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2018-04-05T15:08:16.103" ContentLicense="CC BY-SA 3.0" />
  <row Id="25756" PostTypeId="2" ParentId="25749" CreationDate="2018-04-04T21:18:52.337" Score="4" Body="&lt;p&gt;Since a solar eclipse is a coincidence of a new moon and &#xA;an ascending or descending &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_node&quot; rel=&quot;nofollow noreferrer&quot;&gt;lunar node&lt;/a&gt;,&#xA;the interval between solar eclipses must be a common multiple of&#xA;a whole &lt;a href=&quot;http://moonblink.info/Eclipse/why/months#Synodic&quot; rel=&quot;nofollow noreferrer&quot;&gt;synodic month&lt;/a&gt;&#xA;(or lunation, 29.53 days between same lunar phases)&#xA;and half a &lt;a href=&quot;http://moonblink.info/Eclipse/why/months#Draconic&quot; rel=&quot;nofollow noreferrer&quot;&gt;draconic month&lt;/a&gt;&#xA;(27.21 days between same lunar nodes).&#xA;These line up around &lt;a href=&quot;http://moonblink.info/Eclipse/why/cycles&quot; rel=&quot;nofollow noreferrer&quot;&gt;eclipse seasons&lt;/a&gt;,&#xA;173.3 days apart on average.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The saros cycle of 18.03 years is 223 synodic months, 242 draconic months,&#xA;or 239 &lt;a href=&quot;http://moonblink.info/Eclipse/why/months#Anomalistic&quot; rel=&quot;nofollow noreferrer&quot;&gt;anomalistic months&lt;/a&gt;&#xA;(27.55 days between lunar perigees or apogees).&#xA;This multiple coincidence of phase, node, anomaly, and time of year makes the saros especially useful for prediction.&#xA;However, there are other relationships between eclipses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Between consecutive saros series there is another interval called&#xA;the &lt;a href=&quot;https://en.wikipedia.org/wiki/Inex&quot; rel=&quot;nofollow noreferrer&quot;&gt;inex&lt;/a&gt;:&#xA;358 synodic months, 388.5 draconic months, or about 28.94 years.&#xA;George van den Bergh showed in 1955 that the interval T between any two eclipses can be expressed as T = aI + bS, where I and S are the inex and saros periods, and a and b are integers.&#xA;Some such combinations have names:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                   Synodic  Draconic  Anomalistic  Eclipse&#xA;Name      Formula  Months    Months     Months     Seasons&#xA;&#xA;Semester  5I - 8S      6      6.511       6.43       1.02&#xA;Hepton    5S - 3I     41     44.493      43.94       6.99&#xA;Octon     2I - 3S     47     51.004      50.37       8.01&#xA;Tritos     I - S     135    146.501     144.68      23.00&#xA;Saros      S         223    241.999     238.99      38.00&#xA;Inex       I         358    388.500     383.67      61.00&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Source: &lt;a href=&quot;http://users.skynet.be/felixverbelen/cycles.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;F. Verleben&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Between the eclipse pairs in question, 2422 days = 82 lunations = 10S - 6I or 2 heptons.&#xA;The two pairs also happen to be 3 heptons apart.&#xA;The same hepton family includes southern hemisphere eclipses in Dec 2020, Nov 2030, and Jul 2037.&#xA;Tabulated in a&#xA;&lt;a href=&quot;https://eclipse.gsfc.nasa.gov/SEsaros/SEperiodicity.html#section109&quot; rel=&quot;nofollow noreferrer&quot;&gt;saros-inex panorama&lt;/a&gt;,&#xA;all seven eclipses would fit on a single diagonal line of slope 5/3.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;            Greatest eclipse  Sun      Moon&#xA;   Date      UT     Lon Lat   Decl  EcLat Node  Saros  Inex&#xA;&#xA;2017-08-21  18:26   88W 37N   +12   +0.43 asc    145    39&#xA;2020-12-14  16:13   68W 40S   -23   -0.29 desc   142    44&#xA;2024-04-08  18:17  104W 25N    +7   +0.35 asc    139    49&#xA;2027-08-02  10:06   33E 25N   +18   +0.14 desc   136    54&#xA;2030-11-25  06:50   71E 44S   -21   -0.39 asc    133    59&#xA;2034-03-20  10:17   22E 16N     0   +0.29 desc   130    64&#xA;2037-07-13  02:39  139E 25S   +22   -0.71 asc    127    69&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sources: &lt;a href=&quot;https://eclipse.gsfc.nasa.gov/solar.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;F. Espenak&lt;/a&gt;,&#xA;&lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL HORIZONS&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a Moon-centric point of view, the umbra of a solar eclipse moves straight along the ecliptic, but several factors shape its path on the Earth's surface.&#xA;The terrestrial longitude of a path is directly related to the time of day (UT).&#xA;A path's terrestrial latitude is very sensitive to the Moon's&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Ecliptic_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;ecliptic latitude&lt;/a&gt;&#xA;and is also affected by the Sun's declination.&#xA;Due to the tilt of the Earth's axis, an eclipse path&#xA;slants southwest-northeast near the March equinox,&#xA;bends southward near the June solstice,&#xA;slants northwest-southeast near the September equinox,&#xA;or bends northward near the December solstice.&#xA;The Moon's motion in ecliptic latitude also contributes a slight slant,&#xA;northward at ascending node or southward at descending node,&#xA;as the umbra passes west to east.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both the 2017-2024 and 2027-2034 eclipse pairs occur near opposite equinoxes and at similar times of day, so their paths naturally have opposite slants and similar ranges of longitude.&#xA;The 2030 and 2037 eclipse paths have similar ranges of latitude but barely cross because they occur 4 hours apart in the day.&#xA;Eclipses in the same saros series, if not widely separated in longitude, occur at the same time of year and have nearly parallel paths with few opportunities to cross.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-04-07T01:42:54.633" LastActivityDate="2018-04-07T01:42:54.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25757" PostTypeId="1" AcceptedAnswerId="25790" CreationDate="2018-04-04T21:46:53.877" Score="2" ViewCount="550" Body="&lt;p&gt;I'm having a hard time understanding the Gunn-Peterson effect and the Ly-alpha line and I've written a couple of questions below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trough we see in the spectra of quasars is the Gunn-Peterson trough, right? Does this trough have to start from the Ly-alpha line?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If neutral hydrogen along the line of sight to the distant quasar is few, do we expect the Ly-alpha forest to be sparse?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the main radiative mechanism in the Gunn-Peterson effect?&lt;/p&gt;&#xA;" OwnerUserId="22041" LastActivityDate="2018-04-06T11:29:44.343" Title="What is main radiative mechanism in Gunn-Peterson effect?" Tags="&lt;radiative-transfer&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25758" PostTypeId="1" CreationDate="2018-04-04T22:05:36.177" Score="3" ViewCount="329" Body="&lt;p&gt;What is the temperature under the surface of Mercury?  I know at the surface the temperature varies greatly (-297 F to 801 F).  I also realize that the answer would likely depend on depth.  Do we know the temperature at any depth, except the surface?  I also realize the temperature may vary based on latitude.  I am mainly looking to see if there might be an ideal location / depth for a colony on Mercury for my game Planetary Settlers.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2018-04-04T22:05:36.177" Title="What is the temperature under the surface of Mercury?" Tags="&lt;temperature&gt;&lt;mercury&gt;&lt;heat&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25759" PostTypeId="1" AcceptedAnswerId="25768" CreationDate="2018-04-05T02:49:27.467" Score="5" ViewCount="542" Body="&lt;p&gt;Theoretically, does an evaporating black hole remain a black hole until it evaporates to nothing? Or does it change into something else once it evaporates to below a certain mass?&lt;/p&gt;&#xA;" OwnerUserId="6870" LastActivityDate="2018-04-06T18:26:21.367" Title="Theoretically what happens to a black hole that evaporates?" Tags="&lt;black-hole&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25761" PostTypeId="1" AcceptedAnswerId="25764" CreationDate="2018-04-05T04:36:15.587" Score="2" ViewCount="219" Body="&lt;p&gt;If we look around the cosmos, we will see most of the systems are just flat disks. Either it is solar system, galaxies, or the rings of Saturn. What is the matter, does it depend on how they were formed, or anything else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H4sig.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H4sig.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22552" LastEditorUserId="1891" LastEditDate="2018-04-06T17:55:37.560" LastActivityDate="2018-04-06T17:55:37.560" Title="Flatness of celestial systems like galaxies and Solar System" Tags="&lt;astrophysics&gt;&lt;cosmology&gt;&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2018-04-06T13:45:50.203" ContentLicense="CC BY-SA 3.0" />
  <row Id="25762" PostTypeId="1" AcceptedAnswerId="25763" CreationDate="2018-04-05T05:24:45.247" Score="-2" ViewCount="84" Body="&lt;p&gt;We know black holes emit streams of radiation ,also recently discovered that they emit visible light when they eat any nearby matter .&#xA;        &quot;We find that activity in the vicinity of a black hole can be observed in optical light at low luminosity for the first time,&quot; astronomer and lead researcher, Mariko Kimura from Kyoto University in Japan, told Charles Q. Choi at Space.com. &#xA;         But it is different from what we normally hear that neither matter nor radiation can escape from black holes. &#xA;So what's the reason for this.??&lt;/p&gt;&#xA;" OwnerUserId="22552" LastActivityDate="2018-04-05T06:46:37.830" Title="Strange behaviour of black holes" Tags="&lt;black-hole&gt;&lt;radio-astronomy&gt;&lt;radiation&gt;&lt;black-body&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2018-04-05T12:38:14.927" ContentLicense="CC BY-SA 3.0" />
  <row Id="25763" PostTypeId="2" ParentId="25762" CreationDate="2018-04-05T05:36:47.360" Score="1" Body="&lt;p&gt;Black hole don't allow anything to escape, if the matter/light is inside the event horizon (boundary which describes the point of no return). The emission radiation that is referred in the question is not within the event horizon, so it can escape. This is caused by friction within matter which is swirling in to the black hole. &lt;/p&gt;&#xA;" OwnerUserId="20162" LastActivityDate="2018-04-05T05:36:47.360" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25764" PostTypeId="2" ParentId="25761" CreationDate="2018-04-05T05:40:30.690" Score="0" Body="&lt;p&gt;Flatness is caused by conservation of angular momentum. As solar system/galaxy collapses under its own gravity, its velocity increases to conserve the angular momentum. This high velocity leads to formation of disc shaped structure. The similar manner in which piece of dough becomes flat when it is rotated. &lt;/p&gt;&#xA;" OwnerUserId="20162" LastActivityDate="2018-04-05T05:40:30.690" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25765" PostTypeId="2" ParentId="25759" CreationDate="2018-04-05T05:45:44.007" Score="1" Body="&lt;p&gt;According to theory of general relativity mass warps the spacetime. In black hole, its mass  causes high curvature of spacetime (infinite curvature at its centre called singularity). When black hole evaporates,its mass decreases therefore its ability to wrap spacetime decreases and when it evaporates fully there will be no distortion of spacetime. So it will become normal space without zero/very less curvature of space time. &lt;/p&gt;&#xA;" OwnerUserId="20162" LastEditorUserId="20162" LastEditDate="2018-04-06T18:26:21.367" LastActivityDate="2018-04-06T18:26:21.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25767" PostTypeId="2" ParentId="25762" CreationDate="2018-04-05T06:46:37.830" Score="2" Body="&lt;p&gt;What you need to understand is that a black hole is a region of space defined by an &lt;em&gt;event horizon&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key point about an event horizon is that if you're inside the region it surrounds you cannot get out.  But it's only the stuff inside that is restricted in this absolute way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If something is outside (even just outside) it can, in principle, happily orbit the black hole.  And in fact we know that this is possible because the &lt;a href=&quot;https://en.wikipedia.org/wiki/Sagittarius_A*&quot; rel=&quot;nofollow noreferrer&quot;&gt;super massive black hole at the center of the milky way&lt;/a&gt; is known (and was identified because of) stars that orbit it (and approach very closely).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radiation emitted by material &lt;em&gt;near&lt;/em&gt; a black hole can, therefore, reach us.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2018-04-05T06:46:37.830" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25768" PostTypeId="2" ParentId="25759" CreationDate="2018-04-05T08:13:55.707" Score="2" Body="&lt;p&gt;Utterly unknown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;a href=&quot;http://adsabs.harvard.edu/abs/1974Natur.248...30H&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hawking proposed&lt;/a&gt; radiation, that reduces the mass of the black hole (BH) by black body radiation.  The effective temperature is antiproportional to the mass of the BH.  In the linked paper, Hawking concludes that BHs end in &quot;fairly small explosions by astronomical standards.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are different solutions to the equations of general relativity that produce black holes, not only the common Schwarzschild metric.  Whether the curvature singularity or the event horizon vanish before the BH completely evaporates, depends on the internal structure of the BH.  As any mass, even very small ones, can become BHs if you compress them enough, there might be a singularity until the very end (not taking into account quantum mechanics).  On the other hand a &quot;vanishing event horizon&quot; is not something we should be strictly speaking about, because then it wasn't an event horizon in the first place.  You may see the looming problems with black hole remnants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also: according to the no-hair theorem, BHs are completely characterized by mass, electric charge and angular momentum.  If you can see a remnant, e.g. a clump of matter, could you derive its complete properties from the previous three only?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In research, a remnant seems to be &lt;a href=&quot;https://arxiv.org/abs/1412.8366&quot; rel=&quot;nofollow noreferrer&quot;&gt;generally undesired&#xA; but possible&lt;/a&gt;.  &lt;a href=&quot;https://arxiv.org/abs/1802.04264&quot; rel=&quot;nofollow noreferrer&quot;&gt;This fresh and unreviewed paper&lt;/a&gt; jumps on the bandwagon and proposes white holes as remnants. So the debate about the end of BHs and their information theoretical implications is still a hot topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disclaimer: I'm not a BH specialist.  (And I'm waiting for the time where &quot;black hole specialist&quot; is a craft job designation.)&lt;/p&gt;&#xA;" OwnerUserId="22239" LastEditorUserId="22239" LastEditDate="2018-04-05T18:06:36.690" LastActivityDate="2018-04-05T18:06:36.690" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25769" PostTypeId="1" CreationDate="2018-04-05T08:24:59.517" Score="0" ViewCount="83" Body="&lt;p&gt;If a satellite in an inclined orbit breaks up in orbit and multiple fragments hit a planet, do the individual fragments follow&#xA;1 a sinusoidal path&#xA;2 a geodesic ( or great circle curve)&#xA;3 a straight line&#xA;I'm having real problems trying to visualise this!&#xA;I realise the ground track of an inclined orbit follows a sinusoidal path but what happens when impacting a rotating planet?&lt;/p&gt;&#xA;" OwnerUserId="22032" LastActivityDate="2018-04-06T02:20:13.290" Title="Impact of inclined orbit satellite on a planet" Tags="&lt;planet&gt;" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25770" PostTypeId="2" ParentId="25759" CreationDate="2018-04-05T08:36:44.387" Score="2" Body="&lt;p&gt;Any object whose radius is smaller than its &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schwarzschild radius&lt;/a&gt; is called a black hole. The Schwarzschild radius is given as $$ r_s = \frac{2GM}{c^2}$$&#xA;where G is the gravitational constant, M is the object mass and c is the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's important to note here is that this is a &lt;em&gt;linear&lt;/em&gt; relationship; in other words, the radius decreases in exact proportion to the decrease in mass. We would therefore expect the black hole to continue to behave as a black hole pretty much right up to the point that it completely evaporates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I say &quot;pretty much&quot;, because what happens at the point of final evaporation is that the density and temperature of the black hole approach infinity while its mass and Schwarzschild radius approach zero. The physics at that point is well beyond my modest abilities/knowledge; I suspect our current models break down at a radius below the Planck length.&lt;/p&gt;&#xA;" OwnerUserId="14094" LastActivityDate="2018-04-05T08:36:44.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25771" PostTypeId="2" ParentId="25769" CreationDate="2018-04-05T08:47:37.457" Score="2" Body="&lt;p&gt;Yes, yes and yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your main misconception is that satellites follow a sinusoidal path. That is just a distortion of the map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The path of the satellite is in a single plane. When that plane meets the Earth produces a great circle (which is effectively straight, there are no straight lines on the curved surface of the Earth, great circles are the closest that you get to straight)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you &lt;em&gt;project&lt;/em&gt; that great circle onto a map you may get a sinusoidal path, it depends on the projection. Your statement &quot;The ground track of a satellite follows a sinusoidal path&quot; is not correct, that is just an effect of the transformation from a globe to a plane, not &quot;real&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The path of the fragments is, compared to the size of the Earth is very short, it is on the same great circle that the satellite was orbiting in (except as the fragments get closer to the ground they start to move with the atmosphere). If projected they would be on the same track as the satellite. But a great circle is the closest thing to straight on the surface of the Earth.  The rotation of the Earth is much slower than the orbital speed. By the time the fragments have slowed to terminal velocity then they fall roughly in a parabolic curve to Earth, the same as if you threw a ball off a building.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-04-05T08:47:37.457" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="25772" PostTypeId="1" CreationDate="2018-04-05T08:48:14.713" Score="2" ViewCount="356" Body="&lt;p&gt;The plane of the solar system is at 63 degrees &lt;a href=&quot;https://astronomy.stackexchange.com/questions/546/why-is-our-solar-system-tipped-about-63-with-respect-to-the-plane-of-our-gala/25750#25750&quot;&gt;to the plane of the galaxy&lt;/a&gt;, then either the North or South Pole of Earth would be on the ‘leading edge,’ i.e. pointing roughly in the direction of movement around the galaxy (+/-23.5 degrees depending on the direction of the tilt of the earth at the time). Would there be any effect on the atmosphere at the front of this direction of travel caused by additional collision / friction with space dust or solar wind? For example, if the South Pole was on the leading edge, could something like the ozone hole be exacerbated by interstellar particles, or could the auroras be more spectacular at the pole that points forwards? If so, is there evidence of seasonal variation in these effects as the tilt of the earth increases or decreases that pole’s angle towards the direction of the sun’s movement along the galactic plane?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, would the plane of the planets be shifted slightly to the rear of the midline of the sun because the sun is, in effect, dragging them behind it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, is there any evidence of increase or decrease in the above postulated effect depending on whether the solar system is above or below the galactic plane during its 30 million year &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10267/what-is-the-direction-of-the-movement-of-the-solar-system-in-relation-to-the-gal&quot;&gt;oscillation&lt;/a&gt;? E.g. if there were more interstellar dust while passing through the mid-plane, would that leave a 30 million year signature in polar ice-core samples?&lt;/p&gt;&#xA;" OwnerUserId="22543" LastActivityDate="2019-03-24T06:36:08.473" Title="Direction of Earth’s poles relative to the Sun’s movement around the Galaxy" Tags="&lt;solar-system&gt;&lt;galactic-dynamics&gt;&lt;solar-wind&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25773" PostTypeId="2" ParentId="25772" CreationDate="2018-04-05T09:29:21.140" Score="3" Body="&lt;p&gt;There is no difference in the number of cosmic particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You seem to imagine that there are static clouds of gas in the universe that the sun is moving through.  This is not the case. The gas in the galaxy is orbiting at roughly the same speed as the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun does have some proper motion relative to the interstellar medium, but the interstellar medium doesn't penetrate the bubble formed by the solar wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no dragging. The gravity of the galaxy affects both sun and planets, we all move around the galaxy together. There is no shifting of the plane at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interstellar particles are found in ice cores, &lt;a href=&quot;https://www.researchgate.net/publication/282650791_The_Episodic_Influx_of_Tin-rich_Cosmic_Dust_Particles_During_the_Last_Ice_Age&quot; rel=&quot;nofollow noreferrer&quot;&gt;a paper&lt;/a&gt; suggests that variation in the rate of cosmic radiation may cause hydromagnetic shocks that could transport metal rich dust into the inner solar system. No 30 million year periodicity is mentioned.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-04-05T09:29:21.140" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25775" PostTypeId="1" CreationDate="2018-04-05T10:18:08.807" Score="1" ViewCount="72" Body="&lt;p&gt;Was looking at the sky and I see a satellite , turned around and saw another light , but this light was pulsating like a strobe light , definitely not a plane , next thing I know the light goes out and was not seen again , could this be a dying star , I know what satellites and planes look like in the sky , but I have never seen anything like this , it just went dark after being super bright&lt;/p&gt;&#xA;" OwnerUserId="22555" LastActivityDate="2018-04-05T16:07:04.417" Title="Bright pulsating light" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2018-04-05T20:52:40.420" ContentLicense="CC BY-SA 3.0" />
  <row Id="25780" PostTypeId="2" ParentId="25775" CreationDate="2018-04-05T16:07:04.417" Score="-1" Body="&lt;p&gt;This might be it:&#xA;&lt;a href=&quot;https://gizmodo.com/astronomers-say-giant-disco-ball-in-space-sets-a-bad-pr-1822454265&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomers Say Giant Disco Ball in Space Is Bad for Science&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three foot diameter faceted mirror ball launched late January 2018.&#xA;I don't know either designation or orbit.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2018-04-05T16:07:04.417" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25781" PostTypeId="1" CreationDate="2018-04-05T17:46:09.247" Score="6" ViewCount="174" Body="&lt;p&gt;Ignore the Blender Monkey! (from &lt;a href=&quot;https://blender.stackexchange.com/q/51585/5334&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0SYjI.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0SYjI.gif&quot; alt=&quot;from https://blender.stackexchange.com/q/51585/5334&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;In a close-orbiting binary pair, a small but significant fraction of the light from each star falls upon the other, and the result has to be carefully modeled to simulate the observed spectrum of the system as it rotates.&lt;/p&gt;&#xA;&lt;p&gt;One effect is that each star's radiation will non-uniformly heat the surface of the other star and alter its thermal radiation, but here I'm asking about light from one star being scattered or diffusely reflected by the other.&lt;/p&gt;&#xA;&lt;p&gt;How is this scattered/reflected light calculated, and roughly what is the spectral reflectance?&lt;/p&gt;&#xA;&lt;p&gt;Of course each star will be different, so one example or maybe two would be sufficient.&lt;/p&gt;&#xA;&lt;p&gt;We often treat stars as blackbodies as an approximation, suggesting that stars have low albedo, but at some depth the electron density will be high enough so that the plasma frequency reaches visible wavelengths, increasing reflectivity, though that may be too deep to matter.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2021-04-16T23:25:09.860" LastActivityDate="2021-04-16T23:25:09.860" Title="What is the spectral reflectance of starlight in a close binary?" Tags="&lt;star&gt;&lt;binary-star&gt;&lt;albedo&gt;&lt;eclipsing-star-systems&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25782" PostTypeId="2" ParentId="25769" CreationDate="2018-04-05T22:00:08.003" Score="0" Body="&lt;p&gt;Impact trajectories when plotted in the coordinate frame of a rotating body or planet can best be understood by taking into account the Coriolos effect&lt;/p&gt;&#xA;" OwnerUserId="22032" LastActivityDate="2018-04-05T22:00:08.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25783" PostTypeId="1" CreationDate="2018-04-05T23:06:31.757" Score="2" ViewCount="159" Body="&lt;p&gt;What are six different methods to measure distances in Astronomy in regards of their respective range in terms of astronomical objects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I can name 4: Parallax method, radar ranging, main-sequence fitting, cepheid variables.&lt;/p&gt;&#xA;" OwnerUserId="22102" LastEditorUserId="2365" LastEditDate="2018-04-05T23:27:44.847" LastActivityDate="2018-04-06T07:50:26.860" Title="What are six different methods to measure distances in Astronomy?" Tags="&lt;star&gt;&lt;galaxy&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25784" PostTypeId="1" CreationDate="2018-04-06T01:00:06.123" Score="3" ViewCount="543" Body="&lt;p&gt;A friend of mine tends to ask crazy question about everything. Today he ask me what is the brightest star that is visible on both the hemisphere and in particular from Italy and Argentina. I tried to ask wolframalpha but nothing came up. Thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="11842" LastActivityDate="2018-04-06T01:53:40.743" Title="Brightest star visible from Italy and Argentina" Tags="&lt;star&gt;&lt;night-sky&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25785" PostTypeId="2" ParentId="25784" CreationDate="2018-04-06T01:53:40.743" Score="6" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sirius&quot; rel=&quot;noreferrer&quot;&gt;Sirius&lt;/a&gt; is the brightest star in the sky (excluding the Sun, of course), with a visual apparent magnitude of −1.46, and is visible to everyone in both Argentina and Italy, depending on the season.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, Sirius is visible from anywhere in the Southern Hemisphere, including all of Argentina (which, I might add, is an enormous country stretching from just above the Tropic of Capricorn, all the way south to the Drake Passage separating Tierra del Fuego from Antarctica). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sirius doesn't rise above the horizon for locations in the Northern Hemisphere north of 73°, but this is well beyond the northernmost point of mainland Europe at around 71°N, so Sirius is visible in all of continental Europe. As the northernmost location in Italy is around 47°N, Sirius at its zenith appears well above the horizon to anyone in Italy.&lt;/p&gt;&#xA;" OwnerUserId="14094" LastActivityDate="2018-04-06T01:53:40.743" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25786" PostTypeId="2" ParentId="25769" CreationDate="2018-04-06T02:20:13.290" Score="0" Body="&lt;p&gt;I'm not sure this is a good answer, but I'll give it a shot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with Shoemaker-levy 9, which isn't an ideal example because it broke apart during an orbit about 2 years prior to impact but it's interesting for comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ev4bq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ev4bq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ase.tufts.edu/cosmos/print_images.asp?id=16&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source of picture&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of Shoemaker-levy, while the object broke apart somewhat randomly the pieces all had basically the same orbit, and the ones closer to Jupiter were accelerated a bit faster, so over time the fragments lined up in something close to a straight line.   This happens with debris that falls off a comet as well, the debris tends to orbit in a line, along the same basic orbital path.   Given enough time, and with an elliptical orbit (comets around the sun or Shoemaker around Jupiter), debris tends to line up and follow the same path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4JkBu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4JkBu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same thing happens with comets, though in the case of comets, the debris trail can over time encompass the entire orbit, in a kind of very narrow, elliptical, dusty/tiny asteroid belt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8yFfr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8yFfr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://faculty.virginia.edu/skrutskie/ASTR1210/notes/mshower.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source of picture&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why meteor showers have radiants and they happen at the same time every year.  That's where the Earth intersects the orbiting dust trail from the comet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of meteor showers, any part of the Earth where the radiant is visible in the sky can get meteors.  The spread is significantly larger than the diameter of the Earth (but still a generally narrow band around the sun).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of shoemaker levy, with the breakup happening only one orbit prior to impact, the spread was quite small, and, relative to Jupiter's plane or orbit, the shoemaker-levy objects all hit Jupiter in about the same place, but over a period of about a week, in which time Jupiter was able to make multiple rotations.  Jupiter has almost no axial tilt (3 degrees) so those impacts occurred nearly along a latitude line of Jupiter (3 degrees off).  That's a result of hitting in nearly the same place as the planet rotated below the impact.   In the case of Shoemaker-levy, where the fragments hit in single file over time, the impact site is on a plane perpendicular to the planet's rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WOgaV.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WOgaV.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ifa.hawaii.edu/images/sl9/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source of picture and more images&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your scenario&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If a satellite in an inclined orbit breaks up in orbit and multiple&#xA;  fragments hit a planet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's a little unclear what you mean by orbit.  If the planet is Earth and the satellite is in orbit around the Earth, it mostly wouldn't hit the planet.  Orbits, by their very nature, rarely hit the object they are orbiting.   Now if the satellite was orbiting the sun, it could collide with the Earth if the orbits intersect.   That's the more common type of comet or asteroid impact.   In theory, such a comet or asteroid could break apart as it passed inside Earth's Roche limit.   Impacting objects rarely break apart just passing the Roche limit because they're usually too small and too well held together and if they do, the tidal forces that break them apart would give the pieces a relatively low radial velocity compared to the very high gravitational velocity towards the Earth so the spread would be tiny.  (this doesn't take into account super-heating and exploding in the atmosphere - which can spread out debris pretty far), but you said break apart in orbit, so lets ignore atmospheric spreading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The debris pattern in your scenario would be similar to a shot-gun blast.  Fairly closely concentrated simply because orbital break-ups are low velocity and gravitational impacts are high velocity, but the spread would look similar to, but not precisely the same as a shot-gun pellet spread.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o3sOr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o3sOr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hammernews.blogspot.nl/2006/02/cool-hand-dick-questions-remain-in.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source of image&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some aspects to your question that are unclear such as when the object breaks apart and what it's orbiting, but mostly, comets or meteors don't break up on their way towards impacting a planet, or if they do, the breakup while in orbit is relatively insignificant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are cases where &lt;a href=&quot;https://www.livescience.com/52154-ancient-meteors-struck-simultaneously.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;earth has been hit by a pair of meteors&lt;/a&gt;, that were orbiting each other at the time of impact.  We know that wasn't a &quot;break-up&quot; because the crash sites were too far apart, though in theory, it could have been a shoemaker-levy type of event that may have split the object as it flew past, and then the two objects hit during the next time around.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-04-06T02:20:13.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25788" PostTypeId="2" ParentId="25783" CreationDate="2018-04-06T07:50:26.860" Score="4" Body="&lt;p&gt;Using type Ia supernova light curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The RR Lyrae period-luminosity relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving cluster statistical parallax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surface brightness fluctuations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Redshift-distance relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using merging gravitational wave sources as standard sirens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These and more can be pursued following this link. &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Cosmic_distance_ladder&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Cosmic_distance_ladder&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-04-06T07:50:26.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25790" PostTypeId="2" ParentId="25757" CreationDate="2018-04-06T10:45:25.210" Score="2" Body="&lt;p&gt;The Gunn-Peterson trough is a feature in the spectrum of a (background) source near the Lyman $\alpha$ line, caused by a high density of intervening, neutral hydrogen (HI).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The background source may be a quasar, but could also be e.g. a gamma-ray burst or just a regular galaxy with a strong Ly$\alpha$ line.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Absorption vs. scattering&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The cross section of hydrogen for Ly$\alpha$ photons is quite large, so the mean free path of the photons is very small. Because Ly$\alpha$ excites hydrogen to the first excited level, the electron can only fall back to the ground state again, emitted another Ly$\alpha$ photon, in another direction. Hence, hydrogen doens't &lt;em&gt;absorb&lt;/em&gt; Ly$\alpha$ photons, but rather &lt;em&gt;scatters&lt;/em&gt; them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside the (host) galaxy, the HI density is so large that Ly$\alpha$ is all the time scattered both &lt;em&gt;into&lt;/em&gt; and &lt;em&gt;out of&lt;/em&gt; the LOS. However, when the photons escape the galaxy and enters the circum- and intergalactic medium (IGM), the density decreases. Along the LOS to the galaxy, a photon hitting a hydrogen atom is &quot;lost&quot;, and no new photons are added to the observed flux, so effectively, it becomes an absorption process rather than a scattering process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure below (from &lt;a href=&quot;https://arxiv.org/abs/1012.3175&quot; rel=&quot;nofollow noreferrer&quot;&gt;Laursen 2010&lt;/a&gt;) illustrates this phenomenon: the three photons are initially emitted in random directions, but end up traveling toward the observer on the right. On their way through the IGM, two of them are scattered out of the line of sight (LOS), and the probability of a photon from another galaxy being scattered into the LOS is negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3HGDT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3HGDT.png&quot; alt=&quot;scat&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Reionization&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;In the early days of the (post-&lt;a href=&quot;https://en.wikipedia.org/wiki/Recombination_(cosmology)&quot; rel=&quot;nofollow noreferrer&quot;&gt;recombination&lt;/a&gt;) Universe, hydrogen was mainly neutral, but as star formation emerged, these stars started to &lt;a href=&quot;https://en.wikipedia.org/wiki/Reionization&quot; rel=&quot;nofollow noreferrer&quot;&gt;reionize&lt;/a&gt; the IGM. This process started already at $z\simeq15$–$20$, when the Universe was ~200 million years old , but until $z\simeq6$, when the Universe was ~a billion years old, there was still so much HI left that Ly$\alpha$ photons scatter all the time in the IGM.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Blue wavelengths are eventually redshifted to Lyman $\alpha$&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Now when light leaves a galaxy, it gets redshifted due to the expansion of the Universe. That means that after a while, light &lt;em&gt;blueward&lt;/em&gt; of the Ly$\alpha$ line is redshifted to become Ly$\alpha$, and may hence scatter. As the spectrum travels through the IGM, it redshifts continuously, so that &lt;em&gt;every&lt;/em&gt; wavelength of the &quot;original&quot; Ly$\alpha$ line eventually becomes Ly$\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the HI density in the IGM is high enough, that means that &lt;em&gt;everything&lt;/em&gt; blueward of the Ly$\alpha$ line is wiped out. &lt;strong&gt;This is the Gunn-Peterson trough&lt;/strong&gt;. If the density is &lt;em&gt;low&lt;/em&gt;, but with some neutral clouds dispersed, then an absorption line occurs exactly at the wavelength that, at the location of the cloud, happens to have been redshifted to Ly$\alpha$. When there are many such clouds around, you'll gets lots of absorption lines, forming the so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Lyman-alpha_forest&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lyman $\alpha$ forest&lt;/a&gt;. The fewer the amount of such clouds, the more sparse the forest will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gunn-Peterson trough doesn't &lt;em&gt;have&lt;/em&gt; to start at Ly$\alpha$, but since this is the strongest transition, it is most distinct here. But you will often see a corresponding absorption at Ly$\beta$ (and even Ly$\delta$ and Ly$\gamma$, as well as the stronger metal lines like MgII and CIV). This is visible in the figure below (from &lt;a href=&quot;https://arxiv.org/abs/1510.03368&quot; rel=&quot;nofollow noreferrer&quot;&gt;Becker et al. 2015&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/orRFC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/orRFC.png&quot; alt=&quot;GPT&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see that immediately blueward of Ly$\alpha$, everything is wiped. As you go along the spectrum toward the blue, that part corresponds to later and later epochs, because is takes longer time to redshift that part into Ly$\alpha$. Hence, eventually the Universe becomes less and less neutral, and the absorption is no longer complete.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-04-06T11:29:44.343" LastActivityDate="2018-04-06T11:29:44.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25791" PostTypeId="1" AcceptedAnswerId="25792" CreationDate="2018-04-06T10:56:36.170" Score="10" ViewCount="2225" Body="&lt;p&gt;Hubble's law gives a linear relationship between the distance to a galaxy and it's recessional speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observations of distant type 1a supernovae showed that their red shift (and therefore their recessional velocities) were lower than expect, implying that the rate of expansion of the Universe was lower in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, doesn't this mean that Hubble's law only applies for short distances, as at great distances a plot of recessional velocities against distance would not be linear (as I have crudely tried to show on the graph below)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VqbfE.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VqbfE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22574" LastActivityDate="2018-04-26T16:37:31.580" Title="Does the accelerating expansion of the Universe contradict Hubble's law?" Tags="&lt;expansion&gt;&lt;redshift&gt;&lt;hubble-constant&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25792" PostTypeId="2" ParentId="25791" CreationDate="2018-04-06T11:42:45.470" Score="7" Body="&lt;p&gt;Hubble's law is a bit more subtle than you suppose and an expansion, whether accelerating or decelerating does not invalidate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance and speed that should be used are their values &lt;em&gt;now&lt;/em&gt;. These are known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Comoving_and_proper_distances#Uses_of_the_proper_distance&quot; rel=&quot;noreferrer&quot;&gt;proper distance and velocity&lt;/a&gt; respectively. In that form Hubble's law works just fine, providing that the cosmological principle - that the universe is everywhere homogeneous and isotropic - is satisfied. The problem with that is that neither the recession velocity or distance that we measure are the values that a galaxy has &lt;em&gt;now&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So indeed a plot of a recessional speed versus estimated distance will not be a straight line as we get far enough away that the light takes an appreciable fraction of the lifetime of the universe to travel to us. The curvature of the line can be used to estimate cosmological parameters and this is essentially what has been done with Type Ia supernovae data you refer to. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-04-06T16:09:52.307" LastActivityDate="2018-04-06T16:09:52.307" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25793" PostTypeId="2" ParentId="25791" CreationDate="2018-04-06T11:42:55.790" Score="8" Body="&lt;p&gt;The Hubble law gives the velocity of a distant galaxy &lt;em&gt;right now&lt;/em&gt;. A galaxy at a distance $d$ recedes at a velocity $v = H_0\,d$ right now$^\dagger$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the relation between $d$ and the &lt;em&gt;redshift&lt;/em&gt; — which is the quantity that we &lt;em&gt;observe&lt;/em&gt; — is a non-trivial function of the expansion history of the Universe, obtained by integrating the (inverse) scale factor over time, which in turn depends on the densities of matter, radiation, and dark energy.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Analogy&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Paint dots on a rubber band, hold onto one end, and let your friend walk away with the other end. It doesn't matter whether your friend changes pace, maybe even stops and walks toward you for a while, then starts sprinting away from you. At any given time, the velocity of a given dot with respect to you is simply proportional to its distance from you.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;Strictly speaking, this is only true if our assumption is true that the expansion rate of the Universe is the same at every location, which is a corollary of the Cosmological Principle (i.e. that the Universe is homogeneous and isotropic).&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-04-06T12:22:36.647" LastActivityDate="2018-04-06T12:22:36.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25794" PostTypeId="2" ParentId="12408" CreationDate="2018-04-06T13:48:55.330" Score="1" Body="&lt;p&gt;Earlier answers pointed out that this question is based on a mistake about general relativity, in which gravity neither pushes nor pulls: but it is also based on a mistake about Newton's view and about what he wrote. Newton was emphatic in refusing to commit himself to identify gravitational attraction either as a push or as a pull. He said he was concerned only with the mathematical quantities and relations of the forces, etc. He wrote that he used the words 'attraction' and 'impulse' indifferently, 'one for another' in respect of the general propensity of massive bodies to approach each other, and he expressly abstained from speculation about their physical nature or cause (see quotations below). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accordingly, just as the other answers point out that GRT does not specify gravitation as either a push or a pull, neither did Newton's physics. Whether in GRT or in Newtonian physics, the answer 'push' or 'pull' is unnecessary to an account of the physics. The question in effect poses a false antithesis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the basis so far as what Newton wrote. In the opening parts of the 'Principia', in Definition VIII, Newton wrote (quoted here from the 1729 English translation of his original Latin, available online at &lt;a href=&quot;https://books.google.com/books?id=Tm0FAAAAQAAJ&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://books.google.com/books?id=Tm0FAAAAQAAJ&lt;/a&gt;): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I ... use the words Attraction, Impulse or Propensity of any sort towards a centre, promiscuously, and indifferently, one for another; considering those forces not Physically but Mathematically : Wherefore, the reader is not to imagine, that by those words, I any where take upon me to define the kind, or the manner of any Action, the causes or the physical reason thereof, or that I attribute Forces, in a true and Physical sense, to certain centres (which are only Mathematical points) ...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same vein, his Definition V stated that &quot;a centripetal force is that by which bodies are drawn or impelled, or any way tend, towards a point as to a centre&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later on he emphasised yet again the same abstention from speculation (Principia, Book 1, Section XI, Scholium following Proposition 69):  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I here use the word attraction in general for any endeavour, of what kind soever, made by bodies to approach to each other; whether that endeavour arise from the action of the bodies themselves as tending mutually to, or agitating each other by spirits emitted; or whether it arises from the action of the aether or of the air, or of any medium whatsoever, whether corporeal or incorporeal, any how impelling bodies placed therein towards each other. In the same general sense I use the word impulse, not defining in this treatise the species or physical qualities of forces, but investigating the quantities and mathematical proportions of them; as I observed before in the definitions.&quot; &lt;/p&gt;&#xA;" OwnerUserId="17202" LastEditorUserId="17202" LastEditDate="2018-04-07T08:31:53.907" LastActivityDate="2018-04-07T08:31:53.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25795" PostTypeId="1" AcceptedAnswerId="25807" CreationDate="2018-04-06T14:08:58.573" Score="0" ViewCount="411" Body="&lt;p&gt;The radius of the observable universe is generally taken to be ~45 billion light  years. However, we see distant galaxies as they were many millions of years in the past, so there are two ways I can interpret this radius value:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Light that is reaching Earth for the first time today was emitted by ancient galaxies that were initially a lot closer, but are now estimated to be 45 billion light years away from us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Light that is reaching Earth for the first time today was emitted by ancient galaxies that were initially 45 billion light years away, and are now a lot further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which interpretation, if they're not both wrong, is correct?&lt;/p&gt;&#xA;" OwnerUserId="22574" LastActivityDate="2018-04-07T08:43:59.227" Title="What is the present day radius of the observable universe?" Tags="&lt;expansion&gt;&lt;observable-universe&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25796" PostTypeId="1" CreationDate="2018-04-06T14:37:45.847" Score="1" ViewCount="36" Body="&lt;p&gt;I found an interesting article, claiming very interesting things. But I cannot find any trust-worty sources confirming such bold claims. Maybe someone came across some solid proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Center of gravity of Solar system periodically changes from almost the center of the sun to far towards sun surface (depending on planet movement)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oscillation of the position of Center of gravity influences the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The closer Center of gravity to sun center, the more the intensity of Space rays reaching earth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.1) Intensity of Space rays reaching earth increases water condensation in earth atmosphere (thus we have more clouds), so&#xA;2.2) Higher Intensity of Space rays makes lower temperature in near-earth-surface atmosphere (as we have more clouds), lower Intensity of Space rays makes climate warmer (less clouds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The closer Center of gravity to the sun center, the more space dust (and small meteorites) gets into our atmosphere which decreases its opacity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Position of Center of gravity also influences speed of earth (24-hour) rotation (thru atmospheric effects as well).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://snob.ru/selected/entry/12156&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source (Russian)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22579" LastEditorUserId="22579" LastEditDate="2018-04-06T15:26:28.663" LastActivityDate="2018-04-06T15:26:28.663" Title="Does center of gravity of Solar system influence intensity of space rays reaching earth, space dust, speed of earth rotation?" Tags="&lt;gravity&gt;&lt;solar-system&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25797" PostTypeId="2" ParentId="25572" CreationDate="2018-04-06T16:13:15.653" Score="1" Body="&lt;p&gt;Well, there are three parameters of interest to be evaluated to get intensity of sunlight.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Sinusoidal variation of sun vector on earth's surface througout the year&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The latitude variation. Which is also sinusoidal&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The time variation. Which is also sinusoidal&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$ Intensity = A_0 \sin(\theta - \theta_0) \sin(\pi/2(1 - |t-12|/6))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \theta_0 = A_1\sin(2\pi|d/365.25|)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $d$ is the day and starts from vernal equinox. $t$ is the time from 00:00 - 23:59. $A_0$ and $ A_1$ are constants. Also when Intensity is negative, obviously there it means darkness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is very crude approximation. It assumes circular orbit.&lt;/p&gt;&#xA;" OwnerUserId="22583" LastEditorUserId="22583" LastEditDate="2018-04-06T17:50:00.230" LastActivityDate="2018-04-06T17:50:00.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25798" PostTypeId="1" CreationDate="2018-04-06T18:12:34.660" Score="1" ViewCount="187" Body="&lt;p&gt;If the point in space that belongs to the collapsed star (that formed the black hole) is infinitely dense, why are black holes different sizes? Is it due to them feeding on stars and everything else that they swallow?&lt;/p&gt;&#xA;" OwnerUserId="22586" LastEditorDisplayName="user1569" LastEditDate="2018-04-06T19:19:52.453" LastActivityDate="2018-04-06T20:10:41.467" Title="Are black holes different sizes thanks to the different sized stars that created them?" Tags="&lt;black-hole&gt;" AnswerCount="3" CommentCount="1" ClosedDate="2018-04-08T16:23:17.707" ContentLicense="CC BY-SA 3.0" />
  <row Id="25799" PostTypeId="2" ParentId="25798" CreationDate="2018-04-06T18:37:13.357" Score="0" Body="&lt;p&gt;Black hole have different size eg intermediate mass black hole, supermassive black hole etc. Higher the mass of black hole, larger is the radius of event horizon. All black hole have infinite curvature of space-time at their centre (called singularity), but their ability to warps spacetime away from their respective centre large enough to not allow light to escape depends on their mass. &lt;/p&gt;&#xA;" OwnerUserId="20162" LastActivityDate="2018-04-06T18:37:13.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25801" PostTypeId="2" ParentId="25572" CreationDate="2018-04-06T19:26:14.817" Score="1" Body="&lt;p&gt;The sunlight intensity is the cosine of the sun's elevation angle $\alpha$.  How to calculate the position of the sun is described simply at &lt;a href=&quot;http://pveducation.org/pvcdrom/2-properties-sunlight/suns-position&quot; rel=&quot;nofollow noreferrer&quot;&gt;PVEducation.com&lt;/a&gt;.  To summarize the relevant equations:  $\alpha$ is given by,&#xA;$$\alpha = \arcsin[sin\delta\sin\phi+cos\delta\cos\phi\cos(HRA)]$$&#xA;where $\delta$ is the declination angle of the sun:&#xA;$$\delta = -23.45^\circ \times \cos\left[\frac{360}{365}(d+10)\right]. $$&#xA;$\phi$ is the latitude, d is the day of the year (Jan 1 is 1), and HRA is the hour angle:&#xA;$$HRA = 15^\circ(LST-12).$$  LST is the Local Standard Time.  This should simulate the intensity of light hitting level surfaces throughout the year, however to be really natural one would have to change the position of the lighting (according to these equation) rather than the intensity because early morning and late evening mean a low sun to me rather than a low intensity sun.  &lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2018-04-06T19:26:14.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25802" PostTypeId="1" AcceptedAnswerId="25805" CreationDate="2018-04-06T19:28:54.890" Score="6" ViewCount="285" Body="&lt;p&gt;Recently a Galaxy was discovered that contained no Dark Matter. I was wondering what explanation could be offered for this ? This Galaxy also contains only 1% as many Stars as our Milky Way does . Is there be a connection between these two things, lack of many Stars and lack of Dark Matter ?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastActivityDate="2018-04-18T22:28:36.467" Title="Galaxy discovered lacking Dark Matter" Tags="&lt;dark-matter&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25803" PostTypeId="2" ParentId="25798" CreationDate="2018-04-06T20:08:36.850" Score="0" Body="&lt;p&gt;When the size of a black hole is mentioned, people are talking about the size of the event horizon surrounding the object, not anything about the singularity that might be within.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And since the size of a Schwarzschild black hole depends only on the mass, it's just a different way to describe the object based on the mass estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size (mass) of black holes is basically identical to the mass of the  infalling material.  In the case of stellar black holes, that would mostly be the stellar remnants that formed the hole initially.  For supermassive black holes, the process isn't clear, but it will have &quot;consumed&quot; significant amounts of mass after its creation.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2018-04-06T20:08:36.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25804" PostTypeId="2" ParentId="25798" CreationDate="2018-04-06T20:10:41.467" Score="1" Body="&lt;p&gt;The size of a black hole is not defined by the expanse of its matter content but (usually) its event horizon.  The radius of the event horizon  (of a non-rotating and uncharged black hole) depends on the mass:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_S = \frac{2 G M}{c^2}\\ \small M : \text{black hole mass} \\ \small G: \text{gravitational constant} \\ \small c : \text{speed of light} \\ \small r_S : \text{Schwarzschild radius}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, the size of black holes as final stage of stellar evolution stems from the original star's mass, though some mass is ejected/radiated away during the supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes do grow (increasing $M$ and $r_S$) when they accrete matter from disks or companion stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not all black holes we (kind of) observe are from stars: for example supermassive black holes in galactic centers, which can have &lt;a href=&quot;https://de.wikipedia.org/wiki/Messier_87&quot; rel=&quot;nofollow noreferrer&quot;&gt;billions of solar masses&lt;/a&gt;.  However, it is not yet fully understood, how they form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; The most common black hole descriptions actually don't include any matter.  They are defined in an empty spacetime except for one singular point that is usually interpreted to hold all the matter.  The mass of the black hole is then assigned by identifying the gravitational pull in the Newtonian approximation at some distant point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How the interior of real, non-vacuum black holes looks like, is still up to discussion.&lt;/p&gt;&#xA;" OwnerUserId="22239" LastActivityDate="2018-04-06T20:10:41.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25805" PostTypeId="2" ParentId="25802" CreationDate="2018-04-06T21:28:38.610" Score="9" Body="&lt;p&gt;The paper describing this finding is &lt;a href=&quot;https://arxiv.org/abs/1803.10237&quot; rel=&quot;nofollow noreferrer&quot;&gt;van Dokkum et al. (2018)&lt;/a&gt;. In a galaxy, the ratio of the stellar mass $M_\star$ to dark matter mass $M_\mathrm{DM}$ is normally very small, increasing with mass until Milky Way-sized galaxies where it reaches $\sim1/30$, then decreasing again.&#xA;But the dwarf galaxy &lt;a href=&quot;https://en.wikipedia.org/wiki/NGC_1052-DF2&quot; rel=&quot;nofollow noreferrer&quot;&gt;NGC1052–DF2&lt;/a&gt;, seems to contain little or no dark matter. Galaxies of that mass ($M_\star\sim2\times10^8\,M_\odot$) typically has several hundred times more DM than stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How the galaxy was formed is not known, but the authors speculate on a few different mechanisms that all have to do with the fact that gas, in contrast to DM, may cool and condense and thus form clouds of very high $M_\mathrm{gas}/M_\mathrm{DM}$ ratios: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;NGC1052–DF2 is located neat an large elliptical galaxy (NGC 1052) which could have gone through a merging event, tidally stripping a chunk of gas from one of the mergers. This is consistent with NGC1052–DF2 having a large velocity wrt. the elliptical.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;NGC1052–DF2 could have formed from low-metallicity gas that was swept up in quasar winds (as described in &lt;a href=&quot;https://arxiv.org/abs/astro-ph/9710154&quot; rel=&quot;nofollow noreferrer&quot;&gt;Natarajan et al. 1998&lt;/a&gt;).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Lastly, NGC1052–DF2 could have formed from the fragmentation of gas accreting onto the elliptical, possibly aided by shocks.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; (thanks to @WayfaringStranger): A fourth possibility is that the authors have misinterpreted the data. Shortly after the paper was put out, several other papers criticized the statistical methods used by van Dokkum et al. to infer their result (&lt;a href=&quot;https://arxiv.org/abs/1804.04136&quot; rel=&quot;nofollow noreferrer&quot;&gt;Martin et al. 2018&lt;/a&gt;;&#xA;&lt;a href=&quot;https://arxiv.org/abs/1804.04167&quot; rel=&quot;nofollow noreferrer&quot;&gt;Famaey et al. 2018&lt;/a&gt;;&#xA;&lt;a href=&quot;https://arxiv.org/abs/1804.04139&quot; rel=&quot;nofollow noreferrer&quot;&gt;Laporte et al. 2018&lt;/a&gt;).&#xA;van Dokkum wrote in &lt;a href=&quot;https://www.pietervandokkum.com/ngc1052-df2&quot; rel=&quot;nofollow noreferrer&quot;&gt;a very long reply on his blog&lt;/a&gt; how at least the first of these papers actually confirm, rather than refute, his results. I am not enough of a statistician to comment on who's right and who's wrong, but note that there is currently an ongoing debate on the Facebook group &lt;a href=&quot;https://www.facebook.com/groups/astro.r/&quot; rel=&quot;nofollow noreferrer&quot;&gt;astrostatistics&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You also ask whether the lack of DM is related to its &quot;lack of stars&quot;. I wouldn't say that NGC1052–DF2 &quot;lacks stars&quot;, anymore than the Milky Way &quot;has too many stars&quot; — it's just a small galaxy. But in general, the smaller a galaxy is, the larger the scatter in the $M_\star/M_\mathrm{DM}$ ratio is. Small galaxies, or small clumps of gas, have shallow gravitational potentials, so a small galaxy can more easily get stripped of gas, and a small chunk of gas can more easily escape a galaxy, or accretion stream, without attracting dark matter. In contrast, if would be very hard to conceive a very large galaxy having a considerably different $M_\star/M_\mathrm{DM}$ ratio, and indeed the scatter for massive galaxies is less than a factor of two (e.g. &lt;a href=&quot;https://arxiv.org/abs/1003.3203&quot; rel=&quot;nofollow noreferrer&quot;&gt;More et al. 2010&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-04-18T22:28:36.467" LastActivityDate="2018-04-18T22:28:36.467" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25806" PostTypeId="1" AcceptedAnswerId="25808" CreationDate="2018-04-07T07:08:07.720" Score="11" ViewCount="1301" Body="&lt;p&gt;This &lt;a href=&quot;https://astronomy.stackexchange.com/a/11555/7982&quot;&gt;answer&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Earth's orbital eccentricity varies over time from being nearly circular (low eccentricity of 0.0034) and mildly elliptical (high eccentricity of 0.058). It takes roughly 100,000 years for Earth to undergo a full cycle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why does the Earth's orbital eccentricity oscillate with a period of about 100,000 years? What would be the simplest set of conditions and/or fewest body system that would lead to &lt;em&gt;this particular type&lt;/em&gt; of periodic variation in eccentricity?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-04-07T08:56:05.910" LastActivityDate="2018-04-07T10:31:04.863" Title="Why does the Earth's orbital eccentricity oscillate with a period of about 100,000 years?" Tags="&lt;orbital-mechanics&gt;&lt;mathematics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25807" PostTypeId="2" ParentId="25795" CreationDate="2018-04-07T08:31:25.053" Score="1" Body="&lt;p&gt;The answer is basically #1. The observable universe contains objects whose light has taken up to the age of the universe to travel to us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to the expansion of the universe, the things that emitted that light are about 46 billion light years away &lt;em&gt;now&lt;/em&gt;.&#xA;The exact figure depends on the details of the cosmological parameters and also whether you take the first time that the light can have started out as being the &quot;epoch of recombination&quot; (when the cosmic microwave background was emitted and the universe became transparent to radiation) or the big-bang itself (which would add about 0.5 billion light years to the figure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light emitted just after the epoch of recombination has a redshift of about 1100. That means the scale factor has increased by a similar amount. Thus something that is now 46 billion light years away &lt;em&gt;was&lt;/em&gt; about 42 &lt;em&gt;million&lt;/em&gt; light years away when the light was emitted (of course neither the Earth or our Galaxy existed then).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer cannot be #2 since light from something 45 billion light years away cannot have reached us in the age of the universe.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-04-07T08:43:59.227" LastActivityDate="2018-04-07T08:43:59.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25808" PostTypeId="2" ParentId="25806" CreationDate="2018-04-07T10:22:39.947" Score="13" Body="&lt;p&gt;Saying why gets tricky beyond &quot;because of Jupiter&quot;, but to clarify on the quote, the statement &quot;Earth's eccentricity follows a 100,000 year cycle&quot; is loosely true but it's also an oversimplification.    &lt;a href=&quot;https://en.wikipedia.org/wiki/Milankovitch_cycles#Orbital_shape_(eccentricity)&quot; rel=&quot;noreferrer&quot;&gt;From Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Earth's orbit approximates an ellipse. Eccentricity measures the&#xA;  departure of this ellipse from circularity. The shape of the Earth's&#xA;  orbit varies between nearly circular (with the lowest eccentricity of&#xA;  0.000055) and mildly elliptical (highest eccentricity of 0.0679)&lt;a href=&quot;http://climatica.org.uk/climate-science-information/long-term-climate-change-milankovitch-cycles&quot; rel=&quot;noreferrer&quot;&gt;2&lt;/a&gt; Its geometric or logarithmic mean is 0.0019. The major component of&#xA;  these variations occurs with a period of 413,000 years (eccentricity&#xA;  variation of ±0.012). Other components have 95,000-year and&#xA;  125,000-year cycles (with a beat period of 400,000 years). They&#xA;  &lt;strong&gt;loosely combine into a 100,000-year cycle&lt;/strong&gt; (variation of −0.03 to +0.02).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the highest is 0.0679 and the variations are 0.012 and up to 0.03, that's over 50% variation peak to peak.  That's amplitude, not period, but if we look at the chart it's clearly not a neat and tidy cycle though it's somewhat close to a 100,000 year period (top line on the chart below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yFuAh.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yFuAh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://climatica.org.uk/climate-science-information/long-term-climate-change-milankovitch-cycles&quot; rel=&quot;noreferrer&quot;&gt;Source of image&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cause of Earth's eccentricity variation is the other planets, primarily Jupiter, with Venus and perhaps Mars (small but close) and Saturn also having effects.  Whether a particular planet is responsible for a particular period is a good question.   I'm not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately this is a version of the 3 body problem or, more specifically, &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_perturbation_analysis&quot; rel=&quot;noreferrer&quot;&gt;orbital perturbation&lt;/a&gt;, which is complicated mathematics.   Unfortunately, I can't give a good explanation of precisely why the gravitational perturbations work out to roughly 100,000 years or, more specifically, 413,000, 95,000 and 125,000.   Maybe someone else can. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What would be the simplest set of conditions and/or fewest body system&#xA;  that would lead to this particular type of periodic variation in&#xA;  eccentricity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fewest bodies would be 3.   Sun-Earth-Moon is one example.  The Moon undergoes eccentricity variation on a much faster cycle than any of the planets.   Sun, Earth, Jupiter or any set of Sun and two planets would work too provided the planets were sufficiently large and/or sufficiently close to affect each other's orbits and (I think) you'd want the two planets to not be in orbital resonance.  That might create a different kind of pattern.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-04-07T10:31:04.863" LastActivityDate="2018-04-07T10:31:04.863" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25809" PostTypeId="1" AcceptedAnswerId="25810" CreationDate="2018-04-07T13:03:35.817" Score="2" ViewCount="439" Body="&lt;p&gt;I know the surface of Venus is very hot, but that it gets cooler the higher you go in the atmosphere.  I have heard that there is a section in the atmosphere of Venus that might be somewhat habitable even, that where it is about 1 atmosphere, the temperature is slightly over 100 F.  Not really comfortable, but possibly livable temperature wise (likely with some cooling mechanism).  However, I recently heard that near the poles, the atmosphere is extremely cold: &lt;a href=&quot;http://www.iflscience.com/space/death-plunge-venus-spacecraft-reveals-hottest-planet-not-so-hot/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.iflscience.com/space/death-plunge-venus-spacecraft-reveals-hottest-planet-not-so-hot/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that at the height where the atmospheric pressure is 1 atmosphere, and somewhere between the pole and the equator would lie a comfortable zone of 72 F?  &quot;Extra credit&quot; for suggesting / providing reasoning for a likely latitude where this condition would exist (if it does).  If this does exist, it might make an ideal place for a blimp like sky colony.  I am considering adding this to my Planetary Settlers video game.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2018-04-07T13:37:53.387" Title="Is there a comfortable part of Venus's upper atmosphere?" Tags="&lt;planetary-atmosphere&gt;&lt;temperature&gt;&lt;venus&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25810" PostTypeId="2" ParentId="25809" CreationDate="2018-04-07T13:37:53.387" Score="3" Body="&lt;p&gt;There is a layer of Venus's atmosphere that has bright white clouds, a white sky, no acid rain, 1 atm pressure, a warm tropical climate and 0.9 gravities...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the cloud layers between 50 and 70km altitude above the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dak5S.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dak5S.jpg&quot; alt=&quot;http://faculty.virginia.edu/skrutskie/ASTR1210/notes/venus.html&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22518" LastActivityDate="2018-04-07T13:37:53.387" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25811" PostTypeId="2" ParentId="25802" CreationDate="2018-04-07T16:07:41.310" Score="-1" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Bullet_Cluster&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bullet cluster&lt;/a&gt; We've already seen an example of a large galaxy stripping dark matter from a smaller one.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2018-04-07T16:07:41.310" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="25812" PostTypeId="1" CreationDate="2018-04-08T00:53:58.597" Score="4" ViewCount="147" Body="&lt;p&gt;My favorite thing in this world is looking up at stars but I live in NYC so I can't ever see stars. I was wondering if anyone knew of a device I can attach to a smartphone that would let me look up at the stars and actually see some? I can't afford to buy an expensive telescope. I'm a student and I just want to see some stars. Thanks to anyone who can help.&lt;/p&gt;&#xA;" OwnerUserId="22604" LastEditorUserId="22604" LastEditDate="2018-04-08T01:11:00.123" LastActivityDate="2018-04-26T06:39:10.423" Title="What can a regular person use to view the stars?" Tags="&lt;star&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25813" PostTypeId="1" AcceptedAnswerId="25822" CreationDate="2018-04-08T06:50:43.670" Score="4" ViewCount="226" Body="&lt;p&gt;The color-color diagram is a plot between $U-B$ vs $B-V$ values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a ideal blackbody, &lt;strong&gt;what is the equation of this line?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, &lt;strong&gt;what does the linear relation between $U-B$ and $B-V$ look like?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="15901" LastActivityDate="2018-11-05T09:20:50.233" Title="What is equation of straight line for a ideal blackbody on the color color diagram?" Tags="&lt;star&gt;&lt;observational-astronomy&gt;&lt;astrophysics&gt;&lt;data-analysis&gt;&lt;photometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25814" PostTypeId="2" ParentId="25812" CreationDate="2018-04-08T08:10:11.797" Score="4" Body="&lt;p&gt;Binoculars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't going to attach to a smartphone, not everything in life is best seen through a screen. The camera on a smartphone is remarkable, but your eyes are much more sensitive in low light. Binoculars will make a lot more stars visible, even in heavily light polluted places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pair of binoculars is cheap enough for students. They are categorised as, for example 10x50, which means 10x magnification, and 50mm lenses. You want the second number to be big, which means the binoculars gather more light. 10x50 is often thought to be a good compromise between light gathering and weight. You won't want &quot;zoom&quot; or &quot;night vision&quot; or other such features.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-04-08T10:42:15.373" LastActivityDate="2018-04-08T10:42:15.373" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25816" PostTypeId="1" CreationDate="2018-04-08T10:19:35.777" Score="11" ViewCount="594" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/507OY.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/507OY.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;It is stated that:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;After the main sequence, as fusion weakens or stops in the core, outward radiation weakens. The helium core contracts and heats up. Gravitational energy is converted to thermal energy again!&lt;/p&gt;&#xA;&lt;p&gt;The star will appear to cool slowly and will undergo a modest increase in luminosity. During this phase, the path the star will follow in the HR diagram is almost horizontal to the right of its position on the Main Sequence. Stars in this phase are usually referred to as subgiants.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;But I still can't understand.&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="31410" LastEditDate="2021-04-28T13:17:59.107" LastActivityDate="2021-04-28T13:17:59.107" Title="Why is the track of the subgiant stage almost horizontal on the HR diagram?" Tags="&lt;star&gt;&lt;red-giant&gt;&lt;hr-diagram&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25817" PostTypeId="1" AcceptedAnswerId="25818" CreationDate="2018-04-08T11:58:01.420" Score="6" ViewCount="1632" Body="&lt;p&gt;I know that the Milky Way is a member of the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Local_Group?wprov=sfla1&quot; rel=&quot;noreferrer&quot;&gt;Local Group&lt;/a&gt; and the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Virgo_Supercluster?wprov=sfla1&quot; rel=&quot;noreferrer&quot;&gt;Virgo Supercluster&lt;/a&gt; / &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Laniakea_Supercluster?wprov=sfla1&quot; rel=&quot;noreferrer&quot;&gt;Laniakea Supercluster&lt;/a&gt;. However, I can't seem to find any information about the Galaxy cluster we belong to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the Local Group / Milky Way belong to a galaxy cluster?&lt;/p&gt;&#xA;" OwnerUserId="22609" LastEditorUserId="22609" LastEditDate="2018-04-08T12:03:17.937" LastActivityDate="2020-12-26T18:09:34.733" Title="Does the Milky Way belong to a Galaxy cluster" Tags="&lt;milky-way&gt;&lt;galaxy-cluster&gt;&lt;local-group&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25818" PostTypeId="2" ParentId="25817" CreationDate="2018-04-08T15:53:34.347" Score="6" Body="&lt;p&gt;The &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Local_Group&quot; rel=&quot;noreferrer&quot;&gt;Local Group&lt;/a&gt;&lt;/strong&gt; contains 54 plus galaxies.&#xA;Don't know that that counts as much of a cluster.&#xA;Next up in scale, the Milky way is part of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Laniakea_Supercluster&quot; rel=&quot;noreferrer&quot;&gt;Laniakea Supercluster&lt;/a&gt;&#xA;That contains about 100,000 galaxies, so it's a bit on the large size to call a simple cluster.&#xA;The well known &lt;a href=&quot;https://en.wikipedia.org/wiki/Virgo_Cluster&quot; rel=&quot;noreferrer&quot;&gt;Virgo Cluster&lt;/a&gt; contains about 1300 galaxies.&#xA;So it's clearly a thing of intermediate scale.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2018-04-08T15:53:34.347" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25819" PostTypeId="2" ParentId="25817" CreationDate="2018-04-08T17:31:12.387" Score="6" Body="&lt;p&gt;The milky way is not part of a galaxy cluster. The local group is on a lobe of the Virgo supercluster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not all galaxies are members of large clusters, ours is in a small group. However the only difference between &quot;group&quot; and &quot;cluster&quot; is the size. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-04-08T19:47:55.310" LastActivityDate="2018-04-08T19:47:55.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25820" PostTypeId="2" ParentId="25816" CreationDate="2018-04-08T18:12:15.850" Score="3" Body="&lt;p&gt;I will replace the previous answer to focus on the &quot;subgiant&quot; branch prior to red giant, rather than the pre-main-sequence or the &quot;horizontal branch&quot; of core helium fusion.  Those are other times that the luminosity is constant, but this question is about the subgiant branch, which I missed before. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason the luminosity is nearly constant on the subgiant branch is related to the &quot;mass-luminosity relation&quot; of pre- and main-sequence stars.  It is due to radiative diffusion and how it leads to a luminosity that depends only on mass, for a given composition.  If you compare to pre-main-sequence tracks, you should find that the subgiants more or less retrace that prior evolution, just with a somewhat higher luminosity because many of the electrons have been swallowed into neutrons, reducing the opacity and increasing the all-important rate of radiative diffusion.  It's essentially just a helium-dominated mass-luminosity relation, instead of hydrogen-dominated, as the radius rises due to the details of how the interior is evolving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason the luminosity eventually rises steeply on the red giant branch is that as the degenerate core starts to build in mass, it starts to control the temperature of the fusing region, and this significantly changes the internal structure in ways that get into explaining red giants.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2018-06-09T00:56:21.463" LastActivityDate="2018-06-09T00:56:21.463" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25821" PostTypeId="2" ParentId="25816" CreationDate="2018-04-08T18:37:12.293" Score="-1" Body="&lt;p&gt;From blackbody radiation (&lt;span class=&quot;math-container&quot;&gt;$L = c r^2 T^4$&lt;/span&gt;) the horizontal evolution implies increasing radius and decreasing temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This happens because of the depletion of H at the core:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depletion of H at core -&gt; core contraction -&gt; increasing core temperature + starting H fusion in the envelope (in HR diagram, the star goes up) -&gt; to maintain the equilibrium from the injection of energy of H fusion in the envelope, envelope expands and cools (moving horizontally to the right in HR diagram) -&gt; ...&lt;/p&gt;&#xA;" OwnerUserId="22468" LastEditorUserId="25994" LastEditDate="2019-08-01T11:06:59.990" LastActivityDate="2019-08-01T11:06:59.990" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25822" PostTypeId="2" ParentId="25813" CreationDate="2018-04-08T20:46:09.227" Score="2" Body="&lt;p&gt;Use Planck's law: $$F(\lambda,T) = c  \lambda^{-5}  (e^{d/(\lambda  T)} - 1)^{-1}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since magnitude is defined as, $M = -2.5 \log_{10}(F) + \textrm{constant}$, the color index $A-B$ is&#xA;$$(A - B) = -2.5 \log_{10}(F(\lambda_A,T)/F(\lambda_B,T))$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question, use an appropriate effective wavelength of $U,B,V$. Link $U-B$ and $B-V$ via temperature, you should get $U-B = f(B-V)$.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastEditorUserId="2365" LastEditDate="2018-04-08T22:40:46.507" LastActivityDate="2018-04-08T22:40:46.507" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25831" PostTypeId="1" AcceptedAnswerId="25832" CreationDate="2018-04-09T17:32:42.603" Score="2" ViewCount="188" Body="&lt;p&gt;If I understand correctly, the H-alpha and Ca II K lines are absorption lines of the sun and allow to see the chromosphere. Following the Kirchhoff-Bunsen law, an absorption line is produced by a gas which is cooler than the source of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the chromosphere is hotter than the photosphere and the temperature increases outward (&lt;a href=&quot;https://www.researchgate.net/figure/The-figure-plots-profiles-of-mass-density-dashed-line-and-temperature-solid-line-in_fig6_312376685&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.researchgate.net/figure/The-figure-plots-profiles-of-mass-density-dashed-line-and-temperature-solid-line-in_fig6_312376685&lt;/a&gt;). This is in contradiction with the previous hypothesis. Then how can we see the chromosphere in absorption line? What am I missing in the reasoning? Is the Krichhoff-Bunsen law only valid for a thin medium and here we would be in presence of a thick one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thank you in advance for your help.&lt;/p&gt;&#xA;" OwnerUserId="17423" LastActivityDate="2018-04-09T19:57:05.007" Title="absorption line from the chromosphere" Tags="&lt;the-sun&gt;&lt;spectroscopy&gt;&lt;spectra&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25832" PostTypeId="2" ParentId="25831" CreationDate="2018-04-09T19:32:30.277" Score="2" Body="&lt;p&gt;Most of the absorption line comes from the photosphere in the region where the temperature is still dropping.  Nevertheless, you can also get an absorption line where the temperature is rising, due to scattering effects.  The Kirchoff-Bunsen law you describe is a simplified rule that assumes the lines form in thermal equilibrium with the local temperature, but sometimes scattering produces non-thermal-equilibrium (&quot;non-LTE&quot;) effects.  With scattering, line photons that get absorbed get re-radiated right away without coupling to the local temperature, and their re-emission has a 50% chance of being radiated back downward, where they will be reabsorbed deeper in the atmosphere.  This contributes to absorption, especially near line center (and can cause a feature called &quot;central reversal&quot; there).  But it is true that you can also get emission lines from very strong chromospheres where the Kirchoff-type effect does occur, see the profiles from chromospherically active dwarf stars (called &quot;dMe stars&quot;).&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-04-09T19:32:30.277" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25833" PostTypeId="2" ParentId="25831" CreationDate="2018-04-09T19:37:36.957" Score="1" Body="&lt;p&gt;I think this &lt;a href=&quot;https://www.space.com/17160-sun-atmosphere.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; summarizes really well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should not be confused about photosphere and beyond. From inside to photosphere, the temperature decreases with radius as a result of hydrodynamics; this seems natural. Beyond photosphere, there are chromosphere and corona. These layers are hotter than the photosphere, and the temperature increases with radius. We still do not know for certain why it is so. We can safely say that the plasma effects set in. This requires the better understanding of magnetohydrodynamics (MHD). Kirchhoff's law is invalid in this plasma regime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, chromosphere and corona are seen mainly during the solar eclipse.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastEditorUserId="2365" LastEditDate="2018-04-09T19:57:05.007" LastActivityDate="2018-04-09T19:57:05.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25834" PostTypeId="2" ParentId="25752" CreationDate="2018-04-09T21:44:26.477" Score="1" Body="&lt;p&gt;Surprisingly, we will soon be able to do this well enough to go back several hundred million years and place ourselves, if not in the solar system, inside the Galaxy and roughly in the right part of the Galaxy.   It turns out that we know, with high accuracy, the Earth's motion with respect to the microwave background radiation, which provides a well defined frame of reference for motions.  Therefore, we will be able to work out which direction and about how far the MW has moved with respect to this universal frame and separately the Earth with respect to the center of the galaxy. It requires solving for the growth of what is called the 'microwave dipole velocity' over time.  Some of this velocity develops from nearby objects like the MW and the nearby groups of galaxies.  So, we need to know their masses and relative motions a bit better than we presently do.  But, by far, most of the velocity is from objects quite far away, and this allows us to use the simpler and well understood 'perturbation' techniques. The biggest uncertainty for the Earth's position might be the interactions with local stars back in time and what effects they had on perturbing the Suns orbital motion about the center of the galaxy.  Again we need to know better the masses of local stars and their relative velocities to calculate the Earth's trajectory back in time.  As it happens, the space telescope GAIA, which is about to make its second release of data, will greatly improve our knowledge of the motions and, indirectly, the masses of all these stars and nearby galaxies.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you have got that time machine working, we are nearly ready to go! &lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2018-04-09T21:51:39.493" LastActivityDate="2018-04-09T21:51:39.493" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25835" PostTypeId="1" AcceptedAnswerId="25837" CreationDate="2018-04-10T00:40:13.947" Score="42" ViewCount="15682" Body="&lt;p&gt;Because light can only travel so fast, all of the light we see in the sky was emitted at a previous moment in time.  So if for example we see a supernova or some other great stellar event, by the time we see it, it maybe long over.  That made me kind of curious, what is the most ancient light we can see from earth?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe is supposedly ~13+ billion years old, but we are probably not at the very edge of the known universe so all the light we see is probably less than 13 billion years old.  So what is the oldest light we can see? and as an optional follow-up question how do we know the age of that light? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the light itself may not actually be literally 'old', but its probably obvious what I'm asking here, put another way: what's the longest distance that now earth visible light emitted has traveled to reach the earth?  Though that reformation of the question gets kind of tangled with lensing effects.&lt;/p&gt;&#xA;" OwnerUserId="600" LastEditorUserId="600" LastEditDate="2018-04-10T00:45:47.117" LastActivityDate="2018-12-07T21:55:03.667" Title="How old is the oldest light visible from Earth?" Tags="&lt;earth&gt;&lt;light&gt;" AnswerCount="5" CommentCount="9" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25837" PostTypeId="2" ParentId="25835" CreationDate="2018-04-10T01:28:53.743" Score="65" Body="&lt;p&gt;The oldest light in the universe is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;noreferrer&quot;&gt;cosmic microwave background&lt;/a&gt;. Roughly 380,000 years after the Big Bang, protons and electrons &lt;a href=&quot;https://en.wikipedia.org/wiki/Recombination_(cosmology)&quot; rel=&quot;noreferrer&quot;&gt;&quot;recombined&quot;&lt;/a&gt;&lt;sup&gt;1&lt;/sup&gt; into hydrogen atoms. Before this, any photons scattered off the free electrons in the plasma filling space, and the universe was essentially opaque to light. Once recombination occurred, however, photons were able to &lt;a href=&quot;https://en.wikipedia.org/wiki/Decoupling_(cosmology)&quot; rel=&quot;noreferrer&quot;&gt;&quot;decouple&quot;&lt;/a&gt; from the electrons and move through space unimpeded. This relic radiation is still observable today; it has been redshifted and cooled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can detect light from very distant objects, and we have. It makes more sense to talk about distance in terms of &lt;a href=&quot;https://en.wikipedia.org/wiki/Redshift&quot; rel=&quot;noreferrer&quot;&gt;redshift&lt;/a&gt;; the larger the redshift, the farther away an object is. There are a number of extremely high-redshift objects, some of which have had their measurements confirmed, and others of which have not. Candidates include&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/MACS0647-JD&quot; rel=&quot;noreferrer&quot;&gt;MACS0647-JD&lt;/a&gt; (redshift &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013ApJ...762...32C&quot; rel=&quot;noreferrer&quot;&gt;$z=10.7^{+0.6}_{-0.4}$&lt;/a&gt;), a very distant galaxy.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/UDFj-39546284&quot; rel=&quot;noreferrer&quot;&gt;UDFj-39546284&lt;/a&gt; (redshift from &lt;a href=&quot;http://adsabs.harvard.edu/abs/2011Natur.469..504B&quot; rel=&quot;noreferrer&quot;&gt;$z=10$&lt;/a&gt; to &lt;a href=&quot;http://adsabs.harvard.edu/abs/2013ApJ...765L...2B&quot; rel=&quot;noreferrer&quot;&gt;$z=12$&lt;/a&gt;), another galaxy or protogalaxy (although its distance is uncertain, and its exact nature is disputed).&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/GN-z11&quot; rel=&quot;noreferrer&quot;&gt;GN-z11&lt;/a&gt; (redshift &lt;a href=&quot;http://adsabs.harvard.edu/abs/2016ApJ...819..129O&quot; rel=&quot;noreferrer&quot;&gt;$z=11.09^{+0.08}_{-0.12}$&lt;/a&gt;), a very luminous distant galaxy.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All of these objects would have formed some hundreds of millions of years after the Big Bang, however, so the light we see from them is much &quot;younger&quot; than that of the cosmic microwave background.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1 I've never liked the usage in this context, as this was the &lt;em&gt;first&lt;/em&gt; time they combined; the &quot;re&quot; is kind of misleading.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2018-04-10T01:28:53.743" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25839" PostTypeId="1" CreationDate="2018-04-10T02:42:07.377" Score="1" ViewCount="301" Body="&lt;p&gt;Is our solar system affected by a body unobserved as of yet that causes our solar system to curve through space in an orbit? Is this idea outlandish or us there any evidence to support it? Consider this information: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;In 2007 Voyager 2 unexpectedly reached the edge of the heliosphere at a distance approximately 1.5 billion km less than the distance traveled by Voyager 1, indicating the SS is bullet shaped in the direction of the interstellar magnetic field.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another bit of information is that a binary system with a moving SS was suggested as far back in modern times as 1894 -- with predictions 40x more accurate than an earth wobble to calculate precession of equinoxes (over 100 years). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also unknown to me was precession was observed millennia ago, for which we have some records, but today's accepted notion was formed in the 1850's -- before it was accepted that the solar system even moved. Our current theory of precession takes no account for SS movement. Zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lasly local bodies do not precess at all, only distant stars and -- from what I understand -- all of them. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is a binary star system just bunk? Can it be tested for? &lt;/li&gt;&#xA;&lt;li&gt;Is the axial wobble theory tried tested and come up true every time or Does this theory have predictability issues that have to be tweeked constantly? &lt;/li&gt;&#xA;&lt;li&gt;Why is it that the moons orbit around the earth, as well as the earths orbit around the sun, are increasing and the precessional cycle is speeding up? Shouldn't that have the opposite effect?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="22614" LastEditorUserId="11264" LastEditDate="2018-04-11T16:55:24.147" LastActivityDate="2019-10-08T17:24:43.250" Title="Axial wobble or binary star system" Tags="&lt;the-sun&gt;&lt;the-moon&gt;&lt;precession&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25840" PostTypeId="1" CreationDate="2018-04-10T02:55:42.423" Score="6" ViewCount="163" Body="&lt;p&gt;I'm designing a moon system for a fictional setting, and recently came across the idea of horseshoe orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gist of my question is how many objects can share a horseshoe orbital at a time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to have roughly a dozen Earthlike moons in orbit around a planet. Originally, I was thinking about having them spaced out on the same orbital, or just on different orbitals, but a horshoe setup would be near perfect for multiple reasons I won't get into.&lt;/p&gt;&#xA;" OwnerUserId="22628" LastActivityDate="2018-04-11T20:04:01.747" Title="horseshoe orbits" Tags="&lt;natural-satellites&gt;&lt;gas-giants&gt;&lt;orbital-mechanics&gt;&lt;earth-like-planet&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25842" PostTypeId="1" CreationDate="2018-04-10T08:18:36.927" Score="1" ViewCount="132" Body="&lt;p&gt;What is the canonical method to calculate the surface brightness limit of a 2D image, e.g., in a 1 arcsec2 aperture?&#xA;...if image has erg/s/cm2 units and each pixel has some arcsec2 dimension.&lt;/p&gt;&#xA;" OwnerDisplayName="user19333" LastActivityDate="2018-04-10T08:18:36.927" Title="What is the canonical method to calculate the surface brightness limit of a 2D image, e.g., in a 1 arcsec2 aperture?" Tags="&lt;observational-astronomy&gt;&lt;photometry&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25843" PostTypeId="2" ParentId="25840" CreationDate="2018-04-10T10:47:02.887" Score="4" Body="&lt;p&gt;A horseshoe orbit requires a central body (eg a star) a large orbiting body (a planet) and a small co-orbiting body (an asteroid). If the asteroid has too much mass. The orbit &quot;works&quot; because the large body isn't affected by the small co-orbiting body much. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a central body (a planet) and multiple mid-sized co-orbiting and 12 mutually interacting bodies (moons). This isn't stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some special orbits like that of &lt;a href=&quot;https://en.wikipedia.org/wiki/Janus_(moon)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Janus&lt;/a&gt; and Epimetheus, which can be stable for a while. It may be possible to design orbits with more than two bodies in this form, but the region of stability would become very small, and subject to perturbation from the sun and other planets. 12 major co-orbiting bodies would not be a naturally occurring configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sufficiently large planet could have 12 large moons, providing they are on sufficiently different orbits.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-04-10T10:47:02.887" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25844" PostTypeId="2" ParentId="25835" CreationDate="2018-04-10T14:35:53.930" Score="0" Body="&lt;p&gt;This April 2, 2018, CNN article says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;h2&gt;&lt;a href=&quot;https://www.cnn.com/2018/02/28/asia/big-bang-first-light-universe-intl/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Scientists detect 'fingerprint' of first light ever in the universe&amp;quot;&lt;/a&gt;&lt;/h2&gt;&#xA;&lt;p&gt;Following the Big Bang, physicists believe there was only darkness in the universe for about 180 million years, a period known by scientists as Cosmic “Dark Ages.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So I think your answer might be that Big Bang + 180 million years is the oldest light we can see.&lt;/p&gt;&#xA;" OwnerUserId="22637" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-04-11T18:52:20.850" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25848" PostTypeId="1" CreationDate="2018-04-10T18:33:40.220" Score="1" ViewCount="1325" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TYzX8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TYzX8.jpg&quot; alt=&quot;venus&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took a picture of &lt;strong&gt;Venus&lt;/strong&gt; through my telescope (Celestron Astromaster 130eq) using a 10mm lens with a Barlow lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture isn't really good since I took it with my phone and I still don't have a camera. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it isn't a matter of focus since I fiddled a lot with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what to do to clear the image and reduce the brightness. Should I use a filter? &lt;em&gt;and which one?&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="22644" LastEditorUserId="7472" LastEditDate="2018-04-11T09:05:19.247" LastActivityDate="2018-04-11T15:06:54.397" Title="Photographing Venus with a phone through a telescope" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;photography&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25849" PostTypeId="2" ParentId="25835" CreationDate="2018-04-10T20:46:30.507" Score="11" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What's the oldest light that we can see?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Cosmic Microwave Background is considered to be the oldest E-M radiation detectable to us.  It's in the microwave spectrum, so it can't be seen with the naked eye but is picked-up by &quot;radio telescopes&quot;.  We call it &quot;light&quot; in the broad sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One remarkable aspect about this &lt;strong&gt;background radiation&lt;/strong&gt; is its near-uniformity in all directions.  Astronomers reason that the uniformity is too strong for the source to be a really big thing like a huge balloon... but that would be the case if it was all actually as far apart as it seems to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were really as big as it looks, it would take twice the age of the universe for one side to be affected by the other side!  Instead, astronomers believe that what we see was a very small body, which has become bigger; that's why it looks the same in each direction.  Some of the growth is called &lt;em&gt;metric expansion of space&lt;/em&gt; and has a different meaning than ordinary growth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do we know the age of that light?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The age of the cosmic background light can only be determined &lt;strong&gt;indirectly&lt;/strong&gt;, first by knowing how long ago the Big Bang happened, then by figuring when the light was emitted in the course of the Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By comparing the rate at which everything seems to be getting bigger with how big everything seems to be, in the same way that you might estimate how long it would take to drive to a place given the speed of the road and the distance, we calculate the &lt;strong&gt;Hubble Constant&lt;/strong&gt;.  This helps us calculate how long ago the Big Bang happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there are certain &quot;sound waves&quot; (&lt;strong&gt;baryonic acoustic oscillations&lt;/strong&gt;) where old things we see, including the cosmic microwave background, get brighter and dimmer with a rhythm, like a clock's pendulum.  They can be measured either left-right (for moving things) or by monitoring a video (for stationary things).  Measuring these rhythms and comparing them to the Hubble Constant also helps to calculate how long ago the Big Bang happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the microwave background has physical qualities (like temperature and density) that make it possible for us to &lt;strong&gt;determine when it was emitted&lt;/strong&gt; during the expansion and cooling of the Big Bang.  &lt;strong&gt;&lt;em&gt;Together using all these calculations is how we figure the age of the cosmic microwave background light.&lt;/em&gt;&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomers believe that this combined calculation (called &quot;LCDM&quot;, &quot;Lambda-CDM&quot;, or &quot;Big Bang Cosmology&quot;) is very good because &lt;strong&gt;the different numbers do line up, for the most part*&lt;/strong&gt;.  They were pleased to report more good findings as recently as 2018 when a study called the Dark Energy Survey finished.  Nevertheless, since LCDM includes certain assumptions that may never be validated, and since there are still some unexplained discrepancies, we don't know whether another kind of calculation would be better, provided that it still fits the measurements.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do we know that this is the oldest light?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is only by thinking about the physical qualities of the cosmic microwave background, and thinking about when during the Big Bang it must have emitted its light, that &lt;strong&gt;astronomers identified it as the oldest&lt;/strong&gt; possible light in the universe, older than any stars or galaxies.  It doesn't tell us how old it is by itself; in fact, astronomers are always making sure that it's not in fact just a layer of dust on the telescope!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How far away is the cosmic microwave background?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a really hard question to answer.  According to Big Bang Cosmology, the cosmic microwave background wasn't &quot;somewhere&quot; but instead it was everywhere.  And the distance it has traveled since the Big Bang is different than the time multiplied by light speed, because of metric expansion of space.  This is a result of the relativistic length-contraction due to the speed at which everything is moving.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the observable universe younger than the greater universe, assuming that that exists?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Calculating the amount of time from the &lt;strong&gt;Big Bang&lt;/strong&gt; to now gives the same result whether you consider our observable universe or the greater universe that may exist.  That's why the age of &quot;our&quot; universe is the &lt;strong&gt;same&lt;/strong&gt; as the age of &quot;the&quot; universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;*Some different studies to determine the Hubble constant have given cosmologists pause (&lt;a href=&quot;https://gizmodo.com/expanding-universe-mystery-deepens-1830664922&quot; rel=&quot;nofollow noreferrer&quot;&gt;link 1&lt;/a&gt;, &lt;a href=&quot;https://www.forbes.com/sites/startswithabang/2018/12/06/scientists-cant-agree-on-the-expanding-universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;link 2&lt;/a&gt;); depending on which part of the universe you look, it may be close to 67 or it may be closer to 73 in the standard units.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="22646" LastEditorUserId="22646" LastEditDate="2018-12-07T21:55:03.667" LastActivityDate="2018-12-07T21:55:03.667" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="25850" PostTypeId="1" AcceptedAnswerId="25851" CreationDate="2018-04-10T22:25:27.190" Score="9" ViewCount="1377" Body="&lt;p&gt;After taking a look at some recent papers (examples below) that attempt to derive the value of the Hubble Constant, they all conclude with dramatically different values (speaking in context). Why is this? Each of the techniques seems sufficiently rigorous and equally well rooted in understood physics. Is this simply some form of error or is it due to some yet-to-be-discovered physics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Papers:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.nature.com/nature/journal/v551/n7678/pdf/nature24471.pdf&quot; rel=&quot;noreferrer&quot;&gt;A gravitational-wave standard siren measurement of the Hubble constant&lt;/a&gt;, H&lt;sub&gt;0&lt;/sub&gt; = 70&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://arxiv.org/pdf/1502.01589.pdf&quot; rel=&quot;noreferrer&quot;&gt;Planck 2015 Results&lt;/a&gt;, H&lt;sub&gt;0&lt;/sub&gt; = 67.8&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://arxiv.org/pdf/1801.01120.pdf&quot; rel=&quot;noreferrer&quot;&gt;New Parallaxes of Galactic Cepheids from Spatially Scanning theHubbleSpaceTelescope: Implications for the Hubble Constant&lt;/a&gt;, H&lt;sub&gt;0&lt;/sub&gt; = 73.48&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I do understand that this is a hot topic right now in cosmology, however I just don't yet understand why we're having trouble with getting our measured values of the Hubble Constant to converge.&lt;/p&gt;&#xA;" OwnerUserId="14309" LastActivityDate="2019-03-19T18:29:58.813" Title="Why do we continue to find a discrepancy in the Hubble Constant?" Tags="&lt;cosmology&gt;&lt;hubble-constant&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25851" PostTypeId="2" ParentId="25850" CreationDate="2018-04-11T00:21:20.923" Score="15" Body="&lt;p&gt;To make a long story short, the measurements from Planck and the Hubble Space Telescope disagree, and the reason behind this isn't known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let's look at the values &lt;em&gt;with the uncertainties&lt;/em&gt;. We then have three different results that are, perhaps, not as inconsistent as they originally seemed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$70.0^{+12.0}_{-8.0}\text{ km s}^{-1}\text{ Mpc}^{-1}$&lt;/span&gt; from &lt;a href=&quot;https://www.nature.com/nature/journal/v551/n7678/pdf/nature24471.pdf&quot; rel=&quot;noreferrer&quot;&gt;LIGO&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$67.8\pm0.9\text{ km s}^{-1}\text{ Mpc}^{-1}$&lt;/span&gt; from &lt;a href=&quot;https://arxiv.org/pdf/1502.01589.pdf&quot; rel=&quot;noreferrer&quot;&gt;Planck&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$73.48\pm1.66\text{ km s}^{-1}\text{ Mpc}^{-1}$&lt;/span&gt; from &lt;a href=&quot;https://arxiv.org/pdf/1801.01120.pdf&quot; rel=&quot;noreferrer&quot;&gt;Hubble Space Telescope (HST) observations&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The LIGO measurement is consistent with the others, as the authors happily note in their conclusions. However, there is a discrepancy between the results from Planck and the results from the HST. This isn't a new problem; the Hubble group &lt;a href=&quot;http://adsabs.harvard.edu/abs/2016ApJ...826...56R&quot; rel=&quot;noreferrer&quot;&gt;previously determined a value for the Hubble constant&lt;/a&gt; of &lt;span class=&quot;math-container&quot;&gt;$73.24\pm1.24\text{ km s}^{-1}\text{ Mpc}^{-1}$&lt;/span&gt; in 2016. The uncertainties in the measurements describe one &lt;a href=&quot;https://en.wikipedia.org/wiki/Standard_deviation&quot; rel=&quot;noreferrer&quot;&gt;standard deviation&lt;/a&gt;; it doesn't mean that all values beyond the interval of the measurement plus or minus one standard deviation are impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the difference between Planck results and the 2016 HST data is &lt;span class=&quot;math-container&quot;&gt;$3.4\sigma$&lt;/span&gt;, which is a lot. The group came up with several explanations for the discrepancy:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Systematic errors in the data collection, including in the use of the Cepheid variables used for the measurements.&lt;/li&gt;&#xA;&lt;li&gt;Small-scale variations in the Hubble constant, which appear unlikely to have propagated through the data.&lt;/li&gt;&#xA;&lt;li&gt;Systematic errors in the Planck results.&lt;/li&gt;&#xA;&lt;li&gt;Corrections needed to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lambda-CDM_model&quot; rel=&quot;noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$\Lambda$&lt;/span&gt;CDM model&lt;/a&gt; of cosmology, used by Planck.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the 2018 paper, the authors note that the conflict with the Planck data is even stronger - &lt;span class=&quot;math-container&quot;&gt;$3.7\sigma$&lt;/span&gt; (and we're now getting &lt;a href=&quot;https://arxiv.org/abs/1903.07603&quot; rel=&quot;noreferrer&quot;&gt;results that say it could be &lt;span class=&quot;math-container&quot;&gt;$4.4\sigma$&lt;/span&gt;&lt;/a&gt;; I should check those out). They haven't been able to point a figure at any single problem that could be behind the discrepancy; they do optimistically say that it could be the sign of new physics.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2019-03-19T13:50:48.670" LastActivityDate="2019-03-19T13:50:48.670" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25854" PostTypeId="2" ParentId="25835" CreationDate="2018-04-11T02:47:23.113" Score="3" Body="&lt;p&gt;Scientists have discovered a galaxy, named GN-z11 (already mentioned by &lt;a href=&quot;https://astronomy.stackexchange.com/users/2153/hde-226868&quot;&gt;HDE 226868&lt;/a&gt;), which existed a mere 400 million years after the Big Bang, or about 13.3 billion years ago:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.space.com/32150-farthest-galaxy-smashes-cosmic-distance-record.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Farthest Galaxy Yet Smashes Cosmic Distance Record&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discovery of a 10 billion year old star was announced just last week:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astronomy.com/news/2018/04/hubble-images-farthest-star-ever-seen&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble spots farthest star ever seen&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_the_most_distant_astronomical_objects#List_of_most_distant_objects_by_type&quot; rel=&quot;nofollow noreferrer&quot;&gt;list of distant astronomical objects on Wikipedia&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="22649" LastActivityDate="2018-04-11T02:47:23.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25856" PostTypeId="2" ParentId="25848" CreationDate="2018-04-11T07:54:31.293" Score="5" Body="&lt;p&gt;Phone cameras just aren't designed to take good photographs of astronomical objects. The sensor is too small and the configuration options (like exposure time) are minimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For astronomical photography, a better option is to use (the body of) a &lt;a href=&quot;https://en.wikipedia.org/wiki/Digital_single-lens_reflex_camera&quot; rel=&quot;noreferrer&quot;&gt;DSLR&lt;/a&gt; (or the &lt;a href=&quot;https://en.wikipedia.org/wiki/Single-lens_reflex_camera&quot; rel=&quot;noreferrer&quot;&gt;analog version&lt;/a&gt;); the lens is provided by your telescope. Another option is a specialized &lt;a href=&quot;https://en.wikipedia.org/wiki/Charge-coupled_device#Use_in_astronomy&quot; rel=&quot;noreferrer&quot;&gt;CCD camera&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2018-04-11T07:54:31.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25858" PostTypeId="2" ParentId="25848" CreationDate="2018-04-11T12:22:29.867" Score="2" Body="&lt;p&gt;Your mobile phone will not give you control over exposure and focus and the image will be most likely overexposed if you took it against a dark sky. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overexposure causes saturation of the pixels on the sensor, i.e. pixels that get the same photon count (the maximum possible value) but from points of different light intensity. Detail (Venus has none anyway) is lost on your sensor and cannot be recovered by image manipulation tools. There is nothing you can do in that case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only if you can set the exposure (and focus) manually then you will be able to get a usable image. If that is not possible you could try shooting when the sky is still light to force a shorter exposure time.&lt;/p&gt;&#xA;" OwnerUserId="17367" LastActivityDate="2018-04-11T12:22:29.867" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25859" PostTypeId="2" ParentId="25848" CreationDate="2018-04-11T15:06:54.397" Score="3" Body="&lt;p&gt;It is possible to get usable photos of astronomical objects using a camera phone, but as @Glorfindel and @Jee say, in this case you need to reduce the exposure time.  There are apps that give more control over the camera than the standard app (well that's true for iphones, not sure about others), which are free, or not expensive.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a filter would probably not work because the standard camera app is probably averaging the brightness over an area centred on Venus, so will likely just lengthen the exposure to compensate, getting you back to where you started.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2018-04-11T15:06:54.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25860" PostTypeId="1" CreationDate="2018-04-11T15:59:32.257" Score="-3" ViewCount="117" Body="&lt;p&gt;Any idea about  the design of new better telescope having small length of diameter and having more resolving power.&lt;/p&gt;&#xA;" OwnerUserId="22665" LastEditorUserId="22665" LastEditDate="2018-04-21T15:21:18.263" LastActivityDate="2018-04-21T15:21:18.263" Title="About bigger resolution of telescopes" Tags="&lt;telescope&gt;&lt;space-telescope&gt;&lt;newtonian-telescope&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2018-04-11T21:10:03.723" ContentLicense="CC BY-SA 3.0" />
  <row Id="25861" PostTypeId="2" ParentId="25860" CreationDate="2018-04-11T17:08:50.397" Score="2" Body="&lt;p&gt;It's a large &lt;a href=&quot;https://en.wikipedia.org/wiki/Reflecting_telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;reflecting telescope&lt;/a&gt;, which is more effective than the more familiar lens telescope.    Also, in space, less stabilization is needed.  As long as the rotation is very close to zero, Hubble will stay pointed at the same tiny part of the sky, even as it orbits the Earth.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also avoids atmospheric distortion and light pollution because it's outside of the Earth's atmosphere.  In addition to that, a lot of the images taken by Hubble are timed exposure and digitally put together.   It'll looks at the same part of the sky for several hours to get the best possible image.   One of the more clever things they tried with Hubble is they pointed it at empty space, a region where they'd never seen anything and Hubble saw lots of galaxies - &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble_Deep_Field&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble Deep Field&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's been in orbit since 1990, so, even if it takes several hours to take many of it's images, it's been operational for nearly a quarter of a million hours and counting.   That works out to a lot of images.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I invite additions if I skipped over any key points.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="38" LastEditDate="2018-04-11T18:01:42.757" LastActivityDate="2018-04-11T18:01:42.757" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25862" PostTypeId="2" ParentId="25839" CreationDate="2018-04-11T17:58:10.767" Score="4" Body="&lt;p&gt;There is absolutely no possibility that our Sun is part of a &quot;binary star system&quot;. Even the least luminous star (with a mass of about 8% that of the Sun) would have an apparent brightness and proper motion that would easily have led to its discovery. See for example &lt;a href=&quot;https://astronomy.stackexchange.com/questions/2531/is-sun-a-part-of-a-binary-system&quot;&gt;Is Sun a part of a binary system?&lt;/a&gt; and  &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10124/is-it-possible-that-the-sun-has-a-binary-partner-the-nemesis-theory-that-has-e&quot;&gt;Is it possible that the Sun has a binary partner (the Nemesis Theory) that has eluded detection?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-04-11T17:58:10.767" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25863" PostTypeId="2" ParentId="697" CreationDate="2018-04-11T19:19:11.070" Score="10" Body="&lt;p&gt;Several months after the &lt;a href=&quot;https://astronomy.stackexchange.com/a/831/53&quot;&gt;accepted answer&lt;/a&gt; was posted:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 2014, NASA announced that the WISE survey had ruled out any object with Tyche's characteristics, indicating that Tyche as hypothesized by Matese, Whitman, and Whitmire does not exist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The quotation is from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Tyche_(hypothetical_planet)&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article on Tyche&lt;/a&gt;. The article gives the following references for the statement:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.jpl.nasa.gov/news/news.php?release=2014-075&quot; rel=&quot;noreferrer&quot;&gt;http://www.jpl.nasa.gov/news/news.php?release=2014-075&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://arxiv.org/PS_cache/arxiv/pdf/1004/1004.4584v1.pdf&quot; rel=&quot;noreferrer&quot;&gt;https://arxiv.org/PS_cache/arxiv/pdf/1004/1004.4584v1.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://norwalk.dailyvoice.com/news/does-solar-system-have-giant-new-planet&quot; rel=&quot;noreferrer&quot;&gt;http://norwalk.dailyvoice.com/news/does-solar-system-have-giant-new-planet&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that Tyche is not the same as the (currently) hypothetical &lt;a href=&quot;https://en.wikipedia.org/wiki/Planet_Nine&quot; rel=&quot;noreferrer&quot;&gt;Planet Nine&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="53" LastActivityDate="2018-04-11T19:19:11.070" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25864" PostTypeId="2" ParentId="25840" CreationDate="2018-04-11T20:04:01.747" Score="2" Body="&lt;p&gt;Answering your comment, but it's very related to your question&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To clarify, the original setup was a star, orbitted by a planet, which&#xA;  is, in turn orbitted by moons on a horseshoe around it. What I'm&#xA;  saying is would it be more feasible to have a star, which is orbitted&#xA;  by one large mass + other objects sharing a horshoe orbital with that&#xA;  large planet, rather than being it's satellites? Switch it from them&#xA;  as moons, to a series of planets on orbital with a giant. Usually the&#xA;  horseshoe orbit would only have two objects, but is that necessarily&#xA;  the case? Could it, instead be one large object with multiple smaller&#xA;  ones?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want a planet with a lot of moons, increase the planet's &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere#Solar_System&quot; rel=&quot;nofollow noreferrer&quot;&gt;hill sphere&lt;/a&gt;.  That can be done by increasing the planet's mass or moving the planet further away from it's star.   If you need your planet to be in the Goldilocks zone, make the star bigger, just don't make the star too big because it's life will become too short for habitable planets to happen.  If your star is over 2.5 or 3 solar masses it's life may be to short for the planets to form and moons to cool.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even with a bigger hill sphere, A dozen is pushing it.   Orbits need distance between them for long term stability and 12 probably too many.  If you multiply each distance by 3/2 for resonance and even if you don't have resonance, that's a fair enough ballpark estimate, then the outermost moon is 130 times as far from the planet as the innermost and the innermost moon can't be too close to a large planet, or should undergo fairly rapid tidal forces drawing it inwards or outwards.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you make some of them teeny and distant then 12 moons is easy.   Jupiter has dozens of moons, but just 4 large ones.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sharing orbits in a horseshoe (or Trojan) doesn't help you add moons because a horseshoe orbit would be perturbed by near-by large orbiting objects.   In other words, you need to create extra space around two moons sharing an orbit, so you've not really increased the total number of stable orbits for large moons by putting two in a horseshoe.   Now if you do large and tiny, you can have several tiny, but large and large and some measure of stability, two is the limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A horseshoe orbit is very similar to a Trojan orbit.  If the two moons orbital periods are close enough, they get caught in each other's saddle points and never cross.  If they're a little bit further apart from each other, then the catch-up moon can pass through the saddle point and the two moons can switch places (kinda-sorta), as they undergo the horseshoe dance.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-04-11T20:04:01.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25865" PostTypeId="1" AcceptedAnswerId="27214" CreationDate="2018-04-11T20:38:48.857" Score="2" ViewCount="136" Body="&lt;p&gt;How old are Janus and Epimetheus, and are they believed to have formed out of a collision? (I think would explain their shared orbit).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not the age of their elements, which should be similar to the age of the solar system, but the age of their surface and formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cassini told us that Saturn's rings &lt;a href=&quot;https://www.sciencenews.org/article/saturn-rings-age-young-moons&quot; rel=&quot;nofollow noreferrer&quot;&gt;formed relatively recently&lt;/a&gt;, 100 or 200 million years ago.  Are there similar aging techniques that can be applied to the surface of icy moons and as that been done?   (related, this type of aging technique should apply to Pluto and Charon too, because they are believed to have formed by collision).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  &lt;a href=&quot;https://en.wikipedia.org/wiki/Epimetheus_(moon)#Physical_characteristics&quot; rel=&quot;nofollow noreferrer&quot;&gt;I came across this&lt;/a&gt; which says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The extensive cratering indicates that Epimetheus must be quite old.&#xA;  Janus and Epimetheus may have formed from a disruption of a single&#xA;  parent to form co-orbital satellites, but if this is the case the&#xA;  disruption must have happened early in the history of the satellite&#xA;  system. From its very low density and relatively high albedo, it seems&#xA;  likely that Epimetheus is a very porous icy body.[4] There is a lot of&#xA;  uncertainty in these values, however, and so this remains to be&#xA;  confirmed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, significant uncertainty but probably quite old, early in the history of the satellite system, which would be fairly early in solar-system formation.  Much older than Saturn's rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll leave this question open in case anyone has better info, the above being from Wikipedia, though they reference a NASA article.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-04-11T22:34:58.293" LastActivityDate="2018-08-03T17:10:42.380" Title="How old are Janus and Epimetheus?" Tags="&lt;natural-satellites&gt;&lt;cassini&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25866" PostTypeId="2" ParentId="12645" CreationDate="2018-04-11T21:09:12.927" Score="3" Body="&lt;p&gt;The analogy is false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutron stars consist mainly of neutrons, with a small fraction of protons and electrons in equal numbers. The fraction is density dependent, but is in the range 1%-10%.$^{*}$ Neutron stars are so-named because neutrons dominate both by mass &lt;em&gt;and&lt;/em&gt; number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;normal&quot; star consists of about 75% Hydrogen and 23% Helium and 2% heavier elements (by mass; the fraction varies with age, mass and initial chemical composition).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of this material is completely ionised, so the hydrogen is in the form of protons and an equal number of electrons. Whilst protons (just) dominate the mass, you cannot call it a proton-star because there are equal &lt;em&gt;numbers&lt;/em&gt; of electrons. Actually, because of the contribution from ionised helium and heavier elements there are &lt;em&gt;more&lt;/em&gt; free electrons than there are free protons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^{*}$ Neutron stars also have a crust of neutron-rich nuclei (along with enough electrons to balance their charge), but that is a small fraction in terms of their mass and in terms of particle numbers, so is unimportant to this argument.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-04-11T21:09:12.927" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25867" PostTypeId="2" ParentId="25229" CreationDate="2018-04-11T22:34:49.983" Score="4" Body="&lt;p&gt;No, supernovae have been observed earlier and with better cadence than this one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Kepler Satellite has been observing galaxies at a 30 minute cadence from months before the supernova explosion to months afterwards.  Most of these have been Type Ia which do not have detectable shock breakouts, but there was one SN with a breakout published in the Astrophysical Journal before this one: &lt;a href=&quot;http://adsabs.harvard.edu/abs/2016ApJ...820...23G&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Shock Breakout and Early Light Curves of a Type II-P Supernovae Observed with Kepler&quot;&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Nature paper linked in the question provides 4 other SN that were also detected about as early.  I think the new thing here is that it was discovered so early by an amateur astronomer.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="7982" LastEditDate="2018-04-11T22:42:59.627" LastActivityDate="2018-04-11T22:42:59.627" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25868" PostTypeId="1" CreationDate="2018-04-12T01:29:02.860" Score="3" ViewCount="278" Body="&lt;p&gt;Check out Figure 1 in &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1955PASP...67..154H&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1955PASP...67..154H&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wg48U.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wg48U.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to show that the luminosity of a pre-main-sequence star depends primarily on the mass of the star, and that this basic relation continues to hold, to reasonable accuracy, into the main sequence.  Since there is no fusion in a pre-main-sequence star, is it correct to conclude that the relationship between the mass and luminosity of a pre-main-sequence star does not hinge on fusion physics?&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="10196" LastEditDate="2018-04-12T02:44:30.530" LastActivityDate="2018-11-03T19:38:55.713" Title="What does this figure imply about the luminosity of pre-main-sequence stars?" Tags="&lt;star&gt;&lt;luminosity&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25870" PostTypeId="1" AcceptedAnswerId="42095" CreationDate="2018-04-12T03:11:41.870" Score="5" ViewCount="116" Body="&lt;p&gt;And perhaps display a seasonal variation if the exoplanet has an axial tilt, providing proof of an Earth-like world supporting at least oceans with microbial life?&lt;/p&gt;&#xA;" OwnerUserId="22677" LastEditorUserId="37191" LastEditDate="2021-03-30T12:01:56.890" LastActivityDate="2021-04-08T19:33:11.660" Title="Could diatom blooms affect albedo of an exoplanet?" Tags="&lt;exoplanet&gt;&lt;albedo&gt;&lt;climate&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25871" PostTypeId="2" ParentId="18471" CreationDate="2018-04-12T03:25:24.653" Score="2" Body="&lt;p&gt;Because this has gone unanswered, and using &lt;a href=&quot;https://astronomy.stackexchange.com/questions/24675/is-it-possible-for-two-stars-to-be-in-a-horseshoe-orbit-around-a-much-larger-sta&quot;&gt;this very nice answer&lt;/a&gt; as a reference, I can answer this part way.  &lt;a href=&quot;https://academic.oup.com/mnras/article/426/4/3051/1015282&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source article here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;it was found that if the mass of second body becomes larger than&#xA;  1/1200 of the first body, horseshoe orbits become unstable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So Earth mass orbiting Jupiter (1/318) won't work according to the research above.  You could try Mars mass objects orbiting Jupiter (1/2970) and you might get that to work for a fairly extended number of orbits.   Horseshoe orbits are never stable over the long term the way trojan orbits can be, but they can be stable over numerous orbits if they are properly set up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to work out the specifics for higher mass horseshoe objects, but my attempts at calculations have just sent me in circles so far.   If I make any progress I'll post an update.  You'd have to move the Mars-mass moons quite far from the planet and you'd need to keep the variation in their semi-major axes quite small.   Janus and Epimethius have only 50 km variation in their orbits, which is 1 part in 3000 of their orbital radius.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd explore variations in that range, and I'd start at the outer edge of stable region of Jupiter's hill sphere, and if you can't get it to work, decrease the mass some more, maybe lunar mass.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-04-12T03:25:24.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25873" PostTypeId="1" AcceptedAnswerId="25875" CreationDate="2018-04-12T05:47:03.063" Score="3" ViewCount="1247" Body="&lt;p&gt;I came across few sources that claim Polaris movement cannot be observed with the naked eye. But then, I saw images of the star trails (timelapse), and they clearly show Polaris not quite at the true north and moving around the Earth's axis with a radius that is very visible. See for instance the following images:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PS1Wg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PS1Wg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f6X40.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f6X40.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this image from Wikipedia does not show a trail for Polaris:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ueFfM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ueFfM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing here?&lt;/p&gt;&#xA;" OwnerUserId="22355" LastEditorUserId="7472" LastEditDate="2018-04-12T17:03:02.157" LastActivityDate="2018-04-12T17:03:02.157" Title="Does Polaris have a trail in the sky timelapse?" Tags="&lt;amateur-observing&gt;&lt;rotation&gt;&lt;star-gazing&gt;&lt;pole&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25874" PostTypeId="1" CreationDate="2018-04-12T05:54:35.723" Score="3" ViewCount="510" Body="&lt;p&gt;For mass transfer in a binary system, one star must fill its Roche lobe. What determines whether a star in a binary system will fill its Roche Lobe? How will I calculate it?&#xA;I can't find any mathematical formula that shows any equation of mass transfer.&lt;/p&gt;&#xA;" OwnerUserId="16506" LastActivityDate="2018-05-12T21:29:11.093" Title="Filling the Roche Lobe" Tags="&lt;binary-star&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25875" PostTypeId="2" ParentId="25873" CreationDate="2018-04-12T07:14:59.663" Score="5" Body="&lt;p&gt;Does Polaris have a trail in the sky timelapse? Yes. Every star has a trail, but it's smaller if it's closer to the pole. &lt;a href=&quot;https://en.wikipedia.org/wiki/Polaris&quot; rel=&quot;noreferrer&quot;&gt;Polaris&lt;/a&gt; is close to the pole but it's still 45 (arc)minutes away. I don't think it's visible with the naked eye; &lt;em&gt;no&lt;/em&gt; star trail is visible with the naked eye as you only see one 'instance' of the sky. What perhaps would work, and is more feasible for stars closer to the celestial equator, is trying to see the movement of a star compared to an object on Earth, e.g. a tree, a building or the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exposure times of the photos are different, as @J.Chomel notes in the comments. You can calculate the exposure time by looking at the arc length; in 23 hours and 56 minutes (a &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth%27s_rotation#Stellar_and_sidereal_day&quot; rel=&quot;noreferrer&quot;&gt;'sidereal day'&lt;/a&gt;), a star will make a full circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trails on the last photo have an arc length of about 8 degrees, corresponding to about 30 minutes; the first two photos are more like 9 resp. 5 hours. Also, the zoom factor (of the lens) may play a role; if a photo is zoomed in more, Polaris' movement is more likely to be visible. I'm not sure if this plays a role in this photos, though.&lt;/p&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2018-04-12T07:14:59.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25876" PostTypeId="2" ParentId="25873" CreationDate="2018-04-12T07:23:16.700" Score="3" Body="&lt;p&gt;Polaris is about 1 degree from the celestial North Pole. An image with sufficient magnification and sufficient exposure time will show a (short) trail for Polaris.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between the images is the magnification and the exposure time and Magnification. The first image appears to have been about 10 hour exposure (actually a multiple exposure) The Last one appears to be 30 minutes or less. The first has also been magnified more than the last. The result is that the trail of Polaris is just too short to be visible in the last image.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-04-12T07:23:16.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25877" PostTypeId="2" ParentId="25868" CreationDate="2018-04-12T08:02:39.610" Score="1" Body="&lt;p&gt;It shows that the mass of the (proto)star is the major factor determining bolometric magnitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The brightness of the proto-star increases a little as it contracts and becomes significantly hotter (and the radius decreases), The star moves toward the left of the diagram. As fusion in the core becomes significant the star reaches the maximum point on the curve, this stabilises the core and the star settles into the main sequence, it cools slightly and becomes slightly less bright. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-04-12T08:02:39.610" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25878" PostTypeId="1" CreationDate="2018-04-12T12:15:58.413" Score="5" ViewCount="70" Body="&lt;p&gt;This question is primarily about the mathematics of orbital mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Three-body_problem#Restricted_three-body_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;circular restricted 3 body problem&lt;/a&gt; (in 3 dimensions) assumes two massive bodies in circular orbits around their center of mass, and a third with zero mass. It is usually solved in a rotating frame and in reduced units with the separation equal to unity and the period of rotation of the frame of 2π. Five Lagrange points can be found from the pseudo-potential surface in 2 dimensions (as well as some periodic horizontal Lyapunov orbits), and things like halo, Lissajous, vertical  Lyapunov and other periodic orbits can be found when solving in 3 dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is about a relaxation of some of conditions of the problem, allowing for the third body to have a non-zero mass, at least so that the 2nd and 3rd body have some non-trivial center of mass relationship themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm asking about orbits that would be periodic (or nearly-so). Mathematical stability is not necessary, small displacements can lead to exponential growth of errors, as long as there is at least one mathematically defined periodic orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For visualization purposes, imagine moving Earth's Moon to somewhere in the region associated with the original Sun-Earth L1 point, but that's just to get the conversation going.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, one could think of the &quot;Earth&quot; and &quot;Moon&quot; as both being in heliocentric orbits, but so close that they are in a 1:1 resonance. The &quot;Moon&quot; would orbit the Sun with an average velocity of about 1% slower than that of the Earth so that their heliocentric orbit period were the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has this kind of configuration been explored mathematically? I don't care if it is 3D or only 2D, nor if all six 2-body interactions are considered explicitly or if there are some simplifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, anything that considers some effect of the &quot;Moon's&quot; gravity on the &quot;Earth&quot; so that it is no longer treated as massless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is certainly a body of work on non-unity orbital resonances; 2:1, 3:2 etc. but has 1:1 ever been treated in some way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Considering a 1:1 resonance of two planets around a star in close proximity (e.g. large asteroid or small moon at Sun-Earth L1) - have there been any approximate and/or perturbative mathematical treatments?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;note:&lt;/strong&gt; original motivation is &lt;a href=&quot;https://space.stackexchange.com/a/26619/12102&quot;&gt;this answer&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2020-05-10T02:25:24.290" LastActivityDate="2020-05-10T02:25:53.420" Title="1:1 resonance of two planets around a star in close proximity - approximate perturbative mathematical treatments?" Tags="&lt;orbital-mechanics&gt;&lt;orbital-resonance&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25881" PostTypeId="1" AcceptedAnswerId="25889" CreationDate="2018-04-12T16:08:56.493" Score="1" ViewCount="124" Body="&lt;p&gt;I was doing some research on Alpha Centauri A and B, and thought to myself: 'What would happen if something passed directly through the barycenter of these two stars?' I may theorise that a passing star will offset the barycenter with its gravity, sending the two stars in whatever direction it would go. I am not certain about this, nor am I certain that anyone has the answer. But if they do, please answer, I'd really like to know!&lt;/p&gt;&#xA;" OwnerUserId="22689" LastActivityDate="2018-04-13T00:11:32.220" Title="What would happen if a star passed through the barycenter of two binary stars?" Tags="&lt;star&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25882" PostTypeId="2" ParentId="25881" CreationDate="2018-04-12T17:19:19.777" Score="1" Body="&lt;p&gt;No, a star passing through the barrycenter or anywhere between the stars or nearby wouldn't send the stars flying away nor would it cancel out anything.  The effect would depend on how large your passing through star was, how fast it passed through and from what angle it approached, that is, how close it passed to each of the Alpha Centauri stars.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much of a kick or push it gives each object would depend on the relative direction, speed and distance and the gravitation or mass of your passing star, so it might pull the stars closer together, it might push them father apart, but in that would just be adding a vector to the motion of the stars that was already present.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-04-12T17:19:19.777" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25883" PostTypeId="1" AcceptedAnswerId="25887" CreationDate="2018-04-12T17:23:24.250" Score="0" ViewCount="151" Body="&lt;p&gt;&lt;b&gt;A theory about Blackholes storing the information from the universe.&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being said that Blackholes are condensed mass sufficient enough to break every particle into different elementary particles, broken particles will eventually reach the center(or they break down the time they reach there).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some say that the information is stored on the surface instead, which can be carried by the Hawking's Radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How exactly can Hawking's Radiation carry the information?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How is it possible for information to exist at the event horizon? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(If the mass was swallowed by the blackholes then every single particle should be inside the event horizon as they brought towards the black hole at very high velocity and it's becoming difficult to imagine that some particles were left at the horizon itself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;i&gt;Maybe&lt;/i&gt;: If the information is on the event Horizon then it might not have broken to elementary particles!&lt;/p&gt;&#xA;" OwnerUserId="20655" LastActivityDate="2018-04-12T21:46:50.250" Title="Information Paradox with Hawking's Radiation" Tags="&lt;black-hole&gt;&lt;hawking-radiation&gt;&lt;event-horizon&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25884" PostTypeId="2" ParentId="25874" CreationDate="2018-04-12T20:00:01.023" Score="1" Body="&lt;p&gt;If you're up to deriving the Roche limit, here is something that might be useful to your study. This webpage offers its derivation, based on the Newton's equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jgiesen.de/astro/stars/roche.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.jgiesen.de/astro/stars/roche.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22174" LastActivityDate="2018-04-12T20:00:01.023" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25885" PostTypeId="2" ParentId="25881" CreationDate="2018-04-12T20:23:29.560" Score="1" Body="&lt;p&gt;If you imagine all three stars have similar mass, which would be a normal thing to assume as a first approximation, and assuming the incident speed is larger than the orbital velocity (it likely would be, because it is getting pulled in), then the interaction would seem to be likely to happen with one of the stars in the binary, moreso than the other.  So for a short time, those two stars would behave like two objects undergoing &quot;Rutherford scattering.&quot;  It would indeed depend sensitively on the details (especially the incident angle), but it does seem pretty likely to me that it would disrupt the initial binary.  It's possible that the interloping star could even grab one of the stars into a binary with itself, leaving the other one behind, but if the star comes in fast, it seems pretty likely that the original binary will at the very least be made highly eccentric, and quite likely disrupted altogether.  But yes, there's no substitute for an orbital simulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it occurs to me you might be imagining the third star comes in along the axis of the original binary.  In that special case, the incident star would have &lt;em&gt;no effect at all&lt;/em&gt; if the original binary had two equal mass stars in a circular orbit-- it would simply speed up the whole initial binary toward it, and then return it to its original movement after it passes.  But it wouldn't leave the stars moving-- it would erase its effect as it leaves, and all that would linger is an overall shift in the location of the barycenter.  But it's a very special case there.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2018-04-12T20:28:47.117" LastActivityDate="2018-04-12T20:28:47.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25886" PostTypeId="2" ParentId="25883" CreationDate="2018-04-12T21:00:36.913" Score="0" Body="&lt;p&gt;I hate to answer questions like this because they're very complex ideas that are often attempts at explaining in simpler ways to laymen and the &quot;How/why&quot; questions can be long and not always easy to understand.   See &lt;a href=&quot;https://physics.stackexchange.com/questions/29175/why-is-information-indestructible&quot;&gt;this related question with some very nice long and complex answers&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the mass was swallowed by the blackholes then every single particle&#xA;  should be inside the event horizon as they brought towards the black&#xA;  hole at very high velocity and it's becoming difficult to imagine that&#xA;  some particles were left at the horizon itself&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Particles don't have to be retained at the event horizon for information to be retained just outside.   This gets tricky, but an image of the particle, not the particle itself is retained.  While it's an imperfect analogy, dropping a stone in water creates ripples, which spread out and shrink over time.  By carefully analyzing those ripples, you can determine the mass and size of the stone and when it was thrown in.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How is it possible for information to exist at the event horizon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Information doesn't exist &quot;at&quot; the event horizon.  Nothing can exist at the event horizon in a practical sense, stuff passes through the event horizon, but from the perspective of a person some distance away, they see the object falling into the black hole take forever to get there (in reality the visibility red-shifts beyond recognition very quickly), but if you could see it with some kind of super-long-wave imaging device, and a supercomputer to put the information together, you could in theory see everything that ever fell into the black hole by carefully observing the light that escapes from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, nothing is actually encoded on the event horizon (the video got that point wrong).  It's all images that take a very long time to reach the viewer from just outside the event horizon due to extreme time dilation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How exactly can Hawking's Radiation carry the information?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The precise words in the video are&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;If information is stored on the surface of a black hole, hawking&#xA;  radiation has a chance of learning about the information encoded there&#xA;  and carrying it away&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, you're asking a question about an &quot;if&quot; and a &quot;chance of&quot;, very far from a certainty.   I'm not educated enough in the various theories to say how hawking radiation might pick up information from the black hole's formation and history.  My limited understanding is that it can't.   That Hawking radiation is &quot;new&quot; information so to speak, not carried off bits of old information, but that offends information theory . . . which is a problem for quantum theorists.  Please don't take my word for any of this stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Professor Matt Strassler writes a series of articles that try to explain modern physics to people like me and his stuff gets a top review from me.   &lt;a href=&quot;https://profmattstrassler.com/articles-and-posts/relativity-space-astronomy-and-cosmology/black-holes/black-hole-information-paradox-an-introduction/&quot; rel=&quot;nofollow noreferrer&quot;&gt;His article on this subject is probably worth a look&lt;/a&gt; and he touches on some of the &quot;We don't knows&quot; in better detail.  I remember watching one of professor Susskind's lecture on holograms, black holes and information theory on youtube and enjoyed that as well (there's more than one, easily googled, so I won't link).   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-04-12T21:00:36.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25887" PostTypeId="2" ParentId="25883" CreationDate="2018-04-12T21:46:50.250" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How exactly can Hawking's Radiation carry the information?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It does not, if you mean information about the interior of the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relevant theorem is the &lt;a href=&quot;https://en.wikipedia.org/wiki/No-hair_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;No Hair Theorem&lt;/a&gt;.  Basically you can only get gross statistical properties of black holes, nothing else and no information about the interior structure (meaning any info inside stored in any way, stays there or is lost as randomized radiation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you accept the No Hair Theorem then Hawking radiation would be essentially random in character, giving you no more information about the black hole than the No Hair Theorem allows you to know.  This would mean &quot;information&quot; being lost as a black hole evaporates.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How is it possible for information to exist at the event horizon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't think it's possible for anything to exist &lt;em&gt;at&lt;/em&gt; the event horizon for any time.  It's a surface of zero thickness (as I understand it) and no actual substance.  Nothing can stop there and no real thing could be said to be &quot;on&quot; the horizon, just passing through it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the complication of discussing black holes in relation to things like quantum-scale elementary particles is that while we can't say exactly where a particle is (the uncertainty principle) &quot;classical&quot; general relativity (i.e. without quantum theory) doesn't incorporate this at all and so the idea of a particle having a precise position is perfectly reasonable in classical general relativity.  But that's unlikely to be the case in whatever theory we end up with that links general relativity and quantum field theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect you are hearing things related to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Holographic_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Holographic principle&lt;/a&gt;.  This is pretty much beyond my real understanding in any depth, but note that it is intimately linked with trying to connect String Theory, gravity (and hence a kind of quantum gravity), Thermodynamics and a notion of &quot;information&quot; within those multiple contexts.  I personally don't intend to try and read too much into this as it's beyond cutting edge theory for me (beyond my &quot;event horizon&quot; I suppose :-) ).  However I'd suggest being wary of reading too much into this idea given the present state of the theories involved.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2018-04-12T21:46:50.250" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25888" PostTypeId="2" ParentId="25868" CreationDate="2018-04-12T22:09:50.853" Score="0" Body="&lt;p&gt;By definition, the luminosity of a &lt;strong&gt;pre-main sequence&lt;/strong&gt; star cannot possibly depend on fusion physics, since energetically-speaking, there is no significant fusion occurring!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you had plotted some lower mass tracks from sources a bit more modern than 1955 (the tracks you show are out of date and incorrect, not including the Hayashi phase for example), you would see that there is predicted to be a brief plateau in luminosity (for stars/brown dwarfs less than about 0.3 solar masses) caused by deuterium burning. This has yet to be observationally tested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The luminosity of a convective PMS star (which is how they all begin) is determined by its mass and age.&#xA;To a very close approximation &lt;span class=&quot;math-container&quot;&gt;$L\propto t^{-2/3}$&lt;/span&gt;, with the constant of proportionality being mass dependent. Higher mass stars are more luminous at the same age, with a shallower luminosity-mass relationship than on the main sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If&lt;/em&gt; the star becomes largely radiative, the luminosity becomes roughly independent of time (very roughly) - that luminosity is fixed by the rate at which radiative diffusion can leak the gravitational potential energy of the contracting star. The star follows the radiative horizontal (actually more diagonal for stars lower than about 2 solar masses) Henyey track in the HR diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the star remains convective, then to first order, stars follow vertical Hayashi tracks, almost independent of their mass. The luminosity is given by the rate of change of gravitational potential energy, which is in turn limited by how fast convection can transfer heat to the surface. The Hayashi track terminates when the star becomes radiative (for higher masses) or when nuclear fusion starts (at lower masses), or never (for brown dwarfs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set of more modern tracks is shown below. They are from the Wikipedia page on the Hayashi track, but I believe the tracks are by Palla &amp;amp; Stahler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IEbF4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IEbF4.png&quot; alt=&quot;PMS tracks.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-11-03T19:38:55.713" LastActivityDate="2018-11-03T19:38:55.713" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="25889" PostTypeId="2" ParentId="25881" CreationDate="2018-04-13T00:11:32.220" Score="6" Body="&lt;p&gt;If we assume the binary pair have equal masses, then by symmetry a star passing through their barycenter at right angles to their rotation plane will not change their symmetry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did some simple simulations where a star with mass 0.1 flies through a binary in an originally circular orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first case is with an interloper moving at speed 20. It passes through, losing a bit of momentum to the binary, that now moves slowly in the same direction.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/SCYKF.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SCYKF.png&quot; alt=&quot;Interloper moving at speed 20&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second is an interloper moving more slowly, with speed 1. Now the binary moves along more, and the interaction causes a quasiperiodic oscillation in seimajor axis. &lt;a href=&quot;https://i.stack.imgur.com/UZt7L.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UZt7L.png&quot; alt=&quot;Interloper moving at speed 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just dropping the interloper with zero velocity makes it bound to the binary, oscillating back and forth, driving complex quasiperiodic or chaotic motion. &lt;a href=&quot;https://i.stack.imgur.com/I3IGp.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I3IGp.png&quot; alt=&quot;Interloper moving at speed 0&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to make the pair scatter by throwing in a sufficiently heavy interloper with enough momentum that first drags the binaries close, transfers enough momentum to unbind the system, and then passes along. &lt;a href=&quot;https://i.stack.imgur.com/7PPRw.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7PPRw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if the direction is not exactly symmetric things can turn messy. Note that there are non-chaotic solutions too. &lt;a href=&quot;https://i.stack.imgur.com/Qg9p0.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qg9p0.png&quot; alt=&quot;Misaligned interloper&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum up, interlopers can deposit energy and momentum, split binaries, or create chaotic tangles. But it is not the barycenter that is the important part. &lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2018-04-13T00:11:32.220" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25890" PostTypeId="1" AcceptedAnswerId="25892" CreationDate="2018-04-13T01:03:09.427" Score="3" ViewCount="93" Body="&lt;p&gt;The BBC news podcast Science in Action episode &lt;a href=&quot;https://www.bbc.co.uk/programmes/w3cswmp6&quot; rel=&quot;nofollow noreferrer&quot;&gt;New Planet Hunting Mission&lt;/a&gt; covers the upcoming TESS mission between &lt;code&gt;01:00 and 07:15&lt;/code&gt; with Sara Seager, TESS deputy director of science, Massachusetts Institute of Technology, Cambridge, Massachusetts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After about &lt;code&gt;05:00&lt;/code&gt; Dr. Seager mentions that the TESS spacecraft will measure a special set of order 20,000 stars in each camera's 24x24 degree field at a two minute cadence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How were the stars in each field selected for TESS?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The excellent YouTube video &lt;a href=&quot;https://www.youtube.com/watch?v=mpViVEO-ymc&quot; rel=&quot;nofollow noreferrer&quot;&gt;Transiting Exoplanet Survey Satellite (TESS)&lt;/a&gt; nicely explains how TESS operates and will slowly tile the celestial sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; From see &lt;a href=&quot;https://heasarc.gsfc.nasa.gov/docs/tess/operations.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;TESS Operations&lt;/a&gt;. Click for full size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EhBzM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EhBzMm.jpg&quot; alt=&quot;TESS observations&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-04-13T05:51:46.470" LastActivityDate="2018-04-13T09:58:42.610" Title="How did TESS scientists choose which stars to observe at a &quot;2 minute cadence&quot; in each field?" Tags="&lt;observational-astronomy&gt;&lt;exoplanet&gt;&lt;sky-survey&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25891" PostTypeId="2" ParentId="24576" CreationDate="2018-04-13T01:29:39.177" Score="-1" Body="&lt;p&gt;You guys are all hinting about the effects of a black hole on the solar system, but you are forgetting about the effects on time that gravity creates.  The gravity of a black hole would slow time.  As such, the gravitational time dilation effect would let us see a slow death spiral of the solar system, even though to an external observer, the solar system would already be gone.  The sun would indeed become egg-shaped, but the spiral we see at the end of the Hard Sun miniseries would be the first sign of it.  The five year timetable would be how long it would take for the earth to actually experience the full effects of the black hole, because as we got closer to the black hole, time would continue to slow down and we would feel time dragging out even farther than before.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for it suddenly appearing and nobody noticing, the fact is that someone did notice.  That is the reason the flash drive existed.  Remember, NASA still only monitors NEOs over 140m across.  A black hole is actually much smaller than that - and it is virtually undetectable except in the presence of matter.&lt;/p&gt;&#xA;" OwnerUserId="22692" LastActivityDate="2018-04-13T01:29:39.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25892" PostTypeId="2" ParentId="25890" CreationDate="2018-04-13T07:17:30.303" Score="4" Body="&lt;p&gt;A combination of algorithm and by committee.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The majority of targets are bright, low-mass dwarf stars. The primary goal of TESS is to discover planets by transit, and these targets offer the best chance of that. The way this is done is described by &lt;a href=&quot;https://arxiv.org/abs/1706.00495&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stassun et al. (2017)&lt;/a&gt;. This paper summarises how an input catalogue is constructed (from many data sources) and used to estimate a luminosity, distance, temperature, mass and radius for each star. The catalogue is substantially complete for stars capable of being observed by TESS, since it is based on the 2MASS all-sky catalogue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The candidate target list is designed to meet the priorities of the NASA TESS mission - to find small (rocky) planets in $&amp;lt;10$ day orbits using the transit technique. A subset of targets (near the ecliptic poles) are observed for longer, and there, there is a possibility of finding small planets in longer orbits - up to 120 days. The steps that are taken to prioritise this list are (i) remove giants -- too big for appreciable transits to be detected; (ii) avoid crowded sources -- the TESS point spread function is quite large; (iii) the prioritisation of small stars (which give a large transit signal) and of bright stars (which yield more signal-to-noise in the light curves). The final priorities are artificially boosted for those stars near the ecliptic poles  (the continuous viewing zone) and reduced for stars near the galactic plane, where the de-reddening calculations are more uncertain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, space in the target list was also reserved for targets selected by the &quot;community&quot;.&#xA;There were a number of working groups setup (by science area - I was in the open star clusters group). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was then an open call for proposals last September to bid for which stars to include in the 2-minute cadence list. The bids were mainly from the working groups. A committee (of astronomers) then met to decide which bids/targets were meritorious.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-04-13T09:58:42.610" LastActivityDate="2018-04-13T09:58:42.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25894" PostTypeId="1" CreationDate="2018-04-13T14:12:05.367" Score="1" ViewCount="128" Body="&lt;p&gt;What would happen if you took today's six-parameter &lt;a href=&quot;https://en.wikipedia.org/wiki/Lambda-CDM_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;LCDM&lt;/a&gt; model down to the &lt;strong&gt;&lt;em&gt;5%&lt;/em&gt;&lt;/strong&gt; confidence level?  Understanding that the number of permutations is high, I would be particularly interested just in the present age of the universe t0---- what are t0's higher and lower limits, assuming that all five of the other parameters are leaning all the way to the &lt;strong&gt;&lt;em&gt;plausible floor&lt;/em&gt;&lt;/strong&gt; at the 5% CL either to increase t0 together, or to decrease t0 together?&lt;/p&gt;&#xA;" OwnerUserId="22646" LastEditorUserId="22646" LastEditDate="2018-04-16T13:58:01.383" LastActivityDate="2018-04-16T13:58:01.383" Title="What's the oldest possible age of the universe? (LCDM parameter range)" Tags="&lt;cosmology&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25895" PostTypeId="1" AcceptedAnswerId="25899" CreationDate="2018-04-13T14:26:26.513" Score="2" ViewCount="887" Body="&lt;p&gt;(This question was originally posted via World Building, but I was instructed to post it here)&#xA;&lt;a href=&quot;https://worldbuilding.stackexchange.com/questions/109457/is-a-supernova-powerful-to-destroy-the-other-star-in-a-binary-system?noredirect=1#comment332775_109457&quot;&gt;https://worldbuilding.stackexchange.com/questions/109457/is-a-supernova-powerful-to-destroy-the-other-star-in-a-binary-system?noredirect=1#comment332775_109457&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is basically the same:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can a binary star system have one of the two stars go supernova in the first place, and if so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first scenario would be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Star A of the binary swells to red giant status, and then is either&#xA;touching or consumes star B: what would happen?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Scenario 2:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Star A goes supernova, with maybe 1 AU of distance between it and&#xA;Star B. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What info Id like to know:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can it destroy the other star via supernova? &lt;/li&gt;&#xA;&lt;li&gt;What is the effective    range of the supernova, and can it destroy&#xA;stars a light year or more    away?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="22698" LastEditorUserId="22698" LastEditDate="2018-04-14T02:32:09.823" LastActivityDate="2018-04-14T14:29:49.250" Title="Is a supernova powerful enough to destroy the other star in a binary system?" Tags="&lt;supernova&gt;&lt;binary-star&gt;" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="25896" PostTypeId="1" AcceptedAnswerId="25908" CreationDate="2018-04-13T14:55:00.493" Score="3" ViewCount="1866" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Main_sequence&quot; rel=&quot;nofollow noreferrer&quot;&gt;main sequence&lt;/a&gt; stage for a star begins after it stars burning hydrogen into helium. This is often referred to as the &lt;em&gt;zero age&lt;/em&gt; main sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is confusing since stars have already lived several million years in their pre-main sequence stage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does this come from? Is it a historical designation that just stuck?&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2019-11-26T20:36:02.803" Title="Why is the main sequence often called 'zero age' main sequence?" Tags="&lt;main-sequence&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25897" PostTypeId="2" ParentId="25896" CreationDate="2018-04-13T15:07:08.087" Score="0" Body="&lt;p&gt;Honestly, I hadn't heard the term &lt;em&gt;'zero age main sequence'&lt;/em&gt; until this question came along. The Wikipedia page you linked mentions the zero age term and provides a link to &lt;a href=&quot;http://astronomy.swin.edu.au/cms/astro/cosmos/Z/Zero+Age+Main+Sequence&quot; rel=&quot;nofollow noreferrer&quot;&gt;an explanation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zero Age Main Sequence refers more to a specific point in a star's life. When determining the age of a star, astronomers will sometimes start counting from when it joins the main sequence (so it would be 'zero years old' when it does so). Before this time in this case, the object would be considered a protostar.&lt;/p&gt;&#xA;" OwnerDisplayName="user10106" LastActivityDate="2018-04-13T15:07:08.087" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25898" PostTypeId="2" ParentId="25896" CreationDate="2018-04-13T16:34:20.960" Score="3" Body="&lt;p&gt;The term &quot;zero age Main sequence&quot; is there only because during main-sequence evolution, hydrogen is turned to helium, changing the stellar structure a little (electrons are eaten up, light escapes more easily, so the luminosity rises for example).  Yet all stars along that process are called &quot;main-sequence stars.&quot;  So the &quot;zero-age&quot; subset is simply a term for eliminating confusion about what the rising helium content is doing, it's not intended to be a statement about the actual age of the star.  When astronomers talk about the age of a star, it's the age before the ZAMS, plus the age after the ZAMS.  Usually the distinction doesn't matter, because main-sequence lifetimes tend to be much longer than pre-main-sequence lifetimes.  This is convenient, because it is not possible to identify an exact moment when a star &quot;first forms.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation is made even more unclear when you look at the formal definition of a &quot;star,&quot; which is almost never used in practice but requires that the object be undergoing fusion.  So if you believe that definition, the ZAMS is also the birth of the star, and its age is the age on the main-sequence.  No one actually uses that poor language, because we do actually want to talk about the pre-main-sequence lifetime of a star!  I suppose it would be a bit like counting the 9 months we are in the womb as part of our &quot;lifetime,&quot; but this is the standard for what is done in professional astronomy (though not for human lifetimes).&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-04-13T16:34:20.960" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25899" PostTypeId="2" ParentId="25895" CreationDate="2018-04-13T21:51:24.983" Score="1" Body="&lt;p&gt;The gravitational binding energy of a solar mass star of uniform density is $3GM^2/5R = 2.2774\times 10^{41}$ Joule. The energy of a typical supernova is $10^{44}$ J. So it looks at first glance like the explosion could disperse the other star...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...except that this assumes more than a thousandth of the supernova energy is absorbed and does work on the star. If the other star is more than 16 stellar radii away then it will definitely not be hit by this much energy. And even if it is close enough, converting radiation to mass lift is not going to be efficient. The nova will mainly heat up the surface layer down a few optical depths, meaning a few thousand kilometers get very hot but it is doubtful this does more than push the star away a bit and induces some wobble. &lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2018-04-13T21:51:24.983" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25900" PostTypeId="1" CreationDate="2018-04-13T23:44:33.747" Score="10" ViewCount="1959" Body="&lt;p&gt;I know that a white dwarf is mainly balanced by electron degeneracy pressure and that if it gains more than 1.4 solar masses from any source (such as a companion star or a collision), it explodes as a type I supernova. But is there any possibility that a white dwarf can turn into a neutron star (or possibly a black hole)?&lt;/p&gt;&#xA;" OwnerUserId="19320" LastActivityDate="2021-02-14T14:09:39.870" Title="Is there a possibility that a white dwarf can turn into a neutron star or a black hole?" Tags="&lt;black-hole&gt;&lt;supernova&gt;&lt;neutron-star&gt;&lt;white-dwarf&gt;" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25901" PostTypeId="2" ParentId="25900" CreationDate="2018-04-14T02:45:08.907" Score="3" Body="&lt;p&gt;Not for most white dwarfs, because they are usually full of fusionable material (most notably carbon and oxygen).  When material like that starts to undergo gravitational collapse, it heats up and fuses, creating the type Ia supernova and leaving nothing behind because of the intensity of the explosion.  But some white dwarfs have iron cores, for example, &lt;a href=&quot;https://arxiv.org/abs/astro-ph/9911371&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/astro-ph/9911371&lt;/a&gt;, and it might be expected that the iron core would survive the supernova.  Whether it could gain enough material to collapse into a neutron star I couldn't say, it doesn't sound easy but never say never.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-04-14T02:45:08.907" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25902" PostTypeId="2" ParentId="25900" CreationDate="2018-04-14T04:14:05.377" Score="2" Body="&lt;p&gt;If your white dwarf has a large Iron core . . . maybe.   But probably not.  The rapid type 1a collapse at the Chandrasekhar limit, with the radiative pressure pressing into the Iron core . . . just maybe, but even then I want to say no, it's not possible, only that, that scenario just might has a chance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A black hole is a hard no.   A 2.5 (about) solar mass neutron star can turn into a black hole.  A 1.4 solar mass white dwarf even if made from Iron, even if it somehow manages to collapse into a neutron star, it would be much too light to collapse into a black hole. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-04-14T04:14:05.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25905" PostTypeId="1" CreationDate="2018-04-14T07:28:50.187" Score="2" ViewCount="92" Body="&lt;p&gt;I am planning to conduct a research which involves observing the brightness of a star. So far, I haven't been able to obtain much information as to how I may go forward with the observation process. I'm willing to buy a telescope and other equipment. What telescopes are ideal for this and what other equipment must I require to obtain the apparent brightness value of the star?&lt;/p&gt;&#xA;" OwnerUserId="19320" LastEditorUserId="19320" LastEditDate="2018-04-14T09:50:10.593" LastActivityDate="2018-04-15T08:24:46.563" Title="How do I obtain the brightness of a star in the night sky?" Tags="&lt;star&gt;&lt;telescope&gt;&lt;photometry&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25906" PostTypeId="2" ParentId="25905" CreationDate="2018-04-14T10:12:27.697" Score="1" Body="&lt;p&gt;You probably want to find how bright the star is when calibrated against stars of known brightness. You take a picture of the region around the star, then compare the brightness of the star with the brightness of stars of known brightness in the same field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A standard telescope with camera set-up. Then either a good catalogue or a system such as &lt;a href=&quot;https://github.com/mommermi/photometrypipeline&quot; rel=&quot;nofollow noreferrer&quot;&gt;photometry pipeline&lt;/a&gt; to compare the brightness of the star to others of known brightness.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-04-14T10:12:27.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25907" PostTypeId="2" ParentId="25900" CreationDate="2018-04-14T10:26:31.010" Score="11" Body="&lt;p&gt;The answer is: to a neutron star - possibly; to a black hole, no. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process whereby a neutron star is formed is known as an &lt;em&gt;accretion induced collapse&lt;/em&gt; and is being seriously debated, especially in the case of white dwarfs that are born at the upper end of the &quot;natural mass range&quot; for white dwarfs and then accrete more mass as part of a binary system. An excellent read are the introductory sections of &lt;a href=&quot;https://arxiv.org/abs/1308.4887&quot; rel=&quot;noreferrer&quot;&gt;Taurus et al. (2013)&lt;/a&gt;, who go through the motivation, process and (limited) observational evidence. See also &lt;a href=&quot;https://arxiv.org/abs/1504.05194&quot; rel=&quot;noreferrer&quot;&gt;Schwab et al. (2015)&lt;/a&gt;; &lt;a href=&quot;https://arxiv.org/abs/1802.02437&quot; rel=&quot;noreferrer&quot;&gt;Ruiter et al. (2018)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explosion vs Collapse&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A white dwarf can respond to the accretion of material by exploding or collapsing. It depends on the competition between energy released in fusion reactions and the energy being locked away by endothermic electron capture (also known as neutronisation) reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If thermonuclear reactions are initiated then the likely outcome is a runaway nuclear reaction - the pressure inside the star does not rise rapidly enough to prevent the entire star undergoing fusion. The energy released exceeds the gravitational binding energy and the likely outcome is a type Ia supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the white dwarf is supported by electron degeneracy. If neutronisation begins to occur in the core, then protons (in nuclei) capture electrons to form neutrons. This destabilises the star causing it to collapse. The collapse would proceed (quickly) in a similar way to a core collapse supernova. The nuclei would dissociate, neutronisation would run to near-completion and the collapse would be halted by the formation of a neutron star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is little possibility that a black hole could be formed by such a collapse. The collapsing object would be of order 1.4 solar masses and comfortably smaller than the maximum mass of observed neutron stars (at least 2 solar masses). Therefore the collapse will be halted at the neutron star phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;White dwarfs of moderate mass&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most white dwarfs of moderate mass have a C/O composition. They will need to accrete a lot of mass to get to a density (at about &lt;span class=&quot;math-container&quot;&gt;$4\times 10^{13}$&lt;/span&gt; kg/m&lt;span class=&quot;math-container&quot;&gt;$^3$&lt;/span&gt;, reached at &lt;span class=&quot;math-container&quot;&gt;$1.38M_{\odot}$&lt;/span&gt; in a non-rotating WD) where neutronisation becomes energetically feasible. Before this happens, it is probable that fusion reactions are ignited (due to high density, rather than temperature). The threshold density for ignition is *lower * for nuclei with &lt;em&gt;lower&lt;/em&gt; atomic number (He &amp;lt; C &amp;lt; O) because of lower Coulomb repulsion, and the ignition threshold densities for He and C are also &lt;em&gt;lower than the neutronisation threshold for C&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that in a C/O WD that has accreted a lot of matter, ignition could take place in C at the core, or it could be triggered in He (at even lower densities) at the base of a deep accreted shell of material. The outcome would likely be runaway thermonuclear fusion and the complete destruction of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;More massive white dwarfs&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;O/Ne/Mg WDs are made as the final stages of more massive stars (&lt;span class=&quot;math-container&quot;&gt;$8-10M_{\odot}$&lt;/span&gt;) and are probably born as remnants with much higher mass &lt;span class=&quot;math-container&quot;&gt;$&amp;gt;1.2M_{\odot}$&lt;/span&gt; than typical C/O WDs. More massive WDs are smaller, with higher density. The neutronisation thresholds for O, Ne and Mg are only &lt;span class=&quot;math-container&quot;&gt;$1.9\times10^{13}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$6\times 10^{12}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$3\times 10^{12}$&lt;/span&gt; kg/m&lt;span class=&quot;math-container&quot;&gt;$^3$&lt;/span&gt; respectively (all lower than for C, especially for Ne and Mg). This means that a O/Ne/Mg WD may have to accrete very little mass to reach this central density, begin neutronisation, which leads to collapse. In addition if such densities are insufficient to trigger C burning in a C/O WD, then they certainly won't be high enough to trigger burning in O/Ne/Mg because of stronger coulomb repulsion. Further, if little mass is accreted, then there won't be a deep envelope of accreted material in which to ignite burning off-centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all these reasons, O/Ne/Mg WDs may be more &lt;em&gt;likely&lt;/em&gt; (&lt;a href=&quot;https://arxiv.org/abs/1802.09804&quot; rel=&quot;noreferrer&quot;&gt;Liu et al. 2018&lt;/a&gt;; but see also &lt;a href=&quot;https://arxiv.org/abs/1808.05992&quot; rel=&quot;noreferrer&quot;&gt;Wang 2018&lt;/a&gt;) to collapse than explode (the collapse would cause a type of core-collapse supernova though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Does accretion induced collapse occur?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At present there is only indirect evidence. When we look at recently formed neutron stars - identified as fast-spinning pulsars - we see they generally have very high velocities. These velocities are thought to result from an asymmetric &quot;kick&quot; delivered by a type II core-collapse supernova. This in turn suggests it might be quite difficult to retain a neutron star in a binary system, but many neutron stars &lt;em&gt;are&lt;/em&gt; observed to be in binary systems, and many of these, particularly the millisecond pulsars, are thought to have undergone significant mass transfer in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further evidence comes from the retention of a significant population of neutron stars inside globular clusters. Again, the kicks might have been expected to expel most of these. In addition, there are a number of examples which appear to be &quot;young&quot;, in that the ratio of their spin periods to rate of spin decay indicates they were formed recently. Since there are no high mass stars in globular clusters, and so no possible progenitors for these objects via the core-collapse of massive stars, then accretion induced collapse of a high mass white dwarf is a possibility.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-11-09T19:00:33.590" LastActivityDate="2018-11-09T19:00:33.590" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25908" PostTypeId="2" ParentId="25896" CreationDate="2018-04-14T10:53:32.213" Score="5" Body="&lt;p&gt;The zero age main sequence (ZAMS) is the theoretical locus of points in the Hertzsprung-Russell diagram where the luminosities of young stars (at a range of masses) are mostly supplied by the nuclear reactions that have begun in their cores.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prior to this, the objects are known as pre-main sequence stars and most of their luminosity arises from the release of gravitational potential energy as they contract.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;exact&lt;/em&gt; definition (whether it is 99% or 90% of the luminosity) does vary, but since the nuclear reactions ramp up quite quickly this doesn't materially affect any quantitative analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, the age of a star is not measured from the ZAMS. The reason for this is that what is of physical interest is the time since the star formed and that must include the pre-main sequence lifetime. Whilst this is very short (maybe 1 million years) for stars more massive than the Sun, and certainly shorter than the total main sequence lifetime, it can be much longer (100 million years) for a low-mass M dwarf. This is also the reason that the ZAMS is a theoretical construct. Stars born at the same time will never be all on the ZAMS at the same time. By the time low-mass stars reach the ZAMS, more massive stars will have evolved away from the main sequence!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the origin of the term, I suspect it comes from the early computer/analytic models of stellar evolution. There you would likely miss out the pre main-sequence phase and &lt;em&gt;start&lt;/em&gt; your model with &quot;zero age&quot; at that point where it achieves equilibrium with the luminosity supplied by fusion and with a pristine He/H ratio in its core. Thus in reality, stars of different mass would have different ages at this point, but from the point of view of a theoretical modeller, it would be their starting condition and hence a &quot;zero age&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-04-14T23:13:50.373" LastActivityDate="2018-04-14T23:13:50.373" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25909" PostTypeId="2" ParentId="25905" CreationDate="2018-04-14T14:11:58.920" Score="5" Body="&lt;p&gt;I don't know about the telescope, but I know that you will need the telescope that has CCD technology to be able to convert from counts/second (as brightness) that CCD detects to quantitative values easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the observational procedure, here is the standard regardless of what kind telescope you use:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Take a picture of your object (let's name it &lt;code&gt;ImObj&lt;/code&gt;).&lt;/li&gt;&#xA;&lt;li&gt;Take a picture of a standard star (let's name it &lt;code&gt;ImStd&lt;/code&gt;) or stars (i.e., stars that have real brightness in a database, let's call it &lt;code&gt;MagStdReal&lt;/code&gt;). &lt;/li&gt;&#xA;&lt;li&gt;Take pictures for calibration (let's name it &lt;code&gt;ImCal&lt;/code&gt;). If CCD telescope, these images include dark and flat images.&lt;/li&gt;&#xA;&lt;li&gt;Use &lt;code&gt;ImCal&lt;/code&gt; to calibrate both &lt;code&gt;ImObj&lt;/code&gt; and &lt;code&gt;ImStd&lt;/code&gt; (i.e., removing noise and etc.).&lt;/li&gt;&#xA;&lt;li&gt;Convert image brightness of calibrated ImStd and ImObj to quantitative values (let's call it &lt;code&gt;MagStdFake&lt;/code&gt; and &lt;code&gt;MagObjFake&lt;/code&gt;). Note that you can use any arbitrary zero reference point for &lt;code&gt;MagStdFake&lt;/code&gt; and &lt;code&gt;MagObjFake&lt;/code&gt;, but they use the same reference.&lt;/li&gt;&#xA;&lt;li&gt;Find conversion equation to &lt;code&gt;MagStdReal = f(MagStdFake)&lt;/code&gt;. The simplest form of f() is just linear shift: &lt;code&gt;MagStdReal = a + MagStdFake&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Apply &lt;code&gt;MagObjReal = f(MagObjFake)&lt;/code&gt;. And now you get to do some science.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Also note that, a telescope filters certain wavelengths with certain response function. So, you have to observe everything in one setting of telescope. You cannot mix images or products of different setting of a telescope. Only &lt;code&gt;MagObjReal&lt;/code&gt; or &lt;code&gt;MagStdReal&lt;/code&gt; those are telescope insensitive (ideally).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I provide very short and conceptual procedure to give you the idea. There are a bunch of details that are important. If you use keywords in the list procedure here and search in Google, you should get more relevant hits to each topic.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastEditorUserId="2365" LastEditDate="2018-04-15T08:24:46.563" LastActivityDate="2018-04-15T08:24:46.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25910" PostTypeId="2" ParentId="25895" CreationDate="2018-04-14T14:29:49.250" Score="1" Body="&lt;p&gt;Yes, and it is common that only one star in a binary system goes SN at a time. Besides that, merger that results in SN-like event is also possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first scenario: this is a bit more complicated than you might think. When a star expands, its material reaches Lagrangian point where the material beyond this point is transferred to its companion. The consequence of this depends on parameters. It might end as SN Ia if the companion is white dwarf, or black hole if it is neutron star, etc. Meaning that its companion might not be &quot;consumed.&quot; The case of &quot;consumed&quot; is, if I understand your intention correctly, called common envelope. Normally, common envelope evolution ends at merging. Complication is likely possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scenario 2: Normally, a SN in binary system (at real physical distance) cannot even significantly blow its companion to leave the vicinity (that is why people try to search for possible companion in the vicinity after observing SNe). So, a SN cannot &quot;destroy&quot; its companion. Speaking of that, I don't know for certain the answer to the hypothesized distance of 1AU. For a light year away, that is for certain the explosion will not destroy its companion.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-04-14T14:29:49.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25911" PostTypeId="1" AcceptedAnswerId="25912" CreationDate="2018-04-14T14:43:45.390" Score="4" ViewCount="114" Body="&lt;p&gt;This may be a very stupid question, as I know little about astrophysics, but if solids like Aerogel can float in gasses like sodium hexafluoride, could there be pressures high enough on gas giants to compress a gas to the point where solid particles such as dust collide and create floating landmasses?&lt;/p&gt;&#xA;" OwnerUserId="22710" LastActivityDate="2018-04-14T17:04:43.107" Title="Floating Masses in Gas Giants" Tags="&lt;gas-giants&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25912" PostTypeId="2" ParentId="25911" CreationDate="2018-04-14T17:04:43.107" Score="3" Body="&lt;p&gt;Buoyancy, following &lt;a href=&quot;https://en.wikipedia.org/wiki/Buoyancy#Archimedes&amp;#39;_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Archimedes' law&lt;/a&gt;, depends on the volume the body and the density of displaced material. The force is &#xA;$$ F_{buoy} = \rho_{gas} \cdot V_{body}\cdot g$$&#xA;which has to act against the bodies' own weight $F_{weight} = -m_{body} \cdot g$.&lt;br&gt;&#xA;Here $\rho$ is density, $V$ is Volume, $g$ the local gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're referring to &lt;a href=&quot;https://www.youtube.com/watch?v=HoCAxS4vqwQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; video, then the aerogel that was created in the 1980's is essentially a high-tech balloon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few points that need discussion here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. Balloons the size of continents don't occur naturally anywhere.&lt;/strong&gt;&lt;br&gt;&#xA;I think we can take this as given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. The filling gas&lt;/strong&gt;&lt;br&gt;&#xA;acts to decrease $F_{weight}$ and its pressure must be enough to hold up the body volume. Else the body will collapse and the buoyancy decreases to zero.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A floating continent would need some appropriate filling gas in large quantities, that has lower density than the gas giant atmosphere.&lt;br&gt;&#xA;Impossible is a strong word, but this seems very impossible to me. Also for the continent then be able to uphold its structure under its own weight would be very difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3. Particles either float or experience drag.&lt;/strong&gt;&lt;br&gt;&#xA;If they float, they're coupled to the gas flows. Once they would grow larger, they start experiencing gas drag. When this happens, they loose their momentum and are dragged down into the depths of the planet.&lt;br&gt;&#xA;They then will contribute to the structure of the deep planet, but are lost to any continent-building process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even if your floating continents seem impossible, I think is not a stupid question at all. I might just throw this on our students and test their understanding of buoyancy!&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2018-04-14T17:04:43.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25913" PostTypeId="2" ParentId="25905" CreationDate="2018-04-14T17:24:38.843" Score="2" Body="&lt;p&gt;If the target star varies by half a magnitude or more,&#xA;it's possible to estimate its magnitude visually&#xA;by comparing it to nearby reference stars of similar magnitude.&#xA;&lt;a href=&quot;http://www.skyatnightmagazine.com/feature/how-guide/how-observe-variable-stars&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sky at Night&lt;/a&gt;,&#xA;&lt;a href=&quot;http://www.skyandtelescope.com/observing/celestial-objects-to-watch/the-lure-of-variable-stars/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sky and Telescope&lt;/a&gt;,&#xA;and the &lt;a href=&quot;http://www.rasc.ca/how-observe-variable-stars&quot; rel=&quot;nofollow noreferrer&quot;&gt;Royal Astronomical Society of Canada&lt;/a&gt;&#xA;have articles explaining the procedure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The American Association of Variable Star Observers (AAVSO) provides a&#xA;&lt;a href=&quot;https://www.aavso.org/apps/vsp/&quot; rel=&quot;nofollow noreferrer&quot;&gt;chart plotting tool&lt;/a&gt;&#xA;and comprehensive&#xA;&lt;a href=&quot;https://www.aavso.org/observing-manuals&quot; rel=&quot;nofollow noreferrer&quot;&gt;observing manuals&lt;/a&gt;&#xA;for visual, CCD, DSLR, and other methods.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-04-14T17:24:38.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25914" PostTypeId="1" CreationDate="2018-04-14T18:00:10.987" Score="3" ViewCount="370" Body="&lt;p&gt;Well, I'll start off by saying I am certain I am doing it wrong! Now that I got that out of the way here is my image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://preview.ibb.co/j62Cbn/IMG_0175.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://preview.ibb.co/j62Cbn/IMG_0175.jpg&quot; alt=&quot;Image 1&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://preview.ibb.co/du3kwn/IMG_0163.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://preview.ibb.co/du3kwn/IMG_0163.jpg&quot; alt=&quot;Image 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a feeling I can't see the planet because I need a filter to block out the light surrounding Jupiter, however I honestly didn't expect to have such a profound atmosphere-like disc around Jupiter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My fiance thought it was my own eye being projected, but it was a DSLR doing the actual looking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using the Celestron NexStar 127 SLT which comes stock with a 25mm and 9mm eyepiece. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&#xA;Okay! I focused and was able to see Jupiter, including some aspects of its atmosphere. However, I was not able to take a descent picture of it. Now I'm thinking the need for filters and proper DSLR usage methods will come into play.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WKg7Z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WKg7Z.jpg&quot; alt=&quot;Jupiter 3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21974" LastEditorUserId="21974" LastEditDate="2018-04-16T19:20:10.427" LastActivityDate="2018-04-18T07:28:43.107" Title="My first pictures of Jupiter" Tags="&lt;telescope&gt;&lt;photography&gt;&lt;jupiter&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25915" PostTypeId="2" ParentId="25914" CreationDate="2018-04-14T20:20:51.947" Score="8" Body="&lt;p&gt;Your problem is that Jupiter is completely out of focus.  You need to turn the focus knob so that the bright ring gets smaller and smaller, and then you will find that the bit in the middle is bright, and is Jupiter.  Try turning the focus knob one way, and if the central dark circle gets bigger, then turn the other way, slowly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What your pictures are showing is a magnified image of the central reflector that is located on the front corrector plate of your scope - that's the dark central circle - surrounded by the unfocussed light of Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could start by practising during the day, focus on distant objects (trees etc) and without the camera. Then try at night, again without a camera. Once you can use the telescope to see, try taking a photograph, again you can practise during the day. The moon might then be a good target before trying to get images of planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck&lt;/p&gt;&#xA;" OwnerUserId="11822" LastEditorUserId="11822" LastEditDate="2018-04-15T08:41:03.920" LastActivityDate="2018-04-15T08:41:03.920" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25916" PostTypeId="1" CreationDate="2018-04-14T21:33:39.613" Score="4" ViewCount="123" Body="&lt;p&gt;I would like to know please whether or not the TESS telescope will be much better than the Kepler one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerDisplayName="user22712" LastActivityDate="2018-04-14T21:50:00.143" Title="Will the TESS telescope be much better than the Kepler telescope?" Tags="&lt;space&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25917" PostTypeId="2" ParentId="25916" CreationDate="2018-04-14T21:50:00.143" Score="4" Body="&lt;p&gt;Rather than one being better than the other, I would say that they have different targets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas Kepler focused on a specific area of the sky, TESS will cover an area 350 times bigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, most of the exoplanets discovered by Kepler are far away; TESS will focus on stars within 300 light years away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more info, feel free to watch this video I recently made: &lt;a href=&quot;https://www.youtube.com/watch?list=PL3RiFKfZj3pv1ZqpFxuZinoGtUGEOankw&amp;amp;v=7bsxlG-rh1E&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?list=PL3RiFKfZj3pv1ZqpFxuZinoGtUGEOankw&amp;amp;v=7bsxlG-rh1E&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18421" LastActivityDate="2018-04-14T21:50:00.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25918" PostTypeId="2" ParentId="25894" CreationDate="2018-04-14T23:37:45.140" Score="2" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Q:&lt;/strong&gt; What would happen if you took today's six-parameter LCDM model to the &lt;strong&gt;95%&lt;/strong&gt; confidence level? Understanding that the number of permutations is high ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;NASA's &lt;a href=&quot;https://lambda.gsfc.nasa.gov/education/graphic_history/parameters.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Graphical Parameter Comparisons&lt;/a&gt; webpage has this to say:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Sources in the literature for values of these selected parameters determined between the years 1996 and 2016 are shown in the Data References Table. It would be a formidable task to include all results from all experiments and every paper. Therefore, results presented here are representative, and there are multiple caveats associated with those choices. For example, we do not guarantee that the datapoints are statistically independent of each other, although the choice of data sources is motivated to some extent by a wish to include independent determinations. There also has been no attempt to estimate what fraction of the parameter uncertainties are due to cosmic variance, although this should clearly be a consideration.&lt;/p&gt;&#xA;&lt;p&gt;We discuss the individual parameters in pages that follow, and include an accompanying graphical history for each. Vertical gray lines shown in the graphical histories are weighted averages of the WMAP and Planck data points rather than the entire sample. This is done as a visual aid, rather than a statement that this is the current best value: the WMAP and Planck data values serve as a common thread in each of the plots. In a few cases, liberties were taken in the quoting of asymmetric error bars from published results as symmetric error bars on the plots. For illustrative purposes, the differences are negligible. Quoted uncertainties are &lt;strong&gt;68%&lt;/strong&gt; confidence limits.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://lambda.gsfc.nasa.gov/education/graphic_history/data_ref.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Data references Table&lt;/a&gt; - Contributed by the NASA LAMBDA Archive Team.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The published data isn't accurate enough to reach the confidence level you require.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astro.ucla.edu/%7Ewright/age.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Prof. Wright of UCLA&lt;/a&gt; has a webpage about the age of the universe using actual age measurements, not estimates from cosmological models, but a more authoritative source would be &lt;a href=&quot;https://lambda.gsfc.nasa.gov/education/graphic_history/age.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA's Lambda Archive Team&lt;/a&gt; which has a page on &lt;span class=&quot;math-container&quot;&gt;$t_0$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;There are many ways to tweak the values and make the calculation, the computed value is somewhat dependent of the exact cosmological model assumptions. Within the standard ΛCDM model framework, CMB measurements, combined with BAO and SNIa measurements of the expansion rate, are presently in agreement and tightly constrained. The gray vertical line, representing the weighted average of WMAP and Planck data points, is positioned at &lt;span class=&quot;math-container&quot;&gt;$t_0$&lt;/span&gt; = 13.796 Gyr.&lt;/p&gt;&#xA;&lt;p&gt;This graphic from their website sums it up:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uCpPu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uCpPu.png&quot; alt=&quot;Time zero estimate.&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-04-14T23:37:45.140" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25919" PostTypeId="2" ParentId="25835" CreationDate="2018-04-15T02:01:16.633" Score="2" Body="&lt;p&gt;You've made two questions using semantics:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&amp;quot;How old is the oldest light visible from Earth?&amp;quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;https://astronomy.stackexchange.com/a/13752/18054&quot;&gt;@Pela's answer&lt;/a&gt; to: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/13744/why-is-there-a-difference-between-the-cosmic-event-horizon-and-the-age-of-the-un&quot;&gt;Why is there a difference between the cosmic event horizon and the age of the universe?&lt;/a&gt; - So in ~100M years the most distant light will reach us, from over 116M light years away.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The 16 Gly that the distance to the event horizon is today is sort of a coincidence. It has nothing to do with the age of the Universe. It only depends on the future expansion of the Universe, which in turn depends on the densities of the components of the Universe (Ωb, ΩDM, ΩΛ, etc.). If the Universe has been dominated by matter (or radiation), then there would be no event horizon: No galaxy, ever-so far away would not be visible to us, if we just had the patience to wait. A galaxy is 10,000 billion lightyears away? Just wait long enough (exactly how long depends on the actual density).&lt;/p&gt;&#xA;&lt;p&gt;However, our Universe happens to be dominated by dark energy, which accelerates the expansion without boundaries. This unfortunately means that the light leaving today from a galaxy 17 Gly away will be carried away by the expansion faster than it can travel toward us. In contrast, the light emitted today from a galaxy 15 Gly away will travel in our direction, but will nonetheless initially move away from us due to the expansion. However, its journey toward us makes this expansion rate smaller and smaller (since the expansion rate increases with distance from us), and after a period of time it will have traveled so far that it has overcome expansion and starts decreasing its distance from us and eventually reach us after 100 Gyr or so.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&amp;quot;I guess the light itself may not actually be literally 'old', but its probably obvious what I'm asking here, put another way: &lt;strong&gt;what's the longest distance that now earth visible light emitted has traveled&lt;/strong&gt; to reach the earth? Though that reformation of the question gets kind of tangled with lensing effects?&amp;quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Yes, that's a totally different question ...&lt;/p&gt;&#xA;&lt;p&gt;See one of the earliest papers: &amp;quot;&lt;a href=&quot;https://arxiv.org/abs/astro-ph/0310808&quot; rel=&quot;nofollow noreferrer&quot;&gt;Expanding Confusion: Common Misconceptions of Cosmological Horizons and the Superluminal Expansion of the Universe&lt;/a&gt;&amp;quot; by Davis and Lineweaver (2003).&lt;/p&gt;&#xA;&lt;p&gt;Newer works:&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;&lt;a href=&quot;https://arxiv.org/abs/1305.3943&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Shared Causal Pasts and Futures of Cosmological Events&lt;/a&gt;&amp;quot;, by Friedman, Kaiser, and Gallicchio (2013).&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;&lt;strong&gt;Conclusion:&lt;/strong&gt; ... While the observable spatial densities of galaxies, clusters, and thus quasars are thought to reflect correlations set up during inflation, it remains an open question whether inflationary era events at specific comoving locations — where quasar host galaxies later formed — could yield an observable correlation signal between pairs of eventual quasar emission events at those same comoving locations billions of years after the inflationary density perturbations were imprinted.&lt;/p&gt;&#xA;&lt;p&gt;In closing, we note that all of our conclusions are based on the assumption that the expansion history of our observable universe, at least since the end of inflation, may be accurately described by canonical general relativity and a simply-connected, non-compact FLRW metric. These assumptions are consistent with the latest empirical search for non-trivial topology, which found no observable signals of compact topology for fundamental domains up to the size of the surface of last scattering.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5GRud.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5GRud.png&quot; alt=&quot;Fig. 1&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;FiG. 1. Conformal diagram showing comoving distance, &lt;span class=&quot;math-container&quot;&gt;$R_0χ$&lt;/span&gt; in Glyr, versus conformal time, &lt;span class=&quot;math-container&quot;&gt;$R_0τ /c$&lt;/span&gt; in Gyr, for the case in which events A and B appear on opposite sides of the sky as seen from Earth (α = 180°). The observer sits at Earth at &lt;span class=&quot;math-container&quot;&gt;$χ = 0$&lt;/span&gt; at the present conformal time &lt;span class=&quot;math-container&quot;&gt;$τ = τ0$&lt;/span&gt;. Light is emitted from A at (&lt;span class=&quot;math-container&quot;&gt;$χA, τA$&lt;/span&gt;) and from B at (&lt;span class=&quot;math-container&quot;&gt;$χB, τB$&lt;/span&gt;); both signals reach the Earth along our past lightcone at (&lt;span class=&quot;math-container&quot;&gt;$0,_{τ0}$&lt;/span&gt;). The past-directed lightcones from the emission events (red and blue for A and B, respectively) intersect at (&lt;span class=&quot;math-container&quot;&gt;$χAB, τAB$&lt;/span&gt;) and overlap for &lt;span class=&quot;math-container&quot;&gt;$0 &amp;lt; τ &amp;lt; τAB$&lt;/span&gt; (purple region). For redshifts &lt;span class=&quot;math-container&quot;&gt;$z_A = 1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$z_B = 3$&lt;/span&gt; and a flat ΛCDM cosmology with parameters given in Eq. (11), the events are located at comoving distances &lt;span class=&quot;math-container&quot;&gt;$R_{0χA} = 11.11$&lt;/span&gt; Glyr and &lt;span class=&quot;math-container&quot;&gt;$R_{0χB} = 21.25$&lt;/span&gt; Glyr, with emission at conformal times &lt;span class=&quot;math-container&quot;&gt;$R_{0τA}/c = 35.09$&lt;/span&gt; Gyr and &lt;span class=&quot;math-container&quot;&gt;$R_{0τB}/c = 24.95$&lt;/span&gt; Gyr. The past lightcones intersect at event AB at &lt;span class=&quot;math-container&quot;&gt;$R_{0χAB} = 10.14$&lt;/span&gt; Glyr at time &lt;span class=&quot;math-container&quot;&gt;$R_{0τAB}/c = 13.84$&lt;/span&gt; Gyr, while &lt;strong&gt;the present time is &lt;span class=&quot;math-container&quot;&gt;$R_{0τ0}/c = 46.20$&lt;/span&gt; Gyr&lt;/strong&gt;. Also shown are the cosmic event horizon (line separating yellow and gray regions) and the future-directed lightcones from events A and B (thin dashed lines) and from the origin (0,0) (thick dashed lines). In a ΛCDM cosmology like ours, events in the yellow region outside our current past lightcone are space-like separated from us today but will be observable in the future, while events in the gray region outside the event horizon are space-like separated from observers on Earth forever. Additional scales show redshift (top horizontal axis) and time as measured by the scale factor, &lt;span class=&quot;math-container&quot;&gt;$a(τ)$&lt;/span&gt;, and by proper time, &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt;, (right vertical axis) as measured by an observer at rest at a fixed comoving location.&lt;/p&gt;&#xA;&lt;p&gt;Also see: &amp;quot;&lt;a href=&quot;https://arxiv.org/abs/1711.11395&quot; rel=&quot;nofollow noreferrer&quot;&gt;Causal horizons in a bouncing universe&lt;/a&gt;&amp;quot;, by Bhattacharya, Bari, and Chakraborty (2017):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;&lt;strong&gt;Conclusion:&lt;/strong&gt; The present work shows that the causality problem in bouncing universe is intrinsically related to an understanding of the various phases of the universe during the contraction phase. As our understanding of the contraction phase is purely speculative at present the models we use to figure out the nature of particle horizon remains over simplistic. The present authors believe that although the causality problem in bouncing universe models are far from being solved the present article shows the qualitative and quantitative difficulties one must have to circumvent in the future to produce more meaningful results.&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Short answer: It's &lt;a href=&quot;https://en.wikipedia.org/wiki/Particle_horizon&quot; rel=&quot;nofollow noreferrer&quot;&gt;46.9B light years&lt;/a&gt;. Another Wikipedia page says: &lt;a href=&quot;https://en.wikipedia.org/wiki/Observable_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;46.6B light years&lt;/a&gt;. The experts above calculate 46.2.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-04-15T02:01:16.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25920" PostTypeId="1" AcceptedAnswerId="25921" CreationDate="2018-04-15T11:50:09.537" Score="1" ViewCount="956" Body="&lt;p&gt;I'm thinking of putting a large wall paper with an astronomical motive on a wall. I found quite a number of wall papers by various vendors, but most of them are obviously computer generated images, which is a no-go. It must be an image of an astronomical object, a nebula or a galaxy. (There's also a size requirement, not under 236 x 370 cm, so please do not suggest I use a printed image fitting a common frame format).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I found this wall paper. To my eyes it does not obviously look like CGI. But there are no diffraction artifacts visible for the bright stars; perfect circles it seems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone identify the object? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EPckV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EPckV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6593" LastEditorUserId="6593" LastEditDate="2018-04-15T16:13:09.223" LastActivityDate="2018-04-15T23:18:54.740" Title="Is this an actual nebula or CGI?" Tags="&lt;galaxy&gt;&lt;nebula&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25921" PostTypeId="2" ParentId="25920" CreationDate="2018-04-15T18:47:23.553" Score="6" Body="&lt;p&gt;This appears to be one of Sergii Tsololo's more convincing efforts in the synthetic &quot;space&quot; image genre.&#xA;It is listed on various stock photo sites as&#xA;&quot;&lt;a href=&quot;http://www.alamy.com/stock-photo-blue-purple-space-galaxy-55997572.html&quot; rel=&quot;noreferrer&quot;&gt;Blue and purple space galaxy&lt;/a&gt;,&quot;&#xA;or with added 4-way spikes,&#xA;&quot;&lt;a href=&quot;http://www.alamy.com/stock-photo-blue-purple-space-galaxy-55997572.html&quot; rel=&quot;noreferrer&quot;&gt;Blue space starfield with flares&lt;/a&gt;.&quot;&#xA;A real astrophoto would identify the subject and not be tagged &quot;abstract&quot; or &quot;fantasy.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check it another way, I submitted the image to&#xA;&lt;a href=&quot;http://nova.astrometry.net/&quot; rel=&quot;noreferrer&quot;&gt;Astrometry.net&lt;/a&gt;,&#xA;whose plate solver failed to match the field within their 10 minute time limit.&#xA;When I submitted an amateur photo of the&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Trifid_Nebula&quot; rel=&quot;noreferrer&quot;&gt;Trifid Nebula&lt;/a&gt;&#xA;as a control, Astrometry.net found a solution in less than 1 second.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-04-15T18:47:23.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25922" PostTypeId="2" ParentId="25914" CreationDate="2018-04-15T19:45:03.397" Score="3" Body="&lt;p&gt;The camera's image plane was some distance away from the telescope's focal plane,&#xA;so it captured a cross section of the cone of light from the telescope's mirrors.&#xA;The farther from best focus, the larger and dimmer the &quot;donut.&quot;&#xA;A typical eyepiece field is located near the middle of the eyepiece barrel.&#xA;The image plane of a DSLR is near the back of the camera body.&#xA;When switching from an eyepiece to the camera, you probably need to adjust the telescope's focus knob considerably.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: Now that you found the focus, shorten the exposure or darken the ISO setting so that no pixels are fully saturated.&#xA;Image manipulation software such as &lt;a href=&quot;https://www.gimp.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;GIMP&lt;/a&gt; can brighten a slight underexposure but can't fix an overexposure.&#xA;The red and blue fringes could be due to either off-axis &lt;a href=&quot;https://en.wikipedia.org/wiki/Chromatic_aberration&quot; rel=&quot;nofollow noreferrer&quot;&gt;chromatic aberration&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_refraction&quot; rel=&quot;nofollow noreferrer&quot;&gt;atmospheric refraction&lt;/a&gt;.&#xA;In either case, you can decompose and realign the (already somewhat filtered) RGB channels.&#xA;Finally, &lt;a href=&quot;https://en.wikipedia.org/wiki/Unsharp_masking&quot; rel=&quot;nofollow noreferrer&quot;&gt;unsharp masking&lt;/a&gt; can enhance the small-scale feature contrast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atmospheric turbulence degrades image sharpness to a varying degree, so take several images.&#xA;A standard trick for planetary imaging these days is to record video and use software such as&#xA;&lt;a href=&quot;https://www.astronomie.be/registax/&quot; rel=&quot;nofollow noreferrer&quot;&gt;RegiStax&lt;/a&gt;&#xA;to combine the best frames into a still image.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-04-17T17:51:59.267" LastActivityDate="2018-04-17T17:51:59.267" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25924" PostTypeId="2" ParentId="8227" CreationDate="2018-04-15T21:01:01.490" Score="2" Body="&lt;p&gt;It is very difficult to measure Galactic orbital speeds of stars in our galaxy for several reasons.  We are moving with them and do not have a reliable measurement yet of even our own orbital motion. We only get one component of motion, the radial component (although Gaia is about to change that).  Dust in the galactic plane prevents us from seeing very far in our own galaxy.  So, when we measure Dark Matter in galaxies, it usually done in other galaxies where we can see both sides of the galaxy and, except for inclination angle, directly measure the orbital motion.  &lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2018-04-15T21:01:01.490" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25925" PostTypeId="1" AcceptedAnswerId="25977" CreationDate="2018-04-15T23:08:10.963" Score="6" ViewCount="342" Body="&lt;p&gt;I have recently viewed two news articles announcing a galaxy composed almost entirely of dark matter and one with hardly any. This made me wonder if there is a &quot;continuum&quot; of percentages of dark matter in galaxies, where it has been measured, or if there is a &quot;preferred&quot; range.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="907" LastEditDate="2018-04-16T15:37:11.977" LastActivityDate="2018-04-20T12:20:30.903" Title="How does the amount of dark matter in measured galaxies vary?" Tags="&lt;dark-matter&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25926" PostTypeId="2" ParentId="25920" CreationDate="2018-04-15T23:18:54.740" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm thinking of putting a large wall paper with an astronomical motive on a wall. I found quite a number of wall papers by various vendors, but most of them are obviously computer generated images, which is a no-go. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Buyer beware. Huge printers are expensive, few places have them; you don't want wallpaper that's too skinny or paper that's not actually &quot;wallpaper&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you check that &lt;em&gt;artists&lt;/em&gt; work he photographs a lot of vegetables too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can anyone identify the object? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Getty Images identifies it as: &quot;&lt;a href=&quot;https://www.gettyimages.co.uk/detail/photo/space-galaxy-royalty-free-image/91633822&quot; rel=&quot;nofollow noreferrer&quot;&gt;Space galaxy - Blue and purple space galaxy&lt;/a&gt;&quot;, by 'sololos', Creative #: 91633822. That is most likely the original source. It's also available in a smaller size from the same person at iStock and is called: &quot;&lt;a href=&quot;https://www.istockphoto.com/ca/photo/constellations-cepheus-gm485169533-38602092&quot; rel=&quot;nofollow noreferrer&quot;&gt;Constellations. Cepheus (Cep) - Stock image&lt;/a&gt;&quot;. The image is purchased by various places to use unrestricted, for example the &lt;a href=&quot;https://www.irishtimes.com/news/science/supermassive-irish-scientist-offers-explanation-for-black-holes-1.3008747&quot; rel=&quot;nofollow noreferrer&quot;&gt;Irish Times wrote an article&lt;/a&gt; and used the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;Large&quot; size (assuming that's the size used by the place you plan on purchasing a &lt;strong&gt;copy&lt;/strong&gt; from, to produce their enlarged version) is 4327 x 3245 px (36.64 x 27.47 cm) 300 dpi | 14.0 MP - a 14MP image blown up to 236 x 370 cm will be quite grainy close up. Every pixel will be a 10x10 block, if &lt;em&gt;that&lt;/em&gt; was printed at 300 DPI (don't count on it) then 10 pixels would be 1/30th of an inch (~ 1mm). It's more likely to be 1/4 of that resolution, or 4x4 mm blocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Cepheus_(constellation)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia webpage for Cepheus&lt;/a&gt; and the &lt;a href=&quot;http://www.spacetelescope.org/images/heic0810cf/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble Telescope image&lt;/a&gt; for Cepheus cause me to wonder if the image is correctly titled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uAbdE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uAbdE.jpg&quot; alt=&quot;Cepheus&quot;&gt;&lt;/a&gt;&#xA;&lt;sub&gt;&lt;sub&gt;Credit: NASA, ESA, the Hubble Heritage Team (STScI/AURA)-ESA/Hubble Collaboration and A. Evans (UVA, Charlottesville/NRAO/SBU)&lt;/sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 7.2 megabyte 3600x2749 (9.9MP) .TIF file of the &lt;a href=&quot;http://chandra.harvard.edu/photo/2009/cepb/&quot; rel=&quot;nofollow noreferrer&quot;&gt;constellation Cepheus&lt;/a&gt; is also available from the Chandra X-ray Observatory. By starting with a known source, and using a high quality format, you can find somewhere that makes huge prints, and end up with something that won't devalue your property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, &lt;a href=&quot;http://www.pictorem.com/mural-print.html?width-cm=370&amp;amp;height-cm=236&amp;amp;go=get%20price&quot; rel=&quot;nofollow noreferrer&quot;&gt;one place, wants $400&lt;/a&gt; to create wallpaper. Shop around and check reviews if you only pay a couple of hundred bucks you'll probably be wasting a couple of hundred bucks; it depends upon what you want it to look like &lt;strong&gt;and&lt;/strong&gt; most importantly how close you intend to view it.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastActivityDate="2018-04-15T23:18:54.740" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25930" PostTypeId="2" ParentId="25925" CreationDate="2018-04-16T05:12:35.227" Score="-2" Body="&lt;p&gt;The basic method used to measure dark matter in a galaxy is to observe the extra gravitational pull coming from matter which is not visible. The effect on orbits of stars in a galaxy (determined by observing galaxy rotational curves) and the bending of light by gravitational lensing are the some of the techniques used for measuring the gravitational pull of a galaxy. &lt;/p&gt;&#xA;" OwnerUserId="20162" LastEditorUserId="2365" LastEditDate="2018-04-16T07:28:13.260" LastActivityDate="2018-04-16T07:28:13.260" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25931" PostTypeId="1" AcceptedAnswerId="25998" CreationDate="2018-04-16T09:30:50.213" Score="3" ViewCount="719" Body="&lt;p&gt;Is there an easy way to tell at what angle the solar equator is wrt the horizon at any given time of the day? I would think that it rotates 180 degrees from sunrise to sunset, but not in a linear fashion. But solar noon should always be 0 degrees wrt the horizon right?&lt;/p&gt;&#xA;" OwnerUserId="5286" LastActivityDate="2018-04-23T05:05:45.530" Title="angle to the equator of the Sun on sky at a given time" Tags="&lt;the-sun&gt;&lt;rotation&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25932" PostTypeId="2" ParentId="25931" CreationDate="2018-04-16T10:46:02.007" Score="2" Body="&lt;p&gt;Depends on the precision you need.. The sun's plane of rotation coincides with the plane of rotation of the solar system.&lt;br&gt;&#xA;And for the latter, there is already a name in astronomy, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ecliptic&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ecliptic&lt;/a&gt;. Depending on your latitude on the planet $\delta$, the ecliptic plane cuts through your local horizon in east and west at an angle of $\alpha = 90°-\delta - 23.5°$.&lt;br&gt;&#xA;This is the maximum height of the sun at midday and the sun's height varies thus between 0 and $\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this would then be the angle you're looking for.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another &lt;a href=&quot;https://astronomy.stackexchange.com/questions/11849/what-is-the-angle-between-the-equator-of-the-sun-and-the-plane-of-the-earths-o&quot;&gt;recent answer&lt;/a&gt; pointed out the angle between solar axis of rotation and the Ecliptic is $7.25°$. Translating that into the angle unter which one sees the solar equator, it is still $\alpha$, up to a small correction factor of $- \frac{R_{\odot}}{1 AU} \sin(\delta + 23.5°)$ which can be maximally $-0.05°$.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="2365" LastEditDate="2018-04-20T17:37:23.757" LastActivityDate="2018-04-20T17:37:23.757" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25934" PostTypeId="1" CreationDate="2018-04-16T15:11:27.163" Score="0" ViewCount="80" Body="&lt;p&gt;So, most of you all are probably (well, why wouldn't you be) familiar with The Magellanic Clouds, and hour they are Irregular type galaxies. You should also be aware that they &lt;strong&gt;both&lt;/strong&gt; orbit the Milky Way, so they act sort of like moons to a planet. Since they (most likely) do &lt;strong&gt;not&lt;/strong&gt; have a central Supermassive Black Hole, I wonder, out of all ridiculousness, what are the answer to these two questions?:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What would happen to the Small or Large Magellanic Clouds if a Black Hole (maybe 100 time bigger than the sun) passed though them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What would happen if &lt;strong&gt;Sagittarius A*&lt;/strong&gt; passed through either the Small or Large Magellanic clouds. If you are not aware of Sagittarius A*, it is the Supermassive Black Hole that lies at the center of the Milky Way.&lt;/p&gt;&#xA;" OwnerUserId="22689" LastActivityDate="2018-04-17T16:07:20.240" Title="What if a Black Hole passed through an Irregular type galaxy?" Tags="&lt;sagittarius-a&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25937" PostTypeId="1" AcceptedAnswerId="25995" CreationDate="2018-04-16T16:32:14.947" Score="6" ViewCount="3953" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Surya_Siddhanta&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Surya Siddhanta&lt;/a&gt;, &quot;a Sanskrit treatise in Indian astronomy from the late 4th-century or early 5th-century CE&quot; is truly a great work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how was it possible for the writers to find the exact values of the diameters of different planets and the distance between the sun and the earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, are there any mistakes in this book?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there's any problem in my question please inform me.&lt;/p&gt;&#xA;" OwnerUserId="22722" LastEditorUserId="7982" LastEditDate="2019-04-28T16:42:16.760" LastActivityDate="2021-03-14T18:51:39.247" Title="How did the authors of Surya Siddhanta find the diameters of other planets in the solar system?" Tags="&lt;solar-system&gt;&lt;history&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25939" PostTypeId="2" ParentId="25934" CreationDate="2018-04-16T18:47:31.777" Score="4" Body="&lt;p&gt;Black holes (even SMBs) are actually very small objects on an astronomical scale and in practical terms probably very little would happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There would be some perturbations of stellar orbits within the formation they were in, but unless there was a fairly close approach  very little would happen and on the scale of the entire formation the changes would affect only a small number of objects and not the overall &quot;shape&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put this in perspective, Sag. A* has a mass of about 4,000,000 solar masses.  Sounds big (is big :-) ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the Large Magellanic Cloud has a mass of about 10,000,000,000 solar masses (or about 2500 time larger than Sag. A*).  The dominant gravitational field is still that associated with the LMC itself, not that of Sag A*.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think a good analogy would be a bullet fired into a lake.  Sure, it can cut a path through one isolated region and cause disturbance close to that path, but the bulk of the lake will be undisturbed apart from a slight ripple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Small Magellanic Cloud is still a pretty large mass at about 6,500,000,000 solar masses, so still roughly 1,400 times more massive than Sag. A*.  Again, we're talking bullet in lake territory.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2018-04-16T18:47:31.777" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25940" PostTypeId="1" CreationDate="2018-04-16T21:44:50.873" Score="5" ViewCount="100" Body="&lt;p&gt;In a matter of weeks, I will be performing my first observation using a 25-cm Newton telescope (+CCD camera) and would be grateful for advice regarding an otherwise basic research that I have decided to try out due to this occasion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will be observing eclipsing binary stars (type Algol), yet have never used any program or site. Just to determine which binaries will be visible at my location on a specific date at a specific time - which sites do you use for this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what should be my main goals when taking photos in terms of timing and exposition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later on, I will use APT for photometry and estimate the time length of the eclipse, which will be my objective. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="14653" LastEditorUserId="7982" LastEditDate="2021-04-16T23:28:27.237" LastActivityDate="2021-04-16T23:28:27.237" Title="Observing eclipsing binaries" Tags="&lt;star&gt;&lt;binary-star&gt;&lt;photometry&gt;&lt;eclipsing-star-systems&gt;&lt;time-domain-astronomy&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25941" PostTypeId="2" ParentId="21547" CreationDate="2018-04-17T02:29:43.393" Score="3" Body="&lt;p&gt;I don't have any photos right now, but I think the only photo that really comes close was James K's. The Milky Way looks much like it does in the photos, but with much less magnitude, and virtually no color variation. This is because rods in our eyes allow us to see in dim light, while cones help us differentiate color. If you're in a dark location, in my opinion somewhere blue or darker according to this map: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://darksitefinder.com/maps/world.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://darksitefinder.com/maps/world.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then it is something worth going out, staying up late and seeing and CERTAINLY photographing if you've got a camera that can do long exposures. (some smartphones these days technically can, but I'd find a cheap used APS-C camera like a Sony a5100 or Nikon D3300).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, it's like a huge yet distance milky cloud, and when you look up, you feel incredibly small. It's awesome.&lt;/p&gt;&#xA;" OwnerUserId="16140" LastActivityDate="2018-04-17T02:29:43.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25943" PostTypeId="1" CreationDate="2018-04-17T07:04:11.240" Score="0" ViewCount="119" Body="&lt;p&gt;Is it actually relevant to study objects that are light years far? Because in theories we are actually seeing the past as the light takes years to reach us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say we are studying a asteroid that is going to hit as after several hundred years how can we confirm we are truly studying the actually position and everything of that object as we are seeing object from the past.&lt;/p&gt;&#xA;" OwnerUserId="22733" LastActivityDate="2018-04-17T21:59:03.790" Title="Are we really seeing the asteroids from the past?" Tags="&lt;asteroids&gt;&lt;distances&gt;&lt;space-time&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25944" PostTypeId="2" ParentId="25943" CreationDate="2018-04-17T09:28:18.043" Score="3" Body="&lt;p&gt;Most things in space move very predictably. For reasonably dense objects, like stars, planets or asteroids, gravity far outweighs all of the other forces acting on them, and it's not too hard to take account of gravity and predict their movements (if you have a powerful enough computer. For things like clouds of very thin gas it's a bit harder. So from the positions and movements of all the distant objects that we can see (at the time when the light we see them by left them), we can pretty accurately predict their movements a considerable way into the future, for instance up to possible collisions a few centuries out (for an asteroid) or a few billion years out (for Andromeda). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there is a possibility of something unexpected happening -- between the light we see now being emitted and a possible future collision, a supermassive black hole travelling at close to the speed of light could pass through the Andromeda galaxy and scatter it, for instance, but we don't see things like that happening anywhere else in the universe, so it seems pretty unlikely to happen here.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-04-17T09:28:18.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25946" PostTypeId="2" ParentId="25940" CreationDate="2018-04-17T16:50:59.917" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://www.skymaponline.net&quot; rel=&quot;nofollow noreferrer&quot;&gt;www.skymaponline.net&lt;/a&gt; might be the one you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Main goal when taking images in terms of timing and exposition:&#xA;1. Choose time with good sky condition (e.g., clear sky, no wind)&#xA;2. Expose long enough for good S/N ratio, but not too long that CCD is saturated.&#xA;3. If you want to measure the length of the eclipse, you can follow since it is coming in until going out. Or, you can take a couple of images at different epochs, then assume some trajectory and just doing some math.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-04-17T16:50:59.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25947" PostTypeId="1" AcceptedAnswerId="25957" CreationDate="2018-04-17T17:31:34.277" Score="2" ViewCount="146" Body="&lt;p&gt;I have completely no idea about astronomy, but I am starting growing interest in it as I have been always wondering on what has been discovered out there and since I am studying CS, maybe I will learn more than I think as CS and astronomy as relative to some point. How should I start? Can you suggest me a good website for example or something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I know this is an opinion - based question but still I need to make this question here).&lt;/p&gt;&#xA;" OwnerUserId="22743" LastEditorUserId="1891" LastEditDate="2018-04-18T07:19:10.277" LastActivityDate="2018-04-18T17:31:54.173" Title="Tips for a beginer who wants to learn more about astronomy" Tags="&lt;fundamental-astronomy&gt;" AnswerCount="4" CommentCount="10" FavoriteCount="1" ClosedDate="2018-04-18T22:25:20.007" ContentLicense="CC BY-SA 3.0" />
  <row Id="25948" PostTypeId="1" CreationDate="2018-04-17T17:33:05.903" Score="2" ViewCount="121" Body="&lt;p&gt;I plan to buy a 12&quot; or 14&quot; dobson, the choice depending on my final budget and on the total weight of the instrument. I have several options for the primary mirror. I tried to find a formula that would give the acceptable range for mirror thickness depending on the diameter, maybe also on the material (BK7, pyrex, ....). Is there such a formula, or a back-of-the-enveloppe calculation, which could help me? If not, what do people go with usually (for these diameters, I mean)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your help.&lt;/p&gt;&#xA;" OwnerUserId="10486" LastActivityDate="2018-04-17T23:40:58.123" Title="&quot;Ideal&quot; primary thickness in a Dobson" Tags="&lt;dobsonian-telescope&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25949" PostTypeId="2" ParentId="25948" CreationDate="2018-04-17T17:59:25.540" Score="-2" Body="&lt;p&gt;Since you're just buiding a Dobsonian, you're not going for aspherics and in fact not going for particularly high-quality surface figure at all.  In particular, there's no need to worry about orientation-dependent sag (not that this is a significant factor for such a small diameter in the first place).&lt;br&gt;&#xA;It's not clear whether you are buying or fabbing the primary.  If the latter, just choose a blank such that the center thickness, which is the thinnest spot on the mirror, is at least maybe 5 mm.  You also haven't specified the focal length of your primary, which certainly figures into the thickness and mass calculations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just like having fun :-) ,  buy the blank as thick as you like and use an end mill to lightweight it by removing most of the material from behind, leaving a &quot;web&quot; to support the figured surface.   &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-04-17T17:59:25.540" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25950" PostTypeId="2" ParentId="25940" CreationDate="2018-04-17T21:34:44.417" Score="3" Body="&lt;p&gt;The &lt;a href=&quot;https://www.aavso.org/aavso-eclipsing-binary-section&quot; rel=&quot;nofollow noreferrer&quot;&gt;AAVSO eclipsing binary section&lt;/a&gt; is a good place to start reading.&#xA;Their how-to articles address eclipsing binary specific issues such as predicting times of minimum.&#xA;To get a list of observable Algol-type stars, try their&#xA;&lt;a href=&quot;https://www.aavso.org/observation-planning&quot; rel=&quot;nofollow noreferrer&quot;&gt;observation planner tool&lt;/a&gt;&#xA;with variable type &lt;code&gt;EA&lt;/code&gt;.&#xA;When you've selected a target, you can use their&#xA;&lt;a href=&quot;https://www.aavso.org/apps/vsp/&quot; rel=&quot;nofollow noreferrer&quot;&gt;variable star plotter&lt;/a&gt;&#xA;to make finder and comparison charts.&#xA;Finally their &lt;a href=&quot;https://www.aavso.org/ccd-photometry-guide&quot; rel=&quot;nofollow noreferrer&quot;&gt;guide to CCD photometry&lt;/a&gt; has technical advice on image capture and analysis.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-04-17T21:34:44.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25951" PostTypeId="2" ParentId="25943" CreationDate="2018-04-17T21:59:03.790" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;how can we confirm we are truly studying the actually position and everything of that object as we are seeing object from the past.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All astronomical observations are of things as they &lt;em&gt;were&lt;/em&gt; at a time in the past that depends how far away the object is. We understand that. If we need to calculate the future behaviour of an object, then our extrapolation begins from that past point in time to work out a trajectory through time and space.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-04-17T21:59:03.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25952" PostTypeId="2" ParentId="25948" CreationDate="2018-04-17T23:31:10.267" Score="2" Body="&lt;p&gt;If you're buying the instrument, as opposed to making it from scratch, my advice is to worry about the overall quality, as opposed to a single parameter such as mirror thickness. It boils down to whether you trust the vendor or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, there is no simple formula. You could run the &lt;a href=&quot;http://www.davidlewistoronto.com/plop/&quot; rel=&quot;nofollow noreferrer&quot;&gt;PLOP&lt;/a&gt; simulator and try a few different scenarios, and see what the overall RMS error is in each case. But the values will vary very significantly. The main factors are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the diameter / thickness ratio&lt;/li&gt;&#xA;&lt;li&gt;the number of support points in the mirror cell&lt;/li&gt;&#xA;&lt;li&gt;the relative positions of the support points&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All of the above assumes a traditional mirror with a flat back, and a classic mirror cell with multiple-of-three support points. But the analysis gets even more complex if it's a cell mirror (the back is not flat), or a honeycomb structure (the back is flat but there are empty spaces in the volume of the mirror). And then there are conical mirrors, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's difficult to second-guess the vendor, especially if you've never built an instrument. Just read the reviews, ask other owners, try and gauge the overall build quality.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="616" LastEditDate="2018-04-17T23:40:58.123" LastActivityDate="2018-04-17T23:40:58.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25953" PostTypeId="1" CreationDate="2018-04-17T23:38:09.347" Score="1" ViewCount="151" Body="&lt;p&gt;I am trying to understand how spectrometers or spectroscopy can be used to calculate the surface composition of asteroids for the purposes of asteroid mining. &lt;/p&gt;&#xA;" OwnerUserId="22748" LastActivityDate="2018-04-18T15:25:58.643" Title="How is IR spectroscopy used to determine the composition of asteroids?" Tags="&lt;asteroids&gt;&lt;spectroscopy&gt;&lt;infrared&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25954" PostTypeId="1" CreationDate="2018-04-18T03:47:36.683" Score="5" ViewCount="1282" Body="&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/400457/what-does-general-relativity-say-about-the-relative-velocities-of-objects-that-a&quot;&gt;https://physics.stackexchange.com/questions/400457/what-does-general-relativity-say-about-the-relative-velocities-of-objects-that-a&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;we can actually observe galaxies that are moving away from us at &gt;c&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Um, I think I missed the groundbreaking headline that said scientists have directly measured a galaxy moving away from us at speeds faster than light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the first answer seems to contradict this other answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/107748/how-are-galaxies-receding-faster-than-light-visible-to-observers&quot;&gt;https://physics.stackexchange.com/questions/107748/how-are-galaxies-receding-faster-than-light-visible-to-observers&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If recession velocity at the location of a traveling photon were greater than the speed of light the entire time the photon from a distance galaxy were traveling, we would never observe the photon.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like common sense that at a certain distance, space is expanding faster than light, thus preventing light beyond a certain distance from reaching Earth which means we can't directly measure galaxies traveling faster than light, but we can imply that they do by projecting their velocity beyond the observable universe. But who's right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a galaxy used to be within the observable universe, we can measure photons from that galaxy reaching Earth despite that that galaxy may be currently outside the observable radius of the universe. Galaxies can be projected as travelling faster than light due to the expansion of space, but I haven't seen that we can directly measure them as doing such since the redshift of photons is proportional do the amount of space that is expanding between us and a certain distance. &lt;/p&gt;&#xA;" OwnerUserId="20897" LastEditorUserId="20897" LastEditDate="2018-04-18T15:29:21.060" LastActivityDate="2019-09-06T05:13:25.993" Title="So where are these measurements of galaxies moving faster than light?" Tags="&lt;cosmology&gt;&lt;speed&gt;&lt;hubble-constant&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25955" PostTypeId="1" AcceptedAnswerId="25956" CreationDate="2018-04-18T04:46:36.673" Score="1" ViewCount="760" Body="&lt;p&gt;My textbook (and lots of other sources) shows the earth's longitude of perihelion at the epoch around 103 degrees. I understand that this means (in the neighborhood of J2000) the earth will reach perihelion about 103 degrees after the vernal equinox, however, this would be very close to aphelion (July 6ish) not perihelion (January 3ish).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put another way: If the vernal equinox is March 20th &amp;amp; perihelion is January 3rd, shouldn't the longitude of perihelion be about -76 degrees (or 284 degrees)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help! What am I missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ANSWER--Credit to @Mike_G below, restated here because the final answer is in the comments: &quot;At the March equinox, the alignment is Earth - Sun - first point of Aries.&quot; Beginner mistake: I was trying to find the earth's position in time but I thought the March equinox was when the earth was between the Sun &amp;amp; first point of Aries, which is the wrong arrangement.&lt;/p&gt;&#xA;" OwnerUserId="22750" LastEditorUserId="22750" LastEditDate="2018-04-18T14:21:30.077" LastActivityDate="2018-04-18T14:21:30.077" Title="Longitude of Perihelion at J2000?" Tags="&lt;planet&gt;&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25956" PostTypeId="2" ParentId="25955" CreationDate="2018-04-18T05:17:05.997" Score="1" Body="&lt;p&gt;The Earth's heliocentric ecliptic longitude is opposite to the Sun's geocentric ecliptic longitude. If the Earth's perihelion is at 103 degrees heliocentric longitude, then the geocentric longitude of the Sun at that time is 283 degrees, or 13 degrees past the December solstice.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-04-18T05:17:05.997" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25957" PostTypeId="2" ParentId="25947" CreationDate="2018-04-18T05:59:21.527" Score="4" Body="&lt;p&gt;To me it sounds that you just want to expand your astronomical knowledge and are not actually looking for a career advice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest to just start with reading popular science books. &#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_Evolution_(book)&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Cosmic_Evolution_(book)&lt;/a&gt; , &#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/A_Brief_History_of_Time&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/A_Brief_History_of_Time&lt;/a&gt; for example. &lt;/p&gt;&#xA;" OwnerUserId="22751" LastActivityDate="2018-04-18T05:59:21.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25959" PostTypeId="2" ParentId="25954" CreationDate="2018-04-18T09:27:01.647" Score="1" Body="&lt;p&gt;The situation doesn't even require general relativity, the same thing can happen with ants crawling on a rubber picnic sheet.  Imagine you are sitting in the center of a very stretchy rubber sheet, and ants are crawling toward you at ant-speed from the outer edges of the sheet.  You ask some friends to keep pulling the edges of the sheet away from you at a speed somewhat faster than ants can crawl.  Will that protect you from the ants?  No, believe it or not, the ants will still reach you.  At first they are getting farther from you, but if you mark gridlines on that sheet, you'll see the ants are still advancing across those gridlines all the same.  They get to you eventually, just like the light from a galaxy whose distance from us is increasing faster than c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the second quote is not about the speed at the edge of the sheet, it says the ants don't get to you if all the points on the sheet that the ants access are all moving away from you faster than ant-speed.  That's a very different situation.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-04-18T09:27:01.647" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="25960" PostTypeId="1" AcceptedAnswerId="25978" CreationDate="2018-04-18T09:55:57.683" Score="12" ViewCount="2433" Body="&lt;p&gt;Would you be able to see the curvature of Mars from the top of Olympus Mons? And how much curvature would you see if it was visible, would it be only barely detectable, or would you see a lot of curvature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Olympus Mons is 25km high and the tallest known mountain in the Solar System.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars is just over half the size of Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here on Earth, the curvature of the planet is hardly visible from the height of commerical aeroplane (roughly 10km).  I was wondering, considering the height of Olympus Mons and the size difference of Mars, if the curvature from up there would be very pronounced.&lt;/p&gt;&#xA;" OwnerUserId="2844" LastEditorUserId="9527" LastEditDate="2020-09-24T07:28:07.423" LastActivityDate="2020-09-28T21:00:31.230" Title="Is the curvature of Mars visible from the top of Olympus Mons?" Tags="&lt;mars&gt;&lt;olympus-mons&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25961" PostTypeId="2" ParentId="25954" CreationDate="2018-04-18T10:05:15.150" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So where are these measurements of galaxies moving faster than light?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They're redshift measurements. Check out the Wikipedia &lt;a href=&quot;https://en.wikipedia.org/wiki/Redshift#History&quot; rel=&quot;noreferrer&quot;&gt;redshift&lt;/a&gt; article. It's good stuff. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;we can actually observe galaxies that are moving away from us at &gt;c&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's true. You might think &lt;em&gt;it cannot be&lt;/em&gt;, but it can. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Um, I think I missed the groundbreaking headline that said scientists have measured a galaxy moving away from us at speeds faster than light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not groundbreaking at all. It's simpler than you think.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So the first answer seems to contradict this other answer &lt;a href=&quot;https://physics.stackexchange.com/questions/107748/how-are-galaxies-receding-faster-than-light-visible-to-observers&quot;&gt;https://physics.stackexchange.com/questions/107748/how-are-galaxies-receding-faster-than-light-visible-to-observers&lt;/a&gt; &quot;If recession velocity at the location of a traveling photon were greater than the speed of light the entire time the photon from a distance galaxy were traveling, we would never observe the photon&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, that's a bit unfortunate, because he also says this: &lt;em&gt;&quot;galaxies with red shifts greater than ~3 &lt;strong&gt;were and are&lt;/strong&gt; receding from us faster than light&quot;&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It seems like common sense that at a certain distance, space is expanding faster than light, thus preventing light beyond a certain distance from reaching Earth which means we can't directly measure galaxies traveling faster than light. But who's right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The guy who says we can see galaxies that were and are receding from us faster than light. As per Rob's comment, see &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0310808&quot; rel=&quot;noreferrer&quot;&gt;Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the Universe&lt;/a&gt; by Tamara Davis and Charles Lineweaver. Note this: &lt;em&gt;&quot;We show that we can observe galaxies that have, and always have had, recession velocities greater than the speed of light.&quot;&lt;/em&gt; See page 8: &lt;em&gt;&quot;Amongst those who acknowledge that recession velocities can exceed the speed of light, the claim is sometimes made that objects with recession velocities faster than the speed of light are not observable [App. B: 9–13].&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also have a read of the Wikipedia &lt;a href=&quot;https://en.wikipedia.org/wiki/Ant_on_a_rubber_rope&quot; rel=&quot;noreferrer&quot;&gt;ant on a rubber rope&lt;/a&gt; article. The rubber rope is being stretched as the ant walks along it: &lt;em&gt;&quot;At first consideration it seems that the ant will never reach the end of the rope, but in fact it does (although in the form stated above the time taken is colossal). Whatever the length of the rope and the relative speeds of the ant and the stretching, providing the ant's speed and the stretching remain steady the ant will always be able to reach the end given sufficient time. Once the ant has begun moving, the rubber rope is stretching both in front of and behind the ant, conserving the proportion of the rope already walked by the ant and enabling the ant to make continual progress&quot;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a section on &lt;a href=&quot;https://en.wikipedia.org/wiki/Ant_on_a_rubber_rope#Metric_expansion_of_space&quot; rel=&quot;noreferrer&quot;&gt;the metric expansion of space&lt;/a&gt;. It says this: &lt;em&gt;&quot;By thinking of photons of light as ants crawling along the rubber rope of space between the galaxy and us, we can see that just as the ant can eventually reach the end of the rope, so light from distant galaxies, even some that appear to be receding at a speed greater than the speed of light, can eventually reach Earth, given sufficient time. However, the metric expansion of space is accelerating. An ant on a rubber rope whose expansion increases with time is not guaranteed to reach the endpoint.[3] The &lt;strong&gt;light from sufficiently distant galaxies may still therefore never reach Earth&lt;/strong&gt;&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2018-04-18T10:05:15.150" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25963" PostTypeId="2" ParentId="25524" CreationDate="2018-04-18T13:31:07.833" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the shape of observable universe and shape of space same?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nobody knows for sure. I imagine the answer is &lt;em&gt;yes&lt;/em&gt;, and they're both spherical, but I don't actually know this.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have read that Euclidean space is considered as flat space and Minkowski space is flat space-time. So when we say our observable universe is flat are we saying that the space is flat or space-time is flat? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Space. Space-time is not flat because the universe is expanding. To get a handle on this imagine you dob a big glob of treacle down, and spray a (non-radial) line of cream across it. As the glob of treacle flattens and expands the line of cream will tend to curve. It's similar for a light beam in an expanding universe. If the universe wasn't expanding, the light would go straight. Then space-time would be flat. Note that space-time models space at all times, so there's no motion through space-time. The treacle represents space. Space is expanding, not space-time. Unfortunately people tend to get confused between space and space-time.          &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I know the Planck data has proved the space to be flat but is it only space or space-time? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's space. The paper concerned is &lt;a href=&quot;https://arxiv.org/abs/1303.5086&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck 2013 results. XXVI. Background geometry and topology of the Universe&lt;/a&gt;. They found no evidence of any &quot;higher dimensional curvature&quot;. So if you could go &lt;em&gt;thataway&lt;/em&gt; at some phenomenal speed, you won't end up coming back &lt;em&gt;thisaway&lt;/em&gt;.    &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since in general relativity matter curves space-time and not just space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See what &lt;a href=&quot;http://www-history.mcs.st-andrews.ac.uk/Extras/Einstein_ether.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein said&lt;/a&gt;. A concentration of energy usually in the guise of a massive star &quot;conditions&quot; the surrounding space, rendering it inhomogeneous in a non-linear fashion. Then people call it curved spacetime. Note however that there's no overall gravity in the universe. As userLTK said  the surface of the Earth is &quot;lumpy&quot;, and in similar vein there are  gravitational fields all over the universe. But to the best of our knowledge the universe is not like the surface of the Earth. On the largest scale, to the best of our knowledge, the universe is flat. &lt;/p&gt;&#xA;" OwnerUserId="9772" LastEditorUserId="9772" LastEditDate="2018-04-18T13:37:25.947" LastActivityDate="2018-04-18T13:37:25.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25964" PostTypeId="2" ParentId="25954" CreationDate="2018-04-18T14:03:16.700" Score="0" Body="&lt;p&gt;If I understand the context correctly, the problem of misunderstanding here is the frame of reference. So, the answer is that both are correct because they are talking by using different frames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second diagram in &lt;a href=&quot;http://people.virginia.edu/~dmw8f/astr5630/Topic16/t16_light_cones.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this site&lt;/a&gt; might help you understand better. Y is lookback time, or age of universe. X is comoving distance. The observer is at today. Light cone is red and is not a straight line as one might expect from special relativity because of the universe expansion. However, if you take the vicinity around the observer, the light cone is still approximately the straight line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you extrapolate the straight line out of the observer, you will see stuff staying in the region where v&gt;c (i.e., spacelike). For example, you see the object at comoving distance at 16 Gly today from its emission at about 9 Gyr in the past, and 16 Gly / 9 year is slightly greater than c, as expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to be counter intuitive scenario is because of expressing the distance as comoving frame. Instead, the first diagrame from &lt;a href=&quot;http://people.virginia.edu/~dmw8f/astr5630/Topic16/t16_light_cones.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;the site&lt;/a&gt; shows X as the proper distance, and now all visible locations are in the v&#xA;&#xA;&lt;p&gt;As depicted by the diagrams you will see that we can see only far away as about 46 Gly (comoving distance), a.k.a. horizon. This is because if we go further than that spot, regarding to the history of universe expansion, we at today will stay out of the spacelike.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: all the analysis here still assume speed of light as the constant c in any frame of reference. This might be another misunderstanding, but I will not go into details at this moment.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-04-18T14:03:16.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25965" PostTypeId="1" AcceptedAnswerId="25975" CreationDate="2018-04-18T14:03:17.900" Score="4" ViewCount="445" Body="&lt;p&gt;From my understanding we detect planets by measuring dips in light intensity from the star the habitable planet is orbiting when it passes by it. &#xA;There are 2 things I don’t understand about this method:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Planets in a solar system tend to orbit their star in one disc like plane. Can we only detect planets if this plane is in-alignment with our sensors here on earth/ space. I assume if someone looked at our sun from the ‘bottom’ they would never see a planet cross it. If so this would exclude a big portion of the stars we are looking at? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Our orbit is 1 year. I assume the orbits of habitable plants can vary and having a short/ close to 1 year orbit is not a criteria for habitability. Even so we have to measure 1 dip every year or 10. How can we tell that this a planet with a regular yearly orbit or just anything passing between the star and us. So maybe a better question to ask for number 2 is what can we find out from these dips? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="22760" LastEditorUserId="7472" LastEditDate="2018-04-19T06:46:17.987" LastActivityDate="2018-04-19T06:46:17.987" Title="Detecting habitable planets" Tags="&lt;star&gt;&lt;planet&gt;&lt;exoplanet&gt;&lt;habitable-zone&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25966" PostTypeId="2" ParentId="25965" CreationDate="2018-04-18T14:15:45.507" Score="7" Body="&lt;p&gt;It isn't the only way, Aaron. See the Wikipedia &lt;a href=&quot;https://en.wikipedia.org/wiki/Methods_of_detecting_exoplanets&quot; rel=&quot;noreferrer&quot;&gt;Methods of detecting exoplanets&lt;/a&gt; article. The first method listed is radial velocity. That's where we measure the Doppler shift of the star's &quot;wobble&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second method is transit photography. And as you say it has a drawback. In fact there's more than one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;This method has two major disadvantages. First, planetary transits are observable only when the planet's orbit happens to be perfectly aligned from the astronomers' vantage point. The probability of a planetary orbital plane being directly on the line-of-sight to a star is the ratio of the diameter of the star to the diameter of the orbit (in small stars, the radius of the planet is also an important factor). About 10% of planets with small orbits have such an alignment, and the fraction decreases for planets with larger orbits. For a planet orbiting a Sun-sized star at 1 AU, the probability of a random alignment producing a transit is 0.47%. Therefore, the method cannot guarantee that any particular star is not a host to planets...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The second disadvantage of this method is a high rate of false detections. A 2012 study found that the rate of false positives for transits observed by the Kepler mission could be as high as 40% in single-planet systems...&quot;&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2018-04-18T14:15:45.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25967" PostTypeId="2" ParentId="25953" CreationDate="2018-04-18T14:24:15.367" Score="1" Body="&lt;p&gt;It uses the same fundamental conept of mapping between emission-line wavelengths to certain transitions. Similar as we know that 21cm line is the transition of spin-flip of a neutral hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Infrared (IR) regions is typically where the molecular lines are seen. &lt;a href=&quot;https://en.wikipedia.org/wiki/Infrared#Spectroscopy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; has good and brief discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asteriods are cold so we expect molecules to form. Therefore, we expect to see lines in IR region the most. To be more specific, surface temperature of asteriod is &lt;a href=&quot;https://www.space.com/51-asteroids-formation-discovery-and-exploration.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;~346 K&lt;/a&gt;, which is equivalent to ~83714 A at max emission (assuming blackbody radiation and following Wien's law); this is IR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that, what we see is the surface of the asteriod, not the inside.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-04-18T14:24:15.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25968" PostTypeId="2" ParentId="25947" CreationDate="2018-04-18T14:34:00.143" Score="2" Body="&lt;p&gt;If you want to know about stars, i.e., astronomy or astrophysics, &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/0805304029&quot; rel=&quot;nofollow noreferrer&quot;&gt;Caroll &amp;amp; Ostlie&lt;/a&gt; is somewhat fundamental textbook for undergrad &amp;amp; grad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to know about galaxy and universe, i.e., cosmology, we use the textbook from &lt;a href=&quot;http://carina.fcaglp.unlp.edu.ar/extragalactica/Bibliografia/Ryden_IntroCosmo.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ryden&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My recommendation is to understand stars first, before you deal with galaxy, and then universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will encounter a lot of physics stuff, which you might need to study separately out of just the concepts of astronomy or cosmology, if you want to have deep understanding of the topics.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-04-18T14:34:00.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25969" PostTypeId="2" ParentId="25953" CreationDate="2018-04-18T15:25:58.643" Score="1" Body="&lt;p&gt;To add to KB's answer:&#xA;By and large the method involves looking at the IR absorption spectrum.  If you see a set of lines which correspond to a known mineral, then you conclude that mineral is part of the asteroid's composition.  This assumes you know what's currently illuminating the asteroid (spectral irradiance).  If you're looking at the dark side, then it's a similar analysis but looking for known gaps in the not-quite-black-body exitance for a given mineral specie. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-04-18T15:25:58.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25970" PostTypeId="1" AcceptedAnswerId="25976" CreationDate="2018-04-18T15:35:14.417" Score="2" ViewCount="69" Body="&lt;p&gt;When viewed from above, i.e., looking down from the North Galactic Pole towards the South Galactic Pole, does the sun rotate clockwise or counterclockwise around the Galactic center? &lt;/p&gt;&#xA;" OwnerUserId="22761" LastEditorUserId="616" LastEditDate="2018-04-18T18:16:38.807" LastActivityDate="2018-04-18T19:13:28.603" Title="Direction of movement sun with respect to the NGP" Tags="&lt;milky-way&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25971" PostTypeId="2" ParentId="25954" CreationDate="2018-04-18T16:34:47.733" Score="4" Body="&lt;p&gt;Without going into the technicalities of spacetime diagrams and ants, I think the quickest way to wrap your head around this is to look at it from the distant galaxy's perspective. For instance, let's take &lt;a href=&quot;https://en.wikipedia.org/wiki/GN-z11&quot; rel=&quot;nofollow noreferrer&quot;&gt;GN-z11&lt;/a&gt;, which actually receded from us at $v\simeq4c$ when it emitted the light we see today:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A photon left GN-z11 at $v=c$. Space expands, so although the photon at any time travels &lt;em&gt;locally&lt;/em&gt; at $v=c$, it increased its distance to GN-z11 at an ever-increasing speed $v&amp;gt;c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Milky Way receded from GN-z11 at $v\simeq4c$, but &lt;em&gt;locally&lt;/em&gt; it has $v\simeq0$, and until recently the Universe expanded slower and slower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some point, the photon reached a point where the Universe expanded at, say, $v=0.1c$ wrt. GN-z11, so the photon receded at $v=1.1c$. Later, it reached a point where the Universe expands at $v=c$ wrt. GN-z11, so it receded at $v=2c$ from GN-z11. And so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, it simply &quot;caught up&quot; with the Milky Way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to note, though, that this is only possible because the expansion rate didn't &lt;em&gt;accelerate&lt;/em&gt; until recently. The acceleration puts a limit to how fast galaxies may recede and still be visible. Today, that limit happens to be roughly $v=1.2c$, corresponding to galaxies roughly 17 billion lightyears away. Photons leaving GN-z11 today will never reach us, even though it &quot;only&quot; recedes from us at $v=2.2c$ today.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2018-04-18T16:34:47.733" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25972" PostTypeId="2" ParentId="25947" CreationDate="2018-04-18T17:06:59.427" Score="3" Body="&lt;p&gt;Understanding how things work out in space is only half the fun;&#xA;the other half is seeing them with your own eyes.&#xA;Field guides by&#xA;&lt;a href=&quot;http://web.williams.edu/Astronomy/fieldguide/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pasachoff&lt;/a&gt;&#xA;and others can help bring the two together.&#xA;Magazines like &lt;a href=&quot;http://www.skyandtelescope.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sky &amp;amp; Telescope&lt;/a&gt;&#xA;offer information about temporarily observable objects like comets&#xA;and cover science news at level between mass media and academic journals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hands-on observing experience can motivate programming projects in ways that textbooks can't.&#xA;Given the free availability of such good software as &lt;a href=&quot;http://stellarium.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellarium&lt;/a&gt; and &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL HORIZONS&lt;/a&gt;,&#xA;and libraries like &lt;a href=&quot;http://www.astropy.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astropy&lt;/a&gt;,&#xA;it's wise not to spend much time reinventing wheels.&#xA;On the other hand, implementing a few things yourself will expose you to various astronomy, math, algorithm, and data structure issues.&#xA;You can always check your work against off-the-shelf tools, or better yet, against the real sky.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-04-18T17:06:59.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25973" PostTypeId="2" ParentId="25947" CreationDate="2018-04-18T17:31:54.173" Score="2" Body="&lt;p&gt;If you're like me and you just want lots of interesting facts about space without all of the academic level of complexity, you may enjoy watching the &lt;a href=&quot;https://www.youtube.com/watch?v=0rHUDWjR5gg&amp;amp;list=PL8dPuuaLjXtPAJr1ysd5yGIyiSFuh0mIL&quot; rel=&quot;nofollow noreferrer&quot;&gt;Crash Course Astronomy&lt;/a&gt; series. There are a lot of other good resources on YouTube, such as &lt;a href=&quot;https://www.youtube.com/user/DeepSkyVideos&quot; rel=&quot;nofollow noreferrer&quot;&gt;Deep Sky Videos&lt;/a&gt; or &lt;a href=&quot;https://www.youtube.com/user/tdarnell&quot; rel=&quot;nofollow noreferrer&quot;&gt;Deep Astronomy&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want something more fun and interactive, may I suggest playing &lt;a href=&quot;https://en.wikipedia.org/wiki/Kerbal_Space_Program&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kerbal Space Program&lt;/a&gt;. I've learned a lot about how orbital mechanics, aerodynamics, and engineering just from playing this game. There is a question on the &lt;a href=&quot;https://space.stackexchange.com/questions/22834/what-can-the-ksp-game-actually-teach-about-spaceflight-and-orbital-mechanics-an&quot;&gt;Space Exporation Stack Exchange&lt;/a&gt; that does a good job of covering how much can be learned from playing.&lt;/p&gt;&#xA;" OwnerUserId="14255" LastActivityDate="2018-04-18T17:31:54.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25974" PostTypeId="2" ParentId="25960" CreationDate="2018-04-18T18:09:47.923" Score="6" Body="&lt;p&gt;It would really depend on the local geometry of the volcano.  At the altitude of Olympus Mons here on earth, there are several &lt;a href=&quot;https://earthscience.stackexchange.com/questions/7283/how-high-must-one-be-for-the-curvature-of-the-earth-to-be-visible-to-the-eye&quot;&gt;reports&lt;/a&gt; of being able to see the curvature (U2 and SR-71 pilots).  But that relies on having a clear view of the horizon. Given the smaller radius of mars, if you were on a tower or in a vehicle at that altitude above a flat planet, the curvature would definitely be visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But Olympus Mons is a shield volcano and has a very gentle slope.  I suspect that there's nowhere to stand on the ground that will give you the view you need.  The shallow slope will get in the way of being able to see the horizon at a sufficient distance.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2018-04-18T18:09:47.923" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25975" PostTypeId="2" ParentId="25965" CreationDate="2018-04-18T18:40:32.440" Score="5" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes and yes. Transit detection is only effective for that (small) fraction of planets that pass between the star and our line of sight. Most planets will go undetected by this method.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Also correct. The detection of multiple transits is required to find planets. Even then the dips can be caused by other things (e.g. grazing eclipse binary star systems). This means to find Earthlike planets in Earthlike orbits would require years of observation. But habitable zones around less luminous stars are closer-in and have shorter orbital periods. It is these stars that planet-hunting missions like TESS are focused on.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-04-18T18:40:32.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25976" PostTypeId="2" ParentId="25970" CreationDate="2018-04-18T18:50:25.567" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://doi.org/10.1086/307857&quot; rel=&quot;nofollow noreferrer&quot;&gt;Backer and Sramek 1999&lt;/a&gt;&#xA;observed that relative to more distant radio sources, the galactic center&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sagittarius_A*&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sgr A*&lt;/a&gt;&#xA;had a heliocentric &lt;a href=&quot;https://en.wikipedia.org/wiki/Proper_motion&quot; rel=&quot;nofollow noreferrer&quot;&gt;proper motion&lt;/a&gt;&#xA;of about -6 milliarcseconds per year in galactic longitude.&#xA;&lt;a href=&quot;https://doi.org/10.1086/307855&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reid et al. 1999&lt;/a&gt;&#xA;reported a similar result.&#xA;Looking from NGP to SGP,&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;galactic longitude&lt;/a&gt;&#xA;increases counterclockwise, a decrease in longitude is a clockwise motion, and the Sun revolves clockwise around the galactic center.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-04-18T19:13:28.603" LastActivityDate="2018-04-18T19:13:28.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25977" PostTypeId="2" ParentId="25925" CreationDate="2018-04-18T22:31:56.177" Score="5" Body="&lt;p&gt;&lt;strong&gt;tl;dr&lt;/strong&gt;: In general, smaller galaxies have relatively more dark matter than larger galaxies. The atypical galaxies that you refer to are discussed in the last paragraph.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Mass fractions in the Universe&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Matter in the Universe is dominated by two components; &lt;em&gt;baryons&lt;/em&gt; (i.e. atoms in various forms) and &lt;em&gt;dark matter&lt;/em&gt; (DM). The baryons predominantly exist in two forms: &lt;em&gt;gas&lt;/em&gt; (including plasma) and &lt;em&gt;stars&lt;/em&gt; (some gas condenses to form dust, planets, etc., but that's a minor part).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fraction of the Universe's total energy in the form of matter is $\Omega_\mathrm{M}\simeq0.31$ (most of the rest is dark energy). The baryons and DM have fractions $\Omega_\mathrm{b}=0.05$ and $\Omega_\mathrm{DM}=0.26$, respectively (numbers from &lt;a href=&quot;https://arxiv.org/abs/1502.01589&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck Collaboration et al. 2016&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the baryons and DM comprise fractions&#xA;$f_\mathrm{b}  = \Omega_\mathrm{b}  / \Omega_\mathrm{M} = 16$% and&#xA;$f_\mathrm{DM} = \Omega_\mathrm{DM} / \Omega_\mathrm{M} = 84$% of the total mass, respectively.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Galaxies have more dark matter than the average Universe&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Galaxies formed when DM+gas overdensities decoupled from the Hubble flow and started collapsing, and so were &quot;born&quot; with the &quot;cosmic&quot; fractions. However, observationally it is found that the baryon fraction in galaxies is much smaller; for instance, a &quot;typical&quot; Milky Way-sized galaxy roughly has $f_\mathrm{b}$ only half of the cosmic average, and the DM fraction is correspondingly larger, &gt;$90$% (e.g. &lt;a href=&quot;https://arxiv.org/abs/1403.0947&quot; rel=&quot;nofollow noreferrer&quot;&gt;Werk et al. 2014&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Gas and dark matter dynamics&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The reason is the different mechanisms that affect the dynamics of the particles.&#xA;Whereas the DM is collisionless, the gas particles collide and dissipate their energy, making it easier to cool and contract. Thus, the part of the galaxies that we can observe is much smaller than &quot;whole&quot; galaxy; the  typical &quot;half-light radius&quot; is but a few percent of the emcompassing DM &lt;em&gt;halo&lt;/em&gt; (&lt;a href=&quot;https://arxiv.org/abs/1212.2980&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kravtsov 2013&lt;/a&gt;, &lt;a href=&quot;https://arxiv.org/abs/1701.03526&quot; rel=&quot;nofollow noreferrer&quot;&gt;Somerville et al. 2017&lt;/a&gt;). The half-light radius is the radius within which half of the light is emitted; when you see an image of a galaxy, it typically looks perhaps 4-5 times larger than this. So, something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/95Kd0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/95Kd0.png&quot; alt=&quot;gal+halo&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A significant fraction of the baryons reside in the halo as well. This is the hot gas of millions of Kelvin, and the high temperature and low density makes it difficult to detect, as it glows only in faint X-rays.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Feedback blows out gas, but dark matter stays&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;But various physical processes act so as to try to blow the baryons — but not the DM — out of the galaxies. These processes are known as &lt;em&gt;feedback&lt;/em&gt;, and are due to energy being &quot;injected&quot; in the baryons.&#xA;For massive galaxies with halo masses $M_\mathrm{h}\gtrsim10^{12}\,M_\odot$ this feedback is dominated by their central black hole accreting gas, resulting in extreme luminosities manifesting itself as a &lt;em&gt;quasar&lt;/em&gt;, or &lt;em&gt;active galactic nucleus&lt;/em&gt; (&lt;a href=&quot;https://arxiv.org/abs/astro-ph/9801013&quot; rel=&quot;nofollow noreferrer&quot;&gt;Silk &amp;amp; Rees 1998&lt;/a&gt;; &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0508046&quot; rel=&quot;nofollow noreferrer&quot;&gt;Croton et al. 2006&lt;/a&gt;). For smaller galaxies with $M_\mathrm{h}\lesssim10^{12}\,M_\odot$, the feedback is mostly due to stellar winds and exploding stars depositing thermal and kinetic energy in the surrounding gas (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1986ApJ...303%E2%80%A639D&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dekel &amp;amp; Silk 1986&lt;/a&gt;; &lt;a href=&quot;https://arxiv.org/abs/1110.4638&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hopkins et al. 2012&lt;/a&gt;). At even lower masses, the galaxies are so small that a significant fraction of their gas may be ejected altogether during early epochs of star formation (e.g. &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0002214&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bullock et al. 2000&lt;/a&gt;), resulting in galaxies that consist virtually only of DM and a few stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smaller a galaxy is, the shallower its gravitational potential, and thus the easier it is for the gas to escape the galaxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, in general &lt;strong&gt;the smaller a galaxy is, the larger its amount of dark matter is&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, the larger a halo is, the more its mass fractions converge toward the cosmic fractions. These are only reached for halos of masses $M_\mathrm{h} \gtrsim 10^{15}\,M_\odot$, which are no longer individual galaxies, but rather groups and clusters. This is seen in the figure below (from a paper from last week by &lt;a href=&quot;https://arxiv.org/abs/1804.05064&quot; rel=&quot;nofollow noreferrer&quot;&gt;Henden et al. 2018&lt;/a&gt;), which shows the stellar mass fraction (&lt;em&gt;left&lt;/em&gt;) and gas mass fraction (&lt;em&gt;right&lt;/em&gt;) as a function of halo mass:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LFcj6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LFcj6.png&quot; alt=&quot;fbar&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The baryon fraction is $f_\mathrm{b} = f_\mathrm{stars}+f_\mathrm{gas}$, and the DM fraction — which is what you're asking for — is then given by $f_\mathrm{DM} = 1 - f_\mathrm{stars} - f_\mathrm{gas}$. The stellar fraction is seen to decrease with halo mass, due to the AGN feedback discussed above, but the total star+gas fraction increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note though that although the DM — being collisionless — is not exactly blown out of the galaxy like the baryons, the gravitational attraction between the two components still affects the DM and alters the density profile of the halo (&lt;a href=&quot;https://arxiv.org/abs/1001.3447v2&quot; rel=&quot;nofollow noreferrer&quot;&gt;Duffy et al. 2010&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Galaxies with &lt;em&gt;more&lt;/em&gt; gas than dark matter (?)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Thus, a galaxy without baryons is not a huge mystery, as long as it's small. Even larger galaxies are apparently able to rid themselves of most the baryons; &lt;a href=&quot;https://arxiv.org/abs/1606.06291&quot; rel=&quot;nofollow noreferrer&quot;&gt;van Dokkum et al. (2016)&lt;/a&gt; reported a MW-sized halo with 98% DM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A galaxy &lt;em&gt;with&lt;/em&gt; baryons, but without DM is more spectacular, but was reported recently by &lt;a href=&quot;https://arxiv.org/abs/1803.10237&quot; rel=&quot;nofollow noreferrer&quot;&gt;van Dokkum et al. (2018)&lt;/a&gt; (yes, the same guy).&#xA;In &lt;a href=&quot;https://astronomy.stackexchange.com/questions/25802/galaxy-discovered-lacking-dark-matter/25805#25805&quot;&gt;another answer about DF2&lt;/a&gt; I discuss various processes that might lead to such a galaxy, including misinterpretation of the data.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-04-20T12:20:30.903" LastActivityDate="2018-04-20T12:20:30.903" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25978" PostTypeId="2" ParentId="25960" CreationDate="2018-04-18T23:42:47.190" Score="9" Body="&lt;p&gt;The impression I get from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Olympus_Mons&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page on Olympus Mons&lt;/a&gt; is that from the top of the volcano, you can't even see parts of Mars that aren't Olympus Mons. Quoted from above:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Because of the size of Olympus Mons and its shallow slopes, an observer standing on the Martian surface would be unable to view the entire profile of the volcano, even from a great distance. The curvature of the planet and the volcano itself would obscure such a synoptic view. &lt;strong&gt;Similarly, an observer near the summit would be unaware of standing on a very high mountain, as the slope of the volcano would extend far beyond the horizon, a mere 3 kilometers away.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Emphasis mine.&lt;/p&gt;&#xA;&lt;p&gt;Wikipedia sources the bold line to &lt;a href=&quot;https://web.archive.org/web/20090827141153/http://www.dustymars.net-a.googlepages.com/VOLCANO.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this source in the Way Back Machine&lt;/a&gt;, which works out the math. Near the bottom of the page, there is a section starting with the phrase:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Years ago someone asked me if a person could see the horizon of Mars from the top of Olympus Mons. That is not as simple as one may think. If Mars was a prefect sphere and the slopes of Olympus Mons were smooth then the distance to the horizon could be calculated fairly easy. However, the surface of Mars has hills and valleys, and Tharsis bulges out from the mean spherical surface of Mars several miles, so one must consider that as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;That's pretty much your question. The conclusion is:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Second, if this same [6 foot tall] person stood at the top of Olympus Mons (88704 ft) we would find that he or she could see a lot further to the horizon, some 265 miles away, if nothing was in the way:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\cos A = (11,121,762.53 / (11,121,762.53 + 88704 + 6)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$A = 7.2^\circ = 0.126\ \textrm{radians}  \leftarrow (0.126 \times 11,121,762.53 ft) / 5280 = 265 \textrm{miles}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;At those angles a person 6 feet high would be looking down at an angle of 7.2 degrees in order to see the horizon of Mars, so they would surely look down into the hump positioned at 50 miles away or at the 2.5-degree grade of the slope instead of over it to the horizon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So it would seem that the horizon, as viewed from the top of Olympus Mons, would still be part of Olympus Mons. Indeed, you wouldn't even be able to see most of Olympus Mons from the top.&lt;/p&gt;&#xA;" OwnerUserId="14255" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-04-19T12:22:00.227" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25979" PostTypeId="1" AcceptedAnswerId="26000" CreationDate="2018-04-19T01:41:34.867" Score="10" ViewCount="165" Body="&lt;p&gt;Please excuse an amateur question. &#xA;While trying to think of anything but what was happening during a dental procedure my mind turned to a model of a star close to a rotating black hole and the effects on the drawn in matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it is obvious such matter would be excited to high temperature could the combination of rotation and excitation be enough to induce a sustained fusion reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so would this produce enough energy to maintain a fusion 'ring' at the event horizon - essentially a doughnut star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would there be enough of a reaction to start producing lighter elements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pure curiosity generated by an attempt to distract myself&lt;/p&gt;&#xA;" OwnerUserId="22764" LastActivityDate="2018-04-20T12:35:32.587" Title="Star formation around rotating black holes?" Tags="&lt;black-hole&gt;&lt;star-formation&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25980" PostTypeId="1" CreationDate="2018-04-19T02:46:06.620" Score="4" ViewCount="68" Body="&lt;p&gt;I'm writing a test for a gen-ed level astronomy class that I teach, and would just like a good reliable number to use as the basis for a distance calculation question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, I know that real astronomers use the RR Lyrae stars, but the text were using hasn't introduced variable star standard candles yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at some H-R diagrams in textbook the answer seems to be vaguely around magnitude 0, and I'll use that value in the absence of something more definitive.&lt;/p&gt;&#xA;" OwnerUserId="530" LastActivityDate="2018-04-19T11:30:30.887" Title="Is there an acceptable value for the absolute magnitude of horizontal branch stars?" Tags="&lt;absolute-magnitude&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25982" PostTypeId="1" AcceptedAnswerId="25986" CreationDate="2018-04-19T04:15:10.947" Score="7" ViewCount="2824" Body="&lt;p&gt;I know there are lot of questions on the Hubble constant already, but I am curious to know if it changes with redshift? If at current redshift, &lt;span class=&quot;math-container&quot;&gt;$z=0$&lt;/span&gt;, we know its value to be 0.7, will it be different at higher redshift (&lt;span class=&quot;math-container&quot;&gt;$z=0.1$&lt;/span&gt;)? If so, is there any relationship with redshift?&lt;/p&gt;&#xA;" OwnerUserId="21282" LastEditorUserId="5264" LastEditDate="2020-08-26T10:30:06.177" LastActivityDate="2020-08-26T10:30:06.177" Title="Does the Hubble constant depend on redshift?" Tags="&lt;hubble-constant&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25984" PostTypeId="1" AcceptedAnswerId="25989" CreationDate="2018-04-19T07:34:07.603" Score="3" ViewCount="72" Body="&lt;p&gt;As I understand it most astrophotography is colored using filters and multiple exposures. I would assume that blue filters would be used for higher frequency emissions and red filters for lower frequency emissions, as it at least keeps the &quot;order&quot; of information intact. But besides that, is there any rule or hints to go by when looking at colored astrophotography and deducing the depicted wavelengths? Because a fairly common problem would be, I assume, when something is only visible in either infrared or ultraviolet. How would I differentiate that? Do I just have to know what emission bands certain phenomena are visible in?&lt;/p&gt;&#xA;" OwnerUserId="18954" LastActivityDate="2018-04-19T11:27:39.030" Title="Astrophotography: Is there a rule to the color mapping?" Tags="&lt;photography&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25986" PostTypeId="2" ParentId="25982" CreationDate="2018-04-19T10:15:16.673" Score="18" Body="&lt;p&gt;Yes, definitely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hubble constant describes the expansion rate of the Universe, and the expansion may, in turn, may be &lt;em&gt;decelerated&lt;/em&gt; by &quot;regular&quot; matter/energy, and &lt;em&gt;accelerated&lt;/em&gt; by dark energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's more or less the norm to use the term Hubble &lt;em&gt;constant&lt;/em&gt; $H_0$ for the value today, and Hubble &lt;em&gt;parameter&lt;/em&gt; $H(t)$ or $H(a)$ for the value at a time $t$ or, equivalently, a scale factor $a = 1/(1+z)$, where $z$ is the redshift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value is given by the Friedmann equation:&#xA;$$&#xA;\frac{H^2(a)}{H_0^2} = &#xA;\frac{\Omega_\mathrm{r}}{a^4} +&#xA;\frac{\Omega_\mathrm{M}}{a^3} +&#xA;\frac{\Omega_k}{a^2} +&#xA;      \Omega_\Lambda,&#xA;$$&#xA;where &#xA;$\{&#xA;\Omega_\mathrm{r},&#xA;\Omega_\mathrm{M},&#xA;\Omega_k,&#xA;\Omega_\Lambda&#xA;\} \simeq&#xA;\{&#xA;10^{-3},0.3,0,0.7&#xA;\}&#xA;$&#xA;are the fractional energy densities in radiation, matter, curvature, and dark energy, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, you can solve the above equation at $z=0.1$ and find that the expansion rate was 5% higher than today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since everything but dark energy dilutes with increasing $a$, $H(a)$ will asymptotically converge to a value $H_0\sqrt{\Omega_\Lambda} \simeq 56\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure below shows the evolution of the Hubble parameter with time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Xn2nW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xn2nW.png&quot; alt=&quot;Ht&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted by KenG, the fact that $H$ &lt;em&gt;decreases&lt;/em&gt; with time may seem at odds with the accelerated expansion of the Universe. But $H$ describes how fast a point in space at a given distance recedes. Later, that point will be farther away, and so will recede faster. From the definition of the Hubble parameter, $H\equiv\dot{a}/a$, multiplying by the scale factor shows the acceleration $da/dt$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VbLK0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VbLK0.png&quot; alt=&quot;aHt&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-04-22T21:11:28.147" LastActivityDate="2018-04-22T21:11:28.147" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25987" PostTypeId="2" ParentId="25982" CreationDate="2018-04-19T10:27:42.560" Score="6" Body="&lt;p&gt;What the Hubble constant really depends on is how old was the universe at the time, but if you have a dynamical model of the universe, you can map that into z and come up with a function H(z).  So in that sense, the answer is &quot;yes,&quot; but be careful-- we also think of z as a measure of how far away the objects are, and H does not depend on location it depends on age. What's more, the z we get from a given measurement reflects all the expansion, so all the H's, since that light was emitted, not just the value of that H at that z.  It would be a bit like if you were using your height to talk about your age, and you look at a picture of yourself at 4 feet tall, and say you were growing two inches a year when you were that tall.  That would be like the H that applies in that picture, but the ratio of the height you are now to the height in that picture depends on more than just the rate you were growing in that picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Also-- what do you mean the value of the Hubble constant is 0.7 now?  That sounds like the fraction of total energy that is &quot;dark energy,&quot; so is more about the rate of acceleration of the expansion rather than the rate of expansion itself.  If you are asking about that, then this number has been rising with age of the universe, so can be mapped into a function of z if we are using z to talk about the age of the universe at the time.)&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2018-04-19T10:35:05.360" LastActivityDate="2018-04-19T10:35:05.360" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25988" PostTypeId="1" CreationDate="2018-04-19T10:33:16.327" Score="4" ViewCount="119" Body="&lt;p&gt;ref &lt;a href=&quot;https://www.nature.com/articles/s41467-018-03808-6&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper in Nature&lt;/a&gt; re the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/2008_TC3&quot; rel=&quot;nofollow noreferrer&quot;&gt;Almahata Sitta 2008&lt;/a&gt; meteorite shower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get that they're saying the diamonds could only have been formed within something the size of planet, which subsequently broke apart in the &quot;early days&quot; of the Solar System.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get that Saturn and Jupiter are so massive relative to other Solar System bodies they've marshalled fragments into the Asteroid belt, Kuiper belt, outer disk. And that they're still disturbing those regions enough to occasionally pull things into a cometary orbit/become meteors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This putative planet must have scattered everywhere when it first broke apart/lead to comets and meteors all over the place. But why/how did this fragment wait until 2008 to come to Earth? Where's it been for billions of years? Why didn't it either crash into something a long time ago; or form itself into a stable orbit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/16262/extinct-species-of-meteorite-how-quantitative-is-the-statistical-analysis&quot;&gt;Possibly related&lt;/a&gt;, as in: why haven't meteorites originating from early-formed planets all long gone extinct?&lt;/p&gt;&#xA;" OwnerUserId="22769" LastActivityDate="2018-04-19T12:20:35.567" Title="tiny diamonds in meteorite &quot;signature of a long-lost planet&quot;: why come to Earth in 2008?" Tags="&lt;meteorite&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25989" PostTypeId="2" ParentId="25984" CreationDate="2018-04-19T11:27:39.030" Score="1" Body="&lt;p&gt;Rule in mapping color is just violet (high frequency) &gt;&gt;&gt; red (low one). For example, it was just yesterday thay we had a speaker who who studied in three IR bands: W1 (high frequency = blue), W2 (green), W3 (red).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the image has to provide the description about the mapping to be informative.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-04-19T11:27:39.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25992" PostTypeId="2" ParentId="25988" CreationDate="2018-04-19T12:15:24.857" Score="2" Body="&lt;p&gt;The asteroid has been orbiting the sun for over 4 billion years, either between Mars and Jupiter or between Earth and Mars. Over time, but relatively recently the combination of perturbations of Jupiter, Mars, Earth and other planets slowly changed its orbit eventually bringing the asteroid into an orbit that intersected with that of Earth. And then one day it stopped being &quot;Lucky&quot; and impacted on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key point is that the &lt;em&gt;orbits of asteroids are not fixed&lt;/em&gt; over billions of years. The gravity of planets slowly changes them. Perturbations can move an asteroid from a safe orbit between the planets to an unsafe one that crosses a planet's path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also possible that this asteroid has had one or more close encounters with a planet, this would change the orbit dramatically in a short time.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-04-19T12:15:24.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25993" PostTypeId="2" ParentId="25988" CreationDate="2018-04-19T12:20:35.567" Score="3" Body="&lt;p&gt;Many comets are such lucky ones...&lt;br&gt;&#xA;&lt;strong&gt;That's why we see them with their icy tails evaporating: They have only recently been scattered into the inner solar system.&lt;/strong&gt;&lt;br&gt;&#xA;Just like the well-studied comet &lt;a href=&quot;https://en.wikipedia.org/wiki/67P/Churyumov%E2%80%93Gerasimenko&quot; rel=&quot;nofollow noreferrer&quot;&gt;67P&lt;/a&gt;, you could ask &quot;where has 67P been for all those billion years?&quot;. For at least this comet, we know the answer: It has been orbiting in the outer solar system, until a close encounter in 1840 and a few subsequent later ones with Jupiter brought its' perihelion closer to the sun, where it started its' cometary activity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would place my bet on the origin of those comets being in the outer solar system, because:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The nature article you cite mentions 20 GPa pressure, for which one needs at least a Mercury or Mars type body to generate. This is not necessarily a problem, because Mars-type bodies form quick and early (we know this from &lt;a href=&quot;https://arxiv.org/abs/1205.3030?context=astro-ph&quot; rel=&quot;nofollow noreferrer&quot;&gt;theory&lt;/a&gt; as well as &lt;a href=&quot;https://www.nature.com/articles/nature10077&quot; rel=&quot;nofollow noreferrer&quot;&gt;data&lt;/a&gt;), within the first 3-5 Million years of the solar system, when gas is still there. This is about 10 times faster than an Earth-type body, which has 10 times the mass.&lt;br&gt;&#xA;It is also thought that Mars-type bodies should form the basic building blocks for larger planets - thus there should be many more of them in the early solar system.&lt;/li&gt;&#xA;&lt;li&gt;However even if there are many, now there are essentially none in the inner solar system - the asteroid belt contains roughly 0.1% of a Mars mass. The total mass of the Kuiper belt is estimated to be far larger, by factors of 10-100.&lt;/li&gt;&#xA;&lt;li&gt;If a &lt;a href=&quot;https://en.wikipedia.org/wiki/Nice_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;giant planet instability&lt;/a&gt; occurred, it requires a even much more massive Kuipter belt as trigger, after the gas disappears in the early solar system.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, &lt;strong&gt;Summarizing&lt;/strong&gt;: They're probably out there in the outer solar system, merrily orbiting along.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2018-04-19T12:20:35.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25994" PostTypeId="1" CreationDate="2018-04-19T12:40:49.940" Score="2" ViewCount="83" Body="&lt;p&gt;A short video titled &quot;Why Tess is recording the &quot;sounds of stars&quot; at the bottom of the page of the BBC News article &lt;a href=&quot;http://www.bbc.com/news/science-environment-43790557&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planet-hunter launches from Florida&lt;/a&gt; gives a short description of Asteroseismology by Astronomer Bill Chaplin of the University of Birmingham.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably the sounds played here are some kind of photometric measurements of stars that have been played back a few orders of magnitude faster to make them audible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to listen to more of them, or even download a .wav or similar file and have a look at the spectrum just for fun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I managed to find this University of Birmingham website through searching, and it plays sound clips from four different stars when the cursor is hovered over each one (it takes a while to become responsive I assume due to downloading).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://bison.ph.bham.ac.uk/~miglioa/M4PR/M4_beta0.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://bison.ph.bham.ac.uk/~miglioa/M4PR/M4_beta0.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Roughly how long is the observation time of the star necessary to produce these audio recordings of a few seconds in length?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Where can I find downloadable data; either these .wav sound clips or more traditional raw data files that I could analyze?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kb62N.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kb62N.jpg&quot; alt=&quot;BBC Bill Chaplin U Birmingham&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BcxzU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BcxzU.jpg&quot; alt=&quot;U Birmingham M4 Star Sounds&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-04-19T12:40:49.940" Title="Asteroseismology; recordings of star vibrations duration and data availability?" Tags="&lt;observational-astronomy&gt;&lt;data-analysis&gt;&lt;asteroseismology&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25995" PostTypeId="2" ParentId="25937" CreationDate="2018-04-19T20:21:58.967" Score="17" Body="&lt;p&gt;The authors assume a geocentric universe (first thing that is wrong). They then assume that the planet Mars has the same apparent diameter as a globe 30 &lt;em&gt;yojana&lt;/em&gt; in diameter (about 150 miles) in the same orbit as the moon, from a perspective at the centre of the Earth. This is just stated, and appears to be supposition. It is an incorrect figure. It is conceivable that some kind of system for sighting through a pinhole could be used for this estimate. This method tends to over estimate the apparent size of planets.&lt;/p&gt;&#xA;&lt;p&gt;They next claim that the other planets have apparent sizes that would be the same as globes with diameter 37.5 (Saturn), 45 (Mercury) 52.5 (Jupiter) and 60 &lt;em&gt;yojana&lt;/em&gt; (Venus). These form an arithmetic sequence with a common difference of 7.5, but apart from that, these values seem to be arbitrary (and are wrong). They might have been based on some kind of pinhole observations, but rounded to a simple sequence for the sake of the poetry or for easier memorisation.&lt;/p&gt;&#xA;&lt;p&gt;There is uncertainty about the ancient value of the &lt;em&gt;yojana&lt;/em&gt;. A different conversion to miles will give different values throughout.&lt;/p&gt;&#xA;&lt;p&gt;Using this assumption and the distance already calculated to the moon they calculate the apparent angular diameter of each planet. The actual values they get are wrong, In fact, the angular diameter varies as the planet's distance varies.&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;Planet&lt;/th&gt;&#xA;&lt;th&gt;calculated value&lt;/th&gt;&#xA;&lt;th&gt;modern observed value&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Mars&lt;/td&gt;&#xA;&lt;td&gt;2 arcmin&lt;/td&gt;&#xA;&lt;td&gt;(actually 0.06 - 0.39)&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Saturn&lt;/td&gt;&#xA;&lt;td&gt;2.5 arcmin&lt;/td&gt;&#xA;&lt;td&gt;(0.25 - 0.34)&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Mercury&lt;/td&gt;&#xA;&lt;td&gt;3 arcmin&lt;/td&gt;&#xA;&lt;td&gt;(0.08 - 0.17)&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Jupiter&lt;/td&gt;&#xA;&lt;td&gt;3.5 arcmin&lt;/td&gt;&#xA;&lt;td&gt;(0.51 - 0.83)&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Venus&lt;/td&gt;&#xA;&lt;td&gt;4 arcmin&lt;/td&gt;&#xA;&lt;td&gt;(0.16 - 1.05)&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;As you see the calculated values are much too large (ie wrong). They are comparable with values supposed to have been found by Hipparchus, but without a known observational basis. Tycho Brahe also gave similar values, found by sighting through a pinhole. This simple method doesn't give a good estimate of the angular size. The simple pattern in the calculated values is due to the original assumption that the globes were in an arithmetic sequence.&lt;/p&gt;&#xA;&lt;p&gt;There is then a calculation of the geocentric orbital radii of the planets. They assume that Venus, Mercury and the Sun all orbit at a distance of 3.4 million miles.  This is a significant underestimate, especially for the sun (this is wrong). The distances to Mars, Jupiter and Saturn were also underestimated.&lt;/p&gt;&#xA;&lt;p&gt;One can then combine the apparent angular diameter (which is too large) with the calculated distance (which was too small) to obtain an estimate of the planetary diameter. When this is done, it gives exceedingly good estimates of the diameter of Mercury and Saturn (within 1%). A good estimate for the diameter of Mars (about 10% out), and a poor estimate for Venus and Jupiter (about 50% of the accepted modern values).&lt;/p&gt;&#xA;&lt;p&gt;It is clear that there is much that is wrong. A combination of overestimates of some values and underestimates of others cancels out to give some impressive looking figures for two out of five planets. But do note that the estimates for Jupiter and Venus were a long way out, and all the numbers which the calculations were based on were wrong. In total there are 15 values calcuated here 5 apparent diameters, 5 geocentric distances and 5 actual diameters. Of these only 2 values are impressively accurate.&lt;/p&gt;&#xA;&lt;p&gt;They did better than other pre-telescopic observers, but it isn't clear that they obtained values by anything other than luck.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://web.archive.org/web/20100107161952/http://www.scientificexploration.org/journal/jse_11_2_thompson.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2021-03-14T18:51:39.247" LastActivityDate="2021-03-14T18:51:39.247" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25997" PostTypeId="2" ParentId="25979" CreationDate="2018-04-20T09:35:44.753" Score="0" Body="&lt;p&gt;Heat in accretion disk happens due to friction and friction happens only when there is relative motion. So in that accretion disk a lot of particles are moving relative to each other with high velocities, so fusion should not happen, because for that particle should come together. Even in star (like our sun), mass of star is not enough to produce fusion, and it needs the help of quantum tunneling, so we cannot say that pressure is available within that accretion disk so as to overcome repulsion of nuclear force.  &lt;/p&gt;&#xA;" OwnerUserId="20162" LastActivityDate="2018-04-20T09:35:44.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25998" PostTypeId="2" ParentId="25931" CreationDate="2018-04-20T09:48:06.363" Score="0" Body="&lt;p&gt;I think that if you want to determine the angle between the solar equator and the horizon on Earth at a specific location, there are three components to take into account:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) the angle between the local zenith direction at the specific location and the geocentric polar axis (angle q in the drawing below). It depends on your latitude, it would be 0 degree at the Geographic Pole and 90 degree at the equator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) the angle between the geocentric North Pole and the solar rotation North Pole (angle P) measured eastward  from geocentric north. It can be between -26.31 and +26.31.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) the apparent tilt of the sun's axis of rotation with respect to the ecliptic (angle B0) which can be between -7.25 and +7.25. This angle is also the heliographic latitude of the central point of the solar disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angles P and B0 can be found in an ephemeris table for a given date and time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here an illustration: &#xA;&lt;a href=&quot;https://i.stack.imgur.com/TO0mW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TO0mW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17423" LastEditorUserId="17423" LastEditDate="2018-04-20T10:23:25.520" LastActivityDate="2018-04-20T10:23:25.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26000" PostTypeId="2" ParentId="25979" CreationDate="2018-04-20T11:53:57.343" Score="5" Body="&lt;p&gt;The accretion of material onto (into) black holes (and neutron stars) provides environments that are both very hot and (relatively) dense. Under these circumstances it is possible for nuclear fusion to occur, the question is whether this is significant, both energetically or as a means of producing new chemical elements (nucleosynthesis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to the first of these questions is relatively straightforward. As material falls towards the black hole, its angular momentum forces it to form an accretion disk. Viscous processes heat the disk and provide torques, cause the material to lose energy and angular momentum and eventually allow it to fall into the black hole. Much of the gravitational potential energy (GPE) gained as the material falls towards the black hole does end up heating the material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The innermost stable circular orbit of a black hole is at 3 Schwarzschild radii $=6GM/c^2$, where $M$ is the black hole mass.  The GPE released for material of mass $m$ falling to this radius is $\sim GMmc^2/6GM = mc^2/6$.&#xA;i.e. fully one sixth of the rest mass energy of the material could be released as heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare this with nuclear fusion. The fusion of hydrogen into helium only releases 0.7% of the rest mass as energy that can heat the accretion disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So from the energetic point of view, fusion reactions are negligible, unless they can occur much further out in the disk&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question about nucleosynthesis yields is more complex. The more massive a black hole and the higher the accretion rate, then in general the higher the disk temperature and density and the higher the fusion rate. But it also depends on the details of the cooling processes that are possible and how much material is advected into the black hole. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2008ApJ...681...96H&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hu &amp;amp; Peng (2008)&lt;/a&gt; present some models of accretion onto a 10 solar-mass black hole and suggest that it may be possible to produce certain rare isotopes by this mechanism.  Stellar-sized black holes probably need very substantially super-Eddington accretion rates to achieve the necessary temperatures to sustain nuclear fusion (i.e. much greater accretion rates than are possible by radiation-pressure opposed spherical accretion flows), according to &lt;a href=&quot;http://lup.lub.lu.se/luur/download?func=downloadFile&amp;amp;recordOId=8912003&amp;amp;fileOId=8912097&quot; rel=&quot;nofollow noreferrer&quot;&gt;Frankel (2016)&lt;/a&gt;. Such rates are likely only in the cases where black holes disrupt a binary companion, rather than through a steady accretion flow.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2365" LastEditDate="2018-04-20T12:35:32.587" LastActivityDate="2018-04-20T12:35:32.587" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26002" PostTypeId="1" AcceptedAnswerId="26006" CreationDate="2018-04-20T21:29:45.643" Score="0" ViewCount="58" Body="&lt;p&gt;I need to calculate the eccentricity of a satellite orbit with only the apside altitudes (height above the surface of the Earth), 400 and 1500 km. However, I don't know the semi-major axis ($a$) or semi-minor axis ($b$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{eccentricity} = (1-(b/a)^2)^{1/2}$$&lt;/p&gt;&#xA;" OwnerUserId="22784" LastEditorUserId="7472" LastEditDate="2018-04-22T03:09:23.853" LastActivityDate="2018-04-22T03:09:23.853" Title="Calculate eccentricity from apside altitudes?" Tags="&lt;orbit&gt;&lt;earth&gt;&lt;orbital-elements&gt;&lt;orbital-mechanics&gt;&lt;satellite&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ClosedDate="2018-04-23T19:26:29.563" ContentLicense="CC BY-SA 3.0" />
  <row Id="26003" PostTypeId="1" CreationDate="2018-04-21T00:15:33.187" Score="1" ViewCount="268" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sFTx2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sFTx2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a simple, textbook definition of a binary star orbit.  One low-mass star orbits a higher-mass star in a narrow but definitive center.  For example, the high-mass star would be our own sun--a G-type main sequence star--being orbited by a smaller dwarf, like a red dwarf or a brown dwarf.  But what about binary orbits of two identical stars--like our sun orbiting another G-type main-sequence star?  Is this sort of orbit possible?&lt;/p&gt;&#xA;" OwnerUserId="10085" LastActivityDate="2018-04-22T04:46:28.440" Title="Can Two Stars of One Same Type Orbit Each Other?" Tags="&lt;star&gt;&lt;orbit&gt;&lt;binary-star&gt;&lt;orbital-mechanics&gt;&lt;celestial-mechanics&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26005" PostTypeId="2" ParentId="26003" CreationDate="2018-04-21T05:25:32.963" Score="4" Body="&lt;p&gt;Certainly, the figure only uses different masses because that's the more general possibility (after all, the two masses won't be &lt;em&gt;exactly&lt;/em&gt; the same).  In fact, if you look at the distribution over mass ratio q in binary systems in Figure 1b in &lt;a href=&quot;https://arxiv.org/pdf/1304.3123.pdf&quot; rel=&quot;noreferrer&quot;&gt;https://arxiv.org/pdf/1304.3123.pdf&lt;/a&gt;, it seems that the cumulative number of binaries with mass ratio less than q is more or less equal to q (where q is smaller mass over larger mass).  That means the probability of mass ratio between any q and q+dq is about equal to dq and nearly independent of q.  So the chance of the mass ratio being between 0.9 and 1 is about the same as being between 0 and 0.1, for example.  In some sense, that's like saying all q are equally likely.  And of course if q is near 1, then the stars will have the same age and be in the same evolutionary state, so will be the same stellar type as well.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-04-21T05:25:32.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26006" PostTypeId="2" ParentId="26002" CreationDate="2018-04-21T09:14:28.387" Score="0" Body="&lt;p&gt;The major axis is just the sum of the distance at perigee plus the distance at apogee. This is 400 km + 1500 km + the &lt;em&gt;diameter&lt;/em&gt; (not radius) of the Earth, and the semi-major axis is half of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the semi-minor axis is harder to determine. Therefore, why not use an &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_eccentricity#Calculation&quot; rel=&quot;nofollow noreferrer&quot;&gt;alternative method to calculate the eccentricity&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e = \frac{r_a-r_p}{r_a+r_p}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the $r$'s are the distance to the &lt;em&gt;center&lt;/em&gt; of the Earth, not the apside altitudes. Can you take it from here?&lt;/p&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2018-04-21T09:14:28.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26008" PostTypeId="2" ParentId="26003" CreationDate="2018-04-22T04:46:28.440" Score="1" Body="&lt;p&gt;There are roughly 100 billion stars (quite probably more) in the Milky Way galaxy (ours).  Of those it is thought that about 85% of them are in multi-star systems (binary, triplet, etc.).  There are thought to be of the order of 10 &lt;em&gt;trillion&lt;/em&gt; galaxies in the visible universe (for this I used &lt;a href=&quot;https://www.space.com/26078-how-many-stars-are-there.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article on Space.com&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So something like $100\times 10^9 \times 10 \times 10^{12} = 10^{24}$ stars in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's says about half of the stars in multi-star systems are in binary systems (not triplets, etc.).  That gives us a (ludicrously rough) estimate of about $4\times 10^{23}$ binary systems in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which brings us back to :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But what about binary orbits of two identical stars--like our sun orbiting another G-type main-sequence star? Is this sort of orbit possible?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given the huge number it's hard to imagine not one of those binary systems being, for all intents and purposes, near identical stars and even near identical G-type stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only really similar stars in a binary system I'm personally aware of are &lt;a href=&quot;https://en.wikipedia.org/wiki/Zeta_Ursae_Majoris#Stellar_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mizar Aa and Mizar Ab&lt;/a&gt;.  These are A-type stars, not G-type stars, but they illustrate the idea is possible.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2018-04-22T04:46:28.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26009" PostTypeId="1" AcceptedAnswerId="26010" CreationDate="2018-04-22T14:11:37.313" Score="3" ViewCount="119" Body="&lt;p&gt;So I just derived an expression for the scattered intensity of light from an atmosphere with ideal isotopic scattering: $I/F=\frac{cos(i)}{4(cos(e)+cos(i))}$, where $I$ is intensity, $F$ is flux, $i$ is the angle off incidence, and $e$ is the angle of emission ($i,\ e=0$ are normal to atmosphere).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I plot the scattered intensity as a function of $i$ with $e$ constant, it decreases while the angle increases. Why is this? My theory is that at a large incidence angle, the light encounters a larger surface area, so reflected light will have a higher variation of emission angle, meaning that scattered light at a given emission angle will be lower. Does this make sense? Or is there another reason for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, I plotted scattered intensity as a function of $e$ with $i$ constant. This time, the intensity increases with increases emission angle. I don't have a theory as to why this is the case, nor an intuition to back it up. Why would this be the case?&lt;/p&gt;&#xA;" OwnerUserId="15870" LastEditorUserId="7982" LastEditDate="2021-02-28T11:44:31.963" LastActivityDate="2021-02-28T11:44:31.963" Title="Scattered Intensity of Light vs Angles of Incidence and Emission" Tags="&lt;light&gt;&lt;planetary-atmosphere&gt;&lt;radiation&gt;&lt;atmospheric-effects&gt;&lt;radiative-transfer&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26010" PostTypeId="2" ParentId="26009" CreationDate="2018-04-22T15:42:33.680" Score="3" Body="&lt;p&gt;It is easiest to understand if you fix the incident angle and explain why the emergent intensity is largest at highly oblique angles.  Since the light comes from the outside, it only penetrates so far, and this in turn causes the atmosphere to act like a source of scattered light.  But the source of scattered light is brightest near the top of the atmosphere, since that's where more of the external light penetrates.  Whenever the sources are brighter near the top, it produces what is called &quot;limb brightening,&quot; where if you look from highly oblique angles, you mostly probe those higher, brighter regions.  Looking down the normal is where you see the deepest into the atmosphere, where the external light does not penetrate as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also think about what the individual photons are doing, and ask what is their distribution over emergent angle.  If cos(i)=0, all the photons scatter right at the surface, so it is tantamount to introducing an isotropic radiation field right at the surface.  The photons that go outward will of course have an isotropic distribution, and an isotropic incident radiation field must scatter isotropically (that is a consequence of the principle of reciprocity).  So the emergent distribution is isotropic, but intensity is also per solid angle, so accounts for the foreshortening, and that's where the 1/cos(e) in the intensity comes from.  If cos(i)=1, on the other hand, the incident photons tend to penetrate more, and must diffuse their way out, which gives less of an advantage to low cos(e) after the foreshortening is included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for fixing the angle at which you are looking and altering i, here your result says that the intensity always peaks as cos(i) rises.  Your expression claims that is true at every e, so to me this suggests a normalization error.  You want to keep the incident F fixed, but that requires you must get the same outgoing F if you integrate over all e.  Your result says the I is higher at all e if cos(i) is higher, but that contradicts the idea that you are keeping the incident F the same.  Maybe your result is actually comparing the incident and emergent intensities, not the emergent intensity to incident flux.  Then at higher cos(i), for the same incident I, the incident F is falling, explaining your rise in I/F.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-04-22T15:42:33.680" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26012" PostTypeId="1" CreationDate="2018-04-22T20:03:26.520" Score="0" ViewCount="393" Body="&lt;p&gt;and what should be the minimum focal length of objective lens and focal lens. I've a celestron  which have 300mm focal length of  objective lens and 6/4/12/20 mm of eye piece  but it's hard for me too the jupiter so I asked here &quot;what  should be the minimum magnification of a telescope?&quot;&lt;/p&gt;&#xA;" OwnerUserId="22796" LastEditorUserId="22796" LastEditDate="2018-04-24T14:19:10.400" LastActivityDate="2018-04-24T14:19:10.400" Title="what should be the minimum magnification of the telescope to see properly the cloud(red spot) of jupiter?" Tags="&lt;planet&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2018-04-23T13:30:44.563" ContentLicense="CC BY-SA 3.0" />
  <row Id="26013" PostTypeId="2" ParentId="25931" CreationDate="2018-04-23T05:05:45.530" Score="1" Body="&lt;p&gt;The angle between the observer's zenith, any given celestial object, and the north celestial pole is the object's&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Parallactic_angle&quot; rel=&quot;nofollow noreferrer&quot;&gt;parallactic angle&lt;/a&gt;.&#xA;This angle $q$ depends on the object's&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Hour_angle&quot; rel=&quot;nofollow noreferrer&quot;&gt;hour angle&lt;/a&gt; $H$ and&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Declination&quot; rel=&quot;nofollow noreferrer&quot;&gt;declination&lt;/a&gt; $\delta$,&#xA;and the observer's geographic latitude $\varphi$:&#xA;$$ \tan q = {{\sin H} \over {\tan \varphi \cos \delta - \sin \delta \cos H}} $$&#xA;At local apparent solar noon, $H_\odot$ = 0 and $q_\odot$ = 0 as you expect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun's declination is roughly&#xA;$$ \delta_\odot \approx \varepsilon \sin \left({2\pi {D - 79 \over 365.25}}\right)$$&#xA;where $D$ is the day of the year and $\varepsilon$ is the 23.44$^\circ$ tilt of the Earth's axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Position_angle&quot; rel=&quot;nofollow noreferrer&quot;&gt;position angle&lt;/a&gt; $P$ of the Sun's north pole relative to the north celestial pole depends on both $\varepsilon$ and the Sun's 7.25$^\circ$ axial tilt relative to the ecliptic plane.&#xA;These two components combine somewhat out of phase, roughly&#xA;$$ P \approx -26.3^\circ \sin \left(2\pi {{D - 5} \over 365.25}\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meeus's &lt;a href=&quot;http://www.willbell.com/math/mc1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomical Algorithms&lt;/a&gt; gives more precise formulas for the above in terms of the Sun's ecliptic longitude $\lambda_\odot$, taking the Earth's orbital eccentricity and other factors into account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle of the Sun's equator with respect to the observer's horizon is the same as the angle of the Sun's north pole with respect to the observer's zenith, $P - q_\odot$.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-04-23T05:05:45.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26016" PostTypeId="1" CreationDate="2018-04-23T12:33:58.390" Score="2" ViewCount="1842" Body="&lt;p&gt;Is there a technical term for the &quot;lines&quot; in a constellation or asterism? Alternatively, is there an astronomy related coloquialism or any informal term that refers specifically to these lines, and &lt;em&gt;not&lt;/em&gt; to the constellation itself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To illustrate: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the two examples below, what is the (most?) correct way to describe all of the green lines in the two images.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Argo Navis, an asterism and former constellation.&#xA;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/5/57/Argo_navis_constellation_map.png&quot; alt=&quot;Argo Navis&quot;&gt;&#xA;Source: &lt;a href=&quot;https://commons.wikimedia.org/wiki/Category:Argo_Navis#/media/File:Argo_navis_constellation_map.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The constellation of Draco.&#xA;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/f/f0/Draco_constellation_map.png&quot; alt=&quot;Draco&quot;&gt;&#xA;Source: &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Draco_constellation_map.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Further Background/Details:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was doing some astronomy related browsing and noticed the some of the representations of the same constellation had differences between different sources. The majority were very similar, others had more significant differences. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Version 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lCi4R.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lCi4R.gif&quot; alt=&quot;Version A&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Version 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qpLjq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qpLjq.png&quot; alt=&quot;Version 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Version 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oeiLr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oeiLr.jpg&quot; alt=&quot;Version 3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Version 1 has significantly more lines included in it than Versions 2 and 3.&lt;/li&gt;&#xA;&lt;li&gt;Version 1 has the line between η Cygni and β Cygni, appears to go via φ Cygni.&lt;/li&gt;&#xA;&lt;li&gt;Version 3 has the line between η Cygni and β Cygni, goes via a star to the upper right of χ Cygni on the charts.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I want to search for more examples of the different ways the lines in the constellations have been drawn, and all my google fu has failed to produce any useful search terms. Consequently I'm hoping to find there is a term that describes the lines themselves which should be more helpful at finding  material focused on the ways these lines have been drawn.&lt;/p&gt;&#xA;" OwnerUserId="501" LastEditorUserId="616" LastEditDate="2018-04-23T20:11:41.350" LastActivityDate="2019-04-02T19:46:22.200" Title="What are the &quot;lines&quot; in a constellation or asterism called?" Tags="&lt;observational-astronomy&gt;&lt;constellations&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26017" PostTypeId="1" AcceptedAnswerId="26021" CreationDate="2018-04-23T13:30:42.273" Score="1" ViewCount="212" Body="&lt;p&gt;In this &lt;a href=&quot;https://space.stackexchange.com/a/25121/12102&quot;&gt;answer&lt;/a&gt; I have rearranged an equation from somewhere, and now I can not relocate the source. My re-aranged form is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M_{Abs} = 5 \left(\log_{10}(1329) -\frac{1}{2}\log_{10}(\text{albedo}) -\log_{10}(D_{km})\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the relationship between the absolute magnitude of an object and its diameter and albedo, assuming it is spherical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is a good source for any form of this equation to which I can link in my answer?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-04-23T18:11:22.870" Title="Good source for the relationship between absolute magnitude, diameter, and albedo?" Tags="&lt;asteroids&gt;&lt;absolute-magnitude&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26018" PostTypeId="1" AcceptedAnswerId="26019" CreationDate="2018-04-23T14:15:22.500" Score="5" ViewCount="104" Body="&lt;p&gt;&lt;em&gt;This is being asked from a lay-person point of view. However, please do not hesitate to give more detailed answers that require prior knowledge - if these answer the question more accurately.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My confusion currently comes from the fact that everything in the universe is moving in some way or another - so any reference point we use would also be moving/not completely reliable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, how can we accurately determine that we have completed an exact 360 degree rotation around the sun (and as such, measure exactly how long it took)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;To be clear, this is asking about how we can determine that we have completed a single 360 orbit of our sun, and not the timing aspect of how we know how long a second is. If this situation is different for measuring the orbits of other bodies around their star - it would be great to expand on those differences as well.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="22802" LastActivityDate="2018-04-25T19:29:12.833" Title="How do we measure one complete orbit?" Tags="&lt;orbit&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26019" PostTypeId="2" ParentId="26018" CreationDate="2018-04-23T15:41:38.273" Score="3" Body="&lt;p&gt;When one asks the question, &quot;How is a year defined?&quot;, there are actually a few different answers one can provide. What you're implicitly asking, even if you don't know, is how one defines a &lt;a href=&quot;https://en.wikipedia.org/wiki/Sidereal_year&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sidereal Year&lt;/a&gt;. In effect, the Sidereal Year is the time necessary to complete one, full, 360 degree revolution around the Sun. This is opposed to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Tropical_year&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tropical Year&lt;/a&gt; (or else the Solar Year) which is the time necessary to travel just slightly less than a full 360 degrees (due to things like precession and the definition of the Tropical Year).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your question boils down to, how is a Sidereal year determined? In practice, this is often done by choosing a specific starting point and noting where the Sun is with respect to the stars in the Sky. Then, it will be an entire sidereal year (and a 360 degree revolution) before the Sun returns to the same position as previously observed. Once you've observed the Sun in the &lt;em&gt;exact same position&lt;/em&gt; you know a full revolution has occurred of the Earth around the Sun. It should be noted that the motion of the stars over a year is almost completely inconsequential. The stars are so immensely far away compared to their own motion, that their apparent motion (otherwise known as the proper motion) is basically nothing over a year. In this way, we consider the stars &quot;fixed&quot;.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2018-04-23T15:55:19.483" LastActivityDate="2018-04-23T15:55:19.483" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26020" PostTypeId="1" CreationDate="2018-04-23T17:39:33.473" Score="4" ViewCount="1371" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Metallicity#Relation_between_Z_and_%5BFe/H%5D&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia entry on Metallicity&lt;/a&gt; states that:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\log_{10}\left(\frac{Z/X}{Z_\mathrm{sun}/X_\mathrm{sun}}\right) = [\mathrm{M}/\mathrm{H}]$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$[M/H]$&lt;/span&gt; is the star's total metal abundance (i.e. all elements heavier than helium) defined as a more general expression than the one for [Fe/H]:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$[\mathrm{M}/\mathrm{H}] = \log_{10}{\left(\frac{N_{\mathrm{M}}}{N_{\mathrm{H}}}\right)_\mathrm{star}} - \log_{10}{\left(\frac{N_{\mathrm{M}}}{N_{\mathrm{H}}}\right)_\mathrm{sun}}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The iron abundance and the total metal abundance are often assumed to be related through a constant A as [&lt;em&gt;&lt;strong&gt;citation needed&lt;/strong&gt;&lt;/em&gt;]:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$[\mathrm{M}/\mathrm{H}] = A\times[\mathrm{Fe}/\mathrm{H}]$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; assumes values between 0.9 and 1. Using the formulas presented above, the relation between &lt;span class=&quot;math-container&quot;&gt;$Z$&lt;/span&gt; and [Fe/H] can finally be written as:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\log_{10}\left(\frac{Z/X}{Z_\mathrm{sun}/X_\mathrm{sun}}\right) = A\times[\mathrm{Fe}/\mathrm{H}]$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This all sounds reasonable, but I have two questions:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Where does the &lt;span class=&quot;math-container&quot;&gt;$[\mathrm{M}/\mathrm{H}] = A\times[\mathrm{Fe}/\mathrm{H}]$&lt;/span&gt; relation come from? I could not find a proper source.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1994A%26AS..106..275B&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article&lt;/a&gt; actually says (see Eq 9) that the general relation is &lt;span class=&quot;math-container&quot;&gt;$\log_{10}\left(\frac{Z/X}{Z_\mathrm{sun}/X_\mathrm{sun}}\right) = [\mathrm{Fe}/\mathrm{H}]$&lt;/span&gt; i.e., they seem to equate [M/H] with [Fe/H]. I assume that it is because the aforementioned A parameter is between 0.9 and 1, but again that leaves me with the need for a proper source to state that.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="354" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2020-02-06T18:46:22.653" Title="Correct relation between metallicity (z) and iron content ([Fe/H])" Tags="&lt;metallicity&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26021" PostTypeId="2" ParentId="26017" CreationDate="2018-04-23T18:11:22.870" Score="2" Body="&lt;p&gt;The JPL CNEOS&#xA;&lt;a href=&quot;https://cneos.jpl.nasa.gov/tools/ast_size_est.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;asteroid size estimator&lt;/a&gt;&#xA;and various asteroid albedo papers cite&#xA;&lt;a href=&quot;https://doi.org/10.1006/icar.1996.5664&quot; rel=&quot;nofollow noreferrer&quot;&gt;Harris and Harris 1997&lt;/a&gt;.&#xA;That paper is behind a paywall, but&#xA;&lt;a href=&quot;http://www.boulder.swri.edu/clark/jssrpb04.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stuart and Binzel 2004&lt;/a&gt;&#xA;attribute this formula to it:&#xA;$$ H = C - 5 \log_{10} D - 2.5 \log_{10} p_V$$&#xA;where $H$ is absolute magnitude, $p_V$ is albedo, and $C$ = 15.618.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-04-23T18:11:22.870" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26022" PostTypeId="1" CreationDate="2018-04-23T20:00:40.753" Score="3" ViewCount="90" Body="&lt;p&gt;Hubble's constant is estimated at 15km/s at a distance of million light years, without considering effects of mutual gravity, relativity, etc.&#xA;I am trying to calculate the characteristic expansion time of the universe.&#xA;Its formula is given as reciprocal of the Hubble constant, because time is distance by velocity. But:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v=kx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that velocity and distance are exponential functions of time and so time cannot simply be calculated by dividing 1 million light years by 15km/s.&#xA;So what am I missing? Why is the characteristic expansion time simply the reciprocal of Hubble's constant?&lt;/p&gt;&#xA;" OwnerUserId="22805" LastEditorUserId="7472" LastEditDate="2018-04-24T07:20:07.137" LastActivityDate="2018-04-24T07:20:07.137" Title="calculation of characteristic expansion time of the universe" Tags="&lt;big-bang-theory&gt;&lt;hubble-constant&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26023" PostTypeId="2" ParentId="26022" CreationDate="2018-04-23T21:23:51.193" Score="3" Body="&lt;p&gt;Here is some remarks on the issue, straight from Ryden:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If galaxies are currently moving away from each other, then it implies&#xA;  they were closer together in the past. Consider a pair of galaxies&#xA;  currently separated by a distance $r$, with a velocity $v = H_0r$&#xA;  relative to each other. If there are no forces acting to accelerate or&#xA;  decelerate their relative motion, then their velocity is constant, adn&#xA;  the time that has elapsed since they were in contact is&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$t_0 = \frac{r}{v} = \frac{r}{H_0r} = H_0^{-1}$, &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;independent of their current separation $r$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which part of this are you confused about? I can edit my answer once I better understand your question.&lt;/p&gt;&#xA;" OwnerUserId="13283" LastActivityDate="2018-04-23T21:23:51.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26024" PostTypeId="1" AcceptedAnswerId="26031" CreationDate="2018-04-24T01:27:22.463" Score="4" ViewCount="279" Body="&lt;p&gt;I understand how Solar Parallax calculation determines the Earth-Sun distance using measurements of (a) the baseline distance between two widely-separated terrestrial telescopes and the (b) two angles - one at each telescope - between the baseline and the line from telescope to the centre of the Sun's visible disc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is it that observations of the transit of a planet across the face of the Sun give better results than simply pointing (simultaneously) each telescope at the centre of the Sun's visible disc?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the Solar Transit method also superior (in terms of accuracy) to the method of determining Earth-Planet distances by measuring parallax angles relative to 'fixed' background stars?  If so, why?&lt;/p&gt;&#xA;" OwnerUserId="233" LastActivityDate="2018-04-24T19:00:47.897" Title="How does a Solar Transit give more accurate determination of Earth-to-Sun distance?" Tags="&lt;planetary-transits&gt;&lt;parallax&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26026" PostTypeId="1" CreationDate="2018-04-24T01:55:54.927" Score="1" ViewCount="52" Body="&lt;p&gt;I've been reading about the Earth's analemma from &lt;a href=&quot;https://en.wikipedia.org/wiki/Analemma&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, and I am having trouble visualizing one component. Let's build up to my question.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If the Earth had a circular orbit and no axial tilt, the analemma would be a dot.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the Earth had an eccentric orbit and no axial tilt, the analemma would be a east-west line.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the Earth had a circular orbit and axial tilt, the analemma would be symmetric figure eight.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My question:  In situation 3, I understand the north-south variation of the symmetric analemma, but I don't understand why there would be symmetric east-west variation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the Earth's axial tilt create an east-west variation in the analemma?&lt;/p&gt;&#xA;" OwnerUserId="22808" LastEditorUserId="22808" LastEditDate="2018-04-24T02:01:14.700" LastActivityDate="2018-04-24T02:01:14.700" Title="Why does the Earth's axial tilt create an east-west variation in the analemma?" Tags="&lt;orbit&gt;&lt;earth&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2018-04-24T15:21:50.593" ContentLicense="CC BY-SA 3.0" />
  <row Id="26029" PostTypeId="1" CreationDate="2018-04-24T08:47:24.070" Score="4" ViewCount="52" Body="&lt;p&gt;I use a Meade 200 LX GPS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last night between viewings it was placed in Sleep mode.&#xA;On return the scope was making noises as if the motor was running. The scope was stationary, but had moved to what looked like park position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I pressed MODE and then tried to locate objects using GOTO, but it did not move appropriately either from where it was, or as if from the previous sleep position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then switched everything off and 'started again'...several times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I received a couple of strange messages including that I had changed the location of the scope (it had not moved) and that the batteries were low (it is connected to the mains, not running on batteries).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then decided to leave it overnight and retry in the morning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This morning I started from scratch again and successfully carried out a 2-star alignment, and the scope moved at normal finding speed to the star locations. However, when centring the stars electronically the RA movement was VERY SLOW no matter what speed setting was set. Dec speed is fine!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scope moves to GOTO targets at normal speed, but electronic control of RA is still VERY VERY SLOW.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you please advise what may be the cause of this problem and best options for remediation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FYI we have two hand box controllers and the problem occurs with both. The hand box is not the fault!&lt;/p&gt;&#xA;" OwnerUserId="22812" LastActivityDate="2018-04-24T08:47:24.070" Title="Meade LX200GPS electronic RA movement very slow" Tags="&lt;telescope&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26031" PostTypeId="2" ParentId="26024" CreationDate="2018-04-24T11:44:35.027" Score="8" Body="&lt;p&gt;The problem of determining the solar parallax (and inversely the Sun's distance from Earth) by classical optical astronomy has been in most periods since the 17th century dominated less by principle than by the great difficulties of controlling and reducing the measurement errors of the very small angles involved (the solar parallax is only about 8.794&quot; of arc). A good reference source for the earlier period through the 18th century is Albert van Helden's &quot;Measuring the Universe&quot; (1985, Chicago UP).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method of observing planetary transits across the Sun's disk in order to determine the solar parallax was arguably a method of clear choice mainly or only for the 18th-century solar transits of Venus, which took place only in 1761 and 1769. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The measurement advantage in observing the Venus transits (relative to a difficult direct solar-parallax determination) arises largely because Venus at inferior conjunction is only about a quarter the distance of the Sun, thus its parallax in that configuration is not far off four times as large as the Sun's. What was to be measured directly was the parallax of Venus itself rather than that of the Sun. The size of the parallax angle to be measured was one of a number of factors that it was hoped would allow a smaller relative error in the observational determination. Conversion from the parallax of Venus to that of the Sun was done by using the comparatively much-better-determined relative scale of solar-system distances. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By contrast, Mercury transits, though more frequent in occurrence than those of Venus, only allow measurement of Mercury's parallax, which is not too much bigger at inferior conjunction than the Sun's own parallax itself, thus offering less advantage than a Venus measurement.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the time of the 19th-century Venus transits (again only two, in 1874 and 1882) there was an accumulated experience of the practical measurement difficulties, some of which (like the 'black-drop effect') had not been expected before the 18th-century transits were actually observed. The difficulties meant that advance expectations of measurement accuracy from the 18th-century Venus transits had proved to be over-optimistic.  There was some remaining optimism that the measurement difficulties could be addressed, and expeditions were mounted to observe the 19th-century Venus transits, but still the results were not as much better than before as had been hoped. (A useful reference source for this part of the history is J Ratcliff (2008) &quot;The Transit of Venus Enterprise in Victorian Britain&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the meantime, competing methods of measuring the Sun's parallax began to be devised, and a more successful example was indeed an Earth-planet distance: to Eros, a minor planet whose orbit brought it much closer to the Earth than Venus' distance at solar transit. The crucial aim again was to measure by parallax a distance that should be as close as possible (and well-determined on the relative solar-system scale of distances, like distances of Venus and other planets). The measurements in the Eros case at its close approach to Earth were done by reduction of photographs and of the positions of Eros relative to selected reference stars as seen from a number of observing stations (A R Hinks, 1904, 'Reduction of Photographs of Eros, with a Determination of the Solar Parallax' (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1904MNRAS..64..701H&quot; rel=&quot;noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/1904MNRAS..64..701H&lt;/a&gt;)). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the 1960s onwards, the traditional optical methods of determining solar and planetary distances using solar transits and close approaches of minor planets were entirely superseded by ranging methods, one of the earliest examples being radar-signal reflection from Venus (reviewed along with the Eros measurements e.g. by Eckstein (1963),  &lt;a href=&quot;http://adsabs.harvard.edu/abs/1963AJ.....68..231E&quot; rel=&quot;noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/1963AJ.....68..231E&lt;/a&gt;). More recently many other range measurements have  been obtained e.g. to spacecraft placed in orbit around solar-system bodies. &lt;/p&gt;&#xA;" OwnerUserId="17202" LastEditorUserId="17202" LastEditDate="2018-04-24T12:15:19.863" LastActivityDate="2018-04-24T12:15:19.863" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26033" PostTypeId="1" AcceptedAnswerId="26098" CreationDate="2018-04-24T15:32:22.970" Score="7" ViewCount="737" Body="&lt;p&gt;At innumerable places on the net you can find the claim that leap seconds are inserted into UTC in order to make it track UT1 which again tracks mean solar time at the reference meridian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I tried to locate an actual definition of UT1, the closest I could find was &lt;a href=&quot;http://www.iau.org/static/resolutions/IAU2000_French.pdf#page=23&quot; rel=&quot;noreferrer&quot;&gt;IAU resolution B1.8 from 2000&lt;/a&gt; which seems to define it implicitly from the following relationship with the Earth Rotation Angle:&#xA;$$ \theta = 2\pi\bigl(0.7790572732640+1.00273781191135448({\it JD}_{\rm UT1}-2451545.0)\bigr)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the magic constant 1.00273781191135448 appears to be derived from the ratio between the Earth's rotational period and the sidereal year -- but no provision for adjusting it for observed changes in that ratio seems to be made. (The least significant several digits of the constant must be wrong already).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the Earth's rotation slows down, is UT1 therefore going to drift out of sync with mean solar time at the IERS reference meridian? Or am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="5350" LastEditorUserId="5350" LastEditDate="2018-04-24T16:41:53.843" LastActivityDate="2018-04-30T16:58:32.377" Title="Does Universal Time really track mean solar time?" Tags="&lt;time&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26034" PostTypeId="1" AcceptedAnswerId="26039" CreationDate="2018-04-24T16:18:06.537" Score="9" ViewCount="209" Body="&lt;p&gt;I'm using the term &quot;rogue planet&quot; here loosely to mean anything of planet-level mass that is not orbiting around a star. Popular discussions of these objects always describe them as forming around a star and then being ejected by gravitational interaction with other planets. But I would have thought that, in principle, they could form in interstellar space just by the collapse of a sufficiently small cloud of dust and gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that a small cloud will have less self-gravity, and would therefore take more time to collapse [see the answers for a correction to this assumption] and have a greater chance of being disrupted by other forces. But has a theoretical analysis been done of the minimum mass object that could be expected to form in isolation over the time frame of the current lifetime of the universe? It would be a surprising coincidence of that minimum threshold were the same mass as the mass required for hydrogen fusion.&lt;/p&gt;&#xA;" OwnerUserId="7752" LastEditorUserId="7752" LastEditDate="2018-04-26T03:36:27.047" LastActivityDate="2018-04-26T03:36:27.047" Title="Are rogue planets ever born in isolation?" Tags="&lt;planetary-formation&gt;&lt;rogue-planet&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26035" PostTypeId="2" ParentId="26034" CreationDate="2018-04-24T17:26:09.180" Score="7" Body="&lt;p&gt;Astronomers call the type of object you are describing-- one that condensed from an isolated nebula but was too small to undergo hydrogen fusion-- a &lt;a href=&quot;https://en.wikipedia.org/wiki/Sub-brown_dwarf&quot; rel=&quot;noreferrer&quot;&gt;sub-brown dwarf&lt;/a&gt;.  They are quite difficult to detect as you might imagine, since they &quot;shine&quot; only with the heat of their formation, very dimly in the infrared (the Wikipedia page refers to some that orbit other stars and so could be classified as planets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is theorized that there is a lower-limit on their mass of about 1 Jupiter mass, because smaller masses could not condense from an interstellar cloud in isolation.  A few candidate sub-brown dwarfs have been detected and listed on the linked Wikipedia page; the uncertainty in their mass is high (has to be inferred from their temperature and suspected age).  &lt;a href=&quot;https://en.wikipedia.org/wiki/WISE_0855%E2%88%920714&quot; rel=&quot;noreferrer&quot;&gt;WISE-0855&lt;/a&gt; is the smallest, coldest, and closest sub-brown dwarf in isolation observed so far, estimated 3-10x mass of Jupiter.&lt;/p&gt;&#xA;" OwnerUserId="14434" LastActivityDate="2018-04-24T17:26:09.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26036" PostTypeId="2" ParentId="26024" CreationDate="2018-04-24T19:00:47.897" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://eclipse.gsfc.nasa.gov/transit/HalleyParallax.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Edmond Halley in 1716&lt;/a&gt;&#xA;urged future astronomers to observe the 1761 transit of Venus.&#xA;His key insight was that observers at different geographic latitudes would see Venus cross the Sun along chords of different lengths.&#xA;Especially from temporary remote locations, it would be easier to measure differences in transit duration than to measure parallax directly.&#xA;The disadvantage of Halley's method was that one would have to observe both the beginning and the end of the transit several hours apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ezEmM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ezEmM.png&quot; alt=&quot;Venus transit&quot;&gt;&lt;/a&gt;&#xA;&lt;em&gt;Image source: &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Venus_Transit_%26_Parallax.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikimedia Commons&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Joseph-Nicolas Delisle in 1760 reasoned that observers at different geographic longitudes would see the transit begin or end at different times.&#xA;His method did not require anyone to see the entire transit but did require accurate absolute timekeeping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Black_drop_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;black drop effect&lt;/a&gt; introduced significant timing errors which limited the precision of the distance measurement.&#xA;However, after another transit in 1769, the astronomical community managed to determine the astronomical unit within a few percent of the presently accepted value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further reading:&#xA;&lt;a href=&quot;http://spiff.rit.edu/classes/phys235/venus_t/venus_t.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Michael Richmond&lt;/a&gt;,&#xA;&lt;a href=&quot;http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit4/venussun.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Richard Pogge&lt;/a&gt;,&#xA;&lt;a href=&quot;http://www.open.edu/openlearn/history-the-arts/history/history-science-technology-and-medicine/history-science/history-the-transit-venus&quot; rel=&quot;nofollow noreferrer&quot;&gt;David Hughes&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-04-24T19:00:47.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26037" PostTypeId="1" AcceptedAnswerId="26038" CreationDate="2018-04-24T19:29:18.740" Score="6" ViewCount="306" Body="&lt;p&gt;Does Sun's spectrum expose any red shift at all? Namely, the center of it's visible area (not due to limb effect). If it does, how much is it and does it comply with relativity theory or is it higher/lower than theoretically predicted? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I red about it an article dated 1960 (&lt;a href=&quot;http://adsabs.harvard.edu/full/1960MNRAS.121..421H&quot; rel=&quot;noreferrer&quot;&gt;http://adsabs.harvard.edu/full/1960MNRAS.121..421H&lt;/a&gt;) and not sure if new studies show different results. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="22737" LastActivityDate="2018-04-24T19:47:33.123" Title="Sun's spectrum red shift" Tags="&lt;the-sun&gt;&lt;redshift&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26038" PostTypeId="2" ParentId="26037" CreationDate="2018-04-24T19:47:33.123" Score="2" Body="&lt;p&gt;Light from the solar photosphere is observed to have a gravitational redshift. The expected value, based on General Relativity is equivalent to a doppler redshift of 633 m/s. It is difficult to measure the exact value because it is obscured and blurred by the turbulent motions in the Sun's atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I am aware, the estimated values are entirely consistent with General Relativistic predictions, though the precision is not high enough to make this an especially sensitive test (e.g. $698\pm 113$ m/s &lt;a href=&quot;https://arxiv.org/abs/1207.0177&quot; rel=&quot;nofollow noreferrer&quot;&gt;Takeda &amp;amp; Ueno 2012&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-04-24T19:47:33.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26039" PostTypeId="2" ParentId="26034" CreationDate="2018-04-24T20:15:41.270" Score="10" Body="&lt;p&gt;The minimum mass of a &quot;planet&quot; forming from a gas cloud (definitions of what a planet is are rather slippery, and some would say this is not a planet at all) is &lt;em&gt;not&lt;/em&gt; determined by the time available. The collapse process is rapid - less than a million years. There is a minimum mass though, and what you are referring to is something known as the &lt;strong&gt;fragmentation limit&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cloud becomes unstable and collapses if its mass exceeds the &lt;a href=&quot;https://en.wikipedia.org/wiki/Jeans_instability&quot; rel=&quot;noreferrer&quot;&gt;Jeans mass&lt;/a&gt;. The Jeans mass depends on temperature to the power of 3/2 and inversely on the square root of the cloud density.&#xA;$$ M_J \propto \frac{T^{3/2}}{\rho^{1/2}}$$&#xA;When a cloud of gas collapses, its density increases. If it is able to radiate away heat efficiently, then its temperature can remain more-or-less constant and so the Jeans mass &lt;em&gt;decreases&lt;/em&gt;. This allows the cloud to fragment into smaller pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimum mass that can collapse in isolation will therefore be set by the smallest value that the Jeans mass can attain as the collapse proceeds. This fragmentation limit is in turn set by the cloud becoming opaque to its own radiation, which occurs when the density becomes large enough. At this point, the cloud can no longer efficiently get rid of all the heat that is generated in its interior by the work done by gravity in squashing it. The temperature rises and the Jeans mass stops decreasing. Now, the cloud may still collapse, but it won't break into smaller chunks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fragmentation limit is difficult to calculate with any accuracy, because it depends on the 3D turbulent dynamics of a collapsing cloud and also on whether the cloud is spinning. It is generally thought to be in the range one-to-a-few times the mass of Jupiter (e.g. &lt;a href=&quot;https://www.aanda.org/articles/aa/pdf/2006/42/aa5806-06.pdf&quot; rel=&quot;noreferrer&quot;&gt;Whitworth &amp;amp; Stamatellos 2006&lt;/a&gt;). This is &lt;em&gt;far&lt;/em&gt; below the minimum mass for hydrogen fusion of about 75 Jupiter masses or deuterium fusion of about 13 Jupiter masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evidence that such objects may exist can be found in answers to the related questions &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1715/how-are-rogue-planets-discovered?rq=1&quot;&gt;How are rogue planets discovered?&lt;/a&gt; and &lt;a href=&quot;https://astronomy.stackexchange.com/questions/12536/is-there-any-hard-evidence-that-rogue-planets-exist?rq=1&quot;&gt;Is there any hard evidence that rogue planets exist?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-04-24T22:13:22.140" LastActivityDate="2018-04-24T22:13:22.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26040" PostTypeId="1" CreationDate="2018-04-25T00:24:14.800" Score="1" ViewCount="168" Body="&lt;p&gt;Large optical telescopes want to be located at high altitude (get above as much of the atmosphere as possible). I get the impression that there is also a preference for low latitude (see as much of the sky as possible). And perhaps also for a location with good weather, so that there will be as few cloudy nights in the year as possible? So for example, many large optical telescopes are on the summit of Mauna Kea, Hawaii; there is also McDonald Observatory in Texas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you wanted to put large optical telescopes as far north as possible, but still meet the other requirements, what would be a good location? Looking for high mountains at northern latitudes is an obvious starting point, but how well would that meet the requirement of clear skies? For example, Mount Washington in New Hampshire is the highest peak in the north-eastern United States, but has notoriously terrible weather; this would be an important downside?&lt;/p&gt;&#xA;" OwnerUserId="22818" LastActivityDate="2018-04-25T14:49:25.047" Title="Best locations for large telescopes at high northern latitudes" Tags="&lt;telescope&gt;&lt;night-sky&gt;&lt;weather&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26041" PostTypeId="1" AcceptedAnswerId="26048" CreationDate="2018-04-25T02:31:47.733" Score="5" ViewCount="367" Body="&lt;p&gt;As in taking pix of earth, not just happen to point towards earth during a maneuver or whatever. &lt;/p&gt;&#xA;" OwnerUserId="10117" LastActivityDate="2018-04-25T16:47:31.710" Title="Has the Hubble ever been pointed at earth?" Tags="&lt;hubble-telescope&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26042" PostTypeId="2" ParentId="26041" CreationDate="2018-04-25T03:40:13.013" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The surface of the Earth is whizzing by as Hubble orbits, and the pointing system, designed to track the distant stars, cannot track an object on the Earth. The shortest exposure time on any of the Hubble instruments is 0.1 seconds, and in this time Hubble moves about 700 meters, or almost half a mile. So a picture Hubble took of Earth would be all streaks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hubblesite.org/reference_desk/faq/answer.php.id=78&amp;amp;cat=topten&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://hubblesite.org/reference_desk/faq/answer.php.id=78&amp;amp;cat=topten&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20162" LastEditorUserId="2153" LastEditDate="2018-04-25T12:21:33.430" LastActivityDate="2018-04-25T12:21:33.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26046" PostTypeId="1" CreationDate="2018-04-25T13:05:04.563" Score="1" ViewCount="259" Body="&lt;p&gt;I want to simulate the perihelion precession of Earth. I tried searching  other posts in stack exchange as well relating this question but may be it's my fault that I couldn't understand their answers. I read the history about mathematicians and physicists trying to find a good theory accounting the precession. I read that there are many planetary equation by Newton,Laplace,Hill,Poincare etc. But I am not sure which one gives the more accurate result. I am not even sure if I going on a right track searching the correct theory.&#xA;I'd really appreciate if anyone could help me with the theory which gives fast and more accurate result?&lt;/p&gt;&#xA;" OwnerUserId="6916" LastEditorUserId="616" LastEditDate="2018-04-26T18:08:54.337" LastActivityDate="2018-04-26T18:08:54.337" Title="How to compute the perihelion precession of Earth's orbit?" Tags="&lt;astrophysics&gt;&lt;celestial-mechanics&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26047" PostTypeId="2" ParentId="26040" CreationDate="2018-04-25T14:49:25.047" Score="3" Body="&lt;p&gt;You can (probably) do much better than Mount Washington, both in terms of latitude and air quality. For one, you're right about the weather. According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Mount_Washington_(New_Hampshire)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, it is a convergence point for several storm tracks and next to a breeding ground for low-pressure areas. This not only means a lot of clouds, but very turbulent air, which also lowers the quality of the sky. Also, at just a little shy of $2~km$ in height, you're going to find yourself in the middle of all these clouds, which won't be pleasant. Oh, and you should also consider the difficulties of actually building an observatory in such weather.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, building a telescope at very high latitudes is certainly not a bad idea. On the other side of the globe, there is &lt;a href=&quot;https://en.wikipedia.org/wiki/Ridge_A&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ridge A&lt;/a&gt; on Antarctica, which is agreed to be &quot;the best suited location on the surface of Earth for astronomical research&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, being located on low altitudes allows you to see more of the sky, but you can never see all of the sky at once. Observatories near the equator are great for studying a vast number of different objects over the course of the year. If you're doing a more specific study though (for example, if you want to observe a certain object every night during an extended period of time), you'll have to find yourself a more northern (or southern) observatory. On these higher latitudes, there is a bigger portion of the night sky that is visible throughout the year (constellations in these areas are called 'circumpolar'). This may be desirable for some specific projects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking for good northern spots to build a telescope, we can indeed start by looking at mountains in the northern part of the world. If you take any topographic map of the northern hemisphere, you should immediately notice Greenland sticking out. The big issue with Greenland is that it's covered in a giant layer of ice. You can't build on ice; you'll get in trouble when it melts. We'll disregard that for now though, and we're interested in its highest peak: &lt;a href=&quot;https://en.wikipedia.org/wiki/Gunnbj%C3%B8rn_Fjeld&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gunnbjorn Fjeld&lt;/a&gt;. We're in luck here; Gunnbjorn Fjeld is not covered in ice, but is a rocky peak protruding through the ice. The major difficulty would be actually building there, but I'm going to take a wild guess here and claim that modern building techniques and machines could do it. So this is probably what you're looking for. I'm not an expert on the weather in Greenland though, so I can't say anything about that (I do know that it's cold, which is good, because that means the air is stable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternative locations at northern latitudes would be Iceland (but that's probably a bad idea given its geologic activity) or Siberia. The highest point of Siberia is &lt;a href=&quot;https://en.wikipedia.org/wiki/Klyuchevskaya_Sopka&quot; rel=&quot;nofollow noreferrer&quot;&gt;Klyuchevskaya Sopka&lt;/a&gt; (which is also a bad idea given that it's an active volcano). Apart from that, googling &lt;a href=&quot;https://www.google.be/search?q=mountains%20in%20siberia&amp;amp;rlz=1C1CHBF_enBE792BE792&amp;amp;oq=mountains%20in%20siberia&amp;amp;aqs=chrome..69i57j0l3.3585j0j7&amp;amp;sourceid=chrome&amp;amp;ie=UTF-8&quot; rel=&quot;nofollow noreferrer&quot;&gt;mountains in Siberia&lt;/a&gt; yields lots of peaks which may or may not be well-suited for an observatory. You could probably find a mountain in the Urals as well, but they are not as high as one would ideally want.&lt;/p&gt;&#xA;" OwnerUserId="11824" LastActivityDate="2018-04-25T14:49:25.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26048" PostTypeId="2" ParentId="26041" CreationDate="2018-04-25T16:47:31.710" Score="7" Body="&lt;h1&gt;&lt;a href=&quot;https://what-if.xkcd.com/32/&quot; rel=&quot;noreferrer&quot;&gt;xkcd already did the math on this one!&lt;/a&gt;&lt;/h1&gt;&#xA;&lt;p&gt;Like &lt;a href=&quot;https://astronomy.stackexchange.com/a/26042/14255&quot;&gt;Gauti's answer&lt;/a&gt;, xkcd also links to this question on &lt;a href=&quot;http://hubblesite.org/reference_desk/faq/answer.php.id=78&amp;amp;cat=topten&quot; rel=&quot;noreferrer&quot;&gt;Hubble's website&lt;/a&gt;. Hubble moves too fast for its minimum exposure time to be able to focus on anything on the surface. He goes one step further, linking to this &lt;a href=&quot;http://www.badastronomy.com/mad/2000/hubbleearth.html&quot; rel=&quot;noreferrer&quot;&gt;article on Bad Astronomy&lt;/a&gt; that points out that Hubble frequently points at the sunlight side of Earth to help calibrate its WFPC2 camera.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Still, you may be surprised to find out that Hubble routinely points at the Earth! It uses the bright, daylit Earth to help calibrate one of the cameras on board. The Wide Field Planetary Camera 2 (WFPC2, pronounced ``whiff pick 2'') is an electronic device which detects light, and is similar in principle to a normal digital camera. However, it is far more sensitive, and astronomers are very particular about how well they understand such devices. They want to make sure that every part of the detector is very well calibrated.&lt;/p&gt;&#xA;&lt;p&gt;One way to calibrate the camera is to look at some bright, evenly lit object. If one part of the detector is more sensitive than another, for example, then part of that object will look brighter, even if in reality it isn't. This can be used to correct any later images.&lt;/p&gt;&#xA;&lt;p&gt;In space there aren't any flat fields you can use, but we do have one right here: the Earth. To calibrate WFPC2, sometimes Hubble is pointed straight down at the Earth. As the terrain (or water or whatever) streaks across the field of view it forms a very crude flat field image. The image is called a ``streak flat'', and looks really weird. Trees, houses, all sorts of objects blur across the image. It takes a lot of sophisticated computer processing to turn this into a real flat field for Hubble to use, but luckily there are a lot of smart people working on the problem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;xkcd also shows an approximation of what it would look like if you tried to take a picture of an average desk from orbit. It is just a smear of colors, devoid of any discernible detail. I tried attaching the pictures here, but SE doesn't like the png files for some reason and I don't have time to do the file conversion. Just check out the first link to see what it looks like, plus a more detailed explanation of why.&lt;/p&gt;&#xA;" OwnerUserId="14255" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-04-25T16:47:31.710" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26049" PostTypeId="2" ParentId="26033" CreationDate="2018-04-25T16:53:38.013" Score="4" Body="&lt;p&gt;The rotation of the Earth, UT1 and the Earth Rotation Angle are strictly measured agaist extragalactic light sources. Indeed, nowadays, UT1 is reckoned by using the definition you mentioned, a direct linear relationship with the Earth Rotation Angle between the intermediate celestial and terrestrial frames instead of the actual ficticious mean Sun, because of the impracticality of such a measurement in the context of modern accuracies that are needed by these frames. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a good reference to read about this complex topic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;U. S. Naval Observatory, Precision time and the rotation of the Earth, Dennis D. McCarthy, IAU 2004, &#xA;&lt;a href=&quot;https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S1743921305001377&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S1743921305001377&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a different approach that may also help in the understanding of this topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important thing to keep in mind is that nowadays, with the increasing accuracy needed by many modern applications, we needed reference frames and relationships between them that are as consistent and accurate as possible. For this we reference ourselves on the most stable and easily measurable objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with the Sun was that its location or mean motion cannot feasably be tracked with an accuracy of a few tens of microarcseconds, like quasars. Adopting a definition based on this would have introduced inaccuracies in the definition itself and complexified the relationships, thus undermining the accuracy of the whole system, because UT1-UTC is one of the links between the terrestrial and celestial frames. So the IAU decided to adopt a definition that is &quot;disconnected&quot; from mean solar time, while keeping it reasonably close to the mean Sun by using the adopted ratio. The ratio will be good for decades for civil purposes, and later refinements to keep it reasonably close to the mean Sun while maintaining continuity will be much easier to make anyway. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of imagining the problem is this: say you lived on a boat (Earth) and you needed very accurate measurements of objects around you on the boat and on the sea, say fish, whales, etc (satellites and planets). You see several things around you: the lights of other boats, the lights on a distant shore. Would you prefer to align yourself on another boat's blinding and dancing light (Sun), or on a much more stable and smaller light on the distant shore (quasars), to determine your boat's orientation? Using another moving boat as a reference for coordinate systems would complicate matters and introduce measurement errors while you try to position other objects relative to it. Using the distant shore, however, would make your reference much more stable and accurate, and that does not prevent you from tracking the approximate location of the nearby boat's light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum up, we now have two main reference systems: ITRS (Earth's crust) and ICRS (distant stars), and a link between them (ERA and polar motion) that is as accurate as possible. We keep UT1 close enough to the Sun for all civil purposes, while defining it with a very simple and direct formula related to ERA for accuracy, simplicity and continuity purposes.&lt;/p&gt;&#xA;" OwnerUserId="19814" LastEditorUserId="19814" LastEditDate="2018-04-29T22:27:41.243" LastActivityDate="2018-04-29T22:27:41.243" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26050" PostTypeId="2" ParentId="26018" CreationDate="2018-04-25T19:29:12.833" Score="0" Body="&lt;p&gt;The start and end of a single orbit are when they cross a specified, albeit arbitrary, point.  For satellites, the start of an orbit or rev is when it crosses a boundary like the equator. &lt;/p&gt;&#xA;" OwnerUserId="22828" LastActivityDate="2018-04-25T19:29:12.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26051" PostTypeId="2" ParentId="25812" CreationDate="2018-04-25T19:39:23.320" Score="1" Body="&lt;p&gt;If you live in a city, I'd get binoculars like the others have mentioned.  But I would also plan a trip out into the countryside.  I live 2 hours from a major city and still the sky is dark enough to see thousands of stars, the Milky Way, and even the Andromeda galaxy (faintly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you get binoculars, the first thing I would look at is the Pleiades.  We can see six of the stars with the naked eye, with binoculars, you can see way more in the cluster.  The next thing I would find it he nebula in Orion's belt.&lt;/p&gt;&#xA;" OwnerUserId="22828" LastActivityDate="2018-04-25T19:39:23.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26054" PostTypeId="1" CreationDate="2018-04-26T00:04:44.403" Score="5" ViewCount="165" Body="&lt;p&gt;I'm working on a little side project that traces and identifies galaxies in Hubble's XDF using OpenCV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my current progress with tracing the more prominent galaxies:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/p31Oz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p31Oz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/TVjJw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TVjJw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I was wondering if there was some kind of database for the galaxies that appear in the XDF.&lt;/strong&gt; Like their scientific names, coordinates, distances, and etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a link to the image I am using for the project: &lt;a href=&quot;https://www.nasa.gov/mission_pages/hubble/science/xdf.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.nasa.gov/mission_pages/hubble/science/xdf.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22831" LastEditorUserId="22831" LastEditDate="2018-04-27T08:06:56.030" LastActivityDate="2018-04-28T03:43:12.107" Title="Names of specific galaxies in Hubble's XDF" Tags="&lt;galaxy&gt;&lt;telescope&gt;&lt;hubble-telescope&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26056" PostTypeId="1" CreationDate="2018-04-26T04:55:45.093" Score="2" ViewCount="300" Body="&lt;p&gt;I found the image below in Space.com's article &lt;a href=&quot;https://www.space.com/40406-gaia-release-color-milky-way-map.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This 3D Color Map of 1.7 Billion Stars in the Milky Way Is the Best Ever Made&lt;/a&gt;, although &lt;em&gt;it is not the map mentioned in the title&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The caption for this image reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This radial velocity image shows the movement of 7 billion stars. The colors run from blue (stars moving at 50 km/s toward us) to red (stars moving 50 km/s away from us). The white color shows when, on average, the stars are not moving in the line of sight with respect to us. Stars lagging behind as they orbit the center of the Milky Way appear to be traveling away from us, and those speeding up appear to be traveling toward us. Credit: ESA/Gaia/DPAC&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you imagine a band along the galactic equator the dominant velocity shows two positive and two negative &quot;peaks&quot;, with a zero crossing in the direction of the galactic center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Purely for fun I wanted to see if I could reproduce this behavior with a simple calculation based on a 2D calculation assuming circular motion and a radial density distribution $\rho(r)$ which I could then use to figure out a rotational velocity distribution $v(r)$, bun I swiftly realized that I have no idea what the density profile would look like.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For the purposes of this simple exercise, what would be an analytical expression that roughly matches the Milky Way's radial density profile, projected on to its equatorial plane?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For spherically symmetric distributions, Newton's &lt;a href=&quot;https://en.wikipedia.org/wiki/Shell_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shell theorem&lt;/a&gt; allows one to treat all mass inside a sphere defined by an orbit's radius as if it were at the center, and to ignore all mass in the shell outside of that radius. Is there anything like an analog to this for a radial distribution within a plane?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4Vyep.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4Vyep.jpg&quot; alt=&quot;apparent velocity of the Milky Way seen from Earth (Gaia)&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-04-26T05:01:35.400" LastActivityDate="2018-04-26T16:31:36.620" Title="A rough analytical expression for the Milky Way's radial mass distribution?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;velocity&gt;&lt;mathematics&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26057" PostTypeId="1" AcceptedAnswerId="26059" CreationDate="2018-04-26T05:13:57.747" Score="2" ViewCount="65" Body="&lt;p&gt;I found the image below in Space.com's article &lt;a href=&quot;https://www.space.com/40406-gaia-release-color-milky-way-map.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This 3D Color Map of 1.7 Billion Stars in the Milky Way Is the Best Ever Made&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The caption for this image reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Gaia spacecraft gathered observations for this all-sky view of the Milky Way and neighboring galaxies between July 2014 and May 2016, releasing the data on April 25, 2018. This image shows all the stars' colors and brightness (top), the total density of stars (middle) and the distribution of interstellar gas and dust across the galaxy (bottom). Credit: ESA/Gaia/DPAC&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first two maps reflect data obtained from analysis of GAIA's measurement of starlight as explained in &lt;a href=&quot;https://space.stackexchange.com/a/18237/12102&quot;&gt;this answer&lt;/a&gt;, but the third map is described as &quot;the distribution of interstellar gas and dust across the galaxy.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understand, GAIA does not transmit all complete image data to Earth, but instead does a significant amount of image processing, object detection, classification and analysis, and other data reduction in order to extract positions and &lt;a href=&quot;https://astronomy.stackexchange.com/q/18374/7982&quot;&gt;radial velocity measurements&lt;/a&gt; for point-like objects such as stars, though of course small solar system bodies such as comets and asteroids end up in the raw data as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How is interstellar gas density obtained from GAIA data?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AxdRp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AxdRp.jpg&quot; alt=&quot;various maps of the Milky Way from GAIA&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-04-26T06:27:54.650" Title="How is interstellar gas density mapped from GAIA data?" Tags="&lt;velocity&gt;&lt;interstellar-medium&gt;&lt;gas&gt;&lt;gaia&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26059" PostTypeId="2" ParentId="26057" CreationDate="2018-04-26T06:27:54.650" Score="3" Body="&lt;p&gt;Two techniques immediately spring to mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the stars you detect, you can compare their colours and luminosities (Gaia provides photometric colours and distances) with what you expect for a star of that type at that distance. The difference between what you expect and what you observe tells you the reddening and extinction caused by interstellar dust, integrated in a column towards that star. By going through this process for lots of stars in different directions you build up a 3D picture of the distribution of dust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The measurements are mostly insensitive to gas - simple extrapolation between dust and gas is often assumed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second technique would be just to compare how many stars you can see and with what brightness distribution, with a model for the Galactic stellar population and dust distribution. This is an old-fashioned way of estimating dust extinction, works well locally, but I suspect not what is used in that map, for the simple reasons that Gaia also provides the distances to stars, and one of the purposes of the mission is to improve the Galactic model, not assume it.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-04-26T06:27:54.650" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26060" PostTypeId="2" ParentId="25812" CreationDate="2018-04-26T06:39:10.423" Score="1" Body="&lt;p&gt;I'm too regular person. I find myself in similar situation as I stay in Mumbai. I too don't want to buy any expensive. Here is how I deal with it.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Star Gazing sessions&lt;/strong&gt; away from city.&#xA;I go to star gazing sessions arranged by local astronomy groups every month. I prefer to visit twice a year. One is April and other between Oct to December. This way I cover both summer and winter Sky.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It takes 3 hrs to travel one way to go far away from light pollution. I have been attending this sessions for last 4 years. They have 8 inch and 10 inch Dobsonian reflecting telescopes. Patches of galaxies, nebulae and clusters can be easily seen through it. I made my mom to attend these sessions, she was happy and surprised to see Jupiter and it's moons through telescope. I suggest you to attend these sessions once in a while.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Star Maps or App&lt;/strong&gt;&#xA;Before I goto sleep, I use Star  app to find constellations, asterisms &amp;amp; planets. Thesr apps use real time star map in app for the night. It is helpful to know when they rise and set. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I bought Sky Safari app and been using it for last 2 years. There are amazing free verisons as well like Stellarium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UNQpR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UNQpR.jpg&quot; alt=&quot;Sky Safari App screenshot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Binoculars though can help you looking at stars invisible to naked eye. It's great start but I find using telescopes by attending star gazing sessions more effective way to look stars.&lt;/p&gt;&#xA;" OwnerUserId="21269" LastActivityDate="2018-04-26T06:39:10.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26062" PostTypeId="2" ParentId="25691" CreationDate="2018-04-26T09:23:14.347" Score="0" Body="&lt;p&gt;It is possible the gravity changed locally. Depending on what the asteroid was made of, and the details of what happened in the collision, you might end up with a lump of denser asteroidal material buried in the Earth's crust under the impact crater. That could produce a measurable (although not noticeable to a human) increase in gravity locally.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-04-26T09:23:14.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26063" PostTypeId="2" ParentId="26056" CreationDate="2018-04-26T10:11:27.463" Score="0" Body="&lt;p&gt;@Rob Jeffries mentioned that &quot;You get the density distribution by looking at the velocity data.&quot; I also believe this is what you are looking for, so I will give some calculation details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume spherical symmetry, and circular motion, the gravity equates to the circular motion as&#xA;$$&#xA;\alpha \frac{GMm}{R^2} = \frac{mv^2}{R}&#xA;$$&#xA;where $G$ is gravitational constant, $M$ is the enclosed mass at the radial distance $R$, $v$ is the tangential (not radial) velocity, $m$ is a test mass, and $\alpha$ is a constant for effective potential, which depends on the assumed shape of potential. Then, we can express the mass $M$ as the density profile $\rho$. With the spherical symmetry, the profile $\rho \propto M R^{-3}$. Therefore,&#xA;$$&#xA;\alpha' G \rho R^2 = v^2 .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since observationally we can construct the rotation curve which is $v = f(R)$, the density profile is then a function depending only on $R$: $\rho = g(R)$, i.e., the radial mass distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some notes include i) the mass $M$ includes dark matter; ii) $v$ is the tangential velocity, not the radial velocity as presented in the figure you mentioned.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-04-26T10:11:27.463" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="26065" PostTypeId="2" ParentId="26056" CreationDate="2018-04-26T13:18:19.210" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;For the purposes of this simple exercise, what would be an analytical expression that roughly matches the Milky Way's radial density profile, projected on to its equatorial plane?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The simplest example astronomer's work with is the radial density profile of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Singular_isothermal_sphere_profile&quot; rel=&quot;nofollow noreferrer&quot;&gt;Single Isothermal Sphere&lt;/a&gt; (SIS). It is called this because it is spherically symmetric (and thus applicable to a 2D plane for your purposes) and all objects orbit with the same velocity (and thUs have the same &quot;temperature&quot;, hence isothermal). The density profile takes the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho(r) = \frac{v^2}{4\pi Gr^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $v$ is the rotational velocity. Note that you may see other formulations which use $\sigma_v$ rather than $v$. In this case they're using the &lt;em&gt;velocity dispersion&lt;/em&gt; which is slightly different than the rotational velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other, more realistic density profiles have been found by running simulations of the Universe and matching functional equations to the density profiles of the resulting galaxies. Such popular results are the &lt;a href=&quot;https://en.wikipedia.org/wiki/Navarro%E2%80%93Frenk%E2%80%93White_profile&quot; rel=&quot;nofollow noreferrer&quot;&gt;NFW profile&lt;/a&gt; and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Einasto_profile&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einasto profile&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The NFW profile is a two-parameter function, given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho(r) = \frac{\rho_0}{\frac{r}{R_S}\Big(1+\frac{r}{R_S}\Big)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\rho_0$ and $R_S$ and two, halo-dependent parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Einasto Profile is again a two-parameter model given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho(r) \propto \exp(-Ar^\alpha)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ and $\alpha$ are configurable parameters.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For spherically symmetric distributions, Newton's Shell theorem allows one to treat all mass inside a sphere defined by an orbit's radius as if it were at the center, and to ignore all mass in the shell outside of that radius. Is there anything like an analog to this for a radial distribution within a plane?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Shell theorem for gravity does &lt;em&gt;not&lt;/em&gt; extend to a 2D ring. However, I will say that when talking about the orbits of stars in galaxies, the mass of stars outside a star's orbit are generally considered negligible. The primary reason for this is that it is Dark Matter which comprises most of the mass of a galaxy and contributes the most to defining a star's orbit in a galaxy. The Dark Matter halo is often assumed to be spherically symmetric, in which case Newton's Shell theorem does apply and the mass you're concerned with in determining a star's orbit, is the mass of the Dark Matter halo interior to the star's orbit.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2018-04-26T16:31:36.620" LastActivityDate="2018-04-26T16:31:36.620" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26066" PostTypeId="1" AcceptedAnswerId="26072" CreationDate="2018-04-26T14:12:26.973" Score="3" ViewCount="64" Body="&lt;p&gt;Can a solar flare reach far enough to travel from one ordinary star to another?  If so, how can that solar flare affect the following star?  &lt;/p&gt;&#xA;" OwnerUserId="22339" LastEditorUserId="22339" LastEditDate="2018-04-26T15:27:42.210" LastActivityDate="2018-04-26T17:52:43.770" Title="Is it possible for a solar flare in one star to influence the chances of a solar flare in another star?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;solar-flare&gt;&lt;interstellar&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26067" PostTypeId="1" CreationDate="2018-04-26T14:56:45.587" Score="4" ViewCount="313" Body="&lt;p&gt;The latest info on black holes with regard to their formation is that since space and time go to infinity at the event horizon, from an outside perspective nothing ever enters a black hole, and it also takes an infinite amount of time for the black hole actually to form. Thus, from an outside perspective, no black hole ever is completely finished becoming a black hole, and because of Hawking Radiation, every black hole will evaporate before it fully becomes a black hole, so there will never actually be any fully formed black holes in the universe. Fine, no problem. Question is, since nothing ever enters a black hole, do black holes actually grow in size (that is, the event horizon getting bigger), or are they forever restricted to their original size (apart from two black holes combining to form a larger event horizon)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not about matter accreting on the event horizon; that's the issue, in fact. Rather than crossing the EH and entering the BH, matter seems to accrete on the EH and never goes in. So the BH never grow, the EH never gets larger due to accretion of matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outside observer is not infinitely far away; it's space-time that goes to infinity at the Event Horizon. In fact, it seems to be true that the closer one gets to the EH, the further away one is and the longer it takes to get there, approaching being infinitely far away in both space and time.&lt;/p&gt;&#xA;" OwnerUserId="22841" LastEditorUserId="22841" LastEditDate="2018-04-27T21:20:51.690" LastActivityDate="2018-04-28T11:15:10.570" Title="Black Hole growth" Tags="&lt;black-hole&gt;&lt;event-horizon&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2018-05-01T05:29:05.587" ContentLicense="CC BY-SA 3.0" />
  <row Id="26068" PostTypeId="2" ParentId="26067" CreationDate="2018-04-26T15:58:11.803" Score="3" Body="&lt;p&gt;Good question. I am reminded of Einstein's 1939 paper &lt;a href=&quot;http://www.cscamm.umd.edu/tiglio/GR2012/Syllabus_files/EinsteinSchwarzschild.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;on a stationary system with spherical symmetry consisting of many gravitating masses&lt;/a&gt;. He said &lt;em&gt;“it is easy to show that both light rays and material particles take an infinitely long time (measured in &quot;coordinate time&quot;) in order to reach the point r = μ/2&quot;&lt;/em&gt;. That would suggest black holes can never form, but we have evidence of black holes, see &lt;a href=&quot;https://arxiv.org/abs/0810.4674&quot; rel=&quot;nofollow noreferrer&quot;&gt;Monitoring stellar orbits around the Massive Black Hole in the Galactic Center&lt;/a&gt; by Gillesan et al dating from 2008. So what gives? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I rather think the clue is in the original &quot;frozen star&quot; black hole concept. See Robert Oppenheimer and Hartland Snyder's 1939 paper &lt;a href=&quot;https://journals.aps.org/pr/abstract/10.1103/PhysRev.56.455&quot; rel=&quot;nofollow noreferrer&quot;&gt;on continued gravitational contraction&lt;/a&gt;. Also see the  1971 Physics Today article &lt;a href=&quot;http://authors.library.caltech.edu/14972/1/Ruffini2009p1645Phys_Today.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;introducing the black hole&lt;/a&gt; by Remo Ruffini and John Wheeler. They said in a sense the system is a frozen star, and in another sense the system is not frozen at all. They also said &lt;em&gt;“according to figure 3, the fall of a test particle towards a black hole ends at r=2m as seen by a distant observer. The fall ends at r=0 according to someone falling with the test mass itself. How can two different versions of the truth be compatible?”&lt;/em&gt; They can’t.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some would say that this frozen-star black hole cannot be correct because nothing passes through the event horizon, and therefore black holes cannot grow. After all, that’s more or less what Einstein said. But think in terms of a hailstone. Imagine you're a water molecule. You alight upon the surface of the hailstone. You can't pass through this surface. But you are presently surrounded by other water molecules, and eventually buried by them. So whilst you can't pass through the surface, &lt;em&gt;the surface can pass through you&lt;/em&gt;. So the frozen-star black hole grows like a hailstone. &lt;/p&gt;&#xA;" OwnerUserId="9772" LastEditorUserId="2531" LastEditDate="2018-04-27T23:56:59.327" LastActivityDate="2018-04-27T23:56:59.327" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26070" PostTypeId="2" ParentId="26046" CreationDate="2018-04-26T16:45:19.147" Score="-2" Body="&lt;p&gt;For simulation only, you should be able to show it by simple code as following:&#xA;$$&#xA;\vec{a}_t = - \frac{GM}{r_t^2} \hat{r} \\&#xA;\vec{v}_{t+1} = \vec{v}_t + \vec{a}_t \Delta t \\&#xA;\vec{r}_{t+1} = \vec{r}_t + \vec{v}_t \Delta t &#xA;$$&#xA;where $t$ is time step, $\vec{a}$ is acceleration, $\vec{v}$ is velocity, $\vec{r}$ is position, $\hat{r}$ is a unit vector in radial direction from the origin (set the Sun at the origin), $G$ is gravitational constant, $M$ is Sun mass in this case, and $\Delta t$ is size of time step. You will have to play around with the initial values and step size to get certain orbit representing the case. To make it more real, you derive the initial values from the real parameters, if known.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastEditorUserId="22468" LastEditDate="2018-04-26T16:52:31.933" LastActivityDate="2018-04-26T16:52:31.933" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26071" PostTypeId="1" CreationDate="2018-04-26T17:38:06.827" Score="2" ViewCount="183" Body="&lt;p&gt;I'm working in my final course project and I'm using the recently released &lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR?-source=I/345&quot; rel=&quot;nofollow noreferrer&quot;&gt;GAIA DR2&lt;/a&gt; data. I want to determine open clusters membership and I know this has mainly been done  using proper motions and positions, and maybe photometric data.&lt;br&gt;&#xA;Since now we have access to parallaxes I was wondering if one could also estimate the cluster membership from them, considering all cluster members must have approximately the same parallax value (if they are to be at the same distance from us). I guess the technique can't be much different to the usual one done with proper motions, but I can't work it out in a rigorous way. Any help is appreciated! Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="22595" LastActivityDate="2018-04-26T17:38:06.827" Title="Open cluster membership determination via parallaxes" Tags="&lt;parallax&gt;&lt;star-cluster&gt;&lt;gaia&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26072" PostTypeId="2" ParentId="26066" CreationDate="2018-04-26T17:52:43.770" Score="3" Body="&lt;p&gt;I'd say it's an open question awaiting some observational evidence that it can happen.  It presumably would depend a lot on what kind of situation you are looking for, and what would you call a flare on one star setting off a flare on the other.  Certainly you'd need two stars that both have active atmospheres and are in a close binary, such that one mass ejection could interfere enough with the magnetic configuration in the other star to set off a flare.  Some stars may be so close together that the magnetic configuration of their atmospheres actually connect both stars, so then when get a flare you might see brightenings on both stars but it would be essentially the same flare.  There are even stars called &quot;contact binaries&quot; where it's not even all that clear if you should think of that as two stars that touch (and share a magnetosphere), or just one star that has two separate cores.  But the bottom line is, it sounds like you'd need something unusual going on, involving two stars that were unusually close together and are interacting in some way.  You might want to read up on &quot;RS CVn stars,&quot; for example.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-04-26T17:52:43.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26078" PostTypeId="1" CreationDate="2018-04-27T11:34:18.683" Score="3" ViewCount="533" Body="&lt;p&gt;I was reading about the &lt;a href=&quot;http://earthsky.org/astronomy-essentials/five-petals-of-venus&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pentagram/Flower of Venus&lt;/a&gt;, and how it is caused by a 13:8 ratio between their orbits. I found &lt;a href=&quot;https://i.imgur.com/PqJy8lX.gifv&quot; rel=&quot;nofollow noreferrer&quot;&gt;this GIF particularly fascinating&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, if the ratio changed, would they form other kind of &quot;fractal&quot; shapes? Does this happen between other planets? What would be the condition for the ratio so that the relative motion forms a symmetrical shape like the Cardioid in the case of Venus and Earth?&lt;/p&gt;&#xA;" OwnerUserId="22852" LastActivityDate="2018-04-30T16:12:43.787" Title="Does the relative motion between planets other than Venus and Earth form shapes such as the Flower of Venus?" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;venus&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26079" PostTypeId="1" CreationDate="2018-04-27T13:58:24.347" Score="4" ViewCount="106" Body="&lt;p&gt;I am new to using SPICE and the &lt;a href=&quot;https://naif.jpl.nasa.gov/naif/toolkit.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;SPICE Toolkit&lt;/a&gt;.  My goal is to extract body radius, albedo, and distance from the sun of a number of solar system objects.  The tools provided by NAIF are great and I have been able to get the information items I need from HORIZONS.  I would like to use mice and the right selection of kernels to construct a table of body information.  I am unable to locate the geometric albedo.  It must be in there somewhere since HORIZONS reports it in its body summary (example below).  Please point me to which kernel(s) the information in the example is extracted from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Horizons&gt; 500&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;JPL/HORIZONS                500 Selinur (1903 LA)          2018-Apr-26 15:32:09&#xA;Rec #:     500 (+COV) Soln.date: 2018-Apr-05_11:16:59   # obs: 1803 (1903-2018)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IAU76/J2000 helio. ecliptic osc. elements (au, days, deg., period=Julian yrs): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EPOCH=  2455485.5 ! 2010-Oct-16.00 (TDB)         Residual RMS= .28963&lt;br&gt;&#xA;   EC= .1453222754778464   QR= 2.232822954662918   TP= 2455876.0720004463&lt;br&gt;&#xA;   OM= 290.0072894804399   W=  75.13840817439457   IN= 9.764544808931515&lt;br&gt;&#xA;   A= 2.612473556522465    MA= 268.8351702646443   ADIST= 2.992124158382012&lt;br&gt;&#xA;   PER= 4.22266            N= .233413627           ANGMOM= .02750885&lt;br&gt;&#xA;   DAN= 2.46541            DDN= 2.65631            L= 4.9371039&lt;br&gt;&#xA;   B= 9.434865500000001    MOID= 1.25316           TP= 2011-Nov-10.5720004463  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asteroid physical parameters (km, seconds, rotational period in hours):&lt;br&gt;&#xA;   GM= n.a.                RAD= 20.414             ROTPER= 8.0111&lt;br&gt;&#xA;   H= 9.3                  G= .150                 B-V= n.a.&lt;br&gt;&#xA;                           ALBEDO= .202            STYP= n.a.                  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ASTEROID comments: &#xA;1: soln ref.= JPL#41, OCC=0&#xA;2: source=ORB&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Select ... [A]pproaches, [E]phemeris, [F]tp,[M]ail,[R]edisplay, [S]PK,?,: &lt;/p&gt;&#xA;" OwnerUserId="22757" LastEditorUserId="7982" LastEditDate="2018-04-30T09:02:12.980" LastActivityDate="2018-06-01T20:13:27.070" Title="Which kernel does HORIZONS get the Geometric Albedo from?" Tags="&lt;asteroids&gt;&lt;ephemeris&gt;&lt;albedo&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26080" PostTypeId="1" CreationDate="2018-04-27T14:30:55.697" Score="-4" ViewCount="472" Body="&lt;p&gt;[Edited to remove incorrect geological assumptions]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun is self-rotating, i.e. it spins on its axis. When something spins, there's an apparent force away from the centre, and matter gets flung out. The sun's surface is gas; it follows that beneath the gas surface there could only be void with the density less than the surface. So, is the sun hollow, and if it isn't, what's the reason that the centrifugal force doesn't spin the matter outwards?    &lt;/p&gt;&#xA;" OwnerUserId="17330" LastEditorUserId="14094" LastEditDate="2018-04-28T16:12:34.620" LastActivityDate="2018-04-28T16:12:34.620" Title="Why isn't the Sun hollow?" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;rotation&gt;&lt;density&gt;" AnswerCount="2" CommentCount="5" ClosedDate="2018-04-28T08:50:03.810" ContentLicense="CC BY-SA 3.0" />
  <row Id="26081" PostTypeId="2" ParentId="26080" CreationDate="2018-04-27T14:43:01.807" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;As the surface of the earth is solid the crust,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Correct. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;underneath the crust the mantle which is liquid,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No the mantle is also solid, although more plastic than the crust. There is a liquid core below the mantle&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the liquid density is smaller than the crust. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No the core is made of iron and considerably denser than the mantle rocks above it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This way the earth can exercise self-rotation. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know what you mean. Fully solid bodies, liquid bodies, gaseous bodies can all rotate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Therefore, since the sun also self-rotating, the sun's surface is gas; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Actually plasma, but that is a detail&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;it follows that beneath the gas surface there could only be void which the density smaller than gas, otherwise it wouldn't be possible for the sun to exercise self-rotation. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. This doesn't follow. It is not the case that there is a void, for if there were, the plasma would flow into it. In fact density increases with depth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So, is the sun hollow? Or any reason that the sun can only be hollow inside?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No the sun is not hollow.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-04-27T14:43:01.807" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="26082" PostTypeId="1" AcceptedAnswerId="26088" CreationDate="2018-04-27T15:27:16.610" Score="5" ViewCount="362" Body="&lt;p&gt;I recently watched this video on black holes:&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=ulCdoCfw-bY&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=ulCdoCfw-bY&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that statement should give you an idea of my level of understanding :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if a spinning black hole A could complete a Penrose process through another spinning black hole B's ergosphere. i.e. if it could travel through the ergosphere of another spinning black hole and emerge out of it the other side.  What would happen?  Perhaps this is related to the question of whether black holes can orbit each other (in two or three-body system)?  What happens if black holes orbit each other, but their ergospheres overlap?&lt;/p&gt;&#xA;" OwnerUserId="22854" LastActivityDate="2019-01-16T20:43:08.147" Title="Can a black hole undergo a Penrose process inside another black hole's ergosphere?" Tags="&lt;black-hole&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26083" PostTypeId="1" CreationDate="2018-04-27T15:27:42.817" Score="5" ViewCount="148" Body="&lt;p&gt;Excuse an amateur question, as inspired by this question: &quot;&lt;a href=&quot;https://astronomy.stackexchange.com/q/25835/17330&quot;&gt;How old is the oldest light visible from Earth?&lt;/a&gt;&quot;, how to determine the age of light? Does light get old? Does light have characteristic therefore we know this light &quot;A&quot; is different from that light &quot;B&quot; therefore we can track how it going over time? Like tagging animals to observe their activities? How to differentiate one light from the other?     &lt;/p&gt;&#xA;" OwnerUserId="17330" LastActivityDate="2018-04-28T21:49:12.953" Title="What's the difference of the oldest light and newest light?" Tags="&lt;light&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26084" PostTypeId="2" ParentId="26083" CreationDate="2018-04-27T15:54:39.570" Score="1" Body="&lt;p&gt;According to that context, the old light means the light that travelling from far away in order to reach the observer, and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine the age of light, standard method is to use spectral features (absorption/emission but we will talk about the example of emission only without the loss of generality).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that the observer detects a photon which he is very sure that it must be H$\alpha$ emission (i.e., 656.28 nm at rest frame). But, the observer detects a 700.00-nm photon instead of at that rest-rame wavelength. This is because the photon travelling through the universe and experience time dilation; in general, the longer time travelling, the bigger the observed wavelength is, relative to the rest-frame value. Since we have the relationship $\lambda_{rest} * (1 + z) = \lambda_{obs}$ where $\lambda_{rest}$ ($\lambda_{obs}$) is the rest-frame (observe-frame) wavelength, and $z$ is the redshift, we can find the redshift from the known rest-frame and observe-frame wavelengths. Then, the higher redshift means farther away, means older.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-04-27T15:54:39.570" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26085" PostTypeId="1" CreationDate="2018-04-27T16:21:09.997" Score="1" ViewCount="61" Body="&lt;p&gt;With all of the new NEOs (near-earth objects) are they being pulled in or are they interacting with each other? &lt;/p&gt;&#xA;" OwnerUserId="22856" LastEditorUserId="14094" LastEditDate="2018-04-28T13:02:02.867" LastActivityDate="2018-04-28T13:02:02.867" Title="New NEOs are found daily. Is there any evidence gravity is pulling them to a centralised location?" Tags="&lt;comets&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26086" PostTypeId="2" ParentId="26085" CreationDate="2018-04-27T17:41:16.997" Score="3" Body="&lt;p&gt;No, they are being discovered where they are in their (usually fairly stable) orbits. They are not discovered because they are arriving or going to some particular location. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one exception is long-periodic comets that have fallen in from the Oort cloud and now pass through the inner parts of the solar system, maybe to be captured by the gravity of Jupiter in a closer orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Strictly speaking all these objects are all pulled towards a central location, the Sun, but because they already have a sideways velocity that just means an orbiting motion where they are constantly falling around it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NEOs are too light and small to interact much with each other - the gravitational influence between them is microscopic compared to the influence from the planets which is usually microscopic compared to the sun's influence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason we are seeing a lot more NEOs now is that new instruments are better and more &lt;a href=&quot;https://cneos.jpl.nasa.gov/stats/site_all.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;searches are being conducted&lt;/a&gt;. &lt;a href=&quot;https://en.wikipedia.org/wiki/Lincoln_Near-Earth_Asteroid_Research&quot; rel=&quot;nofollow noreferrer&quot;&gt;LINEAR&lt;/a&gt; for example found  231,082 small Solar System bodies (of which 2,423 are NEOs) in the early 2000s and the Catalina Sky Survey has found 6,515 NEOs since 2009.&lt;/p&gt;&#xA;" OwnerUserId="19816" LastEditorUserId="19816" LastEditDate="2018-04-28T06:47:29.107" LastActivityDate="2018-04-28T06:47:29.107" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26087" PostTypeId="2" ParentId="26083" CreationDate="2018-04-27T17:55:35.957" Score="6" Body="&lt;p&gt;There is no measurable difference between an &quot;old&quot; photon and a &quot;new&quot; one. A photon has a wavelength, a polarisation, and a direction of travel. And no other features. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may ask if it is meaningful to discuss the age of light. It is certainly possible to say that photons were emitted by a certain body 100 years ago, and these photons are being detected now, and so this light is 100 years old. But it is impossible to measure the age as a property of the light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't know of any characteristic that uniquely specifies the &quot;age&quot; of the light, so we just assume it's travelled from as far away as we think the object that generated it is, and call the time to travel that far it's &quot;age&quot;.  In the case of the Cosmic microwave background, this was (we think) released when the universe first became transparent, a few hundred thousand years after the &quot;big bang&quot;. The light from this far back has been &quot;stretched&quot; by the expansion of the universe, but not changed by time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is normally true of many simple objects. They either exist or not. A proton doesn't age. A neutron may decay, but this is an event, not a process. It is only with complex objects made of many particles that they can change with age while maintaining their identity.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-04-28T21:49:12.953" LastActivityDate="2018-04-28T21:49:12.953" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26088" PostTypeId="2" ParentId="26082" CreationDate="2018-04-27T18:11:36.393" Score="5" Body="&lt;p&gt;Black holes can orbit each other (we know this from the gravitational waves from their eventual mergers) and there is no particular reason against a small black hole passing through the ergosphere of a large black hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main problem is avoiding a merger. The the two-black hole spacetime has no analytic solution and numeric simulations are messy, but a merger will definitely happen if the event horizons overlap. In any case, a rotating black hole has an ergosphere with an equatorial radius of $2GM/c^2$ and event horizon somewhere between this and the limiting radius $GM/c^2$. If we use the extremal case, then we can presumably just barely fit in nearly an identical mass and rotation black hole in the ergosphere. More likely a much smaller black hole could fit in. The direction of rotation &lt;em&gt;would&lt;/em&gt; matter for the dynamics, since counter-rotating black holes would tend to coalesce (see &lt;a href=&quot;https://duetosymmetry.com/tool/kerr-calculator-v2/&quot; rel=&quot;noreferrer&quot;&gt;this page&lt;/a&gt; for a plot of the relevant radii and where the &lt;a href=&quot;https://en.wikipedia.org/wiki/Innermost_stable_circular_orbit&quot; rel=&quot;noreferrer&quot;&gt;innermost stable circular orbits&lt;/a&gt; are). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single black hole passing through the ergoregion would not be able to perform the Penrose process, but we can imagine a binary black hole pair that is ripped apart by tidal forces and then presumably the remaining hole would be ejected with a lot of energy. So, yes, we could get fast small black holes this way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice this system would lose a lot of energy through gravitational waves, making it likely that the holes would coalesce anyway. The exact limits require numerical relativity simulations to estimate, and this one would be a tough one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that the gravitational waves also carry away energy, up to 1/8 of the total mass of two equal-sized holes merging, but this energy is very hard to capture. Penrose did note in his classic paper that Misner was first in proposing black hole mining using a variant of this.)&lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2018-04-27T18:11:36.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26089" PostTypeId="2" ParentId="26054" CreationDate="2018-04-27T20:06:12.940" Score="1" Body="&lt;p&gt;After speaking to some people on Astronomy discords, they told me about SINBAD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://simbad.u-strasbg.fr/simbad/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22831" LastActivityDate="2018-04-27T20:06:12.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26090" PostTypeId="2" ParentId="26080" CreationDate="2018-04-27T20:11:46.077" Score="2" Body="&lt;p&gt;With most celestial bodies above a certain size limit, you can safely assume they only become more and more dense the closer you get to the center. This is true for most planets and it's definitely true for the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the aggregation state (solid, liquid, gas) does not matter for the spin (I believe that's what you mean by &quot;self-rotation&quot;). If it's spinning, then it's spinning, no matter what it's made of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What keeps the Sun in one piece is only its gravity, nothing else. The various parts of the Sun are compressed together by the overall gravity of this object. Spin doesn't matter here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spin would matter if the Sun was spinning incredibly fast. But that's not the case, it's not even close; it would have to spin by orders of magnitude faster in order to break up because of that. Like, thousands of times faster, I'm not even sure what the number is but it's huge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, aggregation states are not very important at cosmic scales. Planets like the Earth are technically &quot;solid&quot;, but they behave more like fluids, or more like piles of sand, during an impact. Also, the &quot;solid&quot; parts of the Earth actually do flow like liquids over many millions of years. There are no clearly defined rules here.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2018-04-27T20:11:46.077" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26091" PostTypeId="2" ParentId="26067" CreationDate="2018-04-28T00:30:46.167" Score="2" Body="&lt;p&gt;TL;DR look at the diagram at the end of the post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following the &lt;em&gt;Principle of Relativity,&lt;/em&gt; the laws of physics are the same everywhere and for every observer.  Despite that, measurements of observers can differ, as long as they are consistent within the rules of general relativity.  Furthermore, horizons limit what can be measured by some observer.  In this spirit, let's do a Gedankenexperiment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take an off-the-shelf black hole spacetime: non-rotating, uncharged and spherically symmetric with mass $M$.  Add some matter that is falling towards the black hole: a thin shell of gas to keep the spherical symmetry (very low density gas, such that we can ignore heating and other accretion processes).  Let's throw in a beacon, that follows the gas shell and emits light signals in constant intervals.  What happens?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Freely Falling Observer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This is actually really simple to describe in words:  Everything crashes into the singularity (or whatever garbage dump might be left over from all the spaceships, astronauts, beacons and hypothetical particles that other physicists have thrown in there).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The curvature of spacetime is increasing towards the singularity and &lt;em&gt;there is no other curvature singularity other than the center of the black hole.&lt;/em&gt;  This means, that the gas and the beacon smoothly fall straight through the event horizon and from their point of view there is nothing extraordinary about that (apart from the extreme gravitational lensing and the noticeable &lt;a href=&quot;https://en.wikipedia.org/wiki/Spaghettification&quot; rel=&quot;nofollow noreferrer&quot;&gt;tidal forces&lt;/a&gt;).  Remarkably, they reach the central singularity in &lt;em&gt;finite time,&lt;/em&gt; i.e. the beacon emits a finite number of signals before it hits the singularity.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Outside Observer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;You have specified an outside observer at a fixed position, but then just go to a fixed position that is so far away, that we can call it infinitely far away.  This is mathematically convenient for the description and one of the most common observers chosen, because you don't feel the gravitation of the black hole anymore and your everyday spacetime coordinates are exactly the coordinates of the Schwarzschild metric.  This infinity is only approximated for convenience and has nothing to do with the infinity at the event horizon.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This outside observer sees the gas fall towards the black hole.  The closer it gets to the event horizon, the more it seems to slow down and the fainter it becomes.  The reason is, that the light needs more and more time to leave the gravitational well of the black hole.  Light emitted exactly at the event horizon would need infinitely long to reach this observer.  The signals that the beacon emitted before it crossed the event horizon (in its own proper time of the freely falling observer) are stretched longer and longer into infinity and the total number of signals the observer can ever measure is still finite.  Basically the life of the beacon can be divided into an observable part and an unobservable part.  The observable part is stretched into infinity such that we say that the beacon is slowing down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saying that light takes longer is not entirely correct and in some way similar to saying that there is more space, because the speed of light is constant.  This is described correctly by the Schwarzschild coordinates, which have a &lt;em&gt;coordinate singularity&lt;/em&gt; at the event horizon.  This singularity is &lt;em&gt;not intrinsic to the spacetime&lt;/em&gt; (not a curvature singularity) but arises from the choice of this coordinate system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This description is equally valid and describes the correct physical behavior, just as the freely falling observer.  The apparent paradox of these two different observations can be cleared up by a careful look at the transformation between the observers like many other supposed paradoxa from special relativity.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Growing the Event Horizon&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now we come to the heart of the question: the change of the event horizon.  The black hole has an event horizon at&#xA;$$ R_1 = \frac{2 G M}{c^2}. $$  There is a second &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schwarzschild radius&lt;/a&gt; of the mass of the original black hole $M$ combined with the total gas mass $m$: $$ R_2 = \frac{2 G (M + m)}{c^2} &amp;gt; R_1. $$  The outside observer sees the gas approximating the outer event horizon $R_2$.  To see why, change to the freely falling observer and check what happens to the light emitted by the beacon when it passes the outer event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expansion of the Schwarzschild radius from $R_1$ to $R_2$ is not something you see from the gas shell itself.  However, &lt;strong&gt;other accreting stuff that fell in before is eclipsed!&lt;/strong&gt; Imagine a reckless astronaut who coasted into the black hole before we launched our gas shell.  From our vantage point we see him or her approximating the event horizon $R_1$.  But at some point the gas shell is so close to the old event horizon, that the new outer event horizon hinders some of the light that previously could have reached us to escape.  Still the image of the astronaut doesn't disappear completely, you can only observe less of the astronaut's proper time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might be more easily explained graphically, so I made a sketch.  It is in Schwarzschild coordinates and the most important feature is, that the light cone closes up and tilts towards the singularity the closer you get to the event horizon.  Exactly at the event horizon the light cone is not well defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light from the astronaut emitted at point A is visible to us, but delayed by the gas shell.  Light emitted by Light emitted at point B is extremely delayed and close to the limiting case of which moment will be stretched into infinity.  Light from point C could have reached us without the additional mass, but now it is bound by the additional mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think that there is a very nice description of how the light from point C ends up inside the black hole, because the coordinate singularity is not at a fixed point anymore and non-vacuum black hole solutions are tricky in any case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Vs7Nc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vs7Nc.jpg&quot; alt=&quot;Spacetime diagram in Schwarzschild coordinates&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, you could only see the gas shell forever, if never again any other matter would fall in and conceal it.  The hailstone analogy in the &lt;a href=&quot;https://astronomy.stackexchange.com/a/26068/22239&quot;&gt;answer by John Duffield&lt;/a&gt; illustrates this behavior of the event horizon growing by passing through the accreting matter quite nicely.&lt;/p&gt;&#xA;" OwnerUserId="22239" LastEditorUserId="22239" LastEditDate="2018-04-28T11:15:10.570" LastActivityDate="2018-04-28T11:15:10.570" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26092" PostTypeId="2" ParentId="26054" CreationDate="2018-04-28T03:43:12.107" Score="2" Body="&lt;p&gt;The &lt;a href=&quot;https://archive.stsci.edu/prepds/xdf/&quot; rel=&quot;nofollow noreferrer&quot;&gt;XDF data release&lt;/a&gt;&#xA;has images in FITS format, which includes a world coordinate system to map image coordinates to celestial coordinates.&#xA;Then you can &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-fcoo&quot; rel=&quot;nofollow noreferrer&quot;&gt;query SIMBAD&lt;/a&gt;&#xA;to see what galaxies are near those coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I loaded a WFC 60mas/pixel image into a FITS viewer,&#xA;which indicates that the face-on spiral at center left of the PR image&#xA;is near J2000 3h32m40.8s -27d46m16s.&#xA;(To align the PR image with the equatorial coordinate grid, rotate it about 50 degrees clockwise.)&#xA;SIMBAD found &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=%407814928&amp;amp;Name=%5bRTG2015%5d%2024587&quot; rel=&quot;nofollow noreferrer&quot;&gt;this galaxy&lt;/a&gt; at that location, B-band magnitude 22.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this in Python code, you could try using&#xA;&lt;a href=&quot;http://docs.astropy.org/en/stable/wcs/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;astropy.wcs&lt;/a&gt;&#xA;to get celestial coordinates from the FITS image and&#xA;&lt;a href=&quot;http://astroquery.readthedocs.io/en/latest/simbad/simbad.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;astroquery.simbad&lt;/a&gt;&#xA;to look up objects of interest.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-04-28T03:43:12.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26095" PostTypeId="1" AcceptedAnswerId="26154" CreationDate="2018-04-28T11:31:36.570" Score="2" ViewCount="164" Body="&lt;p&gt;I am using this index page to download GDR2 data files.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cdn.gea.esac.esa.int/Gaia/gdr2/gaia_source/csv/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://cdn.gea.esac.esa.int/Gaia/gdr2/gaia_source/csv/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it keeps giving a different number of links. Sometimes it shows 16,000 files, sometimes 31,000 and the max I have seen is 61,237. What is the correct number I should download? GAIA GDR2 documentation does not list the total number of files available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, if the index lists less than 60k files, it still has the closing /html tag so it would appear I have a complete index file, but the contents are not complete.&lt;/p&gt;&#xA;" OwnerUserId="22867" LastEditorUserId="22867" LastEditDate="2018-04-28T12:52:06.203" LastActivityDate="2019-11-26T11:37:38.623" Title="How many GAIA Data Release 2 files are there?" Tags="&lt;gaia&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26098" PostTypeId="2" ParentId="26033" CreationDate="2018-04-28T20:47:21.070" Score="5" Body="&lt;p&gt;This question asks in effect: Is there a discrepancy, perhaps progressive, between the official time-measure UT1, and some other credible measure of mean solar time? This is harder to answer than it might first seem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;UT1 is the current official representative of mean solar time: it is doubtful that any independent measure of mean solar time (other than direct relatives of UT1 such as UT0 or UT2) has been recently defined or computed to any precision. (The method of the equation of time as normally carried out is only roughly approximate, because it ignores the many solar perturbations discovered since the early 18th-c. and its results may vary from accurate mean solar time by up to about 3 seconds: see Hughes et al. (1989), &quot;The equation of time&quot;,  &lt;a href=&quot;http://adsabs.harvard.edu/abs/1989MNRAS.238.1529H&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/1989MNRAS.238.1529H&lt;/a&gt; .)  But as the question indicates, one reason for questioning UT1 is that solar time naturally depends on two independent variables, related to the angular rates of the earth's axial rotation and of the earth's orbital motion around the sun:  In contrast, the official UT1 calculation appears to condense these into a single constant ratio, thus losing one of the independent variables of the natural physical model. This may well raise suspicion that UT1 deviates or will deviate from a more physically-defined mean solar time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is in fact a (no-longer-official) alternative and physical basis for calculating mean solar time. This can be gathered from the history of 19th-century mean solar time definitions and calculations. At the same time, there is also some documentation from the history to indicate the origin and nature of some of the steps taken in the 20th/21st-century, to change the old 19th-century physical model to the current (and very non-transparent!) basis of calculation for UT1, which is described in: (N Capitaine et al. (2003), &quot;Expressions to implement the IAU 2000 definition of UT1&quot;, Astronomy &amp;amp; Astrophysics 406, 1135-1149, &lt;a href=&quot;http://adsabs.harvard.edu/abs/2003A%26A...406.1135C&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2003A%26A...406.1135C&lt;/a&gt; ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Classical method for determining mean solar time (Nautical Almanac, 19th-c.):&lt;/strong&gt;  (Not discussed here is the method of the equation of time, which grew more cumbersome as perturbations of the apparent solar motion were successively discovered during the 18th-c.; all of them had to be included in precise calculations.)  A simpler equivalent method for mean solar time, to a suitable astronomical precision for its period, was provided by Nevil Maskelyne's tables of 1774 (found at &lt;a href=&quot;http://docs.lib.noaa.gov/rescue/rarebooks_1600-1800/QB12M31774.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://docs.lib.noaa.gov/rescue/rarebooks_1600-1800/QB12M31774.pdf&lt;/a&gt;), especially his table of the Sun's mean longitude converted from degrees to hours and given to a precision of 0.01 time-second. His instructions showed how to use this to convert observed sidereal time to mean solar time. It may be unclear whether Maskelyne intended this method to make a standard out of Greenwich mean time, but that seems to be what happened. Thus initially, according to J J Lalande's 'Astronomie' (3 editions, Paris, 1764, 1771, 1792) methods for mean time still relied usually on the equation of time, but Lalande's third edition of 1792 (vol.1, &lt;a href=&quot;https://archive.org/details/astronomielalande01lala&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://archive.org/details/astronomielalande01lala&lt;/a&gt; , see art.1014-5, p.361) added a mention of Maskelyne's table, commenting that 'in England' it was used to find the mean time of an observation. The Nautical Almanac eventually began (from 1833) to tabulate the relation between sidereal and mean solar time (Sidereal time at (Greenwich) mean noon), and during the rest of the 19th-c. such tabulations provided what became standard observatory methods of finding mean solar from sidereal time (see J C Adams 1884, &quot;On the Definition of Mean Solar Time&quot;, Observatory 7, 42-44, &lt;a href=&quot;http://adsabs.harvard.edu/abs/1884Obs.....7...42A&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/1884Obs.....7...42A&lt;/a&gt; ). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method has a simple physical basis in spherical astronomy, using a division into two parts of the arc of the celestial equator that extends from the equinoctial point eastwards to the point culminating at the meridian -- shown in the diagram below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FRoCb.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FRoCb.gif&quot; alt=&quot;equatorial solar-time diagram&quot;&gt;&lt;/a&gt;&#xA;(The upper part of the diagram represents the celestial equator above the horizon (marked E W) for an observer in the northern hemisphere looking south. M is the culminating point where the equator intersects the (south) meridian.  Q is the equinoctial point 'first point of Aries'. QS represents the sun's mean longitude reckoned from the equinox: it has been (in concept) transferred from the ecliptic to the equator where it now represents the mean right ascension of the fictitious mean sun. For convenience of arrangement, the diagram represents a position in early afternoon (S somewhat west of M) in about June (S is nearly 90° east of Q). (N) is below the horizon.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time-definitions and explanations of J-J de Lalande (1792, 'Astronomie', 3rd ed. vol.1, art.1014, p.361, &lt;a href=&quot;https://archive.org/details/astronomielalande01lala&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://archive.org/details/astronomielalande01lala&lt;/a&gt;) state clearly (expressing right ascensions in degrees, according to Lalande's custom):-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The mean longitude of the sun, plus the mean time converted to degrees, gives the right ascension of the mid-heaven.&quot;   That translates, in terms of the diagram here, to:  QS + MS = QM .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lalande also pointed out (art.1015) that &quot;The Sun's right ascension, or that of the star used to find the right ascension of the midheaven, should be counted from mean equinox like the mean longitude of the sun ....&quot;. Modern readers have to note further that (a) observed and tabulated 'sidereal time' up to about the end of the 19th century always included the effect of nutation in right ascension, i.e. it was 'apparent' sidereal time (see explanation in Nautical Almanac 1864, pp.515, &lt;a href=&quot;https://babel.hathitrust.org/cgi/pt?id=mdp.39015068159329&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://babel.hathitrust.org/cgi/pt?id=mdp.39015068159329&lt;/a&gt; ), and (b) also in those days tabular mean longitudes of the sun still included the effect of mean aberration: Newcomb's tables of 1895-8 were the first to attempt to remove it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lalande (art.1015) continued:  &quot;it follows that the right ascension of the mid-heaven&quot; {i.e. sidereal time} &quot;added to the complement to 24h of the Sun's mean longitude gives the mean time&quot;, and &quot;this method is used in England to find the time of an observation: Mr Maskelyne has given in his tables the Sun's movement in time to hundredths of a second&quot;.&quot;  In terms of the diagram, this translates as:   mean time MS = QM + (24h - QS) = QM - QS . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A modern version of the foregoing classical method for defining mean solar time:&lt;/strong&gt; During the period when methods like the foregoing were current, sources of the solar tables used to compute the annual almanacs were occasionally updated, when newer and more correct theories and tables became available.  Thus, for the (UK) Nautical Almanac, Delambre's mean elements for the Sun were used with minor corrections down to 1833, then Bessel's mean elements from 1834 to 1863, and Leverrier's from 1864 to the end of the century. The principle remained unchanged although slight discontinuities in the tabulations occurred at the points of changeover.  Accordingly, a modernized version of the same method can be implemented using recent accurate data for the sun's mean longitudes. Two modern sources of mean elements/longitudes for the sun (both in terms of ephemeris/dynamical time TT) are available: in J L Simon et al. (1994) &quot;Numerical expressions for precession formulae and mean elements for the Moon and the planets&quot; (derived from the JPL ephemeris DE200 which was the official basis of the Astronomical Almanac from 1984-2002), Astronomy &amp;amp; Astrophysics 282, 663-683, &lt;a href=&quot;http://adsabs.harvard.edu/abs/1994A%26A...282..663S&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/1994A%26A...282..663S&lt;/a&gt; ; and in J Chapront et al., &quot;A new determination of lunar orbital parameters ...&quot; (derived from the JPL successor ephemeris DE405 in official use 2003-14), Astronomy &amp;amp; Astrophysics 387, 700-709, &lt;a href=&quot;http://adsabs.harvard.edu/abs/2002A%26A...387..700C&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2002A%26A...387..700C&lt;/a&gt; .  The solar positions in those two ephemerides hardly differ from each other for present purposes.  Mean solar longitudes reckoned from mean equinox of date are given in Simon et al. The data of Chapront et al. are for a fixed equinox of J2000 to which general precession in longitude has to be applied. Both sets of modern mean longitudes are geometric, leaving mean aberration to be applied as -20&quot;.4955 in arc.  With that adjustment they can then be used like the old mean longitudes, converted to time, treated as right ascensions of the fictitious mean sun, and applied in the formulae given above to relate mean solar time with mean sidereal time.  The results may be compared with current values for UT1. {Some example calculations will be added if requested.}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Modern IERS zero meridian:&lt;/strong&gt; It is worth noting that it appears no allowance is needed for any difference between time at the old transit instrument at Greenwich observatory and time at the new IERS reference (zero) meridian. The ground track of the IERS reference meridian is located about 102m to the east of the meridian line through the old transit instrument. A paper of S Malys et al. (2015), &quot;Why the Greenwich meridian moved&quot; (J Geodesy 89, 1263-1272, at &lt;a href=&quot;http://adsabs.harvard.edu/abs/2015JGeod..89.1263M&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2015JGeod..89.1263M&lt;/a&gt; ) accounts for the difference of ground track as an effect of local gravitational deflection of the vertical. Thus, while the ground track of the IERS zero meridian is appreciably east of the old Airy transit instrument, by a distance which would make a rotation of about 5.3&quot; arc in longitude or about 0.35 time-seconds if it represented a geocentric rotation on a spheroidal earth, there has in fact been no rotation of the reference frame (within measurement error): the deflection of the vertical makes the difference a parallel displacement.  With the nominally meridian (but off-geocentric) plane of the old transit instrument practically parallel to a geocentric meridian plane through the more easterly ground track of the IERS reference meridian, a transit instrument mounted in the IERS geocentric meridian plane would 'see' stellar transits at the same times as those occurring at the Airy transit instrument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Modern changes in mean solar time/UT1:&lt;/strong&gt;  The 19th-century Nautical Almanac standard for mean solar time was altered in several successive ways towards the current standard for UT1. (Not discussed here are the corrections for observed polar wandering and seasonal effects leading to the differences between UT0, UT1 and UT2.)  But there was a history of 19th-c. controversy about one of the changeovers mentioned above from older to newer solar data, worth brief description because it foreshadows and may possibly have influenced some of the 20th/21st-c. departures from the use of mean solar longitudes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At one of the changeovers (from Bessel's data down to the end of 1863, to LeVerrier's data used from 1864 on), a discontinuity of a little over half a second was generated in the tabular relation between sidereal and mean solar time  (shown e.g. by the daily tabular values of sidereal time at mean noon in the Nautical Almanac for '1863 Dec 32' and 1864 Jan 1) (data for Greenwich mean noon 1864 Jan 1 in the 1864 volume, at &lt;a href=&quot;https://babel.hathitrust.org/cgi/pt?id=mdp.39015068159329&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://babel.hathitrust.org/cgi/pt?id=mdp.39015068159329&lt;/a&gt; , and for the same physical day in the 1863 volume where it was designated '1863 Dec 32',  &lt;a href=&quot;https://babel.hathitrust.org/cgi/pt?id=mdp.39015068159006&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://babel.hathitrust.org/cgi/pt?id=mdp.39015068159006&lt;/a&gt; ).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was a brief explanation and apology in the 1864 preface: the step-change arose because LeVerrier's theory had identified new long-term solar periodic perturbations: as long as these remained undiscovered, their effect had been implicit in the old mean elements, and so they had to be removed from the new; the largest of them due to Mars and Jupiter was near its peak at the start of 1864 and responsible for nearly all the time-discrepancy of the changeover. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This changeover and discontinuity attracted little attention for nearly twenty years; but then E J Stone, an astronomer of some repute and director of a university observatory, was trying in the 1880s to account for growing discrepancies between observed places of the moon and positions calculated by Hansen's 1857 'Tables de la Lune'. Stone seized (quite mistakenly), as the cause of the lunar errors, on the 1864 discontinuity of about half a second in calculated mean solar time arising from the changeover from Bessel's mean solar elements to those of Leverrier (E J Stone, 1883 MNRAS 43, 335-345 and 401-407).  The two effects were unrelated and Stone had also made a mistake that exaggerated errors by a factor of about 365. Stone's mistakes were pointed out in print by no less a quartet than John Couch Adams (1883 MNRAS 44, 43-47; 1884 MNRAS 44, 82-84), Arthur Cayley (1883 MNRAS 44, 47-49; 1884 MNRAS 44, 84-85), Simon Newcomb (1884 MNRAS 44, 234-5; 1884 MNRAS 44, 381-3; 1894 MNRAS 54, 286-8) and G B Airy (1883 Observatory 6, 184-5), but Stone stubbornly stuck to his position.  He also wrote a vehement paper for the Royal Society, based on his faulty analysis of the lunar errors, in which he insisted that no revision should ever again be allowed in future to create any discontinuity in the tabular measure of mean solar time (E J Stone (1883), Proc. R. Soc. Lond.  35, 135-7, &lt;a href=&quot;http://rspl.royalsocietypublishing.org/content/35/224-226/135&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://rspl.royalsocietypublishing.org/content/35/224-226/135&lt;/a&gt; ). This was of course a demand incompatible with the use of corrected solar mean longitudes whenever new theories and observations would show the need for correction. But strangely enough, Stone's principle of requiring continuity has actually been followed in the alterations to UT1 made explicitly in 1984 and 2003 (and implicitly in 1960). It is hard to tell whether or not this arose from some descendant of Stone's influence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been four main occasions of alteration away from the 19th-c. practice described above: in 1900, with Newcomb's formula for time quantities; in 1960, and the introduction of Ephemeris Time into many calculations but not into the time quantities; in 1984 with a fresh official definition of UT1 adopted from Aoki et al (1982), &quot;The new definition of Universal Time&quot;, &lt;a href=&quot;http://adsabs.harvard.edu/abs/1982A%26A...105..359A&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/1982A%26A...105..359A&lt;/a&gt; ; and in 2003 with the introduction of the current basis of UT1, described briefly in IAU 2000 resolution B1.8 and more fully in N Capitaine (2003) &quot;Expressions to implement the IAU 2000 definition of UT1&quot;, &lt;a href=&quot;http://adsabs.harvard.edu/abs/2003A%26A...406.1135C&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2003A%26A...406.1135C&lt;/a&gt; . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1900:  Newcomb's time formula made a very small alteration in the time-equivalent of the sun's mean longitude, by increasing the quadratic term. The motivation and effect go back to the work and practice of LeVerrier, summarised by A Gaillot in 1886: &quot;Sur la mesure du temps&quot;, Bulletin Astronomique, Ser. I, v.3, 221-232, &lt;a href=&quot;http://adsabs.harvard.edu/abs/1886BuAsI...3..221G&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/1886BuAsI...3..221G&lt;/a&gt; .  LeVerrier's analysis of precession supposed that the earth's axial rotation was perfectly uniform. He also found that his theory of the precession of the equinoxes in right ascension led to a quadratic term discordant with the quadratic term for the sun's secular acceleration in longitude. By incorporating an increment in the quadratic term in the sun's longitude to remove the 2nd-order discrepancy, the effect was to make increments in mean time closely proportional to those in sidereal time (and to transmit the supposed uniformity of earth rotation into a proportional uniformity of the derived (but altered) calculation for mean solar time).  Newcomb mentioned the point briefly in his book &quot;The elements of the four inner planets and the fundamental constants of astronomy&quot; (1895, at  &lt;a href=&quot;https://archive.org/details/cihm_16774&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://archive.org/details/cihm_16774&lt;/a&gt; , see p.188). This appears to be the historical origin of the still-current practice of making the measure of mean solar time an exact linear function of earth rotation, a point specifically raised by the current question. No-one now supposes that earth- rotation is uniform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1960:  With the introduction of Ephemeris Time (&lt;a href=&quot;https://en.wikipedia.org/wiki/Ephemeris_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Ephemeris_time&lt;/a&gt;) into the official almanacs from 1960, the solar data was effectively corrected, not as had been usual by change of numbers, but by time-shifting it by a displacement DeltaT, the difference (Ephemeris Time - Universal Time). The numbers were maintained unchanged, but they related now to a different timescale. No alteration at all was made in the calculation of nominally mean solar time. The effect was to retain the buildup of the time errors constituted by the difference between the new best estimate of the sun's mean position and the old. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1984: The definition of UT1 was revised in a way that maintained continuity of value and rate at the moment of changeover, the start of 1984 (see S Aoki et al., (1982), link above).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2003: A further revision to UT1 was made, maintaining continuity at the moment of changeover, the start of 2003 (see N Capitaine et al., (2003), link above).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary/conclusion:&lt;/strong&gt;  The description above shows something of how the current calculations for UT1 have departed in principle from classical methods used to define and calculate mean solar time. But it seems no longer to be claimed that UT1 &lt;em&gt;is&lt;/em&gt; mean solar time: thus the IAU 2006 NFA glossary calls UT only &quot;a measure of time that conforms, within a close approximation, to the mean diurnal motion of the Sun&quot;. Current values of UT1 can be compared with results of a modernized version of an old standard method for mean solar time from sidereal time. It seems that differences between this and current UT1 are still well under a second, though they can be expected to increase with increasing DeltaT, the difference in seconds between ((dynamical) TT - UT1). Thus the differences probably are remaining less than allowable differences for example between UT1 and UTC. The IAU gloss seems broadly justified, although it may be of continuing interest to have a method of calculating a more traditional form of mean solar time to check on any accumulating discrepancies.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;{Notice of any mistakes in all of this will be gratefully appreciated!}&lt;/p&gt;&#xA;" OwnerUserId="17202" LastEditorUserId="17202" LastEditDate="2018-04-29T11:50:02.267" LastActivityDate="2018-04-29T11:50:02.267" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="26099" PostTypeId="1" AcceptedAnswerId="29076" CreationDate="2018-04-29T01:38:19.933" Score="6" ViewCount="1429" Body="&lt;p&gt;The &lt;em&gt;tidal radius&lt;/em&gt; is defined in &lt;a href=&quot;http://adsabs.harvard.edu/abs/1962AJ.....67..471K&quot; rel=&quot;noreferrer&quot;&gt;King (1962)&lt;/a&gt; as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the value of r (the radius) at which &lt;em&gt;f&lt;/em&gt; (the density profile) reaches zero...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This also referred by King as the &quot;&lt;em&gt;limiting radius&lt;/em&gt;&quot; written as (Eq 3):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$r_t=R_{GC}\left(\frac{M}{2M_g}\right)^{1/3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$R_{GC}$&lt;/span&gt; is the galactocentric distance, &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is the cluster's mass, and &lt;span class=&quot;math-container&quot;&gt;$M_{g}$&lt;/span&gt; is the galaxy's mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can also be written as &lt;a href=&quot;http://adsabs.harvard.edu/abs/1998MNRAS.299..955P&quot; rel=&quot;noreferrer&quot;&gt;Pinfield et al. (1998) Eq (12)&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$r_t = \left( \frac{GM}{2(A-B)^2} \right)^{1/3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; are the &lt;a href=&quot;https://en.wikipedia.org/wiki/Oort_constants&quot; rel=&quot;noreferrer&quot;&gt;Oort constants&lt;/a&gt; and &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is the total mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I've seen the &quot;&lt;em&gt;tidal radius&lt;/em&gt;&quot; defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$r_t=R_{GC}\left(\frac{M}{3M_g}\right)^{1/3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in &lt;a href=&quot;http://adsabs.harvard.edu/abs/1990ApJ...351..121C&quot; rel=&quot;noreferrer&quot;&gt;Chernoff &amp;amp; Weinberg (1990) Eq 2&lt;/a&gt;. Notice the 3 in the denominator instead of the 2 as above. According to the authors:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The tidal radius depends weakly upon the form of the mass distribution of the halo. The Eq. above is correct for the tidal stress of a Keplerian force field. On the other hand, for constant &lt;span class=&quot;math-container&quot;&gt;$v_g$&lt;/span&gt;, the factor &lt;span class=&quot;math-container&quot;&gt;$3M_g$&lt;/span&gt; becomes &lt;span class=&quot;math-container&quot;&gt;$2M_g$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This same quantity (using &lt;span class=&quot;math-container&quot;&gt;$3M_g$&lt;/span&gt;) is referred to as the &quot;&lt;em&gt;Jacobi radius&lt;/em&gt;&quot; in &lt;a href=&quot;https://arxiv.org/abs/1603.06803&quot; rel=&quot;noreferrer&quot;&gt;Piatti (2015) Eq 4&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also seen the Jacobi radius written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$r_J=\left(\frac{GM}{4\Omega^2-k^2}\right)^{1/3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where &lt;span class=&quot;math-container&quot;&gt;$G, M, \Omega$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; are the gravitational constant, the cluster&#xA;mass, the circular, and the epicyclic frequencies of a near circular&#xA;orbit, respectively) in &lt;a href=&quot;https://www.aanda.org/articles/aa/abs/2011/12/aa18021-11/aa18021-11.html&quot; rel=&quot;noreferrer&quot;&gt;Ernst et al. (2011) Eq (8)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Globular_cluster#Radii&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, both the tidal and the Jacobi radius are the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: are these two radius the same quantity?&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="354" LastEditDate="2019-01-07T21:42:57.257" LastActivityDate="2021-01-18T09:03:51.943" Title="Jacobi vs tidal radius for star cluster" Tags="&lt;tidal-forces&gt;&lt;globular-clusters&gt;&lt;star-cluster&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26100" PostTypeId="1" AcceptedAnswerId="26101" CreationDate="2018-04-29T12:16:07.733" Score="10" ViewCount="6380" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Five billion years from now, the sun will have grown into a red giant star, more than a hundred times larger than its current size.&#xA;  While this metamorphosis into the giant star will change the solar system, scientists are unsure what will happen to the third rock from the sun.We already know that our sun will be bigger and brighter, so that it will probably destroy any form of life on our planet. &#xA;  But whether the Earth's rocky core will survive is uncertain.&#xA;  At the end of its evolution, seven billion years from now, the sun will become a tiny white dwarf star&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(by Leen Decin, professor at the KU Leuven Institute of Astronomy (&lt;a href=&quot;https://www.mirror.co.uk/science/earth-burnt-crisp-sun-not-9416194&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;))&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When our sun runs through its last phase of its life it will expand and tremendous solar flares will flow out of it. Earth will end up getting baked and will lose its atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter is a gas giant planet, a GAS GIANT. So,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What will happen in Jupiter? Will it evolve into a star since it contains hydrogen and helium? How will the atmosphere deal with the extreme heat?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="22623" LastEditorUserId="7472" LastEditDate="2018-04-29T14:32:02.037" LastActivityDate="2021-04-01T12:07:24.887" Title="Fate of Jupiter when our sun dies" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;jupiter&gt;&lt;gas-giants&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26101" PostTypeId="2" ParentId="26100" CreationDate="2018-04-29T13:15:20.827" Score="14" Body="&lt;p&gt;Jupiter won't evolve into a star, it is not big enough. A body would have to have about 80 times the mass of Jupiter for there to be significant fusion occurring in the core. The end of life of the Sun won't change the mass of Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter will continue to orbit the Sun as it evolves into a red giant. Although the solar wind will be much much more powerful, it won't have a significant effect on the overall mass of Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the Sun loses its outer layers it will, in the last million years or so of its life, lose about half its mass. This will have a significant effect on the orbit of Jupiter, causing Jupiter to migrate outwards.  While it is possible that some &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10499/how-much-mass-will-the-sun-have-when-it-becomes-a-white-dwarf&quot;&gt;planet's orbits may become unstable and get ejected from the solar system&lt;/a&gt;, it is more likely that Jupiter will settle into a new, wider orbit around the white dwarf.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2531" LastEditDate="2018-04-30T12:21:16.567" LastActivityDate="2018-04-30T12:21:16.567" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26102" PostTypeId="2" ParentId="26078" CreationDate="2018-04-29T13:27:24.967" Score="1" Body="&lt;p&gt;It boils down to simple remainder mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two planets have a 13:8 resonance, then, for every 1 outer planet orbit, the inner planet moves 1 and 5/8ths times around the sun.  Because both planets move in the same direction, that means the inner planet moves ahead 5/8ths of 1 orbit for every outer planet year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5/8ths in 1 year, means the inner planet passes the outer planet every 8/5ths year, or every 1.6 years.  That's the catch-up time.  The formula is pretty simple:   8/(13-8) or small/(large-small).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That period is known as the synodic period of Venus viewed from Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Throwing out the 13:8, because it's close but not quite right, Venus orbit is 224.701 days and Earth's orbit 365.256 days (sidereal orbits).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;224.701/(365.256 - 224.701)&#xA;or, 224.701 / 140.555 = 1.59867 Earth years, or 583.92 Earth days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.livephysics.com/physical-constants/astronomy-pc/synodic-sidereal-periods-planets/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Venus' synodic period is listed here as 584 days&lt;/a&gt; - close enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All you need for a repeating pattern is a synodic period that's very close to a divisor of an integer and you always get that when the planets are in resonance.    1.59867 is close enough to 1.6 that 5 Venus synodic years comes very close to 8 earth years.   That's what causes the pattern to repeat.   1.59867 times 5 is very close to 8.   I think it misses by maybe 2.6 degrees every 8 years, but it's still close enough to have gotten the attention and admiration of the ancients.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus and Earth aren't really in resonance, just close.  They don't have sufficient tidal forces on each other to maintain a true resonance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flower petal you get &lt;a href=&quot;https://i.imgur.com/PqJy8lX.gifv&quot; rel=&quot;nofollow noreferrer&quot;&gt;in your 2nd link&lt;/a&gt;, the 5 intersection points or meeting points, those happen in in that Spirograph like image when Earth and Venus are on opposite sides of the sun.  They also happen on a 1.6 earth years and move forward 225 degrees every Earth year.  It looks neat, but like a Spirograph it's a cool design based on nothing more than simple fractions.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As far as fractal, flowers are sometimes used as examples of fractal patterns, so I can see how, looking at the flower pattern, one might think it might be fractal.   It's really not.  Fractals are non-repeating.   Kepler Orbits are repeating.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of &lt;a href=&quot;https://en.wikipedia.org/wiki/Milankovitch_cycles&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth's Milankovich cycles&lt;/a&gt; may be fractal in their variation.   That's another topic though.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;@Chappo &lt;/p&gt;&#xA;&#xA;&lt;p&gt;using your 7:4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4/(7-4) = 1.333 years to catch up, so that corresponds to 3 passes every 4 years, or, as you said, a triangle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For resonances and the number of points in the star-pattern, it seems to be a simple matter of subtraction as you said, because the denominator in your example is 7-4, or 3.   Divide by 3, you then need 3 cycles to get back to an integer.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-04-29T13:27:24.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26103" PostTypeId="2" ParentId="26100" CreationDate="2018-04-29T15:13:01.073" Score="6" Body="&lt;p&gt;How will the atmosphere deal with the extreme heat?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &lt;a href=&quot;https://www.universetoday.com/9472/outer-planets-could-warm-up-as-sun-dies/&quot; rel=&quot;noreferrer&quot;&gt;this article as a guide&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;During most of the red giant lifetime, the sun will be only 30 times&#xA;  brighter than its current state. Toward the end of the red giant phase&#xA;  the sun will grow more than 1,000 times brighter, and occasionally&#xA;  release pulses of energy reaching 6,000 times current brightness.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Jupiter is about 5 times further from the Sun than the Earth, so it receives 1/25th the solar energy per area.  If the Sun emits 30 times as much energy during most of its red-giant phase, and if we assume that Jupiter will migrate outwards some, that puts Jupiter (and its moons) probably somewhere in or close to the habitable zone.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen and Helium are pretty inert, but Jupiter has enough Methane and Water vapor in its upper atmosphere to experience a greenhouse effect and gradually heat up, so it will likely grow hot given enough time, even at what we might consider a comfortable distance for a planet.   The Red-giant stage of the Sun will last a couple hundred million years and Jupiter is very large and will take a long time to trap enough solar energy to really start to heat up, but I think that's the outcome.   It would be a good place to put Earth during that time, but Jupiter would probably become a heat trap or run-away greenhouse planet at a certain point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even with heating up, Jupiter is massive enough that it &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_escape&quot; rel=&quot;noreferrer&quot;&gt;probably won't lose much of its hydrogen&lt;/a&gt;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A final point to make is that Jupiter will probably absorb a small percentage of the matter that the sun loses.   Our Sun is estimated to &lt;a href=&quot;http://www.stsci.edu/~jkalirai/ASYarticleJasonKalirai.pdf&quot; rel=&quot;noreferrer&quot;&gt;lose about 54% of its mass&lt;/a&gt; by the time it becomes a white dwarf and much of (some of?) that matter loss will happen during the red giant stage.   That's about 560 Jupiter masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of that ejected material will just form a planetary nebula or leave the solar system.  A tiny percentage of it will be absorbed by Jupiter.   I don't expect it will be much, but Jupiter will probably (could?) add some mass during the red giant stage of the sun.   I don't expect it would be close to enough to become a brown dwarf star, but I think there will be some added hydrogen and helium during that time.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat will make it expand a bit and it will probably get darker.   Losing its lighter bands which are lighter in color due to ice.   But it'll still basically be Jupiter.   Hotter around the surface but otherwise not very different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Europa, however, will probably become an ocean moon instead of an ice moon, at least for a while.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="13437" LastEditDate="2018-04-29T19:57:22.460" LastActivityDate="2018-04-29T19:57:22.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26104" PostTypeId="1" AcceptedAnswerId="26110" CreationDate="2018-04-29T22:56:01.397" Score="6" ViewCount="255" Body="&lt;p&gt;Is there a place on Moon (sure it's near north or south pole) that is always lit by sun? It's to install solar panels if I will choose to make rancho there :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What diameter of Moon Polar circle?&#xA;How high the hill or mast on the pole should be to be always lit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used some online calculators and found next:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Axial tilt to ecliptic is &lt;strong&gt;1.5424°&lt;/strong&gt;, Lunar Radius &lt;strong&gt;1738km&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Polar circle is &lt;strong&gt;46.8km&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Spherical cap height &lt;strong&gt;0.15km&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Are my calculations correct? It shows that it should be near &lt;strong&gt;150 meters high&lt;/strong&gt;. Is any high enough hill exists on a poles? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Where I can check good 3d model of Moon with elevation (sure there is no sea level but ideal sphere)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="22877" LastEditorUserId="22877" LastEditDate="2018-04-29T23:27:47.230" LastActivityDate="2018-04-30T07:19:15.200" Title="Place on Moon that always lit?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26109" PostTypeId="2" ParentId="26033" CreationDate="2018-04-30T06:42:22.290" Score="3" Body="&lt;p&gt;No, not since it has been called Universal Time and calculated with expressions that were internationally agreed has it exactly tracked Mean Solar Time, but it is very close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seago and Seidelmann made several simulations of when the mean sun is over the zero meridian, and they published their result as part of American Astronautical Society meeting Advances in the Astronautical Sciences 148:1789-1807 (meeting was AAS/AIAA Spaceflight Mechanics Meeting Feb. 10-14, 2013, Kauai, HI as seen at &lt;a href=&quot;http://www.univelt.com/book=4186&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.univelt.com/book=4186&lt;/a&gt;&#xA;and paper was AAS 13-486).  This paper is available online at&#xA;&lt;a href=&quot;http://www.agi.com/resources/white-papers/the-mean-solar-time-origin-of-universal-time-and-u&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.agi.com/resources/white-papers/the-mean-solar-time-origin-of-universal-time-and-u&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seago and Seidelmann found that because of assumptions that Simon Newcomb made when constructing his reference frame, UT and MST deviate by an increasing amount that is roughly equal to 0.002738 ∆T&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1896 at a meeting Conférence Internationale des Étoiles Fondamentales in Paris the directors of the major observatories all agreed that Newcomb's expressions in &lt;em&gt;Tables of the Sun&lt;/em&gt; (Astronomical Papers of the American Ephemeris, Vol 6) and other USNO publications should be adopted by all the national ephemerides starting from 1900, and they continued to be used until the IAU 1976 system was implemented in 1984.  Newcomb's expressions are on page 9 of that volume. They were used as the definition of Universal Time, and as noted in other answers all subsequent modifications of the definition of UT1 maintained continuity in value and rate at the date of changeover. Newcomb's expression for Right Ascension of the fictitious mean sun (given as τ) is the same as equation 5 in Seago and Seidelmann.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unknown to Seago and Seidelmann when they wrote that paper in 2013, this deviation had already been understood at the inception of Ephemeris Time and documented by D.H. Sadler in his 1954 monograph in &lt;em&gt;Occasional Notices&lt;/em&gt; of the RAS, 3, p 103 which is transcribed at&#xA;&lt;a href=&quot;https://www.ucolick.org/~sla/leapsecs/twokindsoftime.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.ucolick.org/~sla/leapsecs/twokindsoftime.html&lt;/a&gt;&#xA;The number Sadler had obtained was 0.00273 ΔT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that 1.00273 is the ratio of mean solar and sidereal days, and Sadler's understanding was rooted in his attendance at the 1950 CNRS meeting Constantes Fondamentales de l'Astronomie held at Observatoire de Paris where G.M. Clemence proposed Ephemeris Time in concord with the Astronomer Royal Sir H. Spencer Jones who had documented the nonuniform rotation of the earth in 1939 (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1939MNRAS..99..541S&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/1939MNRAS..99..541S&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, yes, as earth rotation slows the difference between UT=12:00 and mean sun over Greenwich will deviate, but very slowly.&lt;/p&gt;&#xA;" OwnerUserId="14050" LastEditorUserId="14050" LastEditDate="2018-04-30T16:58:32.377" LastActivityDate="2018-04-30T16:58:32.377" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="26110" PostTypeId="2" ParentId="26104" CreationDate="2018-04-30T07:19:15.200" Score="8" Body="&lt;p&gt;This kind of location is called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Peak_of_eternal_light&quot; rel=&quot;noreferrer&quot;&gt;peak of eternal light&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently there is &lt;a href=&quot;http://www.skyandtelescope.com/observing/peaks-of-eternal-light/&quot; rel=&quot;noreferrer&quot;&gt;no certain identification&lt;/a&gt; of such a peak on the Moon, but there are definitely places that get more than 86% sunlight during the lunar year (ignoring eclipses). &lt;a href=&quot;https://agupubs.onlinelibrary.wiley.com/doi/abs/10.1029/2008GL035692&quot; rel=&quot;noreferrer&quot;&gt;Kaguya did not find any permanent peaks&lt;/a&gt;, but there is apeak near Peary at the North Pole that gets 89%, and &lt;a href=&quot;https://linkinghub.elsevier.com/retrieve/pii/S0019103510001375&quot; rel=&quot;noreferrer&quot;&gt;several places near the South Pole that have multi-month illumination&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating the requirements for such a peak is complicated since the moon is definitely not spherical on the hundreds of meter scale; at the very least one needs an ellipsoid model. The south pole situation is complicated because &lt;a href=&quot;http://www.thespacereview.com/article/2423/1&quot; rel=&quot;noreferrer&quot;&gt;there is a large basin around the actual pole&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2018-04-30T07:19:15.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26113" PostTypeId="2" ParentId="26078" CreationDate="2018-04-30T16:12:43.787" Score="0" Body="&lt;p&gt;There is a sequence of orbital resonances between the moons of Jupiter, so tracing the motion of moons relative to one another gives a range of nice patterns&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MLlBe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MLlBe.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this image, the camera follows Ganymede, stationary in the centre of the image. Since the camera moves to follow Ganymede, Jupiter moves relative to the camera, producing the pink circle. Io and Europa have resonant orbits, orange and white, Io has a three petalled &quot;flower&quot;, and Europa has the cardiod.  Calisto orbits outside of Ganymede, and has the pattern of four lobes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image generated by &quot;gravity simulator&quot;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-04-30T16:12:43.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26116" PostTypeId="2" ParentId="1001" CreationDate="2018-04-30T20:25:04.467" Score="4" Body="&lt;p&gt;The completely unscientific explanation...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine the universe to be a balloon.&#xA;Two bodies start close to each other but on opposite surfaces. The expansion of the balloon takes them away from each other at equal speed and such a rate that the light from one at its starting point takes almost the entire history of the universe to reach the other. The distance between the two NOW is not twice the age of the universe - because you cannot travel &quot;through&quot; the balloon - but must instead go round the surface of the balloon... 13.8 * PI billion light years = 43 billion light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not strictly correct, but at least avoids too much worrying about astrophysics and cosmology!&lt;/p&gt;&#xA;" OwnerUserId="22895" LastActivityDate="2018-04-30T20:25:04.467" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26117" PostTypeId="1" CreationDate="2018-04-30T20:52:54.157" Score="2" ViewCount="172" Body="&lt;p&gt;I'm interested in trying to use the data that will be returned by TESS to apply some machine learning/computer sciences concepts. I'm not sure yet what I want to try to do with it (though I have some ideas - see end comments). This because I don't really understand the data structure it will be using. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;So essentially I would like to first some help to get a high-level view of how the TESS data is structured. Then a bit of guidance as to how I could visualize it/get started with some basic example&lt;/em&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Additional info/what I know at this point&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I've checked out &lt;a href=&quot;https://archive.stsci.edu/tess/ete-6.html#ffi&quot; rel=&quot;nofollow noreferrer&quot;&gt;the MAST page&lt;/a&gt;. It details what type of FITS documents will be returned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have downloaded samples of some of these data products and I see they all come in FITS format. On Ubuntu, you can open that in &lt;a href=&quot;http://aladin.u-strasbg.fr/AladinDesktop/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aladin&lt;/a&gt;. However Aladin in itself is a fairly involved piece of software. Do I need Aladin for this, or are there simple alternatives to just view the data quickly to help get a feel for what it means? In other words what kind of navigator do you use for FITS files?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty new to astronomy - I'm coming at this rather from a computer science perspective. Some basic suggestions as to what kind of observation an amateur astronomer might want to look at would be welcomed if you have some.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data linked above (MAST page) are &lt;strong&gt;&lt;em&gt;simulated data&lt;/em&gt;&lt;/strong&gt; (real data will only be available in a couple months). This is fine - mostly interested in being functional with the data when they arrive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Data analysis ideas:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Using kmeans or similar to try to extract  what are some of the &quot;typical&quot; &lt;a href=&quot;http://spaceflight101.com/tess/tess-instrument/&quot; rel=&quot;nofollow noreferrer&quot;&gt;transit light curves&lt;/a&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Generating those transit light curves from the raw data, say from a target pixel&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="22894" LastActivityDate="2018-05-13T14:35:36.510" Title="Understanding expected TESS data structure" Tags="&lt;data-analysis&gt;&lt;tess&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26120" PostTypeId="1" CreationDate="2018-05-01T12:44:41.380" Score="1" ViewCount="323" Body="&lt;p&gt;From what I learned, cosmological event horizon is the horizon that moves away from us at/more than the speed of light. And particle horizon is the horizon that particles inside the horizon can have a causal connection with us. It is the maximum distance from which particles could have traveled to the observer in the age of the universe. This simply indicates a contradiction to me. How is it possible that the cosmological event horizon is closer?&lt;/p&gt;&#xA;" OwnerUserId="22885" LastActivityDate="2018-05-01T16:32:06.647" Title="Why is cosmological event horizon closer to us than the particle horizon?" Tags="&lt;cosmological-horizon&gt;&lt;event-horizon&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26121" PostTypeId="1" AcceptedAnswerId="26125" CreationDate="2018-05-01T12:46:21.783" Score="4" ViewCount="171" Body="&lt;p&gt;It now seems general fact that when a massive star core collapses at the end of its lifecycle it forms a black hole, can a black hole be formed skipping the star phase altogether?&lt;/p&gt;&#xA;" OwnerUserId="16136" LastActivityDate="2018-05-02T18:23:52.000" Title="Is star a necessary phase in black hole lifecycle?" Tags="&lt;star&gt;&lt;black-hole&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26123" PostTypeId="2" ParentId="26121" CreationDate="2018-05-01T16:05:34.540" Score="1" Body="&lt;p&gt;Here's a possible alternative sequence. This is off the top of my head.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose somehow (and quite unlikely in our universe) there's a large cloud of particles all of whom have atomic number greater than &lt;strong&gt;Fe&lt;/strong&gt; .   Over time, this cloud coalesces under the usual laws of gravity.  But due to the high atomic numbers, the pressure and density required for fusion cannot be achieved. (In fact, these elements, even if they fuse, do not release energy but rather absorb it).  The continued coalescence leads to black-hole collapse.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I repeat: I have not the knowledge of the various conditions necessary for collapse, so this may not be possible.  But more or less this is what's described in the link to &lt;a href=&quot;https://academic.oup.com/mnras/article/370/1/289/1026607&quot; rel=&quot;nofollow noreferrer&quot;&gt;oup&lt;/a&gt; provided in the downvoted answer.&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-05-01T16:05:34.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26124" PostTypeId="2" ParentId="26120" CreationDate="2018-05-01T16:25:06.727" Score="2" Body="&lt;p&gt;The &lt;em&gt;particle horizon&lt;/em&gt; marks the region from within which we may have received light. It started out at zero, because light from nowhere had had the time to reach us, and increased as time went by because light from increasingly larger distance reached us. Due to the expansion it will always increase in physical size, but due to the &lt;em&gt;accelerated&lt;/em&gt; expansion, there is a limit to the region of the Universe that we will ever be able to receive light from. Today, the particle horizon is 46 Glyr (billion lightyears) away, and the limit — in the far future — is the part that &lt;em&gt;today&lt;/em&gt; is 63 Glyr away (but in the far future will be much farther away).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;event horizon&lt;/em&gt; can be thought of as sort of the opposite. At any time $t$, it is the distance from which we will be able to receive light. If the Universe contained no dark energy, there would be no limit to how distant a galaxy could be and we would still be able to see it — just wait long enough. However, the accelerated expansion means that an increasingly smaller part of the Universe will be able to emit light toward us with us receiving it, before the expansion carries the photons away faster than they can travel toward us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At Big Bang, the event horizon consisted of particles that today are 63 Glyr away — that's what I meant by &quot;the opposite of the particle horizon&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following figure from &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0310808&quot; rel=&quot;nofollow noreferrer&quot;&gt;Davis &amp;amp; Lineweaver (2004)&lt;/a&gt; may help explain. It show basically three versions of the same thing, namely how the sizes of various horizons evolve with time. The top panel shows time as a function of physical size, while in the middle panel, the $x$ axis show &lt;em&gt;comoving&lt;/em&gt; coordinates, i.e. the coordinates that expand along with the Universe. The lower panel shows not time on the $y$ axis, but conformal time, which can be thought of as the time it would take a photon to travel a distance if we &quot;froze&quot; the Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JdoRt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JdoRt.png&quot; alt=&quot;horizons&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-05-01T16:32:06.647" LastActivityDate="2018-05-01T16:32:06.647" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26125" PostTypeId="2" ParentId="26121" CreationDate="2018-05-01T19:05:44.530" Score="3" Body="&lt;p&gt;There are speculated objects known as &quot;primordial black holes&quot; which could have been present at the origin of the universe, or formed in its early stages via presently unknown processes, but there is no observational evidence that such objects exist. Also, some models for forming supermassive black holes involve the coalescence of such a huge amount of matter that you would not call it a star, unless some kind of supermassive superstar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, the mass could form stars first, and then coalesce into the supermassive black hole, but those stars themselves would not necessarily need to collapse first. Indeed, since much of the mass that goes into stars goes into lower mass stars that don't make black holes, it seems likely that a significant fraction, or even a majority, of the mass in supermassive black holes was not in a black hole prior to falling into the supermassive black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, since we don't know much about dark matter, one presumes it might be possible for dark matter to form black holes via some presently unknown mechanism, but there is no evidence dark matter would do that, and the idea that dark matter might itself be already in the form of black holes has not been supported by observational searches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'd say our current expectation is that supermassive black holes are largely made of matter that did not collapse in a massive star, and stellar-mass black holes could be mostly or entirely matter that did.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14094" LastEditDate="2018-05-02T12:54:05.897" LastActivityDate="2018-05-02T12:54:05.897" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26126" PostTypeId="2" ParentId="26117" CreationDate="2018-05-01T20:21:26.647" Score="0" Body="&lt;p&gt;HEASARC keeps a &lt;a href=&quot;https://fits.gsfc.nasa.gov/fits_viewer.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;list of FITS viewers&lt;/a&gt;.&#xA;Maybe one of those will meet your requirements.&#xA;Of course capabilities and usability vary widely.&#xA;&lt;a href=&quot;https://heasarc.gsfc.nasa.gov/docs/software/ftools/fv/&quot; rel=&quot;nofollow noreferrer&quot;&gt;fv&lt;/a&gt; did enough to help me answer another question here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://archive.stsci.edu/missions/tess/doc/EXP-TESS-ARC-ICD-TM-0014.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;TESS Science Data Products Description Document&lt;/a&gt; gives a one-line description of each of their FITS header fields.&#xA;If Wikipedia and exploratory testing don't close the gaps, ask again here.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-05-01T20:21:26.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26128" PostTypeId="1" AcceptedAnswerId="26130" CreationDate="2018-05-02T03:46:02.487" Score="5" ViewCount="878" Body="&lt;p&gt;Slightly spoilers here, but in the novel &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Three-Body_Problem_(novel)&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Three-Body Problem&lt;/a&gt;, there's a scene in which Civilization #184 is destroyed by the stacked gravitational attraction of the planet's (aptly named Trisolaris) three suns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This destruction is portrayed as things (like a sword) suddenly getting lighter, and people and buildings starting to float up in the air, as the atmosphere of the planet is being sucked into the three aligned suns. This is explained as being a &quot;Trisolar Syzygy&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I don't understand how that is possible. As I understand it, a syzygy happens in any sort of gravitational alignment of some astronomical bodies, and it's apparently &lt;a href=&quot;https://en.wikipedia.org/wiki/Syzygy_(astronomy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;what generates the tidal variation on the earth from the moon's orbit&lt;/a&gt;. But would the tidal forces be enough to cause this effect? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the novel, the planet is in the Alpha Centauri system, which &lt;a href=&quot;https://en.wikipedia.org/wiki/Alpha_Centauri#Planets&quot; rel=&quot;nofollow noreferrer&quot;&gt;may have a planet in the habitable zone&lt;/a&gt;, and is an actual trinary system. Could such an astronomical event actually cause some effect like that in this situation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm asking this here because I thought it wouldn't fit in the Science Fiction Q&amp;amp;A, as I'm more interested in the astronomical side of it, but I'd appreciate a pointer if I don't need to ask this here.&lt;/p&gt;&#xA;" OwnerUserId="10818" LastEditorUserId="10818" LastEditDate="2018-05-02T12:08:59.577" LastActivityDate="2018-05-02T12:53:39.080" Title="Trisolar syzygy and it's effects on the planet" Tags="&lt;planet&gt;&lt;star-systems&gt;&lt;tidal-forces&gt;&lt;earth-like-planet&gt;&lt;n-body-simulations&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26130" PostTypeId="2" ParentId="26128" CreationDate="2018-05-02T06:00:43.060" Score="5" Body="&lt;p&gt;Is it just for the spoiler? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that the 3 suns aligned create a sufficient gravity field for the outermost parts - less well attached - of the planet (atmosphere, people, buildings) to be in the &lt;a href=&quot;https://simple.wikipedia.org/wiki/Roche_limit&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Roche Limit&lt;/em&gt;&lt;/a&gt; of the system, causing them to be sucked out spiraling inwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a funny vid I found to illustrate:  &lt;a href=&quot;https://www.youtube.com/watch?v=6e5yyIcq40o&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=6e5yyIcq40o&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="11264" LastActivityDate="2018-05-02T06:00:43.060" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26131" PostTypeId="1" AcceptedAnswerId="26145" CreationDate="2018-05-02T06:50:56.133" Score="4" ViewCount="385" Body="&lt;p&gt;I made an extract of GAIA DR2 data for parallaxes greater than 500mas. Surprisingly I see a very significant number of sources with parallax greater that 1000, making them much closer than Proxima. This should lead to excited newspaper headlines, but i haven't seen those yet... Am I missing something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see my extract of data here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://gist.github.com/agnes1/1492e94b5e9c0ef71f6ae7944964b666&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://gist.github.com/agnes1/1492e94b5e9c0ef71f6ae7944964b666&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some GAIA source_id rows to look at include:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4039503676370417664&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4042946384760661504&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4042338732771057920&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4046453826887613696&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are many more. The closest source I find is 4062964299525805952 with  parallax of 1851mas. It has a phot_g_mean_mag of 19.63355... Would I be right in assuming this is a brown dwarf?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing I note is that there is a surge of vmag 19/20 sources within 10 parsecs. These are obviously so faint GAIA doe not detect similar sources beyond 10 parsecs. To give some numbers I count 1722 sources (most mag 19/20) within 10 parsecs and only 606 sources in a shell of equal volume beyond that. Again, would these all be (a very large number of) brown dwarfs?&lt;/p&gt;&#xA;" OwnerUserId="22867" LastActivityDate="2018-05-03T16:47:46.353" Title="Very near GAIA DR2 sources - has Proxima Centauri been dethroned?" Tags="&lt;brown-dwarf&gt;&lt;gaia&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26133" PostTypeId="1" CreationDate="2018-05-02T07:58:30.867" Score="4" ViewCount="66" Body="&lt;p&gt;I have decided to do undergraduate thesis on &quot;estimating photometric redshift&quot; or something related to this using machine learning. Reading previous papers, I have come to know that work has been done on this topic using support vector machine, artificial neural network, Bayesian approach and even deep learning. Can you suggest what are the  modifications or additions that I can possibly make?&lt;/p&gt;&#xA;" OwnerUserId="22838" LastEditorUserId="616" LastEditDate="2021-01-23T03:07:47.630" LastActivityDate="2021-01-23T03:07:47.630" Title="Photometric redshift" Tags="&lt;galaxy&gt;&lt;astrophysics&gt;&lt;redshift&gt;&lt;neural-networks&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2019-12-30T19:16:29.430" ContentLicense="CC BY-SA 4.0" />
  <row Id="26135" PostTypeId="2" ParentId="26128" CreationDate="2018-05-02T09:53:32.263" Score="3" Body="&lt;p&gt;In addition to the answer above, passing inside a stars Roche limit would likely cook the planet for anything but the smallest, coldest stars or white dwarfs.   Evaporating the planet's oceans and blowing off much of the atmosphere even before any houses started floating.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Roche limit depends on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Roche_limit#Selected_examples&quot; rel=&quot;nofollow noreferrer&quot;&gt;density and rigidity of the two objects&lt;/a&gt;, but it's on average just a few radii distance, much closer than you'd want a planet to get to any star.  The only natural satellite in our solar-system that orbits within it's Roche limit is Phobos around Mars.  If Phobos was liquid it would drift apart and standing on the Mars side of Phobos, you would experience weightlessness and float off the moon.  Phobos' rigidity and firmness keep the moon itself from drifting apart.  It's also very small and in a nearly circular orbit so the forces are pretty constant.  In your scenario, the planet &quot;passing through&quot; would be very jarring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A planet passing through a Roche limit in a strange, 3 star orbit would experience enormous squeezing and stretching, like Io but much worse.   It would completely resurface with Magma every time it happened.  There's no scenario I can see where people would live long enough to see their houses floating.  They'd be dead long before then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FlsFr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FlsFr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ny.pbslearningmedia.org/resource/112717444-earth-space/a-diagram-explaining-how-tidal-forces-work-on-jupiters-moon-io-earth-and-space/#.WumJIYjwbcc&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source of image&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="10818" LastEditDate="2018-05-02T12:53:39.080" LastActivityDate="2018-05-02T12:53:39.080" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26136" PostTypeId="1" CreationDate="2018-05-02T11:48:18.127" Score="3" ViewCount="139" Body="&lt;p&gt;I have a reflector telescope with a 114mm aperture and 1000mm focal length.  I find that to get the clearest image I can use a 4mm eyepiece, or with my 5x barlow a 20mm however, seeing that I am very interested in astro photography when I try to photograph Jupiter for instance, through the telescope eyepiece, being the size of about a green pea seen through the scope, there is not enough pixel data for my canon eos t2i rebel dlsr to capture a clear photo. Is there some way to enlarge the photo for the camera without blurring the planet's details from over magnification? &lt;/p&gt;&#xA;" OwnerUserId="22913" LastActivityDate="2018-05-09T18:49:40.457" Title="How to improve the quality of an image of the planets when taken through the eyepiece of a telescope with a dlsr camera" Tags="&lt;photography&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26139" PostTypeId="1" CreationDate="2018-05-02T11:54:24.897" Score="7" ViewCount="852" Body="&lt;p&gt;What kind of amateur radioastronomy can any interested citized do with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) A 2 m dish antenna.&#xA;b) A 2,5 m antenna.&#xA;c) A 3 m antenna.&#xA;d) A 5 m antenna. &#xA;e) A &gt;5 m &amp;lt;10 m antenna (likely I can not afford this or the previous but I just want to know...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, space is sometimes a logistic issue for big stuff, but I wondered what kind of objects or sources can someone do with suitable instruments and with aid enough if not an expert. Projects and links would be welcome!&lt;/p&gt;&#xA;" OwnerUserId="2391" LastEditorUserId="2391" LastEditDate="2018-05-02T12:13:57.830" LastActivityDate="2018-06-29T08:59:46.837" Title="Amateur radioastronomy: dish suggestions" Tags="&lt;amateur-observing&gt;&lt;radio-astronomy&gt;&lt;radio-telescope&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26141" PostTypeId="2" ParentId="26139" CreationDate="2018-05-02T13:25:04.233" Score="6" Body="&lt;p&gt;The size of your dish determines two things:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Along with the temperature of your electronics, determines the &lt;strong&gt;signal-to-noise ratio&lt;/strong&gt; of your telescope.&lt;/li&gt;&#xA;&lt;li&gt;The size of your dish determines the &lt;strong&gt;angular resolution&lt;/strong&gt; you can expect. This has an approximate relationship of &lt;span class=&quot;math-container&quot;&gt;$$ R = \lambda / D$$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; is your angular resolution, &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; is your wavelength of light and &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; the width of your aperture (dish).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Instead of talking about dish size lets talk about interesting things to look at and see what the requirements are:&lt;/p&gt;&#xA;&lt;h3&gt;The Sun&lt;/h3&gt;&#xA;&lt;p&gt;The sun is a good source of radio waves in the 10cm wavelength. Since the sun is around half a degree, we would need a dish at least 11.5 meters before you could expect to see the sun as anything more than a point source.&lt;/p&gt;&#xA;&lt;h3&gt;Jupiter&lt;/h3&gt;&#xA;&lt;p&gt;Jupiter has some magnetic field effects that produce radio waves in the 10-100cm wavelength range. Of course Jupiter is 50 arcseconds across, and would need a dish 412 meters in diameter to do any amount of resolution.&lt;/p&gt;&#xA;&lt;h3&gt;Cosmic Microwave Background&lt;/h3&gt;&#xA;&lt;p&gt;The CMB was one of the earliest measurements of radio waves from space. The strongest peak of the CMB is in the 1mm wavelength. However it is very weak. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Discovery_of_cosmic_microwave_background_radiation&quot; rel=&quot;noreferrer&quot;&gt;original investigators&lt;/a&gt; cooled their radio detector with liquid helium, and I imagine something similar would be required.&lt;/p&gt;&#xA;&lt;h3&gt;Terrestrial and Man-made sources&lt;/h3&gt;&#xA;&lt;p&gt;There are plenty of man-made radio sources. Airports have a radio wave emitter in the 10cm range for radar. There are plenty of geosynchronous satellites along the equatorial plane that use the X-band in the ~2-5cm range. These by definition have a strong enough signal to be seen from Earth with even a small dish.&lt;/p&gt;&#xA;&lt;h3&gt;Summary&lt;/h3&gt;&#xA;&lt;p&gt;Resolving power is likely out of the reach of an amateur radio astronomer since the dishes have to be either very large, or you must use a much more sophisticated &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_interferometer&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;radio interferometry&lt;/em&gt;&lt;/a&gt; method to achieve any angular resolution. You can however still see things as point sources in the sky if you are so inclined. An excellent example of what is likely accomplished by most people would be this &lt;a href=&quot;https://youtu.be/aeah3fFYlnA&quot; rel=&quot;noreferrer&quot;&gt;project here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="22897" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-05-02T13:32:49.407" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26142" PostTypeId="2" ParentId="23773" CreationDate="2018-05-02T15:26:03.247" Score="4" Body="&lt;p&gt;Just to provide an analytical formula for @uhoh's correct time-averaged distance, here the derivation of $\langle r\rangle_t=1+\epsilon^2/2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a=1 \qquad c=e\qquad b=\sqrt{1-e^2}\\&#xA;\vec{r}=(\cos \beta-e,\sqrt{1-e^2}\sin \beta)\\&#xA;\vec{r'}=(-\sin \beta,\sqrt{1-e^2}\cos \beta)\\&#xA;|r|^2=\cos^2 \beta -2e\cos \beta+e^2+\sin^2\beta-e^2\sin^2\beta=1-2e\cos \beta+e^2\cos \beta=(1-e\cos \beta)^2\\&#xA;|\vec{r}\times\vec{r'}|=(\cos \beta-e)\sqrt{1-e^2}\cos \beta+\sin \beta\sqrt{1-e^2}\sin \beta=&#xA;\sqrt{1-e^2}(1-e\cos \beta)=2\,\mathrm{d}A/\mathrm{d}\beta=\text{const. }\mathrm{d}t/\mathrm{d}\beta\\\&#xA;\langle|r|\rangle_t=&#xA;\int_0^{T}|r|\mathrm{d}t=\int_0^{2\pi}{|r|\frac{\mathrm{d}t}{\mathrm{d}\beta}}\mathrm{d}\beta&#xA;=\frac{\sqrt{1-e^2}\int_0^{2\pi}{(1-e\cos\beta)^2}\mathrm{d}\beta}&#xA;{\sqrt{1-e^2}\int_0^{2\pi}({1-e\cos\beta})\mathrm{d}\beta}&#xA;=\frac{2\pi+e^2\pi}{2\pi}=1+e^2/2$$&lt;/p&gt;&#xA;" OwnerUserId="13164" LastActivityDate="2018-05-02T15:26:03.247" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26144" PostTypeId="2" ParentId="26121" CreationDate="2018-05-02T18:23:52.000" Score="0" Body="&lt;p&gt;In the very early evolution of the universe, primordial black holes could have formed before there were any stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The early universe was very dense and small variations in density could cause regions to undergo gravitational collapse, forming black holes. These black holes would not have formed from a star, and could be outside the range of masses of a black hole formed from the collapse of the core of a massive star. Depending on the model you use for the scale of inhomogeneities in the Early Universe, these black holes could have masses between a milligram up to several thousand times the mass of the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that some of the supermassive black holes at the centre of galaxies may have been primordial black holes. (but this is speculative)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-05-02T18:23:52.000" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26145" PostTypeId="2" ParentId="26131" CreationDate="2018-05-02T18:45:05.083" Score="4" Body="&lt;p&gt;These are spurious data points. They are likely genuine objects, but the parallax value is incorrect. There are a small number of these in the Gaia data set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 59 object with a parallax greater than Proxima Centuri. These don't represent genuine objects, instead they are when two sources are closely aligned (within about 0.2 arcsec). Most of these are found in the milky way, where there are lots of stars close together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See&#xA;&lt;a href=&quot;https://www.cosmos.esa.int/web/gaia/forum#!/thread/1688565/1688576&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.cosmos.esa.int/web/gaia/forum#!/thread/1688565/1688576&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And section 9 of &lt;a href=&quot;https://arxiv.org/pdf/1804.09365.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/pdf/1804.09365.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22918" LastEditorUserId="2365" LastEditDate="2018-05-03T16:47:46.353" LastActivityDate="2018-05-03T16:47:46.353" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26146" PostTypeId="2" ParentId="26079" CreationDate="2018-05-02T19:15:17.863" Score="2" Body="&lt;p&gt;As barrycarter commented, &lt;a href=&quot;https://naif.jpl.nasa.gov/pub/naif/generic_kernels/spk/asteroids/&quot; rel=&quot;nofollow noreferrer&quot;&gt;codes_300ast_20100725&lt;/a&gt;&#xA;has ephemerides for 300 large asteroids, generated in 2010 and supposedly valid from 1800 to 2199.&#xA;For other asteroids, a &lt;a href=&quot;http://forum.celestialmatters.org/viewtopic.php?f=18&amp;amp;t=629&quot; rel=&quot;nofollow noreferrer&quot;&gt;Celestia developer forum&lt;/a&gt; shows how to how to request an SPK from HORIZONS's telnet interface.&#xA;However, they appear to use the SPK for ephemeris only, adding albedo in Celestia's own SSC file format.&#xA;In SPICE documentation, neither the tutorials nor the &quot;required reading&quot; seems to mention albedo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more direct approach to your goal might be the JPL &lt;a href=&quot;https://ssd.jpl.nasa.gov/sbdb_query.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;Small Body Database Search Engine&lt;/a&gt;.&#xA;For example, if you query for numbered NEO asteroids with absolute magnitude H &amp;lt; 16, you get a table of 169 rows.&#xA;Diameter and albedo are known for some asteroids but not others, even though their orbits and rotational periods may be well established.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-05-02T19:15:17.863" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26147" PostTypeId="2" ParentId="26136" CreationDate="2018-05-02T20:28:22.323" Score="1" Body="&lt;p&gt;If you were using both a camera lens and an eyepiece (afocal method), try eyepiece projection.&#xA;In this method, the eyepiece is the only optical element between the prime focus and the detector.&#xA;You would need an eyepiece projection adapter tube to connect the camera body to the telescope's focuser.&#xA;A longer tube will increase magnification but also increase exposure time.&#xA;Any tube length which reasonably balances those concerns should be OK.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also possible to use a Barlow lens and no eyepiece to extend the effective focal length of the telescope.&#xA;If your adapter tube is wide enough to accommodate the Barlow, you may be able to use it this way too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomy Source has an article with more details about &lt;a href=&quot;http://www.astronomysource.com/2011/10/13/eyepiece-projection/&quot; rel=&quot;nofollow noreferrer&quot;&gt;eyepiece projection&lt;/a&gt;.&#xA;Tele Vue has an overview of that and &lt;a href=&quot;http://www.televue.com/engine/TV3b_page.asp?id=85&quot; rel=&quot;nofollow noreferrer&quot;&gt;other methods&lt;/a&gt;, plugging their own products of course.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-05-09T18:49:40.457" LastActivityDate="2018-05-09T18:49:40.457" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26151" PostTypeId="1" AcceptedAnswerId="26159" CreationDate="2018-05-03T08:22:28.340" Score="3" ViewCount="188" Body="&lt;p&gt;I have 100cm F.L objective and 10cm F.L eyepiece 10cm in diameter and double convex both. &#xA;what are the planetary objects I can see with telescope made with these lenses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also have a double convex and double concave lens of 20cm F.L both having 5cm diameter. will any of the above said lens help the telescope I'm going to make if included somewhere in the design? &lt;/p&gt;&#xA;" OwnerUserId="18486" LastEditorUserId="11264" LastEditDate="2018-05-03T14:09:06.533" LastActivityDate="2018-05-11T21:56:43.690" Title="Need help in creating telescope" Tags="&lt;telescope&gt;&lt;telescope-lens&gt;&lt;telescope-making&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26152" PostTypeId="1" AcceptedAnswerId="26160" CreationDate="2018-05-03T08:53:50.840" Score="2" ViewCount="61" Body="&lt;p&gt;as I was browsing Nasa's Eyes few days ago, I saved a screenshot, &lt;a href=&quot;https://i.stack.imgur.com/Gdh1b.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gdh1b.jpg&quot; alt=&quot;[img]https://i.imgur.com/w1rpBkr.jpg[/img]&quot;&gt;&lt;/a&gt; but it seems text is blurrry, supher-485, first letter is S ends with er-485 from as I can read&#xA;it kinda looked like Cosmic Web, Indra's net, so i saved cool screenshot&#xA;Help me identify this cluster?&lt;/p&gt;&#xA;" OwnerUserId="20219" LastEditorUserId="20219" LastEditDate="2018-05-04T07:59:38.093" LastActivityDate="2018-05-04T07:59:38.093" Title="Help me identify a cluster" Tags="&lt;exoplanet&gt;&lt;globular-clusters&gt;&lt;star-cluster&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2018-05-09T21:05:24.653" ContentLicense="CC BY-SA 4.0" />
  <row Id="26154" PostTypeId="2" ParentId="26095" CreationDate="2018-05-03T11:13:42.300" Score="0" Body="&lt;p&gt;barrycarter provided the answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using curl, I got 61240 lines total, but that includes some headers, so I think 61237 is correct and your browser may just be timing out. In particular, the last file on my list is &quot;GaiaSource_999922404314639104_1000172126596665472.csv.gz&quot;, which seems correct, since the first number increases with each file and seems to finally &quot;wrap around&quot; with the last file. – barrycarter Apr 29 at 3:54&lt;/p&gt;&#xA;" OwnerUserId="22867" LastActivityDate="2018-05-03T11:13:42.300" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26155" PostTypeId="1" AcceptedAnswerId="26162" CreationDate="2018-05-03T11:53:18.760" Score="2" ViewCount="70" Body="&lt;p&gt;I don't know if this question should be here, if it shouldn't just let me know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, I'm programming in Java and I want to get the actual stats of the planet, like orbit ,actual temperature etc... Is there any method, package in any language that makes this?&lt;/p&gt;&#xA;" OwnerUserId="16990" LastActivityDate="2018-05-03T21:02:31.310" Title="Any programme that shows actual planet information?" Tags="&lt;solar-system&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26157" PostTypeId="2" ParentId="26133" CreationDate="2018-05-03T13:32:40.423" Score="2" Body="&lt;p&gt;The process of determining photometric redshift includes, e.g.,:&#xA;1. objects of interest&#xA;2. types of data (e.g., one epoch multiple filters, one filter multiple epochs)&#xA;3. method of analysis (including model, and algorithm)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what you mentioned, those are just algorithms. So, you might take a look around on other aspects as listed.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-05-03T13:32:40.423" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26158" PostTypeId="1" CreationDate="2018-05-03T18:02:23.807" Score="3" ViewCount="49" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZloaJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZloaJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my data and purpose of this assignment is to fit a lorentzian to this data.&#xA;And.. i dont know how to do that....:(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, there is sharply lowered region.&#xA;Couple of hours of googling isn't helpful..&#xA;Second, should I fit lorentzian for 2 times? &#xA;There are two peaks.&lt;/p&gt;&#xA;" OwnerUserId="22931" LastActivityDate="2018-05-03T18:02:23.807" Title="How to fit lorentzian to lyman alpha cross section?" Tags="&lt;star&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26159" PostTypeId="2" ParentId="26151" CreationDate="2018-05-03T18:23:10.483" Score="3" Body="&lt;p&gt;If F is the focal length of the objective, and f the focal length of the ocular, then the magnification of the instrument is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M = F / f&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you'll get 10x with the first combination. This is enough to see craters, mountains and valleys on the Moon. You will probably not see planetary disks or the rings of Saturn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The large diameter of the ocular might be awkward, but it should work. Make sure both lenses are installed on a rigid tube - plastic tubes from water pipes should work. Make sure the lenses are perpendicular to the axis of the tube - small errors there reduce overall performance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will need a way to finely adjust the distance between objective and ocular - the ideal distance is F + f but don't just glue them in place; instead, see if you can make two concentric sliding tubes, you know, like the spyglasses from pirate movies. Then you'll be able to look through the instrument and adjust the distance until the image is sharp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there's any way you could find a used ocular from an old microscope or binoculars, use that as the ocular in your instrument. It will provide higher magnification (smaller focal length) and its performance should be much better than a simple biconvex lens - fewer aberrations, since true oculars have internal corrections (well, most of them). With the higher magnification you will probably see the rings of Saturn, the moons of Jupiter, or even some equatorial belts on Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take it slowly and do it right. This is totally doable with simple materials. This is the design of the first telescope I've built many years ago. You'll be very proud when something you made with your own hands actually works.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Yes, you can stack lenses for extra power in the ocular. If you stack two convergent lenses, so close that they touch each other, having the same focal length, the resulting system has about half the focal length of one lens. If there's some distance in between them, things are a bit more complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will notice that power is not everything, and sometimes more power is beneficial while other times it creates problems. It depends on many factors, which you can discover through experimentation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also read about the various eyepiece designs, which are made for specific purposes - not just magnifying power, but also lack of aberrations and a high quality image. But those require special components to make.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Eyepiece&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Eyepiece&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By all means go ahead and try various combinations yourself. There's a lot to discover. It's a lot of fun.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Telescopes for astronomy cannot be used hand-held. You must use some kind of mount. You could make one, doesn't matter how simple, as long as it's stable and it can turn in both directions (up/down and left/right). E.g. you could make a mount out of a cardboard box:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://aquillam.wordpress.com/2010/01/04/a-cheap-tripod-alternative/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://aquillam.wordpress.com/2010/01/04/a-cheap-tripod-alternative/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K9Zay.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K9Zay.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google &quot;diy refractor mount&quot; or &quot;galileoscope mount&quot; for more ideas.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the eyepiece is convergent, the image is reversed. This is normal for usage in astronomy, where the orientation of the image doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are special devices called prism diagonals that can be inserted between objective and ocular, that fix the reverse image problem. Star diagonals show the image correct-side-up, but left and right are reversed. Correct-image diagonals show the image correct-side-up and also left and right are correct. They achieve this by doing one or multiple reflections in prisms.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2018-05-11T21:56:43.690" LastActivityDate="2018-05-11T21:56:43.690" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26160" PostTypeId="2" ParentId="26152" CreationDate="2018-05-03T18:50:48.630" Score="1" Body="&lt;p&gt;Since this appears to be a screenshot of &lt;a href=&quot;https://eyes.nasa.gov/eyes-on-exoplanets.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eyes on Exoplanets&lt;/a&gt;, I searched for &quot;exoplanet 485&quot; and found &lt;a href=&quot;http://exoplanet.eu/catalog/kepler-485_b/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kepler-485 b&lt;/a&gt;.&#xA;With the colors exaggerated, the label area could be &quot;Kepler-485&quot; in yellow overwritten with &quot;Sun&quot; in orange.&#xA;If this is correct, it's not a cluster, just a star with a planet, plus some artifacts from the instrument which took the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XBTA9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XBTA9.png&quot; alt=&quot;detail of OP&amp;#39;s screenshot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-05-03T18:50:48.630" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26161" PostTypeId="2" ParentId="26151" CreationDate="2018-05-03T20:15:42.097" Score="2" Body="&lt;p&gt;If you use the 100cm FL convex lens as an objective and the 20cm FL concave lens as an eyepiece, the telescope will have only 5x magnification but otherwise be similar to the one Galileo used to discover the &lt;a href=&quot;https://en.wikipedia.org/wiki/Galilean_moons&quot; rel=&quot;nofollow noreferrer&quot;&gt;four largest moons of Jupiter&lt;/a&gt;.&#xA;Here is &lt;a href=&quot;http://galileo.rice.edu/lib/student_work/astronomy96/mtelescope.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;documentation&lt;/a&gt; of such a telescope built by students.&#xA;Beyond 10x or so, you need a steady mount for the telescope, and simple lenses may show &lt;a href=&quot;https://en.wikipedia.org/wiki/Chromatic_aberration&quot; rel=&quot;nofollow noreferrer&quot;&gt;chromatic aberration&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-05-03T20:15:42.097" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26162" PostTypeId="2" ParentId="26155" CreationDate="2018-05-03T21:02:31.310" Score="3" Body="&lt;p&gt;This seems like something that would best be provided by a simple data table which you can open as needed rather than as a feature of the language.  There's one at &lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://nssdc.gsfc.nasa.gov/planetary/factsheet/&lt;/a&gt; that has the basic statistics for the planets (plus Pluto); is that the sort of thing you're after? &lt;/p&gt;&#xA;" OwnerUserId="10914" LastActivityDate="2018-05-03T21:02:31.310" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26163" PostTypeId="1" AcceptedAnswerId="26165" CreationDate="2018-05-03T23:45:28.870" Score="29" ViewCount="3251" Body="&lt;p&gt;I work with astrophysicists and require some basic knowledge of many astronomical sources, however research priorities often demand that most of human knowledge on a subject be taken for granted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am currently researching &lt;a href=&quot;https://en.wikipedia.org/wiki/Nova&quot; rel=&quot;noreferrer&quot;&gt;galactic novae&lt;/a&gt;, and I find it relevant to summarize their history briefly when presenting my research to certain audiences. Unfortunately, I am unable to find any source material which describes &lt;em&gt;how&lt;/em&gt; we know one key aspect of the events: that they are an accreting white dwarf in a stellar binary. This fact appears to be so well founded that no scientific paper feels obligated to cite it when stated, but basic resources like astronomical encyclopedia also make no reference that I've seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How do we know that novae are binary systems?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g., have follow-up observations clearly identified the white dwarf and its companion? Or do other astronomical measurements strongly confirm this binary hypothesis (and make it all but obviously true)? I apologize if it's as simple as &quot;someone looked through a telescope, and it was pretty obvious&quot; -- in my experience no revelation in astrophysics is nearly so simple, but certainly this could be the case.&lt;/p&gt;&#xA;" OwnerUserId="7480" LastActivityDate="2018-05-04T10:15:52.330" Title="We know what a nova is, but how?" Tags="&lt;history&gt;&lt;nova&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26165" PostTypeId="2" ParentId="26163" CreationDate="2018-05-04T05:03:00.363" Score="20" Body="&lt;p&gt;Following a reference to &lt;a href=&quot;https://arxiv.org/abs/1112.2589&quot; rel=&quot;noreferrer&quot;&gt;Darley et al., ApJ &lt;strong&gt;746&lt;/strong&gt;, 61 (2012)&lt;/a&gt; from your Wikipedia link gives a (very technical) discussion of nova progenitors, including distinctions between nova systems where the secondary stars are main sequence or supergiant stars, and distinctions among white dwarfs with different chemistries.&#xA;The first sentence of that paper is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A classical nova (CN) outburst occurs in an interacting&#xA;  binary system comprising a white dwarf (WD, the primary) and&#xA;  typically a late-type main-sequence (MS) star (the secondary)&#xA;  that fills its Roche lobe (&lt;a href=&quot;http://adsabs.harvard.edu/doi/10.1086/146128&quot; rel=&quot;noreferrer&quot;&gt;Crawford &amp;amp; Kraft, 1956&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That suggests the 1956 paper is the original proposal for the Roche overflow model of the classical nova.&#xA;Like many original-idea papers, it's a pretty clear read.&#xA;But for your question, Crawford and Kraft seem to hedge about whether the &quot;blue star&quot; in their particular pair needs to be a white dwarf:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[T]he&#xA;  observed&#xA;  luminosity&#xA;  of&#xA;  the&#xA;  blue&#xA;  star&#xA;  is&#xA;  essentially&#xA;  due&#xA;  to&#xA;  the&#xA;  energy&#xA;  released&#xA;  by&#xA;  the&#xA;  accreted&#xA;  material.&#xA;  This&#xA;  view&#xA;  is&#xA;  strengthened&#xA;  also&#xA;  by&#xA;  the&#xA;  fact&#xA;  that&#xA;  the&#xA;  blue&#xA;  star&#xA;  occupies&#xA;  a&#xA;  peculiar&#xA;  position&#xA;  in&#xA;  the&#xA;  H-R&#xA;  diagram.&#xA;  It&#xA;  lies&#xA;  10.5&#xA;  vis.&#xA;  mag.&#xA;  below&#xA;  the&#xA;  main&#xA;  sequence&#xA;  but&#xA;  about&#xA;  4&#xA;  mag.&#xA;  above&#xA;  the&#xA;  most&#xA;  luminous&#xA;  white&#xA;  dwarfs,&#xA;  whose&#xA;  effective&#xA;  temperature&#xA;  it&#xA;  exceeds&#xA;  by&#xA;  about&#xA;  8000°&#xA;  K.&#xA;  Unless&#xA;  the&#xA;  blue&#xA;  star&#xA;  is&#xA;  essentially&#xA;  degenerate,&#xA;  it&#xA;  can&#xA;  readily&#xA;  be&#xA;  shown&#xA;  that&#xA;  the&#xA;  small&#xA;  radius&#xA;  implies&#xA;  such&#xA;  a&#xA;  high&#xA;  internal&#xA;  temperature&#xA;  that&#xA;  electron&#xA;  scattering&#xA;  is&#xA;  the&#xA;  principal&#xA;  source&#xA;  of&#xA;  opacity.&#xA;  A&#xA;  simple&#xA;  calculation&#xA;  based&#xA;  on&#xA;  the&#xA;  standard&#xA;  model&#xA;  then&#xA;  yields&#xA;  a&#xA;  luminosity&#xA;  8&#xA;  mag.&#xA;  brighter&#xA;  than&#xA;  is&#xA;  observed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, Crawford and Kraft don't come out and say &quot;definitely a WD,&quot; but if it's a non-degenerate star, it's a very strange one.&#xA;More modern observations of novae are compared to detailed models of the dynamics of the surface dynamics, models which &lt;a href=&quot;https://doi.org/10.1086/175741&quot; rel=&quot;noreferrer&quot;&gt;have been debated vigorously for decades&lt;/a&gt;; the current generation of comparisons to data are sensitive to details like &lt;a href=&quot;https://dx.doi.org/10.1086/516838&quot; rel=&quot;noreferrer&quot;&gt;the amount of helium accumulating on the white dwarf's surface&lt;/a&gt; during the nova event. It seems unlikely that such details could even come close if the underlying assumptions about the basic physics of the erupting star were wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that a classical nova system can be thought of as a type of &lt;a href=&quot;https://en.wikipedia.org/wiki/Contact_binary&quot; rel=&quot;noreferrer&quot;&gt;contact binary star&lt;/a&gt;.&#xA;For any reasonable estimate of the size of the giant star, a distance of 10 AU between the two members of the pair seems like an overlarge estimate.&#xA;Ten astronomical units of separation viewed from a distance of 50 parsecs is already a gap of 0.1 seconds of arc. I wouldn't expect to see visible-light photographs showing both the giant star and the white dwarf, but rather that all of the information about the binary systems comes from &lt;a href=&quot;https://en.wikipedia.org/wiki/Binary_star#Spectroscopic_binaries&quot; rel=&quot;noreferrer&quot;&gt;spectroscopy&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1498" LastActivityDate="2018-05-04T05:03:00.363" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26166" PostTypeId="1" CreationDate="2018-05-04T06:54:43.967" Score="8" ViewCount="354" Body="&lt;p&gt;Certainly among the first colonists on Mars there will be a few people interested in Astronomy enough to enjoy the night sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For them, familliar Mars will be missing and there will be a new, unfamiliar planet instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planet by planet, how will their appearance and especially &lt;em&gt;their behavior as seen from Mars&lt;/em&gt; differ from the way they behave in the night sky as seen from Earth?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-05-04T07:11:23.803" LastActivityDate="2020-04-28T22:44:10.243" Title="How will planets behave in the night sky as seen by Mars colonists?" Tags="&lt;mars&gt;&lt;night-sky&gt;" AnswerCount="2" CommentCount="20" ContentLicense="CC BY-SA 4.0" />
  <row Id="26167" PostTypeId="1" AcceptedAnswerId="26168" CreationDate="2018-05-04T09:52:50.560" Score="4" ViewCount="242" Body="&lt;p&gt;Astronomers define the Solar System as the distance under the influence of gravity from the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read that the &lt;a href=&quot;https://en.wikipedia.org/wiki/Oort_cloud&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oort cloud&lt;/a&gt; is probably the farthest object that surrounds the sun. The sphere of this icy cloud is thought to surround the Sun at up to a distance of &lt;strong&gt;50,000 AU&lt;/strong&gt;, making its total diameter nearly two light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How far can the sun's gravity extend to hold celestial bodies around it? The Oort cloud may be the known farthest body but its edge doesn't mark the end of the sun's effective gravitational pull.&lt;/p&gt;&#xA;" OwnerUserId="15324" LastEditorDisplayName="user1569" LastEditDate="2018-05-04T10:15:21.787" LastActivityDate="2018-05-05T18:30:03.003" Title="How far out can the sun keep celestial objects revolving?" Tags="&lt;orbit&gt;&lt;gravity&gt;&lt;oort-cloud&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26168" PostTypeId="2" ParentId="26167" CreationDate="2018-05-04T10:10:18.070" Score="7" Body="&lt;p&gt;The suns gravitational pull extends unlimitedly far and, if the universe were otherwise empty, a body could orbit it (very slowly) at unlimited distance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the universe is NOT otherwise empty. The nearest other stars are about 4 light years from the Sun at the moment, so a body more than two lightyears away will typically feel equally strong attraction from other stars as from the Sun, and so will not really orbit the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a bit of an over-simplification, because (a) the sun is not surrounded by other stars 4 l.y. away in all directions (here is &lt;a href=&quot;http://www.atlasoftheuniverse.com/12lys.html&quot; rel=&quot;noreferrer&quot;&gt;a chart of nearby stars&lt;/a&gt;) so a body might orbit in such a way that it didn't pass between the sun and any of its closest neighbours; (b) not all stars have the same mass, many are smaller than the Sun but some (eg Sirius) much bigger, so their gravitational influences vary and (c) the stars are all moving, so that from time to time one comes quite a bit closer, disrupting orbits.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-05-04T10:10:18.070" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26169" PostTypeId="2" ParentId="26163" CreationDate="2018-05-04T10:15:52.330" Score="7" Body="&lt;p&gt;The real key, I suspect, was that observations of &quot;postnovae&quot; -- classical novae &lt;em&gt;after&lt;/em&gt; the nova outburst, when the light from the outburst itself no long obscured light from the underlying system -- often showed clear characteristics of binary stars. This took the form of periodic dips in the light curve, suggestive of eclipses, or direct spectroscopic evidence for binary motion, or both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is discussed, with references (including the Crawford &amp;amp; Kraft 1956 reference that rob mentions in his answer), in Section 2.2 of the &lt;a href=&quot;http://adsabs.harvard.edu/abs/1978ARA%26A..16..171G&quot; rel=&quot;noreferrer&quot;&gt;1978 review article by Gallagher &amp;amp; Starrfield&lt;/a&gt; in &lt;em&gt;Annual Reviews of Astronomy &amp;amp; Astrophysics&lt;/em&gt;. Section 2.4 discusses some of the evidence then available for the primaries being accreting white dwarfs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you're not already aware of it, review articles in &lt;em&gt;Ann.Rev.A&amp;amp;A&lt;/em&gt; are often a good place to look for answers to questions like this. Sometimes earlier articles are better for certain questions, because they're closer in time to when people were still figuring things out, and so they go over the early evidence in more detail than a later article would.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2018-05-04T10:15:52.330" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26171" PostTypeId="2" ParentId="10896" CreationDate="2018-05-04T14:32:57.220" Score="1" Body="&lt;p&gt;I was about to put this in a comment, but it did not let me because it is too long. So, I will just put it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a certain radius from the center of the black hole, where light inside this radius cannot come out, i.e., event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The light that we see is from materials those are spiralling inwards into the black hole, and they are outside the event horizon. These materials experience frictions, therefore they are heated up, and lose their energy into radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black hole can be formed by various mechanisms, according to theories, with or without star explosions (a.k.a. supernovae, SNe). For star-forming black hole, the size, which is typically referred to the zero-age main sequence (a.k.a. ZAMS) mass, is still under debate. But for sure, the star has to be &quot;massive&quot; $&amp;gt; 8$ solar masses in ZAMS. Note that these massive stars might form neutron stars or black holes after the SNe; also the stars might not leave any compact object if the explosions are pair-instability SNe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a galaxy's black hole, how it is formed is still unknow. One theory is from the accumulation of stuff after SNe, like you said. But, we don't know exactly at this moment, therefore to answer about the &quot;size,&quot; I don't think it can be answered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A black hole is just another massive object possessing huge gravitational potential, like our Sun. Therefore, there are orbits those are stable and are not attracted into the center of the potential.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-05-04T14:32:57.220" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26172" PostTypeId="2" ParentId="26166" CreationDate="2018-05-04T14:36:51.390" Score="3" Body="&lt;p&gt;This should be a relatively simple problem to solve - which is the Academician's way of saying &quot;do it as a homework problem&quot; :-) .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First thing: find some celestial mechanics calculator and determine when each outer and inner planet is visible in Mars' night sky.  That gives you timings.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next thing: compare the distance from each outer planet to Mars vs. Earth. Maybe start with closest and farthest approach as examples. That'll tell you both the angular extent of the planet as well as the relative brightness compared with Terran observations.  Don't forget to add a fudge factor for the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, probably the most obvious difference will be that &lt;em&gt;stars&lt;/em&gt; no longer twinkle. (yeah I'm ignoring during serious dust storms).  You could use a wicked large primary and have no need of adaptive optic corrections.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW there are photos of Earth as seen by the Mars rovers (or maybe just the surveying satellites, I forget) so you can get an idea of the color/features available. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-05-04T14:36:51.390" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26173" PostTypeId="2" ParentId="25781" CreationDate="2018-05-04T19:15:51.440" Score="4" Body="&lt;p&gt;Stars are far from perfect blackbodies due to scattering/reflection.  This is especially true for hotter stars, because of all the free electrons, but even cooler stars can reflect a significant amount.  For example, in aanda.org/articles/aa/pdf/2001/19/aa1009.pdf you will see that they use a reflection albedo of 0.30 for the K star and 1.00 for the F star, but the latter number is not meant to be taken seriously, they simply don't care if the light is reflected or absorbed and re-emitted because it isn't an important term. But the value of 0.30 for the K star might be meant more seriously, though it is still not regarded as a critical parameter because it only affects the color of the light that is reflected, not the total amount of light (given that stars are in radiative equilibrium, so must ultimately return all the incident light, whether it happens by reflection or heating).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, stellar emissions are often characterized by &quot;effective temperature,&quot; to connect the surface flux of a star to the Stefan-Boltzmann formula for the emission of a blackbody by using a T parameter that is not necessarily the actual temperature.  When using this notion, as is quite common for dealing with stars, there is no essential difference between heating of and reflection from the surface of the star in question. The details of the difference have to do with the shape of the spectrum, but that shape is generally not a Planck function anyway, so as soon as one is using the &quot;effective temperature&quot; concept one has already parted company from a detailed understanding of the shape of the spectrum.  (When you do want the details of the spectrum, you will have to model the situation with some care.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are right that the albedo depends on the type of star (hotter stars having high scattering albedo due to all those free electrons), and many stars do show a lot of scattering from their surfaces.  Scattering/reflection elevates the effective temperature above the actual temperature of the layers being looked at, but again, if you stick to effective temperature, it makes no difference if there is scattering or heating of a blackbody, because effective temperature explicitly refers to the outgoing  flux with no claims on the actual temperature.  It is quite common for this distinction to be swept under the rug, and most stellar temperatures are effective temperatures, not actual temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another consequence of radiative equilibrium is that the &quot;reflection effect&quot; cannot alter the total luminosity of the binary system seen from all angles, so if the effective temperature of the surface of the stars is elevated by reflection and/or heating, as seen from some directions, this must be exactly compensated by the reduced brightness due to eclipses seen from other directions.  So the reflection effect is part of the study of eclipse light curves, which are quite useful for understanding things like the sizes of the stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:  a nice paper on the polarization of reflected light was provided in the comments by uhoh, it is  nature.com/articles/s41550-019-0738-7 .  The idea is that if the stars are not too close to each other, there's not a lot of reflected light, but it is still noticeable by virtue of the fact that it is highly linearly polarized in the direction perpendicular to the line between the stars.  So it shows up much better in linearly polarized light, a potentially important new diagnostic of binary systems.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2019-05-01T04:22:07.453" LastActivityDate="2019-05-01T04:22:07.453" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26174" PostTypeId="2" ParentId="26117" CreationDate="2018-05-05T03:12:28.627" Score="2" Body="&lt;p&gt;An insightful option is to open in a python IDE using &lt;a href=&quot;http://docs.astropy.org/en/stable/io/fits/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astropy&lt;/a&gt;. I find it useful because you can manipulate the files as your learn their structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also possible to use matplotlib in conjunction with Astropy and then basically have a FITS file viewer that you can code yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import astropy&#xA;import matplotlib.pyplot as plt&#xA;from astropy.visualization import astropy_mpl_style&#xA;&#xA;&#xA;    def open_fits_file(fits_path):&#xA;        for file in Path(fits_path).iterdir():&#xA;            #Not really using the with, but it'll manage closing the file...&#xA;            with fits.open(file) as opened_fits:&#xA;                #printing header infos for that file&#xA;                opened_fits.info()&#xA;                plot_image(fits.getdata(file, ext=2))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then to plot the image:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def plot_image(fit_image_data):&#xA;    #astropy has better plot settings for this use&#xA;    plt.style.use(astropy_mpl_style)&#xA;    plt.figure()&#xA;    plt.imshow(fit_image_data, cmap=&quot;gray&quot;)&#xA;    plt.colorbar()&#xA;    plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I used that to plot the pixels  from these simulated data linked in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned by Mike G, HEASARC as a list of viewer, including FV (fits viewer). It works fine and it has some built-in support to link up with other astronomy database. It has an apollo-era feel in terms of appearance, which is nice... I guess?&lt;/p&gt;&#xA;" OwnerUserId="22894" LastEditorUserId="22894" LastEditDate="2018-05-13T14:35:36.510" LastActivityDate="2018-05-13T14:35:36.510" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26177" PostTypeId="1" AcceptedAnswerId="26178" CreationDate="2018-05-05T07:14:39.457" Score="5" ViewCount="462" Body="&lt;p&gt;@PM2Ring's &lt;a href=&quot;https://astronomy.stackexchange.com/questions/26166/how-will-planets-behave-in-the-night-sky-as-seen-by-mars-colonists#comment45224_26166&quot;&gt;comment&lt;/a&gt; mentions the story &lt;em&gt;Transit of Earth&lt;/em&gt; written by the famous British science fiction writer, science writer and futurist,&lt;a href=&quot;https://www.youtube.com/watch?v=csxT8Y9MaSs&quot; rel=&quot;nofollow noreferrer&quot;&gt;3&lt;/a&gt; inventor, undersea explorer, television series host, Fellow of the Royal Astronomical Society and Commander of the Order of the British Empire and celebrated polymath &lt;a href=&quot;https://en.wikipedia.org/wiki/Arthur_C._Clarke&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arthur C. Clarke&lt;/a&gt; in which the character's task is to record a transit of the Earth across the Sun's disk as seen from Mars.&lt;/p&gt;&#xA;&lt;p&gt;According to the story, this happens on May 11, 1984.&lt;/p&gt;&#xA;&lt;p&gt;The story also mentions that the previous occurrence would have been in 1905, and the following one would be 100 years following this one, presumably in 2084.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The radio has just printed a message from Earth, reminding me the transit starts in two hours, as if I’m likely to forget, when four men have already died so that I can be the first human being to see it, and the only one, for exactly a hundred years.&lt;/p&gt;&#xA;&lt;p&gt;It isn’t often that Sun, Earth and Mars line up neatly like this. The last time was in 1905, when poor old Lowell was still writing his beautiful nonsense about the canals and the great dying civilization that built them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Were there in fact transits of the Earth as seen from Mars on these dates (in 1905, on 11-May-1984, and in 2084)? If they do occur at roughly 100 year intervals as suggested, there may in fact be people on Mars to see the next one presuming that Sir Clarke did his maths right (did he ever not?). On what day would that be?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;You can listen to Sir Clarke read the story himself in the YouTube video &lt;a href=&quot;https://www.youtube.com/watch?v=csxT8Y9MaSs&quot; rel=&quot;nofollow noreferrer&quot;&gt;'Transit Of Earth' by Arthur C. Clarke read by himself&lt;/a&gt;. You can also listen here: &lt;a href=&quot;http://recordbrother.typepad.com/imagesilike/files/transit_of_earth.mp3&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://recordbrother.typepad.com/imagesilike/files/transit_of_earth.mp3&lt;/a&gt; and it seems download &lt;code&gt;HMIcomposite_2Kx2K.mov&lt;/code&gt; from NASA here (though I haven't verified it yet): &lt;a href=&quot;https://svs.gsfc.nasa.gov/vis/a000000/a003900/a003941/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://svs.gsfc.nasa.gov/vis/a000000/a003900/a003941/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-08-11T01:43:09.473" Title="When will the next transit of Earth be visible from Mars? Was the last one really on May 11, 1984?" Tags="&lt;mars&gt;&lt;planetary-transits&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26178" PostTypeId="2" ParentId="26177" CreationDate="2018-05-05T07:27:38.530" Score="5" Body="&lt;p&gt;Yes. The Clarke story is quite correct. Earth (and Moon) would transit the sun from the viewpoint of Mars on 8 May 1905, 11 May 1984 and again on 10 Nov 2084.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;full list over a period of about 1000 years, from &lt;a href=&quot;https://en.wikipedia.org/wiki/Transit_of_Earth_from_Mars&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia Transit of Earth from Mars&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;November 10, 1595   May 5, 1621     May 8, 1700     November  9, 1800   &#xA;November 12, 1879   May 8, 1905     May 11, 1984    November 10, 2084   &#xA;November 15, 2163   May 10, 2189    May 13, 2268    November 13, 2368   May 10, 2394&#xA;November 17, 2447   May 13, 2473    May 16, 2552    November 15, 2652   May 13, 2678&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;They follow a pattern of occurring after 26, 79 then 100 years.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="7982" LastEditDate="2019-08-11T01:43:09.473" LastActivityDate="2019-08-11T01:43:09.473" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26181" PostTypeId="1" CreationDate="2018-05-05T13:12:00.800" Score="-1" ViewCount="121" Body="&lt;p&gt;does Keplers third law work if one were to, in some way use the semi minor axis instead of the semi major axis? and does the difference between the arithmetic, geometric and harmonic means play any role?&lt;/p&gt;&#xA;" OwnerUserId="7391" LastActivityDate="2018-05-05T21:34:18.303" Title="keplers third law" Tags="&lt;kepler&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26183" PostTypeId="2" ParentId="26181" CreationDate="2018-05-05T15:41:37.377" Score="1" Body="&lt;p&gt;No, the relationship that Kepler discovered was between the length of the long axis of the ellipse, and the period of the planet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible for a comet with a very long skinny orbit, and an asteroid with a nearly circular orbit to have the same minor axis length, but obviously, the comet will have a much longer orbital period. For example, Halley's comet has a semiminor axis of about 3.5AU. An asteroid in a circular orbit with the same semi-minor axis would have a period of about 6 years, compared to Halley's period of 76 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know the semi-minor axis, and the perihelion distance (for example), then it would be possible to work out the semi-major axis, and the relative orbital period, but you would need more information than just the semi-minor axis.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-05-05T21:34:18.303" LastActivityDate="2018-05-05T21:34:18.303" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26186" PostTypeId="1" CreationDate="2018-05-05T20:02:49.753" Score="0" ViewCount="93" Body="&lt;p&gt;In all these years of listening, why hasn't SETI detected anything ? Is there no intelligent life out there or is it too far away to detect ?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastActivityDate="2018-05-05T23:20:57.810" Title="SETI search for extraterrestrial life" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2018-05-06T08:46:35.530" ContentLicense="CC BY-SA 4.0" />
  <row Id="26187" PostTypeId="2" ParentId="26186" CreationDate="2018-05-05T23:20:57.810" Score="1" Body="&lt;p&gt;There are many reasons why nothing has been detected. We could be too far away, no one is listening, listened at wrong time, and many more. But this does not mean there is no life on other planets. We just haven't discovered them yet.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastActivityDate="2018-05-05T23:20:57.810" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26188" PostTypeId="1" CreationDate="2018-05-06T00:55:31.850" Score="5" ViewCount="144" Body="&lt;p&gt;I'm not an astronomer but I enjoy looking up at the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So tonight I saw &lt;em&gt;something&lt;/em&gt; moving in the sky. It looked like a star, and it was definitely not a plane (I checked in flightradar).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an app, similar to flightradar for planes, that will tell me what's traveling above me in the sky?&lt;/p&gt;&#xA;" OwnerUserId="17849" LastEditorUserId="616" LastEditDate="2018-05-08T14:10:35.683" LastActivityDate="2018-05-28T14:13:33.223" Title="App that shows what's passing in sky?" Tags="&lt;night-sky&gt;&lt;artificial-satellite&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26189" PostTypeId="2" ParentId="26188" CreationDate="2018-05-06T03:28:55.390" Score="5" Body="&lt;p&gt;Chris Peat maintains a very sophisticated database of satellites and their visibilites at &lt;a href=&quot;http://heavens-above.com&quot; rel=&quot;noreferrer&quot;&gt;http://heavens-above.com&lt;/a&gt;, and there is an app available.&lt;/p&gt;&#xA;" OwnerUserId="1498" LastEditorUserId="1498" LastEditDate="2018-05-06T07:33:50.567" LastActivityDate="2018-05-06T07:33:50.567" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26190" PostTypeId="2" ParentId="26188" CreationDate="2018-05-06T07:11:48.180" Score="1" Body="&lt;p&gt;Try something like Stellarium (&lt;a href=&quot;https://play.google.com/store/apps/details?id=com.noctuasoftware.stellarium&quot; rel=&quot;nofollow noreferrer&quot;&gt;apk&lt;/a&gt; or &lt;a href=&quot;http://stellarium.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;PC&lt;/a&gt;), or Celestron's Starry Night &lt;a href=&quot;https://play.google.com/store/apps/details?id=com.celestron.skyportal&quot; rel=&quot;nofollow noreferrer&quot;&gt;apk&lt;/a&gt; for your phone (if you don't want to pay).&#xA;Hope that helps!&#xA;[Edit: I'm sorry I haven't mentioned anything on the Appstore; Stellarium is also available for &lt;a href=&quot;http://stellarium.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;MacOS&lt;/a&gt; and for &lt;a href=&quot;https://itunes.apple.com/in/app/stellarium-mobile-sky-map/id643165438?mt=8&quot; rel=&quot;nofollow noreferrer&quot;&gt;iOS&lt;/a&gt;, and also Celestron's Starry Night for &lt;a href=&quot;https://itunes.apple.com/in/app/celestron-skyportal/id877780544?mt=8&quot; rel=&quot;nofollow noreferrer&quot;&gt;iOS&lt;/a&gt;. This answer seems to help out on the iOS side: &#xA;&lt;a href=&quot;https://astronomy.stackexchange.com/a/26205/22907]&quot;&gt;https://astronomy.stackexchange.com/a/26205/22907]&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22907" LastEditorUserId="22907" LastEditDate="2018-05-16T05:23:26.863" LastActivityDate="2018-05-16T05:23:26.863" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26193" PostTypeId="1" CreationDate="2018-05-06T17:14:26.467" Score="3" ViewCount="209" Body="&lt;p&gt;As far as I understand the wave-optics effects,such e.g.diffraction at a caustic, are hardly observable in the gravitational lensing. Are there any observations where the diffraction paterns of the visible light have been seen? If not visible, which wavelength?&lt;/p&gt;&#xA;" OwnerUserId="18255" LastActivityDate="2019-01-02T21:01:30.310" Title="Is Diffraction of light observed in gravitational lensing/microlensing?" Tags="&lt;gravitational-lensing&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26195" PostTypeId="1" AcceptedAnswerId="26198" CreationDate="2018-05-07T00:57:38.400" Score="3" ViewCount="683" Body="&lt;p&gt;If I’m correct, the International Space Station is in free fall, but is moving fast enough to keep it’s position. Assuming that is true, does that mean that some/all planets orbiting the Sun are in free fall of the Sun?&lt;/p&gt;&#xA;" OwnerUserId="22740" LastActivityDate="2018-05-07T06:17:08.873" Title="Are the planets in our solar system in free fall?" Tags="&lt;orbit&gt;&lt;gravity&gt;&lt;solar-system&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26198" PostTypeId="2" ParentId="26195" CreationDate="2018-05-07T06:03:43.977" Score="8" Body="&lt;p&gt;Yes. Free fall means the only force acting on a body is gravity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only force acting on the Earth and every other body in orbit is gravity, so all the planets are in free fall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There are very small forces from, for example, solar wind but these are minute in comparison to the gravitational forces)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-05-07T06:17:08.873" LastActivityDate="2018-05-07T06:17:08.873" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26200" PostTypeId="1" AcceptedAnswerId="26215" CreationDate="2018-05-07T10:19:25.147" Score="5" ViewCount="726" Body="&lt;p&gt;With Astronomical Algorithms by J. Meeus (1998)&#xA;I have obtained the relevant illuminated fraction of the disk and position angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But they're numbers (all Greek to me!) I know, Meeus explains that they're ratios of the lengths of the illuminated limb and areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, so I have a ratio.  How do I implement that together with the position angle to work out the direction of the illuminated limb and the amount illuminated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming that the ratio 1.0 is totally illuminated (i.e. Full Moon)&#xA;and 0.0 is nothing (i.e. New Moon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then there is the position angle, I'm assuming that the direction in degrees is starting at 0 pointing upwards.  Is that the crescent shape direction i.e. the horns are pointing away from the directional angle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to be able plot this so that I can show visually the moon's illuminated fraction and direction for any day in any given year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said, I have the numbers, I'm trying to translate that into a more visually appealling and meaningful result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meeus says this is the selenocentric elongation of the Earth from the Sun (Phase Angle) - i.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the formula for illuminated fraction k is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k = ( 1 + cos i ) / 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;tan i = (R sin l) / (D - (R cos l))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R = distance Earth-Sun&#xA;D = distance Earth-Moon&#xA;l = geocentric elongation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meeus calls k the ratio of the lengths BC:AC&#xA;the same as the ratio of the areas: NBSC:NASC&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rOTD1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rOTD1.jpg&quot; alt=&quot;Illuminated Fraction of Moon&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22963" LastEditorUserId="22963" LastEditDate="2018-05-08T10:02:20.220" LastActivityDate="2018-05-08T14:07:12.860" Title="Implementation of formulae for Meeus for the Moon" Tags="&lt;the-moon&gt;&lt;observational-astronomy&gt;&lt;celestial-mechanics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26201" PostTypeId="2" ParentId="23794" CreationDate="2018-05-07T15:35:05.877" Score="1" Body="&lt;p&gt;I am interested in this topic as well and have found a news item describing another survey program that may contribute to the answer to this question: &#xA;&lt;a href=&quot;http://www.neoshield.eu/protecting-earth-asteroids-comets-neos/meteor-asteroid-neo-near-earth-object-comet/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.neoshield.eu/protecting-earth-asteroids-comets-neos/meteor-asteroid-neo-near-earth-object-comet/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22965" LastEditorUserId="22965" LastEditDate="2018-05-07T15:45:01.933" LastActivityDate="2018-05-07T15:45:01.933" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26202" PostTypeId="2" ParentId="26193" CreationDate="2018-05-07T17:10:12.630" Score="1" Body="&lt;p&gt;I'm not sure how to define the &quot;edges&quot; or obscurations in a case of gravitational lensing.  The image we produce on Earth is the combination of our local telescope lens elements and the &quot;grav lens&quot; a zillion Kessel Runs away from us.  Even if you use the premise that the black hole itself is a central obscuration which should cause a Spot of Arago, I don't think any other diffraction-like effect is observable from such a great distance.  All the wavefronts of  light which passes thru the GravLens region continue to expand, ending up near-parallel from Earth's viewpoint. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-05-07T17:10:12.630" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26203" PostTypeId="2" ParentId="23794" CreationDate="2018-05-07T18:29:51.403" Score="4" Body="&lt;p&gt;The &quot;small comets&quot; buzz was temporary and has been completely superceded by subsequent evidence that it was all noise in the detectors.  A careful investigation that produced a null result looking for the comets is published in &lt;a href=&quot;http://adsabs.harvard.edu/abs/2000JGR...10524907M&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2000JGR...10524907M&lt;/a&gt;.  The moral is, beware of the &quot;1 sigma effect,&quot; where a researcher sees a hint of a signal and is beguiled into seeing it as something real.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-05-07T18:29:51.403" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26204" PostTypeId="2" ParentId="18798" CreationDate="2018-05-07T19:58:12.360" Score="2" Body="&lt;p&gt;I'm not sure where in the above, the eccentricity of 0.0679 came from, maybe he meant 0.0167. Also, the Earths orbit does not come that close to the sun as the diagram shows. The orbit is near circular, with an eccentricity of 0.01671123, which means a and b (which isn't show in the above) are very close values to one another. The Focus points can be given be $F_1 = (-ae, 0)$ and $F_2 = (ae, 0)$ and eccentricity being $e = \sqrt{1-(b^2/a^2)}$. Use Pythagoras to figure the rest out.&lt;/p&gt;&#xA;" OwnerUserId="22968" LastEditorUserId="7472" LastEditDate="2018-05-08T13:29:23.830" LastActivityDate="2018-05-08T13:29:23.830" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26206" PostTypeId="1" CreationDate="2018-05-07T23:15:43.380" Score="5" ViewCount="550" Body="&lt;p&gt;Establishing the position of celestial bodies (stars, and other objects) is based on imaging technology. Depending on the resolution of the imaging sensor used, an object is identified by a group of neighboring (bright) pixels. These pixels surround the center of gravity of the detected celestial body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In image processing, the central position of a round object (a sphere) is generally determined by computing the center of gravity of the pixels surrounding the detected object. The weights are the intensities of the pixels in the vicinity of the brightest (central) pixel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: how is the &lt;em&gt;variance of this center-of-gravity estimate&lt;/em&gt; being calculated, within astronomy?&lt;/p&gt;&#xA;" OwnerUserId="22970" LastEditorUserId="2365" LastEditDate="2018-05-08T18:45:38.810" LastActivityDate="2018-05-17T19:22:46.990" Title="Measurement uncertainty of the computed center of gravity from pixel image" Tags="&lt;photography&gt;&lt;software&gt;&lt;astrometry&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26210" PostTypeId="1" AcceptedAnswerId="26212" CreationDate="2018-05-08T05:36:06.473" Score="4" ViewCount="192" Body="&lt;p&gt;I'm not really a physicist so I don't know if it's possible to get the real orbital velocity of a star from its radial velocity. I'm working on a paper for school and I want to get the mean velocity of selected open clusters to graph the galaxy's rotational curve, but in the databases I only encounter radial velocities. How do I get the real orbital velocity from those parameters? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="22975" LastActivityDate="2018-06-30T20:19:27.313" Title="Calculating obital velocity from radial velocity" Tags="&lt;astrophysics&gt;&lt;orbital-mechanics&gt;&lt;orbital-elements&gt;&lt;velocity&gt;&lt;star-cluster&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26212" PostTypeId="2" ParentId="26210" CreationDate="2018-05-08T06:37:24.880" Score="5" Body="&lt;p&gt;You can't without assuming something about the overall velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radial velocity is one component of a velocity vector; you are missing the other two components, which could in principle be anything. However, you could &lt;em&gt;assume&lt;/em&gt; that as &lt;strong&gt;open&lt;/strong&gt; clusters are mostly quite young and members of the Galactic disc population, that they are moving in the disc on roughly circular orbits. NB. This cannot possibly work for &lt;strong&gt;globular&lt;/strong&gt; clusters, which are an entirely different population, many with non-circular orbits and orbits out of the galactic plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having made that assumption, your radial velocity is the orbital velocity resolved through an angle that is formed between your line of sight to the cluster and a tangent to the Galactic centre at the cluster's Galactic position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method will not work well for clusters with a Galactic latitude much different to zero, or clusters with small Galactic longitude (which should have zero radial velocity in this model).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all this to work you need the radial velocity, the Galactic latitude and longitude of the cluster, the distance to the cluster and an assumed distance to the Galactic centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've had a go at sketching the situation. The cluster is on a circular orbit around the Galactic centre with a true orbital velocity $v$ (shown as a vector). We measure the component of this in our line of sight as the radial velocity, where &#xA;$$v_r = v \cos \alpha.$$&#xA;The trick is to find $\alpha$ using the triangle formed by: us, the Galactic centre and the cluster. If the distance to the galactic centre is $d_{gc}$ and the distance to the cluster is $D$ and the galactic longitude of the cluster (angle measured from the direction of the Galactic centre) is $l$, then I think that the distance of the cluster from the Galactic centre $d_{Cl}$ is given by the cosine rule as&#xA;$$ d_{Cl} = \left( d_{gc}^{2} + D^2 - 2d_{gc}D \cos l \right)^{1/2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the angle $\beta$ at the apex defined by the cluster id given by&#xA;$$\cos \beta = \frac{d_{Cl}^2 + D^2 - d_{gc}^2}{2d_{Cl}D}$$&lt;br&gt;&#xA;which leads to&#xA;$$ \cos \beta = \frac{D - d_{gc} \cos l}{d_{Cl}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then look at the diagram and note that $\beta = 90 - \alpha$, so that&#xA;$\cos \beta = \sin \alpha$. Hence you can calculate $\alpha$ and hence calculate $v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MO5dL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MO5dL.png&quot; alt=&quot;Orbital geometry&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-06-30T20:19:27.313" LastActivityDate="2018-06-30T20:19:27.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26215" PostTypeId="2" ParentId="26200" CreationDate="2018-05-08T12:46:37.250" Score="5" Body="&lt;p&gt;Your understanding of the illumination is correct. k is the ratio of the illuminated length BC to the diameter AC. New Moon has an illumination of 0, Full Moon an illumination of 1, and the quarter phase an illumination of 0.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position angle is measured counterclockwise from north (celestial north, along a line of right ascension) to the bright limb C. You need to calculate the parallactic angle, q, which is the angle from &quot;up&quot; to celestial north. From chapter 14 (in the 2009 version)&#xA;$$&#xA;\tan q = \frac{\sin H}{\tan \phi \; \cos \delta - \sin \delta \; \cos H}&#xA;$$&#xA;where $\phi$ is the geographical latitude, $\delta$ is the declination of the Moon, and H is the hour angle. The diagram below shows a positive value for both q and PA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zQevl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zQevl.png&quot; alt=&quot;Parallactic and Position Angles&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17355" LastEditorUserId="616" LastEditDate="2018-05-08T14:07:12.860" LastActivityDate="2018-05-08T14:07:12.860" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26217" PostTypeId="2" ParentId="26206" CreationDate="2018-05-08T15:31:41.640" Score="1" Body="&lt;p&gt;Dunno about &quot;...generally determined...&quot; but in similar work I've performed (and witnessed) a 2-dimensional spline fit is performed on the pixel intensity data to determine the power centroid to sub-pixel resolution.   As you noted, we make the reasonable assumption that the object is close to spherical and close to having azimuthal constant density (that is, density may vary with radius but not angle).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The uncertainty (variance) in this calculation is usually calculated by applying standard statistical methods to the observed variances in the received signal in each pixel, after accounting for jitter in the line-of-sight position.  (and of course accounting for electronics noise, etc).   Essentially, one can look at the variance in the spline-calculated peak over N frames, or look at the variance in all the pixels, determine their contribution to the spline fit, and weight their contributions accordingly.&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-05-08T15:31:41.640" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26218" PostTypeId="1" AcceptedAnswerId="26220" CreationDate="2018-05-08T15:43:05.230" Score="3" ViewCount="4279" Body="&lt;p&gt;From gcseastronomy.co.uk:&#xA;&lt;em&gt;A star culminates at 50°. It has a declination of +20°. What is the latitude from where it is observed?&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the relationship between a culmination, declination, and latitude? How would one work this out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the answer:&#xA;50 - 30 = 20°&#xA;Latitude = 20°&lt;/p&gt;&#xA;" OwnerUserId="22985" LastActivityDate="2018-05-08T18:44:48.433" Title="How to calculate observer's latitude from declination and culmination of a star?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26219" PostTypeId="2" ParentId="26206" CreationDate="2018-05-08T17:38:52.833" Score="4" Body="&lt;p&gt;Every pixel value $S_i$ on the detector at $\vec x_i$ has some error $N_i$: CCDs for example have a background noise $N_\text{bkg}$ from read-out electronics, thermal noise, and sky background, plus a Poissonian photon noise $N_S=\sqrt{S}$.  In many cases this noise follows a Gaussian distribution reasonably well.  After subtracting the background, a position measurement&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec x = \frac{\sum{S_i \vec x_i}}{\sum{S_i}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has the uncertainty&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \operatorname{std} \vec x&#xA;= \left[\sum_i \left(\frac{\partial x}{\partial S_i} N_i\right)^2 \right]^{1/2}&#xA;= \frac{\left[\sum_i&#xA;    \left(&#xA;      \sum_j\frac{(x_i - x_j) S_j}{\sum_k S_k} N_i&#xA;    \right)^2&#xA;  \right]^{1/2}}{\sum_k S_k} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ N_i^2 = N_\text{bkg}^2 + N_{S_i}^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using Gaussian error propagation ... I hope I have run the math correctly.  The way I've written it is a bit weird, but shows one important property of this method: If you look at the sum over $j$, you can see how the noise of a pixel is basically weighted by the pixel's distance to the centroid result.  The same error on a pixel value has more impact if the pixel is farther away from the centroid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Better methods account for the uncertainty $N_i$ on pixel values in the first equation already.  You can do this by introducing additional weights in your centroid or go to fitting models, which is the “usual” method in astrometry, as far as I have seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This more involved approach to measuring positions uses the point spread function $P(\vec x_i - \vec x_\text{obj})$ of the instrument under given observing conditions.  The model can be an approximation, e.g. a Moffat function, or an empirical model build from unconfused bright stars in the image, e.g. a spline interpolation.  For point sources, the typical least-squares fit of position and flux of the model to an image easily produces results close to the statistical optimum of parameter estimation in terms of uncertainty.  With our modern computing power the easiest way to get the uncertainty for a given model and data noise is often a &lt;a href=&quot;https://en.wikipedia.org/wiki/Bootstrapping_(statistics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;bootstrapping algorithm&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course extended objects require a bit more work in the model, for example assumptions on their shape, as you have given in your question.&lt;/p&gt;&#xA;" OwnerUserId="22239" LastEditorUserId="22239" LastEditDate="2018-05-08T17:46:11.227" LastActivityDate="2018-05-08T17:46:11.227" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26220" PostTypeId="2" ParentId="26218" CreationDate="2018-05-08T18:44:48.433" Score="5" Body="&lt;p&gt;Let $\delta$ be the star's declination and $\phi$ be the observer's terrestrial latitude.&#xA;Neglecting atmospheric refraction, the altitude at culmination is&#xA;$$\mathsf{alt_{max}} = 90^\circ - |\phi - \delta| $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some special cases:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A star whose declination equals the observer's latitude ($\delta = \phi$) culminates at the zenith (altitude 90$^\circ$).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A star on the celestial equator ($\delta$ = 0) culminates at altitude 90$^\circ$ - $|\phi|$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the example in the question, if we solve for $\phi$,&#xA;$$\phi = \delta \pm (90^\circ - \mathsf{alt_{max}}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the observer's latitude is 20$^\circ$ $\pm$ 40$^\circ$, either 20$^\circ$S or 60$^\circ$N.&#xA;GCSE Astronomy's answer is incorrect.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-05-08T18:44:48.433" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26221" PostTypeId="1" CreationDate="2018-05-08T19:52:36.473" Score="4" ViewCount="105" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Wow!_signal&quot; rel=&quot;nofollow noreferrer&quot;&gt;WOW! signal&lt;/a&gt;'s measurement took the average signal strength over 12-second intervals. The average intensity's peak was just over 30-1 signal to noise, meaning that the 12 second interval had an average strength of 30-1. But evidence suggests the signal came from a point source. Has anyone ever done the math to estimate the point source strength? Is this even possible? &lt;/p&gt;&#xA;" OwnerUserId="22988" LastEditorUserId="616" LastEditDate="2018-05-09T12:42:36.493" LastActivityDate="2018-05-09T12:42:36.493" Title="WOW! signal calculation of point source intensity" Tags="&lt;seti&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26222" PostTypeId="2" ParentId="20415" CreationDate="2018-05-08T20:39:33.230" Score="4" Body="&lt;p&gt;I'm going to post answers for both hemispheres, because there are good points from both sets of directions I've found.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;For the northern hemisphere:&lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The &quot;declination drift method&quot; is the most accurate way to accomplish this. The method is straightforward, but it does require some time and patience.&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;First, aim the mount's polar axis roughly at Polaris. Now point the telescope at a star that's somewhat above the celestial equator and as close to south as you can judge by looking opposite Polaris. Put in a high-power eyepiece. If the eyepiece has cross hairs, center the star on them. Otherwise put the star on the north or south edge of the field and defocus it a little. Turn on the clock drive, and ignore any east-west drift.&lt;/li&gt;&#xA;  &lt;li&gt;If the star drifts south in the eyepiece, the polar axis is pointing too far east. &lt;/li&gt;&#xA;  &lt;li&gt;If the star drifts north, the polar axis is too far west.&lt;/li&gt;&#xA;  &lt;li&gt;Shift the polar axis left or right accordingly, until there is no more drift.&lt;/li&gt;&#xA;  &lt;li&gt;Now aim at a star that's near the celestial equator low in the eastern sky.&lt;/li&gt;&#xA;  &lt;li&gt;If the star drifts south, the polar axis points too low.&lt;/li&gt;&#xA;  &lt;li&gt;If the star drifts north, the polar axis points too high.&lt;/li&gt;&#xA;  &lt;li&gt;Again, shift the polar axis accordingly.&lt;/li&gt;&#xA;  &lt;li&gt;Now go back and repeat from the beginning, because each adjustment throws the previous one slightly off. When all visible drift is eliminated the telescope is very accurately aligned, and you can take long deep-sky exposures.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Via &lt;a href=&quot;http://www.skyandtelescope.com/astronomy-resources/accurate-polar-alignment/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sky &amp;amp; Telescope&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;For the southern hemisphere:&lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Part 1&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Direct your telescope to a bright star that's low-ish (about 20 degrees) on the Eastern Horizon, near the Celestial Equator (i.e. 0 degrees DEC). The Orion nebula would be good to look at in mean time (given the right time of the year - as it may be too high or not visible).&lt;/li&gt;&#xA;  &lt;li&gt;At this point, the east/west rotation of your mount has little effect, leaving you to correct the angle of elevation of the mount.&lt;/li&gt;&#xA;  &lt;li&gt;Now, if the star drifts NORTH, your polar axis is too low - change the axis to angle further up in the sky (i.e. the latitude knob on your eq mount)&lt;/li&gt;&#xA;  &lt;li&gt;But if the star drifts SOUTH, your plar axis is too high - change the axis to angle further down toward land (i.e. the latitude knob on your eq mount).&lt;/li&gt;&#xA;  &lt;li&gt;You can tell if the star is drifting north or south by allowing the star to drift for a little bit - now, move your telescope to &quot;catch up with the star&quot; using the DEC control. If you see it heading north, then it's drifting north and vice-versa!&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Part 2&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Now, point the telescope to a bright star on the Meridian and on the celestial equator i.e. right above your head somewhere. This is so the elevation of the mount (as corrected in the steps above) will have minumal effect, allowing you to accurately correct the east/west rotation of the mount.&lt;/li&gt;&#xA;  &lt;li&gt;Now, if the star drifts NORTH, your polar axis is too far EAST - so rotate your mount further West (i.e. rotate your mount clockwise while looking down at it).&lt;/li&gt;&#xA;  &lt;li&gt;Now, if the star drifts SOUTH, your polar axis is too far WEST - so rotate your mount further East (i.e. rotate your mount anti-clockwise while looking down at it).&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Part 3&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Repeat these two steps a few times to increase the accuracy of your plar alignment - repeating Part 2 will allow you to get Part 1 more accurate.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Via &lt;a href=&quot;https://www.ozscopes.com.au/How-to-Polar-Align-Equatorial-Mount-Southern-Hemisphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;OZScopes&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Edit to provide less-precise-but-more-prompt method&lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Polar alignment for southern hemisphere observers is very similar to techniques used for the northern hemisphere. The major difference is the lack of a bright pole star like Polaris near the south celestial pole (SCP) to aid orientation used in several alignment procedures. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Rough polar alignment using the latitude scale of your scope is identical to the procedure used in the northern hemisphere.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Either look on a map, use Google Earth, or consult an almanac to find your observing site's latitude. Unlock any latitude adjustment screws on the sides of the mount and turn the latitude adjustment screws until the index on the polar axis reads your latitude. Tighten the adjustment screws if needed to secure the latitude setting. (You may also need to loosen the center pivot bolt by turning the hex nut to allow the equatorial mount head to be tilted.)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/naEI4.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/naEI4.gif&quot; alt=&quot;Latitude adjustment of the mount&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Latitude adjustment of the mount&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now complete the polar alignment by turning the entire mount (not either axis; both should be clamped tightly) to align the upwards end of the polar axis with south on the horizon. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Another more accurate rough method is to point to Sigma Octantis. This star is located only one degree from the SCP, the point in the sky around which all the other stars appear to rotate, and where the polar axis of a properly aligned equatorial mount should point.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;First, set up the mount so that the polar axis is pointing south.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Second, unlock the declination clamp and move the scope in declination so that the tube is parallel to the polar axis. Your declination setting circles should read 90° in this orientation. Clamp the declination lock.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Note: These last steps involve moving the entire mount. Don't use either the RA or Dec motions to change the position of the tube.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Move the mount in altitude and azimuth until Sigma Octantis is in your finder's field of view or centered in your finderscope. At magnitude 5.5, it is a fainter star--just slightly brighter than the unaided eye limit--so your finder will definitely be helpful here.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OGjws.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OGjws.gif&quot; alt=&quot;Vicinity of south celestial pole&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Sigma Octantis and the SCP&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Tweak the position of the mount by again moving the mount, this time centering Sigma in the eyepiece field of view. Altitude can be adjusted using the latitude adjustment screw or shortening-lengthening tripod legs.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For greater accuracy, you can point directly at the SCP using an offset from Sigma Octantis. (This can be done after aligning the optical axis of your finderscope with the polar axis.) Use the chart above or these patterns of southern circumpolar stars to find the SCP. Draw an imaginary line toward the SCP through the Southern Cross stars Gamma Crucis and Alpha Crucis (down the long axis of the cross). Draw another imaginary line toward the SCP at a right angle to a line connecting Alpha Centauri and Beta Centauri. The intersection of these two imaginary lines will point you close to the south celestial pole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Via &lt;a href=&quot;https://www.celestron.com/blogs/knowledgebase/how-do-i-polar-align-my-telescope-if-im-in-the-southern-hemisphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;Celestron&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22986" LastEditorUserId="22986" LastEditDate="2018-05-09T15:06:15.860" LastActivityDate="2018-05-09T15:06:15.860" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26223" PostTypeId="2" ParentId="26221" CreationDate="2018-05-09T01:07:59.667" Score="2" Body="&lt;p&gt;The plot is pretty straight forward.   6EQUJ5 puts the peak signal between Q and U, closer to U.   The letters represent two digit numbers, 1-9, then A-Z.  A being 10, Z being 35.   U was the highest signal strength received equivalent to a 30.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Wow!_signal#/media/File:Wow_signal_profile.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;A plot fits rather neatly&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia has a nice summary of the event.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Wow!_signal#Celestial_location&quot; rel=&quot;nofollow noreferrer&quot;&gt;The location is discussed here&lt;/a&gt; with some margin for error due to there being two listening devices and the design being, as I understand it, somewhat less than ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the signal remains unexplained, it was some 30 times stronger than baseline, but 3s and 4s were fairly common.  It was only 7-10 times stronger than signal strengths they received regularly.   I lean towards it being a natural but unexpected event, but there's not sufficient data to close the case, so to speak.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-05-09T01:07:59.667" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26227" PostTypeId="1" AcceptedAnswerId="26233" CreationDate="2018-05-09T11:39:57.117" Score="4" ViewCount="291" Body="&lt;p&gt;Everywhere I read that to view the Sun in the telescope, one has to use a solar filter. The filter must be handled carefully to avoid any holes in it and even mustn't be stretched so as to avoid reducing its reflectivity in some spots. Otherwise it becomes dangerous due to possibility of overexposing/burning observer's retina.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, if all we need is to reduce illuminance, and covering part of the telescope's aperture won't obscure the image (since the aperture is far from being in focus), why not simply reduce the aperture to a very small hole? Photo cameras do this all the time, why wouldn't this work for telescopes? Or does a solar filter block IR/UV light, in addition to reducing visible illuminance?&lt;/p&gt;&#xA;" OwnerUserId="735" LastActivityDate="2018-05-11T07:08:09.310" Title="Why do we use solar filters instead of simply reducing aperture of the telescope?" Tags="&lt;the-sun&gt;&lt;telescope&gt;&lt;observational-astronomy&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26228" PostTypeId="1" CreationDate="2018-05-09T13:19:45.103" Score="2" ViewCount="235" Body="&lt;p&gt;Would things be the same if the Sun was orbiting Earth and the Earth was in the center of the Solar System, or would there be any big difference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like why people before thought the Earth was the center of the Solar System since it didn't conflict with their experience.&lt;/p&gt;&#xA;" OwnerUserId="22996" LastEditorUserId="2153" LastEditDate="2018-05-09T13:56:58.150" LastActivityDate="2018-07-09T08:26:22.957" Title="Would things be exactly the same if the Sun was orbiting the Earth and the Earth was in the center of the Solar System?" Tags="&lt;the-sun&gt;&lt;orbit&gt;&lt;solar-system&gt;&lt;earth&gt;" AnswerCount="6" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26229" PostTypeId="2" ParentId="26228" CreationDate="2018-05-09T13:32:58.650" Score="2" Body="&lt;p&gt;It depends on what you mean by &quot;in the center&quot; or &quot;switch places.&quot;  If you physically moved the Earth to where the Sun is, and the Sun to where the Earth is, and gave each the speed that the other had while keeping the same reference frame, this would do tremendous violence to all the other planets in the solar system (though the Earth would continue to orbit more or less the same).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if what you are talking about is simply the language of what you call the center, then you are talking about changing to a different set of coordinates where the Earth is at the stationary center of your coordinate system.  Doing that presents no problems, and nothing is different except the language you are using to talk about what is going on.  One of the great things Einstein did was to reframe the laws of physics in a form that was independent of coordinates, allowing us to forever separate what the laws say is happening from the language we use to talk about what is happening (i.e., our chosen coordinates).  So it all comes down to whether you think &quot;being in the center&quot; is a real thing, or just the language of some coordinate system.  If the former, then no, things would not be at all the same, if the latter, then yes, they'd be exactly the same, if you are careful to use coordinate-independent laws like general relativity.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2018-05-09T15:26:26.510" LastActivityDate="2018-05-09T15:26:26.510" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="26230" PostTypeId="2" ParentId="26228" CreationDate="2018-05-09T15:01:13.297" Score="0" Body="&lt;p&gt;Let us suppose that &quot;a wizard did it&quot;. The law of gravity no longer applies and the motion of the sun is entirely due to magic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off what about the other planets? If they continue to orbit about the centre of the solar system, they would not move across the sky in the looping orbits that they seem to take. Venus and Mercury would not remain near the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the planets were to move in exactly the same way across our sky, then their routes through space would have to be complex looping affairs. The wizard would be busy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the stars? If the Earth is not orbiting the sun then they wouldn't have the yearly wobble in their position. The wobble is tiny but important in measuring the distance of the stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A wizard whold have to move the whole universe around in a 150million km circle each year for things to be exactly the same&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-05-09T15:01:13.297" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26231" PostTypeId="2" ParentId="26227" CreationDate="2018-05-09T15:06:17.667" Score="1" Body="&lt;p&gt;You would have to make the aperture so small diffraction would wreck the view. Also, resolution of a telescope is proportional to aperture, so the telescope with reduced aperture would also have reduced resolving power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As indicated in the comments, you should NEVER look at the sun with your naked eye or any optical device without proper solar filtering, e.g. Baader solar foil for a telescope.&lt;/p&gt;&#xA;" OwnerUserId="22867" LastEditorUserId="22867" LastEditDate="2018-05-11T07:08:09.310" LastActivityDate="2018-05-11T07:08:09.310" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26232" PostTypeId="2" ParentId="26227" CreationDate="2018-05-09T15:45:43.070" Score="0" Body="&lt;p&gt;If you try (&lt;strong&gt;do not do this!)&lt;/strong&gt; stopping down a digital camera and aiming it at the sun, you will almost certainly discover that, for anything larger than a  pinhole, the sun will be imaged in a small enough spot to damage the pixels involved.  Further, unless the pinhole stop is placed prior to all intermediate images (which depends heavily on the lens design), it's likely the sunlight could damage other optical surfaces.  Ditto for DoNotLookAtTheSun thru a tiny pinhole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible in theory to design a telescope stop which is not a small hole but rather a strategic radially adjusted pattern to limit the total light transmitted without losing the high-frequency phase information necessary to produce a clean image, but that's not worth the pain and effort.  It is much easier to use a uniform filter and avoid introducing any spatial frequency (diffraction) losses. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-05-09T15:45:43.070" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26233" PostTypeId="2" ParentId="26227" CreationDate="2018-05-09T15:55:14.240" Score="8" Body="&lt;p&gt;Suppose we want to make the Sun appear only as bright as the full Moon does in a 5-inch telescope.&#xA;A difference of 14 magnitudes is a factor of 400 000; we can either add a filter transmitting 2.5 millionths of the light (including UV and IR), or reduce the aperture diameter by a factor of 630.&#xA;We can achieve that by covering the 5-inch aperture with either a large #14 welding shade, a similarly dark &lt;a href=&quot;http://www.mreclipse.com/Special/filters.html&quot; rel=&quot;noreferrer&quot;&gt;solar filter&lt;/a&gt;, or a mask with a 0.2 mm pinhole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_resolution&quot; rel=&quot;noreferrer&quot;&gt;angular resolution&lt;/a&gt; is inversely related to aperture.&#xA;For a wavelength of 500 nm and an aperture of 0.2 mm, the Rayleigh criterion gives 1.2 * (500 nm / 0.2 mm) = 0.003 radian = 10 arcminutes, about 1/3 the angular diameter of the Sun.&#xA;Even large sunspots would be blurred beyond recognition.&#xA;A 1.0 mm hole would give 2 arcmin resolution but make the Sun too bright for direct viewing.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-05-09T15:55:14.240" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26234" PostTypeId="1" CreationDate="2018-05-09T19:59:18.227" Score="2" ViewCount="552" Body="&lt;p&gt;I watch the discovery science channel a lot &amp;amp; enjoy the material about space, stars, planets, etc.  In one episode the narrator states our Sun will one day become a white dwarf star.  The size of our Sun will double and ultimately the Earth will be inside this enlarged Sun. This the end of all life on Earth.  Then I read a recent Snapple bottle cap that stated our Sun is really white.  I read it again and chuckled: is it already a White dwarf star at this point?  I take it that the original color of the star happens to be white.  How will we know when the Sun will become a White dwarf and increase in size the engulf the Earth? Basically what is the time frame in this context.&lt;/p&gt;&#xA;" OwnerUserId="22998" LastEditorUserId="2365" LastEditDate="2018-05-09T21:33:06.130" LastActivityDate="2018-05-10T20:35:12.650" Title="Our Sun already a White dwarf?" Tags="&lt;star&gt;&lt;dwarf-star&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26235" PostTypeId="2" ParentId="26234" CreationDate="2018-05-09T20:43:50.227" Score="5" Body="&lt;p&gt;First. In about 6 billion years the Sun will become much larger than it is now, maybe a factor of 20. It will be a red giant. Then a little while later it will get really big, as large as the earth's orbit and perhaps the Earth will be engulfed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second. Only &lt;em&gt;after&lt;/em&gt; this does the Sun lose about half of its mass and the remainder is exposed as a &lt;em&gt;small&lt;/em&gt;, hot, white dwarf star, about the &lt;em&gt;same size&lt;/em&gt; as the Earth but half the mass of the Sun. It will have a density about 1 million times that of water and it is held up by a pressure that only arises because of the quantum mechanical nature of the electrons within it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun is not currently a white dwarf. It is a G-type dwarf star, supported by &quot;normal&quot; gas pressure and will remain reasonably stable, though will become about 50% more luminous over the next 5 billion years or so.  &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-05-09T20:43:50.227" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26237" PostTypeId="1" CreationDate="2018-05-09T22:37:16.080" Score="3" ViewCount="80" Body="&lt;p&gt;I was reading about how powerful supernova were and found out there are even bigger supernovae also called hyper nova.  A large explosion was discovered in 2016 I believe (or was it 2017).   I would like to know on a human scale how much power that is.  Suppose an inventor produced an engine technology that has the output of the largest superluminous super nova to project a space shuttle.  How fast would the space ship be projected forward?  That is the output pushes air to project the space ship forward.  What speed would it reach.  Part of me wants to say faster than light.  I imagine the ship will travel so fast it would disintegrate since matter cannot exceed the speed of light. &lt;/p&gt;&#xA;" OwnerUserId="22998" LastActivityDate="2018-05-10T14:53:58.453" Title="Superluminous super nova output" Tags="&lt;supernova&gt;&lt;speed&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26238" PostTypeId="2" ParentId="26234" CreationDate="2018-05-10T03:50:24.157" Score="2" Body="&lt;p&gt;The Sun is white, in the sense that you can hold up a white piece of paper to it and see no color, it's what our eyes have adapted to.  And it is a dwarf, in the sense that it is a main-sequence star, and all such stars are called dwarfs.  But it is definitely not a white dwarf, that's a much smaller (size of Earth) and much older star.  The names are not always very clear!&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-05-10T03:50:24.157" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26240" PostTypeId="2" ParentId="26237" CreationDate="2018-05-10T05:21:12.693" Score="4" Body="&lt;p&gt;This puts us in the realm of silly numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A supernova has enough energy to rip a massive star apart. Putting all that energy into one small rigid body isn't going to end well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hypernova gives off about $10^{46}$ Joules. We will need the formula for relatavistic kinetic energy $E=mc^2(\gamma-1)$ where $\gamma$ depends on velocity. The mass of the shuttle is about $2\times 10^6$ kg, and $c^2=9\times 10^{16}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $10^{46} = (2\times 10^6) \times (9\times 10^{16}) (\gamma - 1)$. Rearranging, $\gamma = 5.5\times10^{22}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In relativity $\gamma = 1/\sqrt{1-v^2/c^2}$ So even this massive amount of energy won't make the shuttle go faster than light. It will make it go at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;99.99999999999999999999999999999999999999999998% &lt;/p&gt;&#xA;&#xA;&lt;p&gt;of the speed of light.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-05-10T05:21:12.693" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26241" PostTypeId="1" AcceptedAnswerId="26243" CreationDate="2018-05-10T11:03:05.290" Score="14" ViewCount="1162" Body="&lt;p&gt;Cosmic microwave background radiation is 2.7 K today. When would it have been between 0 and 100 C?&lt;/p&gt;&#xA;" OwnerUserId="23002" LastEditorUserId="7472" LastEditDate="2018-05-10T16:07:22.883" LastActivityDate="2018-05-10T20:58:36.847" Title="How long after the Big Bang would CMB radiation have been between 273 and 373 K?" Tags="&lt;cmb&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26243" PostTypeId="2" ParentId="26241" CreationDate="2018-05-10T11:45:03.330" Score="17" Body="&lt;p&gt;The temperature of the cosmic microwave background scales as the inverse of the cosmic scale factor $a$. i.e. When everything was at half the separation it is now, then the CMB was twice the (absolute) temperature. The scale factor in turn is reciprocally related to the redshift by $a/a_0 = (1 +z)^{-1}$, where $z$ is the redshift and $a_0$ is the present-day scale factor, usually taken to be 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this we see that&#xA;$$T(z) = T_0 (1+z),$$&#xA;where $T(z)$ is the temperature at redshift $z$ at some time in the past and  $T_0 = 2.73$K is the temperature of the CMB now. For the range of temperatures you specify requires redshifts in the range $99&amp;lt; z&amp;lt; 136$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting from a redshift to a time in the universe since the big bang is not so simple. The answer depends on what you take to be the &quot;cosmological parameters&quot; - i.e. the values of the cosmic matter density, the dark energy density and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we can avoid looking &quot;under the hood&quot; and use a &lt;a href=&quot;http://home.fnal.gov/~gnedin/cc/&quot; rel=&quot;noreferrer&quot;&gt;cosmology calculator&lt;/a&gt;. The one I have linked to is for a &quot;flat&quot; universe and has default values for what are currently good estimates for the matter density and present-day Hubble parameter. For the range of $z$ I found above, this corresponds to an age of &lt;strong&gt;between 10.8 and 17.3 million years after the big bang&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess you are perhaps thinking about &lt;a href=&quot;https://www.nature.com/news/life-possible-in-the-early-universe-1.14341&quot; rel=&quot;noreferrer&quot;&gt;these ideas&lt;/a&gt; that life could have been around about 15 million years after the big bang.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-05-10T20:58:36.847" LastActivityDate="2018-05-10T20:58:36.847" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26244" PostTypeId="2" ParentId="26237" CreationDate="2018-05-10T14:53:58.453" Score="1" Body="&lt;p&gt;This comment will not answer about the speed of a rocket, because I don't know other factors involving. But this comment will give you energy scales, which you can work on later for the speed of a rocket. I cannot put this comment in the &quot;comment&quot; because it will be too long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hypernovae (a.k.a. supernovae Ic broad line (SNe Ic-BL) and superluminous supernovae (SLSNe) are core-collapse supernovae (CCSNe).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical CCSNe are believed to be delayed neutrino-driven explosions, where total explosion energy is $10^{53}$ erg. For typical SNe (i.e., not SNe Ic-BL nor SLSNe), only $\sim 1$% of the total explosion energy is converted into radiation, i.e., $\sim 10^{51}$ erg. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;SNe Ic-BL and SLSNe radiate $\gtrsim 10^{51} - 10^{52}$ erg, more than those of the typical SNe. Because they are more energetic than $\sim 10^{51}$ erg predicted from typical CCSNe, their explosion mechanisms or energy sources are under debate. Currently, the power from a spinning down magnetar is the most favored candidate. The magnetar spin-down can supply $\sim 10^{52}$ erg, with peak $\lesssim 10^{45}$ erg/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Talking in term of power, i.e., luminosity, typical SNe has peaks UV/optical/NIR outputs $\lesssim 10^{43}$ erg/s. SNe Ic-BL are about the same scale, but they are more energetic because spectral evidence implies fast moving ejecta. SLSNe are $\gtrsim 10^{44}$ erg/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To illustrate how powerful these SNe are, we normally compare them to solar radiation $\sim 10^{33}$ erg/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there are even more powerful explosions predicted such as pair-instability supernovae (PISNe). PISNe can supply $\sim 10^{53}$ erg. But because of very massive ejecta, the peak luminosity is predicted to be comparable to those of SLSNe, but with very long timescale. Some slow-evolving SLSNe are candidates for PISNe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this is helpful for you to use in calculating the speed of a rocket.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-05-10T14:53:58.453" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26245" PostTypeId="1" CreationDate="2018-05-10T16:17:51.380" Score="3" ViewCount="108" Body="&lt;p&gt;If we want to observe a galaxy for which we have optical data, what parameters we need to know to calculate its total radio flux at 3 cm? &#xA;Specially I am interested to know how to differentiate among the different process in radio, like synchroton, free-free and inverse compton? Specially when I know its Hydrogen recombination line luminosity is in the power of 42, and electron temperature 10000 K with distance 20 Mpc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From literature, I assume 3 cm is 10GHz, which should fall under either synchroton or free-free emission. But how do I know exactly?&lt;/p&gt;&#xA;" OwnerUserId="21282" LastActivityDate="2018-06-10T14:25:48.970" Title="Different radio process band ranges" Tags="&lt;radio-astronomy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26246" PostTypeId="2" ParentId="26234" CreationDate="2018-05-10T20:35:12.650" Score="1" Body="&lt;p&gt;There's a bit of confusion in the nomenclature here. This is very important: The nomenclature for labeling largish objects in space was developed before scientists knew anything about fusion. Stars on the main sequence fuse hydrogen to form helium. Except for extremely massive stars, every star on the main sequence is a so-called dwarf star. These main sequence stars are called dwarfs because they aren't anywhere near as large as are the giant stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our Sun is on the main sequence (i.e., it's fusing hydrogen into helium) and will remain so for another five to seven billion years. Star-sized objects leave the main sequence when they've consumed all available hydrogen. Stars that aren't too small will eventually start fusing helium after they leave the main sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite the common word &quot;dwarf&quot;, white dwarfs form a category of objects that is very distinct from dwarf stars. White dwarfs are objects that not only have left the main sequence, but have also fused every last dreg of fusible material available to them. This necessarily makes white dwarfs very compact, and temporarily, extremely hot. Note well: &quot;Temporarily&quot; to an astronomer or astrophysicist means something very different than it does to everyone else. Once again, the nomenclature was developed over a hundred years ago, well before nuclear fusion was known to exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would perhaps be nice if the International Astronomical Union would come up with a working definition of what constitutes a &quot;star&quot;. However, I suspect they have had their fill with naming conventions given the huge controversy over their definition of what qualifies as a &quot;planet&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2018-05-10T20:35:12.650" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26247" PostTypeId="1" CreationDate="2018-05-10T21:04:40.533" Score="4" ViewCount="102" Body="&lt;p&gt;I've read articles that state the age of the solar system has been determined by measuring the amount of decay of isotopes in some meteorite, which has demonstrated that  the lead isotopes in the meteorite have cooled and locked into place some number of years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if a yellow sun can not create heavier elements, wouldn't it be correct to say the age of an meteorite dated this way is how long it's been since the asteroid was ejected from a blue or heavier star? The sun may have started burning many years after that.&lt;/p&gt;&#xA;" OwnerUserId="23008" LastEditorDisplayName="user1569" LastEditDate="2018-05-13T20:08:34.200" LastActivityDate="2018-05-13T20:08:34.200" Title="Is the age of the solar system equal to the age of meteorites?" Tags="&lt;star&gt;&lt;planet&gt;&lt;solar-system&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26248" PostTypeId="2" ParentId="26247" CreationDate="2018-05-10T21:43:00.963" Score="4" Body="&lt;p&gt;The radiometric dating of &lt;strong&gt;meteorites&lt;/strong&gt; tells you when certain radioactive species were incorporated into that body. It is extremely difficult to see how that material could have got there at a later time, so the age you get is the age of formation of that body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meteorites do not form in stars. They require considerably cooler conditions, but they also require a certain density to stick together. Those conditions are found in a protoplanetary disk. So you might well ask, well maybe the meteorites were formed around another star and then somehow they have ended up in orbit around our Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is quite unlikely. Very similar ages have been determined for some tens of meteorites and that age happens to be just a touch older than the oldest rocks that can be found on the Earth and the Moon. There is no big age spread among meteorites. That age also coincides with the age for the Sun deduced from physical models (though this age is less accurate because of some remaining uncertainties in the physics, but it is not hundreds of millions of years out).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious inference is that these meteorites were among the first solid bodies formed in the protoplanetary disk that ultimately formed the Earth and Moon and hence represents a &lt;em&gt;lower limit&lt;/em&gt; to the age of the solar system. It depends where you wish to start the clock; the formation of the first solid bodies seems as good a place as any and models for disk formation and evolution suggest that these solid bodies do form within about a million years of the molecular cloud that formed the Sun beginning to collapse.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-05-13T18:54:34.457" LastActivityDate="2018-05-13T18:54:34.457" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26249" PostTypeId="2" ParentId="26245" CreationDate="2018-05-11T00:45:20.903" Score="1" Body="&lt;p&gt;Without a lot of expensive equipment the best you can do is make a rough estimate. By incorrectly assuming that the light that you see describes the black body temperature of the object, by assuming that the system is in &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermal_equilibrium&quot; rel=&quot;nofollow noreferrer&quot;&gt;thermal equilibrium&lt;/a&gt; with itself and that the temperature within the system is spatially and temporally uniform, you can use &lt;a href=&quot;https://en.wikipedia.org/wiki/Planck%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck's law&lt;/a&gt; to calculate the spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dTbvu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dTbvu.png&quot; alt=&quot;Planck&amp;#39;s Law Blackbody Radiation&quot;&gt;&lt;/a&gt;&#xA;Planck's law describes the spectral density of electromagnetic radiation emitted by a black body in thermal equilibrium at a given temperature $T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Radiance&quot; rel=&quot;nofollow noreferrer&quot;&gt;spectral radiance&lt;/a&gt; of a body, $B_ν$, describes the amount of energy it gives off as radiation of different frequencies. It is measured in terms of the power emitted per unit area of the body, per unit solid angle that the radiation is measured over, per unit frequency. &lt;a href=&quot;https://en.wikipedia.org/wiki/Max_Planck&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck&lt;/a&gt; showed that the spectral radiance of a body for &lt;a href=&quot;https://en.wikipedia.org/wiki/Frequency&quot; rel=&quot;nofollow noreferrer&quot;&gt;frequency&lt;/a&gt; $ν$ at &lt;a href=&quot;https://en.wikipedia.org/wiki/Absolute_temperature&quot; rel=&quot;nofollow noreferrer&quot;&gt;absolute temperature&lt;/a&gt; $T$ is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ B_{\nu }(\nu ,T)={\frac {2h\nu ^{3}}{c^{2}}}{\frac {1}{e^{\frac {h\nu }{k_{\mathrm {B} }T}}-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $k_B$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Boltzmann_constant&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boltzmann constant&lt;/a&gt;, $h$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Planck_constant&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck constant&lt;/a&gt;, and $c$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Speed_of_light&quot; rel=&quot;nofollow noreferrer&quot;&gt;speed of light&lt;/a&gt; in the medium, whether material or vacuum. The spectral radiance can also be measured per unit &lt;a href=&quot;https://en.wikipedia.org/wiki/Wavelength&quot; rel=&quot;nofollow noreferrer&quot;&gt;wavelength&lt;/a&gt; $\lambda$ instead of per unit frequency. In this case, it is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_{\lambda }(\lambda ,T)={\frac {2hc^{2}}{\lambda ^{5}}}{\frac {1}{e^{\frac {hc}{\lambda k_{\mathrm {B} }T}}-1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The law may also be expressed in other terms, such as the number of photons emitted at a certain wavelength, or the energy density in a volume of radiation. The SI units of $B_ν$ are $W·sr^{−1}·m^{−2}·Hz^{−1}$, while those of $B_\lambda$ are $W·sr^{−1}·m^{−3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;&lt;a href=&quot;https://ncc.nesdis.noaa.gov/data/planck.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck Calculator for Infrared Remote Sensing&lt;/a&gt;&quot; can be used to make some calculations:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Temperature T(K)                 10,000&#xA;Center Wavelength μm             11.0&#xA;Spectral Radiance (W/m^2-sr-μm)  5292.388800699331&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;With a temperature of 10K the center frequency is 11 $\mu m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the oversimplified (inaccurate) layperson's calculation. More accurate calculations are made by allowing for everything from: relative velocity, the composition of the object's atmosphere (and our own), the effect of distance, even other elements, upon the spectral distribution (short list).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, after correcting for light travel time effects the response of the Balmer and optical helium lines should generally be strongest during low continuum luminosity states. Responsivity that depends upon photon flux and continuum state may explain a number of outstanding problems currently under investigation in broad line variability studies of these and other emission lines. [Abridged] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once calculations are made for the blackbody curve in accordance with Planck's law it is necessary to transpose that to Rayleigh–Jeans law which is applicable at large wavelengths (low frequencies) but strongly disagrees at short wavelengths (high frequencies) where Planck's law is more useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Wikipedia's &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Rayleigh%E2%80%93Jeans_law#Comparison_to_Planck&amp;#39;s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Comparison [of Rayleigh–Jeans law] to Planck's law&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YSSaI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YSSaI.png&quot; alt=&quot;Comparison of Rayleigh–Jeans law with Planck&amp;#39;s law.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further reading:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;&lt;a href=&quot;https://arxiv.org/abs/0903.0562&quot; rel=&quot;nofollow noreferrer&quot;&gt;Introduction to Millimeter/Sub-Millimeter Astronomy&lt;/a&gt;&quot; 3 Mar 2009, by Wilson&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;&lt;a href=&quot;https://arxiv.org/abs/astro-ph/0403689&quot; rel=&quot;nofollow noreferrer&quot;&gt;Free-Free Spectral Energy Distributions of Hierarchically Clumped HII Regions&lt;/a&gt;&quot; 30 Mar 2004, by Ignace and Churchwell&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;&lt;a href=&quot;https://arxiv.org/abs/astro-ph/0402506&quot; rel=&quot;nofollow noreferrer&quot;&gt;What the Optical Recombination Lines Can Tell Us About the Broad Line Regions of Active Galactic Nuclei&lt;/a&gt;&quot; 17 Jun 2005, by Korista and Goad.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you want to include the effect of quantum gravity in your calculations a recent paper is: &quot;&lt;a href=&quot;https://arxiv.org/abs/1206.5631&quot; rel=&quot;nofollow noreferrer&quot;&gt;Black body radiation in a model universe with large&#xA;extra dimensions and quantum gravity effects&lt;/a&gt;&quot; 8 Jan 2018, by Nozari, Anvari, and Sefiedgar&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Abstract:&lt;/strong&gt; &quot;We analyze the problem of black body radiation in the presence of quantum gravity effects encoded in modified dispersion relations and in a model universe with large extra dimensions. In this context, modified form of Planck distribution, Jeans number, equipartition theorem, spectral energy density, Stefan-Boltzmann law and Wien's law are found and the corresponding results are interpreted. Then, entropy and specific heat of black body radiation are obtained in this setup. Finally, the modified form of Debye law and Dulong-Petit law are investigated in this framework.&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This article from the PrimaLuce Lab's website: &quot;&lt;a href=&quot;https://www.primalucelab.com/radioastronomy/blog/radioastronomia-a-scuola-radio-astronomy-at-school&quot; rel=&quot;nofollow noreferrer&quot;&gt;Radio astronomy at school with SPIDER radio telescopes&lt;/a&gt;&quot; explains backyard radio astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their 3M diameter SPIDER 500A's H142-One receiver has a 1.42 GHz superheterodyne type radiometer/ spectrometer, double conversion (type UP/DOWN) with 50 MHz bandwidth (RF=1.395MHz-1.445MHz) and 14-bit analog to digital converter. It has a spectrometer with 1024 channels (each 61 KHz) and is able to record many radio sources in the Universe, with a theoretical flow of at least 5 Jy. Price available upon request.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their SPIDER 230C Compact Radio Telescope measures 2.3 meters and is priced at U$16K, from what I can determine from reading the literature it features the same receiver and software; dish diameter and mount capacity being the only difference, along with beam width and sensitivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mesh on their dishes is good up to 5 GHz but the feedhorn normally supplied is for 1.42 GHz, I imagine that they are willing to do some amount of customization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An explanation of the atmospheric window can be found here: &quot;&lt;a href=&quot;https://radiojove.gsfc.nasa.gov/education/educ/radio/tran-rec/exerc/iono.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Effects of Earth's Upper Atmosphere on Radio Signals&lt;/a&gt;&quot;, this image is from NASA's website:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/otdxp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/otdxp.jpg&quot; alt=&quot;Atmospheric Windows&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; From literature, I assume 3 cm is 10GHz, which should fall under either synchroton or free-free emission. But how do I know exactly?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3cm is 10 GHz and can be received with a 10 GHz receiver. The difference between synchrotron and free-free is explained on NRAO's webpages  &lt;a href=&quot;https://www.cv.nrao.edu/~sransom/web/Ch4.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Free–Free Radiation&lt;/a&gt; and &lt;a href=&quot;https://www.cv.nrao.edu/~sransom/web/Ch5.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Synchrotron Radiation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acceleration by a magnetic field produces magnetobremsstrahlung, the German word for “magnetic braking radiation.” The character of magnetobremsstrahlung depends on the speeds of the electrons, so these somewhat different types of radiation are given specific names. Gyro radiation comes from electrons whose velocities are much smaller than the speed of light: $v≪c$. Mildly relativistic electrons whose kinetic energies are comparable with their rest mass $m_e⁢c^2$ emit cyclotron radiation, and ultrarelativistic electrons (kinetic energies $≫m_e⁢c^2$) produce synchrotron radiation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acceleration by an electric field accounts for free–free radiation, the resulting emission is called free–free radiation because the electron is free both before and after the interaction; it is not captured by the ion. If the ionized interstellar cloud is reasonably dense, the electrons and ions interact often enough that they quickly come into local thermodynamic equilibrium (LTE) at some common temperature, so free–free radiation is usually thermal emission. &lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="18054" LastEditDate="2018-05-11T14:03:42.590" LastActivityDate="2018-05-11T14:03:42.590" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26250" PostTypeId="1" CreationDate="2018-05-11T00:47:53.600" Score="6" ViewCount="2210" Body="&lt;p&gt;The Gaia data release 2 dataset has some parallaxes listed as negative. What is the proper interpretation of this? Should they be discarded as bad data? Use their absolute value to calculate the distance? Use them as-is and treat them as negative distances along the right ascension/declination vector?&lt;/p&gt;&#xA;" OwnerUserId="23009" LastActivityDate="2020-05-17T15:23:28.870" Title="What is the proper interpretation of a negative parallax?" Tags="&lt;parallax&gt;&lt;gaia&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26251" PostTypeId="2" ParentId="26250" CreationDate="2018-05-11T02:57:55.993" Score="7" Body="&lt;p&gt;Did you read this section of the documentation?  It suggests there are ways to deal with it, but I have not examined the paper it refers to.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;• For closely aligned sources (separated by 0.2–0.3 arcsec), which are only occasionally resolved in the Gaia observations, confusion in the observation-to-source matching can lead to spurious parallax values which are either very large or have a negative value very far away from zero in terms of the formal parallax uncertainty quoted in the catalogue. These sources tend to be faint and located in crowded regions and are also associated with unreliable (large) proper motions (Gaia Collaboration et al. 2018b). Guidance on how to clean samples from spurious parallax values is provided in Lindegren et al. (&lt;a href=&quot;https://gea.esac.esa.int/archive/documentation/GDR2/bib.html#bib2&quot; rel=&quot;noreferrer&quot;&gt;2018&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It also says there's a small bias, but it shouldn't be universally removed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;• The systematic errors in the parallaxes are estimated to be below the 0.1 mas level (Lindegren et al. 2018) but the following systematics remain. There is an overall parallax zeropoint which, from an examination of QSO parallaxes, is estimated to be around −0.03 mas (in the sense of the Gaia DR2 parallaxes being too small). The estimated parallax zeropoint depends on the sample of sources examined (Arenou et al. 2018) and the value above should not be used to ’correct’ the catalogue parallax values....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2018-05-11T02:57:55.993" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26252" PostTypeId="2" ParentId="26250" CreationDate="2018-05-11T06:45:57.890" Score="3" Body="&lt;p&gt;It depends how negative the parallax is and what your &quot;prior&quot; knowledge is of the distance to the star is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As another answer suggests, there are some spurious large negative (and positive) parallaxes for faint, crowded sources. If possible, these should be removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is not the case, and the parallax is negative, but close to zero within its uncertainty, then it is telling you that you have a lower limit to the distance of the object (i.e. measurement uncertainties have led to a small negative parallax). Crudely speaking, you could add a couple of error bars to the parallax and treat that as a 95% upper limit (remember the 0.1 mas possible systematic error too).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under no circumstances should you &quot;use them as is&quot;, since there is no physical basis for a negative parallax or negative distance.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-05-11T06:45:57.890" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26253" PostTypeId="1" AcceptedAnswerId="40781" CreationDate="2018-05-11T07:13:58.947" Score="6" ViewCount="201" Body="&lt;p&gt;I understand the Moon was perhaps five Earth radii away when it first formed (assuming it was formed by a giant impact of a Mars-sized body), and Earth has since transferred its rotational energy into the moon's orbital energy. How fast would Earth have been spinning at the point the Moon was formed? Assuming it was spinning very fast (day under 10 hours) would this have caused a noticeable decrease in effective &quot;gravity&quot; at the equator? Presumably the Precambrian microbes could have reached equatorial orbit really easily, assuming they had a space program.&lt;/p&gt;&#xA;" OwnerUserId="22867" LastEditorUserId="22867" LastEditDate="2018-05-11T15:24:20.247" LastActivityDate="2021-01-25T03:54:15.287" Title="How fast was the Earth spinning directly after the Moon formed?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26256" PostTypeId="1" AcceptedAnswerId="26257" CreationDate="2018-05-11T15:18:23.643" Score="4" ViewCount="965" Body="&lt;p&gt;I've come across multiple answers to this question. So, the moon is moving away from the Earth a couple of centimeters per year and the Earth's rotation is slowing down. &lt;a href=&quot;http://astro-interest.com/2014/10/the-speed-of-rotation-of-earth-is-decreasing-due-to-revolution-of-moon/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; it says the moon's revolution is speeding up, another place it says it is slowing down. So, due to tidal effects, is the moon's revolution slowing down or speeding up?&lt;/p&gt;&#xA;" OwnerUserId="16048" LastEditorUserId="16048" LastEditDate="2018-05-11T16:21:31.570" LastActivityDate="2018-05-12T10:28:54.307" Title="Is the moon's revolution speed increasing or decreasing" Tags="&lt;the-moon&gt;&lt;tidal-locking&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26257" PostTypeId="2" ParentId="26256" CreationDate="2018-05-11T15:48:47.920" Score="2" Body="&lt;p&gt;The moon's rotation is very slowly decreasing. The page you link to is correct but confusing, and you need to understand one of the weird things about orbits to understand it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If you speed up in an orbit you go slower&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is how this apparent paradox works.  If you are orbiting the Earth and you fire your rockets to speed up, you will lift yourself to a higher orbit, where you will be moving slower. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tidal bulge is a large mass and this large mass is in front of the moon. Any large mass has a gravitational effect and there is a gravitational force on the moon by the bulge, and an equal and opposite force on the bulge by the moon. The bulge is in front of the moon, so the gravitational action of the bulge pulls the moon forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the tidal bulge acts to pull the moon forward in its orbit, which makes it slow down, and since the moon is tidally locked it also slows its rotation speed. However the moon is already spinning quite slowly, only one complete rotation per month, so the amount of slow down is very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article writes&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The slowing of the speed of rotation of Earth also results in the increase in speed of revolution of moon around the Earth. [...] The increase in speed of revolution of moon around the Earth also results in the increase in radius of orbit of moon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You should understand that when it says &quot;increase in speed of the moon&quot; it means that the moon is actually slowing down due to the increase in radius.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-05-11T16:20:46.947" LastActivityDate="2018-05-11T16:20:46.947" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="26258" PostTypeId="1" AcceptedAnswerId="26260" CreationDate="2018-05-11T19:12:41.120" Score="5" ViewCount="63" Body="&lt;p&gt;Mercury and Venus have no satellites. Then earth has moon (synchronized orbit). For Mars - both its satellites (Fobos and Deimos) also synchronized. Analyzing other planets is hard - too many satellites. Maybe someone knows percentage - all of them synchronized or which ones are not synchronized (if just a few ones are not synchronized). All 4 largest Jupiter satellites are also synchronized.&lt;/p&gt;&#xA;" OwnerUserId="23018" LastEditorUserId="23018" LastEditDate="2018-05-11T19:28:45.227" LastActivityDate="2018-05-11T20:17:15.133" Title="How many (or percentage) satellites of Solar system planets have their orbits synchronized like the Moon (always facing earth)?" Tags="&lt;satellite&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26259" PostTypeId="2" ParentId="26256" CreationDate="2018-05-11T20:15:53.647" Score="5" Body="&lt;p&gt;From the cited article,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The slowing of the speed of rotation of Earth also results in the increase in speed of revolution of moon around the Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I suspect that this is the statement that is causing the OP to be confused, and rightly so. This statement is incorrect. Whether &quot;speed of rotation&quot; means orbital velocity or angular velocity, both are decreasing as the Moon slowly retreats from the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon is retreating very slowly from the Earth, currently about 3.78 cm per year. Dividing by 385000 km (the Earth-Moon mean orbital radius) yields about 10&lt;sup&gt;-10&lt;/sup&gt; parts per year. That qualifies as &quot;very slowly&quot;. For simplicity, I'll assume the Moon's orbit is essentially circular, plus this very slow spiraling out. For a circular orbit, the relation between orbital radius and angular velocity is $$r^3\omega^2 = G(M+m)$$&#xA;This essentially is Kepler's third law. Assuming $G$, $M$, and $m$ are constant, differentiating both sides with respect to time yields&#xA;$$r^2\omega (3\dot r \omega + 2 r \dot \omega) = 0$$&#xA;meaning&#xA;$$r \dot \omega = -\frac32 \dot r \omega$$&#xA;Suppose some quantity $q$ is of the form to $q=r^n\omega^m$. Differentiating with respect to time yields $\dot q = r^{n-1}\omega^{m-1}\left(n\dot r\omega + m r \dot\omega\right)$. Since $r\dot\omega = -\frac32 \dot r \omega$, this can be rewritten as $$\dot q = \left(n - \frac 3 2 m\right) \dot r r^{n-1}\omega^m$$&#xA;Quantities of this form include angular velocity ($q=\omega$, so $n=0, m=1$ in this case), orbital velocity $(q=r\omega$, so $n=m=1$), and angular momentum (q$=r^2\omega$, so $n=2, m=1$).  Note that $(n-\frac 32m)$ is negative for angular and orbital velocity but positive for orbit angular momentum. Orbit angular momentum thus has the same sign as does $\dot r$ while angular velocity and orbital velocity have the opposite sign as does $\dot r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a sanity check, look to centripetal acceleration. This is given by $a = r\omega^2$. In terms of the above, $n=1$ and $m=2$. The term $n-\frac32m$ is clearly negative in this case, so centripetal acceleration decreases as the Moon retreats from the Earth. Centripetal acceleration is also give by Newton's law of gravitation: It is proportional to the inverse of the square of the distance between the Earth and the Moon, so this does indeed decrease as the Moon retreats from the Earth.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2018-05-11T22:46:38.097" LastActivityDate="2018-05-11T22:46:38.097" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26260" PostTypeId="2" ParentId="26258" CreationDate="2018-05-11T20:17:15.133" Score="6" Body="&lt;p&gt;This is not really known; while the rotation period of most of the larger satellites is known, and they're mostly synchronized, for the smaller ones it's harder to determine and they're &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_locking#Timescale&quot; rel=&quot;noreferrer&quot;&gt;less likely to be synchronized&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One conclusion is that, other things being equal ... a large moon will lock faster than a smaller moon at the same orbital distance from the planet. ... A possible example of this is in the Saturn system, where Hyperion is not tidally locked, whereas the larger Iapetus, which orbits at a greater distance, is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wikipedia has a &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_locking#List_of_known_tidally_locked_bodies&quot; rel=&quot;noreferrer&quot;&gt;list of synchronized satellites&lt;/a&gt; and 'suspects', if you want to investigate further.&lt;/p&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2018-05-11T20:17:15.133" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26262" PostTypeId="1" CreationDate="2018-05-11T22:25:15.873" Score="2" ViewCount="81" Body="&lt;p&gt;Today in the tech age we have many communication technologies cell tower for cell  phones,  wifi, nuclear subs have some classified method of communication,  etc.  What communication technology is being used if a satellite probe is sent past Jupiter &amp;amp; transmits a signal back to scientists to monitor progress or review high definition detailed images?&lt;/p&gt;&#xA;" OwnerUserId="22998" LastActivityDate="2018-05-11T23:00:21.313" Title="Communication for far away satellites" Tags="&lt;communication&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26263" PostTypeId="2" ParentId="26262" CreationDate="2018-05-11T23:00:21.313" Score="4" Body="&lt;p&gt;That's the job of NASA's &lt;a href=&quot;https://deepspace.jpl.nasa.gov&quot; rel=&quot;nofollow noreferrer&quot;&gt;Deep Space Network&lt;/a&gt;, plus sites in other countries that perform similar functions. NASA currently has facilities in California, Spain, and Australia whose primary function is to communicate with devices sent beyond low Earth orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to these Deep Space Network sites, NASA and the US Department of Defense jointly operate the &lt;a href=&quot;https://en.wikipedia.org/wiki/Tracking_and_Data_Relay_Satellite_System&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tracking and Data Relay Satellite System&lt;/a&gt; network. This system comprises geosynchronous satellites outfitted with communications equipment, relay stations in White Sands, New Mexico and in Guam, plus a control station in Greenbelt, Maryland. The European Space Agency has developed a similar system that takes advantage of recent developments in long distance laser communications. These satellite-based systems are intended for use by vehicles near the Earth. More remote vehicles such as probes to Mars, Venus, and beyond generally use the Deep Space Network.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2018-05-11T23:00:21.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26264" PostTypeId="2" ParentId="26256" CreationDate="2018-05-12T02:15:28.597" Score="4" Body="&lt;p&gt;Some confusion may have arisen in parts of the literature due to the facts that (a) the moon's mean rate of angular motion (and revolution speed) does appear to be speeding up if it is measured by mean solar time or by sidereal time, but also (b) the mean rate appears to be slowing down if it is measured by an atomic or dynamical time-scale.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is because there are (at least) three real effects in operation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) an &lt;em&gt;apparent&lt;/em&gt; increase in the moon's rate of motion caused by the long-term slowing-down of earth rotation (when the day is longer the moon would appear to go further even if its real rate is unchanged),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) a &lt;em&gt;real&lt;/em&gt; increase in the moon's rate due to the long-term reduction in the eccentricity of the earth-sun orbit, and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) a &lt;em&gt;real&lt;/em&gt; decrease in rate due to tidal retardation of the moon's motion itself.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of all three effects is an apparent net positive acceleration. (When effect (1) is included, that of course implies the use of a mean solar or sidereal clock and time-scale.) But if apparent effect (1) is taken out, by using an atomic timescale instead, the remaining real effect (3) more than offsets real effect (2), leaving a net negative acceleration (retardation). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Historically these were discovered in the order, first the aggregate of (1)+(2)+(3) from the 1690s onwards, then effect (2) from the late 1700s, and finally effects (1) and (3) were suspected in the later 1800s and after much work separately verified during the 20th century. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edmond Halley in the 1690s suspected a net positive acceleration; but he could not evaluate it quantitatively, for lack of good geographical longitudes of the places where the ancient moon observations had been taken. He appealed for such measurements in the last two pages (174-5) of a paper of 1693 (&lt;a href=&quot;http://rstl.royalsocietypublishing.org/content/19/218/160&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://rstl.royalsocietypublishing.org/content/19/218/160&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 1740s Richard Dunthorne had better data, and by evaluating ancient moon observations in the near East he estimated the mean acceleration as +10 arc&quot;/cy/cy (arc-seconds, centuries) (sources cited in &lt;a href=&quot;https://en.wikipedia.org/wiki/Richard_Dunthorne&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Richard_Dunthorne&lt;/a&gt;). A closer estimate is about +12.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then Laplace in 1788 published results of theoretical calculations of celestial mechanics to identify the real accelerative effect on the moon caused by the (by-then established) slow and slight reduction in eccentricity of the earth-sun orbit (&quot;Sur l'equation seculaire de la Lune&quot;, (&lt;a href=&quot;http://gallica.bnf.fr/ark:/12148/bpt6k77599c/f248&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://gallica.bnf.fr/ark:/12148/bpt6k77599c/f248&lt;/a&gt;)). Unfortunately Laplace treated as negligible too many of the smaller components of the infinite series that he had to deal with. His result overestimated the real effect by about double. By chance this seemed to match the whole of the observed apparent acceleration. The apparent agreement between theory and observation quietened further enquiry for about half a century, until in 1853 J C Adams established that a correct calculation of this accelerative effect accounted only for about half of the apparent acceleration observed (&quot;On the secular variation of the moon's mean motion&quot;, Phil. Trans. R. Soc. Lond. 1853 vol.143 397-406) (&lt;a href=&quot;http://rstl.royalsocietypublishing.org/content/143/397&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://rstl.royalsocietypublishing.org/content/143/397&lt;/a&gt;). Adams' finding caused a controversy amongst astronomers that lasted some years, but it was finally accepted that Adams (and Delaunay, who confirmed his calculations) were in the right. But that left about half of the observed apparent lunar acceleration still to be accounted for. The size of effect (2) (real positive acceleration due to changing solar eccentricity) was calculated at about +6 arc&quot;/cy/cy from the later 19th-c. onwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Irregular slowing-down of earth rotation was established more recently, in the 1920s-1930s: previous suspicions were converted to practical certainty by works of E W Brown (1926), W de Sitter (1929) and a more complete quantitative study by H Spencer Jones (1939) (&quot;The rotation of the earth, and the secular accelerations of the sun, moon and planets&quot;) (&lt;a href=&quot;http://articles.adsabs.harvard.edu/full/1939MNRAS..99..541S&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://articles.adsabs.harvard.edu/full/1939MNRAS..99..541S&lt;/a&gt;). This last paper set the basis for a practical time-scale freed from irregular earth-rotation effects, and led via Ephemeris Time (&lt;a href=&quot;https://en.wikipedia.org/wiki/Ephemeris_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Ephemeris_time&lt;/a&gt;) to the later varieties of atomic time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tidal deceleration was first well-measured at about -26 arc&quot;/cy/cy as recently as 1975 by Morrison and Ward (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1975MNRAS.173..183M&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/1975MNRAS.173..183M&lt;/a&gt;) by an analysis of traditional optical results (perhaps surprisingly, by a long series of Mercury solar transit timings). As the period of lunar laser-ranging measurements has grown longer and the technique more precise, the 1975 value has been refined more recently to about -25.858 (Chapront et al., 2002 (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2002A%26A...387..700C&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2002A%26A...387..700C&lt;/a&gt;)).  (The 3.8cm or so of annual lunar recess from the earth established by lunar laser-ranging is believed to be a result of the tidal retardation, with relatively small effect on velocity.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net deceleration in modern estimates of lunar motion on atomic timescales amounts in aggregate to about -11.5 &quot;/cy/cy (relative to equinox of date).  &lt;/p&gt;&#xA;" OwnerUserId="17202" LastEditorUserId="17202" LastEditDate="2018-05-12T10:28:54.307" LastActivityDate="2018-05-12T10:28:54.307" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26267" PostTypeId="1" AcceptedAnswerId="26269" CreationDate="2018-05-12T11:23:20.037" Score="5" ViewCount="149" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Hyperion_(moon)&quot; rel=&quot;noreferrer&quot;&gt;wikipedia&lt;/a&gt; (see info below image) Hyperion's rotation period around Saturn is &quot;chaotic&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does that mean? The period (in days/hours) is different every rotation without any law helping to define it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can it be?&lt;/p&gt;&#xA;" OwnerUserId="23018" LastEditorUserId="2618" LastEditDate="2018-05-12T12:53:26.210" LastActivityDate="2018-05-12T14:07:42.413" Title="How can a rotation period of a planet's satellite/moon be &quot;chaotic&quot; (like Hyperion around Saturn)?" Tags="&lt;rotation&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26268" PostTypeId="2" ParentId="12590" CreationDate="2018-05-12T13:41:28.107" Score="1" Body="&lt;p&gt;It is an effect caused by the inclination of the precession axis. This axis takes about 23,000 years to complete. If we could live so long we would see the anemalema spinning in the sky&lt;/p&gt;&#xA;" OwnerUserId="23023" LastActivityDate="2018-05-12T13:41:28.107" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26269" PostTypeId="2" ParentId="26267" CreationDate="2018-05-12T14:07:42.413" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;According to wikipedia (see info below image) Hyperion's rotation period around Saturn is &quot;chaotic&quot;. What does that mean?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The wikipedia article refers to Hyperion's rotation being chaotic rather than its orbit about Saturn. I'll assume you are asking how Hyperion's axial rotation can be chaotic. (Its orbit is not.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An object that is not subject to any external torques will have a constant angular momentum. This does not mean that such an object will have a constant angular velocity. Place a rubber band about a book and toss it in the air, giving it a nice spin rate before letting it go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yzD8c.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yzD8c.png&quot; alt=&quot;Canonical experiment regarding stability of rotations using a book on classical mechanics&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the question &lt;a href=&quot;https://physics.stackexchange.com/questions/67957/stability-of-rotation-of-a-rectangular-prism&quot;&gt;Stability of rotation of a rectangular prism&lt;/a&gt; for details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hyperion is small enough and rigid enough that it's not quite capable of pulling itself into a roundish shape by its self gravitation. It instead has a markedly triaxial shape. Designating the principal moments of inertia as $A &amp;lt; B &amp;lt; C$ (corresponding respectively to the rotations in subfigures &lt;em&gt;b&lt;/em&gt;, &lt;em&gt;c&lt;/em&gt;, and &lt;em&gt;a&lt;/em&gt; in the above image), the longitudinal libration parameter $(B-A)/C$ is about 0.21 for Hyperion. That's huge, and is a key factor that makes Hyperion's rotation chaotic. If Hyperion was a perfectly rigid body that wasn't subject to external torques and was instantaneous rotating about an axis close to that depicted as &lt;em&gt;c&lt;/em&gt; in the above figure, it's rotation would essentially be chaotic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of the above conditions is true. Hyperion is not perfectly rigid and it is subject to external torques. Almost every regular moon in the solar system exhibits synchronous rotation and revolution. This is a result of tidal torques on the moon by the parent planet and the moon exhibiting a bit of plasticity (non rigid behavior) in response to those torques. The torques coupled with the plasticity are the mechanism by which those moons eventually became tidally locked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other possible spin:orbit couplings than the 1:1 coupling exhibited by those tidally-locked moons. Mercury, for example, is gravitationally locked with the Sun in a 3:2 spin-orbit resonance. Mercury has a rather low longitudinal libration parameter of about $10^{-8}$. The high eccentricity of Mercury's orbit together with its nearly spheroidal nature makes it more likely to have been captured in the observed 3:2 spin-orbit resonance than in a 1:1 resonance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With most moons and planets/exoplanets that orbit sufficiently close to the parent planet or star, the torque, plasticity, and low longitudinal libration parameter will eventually result in the orbiting body slipping into one of those small integer resonances. This is very unlikely to happen with Hyperion thanks to that very large longitudinal libration parameter. When Hyperion's spin and orbit happen to get close to a small integer ratio, the torque from Saturn is more likely to kick the moon into a rotation that is further away from the two stable axes than toward either one. It tumbles away from the resonant spin.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2018-05-12T14:07:42.413" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26270" PostTypeId="2" ParentId="25874" CreationDate="2018-05-12T21:29:11.093" Score="2" Body="&lt;p&gt;One should distinguish the &quot;Roche limit&quot; from the &quot;Roche lobe,&quot; as the latter is what you need when you have two objects of similar mass, like a binary star.  The Roche lobe is an effective equipotential in a frame that cororates with the orbit of the binary (assumed circular).  The equipotential is &quot;effective&quot; because it includes both stellar gravities, and also the centrifugal force from the rotating frame.  The Roche lobe is not only an effective equipotential, it is the last one that is closed-- any equipotential further out connects to infinity and cannot correspond to material in equilibrium that is corotating with the orbit.  From the side, this last equipotential looks like a kind of infinity sign.  Material outside that &quot;infinity sign&quot; has to start moving in the rotating frame, bringing in coriolis forces and in generally leading to complicated motion.  However, if close to the Roche lobe, the gas tends to flow through the nodal point at the center of the &quot;infinity&quot;, so tends to be transferred from the domain of gravitational attraction of one star to the other.  That's &quot;mass transfer.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting up the calculation of the Roche lobe is fairly easy because you only need the two gravities and the centrifugal force at the orbital period.  But it is a difficult set of equations to solve, and even at the research frontier, is usually approximated rather than calculated exactly.  No doubt this is the source of your difficulty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why one star fills its Roche lobe is a consequence of stellar evolution, often as one star attempts to evolve into a red giant or asymptotic giant.  Before it can reach its giant radius, its outer layers encounter the Roche lobe, and pass along it through the central node (the &quot;Roche point&quot;) and may ultimately end up attracted to the other star.  This is what is thought to have happened in Algol, for example, which is a binary where the star with the greater mass now started out as the star with the lesser mass (because it gained from its companion).  The rate of mass transfer is also difficult to compute, because it depends on how fast the mass-loser is evolving.  It is only its evolutionary rate that causes the &quot;Roche overflow,&quot; so the mass transfer rate is pegged to the evolutionary rate, which is largely due to rate of fusion that adds mass to its core.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-05-12T21:29:11.093" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26271" PostTypeId="1" CreationDate="2018-05-12T21:33:05.497" Score="3" ViewCount="117" Body="&lt;p&gt;Recently I took pictures of the NGC 884 cluster in the B and I bands and selected some 30-what stars from them. I've reduced the data to a colour-magnitude diagram (absolute magnitude in the I band $M_I$ vs. (B-I) colour index). I am now asked to designate where the stars I've selected from this cluster are located on the Hertzsprung-Russell diagram and the approximate age of the stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is the following: any HR diagram that I can find is either expressed in effective temperature instead of colour index, or is in the colour index (B-V). Also I can only seem to find empirical formulas for the effective temperature that use the (B-V) colour index, not the (B-I) one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know where to find an HR diagram that uses the (B-I) colour index?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/izyMF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/izyMF.png&quot; alt=&quot;absolute mag vs. (B-I) colour index&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22839" LastEditorUserId="31410" LastEditDate="2021-05-27T13:56:01.590" LastActivityDate="2021-05-27T13:56:01.590" Title="Locating a star cluster on a Hertzsprung-Russell diagram with other color index" Tags="&lt;amateur-observing&gt;&lt;data-analysis&gt;&lt;hr-diagram&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26272" PostTypeId="2" ParentId="26206" CreationDate="2018-05-12T23:44:19.200" Score="1" Body="&lt;p&gt;This interesting problem is clearly relevant in astronomy as well. Within medical imaging, determining the center of an object also comes into play in several, special application areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I posed the question here to become informed as to which formula's are being used within astronomy: How is the uncertainty of the 'COG' being estimated, in the presence of (additive) measurement noise?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 2002, we derived a general formula for the variance of the estimated center of gravity, in the presence of normally distributed additive noise, associated with each pixel value. The reference is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H.C. van Assen, M. Egmont-Petersen, J.H.C. Reiber. &quot;Accurate object localization in gray level images using the center of gravity measure: accuracy versus precision&quot;, &lt;em&gt;IEEE Transactions on Image Processing&lt;/em&gt;, Vol. 11, No. 12, pp. 1379-1384, 2002.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will first give the general formula, which only assumes that the center of the window surrounding the object is $(0,0)$. The object &lt;em&gt;does not have to be positioned centrally&lt;/em&gt; for this formula to hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the additive measurement noise associated with each pixel as $\epsilon \sim U(0,\sigma^2)$, with $\sigma$ being its standard deviation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define a squared (sub)image ${\cal W}$ of dimensions $(d+1) \times (d+1)$ pixels, with coordinates, $x=-\frac{d}{2},\frac{d}{2}+1,\ldots,\frac{d}{2}$, and $y=-\frac{d}{2},\frac{d}{2}+1,\ldots,\frac{d}{2}$. Let ${\bf w}_{x,y}$ be the true pixel intensity of pixel $(x,y)$ in absense of any noise, in ${\cal W}$. Define the signal-plus-noise image $W$ as: $w_{x,y}={\bf w}_{x,y} + \epsilon$. Define the total number of pixels in the (sub)image as $N=(d+1)^2$, which includes the central 0-th row and 0-th column of $W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The values of $w_{x,y}$, with ($w_{x,y} \geq 0$), are the ones actually being observed. Close to the center $(x=0,y=0)$ a &lt;strong&gt;bright object&lt;/strong&gt; of interest has been located.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The estimated center of gravity $cog$ of this object is computed as follows:&#xA;$$&#xA;\widehat{cog}(x,y)=\left(\dfrac{\sum_{x,y} \; x\,w_{x,y}}{\sum_{x,y} \; w_{x,y}},\dfrac{\sum_{x,y} \; y\,w_{x,y}}{\sum_{x,y} \; w_{x,y}}\right)&#xA;$$&#xA;where $x$ and $y$ are running indices such that each pixel in $W$ enters the calculation of a sum (sigma) exactly one time. $cog$ is the measurement that is influenced by measurement noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the delta rule two successive times, we derived a general approximate formula for the variance of the cog, given a known noise level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $x2$ as:&#xA;$$&#xA;x2 = \sum_{x} \; \sum_{y} \; x^2&#xA;$$&#xA;and similarly $y2$ as:&#xA;$$&#xA;y2 = \sum_{x} \; \sum_{y} \; y^2&#xA;$$&#xA;Finally, the average 'weight' (average pixel intensity) is given by:&#xA;$$&#xA;\hat{\mu}_w = N^{-1} \; \sum_{x} \; \sum_{y} \; w_{x,y}&#xA;$$&#xA;Let $\widehat{cog}(x)$ denote the estimated center of gravity $x$-coordinate and $\widehat{cog}(y)$ the estimated center of gravity $y$-coordinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The MLE derived variance estimates of $\widehat{cog}(x)$ and $\widehat{cog}(y)$ are:&#xA;$$&#xA;\text{var}(\widehat{cog}(x))=\left(\frac{\sigma^2 \; x2}{\left[N \; \hat{\mu}_w \; \widehat{cog}(x)\right]^2} + \frac{\sigma^2}{N \; (\hat{\mu}_w)^2} \right) \; (\widehat{cog}(x))^2&#xA;$$&#xA;and &#xA;$$&#xA;\text{var}(\widehat{cog}(y))=\left(\frac{\sigma^2 \; y2}{\left[N \; \hat{\mu}_w \; \widehat{cog}(y)\right]^2} + \frac{\sigma^2}{N \; (\hat{\mu}_w)^2} \right) \; (\widehat{cog}(y))^2&#xA;$$&#xA;It turns out that when the &lt;strong&gt;true cog&lt;/strong&gt; is precisely $(0,0)$, then the limit (simplified) formulas for the cog-variance hold:&#xA;$$&#xA;\lim_{cog(x) \to 0} \; \; \lim_{cog(y) \to 0} \; \text{var}(\widehat{cog}(x)) = \frac{\sigma^2 \; x2}{(N \; \hat{\mu}_w)^2}&#xA;$$ &#xA;and &#xA;$$&#xA;\lim_{cog(x) \to 0} \; \; \lim_{cog(y) \to 0} \; \text{var}(\widehat{cog}(y)) = \frac{\sigma^2 \; y2}{(N \; \hat{\mu}_w)^2}&#xA;$$&#xA;as the second (additive) variance term within the outer parentheses then vanishes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My recent simulations show that above $95\%$ of the actual variance results from out defined variance formula, as presented here. This simulation result also holds when the true &lt;code&gt;cog&lt;/code&gt; deviates more than one coordinate from the central position $(0,0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simulation graphic showing the accuracy of the variance estimation will be added to this answer, one of the coming days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Multiplicative noise&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Poisson noise occurs in CCD-cameras, its influence can especially be observed in gradient transitions of image intensity from very dark areas to really bright ones. The magnitude of Poisson noise is known to be proportional to signal intensity. In this case, &lt;em&gt;multiplicative noise&lt;/em&gt; is present in the observed image:&#xA;$$&#xA;w_{x,y}={\bf w}_{x,y} \, \, \cdot \alpha \, \cdot  \, \epsilon&#xA;$$&#xA;with $\alpha$ being a proportionality constant and $\epsilon$ the normally distributed noise term (which approximates a Poisson distribution well for moderate $w_{i,j} &amp;gt;0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Performing a simple $\ln(\cdot)$ transformation, $lw_{i,j}=\ln(w_{i,j})$ yields a transformed image $lw$ perturbed by &lt;em&gt;additive&lt;/em&gt; noise. Subsequently, the cog-variance estimator can be applied to the $lw$-image.&lt;/p&gt;&#xA;" OwnerUserId="22970" LastEditorUserId="22970" LastEditDate="2018-05-17T19:22:46.990" LastActivityDate="2018-05-17T19:22:46.990" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26274" PostTypeId="1" AcceptedAnswerId="26275" CreationDate="2018-05-13T06:12:50.807" Score="4" ViewCount="148" Body="&lt;p&gt;&lt;em&gt;I read somewhere that an exoplanet in the name of HAT-P-7B orbits its star backwards.Why?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J1Ggz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J1Ggz.jpg&quot; alt=&quot;HAT P 7B&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;And why did we consider our orbital movement to be forward?Is there some scientific reason to consider the movement as forward?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="22623" LastEditorUserId="22623" LastEditDate="2018-06-08T18:47:52.883" LastActivityDate="2018-06-08T18:47:52.883" Title="What makes an exoplanet to orbit around its star backwards?And why did we consider our orbital movement to be forward?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;orbit&gt;&lt;exoplanet&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26275" PostTypeId="2" ParentId="26274" CreationDate="2018-05-13T08:05:19.010" Score="7" Body="&lt;p&gt;All the planets in the solar system orbit the same way. This is not a coincidence. Planets all form from the same protoplanetary disc that surrounded the young sun. The particles in the disc orbited the same way as the sun rotates, and the planets continue to orbit in the same direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this context &lt;em&gt;forward means orbiting in the same direction as the star rotates&lt;/em&gt;. We would expect most planets to orbit &quot;forwards&quot; for the same reason that all the planets in the solar system orbit forwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What probably happened with HAT-P-7b is that it was thrown into an odd orbit following a close encounter with another planet, or perhaps another star. It is also conceivable that it is a captured rogue planet. See a &lt;a href=&quot;https://www.newscientist.com/article/dn17613-second-backwards-planet-found-a-day-after-the-first/&quot; rel=&quot;noreferrer&quot;&gt;New Scientist article about the planet&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-05-13T08:05:19.010" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26276" PostTypeId="2" ParentId="26271" CreationDate="2018-05-13T19:03:19.553" Score="2" Body="&lt;p&gt;With an absolute magnitude of zero and a colour (any colour) of zero, your stars are around spectral type A0 (by definition). The only thing that can confuse this is reddening/extinction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is appreciable reddening/extinction, then your stars could be more luminous and hotter (O/B stars), but there isn't any way to tell from your data.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-05-13T19:03:19.553" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26277" PostTypeId="1" AcceptedAnswerId="26278" CreationDate="2018-05-13T19:09:16.627" Score="6" ViewCount="197" Body="&lt;p&gt;Apologies in advance for my over-exposure to pop science.  I want to propose a telescope powerful enough to collect imagery from 10^-32 seconds after the “Big Bang”, or as soon as the end of the inflationary epoch.  With light coming from the universe when roughly 10 ly in diameter, along with the expansion to our current size, I would only see the tiniest fraction of light from so long ago with a telescope looking at that far of a distance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming my telescope has an identical view field as the Hubble Space Telescope, what fraction of the total volume of the universe at end of the inflationary epoch would I actually be viewing?&lt;/p&gt;&#xA;" OwnerUserId="10234" LastActivityDate="2018-05-14T15:47:49.513" Title="Portion of universe visible if gathering image from inflationary epoch" Tags="&lt;hubble-telescope&gt;&lt;early-universe&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26278" PostTypeId="2" ParentId="26277" CreationDate="2018-05-13T21:13:28.183" Score="11" Body="&lt;p&gt;&lt;strong&gt;tl;dr&lt;/strong&gt;: Your field of view would cover roughly one square centimeter of the sky at that time, and you would observe roughly 50 billionths of the observable Universe.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;You can't really…&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;With photons, you will never be able to see further back than to &lt;a href=&quot;https://en.wikipedia.org/wiki/Recombination_(cosmology)&quot; rel=&quot;noreferrer&quot;&gt;recombination&lt;/a&gt;, when the Universe was 380,000 yr old, because until then, the free electrons made is opaque to radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With neutrinos, you can (in principle) see back to 1 sec after Big Bang, but to &quot;see&quot; all the way back to inflation, you will probably need gravitational waves.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;…but let's assume you can anyway&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Anyway, lets's do a thought experiment and assume that you have a RLT (Ridiculously Large Telescope) with a field of view (FoV) of the HST. What fraction of the Universe can you see? The answer might surprise you:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The farther an object is from you, the smaller an angle it spans on the sky, i.e. the smaller it looks. This is true for birds and planets and even for the nearest galaxies. However, for distant galaxies a strange effect counteracts this: Because of 1) the finite speed of light and 2) the expansion of the Universe, distant galaxies were closer to you when they emitted the light you see today, and therefore spanned a larger angle. Hence, if you compare galaxies of the same physical size, they look smaller and smaller until around 15 Glyr (giga-lightyears), after which they'll look larger and larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hubble Ultra Deep Field has FoV of 2.8 by 2.5 arcminutes. The region of the Universe that this FoV spans is thus largest at a distance of 15 Glyr, where it span $4.8\times4.3\,\mathrm{Mlyr}^2$ (i.e. square-mega-lightyears). But then your FoV begins to span an increasingly smaller region. At recombination, it spans only $34\times30\,\mathrm{klyr}^2$ — in other words, if a Milky Way-sized galaxy were present there (it weren't), it would be larger than the FoV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It you could look all the way back to 4 hours after Big Bang, your FoV would span roughly a square-lightyear. Three minutes after BB, when all the Universe's hydrogen and helium had just been created, it would span $\sim0.1\times0.1\,\mathrm{lyr}^2$. A few microseconds after BB, it would span roughly a square-AU (the distance from Earth to Sun). Approximately $10^{-22}\,\mathrm{s}$ after BB, it would span a square kilometer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And at $10^{-32}\,\mathrm{s}$ after BB, your FoV would span $0.8\times0.7\,\mathrm{cm}^2$, i.e. roughly one square-centimeter!&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Fraction of the Universe observed&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The part of the observable Universe that you would observe is the volume &quot;behind&quot; your FoV. That can be obtained by integrating the area along the distance, but an easier way is to simply note the fact that, if your FoV were the whole sky, you'd observe the whole Universe. Since your FoV covers a fraction of the sky of&#xA;$$&#xA;\begin{array}\\&#xA;f &amp;amp; = &amp;amp; \frac{2.8'\times2.5'}{\mathrm{full\,sky}}\\&#xA;  &amp;amp; = &amp;amp; \frac{8\times10^{-4}\,\mathrm{rad}\,\,\times\,\,7\times10^{-4}\,\mathrm{rad}}{4\pi\,\mathrm{rad}^2}\\&#xA;  &amp;amp; = &amp;amp; 5\times10^{-8},&#xA;\end{array}&#xA;$$&#xA;this is the fraction of the Universe that you would observe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At $t=10^{-32}\,\mathrm{s}$, the part of the Universe later became our observable Universe &lt;em&gt;today&lt;/em&gt;, was about ten meters in radius. Another way to calculate the fraction is to realize that the above mentioned observed square-centimeter roughly comprises a fraction $f$ of the surface area of a sphere with a radius of ten meters.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-05-14T15:47:49.513" LastActivityDate="2018-05-14T15:47:49.513" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26279" PostTypeId="1" CreationDate="2018-05-13T22:00:44.887" Score="2" ViewCount="159" Body="&lt;p&gt;If I am standing looking towards the East from the Northern Hemisphere, do the constellations appear to rotate towards the West via the North or the West via the South or am I completely off target all together? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do we say Capricorn is in December at the Solstice? Where do we see it in the sky in the Northern Hemisphere in December and likewise would we see Aquarius in that same position in February etc..? &lt;/p&gt;&#xA;" OwnerUserId="23030" LastEditorDisplayName="user1569" LastEditDate="2018-05-14T07:37:34.383" LastActivityDate="2018-05-14T07:37:34.383" Title="Which direction do the constellation revolve in?" Tags="&lt;rotation&gt;&lt;constellations&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26280" PostTypeId="2" ParentId="26279" CreationDate="2018-05-13T22:33:25.133" Score="1" Body="&lt;p&gt;The sky appears to rotate around the Poles. Stars that are in the northern Hemisphere will move in circles around the Pole. Some Southern hemisphere stars are also visible from the North, the move in circle around the southern pole (Although the star is sometimes visible the southern pole is always below the horizon from the Northern hemisphere)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A star that rises due East, will reach its highest position in the sky when it it is due South. It won't enter the Northern half of the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomers don't tend to talk much of Capricorn in December. (That leads to astrological nonsense) But it means that the sun in in the same region of the sky as Capricorn at that time (which means that you certainly can't see it) In fact the best time to see Capricorn is in Early Summer, when it is as far from the sun as possible.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="13426" LastEditDate="2018-05-13T22:44:09.300" LastActivityDate="2018-05-13T22:44:09.300" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26281" PostTypeId="1" CreationDate="2018-05-14T06:31:09.963" Score="18" ViewCount="2855" Body="&lt;p&gt;The energy of electromagnetic radiation is related to frequency; higher the frequency, higher the energy level. If electromagnetic waves have lower frequency when they arrive on Earth than originally emitted due to the Doppler effect, where, according to the conservation of energy, does the excess energy go?&lt;/p&gt;&#xA;" OwnerUserId="22988" LastActivityDate="2018-05-15T10:58:13.070" Title="Energy loss from Doppler effect" Tags="&lt;doppler-effect&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26282" PostTypeId="2" ParentId="26281" CreationDate="2018-05-14T07:33:15.170" Score="13" Body="&lt;p&gt;It's spread out in time. If a source emits a 1W pule of energy lasting 1 second and the receiver is receding so fast that it is Doppler shifted to a frequency which means the power is just 0.5W, then the pulse will take 2seconds to arrive (since the end of it had further to travel).&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-05-14T07:33:15.170" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26283" PostTypeId="1" CreationDate="2018-05-14T08:21:03.103" Score="1" ViewCount="314" Body="&lt;p&gt;I have a template spectral energy distribution (SED) at z=0. I want to shift the template SED to an arbitrary redshift (z&gt;0), how to do this in python?&lt;/p&gt;&#xA;" OwnerUserId="23032" LastEditorUserId="9527" LastEditDate="2020-02-26T07:56:30.063" LastActivityDate="2020-03-27T08:02:20.330" Title="Redshifting restframe SED" Tags="&lt;observational-astronomy&gt;&lt;galaxy&gt;&lt;redshift&gt;&lt;spectroscopy&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26284" PostTypeId="1" AcceptedAnswerId="26285" CreationDate="2018-05-14T08:40:59.627" Score="5" ViewCount="318" Body="&lt;p&gt;Can a black dwarf be suitable for manned exploration, once it cools down? And potentially used for space mining?&lt;/p&gt;&#xA;" OwnerUserId="23033" LastActivityDate="2018-05-14T15:02:06.030" Title="Black Dwarf Capabilities" Tags="&lt;star&gt;&lt;star-formation&gt;&lt;dwarf-star&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26285" PostTypeId="2" ParentId="26284" CreationDate="2018-05-14T08:50:18.390" Score="7" Body="&lt;p&gt;A 1 solar mass, Earth sized black dwarf would have a surface gravity of about 360 000 g which probably rules out manned exploration by anything we would normally think of as human. For similar reasons, mining would be quite challenging. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another obstacle is that the Universe is not old enough to have produced any black dwarves yet. The oldest white dwarves still have a surface temperature of thousands of Kelvins. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, if humanity is still around when they have formed, visiting and mining them will probably be easy.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="21928" LastEditDate="2018-05-14T10:17:55.757" LastActivityDate="2018-05-14T10:17:55.757" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26287" PostTypeId="1" AcceptedAnswerId="26291" CreationDate="2018-05-14T09:29:59.983" Score="6" ViewCount="71" Body="&lt;p&gt;Simply put, would, say, rocky planets orbiting red dwarfs have a noticably different chemical makeup than rocky planets orbiting B-type main sequence stars? &lt;em&gt;Mutatis mutandis&lt;/em&gt; gas giants and the other types of host stars.&lt;/p&gt;&#xA;" OwnerUserId="23035" LastActivityDate="2018-05-14T14:57:38.357" Title="Is there any difference among exoplanets in composition due to host star?" Tags="&lt;star&gt;&lt;planet&gt;&lt;exoplanet&gt;&lt;astrochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26290" PostTypeId="2" ParentId="26281" CreationDate="2018-05-14T12:48:36.493" Score="12" Body="&lt;p&gt;Completely ignoring relativistic effects, it depends on which reference frame you are using; the &quot;missing&quot; energy is seen as kinetic energy in either the emitting or receiving atom as recoil depending on which one you view as moving. Energy is not conserved between reference frames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am travelling at a velocity away from you, and I emit a photon at you which I observe to have a frequency &lt;em&gt;f&lt;/em&gt;, then I will assume that photon has an energy E = hf, where h is the Planck constant. I will never observe a different energy for that photon - in my reference frame, energy is conserved. &#xA;You, however, will observe a different frequency &lt;em&gt;f&lt;/em&gt;, and therefore a different energy E. This energy remains constant for you - energy is conserved in your reference frame - but the energy I observe and the energy you observe differ - energy is not conserved between our reference frames; that is to say, the energy is &lt;em&gt;conserved&lt;/em&gt; but not &lt;em&gt;invariant&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider - I drive past stationary you in a car and throw a tennis ball at you. In my perspective, the tennis ball has greater kinetic energy (it is moving at my speed, plus the speed of the ball) than in your perspective. Energy is not invariant in this circumstance either!&lt;/p&gt;&#xA;" OwnerUserId="23039" LastEditorUserId="23039" LastEditDate="2018-05-14T12:54:58.463" LastActivityDate="2018-05-14T12:54:58.463" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26291" PostTypeId="2" ParentId="26287" CreationDate="2018-05-14T14:57:38.357" Score="5" Body="&lt;p&gt;The chemical makeup of space dust -- the stuff  which agglomerates into planets -- is dependent on the history of supernovas and neutron stars and the like which created the atoms in a given region of space.  This is unlikely to bear any relationship to the size or age of the star around which said items revolve. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-05-14T14:57:38.357" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26292" PostTypeId="2" ParentId="10108" CreationDate="2018-05-14T15:40:38.570" Score="3" Body="&lt;p&gt;I checked the &lt;a href=&quot;http://www.ctio.noao.edu/%7Eatokovin/stars/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Multiple Star Catalog&lt;/a&gt; for data on these systems. It has hierarchical information on &lt;a href=&quot;http://www.ctio.noao.edu/%7Eatokovin/stars/stars.php?cat=HIP&amp;amp;number=79374&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nu Scorpii&lt;/a&gt; and &lt;a href=&quot;http://www.ctio.noao.edu/%7Eatokovin/stars/stars.php?cat=HIP&amp;amp;number=115990&quot; rel=&quot;nofollow noreferrer&quot;&gt;AR Cassiopeiae&lt;/a&gt;, which I also used for &lt;a href=&quot;https://astronomy.stackexchange.com/a/24312/2153&quot;&gt;this answer&lt;/a&gt;. I'll try to explain the orbital structure of both systems.&lt;/p&gt;&#xA;&lt;h2&gt;Nu Scorpii&lt;/h2&gt;&#xA;&lt;p&gt;Think of Nu Scorpii as two subsets of stars, Nu Scorpii AB and Nu Scorpii CD, orbiting each other with a period of 105,000 years. AB contains four stars:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Nu Scorpii Aa, Ab and Ac:&lt;/strong&gt; Aa and Ab are a binary system (period: 5.5 days) orbited by Ac (period: 8.3 years).&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Nu Scorpii B:&lt;/strong&gt; This star orbits Nu Scorpii A - the three stars above - with a period of 675 years.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Nu Scorpii CD contains three stars:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Nu Scorpii C:&lt;/strong&gt; C is a single star.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Nu Scorpii Da and Db:&lt;/strong&gt; Da and Db are a binary system (period: unknown), which together orbit C over 3,100 years.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h2&gt;AR Cassiopeiae&lt;/h2&gt;&#xA;&lt;p&gt;AR Cassiopeiae is more complicated; it contains three main subsets:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;AR Cassiopeiae AB:&lt;/strong&gt; The contains three stars: Aa and Ab, a binary orbiting each other every 6 days, and B, orbiting A every 545 years.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;AR Cassiopeiae CD:&lt;/strong&gt; These pair of binary stars orbit each other once every 2,200 years, and orbit component AB in 438,000 years.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;AR Cassiopeiae FG:&lt;/strong&gt; This is another binary, with a period of 58,000 years. It orbits AB every 381,000 years.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Multiple star systems are usually nested like this - binaries and triple systems inside more binaries. I would image it's rare to have any other setup, like four stars orbiting together, because such a system would be highly unstable. You really need to nest the stars, with large separations between components (i.e. the distance between Nu Scorpii AB and CD must be much larger than the distance between Nu Scorpii A and Nu Scorpii B).&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-05-14T15:40:38.570" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26295" PostTypeId="1" AcceptedAnswerId="26296" CreationDate="2018-05-14T19:18:36.243" Score="6" ViewCount="304" Body="&lt;p&gt;I wonder if there a difference between horizon and comoving distances in cosmology ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean both correspond to a null geodesic $(ds^2=0)$ right ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_{horizon} (t) =a(t) \int\limits_{0}^{r} \frac{{\rm d}r'}{\sqrt{1-kr'^2}}= a(t) \int\limits_{0}^{t}\frac{ c \, {\rm d}t'}{a(t')}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $a(t)$ the scale factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the difference just that for a comoving distance the integration bounderies changes ? like $r_e$,  $r_0$  and $t_0$, $t_e$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for help&lt;/p&gt;&#xA;" OwnerUserId="23047" LastActivityDate="2018-05-17T13:55:47.120" Title="What is difference between horizon and comoving distances?" Tags="&lt;cosmology&gt;&lt;distances&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26296" PostTypeId="2" ParentId="26295" CreationDate="2018-05-14T19:44:04.430" Score="4" Body="&lt;p&gt;The comoving distance is a distance. You can calculate the comoving distance from your head to your toes (though it's more useful in cosmological contexts). It is defined such that &lt;em&gt;today&lt;/em&gt; it coincides with the actual, physical distance you would measure if you froze the Universe and laid out meter sticks. In 11 billion years from now, when the Universe has doubled in size, physical (cosmological) distances have increased by a factor of two, but the comoving distances are, by definition, the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;horizon&quot; is a term used for a (non-physical) boundary, e.g. the &quot;particle horizon&quot;, which is the boundary between the observable and the un-observable Universe. Or the &quot;(cosmological) event horizon&quot;, which marks the boundary between regions from which we may at some point receive a signal emitted today, and regions from which we cannot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like on Earth it makes sense to talk about &quot;the distance to the horizon&quot;, you can say, e.g., &quot;&lt;em&gt;The comoving distance to the particle horizon is 46 billion lightyears&lt;/em&gt;&quot;, or &quot;&lt;em&gt;The comoving distance to the event horizon is 17 billion lightyears&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you say, to calculate the comoving distance you use the Robertson-Walker metric (assuming here a flat Universe for simplicity)&#xA;$$&#xA;ds^2 = -c^2 dt^2 + a(t)^2 \big[ dr^2 + r^2 d\Omega^2 \big],&#xA;$$&#xA;and set $ds=d\Omega=0$ to measure along a radial null geodesic. Then $dr = c\,dt/a(t)$, or&#xA;$$&#xA;r = c \int_{t_\mathrm{em}}^{t_\mathrm{obs}} \frac{dt}{a(t)},&#xA;$$&#xA;where $t_\mathrm{em}$ and $t_\mathrm{obs}$ is the time when the light is emitted and observed, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the comoving distance to the particle horizon, you set $t_\mathrm{em} = 0$ and $t_\mathrm{obs} =$ today, because the particle horizon corresponds to light emitted at Big Bang and observed today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the comoving distance to the event horizon, you set $t_\mathrm{em} =$ today and $t_\mathrm{obs} = \infty$, because the event horizon corresponds to light emitted today and observed in the (almost) infinitely distant future.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-05-17T13:55:47.120" LastActivityDate="2018-05-17T13:55:47.120" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26297" PostTypeId="1" AcceptedAnswerId="26302" CreationDate="2018-05-14T20:24:00.250" Score="4" ViewCount="95" Body="&lt;p&gt;I was wondering what one would see if they were to look through a telescope tracking a meteor as it shot through the sky. Technology aside, would one be able to see the object tumbling or would it always be obscured by the light/dust/plasma?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically would there be any advantage to getting a high shutter-speed video with high magnification tracking the leading edge of a meteor or would it not offer any greater detail than just a bright streak in the sky?&lt;/p&gt;&#xA;" OwnerUserId="23049" LastActivityDate="2018-05-15T10:52:42.860" Title="Is it possible to image a meteor so the rock is visible?" Tags="&lt;observational-astronomy&gt;&lt;photography&gt;&lt;meteor&gt;&lt;meteor-shower&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26299" PostTypeId="1" AcceptedAnswerId="26300" CreationDate="2018-05-15T03:42:04.270" Score="1" ViewCount="118" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tTSij.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tTSij.gif&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/262813/why-does-the-sun-have-to-be-nearly-fully-covered-to-notice-any-darkening-in-an-e&quot;&gt;This question on Physics SE&lt;/a&gt; about apparent brightness during a solar eclipse has a log-scale graph of light intensity versus time.  Parts of it are the shape I'd expect: a nearly-smooth curve going from &quot;extremely bright&quot; to &quot;very dim&quot;.  However, there's a great deal of apparent noise in the graph, with jumps of nearly a full division in the graph (corresponding to about a two-fold or eight-fold change in brightness, depending on if it's a natural or base-10 log).  What causes these jumps?&lt;/p&gt;&#xA;" OwnerUserId="1559" LastActivityDate="2018-05-15T07:58:10.573" Title="What would cause large jumps in brightness during a solar eclipse?" Tags="&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26300" PostTypeId="2" ParentId="26299" CreationDate="2018-05-15T07:58:10.573" Score="6" Body="&lt;p&gt;Eric Pauer, who produced the chart, has an interesting web site that covers this eclipse.  It is evident that the weather was not great at the time of the eclipse, with intermittent cloud, and it will have been passing clouds that caused the variations in intensity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see the evidence from this page &#xA;&lt;a href=&quot;http://www.pauerhome.com/eclipse/eclipse99/1contact/1contact.htm&quot; rel=&quot;noreferrer&quot;&gt;http://www.pauerhome.com/eclipse/eclipse99/1contact/1contact.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2018-05-15T07:58:10.573" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26302" PostTypeId="2" ParentId="26297" CreationDate="2018-05-15T10:34:51.317" Score="1" Body="&lt;p&gt;Meteors are very small, typical ones are the size of a grain of sand, and bright fireballs are only a few grams. So they are nowhere near as big as &quot;rocks&quot; and there is no way to image them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Meteoroid&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Meteoroid&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://curious.astro.cornell.edu/physics/73-our-solar-system/comets-meteors-and-asteroids/meteors/303-what-is-the-typical-size-of-a-visible-shooting-star-intermediate&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://curious.astro.cornell.edu/physics/73-our-solar-system/comets-meteors-and-asteroids/meteors/303-what-is-the-typical-size-of-a-visible-shooting-star-intermediate&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.amsmeteors.org/fireballs/faqf/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amsmeteors.org/fireballs/faqf/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let's say you could fly next to a shooting star in a UFO and image it that way. The next problem you would face is that the meteor is incandescent and and it would be difficult to get anything but an overexposed blob in a regular camera.&lt;/p&gt;&#xA;" OwnerUserId="22867" LastEditorUserId="22867" LastEditDate="2018-05-15T10:52:42.860" LastActivityDate="2018-05-15T10:52:42.860" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26303" PostTypeId="2" ParentId="26281" CreationDate="2018-05-15T10:58:13.070" Score="1" Body="&lt;p&gt;Conservation of energy doesn't apply to this situation because the energy you measure when at rest with respect to the source and the energy you measure when moving with respect to the source are in different reference frames. Energy is not conserved between different reference frames; in other words, if you're going to use conservation of energy, you have to make all your measurements without changing velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information, have a look at &lt;a href=&quot;https://physics.stackexchange.com/questions/1368/is-kinetic-energy-a-relative-quantity-will-it-make-inconsistent-equations-when-a&quot;&gt;https://physics.stackexchange.com/questions/1368/is-kinetic-energy-a-relative-quantity-will-it-make-inconsistent-equations-when-a&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;varbatim from &lt;a href=&quot;https://physics.stackexchange.com/a/15280/168619&quot;&gt;David Z's answer for a question on physics SE&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="23042" LastActivityDate="2018-05-15T10:58:13.070" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26304" PostTypeId="2" ParentId="26283" CreationDate="2018-05-15T13:06:55.483" Score="0" Body="&lt;p&gt;$\lambda_{rest} \times (1 + z) = \lambda_{obs}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\lambda_{rest}$ is your restframe wavelength template.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-05-15T13:06:55.483" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26306" PostTypeId="1" CreationDate="2018-05-15T21:46:43.260" Score="3" ViewCount="117" Body="&lt;p&gt;The Sun, in its apparent motion along the ecliptic through the year, passes over the background of distant stars. However, given that we can't observe those stars due to the brightness of the Sun, how were the ancient astronomers able to identify which star the Sun appears to be over or near to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only possible explanation I can come up with is that they observed the morning sky just before dawn (while stars were still visible) or the night sky just after sunset and so they couldn't really identify the precise location of the Sun on the celestial sphere but they could describe its position in rather general terms, which is why they might say, e.g. that the Sun is in the constellation of the Capricorn. Am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="522" LastEditorUserId="522" LastEditDate="2018-05-17T13:27:22.187" LastActivityDate="2018-05-17T13:27:22.187" Title="location of the sun on the celestial sphere" Tags="&lt;ecliptic&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26309" PostTypeId="1" CreationDate="2018-05-16T12:55:45.470" Score="1" ViewCount="111" Body="&lt;p&gt;Can someone help me? I have a case, like, we have H-R diagram and two stars on it. One of them is a blue star on main sequence, another is a red giant. They have equal luminosity. If they are in a double star system and seen from Earh as one star, then where on the diagram should I place this star? What luminosity amd temperature will it have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe I should sum up their luminosity and then, using Stefan Boltzmann law, calculate the temperature? But does the law work for red giants? And can I sum the luminosity?&lt;/p&gt;&#xA;" OwnerUserId="23070" LastEditorUserId="1891" LastEditDate="2018-05-17T06:54:12.557" LastActivityDate="2018-05-17T17:33:23.960" Title="H-R diagram, Double stars" Tags="&lt;temperature&gt;&lt;luminosity&gt;&lt;star-systems&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26310" PostTypeId="1" AcceptedAnswerId="26311" CreationDate="2018-05-16T14:30:49.820" Score="3" ViewCount="870" Body="&lt;p&gt;As an enthusiastic amateur scientist, I'm plowing my way through a cool &lt;a href=&quot;http://sci.esa.int/gaia/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gaia&lt;/a&gt; &lt;a href=&quot;https://www.cosmos.esa.int/web/gaia/dr2&quot; rel=&quot;nofollow noreferrer&quot;&gt;DR2&lt;/a&gt;-based paper, &lt;a href=&quot;https://arxiv.org/pdf/1805.05849.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;An Empirical Measurement of the Initial-Final Mass Relation with Gaia White Dwarfs&quot;&lt;/a&gt;. I'm doing all right, but have hit a roadblock: understanding the meaning of the stellar color value G&lt;sub&gt;BP&lt;/sub&gt; - G&lt;sub&gt;RP&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G&lt;sub&gt;BP&lt;/sub&gt; is the blue photometry value, and G&lt;sub&gt;RP&lt;/sub&gt; is the red photometry value, so G&lt;sub&gt;BP&lt;/sub&gt; - G&lt;sub&gt;RP&lt;/sub&gt; is the difference between the two, giving a number broadly encoding the star's color. Here's a graph from the paper that uses this value:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yvmaD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yvmaD.png&quot; alt=&quot;Gaia graph; absolute magnitude versus GBP - GRP&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a graph of white dwarf (&quot;WD&quot;) color versus absolute magnitude. On the left, the absolute magnitude M&lt;sub&gt;G&lt;/sub&gt; increases (downwards) as the star grows dimmer, and across the bottom the color G&lt;sub&gt;BP&lt;/sub&gt; - G&lt;sub&gt;RP&lt;/sub&gt; increases (rightwards) as the star grows redder. Hence the slope of the curve; as a WD cools, it gets smoothly dimmer and redder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But wait! as a star gets redder, the the &quot;blue&quot; brightness should become relatively smaller than the &quot;red&quot; brightness, right? So, the G&lt;sub&gt;BP&lt;/sub&gt; - G&lt;sub&gt;RP&lt;/sub&gt; value should get smaller and smaller, right? But that's the opposite of what this graph (and other graphs in the paper) apparently shows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should G&lt;sub&gt;BP&lt;/sub&gt; - G&lt;sub&gt;RP&lt;/sub&gt; get larger as a star cools and gets redder? Where am I confused? (I'm assuming the professionals aren't all confused...)&lt;/p&gt;&#xA;" OwnerUserId="9889" LastEditorDisplayName="user10106" LastEditDate="2018-05-16T15:44:43.273" LastActivityDate="2018-05-18T07:03:33.160" Title="Gaia Gbp - Grp: why does it get larger as star gets redder?" Tags="&lt;temperature&gt;&lt;white-dwarf&gt;&lt;gaia&gt;&lt;absolute-magnitude&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26311" PostTypeId="2" ParentId="26310" CreationDate="2018-05-16T15:07:05.123" Score="3" Body="&lt;p&gt;But you demonstrate that you know the answer to this question! Fainter objects have larger magnitudes. So as the white dwarf becomes cooler and redder, its blue magnitude $(G_{BP})$ grows by more than its red magnitude $(G_{RP})$ and the colour-index $(G_{BP} - G_{RP})$ becomes larger. This is very similar to the commonly used $B-V$ colour, which is larger for cooler stars.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-05-18T07:03:33.160" LastActivityDate="2018-05-18T07:03:33.160" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26312" PostTypeId="2" ParentId="26206" CreationDate="2018-05-16T15:17:43.733" Score="3" Body="&lt;p&gt;You might take a look at &lt;a href=&quot;https://academic.oup.com/mnras/article/371/1/323/980402&quot; rel=&quot;nofollow noreferrer&quot;&gt;this 2006 paper by Thomas et al.&lt;/a&gt; on centroiding algorithms for astronomical adaptive-optics (AO) systems, which includes a detailed discussion of error estimates for the centroid position using different algorithms. The approach you describe in your answer corresponds to what they call &quot;simple centroid&quot; (Section 3); they refer to a book chapter by Rousset (1999) for the detailed analysis (which I believe includes Poisson and readout-noise contributions, and so is not identical to your result).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, the &quot;center-of-gravity&quot; approach seems to be used in situations where one needs fast, computationally cheap estimates, such as in AO systems (where a stellar centroid needs to be determined many times a second), or as a crude first guess to provide a starting point for a more sophisticated analysis. Post-observing analysis of astronomical images generally uses more complex/sophisticated approaches, depending on whether the source is a star or other point sources versus an extended source, whether you have an accurate model of the point spread function (which may be non-circular), deblending of neighboring sources, what the noise characteristics of your data are, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, I'd guess that most such analyses use some kind of nonlinear least-squares or maximum-likelihood analysis that involves fitting a model to the data. The errors in the fitted model parameters (including the centroid position) can be derived from simplistic assumptions about the fit landscape (e.g., Levenberg-Marquardt and other gradient-based minimization algorithms sometimes provide covariance matrices based on treating the local $\chi^{2}$ landscape as a parabola), from bootstrap resampling, or from Markov Chain Monte Carlo approaches. This may be supplemented by running simulations of the fitting process on artificial images of simple models of stars or galaxies, to get some quasi-empirical estimates or corrections for the centroid (and other parameter) uncertainties.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2018-05-17T08:03:52.253" LastActivityDate="2018-05-17T08:03:52.253" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26315" PostTypeId="1" CreationDate="2018-05-17T07:03:05.150" Score="2" ViewCount="693" Body="&lt;p&gt;Using Equation 28 in Section 3.8.1 of &lt;a href=&quot;https://www.nrel.gov/docs/fy08osti/34302.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solar Position Algorithm&lt;/a&gt; (Reda, I. and A. Andreas, Jan. 2008), I computed the mean sidereal time at Greenwich for 2003 October 17 for four different times (UT), using a Casio fx-991MS. The computed values were cross-schecked online using the URL &lt;a href=&quot;http://dc.zah.uni-heidelberg.de/apfs/times/q/form#_queryForm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Computation of GAST, GMST, and ERA&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation 28 used to compute GMST is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;v0=280.46061837 + 360.98564736629*(JD - 2451545) + 0.000387933*JC^2 - JC^3/38710000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;JC is computed as:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;JC = (JD - 2451545)/36525&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The computed values of JD and JC are as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1. 00:00:00 UT: 2452929.5   (JD) &amp;amp; 0.037905544 (JC)&#xA;2. 05:30:00 UT: 2452929.729 (JD) &amp;amp; 0.037911818 (JC)&#xA;3. 11:00:00 UT: 2452929.958 (JD) &amp;amp; 0.037918083 (JC)&#xA;4. 19:21:00 UT: 2452930.306 (JD) &amp;amp; 0.037927611 (JC)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However, there is a clear offset between the computed values and the values provided by the aforementioned URL. For example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1. 00:00:00 UT: 22.32738 hour    (Computed) &amp;amp; 1.6726265 hour (URL)&#xA;2. 05:30:00 UT: 16.7231 hour     (Computed) &amp;amp; 7.1960077 hour (URL)&#xA;3. 11:00:00 UT: 11.30528667 hour (Computed) &amp;amp; 12.702743 hour (URL)&#xA;4. 19:21:00 UT: 2.93042 hour     (Computed) &amp;amp; 21.075605 hour (URL)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Moreover, if I subtract 22.32738 hour from 24.0, I will get 1.67262, which is the value provided by the URL.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same time, using the same set of equation , if I compute GMST for 1987 April 10 at 00:00:00, the computed and URL values are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;April 10 at 00:00:00:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;JD=2446895.5&#xA;JC=-0.127296372&#xA;GMST = 13.18 hour (Computed) &amp;amp; 13.17954 hour (URL)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Kindly help me to understand where exactly lies the problem. If the problem lies with the concept regarding the interpretation, please help for a better understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Knew gmst is now deprecated , but since i am a beginner thought of understanding gmst foremost and then will dwell into current scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="23080" LastEditorUserId="616" LastEditDate="2018-05-17T16:47:20.117" LastActivityDate="2018-05-20T22:47:10.470" Title="Computation of mean sidereal time at Greenwich using solar position algorithm" Tags="&lt;the-sun&gt;&lt;time&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26317" PostTypeId="2" ParentId="26315" CreationDate="2018-05-17T16:45:49.650" Score="1" Body="&lt;p&gt;The formula 28 from SPA that you linked is given with units and in a form that do not match official versions. It takes a little work to 'translate' it: also, it is needlessly susceptible to roundoff error (because all significant figures for both the date and time-of-day are strung together in one very long number).  The authors of SPA do not appear to have been striving for full official accuracy, probably their application did not require that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't quite work out from the data in the question what might have happened to the calculations, online or otherwise, but maybe the following reference data could be of help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GMST for 0h UT is still tabulated in the Astronomical Almanac, and there are also two rather easy formulae derived from the official definitions that you can use (see below), they make somewhat better preservation of end-figures than a calculation in the SPA form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the meantime, just for reference, the Astronomical Almanac for 2003 gave the GMST for 0h UT on 2003 Oct 17 as 1h 40m 21.4554s (1.6726265 h).  That is for 0h UT, and during the following 24h, the GMST for x hr UT is is easily found, it is very close to  $ [ (GMST for 0h)  +  x * 1.00273791 ] $ . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2003 was a changeover year for the official (IAU) calculations that relate GMST and UT1. Before 2003, the formula of Aoki et al (1982), &quot;The new definition of universal time&quot; (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1982A&amp;amp;A...105..359A&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/1982A&amp;amp;A...105..359A&lt;/a&gt;) was the one adopted. From 2003 Jan 1 the formula of N Capitaine et al. (2003), &quot;Expressions to implement the IAU 2000 definition of UT1&quot;, Astronomy &amp;amp; Astrophysics 406, 1135-1149 (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2003A%26A...406.1135C&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2003A%26A...406.1135C&lt;/a&gt;)  was (and still is) adopted. The difference between the results during 2003 was small (+0.0003 time-secs to get the result of the newer calculation), but it looks as if AA for 2003 was still using the older 1982 formula. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Aoki et al (1982) formula gave :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GMST of 0h UT (in time-secs) = &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 24110.54841 + 8640184.812866 *Tcy + 0.093104 *Tcy^2 -0.0000062*Tcy^3 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;plus a &lt;em&gt;very&lt;/em&gt; small correction for DeltaT: this can be had from the version of the Aoki formula given in the Capitaine reference at B.3 on page 1149, (which seems to be the one that was used in official calculations) but it can practically be left out at the precision of the data in the question. Tcy means the time interval from J2000 (Julian Date 2451545.0) reduced to centuries of 36525 days. (The JD for 0h on 2003 Oct 17 0h UT was JD(UT) 2452929.5 days and DeltaT was close to +64.55 secs.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current formula from the Capitaine (2003) reference (most conveniently from formula B.2 on page 1149) gives &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 24110.5493771 + 8640184.7945360 *Tcy + 0.0931118*Tcy^2 -0.0000062*Tcy^3 $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;plus a very small fourth-order term (given in the reference but omitted here) and again a very small correction for the slight effect of DeltaT which can be left out or included if desired as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ +307.4771600 * DeltaT/(86400.0*36525.0) $ (time-secs), where DeltaT is still in seconds.&lt;/p&gt;&#xA;" OwnerUserId="17202" LastEditorUserId="17202" LastEditDate="2018-05-17T16:55:45.757" LastActivityDate="2018-05-17T16:55:45.757" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26318" PostTypeId="1" CreationDate="2018-05-17T17:13:18.913" Score="6" ViewCount="346" Body="&lt;p&gt;The Moon receives most of its light from the Sun, either directly through sunlight or indirectly through Earthshine. However, there are many other dimmer sources of illumination. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Starlight.&lt;/strong&gt; This is probably the most important one. Under exceptional conditions, it &lt;a href=&quot;https://apod.nasa.gov/apod/ap100823.html&quot; rel=&quot;noreferrer&quot;&gt;creates shadows&lt;/a&gt; here on Earth.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Artificial light.&lt;/strong&gt; Mostly city lights and gas flares.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Meteorological activity.&lt;/strong&gt; Lightning storms on Earth and other planets are visible from space, and here on Earth, they can trigger other sources of light such as firestorms.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Bioluminescence.&lt;/strong&gt; Light-emitting microorganisms are bright enough and numerous enough to be &lt;a href=&quot;http://www.pnas.org/content/102/40/14181&quot; rel=&quot;noreferrer&quot;&gt;seen from space&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Transient bright lights.&lt;/strong&gt; Events such as supernovas and asteroid impacts are rare, but when they happen, they are a significant source of illumination. Over long timescales, for example a billion years, this may be the biggest contributor of non-solar light.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How much do these light sources illuminate the Moon? If there was no Sun, could we still see the Moon with these lights? And of the three Earth-based sources (artificial light, meteorological activity, bioluminescence), which one illuminates the best?&lt;/p&gt;&#xA;" OwnerUserId="23085" LastActivityDate="2018-05-21T15:22:55.227" Title="How much do starlight and artificial light illuminate the Moon?" Tags="&lt;the-moon&gt;&lt;light&gt;&lt;observational-astronomy&gt;&lt;weather&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26319" PostTypeId="2" ParentId="26309" CreationDate="2018-05-17T17:26:59.037" Score="1" Body="&lt;p&gt;&quot;Maybe I should sum up their luminosity and then, using Stefan Boltzmann law, calculate the temperature? But does the law work for red giants? And can I sum the luminosity?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer is not about placing the object on HR diagram, but is about the characteristic temperature as a blackbody.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like you have a point source which you know that it is a combination of two stars: one blue, and the other redder. You can apply two-component blackbody: $F_{\lambda,total} = C_1 \times F_{\lambda,1}(T_1) + C_2 \times F_{\lambda,2}(T_2) ; T_1 \ne T_2$ where $F_{\lambda,total}$ is the observed specific fluxes from the point source, $F_{\lambda,1}, F_{\lambda,2}$ are blackbodies with temperature $T_1, T_2$ respectively, $C$s are normalization constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blackbody approximately (to the first order) works with stars including red giants as long as you include a portion of the SED where the line absorption/emission is negligible. For example, some stars have strong absorption at less than 4000 A due to metals, therefore that portion deviates significantly from the blackbody.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can sum fluxes, since fluxes are normalized for the luminosity distance. Since luminosity $L$ is $L = F \times 4 \pi D_L^2$ where $D_L$ is the luminosity distance, the total luminosity is $L_1 + L_2 = 4 \pi D_{L,1}^2 \times F_1 + 4 \pi D_{L,2}^2 \times F_2 = C_1 \times F_1 + C_2 \times F_2$. Therefore, you can sum the luminosity when $D_{L,1} \approx D_{L,2}$.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastEditorUserId="22468" LastEditDate="2018-05-17T17:33:23.960" LastActivityDate="2018-05-17T17:33:23.960" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26321" PostTypeId="1" AcceptedAnswerId="26338" CreationDate="2018-05-18T07:08:49.987" Score="3" ViewCount="133" Body="&lt;p&gt;What kind of stability it provides to each and every celestial body that each one them are round in shape? &lt;/p&gt;&#xA;" OwnerUserId="22623" LastActivityDate="2018-05-20T16:52:56.453" Title="Why planets and stars are always round/oval in size?" Tags="&lt;star&gt;&lt;planet&gt;&lt;celestial-mechanics&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" ClosedDate="2018-06-03T00:28:09.050" ContentLicense="CC BY-SA 4.0" />
  <row Id="26325" PostTypeId="2" ParentId="25524" CreationDate="2018-05-18T16:11:26.223" Score="3" Body="&lt;p&gt;When talking about curvature in (General) Relativity one usually refers to the metric interval and how this depends on the space-time coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the metric interval in Minkowski space can be written&#xA;$$ ds^2 = c^2\ dt^2 - dr^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an example of both flat space and flat space-time. It is flat space because if I want to find the distance between two events happening at the same time $t$, it is simply the integral of $dr$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly I can ask what the time interval is between two events and this is just given by the integral of $dt$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A counterexample is seen when using the Schwarzschild metric, which is appropriate in the space outside a spherically symmetric mass. Considering only the radial and time components, the metric is&#xA;$$ds^2 = c^2 (1 - r_s/r)\ dt^2 - (1- r_s/r)^{-1}\ dr^2,$$&#xA;where $r_s = 2GM/c^2$ is the Schwarzschild radius associated with the mass $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there is no simple relationship between $dr$ and the spatial separation between two events that occur at the same $t$, we must do an integral. Similarly, to work out the time interval between two events at the same $r$ we must do a more complicated integral. Both space and space-time are curved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The evolution of the universe as a whole is governed by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann%E2%80%93Lema%C3%AEtre%E2%80%93Robertson%E2%80%93Walker_metric&quot; rel=&quot;nofollow noreferrer&quot;&gt;Robertson-Walker metric&lt;/a&gt;.&#xA;$$ds^2 = c^2\ dt^2 - a(t)^2\frac{dr^2}{1 - kr^2},$$&#xA;where $k$ is a curvature parameter that may be -1, +1 or zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment, the best measurements are consistent with a zero curvature - that the universe is spatially flat. That does not mean it is flat on much large scales than we can observe - that is still possible. A curved universe that was exponentially &quot;inflated&quot; shortly after the big bang may well mimic a flat universe even though the curvature is finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note though that because the universe is inherently &quot;lumpy&quot;, this spatial flatness is only true on scales much larger than planets, stars, galaxies and even clusters of galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether you could call space-time flat in a spatially flat Robertson-Walker universe I doubt. The reason for this is the time-dependent scale factor $a(t)$, which ensures there is a coupling between the $r$ and $t$ coordinates. It is this coupling that is responsible, for example, for the cosmological redshift of light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A technical answer would need to look at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Einstein_tensor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein Tensor&lt;/a&gt; which characterises the curvature of space-time caused by mass and energy. In the Robertson-Walker metric, this tensor definitely has non-zero components, so you would have to say that even if space is flat, space-time is curved.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-05-18T16:11:26.223" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26326" PostTypeId="1" CreationDate="2018-05-18T17:06:32.130" Score="3" ViewCount="105" Body="&lt;p&gt;I'm coming across the concept of &quot;period doubling&quot; in  RR Lyrae stars.  When I look up what period doubling is on the Internet, I find general-level explanations of bifurcations of dynamical systems and transitioning to chaos. However, I am unable to apply these things to stars in a way I understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is period doubling, and what is physically happening in a star to lead to period doubling?&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2018-05-19T00:57:54.187" Title="What is period doubling in a variable star?" Tags="&lt;variable-star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26327" PostTypeId="2" ParentId="26326" CreationDate="2018-05-19T00:52:35.887" Score="3" Body="&lt;p&gt;I think part of the confusion comes from the fact that &quot;period doubling&quot; can mean two rather different things, which are probably related but this is not completely known.  The first is a purely observational meaning, which happens when you have a clear periodic pulsation in a star, but the amplitude shows a higher-lower-higher-lower pattern that implies a &quot;double&quot; period to see the full cycle.  Then there's the theoretical meaning, which is a step in the approach to chaos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way the approach to deterministic chaos often works is, you have a simple system that shows a simple period.  But then you increase the degree of nonlinearity in the system, and you see bifurcations into something that requires double the number of iterations to be cyclic.  If each iteration represents a cycle time, like the time to go around an orbit or the time for a pulsation, then that means double the period.  Maybe you started with an interative mapping that converges to a fixed point, but then you increased the nonlinearity and it turned into a limit cycle where the iterations go back and forth between two points.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it's not immediately obvious the connection between that and the amplitude modulation of a simple cycle, and I think many of the sources are purposefully ambiguous about that connection because nonlinear dynamics is still a mysterious field.  But the basic idea seems to be, if you have a linear system, it has uncoupled pulsation modes that are higher harmonics of the lowest frequency &quot;fundamental&quot; mode.  But if you have a nonlinear system, you can have coupling between those modes, which end up looking like amplitude modulation in a lower harmonic due to resonant coupling with a higher harmonic.  If the coupling has an odd-integer relationship, like a frequency of 2 coupling to a frequency of 3, then if the first oscillation in the frequency 2 mode couples resonantly with the frequency 3 harmonic, then the second oscillation will not couple well, it will hit the harmonic when it is 3/2 of the way through a cycle, so is out of phase.  Then the third cycle of the frequency 2 mode will hit the frequency 3 mode when it has gone through 3 full cycles and is back into resonance.  So that gives the &quot;period doubling&quot; where the amplitude of the frequency 2 harmonic gets a boost every second time around, owing to its ability to receive energy from the higher mode as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The connection between that and period doubling bifurcations is not entirely obvious either, but it seems that the idea is that if you have nonlinearity, the period doubling you find on the path to chaos might be related to this kind of resonant interaction between the simple behavior you see, and higher frequency behaviors that could exist but only make their presence known as a kind of residue from how they interact with the behavior you do see.  In other words, if the linearity is weak, some simple oscillation gets excited, but if you crank up the nonlinearity, that simple oscillation starts to overlap with more complicated behaviors that require more cycles to return to the original resonance relationship.  Crank up the nonlinearity even more, and all you see are the complicated behaviors, you have complete chaos.  The period doubling in variable stars represents only a small amount of nonlinearity, it's not chaotic and the relatively simple nature of the resonances allow for a large amplitude to build up and allow the stars to be classified as variable in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do more research into this, you might look up the &quot;Blazhko effect,&quot; or how period doubling also applies to other variable stars.  It sounds like an active area of research where there are still more questions than answers, but Kepler data has the promise of producing the necessary information to solve it.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2018-05-19T00:57:54.187" LastActivityDate="2018-05-19T00:57:54.187" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26328" PostTypeId="2" ParentId="20840" CreationDate="2018-05-19T03:16:42.730" Score="1" Body="&lt;p&gt;It's a bit subtle, the key thing the partial ionization does is to keep the temperature from changing much.  What you really want is an increase in density, not temperature.  The reason the kappa mechanism is important is that it allows heat to be added to the gas when it is compressed, and removed when the gas is expanded, that's what allows for the pulsation to have energy added to it.  Heat is added when the opacity is high, and removed when low, so you need the opacity to go up when the gas is compressed, and go down when the gas is expanded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, normally opacity goes down when gas is compressed, because the rise in temperature causes ionization, which actually hurts metal opacity by reducing the number of bound electrons (metal electrons have higher opacity in general when bound, due to resonances, than when free, though the hydrogen and helium we had in the Big Bang are an exception that works the other way-- in a star, it is usually metal opacity that is important and variable).  But increases in density also tend to produce more bound metal electrons, so there's a competition there that temperature usually wins and stars are pulsationally stable.  But in partial ionization zones, temperature changes are reduced by the phase changes eating up the excess energy, and that gives the density increase the edge that allows opacity to rise when the gas is compressed.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-05-19T03:16:42.730" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26332" PostTypeId="1" AcceptedAnswerId="26333" CreationDate="2018-05-19T12:29:27.460" Score="2" ViewCount="116" Body="&lt;p&gt;Why is there so much excitement over &lt;a href=&quot;https://www.nature.com/articles/d41586-018-04969-6&quot; rel=&quot;nofollow noreferrer&quot;&gt;helium having been discovered on an exoplanet&lt;/a&gt;? Isn't it oxygen that we should be looking for, since this is what is needed for life? What do we need helium for?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastEditorUserId="14094" LastEditDate="2018-05-24T07:44:40.867" LastActivityDate="2018-05-24T07:44:40.867" Title="Helium discovered on an exo planet" Tags="&lt;exoplanet&gt;&lt;planetary-atmosphere&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26333" PostTypeId="2" ParentId="26332" CreationDate="2018-05-19T12:51:52.437" Score="4" Body="&lt;p&gt;You are right that helium around a gas giant is no kind of surprise, and oxygen on an Earthlike planet would be spectacularly more important.  However, the excitement is simply that if you can detect one element, it means you might be able to detect another.  It just gives you more ways to learn about the planet, and offers promise for more new ways to learn.  One might find analogy with the first detection of exoplanets, which were &quot;hot Jupiters.&quot;  No one was surprised stars can have planets, and no one thinks hot Jupiters are conducive to life, but if you can discover one kind of planet, it bodes well for the later discovery of others.  Anything we can learn about composition is an important step in the right direction.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-05-19T12:51:52.437" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26334" PostTypeId="1" AcceptedAnswerId="26369" CreationDate="2018-05-19T18:10:01.673" Score="1" ViewCount="479" Body="&lt;p&gt;Can somebody tell me why the constellations I've marked here have areas that resemble pieces of a puzzle, with very varying width and height and shapes, instead of being logically/evenly laid out on the sky?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Virgo is much bigger than Cancer, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image is a screenshot from Stellarium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6zgo2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6zgo2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23110" LastEditorUserId="616" LastEditDate="2018-06-01T18:43:15.493" LastActivityDate="2018-06-01T18:43:15.493" Title="Why oddly shaped constellation areas?" Tags="&lt;constellations&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26335" PostTypeId="2" ParentId="26334" CreationDate="2018-05-19T23:11:52.717" Score="3" Body="&lt;p&gt;The constellations were devised in Greece. They grouped the stars in ways that told stories. They put a structure onto the sky. There was never any attempt to divide the sky into equal sized parts. Some of the constellations were just bigger than others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stars are not logically or evenly laid out, and the stories that are told about the stars did not require even sized constellations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From ancient times to now the constellations have proved a convenient, if somewhat quirky, way of arranging the sky. Astronomers divided the sky into regions, each named after the constellation it contained although with some variation, for example splitting the claws of the scorpion off and creating &quot;Libra&quot;. Astrologers split the year into 12 months based on the general position of the sun. The astrological periods don't match up very well with the actual borders of the constellations. For example, the sun is within the modern borders of Libra between Oct 31 and Nov 22. Whereas astrologers will say that the sun birth sign is from September 23 to October 22. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-05-19T23:11:52.717" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="26336" PostTypeId="1" CreationDate="2018-05-20T14:12:06.400" Score="1" ViewCount="74" Body="&lt;p&gt;Is Earth the perfect place for life ? If it isn't then how could conditions be better here ? For example, the size of the Planet, our atmosphere, distanace from our Sun etc.&lt;/p&gt;&#xA;" OwnerUserId="2562" LastActivityDate="2018-05-20T15:49:36.683" Title="Ideal conditions for life" Tags="&lt;life&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2018-05-22T11:13:36.117" ContentLicense="CC BY-SA 4.0" />
  <row Id="26337" PostTypeId="2" ParentId="26336" CreationDate="2018-05-20T15:49:36.683" Score="3" Body="&lt;p&gt;It is a very good place for Earth Life, but that is because we have evolved and adapted to this planet. For example, we like to have an atmospheric pressure of about 1000 mBar, and a partial O2 pressure of about 200 mBar, which is exactly what the atmosphere provides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth would be a very bad place for a species to which oxygen was toxic. A species that hadn't evolved here would probably find conditions challenging in one way or another.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-05-20T15:49:36.683" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26338" PostTypeId="2" ParentId="26321" CreationDate="2018-05-20T16:29:28.197" Score="1" Body="&lt;p&gt;The gravitational is radial only. That means if you tried a different shape then you would have regions, like hills, where a change in the angle would experience a much less attraction towards the center. This region would, over time, be attracted to a different angle with a greater force of attraction. Or spherical. This results in a spherical-like shape.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastActivityDate="2018-05-20T16:29:28.197" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26340" PostTypeId="2" ParentId="13356" CreationDate="2018-05-20T17:16:46.547" Score="-1" Body="&lt;p&gt;Since any form of contact, would be next to impossible, because of the distance involved, I think the only importance of such a discovery would be for our curiosity .&lt;/p&gt;&#xA;" OwnerUserId="2562" LastActivityDate="2018-05-20T17:16:46.547" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26341" PostTypeId="2" ParentId="24670" CreationDate="2018-05-20T17:18:10.593" Score="0" Body="&lt;p&gt;Globular clusters vary in size quite a bit. They contain from a few hundred thousand to over a million stars. Probably best way to determine size is to measure its extent across the sky by telescope and then calculate that distance in km. One could also model the size by assuming the number of stars, the average size of the stars, and the spacing based on this size.  One could build a globular cluster of 300,000 stars or so and determine the approximate size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You ask if 120-150 ly is the average diameter of such globular clusters. I'm confused by exactly what you mean but you have sampled a small range of sizes considering globular clusters can be well over 300 ly in diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Wikipedia, and based on comment from @AtmosphericPrisonEscape, you have &quot;Almost all globular clusters have a half-light radius of less than 10 pc, although there are well-established globular clusters with very large radii (i.e. NGC 2419 (Rh = 18 pc) and Palomar 14 (Rh = 25 pc)).[64]&quot;&lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="16926" LastEditDate="2018-06-19T17:55:54.227" LastActivityDate="2018-06-19T17:55:54.227" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26342" PostTypeId="2" ParentId="13356" CreationDate="2018-05-20T18:13:14.323" Score="1" Body="&lt;p&gt;Another point to bear in mind is that very few topics in astronomy are &quot;important&quot; in any kind of absolute or demonstrable way, astronomy is more or less the study of all the things that &lt;em&gt;don't&lt;/em&gt; directly affect us (with a few exceptions that no longer motivate or justify astronomy).  Instead, one of the most important reasons to do astronomy is to create a sense of place, a sense of where we stand in the grand scheme-- it is the way astronomy gives us a context we can't get from anything else.   So I think discovery of life elsewhere would have a transformative effect in that respect, because it would change how we understand our own story.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By that I mean, right now we really don't even know if we are looking for a story of life on Earth that sounds like a spectacularly unlikely sequence of events that could only occur on one planet in a trillion, or if we are looking for a common, even inevitable, story that happens almost everywhere.  And there is no way we could ever know which story we are looking for by looking only at life on Earth.  No study of evolution, no study of the history of our weather, no understanding of the reasons for mass extinctions or even if mass extinctions require any reason beyond chance fluctuations, could ever really tell us if the story we are telling about our own appearance on this planet is a common or inevitable kind of story, or a spectacularly unlikely story that could only play out at all in a truly vast universe.  So that's one of the most important elements of discovering life elsewhere, to understand the nature of life itself, including our own.  It's the sense of &lt;em&gt;context&lt;/em&gt; that astronomy provides.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-05-20T18:13:14.323" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26343" PostTypeId="2" ParentId="13356" CreationDate="2018-05-20T18:31:08.893" Score="0" Body="&lt;p&gt;To be frank, mate, your viewpoint is quite odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find life - intelligent life - on another planet would be THE most amazing, world-changing, breathtaking, scary, exciting discovery in the history of mankind (and that's an obscene understatement).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just imagine it, man. We'd know we weren't alone. We'd then establish a communication with the beings, eventually, and start learning about their civilisation etc. The implications of it are endless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if, by life on other planets you mean microbes, bacteria etc, then yes, I agree that's boring and isn't something that would change anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I answered this with the belief it was a new question. I've only just realised how old it is.&lt;/p&gt;&#xA;" OwnerUserId="22083" LastEditorUserId="22083" LastEditDate="2018-05-20T18:40:01.167" LastActivityDate="2018-05-20T18:40:01.167" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26344" PostTypeId="2" ParentId="26318" CreationDate="2018-05-20T19:57:04.737" Score="3" Body="&lt;p&gt;To reverse the question: if you were on the Moon, what sources of light would you see?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The Sun would be the major source of light.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Although the Moon's albedo is only about 12%, reflected light from the ground would be the second major source.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Earth would be in a poor 3rd place. You could see the general shape of the continents, but not storms or any other light-emitting meteorological or biological activity (it's there, but much too faint to be seen from the Moon)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The other planets would be poor 4th place.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Starlight would be in a poor 5th place.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It would be similar to standing on Earth, except:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The Moon has no atmosphere, so:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the Sun would be brighter&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;there'd be no light diffusion, so you could see the Earth/planets/stars in the daytime&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Meteors would not leave a fiery trail -- they would simply hit the moon silently. Asteroid impacts on the Moon would also produce no sound.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There wouldn't be a moon in the sky, but there would be an Earth instead.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You add a twist to the question by asking about past and future light:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In about 7.6 billion years, the Sun will expand to (probably) engulf the Earth and Moon, which will probably be the most light the Moon will see before it's reduced to nothingness.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Moon possibly saw a lot of light when it was created through planetary collision. Not sure this counts, because the Moon didn't really exist until sometime after this.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Any other major source of light (such as supernovas) would be visible from Earth as well. That doesn't mean there haven't been any super-bright supernovas, but if they were bright enough and lasted long enough to generate more light than the Sun has during its lifetime, they would probably have a major impact on Earth (who knows, maybe that's why the dinosaurs died ), and there's really no evidence anything that liked happened.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user21" LastEditorUserId="23100" LastEditDate="2018-05-21T15:22:55.227" LastActivityDate="2018-05-21T15:22:55.227" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26345" PostTypeId="2" ParentId="26315" CreationDate="2018-05-20T21:06:45.003" Score="0" Body="&lt;p&gt;Reda and Andreas 2008 equation 28 comes from Jean Meeus, &lt;a href=&quot;http://www.willbell.com/math/mc1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Astronomical Algorithms&lt;/em&gt;&lt;/a&gt;, equation 12.4.&#xA;He derived it from &lt;a href=&quot;http://adsabs.harvard.edu/abs/1982A%26A...105..359A&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aoki et al. 1982&lt;/a&gt; equation 13, which IAU used from 1984 through 2002.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's break it down:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The constant term is GMST in degrees at 2000-01-01 12:00 UT, also known as JD 2451545.0 or J2000.0.&lt;/li&gt;&#xA;&lt;li&gt;The linear term is Earth's mean sidereal rotation rate in degrees per solar day, multiplied by the number of days since JD 2451545.&lt;/li&gt;&#xA;&lt;li&gt;The JC^2 and JC^3 terms approximate the long-term effects of Earth's precession and change in rotation rate. Between 1900 and 2100, they amount to less than 0.1 second of time and may be omitted unless you require high precision.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The set of reference frames which IAU adopted in 2000 and implemented in 2003 was motivated by the extreme precision requirements of very long baseline radio interferometry (VLBI).&#xA;The difference between the old and new expressions for GMST is zero on 2003-01-01 and negligible within a century of that date unless you are doing VLBI yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The results you previously got for 2003-10-17 could be due to inverting an addition or subtraction somewhere, maybe in normalizing to [0, 360].&#xA;If you try again, it may come out right as it did for 1987-04-10.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-05-20T22:47:10.470" LastActivityDate="2018-05-20T22:47:10.470" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26347" PostTypeId="1" AcceptedAnswerId="26361" CreationDate="2018-05-21T09:38:00.487" Score="3" ViewCount="205" Body="&lt;p&gt;I'm trying to make a simple app for my own use to be able to check which of my favorit restaurants have outdoor seating sun and for how long etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the data in the title (sunrise, noon, sunset, longitude and latitude) I can draw a path where the sun should be at any given time, but I want to display the hight as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I live in sweden where the sun changes &quot;hight angle&quot; (sorry I don't know the exact term for it) which make the angle relevant when deciding on seatings amongst high buildings.&#xA;My end goal is to be able to use google maps 3D data to actually cast shadows on the map in a realistic way!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no Idea if I'm in the right place but I figure I'll give it a shot :)&lt;/p&gt;&#xA;" OwnerUserId="23119" LastActivityDate="2018-05-22T15:06:35.070" Title="Given sunrise, noon, sunset, longitude and latitude can I calculate the &quot;hight&quot; or &quot;ascension&quot; of the sun?" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;orbital-elements&gt;&lt;positional-astronomy&gt;&lt;mathematics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26348" PostTypeId="1" CreationDate="2018-05-21T09:39:11.807" Score="2" ViewCount="53" Body="&lt;p&gt;I am very much interested in becoming an astronomer as a career. I know that my main focus should be in mathematics and science but; are there any specific degrees or competitive examinations I must obtain or attend, respectively? I am writing this question because I didn't obtain the desired answer.Please see my comment below for further information.&lt;/p&gt;&#xA;" OwnerUserId="23120" LastEditorUserId="23120" LastEditDate="2018-06-03T17:25:20.997" LastActivityDate="2018-06-03T17:25:20.997" Title="How to become an astonomer?" Tags="&lt;amateur-observing&gt;&lt;observational-astronomy&gt;" AnswerCount="0" CommentCount="6" FavoriteCount="1" ClosedDate="2018-05-21T18:41:35.837" ContentLicense="CC BY-SA 4.0" />
  <row Id="26349" PostTypeId="1" AcceptedAnswerId="36967" CreationDate="2018-05-21T10:33:09.897" Score="8" ViewCount="350" Body="&lt;p&gt;I'm looking at the Mars topography map: &#xA;&lt;a href=&quot;https://i.stack.imgur.com/LP81v.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LP81v.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Mars_topography_(MOLA_dataset)_with_poles_HiRes.jpg&quot; rel=&quot;noreferrer&quot;&gt;original link&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...and I can't shake the impression that someone just messed up the placement of the center point of the planet, shifting it about 8km north from where it should be. If the center point was ~8km further south, the north and south altitudes would be roughly the same. Moreover, it's even hard to argue about placement of equator, as the northern pole is about 16km closer to the equatorial plane than the southern - due to the altitude difference. Move the equator and it all evens out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's fairly hard to believe such a mistake to have persisted so long, so... is there any deeper reason behind this strong asymmetry of the planet, other than just a topographical snafu? &lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2021-05-06T08:12:21.990" Title="Why are Mars northern/southern hemisphere altitudes so different?" Tags="&lt;mars&gt;&lt;topographic-map&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26351" PostTypeId="2" ParentId="26349" CreationDate="2018-05-21T11:36:56.870" Score="10" Body="&lt;p&gt;One part of your question could be paraphrased &quot;why do we measure altitude on Mars the way we do&quot;. As you observe we could redefine the ellipsoid of zero altitude to much more closely fit the planet by moving it South a bit. The &lt;a href=&quot;https://www.quora.com/How-do-we-calculate-altitude-on-Mars-since-theres-no-large-body-of-water-to-use-as-reference&quot; rel=&quot;noreferrer&quot;&gt;answer&lt;/a&gt; is based on the gravity. 0 altitude is on a surface of constant gravitational potential energy, which was originally determined in terms of atmospheric pressure, but more recently by direct gravity measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other part might ask why this is in terms of the history of Mars. I don't think there is a definitive accepted answer, but I'll leave that for someone else.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-05-21T11:36:56.870" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26352" PostTypeId="1" AcceptedAnswerId="26353" CreationDate="2018-05-21T14:42:47.443" Score="2" ViewCount="44" Body="&lt;p&gt;I'm reading about Antikythera machine and its functions.&#xA;So I read about solar cycle and moon cycle and how they used 19 solar years to compare with 235 lunar cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is my question: They know about leap day at the time (~ 200 BC)?&lt;/p&gt;&#xA;" OwnerUserId="21949" LastEditorUserId="21949" LastEditDate="2018-05-21T17:40:36.033" LastActivityDate="2018-05-21T17:40:36.033" Title="Ancient greeks know about leap day?" Tags="&lt;the-sun&gt;&lt;the-moon&gt;&lt;moon-phases&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2018-05-21T17:50:55.943" ContentLicense="CC BY-SA 4.0" />
  <row Id="26353" PostTypeId="2" ParentId="26352" CreationDate="2018-05-21T14:52:04.017" Score="2" Body="&lt;p&gt;Well, they know according to the Metonic Calendar created in ~430 B.C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To know more about &lt;a href=&quot;https://en.wikipedia.org/wiki/Metonic_cycle&quot; rel=&quot;nofollow noreferrer&quot;&gt;click here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="21949" LastActivityDate="2018-05-21T14:52:04.017" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26355" PostTypeId="1" CreationDate="2018-05-21T19:32:16.463" Score="1" ViewCount="120" Body="&lt;p&gt;I am wanting to plot a velocity profile (a plot of velocity against the star's distance from the galactic core) from the Gaia DR2, but so far my efforts have been disappointing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The information I have for each star are the declination ($\delta$), right ascension ($\alpha$), radial velocity ($v_r$), parallax ($\theta_\text{parallax}$) and proper motion ($\mu$) (ra and dec).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to calculate the star's distance from the galactic core using the cosine rule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ d = \frac{1000}{\theta_{\text{parallax}}}$$&#xA;$$\theta = \arccos{[\sin{\delta_\text{star}}\sin{\delta_\text{core}} + \cos{\delta_\text{star}}\cos{\delta_\text{core}}\cos{(\alpha_\text{star}-\alpha_\text{core})}]}$$&#xA;$$ r = \sqrt{d_\text{core}^2+ d^2 - 2d \times d_\text{core} \times \cos{\theta}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $d$ is the distance from the Sun to the star. $d_\text{core}$ is the distance from the Sun to the galactic core. $\theta$ is the angular separation of the star and the galactic core. $r$ is the distance from the star to the core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the star's speed using the equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu = \sqrt{\mu_\text{dec}^2 + \mu_\text{ra}^2}$$&#xA;$$ v_t = 4.7 \mu d $$&#xA;$$ v = \sqrt{v_r^2 + v_t^2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the plot doesn't show any kind of trend (just a kind of blob). Have I calculated correctly and, if not, how could I calculate the tangential velocities and radius from the galactic core?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, I want to make the kind of plot that helps us identify dark matter in galaxies.&lt;/p&gt;&#xA;" OwnerUserId="1494" LastEditorUserId="1494" LastEditDate="2018-05-22T14:22:15.257" LastActivityDate="2018-05-22T14:22:15.257" Title="How to calculate the various properties of a star?" Tags="&lt;data-analysis&gt;&lt;astrometry&gt;&lt;gaia&gt;" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26357" PostTypeId="2" ParentId="10490" CreationDate="2018-05-21T20:39:36.047" Score="6" Body="&lt;p&gt;The modern 7-day week was adopted in the late Roman Empire, replacing an 8-day &quot;nundinal week&quot; that had previously been in use mainly for market intervals. As far as I know, there was no particular reason for the 8-day length other than that it was a convenient size to separate market days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the early Empire, Christians and Jews both used a 7-day week for religious purposes and as they grew to be a major part of the Empire's population, it became convenient to link the market week with the religious week. This worked because 7 worked about as well as 8 for markets, and so by the late Empire, it was a done deal.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://en.wikipedia.org/wiki/Nundinae&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Nundinae&lt;/a&gt; for a good discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where did the Jewish 7-day cycle come from?  It's not certain.  A practical market cycle can be anywhere in the 5-10 day range -- it doesn't matter a lot -- so if early Jews preferred a 7-day religious week (from Genesis, Chapter 1, probably) then the market cycle would have synched up just like in the Roman Empire.  It's also possible they inherited it from some previous mid-Eastern culture.  I doubt anyone knows for sure any more, so it must remain to some degree speculation.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2018-05-21T22:20:30.443" LastActivityDate="2018-05-21T22:20:30.443" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26359" PostTypeId="1" CreationDate="2018-05-22T01:32:27.227" Score="4" ViewCount="65" Body="&lt;p&gt;My rudimentary experience of photography and telescopes suggests that with commonly affordable optics any object beyond the range of hundreds to thousands of feet will be in adequate focus at the proper &quot;infinity&quot; position. For example the advice here for overcoming difficulties in seeing the focus of a star on a DLSR screen to instead focus on a brighter and distant city light: &lt;a href=&quot;https://photo.stackexchange.com/a/23973&quot;&gt;https://photo.stackexchange.com/a/23973&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However an interlocutor recently offered this advice (collated from a series of comments):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Apparently you haven't actually used a camera or a telescope. The focus adjustment is there for a reason. Any astronomer will tell you focus changes are needed between stars, stars farther away,  planets, and the Moon. ... If focused on the Moon, you need to refocus slightly when looking at stars. When looking at Jupiter you need to refocus if the target is the Moon. Close stars vs far away stars, the same thing, a need to refocus. ...  To the tune of: Jupiter and it's moons, which are close to Jupiter , need focus adjustment between them. Either Jupiter or it's moons will be in focus. Ask any astronomer. That's a difference and it's well past the supposed &quot;infinity focus&quot; point.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't be wholly surprised if retargeting a telescope could require a focus tweak to correct for slight changes of the mirror distances (etc), or from thermal shrinkage across a cooling evening, but such effects would probably be invariant across objects, whereas as described in that quote targetting back on the original object would presumably require the focus to be restored to the original setting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a distance-dependent component for focus in astronomical observation, and if so what sort of equipment does it affect in practice?&lt;/p&gt;&#xA;" OwnerUserId="23130" LastActivityDate="2018-05-22T01:46:34.347" Title="Does the focus of terrestial telecopes need adjustment for the distance of astronomical objects?" Tags="&lt;photography&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26360" PostTypeId="2" ParentId="26359" CreationDate="2018-05-22T01:46:34.347" Score="4" Body="&lt;p&gt;This is incorrect in any operational sense, and no astronomer will tell you that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Any astronomer will tell you focus changes are needed between stars, stars farther away, planets, and the Moon. ... If focused on the Moon, you need to refocus slightly when looking at stars. When looking at Jupiter you need to refocus if the target is the Moon. Close stars vs far away stars, the same thing, a need to refocus. ... To the tune of: Jupiter and it's moons, which are close to Jupiter , need focus adjustment between them. Either Jupiter or it's moons will be in focus. Ask any astronomer. That's a difference and it's well past the supposed &quot;infinity focus&quot; point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's true that Jupiter and the Moon are at different distances and thus geometrical optics says that the focus needs to be adjusted, but this is &lt;em&gt;only&lt;/em&gt; true in mathematically in an idealize instrument. In any real instrument, it is not possible to discern any difference between the various focus points.  They are all at infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one size really does fit all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With exceptional quality optics, it's possible that a distant streetlight might require a very, very slightly different focus than the Moon or Jupiter. Nearby streetlights, sure.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-05-22T01:46:34.347" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26361" PostTypeId="2" ParentId="26347" CreationDate="2018-05-22T03:18:21.497" Score="3" Body="&lt;p&gt;To the precision that you need, you can calculate the azimuth A and altitude h of the Sun from the following equations:&#xA;$$&#xA;\tan(A) = \frac{\sin(H)}{\cos(H)\sin(\phi)-\tan(\delta)\cos(\phi)}\\&#xA;\sin(h)=\sin(\phi)\sin(\delta)+\cos(\phi)\cos(\delta)\cos(H)&#xA;$$&#xA;where&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A is the azimuth measured from due south (positive towards the west)&lt;/li&gt;&#xA;&lt;li&gt;H is the hour angle which is the number of degrees before (negative angle) or after (positive) solar noon. Calculate this as 15*1.00274*(time of day - time of noon).&lt;/li&gt;&#xA;&lt;li&gt;$\phi$ is the latitude (positive for northern hemisphere)&lt;/li&gt;&#xA;&lt;li&gt;$\delta$ is the declination of the Sun. This can be calculated from the second equation by assuming the altitude of the Sun h=0 when it is rising or setting, and H=15*1.00274*time from sunrise to solar noon. The assumption here is that the time from solar noon to sunset is the same.&lt;/li&gt;&#xA;&lt;li&gt;h is the altitude of the Sun above the horizon.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All of the above quantities are in degrees. You may need to convert those to radians to perform the trigonometry calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more accurate calculation, you would calculate the position of the Sun (its right ascension $\alpha$ and declination $\delta$) and Greenwich Mean Sidereal Time (GMST) based on the date and time throughout the day. Then using the latitude and longitude, you would calculate H, A, and h for various times of the day.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastEditorUserId="17355" LastEditDate="2018-05-22T15:06:35.070" LastActivityDate="2018-05-22T15:06:35.070" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26363" PostTypeId="1" AcceptedAnswerId="26365" CreationDate="2018-05-22T08:42:42.137" Score="1" ViewCount="60" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4gDNG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4gDNG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find this image in the following linked index.&lt;br&gt;&#xA;&lt;a href=&quot;https://apod.nasa.gov/apod/archivepix.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://apod.nasa.gov/apod/archivepix.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23136" LastActivityDate="2018-05-22T14:15:32.403" Title="Does anybody know date of publishing of following spirit rover image in NASA apod?" Tags="&lt;nasa&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26364" PostTypeId="1" AcceptedAnswerId="27035" CreationDate="2018-05-22T11:14:47.463" Score="2" ViewCount="56" Body="&lt;p&gt;I need some help obtaining IR background diffuse data from the WISE observation. There is already some work done in this regard; see: &lt;a href=&quot;http://wise.skymaps.info/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://wise.skymaps.info/&lt;/a&gt; . These are .fits files. How do I find out the photon irradiance for a range (say glat b/w 0 and 30) from these files? Any help is appreciated. Also, how do I correlate this with UV flux for the same range using python?&lt;/p&gt;&#xA;" OwnerUserId="23137" LastActivityDate="2018-10-19T08:00:31.883" Title="Obtain IR diffuse data from WISE" Tags="&lt;fundamental-astronomy&gt;&lt;data-analysis&gt;&lt;python&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26365" PostTypeId="2" ParentId="26363" CreationDate="2018-05-22T14:15:32.403" Score="4" Body="&lt;p&gt;It was posted on &lt;a href=&quot;https://apod.nasa.gov/apod/ap021126.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;November 26, 2002&lt;/a&gt;, with the caption &quot;Name This Martian Robot.&quot;&#xA;There was an &lt;a href=&quot;https://mars.nasa.gov/mer/newsroom/pressreleases/20021104.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;essay contest&lt;/a&gt; to name the two Mars Exploration Rovers to be launched in 2003.&#xA;The winner was &lt;a href=&quot;https://mars.nasa.gov/mer/newsroom/pressreleases/20030608a.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sofi Collis&lt;/a&gt;, a third grader in Arizona.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-05-22T14:15:32.403" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26366" PostTypeId="1" CreationDate="2018-05-22T15:05:47.913" Score="3" ViewCount="421" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zv4fS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why the moon looks like the cloud-structure on the blue sky among the clouds?&lt;/p&gt;&#xA;" OwnerUserId="23141" LastEditorUserId="1891" LastEditDate="2018-05-22T20:26:34.203" LastActivityDate="2018-05-23T09:55:05.613" Title="Why is the color of the moon so similar to the clouds?" Tags="&lt;the-moon&gt;&lt;naked-eye&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26367" PostTypeId="2" ParentId="5869" CreationDate="2018-05-22T16:00:16.803" Score="6" Body="&lt;p&gt;Updated 2018: @Stan Liou's answer is excellent, but since he wrote that answer, we have accurately measured the speed of gravitational waves and confirmed to a very high degree of accuracy that it is the speed of light in vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In August 2017, LIGO observed GW170817, a neutron star inspiral. X-ray observatories in orbit detected x-rays from the collision with at most a 2-second difference in travel time, so speed-of-light == speed-of-gravitation is confirmed.  &lt;a href=&quot;https://www.nextbigfuture.com/2017/11/speed-of-gravitational-waves-is-very-near-the-speed-of-light.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's a decent summary of the results.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2018-05-26T11:16:57.267" LastActivityDate="2018-05-26T11:16:57.267" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26368" PostTypeId="2" ParentId="26366" CreationDate="2018-05-22T19:07:50.903" Score="1" Body="&lt;p&gt;The moon has dark and bright areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it is visible in the daytime the bright areas appear roughly as bright as clouds, but the dark areas have no light to add to the scene and we are just left with the blue light from the sky generally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sky blue is not as bright as the clouds or the moon, at least when it's visible like this (typically near the end of day and early morning) so it does not conceal the bright clouds or the moon's brighter parts.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2018-05-22T19:07:50.903" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26369" PostTypeId="2" ParentId="26334" CreationDate="2018-05-22T19:59:56.113" Score="8" Body="&lt;p&gt;The constellations are artifacts of visual pattern recognition, many dating back to antiquity.&#xA;As the stars are distributed unevenly, naturally some constellations are larger than others.&#xA;Ptolemy catalogued 48 of them around 150 CE.&#xA;The other 40 were added between 1596 and 1763, either to cover the far southern sky or to fill gaps between the older figures.&#xA;Ian Ridpath gives a comprehensive history in &lt;a href=&quot;http://www.ianridpath.com/startales/contents.htm&quot; rel=&quot;noreferrer&quot;&gt;Star Tales&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The modern constellation boundaries were &lt;a href=&quot;http://www.ianridpath.com/boundaries.htm&quot; rel=&quot;noreferrer&quot;&gt;standardized in the 1920s&lt;/a&gt; by an IAU committee led by Eugène Delporte.&#xA;Their goals were to assign each square degree of sky unambiguously to one constellation, to be consistent with scientific literature already published, and to preserve the traditional figures where possible.&#xA;Equal size wasn't on the agenda.&#xA;The boundaries align with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Equatorial_coordinate_system&quot; rel=&quot;noreferrer&quot;&gt;equatorial grid&lt;/a&gt; as of 1875:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mBfnG.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mBfnG.png&quot; alt=&quot;IAU constellation boundaries&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The zodiac of 12 equal parts was invented by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Zodiac#Early_history&quot; rel=&quot;noreferrer&quot;&gt;Babylonians around 400 BCE&lt;/a&gt; and later refined by Ptolemy.&#xA;Each sign corresponds to a box 30 degrees wide and about 20 degrees high.&#xA;It's important to distinguish between a constellation (a visual group of bright stars) and a sign (a uniform box along the ecliptic).&#xA;For example, Virgo the &lt;em&gt;sign&lt;/em&gt; spans &lt;a href=&quot;https://en.wikipedia.org/wiki/Ecliptic_coordinate_system&quot; rel=&quot;noreferrer&quot;&gt;ecliptic longitudes&lt;/a&gt; 150$^\circ$ to 180$^\circ$, whereas Virgo the &lt;em&gt;constellation&lt;/em&gt; is somewhat wider and farther east.&#xA;In Ptolemy's time the signs and constellations were not far apart:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0kXwo.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0kXwo.png&quot; alt=&quot;Zodiac in the time of Ptolemy&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ptolemy defined the &lt;a href=&quot;https://en.wikipedia.org/wiki/Sidereal_and_tropical_astrology&quot; rel=&quot;noreferrer&quot;&gt;tropical zodiac&lt;/a&gt; in reference to the equinoxes and solstices; for example, the boundary between the signs of Gemini and Cancer lies at the June solstice, wherever that may be among the stars.&#xA;Consequently, &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_precession&quot; rel=&quot;noreferrer&quot;&gt;precession&lt;/a&gt; makes the signs drift westward about 1.4$^\circ$ per century relative to the constellations.&#xA;In the present day, the zodiac is roughly one sign ahead of its constellations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dQp9i.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dQp9i.png&quot; alt=&quot;Zodiac in the present day&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-05-28T13:45:11.247" LastActivityDate="2018-05-28T13:45:11.247" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26371" PostTypeId="2" ParentId="26366" CreationDate="2018-05-22T20:51:20.700" Score="2" Body="&lt;p&gt;The Moon is a dark gray in color -- it reflects light poorly but uniformly. (Its color is about the same  as fresh asphalt!  It's albedo is about 12%)  Clouds are (usually) light gray with an albedo ranging from 10% to 90%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both the Moon and the clouds shine by reflected sunlight, so they're both the same color. (Clouds usually have a greater surface brightness owing to their lower albedo.)&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-05-22T20:51:20.700" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26373" PostTypeId="1" AcceptedAnswerId="26379" CreationDate="2018-05-23T01:39:58.597" Score="4" ViewCount="1679" Body="&lt;p&gt;The angular resolution of a telescope is given by $\frac{1.22\lambda}{D}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should mean that the smaller the aperture D, the higher the angular resolution. But it is exactly the opposite - why?&lt;/p&gt;&#xA;" OwnerUserId="23146" LastEditorUserId="907" LastEditDate="2018-05-23T14:37:06.463" LastActivityDate="2018-05-23T19:30:34.180" Title="Relation between angular resolution and aperture?" Tags="&lt;telescope&gt;&lt;fundamental-astronomy&gt;&lt;telescope-lens&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26374" PostTypeId="2" ParentId="26373" CreationDate="2018-05-23T02:34:23.617" Score="6" Body="&lt;p&gt;The angular resolution only tells you your ability to distinguish (&quot;resolve&quot;) two separate points of light. For example, your ability to distinguish the two distinct headlamps on a car from being a single light in the distance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angular &lt;em&gt;magnification&lt;/em&gt; has nothing to do with the angular resolution: it is determined by the focal lengths of your primary and objective, and the eyepiece if applicable. &lt;/p&gt;&#xA;" OwnerUserId="20489" LastActivityDate="2018-05-23T02:34:23.617" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26375" PostTypeId="2" ParentId="26373" CreationDate="2018-05-23T06:12:06.777" Score="6" Body="&lt;p&gt;Your formula is correct. In the absence of instrumental effects or blurring of the atmosphere, it gives (approximately) the minimum &lt;em&gt;angular separation&lt;/em&gt; between two objects that can be resolved. The larger this is, the &lt;em&gt;worse&lt;/em&gt; your telescope is. You want the resolution to be &lt;em&gt;small&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-05-23T19:30:34.180" LastActivityDate="2018-05-23T19:30:34.180" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26376" PostTypeId="1" CreationDate="2018-05-23T06:37:57.487" Score="1" ViewCount="125" Body="&lt;p&gt;I want to create a semi realistic star system generator but I am not sure where to &quot;put in&quot; gas giants. My guess is they can form pretty much anywhere within the sphere of influence of a star at the right circumstances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am still debating how to exactly go about generating the system but it won't be super complicated. I probably have plop in a star with a random mass and based on modifiers like distance, reflection, greenhouse and random chance plop in certain types of planets. It does not need to be realistic but it needs to feel realistic for none astronomy experts.&lt;/p&gt;&#xA;" OwnerUserId="23152" LastEditorUserId="23152" LastEditDate="2018-05-23T08:55:17.320" LastActivityDate="2018-05-23T15:05:38.367" Title="Where do gas giants end up?" Tags="&lt;planet&gt;&lt;gas-giants&gt;&lt;star-systems&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26377" PostTypeId="2" ParentId="26366" CreationDate="2018-05-23T09:55:05.613" Score="-1" Body="&lt;p&gt;Sry Guys ;) english is not my main language so maybe i mesed in question :) still  i think i found the answer , it serms  everything on a blue sky would look cloudish ? My main concern was why moon looks more cloudish than clouds moonish ;) but now i think its just the contecst ;)&#xA;Thx&lt;/p&gt;&#xA;" OwnerUserId="23141" LastActivityDate="2018-05-23T09:55:05.613" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26379" PostTypeId="2" ParentId="26373" CreationDate="2018-05-23T12:31:51.303" Score="2" Body="&lt;p&gt;Part of the confusion stems from the words used.  &quot;higher&quot; resolution doesn't mean better, it means only bigger objects can be resolved; but we want to be able to resolve small objects, such as the angular distance between a pair of double stars.  So small is better in this situation.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2018-05-23T12:31:51.303" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26381" PostTypeId="2" ParentId="26376" CreationDate="2018-05-23T15:05:38.367" Score="2" Body="&lt;p&gt;What you're asking about is not known at the moment, but I can briefly comment on how flawed our knowledge is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) &lt;strong&gt;Theoretical predictions&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As one comment mentioned, the question &quot;where do planets end up?&quot; is dependent on the answer to &quot;where do planets form?&quot; and the answer to &quot;how fast do planets migrate?&quot; which is not well understood at the moment.&lt;br&gt;&#xA;There are of course predictions for both those processes, but those essentially don't agree on anything and especially not with the observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;https://arxiv.org/abs/1610.07202&quot; rel=&quot;nofollow noreferrer&quot;&gt;recent overview&lt;/a&gt; and &lt;a href=&quot;https://www.youtube.com/watch?v=XOVkkD71gWQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;a very short video&lt;/a&gt; by the authors of that paper summarizing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) &lt;strong&gt;Observations&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have only very few multi-planet systems, and not a single one where we can claim we know all planets down to a certain size or mass limit.&lt;br&gt;&#xA;I suggest you take a look at the data on &lt;a href=&quot;http://exoplanets.org/table&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://exoplanets.org/table&lt;/a&gt; and use the filter &quot;NCOMP &gt;= 5&quot; (also sort by name), this way you get current candidate systems with equal to or more than 5 planets.&lt;br&gt;&#xA;You will see there are not many systems that we can base any systematic understanding on.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plots like the first one in the answer to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/12456/estimates-of-exoplanets-distribution-consistent-with-current-data/12462&quot;&gt;this question&lt;/a&gt; that seem to indicate that we know quite many systems, hide the fact that those systems are all single or double planet-systems.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This prohibits basing your generated systems based on data or planet formation theory. But I suggest that you can use a different method:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) &lt;strong&gt;Long-term stability&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't know all systems that are possible, but we certainly know that all systems under certain conditions become unstable. &lt;a href=&quot;https://slidex.tips/download/the-long-term-dynamical-evolution-of-planetary-systems&quot; rel=&quot;nofollow noreferrer&quot;&gt;There is an overview article&lt;/a&gt; or also a &lt;a href=&quot;https://www.youtube.com/watch?v=3wKY0LuDuIk&quot; rel=&quot;nofollow noreferrer&quot;&gt;long video&lt;/a&gt; outlining that there should be a simple stability criterion for planetary systems:&lt;br&gt;&#xA;If any two planets come closer to each other than ~10 mutual Hill-radii, their orbits will become unstable and possible send the whole system into chaos.&lt;br&gt;&#xA;The mutual Hill radius of two planets with masses $m_1$ and $m_2$, and central star $m_s$ and semi-major axis distances $a_1$ and $a_2$ from the star is  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_{hill,mutual} = \left( \frac{m_1+m_2}{3 \cdot m_s} \right)^{1/3} \cdot \frac{\left( a_1 + a_2 \right)}{2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my idea of a simple solar-system generator would be to choose some first planetary mass, place it somewhere.&lt;br&gt;&#xA;Then build up the star system from the inside out planet by planet, by chosing the next planetary mass from one of the above mentioned 1-or-2-planet distributions (this one could be a realistic one, like in &lt;a href=&quot;https://arxiv.org/abs/1109.2497&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mayor et al. 2011&lt;/a&gt;, if you know how to re-shape random distributions).&#xA;You calculate $a_2$ for them such that their mutual distance is $&amp;gt; 10 \cdot R_{Hill,mutual}$.&lt;br&gt;&#xA;In this way you'll probably get pretty packed systems, but you can of course take their mutual distances to be larger than $10 \cdot R_{mutual, Hill}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this way the properties of your generated star systems would be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mutual distances: Stable physical foundation&lt;/li&gt;&#xA;&lt;li&gt;Planetary mass distribution: Would be on shaky feet, as we don't know enough about multi-planet systems, and our data on low-multi systems is heavily biased towards the methods used to detect them.&lt;/li&gt;&#xA;&lt;li&gt;System architecture / order of the planets: So far we've seen many unexpected things and there is no data or reason to assume there should be any strong constraints. I think here you can let your fantasy or the random masses generator roam free.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You could even go one step further and include one of the (half-way) realistic &lt;a href=&quot;https://en.wikipedia.org/wiki/Initial_mass_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;mass distributions for stars&lt;/a&gt;, to generate your $m_s$ values.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your code (e.e. if you code in C) will only be able to give you uniformly distributed random numbers. If you want them to follow a specific distribution, you need to transform them. Google &quot;transform probability density function&quot; or ask someone with math education that you know, if you don't know how that works.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2018-05-23T15:05:38.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26385" PostTypeId="1" AcceptedAnswerId="26392" CreationDate="2018-05-23T20:53:28.263" Score="5" ViewCount="93" Body="&lt;p&gt;If it had been detected gravitational waves from merging of black holes (BH), and neutron stars (NS), why there are no detection of black holes merging with neutron stars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why a BH-NS merging should be a rarer event than BH-BH and NS-NS?&lt;/p&gt;&#xA;" OwnerUserId="23166" LastActivityDate="2018-05-24T16:01:10.287" Title="GW from merging of neutron stars and black holes" Tags="&lt;astrophysics&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26386" PostTypeId="1" CreationDate="2018-05-24T01:29:20.110" Score="1" ViewCount="107" Body="&lt;p&gt;Every 39.88 minutes 10 degrees ($\frac{360^\circ}{36}$) of arc passes over the zenith. It can also be stated that the 36th harmonic of the rotation period of the earth (1435.8 min) is equal to 39.88 min. Note the similar relationships with the rapidly rotating planets:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mars, rotation period of $\frac{1477.2 min}{\text{harmonic of }37} = 39.92$&lt;/li&gt;&#xA;&lt;li&gt;Jupiter: $\frac{595.8}{15} = 39.72$&lt;/li&gt;&#xA;&lt;li&gt;Saturn: $\frac{639.6}{16} = 39.98$&lt;/li&gt;&#xA;&lt;li&gt;Uranus: $\frac{1035}{26} = 39.81$&lt;/li&gt;&#xA;&lt;li&gt;Neptune: $\frac{966.6}{24} = 40.28$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the slowly rotating planets, and the Moon, the numbers are larger, but still very close to 39.88:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mercury: $\frac{3520.2}{88} = 40.00 \text{     } (88 = 22 \cdot 4)$&lt;/li&gt;&#xA;&lt;li&gt;Venus: $\frac{14582.4}{368} = 39.63 \text{     } (368 = 23 \cdot 16)$&lt;/li&gt;&#xA;&lt;li&gt;Pluto: $\frac{9223.2}{232} = 39.76 \text{     } (232 = 29 \cdot 8)$&lt;/li&gt;&#xA;&lt;li&gt;Moon: $\frac{39343.68}{992} = 39.66 \text{     } (992 = 31 \cdot 32)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The questions are: Are these relationships well known? Are they factors in other planetary phenomena?&lt;/p&gt;&#xA;" OwnerDisplayName="user23167" LastEditorUserId="1891" LastEditDate="2018-05-26T07:46:04.853" LastActivityDate="2018-05-26T07:46:04.853" Title="Are harmonic relationships among the rotation periods of the planets well known? Are they factors in other planetary phenomena?" Tags="&lt;rotation&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2018-06-03T00:28:41.550" ContentLicense="CC BY-SA 4.0" />
  <row Id="26387" PostTypeId="1" AcceptedAnswerId="26400" CreationDate="2018-05-24T01:39:07.693" Score="3" ViewCount="709" Body="&lt;p&gt;Do those few thousand kilometers of difference going around the earth significantly affect the amount of thermal energy the moon receives from the sun to warm up its surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, are there more significant differences between these variables when the earth goes from perihelion to aphelion and vice-versa?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.imgur.com/dvZqWV2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/dvZqWV2.png&quot; alt=&quot;Moon&amp;#39;s orbit around earth around Sun, not to scale.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22180" LastEditorUserId="22180" LastEditDate="2018-05-25T04:15:14.303" LastActivityDate="2018-05-25T12:45:17.810" Title="Is there a significant temperature difference on the moon from apogee to perigee or vice-versa?" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;natural-satellites&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26388" PostTypeId="1" CreationDate="2018-05-24T07:34:38.577" Score="3" ViewCount="75" Body="&lt;p&gt;When our sun expands earthlings, if still alive, will have to leave. The Galilean moons offer a lot of water but we would also need land.  As far as I can tell there will be no land for earthlings to live on when the oceans of these moons melt. Is anyone aware of &lt;em&gt;land&lt;/em&gt;  masses on these moons and if so how large? &lt;/p&gt;&#xA;" OwnerUserId="22957" LastActivityDate="2018-05-24T09:49:50.503" Title="Apart from Io, will any Galilean moons have oceans and land if their ice melts" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26389" PostTypeId="2" ParentId="26388" CreationDate="2018-05-24T09:49:50.503" Score="2" Body="&lt;p&gt;We're not even sure the Galilean moons will still be orbiting Jupiter in 5 billion years.   They probably will, but it's hard to say with certainty.   It's difficult to model their orbital stability over that long a time-frame.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the spirit of your question, it takes much more than oceans and land to make a planet or moon livable.   You also need atmosphere, proper gravity (Jupiter's moons have very low gravity) and the right mix of elements.   If, somehow, one of the Galilean moons had continents, it's &quot;ocean water&quot; would be full of ammonia and other elements and we'd find it both stinky and likely toxic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the answer to your question is no, there are no continents.   The ice/water to silicate and metal-oxide ratio for the Galilean moons is much too high for it to ever form continents, and even if it could, a moon that light would have some difficulty retaining it's oceans.   Some would boil of into an atmosphere and with that low gravity, the atmosphere would be stripped from the moon over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a practical standpoint, there aren't any good options for mankind to move to in 1-4 billion years.   Terra-forming mars is possible but difficult and only a temporary solution.  In time the Sun would get too hot for Mars as well.   Titan is an interesting option because it has about the right atmospheric pressure.   That's actually very useful, though living on Titan would be kind of dreary.  It's under a constant thick blanket of clouds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The good news is, there's a lot of time to find a solution and living in space doesn't require living on the surface.   Living under the surface of an icy body (like Pluto) would be workable too.  In fact, there's several advantages to living inside an icy world rather than trying to construct one where we live on the surface.   Much less work to set it up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If someone wants to move this to world-building, feel free, but I felt the question was asked in good faith.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-05-24T09:49:50.503" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26391" PostTypeId="1" CreationDate="2018-05-24T10:31:36.613" Score="3" ViewCount="118" Body="&lt;p&gt;To my understanding, a Kepler telescope is designed to conjugate one plane to another. For example scale down/up the incoming wavefront onto a wavefront sensor for measurement. It works well for small wavefronts, i.e. the wavefront is relatively flat (e.g. $&amp;lt;10 \lambda$ wavefront RMS).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would happen if a large aspherical wavefront is impinging onto the same Kepler telescope system? Paper &lt;a href=&quot;https://www.spiedigitallibrary.org/conference-proceedings-of-spie/5252/0000/Calibration-issues-with-Shack-Hartmann-sensors-for-metrology-applications/10.1117/12.513462.full&quot; rel=&quot;nofollow noreferrer&quot;&gt;Calibration issues with Shack-Hartmann sensors for metrology applications&lt;/a&gt; suggests to place the aperture at the shared focal plane for performance improvement, as below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Cid9R.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cid9R.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to understand the intuition behind? Why would Kepler telescope fail for large aspherical wavefronts?&lt;/p&gt;&#xA;" OwnerUserId="16553" LastActivityDate="2018-10-22T17:01:04.350" Title="How to understand Kepler telescope under large aspherical wavefront" Tags="&lt;telescope&gt;&lt;optics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26392" PostTypeId="2" ParentId="26385" CreationDate="2018-05-24T16:01:10.287" Score="2" Body="&lt;p&gt;I don't think we have a clear answer yet.  LIGO hoped to detect black hole (BH) mergers and expected to detect neutron star (NS) mergers, yet the BH mergers are observed more frequently, partly because they are louder and thus LIGO can see them from further away and partly because the 20-40 solar mass BHs LIGO observed are more common than we expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BHs this size can be formed by the supernova implosion of a massive star, but BHs formed this way should be quite rare  They can also be formed by the merger of two smaller BHs -- which is exactly what LIGO was observing!  It seems likely that these larger (but still stellar mass) BHs were produced by previous mergers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the end, the frequency of LIGO detections depends on two things: The real distribution of objects in binaries: NS-NS, NS-BH, and BH-BH and the loudness/brightness of the gravitational waves emitted by the merger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, black holes form from more massive stars than neutron stars, and that means rarer stars. (The moreso since the more massive a star is, the quicker it sheds mass through stellar winds.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So NS-NS systems come from lesser-massive stellar pairs.  NS-BH systems form the same way, but with one of the original stars being more massive than would lead to a NS.  On the other hand, the BH-BH systems we're detecting probably come from BHs that themselves were the result of mergers in very dense stellar environments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, we wouldn't expect that NS-BH mergers would be especially correlated with BH-BH mergers, since they precursor systems are formed through different mechanisms.  But we would expect NS-BH mergers to be rarer than NS-NS mergers since the precursor systems are rarer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a bit of handwaving in that!  I'll try to find some papers on the subject -- there &lt;em&gt;must&lt;/em&gt; be a bunch.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-05-24T16:01:10.287" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26393" PostTypeId="1" AcceptedAnswerId="26394" CreationDate="2018-05-24T18:00:35.487" Score="3" ViewCount="4089" Body="&lt;p&gt;I'm looking for some form of &quot;rough and ready&quot; formula to convert between redshift &lt;em&gt;z&lt;/em&gt; value, years since BB, and distance, so that when I read an astronomy paper and it discusses an event that occurred at z=10+/-0.5, or a quasar at z=7, I can get a rough idea when the event occurred and how far away they are saying the quasar is, as context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware that this is an inherently vague question, since the times and distances depend on the model and convention chosen, and also that I'm not specifying proper distances/comoving coordinates, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My assumption is that most current models and their parameters,  that are mainstream accepted and used, will be based on very similar parameters of very similar models (if not the same model), such as latest Planck parameters, or similar. They might differ more due to assumptions about very early times, but redshift is inherently &gt;~380 k years, and if they did vary a lot for times beyond that timeframe, we'd have real issues.   In other words, I suspect/hope that any differences in answers due to model variation, won't significantly change the answer to &lt;em&gt;this&lt;/em&gt; question. So the only question will (hopefully) be, which convention or type of metric/distance will be appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an assumption needs making (such as the meaning of 'distance' to be applied, or the zero point used for time: BB or end of inflation etc) please make an assumption  most likely to help me, and I'll edit the question to clarify if needed, once I see what is too vague in the question as initially worded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you !&lt;/p&gt;&#xA;" OwnerUserId="18601" LastEditorUserId="18601" LastEditDate="2018-05-24T18:17:39.580" LastActivityDate="2018-06-02T14:31:03.163" Title="Approximate conversion of redshift 'z' to a time and/or distance, when reading papers?" Tags="&lt;cosmology&gt;&lt;redshift&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26394" PostTypeId="2" ParentId="26393" CreationDate="2018-05-24T19:15:39.587" Score="2" Body="&lt;p&gt;Use one of the &quot;cosmology calculators&quot;. The conversions depend on what you assume for the cosmological parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's one that will do what you want. &lt;a href=&quot;http://home.fnal.gov/~gnedin/cc/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://home.fnal.gov/~gnedin/cc/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. For the default parameters, $z=7$ corresponds to a look back time of 13.01 billion years, whilst $z=10.5$ corresponds to a look back time of 13.33 Gyr, with the current age of the universe being 13.78 Gyr. Choosing the &quot;luminosity distance&quot; option, we that the object has a luminosity distance of 112 Gpc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the proper distance you could use &lt;a href=&quot;http://www.astro.ucla.edu/~wright/CosmoCalc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.astro.ucla.edu/~wright/CosmoCalc.html&lt;/a&gt;&#xA;According to this calculator the $z=10.5$ object is currently at a proper distance of 9.76 Gpc (31.8 billion light years).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="7411" LastEditDate="2018-05-25T12:48:27.107" LastActivityDate="2018-05-25T12:48:27.107" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26395" PostTypeId="2" ParentId="26386" CreationDate="2018-05-24T20:07:22.860" Score="4" Body="&lt;p&gt;I'm very sorry for you, but this is just (mathematical) probability at work and no (astronomical) pattern. I'm on mobile right now so I can't do a full analysis, but note that if you divide Mercury's period (3520.2) by 89 instead of 88, the result is 39.55. Still pretty close to 39.88. In other words, the larger the numbers get, the easier is to find a 'divisor' close to 39.88.&lt;/p&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2018-05-24T20:07:22.860" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26396" PostTypeId="1" CreationDate="2018-05-25T00:12:42.097" Score="0" ViewCount="410" Body="&lt;p&gt;How do you determine the temperature of a star by color?&lt;/p&gt;&#xA;" OwnerUserId="23179" LastEditorUserId="6" LastEditDate="2019-01-29T14:43:48.263" LastActivityDate="2019-01-29T14:43:48.263" Title="How can the temperature of stars can be determined by their color?" Tags="&lt;star&gt;&lt;fundamental-astronomy&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="26399" PostTypeId="2" ParentId="26386" CreationDate="2018-05-25T09:59:36.783" Score="2" Body="&lt;p&gt;So you divide the rotation periods of the planets by any number you need it to be for the end result to be 40, and wonder why the result is 40?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Circular argumentation works because circular argumentation works. It's just not meaningful in any way.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2018-05-25T09:59:36.783" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26400" PostTypeId="2" ParentId="26387" CreationDate="2018-05-25T12:45:17.810" Score="3" Body="&lt;p&gt;Roughly speaking, the moon's orbit has a semi-major axis of about 400,000 km. The semi-major axis of the Earth's orbit is 149.6 million km. The intensity of sunlight received by the Moon depends on the inverse of the square of the distance, so the difference amounts to $((1.496e8 + 4e5)/(1.496e8 - 4e5))^2$ = 1.011, or about 1%. So, not very much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variation in Earth's distance from the Sun is about five &lt;i&gt;million&lt;/i&gt; km, so that's going to have more of an effect than the lunar orbit itself (as Chappo pointed out in their comment). That works out to a variation of about 6.9% in insolation from perihelion to aphelion. The maximum possible variation -- full moon at aphelion versus new moon at perihelion -- would be $((1.521e8 + 4e5)/(1.471e8 - 4e5))^2$, which works out to about 8.1%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference in insolation from new moon to full moon &lt;em&gt;at perihelion&lt;/em&gt; is about 1.09%, versus a difference of 1.06% at aphelion.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2018-05-25T12:45:17.810" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26402" PostTypeId="2" ParentId="26396" CreationDate="2018-05-25T13:10:09.167" Score="3" Body="&lt;p&gt;I think you are talking about color index and temperature. So, I will briefly explain under that circumstance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Color index, e.g., B-V is the difference in brightness (magnitude) of the light observed by filters B and V. (The following discussion can be generalized to other color indeces, e.g., U-B or g-i). Filters B and V have effective wavelengths 4450 (bluer) and V 5510 (redder) A (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Photometric_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;), respectively, therefore B-V increases when the object becomes red. (Recall that magnitude is more negative for being brighter). The temperature which is derived from the color index is called &quot;color temperature;&quot; this is different to blackbody temperature, or kinetic temperature, or other types of inferred temperatures (see [this &lt;a href=&quot;https://www.physics.byu.edu/faculty/christensen/Physics%20427/FTI/Measures%20of%20Temperature.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the color temperature, this is very observational inference. To map from the color index to the color temperature, it is inferred from other well-known object. See &lt;a href=&quot;https://en.wikipedia.org/wiki/Color_index&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to work around is to transform the color index into the flux ratio, then applying the blackbody function as the inference. For example, from B-V &gt;&gt;&gt; flux(4450A)/flux(5510A) &gt;&gt;&gt; blackbody temperature corresponding to the flux ratio.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-05-25T13:10:09.167" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26403" PostTypeId="2" ParentId="2221" CreationDate="2018-05-25T13:20:11.733" Score="3" Body="&lt;p&gt;First, I'd like to point out and commend @user83692435's reply which came first and is correct.  Expanding on it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image of a virtual particle/anti-particle pair being created and then one of the pair being swallowed by the event horizon leaving the other as real is an analogy which provides a picture of what is happening, but is definitely not correct. Popularizers continue to use it because what is really going on is extremely complex and not easily explainable in words. (And I won't try!)  But here's a link to a &lt;a href=&quot;https://arxiv.org/abs/1511.08221&quot; rel=&quot;nofollow noreferrer&quot;&gt;technical paper on the subject&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But perhaps the most telling point against the simple explanation is that Hawking radiation &lt;em&gt;doesn't&lt;/em&gt; come from the event horizon which the analogy requires, but from the space outside it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second point against the Hawking analogy is that the event horizon is &lt;em&gt;enormously&lt;/em&gt; deep in the black hole's potential well. For a particle or photon to &lt;em&gt;escape&lt;/em&gt; the BH (which Hawking radiation must) it must be created with enough additional energy to escape the BH -- and a BH can be thought of an an object with an escape velocity greater than the speed of light. Wimpy little virtual particles which lost their partner to the BH  would never make it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you'd like to dig a bit deeper, I recommend Sabine Hossenfelder's blog Backreaction which has a &lt;a href=&quot;http://backreaction.blogspot.com/2015/12/hawking-radiation-is-not-produced-at.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;long post with many links&lt;/a&gt;  to further information. Backreaction is one of the best of the frontier physics popularizing blogs going these days, in significant part because Hossenfelder is an active researcher and a good writer.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2018-05-25T17:19:39.017" LastActivityDate="2018-05-25T17:19:39.017" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26404" PostTypeId="2" ParentId="26391" CreationDate="2018-05-25T14:52:27.523" Score="1" Body="&lt;p&gt;Not sure what you're asking here.  Pretty much any telescope will be unable to simultaneously produce infinity (magnified) output &lt;strong&gt;and&lt;/strong&gt; an aberration-free intermediate focus.  Further, an aspherical wavefront can't be produced purely by a radiating point source (all objects are collections of point sources), so no optical system can refocus an ugly wavefront.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way a Shack-Hartmann WFS works is basically not to care about aberrated images. Each subaperture's image is unresolved to begin with, and all the sensor needs is to determine the approximate centroid location. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-05-25T14:52:27.523" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26406" PostTypeId="1" CreationDate="2018-05-25T18:41:32.267" Score="1" ViewCount="68" Body="&lt;p&gt;Is there a way to take the red shift of an object and, from that, calculate a rough estimate of its distance from us (in parsecs, preferably)?&lt;/p&gt;&#xA;" OwnerUserId="1494" LastActivityDate="2018-05-26T15:55:20.850" Title="How can you estimate distance of an object by its redshift?" Tags="&lt;distances&gt;&lt;redshift&gt;&lt;astrometry&gt;" AnswerCount="0" CommentCount="8" ClosedDate="2018-05-28T22:12:16.113" ContentLicense="CC BY-SA 4.0" />
  <row Id="26407" PostTypeId="1" CreationDate="2018-05-25T18:54:50.543" Score="3" ViewCount="30" Body="&lt;p&gt;I am looking to create a function to correct for telluric interference and to do so I need to find data on Ozone interference. &#xA;It has been suggested to me that Astropy may have a function that does a similar thing and it would probably retrieve data of a high quality. &#xA;Does anyone know if such a thing exists and if so, how to determine where it is looking to get that data?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&#xA;(I am new to python and astronomy as a whole so please take it easy on me if that's a bad question)&lt;/p&gt;&#xA;" OwnerUserId="23185" LastActivityDate="2018-05-25T18:54:50.543" Title="Telluric Interference data" Tags="&lt;astropy&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26408" PostTypeId="1" AcceptedAnswerId="26409" CreationDate="2018-05-25T19:40:34.097" Score="2" ViewCount="872" Body="&lt;p&gt;How was the Earth to sun distance measured. How do we know how far from the Earth the Sun actually is.&lt;/p&gt;&#xA;" OwnerUserId="23178" LastEditorUserId="2365" LastEditDate="2018-05-25T21:05:44.617" LastActivityDate="2021-03-27T08:20:22.033" Title="How was the distance from the Earth to the Sun determined" Tags="&lt;the-sun&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26409" PostTypeId="2" ParentId="26408" CreationDate="2018-05-25T19:54:41.967" Score="4" Body="&lt;p&gt;There are a bunch of ways.  The simplest is triangulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a transit of Mercury or Venus, when the planet gets between the Earth and the Sun. If you observe the planet's path across the face Sun from two different places, measuring the chords across the Sun's disk and the timings, using no more than trigonometry, you can get an accurate distance to the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a &lt;a href=&quot;https://physics.stackexchange.com/questions/29363/how-did-halley-calculate-the-distance-to-the-sun-by-measuring-the-transit-of-ven&quot;&gt;good discussion of the method on the Physics SE&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It's worth noting that it is easy to measure the distance to, say, Mars in terms of the &lt;em&gt;ratio&lt;/em&gt; of the distance to Mars compared with the Earth-Sun distance, than it is to measure the Earth-Sun distance in miles.  Hence the Astronomical Unit.  We accurately knew many astronomical distances in AU before we knew the length of the AU equally accurately.)&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-05-25T19:54:41.967" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26410" PostTypeId="1" CreationDate="2018-05-26T01:10:27.457" Score="2" ViewCount="207" Body="&lt;p&gt;I'm a beginner in astronomy and I'm trying to figure out how to model the positions of stars in the night sky mathematically (ie a function which spits out azimuth and altitude as a function of time). I was wondering if anyone knew of any such function, I understand it would vary with your position on earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, does anyone know anything about the mathematics of star movements, I'm trying to learn all I can. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="23187" LastEditorUserId="616" LastEditDate="2018-05-26T22:02:17.047" LastActivityDate="2018-08-29T22:59:02.170" Title="Position of specific stars in the sky as a function of time" Tags="&lt;star&gt;&lt;coordinate&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26413" PostTypeId="1" AcceptedAnswerId="26467" CreationDate="2018-05-27T00:41:48.613" Score="7" ViewCount="456" Body="&lt;p&gt;I've seen standard uv plots used discussions of interferometric array designs regularly, and I have a vague idea that each arc may correspond to a pair of elements within the array, and the coordinates $u, v$ in the plot may be something like the path length difference between the source and the two elements projected into some cartesian coordinates, and that the length of the arcs corresponds to the total duration of the observation, but it's all still fuzzy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the elements may be on topography or be so far apart (as in the example below) that the curvature of the Earth is even larger, the formal definition of the $uv$ plane should take this into consideration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several more examples of uv plots and comparisons to point spread functions in the SlidePlayer presentation &lt;a href=&quot;http://slideplayer.com/slide/3602170/&quot; rel=&quot;noreferrer&quot;&gt;Array Design and Simulations&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the math behind the construction of $uv$ plots in radio astronomy? A source that explains it and shows how one can construct them would be extremely helpful, as will any intuitive hints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; From the recent ArXiv preprint &lt;a href=&quot;https://arxiv.org/abs/1805.09223&quot; rel=&quot;noreferrer&quot;&gt;Detection of Intrinsic Source Structure at ~3 Schwarzschild Radii with Millimeter-VLBI Observations of Sagittarius A*&lt;/a&gt; (also &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/1538-4357/aabe2e&quot; rel=&quot;noreferrer&quot;&gt;ApJ Open Access&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/L8MZy.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L8MZy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; From the 2016 ArXiv preprint &lt;a href=&quot;http://inspirehep.net/record/1254036/plots&quot; rel=&quot;noreferrer&quot;&gt;High-Angular-Resolution and High-Sensitivity Science Enabled by Beamformed ALMA&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/I0M8I.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I0M8I.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-05-27T00:47:20.027" LastActivityDate="2018-05-30T21:13:04.487" Title="Math behind a uv plot in interferometry?" Tags="&lt;radio-astronomy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26414" PostTypeId="2" ParentId="5869" CreationDate="2018-05-27T09:35:09.340" Score="0" Body="&lt;p&gt;Gravitational waves have been detected over the past couple of years, and these are simply ripples in gravitational fields that propagate at the speed of light. We know they propagate at the speed of light because that's how gravity wave detectors like LIGO work, and because we have correlated some of the gravity signals to light signals that arrive at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.ligo.caltech.edu/page/press-release-gw170817&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.ligo.caltech.edu/page/press-release-gw170817&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22867" LastActivityDate="2018-05-27T09:35:09.340" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26415" PostTypeId="2" ParentId="26410" CreationDate="2018-05-27T16:22:33.047" Score="1" Body="&lt;p&gt;It's difficult to answer this question with any confidence, since it is so wide ranging, and the original poster has not (yet) provided any clarification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it seems likely that the book &quot;Astronomical Algorithms&quot; by Jean Meeus, published by Willmann-Bell, will provide the answer(s).  Failing that, the publisher's website contains several other reference works on this topic. I have no affiliation with the publisher or author.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, as the author of the question is a beginner, I wonder whether in fact a simple introduction to astronomy, of which there are hundreds, might be a better place to start.  Jumping in by trying to model the positions of stars looks hard, particularly as there are plenty of good applications already available, some of which are free.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2018-05-27T16:22:33.047" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26416" PostTypeId="1" AcceptedAnswerId="26420" CreationDate="2018-05-28T02:16:27.747" Score="7" ViewCount="3555" Body="&lt;p&gt;Hypothetically speaking, if Saturn had a solid surface that we could stand on, would we be able to see our Moon from the North Pole of Saturn? Would the rings of Saturn obstruct the view?&lt;/p&gt;&#xA;" OwnerUserId="22180" LastActivityDate="2018-05-29T04:04:29.300" Title="Hypothetically, would we be able to see the moon from Saturn's North Pole?" Tags="&lt;the-moon&gt;&lt;saturn&gt;&lt;planetary-ring&gt;&lt;hypothetical&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26417" PostTypeId="2" ParentId="26416" CreationDate="2018-05-28T02:56:29.573" Score="14" Body="&lt;p&gt;First of all, at that distance seeing the Moon and seeing the Earth amounts to the same thing. At its closest, Saturn is around 3000 times as far from Earth as the Moon is, so viewed from Saturn, the Moon is never more than about a minute of arc away from the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can see Earth from the North Pole of Saturn, you can see the Moon, also.  (Though it would be a lot dimmer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But&lt;/em&gt; if you can See the Earth from Saturn's pole, then you can also see Saturn's pole from Earth.  So your question can equally well be phrased as &quot;Can Saturn's north pole be seen from Earth and if so, when?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you &lt;a href=&quot;https://www.google.com/search?hl=en&amp;amp;authuser=0&amp;amp;tbm=isch&amp;amp;source=hp&amp;amp;biw=2133&amp;amp;bih=1000&amp;amp;ei=pW4LW8bpDejP5gLkoYiYAQ&amp;amp;q=saturn%20from%20aerth&amp;amp;oq=saturn%20from%20aerth&amp;amp;gs_l=img.3...979.3763.0.4137.0.0.0.0.0.0.0.0..0.0....0...1ac.1.64.img..0.0.0....0.H_HbqnDBOaw&quot; rel=&quot;noreferrer&quot;&gt;look at pictures of Saturn from Earth&lt;/a&gt;, the rings are sometimes tilted so far that they cover one polar region, though it's also true at those times that polar region tilted away from Earth, anyway!  Most of the time the rings are either not tilted enough to cover the poles or tilted to cover the south pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is that during most of Saturn's orbit around the Sun, the rings would not obscure the sightline from Saturn's north pole to the Moon.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-05-28T02:56:29.573" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26418" PostTypeId="1" AcceptedAnswerId="26419" CreationDate="2018-05-28T03:06:08.570" Score="3" ViewCount="182" Body="&lt;p&gt;&lt;a href=&quot;https://worldbuilding.stackexchange.com/a/99325/760&quot;&gt;An answer to a question on Worldbuilding.SE&lt;/a&gt; opined that if you were suddenly transported an unknown distance and direction away &lt;em&gt;by magic&lt;/em&gt; you could use the Cosmic Microwave Background to determine your position relative to the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This lead me to wonder, what does the CMB look like at different positions in the universe? Can changes in how the CMB looks be used to determine your position, direction, or velocity relative to Earth?&lt;/p&gt;&#xA;" OwnerUserId="2019" LastEditorUserId="7982" LastEditDate="2019-08-26T14:49:56.323" LastActivityDate="2019-08-26T14:49:56.323" Title="What would the CMB look like somewhere else in the universe?" Tags="&lt;cosmology&gt;&lt;cmb&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26419" PostTypeId="2" ParentId="26418" CreationDate="2018-05-28T03:15:50.460" Score="4" Body="&lt;p&gt;The CMB is nearly featureless, but the operative word is &lt;em&gt;nearly&lt;/em&gt;. The tiny temperature fluctuations that we've measured should look about the same anywhere within tens to hundreds of millions of light years of Earth or so. This would allow you to get an absolute orientation relative to the CMB and thus relative to Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's fairly easy to measure velocity relative to the CMB, so if you measure your velocity relative to the CMB somewhere else (within that 10-100 million lightyears of Earth, anyway) and if you know your orientation, you can compute your velocity relative to Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't see any way to easily measure the distance from Earth except very approximately. (The CMB that you see from any point is different because it comes from a different sphere. If you move far enough from Earth, the temperature fluctuations you see are no longer the ones we see from Earth because you're sampling a different part of the 300-year-old universe.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Later:&lt;/strong&gt;&#xA;I've been thinking about the position problem and there's a solution, but it's complicated.  The CMB samples the universe of 13.7 billion years ago and from any point in it see the distribution of matter in a shell that old. If we move to another point, in effect, the shell shifts also and samples different parts of the gas. The key point is that the faint ripples in the gas's temperature that we see as mottling in the CMB are three-dimensional bubbles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you shift a short distance in space, the CMB will shift the most in the direction of motion (both in front and behind) and the least at right angles. Furthermore, the amount of shift will depend on the angular size of the fluctuation you're looking at. Low-angular-frequency fluctuations come from physically large ripples int he gas and you need to move a long distance to move the shell you are sampling out of the bubble. High-angular-frequencies ripples come from smaller bubbles and shorter shifts in position will take the shell you observe out of one bubble and into another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Waving my hands just a tiny bit, it should be possible to use statistical methods to estimate the amount and direction of movement by comparing two detailed CMB maps. A major limitation of the method is that the finer the angular resolution, the sooner you lose all correlation between the two views fo the CMB, but the coarser the angular resolution, the coarser your measurement will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numbers? As long as we're not talking about jaunteing many billions of light years, we can ignore general relativistic issues. A fluctuation with an angular size of 1 degree corresponds to a CMB fluctuation which appears to us right now to be around 250 million light years across and consequently ought have a good chance to be visible for moves within roughly that distance. This suggests that the CMB method ought to be able to give you a rough position up to maybe a half-billion or even a billion light years, but will drop off rapidly in accuracy after that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that for shorter hops, it ought to be possible to actually recognize galaxy super-clusters and get a position that way.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2018-05-28T13:38:40.640" LastActivityDate="2018-05-28T13:38:40.640" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="26420" PostTypeId="2" ParentId="26416" CreationDate="2018-05-28T03:37:22.453" Score="12" Body="&lt;p&gt;Yes, if you observe Earth and the Moon at a favorable time.&#xA;Near a Saturn summer solstice, e.g. between 2012 and 2022, Earth appears well above the horizon from Saturn's north pole.&#xA;If the planet body is out of the way, so are the rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other observability issue is Earth's elongation from the Sun.&#xA;This reaches a maximum of about 6$^\circ$ at intervals of about 189 days, e.g. late September 2018 or early April 2019.&#xA;Even then, you would need to shade your optics from the Sun and be above Saturn's atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-05-28T03:37:22.453" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26423" PostTypeId="2" ParentId="26416" CreationDate="2018-05-28T08:52:13.367" Score="9" Body="&lt;p&gt;For these kind of questions, you might want to use &lt;a href=&quot;http://stellarium.org/&quot; rel=&quot;noreferrer&quot;&gt;Stellarium&lt;/a&gt;, a free open source planetarium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can specify the location of the observer on many celestial bodies, including Saturn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/y9mZl.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y9mZl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Any time between 2011 and 2023&lt;/h1&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;With this tool, you can see that the moon will be in the northern saturnian sky non-stop between 2011 and 2023.&lt;/li&gt;&#xA;&lt;li&gt;You can also see that the moon will have an &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_magnitude&quot; rel=&quot;noreferrer&quot;&gt;apparent magnitude&lt;/a&gt; of $5$ when full.&lt;/li&gt;&#xA;&lt;li&gt;The sun has a magnitude of approximately $-20$ as seen from Saturn. It is $10^{10}$ times brighter than the full moon!&lt;/li&gt;&#xA;&lt;li&gt;Saturn is an outer planet, so Earth and Luna can never be in &lt;a href=&quot;https://en.wikipedia.org/wiki/Opposition_%28planets%29&quot; rel=&quot;noreferrer&quot;&gt;opposition&lt;/a&gt; as seen from Saturn.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h1&gt;During a Titan transit&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;What we call a &lt;a href=&quot;http://hubblesite.org/image/2508/news_release/2009-12&quot; rel=&quot;noreferrer&quot;&gt;Titan transit&lt;/a&gt; is basically a solar eclipse for Saturnians. During this time, it should be much easier to observe Luna from Saturn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MJ0Jo.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MJ0Jo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ljT3c.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ljT3c.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;2nd November 2038&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yoUDt.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yoUDt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the 2nd of November 2038 at 15:45 UTC, a Titan transit will occur in the northern hemisphere. Luna will be almost full as seen from Saturn.&lt;/p&gt;&#xA;" OwnerUserId="18474" LastEditorUserId="18474" LastEditDate="2018-05-28T12:15:49.180" LastActivityDate="2018-05-28T12:15:49.180" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26424" PostTypeId="1" AcceptedAnswerId="26441" CreationDate="2018-05-28T14:11:07.317" Score="1" ViewCount="75" Body="&lt;p&gt;I haven't been able to find a reference to the first who showed that the small density fluctuations in the cosmic microwave background led to the formation of large scale structure as we know it.&lt;/p&gt;&#xA;" OwnerUserId="23047" LastEditorUserId="14094" LastEditDate="2018-05-29T08:18:18.530" LastActivityDate="2018-05-29T12:30:05.437" Title="Who discovered the relation between CMB fluctuations and large-scale structure formation?" Tags="&lt;cosmology&gt;&lt;cmb&gt;&lt;early-universe&gt;&lt;formation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26426" PostTypeId="1" AcceptedAnswerId="26430" CreationDate="2018-05-28T16:34:33.430" Score="2" ViewCount="54" Body="&lt;p&gt;When black holes forms out of a supernova do they have a very high velocity ?? Or do they remains at the position of parent star ??&lt;/p&gt;&#xA;" OwnerUserId="22552" LastActivityDate="2018-05-28T18:44:04.530" Title="When black holes forms out of a supernova do they have a very high velocity?" Tags="&lt;black-hole&gt;&lt;astrophysics&gt;&lt;fundamental-astronomy&gt;&lt;stellar-astrophysics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26427" PostTypeId="1" AcceptedAnswerId="26429" CreationDate="2018-05-28T16:45:20.023" Score="9" ViewCount="3970" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;The most massive black holes in the Universe, the supermassive black&#xA;holes with millions of times the math [sic] of the Sun will have a&#xA;temperature of 1.4 x &lt;span class=&quot;math-container&quot;&gt;$10^{-14}$&lt;/span&gt; Kelvin. That's low. Almost absolute zero,&#xA;but not quite. A solar mass black hole might have a temperature of&#xA;only 0.00000006 Kelvin.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://phys.org/news/2016-09-cold-black-holes.html&quot; rel=&quot;noreferrer&quot;&gt;September 5, 2016 by Fraser Cain, Universe Today&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Black holes absorb every form of energy, even light. Absorption of energy should raise its temperature but still it is extremely cold, why?&lt;/p&gt;&#xA;" OwnerUserId="22623" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-10-31T12:27:52.787" Title="Why black holes are extremely cold?" Tags="&lt;black-hole&gt;&lt;universe&gt;&lt;supermassive-black-hole&gt;&lt;heat&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26428" PostTypeId="1" AcceptedAnswerId="26432" CreationDate="2018-05-28T17:30:21.187" Score="3" ViewCount="131" Body="&lt;p&gt;I've read in Hawking's book &lt;em&gt;The Theory of Everything&lt;/em&gt; that total energy of the Universe is zero. He explained this by saying that if there are two bodies separated by a distance and having some mass, they possess potential energy and the same amount of negative energy  with the gravitational field. Thus he concluded that total energy of the Universe  is zero. But here I've a question: how do we account for energy which is in the form of radiation?&lt;/p&gt;&#xA;" OwnerUserId="22552" LastEditorUserId="14094" LastEditDate="2018-05-29T08:18:30.383" LastActivityDate="2018-05-30T14:21:25.040" Title="Account for radiation energy in the statement &quot;total energy of universe is zero&quot;" Tags="&lt;universe&gt;&lt;astrophysics&gt;&lt;fundamental-astronomy&gt;&lt;radiation&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26429" PostTypeId="2" ParentId="26427" CreationDate="2018-05-28T18:35:21.480" Score="16" Body="&lt;p&gt;Under General Relativity (GR) alone, a Black Hole's (BH's) &lt;a href=&quot;https://en.wikipedia.org/wiki/Event_horizon&quot; rel=&quot;noreferrer&quot;&gt;event horizon&lt;/a&gt; is a point of no return -- anything that passes through the event horizon is lost and gone forever, and &lt;em&gt;nothing&lt;/em&gt; comes out. Hence, under GR alone, BHs are utterly black and don't have a temperature at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why the absorption of radiation (or anything else) by a BH doesn't raise its temperature -- it just gets swallowed up and lost. (It's mass, angular momentum and charge do remain, but that's all -- see the &lt;a href=&quot;https://en.wikipedia.org/wiki/No-hair_theorem&quot; rel=&quot;noreferrer&quot;&gt;No Hair Theorem&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: The &lt;a href=&quot;https://en.wikipedia.org/wiki/Accretion_disk&quot; rel=&quot;noreferrer&quot;&gt;accretion disk&lt;/a&gt; that &lt;em&gt;surrounds&lt;/em&gt; a BH can be very hot indeed, but that's another thing entirely.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stephen Hawking discovered that applying quantum mechanics to BHs showed that BHs &lt;em&gt;would&lt;/em&gt; emit a random spray of radiation, and that that radiation was precisely what a black body would emit -- &lt;a href=&quot;https://en.wikipedia.org/wiki/Black-body_radiation&quot; rel=&quot;noreferrer&quot;&gt;black body radiation&lt;/a&gt;.  This is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;noreferrer&quot;&gt;Hawking radiation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black body radiation is simply the thermal emission of a perfect absorber of radiation, and leads to the inescapable conclusion that a BH &lt;em&gt;does&lt;/em&gt; have a non-zero temperature.  Interestingly, Hawking's analysis showed that the effective temperature of the BH is inversely proportional to its mass and that solar-mass BHs (which are the smallest for which we have actual evidence) would have a temperature of about 0.00000006 K. Kinda cold, but still not zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, unintuitively, a solar mass BH get &lt;em&gt;colder&lt;/em&gt; as it absorbs radiation. Because any radiation (or anything else) it absorbs increases its mass, and since higher mass BHs are colder, the more energy you dump into one, the colder it gets!&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-05-28T18:35:21.480" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26430" PostTypeId="2" ParentId="26426" CreationDate="2018-05-28T18:44:04.530" Score="7" Body="&lt;p&gt;There is no specific answer to this -- anything from &quot;just sits there&quot; to flys away at high speed is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It all depends on the symmetry of the supernova (SN) explosion. &lt;a href=&quot;https://www.bing.com/videos/search?q=modelling%20supernova%20explosions&amp;amp;qpvt=modelling%20supernova%20elplosions&amp;amp;FORM=VDRE&quot; rel=&quot;noreferrer&quot;&gt;Extensive modelling&lt;/a&gt; shows that the explosions can be quite asymmetrical, and if they are the gravitational waves created can give the new black hole (BH) quite a kick.  If the explosion is relatively symmetric, the new BH can more or less stay right where it was formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if the exploding star is a member of a close binary, the SN explosion will usually give the companion star a high velocity. Before the explosion, a close pair may be rotating around their joint center of mass at 1000 kps. When one star goes SN, it blows most of its mass right out of the system and high velocity. This doesn't change the velocity of the companion star which suddenly is no longer bound and has escape velocity from the star system -- so it more-or-less goes flying off in a straight line at its orbital velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the &lt;a href=&quot;http://www.esa.int/Our_Activities/Space_Science/Gaia/Artificial_brain_helps_Gaia_catch_speeding_stars&quot; rel=&quot;noreferrer&quot;&gt;runaway stars discovered by Gaia&lt;/a&gt; are probably companion stars to SNs.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-05-28T18:44:04.530" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26431" PostTypeId="1" CreationDate="2018-05-28T22:13:38.747" Score="4" ViewCount="851" Body="&lt;p&gt;I'm currently writing a code to compute the two-point correlation function of a set of galaxies. As I work with a loooot of galaxies, it was suggested to me not to do it in real space, but to compute the power spectrum in Fourier space and get the correlation from the power spectrum, as is explained e.g. in appendix B &lt;a href=&quot;https://arxiv.org/abs/1507.01948&quot; rel=&quot;nofollow noreferrer&quot;&gt;in this paper (arXiv:1507.01948)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I need are some very simple test cases with known results to test my code at each step in order to ensure that I really get what I want. Can anybody point me to any such simple tests? Or at least to a already existing code (preferably in Python or Fortran) that does exactly this, so I can cross-compare the results?&lt;/p&gt;&#xA;" OwnerUserId="20403" LastActivityDate="2018-06-18T22:03:14.253" Title="Tests for code that computes two-point correlation function of galaxies" Tags="&lt;galaxy&gt;&lt;cosmology&gt;&lt;data-analysis&gt;&lt;python&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26432" PostTypeId="2" ParentId="26428" CreationDate="2018-05-28T22:48:17.890" Score="-1" Body="&lt;p&gt;It must be the same situation as if you had bits of mass flying off the surface of the two masses in the simple model. It would not take much kinetic energy, at the huge $mc^2$ scale, to get those bits to fly off to infinity in a Newtonian potential energy, but you'd have a huge $mc^2$ lost from your system. So you must also have a similar loss in the general relativistic gravitational potential energy, if the energy was zero originally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the lost radiation must also represent a significant loss of potential energy, not just the tiny gravitational redshift. I don't know how Hawking is counting potential energy to get the energy to be zero, but it can't be the simple Newtonian potential. Remember that the zero in a potential energy scale is not usually of physical consequence, because it's only changes in potential that matter, but to get the total energy to be zero, he must have some way to get a relevant scale.  Much more formal mathematical arguments are available, such as &lt;a href=&quot;http://www.prespacetime.com/index.php/pst/article/view/81&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.prespacetime.com/index.php/pst/article/view/81&lt;/a&gt;, but even these are debated by experts, who each attribute different degrees of significance to the ability to make the total energy zero.  It is not at all clear that total energy is a relevant physical quantity in general relativity, but for those who wish to make it so, as Hawking does, it may indeed be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that the tiny energy in the radiation today is not really relevant to this question, because the question is about originating a universe.  When the universe originated, radiation was extremely important, so the question does need to be addressed for anyone who wishes the energy to be zero.  The real issue appears to be, whether or not that is an important goal in the first place.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2018-05-30T14:21:25.040" LastActivityDate="2018-05-30T14:21:25.040" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26433" PostTypeId="2" ParentId="26428" CreationDate="2018-05-29T01:57:04.733" Score="4" Body="&lt;p&gt;The total energy of the universe consists of the mass energy of all the matter (both normal and dark), the mass-energy of the radiation and of the dark energy &lt;em&gt;plus&lt;/em&gt; the gravitational potential energy of the universe (which is negative).  The &lt;em&gt;hypotheses&lt;/em&gt; is that this sums to zero.  As far as I know, there is no experimental evidence for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is good evidence for the size of the various mass-energy terms.  The best current estimates are that the mass-energy of the universe is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;73% -- Dark energy &lt;/li&gt;&#xA;&lt;li&gt;23% -- Dark matter&lt;/li&gt;&#xA;&lt;li&gt;3.6% -- Interstellar gas &lt;/li&gt;&#xA;&lt;li&gt;0.4% -- Stars (includes black holes) &lt;/li&gt;&#xA;&lt;li&gt;v. small -- Relativistic neutrinos&lt;/li&gt;&#xA;&lt;li&gt;v. small -- Radiation (CMB and stellar radiation combined)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This balance changes.  As the universe expands, the density of matter (dark and normal) varies as the inverse 3rd power of the size of the universe. (Matter stays the same and the volume expands as the cube of the radius.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the density of dark energy stays constant, since it is a property of empty space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The density of radiation is proportional to the inverse &lt;strong&gt;4th&lt;/strong&gt; power of the radius, because not only does the increased volume dilute the radiation like it dilutes the matter, but also the radiation gets red-shifted, contributing an extra power of the radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Incidentally, that means that the fraction of the total mass-energy of the universe in radiation decreases with time, and is estimated presently to be 1000:1 matter:radiation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So radiation seems simply to be too small to matter.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-05-29T01:57:04.733" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="26434" PostTypeId="2" ParentId="26416" CreationDate="2018-05-29T04:04:29.300" Score="0" Body="&lt;p&gt;Depends on whether you mean &quot;unobstructed line of sight&quot; or &quot;can actually make out with the naked eye&quot;. The other answers have answered the first question, but I think it's worth pointing out the human eye has a resolution of about 10^-4 radians. Saturn is about 1.2 x 10^12 meters away, which means that the moon is about 1.2 x 10^12 meters away from Saturn. The diameter of the moon is about 3.4 x 10^6 m, so the moon is about 3 x 10^5 moon diameters away from Saturn. We can therefore approximate the apparent size of the moon as 1/(3 x 10^5), or 3 x 10^-6, radians, making it about 1/30 the size necessary to be seen from Saturn. Which is not as large of a margin as I thought, so I'm wondering whether there's an error somewhere.&lt;/p&gt;&#xA;" OwnerUserId="19010" LastActivityDate="2018-05-29T04:04:29.300" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26436" PostTypeId="2" ParentId="10344" CreationDate="2018-05-29T06:07:24.200" Score="0" Body="&lt;p&gt;Friedmann made two very simple assumptions about universe,one is that the universe looks identical in which ever direction,the second one is that this would be true if we are observing the universe from anywhere else&lt;/p&gt;&#xA;" OwnerUserId="23209" LastActivityDate="2018-05-29T06:07:24.200" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26437" PostTypeId="2" ParentId="8656" CreationDate="2018-05-29T08:33:33.790" Score="0" Body="&lt;p&gt;The stars do orbit the center every 250 Million years according to a &lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_year&quot; rel=&quot;nofollow noreferrer&quot;&gt;galactic year&lt;/a&gt;, That's 20-25 times since the formation of the solar system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The complex shape of the galaxy, with a disk, bulge, bar and arms adds many non concentric gravity poles to the equation. Easiest to imagine is a wave of movement relative to the disk. It would be nice to see a gravity map of a galaxy in 3D to see that it is not a 1 body at the center kind of gravity map. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some (many? n%) stars can result from the pressure front of supernovae that happen inside star nurseries, which gives them erratic vectors compared to the disk and the arms. Perhaps if there wasn't the constant energy of supernovae affecting the birth and movement of objects in the galaxy, there would be rings as seen for Saturn. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a parallel with mapping rocks in the solar system. If you measure the denser zones to try to figure out where every rock was moving, you'd find your predictions soon become chaos of spirals and local deflection, rather than simple trajectories of a predictable fashion. &lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2018-05-30T22:46:13.657" LastActivityDate="2018-05-30T22:46:13.657" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26439" PostTypeId="2" ParentId="26427" CreationDate="2018-05-29T11:14:12.713" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Black holes absorb every form of energy, even light. Absorption of energy should raise its temperature but still it is extremely cold, why?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because of the infinite gravitational time dilation. The thing to understand is that temperature is a measure of motion. A hot gas is one where the molecules are, on average, moving faster than in a cold gas. See the Wikipedia &lt;a href=&quot;https://en.wikipedia.org/w/index.php?title=Temperature&amp;amp;oldid=843448637&quot; rel=&quot;nofollow noreferrer&quot;&gt;temperature&lt;/a&gt; article and note this: &lt;em&gt;&quot;The coldest theoretical temperature is absolute zero, at which the thermal motion of all fundamental particles in matter reaches a minimum&quot;&lt;/em&gt;. Gravitational time dilation means things are moving slower. When the gravitational time dilation is infinite, things aren't moving at all. This is why the black hole was originally known as the frozen star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stephen Hawking wrote a &lt;a href=&quot;https://projecteuclid.org/euclid.cmp/1103858973&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper&lt;/a&gt; in 1972 with Brandon Carter and Jim Bardeen where they said &lt;em&gt;&quot;It should however be emphasized that κ/8π and A are distinct from the temperature and entropy of the black hole. In fact the effective temperature of a black hole is absolute zero”&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert Wald said much the same in &lt;a href=&quot;https://books.google.co.uk/books?id=Ry7pCAAAQBAJ&amp;amp;pg=PA54&amp;amp;lpg=PA54#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;black hole physics&lt;/a&gt;. On page 69 he said in classical black hole physics &lt;em&gt;“κ has nothing whatever to do with the physical temperature of a black hole, which is absolute zero by any reasonable criteria”&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black hole is said to have an effective temperature by virtue of Hawking radiation, but as Fraser Cain said, it's very low. And as Mark said, &lt;em&gt;&quot;under GR alone, BHs are utterly black and don't have a temperature at all&quot;&lt;/em&gt;. More importantly, Hawking radiation is said to be emitted from outside the event horizon. So it isn't actually the temperature of the black hole. Just as &lt;em&gt;&quot;the accretion disk that surrounds a BH can be very hot indeed&quot;&lt;/em&gt; but it isn't actually the temperature of the black hole.  &lt;/p&gt;&#xA;" OwnerUserId="9772" LastEditorUserId="9772" LastEditDate="2018-05-29T11:30:27.023" LastActivityDate="2018-05-29T11:30:27.023" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26440" PostTypeId="2" ParentId="13081" CreationDate="2018-05-29T11:39:49.957" Score="-3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the temperature inside a Black Hole?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Absolute zero. That's because of the infinite gravitational time dilation. &lt;a href=&quot;https://en.wikipedia.org/wiki/Temperature&quot; rel=&quot;nofollow noreferrer&quot;&gt;Temperature&lt;/a&gt; is a measure of motion of sorts. Gravitational time dilation means things are moving slower. When the gravitational time dilation goes infinite there is no motion, and so no temperature. This is why the black hole was originally known as the frozen star. See the 1939 paper by Robert Oppenheimer and Hartland Snyder &lt;a href=&quot;https://journals.aps.org/pr/abstract/10.1103/PhysRev.56.455&quot; rel=&quot;nofollow noreferrer&quot;&gt;on continued gravitational contraction&lt;/a&gt;. People often say this is outdated, see for example Brown’s mathpages article &lt;a href=&quot;http://mathpages.com/rr/s7-02/7-02.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;the formation and growth of black holes&lt;/a&gt;, but I don't think it is. I say that because I've read the &lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol7-trans/156?highlightText=%22spatially%20variable%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein digital papers&lt;/a&gt;, and I like to think I know &lt;a href=&quot;http://physicsdetective.com/how-gravity-works/&quot; rel=&quot;nofollow noreferrer&quot;&gt;how gravity works&lt;/a&gt;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless of that, in 1972 Stephen Hawking, Brandon Carter, and Jim Bardeen wrote a &lt;a href=&quot;https://projecteuclid.org/euclid.cmp/1103858973&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper&lt;/a&gt; and said &lt;em&gt;&quot;the effective temperature of a black hole is absolute zero”&lt;/em&gt;. Robert Wald said much the same in &lt;a href=&quot;https://books.google.co.uk/books?id=Ry7pCAAAQBAJ&amp;amp;pg=PA54&amp;amp;lpg=PA54#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;black hole physics&lt;/a&gt;: &lt;em&gt;“κ has nothing whatever to do with the physical temperature of a black hole, which is absolute zero by any reasonable criteria”&lt;/em&gt;.  Note that Hawking radiation is nothing to do with the temperature &lt;em&gt;inside&lt;/em&gt; a black hole.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it insanely hot? Cause of so much gravity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. Or should I say &lt;em&gt;I don't think so&lt;/em&gt;. I don't actually know for certain. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it close to absolute zero? Cause the matter is so closely packed, there is hardly any space for particles to move?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes AFAIK it's close to absolute zero. Or it's actually absolute zero. But not because matter is so closely packed. In fact there might not even &lt;em&gt;be&lt;/em&gt; any particles inside a black hole. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it room temperature? Cause Cooper didn't die in Interstellar when he went into a Black Hole?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, it isn't room temperature. &lt;a href=&quot;https://www.google.co.uk/search?source=hp&amp;amp;ei=EzoNW9m7E8-YsAftj6TYCA&amp;amp;q=interstellar+psuedoscince&amp;amp;oq=interstellar+psuedoscince&amp;amp;gs_l=psy-ab.3..0i13i30k1.1040.6320.0.6742.25.22.0.3.3.0.96.1624.22.22.0..2..0...1c.1.64.psy-ab..0.24.1605...0j0i131k1j0i22i30k1j0i22i10i30k1j33i160k1j33i13i21k1.0.VE-bvf5BR9A&quot; rel=&quot;nofollow noreferrer&quot;&gt;Interstellar&lt;/a&gt; is just science fiction I'm afraid, even though it was portrayed as science fact. &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2018-05-29T11:39:49.957" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26441" PostTypeId="2" ParentId="26424" CreationDate="2018-05-29T12:30:05.437" Score="1" Body="&lt;p&gt;I'm not an expert in this, but I'll give it a try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you've got the argument the other way round here. This was never discovered. It was assumed. But why?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We know for the CMB being a redshifted blackbody-spectrum. This means the CMB consists of photons that were emitted in interaction with matter. &lt;/li&gt;&#xA;&lt;li&gt;The CMB fluctuations inform us about the depths of gravity wells that those photons had to escape from (amongst other things). Thus we get a picture of the baryonic and dark matter density at the time of photon-matter decoupling.&lt;/li&gt;&#xA;&lt;li&gt;This allows us to set up linear perturbation theory + simulations to look what happens to a uniform gas with the measured fluctuations on top. That's the rough idea.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You could now of course ask &quot;who was the first to implement this?&quot;.&#xA;The earliest theoretical work that I'm aware of &lt;a href=&quot;http://adsabs.harvard.edu/abs/1965ApJ...142.1317P&quot; rel=&quot;nofollow noreferrer&quot;&gt;(Peebles 1965)&lt;/a&gt; linking &quot;a possible universal blackbody radiation&quot; with the formation of galaxies came in the year of the discovery of background radiation by &lt;a href=&quot;http://adsabs.harvard.edu/abs/1965ApJ...142..419P&quot; rel=&quot;nofollow noreferrer&quot;&gt;Penzias &amp;amp; Wilson 1965&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2018-05-29T12:30:05.437" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26442" PostTypeId="2" ParentId="13081" CreationDate="2018-05-29T14:33:26.127" Score="2" Body="&lt;p&gt;The temperature of a black hole as seen from the outside is the temperature of the Hawking radiation, which @jyoti has discussed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature you would experience as you fell freely through the event horizon, is a different question. I think the answer depends on what else is falling into the black hole at the same time. From its own viewpoint, everything that falls into a black hole reaches the singularity in a fixed maximum time (a few days for the very largest supermassive black holes) so you could only exchange energy with things that fell in around the same time as you did. If there was very little such stuff, then I don't think you'd notice anything out of the ordinary -- you'd be exposed to the cold of space, just as you were before (assuming the black hole is big enough that you haven't been heated up the tides of the black hole distorting you. On the other hand if lots of stuff was falling into the black hole with you then the whole lot would get churned up, and you'd be falling in as part of a mass of very hot plasma. Eventually, at least in a non-spinning black hole, you'd approach the singularity and be ripped apart by tides.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="21928" LastEditDate="2018-05-31T19:30:50.650" LastActivityDate="2018-05-31T19:30:50.650" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26443" PostTypeId="2" ParentId="26428" CreationDate="2018-05-29T15:41:06.763" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I've read in Hawking's book The Theory of Everything that total energy of the Universe is zero. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It isn't true. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;He explained this by saying that if there are two bodies separated by a distance and having some mass, they possess potential energy and the same amount of negative energy with the gravitational field. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That isn't true either. If you let these two bodies fall towards each other,  their gravitational potential energy is converted into kinetic energy. Then when the bodies collide some of this kinetic energy gets radiated away, and you're left with a &lt;a href=&quot;https://en.wikipedia.org/wiki/Binding_energy#Mass-energy_relation&quot; rel=&quot;nofollow noreferrer&quot;&gt;mass deficit&lt;/a&gt;. But note that it's only a mass deficit. The two bodies don't disappear and end up as some zero-energy nothing. Besides, gravitational field energy isn't negative, it's positive. That's why &lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol6-trans/197?highlightText=gravitatively&amp;amp;backToSearch=https%3A%2F%2Feinsteinpapers.press.princeton.edu%2F~searchResults%3FsearchMode%3Dapp%26searchParam-titles%3Dfull%26searchParam-versions%3Dall%26searchParam-years%3D%257B%2522operator%2522%253A%2522%2522%257D%26searchParam-sort%3Ddate%26searchParam-terms%3Dgravitatively%26searchParam-start%3D0%26searchParam-page%3D20&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein said&lt;/a&gt; &lt;em&gt;&quot;the energy of the gravitational field shall act gravitatively in the same way as any other kind of energy&quot;&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point to note is that when you throw a 1kg brick up in the air you do work on it. You give it kinetic energy*. Gravity then converts this kinetic energy into gravitational potential energy, and the brick slows down. Note that this gravitational potential energy is in the brick, not in the gravitational field, or anywhere else. You did work on the brick, and when the brick is at the top of its arc, its mass is greater. Hence when the brick falls to Earth you end up with the mass deficit. Hence when you throw the brick upwards at 11.7 km/s, it's got escape velocity. It leaves the system, taking the original 1kg worth of mass-energy away, along with the mass-equivalence of the 11.7 km/s of kinetic energy.      &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thus he concluded that total energy of the Universe is zero. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's popscience rubbish I'm afraid. Hawking was always coming out with things like that, and few would criticise him because of the wheelchair. Those that did tended not to get any publicity.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But here I've a question: how do we account for energy which is in the form of radiation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By measuring the CMBR. There isn't much radiation compared to other stuff. The universe review &lt;a href=&quot;https://universe-review.ca/F02-cosmicbg08.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;dark energy&lt;/a&gt; article says radiation comprises only about 0.005% of the mass-energy of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;* Whilst conservation of p=mv momentum means there was an effect on the Earth, it's very slight. The KE=½mv² kinetic energy is not shared equally because the Earth didn't move in any detectable fashion. &lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2018-05-29T15:41:06.763" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26444" PostTypeId="1" AcceptedAnswerId="26448" CreationDate="2018-05-29T16:17:32.960" Score="12" ViewCount="6604" Body="&lt;p&gt;&lt;a href=&quot;https://www.livescience.com/32716-whats-at-the-center-of-the-milky-way.html&quot; rel=&quot;noreferrer&quot;&gt;What's at the Center of the Milky Way?&lt;/a&gt;&#xA; In this article it is said that a supermassive black hole lies in the center of milky way galaxy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At its center, surrounded by 200-400 billion stars and undetectable to the human eye and by direct measurements, lies a supermassive black hole called Sagittarius A*, or Sgr A* for short.&#xA;  The Milky Way has the shape of a spiral and rotates around its center, with long curling arms surrounding a slightly bulging disk. It's on one of these arms close to the center that the sun and Earth are located. Scientists estimate that the galactic center and Sgr A* are around 25,000 to 28,000 light-years away from us. The entire galaxy is around 100,000 light-years across.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We revolve around the center every 250 million years.Presumbably we rotate beacuse of the BH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;When the black hole dies in our galaxy will we be thrown out of the revolving orbit?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;The shape of the galaxy is expected to change right?It will be some irregular shape not spherical?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="22623" LastEditorUserId="22623" LastEditDate="2018-05-29T16:30:39.760" LastActivityDate="2018-05-31T15:06:36.340" Title="What will happen to the shape of a galaxy when a super massive black hole lying in its center dies(evaporates out)?" Tags="&lt;black-hole&gt;&lt;galaxy&gt;&lt;milky-way&gt;&lt;supermassive-black-hole&gt;" AnswerCount="6" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26445" PostTypeId="2" ParentId="26444" CreationDate="2018-05-29T16:28:43.840" Score="22" Body="&lt;p&gt;Answer: Not much&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Milky Way's central black hole (BH) masses about 5 million suns, while the galaxy masses 100 billion to a trillion suns.  Consequently, the central BH is pretty much irrelevant to the dynamics of stellar orbits except very close to the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what do you mean by &quot;the black hole dies&quot;? Do you mean evaporates through Hawking radiation? (That's the only process we know of that can do away with a BH, and it is so slow that the galaxy will long since have disappeared before the central black hole evaporates.)&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2018-05-29T17:10:45.587" LastActivityDate="2018-05-29T17:10:45.587" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="26446" PostTypeId="1" CreationDate="2018-05-29T16:50:38.607" Score="6" ViewCount="561" Body="&lt;p&gt;We know about events like two neutron star colliding and resulting in a black hole, also collision of black holes, and collision of galaxies. But we never see a satellite such as a moon colliding with a planet, or a planet colliding with a star, or an artificial satellite colliding with Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can't such celestial bodies (like black holes or neutron stars) remain revolving each other? What makes them collide? Is anything other than gravity responsible for this?&lt;/p&gt;&#xA;" OwnerUserId="22552" LastEditorUserId="616" LastEditDate="2018-05-30T04:17:10.123" LastActivityDate="2018-05-30T04:17:10.123" Title="Why do neutron stars collide instead of just revolving around each other like planets revolve around the Sun?" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;astrophysics&gt;&lt;neutron-star&gt;&lt;stellar-astrophysics&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26447" PostTypeId="2" ParentId="26444" CreationDate="2018-05-29T17:00:54.960" Score="-3" Body="&lt;p&gt;The evaporation of the super massive black holes will take a billions of years and thus the gravitational attraction will get weaken in a very long course of time . This will result in  expansion of the the galaxy and all the stars system and gases will spread in the universe .&#xA; But Hawking radiation is very slow process , even it is possible that till that time all the fuel of the stars will get burnt up (hydrogen) resulting into total darkness .&lt;/p&gt;&#xA;" OwnerUserId="22552" LastEditorUserId="22552" LastEditDate="2018-05-29T17:06:04.740" LastActivityDate="2018-05-29T17:06:04.740" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26448" PostTypeId="2" ParentId="26444" CreationDate="2018-05-29T17:45:34.690" Score="27" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Presumbably we rotate beacuse of the BH.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. The galaxy is being held in one piece due to its own total gravity. The black hole is only a small fraction of that. Basically, the BH doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When the black hole dies in our galaxy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The BH will probably be the last thing left of our galaxy at the end. And even then it will take some incredibly long time for it to evaporate. BH evaporation for very large BHs is basically the slowest process you could imagine.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It will be some irregular shape not spherical?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The galaxy is not spherical. Its shape is rather more like a round disk (with some irregularities and some features like arms, etc).&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2018-05-29T17:45:34.690" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26449" PostTypeId="2" ParentId="26446" CreationDate="2018-05-29T17:47:18.537" Score="3" Body="&lt;p&gt;It's just random. Celestial bodies at that size typically don't collide with each other. They just go on and on for a very long time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A typical collision mechanism is when two neutron stars orbit each other. Their orbits actually do decay slowly over time, because they emit gravitational waves - it's a very slow process and it takes a very long time. When they get close enough to each other, a collision occurs - or rather they merge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind this is a rare event.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2018-05-29T23:36:52.173" LastActivityDate="2018-05-29T23:36:52.173" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="26450" PostTypeId="2" ParentId="26427" CreationDate="2018-05-29T17:54:12.673" Score="3" Body="&lt;p&gt;You got some very good answers already. I just want to point out this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;temperature&quot; of a black hole is more like just &quot;a way of speaking&quot;. It's not temperature as normally understood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is this process called the Hawking radiation where vacuum near a black hole produces a stream of particles, borrowing energy from the black hole's gravity to create those particles - and so it appears like the black hole &quot;emits&quot; radiation. Since this is radiation, you could in theory measure its temperature. But that's just the temperature of the Hawking radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously you couldn't stick a thermometer into a black hole.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2018-05-29T17:54:12.673" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26451" PostTypeId="2" ParentId="26446" CreationDate="2018-05-29T18:06:14.990" Score="7" Body="&lt;p&gt;Neutron stars are one of the possible end products of the evolution of stars greater than around 8 solar masses.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you start out with a close binary pair of these fairly massive stars -- not common, but not rare, either -- the more massive star will evolve to a red giant and tides (or even friction) in the extended envelope will pull them closer together.  The red giant will eventually supernova and may produce a neutron star. Later the originally less massive star will do the same, sometimes resulting in a very close binary pair of neutron stars. (Or sometimes a &lt;a href=&quot;https://en.wikipedia.org/wiki/Binary_pulsar&quot; rel=&quot;noreferrer&quot;&gt;neutron star/white dwarf pair&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The close binary will emit significant orbital energy in gravitational waves, causing the neutron stars to spiral towards each other and, eventually to coalesce.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason they don't just sit there orbiting stably like planets do is because they and very massive and very close together, and consequently orbit each other &lt;em&gt;very&lt;/em&gt; rapidly.  Since the intensity of gravitational wave emission depends on the radiating body's mass and its acceleration, close neutron star pairs can radiate away their orbital energy in &quot;mere&quot; hundreds of millions of years.  The exact same effect happens for planets, but because of the lower mass and lower acceleration, the time to radiate significant energy is far longer than the current age of the universe.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-05-29T18:06:14.990" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26452" PostTypeId="2" ParentId="26444" CreationDate="2018-05-30T01:31:56.200" Score="13" Body="&lt;p&gt;Absolutely nothing left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time for stellar black holes to evaporate is said to exceed the proton half life. How much more the galactic black holes. And by the way, this time is currently increasing as even stellar black holes are currently growing from the cosmic background radiation alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe must pass through the intermediate phase of black holes and empty space before this happens.&lt;/p&gt;&#xA;" OwnerUserId="8056" LastActivityDate="2018-05-30T01:31:56.200" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26453" PostTypeId="1" AcceptedAnswerId="26460" CreationDate="2018-05-30T02:15:18.887" Score="4" ViewCount="213" Body="&lt;p&gt;Hypothetically speaking (once again), if the Sun stays the same size forever (i.e. it somehow maintains a stable hydrogen supply and the helium product just disappears), how long would it take before the decay of the Earth's orbit sends it plummeting into the Sun and disintegrating?&lt;/p&gt;&#xA;" OwnerUserId="22180" LastEditorUserId="2365" LastEditDate="2018-05-30T09:58:05.690" LastActivityDate="2018-05-31T10:13:54.383" Title="Hypothetically, if the sun stays the same size forever, when will the Earth fall into it?" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;orbital-migration&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26454" PostTypeId="2" ParentId="26453" CreationDate="2018-05-30T02:32:28.957" Score="3" Body="&lt;p&gt;This is unpredictable.  The solar system is a chaotic system and it's long-term behavior can't be predicted.  But it's also a &lt;em&gt;very stable&lt;/em&gt; chaotic system and the planetary orbits have been stable for 4 billion years and will remain stable for some millions of years into the future.  See the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stability_of_the_Solar_System&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article for an overview.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we numerically integrate the orbits of the planets into the future in an ensemble (doing many integrations, each with small differences in starting values that are still within experimental error), a small fraction of the ensemble members diverge chaotically. (E.g., Mercury is eventually flung into a wild orbit which may impact Venus.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key point here is that it is intrinsically unpredictable more than some millions of years into the future -- we're definitely safe for much more than a thousand human lifetimes -- other than statistically.  But base don those statistics, it seems likely that the solar system will be disrupted due to chaotic dynamics before the Earth's orbit decays due to tidal forces or anything like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Another presently unpredictable future is an interloping star coming close enough to disrupt planetary orbits. Stars pass through the Oort cloud on a regular basis -- one each million years or so -- but getting inside the orbit of Pluto is &lt;em&gt;much&lt;/em&gt; less likely and hasn't happened yet as far as we can tell.  Nonetheless, this &lt;em&gt;could&lt;/em&gt; happen any time.)&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-05-30T02:32:28.957" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="26455" PostTypeId="2" ParentId="26446" CreationDate="2018-05-30T04:07:09.307" Score="2" Body="&lt;p&gt;The previous answers responded to the main point &lt;em&gt;how&lt;/em&gt; mergers between massive compact objects (i.e. black holes with black holes, neutron stars with neutron stars, etc) occur. We haven't yet observed a merger of a black hole with a neutron star but no doubt such mergers do occur. My answer provides some additional information about the history of detecting them, and also responds to the question about satellites.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;LIGO&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As Florin notes, all mergers of massive compact bodies are rare events. However, the number of stars in the Universe is mind-bogglingly huge, so &quot;rare event&quot; x &quot;huge population&quot; means there are many such events happening across the Universe all the time, but we are &lt;em&gt;only just beginning&lt;/em&gt; to detect them with our current level of technology. It's a very exciting time to be an astronomer!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first &lt;a href=&quot;https://en.wikipedia.org/wiki/LIGO&quot; rel=&quot;nofollow noreferrer&quot;&gt;Laser Interferometer Gravitational-Wave Observatory&lt;/a&gt; (LIGO) started operating in 2002, looking for the telltale gravitational wave signature from the merger of two black holes. After after 8 years it still hadn't detected a single g-wave. An advanced LIGO (&quot;aLIGO&quot;) was then developed with about four times the sensitivity of the previous LIGO, and started collecting data in September 2015. Almost immediately (in fact, while they were still running tests on it prior to it formally being launched!) aLIGO detected the gravitational wave signature of the &lt;a href=&quot;https://en.wikipedia.org/wiki/First_observation_of_gravitational_waves&quot; rel=&quot;nofollow noreferrer&quot;&gt;merger of two stellar-mass black holes&lt;/a&gt; over a billion light years from Earth. Three months later, aLIGO detected a second merger, and since then a further three black hole mergers have been detected. And in August 2017 aLIGO detected the &lt;a href=&quot;https://en.wikipedia.org/wiki/GW170817&quot; rel=&quot;nofollow noreferrer&quot;&gt;collision of two neutron stars&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further retrofitted enhancements to the LIGO network (the &quot;A+&quot; proposals) over the next decade are expected to nearly double aLIGO's sensitivity, so we can expect significantly more compact object mergers/collisions to be detected in future.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Earth satellites&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Regarding artificial satellites &quot;colliding&quot; with the Earth, we see this happen all the time - sometimes spectacularly! According to &lt;a href=&quot;https://www.nasa.gov/news/debris_faq.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA&lt;/a&gt;, &quot;During the past 50 years an average of one cataloged, or tracked, piece of debris fell back to Earth each day&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About half of Earth's artificial satellites are in orbit either within the &lt;a href=&quot;https://en.wikipedia.org/wiki/Exosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;exosphere&lt;/a&gt; - which extends from roughly 700 km to 10,000 km above sea level - or tens of thousands of km above the exosphere in geosynchronous orbit (such as some communications and GPS satellites). The exosphere is so extremely thin that the molecules - mostly hydrogen - rarely interact, so it no longer behaves as a gas, there is no &quot;weather&quot;, and drag is minimal (although not zero). The orbits of satellites in the exosphere and above are mostly affected by solar wind, radiation pressure, variations in the Earth's gravitational field (e.g. from tall mountain ranges), and the gravitational influence of the Sun and the Moon, all of which can lead to an orbit slowly decaying (losing height).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since 2002, the USA has required geostationary satellites to be moved to a &lt;a href=&quot;https://en.wikipedia.org/wiki/Graveyard_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;graveyard orbit&lt;/a&gt; at the end of their operational life, to clear them away from operational satellites. This requires a small rocket boost to achieve an additional 300km in altitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, about 500 operational satellites are instead in &lt;a href=&quot;https://en.wikipedia.org/wiki/Low_Earth_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;low earth orbit&lt;/a&gt; (&quot;LEO&quot;) with an altitude of 2,000 km or less. LEO is the simplest and cheapest option for satellite placement, but the downside is that atmospheric drag becomes increasingly relevant once you're  in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;thermosphere&lt;/a&gt; from 80km to about 1000 km above sea level. The International Space Station orbits in this layer, at between 350 and 420 km; the Iridium communications satellites orbit within the upper thermosphere at 780km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For LEO satellites, boosting to a graveyard orbit may be impractical. Satellites in lower orbits will lose altitude relatively quickly, and below 160km in altitude the orbit decays rapidly, eventually plummeting towards Earth. Small satellites will burn up on re-entry, but larger ones may not burn up completely and can hit the ground intact. Ideally, these larger satellites undergo a planned de-orbit (a controlled re-entry) to crash into a remote part of the ocean. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_reentering_space_debris&quot; rel=&quot;nofollow noreferrer&quot;&gt;lists the largest space debris&lt;/a&gt; that has re-entered Earth's atmosphere. For example, the Salyut and Mir space stations were successfully de-orbited into the Pacific Ocean. The Skylab space station was slightly less successful, crashing spectacularly into the Australian outback in 1979.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And earlier this year, on 2 April 2018, China's &lt;a href=&quot;https://en.wikipedia.org/wiki/Tiangong-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tiangong-1&lt;/a&gt; space station made news headlines with its uncontrolled re-entry as a result of China losing its telemetry link with the spacecraft. The segments that failed to burn up on re-entry crashed into a remote part of the South Pacific, but this was entirely by accident rather than design.&lt;/p&gt;&#xA;" OwnerUserId="14094" LastActivityDate="2018-05-30T04:07:09.307" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26456" PostTypeId="2" ParentId="24842" CreationDate="2018-05-30T04:11:48.300" Score="1" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;To what extent do the spin of smaller celestial bodies reflect the spin of the larger system’s of which they are a part?&lt;/p&gt;&#xA;&lt;p&gt;What about the rotations at other scales of the larger systems we are in, such as our local star cluster, our local galactic cluster, and so forth? Thank you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;There's an amusing explanation here: &lt;a href=&quot;https://space.stackexchange.com/questions/1762/changing-the-rotational-rate-of-a-natural-body&quot;&gt;Changing the rotational rate of a natural body&lt;/a&gt;, it's about &lt;a href=&quot;https://xkcd.com/162/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this much&lt;/a&gt; (as far as &lt;strong&gt;most&lt;/strong&gt; people are concerned):&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VVlE5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VVlE5.jpg&quot; alt=&quot;xkcd - Angular Momentum&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Simple and incorrect answer&lt;/strong&gt;: In &lt;a href=&quot;https://en.wikipedia.org/wiki/Newtonian_mechanics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Newtonian mechanics&lt;/a&gt; the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_field&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravitational field&lt;/a&gt; of a body depends only on its mass, not on it's rotation. If you go a bit further and assume that the spinning object has perfectly uniform density then it matters not whether it spins nor the direction.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;More precise answer&lt;/strong&gt;: The &lt;a href=&quot;https://en.wikipedia.org/wiki/Lense%E2%80%93Thirring_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lense–Thirring effect&lt;/a&gt; is very small—about one part in a few trillion. To detect it, it is necessary to examine a very massive object, or build an instrument that is very sensitive. The non-static stationary distributions of mass–energy causes &lt;a href=&quot;https://en.wikipedia.org/wiki/Frame-dragging&quot; rel=&quot;nofollow noreferrer&quot;&gt;frame dragging&lt;/a&gt; resulting in mass–energy currents and what is known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitoelectromagnetism#Background&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravitomagnetism&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;This approximate reformulation of &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitation&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravitation&lt;/a&gt; as described by &lt;a href=&quot;https://en.wikipedia.org/wiki/General_relativity&quot; rel=&quot;nofollow noreferrer&quot;&gt;general relativity&lt;/a&gt; in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Weak_field_approximation&quot; rel=&quot;nofollow noreferrer&quot;&gt;weak field limit&lt;/a&gt; makes an apparent field appear in a &lt;a href=&quot;https://en.wikipedia.org/wiki/Frame_of_reference&quot; rel=&quot;nofollow noreferrer&quot;&gt;frame of reference&lt;/a&gt; from that of a freely moving inertial body. This apparent field may be described by two components that act respectively like the electric and magnetic fields of electromagnetism, and by analogy these are called the gravitoelectric and gravitomagnetic fields, since these arise in the same way around a mass that a moving electric charge is the source of electric and magnetic fields. The main consequence of the gravitomagnetic field, or velocity-dependent acceleration, is that a moving object near a massive rotating object will experience acceleration not predicted by a purely Newtonian (gravitoelectric) gravity field. More subtle predictions, such as induced rotation of a falling object and precession of a spinning object are among the last basic predictions of general relativity to be directly tested.&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitoelectromagnetism#Equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;equations for calculating the effect&lt;/a&gt; are straightforward but perhaps more than you wanted to know. Modelling this complex behaviour as a curved spacetime problem &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitoelectromagnetism#Higher-order_effects&quot; rel=&quot;nofollow noreferrer&quot;&gt;has yet to be done&lt;/a&gt; and is believed to be very difficult.&lt;/p&gt;&#xA;&lt;p&gt;If you're trying to point a camera at a distant object it's a big deal:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MYVZP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MYVZP.jpg&quot; alt=&quot;Frame Dragging&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-05-30T04:11:48.300" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26457" PostTypeId="1" AcceptedAnswerId="26461" CreationDate="2018-05-30T07:18:13.220" Score="0" ViewCount="1076" Body="&lt;p&gt;Could a planet like the Earth or any other explode or be destroyed and did we see a planet explosion with Hubble?&lt;/p&gt;&#xA;" OwnerUserId="23220" LastEditorUserId="2365" LastEditDate="2018-05-30T09:57:03.897" LastActivityDate="2019-11-09T15:22:35.593" Title="Could planets explode in space?" Tags="&lt;planet&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26458" PostTypeId="2" ParentId="26457" CreationDate="2018-05-30T08:21:29.310" Score="4" Body="&lt;p&gt;We never saw a planet explosion with Hubble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make something explode it has to produce a lot of energy in a very short time. Gravitational collapse and thermonuclear fusion in degenerate matter are two possible ways of producing enough energy to produce a supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These mechanisms can not cause a planet to explode. Planets are solid and supported by their own matter, so can't collapse, and they don't contain large amounts of hot degenerate matter, so don't have any kind of fusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets don't produce much of their own energy, and there is no mechanism known that could cause a planet to explode. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest you may get is planetary collisions, such as that which likely formed the moon. But these have never been directly observed.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-05-30T08:21:29.310" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26460" PostTypeId="2" ParentId="26453" CreationDate="2018-05-30T11:23:19.557" Score="5" Body="&lt;p&gt;As has already been said, the major sources of change to Earth's orbit are interactions with other planets and passing stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're ruling out mass loss of the Sun, so the next consideration is probably tidal interactions between the Earth and Sun. &lt;a href=&quot;https://arxiv.org/abs/0905.3008&quot; rel=&quot;nofollow noreferrer&quot;&gt;This paper&lt;/a&gt; suggests that the Earth is receding from the Sun by about 15cm/yr for this reason. This is 150 km per million years, so over a trillion years or so, would move Earth quite a bit further from the Sun. The effect would reduce as the Earth got further away from the Sun, but I think it would always be slowly receding. I haven't done the calculation, but I'm not sure it would ever reach tidal lock -- the angular momentum of the Sun is too high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Independently of this, though, orbital energy and angular momentum would slowly be lost by gravitational radiation. &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_wave#Compact_binaries&quot; rel=&quot;nofollow noreferrer&quot;&gt;This wikipedia page&lt;/a&gt; gives a formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t={\frac {5}{256}}\,{\frac {c^{5}}{G^{3}}}\,{\frac {r^{4}}{(m_{1}m_{2})(m_{1}+m_{2})}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the time to collision ignoring tidal effects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can plug in known values for the parameters $r = 1.5 \times 10^{11} m$, $c = 3\times 10^8 m/s$, $G = 6.7 \times 10^{-11} m^3 kg^{-1} s^{-2}$, $m_1 = 6\times 10^{24} kg$, $m_2 = 2 \times 10^{30} kg$ and get  about $3.3\times 10^{30} s$ as the time to collision. I suspect that's the time until the centres of the Earth and Sun coincide, so the actual collision would be a bit earlier, but probably not enough to notice at this precision. So that's about 10 million million times the lifetime of the universe so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't find enough information to work out how these two effects would interact. Clearly in the first $10^{12}$ years tidal evolution would be more important, but that conserves angular momentum and slows as the Suns rotation slows, whereas gravitational radiation actually carries angular momentum away, so might dominate in the long term. On the other hand the radiation drops as a high power of the separation, so we might have the Earth moving ever more slowly away from the Sun, while radiating gravitational waves of ever weaker strength, and never reaching tidal lock or starting to move back in. &lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="21928" LastEditDate="2018-05-31T10:13:54.383" LastActivityDate="2018-05-31T10:13:54.383" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26461" PostTypeId="2" ParentId="26457" CreationDate="2018-05-30T11:32:12.680" Score="2" Body="&lt;p&gt;Of course if you supply enough energy, you could destroy a planet. If it's applied efficiently enough then the energy you need is basically the gravitational binding energy of the planet. For the Earth that is about $2\times 10^{32} J$ (&lt;a href=&quot;http://www.astronomy.ohio-state.edu/~pogge/Ast162/Intro/gravity.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;) which is perhaps better though of as the energy equivalent of about 2 million million tons of mass. So if you could supply a million million tons of antimatter (roughly a 1km diameter asteroid) and get it into the centre of the Earth so that it could annihilate with the matter there, that would do it and you'd be left with a (hot, radioactive) asteroid belt where the Earth used to be. (see Greg Bear's &quot;The Forge of God&quot; for a fictionalised account of this. Alternatively, that is twelve days of the Sun's total output if you could somehow focus it on the Earth. &lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-05-30T11:32:12.680" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26464" PostTypeId="1" CreationDate="2018-05-30T17:29:09.360" Score="0" ViewCount="1525" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;When observing from a latitude of 50 degrees North on Earth, the Celestial Equator will have a maximum altitude of:&#xA;  &lt;em&gt;Answer: 40 degrees&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've looked this up and it says something about the observer's co-latitude. However, I do not understand why this is the case. Could someone explain this please?&lt;/p&gt;&#xA;" OwnerUserId="22985" LastActivityDate="2018-06-05T02:43:01.400" Title="How to calculate the altitude of the Celestial Equator from an observer's latitude? Exam question" Tags="&lt;earth&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26465" PostTypeId="1" AcceptedAnswerId="26466" CreationDate="2018-05-30T19:37:45.543" Score="5" ViewCount="358" Body="&lt;p&gt;Assume two black holes collide head-on.  In other words, they were not orbiting one another before the collision.  I know this is unlikely.  Further assume that their sizes and distance from Earth are similar to past collisions detected by LIGO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that detectable gravitational waves are caused by massive objects undergoing massive accelerations and that detections so far have been of colliding objects orbiting one another many times within a fraction of a second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that a head-on collision would create a single gravitational &quot;pulse&quot; instead of a wave.  I don't know if &quot;pulse&quot; is the right word for this.  Would this pulse be detectable by LIGO?&lt;/p&gt;&#xA;" OwnerUserId="20370" LastEditorUserId="37191" LastEditDate="2021-02-12T21:29:20.687" LastActivityDate="2021-02-12T21:29:20.687" Title="Would LIGO Detect Head-On Collision?" Tags="&lt;black-hole&gt;&lt;gravitational-waves&gt;&lt;ligo&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26466" PostTypeId="2" ParentId="26465" CreationDate="2018-05-30T20:05:05.183" Score="8" Body="&lt;p&gt;There are two issues: Would there be gravitational waves to be detected and would LIGO detect them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the first issue, gravitational waves are quadrupolar, and a cylindrically symmetric system will not produce any. (Specifically, the second time derivative of the quadrupole moment of an isolated system's stress–energy tensor must be non-zero in order for it to emit gravitational radiation.) So a head-on collision of equally massive black holes would not produce significant gravitational waves.  If the BHs were of different masses, or the collision not quite head-on, gravitational waves would be produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the second issue there are at least three ways LIGO might might miss them.  First, they might simply be too faint.  A merger through rotation is one of the best ways to produce gravitational waves, and a head-on collision is one of the worst (during a BH merger, of course). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second way it might miss is if the merger is in one of its blind spots. Each LIGO detector has blind spots and there's no rule that says a BH merger can't be in one of them. (In fact, the neutron star merger was located by use of the blind spots. The third gravitational wave (GW) detector, Virgo, in Italy, didn't pick up the neutron star merger.  It was bright/loud enough that it should have, which meant that the merger was in one of Virgo's blind spots, which helped to narrow down its location.  A GW detected in only one detector would be listed as a possible GW, but not as a detection, because it might simply be terrestrial interference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But another way it might miss is if LIGO had not been looking for the waveform of that kind of merger. General Relativity allows us to compute in very good detail the GW signal from any type of collision that we think likely. Because the GW signals are so faint, the first step of detection is matching computed GW waveforms to the data and looking for matches. Once a match is found, the signal is analyzed in more detail, but they have to know where to look. If no one expected the waveform and it was not strong enough to pop out of the data and yell &quot;Here I am, dummy!&quot; it might be missed. Since head-on collisions are expected to be &lt;em&gt;very&lt;/em&gt; rare, it's possible that they are not in the library of waveforms that LIGO checks.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-05-30T20:05:05.183" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26467" PostTypeId="2" ParentId="26413" CreationDate="2018-05-30T21:13:04.487" Score="4" Body="&lt;p&gt;Rick Perley gives a lecture about interferometry to a biannual &lt;a href=&quot;http://www.cvent.com/events/15th-synthesis-imaging-workshop/event-summary-e34ab08b2a854a48bae723c8fe86d2d4.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;NRAO Synthesis Imaging Workshop&lt;/a&gt;.&#xA;&lt;a href=&quot;https://science.nrao.edu/science/meetings/2016/15th-synthesis-imaging-workshop/SISS15Advanced.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Part 2 of his 2016 talk&lt;/a&gt; discusses u,v plane coverage.&#xA;The 2016 &lt;a href=&quot;http://www.iram-institute.org/EN/content-page-331-7-67-331-0-0.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IRAM Millimeter Interferometry School&lt;/a&gt; lectures by &lt;a href=&quot;http://www.iram.fr/IRAMFR/IS/IS2016/presentations/pety-basis.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;J. Pety&lt;/a&gt; and &lt;a href=&quot;http://www.iram.fr/IRAMFR/IS/IS2016/presentations/gueth-mm-interferometers.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;F. Gueth&lt;/a&gt; may also help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\hat{u}, \hat{v}, \hat{w}$ be orthogonal unit vectors with $\hat{w}$ pointing from Earth to the radio source, $\hat{v}$ oriented northward, and $\hat{u}$ oriented eastward.&#xA;Let $B_{ij}$ be the baseline vector between antennas $i$ and $j$ in the array.&#xA;Then $(B_{ij} \cdot \hat{u}, B_{ij} \cdot \hat{v})$&#xA;and $(-B_{ij} \cdot \hat{u}, -B_{ij} \cdot \hat{v})$&#xA;are points in the u,v plot, and&#xA;$B_{ij} \cdot \hat{w} / c$ is a time offset between the signals at $i$ and $j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If both antennas observe the source for several hours, $B_{ij}$ rotates with the Earth, and the two points in the u,v plot grow into elliptical arcs.&#xA;The shape of those ellipses depends on the declination of the source, so&#xA;the same array will have different u,v coverage for sources at different declinations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interfering signals from the array are combined in a visibility function $V(u, v)$.&#xA;The image of the source is an inverse Fourier transform of $V$, typically deconvolved using an algorithm called CLEAN.&#xA;To minimize image artifacts, a diverse set of baseline lengths and orientations, without large gaps in u,v coverage, is desirable.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-05-30T21:13:04.487" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26468" PostTypeId="2" ParentId="26464" CreationDate="2018-05-30T21:17:01.153" Score="3" Body="&lt;p&gt;Perhaps you have heard that the altitude of the celestial pole (3 in the figure below) is equal to your latitude. In other words, angle 5 is the same as your latitude. Of course, the angle between the celestial pole (3) and the celestial equator (4) is 90 degrees (angle 6). From this, you should be able to calculate the angle of the equator (angle 7).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Item 1 is the horizon, and 2 is the celestial sphere. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/j1NME.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j1NME.png&quot; alt=&quot;horizon and sky for northern observer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2018-05-30T21:17:01.153" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26469" PostTypeId="1" AcceptedAnswerId="26475" CreationDate="2018-05-30T22:09:30.480" Score="2" ViewCount="35" Body="&lt;p&gt;I have read that the lifetime of stars is much less than the period of rotation for a spiral galaxy. For example &lt;a href=&quot;https://www.coursera.org/learn/evolvinguniverse/lecture/wlYhz/the-origin-of-spiral-arms&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.coursera.org/learn/evolvinguniverse/lecture/wlYhz/the-origin-of-spiral-arms&lt;/a&gt; the professor uses the existence of young blue stars whose creation is triggered by a persistent density wave to claim that the stars lifetime is less than the rotation period. I am not satisfied with this. Given the enormous timescale, how do we know this?&lt;/p&gt;&#xA;" OwnerUserId="23225" LastActivityDate="2018-05-31T07:08:34.127" Title="Relationship between galactic rotation and stellar lifetime" Tags="&lt;star&gt;&lt;galaxy&gt;&lt;rotation&gt;&lt;spiral-arms&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26470" PostTypeId="2" ParentId="26444" CreationDate="2018-05-30T23:51:39.213" Score="2" Body="&lt;p&gt;A macroscopic black hole cannot shrink as long as any radiation source (like other galaxies) is within sight. The Hawking radiation is very weak; black holes are called black for a reason. In fact the Hawking radiation is already outweighed by the cosmic microwave background radiation alone for black holes heavier than the moon. This is just a function of temperature: The universe background has a temperature of 2.72 K &amp;mdash; in order to emit more radiation than it absorbs the black hole must be hotter, &lt;a href=&quot;https://en.wikipedia.org/wiki/Hawking_radiation#Black_hole_evaporation&quot; rel=&quot;nofollow noreferrer&quot;&gt;which requires a mass smaller than the moon's&lt;/a&gt;. Solar mass black holes have a low temperature in the 6E-8 K order of magnitude. That means that even absent any matter it could absorb and absent any specific source of radiation a massive black hole would still grow, not shrink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of Sagittarius A* there is a lot of matter and radiation around, namely our galaxy, which will eventually fall into the black hole, if left undisturbed for a sufficiently long time. The resulting super duper massive black hole would be super duper cold (around E-19K, give or take a few orders of magnitude) and could feed even from an ever-cooler microwave background for a long time. Only when everything has been absorbed or disappeared beyond the event horizon can it even start shrinking at all. And because it is very very cold it will shrink very very slowly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is more likely that other events will precede this evaporation though. &lt;a href=&quot;https://arxiv.org/pdf/0704.0221.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This paper&lt;/a&gt; describes how in the far future &amp;mdash; say, 100 billion years &amp;mdash; the accelerating expansion of the universe will leave us stranded on the gravitationally bound island of our local group, because everything else &quot;expands away&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some point the black holes in this island will have absorbed all surrounding matter until only orbiting black holes are left. They will eventually fall into each other because they lose kinetic energy through gravitational waves. The end scenario is a single giant black hole which rotates enormously fast (making temperature estimates more difficult). It is conceivable that at some point in this process the background radiation will become colder than the black hole(s) so that the ever more massive black holes indeed, finally, start evaporating. Very &lt;em&gt;very&lt;/em&gt; very slowly though.&lt;/p&gt;&#xA;" OwnerUserId="11190" LastEditorUserId="11190" LastEditDate="2018-05-31T15:06:36.340" LastActivityDate="2018-05-31T15:06:36.340" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26471" PostTypeId="1" AcceptedAnswerId="26474" CreationDate="2018-05-31T02:51:43.280" Score="-1" ViewCount="82" Body="&lt;p&gt;LIGO detected a black hole merger for bh's of about 30 solar masses, with a schwartzchild radii of about 20km. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Were the gravity waves generated by masses distanced by a few microns or a few kilometers? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A LIGO &lt;a href=&quot;https://physics.aps.org/featured-article-pdf/10.1103/PhysRevLett.116.061102&quot; rel=&quot;nofollow noreferrer&quot;&gt;article states&lt;/a&gt;: &#xA;&lt;em&gt;The Keplerian effective black hole separation in units of Schwarzschild radii&#xA;(RS ¼ 2GM=c2) and the effective relative velocity given by the&#xA;post-Newtonian parameter v=c ¼ ðGMπf=c3Þ1=3, where f is the&#xA;gravitational-wave frequency calculated with numerical relativity&#xA;and M is the total mass&lt;/em&gt; (FIG. 2.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They state a relative velocity of 0.4 c for 100ms, can't we compare that to electron orbits or something practical? Electrons travel at 0.01 c. Can we say that the merging masses are 40 times smaller than an H atom?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we state that the signal does not continue for an entire minute up to 90Khz, perhaps forever, except that the rest is too faint to detect on LIGO For some reason? &lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2018-05-31T22:02:00.500" LastActivityDate="2018-05-31T22:02:00.500" Title="Physical measurements and celerity of black hole mergers?" Tags="&lt;black-hole&gt;&lt;gravity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26472" PostTypeId="2" ParentId="26465" CreationDate="2018-05-31T03:22:58.507" Score="2" Body="&lt;p&gt;The waveform of binary inspiral was about 100ms with a peak for every rotation, about 10 waves/rotations were measured, ranging from 30 to 200Hz. The average female voice has a fundamental frequency of about 200hz, to have some idea, and a typical pitchfork is about 440Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The waveform for a head on collision would about 5ms, probably less, kindof like a snap sound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be indistinguishable from background noise. Sounds that consist of a single peak are like snaps on a vynil. you would challenge LIGO to detect a snap on a vynil record every hextillion years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A direct collision of solar mass bh's is something like 1 in a hextillion years for the entire universe, because the singularities are as big as atoms, distanced by 100ds of kms travelling in the 20kmps kind of range.&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2018-05-31T03:28:03.097" LastActivityDate="2018-05-31T03:28:03.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26473" PostTypeId="2" ParentId="15023" CreationDate="2018-05-31T03:57:29.893" Score="4" Body="&lt;p&gt;There's a detailed graph of it here:&lt;a href=&quot;https://i.stack.imgur.com/fjuSZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fjuSZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/First_observation_of_gravitational_waves&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/First_observation_of_gravitational_waves&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the most precise reference for your query is the Time-Frequency graphs in black and green, also called waterfall graphs, spectrographs, sonograms and so forth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the Y axis, the width of the green smudge is about 30-50Hz, That's the frequency-uncertainty for the TF graph for every wave of the roughly 10 waves that make up the signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the actual real measurement that LIGO returned, and there are idealized versions &lt;a href=&quot;https://www.google.fr/search?q=ligo%20spectrogram&amp;amp;num=100&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ved=0ahUKEwil7NKMhK_bAhXEbhQKHYTVDtkQ_AUICigB&amp;amp;biw=1269&amp;amp;bih=656#imgrc=7fHKhmhfDItCoM:&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; with an uncertainty range of about 5Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in sound theory, you can research about Wigner-Ville TF graph, which is the most precise sound analysis mathematics known, and which uses quantum uncertainty estimates to minimize the quantum uncertainty of placing single audio waves in time at a given known frequency. If you have java JVM you can compute a more deeply processed TF graph using sonogram 4 visible speech demo version. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an synth-reverse-engineering sound scientist, I have to say that the precision of the wave is incredibly clear, it's the same as recording an individual bird chirp from 20 meters away on a clear microphone. Gravity actually has sound, like a bird call, except it's not carried by air compression it's carried by space time at &lt;em&gt;c&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://losc.ligo.org/s/images/omega_sample.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://losc.ligo.org/s/images/omega_sample.png&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2018-06-05T07:40:38.867" LastActivityDate="2018-06-05T07:40:38.867" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26474" PostTypeId="2" ParentId="26471" CreationDate="2018-05-31T06:35:25.487" Score="2" Body="&lt;p&gt;A singularity is not a object. it is a time and position at which infinities appear in the solution of general relativity. When you speak of a &quot;non-zero size singularity&quot; I assume you mean a physical object inside a black hole (as could be predicted by some forms of string theory) Such objects are not part of General relativity, which models black holes as a region of spacetime including a point singularity (for a simple black hole) or a ring singularity (for a rotating black hole)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These results are consistent with General Relativity and two merging Kerr black holes (which have a ring singularity in GR).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They would also be consistent with other models of gravity, including models in which a physical singularity doesn't form. But there is no positive evidence here for there being an object that is a millionth of an angstrom behind the event horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what you mean by &quot;binary singularities distanced by a few microns&quot;  The amplitude of the gravitational waves detected was much smaller than that, and there is nothing about microns in the linked video.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would not expect the signal to continue to increase in frequency up to 90KHz. A new single event horizon forms at some point, in this case when the gravitational waves have reached about 200Hz. A short &quot;ring down&quot; period follows the merger as a single black hole settles into a new stable state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objects measured by LIGO are black holes, with an event horizon about 20km in diameter. The measurements at LIGO don't suggest that any object exists inside the black hole. Nothing in these data suggest the objects were the size of electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass estimates have quite large error bars, for the first Gravitational wave detected, the masses were about 30 solar masses +- 4solar masses, with similar uncertainties in the Schwartzchild radii&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the orginal LIGO article &lt;a href=&quot;https://physics.aps.org/featured-article-pdf/10.1103/PhysRevLett.116.061102&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://physics.aps.org/featured-article-pdf/10.1103/PhysRevLett.116.061102&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-05-31T06:35:25.487" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26475" PostTypeId="2" ParentId="26469" CreationDate="2018-05-31T07:08:34.127" Score="3" Body="&lt;p&gt;We know how long stars take to orbit in a spiral galaxy, because we know how big galaxies are (we can measure the size by knowing the distance using various methods such as Cepheid variables, and type Ia supernovae) and we know how fast the stars are moving. (We can measure the velocity of stars using the doppler method)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know how long stars last for because we have a good model of how stars work. We can use this model to predict that large, hot stars will burn themselves out in a short time. We can verify this model by looking at the numbers of blue stars, red supergiants and supernovae. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the first part we know that stars take about 250 million years to complete an orbit of the Milky way (which is typical for similar sized galaxies) From the second part, we know that giant blue stars live for only tens of million years, and so could not complete a full orbit of the milky way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, smaller stars like the sun live for billions of years and so make multiple orbits.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-05-31T07:08:34.127" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26476" PostTypeId="2" ParentId="26444" CreationDate="2018-05-31T08:51:03.187" Score="2" Body="&lt;p&gt;To answer this, let's look at the next few billion/trillion/quadrillion/? years, and get a sense of the size of our galaxy and its central black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing that happens related to your question, is that our galaxy and Andromeda collide and merge. This happens in some billions of years. When galaxies merge, the combined galaxy exists, but may have a different form, merged central black holes, and stars (or in some cases even one or both black holes) may be flung out of the combined galaxy. But the galaxy will endure, in some form or other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's because a galaxy isn't held together by its central black hole. &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;A sense of scale: mass&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In our galaxy, the central BH has a mass of about &lt;strong&gt;4 -4.5 million suns&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bigger part is the stars, gas, and other ordinary baryonic matter (some hundreds of billions of stars, although many are red dwarfs and smaller than our sun). The ordinary matter is estimated to be about &lt;strong&gt;600 billion suns&lt;/strong&gt;,or about 150,000 times the mass of the central black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the biggest part is dark matter. Explained simply, even taking into account all of the above mass, the galaxy still wouldn't be massive enough to rotate as it does. Calculations show that about 85% of all matter in our galaxy is &quot;dark matter&quot; - a type of matter that isn't made of ordinary atoms, but is suspected to be made of particles that can't interact much except through gravity (so we can't detect it through radiation, it doesn't form planets, stars or black holes, etc). Dark matter would be about &lt;strong&gt;3.5 trillion suns&lt;/strong&gt;, or about 850,000 times the mass of the central BH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total mass (ordinary + dark matter) is about &lt;strong&gt;4 trillion suns&lt;/strong&gt; or about &lt;strong&gt;a million times the mass of the central black hole&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;A sense of scale: diameter&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Considering size rather than mass, the central BH is perhaps the size of Uranus' orbit (about &lt;strong&gt;12 light hours diameter&lt;/strong&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The visible galaxy is about &lt;strong&gt;100,000 light &lt;em&gt;years&lt;/em&gt; diameter&lt;/strong&gt;, or about 70 million times the BH size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extent of the dark matter halo is less certain (and has less of a defined edge), but depending on which research is right, may extend between &lt;strong&gt;500,000 and 1 million light years diameter&lt;/strong&gt;, or something along those lines (from memory), or a little under half a billion times the BH size.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Summary&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The central BH contains about &lt;strong&gt;a millionth (0.0001%)&lt;/strong&gt; of the galaxy's mass, and about &lt;strong&gt;2 billionths (0.0000002%)&lt;/strong&gt; of its diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the central black hole is actually, and oddly, almost insignificant in terms of our galaxy's present-day structure. It might have been crucial for the formation of the galaxy, but that was long, long ago. It's not the current reason we rotate, and its not the reason we stay in galactic orbit. If it vanished or was ejected tomorrow, nothing at all would change except for a comparatively few stars in the galactic centre that directly orbit the BH. We're nowhere near there. We are in a spiral arm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line is, if the central BH vanished or left our galaxy, we and our descendants wouldn't ever notice, except for a change in X-ray emissions from that region (as detected by radio telescopes), and a few very faint stars in that region moving slightly differently over the millennia. That's all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as other answers explain, a black hole takes an immense time to evaporate, so in reality, two things will happen: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;On a timescale of billions to trillions of years&lt;/strong&gt; At some point the merging Milky Way/Andromeda galaxy (or a successor galaxy) will keep, merge or eject its central BH. This event won't be an 'end' to the galaxy or the stars in them, although the combined galaxy probably won't be a spiral shape;  merged galaxies are common. The combined galaxy will settle down and things will continue.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;On a timescale beyond human comprehension (quadrillions upon quadrillions of years)&lt;/strong&gt; If our universe still exists in its present structure and the standard model and standard cosmology are about right, the central BH &lt;em&gt;will&lt;/em&gt; eventually evaporate. But the galaxy (and all galaxies, and most matter) will have decomposed long, long, long before that can happen.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="18601" LastEditorUserId="18601" LastEditDate="2018-05-31T14:17:42.137" LastActivityDate="2018-05-31T14:17:42.137" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26478" PostTypeId="1" AcceptedAnswerId="26481" CreationDate="2018-05-31T15:46:44.500" Score="0" ViewCount="103" Body="&lt;p&gt;What would be the (theoretical) apparent visual magnitudes of Spica A and Spica B (since they are too close together to be resolved telescopically)?  The Spica System is 77 parsecs away, and the combined apparent visual magnitude of the two stars is 0.97 at maximum (presumably when the stars are not eclipsing each other).  Their orbital parameters and respective luminosities, temperatures, and radii are given on Wikipedia: &lt;a href=&quot;https://en.wikipedia.org/wiki/Spica&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Spica&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="23235" LastActivityDate="2018-05-31T16:33:17.767" Title="What are the apparent visual magnitudes of Spica A and Spica B?" Tags="&lt;binary-star&gt;&lt;apparent-magnitude&gt;&lt;magnitude&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26479" PostTypeId="1" CreationDate="2018-05-31T16:01:48.507" Score="1" ViewCount="35" Body="&lt;p&gt;My goal is to learn about an example of a Dwarf Spheroidal or a Dwarf Irregular Galaxy which is not a Satellite / Companion Galaxy (or is at least 500 000 L.y. away from another Galaxy), and doesn't measure more than 500 Light years in Diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must say that merely learning whether such Galaxies exist or not, would also be extremely helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another great alternative would be a Galaxy of any shape as long as it's less than 500 L.y. in Diameter and at least 500 000 L.y. away from another Galaxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The perfect example however would be a Dwarf Spheroidal or a Dwarf Irregular below 500 L.y. in Diameter, at about the average distance away from another Galaxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a few lists of Dwarf Galaxies, but they are lacking details regarding the sizes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="23236" LastActivityDate="2018-05-31T16:01:48.507" Title="Looking for a Dwarf Galaxy &lt; 500 L.y. in Diameter, which is not a Satellite / Companion, or is at least 500 000 L.y. away from another Galaxy" Tags="&lt;galaxy&gt;&lt;galactic-dynamics&gt;&lt;intergalactic-space&gt;&lt;dwarf-galaxy&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26480" PostTypeId="2" ParentId="12868" CreationDate="2018-05-31T16:17:50.793" Score="3" Body="&lt;p&gt;The statement that a Pulsar will look like a black body with a high temperature does is not supported by the evidence. Optical measurements of the Crab Pulsar show a flat spectrum see &lt;a href=&quot;https://www.eso.org/sci/publications/messenger/archive/no.83-mar96/messenger-no83-37-40.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;. This is a result of the optical emission being from synchrotron radiation rather than the hot surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The recent Gaia DR2 results include the Crab Pulsar as DR23403818172572314624 this has a BP-RP colour of 1.0494 which equates to a temperature of around 5,100 K from the DR2 HR diagram. This is very similar to the temperature shown in the DR2 data. This needs to be used with caution as the calibration is for a star with a 'Black Body' atmosphere rather than an 'atmosphere' radiating due to synchrotron Radiation. See &lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR-5?-info=XML&amp;amp;-out.add=.&amp;amp;-c=83.63306103515%2022.01449702667&amp;amp;-c.rs=0.1&amp;amp;-source=I/345/gaia2&amp;amp;Source===3403818172572314624&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; for the full DR2 data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't know how large the radiating 'atmosphere' is but a rough idea could be calculated from the DR2 data in the link above. However the parallax (distance) uncertainty is quite large so would need a better distance measurement.&lt;/p&gt;&#xA;" OwnerUserId="23237" LastEditorUserId="1891" LastEditDate="2018-06-01T02:23:17.967" LastActivityDate="2018-06-01T02:23:17.967" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26481" PostTypeId="2" ParentId="26478" CreationDate="2018-05-31T16:33:17.767" Score="1" Body="&lt;p&gt;The Wikipedia article says that Spica A has 8 times the luminosity of Spica B (presumably that's at maximum.)  Magnitude is -2.5*log&lt;sub&gt;10&lt;/sub&gt;(I/I&lt;sub&gt;0&lt;/sub&gt;). So if the combined magnitude is 0.97, the magnitude of component A is ~1.1 and the magnitude of component B is ~3.3.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-05-31T16:33:17.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26482" PostTypeId="2" ParentId="13081" CreationDate="2018-05-31T16:47:36.597" Score="0" Body="&lt;p&gt;The temperature of radiation emitted by a static black hole (without an accretion disc) is from pair production.  The black hole loses mass when a virtual (negative-energy) particle-antiparticle pair forms exactly on the event horizon.  One of the virtual particles falls into the black hole, separating the pair.  The other particle becomes real and is emitted as positive mass-energy, and the particle that falls into the singularity remains virtual as negative mass-energy.  The net result is that mass-energy conservation is preserved, and the black hole loses mass-energy by emitting a real particle.  Since the escape velocity below the event horizon is beyond the speed of light, any particles created inside the black hole (real or otherwise) would always immediately fall into the singularity.  So if you were able to safely cross the event horizon (which is speculative—some recent studies have hypothesized that there many be &quot;wall of fire&quot; at the event horizon), I would think that you should only be able to register absolute zero, with the exception of any incoming radiation raining down from above you.  The virtual negative-mass component of Hawking radiation should presumably be undetectable, and even if it was, it would register as being below absolute zero, in violation of thermodynamics (and energy conservation).  Note though that the &quot;temperature&quot; of the singularity itself can be thought of as being measured in Planck temperatures (10^32 kelvins), since spacetime breaks down at that point (the Planck scale of M-theory, where one Planck temperature is the equivalent of one Planck energy, or the Planck mass).&lt;/p&gt;&#xA;" OwnerUserId="23235" LastActivityDate="2018-05-31T16:47:36.597" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26483" PostTypeId="2" ParentId="26410" CreationDate="2018-05-31T17:06:59.853" Score="0" Body="&lt;p&gt;The planet beneath us is providing the majority of movement.  The simplest possible model to figure out where a star should be is to just rotate the celestial sphere by 360 degrees every 24 hours.  This is what a clockdrive on an equatorial mount does, to track a particular star's position as the planet rotates.  If you know a star's coordinates in right ascension and declination, you can convert this to altitude and azimuth by rotating the spherical coordinate system.  Over longer periods of time though, there are many other motions that have to be included for better accuracy.&lt;/p&gt;&#xA;" OwnerUserId="23235" LastActivityDate="2018-05-31T17:06:59.853" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26484" PostTypeId="2" ParentId="14475" CreationDate="2018-05-31T19:04:18.110" Score="1" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Is Io's orbit or rotation affected by its volcanism?&lt;/p&gt;&#xA;&lt;p&gt;The rotations of comets and asteroids are affected by outgassing volatilities. Io is very volcanically active. Does this give Io a varying orbit and a slow rotation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Jupiter's moon &lt;a href=&quot;https://photojournal.jpl.nasa.gov/catalog/?IDNumber=PIA09257&quot; rel=&quot;nofollow noreferrer&quot;&gt;Io&lt;/a&gt; has a &lt;a href=&quot;https://pubs.usgs.gov/sim/3168/&quot; rel=&quot;nofollow noreferrer&quot;&gt;lot of volcanoes&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;A similar answer to your question is provided at Quora's question: &amp;quot;&lt;a href=&quot;https://www.quora.com/Could-the-eruptions-of-all-the-volcanoes-on-Earth-simultaneously-knock-it-off-its-rotation-or-change-its-orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Could the eruptions of all the volcanoes on Earth simultaneously knock it off its rotation or change its orbit?&lt;/a&gt;&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;There are two differences: Eruptions on Io can reach &lt;a href=&quot;https://en.wikipedia.org/wiki/Escape_velocity&quot; rel=&quot;nofollow noreferrer&quot;&gt;escape velocity&lt;/a&gt; (&lt;a href=&quot;https://en.wikipedia.org/wiki/Io_(moon)&quot; rel=&quot;nofollow noreferrer&quot;&gt;2.558 km/s&lt;/a&gt;), and matter is subject to gravity as it is propelled away from the moon allowing some to escape and some to return; so it is a source of wobbling and rotational speed changes.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Jupiter's rocky moon Io is the &lt;a href=&quot;https://solarsystem.nasa.gov/moons/jupiter-moons/io/in-depth/&quot; rel=&quot;nofollow noreferrer&quot;&gt;most volcanically active&lt;/a&gt; world in the solar system, with hundreds of volcanoes, some erupting lava fountains dozens of miles (or kilometers) high. Io’s remarkable activity is the result of a tug-of-war between Jupiter's powerful gravity and smaller but precisely timed pulls from two neighboring moons that orbit farther from Jupiter – Europa and Ganymede.&lt;/p&gt;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Orbit and Rotation&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Although Io always points the same side toward Jupiter in its orbit around the giant planet, the large moons Europa and Ganymede perturb Io's orbit into an irregularly elliptical one. Thus, in its widely varying distances from Jupiter, Io is subjected to tremendous tidal forces.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;These forces cause Io's surface to bulge up and down (or in and out) by as much as 330 feet (100 meters). Compare these tides on Io's solid surface to the tides on Earth's oceans. On Earth, in the place where tides are highest, the difference between low and high tides is only 60 feet (18 meters), and this is for water, not solid ground.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Io's orbit, keeping it at more or less a cozy 262,000 miles (422,000 kilometers) from Jupiter, cuts across the planet's powerful magnetic lines of force, thus turning Io into a electric generator. Io can develop 400,000 volts across itself and create an electric current of 3 million amperes. This current takes the path of least resistance along Jupiter's magnetic field lines to the planet's surface, creating lightning in Jupiter's upper atmosphere.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Surface&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The tidal forces generate a tremendous amount of heat within Io, keeping much of its subsurface crust in liquid form seeking any available escape route to the surface to relieve the pressure. Thus, the surface of Io is constantly renewing itself, filling in any impact craters with molten lava lakes and spreading smooth new floodplains of liquid rock. ...&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Two-body_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;two-body problem&lt;/a&gt; explains how multiple interacting objects affect the orbit of each other. While the &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_eccentricity&quot; rel=&quot;nofollow noreferrer&quot;&gt;orbital eccentricity&lt;/a&gt; of Io is quite low (&lt;a href=&quot;https://en.wikipedia.org/wiki/Io_(moon)&quot; rel=&quot;nofollow noreferrer&quot;&gt;0.0041&lt;/a&gt;) the orbits of three of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Galilean_moons#Orbit_animation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Galilean moons&lt;/a&gt; (Ganymede, Europa and Io, excluding Callisto) have a &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Orbital_resonance&quot; rel=&quot;nofollow noreferrer&quot;&gt;1:2:4 resonance&lt;/a&gt;. The &amp;quot;pushing&amp;quot; will act in periodic repetition to have a cumulative effect on the motion.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vFgvf.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vFgvf.gif&quot; alt=&quot;Orbits of the Galilean moons&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Orbital resonances greatly enhance the mutual gravitational influence of the bodies, i.e. their ability to alter or constrain each other's orbits. In most cases, this results in an unstable interaction, in which the bodies exchange momentum and shift orbits until the resonance no longer exists. Under some circumstances, a resonant system can be stable and self-correcting, so that the bodies remain in resonance.&lt;/p&gt;&#xA;&lt;p&gt;The small effect of the volcanoes is corrected by the much greater effect of resonance. The website space.com estimates Io's age at about &lt;a href=&quot;https://www.space.com/16419-io-facts-about-jupiters-volcanic-moon.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;4.5 billion years&lt;/a&gt; old, the same age as Jupiter. If the orbit was unstable and changing it may well have been destroyed long ago.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-05-31T19:04:18.110" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26486" PostTypeId="1" CreationDate="2018-06-01T01:21:16.303" Score="1" ViewCount="192" Body="&lt;p&gt;I have two column data as magnitude and period of several stars. I have plotted magnitude vs. period, but I need to add a column as error bar for Y values. How can I write the code to put the error bars on data points?&lt;/p&gt;&#xA;" OwnerUserId="23239" LastEditorUserId="1891" LastEditDate="2018-06-01T03:56:41.417" LastActivityDate="2018-06-01T03:56:41.417" Title="Error bar in Python" Tags="&lt;python&gt;&lt;absolute-magnitude&gt;" AnswerCount="0" CommentCount="6" ClosedDate="2018-06-02T04:23:36.117" ContentLicense="CC BY-SA 4.0" />
  <row Id="26487" PostTypeId="2" ParentId="8675" CreationDate="2018-06-01T04:23:39.833" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Q: &quot;Is there a cosmic, rather than technological, upper limit to what a telescope can resolve?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Space radio interferometers could have a baseline of millions of kilometers, but is there a point where a larger baseline doesn't improve the resolution anymore because the photons observed are distorted before they arrive?'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Deviation of the wavefront from a blackbody spectrum and the processes that create them is reasonably well understood, the phases of the incoming wavefronts can be measured to enhance the directionality and reduce the distortion. Correlation of the different frequencies allows detection of even small amounts of distortion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See: &quot;&lt;a href=&quot;https://arxiv.org/abs/1109.6552&quot; rel=&quot;nofollow noreferrer&quot;&gt;The evolution of CMB spectral distortions in the early Universe&lt;/a&gt;&quot; (29 Sept 2011), by J. Chluba, R.A. Sunyaev and the Video: &quot;&lt;a href=&quot;https://www.perimeterinstitute.ca/videos/spectral-distortions-cmb-and-what-we-might-learn-about-early-universe-physics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Spectral distortions of the CMB and what we might learn about the early universe&lt;/a&gt;&quot;, by Jens Chluba, and &quot;&lt;a href=&quot;https://arxiv.org/abs/1405.6938&quot; rel=&quot;nofollow noreferrer&quot;&gt;Science with CMB spectral distortions&lt;/a&gt;&quot; (27 May 2014), by Jens Chluba.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method is explained in the paper &quot;&lt;a href=&quot;https://arxiv.org/abs/astro-ph/0008134&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cosmic Microwave Background constraint on residual annihilations of relic particles&lt;/a&gt;&quot; (9 Aug 2000), by McDonald, Scherrer, and Walker, page 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;In this paper we compute the energy injected into the CMB by annihilating particles as a function of their mass and annihilation rate (i.e., the product of cross section and abundance squared). We derive constraints on the particle properties by comparison with the observed limits on chemical potential ($\mu$) distortions, and Compton-$y$ distortions (§II). We compare these constraints to similar constraints obtained from the production of deuterium by photodissociation of primordial helium (§III.A), and from the diffuse photon background produced after recombination by extragalactic annihilations (§III.B), and annihilations in the Milky Way halo (§III.C).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;II. DISTORTIONS OF THE CMB ENERGY SPECTRUM&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We consider first the effect of annihilation products on the CMB energy spectrum. The distortion of the spectrum takes place in two steps: first the high energy annihilation products rapidly dissipate their energy into the background photons and electrons, and then the low energy background evolves more slowly in an effort to restore the Planck spectrum. The permanence of distortions produced after $z ≃ 10^6$ is simple to understand in the following way: A Planck spectrum with a given photon number density must have a specific energy density. For $z ≃ 10^6$, photon nonconserving processes (double Compton scattering and bremsstrahlung) are inefficient&#xA;  in the background plasma. Therefore, if energy is injected into the CMB but not the the correct number of photons, a Planck spectrum cannot be restored. We now discuss in more detail the form of the distortions pro-&#xA;  duced in different redshift intervals. ...'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See: &quot;&lt;a href=&quot;https://public.nrao.edu/news/2018-cdl-paf/#PRimage0&quot; rel=&quot;nofollow noreferrer&quot;&gt;New Technology Offers to Broaden Vision for Radio Astronomy&lt;/a&gt;&quot;. Even a tiny antenna is very capable, this &lt;em&gt;tiny&lt;/em&gt; antenna can see in seven directions simultaneously:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://public.nrao.edu/wp-content/uploads/2018/04/DSC01858.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BY6Gv.jpg&quot; alt=&quot;19 Element Phased Array Feed&quot;&gt;&lt;/a&gt;&#xA;&lt;sup&gt;[Click image to zoom]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adaptive optics are used with some optical telescopes, no reason not to apply the same algorithm to VLBI.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;a href=&quot;https://www.laserfocusworld.com/articles/print/volume-52/issue-07/features/adaptive-optics-eso-s-very-large-telescope-sees-four-times-first-laser-light.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Adaptive Optics: ESO's Very Large Telescope sees four times first (laser) light&lt;/a&gt;&quot; (07/11/2016):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;... even at the best optical sites on Earth such as the 2600 m summit of Cerro Paranal or the 4205 m summit of Mauna Kea in Hawaii, refractive-index variations in the atmosphere lead to a &lt;strong&gt;seeing-limited resolution of 0.4 arcsec in contrast to the theoretical diffraction-limited resolution of less than 0.02 arcsec for the 8.2 m VLT telescopes&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;a href=&quot;https://arxiv.org/abs/1705.02067&quot; rel=&quot;nofollow noreferrer&quot;&gt;PKS 1954-388: RadioAstron Detection on 80,000 km Baselines and Multiwavelength Observations&lt;/a&gt;&quot; (5 May 2017), by Edwards, Kovalev, Ojha, An, Bignall, Et al.:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;strong&gt;1 INTRODUCTION&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A major challenge in astronomy is the struggle to observe objects with an angular resolution sucient to probe the underlying physical mechanisms. The longer wavelengths of radio-astronomy initially made the quest for high angular resolution more difficult, but the relative ease of preserving phase information enabled the technique of Very Long Baseline Interferometry (VLBI). Intercontinental VLBI &lt;strong&gt;routinely achieves milliarcsecond-scale angular resolutions&lt;/strong&gt;, and extending the baselines between telescopes into space, with satellitebased telescopes, currently yields the highest angular resolution achieved in astronomy.&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's a newer paper than the one you linked to, using a longer baseline, with better results. See: &quot;&lt;a href=&quot;https://arxiv.org/abs/1609.04008&quot; rel=&quot;nofollow noreferrer&quot;&gt;PSR B0329+54: Substructure in the scatter-broadened image discovered with RadioAstron on baselines up to 330,000 km&lt;/a&gt;&quot; (13 Sept 2016), by Popov, Bartel, Gwinn, Johnson, Andrianov, Fadeev, Et al., the conclusion on page 8:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;strong&gt;7 SUMMARY AND CONCLUSIONS&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Here we summarize our observations and results and give our conclusions.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(i) We made VLBI observations of PSR B0329+54 with RadioAstron at 324 MHz on projected baselines up to 330,000 km or 350 M$\lambda$. Our goal was to investigate scattering properties of the ISM which affect radio observations of all celestial sources. While the results of such observations are in general influenced by the convolution of source structure with the scattering processes, pulsars are virtually&#xA;  point-like sources and signatures in the observational results can be directly related to the ISM scattering properties.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(ii) Visibility function at short ground-ground baselines manifests a single bright spike in delay-rate space that vanishes on long space-ground baselines. Thus, the scattering disk of PSR B0329+54 was completely resolved on ground-space baselines of 15,000 to 30,000 km. The FWHM of the&#xA;  angular diameter is 4.8 ± 0.8 &lt;a href=&quot;https://en.wiktionary.org/wiki/milliarcsecond&quot; rel=&quot;nofollow noreferrer&quot;&gt;mas&lt;/a&gt; at 324 Hz.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(iii) The diffractive length scale or size of the diffraction spot near Earth is 17, 000 ± 3, 000 km.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(iv) With the assumption of turbulent and large-scale irregularities in the plasma, the effective scattering screen is located at d/D = 0.6 ± 0.1 or somewhat more than half of the distance from Earth to the pulsar.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(v) &lt;strong&gt;At longer projected baselines, up to 330,000 km, significant visibility amplitudes were detected, although none were expected from the scattering disk&lt;/strong&gt;. They are scattered around a mean which stays approximately constant up to the longest baselines. &lt;strong&gt;This result indicates that substructure was discovered in the scatter-broadened image of PSR B0329+54&lt;/strong&gt;.&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your last question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Q: &quot;I'm instead asking about cosmic limitations due to for example interstellar and extragalactic gas which scatters light.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because you are looking at so many frequencies from so many angles for an extended period of time it's possible to use a supercomputer to see through (around) atoms, fog, dust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some layperson's explanations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ted Talks video: &quot;&lt;a href=&quot;https://www.ted.com/talks/michael_rubinstein_see_invisible_motion_hear_silent_sounds_cool_creepy_we_can_t_decide&quot; rel=&quot;nofollow noreferrer&quot;&gt;See invisible motion, hear silent sounds&lt;/a&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CNN YouTube: &quot;&lt;a href=&quot;https://www.youtube.com/watch?v=Rqi3jpBSyCc&quot; rel=&quot;nofollow noreferrer&quot;&gt;New technology makes troops invisible&lt;/a&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ascendent Technology YouTube &quot;&lt;a href=&quot;https://www.youtube.com/watch?v=pF85ALXyNDw&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thermal Infrared &amp;amp; Visible CCTV PTZ Camera Fire Monitoring &amp;amp; Detecting See Through Smoke &amp;amp; Haze&lt;/a&gt;&quot;&lt;/p&gt;&#xA;" OwnerUserId="18054" LastActivityDate="2018-06-01T04:23:39.833" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26488" PostTypeId="1" CreationDate="2018-06-01T05:04:31.803" Score="0" ViewCount="129" Body="&lt;p&gt;ok so I am a total amateur when it comes to telescopes and lens, and this is probably a pretty stupid question, but I am trying to find a lens/eyepiece that is good for viewing Jupiter and Saturn up close, and I have no idea what I am looking for, I have an Orion Sky Quest XT8 plus reflector telescope, (not sure if that matters) but if someone knows what I am looking for pls lmk, thx Noelle:)  &lt;/p&gt;&#xA;" OwnerUserId="23241" LastActivityDate="2018-06-01T07:41:23.227" Title="what kind of telescope eyepieces is best for viewing planets up close" Tags="&lt;telescope-lens&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26489" PostTypeId="2" ParentId="26488" CreationDate="2018-06-01T07:34:36.643" Score="1" Body="&lt;p&gt;To get good views of Jupiter and Saturn you need quite high magnification - probably 100 times or more. The XT8 has a focal length of 1200 mm and so to get that magnification will need an eyepiece of 12 mm focal length. I believe the scope comes with a 10mm eyepiece which will give 120 times magnification, which should give decent views. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could go for higher magnification with shorter focal length eyepieces, but if you do you will find that the object you are observing scoots across the field of view rather quickly and you will spend more time moving the telescope than actually observing.  As you get more experienced you may find you get more adept at moving the scope, and so could use more magnification. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It’s important that the mirrors of the scope are properly aligned (collimated) so as to get the sharpest views. The manual should tell you how to do that. &lt;/p&gt;&#xA;" OwnerUserId="11822" LastEditorUserId="11822" LastEditDate="2018-06-01T07:41:23.227" LastActivityDate="2018-06-01T07:41:23.227" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26490" PostTypeId="1" CreationDate="2018-06-01T11:34:30.760" Score="3" ViewCount="668" Body="&lt;p&gt;The exact question goes like this:&#xA;In the real Universe the expansion is not completely uniform. Rather, galaxies exhibit&#xA;some random motion relative to the overall Hubble expansion, known as their&#xA;peculiar velocity and caused by the gravitational pull of their near neighbours. Supposing&#xA;that a typical (e.g. root mean square) galaxy peculiar velocity is 600 km s-1,&#xA;how far away would a galaxy have to be before it could be used to determine the&#xA;Hubble constant to ten per cent accuracy, supposing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) The true value of the Hubble constant is 100 km s-1 Mpc-1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) The true value of the Hubble constant is 50 km s-1 Mpc-1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume in your calculation that the galaxy distance and redshift could be measured&#xA;exactly. Unfortunately, that is not true of real observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started off by writing the hubbles equation and as they mentioned it should be accurate to 10% so its peculiar velocity (given as about 600 km/s) is less than or equal to 10% of its expansion velocity. But this gives me different answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual answer to the problem as stated in the book is&#xA;&lt;em&gt;Slightly different answers are possible depending how you deal with the rms velocity.&#xA;You should get something like r &gt; 35 Mpc for H0 = 100km s-1 and r &gt; 70Mpc&#xA;for Ho = 50kms-1&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing here? Has it got to do with the RMS velocity as mentioned? Please explain. &lt;/p&gt;&#xA;" OwnerUserId="23244" LastEditorUserId="23748" LastEditDate="2019-06-20T14:39:24.990" LastActivityDate="2019-06-20T14:39:24.990" Title="Question from Introduction to Modern Cosmology by Andrew Liddle" Tags="&lt;galaxy&gt;&lt;cosmology&gt;&lt;distances&gt;&lt;redshift&gt;&lt;hubble-constant&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26491" PostTypeId="2" ParentId="26490" CreationDate="2018-06-01T16:12:32.340" Score="3" Body="&lt;p&gt;From the definition of the rms (e.g. &lt;a href=&quot;https://en.wikipedia.org/wiki/Root_mean_square#Relationship_to_other_statistics&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;),&#xA;$$&#xA;\mathrm{rms}^2(x) = \langle x \rangle^2 + \sigma_x^2,&#xA;$$&#xA;where $\langle x \rangle$ is the mean value, and $\sigma_x$ is the dispersion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For galaxies with random velocities, the &lt;em&gt;mean&lt;/em&gt; velocity should be $\langle x \rangle = 0$, unless they drift in some direction. Hence, the velocity dispersion should be $600\,\mathrm{km}\,\mathrm{s}^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For randomly distributed velocities, the speed $V_{||}$ you measure along your line of sight (LOS) is a factor $\sqrt{3}$ smaller, i.e. $\sim350\,\mathrm{km}\,\mathrm{s}^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, to get a 10% accuracy, the mean value should be 10 times larger, i.e. $\sim 3500\,\mathrm{km}\,\mathrm{s}^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a Hubble constant of $H_0 = 100\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$, that corresponds to a distance of&#xA;$$&#xA;d = \frac{v_{||}}{H_0} \simeq 35\,\mathrm{Mpc},&#xA;$$&#xA;while for $50 \,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$ you get $70\,\mathrm{Mpc}$.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-06-03T12:56:00.767" LastActivityDate="2018-06-03T12:56:00.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26492" PostTypeId="1" CreationDate="2018-06-02T10:09:50.237" Score="0" ViewCount="83" Body="&lt;p&gt;&lt;strong&gt;Poetic human centric approach:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Universe/Laniakea/Virgo/MilkyWay/Sol/Earth/Humans/AndreasT&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Little bit more technical:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Universe/0/SuperCluster/0/GalaxyCluster/0/Galaxy/0/StarCluster/0/Star/0/Planet/3/../HomoSapiensSapiens/2834929849289&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;More ... multiversal?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is there something in the spirit of the rebasing of SI units to universal concepts that makes us not be the origin(hah!) of the system?&lt;/p&gt;&#xA;" OwnerUserId="23251" LastActivityDate="2018-06-03T20:30:49.580" Title="What is the official classification and naming system of objects in the universe(, or what would my URI look like)?" Tags="&lt;star-systems&gt;&lt;naming&gt;&lt;ephemeris&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26493" PostTypeId="1" CreationDate="2018-06-02T10:17:38.647" Score="2" ViewCount="123" Body="&lt;p&gt;After the Big bang occurred, was the space filled with hot gas totally?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there's any problem in my question please inform me. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="22722" LastEditorUserId="22722" LastEditDate="2018-06-02T10:55:31.173" LastActivityDate="2018-06-03T21:14:17.030" Title="Just after the Big Bang, was the space, in its entirety, filled with Hot gas?" Tags="&lt;big-bang-theory&gt;&lt;plasma-physics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26494" PostTypeId="2" ParentId="26492" CreationDate="2018-06-02T10:48:35.580" Score="2" Body="&lt;p&gt;Objects aren't named like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either objects are given &quot;names&quot; that have no structure. For example, an asteroid may be named &quot;Vesta&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or they are given essentially meaningless catalogue numbers. For example, a star might be catalogued as HD 138987. It is number 138987 in the Henry Draper catalogue. Some major stars do have their constellation as part of the name. Alpha Centauri is the brightest star in the Centaurus constellation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comets are often just named according to the year of discovery &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other objects are named by their position in the sky, using polar coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no URI-like system in use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this is that there is no natural hierarchy. When naming species of animal, they naturally group into genus, family, order, class as a result of evolution. No such structure is found among astronomical objects, so a hierarchical approach is not appropriate.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-06-02T10:48:35.580" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26495" PostTypeId="2" ParentId="26493" CreationDate="2018-06-02T10:51:31.863" Score="3" Body="&lt;p&gt;I'm not sure what you mean by &quot;nebulae&quot;.  You could mean &quot;matter in a gaseous or plasma state&quot; or &quot;cloud of low density gas and dust in space&quot; (which is what astronomers mean by nebulae.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The former case is believed to be correct, and it &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Bang&quot; rel=&quot;nofollow noreferrer&quot;&gt;pretty well described in Wikipedia&lt;/a&gt;. I'll summarize. Immediately after the Big Bang (10&lt;sup&gt;-20&lt;/sup&gt; seconds or so) the universe was full of extremely hot and extremely dense matter -- hot enough that not only did atoms come apart, and nuclei come apart, but even protons and neutrons came apart into a quark-gluon plasma. This cooled into a stew of elementary particles, which cooled into mostly protons, neutrons and electrons, which cooled partly into nuclei. At this point we're maybe a minute into the expansion and the universe was full of a plasma that was still hotter than the center of a star. After around 300,000 years, the plasma cooled enough to recombined into atoms. Stars eventually started to form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No astronomer would call &lt;em&gt;any&lt;/em&gt; stage of this expansion a &quot;nebula&quot; -- they first formed later after the atomic gas left a milion years after the Big Bang started to condense into stars and patches of gas and patches close to vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line: Nebulae did not exist until millions of years after the Big Bang, but the universe was filled with hot gas much earlier in the Big Bang.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-06-02T10:51:31.863" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26496" PostTypeId="2" ParentId="24275" CreationDate="2018-06-02T14:15:35.463" Score="1" Body="&lt;p&gt;The measurements are of distance from one particular place on Earth, to one particular place on the Moon. But these places move around relative to the center of the Earth or moon ... the Earth surface &quot;flexes&quot; due to various tidal forces from the Moon, sun, and other planets, and so does the moon to a lesser degree. In addition, the moon doesn't orbit the Earth in a nice perfect circle. Then you need to consider that the speed of light through the atmosphere is not constant (varies with weather).  Also, the measurement instruments have a lot of noise and jitter (it's a lot compared to the precision &amp;amp; accuracy of the measurements).&lt;/p&gt;&#xA;" OwnerUserId="23252" LastActivityDate="2018-06-02T14:15:35.463" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26497" PostTypeId="2" ParentId="26393" CreationDate="2018-06-02T14:31:03.163" Score="0" Body="&lt;p&gt;I suspect that converting red-shift to distance, for any &lt;em&gt;particular&lt;/em&gt; object, is going to have a lot of error. Especially for z &gt; 1. Even for 0.05 &amp;lt; z &amp;lt; 1, the errors are often over 10%.&lt;/p&gt;&#xA;" OwnerUserId="23252" LastActivityDate="2018-06-02T14:31:03.163" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26498" PostTypeId="1" CreationDate="2018-06-02T14:39:29.323" Score="5" ViewCount="176" Body="&lt;p&gt;There are various good models for solid-body tides on the Earth. Are there any recommendations for tides on the Moon?&lt;/p&gt;&#xA;" OwnerUserId="23252" LastActivityDate="2018-06-04T09:15:11.607" Title="Solid-body tides on the Moon" Tags="&lt;the-moon&gt;&lt;tidal-forces&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26500" PostTypeId="1" AcceptedAnswerId="26502" CreationDate="2018-06-03T07:58:18.237" Score="8" ViewCount="564" Body="&lt;p&gt;As far as I was aware, stars don't usually have lithium, but it has been observed that there is plenty of lithium on our star's surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, a team lead by Garik Israelian took a census of &lt;em&gt;500 stars&lt;/em&gt;, &lt;em&gt;70&lt;/em&gt; of which are &lt;em&gt;known to host planets&lt;/em&gt; and about a &lt;em&gt;quarter&lt;/em&gt; of which are &lt;em&gt;particularly Sun-like&lt;/em&gt;. Using ESO’s HARPS spectrograph, this team of astronomers (see &lt;a href=&quot;https://doi.org/10.1038/nature08483&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;) found that the Sun-like stars which host planets have destroyed their lithium much more efficiently than “planet-free” stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do Sun-like stars have lithium on their surfaces and why do stars that host planets destroy their lithium more efficiently?&lt;/p&gt;&#xA;" OwnerUserId="23217" LastEditorUserId="6" LastEditDate="2018-06-04T14:04:58.150" LastActivityDate="2018-06-04T14:04:58.150" Title="Why does the Sun's surface contain lithium?" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26502" PostTypeId="2" ParentId="26500" CreationDate="2018-06-03T14:13:41.120" Score="13" Body="&lt;p&gt;The Sun has lithium because, like all the elements that it has heavier than helium, it was born with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many stars, particularly older metal-rich stars or cool stars with deep convective envelopes show very little Li in their atmosphere - and that is true for the Sun too. But this trace of lithium is still detectable (just).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By comparison with meteorites, we know that the Sun was born with about 100 times the abundance of Li that is seen in its photosphere now. Roughly speaking, about 20% of this Li was produced by big-bang nucleosynthesis and the rest was likely produced, before the Sun was born, in (giant) stars and (classical) novae via rather complicated synthesis and mixing routes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the Sun's Li has been destroyed in its interior. It reacts with protons to form helium at temperatures lower than required for hydrogen fusion. However, the exterior of the Sun has not been completely mixed with the interior, so a little still remains in the photosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to why stars with planets deplete their Li more efficiently, well there is still a bit of a question mark over whether that is true or not. If it is, then it probably has something to do with the history of rotation and angular momentum loss in the star. Stars with long-lived, planet-forming discs may have spun down more readily, creating an internal differential rotation that aids the mixing process. It is a clear observational result, that slower-rotating stars in young clusters have depleted more Li than their faster rotating siblings.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-06-03T14:19:00.757" LastActivityDate="2018-06-03T14:19:00.757" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26504" PostTypeId="1" AcceptedAnswerId="26506" CreationDate="2018-06-03T20:20:16.600" Score="5" ViewCount="368" Body="&lt;p&gt;I'm new to amateur astronomy, and I recently acquired my first small telescope. I've just learned a bit about surface brightness after an unsuccessful night of trying to locate a few nebulae that were beyond my scope's abilities, but I'm wondering if it's as cut-and-dry as I've understood so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I'm asking is that when I look up the surface brightness of the Great Orion Nebula on Stellarium, it shows up at 13.07, a bit beyond the limiting stellar magnitude of my scope at 11.7. However, I am pretty confident that I was able to see the Orion Nebula with my scope on another occasion. It's a large enough object that I'm pretty sure I observed it, and the features looked correct (but faint and colorless) when compared to high-detail photos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to get to the point of my question, are there other things I should keep in mind when determining whether I'll be able to see an object with my scope? I would like to avoid more frustrating nights in the future by not picking impossible targets. I also recognize that as an amateur I might have seen something completely different than the Orion Nebula.&lt;/p&gt;&#xA;" OwnerUserId="23256" LastActivityDate="2018-06-04T05:19:31.877" Title="Surface brightness of deep-sky objects and telescope specs" Tags="&lt;telescope&gt;&lt;deep-sky-observing&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26505" PostTypeId="2" ParentId="26492" CreationDate="2018-06-03T20:22:18.497" Score="1" Body="&lt;p&gt;There have been a few discussions and opinions offered:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://sherwoodharrington.com/CosAddr.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;DeAnza College Planetarium&lt;/a&gt; gives it's &lt;em&gt;cosmic address&lt;/em&gt; as: Fujitsu Planetarium,  Anza College, Stevens Creek Boulevard, Cupertino, California, U.S.A., Planet Earth, Earth-Moon System, Solar System, Milky Way Galaxy, Local Group, Virgo Supercluster, Observable Universe. - That's somewhat commonly used (or variations thereof), and not a recent / accurate location reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more creditable, certainly shorter, take is offered by Nature.com in their article: &quot;&lt;a href=&quot;https://www.nature.com/news/earth-s-new-address-solar-system-milky-way-laniakea-1.15819&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth's new address: &lt;strong&gt;'Solar System, Milky Way, Laniakea'&lt;/strong&gt;&lt;/a&gt;&quot; and their paper &quot;&lt;a href=&quot;https://www.nature.com/articles/nature13674&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Laniakea supercluster of galaxies&lt;/a&gt;&quot; (4 Sept 2014), by Tully, Courtois, Hoffman and Pomarède - Free @ &lt;a href=&quot;https://arxiv.org/abs/1409.0880&quot; rel=&quot;nofollow noreferrer&quot;&gt;arXiv&lt;/a&gt;, 2nd &lt;a href=&quot;https://www.researchgate.net/publication/265337589_The_Laniakea_supercluster_of_galaxies&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Opinion supported by this National Radio Astronomy Observatory news release: September 3, 2014 at 5:00 pm EDT, &quot;&lt;a href=&quot;https://public.nrao.edu/news/supercluster-gbt/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Newly Identified Galactic Supercluster Is Home to the Milky Way&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 'more technical' approach portion of your question involves &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Human_taxonomy&quot; rel=&quot;nofollow noreferrer&quot;&gt;human taxonomy&lt;/a&gt;, we are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{array}{ll}&#xA;Kingdom: &amp;amp; Animalia \\&#xA;Phylum: &amp;amp; Chordata \\&#xA;Class: &amp;amp; Mammalia \\&#xA;Order: &amp;amp; Primates \\&#xA;Suborder: &amp;amp; Haplorhini \\&#xA;Infraorder: &amp;amp; Simiiformes \\&#xA;Family: &amp;amp; Hominidae \\&#xA;Subfamily: &amp;amp; Homininae \\&#xA;Tribe: &amp;amp; Hominini \\&#xA;Genus: &amp;amp; Homo \; (Linnaeus, 1758) \\&#xA;\end{array}$&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="18054" LastEditDate="2018-06-03T20:30:49.580" LastActivityDate="2018-06-03T20:30:49.580" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26506" PostTypeId="2" ParentId="26504" CreationDate="2018-06-03T20:51:24.387" Score="3" Body="&lt;p&gt;Although you have not given any information about your telescope, the Orion nebula is easy to see with almost any telescope, even under urban light polluted skies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surface brightness is almost always greater (ie fainter) than magnitude - so for example the total magnitude of the Orion nebula is about 4 (from memory) but you say surface brightness of 13.  The reason is that the area of the Orion nebula is quite large, at least when measured in arc seconds squared, and the eye/brain has some ability to integrate the light.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works both ways: there are objects, eg M33 and M101 in my experience, which have a magnitude that would make you think they are easy targets for small telescopes, but because the brightness is spread over a large area, they are difficult to see from a typical urban or suburban location.  The important measure in these cases is to compare the surface brightness with that of the background sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's quite a complex topic, and Roger Clark's book &quot;Visual Astronomy of the Deep Sky&quot; has some useful discussion of this, and much more.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastEditorUserId="11822" LastEditDate="2018-06-04T05:19:31.877" LastActivityDate="2018-06-04T05:19:31.877" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26508" PostTypeId="1" CreationDate="2018-06-04T02:52:34.663" Score="1" ViewCount="34" Body="&lt;p&gt;A few days ago I saw more than 50 stars moving in a group at night in SW of Florida, what this could be satellites or something else?&lt;/p&gt;&#xA;" OwnerUserId="23259" LastEditorUserId="10196" LastEditDate="2018-06-04T12:10:18.567" LastActivityDate="2018-06-04T12:10:18.567" Title="Stars shape moving at night" Tags="&lt;identify-this-object&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2018-06-04T14:34:39.093" ContentLicense="CC BY-SA 4.0" />
  <row Id="26509" PostTypeId="2" ParentId="240" CreationDate="2018-06-04T08:29:19.743" Score="-4" Body="&lt;p&gt;What we consistantly overlook and are fooled by, is our earths power and own quirky magnetic self. Earth also has a pull. Therfor, your carefully calculated 29+ day moon cyle which varies-is incorrrct. By about 4 more fluctuating days. We dont see it from earth and it appears normal because the moln takes its usual time to appear, that is why we &quot;think&quot; we understand it.   :)&lt;/p&gt;&#xA;" OwnerUserId="23262" LastActivityDate="2018-06-04T08:29:19.743" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26510" PostTypeId="2" ParentId="26498" CreationDate="2018-06-04T08:54:14.030" Score="4" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Solid-body tides on the Moon&lt;/p&gt;&#xA;&lt;p&gt;There are various good models for solid-body tides on the Earth. Are there any recommendations for tides on the Moon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This answer starts with simplistic explanations and links with trivial calculations and progresses to links that, along with their references, provide exacting and lengthy approaches. @scb, you might want to skip halfway through the answer.&lt;/p&gt;&#xA;&lt;p&gt;To make accurate calculations one needs to know about various parameters of the moon: composition of the crust and &lt;a href=&quot;https://en.wikipedia.org/wiki/Internal_structure_of_the_Moon&quot; rel=&quot;nofollow noreferrer&quot;&gt;internal structure of the moon&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Love_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;Love numbers&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;precession&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_acceleration&quot; rel=&quot;nofollow noreferrer&quot;&gt;tidal acceleration&lt;/a&gt;, even &lt;a href=&quot;https://en.wikipedia.org/wiki/Time_dilation&quot; rel=&quot;nofollow noreferrer&quot;&gt;time dilation&lt;/a&gt; - each results in a tiny error if omitted, a larger error if you skip them all.&lt;/p&gt;&#xA;&lt;p&gt;An explanation of the anomalies of the Moon's crust is offered in Space.com's article: &amp;quot;&lt;a href=&quot;https://www.space.com/21364-moon-gravity-mascons-mystery.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mystery of Moon's Lumpy Gravity Explained&lt;/a&gt;&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;A particularly poor calculation is offered on this NASA webpage &amp;quot;&lt;a href=&quot;https://history.nasa.gov/SP-345/ch9.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;SP-345 Evolution of the Solar System - 9. SPIN AND TIDES&lt;/a&gt;&amp;quot;. It leaves out a lot (like &lt;a href=&quot;https://en.wikipedia.org/wiki/Love_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;Love numbers&lt;/a&gt;, precession, &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_acceleration&quot; rel=&quot;nofollow noreferrer&quot;&gt;tidal acceleration&lt;/a&gt;, etc.) providing a quick and easy calculation along with a bit of unknown inaccuracies.&lt;/p&gt;&#xA;&lt;p&gt;A short and quick explanation, less than you asked for, is provided by Space.com's article: &amp;quot;&lt;a href=&quot;https://www.space.com/26246-lunar-tide-seen-from-space.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Moon Bumps: Earth's Gravity Creates Lunar Bulges&lt;/a&gt;&amp;quot; which explains:&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Earth's gravitational pull is so powerful that it creates a small bulge on the surface of the moon. ... The corresponding distorting effect on the moon, called the lunar body tide, is more difficult to see, because the moon is solid except for a molten core. But &lt;strong&gt;Earth's pull raises a small bulge about 20 inches (50 centimeters) from the surface on the near side of the moon and a matching bulge on the far side&lt;/strong&gt;. [Rob's additional comment: they won't be equal.]&lt;/p&gt;&#xA;&lt;p&gt;The same side of the moon always faces Earth, but &lt;strong&gt;the bulge does move around a few inches over time, wobbling and following Earth's pull like a magnet&lt;/strong&gt;, as the moon shifts slightly during its orbit.&lt;/p&gt;&#xA;&lt;p&gt;Scientists observed the lunar body tide using &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_Reconnaissance_Orbiter#Results&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA's Lunar Reconnaissance Orbiter satellite&lt;/a&gt;, which is mapping the height of features on the moon's surface, and &lt;a href=&quot;https://en.wikipedia.org/wiki/GRAIL#Objectives&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA's Gravity Recovery and Interior Laboratory satellite&lt;/a&gt;, which is mapping the moon's gravitational field. &lt;strong&gt;The satellites measured the height of 350,000 locations spread across areas of the moon closest to Earth and areas of the moon on the opposite side from Earth. The satellites passed over each location several times, so scientists could compare the height of each spot from one satellite pass to the next. By identifying which spots changed height, the researchers could track the lunar tide.&lt;/strong&gt;&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;A detailed calculation of the shape of the moon with plenty of math is covered in: &amp;quot;&lt;a href=&quot;https://agupubs.onlinelibrary.wiley.com/doi/full/10.1002/2014GL061195&quot; rel=&quot;nofollow noreferrer&quot;&gt;Evidence for lunar true polar wander and a past low‐eccentricity, synchronous lunar orbit&lt;/a&gt;&amp;quot;. While covering a vast number of factors affecting the shape of the moon today body tides is amongst the points covered:&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;&lt;strong&gt;Abstract&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;As first noted 200 years ago by Laplace, the Moon's rotational and tidal bulges are significantly larger than expected, given the Moon's present orbital and rotational state. This excess deformation has been ascribed to a fossil figure, frozen in when the Moon was closer to the Earth. However, the observed figure is only consistent with an eccentric and non‐synchronous orbit, contrary to our understanding of the Moon's formation and evolution. Here, we show that lunar mascons and impact basins have a significant contribution to the observed lunar figure. Removing their contribution reveals a misaligned fossil figure consistent with an early epoch of true polar wander (driven by the formation of the South Pole‐Aitken Basin) and an early low‐eccentricity, synchronous lunar orbit. This new self‐consistent model solves a long‐standing problem in planetary science and will inform future studies of the Moon's dynamical evolution and early dynamo.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;1 Introduction&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The long‐wavelength lunar figure is triaxial, due to the combination of rotational deformation (which creates a rotational bulge) and tidal deformation (which creates a tidal bulge along the Earth‐Moon axis). However, it has been known since Laplace [1878] that the observed deformation is significantly larger than the predicted deformation assuming hydrostatic equilibrium and the Moon's current orbital and rotational state (Figure S1). This difference has been ascribed to the presence of a fossil figure, frozen in when the Moon was closer to the Earth and the rotational and tidal potentials were larger [Sedgwick, 1898; Jeffreys, 1915; Lambeck and Pullan, 1980]. More recent work by Garrick‐Bethell et al. [2006] has shown that the observed deformation not only requires freezing the fossil figure while on an orbit with a smaller semimajor axis, but also with significant eccentricity. Matsuyama [2013] extended this work by taking into account the effect of finite lithospheric rigidity; however, the required initial orbital eccentricity remains large.&lt;/p&gt;&#xA;&lt;p&gt;...&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;That article includes &lt;strong&gt;many&lt;/strong&gt; references which explain how the numerous equations were derived. As this illustration shows the article covers the development of the Moon's shape throughout history and attempts to calculate the minute bulging more exactly than the links referred to at the start of this answer.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dPGJs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dPGJs.jpg&quot; alt=&quot;Figure 3.&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;&lt;strong&gt;Figure 3.&lt;/strong&gt; Our proposed model for the formation of the Moon's fossil figure. (a) The Moon coalesced from the debris of giant impact. (b) Under the action of tides, the Moon migrated outward and cooled from a magma ocean, perhaps experiencing periods of non‐synchronous rotation or high eccentricity. (c) As the moon cools, it forms an elastic lithosphere capable of supporting long‐term deformation, resulting in the fossil figure. (d–e) The South Pole‐Aitken (SPA) basin forming impact occurs, resulting in ~15° of reorientation, placing SPA closer to the south pole. (f) Subsequent large impacts and mare volcanism occur but do not significantly alter the lunar figure. (g) Ultimately, the Moon migrates to its current orbital configuration.&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;Additional references:&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;&lt;a href=&quot;https://arxiv.org/abs/gr-qc/0411095&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lunar Laser Ranging Science&lt;/a&gt;&amp;quot; (18 Nov 2004), by Williams, Boggs, Turyshev, and Ratcliff.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.594.6995&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lunar Geophysics, Geodesy, and Dynamics&lt;/a&gt;&amp;quot; (&lt;a href=&quot;https://cddis.nasa.gov/lw13/lw_proceedings.html#lunar&quot; rel=&quot;nofollow noreferrer&quot;&gt;Homepage&lt;/a&gt;, direct &lt;a href=&quot;https://cddis.nasa.gov/lw13/docs/papers/sci_williams_1m.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;.PDF&lt;/a&gt; link) (2003), by Williams and Dicky. That paper contains this useful text:&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;The present lunar rotation model involves elastic solid body and fluid core. It will become more complicated in&#xA;the future. Numerically integrated physical librations (no fluid core) are available with DE403 at the JPL ephemeris&#xA;web site (&lt;a href=&quot;http://ssd.jpl.nasa.gov/horizons.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ssd.jpl.nasa.gov/horizons.html&lt;/a&gt;).&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;That website has moved, it's current address is: &lt;a href=&quot;https://ssd.jpl.nasa.gov/?horizons&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ssd.jpl.nasa.gov/?horizons&lt;/a&gt; - a bit of searching there turns up this URL: &lt;a href=&quot;https://ssd.jpl.nasa.gov/?lunar_doc&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lunar Constants and Models&lt;/a&gt; which says:&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Title:     Lunar Constants and Models Document&lt;/p&gt;&#xA;&lt;p&gt;Author:     Ralph B. Roncoli&#xA;Date:   September 23, 2005&lt;/p&gt;&#xA;&lt;p&gt;Reference:      JPL Technical Document D-32296&lt;/p&gt;&#xA;&lt;p&gt;Summary: &lt;strong&gt;The primary purpose of this document is to provide a single source for the constants and models to be used in the trajectory and navigation design of missions whose objective is to orbit or land on the Moon.&lt;/strong&gt; A secondary objective is to provide the mission analyst with some basic background information about the Moon, its orbit, and the previous missions that have explored the Moon. As a result, &lt;strong&gt;this document contains more information than the typical constants and models document&lt;/strong&gt;. Some of the data are required for mission studies while other data are simply provided for &amp;quot;educational purposes&amp;quot;. &lt;strong&gt;this document provides only brief descriptions of the constants and models. The user should consult the references if more detailed information is desired.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Format:     PDF&#xA;Link:   &lt;a href=&quot;https://ssd.jpl.nasa.gov/dat/lunar_cmd_2005_jpl_d32296.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;download here&lt;/a&gt;&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;So that last list of references is &amp;quot;build your own rocket and land on the moon&amp;quot; accurate.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-06-04T09:15:11.607" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26513" PostTypeId="1" CreationDate="2018-06-04T11:45:54.343" Score="4" ViewCount="335" Body="&lt;p&gt;I have searched in Google and found interesting articles for atmospheric refraction influence on sun's elevation above the horizon (from 0° till 90°).&#xA;eg. Bennett G.G. etc.&#xA;I would like to know if there is a formula that approximates the refraction when the sun's center is below the horizon (from 0° to ca. -3°) and taking in account for temperature and air pressure.&lt;/p&gt;&#xA;" OwnerUserId="16985" LastEditorDisplayName="user10106" LastEditDate="2018-06-04T12:31:58.317" LastActivityDate="2018-11-09T00:01:10.170" Title="Formula for sun's atmospheric refraction below the horizon" Tags="&lt;positional-astronomy&gt;&lt;atmospheric-effects&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26514" PostTypeId="1" CreationDate="2018-06-04T14:30:11.277" Score="5" ViewCount="148" Body="&lt;p&gt;I was reading &lt;a href=&quot;http://adsabs.harvard.edu/abs/2005ApJ...627..953W&quot; rel=&quot;nofollow noreferrer&quot;&gt;a paper&lt;/a&gt; on the differential emission measures of a set of hot O- and B- type stars. As the authors discuss in Section 3 (page 959), two stars, &lt;a href=&quot;https://en.wikipedia.org/wiki/Tau_Scorpii&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\tau$ Sco&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Theta1_Orionis_C&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\theta^1$ Ori C&lt;/a&gt;, have narrow emission lines compared to other stars of the same spectral type. There is a good amount of evidence that both stars are young and have magnetic fields; the stellar winds travel along the field lines and collide at the magnetic equator, producing shocks and, indirectly, X-ray emission. This is in contrast to the mechanism behind shock formation in other O- and B- type stars, where shocks come from the line instability transition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the plasma at the magnetic equator is roughly stationary because of the collision; this means - if I'm interpreting things correctly - that there is little to no broadening of the lines, and so they are comparatively narrow. Other spectral characteristics and observations support this model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Has the presence of such comparatively narrow lines in the spectra of O- and B- type stars been observed in cases where there is no magnetic field present? Very few O-type stars have significant magnetic fields, and theory predicts that none should, so these stars are the few exceptions to the rule.&lt;/li&gt;&#xA;&lt;li&gt;If so, are there other mechanisms that could be responsible for this in cases without magnetic fields? I'm grasping at straws for ideas, like low collisional broadening.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I talked with my advisor, and we briefly discussed colliding wind shocks (CWS), which form at the interface between the stellar winds in a binary system. It's thought that these shocks contribute only a small amount to the total X-ray production of the system (see &lt;a href=&quot;http://adsabs.harvard.edu/abs/2011ApJS..194....5G&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gagné et al. (2011)&lt;/a&gt;), though, so any narrow lines produced in that plasma - if any exist - wouldn't have much influence over the measured spectra. So that seems to be ruled out.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2018-06-13T14:54:27.937" LastActivityDate="2018-06-13T14:54:27.937" Title="Do narrow lines in the spectra of O- and B- type stars always indicate magnetic fields?" Tags="&lt;star&gt;&lt;magnetic-field&gt;&lt;spectra&gt;&lt;solar-wind&gt;&lt;x-ray&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26515" PostTypeId="2" ParentId="11835" CreationDate="2018-06-04T15:38:09.893" Score="0" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;The battle over Gliese 581 and Gliese 667C?&lt;/p&gt;&#xA;&lt;p&gt;Is there a direction that a majority of the professional community is leaning?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The &lt;a href=&quot;http://exoplanet.eu/team/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Exoplanets Team&lt;/a&gt; maintains the &amp;quot;&lt;a href=&quot;http://exoplanet.eu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Extrasolar Planets Encyclopaedia&lt;/a&gt;&amp;quot; which is updated daily.&lt;/p&gt;&#xA;&lt;p&gt;The UI is a bit fiddly on a mobile device but the raw data can be downloaded to desktop applications that are able to work with it. Links expire quickly so I can only post 'root URLs'.&lt;/p&gt;&#xA;&lt;p&gt;If you use the variable &lt;code&gt;star_name=&amp;quot;GJ 667 C&amp;quot;&lt;/code&gt; for the search term on the webpage: &lt;a href=&quot;http://exoplanet.eu/diagrams/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://exoplanet.eu/diagrams/&lt;/a&gt; you can create this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fIMX1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fIMX1.png&quot; alt=&quot;Plot of GJ 667 C Planet's Mass and Radius&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Use the search bar at the URL: &lt;a href=&quot;http://exoplanet.eu/catalog/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://exoplanet.eu/catalog/&lt;/a&gt; for this result:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FjP6Q.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FjP6Q.jpg&quot; alt=&quot;Catalog of GJ 667 C's Planets&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Wikipedia's information page on the: &lt;a href=&quot;https://en.wikipedia.org/wiki/Extrasolar_Planets_Encyclopaedia&quot; rel=&quot;nofollow noreferrer&quot;&gt;Extrasolar Planets Encyclopaedia&lt;/a&gt; website.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-06-04T15:38:09.893" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26517" PostTypeId="1" AcceptedAnswerId="26518" CreationDate="2018-06-04T16:17:15.217" Score="4" ViewCount="326" Body="&lt;p&gt;The interstellar asteroid 'Oumuamua was discovered on October 19, 2017. Once discovered, less than a month's worth of precovery observations were found on various sky surveys. By mid-November, scientists had worked out its orbital eccentricity, and soon after its probable origin and destination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With such a quick visit and so little data, how did scientists figure out so quickly where it came from and where it was going? Did using space probes for observations impart enough parallax to easily determine that, or did Earth-based observation have enough parallax themselves?&lt;/p&gt;&#xA;" OwnerUserId="17585" LastActivityDate="2018-06-06T23:50:21.653" Title="How did astronomers determine the path of 'Oumuamua so quickly?" Tags="&lt;asteroids&gt;&lt;interstellar&gt;&lt;oumuamua&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26518" PostTypeId="2" ParentId="26517" CreationDate="2018-06-04T18:21:28.353" Score="8" Body="&lt;p&gt;&lt;a href=&quot;https://www.stardust2013.eu/Portals/63/Images/Training/OTS%20Repository/gronchi_OTS2013.pdf&quot; rel=&quot;noreferrer&quot;&gt;Three accurate observations are sufficient to fix a Keplerian orbit&lt;/a&gt; (ie an elliptical or hyperbolic orbit with the sun at the focus)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, observations are not perfectly accurate due to limitations of the equipment and observations over a short time are particularly prone to observational error being magnified. Moreover the orbit will be perturbed by the gravity of the planets, so won't be perfectly Keplerian. For this reason pre-discovery images will help fix the exact orbit more accurately.&lt;br&gt;&#xA;As the time difference between first and last observation is critical in the quality of the orbital determination, we talk about the length (in days) of the observation arc as a measure of how well defined the orbit is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiple observations can reduce error by an averaging effect (the Gaussian curve is so named from Gauss's use of it in orbital determination) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However once you have three or more observations of a body you can determine its orbit almost immediately, using a computer to do the calculations for you. It would have been immediately obvious that the object was on a very unusual trajectory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no need to use space probes for the observations. The relative motion of the Earth and the Object provide sufficient parallax to determine distance, position and velocity.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-06-04T18:21:28.353" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26519" PostTypeId="1" AcceptedAnswerId="26520" CreationDate="2018-06-04T19:18:41.830" Score="5" ViewCount="151" Body="&lt;p&gt;Where do asteroids impact upon the Earth most often?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are impact zones primarily along the equator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an average area where impacts are more likely to occur?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a zone, is there a way to determine if it's changed? More to the point, if there is an impact zone, are there any similar areas that might indicate the zone has shifted?&lt;/p&gt;&#xA;" OwnerUserId="23274" LastEditorUserId="2365" LastEditDate="2018-06-04T20:10:27.090" LastActivityDate="2018-06-04T23:19:04.257" Title="Where do asteroids impact upon the Earth most often?" Tags="&lt;asteroids&gt;&lt;crater&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26520" PostTypeId="2" ParentId="26519" CreationDate="2018-06-04T20:08:02.157" Score="4" Body="&lt;p&gt;The occur all over the world.  On this image each yellow circle is an recent impact of a small asteroid the size of the circle is related to the energy of the fireball (the sizes of the circles don't represent crater areas). You will note that there is no region that get more impacts, and similarly, nowhere is safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The very large yellow circle is the Chelyabinsk meteor in Russia. It was the biggest such impact over land in recent times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vZM5n.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vZM5n.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-06-04T23:19:04.257" LastActivityDate="2018-06-04T23:19:04.257" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="26521" PostTypeId="2" ParentId="26464" CreationDate="2018-06-04T21:32:13.367" Score="2" Body="&lt;p&gt;In addition to @JohnHoltz correct answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've always found it useful to answer this type of question by considering extremes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're at the equator ($0$ latitude), where is the Celestial equator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're at the Pole ($90$ latitude), where, the Celestial equator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you move each degree from equator to pole, what happens to the elevation of the equator, and how fast?&lt;/p&gt;&#xA;" OwnerUserId="420" LastEditorUserId="420" LastEditDate="2018-06-05T02:43:01.400" LastActivityDate="2018-06-05T02:43:01.400" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26522" PostTypeId="2" ParentId="490" CreationDate="2018-06-04T22:32:46.857" Score="2" Body="&lt;p&gt;An important factor is that the comet had fallen through Jupiter's gravitational field, picking up Jupiter's escape velocity of about 60 &lt;span class=&quot;math-container&quot;&gt;${\rm km/s}$&lt;/span&gt; in addition to whatever velocity it had as it approached the Jupiter system. &lt;a href=&quot;https://www.lpi.usra.edu/meetings/lpsc97/pdf/1351.PDF&quot; rel=&quot;nofollow noreferrer&quot;&gt;A paper by D. A. Crawford from the Sandia National Laboratories&lt;/a&gt; gives detailed estimates of the sizes of all the fragments at impact. Three of them are estimated at 1km diameter or more, so they were not really small.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="37191" LastEditDate="2021-01-29T09:12:30.390" LastActivityDate="2021-01-29T09:12:30.390" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26524" PostTypeId="1" CreationDate="2018-06-05T08:03:45.793" Score="2" ViewCount="55" Body="&lt;p&gt;If meteorites and bolides are aligned to the solar system,  then do they have twice as much atmosphere to go through at high latitudes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wouldn't we expect a higher percentage of metal meteorites further from the equator,  because they disintegrate slower in prolonged atmospheric traversal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image from:&#xA;&lt;a href=&quot;https://academic.oup.com/mnras/article/359/2/551/987238&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://academic.oup.com/mnras/article/359/2/551/987238&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.skyandtelescope.com/observing/random-meteors-arent-random02112015/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.skyandtelescope.com/observing/random-meteors-arent-random02112015/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://oup.silverchair-cdn.com/oup/backfile/Content_public/Journal/mnras/359/2/10.1111/j.1365-2966.2005.08918.x/2/359-2-551-fig001.gif?Expires=1528360597&amp;amp;Signature=ASl7-HtSQRnhGpAZ60T8QPqoEEhSIZQ46aF58CVmWncph76O8PqY2pG0pbDEaQhO6RzVFY3NU0dNp3pgM43a3E~F2rymisXef~tWYmSoJnQL1D5Z0H6Q90-BTasw093GObwuS1OpJEULcnpmHaHKMNyTqhHX5-mYh6jJ~uWqzxX7cifjJLqx1E4L-lCaxoWCpKBCJF~FteWFK4TIPmz4JpqdnKj6IuMAEMY4V46HgKopk2-jX99Cnt8HaKYOpod-RJIsLMTODX6QN1AT~5ZeOfJhm~bHShfob9bgXXjrghTRSqX33XFw6E5UmOv7Bzbj6qDYAOIc9zEoKkldQPpqwA__&amp;amp;Key-Pair-Id=APKAIE5G5CRDK6RD3PGA&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2018-06-06T09:59:32.540" LastActivityDate="2018-06-06T09:59:32.540" Title="Is thermal disintegration of meteorites variable with lattitude?" Tags="&lt;planetary-atmosphere&gt;&lt;meteorite&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26525" PostTypeId="1" CreationDate="2018-06-05T08:44:45.847" Score="2" ViewCount="81" Body="&lt;p&gt;Now that we know that gravity waves exist,  we can consider to design a new unit of measurement,  i.e.  deciGravs (from dB) and we can predict what the backround noise of dG's is locally and in space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a theoretical high-res LIGO that can graph gravity waves here and far away,  we can ffwd the record to represent the dG's at human timeframes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the loudest local waves? Can we see the moon as the loudest local wave,  followed by other planets and moons,  are local bodies much louder than local bh waves? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an expected 3D wave map type locally and in deep space that we can model with graphical animations? Can we draw it easily in 2D/3Dusing 5 lines of code for every body and lots of processing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps it makes sense to define a frequency chart of dG's and astroHz and a bit like the electromagnetic waves,  and chart gravity-waves,  amplitude and frequency trends and types?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps supernovas and solar flares and many background waves appear on graviCharts.  Can we know what graviCharts and backround noise of gravity resembles here and far away?&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2018-06-05T09:14:40.813" LastActivityDate="2018-06-05T20:48:33.740" Title="What does gravity-wave noise resemble locally?" Tags="&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26526" PostTypeId="1" CreationDate="2018-06-05T09:29:09.057" Score="2" ViewCount="55" Body="&lt;p&gt;&lt;a href=&quot;https://www.spacetelescope.org/images/opo0109f/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This web page&lt;/a&gt; from ESA/Hubble, shows the size of the universe over time. (I know that &quot;size of the universe&quot; is a gross simplification and open to multiple definitions and ambiguity, but this kind of graph is roughly what I'm after).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But no source data is cited. This is the closest I can find to what I'm after - where can I find source data to match this graph?&lt;/p&gt;&#xA;" OwnerUserId="18601" LastActivityDate="2018-08-07T15:02:03.597" Title="Where can I find source data for the graph of metric expansion over time, as shown in this ESA web page?" Tags="&lt;universe&gt;&lt;expansion&gt;&lt;data-analysis&gt;&lt;cosmological-inflation&gt;&lt;space-telescope&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26527" PostTypeId="1" CreationDate="2018-06-05T11:00:24.380" Score="1" ViewCount="220" Body="&lt;p&gt;This is the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;By considering both the Friedmann and acceleration equations, and assuming a pressureless&#xA;Universe, demonstrate that in order to have a static Universe we must have&#xA;a closed Universe with a positive vacuum energy. Using either physical arguments&#xA;or mathematics, demonstrate that this solution must be unstable.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started off with the friedmann equation and equated H^2=0 and K&gt;0. How to proceed further and prove that lamda is positive?&lt;/p&gt;&#xA;" OwnerUserId="23244" LastActivityDate="2018-06-07T13:42:39.643" Title="Question on Friedman equation involving cosmological constant" Tags="&lt;universe&gt;&lt;astrophysics&gt;&lt;cosmology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26528" PostTypeId="2" ParentId="26525" CreationDate="2018-06-05T12:18:59.080" Score="1" Body="&lt;p&gt;For orbiting objects, gravitational wave intensity depends on the second time derivative of the quadrupole moment of an isolated system's stress–energy tensor (which for the wimpy objects in our vicinity is proportional to the mass times the line of sight acceleration) and drop off with the distance. So you can easily estimate the GW intensity of any object.  Hint: It's &lt;em&gt;small&lt;/em&gt; for nearby objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As requested, I have tried to estimate the GW amplitude and I found the numbers somewhat surprising.  I looked at the Moon, Jupiter, Io, the LIGO neutron star and a pair of supermassive BHs if they were at the Galactic center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got the mass of each object in Suns, the orbital radius in miles, the orbital period in years and the distance in AU.  The angular acceleration is computed as r * omega&lt;sup&gt;2&lt;/sup&gt;. Multiply by mass to get GW radiation at source. Divide by distance to get GW amplitude observed at Earth. (All units relative and arbitrary.)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Object             Acceleration   Mass*Acceleration  Est. amplitude&#xA;Moon               3 x 10**7      1                  4 x 10**2&#xA;Jupiter            3 x 10**6      3 x 10**3          7 x 10**2&#xA;Io                 1 x 10**10     5 x 10**2          1 x 10**2&#xA;LIGO Neutron star  1 x 10**19     3 x 10**16         4 x 10**6&#xA;Supermassive BHs   1 s 10**9      2 x 10**15         3 x 10**6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(The supermassive BH numbers are really quite arbitrary, since you can assume any separation you like and pretty much any mass.  I picked 1 AU and 1,000,000 suns mass.  You can easily get much larger numbers, but the frequency will typically be too low to be seen by LIGO.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other &lt;strong&gt;really important&lt;/strong&gt; thing to remember is that in spite of Io being about as bright as the sources LIGO &lt;em&gt;has&lt;/em&gt; observed, the frequency of Io's radiation is about 10&lt;sup&gt;-5&lt;/sup&gt; Hz. LIGO is only really sensitive in the 10-1000Hz range and has no chance at all of ever seeing IO or Jupiter or the Moon or even a pair of supermassive BHs that are not in the last milliseconds of merger.  If eLISA gets built, it can probably see at least some supermassivs BH pairs.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2018-06-05T20:48:33.740" LastActivityDate="2018-06-05T20:48:33.740" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26529" PostTypeId="1" AcceptedAnswerId="26547" CreationDate="2018-06-05T15:05:06.883" Score="9" ViewCount="185" Body="&lt;p&gt;I was thinking, as CMB is a snapshot of a slice of the Universe during recombination, how much the cmb is changing with time ?&#xA;I mean as we progress in time we look a CMB a bit more faraway each time, could this effect be seen and could we watch the pattern of CMB anisotropies changing with time (on a &quot;human&quot; timescale I mean) ? &lt;/p&gt;&#xA;" OwnerUserId="23047" LastActivityDate="2018-06-07T15:14:47.710" Title="Does the CMB pattern evolve in a human lifetime?" Tags="&lt;universe&gt;&lt;cosmology&gt;&lt;observable-universe&gt;&lt;cmb&gt;&lt;early-universe&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26530" PostTypeId="1" CreationDate="2018-06-05T15:37:50.873" Score="2" ViewCount="115" Body="&lt;p&gt;Can anyone show the steps involved in deriving this equation from the friedmann equation with the cosmological constant involved&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RY4AW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RY4AW.jpg&quot; alt=&quot;equation of the age of the universe&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23244" LastActivityDate="2018-06-05T15:37:50.873" Title="How to derive the equation for the age of a flat universe with a positive cosmological term?" Tags="&lt;universe&gt;&lt;cosmology&gt;&lt;age&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26531" PostTypeId="2" ParentId="21739" CreationDate="2018-06-05T23:59:10.443" Score="2" Body="&lt;p&gt;Experimental answer, at least with a 40mm PST: No. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't worry, I didn't miss much of it :)&lt;/p&gt;&#xA;" OwnerUserId="17585" LastEditorUserId="17585" LastEditDate="2018-06-06T18:55:01.167" LastActivityDate="2018-06-06T18:55:01.167" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26534" PostTypeId="2" ParentId="23631" CreationDate="2018-06-06T13:31:58.867" Score="1" Body="&lt;p&gt;I did this exercise with IRAF by the functions geomap and geotran. There is PYRAF version of the IRAF, but I have never tried the python version. Using these routines, you can find the tranformation equation regardless of the header. What you need to know is some standard objects of known WCS and pixel coordinates.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-06-06T13:31:58.867" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26535" PostTypeId="1" AcceptedAnswerId="26538" CreationDate="2018-06-06T16:45:16.937" Score="4" ViewCount="245" Body="&lt;p&gt;Similar to the question asked &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1018/how-many-stars-and-galaxies-can-be-seen-by-the-naked-eye&quot;&gt;here&lt;/a&gt;, but my question is not about what can be seen with the naked eye but rather what might be seen if pointing my Newtonian reflector (full specs outlined below) at an average point in the sky. (This is why my questions is different from the potential duplicate...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was doing this last night because it was cloudless and ok seeing, and whenever I point the scope at a given patch of sky I always see a much larger amount of objects (obviously) but how many of these additional dots are galaxies and how many are just nearby red dwarfs or average but more distant stars in the milky way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do galaxies ever appear as points of light similar to faint stars, or are they always more diffuse like Andromeda is through binoculars or my telescope? To be clear here, I'm not talking about the galaxies in our local group that are close enough to be discernable as actually fuzzy objects, but rather more distant galaxies that might be very bright (i.e. large elliptical galaxies). I feel as though I'm answering my own question here and that galaxies are only seen as diffuse objects and not points of light, but I can't think of a good way to test this hypothesis with my equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume ideal conditions (low light pollution, perfect seeing, etc.) although I am also curious to know how much this changes if the light pollution conditions worsen, particularly in my backyard. For reference, I can somewhat make out the milky way on a cloudless, moonless night. Andromeda is not visible without small binoculars on any night here (or it is and my eyes just aren't trained enough to find it without them).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better way to phrase my question might be this. Imagine if I were to remove all the stars in the milky way so that the only objects in the night sky were other galaxies. How densely populated with the sky be to the naked eye, and how densely populated would it be through my telescope?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for taking to the time to read my question(s)! I hope to be able to give my own answers one day.&#xA;Clear skies everyone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Telescope specs:&#xA;NexStar 130mm&#xA;Aperature: 130mm&#xA;Focal Length: 650mm&#xA;Eyepiece: 9mm, 1.25&quot; Plossl&lt;/p&gt;&#xA;" OwnerUserId="23285" LastEditorUserId="23285" LastEditDate="2018-06-08T00:57:59.513" LastActivityDate="2018-06-08T00:57:59.513" Title="How many galaxies can be seen in a given patch of sky" Tags="&lt;galaxy&gt;&lt;telescope&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26536" PostTypeId="1" AcceptedAnswerId="26541" CreationDate="2018-06-06T19:39:04.553" Score="4" ViewCount="979" Body="&lt;p&gt;China's FAST (Five-hundred-meter Aperture Spherical radio Telescope) is the largest radio telescope in the world.(This is huge!!!)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/uPEdm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uPEdm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we all know Space Telescopes are better than Earth-Based telescopes always.(Hubble space telescope was such a tremendous success)Even though ground-based observatories are usually located in highly elevated areas with minimal light pollution, but still they contend with atmospheric turbulence, which limits the sharpness of images taken from this vantage point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What if all the space agencies collaborate together to build a telescope half the size of China's FAST in moon(Even smaller than half),even then it will be more powerful than the &quot;FAST&quot; right(because the atmosphere of moon is very less dense)?Will it be impractical?It may take decades to build one but still the outcome will be huge(We may even look and find aliens!!)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;(Taking in account that Elon Musk is already planning Colonization of Mars)This project looks somewhat tamable than that,right?  Is there a possibility of this happening? Is it feasible enough?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="22623" LastEditorUserId="22623" LastEditDate="2018-06-06T21:38:32.140" LastActivityDate="2018-06-07T11:55:49.270" Title="Is it practical to build a giant telescope in Moon? (Considering the fact that the atmosphere of moon is very rare)" Tags="&lt;the-moon&gt;&lt;telescope&gt;&lt;observational-astronomy&gt;&lt;space-telescope&gt;&lt;deep-sky-observing&gt;" AnswerCount="4" CommentCount="14" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26537" PostTypeId="2" ParentId="26536" CreationDate="2018-06-06T22:10:59.997" Score="1" Body="&lt;p&gt;Telescopes on the moon are a &lt;strong&gt;fantastic idea&lt;/strong&gt;. If you &quot;thought it up yourself&quot; congratulations, because it's a great idea for the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you may know we already have many fantastic telescopes in space (that is to say, generally orbiting or near our Earth). The &lt;strong&gt;next step&lt;/strong&gt; would be telescopes on the Moon. (It is far, far, harder to build one on the moon, than simply lift one up to orbit.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can easily find many beginning studies about it; for example &lt;a href=&quot;https://www.space.com/5628-nasa-envisions-huge-lunar-telescope.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; is one by NASA on your great idea, from ten years ago!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your idea is a good one, and hopefully it comes sooner rather than later. I bet we are 30 years away from telescopes on the moon. But they will come.&lt;/p&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="7472" LastEditDate="2018-06-07T11:31:10.363" LastActivityDate="2018-06-07T11:31:10.363" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26538" PostTypeId="2" ParentId="26535" CreationDate="2018-06-06T22:18:49.113" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Do galaxies ever appear as points of light similar to faint stars&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;sure, just google for photos&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Telescope specs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;it depends on your exposure time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many galaxies does Hubble see?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hellova lot!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble_Ultra-Deep_Field&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Hubble_Ultra-Deep_Field&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Mck4Y.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mck4Y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13071" LastActivityDate="2018-06-06T22:18:49.113" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26540" PostTypeId="2" ParentId="26517" CreationDate="2018-06-06T22:36:17.840" Score="5" Body="&lt;p&gt;The Minor Planet Center (MPC) routinely gives &lt;a href=&quot;https://www.minorplanetcenter.net/mpec/RecentMPECs.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;preliminary orbits for newly discovered asteroids&lt;/a&gt;  with observations spanning only 48 hours.&#xA;Ordinary elliptical orbits are computed automatically.&#xA;After further observations are reported, MPC may issue a more precise orbit estimate.&#xA;The uncertainty never reaches zero, even if we track an asteroid for 100 years or longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ssd.jpl.nasa.gov/sbdb.cgi?spk=3788040&quot; rel=&quot;nofollow noreferrer&quot;&gt;'Oumuamua's trajectory&lt;/a&gt; took longer to determine because it didn't fit the usual elliptical model.&#xA;&lt;a href=&quot;https://www.eso.org/public/archives/releases/sciencepapers/eso1737/eso1737a.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Meech et al. 2017&lt;/a&gt; tell in their Methods section how they tried and rejected two different elliptical orbits before settling on a hyperbolic trajectory.&#xA;Even with observations spanning 6 days, MPC issued their &lt;a href=&quot;https://minorplanetcenter.net/mpec/K17/K17UI1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;preliminary trajectory&lt;/a&gt; with conditions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Further observations of this object are very much desired.&#xA;  &lt;em&gt;Unless there are serious problems with much of the astrometry listed below&lt;/em&gt;, strongly hyperbolic orbits are the only viable solutions.&#xA;  Although it is probably not too sensible to compute meaningful original and future barycentric orbits, given the very short arc of observations, the orbit below has e ~ 1.2 for both values.&#xA;  &lt;em&gt;If further observations confirm the unusual nature of this orbit&lt;/em&gt;, this object may be the first clear case of an interstellar comet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-06-06T23:50:21.653" LastActivityDate="2018-06-06T23:50:21.653" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26541" PostTypeId="2" ParentId="26536" CreationDate="2018-06-07T01:18:38.073" Score="7" Body="&lt;p&gt;It's very unlikely that large optical telescopes will ever be built on the Moon, because the Moon is almost the worst possible place to build them. (The surfaces any of the  planets other than Earth are worse.)  It has no particular advantages over orbit and costs a lot more to build there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon looked like a good location when observatory technology was film-based, because observatories required people and people needed a base to live in and work best under gravity.  But we now have the technology to shade an orbiting telescope from the Sun, point it exquisitely accurately, and take endless photos without changing the film -- all in orbit.  Orbit is easier to get to than the Lunar surface -- even the Webb (should it ever launch) which will be parked a million miles from Earth, is easier to get to than the surface of the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we had the space capabilities to build a giant reflector on the Moon, we could build an even bigger and better one in orbit for the same cost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You spoke of FAST. FAST is a radio telescope. It's likewise much cheaper to build radio telescopes in orbit than on the Lunar surface. The one possible advantage of the Moon is that the far side of the Moon might have advantages of being especially radio quiet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even there, it seems likely that any space radio telescope would be a large array of radio telescopes joined to form a giant synthetic aperture, and that would almost certainly want to be much larger than earth-based instruments -- for which the Moon is already too small.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2018-06-07T11:55:49.270" LastActivityDate="2018-06-07T11:55:49.270" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26542" PostTypeId="2" ParentId="26536" CreationDate="2018-06-07T02:05:08.117" Score="1" Body="&lt;p&gt;There are advantages and disadvantages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth based telescopes have disadvantages because of the atmosphere, which blocks some light and it blocks some wavelengths almost completely and it also bends the incoming light.   Some of the atmospheric displacement can be fixed with modern computers, so Earth based telescopes have improved significantly over the last 20-30 years, but they still have the disadvantage Earth's atmosphere blocking some of the light coming in from space.  Today's Earth based telescopes are as good as Hubble (I read that somewhere) at least in the visible spectrum, in large part because of computerized imaging.  When Hubble was launched (and fixed) at the time if was lightyears better than anything on Earth.  Today it's close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.space.com/6716-major-space-telescopes.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chandra and Spitzer&lt;/a&gt; see in wavelengths that don't make it to Earth, so they need to be in orbit to work.   James Webb will also view in the infra-red as well as visible wavelengths, so it needs to be in space to take advantage of it's infrared imaging. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space is much better because it gets a much cleaner picture and it receives all wavelengths.   There are disadvantages to being in space too, like cosmic rays, solar storms and the occasional near-light-speed proton but on average, space is better than Earth and the surface of the Moon is similar to being in space due to there being essentially no atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A big problem with telescopes in space is that launching that much material is very expensive.  It costs about $10,000 per pound to launch something into space.  That's why there's so few you can count space telescopes on your fingers (fingers and toes actually, &lt;a href=&quot;https://www.space.com/6716-major-space-telescopes.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;I think there's 13&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also important to keep large telescopes somewhat cold.  (&lt;a href=&quot;https://jwst.nasa.gov/sunshield.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;James Webb will have a large sun shield&lt;/a&gt;).   That's especially important for infrared imaging. Provided the area is shaded, the moon will remain sufficiently cold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big advantage the Moon has is that it's full of silicon, which is what mirrors are made of, so by using lunar material, much less stuff needs to be launched and that saves a ton of money and increases the size of this theoretical telescope.   That's the real advantage.  We can't launch a 100 or 200 foot telescope from Earth because it's too big, but we could in theory, built one that size on the moon because there's lots and lots of stuff to use.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A smaller advantage is that the body of the moon acts as a shield for what's below it, but the real advantage is that the Moon is made of stuff that can be turned into a telescope.   Really any large body a few miles in diameter would work, though working on a surface with gravity might be easier.   Personally I think Ceres would be a great place to build.  Very low gravity (but enough where you can set things down and not worry about them bouncing away) farther from the sun so a smaller sun-shield.   Any reasonably large body in space could theoretically be used to build a telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon, however, is closer, so it's both more easily repaired and it's close enough that it's easier to transmit images back to Earth as opposed to a Phobos based or Ceres based telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shouldn't be mistaken for easy though.  It's never been done and might run into unanticipated problems and end up being quite difficult, but it's a nice idea.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-06-07T02:05:08.117" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26544" PostTypeId="1" CreationDate="2018-06-07T09:00:07.343" Score="1" ViewCount="1165" Body="&lt;p&gt;Do we see planet Venus every day? (Just before sun sets or Just after sun sets) throughout year? or some days we don't see it?&lt;/p&gt;&#xA;" OwnerUserId="23290" LastEditorUserId="907" LastEditDate="2018-06-08T14:42:15.780" LastActivityDate="2018-06-10T01:14:08.693" Title="Do we see planet venus everyday?" Tags="&lt;venus&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26545" PostTypeId="2" ParentId="26544" CreationDate="2018-06-07T09:45:20.463" Score="6" Body="&lt;p&gt;No. Some days Venus is an evening star, visible for up to a few hours after the Sun sets. Some days it is a morning star, visible for a few hours before the Sun rises. Some days it is so close to the Sun in the sky that we can't see it at all (or even actually behind the Sun).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus can also be seen in daylight, if it is above the horizon, not too close to the Sun and the sky is clear. You need to know where to look for it, though, it's only a little brighter than the sky.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="2365" LastEditDate="2018-06-07T09:50:34.360" LastActivityDate="2018-06-07T09:50:34.360" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26547" PostTypeId="2" ParentId="26529" CreationDate="2018-06-07T10:22:17.863" Score="5" Body="&lt;p&gt;In principle yes, in practice no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As seen in the temperature power spectrum below, the Planck satellite detects power (i.e. &quot;a signal&quot;) even on the smallest probed scales, which is a few arcminutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gW7Di.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gW7Di.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;sup&gt;Fig. 1 from &lt;a href=&quot;https://arxiv.org/abs/1502.01589&quot; rel=&quot;noreferrer&quot;&gt;Planck Collaboration et al. (2016)&lt;/a&gt; with my own approximate corresponding angles annotated in red.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smallest scale shown in the figure is $\ell = 2500$, corresponding to an angle of roughly $4'$. At the redshift of the CMB ($z=1100$), this angle subtends 57 million lightyears (Mlyr) &lt;em&gt;today&lt;/em&gt;. When the CMB was emitted, such a region was only 52 &lt;em&gt;thousands&lt;/em&gt; lightyears (klyr) across).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, we can see that at least down to regions that were roughly 50 klyr across, the CMB pattern is anisotropic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As time goes, the CMB we see will come from an increasingly larger distance. Hence, when this distance has increased by roughly 50 klyr, the CMB pattern should have evolved. This will take a bit more than 50 thousand years$^\dagger$(kyr), i.e. much longer than a human lifetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be able to see changes within 100 years, we would need an angular resolution of 0.5 arcseconds. But angular resolution is not enough — the power seems to fall off as a power law on these scales, so a higher sensitivity would also be needed.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;A bit more, because space expands somewhat during the time that light travels that distance, so in total roughly 55 kyr.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-06-07T15:14:47.710" LastActivityDate="2018-06-07T15:14:47.710" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26548" PostTypeId="2" ParentId="26536" CreationDate="2018-06-07T10:34:08.780" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://www.spiedigitallibrary.org/conference-proceedings-of-spie/5487/1/Buyers-guide-to-telescopes-at-the-best-sites--Dome/10.1117/12.596320.full?SSO=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;This 2004 article&lt;/a&gt; although a little dated now, discusses the relative merits of three possible novel sites for new major optical/IR telescopes: Dome A in Antarctica, the South pole of the Moon, and the Earth-Sun L2 point (or an orbit around it). Each is best for different purposes. A particular attraction of the Lunar site is the ability to stare at the same part of the sky for very long periods to detect very faint objects.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-06-07T10:34:08.780" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26550" PostTypeId="2" ParentId="26527" CreationDate="2018-06-07T13:35:17.357" Score="1" Body="&lt;p&gt;When including a cosmological constant, the acceleration equation reads&#xA;$$\frac{\ddot{a}}{a}=-\frac{4\pi G}{3c^2}(\epsilon+3P)+\frac{\Lambda}{3}$$&#xA;In a pressureless universe, $P=0$, and in a static universe $\frac{\ddot{a}}{a}=0$. The equation becomes:&#xA;$$\frac{4\pi G}{3c^2}\epsilon=\frac{\Lambda}{3}$$Energy density is always positive and non-zero, so $\Lambda&amp;gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is unstable since after a slight perturbation, if $a$ grows, $\epsilon$ will go down ($\epsilon_m\propto a^{-3}$ for matter for example), which will result in an increase of the acceleration, making $a$ go up even more.&lt;/p&gt;&#xA;" OwnerUserId="22391" LastEditorUserId="22391" LastEditDate="2018-06-07T13:42:39.643" LastActivityDate="2018-06-07T13:42:39.643" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26556" PostTypeId="1" AcceptedAnswerId="26558" CreationDate="2018-06-08T11:50:13.933" Score="3" ViewCount="107" Body="&lt;p&gt;In &lt;a href=&quot;http://www.spiegel.de/fotostrecke/curiosity-marsrover-entdeckt-organisches-material-fotostrecke-161363-2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this diagram&lt;/a&gt; about the methane concentration in the martian atmosphere, there are data points labeled &quot;Mars year 32&quot;, &quot;Mars year 33&quot;, and &quot;Mars year 34&quot;. How are Mars years counted? Is there a &quot;Mars year 0&quot;? What is the event that defines the start of the time scale?&lt;/p&gt;&#xA;" OwnerUserId="11617" LastEditorUserId="11617" LastEditDate="2018-06-08T12:27:27.077" LastActivityDate="2018-06-08T13:38:14.490" Title="How are mars years counted?" Tags="&lt;mars&gt;&lt;time&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26557" PostTypeId="2" ParentId="26526" CreationDate="2018-06-08T13:27:02.463" Score="1" Body="&lt;p&gt;I was hoping a physicist might answer this, as I'm just an enthusiastic armchair scientist whose undergrad calculus was 30 years ago. But as no one else has had a go, let me try an answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graph in the web page is a simplistic and somewhat exaggerated representation of the evolution of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Scale_factor_(cosmology)&quot; rel=&quot;nofollow noreferrer&quot;&gt;cosmic scale factor&lt;/a&gt; &lt;em&gt;a(t)&lt;/em&gt; over time. You can derive the &quot;source data&quot; yourself using the following metrics for the three distinct periods in the expansion of the Universe, since the only variable is &lt;em&gt;t&lt;/em&gt; which in this case is the age of the Universe. The scale factor &lt;em&gt;a(t)&lt;/em&gt; is dimensionless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the earliest period of the Universe up to about 47,000 years after the Big Bang, the Universe was dominated by radiation. The evolution of the scale factor in the FLRW metric for this period is given by: $$a(t) \propto \sqrt t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that very brief time (in cosmological terms), matter took over as the dominant factor in the Universe for the next 9.8 billion years, and the evolution of the scale factor is given by: $$a(t) \propto t^{2/3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the Universe in the current period is dominated by dark energy, and the evolution of the scale factor is given by: $$a(t) \propto exp (Ht)$$&#xA;where H is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble_constant&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble constant&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practical terms, if you'd like to generate a graph similar to that in the web page, you can set $t_1$ = 13.79 Gyr (i.e. the current age of the Universe). I found that the first era of radiation domination is so short that it's not worth including in the graph. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my simplistic rendition:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/cYbvH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cYbvH.png&quot; alt=&quot;evolution of scale factor over time&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14094" LastEditorUserId="14094" LastEditDate="2018-06-08T13:33:54.300" LastActivityDate="2018-06-08T13:33:54.300" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26558" PostTypeId="2" ParentId="26556" CreationDate="2018-06-08T13:38:14.490" Score="4" Body="&lt;p&gt;Given that &lt;a href=&quot;https://www.space.com/40819-mars-methane-organics-curiosity-rover.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;the methane variations are seasonal&lt;/a&gt;, the diagram is labeled in Martian years - the amount of time it take Mars to travel around the Sun once (about 1.88 Earth years). This makes it much easier to see the cycles rather than using Earth years. In particular, the x-axis has a length of one Martian year; the data point show seasonal methane measurements for that year. So, for instance, the red data point was take early in the spring of Mars year 34, the first blue point was taken at the start of summer of Mars year 32 (earlier!), and so on and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific year numbering is &lt;a href=&quot;http://www.planetary.org/explore/space-topics/mars/mars-calendar.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;an arbitrary convention&lt;/a&gt; begun on April 11, 1955 (the &lt;a href=&quot;https://en.wikipedia.org/wiki/Timekeeping_on_Mars#Martian_year&quot; rel=&quot;nofollow noreferrer&quot;&gt;northern spring equinox&lt;/a&gt;). The &lt;a href=&quot;http://www.daviddarling.info/encyclopedia/S/solar_longitude.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar longitude&lt;/a&gt; system is applied to Mars; it's a geometric way of describing where in its orbit a planet is. The solar longitude is denoted $L_2$; on April 11, 1955, $L_s=0^\circ$; it increases to $L_s=360^\circ$, at which point a new Martian year begins.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2018-06-08T13:38:14.490" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26559" PostTypeId="2" ParentId="26228" CreationDate="2018-06-08T16:19:18.270" Score="0" Body="&lt;p&gt;Laws of gravity would require a massive rewrite, and Earth would likely in turn be orbiting Sputnik.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2018-06-08T16:19:18.270" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26560" PostTypeId="1" AcceptedAnswerId="26569" CreationDate="2018-06-08T17:20:27.013" Score="4" ViewCount="252" Body="&lt;p&gt;Querying GAIA DR2 for Polaris (for example using &lt;a href=&quot;https://gea.esac.esa.int/archive/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://gea.esac.esa.int/archive/&lt;/a&gt;) yields empty parallax columns. Anyone knows why it was not measured/published?&lt;/p&gt;&#xA;" OwnerUserId="23303" LastActivityDate="2021-04-03T09:17:02.470" Title="Why no Polaris distance in GAIA DR2?" Tags="&lt;parallax&gt;&lt;gaia&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26561" PostTypeId="1" AcceptedAnswerId="26570" CreationDate="2018-06-08T18:33:27.553" Score="10" ViewCount="2124" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XZEGJ.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XZEGJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do planets have an axial tilt? From the above image we can see that each planet's axial tilt angle varies and differs from the others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was the cause of this, was this from the beginning of our solar systems birth? Or did it happen later?  &lt;/p&gt;&#xA;" OwnerUserId="22623" LastEditorUserId="616" LastEditDate="2019-07-21T21:27:06.933" LastActivityDate="2019-07-21T21:27:06.933" Title="Why does each and every planet of our solar system have an unique axial tilt angle?" Tags="&lt;planet&gt;&lt;solar-system&gt;&lt;earth&gt;&lt;gas-giants&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26562" PostTypeId="1" CreationDate="2018-06-08T20:09:26.253" Score="7" ViewCount="167" Body="&lt;p&gt;Let's take &lt;a href=&quot;https://en.wikipedia.org/wiki/Europa_(moon)&quot; rel=&quot;noreferrer&quot;&gt;Europa&lt;/a&gt; for example:&lt;br /&gt;&#xA;The predominant model suggests that heat from tidal flexing causes the ocean to remain liquid and drives ice movement similar to plate tectonics.&lt;/p&gt;&#xA;&lt;p&gt;The heat generated from the tidal flexing has to go somewhere. Is it all carried away by the water vapor plumes we recently detected? Is more water from the ice cover melting away? Is it just radiated away?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;I would expect it to cause sublimation as well&lt;/strong&gt;: There is hardly an atmosphere (&lt;a href=&quot;https://phys.org/news/2015-09-jupiter-moon-europa.html&quot; rel=&quot;noreferrer&quot;&gt;molecular oxygen at 0.1 micro Pascals, or 10&lt;sup&gt;-12&lt;/sup&gt; times that of the Earth&lt;/a&gt;), and a tenuous ionosphere.&lt;br /&gt;&#xA;And that same article suggest that the surface ice is &lt;em&gt;also&lt;/em&gt; broken down by radiation:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;the oxygen [...] is formed through the process of radiolysis, where ultraviolet radiation from the Jovian magnetosphere collides with the icy surface, splitting water into oxygen and hydrogen. The same radiation also creates collisional ejections of these products from the surface, and the balance of these two processes forms an atmosphere.&lt;/p&gt;&#xA;&lt;p&gt;The hydrogen, meanwhile, lack the mass needed to be retained as part of the atmosphere and most is lost to space. This escaped hydrogen, along with portions of atomic and molecular oxygen that are ejected, forms a gas torus in the vicinity of Europa's orbit around Jupiter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Or do we have reason to believe that any losses are compensated by ice being deposited from external sources?&lt;/p&gt;&#xA;" OwnerDisplayName="user1569" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-06-14T09:17:48.190" Title="Icy moons: why hasn't the ice sublimated away?" Tags="&lt;europa&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26563" PostTypeId="1" AcceptedAnswerId="26564" CreationDate="2018-06-08T20:26:50.960" Score="2" ViewCount="1577" Body="&lt;p&gt;I saw reports that only Andromeda Galaxy and Big/Small Magellanic Cloud galaxy stars could be visible with the naked eye (how many stars would be from them out of a few thousand visible?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question appeared because there are some very &quot;sparse&quot; galaxies (very few stars in them) and scientists say night sky in them is practically all black (starless)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So is it true that 99,9% visible stars in the night sky (naked eye) are from the Milky Way only?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="23305" LastActivityDate="2018-06-08T20:45:57.837" Title="Do we see stars from other galaxies (not Milky way) in the night sky with the naked eye?" Tags="&lt;galaxy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26564" PostTypeId="2" ParentId="26563" CreationDate="2018-06-08T20:45:57.837" Score="5" Body="&lt;p&gt;Stars in other galaxies are too dim for our eyes to see them as individual points of light. The brightest stars in the Magellanic clouds are magnitude 10, and we can only see magnitude 6 stars (bigger magnitude means dimmer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So instead we see the Magellanic Clouds and the Andromeda galaxy as patches of light. The galaxy is visible, but its individual stars are not. The only time a &quot;star&quot; was visible in the Large Magellanic cloud was a supernova in 1987.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a very sparse galaxy the night sky would be very dark, with few visible stars. Other galaxies might be visible, but they would be dim patches of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;100% of the naked-eye stars you see are in the Milky Way. You can see Andromeda, but you can't see any individual stars in it.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-06-08T20:45:57.837" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26565" PostTypeId="1" AcceptedAnswerId="26571" CreationDate="2018-06-08T20:59:53.687" Score="1" ViewCount="349" Body="&lt;p&gt;Can a red dwarf star or a white dwarf have enough energy to support life on earth? How it will change the climate and the seasons of earth if it is our only star? (energy source)(Expecting it to change since earths orbital mechanics will also change) &lt;/p&gt;&#xA;" OwnerUserId="22623" LastEditorUserId="22623" LastEditDate="2018-06-08T21:10:48.040" LastActivityDate="2019-04-01T08:00:01.000" Title="Can life on earth be supported by only a Red Dwarf star or a White Dwarf(state still hot not cooled enough yet)?(Provided it is near to the star)" Tags="&lt;star&gt;&lt;life&gt;&lt;earth-like-planet&gt;&lt;white-dwarf&gt;&lt;dwarf-star&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26566" PostTypeId="1" CreationDate="2018-06-08T21:11:35.540" Score="0" ViewCount="758" Body="&lt;p&gt;What other galaxies can be seen with the naked eye in the night sky? Can we see any other galaxy besides Andromeda and Magellanic Clouds?&lt;/p&gt;&#xA;" OwnerUserId="23305" LastActivityDate="2018-06-11T23:06:12.123" Title="What other galaxies can be seen with the naked eye in the night sky?" Tags="&lt;galaxy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26567" PostTypeId="2" ParentId="26561" CreationDate="2018-06-08T21:19:43.260" Score="5" Body="&lt;p&gt;You could also ask:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is each planet a different size.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is each planet a different colour.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or even&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why are the apples in my fruit bowl pointing in different directions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason is that, unless I make the effort to point the apples in the same direction they will point in random directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The planets were formed independently, (though at about the same time) and the environment of each planet, was slightly different. After their initial formation, they have been subject to different histories of perturbation and collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall effect is that the axial tilt of each planet is essentially a random number (though not uniformly distributed). If you choose 8 random numbers you would be surprised if two came out exactly the same.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-06-08T21:19:43.260" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26568" PostTypeId="2" ParentId="26566" CreationDate="2018-06-08T21:25:12.310" Score="3" Body="&lt;p&gt;We can see the Magellenic clouds easily, and we can see Andromeda on very dark, clear nights, but only when we are far from the city.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people claim to have seen the Triangulum galaxy (the third largest member of the local group, after Andromeda and the Milky way). Exceptional conditions, and very well trained eyes would be required to see it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No other galaxies are visible, even under the best conditions.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-06-08T21:25:12.310" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26569" PostTypeId="2" ParentId="26560" CreationDate="2018-06-08T21:50:57.830" Score="5" Body="&lt;p&gt;My query for stars north of +75° and brighter than magnitude 4 returns only 6 stars, brightest of which is γ Cephei.&#xA;Polaris may be &lt;a href=&quot;https://www.cosmos.esa.int/web/gaia/faqs#BrightStar&quot; rel=&quot;nofollow noreferrer&quot;&gt;too bright for Gaia&lt;/a&gt;.&#xA;Even if it doesn't saturate the detector, a bright star will probably have larger astrometric errors than a star between magnitudes 6 and 12.&#xA;DR2 coverage of magnitude 5 and brighter stars is better than DR1 but still rather sparse.&#xA;Section 3.1 of &lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2018A%26A...616A...1G/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt; plots parallax coverage vs. magnitude and explains why some stars have positions but not parallaxes.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2021-04-03T09:17:02.470" LastActivityDate="2021-04-03T09:17:02.470" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26570" PostTypeId="2" ParentId="26561" CreationDate="2018-06-08T22:19:04.443" Score="10" Body="&lt;p&gt;In the early stages of the formation of the solar system, &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Planetesimal&quot; rel=&quot;nofollow noreferrer&quot;&gt;planetesimals&lt;/a&gt; start condensing and everything rotates with angular momentum inherited from the collapsing cloud of gas and dust, so the planetesimals all have their orbit and spin axes closely aligned with that of the proto-Sun. And while a planetesimal continues to grow by attracting nearby material its spin orientation isn't affected very much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But eventually the planetesimals get large enough that they start attracting one another, and collisions occur. These collisions can disturb the spin axis of a planetesimal, especially if the collision is off-centre, and if the two bodies have roughly similar momentum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Big planetesimals grow faster than small ones, and are less likely to have their spin axis disturbed. So proto-Jupiter manages to capture most of the matter that doesn't end up in the proto-Sun, and suffers little disruption to its spin axis. The smaller planets aren't so lucky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the rough picture, and it misses a &lt;em&gt;lot&lt;/em&gt; of details. Some of the details are still unclear, and more work is needed to clarify them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest mystery is Uranus. We don't know why Uranus is rolling around on its side, but we presume that was caused by some interaction fairly late in its history because its major moons orbit in its equatorial plane, pretty much. One theory is that it suffered collateral damage in whatever event upset Saturn's spin axis. Note that Saturn's rings and large moons orbit in Saturn's equatorial plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury has a short orbital period and hence a high orbital speed. It's tidally locked to the Sun, so that it spins 3 times on its axis in 2 revolutions around the Sun; that tidal locking also locks it's spin axis fairly closely to its revolution axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus is also a bit of a mystery. It was formerly believed to be tidally locked, with its year equal to its day, but actually its rotational period is about 243 (Earth) days, about 20 days longer than its year, and it's spinning backwards! We don't know why, but suspect it may be due to a strong coupling between it's surface and very dense atmosphere. Also note that Venus experienced planet-wide severe tectonic activity not so long ago (on an astronomical timescale), so its surface is relatively new, and a lot of the planet's angular momentum may have been transferred to its atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="16685" LastEditDate="2019-07-20T08:38:00.873" LastActivityDate="2019-07-20T08:38:00.873" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26571" PostTypeId="2" ParentId="26565" CreationDate="2018-06-08T22:52:33.787" Score="5" Body="&lt;p&gt;Certainly a red-dwarf star can have enough energy for a planet around it to be in the goldilocks zone.   There are some difficulties with red-dwarf stars and Earth like planets.   The planet would need to be very close to the star and as a result, tidally locked.   The orbital period would be quite short, so there would be no seasons and one side of the planet would be in permanent sun the other side in permanent shade.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Red-Dwarf stars, unlike our sun, grow colder with time and they can be quite magnetically active, sometimes called flare stars, though not all red-dwarf stars are flare stars, many start out that way.   The proximity to that much solar-ejection could strip planets of their atmosphere early in the planet's life.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of that makes it impossible.  It's certainly possible for a Red-dwarf star to host an Earth-like planet, but there's some difficulty.  Orange or Yellow stars are, on average, considered better candidates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A white dwarf, however, likely not possible.   A white dwarf follows the red-giant stage of a star.  It would be too small and give off too little energy and any planets that survived the red-giant phase would likely be very distant.   Life and/or an Earth-like planet around a white dwarf is virtually impossible.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-06-08T22:52:33.787" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26572" PostTypeId="1" CreationDate="2018-06-09T01:59:07.810" Score="3" ViewCount="888" Body="&lt;p&gt;What are the elements of a black hole it just says what a black hole is? There must be elements that the sun consists of, right?&lt;/p&gt;&#xA;" OwnerUserId="23308" LastEditorUserId="37191" LastEditDate="2021-02-12T21:01:39.843" LastActivityDate="2021-02-12T21:01:39.843" Title="What does a black hole elements consists of?" Tags="&lt;black-hole&gt;&lt;astrochemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26573" PostTypeId="1" CreationDate="2018-06-09T04:33:01.957" Score="-1" ViewCount="141" Body="&lt;p&gt;I've heard somewhere that when we observe distant galaxies, we actually look into the past. For example, if a planet is 3 million light years away from us, it means that we observe its state as it was 3 million years ago, am I right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I think this explains why we haven't encountered any aliens yet. Because we see their planets in their ancient times, their civilizations haven't formed yet. So we think that nobody's around us. But instead we can actually be surrounded by alien civilizations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this idea plausible?&lt;/p&gt;&#xA;" OwnerUserId="16374" LastEditorUserId="14094" LastEditDate="2018-06-11T10:59:56.080" LastActivityDate="2018-06-11T10:59:56.080" Title="Can aliens possibly be around us?" Tags="&lt;exoplanet&gt;&lt;life&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26574" PostTypeId="2" ParentId="26573" CreationDate="2018-06-09T07:24:58.377" Score="5" Body="&lt;p&gt;There are a number of misconceptions here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, most of the visible stars that we see are much closer than you think. Most are within a few hundred light-years. With telescopes you can see more distant stars, but also lots of dim nearby ones. So the &quot;3 million&quot; value is wrong. &quot;3 hundred&quot; would be better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly stars formed at different times. If the star formed before the sun, the planet would have had more time to develop. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, there is no reason to think that life on all planets develops at the same rate or in the same way.  For 3 billion years, life on Earth was mostly &quot;slime&quot; then suddenly complex animals evolved.  Perhaps on another planet the &quot;slime&quot; period took longer, or shorter, or the complex life never happened at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is something of a mystery on why we don't see other intelligent life, and your notion of &quot;we see their past&quot; doesn't explain it.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-06-09T07:24:58.377" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26575" PostTypeId="1" CreationDate="2018-06-09T09:40:09.063" Score="-2" ViewCount="116" Body="&lt;p&gt;This &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_nearest_stars_and_brown_dwarfs&quot; rel=&quot;nofollow noreferrer&quot;&gt;list&lt;/a&gt; says that Tau Ceti is very like the Sun (spectrally). Others are predominantly red dwarfs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is Tau Ceti closest star which is quite like the sun according to some criteria?&lt;/p&gt;&#xA;" OwnerUserId="23305" LastActivityDate="2018-06-15T09:26:28.073" Title="What sun-like star is closest to earth/sun?" Tags="&lt;star&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="0" ClosedDate="2018-06-21T16:55:11.620" ContentLicense="CC BY-SA 4.0" />
  <row Id="26578" PostTypeId="2" ParentId="26575" CreationDate="2018-06-09T15:08:39.643" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;According to their system of classification, the Sun is known as a yellow dwarf star. This group of stars are relatively small, containing between 80% and 100% the mass of the Sun. So the Sun is at the higher end of this group. The official designation is as a G 2V star.      &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(by Fraser Cain)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is Tau Ceti closest star which is quite like the sun according to some criteria?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No,Tau Ceti's stellar class is G 8Vp. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest star bearing our Sun's stellar class is Alpha Centauri A (HD 128620)&lt;/p&gt;&#xA;" OwnerUserId="22623" LastActivityDate="2018-06-09T15:08:39.643" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26579" PostTypeId="2" ParentId="26572" CreationDate="2018-06-09T18:35:58.393" Score="3" Body="&lt;p&gt;If your question is &amp;quot;What elements are in a black hole?&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;The answer is: The gravity is too great for there to be &amp;quot;elements&amp;quot; (atoms or molecules) &lt;strong&gt;inside&lt;/strong&gt; or near the surface of a black hole. The extreme gravity tears apart everything and compresses it.&lt;/p&gt;&#xA;&lt;p&gt;Some distance away (especially far away, here on Earth) the black hole doesn't alter matter, but as things get closer they fall at increasing speeds to the surface, becoming a part of the black hole. Everything becomes crushed, becoming &lt;a href=&quot;https://en.wikipedia.org/wiki/Elementary_particle&quot; rel=&quot;nofollow noreferrer&quot;&gt;elementary particles&lt;/a&gt;; what atoms are made of.&lt;/p&gt;&#xA;&lt;p&gt;NASA's webpage: &amp;quot;&lt;a href=&quot;https://www.nasa.gov/feature/jpl/nustar-probes-black-hole-jet-mystery&quot; rel=&quot;nofollow noreferrer&quot;&gt;NuSTAR Probes Black Hole Jet Mystery&lt;/a&gt;&amp;quot; explains that plasma, x-rays, and eventually light can leave a black hole at the poles. See also: &amp;quot;&lt;a href=&quot;https://arxiv.org/abs/1710.09838&quot; rel=&quot;nofollow noreferrer&quot;&gt;An elevation of 0.1 light-seconds for the optical jet base in an accreting Galactic black hole system&lt;/a&gt;&amp;quot; (26 Oct 2017) which says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;Relativistic plasma jets are observed in many accreting black holes. According to theory, coiled magnetic fields close to the black hole accelerate and collimate the plasma, leading to a jet being launched. Isolating emission from this acceleration and collimation zone is key to measuring its size and understanding jet formation physics. But this is challenging because emission from the jet base cannot be easily disentangled from other accreting components.&lt;/p&gt;&#xA;&lt;p&gt;Here, we show that rapid optical flux variations from a Galactic black-hole binary are delayed with respect to X-rays radiated from close to the black hole by ~0.1 seconds, and that this delayed signal appears together with a brightening radio jet. The origin of these sub-second optical variations has hitherto been controversial.&lt;/p&gt;&#xA;&lt;p&gt;Not only does our work strongly support a jet origin for the optical variations, it also sets a characteristic elevation of &amp;lt;~&lt;span class=&quot;math-container&quot;&gt;$10^3$&lt;/span&gt; Schwarzschild radii for the main inner optical emission zone above the black hole, constraining both internal shock and magnetohydrodynamic models.&lt;/p&gt;&#xA;&lt;p&gt;Similarities with blazars suggest that jet structure and launching physics could potentially be unified under mass-invariant models. Two of the best-studied jetted black hole binaries show very similar optical lags, so this size scale may be a defining feature of such systems.&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So, what goes in is not what comes out.&lt;/p&gt;&#xA;&lt;p&gt;See also this article about &lt;a href=&quot;https://en.wikipedia.org/wiki/Active_galactic_nucleus&quot; rel=&quot;nofollow noreferrer&quot;&gt;AGN&lt;/a&gt;: &amp;quot;&lt;a href=&quot;https://arxiv.org/abs/1802.10308&quot; rel=&quot;nofollow noreferrer&quot;&gt;The largely unconstrained multiphase nature of outflows in AGN host galaxies&lt;/a&gt;&amp;quot; (28 Feb 2018).&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://hubblesite.org/image/968/news_release/2000-20&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PUNMb.jpg&quot; alt=&quot;Galaxy M87&quot; /&gt;&lt;/a&gt; &lt;sup&gt;Black Hole-Powered Jet of Electrons and Sub-Atomic Particles Streams From Center of Galaxy M87&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Frame-dragging#Astronomical_evidence&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomical evidence&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Relativistic_jet&quot; rel=&quot;nofollow noreferrer&quot;&gt;Relativistic jets&lt;/a&gt; may provide evidence for the reality of frame-dragging. Gravitomagnetic forces produced by the Lense–Thirring effect (frame dragging) within the ergosphere of rotating black holes combined with the energy extraction mechanism by Penrose have been used to explain the observed properties of relativistic jets. The gravitomagnetic model developed by Reva Kay Williams predicts the observed high energy particles (~GeV) emitted by quasars and active galactic nuclei; the extraction of X-rays, γ-rays, and relativistic e&lt;span class=&quot;math-container&quot;&gt;$^-$&lt;/span&gt;—&lt;span class=&quot;math-container&quot;&gt;$\,e^+$&lt;/span&gt; pairs; the collimated jets about the polar axis; and the asymmetrical formation of jets (relative to the orbital plane).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;See also:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Penrose_process&quot; rel=&quot;nofollow noreferrer&quot;&gt;Penrose process&lt;/a&gt; (also called Penrose mechanism) is a process theorised by &lt;a href=&quot;https://en.wikipedia.org/wiki/Roger_Penrose&quot; rel=&quot;nofollow noreferrer&quot;&gt;Roger Penrose&lt;/a&gt; wherein &lt;strong&gt;energy can be extracted&lt;/strong&gt; from a &lt;a href=&quot;https://en.wikipedia.org/wiki/Rotating_black_hole&quot; rel=&quot;nofollow noreferrer&quot;&gt;rotating black hole&lt;/a&gt;. That extraction is made possible &lt;strong&gt;because [some of] the rotational energy&lt;/strong&gt; of the black hole &lt;strong&gt;is located&lt;/strong&gt; not inside the &lt;a href=&quot;https://en.wikipedia.org/wiki/Event_horizon&quot; rel=&quot;nofollow noreferrer&quot;&gt;event horizon&lt;/a&gt; of the black hole, but on the &lt;strong&gt;outside&lt;/strong&gt; of it in a region of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kerr_metric&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kerr&lt;/a&gt; metric &lt;a href=&quot;https://en.wikipedia.org/wiki/Spacetime&quot; rel=&quot;nofollow noreferrer&quot;&gt;spacetime&lt;/a&gt; called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ergosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;ergosphere&lt;/a&gt;, a region in which a particle is necessarily propelled in locomotive concurrence with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Frame-dragging&quot; rel=&quot;nofollow noreferrer&quot;&gt;rotating spacetime&lt;/a&gt;. In the process, a lump of matter enters into the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ergosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;ergosphere&lt;/a&gt; of the black hole, and once it enters the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ergosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;ergosphere&lt;/a&gt;, it is forcibly split into two parts.&lt;/p&gt;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&lt;p&gt;Although &lt;a href=&quot;https://en.wikipedia.org/wiki/Conservation_of_momentum&quot; rel=&quot;nofollow noreferrer&quot;&gt;momentum is conserved&lt;/a&gt; the effect is that &lt;strong&gt;more energy can be extracted than was originally provided&lt;/strong&gt;, the difference being &lt;strong&gt;provided by the black hole&lt;/strong&gt; itself. In summary, the process results in a slight decrease in the angular momentum of the black hole, which corresponds to a transference of energy to the matter. The &lt;strong&gt;momentum lost is converted to energy extracted&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The maximum amount of energy gain possible for a single particle via this process is 20.7%. The process obeys the &lt;a href=&quot;https://en.wikipedia.org/wiki/Black_hole_thermodynamics&quot; rel=&quot;nofollow noreferrer&quot;&gt;laws of black hole mechanics&lt;/a&gt;. A consequence of these laws is that if the process is performed repeatedly, the black hole can eventually lose all of its angular momentum, becoming non-rotating, i.e. a &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_black_hole&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schwarzschild black hole&lt;/a&gt;. In this case &lt;strong&gt;the theoretical maximum energy that can be extracted from a black hole is 29% its original mass&lt;/strong&gt;. Larger efficiencies are possible for charged rotating black holes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/1/10/Ergosphere_and_event_horizons_of_a_rotating_black_hole.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UqJoi.gif&quot; alt=&quot;Rotating Black Hold.&quot; /&gt;&lt;/a&gt;&#xA;Click to animate.&lt;/p&gt;&#xA;&lt;p&gt;An &lt;strong&gt;unrelated&lt;/strong&gt; effect is described as &lt;a href=&quot;https://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hawking radiation&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Physical insight into the process may be gained by imagining that particle–antiparticle radiation is emitted from just beyond the event horizon. This radiation does not come directly from the black hole itself, but rather is a result of virtual particles being &amp;quot;boosted&amp;quot; by the black hole's gravitation into becoming real particles. As the particle–antiparticle pair was produced by the black hole's gravitational energy, the escape of one of the particles lowers the mass of the black hole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;That shouldn't be confused with the &lt;a href=&quot;https://physics.stackexchange.com/questions/66064/can-anyone-please-explain-hawking-penrose-singularity-theorems-and-geodesic-inco&quot;&gt;Hawking-Penrose Singularity Theorem&lt;/a&gt; either.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-06-10T02:02:30.040" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26580" PostTypeId="2" ParentId="8605" CreationDate="2018-06-09T20:41:50.983" Score="1" Body="&lt;p&gt;Neal Turner's answer from the &lt;a href=&quot;http://www.ustream.tv/recorded/72967023#to01:02:14&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;The Birth of Planets&quot; von Karman Lecture&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How would we detect planets around a far-flung, identical copy of our solar system? Would our planets need to be detected using the transit method?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the whole, yes. Jupiter you could probably detect by the radial velocity method if you're willing to wait one orbit or maybe two to be sure, so 12 years for Jupiter to go around the sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other planets would be really tough. If they transited you could detect them with technology similar to ours. You'd have to be lucky because our solar system is not compact like [others discovered by Kepler]; it's quite spread out. If you have a planet very close to its star, you have a decent chance, if it has a random orientation, that it'll be along your line of sight. If it's very far away there's many more possibilities for its orientation and there's a much smaller probability if things are random that you'll get it along your line of sight exactly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for someone to see our Jupiter from a nearby star is much less likely than for us to see a hot Jupiter. There's only a small number of aliens looking at our solar system and seeing it through transits right now.&lt;/p&gt;&#xA;" OwnerUserId="2098" LastActivityDate="2018-06-09T20:41:50.983" CommentCount="0" CommunityOwnedDate="2018-06-09T20:41:50.983" ContentLicense="CC BY-SA 4.0" />
  <row Id="26581" PostTypeId="1" AcceptedAnswerId="26595" CreationDate="2018-06-09T21:26:52.140" Score="4" ViewCount="151" Body="&lt;p&gt;I sincerely apologize if this question is stupid, however after all the articles I read so far, I could not confirm this assumption, which I think is actually a fact, but I'm not completely confident.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm writing on the topic of proportions of the universe, and in one of the statements I claim the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Just one small void inside a supercluster of galaxies can span across 10 to 20 megaparsecs (32 to 65 million light-years ) in diameter.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me state that I understand that a void within a galaxy supercluster may not necessarily be completely surrounded by galaxies of that supercluster, and can extend far out from the supercluster it partially occupies, so to clarify what I mean by my question on a practical example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to draw a straight line across a map of a galaxy supercluster, can that line start inside a group of galaxies of that supercluster, go across a void of 10 - 20 megaparsecs, and end inside another group of galaxies of that same supercluster?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am basing my &quot;small&quot; classification of a cosmic void 10 - 20 megaparsecs in diameter on the Wikipedia article &quot;Void (Astronomy)&quot; &lt;a href=&quot;https://en.wikipedia.org/wiki/Void_(astronomy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Void_(astronomy)&lt;/a&gt; in which it says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot; ...Voids typically have a diameter of 10 to 100 megaparsecs; particularly large voids, defined by the absence of rich superclusters, are sometimes called supervoids. ...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My only goal is basically to find out whether I would be wrong or not if I state that small voids of 10 - 20 megaparsecs, really actually exist within, and not only between superclusters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any eventual help on this topic, I would be sincerely grateful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kind regards&lt;/p&gt;&#xA;" OwnerUserId="23307" LastActivityDate="2018-06-11T04:57:08.820" Title="Are there small voids of 10 - 20 megaparsecs also within, and not only between superclusters?" Tags="&lt;galaxy-cluster&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26582" PostTypeId="2" ParentId="26544" CreationDate="2018-06-10T01:14:08.693" Score="2" Body="&lt;p&gt;Venus goes through a 19 month synodic cycle as seen from Earth, roughly:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;8 months as an evening star&lt;/li&gt;&#xA;&lt;li&gt;1 month passing in front of the Sun&lt;/li&gt;&#xA;&lt;li&gt;8 months as a morning star&lt;/li&gt;&#xA;&lt;li&gt;2 months passing behind the Sun&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In any given month, Venus can be an evening star all month, be a morning star all month, be too close to the Sun, or pass from one state to the next.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-06-10T01:14:08.693" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26583" PostTypeId="1" AcceptedAnswerId="26584" CreationDate="2018-06-10T05:53:34.917" Score="2" ViewCount="1291" Body="&lt;p&gt;More massive the body is more is the gravity.Still often smaller black holes are considered to be very dangerous than the bigger ones,why?&lt;/p&gt;&#xA;" OwnerUserId="22623" LastEditorUserId="616" LastEditDate="2018-06-10T17:25:28.477" LastActivityDate="2018-06-10T17:25:28.477" Title="Why smaller black holes are considered to be more dangerous than bigger black holes?" Tags="&lt;black-hole&gt;&lt;astrophysics&gt;&lt;supermassive-black-hole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26584" PostTypeId="2" ParentId="26583" CreationDate="2018-06-10T06:37:05.220" Score="2" Body="&lt;p&gt;All black holes are dangerous if you get close enough, but smaller ones are more dangerous because they cause stronger tidal effects, so they can cause spaghettification well before you reach the event horizon, whereas you can cross the event horizon of a supermassive black hole without the tidal effects becoming noticeable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very small black holes would also be dangerous due to the large amount of Hawking radiation they'd emit. E.g., a 1 million metric ton hole has a luminosity of over 356 terrawatts. However, there's no known mechanism to produce black holes that small, apart from waiting for a large hole to get small by evaporating via Hawking radiation, and that won't even start to happen for a &lt;em&gt;very&lt;/em&gt; long time, since the cosmic microwave background is currently around a billion times hotter than even the smallest stellar black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, the main danger in the neighborhood of a black hole is if the hole is active, that is, if it is currently surrounded by matter that it's absorbing. That matter gets accelerated to very high speeds due to the gravitational acceleration and when it collides with other matter large amounts of dangerous radiation is released.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2018-06-10T06:37:05.220" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26585" PostTypeId="1" AcceptedAnswerId="26586" CreationDate="2018-06-10T06:51:22.420" Score="1" ViewCount="128" Body="&lt;p&gt;Certainly all stars have enough energy for a planet to host life. But a more magnetically active star is not preferred because solar ejection could strip planets of their atmosphere. What's the right balance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Which stellar category of a star is suitable for hosting life excluding G2V class (our Sun's category)?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="22623" LastEditorUserId="22623" LastEditDate="2018-06-10T12:00:47.470" LastActivityDate="2018-06-10T12:00:47.470" Title="What types of star do we know are hostile to life (like ours)?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;astrophysics&gt;&lt;life&gt;&lt;earth-like-planet&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26586" PostTypeId="2" ParentId="26585" CreationDate="2018-06-10T09:21:02.483" Score="4" Body="&lt;p&gt;We so far have a sample size of 1. We know that a largish star, with a surface temperature at about 4.5 billion years of 5800K can be stable enough to support life for long enough for intelligence to evolve. We are fairly sure that very large stars don't last long enough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With only one data point to extrapolate there is a tendency to think that stars that are like the sun offer the best environment for life. We are fairly sure that very large stars don't last long enough. It seems reasonable to exclude highly variable stars, or ones whose outbursts would strip the atmosphere from any planet. Intuitively these would seem to be less likely sites for life. But there is no strong evidence in the form of multiple discoveries of extra-terrestrial life to support this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short we don't know what stars can support life.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-06-10T09:21:02.483" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26587" PostTypeId="1" AcceptedAnswerId="26591" CreationDate="2018-06-10T17:57:21.700" Score="3" ViewCount="189" Body="&lt;p&gt;In the Astronomy magazine I read that the stellar mass black holes are only the size of Rhode Island. Can they swallow Earth?&lt;/p&gt;&#xA;" OwnerUserId="23317" LastActivityDate="2018-06-11T00:18:06.230" Title="Can a stellar mass black hole swallow Earth?" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26588" PostTypeId="2" ParentId="14475" CreationDate="2018-06-10T18:05:40.340" Score="0" Body="&lt;p&gt;If you read the Astronomy magazine, you should have seen the star charts, and the article about Jupiter and the moons. So, the answer is no, the activity does &lt;strong&gt;not&lt;/strong&gt; affect it's orbit.&lt;/p&gt;&#xA;" OwnerUserId="23317" LastActivityDate="2018-06-10T18:05:40.340" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26589" PostTypeId="2" ParentId="26565" CreationDate="2018-06-10T18:16:55.393" Score="-1" Body="&lt;p&gt;It is possible that the red dwarfs can support life, but we would need a strong magnetic field and a path that would take us closer and closer because that dwarf will cool overtime. A white dwarf would not be possible because they are the remnants of dead stars, so Earth would not have survived the blast (the other planets wouldn't have either).&lt;/p&gt;&#xA;" OwnerUserId="23317" LastActivityDate="2018-06-10T18:16:55.393" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26590" PostTypeId="1" CreationDate="2018-06-10T21:39:22.190" Score="4" ViewCount="62" Body="&lt;p&gt;Do the cometary gas ions which make up the plasma tail of a comet reach the velocity of the Sun's solar wind?&lt;/p&gt;&#xA;" OwnerDisplayName="user23319" LastEditorDisplayName="user23319" LastEditDate="2018-06-10T21:48:06.160" LastActivityDate="2018-06-11T13:18:08.937" Title="Do the cometary gas ions which make up the plasma tail of a comet reach the velocity of the Sun's solar wind?" Tags="&lt;comets&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26591" PostTypeId="2" ParentId="26587" CreationDate="2018-06-10T22:03:57.500" Score="5" Body="&lt;p&gt;There's a fairly common misconception in this question that black holes &quot;suck everything in&quot;.   They do - at distances close to their event horizon but as noted, the event horizon's diameter is about the distance across rhode island (different masses have different sizes), but many stellar mass black holes would fit inside the border of rhode island.  Virtually all of them would fit inside pennsylvania for example.   They're tiny.    I'd rate my chances of survival as much higher in a space-ship 1 million miles from a stellar mass black hole than in a space-ship 1 million miles from our Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At normal planetary distances, a planet would just orbit a black hole and likely maintain a stable orbit for billions of years.   Gravitationally, they're no different than a star at those distances.  The orbital mechanics is based on mass and a stellar mass black hole has stellar mass.   What makes it different is it's shrunk inside it's schwarzchild radius, so compared to stars, it's incredibly small and very dense.   The &quot;weird stuff&quot; only happens at very close distances - basically at distances you can't reach around a star because you'd be inside the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So . . . what would happen if you put the Earth next to a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure why you'd want to do that, but the Earth would begin to break apart form tidal forces before it got that close.  The gravity would be stronger on the near side than the far side and the Earth would stretch.  This is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Spaghettification&quot; rel=&quot;nofollow noreferrer&quot;&gt;spagettification&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you theoretically made a black hole just appear near the Earth then it would begin to eat the Earth.   Think of a black hole as a drain, a drain some 20-30 miles across, and think of the Earth as a liquid trying to get down that drain.   It's a lot of material squeezing into a comparatively tiny space and in the process the Earth material would get very hot, and likely form into an &lt;a href=&quot;https://en.wikipedia.org/wiki/Accretion_disk&quot; rel=&quot;nofollow noreferrer&quot;&gt;accretion disk&lt;/a&gt; trying to get inside the black hole and the formation of the accretion disk would eject two cosmic jets along the axes.   &lt;a href=&quot;http://www.sciencemag.org/news/2011/03/scienceshot-black-holes-are-messy-eaters&quot; rel=&quot;nofollow noreferrer&quot;&gt;Black holes are sloppy eaters&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't tell you the precise amount of Earth that would be absorbed if they were placed right next to each other, but that's also an impossible scenario.   Everything in space is moving and gravity guarantees that if the Earth did meet or closely orbit a stellar mass black hole, that it would be moving enormously fast.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth could in theory, but extremely unlikely, either orbit a black hole or collide with one, or pass very close to one, close enough to break apart.  Each scenario would have a different outcome with a different percentage of Earth falling into the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't feel like going through each scenario individually, but the short answer is that the Earth would get quickly ripped apart and some of it, but not all, would fall into the black hole.  Some would be ejected as high speed charged particles.   The matter spiralling in would create a very strong magnetic field and because protons and electrons are charged, much of the matter wouldn't entirely follow the gravitational forces.  Some would get shot out as &lt;a href=&quot;https://en.wikipedia.org/wiki/Accretion_disk#Magnetic_fields_and_jets&quot; rel=&quot;nofollow noreferrer&quot;&gt;jets of charged particles&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a rough answer anyway.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-06-11T00:18:06.230" LastActivityDate="2018-06-11T00:18:06.230" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26593" PostTypeId="2" ParentId="26573" CreationDate="2018-06-11T01:36:12.933" Score="1" Body="&lt;p&gt;In addition to James K's answer, you might find the following interesting and of relevance to your question about observing distant planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can't normally detect exoplanets (i.e. planets beyond our own solar system) if they're very far away. It's hard enough to detect dim stars at a distance, and planets capable of supporting intelligent life are &lt;em&gt;much&lt;/em&gt; smaller than stars and emit extremely little light of their own (mostly as infrared, i.e. heat), so this makes it almost impossible to see far-distant planets. The most distant exoplanet we've observed &lt;em&gt;directly&lt;/em&gt; is &lt;a href=&quot;https://en.wikipedia.org/wiki/CVSO_30&quot; rel=&quot;nofollow noreferrer&quot;&gt;CVSO 30 c&lt;/a&gt;, a gas giant 4.7 times the mass of Jupiter and located 1,200 light years from Earth, in the constellation of Orion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We mainly detect exoplanets &lt;em&gt;indirectly&lt;/em&gt;, by observing either a pattern of wobbles in a star's position (the &lt;a href=&quot;https://en.wikipedia.org/wiki/Doppler_spectroscopy&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;radial-velocity method&quot;&lt;/a&gt;), or a pattern of changes in a star's light due to an exoplanet passing in front of the star (the &lt;a href=&quot;https://en.wikipedia.org/wiki/Methods_of_detecting_exoplanets#Transit_photometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;transit method&quot;&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most distant &lt;em&gt;potentially habitable&lt;/em&gt; planet so far confirmed is &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler-443b&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kepler-443b&lt;/a&gt; in the constellation Cygnus, at 2,540 light years from Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most distant of all exoplanets discovered (as at June 2018) is &lt;a href=&quot;https://en.wikipedia.org/wiki/SWEEPS-11&quot; rel=&quot;nofollow noreferrer&quot;&gt;SWEEPS-11&lt;/a&gt;, a &lt;a href=&quot;https://en.wikipedia.org/wiki/Hot_Jupiter&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;hot Jupiter&quot;&lt;/a&gt; with a mass 9.7 times that of Jupiter and orbiting so close to its parent star that its &quot;year&quot; is only 43 hours. It's around 27,700 light years away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these exoplanets are within our own Milky Way galaxy. To give you an idea of distance, the centre of the Milky Way is about 26,000 light years from Earth, and from the centre to the outer edge of the galactic disk is up to 90,000 light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mention a distance of 3 million light years away. This is well beyond the &lt;a href=&quot;https://en.wikipedia.org/wiki/Andromeda_Galaxy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Andromeda Galaxy&lt;/a&gt;, the nearest major galaxy to our own, at around 2.5 million light years. Detection of exoplanets at that distance isn't currently possible with the transit or radial-velocity methods, but &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_microlensing&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravitational microlensing&lt;/a&gt; provides a method of detecting extremely distant objects. Analysis of the 1999 microlensing event &lt;a href=&quot;https://en.wikipedia.org/wiki/PA-99-N2&quot; rel=&quot;nofollow noreferrer&quot;&gt;PA-99-N2&lt;/a&gt; &lt;em&gt;suggests&lt;/em&gt; the presence of a planet orbiting a star in the Andromeda Galaxy. However, intelligent aliens would have to be able to send signals of inconceivable strength for us to be able to detect them at that distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even more interesting is the claim, &lt;a href=&quot;https://news.nationalgeographic.com/2018/02/exoplanets-discovery-milky-way-galaxy-spd/&quot; rel=&quot;nofollow noreferrer&quot;&gt;published in February this year&lt;/a&gt;, of the microlens discovery of a collection of about 2,000 &lt;a href=&quot;https://en.wikipedia.org/wiki/Rogue_planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;rogue planets&lt;/a&gt; in the amazing quasar &lt;a href=&quot;https://en.wikipedia.org/wiki/RX_J1131-1231&quot; rel=&quot;nofollow noreferrer&quot;&gt;RX J1131-1231&lt;/a&gt;, which is 3.8 &lt;em&gt;billion&lt;/em&gt; light-years distant.&lt;/p&gt;&#xA;" OwnerUserId="14094" LastActivityDate="2018-06-11T01:36:12.933" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26594" PostTypeId="1" CreationDate="2018-06-11T02:12:56.233" Score="7" ViewCount="218" Body="&lt;p&gt;What is the value of Saturn's &quot;tropical year&quot; (as opposed to its &quot;sidereal year&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am seeing it listed as 10 days shorter (10,747 days) than the sidereal year (10,759.22 days).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could this be if the precession of the &lt;strong&gt;axis&lt;/strong&gt; (equinox) is so slow &lt;a href=&quot;https://www.astro.umd.edu/~dphamil/research/preprints/WardHam04.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;(~1.86M years)&lt;/a&gt;? Is this an &quot;transient/osculating&quot; value? Frame of reference (e.g. J2000)? Am I completely missing something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;em&gt;think&lt;/em&gt; I can calculate the tropical year by (close to, this works for Earth):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_{tropical} = P_{sidereal} \cdot (1-\frac{P_{sidereal} }{P_{precession}})$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Using &lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/saturnfact.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA planetary facts data on Saturn&lt;/a&gt;&#xA;(these figures are re-used on many websites):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$10759.22\cdot(1-\frac{10759.22}{1.86\times10^6 \space\times\space 365.256})=10759.05$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Using &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL Horizons&lt;/a&gt; data:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$10755.698\cdot(1-\frac{10755.698}{1.86\times10^6 \space\times\space 365.256})=10755.528$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;about 0.170 days less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, interestingly, see the &lt;em&gt;instantaneous&lt;/em&gt; &lt;strong&gt;sidereal&lt;/strong&gt; year of Saturn from the graph of JPL Horizon's data, using the sun's centre (blue) and solar system barycentre (red). I think the &lt;em&gt;instantaneous&lt;/em&gt; &lt;strong&gt;tropical&lt;/strong&gt; year (at epoch) would always be 0.170 days less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is where the confusion is: texts have used particular values, and further the &lt;em&gt;same&lt;/em&gt; text has picked a sidereal from one epoch (or average), and a tropical from another...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Y8GdD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y8GdD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn's instantaneous length of sidereal year in days, at epoch.&lt;/p&gt;&#xA;" OwnerUserId="23322" LastEditorUserId="23322" LastEditDate="2018-06-14T22:02:37.990" LastActivityDate="2018-06-14T22:02:37.990" Title="Saturn's tropical year?" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26595" PostTypeId="2" ParentId="26581" CreationDate="2018-06-11T04:57:08.820" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My only goal is basically to find out whether I would be wrong or not if I state that small voids of 10 - 20 megaparsecs, really actually exist within, and not only between superclusters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The largest structure is called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Galaxy_filament&quot; rel=&quot;nofollow noreferrer&quot;&gt;filament&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are in the supercluster &lt;a href=&quot;https://en.wikipedia.org/wiki/Laniakea_Supercluster&quot; rel=&quot;nofollow noreferrer&quot;&gt;Laniakea&lt;/a&gt;, in the smaller supercluster &lt;a href=&quot;https://en.wikipedia.org/wiki/Virgo_Supercluster&quot; rel=&quot;nofollow noreferrer&quot;&gt;Virgo&lt;/a&gt;, in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Milky_Way&quot; rel=&quot;nofollow noreferrer&quot;&gt;Milky Way&lt;/a&gt; galaxy, in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_System&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solar&lt;/a&gt; system, on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/f/f4/Earth%27s_Location_in_the_Universe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uW2lg.jpg&quot; alt=&quot;Earth to Laniakea&quot;&gt;&lt;/a&gt;$$\text{Click to zoom in - 23,624 × 2,953 pixels}$$&#xA;&lt;a href=&quot;https://i.stack.imgur.com/I8PgS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I8PgS.jpg&quot; alt=&quot;Laniakea&quot;&gt;&lt;/a&gt;$$\text{Laniakea}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AFi0Z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AFi0Z.jpg&quot; alt=&quot;Our Local Group&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Supercluster#Distribution:_cosmic_voids_and_sheets&quot; rel=&quot;nofollow noreferrer&quot;&gt;Distribution: cosmic voids and sheets&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Research has tried to understand the way superclusters are arranged in space. Maps are used to display the positions of 1.6 million galaxies. Three-dimensional maps are used to further understand the positions of these superclusters. To map them three-dimensionally, the position of the galaxy in the sky as well as the galaxy's redshift are used for calculation. The galaxy's redshift is used with Hubble's law to determine its position in three-dimensional space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was discovered from those maps that superclusters of galaxies are not spread uniformly across the universe but they seem to lie along filaments. Maps reveal huge voids where there are extremely few galaxies. Some dim galaxies or hydrogen clouds can be found in some voids, but most galaxies are found in sheets between the voids. The voids themselves are often spherical but the superclusters are not. They can range from being &lt;strong&gt;100 million to 400 million light-years&lt;/strong&gt; [that's &lt;strong&gt;30.660139381 to 122.64055752 megaparsec&lt;/strong&gt;] in diameter. The pattern of sheets and voids contains information about how galaxy clusters formed in the early universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a sponge analogy used often that compares a sponge to the pattern of clusters of galaxies in the universe – the holes are the voids and the other parts are the locations of the superclusters.&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, yes - there are small and large voids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also this Wikipedia page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Virgo_Supercluster#Galaxy_distribution&quot; rel=&quot;nofollow noreferrer&quot;&gt;Galaxy distribution&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number density of galaxies in the LS falls off with the square of the distance from its center near the Virgo Cluster, suggesting that this cluster is not randomly located. Overall, the vast majority of the luminous galaxies (less than absolute magnitude −13) are concentrated in a small number of clouds (groups of galaxy clusters). Ninety-eight percent can be found in the following 11 clouds (given in decreasing order of number of luminous galaxies): Canes Venatici, Virgo Cluster, Virgo II (southern extension), Leo II, Virgo III, Crater (NGC 3672), Leo I, Leo Minor (NGC 2841), Draco (NGC 5907), Antlia (NGC 2997) and NGC 5643. Of the luminous galaxies located in the disk, one third are in the Virgo Cluster, while the remainder are found in the Canes Venatici Cloud and Virgo II Cloud, plus the somewhat insignificant NGC 5643 Group. The luminous galaxies in the halo are also concentrated in a small number of clouds (94% in 7 clouds). This distribution indicates that &quot;most of the volume of the supergalactic plane is a great void.&quot; A helpful analogy that matches the observed distribution is that of soap bubbles. Flattish clusters and superclusters are found at the intersection of bubbles, which are large, roughly spherical (on the order of &lt;strong&gt;20–60 Mpc in diameter&lt;/strong&gt;) voids in space. Long filamentary structures seem to predominate. An example of this is the Hydra-Centaurus Supercluster, the nearest supercluster to the LS, which starts at a distance of roughly 30 Mpc and extends to 60 Mpc.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastActivityDate="2018-06-11T04:57:08.820" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26596" PostTypeId="1" AcceptedAnswerId="26597" CreationDate="2018-06-11T09:41:30.423" Score="1" ViewCount="228" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1lUFM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1lUFM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;em&gt;(Our sun compared to some of the known stars)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;I know that the star is born in a nebula.Do only a extremely gigantic nebula give rise to large-radius star or is there any other factors related?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="22623" LastEditorUserId="907" LastEditDate="2018-06-11T14:24:34.903" LastActivityDate="2018-06-11T14:24:34.903" Title="What factors makes a star big in size(Physically)?Is the size of the nebula a relevant factor?" Tags="&lt;star&gt;&lt;stellar-evolution&gt;&lt;nebula&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26597" PostTypeId="2" ParentId="26596" CreationDate="2018-06-11T10:28:51.740" Score="3" Body="&lt;p&gt;The size of the nebula doesn't matter at all, what matters to the size of a star is the mass and age of that star.  Even the mass of the nebula doesn't really matter, because giant molecular clouds fragment dramatically as they make many many stars.  It's a very complicated and detailed process that controls how they fragment, so usually one merely assumes a statistical distribution of stellar masses, called the &quot;initial mass function,&quot; which applies more or less universally to any sized nebula. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the mass of the star is determined (statistically), the star contracts with age, so very young stars have all kinds of different sizes while they contract.  Eventually fusion initiates in the core (which is sometimes considered the &quot;birth&quot; of the star, though it was pretty much already a star before that), and then the size of the star is determined by its mass because a higher mass star doesn't need to contract as small to reach fusion temperatures.  Later, when core fuel runs out, stellar envelopes expand dramatically into giants or supergiants, which to a point reverses the process for the envelope as what happened in the original contraction of the star.  The size varies a lot after that.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2018-06-11T10:34:28.837" LastActivityDate="2018-06-11T10:34:28.837" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26598" PostTypeId="2" ParentId="26590" CreationDate="2018-06-11T10:41:13.253" Score="1" Body="&lt;p&gt;Yes, the ions are said to be &quot;swept out&quot; by the solar wind, such that they reach the speed of the solar wind.  That speed varies a fair amount, however, because there is both a &quot;fast&quot; (about 600 km/s) and &quot;slow&quot; (about 300 km/s) solar wind.  It is really the magnetic fields entrained in the wind that sweep out the charged ions, since each ion has such a low mass it is easily affected by those fields.  Note the dust particles from a comet, which is the tail that we see, are much more massive and are not affected by the solar wind, so they are moving out much more slowly and often make a noticeably curved shape.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-06-11T10:41:13.253" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26600" PostTypeId="2" ParentId="26594" CreationDate="2018-06-11T14:38:22.643" Score="1" Body="&lt;p&gt;A quick skim of a few papers on Earth's tropical year, e.g. &lt;a href=&quot;http://www.public.asu.edu/~mjwhite/Sidereal%20and%20tropical%20years.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;M.J.White&lt;/a&gt;, suggests to me that existing formulas are best-fit to data, not analytic.  How Saturn's ellipticity of orbit &amp;amp; axial precession(s) play into this is beyond my learning, but I'm getting the feeling that &quot;it just is,&quot; and that any number of cofactors such as planetary layers, densities, and so on might affect what's being observed.&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-06-11T14:38:22.643" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26601" PostTypeId="2" ParentId="18588" CreationDate="2018-06-11T21:51:34.190" Score="1" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;I suppose it is the Saturn system's barycenter which has been located gravitationally. At this level of precision I suppose that the moons', especially Titan's tidal effects must be considered. ..., so I wonder if these too are being considered at the 4 km precision level, and overall exactly how the point located is defined?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Short answer:&lt;/strong&gt; No.&lt;/p&gt;&#xA;&lt;p&gt;As pointed out in the comments by @AtmosphericPrisonEscape (Oct 5 '16 at 14:42) and @userLTK (Oct 6 '16 at 9:25) - it is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Barycenter&quot; rel=&quot;nofollow noreferrer&quot;&gt;barycentric&lt;/a&gt; measurement.&lt;/p&gt;&#xA;&lt;p&gt;Other references:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/26510/18054&quot;&gt;Solid-body tides on the Moon&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/26484/18054&quot;&gt;Is Io's orbit or rotation affected by its volcanism?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The Barycenter coordinates are &lt;strong&gt;an average, a mean&lt;/strong&gt;, small &lt;a href=&quot;https://en.wikipedia.org/wiki/Perturbation_(astronomy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;perturbations&lt;/a&gt; are smoothed in the calculations by various factors such as: measurement accuracy, &lt;a href=&quot;http://adsabs.harvard.edu/abs/2016DPS....4842123J&quot; rel=&quot;nofollow noreferrer&quot;&gt;use of the International Celestial Reference Frame&lt;/a&gt; (ICRF), even the &lt;a href=&quot;http://joelouismarino.github.io/blog_posts/blog_whitening.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cholesky whitening&lt;/a&gt; (Source: &amp;quot;&lt;a href=&quot;https://arxiv.org/abs/1008.3607&quot; rel=&quot;nofollow noreferrer&quot;&gt;Measuring the mass of solar system planets using pulsar timing&lt;/a&gt;&amp;quot; (21 Aug 2010)) used in some of the calculations.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;In astronomy, &lt;a href=&quot;https://en.wikipedia.org/wiki/Barycenter&quot; rel=&quot;nofollow noreferrer&quot;&gt;barycentric coordinates&lt;/a&gt; are non-rotating coordinates with the origin at the center of mass of two or more bodies. The International Celestial Reference System is a barycentric one, based on the barycenter of the Solar System.&lt;/p&gt;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Relativistic corrections&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;In classical mechanics, this definition simplifies calculations and introduces no known problems. In general relativity, problems arise because, while it is possible, within reasonable approximations, to define the barycenter, the associated coordinate system does not fully reflect the inequality of clock rates at different locations&lt;/strong&gt;. Brumberg explains how to set up barycentric coordinates in general relativity in his book &amp;quot;&lt;a href=&quot;https://books.google.ca/books?id=yA1oLkHkNGMC&amp;amp;printsec=frontcover&amp;amp;dq=0750300620&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwiP8sHHv8zbAhUFNX0KHa5ACMIQ6AEICjAA#v=onepage&amp;amp;q=0750300620&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;Essential Relativistic Celestial Mechanics&lt;/a&gt;&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;The coordinate systems involve a world-time, i.e. a global time coordinate that could be set up by telemetry. Individual clocks of similar construction will not agree with this standard, because they are subject to differing gravitational potentials or move at various velocities, so the world-time must be slaved to some ideal clock that is assumed to be very far from the whole self-gravitating system. This time standard is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Barycentric_Coordinate_Time&quot; rel=&quot;nofollow noreferrer&quot;&gt;Barycentric Coordinate Time&lt;/a&gt;, or TCB.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JeNel.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JeNel.png&quot; alt=&quot;SSB - Solar System Barycenter&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Example: Motion of the &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Solar_system_barycenter.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;barycenter of the Solar System&lt;/a&gt; relative to the Sun, 1945–1995.&lt;/p&gt;&#xA;&lt;p&gt;In the paper &amp;quot;&lt;a href=&quot;https://www.researchgate.net/publication/268360163_UNCERTAINTIES_IN_THE_JPL_PLANETARY_EPHEMERIS&quot; rel=&quot;nofollow noreferrer&quot;&gt;Uncertainties in the JPL Planetary Ephemeris&lt;/a&gt;&amp;quot;, by Folkner, on page 47, is shown this chart of the uncertainty in right ascension, declination, and distance of the barycenters of the Earth and Saturn systems calculated for the years 1950 to 2050:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tEyFH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tEyFH.jpg&quot; alt=&quot;Saturn's Barycenter Error&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Celestial mechanics and the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/N-body_problem#Many_bodies&quot; rel=&quot;nofollow noreferrer&quot;&gt;n-body problem&lt;/a&gt; take a long time to solve.&lt;/p&gt;&#xA;&lt;p&gt;The paper associated with these measurements of Saturn is titled: &amp;quot;&lt;a href=&quot;https://arxiv.org/abs/1012.0264&quot; rel=&quot;nofollow noreferrer&quot;&gt;VLBA Astrometric Observations of the Cassini Spacecraft at Saturn&lt;/a&gt;&amp;quot; (1 Dec 2010), on page 2 it says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;These observations provide positions for the center of mass of Saturn in the International Celestial Reference Frame (ICRF) with accuracies ∼0.3 milli-arcsecond (1.5 nrad), or about 2 km at the average distance of Saturn.&lt;/p&gt;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&lt;p&gt;The DE 422 post-fit residuals for Saturn with respect to the VLBA data are generally 0.2 &lt;a href=&quot;https://en.wikipedia.org/wiki/Minute_and_second_of_arc#Symbols_and_abbreviations&quot; rel=&quot;nofollow noreferrer&quot;&gt;mas&lt;/a&gt;, but additional observations are needed to improve the positions of all of our phase reference sources to this level. Over time we expect to be able to improve the accuracy of all three coordinates in the Saturn ephemeris (latitude, longitude, and range) by a factor of at least three. This will represent a significant improvement not just in the Saturn ephemeris but also in the link between the inner and outer solar system ephemeredes and in the link to the inertial ICRF.&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;sub&gt;I might return for an edit if there is interest in this Q&amp;amp;A.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-06-11T22:21:46.993" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26602" PostTypeId="2" ParentId="26566" CreationDate="2018-06-11T23:06:12.123" Score="0" Body="&lt;p&gt;According to &lt;a href=&quot;https://starchild.gsfc.nasa.gov/docs/StarChild/questions/question15.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; the main galaxy you can see besides the milky way is Andromeda. However, the Large and Small Magellanic clouds can be southern hemisphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find Andromeda, Start at the northeast corner of the Great Square of Pagasus - the constellation Andromeda forms a &quot;handle&quot; attached to the northeast corner of the &quot;bowl&quot; formed by the Great Square. Find the second bright star in the handle (the last star before the end of the handle) and from that star, make a 90 degree turn to the line that joins the two handle stars, pass the first star you see and look at the 2nd &quot;star&quot;. It appears to be a &quot;fuzzy star&quot;. That is the Andromeda Galaxy. (Source: See above)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A chart to help you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H1sNN.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H1sNN.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll probably need a telescope to view other galaxies. Other than that, that is pretty much it!&lt;/p&gt;&#xA;" OwnerUserId="16048" LastActivityDate="2018-06-11T23:06:12.123" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26603" PostTypeId="2" ParentId="26513" CreationDate="2018-06-11T23:46:35.730" Score="2" Body="&lt;p&gt;Without taking temperature and pressure into account, from my old code I found this approximate formula: &lt;em&gt;(angles in degrees)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta \alpha_{reflect} = tan[(90°-\alpha_{technical}) \times 0.9679] \times 0.023626$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this calculated reflection angle exceeds $\frac{34.9}{60}$ degrees, use that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observed elevation would be $\alpha_{observed}=\alpha_{technical}+\Delta\alpha_{reflect}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Example&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a point technically (without atmosphere) 34.9' (minutes of arc) below horizon, &lt;em&gt;appears&lt;/em&gt; on the horizon.&lt;/li&gt;&#xA;&lt;li&gt;a point technically on the horizon, &lt;em&gt;appears&lt;/em&gt; 28.1' above the horizon.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="23322" LastActivityDate="2018-06-11T23:46:35.730" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26604" PostTypeId="2" ParentId="79" CreationDate="2018-06-12T00:02:17.320" Score="3" Body="&lt;p&gt;The hottest stars - and here, I assume that &quot;star&quot; excludes stellar remnants such as white dwarfs, neutron stars, and other exotic compact objects - are likely &lt;a href=&quot;https://en.wikipedia.org/wiki/Wolf%E2%80%93Rayet_star&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wolf-Rayet stars&lt;/a&gt;, a class of hot, hydrogen-deficient stars characterized by hydrogen depletion and noticeable carbon, nitrogen and oxygen lines. The massive Population I sub-type are likely former O-type high-mass main sequence stars with exceptionally strong stellar winds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/91/2153&quot;&gt;Guillochon's answer&lt;/a&gt; mentions that O-type stars often have surface temperatures of about 30,000 K. Many, - if not most - Wolf-Rayet stars exceed that by drastic amounts. Some of the hottest may be the Wolf-Rayet components of the binaries &lt;a href=&quot;https://en.wikipedia.org/wiki/AB7&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;AB7&lt;/strong&gt;&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/SMC_AB8&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;AB8&lt;/strong&gt;&lt;/a&gt;, in the Small Magellanic Cloud. Both have normal O-type companions, which are also exceedingly hot. However, the maximum temperatures for the Wolf-Rayet components may be 105,000 K and 141,000 K, respectively (Wikipedia cites &lt;a href=&quot;http://adsabs.harvard.edu/abs/2016A&amp;amp;A...591A..22S&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shenar et al. (2016)&lt;/a&gt; here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here's the problem. &lt;a href=&quot;http://pacrowther.staff.shef.ac.uk/IAUS250.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;It's notoriously difficult to determine the temperatures of Wolf-Rayet stars&lt;/a&gt; to the desired accuracy. Why? Well, it's largely because of their stellar winds and high mass-loss rates. Parts of the atmospheres and winds are optically thick, meaning that we can't necessarily observe where the &quot;surface&quot;, as normally described in stellar astrophysics, lies. Therefore, let's bear in mind that the listed temperatures may be off a bit - although Wolf-Rayet stars are still quite clearly hotter than normal O-type stars.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2018-06-12T00:02:17.320" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26605" PostTypeId="1" AcceptedAnswerId="26606" CreationDate="2018-06-12T00:09:11.123" Score="3" ViewCount="89" Body="&lt;p&gt;I heard and read that a wormhole will collapse if they transport any positively charged matter. So, if wormholes exist, they probably close as soon as normal matter enters it, how can we access them without negative matter?&lt;/p&gt;&#xA;" OwnerUserId="23317" LastEditorUserId="23317" LastEditDate="2018-06-15T01:56:51.583" LastActivityDate="2018-06-15T01:56:51.583" Title="If wormholes exist and behave like we think they do, can we cross them without &quot;negatively charged matter &quot;?" Tags="&lt;wormhole&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2018-06-14T10:49:04.140" ContentLicense="CC BY-SA 4.0" />
  <row Id="26606" PostTypeId="2" ParentId="26605" CreationDate="2018-06-12T00:51:09.210" Score="7" Body="&lt;p&gt;The best current answer is &quot;We can't&quot;.  We do not presently know of any way to stabilize wormholes without large negative energy densities. (Some argue that even that isn't enough, but I don't believe that the matter is settled, even on a theoretical level.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) No wormhole has &lt;em&gt;ever&lt;/em&gt; been observed, nor have we observed anything which is a predicted effect of a wormhole: Wormholes are entirely theoretical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Having said that, Wormholes are a prediction of General Relativity, which is a very well-tested and well-established theory, so it's unwise to bet against them existing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) Macroscopic wormholes are a purely GR phenomenon and while the fields involved are strong, they are not so strong that they are in the realm where we expect the eventual unified theory of quantum gravity to give markedly different results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d) And by the way, we haven't a clue how to create large negative energy densities.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-06-12T00:51:09.210" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26610" PostTypeId="1" AcceptedAnswerId="26611" CreationDate="2018-06-12T11:55:54.397" Score="39" ViewCount="8881" Body="&lt;p&gt;This question is a follow-up to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/11743/do-bigger-telescopes-equal-better-results&quot;&gt;Do bigger telescopes equal better results?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much bigger does a ground-based mirror have to be to match what a space-based one can do?  I guess I'm asking primarily for visible light, but I'm interested in general too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess on the ground, you're safe from micrometeorites, so it will probably last longer.  At what point does it become cheaper to build a telescope on the moon or something?&lt;/p&gt;&#xA;" OwnerUserId="23338" LastActivityDate="2018-06-15T00:34:49.293" Title="Why are we building larger land-based telescopes instead of launching larger ones into space?" Tags="&lt;telescope&gt;&lt;space-telescope&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26611" PostTypeId="2" ParentId="26610" CreationDate="2018-06-12T12:55:37.173" Score="72" Body="&lt;p&gt;It's cheaper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) With adaptive optics you can get 0.1 arc second resolution on the ground (admittedly only on a mountain top with particularly good air flow, but still!). This eliminates one of the major advantages of space until you get above several meters mirror diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Rocket fairings are the shrouds which protect payloads during the supersonic atmospherics speeds reached during launch. A 5 meter fairing is about the largest that can be flown, which limits the size of the one-piece mirrors which can be launched.  (The Dreaded Webb Telescope's mirror is in pieces which will assemble themselves in space -- a &lt;em&gt;very&lt;/em&gt; scary and &lt;em&gt;very&lt;/em&gt; expensive piece of design.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Servicing a telescope on the top of Mauna Kea or in the high Chilean Andes is a difficult and expensive process. Servicing a telescope in orbit makes that look like small change.  (Cost comparable to the cost of &lt;em&gt;building&lt;/em&gt; a new giant scope on Earth.) And in-orbit servicing can't even be done with current technology except in low earth orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) While high resolution is one frontier in astronomy, going deep is another, and going deep requires &lt;em&gt;big&lt;/em&gt; mirrors.  A 30 meter mirror on Earth gathers much more light than a 5 meter mirror in space.  The giant terrestrial telescopes simply do a better job of being light buckets for spectroscopy than anything we can yet put into space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line is that with the development of adaptive optics, space-based telescopes of currently buildable and launchable size lost their main advantage over ground-based telescopes.  And since they're 10x to 100x the cost, they are simply not worth building for many purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space based telescopes still hold a significant edge in parts of the spectrum blocked by the atmosphere such as UV and IR (Webb), and for certain tasks involving long-term high accuracy photometry (Kepler) and astrometry (Gaia).  But for general purpose use, the balance seems firmly on the ground's side for large telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will change if space flight becomes cheaper -- the SpaceX BFR, for example, with its 9 meter fairing and dramatically lower launch costs, offers great hope for space telescopes.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2018-06-12T15:03:52.830" LastActivityDate="2018-06-12T15:03:52.830" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="26612" PostTypeId="2" ParentId="26610" CreationDate="2018-06-12T15:18:11.530" Score="9" Body="&lt;p&gt;Answering your subquestion about building on the moon:&#xA;This is subject to the same launch costs and restrictions as a space-based 'scope, plus you have to deal with landing and with gravitational sag.  So the first thing you need is a functioning moon base that can manufacture all components from local raw materials.   Once that's in place (insert large laugh here), you'll still need adaptive optics (as do multi-element scopes like JWebb) for alignment and gravitational sag offsets, but as those are static, you don't need the high-frequency response required on Earth to handle atmospheric aberrations.  You'll want to build on the &quot;dark side&quot; so that Terran light doesn't foul everything up.  &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-06-12T15:18:11.530" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26613" PostTypeId="1" AcceptedAnswerId="26615" CreationDate="2018-06-12T18:42:36.270" Score="-1" ViewCount="80" Body="&lt;p&gt;What will happen?Will our atmosphere head to the next gravity dominant body?Will our moon hold a part of our atmosphere or is too weak to hold it?Where all those gases are supposed to go?&lt;/p&gt;&#xA;" OwnerUserId="22623" LastEditorUserId="616" LastEditDate="2018-06-12T19:36:10.740" LastActivityDate="2018-06-12T21:20:49.407" Title="Where will the atmosphere of our Earth go if our sun blows it away completely?" Tags="&lt;the-sun&gt;&lt;planet&gt;&lt;planetary-atmosphere&gt;&lt;solar-wind&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26614" PostTypeId="2" ParentId="26610" CreationDate="2018-06-12T20:36:53.467" Score="17" Body="&lt;p&gt;In addition to Mark's great answer ...&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Why  are we building larger land-based telescopes instead of launching larger ones into space?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If you had money for two homes, one near work and a 'summer cottage' in the woods, how would you divide your budget?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;This question is a follow-up to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/11743/do-bigger-telescopes-equal-better-results&quot;&gt;Do bigger telescopes equal better results?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Yes, and I'm not a fan of those answers, perhaps @MarkOlson is also not impressed.&lt;/p&gt;&#xA;&lt;p&gt;Those answers &lt;em&gt;miss&lt;/em&gt; adaptive optics (dismissing it as expensive and not particularly effective) and the ability to &lt;em&gt;easily&lt;/em&gt; upgrade everything except the size of the building and main mirror.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;How much bigger does a ground-based mirror have to be to match what a space-based one can do? I guess I'm asking primarily for visible light, but I'm interested in general too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It's not so much &amp;quot;how much bigger&amp;quot;, it's &amp;quot;effectively market your idea, secure as much funding as possible, and build the largest building with the biggest main mirror possible&amp;quot;. Dig deep and build what you can, not be upgraded as big as you can - sensors and supercomputers can fix the rest.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I guess on the ground, you're safe from micrometeorites, so it will probably last longer. At what point does it become cheaper to build a telescope on the moon or something?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Ground and space-based telescopes are useful, moon-based less so.&lt;/p&gt;&#xA;&lt;p&gt;When we have &amp;quot;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Acme_Corporation&quot; rel=&quot;noreferrer&quot;&gt;Acme&lt;/a&gt; Telescope Company&amp;quot; open their first store on the moon the price to purchase will fall, until then Earth and space-based will be cheaper. With space-based it can meet you halfway for repairs, with ground based (even on the top of a mountain) a repair facility is often close at hand.&lt;/p&gt;&#xA;&lt;p&gt;At Paranal the &lt;a href=&quot;https://www.eso.org/public/images/mmb-coatn-tank-pano4/&quot; rel=&quot;noreferrer&quot;&gt;mirror maintenance building&lt;/a&gt; is located on the mountain top, near the mirrors.&lt;/p&gt;&#xA;&lt;p&gt;The Scientific America article: &lt;a href=&quot;https://www.scientificamerican.com/article/is-the-james-webb-space-telescope-too-big-to-fail/&quot; rel=&quot;noreferrer&quot;&gt;Is the James Webb Space Telescope &amp;quot;Too Big to Fail?&amp;quot;&lt;/a&gt; explains:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;“Assuming we make it to the injection trajectory to Earth-Sun L2, of course the next most risky thing is deploying the telescope. And unlike Hubble we can’t go out and fix it. Not even a robot can go out and fix it. So we’re taking a great risk, but for great reward,” Grunsfeld says.&lt;/p&gt;&#xA;&lt;p&gt;There are, however, &lt;strong&gt;modest efforts being made to make JWST “serviceable” like Hubble,&lt;/strong&gt; according to Scott Willoughby, JWST’s program manager at Northrop Grumman Aerospace Systems in Redondo Beach, California. The aerospace firm is NASA’s prime contractor to develop and integrate JWST, and has been tasked with provisioning for a “launch vehicle interface ring” on the telescope that could be “grasped by something,” whether astronaut or remotely operated robot, Willoughby says. If a spacecraft were sent out to L2 to dock with JWST, it could then attempt repairs—or, if the observatory is well-functioning, simply top off its fuel tank to extend its life. But presently no money is budgeted for such heroics. In the event that JWST suffers what those in spaceflight understatedly call a “bad day,” whether due to rocket mishap or deployment glitch or something unforeseen, Grunsfeld says there’s presently an ensemble of in-space observatories, including Hubble, and an ever-expanding collection of powerful ground-based telescopes that would offset such misfortune.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QVO0C.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QVO0C.png&quot; alt=&quot;LVIR&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Launch Vehicle Interface Ring (LVIR) Forgings (2) delivered&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Quote from the &amp;quot;&lt;a href=&quot;https://www.nasa.gov/vision/universe/starsgalaxies/mirror_size.html&quot; rel=&quot;noreferrer&quot;&gt;James Webb Space Telescope&lt;/a&gt;&amp;quot; (JWST) website:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The completed primary mirror will be over 2.5 times larger than the diameter of the Hubble Space Telescope's primary mirror, which is 2.4 meters in diameter, but will weigh roughly half as much.&lt;/p&gt;&#xA;&lt;p&gt;The James Webb Space Telescope will collect light approximately 9 times faster than the Hubble Space Telescope when one takes into account the details of the relative mirror sizes, shapes, and features in each design,&amp;quot; said Eric Smith, JWST program scientist at NASA Headquarters, Washington. The increased sensitivity will allow scientists to see back to when the first galaxies formed just after the Big Bang. The larger telescope will have advantages for all aspects of astronomy and will revolutionize studies of how stars and planetary systems form and evolve.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;See also: &amp;quot;&lt;a href=&quot;https://jwst.nasa.gov/comparison_about.html&quot; rel=&quot;noreferrer&quot;&gt;Webb vs Hubble Telescope&lt;/a&gt;&amp;quot;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;... more distant objects are more highly redshifted, and their light is pushed from the UV and optical into the near-infrared. Thus observations of these distant objects (like the first galaxies formed in the Universe, for example) requires an infrared telescope.&lt;/p&gt;&#xA;&lt;p&gt;This is the other reason that Webb is not a replacement for Hubble is that its capabilities are not identical. Webb will primarily look at the Universe in the infrared, while Hubble studies it primarily at optical and ultraviolet wavelengths (though it has some infrared capability). Webb also has a much bigger mirror than Hubble. This larger light collecting area means that Webb can peer farther back into time than Hubble is capable of doing. Hubble is in a very close orbit around the earth, while Webb will be 1.5 million kilometers (km) away at the second Lagrange (L2) point.&lt;/p&gt;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;How far will Webb see?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Because of the time it takes light to travel, the further away an object is, the further back in time we are looking.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FZT5c.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FZT5c.jpg&quot; alt=&quot;Seeing backwards in time.&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;This illustration compares various telescopes and how far back they are able to see. Essentially, Hubble [HST] can see the equivalent of &amp;quot;toddler galaxies&amp;quot; and Webb Telescope [JWST] will be able see &amp;quot;baby galaxies&amp;quot;. One reason Webb will be able to see the first galaxies is because it is an infrared telescope. The universe (and thus the galaxies in it) is expanding. When we talk about the most distant objects, Einstein’s General Relatively actually comes into play. It tells us that the expansion of the universe means it is the space between objects that actually stretches, causing objects (galaxies) to move away from each other. Furthermore, any light in that space will also stretch, shifting that light's wavelength to longer wavelengths. This can make distant objects very dim (or invisible) at visible wavelengths of light, because that light reaches us as infrared light. Infrared telescopes, like Webb, are ideal for observing these early galaxies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Updates in adaptive optical techniques are ongoing, see: &amp;quot;&lt;a href=&quot;https://arxiv.org/abs/1806.02881&quot; rel=&quot;noreferrer&quot;&gt;Fast Coherent Differential Imaging on Ground-Based Telescopes using the Self-Coherent Camera&lt;/a&gt;&amp;quot; (7 Jun 2018), by Benjamin L. Gerard, Christian Marois, and Raphaël Galicher:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;We develop the framework for one such method based on the self-coherent camera (SCC) to be applied to ground-based telescopes, called Fast Atmospheric SCC Technique (FAST). We show that with the use of a specially designed coronagraph and coherent differential imaging algorithm, recording images every few milliseconds allows for a subtraction of atmospheric and static speckles while maintaining a close to unity algorithmic exoplanet throughput. Detailed simulations reach a contrast &lt;strong&gt;close to the photon noise limit after 30 seconds for a 1 % bandpass in H band on both 0th and 5th magnitude stars. For the 5th magnitude case, this is about 110 times better in raw contrast than what is currently achieved from ExAO instruments if we extrapolate for an hour of observing time&lt;/strong&gt;, illustrating that sensitivity improvement from this method could play an essential role in the future detection and characterization of lower mass exoplanets.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In short, sometimes they can completely remove the atmosphere. Improvements are coming.&lt;/p&gt;&#xA;&lt;p&gt;ESO 4LGSF - &lt;a href=&quot;http://www.eso.org/sci/facilities/develop/4LGSF.html&quot; rel=&quot;noreferrer&quot;&gt;Laser Guide Stars Facility&lt;/a&gt; - Four lasers are used to create guide stars for the AO.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-06-14T17:36:21.550" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="26615" PostTypeId="2" ParentId="26613" CreationDate="2018-06-12T21:20:49.407" Score="5" Body="&lt;p&gt;The Sun erodes a planetary atmosphere through its solar wind.  Presently, the solar wind passes Earth at 250-750 kps, which is roughly ten times the Sun's escape velocity from earth orbit.  So atoms and molecules of gas eroded by the solar wind are carried right out of the Solar System into the beginnings of interstellar space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At that point (ca. 100 AU) the solar wind starts interacting with the interstellar medium (IM) which is a very low density plasma. The Sun is plowing through the IM at a speed around 25 kps, and there's a bow shock ahead of it and a long tail &quot;downstream.&quot;  The Sun's magnetic field also plays a big role at extreme distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems likely that some of the solar wind escapes altogether, while some of it piles up out there. (Note that at 100 AU, there's a &lt;em&gt;lot&lt;/em&gt; of space and Earth's entire atmosphere and all its water too, wouldn't make a noticeable impact on gas density out there.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, best answer is that as Earth's gasses are stripped away, it gets carried out to around 100 AU by the solar wind where it stagnates, some staying there and some leaking off into the IM.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-06-12T21:20:49.407" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26617" PostTypeId="1" AcceptedAnswerId="27025" CreationDate="2018-06-12T22:32:17.760" Score="7" ViewCount="160" Body="&lt;p&gt;I'm currently trying to determine the position of the sun and the moon from jpl ephemeris DE200 which is referred to the dynamical equator and equinox of 2000 and uses tdb time. I am using the j2000  system ECI.The things that trouble me are the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.Is the moon position given with respect to Earth or with respect to the barycenter? Also when we find the sun position by using the earth-moon barycenter position and lunar coordinate , will the sun position be with respect to Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.If I'm using UTC time  , how exactly to obtain the TDB or TAi, since it's oscillating?( After that I want to obtain the JD for TDB)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.If I'm using  J2000 ECI coordinate system , do I need to take in acount the effect of nutation and precession? And how to do that? &lt;/p&gt;&#xA;" OwnerUserId="22489" LastActivityDate="2018-07-20T20:49:56.853" Title="What are the problems that need to be taken in consideration when trying to find planetary position by using jpl ephemerides?" Tags="&lt;solar-system&gt;&lt;positional-astronomy&gt;&lt;time&gt;&lt;ephemeris&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26618" PostTypeId="1" CreationDate="2018-06-13T03:11:21.640" Score="3" ViewCount="132" Body="&lt;p&gt;I heard that stars would live longer if they consumed more hydrogen and material from their 'environment'. Is that true? For example, if the sun consumed all of the gas giants, would it live a little bit longer?&lt;/p&gt;&#xA;" OwnerUserId="23317" LastEditorDisplayName="user1569" LastEditDate="2018-06-13T09:57:30.550" LastActivityDate="2018-06-13T09:57:30.550" Title="Will stars live longer if they consume more hydrogen?" Tags="&lt;the-sun&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26619" PostTypeId="2" ParentId="26618" CreationDate="2018-06-13T03:28:22.287" Score="2" Body="&lt;p&gt;The sun has a mass of $1.989*10^{30}$ kg. Jupiter's mass is $1.898*10^{27}$ kg. That's only about 0.1% of the sun's mass. From a simple estimation, &lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2014/07/14/solar_wind_versus_fusion_how_does_the_sun_lose_mass.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;the sun consumes&lt;/a&gt; about $4.4*10^9$ kg of hydrogen per second via nuclear fusion. Therefore, it burns about 1 Jupiter Mass every 1.4 million years. Considering the fact that &lt;a href=&quot;http://curious.astro.cornell.edu/about-us/51-our-solar-system/the-sun/birth-death-and-evolution-of-the-sun/167-how-do-you-calculate-the-lifetime-of-the-sun-advanced&quot; rel=&quot;nofollow noreferrer&quot;&gt;the sun is about 5 billion years old&lt;/a&gt;, and will live another 5 billion years, this would increase the sun's lifetime by about 0.1%. This is ultimately a negligible contribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn and the other gas giants are all an order of magnitude smaller than Jupiter, so their contributions would be even smaller - just a few hundred thousand years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, this is all assuming that all of the mass added by the planets we've added is being used for fusion. In reality, Jupiter is &lt;a href=&quot;https://www.space.com/18388-what-is-jupiter-made-of.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;only about 90% Hydrogen&lt;/a&gt;. Also, adding more mass to the sun would cause it to fuse hydrogen a little more quickly, to counteract the increased gravitational pressure its core would experience. This is why &lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/M/Main+Sequence+Lifetime&quot; rel=&quot;nofollow noreferrer&quot;&gt;giant stars die more quickly&lt;/a&gt; than smaller stars. So this would slightly counteract the increased lifetime from the added fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TL;DR smashing all the gas giants into the sun wouldn't meaningfully change its lifetime.&lt;/p&gt;&#xA;" OwnerUserId="13426" LastActivityDate="2018-06-13T03:28:22.287" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26620" PostTypeId="2" ParentId="26514" CreationDate="2018-06-13T05:03:34.300" Score="-1" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Do narrow lines in the spectra of O- and B- type stars always indicate magnetic fields?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The narrow lines don't necessarily require a magnetic field, but any star that large will have one.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I'm wondering two things:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Has the presence of such comparatively narrow lines in the spectra of O- and B- type stars been observed in cases where there is no magnetic field present?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/O-type_main-sequence_star&quot; rel=&quot;nofollow noreferrer&quot;&gt;O-type main-sequence stars&lt;/a&gt; have high metallicity, they will have a magnetic field. &lt;a href=&quot;https://en.wikipedia.org/wiki/Tau_Scorpii&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tau Scorpii&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Theta1_Orionis_C&quot; rel=&quot;nofollow noreferrer&quot;&gt;Theta&lt;span class=&quot;math-container&quot;&gt;$^1$&lt;/span&gt; Orionis C&lt;/a&gt; info at their links. The Phys.Org article: &amp;quot;&lt;a href=&quot;https://m.phys.org/news/2016-01-strong-magnetic-fields-majority-stars.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Strong magnetic fields discovered in majority of stars&lt;/a&gt;&amp;quot; says:&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;An international group of astronomers led by the University of Sydney has discovered strong magnetic fields are common in stars, not rare as previously thought, which will dramatically impact our understanding of how stars evolve.&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;This Wikipedia article: &amp;quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_dynamo&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solar Dynamo&lt;/a&gt;&amp;quot; could use some citations, but here's what it says:&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;The solar dynamo is the physical process that generates the Sun's magnetic field. A dynamo, essentially a naturally occurring electric generator in the Sun's interior, produces electric currents and a magnetic field, following the laws of Ampère, Faraday and Ohm, as well as the laws of hydrodynamics, which together form the laws of magnetohydrodynamics. The detailed mechanism of the solar dynamo is not known and is the subject of current research.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Mechanism&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;A dynamo converts kinetic energy into electric-magnetic energy. An electrically conducting fluid with shear or more complicated motion, such as turbulence, can temporarily amplify a magnetic field through Lenz's law: fluid motion relative to a magnetic field induces electric currents in the fluid that distort the initial field. If the fluid motion is sufficiently complicated, it can sustain its own magnetic field, with advective fluid amplification essentially balancing diffusive or ohmic decay. Such systems are called self-sustaining dynamos. The Sun is a self-sustaining dynamo that converts convective motion and differential rotation within the Sun to electric-magnetic energy.&lt;/p&gt;&#xA;&lt;p&gt;...&amp;quot;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If so, are there other mechanisms that could be responsible for this in cases without magnetic fields? I'm grasping at straws for ideas, like low collisional broadening.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Our Q&amp;amp;A: &amp;quot;&lt;a href=&quot;https://physics.stackexchange.com/questions/257375/different-width-of-spectral-lines-for-different-groups-of-stars&quot;&gt;Different width of spectral lines for different groups of stars&lt;/a&gt;&amp;quot; has &lt;a href=&quot;https://physics.stackexchange.com/a/257471/170832&quot;&gt;Rob Jefferies answer&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;This seems well explained in section 3.4 of &amp;quot;&lt;a href=&quot;https://arxiv.org/abs/1509.06482&quot; rel=&quot;nofollow noreferrer&quot;&gt;Magnetically Confined Wind Shocks in X-rays - a Review&lt;/a&gt;&amp;quot; (22 Sept 2015), by Asif ud-Doula and Yael Nazé, on page 10:&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;&lt;strong&gt;3.4. Structure of confined winds, as revealed by high-resolution spectra&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;High-resolution spectra can yield a wealth of information. With current instrumentation, line widths and shifts can be evaluated with precisions down to a few tens km s&lt;span class=&quot;math-container&quot;&gt;$^{−1}$&lt;/span&gt; in the most favorable cases (a few hundreds km s&lt;span class=&quot;math-container&quot;&gt;$^{−1}$&lt;/span&gt; more typically). Furthermore, the comparison of lines from H-like and He-like ions and of components of &lt;span class=&quot;math-container&quot;&gt;$f ir$&lt;/span&gt; triplets of He-like ions constrain the temperature and location of the emitting plasma. However, such measurements are currently only possible for the brightest X-ray sources, so that few magnetic massive stars have been investigated&#xA;in this respect (τ Sco - Mewe et al. 2003; Cohen et al. 2003, θ&lt;span class=&quot;math-container&quot;&gt;$^1\!$&lt;/span&gt; Ori C - Schulz et al. 2000; Gagné et al. 2005a, b, HD 191612 - Nazé et al. 2007, HD 148937 - Nazé et al. 2008, 2012, β Cep - Favata et al. 2009, IQ Aur - Robrade and Schmitt 2011).&lt;/p&gt;&#xA;&lt;p&gt;Within noise limitations, the X-ray lines of magnetic massive stars were found to be symmetric, and globally unshifted. This agrees well with MHD models. In the case of θ&lt;span class=&quot;math-container&quot;&gt;$^1\!$&lt;/span&gt; Ori C, global fitting however suggests small variations in velocity (Gagné et al., 2005a): from −75 km s&lt;span class=&quot;math-container&quot;&gt;$^{−1}$&lt;/span&gt; when the star is seen pole-on to about 100 km s&lt;span class=&quot;math-container&quot;&gt;$^{−1}$&lt;/span&gt; when seen edge-on. This change needs to be confirmed as the errors are large but also because one cannot exclude a stochastic variation when only a single observation per phase is available. If further observations provide evidence that velocity varies with phase, then refinement of models will be needed, as no such changes are currently predicted (Gagné et al., 2005a).&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Reported widths of X-ray lines largely depend on object and ion considered. The narrowest widths, so far, were found for β Cep, whose lines are dominated by instrumental resolution, yielding only upper limit on intrinsic widths (&amp;lt;600 km s&lt;span class=&quot;math-container&quot;&gt;$^{−1}$&lt;/span&gt;, Favata et al., 2009). Larger widths, full width at half-maximum (FW HM) ∼ 600 − 800 km s&lt;span class=&quot;math-container&quot;&gt;$^{−1}$&lt;/span&gt;, were reported for ions with high ionization potential (Mg, Si, S) in τ Sco, θ&lt;span class=&quot;math-container&quot;&gt;$^1\!$&lt;/span&gt; Ori C, and HD 148937, three stars with more rapid winds than β Cep. Such widths are much smaller than observed for “normal” O-type stars (FW HM ∼ v∞), where lines arise in embedded wind shocks distributed all over the wind, hence cover a larger velocity range. They indicate formation in slowly moving plasma, in agreement with the confined winds scenario. However, most MHD models predict even narrower lines (Gagné et al., 2005a).&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Furthermore, lines from ions with lower ionization potential, notably oxygen, appear broader (FW HM ∼ 1800−2000 km s&lt;span class=&quot;math-container&quot;&gt;$^{−1}$&lt;/span&gt; Gagné et al., 2005a; Nazé et al., 2007, 2008). These lines are associated with cooler plasma, which could have a different origin than hotter plasma. For example,&#xA;the dominant hot plasma in θ&lt;span class=&quot;math-container&quot;&gt;$^1\!$&lt;/span&gt; Ori C is thought to arise in confined winds while the cooler one could arise in embedded wind shocks as in normal O-stars (Gagné et al., 2005a). This dual origin could be supported by the different temperatures derived from the different ions (Schulz et al., 2000). In &lt;a href=&quot;https://arxiv.org/abs/1801.06152&quot; rel=&quot;nofollow noreferrer&quot;&gt;Of?p&lt;/a&gt; stars, however, the spectra are dominated by the cooler component, i.e. confined winds emit soft X-rays in these objects (see above), but it cannot be excluded that current errors, which are large, are somewhat blurring the picture.&lt;/p&gt;&#xA;&lt;p&gt;In He-like triplets, the forbidden line is suppressed when the density is high or the UV radiation is intense. In the case of massive stars, the latter effect is the most important one and, thanks to dilution with distance, enables us to locate the emitting region. In &lt;strong&gt;τ Sco&lt;/strong&gt;, &lt;strong&gt;θ&lt;span class=&quot;math-container&quot;&gt;$^1\!$&lt;/span&gt; Ori C&lt;/strong&gt;, HD 148937, and &lt;strong&gt;β Cep&lt;/strong&gt;, the start of the emitting region is found to be close to the photosphere, at radii r ∼ 1.5 − 3 R&lt;span class=&quot;math-container&quot;&gt;$_∗$&lt;/span&gt; for the first three stars and r ∼ 4 − 6 R&lt;span class=&quot;math-container&quot;&gt;$_∗$&lt;/span&gt; for the latter case. These values are only slightly lower than the corresponding &lt;a href=&quot;https://en.wikipedia.org/wiki/Alfv%C3%A9n%27s_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Alfvén&lt;/a&gt; radii of these stars, they thus appear qualitatively compatible with MHD simulations. In IQ Aur, the forbidden line was found to be normal, suggesting a formation radius larger than 7R&lt;span class=&quot;math-container&quot;&gt;$_∗$&lt;/span&gt; (Robrade and Schmitt, 2011) - despite a large value, this also appears compatible with the supposed location of confined winds in this star.&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;Other reading:&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;&lt;a href=&quot;https://arxiv.org/abs/astro-ph/0606156&quot; rel=&quot;nofollow noreferrer&quot;&gt;The surprising magnetic topology of &lt;span class=&quot;math-container&quot;&gt;$\tau$&lt;/span&gt; Sco: fossil remnant or dynamo output?&lt;/a&gt;&amp;quot; (7 Jun 2006), by JF Donati, ID Howarth, MM Jardine, P Petit, C Catala, JD Landstreet, JC Bouret, E Alecian, JR Barnes, T Forveille, F Paletou, and N Manset&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;&lt;a href=&quot;https://arxiv.org/abs/astro-ph/0601623&quot; rel=&quot;nofollow noreferrer&quot;&gt;The magnetic field and confined wind of the O star θ&lt;span class=&quot;math-container&quot;&gt;$^{1\!}$&lt;/span&gt; Orionis C&lt;/a&gt;&amp;quot; (26 Jan 2006), by G.A. Wade, A.W. Fullerton, J.-F. Donati, J.D. Landstreet, P. Petit, S. Strasser&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;&lt;a href=&quot;http://adsabs.harvard.edu/full/2006IAUS..230...45M&quot; rel=&quot;nofollow noreferrer&quot;&gt;X-ray hardness ratios for stars of different spectral types&lt;/a&gt;&amp;quot; (2006), by Meurs, E. J. A., Casey, P., &amp;amp; Norci, L.&lt;/p&gt;&#xA;&lt;p&gt;Wikipedia - &lt;a href=&quot;https://en.wikipedia.org/wiki/Zeeman_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zeeman Effect&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;&lt;a href=&quot;http://adsabs.harvard.edu/full/1958ApJ...128..228B&quot; rel=&quot;nofollow noreferrer&quot;&gt;Magnetic Fields of the A-Type Stars&lt;/a&gt;&amp;quot; (30 Apr &lt;strong&gt;1958&lt;/strong&gt;), by H. W. Babcock&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-06-13T07:51:27.570" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26621" PostTypeId="2" ParentId="26618" CreationDate="2018-06-13T07:40:25.053" Score="5" Body="&lt;p&gt;Adding mass of similar composition to the Sun would effectively turn the Sun into a higher-mass main-sequence star.  That would not increase the gravitational pressure, it would lower it, because it would cause the Sun to expand to lower density.  It would also cause the Sun to fuse hydrogen faster, to the point that it would actually reduce the lifetime of the Sun, not increase it.  All this can be seen by consulting models of higher-mass main-sequence stars.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it seems counterintuitive that adding mass would weaken the gravity, this is because the core must self-regulate its fusion rate to match the rate that light is escaping, and simply adding mass would upset that balance unless the Sun expanded to recover it.  Note that main-sequence star models do not particularly care about the history of how the mass got there, they are simply the only equilibrium models that have the appropriate mass and composition and are undergoing core fusion.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-06-13T07:40:25.053" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26622" PostTypeId="1" AcceptedAnswerId="26623" CreationDate="2018-06-13T07:42:13.627" Score="17" ViewCount="2036" Body="&lt;p&gt;&lt;a href=&quot;https://www.quora.com/Why-does-a-star-expand-as-it-becomes-a-red-giant#__w2_TNbVPRD_link&quot; rel=&quot;noreferrer&quot;&gt;Please refer to this answer from Quora:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... a star will become a red giant before it begins burning helium. In fact, it will bloat into a red giant while still burning hydrogen in a shell on the surface of the helium core. Shell burning, however, does release more energy than core-burning, but even that alone wouldn’t cause the star to become a red giant, since it could just shine brighter. The real culprit is the combination of the higher energy production rate AND that &lt;strong&gt;the star has a higher opacity at that stage in life&lt;/strong&gt;. This produces an energy crisis where the energy cannot escape radiatively fast enough, and convection would have to be supersonic to rebalance the star. Since supersonic convection is highly disfavored (i.e. impossible), the star greatly expands to the point that the energy flux at the burning shell matches the energy flux at the now much larger surface again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is one of the best explanations I've found for how stars transition to red giants. However, the section I've bolded is confusing to me: why would the star become more opaque, and what causes that to occur?&lt;/p&gt;&#xA;" OwnerUserId="23353" LastActivityDate="2018-06-13T09:02:34.613" Title="When a star reaches the red giant phase, why does it become more opaque?" Tags="&lt;star&gt;&lt;red-giant&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26623" PostTypeId="2" ParentId="26622" CreationDate="2018-06-13T08:13:53.593" Score="8" Body="&lt;p&gt;There are some answers in &lt;a href=&quot;https://www.ast.cam.ac.uk/~pettini/STARS/Lecture12.pdf&quot; rel=&quot;noreferrer&quot;&gt;this lecture&lt;/a&gt; and &lt;a href=&quot;https://www.ast.cam.ac.uk/~pettini/STARS/Lecture13.pdf&quot; rel=&quot;noreferrer&quot;&gt;this one&lt;/a&gt; although even there the author admits that the whole story is complicated, and not perfectly known. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that shell burning and contraction of the no-longer-burning core of the star causes some expansion of the outer layers (this is described as the &quot;mirror principle&quot; and partly explained in lecture 12 (section 12.4). This causes the outer layers to cool, until H- ions start to form in them (around 5000K). These interact with light much more than neutral hydrogen atoms or bare protons (H+) which accounts for the increase in opacity.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-06-13T08:13:53.593" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26624" PostTypeId="2" ParentId="26622" CreationDate="2018-06-13T09:02:34.613" Score="6" Body="&lt;p&gt;The real reason stars bloat into red giants is not due to opacity changes, it is due to the creation of a degenerate core of non-fusing helium at the center.  This degenerate core has a strong gravity, which dictates a high temperature to the hydrogen fusing shell that sits atop it.  This is quite important, because when a star is undergoing core fusion like our Sun, the fusion rate self-regulates by adjusting the temperature.  But when the temperature is dictated by the gravity of the degenerate core, it tends to be quite high, and the fusion rate, unable to self-regulate, goes nuts.  Something has to give, because as said above, there is an energy crisis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What gives is that the energy generation rate causes the envelope to expand, which lifts weight off the fusing shell.  That reduces its density, and the amount of gas undergoing fusion.  So unlike core fusion, which self-regulates its temperature, shell fusion self-regulates its amount of mass, by lifting off weight.  But to lift off such a large amount of weight requires a very significant expansion of the envelope, ergo a red giant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The H minus opacity near the surface does control exactly how large the star gets, because once the mass has been lifted off it doesn't matter so much to the shell exactly how far out it gets lifted.  But it does matter to the star's ability to radiate that energy into space, so that controls the radius of the star, it just isn't what makes that radius large in the first place.  So the reason the envelope expands in the first place has nothing to do with opacity, the opacity changes come into play after significant expansion has already occurred, and controls where the stellar surface ends up.  If the stellar opacity did not change at all, it would still be a giant star, just not quite the same radius.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-06-13T09:02:34.613" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26625" PostTypeId="1" AcceptedAnswerId="26627" CreationDate="2018-06-13T14:33:47.290" Score="0" ViewCount="1794" Body="&lt;p&gt;I am interested in calculating the angles formed between planets like Saturn opposite to Neptune - 180 degrees, PlanetX opposite/trine/square to PlanetY. How to find such degrees?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use Stellarium software to find positions of planets. Here I am adding positions of Saturn, Jupiter and Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/P4Dj4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P4Dj4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/P0AeX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P0AeX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eGU5c.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eGU5c.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stellarium shows information about selected objects, can it be used to calculate angle formed, for example, angle formed between Saturn and Jupiter taking Sun as reference point?&lt;/p&gt;&#xA;" OwnerUserId="19873" LastEditorUserId="19873" LastEditDate="2018-06-14T16:47:04.383" LastActivityDate="2018-06-14T19:28:31.387" Title="How to calculate the angle formed between 2 planets?" Tags="&lt;planetary-transits&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26626" PostTypeId="1" AcceptedAnswerId="26629" CreationDate="2018-06-13T16:36:28.877" Score="1" ViewCount="120" Body="&lt;p&gt;If a rogue planet disrupts Earth's orbit slightly to cause a long winter or summer will the Earth's orbit normalize back close its original orbit over time? Could a Planet X or some other elliptical orbit large body revolving around the sun effect Earth's orbit by being inside the inner solar system? &lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-06-14T02:58:47.247" LastActivityDate="2018-06-14T02:58:47.247" Title="Are large body orbits self correcting?" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;solar-system&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26627" PostTypeId="2" ParentId="26625" CreationDate="2018-06-13T17:02:18.847" Score="1" Body="&lt;p&gt;The planets are moving along elliptical paths, and we are viewing them from another planet that is also moving along an elliptical path, this makes the motion that the planets make in the sky (relative to distant stars) seem to loop and move with an inconsistent rate. The planets don't all orbit in the same plane, and so may be above or below the ecliptic (the apparent path in the sky of the sun)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is no shortcut to finding the angle between two planets: You have to calculate the sky position of the planets and then find the angle between those positions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step is called calculating an ephemeris. The details are beyond the scope of this answer, but NASA have a good ephemeris calculator at &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ssd.jpl.nasa.gov/horizons.cgi&lt;/a&gt;. Alternatively, software like Stellarium will be able to calculate the position of a planet on any given date in the near future or past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have the position of the planets as RA and Dec, and you have converted the units to decimal degrees, you can calculate the angle $A$ between them using&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos(A) = \sin(\mathrm{Dec_1})\sin(\mathrm{Dec_2}) + \cos(\mathrm{Dec_1})\cos(\mathrm{Dec_2})\cos(\mathrm{RA_1} - \mathrm{RA_2})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further details and a calculator is found at &lt;a href=&quot;http://www.gyes.eu/calculator/calculator_page1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.gyes.eu/calculator/calculator_page1.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-06-13T17:02:18.847" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26628" PostTypeId="1" AcceptedAnswerId="26632" CreationDate="2018-06-13T17:41:01.773" Score="7" ViewCount="173" Body="&lt;p&gt;After reading &lt;a href=&quot;https://weather.com/science/space/news/2018-06-13-mars-brighter-closer-since-2003&quot; rel=&quot;noreferrer&quot;&gt;this article&lt;/a&gt;, I was inspired to run some simulations at NASA's JPL &lt;a href=&quot;https://space.jpl.nasa.gov/&quot; rel=&quot;noreferrer&quot;&gt;Solar System Simulator&lt;/a&gt; to see Earth's and Mars' orbits from above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I could visually determine (from the JPEG images), &lt;a href=&quot;https://space.jpl.nasa.gov/cgi-bin/wspace?tbody=1000&amp;amp;vbody=1001&amp;amp;month=7&amp;amp;day=26&amp;amp;year=2018&amp;amp;hour=00&amp;amp;minute=00&amp;amp;fovmul=1&amp;amp;rfov=2&amp;amp;bfov=30&amp;amp;porbs=1&amp;amp;brite=1&amp;amp;showac=1&quot; rel=&quot;noreferrer&quot;&gt;Mars is at opposition on July 26&lt;/a&gt;, but its perihelion is &lt;a href=&quot;https://space.jpl.nasa.gov/cgi-bin/wspace?tbody=1000&amp;amp;vbody=1001&amp;amp;month=8&amp;amp;day=31&amp;amp;year=2018&amp;amp;hour=00&amp;amp;minute=00&amp;amp;fovmul=1&amp;amp;rfov=2&amp;amp;bfov=30&amp;amp;porbs=1&amp;amp;brite=1&amp;amp;showac=1&quot; rel=&quot;noreferrer&quot;&gt;sometime around September&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When (and where in their orbits) is Mars brightest from Earth this year?&lt;/p&gt;&#xA;" OwnerUserId="13976" LastActivityDate="2018-06-14T04:18:32.220" Title="Is Mars brightest from Earth at opposition, perihelion or somewhere in between?" Tags="&lt;planet&gt;&lt;observational-astronomy&gt;&lt;mars&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26629" PostTypeId="2" ParentId="26626" CreationDate="2018-06-13T17:58:35.953" Score="9" Body="&lt;p&gt;No. Basically, there is no such thing as a stable orbit in a realistic solar system.  &lt;em&gt;All&lt;/em&gt; planetary orbits are unstable due to gravitational interactions between planets, but how long it takes the instability to make a big change in a planet's orbit varies a &lt;em&gt;lot&lt;/em&gt;.  It can be a few of the planet's years, thousands, millions or billions. (Earth's orbit has been stable, so far, for 4+ billion years.)  But stability going forward can't be guaranteed beyond at most a hundred million or so years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main source of instability is resonances. A resonance is where two planets are in orbits where the ratio of their periods is a simple fraction like 1/2 or 2/3 or 5/3 or whatever.  Because of the simple relationship between periods, planets repeat the same configuration in space repeatedly, and gravitational interactions can build up. (Without the resonance, the interactions are just as strong, but in random directions and thus mostly cancel.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resonances can disrupt an orbit or can stabilize it. The Hilda group of asteroids are in a stable 3:2 resonance with Jupiter, for example, and Pluto is in a 2:3 resonance with Neptune.  Both of these resonances act to stabilize the orbits.  But the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kirkwood_gap&quot; rel=&quot;noreferrer&quot;&gt;Kirkwood Gap&lt;/a&gt; in the asteroid belt is a 2:1 resonance with Jupiter which tends to keep the gap empty by disrupting the orbits of any asteroids that stray into the gap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the interaction, Earth will be in a new orbit and that orbit may also be stable over an extended period or it may be one where interactions with other planets quickly render the new orbit unstable.  There's no way to tell without &lt;em&gt;detailed&lt;/em&gt; orbital information on the encounter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any event, there is no restoring force to push Earth back to its old orbit.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2018-06-13T19:26:14.670" LastActivityDate="2018-06-13T19:26:14.670" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26630" PostTypeId="2" ParentId="26628" CreationDate="2018-06-13T18:58:53.610" Score="2" Body="&lt;p&gt;The two main factors affecting the brightness of Mars are its distance from Earth, and the phase (ie how much of the planet is illuminated from the point of view of Earth. Both of these are optimal when Mars is at opposition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stellarium has Mars reaching maximum brightness on the 28th of July with a magnitude of -2.78, very close to opposition. The distance of Mars from the sun has only a secondary effect on the brightness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On that date it will be close to the full moon, but low in the sky for Northern Hemisphere observers.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-06-13T18:58:53.610" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26632" PostTypeId="2" ParentId="26628" CreationDate="2018-06-13T20:55:59.747" Score="6" Body="&lt;p&gt;&lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;noreferrer&quot;&gt;JPL HORIZONS&lt;/a&gt; gives these values for Mars as seen from Earth.&#xA;&lt;code&gt;r&lt;/code&gt; is the Sun-Mars distance in AU, &lt;code&gt;delta&lt;/code&gt; is the Earth-Mars distance in AU, and &lt;code&gt;S-O-T&lt;/code&gt; is the Sun-Earth-Mars angle in degrees.&#xA;I added &lt;code&gt;[]&lt;/code&gt; to indicate minimum or maximum values.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Date__(UT)__HR:MN   APmag         r     delta    S-O-T&#xA;&#xA;2018-Jul-23 16:00   -2.75   1.40246   0.38911   172.30&#xA;2018-Jul-25 14:00   -2.77   1.40106   0.38730   173.27&#xA;2018-Jul-27 12:00  [-2.78]  1.39971   0.38600  [173.51]&#xA;2018-Jul-29 10:00  [-2.78]  1.39840   0.38522   172.94&#xA;2018-Jul-31 08:00   -2.77   1.39714  [0.38497]  171.73&#xA;2018-Aug-02 06:00   -2.75   1.39592   0.38523   170.11&#xA;&#xA;2018-Sep-14 12:00   -1.74   1.38147   0.50640   128.93&#xA;2018-Sep-16 12:00   -1.69  [1.38144]  0.51602   127.49&#xA;2018-Sep-18 12:00   -1.64   1.38147   0.52588   126.08&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Mars is at opposition on Jul 27, closest to Earth on Jul 31, and at perihelion on Sep 16.&#xA;Mars is at peak brightness for several days between opposition and closest approach to Earth.&#xA;A difference of 0.01 magnitude is imperceptible to the eye and smaller than the uncertainty of most ground-based CCD photometry.&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Classical_albedo_features_on_Mars&quot; rel=&quot;noreferrer&quot;&gt;Albedo features&lt;/a&gt; and dust storms probably contribute at least that much variation in magnitude anyway.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-06-13T20:55:59.747" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26633" PostTypeId="1" AcceptedAnswerId="26646" CreationDate="2018-06-13T23:05:41.723" Score="2" ViewCount="1318" Body="&lt;p&gt;If we could hear black holes, how loud would they be? I know that this is basically the same question as the question &quot;How loud would the Sun be?&quot;, except this time there is a black hole. But since the black hole sucks in stuff like a passenger jet engine sucks in air, about how many decibels would it be? Also, would it lethal to humans if we were 1 AU away and what would it do to our planet?&lt;/p&gt;&#xA;" OwnerUserId="23317" LastEditorUserId="37191" LastEditDate="2021-03-26T09:47:48.750" LastActivityDate="2021-03-26T09:47:48.750" Title="How loud would a black hole be?" Tags="&lt;black-hole&gt;&lt;sound&gt;" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="26634" PostTypeId="2" ParentId="20466" CreationDate="2018-06-14T00:44:59.633" Score="1" Body="&lt;p&gt;The simple answer is: the larger units such as AU or light years are easier for the human brain to remember. And, we should avoid putting units with to many zeros trailing after the first few digits, for example:1,000,000,000,000,000,000,000,000,000,000 meters. we could use AU or for even higher distances, light years. If it was shorter, well we cold still use meters but with an exponent.&lt;/p&gt;&#xA;" OwnerUserId="23317" LastEditorUserId="23317" LastEditDate="2018-06-17T17:55:56.693" LastActivityDate="2018-06-17T17:55:56.693" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26635" PostTypeId="2" ParentId="26628" CreationDate="2018-06-14T04:18:32.220" Score="2" Body="&lt;p&gt;The key is in &quot;as seen from Earth&quot;. Mars is illuminated by the Sun and the distance from the Sun varies only slightly. Therefore, it is always the same brightness. A photo of Mars through a telescope will always use the same exposure no matter how far it is from Earth. The planets do not behave like point sources of light. When Mars is twice as close, it fills 4 times the area in the sky. This exactly counters the inverse square spreading of the light from Mars. The same is true for all the planets. If they got brighter as they get closer, they would get brighter as YOU got closer. The Apollo astronauts would have boiled then vaporized as they approached the Moon. It is a common error. N. D. Tyson makes it with a truly dumb statement near the beginning of his Cosmos series. Something about the brightness of the Moon long ago when it was closer to the Earth. Think of how many people read that script and it still slipped by.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The brightness of an extended object as viewed, is not the same as the illumination it provides. If the Moon filled the sky, it would not look brighter. You would not have to wear welder's goggles to look at it. But the total amount of light reflected on the Earth would be much greater and more like the illumination on an overcast day (and there would never be a full Moon because the Earth would get in the way).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can easily calculate a brightness compared to the Earth (or Moon) daylight brightness and the ratio of the squares of the distance from the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you add that you want brightness as perceived from Earth, it depends on distance and phase. If you think of it as wanting to read by Mars light, it is &quot;brighter&quot; if it is closer and it is &quot;brightest&quot; at &quot;full Mars&quot; or opposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a gaggle of terms used in optics to describe various ways of describing brightness and intensity and illumination and they get very confusing.&lt;/p&gt;&#xA;" OwnerUserId="5526" LastActivityDate="2018-06-14T04:18:32.220" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26636" PostTypeId="1" AcceptedAnswerId="26637" CreationDate="2018-06-14T05:52:29.100" Score="2" ViewCount="218" Body="&lt;p&gt;A hopefully simple terminology question:  When you have an object that orbits another object in space, what do you call the object being orbited in relation to the object orbiting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, you might call the object orbiting the &quot;satellite&quot; of the orbited object, but what would you call the object being orbited?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbitee is all i've been able to come up with but it doesn't seem quite right ;)&lt;/p&gt;&#xA;" OwnerUserId="23363" LastActivityDate="2018-06-14T10:28:35.973" Title="Terminology of Orbits" Tags="&lt;orbit&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26637" PostTypeId="2" ParentId="26636" CreationDate="2018-06-14T06:40:16.103" Score="5" Body="&lt;p&gt;When one body has a much larger mass, so it's motion is negligable, it can be called the &lt;em&gt;central body&lt;/em&gt; (even for eccentric orbits, where the central body is a the focus of the ellipse, and far from the centre)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the terminology used by Wikipedia editors&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The central body in an orbital system can be defined as the one whose mass (M) is much larger than the mass of the orbiting body (m)...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also by &lt;a href=&quot;http://radio.astro.gla.ac.uk/a1dynamics/twobody.pdf&quot; rel=&quot;noreferrer&quot;&gt;Glasgow university&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The most straightforward orbit calculations occur when the central body is much more massive than the orbiting body.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;a href=&quot;https://books.google.co.uk/books?id=COgDAAAAMBAJ&amp;amp;pg=PA1582&amp;amp;lpg=PA1582&amp;amp;dq=orbit%20central%20body&amp;amp;source=bl&amp;amp;ots=M6Qwsa8yzQ&amp;amp;sig=eBD-uyvTJCsf_3mElv-3nlOokcM&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwiHs9q5w9LbAhVOYsAKHQ3pBT8Q6AEInQEwDQ&quot; rel=&quot;noreferrer&quot;&gt;elsewhere&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The angular momentum of a satellite depends on both the mass of orbiting satellite and central body as well as the radius of the orbit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Normally, in a specific situation &quot;planet&quot;, or &quot;star&quot; can be used.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-06-14T06:40:16.103" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26638" PostTypeId="1" CreationDate="2018-06-14T07:06:58.867" Score="2" ViewCount="56" Body="&lt;p&gt;Since decades it's well known that a correlation between stellar mass and metallicity (Z; both stellar and gas-phase Z, but here I focus on the gas-phase Z) exists (e.g., Tremonti+04; analysis of &gt;53000 galaxies from SDSS).&#xA;More recent results have shown that the scatter of this &quot;mass-Z&quot; relation is mainly driven by the star formation rate (SFR; e.g., Mannucci+10), i.e., at a fixed stellar mass, galaxies with higher SFRs have lower Z. This happen systematically over a wide range of stellar masses.&lt;br&gt;&#xA;What's the physical mechanism(s) responsible for that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, even more recently (e.g., Bothwell+16) it has been shown that the &quot;real&quot; driver of the &quot;mass-Z&quot; relation's scatter is the molecular gas mass (the previously found dependence on SFR may be only a by-product of the Schmidt-Kennicutt relation between gas and SFR). &#xA;How this would change the interpretation of the first problem?&lt;/p&gt;&#xA;" OwnerDisplayName="user19333" LastActivityDate="2018-06-14T07:06:58.867" Title="What's the profound reason(s) why the star formation rate drives the scatter of the mass-metallicity relation in galaxies?" Tags="&lt;galaxy&gt;&lt;astrophysics&gt;&lt;observational-astronomy&gt;&lt;star-formation&gt;&lt;metallicity&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26640" PostTypeId="2" ParentId="26562" CreationDate="2018-06-14T08:53:24.200" Score="4" Body="&lt;p&gt;I'll pick up on your example as it can be used very well for illustrating the main points:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First: The solar intensity in Jupiters orbit is less than 4 percent compared to Earths. So much less energy comes from the Sun which limits possible sublimation significantly&lt;/li&gt;&#xA;&lt;li&gt;Second: Sure tidal flexing leads to fountains. These fountains blow material into space. But this does not mean that all the material will leave Europas surface forever. It can fall back or it stays in Europas orbit and returns to the surface due to the effects of Jupiters magnetic field or gravitational forces of the other moons.&lt;/li&gt;&#xA;&lt;li&gt;Third: Sure the heat of the tidal flexing has to go somewhere. And it is largely irratiated away. As Europas surface is quite bright it reflects most sunlight away instead of absorbing it. This makes Europas surface quite cool (50-125K). Compare it with the darker neighboring moon Ganymede (70-152K). As you can see the albedo (which means the reflectivity of the surface) is far more important than the tidal heating. &lt;/li&gt;&#xA;&lt;li&gt;Fourth: Jep some of the water is split up by the radiation. This leads to an oxygen 'atmosphere' on Europa with 0.1 µPa surface pressure. While this is detectable it still is practically a vacuum. There is very little oxygen there, probably not more than a few hundred or thousand tons on the whole moon. Not enough to threaten the existence of moon with a mass of 4.8 * 10^22 kg&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In summary: Yes, icy moons lose some of their material. But for exactly this reason they only appear in the outer solar system where the solar radiation is low enough to make this effect negligble.&lt;/p&gt;&#xA;" OwnerUserId="22927" LastEditorUserId="22927" LastEditDate="2018-06-14T09:17:48.190" LastActivityDate="2018-06-14T09:17:48.190" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26642" PostTypeId="2" ParentId="26636" CreationDate="2018-06-14T10:28:35.973" Score="4" Body="&lt;p&gt;&quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Primary_(astronomy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Primary&lt;/a&gt;&quot; is quite a common term for this.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-06-14T10:28:35.973" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26643" PostTypeId="1" AcceptedAnswerId="26645" CreationDate="2018-06-14T15:35:56.737" Score="11" ViewCount="1503" Body="&lt;p&gt;I know for most of you the question may look silly, but Kerala is a state in India where the moon (Ramadan moon) will be sighted but not in rest of India.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was reading the news article where I got a little hint but was not clear for me since I am newbie, tried to google but couldn't find much info :) can someone explain me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Eid is likely to be celebrated in Kerala tomorrow. Due to the&#xA;  geographical location of the coastal state, the lunar moon is sighted&#xA;  in the state a day before the rest of India.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is the article link &lt;a href=&quot;http://www.india.com/news/india/eid-2018-in-india-moon-sighting-or-chand-raat-live-news-updates-3101209/&quot; rel=&quot;noreferrer&quot;&gt;Moon sighted in Kerala&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15407" LastEditorUserId="616" LastEditDate="2018-06-15T16:20:27.347" LastActivityDate="2018-06-15T22:29:49.477" Title="Why moon sighted only in Kerala but not in rest of India?" Tags="&lt;the-moon&gt;&lt;observational-astronomy&gt;&lt;moon-phases&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26644" PostTypeId="1" AcceptedAnswerId="26647" CreationDate="2018-06-14T16:35:16.113" Score="-1" ViewCount="51" Body="&lt;p&gt;Could two objects be caught into an orbit around some kind of dense object, perhaps a rogue planet. Could they share an atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lpfjk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lpfjk.png&quot; alt=&quot;planetesimals&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23371" LastEditorUserId="23371" LastEditDate="2018-06-14T20:10:15.970" LastActivityDate="2018-06-14T20:10:15.970" Title="Could two planetesimals orbit a dense object and share an atmosphere?" Tags="&lt;gravity&gt;&lt;hypothetical&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26645" PostTypeId="2" ParentId="26643" CreationDate="2018-06-14T16:50:42.860" Score="15" Body="&lt;p&gt;Think of it like this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each hour after New Moon, the Moon is moving away from the Sun. Depending on the position of the Sun, Moon, and observer, it is unlikely to be visible until at least 15 hours after New Moon, and in some cases not until 24 hours after New Moon.&lt;/li&gt;&#xA;&lt;li&gt;Someone will be the first to see the thin crescent after the New Moon. This occurs when the Sun is far enough below the horizon to make the sky &quot;dark&quot; and the Moon is far enough above the horizon (and away from the Sun) to be &quot;clear&quot; of the sky's glow. (&quot;dark&quot; and &quot;clear&quot; are relative. The sky will still be quite bright but not like the middle of the day.)&lt;/li&gt;&#xA;&lt;li&gt;Observers that are farther east than the first person would not be able to see the Moon because the time would have been earlier. That is, the Moon would have been closer to the Sun and not visible.&lt;/li&gt;&#xA;&lt;li&gt;Observers that are farther west should be able to see the Moon more easily because it is later, and therefore the Moon is farther from the Sun. In other words, the longitude affects how easy it is to see the new Moon.&lt;/li&gt;&#xA;&lt;li&gt;Whether it is easy or difficult to see the new Moon also depends on the latitude because this can affect where the Moon is relative to the Sun and horizon. This variation changes throughout the year.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It looks like Kerala is on the west coast, so they may have been in the position to see the Moon in that 15 to 24 hour time range whereas it would have been too early for people further to the east to see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kerala is also farther south than the rest of India. I have not done the calculations for this New Moon, but it is possible that being south and west increased the possibility of seeing the Moon before people further north or east would see it. If this is the case, then Kerala would have been in the best place to be the first to see the new Moon -- assuming the sky was clear :-).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/09cqS.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/09cqS.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image shows that, according to the calculations by &lt;a href=&quot;https://www.moonsighting.com/visibility.html&quot; rel=&quot;noreferrer&quot;&gt;https://www.moonsighting.com/visibility.html&lt;/a&gt;, the new moon might be visible from only the far South-Western coast of India.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastEditorUserId="2365" LastEditDate="2018-06-14T19:11:42.560" LastActivityDate="2018-06-14T19:11:42.560" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26646" PostTypeId="2" ParentId="26633" CreationDate="2018-06-14T17:26:22.740" Score="3" Body="&lt;p&gt;Nothing comes out from the BH (well, with a single known, very small &lt;a href=&quot;https://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;exception&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Accretion_disk&quot; rel=&quot;nofollow noreferrer&quot;&gt;accretion disk&lt;/a&gt; of the Black Holes most likely has various streams, pressure differences and so in it. Calculating it or measuring it is surely a very hard task, if you have luck, a professional explains the results in a better answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The accretion disc of the typical, star-sized black holes is much smaller as 1AU. And sound doesn't propagates in vacuum. Thus, from 1AU distance from the Black Hole we most likely wouldn't hear anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other radiations (mainly gamma from the disc) may be deadly even from 1AU, but not its sound.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="1891" LastEditDate="2018-06-18T22:31:35.017" LastActivityDate="2018-06-18T22:31:35.017" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26647" PostTypeId="2" ParentId="26644" CreationDate="2018-06-14T17:54:26.340" Score="1" Body="&lt;p&gt;So how much do you know about drag? Orbits in atmosphere are short lived - as reference see any low earth orbit satellite. They need regular boosts or else they fall in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your objects could orbit... Briefly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But not in the way you have drawn in the diagram, and not for any length of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically it's a no, in any real terms.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2018-06-14T17:54:26.340" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26648" PostTypeId="2" ParentId="26633" CreationDate="2018-06-14T19:43:26.550" Score="2" Body="&lt;p&gt;Physicists talk about phonons to describe energy transmission in elastic media similar to sound. There hasn't been much research about phonon transmission through the void of space because it is a forbidden event in physics. There is no sound in space. Some scientists have studied phonon tunneling through few angstroms of vacuum. &lt;a href=&quot;https://physicsworld.com/a/phonons-tunnel-across-the-vacuum/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://physicsworld.com/a/phonons-tunnel-across-the-vacuum/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find out more, you can read about phonons and sonic black holes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ears and microphones require air pressure and displacement of mass onto the auditory diaphragm. The mass and pressure of the interstellar medium is too low to actively move sound-sensing diaphragms, unless you have a lot of ions/gas massive objects to do that. In space there is a vacuum. if an astronaut played drums beside you in space, they would not displace any air pressure. there would be no noise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had a 1000Watt studio speaker beside you in space, it's quite possible that you would receive less than a single electron worth of sound pressure energy from that speaker per m2. something like 10^-20 watts. 100 dB is about 0.01W/m2 and is at the limit of hearing impairment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know the warning: &quot;In Space, No one can hear you scream&quot;... the film aliens...&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2018-06-16T09:33:08.660" LastActivityDate="2018-06-16T09:33:08.660" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="26649" PostTypeId="1" AcceptedAnswerId="26679" CreationDate="2018-06-14T20:55:58.423" Score="5" ViewCount="555" Body="&lt;p&gt;The galaxy correlation function, when computed directly from a galaxy catalog, relies on computing a randomly generated counterpart catalog.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When dealing with a simulated galaxy mock-catalog, where the sample of galaxies is roughly uniform over a large box (e.g. the galaxy mocks created from Millennium/Millennium II), the random catalog can be just be a uniform distribution of points in the same volume as the real data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, real-world galaxy catalogs are not homogenous boxes, and have very irregular shapes and observationally induced redshift and angular distributions. Even some mock-catalogs are endowed with these properties to better represent real data. In these cases, how is a random catalog best produced? With some sort of probability distribution in coordinates which attempts to mimic the &quot;non-physical&quot; (i.e. not due to observational constraints) part of the distribution? Or a simple binary mask which filters out empty regions, but otherwise leaves the random catalog uniform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the answer is that some nonuniform distribution should be used, is there a usual method for doing so? The literature I have found (e.g. for the SDSS published data, and for galaxy mocks mimicking SDSS's constraints) does not discuss this in any way. Note, if this is a highly involved process beyond a brief summary and a source or two, I do not expect a full answer here. Rather, I am trying to determine if there is an existing set of standard methods which are documented but have eluded me thus far.&lt;/p&gt;&#xA;" OwnerUserId="23373" LastActivityDate="2018-06-19T05:04:32.330" Title="Generating random galaxy catalogs for correlation functions" Tags="&lt;galaxy&gt;&lt;cosmology&gt;&lt;data-analysis&gt;&lt;galaxy-cluster&gt;&lt;python&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26650" PostTypeId="1" CreationDate="2018-06-14T22:40:27.907" Score="12" ViewCount="326" Body="&lt;p&gt;A main sequence star will fuse some of its hydrogen, but not all. In massive stars ($&amp;gt;1.5M_\odot$) the core is convective but the rest of the atmosphere radiative and hence does not mix much: as it undergoes shell fusion it will produce an onion-like structure with unused hydrogen on top. Solar mass stars only do this up to helium, but again leave a mantle of unused hydrogen. Stars less than $0.35M_\odot$ are fully convective and can in principle use up all hydrogen. However, I suspect this is not complete except for very low-mass M dwarves that have trillions of years to mature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at planetary nebulae, I have seen statements that they are about 90 percent hydrogen, 10 percent helium. This seems to fit with papers I have found (&lt;a href=&quot;https://www.aanda.org/articles/aa/pdf/2007/47/aa8655-07.pdf&quot; rel=&quot;noreferrer&quot;&gt;example&lt;/a&gt;, &lt;a href=&quot;https://arxiv.org/pdf/0709.0711.pdf&quot; rel=&quot;noreferrer&quot;&gt;example&lt;/a&gt;) although in some cases helium may reach 29% (&lt;a href=&quot;http://outters.fr/wp/wp-content/uploads/2015/12/Frew_thesis_2008_final.pdf&quot; rel=&quot;noreferrer&quot;&gt;example&lt;/a&gt;). Given that for a G star about half of the mass is ejected, that would suggest a fraction $0.9\times 0.5 = 0.45$ of unused hydrogen. But surely the fractions will be different for other masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to sum up, what is known about the fraction of hydrogen that is never fused over a stellar lifetime? &lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2018-06-14T22:40:27.907" Title="What fraction of a star's hydrogen store will be fused over its lifespan?" Tags="&lt;stellar-evolution&gt;&lt;hydrogen&gt;" AnswerCount="0" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26651" PostTypeId="2" ParentId="26610" CreationDate="2018-06-15T00:34:49.293" Score="1" Body="&lt;p&gt;Answer to telescopes on the moon. Being on the surface of the moon creates problems as compared to floating freely in space a long way from any planet/moon. Gravity distorts mirrors/mechanics, requires extra engineering to support weight, half sky is blocked by moon at any time, thermal emission from ground, temperature changes with day-night cycle, dust  ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The far side of the moon would be the best place to do low frequency radio observations. The moon blocks all the emissions from earth.&lt;/p&gt;&#xA;" OwnerUserId="10761" LastActivityDate="2018-06-15T00:34:49.293" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26653" PostTypeId="2" ParentId="26575" CreationDate="2018-06-15T09:26:28.073" Score="1" Body="&lt;p&gt;As was pointed out by @JamesK in a comment, one has to define how similar to the sun your star should be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want only the same spectral class, then Alpha Centauri A is your pick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it should be a &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_analog&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar analog&lt;/a&gt;, it has to have similar spectral class, and similar amount of &lt;a href=&quot;https://en.wikipedia.org/wiki/Metallicity&quot; rel=&quot;nofollow noreferrer&quot;&gt;heavy elements&lt;/a&gt; and a similar planetary system (as far as is possible to detect at the moment, which is not very).&#xA;Going with this an reading off-of the wikipedia list, your pick would then be &lt;a href=&quot;https://en.wikipedia.org/wiki/Sigma_Draconis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sigma Draconis&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, there's a even stricter class of similarity, namely the solar twin. Those aren't exact twins, as no such thing has been found to date. But by all other relevant parameters, particularly the spectroscopically determined element abundances, they're very much like the sun.&lt;br&gt;&#xA;Here the closest one would be &lt;a href=&quot;https://en.wikipedia.org/wiki/18_Scorpii&quot; rel=&quot;nofollow noreferrer&quot;&gt;18 Scorpii&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is Tau Ceti closest star which is quite like the sun according to some criteria?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So you see, here you can pick your criteria and choose your closest star according to those.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2018-06-15T09:26:28.073" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26654" PostTypeId="1" AcceptedAnswerId="26658" CreationDate="2018-06-15T21:22:20.303" Score="2" ViewCount="157" Body="&lt;p&gt;&lt;a href=&quot;http://chandra.harvard.edu/photo/2017/arp299/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arp 299&lt;/a&gt; is in the news, with most sources reporting it as 140 or 150 million light-years away. But what kind of distance is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://science.sciencemag.org/content/early/2018/06/13/science.aao4669&quot; rel=&quot;nofollow noreferrer&quot;&gt;The paper&lt;/a&gt; which is trhe source for the news item has &lt;a href=&quot;http://www.sciencemag.org/cgi/content/full/science.aao4669/DC1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Supplementary Materials&lt;/a&gt; which mention the distance but don't appear to say what kind or give details:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The merging pair of galaxies Arp 299 (Fig. 1) at a 44.8 Mpc distance (33; corrected to the cosmic microwave background reference frame; H0 = 73 km s^-1 Mpc^-1) is one of the brightest nearby LIRGs with an IR luminosity LIR ~ 7x10^11 L_⊙ (34; L_⊙ = 3.833×10^26 W).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it obvious what method is used (and thus what type of distance this is) to measure the distance to this galaxy? Is there a standard reference/database/etc. giving basic information like distances of astronomical objects like galaxies? I'd like to know the answer for this galaxy, but answers that generalize will naturally be more interesting.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edward L. Wright has &lt;a href=&quot;http://www.astro.ucla.edu/~wright/CosmoCalc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;a cosmology calculator&lt;/a&gt; that is useful in converting between different types of distances. His &lt;a href=&quot;http://www.astro.ucla.edu/~wright/distance.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;page on distances&lt;/a&gt; gives many methods for measuring distances; I feel that if I knew which type (if any) of these methods was used to find the distance I'd have a decent chance of figuring out what kind of distance it was (and could find the other distances by bisection).&lt;/p&gt;&#xA;" OwnerUserId="10111" LastActivityDate="2018-06-16T10:59:32.280" Title="How far away is this galaxy?" Tags="&lt;light&gt;&lt;cosmology&gt;&lt;distances&gt;&lt;redshift&gt;&lt;mathematics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26655" PostTypeId="2" ParentId="26654" CreationDate="2018-06-15T21:39:03.027" Score="3" Body="&lt;p&gt;It's the good old physical distance — also called &lt;em&gt;proper distance&lt;/em&gt; — i.e. what you'd measure if you stopped the expansion of the Universe and patiently laid out meter sticks from us to Arp 299. It's also the &lt;a href=&quot;https://en.wikipedia.org/wiki/Comoving_and_proper_distances&quot; rel=&quot;nofollow noreferrer&quot;&gt;comoving distance&lt;/a&gt; — i.e. the distance measured in coordinates that expand with the Universe and hence don't change with time — since that distance is defined to coincide the physical distance today. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Luminosity_distance&quot; rel=&quot;nofollow noreferrer&quot;&gt;Luminosity distances&lt;/a&gt; are used mostly in relations between flux, magnitudes, and distances, and are rarely quoted as a distance without explicitly saying that it's the luminosity distance. This distance is defined such that it satisfies the inverse-square law, and takes into account the expansion of the Universe and the redshift of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same goes for &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_diameter_distance&quot; rel=&quot;nofollow noreferrer&quot;&gt;angular diameter distances&lt;/a&gt;, which is defined such that it takes into account the fact that an observed object spanned a larger angle on the sky when it emitted the light we observe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In popular science, the physical distance is usually what is quoted. Sometimes, however, journalists use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Distance_measures_(cosmology)#Light-travel_distance&quot; rel=&quot;nofollow noreferrer&quot;&gt;light travel distance&lt;/a&gt;, which is the time the light has traveled, divided by the speed of light. That means that a photon emitted at the Big Bang, 13.8 billion years ago, has a light travel distance of 13.8 billion lightyears. But that's sort of a misleading number, since in fact the particle that emitted the photon is now much farther away, due to the expansion of the Universe (in fact 46.3 billion lightyears), and as &lt;a href=&quot;http://www.astro.ucla.edu/~wright/Dltt_is_Dumb.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ned Wright says&lt;/a&gt; the term should be avoided.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-06-16T07:19:06.560" LastActivityDate="2018-06-16T07:19:06.560" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26656" PostTypeId="2" ParentId="26643" CreationDate="2018-06-15T22:29:49.477" Score="7" Body="&lt;p&gt;The visibility of a narrow crescent moon depends not only on the width of the crescent, but also on the difference between the moon's and sun's altitudes relative to the local horizon.&#xA;The greater this difference, the darker the sky near the moon can be before it sets.&#xA;Estimated moon visibility maps for 2018-06-14 by&#xA;the &lt;a href=&quot;http://astro.ukho.gov.uk/nao/miscellanea/ramadan_2018/F2018Jun14.pdf&quot; rel=&quot;noreferrer&quot;&gt;UK Nautical Almanac Office&lt;/a&gt;&#xA;and &lt;a href=&quot;http://www.staff.science.uu.nl/~gent0113/islam/1439/1439g_10b.pdf&quot; rel=&quot;noreferrer&quot;&gt;R H van Gent&lt;/a&gt;,&#xA;based on a &lt;a href=&quot;http://astro.ukho.gov.uk/download/NAOTN69.pdf&quot; rel=&quot;noreferrer&quot;&gt;1997 paper by B D Yallop&lt;/a&gt;,&#xA;do not guarantee visibility anywhere in India but do indicate better chances to see it from Kerala.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's compare Stellarium screen shots for Kozhikode (11.2N 75.8E), where the &lt;a href=&quot;https://timesofindia.indiatimes.com/india/eid-to-be-celebrated-on-saturday-today-in-kerala/articleshow/64594922.cms&quot; rel=&quot;noreferrer&quot;&gt;Times of India reported&lt;/a&gt; a June 14 sighting, and New Delhi (28.6N 77.2E), at Yallop's &quot;best time&quot; 4/9 of the way from sunset to moonset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9ocN5.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9ocN5.png&quot; alt=&quot;Moon from Kozhikode&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Kozhikode (above) at 19:04 IST, the moon is 5 degrees above the horizon, and the sun is 4 degrees below the horizon, making the sky just dark enough for an experienced observer in good weather conditions to see the moon faintly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In New Delhi (below), the sun in June sets half an hour later but descends at a shallower angle.&#xA;At 19:31 IST, the moon is only 4 degrees above the horizon, and the sun is only 3 degrees below the horizon. The slightly older moon is slightly farther elongated from the sun and has a slightly wider crescent, but the lower altitude and brighter sky make it harder to see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JZwNi.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JZwNi.png&quot; alt=&quot;Moon from New Delhi&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-06-15T22:29:49.477" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26658" PostTypeId="2" ParentId="26654" CreationDate="2018-06-16T10:59:32.280" Score="4" Body="&lt;p&gt;The article references a paper by Huo et al. (2004; that's the reference number &quot;33&quot; in your quoted text). Huo et al. merely list a redshift of &quot;0.0103&quot; for Arp 299 in their Table 1 (without any apparent sources); combined with the fact that the authors of the &lt;i&gt;Science&lt;/i&gt; paper say what Hubble constant they assumed (73 km/s/Mpc), this tells us that the distance is redshift-based. That is, they used the redshift plus the Hubble constant to derive the (comoving) distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look up Arp 299 in the &lt;a href=&quot;http://ned.ipac.caltech.edu&quot; rel=&quot;noreferrer&quot;&gt;NASA Extragalactic Database&lt;/a&gt; and scroll down &lt;a href=&quot;http://ned.ipac.caltech.edu/cgi-bin/objsearch?objname=arp%20299&amp;amp;extend=no&amp;amp;list_limit=5&quot; rel=&quot;noreferrer&quot;&gt;the resulting page&lt;/a&gt;, you can see a list of possible redshifts (&quot;Calculated and Corrected Velocities&quot;) for this galaxy, ranging from 3088 km/s (which is in fact $z = 0.0103$) to 3511 km/s. Immediately below that, you can see the derived distances (&quot;Hubble Flow Distance&quot;), assuming $H_0 = 73$, the same value the authors say they used. The closest number to their value is the 44.4 Mpc, assuming &quot;3K CMB&quot; correction (which seems to match the authors' description). Why this isn't exactly the same as the authors' value of 44.8 Mpc I don't know; possible a slightly different value of the redshift was used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this relatively small redshift, the differences between the different distances that pela mentions are pretty minor. E.g., plugging $z = 0.0108$ (NED's &quot;3K CMB&quot; corrected redshift) and $H_0 = 73$ into Ned Wright's cosmology calculator gives you a comoving distance of 44.2 Mpc, a luminosity distance of 44.7 Mpc, and an angular-diamter distance of 43.8 Mpc. The differences between these are all smaller than any plausible uncertainty based on using the redshift.[*]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to summarize, the distance in the paper is (as pela says) the comoving distance; the method used to derive it was the redshift of the galaxy (corrected to a CMB-relative value) plus a Hubble constant of 73 km/s/kpc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[*] E.g., if we assume a redshift uncertainty of 300 km/s, based on possible peculiar motion of Arp 299, then the distances are uncertain by +/- 4 Mpc.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2018-06-16T10:59:32.280" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26659" PostTypeId="1" AcceptedAnswerId="26667" CreationDate="2018-06-16T15:14:13.597" Score="4" ViewCount="2112" Body="&lt;p&gt;The definition of the angular diameter distance is the ratio of an object's physical transverse size to its angular size. However when I was reading my textbook, &lt;em&gt;Astrophysics in a Nutshell by Dan Maoz pp.220-221&lt;/em&gt;, I am having some trouble trying to understand the notion of &lt;strong&gt;angular diameter distance to the last scattering surface&lt;/strong&gt;. The text calculates the angular diameter distance to the last scattering surface $D_A$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider flat cosmology (k=0) with no cosmological constant. We wish to calculate the angular size on the sky, as it appears today of a region of physical size&#xA;  $$D_s=\frac{2ct_{rec}}{\sqrt{3}}=140 kpc$$ from which light was emitted at time $t_{rec}$. Between recombination and the present time, the Universal expansion is matter-dominated, with $R\propto t^{2/3}$ for this model&#xA;  $$\frac{R}{R_0}=\left( \frac{t}{t_0} \right)^{2/3} = \frac{1}{1+z}$$&#xA;  and hence we can write&#xA;  $$D_s=\frac{2ct_0}{\sqrt{3}}(1+z_{rec})^{-3/2}$$&#xA;  The angle subtended by the region equals its size, divided by its distance to us at the&#xA;  &lt;strong&gt;time of emission&lt;/strong&gt; (since that is when the angle between rays emanating from two&#xA;  sides of the region was set).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what does the last line actually mean..Can someone please elaborate more on this? I just simply take the $D_s$ as the &quot;physical transverse size&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As we are concerned with observed angles, the type&#xA;  of distance we are interested in is the distance which, when squared and multiplied&#xA;  by 4π, will give the area of the sphere centered on us and passing through the&#xA;  said region. If the comoving radial coordinate of the surface of last scattering is r, the required distance is &lt;strong&gt;currently&lt;/strong&gt; just $r\times R_0$ and is called the proper motion distance. The proper motion distance can be solved using null geodesic in the FRW metric&#xA;  $$\int_{t_{rec}}^{t_0} \frac{c dt}{R(t)} = \int_{0}^{r}\frac{dr}{\sqrt{1-kr^2}}$$&#xA;  Setting k = 0, and substituting&#xA;  $$R(t)=R_0 \left( \frac{t}{t_0} \right)^{2/3}$$ and integrate&#xA;  $$rR_0=3ct_0[1-(1+z_{rec})^{-1/2}]$$   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I take this as the physical distance of the region from us. The next part is what confuse me:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, at the time of emission, the scale factor of the Universe was 1 + z times smaller. The so-called &lt;strong&gt;angular diameter distance&lt;/strong&gt; to the last scattering surface is therefore&#xA;  $$D_A=\frac{rR_0}{1+z}=3ct_0[(1+z_{rec})^{-1}-(1+z_{rec})^{-3/2}]$$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How does a physical distance $rR_0$ comes into play in the angular diameter distance, because from its definition it is just $$D_A=\frac{\text{physical transverse size}}{\text{angular size}}$$??&lt;/p&gt;&#xA;" OwnerUserId="23387" LastActivityDate="2018-06-18T01:30:30.500" Title="Angular Diameter Distance" Tags="&lt;astrophysics&gt;&lt;cosmology&gt;&lt;expansion&gt;&lt;angular-diameter&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26660" PostTypeId="1" AcceptedAnswerId="26664" CreationDate="2018-06-16T20:48:49.077" Score="1" ViewCount="108" Body="&lt;p&gt;We all know that solar flares can rip a planet's atmosphere if the magnetic field is weak.Solar flares travel at tremendous speed so when they hit the gas molecules of a atmosphere the molecules are ejected from the gravitational pull and thrown away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Can a giant planet's gravity pull out its moons atmosphere?Is gravity enough to pull its atmosphere or do we need an instantaneous force/impact to blow away an atmosphere?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="22623" LastActivityDate="2018-06-18T17:51:49.103" Title="Can a planet's gravity rip out its moon's atmosphere?" Tags="&lt;planet&gt;&lt;planetary-atmosphere&gt;&lt;natural-satellites&gt;&lt;solar-flare&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26661" PostTypeId="1" AcceptedAnswerId="26665" CreationDate="2018-06-16T21:26:10.050" Score="4" ViewCount="762" Body="&lt;p&gt;Suppose two SMBHs (supermassive black holes) are brought close (e.g. during a galactic merger).&#xA;Suppose that a &quot;normal&quot; black hole drifts into the space between the SMBHs.&#xA;Could the gravitational pull of the SMBHs be sufficient to tear the normal black hole to pieces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, will we be able to examine the fragments, free from being obscured by the black hole's event horizon?&lt;/p&gt;&#xA;" OwnerUserId="23390" LastEditorUserId="53" LastEditDate="2018-06-16T22:46:48.460" LastActivityDate="2018-06-17T12:26:13.297" Title="Can a black hole be torn apart by SMBHs?" Tags="&lt;black-hole&gt;&lt;supermassive-black-hole&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26662" PostTypeId="2" ParentId="26661" CreationDate="2018-06-16T21:42:52.547" Score="3" Body="&lt;p&gt;No. The total mass of the Black Hole is in the point in its center, but there is no return from passing the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Black_hole_thermodynamics#The_laws_of_black_hole_mechanics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Area theorem&lt;/a&gt; (from &lt;a href=&quot;https://en.wikipedia.org/wiki/Stephen_Hawking&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hawking&lt;/a&gt;) says, that the total area of the event horizons never decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't break up the hole, there is nothing on the event horizon, only the geometry of the spacetime looks so, that there is no way out any more. The mass of the black hole is (probably) in a small point in the centre. What you call &quot;black hole&quot;, is mainly vacuum. It doesn't matter, how do we play snooker with the black holes, the result will be at most some black hole merging, and not a broken up hole.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="1891" LastEditDate="2018-06-17T12:26:12.493" LastActivityDate="2018-06-17T12:26:12.493" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26663" PostTypeId="2" ParentId="26661" CreationDate="2018-06-17T04:02:30.943" Score="1" Body="&lt;p&gt;Taking this question from a different perspective.   Lets forget the black hole.  Black holes run into the problem with having unknown interiors inside their inescapable event horizons with the possibility of a near singularity in size at their center due to no known force being able to prevent their continued collapse, so forget black holes.  A neutron star is easier to work with as a theoretical scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine, instead, that we have a very high density Neutron star on the verge of sufficient mass and density to collapse.   The Neutron star equivalent of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Chandrasekhar_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chandrasekhar limit&lt;/a&gt;, aka the &lt;a href=&quot;https://en.wikipedia.org/wiki/Tolman%E2%80%93Oppenheimer%E2%80%93Volkoff_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tolman–Oppenheimer–Volkoff limit&lt;/a&gt;.  This isn't a black hole, but it's as close as you can get to a black hole without being a black hole, with an &lt;a href=&quot;https://en.wikipedia.org/wiki/Neutron_star#Formation&quot; rel=&quot;nofollow noreferrer&quot;&gt;escape velocity approaching to the speed of light&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets take your scenario:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose two SMBHs (supermassive black holes) are brought close (e.g.&#xA;  during a galactic merger). Suppose that a &quot;normal&quot; black hole drifts&#xA;  into the space between the SMBH&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I like your scenario because galaxies and their supermassive central black holes do merge and as a result it's theoretically possible that this could happen.  Keep in mind, everything in space is in free-fall and objects in free fall don't experience the gravity of the object they orbit.  What they experience are tidal forces and the tidal forces of super-massive black holes is comparatively pretty weak.  You get the highest tidal forces (just outside the event horizon) from the smallest stellar black holes not supermassive ones.  &lt;a href=&quot;https://physics.stackexchange.com/questions/26515/what-is-exactly-the-density-of-a-black-hole-and-how-can-it-be-calculated&quot;&gt;See here&lt;/a&gt; &lt;a href=&quot;http://curious.astro.cornell.edu/86-the-universe/black-holes-and-quasars/general-questions/423-what-is-the-density-of-a-black-hole-advanced&quot; rel=&quot;nofollow noreferrer&quot;&gt;or here&lt;/a&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Spaghettification#Examples_of_weak_and_strong_tidal_forces&quot; rel=&quot;nofollow noreferrer&quot;&gt;or here&lt;/a&gt;.  A Neutron star would withstand any tidal forces from two colliding supermassive black holes with ease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would happen is some tidal stretching like Europa experiences from Io and Ganymede as it orbits Jupiter.   In the case of Io, &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_heating_of_Io&quot; rel=&quot;nofollow noreferrer&quot;&gt;this is often called tidal heating&lt;/a&gt;, but the heat comes from stretching and squashing and it's continuously generated every orbit due to the 3 body dynamics.  Super-dense objects like Neutron stars are much more resistant to stretching and squashing, so the gravitational tidal forces would need to be enormous to have any effect.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 body tidal stretching outside the event horizon of super-massive black holes is comparatively weak.  Much too weak to pull apart a neutron star.   You'd get the strongest theoretical tidal stretching from stellar mass black holes and the strongest tidal forces possible from the smallest stellar mass black holes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this theoretical scenario to work, you'd need two stellar mass black holes about as small as black holes can form, which would have maximum density in relation to the volume of their event horizon and a Neutron star to fall perfectly in between them, nearly touching them as it falls between them and in such a scenario, the tidal forces might pull apart the high density Neutron star . . . if it was perfectly timed.   The improbability of a 3 body scenario like this actually happening is essentially zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should add, I'm not sure how the relativistic factors would affect the tidal forces.  Weird things happen with relativity, for example, &lt;a href=&quot;https://physics.stackexchange.com/questions/207816/what-is-the-relation-between-orbital-velocity-and-escape-velocity-in-strongly-re&quot;&gt;the escape velocity becomes lower than the orbital velocity&lt;/a&gt; and precisely how the relativistic effects would affect orbital velocity and tidal stretching, if they would affect it at all, I'm not sure.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pulling apart a black hole raises a whole set of impossibilities and to my understanding, covered by @Peterh answer.  Pulling apart a dense Neutron star between two small, close to maximum density stellar mass black holes if the Neutron star threads the needle between them as they tightly orbit each other, just barely, maybe,  but as noted, it's a prohibitively improbable scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's not the question you asked, but I think it touches on the idea behind your question.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-06-17T04:02:30.943" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26664" PostTypeId="2" ParentId="26660" CreationDate="2018-06-17T04:42:09.043" Score="4" Body="&lt;p&gt;It's probably rare that moons have atmospheres to begin with. Planets don't generate solar flares. Planets emit thermal radiation as a result of their heat and during formation and collision planets can get significantly hot.   Earth is believed to have glowed red hot following the theoretical giant impact and that heat actually may have helped give the Moon its lopsided crust.  &lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2014/07/01/the_moon_s_two_faces_why_are_they_so_different.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;See the Earthshine part of this article&lt;/a&gt;.  The Moon, which also likely glowed red hot after it formed, was unlikely to have ever had an atmosphere, but in theory, a hot planet could thermally strip a nearby moon if that moon had an atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A planet with a surface temperature of a couple thousand degrees could heat the near side of its moon more than the star that they both orbit, perhaps giving off enough heat to cause the loss of the moon's atmosphere through &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_escape#Thermal_escape_mechanisms&quot; rel=&quot;nofollow noreferrer&quot;&gt;thermal or Jeans escape&lt;/a&gt; which is basically a product of upper atmospheric temperature and escape velocity.  It might not even be that uncommon for moons to receive more heat from the planets they orbits than the star they both orbit when the planet is undergoing sufficient bomboardment.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second way that a planet might strip a moon's atmosphere is if the moon was close to the planet's &lt;a href=&quot;https://en.wikipedia.org/wiki/Roche_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Roche limit&lt;/a&gt;.  The atmosphere is less gravitationally bound and even if the moon comfortably holds itself together outside the Roche limit, its atmosphere would extend miles above its surface and be less gravitationally bound where gradual atmospheric stripping would be possible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moons rarely have atmospheres to begin with, but ice moons at the right temperature can generate atmospheres through outgassing.   Pluto, for example, has a &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Pluto&quot; rel=&quot;nofollow noreferrer&quot;&gt;tenuous atmosphere from the thawing of frozen nitrogen&lt;/a&gt; and other ices on its surface and Pluto's low gravity means this atmosphere is very dispersed and it can extend quite far from the planet, far enough that Pluto and its moon Charon &lt;a href=&quot;https://www.newscientist.com/article/dn25684-pluto-and-its-moon-snuggle-under-a-shared-atmosphere/&quot; rel=&quot;nofollow noreferrer&quot;&gt;may exchange atmosphere&lt;/a&gt; (grain of salt warning with that article, to say they share an atmosphere is an interpretation that I'm not 100% comfortable with, but the article does make the point that an atmosphere can be tenuously held by a comparatively low gravity object like Pluto, and as a result, affected and even shared with a nearby object in a close orbit). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A third theoretical way a planet could strip a moon of its atmosphere would be through magnetism.   Jupiter's enormous magnetic field generates a radiation belt that extends beyond its inner moons, &lt;a href=&quot;https://www.astrobio.net/news-exclusive/hiding-from-jupiters-radiation/&quot; rel=&quot;nofollow noreferrer&quot;&gt;certainly beyond the inner 3&lt;/a&gt; of its 4 Galilean moons.   The high temperature of the charged particles in a planet's radiation belt could strip a moon of its atmosphere in a similar way that the solar wind can strip planets of theirs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You just asked about gravity and I gave you three possible scenarios; radiation belts and thermal radiation aren't gravity but they could theoretically strip a moon of its atmosphere.  The Roche limit scenario is essentially driven by gravity, so the answer is yes, it's possible.  It's not actually the gravity but the tidal forces, which is a product of the gravity that can strip the atmosphere of a moon close to its planet's roche limit.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moons with atmospheres are probably pretty rare and moons that orbit close to their planet's Roche limits are also fairly rare.  So it might be a rare scenario where the two situations coincide.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="14094" LastEditDate="2018-06-18T07:14:54.407" LastActivityDate="2018-06-18T07:14:54.407" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26665" PostTypeId="2" ParentId="26661" CreationDate="2018-06-17T12:26:13.297" Score="5" Body="&lt;p&gt;@peterh's answer contains the essential point, but I think lacks enough background. Let me try to go a bit deeper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, let's be clear that we're talking about the classical Black Holes (BHs) of General Relativity (GR), and not the true BHs of the as-yet unknown theory of quantum gravity (TQG). To answer your question, TQG does not have a role, because your question is about a regime which is far from where TQG effects would be relevant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's contrary to intuition, but the vicinity of a SMBH is a &lt;em&gt;less&lt;/em&gt; exotic and &lt;em&gt;less&lt;/em&gt; stressed environment than the vicinity of an ordinary BH. If you fell toward a solar-mass BH, the tidal forces would have torn you apart well before you got to the event horizon, while in a SMBH, it happens well after you pass the event horizon. Likewise the Hawking radiation (which can be taken as a measure of the stress in spacetime) is greater outside a solar mass BH than it is outside a SMBH. &lt;a href=&quot;https://en.wikipedia.org/wiki/Spaghettification&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; has a decent article on this which, while it contains at least one howler, on the whole seems to be reliable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line: the tidal forces which tear things apart are &lt;em&gt;weaker&lt;/em&gt; outside a SMBH than outside a smaller BH. But we've already observed several smaller BHs collide using LIGO, and they don't shred each other, but merge. So we can confidently say that a SMBH will not shred a smaller BH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to understand why a BH does not shred is to think about what's actually there. The answer is &lt;em&gt;nothing&lt;/em&gt;. If you could survive the tidal forces (perhaps by falling into a SMBH) and could avoid the matter which generally forms an accretion disk around a BH (maybe by finding one in intergalactic space where there's no matter to pick up) and fall through the event horizon (EH), you'd notice nothing at all. It's just ordinary empty space outside the hole, at the EH, and inside the EH. &lt;em&gt;Locally&lt;/em&gt; (i.e., it the cabin of your (now doomed) spaceship), there's nothing to see and nothing to detect, other than the ever-increasing tidal forces which will eventually tear you apart.  The BH is nothing but curved spacetime and there's nothing else there. What's to tear apart?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using words. The mathematics of GR, which is very difficult but is also well-understood, describes exactly what happens when to BHs approach one another and Hawking's Area Theorem does prove that under GR, merger is the inevitable result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the Theory of Quantum Gravity? Obviously we don't know much about it, since it hasn't been discovered yet, but one of the few things we're sure of is that it will give results exactly* the same as GR at &quot;large&quot; distances. And &quot;large&quot; is this case is probably anything bigger than an atom. Even if the limit is &quot;bigger than a bread box&quot;, the two BHs are far too large for their collision to have any TQG effects. The TQG effects all happen down at the singularity the GR predicts at the center of the BH -- but whatever happens there is lost to the outside world.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-06-17T12:26:13.297" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26666" PostTypeId="1" AcceptedAnswerId="26668" CreationDate="2018-06-17T15:28:08.437" Score="2" ViewCount="123" Body="&lt;p&gt;If you were to place a long cylinder, open at one end and closed at the other end, and position this cylinder so that the open end is facing the Sun, and then you were to inject plasma ions, such as oxygen ions, into the cylinder near the closed end, would the solar wind entering the open end of the cylinder be strong enough to keep the oxygen ions pinned against the closed end of the cylinder, effectively trapping them there? How close to the Sun would this cylinder need to be for the solar wind to be strong enough to 'entrap' these oxygen ions?&lt;/p&gt;&#xA;" OwnerDisplayName="user23327" LastEditorDisplayName="user23327" LastEditDate="2018-06-18T19:25:15.047" LastActivityDate="2018-09-22T14:01:15.817" Title="Would the dynamic pressure of the Sun’s solar wind be strong enough to trap plasma ions within an open container?" Tags="&lt;astrophysics&gt;&lt;comets&gt;&lt;space-probe&gt;&lt;astrochemistry&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="26667" PostTypeId="2" ParentId="26659" CreationDate="2018-06-17T17:09:49.623" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How does a physical distance $rR_0$ comes into play in the angular diameter distance, because from its definition ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The manner you choose to calculate it is explained on Wikipedia's &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_diameter_distance&quot; rel=&quot;nofollow noreferrer&quot;&gt;Angular diameter distance&lt;/a&gt;&quot; webpage:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angular diameter distance is a distance measure used in astronomy. It is defined in terms of an object's physical size, $x$, and $\theta$ the angular size of the object as viewed from earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_{A}={\frac {x}{\theta}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angular diameter distance depends on the assumed cosmology of the universe. The angular diameter distance to an object at &lt;a href=&quot;https://en.wikipedia.org/wiki/Redshift&quot; rel=&quot;nofollow noreferrer&quot;&gt;redshift&lt;/a&gt;, $z$, is expressed in terms of the comoving distance, $r$ as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_{A}={\frac {S_{k}(r)}{1+z}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $S_{k}(r)$ is the FLRW coordinate defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_{k}(r)={\begin{cases}\sin \left({\sqrt {-\Omega _{k}}}H_{0}r\right)/\left(H_{0}{\sqrt {|\Omega _{k}|}}\right)&amp;amp;\Omega _{k}&amp;lt;0\\r&amp;amp;\Omega _{k}=0\\\sinh \left({\sqrt {\Omega _{k}}}H_{0}r\right)/\left(H_{0}{\sqrt {|\Omega _{k}|}}\right)&amp;amp;\Omega _{k}&amp;gt;0\end{cases}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\Omega _{k}$ is the curvature density and $H_{0}$ is the value of the Hubble parameter today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the currently favoured geometric model of our Universe, the &quot;angular diameter distance&quot; of an object is a good &lt;strong&gt;approximation&lt;/strong&gt; to the &quot;real distance&quot;, i.e. the proper distance when the light left the object. &lt;strong&gt;Note that beyond a certain redshift, the angular diameter distance gets smaller with increasing redshift.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also Wikipedia's &quot;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Distance_measures_(cosmology)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Distance measures (cosmology)&lt;/a&gt;&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0qzVH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0qzVH.png&quot; alt=&quot;Comparison of cosmological distance measures&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{l}&#xA;\text{Comparison of cosmological distance $\qquad\quad\;\,$ Comparison of cosmological distance} \\&#xA;\text{measures, from redshift 0 to redshift 0.5. $\qquad$ measures, from redshift 0 to redshift 10K.} \\&#xA;\hline&#xA;\qquad\text{The background cosmology is Hubble parameter 72 km/s/Mpc, $\Omega_\Lambda=0.732$,} \\&#xA;\qquad\text{$\Omega_{\rm matter}=0.266$, $\Omega_{\rm radiation}=0.266/3454$, and $\Omega_k$ chosen so that the sum of} \\&#xA;\qquad\text{Omega parameters is 1.} &#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how, even at a small $z$, the choice of &lt;a href=&quot;http://inspirehep.net/record/1641147?ln=en&quot; rel=&quot;nofollow noreferrer&quot;&gt;cosmological model&lt;/a&gt; is important for full accuracy.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The book &quot;Astrophysics in a Nutshell: &lt;strong&gt;Second Edition&lt;/strong&gt;&quot; was published &lt;a href=&quot;https://www.chapters.indigo.ca/en-ca/books/product/9780691164793-item.html?s_campaign=goo-Shopping_Books&amp;amp;ds_rl=1254699&amp;amp;gclid=CjwKCAjwjZjZBRAZEiwAPeLSK8581OpCWDReUl_4g_3e55HS9GBVhKF4hMknC3JwUtW849A79Wa4LRoCo6YQAvD_BwE&quot; rel=&quot;nofollow noreferrer&quot;&gt;Feb 23, 2016&lt;/a&gt;, the first edition was published on &lt;a href=&quot;https://books.google.ca/books/about/Astrophysics_in_a_Nutshell.html?id=yWidGMhCNX8C&amp;amp;redir_esc=y&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dec 4, 2011&lt;/a&gt;. The first edition is &lt;strong&gt;&lt;em&gt;old&lt;/em&gt;&lt;/strong&gt; (and 25% of the cost of), the second edition is not new.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia explains the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lambda-CDM_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lambda-CDM model&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The ΛCDM (Lambda cold dark matter) or Lambda-CDM model is a parametrization of the Big Bang cosmological model in which the universe contains a cosmological constant, denoted by Lambda (Greek Λ), associated with dark energy, and cold dark matter (abbreviated CDM). It is frequently referred to as the standard model of Big Bang cosmology because it is the &lt;strong&gt;simplest model&lt;/strong&gt; that provides a &lt;strong&gt;reasonably good&lt;/strong&gt; account of the following properties of the cosmos:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;the existence and structure of the cosmic microwave background&lt;/li&gt;&#xA;  &lt;li&gt;the large-scale structure in the distribution of galaxies&lt;/li&gt;&#xA;  &lt;li&gt;the abundances of hydrogen (including deuterium), helium, and lithium&lt;/li&gt;&#xA;  &lt;li&gt;the accelerating expansion of the universe observed in the light from distant galaxies and supernovae&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;The model assumes that general relativity is the correct theory of gravity on cosmological scales. &lt;strong&gt;It emerged in the late 1990s&lt;/strong&gt; as a &lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/c/concordance+model&quot; rel=&quot;nofollow noreferrer&quot;&gt;concordance cosmology&lt;/a&gt;, after a period of time when &lt;strong&gt;disparate observed properties of the universe appeared mutually inconsistent, and there was no consensus on the makeup of the energy density of the universe&lt;/strong&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The ΛCDM model &lt;strong&gt;can be extended&lt;/strong&gt; [tweaked to &lt;em&gt;fix&lt;/em&gt; it, depending on what you are doing] by adding cosmological inflation, quintessence and other elements that are current areas of speculation and research in cosmology.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Some alternative models&lt;/strong&gt; &lt;strong&gt;&lt;em&gt;challenge&lt;/em&gt;&lt;/strong&gt; &lt;strong&gt;the assumptions of the ΛCDM model&lt;/strong&gt;. Examples of these are modified Newtonian dynamics, modified gravity, theories of large-scale variations in the matter density of the universe, and &lt;strong&gt;scale invariance&lt;/strong&gt; of empty space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason to nitpick about a &lt;em&gt;tiny&lt;/em&gt; difference of opinion is because the distances involved are so enormous, and the difference really isn't so small either. Depending on the distance the amount of time that has passed means that the space through which the light is passing changes somewhat throughout it's lifetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also: &quot;&lt;a href=&quot;https://arxiv.org/abs/1605.06314&quot; rel=&quot;nofollow noreferrer&quot;&gt;Scale invariant cosmology III: dynamical models and comparisons with observations&lt;/a&gt;&quot; (19 May 2016) by André Maeder:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The basic equations of cosmology are modified, showing an acceleration of the expansion after a certain initial period, the duration of which depends on the mean density of the Universe. Another major consequence of the scale&#xA;  invariance is that the laws of conservation of matter-energy show some dependence on the cosmic time. This dependence is very weak for models with a non-zero matter density, but at the conceptual level this is not a minor effect.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We do think it is worth to undertake the present exploration for &lt;strong&gt;two&lt;/strong&gt; main reasons. One is that the recent cosmological results suggest that a totally unknown form of matter-energy, the dark energy, dominates the energy content of the Universe. This is a major problem. The other main reason is that &lt;strong&gt;scale invariance is not a kind of adjusted trick to make things work&lt;/strong&gt;. But it is a basic physical change, that responds to the fundamental wish (Dirac 1973) that the equations expressing basic laws should be invariant under the widest group of transformations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The newer paper &quot;&lt;a href=&quot;https://arxiv.org/abs/1701.03964&quot; rel=&quot;nofollow noreferrer&quot;&gt;An alternative to the LCDM model: the case of scale invariance&lt;/a&gt;&quot; (14 Jan 2017), by André Maeder contains the calculations ($d_M = R_0 r_1$, $d_A = d_M/(1+z)$, on pages 13 and 14) and the following graph:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FFl4Z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FFl4Z.jpg&quot; alt=&quot;Lambda vs. Scale Invariant&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Figure 5. The angular diameter distance $d_A$ vs. redshift $z$ for flat scale invariant models (continuous red lines) compared to flat ΛCDM&#xA;models (broken blue lines). The curves are given for Ωm = 0, 0.1, 0.3, 0.99, from the upper to the lower curve in both cases (at $z$ &gt; 3).&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please refer to that paper for details on the derivation of the calculations.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="18054" LastEditDate="2018-06-18T01:30:30.500" LastActivityDate="2018-06-18T01:30:30.500" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26668" PostTypeId="2" ParentId="26666" CreationDate="2018-06-17T18:01:32.237" Score="0" Body="&lt;p&gt;So you basically said that we could use the Sun to propel a rocket. We already have rockets like this equipped with solar sails. These solar sails have a &quot;sail&quot; that catches the stuff thrown out of the Sun. I'm no expert on this way of travel, but that is the basic description of how it works.&lt;/p&gt;&#xA;" OwnerUserId="23317" LastActivityDate="2018-06-17T18:01:32.237" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26669" PostTypeId="1" CreationDate="2018-06-18T06:05:32.030" Score="4" ViewCount="178" Body="&lt;p&gt;I have the telescope in the title, which camps with 10mm and 25mm eyepieces. Both cannot really see anything in detail past the moon. All planets look like stars. What are good eyepieces for seeing all the other planets in detail, respectively?&lt;/p&gt;&#xA;" OwnerUserId="23395" LastEditorUserId="11822" LastEditDate="2018-06-18T07:52:41.027" LastActivityDate="2018-06-18T08:19:37.367" Title="Orion Skyquest xt8i Intelliscope Dobsonian Telescope Lens" Tags="&lt;telescope&gt;&lt;orion&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26671" PostTypeId="2" ParentId="19092" CreationDate="2018-06-18T07:58:14.857" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/13330/why-hasnt-the-9th-planet-been-detected-already&quot;&gt;This&lt;/a&gt; question already discusses the challenges of discovering planet 9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main problem of the hypothetical planet 9 as compared to Sedna and 2012 VP113 is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sedna and 2012 VP113 have orbits that reach far out but currently they are near their perihelion so it was possible to observe them in the visible light spectrum. Planet 9's orbit is far out or at least it is currently near its aphelion. Otherwise it probably would have already been discovered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the planet's orbit is probably tilted towards the ecliptic and it moves very slowly compared to the stars in the background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this position the maximum of its radiation would be emitted in the far infrared and it moves before the background of the Milky Way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to summarize:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would need a long term observation with an far-infrared capable, spaceborne, telescope (because our atmosphere absorbs far infrared radiation) which has to include the background of the Milky Way. The telescope has to be able to detect faint infrared sources. We could be lucky and planet 9 just happens to pass in front of a star as we are observing it. Then we could get our first hint with an optical telescope. But that's pure luck and we can't plan towards that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Wide-field_Infrared_Survey_Explorer&quot; rel=&quot;nofollow noreferrer&quot;&gt;WISE&lt;/a&gt; already tried to find planet 9 but did not find it. The &lt;a href=&quot;https://en.wikipedia.org/wiki/James_Webb_Space_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;James Webb Telescope&lt;/a&gt; may be our sole instrument capable of discovering planet 9 directly. But it's not built to permanently survey an area of the sky although it could confirm a finding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Gaia_(spacecraft)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gaia&lt;/a&gt; could probably detect planet 9 if it happens to pass in front a star.&lt;/p&gt;&#xA;" OwnerUserId="22927" LastActivityDate="2018-06-18T07:58:14.857" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26672" PostTypeId="2" ParentId="25007" CreationDate="2018-06-18T08:16:21.690" Score="2" Body="&lt;p&gt;Well I can't give a mathematically based answer. I can only tell that some man-made objects were recorded by sky surveys. E.g. the &lt;a href=&quot;https://en.wikipedia.org/wiki/Rosetta_(spacecraft)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rosetta&lt;/a&gt; spacecraft was marked as a newly found asteroid by the Catalina Sky survey although the error was corrected quite quickly.&#xA;Rosetta made an earth flyby at this time so it was pretty close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The SpaceX Tesla will probably not come close to earth for quite a time and as such it probably won't show up in any survey.&lt;/p&gt;&#xA;" OwnerUserId="22927" LastActivityDate="2018-06-18T08:16:21.690" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26673" PostTypeId="2" ParentId="26669" CreationDate="2018-06-18T08:19:37.367" Score="7" Body="&lt;p&gt;The telescope you have is quite well regarded, and the eyepieces that come with it are of reasonable quality.  If you can see the moon clearly, including craters and other features, then you should be able to see that the planets, particularly Jupiter and Saturn,  are not like stars.  At the very least they will look like disks, not points of light, and saturn's rings will be unmistakable. The 10mm lens will give magnification of 120, which is more than adequate.  My guess is that you are not looking at the planets, but stars close to them.&lt;br&gt;&#xA;Make sure sure the finder scope is properly aligned - use the moon for that - and then try again.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2018-06-18T08:19:37.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26674" PostTypeId="2" ParentId="22406" CreationDate="2018-06-18T09:24:14.997" Score="0" Body="&lt;p&gt;Easy. Bounce a laser off the moon at a specific time of day and do this everyday for years and years. You could in fact get the distance it moves away from earth down to the day just by the time it takes for the laser light to come back to your device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the time it takes for the light to come back with the speed of light and divide by 2. Note down the time each day and check the differences over time. You'll be able to get the change of rate down to the day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heck, we could even fly back to the moon, house a computer with some batteries and a few solar panels to charge them and ping it. On the trip back just drop off a few satellites 22,000 miles apart back to earth to boost signal so it could connect to the internet lol. Fun aside.&lt;/p&gt;&#xA;" OwnerUserId="23398" LastActivityDate="2018-06-18T09:24:14.997" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26675" PostTypeId="1" AcceptedAnswerId="26779" CreationDate="2018-06-18T11:14:51.927" Score="2" ViewCount="1078" Body="&lt;p&gt;I have 4000 fits-files images from SPITZER, each containing the same star over a time period. Each image is a photmetry cube of 32 pixel * 32 pixel * 64 consequent time frames. I need to look through all the single frames and order the pixel by brightness. Not hard to do with Python, but sometimes I have in the frames short-lived events/artefacts which are brighter than the brightest pixel of the star (see below). These artefacts last only over one or two consequent time frames&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I easily correct for the artefacts? Is there an astropy/pyfits package that does this? Or some kind of bad pixel flagging in SPITZER (I know that Kepler-images have flagging of bad pixel)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Optional: It would be convenient to have the code in vectorized form, avoiding loops, so that I save computtional time &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Fig: In the left frame everything is normal, in the right frame, in the top right corner you see one of the artefacts, having a value below a saturation treshhold&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mgTV7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mgTV7.png&quot; alt=&quot;In the left panel everything is normal, in the right panel, in the top right corner you see one of the artefacts, having a value below a saturation treshhold&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23399" LastEditorUserId="23399" LastEditDate="2018-06-20T17:43:12.797" LastActivityDate="2018-06-27T11:24:19.587" Title="Python: Find brightest pixel in fits-file and ignore brighter artefacts" Tags="&lt;python&gt;&lt;light-curve&gt;&lt;fits-header&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26676" PostTypeId="2" ParentId="26675" CreationDate="2018-06-18T17:48:09.873" Score="0" Body="&lt;p&gt;Apart from the solution suggested in the comments (PSF fit), which is probably the proper way of doing further analyses, several solutions come to mind:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Check if the artefacts are in fact saturated pixels (sometimes the bias is subtracted, so you may need to check for saturation value minus bias). Ignore any pixels with that value.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If that doesn't work:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;Get the mean value for each pixel across the time series, reject all pixels that are more that $x$ (%) away from it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For each pixel across your time series, check if its brightness changes by more than $x$ (%). Note previous value, reject as long as subsequent values are also above the threshold.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;All these methods will have problems if the bright pixel is at the position of the star.&lt;/p&gt;&#xA;" OwnerUserId="13681" LastActivityDate="2018-06-18T17:48:09.873" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26677" PostTypeId="2" ParentId="26660" CreationDate="2018-06-18T17:51:49.103" Score="1" Body="&lt;p&gt;Possibly an extreme case of the Roche limit (described aptly by userLTK) a moon with atmosphere in a low orbit could lead to tidal forces deforming the atmosphere to the point that at least some of the gas is pulled free from the moon's grav. field.  I wouldn't count on it,  since it's a bit hard to imagine the planet itself being atmosphere-free.  &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-06-18T17:51:49.103" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26678" PostTypeId="1" AcceptedAnswerId="26685" CreationDate="2018-06-18T19:53:34.413" Score="4" ViewCount="455" Body="&lt;p&gt;Is it possible to use a standard Newtonian laser collimator, &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B01D8HMBCQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;such as this one&lt;/a&gt; to collimate an RCT?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, is the procedure any different than collimating a Newtonian (i.e. make the laser go into the hole by adjusting the mirrors)?&lt;/p&gt;&#xA;" OwnerUserId="23402" LastEditorDisplayName="user1569" LastEditDate="2018-06-19T12:12:36.470" LastActivityDate="2018-06-19T12:12:36.470" Title="Can a Ritchey–Chrétien telescope be collimated using a standard Newtonian laser collimator?" Tags="&lt;collimation&gt;&lt;rc-telescope&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26679" PostTypeId="2" ParentId="26649" CreationDate="2018-06-18T20:36:24.750" Score="2" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Generating random galaxy catalogs for correlation functions&lt;/p&gt;&#xA;&lt;p&gt;... the random catalog can be just be a uniform distribution of points in the same volume as the real data.&lt;/p&gt;&#xA;&lt;p&gt;However, real-world galaxy catalogs are not homogenous boxes, and have very irregular shapes and observationally induced redshift and angular distributions. Even some mock-catalogs are endowed with these properties to better represent real data. In these cases, how is a random catalog best produced? ...&lt;/p&gt;&#xA;&lt;p&gt;If the answer is that some nonuniform distribution should be used, is there a usual method for doing so? The literature I have found (e.g. for the SDSS published data, and for galaxy mocks mimicking SDSS's constraints) does not discuss this in any way. &lt;strong&gt;Note, if this is a highly involved process beyond a brief summary and a source or two, I do not expect a full answer here&lt;/strong&gt;. Rather, I am trying to determine if there is an existing set of standard methods which are documented but have eluded me thus far.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;For SDSS see:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&amp;quot;&lt;a href=&quot;https://www.aanda.org/articles/aa/full/2006/08/aa3902-05/aa3902-05.right.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Large scale correlations in galaxy clustering from the two degree field galaxy redshift survey&lt;/a&gt;&amp;quot; (2006), by N. L. Vasilyev, Yu. V. Baryshev and F. Sylos Labini.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&amp;quot;&lt;a href=&quot;https://arxiv.org/abs/1303.2432&quot; rel=&quot;nofollow noreferrer&quot;&gt;The SDSS Galaxy Angular Two-Point Correlation Function&lt;/a&gt;&amp;quot; (12 Mar 2013), by Y. Wang, R. J. Brunner, J. C. Dolence. The include their SQL query to the SDSS catalog archive server:&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;1 INTRODUCTION&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;One of the most powerful and simplest probes of the galaxy distribution is the two-point angular correlation function, which quantifies the excess probability above a random distribution of finding one galaxy within a specified angle of another galaxy. For the case of a Gaussian random field, the two-point angular correlation function and its Legendre transform pair provide a complete statistical characterization of the galaxy clustering (see, e.g., Peebles 1980). Even for the case of non-Gaussianity, the two-point angular correlation function provides a simple and important statistical test of galaxy formation models (Tegmark et al. 2004). The two-point angular correlation function has been studied at bright magnitudes from the data releases from the Sloan Digital Sky Survey (SDSS) such as the Early Data Release (EDR; Connolly et al. 2002).&lt;/p&gt;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;To accurately calculate the galaxy two-point angular correlation function, we must first minimize potential systematic effects in the galaxy catalog used to measure the correlation function&lt;/strong&gt;. The systematics of the SDSS EDR were thoroughly studied by Scranton et al. (2002). To minimize the systematic effects of seeing and Galactic extinction, they determined that the SDSS EDR galaxy sample had to be masked to exclude regions with seeing greater than 1.&amp;quot;75 and reddening &amp;gt; 0.2 magnitudes. Given the importance of minimizing the impact of systematic effects on the galaxy two-point angular correlation function and the significant changes that were made in the SDSS data processing pipeline between the SDSS EDR and the SDSS DR7, we have repeated many of the tests presented in Scranton et al. (2002) by using the SDSS DR7 data. &lt;strong&gt;In this paper we present the methods used to contain these systematic effects, the results of these systematic tests, the actual galaxy two-point angular correlation function for the SDSS DR7, and our massively parallel implementation that rapidly calculates correlation functions for large data sets&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Also: &amp;quot;The galaxy correlation function and power spectrum&amp;quot; (&lt;a href=&quot;http://www.uio.no/studier/emner/matnat/astro/AST4320/h12/undervisningsmateriale/galaxyclustering.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;.PDF&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;p&gt;Wikipedia's &amp;quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Correlation_function_(astronomy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Correlation function (astronomy)&lt;/a&gt;&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;CalTech: &amp;quot;&lt;a href=&quot;https://ned.ipac.caltech.edu/level5/March04/Jones/Jones5.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Measurements of Clustering&lt;/a&gt;&amp;quot; (Use [Next] or [Contents] buttons to view all the pages).&lt;/p&gt;&#xA;&lt;p&gt;There are fixups applied to the raw data to &lt;a href=&quot;https://astronomy.stackexchange.com/a/26667/18054&quot;&gt;correct for redshift and angular diameter distance&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;For example catalogs and a description of the procedures see Shaun Cole's website: &amp;quot;&lt;a href=&quot;http://star-www.dur.ac.uk/%7Ecole/mocks/main.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mock Galaxy Redshift Catalogues&lt;/a&gt;&amp;quot; where software and full details of the N-body simulations and the contruction and properties of the mock catalogues is provided for both 2dF Galaxy Redshift Survey (2dFGRS)  and SDSS galaxy redshift surveys.&lt;/p&gt;&#xA;&lt;p&gt;Halo bias relation to dark matter and N-body simulations are discussed in &amp;quot;&lt;a href=&quot;https://arxiv.org/abs/1806.05870&quot; rel=&quot;nofollow noreferrer&quot;&gt;BAM: Bias Assignment Method to generate mock catalogs&lt;/a&gt;&amp;quot; (15 June 2018), by A. Balaguera-Antolínez, Francisco-Shu Kitaura, Marcos Pellerejo-Ibañez, Cheng Zhao, and Tom Abel.&lt;/p&gt;&#xA;&lt;p&gt;Also check out the (under construction, and too busy working and writing papers) website: &amp;quot;&lt;a href=&quot;https://cgs.obs.carnegiescience.edu/CGS/Publications.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Carnegie-Irvine Galaxy Survey (CGS)&lt;/a&gt;&amp;quot; - with missing papers available on arXix:&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;&lt;a href=&quot;https://arxiv.org/abs/1111.4605&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Carnegie-Irvine Galaxy Survey. I. Overview and Atlas of Optical Images&lt;/a&gt;&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;&lt;a href=&quot;https://arxiv.org/abs/1111.4606&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Carnegie-Irvine Galaxy Survey. II. Isophotal Analysis&lt;/a&gt;&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;&lt;a href=&quot;https://arxiv.org/abs/1212.2639&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Carnegie-Irvine Galaxy Survey. III. The Three-Component Structure of Nearby Elliptical Galaxies&lt;/a&gt;&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;&lt;a href=&quot;https://arxiv.org/abs/1603.00876&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Carnegie-Irvine Galaxy Survey. IV. A Method to Determine the Average Mass Ratio of Mergers That Built Massive Elliptical Galaxies&lt;/a&gt;&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;&lt;a href=&quot;https://arxiv.org/abs/1707.03570&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Carnegie-Irvine Galaxy Survey. V. Statistical study of bars and buckled bars&lt;/a&gt;&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;&lt;a href=&quot;https://arxiv.org/abs/1806.06591&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Carnegie-Irvine Galaxy Survey. VI. Quantifying Spiral Structure&lt;/a&gt;&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;&lt;a href=&quot;https://arxiv.org/abs/1806.06350&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Carnegie-Irvine Galaxy Survey. VII. Constraints on the Origin of S0 Galaxies from Their Photometric Structure&lt;/a&gt;&amp;quot;&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-06-19T05:04:32.330" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26680" PostTypeId="2" ParentId="26431" CreationDate="2018-06-18T22:03:14.253" Score="1" Body="&lt;p&gt;A very simple example is to fill a 3d density with a regular plane wave, which should result in a single peak in the power spectrum. Here is a simple python scripts that does exactly that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computing the two point correlation function should be only one inverse fourier transform of the Power spectrum away. (Depending on which FFT library you use, you might need to renormalize the results. FFTW and numpy.fft for example have unnormalized fourier transforms: $F^{-1}[F[f(x)] = Nf(x) $, where $N$ is the number of samples.)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#!/usr/bin/python3&#xA;&#xA;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;&#xA;&#xA;#==================================&#xA;def main():&#xA;#==================================&#xA;&#xA;&#xA;    nc = 128                # define how many cells your box has&#xA;    boxlen = 50.0           # define length of box&#xA;    Lambda = boxlen/4.0     # define an arbitrary wave length of a plane wave&#xA;    dx = boxlen/nc          # get size of a cell&#xA;&#xA;    # create plane wave density field&#xA;    density_field = np.zeros((nc, nc, nc), dtype='float')&#xA;    for x in range(density_field.shape[0]):&#xA;        density_field[x,:,:] = np.cos(2*np.pi*x*dx/Lambda)&#xA;&#xA;    # get overdensity field&#xA;    delta = density_field/np.mean(density_field) - 1&#xA;&#xA;    # get P(k) field: explot fft of data that is only real, not complex&#xA;    delta_k = np.abs(np.fft.rfftn(delta).round())&#xA;    Pk_field =  delta_k**2&#xA;&#xA;    # get 3d array of index integer distances to k = (0, 0, 0)&#xA;    dist = np.minimum(np.arange(nc), np.arange(nc,0,-1))&#xA;    dist_z = np.arange(nc//2+1)&#xA;    dist *= dist&#xA;    dist_z *= dist_z&#xA;    dist_3d = np.sqrt(dist[:, None, None] + dist[:, None] + dist_z)&#xA;&#xA;    # get unique distances and index which any distance stored in dist_3d &#xA;    # will have in &quot;distances&quot; array&#xA;    distances, _ = np.unique(dist_3d, return_inverse=True)&#xA;&#xA;    # average P(kx, ky, kz) to P(|k|)&#xA;    Pk = np.bincount(_, weights=Pk_field.ravel())/np.bincount(_)&#xA;&#xA;    # compute &quot;phyical&quot; values of k&#xA;    dk = 2*np.pi/boxlen&#xA;    k = distances*dk&#xA;&#xA;    # plot results&#xA;    fig = plt.figure(figsize=(9,6))&#xA;    ax1 = fig.add_subplot(111)&#xA;    ax1.plot(k, Pk, label=r'$P(\mathbf{k})$')&#xA;&#xA;    # plot expected peak:&#xA;    # k_peak = 2*pi/lambda, where we chose lambda for our planar wave earlier&#xA;    ax1.plot([2*np.pi/Lambda]*2, [Pk.min()-1, Pk.max()+1], label='expected peak')&#xA;    ax1.legend()&#xA;    plt.show()&#xA;&#xA;&#xA;&#xA;&#xA;&#xA;&#xA;&#xA;&#xA;#==================================&#xA;if __name__ == &quot;__main__&quot;:&#xA;#==================================&#xA;&#xA;    main()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="20403" LastActivityDate="2018-06-18T22:03:14.253" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26682" PostTypeId="1" AcceptedAnswerId="26686" CreationDate="2018-06-18T23:54:08.657" Score="2" ViewCount="1911" Body="&lt;p&gt;If black holes swallow stuff, will a big black hole eventually swallow the universe? The universe will eventually go dark, because there will not be enough matter to make stars, and all that will be left is a bunch of black holes and they can merge. So will the universe get swallowed by a big black hole someday?&lt;/p&gt;&#xA;" OwnerUserId="23317" LastActivityDate="2018-06-19T18:46:35.307" Title="Will a black hole swallow the universe?" Tags="&lt;black-hole&gt;&lt;universe&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26683" PostTypeId="1" CreationDate="2018-06-19T00:01:52.037" Score="1" ViewCount="304" Body="&lt;p&gt;I'm currently writing a code to compute the two-point correlation function of a set of galaxies. As I work with a loooot of galaxies, it was suggested to me not to do it in real space, but to compute the power spectrum in Fourier space and get the correlation from the power spectrum, as is explained e.g. in &lt;a href=&quot;https://arxiv.org/pdf/1507.01948.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;appendix B in this paper (arXiv:1507.01948)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, it seems like I'm doing something wrong, and I don't know exactly what. For the FFT, I use the FFTW library, which as far as I see follows the same conventions concerning indexing and definition of the DFT as the numpy.fft library. If necessary, I'll gladly supply my full code and galaxy catalogues, but meanwhile, I'll describe it step by step:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First read in the galaxy data (Stellar mass, x,y,z positions) and compute the density contrast field $$ \delta(\mathbf{x}) = \frac{\rho(\mathbf{x})}{\bar{\rho}} -1 $$&#xA;where $\rho(\mathbf{x})$ is the density field in units of $M_{sol}/Mpc^3$ and $\bar{\rho}$ is the mean density of the entire simulation box, stored in a 3d-array of size $N^3$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;compute $\delta(\mathbf{k})$ , the fourier transform  of $\delta(\mathbf{x})$ [meaning just calling the fftw_execute_dft_r2c() routines, nothing else added at this step]&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;compute the &quot;power spectrum field&quot; $P(\mathbf{k}) = \delta(\mathbf{k})\bar{\delta}(\mathbf{k}) / (2\pi)^3$ , where $\bar{\delta}(\mathbf{k})$ is the complex conjugate of $\delta(\mathbf{k})$ . Note that at this point, it is not averaged yet, as this is only an intermediate step.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;compute $k = | \mathbf{k} | = \sqrt{k_x^2 + k_y^2 + k_z^2}$ for every $(k_x, k_y, k_z)$ of the available $P(\mathbf{k})$. &#xA;I also take into account the convention that $\mathbf{k} = 0$ is at index $(0,0,0)$ in python/C and (1,1,1) for Fortran, respectively, and that for $j &amp;gt; N/2+1$, $k_j \neq k_j \Delta k$, but $k_j = (-N + j)\Delta k$. &#xA;(For a real-to-complex FFT of a 3d-array, this holds for 2 of the 3 dimensions. Using the fact that $\tilde{f}(\mathbf{k}) = \bar{\tilde{f}}(\mathbf{-k})$ for the Fourier transform $\tilde{f}$ of a real function $f$, one array dimension will be truncated to length $N/2+1$ to avoid redundant information. For Fortran, it's the first index; for C/python, it's the last index.) $\Delta k$ is given by $\frac{2\pi}{boxlength}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;histogrammize $P(\mathbf{k})$ to obtain $P(k)$ : Set up $n_b$ number of bins for $k$ between $k_{min}=0$ and $k_{max} = N/2 \Delta k$. I chose $k_{max}$ this way to ensure that the samples are equal in the sense that that is the maximal radius from $\mathbf{k} = 0$ which contains a full sphere inside the box without intersecting with already used samples. &#xA;I histogrammize both weighted $P_{hist,weight}$ with the value of each $P(\mathbf{k})$, as well as unweighted, $P_{hist,count}$ to obtain the number of counts in each bin, in order to compute the average value in each bin at the end: &#xA;$$P(k_i) = \frac{P_{hist,weigh}(k_i)}{P_{hist,count}(k_i)}$$&#xA;This concludes the computation of the Power spectrum. Next comes the correlation function:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Compute the inverse Fourier transform of the &quot;power spectrum field&quot; $P(\mathbf{k}) = \delta(\mathbf{k})\bar{\delta}(\mathbf{k}) / (2\pi)^3$ to obtain the &quot;correlation field&quot; $\xi(\mathbf{x})$. This time, a normalisation is necessary, because the FFTW transforms are unnormalised: $\mathcal{F}^{-1}[\mathcal{F}[f(x)] = N f(x) $, with $N$ being the array size for a 1d array. In my 3d-case, I divide the values by $N^3$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Histogrammize $\xi(\mathbf{x})$ in the same way I histogrammize $P(\mathbf{k})$ to obtain $\xi(r)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the results are orders of magnitude off of observational values. I took the observational values for the power spectrum from &lt;a href=&quot;https://arxiv.org/pdf/astro-ph/0111575.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, and a best-fit function for the galaxy correlation function from &lt;a href=&quot;https://arxiv.org/pdf/astro-ph/0301280.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a plot of my results, the green lines being the observational data, while the blue/orange line are two different cases obtained from the mock galaxy catalogue (whether orphan galaxies are included or not). This particular image was created with an array of length $N=512$ per dimension.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/7A5Vm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7A5Vm.png&quot; alt=&quot;my results for the power spectrum and galaxy correlation function&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now clearly something is way off here, even though the general shape doesn't seem too bad. I have 2 suspicions for the reason behind this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) I must be doing something wrong with the normalisation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I do something wrong when using different conventions of Fourier transforms: As far as I see, the Power spectra are computed using the convention &#xA;$$ \mathcal{F}[f(x)] = \int f(x) e^{-i kx} dx $$&#xA;while the FFTW/numpy libraries use &#xA;$$ \mathcal{F}[f(x)] = \int f(x) e^{-i 2\pi kx} dx $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check those suspicions, I checked when the normalisation is required by transforming and inverse-transforming a known function. Specifically:&#xA; $$ \delta_D(x-1)/2 - \delta_D(x+1)/2 \leftrightarrow - i \sin(k) $$&#xA;following the convention $ \mathcal{F}[f(x)] = \int f(x) e^{-i kx} dx $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get exactly correct results (nice peaks and a sinus wave) for both the Fourier transform and the inverse transform by following these rules:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 ) To change between transform conventions, substitute $k \rightarrow 2 \pi k$ : a $k$ in the FFTW convention corresponds to $2\pi k$ in the other convention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 ) Normalise by dividing by $N$ after the inverse transform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If necessary, I'll gladly provide the code I tested this with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the problem remains. I still think that I am doing something wrong with the normalisation, because  I verified that the value of the Power spectrum $P(k)$ indeed does vary by orders of magnitude using different numbers of samples $N$, the correlation function however doesn't. Maybe this is somehow connected to the fact that I compute the square value $\delta(\mathbf{k})\bar{\delta}(\mathbf{k})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody please please help me? I've been stuck on this for weeks now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I've uploaded my code here: &lt;a href=&quot;https://bitbucket.org/mivkov/sharing/src/master/computing_power_spectrum/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://bitbucket.org/mivkov/sharing/src/master/computing_power_spectrum/&lt;/a&gt; if anyone wants to see it directly.&lt;/p&gt;&#xA;" OwnerUserId="20403" LastEditorUserId="20403" LastEditDate="2018-06-19T15:22:35.157" LastActivityDate="2018-06-19T15:22:35.157" Title="Correctly scaling Power Spectrum and Correlation Function of Galaxies from Mock Catalogues" Tags="&lt;galaxy&gt;&lt;cosmology&gt;&lt;data-analysis&gt;&lt;spectra&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26685" PostTypeId="2" ParentId="26678" CreationDate="2018-06-19T07:20:23.947" Score="6" Body="&lt;p&gt;No, the fundamental designs of the two types of telescope are in this respect quite different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Newtonian laser collimator works by shining a laser through the eyepiece holder to the secondary mirror which reflects the beam onto the primary. The primary, usually a parabolic mirror but always concave, is adjusted so that the beam is reflected back to the secondary mirror such that the reflected beam returns to its point of origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A RCT, or other Cassegrain-like telescopes have a convex, possibly hyperbolic, mirror as the secondary. While it is possible to shine the laser through the eyepiece holder to the secondary mirror, and the mirror can be adjusted to reflect the laser back to the point of origin, it cannot be reflected onto the primary and then back to the secondary*. Any adjustment of the secondary to reflect the laser back to the origin only ensures that the incident surface of the secondary mirror is perpendicular to the line-of-sight through the eyepiece holder. It &lt;strong&gt;&lt;em&gt;does not&lt;/em&gt;&lt;/strong&gt; ensure that the secondary mirror is aligned correctly with respect to line-of-sight through the eyepiece, and neither with respect to the primary mirror.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nTG9z.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nTG9z.png&quot; alt=&quot;Cassegrain&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/ridPU.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ridPU.png&quot; alt=&quot;Newtonian&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* Notwithstanding that the two mirrors could be misaligned so badly that this could occur at some particular point.&lt;/p&gt;&#xA;" OwnerUserId="18643" LastEditorUserId="18643" LastEditDate="2018-06-19T08:52:52.800" LastActivityDate="2018-06-19T08:52:52.800" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26686" PostTypeId="2" ParentId="26682" CreationDate="2018-06-19T11:50:53.080" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If black holes swallow stuff, will a big black hole eventually swallow the universe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As far as I know, no. Because the universe is expanding, and we have good evidence that the expansion is &lt;a href=&quot;https://en.wikipedia.org/wiki/Accelerating_expansion_of_the_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;accelerating&lt;/a&gt;. Yes,  some people talk about the &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Crunch&quot; rel=&quot;nofollow noreferrer&quot;&gt;Big Crunch&lt;/a&gt;, but there's no evidence for it. I don't think there ever will be. I say that because it's space that's expanding, and a gravitational field isn't collapsing space. The universe didn't collapse when it was smaller and denser, so I see no way it can collapse in future. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The universe will eventually go dark, because there will not be enough matter to make stars, and all that will be left is a bunch of black holes and they can merge. So will the universe get swallowed by a big black hole someday?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As above. Even if star formation ceased and a whole lot of matter ended up in black holes, the universe would still be expanding. As far as we know. The expansion would have to reverse for the black holes to coalesce. Mind you, it might be a good thing if it did. As Stephen Hawking &lt;a href=&quot;https://www.google.co.uk/search?source=hp&amp;amp;ei=i-woW9bqNofWgAaCo4bQBw&amp;amp;q=hawking+bbc+universe+black+hole&amp;amp;oq=hawking+bbc+universe+black+hole&amp;amp;gs_l=psy-ab.3..33i160k1.1454.10013.0.10514.32.31.0.1.1.0.230.3513.7j19j2.28.0....0...1c.1.64.psy-ab..3.24.2921...0j0i131k1j0i22i30k1j33i21k1.0.tklGibOyRdI&quot; rel=&quot;nofollow noreferrer&quot;&gt;said&lt;/a&gt;, the universe is like a black hole in reverse. If the universe ended up as one big black hole, it would be ending up the way (we think) it started out. Then you'd have a &lt;a href=&quot;https://en.wikipedia.org/wiki/Cyclic_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;cyclical universe&lt;/a&gt;.   &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2018-06-19T11:50:53.080" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26687" PostTypeId="1" AcceptedAnswerId="26689" CreationDate="2018-06-19T12:56:36.530" Score="23" ViewCount="10372" Body="&lt;p&gt;Sorry for the absolute begginer question here, but is our sun a part of some globular cluster?&#xA;It is something related to Virgo supercluster?&lt;/p&gt;&#xA;" OwnerUserId="21949" LastEditorUserId="16763" LastEditDate="2018-06-19T14:47:49.887" LastActivityDate="2018-06-20T15:28:14.670" Title="Is our sun in a star cluster?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;globular-clusters&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26688" PostTypeId="2" ParentId="26687" CreationDate="2018-06-19T13:16:25.170" Score="15" Body="&lt;p&gt;It is very likely that most stars are born in clusters, ranging in size from 100 stars to a million or more (&lt;a href=&quot;https://arxiv.org/abs/astro-ph/0301540&quot; rel=&quot;noreferrer&quot;&gt;Lada &amp;amp; Lada 2003&lt;/a&gt;). It has been suggested (from indirect evidence) that our Sun &lt;em&gt;was&lt;/em&gt; born in a cluster of somewhere between 1000 and 10,000 siblings (&lt;a href=&quot;https://arxiv.org/abs/1001.5444&quot; rel=&quot;noreferrer&quot;&gt;Adams 2010&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, &lt;em&gt;most&lt;/em&gt; clusters do not survive beyond about 10 million years from their main star formation event. Various processes (gas expulsion, galactic tidal fields, dynamical interactions) pull these clusters apart and scatter their contents (i.e. the stars that were born in them) amongst the galactic field stars. The identity and location of our Sun's birth cluster and even the identity of its siblings can only be pieced together with a forensic examination of stellar kinematics and chemistry. That study has begun and is being greatly aided by massive spectroscopic surveys like Gaia-ESO, SDSS and LAMOST and with new astrometric/kinematic information from Gaia DR2. But at the moment we cannot be sure what cluster, if any, Sun was born in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Virgo supercluster refers to a cluster of &lt;em&gt;galaxies&lt;/em&gt; and has nothing to do with the birth of stars in our Galaxy or of the Sun.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-06-19T15:20:26.867" LastActivityDate="2018-06-19T15:20:26.867" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26689" PostTypeId="2" ParentId="26687" CreationDate="2018-06-19T13:17:07.133" Score="38" Body="&lt;p&gt;No the sun is not part of a cluster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several types of clusters that we see in the sky. The most familiar is the &quot;open cluster&quot;, like the Pleiades. These are a group of stars that formed together and have remained close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the stars drift apart they can become part of a &quot;moving group&quot;, a collection of stars that don't appear to be a cluster, but since they share the same age and direction of motion we can tell they used to be a cluster.  Many nearby stars are part of the Ursa Major moving group, but the sun is not one. It just happens to be in the same part of the Milky way. The sun was probably part of a cluster shortly after it formed (4.6 billion years ago) but that cluster has long ago broken up. We don't (yet) know of any other stars that seem to have come from the same cluster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Globular clusters, like M13 and Omega Centauri, are larger and have many more stars tightly packed together. They are all rather distant, and the brightest look like slightly fuzzy stars (in fact Omega Centauri was originally thought to be a star).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course a galaxy is a group of 100 billion stars. We don't normally think of galaxies as a star cluster, because they are so much bigger and the stars in them don't form at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there are clusters of galaxies, the Virgo cluster is a cluster of galaxies, and the local group of galaxies is on the edge of this cluster. But the Virgo cluster is not a star cluster.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-06-20T06:20:22.813" LastActivityDate="2018-06-20T06:20:22.813" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26690" PostTypeId="2" ParentId="14525" CreationDate="2018-06-19T13:25:29.073" Score="2" Body="&lt;p&gt;I stumbled across this page while searching for pointing accuracy information for other telescopes. I currently work at Jodrell Bank and I'm afraid most of the information presented about the Lovell Telescope, at least, is basically incorrect or at least misleading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The telescope definitely does not have pointing accuracy to within 10arcsec rms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first point here is that there's a difference between &lt;em&gt;absolute&lt;/em&gt; pointing accuracy and relative accuracy while a source is being tracked. As touched on by others, diurnal temperature variations along with wind have the largest effect on absolute accuracy and we regularly find telescopes are arcminutes away from where they should be, before observations are started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The technique is to add &quot;offsets&quot; into the pointing model by tracking bright calibrator source for a short time. This is repeated roughly hourly, depending on what frequency the telescope is operating at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the most popular answer claims 10arcsec rms, which might be true on short timescales, but definitely not in absolute terms without current offsets. Further, the pointing model itself -- the thing which tells the telescope how and where to track -- is only quite basic for the Lovell at the moment. This is in contrast to many other observatories and one of the top priorities for improvement at the moment. The pointing model can lead the telescope away from the source over long periods of time, which is why regularly resetting the pointing is so important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is correct in saying that the biggest issue in terms of moment-to-moment pointing accuracy is Servo control -- the system which turns computer requests for a particular direction into mechanical reality. Actually, the dish sagging under its own weight only really affects the beam shape and sensitivity, not the pointing. The Lovell telescope, and many others, use motors working against each other to achieve a consistent drive rate and this works remarkably well. The lurches and wobbles are of order 1/100 the beam width, which makes them basically negligible.&lt;/p&gt;&#xA;" OwnerUserId="23409" LastActivityDate="2018-06-19T13:25:29.073" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26691" PostTypeId="1" AcceptedAnswerId="26694" CreationDate="2018-06-19T15:33:21.343" Score="6" ViewCount="153" Body="&lt;p&gt;Last weekend, I had the pleasure of visiting an observatory where there was a series of lectures to cover up the fact that there was too much cloud cover to actually observe much at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of these lectures involved the life-cycles of stars, which mainly involves Hydrogen and Helium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out of curiosity, I raised the question&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How old is the matter that make us and where does it come from?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &quot;where&quot; is covered by &lt;a href=&quot;https://en.wikipedia.org/wiki/Nucleosynthesis&quot; rel=&quot;noreferrer&quot;&gt;Nucleosynthesis&lt;/a&gt;, but the &quot;how old&quot; was shrugged away with &quot;maybe 2-3 sun lifetimes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to more accurately estimate the age of the matter that made up our solar system's accretion disk?  Or is the likely source of the supernova/collisions involved unlikely to be known?&lt;/p&gt;&#xA;" OwnerDisplayName="user20278" LastActivityDate="2018-06-19T19:22:19.277" Title="Do we know how old the matter that makes us is?" Tags="&lt;earth&gt;&lt;matter&gt;&lt;nucleosynthesis&gt;&lt;age&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26692" PostTypeId="2" ParentId="14525" CreationDate="2018-06-19T16:24:20.337" Score="1" Body="&lt;p&gt;This is a complicated question with a number of good answers already posted.  It's complicated since there are (broadly) two kinds of radio telescopes -- single dishes and interferometers -- and (even more broadly) two kinds of observation -- imaging, and spectroscopy/photometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most important thing to remember is that to a good first approximation &quot;all&quot; you need to do is get the source in the telescope's beam and avoid any conflicting sources.  As @Hobbes noted, the beamwidth (in radians) is the wavelength/dish diameter.  (Specifically, it's 1.2*wavelength/diameter -- see &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_resolution&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;.) Even for big dishes this can be large: multiple degrees, though it gets down into the arc minute range for higher radio frequencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Individual dishes are steered by moving something: usually the dish, but in the case of one like Arecibo, by moving the feed horn.  There's no real need to steer more accurately than a fraction -- a quarter, say -- of the beam width. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even a big dish like Green Bank's isn't used much for imaging (its images would have a resolution of many arc minutes at best), but to measure the source's time variation at a variety of wavelengths.  And for this it just needs the source to be somewhere near the center of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for this kind of telescope, you need pointing accuracy ranging from several degrees down to maybe ten arc minutes.  Somewhere in the system, there's a gearbox and motors which drive the dish, and there are indicators which show position. Once this is calibrated -- probably by making accurate measurements of the dish's actual position as a function of the gearbox readings -- this accuracy of pointing can be done by driving the dish to the desired position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are doing interferometry, things are much more complicated!  The VLA works in centimeter wavelengths with 25-meter dishes, so the beam width of an individual dish is about a minute of arc.  Aiming an individual dish is done in the same way as any other individual dish, though the mechanisms are typically more precise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the resolution of an array like the VLA is &lt;em&gt;much&lt;/em&gt; better than the resolution of the individual dishes. (The &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_interferometer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; on the subject is mediocre, but tracing down some of its references will help a lot.)  Basically, a single very large dish is constructed mathematically by combining together the signals received at each individual dish (both intensity and time) and the &lt;em&gt;precise&lt;/em&gt; location of each dish (but not so much where each was pointed!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ultra-precise pointing that's possible from an interferometer comes from the mathematical processing of the data, not from the mechanics of the pointing of the individual dishes.  (In fact, there are a number of useful interferometers -- &lt;a href=&quot;https://en.wikipedia.org/wiki/LOFAR&quot; rel=&quot;nofollow noreferrer&quot;&gt;LOFAR&lt;/a&gt; in the Netherlands is a good example -- which don't point at all,, but basically are comprised of omnidirectional antennas.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line: Individual dishes point using mechanics;  Interferometric arrays point using mathematics.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-06-19T16:24:20.337" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26693" PostTypeId="2" ParentId="26682" CreationDate="2018-06-19T18:46:35.307" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;all that will be left is a bunch of black holes and they can merge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is not clear actually. Keep in mind, black holes do evaporate via Hawking radiation. It just takes a very long time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the ultimate fate of the universe might be a uniform soup of cold photons instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is all highly speculative.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2018-06-19T18:46:35.307" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26694" PostTypeId="2" ParentId="26691" CreationDate="2018-06-19T19:07:34.687" Score="9" Body="&lt;p&gt;The material (heavier than helium) that makes up the solar system was made in millions, if not hundreds of millions of stars that lived and died  in the ~7 billion years between the formation of the Galaxy and the birth of the Sun. This material has been thoroughly mixed in the interstellar medium and so the heavier elements arise from countless individual stars of different ages. You can say something probabilistically, based on what we understand about the build up of chemical enrichment in our Galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with hydrogen - the vast majority of hydrogen is primordial. That is, it hasn't been inside another star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The carbon in the solar system is made in a variety of stars. Some will come from high mass stars and dispersed either when they exploded as supernovae or via winds from their surfaces during the Wolf-Rayet phase; some will have been manufactured in lower mass asymptotic giant branch stars and expelled in their winds. The distinction is important - massive stars enrich the Galaxy very promptly (see below, when discussing oxygen), near the beginning of the life of the Galaxy, long before the Sun was born. The enrichment rate by supernovae and massive stars tailed off considerably at later times. On the other hand, enrichment by lower mass stars takes longer because of their longer lifetimes. Thus the carbon in the solar system will have a spread of ages from about 13 billion (the oldest stars in the Galaxy), right up to the age of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elements like oxygen, phosphorus, magnesium, sulphur and silicon are primarily made and dispersed by massive stars and supernovae. These are likely to be at the older end of the range discussed above. On the other hand, nitrogen is also made in a mixture of stars, both high and low masses, with a consequent range of ages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron and nickel are mostly dispersed into the interstellar medium by type Ia supernovae. These are exploding white dwarfs, which in turn are the remnants of comparatively long-lived stars of a couple of solar masses or so. Most likely, these elements were made not long before the Sun was born. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;https://astronomy.stackexchange.com/questions/22694/parent-stars-of-our-sun-where-are-its-remains&quot;&gt;Parent stars of our Sun - Where are its remains?&lt;/a&gt; and &lt;a href=&quot;https://astronomy.stackexchange.com/questions/16311/how-can-there-be-1-000-stellar-ancestors-before-our-sun/16313#16313&quot;&gt;How can there be 1,000 stellar ancestors before our Sun?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more technical description of this differential enrichment history can be found in &lt;a href=&quot;https://arxiv.org/abs/0811.2258&quot; rel=&quot;noreferrer&quot;&gt;Pippino &amp;amp; Matteucci (2008)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-06-19T19:22:19.277" LastActivityDate="2018-06-19T19:22:19.277" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26695" PostTypeId="1" CreationDate="2018-06-19T20:25:07.037" Score="1" ViewCount="242" Body="&lt;p&gt;From the simple perspective of someone like myself, it appears that our sun is &quot;fixed in place&quot; (from the perspective of the solar system itself) and that everything else of lower mass orbits around it.  I suspect that the sun wobbles due to the other objects (that the barycenter is either near it or within its radius).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears from reading about binary stars of more-or-less equal mass that they cannot have one &quot;fixed in place&quot; and the other orbiting it.  The barycenter is basically between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that perspective correct?  Is it ever possible for a low mass star to be the &quot;fixed in place&quot; center with a large mass star orbiting it?  Or does the barycenter always shift toward the more massive object?&lt;/p&gt;&#xA;" OwnerUserId="18065" LastActivityDate="2018-06-19T22:19:46.573" Title="Do all orbiting objects have barycenters?" Tags="&lt;orbital-mechanics&gt;&lt;star-systems&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26696" PostTypeId="2" ParentId="26695" CreationDate="2018-06-19T20:35:01.673" Score="4" Body="&lt;p&gt;The barycenter of a system consisting of two objects of mass m1 and m2 with their centers separated by a distance d is always along the line connecting the two centers and m1/(m1+m2) of the way along the line from the center of m1 to the center of m2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for the Sun and the Earth, where the Sun weighs in at about 333,000 Earth masses, the barycenter will be about 1/333001 of the way from the center of the Sun to the center of the Earth, or about 280 miles from the center of the Sun in the direction of Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(So it might as well be fixed in place for all you or I could tell without high-quality instruments, but it does wobble in a 270-mile orbit once each year.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The presence of the rest of the planets makes this all a lot more complicated, of course and the Sun actually traverses a much more complicated path than it would if the Solar System was a simple two-body system. In this case, the barycenter is possibly better thought of as the center of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OTOH, since Jupiter is so much heavier than everything else, to a decent first approximation the Solar System &lt;em&gt;is&lt;/em&gt; a binary system. The barycenter for the Sun-Jupiter pair is actually &lt;em&gt;outside&lt;/em&gt; the Sun, at about 1.09 Solar radii.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See my answer at &lt;a href=&quot;https://astronomy.stackexchange.com/questions/18169/what-point-does-earth-actually-orbit&quot;&gt;What point does Earth actually orbit?&lt;/a&gt; for more.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2018-06-19T22:19:46.573" LastActivityDate="2018-06-19T22:19:46.573" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26697" PostTypeId="2" ParentId="26695" CreationDate="2018-06-19T20:39:13.817" Score="2" Body="&lt;p&gt;&quot;Barycenter&quot; is really just a fancy term for the center of mass of a system. You can compute it quite easily if you know the positions of all the objects in a system at a given time. This is usually quite handy, at least when dealing with either a pair of similar-mass objects, such as a binary star, or in systems where one object is gravitationally dominant and most other objects orbit it, like in a planetary system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A barycenter isn't really &quot;fixed&quot;, because no reference frame is absolute; we know this from the principle of relativity, which discards the idea of privileged inertial reference frames. However, when discussing orbital mechanics, it's often quite convenient to use &lt;em&gt;barycentric coordinates&lt;/em&gt;. In a binary star system, for instance, the barycenter is a focus of the orbits of both bodies, and so it is an obvious choice for designating as a reference point, as we can then calculate the stars' positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, things get complicated if we look at more complex systems. For instance, &lt;a href=&quot;https://astronomy.stackexchange.com/q/18169/2153&quot;&gt;Earth doesn't really orbit the Solar System's barycenter&lt;/a&gt;; it orbits the Sun-Earth barycenter - which is, for &lt;em&gt;most&lt;/em&gt; purposes, the center of the Sun. Finding the barycenter of the Solar System isn't too important if you're computing Earth's orbit approximately (for more exact calculations, you might want to take other bodies into account, namely, the giant planets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the equation for the center of mass of two bodies of masses $m_1$ and $m_2$ and positions $\mathbf{x}_1$ and $\mathbf{x}_2$ is&#xA;$$\mathbf{x}_{\text{cm}}=\frac{m_1\mathbf{x}_1+m_2\mathbf{x}_2}{m_1+m_2}$$&#xA;and if $m_1\ll m_2$,&#xA;$$\mathbf{x}_{\text{cm}}=\frac{m_1\mathbf{x}_1}{m_1+m_2}+\frac{m_2\mathbf{x}_2}{m_1+m_2}\approx\frac{m_1}{m_2}\mathbf{x}_1+\frac{m_2}{m_2}\mathbf{x}_2\approx\mathbf{x}_2$$&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2018-06-19T20:39:13.817" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26698" PostTypeId="2" ParentId="18169" CreationDate="2018-06-19T21:21:26.250" Score="4" Body="&lt;p&gt;@Incnis Mrsi's answer is correct: The Earth orbits around the center of mass of the Solar System, which is dominated by the Sun with a small correction due to the Jupiter and smaller still by the other giant planets.&lt;/p&gt;&#xA;&lt;p&gt;See &lt;a href=&quot;https://spaceplace.nasa.gov/barycenter/en/&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA's Space Place&lt;/a&gt; for a nice short, simple explanation.&#xA;See &lt;a href=&quot;https://zidbits.com/2011/09/the-earth-doesnt-actually-orbit-the-sun/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zidbits&lt;/a&gt; for a description with &lt;a href=&quot;https://zidbits.com/wp-content/uploads/2011/09/solar-system-barycenter.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;a very nice graphic&lt;/a&gt;&#xA;WMU has a &lt;a href=&quot;http://homepages.wmich.edu/%7Ekorista/solarsystem_barycenter.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;nice article&lt;/a&gt; showing the effects of the different planets&lt;/p&gt;&#xA;&lt;p&gt;You can easily see that this must be true in several ways.&lt;/p&gt;&#xA;&lt;p&gt;First of all, let's consider just the Sun-Earth-Jupiter triplet. If each planet orbited around its barycenter with the Sun as if the other planets didn't exist, you'd have the sun moving in a small circle with the Earth and a much larger and 12x slower circle with Jupiter &lt;em&gt;at the same time&lt;/em&gt; without one affecting the other.&lt;/p&gt;&#xA;&lt;p&gt;Because the Earth's mass is so small, to a pretty good approximation, you can look at the Earth as orbiting once a year around a Sun which is itself orbiting around the Sun-Jupiter barycenter in twelve years.&lt;/p&gt;&#xA;&lt;p&gt;A second experimental point in support of this comes from extra solar planet discovery through Doppler measurements. When a star has multiple planets big enough to cause stellar motions big enough to be detected by our instruments, we see the star's light-of-sight velocity changing in a way consistent with it rotating around the center of mass of the star and several planets. Otherwise, we could not detect the other planets.&lt;/p&gt;&#xA;&lt;p&gt;A third way is a gedanken experiment. Consider Tatooine, and early-like planet orbiting a double star. Does Tatooine orbit just one of them, ignoring the other?  How could that be?  (How does it pick which one to ignore?) It &lt;em&gt;must&lt;/em&gt; orbit their common center of mass.&lt;/p&gt;&#xA;&lt;p&gt;Bottom line: The Sun and all the planets orbit around the Solar System's center of mass.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2020-10-27T01:04:34.010" LastActivityDate="2020-10-27T01:04:34.010" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26699" PostTypeId="1" AcceptedAnswerId="26742" CreationDate="2018-06-20T00:39:32.417" Score="2" ViewCount="82" Body="&lt;p&gt;I am interested in the composition of asteroids from the perspective of determining potential targets for mining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect there already exists some data that can be used to identify reasonable targets for mining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific mining method I am looking to evaluate is to send a &quot;collector&quot; out and &quot;grab&quot; a target, bringing it back into LEO for mineral extraction.  Because of this I would like to identify targets that are &amp;lt; 20m diameter. The ideal material to mine/extract would be a precious metal of some kind. This would give the most &quot;bang for the buck&quot;. (The buck in this case being the effort rather than Delta V).&lt;br /&gt;&#xA;Considering the ideal material, I suspect a list of &quot;M&quot; type asteroids and their properties would be the data I am after.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw a question that asked about &lt;a href=&quot;https://astronomy.stackexchange.com/questions/25953/how-is-ir-spectroscopy-used-to-determine-the-composition-of-asteroids&quot;&gt;using IR Spectroscopy&lt;/a&gt; to determine the surface composition of asteroids.  That gave some good insight - but no indication of where such data currently exists. A table of this data would be a reasonable starting point.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also wondering if we can determine anything about the relative density of an asteroid by the gravity it exerts on nearby objects.  If that data is available anywhere that would also be an interesting place for me to look.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also interested in trying to determine the orbits of any target asteroids, so where would I look for data on the orbits of those that are selected?&lt;/p&gt;&#xA;" OwnerUserId="23414" LastEditorDisplayName="user1569" LastEditDate="2018-06-21T11:00:07.093" LastActivityDate="2018-06-24T04:07:35.093" Title="Data on the composition and orbits of asteroids" Tags="&lt;asteroids&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26700" PostTypeId="2" ParentId="669" CreationDate="2018-06-20T04:38:28.713" Score="3" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;What is in the center of the universe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This question over at Physics.SE: &amp;quot;&lt;a href=&quot;https://physics.stackexchange.com/a/136861/170832&quot;&gt;Did the Big Bang happen at a point?&lt;/a&gt;&amp;quot;, which has an answer with over 300 UpVotes, explains:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;The simple answer is that no, the Big Bang did not happen at a point. Instead it happened everywhere in the universe at the same time. Consequences of this include:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The universe doesn't have a centre: the Big Bang didn't happen at a point so there is no central point in the universe that it is expanding from.&amp;quot;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The universe isn't expanding into anything: because the universe isn't expanding like a ball of fire, there is no space outside the universe that it is expanding into.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;We are less than a spec in our &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Laniakea_Supercluster&quot; rel=&quot;nofollow noreferrer&quot;&gt;supercluster&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mCz0I.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mCz0I.jpg&quot; alt=&quot;Laniakea&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;There is a Wikipedia webpage: &amp;quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/History_of_the_center_of_the_Universe#The_nonexistence_of_a_center_of_the_Universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;History of the Center of the Universe - The nonexistence of a center of the Universe&lt;/a&gt;&amp;quot; which explains:&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;A homogeneous, isotropic universe does not have a center.&amp;quot; - Source: Livio, Mario (2001). &lt;a href=&quot;https://books.google.com/books?id=4EidS6_VVNYC&amp;amp;pg=PA53&amp;amp;dq=cosmological%20principle%20%22center%20of%20the%20universe%22&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=Eyt3T5HFGuzy2gXsm_C1Bg&amp;amp;ved=0CFQQ6AEwBjgU&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Accelerating Universe: Infinite Expansion, the Cosmological Constant, and the Beauty of the Cosmos&lt;/a&gt;. John Wiley and Sons. p. 53. Retrieved 31 March 2012.&lt;/p&gt;&#xA;&lt;p&gt;See also this CalTech video: &amp;quot;&lt;a href=&quot;http://www.spitzer.caltech.edu/video-audio/664-ask2009-001-Where-Is-the-Center-of-the-Universe-&quot; rel=&quot;nofollow noreferrer&quot;&gt;Where is the center of the universe?&lt;/a&gt;&amp;quot;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If the universe has formed &amp;amp; originated by a Big Bang Explosion, then there must be empty space left in the center of the explosion site, as all the matter is travelling at tremendous speeds away from the center, and there must be more matter, stars, galaxies and dust, etc near the present periphery or circumference or horizon of the present universe. As that big explosion has taken place about 13.7 billion years back, then the outer boundaries of our universe are 13.7 billion light years away from the centre of the explosion of Big Bang.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NDdAE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NDdAE.jpg&quot; alt=&quot;Before, during, and after - The Big Bang&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Have our astronomers discovered hollowness or emptiness anywhere in the centre of the universe or not?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Zooming in to the Milky Way (center of this image, but not the center of the universe) we see:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UQwZj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UQwZj.jpg&quot; alt=&quot;Milky Way&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The &lt;strong&gt;blue areas near us are the &lt;a href=&quot;https://en.wikipedia.org/wiki/Local_Void&quot; rel=&quot;nofollow noreferrer&quot;&gt;local void&lt;/a&gt;&lt;/strong&gt;, while the area to the left is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Great_Attractor&quot; rel=&quot;nofollow noreferrer&quot;&gt;great attractor&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The shape of the universe, that we can detect/see, is complicated - it not a simple sphere or football shaped, radiating from a central point. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Planck_(spacecraft)#2015_data_release&quot; rel=&quot;nofollow noreferrer&quot;&gt;current measurement&lt;/a&gt; of the age of the universe is 13.799±0.021 billion (&lt;span class=&quot;math-container&quot;&gt;$10^9$&lt;/span&gt;) years within the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lambda-CDM_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lambda-CDM concordance model&lt;/a&gt;. We can only see and measure so far, and during the past nearly 14 billion year parts of the universe have grown denser and parts have spread apart.&lt;/p&gt;&#xA;&lt;p&gt;See these Wikipedia webpages: &amp;quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Observable_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;Observable universe&lt;/a&gt;&amp;quot; and &amp;quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Observational_cosmology&quot; rel=&quot;nofollow noreferrer&quot;&gt;Observational cosmology&lt;/a&gt;&amp;quot;, this is from &amp;quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Universe#Size_and_regions&quot; rel=&quot;nofollow noreferrer&quot;&gt;Size and regions&lt;/a&gt;&amp;quot;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The size of the Universe is somewhat difficult to define. According to the general theory of relativity, some regions of space may never interact with ours even in the lifetime of the Universe due to the finite speed of light and the ongoing expansion of space. For example, radio messages sent from Earth may never reach some regions of space, even if the Universe were to exist forever: space may expand faster than light can traverse it.&lt;/p&gt;&#xA;&lt;p&gt;Distant regions of space are assumed to exist and to be part of reality as much as we are, even though we can never interact with them. The spatial region that we can affect and be affected by is the observable universe.&lt;/p&gt;&#xA;&lt;p&gt;The observable universe depends on the location of the observer. By traveling, an observer can come into contact with a greater region of spacetime than an observer who remains still. Nevertheless, even the most rapid traveler will not be able to interact with all of space. Typically, the observable universe is taken to mean the portion of the Universe that is observable from our vantage point in the Milky Way.&lt;/p&gt;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Comoving_distance&quot; rel=&quot;nofollow noreferrer&quot;&gt;proper distance&lt;/a&gt;—the distance as would be measured at a specific time, including the present—between Earth and the edge of the observable universe is 46 billion light-years (14 billion &lt;a href=&quot;https://en.wikipedia.org/wiki/Parsec&quot; rel=&quot;nofollow noreferrer&quot;&gt;parsecs&lt;/a&gt;), making the diameter of the observable universe about 91 billion light-years (&lt;span class=&quot;math-container&quot;&gt;$28×10^9$&lt;/span&gt; pc). The distance the light from the edge of the observable universe has travelled is very close to the age of the Universe times the speed of light, 13.8 billion light-years (&lt;span class=&quot;math-container&quot;&gt;$4.2×10^9$&lt;/span&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Parsec&quot; rel=&quot;nofollow noreferrer&quot;&gt;parsecs&lt;/a&gt;), but this does not represent the distance at any given time because the edge of the observable universe and the Earth have since moved further apart. For comparison, the diameter of a typical galaxy is 30,000 light-years (9,198 &lt;a href=&quot;https://en.wikipedia.org/wiki/Parsec&quot; rel=&quot;nofollow noreferrer&quot;&gt;parsecs&lt;/a&gt;), and the typical distance between two neighboring galaxies is 3 million light-years (919.8 &lt;a href=&quot;https://en.wikipedia.org/wiki/Parsec#Parsecs_and_kiloparsecs&quot; rel=&quot;nofollow noreferrer&quot;&gt;kiloparsecs&lt;/a&gt;). As an example, the Milky Way is roughly 100,000–180,000 light years in diameter, and the nearest sister galaxy to the Milky Way, the Andromeda Galaxy, is located roughly 2.5 million light years away.&lt;/p&gt;&#xA;&lt;p&gt;Because we cannot observe space beyond the edge of the observable universe, it is unknown whether the size of the Universe in its totality is finite or infinite.&lt;/p&gt;&#xA;&lt;p&gt;Estimates for the total size of the universe, if finite, reach as high as &lt;span class=&quot;math-container&quot;&gt;$10^{{10}^{{10}^{122}}}$&lt;/span&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Parsec#Megaparsecs_and_gigaparsecs&quot; rel=&quot;nofollow noreferrer&quot;&gt;megaparsecs&lt;/a&gt;, implied by one resolution of the No-Boundary Proposal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;According to the proposal &lt;a href=&quot;https://en.wikipedia.org/wiki/Hartle%E2%80%93Hawking_state&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hartle–Hawking state&lt;/a&gt;: &amp;quot;The universe has no initial boundaries in time nor space&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;Dr. &lt;a href=&quot;http://www.ifa.hawaii.edu/users/tully/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brent Tulley&lt;/a&gt; published an article: &amp;quot;&lt;a href=&quot;https://www.nature.com/articles/nature13674&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Laniakea supercluster of galaxies&lt;/a&gt;&amp;quot; (free &lt;a href=&quot;https://arxiv.org/abs/1409.0880&quot; rel=&quot;nofollow noreferrer&quot;&gt;arXiv preprint&lt;/a&gt;) and associated supplimentary &lt;a href=&quot;https://media.nature.com/original/nature-assets/nature/journal/v513/n7516/extref/nature13674-sv1.mp4&quot; rel=&quot;nofollow noreferrer&quot;&gt;video&lt;/a&gt;, along with &lt;a href=&quot;https://vimeo.com/pomarede&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dr. Daniel Pomarède's Vimeo directory&lt;/a&gt;, specifically this video: &lt;a href=&quot;https://vimeo.com/64868713&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cosmography of the Local Universe (FullHD version)&lt;/a&gt; from which these images were drawn, which shows the shape of part of the universe as we know it:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Take the WMAP data and project all galaxies within 8K km/s (1:18 on the video) onto a 3D space:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9nyST.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QsoCU.jpg&quot; alt=&quot;WMAP -&amp;gt; 3D&quot; /&gt;&lt;/a&gt;&#xA;Click image to animate&lt;/p&gt;&#xA;&lt;p&gt;A close up of our location shows the large &lt;a href=&quot;https://en.wikipedia.org/wiki/Local_Void&quot; rel=&quot;nofollow noreferrer&quot;&gt;local void&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fBOk4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fBOk4.jpg&quot; alt=&quot;Milky Way&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Zooming out reveals part of the universe, see the video linked above for more info:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FP0yZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FP0yZ.jpg&quot; alt=&quot;10K km/s&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-06-20T04:38:28.713" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26702" PostTypeId="1" AcceptedAnswerId="26703" CreationDate="2018-06-20T07:07:21.580" Score="3" ViewCount="145" Body="&lt;p&gt;I am setting up a 2.4 metre parabola pointing straight up to observe the 21cm hydrogen line at 1420MHz with meridian drift scans.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm planning to place the dish on a steel pole embedded in the ground and protruding 2 metres from the ground surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a tree some metres away that has a few branches well overhead of the antenna.  It's a Eucalyptus tree, so not heavy with green leaves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:  Should I be concerned about interference from a few overhanging branches (say about 15 metres above the dish)?  My understanding is that shorter lines in the cm range should be less affected by interference (e.g weather), but I'm not sure about trees.&lt;/p&gt;&#xA;" OwnerUserId="23153" LastEditorUserId="23153" LastEditDate="2018-06-20T08:41:24.240" LastActivityDate="2018-06-20T12:57:45.413" Title="Hydrogen Line observation - Interference from trees" Tags="&lt;radio-astronomy&gt;&lt;hydrogen&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26703" PostTypeId="2" ParentId="26702" CreationDate="2018-06-20T12:38:26.240" Score="3" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrogen_line&quot; rel=&quot;nofollow noreferrer&quot;&gt;21 cm line&lt;/a&gt; is from &lt;em&gt;atomic hydrogen&lt;/em&gt; which is going to be free, neutral atoms in vacuum, where the electron is bound only to the proton. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In water, or any other hydrogen containing molecule, the electron's orbit will be dramatically modified (or even &quot;stolen&quot;) by the atom to which it is attached, and so the transition will no longer exist as a narrow line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly there might be all kinds of RF interference from objects encroaching on your antenna's reception pattern, which may have side-lobes as well as the main beam pointing upwards, so like @Kozaky's &lt;a href=&quot;https://astronomy.stackexchange.com/questions/26702/hydrogen-line-observation-interference-from-trees#comment46270_26702&quot;&gt;comment&lt;/a&gt; suggests, you should to some experimentation if possible to understand contributions to background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your sensitivity will depend on the bandwidth you choose for your receiver, as well as the thermal and Doppler broadening of of the source that passes by, so make sure you have an understanding of what the bandwidth should be to see a peak, and also if there are significant offsets in frequency due to a &lt;em&gt;Doppler shift&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G2GZe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G2GZe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; From &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Hydrogen-SpinFlip.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ground state hyperfine levels of hydrogen (parallel and antiparallel) with the spin-flip transition, emitting radiation at 1420 MHz. The corresponding wavelength is 21 cm. (21-cm line, hydrogen line)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-06-20T12:44:27.297" LastActivityDate="2018-06-20T12:44:27.297" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26704" PostTypeId="2" ParentId="26702" CreationDate="2018-06-20T12:57:45.413" Score="1" Body="&lt;p&gt;The GPS system uses frequencies either side of the 1420 MHz frequency of neutral hydrogen.  The GPS system works in woods and forests, so my guess is that you should be ok.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2018-06-20T12:57:45.413" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26706" PostTypeId="2" ParentId="26699" CreationDate="2018-06-20T17:44:10.053" Score="3" Body="&lt;p&gt;The &lt;a href=&quot;https://ssd.jpl.nasa.gov/sbdb_query.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL Small-Body Database Search Engine&lt;/a&gt; is a good place to start.&#xA;If diameter &amp;lt; 20 m, then probably &lt;a href=&quot;https://cneos.jpl.nasa.gov/tools/ast_size_est.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;absolute magnitude&lt;/a&gt; H &gt; 25.&#xA;Fortunately we know of over 4000 asteroids in that size range, mostly after close approaches to Earth.&#xA;Unfortunately the orbit and magnitude are all we know about most of those, due to limited observational data.&#xA;JPL SBDB lists only about 40 asteroids with known Tholen spectral type M, all with perihelion q &gt; 1.5 AU and diameter &gt; 30 km.&#xA;The picture is a little brighter for type S, with over 300 known, a few with diameter &amp;lt; 2 km and &lt;a href=&quot;https://en.wikipedia.org/wiki/Minimum_orbit_intersection_distance&quot; rel=&quot;nofollow noreferrer&quot;&gt;minimum orbit intersection distance&lt;/a&gt; &amp;lt; 50 LD.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-06-20T17:44:10.053" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26707" PostTypeId="1" CreationDate="2018-06-21T06:32:06.113" Score="8" ViewCount="1017" Body="&lt;p&gt;The Phys.org article &lt;a href=&quot;https://phys.org/news/2018-06-universe-ordinary.html&quot; rel=&quot;noreferrer&quot;&gt;Researchers find last of universe's missing ordinary matter&lt;/a&gt; says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Ordinary matter, or &amp;quot;baryons,&amp;quot; make up all physical objects in existence, from stars to the cores of black holes. But until now, astrophysicists had only been able to locate about two-thirds of the matter that theorists predict was created by the Big Bang.&lt;/p&gt;&#xA;&lt;p&gt;In the new research, an international team pinned down the missing third, finding it in the space between galaxies. That lost matter exists as filaments of oxygen gas at temperatures of around 1 million degrees Celsius, said CU Boulder's Michael Shull, a co-author of the study.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;and refers to F. Nicastro et al, Observations of the missing baryons in the warm–hot intergalactic medium, Nature (2018). &lt;a href=&quot;http://dx.doi.org/10.1038/s41586-018-0204-1&quot; rel=&quot;noreferrer&quot;&gt;DOI: 10.1038/s41586-018-0204-1&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The Wikipedia page &lt;a href=&quot;https://en.wikipedia.org/wiki/Warm%E2%80%93hot_intergalactic_medium&quot; rel=&quot;noreferrer&quot;&gt;Warm–hot intergalactic medium&lt;/a&gt; says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Part of the gravitational energy supplied by these effects is converted into thermal emissions of the matter by collisionless shock heating&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;but the link to &amp;quot;collisionless&amp;quot; is ambiguous and I haven't a clue how a shock wave can get hot without collisions.&lt;/p&gt;&#xA;&lt;p&gt;Is there a way to understand why is the Warm-Hot Intergalactic Medium (WHIM) is so hot, and what the &amp;quot;collisionless shock heating&amp;quot; process is that appears to be heating it?&lt;/p&gt;&#xA;&lt;p&gt;Are the temperatures associated with the motion of the atoms and with the excited states of the atoms similar? With only of order 1 atom per cubic meter, collisions must be so infrequent that it seems to me that it might not be so hard for these two temperatures to diverge.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-06-21T12:28:00.773" Title="Why is the Warm-Hot Intergalactic Medium (WHIM) so hot, and what is &quot;collisionless shock heating&quot;?" Tags="&lt;temperature&gt;&lt;interstellar-medium&gt;&lt;intergalactic-space&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26708" PostTypeId="2" ParentId="26707" CreationDate="2018-06-21T09:22:16.593" Score="5" Body="&lt;p&gt;The answer is almost certainly magnetic fields. A collisionless shock occurs when you try to propagate an increase/density in pressure through a sparse plasma at faster than the sound speed. Whilst the ions and electrons don't collide (very frequently), the magnetic fields that thread the plasma do accelerate the charged particles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a collisionless shock the magnetic fields will become compressed and turbulent leading to the dissipation of energy. The energy is transferred into the motion of particles in the plasma (and radiation). This can be characterised with a temperature, but the plasma can have a significantly non-thermal component and indeed collisionless shock heating is one way to produce very energetic particles. I would guess that when a &quot;temperature&quot; is discussed with respect to the intergalactic medium it is a rough characterisation - indeed the review by &lt;a href=&quot;http://ttps://arxiv.org/abs/0801.1008&quot; rel=&quot;noreferrer&quot;&gt;Bykov et al. (2008)&lt;/a&gt; provides more detail, referring to different temperatures for the ions and electrons, anisotropic non-Maxwellian velocity distributions etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course some sort of thermal relaxation moderated by collisions between ions and electrons does then take place post-shock. How long it takes to reach a  thermal equilibrium depends on the density and &quot;temperature&quot; of the plasma. At very low densities and high temperatures they may not do so in a Hubble time. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-06-21T09:22:16.593" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26709" PostTypeId="1" AcceptedAnswerId="26712" CreationDate="2018-06-21T12:05:30.120" Score="1" ViewCount="220" Body="&lt;p&gt;I always thought the only fuel for a star was hydrogen, which is fused into helium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But while reading some &lt;a href=&quot;https://astronomy.stackexchange.com/questions/8460/can-our-sun-become-a-black-hole?rq=1&quot;&gt;questions and answers here in ASO,&lt;/a&gt; I saw phrases like &quot;This balance stays relatively stable until the star runs out of whatever its current fuel is&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides hydrogen, what are the possible fuels in a star’s core?&lt;/p&gt;&#xA;" OwnerUserId="21949" LastEditorUserId="23402" LastEditDate="2018-06-22T13:17:14.347" LastActivityDate="2018-07-04T17:38:00.060" Title="What are the possible star fuels?" Tags="&lt;star&gt;&lt;core&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26710" PostTypeId="2" ParentId="24830" CreationDate="2018-06-21T12:18:58.773" Score="2" Body="&lt;p&gt;That's very, very difficult to answer with a lot of if's. If the earth had no atmosphere it would be 255K-277K (-18-3°C) on the day side and a lot colder on the nightside. But if we include the atmosphere into our calculations we can get widely differing results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sake of simplicity let's assume that earth tidal locks and nothing else happens. No change of atmosphere because many plants die or human population and therefore CO2 emmissions go down or something like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case I guess that the temperature would stay about the same. While one side of the earth enjoys permanent sunlight, the air on this side heats up and therefore rise to higher layers of the atmosphere. Cooler air from the other side of the earth streams in and the hot air in exchange streams to the cooler side where it cools down and returns to the bright side. So while one side of the earth is permanently sunny it is also permanently cooled down by air from the other side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dark side would not cool out because it is permanently heated by the hot air from the sunny side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the whole we would live on a very windy planet. But thats it.&lt;/p&gt;&#xA;" OwnerUserId="22927" LastActivityDate="2018-06-21T12:18:58.773" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26711" PostTypeId="2" ParentId="26707" CreationDate="2018-06-21T12:19:47.150" Score="6" Body="&lt;p&gt;It should probably be added that the article includes a glaring error of the type you often see when the science writer apparently did not take an elementary astronomy class (this is why we have such classes!).  When the article states that the &quot;lost matter exists as filaments of oxygen gas&quot;, you can be sure that Michael Shull never said any such thing, because that language clearly asserts that the &quot;lost&quot; matter is oxygen.  Of course what is meant is that the previously undetected baryons exist primarily as free protons (and electrons), which is merely &lt;em&gt;evidenced by&lt;/em&gt; the presence of detectable oxygen that is simply assumed to be in the normal ratio we find between oxygen and hydrogen.  We could also quibble that calling it &quot;lost&quot; is hyperbole, a bit like saying the brain matter of the writer was heretofore &quot;lost&quot; because it couldn't be seen by looking into his/her ears, and was only recently &quot;found&quot; in some recent MRI.   Astronomers have long expected the WHIM to be there, and knew it would be hard to see, so it was hardly &quot;lost.&quot;  But it is certainly nice to have greater certainty that it is there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And not only is there a crucial assumption of the oxygen-to-hydrogen ratio, the OP is astute to ask how well we can even know the oxygen content itself.  All we see is the light from the oxygen, so we have to model how emissive it is to know how much oxygen is there.  That's not always easy at low densities where the emissivity can differ from what you'd get in local thermodynamic equilibrium, as was pointed out in the question and the first answer also.  So not all the uncertainties have been resolved, but it's certainly a nice step forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One final point regards why the gas is so hot.  This is related to the fact that low density gas cools itself very inefficiently, because cooling generally requires collisions between the particles (which generates the light).  Heating, on the other hand, can occur in various ways that might involve interactions with something other than the particles, say with cosmic rays or with magnetic fields.  Those latter elements can be present even if the density is very low, leading to high temperatures.  There is also a radiative cooling instability above about 10,000 K, where gas that is heated above that starts to cool less efficiently (as hydrogen gets all ionized), and responds by being compressed by the ambient pressure, which drives the temperature up even more.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2018-06-21T12:28:00.773" LastActivityDate="2018-06-21T12:28:00.773" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26712" PostTypeId="2" ParentId="26709" CreationDate="2018-06-21T12:52:17.903" Score="6" Body="&lt;p&gt;As a star begins to age and exhaust its hydrogen supply, pressure will build in its interior until temperatures rise enough to begin fusing helium into carbon. If the star is massive enough, this process will continue, using heavier elements each time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Referring to this link on &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_nucleosynthesis#Key_reactions&quot; rel=&quot;nofollow noreferrer&quot;&gt;stellar fusion&lt;/a&gt;, depending on their mass, stars use a variety of elements from hydrogen to silicon as 'fuel'. Stellar cores of stars towards the end of their giant stages will have cores of iron, but iron doesn't fuse into anything heavier in a stellar core.&lt;/p&gt;&#xA;" OwnerDisplayName="user10106" LastEditorDisplayName="user10106" LastEditDate="2018-06-21T14:56:56.623" LastActivityDate="2018-06-21T14:56:56.623" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26713" PostTypeId="1" AcceptedAnswerId="26746" CreationDate="2018-06-21T16:18:43.473" Score="3" ViewCount="247" Body="&lt;p&gt;If you were place a strong permanent magnet into the solar wind, like a 1 inch cube neodymium magnet, would the charged particles of the solar wind colliding with the magnet's magnetic field push the magnet out of our solar system, that being if it does not encounter any planets or objects in its path? &lt;/p&gt;&#xA;" OwnerDisplayName="user23327" LastActivityDate="2018-06-24T10:10:00.260" Title="Will the Sun's solar wind push a permanent magnet out of our solar system?" Tags="&lt;the-sun&gt;&lt;astrophysics&gt;&lt;magnetic-field&gt;&lt;solar-wind&gt;&lt;plasma-physics&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26714" PostTypeId="1" CreationDate="2018-06-21T20:27:59.060" Score="2" ViewCount="78" Body="&lt;p&gt;I am dealing with a problem which requires calculating either the correlation function or the power spectrum. At times, the 2-point-correlator seems the better route, at others, the PS. Right now I am having some questions about how to practically calculate a PS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the basic setup: I have a mock galaxy catalog. Let's say I want the PS. I could Fourier transform the 2PC, but let's say I want to avoid working with a 2PC code. So instead, I make the assumption that the galaxies trace the density field (it helps that I have the DM halo masses from the underlying simulation which produced the mock).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I bin the galaxy catalog, giving me a coarse-grained list of masses for each bin, $\{m_{ijk}\}$. It is trivial to turn this into a density contrast list $\{\delta_{ijk}\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a couple of options now to proceed to calculating the PS. Formally, the PS is given by&#xA;$$&#xA;P(k) \sim \langle \, |\tilde \delta(\vec{k})|^2 \,\rangle&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;up to some normalization, with $\tilde \delta$ being the fourier transform of $\delta(\vec{x})$, the full density contrast &lt;em&gt;field&lt;/em&gt;, and the angle brackets denoting volume averaging. The seemingly natural thing to do would then be&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;FFT my binned list $\{\delta_{ijk}\}$&lt;/li&gt;&#xA;&lt;li&gt;get list of values $\{ P_{ijk} \}$&lt;/li&gt;&#xA;&lt;li&gt;average this list to get the PS $\{ P_k \}$, now just a function of $k$ instead of $k$-space position.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is potentially problematic for me, as unless I bin over large scales, the memory required to store the zeros present in the array of binned values is too large. With the data I have, binning at $&#xA;\sim 10$ Mpc$/h$ spaces or greater is needed to get a managable array.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My solution for calculating the 2PC was to avoid calculating an array entirely, and instead performing a regrouping operation to bin the catalog. This keeps the memory usage proportional to the catalog length instead of to the binning resolution cubed. However, it effectively means I am dealing with a sparse density array, which AFAIK cannot be used directly in most existing FFT codes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves me with a dilemma. I can either coarse-grain my density array $\{ \delta_{ijk} \}$ even more, losing information at $\sim 1$ Mpc scales, or else I suppose I could interpolate the density, giving me a full function $\delta(\vec{x})$ that I could pass some coarser mesh into, and then pass into an FFT. The problem with interpolating is that this operation requires the (now implicit) knowledge of the empty cells in the sparse array.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, one &lt;a href=&quot;https://stackoverflow.com/questions/37127765/ifft-and-fft-for-sparse-vectors&quot;&gt;SO post&lt;/a&gt; has a comment suggesting to merely calculate the Fourier integral instead of using FFT methods. I fear this would still present a problem, as this takes the time complexity from $O(n\log(n))$ to $O(n^2)$, and $n$ for me is $\sim 10^6$ or higher (becoming as high as $10^8$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I missing something? I feel like people calculate PS and 2PC's all the time for catalogs this large, and like my attempts at solving this issue could be flawed. My precise question is, are my options just as I have laid out, or is there another way of handling this calculation I have not described?&lt;/p&gt;&#xA;" OwnerUserId="23373" LastEditorUserId="23373" LastEditDate="2018-06-26T16:00:03.233" LastActivityDate="2018-06-26T16:00:03.233" Title="Efficiently calculating power spectra from galaxy catalogs" Tags="&lt;galaxy&gt;&lt;cosmology&gt;&lt;data-analysis&gt;&lt;galaxy-cluster&gt;&lt;python&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26715" PostTypeId="2" ParentId="26713" CreationDate="2018-06-21T21:26:21.680" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If you were place a strong permanent magnet into the solar wind, like a 1 inch cube neodymium magnet, would the charged particles of the solar wind colliding with the magnet's magnetic field push the magnet out of our solar system ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Technically speaking, yes. Practically speaking, &lt;a href=&quot;https://image.gsfc.nasa.gov/poetry/ask/a11381.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;it would take an extremely long time&lt;/a&gt;. The solar wind consists of &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_wind#Pressure&quot; rel=&quot;nofollow noreferrer&quot;&gt;5 atoms per cubic centimeter per second&lt;/a&gt;. Solar sails are usually huge and &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Solar_sail#Solar_radiation_pressure&quot; rel=&quot;nofollow noreferrer&quot;&gt;work by reflection&lt;/a&gt;, simply pushing on a tiny object is inefficient, being magnetic helps very little.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastActivityDate="2018-06-21T21:26:21.680" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26716" PostTypeId="1" CreationDate="2018-06-21T22:57:14.053" Score="2" ViewCount="106" Body="&lt;p&gt;I'm assuming that when retrieving a meteor for study and later display, there are precautions taken against possible radiation coming from it. Is my assumption correct and are there other precautions taken such as protection from biohazards? (e.g. is it not possible for a meteor to carry a foreign viral agent?) &lt;/p&gt;&#xA;" OwnerUserId="23441" LastEditorUserId="2365" LastEditDate="2018-06-22T05:59:34.467" LastActivityDate="2018-06-22T05:59:34.467" Title="Precautions when retrieving material originated from space" Tags="&lt;earth&gt;&lt;meteor&gt;&lt;meteorite&gt;&lt;impact&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26717" PostTypeId="2" ParentId="26716" CreationDate="2018-06-21T23:16:01.287" Score="3" Body="&lt;p&gt;Meteorites are composed of rock and iron. They are not radioactive, at least no more so than any other stone.  A banana is probably more radioactive than a meteorite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You do not need protection for foreign viral agents either. Meteorites are stones that have been orbiting in space, usually for billions of years. They do not contain any living things. The have also passed through the atmosphere so fast that their surface can begin to melt from the friction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can pick up a meteorite with you unprotected hands&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;For pedants: A meteorid is &quot;space stone&quot;. Meteor is the flash of light in the sky as a meteoroid passes through the atmosphere. Meteorite is the stone from space that you find on the ground. It is the meteorite that you retrieve for study.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a suggestion that ancient meteorites bought the ingredients for life to Earth, or even seeded life on the planet, but it is speculative as we don't have the evidence to prove it.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-06-21T23:16:01.287" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26718" PostTypeId="2" ParentId="22673" CreationDate="2018-06-22T08:43:39.113" Score="1" Body="&lt;p&gt;Maybe too late but probably he is referring to only take into account pixels with an intensity greater than 3*sigma. If not, you will be counting random noise as continuum. Obviously you will lose a lot of pixels, but those pixels doesn't carry any information at all.&lt;/p&gt;&#xA;" OwnerUserId="23447" LastActivityDate="2018-06-22T08:43:39.113" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26719" PostTypeId="1" AcceptedAnswerId="27071" CreationDate="2018-06-22T09:39:59.623" Score="0" ViewCount="101" Body="&lt;p&gt;What are the axioms (if any) behind the LCDM model of cosmology?&#xA;NB: axioms, not postulates (e.g., inflation)&lt;/p&gt;&#xA;" OwnerDisplayName="user19333" LastEditorUserId="11143" LastEditDate="2018-06-22T15:42:06.057" LastActivityDate="2018-08-23T14:01:09.367" Title="Fundamental axioms in LCDM" Tags="&lt;observational-astronomy&gt;&lt;astrophysics&gt;&lt;cosmology&gt;&lt;big-bang-theory&gt;&lt;cosmological-inflation&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26720" PostTypeId="2" ParentId="26709" CreationDate="2018-06-22T10:12:12.533" Score="2" Body="&lt;p&gt;Stars can consume everything as fuel in the periodic table up to iron, which is endothermic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When they consume other elements different from Hydrogen, their radiation pressure gets higher and the star expands its volume, sometimes even it surpasses the gravitation force of the star and starts to lose material. &lt;/p&gt;&#xA;" OwnerUserId="23447" LastActivityDate="2018-06-22T10:12:12.533" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26721" PostTypeId="1" CreationDate="2018-06-22T10:16:39.437" Score="4" ViewCount="2713" Body="&lt;p&gt;Maybe its a dumb question, but to convert Jy/beam to Jy, I just have to multiply it by the beam size in sr right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being $\Omega$ the beam size: &#xA;$\Omega = \frac{\pi \theta_{maj} ~~ \theta_{min}}{4 \ln{2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jy/beam $\cdot \ \Omega$ = Jy   ?&lt;/p&gt;&#xA;" OwnerUserId="23447" LastEditorUserId="616" LastEditDate="2018-06-22T13:31:52.950" LastActivityDate="2020-09-01T12:18:44.770" Title="Converting Jy/beam to Jy?" Tags="&lt;radio-astronomy&gt;&lt;units&gt;&lt;interferometry&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26722" PostTypeId="2" ParentId="26713" CreationDate="2018-06-22T12:10:40.910" Score="1" Body="&lt;p&gt;The Sun's gravity would pull in any large object not already in orbit, that would be dominant over magnetic forces from the very weak wind.  If the object was already in orbit, the situation would be akin to dust particles leaving a comet-- their path is more affected by light pressure than magnetic pressure, but for a large object, neither of those could rival gravity.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-06-22T12:10:40.910" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26724" PostTypeId="2" ParentId="26721" CreationDate="2018-06-22T15:38:39.367" Score="3" Body="&lt;p&gt;So long as you accurately know the beam size, then yes multiplying your Jy/beam measurement (effectively flux density) by the beam size (effectively area of flux) will give you the total Jy (effectively the flux).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://tsrice-research.blogspot.com/2016/05/how-to-manually-extract-flux-in-jy-from.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;See this source as an example.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2018-06-22T15:38:39.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26725" PostTypeId="1" AcceptedAnswerId="26732" CreationDate="2018-06-22T15:41:57.143" Score="3" ViewCount="139" Body="&lt;p&gt;I am an undergraduate doing summer project at USO. I am working on Sun's Magnetic field model. I got spiral graph for that, which looks like any spiral Galaxy. Do you think that there is some correlation between them. Or let me put this way, is magnetic field of Galaxy spiral?&#xA;Thank You in advance.&lt;img src=&quot;https://i.stack.imgur.com/BOTF4.png&quot; alt=&quot;LHS is magnetic field lines I got for Sun, RHS is some spiral Galaxy!&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="23450" LastActivityDate="2018-06-25T22:11:12.787" Title="Are magnetic field lines Spiral for Spiral Galaxies? If no, then how it look like topologically?" Tags="&lt;galaxy&gt;&lt;magnetic-field&gt;&lt;spiral-arms&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26726" PostTypeId="2" ParentId="26725" CreationDate="2018-06-22T15:56:18.700" Score="3" Body="&lt;p&gt;Take a look yourself! Below is an image detailing the Milky Way's magnetic field structure &lt;a href=&quot;https://www.ias.u-psud.fr/soler/planckhighlights.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;as measured from the Planck satellite&lt;/a&gt; (and thus necessarily from within the galaxy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Fouh6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fouh6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although of course this has some projection effects going on and is integrated along the line of sight, so it's not the same as a view of another galaxy or as a 3D model, but it gives you some insight into the general structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, magnetic fields of galaxies are messy and unpredictable. This is due to the fact that mostly, there isn't a source for magnetic generation within the galaxy that is of a galactic size. The magnetic field of the galaxy is really just the sum total of magnetic fields produced by all the gas/dust/stars/etc. and all of these objects are doing their own thing and producing their own field, irrespective of what everything else is doing. This means there's going to be no meaningful, coherent contribution to a galactic magnetic field such that a nice structure arises.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your particular field from the Sun, I can't be sure what your dimensions or units are or how you're viewing the field so its hard to process your image. Overall, our Sun's magnetic field as an archimedian spiral like structure to it when viewed in the plane of the ecliptic and a &quot;squashed&quot; dipolar structure on a large scale (due to it hitting the ISM).&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2018-06-22T15:56:18.700" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26727" PostTypeId="1" CreationDate="2018-06-22T16:18:49.920" Score="-1" ViewCount="174" Body="&lt;p&gt;I posit that the giant planets comprise methane gas hydrate (CH4)8(H2O)46). If so, would not the Sun have to have a large composition of oxygen?&lt;/p&gt;&#xA;" OwnerUserId="23451" LastActivityDate="2018-06-23T14:15:27.480" Title="The Sun's composition is based on the currently assumed composition of the giant planets" Tags="&lt;solar-system&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2018-06-23T14:12:51.493" ContentLicense="CC BY-SA 4.0" />
  <row Id="26728" PostTypeId="1" AcceptedAnswerId="26768" CreationDate="2018-06-22T18:09:21.683" Score="4" ViewCount="673" Body="&lt;p&gt;I’m a hobbyist programmer with a profound love for procedural content generation.  I’m working on a 2D space exploration game and while it’s not meant to simulate anything real, I’d like to give it a natural feel.  I have no real knowledge in astronomy, so I figured someone here should have an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if there are a few simple mathematical models I could use to determine some parameters for my generation algorithm.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the orbital radius of the planets&lt;/li&gt;&#xA;&lt;li&gt;the size of the planets (is it more likely that gas giants and the likes will show up closer to the sun or further away, etc.)&lt;/li&gt;&#xA;&lt;li&gt;the number of planets orbiting a star&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Also, a few questions come to my mind too:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;is it a common occurence to have a star with nothing orbiting it?  Like just a single lone star sitting there in the middle of space.&lt;/li&gt;&#xA;&lt;li&gt;IIRC, Earth is kind of a weird thing (besides the fact that it hosts life) because our moon is way bigger than the average moons and also the only satellite of the planet.  Is there any usual correllation between the planet size/density and the size or number of moons around it?&lt;/li&gt;&#xA;&lt;li&gt;Is an asteroid belt something common in solar systems?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot for your help!&lt;/p&gt;&#xA;" OwnerUserId="23452" LastActivityDate="2018-06-26T10:35:12.937" Title="Simple mathematical models of solar systems" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;solar-system&gt;&lt;asteroids&gt;&lt;mathematics&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26729" PostTypeId="2" ParentId="26727" CreationDate="2018-06-22T19:32:39.243" Score="2" Body="&lt;p&gt;Oxygen is the third most abundant element in the Sun, hydrogen and helium being first and second.&#xA;Most sources estimate solar oxygen abundance between 0.6% and 1.0% by mass.&#xA;The astronomical literature expresses element abundance as a logarithm of atom count relative to hydrogen:&#xA;$$A(x) = \log{n(x)\over{n(\mathrm{H})}} + 12$$&#xA;&lt;a href=&quot;https://arxiv.org/abs/1010.2746&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lodders 2010&lt;/a&gt;, based on several contemporary studies, recommends a photospheric oxygen abundance value $A(\mathrm{O})$ = 8.73 $\pm$ 0.07, or 0.046% to 0.063% as many oxygen atoms as hydrogen atoms.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-06-22T20:13:19.237" LastActivityDate="2018-06-22T20:13:19.237" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26730" PostTypeId="2" ParentId="18169" CreationDate="2018-06-22T21:47:50.107" Score="1" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;What point does Earth actually orbit?&lt;/p&gt;&#xA;&lt;p&gt;Neverthess, there is still a barycenter (often outside the limb of the Sun) about which the Sun does a complex dance.&lt;/p&gt;&#xA;&lt;p&gt;Now, &lt;strong&gt;what is Earth actually orbiting?&lt;/strong&gt; The Earth's orbit is described to be an ellipse; the &amp;quot;center&amp;quot; of an elliptical orbit occurs at one of the focal points of the ellipse. What is at the focal point of the elliptical path the Earth follows? Is it the solar system barycenter or is it the Sun's (wobbling) center of mass? In other words, is the Earth's orbit shifting as the Sun is tugged around, our distance from the Sun never varying by more than the eccentricity of the ellipse, or do we orbit the solar system barycenter, and ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Everything is relative, and relative to what.&lt;/p&gt;&#xA;&lt;p&gt;See NASA's article &amp;quot;&lt;a href=&quot;https://spaceplace.nasa.gov/barycenter/en/&quot; rel=&quot;nofollow noreferrer&quot;&gt;What is a barycenter?&lt;/a&gt;&amp;quot;, the section &amp;quot;Barycenters in our solar system&amp;quot;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Where is the barycenter between Earth and the sun? Well, the sun has lots of mass. In comparison, Earth's mass is very small. That means the sun is like the head of the sledgehammer. So, the barycenter between Earth and the sun is very close to the center of the sun.&lt;a href=&quot;https://en.wikipedia.org/wiki/Barycenter#Primary%E2%80%93secondary_examples&quot; rel=&quot;nofollow noreferrer&quot;&gt; [It orbits 449 km from the sun's center&lt;/a&gt;.]&lt;/p&gt;&#xA;&lt;p&gt;Jupiter is a lot larger than Earth. It has 318 times more mass. As a result, the barycenter of Jupiter and the sun isn’t in the center of the sun. It’s actually just outside the sun's surface!&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XFK0s.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XFK0s.jpg&quot; alt=&quot;Sun - Jupiter Barycenter&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The barycenter of our solar system is the combination of all it's masses, this is how much the sun wobbles:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WhVXV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WhVXV.png&quot; alt=&quot;Solar System Barycenter&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Last but not least, user &lt;a href=&quot;https://www.physicsforums.com/threads/orientation-of-the-earth-sun-and-solar-system-in-the-milky-way.888643/page-2&quot; rel=&quot;nofollow noreferrer&quot;&gt;&amp;quot;fizixfan&amp;quot; at the PhysicsForum&lt;/a&gt; has provided this interesting graphic:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3zuad.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3zuad.jpg&quot; alt=&quot;Orbit of the Solar System Through Space, Around the Supergalactic Plane.&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Our orbit around the &lt;a href=&quot;https://en.wikipedia.org/wiki/Celestial_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;supergalactic plane&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-06-22T21:47:50.107" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26732" PostTypeId="2" ParentId="26725" CreationDate="2018-06-23T12:06:53.947" Score="4" Body="&lt;p&gt;The short answer is: yes, magnetic fields in spiral galaxies are generally spiral in orientation. (But it probably isn't directly correlated with your solar model.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, this is a map of radio polarization (a reflection of magnetic field direction) in the galaxy IC 342 by Beck et al. (2015):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WA2W6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WA2W6.jpg&quot; alt=&quot;Magnetic field lines for IC 342&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The short yellow lines are indicators of polarization, and thus magnetic field, direction. See &lt;a href=&quot;https://www.mpg.de/9283089/magnetic-field-galaxy-IC342&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for details of the observation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very good, if somewhat technical, reference is Rainer Beck's recent (2015) review article &lt;a href=&quot;http://arxiv.org/abs/1509.04522&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Magnetic Fields in Spiral Galaxies&quot;&lt;/a&gt;; this discusses both observations of magnetic fields in nearby galaxies (and the Milky Way) and the models for how they might be generated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He also discusses, for example, evidence for magnetica fields in the halos of galaxies; these are (seen in projection) sometimes radial and sometimes more X-shaped, as in the case of NGC 891:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/R0SMN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R0SMN.jpg&quot; alt=&quot;magnetic field lines for NGC 891&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(From &lt;a href=&quot;https://arxiv.org/abs/0806.2060&quot; rel=&quot;nofollow noreferrer&quot;&gt;Krause 2009&lt;/a&gt;; the short white lines indicate magnetic field line directions, based on radio polarization.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The evidence for the Milky Way is somewhat ambiguous, for the usual reasons that it's hard to get a global picture of something you're inside of. One intriguing finding is that the magnetic field sometimes reverses direction as a function of radius; this hasn't been seen in other galaxies (though this may just be a function of resolution or the fact that we are measuring the Milky Way's magnetic fields primarily in a narrow layer of gas near the Galactic plane).&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2018-06-25T22:11:12.787" LastActivityDate="2018-06-25T22:11:12.787" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26733" PostTypeId="1" CreationDate="2018-06-23T13:40:12.793" Score="5" ViewCount="2510" Body="&lt;p&gt;if we are in the milky way, and people are able to take photographs from Earth that look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LXNST.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LXNST.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then how come almost all of Earth's photos look like this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/O44Ry.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O44Ry.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first photo, we are seeing what is &quot;outside Earth&quot;, and so if we actually went &quot;outside Earth&quot;, why don't we see these gases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because of the Sun blinding our vision? Suppose a satellite was hovering over the part of Earth where its night time (very little sun interference). If it &quot;turned around and faced the opposite direction from earth&quot;, it would see the void, but would it see all these gases, or just the black void.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;" OwnerUserId="17327" LastActivityDate="2018-06-24T02:44:28.177" Title="Why do photos of Earth always have a black background?" Tags="&lt;star&gt;&lt;galaxy&gt;&lt;earth&gt;&lt;amateur-observing&gt;&lt;photography&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26734" PostTypeId="2" ParentId="26733" CreationDate="2018-06-23T14:01:49.133" Score="5" Body="&lt;p&gt;As viewed from nearby, the sunlit Earth is &lt;em&gt;far&lt;/em&gt; brighter per unit area than the starry sky because the Earth fills the sky as a solid body, but the stars are tiny, tiny dots filling only a very small portion of the sky -- much less than a pixel. So a properly exposed picture showing the sunlit Earth against space will completely underexpose the stars, leaving space looking black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't happen in a nighttime picture of the moonless sky from Earth, because there is no sunlit landscape to drown out the stars. (Though just the dim nighttime skyglow does a pretty good job.) If the Moon is int he shy, once again, all but the brighter the stars will be invisible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are in space and aim your camera away from the Earth, Sun and Moon, and if your camera is well designed so that the len is properly shaded from all three, you can get wonderful images of the starry sky -- better than from on Earth.  Hubble does this all the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the things that look like gasses in the picture of the Milky Way are actually almost entirely swarms of stars that the eye can't resolve.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-06-23T14:01:49.133" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26735" PostTypeId="2" ParentId="26727" CreationDate="2018-06-23T14:05:02.990" Score="2" Body="&lt;p&gt;Although the Sun and planets were born from the same cloud of gas and dust, it does not follow that the chemical abundance mixture in the Sun and planets will be the same. That is trivially obvious if you consider the case of the Earth, which contains comparatively little of the lightest elements - hydrogen and helium - which actually make up the bulk of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true to a lesser extent with the giant planets Jupiter and Saturn. Their bulk abundances are reasonably close to that of the Sun but not exactly the same. The reasons behind this are complex but are to do with the formation mechanism of the planets (which took some time after the Sun was formed) and chemical fractionation and condensation within the disc of gas and dust from which the planets came.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chemical abundance mixture of the Sun is &lt;em&gt;not&lt;/em&gt; derived from the abundances of chemical elements in the giant planets (which are more uncertain), but from a combination of analysis of the Sun's spectrum, which tells you about the composition of the photosphere, and by a detailed study of the composition of meteorites, which are thought to have formed very early in the life of the solar system and contain a pristine record of the abundances bar the absence of volatiles like hydrogen and helium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your reference to methane hydrates (also known as clathrates) is puzzling. These clathrates are well known to exist in the ices of the satellites of Jupiter and Saturn. Clathrates can form early in the life of the solar system and were probably incorporated into many of the icy and rocky bodies of the outer solar system. I'm not aware of any particular evidence that they occur in great abundances in the atmospheres of Jupiter and Saturn (the outer atmospheres are all we can see) and they certainly would not survive the temperatures and pressure in the interiors of these planets.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-06-23T14:15:27.480" LastActivityDate="2018-06-23T14:15:27.480" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26736" PostTypeId="2" ParentId="26733" CreationDate="2018-06-23T15:12:40.543" Score="6" Body="&lt;p&gt;The photo of the Milky Way is an exposure that was 30 seconds or longer in order to capture the stars. The photo of the Earth was an exposure that was probably 1/100 of a second. (It is daytime!) The stars will not appear in such a short exposure.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2018-06-23T15:12:40.543" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26737" PostTypeId="2" ParentId="6316" CreationDate="2018-06-23T18:04:46.070" Score="1" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;... raise questions about how much we know about the &lt;strong&gt;true&lt;/strong&gt; (&lt;a href=&quot;https://en.wikipedia.org/wiki/IAU_definition_of_planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;IAU definition&lt;/a&gt;) &lt;strong&gt;number&lt;/strong&gt; of planets in our solar system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Planet#Solar_System&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eight&lt;/a&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Question: Does the discovery of &lt;a href=&quot;https://en.wikipedia.org/wiki/2012_VP113&quot; rel=&quot;nofollow noreferrer&quot;&gt;2012 VP&lt;span class=&quot;math-container&quot;&gt;$_{113}$&lt;/span&gt;&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/90377_Sedna&quot; rel=&quot;nofollow noreferrer&quot;&gt;90377 .&lt;/a&gt; increase the odds of some kind of planetary object on the edge of the known solar system (the &lt;a href=&quot;https://en.wikipedia.org/wiki/Oort_cloud&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oort cloud&lt;/a&gt;)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/2012_VP113&quot; rel=&quot;nofollow noreferrer&quot;&gt;2012 VP&lt;span class=&quot;math-container&quot;&gt;$_{113}$&lt;/span&gt;&lt;/a&gt; is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Minor_planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;minor planet&lt;/a&gt;, that does not imply that it is a &amp;quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;planet&lt;/a&gt;&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/90377_Sedna&quot; rel=&quot;nofollow noreferrer&quot;&gt;90377 Sedna&lt;/a&gt; is a large minor &lt;a href=&quot;https://en.wikipedia.org/wiki/Minor_planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;minor planet&lt;/a&gt;, also not a &amp;quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;planet&lt;/a&gt;&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Oort_cloud&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oort cloud&lt;/a&gt; has so many tiny objects that they upset each other's orbit, occasionally sending comets into our solar system. Part of being a planet is having an orbit &lt;strong&gt;and&lt;/strong&gt; clearing your orbit; these objects do neither.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FYbQX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FYbQX.jpg&quot; alt=&quot;Oort Cloud - close-up slice&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/R7lTM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R7lTM.jpg&quot; alt=&quot;Oort Cloud - distant view&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;How could we find out if it is very cold and dark?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Every object in space affects the movement of every other object.&lt;/p&gt;&#xA;&lt;p&gt;We can detect tiny black holes at a distance, they are cold and dark; detecting a new planet in our solar system isn't beyond our ability, there simply isn't one there.&lt;/p&gt;&#xA;&lt;p&gt;It's not impossible for a large object to begin orbit of the sun and establish that it will orbit a reasonably fixed path (some wobble is OK, drastic changes in direction or too close to something large enough isn't going to maintain it's orbit (and clearing ability)).&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Pluto&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pluto&lt;/a&gt; takes 248.00 years (90,560 days) to complete an orbit. Discovered in 1930 it lost its status as a planet in 2006. If something appears out of nowhere it will take time for it to be approved, so no surprises in the next few years.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I mean, how many &lt;a href=&quot;https://en.wikipedia.org/wiki/90377_Sedna&quot; rel=&quot;nofollow noreferrer&quot;&gt;90377 Sedna&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/2012_VP113&quot; rel=&quot;nofollow noreferrer&quot;&gt;2012 VP&lt;span class=&quot;math-container&quot;&gt;$_{113}$&lt;/span&gt;&lt;/a&gt; objects would we need in order to guess its position and location?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Objects very distant to each other affect each other to a much smaller extent that objects which are closer to each other. As time passes, observations and new calculations are made, results are confirmed, &lt;strong&gt;and&lt;/strong&gt; it passes IAU committee approval something might be discovered and become a new planet, instead scientists are looking outside our solar system for exoplanets.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Beta_Pictoris&quot; rel=&quot;nofollow noreferrer&quot;&gt;Beta Pictoris&lt;/a&gt; (β Pic) is the second brightest star in the constellation &lt;a href=&quot;https://en.wikipedia.org/wiki/Pictor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pictor&lt;/a&gt;. It is located 63.4 light years from the Solar System. Scientists believe there are a few planets in orbit around the star.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;&lt;a href=&quot;https://www.cambridge.org/core/journals/proceedings-of-the-international-astronomical-union/article/planet-companion-around-pictoris/4516DB8AC8CE30CC549B9236C78D6A39&quot; rel=&quot;nofollow noreferrer&quot;&gt;The planet companion around β Pictoris&lt;/a&gt;&amp;quot; (Oct 2010), by Anne-Marie Lagrange, Mickaël Bonnefoy, Gael Chauvin, Daniel Apai, Et al.:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Abstract&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The β Pic disk of dust and gas has been regarded as the prototype of young planetary systems since the 1980s and has revealed over the years an impressive amount of indirect signs pointing toward the presence of at least one giant planet. We present here the recently detected first giant planet around this star. We show how this planet could explain some very peculiar features of the star environment (disk, spectroscopic variability), and how it constrains the scenarios of planetary system formation (timescales, mechanisms).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Scientists can see far and do look for planets, it's unlikely that there's a 11th planet that's been unobserved. They have nicknamed β Pic's &lt;a href=&quot;http://www.solstation.com/stars2/beta-pic.htm#planetb&quot; rel=&quot;nofollow noreferrer&quot;&gt;planet &amp;quot;b&amp;quot;&lt;/a&gt;. Here's its photo:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mmyVE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mmyVE.jpg&quot; alt=&quot;Planet &amp;quot;b&amp;quot;&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Reading about this topic, I don't know how to quantify the odds of finding:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;PX (Planet X): An Earth sized body perturbing the orbits of &lt;a href=&quot;https://en.wikipedia.org/wiki/90377_Sedna&quot; rel=&quot;nofollow noreferrer&quot;&gt;90377 Sedna&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/2012_VP113&quot; rel=&quot;nofollow noreferrer&quot;&gt;2012 VP&lt;span class=&quot;math-container&quot;&gt;$_{113}$&lt;/span&gt;&lt;/a&gt; and other similar objects.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;There is &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/0004-6256/151/2/22&quot; rel=&quot;nofollow noreferrer&quot;&gt;support for the existence of the object&lt;/a&gt;, but under the IAU definition imposed by your question it hasn't yet met the requirements to become a planet. &lt;strong&gt;The announcement does not mean there is a new planet in our solar system. The existence of this distant world is only theoretical at this point and no direct observation of the object nicknamed have been made&lt;/strong&gt;. The mathematical prediction of a planet could explain the unique orbits of some smaller objects in the Kuiper Belt, a distant region of icy debris that extends far beyond the orbit of Neptune. Astronomers are now searching for the predicted planet.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Talk:Telisto_(hypothetical_planet)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thelistos&lt;/a&gt;: A super-Earth (about 5-10 Earth masses times a O(1) factor)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Misnomer. Deleted from Wikipedia, redirects to planet X.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Tyche_(hypothetical_planet)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tyche&lt;/a&gt;: (Now disfavoured by WISE data, I think it can not be excluded if it is extremely cold and &amp;quot;dark&amp;quot;) Gas giant (Neptune-like, Saturn-like, etc.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Disproven. Does not exist.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Nemesis: A &amp;quot;dark star&amp;quot;/brown dwarf object.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/128_Nemesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;128 Nemesis&lt;/a&gt; is an asteroid, the &lt;em&gt;other&lt;/em&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Nemesis_(hypothetical_star)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nemesis&lt;/a&gt; is a hypothetical red dwarf or brown dwarf, originally postulated in 1984 to be orbiting the Sun at a distance of about 95,000 AU (1.5 light-years). Thought to be somewhat beyond the Oort cloud, it's &lt;a href=&quot;https://www.google.com/search?q=postulated&quot; rel=&quot;nofollow noreferrer&quot;&gt;postulated&lt;/a&gt; to explain a perceived cycle of mass extinctions in the geological record, which seem to occur more often at intervals of 26 million years.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Which objects are defined as planets is precisely defined, there's only some overlap for non-planets:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z786z.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z786z.png&quot; alt=&quot;Planet or not.&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here is the text of the &lt;a href=&quot;https://www.iau.org/static/resolutions/Resolution_GA26-5-6.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;IAU’s Resolution B5: Definition of a Planet in the Solar System&lt;/a&gt; (.PDF):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The IAU therefore resolves that planets and other bodies, except satellites, in our Solar System be defined into three distinct categories in the following way:&lt;/p&gt;&#xA;&lt;p&gt;(&lt;strong&gt;1&lt;/strong&gt;) A planet&lt;span class=&quot;math-container&quot;&gt;$^1$&lt;/span&gt; is a celestial body that&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\quad$&lt;/span&gt; (&lt;strong&gt;a&lt;/strong&gt;) is in orbit around the Sun,&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\quad$&lt;/span&gt; (&lt;strong&gt;b&lt;/strong&gt;) has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape, and&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\quad$&lt;/span&gt; (&lt;strong&gt;c&lt;/strong&gt;) has cleared the neighbourhood around its orbit.&lt;/p&gt;&#xA;&lt;p&gt;(&lt;strong&gt;2&lt;/strong&gt;) A &amp;quot;dwarf planet&amp;quot; is a celestial body that&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\quad$&lt;/span&gt; (&lt;strong&gt;a&lt;/strong&gt;) is in orbit around the Sun,&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\quad$&lt;/span&gt; (&lt;strong&gt;b&lt;/strong&gt;) has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt;,&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\quad$&lt;/span&gt; (&lt;strong&gt;c&lt;/strong&gt;) has &lt;em&gt;&lt;strong&gt;not&lt;/strong&gt;&lt;/em&gt; cleared the neighbourhood around its orbit, and&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\quad$&lt;/span&gt; (&lt;strong&gt;d&lt;/strong&gt;) is not a satellite.&lt;/p&gt;&#xA;&lt;p&gt;(&lt;strong&gt;3&lt;/strong&gt;) All other objects&lt;span class=&quot;math-container&quot;&gt;$^3$&lt;/span&gt;, except satellites, orbiting the Sun shall be referred to  collectively as &amp;quot;Small Solar System Bodies&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;code&gt;1&lt;/code&gt;&lt;/strong&gt; &lt;code&gt;The eight planets are: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune.&lt;/code&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;code&gt;2&lt;/code&gt;&lt;/strong&gt; &lt;code&gt;An IAU process will be established to assign borderline objects to the dwarf planet or to another category.&lt;/code&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;code&gt;3&lt;/code&gt;&lt;/strong&gt; &lt;code&gt;These currently include most of the Solar System asteroids, most Trans-Neptunian Objects (TNOs), comets, and other small bodies.&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-06-23T18:04:46.070" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26738" PostTypeId="1" AcceptedAnswerId="26743" CreationDate="2018-06-23T22:59:11.757" Score="5" ViewCount="403" Body="&lt;p&gt;I'm writing a sci-fi novel in which humans venture to the stars for the first time. Their &quot;shake down&quot; cruise takes them to Proxima Centauri. I'm trying to understand where Proxima Centauri is located relative to the sun rather than Earth, as that seems a more logical reference for navigation to nearby stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The location of Proxima Centauri in Earth's night sky is described with the Earth-relevant coordinates&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Right ascension   14h 39m 36.49400&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Declination –60° 50′ 02.3737&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, the Earth's orbit is inclined 7.155° to the Sun's equator and our axis is tilted at 23.4392811°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a reference indicating Proxima Centauri's sol-relative position, or can that position be calculated based on the above information?&lt;/p&gt;&#xA;" OwnerUserId="1315" LastActivityDate="2018-06-26T02:00:45.220" Title="Directions to Proxima Centauri" Tags="&lt;fundamental-astronomy&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26739" PostTypeId="2" ParentId="26733" CreationDate="2018-06-24T00:34:51.513" Score="5" Body="&lt;p&gt;Photos of Earth from space do not always have a black background (the other answers explain why usually, they do).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://apod.nasa.gov/apod/ap151107.html&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://apod.nasa.gov/apod/image/1511/iss044e045565a.jpg&quot; alt=&quot;Milky way from space&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NASA Astronomy Picture of the Day, Milky Way seen from ISS, taken by Astronaut Scott Kelly on August 9, 2015.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;https://www.youtube.com/watch?v=9kkwKW28PRU&quot; rel=&quot;noreferrer&quot;&gt;this timelapse on youtube&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1462" LastActivityDate="2018-06-24T00:34:51.513" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26742" PostTypeId="2" ParentId="26699" CreationDate="2018-06-24T04:07:35.093" Score="0" Body="&lt;p&gt;So I have just found this &lt;a href=&quot;http://www.asterank.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;database&lt;/a&gt; - which looks like exactly what I was after!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has over ½ a million asteroids listed - giving their value and cost to get them as well as a graphical representation of their orbit!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the  Website's description:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Details on orbits and basic physical parameters are sourced from the&#xA;  Minor Planet Center and NASA JPL. Composition data is based on&#xA;  spectral classification and size.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="23414" LastActivityDate="2018-06-24T04:07:35.093" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26743" PostTypeId="2" ParentId="26738" CreationDate="2018-06-24T06:16:02.153" Score="4" Body="&lt;p&gt;A system of coordinates requires a fixed plane to define &quot;zero latitude&quot; and a direction chosen in that plane to define &quot;zero longitude&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For directions to the stars, the distances are so great that it matters not if you base your coordinates at the Sun or at the Earth, because the distance between the Sun and Earth is so small in relation to the distances to the stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few coordinate systems in use.  The first uses the Earth's Equator as the fixed plane and the direction of the sun on March 21 (vernal equinox) as the direction of zero longitude. For historical reasons the longitude in this system is called Right Ascension and latitude is called Declination. It would be quite suitable for your adventurers to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also use the plane of the solar system, or the plane of the galaxy as the fixed plane. In these coordinates systems &lt;a href=&quot;https://lambda.gsfc.nasa.gov/toolbox/tb_coordconv.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Proxima is located at&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-44:45:47.62 Latitude(deg)   +239:06:53.08 Longitude(deg) Ecliptic&#xA;-01:55:37.90 Latitude(deg)   +313:56:23.96 Longitude(deg) Galactic&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But remember that Proxima is moving, so these figures are not fixed.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="616" LastEditDate="2018-06-24T11:43:09.440" LastActivityDate="2018-06-24T11:43:09.440" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26744" PostTypeId="1" CreationDate="2018-06-24T09:34:54.733" Score="3" ViewCount="29" Body="&lt;p&gt;In optical astronomy, much (most?) electromagnetic radiation is emitted at well defined frequencies, and this can be used to infer a redshift for the source, and hence its recessional velocity, age, or distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't read anything specifically suggesting well defined source frequencies for gravitational wave astronomy, which would allow a redshift to be calculated the same way. I assume there would be a redshift effect since like EM radiation a gravitational wave propagates at (or exceedingly close to) the speed of light. But then again it propagates as a &lt;em&gt;ripple in&lt;/em&gt; spacetime rather than a phenomenon &lt;em&gt;embedded in&lt;/em&gt; spacetime, so perhaps not.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Do gravitational waves exhibit a frequency shifting effect akin to a redshirt, due to cosmic expansion, and which can be used to infer distance/time? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If so, do they also exhibit some kind of discreet originating frequencies, or some way that we can infer their originating frequencies, to compare the received signal and deduce the amount of redshifting they have undergone?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="18601" LastActivityDate="2018-06-24T09:34:54.733" Title="Do gravitational waves have distinct bands or parameters from which a source redshift can be inferred?" Tags="&lt;redshift&gt;&lt;gravitational-waves&gt;&lt;special-relativity&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2018-06-25T11:05:13.950" ContentLicense="CC BY-SA 4.0" />
  <row Id="26746" PostTypeId="2" ParentId="26713" CreationDate="2018-06-24T10:02:17.617" Score="2" Body="&lt;p&gt;No. The magnetic field has two poles, the force on the two pole of the magnet is equal and opposite. This is why a compass needle will align in the North South direction, but is not pulled towards the North or the South.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two magnets will attract or each other because the North pole of one is nearer the South pole of the other. You can get a net force on a magnet when the field-lines are getting bunched together. But the magnetic field in the plasma is not like this. To a first approximation they are parallel to each other and perpendicular to the equator of the sun. In fact the magnetic field is complex as it is carried with the flow of solar wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There would be no net force on the magnet, but it may tend to align with the magnetic field. This might cause a turning force, but no acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There would be some force due to solar wind particles impinging on the object, but since there is only something like 5 atoms per cm3, the force isn't very big, and it has nothing to do with the object being a magnet.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-06-24T10:10:00.260" LastActivityDate="2018-06-24T10:10:00.260" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26747" PostTypeId="2" ParentId="22673" CreationDate="2018-06-24T13:19:10.500" Score="1" Body="&lt;p&gt;The professor might also mean to subtract the background region fluxes for the noise subtraction. Then, calculating the statistics for the subtracted flux value and error of the measurement, say A +/- dA. Then, applying the 3-sigma criteria to determine whether there is a radio source detected at the location, i.e., if A - 3*dA &gt; 0, there is a source, otherwise there is no source. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is no source, you will want to quote the value for 3-sigma upper limit instead, e.g., A + 3*dA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, note that for a faint source, regular poisson or Gaussian statistics are not good representative. In X-ray, we use Cash statistics for low-count cases. I think the same statistics might be applicable with radio observations.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-06-24T13:19:10.500" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26748" PostTypeId="2" ParentId="26738" CreationDate="2018-06-24T19:51:01.673" Score="4" Body="&lt;p&gt;Equatorial coordinates do refer to Earth's rotation and orbit, but the related &lt;a href=&quot;http://aa.usno.navy.mil/faq/docs/ICRS_doc.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;International Celestial Reference System&lt;/a&gt; (ICRS) can also be used in deep space.&#xA;The ICRS origin is the solar system &lt;a href=&quot;https://en.wikipedia.org/wiki/Barycenter&quot; rel=&quot;nofollow noreferrer&quot;&gt;barycenter&lt;/a&gt;, and the coordinate axes are aligned with the Earth equator and equinox as of the year 2000.&#xA;The International Celestial Reference Frame (ICRF) is a catalog of precise ICRS positions of a few hundred extragalactic radio sources, mostly quasars.&#xA;In optical wavelengths, a Hipparcos reference frame links stars in this galaxy to ICRS, and a &lt;a href=&quot;https://arxiv.org/abs/1804.09377&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gaia reference frame&lt;/a&gt; of quasars is in development.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A complete star catalog includes not only equatorial positions at the reference epoch but also parallaxes and proper motions.&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_parallax&quot; rel=&quot;nofollow noreferrer&quot;&gt;Parallax&lt;/a&gt;, inversely related to distance, is the annual wobble of a star's geocentric position around the heliocentric average, less than 1 arcsecond even for the nearest stars.&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Proper_motion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Proper motion&lt;/a&gt; is the change in apparent position over time as the Sun and other stars move around the galaxy.&#xA;With this information plus &lt;a href=&quot;https://en.wikipedia.org/wiki/Radial_velocity&quot; rel=&quot;nofollow noreferrer&quot;&gt;radial velocity&lt;/a&gt;, the crew could predict and analyze the apparent positions of nearby stars as the spacecraft travels among them.&#xA;The ICRF quasars' great distance fortunately makes their parallaxes and proper motions negligible but unfortunately makes them rather faint in the optical band.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to on-board &lt;a href=&quot;https://en.wikipedia.org/wiki/Astrometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;astrometry&lt;/a&gt;, I suppose the spacecraft would use &lt;a href=&quot;https://en.wikipedia.org/wiki/Star_tracker&quot; rel=&quot;nofollow noreferrer&quot;&gt;star trackers&lt;/a&gt; for navigation.&#xA;Satellites and planetary probes commonly track Canopus as it is very bright and fairly distant.&#xA;Other distant bright stars such as Rigel or Deneb might help with attitude control.&#xA;Nearby stars such as Sirius and Vega would appear to shift by a few degrees as the spacecraft progressed toward its destination.&#xA;By the end of the trip, the Sun would fade to magnitude +0.4 but remain a useful reference point.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-06-24T20:06:02.327" LastActivityDate="2018-06-24T20:06:02.327" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26749" PostTypeId="1" CreationDate="2018-06-24T23:21:22.537" Score="5" ViewCount="185" Body="&lt;p&gt;I've heard one of the reason why Titan has such a thick atmosphere is because of its really low temperature. So say that were were to magically bring Titan to the Same orbit as Earth, and the moon starts to rapidly heating up. How much of its atmosphere would it lose? Would we still be able to stand there without a pressurized suit? Is it possible to terraform Titan to the extent that we can heat up its surface while still retaining its thick atmosphere? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks &lt;/p&gt;&#xA;" OwnerUserId="22660" LastActivityDate="2018-06-25T10:48:47.187" Title="Say we were to bring Titan to Earth's orbit. How much would its atmospheric pressure change?" Tags="&lt;planetary-atmosphere&gt;&lt;natural-satellites&gt;&lt;titan&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26750" PostTypeId="2" ParentId="26749" CreationDate="2018-06-25T07:49:05.430" Score="2" Body="&lt;p&gt;As Temperature rises, so would the pressure, this would work immediately, as heat from the sun reaches Titan's surface.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atmospheric escape (and thus, pressure loss, as you describe it) would work on a much longer timescale.&lt;br&gt;&#xA;We can make a few estimates to determine this timescale: On the most simplified level, the composition if Titan's atmosphere is that of Earth's - it's mostly molecular Nitrogen.&lt;br&gt;&#xA;So the mean molecular weight $\mu$ would be the same. The temperature would be similar compared to Earth by assumption of your experiment, then the only important difference is the mass and radius of the two bodies, reducing the escape speed by a factor of ~10.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Titan has 2% of the mass of Earth, so the ratio of escape speed to thermal molecular speed at the exobase, from where molecules are lost, would approximately scale like $\sim 1/\sqrt{10}\approx 3.1$&lt;br&gt;&#xA;To determine escape rates, those go roughly like $\sim \exp(v_{th}/v_{esc})$ and would thus increase by $\sim \exp(3.1)\approx 22 $, compared to today's level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know present estimates for the lifetime of Titan's atmosphere, but this factor would e.e. bring it down from 5 billion years to only 250 million years. That's really short, geologically speaking!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course it's not enough to affect you in your pressure suit on the surface. You'd rather be concerned about an overpressure than underpressure.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2018-06-25T07:49:05.430" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26751" PostTypeId="1" CreationDate="2018-06-25T08:07:47.973" Score="6" ViewCount="164" Body="&lt;p&gt;In a broad sense I'm trying to understand how a binary black hole or neutron star systems can form from isolated binary massive star systems (i.e., other than from migration within rich BH/NS clusters), and especially, how these can form within the constraints that allow for observed merger rates. I was hoping that &lt;a href=&quot;https://astronomy.stackexchange.com/questions/13681/the-black-hole-binary-that-was-detected-by-advanced-ligo-how-do-such-hypergian&quot;&gt;this question&lt;/a&gt; would answer me, but it doesn't, because I realised I've never seen an explanation of how a binary system could survive a SN of one of its stars in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understand, we think one route to observed BH/NS mergers is a binary system of massive stars. This gives a scenario of 2 massive stars in an isolated binary system, one of which undergoes core collapse and type 2 SN. The two stars must be quite close (or later perturbed) and not greatly &quot;kicked&quot; on formation of either compact object, otherwise they couldn't later merge in the observed time. They probably also have similar ages (formed in the same cloud).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in this scenario&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;How does the second-to-collapse star, survive being ripped apart by a nearby SN at such short distance, and remain in a small enough orbit, which we know happens since we later observe the merger of two compact objects? How does it survive structurally and regain stability in a form capable of future SN, given the intense flood of neutrinos and other energy/particles, some of which can penetrate to its core? Or does the first SN directly trigger the second SN?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Even assuming the second star survives with enough mass and undisrupted structure for SN (or is triggered directly into SN), wouldn't the transferred momentum from the first SN ejecta &quot;on its doorstep&quot; be sufficient to simultaneously unbind the binary system, or the momentum transferred from the second SN be enough to eject the first NS/BH, so that the two resulting compact objects no longer form a binary system?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="18601" LastEditorUserId="18601" LastEditDate="2018-06-25T08:19:14.927" LastActivityDate="2018-06-27T02:55:25.047" Title="How do isolated binary compact systems survive their formational supernovae?" Tags="&lt;black-hole&gt;&lt;supernova&gt;&lt;binary-star&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26752" PostTypeId="2" ParentId="26749" CreationDate="2018-06-25T10:48:47.187" Score="1" Body="&lt;p&gt;There's a basic problem that I don't see addressed.   Titan has oceans of methane.   If you warm up it's surface, those oceans will evaporate and the atmospheric pressure will increase rapidly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, no, you can't warm up titan to teraform it because, at least for a while, the atmospheric pressure would become too great.   In the long run it would lose it's atmosphere.  It will probably lose it's atmosphere anyway given enough time and certainly as the sun goes red giant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as settlements go, the cold weather of Titan is probably easier for a human colony than the near vacuum and fine/toxic dust on Mars.   A vacuum requires much more technology to survive in than extreme cold.  (to my limited knowledge anyway).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars has more metals and it's better for solar energy and its gravity is more livable.    Solar energy would be basically useless on Titan and the gravity is too low for humans to maintain a good sense of up and down.   There's pros and cons to both.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-06-25T10:48:47.187" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26753" PostTypeId="1" AcceptedAnswerId="26757" CreationDate="2018-06-25T12:02:02.363" Score="1" ViewCount="36" Body="&lt;p&gt;I am trying to work on data from a .fits file and it has an IVAR header. How can I use inverse variation information along with the actual signal data to get better results or reduce noise?&lt;/p&gt;&#xA;" OwnerUserId="23468" LastActivityDate="2018-06-25T13:23:20.287" Title="In sky surveys, what does inverse variation really mean? How do we use inverse variation to remove noise?" Tags="&lt;fits-header&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26754" PostTypeId="1" AcceptedAnswerId="26756" CreationDate="2018-06-25T12:06:01.610" Score="2" ViewCount="325" Body="&lt;p&gt;I'm wondering that what would be the &lt;strong&gt;escape velocity&lt;/strong&gt; of our Galaxy?&lt;/p&gt;&#xA;" OwnerUserId="23217" LastActivityDate="2018-06-25T12:34:33.460" Title="Escape velocity of Milky Way galaxy" Tags="&lt;galaxy&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2018-06-25T13:15:00.303" ContentLicense="CC BY-SA 4.0" />
  <row Id="26756" PostTypeId="2" ParentId="26754" CreationDate="2018-06-25T12:34:33.460" Score="2" Body="&lt;p&gt;500+ kps.  There's a pretty good answer over at &lt;a href=&quot;https://space.stackexchange.com/questions/3948/escape-velocity-for-our-galaxy&quot;&gt;https://space.stackexchange.com/questions/3948/escape-velocity-for-our-galaxy&lt;/a&gt;, but the question appears to fit here as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a mass all located inside a circular orbit, escape velocity is a simple function of orbital velocity, but that ignores most of the mass outside that circular orbit (all of it if its distribution is spherically symmetric).  The Milky Way has a &lt;em&gt;lot&lt;/em&gt; of mass outside the Sun's orbit, though.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-06-25T12:34:33.460" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26757" PostTypeId="2" ParentId="26753" CreationDate="2018-06-25T13:23:20.287" Score="1" Body="&lt;p&gt;It is inverse &lt;em&gt;variance&lt;/em&gt;. A larger inverse variance means the value in that pixel is less uncertain (in a relative sense).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inverse variance can be used as a weighting factor - it directly gives you level of significance for the measurement. It can also be set to zero for measurements that should not be used (for whatever reason).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example would be finding the mean value of a set of pixels. A standard mean would apply equal weight to all pixels. A &lt;a href=&quot;https://en.wikipedia.org/wiki/Weighted_arithmetic_mean&quot; rel=&quot;nofollow noreferrer&quot;&gt;weighted mean&lt;/a&gt; would be calculated as&#xA;$$ \bar{x} = \frac{\sum_i x_i w_i}{\sum_i w_i},$$&#xA;where $w_i$ here would be the inverse variance of each pixel.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-06-25T13:23:20.287" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26759" PostTypeId="2" ParentId="20754" CreationDate="2018-06-25T18:34:12.503" Score="1" Body="&lt;p&gt;So to estimate the answer, you need to convert the mass density to number density by $n_H = \rho_H / m_H$. After getting the current number density you need to redshift it to finite $z$, here $z=6$. Keep in mind the density of hydrogen evolves $\sim (1+z)^3$ due to the expansion of the universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, the assumption 'let universe be completely made from hydrogen' renders this question highly unrealistic, hence I agree with @James K that this is not real world astronomy. In the real world, only about 5% of the total energy is made of baryons, among which Hydrogen abundance is $\sim 75\%$. Also, at redshift $z\sim 6$, star formation etc. lowers the neutral hydrogen gas density, c.f. recent observations &lt;a href=&quot;https://arxiv.org/pdf/1506.02037.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://arxiv.org/pdf/1503.05553.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="23472" LastActivityDate="2018-06-25T18:34:12.503" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26760" PostTypeId="1" CreationDate="2018-06-25T18:35:29.573" Score="2" ViewCount="1416" Body="&lt;p&gt;I've owned my AstroMaster 114EQ for about a year now, and feel like I know my way around pretty well, but I'm only just becoming aware that collimation is recommended. The &lt;a href=&quot;https://www.manualslib.com/manual/356435/Celestron-Astromaster-114eq.html?page=26#manual&quot; rel=&quot;nofollow noreferrer&quot;&gt;user manual has instructions on basic collimation&lt;/a&gt;. Does this look sufficient? Or should I invest in a &lt;a href=&quot;https://www.opticsplanet.com/celestron-collimation-tool.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;celestron 94183 collimation cap&lt;/a&gt;, a &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B00009R7RJ&quot; rel=&quot;nofollow noreferrer&quot;&gt;celestron collimation eyepiece&lt;/a&gt;, or something fancy like an &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B008XEGXMO&quot; rel=&quot;nofollow noreferrer&quot;&gt;Orion LaserMate Deluxe II&lt;/a&gt;? The telescope itself cost less than $200 so I can't help but wonder if the fancier collimators are overkill. I've watched &lt;a href=&quot;https://www.youtube.com/watch?v=MwMUgBG5qkg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this longish video&lt;/a&gt; and one must apparently perform some disassembly on the telescope to user the laser collimator. This doesn't seem like the easy 'couple of minutes' adjustment I've seen described.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also confused about the process of adjusting the secondary mirror. If anyone has good instructions about my specific model, I would greatly appreciate it. I'd really like to get it tuned up now that Mars is looking so good.&lt;/p&gt;&#xA;" OwnerUserId="16705" LastActivityDate="2018-07-18T19:47:18.353" Title="collimating AstroMaster 114" Tags="&lt;collimation&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26761" PostTypeId="2" ParentId="23257" CreationDate="2018-06-25T20:16:50.767" Score="1" Body="&lt;p&gt;Yes, mathematics is essential for all branches of physics and critical for astrophysics. It is required that you improve your mathematical skills throughout school and university.&lt;/p&gt;&#xA;" OwnerUserId="23468" LastActivityDate="2018-06-25T20:16:50.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26762" PostTypeId="2" ParentId="23257" CreationDate="2018-06-25T20:52:53.217" Score="1" Body="&lt;p&gt;I agree with the answers here suggesting that geometry, algebra, trigonometry, and calculus are all highly helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if I were you, I would reframe your approach a bit. Instead of asking yourself &quot;I want to read more advanced sources on astrophysics, what math should I read first&quot;, I'd instead try to move towards this kind of process:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What particular subject do you want to start with (e.g. how stars form, or what is a black hole, etc.)&lt;/li&gt;&#xA;&lt;li&gt;Read a general resource about this (e.g. Wikipedia articles, popular science pages online - long form books like Hawking's are useful, but can be hard to probe for just one topic)&lt;/li&gt;&#xA;&lt;li&gt;Ask yourself how much deeper you want to go; what didn't you understand, what holes are left unfilled, etc.&lt;/li&gt;&#xA;&lt;li&gt;Now, go find something more technical (a textbook, that mathy section of the Wikipedia article, whatever you can find)&lt;/li&gt;&#xA;&lt;li&gt;If you start seeing math/physics you don't understand, you may have to dig around to find out what you are missing. You can ask questions on SE about this kind of &quot;mathematics/physics orientation&quot;.&lt;/li&gt;&#xA;&lt;li&gt;Learn what you need to progress, and then keep going. Repeat the above steps as necessary.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The main difference in this approach is that you spend less time moving through your &quot;prerequisites&quot; list, and more time reading about what you want to learn. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there won't be any skipping algebra, trig, or calculus if you want to dig much deeper than &lt;em&gt;A Brief History of Time&lt;/em&gt;. You will have to learn a lot of that stuff to understand basically any textbooks on astrophysics. It could be worth your time to go through some of the math and physics exercises on Khan Academy or another online learning resource.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The route I am recommending is filled with plenty of frustration, and might be annoying at times. But speaking as someone who's been learning a lot on-the-side of undergrad and graduate school for the last five years, I think it works better than trying to systematically learn everything in order. And it has the advantage of whetting your apetite with the &quot;good stuff&quot; you're trying to learn more frequently. YMMV, however.&lt;/p&gt;&#xA;" OwnerUserId="23373" LastActivityDate="2018-06-25T20:52:53.217" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26764" PostTypeId="2" ParentId="26738" CreationDate="2018-06-26T02:00:45.220" Score="1" Body="&lt;p&gt;I think I have a better way for you to visualize locations of stars relative to the sun. I am also writing a science-fiction book/series so I would be glad to share ideas. I have finished one book of what I hope to one day be an eight book series. Over the course of my eight books, I am having a generation starship visit 18 star systems then return to Earth over a 600 year period. I wanted to visualize the shortest route so I actually built a 3D model out of foam balls and stiff wire. It takes about a half an hour to make this if you have the following information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GALACTIC (CARTESIAN) COORDINATES&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Create an X,Y,Z coordinate system with the following directions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X,Y plane is the &quot;galactic plane&quot;: this is not well defined, but ideally&#xA;there is an equal amount of mass (or star count) above and below the plane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X: line pointing in the direction of the center of the galaxy from the sun&#xA;   (distance of sun &quot;above&quot; the plane is negligible compared to the distance &#xA;   to the galactic center/ we arbitrarily choose the sun to be &quot;above&quot; the&#xA;   galactic plane (about 67 +/- 11 Light Years according to Humphrey &amp;amp; Larsen&#xA;   1995AJ Vol 110 p 2183)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Y: line perpendicular to X, positive in the direction of galactic rotation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Z: line perpendicular to the galactic plane that passes through the sun, &#xA;   positive in the &quot;Galactic North&quot; direction (farther &quot;above&quot; the galactic &#xA;   plane)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now set the sun at (X,Y,Z) = (0,0,0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these coordinates Proxima was at (2.946,-3.021,-0.169) in the year 2000 per:&#xA;&lt;a href=&quot;http://www.stellar-database.com/Scripts/search_star.exe?Name=alpha+centauri&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.stellar-database.com/Scripts/search_star.exe?Name=alpha+centauri&lt;/a&gt;&#xA;Of course stars move, but if your book is near future this is close enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method would just be for visualization and you would probably need to put in a few more nearby stars to get relative locations. For navigation, answers above would work (if you used a computer to calculate where Proxima will be at arrival time, you point that way and make small corrections when you get close).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out this cool 3D plot:&#xA;&lt;a href=&quot;http://www.atlasoftheuniverse.com/50lys.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.atlasoftheuniverse.com/50lys.html&lt;/a&gt;&#xA;At this scale, Proxima and Alpha Centauri are in the same place.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastActivityDate="2018-06-26T02:00:45.220" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26768" PostTypeId="2" ParentId="26728" CreationDate="2018-06-26T10:35:12.937" Score="1" Body="&lt;p&gt;I &quot;grew up&quot; with the old roleplaying game 2300AD's system creation rules, which were based on fairly hard science (and used the Gliese star catalog as a map!) They have stood the test of time fairly well (&lt;a href=&quot;http://evildrganymede.net/wp/2014/01/17/gdw-2300ad-worldbuilding-system-part-1/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here is an astronomer friend's take on it&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a rough sketch of how I typically make my system generators, which is based on the same logic:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Generate a number $N$ of planets in the system. At present there is no clear data constraining this, although we may imagine high metallicity stars could have more planets (but maybe not).&lt;/li&gt;&#xA;&lt;li&gt;Generate orbits $a_i$: start with a random innermost orbit $a_1$, and then make $a_{n+1}=(1+r_n)a_n$ where $r_n \sim U(K_1,K_2)$ is a random number. This imitates the Titus-Bode law pattern and looks roughly like observed systems. I tend to use $K_1=0.1, K_2=2$ but this is guesswork. Note that this process has problems handling hot Jupiters really close - you may want to add them by hand. It also ignore resonances etc. &lt;/li&gt;&#xA;&lt;li&gt;Generate a planetary seed size/mass. This can be done in an elaborate way by looking at exoplanet $(a,M)$ distributions and extrapolating to unobserved corners, or using some suitable skew random number distribution.&lt;/li&gt;&#xA;&lt;li&gt;From the mass and distance to the star, estimate planetary and atmospheric composition. Basically, outside the snowline ($\approx 1.4\sqrt{L/L_\odot}$ where $L$ is star luminosity) planets will start accumulating volatiles, and having ice crusts become possible. Essentially, decide on a core density randomly.&lt;/li&gt;&#xA;&lt;li&gt;Calculate the approximate minimum molecular weight retained based on the surface gravity. This will determine the atmospheric density. Especially, if the planet retains hydrogen it will become a gas giant: multiply up the mass a lot (the overall radius of a planet scales as $M^{1/3}$ for solid planets up to a few earth masses, and up to Jupiter size for gas giants - then degeneracy pressure will make the size level off and they become denser instead). Note that this is going to be the most fudgy step, since actually estimating a consistent atmosphere (especially exosphere) temperature, composition and pressure is really involved (and nontrivial even in reality). &lt;/li&gt;&#xA;&lt;li&gt;Given the size, temperature zone and atmosphere classify it suitably (&quot;ice ball&quot;, &quot;gas giant&quot;, &quot;venusian hothouse&quot;, &quot;super-earth&quot;, ...).&lt;/li&gt;&#xA;&lt;li&gt;Add random moons, rotation periods, eccentricities and whatnot, calculating their effects (like tide sizes, estimated magnetic fields, temperature variations, number of Hadley cells...) - &lt;em&gt;lots&lt;/em&gt; of handwaving here, even when you base it on real astronomy and atmospheric physics papers.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is the simple version. It does not try to actually simulate planet formation, where important processes like migrating gas giants can really affect systems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You extra questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now it looks like planets are very common, so a system with no planets is likely pretty unusual. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Big moons might be uncommon, or they might be ordinary. In the solar system Pluto/Charon is also a pretty even pair. I personally think double-planets may be more common than people expect. We'll see. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asteroid belts are likely pretty common. In a way we have two, the main belt and Kuiper belt. I would expect most systems to at least have a bit of debris that has not cohered in the outer regions. &lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2018-06-26T10:35:12.937" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26769" PostTypeId="1" AcceptedAnswerId="26770" CreationDate="2018-06-26T16:49:03.783" Score="7" ViewCount="152" Body="&lt;p&gt;Apparently EGSY8p7 is the object with the longest light travel distance, 13.2 gly or a redshift of z = 8.68 (Wikipedia). &#xA;So the light took 13.2 billion years to travel to us from that object, but we know that the universe has expanded since (I believe with an expansion rate that varied over time). So given our current theories of how the universe expanded, is it possible to calculate the theoretical distance of that object to us right now? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume it must be close to what is believed the radius of the observable universe (46.5 Gly) but how would one go about to calculate this? &lt;/p&gt;&#xA;" OwnerUserId="23482" LastActivityDate="2018-06-26T19:23:07.220" Title="How far would EGSY8p7 be away now?" Tags="&lt;galaxy&gt;&lt;expansion&gt;&lt;age&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26770" PostTypeId="2" ParentId="26769" CreationDate="2018-06-26T19:23:07.220" Score="10" Body="&lt;p&gt;30.4 billion lightyears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current distance — i.e. the distance that one would measure if we froze the Universe and started laying out measuring rods — is called the &lt;em&gt;proper distance&lt;/em&gt;, or &lt;em&gt;physical distance&lt;/em&gt;, in astronomy. By definition, it corresponds &lt;em&gt;today&lt;/em&gt; to another often-used term in astronomy, namely the &lt;em&gt;comoving distance&lt;/em&gt;. While the former increases with time, the latter is defined in a coordinate system that expands with the Universe, and so is constant at all times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the comoving distance to EGSY8p7 was the same when the Universe was half the size of what it is today, but the proper distance was half of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For galaxies so distant as EGSY8p7, we don't measure distances directly. Instead we measure the redshift of some spectral absorption or emission lines, and then use a model of the expansion of the Universe to convert this redshift to a distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann%E2%80%93Lema%C3%AEtre%E2%80%93Robertson%E2%80%93Walker_metric&quot; rel=&quot;noreferrer&quot;&gt;FLRW metric&lt;/a&gt;, the resulting distance today of an object at redshift $z$ is given by&#xA;$$&#xA;d_C = \frac{c}{H_0} \int_0^z dz' \frac{1}{\sqrt{\Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_k(1+z)^2 + \Omega_\Lambda}},&#xA;$$&#xA;where $c$, $H_0$, and $\{\Omega_r,\Omega_m,\Omega_k,\Omega_\Lambda\}$ are the speed of light, Hubble constant, and density parameters of the various components of the Universe, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $z=8.68$ and with the latest cosmological parameters from &lt;a href=&quot;https://arxiv.org/abs/1502.01589&quot; rel=&quot;noreferrer&quot;&gt;Planck Collaboration et al. (2016)&lt;/a&gt;, this integral evaluates to $d_C = 30.4\,\mathrm{Glyr}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $z\rightarrow\infty$, you get $d_C = 46.3\,\mathrm{Glyr}$, which is &quot;the edge of the observable Universe&quot;, also known as &lt;em&gt;the particle horizon&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note though that EGSY8p7 no longer holds the redshift record. The galaxy GN-z11 (&lt;a href=&quot;https://arxiv.org/abs/1603.00461&quot; rel=&quot;noreferrer&quot;&gt;Oesch et al. 2016&lt;/a&gt;) has a redshift of $z=11.09$, corresponding to a distance of $d_C = 32.2\,\mathrm{Glyr}$.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2018-06-26T19:23:07.220" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26771" PostTypeId="2" ParentId="26760" CreationDate="2018-06-26T19:42:19.230" Score="1" Body="&lt;p&gt;With a tube length of 46 cm and a focal length of 100 cm, probably the AstroMaster 114EQ is not a classical Newtonian but uses a fast (e.g. f/4) spherical primary mirror and a corrector lens built into the base of the focuser to reduce spherical aberration and increase the effective focal length (e.g. to f/8.8).&#xA;This is a variant of the Jones-Bird design of some &lt;a href=&quot;http://www.asemonline.org/Home/telescope-and-equipment-resources/40cm-jones-bird-telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;custom telescopes&lt;/a&gt;.&#xA;There is &lt;a href=&quot;http://www.astronomyforum.net/reflector-telescope-forum/228369-jones-bird-telescope-design-isnt.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;some doubt&lt;/a&gt; about how well the corrector actually works in budget-priced telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most collimation instructions (e.g. &lt;a href=&quot;http://garyseronik.com/a-beginners-guide-to-collimation/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Seronik&lt;/a&gt; or &lt;a href=&quot;https://www.skyandtelescope.com/astronomy-resources/how-to-align-your-newtonian-reflector-telescope/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carlin&lt;/a&gt;) assume a classical Newtonian design.&#xA;Jones-Bird collimation is similar in theory but more complicated in practice.&#xA;In particular, the corrector hinders the use of a laser device.&#xA;Try a collimation cap; anything that fits a 1.25 inch focuser is fine.&#xA;You can even get decent results with a homemade one.&#xA;If you decide to invest in a Cheshire sight tube, it can be used with any Newtonian reflector.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-07-18T19:47:18.353" LastActivityDate="2018-07-18T19:47:18.353" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26772" PostTypeId="2" ParentId="26751" CreationDate="2018-06-26T22:30:03.180" Score="2" Body="&lt;p&gt;OK, here's what I've been able to piece together with a little research (not really my field, so I may get a few things wrong).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short answer is a combination of: A) It's pretty hard to actually disrupt a massive star via a neighboring supernova, and the two stars aren't intially that close to each other; and B) Lots of complicated effects due to the binary interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll focus on what seems to be the best current model for forming a compact double neutron-star (DNS) system -- one with a small enough separation in the resulting neutron stars so that gravitational-wave emission can cause them to merge within a few billion years -- which is discussed in this 2017 article by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2017ApJ...846..170T&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tauris et al.&lt;/a&gt;. It looks a bit like this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Massive binary star system forms, with moderate separation (&lt;em&gt;not&lt;/em&gt; as close as the final DNS binary will be).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Primary star (initially more massive) evolves faster, to the point where it expands and fill its Roche lobe, leading to mass transfer to the secondary.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This means the primary loses mass while the secondary gains mass. So when the primary reaches the supernova stage, it has less mass in its envelope to eject -- &lt;em&gt;and&lt;/em&gt; the companion star may now be the more massive of the two.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Primary's core collapses, producing first SN. This probably won't destroy the companion; it &lt;em&gt;might&lt;/em&gt; disrupt the system -- not so much because the companion is kicked a lot but because the NS remnant is lower mass than the pre-SN primary star, and it probably gets a kick from explosion asymmetries. But recall that the companion star may now be more massive, so the gravitational disruption is less.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Companion star evolves, expanding to overfill its Roche lobe, leading to a &quot;common envelope&quot; phase where the NS is actually inside the companion star's outer envelope.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Common envelope phase leads to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. Significant shrinking of the orbit as the NS experiences dynamical friction;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. Ejection of much of the companion star's outer envelope.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;System is now much more compact, with a NS and a stripped helium-star companion.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Possible further mass transfer from companion onto neutron star, leading to even less of an envelope around the companion and an even lower mass.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Companion's core collapses, causing second SN. This has little effect on the first NS (i.e., the former primary star), since the NS is a very small target with extreme density and surface gravity: &lt;em&gt;very&lt;/em&gt; hard to disrupt. In addition, the companion has very little envelope left to eject at this point.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Some key points:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The binary starts off with a significantly wider separation than the final DNS will have, so the effects of the first SN on the companion will be less extreme than you might imagine;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mass transfer (and ejection) during the Roche overflow and common envelope stages removes mass from the stars, which means there's less envelope to eject when a SN kicks off (so it does less damage to the other star/NS) and also that the SN ejection of a star's envelope doesn't have a strong effect on the gravitational binding energy, since there's less of an envelope to eject.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This paper by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2015A&amp;amp;A...584A..11L&quot; rel=&quot;nofollow noreferrer&quot;&gt;Liu et al. (2015)&lt;/a&gt; presents some modern simulations of the effect of a SN blast wave on a nearby companion star. It turns out to be pretty hard to disrupt the companion: even for relatively small separations, the companion only loses about 10% of its mass, and this goes down to a few percent or less as the separation increases. (Remember that the companion is itself a massive star most likely still on the main sequence, and so relatively compact, with high surface gravity.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the explosion &lt;em&gt;push&lt;/em&gt; the companion star away? Probably not very much: again, the companion star is itself quite massive -- maybe &lt;em&gt;more&lt;/em&gt; massive than the original primary because of the mass transfer in stage 2 (above) -- so it's hard to push it around.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2018-06-26T22:30:03.180" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26773" PostTypeId="1" CreationDate="2018-06-26T22:51:45.537" Score="2" ViewCount="154" Body="&lt;p&gt;If you go to the following video, at this certain time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/Iy7NzjCmUf0?t=435&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/Iy7NzjCmUf0?t=435&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you will see that the narrator says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;We simply don't know what else is out there because the light from these incredibly distant places has not yet had enough time in the universes history to reach us yet back on Earth. And the light from some places might never reach us at all. Because some parts of space, very far from Earth, are expanding away from us, &lt;strong&gt;faster then the speed of light&lt;/strong&gt;, that means that the light from these places will never, in an infinite amount of time reach Earth&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am I understanding this correctly? Some parts in the universe are expanding faster then the speed of light? Is it yet proven that there are definitely things faster then the speed of light?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like I am misunderstanding the above statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S, try watching the entire video, it will be 10 minutes well spent.&lt;/p&gt;&#xA;" OwnerUserId="17327" LastEditorUserId="16354" LastEditDate="2018-06-27T21:03:40.393" LastActivityDate="2018-06-28T02:39:31.107" Title="Are galaxies expanding away from us at more than the speed of light?" Tags="&lt;galaxy&gt;&lt;light&gt;&lt;expansion&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2018-06-27T21:05:49.540" ContentLicense="CC BY-SA 4.0" />
  <row Id="26774" PostTypeId="2" ParentId="26773" CreationDate="2018-06-27T00:20:30.403" Score="4" Body="&lt;p&gt;You've been confused by a subtle error that seems always to be made when trying to express the speed of light limitation of Relativity in words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's fundamental is something called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lorentz_covariance&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lorentz Symmetry AKA Lorentz Invariance&lt;/a&gt; which is the basis of Relativity. This symmetry is a &lt;em&gt;local&lt;/em&gt; symmetry and it's a strong symmetry of of the physical universe. Basically, what it says is that the laws of physics are independent of velocity. A &lt;em&gt;consequence&lt;/em&gt; of this symmetry is that two objects can't pass each other at a speed greater than that of light. (The reason why is not especially difficult, but best explained using mathematics rather than words.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General Relativity is not a local theory -- it's a theory of the structure of spacetime -- but its local structure is that of Special Relativity, and hence it has Lorentz symmetry.  One of the surprising (utterly surprising -- Einstein himself didn't believe it for many years) consequences of General Relativity is that spacetime isn't stable, but it must either expand or contract.  This expansion is not the galaxies flying further and further into the void like the bits of an exploded hand grenade, but is spacetime itself expanding everywhere (there is no center of expansion) and the galaxies everywhere more-or-less sorta sitting still in the expanding spacetime and being carried along with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you pick two points which are far enough apart, they can be receding from each other at a speed greater than that of light -- or, alternatively and just as correctly -- more than 300,000 km of new space can be created between them each second.  Because nothing is moving past each other at FTL speeds, local Lorentz symmetry is preserved and Einstein is happy and his equations are satisfied and all's well with the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that everything I've said is wrong, but I hope it's not misleadingly wrong. &#xA;You really need to do the math to see what's actually happening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the upshot is that we can't fly to the Moon (or Mars or Alpha Centauri or wherever) and come back at a speed faster than light, but many distant galaxies -- which we will never see or hear from again -- have just passed c moving away from us, and more do so every year.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="616" LastEditDate="2018-06-27T01:10:42.453" LastActivityDate="2018-06-27T01:10:42.453" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26775" PostTypeId="2" ParentId="26773" CreationDate="2018-06-27T03:26:05.180" Score="3" Body="&lt;p&gt;And indeed there is another error in the quoted statement, which is that the statement implies there is something significant about objects whose distance from us is increasing faster than the speed of light, as though the light from those objects would never reach us.  This is incorrect, we routinely see galaxies that have always been separating from us at a rate larger than c-- they were separating faster than that when the light was emitted, and they have only been speeding up since then, yet we see them clearly because a model where space itself is expanding at that rate makes it possible.  It's subtle but routine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, it should also be mentioned that because the expansion is accelerating, light emitted now (meaning at this same cosmological age) from  galaxies that are separating from us faster than c now will never be able to be seen.  But again, there's nothing special about c-- light from galaxies separating from us at a rate less than c right now will also never be seen, it just depends on how fast the acceleration goes from here on.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2018-06-28T02:39:31.107" LastActivityDate="2018-06-28T02:39:31.107" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26777" PostTypeId="2" ParentId="20906" CreationDate="2018-06-27T04:41:31.237" Score="1" Body="&lt;p&gt;I have compiled exactly the kind of list that you're looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stars are identified by their &lt;a href=&quot;http://tdc-www.harvard.edu/catalogs/bsc5.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Yale Bright Star Catalogue&lt;/a&gt; HR designation. This is the best catalogue to use to define the IAU constellations because it is conveniently-sized (only 9,110 stars, down to a magnitude of about 7.5) and all the IAU constellation patterns can be defined in terms of these stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is based on &lt;a href=&quot;https://www.iau.org/public/themes/constellations/&quot; rel=&quot;nofollow noreferrer&quot;&gt;canonical IAU constellations&lt;/a&gt;. I also throw in J2000 positions, magnitudes and names for each of the stars in each pair of the constellation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, Cassiopeia appears as 4 lines (below shown wrapped)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Cas 21  2.294583    59.149722   2.27    Caph|bet Cas|11 Cas &#xA;    168 10.127083   56.537222   2.23    Schedar|alf Cas|18 Cas&#xA;Cas 168 10.127083   56.537222   2.23    Schedar|alf Cas|18 Cas  &#xA;    264 14.177083   60.716667   2.47    BU 499A|BU 1028|gam Cas|27 Cas&#xA;Cas 264 14.177083   60.716667   2.47    BU 499A|BU 1028|gam Cas|27 Cas  &#xA;    403 21.454167   60.235278   2.68    Ruchbah|BUP 19A|del Cas|37 Cas&#xA;Cas 403 21.454167   60.235278   2.68    Ruchbah|BUP 19A|del Cas|37 Cas  &#xA;    542 28.598750   63.670000   3.38    Segin|eps Cas|45 Cas&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mkweb.bcgsc.ca/universe-voids-and-superclusters/constellation.shapes.iau.txt&quot; rel=&quot;nofollow noreferrer&quot;&gt;Download my list here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the asterisms defined by the IAU may be slightly different from what you see in planetarium programs or star charts. However, the IAU should be seen as an authoritative source and any variation on the patterns is up to the &quot;poetic license&quot; of the user.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of these stars (possibly all, I haven't actually checked) will have a &lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR-3?-source=I/239/hip_main&amp;amp;-out.max=50&amp;amp;-out.form=HTML%20Table&amp;amp;-out.add=_r&amp;amp;-out.add=_RAJ,_DEJ&amp;amp;-sort=_r&amp;amp;-oc.form=sexa&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hipparcos&lt;/a&gt; (HIP) and/or &lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR-3?-source=I/259/tyc2&amp;amp;-out.max=50&amp;amp;-out.form=HTML%20Table&amp;amp;-out.add=_r&amp;amp;-out.add=_RAJ,_DEJ&amp;amp;-sort=_r&amp;amp;-oc.form=sexa&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tycho&lt;/a&gt; (TYC) index entry as well and also&lt;/p&gt;&#xA;" OwnerUserId="23484" LastActivityDate="2018-06-27T04:41:31.237" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26778" PostTypeId="2" ParentId="26139" CreationDate="2018-06-27T09:02:10.547" Score="4" Body="&lt;p&gt;I have a used 2.4 metre mesh C band dish that I picked up for free that I will be converting for observing the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrogen_line&quot; rel=&quot;nofollow noreferrer&quot;&gt;21cm hydrogen line @ 1420MHz&lt;/a&gt;. I was lucky with this dish as it's in as new condition, but you need to be aware of any rusting and damage to the mesh that will distort your data.  I'm mounting mine in my backyard on a 75mm steel pole 2 metres above the ground pointing straight up.  This is to do &lt;a href=&quot;https://www.assa.org.au/resources/radio-astronomy/meridian-drift-scanning/&quot; rel=&quot;nofollow noreferrer&quot;&gt;meridian drift scans&lt;/a&gt;, so you don't need to have a movable dish, the planet's rotation does this for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydrogen line can be used to observe deep space objects which emit strong hydrogen line signals.  Observing this spectrum from the Milky Way is one example, as well as say, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cygnus_A&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cygnus A galaxy&lt;/a&gt; and others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The old large (2 to 3 metres or so) C band dishes once used for satellite TV, now largely replaced by smaller Ku band dishes, can be picked up for about AU$100 or even for free from household backyards.  These are used in a number of amatuer radio astronomy projects, especially for the 21 cm line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have the dish, you will typically need to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Replacing the Low Noise Block (LNB)&lt;/strong&gt;. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Low-noise_block_downconverter&quot; rel=&quot;nofollow noreferrer&quot;&gt;LNB&lt;/a&gt; is attached to the top of the dish struts.  You will need to replace this with one made specifically for 1420MHz.  You can make your own LNB with details at &lt;a href=&quot;http://www.setileague.org/hardware/feedchok.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.setileague.org/hardware/feedchok.htm&lt;/a&gt;.  On that site there is an Excel spreadsheet with variables for adjusting some of the measurements of the LNB.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The LNB for the hydrogen line basically consists of an aluminium tube (waveguide) capped at one end.  Inside the tube is the antenna probe, which is just a brass rod - length and placement varies according to the projects I have seen, but the SETI guide above should be OK.  The probe is soldered to the centre pin of the coaxial cable connector fitted to the waveguide.  Again, refer to the SETI page above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also purchase one ready made from &lt;a href=&quot;https://www.radioastronomysupplies.com/store/p22/1420_MHz._CYLINDRICAL_FEEDHORN_AND_CHOKE.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.radioastronomysupplies.com/store/p22/1420_MHz._CYLINDRICAL_FEEDHORN_AND_CHOKE.html&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;You will need a Low Noise Amplifier (LNA) for 1420MHz.&lt;/strong&gt; The LNA will need a Gain of &gt; 30dB and a Noise Figure (NF) of somewhere around 0.3dB or lower.  The higher the Gain (sensitivity) and the lower the NF the better, though obviously at a price.  The LNA should be mounted on the coaxial cable connected to the LNB antenna probe inside the LNB waveguide. The closer the better. I have no connections with Radio Astronomy Supplies, but they also have what appears to be a decent LNA for the hydrogen line:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.radioastronomysupplies.com/store/p9/1420_MHz._HIGH_PERFORMANCE_LNA.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.radioastronomysupplies.com/store/p9/1420_MHz._HIGH_PERFORMANCE_LNA.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.ebay.com.au/itm/LNA-21cm-1420-MHz-N-female-/113010650218?hash=item1a4ff5ac6a%3Ag%3ACuwAAOSwDMNbBXhA&amp;amp;nma=true&amp;amp;si=gILuRaA7xRatFaHflv7C6kI31Pw%253D&amp;amp;orig_cvip=true&amp;amp;rt=nc&amp;amp;_trksid=p2047675.l2557&quot; rel=&quot;nofollow noreferrer&quot;&gt;Another LNA made for 1420MHz&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A receiver.&lt;/strong&gt;  The receiver allows you to interpret the signal coming down from the LNA.  I have purchased a cheapish (AU$30) Software-defined Radio (SDR) USB dongle for my setup which will act as the receiver.  In particular a &lt;a href=&quot;https://www.rtl-sdr.com/buy-rtl-sdr-dvb-t-dongles/&quot; rel=&quot;nofollow noreferrer&quot;&gt;RTL-SDR Blog R820T2 RTL2832U 1PPM TCXO SMA Software Defined Radio&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One example of such usage is at &lt;a href=&quot;https://www.rtl-sdr.com/hydrogen-line-observation-with-an-rtl-sdr/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.rtl-sdr.com/hydrogen-line-observation-with-an-rtl-sdr/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More discussion on SDR for observing the hydrogen line is at h&lt;a href=&quot;https://www.rtl-sdr.com/rtl-sdr-for-budget-radio-astronomy/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ttps://www.rtl-sdr.com/rtl-sdr-for-budget-radio-astronomy/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The SDR dongle connects to the coaxial line from the LNA. You can then plug the SDR dongle into your computer's USB port. Beware of the length of coaxial line as longer lines will lose data.  An alternative is discussed below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Software to observe the data.&lt;/strong&gt;  There are a number of open-source applications for SDR reception.  Possibly the most popular for SDR - radio astronomy is &lt;a href=&quot;https://www.rtl-sdr.com/sdrsharp-users-guide/&quot; rel=&quot;nofollow noreferrer&quot;&gt;SDR#&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Using a Raspberry Pi 3 B+ as a server from the dish.&lt;/strong&gt;  Alternative to using coaxial cable from the LNA to SDR at the computer is to have a &lt;a href=&quot;https://www.raspberrypi.org/blog/raspberry-pi-3-model-bplus-sale-now-35/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Raspberry Pi 3 B+&lt;/a&gt; (RPi) act as a server to send the data to computer via Ethernet cable, rather than coaxial cable.  This has a number of possible advantages including much less or no data loss depending on the cable and length.  I will be using Cat6 cable up to around 20 to 30 metres.  The cable plugs into the RPi RJ45 Ethernet port.  The SDR dongle plugs into a RPi USB port.  The LNA attaches to the SDR dongle directly via the coaxial connectors/adapters.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This setup can be mounted on the mounting pole for the dish contained in a weatherproof and ventilated box, something like &lt;a href=&quot;https://www.desertcart.ae/products/23590912-outdoor-poe-interface-enclosure-box-with-cooling-fan-48-volt-dc-hyperlink-poe-passive-injectors-weatherproof-cabinet-box-waterproof&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;  You would then need to think about powering this setup.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently I'm looking at &lt;a href=&quot;https://en.wikipedia.org/wiki/Power_over_Ethernet&quot; rel=&quot;nofollow noreferrer&quot;&gt;Power Over Ethernet&lt;/a&gt; (POE) to the RPi 3 B+, possibly using the &lt;a href=&quot;https://www.raspberrypi.org/products/poe-hat/&quot; rel=&quot;nofollow noreferrer&quot;&gt;RPi POE HAT&lt;/a&gt; when it's released this year.  Then you could take power from the RPi and use &lt;a href=&quot;https://core-electronics.com.au/pololu-9v-step-up-voltage-regulator-u3v12f9.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;boost converter&lt;/a&gt; to 9v or 12v to power the LNA of choice. as well as any 5V cooling fans you have in your box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then when connect to the RPi from your computer (e.g. using SSH) you should be set to receive data.  The other advantage to this setup is that since the RPi is acting as a server connected to the internet, you can access your dish from anywhere in the world with an internet connection.  There is some discussion of this &lt;a href=&quot;https://roughlea.wordpress.com/raspberry-pi-experiences/configuring-the-raspberry-pi-as-an-ssh-server/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, &lt;a href=&quot;https://thepi.io/how-to-ssh-into-the-raspberry-pi/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://radioadastra.blogspot.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23153" LastEditorUserId="23153" LastEditDate="2018-06-29T08:59:46.837" LastActivityDate="2018-06-29T08:59:46.837" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26779" PostTypeId="2" ParentId="26675" CreationDate="2018-06-27T11:24:19.587" Score="1" Body="&lt;p&gt;Following Mick's comment, I found an astropy-function - DAOStarFinder - that does what I was looking for. It scans an image for light-sources and one can set threshhold for a detection, e.g. the &quot;roundness&quot;, &quot;sharpness&quot; or brightness of an object have to be between [-0.05,0.05], [0.6,0.8] or &gt;5*standard_deviation, respectively. I must say, that I have the impression the roundness and sharpness measure don't work consistently and sometimes values for the same object change enormously from one frame to the other. But DAOStarFinder gives you also a rough estimate of the flux density of the object, which I could use for sorting the sources and filtering out the relevant one. Here is a link to &lt;a href=&quot;http://photutils.readthedocs.io/en/stable/detection.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;example script&lt;/a&gt; and to the &lt;a href=&quot;http://photutils.readthedocs.io/en/stable/api/photutils.DAOStarFinder.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;documentation&lt;/a&gt; of DAOStarFinder. And here the example script explicitly:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt;&amp;gt;&amp;gt; from astropy.stats import sigma_clipped_stats&#xA;&amp;gt;&amp;gt;&amp;gt; from photutils import datasets&#xA;&amp;gt;&amp;gt;&amp;gt; from photutils import DAOStarFinder&#xA;&#xA;&amp;gt;&amp;gt;&amp;gt; hdu = datasets.load_star_image()    &#xA;&amp;gt;&amp;gt;&amp;gt; data = hdu.data[0:400, 0:400]    &#xA;&amp;gt;&amp;gt;&amp;gt; mean, median, std = sigma_clipped_stats(data, sigma=3.0, iters=5)&#xA;&amp;gt;&amp;gt;&amp;gt; daofind = DAOStarFinder(fwhm=3.0, threshold=5.*std)    &#xA;&amp;gt;&amp;gt;&amp;gt; sources = daofind(data - median)    &#xA;&#xA;&amp;gt;&amp;gt;&amp;gt; print(sources)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="23399" LastActivityDate="2018-06-27T11:24:19.587" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26780" PostTypeId="1" CreationDate="2018-06-27T15:10:42.440" Score="52" ViewCount="14680" Body="&lt;p&gt;The light from the moon is light being reflected from the sun. The sun, in space, is white. But on Earth, when the light is filtered through an atmosphere, the light appears yellow. So then, why is moonlight white through the atmosphere?&lt;/p&gt;&#xA;" OwnerUserId="23524" OwnerDisplayName="Andrew Johnson" LastEditorUserId="7982" LastEditorDisplayName="user14720" LastEditDate="2021-02-28T11:41:36.677" LastActivityDate="2021-02-28T11:41:36.677" Title="Why is moon light not the same color as sunlight?" Tags="&lt;the-sun&gt;&lt;the-moon&gt;&lt;atmospheric-effects&gt;" AnswerCount="5" CommentCount="8" FavoriteCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26781" PostTypeId="1" CreationDate="2018-06-27T16:11:26.167" Score="1" ViewCount="53" Body="&lt;p&gt;I have an analog question of this one : &lt;a href=&quot;https://astronomy.stackexchange.com/questions/22388/generate-an-uniform-distribution-on-the-sky&quot;&gt;Generate an uniform distribution on the sky&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I generate an uniform distribution on the sky, but just a patch of it, with $$\alpha_{min} \leq \alpha \leq \alpha_{max}$$ and &#xA;$$\delta_{min} \leq \delta \leq \delta_{max}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for Right Ascention, I just have to take uniform distribution between $\alpha_{min}$and $ \alpha_{max}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For declination it works if I take $$\delta_{random} = \sin^{-1}\left( \mathrm{Uniform}(\sin\delta_{min},\, \sin\delta_{max}) \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is that I don't know how to demonstrate this formula for the generation of delinations. Any idea how to make the proof of this formula ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw in &lt;a href=&quot;https://astronomy.stackexchange.com/questions/22388/generate-an-uniform-distribution-on-the-sky&quot;&gt;Generate an uniform distribution on the sky&lt;/a&gt; @RobJeffries post which gives an expression of $P(\delta)$ and demonstrates that $\delta = \sin^{-1}(2P-1)$ where $P$ is a random number between 0 and 1, which is compatible to my formula with $\delta_{min} = -\frac{\pi}{2}$ and $ \delta_{max} = +\frac{\pi}{2}$, but I was not able to generalize his demonstration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for your help!&lt;/p&gt;&#xA;" OwnerUserId="23047" LastEditorUserId="23047" LastEditDate="2018-06-28T12:54:58.607" LastActivityDate="2018-06-28T12:54:58.607" Title="Generate an uniform distribution on the sky between limits" Tags="&lt;coordinate&gt;&lt;positional-astronomy&gt;&lt;mathematics&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26782" PostTypeId="1" AcceptedAnswerId="26791" CreationDate="2018-06-27T16:47:46.853" Score="1" ViewCount="671" Body="&lt;p&gt;Our solar system travels at an average speed of 515,000 mph and our galaxy at 1.3 million mph through space. What is the fastest moving body in space recorded and the fastest can a black hole or galaxy travel through the universe?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="2531" LastEditDate="2018-06-30T00:27:44.870" LastActivityDate="2018-07-03T13:44:26.890" Title="What is the fastest a galaxy or other body has been measured to (or can) travel?" Tags="&lt;black-hole&gt;&lt;galaxy&gt;&lt;astrophysics&gt;&lt;space-time&gt;&lt;velocity&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26783" PostTypeId="1" CreationDate="2018-06-27T17:05:20.090" Score="8" ViewCount="281" Body="&lt;p&gt;Could two galaxies (one big and one small)intersect at a velocity to allow the smaller black hole to escape but not the galaxy around it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mheGW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mheGW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="6" LastEditDate="2018-06-28T19:19:22.293" LastActivityDate="2018-06-30T11:49:57.587" Title="Can &quot;rogue&quot; supermassive black holes be made this way?" Tags="&lt;black-hole&gt;&lt;galaxy&gt;&lt;mass&gt;&lt;velocity&gt;&lt;density&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26786" PostTypeId="2" ParentId="26780" CreationDate="2018-06-27T17:13:39.853" Score="63" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The light from the moon is light being reflected from the sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is at least one reason you should &lt;em&gt;not&lt;/em&gt; expect the Moon to have the same color.  Sunlight hitting an e.g. blue object would appear predominantly blue and similar.  So the color we might expect to see from the Moon is going to be adjusted by the color properties of the Moon's surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it happens &lt;a href=&quot;http://meteorites.wustl.edu/lunar/apollo_regolith_breccias.htm&quot; rel=&quot;noreferrer&quot;&gt;lunar regolith&lt;/a&gt; is essentially gray in color (i.e. neutral), so it more or less reflects the spectrum it receives.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The sun, in space, is white&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No it is not.  There are a couple of reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun is an exceptionally bright object (for us) and will saturate our color vision so that it seems like it's white when we look at it directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Color_index&quot; rel=&quot;noreferrer&quot;&gt;Sun is actually a yellow star&lt;/a&gt;.  It's spectrum peaks in the (visual) blue-green range.  There are stars that have a red or blue color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;White is a tricky one.  White is a perceived color.  There is no &quot;white&quot; wavelength in the EM spectrum.  White is a balance between the three different color sensors your human vision uses.  The Sun itself is not white because the color combination &quot;white&quot; we're balanced for is based not on the light from the Sun in the vacuum of space, but on the light that reaches the surface of the Earth and also what is reflected from the Earth (which is part of the ambient lighting we normally experience).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;why is moonlight white through the atmosphere?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seemed a nice clear sky last night around my neck of the woods and the Moon looked yellow to me.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Sunlight#Spectral_composition_of_sunlight_at_Earth&amp;#39;s_surface&quot; rel=&quot;noreferrer&quot;&gt;This will vary&lt;/a&gt; with your location and atmospheric conditions, but also with your personal perception of color.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2018-06-27T17:13:39.853" CommentCount="15" ContentLicense="CC BY-SA 4.0" />
  <row Id="26787" PostTypeId="2" ParentId="26780" CreationDate="2018-06-27T17:51:12.887" Score="42" Body="&lt;p&gt;Reflected moonlight is actually slightly reddened compared with the incident solar spectrum (&lt;a href=&quot;http://iopscience.iop.org/0031-9120/48/3/360/article&quot; rel=&quot;noreferrer&quot;&gt;Ciocca &amp;amp; Wang 2013&lt;/a&gt;). That same light is then transmitted through out atmosphere in exactly the same way as sunlight.&lt;/p&gt;&#xA;&lt;p&gt;Any phenomenology (which appears disputed) as per the claim in the question is purely down to the nuances of our colour perception.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/72dkh.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/72dkh.png&quot; alt=&quot;The reflected moonlight spectrum&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4wtWq.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4wtWq.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Figure 8: Averaged geometrical moon albedos measured by GOME from July 1995, November 1995, and September 1996.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;From ESA's &lt;a href=&quot;https://earth.esa.int/workshops/ers97/papers/dobber/index-2.html&quot; rel=&quot;noreferrer&quot;&gt;GOME moon measurements, including instrument characterisation and moon albedo&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-01-25T08:04:56.077" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26788" PostTypeId="2" ParentId="26783" CreationDate="2018-06-27T19:44:55.457" Score="4" Body="&lt;p&gt;Yes, and in fact a mechanism somewhat like this has probably dumped a large number of BHs into intergalactic space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes tend to settle towards the center of galaxies (an effect of dynamical friction). As they settle, they &quot;cool&quot; by evaporation. The chaos of BHs orbiting the center of mass all interact, especially when two of them approach closely. Depending on the geometry of the near-miss, one BH can gain energy at the expense of the other. One swings into a larger orbit and the other goes into a smaller orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometime the larger orbit is hyperbolic and the BH is thrown right out of the galaxy.  This removes orbital energy from the assemblage of BHs and the whole thing shrinks a bit and encounters become a bit more common.  In the end, many of the original set of BHs are thrown out into intergalactic space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many?  No one knows yet. We have good evidence of a single very large BH (&gt;10&lt;sup&gt;6&lt;/sup&gt; solar masses) at the center of the Milky Way, but recent results have suggested there may be as many as 10,000 smaller BHs (~10 solar masses each) in orbit around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the latter is right, there may be a &lt;em&gt;lot&lt;/em&gt; of BHs wandering intergalactic space!&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-06-27T19:44:55.457" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26789" PostTypeId="2" ParentId="26780" CreationDate="2018-06-27T21:15:51.397" Score="25" Body="&lt;p&gt;Surprisingly, the moonlight is actually &lt;strong&gt;slightly warmer color&lt;/strong&gt; than sunlight, as the moon reflectance is higher for longer wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet, on clear nights, with the full moon high in the sky (as little atmospheric influence as possible) the landscape around us appears &lt;strong&gt;blueish&lt;/strong&gt;, because of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Purkinje_effect&quot; rel=&quot;noreferrer&quot;&gt;Purkinje effect&lt;/a&gt;: at low illumination levels our red color sensitivity decreases (as our vision system gradually switches from daylight (cones) to night time vision (using rod cells)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, the moon surface is bright enough and its color can be seen correctly (not distorted by the Purkinje effect). &lt;strong&gt;So how how does it relate to the question&lt;/strong&gt;, where we ask about the moonlight color? In my opinion, we tend to perceive the general impression of the color by looking at the surroundings, not just the light source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;That's why the sunlight looks &quot;warm&quot; (more yellow) and the moonlight looks &quot;cold&quot;&lt;/strong&gt; (more blue - because our color perception is changed by the insufficient light level), even though the &lt;em&gt;real colors&lt;/em&gt; are pretty much the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this is can even become a real problem when taking very long exposure photos at night! They look almost like daylight photos, destroying the intended atmosphere of mystery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objectively, the photograph is correct, but it is &lt;strong&gt;not what we see with our own eyes&lt;/strong&gt;. Add some blue tint and the night feeling is back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: (note the stars, this is certainly not a daylight photo!)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/qh5gD.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qh5gD.jpg&quot; alt=&quot;full moon&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.flickr.com/photos/thundi/5371654787/in/photostream/&quot; rel=&quot;noreferrer&quot;&gt;Abrahim Asad - Mathimago fanni Athiri, Fuvahmulah Beach on Full Moon Night LongExposure&lt;/a&gt; (CC BY-ND 2.0)&lt;/p&gt;&#xA;" OwnerUserId="7654" LastEditorUserId="7654" LastEditDate="2018-07-01T13:09:04.303" LastActivityDate="2018-07-01T13:09:04.303" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26790" PostTypeId="2" ParentId="26783" CreationDate="2018-06-27T21:36:57.390" Score="10" Body="&lt;p&gt;I assume you're asking about central supermassive black holes (SMBHs, one per galaxy), not stellar-mass black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is yes, but what actually happens is the two SMBHs have to merge first, and then the resulting &lt;em&gt;combined&lt;/em&gt; SMBH can sometimes be ejected from the combined (merged) galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edited to add: Since you've updated the question with a series of diagrams, I should state explicitly that the scenario suggested by the diagrams -- stars in smaller galaxy merge into big galaxy, but SMBH continues on almost unaffected -- is not physically possible. Most of the stars from the smaller galaxy will not end up in the center of the big galaxy, but because of &lt;a href=&quot;https://en.wikipedia.org/wiki/Dynamical_friction&quot; rel=&quot;nofollow noreferrer&quot;&gt;dynamical friction&lt;/a&gt;, the SMBH &lt;em&gt;will&lt;/em&gt;.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.nasa.gov/feature/goddard/2017/gravitational-wave-kicks-monster-black-hole-out-of-galactic-core&quot; rel=&quot;nofollow noreferrer&quot;&gt;This NASA press release&lt;/a&gt; from 2017 describes the discovery of a quasar apparently ejected from a recently merged galaxy. I'll go ahead and quote their description of the suggested mechanism (this possibility has been suggested by theoretical studies going back at least ten or fifteen years):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;According to their theory, two galaxies merge, and their black holes&#xA;  settle into the center of the newly formed elliptical galaxy. As the&#xA;  black holes whirl around each other, gravity waves are flung out like&#xA;  water from a lawn sprinkler. The hefty objects move closer to each&#xA;  other over time as they radiate away gravitational energy. If the two&#xA;  black holes do not have the same mass and rotation rate, they emit&#xA;  gravitational waves more strongly along one direction. When the two&#xA;  black holes collide, they stop producing gravitational waves. The&#xA;  newly merged black hole then recoils in the opposite direction of the&#xA;  strongest gravitational waves and shoots off like a rocket.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since most massive galaxies -- including those that have undergone major mergers in the past -- have a SMBH in their center, the gravitational recoil usually isn't strong enough to eject the SMBH; instead, the SMBH loses energy to the stars in the inner part of the merged galaxy via dynamical friction, and settles back into the center. But it appears that sometimes there's enough of a kick to allow the SMBH to escape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is that if two galaxies merge and their SMBHs form a binary, and then &lt;em&gt;another&lt;/em&gt; galaxy (with its own SMBH) merges &lt;em&gt;before&lt;/em&gt; the previous two SMBHs have actually merged, then you can have a three-body interaction between the late-arrival SMBH and the binary SMBH, which could result in one of the SMBHs being ejected. But this requires the right timing, and probably doesn't happen very often.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2018-06-30T11:49:57.587" LastActivityDate="2018-06-30T11:49:57.587" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26791" PostTypeId="2" ParentId="26782" CreationDate="2018-06-27T22:14:07.400" Score="4" Body="&lt;h3&gt;Hubble velocities&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The Universe expands and carries galaxies away from each other with a relative velocity proportional to the distance between them. This is Hubble's law, and if the Universe is infinitely large, there is no limit to how fast two galaxies may recede from each other. In our observable Universe, the most distant galaxy observed (GN-z11; &lt;a href=&quot;https://arxiv.org/abs/1507.02679&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oesch et al. 2016&lt;/a&gt;) recedes from the Milky Way at more than twice the speed of light ($c\simeq300\,000\,\mathrm{km}\,\mathrm{s}^{-1}$).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Peculiar velocities&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;However, this is not a motion &lt;em&gt;through space&lt;/em&gt;, which seems to be what you are asking about. Nothing is able to travel through space faster than the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxies generally move through space with velocities of the order of a few $100\,\mathrm{km}\,\mathrm{s}^{-1}$; in massive clusters, they may acquire velocities of a few $1000\,\mathrm{km}\,\mathrm{s}^{-1}$ (e.g. &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0611945&quot; rel=&quot;nofollow noreferrer&quot;&gt;Karachentsev et al. 2006&lt;/a&gt;). These so-called &lt;em&gt;peculiar&lt;/em&gt; velocities are wrt. to the &quot;global&quot; frame, i.e. the frame in which an observer that follows the cosmic expansion lies still. This is also the frame in which the cosmic microwave background is (statistically) the same in all directions.$^\dagger$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes, stars, planets, and gas cloud move around in the reference frame of their galaxies with characteristic velocities of the order of $100\,\mathrm{km}\,\mathrm{s}^{-1}$, so this will typically be smaller than that of their host galaxy, if the galaxy lies in a large cluster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When massive celestial bodies such as neutron stars or black holes merge, they reach velocities of the order of half the speed of light, as discussed further in Rob Jeffries' answer. This is as measured in the frame of the center of mass.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Velocities of particles&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;EDIT: The following was written before the question was changed to emphasize &lt;em&gt;celestial&lt;/em&gt; bodies:&#xA;Much higher velocities are found for small particles such as cosmic rays; these are massive particles accelerated to extreme energies, e.g. by supernova explosions. The record holder of such particles is still (I think) the so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Oh-My-God_particle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oh-My-God particle&lt;/a&gt; which traveled at $0.9999999999999999999999951c$ (in Earth's frame), much faster than the velocities reached in the Large Hadron Collider.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^\dagger$&lt;sub&gt;Of course you can always define a frame that moves with almost the speed of light, such that all object acquire that speed &quot;artificially&quot;, but then the Universe isn't isotropic anymore.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-06-30T09:53:03.467" LastActivityDate="2018-06-30T09:53:03.467" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26792" PostTypeId="2" ParentId="26782" CreationDate="2018-06-27T22:32:48.787" Score="3" Body="&lt;p&gt;If we are talking about any significant celestial bodies, then it will be the relative velocities of a pair of merging black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The typical relative velocity due to the orbital speed of the black hole components just before merger is greater than half the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing comes close to that on a galactic scale. Typical peculiar velocities for galaxies in a cluster are a few hundred to a few thousand km/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The plot below shows the inferred relative velocities (bottom) of merging black holes from the first LIGO gravitational wave detection.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qQwUS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qQwUS.png&quot; alt=&quot;Merging black holes&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-06-27T22:32:48.787" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26793" PostTypeId="1" CreationDate="2018-06-28T00:26:15.837" Score="1" ViewCount="156" Body="&lt;p&gt;If there was a massive enough planet outside of earth's orbit to affect it, how would the earth's velocity change? I know it would be small, but would the velocity increase or decrease?&lt;/p&gt;&#xA;" OwnerUserId="18850" LastActivityDate="2018-08-06T07:58:06.833" Title="How does the gravity of one planet affect the velocity of other planets?" Tags="&lt;gravity&gt;&lt;solar-system&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26794" PostTypeId="2" ParentId="26793" CreationDate="2018-06-28T02:47:24.240" Score="1" Body="&lt;p&gt;Sometimes it would increase and sometimes it would decrease, depending on where the other planet was.  But we do have examples already of planets affecting other planets, for example the orbit of Mercury precesses around the Sun due to the gravity of the other planets (largely Jupiter), as well as due to general relativity.  The Earth's orbit is also already affected by the other planets, but even Jupiter has 1/1000 the mass of the Sun and is much farther away, so the effects are tiny.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-06-28T02:47:24.240" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26795" PostTypeId="1" AcceptedAnswerId="26820" CreationDate="2018-06-28T08:16:10.250" Score="3" ViewCount="274" Body="&lt;p&gt;I found a &lt;a href=&quot;http://www.isthe.com/chongo/tech/astro/HR-temp-mass-table-byhrclass.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;table&lt;/a&gt; that was based on data from 2003 for Stellar Classification, and I can't get to the given data source. I plan on using the table as a basis for a program that spits out random star systems with stars +-5% of the given data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know type R was renamed to C so I am already throwing that out. Is there anything else I should ignore from this table or add? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or this table is crap and should not be used as there are glaring errors. &lt;/p&gt;&#xA;" OwnerUserId="23504" LastEditorUserId="23504" LastEditDate="2018-10-31T03:26:00.783" LastActivityDate="2018-10-31T03:26:00.783" Title="Stellar Classification Table and Star RNG" Tags="&lt;star&gt;&lt;mathematics&gt;&lt;resource-request&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26796" PostTypeId="2" ParentId="26793" CreationDate="2018-06-28T09:29:24.390" Score="2" Body="&lt;p&gt;The most famous example of this lead to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Discovery_of_Neptune#Irregularities_in_Uranus&amp;#39;s_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;discovery of Neptune&lt;/a&gt;.   Planetary orbits were neatly worked out by Kepler and Newton, so they had a good idea of how fast planets should be moving.   Uranus wasn't moving as fast as it should have, and that lead to the conclusion that there was an 8th planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the two planets orbit in a similar plane and don't cross, the inner planet moves faster so it's always catching up to and passing the outer planet.   When it's behind the planet and catching up, it moves faster, when it's ahead, it moves slower.   The effect is, as you stated, very very small but in the case of Uranus, with an orbital period of 84 years, so it spends a couple decades being accelerated by Neptune until it passes it, then it spends a similar amount of time being slowed down by it.   The observed speeding up and slowing down over years was enough to lead to a calculation of the approximate location of, and shortly after, the discovery of Neptune.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 3-body problem is complicated and the math involved is somewhat advanced, but you can do a rough approximation of acceleration and altered distance from the expected orbit using Newton's laws of gravitational acceleration g=GM/r^2 over roughly a fraction of the orbital period.  It's surprisingly small, but over a couple months (for Earth) or a couple decades (for Uranus) it works out to about what you might expect.  A noticeable but tiny fraction of the orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over time, these variations balance out, so the orbits remain largely unchanged over thousands, millions and even billions of years, unless the planets happen to fall into resonance.  Resonance can lead to the gradual destabilizing of orbits. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-06-28T09:29:24.390" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26797" PostTypeId="2" ParentId="13375" CreationDate="2018-06-28T10:01:35.493" Score="3" Body="&lt;p&gt;Look at J2000 Solar System Barycentric (spatial reference) and Barycentric Dynamical Time (TDB).  Together they are good space and time coordinate system.  NASA/JPL has some good info and data on these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of time, TDB is rescaled so that from the Earth, it appears to be close to the same as TT (roughly ~UTC). The rescaling is due to the fact we are in a gravitational well, as well as moving at 30 km/s with respect to the solar system barycentre, due to relativity. The original unscaled is called Barycentric Coordinate Time (TCB) and differs by ~0.5seconds / year.&lt;/p&gt;&#xA;" OwnerUserId="23322" LastActivityDate="2018-06-28T10:01:35.493" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26798" PostTypeId="2" ParentId="13375" CreationDate="2018-06-28T11:29:37.300" Score="1" Body="&lt;p&gt;We need a system to describe &quot;where things are in the sky&quot;. Even a cursory glance at the sky will find that &quot;things move across the sky daily&quot;. So instead of describing where something is directly, we will describe where it is relative to the stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However some of the stars move (due to their actual motion relative to the sun) and appear to wobble (due the motion of the Earth around the sun). So let us consider those objects that are so far that any such motion is undetectable. For example, quasars. Other distant stars are also suitable as they don't have a measurable motion. I'll call these sources &quot;fixed stars&quot;. The goal is to describe a system of coordinates in which the fixed stars don't move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For our coordinate system we will use the plane of the Earth's equator on the March Equinox (chosen in part so the plane passes through the sun). Declension is defined as the angle relative to this plane. Right Ascension is then the angle between the line through the Earth and Sun, and the line formed by projecting the object onto the plane. For very distant objects, it doesn't matter if we use the sun or the Earth as the centre as the angle will be the same to any reasonable level of accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, choosing the March equinox in this way causes a problem, because the plane of the Earth's equator is slowly changing, and this means that the position relative to this plane will also slowly change. The RA and Dec of a quasar will slowly change due to this precession.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution to this issue is to define the coordinate system on a particular date &quot;Jan 1st 2000&quot;. With this convention we can assign a position to the quasar, and it won't change. This is a coordinate system that can describe the position of any object relative to the fixed stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can now define the position of any object in the same coordinates. For nearby stars we can describe their proper and apparent motion relative to this coordinate system. For planets the position relative to the fixed stars varies from day to day, due to the relative motion of the planets. It also depends on the location of the observer. So I can talk about the location of Jupiter at midnight on 28 June 2018, from Perth, WA, using the J2000.0 coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 2000.0 defines exactly which fixed coordinate system we are using. But to describe the location of Jupiter in the sky we also need to use an observation time and date and location. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-06-28T11:29:37.300" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26799" PostTypeId="1" AcceptedAnswerId="26804" CreationDate="2018-06-28T16:01:57.150" Score="1" ViewCount="383" Body="&lt;p&gt;I've been playing Universe Sandbox for a while now and I have never been able to make a near Earth asteroid collide with it. By &quot;&lt;em&gt;make&lt;/em&gt;&quot; I mean, simulating billions of years of known, near Earth objects until one of their orbits intersect. In &lt;a href=&quot;https://astronomy.stackexchange.com/questions/16395/is-universe-sandbox-2-realistic&quot;&gt;this explanation&lt;/a&gt; Sir Cumference describes that Universe Sandbox 2 uses Newtonian physics to in its simulations. But wouldn't that calculate enough pull from Earth to slightly change the trajectory of the NEO to eventually impact?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know, with almost 100% certainty that a LARGE NEO will impact Earth (probably sooner than later), otherwise why would we be investing so heavily on &lt;a href=&quot;https://www.jpl.nasa.gov/asteroidwatch/&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Jet Propulsion Lab's Asteroid Watch program&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the game not accurately simulate gravitational pull on those large scales or is there something else (probably a lot) I'm missing? Are all known NEOs not projected to have a relevant collision intersection with Earth?&lt;/p&gt;&#xA;" OwnerUserId="23510" LastEditorUserId="7982" LastEditDate="2021-03-08T00:02:35.510" LastActivityDate="2021-03-08T00:02:35.510" Title="Universe Sandbox 2 accurate representation of gravity" Tags="&lt;gravity&gt;&lt;near-earth-object&gt;&lt;n-body-simulations&gt;&lt;newtonian-gravity&gt;&lt;universe-sandbox&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26800" PostTypeId="1" CreationDate="2018-06-28T20:05:17.850" Score="0" ViewCount="174" Body="&lt;p&gt;Is &lt;a href=&quot;https://en.wikipedia.org/wiki/Limb_darkening&quot; rel=&quot;nofollow noreferrer&quot;&gt;limb darkening&lt;/a&gt; of the Sun because at centre we are directly looking the core and different zones, whereas at limbs we are seeing only the photospheric part (and less beneath photospheric part)? Or there is some other physics behind this phenomenon?&lt;/p&gt;&#xA;" OwnerUserId="23450" LastEditorUserId="616" LastEditDate="2018-06-30T00:36:16.913" LastActivityDate="2018-06-30T00:36:16.913" Title="Is Limb Darkening because at centre we are directly looking core, whereas at limbs we are seeing only photospheric part?" Tags="&lt;the-sun&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26801" PostTypeId="2" ParentId="26800" CreationDate="2018-06-28T20:24:45.430" Score="2" Body="&lt;p&gt;We can't see the core, but we can see deeper into the photosphere of the sun. The deeper layers of the photosphere are hotter and brighter, and this is the cause of limb darkening.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-06-28T20:24:45.430" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26802" PostTypeId="2" ParentId="26799" CreationDate="2018-06-28T22:52:46.393" Score="1" Body="&lt;p&gt;Impact probabilities are &lt;em&gt;probabilities&lt;/em&gt; because there are uncertainties in the orbits themselves, as well as in the simulation or propagation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words the measurements used to produce the starting points have plenty of observational uncertainties, and the error in the simulation grows due to these uncertainties grows the longer you run it for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even if you had a &quot;perfect&quot; simulator, you'd have to run the simulation &lt;em&gt;&lt;a href=&quot;https://neologisms.rice.edu/index.php?a=term&amp;amp;d=1&amp;amp;t=2562&quot; rel=&quot;nofollow noreferrer&quot;&gt;kerjillions&lt;/a&gt;&lt;/em&gt; of times to include a thousand slightly different staring point for each of the bodies in the simulation. Since everything interacts with everything (in the case of gravity) that becomes a huge problem. Real calculations use all kinds of techniques to try to include uncertainties in their propagator, hopefully another answer here will expand on that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there are uncertainties in the non-gravitational forces such as outgassing and pressure from sunlight and the solar wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To read more on that, see the new BBC News article &lt;a href=&quot;https://www.bbc.com/news/science-environment-44630125&quot; rel=&quot;nofollow noreferrer&quot;&gt;Interstellar visitor's identity solved&lt;/a&gt; and also scroll down to the link that says &lt;strong&gt;The full study is published in Nature&lt;/strong&gt; where an &lt;em&gt;open access&lt;/em&gt; version of the Nature paper &lt;a href=&quot;https://www.nature.com/articles/s41586-018-0254-4#Sec19&quot; rel=&quot;nofollow noreferrer&quot;&gt;Non-gravitational acceleration in the trajectory of 1I/2017 U1 (‘Oumuamua)&lt;/a&gt; is available for you to read.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-06-29T23:09:24.793" LastActivityDate="2018-06-29T23:09:24.793" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26803" PostTypeId="1" AcceptedAnswerId="26841" CreationDate="2018-06-28T23:00:12.723" Score="1" ViewCount="797" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/16396/7982&quot;&gt;This answer&lt;/a&gt; points out that including General Relativity in a simulation can be computationally intensive and may not be present in Universe Sandbox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, answers to &lt;a href=&quot;https://space.stackexchange.com/q/23408/12102&quot;&gt;How to calculate the planets and moons beyond Newtons's gravitational force?&lt;/a&gt; discuss approximations that go most of the way toward including main GR effects in a &lt;em&gt;typical&lt;/em&gt; solar system simulation where the effects are small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any options to &quot;turn on&quot; an approximation to GR in Universe Sandbox and/or Universe Sandbox 2? Or plans to add it later? And to double check, is it absolutely certain that only Newtonian mechanics is used currently for propagating orbits?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2021-03-08T00:04:03.940" LastActivityDate="2021-03-08T00:04:03.940" Title="Is a General Relativity approximation available in Universe Sandbox (1 or 2)?" Tags="&lt;space&gt;&lt;n-body-simulations&gt;&lt;universe-sandbox&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26804" PostTypeId="2" ParentId="26799" CreationDate="2018-06-29T01:58:09.327" Score="2" Body="&lt;p&gt;A NEO only impacts Earth if both bodies are in the orbit intersection zone at the same time, closely enough for gravity $(F \propto 1/r^2)$ to bring them in contact.&#xA;In a simulation, the timestep should also be short enough to detect a collision.&#xA;An object at a moderate relative speed of 10 km/s crosses Earth's diameter in about 20 minutes; a longer timestep could turn a hit into a miss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're already modeling real asteroids, try some which have hit Earth:&#xA;&lt;a href=&quot;https://ssd.jpl.nasa.gov/sbdb.cgi?spk=3430291;orb=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;2008 TC3&lt;/a&gt;,&#xA;&lt;a href=&quot;https://ssd.jpl.nasa.gov/sbdb.cgi?spk=3655730;orb=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;2014 AA&lt;/a&gt;,&#xA;&lt;a href=&quot;https://ssd.jpl.nasa.gov/sbdb.cgi?spk=3825058;orb=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;2018 LA&lt;/a&gt;.&#xA;The short observation arcs make their pre-impact orbits highly uncertain, so a few small bodies spread within $\pm\sigma$ of each orbital element value may improve your chances.&#xA;Alternatively you could set an object up for immediate collision with Earth at a relative speed of 12 to 20 km/s in any direction, and run the simulator backward to see how it would get there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The asteroid risk is real, but we should not overstate it.&#xA;If you visit JPL CNEOS &lt;a href=&quot;https://cneos.jpl.nasa.gov/sentry/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sentry&lt;/a&gt; and use unconstrained settings, it shows only a few objects with more than 1 chance in 1000 to hit Earth in the next 100 years, and most of those are small.&#xA;Few NEO orbits are known precisely enough to make meaningful predictions beyond that.&#xA;The &lt;a href=&quot;https://cneos.jpl.nasa.gov/sentry/palermo_scale.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Palermo scale&lt;/a&gt; quantifies a given object's threat to Earth in the foreseeable future.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-06-30T17:19:11.073" LastActivityDate="2018-06-30T17:19:11.073" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26806" PostTypeId="2" ParentId="26800" CreationDate="2018-06-29T07:30:52.327" Score="1" Body="&lt;p&gt;The visible surface of the Sun is defined by that location within the Sun's atmosphere from which photons produced in the gas can escape towards the Earth. That location is determined by the amount of gas &lt;em&gt;between&lt;/em&gt; it and the Earth, since that gas would reabsorb photons if there were enough of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you need to think about the geometry of a roughly, spherically symmetric Sun.  When we look at the solar centre, photons can escape from a certain layer in its interior. That layer is nowhere near the core of the Sun, it is the solar photosphere at a temperature of approximately 6000K and at a radius of $6.96\times 10^8$ m from the centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine looking towards the limb. The same quantity of intervening gas is reached by the time you get to a layer in the Sun that is at a larger radius than at the solar centre. Because the temperature decreases as you move outwards in the solar photosphere, this layer is cooler and has a lower surface brightness (surface brightness is proportional to $T^4$). As a result, the solar centre looks hotter and brighter than the limb of the Sun.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-06-29T07:30:52.327" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26809" PostTypeId="1" CreationDate="2018-06-29T14:01:56.847" Score="3" ViewCount="305" Body="&lt;p&gt;How can I get heliocentric distance to a star when I have galactic x y and z coordinates known ?&lt;/p&gt;&#xA;" OwnerUserId="23522" LastActivityDate="2018-06-29T14:45:58.563" Title="Heliocentric distance from galactic x y and z coordinates" Tags="&lt;positional-astronomy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26810" PostTypeId="2" ParentId="26809" CreationDate="2018-06-29T14:45:58.563" Score="2" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;galactic coordinate system&lt;/a&gt; is centred on the sun with the xy plane in the plane of the galaxy and the x coordinate pointed towards the centre of the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it is a heliocentric system the Euclidian distance is $\sqrt{x^2+y^2+z^2}$.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-06-29T14:45:58.563" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26811" PostTypeId="1" AcceptedAnswerId="26823" CreationDate="2018-06-29T20:01:18.623" Score="5" ViewCount="1581" Body="&lt;p&gt;I have seen models of 2 black hole mergers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TD19z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TD19z.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.theverge.com/2017/11/16/16665132/gravitational-waves-ligo-virgo-black-holes-merger-multi-messenger-astronomy&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.theverge.com/2017/11/16/16665132/gravitational-waves-ligo-virgo-black-holes-merger-multi-messenger-astronomy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vNaFX.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vNaFX.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we see a space ship appear to be frozen in time as it approaches the event horizon then would a back hole merger be faster then perceived in our time? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;John put it well in the comments:  People say merging black holes end up moving at half the speed of light, but don't say anything about the gravitational time dilation wherein the &quot;coordinate&quot; speed of light goes to zero at the event horizon. Which would mean black holes would have to be moving faster than the local speed of light. How can this be right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/11673/can-things-move-faster-than-light-inside-the-event-horizon-of-a-black-hole?rq=1&quot;&gt;Can things move faster than light inside the event horizon of a black hole?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-06-30T15:47:57.067" LastActivityDate="2018-07-02T03:38:37.967" Title="What is the actual black hole merger speed?" Tags="&lt;black-hole&gt;&lt;astrophysics&gt;&lt;amateur-observing&gt;&lt;space-time&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26812" PostTypeId="2" ParentId="26250" CreationDate="2018-06-29T23:17:25.157" Score="3" Body="&lt;p&gt;Although should not use the negative parallaxes, you should not ignore them either. If you are looking at populations of objects, removing those with negative parallaxes will lead to significant bias in your results, as &lt;a href=&quot;https://arxiv.org/abs/1804.09376&quot; rel=&quot;nofollow noreferrer&quot;&gt;Luri et al. 2018&lt;/a&gt; has shown.&lt;/p&gt;&#xA;" OwnerUserId="23530" LastActivityDate="2018-06-29T23:17:25.157" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26813" PostTypeId="1" AcceptedAnswerId="26879" CreationDate="2018-06-30T02:39:02.797" Score="1" ViewCount="174" Body="&lt;p&gt;Magnetic field strength is measured using Zeeman splitting. This is the one of the way Sun's magnetic field strength is measured. Now, how one measure vector magnetic field?&#xA;Vector magnetic field = Magnetic field and it's components.&lt;/p&gt;&#xA;" OwnerUserId="23450" LastActivityDate="2018-07-06T02:18:24.057" Title="How does one measure vector magnetic field of astrophysical object?" Tags="&lt;magnetic-field&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26815" PostTypeId="1" AcceptedAnswerId="26816" CreationDate="2018-06-30T04:09:10.413" Score="2" ViewCount="57" Body="&lt;p&gt;Dynamo theory suggest how magnetic field generated on Earth. Due to flowing magma which contains molten iron generates magnetic field on earth. But how magnetic field is generated over sun? &lt;/p&gt;&#xA;" OwnerUserId="23450" LastActivityDate="2018-06-30T04:37:58.183" Title="How is magnetic field is generated over the sun?" Tags="&lt;the-sun&gt;&lt;magnetic-field&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26816" PostTypeId="2" ParentId="26815" CreationDate="2018-06-30T04:37:58.183" Score="4" Body="&lt;p&gt;The Sun's interior is partly or completely ionised and therefore supports the flow of electric currents. The solar magnetic field is explained by a &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_dynamo&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar dynamo&lt;/a&gt; that converts the kinetic energy of electrically charged fluid flows in the Sun (due to differential rotation and convection) into electromagnetic energy.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-06-30T04:37:58.183" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26817" PostTypeId="1" AcceptedAnswerId="26819" CreationDate="2018-06-30T12:58:58.640" Score="1" ViewCount="154" Body="&lt;p&gt;Recently i saw a post comparing sizes of different galaxies with the Milky Way. The largest one being the Ic 1011.What makes this galaxy so large? What factors decide the size of a galaxy? &lt;a href=&quot;https://i.stack.imgur.com/S7He7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S7He7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22623" LastActivityDate="2018-06-30T15:22:59.380" Title="What factors determines the size of a galaxy?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26818" PostTypeId="1" AcceptedAnswerId="26822" CreationDate="2018-06-30T13:00:32.313" Score="2" ViewCount="1958" Body="&lt;p&gt;If anyone has a derivation of this result, it would be appreciated greatly.&lt;/p&gt;&#xA;" OwnerUserId="23536" LastActivityDate="2018-06-30T19:15:46.363" Title="Why is the Innermost Stable Circular Orbit (ISCO) of a non-rotating black-hole 3 times the Schwarzschild radius?" Tags="&lt;orbit&gt;&lt;black-hole&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26819" PostTypeId="2" ParentId="26817" CreationDate="2018-06-30T15:22:59.380" Score="2" Body="&lt;p&gt;Galaxies reside in dark matter halos, and the size of the galaxy is more or less proportional to the size if its halo. The distribution of the sizes of halos is given by the &lt;em&gt;halo mass function&lt;/em&gt; (HMF), which is a declining function with mass — that is, the smaller the mass, the larger the number. At low masses, the HMF falls off as a power law, while at higher masses there is an exponential cut-off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, a galaxy such as IC 1011 is much rarer than a galaxy like M87, which in turn is rarer than Andromeda and the Milky Way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An analogy is the size distribution of towns and cities. Capitals are rare, towns are common, and villages are even more common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;reason&lt;/em&gt; the galaxy masses follow this distribution is first given by how dark matter in the early Universe fragments into clumps, and start collapsing and thus detach themselves from the Hubble flow. The first analytical attempt to solve the problem of structure formation assumed spherical collapse, and was presented by &lt;a href=&quot;http://adsabs.harvard.edu/abs/1974ApJ...187..425P&quot; rel=&quot;nofollow noreferrer&quot;&gt;Press &amp;amp; Schechter (1974)&lt;/a&gt;. Later, assuming ellipsoidal collapse, &lt;a href=&quot;https://arxiv.org/abs/astro-ph/9901122&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sheth &amp;amp; Tormen (1999)&lt;/a&gt; gave an improved solution. More realistic forms cannot be calculated analytically, but must be estimated using large-scale $N$-body simulations (e.g. &lt;a href=&quot;https://arxiv.org/abs/0803.2706&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tinker et al. 2008&lt;/a&gt;, &lt;a href=&quot;https://arxiv.org/abs/1002.3660&quot; rel=&quot;nofollow noreferrer&quot;&gt;Klypin et al. 2011&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the Universe evolves, galaxies merge to form larger galaxies. Thus, the distribution depends on the time one considers. The figure below (from &lt;a href=&quot;https://arxiv.org/abs/1002.3660&quot; rel=&quot;nofollow noreferrer&quot;&gt;Klypin et al. 2011&lt;/a&gt;) shows how the HMF evolves from early times ($z=10$) to now ($z=0$). You'll see that the amplitude grows with time, because more galaxies are formed. From $z=2.5$ to $z=0$ you can also see that the number density of small halos actually &lt;em&gt;de&lt;/em&gt;crease a little, because small galaxies are &quot;eaten&quot; by larger ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7ga1J.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7ga1J.png&quot; alt=&quot;HMF&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Halo mass function, i.e. number density of dark matter halos as a function of mass. Points are from an $N$-body simulation, while curves show the Sheth-Tormen approximation, which provides a very accurate fit at $z = 0$, but overpredicts the number of halos at higher redshifts.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2018-06-30T15:22:59.380" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26820" PostTypeId="2" ParentId="26795" CreationDate="2018-06-30T16:17:43.863" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I found a table that was based on data from 2003 for Stellar Classification, and I can't get to the given data source ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The website that was the source of the data was last active  in 2006. I've replaced the broken URLs with working URLs. I wouldn't recommend relying on data that old, there's likely a newer source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Original (404) Text and URLs&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Credit:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The table data was obtained from &lt;a href=&quot;http://www.spacegear.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;SpaceGear.Org&lt;/a&gt;'s &lt;a href=&quot;http://www.spacegear.org/data/SCTable/sctable.3.0.csv&quot; rel=&quot;nofollow noreferrer&quot;&gt;SCTable&lt;/a&gt; version 3.0 09/23/2003. The following data source was obtained from their &lt;a href=&quot;http://www.spacegear.org/data/SCTable/&quot; rel=&quot;nofollow noreferrer&quot;&gt;SCTable web page&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Original Text and repaired URLs:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Credit:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The table data was obtained from &lt;a href=&quot;https://web.archive.org/web/20060410153507/http://www.spacegear.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;SpaceGear.Org&lt;/a&gt;'s &lt;a href=&quot;https://web.archive.org/web/20061012065012/http://www.spacegear.org/data/SCTable/sctable.3.0.csv&quot; rel=&quot;nofollow noreferrer&quot;&gt;SCTable&lt;/a&gt; version 3.0 09/23/2003. The following data source was obtained from their &lt;a href=&quot;https://web.archive.org/web/20061012065004/http://www.spacegear.org/data/SCTable/&quot; rel=&quot;nofollow noreferrer&quot;&gt;SCTable web page&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The 3rd links says that they derived the data from:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Absolute magnitude was derived from the standard Hertzsprung-Russell (HR) diagram. The book &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/0521899540&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stars and Their Spectra by James B. Kaler&lt;/a&gt; was used as a resource for deriving the absolute magnitude for each possible luminosity class. Classifications N, S, WC, WN, C, and R are mapped as follows:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N and S are equivalent to M.&#xA;WC and WN are equivalent to O.&#xA;C spans G4 to M9.&#xA;R spans G4 to M0.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Temperature was derived from the relationship between luminosity class and spectral class. The book &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/0521899540&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stars and Their Spectra by James B. Kaler&lt;/a&gt; and &lt;a href=&quot;http://www.vendian.org/mncharity/dir3/starcolor/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mitchell Charity's work on star colors&lt;/a&gt; was used as a resource for deriving temperature for Main Sequence (V), Supergiant (Ia), Giant (II), and Dwarf (D) luminosity classes. Luminosity classes Ia0, Ib, III, IV, and VI are mapped as follows:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Temperatures of Ia0 and Ib are equivalent to Ia.&#xA;Temperatures of II is equivalent to III.&#xA;Temperatures of IV and VI are equivalent to V.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Color was derived from &lt;a href=&quot;http://www.vendian.org/mncharity/dir3/starcolor/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mitchell Charity's work on star colors&lt;/a&gt;. Each designated color is defined with at least one field (R, G, or B) at full saturation (value equals 255). Actual color values for display would be attenuated across all color components based on the star's apparent magnitude in a given situation.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Mass was derived from many sources including &lt;a href=&quot;https://www.springer.com/us/book/9783662216415&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astrophysical Formulae by Kenneth R. Lang&lt;/a&gt;. Many rows are heavily interpolated from known data on star mass. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;See how I fixed all the URLs so they work correctly. If there is more from that webpage that you wanted sources for then you can use the same method. Since I was able to get to the raw CSV data that should be easier to use in your program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also Wikipedia's webpage on &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Stellar_classification&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellar Classification&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of a website providing similar data (from 2015) is: &quot;&lt;a href=&quot;http://www.johnpratt.com/items/astronomy/mag_5_stars.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Mag 5 Star Catalog&lt;/a&gt;&quot; or see: &quot;&lt;a href=&quot;http://www.astronexus.com/hyg&quot; rel=&quot;nofollow noreferrer&quot;&gt;The HYG Database&lt;/a&gt;&quot;, his latest update to &lt;a href=&quot;https://github.com/astronexus/HYG-Database&quot; rel=&quot;nofollow noreferrer&quot;&gt;GitHub&lt;/a&gt; is a couple of years ago.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastActivityDate="2018-06-30T16:17:43.863" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26821" PostTypeId="1" CreationDate="2018-06-30T18:47:20.060" Score="3" ViewCount="266" Body="&lt;p&gt;The plot of &lt;a href=&quot;https://en.wikipedia.org/wiki/Thundarr_the_Barbarian#Plot&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thundarr the Barbarian&lt;/a&gt; got me thinking about the science behind &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the passage of a runaway planet between the Earth and the Moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and its effect on the Earth and Moon themselves, as well as their orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the following scenario:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;planet has half the mass of Earth, and is moving at 16km/s&lt;/li&gt;&#xA;&lt;li&gt;moving perpendicular to the plane of the ecliptic&lt;/li&gt;&#xA;&lt;li&gt;passes directly between the Earth and the Moon with closest distance to Earth of 192,500 km&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What sort of effects, if any, might this have on Earth and those of us residing there during the passage in later months and years?&lt;/p&gt;&#xA;" OwnerUserId="23538" LastEditorUserId="7982" LastEditDate="2018-07-01T07:35:31.110" LastActivityDate="2018-07-01T19:38:32.040" Title="Effects of a rogue planet passing between the Earth and the Moon?" Tags="&lt;astrophysics&gt;&lt;rogue-planet&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26822" PostTypeId="2" ParentId="26818" CreationDate="2018-06-30T19:15:46.363" Score="2" Body="&lt;p&gt;There's a fairly full treatment in &lt;a href=&quot;https://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll7.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;these lecture notes&lt;/a&gt; but it's not really amenable to a summary in a stackexchange answer. However they end up with the following equation for the radius of a circular orbit of a particle with angular momentum $L$ and mass (ie not a photon). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$GMr^2 - L^2r + 3GML^2 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For suitable values of $L$ this has two solutions for $r$ and, one inside the ISCO radius (which is $6GM$ in these units) and one outside. Further analysis shows that the inner orbit is unstable and the outer one stable. See around equations 7.55 for more details. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photons have an unstable circular orbit at $3GM$ and the event horizon is at $2GM$.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-06-30T19:15:46.363" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26823" PostTypeId="2" ParentId="26811" CreationDate="2018-06-30T19:43:05.913" Score="3" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;What is the actual black hole merger speed?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It varies per object, and throughout the merger (and doesn't neatly increase linearly).&lt;/p&gt;&#xA;&lt;p&gt;An object with a lot mass can not come close to the speed of light, that would require almost infinite energy. Extremely tiny objects, particles, can come &lt;a href=&quot;https://physics.stackexchange.com/a/227930/170832&quot;&gt;close to the speed of light&lt;/a&gt; but not actually reach it. Light can go the speed of light, &lt;strong&gt;only&lt;/strong&gt; in a vacuum. Gravity &lt;strong&gt;always&lt;/strong&gt; travels at the speed of light, unlike &lt;strong&gt;everything&lt;/strong&gt; else which goes slower.&lt;/p&gt;&#xA;&lt;p&gt;The first animation shows what happens when two large objects approach each other, miss, and then don't &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravity_assist&quot; rel=&quot;nofollow noreferrer&quot;&gt;slingshot&lt;/a&gt; away from each other - they become trapped in an &lt;a href=&quot;https://www.soundsofspacetime.org/elliptical-binaries.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;extreme elliptical orbit&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The second animation shows what happens when the orbits are relatively close, &amp;gt; ~0.01 parsecs - instead of an elliptical orbit they have a somewhat circular-like orbit, around the &lt;a href=&quot;https://en.wikipedia.org/wiki/Two-body_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;barycenter&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;There are many different trajectories and resulting orbits, the binaries are also affected by: inflow, outflow, and other objects entering and leaving the area of their orbits. No one answer, no two are alike.&lt;/p&gt;&#xA;&lt;p&gt;The paper &amp;quot;&lt;a href=&quot;https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.116.061102&quot; rel=&quot;nofollow noreferrer&quot;&gt;Observation of Gravitational Waves from a Binary Black Hole Merger&lt;/a&gt;&amp;quot; (Feb 11 2016), by B. P. Abbott et al. offers &lt;strong&gt;one study of one merger&lt;/strong&gt;, they provide this illustration:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VnDOz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VnDOz.png&quot; alt=&quot;Binary Merger&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Top:&lt;/strong&gt; Estimated gravitational-wave strain amplitude from &lt;a href=&quot;https://en.wikipedia.org/wiki/First_observation_of_gravitational_waves&quot; rel=&quot;nofollow noreferrer&quot;&gt;GW150914&lt;/a&gt; projected onto H1. This shows the full bandwidth of the waveforms, without the filtering used for Fig. 1. The inset images show numerical relativity models of the black hole horizons as the black holes coalesce. &lt;strong&gt;Bottom:&lt;/strong&gt; The Keplerian effective black hole separation in units of Schwarzschild radii (&lt;span class=&quot;math-container&quot;&gt;$R_S=2GM/c^2$&lt;/span&gt;) and the effective relative velocity given by the post-Newtonian parameter &lt;span class=&quot;math-container&quot;&gt;$v/c=(GMπf/c^3)^{1/3}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt; is the gravitational-wave frequency calculated with numerical relativity and &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is the total mass (value from Table 1).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In-between those distances occurs a situation referred to as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Binary_black_hole#Final_parsec_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;final parsec problem&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Final parsec problem&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;When two galaxies collide, the supermassive black holes at their centers do not hit head-on, but would shoot past each other if some mechanism did not bring them together. The most important mechanism is &lt;a href=&quot;https://en.wikipedia.org/wiki/Dynamical_friction&quot; rel=&quot;nofollow noreferrer&quot;&gt;dynamical friction&lt;/a&gt;, which brings the black holes to within a few parsecs of each other. At this distance, they form a bound, binary system. The binary system must lose &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;orbital energy&lt;/a&gt; somehow, for the black holes to orbit more closely or merge.&lt;/p&gt;&#xA;&lt;p&gt;Initially, the explanation is easy. The black holes transfer energy to gas and stars between them, ejecting matter at high speed via a gravitational slingshot and thereby losing energy. However, the volume of space subject to this effect shrinks as the orbits do, and &lt;strong&gt;when the black holes reach a separation of about one parsec, there is so little matter left between them that it would take billions of years to orbit closely enough to merge&lt;/strong&gt; - more than the age of the universe. Gravitational waves can be a significant contributor, but not until the separation shrinks to a much smaller value, roughly 0.01–0.001 parsec.&lt;/p&gt;&#xA;&lt;p&gt;Nonetheless, supermassive black holes appear to have merged, and what appears to be a pair in this intermediate range has been observed, in &lt;a href=&quot;https://en.wikipedia.org/wiki/PKS_1302-102&quot; rel=&quot;nofollow noreferrer&quot;&gt;PKS 1302-102&lt;/a&gt;. The question of how this happens is the &amp;quot;final parsec problem&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;A number of solutions to the final parsec problem have been proposed. Most involve the interaction of the massive binary with surrounding matter, either stars or gas, which can extract energy from the binary and cause it to shrink. For instance, if enough stars pass close by to the orbiting pair, their gravitational ejection can bring the two black holes together much quicker than would otherwise be the case.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;See the &amp;quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Binary_black_hole#Final_parsec_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;final parsec problem&lt;/a&gt;&amp;quot; link for the whole webpage.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;... the &lt;a href=&quot;https://en.wikipedia.org/wiki/Time_dilation#Gravitational_time_dilation&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravitational time dilation&lt;/a&gt; ...&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Means: In the black hole's timeframe time moves forward normally for both the black hole and the observer &lt;strong&gt;in their own frames&lt;/strong&gt;. In each other's frames the outside observer see the black hole's time seemingly come to a standstill while the black hole observer sees the outside observer's time pass very rapidly.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-06-30T19:51:19.580" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26825" PostTypeId="1" AcceptedAnswerId="26857" CreationDate="2018-07-01T16:24:26.840" Score="1" ViewCount="198" Body="&lt;p&gt;Beside that, is that possible our sun became &lt;strong&gt;pulsar&lt;/strong&gt;? And if not (and i know likely) but how much more mass does it need to become one ?&lt;/p&gt;&#xA;" OwnerUserId="23217" LastActivityDate="2018-07-03T13:33:19.990" Title="How much mass does our sun need for exploding like supernova?" Tags="&lt;pulsar&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26826" PostTypeId="1" CreationDate="2018-07-01T16:59:22.727" Score="2" ViewCount="719" Body="&lt;p&gt;Is there significant time dilation on the other planets orbiting the sun?  Would an 'hour' experienced on earth actually be 59 or 61 minutes on other planets, depending on their distance to the sun?  What about the moons orbiting Jupiter?&lt;/p&gt;&#xA;" OwnerUserId="23548" LastActivityDate="2018-07-01T18:45:57.423" Title="Is there any significant time dilation on other planets orbiting the sun?" Tags="&lt;astrophysics&gt;&lt;general-relativity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26827" PostTypeId="2" ParentId="26826" CreationDate="2018-07-01T17:49:53.203" Score="3" Body="&lt;p&gt;It depends on what you mean by &quot;significant.&quot;  None of the orbits of other planets exhibit relativistic effects at the level you specify.  You can estimate the overall size of relativistic corrections by squaring the ratio of v/c, where v is the speed of the orbit.  For the Earth, that number is .00000001, and it is not more than about a factor of 3 higher for any other orbit.  So it's pretty negligible, but that doesn't mean it has no consequences-- over many orbits, for example, Mercury precesses by a larger amount than can be understood without relativity.  Also, if one takes the relativistic perspective, the reason the planets follow the orbits they do is because of that tiny time dilation effect-- that small effect is enough to select the orbits we see over other possible ones.  So that's &quot;significant,&quot; but not like a 59 minute hour.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2018-07-01T18:45:57.423" LastActivityDate="2018-07-01T18:45:57.423" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26828" PostTypeId="2" ParentId="26825" CreationDate="2018-07-01T18:53:30.073" Score="2" Body="&lt;p&gt;The smallest stars that could explode in supernova have a mass of at least about 8 times the mass of the sun. So it is impossible for the sun to become a pulsar at the end of its life. It will form a red giant which will then lose its outer layers and form a white dwarf.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-07-01T18:53:30.073" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26829" PostTypeId="2" ParentId="26825" CreationDate="2018-07-01T18:56:00.330" Score="0" Body="&lt;p&gt;What's more, there is a path to supernova involving less added mass, and that is if the Sun is first allowed to become a white dwarf, then it would only require one more solar mass to be added to create a type Ia supernova.  But that would not make a pulsar, and it isn't any more likely to happen than adding 7 solar masses.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-07-01T18:56:00.330" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26830" PostTypeId="2" ParentId="26821" CreationDate="2018-07-01T19:38:32.040" Score="2" Body="&lt;p&gt;It doesn't look good. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tidal effects. The passing planet will certainly cause disruptive tides as it passes. The exact height of the tides would be hard to calculate as it depends on how the tides flow around the land masses, but the effects would be much greater than lunar tides. The tidal effect could also have consequences for the earthquakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disruption to the orbit of the moon. The moon's orbit will be significantly altered. It could cause the moon to enter an elliptical orbit (with consequences for tides).  There would not be enough disruption to cause the moon to leave Earth orbit immediately&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disruption to Earth's orbit.  The orbit of Earth would also be significantly changed, it could enter a significantly elliptical orbit, this would change the amount insolation over the year and greatly change the Earth's climate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the disruption were significant enough, it could bring the Earth's orbit uncomfortably close to that of Venus, this could result in further disruption to the moon's orbit&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately space is big, and even if there are a lot of rogue planets out there, it would be a very rare event for one to pass so close to star&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-07-01T19:38:32.040" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26831" PostTypeId="1" AcceptedAnswerId="26837" CreationDate="2018-07-01T20:48:03.057" Score="6" ViewCount="1341" Body="&lt;p&gt;We take observations from earth but to define radial velocity, transverse velocity and the proper motion of stars, why we consider them with respect to the sun? and then we do some corrections due to both rotational and orbital motion of the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why can't we directly consider these motions with respect to earth instead of Sun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what is the whole procedure of correction?&lt;/p&gt;&#xA;" OwnerUserId="23550" LastActivityDate="2018-07-01T23:43:42.963" Title="Why we define Stellar motions with respect to sun?" Tags="&lt;stellar-astrophysics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26832" PostTypeId="2" ParentId="26831" CreationDate="2018-07-01T21:15:02.603" Score="6" Body="&lt;p&gt;We use the position of the sun (or more accurately the centre of mass of the solar system) as this gives us a very nearly inertial frame of reference. An inertial frame is one which is not changing its velocity. The surface of the Earth does not define an inertial frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth orbits the sun and so it is moving in opposite directions in summer and winter. If we were to use the radial velocity (for example) relative to the Earth, we would get a different value in summer from winter. The orbital speed of the Earth is about 30km/s, and the radial velocity of stars tends to be in the 10s of kilometers per second, so the Earth's motion is a significant amount. If, however, we use the centre of mass of the solar system (which is always very close to the sun) then we can record the actual motion of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process of finding the relative motion is to measure the radial motion relative to the telescope doing the observations, and then subtract the velocity of the Earth around the sun, and the velocity of the telescope relative to the centre of the Earth. Since these two velocities are well known, compensating for them is essentially no more than subtraction.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-07-01T23:23:17.150" LastActivityDate="2018-07-01T23:23:17.150" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26834" PostTypeId="1" CreationDate="2018-07-01T21:41:54.127" Score="0" ViewCount="620" Body="&lt;p&gt;I'm currently stuck on a problem. &#xA;I have the RA and DEC for JD2000:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RA = 17 57 48.49980&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DEC = + 04 41 36.1114&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to calculate the RA and DEC for 01.01.2017, how can I do it? &#xA;The solution might be easy but I'm totally stuck right now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="23551" LastEditorUserId="616" LastEditDate="2018-07-01T23:03:16.510" LastActivityDate="2018-07-01T23:34:55.520" Title="Calculation of RA and DEC" Tags="&lt;coordinate&gt;&lt;precession&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26835" PostTypeId="2" ParentId="26834" CreationDate="2018-07-01T23:11:47.060" Score="4" Body="&lt;p&gt;Simon Littlefair of Sheffield University gives &lt;a href=&quot;http://slittlefair.staff.shef.ac.uk/teaching/phy115/session2/moreCel/moreCel.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;formulae for adjusting for precession&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;Δα = M + N \sinα \tanδ&#xA;$$&#xA;$$&#xA;Δδ = N \cosα&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where M and N are given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M = 1°.2812323 T + 0°.0003879 T^2 + 0°.0000101 T^3$$&#xA;$$N = 0°.5567530 T - 0°.0001185 T^2 + 0°.0000116 T^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $T$ is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;T = (t-2000.0)/100&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where t is the observation date, in years. $Δα$ is the change in RA that you would need to add, in degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(formulae quoted directly from the website)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you would have T = 0.17, M= 0.21782075093,N=0.09464464234&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual changes in RA and Dec can be calculated from the trig formulae above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately caltech provide an &lt;a href=&quot;https://ned.ipac.caltech.edu/forms/calculator.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;online calculator&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-07-01T23:11:47.060" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26837" PostTypeId="2" ParentId="26831" CreationDate="2018-07-01T23:43:42.963" Score="7" Body="&lt;p&gt;The Earth is a moving (actually, accelerating) platform from which we make our observations. If you want to describe the motion of a distant celestial body, then it does not make much sense to provide a geocentric velocity, because this will depend on exactly when the observations were taken (because the Earth orbits with a &lt;em&gt;speed&lt;/em&gt; of about 30 km/s, but the &lt;em&gt;velocity&lt;/em&gt; changes throughout the year), the position on the Earth that the measurement was taken (because the Earth's equator revolves at around 0.5 km/s, but this is different at different latitudes) and the position of the star in the sky (because the previous two effects have to be resolved in the direction of the star).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that, then it is &lt;em&gt;far&lt;/em&gt; easier to quote a velocity in a frame of reference that does not depend on these things. One can use the heliocentric frame in most circumstances - that is the frame of reference in which the Sun is stationary. To do this, one has to remove the components of the Earth's orbital and rotational velocity. However, if very precise measurements are being done then it is better to use the &lt;em&gt;barycentric&lt;/em&gt; reference frame, which is with respect to the centre of mass of the solar system (since the Sun orbits this with a speed of around 0.1 m/s).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-07-01T23:43:42.963" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26838" PostTypeId="1" AcceptedAnswerId="26842" CreationDate="2018-07-02T02:56:44.060" Score="2" ViewCount="133" Body="&lt;p&gt;In billions of stars, how space scientist keeps the track of the stars ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a star called &lt;a href=&quot;https://en.wikipedia.org/wiki/KIC_8462852&quot; rel=&quot;nofollow noreferrer&quot;&gt;KIC 8462852&lt;/a&gt;, I saw that picture, star is so small from the earth. It has higher chance to miss that star in the universe.&lt;/p&gt;&#xA;" OwnerUserId="23555" LastEditorUserId="23555" LastEditDate="2018-07-02T03:03:41.587" LastActivityDate="2018-07-02T08:08:08.607" Title="How space scientist keeps the track of the stars?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26839" PostTypeId="2" ParentId="12824" CreationDate="2018-07-02T03:04:42.210" Score="2" Body="&lt;p&gt;OK, lets start with the simplest mathematical approach to illustrate the path to a fully analytical answer.&#xA;The sun presents an angular width of 32 arcminutes to any point on earth.  That is 32/60 or 0.533 degrees of arc or angular span. &#xA;Lets assume the Earth does not have its 23 degrees of tilt, for this first approximation.&#xA;Then as a second approximation lets assume the Earth rotates around the sun in 24 hours, you are still on the equator.&#xA;Our calculation is as follows;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.533 degrees/360 degrees) = (hours sunset/24 hours).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve for hours sunset and you get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;24 hrs X (0.533/360) = 0.0355 hrs, which is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.0355 hrs X 60 min/hrs = 2.13 minutes, which is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.13 min X 60 secs/min = 128 seconds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, now that is the first order approximation only and explains the minima of the nice charts previously provided.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first and trivial correction would be to notice that the 24 hour assumption is not accurate, hence leap years!  Beyond that we have actually 23:56 per year.  That will get you 127.56 seconds for sunset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real solution for the deep divers out there is to understand that the angular width of the sun in the sky is 32 arcminutes but only for any one instant in time for any one point on the Earth.  So the next calculation would be to integrate over the diameter of the earth to incorporate the angular width of you are traversing during the sunset traverse time.  You the observer are moving, rotating with the surface of the Earth, and hence you are spreading out the apparent angular size of the sun to the extent you are traversing during that sunset period, and this will add time to the sunset period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that is the easier side of all of this.  The next calculation would add the geometric correction for latitude which the observer is located in.  This introduces horizontal relative component of motion of the sun to the observer, greatly increasing the time when one is not at the summer or winter equinoxes. (The prior calculations had the sun directly perpendicular to the rotation of the Earth.)  In the titled Earth Sun systems, this effect is minimized at the equinox positions of the earth sun system and asymptotes toward the prior calculation if one is on the equator and on the equinox twice per year.  Again, this is seen nicely in the charts of the previous answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that help folks understand some of basic underpinnings of the math and geometry which the actual calculations must take into effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No calculators allowed and you can still get there.&lt;/p&gt;&#xA;" OwnerUserId="23556" LastEditorUserId="23556" LastEditDate="2018-07-02T03:24:43.453" LastActivityDate="2018-07-02T03:24:43.453" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26840" PostTypeId="1" CreationDate="2018-07-02T08:00:11.557" Score="1" ViewCount="178" Body="&lt;p&gt;Local standard Of Rest is the hypothetical origin of a reference system with respect to which the motions of all stars in some neighborhood of the sun average out to be zero&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which type of motions of stars average out to be zero and how can I justify that their average will be zero?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is Peculiar Motion? How does it look like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How we measure the proper motions of stars as seen from the sun as we can only observe from earth?&lt;/p&gt;&#xA;" OwnerUserId="23550" LastActivityDate="2018-07-02T08:39:04.127" Title="The Solar Motion and the peculiar velocities of stars" Tags="&lt;fundamental-astronomy&gt;&lt;observational-astronomy&gt;&lt;stellar-astrophysics&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26841" PostTypeId="2" ParentId="26803" CreationDate="2018-07-02T08:05:02.373" Score="4" Body="&lt;p&gt;As noted by @MikeG in this &lt;a href=&quot;https://astronomy.stackexchange.com/questions/26803/is-a-general-relativity-approximation-available-in-universe-sandbox-1-or-2#comment46449_26803&quot;&gt;comment&lt;/a&gt;, the Universe Sandbox 2 &lt;a href=&quot;http://universesandbox.com/faq/#sim&quot; rel=&quot;nofollow noreferrer&quot;&gt;FAQ&lt;/a&gt; addresses the question, but not in a satisfying or helpful way:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Does it account for relativity?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;No, the physics in Universe Sandbox² is currently only &lt;a href=&quot;https://en.wikipedia.org/wiki/Classical_mechanics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Newtonian&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Why? The short answer is that &lt;strong&gt;you need a supercomputer to accurately simulate general relativity&lt;/strong&gt;. (emphasis added)&lt;/p&gt;&#xA;&lt;p&gt;Jenn, astrophysicist and Universe Sandbox² developer, &lt;a href=&quot;http://universesandbox.com/blog/2016/02/gravitational-waves/&quot; rel=&quot;nofollow noreferrer&quot;&gt;explains more in a blog&lt;/a&gt; post: &amp;quot;General relativity requires simulating the spacetime itself. That is, taking your simulation space, discretizing it to a hi-res 3-D grid and checking the effect that each and every point in that grid has on all neighboring points at every timestep. Instead of simulating N number of bodies, you are simulating a huge number of points. You start with some initial data of the shape of your spacetime and then see how it evolves according to the Einstein equations, which are 10 highly non-linear partial differential equations.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Why do I say this is not very satisfying or helpful?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Because it's quite easy to &lt;em&gt;nearly completely account&lt;/em&gt; for General Relativity (GR) effects in the orbits of solar system bodies just by adding some terms to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ordinary_differential_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;ODE&lt;/a&gt;s that are integrated to propagate orbits.&lt;/p&gt;&#xA;&lt;p&gt;Answers to the question &lt;a href=&quot;https://space.stackexchange.com/q/23408/12102&quot;&gt;How to calculate the planets and moons beyond Newtons's gravitational force?&lt;/a&gt; (including mine) show how this is done all the time. I reproduced what's normally done in Python rather than in a compiled language for transparancy, it would be much faster if included in a reasonable simulator implementation.&lt;/p&gt;&#xA;&lt;p&gt;As &lt;a href=&quot;https://space.stackexchange.com/a/23409/12102&quot;&gt;explained here&lt;/a&gt;, to the Newtonian acceleration given by:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathbf{a_{Newton}} = -GM \frac{\mathbf{r}}{|r|^3},$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;You can just add this expression:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathbf{a_{GR}} = GM \frac{1}{c^2 |r|^3}\left(4 GM \frac{\mathbf{r}}{|r|} - (\mathbf{v} \cdot \mathbf{v}) \mathbf{r} + 4 (\mathbf{r} \cdot \mathbf{v}) \mathbf{v} \right),$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;To get close, you certainly don't need to evaluate these GR terms for every pair of objects. Interaction between the Moon and Ceres for example could still be Newtonian.&lt;/p&gt;&#xA;&lt;p&gt;All approximations are approximate. The FAQ's suggestion that they don't do it because you need a supercomputer to be accurate is slightly misleading. Using this straightforward approximation is &lt;em&gt;much better&lt;/em&gt; than not using it in this context.&lt;/p&gt;&#xA;&lt;p&gt;Yes for &lt;em&gt;Gravitational Waves&lt;/em&gt;, collapses, and other effects &amp;quot;requires simulating the spacetime itself. That is, taking your simulation space, discretizing it to a hi-res 3-D grid and checking the effect that each and every point in that grid has on all neighboring points at every timestep&amp;quot; but just adding the major GR effects to the propagation of solar system bodies certainly doesn't need anything more than  whatever computer you are already running your copy of Universe Sandbox 2 on already.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The FAQ answer continues:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;We are, however, interested in adding in a few features which would address some effects of relativity. One example is setting gravity to travel at the speed of light, instead of instantaneously taking effect as it currently does. You can read more about these in Jenn's blog post: &lt;a href=&quot;http://universesandbox.com/blog/2016/02/gravitational-waves/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gravitational Waves &amp;amp; Universe Sandbox².&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I had thought so as well, but according to @DavidHammen's &lt;a href=&quot;https://space.stackexchange.com/a/21472/12102&quot;&gt;thoughtful answer&lt;/a&gt; this is probably not so. I haven't tested this yet, but since this comes from a reliable source, I tend to believe it. Retarding gravity alone, without these other terms, seems to be the &lt;em&gt;wrong thing to do&lt;/em&gt; in this case.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-07-02T08:14:21.937" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26842" PostTypeId="2" ParentId="26838" CreationDate="2018-07-02T08:08:08.607" Score="2" Body="&lt;p&gt;Scientists make lists of stars in a catalogue. In the past this would be in a book nowadays it is more likely to be a computer database.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stars don't move (much) so it is possible to make a list of them by position Modern catalogues are online (see &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/&quot; rel=&quot;nofollow noreferrer&quot;&gt;SINBAD&lt;/a&gt; older ones were on paper (see the &lt;a href=&quot;https://archive.org/stream/henrydrapercata00obsegoog#page/n28/mode/2up&quot; rel=&quot;nofollow noreferrer&quot;&gt;1918 Henry Draper catalogue&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Henry Draper catalogue is typical and easier to follow. It is just a table, the first column is the catalogue number (used instead of a &quot;name&quot;)  the second column is a cross reference to another database. Then there is information on the location of the star in the sky using the RA and Dec system. Next comes information on the brightness of the star (estimated in two ways). Some letters come next to describe the spectrum (or &quot;colour&quot;) of the star and finally some short &quot;remarks&quot;, and a &quot;plate number (the number of the photograph that contains an image of the star&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scientists need to be systematic and careful to avoid missing anything.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-07-02T08:08:08.607" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26843" PostTypeId="2" ParentId="26840" CreationDate="2018-07-02T08:39:04.127" Score="1" Body="&lt;p&gt;When you talk about the motion of anything, you need to first define what &quot;not moving&quot; means.  For example I may be sitting in a chair (and not moving relative to the chair), but I know that the Earth is rotating, and orbiting the sun, and the whole solar system is orbiting in the Milky Way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I look at the motion of stars, I know that (nearly?) all the nearby stars are also orbiting the Milky Way. But I want to ignore that overall motion, and only consider the motion of the stars relative to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to measure the motion of a star relative to the observer. We also know the motion of the observer relative to the sun (since we know how fast the Earth spins and how fast it orbits the sun) So by simply subtracting one motion from the other we can find the motion of the star relative to the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But instead of using the sun as a fixed frame of reference, I can look at the local stars, find their average motion relative to the sun, and subtract that motion from each individual velocity. This will give a motion relative to the motion of the stars around the Milky Way. I know the average will be zero because I have chosen to subtract a certain amount from each star's velocity to make the average zero. It is zero &quot;by construction&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(simple example Consider the numbers &quot;3,7,2,3,5&quot;. The average is 4. If I subtract 4 from each number I get (-1, 3, -2 -1, 1) with average = 0)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motion of a star, when you have subtracted the average motion of all the local stars, is called the peculiar motion&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-07-02T08:39:04.127" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26844" PostTypeId="2" ParentId="26780" CreationDate="2018-07-02T09:17:14.003" Score="2" Body="&lt;p&gt;There are lots of good an sophisticated answers here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One important one I've not seen yet is this: Colour is relative. I don't mean this in a scientific or general way as others have listed (correctly). But simply: the immediate background colour is very important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try it with any colour sample: place a dark and then a light colour next to the sample. The sample will appear to change, though of course it has not. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Checker_shadow_illusion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Checker Board Illusion&lt;/a&gt; is a black/white and grey version of this. But it also works with colours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The colour the Moon is right next to will change how we see it. At night, the sky is black or a dark cool shade. So even when it's grey it seems bright at white in contrast. When we see the Moon in the day, with a blue sky, it also looks white, but it's fainter. And we're seeing the moon through the &lt;a href=&quot;https://www.explainxkcd.com/wiki/index.php/1818:_Rayleigh_Scattering&quot; rel=&quot;nofollow noreferrer&quot;&gt;blue atmosphere&lt;/a&gt; So that's going to make the background contrast different but also make the light more blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But seeing the Moon in the daytime is imagined to be unusual. The Moon is supposed to represent night, and cold, and dark. We are social animals. Our perception is not objective but also social. If we only saw the Moon in the Day we would probably think of it as grey blue. But since we consider the Moon mostly at night and draw it next to black - then it becomes white.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The really weird question that I still haven't figured out is: why is the Sun so yellow? Our perception has evolved white light to match the Sun's output, but still we call the sun yellow or Orange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Colour is weird.&lt;/p&gt;&#xA;" OwnerUserId="23559" LastActivityDate="2018-07-02T09:17:14.003" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26847" PostTypeId="1" AcceptedAnswerId="26848" CreationDate="2018-07-02T17:28:54.497" Score="1" ViewCount="154" Body="&lt;p&gt;Wouldn't it be easier with geostationary satellites? Has is to do with the monitoring of gamma-ray bursts?&lt;/p&gt;&#xA;" OwnerUserId="23564" LastActivityDate="2018-07-02T17:49:09.993" Title="Why do GPS satellites move in relation to earth's surface?" Tags="&lt;artificial-satellite&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2018-07-02T18:23:04.103" ContentLicense="CC BY-SA 4.0" />
  <row Id="26848" PostTypeId="2" ParentId="26847" CreationDate="2018-07-02T17:49:09.993" Score="1" Body="&lt;p&gt;Geostationary orbits (GEO) are necessarily over the equator, and satellites in GEO are not visible at all poleward of about 80 degrees -- they are always below the horizon (There's a &lt;a href=&quot;https://physics.stackexchange.com/questions/71582/is-it-possible-to-have-a-geostationary-satellite-over-the-poles&quot;&gt;good explanation&lt;/a&gt; on physics.stackexchange.com).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a problem because for GPS to work, you need at least four satellites to be visible &lt;em&gt;and&lt;/em&gt; they need to be well-separated in the sky.  Because it's rare for the whole sky to be &quot;visible&quot; all the time -- parts are blocked by buildings, metal vehicles, hills and suchlike -- you in fact need more than four in the sky as seen from any point on Earth to have a minimum of four actually visible all the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even at lower latitudes where GEO satellites are visible, they still all are in the southern (or all in the northern) sky and thus won't give a good GPS fix.  And the military (who were paying for this) wanted to be able to use GPS in the north, not just in the equatorial regions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, possible GEO satellite locations trace a gently curved line across the sky, and even if there were a dozen visible and even at low latitude, they'd give very poor accuracy perpendicular to that line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution was to put the GPS satellites into many different orbits which were strongly inclined to the equator. They wanted them to be fairly slow-moving as viewed from Earth, so LEO wasn't viable.  (LEO also required too many satellites to get adequate coverage.)   They needed to be above the Van Allen belts for longevity. But they also needed them to be as low as possible to maximize signal strength. 10,000 miles up turned out to be the right engineering compromise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gamma ray bursts were lagniappe.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-07-02T17:49:09.993" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26850" PostTypeId="2" ParentId="26780" CreationDate="2018-07-02T22:29:21.313" Score="6" Body="&lt;p&gt;I agree with Jack's comment, it's partly misconception and party due to the way rods and cones work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &quot;&lt;a href=&quot;http://www.academia.edu/15609535/By_the_light_of_the_silvery_Moon_fact_and_fiction&quot; rel=&quot;noreferrer&quot;&gt;By the light of the silvery Moon: fact and fiction&lt;/a&gt;&quot; (from where the illustration and quotes were obtained), by Marco Ciocca and Jing Wang, they explain:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Page 365: &quot;The rods are effectively colour blind. Scotopic vision has a higher sensitivity to light levels than photopic, but is effectively colour blind. We convoluted the sunlight spectrum shown in ﬁgure 5 with the photopic response above and the result is shown in ﬁgure 8. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1hDWb.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1hDWb.jpg&quot; alt=&quot;Figure 8&quot;&gt;&lt;/a&gt;&#xA;  Figure 8: Perception of sunlight and moonlight due to our eyes.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;During the daytime photopic vision clearly dominates, and the resulting spectral distribution of perceived light, even without considering the lack of blue components of the solar spectrum due to atmospheric scattering, is clearly shifted towards yellow-green (560 nm) because cone sensitivity is biased towards that part of the visible spectrum $^{[20]}. There is little sensitivity below 450 nm, with the peak at 560 nm returning to zero at 700 nm. Therefore sunlight appears strongest around 560 nm (yellow-green, the colour of tennis balls, unsurprisingly), and that is why we think of sunlight as golden. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;On the other hand, rods show a much more sensitive response towards shorter wavelengths (rod vision shows a peak around 510 nm), but are unable to detect colour. To show the peak response of the rods we convoluted the Moonspectrum of ﬁgure 5 with the rods’ response. The result is also shown in ﬁgure 8. If the light was so low that only the rods were detecting it, the Moon would appear white because the rods are colour blind and our convolution would not lead to a colour bias. In reality, however, at the light level of moonlit landscapes human eyes are working in a condition that has been dubbed as mesopic vision, a combination between photopic and scotopic vision in dim lighting $^{[16, 17]}$. Cones are still furnishing information but at a reduced rate, while the rods, with their superior light-level detections, allow low light-level sensitivity. Under these conditions, the rods couple to blue receptors in the cones to show blue light $^{[3]}$. There seems to be be, therefore, physiological reasons why moonlight appears bluish.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Page 366: &quot;Our perception, then, of moonlight as a colder, silvery light with a blue tinge is acombination of two effects: a physical one due to diminishing Rayleigh scattering as the Moon gets higher in the sky and a physiological one (knownas the Purkinje shift $^{[6]}$). The physiological shift is because cones are more sensitive to yellow-green light, whereas the rods respond best to green-blue light $^{[20]}$ and couple back to blue cones $^{[3]}$. This bias of our eyes towards shorter wavelengths in dim light creates the illusion of a colder bluer moonlight, even though moonlight is actually overall redder than sunlight.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Figure 5 on it's own is misleading due to the above explanation and this text in the paper which explains figure 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Page 363: &quot;We also compared the Moon at altitude 57° spectrum with sunlight in ﬁgure 5, which is normalized the same way as in ﬁgure 4. Since direct sunlight will damage the spectrometer, the data were collected with the telescope–spectrometer pointed in a direction at as small an angular distance away from the Sun as prudent. We did so to protect the instrument while at the same time trying to minimize atmospheric effects. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The data presented in figure 5 needs to be compensated and convolved with the spectrum of eyesight under bright and dark conditions, as shown in figure 8.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;[3]&lt;/code&gt; Khann S M and Pattanaik S N 2004 Modeling blue shift in moonlit scenes by rod cone interaction J. Vision 4 316&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;[6]&lt;/code&gt; Naylor J 2002 Out of the Blue, a 24-hour Skywatcher’s Guide&#xA;   (New York: Cambridge University Press) pp 7–11, 86–87, 195–197&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;[16]&lt;/code&gt; Stockman A and Sharpe L T 2006 Into the twilight zone: the complexities of mesopic vision and luminous efﬁciency Ophthalmic Physiol. Opt. 26 225–39&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;[17]&lt;/code&gt; Pokorny J and Cao D 2010 Rod and cone contributions to mescopic vision Proc. CIE (Int. Commission on Illumination) Conf. Lighting Quality and Energy Efﬁciency, (Vienna)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;[20]&lt;/code&gt; Cornsweet T N 1970 Visual Perception (New York: Academic) pp 145-8&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18054" LastActivityDate="2018-07-02T22:29:21.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26851" PostTypeId="1" CreationDate="2018-07-02T23:15:39.713" Score="2" ViewCount="1187" Body="&lt;p&gt;First of all, I have very basic knowledge of astronomy so sorry if I make some mistakes in my terminology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to plot the Southern Cross. I can find the name of its 5 stars. Also, I have some knowledge of Python and I've tried to use &lt;a href=&quot;http://rhodesmill.org/pyephem/&quot; rel=&quot;nofollow noreferrer&quot;&gt;PyEphem&lt;/a&gt; and &lt;a href=&quot;http://www.astropy.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;AstroPy&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't know is: what kind of information do I need to plot the cross on the circle? I mean, I'd like to plot it exactly at the right position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most difficult part is: the observer should be in &lt;code&gt;Rio de Janeiro&lt;/code&gt; and the date should be &lt;code&gt;1889-11-15 08:30:00 UTC-03&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to collect some info with the Astropy function &lt;code&gt;FixedTarget.from_name()&lt;/code&gt; but, if I understood correctly, the date is too far in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anybody could give me some guidelines (maybe an online tool) to create a sky map with the cross in the right position, I'd really appreciate it. &lt;/p&gt;&#xA;" OwnerUserId="23567" LastEditorUserId="23547" LastEditDate="2018-07-05T19:51:41.377" LastActivityDate="2018-07-08T19:14:56.677" Title="How to construct a sky map" Tags="&lt;star&gt;&lt;amateur-observing&gt;&lt;positional-astronomy&gt;&lt;python&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26852" PostTypeId="1" AcceptedAnswerId="26853" CreationDate="2018-07-03T00:54:31.947" Score="4" ViewCount="195" Body="&lt;p&gt;Our sun is said to be incapable of producing heavier elements, and these are thought to originate from older stars dying and going supernova. If our sun is 4-5 billion years old, and has a lifetime of 10 billion years, how could an earlier sun have lived and died in the short time between the creation of the universe and the birth of our fine solar system?&lt;/p&gt;&#xA;" OwnerUserId="23569" LastEditorUserId="23569" LastEditDate="2018-07-03T03:01:20.237" LastActivityDate="2018-07-03T03:01:20.237" Title="How can heavy elements exist when the universe is young compared to the lifetime of a star?" Tags="&lt;the-sun&gt;&lt;universe&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2018-07-03T15:43:58.373" ContentLicense="CC BY-SA 4.0" />
  <row Id="26853" PostTypeId="2" ParentId="26852" CreationDate="2018-07-03T01:17:28.350" Score="6" Body="&lt;p&gt;First, stars evolve at &lt;em&gt;radically&lt;/em&gt; different rates. The higher the star's mass, the faster it evolves.  The more massive stars had plenty of time to go through multiple generations before the Sun started to coalesce.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It went even faster than it does today, because the collapse of interstellar clouds to form stars is halted mainly by radiation pressure from the new star blowing the rest of its birth cloud away and ending accretion. Since hydrogen and helium are transparent to most of the new star's radiation, the very first stars could grow larger than stars do today. (They were pretty much made of nothing but H and He (with a bit of Li), since very little of the heavy elements had yet formed.) They went through their lifetimes and turned into supernovae and neutron stars very quickly. The largest had lifetimes on the order of just a million years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So by a &lt;em&gt;very&lt;/em&gt; short time after the first stars lit, they were already spewing heavy elements back into space and later stars already had some heavier elements in them at formation.  The fraction of heavy elements has increased steadily since then, but the Sun formed many billions of years after heavy elements became common.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-07-03T01:17:28.350" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26854" PostTypeId="1" CreationDate="2018-07-03T01:36:57.760" Score="2" ViewCount="111" Body="&lt;p&gt;The earliest life on Earth is thought to have began shortly after its formation (a few hundred million years after, by fossil evidence, and this while the late heavy bombardment was raging). With the next discernible leap (eukaryotes) happening billions of years later, That seems like a very short timeframe for life to spontaneously develop. Is it possible that life was seeded on an early Earth by a supernova of an earlier, life-bearing star, blasting living material across the cosmos?&lt;/p&gt;&#xA;" OwnerUserId="23569" LastActivityDate="2018-07-03T11:57:51.530" Title="Could Earth have been seeded with life?" Tags="&lt;earth&gt;&lt;life&gt;" AnswerCount="1" CommentCount="7" ClosedDate="2018-07-05T03:34:15.050" ContentLicense="CC BY-SA 4.0" />
  <row Id="26856" PostTypeId="2" ParentId="26854" CreationDate="2018-07-03T11:57:51.530" Score="0" Body="&lt;p&gt;It seems &lt;em&gt;very&lt;/em&gt; unlikely because while a supernova which sheds enough mass int he explosion can scatter matter from the planets of its system into interstellar space, it also tends to irradiate that matter and to heat it to a very high temperature, so at least as far as a supernova's planets being the source of life, it's difficult to see how that might work.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-07-03T11:57:51.530" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26857" PostTypeId="2" ParentId="26825" CreationDate="2018-07-03T13:33:19.990" Score="1" Body="&lt;p&gt;There are at least two types of supernovae (SNe): thermonuclear supernovae (TSNe) and core-collapse supernovae (CCSNe).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TSNe come from the explosions of white dwarves (WDs) in binary systems. A star with mass at zero age about &amp;lt; 8 solar mass, it normally evolves and ends up being a WD. Our sun is in this category. So, we expect our sun to evolve to be a WD and give no SN explosion without further perturbation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Massive stars (&gt; 8 solar mass), by themselves, evolve to CCSNe. Some of these events leave compact objects either neutron stars or black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pulsar is a rapidly-rotating and magnetized neutron star (or white dwarf but very rare) with misalignment between the rotational axis and the magnetic dipole, and it beams the EM to the line of sight. Speaking of that, if we consider the pulsar as a special case of a neutron star, the progenitor star should have its zero age main sequence mass &gt; 8 solar mass, for ending up in CCSN and hopefully leaving a neutron star as the compact object. Therefore, it is less likely for the sun to be a pulsar at the end of the tunnel. How much more mass? This is even harder to tell because that means the sun has to evolve in a binary system, where the evolutionary path is even more diverse and harder to predict.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-07-03T13:33:19.990" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26859" PostTypeId="2" ParentId="26851" CreationDate="2018-07-03T16:02:18.070" Score="2" Body="&lt;p&gt;There are a few details here that make this non-trivial. The year doesn't matter much, since the same stars are visible each year. Further, in Stellarium, you can set the date to 1889 so this is not an issue but we have to turn off atmospheric effects to make the stars visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More tricky is that the Brazilian flag is designed as if viewed &quot;from above&quot;. We can flip the image from Stellarium to simulate this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the stars are on a globe and the flag is flat, so a projection is needed to represent it. It's not clear what projection the flag designers used, but perhaps it is &quot;orthographic&quot;. Stellarium can do that projection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The projection also seems to be slanted, so &quot;north-south&quot; isn't vertical, although it's not clear by how much. We can estimate this slant from the position of the equator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here is an image of the sky from Rio. The Southern Cross is visible towards the bottom of the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7SeZ0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7SeZ0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to do this yourself in Pyephem you can create a &quot;fixed body&quot; object for each star (Pyephem has a catalogue of brighter stars), create an observer in Rio, and call &quot;body.compute(observer)&quot; which will calculate the azimuth and altitude of the star. Their orthographic projection is then the application of the formulas found on Wikipedia, and you can plot that using any plotting package. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="23547" LastEditDate="2018-07-08T19:14:56.677" LastActivityDate="2018-07-08T19:14:56.677" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26860" PostTypeId="1" AcceptedAnswerId="26865" CreationDate="2018-07-03T19:42:36.993" Score="3" ViewCount="137" Body="&lt;p&gt;As detailed in &lt;a href=&quot;https://physics.stackexchange.com/questions/41450/is-there-a-simple-accurate-formula-for-calculating-transit-times-from-rise-and&quot;&gt;https://physics.stackexchange.com/questions/41450/is-there-a-simple-accurate-formula-for-calculating-transit-times-from-rise-and&lt;/a&gt;, the sunrise, and set times can accurately be used to compute the sun's transit. Yet, in some cases, these do not exist, e.g. in Finland in the summer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an easier and/or more general way to compute the transit time (and only the transit time) given the UTC and observer position (latitude, longitude, height). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;The library at &lt;a href=&quot;http://www.jstott.me.uk/jsuntimes&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.jstott.me.uk/jsuntimes&lt;/a&gt; gives sunrise and sunset, but not the transit time. &lt;code&gt;find_transit&lt;/code&gt; at &lt;code&gt;riset_cir.c&lt;/code&gt; in &lt;a href=&quot;https://github.com/brandon-rhodes/pyephem/tree/master/libastro-3.7.7&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/brandon-rhodes/pyephem/tree/master/libastro-3.7.7&lt;/a&gt; might do this, but seems too general and complicated.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="23386" LastEditorUserId="23386" LastEditDate="2018-07-04T10:22:54.520" LastActivityDate="2018-07-04T13:26:54.510" Title="How to compute the sun's (upper) transit" Tags="&lt;the-sun&gt;&lt;planetary-transits&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="26861" PostTypeId="1" AcceptedAnswerId="26862" CreationDate="2018-07-03T22:07:02.770" Score="14" ViewCount="1031" Body="&lt;p&gt;This page on &lt;a href=&quot;https://en.wikipedia.org/wiki/Quasar&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia -- Quasars&lt;/a&gt; mentions that the &quot;The largest known [quasar] is estimated to consume matter equivalent to 600 Earths per minute&quot;. However, there is no citation for this comment. How can I find out where this information came from? I've commented in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Talk:Quasar&quot; rel=&quot;noreferrer&quot;&gt;Talk&lt;/a&gt; section for the page.&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2018-07-06T04:12:54.213" Title="Quasar mass and accretion rates" Tags="&lt;quasars&gt;&lt;accretion-discs&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26862" PostTypeId="2" ParentId="26861" CreationDate="2018-07-03T23:27:33.210" Score="12" Body="&lt;p&gt;Tricky to say for sure, but I would imagine it comes about from measurements of the luminosity and inference of the black hole mass in such systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most extreme objects radiate at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Eddington_luminosity&quot; rel=&quot;noreferrer&quot;&gt;Eddington luminosity&lt;/a&gt;, where gravitational forces on matter falling into the black hole are balanced by radiation pressure from the heated material closer in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If infalling mass is converted to luminosity at a rate of &#xA;$$ L = \epsilon \dot{M} c^2,$$&#xA;where $\dot{M}$ is the mass accretion rate, $L$ is the luminosity and $\epsilon$ is an efficiency factor, which should be of order 0.1; then the mass accretion rate at the Eddington limit is given by&#xA;$$ \dot{M} = \frac{4\pi G M m_p}{\epsilon c \sigma_T} \simeq 1.4\times 10^{15}\frac{M}{M_{\odot}}\ {\rm kg/s},$$&#xA;where $M$ is the black hole mass, $m_p$ the mass of a proton and $\sigma_T$ is the Thomson scattering cross-section for free electrons (the major source of opacity in the infalling hot gas).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest supermassive black holes in the universe have $M \simeq 10^{10}M_{\odot}$ and thus the Eddington accretion rate for such objects is about $1.4\times 10^{25}$ kg/s or about 2.3 Earths/second or 140 Earths per minute. The difference between this estimate and the one on the wikipedia page could be what is assumed for the biggest $M$ or that $\epsilon$ is a bit smaller than 0.1 or indeed that the luminosity could exceed the Eddington luminosity (because the accretion isn't spherical).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps a simpler way to get the answer is to find the most luminous quasar and divide by $\epsilon c^2$. The most luminous quasar ever seen is probably something like &lt;a href=&quot;http://www.mdpi.com/2075-4434/5/1/3/pdf&quot; rel=&quot;noreferrer&quot;&gt;3C 454.3&lt;/a&gt;, which reaches $\sim 5\times 10^{40}$ Watts in its highest state. Using $\epsilon = 0.1$ yields about an Earth mass per second for the accretion rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So perhaps the number on the wikipedia page is a &lt;em&gt;little&lt;/em&gt; exaggerated.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-07-03T23:38:29.803" LastActivityDate="2018-07-03T23:38:29.803" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26864" PostTypeId="2" ParentId="26861" CreationDate="2018-07-04T06:45:46.013" Score="2" Body="&lt;p&gt;Here is a study from 2012 for the largest recorded quasar which quotes an output of 400 times the mass of the sun per year, which is 253 earth masses per minute (133178400 M ⊕ / 525600 mins) at 2.5 percent the speed of light, located 1 billion light years away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://vtnews.vt.edu/articles/2012/11/112912-science-quasar.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://vtnews.vt.edu/articles/2012/11/112912-science-quasar.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the largest recorded quasar, I don't know the figure for the largest theoretical quasar, there are apparently &lt;a href=&quot;http://www.dailygalaxy.com/my_weblog/2012/11/monster-quasar-discovered-with-power-outflow-100-times-milky-way-galaxy.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;hundreds of people&lt;/a&gt; theorizing and debating the theoretical maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/t2wst.jpg&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;SDSS J1106+1939&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t2wst.jpg&quot; alt=&quot;SDSS J1106+1939&quot; title=&quot;SDSS J1106+1939&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2018-07-06T04:12:54.213" LastActivityDate="2018-07-06T04:12:54.213" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26865" PostTypeId="2" ParentId="26860" CreationDate="2018-07-04T09:12:07.560" Score="1" Body="&lt;p&gt;Here is the page code for &lt;a href=&quot;http://sionnagh.com/mtdistcalc.php&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;my BASIC-to-php implementation&quot;&gt;my BASIC-to-php implementation&lt;/a&gt; of the algorithms to calculate the  (geocentric) RA &amp;amp; Dec of the Sun for a given date from Jean Meeus' book &lt;em&gt;Astronomical Algorithms&lt;/em&gt; as well as calculating the angular distance between two celestial objects.  I can't remember the accuracy of the calculations and I didn't make the corrections for observer lat &amp;amp; long.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt;!DOCTYPE HTML PUBLIC &quot;-//W3C//DTD HTML 4.01 Transitional//EN&quot;&amp;gt;&#xA;&amp;lt;html&amp;gt;&#xA;&amp;lt;head&amp;gt;&#xA;&amp;lt;META HTTP-EQUIV=&quot;Content-Type&quot; CONTENT=&quot;text/html; charset=ISO-8859-1&quot; /&amp;gt;&#xA; &amp;lt;base href=&quot;sionnagh.com&quot; /&amp;gt;&#xA;&amp;lt;title&amp;gt;Mick Todd Distance Calculator&amp;lt;/title&amp;gt;&#xA;&amp;lt;STYLE TYPE=&quot;text/css&quot;&amp;gt;&#xA;&amp;lt;!--&#xA;BODY&#xA;   {&#xA;   font-family:sans-serif;&#xA;   }&#xA;A:link{color:white}&#xA;A:visited{color:yellow}&#xA;--&amp;gt;&#xA;&amp;lt;/STYLE&amp;gt;&#xA;&amp;lt;/head&amp;gt;&#xA;&amp;lt;body&amp;gt;&#xA;&amp;lt;?php include(&quot;mtheader.php&quot;); ?&amp;gt;&#xA;&amp;lt;h1&amp;gt;Mick Todd Distance Calculator&amp;lt;/h1&amp;gt;&#xA;&amp;lt;?php &#xA;$earthincl=23.44;	//Earth's axial tilt&#xA;$sollat=0;&#xA;&#xA;$fdate = $_POST[&quot;fdate&quot;];&#xA;if ($fdate==&quot;&quot;) {&#xA;	$fdate=getdate(date(&quot;U&quot;)-date(&quot;Z&quot;));&#xA;	$fday=$fdate[mday];&#xA;	if (strlen($fday)==1) {$fday=&quot;0&quot;.$fday;}&#xA;	$fmonth=$fdate[mon];&#xA;	if (strlen($fmonth)==1) {$fmonth=&quot;0&quot;.$fmonth;}&#xA;	$fyear=$fdate[year];&#xA;	//$fdate=$fdate[mday].&quot;/&quot;.$fdate[mon].&quot;/&quot;.$fdate[year];&#xA;	$fdate=$fday.&quot;/&quot;.$fmonth.&quot;/&quot;.$fyear;&#xA;}&#xA;else {&#xA;	$fday=substr($fdate,0,2);&#xA;	$fmonth=substr($fdate,3,2);&#xA;	$fyear=substr($fdate,6,4);&#xA;}&#xA;$fra1 = $_POST[&quot;fra1&quot;];&#xA;$fra1 = strtolower($fra1);&#xA;if ($fra1&amp;lt;&amp;gt;&quot;&quot;) {&#xA;  if (strpos($fra1,&quot;h&quot;)==0) {&#xA;  $frad1=$fra1;&#xA;  $fra1=converttohms($fra1);&#xA;  }&#xA;  else {&#xA;    $frad1=converthtod($fra1);&#xA;  }&#xA;}&#xA;$fdec1 = $_POST[&quot;fdec1&quot;];&#xA;$fdec1 = strtolower($fdec1);&#xA;if ($fdec1&amp;lt;&amp;gt;&quot;&quot;) {&#xA;  if (strpos($fdec1,&quot;d&quot;)==0) {&#xA;    $fdecd1=$fdec1;&#xA;    $fdec1=converttodms($fdec1);&#xA;  }&#xA;  else {&#xA;    $fdecd1=convertdtod($fdec1);&#xA;  }&#xA;}&#xA;$fra2 = $_POST[&quot;fra2&quot;];&#xA;$fra2 = strtolower($fra2);&#xA;if ($fra2&amp;lt;&amp;gt;&quot;&quot;) {&#xA;  if (strpos($fra2,&quot;h&quot;)==0) {&#xA;    $frad2=$fra2;&#xA;    $fra2=converttohms($fra2);&#xA;  }&#xA;  else {&#xA;    $frad2=converthtod($fra2);&#xA;  }&#xA;}&#xA;$fdec2 = $_POST[&quot;fdec2&quot;];&#xA;$fdec2 = strtolower($fdec2);&#xA;if ($fdec2&amp;lt;&amp;gt;&quot;&quot;) {&#xA;  if (strpos($fdec2,&quot;d&quot;)==0) {&#xA;    $fdecd2=$fdec2;&#xA;    $fdec2=converttodms($fdec2);&#xA;  }&#xA;  else {&#xA;    $fdecd2=convertdtod($fdec2);&#xA;  }&#xA;}&#xA;&#xA;function JulianDay ($fday, $fmonth, $fyear, $fUT) {&#xA;    $JDpartsign=(100*$fyear+$fmonth-190002.5)/abs(100*$fyear+$fmonth-190002.5);&#xA;	$JD = 367*$fyear - floor((7*($fyear+floor(($fmonth+9)/12)))/4) + floor((275*$fmonth)/9) + $fday + 1721013.5 + $fUT/24 - 0.5*$JDpartsign + 0.5;&#xA;	return $JD;&#xA;}&#xA;function converttodms($inval) {&#xA;	if ($inval==0) {&#xA;	  $fdsign=1;&#xA;	}&#xA;	else {&#xA;	  $fdsign=$inval/abs($inval);&#xA;	}&#xA;	$inval=abs($inval);&#xA;	$fddeg=floor($inval);&#xA;	$remain=($inval-$fddeg)*60;&#xA;	$fdmin=floor($remain);&#xA;	if (strlen($fdmin)==1) { $fdmin=&quot;0&quot;.$fdmin; }&#xA;	$remain=($remain-$fdmin)*60;&#xA;	$fdsec=round($remain*100)/100;&#xA;	if (strpos($fdsec,&quot;.&quot;)==1) { $fdsec=&quot;0&quot;.$fdsec; }&#xA;	$fddeg=$fdsign*$fddeg;&#xA;	$fdms=$fddeg.&quot;d&quot;.$fdmin.&quot;m&quot;.$fdsec.&quot;s&quot;;&#xA;&#xA;    return $fdms;&#xA;}&#xA;function converttohms($inval) {&#xA;    $inval=$inval/15;&#xA;    $fddeg=floor($inval);&#xA;    $remain=($inval-$fddeg)*60;&#xA;	$fdmin=floor($remain);&#xA;	if (strlen($fdmin)==1) { $fdmin=&quot;0&quot;.$fdmin; }&#xA;    $remain=($remain-$fdmin)*60;&#xA;	$fdsec=round($remain*100)/100;&#xA;	if (strpos($fdsec,&quot;.&quot;)==1) { $fdsec=&quot;0&quot;.$fdsec; }&#xA;    $fdms=$fddeg.&quot;h&quot;.$fdmin.&quot;m&quot;.$fdsec.&quot;s&quot;;&#xA;&#xA;    return $fdms;&#xA;}&#xA;function converthtod($inval) {&#xA;  $posh=strpos($inval,&quot;h&quot;);&#xA;  $posm=strpos($inval,&quot;m&quot;);&#xA;  $poss=strpos($inval,&quot;s&quot;);&#xA;  $temph=substr($inval,0,$posh);&#xA;  if ($posm) {&#xA;    $tempm=substr($inval,$posh+1,$posm-($posh+1));&#xA;  }&#xA;  if ($poss) {&#xA;    $temps=substr($inval,$posm+1,$poss-($posm+1));&#xA;  }&#xA;&#xA;  $fdms=($temph+$tempm/60+$temps/3600)*15;&#xA;&#xA;  return $fdms;&#xA;}&#xA;function convertdtod($inval) {&#xA;  $posd=strpos($inval,&quot;d&quot;);&#xA;  $posm=strpos($inval,&quot;m&quot;);&#xA;  $poss=strpos($inval,&quot;s&quot;);&#xA;  $tempd=substr($inval,0,$posd);&#xA;  if ($posm) {&#xA;    $tempm=substr($inval,$posd+1,$posm-($posd+1));&#xA;  }&#xA;  if ($poss) {&#xA;    $temps=substr($inval,$posm+1,$poss-($posm+1));&#xA;  }&#xA;  $fdms=$tempd+$tempm/60+$temps/3600;&#xA;&#xA;  return $fdms;&#xA;}&#xA;&#xA;function SolarCoords($fday, $fmonth, $fyear, $fUT) {&#xA;//Julian Day of 1991/ 5/19 at 13 UT         JD = 2448396.04167&#xA;//Julian day of 2000/01/01 at 12 UT         JD = 2451545.0&#xA;//number of Julian days since 2000/01/01 at 12 UT               -3148.95833&#xA;//number of Julian centuries since 2000/01/01 at 12 UT  T = - 3148.95833/36525&#xA;//used by the algorithm for L                                                           = 0.086213780&#xA;&#xA;    $k = 2*pi()/360;&#xA;  $T = (JulianDay($fday, $fmonth, $fyear, $fUT)-2451545.0)/36525;&#xA;    //echo &quot;\$T: &quot;.$T.&quot;&amp;lt;br /&amp;gt;&quot;;&#xA;	//mean anomaly, degree&#xA;	$M = 357.52910 + 35999.05030*$T - 0.0001559*$T*$T - 0.00000048*$T*$T*$T;&#xA;    //echo &quot;\$M: &quot;.$M.&quot;&amp;lt;br /&amp;gt;&quot;;&#xA;&#xA;    // mean longitude, degree&#xA;    $L0 = 280.46645 + 36000.76983*$T + 0.0003032*$T*$T;&#xA;    //echo &quot;\$L0: &quot;.$L0.&quot;&amp;lt;br /&amp;gt;&quot;;&#xA;&#xA;    $DL = (1.914600 - 0.004817*$T - 0.000014*$T*$T)*sin($k*$M) + (0.019993 - 0.000101*$T)*sin($k*2*$M) + 0.000290*sin($k*3*$M);&#xA;	//echo &quot;\$DL: &quot;.$DL.&quot;&amp;lt;br /&amp;gt;&quot;;&#xA;&#xA;    // true longitude, degree&#xA;    $L = $L0 + $DL;&#xA;	if (abs($L)&amp;gt;360) {&#xA;        $divsign=($L/360)/abs($L/360);&#xA;		$div=floor(abs($L/360));&#xA;		$L=$L-$divsign*$div*360;&#xA;		if ($L&amp;lt;0) { $L=$L+360; }&#xA;    }&#xA;    $L=round($L*10000)/10000;&#xA;&#xA;    return $L;&#xA;&#xA;}&#xA;&#xA;?&amp;gt;&#xA;&amp;lt;script type=&quot;text/javascript&quot;&amp;gt;&#xA;    &amp;lt;!--&#xA;    function settoday() {&#xA;        var d=new Date();&#xA;        dday=d.getDate();&#xA;        if (dday&amp;lt;10) { dday=&quot;0&quot; + dday; }&#xA;        dmonth=d.getMonth()+1;&#xA;        if (dmonth&amp;lt;10) { dmonth=&quot;0&quot; + dmonth; }&#xA;        dyear=d.getFullYear();&#xA;&#xA;        ddate= dday + &quot;/&quot; + dmonth + &quot;/&quot; + dyear;&#xA;&#xA;        return ddate;&#xA;&#xA;    }&#xA;&#xA;    //--&amp;gt;&#xA;&amp;lt;/script&amp;gt;&#xA;&amp;lt;form METHOD=&quot;POST&quot; ACTION=&quot;mtdistcalc.php&quot; name=&quot;mtdistcalc&quot;&amp;gt;&#xA;&#xA;&#xA;&amp;lt;p&amp;gt;Date &amp;lt;input type=&quot;text&quot; name=&quot;fdate&quot; id=&quot;fdate&quot; size=10 value=&quot;&amp;lt;?php echo $fdate; ?&amp;gt;&quot;&amp;gt; &amp;lt;input type=&quot;button&quot; value=&quot;Now&quot; onclick=&quot;fdate.value=settoday();&quot;&amp;gt;&amp;lt;/p&amp;gt;&#xA;&amp;lt;p&amp;gt;Object coordinates (#1) RA &amp;lt;input type=&quot;text&quot; name=&quot;fra1&quot; id=&quot;fra1&quot; size=15 value=&quot;&amp;lt;?php echo $fra1; ?&amp;gt;&quot;&amp;gt;  Decl &amp;lt;input type=&quot;text&quot; name=&quot;fdec1&quot; id=&quot;fdec1&quot; size=15 value=&quot;&amp;lt;?php echo $fdec1; ?&amp;gt;&quot;&amp;gt;&amp;lt;/p&amp;gt;&#xA;&amp;lt;p&amp;gt;Object coordinates (#2) RA &amp;lt;input type=&quot;text&quot; name=&quot;fra2&quot; id=&quot;fra2&quot; size=15 value=&quot;&amp;lt;?php echo $fra2; ?&amp;gt;&quot;&amp;gt;  Decl &amp;lt;input type=&quot;text&quot; name=&quot;fdec2&quot; id=&quot;fdec2&quot; size=15 value=&quot;&amp;lt;?php echo $fdec2; ?&amp;gt;&quot;&amp;gt;&amp;lt;/p&amp;gt;&#xA;  &amp;lt;p&amp;gt;&amp;lt;input TYPE=SUBMIT VALUE=&quot;Calculate distances&quot;&amp;gt;&amp;lt;/p&amp;gt;&#xA;&#xA;&#xA;&amp;lt;/form&amp;gt;&#xA;&amp;lt;?php&#xA;echo &quot;&amp;lt;table&amp;gt;&amp;lt;tr&amp;gt;&amp;lt;td&amp;gt;Date:&amp;lt;/td&amp;gt;&amp;lt;td&amp;gt;&quot;.$fday.&quot;/&quot;.$fmonth.&quot;/&quot;.$fyear.&quot; &quot;.&quot;0000UT&amp;lt;/td&amp;gt;&amp;lt;/tr&amp;gt;&quot;;&#xA;echo &quot;&amp;lt;tr&amp;gt;&amp;lt;td&amp;gt;Julian Day: &amp;lt;/td&amp;gt;&amp;lt;td&amp;gt;&quot;.JulianDay($fday,$fmonth,$fyear,&quot;0&quot;).&quot;&amp;lt;/td&amp;gt;&amp;lt;/tr&amp;gt;&quot;;&#xA;echo &quot;&amp;lt;tr&amp;gt;&amp;lt;td&amp;gt;Ecliptic latitude of the Sun: &amp;lt;/td&amp;gt;&amp;lt;td&amp;gt;&quot;.$sollat.&quot;&amp;amp;deg; (assumed)&amp;lt;/td&amp;gt;&amp;lt;/tr&amp;gt;&quot;;&#xA;$sollong = SolarCoords($fday,$fmonth,$fyear,&quot;0&quot;);&#xA;echo &quot;&amp;lt;tr&amp;gt;&amp;lt;td&amp;gt;Ecliptic longitude of the Sun: &amp;lt;/td&amp;gt;&amp;lt;td&amp;gt;&quot;.$sollong.&quot;&amp;amp;deg; (&quot;.converttodms($sollong).&quot;)&amp;lt;/td&amp;gt;&amp;lt;/tr&amp;gt;&quot;;&#xA;echo &quot;&amp;lt;tr&amp;gt;&amp;lt;td&amp;gt;Earth's axial inclination: &amp;lt;/td&amp;gt;&amp;lt;td&amp;gt;&quot;.$earthincl.&quot;&amp;amp;deg;&amp;lt;/td&amp;gt;&amp;lt;/tr&amp;gt;&amp;lt;/table&amp;gt;&amp;lt;br /&amp;gt;&quot;;&#xA;&#xA;//echo &quot;\$feclong &quot;.$feclong.&quot;&amp;lt;br /&amp;gt;&amp;lt;br /&amp;gt;&quot;;&#xA;&#xA;$deltaSol = round(asin(sin($sollat*M_PI/180)*cos($earthincl*M_PI/180) + cos($sollat*M_PI/180)*sin($earthincl*M_PI/180)*sin($sollong*M_PI/180))*180/M_PI*10000)/10000;&#xA;$alphaSol = round(acos(cos($sollong*M_PI/180)*cos($sollat*M_PI/180) / cos($deltaSol*M_PI/180))*180/M_PI*10000)/10000;&#xA;if ($sollong&amp;gt;180) {&#xA;	$alphaSol = 360-$alphaSol;&#xA;}&#xA;echo &quot;&amp;lt;table&amp;gt;&amp;lt;tr&amp;gt;&amp;lt;td colspan=2&amp;gt;Sol&amp;lt;/td&amp;gt;&amp;lt;/tr&amp;gt;&amp;lt;tr&amp;gt;&amp;lt;td&amp;gt;RA: &amp;lt;/td&amp;gt;&amp;lt;td&amp;gt;&quot;.$alphaSol.&quot;&amp;amp;deg; (&quot;.converttohms($alphaSol).&quot;)&amp;lt;/td&amp;gt;&amp;lt;/tr&amp;gt;&amp;lt;tr&amp;gt;&amp;lt;td&amp;gt;Dec:&amp;lt;/td&amp;gt;&amp;lt;td&amp;gt;&quot;.$deltaSol.&quot;&amp;amp;deg; (&quot;.converttodms($deltaSol).&quot;)&amp;lt;/td&amp;gt;&amp;lt;/tr&amp;gt;&amp;lt;/table&amp;gt;&amp;lt;br /&amp;gt;&quot;;&#xA;&#xA;//spherical angular distance&#xA;//cos(theta) = sin(delta_a) sin(delta_b) + cos(delta_a) cos(delta_b) cos(alpha_a-alpha_b)&#xA;if ($fra1&amp;lt;&amp;gt;&quot;&quot; AND $fdec1&amp;lt;&amp;gt;&quot;&quot;) {&#xA;  $theta1 = round(acos(sin($deltaSol*M_PI/180)*sin($fdecd1*M_PI/180) + cos($deltaSol*M_PI/180)*cos($fdecd1*M_PI/180)*cos(($alphaSol-$frad1)*M_PI/180))*180/M_PI*1000)/1000;&#xA;}&#xA;if ($fra2&amp;lt;&amp;gt;&quot;&quot; AND $fdec2&amp;lt;&amp;gt;&quot;&quot;) {&#xA;  $theta2 = round(acos(sin($deltaSol*M_PI/180)*sin($fdecd2*M_PI/180) + cos($deltaSol*M_PI/180)*cos($fdecd2*M_PI/180)*cos(($alphaSol-$frad2)*M_PI/180))*180/M_PI*1000)/1000;&#xA;}&#xA;&#xA;if ($fra1&amp;lt;&amp;gt;&quot;&quot; AND $fdec1&amp;lt;&amp;gt;&quot;&quot; AND $fra2&amp;lt;&amp;gt;&quot;&quot; AND $fdec2&amp;lt;&amp;gt;&quot;&quot;) {&#xA;  $theta3 = round(acos(sin($fdecd1*M_PI/180)*sin($fdecd2*M_PI/180) + cos($fdecd1*M_PI/180)*cos($fdecd2*M_PI/180)*cos(($frad1-$frad2)*M_PI/180))*180/M_PI*1000)/1000;&#xA;  echo &quot;&amp;lt;p&amp;gt;&amp;lt;font color=\&quot;#0000FF\&quot;&amp;gt;Angular distance between Object #1 and Object #2 is: &quot;.$theta3.&quot;&amp;amp;deg;&amp;lt;/font&amp;gt;&amp;lt;/p&amp;gt;&quot;;&#xA;}&#xA;&#xA;if (($fra1&amp;lt;&amp;gt;&quot;&quot; and $fdec1&amp;lt;&amp;gt;&quot;&quot;) or ($fra2&amp;lt;&amp;gt;&quot;&quot; and $fdec2&amp;lt;&amp;gt;&quot;&quot;)) {&#xA;  echo &quot;&amp;lt;table&amp;gt;&amp;lt;tr&amp;gt;&amp;lt;td&amp;gt;&amp;amp;nbsp;&amp;lt;/td&amp;gt;&amp;lt;td align=center&amp;gt;&amp;lt;b&amp;gt;RA&amp;lt;/b&amp;gt;&amp;lt;/td&amp;gt;&amp;lt;td align=center&amp;gt;&amp;lt;b&amp;gt;Dec&amp;lt;/b&amp;gt;&amp;lt;/td&amp;gt;&amp;lt;td&amp;gt;&amp;lt;b&amp;gt;Solar elong.&amp;lt;/b&amp;gt;&amp;lt;/td&amp;gt;&amp;lt;/tr&amp;gt;&quot;;&#xA;}&#xA;if ($fra1&amp;lt;&amp;gt;&quot;&quot; AND $fdec1&amp;lt;&amp;gt;&quot;&quot;) {&#xA;  echo &quot;&amp;lt;tr&amp;gt;&amp;lt;td&amp;gt;Object #1:&amp;lt;/td&amp;gt;&amp;lt;td&amp;gt;&amp;lt;font color=Maroon&amp;gt;&quot;.$fra1.&quot;&amp;lt;/font&amp;gt;&amp;lt;/td&amp;gt;&amp;lt;td&amp;gt;&amp;lt;font color=Maroon&amp;gt;&quot;.$fdec1.&quot;&amp;lt;/font&amp;gt;&amp;lt;/td&amp;gt;&amp;lt;td&amp;gt;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;lt;font color=Maroon&amp;gt;&quot;.$theta1.&quot;&amp;amp;deg;&amp;lt;/font&amp;gt;&amp;lt;/td&amp;gt;&amp;lt;/tr&amp;gt;&quot;;&#xA;}&#xA;if ($fra2&amp;lt;&amp;gt;&quot;&quot; AND $fdec2&amp;lt;&amp;gt;&quot;&quot;) {&#xA;  echo &quot;&amp;lt;tr&amp;gt;&amp;lt;td&amp;gt;Object #2:&amp;lt;/td&amp;gt;&amp;lt;td&amp;gt;&quot;.$fra2.&quot;&amp;lt;/td&amp;gt;&amp;lt;td&amp;gt;&quot;.$fdec2.&quot;&amp;lt;/td&amp;gt;&amp;lt;td&amp;gt;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&quot;.$theta2.&quot;&amp;amp;deg;&amp;lt;/td&amp;gt;&amp;lt;/tr&amp;gt;&quot;;&#xA;}&#xA;&#xA;&#xA;?&amp;gt;&#xA;&amp;lt;/body&amp;gt;&#xA;&amp;lt;/html&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2018-07-04T09:12:07.560" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26868" PostTypeId="2" ParentId="26709" CreationDate="2018-07-04T17:38:00.060" Score="1" Body="&lt;p&gt;Stars use pretty much every element from Hydrogen to Iron as fuels. The three main ones are Hydrogen, Helium and Carbon. Hydrogen is a light element so it produces much energy as it fuses into deuterium or Helium. The heavier the element is the less energy it produces. Heavier element shortens the supply of fuels as the rate of reaction keeps on increasing. For example Carbon takes 600 years to fuse into neon and silicon takes one day to fuse into Iron. After iron it just becomes counter efficient for the star to fuse it into heavier elements as it would need more energy to fuse into heavier elemnets than they would produce. So after iron it comes to a slow stop.So anyting from hydrogen to iron could be use as fuel&#xA;Reference: &lt;a href=&quot;http://abyss.uoregon.edu/~js/ast122/lectures/lec18.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://abyss.uoregon.edu/~js/ast122/lectures/lec18.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23582" LastActivityDate="2018-07-04T17:38:00.060" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26869" PostTypeId="2" ParentId="24196" CreationDate="2018-07-04T18:38:19.823" Score="4" Body="&lt;p&gt;There's no maximum theoretical size short of the size of the universe.  But there's no way in nature that a Black Hole that big could form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More interesting, perhaps, are the BHs which formed naturally. The largest natural BHs we know of the the supermassive BHs in galactic centers.  There is a very interesting relationship that we have observed: &lt;em&gt;The mass of Galactic BHs is never more than about 0.001 of the mass of their home galaxy's central bulge&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't understand -- yet -- in any detail why this is so, but it is strongly suspected that a galactic bulge an its central BH grow at the same time and stop growing together, also. Modelling work has been done which makes this plausible, but as far as I know no one has a fully satisfactory theory explaining their origin and growth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supermassive BHs sometimes merge after their host galaxies merge, though in most cases there probably hasn't been enough time for that to happen. (We do know of galaxies with multiple SMBHs in them.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting this all together, the largest BHs found in nature will be a few times more than 0.001 of the mass of the largest galactic bulges, and will be formed when  several especially large galaxies merge and their SMBHs eventually also merge.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-07-04T18:38:19.823" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26870" PostTypeId="1" CreationDate="2018-07-05T00:54:47.623" Score="5" ViewCount="335" Body="&lt;p&gt;Could some influence from outside the observable universe be causing the expansion?&lt;/p&gt;&#xA;" OwnerUserId="23586" LastEditorUserId="14094" LastEditDate="2018-07-27T03:25:39.510" LastActivityDate="2018-08-20T12:19:58.523" Title="Influences from outside the observable universe explaining dark energy and expansion?" Tags="&lt;universe&gt;&lt;cosmology&gt;&lt;expansion&gt;&lt;dark-energy&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26871" PostTypeId="1" AcceptedAnswerId="26876" CreationDate="2018-07-05T02:20:59.253" Score="6" ViewCount="102" Body="&lt;p&gt;In this picture of 3C273, we can see a jet&amp;mdash;estimated to be around 200,000 light-years long&amp;mdash;being emitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fhLVR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fhLVR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to find an estimate for the mass of the jet. It's a stream of electrons, accelerated through synchrotron radiation (see &lt;a href=&quot;https://ned.ipac.caltech.edu/level5/March02/Courvoisier/Cour6.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;). There seems to be lots of information on its spectroscopic properties and the mass of the quasar (886 million solar masses), but I can't find an estimate for the mass of the jet itself. Does anyone know of a source I could look at to give me some idea?&lt;/p&gt;&#xA;" OwnerUserId="17535" LastEditorUserId="37191" LastEditDate="2021-04-01T07:48:45.383" LastActivityDate="2021-04-01T07:48:45.383" Title="3C273 jet mass estimate" Tags="&lt;quasars&gt;&lt;resource-request&gt;&lt;astronomical-jet&gt;&lt;estimate&gt;&lt;synchrotron-radiation&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26873" PostTypeId="2" ParentId="26870" CreationDate="2018-07-05T08:00:19.980" Score="3" Body="&lt;p&gt;EDIT - The original question was edited to enable it to be reopened, but my answer below responded to the way the question was initially framed. &lt;em&gt;I've left my answer intact as it may be useful in addressing confusion about the difference between &quot;the universe&quot; and &quot;the observable universe&quot;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Could some pressure from outside of our universe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;Universe&lt;/a&gt; is &quot;all of space and time and their contents&quot;. So it makes no sense to talk of something &quot;outside&quot; all of space and time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might, of course, be referring just to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Observable_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;observable universe&lt;/a&gt;, &quot;a spherical region of the Universe comprising all matter that can be observed from Earth at the present time.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the conditions beyond the observable universe are what is causing expansion, then the Universe is neither homogenous nor isotropic. While we have no way of disproving this idea, all the evidence points to a homogenous and isotropic Universe and the vast majority of physicists subscribe to this model.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;causing the expansion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Expansion_of_the_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;metric expansion of the Universe&lt;/a&gt; is a generic property of the Universe under the &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann%E2%80%93Lema%C3%AEtre%E2%80%93Robertson%E2%80%93Walker_metric&quot; rel=&quot;nofollow noreferrer&quot;&gt;FLRW metric&lt;/a&gt;, an exact solution of Einstein's field equations of general relativity. There is no need to propose an external cause for the expansion.&lt;/p&gt;&#xA;" OwnerUserId="14094" LastEditorUserId="14094" LastEditDate="2018-08-20T12:19:58.523" LastActivityDate="2018-08-20T12:19:58.523" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26874" PostTypeId="2" ParentId="26870" CreationDate="2018-07-05T12:02:44.793" Score="2" Body="&lt;p&gt;Yes, it's possible, but there's absolutely no evidence for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;String theory -- which is, I emphasize, a theory for which there is &lt;em&gt;no&lt;/em&gt; evidence -- has many incompatible variations, and some of them (e.g., the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ekpyrotic_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ekpyrotic theories&lt;/a&gt;) suggest that our 4D universe may be one of many that exist on &quot;branes&quot; in a higher-dimensional space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Collisions between the branes may be possible and would have an impact (pun unavoidable) on our universe -- and may even have been the cause of the Big Bang.  Such collisions ought to have left an imprint on spacetime, but searches for the expected fingerprints have come up negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None-the-less, String theory and all its many children are flexible enough and unconstrained by observation enough that a solution where the cosmological constant is affected by other brane universes is plausible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I wouldn't believe &lt;em&gt;any&lt;/em&gt; of that until there's some observational evidence for String theory.  At the moment it's basically a modern version of &lt;a href=&quot;https://en.wikipedia.org/wiki/Mysterium_Cosmographicum&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kepler's Mysterium Cosmographicum&lt;/a&gt; -- theory running wild.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-07-05T12:02:44.793" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26875" PostTypeId="1" CreationDate="2018-07-05T14:38:07.200" Score="5" ViewCount="361" Body="&lt;p&gt;For a given &lt;strong&gt;date&lt;/strong&gt;, &lt;strong&gt;longitude&lt;/strong&gt;, and &lt;strong&gt;latitude&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Popular apps and websites give the time when the sun's center or edge is on the horizon (i.e., the time of sunrise and sunset) or its transit (i.e., noontime), but I want to ask how to calculate the time needed for the &lt;strong&gt;solar disk&lt;/strong&gt; to pass the &lt;strong&gt;horizon's&lt;/strong&gt; line and the &lt;strong&gt;transit's&lt;/strong&gt; line (preferably in c# or java)?&lt;/p&gt;&#xA;" OwnerUserId="23589" LastEditorUserId="23547" LastEditDate="2018-07-05T19:51:18.523" LastActivityDate="2018-07-08T15:24:51.590" Title="How to calculate the time for the solar disk to pass the horizon and transit's lines" Tags="&lt;the-sun&gt;&lt;mathematics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26876" PostTypeId="2" ParentId="26871" CreationDate="2018-07-05T15:15:34.760" Score="2" Body="&lt;p&gt;Trigger Warning:  Major hacking going on here.&#xA;[edited from 1 ly to 1000 ly diameter]&#xA;Scaling from the image, I'll approximate the jet diameter as 1000 ly . So there's a volume of roughly 2e4*1e6 * pi/4 cubic ly.   For absolutely no good reason, I'll propose an electron ejection density of 1E6 per cubic km -- at least those are round numbers and easily scalable if anyone can find references to actual calculations/measurements --, so using 1 cubic ly = 8.468e+38 cubic km, I get 1.33e+56 electrons in the jet.  &lt;/p&gt;&#xA;" OwnerUserId="907" LastEditorUserId="907" LastEditDate="2018-07-06T13:22:59.717" LastActivityDate="2018-07-06T13:22:59.717" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26877" PostTypeId="2" ParentId="12576" CreationDate="2018-07-05T17:18:50.597" Score="0" Body="&lt;p&gt;Think of the universe as a birthday cake. If you had a cake that was a million light years across, you could fit at least a hundred thousand candles on it, right? Probably many more. But just think how many candles you could fit on a cake that was infinite in size - millions, if not billions. But if you put this cake back in the oven and bake a smaller cake onto the side of it, then you could say you've &quot;expanded the cake&quot;, despite it being infinite in size. In this way, you could keep adding to your cake until it truly was infinite.&lt;/p&gt;&#xA;" OwnerUserId="23594" LastActivityDate="2018-07-05T17:18:50.597" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26878" PostTypeId="1" AcceptedAnswerId="26890" CreationDate="2018-07-05T22:08:24.673" Score="2" ViewCount="138" Body="&lt;p&gt;Conservation of angular momentum doesn't seem correct to me. I think this because radiation is present everywhere, no space is completely void, and no object perfect. Thus all objects will have uneven radiation pressure. My question is there anything in space that is not increasing or decreasing in rotation or is fixed in space? Is the rate in which an object spins is determined by surrounding celestial bodies? Or in other words any object in space depending on its position in the galaxy will slow or increase to an ideal rotation?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-07-07T16:11:59.147" LastActivityDate="2018-07-07T16:11:59.147" Title="Is it possible to define a state of zero absolute angular momentum anywhere in the universe?" Tags="&lt;rotation&gt;&lt;radio-astronomy&gt;&lt;positional-astronomy&gt;&lt;galactic-dynamics&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26879" PostTypeId="2" ParentId="26813" CreationDate="2018-07-06T02:18:24.057" Score="0" Body="&lt;p&gt;Using spectropolarimeter one can measure vector Magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A spectropolarimeter based on a ferroelectric liquid-crystal modulator is described. An optical system with spatial modulation of the positions of the components of Zeeman splitting is a specific feature of this instrument. In comparison to the familiar instruments, the developed spectropolarimeter utilizes the light flux more efficiently and contains only one photodetector array. An operating spectropolarimeter developed at the Sayan Solar Observatory is considered as an example. Comparative estimates of noises in different operating modes are presented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Ref: Kolobov, D.Y., Kobanov, N.I. &amp;amp; Grigoryev, V.M. Instrum Exp Tech (2008) 51: 124. &lt;a href=&quot;https://doi.org/10.1134/S0020441208010156&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://doi.org/10.1134/S0020441208010156&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="23450" LastActivityDate="2018-07-06T02:18:24.057" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26880" PostTypeId="1" CreationDate="2018-07-06T11:36:46.690" Score="19" ViewCount="1453" Body="&lt;p&gt;Yesterday my sister and I have discussed the recent efforts of SpaceX to create a permanent colony on mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When coming to the several dangers of creating a habitat on a different planet, we got stuck on the probability of mars getting hit by an astronomical object thats large enough to wipe out life anywhere on the planet.&#xA;I assume the probability is higher than on Earth, due to Mars being closer to the Asteroid Belt, its size and thinner atmosphere. There are more potential objects to enter martian atmosphere and can be smaller than on Earth to create the same hazardous impact for life. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to research the numbers, but the only calculations I found were for specific asteroids and their probability to hit Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On this StackExchange there are some questions concerningthe probability of an &lt;a href=&quot;https://astronomy.stackexchange.com/questions/11418/what-is-the-probability-of-an-astronomical-body-hitting-you&quot;&gt;individual  (on earth)&lt;/a&gt; or a &lt;a href=&quot;https://astronomy.stackexchange.com/questions/21372/is-there-any-mathematical-formula-paper-or-anything-that-shows-the-probability&quot;&gt;specific spot (on the moon)&lt;/a&gt; getting hit, but I found none for the hit of an object with a certain amount of impact (I hope thats the right word here). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the probability of an astronomical object hitting mars anywhere on its surface and kill all people currently living there?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any impact destroying life-support systems and thus wiping out life on mars indirectly can also be included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that there is only one habitat, it is built directly on the surface and the first humans have arrived with the latest flight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are any questions about the scope of this question, don't hesitate to ask.&lt;/p&gt;&#xA;" OwnerUserId="23598" LastActivityDate="2018-07-07T02:03:47.373" Title="Whats the probability of a hazardous impact for human life on Mars?" Tags="&lt;asteroids&gt;&lt;mars&gt;&lt;life&gt;&lt;impact-probability&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26881" PostTypeId="2" ParentId="26880" CreationDate="2018-07-06T12:03:58.287" Score="15" Body="&lt;p&gt;We actually have a very good idea of this because the &lt;a href=&quot;https://en.wikipedia.org/wiki/Mars_Reconnaissance_Orbiter&quot; rel=&quot;noreferrer&quot;&gt;Mars Reconnaissance Orbiter&lt;/a&gt; has been orbiting Mars for over a decade. The MRO is, basically, a spy satellite around Mars and is continually taking high-resolution photos of the surface. It has revisted much of the surface, taking pictures multiple times over the years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, we have a very good idea of the rate of formation of craters on Mars. We have seen numerous craters that have formed between visits to a spot, but none are very large. The key thing is that we now have a very solid measurement of the flux of smaller objects onto Mars and can say with certainty, that that sort of impact is way down on the list of perils. (It's probably less than the odds of being killed by a tornado in the Midwest.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Larger impacts are a bit harder since we haven't seen any, but we have two lines of evidence. First, space debris follows a distribution in size vs. frequency that we've measured, and it's not much different around Mars' orbit than it is around Earth's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the &lt;em&gt;extensive&lt;/em&gt; photography we have of Mars' surface (which we now have mapped better than Earth's!) allows us to count craters and to make estimates of their age. We know that small craters a few yards to tens of yards across are much more common and we know the rate at which they hit, so astronomers count the number of small craters on the floors of larger craters and can make a very good estimate of the age of the larger crater. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this we have developed measurements of the rate of cratering at all sizes, and, again, the hazard isn't that great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd note that the risk of what on Earth would be civilization-killers is actually &lt;em&gt;lower&lt;/em&gt; on Mars because of its lack of an atmosphere. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Chicxulub_impactor&quot; rel=&quot;noreferrer&quot;&gt;Chicxulub&lt;/a&gt; impact that killed the dinosaurs killed most of them through the superheated steam and air shockwave that went around the Earth followed by the years of Asteroid Winter that followed.  Mars' atmosphere isn't heavy enough to support either, so &quot;all&quot; you'd have to do is hunker down and wait out the fall of rock fragments... (Still, it's an event better watched from orbit than from the ground.)&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-07-06T12:03:58.287" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="26882" PostTypeId="1" CreationDate="2018-07-06T12:31:25.883" Score="3" ViewCount="48" Body="&lt;p&gt;Following on from a query concerning &lt;a href=&quot;https://astronomy.stackexchange.com/questions/26861/quasar-mass-and-accretion-rates&quot;&gt;quasar mass&lt;/a&gt;, what fraction of matter is ejected compared to matter that is lost to the BH? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a graph that describes the fractions of the mass of quasars which contribute to disk radiation, plume radiation and black-hollowed? (sorry I can't think of a word because accretion and accretion disk is confusing) &lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2018-07-07T09:52:44.333" LastActivityDate="2018-07-07T09:52:44.333" Title="What fractions of matter are retained and displaced by quasars?" Tags="&lt;black-hole&gt;&lt;quasars&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26883" PostTypeId="1" AcceptedAnswerId="26884" CreationDate="2018-07-06T13:23:53.467" Score="2" ViewCount="672" Body="&lt;p&gt;If the energy/matter that was contained in the singularity before the Big Bang had always existed, and if it’s true that energy/matter cannot be created or destroyed, then what explains its existence?&lt;/p&gt;&#xA;" OwnerDisplayName="user23327" LastActivityDate="2018-10-31T03:42:53.103" Title="What explains the existence of energy/matter if it cannot be created or destroyed?" Tags="&lt;universe&gt;&lt;astrophysics&gt;&lt;cosmology&gt;&lt;cosmological-inflation&gt;&lt;matter&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="26884" PostTypeId="2" ParentId="26883" CreationDate="2018-07-06T13:53:33.783" Score="7" Body="&lt;p&gt;That's a very complicated question!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let's remember that Moses didn't bring the Law of Conservation of Energy down from Sinai on stone tablets -- it's something that we've observed to be true of the universe today, and which our theories use to make very accurate predictions. But this doesn't mean that it was always true, and particularly that it was true early enough in the Big Bang that our current theories don't apply.  We &lt;em&gt;know&lt;/em&gt; that our current theories fail during the &lt;a href=&quot;https://en.wikipedia.org/wiki/Planck_units#Cosmology&quot; rel=&quot;noreferrer&quot;&gt;Planck Era&lt;/a&gt;, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, &lt;a href=&quot;https://en.wikipedia.org/wiki/Noether%27s_theorem&quot; rel=&quot;noreferrer&quot;&gt;Noether's Theorem&lt;/a&gt; says that energy conservation exists because the Lagrangians of mass-energy are time-symmetric. Time symmetry is a very good symmetry today in regionals of space small compared with the entire universe, but there may well have been eras in which it wasn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, there probably were such eras, just as time-symmetry is not a good symmetry even today at very large scales.  (The universe is expanding, and on the face of it, this causes a non-conservation of energy.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thirdly, if you try to save the law of conservation of energy by calculating the total energy of the entire universe including the gravitational energy, one reasonable set of arguments (energy calculations in General Relativity can be very difficult, since gravitational energy tends to be non-local) yields a total energy of zero! The mass-energy in the universe is exactly balanced by the gravitational potential energy.  If this is correct, the total energy has always been zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is all speculative!  Remember, that real physics is an experimental science and if you can't test a theory it may be interesting, but it isn't physics.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-07-06T13:53:33.783" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26885" PostTypeId="2" ParentId="13213" CreationDate="2018-07-06T14:33:56.340" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What percentage of &lt;a href=&quot;https://en.wikipedia.org/wiki/Spiral_galaxy#Origin_of_the_spiral_structure&quot; rel=&quot;nofollow noreferrer&quot;&gt;galaxies rotate with trailing arms&lt;/a&gt;?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I am looking for the percentage of galaxies that rotate with trailing arms. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Do you know of any newer publication on this question? More recent than Pasha &amp;amp; Smirnov (1982) where the sample is quite small?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think the best answer is that exact numbers are very difficult to determine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take for example page 5 of &quot;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1987ApJ...314....3E&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arm classifications for spiral galaxies&lt;/a&gt;&quot; (1987) by D. M. Elmegreen and B. G. Elmegreen, cited over 200 times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Affiliation: IBM Thomas J. Watson Research Center, Yorktown Heights, NY&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Publication: Astrophysical Journal, Part 1 (ISSN 0004-637X), vol. 314, March 1, 1987, p. 3-9.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Page 5:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;&lt;strong&gt;II. THE CATALOGS&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;a) Selection Criteria&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Table 2 includes all &lt;strong&gt;708&lt;/strong&gt; spiral galaxies listed in the Second&#xA;  &lt;a href=&quot;https://heasarc.nasa.gov/W3Browse/all/rc3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reference Catalogue of Bright Galaxies - [3rd] Edition link&lt;/a&gt; (A. de Vaucouleurs, &lt;a href=&quot;https://en.wikipedia.org/wiki/Galaxy_morphological_classification#De_Vaucouleurs_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;G. de Vaucouleurs&lt;/a&gt;, and H. G. Corwin 1976), with declinations $\delta \gt -35°$, inclination-corrected radii at 25 mag arcsec$^{-2}$, R$_{25}$, greater than $1^\prime$, and inclinations less than 60°; &lt;strong&gt;654&lt;/strong&gt; galaxies could be classified, &lt;strong&gt;54&lt;/strong&gt; galaxies which met the criteria but &lt;strong&gt;were too faint&lt;/strong&gt;, &lt;strong&gt;had no details&lt;/strong&gt; (were overexposed on the POSS or were too early in Hubble type to distinguish structure) or were &lt;strong&gt;too inclined&lt;/strong&gt; for reliable classification are also included in Table 2, but with designations f n, or i, respectively, instead of arm classes. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;b) Self-Consistency Checks&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Most galaxies in Table 2 were classified at least 3 times, separated by intervals of several months, as a consistency check. They were classified from both paper and glass copies of the POSS. In the de Vaucouleurs sample, &lt;strong&gt;~10%&lt;/strong&gt; of the galaxies were given different arm classes during independent examinations on the paper prints. An &lt;strong&gt;8%&lt;/strong&gt; variation occurred when the glass images were subsequently examined. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For the high-resolution galaxies listed in Table 3, again &lt;strong&gt;~10%&lt;/strong&gt; of the arm classes changed upon reclassification. In these cases the large amount of detail presented confusion; irregular features were given more or less weight at different times of classification. Arm classifications become increasingly difficult for very small, faint galaxies. The cutoff of $1 ^\prime$ radius for the galaxies in Table 2 represents a minimum practical size for classification from the POSS.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you can see, even with expert examination multiple times of the same data, determination of exact numbers with any accuracy is impossible at the present time. A guess is &quot;most of them&quot;, an exception example is: &lt;a href=&quot;https://en.wikipedia.org/wiki/NGC_4622&quot; rel=&quot;nofollow noreferrer&quot;&gt;NGC 4622&lt;/a&gt; (also called a backward galaxy).&lt;/p&gt;&#xA;" OwnerUserId="18054" LastActivityDate="2018-07-06T14:33:56.340" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26886" PostTypeId="1" AcceptedAnswerId="26887" CreationDate="2018-07-06T20:28:44.850" Score="1" ViewCount="2002" Body="&lt;p&gt;I have heard theories for the big bang like: of another universe existing before the Big Bang, or an unstable white hole that blew up, or our universe detaching from our mother universe, or our universe coming out of a black hole. Most of these theories don't make sense, and I want to know the most-likely-to-be-true theory for how the singularity was made.&lt;/p&gt;&#xA;" OwnerUserId="23317" LastActivityDate="2021-05-19T16:56:06.000" Title="What created the big bang's singularity?" Tags="&lt;big-bang-theory&gt;&lt;singularity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26887" PostTypeId="2" ParentId="26886" CreationDate="2018-07-06T20:50:24.230" Score="13" Body="&lt;p&gt;No one knows what came before the Big Bang if, indeed, anything did.  Theories include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;http://Ekpyrotic%20universe&quot; rel=&quot;noreferrer&quot;&gt;Ekpyrotic universe&lt;/a&gt; theory where the BB was the result of the&#xA;collision of branes.&lt;/li&gt;&#xA;&lt;li&gt;Various &lt;a href=&quot;https://en.wikipedia.org/wiki/Cyclic_model&quot; rel=&quot;noreferrer&quot;&gt;oscillating universe&lt;/a&gt; theories where&#xA;the expansion of the universe ultimately reverses into a Big Crunch&lt;/li&gt;&#xA;&lt;li&gt;Quantum fluctuation models where a zero-net-energy universe is a&#xA;quantum fluctuation from nothing.&lt;/li&gt;&#xA;&lt;li&gt;Various &lt;a href=&quot;https://en.wikipedia.org/wiki/Multiverse&quot; rel=&quot;noreferrer&quot;&gt;multiverse models&lt;/a&gt;&#xA;where the universe we see is only an infinitesimal bubble in a much,&#xA;much, much larger universe.&lt;/li&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Hartle%E2%80%93Hawking_state&quot; rel=&quot;noreferrer&quot;&gt;Hartle-Hawking&lt;/a&gt; idea where&#xA;spacetime gets a bit scrambled and there &lt;em&gt;is&lt;/em&gt; no before. (Really:&#xA;time turns into a space dimension if you go far enough back, and&#xA;there is no time.)&lt;/li&gt;&#xA;&lt;li&gt;Stringy stuff. Very difficult math.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I've only scratched the surface -- making models like this is an avocation to some physicists, while others have discovered that it sells books and generates publicity.  All of the theories have one thing in common: &lt;strong&gt;There is no experimental evidence for any of them.&lt;/strong&gt; Additionally, many of the theories have been shown to contradict currently-understood physical law. (Especially the 2nd Law of Thermodynamics.) The authors generally just shrug and write another paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(On the religious side, speculation about what happened before Creation has been around as far back as written history, with roughly equal success. In the 4th century St. Augustine, after being pushed to answer that question retorted that before the Creation, God was creating a Hell for people who ask such questions. &#xA; That didn't stop the speculation either.)&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-07-06T20:50:24.230" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="26888" PostTypeId="2" ParentId="26880" CreationDate="2018-07-07T02:03:47.373" Score="3" Body="&lt;p&gt;Further to Mark Olson's answer, while an asteroid impact is popularly thought of as the reason for &lt;a href=&quot;https://en.wikipedia.org/wiki/Extinction_event&quot; rel=&quot;nofollow noreferrer&quot;&gt;extinction events on Earth&lt;/a&gt;, the reality is more prosaic, with volcanism and glaciation being the most common causes. In all cases, including impact events, the loss of species is primarily the result of a dramatic change to the global &lt;em&gt;environment&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The K–Pg extinctions were mostly plants (from loss of photosynthesis), pure herbivores (from loss of plant food) and pure carnivores (from loss of prey animals especially herbivores). A Mars base would have to find solutions to a temporary reduction in solar energy, e.g. to sustain its hydroponics, maintain indoor warmth, etc - but hopefully it would have alternative energy sources (generation or storage) anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Excluding an impact in the local vicinity, a Mars base is less at risk from the immediate aftermath than if a similar impact occurred on Earth. There won't be a giant tidal wave or fire front to contend with. There won't the same global infrared atmospheric firestorm as that of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cretaceous%E2%80%93Paleogene_extinction_event#Effects_of_impact&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chicxulub impact&lt;/a&gt;, since the Mars atmosphere is much thinner and has little free oxygen. There will be no burnt forests and no vaporisation of combustible hydrocarbons and sulphur in undersea carbonate rock, each of which contributed to the &quot;impact winter&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Debris impacts might be an issue, but remember the lower Mars gravity will affect the ejecta differently. Now &lt;em&gt;that&lt;/em&gt; would make an interesting question in its own right.&lt;/p&gt;&#xA;" OwnerUserId="14094" LastActivityDate="2018-07-07T02:03:47.373" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26890" PostTypeId="2" ParentId="26878" CreationDate="2018-07-07T08:02:40.080" Score="2" Body="&lt;p&gt;Angular momentum does not change unless a torque acts upon a body. So your question is just whether the torque on a body could ever be exactly zero. The answer is of course that it can be arbitrarily small but we can never be sure it is exactly zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More interesting questions would have been &quot;does the CMB exert a torque on an object?&quot;, or &quot;is it possible to define a state of zero absolute angular momentum?&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-07-07T08:02:40.080" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26891" PostTypeId="2" ParentId="8782" CreationDate="2018-07-07T11:12:42.197" Score="4" Body="&lt;p&gt;in the early universe there was more diffuse matter in gaseous form available to the black holes. The galaxies were more foggy with diffuse matter and less void, so a BH travelling through a galaxy could catch a continuous cloud of matter as it went along, resulting in huge capture rates. Back then, most gas had not yet collapsed to form stars, there was more fuel available for both the feeding of black holes and the forming of new stars. Much of that fuel was subsequently consumed in the formation of stars during the first few billion years after the big bang.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MQm6V.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MQm6V.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2020-06-27T06:30:47.173" LastActivityDate="2020-06-27T06:30:47.173" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26892" PostTypeId="1" AcceptedAnswerId="28644" CreationDate="2018-07-07T13:51:17.437" Score="2" ViewCount="99" Body="&lt;p&gt;There is a phrase in a &lt;a href=&quot;https://www.sciencedaily.com/releases/2018/07/180703105953.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Science Daily&lt;/a&gt; article which appears to be wrong: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The Milky Way has a very high content of carbonaceous dust, which has been shown to be very rare in other galaxies.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a theory that explains these observations?&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2018-07-08T07:47:25.087" LastActivityDate="2018-12-06T14:37:23.537" Title="carbonaceous content of galaxies?" Tags="&lt;interstellar-medium&gt;&lt;dust&gt;&lt;elemental-abundances&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26895" PostTypeId="2" ParentId="12929" CreationDate="2018-07-07T17:15:23.860" Score="2" Body="&lt;p&gt;The key issue is the opacity of the atmosphere, because I presume the question is about the temperature at the solid surface of the Earth.  The atmospheric opacity can be seen from &lt;a href=&quot;https://physics.stackexchange.com/questions/135260/can-someone-explain-to-me-the-concept-of-atmosphere-opacity&quot;&gt;https://physics.stackexchange.com/questions/135260/can-someone-explain-to-me-the-concept-of-atmosphere-opacity&lt;/a&gt;, where you can see that the &quot;rainbow&quot; of maximum heat flux from the Sun happens to hit a kind of hole in atmospheric opacity.  That has a significant warming effect on the Earth, and is exacerbated by the Greenhouse effect.  If sunlight was further into the infrared, the graph shows that much more of it would be intercepted in the atmosphere.  That would make the surface significantly colder, though certainly not a factor of 2 colder.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No doubt the question is of more than passing interest, because M dwarfs are the most numerous main-sequence stars and are therefore interesting for life.  To have life near an M dwarf, the planet would need to be closer than Earth is to the Sun, but the effect of moving the planet closer and shrinking and cooling the star would be similar to leaving Earth where it is and making the star cooler and larger.   So the nature of atmospheric opacity for wet atmospheres must be of great significance for understanding the prospects for life around M dwarfs.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-07-07T17:15:23.860" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26896" PostTypeId="1" CreationDate="2018-07-07T19:26:43.520" Score="3" ViewCount="203" Body="&lt;p&gt;I would like to know please whether or not we have sent radio signals to exoplanets that may be habitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance !&lt;/p&gt;&#xA;" OwnerDisplayName="user22712" LastEditorUserId="43" LastEditDate="2020-09-21T13:03:01.157" LastActivityDate="2020-09-21T13:03:01.157" Title="Have we sent radio signals to exoplanets that may be habitable?" Tags="&lt;exoplanet&gt;&lt;seti&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26898" PostTypeId="1" CreationDate="2018-07-08T00:36:14.203" Score="2" ViewCount="17481" Body="&lt;p&gt;Some friends and I were backpacking in the mountains, and at night we saw a very bright orange object in the sky. It looked brighter than Venus. After a couple minutes we realized it was moving horizontally very slowly. Its movement wasn't noticeable unless looking at it next to a reference point like a tree. This was way brighter and slower than any satellite I've seen. Anyone know what this was?&lt;/p&gt;&#xA;" OwnerUserId="23610" LastEditorUserId="616" LastEditDate="2018-07-16T15:32:26.640" LastActivityDate="2018-07-16T15:32:26.640" Title="Saw a bright orange object in the sky. What was it?" Tags="&lt;planet&gt;&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="5" ClosedDate="2018-07-12T15:06:41.563" ContentLicense="CC BY-SA 4.0" />
  <row Id="26899" PostTypeId="1" CreationDate="2018-07-08T04:10:15.673" Score="2" ViewCount="183" Body="&lt;p&gt;I have Redshift (&lt;code&gt;z&lt;/code&gt;), &lt;code&gt;RA&lt;/code&gt; and &lt;code&gt;Dec&lt;/code&gt; positions for two points in a galaxy catalogue (say &lt;strong&gt;P1&lt;/strong&gt; = [&lt;code&gt;z1&lt;/code&gt;=4, &lt;code&gt;RA1&lt;/code&gt;=140, &lt;code&gt;Dec1&lt;/code&gt;=2] and &lt;strong&gt;P2&lt;/strong&gt; = [&lt;code&gt;z2&lt;/code&gt;=6, &lt;code&gt;RA2&lt;/code&gt;=150, &lt;code&gt;Dec2&lt;/code&gt;=5], where &lt;code&gt;RA&lt;/code&gt; and &lt;code&gt;Dec&lt;/code&gt; are given in degrees). How can I find the &lt;strong&gt;comoving distance&lt;/strong&gt; between these two points using any package in &lt;strong&gt;python&lt;/strong&gt; (like &lt;strong&gt;AstroPy&lt;/strong&gt;, &lt;strong&gt;CosmoloPy&lt;/strong&gt; etc)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am assuming a Flat-LCDM model with the cosmology &lt;code&gt;H0=70, Om0=0.3, Ob0=0.05&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presently, I am using this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import cosmolopy.distance as cd&#xA;cosmo = {'omega_M_0' : 0.3, 'omega_lambda_0' : 0.7, 'h' : 0.70}&#xA;cosmo = cd.set_omega_k_0(cosmo)&#xA;z1 = 4.0&#xA;z2 = 6.0&#xA;d_co = cd.comoving_distance(z2, **cosmo)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Specifically, my questions are these:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) In the said example, how do I use information of the starting redshift &lt;code&gt;z1&lt;/code&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) Do I need to use information of [&lt;code&gt;RA1&lt;/code&gt;, &lt;code&gt;Dec1&lt;/code&gt;] and [&lt;code&gt;RA2&lt;/code&gt;, &lt;code&gt;Dec2&lt;/code&gt;] for an accurate calculation of the comoving separation between &lt;code&gt;P1&lt;/code&gt; and &lt;code&gt;P2&lt;/code&gt; (I need to find the comoving distance to a great degree of accuracy)? &lt;/p&gt;&#xA;" OwnerUserId="23612" LastActivityDate="2018-07-08T04:10:15.673" Title="Comoving distance between two points [(RA1, Dec1, z1) and (RA2, Dec2, z2)]" Tags="&lt;distances&gt;&lt;python&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26900" PostTypeId="2" ParentId="26898" CreationDate="2018-07-08T11:55:54.327" Score="6" Body="&lt;p&gt;If the bright orange object was low in the southeast around midnight, it may have been &lt;a href=&quot;http://earthsky.org/astronomy-essentials/why-is-mars-sometimes-bright-and-sometimes-faint&quot; rel=&quot;noreferrer&quot;&gt;Mars&lt;/a&gt;.&#xA;In late July and early August 2018, it will be only 0.39 &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_unit&quot; rel=&quot;noreferrer&quot;&gt;AU&lt;/a&gt; away from Earth and shine with apparent magnitude -2.7, slightly brighter than Jupiter.&#xA;This is not as bright as Venus but may stand out more strongly in a midnight sky than Venus does in twilight.&#xA;For most of Mars's 2.1 year &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_period&quot; rel=&quot;noreferrer&quot;&gt;synodic period&lt;/a&gt;, it is farther from Earth and appears fainter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the Earth rotates, celestial objects appear to move about 1/4 degree per minute from east to west.&#xA;For observers in the northern hemisphere, objects near the southern horizon appear to move horizontally left to right.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-07-08T11:55:54.327" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26901" PostTypeId="1" AcceptedAnswerId="41486" CreationDate="2018-07-08T12:45:50.480" Score="2" ViewCount="57" Body="&lt;p&gt;&amp;quot;Give a man a watch and he will know the time; give him two and he will never be sure&amp;quot; - this was my impression when reading the Wikipedia article on &lt;a href=&quot;https://en.wikipedia.org/wiki/162173_Ryugu&quot; rel=&quot;nofollow noreferrer&quot;&gt;162173 Ryugu&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Mean diameter:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;0.865±0.015 km[3]&lt;/li&gt;&#xA;&lt;li&gt;0.87 km[4]&lt;/li&gt;&#xA;&lt;li&gt;0.90±0.14 km[5]&lt;/li&gt;&#xA;&lt;li&gt;0.92±0.12 km[6]&lt;/li&gt;&#xA;&lt;li&gt;0.980±0.029 km[7]&lt;/li&gt;&#xA;&lt;li&gt;1.13±0.03 km[8]&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And no mass estimates at all.&lt;/p&gt;&#xA;&lt;p&gt;Now that Hayabusa2 has arrived, do we have solid measurements of Ryugu's size and mass yet?&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2021-02-17T08:18:07.590" Title="Do we finally have a decent set of parameters of 162173 Ryugu?" Tags="&lt;asteroids&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26902" PostTypeId="2" ParentId="26875" CreationDate="2018-07-08T15:24:51.590" Score="4" Body="&lt;p&gt;The time it takes the sun to cross the horizon is given at:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/24304/&quot;&gt;Expression for length of sunrise/sunset as function of latitude and day-of-year&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/12824/&quot;&gt;How long does a sunrise or sunset take?&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Re the time it takes to cross the transit line, this doesn't really answer your question, but, to a good approximation, the time varies between 128 seconds at the equinoxes and 140 seconds at the solstices, regardless of latitude or longitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, the formula is $\frac{128}{\cos (\text{dec})}$ seconds, where $\text{dec}$ is the Sun's declination. You can calculate the Sun's declination itself using the formulas at &lt;a href=&quot;https://en.wikipedia.org/wiki/Position_of_the_Sun&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Position_of_the_Sun&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculation here is relatively simple: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The Sun travels $360 \cos (\text{dec})$ degrees in a 24-hour day, where $\text{dec}$ is the Sun's declination&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When the Sun is transiting, the motion is perpendicular to the transit line (the Sun's motion is entirely in azimuth, not in altitude)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Therefore, all of the Sun's angular motion translates to motion across the transit line; in contrast, the Sun rises and sets a (non-perpendicular) angle (except at tropical latitudes on the two days where the Sun passes directly overhead), so sunsets and sunrises take longer than $\frac{128}{\cos (\text{dec})}$ seconds&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since the Sun has an angular diameter of 32 minutes or $\frac{8}{15}$ degrees, it takes $\frac{\frac{8}{15}}{360 \cos (\text{dec})}$ of a day for the Sun to cross the transit line&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since a day is 86400 seconds, this works out to $\frac{128}{\cos (\text{dec})}$ seconds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Caveats and nitpicks:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I assume there are 86400 seconds between successive noons. This is incorrect for two reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The time between noons is not exactly one day. The cumulative difference forms the &lt;a href=&quot;https://en.wikipedia.org/wiki/Equation_of_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;Equation of Time&lt;/a&gt; but the day to day difference is quite small&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Earth's day is slightly longer than 86400 seconds, which is why we need &lt;a href=&quot;https://en.wikipedia.org/wiki/Leap_second&quot; rel=&quot;nofollow noreferrer&quot;&gt;leap seconds&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Sun's angular diameter actually varies based on Earth's distance from the Sun, but 32 minutes is a good approximation&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I assume the Sun's altitude doesn't change while it's transiting. While this is a good approximation, the altitude does change slightly&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I assume the Sun's declination doesn't change while it's transiting. The change in declination is very small, so this is a reasonable assumption&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are probably other assumptions I made implicitly that I am not noting here.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Refraction is not an issue, since the motion we are discussing is azimuthal and not in altitude.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I used the &quot;one over cosine&quot; form above to make things easier for non-mathematicians. The more compact form would use &quot;secant&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The calculations I did for this problem are disorganized, but available at: &lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/STACK/bc-solar-transit.m&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/STACK/bc-solar-transit.m&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2018-07-08T15:24:51.590" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26903" PostTypeId="2" ParentId="26228" CreationDate="2018-07-08T19:28:32.717" Score="1" Body="&lt;p&gt;As @Ken G noted, it all depends on what you are asking.  I &lt;em&gt;think&lt;/em&gt; the question is, &quot;What would happen all apparent motions of the heavenly bodies and all phenomena we see on Earth remained unchanged and yet the Earth was stationary and the Sun (and everything else, I guess) revolved around it?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is, nothing need change except that the laws of physics would get hugely more complicated -- unrecognizably more complicated -- or else they would turn out to be essentially arbitrary.  It's perfectly possible to rewrite Newtonian (or Einsteinian) mechanics so that there's an arbitrary non-moving and non-accelerated body, but the force laws get very complicated and there are messy &quot;extra&quot; forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have an inbuilt prejudice (which I should make clear I share) that, their effectiveness at describing nature being equal, the simpler the laws of physics are, the more likely they are to be true.  The Earth going around the Sun yields simple laws of nature.  Having the Sun going around the Earth requires contorted, ugly laws of nature, but theories can be constructed each why that yield the same result.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-07-08T19:28:32.717" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26904" PostTypeId="1" CreationDate="2018-07-09T00:00:14.010" Score="3" ViewCount="110" Body="&lt;p&gt;I'm trying to code a piece of software which will calculate some times for me that would greatly help me in my day to day but none of my formulae work if the sun doesn't go at least 18 degrees below the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, I have different formulae to use but I need to program the software to be able to use this other formula on these days. I have no way of determining the maximum the sun will depress below the horizon on any specific day or location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not an expert in any sense of the word in astronomy and literally only know what is relevant to what I need, but I think what I am trying to ask is if it is possible to determine the if the minimum solar elevation / altitude angle is less than -18 degrees on a day / location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have attempted to search but was unsuccessful and am even unsure if what I am asking for is even possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help!&lt;/p&gt;&#xA;" OwnerUserId="23620" LastEditorUserId="23620" LastEditDate="2018-07-09T00:27:17.453" LastActivityDate="2020-11-26T01:21:11.300" Title="Is there any formulae to calculate whether the sun will depress 18 degrees or more below the horizon on a certain day / location?" Tags="&lt;the-sun&gt;&lt;observational-astronomy&gt;&lt;horizon&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26905" PostTypeId="2" ParentId="26228" CreationDate="2018-07-09T03:39:34.710" Score="-1" Body="&lt;p&gt;So, to answer your question let's go to 322 BCC where Earth was at the center and all stars and planets revolves around it. But planet used to show unusual behavior. They use to retrograde (&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Apparent_retrograde_motion&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Apparent_retrograde_motion&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in retrogration, unlike stars, motion the planets seems to move forward then backward and again forward. To understand this go to link above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later to explain retrogration of planets let's go to 3rd century of BC where in the Hipparchian and Ptolemic Astronomy they used concept of epicycle to explain retrogration of planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Deferent_and_epicycle&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Deferent_and_epicycle&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in epicycle, there is one orbit where some invisible thing is revolving, now on this invisible thing, imagine another orbit. It is like orbit of moon in our present solar system, but we can't see earth, and in this model the orbit of moon is similar to orbit of planets. (See link for better explanation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, with this we can explain retrogration motion of planets. But it lack precision. So what we can do is, introduce epicycle within an epicycle, and epicycle within it... And so on, and use FFT (Fast Fourier Transformation) to decide radius of orbits of various epicycles to match observation data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have a system where Earth is at center, stars move around it. So, do the Sun. And planets moves in FFT epicycles.&lt;/p&gt;&#xA;" OwnerUserId="23450" LastActivityDate="2018-07-09T03:39:34.710" CommentCount="0" CommunityOwnedDate="2018-07-09T03:39:34.710" ContentLicense="CC BY-SA 4.0" />
  <row Id="26907" PostTypeId="2" ParentId="26228" CreationDate="2018-07-09T08:26:22.957" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Would things be the same if the Sun was orbiting Earth and the Earth was in the center of the Solar System, or would there be any big difference?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Such a model is impossible when combined with visual observations, much as the Earth could not be flat and coincide with the effects of gravity and the observations made when traveling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Geocentric_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;geocentric model&lt;/a&gt; of the universe was first proposed in the 6th century BC by &lt;a href=&quot;https://en.wikipedia.org/wiki/Anaximander&quot; rel=&quot;nofollow noreferrer&quot;&gt;Anaximander&lt;/a&gt;. In the 2nd century AD the &lt;a href=&quot;https://en.wikipedia.org/wiki/Geocentric_model#Ptolemaic_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ptolemaic model&lt;/a&gt; was developed to explain the observed motion of the moon and planets using an &lt;a href=&quot;https://en.wikipedia.org/wiki/Equant&quot; rel=&quot;nofollow noreferrer&quot;&gt;equant&lt;/a&gt; (a sort of &lt;a href=&quot;https://en.wikipedia.org/wiki/Barycenter&quot; rel=&quot;nofollow noreferrer&quot;&gt;barycenter&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Geocentric_model#Islamic_astronomy_and_geocentrism&quot; rel=&quot;nofollow noreferrer&quot;&gt;By the 10th century AD&lt;/a&gt; scholars started to doubt the Ptolemaic model and from the late 16th century onward, it was gradually superseded by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Heliocentrism&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heliocentric model&lt;/a&gt; of Copernicus, Galileo and Kepler. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;https://en.wikipedia.org/wiki/William_Herschel#Further_discoveries&quot; rel=&quot;nofollow noreferrer&quot;&gt;1783&lt;/a&gt; the observations of William Herschel, Friedrich Bessel, and other astronomers showed that the sun, while near the barycenter of the solar system, was not at the center of the universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a video called &quot;&lt;a href=&quot;https://www.youtube.com/watch?v=wGjlT3XHb9A&amp;amp;feature=youtu.be&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ptolemaic Planetary model&lt;/a&gt;&quot; by YouTube user rufjunk which demonstrates how the geocentric and heliocentric models would appear.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastActivityDate="2018-07-09T08:26:22.957" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26909" PostTypeId="1" CreationDate="2018-07-09T13:17:25.767" Score="25" ViewCount="14223" Body="&lt;p&gt;What are the simplest experiments or calculations that give evidence that the earth revolves around the sun? Can you please explain them and reference the history? Many simple explanations &lt;a href=&quot;https://io9.gizmodo.com/5464810/the-earth-revolves-around-the-sun--prove-it&quot; rel=&quot;noreferrer&quot;&gt;such as this&lt;/a&gt; cite observations such as that relative position of two stars are observed from earth vary every night - which would not be true if the stars orbited the earth. But isn't the observation also consistent with a model where the stars orbit the earth but do so at different speeds, while the earth still orbits the sun? Simple explanations would be helpful.&lt;/p&gt;&#xA;" OwnerUserId="23596" LastEditorUserId="13850" LastEditDate="2018-07-10T11:36:19.477" LastActivityDate="2019-03-27T11:58:19.650" Title="Simple experimental evidence that Earth revolves around Sun" Tags="&lt;the-sun&gt;&lt;orbit&gt;&lt;earth&gt;&lt;history&gt;" AnswerCount="9" CommentCount="9" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26910" PostTypeId="2" ParentId="26909" CreationDate="2018-07-09T13:34:41.860" Score="41" Body="&lt;p&gt;The answer is ironic: Without good instruments, &lt;em&gt;there is no evidence&lt;/em&gt;. The people who thought that the Sun went around the Earth were perfectly correct as far as the actual evidence went until the early 1700s and mid-1800s when two lines of evidence opened up that showed that the Earth moved.&lt;/p&gt;&#xA;&lt;h3&gt;Aberration of Starlight&lt;/h3&gt;&#xA;&lt;p&gt;Wikipedia has a &lt;a href=&quot;https://en.wikipedia.org/wiki/Aberration_of_light&quot; rel=&quot;noreferrer&quot;&gt;correct but over-complicated explanation&lt;/a&gt;. The easiest way to think about it is to imagine yourself at a stop sign in a car in the rain, and the rain is falling straight down. When you start moving, the rain's apparent direction of fall changes so that it appears to be falling from ahead of you and slanting down towards you. That's aberration.&lt;/p&gt;&#xA;&lt;p&gt;In the early 1700s, the stars were discovered to be shifting position, and in 1727, James Bradley correctly identified it as abberation of starlight due to the motion of the Earth around the Sun. (For any star in the ecliptic, the Earth is moving towards it at some time of the year and away from it six months later.)&lt;/p&gt;&#xA;&lt;h3&gt;Parallax&lt;/h3&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Parallax&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia's article on parallax&lt;/a&gt; is better, and I refer you to it for details. Basically, if you hold your finger up before you and look at it with your left eye closed, and then with your right eye closed, it appears to jump with respect to the background -- the wall beyond or the trees outside or whatever. Switch back and forth between your eyes quickly to see it clearly.&lt;/p&gt;&#xA;&lt;p&gt;As the Earth circles around the Sun, nearby stars also appear to shift their position relative to the more distant stars.  A key point here is that there were good scientific reasons to suppose that the stars were much smaller than the Sun. Seen through a telescope, stars showed disks and if they were like the Sun, their distance could be deduced from those disks. And they were close enough that if the Earth really went around the Sun, parallax should have been observed. But it wasn't and the lack of any noticeable parallax was a strong empirical argument against Heliocentric theories.&lt;/p&gt;&#xA;&lt;p&gt;In reality of course parallax exists, but the parallax of all stars is small, because they are much further away than was estimated from their disks. (The visible disks were actually &lt;a href=&quot;https://en.wikipedia.org/wiki/Diffraction&quot; rel=&quot;noreferrer&quot;&gt;diffraction disks&lt;/a&gt; and not true disks at all -- but it was not until nearly a century later that diffraction began to be understood.)  Friedrich Bessel first measure the real parallax of a star in 1838.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-03-27T11:58:19.650" CommentCount="25" ContentLicense="CC BY-SA 4.0" />
  <row Id="26911" PostTypeId="1" AcceptedAnswerId="26912" CreationDate="2018-07-09T16:53:01.980" Score="2" ViewCount="284" Body="&lt;p&gt;Do planets like Jupiter have a system of latitudes and longitudes (I believe yes)? So if yes, what purpose do they serve? Is there anything different in their purpose from Earth lat-longs? and What would be the average distance between each 1-degree latitude on Jupiter?&lt;/p&gt;&#xA;" OwnerUserId="23627" LastActivityDate="2018-07-09T17:11:36.617" Title="What would be the average distance between each 1-degree latitude on Jupiter?" Tags="&lt;distances&gt;&lt;jupiter&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26912" PostTypeId="2" ParentId="26911" CreationDate="2018-07-09T17:05:57.643" Score="5" Body="&lt;p&gt;Yes, they have coordinate systems as well. For example, the Wikipedia page on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Great_Red_Spot&quot; rel=&quot;noreferrer&quot;&gt;Great Red Spot&lt;/a&gt; mentions it lies 22 degrees south of the planet's equator. Jupiter has a polar radius of 66,854 km; a latitude degree is $2 \pi / 360$ times that, so almost 1,167 km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that because Jupiter lacks a solid surface, longitudes are less well-defined. In fact, there are &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Jupiter#Great_Red_Spot&quot; rel=&quot;noreferrer&quot;&gt;three different systems&lt;/a&gt; for this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because Jupiter's visible features do not rotate uniformly at all latitudes, astronomers have defined three different systems for defining the longitude.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rocky planets like Mars do have a well-defined coordinate system, as shown in &lt;a href=&quot;https://en.wikipedia.org/wiki/Mars#/media/File:USGS-MarsMap-sim3292-20140714-crop.png&quot; rel=&quot;noreferrer&quot;&gt;this map&lt;/a&gt; and on &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_quadrangles_on_Mars&quot; rel=&quot;noreferrer&quot;&gt;this page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These coordinates are used to mark and find interesting places on either the surface (like on Earth) or in the atmosphere of a planet. There are also tables available showing which latitude/longitude is directed towards us at a certain moment, so that you can compute whether a particular region of the planet is visible from Earth or not.&lt;/p&gt;&#xA;" OwnerUserId="7472" LastEditorUserId="7472" LastEditDate="2018-07-09T17:11:36.617" LastActivityDate="2018-07-09T17:11:36.617" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26914" PostTypeId="1" CreationDate="2018-07-09T19:24:39.097" Score="2" ViewCount="300" Body="&lt;p&gt;We know that with the passage of time, hydrogen inside (near or at the core) a star is converted to helium and other heavy elements via fusion. We can then deduce that the old sun contains heavier elements in comparison to the new one. Does that mean that the mass of the old sun is more than the new sun? &lt;/p&gt;&#xA;" OwnerUserId="23550" LastEditorUserId="14094" LastEditDate="2018-07-14T18:08:35.050" LastActivityDate="2018-07-14T18:08:35.050" Title="Old star vs New star" Tags="&lt;the-sun&gt;&lt;stellar-astrophysics&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26915" PostTypeId="2" ParentId="26914" CreationDate="2018-07-09T19:35:50.630" Score="4" Body="&lt;p&gt;The main fusion reaction converts four atoms of hydrogen into one atom of helium. The total mass of for hydrogen atoms is a little more than the mass of one helium atom, the difference being converted to energy. Thus the mass of the star goes down a bit as it fuses hydrogen. It actually goes down more due to matter being lost in the stellar wind, but both effects reduce the mass over time&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-07-09T19:35:50.630" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26916" PostTypeId="2" ParentId="26914" CreationDate="2018-07-09T20:14:55.993" Score="8" Body="&lt;p&gt;Let me try to add some numbers to &lt;a href=&quot;https://astronomy.stackexchange.com/a/26915/2153&quot;&gt;Steve's answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun's luminosity is about $L_{\odot}=4\times10^{26}\text{ J/s}$. Now, if we assume that the majority of that energy comes from nuclear fusion, we have&#xA;$$L_{\odot}=\frac{dE}{dt}=\frac{d(mc^2)}{dt}=c^2\frac{dm}{dt}=c^2\dot{M}_{\text{nuc}}$$&#xA;Therefore, we can write the mass-loss due to nuclear fusion as $\dot{M}_{\text{nuc}}=L_{\odot}/c^2\sim4\times10^{9}\text{ kg/s}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the Sun also loses mass via the &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_wind&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar wind&lt;/a&gt;. I believe the mass loss rate is a few times about $\dot{M}_{\text{wind}}\sim3\times10^{-14}M_{\odot}/\text{yr}\sim2\times10^9\text{ kg/s}$. Coronal mass ejections, solar flares and similar phenomena are often accounted for in this figure; at any rate, they don't add a significant amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This figure should increase as the Sun moves into the red giant (and especially asymptotic giant branch) phase of its life; it winds will increase to become much more significant that they are today. However, fusion will also increase, and the Sun's luminosity will increase by a factor of 1000 or 10000; therefore, the mass loss due to fusion will also increase. However, I &lt;em&gt;think&lt;/em&gt; mass loss due to winds will increase more significantly, possibly by a factor of $\sim10^6$ - but I'll check that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, an older star will have a lower mass than a younger star. For a Sun-like star, this would be hard to measure over the span of a human lifetime - remember that it only loses about $10^{-13}$ solar masses each year. Long-term records could make it possible, though, through various observations of the Sun and its effects. Centuries or millennia of data-gathering do tend to yield results.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2018-07-10T17:00:24.410" LastActivityDate="2018-07-10T17:00:24.410" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26917" PostTypeId="2" ParentId="26909" CreationDate="2018-07-09T20:35:15.047" Score="19" Body="&lt;p&gt;You cannot prove that the Earth orbits the Sun rather than vice versa because this goes very much against the grain of all frames of reference being equally valid (but some make a lot more sense than others). For example, it makes much more sense to use an Earth-centered, Earth-fixed point of view rather than a non-rotating geocentric, heliocentric, barycentric, or galactocentric point of view when modeling the weather or the tides. One could, for example, use a heliocentric  or even galactocentric point of view to model the Earth's weather, but doing so would be beyond stupid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, when modeling the behavior solar system it makes much more sense to use a heliocentric, or even better, a solar system barycentric point of view. One could however use an Earth-centered, Earth-fixed point of view because &lt;strong&gt;all frames of reference are equally valid&lt;/strong&gt; (in theory). Doing so would of course make the equations of motion quite ugly, and uglier yet on trying to make those equations of motion relativistically correct. A geocentric point of view nonetheless remains theoretically valid -- even for modeling the behavior of the Milky Way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with a geocentric point of view isn't that it's invalid (which it isn't). The problem is that advocates of geocentricism argued (and sadly, continue to argue) that this is the one and only valid point of view. This argument is invalid, because once again, all frames of reference are equally valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note well: Just because inertial frames are special in some sense does not mean that non-inertial frames are invalid.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2018-07-10T11:17:38.207" LastActivityDate="2018-07-10T11:17:38.207" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="26918" PostTypeId="2" ParentId="26909" CreationDate="2018-07-10T00:51:51.630" Score="12" Body="&lt;p&gt;If you start with the idea that the planets, the sun, the moon &lt;em&gt;and the earth&lt;/em&gt; are all bodies that all move through space, exclude the apparently fixed stars, and then see what evidence there is as to &lt;em&gt;how&lt;/em&gt; they move relative to each other, then in that context there is some evidence to be found in naked-eye astronomy aided by navigational instruments available even to the ancients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The patterns of observed movement of the planets is evidence of heliocentric orbit.  The visible planets follow certain patterns.  First, Mercury and Venus:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;They are always seen in the vicinity of the sun.&lt;/li&gt;&#xA;&lt;li&gt;The observed angular separations of both Mercury and Venus from the sun have a regular pattern.&lt;/li&gt;&#xA;&lt;li&gt;Mercury has a much closer maximum separation than Venus, and its angular separation changes at a much faster rate.&lt;/li&gt;&#xA;&lt;li&gt;Both planets stay close to the ecliptic, and never oscillate normal to it.&lt;/li&gt;&#xA;&lt;li&gt;Both planets' orbits around the sun can be documented and predicted with relative ease.  This can be done imprecisely even without a telescope, though it is much harder for Mercury, being so close to the sun.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Beginning with the premise of bodies moving through the heavens, I believe the evidence is there for Mercury and Venus having a heliocentric orbit.  Kepler described it precisely, but the ancient Greeks were able to model their motion very well without telescopes in &lt;a href=&quot;https://en.wikipedia.org/wiki/Antikythera_mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Antikythera Mechanism&lt;/a&gt; in &lt;em&gt;geocentric&lt;/em&gt; terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an ancient Greek astronomer had wanted to precisely model the motion of the inner planets in &lt;em&gt;heliocentric&lt;/em&gt; terms, he could have.  The way to do it is to assume the fixed stars are rigidly fixed, and measure the angular distances between them all, and then plot the motions of the moving planets among them.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Sextant&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sextants&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Mariner%27s_astrolabe&quot; rel=&quot;nofollow noreferrer&quot;&gt;other devices&lt;/a&gt; were used by ancient mariners who were highly skilled even with &lt;a href=&quot;https://arstechnica.com/science/2018/04/mysterious-sunstones-in-medieval-viking-texts-could-really-have-worked/&quot; rel=&quot;nofollow noreferrer&quot;&gt;primitive ones&lt;/a&gt;.  So this could have been done to realize the &quot;simple experiement or calculation&quot; you are asking for.  Whether it ever &lt;em&gt;was&lt;/em&gt; done, with that question in mind, is a bit different issue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the earth itself.  Even in the ancient world the relationship between the &lt;a href=&quot;https://en.wikipedia.org/wiki/Sidereal_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;sidereal day&lt;/a&gt; and the solar day has been &lt;a href=&quot;https://en.wikipedia.org/wiki/Astrological_sign&quot; rel=&quot;nofollow noreferrer&quot;&gt;well understood&lt;/a&gt;.  The precession of the sun around the ecliptic plane is evidence of a heliocentric orbit.  One just has to &lt;a href=&quot;https://en.wikipedia.org/wiki/Orrery&quot; rel=&quot;nofollow noreferrer&quot;&gt;model it&lt;/a&gt; to make this clear.  Ancient calculations relating to sidereal time and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Metonic_cycle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Metonic cycle&lt;/a&gt; reveal that the earth's heliocentric motion could have been mathematically modeled, if conceived of and desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the outer planets, to my mind this is the least intuitive, but there is evidence for a heliocentric orbit for them too, but only by building on the idea that earth and the inner planets orbit the sun.  This comes from observing their &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_retrograde_motion&quot; rel=&quot;nofollow noreferrer&quot;&gt;retrograde motion&lt;/a&gt;.  These planets will move retrograde against the &quot;fixed background stars&quot; at certain times, and those times can be correlated to their angular separation from the sun.  Also the different planets move through the zodiac at different speeds, which also correlate with the amplitude of retrograde motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you simulate all this with a heliocentric orrery, it is very plainly evident that we on an inner, faster planet observe an outer, slower planet in its orbit.  The ancient Greeks had enough skill to model the motions of Mars, Jupiter and Saturn in their &lt;a href=&quot;https://en.wikipedia.org/wiki/Antikythera_mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;Antikythera Mechanism&lt;/a&gt; in &lt;em&gt;geocentric&lt;/em&gt; terms.  So it follows that a precise, mathematical model of heliocentric motion for the outer planets was within their reach, if they ever reached for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also some evidence that at least some ancient thinkers &lt;em&gt;were&lt;/em&gt; able to decode all this into a heliocentric model.  The ancient Greek &lt;a href=&quot;https://en.wikipedia.org/wiki/Aristarchus_of_Samos&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aristarchus of Samos&lt;/a&gt; had a heliocentric model.  However, Plato and others seemed to disfavor it, and &lt;a href=&quot;https://www.youtube.com/watch?v=UpLcnAIpVRA&quot; rel=&quot;nofollow noreferrer&quot;&gt;this reconstruction&lt;/a&gt; of &lt;a href=&quot;https://en.wikipedia.org/wiki/Antikythera_mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Antikythera Mechanism&lt;/a&gt; which is believed to come well after Aristarchus' day features a &lt;em&gt;geocentric&lt;/em&gt; orrery which models planetary retrograde motion.  And heliocentric thinking stayed &lt;a href=&quot;https://en.wikipedia.org/wiki/Galileo_affair&quot; rel=&quot;nofollow noreferrer&quot;&gt;within the minority&lt;/a&gt; in the west until the modern age.  Perhaps the obvious geocentric orbit of the moon, or the question of the stars (whether they should be included in any correct model or not), or the lack of a universal theory of gravity, sufficiently obscured for them what to us is clear.&lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorUserId="1160" LastEditDate="2018-07-12T04:23:48.373" LastActivityDate="2018-07-12T04:23:48.373" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26920" PostTypeId="1" AcceptedAnswerId="26921" CreationDate="2018-07-10T11:31:27.337" Score="7" ViewCount="173" Body="&lt;p&gt;As I watch the &lt;a href=&quot;https://youtu.be/ONUSP23cmAE&quot; rel=&quot;noreferrer&quot;&gt;video&lt;/a&gt; of the asteroids being discovered, all I see is prograde orbits. Their prevalence is fully understandable and justified, but the Sun gets interstellar visitors occasionally and I see no specific reason why it couldn't catch one into a retrograde orbit. They would obviously be a minority and likely not nearly as long-lasting as these in prograde orbits, but they should exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did we observe any? If so, what sort of percentage of the whole are they? If not, why? &lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2018-07-10T13:54:46.040" Title="Are there any natural satellites in retrograde Sun orbit?" Tags="&lt;solar-system&gt;&lt;asteroids&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26921" PostTypeId="2" ParentId="26920" CreationDate="2018-07-10T13:49:10.763" Score="12" Body="&lt;p&gt;Yes, there are, but they are very rare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_exceptional_asteroids#Retrograde&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia has a list of retrograde asteroids&lt;/a&gt;, and mentions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As of September 2016, of the more than 700,000 minor planets known, there are only 95 known retrograde minor planets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That means only 0.014% of known minor planets are in a retrograde orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The asteroid with the highest known inclination is &lt;a href=&quot;https://en.wikipedia.org/wiki/2013_LA2&quot; rel=&quot;noreferrer&quot;&gt;2013 LA2&lt;/a&gt;, with an inclination of reportedly (175.189±0.007)° according to the &lt;a href=&quot;https://ssd.jpl.nasa.gov/sbdb.cgi?sstr=2013LA2;cad=1&quot; rel=&quot;noreferrer&quot;&gt;JPL small-body database browser&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1462" LastEditorUserId="1462" LastEditDate="2018-07-10T13:54:46.040" LastActivityDate="2018-07-10T13:54:46.040" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26922" PostTypeId="1" CreationDate="2018-07-10T13:58:26.883" Score="4" ViewCount="42" Body="&lt;p&gt;In the JPL small body browser, orbital elements are all provided with uncertainties and even a covariance matrix, such as for &lt;a href=&quot;https://ssd.jpl.nasa.gov/sbdb.cgi?sstr=2013LA2;cad=1;old=0;orb=0;cov=1;log=0#elem&quot; rel=&quot;nofollow noreferrer&quot;&gt;2013 LA2&lt;/a&gt;.  Very good.  But underneath, the absolute magnitude is reported with a sigma of &quot;n/a&quot;.  Why would the uncertainty for an absolute magnitude be not applicable?  Is it not a measurement but somehow given by definition?&lt;/p&gt;&#xA;" OwnerUserId="1462" LastActivityDate="2018-07-10T13:58:26.883" Title="In the JPL small body browser, why is uncertainty on absolute magnitude &quot;N/A&quot;?" Tags="&lt;observational-astronomy&gt;&lt;absolute-magnitude&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26924" PostTypeId="2" ParentId="26909" CreationDate="2018-07-11T02:01:17.373" Score="5" Body="&lt;p&gt;The best experimental evidence is probably &lt;strong&gt;retrograde motion&lt;/strong&gt;. The data is not easily acquired: it takes a long time to collect, not to mention an astronomer would have to stay up every night keeping painstaking measurements of the positions of each object. But it can be done (ancient Greeks were aware of it) and in the modern world you can simply use a simulator like &lt;a href=&quot;http://stellarium.org/&quot; rel=&quot;noreferrer&quot;&gt;Stellarium&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Download Stellarium, start it up, and navigate to your local position. Then set the simulation running and speed it up many times. You should see the sun and stars rotate around you. Then turn the ground off (so you can see through the Earth), turn the atmosphere off (so you can see stars during the day), switch to the equatorial mount (Ctrl + M; this is the mount where most of the sky is stationary), and zoom out until the Sun, the Moon, and all the planets appear to move in a circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now look carefully at the motions of all the planets. You should see that the Moon (and the Sun) goes in circles without ever slowing down. This is what you'd expect if they went around the Earth. However Mercury does not follow this motion - it visibly disappears around the Sun. Mars behaves differently as well: it goes round and round, then stops, goes backwards, and then goes round and round again. This last behavior is called retrograde motion and its explanation occupied a lot of ancient astronomy. Ancient Greeks came up with a complicated theory of epicycles to explain it, given that the planets orbited the Earth and moved in perfect circles (neither of these are true in modern knowledge).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However retrograde motion can be easily explained if Mars didn't go around the Earth, but went around the Sun instead. This would simply mean that Mars goes retrograde when we overtake it on its orbit. In addition, this also explains how each time Mars goes retrograde, it is at its brightest, plus it is on the opposite side of the sky relative to the Sun. It also explains why Mercury does its loops around the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't mean that the geocentric model is not able to account for the same observations, but it's drastically simpler. In the heliocentric model, every planet goes round the Sun on a simple path, an ellipse. In the geocentric model, every planet goes round the Earth, but on epicycle after epicycle. That's when we apply Occam's Razor and conclude that the simpler explanation is correct.&lt;/p&gt;&#xA;" OwnerUserId="20348" LastActivityDate="2018-07-11T02:01:17.373" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26925" PostTypeId="1" AcceptedAnswerId="26931" CreationDate="2018-07-11T02:17:25.713" Score="3" ViewCount="90" Body="&lt;p&gt;Excluding gas giants, roughly what percentage of planets in the habitable zones of their stars would have large moons like Earth does, as opposed to Mars and its small moons/captured asteroids?&lt;/p&gt;&#xA;" OwnerUserId="23652" LastActivityDate="2018-07-12T15:12:18.133" Title="Number of planets in a habitable zone with large moons" Tags="&lt;exoplanet&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2018-07-15T23:29:19.437" ContentLicense="CC BY-SA 4.0" />
  <row Id="26927" PostTypeId="1" CreationDate="2018-07-11T08:07:23.983" Score="3" ViewCount="123" Body="&lt;p&gt;I understand that the Universe could be either finite or infinite and that it is, at the least, &quot;very large&quot;. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, why doesn't the math involved in making the CMB and other predictions, in effect, tell us the size of the Universe? We know how hot it was then and how cool it is now. Does scale not affect these calculations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words if a given volume is a million degrees on &quot;day&quot; one and it's now 2.5 degrees don't those temperatures change if you make the volume 200 times larger?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculations make the Observable Universe, the CMB and the temperatures then and now all come out right and validate the Big Bang theory. How can they do that with both a smaller Universe (Observable Universe) and the much larger actual Universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it just because whatever the size of the Universe is, it was all 1 million degrees (example) at one point? Otherwise, it seems the predictive math involved would produce different numbers for different volume.&lt;/p&gt;&#xA;" OwnerUserId="23655" LastActivityDate="2018-07-11T16:54:04.333" Title="Methodology for size of Universe" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26928" PostTypeId="2" ParentId="26909" CreationDate="2018-07-11T12:40:14.397" Score="-1" Body="&lt;p&gt;Very simply: because of relative motion, no proof exists.  Any situation that you come up with can be explained by a tweaked geocentric module. Albert Einstein came to the same conclusion when he said   &quot;I have come to believe that the motion of the Earth cannot be detected by any optical experiment.&quot;  and &quot;...to the question whether or not the motion of the Earth in space can be made perceptible in terrestrial experiments. We have already remarked... that all attempts of this nature led to a negative result. Before the theory of relativity was put forward, it was difficult to become reconciled to this negative result.&quot; &lt;/p&gt;&#xA;" OwnerUserId="23658" LastEditorUserId="7472" LastEditDate="2018-07-11T18:11:25.080" LastActivityDate="2018-07-11T18:11:25.080" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26929" PostTypeId="2" ParentId="26909" CreationDate="2018-07-11T12:56:24.557" Score="1" Body="&lt;p&gt;Well... the seasonal cycle is evidence enough that the Earth and Sun are orbiting each other. Whether A orbits B or B orbits A is an argument about relative mass. If you find that the movement of all the other planets are consistent with them orbiting the Sun but not the Earth, you can conclude that the Sun's mass is enormous and therefore barely affected by the pull of the Earth.&lt;/p&gt;&#xA;" OwnerUserId="23659" LastActivityDate="2018-07-11T12:56:24.557" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26930" PostTypeId="2" ParentId="26927" CreationDate="2018-07-11T12:57:47.107" Score="2" Body="&lt;p&gt;There's a subtle misunderstanding there, which is rarely made clear in discussions of the Big Bang and the current model of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Big Bang &lt;em&gt;did not&lt;/em&gt; explode to fill the universe. Matter and energy has always filled the entire universe pretty uniformly, but 13+ GYears ago, the universe was &lt;em&gt;very&lt;/em&gt; small and its energy density was &lt;em&gt;very&lt;/em&gt; high. But it was still completely filled. Our theories combined with our observations don't tell us whether it was infinite or finite, just that whatever it was, it was was very dense and very energetic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From that point -- what came before, if anything is utterly beyond the current scope of physics and is pure speculation -- space itself started to expand very, very rapidly and since the mass-energy was conserved, the density of mass-energy rapidly decreased.  This hasn't stopped, though it has slowed.  Our best measurements to date suggest that it will never stop and probably is speeding up some.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we look at the CMB, we're looking back in time to about 380,000 years after the Big Bang, to the point where the universe had cooled so that it was no longer a plasma and became transparent to light. Prior to that point the plasma was essentially opaque and there is no way to use electromagnetic radiation to see earlier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see the CMB 13 GYears away in every direction, but we're not seeing to the &quot;edge&quot; of the universe (if there is one). We're seeing to the most distant point from which light has has &lt;em&gt;time&lt;/em&gt; to get to us travelling at lightspeed. So 13 GYears back is 13 GLightyears out.  There's &lt;em&gt;stuff&lt;/em&gt; (probably stars and planets and galaxies and all the stuff we know and love) further away, but we can't see it because light hasn't had time to get here from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the universe that we can &lt;em&gt;in principle&lt;/em&gt; observe is a 26 GLightyear bubble in a much larger universe. (BTW, all these measurements are light-travel distances. When you take GR into account it gets more complicated, but those complications are not critical to this discussion.  See &lt;a href=&quot;https://en.wikipedia.org/wiki/Observable_universe#Misconceptions_on_its_size&quot; rel=&quot;nofollow noreferrer&quot;&gt;Observable Universe&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Distance_measures_(cosmology)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Distance Measures&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Comoving_and_proper_distances&quot; rel=&quot;nofollow noreferrer&quot;&gt;Comoving and Proper Distances&lt;/a&gt; for typically Wikipedian incomplete-and-confusing but still useful articles.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one thing based on observation that we can say is that the universe is probably at least ten times the diameter of what we can observe. Assuming General Relativity and assuming that the larger universe is about as homogeneous as what we can see, the fact that we observe &lt;em&gt;our&lt;/em&gt; bit of it to be very flat means that the edge or the radius of curvature or whatever makes the whole thing finite has to be at least ten times further away -- otherwise we'd see curvature in our own observable patch of space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the problem is that the part of the universe that we can see is &lt;em&gt;observed&lt;/em&gt; to be too small a part of the whole shebang to make any kind of estimate of how big the entire universe is.  (It's a whole lot like trying to survey an area of a few square miles accurately enough to estimate the size of the Earth. There's so much noise in the measurements due to local topography, that all you can do is say &quot;It's &lt;em&gt;a lot&lt;/em&gt; bigger than a few miles.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Caveat: If I understood you correctly, your question is about the size of the whole universe, not just the visible universe.  If I'm wrong, let me know and I'll frantically re-write this!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that even the parts which rely on observation necessarily rely on those observations filtered through the lenses of theory. I've tried to use only the parts of cosmology which seem to be solid and to avoid the wild speculation that produces popular science books and TV specials.  But it's still quite possible that it will all be overturned as we come to learn more. (I'll be surprised, but &lt;em&gt;that's&lt;/em&gt; happened more than once!)&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2018-07-11T16:54:04.333" LastActivityDate="2018-07-11T16:54:04.333" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26931" PostTypeId="2" ParentId="26925" CreationDate="2018-07-11T16:23:05.087" Score="3" Body="&lt;p&gt;It's not quite the drake equation in terms of your inputs determining your estimate but it's not too far off.   Our telescopes can't get a good look at existing or forming solar systems.  They've spotted several thousand planets but only a &lt;a href=&quot;https://www.nasa.gov/jpl/multimedia/exomoon-pia17998&quot; rel=&quot;nofollow noreferrer&quot;&gt;tiny number of possible moons&lt;/a&gt; and I don't think any of them are confirmed.    It's also not always possible to tell is some of the &quot;earth-like&quot; planets that have been observed are more like sub-neptunes or perhaps water worlds.   Any kind of gaseous/water or ice world planet would have a hard time making an impact moon.  Two rocky bodies colliding are best for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Young solar systems are full of collisions during their planet formation period so there are probably enough collisions with the approximate size/ratio of &lt;a href=&quot;https://en.wikipedia.org/wiki/Giant-impact_hypothesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Theia-Earth&lt;/a&gt;.  At least, it's probably not uncommon, but size isn't the only factor.   The impact has to be at the right angle to generate sufficient angular momentum, because the main body ideally rotates faster than its Moon orbits.  This faster rotation and the tidal force of the moon on the planet pushes the moon outward over time and stabilizes the Moon's orbit (provided &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;the hill sphere is larger than the Moon's distance over time&lt;/a&gt;).   If the main body rotates slower than the moon orbits, the moon is drawn towards the planet and would likely destabilize before long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An impact too straight on is bad for moon formation and too glancing is also bad.  The impact needs to be within a proper range in size, ratio between the two objects and angle of impact and without a surface with lots of ice and/or water.   The requirements are pretty precise, so stable large collision moon formation should at least be somewhat rare because of all the necessary conditions.   It's also not known with accuracy how many of the &lt;a href=&quot;http://www.washington.edu/news/2015/01/28/some-potentially-habitable-planets-began-as-gaseous-neptune-like-worlds/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;habitable&quot; planets started out rocky&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Migrating jupiters and other planets are another problem.   It's fairly common for Jupiters to migrate which should destabilize a percentage of inner planets with stable collision moons.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest issue is the size of the star.   The habitable zone of red dwarf stars is very close to the star and moons wouldn't be stable around planets in those systems.    Those planets would likely be tidally locked, which means they'd rotate slower than their moon orbits and that would likely lead to destabilizing the moons orbit and the moon crashing into the planet, that is, if the planet had a moon at all, it would need to form in a very tight orbit.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there's a few estimates to make a guess at an input.   I don't think an accurate estimate is possible but if I was to guess, I think large collision moons will be found.  I don't think they're super rare, and I base that guess on a reasonable abundance of rocky material in solar system formation and a very large number of collisions, so even if the Earth-Theia collision was one in a million, there's enough collisions happening that other two body systems like ours probably create long term stable moons from time to time.  So, rare but I think, not super-rare.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We probably won't have an answer for this until we get a better look with the &lt;a href=&quot;https://www.jwst.nasa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;James Webb Telescope&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-07-12T15:12:18.133" LastActivityDate="2018-07-12T15:12:18.133" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26933" PostTypeId="2" ParentId="26909" CreationDate="2018-07-12T06:42:11.720" Score="1" Body="&lt;p&gt;Detailed observations of any star in the sky reveal that the Earth moves in an elliptical orbit with a speed of approximately 30 km/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the line of sight velocities of stars are measured using the Doppler effect, they have to be corrected for the motion of the Earth. If they are not, then one would see an unexplained modulation of the velocities, with a period of 1 year and an amplitude of up to 30 km/s that would differ depending on the direction of the star with respect to the Earth-Sun orbital plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, a geocentric model fails to explain why an observer on the Earth sees the positions of stars on the sky execute periodic ellipses on the sky with amplitudes (a.k.a. the trigonometric parallax) that appear to be inversely correlated with how far away they are, but all with a period of one year. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps these are not the&quot;simple&quot; experiments that you were thinking of, but the universe cannot always be understood with what is visible to the naked eye and common sense.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-07-12T06:51:59.000" LastActivityDate="2018-07-12T06:51:59.000" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26934" PostTypeId="2" ParentId="10979" CreationDate="2018-07-12T07:01:47.370" Score="4" Body="&lt;p&gt;This is a very old question, and already has great answers, including a diagram drawn to scale. I just want to add a very simple analogy that shows how both pictures in the question are wrong (the second one is less so than the first, if we admit the dubious &lt;em&gt;degree of wrong&lt;/em&gt;). Below is how I explained the Moon-Sun motion to some friends, and they grasped it right away, so I still hope my addtion may be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of the Olympic sport of long track speed-skating. Two competitors are running close to each other at a great speed on almost concentric circles (let's make this track round for simplicity, not oval like in real sport). Since one of the skaters is running a bit longer track, the rules require they switch lanes on every lap, so no one of the two is at a disadvantage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/A9abP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/A9abP.jpg&quot; alt=&quot;Jan Smeekens (NED) at a world cup speedskating in Heerenveen, the Netherlands.&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sup&gt;Image credit: &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Jan_Smeekens_(23-02-2008).jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia File:Jan_Smeekens_(23-02-2008).jpg&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suddenly, the Olympic committee changes the rules. The skaters must now run not a 400m track, but a whopping 10km long one (a circle about 3.2km in diameter), and change tracks not once per lap, but 13 times. The stadium is large, and the curve of the track is so slight, that the outer runner during the change still traces a convex curve when moving towards the inner track at 0.2 m/s while maintaining 15 m/s forward speed along the track [Note: these two figures, unlike others, are inconsequential for the analogy].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, say one of the guys is not quite consistent. He speeds up a bit, then slows down a bit, but still catches up with the second. In fact, it happens so that when he moves outside, he's behind the second, but when moving to the inside track, he's ahead. Imagine what you would see from a drone camera following the runners from above. &lt;em&gt;Wow, they orbit each other&lt;/em&gt;! And neither of the paths is concave, like on the second picture, nor are there any backward-going loops like that on the first picture. They always go forward, to bring home that medal!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numbers above are approximating Earth and Moon orbits. Earth orbit around the Sun is ~400 times larger than that of the Moon around the Earth, as is a standard 2-lane track width of 4m is 1/400 of the 1.6 km radius of our supertrack. The Moon makes about 13 turns around the Earth in one Earth's year. Of course, in a better analogy, the Earth runs in the middle of the track, deviating only slightly, and the Moon is rhythmically bobbing between the innermost and outermost edges. There are also no abrupt lane changes at set points for the Moon (and this smooths out the curve even more away from concavity compared to skaters paths). But since we are in a thought experiment, let's pretend the Olympic committee has an exception for this peculiar pair.&lt;/p&gt;&#xA;" OwnerUserId="16781" LastActivityDate="2018-07-12T07:01:47.370" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26935" PostTypeId="1" AcceptedAnswerId="26947" CreationDate="2018-07-12T08:03:25.897" Score="3" ViewCount="277" Body="&lt;p&gt;I'm not sure, whether there is any other observable super cluster like our local one, Laniakea, and if they do exists, do they also get pulled by the Great Attractor? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a little confused about superclusters and The Great Attractor, how they do exists relative to each other and how we can see/know from our galaxy, specifically from the Earth. &lt;/p&gt;&#xA;" OwnerUserId="23665" LastEditorUserId="23678" LastEditDate="2018-07-13T15:01:28.943" LastActivityDate="2018-07-13T16:10:34.057" Title="Do other supercluster like Laniakea also get pulled by the Great Attractor?" Tags="&lt;redshift&gt;&lt;galaxy-cluster&gt;&lt;local-group&gt;&lt;apparent-motion&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26936" PostTypeId="1" AcceptedAnswerId="27048" CreationDate="2018-07-12T10:08:12.813" Score="5" ViewCount="131" Body="&lt;p&gt;I suspect our best bet for finding advanced civilisations is to look for gamma rays characteristic of nuclear/antimatter propulsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an alien civilisation was using something similar to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Project_Orion_(nuclear_propulsion)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Orion drive&lt;/a&gt;, how close would they have to be for us to be able to detect the gamma rays?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my mind, this question has two parts:&#xA;What are the magnitude/frequency limits on the gamma ray telescopes in orbit? (Specifically Fermi, assuming that has the best detection capabilities of an orbital gamma ray observatory)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the magnitude and frequency profile on, say, a 1 megaton nuclear bomb? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've had a bit of trouble finding these out myself from a casual search, and am wondering if maybe there are military secret issues surrounding the detection of gamma rays. &lt;/p&gt;&#xA;" OwnerUserId="17655" LastEditorUserId="17655" LastEditDate="2018-07-22T07:10:43.580" LastActivityDate="2018-07-23T03:51:28.210" Title="Gamma ray detection for SETI. How far away could we detect an alien nuclear powered rocket?" Tags="&lt;gamma-ray-bursts&gt;&lt;seti&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26937" PostTypeId="2" ParentId="11296" CreationDate="2018-07-12T13:47:14.413" Score="0" Body="&lt;p&gt;I'm using my own web app for this: &lt;a href=&quot;https://deep-skies.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://deep-skies.com&lt;/a&gt; . You can manually create an observation log or import one saved in a .skylist file (default format in SkySafari app 4 &amp;amp; 5 versions). You also have a nice overview of all of your observing sessions. &lt;/p&gt;&#xA;" OwnerUserId="23668" LastEditorUserId="23668" LastEditDate="2018-07-16T13:18:56.127" LastActivityDate="2018-07-16T13:18:56.127" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26938" PostTypeId="2" ParentId="26898" CreationDate="2018-07-12T14:12:07.010" Score="0" Body="&lt;p&gt;I came here to ask the same question on July 8 but I found you had asked already 3 hours ago. So I waited for answers. When Mike G assured it's mars, I felt better, because when I saw the object I felt it's never a satellite or UFO like something. But we are in different Geo location, right? So how can that answer be authentic?(although, Mike has provided some technical info for Geo locations, but I'm a bit lazy to measure and find that way) So I had to search for any reliable source. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yesterday, searching on google, I found &lt;a href=&quot;https://www.fourmilab.ch/cgi-bin/Yourhorizon&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.fourmilab.ch/cgi-bin/Yourhorizon&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I corrected Latitude and Longitude with my location( the city I lived), which I found in wikipedia or you can use google maps( but that case you have to convert the decimal coordinates to degree). And then Updating the data with real time, it's just almost accurate (with eye level measurement) for the red planet and else. Yes, it's mars. But wait, there are more, I found Saturn, Jupiter- I was blown away, identifying them as what they are for the first time in my life. (planets are shown with symbols- the legends you can find on &lt;a href=&quot;https://www.fourmilab.ch/yoursky/help/icons.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.fourmilab.ch/yoursky/help/icons.html&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think as it's relative with Geo location, you can get the actual answer, trying this site, the way I found my answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks &lt;/p&gt;&#xA;" OwnerUserId="23613" LastEditorUserId="23613" LastEditDate="2018-07-13T00:46:08.133" LastActivityDate="2018-07-13T00:46:08.133" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26939" PostTypeId="2" ParentId="26909" CreationDate="2018-07-12T15:53:10.437" Score="1" Body="&lt;p&gt;This might oversimplify things but here's my go:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Create a flat surface (the larger the better as long as it stays flat), e.g. by placing a board on a still surface of water.&lt;/li&gt;&#xA;&lt;li&gt;Put up a long pole (the longer the better) vertically on that surface at noon.&lt;/li&gt;&#xA;&lt;li&gt;Measure its shadow (direction and length), which needs to be completely on the flat surface.&lt;/li&gt;&#xA;&lt;li&gt;Have someone do the same (esp. same length of pole) at the same time far to the north of you (the further the better).&lt;/li&gt;&#xA;&lt;li&gt;Have a third of the exact same measurements far to the south of you.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Evaluating the measurements should establish:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Earth's surface is roughly spherical (actually earth is an oblate ellipsoid but you need more than 3 measurements to confirm that)&lt;/li&gt;&#xA;&lt;li&gt;Earth diameter is within reported values (+/- expected deviation for measurement error and the fact that you only measured a very rough estimation)&lt;/li&gt;&#xA;&lt;li&gt;Rough estimation of earth-sun distance by triangulating&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Using a pinhole camera you can now achieve a rough estimation of sun's actual diameter by its apparent diameter and the distance estimation from above. Even accumulating all the measurement errors, the difference in size between sun and earth should be some orders of magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attach two balls to the opposite ends of a rod (the lighter the rod compared to the balls the better). The balls need to be rough approximations of the above established measurements (e.g. you could guess the sun is pure hydrogen and the earth is pure iron to achieve an estimation of mass). Attach a string to the rod and find the point of balance. Most likely it will be way to the ball representing the sun (you need to accommodate for weight of the rod).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can now make the two balls circle each other while hanging from the string.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which one revolves around the other?&lt;/p&gt;&#xA;" OwnerUserId="23670" LastActivityDate="2018-07-12T15:53:10.437" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26940" PostTypeId="1" CreationDate="2018-07-13T02:23:55.897" Score="2" ViewCount="148" Body="&lt;p&gt;Has the moon’s velocity changed since ‘shortly’ after it’s formation due to impacts? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wouldn't it be likely that a significant collision would have occurred over such a long period of time? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would seem an object with enough momentum to change the velocity of the Moon would have made an impact on its velocity at some point over the millennia. &lt;/p&gt;&#xA;" OwnerUserId="14603" LastEditorUserId="7982" LastEditDate="2018-07-19T01:53:21.437" LastActivityDate="2018-07-19T12:51:04.973" Title="Changes of the velocity of the Moon due to impacts from other objects?" Tags="&lt;the-moon&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26941" PostTypeId="2" ParentId="26940" CreationDate="2018-07-13T02:56:21.563" Score="4" Body="&lt;p&gt;Our best current understanding of the formation of the Moon has it forming quite close to Earth.  Because even then its orbit around the Earth took more than one day (Earth was spinning faster, also), the effect of tides has always been for the Earth to transfer some of its rotational energy to the Moon. This slows the Earth's rotation  and makes the Moon move outwards into a higher (and slower) orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's been around four billion years since the Moon last suffered a major impact, so collisions have not played a significant role since then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expanding (a lot) on this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we measure Lunar ages? Two ways: Cratering and isotope dating. The cratering technique was developed by William K. Hartmann of the University of Arizona and makes use of the amazingly simple fact that when two craters overlap, you can tell which is older. Combine that with the observed fact that while the cratering rate varies with time, it changes smoothly and the distribution of craters is essentially random. Using that and &lt;em&gt;a lot&lt;/em&gt; of painstaking analysis of photography, the &lt;em&gt;order&lt;/em&gt; of events on the Moon can be determined quite closely (E.g., &lt;em&gt;this&lt;/em&gt; crater, &lt;em&gt;then&lt;/em&gt; this mare, then &lt;em&gt;those&lt;/em&gt; craters, etc.) as well as getting rough estimates of the actual time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isotope dating of Lunar meteors (bits of Lunar rock blasted into space by impacts) and Apollo samples lets us put precise times to various bits of the sequence resulting in a reliable chronology going back nearly to the Moon's formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arguably there were four stages in the Moon's history:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stage 1: &lt;a href=&quot;https://en.wikipedia.org/wiki/Giant-impact_hypothesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Formation of the Moon by Giant Impact&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon forms out of the debris in orbit after a Mars-sized impactor strikes the Earth about 4.5 billion years ago during the formation of the Solar System.  This was the final giant impact involving either the Earth or the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stage 2: The Early Years&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When formed, the early Moon was around 25,000 miles from Earth and the huge tides raised by the Moon in the Earth and by the Earth in the moon rapidly (by astronomical standards, anyway) further out.  The Moon (and Earth) continued to be hit by big rocks (dinosaur-killer-sized and bigger) but not by things in the many hundreds of miles in diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stage 3: The &lt;a href=&quot;https://en.wikipedia.org/wiki/Late_Heavy_Bombardment&quot; rel=&quot;nofollow noreferrer&quot;&gt;Late Heavy Bombardment&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can tell from cratering patterns and from dating Lunar samples, that 4.1 to 3.8 billion years ago there was an upsurge in big rocks (dinosaur-killer-sized) hitting the Moon. We also see evidence of this on other planets, though not on Earth since so little remains of the Earth's crust from that era.) We don't know why the LHB happened, though there are &lt;em&gt;lots&lt;/em&gt; of theories.  (There are also scientists who think the LHB was not as heavy and more spread out, but they are currently in a minority.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stage 4: Retirement&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the end of the LHB, the Moon has had few new craters created and has simply slowly receded from Earth (enjoying a serene and fairly uneventful old age.)  There are still craters being formed now and then -- the freshest big one, &lt;a href=&quot;https://en.wikipedia.org/wiki/Tycho_(lunar_crater)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tycho&lt;/a&gt;, is only 108 million years old!  Another recent one, &lt;a href=&quot;https://en.wikipedia.org/wiki/Aristarchus_(crater)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aristarchus&lt;/a&gt;, is around 400 million years old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are typical of the last few billion years: &lt;em&gt;Very&lt;/em&gt; occasional craters caused by rocks in the 10-50 mile rage. They make a big hole, but happen rarely (&lt;em&gt;ca.&lt;/em&gt; even hundred million years) and have had no significant impact (pun intended) on the Moon's orbit.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2018-07-19T12:51:04.973" LastActivityDate="2018-07-19T12:51:04.973" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26943" PostTypeId="1" CreationDate="2018-07-13T06:05:04.190" Score="6" ViewCount="152" Body="&lt;p&gt;There is a famous story of an experienced astronomer setting his beard on fire because he left the cap off his telescopes finder. I know it's said to be true of Galileo, but I thing it's relatively recent - i.e. 20th century.&#xA;Does anyone know his name? Or can someone cite a source? &lt;/p&gt;&#xA;" OwnerUserId="23674" LastEditorUserId="11264" LastEditDate="2018-07-13T11:35:25.630" LastActivityDate="2020-06-02T13:41:30.213" Title="Which astronomer set his beard on fire because he left the cap off his telescopes finder?" Tags="&lt;the-sun&gt;&lt;observational-astronomy&gt;&lt;history&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26944" PostTypeId="1" CreationDate="2018-07-13T06:16:49.780" Score="1" ViewCount="35" Body="&lt;p&gt;did anyone watch the sky in the west at 1:15 am.On 7-13-18. a star moving around in the greenup ky UT WOULD go back and forth&lt;/p&gt;&#xA;" OwnerUserId="23675" LastActivityDate="2018-07-13T06:16:49.780" Title="a star in the sky thay moves back and forth way up in the sky" Tags="&lt;star&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2018-07-13T13:36:04.403" ContentLicense="CC BY-SA 4.0" />
  <row Id="26945" PostTypeId="1" AcceptedAnswerId="26949" CreationDate="2018-07-13T13:17:25.657" Score="3" ViewCount="95" Body="&lt;p&gt;I`m wondering if the gravitational pull from earth that should be constant around the planet, will get amplified by a celestial body like the moon orbiting the earth? &lt;/p&gt;&#xA;" OwnerUserId="23678" LastActivityDate="2018-07-13T17:40:21.840" Title="Can gravity be amplified by other celestial bodies in its vicinity?" Tags="&lt;gravity&gt;&lt;newtonian-gravity&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="26946" PostTypeId="2" ParentId="26935" CreationDate="2018-07-13T13:57:49.557" Score="1" Body="&lt;p&gt;The method used to see the movement of galaxy&lt;code&gt;s relative movement to each other they measure the redshift of the galaxy&lt;/code&gt;s&#xA;&lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/Astro/redshf.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/hbase/Astro/redshf.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked in to you question and found some answers for you in another topic on this site.Seems like the super cluster Laniekea is mowing towards the Great Attractor and accelerating as a whole at different speeds due to there diffrent masses.&#xA; &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6288/when-will-the-milky-way-arrive-at-the-great-attractor-and-what-all-happen-the?rq=1&quot;&gt;When will the Milky Way &amp;quot;arrive&amp;quot; at the Great Attractor, and what all happen then?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23678" LastActivityDate="2018-07-13T13:57:49.557" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26947" PostTypeId="2" ParentId="26935" CreationDate="2018-07-13T16:10:34.057" Score="2" Body="&lt;p&gt;In short, not much gravitational pull at all, not relative to the distance and relative velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything in the observable universe is gravitationally attracted to everything else, at least &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_cosmological_horizons&quot; rel=&quot;nofollow noreferrer&quot;&gt;within the appropriate horizon&lt;/a&gt;.  There should be some objects far enough apart, taking into account cosmic inflation, that gravity can't travel that far over the age of the Universe.  For example the farthest galaxy we can see in one direction and the farthest one we can see in the opposite direction, those two galaxies can't see each other and they likely experience no gravitational pull from each other, but that's not really relevant to your question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Great_Attractor&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Great Attractor&lt;/a&gt; is a comparatively small but very dense part of Laniakea, with an estimated mass thousands of times the mass of the Milky Way.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Laniakea_Supercluster&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Laniakea supercluster&lt;/a&gt; has a mass of about 100,000 Milky Ways, so any gravitational attraction one supercluster experiences would primarily be from other entire superclusters, and the great attractor would be a small percentage of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Superclusters experience gravity from other superclusters but the Universe is very uniform, so that gravitational pull is largely balanced out because it's coming from all sides.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This touches on one of the modern puzzles in physics, &lt;a href=&quot;https://en.wikipedia.org/wiki/Horizon_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;called the Horizon problem&lt;/a&gt;. This question goes back to the big bang and why the young universe was very close to, but not precisely uniform.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be a fun exercise to calculate the gravitational pull neighboring superclusters have on each other.   I imagine it's pretty negligible compared to the expansion of space between them.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-07-13T16:10:34.057" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26948" PostTypeId="1" AcceptedAnswerId="26964" CreationDate="2018-07-13T16:39:44.543" Score="3" ViewCount="429" Body="&lt;p&gt;I have a distant $z\sim3$ galaxy with absolute AB magnitudes $M_B$ and $M_I$, where the central wavelengths of the $B$ and $I$ passbands are $\sim0.4\,\mu\text{m}$ and $\sim0.8\,\mu\text{m}$, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this galaxy was to experience $1\,\text{mag}$ of $V$-band extinction due to dust &lt;strong&gt;within it&lt;/strong&gt;, where the central wavelength of the $V$ passband is $\sim0.5\,\mu\text{m}$, how do I work out what the absolute magnitudes $M_B$ and $M_I$ would be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help with this issue.&lt;/p&gt;&#xA;" OwnerUserId="23614" LastEditorUserId="23614" LastEditDate="2018-07-15T20:47:24.733" LastActivityDate="2018-07-15T20:47:24.733" Title="Absolute magnitudes with dust extinction" Tags="&lt;dust&gt;&lt;absolute-magnitude&gt;&lt;extinction&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26949" PostTypeId="2" ParentId="26945" CreationDate="2018-07-13T17:40:21.840" Score="1" Body="&lt;p&gt;Gravity itself cannot be amplified but the forces are additive. The earth is held in orbit by the Sun's gravity but earth is also affected by the moon's gravity. Depending on their position the moon's gravity could be aligned with Sun's gravity, increasing the gravitational pull. Or the moon could reduce the effect of suns gravity if on opposite side of the earth relative to the sun.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastActivityDate="2018-07-13T17:40:21.840" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26950" PostTypeId="1" AcceptedAnswerId="26954" CreationDate="2018-07-13T20:18:36.620" Score="3" ViewCount="122" Body="&lt;p&gt;I've just found out that moon's dark side is 30 miles thicker then the closer side to the Earth. So I'm wondering about moon's mass. How come it's still balanced? Supposing the far side is indeed heavier, considering More Mass More Gravity, wouldn't it have to be closer to Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if I'm making stupid assumptions, I'm just a curious fela. &lt;/p&gt;&#xA;" OwnerUserId="23680" LastActivityDate="2018-07-14T03:45:59.963" Title="How come moon's difference in thickness between sides does not create an unbalance on its mass?" Tags="&lt;the-moon&gt;&lt;gravity&gt;&lt;mass&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26951" PostTypeId="1" CreationDate="2018-07-13T22:12:38.043" Score="5" ViewCount="611" Body="&lt;p&gt;There were theories that Pluto's atmosphere collapses when it gets too far from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I've read &lt;a href=&quot;https://arxiv.org/abs/1601.05672&quot; rel=&quot;noreferrer&quot;&gt;papers&lt;/a&gt; that says that it &lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/2013/09051420-plutos-atmosphere-does-not-collapse.html&quot; rel=&quot;noreferrer&quot;&gt;doesn't&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;a href=&quot;https://www.nasa.gov/feature/new-horizons-reveals-pluto-s-atmospheric-pressure-has-sharply-decreased&quot; rel=&quot;noreferrer&quot;&gt;NASA&lt;/a&gt; says that the surface pressure went down recently (implying that it does).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any clear data as to what's going on?&lt;/p&gt;&#xA;" OwnerUserId="23681" LastActivityDate="2018-07-15T11:58:06.257" Title="Does Pluto's atmosphere collapse?" Tags="&lt;planetary-atmosphere&gt;&lt;pluto&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26952" PostTypeId="2" ParentId="26951" CreationDate="2018-07-13T23:00:43.827" Score="4" Body="&lt;p&gt;Found this link by Emily Lakdawalla, she wrote a detailed explanation about Pluto`s seasons and its movement thru the solar system. Here research seems to contradict the atmosphere collapse and and here 3 models and research on Pluto. Seems the most likely model gives the highest pressure in the atmosphere between 2020 ad 2040, but none a atmosphere collapse. &lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/2013/09051420-plutos-atmosphere-does-not-collapse.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23678" LastEditorUserId="5278" LastEditDate="2018-07-15T11:58:06.257" LastActivityDate="2018-07-15T11:58:06.257" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26954" PostTypeId="2" ParentId="26950" CreationDate="2018-07-14T03:45:59.963" Score="4" Body="&lt;p&gt;As is the case on Earth, the Moon's crust is less dense than the material that comprises the Moon's mantle or core. On the Earth, this means that areas with thick crust rise above sea level. (These are our continents.) On the Moon, this means that the Moon's center of figure (it's center based on surface topography) is offset by a couple of kilometers from Moon's center of mass, with the center of mass closer to the Earth than the center of figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This center of mass to center of figure offset does not mean much with regard to the Moon's stability. What matters much more is the distribution of mass top to bottom vs side to side vs front to back. To be stable in a tidally locked orientation, a moon's principal axes deduced from the moment of inertia tensor should be very close to the set of axes as seen from the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, a very specific ordering of the moments about the principal axes as deduced from the inertia tensor needs to exist. For the moon to be in a stable tidally locked configuration, the axis of rotation needs to be more or less orthogonal to the moon's orbital plane, the moment of inertia about the rotation axis needs to be the largest, and the moment of inertia about the line from the center of the planet to the center of the moon needs to be the smallest, leaving the moment of inertia about the side-to-side as the intermediate axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This most definitely is the case with the Earth's Moon. The angular difference between the Moon's geometric axes as seen from the Earth and the Moon's principal axes deduced from its inertia tensor (in technical terms, the angular difference between the Moon-Mean Earth frame and the Moon's principal axis frame) is a few hundredths of a degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That the Moon's far side has a much thicker crust than does the near side is a third order effect. The Moon would be nearly as stable as it is now if it was the Moon's far side with its much thicker crust that faced the Earth rather than facing empty space.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2018-07-14T03:45:59.963" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26955" PostTypeId="1" AcceptedAnswerId="26957" CreationDate="2018-07-14T03:57:39.033" Score="15" ViewCount="1570" Body="&lt;p&gt;Do any of the planets have a Roche limit that is strong enough to be felt by an astronaut whilst in orbit?  &lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-12-27T19:14:56.223" LastActivityDate="2021-02-04T21:05:10.693" Title="Is there any orbit at which the Roche limit can be &quot;felt&quot;?" Tags="&lt;gravity&gt;&lt;jupiter&gt;&lt;roche-limit&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26957" PostTypeId="2" ParentId="26955" CreationDate="2018-07-14T08:22:59.133" Score="24" Body="&lt;p&gt;Roche limit happens where the gravity of the object, trying to pull the object together, becomes smaller than the tidal force (trying to pull the object apart).&lt;/p&gt;&#xA;&lt;p&gt;But the astronaut is bound by not gravity, rather by the electromagnetic interaction between his/her atoms. The own gravity of the astronaut is negligible, compared to the electromagnetic interaction.&lt;/p&gt;&#xA;&lt;p&gt;However, the tidal force affecting an astronaut, should require a little calculation. We can derive the formula of the gravitational acceleration around a point-like body (&lt;span class=&quot;math-container&quot;&gt;$F=\frac{GM}{r^2}$&lt;/span&gt;), we get&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{dF}{dr}=\frac{2GM}{r^3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;(We can ignore the sign on obvious reasons.)&lt;/p&gt;&#xA;&lt;p&gt;Here &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; is the gravitational constant, &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is the mass of the body, and &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; is the distance.&lt;/p&gt;&#xA;&lt;p&gt;Substituting the values of the Sun, we get &lt;span class=&quot;math-container&quot;&gt;$\frac{2\cdot 6.67 \cdot 10^{-11} \cdot 2 \cdot 10^{30}}{(7\cdot 10^8)^3} \approx 7.78\cdot 10^{-7} \frac{\mathrm{m/s^2}}{\mathrm{m}} \approx \underline{\underline{8 \cdot 10^{-8} \frac{g}{\mathrm{m}}}}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;More clearly, if we are orbiting the Sun just above its surface, a roughly 2m long astronaut feel that his head and foot are pulled apart by around &lt;span class=&quot;math-container&quot;&gt;$1.6\cdot 10^{-7}g$&lt;/span&gt; weight. In the case of a &lt;span class=&quot;math-container&quot;&gt;$70\:\mathrm{kg}$&lt;/span&gt; astronaut, it is around the weight of &lt;span class=&quot;math-container&quot;&gt;$0.0112$&lt;/span&gt; gram on the Earth.&lt;/p&gt;&#xA;&lt;p&gt;The astronaut wouldn't feel it, but not very sensitive sensors could already measure it.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;sub&gt;This calculation sometimes used &lt;span class=&quot;math-container&quot;&gt;$\mathrm{g}$&lt;/span&gt; for &amp;quot;gram&amp;quot;, as unit of mass, and &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; as the (non-standard) unit of acceleration.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="1891" LastEditDate="2021-02-04T21:05:10.693" LastActivityDate="2021-02-04T21:05:10.693" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26958" PostTypeId="2" ParentId="26955" CreationDate="2018-07-14T08:23:10.280" Score="14" Body="&lt;p&gt;The Roche limit is where the tidal forces exerted on an orbiting object are sufficient to overcome the &lt;em&gt;self-gravity&lt;/em&gt; of that object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;self-gravity&quot; of an astronaut is tiny. We can estimate it as something like &lt;span class=&quot;math-container&quot;&gt;$$F_{\rm grav} \sim \frac{Gm^2}{(h/2)^2},$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; is the mass of the astronaut (+ equipment) and &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; is their size (height). Assuming &lt;span class=&quot;math-container&quot;&gt;$m=100$&lt;/span&gt; kg and &lt;span class=&quot;math-container&quot;&gt;$h=2$&lt;/span&gt; m, then the self-gravity  force is &lt;span class=&quot;math-container&quot;&gt;$2.7\times 10^{-6}$&lt;/span&gt; N. This is a force that is too small to feel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tidal force on the astronaut a distance &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; from a body of mass &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is approximately&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ F_{\rm tidal} \simeq 2 \frac{GMm}{R^3} h,$$&lt;/span&gt;&#xA;assuming they are pointing feet-first towards the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Roche limit is where &lt;span class=&quot;math-container&quot;&gt;$F_{\rm grav} &amp;lt; F_{\rm tidal}$&lt;/span&gt;, so where&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{Gm^2}{(h/2)^2}&amp;lt; 2 \frac{GMm}{R^3} h$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ R &amp;lt; h\left(\frac{M}{2m}\right)^{1/3}$$&lt;/span&gt;&#xA;For example if &lt;span class=&quot;math-container&quot;&gt;$M=M_{\rm Earth}$&lt;/span&gt; and for the astronaut above, then &lt;span class=&quot;math-container&quot;&gt;$R &amp;lt; 60,000$&lt;/span&gt; km for tidal break up. Which seems odd, because astronauts work quite happily in low-Earth orbit where the tidal forces are much stronger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with this calculation is that astronauts are not held together by self-gravity and a tidal field at the Roche limit has a negligible effect on a small body that is actually held together by atomic forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to experience a tidal field that can be felt on astronaut scales, let's say larger than 10 N (imagine hanging a 1 kg weight from your ankle on Earth), you would have to get much closer to the source of gravity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a fixed mass equal to that of the Earth, we can work out you would need to get to within 500 km &lt;em&gt;of the centre of mass&lt;/em&gt; in order to feel the tidal force. For the Earth (and also for other solar system bodies you could do the same calculation for) this would put you well &lt;em&gt;inside&lt;/em&gt; the Earth, which isn't possible and in any case we could not assume that &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; was fixed in that case, because it is the mass interior to &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; that counts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way that an astronaut could &quot;feel&quot; a tidal force would be to approach a compact star - a high density neutron star, white dwarf or black hole. There you can generate a very strong tidal field and, because they are compact, an astronaut could get close enough to feel it.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2020-04-27T18:05:54.453" LastActivityDate="2020-04-27T18:05:54.453" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26960" PostTypeId="5" CreationDate="2018-07-14T15:22:24.260" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2018-07-14T15:22:24.260" LastActivityDate="2018-07-14T15:22:24.260" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26961" PostTypeId="4" CreationDate="2018-07-14T15:22:24.260" Score="0" Body="Questions about positions of objects in the sky, either relative to a fixed coordinate system like the equatorial and ecliptic coordinate systems, or relative to an observer's point of view." OwnerUserId="7472" LastEditorUserId="7472" LastEditDate="2018-07-14T18:08:46.273" LastActivityDate="2018-07-14T18:08:46.273" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26962" PostTypeId="2" ParentId="26955" CreationDate="2018-07-14T20:57:46.863" Score="5" Body="&lt;p&gt;Expanding on Peterh's answer, we could try to find how should be an astronomical object for the tidal forces be felt by an astronaut orbiting it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have any reliable data on how strong need the tidal force to be felt. However, with a big simplification, we can very roughly model the upper and lower body of an astronaut as two masses placed about 1 meter apart. For a 70 kg astronaut under a tidal force of &lt;span class=&quot;math-container&quot;&gt;$0.1·g$&lt;/span&gt; per meter (where &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; is the acceleration of gravity), the difference of pull between those two 35 kg masses would be &lt;span class=&quot;math-container&quot;&gt;$0.1·35 kg = 3.5 kg$&lt;/span&gt;. Those force would stretch the astronaut's waist and would be clearly noticeably (maybe a force ten times weaker would be noticeable, too, but I'll will stick to &lt;span class=&quot;math-container&quot;&gt;$0.1m^{-1}·g$&lt;/span&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Peterh's formulas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$r=\sqrt[3]{\frac{2·G·M}{0.1m^{-1}·g}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a 1 solar mass object:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$r=\sqrt[3]{\frac{2·6.67·10^{-11}·2·10^{30}}{0.1·g}}=6481168 m = 6481 km$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, an astronaut orbiting a sun-sized mass at a distance similar to Earth's radius would clearly feel tide forces when their head or feet point to the object. Of course the object would need to be a black hole or a neutron star to fit inside of the orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a more massive object the orbit could be larger, but given that mass is inside a cubic root, radius would grow very slowly.&lt;/p&gt;&#xA;" OwnerUserId="14097" LastEditorUserId="14097" LastEditDate="2020-04-27T17:25:50.313" LastActivityDate="2020-04-27T17:25:50.313" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26963" PostTypeId="2" ParentId="8367" CreationDate="2018-07-15T18:14:56.660" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://mesa.sourceforge.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;MESA&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is designed specifically for stellar evolution. There are many modules that you can play around with, including binary interaction.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-07-15T18:14:56.660" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26964" PostTypeId="2" ParentId="26948" CreationDate="2018-07-15T18:32:04.997" Score="1" Body="&lt;p&gt;Check &lt;a href=&quot;http://astro.pas.rochester.edu/~aquillen/ast142/Lecture/Lecture12.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this lecture note&lt;/a&gt;. I believe there is everything you need to understand how to proceed. In brief, you should have extra information about the relative absorption, shown in the table on p.18. Given the correction in one band (which you already have $A_V = 1)$, you can find the correction in other bands using the table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the relative absorption is different for different scenario, e.g., different RADEC, different host, different type of stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that there are two types of extinction: galactic and extragalactic.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-07-15T18:32:04.997" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26965" PostTypeId="1" AcceptedAnswerId="26978" CreationDate="2018-07-15T18:46:02.713" Score="2" ViewCount="132" Body="&lt;p&gt;A galaxy 13 billion light-years distant is 94% of the way to the big bang. The universe was 800 million years old when the light was emitted. That galaxy today is traveling away from us at 94% of the speed of light roughly. So 13 billion years ago it was 750 million light years away from us. Assuming the expansion is NOT greater than light speed, that photon did NOT go backward. This is also evidenced by the fact that the photon arrived at earth. It seems to me the photon took 13 billion years to cross 750 million light years. What am I doing wrong?&#xA;   Question two. If this is right then please explain how we go from 750 (or 800) million light years to 46.5 billion light years.&lt;/p&gt;&#xA;" OwnerUserId="23697" LastActivityDate="2018-07-17T17:50:54.807" Title="Yet another Question about the expansion of universe" Tags="&lt;universe&gt;&lt;observable-universe&gt;&lt;size&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26967" PostTypeId="2" ParentId="25030" CreationDate="2018-07-15T20:40:48.340" Score="1" Body="&lt;p&gt;So in this model, all of the observable universe is expanding. It is simply that it is part of a larger universe which is being locally stretched, as part of a an accelerating convergence to a central point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the answer is that we cannot distinguish this model from an expanding universe (assuming the structure of the convergence is very carefully tuned to look like what we see &quot;locally&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it makes no difference as far as the observable universe is concerned, so we have to resort to Occam's razor. The expanding universe is simpler, and arises naturally from our theories of gravity, etc.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-07-15T20:40:48.340" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26969" PostTypeId="1" CreationDate="2018-07-16T11:53:14.443" Score="0" ViewCount="97" Body="&lt;p&gt;Space is dark. But if we direct telescopes to seemingly dark portions of sky, we see that it is filled with galaxies. If the universe is theoretically infinite, wouldn't an infinite number of photons be reaching the earth and illuminating the night sky brighter than it appears?&lt;/p&gt;&#xA;" OwnerUserId="23569" LastActivityDate="2018-07-16T15:05:22.823" Title="With all the stars in the universe, why isn't the sky saturated with light?" Tags="&lt;light&gt;&lt;space&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2018-07-16T17:49:25.247" ContentLicense="CC BY-SA 4.0" />
  <row Id="26970" PostTypeId="2" ParentId="26969" CreationDate="2018-07-16T15:05:22.823" Score="-2" Body="&lt;p&gt;Logic1: If the universe is infinite AND static, the sky will be saturated with light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note &quot;AND.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Logic2: If the sky is not saturated with light, the universe is not infinite or not static.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note &quot;OR.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Logic1 = Logic2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observation: the sky is not saturated with light.&#xA;Therefore: the universe is not infinite or not static.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the universe is dynamic, at least.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-07-16T15:05:22.823" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26971" PostTypeId="2" ParentId="25416" CreationDate="2018-07-16T18:50:44.923" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does Uranus and Neptune have more methane than Jupiter and Saturn?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's a combination of equations of state (&lt;a href=&quot;https://en.wikipedia.org/wiki/Equation_of_state&quot; rel=&quot;noreferrer&quot;&gt;EOS&lt;/a&gt;), &lt;a href=&quot;https://en.wikipedia.org/wiki/Serpentinite#Abiotic_methane_production_on_Mars_by_serpentinization&quot; rel=&quot;noreferrer&quot;&gt;serpentinization&lt;/a&gt;, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmological_principle#Criticism&quot; rel=&quot;noreferrer&quot;&gt;mixing&lt;/a&gt; (rotational and convective) that favors a preference for some reactions (and resulting compounds) over others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the references below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The giant planets are all mostly hydrogen and helium, but Uranus and Neptune have relatively large amounts of hydrogen compounds like methane (that's what gives them their color).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Jupiter and Saturn are &lt;a href=&quot;https://en.wikipedia.org/wiki/Gas_giant&quot; rel=&quot;noreferrer&quot;&gt;gas giants&lt;/a&gt;, Uranus and Neptune are &lt;a href=&quot;https://en.wikipedia.org/wiki/Ice_giant&quot; rel=&quot;noreferrer&quot;&gt;ice giants&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kSWhx.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kSWhx.jpg&quot; alt=&quot;Cutaway view of Jupiter, Saturn, Uranus and Neptune.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My question is why did that happen? How did Uranus and Neptune get their methane? My impression is that all the gas giants were far enough out for methane to condense into ice, so how did Uranus and Neptune end up preferentially with methane?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See Wikipedia's &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Extraterrestrial_atmosphere#Gas_giants&quot; rel=&quot;noreferrer&quot;&gt;Extraterrestrial Atmosphere&lt;/a&gt;&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QRtzr.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QRtzr.png&quot; alt=&quot;Graphs of Escape Velocity vs. Gas Composition.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Graphs of escape velocity against surface temperature of some Solar System objects showing which gases are retained. The objects are drawn to scale, and their data points are at the black dots in the middle. Data is based on &quot;&lt;a href=&quot;http://ircamera.as.arizona.edu/astr_250/Lectures/Lec_05sml.htm&quot; rel=&quot;noreferrer&quot;&gt;Lecture 5: Overview of the Solar System, Matter in Thermodynamc Equilibrium&lt;/a&gt;&quot; and &quot;&lt;a href=&quot;http://www-spof.gsfc.nasa.gov/stargaze/StarFAQ12.htm&quot; rel=&quot;noreferrer&quot;&gt;Stargazer's FAQ - How exactly are atmospheres held?&lt;/a&gt;&quot;.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia says little about the atmosphere of these planets, and the least about Uranus and Neptune:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Jupiter#Chemical_composition&quot; rel=&quot;noreferrer&quot;&gt;Atmosphere of Jupiter&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;There are no methane clouds as the temperatures are too high for it to condense.&quot; - Source: &quot;&lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/S0032063304002442?via%3Dihub&quot; rel=&quot;noreferrer&quot;&gt;Jupiter's ammonia clouds — localized or ubiquitous?&lt;/a&gt;&quot; (April 9 2004), by S.K.Atreya, A.S.Wong, K.H.Baines,  M.H.Wong, and T.C.Owen.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Quotes from the paper:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Page 502: &quot;For the production of polycyclic aromatic hydrocarbons (PAHs), chemistry begins with the destruction of methane (CH$_4$) by solar UV photons at $\lambda \le$160 nm, ultimately leading to the formation of benzene ($c$-C$_6$H$_6$, or A$_1$) and other complex hydrocarbons (Fig. 3). In the polar auroral regions where energetic particles also break down methane, ion chemistry becomes dominant in the production of benzene and heavy hydrocarbons (Wong et al., 2003, and Fig. 3).&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Saturn#Atmosphere&quot; rel=&quot;noreferrer&quot;&gt;Atmosphere of Saturn&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Ultraviolet radiation from the Sun causes methane photolysis in the upper atmosphere, leading to a series of hydrocarbon chemical reactions with the resulting products being carried downward by eddies and diffusion. This photochemical cycle is modulated by Saturn's annual seasonal cycle.&quot;. - Source: &quot;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2008sf2a.conf..405G&quot; rel=&quot;noreferrer&quot;&gt;Ethane, acetylene and propane distribution in Saturn's stratosphere from Cassini/CIRS limb observations&lt;/a&gt;&quot; (Nov. 2008), by S. Guerlet, T. Fouchet, and B. Bézard.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Quotes from the paper:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Page 406: &quot;&lt;strong&gt;3 Method&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We used a line-by-line radiative transfer model to calculate synthetic spectra. It included opacity from &lt;strong&gt;CH$_4$&lt;/strong&gt;, CH$_3$D, C$_2$H$_6$, C$_2$H$_2$, C$_3$H$_8$, C$_3$H$_4, C$_4$H$_2 and collision-induced opacity from H2-He and H2-H2 . The atmospheric grid consisted in [of] 360 layers from 10 bar to 10−8 bar. It was coupled with an iterative inversion algorithm adapted from Conrath et al. (1998), in order to retrieve the atmospheric state (temperature, hydrocarbon vertical profiles) from the measured spectra.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;As a molecular emission intensity depends on both its abundance and temperature, we proceeded in two steps. First, we retrieved the temperature vertical profile from the methane ν4 emission band at 1305 m$^{−1}$ (assuming it is uniformly mixed with a vmr of 4.5 x10$^{−3}$ (Flasar et al. 2005)), providing information in the 1 mbar - 2 $\mu$bar region. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Figure 1 shows an example of a comparison between synthetic and observed emission bands of ethane, acetylene and propane at two given pressure levels (all the different pressure levels probed by CIRS have not&#xA;  been plotted for the sake of clarity) and Fig. 3 the Corresponding retrieved profiles.&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What that means is that more complex compounds than methane are favored by the conditions, see comments above concerning &quot;equations of state&quot;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Ice_giant#Atmosphere_and_weather&quot; rel=&quot;noreferrer&quot;&gt;Atmosphere of Uranus and Neptune&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The  gaseous outer layers of the ice giants have several similarities to those of the gas giants. These include long-lived, high-speed equatorial winds, polar vortices, large-scale circulation patterns, and complex chemical processes driven by ultraviolet radiation from above and mixing with the lower atmosphere.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Studying the ice giants' atmospheric pattern also gives insights into atmospheric physics. Their compositions promote different chemical processes and they receive far less sunlight in their distant orbits than any other planets in the Solar System (increasing the relevance of internal heating on weather patterns).&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/&quot; rel=&quot;noreferrer&quot;&gt;NASA Factsheets&lt;/a&gt; - Atmospheric composition (by volume, uncertainty in parentheses):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/jupiterfact.html&quot; rel=&quot;noreferrer&quot;&gt;Jupiter&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Major: Molecular hydrogen (H$_2$) - 89.8% (2.0%); Helium (He) - 10.2% (2.0%)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Minor (ppm): Methane (CH$_4$) - 3000 (1000); Ammonia (NH$_3$) - 260 (40); Hydrogen Deuteride (HD) - 28 (10); Ethane (C$_2$H$_6$) - 5.8 (1.5); Water (H$_2$O) - 4 (varies with pressure)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Aerosols: Ammonia ice, water ice, ammonia hydrosulfide&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/saturnfact.html&quot; rel=&quot;noreferrer&quot;&gt;Saturn&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Major: Molecular hydrogen (H$_2$) - 96.3% (2.4%); Helium (He) - 3.25% (2.4%) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Minor (ppm): Methane (CH$_4$) - 4500 (2000); Ammonia (NH$_3$) - 125 (75); Hydrogen Deuteride (HD) - 110 (58); Ethane (C$_2$H$_6$) - 7 (1.5) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Aerosols: Ammonia ice, water ice, ammonia hydrosulfide&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/uranusfact.html&quot; rel=&quot;noreferrer&quot;&gt;Uranus&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Major: Molecular hydrogen (H$_2$) - 82.5% (3.3%); Helium (He) - 15.2% (3.3%) Methane (CH$_4$) - 2.3% &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Minor (ppm): Hydrogen Deuteride (HD) - 148 &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Aerosols: Ammonia ice, water ice, ammonia hydrosulfide, methane ice(?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/neptunefact.html&quot; rel=&quot;noreferrer&quot;&gt;Neptune&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Major: Molecular hydrogen (H$_2$) - 80.0% (3.2%); Helium (He) - 19.0% (3.2%); Methane (CH$_4$) 1.5% (0.5%)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Minor (ppm): Hydrogen Deuteride (HD) - 192; Ethane (C$_2$H$_6$) - 1.5&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Aerosols: Ammonia ice, water ice, ammonia hydrosulfide, methane ice(?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Additional references:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;a href=&quot;http://www.scielo.org.mx/scielo.php?pid=S1405-33222015000300002&amp;amp;script=sci_abstract&quot; rel=&quot;noreferrer&quot;&gt;Methane in the Solar System&lt;/a&gt;&quot; in English, (Bol. Soc. Geol. Mex [online]. 2015, vol.67, n.3, pp.377-385.), by Andrés Guzmán-Marmolejo and Antígona Segura.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;a href=&quot;https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3689174/&quot; rel=&quot;noreferrer&quot;&gt;Abiotic Production of Methane in Terrestrial Planets&lt;/a&gt;&quot; (Astrobiology. 2013 Jun; 13(6): 550–559), by Andrés Guzmán-Marmolejo, Antígona Segura, and Elva Escobar-Briones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;a href=&quot;https://www.ncbi.nlm.nih.gov/pubmed/25774974&quot; rel=&quot;noreferrer&quot;&gt;Methane clathrates in the solar system&lt;/a&gt;&quot; (Astrobiology. 2015 Apr;15(4):308-26), by Mousis O, Chassefière E, Holm NG, Bouquet A, Waite JH, et al.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NASA - &quot;&lt;a href=&quot;https://www.nasa.gov/centers/goddard/news/topstory/2007/earthsized_planets.html&quot; rel=&quot;noreferrer&quot;&gt;Scientists Model a Cornucopia of Earth-sized Planets&lt;/a&gt;&quot; (Sept. 24 2007).&lt;/p&gt;&#xA;" OwnerUserId="18054" LastActivityDate="2018-07-16T18:50:44.923" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26973" PostTypeId="1" CreationDate="2018-07-17T11:22:34.180" Score="5" ViewCount="131" Body="&lt;p&gt;I found &lt;a href=&quot;https://www.space.com/16943-supernova-explosion-solar-system-formation.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;an article&lt;/a&gt; suggesting a supernova injected debris which helped nudge the formation of the Solar System 4.5 billion years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't find a name for this hypothetical star/supernova after some research. Is there one? What is it?&lt;/p&gt;&#xA;" OwnerUserId="23008" LastEditorUserId="2153" LastEditDate="2018-07-17T13:35:24.830" LastActivityDate="2018-07-17T14:07:36.790" Title="Is there a name for the star/supernova hypothesized to have been a precursor to the Solar System?" Tags="&lt;the-sun&gt;&lt;solar-system&gt;&lt;supernova&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26974" PostTypeId="2" ParentId="14537" CreationDate="2018-07-17T13:01:54.560" Score="2" Body="&lt;p&gt;This is an old question, but I'd like to answer it anyway. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does a star, when collapsing into a neutron star, or even a black hole, pick up any significant relativistic mass due to its rapid collapse rate? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. Conservation of energy applies. When you drop a 100000kg asteroid from some high elevation above the Earth, some of the asteroid's gravitational potential energy, which is mass-energy, is converted into kinetic energy. The mass-equivalence of the system does not increase as the asteroid falls. Instead the asteroid's rest mass reduces as the relativistic mass associated with its kinetic energy increases*.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the asteroid hits the Earth (BOOM!) some of the kinetic energy will get dissipated and radiated out into space, and then you're left with a mass deficit. The mass of the system is then less than it was. See &lt;a href=&quot;https://en.wikipedia.org/w/index.php?title=Binding_energy&amp;amp;oldid=850260579#Mass-energy_relation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;: &lt;em&gt;&quot;This missing mass may be lost during the process of binding as energy in the form of heat or light, with the removed energy corresponding to the removed mass through Einstein's equation E = mc²&quot;&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same applies to a collapsing star. There is no &quot;pick up&quot; of relativistic mass. Conservation of energy applies. But some of this energy is typically radiated away, so you're left with a reduced mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same also applies to a brick. When you drop a brick, gravity converts some of its rest mass into kinetic energy. This gets dissipated, and you're left with a mass deficit. Then when you lift the brick you do work on it. You add energy to it, thereby increasing its mass. Then you're back where you started. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same also applies to an electron. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;* this also applies to the Earth, but to a much lesser extent, so we tend to forget about it.&lt;/sub&gt; &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2018-07-17T13:01:54.560" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26975" PostTypeId="2" ParentId="26973" CreationDate="2018-07-17T13:25:27.077" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I couldn't find a name for this hypothetical star/nova after some research.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First off, the existence of this former object is hypothetical. Others working in this field have different hypotheses / conjectures; the science is far from settled. Secondly, if true, this star formed a supernova at the time the solar system was born and hence it no longer exists as a star. Thirdly, what's the point? Naming this hypothetical object would add zero credence to the hypothesis, and arguably would detract as being a bit premature.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2018-07-17T13:25:27.077" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26976" PostTypeId="2" ParentId="26973" CreationDate="2018-07-17T13:49:19.723" Score="4" Body="&lt;p&gt;Names of stars can serve a variety of purposes. Let's look at the different names and designations of &lt;a href=&quot;https://en.wikipedia.org/wiki/Betelgeuse&quot; rel=&quot;nofollow noreferrer&quot;&gt;Betelgeuse&lt;/a&gt;, for example.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A star's colloquial name, for lack of a better word, &lt;a href=&quot;https://en.wikipedia.org/wiki/IAU_Working_Group_on_Star_Names&quot; rel=&quot;nofollow noreferrer&quot;&gt;is sometimes chosen to recognize past cultural value&lt;/a&gt;. In the case of Betelgeuse, this name is . . . Betelgeuse. In the case of the hypothetical star, though, there is no name from antiquity, as it was not around in antiquity.&lt;/li&gt;&#xA;&lt;li&gt;A star's &lt;a href=&quot;https://en.wikipedia.org/wiki/Bayer_designation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bayer designation&lt;/a&gt; describes what constellation it is in and its apparent brightness. Betelgeuse appears to us as one of the brightest stars in Orion, so its Bayer designation is $\alpha$ Orionis. As we know nothing about the location or absolute magnitude of the hypothetical star (and heck, the constellations wouldn't have been the same 4.6 billion years ago), giving it a Bayer designation would be pointless.&lt;/li&gt;&#xA;&lt;li&gt;The name of a star in various catalogs may be arbitrary or may tell you something about its current location in the sky (see for instance the &lt;a href=&quot;https://en.wikipedia.org/wiki/Henry_Draper_Catalogue&quot; rel=&quot;nofollow noreferrer&quot;&gt;Henry Draper Catalogue&lt;/a&gt;. However, as I said above, things change on astronomical timescales, and a star will wander from its celestial coordinates on timeframes of hundreds, thousands, or millions of years. And there's no point in assigning this star an arbitrary number.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Finally, &lt;a href=&quot;https://astronomy.stackexchange.com/a/26975/2153&quot;&gt;as David Hammen said&lt;/a&gt;, the existence of this star is still very much hypothetical. Assuming that &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/2041-8205/756/1/L9/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;the paper he found&lt;/a&gt; is the right one (and I think it is), the authors don't claim to identify the supernova progenitor with any supernova remnant, location in the sky, or astronomical object. They simply attempt to support the supernova hypothesis with more modeling, something which itself isn't new in the context of this theory. It's certainly nowhere near compelling enough to name the hypothetical object, nor, I think, is the overall evidence for the theory.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2618" LastEditDate="2018-07-17T14:07:36.790" LastActivityDate="2018-07-17T14:07:36.790" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26978" PostTypeId="2" ParentId="26965" CreationDate="2018-07-17T17:50:54.807" Score="3" Body="&lt;p&gt;Neither the amount of time it took for light to reach us emitted long ago from a distant galaxy nor the distance to that galaxy are directly observable. What is observable is the redshift, how much light has been shifted toward longer wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of time it took for that redshifted light to reach us can be calculated from the redshift $z$ by assuming various parameters regarding the universe (regular matter vs dark matter, the Hubble constant, whether the universe is open, flat, or closed, etc.), and by assuming that that galaxy was and our galaxy is more or less moving with the local Hubble flow. A distance value can be calculated simply by multiplying the calculated time by the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though the second calculation is much simpler than the first, it is this second calculation that is rather misleading. There's an underlying assumption in this calculation that the universe is not expanding. The universe is expanding, so it's not quite valid to multiple very long times such as 13 billion years by the speed of light to get 13 billion light years. In fact, that distant galaxy is about 29 billion light years from us now and was only about 3.5 billion light years away when the light was emitted.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2018-07-17T17:50:54.807" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26979" PostTypeId="1" AcceptedAnswerId="26984" CreationDate="2018-07-17T22:12:42.250" Score="0" ViewCount="745" Body="&lt;p&gt;The Universe is expanding faster than the speed of light then does that mean matter at the edge of the universe is also traveling at the speed of light or are ether opposite sides of the universe are expanding over half the speed of light equaling more than the speed of light ? Looking from outside of the universe to the nearest edge would light proceed slower or faster combined with the speed of matter traveling on the edge of the universe?  &lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-07-17T22:18:33.090" LastActivityDate="2018-07-18T13:07:25.997" Title="Does the edge of the Universe travel faster than the speed of light?" Tags="&lt;universe&gt;&lt;observational-astronomy&gt;&lt;light&gt;&lt;expansion&gt;&lt;speed&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26980" PostTypeId="1" AcceptedAnswerId="27096" CreationDate="2018-07-17T22:23:47.567" Score="1" ViewCount="783" Body="&lt;p&gt;Are there stars, galaxies etc that we cannot see because they are traveling too fast and their spectrum is shifted below our visible range? From what I understand, red shift is caused by stars and such moving away from the viewer. At what speed does a star etc have to travel to be invisible to us?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if you are in one galaxy on one side of the universe that is expanding looking at a galaxy on the other expanding edge side. They would be traveling faster than the speed of light away from each other. Would they be invisible to each other?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-07-26T20:35:11.833" LastActivityDate="2018-07-27T11:23:41.540" Title="At what speed does something have to travel away from us for it to red shift enough that it becomes invisible to the human eye?" Tags="&lt;star&gt;&lt;redshift&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26981" PostTypeId="2" ParentId="26979" CreationDate="2018-07-17T22:40:15.853" Score="2" Body="&lt;p&gt;Nothing travels through space &quot;faster than the speed of light&quot;. Distant galaxies can appear to be receding &quot;faster than the speed of light&quot; because the space between they and us is expanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no physics that prevents a distant object &lt;em&gt;appearing&lt;/em&gt; to travel faster than the speed of light. The speed of light limit and constancy applies to &lt;em&gt;local&lt;/em&gt; measurements. The speed of light is not a constant when it is measured &quot;elsewhere&quot; (by this, I mean somewhere different to where you are).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot &quot;look from outside the universe&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-07-18T13:07:25.997" LastActivityDate="2018-07-18T13:07:25.997" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26982" PostTypeId="1" CreationDate="2018-07-17T22:53:05.587" Score="-7" ViewCount="178" Body="&lt;p&gt;Some questions have been posed about black holes becoming stars again once they attract enough hydrogen/atomic matter from another star. Some answers were a forceful NO. Never. However - let's ponder this again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: Once material is absorbed, is there a physical reason why fusion cannot take place once more? Does the process of entering the black hole strip out all the nuclear energy contained in mass/matter? Or is it that matter that makes it past the event horizon has potential to fuse but because of that time dilation due to the already present dense mass make this process take billions of years to resolve? &lt;/p&gt;&#xA;" OwnerUserId="23710" LastActivityDate="2018-07-17T23:31:49.897" Title="Black-hole becoming a star again - follow-up question" Tags="&lt;star&gt;&lt;black-hole&gt;&lt;astrophysics&gt;&lt;star-formation&gt;&lt;time-dilation&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2018-07-19T17:24:12.727" ContentLicense="CC BY-SA 4.0" />
  <row Id="26983" PostTypeId="2" ParentId="26982" CreationDate="2018-07-17T23:31:49.897" Score="2" Body="&lt;p&gt;Anything that's inside the event horizon has vanished forever and ever from this universe. There is no way back, period. Once inside, any direction you're turning, you're looking at the center of the BH, there is no direction pointing &quot;out&quot; anymore - that's the real reason why you can't get back out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physics inside the event horizon are radically different from the physics in normal space. It's pointless to speculate about fusion, etc. Regardless, what happens inside the event horizon, stays inside the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything falling towards the event horizon will take an infinite amount of time, from your perspective (the distant observer) to even reach the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just accept it. Once a black hole, always a black hole.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2018-07-17T23:31:49.897" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26984" PostTypeId="2" ParentId="26979" CreationDate="2018-07-17T23:44:22.420" Score="5" Body="&lt;p&gt;The universe has no edge, as far as we can tell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the &lt;em&gt;observable&lt;/em&gt; universe does have an &quot;edge&quot; - it's the most distant places we can see. But that's just us and our limited instruments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the expansion of the universe, it is true that two objects, far enough from each other, could actually move faster than light relative to each other. That means light sent from one of them would never reach the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the speed of light, no matter how you measure it, is always the same. It does not matter how fast or slow its source was moving - when you measure the speed of that light, it's always, ALWAYS the same, for you, locally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how can that be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, it's because the universe is expanding because space itself is growing. There's more space all the time in between distant objects. And the further apart two objects are, the faster the space between them is growing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's like a canvas with galaxies painted on it, but the canvas itself is growing. However, when light is traveling across the canvas, its speed is always the same relative to whatever galaxies it may find along the way.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2018-07-18T00:44:52.800" LastActivityDate="2018-07-18T00:44:52.800" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26987" PostTypeId="1" AcceptedAnswerId="27021" CreationDate="2018-07-18T09:42:22.370" Score="6" ViewCount="203" Body="&lt;p&gt;Do we know at which angle the Event Horizon Telescope will look at the accretion disk of Sagittarius A*? Additionally, would it matter if we are exactly in the plane with the accretion disk, i.e. could we still observe the event horizon?&lt;/p&gt;&#xA;" OwnerUserId="23717" LastEditorUserId="2153" LastEditDate="2018-07-18T13:17:15.890" LastActivityDate="2018-07-20T15:19:15.110" Title="Do we know at which angle the Event Horizon Telescope will look at the accretion disk of Sagittarius A*?" Tags="&lt;telescope&gt;&lt;supermassive-black-hole&gt;&lt;sagittarius-a&gt;&lt;galactic-center&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26988" PostTypeId="2" ParentId="26987" CreationDate="2018-07-18T13:10:31.700" Score="7" Body="&lt;p&gt;We don't know the answer to this (or if someone does, I'd really like to hear that answer). Our galaxy's Super-Massive Black Hole (SMBH) is an unusually quiet one, with little to no accretion disk. If one indeed exists, and it almost assuredly exists, even if it is tenuous and small, we have not yet seen it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Around 2013, there was a &lt;a href=&quot;https://en.wikipedia.org/wiki/Sagittarius_A*#Discovery_of_G2_gas_cloud_on_an_accretion_course&quot; rel=&quot;noreferrer&quot;&gt;gas cloud designated G2&lt;/a&gt; that was expected to be accreted which would have given us a lot of insight into the black hole and its structure. Unfortunately there was little to no activity observed. &lt;a href=&quot;https://www.nature.com/articles/nature10767&quot; rel=&quot;noreferrer&quot;&gt;A Nature article&lt;/a&gt; about this gas cloud goes into more detail and talks about the accretion disk, as quoted below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The cautiously predicted flare-up would be a valuable probe of the environment immediately surrounding the black hole, about which little is known. What we do know is that the average rate of mass accretion driving the emission from Sgr A* is exceptionally small, thus explaining why our Galactic black hole is extremely dim. That we can detect it at all is due largely to its proximity — it is about 100 times closer to Earth than the nearest supermassive black hole in another galactic nucleus, that of the Andromeda galaxy. We also know that the accreting gas is heated so much that the individual gas particles almost all approach the speed of light, and that the observed emission emanates from a magnetohydrodynamic maelstrom. But the basic parameters of this small-accretion-rate regime are still being sought: does the accreting gas form a well-defined disk or an outflowing jet, as it often does in the more prominent cases observed in many other galaxies? Or is it better described as a more isotropic inflow comprising stochastically distributed density fluctuations?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2018-07-18T13:17:17.960" LastActivityDate="2018-07-18T13:17:17.960" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26989" PostTypeId="1" AcceptedAnswerId="26993" CreationDate="2018-07-18T16:44:55.573" Score="6" ViewCount="281" Body="&lt;p&gt;This image of Neptune taken with the VLT is really impressive. The resolution is achieved by recent improvements in the adaptive optics.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gizmodo: &lt;a href=&quot;https://gizmodo.com/new-super-crisp-images-of-neptune-show-how-far-our-tele-1827683475&quot; rel=&quot;noreferrer&quot;&gt;New Super-Crisp Images of Neptune Show How Far Our Telescopes Have Come&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Space.com &lt;a href=&quot;https://www.space.com/41202-new-adaptive-optics-produces-stunning-views.html&quot; rel=&quot;noreferrer&quot;&gt;Telescope Upgrade Produces Stunningly Clear Views of Space&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But I don't really understand, is this coming from just one 8 meter mirror, or is this using more than one of the four that make up the VLT? Also, is this really using adaptive optics &lt;em&gt;that work in the visible, including blue wavelengths?&lt;/em&gt; Or is this a false-color image? See the excellent answer(s) to &lt;a href=&quot;https://astronomy.stackexchange.com/q/14795/7982&quot;&gt;Why (actually) aren't ground-based observatories using adaptive optics for visible wavelengths?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/me10Y.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/me10Y.jpg&quot; alt=&quot;Neptune VLT new adaptive optics&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-07-18T19:41:51.663" Title="How did VLT's adaptive optics obtain this resolution for Neptune? Is it really working in visible wavelengths?" Tags="&lt;angular-resolution&gt;&lt;neptune&gt;&lt;adaptive-optics&gt;&lt;very-large-telescope&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26990" PostTypeId="1" CreationDate="2018-07-18T17:15:09.470" Score="2" ViewCount="261" Body="&lt;p&gt;Does the size of the atom place a theoretical limit on a telescope's focal length (and thus, resolution)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that that the allowed margin of error for parabolic mirrors drops with an increasing focal length. It's already like 25 nm for advanced Earth-based telescopes - which is just 50-250 atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we try building mirrors with an ever increasing focal length (e. g. as a part of a huge space-based interferometer), will we eventually reach a point where the precision required will be less than the size of the atom and thus no longer achievable? If yes, how far away are we from this limit now?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we had unlimited resources, would it be possible to construct a space-based telescope for imagining extrasolar planets in great detail, or would this be not even theoretically possible due to the fixed atom size?&lt;/p&gt;&#xA;" OwnerUserId="23720" LastActivityDate="2021-04-05T23:56:01.047" Title="Does the size of the atom limit the focal length of telescopes?" Tags="&lt;telescope&gt;&lt;mirror&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26991" PostTypeId="1" AcceptedAnswerId="26992" CreationDate="2018-07-18T18:36:24.010" Score="2" ViewCount="55" Body="&lt;p&gt;Type Ia supernovae (SNIa) are thermonuclear supernovae that generates a lot of Nickel 56 which emits photons while disintegrating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is the secondary maximum in lightcurves for reddish bands (see I, J, H, K bands in the figure below) explained ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fDnXT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fDnXT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23047" LastEditorUserId="23047" LastEditDate="2018-07-18T18:44:03.570" LastActivityDate="2018-07-18T19:14:31.007" Title="Secondary maximum in SNIa lightcurves for near infrared" Tags="&lt;astrophysics&gt;&lt;supernova&gt;&lt;stellar-evolution&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26992" PostTypeId="2" ParentId="26991" CreationDate="2018-07-18T19:08:48.473" Score="2" Body="&lt;p&gt;The main power driving a supernova (of type Ia) light curve is the explosive detonation of a carbon/oxygen white dwarf. This produces a large amount of iron-peak elements (particularly Ni) via nuclear fusion of carbon, oxygen and alpha particles via magnesium, neon, sulphur and silicon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The decay of the light curve is then largely governed by the radioactive decay of $^{56}$Ni via $^{56}$Co to $^{56}$Fe, which releases gamma rays that &quot;reheat&quot; the ejecta. The half-lives of these decays are about 6 and 77 days respectively, and this is the basic reason for the decay of the light curve happening on timescales of tens of days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as you have noticed, in the (rest-frame) infrared part of the spectrum, there is a secondary peak. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, whilst the emission in the visible part of the spectrum is largely of a thermal continuum nature (the ejecta are typically at temperatures of $\sim$10,000K during the early parts of the decay), the infrared portion can be dominated (and certainly is after $&amp;gt;20$ days) by line emission (that is then smeared out by turbulent doppler broadening). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The species that emit this radiation are ionised iron-peak elements (mainly cobalt and iron). As the ejecta cools, the species that are present recapture electrons, becoming less ionised. At around 7000K there is a fairly abrupt transition between the gas consisting of mainly doubly ionised species to mainly singly ionised species. This results in an order of magnitude increase in emissivity of the gas due to radiative recombination of the ions and electrons. It turns out that this radiation emerges mainly at near infrared wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the gas cools further and becomes wholly singly ionised, or even neutral, then the emissivity falls drastically again. There is therefore a &quot;sweet spot&quot; where the infrared emissivity of the ejecta is maximised, that occurs about 20-30 days after the explosion, that manifests itself as a secondary bump in the infrared light curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much more detail can be found in &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/506588/pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kasen (2006)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-07-18T19:14:31.007" LastActivityDate="2018-07-18T19:14:31.007" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26993" PostTypeId="2" ParentId="26989" CreationDate="2018-07-18T19:34:10.340" Score="4" Body="&lt;p&gt;It is an image taken with the new narrow field mode of the MUSE instrument using the GALACSI Adaptive optics module on a single (UT4) VLT telescope using laser guide stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having a great deal of difficulty (e.g. from &lt;a href=&quot;http://www.eso.org/public/news/eso1824/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this press release&lt;/a&gt;) in working out at what wavelength(s) this image was taken. I do not believe that the AO system is working at blue wavelengths and therefore the comparison with HST (which certainly does) is somewhat misleading. I think (&lt;a href=&quot;https://www.eso.org/sci/facilities/paranal/instruments/muse/inst.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;from ESO's own instrumentation pages&lt;/a&gt;) that the MUSE narrow field mode has zero efficiency at blue wavelengths and the description of &lt;a href=&quot;https://www.eso.org/sci/facilities/develop/ao/sys/galacsi.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;GALACSI&lt;/a&gt; only talks about AO correction approaching diffraction limited images at 650 nm (i.e. red wavelengths).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The captioning to the press release images found &lt;a href=&quot;http://www.eso.org/public/images/eso1824a/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; suggest that the shortest wavelengths used in this &lt;strong&gt;false-colour image&lt;/strong&gt; were 550nm and that most of the detail you see is coming from redder wavelengths (600-920nm).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-07-18T19:41:51.663" LastActivityDate="2018-07-18T19:41:51.663" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26994" PostTypeId="2" ParentId="26990" CreationDate="2018-07-18T23:47:17.897" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the size of the atom limit the focal length of telescopes?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Does the size of the atom place a theoretical limit on a telescope's focal length (and thus, resolution)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, the size of an atom does not limit the focal length of a telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Halfway through this answer I further explain why the size of an atom isn't relevant and that instead we can use the size of an electron. Hydrogen is the smallest atom, it has one &lt;a href=&quot;https://en.wikipedia.org/wiki/Electron&quot; rel=&quot;nofollow noreferrer&quot;&gt;electron&lt;/a&gt;, one &lt;a href=&quot;https://en.wikipedia.org/wiki/Proton&quot; rel=&quot;nofollow noreferrer&quot;&gt;proton&lt;/a&gt; and no &lt;a href=&quot;https://en.wikipedia.org/wiki/Neutron&quot; rel=&quot;nofollow noreferrer&quot;&gt;neutrons&lt;/a&gt;. An electron is 1/1836th the size of a proton; thus it is much smaller than an atom of $^1$H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Proton-to-electron_mass_ratio&quot; rel=&quot;nofollow noreferrer&quot;&gt;proton-to-electron mass ratio&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\mu = m_p/m_e = 1836.15267389(17).$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The number enclosed in parentheses is the measurement uncertainty on the last two digits. The value of $\mu$ is known to about 0.1 parts per billion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For an explanation of what actually does limit the focal length see: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/498/what-limits-the-usable-focal-length-of-telescopes-currently&quot;&gt;What limits the usable focal length of telescopes currently?&lt;/a&gt;, to see why &lt;strong&gt;an atom does not limit the resolution of a telescope&lt;/strong&gt; skip halfway through this answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/498/what-limits-the-usable-focal-length-of-telescopes-currently&quot;&gt;Arne wrote&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Visual resolution of a telescope is directly proportional to the aperture of the telescope. The focal length, and hence the magnification that can be achieved, is then just following on the visual resolution.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The telescopes today are usually so well build that they are diffraction limited, which means optical resolution due to diffraction is the limiting factor. If you want to have &quot;higher magnification&quot; in a telescope, you always want to have a larger aperture. The longer focal length may help, but is not quite necessary.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;And, as Jeremy said, the limiting resource in this is money. There are some engineering problems with building extremely large telescopes, but most of these can be solved, given enough money, time and resources.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is &lt;a href=&quot;https://en.wikipedia.org/wiki/Diffraction-limited_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;diffraction&lt;/a&gt; that limits resolution of a telescope, not the size of an atom. That part of your question is essentially what differentiates your question from the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Wikipedia's page &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Diffraction-limited_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;diffraction-limited system&lt;/a&gt;&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The resolution of a given instrument is proportional to the wavelength of the light being observed, and inversely proportional to the size of its objective. For telescopes with circular apertures, the size of the smallest feature in an image that is diffraction limited is the size of the Airy disk. As one decreases the size of the aperture in a lens, diffraction increases. At small apertures, such as f/22, most modern lenses are limited only by diffraction.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In astronomy, a diffraction-limited observation is one that is limited only by the optical power of the instrument used. However, most observations from Earth are seeing-limited due to atmospheric effects. Optical telescopes on the Earth work at a much lower resolution than the diffraction limit because of the distortion introduced by the passage of light through several kilometres of turbulent atmosphere. Some advanced observatories have recently started using adaptive optics technology, resulting in greater image resolution for faint targets, but it is still difficult to reach the diffraction limit using adaptive optics&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gaRtb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gaRtb.png&quot; alt=&quot;xxxxxxxx&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Log-log plot of aperture diameter vs angular resolution at the diffraction limit for various light wavelengths compared with various astronomical instruments. For example, the blue star shows that the Hubble Space Telescope is almost diffraction-limited in the visible spectrum at 0.1 arcsecs, whereas the red circle shows that the human eye should have a resolving power of 20 arcsecs in theory, though normally only 60 arcsecs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why is the size of an atom not a consideration?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Wikipedia's page: &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Calorimetric_Electron_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Calorimetric Electron Telescope&lt;/a&gt;&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The CALorimetric Electron Telescope (CALET) is a space telescope being mainly used to perform &lt;strong&gt;high precision observations of electrons&lt;/strong&gt; and gamma rays. It &lt;strong&gt;tracks the trajectory of electrons&lt;/strong&gt;, protons, nuclei, and gamma rays and &lt;strong&gt;measures their direction, charge and energy&lt;/strong&gt;, which may help understand the nature of dark matter or nearby sources of high-energy particle acceleration.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For more information see: &quot;&lt;a href=&quot;http://iopscience.iop.org/article/10.1088/1742-6596/632/1/012023/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;The CALorimetric Electron Telescope (CALET) for high-energy astroparticle physics on the International Space Station&lt;/a&gt;&quot; (O Adriani et al 2015 J. Phys.: Conf. Ser. 632 012023):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7GXXA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7GXXA.jpg&quot; alt=&quot;Figure 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Figure 1. Drawing of the CALET payload. On the ISS the payload will be installed in such a way to have open sky on top, Earth on bottom.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hKdbb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hKdbb.png&quot; alt=&quot;Figure 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Figure 2. Drawing of the CALET-CAL instrument, with a typical shower of secondary particles generated by a 1 TeV electron incident from top.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Page 2: &quot;&lt;strong&gt;2.1. CALET-CAL instrument&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The CALET calorimeter (see figure 2) is composed of a charge detector (CHD), a pre-shower imaging calorimeter (IMC) and a total absorption calorimeter (TASC). It is optimized for particle identification and energy measurement of several cosmic-ray species:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;electrons and positrons in the 1 GeV - 20 TeV energy range (with no charge sign discrimination);&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;photons from few GeV to ∼ 10 TeV;&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;nuclei up to the Fe region, with energy from tens of GeV to ∼ 1 PeV;&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;ultra-heavy (Z &gt; 28) nuclei with E &gt; 600 MeV/nucleon (in this case, with no energy measurement).&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;The CHD, positioned on top of the IMC structure, is composed of two layers of plastic scintillator with mutually orthogonal segmentation in 14 bars (SciBars), each of dimensions 3.2 × 1.0 × 44.8 cm$^3$ and read by a photomultiplier tube (PMT, Hamamatsu Photonics R7400U-06) and front-end circuit (FEC) with charge sensitive amplifier, for a total of 28 channels. The CHD determines the charge absolute value |Z| of the incoming charged particle, through the Z$^2$ dependence of the specific ionization loss. The very low uncertainty in the Z measurement (0.1 for light nuclei up to B, 0.3 in the Fe region) allows for resolving individual chemical elements with Z from 1 to 40.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The IMC is a finely segmented sampling calorimeter, with surface area of 45 × 45 cm$^2$ and total thickness of 3 radiation lengths X$_0$; internally, 8 double layers of scintillating fibres (SciFi, 1 mm$^2$ cross-section) are interleaved with a sequence of 7 tungsten plates: 5 of thickness 0.2 X$_0$ and 2 of thickness 1.0 X$_0$. The fibres of each double layer are mutually orthogonal and arranged in belts, each read by a 64-channel multi-anode PMT (MAPMT, Hamamatsu Photonics R7600-M64) and VA front-end ASIC circuit, for a total of 7168 channels. The IMC fine granularity allows for &lt;strong&gt;precise determination of the incoming particle trajectory&lt;/strong&gt;, localization of the starting point of the secondary shower possibly generated, discrimination of the primary incident particle against possible backscattering from the shower developing in the pre-shower&#xA;  imaging calorimeter (IMC) and underlying total absorption calorimeter (TASC).&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus we can see a tiny particle, if it has enough energy, from as far as it can travel.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastActivityDate="2018-07-18T23:47:17.897" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26995" PostTypeId="1" AcceptedAnswerId="26996" CreationDate="2018-07-19T01:23:39.980" Score="8" ViewCount="478" Body="&lt;p&gt;The question &lt;a href=&quot;https://astronomy.stackexchange.com/q/26987/7982&quot;&gt;Do we know at which angle the Event Horizon Telescope will look at the accretion disk of Sagittarius A*?&lt;/a&gt; about just how close to edge-on we would be viewing any accretion disk around the black hole in the center of our galaxy, and @Zephyr's &lt;a href=&quot;https://astronomy.stackexchange.com/a/26988/7982&quot;&gt;excellent answer&lt;/a&gt; have got be thinking about the geometry and angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the angular momentum vector of a black hole will likely reflect that of all the matter that fell into it , would the plane of any accretion disk around a black hole reflect the black hole's &lt;em&gt;current equatorial plane&lt;/em&gt;, or a plane defined by rotation of whatever material is currently falling in to it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to formulate this question might be to propose an incredibly unlikely &lt;a href=&quot;https://en.wikipedia.org/wiki/Thought_experiment&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gedankenexperiment&lt;/a&gt;. If I inserted a black hole into the center of a rotating disk of material so that the black hole's axis &lt;em&gt;was in the plane of the disk of material&lt;/em&gt; rather than parallel to its rotational axis, would a well-defined accretion disk form all all, and if so, what would determine the orientation of that plane?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="1437" LastEditDate="2019-04-11T22:23:22.840" LastActivityDate="2019-04-11T22:23:44.203" Title="What defines the plane of an accretion disk around a black hole?" Tags="&lt;black-hole&gt;&lt;supermassive-black-hole&gt;&lt;accretion-discs&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26996" PostTypeId="2" ParentId="26995" CreationDate="2018-07-19T06:55:52.623" Score="6" Body="&lt;p&gt;Let me give an answer based on my intuitive understanding of the thing...&lt;/p&gt;&#xA;&#xA;&lt;h3&gt; Initial conditions&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The initial accretion disk movement is defined by the objects falling into it, the black hole's mass and its spin of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take the situation where a single star falls on a quiet black-hole, the accretion will appear along the orbit of the infalling object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/opCmx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/opCmx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;sup&gt;Credit: NASA, S. Gezari (The Johns Hopkins University, Baltimore, Md.), A. Rest (Space Telescope Science Institute, Baltimore, Md.), and R. Chornock (Harvard-Smithsonian Center for Astrophysics, Cambridge, Ma.)&lt;/sup&gt; and live on &lt;a href=&quot;https://www.youtube.com/watch?v=CpaVzOKihBQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;youtube&lt;/a&gt;...&lt;/p&gt;&#xA;&#xA;&lt;h3&gt; Equilibrium in an active galactic nucleus&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Now on the long run, if additional matter is constantly coming in, at first thought I would think &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the accretion disk momentum will be exchanged with the one of the black hole through the &lt;a href=&quot;https://en.wikipedia.org/wiki/Penrose_process&quot; rel=&quot;nofollow noreferrer&quot;&gt;Penrose process&lt;/a&gt;, &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Infalling matter momentum will reflect the overall host galaxy momentum - especially if the galaxy is a spiral &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11264" LastEditorUserId="1437" LastEditDate="2019-04-11T22:23:44.203" LastActivityDate="2019-04-11T22:23:44.203" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26997" PostTypeId="1" CreationDate="2018-07-19T10:04:58.427" Score="-1" ViewCount="54" Body="&lt;p&gt;We know that on a longer time scale (millions of years) the orbits of planes are unstable and it is impossible to make predictions about eclipses millions of years into the future. But on a shorter scale, the behaviour of orbits can be considered to be fairly stable. I was studding the distribution of the number of solar eclipses in short time scale (thousands of years as opposed to millions of years)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://eclipse.gsfc.nasa.gov/SEcat5/SEcatalog.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA has published the list of the count of solar eclipses&lt;/a&gt; (total + annular  + partial) every century from 1999 BC to 3000 AD. When we plot count of eclipses against the century (-19 for 1999 BC to 1900 BC, +30 for 2900 AD to 3000 AD) we get a curve which roughly looks like a sine wave. I fitted a sine curve on this data and obtained a good fit with $R^2 = 0.9$. Since a sine function is period function with a period of $2\pi$ it means that the number of solar eclipses in a century should roughly repeat every $2\pi$ century or 6th century (6.28th century to be precise). Example: For the 6 century cycle starting from the first century AD we have (1AD, 248), (7AD, 251), (13AD, 246), (19AD, 242) etc. Generalising this, for any time period $T$ instead a century, we have the hypothesis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hypothesis&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The number of solar eclipses in a time interval of length $T$ has a quasi-period of $2\pi T$.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Is it possible to provide a theoretical justification to this empirically observed trend? I am more interested in why should the quasi-period be $2\pi T$ rather than just why there should be a quasi-period.&lt;/p&gt;&#xA;" OwnerUserId="23712" LastActivityDate="2018-07-19T10:32:43.397" Title="The total number of solar eclipses in a century repeats roughly every 6th century" Tags="&lt;the-sun&gt;&lt;orbit&gt;&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26998" PostTypeId="2" ParentId="26997" CreationDate="2018-07-19T10:32:43.397" Score="2" Body="&lt;p&gt;From the same &lt;a href=&quot;https://eclipse.gsfc.nasa.gov/SEcat5/SEcatalog.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA page&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The recurrence of solar eclipses is governed by the &lt;a href=&quot;https://eclipse.gsfc.nasa.gov/SEsaros/SEsaros.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Saros&lt;/a&gt; cycle. For more information, see &lt;a href=&quot;https://eclipse.gsfc.nasa.gov/SEsaros/SEperiodicity.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Periodicity of Solar Eclipses&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which then goes on to say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The periodicity and recurrence of eclipses is governed by the Saros cycle, a period of approximately 6,585.3 days (18 years 11 days 8 hours). It was known to the Chaldeans as a period when lunar eclipses seem to repeat themselves, but the cycle is applicable to solar eclipses as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2018-07-19T10:32:43.397" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26999" PostTypeId="1" AcceptedAnswerId="27000" CreationDate="2018-07-19T12:15:48.280" Score="19" ViewCount="3227" Body="&lt;p&gt;I saw multiple news sites reporting that a team discovered 12 new Jupiter's moons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.cnet.com/videos/twelve-new-jupiter-moons-found-including-one-reckless-one/&quot; rel=&quot;nofollow noreferrer&quot;&gt;c|net - Twelve new Jupiter moons found, including one reckless one&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://blogs.discovermagazine.com/d-brief/2018/07/17/new-jupiter-moons/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Discover - Jupiter’s Got Twelve New Moons — One is a Bit of a Problem Child&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;While some other news sites claim that they found only 10 new moons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.nature.com/articles/d41586-018-05725-6&quot; rel=&quot;nofollow noreferrer&quot;&gt;nature - Jupiter has 10 more moons we didn't know about — and they're weird&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://earthsky.org/space/10-new-moons-discovered-jupiter-1-oddball&quot; rel=&quot;nofollow noreferrer&quot;&gt;EarthSky - Astronomers discover 10 new moons for Jupiter&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Why does the number of moons differ in the different articles? Did the team now found 10 or 12 moons?&lt;/p&gt;&#xA;" OwnerUserId="23730" OwnerDisplayName="Yannjoel" LastEditorUserId="23730" LastEditDate="2018-07-23T12:46:53.420" LastActivityDate="2018-07-23T12:46:53.420" Title="Did we discover 10 or 12 new moons of Jupiter?" Tags="&lt;jupiter&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="27000" PostTypeId="2" ParentId="26999" CreationDate="2018-07-19T13:13:45.057" Score="20" Body="&lt;p&gt;Per the &lt;a href=&quot;https://carnegiescience.edu/news/dozen-new-moons-jupiter-discovered-including-one-%E2%80%9Coddball%E2%80%9D&quot; rel=&quot;noreferrer&quot;&gt;Carnegie Science article&lt;/a&gt; that Magic Octopus Urn linked from NASA in the comments, a Carnegie Science team led by Scott S. Sheppard noticed something new in spring of 2017 (though some observations occurred as early as 2016). It took a year to confirm the discovery of the new moons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ten of the moons orbit in the outer swarm of moons, which is one of the ways to divide the groups of moons. Nine of these follow the pattern of the other moons which orbit Jupiter in retrograde (the opposite direction of the planet's rotation). One they call &quot;oddball&quot; because it orbits prograde (the same direction as the planet's rotation) and has a more inclined orbit than the inner prograde moons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two of the moons orbit in the inner group, which is what brings the total number of discoveries up to twelve, as referenced in the first set of articles in your post. Like the other inner moons, these orbit prograde.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the retrograde moons and the &quot;oddball&quot; were first noticed in 2016. One of the inner group and the rest of the retrograde moons were discovered in 2017. The second inner group moon was discovered in 2018, then most were announced together on July 17, 2018. (Dates collected from &lt;a href=&quot;https://en.wikipedia.org/wiki/Moons_of_Jupiter#List&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The retrograde moon discovered in 2016 and one of the retrograde moons discovered in 2017 were announced in 2017, this means only ten moons were announced on July 17, 2018, which is why the Nature article refers to ten new moons. The EarthSky article talks some about the ten and the twelve, and it does clarify both groups (divided by inner/outer and year of announcement).&lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorUserId="6" LastEditDate="2018-07-19T14:30:21.577" LastActivityDate="2018-07-19T14:30:21.577" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27001" PostTypeId="2" ParentId="22903" CreationDate="2018-07-19T13:21:43.487" Score="2" Body="&lt;p&gt;It is quite beautiful to imagine such a condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First thing is that we can't observe them with eyes, as no reflection of light will occur from black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the observer is in rest relative to merging black holes: now when the black holes are merging, the surrounding space-time's gravitational waves will start increasing, and their amplitude will also increasing. It affects the observer, because the waves start pushing away the observer from the black holes. But it doesn't effect the images of the observation, because light will also bend in the gravitational waves. Also time dilation will occur due to the gravitational waves, but what actually happening when they were merging, I can't imagine.&lt;/p&gt;&#xA;" OwnerDisplayName="user23728" LastEditorUserId="1891" LastEditDate="2018-07-19T19:20:30.723" LastActivityDate="2018-07-19T19:20:30.723" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27002" PostTypeId="1" CreationDate="2018-07-19T12:36:40.210" Score="2" ViewCount="378" Body="&lt;p&gt;I just &lt;a href=&quot;https://space.stackexchange.com/questions/29544/new-jupiter-moons&quot;&gt;saw this question&lt;/a&gt; and that got me to &lt;a href=&quot;https://solarsystem.nasa.gov/moons/jupiter-moons/overview/?page=0&amp;amp;per_page=40&amp;amp;order=name+asc&amp;amp;search=&amp;amp;condition_1=9%3Aparent_id&amp;amp;condition_2=moon%3Abody_type%3Ailike&quot; rel=&quot;nofollow noreferrer&quot;&gt;this NASA page&lt;/a&gt; which led me &lt;a href=&quot;https://carnegiescience.edu/news/dozen-new-moons-jupiter-discovered-including-one-%E2%80%9Coddball%E2%80%9D&quot; rel=&quot;nofollow noreferrer&quot;&gt;to this scientific article&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wMgFe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wMgFe.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;“Our other discovery is a real oddball and has an orbit like no other known Jovian moon,” Sheppard explained. “It’s also likely Jupiter’s smallest known moon, being less than one kilometer in diameter”.&lt;/p&gt;&#xA;&lt;p&gt;This new “oddball” moon is more distant and more inclined than the prograde group of moons and takes about one and a half years to orbit Jupiter.  So, unlike the closer-in prograde group of moons, this new oddball prograde moon has an orbit that crosses the outer retrograde moons.&lt;/p&gt;&#xA;&lt;p&gt;As a result, head-on collisions are much more likely to occur between the “oddball” prograde and the retrograde moons, which are moving in opposite directions.&lt;/p&gt;&#xA;&lt;p&gt;“This is an unstable situation,” said Sheppard. “Head-on collisions would quickly break apart and grind the objects down to dust.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;My question is, why are most of the moons that are further away from Jupiter retrograde orbits, while the ones extremely close to Jupiter are prograde orbits? What possible situations were required for Valetudo to end up where it is opposed to where the rest of the pro-grade moons are clustered? Pure chance / a series of unlikely events?&lt;/p&gt;&#xA;" OwnerUserId="23729" OwnerDisplayName="Magic Octopus Urn" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-07-20T18:47:13.300" Title="How did &quot;oddball&quot; Valetudo, Jupiter's new prograde moon, end up in a wider orbit?" Tags="&lt;orbit&gt;&lt;jupiter&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27004" PostTypeId="1" AcceptedAnswerId="27005" CreationDate="2018-07-19T15:48:24.430" Score="2" ViewCount="162" Body="&lt;p&gt;I know that viewing Mercury is a challenge, but I'm fairly close to the equator and Mercury was at greatest elongation just a week ago.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://in-the-sky.org/data/planets.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;In-the-sky.org&lt;/a&gt; says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;All around the world, Mercury is visible as an evening object, having recently passed greatest elongation east. &lt;strong&gt;From Taipei, however, it is not observable&lt;/strong&gt; – it will reach its highest point in the sky during daytime and is no higher than 10° above the horizon at dusk.&lt;/p&gt;&#xA;&lt;p&gt;12 Jul 2018, 12:00 CST   –  Mercury at greatest elongation east&#xA;27 Aug 2018, 05:48 CST   –  Mercury at greatest elongation west&#xA;06 Nov 2018, 22:59 CST   –  Mercury at greatest elongation east&#xA;15 Dec 2018, 23:17 CST   –  Mercury at greatest elongation west&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Of course, since Mercury stays near the Sun it will always reach its highest point in the sky during daytime, &lt;em&gt;probably around lunchtime!&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;But I don't understand why it would be viewable all around the world (presumably just after sunset since it is currently near maximum elongation East), but not from Taipei.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Why does in-the-sky.org say that Mercury is not visible from Taipei these days?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GYXXG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GYXXG.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-07-19T16:13:48.803" Title="Why does in-the-sky.org say that Mercury is not visible from Taipei these days?" Tags="&lt;planet&gt;&lt;night-sky&gt;&lt;mercury&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27005" PostTypeId="2" ParentId="27004" CreationDate="2018-07-19T16:13:48.803" Score="6" Body="&lt;p&gt;The logic that In-The-Sky.org uses to decide whether events are observable or not doesn't currently work very well for Mercury and Venus. I have some plans to fix this in the long term, but always have far too many projects on the go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you correctly say, it's blazingly obvious that Mercury is highest in the sky at lunchtime, and the code I wrote to generate these descriptions should realise that (but doesn't, currently).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement that Mercury is going to be very challenging to observe from Taipei is basically correct. At this time of year, Mercury doesn't venture very far from the Sun, even at greatest elongation. The best time to see it in the evening sky is around late September. This is true regardless of where you live in the world, so the statement &quot;all around the world...&quot; in the web page is basically wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I say, I have plans to improve all of this, but the website is a one-man effort, done entirely in my spare time, so things don't always get fixed quickly :-)&lt;/p&gt;&#xA;" OwnerUserId="23733" LastActivityDate="2018-07-19T16:13:48.803" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27007" PostTypeId="2" ParentId="22903" CreationDate="2018-07-19T20:34:39.463" Score="3" Body="&lt;p&gt;It could do serious damage if things were just right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three solar masses of energy is about 5x10&lt;sup&gt;47&lt;/sup&gt; Joules.  1 AU is about 1.5x10&lt;sup&gt;13&lt;/sup&gt; cm. The area of the sphere is thus about 4x10&lt;sup&gt;27&lt;/sup&gt; cm&lt;sup&gt;2&lt;/sup&gt;.  That's a flux of 10&lt;sup&gt;20&lt;/sup&gt; joules cm&lt;sup&gt;-2&lt;/sup&gt;.  That's plenty of energy to do damage...if it can couple to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strain of the GW measured at Earth is around 10&lt;sup&gt;-22&lt;/sup&gt;.  Strain drops off as the inverse distance, and the ratio of a gigaparsec to an AU is about 10&lt;sup&gt;14&lt;/sup&gt;, so the strain at 1AU from the merger would be about 10&lt;sup&gt;-8&lt;/sup&gt;. By itself, this would probably not harm you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's not a simple one-time strain, it is oscillating with the orbital frequency of the merging black holes, and anything which had a physical resonance at that frequency could pick up energy, possibly significant energy, from the GW fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that this depends on orientation. The mechanical resonance needs to align with the GW, and GWs are not emitted in all directions, so you'd want (or, perhaps, not want!) to be in the plane of the merger.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument that GWs can't deposit energy is plainly wrong -- if it were true, LIGO couldn't detect them! -- but if you believed that, you were in superb company: It took Einstein decades before he became convinced that GWs actually carried energy!&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2019-09-02T01:03:04.190" LastActivityDate="2019-09-02T01:03:04.190" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27008" PostTypeId="1" CreationDate="2018-07-19T21:18:25.903" Score="2" ViewCount="3273" Body="&lt;p&gt;I was out with my dad on our porch watching the stars with our telescope. after awhile of looking and taking pictures I noticed something strange in the sky. There were two airplanes flying close to each other, but next to those two airplanes was a very bright light (see how the stars are bright? Well that one was a lot bigger and I lot brighter, and to give you an example it was moving as fast as the planes, I showed my dad and he as well saw it, so I knew I wasn’t dreaming, it wasn’t moving straight either, both me and my dad could see it was going in zigzags, not a straight line, it desapeared after awhile behind he trees, we didn’t get to look at it through the telescope since it was moving so fast...what could it be...? Both me and my dad agree it was going as fast as the planes next to it, it was brighter then both the planes and the stars and it could not have been a third plane since broth planes were just so small and it looked like a very big star. FINALY it was not moving in a straight line...so the what would it be...?  &lt;/p&gt;&#xA;" OwnerUserId="23734" LastActivityDate="2018-07-20T01:02:36.290" Title="Strange bright light in the sky flying fast" Tags="&lt;planet&gt;&lt;light&gt;&lt;space&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2018-07-20T17:56:58.647" ContentLicense="CC BY-SA 4.0" />
  <row Id="27009" PostTypeId="2" ParentId="21437" CreationDate="2018-07-19T21:51:08.667" Score="1" Body="&lt;p&gt;What the mass increase from special relativity tells you is that if you see a physical system (say, a rocket with a physics laboratory on board for convenience) flying past your location at a certain relative velocity and you measure the mass of an object in that laboratory, for instance by measuring its gravitational effect on another mass in the same laboratory (using lasers to measure distances and velocities and so on), you will get an answer which is higher than the one an observer in that laboratory would get.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In and of itself this doesn't even tell you much about what you'd measure about a far distant laboratory, especially one far enough away that the expansion of the universe significantly effects your measurements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, since these two galaxies on opposite sides of the observable universe cannot actually  exchange messages and observe one another, the mass increase is not really a consideration -- you can't measure the mass of an object in the other laboratory, because you can't get a signal there and back.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-07-19T21:51:08.667" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27011" PostTypeId="2" ParentId="26990" CreationDate="2018-07-19T23:45:15.760" Score="6" Body="&lt;p&gt;Ok, finally I can explain it. Not sure why it took so long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line, for visible light atoms don't matter. Let's say the blue side of the visible spectrum of light has a wavelength of 400 nm (round number for simplicity). Let's assume the distance between two atoms in glass is 0.2 nm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is, light is a wave. Anything smaller than about 1/4 of a wavelength becomes essentially invisible to that wave. It's not a sharp threshold, but as things get smaller and smaller, the wave &quot;sees&quot; them less and less. It just goes around the object as if it's not even there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/4 wave of visible light is 100 nm, which is 500 glass atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I make telescope mirrors. I can still see the effects of surface errors down to about 1/20 ... 1/25 of a wave. Beyond that the image is not really distinguishable from a geometrically perfect mirror, either in real-life tests under the stars, or in the lab using the equipment. Maybe an interferometer could push that limit a bit further, but not a whole lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say the absolute measurable limit is 1/50 wave, which is 8 nm, which is 40 atoms. That's plenty of atoms. Any defects smaller than that simply do not change the image created by the mirror - when talking about regular telescope mirrors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface level difference from a single layer of atoms does not even register in visible light. It takes far more thickness than that to make a visible difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your ultra-long focal length mirror would be almost flat, slowly adding extra atomic layers towards the edge. As long as it approximates the desired ideal surface (paraboloid, etc), then it will be fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see one exception. If the focal ratio is indeed so big that there isn't enough curvature in the surface to require even a single extra layer of atoms at the edge. Then, indeed, the mirror would not work, because it's flat down to the atomic level. Keep in mind though, that would be a mirror with a humongous focal ratio - and indeed with a cosmic-scale focal length. There are many issues with such a system before you even begin thinking about surface errors. Probably not feasible with current technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then what you would do is make a segmented mirror, increasing the apparent diameter and reducing the focal ratio. You could even have multiple flat segments (flat down to the atomic level) but far from each other to massively increase the total diameter, and slightly tilted according to the shape of the ideal surface. If you can afford a cosmic-scale focal length, then you can surely afford a cosmic scale diameter, and that's how you reduce the focal ratio. Maintaining that precise tilt angle between segments would be a tremendous feat of engineering.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2018-07-20T00:05:55.783" LastActivityDate="2018-07-20T00:05:55.783" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27012" PostTypeId="2" ParentId="27008" CreationDate="2018-07-20T01:02:36.290" Score="2" Body="&lt;p&gt;This sounds kind of like a &lt;a href=&quot;https://en.wikipedia.org/wiki/Foo_fighter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Foo Fighter&lt;/a&gt;. &#xA;There have been various reports of weird glowing orbs following aircraft and zig-zagging around them ever since World War two. In fact both sides assumed they were secret weapons belonging to the other side up until the end of the war.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to what they actually are exactly, well &lt;em&gt;nobody really knows&lt;/em&gt;. The most plausible answer is that these are examples of &lt;a href=&quot;https://en.wikipedia.org/wiki/Ball_lightning&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ball lightning&lt;/a&gt; or some other electrostatic phenomenon. As far as I know, it has never been conclusively replicated in the lab.&lt;/p&gt;&#xA;" OwnerUserId="17655" LastActivityDate="2018-07-20T01:02:36.290" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27013" PostTypeId="1" AcceptedAnswerId="27032" CreationDate="2018-07-20T01:40:01.753" Score="1" ViewCount="103" Body="&lt;p&gt;I came across an &lt;a href=&quot;https://www.technologyreview.com/s/536091/spacecraft-traveling-close-to-light-speed-should-be-visible-with-current-technology-say/&quot; rel=&quot;nofollow noreferrer&quot;&gt;interesting article&lt;/a&gt; that speculates that we could detect an object of significant mass (&quot;space-ship size&quot;) moving through our galactic neighborhood at a substantial fraction of the speed of light with existing technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That got me wondering (for a sci-fi novel I'm writing) whether there would be evidence of such passage &lt;em&gt;near the point in space where it occurred&lt;/em&gt; years or decades after such passage far from any star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CMB interactions mentioned in the article would spread from the point of passage at the speed of light so would have long departed the area. Is it likely that some artifact of passage would still be present, such as a change to the interstellar medium (change to ionization rates, fusion of lighter elements, etc)? How quickly would any such changes disperse? For purposes of the question, assume that scientific observations are possible at the point of suspected passage.&lt;/p&gt;&#xA;" OwnerUserId="1315" LastEditorUserId="1315" LastEditDate="2018-07-21T00:30:34.193" LastActivityDate="2018-07-27T20:07:36.460" Title="Detect Passage of Massive Relativistic Object" Tags="&lt;astrophysics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27014" PostTypeId="1" AcceptedAnswerId="27015" CreationDate="2018-07-20T02:10:30.627" Score="1" ViewCount="57" Body="&lt;p&gt;Whether it be the moon (especially when full) or tonight, as Mars is closer than it has been in decades, it appears that these bodies are larger when close to the horizon than overhead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this an optical illusion (i.e. the actual appearance is no larger), or is there some refraction or other effect that actually makes the appearance larger? &lt;/p&gt;&#xA;" OwnerUserId="22212" LastActivityDate="2018-07-20T02:32:31.060" Title="Do celestial bodies actually appear larger along the horizon?" Tags="&lt;the-moon&gt;&lt;size&gt;&lt;terrestrial-planets&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2018-07-22T03:42:09.650" ContentLicense="CC BY-SA 4.0" />
  <row Id="27015" PostTypeId="2" ParentId="27014" CreationDate="2018-07-20T02:32:31.060" Score="3" Body="&lt;p&gt;No, it's an illusion. Probably the ancient one. Simple experiment you can do is, set grid on telescope, measure the angle subtended when moon is at horizon and when moon is atop. You will see angle subtended by moon is same, hence the size are same. It's merely an illusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details:&#xA;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Moon_illusion&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Moon_illusion&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23450" LastActivityDate="2018-07-20T02:32:31.060" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27016" PostTypeId="1" AcceptedAnswerId="27020" CreationDate="2018-07-20T05:01:45.933" Score="8" ViewCount="917" Body="&lt;p&gt;Recently some new moons of Jupiter were discovered. However the moons are pretty damn small indeed - only a few kilometers wide. That makes me wonder if they should even be called moons in the first place. The Wikipedia article on &lt;a href=&quot;https://en.wikipedia.org/wiki/Natural_satellite&quot; rel=&quot;noreferrer&quot;&gt;natural satellites&lt;/a&gt; says there's no clear lower limit (or upper limit for that matter). But if we accept that, then Jupiter already has countless numbers of moons in its ring system, and even the Earth has multiple moons since the International Space Station is a &quot;moon&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exactly is a moon? Is it really a vague notion of &quot;something that goes around a planet&quot;? If so, why hasn't the IAU clarified what exactly a moon is? They've already done so for planets; it seems natural to do so for moons as well.&lt;/p&gt;&#xA;" OwnerUserId="20348" LastEditorUserId="9527" LastEditDate="2020-02-20T14:35:36.433" LastActivityDate="2020-02-20T14:35:36.433" Title="What exactly is a &quot;moon&quot;?" Tags="&lt;natural-satellites&gt;&lt;definition&gt;&lt;moonlet&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27017" PostTypeId="2" ParentId="19482" CreationDate="2018-07-20T05:33:07.013" Score="1" Body="&lt;p&gt;I might be answering out of turn, but nobody (so far as I noticed) has mentioned the &quot;Dobsonian Sun Telescope&quot;, which a Newtonian design that uses a full-aperture plane two-way mirror as secondary set at 45 degrees, that reflects 95% of the sunlight away from the optical path. This is a safety feature, should it be broken, the secondary mirror will also be broken as the optical path is disrupted. The Primary mirror is unsilvered / unaluminised. There is also a welders' glass in the optical path.&#xA;&lt;a href=&quot;http://planitarium.net/sfsa/sunscope/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://planitarium.net/sfsa/sunscope/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23674" LastActivityDate="2018-07-20T05:33:07.013" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27018" PostTypeId="2" ParentId="27016" CreationDate="2018-07-20T07:15:48.703" Score="2" Body="&lt;p&gt;According to &lt;a href=&quot;https://solarsystem.nasa.gov/moons/overview/&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;NASA&quot;&gt;NASA - What is a Moon?&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Planets and asteroids orbit the Sun. Moons—also known as natural satellites—orbit planets and asteroids. Moons come in many shapes, sizes and types. Most are airless, but a few have atmospheres and even hidden oceans. There are hundreds of moons in our solar system—even a few asteroids have small companion moons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The same question on &lt;a href=&quot;https://www.universetoday.com/60072/what-is-a-moon/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Universe Today - What is a Moon?&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A moon is defined to be a celestial body that makes an orbit around a planet, including the eight major planets, dwarf planets, and minor planets. A moon may also be referred to as a natural satellite, although to differentiate it from other astronomical bodies orbiting another body, e.g. a planet orbiting a star, the term moon is used exclusively to make a reference to a planet’s natural satellite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many other sites have similar explanations. By any of the definitions, though, the International Space Station is &lt;em&gt;not&lt;/em&gt; a moon, as a moon is defined as a &lt;em&gt;natural satellite&lt;/em&gt; orbiting a planet, dwarf planet or minor planet.&lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2018-07-20T07:15:48.703" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27020" PostTypeId="2" ParentId="27016" CreationDate="2018-07-20T12:32:01.700" Score="11" Body="&lt;p&gt;Unlike &quot;planet&quot; the IAU hasn't attempted to precisely define &quot;moon&quot;. General usage requires that a &quot;moon&quot; is a natural satellite of a planet (or dwarf planet, asteroid, or perhaps even of another moon?) and it is big enough for us to have seen it as an independent body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This contrasts with the &quot;moonlets&quot; that have been detected in Saturn's rings by their graviational effects on the even smaller ring particles. We know that a body exists there, but it is too small to be directly imaged. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such the question &quot;How many moons does Jupiter have&quot; is probably unanswerable, since it depends on the sensitivity of your telescope and how close you are to the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smallest moons are about 10km across, but moonlets in the rings exist that are only a few hundred metres across. So in current usage, 10k might be roughly where people stop saying &quot;moon&quot; and start saying &quot;moonlet&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Language exists to serve us, the definition of planet was only made to solve a particular naming problem.  No such problem has arisen for moons. So there is no authoritative answer.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-07-20T12:32:01.700" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="27021" PostTypeId="2" ParentId="26987" CreationDate="2018-07-20T14:04:01.267" Score="1" Body="&lt;p&gt;After some research, I think this answers my question: The orientation of the disk seems not to be known with certainty, however, &#xA;&lt;a href=&quot;https://dx.doi.org/10.1051/0004-6361:20078009&quot; rel=&quot;nofollow noreferrer&quot;&gt;Meyer et al. (2007)&lt;/a&gt; gives estimates with $3\sigma$ confidence that &quot;the position angle is ∼60◦−108◦ (east of north) in combination with a large inclination angle&quot;, meaning that the disk is seen more or less edge on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/0909.0267&quot; rel=&quot;nofollow noreferrer&quot;&gt;This 2009 paper&lt;/a&gt; indicates that the orientation of the disk will of course influence the pictures, but in any case one will see the event horizon. I assume that this has to do with the strong gravitational lensing, basically projecting the plane onto the sky.&lt;/p&gt;&#xA;" OwnerUserId="23717" LastEditorUserId="23717" LastEditDate="2018-07-20T15:19:15.110" LastActivityDate="2018-07-20T15:19:15.110" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27022" PostTypeId="2" ParentId="27013" CreationDate="2018-07-20T15:34:02.070" Score="0" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Detect Passage of Massive Relativistic Object&lt;/p&gt;&#xA;&lt;p&gt;... we could detect a massive object moving through our galactic neighborhood at a substantial fraction of the speed of light with existing technology.&lt;/p&gt;&#xA;&lt;p&gt;... wondering ... whether there would be evidence of such passage years or decades after such passage far from any star.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Yes.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A large object passes through a galaxy over 10 light years away.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We begin observation of that portion of the sky, 9 years 11 months later (or sooner).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The following month we start to notice a change in the orbits of objects in that area.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;If the massive object was a star we would see it as though it's travel through that system was occuring in &lt;em&gt;real time&lt;/em&gt; (our time, of course all times and their frames are real; this is a simplified explanation) even though what we're seeing occurred over a decade ago.&lt;/p&gt;&#xA;&lt;p&gt;If the massive object was a black hole we would not see it but we would see the movement of other objects in an unexpected manner, and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Weak_gravitational_lensing&quot; rel=&quot;nofollow noreferrer&quot;&gt;lensing&lt;/a&gt; of light.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pvYdU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pvYdU.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The effects of foreground galaxy cluster mass on background galaxy shapes. The upper left panel shows (projected onto the plane of the sky) the shapes of cluster members (in yellow) and background galaxies (in white), ignoring the effects of weak lensing. The lower right panel shows this same scenario, but includes the effects of lensing. The middle panel shows a 3-d representation of the positions of cluster and source galaxies, relative to the observer. Note that the background galaxies appear stretched tangentially around the cluster.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In the case of &lt;a href=&quot;https://en.wikipedia.org/wiki/Strong_gravitational_lensing&quot; rel=&quot;nofollow noreferrer&quot;&gt;very large masses&lt;/a&gt; well positioned to deflect the background the effect is more pronounced.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gP3RN.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QePWb.png&quot; alt=&quot;Gravitational Lensing&quot; /&gt;&lt;/a&gt;&#xA;&lt;sup&gt;Click to animate&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A remote light source passing behind a gravitational lens. There is a large point mass in the center acting as a lens. The aqua circle is how we would see the light source if there was no lens, while the white spots/circle is the light source as seen through the lens. If the light source is collinear with the earth and lens, the image is an &amp;quot;Einstein ring&amp;quot;. When the source is off this line we see a double image. As it moves far away, one of the images gets fainter while the other one is almost not affected by the lens any more (thus coinciding with cyan circle).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-07-20T15:34:02.070" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27023" PostTypeId="1" CreationDate="2018-07-20T18:42:12.527" Score="3" ViewCount="54" Body="&lt;p&gt;On July 13, 2018, while camping in Southwest Oregon, a bright light appeared in the sky at sunset. It was much brighter than Jupiter, also visible. It was moving slowly, west to east. I estimate it took 1/2 hour to cross the sky, though I didn't watch it all the way. The brightness seemed to vary, as did the coloration. It seemed it was reflecting the sunlight. It was higher than the clouds.&#xA;   I had a spotting scope with me and it was not a point nor a disc, but rather circular, bright around the edge with a less-bright center. It kind of looked like a planetary nebula, only quite bright. I thought maybe it was a weather balloon, but I really have no idea.&lt;/p&gt;&#xA;" OwnerUserId="23747" LastActivityDate="2018-07-20T18:42:12.527" Title="Unknown moving light in sky" Tags="&lt;amateur-observing&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2018-07-23T13:47:05.793" ContentLicense="CC BY-SA 4.0" />
  <row Id="27024" PostTypeId="2" ParentId="27002" CreationDate="2018-07-20T18:47:13.300" Score="4" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;My questions are: &lt;strong&gt;1&lt;/strong&gt;.:  why are most of the moons that are further away from Jupiter retrograde orbits, while the ones extremely close to Jupiter are prograde orbits? &lt;strong&gt;2&lt;/strong&gt;.: What possible situations were required for Valetudo to end up where it is opposed to where the rest of the pro-grade moons are clustered? Pure chance / a series of unlikely events?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Answer 1.&lt;/strong&gt; Inner moons usually orbit in the same direction as the rotation of their parent planet, this is favored due to one means of their formation, and the effect of tidal forces if they were close to the planet and in a retrograde orbit.&lt;/p&gt;&#xA;&lt;p&gt;Source 1: Wikipedia: &amp;quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Moons_of_Jupiter#Regular_satellites&quot; rel=&quot;nofollow noreferrer&quot;&gt;Moons of Jupiter - Regular Satellites&lt;/a&gt;&amp;quot;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;The inner moons are in a 1:2:4 orbital resonance. Models suggest that they formed by slow accretion in the low-density Jovian subnebula—a disc of the gas and dust that existed around Jupiter after its formation—which lasted up to 10 million years in the case of Callisto.&lt;span class=&quot;math-container&quot;&gt;$^{[39]}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;[39]: &amp;quot;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2002AJ....124.3404C&quot; rel=&quot;nofollow noreferrer&quot;&gt;Formation of the Galilean Satellites: Conditions of Accretion&lt;/a&gt;&amp;quot; (Dec 2002), by Robin M. Canup and William R. Ward.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Source 2: Wikipedia - &amp;quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere#True_region_of_stability&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hill Sphere - True Region of Stability&lt;/a&gt;&amp;quot;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;The region of stability for retrograde orbits at a large distance from the primary, is larger than the region for prograde orbits at a large distance from the primary. This was thought to explain the preponderance of retrograde moons around Jupiter; however, Saturn has a more even mix of retrograde/prograde moons so the reasons are more complicated.&lt;span class=&quot;math-container&quot;&gt;$^{[3]}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;[3]: &amp;quot;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2003Natur.423..264A&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chaos-assisted capture of irregular moons&lt;/a&gt;&amp;quot; (May 2003), by Sergey A. Astakhov, Andrew D. Burbanks, Stephen Wiggins, and David Farrelly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Answer 2.&lt;/strong&gt; It is possible that Valetudo was a &lt;a href=&quot;https://en.wikipedia.org/wiki/Binary_system#Binary_companion_(minor_planets)&quot; rel=&quot;nofollow noreferrer&quot;&gt;binary companion&lt;/a&gt; and was captured in an exchange reaction.&lt;/p&gt;&#xA;&lt;p&gt;Source 1: Wikipedia: &amp;quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Moons_of_Jupiter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Moons of Jupiter&lt;/a&gt;&amp;quot;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;These moons were probably captured by Jupiter from solar orbits.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Source 2: &amp;quot;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2005AGUFM.P11B0120A&quot; rel=&quot;nofollow noreferrer&quot;&gt;Satellite Capture via Binary Exchange Reactions: Application to Triton&lt;/a&gt;&amp;quot; (Dec 2005), by C. Agnor and D.P. Hamilton:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;... exchange reactions during planet-binary encounters (i.e. ~ an encounter in which one member of a binary is expelled and its place taken by the planet) represent a new pathway for the capture of planetary satellites. This new capture mechanism may be relevant to origin and evolution of several small body populations in the solar system (e.g.~the icy and irregular satellites of the giant planets).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Additional Info: &lt;a href=&quot;https://www.science20.com/robert_inventor/so_you_thought_you_knew_what_planets_look_like_shapes_of_rapidly_spinning_planets-155538&quot; rel=&quot;nofollow noreferrer&quot;&gt;Robert Walker explains&lt;/a&gt; and offers this image showing how contact and overcontact binaries, ternaries, quarternaries, etc. can form stable and unstable systems. Those can be torn apart by the tidal forces of larger (even smaller) objects.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XYevy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XYevy.png&quot; alt=&quot;Binary, ternary and quarternary objects&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-07-20T18:47:13.300" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27025" PostTypeId="2" ParentId="26617" CreationDate="2018-07-20T20:49:56.853" Score="2" Body="&lt;p&gt;You don't mention what you are trying to access this with but I would strongly recommend AstroPy for handling the ephemeris and handling the co-ordinates and time systems. For your questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The Moon and the Earth are both given relative to the Earth-Moon barycenter (center of mass). Separately you can get the vectors from the Solar System Barycenter to the Earth-Moon Barycenter. So if you want SSB-&gt;Moon you will need to do two calls to the ephemeris and do vector addition.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It depends how accurately you want things and how fast the thing you want is moving. You can probably use Terrestrial Time (TT) with the ephemeris and you want be horribly off for anything other than Moon. The offset TT-UTC (currently 67.184 seconds) consists of two parts; a variable part from UTC to TAI, currently 37 seconds, which consists (mostly) of the number of leapseconds and a fixed part (32.184 seconds) from TAI to TT. TDB varies around TT by about 2 milliseconds and it's somewhat periodic (dominated by Earth's orbit ellipticity and Jupiter and Saturn) but doing it properly requires a 791 term expansion (see e.g. &lt;a href=&quot;http://star-www.rl.ac.uk/docs/sun67.htx/sun67ss154.html#Q1-158-946&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://star-www.rl.ac.uk/docs/sun67.htx/sun67ss154.html#Q1-158-946&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes, if you want to do anything related from the Earth such as predicting positions of something in the sky, you will need to take the frame bias, precession and nutation into account along with Earth rotation/sidereal time to align the celestial reference frame (ICRS/J2000) with that of the Earth (ITRS).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I recommend USNO Circular 179 (&lt;a href=&quot;http://aa.usno.navy.mil/publications/docs/Circular_179.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://aa.usno.navy.mil/publications/docs/Circular_179.php&lt;/a&gt;) which has a lot of details of how to do this in the new IAU framework. Meeus's book is very good (I have it also) but is based on the old IAU1984 equator and equinox of date methods.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-07-20T20:49:56.853" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27026" PostTypeId="2" ParentId="24366" CreationDate="2018-07-20T21:08:09.147" Score="2" Body="&lt;p&gt;The DE430 JPL ephemeris &quot;memo&quot; (&lt;a href=&quot;https://ipnpr.jpl.nasa.gov/progress_report/42-196/196C.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ipnpr.jpl.nasa.gov/progress_report/42-196/196C.pdf&lt;/a&gt;) has details of how it is constructed and what from. Mercury and Venus's positions are tracked by tracking the spacecraft orbiting them (to sub-km accuracy). Mars, Jupiter and Saturn are also measured from the orbiting spacecraft such as Galileo and Cassini. The outer planets plus Pluto positions are primarily done from astrometry so measuring the positions of the planets relative to the background stars, in the past through photographic plates or with CCDs in more modern times. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Positions of stars and planets are all measured in something called the International Celestial Reference System (ICRS) which is based on measuring very distant and fixed on the sky objects called quasars using a precise radio technique called VLBI.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-07-20T21:08:09.147" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27027" PostTypeId="2" ParentId="21541" CreationDate="2018-07-20T21:57:04.840" Score="5" Body="&lt;p&gt;It depends on your definition of &quot;layman&quot; but Patrick Wallace has a slide presentation on what changed and why with the new &lt;a href=&quot;https://www.slideserve.com/harry/the-iau-2000-06-changes-to-celestial-reference-systems-a-bluffer-s-guide&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference systems&lt;/a&gt;. There is also the &lt;a href=&quot;http://aa.usno.navy.mil/publications/docs/Circular_179.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;USNO Circular 179&lt;/a&gt; which covers all of this and is pretty readable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a lot of ways it boils down to the realization that the wobbly (in space and rotation) Earth was a lousy thing to base co-ordinate systems on. The new systems are what's known as &quot;kinematically non-rotating&quot; which mean they are fixed in space and don't move. This makes it much easier to make more precise theories (which were needed) to describe how things like the orientation of Earth's pole (the precession and nutation you may have heard about) move in the Solar System without having to reference it to something like the equator and ecliptic which are always moving about.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastEditorUserId="23748" LastEditDate="2018-07-20T23:43:35.980" LastActivityDate="2018-07-20T23:43:35.980" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27028" PostTypeId="2" ParentId="14412" CreationDate="2018-07-20T22:21:09.090" Score="1" Body="&lt;p&gt;This doesn't need programming but the &lt;a href=&quot;http://www.astromatic.net/software/skymaker&quot; rel=&quot;nofollow noreferrer&quot;&gt;SkyMaker&lt;/a&gt; package will generate very realistic astronomical images from catalogs produced by &lt;a href=&quot;http://www.astromatic.net/software/stuff&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stuff&lt;/a&gt;. It includes realistic models of stars and galaxies and the point spread function including e.g. atmospheric blurring and noise effects in the CCD (blooming, saturation etc)&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-07-20T22:21:09.090" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27029" PostTypeId="2" ParentId="25227" CreationDate="2018-07-20T23:03:35.963" Score="1" Body="&lt;p&gt;Most professional astronomers tend to use whatever package they are used to or got from somewhere else and don't look &quot;under the hood&quot;. Very few astronomers I've met tend to write the own code that needs to deal with the multiple time and co-ordinate systems needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most tend to use AstroPy these days which hides a lot of details but uses the SOFA/ERFA routines underneath which do indeed use the modern IAU2006/2000 framework and Earth Rotation Angle. However most textbooks, webpages etc still use the old methods to compute an apparent place and then sidereal time to rotate and relate that to the terrestrial system. So a lot of people still use the old methods because that's what they learnt initially or read recently. Until there is an update of a readable book like Jean Meeus's &lt;em&gt;Astronomical Algorithms&lt;/em&gt; to the new system, I suspect sidereal time etc will persist for a good while longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two methods give the same answers for positions to more precision than most astronomers can measure for most objects; exceptions would be pulsars, quasars, spacecraft (all three mostly done in the radio and/or with VLBI techniques) and GAIA-related measurements.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-07-20T23:03:35.963" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27030" PostTypeId="2" ParentId="22466" CreationDate="2018-07-20T23:42:03.310" Score="2" Body="&lt;p&gt;Virtually all KBO searches use mosaic CCD cameras operating in the visible part of the spectrum, usually with a red filter like SDSS-r' (centered at approx. 622nm &lt;a href=&quot;http://skyserver.sdss.org/dr2/en/proj/advanced/color/sdssfilters.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;SDSS filters&lt;/a&gt;) on big (~4m telescopes). This is what we did with the OSSOS survey I was (lightly) involved with (&lt;a href=&quot;http://www.ossos-survey.org/about.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.ossos-survey.org/about.html&lt;/a&gt;). Although cold KBOs are brightest in the infrared, the background from the atmosphere is much higher which means you are less sensitive, unless you go to space (very expensive). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also IR detectors are smaller and less sensitive than optical CCDs and much harder to make into mosaic cameras and keep cold (to increase sensitivity). This means you can't cover as much sky area per night which makes finding rare and faint KBOs much slower. A CCD and red filter is a good balance between brightness of the objects you are looking for, the sensitivity of your detector (CCDs get less sensitive as you move more to the red or blue of the green-red central part of the optical spectrum) and a low background from the atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-07-20T23:42:03.310" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27032" PostTypeId="2" ParentId="27013" CreationDate="2018-07-21T02:11:57.117" Score="2" Body="&lt;p&gt;Probably not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that your question can be stated &quot;Can we detect a lingering effect on the local environment in interstellar space a decade or two after the passage of a relativistic starship?&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step is to ask &lt;em&gt;just how relativistic?&lt;/em&gt; If it's just &quot;really fast&quot; (say, 0.99c) there would be one set if effects, but if it was much fast -- 0.99999c, for instance -- some new effects would come into play.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second is to ask about the density of gas in the vicinity. If it's like normal interstellar space (typically a million atoms per cubic meter) its passage will be much harder to detect than if we're in something like a Giant Molecular Cloud where densities can be 100 to a million times greater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, you won't see any lingering electromagnetic radiation: That will necessarily be light years off by a decade later and you specified local effects.  That leaves two things that I can see: (a) Effects on the pre-existing gas, and (b) bits of ship that remain behind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Interstellar_medium&quot; rel=&quot;nofollow noreferrer&quot;&gt;interstellar medium&lt;/a&gt; is nearly entirely atomic hydrogen, either neutral or ionized. The temperature is typically 100-1000 K. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Mean_free_path&quot; rel=&quot;nofollow noreferrer&quot;&gt;mean free path&lt;/a&gt; of these atoms is very large: 10&lt;sup&gt;5&lt;/sup&gt; miles up to a million times that. The &lt;a href=&quot;http://www.wolframalpha.com/input/?i=speed%20of%20100K%20hydrogen%20atoms&quot; rel=&quot;nofollow noreferrer&quot;&gt;mean speed of the atoms&lt;/a&gt; is about 1 kps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;That means&lt;/em&gt; that any effect the passage of the ship has on the local interstellar medium will be entirely erased by the diffusion of the interstellar medium (IM) in a short time.  Imagine that the ship blows a tube of vacuum in the IM a million miles across. Neutral hydrogen atoms from outside of this region (just a few mean free paths away) will move in at 1 kps. In a few million seconds, the void created by the ship's passage will be refilled. (A  million seconds is about 10 days, so a single year is plenty of time for the IM to diffuse back to equilibrium.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the IM is &lt;em&gt;much&lt;/em&gt; denser, the effect may linger as long as a decade, but we're not talking about the IM anywhere near us.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second potential source is bits of the ship left behind. Hitting the IM at .99c is precisely the same as the IM hitting the ship at .99c, and that's &lt;a href=&quot;http://www.wolframalpha.com/input/?i=energy%20of%20hydrogen%20at%20.99c&quot; rel=&quot;nofollow noreferrer&quot;&gt;quite hard radiation&lt;/a&gt; -- the IM's protons are hitting at 5 GeV. Unless the ship has force shields of some sort, this is likely to cause some spallation -- atoms of the ship's front end eroded into space. This &lt;em&gt;might&lt;/em&gt; be detectable, but the problem I see is that those bits are unlikely to be dumped into space with zero transverse velocity and will thus quickly leave the local area and be undetectable a decade later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ship it may dump enough energy into the IM to heat it -- creating a long tube of hotter than normal IM along its path. The problem -- again -- is that the mean free path is so long that the hot atoms would immediately decamp for distant places and the colder atoms outside the ships's range would diffuse in, erasing the tracks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the speed were insanely high (as if .99c wasn't insane enough), possibly you might see some sort of effect where a bow wave of plasma builds up big enough to clear a &lt;em&gt;really&lt;/em&gt; wide tube along the path, long enough that it can't fill in in a decade, or enough energy is dumped into the IM that it's still excited a decade later.  The problem is that that energy -- which is large -- is coming from the ship's kinetic energy through &lt;em&gt;friction&lt;/em&gt; with the IM. (a) This slows the ship and (b) it probably does very bad things to the ship's structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last effect is that we might see some side effect of the ship's drive or shields. But since they're basically magic (&lt;a href=&quot;https://en.wikipedia.org/wiki/Clarke%27s_three_laws&quot; rel=&quot;nofollow noreferrer&quot;&gt;Clarke's Third Law&lt;/a&gt;) I don't see how anyone can give you a scientific answer.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="1315" LastEditDate="2018-07-27T20:07:36.460" LastActivityDate="2018-07-27T20:07:36.460" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27033" PostTypeId="1" CreationDate="2018-07-21T03:48:28.433" Score="5" ViewCount="239" Body="&lt;p&gt;Why is the sunset not the latest on 21 June, but only on 27 June. Purely geometrically regarding the orbit of the two celestial bodies this should not be like this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there somewhere a good source, that explains this situation ?&lt;/p&gt;&#xA;" OwnerUserId="19586" LastEditorUserId="2365" LastEditDate="2018-07-21T06:40:40.530" LastActivityDate="2018-07-23T19:49:49.413" Title="Why is sunset on 21st of June not the latest?" Tags="&lt;the-sun&gt;&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27034" PostTypeId="2" ParentId="14041" CreationDate="2018-07-21T04:38:36.493" Score="5" Body="&lt;p&gt;Given that the uncertainty in deltaT, the difference between the uniform dynamical time needed to predict the positions of the Sun and the Moon and civil (solar) time was already 4.5 hours by year -4000 (&lt;a href=&quot;https://eclipse.gsfc.nasa.gov/SEcat5/uncertainty.html&quot; rel=&quot;noreferrer&quot;&gt;https://eclipse.gsfc.nasa.gov/SEcat5/uncertainty.html&lt;/a&gt;), then I would say by 65 million years ago we would have no idea of which way the Earth was facing at the time. Unless the dinosaurs left eclipse records that would let us calibrate the much faster spinning Earth back then, we would have no means to work out what time of day the impact happened.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-07-21T04:38:36.493" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27035" PostTypeId="2" ParentId="26364" CreationDate="2018-07-21T05:41:43.537" Score="1" Body="&lt;p&gt;Assuming you have downloaded their software from the link you provided and have astropy installed, you should be able to do:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from wssa_utils import wssa_getval&#xA;from astropy import units as u&#xA;from astropy.coordinates import SkyCoord&#xA;&#xA;# Declare your wanted point in Galactic co-ordinates&#xA;c = SkyCoord(l=0, b=30, unit=(u.deg,u.deg), frame='galactic')&#xA;ra = c.icrs.ra&#xA;dec = c.icrs.dec&#xA;vals = wssa_getval(ra, dec)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The results look like they will be in MegaJanskys per steradian, which would be &lt;code&gt;u.MJy/u.sr&lt;/code&gt; in astropy units.&#xA;For UV data, I'm not sure what is best, maybe &lt;a href=&quot;http://galex.stsci.edu/GR6/&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;GALEX&quot;&gt;GALEX&lt;/a&gt; ?&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-07-21T05:41:43.537" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27036" PostTypeId="2" ParentId="27033" CreationDate="2018-07-21T08:33:23.343" Score="5" Body="&lt;p&gt;It's explained in &lt;a href=&quot;https://en.wikipedia.org/wiki/Analemma#Earliest_and_latest_sunrise_and_sunset&quot; rel=&quot;noreferrer&quot;&gt;this Wikipedia article&lt;/a&gt;. Basically, the length of a day is defined by both the &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth%27s_rotation&quot; rel=&quot;noreferrer&quot;&gt;Earth's rotation&lt;/a&gt; itself and Earth orbiting around the sun. Because the Earth's orbit around the Sun isn't a circle but an ellipse (making it moving faster when it's closer to the Sun), and the Earth rotates in a plane different than it orbits the Sun, an &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_time#Apparent_solar_time&quot; rel=&quot;noreferrer&quot;&gt;apparent solar day&lt;/a&gt; isn't always 24 hours; in June, it's slightly longer. That means every day, &quot;noon&quot; (the time of day when the Sun is at its highest point) happens 'later' (compared to the time on your watch), so &lt;em&gt;sunset&lt;/em&gt; (and sunrise) are also later. The first few days after June 21st, this effect is more significant than the effect of the Sun moving south (causing the day* to become shorter and the sunrise falling earlier). The same effect causes the earliest sunrise (in your location) to be on June 15th, so before the solstice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;sub&gt;: day as in day vs. night, the period between sunrise and sunset, so not 24 hours&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="7472" LastEditorUserId="7472" LastEditDate="2018-07-21T08:47:41.730" LastActivityDate="2018-07-21T08:47:41.730" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27038" PostTypeId="2" ParentId="27033" CreationDate="2018-07-21T10:04:41.400" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is the sunset not the latest on 21 June, but only on 27 June?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When the latest sunset occurs depends very much on latitude. For people who live in the far north (e.g., Anchorage, Alaska or Reykjavik, Iceland), the latest sunset occurs on the day of or a few days after the solstice. For people who live just north of the equator, the latest sunset occurs in late July.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason the latest sunset occurs in late July just north of the equator has nothing to do with the change in the time from sunrise to sunset; there is very little variation in the portion of a day that is sunlit near the equator. The reason the latest sunset is in late July is instead because that is when solar noon is at its latest. Solar noon occurs later and later in the day in the weeks preceding late July, but then it reverses direction. There are four extrema of the timing of solar noon over the course of a year. Late July represents one of those four extrema.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time at which solar noon occurs varies over the course of a year. This variation is a consequence of the Earth's axial tilt and the eccentricity of its orbit. Mathematically, this is described by the &lt;a href=&quot;http://aa.usno.navy.mil/faq/docs/eqtime.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;equation of time&lt;/a&gt;, depicted in the two graphs below. The first shows the cumulative effect; this is the &quot;equation of time.&quot; The second graph shows how much local noon changes on a day-per-day basis. Negative values such as in mid May to late July indicate when solar noon occurs later than it did on the previous day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WWuIN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WWuIN.jpg&quot; alt=&quot;Plot of the equation of time, showing the variance of solar noon vs clock noon over the course of a year.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rVrHa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rVrHa.jpg&quot; alt=&quot;Plot of the rate of change of the equation of time, showing how much solar noon changes per day. Negative values indicate that solar noon is getting later.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length of time that a given place on the Earth north of the equator is sunlit decreases in the days after the solstice, with how much depending on latitude. Solar noon occurs later and later in the days after the solstice, and this change is the same for every location where the Sun is visible. This creates a latitudinal tension regarding when the latest sunset occurs. In far northern latitudes, the rapidly decreasing length of sunlight hours after the solstice quickly overcomes the slowly changing time at which solar noon occurs. Closer to the equator, the small seasonal changes in daylight make the latest solar noon mark the day with the latest sunset. When the latest sunset occurs is somewhere between these two extremes for intermediate northern latitudes. For your locale, the latest sunset was on June 27. Where I live, it occurred a few days later than that.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2018-07-23T19:49:49.413" LastActivityDate="2018-07-23T19:49:49.413" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27039" PostTypeId="1" AcceptedAnswerId="27040" CreationDate="2018-07-21T13:53:32.600" Score="0" ViewCount="69" Body="&lt;p&gt;Can a Black Hole enter inside a Worm hole? If yes then what will a observer observe if it present at the another side of Worm hole?&lt;/p&gt;&#xA;" OwnerUserId="23742" LastEditorUserId="2365" LastEditDate="2018-07-21T17:08:18.330" LastActivityDate="2018-07-21T17:26:44.900" Title="Black Holes and Worm Holes" Tags="&lt;black-hole&gt;&lt;astrophysics&gt;&lt;wormhole&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ClosedDate="2018-07-23T18:18:16.527" ContentLicense="CC BY-SA 4.0" />
  <row Id="27040" PostTypeId="2" ParentId="27039" CreationDate="2018-07-21T17:26:44.900" Score="4" Body="&lt;p&gt;Wormholes are a theoretical solution to Einstein's equations, but, unlike black holes, they have never been observed and probably don't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they did exist, they would almost immediately close due to the presence of matter in them. To stop them from closing you would need some &quot;exotic matter&quot; with remarkable properties, like &quot;negative mass&quot;. Exotic matter has never been observed, and probably doesn't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to push something massive, like a black hole close to a wormhole would almost certainly cause the wormhole to close. Perhaps if the wormhole were sufficiently big with enough negative mass to stabalise it it would be possible for the black hole to traverse it. The black hole would pass through the worm hole just like anything else. At the other end you would see the black hole coming through the wormhole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But remember, fundamentally, wormholes don't exist.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-07-21T17:26:44.900" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27041" PostTypeId="2" ParentId="18444" CreationDate="2018-07-21T18:27:19.013" Score="2" Body="&lt;p&gt;For all of these direct imaging results, the critical parameter is the contrast as a function of separation. This lets you know how much fainter an object you can see around the much brighter primary object whose light has been suppressed by the coronograph (the black circle in the center of the star). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the change in eclipse timings (Figure 1 in their paper) you can predict the expected mass and separation of the third object (proposed brown dwarf) using their equation 1 since you know the mass of the binary. From evolutionary models of brown dwarfs, you can make predictions of how bright the proposed brown dwarf should be for its predicted mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then look for an object at the right separation in the SPHERE images. This is shown in their Figure 3 (right hand figure of the 2) where they show how bright an object they could see as a function of separation from the primary star; everything above the solid curve is something they should see. The vertical dotted lines show the boundary on the separation of the brown dwarf, predicted from the eclipse timings. The diamond symbol is the predicted brightness of the brown dwarf from the evolutionary models given its mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is well above the contrast curve (by a factor of about 15x) and nothing is seen in the images at that separation, this strongly suggests that the proposed brown dwarf does not exist. The only &quot;get-outs&quot; are if the SPHERE team have measured their contrast curve incorrectly (unlikely at the level needed) or our models of how bright brown dwarfs should be are also wrong by about a factor of 15x and the brown dwarf is much, much fainter than predicted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Applegate mechanism is a bit weird. The idea is that as the star goes through magnetic activity cycles (as the Sun and a lot of other stars do) and the magnetic field strength grows and shrinks, this causes the shape of the star to change, bulging more or less at the star's equator as the cycle progresses. This changes the amount of angular momentum in the star, which has to couple into the binary orbit as the angular momentum has to be conserved, causing the binary orbit to shrink or expand. This would then explain the change in timings of the eclipses without the need for the third body brown dwarf tugging on binary. Unfortunately the Applegate mechanism, although it may work in V471 Tau, can't explain the variations in a bunch of other binary systems (magnetic field strength not strong enough) so it can't be the only solution for eclipse time variations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latest paper I could find that references the Vaccaero et al. 2015 paper, is &lt;a href=&quot;http://adsabs.harvard.edu/abs/2017ASPC..509..571V&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vanderbosch et al. 2017&lt;/a&gt;. They argue that because of the timing of the eclipses change but the spin period of the white dwarf secondary doesn't change, a third body such as a brown dwarf can't explain the changes as it should change the timing of both &quot;clocks&quot;.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-07-21T18:27:19.013" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27042" PostTypeId="1" AcceptedAnswerId="27047" CreationDate="2018-07-22T02:56:08.203" Score="1" ViewCount="104" Body="&lt;p&gt;Given the local sidereal time (LST) of a star when it rises, how can I find out the time when it will rise (according to my watch) on a specific date? &lt;/p&gt;&#xA;" OwnerUserId="16506" LastActivityDate="2018-07-22T20:36:00.903" Title="Rising &amp; Setting time of a star on a specific date" Tags="&lt;amateur-observing&gt;&lt;coordinate&gt;&lt;time&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27043" PostTypeId="1" CreationDate="2018-07-22T06:42:16.040" Score="3" ViewCount="463" Body="&lt;p&gt;I recently got my first telescope, a Bushnell reflector telescope. I spent a good while setting up my alignment. And I can see the moon perfectly! I went to look at another, Mars. I got it aligned in my finderscope. And i go to focus to see. And i see....basically the mirror in my scope with the lines. Its clearly getting light from the star....but i can't get it to focus on the star at all! Help!!&lt;img src=&quot;https://i.stack.imgur.com/BCTsL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="23757" LastActivityDate="2018-09-21T11:55:17.143" Title="Struggling to use my new telescope" Tags="&lt;star&gt;&lt;planet&gt;&lt;telescope&gt;&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27044" PostTypeId="2" ParentId="27043" CreationDate="2018-07-22T06:53:46.147" Score="3" Body="&lt;p&gt;It looks to me like you have two problems.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You need to adjust the focus. All telescopes come with a knob that alters the focus. This is pretty easy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the shadow you see is from the secondary mirror, and it is off centre. I don't know what the word for this is (off-collimation?), but I had it the other night and I fixed it by wiggling the secondary mirror and making sure it was flush with the hole it's supposed to fit in.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Try using the focus and if it's still bad, take another shot and post it here and maybe someone with more experience than me can help.&lt;/p&gt;&#xA;" OwnerUserId="17655" LastActivityDate="2018-07-22T06:53:46.147" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27046" PostTypeId="1" AcceptedAnswerId="27050" CreationDate="2018-07-22T14:36:38.147" Score="1" ViewCount="293" Body="&lt;p&gt;I have tried to compare the Ephemeris data from Nasa's Horizons with ephemeris &#xA;tables from astro.com. And they do not match. Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Date: 22th of July 2018. Body: Mars. Time 00:00 UT&#xA;Horizons: 309 degrees (Ephemeris type=Observer)&#xA;Astro.com: 305 degrees (&lt;a href=&quot;http://www.astro.com/swisseph/ae/2000/ae_2018.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.astro.com/swisseph/ae/2000/ae_2018.pdf&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But some bodies are matching. Like Mercury and Saturn. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I doing something wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources:&#xA;Nasa Horizons: &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi#top&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ssd.jpl.nasa.gov/horizons.cgi#top&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astro.com: &lt;a href=&quot;http://www.astro.com/swisseph/swepha_e.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.astro.com/swisseph/swepha_e.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23761" LastActivityDate="2018-07-23T05:26:19.790" Title="Ephemeris data mismatch between Nasa Horizonz and Astro.com" Tags="&lt;ephemeris&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="27047" PostTypeId="2" ParentId="27042" CreationDate="2018-07-22T20:36:00.903" Score="1" Body="&lt;p&gt;You'll need to convert local sidereal time to local time on the specific date, using the longitude of your location. This can be done with the formula described &lt;a href=&quot;https://astronomy.stackexchange.com/q/20296/7472&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$&#xA;UT = \frac{LST - \frac{L}{15} - 6.697374558 - 2400.051336*T - 0.000025862*T^2)}{1.0027379093}&#xA;$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;where $UT$ is the time in Universal Coordinated Time, $LST$ is the Local Sidereal Time, $L$ is the longitude, and $T$ is:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$&#xA;T = \frac{JD-2451545.0}{36525.0}&#xA;$$&#xA;  where JD is the Julian Date at noon UTC.  Once I've calculated the time in UT, I then add or subtract 24 until the result is within 0 - 24 hours.  Then I can convert from UT to local time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2018-07-22T20:36:00.903" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27048" PostTypeId="2" ParentId="26936" CreationDate="2018-07-23T01:22:37.707" Score="1" Body="&lt;p&gt;It's unclear if nuclear energy would be a choice propulsion for a space faring civilization.   It's possible, but it's not the only option.   That said, if it was the method of choice, it's not at all clear to me that it would be easily detected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me preface this with stating that I'm not an expert.   There's two kinds of basic nuclear engine that could be used for space travel, closed or open.  An open nuclear propulsion is much simpler.   Basically explosions are made outside the ship that propel the ship by force from the explosion.   Tiny nuclear pelots are detonated and that's what pushes the ship. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other method would be an internal nuclear engine, well insulated that generates electricity and electricity generates magnetic fields and presumably ion thrusters that push the ship.   We wouldn't detect nuclear energy if it's internal and insulated.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Orion ship that you mention in your question wouldn't make one megaton explosions, but many smaller explosions.  It would carry more fuel than 1 megaton, but it would use it for tiny explosions over it's journey.    Space is also full of Gamma radiation, so I don't think gamma radiation would stand out.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't look for gamma-rays from Earth's surface because the Atmosphere blocks most of them, so you need a space telescope and our best space telescope for that is the Fermi Gamma-ray Telescope.  &lt;a href=&quot;https://fermi.gsfc.nasa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://fermi.gsfc.nasa.gov/&lt;/a&gt;  (having trouble with linking, not sure why).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Fermi scans the entire sky, so it's really more of a survey than a telescope, and gamma ray sources are still being studied, with 31% (per chart) listed as unknown sources.   &lt;a href=&quot;https://www.nasa.gov/mission_pages/GLAST/news/gamma-ray-census.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.nasa.gov/mission_pages/GLAST/news/gamma-ray-census.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think our current equipment is close to detecting a 1 megaton nuclear bomb a few star systems away, much less an orion type spaceship.   I think it's too few gamma rays at several light years distance.  Perhaps with better equipment we'd have a shot, but I don't think we're anywhere close to that now.    That's largely a guess though, I'm not an expert on the subject.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's something wrong with my account, I can't comment either, so I'll post it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your assumption is that 1 megaton is large enough to be visible from lightyears away - by gamma ray telescope.   I'm not sure that's true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 Megaton = 4.18 x 10^15 Joules.&#xA;1 Joule = 6.24 x 10^18 electronvolts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, 1 Megaton in explosive energy (and not all of it would go to gamma rays, but a decent fraction).   But ignoring that fraction, 1 Megaton = 2.61 x 10^34 Electron volts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface area of a sphere 4.1 light years in radius (the distance to the closest star), 4 Pi r^2, where the radius is 3.88 x 10^16 meters, is 1.89 x 10^34 square meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a single megaton blast 4.1 light years away would average out to a bit over 1 electron volt on a 1 meter gamma-ray telescope.    A typical gamma ray is thousands of electron volts, so unless you have a very large telescope, there's a good chance you wouldn't pick up a single gamma ray from a one megaton blast 4.1 light year away, or maybe you'd pick up one if you were lucky.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.1 light years is very far.  Far enough that a telescope, unless it was enormous, would have a hard time seeing a nuclear blast that far away, and that's the closest star system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corrections are welcome but that's what the math looks like to me.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-07-23T03:51:28.210" LastActivityDate="2018-07-23T03:51:28.210" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27050" PostTypeId="2" ParentId="27046" CreationDate="2018-07-23T04:56:27.370" Score="1" Body="&lt;p&gt;You might be comparing an &lt;a href=&quot;https://en.wikipedia.org/wiki/Ecliptic_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;ecliptic longitude&lt;/a&gt; to a &lt;a href=&quot;https://en.wikipedia.org/wiki/Equatorial_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;right ascension&lt;/a&gt;.&#xA;These are measured around different circles, the ecliptic and the equator respectively.&#xA;Their values coincide at the solstices ($\lambda$ 90$^\circ$ or 270$^\circ$, RA 6h or 18h) but generally not elsewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With target body Mars, a geocentric observer location, and output quantities 2 (apparent RA &amp;amp; Dec) and 31 (observer ecliptic lon &amp;amp; lat) selected, JPL HORIZONS gives:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Date__(UT)__HR:MN     R.A.__(a-apparent)__DEC    ObsEcLon    ObsEcLat&#xA;2018-Jul-22 00:00     20 38 23.96 -24 58 44.6 305.5424192  -6.2964464&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;RA 20h 38m 24s is 309.60$^\circ$, but the quantity to compare is the ecliptic longitude, 305.54$^\circ$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Astrodienst ephemeris table, &lt;a href=&quot;http://www.astro.com/swisseph/swephinfo_e.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;based on JPL DE431&lt;/a&gt;, gives ♒ 5$^\circ$ 33', i.e. ecliptic longitude 300$^\circ$ + 5.55$^\circ$ = 305.55$^\circ$, for Mars on 2018-07-22.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-07-23T05:26:19.790" LastActivityDate="2018-07-23T05:26:19.790" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27051" PostTypeId="1" AcceptedAnswerId="27055" CreationDate="2018-07-23T05:35:52.967" Score="15" ViewCount="2399" Body="&lt;p&gt;This is a follow up question to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/27016/what-exactly-is-a-moon/&quot;&gt;What exactly is a &quot;moon&quot;?&lt;/a&gt; The conclusions I draw from James K's answer is that the IAU should define what a moon is. They haven't done so yet, but they should.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious way to go about defining &quot;moon&quot; would be to build on the definition of a planet:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Planets must be in orbit around the Sun&lt;/li&gt;&#xA;&lt;li&gt;Planets must be round due to their own gravity&lt;/li&gt;&#xA;&lt;li&gt;Planets must have &quot;cleared the neighbourhood&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This immediately suggests the equivalent definition of a moon:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Moons must be in orbit around the host planet&lt;/li&gt;&#xA;&lt;li&gt;Moons must have [minimum size]. Unfortunately being round doesn't work since not all moons are round (c.f. Mars' moons, unless one reclassifies those as not moons)&lt;/li&gt;&#xA;&lt;li&gt;Moons must have &quot;cleared the neighbourhood&quot; of their own orbit, i.e. in their particular orbit around the host they are by far the most massive body&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Would such a definition work? If so, why hasn't the IAU also defined moons? It seems so natural after all. If not, what's the catch?&lt;/p&gt;&#xA;" OwnerUserId="20348" LastEditorUserId="20348" LastEditDate="2018-07-23T07:31:17.393" LastActivityDate="2018-07-24T14:28:57.637" Title="Would an analogue of the definition for planets also work for moons?" Tags="&lt;natural-satellites&gt;&lt;definition&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27052" PostTypeId="2" ParentId="27051" CreationDate="2018-07-23T07:03:43.143" Score="19" Body="&lt;p&gt;In 2006 the IAU had a trilemma. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;They could decide that Eris was a planet, and potentially allow for future discoveries of tens of new planets. &lt;/li&gt;&#xA;&lt;li&gt;They could be inconsistent, declare that Pluto was a planet, but Eris (and Ceres) wasn't&lt;/li&gt;&#xA;&lt;li&gt;They could come up with a definition of &quot;planet&quot; that would exclude Eris, and consequently also exclude Pluto.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Each option is problematic: We teach eight-year-olds to chant the names of the planets, would we want to make them chant the names of 20 or thirty Kuiper belt objects?  Being willfully inconsistent would lead to further argument. Re-describing Pluto would go in the face of 75 years of tradition. There was no option that would keep everyone happy, but they chose the third. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't have a problem with a more relaxed defintion of &quot;moon&quot; we are used to new &quot;moons&quot; being discovered. Eight-year-olds don't need to chant their names.  The casual definition of &quot;if you find it, you can name it&quot; works. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issues with planets are particular to our expectations that &quot;planets&quot; are important. We have no such expectation with moons.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-07-24T14:28:57.637" LastActivityDate="2018-07-24T14:28:57.637" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27055" PostTypeId="2" ParentId="27051" CreationDate="2018-07-23T09:49:44.353" Score="10" Body="&lt;p&gt;Singling out one point:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;Moons must have &quot;cleared the neighbourhood&quot; of their own orbit, i.e. in their particular orbit around the host they are by far the most massive body&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This would mean that &lt;a href=&quot;https://en.wikipedia.org/wiki/Co-orbital_configuration&quot; rel=&quot;noreferrer&quot;&gt;co-orbital moons&lt;/a&gt; won't exist anymore; e.g. &lt;a href=&quot;https://en.wikipedia.org/wiki/Telesto_(moon)&quot; rel=&quot;noreferrer&quot;&gt;Telesto&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Calypso_(moon)&quot; rel=&quot;noreferrer&quot;&gt;Calypso&lt;/a&gt; share their orbit around Saturn with the much more massive &lt;a href=&quot;https://en.wikipedia.org/wiki/Tethys_(moon)&quot; rel=&quot;noreferrer&quot;&gt;Tethys&lt;/a&gt;. With the proposed definition, only Tethys is a moon, but if Telesto and/or Calypso would be bigger (so that Tethys is no longer &lt;em&gt;by far the most massive body&lt;/em&gt;) none of them would qualify as a moon, not even Tethys, no matter how big they are compared to other (potential) moons orbiting Saturn.&lt;/p&gt;&#xA;" OwnerUserId="7472" LastEditorUserId="7472" LastEditDate="2018-07-23T11:01:34.937" LastActivityDate="2018-07-23T11:01:34.937" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27056" PostTypeId="2" ParentId="27051" CreationDate="2018-07-23T12:27:16.897" Score="5" Body="&lt;p&gt;I'd like to expand on @James K's good answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question you need to answer before your question can be answered is: &lt;em&gt;why&lt;/em&gt; do you want to have a definition of &quot;moon&quot;?  What purpose does it serve other than pedantry?  (Remember that there is a continuum of objects from the largest stars down to the smallest bits of space dust -- whatever you do, you're drawing lines dividing &lt;em&gt;this&lt;/em&gt; patch of sand from &lt;em&gt;that&lt;/em&gt; one. Giving groups of similar objects a collective name is useful, but why is it important to precisely define the line's position down to individual grains?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because there are some laws which apply to moons and not to smaller bodies?  Because it's a step to making your organization the definitive source of naming policy?  Because full-fledged moons are more attractive tourist destinations?   To get publicity?  Because there's a turf fight between the Department of Small-Body Astronomy and the Department of Lunar Astronomy that needs to be settled?  Because several of your colleagues like making lists?  Because the name will make a difference in the scientific study of the bodies?  Because you're at this meeting and bored?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the reasons may seem flippant, but many of them (most of them, I suspect) played a role in the IAU's definition of planet.  So if you can answer &lt;em&gt;why&lt;/em&gt;, you'll probably also be able to answer your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(My own guess is that having a formal definition is important to people who like to make lists, but not otherwise scientifically important.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For myself, I'd say that a satellite is an object that goes around a larger non-stellar object.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-07-23T12:27:16.897" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27057" PostTypeId="1" AcceptedAnswerId="27059" CreationDate="2018-07-23T13:16:18.553" Score="3" ViewCount="1705" Body="&lt;p&gt;I know that Earth is always falling towards the Sun, but due to its sideways momentum it just misses always.  But how did it all start?  To go into detail, when the gas cloud with zero angular momentum collided, our solar system was made with planets and sun having their own angular momentum which makes them rotate.  But how did the revolution start in the beginning?  Were  they placed in such a position and distance that they started attracting each other but how ? .They were having only rotation .  But a revolution started which forced them to follow a elleptical path  so how this happened in the beginning?  Did some external force act on them and revolution started?  How was  this revolution started?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i have assumed dust cloud was having zero angular momentum but it can have any value that does not affect the question i think ? &lt;/p&gt;&#xA;" OwnerUserId="19788" LastEditorUserId="19788" LastEditDate="2018-07-24T06:05:19.100" LastActivityDate="2018-07-24T11:12:08.450" Title="How did the solar system get its angular momentum? Why did Earth not fall into the Sun at the start?" Tags="&lt;the-sun&gt;&lt;earth&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" ClosedDate="2018-07-23T18:09:20.073" ContentLicense="CC BY-SA 4.0" />
  <row Id="27058" PostTypeId="2" ParentId="27057" CreationDate="2018-07-23T14:39:46.073" Score="2" Body="&lt;p&gt;Imagine the initial gas and dust cloud before it collapsed. It was enriched in heavy elements by supernovaes from other stars. As we could simplify a supernova as an &quot;explosion&quot;, the shock waves of the novae alone will provide some momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally the sun did not form alone but is surrounded by other stars, which will also exert some force on the initial dust cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other fact to bear in mind is that even if the cloud collapsed towards it's center, the individual particles and gas molecules can and will have their own path through this cloud. The will collide with each other, be pushed around by radiation and gravity of other bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we set aside these forces and trace the path of one of this particles towards the center there are two things that could happen. It &quot;hits&quot; the center or it doesn't. If it doesn't hit the center it will continue it's path on a elliptical path around the center. As the gas cloud already has an initial momentum by the forced described above, all particles that do not follow this momentum are more likely to suffer a head-on collision, loose some or all of their momentum and therefore fall towards the center. The other particles will be more likely to follow their path. This will lead to a situation where all particles, that do not follow the overall momentum of the cloud, will fall towards the center and unify with our Proto-Sun. Therefore only the particles that followed the initial momentum of the cloud had the cance to form planets and asteroids.&lt;/p&gt;&#xA;" OwnerUserId="22927" LastActivityDate="2018-07-23T14:39:46.073" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27059" PostTypeId="2" ParentId="27057" CreationDate="2018-07-23T14:43:29.297" Score="11" Body="&lt;p&gt;A particle's angular momentum around a point is the particle's mass times its distance from the point times its velocity around the point.  So a small velocity at a large distance can still mean a lot of angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bit of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Star_formation&quot; rel=&quot;noreferrer&quot;&gt;Giant Molecular Cloud which collapses to form one solar system&lt;/a&gt; is typically on the order of a light-year across or somewhat less. It normally possesses some angular momentum -- just from random motions if nothing else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Star_formation&quot; rel=&quot;noreferrer&quot;&gt;When it collapses&lt;/a&gt; to form a solar system, its size drops from maybe a light-year down to maybe 10-100 AUs. A light year is about 60,000 AUs (astronomical units), so it is collapsing to about &lt;sup&gt;1&lt;/sup&gt;/&lt;sub&gt;1000&lt;/sub&gt; the size. To conserve angular momentum, the circular velocity of the matter must &lt;em&gt;increase&lt;/em&gt; by a factor of a thousand.  And the fraction of the matter which will form the star collapses by &lt;em&gt;another&lt;/em&gt; factor of a thousand or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only does this give you plenty of speed, but for the star to form it actually needs to &lt;em&gt;shed&lt;/em&gt; angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_momentum&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article on angular momentum&lt;/a&gt; for a lot of good details.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-07-23T14:43:29.297" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27060" PostTypeId="2" ParentId="27057" CreationDate="2018-07-23T15:41:27.620" Score="1" Body="&lt;p&gt;Ontological/statistical answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume a great many clouds of matter, each of which collapses. They can start out in any configuration (speed, rotation, momentum; being more homogenous or more disturbed etc.) whatsoever. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are comparatively few special overall configurations: the ones where there is no noticable net rotational momentum, and no appreciable local clumps with their own momentum. &lt;em&gt;These&lt;/em&gt; few configurations will lead to not forming a solar system with planets; if all turns out well (highly unlikely), everything will just smack together in the middle and create a very lonely star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All other cases will lead to some form of momentum, which will start to spin faster as matter gets drawn into a smaller volume (compare dancers who tuck their arms in).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: lots and lots of statistical probability for a random start configuration to end up with some momentum; very very small probability for that being not the case.&lt;/p&gt;&#xA;" OwnerUserId="14297" LastEditorUserId="14297" LastEditDate="2018-07-24T11:12:08.450" LastActivityDate="2018-07-24T11:12:08.450" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27062" PostTypeId="1" AcceptedAnswerId="27068" CreationDate="2018-07-23T20:45:28.743" Score="3" ViewCount="125" Body="&lt;p&gt;Could a lens or mirror be placed in orbit on the same flight path with other parts to make a telescope or on the horizon of the moon to see farther with the Hubble? &lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-07-25T02:19:55.497" LastActivityDate="2018-07-25T02:19:55.497" Title="Can a telescope image be amplified to see more by adding distant lenses or mirrors?" Tags="&lt;amateur-observing&gt;&lt;photography&gt;&lt;hubble-telescope&gt;&lt;telescope-lens&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27063" PostTypeId="1" AcceptedAnswerId="27064" CreationDate="2018-07-23T20:47:53.810" Score="2" ViewCount="632" Body="&lt;p&gt;We constantly read about &quot;clouds of dust and gas&quot; surrounding a (or being the initial state of a) star, but what elements constitute such dust and gas?&lt;/p&gt;&#xA;" OwnerUserId="23775" LastEditorUserId="38" LastEditDate="2018-07-31T10:49:11.877" LastActivityDate="2018-07-31T10:49:11.877" Title="What are the so called clouds of dust and gas made of?" Tags="&lt;star&gt;&lt;dust&gt;&lt;gas&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27064" PostTypeId="2" ParentId="27063" CreationDate="2018-07-23T21:08:43.427" Score="4" Body="&lt;p&gt;The clouds of gas and dust that form stars are usually what are called Molecular Clouds and &lt;a href=&quot;https://en.wikipedia.org/wiki/Molecular_cloud&quot; rel=&quot;nofollow noreferrer&quot;&gt;Giant Molecular Clouds&lt;/a&gt; (GMCs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;Molecular&quot; means that most of the atoms are combined into molecules rather than being free atoms.  So the hydrogen is mostly H&lt;sub&gt;2&lt;/sub&gt; and not H.  This is really important, since a cloud comprised of molecules can radiate away the heat of collapse much easier than a cloud of atoms can. If that heat isn't radiated away, the cloud won't collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the contents go, clouds int he solar neighborhood today have a composition very similar to that of the outer layers of the Sun: around 90% (71%) hydrogen, 9% (27%) helium .08% (1%) oxygen, .04% (0.5%) carbon and lesser amounts of things like nitrogen, silicon, magnesium, neon, iron and sulfur.  Note that these are atom% followed by (mass%).  See &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/suncomp.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; for a table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomers would call it 90% (71%) hydrogen, 9% (27%) helium and less than 1% (2%) &quot;metals&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much earlier in the universe's history the percentage of hydrogen was higher and the percentage of everything else was lower.   Since the Sun is a relatively new star, the gas it condensed from was &quot;polluted&quot; by the gas emitted by generations of older stars which enriched it with helium and &quot;heavy metals&quot;.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="7411" LastEditDate="2018-07-24T06:35:21.983" LastActivityDate="2018-07-24T06:35:21.983" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27065" PostTypeId="2" ParentId="27062" CreationDate="2018-07-23T23:43:56.710" Score="6" Body="&lt;p&gt;Speaking in general, telescopes are designed as complete systems from the beginning. It's not generally feasible to improve performance by just adding yet another lens or mirror to the assembly. The reasons for this are not simple to explain, since this is basically the whole theory of instrument design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually you start with a set of requirements (limiting magnitude, angular resolution, linear resolution in prime focus, etc) and with a set of constraints (cost, size, weight, materials).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You settle on a given architecture (newtonian reflector, Ritchey-Chrétien, doublet / triplet / quadruplet refractor, etc), and then start to work out the main parameters: aperture, focal length, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's an iterative process involving computer aided design to work out the details of each optical element; at each step you need to solve problems such as eliminating internal reflections, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, what you have is the design of the whole optical stack: say a Dall-Kirkham reflector, 1 meter aperture, f/8, with mirrors of whatever radiuses of curvature, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once that's set in stone, it's hard to change in a major way. Minor upgrades might be possible (add a corrector lens, or a different secondary mirror), but not huge modifications.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;TLDR: No. Start with a bigger instrument from scratch if that's what you want.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2018-07-23T23:43:56.710" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27066" PostTypeId="1" CreationDate="2018-07-24T03:34:10.713" Score="3" ViewCount="90" Body="&lt;p&gt;According to the Wikipedia article &lt;a href=&quot;https://en.wikipedia.org/wiki/Elliptical_galaxy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Elliptical galaxy&lt;/a&gt;, elliptical galaxies have a sparse interstellar medium. I know that probably sometimes two spiral galaxies collide then exponentially approach the state of being an elliptical galaxy. For a spiral galaxy that will never collide because if it hasn't collided before all the other galaxies went past its cosmic horizon, I wonder whether its interstellar medium will slowly get thinner until it's so thin that stars are no longer confined to the plane of the galaxy by friction and then as soon as a few of them start to go slightly outside the plane, they will bring with them their gravitational field causing stars to go further outside the plane and then after it gets wide enough, the interstellar medium will be even faster to get consumed by stars moving in random directions.&lt;/p&gt;&#xA;" OwnerUserId="13685" LastActivityDate="2018-07-24T09:40:25.170" Title="Will all galaxies eventually become elliptical galaxies?" Tags="&lt;astrophysics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27068" PostTypeId="2" ParentId="27062" CreationDate="2018-07-24T06:45:34.920" Score="4" Body="&lt;p&gt;Far from &quot;orbit&quot; some of the longest telescopes in history are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Tubed: &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Aerial_telescope#Very_long_%22tubed%22_telescopes&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hevelius' Longest Telescope from Machinae Coelestis&lt;/a&gt;&quot; - 150 feet&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yxAOP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yxAOP.jpg&quot; alt=&quot;Hevelius&amp;#39; Telescope&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Tubeless: &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Aerial_telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Huygen's Telescope&lt;/a&gt;&quot; - 210 feet&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/W1BT9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W1BT9.jpg&quot; alt=&quot;Huygen&amp;#39;s Telescope&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Adrien_Auzout&quot; rel=&quot;nofollow noreferrer&quot;&gt;Adrian Auzout&lt;/a&gt; and others made telescopes of from &lt;a href=&quot;https://en.wikipedia.org/wiki/Aerial_telescope#Tubeless_%22aerial%22_telescopes&quot; rel=&quot;nofollow noreferrer&quot;&gt;300 to 600 ft&lt;/a&gt; (90 to 180 m) focal length, but I've not found a drawing from the late 1600's of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory you could have a distant &lt;em&gt;objective lens&lt;/em&gt; but either it would need to be huge (costly) or your field of view would need to be extremely narrow (only useful for extremely distant objects). Then there's the cost of stabilizing the lens and maintaining alignment.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastActivityDate="2018-07-24T06:45:34.920" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27070" PostTypeId="2" ParentId="27066" CreationDate="2018-07-24T09:40:25.170" Score="5" Body="&lt;p&gt;First of all, stars are not confined to the plane of a spiral galaxy's disk by friction with the interstellar medium. Stars are very massive objects and the interstellar medium is extremely thin (comparable to the best laboratory vacuums on Earth), so there is effectively no friction involved. (And if there were, it would cause stars to spiral into the center of the galaxy.) Collisions between gas clouds do tend to keep the &lt;em&gt;gas&lt;/em&gt; confined to a thin disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, an isolated spiral galaxy will gradually turn into a &lt;a href=&quot;https://en.wikipedia.org/wiki/Lenticular_galaxy&quot; rel=&quot;noreferrer&quot;&gt;lenticular (a.k.a. S0) galaxy&lt;/a&gt;, not into an elliptical. As it uses up its gas in forming stars, star formation will then gradually cease, and the visible spiral arms will weaken and fade away; you will be left with a smooth disk of stars. (This process can happen more quickly in galaxy clusters, where the high-speed motion of a spiral through the hot, high-pressure gas of the cluster rapidly strips out the gas, leaving behind just the stars.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars in spiral-galaxy disks &lt;em&gt;do&lt;/em&gt; tend to increase their vertical motions a bit over time, due to the gravitational effects of minor mergers, spiral arms, and massive molecular clouds (&lt;em&gt;not&lt;/em&gt; because of friction, but because the gravity of the molecular cloud shifts the orbits of nearby stars a bit). Ironically, this means that the thinning out of the interstellar medium will probably &lt;em&gt;weaken&lt;/em&gt; the tendency of stars to increase their vertical motions, because there won't be any more molecular clouds to perturb the stellar orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given sufficient time, close encounters between pairs of stars will tend to randomize their orbits, and the lenticular galaxy &lt;em&gt;will&lt;/em&gt; eventually evolve into something more like an elliptical galaxy. However, because the average distances between stars are so large, the frequency of close encounters is very low, and the whole process will take &lt;em&gt;trillions&lt;/em&gt; of years.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2018-07-24T09:40:25.170" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27071" PostTypeId="2" ParentId="26719" CreationDate="2018-07-24T12:00:25.937" Score="0" Body="&lt;p&gt;The fundamental assumptions of LCDM cosmology are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;General relativity is valid on cosmological scales.&lt;/li&gt;&#xA;&lt;li&gt;Universe is dominated by &quot;cold&quot; dark matter (origin, composition unknown).&lt;/li&gt;&#xA;&lt;li&gt;The metric of the universe is given by the  Friedmann–Lemaître–Robertson–Walker metric.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Reference: &lt;a href=&quot;https://en.wikipedia.org/wiki/Lambda-CDM_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Lambda-CDM_model&lt;/a&gt;&#xA; A combination of 1 and 2 helped explain large scaled structure formation, which is smaller masses merging to become bigger systems while 3 helped explain the expansion of space depending on the matter density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lambda parameter comes from invoking general relativity. Although Einstein added it to support his own view of a static universe (disproved a decade later by Hubble's observations), it was later re-interpreted as a negative energy density of space (see Harvey, Alex 2012 &lt;a href=&quot;https://arxiv.org/pdf/1211.6338.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/pdf/1211.6338.pdf&lt;/a&gt; and/or Carroll, Sean 2001 &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0004075&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/astro-ph/0004075&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helped.&lt;/p&gt;&#xA;" OwnerUserId="23557" LastActivityDate="2018-07-24T12:00:25.937" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27072" PostTypeId="2" ParentId="2241" CreationDate="2018-07-24T12:25:00.937" Score="0" Body="&lt;p&gt;&quot;How&quot; can the universe be infinite? So I am inferring that there must be a 'procedure' of sorts to make it infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space can exist by itself without having anything. Vacuum, Intergalactic space, the huge voids seen in the large scale structure maps are examples. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is means no matter how much the universe expands, space cannot break apart like atoms. One may say it has no elastic modulus. One may use this as a basic premise to explain how there can be an infinite universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, theories of cosmic topology are in good agreement with the predictions of our universe's matter density being very close to 1 which points to a universe expanding forever. (see &lt;a href=&quot;http://www.scholarpedia.org/article/Cosmic_Topology#fabre:2013&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.scholarpedia.org/article/Cosmic_Topology#fabre:2013&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It cannot be infinite in a finite time because time is space and time are more or less the same. They are one entity called spacetime.  &lt;/p&gt;&#xA;" OwnerUserId="23557" LastActivityDate="2018-07-24T12:25:00.937" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27073" PostTypeId="2" ParentId="27016" CreationDate="2018-07-24T15:04:10.550" Score="0" Body="&lt;p&gt;I wont attempt to propose a complete definition, but only suggest a few criteria, that could be meaningful when defining a moon. As I see it, a formal definition, should either serve a very specific purpose (none given in question) or clarify a term already in general use. as a clarification of an everyday term, the formal definition should strive to be consistent with how, the word is already in common use, and not break drastically with existing understandings of the word.&#xA;I would propose, that moons are first and foremostnatural satelites orbiting around planets, dwarf-planets and exo-planets&#xA;(a satelite orbiting smaller objects, such as an asteroid, would probably not meet most people's idea of a real moon.&#xA;As for size, and clearing it's orbit, I would suggest, that the satelite, should be large enough and distinct enough from other objects in its orbital neighbourhood, to be clearly visible, as a distinct object, with the naked eye of a hypothetical human being, standing on the surface of the orbited (parent) object(at least, during part of its orbit).&lt;/p&gt;&#xA;" OwnerUserId="23787" LastActivityDate="2018-07-24T15:04:10.550" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27074" PostTypeId="1" CreationDate="2018-07-24T16:37:33.207" Score="1" ViewCount="131" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/706/why-is-the-earths-center-still-hot-after-millions-of-years&quot;&gt;This question&lt;/a&gt; poses the question of why the Earth is still hot after 4.5 billion years, and the primary reasons were:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(1) heat from when the planet formed and accreted, which has not yet&#xA;  been lost; (2) frictional heating, caused by denser core material&#xA;  sinking to the center of the planet; and (3) heat from the decay of&#xA;  radioactive elements.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With that in mind, would it be possible for a planet of Earth's size and age to have a cold core, mantle, and crust?&lt;/p&gt;&#xA;" OwnerUserId="23788" LastActivityDate="2018-07-25T17:50:29.710" Title="Could an Earth-sized planet be cold?" Tags="&lt;planet&gt;&lt;geology&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27076" PostTypeId="1" CreationDate="2018-07-24T18:25:11.213" Score="1" ViewCount="66" Body="&lt;p&gt;On July 13, 2018, just after the clouds had lost their color, a bright object became visible about 20 degrees above the western horizon (there was a hill).&#xA;It moved slowly east and had colors similar to what the clouds had recently.&#xA;In my spotting scope it was circular, sort of looked like the ring nebula, with an apparent diameter bigger than Jupiter.&#xA;It wasn't a satellite, nor an airplane. Weather balloon? Any ideas?&#xA;I know this isn't exactly an astronomy question, but I couldn't find another forum interested in such phenomina.&lt;/p&gt;&#xA;" OwnerUserId="23747" LastActivityDate="2018-07-24T18:25:11.213" Title="What was that object?" Tags="&lt;amateur-observing&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2018-07-25T08:00:32.407" ContentLicense="CC BY-SA 4.0" />
  <row Id="27077" PostTypeId="1" CreationDate="2018-07-25T04:14:27.037" Score="3" ViewCount="1021" Body="&lt;p&gt;Using the library &lt;a href=&quot;http://www.zeptomoby.com/satellites/&quot; rel=&quot;nofollow noreferrer&quot;&gt;OrbitTools&lt;/a&gt; I can input a TLE and get back ECI coordinates. I want to find the RA DEC though, so I use &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;right_ascension = atan(y / x) * 180.0 / PI;&#xA;&#xA;declination = acos(z/ sqrt(x* x + y * y + z * z)) * 180.0 / PI;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I figure there is likely some constant I need to add (PI/2, PI, etc.) but comparing the calculated RA to that in the TLE, there is no constant that would fix it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also noticed x,y,z can be very low (on the order of ~100 km in some cases) which would suggest its measured from earth's surface, so I would need to add the earth's radius to x,y,z, but this also doesn't fix my problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, I tried using the built in look angle function from OrbitTools, assuming that (0,0,0) (latitude, longitude, height) would be equivalent to RA DEC but... its not. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="23792" LastEditorUserId="907" LastEditDate="2018-07-25T15:20:05.480" LastActivityDate="2018-07-26T20:27:38.863" Title="Help converting ECI to RA DEC" Tags="&lt;satellite&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27079" PostTypeId="1" AcceptedAnswerId="27081" CreationDate="2018-07-25T09:49:52.877" Score="2" ViewCount="123" Body="&lt;p&gt;I'm just done reading &lt;a href=&quot;http://tofspot.blogspot.com/2013/10/the-great-ptolemaic-smackdown-table-of.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Great Ptolemaic Smackdown&lt;/a&gt;, and it provided a good timeline of establishing heliocentrism, broad acceptation of Kepler's system, and the initial (failed) Galileo's experiment attempting measurement of parallax of distant stars (which was based on assumption of two near stars of different brightness being a long distance apart; Galileo unfortunately picked a binary star) - all the models described within assume a fixed, mostly immutable &quot;celestial sphere&quot; of distant stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When, and how was this concept disproven? &lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2018-07-25T13:07:49.997" Title="When did the concept of &quot;celestial sphere&quot; fall?" Tags="&lt;history&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27080" PostTypeId="2" ParentId="2241" CreationDate="2018-07-25T10:52:56.073" Score="-1" Body="&lt;p&gt;It would be cool if Hubble saw a signpost marked: &quot;please do not go beyond this point, this is the edge of the universe.&quot; perhaps even a wall or a balustrade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can the universe be infinite? infinite size, infinite age, infinitely small quantum constructs? The main problem we have with infinity is that it defies human understanding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spherical and circular equations pervade the universe: Stars, CMB, planet and moon orbits, atoms, photons, Pi, they all can use sine-cosine equations. Rotations are infinite. Pi and sine-cosine are infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the cosmos is made up of infinite maths functions, why would it not be infinite? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many angles does space have? infinity. how long does a galaxy turn around? eternity. The circles of angular momentum go on ad-eternum. rotations are infinite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why would there be a spacial limit to infinitely large or infinitely small space? Because it defies human understanding? that's not a good enough reason. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything in the known universe depends on mathematical principles which are infinite, numbers are infinite, angles are infinite, generally sine-cosine, linear, circular and spherical objects have infinite turns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mandelbrot set is infinite. It's a much simpler equation than the objects of the universe, and it is infinite. So, i'd encourage you to study the mandelbrot and ask yourself: how can it be infinite? Then you will have some advantage to apply that study to objects of the universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a curious thing that bridges the infinitely large and small objects in the universe: Black hole singularities are nearly entirely composed of sine-cosine maths descriptions, but they withhold millions of suns of atoms in the space the size a money coin or a CD or a nano-meter, all inside a sine-cosine disk, which is a circular phenomenon weighing billions of stars. If suns can fit into basketballs, and perhaps atoms, how can we judge what is big or small at all? If very big things are reborn to very small, and grow to very big again, it's kind of like the circle of life, and it means that you are the center of the universe, which is a good feeling. Atoms and Stars don't matter compared to you. &lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2018-07-25T12:35:44.797" LastActivityDate="2018-07-25T12:35:44.797" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27081" PostTypeId="2" ParentId="27079" CreationDate="2018-07-25T13:07:49.997" Score="3" Body="&lt;p&gt;Friedrich Bessel's 1838 observation of stellar parallax disproved the notion that all stars were equally distant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, by this time, the idea that the stars were on a sphere surrounding the solar system had already been discounted (but not disproven) by the application of the Copernican principle (The Earth does not have a privileged position) and Occam's Razor (Adopt the model that has the fewest assumptions)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Giordano Bruno is particularly noted with the idea (c. 1580) that the sun is just one star among many which are distributed throughout the universe.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-07-25T13:07:49.997" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27082" PostTypeId="2" ParentId="27074" CreationDate="2018-07-25T15:30:18.897" Score="1" Body="&lt;p&gt;If you peruse Google for Earth cooling rates, you might find what you're looking for. Here's &lt;a href=&quot;http://apps.usd.edu/esci/creation/age/content/failed_scientific_clocks/kelvin_cooling.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;one&lt;/a&gt; of many sources.  I strongly recommend reading the entire page,as it discusses some of the geology-based estimates as well.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Kelvin's primary attack on geologic dating was that measurements of&#xA;  the rates of geologic processes were highly uncertain, if it was&#xA;  possible to measure them at all (Hallam, 110). Accordingly, Kelvin&#xA;  sought to apply the knowledge of physics to the problem. Given that&#xA;  temperature increases the further one descends below Earth's surface,&#xA;  Kelvin concluded that Earth was slowly cooling. He set out to&#xA;  calculate the time required for the Earth to cool, and thereby&#xA;  solidify, from an initially molten state. The idea that Earth had&#xA;  begun as an incredibly hot sphere of liquid dates back to Descartes&#xA;  and Leibnitz. This assumed initial condition was the linchpin for&#xA;  Kelvin's entire method (Hallam, 110). Bits of material at the surface&#xA;  would sink before solidifying, creating convection currents that kept&#xA;  the Earth at a uniform temperature until solidification began at the&#xA;  core (Hallam, 110; Knopf, 445). Kelvin needed to know: (1) the&#xA;  temperature at Earth's core, (2) the temperature gradient with regard&#xA;  to depth below the surface, and (3) the thermal conductivity of rocks.&#xA;  The gradient was established to be around one degree Fahrenheit for&#xA;  every fifty feet. Kelvin made his own measurements of conductivity.&#xA;  The problem was determining the temperature at the core. This is where&#xA;  Kelvin's theory of solidification enters the picture. Because the core&#xA;  was thought to be solid rock, its temperature could not exceed the&#xA;  melting point of rocks (Hallam, 110). He constructed the following&#xA;  equation:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;dθ / dx = S / h √(π t)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The temperature gradient is expressed by dθ / dx; h is the thermal&#xA;  conductivity; x is the distance below the surface, and θ is the&#xA;  temperature (Holmes, 445). In 1862, Kelvin arrived at a likely age of&#xA;  100 million years. Because of uncertainties in the data, the lower and&#xA;  upper limits were 20 million and 400 million years (Dalyrmple, 26;&#xA;  Hallam, 111).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That said, there are plenty of more recent papers which estimate the cooling rates (with and without the help of the radioactive elements) based on the kinds of data now available as to the current and historical makeup of the core, mantle, and outer layer(s), heat transfer cycles, etc.  &lt;/p&gt;&#xA;" OwnerUserId="907" LastEditorUserId="6" LastEditDate="2018-07-25T17:50:29.710" LastActivityDate="2018-07-25T17:50:29.710" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27083" PostTypeId="1" AcceptedAnswerId="27084" CreationDate="2018-07-25T16:54:44.230" Score="4" ViewCount="204" Body="&lt;p&gt;I have done a lot of searching online and while I have been able to find out that theoretically a moon could exist in a near-synchronous/stationary orbit, I am wondering if it is possible for multiple moons to exist in this state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if I've explained appropriately or not, so if this needs to be a little clearer, please let me know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of this question is that I am writing a novel and I had this idea that the planet has 9 moons and that the area of land that the story takes place in the moons only travel across the sky over the course of many days.  One moon might take 60 days to get from one horizon to the other, while another might take 15 days, and others would be varying times in between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my limited knowledge, I have surmised that a moon closer to the planet would take less time to cross the sky than a moon further away.  Obviously, for this to work, the moons would have to be in a supersynchronous orbit, not fully synchronous.  Would the range of distance for supersynchronous orbits allow multiple moons to exist in staggered distances and still be close enough to remain visible from one spot continuously for multiple days at a time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disclaimer:&#xA;I am not sure that this is the appropriate place to ask this question, since it is mostly theoretical.  I also want to apologize if my English is confusing.  Finally, weird letters and symbols in math make my head hurt, so I am sorry I cannot follow complicated mathematical formulas, but feel free to post them if you believe it may help answer the question.&lt;/p&gt;&#xA;" OwnerUserId="23801" LastActivityDate="2018-07-26T09:09:25.013" Title="Is it possible for a planet to have multiple moons in a nearly stationary orbit?" Tags="&lt;orbit&gt;&lt;natural-satellites&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27084" PostTypeId="2" ParentId="27083" CreationDate="2018-07-25T17:44:15.130" Score="4" Body="&lt;p&gt;You may want to try the Worldbuilding site &lt;a href=&quot;https://worldbuilding.stackexchange.com/&quot;&gt;https://worldbuilding.stackexchange.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, what you're describing is more or less possible, with some caveats. A major question is that of stability - how long those configurations remain in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several notions you need to think about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geostationary orbit: it's the orbit where satellites have an orbital period equal to the planet's rotational period, so they appear stationary in the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roche limit: the lowest orbit where natural satellites remain in one piece. Any lower than that and they break up into pieces due to tidal forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hill sphere: is the region where the planet's attraction dominates and satellite orbits remain stable (as opposed to being stolen by the star, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually Roche limit &amp;lt; Geostationary orbit &amp;lt; Hill sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Satellites at the geostationary orbit appear fixed. Those close to the geostationary orbit would indeed appear to move extremely slowly across the sky. Orbits higher than geostationary would appear to move backwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regards to the long-term stability of multiple moons in various orbits, they would be stable if the orbits are far enough from each other, or if they're sitting in each other's Lagrange points. Also see here &lt;a href=&quot;https://worldbuilding.stackexchange.com/questions/36929/stability-of-multiple-moons&quot;&gt;https://worldbuilding.stackexchange.com/questions/36929/stability-of-multiple-moons&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2018-07-25T17:44:15.130" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27086" PostTypeId="1" CreationDate="2018-07-25T23:49:24.373" Score="2" ViewCount="33" Body="&lt;p&gt;I suppose this applies to all solar events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know centuries ago they knew the quantity of these things fluxuated in  solar cycles. Apart from that I know there has been advancement in solar weather forecasting but how accurate is it? Could, say ESA, have enough foreknowledge to launch a satellite or have one poised in solar orbit to 'go fishing' to investigate the emissions in someway? ?composition of particles- for example?&lt;/p&gt;&#xA;" OwnerUserId="23805" LastActivityDate="2020-11-25T03:03:27.363" Title="how accurate are the predictions for coronal mass ejections-(location, timing and magnitude)?" Tags="&lt;coronal-mass-ejection&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27087" PostTypeId="1" CreationDate="2018-07-26T00:47:44.737" Score="3" ViewCount="166" Body="&lt;p&gt;Can anyone help with this question? My question arose from an eclipse related observation that I made while looking into lunar Saros series 129. It is the Saros series number for the lunar eclipse that occurred on this past July 27, 2018. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I saw that this Saros 129 series was &quot;born&quot; June 10, 1351 and the series will end June 24, 2613 (70 saros cycles of 18 years = roughly 1260 years). So, I became curious to see if another Saros series had ended just before Saros 129 began. I saw that Saros 89 (less by 40 in series numbering) had ended just four years earlier on July 23, 1347. Investigating further, I learned that there are roughly 40 different Saros series that are &quot;active&quot; at any one given time. So then I looked back further to lunar Saros 49 and observed that there does indeed seem to be a pattern involving roughly &quot;40&quot; less in the numbering (i.e. yes indeed, series 49 also ended just as series 89 began). These observations do seem to me as more than mere coincidences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So, my question now is - Are these particular lunar Saros series (&lt;em&gt;separated by &quot;40&quot; in numbering&lt;/em&gt;) related in any way? - series 49, 89, and 129. Is there another cycle or a &quot;greater&quot; cycle also at work here that I am unaware of?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please see the comments below if what I am asking seems unclear - my comments might rephrase the question for clarity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; - I checked several other Saros series numbers separated by 40. I am adding additional information below - hoping that it might help find an answer to this question about relatedness of the &quot;40&#xA;  apart&quot; series.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Checking other series spaced by 40 numbering, these Saros series (that are numbered 40 apart – i.e. 40, 80, 120 etc.) always contain eclipses that are separated by &lt;strong&gt;~1418 days&lt;/strong&gt; between eclipses in each respective series (i.e. exactly 4 lunar years/48 lunar months).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Small Saros series&lt;/strong&gt; (those containing fewer cycles - i.e. 70-74 saros cycles in the series) contain saros cycles that &lt;strong&gt;overlap minimally&lt;/strong&gt; (Saros 129 which is only &lt;strong&gt;70 eclipse cycles in length has no overlap&lt;/strong&gt; and began just after Saros 89 came to an end &lt;strong&gt;1418 days earlier&lt;/strong&gt;). Note - zero eclipse cycles between them overlap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Large Saros series&lt;/strong&gt; (those containing more cycles – i.e. 80-84 saros cycles in the series) contain saros cycles that &lt;strong&gt;overlap accordingly&lt;/strong&gt; (Saros 120 which is 83 cycles long thus began 13 cycles before Saros 80 came to an end with the overlapping eclipses each gapped &lt;strong&gt;by 1418 days&lt;/strong&gt;). Note - 13 eclipse cycles overlap accordingly - again about &lt;strong&gt;70 eclipse cycles with no overlap&lt;/strong&gt; between these series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like about 70 eclipse cycles (constituting ~1262 solar years or ~1300 lunar years) is the standard length of each adjacent Saros series – considering the &lt;strong&gt;non-overlapping portion only&lt;/strong&gt; – each non-overlapping portion separated by 4 lunar years.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; - Thanks to anyone who can shed any additional light on this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="22542" LastEditorUserId="22542" LastEditDate="2018-08-14T20:12:10.783" LastActivityDate="2018-08-14T20:12:10.783" Title="Are Saros series separated by &quot;40&quot; in numbering related in any way?" Tags="&lt;the-moon&gt;&lt;solar-system&gt;&lt;observational-astronomy&gt;&lt;solar-eclipse&gt;&lt;lunar-eclipse&gt;" AnswerCount="0" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27088" PostTypeId="1" CreationDate="2018-07-26T01:37:17.440" Score="2" ViewCount="171" Body="&lt;p&gt;I have a Bushnell Voyager Reflector telescope. It came with 2 eyepieces: k4mm and k20mm. And I am trying to find additional eyepieces to zoom in further on planets and stars, but the bushnell website doesnt offer any. I am unsure if other brands will work with the telescope. Or is my teleacope not powerful enough to look further out.&lt;/p&gt;&#xA;" OwnerUserId="23757" LastActivityDate="2018-07-26T07:28:55.110" Title="Beginner looking for extra eyepieces for telescope" Tags="&lt;planet&gt;&lt;telescope&gt;&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27090" PostTypeId="1" AcceptedAnswerId="34452" CreationDate="2018-07-26T05:33:23.593" Score="2" ViewCount="154" Body="&lt;p&gt;I wrote a &lt;a href=&quot;https://space.stackexchange.com/a/29697/12102&quot;&gt;quickie answer&lt;/a&gt; where I estimate the total rotational angular momentum of Jupiter as about 7E+38 kg m^2/s. However it bothers me because one of the sources gets this number assuming uniform density, and the other two cite 6.9E+38 kg m^2/s but it's not clear to me where this comes from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there better estimates that use reasonable models for Jupiter's density distribution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;update:&lt;/strong&gt; If not answerable due to uncertainties in density distribution and/or gradients in rotation rate (i.e. non rigid-body rotation) which I sort-of remember reading about recently, then an explanation why a better value isn't available would be an acceptable answer.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-12-24T01:43:19.440" LastActivityDate="2020-01-01T03:52:33.227" Title="Total rotational angular momentum estimates for Jupiter?" Tags="&lt;jupiter&gt;&lt;planetary-science&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27091" PostTypeId="2" ParentId="27088" CreationDate="2018-07-26T07:28:55.110" Score="2" Body="&lt;p&gt;First of all, any eyepiece of the correct diameter will work with your telescope. Typical diameters are 1.25&quot; and 2&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The eyepieces you have listed tell you the focal length of the lens. This relates to the magnification - the focal length of the telescope divided by the focal length of the lens will tell you the magnification. So the 20 mm eyepiece will give you lower magnification than the 4 mm eyepiece. Thus to work out the magnication you will need to know the size and focal length of the telescope for starters. A Google search showed there are at least a couple of models but your owner's manual will give you this information. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a caveat: Forget (at least a bit) about magnification. Many telescope brands will [try to] sell the magnification, because this is a big selling point. BUT higher magnification might just turn a small fuzzy blob into a bigger fuzzy blob. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most important features of a telescope are the aperture (diameter) and lens/mirror quality. Bigger aperture means more light means easier to see faint stuff. And a quality lens means the difference between a fuzzy blob and a crystal-clear image. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will probably find that the mirrors in your telescope are reasonable quality; the lenses are usually where the sellers save on cost. I am confident that if you buy a good quality lens in the 10-15 mm range and one in the 20-25 mm range you will see very nice, clear images of the Moon, the planets out to Saturn, nebulae and star clusters.&lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2018-07-26T07:28:55.110" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27092" PostTypeId="1" AcceptedAnswerId="27097" CreationDate="2018-07-26T07:47:15.483" Score="1" ViewCount="70" Body="&lt;p&gt;Soon a lunar eclipse will take place soon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if I will be able to see the moon during the eclipse (from 9:30 PM to 11:00 PM), because, I have some building, mountains... in my field of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I wonder:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On lunar eclipes, will the Moon position at 9:30 PM be the same than the Sun position at 9:30 AM ?&lt;/p&gt;&#xA;" OwnerUserId="23809" LastActivityDate="2018-07-26T12:28:15.317" Title="Determine full Moon position into sky from Sun position" Tags="&lt;the-moon&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27093" PostTypeId="2" ParentId="27083" CreationDate="2018-07-26T09:09:25.013" Score="2" Body="&lt;p&gt;To add to the answer above, (and this is kind of a world building answer), but your scenario, at the very least, would be unusual and difficult but maybe not impossible.   The answer is a little surprising, but the way to do it is to make the moons smaller and bring them in closer to the planet.  They'd have less gravitational interaction with each other that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not difficult to have a moon with a 60 day orbital period.   Our moon, for example has a 27.3 day period (sidereal) with an apparent &lt;a href=&quot;https://community.dur.ac.uk/john.lucey/users/lunar_sid_syn.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;29.5 day synodic cycle&lt;/a&gt;, full moon to full moon, but the moon basically circles the sky in a little bit over 24 hours because of the Earth's rotation.   Any relatively distant moon will orbit it's planet slowly.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that's not really the issue.  If the planet has normal days, similar to 24 hour days on Earth, and I'm guessing by your question that you don't want the day/night period on your planet to be huge, then any distant moon's apparent orbital period will be determined by the planet's rotation period, similar to our moon which appears to orbit the Earth about once a day (25 hours or so).   If your planet has what we might consider regular days, 12 hours, 16 hours, 24 hours, what have you, then your 9 moons would need to crowd around the geostationary radius and crowding an orbit with 9 objects that would be visible from the planet's surface is never a good idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9 visible moons from the surface along is a lot.   Jupiter has over 60 known moons and more waiting to be discovered, but most of those wouldn't be distinguishable as moons from the planet's surface.   Only the big 4 would be clearly identifiable as moons.  The rest are too small.   9 visible moons is a huge number and crowding those 9 moons around the geostationary orbital radius is nuts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is a way to do it and it's straight forward math.   Bring the geostationary orbit closer to the planet by making the planet less massive and by increasing it's rotation period, shortening it's days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take Earth as a starting point.   &lt;a href=&quot;https://en.wikipedia.org/wiki/Geostationary_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth's geostationary orbit&lt;/a&gt; is about 35,786 km above the equator, and the equator's radius is 6,378 km, so for orbital purposes, it's 42,164 km from Earth's center of mass.  Both numbers are relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you reduce the mass of your planet to 1/2 the mass of the Earth (&lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6&quot; rel=&quot;nofollow noreferrer&quot;&gt;orbital period formula is proportional to (r^3/m)^.5&lt;/a&gt;), r^3 needs to equal 1/2, so r is reduced to 79.3%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if you increase the rate of rotation of your planet from 24 hours to 16 hours.   The period is now 2/3rds and, same formula, the ratio of the period to the radius the power of 3/2.   2/3rds the period equates to a 76.3% of the initial radius.   Combined, with these factors, the new geostationary radius, from the center of mass is 60.6% of 41,164, or about 24,933 km from the planet's center, or about (assuming 5,100 km radius for the smaller planet).   19,833 km from the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you make the planet even smaller or the planet's rotation period even faster, you can bring the moons even closer.    So, why does this help you?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you bring the moons closer to the planet's surface you can make them smaller and they can still be visible.    As you make a moon smaller, say you reduce it's radius by 50%.   It's mass is reduced by a factor of 8 but it's angular diameter is only reduced by a factor of 2 and it's apparent brightness by a factor of 4.   By shrinking your planet and by increasing it's rate of rotation you move the geostationary orbital radius closer in and you can make the moons smaller but they can still be visible.   By making them smaller, they would have less gravitational interaction with each other and you could theoretically have 9 small moons in tight orbits, visible from the planet's surface, with relatively long rotation periods.   You could explain the collection of small moons as the result of a collision between two moons.   It's a stretch, but not impossible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want astronomical accuracy though, moons that close to a planet would pass through the planet's shadow fairly often and given their size, they'd probably disappear from view.   You could offset that somewhat by giving the planet a sizable axial tilt and the moons orbit the planet's equator, but they could still pass through the planet's shadow as the planet approached it's spring and fall equinox.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that said, most of the Moons should still have a not that slow orbit, 1-2-3 days.  Getting a 60 day orbit requires the moon be very near to the geostationary radius and another with 15 days, that's also quite close.   You shouldn't have to many moons with long period orbits because there's no realistic way to do it, but you could have 2 maybe 3 in the 10-60 day range, and I'd push the others out into the 1-2-3-5 day range.  You could even have a retrograde moon if you wanted.  It wouldn't really be retrograde, it would just appear that way because it orbits the planet faster than the planet rotates.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to get creative, you could have a &lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/2006/janus-epimetheus-swap.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Janus/Epimetheus dance&lt;/a&gt;.   :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A final point.  Your moons in this scenario shouldn't be round.  They would be small enough to be lumpy potato like objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(too long?)&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-07-26T09:09:25.013" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27095" PostTypeId="2" ParentId="27092" CreationDate="2018-07-26T11:21:15.367" Score="1" Body="&lt;p&gt;Visibility of this eclipse will depend on many things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 Is the moon visible at your location during the eclipse - your profile shows you live in France, so yes, but the eclipse may have already started at the time of moon rise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 What is the weather going to be like -ie will it be clear?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 Are there local obstructions? To get to your question, where will the moon be at any time?  It will not be where the sun is 12 hours earlier; that would only happen if the eclipse happened on an equinox.&lt;br&gt;&#xA;I think your best bet is to download one of the several free sky map programs, eg Stellarium, and set it to your location.  In any event the eclipse will last several hours, so even if you cant see it at first, you may be able to see it later.  Alternatively go out to some other location.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2018-07-26T11:21:15.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27096" PostTypeId="2" ParentId="26980" CreationDate="2018-07-26T11:42:31.903" Score="3" Body="&lt;p&gt;The following deals only with redshift caused by motion (Doppler effect).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wavelength (wl) shift for an object moving away from a stationary observer is calculated by following formula: shift = wl x V/C&#xA;with V speed of the moving object and C speed of light&#xA;(source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Doppler_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Doppler_effect&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we want a Sun-like star to become invisible to the human eye. Our Sun's emission starts around 250 nm and human vision ends around 700 nm (source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Sunlight#/media/File:Solar_spectrum_en.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Sunlight#/media/File:Solar_spectrum_en.svg&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we want a minimum shift of 700 - 250 = 450 nm for wl 250 nm.&#xA;Formula yields V/C = 450 / 250 = 1,8&#xA;which&#xA;(1) is impossible because nothing moves faster than light&#xA;(2) makes the classical formula irrelevant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relativistic formula (for objects moving at speeds &gt; C/10) is:&#xA;shift / wl = SQRT( (1 + V/C) / (1 - V/C) ) -1&#xA;with SQRT the square root&#xA;(source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Relativistic_Doppler_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Relativistic_Doppler_effect&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rounding (shift / wl) up to 2, the relativistic formula yields V/C = 8/10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed V would have to be even higher for a hotter (white to bluish) star with UV emission starting below 250 nm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my answer to the question is:&#xA;very close to the speed of light (80% in above calculation)&#xA;and therefore not a real-life scenario (see below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Among the nearby stars, the largest radial velocities with respect to the Sun are +308 km/s&quot; (source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Doppler_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Doppler_effect&lt;/a&gt;), that is 1000 times smaller than C = 3e5 km/s. Rounding down to 300 km/s, it would produce a maximum shift of 700 x 300 / 3e5 = 0,7 nm, several hundred times smaller than required for invisibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides, the star would still appear on an infrared image.&lt;/p&gt;&#xA;" OwnerUserId="23782" LastEditorUserId="23782" LastEditDate="2018-07-27T11:23:41.540" LastActivityDate="2018-07-27T11:23:41.540" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27097" PostTypeId="2" ParentId="27092" CreationDate="2018-07-26T12:28:15.317" Score="2" Body="&lt;p&gt;The Sun's and Moon's &lt;a href=&quot;https://en.wikipedia.org/wiki/Hour_angle&quot; rel=&quot;nofollow noreferrer&quot;&gt;hour angles&lt;/a&gt; at those times will be similar (both about 48$^\circ$ east of the meridian), but their &lt;a href=&quot;https://en.wikipedia.org/wiki/Declination&quot; rel=&quot;nofollow noreferrer&quot;&gt;declinations&lt;/a&gt; will be opposite (Sun 19$^\circ$ north of the equator, Moon 19$^\circ$ south).&#xA;The Sun in the morning will be over 30$^\circ$ higher than the Moon in the evening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the Moon rises and sets about 50 minutes later each day, a better predictor for the Moon's position at 21:30 on July 27 is the Moon's position at 20:40 on July 26, about 1$^\circ$ lower than the eclipse position if observing from a mid-northern latitude.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-07-26T12:28:15.317" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27098" PostTypeId="1" CreationDate="2018-07-26T12:33:25.377" Score="5" ViewCount="127" Body="&lt;p&gt;I was reading on how Type II Supernovae release as much as 99% of their energy as neutrinos. So I was thinking: suppose there was a large star and black hole binary system, but the two were not close enough for the BH to draw mass out of the star, so the star aged naturally instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one were close enough to observe when the star's core collapses - but before the supernova bursts through the star's surface - would it be possible to see the black hole immediately grow in size for no apparent reason as it eats up the neutrinos from the companion star's collapse?&lt;/p&gt;&#xA;" OwnerUserId="23811" LastEditorUserId="1891" LastEditDate="2018-07-26T12:49:29.050" LastActivityDate="2018-07-26T14:42:29.827" Title="Star - Black Hole binary. Star core Collapses. Does the black hole immediately grow from neutrinos?" Tags="&lt;black-hole&gt;&lt;supernova&gt;&lt;neutrinos&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27099" PostTypeId="2" ParentId="27098" CreationDate="2018-07-26T12:56:16.603" Score="4" Body="&lt;p&gt;Yes. Neutrinos which hit the BH will add to its mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you may be thinking that the neutrinos pass through the nascent supernova without interaction. In a dense enough environment, neutrinos &lt;em&gt;do&lt;/em&gt; interact, and absorption of neutrinos by the layers beyond the core is a major source of the energy that disrupts the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a bit more detail, the core collapse happens through inverse beta decay, where a proton and an electron react to form a neutron and a neutrino. This basically turns the star's iron core to neutrons. The energy is carried away by the neutrinos and the absorption of this &lt;em&gt;enormous&lt;/em&gt; flux of neutrinos by the outer layers drives the SN.  The &lt;a href=&quot;https://en.wikipedia.org/wiki/Type_II_supernova&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; has a good (but IMO somewhat incoherent) longer discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still the &lt;em&gt;many&lt;/em&gt; neutrinos which do escape do so before the shock wave blows the star apart and the energy they carry will bulk up a nearby BH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I doubt that it would easily be observable, though. Stellar-mass BHs are &lt;em&gt;small&lt;/em&gt; in cross-section and would only absorb neutrinos from a patch of the sky proportional to their area. If a 10km BH was orbiting a million km star right at its surface, it would only absorb one part in 10&lt;sup&gt;10&lt;/sup&gt; of the neutrinos, producing roughly a 10&lt;sup&gt;-9&lt;/sup&gt; fractional increase in size.  &lt;em&gt;Very&lt;/em&gt; difficult to measure!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're using this to provide an early warning of the core-collapse (perhaps so you can start you FTL drive and get outta there) it's fairly useless. First, the neutrino flux is probably high enough to fry people who are close enough to need the early warning.  And second, if you're far enough away not to be fried by the neutrinos, you have plenty of time to see the star rapidly swelling  and get moving before the comparatively slow shock wave gets near you.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2018-07-26T14:42:29.827" LastActivityDate="2018-07-26T14:42:29.827" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27100" PostTypeId="1" CreationDate="2018-07-26T18:42:46.380" Score="1" ViewCount="2462" Body="&lt;p&gt;If a satellite is revolving in a GEO orbit, then what will be the eclipse duration for it? How to calculate when the satellite will enter the dark side of the Earth? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any formula or online tool would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="23810" LastEditorUserId="7472" LastEditDate="2018-07-26T20:27:33.383" LastActivityDate="2019-04-14T11:24:53.713" Title="Calculate eclipse days and duration of satellite" Tags="&lt;earth&gt;&lt;solar-eclipse&gt;&lt;satellite&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27102" PostTypeId="2" ParentId="27077" CreationDate="2018-07-26T20:27:38.863" Score="1" Body="&lt;p&gt;Assuming you are calling &lt;code&gt;cEci&lt;/code&gt; and have the units and sign conventions correct (longitudes are notoriously inconsistent as to which way is positive) then you should be able to calculate it almost as you have as:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;distance = sqrt(x*x + y*y + z*z)&#xA;RA = atan2(y/x)&#xA;# Normalize right ascension into the 0..2*PI range&#xA;if ( RA &amp;lt; 0. ) &#xA;  RA += PI + PI&#xA;Dec = asin(z/distance)&#xA;# Convert radians to hms and or degrees to taste&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This code was taken from Bill Gray's &lt;a href=&quot;https://github.com/Bill-Gray/sat_code&quot; rel=&quot;nofollow noreferrer&quot;&gt;sat_code&lt;/a&gt;, specifically &lt;code&gt;observe.cpp&lt;/code&gt;. Other bits of that code handle the conversion from a latitude, longitude, height (geodetic coordinates) to a X,Y,Z vector from the Earth center to the observing site (geocentric co-ordinates) and then the rotation by the Local Mean Sidereal Time (site longitude + GMST, where GMST is the Greenwich Mean Sidereal Time) and then the 2 vectors (Geocenter-&gt;Site and Geocenter-&gt;Sattelite) are added. However it looks &lt;code&gt;cEci&lt;/code&gt; already has this as I see references to the flattening factor and LMST.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing to note with TLEs is they are not a very good representation of the satellite motion/orbit over long periods of times. You should make sure your TLEs are up-to-date or use something like SGP4 to apply the perturbations and update the TLEs.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-07-26T20:27:38.863" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27104" PostTypeId="1" CreationDate="2018-07-26T21:04:23.710" Score="2" ViewCount="1100" Body="&lt;p&gt;My husband has told me that there are nights when there is no moon in the sky, because it has risen in the morning and set before dusk. I don’t believe him. Does this happen?&lt;/p&gt;&#xA;" OwnerUserId="23816" LastActivityDate="2018-08-26T04:03:21.680" Title="Moonless nights?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27105" PostTypeId="1" CreationDate="2018-07-26T21:45:50.633" Score="3" ViewCount="194" Body="&lt;p&gt;So I'm interested in space scientifically. I do have a firm belief that there could be intelligent civilizations in our universe. Just wanted to get that out of the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been spending the evenings in the garden due the weather. Past two days I've seen the same similar thing and I'm not sure what it is. I have observed many 'shooting stars', planets, satellites, and aircraft. It's just one solid bright white light. I actually checked it with binoculars I had in the kitchen an yeah all I can see is bright white light. It is much brighter and larger to my eye than satellites appear. It is about 22:00 when I see these it is pretty dark but I wouldn't say peak darkness. It moves steadily faster than large aircraft like commercial flights that I can see. It moves in a straight line I think and is fully visible but just as that bright light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could possibly go to lengths and get images of it if anyone wants I took video on both occasions. They are pretty clear and I tried adjusting stuff on the camera to improve the quality while filming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, would be interested in what it could be obviously. Responses will be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDITS:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, so the town I'm in is called Colchester, United Kingdom.&#xA;I was facing West, it came from above the cloud, was to the right side of Bootes. Carried on in a straight line across the whole sky travelling East until it was no longer visible. The light remained bright the entire time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, and 22:00, GMT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am uploading the videos and images, I will link to it.&lt;/p&gt;&#xA;" OwnerUserId="23817" LastEditorUserId="23817" LastEditDate="2018-07-27T08:56:05.503" LastActivityDate="2018-07-30T11:27:12.090" Title="What is it that I have been seeing in the sky?" Tags="&lt;identify-this-object&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27106" PostTypeId="1" AcceptedAnswerId="27111" CreationDate="2018-07-26T21:57:52.957" Score="4" ViewCount="142" Body="&lt;p&gt;I have always been fascinated to see that with so little knowledge, scientists have succeeded in discovering the planets of the solar system, and have succeeded in defining them by their masses as well as by their orbits, temperatures, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to try to trace this process myself but I do not know how to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My starting postulate is simple, I have at my disposal a simple telescope and DSLR camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal is to define the solar planets, ie to know their masses, their sizes, their orbits, their revolutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to achieve it? Do you have any leads?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT :&#xA;Thank you for your awesome answers and sorry for my late reply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that I have to separate this problem into 3 different points :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;earth's variable : I can find it by reproducing the Greek calculus  &lt;/li&gt;&#xA;&lt;li&gt;Sun's variable : with the two-body problem   &lt;/li&gt;&#xA;&lt;li&gt;planets' variable : with the Gauss' method &amp;amp; observation  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I keep you update about my advancements ;)&lt;/p&gt;&#xA;" OwnerUserId="23812" LastEditorUserId="2365" LastEditDate="2018-08-15T12:53:00.007" LastActivityDate="2018-08-15T12:53:00.007" Title="How to know the orbital and physical characteristics of the solar planets with only a non-professional telescope?" Tags="&lt;planet&gt;&lt;solar-system&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27107" PostTypeId="2" ParentId="27106" CreationDate="2018-07-26T22:31:00.447" Score="1" Body="&lt;p&gt;Some parts of this are easier than others!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By observing the positions of the planets among the stars from night to night over an extended period of time, &lt;em&gt;and then applying a lot of not especially complicated -- but&lt;/em&gt; quite &lt;em&gt;tedious -- calculation&lt;/em&gt; you can determine the planets' orbits around the Sun with good accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, with careful observation, you can measure the &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_diameter&quot; rel=&quot;nofollow noreferrer&quot;&gt;angular diameter&lt;/a&gt; of the planets as observed from Earth. Combining those angular measurements with the orbits you have determined lets you determine the planets' actual sizes. This will be quite accurate in some cases (e.g., Jupiter, where the error should be less than one percent), and much higher in others cases (e.g., Uranus). This is because Jupiter can be as big as 50 arc seconds, while Uranus is never bigger than just over two arc seconds.  Since the practical resolution of an amateur telescope is around a half arc second when the seeing is very good, your estimated diameters will be pretty rough for the ice giants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measuring the masses of the planets is much harder. The only way I can think of the do it is to continue to measure the planets' positions as accurately as possibly for as long as possible. Applying a lot of tedious arithmetic (&lt;em&gt;much&lt;/em&gt; better done on a computer) you can show that their orbits are affected by more than just the Sun's gravity and can make estimates of each planet's mass based on how their gravity perturbs the other planets.  Close, massive planets are easier to measure this way than distant, light planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotation period is easy for planets like Mars and Jupiter which have large disks with easily visible markings and &lt;em&gt;much&lt;/em&gt; harder for featureless planets like Venus and hard-to-observe ones like Mercury. (In fact, we really didn't know the rotation periods of either prior to sending probes. Some astronomers got it right, but they did not convince their colleagues.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Temperature is hard. I don't think you can measure it given your specified instruments, but you can certainly make a rough estimate using the known distance from the Sun.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-07-26T22:31:00.447" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27108" PostTypeId="1" AcceptedAnswerId="27109" CreationDate="2018-07-26T23:49:56.787" Score="4" ViewCount="57" Body="&lt;p&gt;Very novice astronomer here. I keep seeing terms like &quot;S-Complex&quot; thrown around when talking about asteroids, but no matter how much Googling I do I can't figure out what that exactly means (every article just writes about it assuming the reader knows the definition).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally I looked up a free textbook and read the section on asteroids. In &lt;a href=&quot;https://cnx.org/contents/LnN76Opl@14.4:chtGdUPl@6/Asteroids&quot; rel=&quot;nofollow noreferrer&quot;&gt;Composition and Classification&lt;/a&gt; it talks about there being three &quot;types&quot; of asteroids (another term I have seen thrown around): C-type, S-type, and M-type. (I've also seen them referred to as Type C, Type S, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So are the words &quot;type&quot; and &quot;complex&quot; just synonymous words to classify asteroids? If not, what does &quot;_ -Complex&quot; mean when talking about things like S-Complex asteroids?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="19852" LastActivityDate="2018-07-27T01:26:41.383" Title="Are asteroid complexes and types the same?" Tags="&lt;asteroids&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27109" PostTypeId="2" ParentId="27108" CreationDate="2018-07-27T01:26:41.383" Score="3" Body="&lt;p&gt;Yes they are basically the same. The original asteroid taxonomies were defined from photometry (brightness) in a few filters such as the Eight Color Asteroid Survey of Tholen (1984). This resulted in asteroids with similar slopes being grouped together into the main complexes or classes. Once we got more detailed spectra of asteroids subtle differences between sub-members of e.g. the S-complex showed up and so the large complex was broken into more sub-classes such as Sc, Sq etc.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-07-27T01:26:41.383" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27110" PostTypeId="2" ParentId="27104" CreationDate="2018-07-27T03:45:33.443" Score="4" Body="&lt;p&gt;The short answer is yes: there are hours of the night when the Moon is not visible. As others have indicated, the New Moon rises around the time of sunrise and sets around the time of sunset; therefore, the New Moon is not visible in the nighttime sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon takes approximately 27 days to move through the zodiac, so its eastward movement of 360 degrees over 27 days leads to the Moon rising approximately 50 minutes later each day. (That is an average; the details are more complex.) If it rises at 9 pm today, it will rise around 9:50 pm tomorrow, and so on. After enough days, it will be dark for many hours before the Moon rises.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, when you see the First Quarter Moon in the evening sky, it sets in the middle of the night. The remainder of the night is thus without a Moon.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2018-07-27T03:45:33.443" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27111" PostTypeId="2" ParentId="27106" CreationDate="2018-07-27T04:12:37.707" Score="3" Body="&lt;p&gt;This is a fascinating goal, and very difficult. It would make a book on recreating the history of science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It took many people spending lifetimes collecting data to get anywhere. However, you don't need to rediscover that the planets go around the Sun I would think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some key milestones that need to be reached. For example, you can distances to planet orbits in terms of ratios of a reference distance, like the Earth Sun distance, WITHOUT knowing the actual Earth-Sun distance. You only know that Mars' orbit is x times bigger than Earth-Sun orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first calculation of Earth-Moon distance was done by the Greeks. Aristarchus or Eratosthenes perhaps. One of them found the diameter of the Earth using shadows  of monuments and in wells at noon, and the other estimated the arc of the Earth by Its shadow on the Moon, and thus the size of the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think then it goes something like Newton measure the mass of the Earth by measuring g for a dropped object. Then using estimates of the Moon's distance and its orbital period he could determine the constant G in his law of universal gravitation. Then you can calculate the Earth-Sun distance and the rest of the planets fall into place based on their periods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the other challenge is collecting useful data for determining orbits of planets. There are a couple of ways to &quot;determine an orbit from observations&quot;. And this is a particularly useful project. Getting a mental picture of what to do can be difficult. But basically, you record the time and the position of a planet in some Earth centered coordinate system, like Right Ascension and Declination. Or relative to the know position of nearby stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the orbits are ellipses with the Sun at one focus, so the Earth based system is not much good. You convert all your observations to Sun centered coordinates in the form of lines or vectors that point to the planet. But they point to the Planet from where the Earth was when you took the measurements. These are a set of lines in space and they don't go through the Sun or anything like that. They are just from some point in space (where the Earth was) pointing to the Planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C.F. Gauss (I think Laplace improved on this - not sure) showed that you only need three observations to find an orbit because only one ellipse will have points on the three or more lines and have a focus at the Sun. Two points are enough if you have other information like is it getting closer or further and which way is it going (it might be in retrograde so you can't tell).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More measurements taken further apart will increase accuracy and refine the orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the distance to Jupiter or Saturn could be found by watching their moon's orbits and noticing how they stray from prediction with distance due to the speed of light at opposition versus being further away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you can see that the steps are not too daunting if you don't need to wait for a lunar eclipse and take a few other things as known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you could do the temperatures with a small reflecting telescope. It is by detecting the infra-red radiation from the planet. A reflector is used so as not to absorb the IR. The instruments used were a bolometer that responds to heat and a galvanometer. The galvanometer is an incredible precise and sensitive instrument with no electronics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Search on &quot;calculate orbits from observations&quot; and there are bunch of PDFs from good sources.&lt;/p&gt;&#xA;" OwnerUserId="5526" LastActivityDate="2018-07-27T04:12:37.707" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27112" PostTypeId="1" CreationDate="2018-07-27T05:01:57.743" Score="2" ViewCount="135" Body="&lt;p&gt;Approximately what percent of exoplanet gas giants discovered so far are in the habitable zones of their stars? I hear that most of them are &quot;Hot Jupiters&quot;, orbiting so close to their stars that the heat causes them to swell up like a hot air balloon. &lt;/p&gt;&#xA;" OwnerUserId="23652" LastEditorUserId="2153" LastEditDate="2018-07-31T14:25:46.287" LastActivityDate="2018-07-31T14:25:46.287" Title="What percentage of gas giant exoplanets are in the habitable zone?" Tags="&lt;exoplanet&gt;&lt;gas-giants&gt;&lt;habitable-zone&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27113" PostTypeId="1" CreationDate="2018-07-27T10:37:09.487" Score="6" ViewCount="1088" Body="&lt;p&gt;According to this article:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.theguardian.com/science/2018/jul/26/star-spotted-speeding-near-milky-way-black-hole-for-first-time&quot; rel=&quot;noreferrer&quot;&gt;https://www.theguardian.com/science/2018/jul/26/star-spotted-speeding-near-milky-way-black-hole-for-first-time&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the Very Large Telescope is &quot;powerful enough to see a tennis ball on the moon from Earth&quot;.  I guess that the moon is much to bright and moves too fast for the VLT to take good pictures of it, but it got me wondering: what is the most magnified picture of the moon ever taken from Earth?&lt;/p&gt;&#xA;" OwnerUserId="23822" LastEditorUserId="907" LastEditDate="2018-07-27T17:52:40.243" LastActivityDate="2018-07-27T17:52:40.243" Title="What is the highest resolution image of the moon taken from Earth's surface?" Tags="&lt;the-moon&gt;&lt;telescope&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27114" PostTypeId="2" ParentId="27112" CreationDate="2018-07-27T10:42:43.807" Score="3" Body="&lt;p&gt;Of the gas giants, the number of &quot;hot Jupiters&quot; is about the same as the number of &quot;hot Neptunes.&quot; A little under 9% of known gas giants are in the habitable zones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An &lt;a href=&quot;https://www.popularmechanics.com/space/deep-space/a13733860/all-the-exoplanets-weve-discovered-in-one-small-chart/&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; by John Wenz has the discoveries thus far collated into a chart:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AMFTl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AMFTl.jpg&quot; alt=&quot;All the Exoplanets We’ve Discovered in One Small Chart&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The claim, &quot;most of them are &quot;Hot Jupiters&quot;, orbiting so close to their stars that the heat causes them to swell up like a hot air balloon,&quot; is at best a bit of a stretch - the only reference that I found for this was a &lt;em&gt;Gizmodo&lt;/em&gt; article about Kelt-9b. The &lt;a href=&quot;https://arxiv.org/abs/1706.06723&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Nature&lt;/em&gt; paper&lt;/a&gt;, &lt;em&gt;A giant planet undergoing extreme ultraviolet irradiation by its hot massive-star host&lt;/em&gt;, that discusses Kelt-9b says, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The planet is also extremely inflated relative to theoretical models, with a radius of $\sim 1.9\ R_J$. Poor redistribution of&#xA;  heat and radius inflation (both of which are also observed in WASP-33b), have been linked to high stellar insolation, although the exact physical mechanisms remain uncertain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So they did observe that Kelt-9b had a larger radius than is normally expected from a planet of that mass, and it receives massive amounts of radiation from its star, but it cannot be concluded from that that it is the same as putting a balloon filled with air near a heat source.&lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2018-07-27T10:42:43.807" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27115" PostTypeId="1" AcceptedAnswerId="27116" CreationDate="2018-07-27T13:02:49.617" Score="1" ViewCount="106" Body="&lt;p&gt;Does Dark Matter affect Space Time Curvature? If so, then what is the formula to calculate space time curvature due to dark matter?&lt;/p&gt;&#xA;" OwnerUserId="23742" LastEditorUserId="2365" LastEditDate="2018-07-27T14:31:46.330" LastActivityDate="2018-07-27T18:32:43.807" Title="Dark Matter Curvature" Tags="&lt;space-time&gt;&lt;general-relativity&gt;&lt;dark-matter&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27116" PostTypeId="2" ParentId="27115" CreationDate="2018-07-27T14:51:15.693" Score="2" Body="&lt;p&gt;This answer comes with a small warning.  Actual dark matter has not been discovered, so this is theoretical, but a very well understood theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All matter, dark or light, affects the curvature of spacetime in the same way. The formula for the curvature is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Einstein_field_equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein Field Equation.&lt;/a&gt; The curvature is defined by the value of a 4x4 matrix, called the &quot;stress energy tensor&quot;. Which describes the density of energy and momentum in space-time. The formula doesn't care if the energy and momentum comes from regular matter or dark matter.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-07-27T14:51:15.693" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27117" PostTypeId="1" CreationDate="2018-07-27T15:36:55.050" Score="4" ViewCount="55" Body="&lt;p&gt;I am analyzing data from an instrument that measures low energy electron flux at Geo-Stationary orbit.  There are sharp oscillations in the particle flux of 30 eV electrons occurring from about noon to 8 pm (space-craft local).  The period of this activity is relatively consistent at about one cycle every 13 minutes.  I am trying to determine if this is a natural phenomenon or systemic contamination.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;  I have added a plot of data containing the oscillations.  The Y-axis represents counts per interval and X axis is time.  The purple line represents the south-most zone and the red the north-most.  I have intentionally removed the title/legend/labels.  The oscillations are obvious in the right side of the plot.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RVmBI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RVmBI.png&quot; alt=&quot;30 eV electron counts for all zones.  Purple and Red lines are South-most and North-most zones respectively.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are there any natural cycles within the solar-wind or radiation belts that have a similar frequency?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are there evenly spaced and distinct longitudes containing higher densities of electron cycling that the space-craft might be passing through?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other ideas as to the cause of the oscillation are welcome.&lt;/p&gt;&#xA;" OwnerUserId="23824" LastEditorUserId="23824" LastEditDate="2018-07-31T15:36:54.210" LastActivityDate="2018-07-31T15:36:54.210" Title="Cause of ~10 minute Oscillations in Low Energy Electron Flux in Earth's Magnetosphere" Tags="&lt;magnetic-field&gt;&lt;solar-wind&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27119" PostTypeId="2" ParentId="27112" CreationDate="2018-07-27T18:15:10.820" Score="3" Body="&lt;p&gt;I was curious about the figure given in &lt;a href=&quot;https://astronomy.stackexchange.com/a/27114/2153&quot;&gt;Mick's answer&lt;/a&gt; (~9%), so I did some data-crunching of my own. I looked at the &lt;a href=&quot;http://exoplanetarchive.ipac.caltech.edu&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA Exoplanet Archive&lt;/a&gt;, in particular the table of confirmed planets. I was interested in four parameters:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt;, the planet's semi-major axis&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$M\sin i$&lt;/span&gt;, the planet's minimum mass&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;, the temperature of the star&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt;, the radius of the star&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;From the final two, I could approximate the luminosity of the star by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stefan-Boltzmann law&lt;/a&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$L=4\pi\sigma R^2T^4$$&lt;/span&gt;&#xA;I downloaded the dataset (3774 exoplanets) including only those parameters, then filtered out all the planets that lacked all four (leaving 932). I then created lists of all four parameters. Going back through this list, I removed planets of less than &lt;span class=&quot;math-container&quot;&gt;$\sim0.05$&lt;/span&gt; Jupiter masses (i.e. Neptune-mass or less) and objects of more than &lt;span class=&quot;math-container&quot;&gt;$\sim20$&lt;/span&gt; Jupiter masses (which are really essentially brown dwarfs - the cutoff's fuzzy). This left me with 751 probable giant planets, including ice giants.&lt;/p&gt;&#xA;&lt;p&gt;I estimated the habitable zone by &lt;a href=&quot;https://en.wikipedia.org/wiki/Effective_temperature#Blackbody_Temperature&quot; rel=&quot;nofollow noreferrer&quot;&gt;using a black body approximation&lt;/a&gt; for the surface temperature of a planet:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$T_p=\left(\frac{L(1-a)}{16\pi\sigma d^2}\right)^{1/4}$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is albedo and &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; is the mean distance to the star. I set &lt;span class=&quot;math-container&quot;&gt;$a\approx0$&lt;/span&gt;, for simplicity. Then, for each planet, I calculated the orbital radii at which water would freeze and boil, and finally compared the measured semi-major axis to see if it fit inside the two. I ended up with 72 giant planets in the habitable zone - a total of 9.6% of the giant planets I considered, similar to the result given in Mick's answer.&lt;/p&gt;&#xA;&lt;p&gt;In summary, according to the NASA data I looked at, around 10% of giant planets lie in the habitable zone, although uncertainties in planetary and stellar parameters could change this.&lt;/p&gt;&#xA;&lt;h3&gt;Notes:&lt;/h3&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Increasing the minimum mass even up to &lt;span class=&quot;math-container&quot;&gt;$1M_J$&lt;/span&gt; changed my result by no more than 1% of the population; lowering it down to &lt;span class=&quot;math-container&quot;&gt;$0.01M_J$&lt;/span&gt; also changed it by only about 1% of the population.&lt;/li&gt;&#xA;&lt;li&gt;Varying the albedo from &lt;span class=&quot;math-container&quot;&gt;$0.1$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$0.9$&lt;/span&gt; changed the results by about 1.5% of the population, going in both directions.&lt;/li&gt;&#xA;&lt;li&gt;I assumed no greenhouse effect when calculating the effective surface temperature.&lt;/li&gt;&#xA;&lt;li&gt;Removing the lower mass restriction entirely (i.e. including terrestrial planets) brought the result down by about 1% of the population.&lt;/li&gt;&#xA;&lt;li&gt;I used Python after downloading the &lt;code&gt;.csv&lt;/code&gt; file from the Archive (&lt;a href=&quot;https://github.com/HDE226868/Gas-giant-distribution&quot; rel=&quot;nofollow noreferrer&quot;&gt;code available on GitHub&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-07-27T19:59:01.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27120" PostTypeId="2" ParentId="27115" CreationDate="2018-07-27T18:32:43.807" Score="3" Body="&lt;p&gt;Yes, the same &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_lens#Explanation_in_terms_of_space%E2%80%93time_curvature&quot; rel=&quot;nofollow noreferrer&quot;&gt;equation&lt;/a&gt; as is used for other matter.&lt;/p&gt;&#xA;&lt;p&gt;Marc Postman, Ray Villard and Donna Weaver of the &lt;a href=&quot;http://www.stsci.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Space Telescope Science Institute&lt;/a&gt; posted the results of a &lt;a href=&quot;http://hubblesite.org/images/news/44-clash&quot; rel=&quot;nofollow noreferrer&quot;&gt;CLASH study&lt;/a&gt;: &amp;quot;&lt;a href=&quot;http://hubblesite.org/image/2889/news_release/2011-25&quot; rel=&quot;nofollow noreferrer&quot;&gt;News Release 2011-25&lt;/a&gt;&amp;quot;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QNdOl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QNdOl.jpg&quot; alt=&quot;Galaxy Cluster MACS 1206&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;This image of galaxy cluster &lt;a href=&quot;https://translate.googleusercontent.com/translate_c?depth=1&amp;amp;nv=1&amp;amp;rurl=translate.google.com&amp;amp;sl=auto&amp;amp;sp=nmt4&amp;amp;tl=en&amp;amp;u=https://it.wikipedia.org/wiki/MACS_J1206.2-0847&amp;amp;xid=25657,15700022,15700122,15700124,15700149,15700168,15700186,15700191,15700201,15700208&amp;amp;usg=ALkJrhhQO6EpahOSD2p3iLaT6i-6Lu6QcA&quot; rel=&quot;nofollow noreferrer&quot;&gt;MACS J1206.2-0847&lt;/a&gt; (or MACS 1206 for short) is part of a broad survey with NASA's Hubble Space Telescope.&lt;/p&gt;&#xA;&lt;p&gt;The distorted shapes in the cluster are distant galaxies from which the light is bent by the gravitational pull of an invisible material called dark matter within the cluster of galaxies. This cluster is an early target in a survey that will allow astronomers to construct the most detailed dark matter maps of more galaxy clusters than ever before.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;On the NASA website: &amp;quot;&lt;a href=&quot;https://www.nasa.gov/image-feature/a-clash-of-clusters-provides-new-clue-to-dark-matter&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Clash of Clusters Provides New Clue to Dark Matter&lt;/a&gt;&amp;quot;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7lFF0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7lFF0.jpg&quot; alt=&quot;Formations of Dark Matter&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;... this composite image, made using data from the Hubble Space Telescope Chandra X-Ray Observatory, researchers have discovered clues to increase our knowledge of dark matter. This image is a powerful collision of clusters. These clusters show formations of dark matter, where it separated from the ordinary matter. The different color disparities, along with the interesting evidence for dark matter ...&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Another article with the same photo: &amp;quot;&lt;a href=&quot;https://www.spacetelescope.org/news/heic0818/&quot; rel=&quot;nofollow noreferrer&quot;&gt;heic0818 — Science Release -- Clash of clusters provides new dark matter clue&lt;/a&gt;&amp;quot;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;New Hubble and Chandra observations of the cluster known as MACSJ0025.4-1222 indicate that a titanic collision has separated dark from ordinary matter. This provides independent confirmation of a similar effect detected previously in a target dubbed the &lt;a href=&quot;https://en.wikipedia.org/wiki/Bullet_Cluster&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bullet Cluster&lt;/a&gt;, showing that the Bullet Cluster is not an anomalous case.&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Wikipedia webpage on CLASH - &lt;a href=&quot;https://en.wikipedia.org/wiki/Cluster_Lensing_and_Supernova_survey_with_Hubble&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cluster Lensing and Supernova survey with Hubble&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-07-27T18:32:43.807" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27121" PostTypeId="1" AcceptedAnswerId="27123" CreationDate="2018-07-27T19:36:17.710" Score="11" ViewCount="722" Body="&lt;p&gt;According to the &lt;a href=&quot;https://www.bbc.co.uk/news/science-environment-44945452&quot; rel=&quot;noreferrer&quot;&gt;BBC&lt;/a&gt; the lunar eclipse on July 27th/28th 2018 is the &quot;longest this century&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longest isn't a very good way of describing things, since it's a relative measure without reference to actual numbers, and implies it's somehow longer in some way that might actually matter in a practical way.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;How much longer is this lunar eclipse than:&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;An &quot;average&quot; lunar eclipse? &lt;/li&gt;&#xA;&lt;li&gt;The next lunar eclipse? &lt;/li&gt;&#xA;&lt;li&gt;The shortest lunar eclipse this century? &lt;/li&gt;&#xA;&lt;li&gt;The 2nd longest lunar eclipse this century?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(Feel free to add in any questions/answers I might have missed that may explain &quot;longest&quot; in a better context.)&lt;/p&gt;&#xA;" OwnerUserId="23827" LastEditorUserId="7472" LastEditDate="2018-07-28T16:37:58.507" LastActivityDate="2018-07-28T16:37:58.507" Title="How much longer is this (July 27th 2018) lunar eclipse than others?" Tags="&lt;lunar-eclipse&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27122" PostTypeId="2" ParentId="27121" CreationDate="2018-07-27T20:12:54.480" Score="9" Body="&lt;p&gt;The NASA &lt;a href=&quot;https://eclipse.gsfc.nasa.gov/LEcat5/LEcatalog.html&quot; rel=&quot;noreferrer&quot;&gt;NASA Five Millenium Catalog of Lunar Eclipse&lt;/a&gt; says that the longest total lunar eclipse was 106.6 minutes or 1 hr 46.6 minutes. The link for &lt;a href=&quot;https://eclipse.gsfc.nasa.gov/LEcat5/LE2001-2100.html&quot; rel=&quot;noreferrer&quot;&gt;this century&lt;/a&gt; shows that the July 27th eclipse is indeed the longest at 1hr 43 minutes (rounding up). So it is definitely up towards the maximum length at ~96% of the longest known. Two times can be given for eclipse duration; the full duration measured from first to fourth contacts i.e. when the shadow of the Earth just touches the edge (limb) of the Moon to when the shadow just leaves the Moon on the other side. The second duration is that of totality from 2nd to 3rd contacts during which time the Earth's shadow completely covers the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; Corrected eclipse timings discussion&lt;/p&gt;&#xA;" OwnerUserId="23748" LastEditorUserId="23748" LastEditDate="2018-07-27T20:32:06.830" LastActivityDate="2018-07-27T20:32:06.830" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27123" PostTypeId="2" ParentId="27121" CreationDate="2018-07-27T20:24:40.363" Score="16" Body="&lt;p&gt;The NASA has a &lt;a href=&quot;https://eclipse.gsfc.nasa.gov/eclipse.html&quot; rel=&quot;noreferrer&quot;&gt;lot of data&lt;/a&gt; on eclipses. I've pulled the &lt;a href=&quot;https://eclipse.gsfc.nasa.gov/LEcat5/LE2001-2100.html&quot; rel=&quot;noreferrer&quot;&gt;table for the 21st century&lt;/a&gt; in Excel. There are 228 eclipses of which 85 are total. The durations below are for the &lt;em&gt;totality phase&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;duration of today's eclipse&lt;/strong&gt; is 103 minutes.&lt;br&gt;&#xA;The &lt;strong&gt;average duration&lt;/strong&gt; is 68.7 minutes.&lt;br&gt;&#xA;The &lt;strong&gt;duration of the next total eclipse&lt;/strong&gt;, January 21st, 2019 is 62 minutes.&lt;br&gt;&#xA;The &lt;strong&gt;shortest eclipse this century&lt;/strong&gt;, April 4th, 2015, is 4.7 minutes.&lt;br&gt;&#xA;The &lt;strong&gt;second longest eclipse this century&lt;/strong&gt;, June 26th, 2029, is 101.9 minutes.&lt;br&gt;&#xA;The &lt;strong&gt;median duration&lt;/strong&gt;, as requested by @FooBar is 71.7 minutes (slightly more than the average).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;For what it's worth, here in the Netherlands it's probably also the &lt;em&gt;warmest&lt;/em&gt; lunar eclipse ever; at sunset, when the moon rose, it was 31&amp;deg;C.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="7472" LastEditorUserId="7472" LastEditDate="2018-07-28T13:27:00.300" LastActivityDate="2018-07-28T13:27:00.300" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27124" PostTypeId="2" ParentId="27105" CreationDate="2018-07-27T20:54:16.890" Score="5" Body="&lt;p&gt;The International Space Station can appear as bright as Venus, typically crosses the sky in about 6 minutes, and passes directly over the UK several evenings this week.&#xA;Heavens Above provides a &lt;a href=&quot;https://www.heavens-above.com/PassSummary.aspx?satid=25544&quot; rel=&quot;nofollow noreferrer&quot;&gt;table of ISS passes&lt;/a&gt; visible from a given location.&#xA;Their &lt;a href=&quot;https://www.heavens-above.com/passdetails.aspx?lat=51.8959&amp;amp;lng=0.8919&amp;amp;loc=Colchester&amp;amp;alt=8&amp;amp;tz=GMT&amp;amp;satid=25544&amp;amp;mjd=58324.9287112233&amp;amp;type=V&quot; rel=&quot;nofollow noreferrer&quot;&gt;chart of a July 25 pass&lt;/a&gt; is consistent with your revised description.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-07-28T02:00:56.547" LastActivityDate="2018-07-28T02:00:56.547" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27125" PostTypeId="1" AcceptedAnswerId="27126" CreationDate="2018-07-27T20:54:50.493" Score="4" ViewCount="172" Body="&lt;p&gt;I have theoretical question.&#xA;Let's assume that it is full moon and two different humans are looking at it at the same time. One human is for example in London and one in Beijing. It doesn't really matter where exactly, important is that they are almost on opposite side of Earth but they both see the moon at the same time.&#xA;I assume they see the moon from different angle. Is it possible that one of them see different part of the moon because of this angle. It could be just the border, tiny part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you please tell me, how big is this tiny part of the moon which one see and the other don't? Few meters? Kilometers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks everyone for help.&lt;/p&gt;&#xA;" OwnerUserId="23829" LastActivityDate="2018-07-27T21:23:58.840" Title="Different full moon face on different places on Earth same time" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27126" PostTypeId="2" ParentId="27125" CreationDate="2018-07-27T21:23:58.840" Score="9" Body="&lt;p&gt;Yes, they do see a slightly different face of the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to calculate how different. If they're on nearly opposite sides of the Earth so that one sees the Moon just rising and the other sees the Moon just setting, they are seeing it from about 1.9 degrees apart. (A triangle with a base of 8000 miles (the distance they are apart) with sides of 240,000 miles (the distance to the Moon) yields an angle of 1.9 degrees (8/240 radians).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So they'll be viewing the Moon from 1.9 degrees apart and will see a full moon displaced by 1.9 degrees, or about 36 miles on the surface. So one will see 36 miles of Moon more on the Lunar east side than the  other and the other will see about 32 miles more on the Lunar west side.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-07-27T21:23:58.840" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27128" PostTypeId="1" CreationDate="2018-07-27T23:21:27.013" Score="1" ViewCount="52" Body="&lt;p&gt;Sometimes when I look up at the sky at night, I can see something moving. It’s so small that I barely can see it, but I can’t see it when I look directly at it. I need to look beside it to see it. It slowly moves across the sky, but it switches sometimes directions. It sometimes disappears and after a while, it’s back. It’s not a strong light at all so it’s easy to miss.&lt;/p&gt;&#xA;" OwnerUserId="23832" LastActivityDate="2018-07-27T23:21:27.013" Title="Almost unseeable light moving over nightsky. What is it?" Tags="&lt;star&gt;&lt;light&gt;&lt;movement&gt;" AnswerCount="0" CommentCount="6" ClosedDate="2018-09-18T07:00:56.000" ContentLicense="CC BY-SA 4.0" />
  <row Id="27129" PostTypeId="1" AcceptedAnswerId="27130" CreationDate="2018-07-28T03:23:54.547" Score="4" ViewCount="103" Body="&lt;p&gt;I was just wondering if it had anything to do with the fact that last year we were blessed with the opportunity to see the total solar eclipse in our neck of the woods. So this is the universe's way of making things 'even'. Or is it pure chance and luck?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jokes aside, there is probably a physics explanation regarding tidal forces or angular momentum, so I am curious how these factors affect which parts of the world get to witness the event. &lt;/p&gt;&#xA;" OwnerUserId="23833" LastActivityDate="2018-07-28T06:35:24.317" Title="Why was the July 27th 2018 lunar eclipse not visible in North America?" Tags="&lt;the-moon&gt;&lt;lunar-eclipse&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27130" PostTypeId="2" ParentId="27129" CreationDate="2018-07-28T06:35:24.317" Score="3" Body="&lt;p&gt;Nothing so romantic as the universe being fair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The path of the moon is tilted by about 5 degrees from the path of the sun. but twice a month the moon must cross the path of the sun.  When this moment coincides exactly with full moon there is a lunar eclipse. Since the sun is on the opposite side of the Earth, lunar eclipses are only visible during the night, and &lt;em&gt;this one happened to occur during the day for the USA&lt;/em&gt;. When it is day in the USA it is night in China, and the eclipse was visible from the other side of the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, these things aren't random, but there is no simple formula like &quot;if you see a solar eclipse, then the next lunar eclipse won't be visible.&quot;  You can just go to any of the eclipse prediction websites to find when the next one will be visible in your area.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-07-28T06:35:24.317" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27131" PostTypeId="2" ParentId="897" CreationDate="2018-07-28T07:38:01.833" Score="1" Body="&lt;p&gt;This is an old question but I thought it good to add an update since more information has become available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ceres &lt;a href=&quot;https://www.universetoday.com/134922/confirmed-ceres-transient-atmosphere/&quot; rel=&quot;nofollow noreferrer&quot;&gt;has indeed been confirmed by Dawn&lt;/a&gt; to have a tenuous atmosphere created by ice sublimating off the surface.&lt;/p&gt;&#xA;" OwnerUserId="14969" LastActivityDate="2018-07-28T07:38:01.833" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27132" PostTypeId="2" ParentId="12646" CreationDate="2018-07-28T08:03:54.000" Score="2" Body="&lt;p&gt;A neutron star will form if you have roughly 1.4-3 solar masses of matter which is not producing enough energy to hold itself up through radiation pressure. So in principle you could assemble 2 solar masses of iron and it would collapse under its own gravity and form a neutron star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the vast majority of the universe is mixed hydrogen and helium, and if you assemble 2 solar masses of those materials, fusion will start well before it collapses and you get a normal star. So it seems likely that every natural neutron star in the real universe will have been through a nuclear burning phase, and so be a &quot;stellar&quot; neutron star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is just conceivable that a density fluctation in the very early universe might have been just the right size that the matter there cooled to neutronium rather than just a denser cloud of normal matter, but it would have been surrounded by fairly dense normal matter, so would probably grow into a black hole. &lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="21928" LastEditDate="2018-07-29T08:14:27.270" LastActivityDate="2018-07-29T08:14:27.270" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27134" PostTypeId="1" AcceptedAnswerId="27146" CreationDate="2018-07-28T19:08:26.863" Score="4" ViewCount="199" Body="&lt;p&gt;NASA has a page dedicated to lunar eclipses and provides &lt;a href=&quot;https://eclipse.gsfc.nasa.gov/LEplot/LEplot2001/LE2018Jul27T.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;charts&lt;/a&gt;. How are these charts to be read? What do P1, P4, U1,... mean? What about the world map below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SOBLp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SOBLpm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click for full size view&lt;/p&gt;&#xA;" OwnerUserId="9527" LastEditorUserId="7982" LastEditDate="2018-07-31T02:37:51.890" LastActivityDate="2018-07-31T02:37:51.890" Title="How to read an eclipse chart?" Tags="&lt;lunar-eclipse&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27135" PostTypeId="2" ParentId="27134" CreationDate="2018-07-28T19:58:40.490" Score="5" Body="&lt;p&gt;'P' is the penumbra, the zone where the Earth blocks some, but not all, of the Sun's light from reaching the Moon. 'U' is the umbra, the zone where the Earth blocks all of the light from the Sun (see &lt;a href=&quot;http://www.mreclipse.com/Special/LEprimer.html&quot; rel=&quot;noreferrer&quot;&gt;this primer&lt;/a&gt;). As the Moon moves across the two zones then you will get various times as the Moon touches those zones. So P1 is where the Moon just starts to touch the penumbral shadow, U1 is where the Moon just touches the umbral shadow cone, U2 is where all the Moon is inside the umbral shadow and the total eclipse begins. Similarly U4 and P4 are where the Moon last touches the umbral and penumbral shadows. Note that the times in the chart are in Universal Time so you need to account for your timezones' difference from UTC.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-07-28T19:58:40.490" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27136" PostTypeId="1" CreationDate="2018-07-29T02:57:40.357" Score="1" ViewCount="196" Body="&lt;p&gt;I have uploaded a h alpha photo in SAO Image DS9, but want to locate a particular region for which I have the galactic coordinates. How to input the coordinates in DS9 to view the particular zone?&lt;/p&gt;&#xA;" OwnerUserId="23843" LastEditorUserId="7472" LastEditDate="2018-07-29T09:41:41.500" LastActivityDate="2018-07-29T09:41:41.500" Title="Galactic coordinates input in DS9" Tags="&lt;ds9&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27137" PostTypeId="1" CreationDate="2018-07-29T03:01:40.717" Score="1" ViewCount="156" Body="&lt;p&gt;How to calculate the approximate angular (how many degrees across?) from an image in DS9 rendering? &lt;/p&gt;&#xA;" OwnerUserId="23843" LastEditorUserId="7472" LastEditDate="2018-07-29T09:41:58.063" LastActivityDate="2018-07-29T09:41:58.063" Title="How to calculate angular size in ds9?" Tags="&lt;ds9&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27138" PostTypeId="1" CreationDate="2018-07-29T08:10:39.597" Score="2" ViewCount="791" Body="&lt;p&gt;My Model 36050 telescope came with the H6mm, H20mm and the green filtered moon lenses. It has a 360mm focal length and a 50mm diamiter.  What can be seen with it?&lt;/p&gt;&#xA;" OwnerUserId="23845" LastActivityDate="2018-07-30T18:36:47.743" Title="I have a Model 36050 telescope" Tags="&lt;planet&gt;&lt;telescope&gt;&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27139" PostTypeId="1" CreationDate="2018-07-29T14:45:35.987" Score="1" ViewCount="166" Body="&lt;p&gt;The value of the surface brightness is in what unit? I hope it is in Rayleighs. Is that true?&lt;/p&gt;&#xA;" OwnerUserId="23843" LastEditorUserId="7472" LastEditDate="2018-07-29T20:02:58.397" LastActivityDate="2018-07-29T22:19:38.727" Title="What unit is used for region Surface brightness in DS9?" Tags="&lt;light&gt;&lt;ds9&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27140" PostTypeId="2" ParentId="27105" CreationDate="2018-07-29T16:41:32.197" Score="2" Body="&lt;p&gt;Very likely the ISS. The time it takes to cross the sky varies depending on the height above the horizon, up to a maximum of about 6 minutes. Nasa provides a site called  &lt;a href=&quot;http://spotthestation.nasa.gov&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Spot The Station&quot;&lt;/a&gt; where you can sign up to get a notification email whenever an ISS pass is visible from your location.&lt;/p&gt;&#xA;" OwnerUserId="23848" LastActivityDate="2018-07-29T16:41:32.197" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27141" PostTypeId="1" CreationDate="2018-07-29T17:21:51.283" Score="0" ViewCount="130" Body="&lt;p&gt;I have a question relating to the hot temperature this summer. There are 3 reasons that I can come up with:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Chemical, greenhouse effect.&lt;/li&gt;&#xA;&lt;li&gt;Increased solar activity.&lt;/li&gt;&#xA;&lt;li&gt;Orbital Resonance. The Earth is closer to the Sun, due to gravitational pull from other planets.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Anybody know something about Orbital Resonance? The orbits are affected by one another as they line up? Particularly the larger ones like Jupiter, Neptune and Saturn?&lt;/p&gt;&#xA;" OwnerUserId="23849" LastEditorUserId="7472" LastEditDate="2018-07-29T20:02:20.520" LastActivityDate="2018-07-31T15:06:17.910" Title="What might be the cause of the hot summer on Earth in 2018?" Tags="&lt;orbit&gt;" AnswerCount="3" CommentCount="5" ClosedDate="2018-07-31T15:50:15.443" ContentLicense="CC BY-SA 4.0" />
  <row Id="27142" PostTypeId="1" AcceptedAnswerId="27152" CreationDate="2018-07-29T17:31:06.577" Score="1" ViewCount="706" Body="&lt;p&gt;How did this 'black fabric' that we call space start? How was it made? How was the initial singularity formed? How did the first atom come into existence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it all started from a tiny particle that expanded and is now still expanding, how did that tiny particle come into existence ?&lt;/p&gt;&#xA;" OwnerUserId="19788" LastEditorDisplayName="user1569" LastEditDate="2018-07-31T07:54:00.173" LastActivityDate="2018-07-31T08:49:03.127" Title="How did space start?" Tags="&lt;space&gt;&lt;big-bang-theory&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27143" PostTypeId="1" CreationDate="2018-07-29T18:30:16.037" Score="2" ViewCount="124" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kjIy9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kjIy9.png&quot; alt=&quot;Image of light curve of an eclipsing binary with description&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This light curve seems really weird to me, I'm used to seeing light curves where the barycenter is a straight line. It either stays at a place, moves at a constant speed or slowly increases its speed. But here it looks like the barycenter periodically changes its speed. Why is this happening? Can someone please explain the details of this graph.&lt;/p&gt;&#xA;" OwnerUserId="16506" LastEditorUserId="7982" LastEditDate="2021-04-16T23:24:32.823" LastActivityDate="2021-04-16T23:24:32.823" Title="Sinusoidal Light curve of an eclipsing binary" Tags="&lt;light-curve&gt;&lt;time-domain-astronomy&gt;&lt;eclipsing-star-systems&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2021-04-21T20:56:10.477" ContentLicense="CC BY-SA 4.0" />
  <row Id="27144" PostTypeId="1" AcceptedAnswerId="27154" CreationDate="2018-07-29T19:09:00.907" Score="3" ViewCount="182" Body="&lt;p&gt;Looking at &lt;a href=&quot;https://www.nasa.gov/image-feature/jpl/swirling-jovian-storm-natural-color&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jupiter image &lt;/a&gt;  the atmosphere edge of planet is very sharp as if it has solid surface which is not as it’s gases giant and there should be some thinning of atmosphere like we see around earth . Or there and could be seen on higher resolution images?&lt;/p&gt;&#xA;" OwnerUserId="16136" LastEditorUserId="16136" LastEditDate="2018-07-30T03:51:56.033" LastActivityDate="2018-07-30T15:37:16.457" Title="Why does Jupiter atmosphere does not get thinner on edges" Tags="&lt;planet&gt;&lt;solar-system&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27146" PostTypeId="2" ParentId="27134" CreationDate="2018-07-29T20:12:45.453" Score="4" Body="&lt;p&gt;astrosnapper already explained the chart and the Us and Ps, so I'm going to focus on the map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The map shows you which phase of the eclipse is visible from where. For example Mauritius gets to see the whole eclipse while North America is out of luck and doesn't get to see any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my vantage point in Germany I should have been able to see from somewhere between U1 (first contact with core shadow) and U2 (start of totality) till the end.&#xA;I cannot see P1, P2 and U1 because the moon is still below the horizon from my point of view.&#xA;In the end, some low hanging clouds got in the way of my seeing U2 but I got a very good view of U3 (end of totality).&lt;/p&gt;&#xA;" OwnerUserId="6822" LastActivityDate="2018-07-29T20:12:45.453" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27147" PostTypeId="2" ParentId="27141" CreationDate="2018-07-29T20:18:03.440" Score="2" Body="&lt;p&gt;Seen globally, the Earth is 0.75—1.0 degrees warmer than the average. This is what is modelled by warming due to raised levels of greenhouse gases.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently the sun is close to a solar minimum. Perturbations of the Earth's orbit by other planets do not significantly change the level of solar insolation. Also the ENSO is currently close to neutral. The hot temperatures across western Europe and East Asia are a combination of the effects of greenhouse gases, and the random influence of weather.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-07-29T20:18:03.440" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27149" PostTypeId="2" ParentId="27139" CreationDate="2018-07-29T22:19:38.727" Score="4" Body="&lt;p&gt;ds9 is a common image analysis tool used in astronomy (&lt;a href=&quot;http://ds9.si.edu/site/Home.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ds9.si.edu/site/Home.html&lt;/a&gt;). &#xA;The units of surface brightness are simply detector counts per square arcsecond as that is all that can easily be deduced from an image. If you want to turn that into more physical units, you will need to know the gain of the CCD (how many photons produce each count) and you will need to make some assumptions about the wavelength of light that is producing the surface brightness of the emission &lt;a href=&quot;http://www.astronomy.ohio-state.edu/~pogge/Ast871/Notes/Rayleighs.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Note on Rayleighs may help&lt;/a&gt;; if all of this emission is coming from Halpha, then this is a reasonable assumption.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-07-29T22:19:38.727" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27150" PostTypeId="2" ParentId="27142" CreationDate="2018-07-29T22:41:48.833" Score="2" Body="&lt;p&gt;Your confusion comes from the words &quot;singularity&quot; and perhaps &quot;atom&quot; and &quot;particle&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An equation like y=1/x is singular when x=0. For a physical object modelled by the equation y=1/x the model can't be applied to understand the object at x=0, and it may or may not make sense to use the model for negative values of x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The singularity is not an object. It would be better if people said &quot;the equations that model the universe become singular at a point in time. The models that we have for the universe don't work at time  t = 0, and it is not clear if negative values of &quot;t&quot; are physically meaningful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The singularity &lt;em&gt;was not a tiny particle that expanded&lt;/em&gt;. Since the singularity is not an object, it doesn't make much sense to ask &quot;how did it form&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atoms are another thing altogether.  We know with great detail how protons and electrons formed. How most of the protons that formed were annihilated by antiprotons, but a few remained, and as the universe cooled, electrons bound with the protons to form atoms.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-07-29T22:41:48.833" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27151" PostTypeId="1" CreationDate="2018-07-30T00:34:32.283" Score="4" ViewCount="631" Body="&lt;p&gt;I was reading up on &lt;a href=&quot;https://en.wikipedia.org/wiki/Bo%C3%B6tes_void&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bootes void&lt;/a&gt; and came across &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_largest_cosmic_structures#List_of_largest_voids&quot; rel=&quot;nofollow noreferrer&quot;&gt;this list&lt;/a&gt; of the largest voids in the visible universe, and apparently &lt;a href=&quot;https://en.wikipedia.org/wiki/KBC_Void&quot; rel=&quot;nofollow noreferrer&quot;&gt;the one we're in, the KBC void&lt;/a&gt;, is not only the biggest, but damned-near spherical.  The Wikipedia page says that the Milky Way is within a few hundred million light-years of the center of the KBC void, but I'm unable to find out what, if anything, is closest to the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this information available anywhere?  Does anyone know what's at the center of the KBC void?&lt;/p&gt;&#xA;" OwnerUserId="23852" LastEditorUserId="2153" LastEditDate="2018-07-30T04:12:31.463" LastActivityDate="2018-07-30T04:12:31.463" Title="Which galaxy is closest to the center of the KBC void?" Tags="&lt;galaxy&gt;&lt;galaxy-cluster&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27152" PostTypeId="2" ParentId="27142" CreationDate="2018-07-30T03:43:55.410" Score="4" Body="&lt;p&gt;I'd like to add onto what James K said, but I'll try and simplify things too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A singularity may sound like everything was at a &quot;single point&quot;, but that's not what the word means in physics/math. A singularity, loosely speaking, is just a point where a function reaches infinity. It's not a physical object, but an input to a function. As James mentioned, 1/x has a singularity at x=0. At the moment of the Big Bang, several functions have singularities, as I discuss below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So with that set aside, let's explain the Big Bang. Our modern theory of gravity, general relativity, indicates that the universe is currently expanding (i.e. space is being created between all matter). By that logic, we expect the matter in the universe to be closer together as we look further back in time. Our physical theories indicate that at earlier periods, temperatures were so high that even atoms couldn't form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our theories can handle these conditions pretty well, but eventually they start to fall apart. At a point around 13.8 billion years in the past, our calculations predict that the amount of space in the Universe should be zero. As a result, density, gravity and temperature would skyrocket to infinity — in other words, they have singularities at that moment. This moment is known as the Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These infinities are a problem, since our theories break down at such extreme conditions. The Big Bang is an unsolved problem that modern physics has yet to explain. It's not the beginning of the universe per se, but rather we do not know what happened around, or before, that moment.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2018-07-30T03:43:55.410" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27154" PostTypeId="2" ParentId="27144" CreationDate="2018-07-30T06:42:04.380" Score="3" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Why does Jupiter's atmosphere not get thinner at the edges?&lt;/p&gt;&#xA;&lt;p&gt;Looking at an image of the atmosphere's edge it is very sharp as if it has a solid surface, which it has not since it’s a gas giant. There should be some thinning of the atmosphere like we see around Earth. Would this be seen on higher resolution images?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;When you look at a spherical object from a distance it appears round, sometimes when you view it from close up it doesn't appear quite so smooth (or round). The wavelength that you look at an object with also affects what you are able to see.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wXx5T.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wXx5T.jpg&quot; alt=&quot;High resolution SEM image of CN ball bearing&quot; /&gt;&lt;/a&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\scriptsize \text{Figure 6. High resolution SEM image of CN ball bearing inner ring fabricated from 52100 steel showing} \\ \scriptsize \text{fine globular cementite. Presence of fine precipitates of carbides and nitrides uniformly aggregated in the} \\ \scriptsize \text{fine martensitic microstructure, which act as obstacles to the motion of dislocations.}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The above scanning electron microscope image of a smoothly ground ball bearing illustrates this, from a relatively far distance it appears smooth; from much closer the particles constituting its makeup become visible. Image Source: &amp;quot;&lt;a href=&quot;http://file.scirp.org/Html/3-1180152_34218.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Effect of Carbonitriding on Endurance Life of Ball Bearing Produced from SAE 52100 Bearing Steels&lt;/a&gt;&amp;quot;, by Karthikeyan Rajan, Vinod Joshi, and Arindam Ghosh.&lt;/p&gt;&#xA;&lt;p&gt;A similar principle applies to viewing Jupiter from so far away. While its clouds are &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Atmosphere_of_Jupiter#Vertical_structure&quot; rel=&quot;nofollow noreferrer&quot;&gt;far greater than 2000 km thick&lt;/a&gt; most of the gases are opaque thus not allowing you to see very far into the atmosphere using visible light.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yJzCb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yJzCb.jpg&quot; alt=&quot;Images of Jupiter&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The upper left image shows Jupiter from a distance using visible light. Source: &amp;quot;&lt;a href=&quot;https://www.missionjuno.swri.edu/junocam/processing?id=5150&quot; rel=&quot;nofollow noreferrer&quot;&gt;PJ14 Image 26 - Hot Spot (Enhanced)&lt;/a&gt;&amp;quot;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The upper right image  shows the different layers of Jupiter's atmosphere. Source: &amp;quot;&lt;a href=&quot;https://www.missionjuno.swri.edu/jupiter/atmosphere?show=hs_jupiter_atmosphere_story_how-deep-is-the-atmosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;How deep is the atmosphere?&lt;/a&gt;&amp;quot;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The lower left image from NASA's Juno spacecraft approaching Jupiter on Aug. 27, 2016, the Jovian Infrared Auroral Mapper (JIRAM) instrument captured the planet's glow in infrared light. Source: &amp;quot;&lt;a href=&quot;https://www.jpl.nasa.gov/spaceimages/details.php?id=PIA21036&quot; rel=&quot;nofollow noreferrer&quot;&gt;Juno Captures Jupiter's Glow in Infrared Light&lt;/a&gt;&amp;quot;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The last two images are from: &amp;quot;&lt;a href=&quot;https://www.jpl.nasa.gov/news/news.php?feature=7096&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA's Juno Mission Provides Infrared Tour of Jupiter's North Pole&lt;/a&gt;&amp;quot;, there is also a &lt;strong&gt;close-up video flythrough of the atmosphere&lt;/strong&gt; created from infrared imaging. A second video on that webpage, titled: &amp;quot;Jupiter's Dynamo&amp;quot;, shows magnetometer imaging overlaid on visible spectrum imaging. The YouTube links for those videos are: &amp;quot;&lt;a href=&quot;https://www.youtube.com/watch?v=By6sZ6RGCEQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;Low 3-D Flyover of Jupiter’s North Pole in Infrared&lt;/a&gt;&amp;quot; and &amp;quot;&lt;a href=&quot;https://www.youtube.com/watch?v=LPvfeOiKbm8&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jupiter's Dynamo&lt;/a&gt;&amp;quot;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-07-30T06:42:04.380" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27156" PostTypeId="2" ParentId="27142" CreationDate="2018-07-30T08:31:13.160" Score="0" Body="&lt;p&gt;I guess here there is an assumption that there was a singularity, where as Sir Circumference points out, both relativity and quantum mechanics break down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Einstein even considered that the universe goes through eternal oscillations of expansion, crunch and bounce before Tolman showed that entropy can only increase, thereby resulting in progressively slower expansion. Which means through extrapolation there must have been some sort of beginning.  Entropy is the problem in any cyclical model of the universe that involves matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recent theorists such as Paul Steinhardt (Princeton) and Neil Turok (Cambridge) have proposed another theory of cyclical epochs where there is a limit on the net expansion phase brought about by colliding branes (membranes) and dark energy limiting entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative theory is the Baum-Frampton Cyclic Model which also proposes cyclical phases in regard to expansion, turnaround, contraction and bounce.  Expansion results in the fragmentation of all matter into patches, or separate universes.  Just before the end of time, at the turnaround, the patches individually contract and then bounce to infinitely repeat the cycle.  Entropy becomes a non-player as the theory contends that contraction of each universe results in an empty state, that is, only dark energy and no matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steinhardt–Turok model&#xA;&lt;a href=&quot;https://arxiv.org/abs/hep-th/0111030&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/hep-th/0111030&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Baum-Frampton model&#xA;&lt;a href=&quot;http://Baum-Frampton%20Cyclic%20Model&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/hep-th/0703162&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23153" LastEditorUserId="23153" LastEditDate="2018-07-31T08:49:03.127" LastActivityDate="2018-07-31T08:49:03.127" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27157" PostTypeId="2" ParentId="27141" CreationDate="2018-07-30T09:26:40.577" Score="2" Body="&lt;p&gt;Short answer:   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Local seasonal weather variation is statistical noise, driven in part by ocean currents, but year to year, it's generally considered natural variation or statistical noise.   Greenhouse gas seems to play a role in greater variation, so that's probably part of it (as are Ocean currents), but statistical noise/natural variation is almost certainly the bigger reason why a single season would be unusually hot or cold or wet or dry in a particular location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's kind of an astronomy question with two of your suggestions, but the astronomy part is mostly irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://data.giss.nasa.gov/gistemp/maps/&quot; rel=&quot;nofollow noreferrer&quot;&gt;The NASA June 2018 Temperature anomaly map&lt;/a&gt; shows more than 4 degrees C above normal in parts of Russia/maybe Northern China and parts of Antarctica.   July 2018 isn't out yet but you can check that link in a couple weeks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precise reasons for local warm spells is a little complicated and more Earth Science than astronomy, Earth has warmed, as noted in the other answer about 0.75-1.00 degrees C, but that's global average.   Over oceans the number is smaller and over land it's considerably larger than 1 degree and some regions are warming faster than others.  The greenhouse effect may also play a role in greater temperature variation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Increased solar activity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As noted in the other answer, solar activity is actually down this year, but the sun is very consistent.  The variation in radiative energy during an active solar cycle and a quiet one is about 0.1%.  Not enough to make a significant difference in local temperature on Earth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Orbital Resonance. The Earth is closer to the Sun, due to&#xA;  gravitational pull from other planet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Orbital resonance has probably an even smaller effect year to year.  It's true that if Earth is between the Sun and Jupiter, there will be a slight tidal effect that pulls Earth slightly further away from the sun, like a tidal bulge, The overall effect is pretty tiny on a year to year basis and I'm tempted to say virtually irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Anybody know something about Orbital Resonance? The orbits are&#xA;  affected by one another as they line up? Particularly the larger ones&#xA;  like Jupiter, Neptune and Saturn?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a good astrology question, though I'm not sure it's specific enough to be a good question here.  You're really asking about orbital perturbations from other planets.   (&lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_resonance&quot; rel=&quot;nofollow noreferrer&quot;&gt;orbital resonance&lt;/a&gt; is something else).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Math behind &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_perturbation_analysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;orbital perturbation analysis&lt;/a&gt; gets pretty intense.  Here's a &lt;a href=&quot;https://astronomy.stackexchange.com/questions/26793/how-does-the-gravity-of-one-planet-affect-the-velocity-of-other-planets?rq=1&quot;&gt;question where it's answered somewhat on a single orbit&lt;/a&gt;.   Over tens of thousands or hundreds of thousands of Earth orbits, &lt;a href=&quot;https://en.wikipedia.org/wiki/Milankovitch_cycles#Orbital_shape_(eccentricity)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth's eccentricity wobbles&lt;/a&gt;, mostly influenced by Jupiter and Saturn, and that eccentricity variation is thought to play a key role in the 100,000 year ice age cycle.  But over single orbits the effect is tiny and not likely to affect seasonal weather.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is, ofcourse, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Butterfly_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;butterfly flaps its wings theory&lt;/a&gt;, in which case, small changes like how much Jupiter moves Earth can have an effect in the sense that small input variations can lead to large changes down the line, but that's a local effect.   Butterflies don't put more energy into the system, nor does every butterfly cause a hurricane, but they may be able to move them around, at least, that's what the non-repeating mathematical analysis tells us, so according to the butterfly effect theory, Jupiter might have an effect, but not a predictable or repeating one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that said, statistical noise is probably the real answer to your question.  perhaps enhanced by greenhouse and affected by El Nino or La Nina, and for the most part, Jupiter and the Sun or not really factors, but it just might be that pesky butterfly in Africa.   ;-)&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-07-30T09:26:40.577" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27158" PostTypeId="2" ParentId="27138" CreationDate="2018-07-30T10:17:24.120" Score="3" Body="&lt;p&gt;With this scope you will be able to see quite a few things. Unfortunately, you won't be able to resolve much detail with some objects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 50mm aperture, the light gathering ability is not fantastic. That being said, some brighter objects will be easily seen. Your max power (zoom) is approximately 2*aperture, and the power is your objective focal length (360) / eyepiece focal length. Say you have a 6mm eyepiece in, your power (zoom) will be 360/6 = 60&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PLANETS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury and Venus will be easily seen with your telescope, at the highest power, you may be able to resolve some of the phases of Venus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars will be visible as a reddish round blob, with hazy edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter will be visible along with the 4 Galilean moons (Io, Ganymeade, Europa, Callisto), although cloud bands will be very hard to resolve. Perhaps if you are at a dark sky site with perfect seeing, you may be able to make out a faint cloud band, but you will likely be looking at a large bright dot, with 4 smaller ones near it. A handy app for your phone called 'Moons of Jupiter' will allow you to tell which ones you are looking at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn will be a nice sight as you will be able to see the rings. This is always a spectacular sight for newcomers to astronomy. Although it will be rather small (even at maximum power), it will still be a nice sight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uranus and Neptune will likely be too dim for you to see. In a dark sky site in great conditions, you may be able to make out a tiny dim dot for each, but that is unlikely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DSO's (Deep Sky Objects)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For DSO's, your best bet is to go for clusters and nebula at around mag +6.5-+7 max. Any dimmer than that and you will struggle. This means you will be able to see such globular clusters as the great cluster in Hercules, the Andromeda galaxy and the Orion nebula (these 3 were picked as they are the brightest in their categories) although they will look like just fain smudges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stars&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the brighter Carbon stars are a nice sight as they are often pleasing shades of red, which distinguish them from background stars. Binary stars are also a popular sight. There are a lot of good lists of binary stars online, although you will want to go for ones with a large separation between them, so that you will be able to see both. Mizar and Alcor are a good starting point for binary stars, these are easy to locate in the handle of the 'Big Dipper' in Ursa Major. If you want a bit more from your binary stars, you could look for some colour contrasting binaries, such as Albiero in Cygnus, which is a lovely binary star with one blue and one gold star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some open star clusters may be nice to go for, such as the Beehive cluster in Cancer, or the Pleiades in Taurus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best thing to do is to download a few apps, get yourself a star chart and get to a local dark sky sight and see what you can see. There are hundreds of objects you will be able to see. You could also look for a local Astronomy group based in your location for further guidance and also a chance to look through some other telescopes too!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh and let us not forget the Moon! That is always a beautiful sight!&lt;/p&gt;&#xA;" OwnerUserId="17096" LastEditorUserId="616" LastEditDate="2018-07-30T18:36:47.743" LastActivityDate="2018-07-30T18:36:47.743" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27159" PostTypeId="2" ParentId="27105" CreationDate="2018-07-30T11:27:12.090" Score="2" Body="&lt;p&gt;The ISS can be very bright. I think that may be what you were looking at. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am in South England, so would see a very similar sight to you. That week the ISS was passing W to E, and it would have appeared to come from the 'right' of Bootes if facing W. I observed it myself that week. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, my astronomy group chat shows that on the night of the 25th we observed the ISS pass at 22.30ish and again at 00.05ish. Judging by the date of your post and mentioning you had seen it the 'past few days' I am 99% certain that your observations match my own. You were looking at the ISS. &lt;/p&gt;&#xA;" OwnerUserId="17096" LastActivityDate="2018-07-30T11:27:12.090" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27160" PostTypeId="1" CreationDate="2018-07-30T11:47:36.300" Score="1" ViewCount="160" Body="&lt;p&gt;When I read about the universe having always expanded from a 'singularity', I can't help but wonder if that is perhaps not the case. If a singularity is a point at which everything is infinite and is, theoretically, impossible, why are we quite so convinced it existed? What evidence is there that the universe didn't start out as a single square centimetre or even perhaps half the size it is now, because weve only known it's been expanding for the last century or so because of technological advances? Perhaps it was sizeable to begin with but the pressure of more atoms forming has pushed it outwards? Probably a ridiculous theory but Im intrigued to hear what others think about the whole subject because I just don't think the current presumption is based on any real evidence and is likely to be totally wrong.&lt;/p&gt;&#xA;" OwnerUserId="23857" LastActivityDate="2018-08-01T01:37:01.440" Title="Why do we presume that the universe has always expanded?" Tags="&lt;universe&gt;&lt;early-universe&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27161" PostTypeId="1" AcceptedAnswerId="41197" CreationDate="2018-07-30T13:33:21.523" Score="7" ViewCount="362" Body="&lt;p&gt;The article &lt;a href=&quot;https://www.rmg.co.uk/discover/behind-the-scenes/blog/first-light&quot; rel=&quot;nofollow noreferrer&quot;&gt;First Light: a new era for the Royal Observatory&lt;/a&gt; says (in part):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;25 June 2018&lt;/p&gt;&#xA;&lt;p&gt;The first modern, research-grade telescopes have just been installed at the Royal Observatory Greenwich, making the institution a working observatory once again after 60 years. Marek Kukula, Brendan Owens and Tom Kerss explain what this will mean for them, researchers, and the public.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;&lt;em&gt;Modern equipment has allowed us to bypass a lot of the disadvantages that drove astronomy away from Greenwich in the middle of the twentieth century.&amp;quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;How exactly will &amp;quot;modern equipment&amp;quot; do this? A blocking filter for the bright yellow sodium line of some streetlights might have some effect, but with high pressure mercury lights and now broadband LED illumination, what other tricks do Greenwich astronomers have up their sleeves?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BamuK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BamuK.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;AMAT’s dual role is aptly matched to its name-sake. Annie Maunder explored the night sky and its phenomena as an astronomer, astrophotographer and science communicator in Greenwich during the 1890s and early 20th Century. She was hired as a human computer at the Royal Observatory in 1891, and though, as a government employee, she was required to give up her position when she married fellow Greenwich astronomer Edward Walter Maunder, she continued to participate in science, publishing books to inform and inspire others and returning to work at the Observatory during the First World War. She was also one of the first women to be elected as a Fellow of the Royal Astronomical Society in 1916. By the end of her life she had gained an international reputation, particularly for her work examining the Sun during eclipses for which she designed her own specialised cameras.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2021-02-03T08:08:27.267" Title="How will &quot;modern equipment&quot; allow the Royal Observatory to now avoid some of the effects of light pollution at Greenwich?" Tags="&lt;observatory&gt;&lt;light-pollution&gt;&lt;atmospheric-effects&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27162" PostTypeId="2" ParentId="27161" CreationDate="2018-07-30T13:52:55.530" Score="4" Body="&lt;p&gt;It's a little hard to guess what research can be done there -- Today, Greenwich is close to the worst place on Earth to have an observatory: At sea level, in the light dome of one of the largest cities in the world, in a country famous for its clouds and fogs.&lt;/p&gt;&#xA;&lt;p&gt;The Wikipedia article on the observatory mentions the new telescope and says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In 2018 the Annie Maunder Astrographic Telescope (AMAT) was installed at the ROG in Greenwich. AMAT is a cluster of four separate instruments, to be used for astronomical research; it had achieved first light by June 2018:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A 14-inch reflector can take high resolution images of the sun, moon and planets.&lt;/li&gt;&#xA;&lt;li&gt;An instrument dedicated to observing the sun.&lt;/li&gt;&#xA;&lt;li&gt;An instrument with interchangeable filters to view distant nebulae at different optical wavelengths.&lt;/li&gt;&#xA;&lt;li&gt;A general-purpose telescope.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;At a guess, they hope to do some sort of star position measurements which is one sort of astronomy that is less affected by haze and light pollution (though it needs steady air -- I have no idea what the air steadiness is like there). But even that is still better done with larger instruments in better locations, and by satellites such as Gaia.&lt;/p&gt;&#xA;&lt;p&gt;It really sounds like a public outreach facility or an undergraduate training site more than a research facility.  It's also worth noting that the pictures show equipment that is at the serious amateur level -- I certainly know amateurs who have better equipment.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-07-30T13:52:55.530" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27163" PostTypeId="1" CreationDate="2018-07-30T15:15:23.353" Score="1" ViewCount="56" Body="&lt;p&gt;Suppose somebody went on an island and don't have sense about time, dates, months. The only thing he has is seeing the sky. Is it possible to recognise which lunar month it is by just looking the sky? I ask this because I know how to see beginning of the lunar month, but how to recognise which month is it? So we can look at the Sun, Moon, the stars. Is there a way to figure that out? &lt;/p&gt;&#xA;" OwnerUserId="23859" LastActivityDate="2018-07-30T17:51:50.610" Title="Recognising lunar month only by looking the sky" Tags="&lt;the-sun&gt;&lt;the-moon&gt;&lt;star-systems&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27164" PostTypeId="2" ParentId="27144" CreationDate="2018-07-30T15:24:26.277" Score="2" Body="&lt;p&gt;There certainly is thinning. As the other answer points out, Jupiter is really big, so just about anything else is dwarfed by it's scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scale height can be though of as the distance that the atmospheric density drops by a factor of 1/e, so a few scale heights can be though of as the &quot;thickness&quot; of the atmosphere loosely speaking. It's a loose concept because both temperature and composition can be changing with altitude, it's a good first approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Scale_height#Planetary_examples&quot; rel=&quot;nofollow noreferrer&quot;&gt;scale heights&lt;/a&gt; of Jupiter and Saturn are in fact much longer than that of Earth's, Wikipedia lists them at about 27 and 60 kilometers, versus Earth's 8 (all rough values). Scale height can be estimated with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H = \frac{kT}{Mg}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where M is a measure of the masses of the body's atmosphere's constituent molecules, and $g$ is the surface gravity (acceleration) of the body in question.  While Jupiter's surface gravity is higher than that of Earth (about 25 m/s^2 versus 9.8 m/s^2), the atmosphere is mostly hydrogen as H2, which per molecule or per mole is fourteen times lighter than Earth's mostly nitrogen as N2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the definition of the &quot;edge&quot; of Jupiter, which is somewhat arbitrary, you can read @MarkAdler's &lt;a href=&quot;https://space.stackexchange.com/a/17056/12102&quot;&gt;excellent answer&lt;/a&gt; to the Space SE question &lt;em&gt;&quot;What defines the radius of a ball of gas like Jupiter?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For a visualization of &lt;em&gt;something similar&lt;/em&gt; the ill-defined edge of &lt;em&gt;Saturn's&lt;/em&gt; atmosphere (with a scale height double that of Jupiter's) I can steer you to the image in the question &lt;a href=&quot;https://astronomy.stackexchange.com/q/16703/7982&quot;&gt;Refraction by Saturn's atmosphere - how dense is it here?&lt;/a&gt; There are additional links there, but I'l also shown it below, where you can see the appearance of Saturn's rings bend because they are being viewed through its atmosphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can imagine, by the way that the refraction increases closer to the planet that the atmosphere is becoming thicker. There are geometrical effects as well, so this is not a &quot;proof&quot;, but only a visual aid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1Q4rY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1Q4rY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; cropped portion of the NASA's Cassini image &lt;a href=&quot;http://photojournal.jpl.nasa.gov/catalog/PIA20491&quot; rel=&quot;nofollow noreferrer&quot;&gt;from here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://twitter.com/CassiniSaturn/status/757599817316966400&quot; rel=&quot;nofollow noreferrer&quot;&gt;This tweet&lt;/a&gt; from CassiniSaturn shows an image of Saturn's rings being optically distorted by the dense atmosphere. It is explained more fully &lt;a href=&quot;https://saturn.jpl.nasa.gov/resources/7394/&quot; rel=&quot;nofollow noreferrer&quot;&gt;on this page&lt;/a&gt; and is available &lt;a href=&quot;http://photojournal.jpl.nasa.gov/catalog/PIA20491&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-07-30T15:37:16.457" LastActivityDate="2018-07-30T15:37:16.457" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27165" PostTypeId="1" AcceptedAnswerId="27185" CreationDate="2018-07-30T16:45:29.487" Score="6" ViewCount="641" Body="&lt;p&gt;The above emission line, as I understand, is a useful probe of early star-forming galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I do not understand what the ']' to the right of the 'CIII' means. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could not find any online reference that can help me decipher this and other spectral line names (for example what does the double lambda in 'O III] λλ1661' mean. Any good reference is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="23860" LastActivityDate="2018-08-01T01:57:47.540" Title="What does the ']' in the spectral line &quot;CIII] 1909 Å&quot; mean?" Tags="&lt;galaxy&gt;&lt;spectroscopy&gt;&lt;star-formation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27166" PostTypeId="2" ParentId="27165" CreationDate="2018-07-30T17:08:04.960" Score="0" Body="&lt;p&gt;&quot;O III]&quot; looks a lot like a typo for [O III], a notation which is &lt;a href=&quot;https://en.wikipedia.org/wiki/Doubly_ionized_oxygen&quot; rel=&quot;nofollow noreferrer&quot;&gt;widely used&lt;/a&gt;.  A [a recent paper in ApJ] uses both &quot;[O III]&quot; and &quot;O III]&quot; and I don't see a pattern.  I can't believe that journal standards have slipped &lt;em&gt;that&lt;/em&gt; badly, though, so I'm not certain it's a typo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've never run across the &quot;λλ&quot; notation before, but in &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0004-637X/719/2/1168/pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;that recent paper in ApJ&lt;/a&gt; I notice that &quot;λλ&quot; always precedes a pair of wavelengths (usualy a doublet) while a single λ proceeds single wavelengths. It appears it is either a symbol for &quot;doublet&quot; or the plural of λ&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-07-30T17:08:04.960" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27167" PostTypeId="2" ParentId="27163" CreationDate="2018-07-30T17:51:50.610" Score="1" Body="&lt;p&gt;Not easily. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lunar Islamic calendar is based on a strict succession of months, with each month based only on the position of the moon, and no intercalated months or months extended to 31 days in order to keep the cycle of months aligned with the solar year. The start of the months are determined by the moon, the names of the months are not defined by any astronomical feature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to determine which Islamic month it is you need to know the actual amount of time that has passed. Now the positions of the planets change day by day, and by careful observation of the planets, and given a detailed ephemeris of their positions, one could work out the exact date, and therefore the Islamic month. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is similar to determining the day of the week by astronomical observations, since the start of the day is based on the motion of the sun, but the name of the day is not defined on any astronomical feature.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-07-30T17:51:50.610" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27168" PostTypeId="1" AcceptedAnswerId="27169" CreationDate="2018-07-30T18:24:34.720" Score="1" ViewCount="156" Body="&lt;p&gt;Our solar system is moving at approx. 52,000 mph, so where is the Alpha Centauri system, which is assumed to be also moving through interstellar space at a similar velocity since it is part of our Galaxy. Is the A. Centauri system moving in front of us at 12 o’clock, parallel to us at 3 o’clock, or some other configuration?&lt;/p&gt;&#xA;" OwnerUserId="23862" LastEditorUserId="14094" LastEditDate="2018-07-31T07:54:06.643" LastActivityDate="2018-07-31T07:54:06.643" Title="Our Solar System’s interstellar movement in relation to the Alpha Centauri System" Tags="&lt;movement&gt;&lt;interstellar&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27169" PostTypeId="2" ParentId="27168" CreationDate="2018-07-30T19:23:06.993" Score="1" Body="&lt;p&gt;If you were looking down on the galaxy from above with 12 o'clock pointing from the Sun to the Galactic center, then using &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=alpha%20cen&amp;amp;NbIdent=1&amp;amp;Radius=2&amp;amp;Radius.unit=arcmin&amp;amp;submit=submit%20id&quot; rel=&quot;nofollow noreferrer&quot;&gt;SIMBAD&lt;/a&gt; we can get the galactic co-ordinates as approx (315,-0.6) in longitude, latitude respectively. Longitude is measured anti-clockwise from the line from the Sun to the galactic center, so Alpha Cen would be approximately between 1 and 2 o'clock. ESO have a map &lt;a href=&quot;https://www.eso.org/public/images/eso0303c/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; showing the nearby stars within 12.5 light years.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-07-30T19:23:06.993" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27170" PostTypeId="2" ParentId="13733" CreationDate="2018-07-30T22:01:04.320" Score="3" Body="&lt;p&gt;While the original post inquires specifically about gravitational lensing, I'd like to suggest that this is not the only criteria for &quot;seeing&quot; a black hole. Clearly, since no light can escape from a black hole, the actual hole itself cannot be seen. However, there are a variety of epiphenomena that can be observed, and gravitational influence on other celestial bodies (e.g., stars) can certainly be observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The European Southern Observatory (ESO) released a video of the center of the Milky Way galaxy made from 20 years of observations of this area by the &lt;a href=&quot;https://www.eso.org/public/videos/eso1825e/&quot; rel=&quot;nofollow noreferrer&quot;&gt;NACO instrument on ESO's Very Large Telescope in Chile&lt;/a&gt;. This seems to me like extremely compelling evidence that an enormous black hole exists at the center of the Milky Way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for gravitational lensing, &lt;a href=&quot;https://www.space.com/8830-massive-black-hole-bends-light-magnify-distant-galaxy.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Space.com reported just such an image attributed to Courbin, Djorgovski, &amp;amp; Meylan&lt;/a&gt;. This image may satisfy the exact criteria requested by the OP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT/UPDATE: This morning (April 10, 2019), the &lt;a href=&quot;https://eventhorizontelescope.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Event Horizons Telescope&lt;/a&gt; held press releases on four continents &lt;a href=&quot;https://aasnova.org/2019/04/10/first-images-of-a-black-hole-from-the-event-horizon-telescope/&quot; rel=&quot;nofollow noreferrer&quot;&gt;announcing&lt;/a&gt; that they have indeed imaged M87*, a black hole of an estimated 6.5 billion solar masses in the Messier 87 galaxy, some 55 million light years away. Strictly speaking, we still cannot see the black hole itself but rather its epiphenomena such as the accretion disk and the turbulence caused by the presence of the black hole. This image, created by &lt;a href=&quot;https://eventhorizontelescope.org/infographics&quot; rel=&quot;nofollow noreferrer&quot;&gt;very long baseline radio interferometry&lt;/a&gt;, is not a photograph in the traditional sense, but rather a false-color image composite generated by converting enormous volumes of radio wave observations with a supercomputer into an image visible to the human eye.&lt;a href=&quot;https://i.stack.imgur.com/jKQ24.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jKQ24.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16705" LastEditorUserId="16705" LastEditDate="2019-04-10T18:09:58.783" LastActivityDate="2019-04-10T18:09:58.783" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27171" PostTypeId="2" ParentId="27160" CreationDate="2018-07-30T23:00:11.013" Score="4" Body="&lt;p&gt;We can, after making some approximations, trace the movements and locations of the matter in the universe back in time, using the laws of physics (mainly general relativity). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment we see a certain density of matter which is, on a large enough scale, expanding uniformly -- as if space were being blown up so that every bit of it doubles in size every seven billion or so years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we extrapolate that back we find that long ago the matter density was much higher, but it was still expanding in a similar kind of way, and this would seem to hold true back to a state of very high density indeed, at which point our existing understanding of the physical laws stops giving us an extrapolation that makes any kind of sense and we have to stop.  Up to that point though, the extrapolation gives us a picture which is pretty consistent with all the various ways we have of observing it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as someone has said, the universe might have been created moments ago with everything, including our memories set up just so, but this doesn't seem a very useful line of speculation. Instead we say that, based on the physics we understand, and everything we can see and measure in the universe, it looks like its been expanding ever since it was incredibly hot and dense about 13.8 billion years ago. &lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="21928" LastEditDate="2018-07-31T06:49:21.010" LastActivityDate="2018-07-31T06:49:21.010" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27172" PostTypeId="2" ParentId="27160" CreationDate="2018-07-31T05:24:05.110" Score="1" Body="&lt;p&gt;The problem with a singularity Big Bang theory is that both relativity and quantum mechanics break down as you extrapolate back to the singularity event.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other theoretical models propose that the universe goes through stages of expansion and contraction, that is no Big Bang nor singularity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main issue that cyclical theories must address is that the universe tends toward high entropy as per the second law of thermodynamics.  This means that the rate of expansion progressively slows down as energy is depleted resulting in the Heat Death, Big Rip, Big Crunch etc depending on your leaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two models of cyclical theories are the Steinhardt–Turok model and the Baum-Frampton Cyclic model.  Both models address the issue of entropy in different ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steinhardt–Turok model &lt;a href=&quot;https://arxiv.org/abs/hep-th/0111030&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/hep-th/0111030&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Baum-Frampton model &lt;a href=&quot;https://arxiv.org/abs/hep-th/0703162&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/hep-th/0703162&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23153" LastEditorUserId="23153" LastEditDate="2018-08-01T01:37:01.440" LastActivityDate="2018-08-01T01:37:01.440" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27174" PostTypeId="1" AcceptedAnswerId="27176" CreationDate="2018-07-31T09:33:26.680" Score="10" ViewCount="1278" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ssSp5.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ssSp5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture was taken with a smartphone camera at Latitude 48.860045 | Longitude: 2.366202 the 31th of July at 01h07min15s Paris time. On the left is obviously the Moon, and the lower red circle points the point of Latitude: 48.852886 | Longitude: 2.369167. (It's the angel of liberty, place de la Bastille, Paris.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does the upper circle point to ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the beginning I thought it was Venus, but as far as I remember, it disappears quite early in the evening. I don't think it is Mars : the light reflection on it isn't red enough.&lt;/p&gt;&#xA;" OwnerUserId="23868" LastEditorUserId="23868" LastEditDate="2018-07-31T09:59:53.667" LastActivityDate="2018-07-31T22:09:22.537" Title="Which planet or star is this?" Tags="&lt;amateur-observing&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27175" PostTypeId="2" ParentId="27174" CreationDate="2018-07-31T12:21:03.420" Score="2" Body="&lt;p&gt;At Paris, Jupiter sets at 12:51 am and Saturn sets at 3:37 am.  Both are bright and both set in the southwest.  At 1:05 AM, Saturn would have been at 15 degrees elevation in the SSW.  The Moon would have been to the SW at about the same altitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it looks like this was Saturn.  (Someone with a full-up planetarium program could confirm.)&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-07-31T12:21:03.420" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27176" PostTypeId="2" ParentId="27174" CreationDate="2018-07-31T13:25:42.253" Score="12" Body="&lt;p&gt;I think it is Mars.  Although it's difficult to tell, just looking at the image, but if it were Saturn then you would also be able to see Mars.  Mars is much brighter than Saturn at the moment, so if Saturn were visible, Mars would be even brighter.  The fact that there is only one bright object (other than the moon) is, I think, conclusive evidence that this must be Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking over the city, the large brick building on the Rue St Gille is visible, and the object is visible directly in line with the phone mast on that building. That gives a direction just East of South. Checking with Stellarium at the given time shows Mars at exactly that position, at an azimuth of 171. &lt;/p&gt;&#xA;" OwnerUserId="11822" LastEditorUserId="11822" LastEditDate="2018-07-31T22:09:22.537" LastActivityDate="2018-07-31T22:09:22.537" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="27177" PostTypeId="2" ParentId="27141" CreationDate="2018-07-31T14:55:26.737" Score="0" Body="&lt;p&gt;The orbital calculations is way above my head . Since the question is stated &quot;summer&quot; I guess its locally considering the northern globe experiencing summer, this time of year . I live in the northern gulf stream area (norway) and this summer has been exceptionally hot . Ive heared that the Earth is closer to the Sun in December, but due to the Earth tilt, its winter in the Northern globe at that time. That tilt must however contribute to just an increase of distance from the Sun, to make a snowy winter. Or a temprature differance of + 30C summer to - 10C winter ( differance of 40C) . And I thought the &quot;thugging&quot; of planets might do the same. Orbital Resonance might be negligible, and might as well be a butterfly . Maybe the heat just has been a lack of clouds ! And the Sun has had free access to the surface ? &lt;/p&gt;&#xA;" OwnerUserId="23849" LastEditorUserId="23849" LastEditDate="2018-07-31T15:06:17.910" LastActivityDate="2018-07-31T15:06:17.910" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27178" PostTypeId="1" CreationDate="2018-07-31T16:01:25.757" Score="2" ViewCount="84" Body="&lt;p&gt;While messing around in &lt;em&gt;&lt;a href=&quot;https://dan-ball.jp/en/javagame/planet/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://dan-ball.jp/en/javagame/planet/&lt;/a&gt;&lt;/em&gt;. I came across a pretty stable orbit of 2 bodies around a star&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://imgur.com/NpxHJBw.gif&quot; alt=&quot;gif&quot; title=&quot;weird orbit&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbit yellow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://imgur.com/ZW8bqSH.png&quot; alt=&quot;yellow&quot; title=&quot;yellow orbit&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbit Blue:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://imgur.com/wv1NA7M.png&quot; alt=&quot;blue&quot; title=&quot;blue orbit&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;full album: &lt;a href=&quot;https://imgur.com/a/ruyenmY&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://imgur.com/a/ruyenmY&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a know orbit type? if so what is the maximum distance between the inner circular orbit and outer circular orbit?&lt;/p&gt;&#xA;" OwnerUserId="23870" LastActivityDate="2018-07-31T16:01:25.757" Title="Weird 2 body swapping orbit" Tags="&lt;orbit&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27179" PostTypeId="2" ParentId="27174" CreationDate="2018-07-31T16:12:12.463" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hxgt6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hxgt6.png&quot; alt=&quot;![enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a &lt;a href=&quot;http://stellarium.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellarium&lt;/a&gt; simulation of the Paris sky at the time you indicate. Mars and the Moon are at about the same altitude, as in your photo, while Saturn is further to the west, a little higher, and two or three magnitudes dimmer.&lt;/p&gt;&#xA;" OwnerUserId="1498" LastActivityDate="2018-07-31T16:12:12.463" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27180" PostTypeId="1" AcceptedAnswerId="27182" CreationDate="2018-07-31T16:30:49.927" Score="30" ViewCount="6314" Body="&lt;p&gt;As I see &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_population#Population_I_stars&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;, the Sun belongs to the Population I group of stars, which is the 3rd generation of the stars in our universe. 1st generation stars are Population III, 2nd generation are Population II, and 3rd generation are Population I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the 1st generation (Population III) of stars died, that means most of the hydrogen was burned to helium. Stars die when there is no hydrogen left. Later, the 2nd generation of stars (Population II) appeared and they fuse another portion of hydrogen into heavier elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 1st and 2nd star generations burned hydrogen to helium and more heavier elements, then shouldn't like 90% of all universe hydrogen already be converted to helium and something else? If yes, then there should not be enough hydrogen to make the Sun.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;UPDATE 1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Thanks for all your answers. They are very useful. Now a new subquestion appeared.&#xA;When the star dies, like our Sun, it sends out external layers and core becomes white/other dwarf. In this case, new star can be formed only from the hydrogen from the external layer. The questions what is the percentage of initial star hydrogen after burning it to helium goes from this external layer to outer space?&lt;/p&gt;&#xA;" OwnerUserId="15004" LastEditorUserId="15004" LastEditDate="2018-08-02T11:40:58.007" LastActivityDate="2018-08-02T11:40:58.007" Title="Where did the Sun get hydrogen to work with if it is in the 3rd generation of stars?" Tags="&lt;the-sun&gt;&lt;stellar-evolution&gt;&lt;hydrogen&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27181" PostTypeId="2" ParentId="27180" CreationDate="2018-07-31T16:59:56.783" Score="21" Body="&lt;p&gt;I think you've answered your own question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;if 1st and 2nd stars generation burned hydrogen to helium and more heavier elements, then should it be like 90% of all universe hydrogen already converted to helium and something else? If yes, then there should not be enough hydrogen to make the Sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clearly the Sun does have enough Hydrogen to form and the universe is not 90% Helium and heavier elements (it is in fact &lt;a href=&quot;https://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements&quot; rel=&quot;noreferrer&quot;&gt;~74% Hydrogen, ~24% Helium, and a fraction of heavier elements&lt;/a&gt;). That means the first and second generations of stars have &lt;em&gt;not&lt;/em&gt; burned up most of the Hydrogen and your basic assumptions are wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your main incorrect assumption comes from the statement&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[A] star dies when there is no hydrogen left.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A more correct statement would be &quot;A star dies when there is no hydrogen left in its core&quot;&lt;sup&gt;1&lt;/sup&gt;. Once the core runs out of hydrogen for fusing, it generally can't support the gravitational pressure trying to compact it and it begins the stages of death. However, the outer shell around the core, which can make up 50-70% of a star's mass, never gets fused, thus remaining Hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; Technically its more complicated than that, and the notion of a when a star &quot;dies&quot; isn't well demarcated. But that's another question for another day.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2018-07-31T19:46:24.637" LastActivityDate="2018-07-31T19:46:24.637" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27182" PostTypeId="2" ParentId="27180" CreationDate="2018-07-31T17:01:15.997" Score="29" Body="&lt;p&gt;Most of the galaxy's gas is not incorporated into stars and remains as gas and dust. This is not really my area of expertise, but papers such as &lt;a href=&quot;http://adsabs.harvard.edu/abs/2009ApJS..181..321E&quot; rel=&quot;noreferrer&quot;&gt;Evans et al. 2008&lt;/a&gt; and &lt;a href=&quot;https://arxiv.org/pdf/1805.12039.pdf&quot; rel=&quot;noreferrer&quot;&gt;Matthews et al. 2018&lt;/a&gt; seem to suggest that in the Giant Molecular Clouds where most stars in the Milky Way Galaxy form, the star formation efficiency is about 3-6%. So the vast majority of the gas (94-97%) is not made into stars. In very dense environments such as globular clusters, which were formed much earlier in the Milky Way's history, the star formation efficiency get as high as approx. 30%. The canonical quoted rate for &quot;regular&quot; spiral galaxies like the Milky Way is about 1 solar mass of new stars are made per year, which is very low summed across the whole galaxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars also give off a fair amount of their outer, hydrogen rich outer layers during the later red giant phases when the stellar wind is stronger and the atmosphere expands a huge amount (radius of the Sun during the red giant phase will be about what the Earth's orbit is now). Also in the end state when the white dwarf is formed, it's only the core and inner layers that form the white dwarf. The typical white dwarf mass is about 0.6 times the mass of the Sun (&lt;a href=&quot;https://arxiv.org/abs/astro-ph/0612277&quot; rel=&quot;noreferrer&quot;&gt;S. Kepler et al. 2006&lt;/a&gt;) and so there will be a fair amount of unfused hydrogen-rich outer atmosphere left over after the star dies. For higher mass stars, even more of the mass goes into the (ejected at high speed) envelope than goes into the remaining neutron star. These high mass stars are much rarer though; most of the Milky Way's stars are faint, cool M dwarfs.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastEditorUserId="5350" LastEditDate="2018-07-31T20:57:43.633" LastActivityDate="2018-07-31T20:57:43.633" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27183" PostTypeId="1" CreationDate="2018-07-31T17:03:38.780" Score="1" ViewCount="34" Body="&lt;p&gt;Another trouble in SAOImae DS9: with the Rotation Measure catalogue I have just downloaded: “Rotation measure image of the sky (Taylor+2009)”, even if I have set up the WCS to galactic before the catalogue loadind, I obtain the table file with another WCS (and the option to chane WCSto another one is inactivated); then I can not obtain from it the galactic longitude and altitude needed for my report..&lt;/p&gt;&#xA;" OwnerUserId="23843" LastActivityDate="2018-07-31T17:03:38.780" Title="help needed with inactivated option for wcs in SAO Image DS9" Tags="&lt;ds9&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27185" PostTypeId="2" ParentId="27165" CreationDate="2018-08-01T01:57:47.540" Score="6" Body="&lt;p&gt;Under conditions of very high temperature and very low pressure &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Forbidden_mechanism#In_astrophysics_and_atomic_physics&quot; rel=&quot;noreferrer&quot;&gt;forbidden&lt;/a&gt;&lt;/em&gt; conditions can occur, they shift the color of the excitation radiation from the ultraviolet into the visible spectrum giving nebula their beautiful colors. These are represented using brackets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Wikipedia's Selection Rules, section &lt;a href=&quot;https://en.wikipedia.org/wiki/Selection_rule#Angular_momentum&quot; rel=&quot;noreferrer&quot;&gt;Angular Momentum&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Semi-forbidden transitions (resulting in so-called intercombination lines) are electric dipole (E1) transitions for which the selection rule that the spin does not change is violated. This is a result of the failure of &lt;a href=&quot;https://en.wikipedia.org/wiki/LS_coupling&quot; rel=&quot;noreferrer&quot;&gt;LS coupling&lt;/a&gt;.&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That notation would read: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Element.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ionization state.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Forbidden_mechanism#Beta_decay&quot; rel=&quot;noreferrer&quot;&gt;Forbiddenness&lt;/a&gt; - superallowed, allowed, first forbidden, second forbidden, third forbidden, fourth forbidden, etc. Think of &quot;forbidden&quot; as meaning &quot;likelihood&quot;. Represented with one or two brackets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;See note below&lt;/strong&gt;: &quot;In many-electrons systems &lt;strong&gt;transitions that violate rule 4&lt;/strong&gt; change the total spin (&lt;strong&gt;semi-forbidden&lt;/strong&gt; transitions) and are called &lt;strong&gt;intercombination lines&lt;/strong&gt; (e.g. &lt;strong&gt;CIII]&lt;/strong&gt;). Transitions which violate the propensity rule 5 and/or 6 are strictly &lt;strong&gt;forbidden&lt;/strong&gt;, and &lt;strong&gt;are labelled by&lt;/strong&gt; &lt;strong&gt;&lt;em&gt;two&lt;/em&gt;&lt;/strong&gt; &lt;strong&gt;square brackets&lt;/strong&gt; (e.g. &lt;strong&gt;[CIII]&lt;/strong&gt;). Table 1 gives examples of the 3 types of transitions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Do not&lt;/strong&gt; confuse the use of two brackets around an element with their usage around abbreviations for massive &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_classification#Hot_blue_emission_star_classes&quot; rel=&quot;noreferrer&quot;&gt;hot blue emission stars&lt;/a&gt;. With massive Wolf-Rayet stars, their spectra groups are enclosed in square brackets: e.g. [WC]. Most of these show [WC] spectra, some [WO], and very rarely [WN]. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &quot;double Lambda&quot;, in the example you provided, is a 'half doublet'. A doublet for that element would be OIII]λλ1661,1666 another example would be SiIII]λλ1882,1892. One paper where this is discussed is: &quot;&lt;a href=&quot;https://arxiv.org/abs/1612.06866&quot; rel=&quot;noreferrer&quot;&gt;CIII] Emission in Star-forming Galaxies at z ~ 1&lt;/a&gt;&quot; (Mar 4 2017), by Xinnan Du, Alice E. Shapley, Crystal L. Martin, and Alison L. Coil. A doublet is two distinct spectral lines formed independently but which are so close that previously inaccurate instruments and poorer understanding of these mechanisms led many to believe a single line at a particular location. See: &quot;&lt;a href=&quot;https://books.google.ca/books?id=BnPZ249ArwoC&amp;amp;pg=PA29&amp;amp;lpg=PA29&amp;amp;dq=%CE%BB%CE%BB%20doublet&amp;amp;source=bl&amp;amp;ots=KF_FT1pgsx&amp;amp;sig=o3SVMEXXVMqNGi8L9IePKoWsLog&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwjuxsKA18rcAhUNGjQIHbv9ABIQ6AEIHzAI#v=onepage&amp;amp;q=%CE%BB%CE%BB%20doublet&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;Performance Analysis of Standard Fourier-Transform Spectrometers&lt;/a&gt;&quot; 2007 by Douglas Cohen.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Å (&lt;code&gt;$\unicode{x212B}$&lt;/code&gt;$\unicode{x212B}$ &lt;strong&gt;or&lt;/strong&gt; &lt;code&gt;$\unicode{x00C5}$&lt;/code&gt;$\unicode{x00C5}$, a Latin capital letter &quot;A&quot; with a ring above it) is a Swedish letter, the symbol for the &lt;a href=&quot;https://en.wikipedia.org/wiki/%C3%85ngstr%C3%B6m&quot; rel=&quot;noreferrer&quot;&gt;ångström&lt;/a&gt;, a unit of length equal to 10$^{−10}$ m (one ten-billionth of a metre) or 0.1 nanometre.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The Wikipedia webpage: &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Spectroscopic_notation&quot; rel=&quot;noreferrer&quot;&gt;Spectroscopic notation&lt;/a&gt;&quot; should be helpful:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Ionization states&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Spectroscopists&quot; rel=&quot;noreferrer&quot;&gt;Spectroscopists&lt;/a&gt; customarily refer to the spectrum arising from a given ionization state of a given element by the &lt;strong&gt;element's symbol&lt;/strong&gt; followed by a &lt;strong&gt;Roman numeral&lt;/strong&gt;. The numeral &lt;strong&gt;I&lt;/strong&gt; is used for spectral lines associated with the neutral element, &lt;strong&gt;II&lt;/strong&gt; for those from the first ionization state, III for those from the second ionization state, and so on.&lt;a href=&quot;https://en.wikipedia.org/wiki/Spectroscopists&quot; rel=&quot;noreferrer&quot;&gt;10&lt;/a&gt; For example, 'He I' denotes lines of neutral helium, and 'C IV' denotes lines arising from the third ionization state, C3+, of carbon. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;See also: Wikipedia's Ionization energy, section &lt;a href=&quot;https://en.wikipedia.org/wiki/Ionization_energy#Values_and_trends&quot; rel=&quot;noreferrer&quot;&gt;Values and Trends&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;More about doublets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;a href=&quot;https://escholarship.org/uc/item/6c20s0ft&quot; rel=&quot;noreferrer&quot;&gt;The Grism Lens-Amplified Survey from Space (GLASS). XI. Detection of C IV in Multiple Images of the z = 6.11 Lyα Emitter behind RXC J2248.7-4431&lt;/a&gt;&quot; (2017), by Schmidt, KB; Huang, KH; Treu, T; Hoag, A; Bradač, M; Henry, AL; Jones, TA, et al.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;a href=&quot;https://arxiv.org/abs/astro-ph/9706264&quot; rel=&quot;noreferrer&quot;&gt;The UV Properties of the Narrow Line Quasar I Zwicky 1&lt;/a&gt;&quot; (Jun 26 1997), by Ari Laor (Technion), Buell T. Jannuzi, Richard F. Green, Todd A. Boroson (NOAO).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &quot;Spectroscopy - IPAG Grenoble&quot; (&lt;a href=&quot;http://ipag.obs.ujf-grenoble.fr/~hilyblap/teaching/m2astro/ism/notes3.pdf&quot; rel=&quot;noreferrer&quot;&gt;.PDF&lt;/a&gt;), page 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;strong&gt;2.3 Selection rules for Hydrogenic atoms&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In general, there is always a non-zero probability for any transition between two states to occur. However, in some instances, the probability is exceedingly small. Under some approximations (e.g. L-S coupling, dipole approximation, ...) the matrix element might be strictly zero. However, higher-order terms generally lead to non-zero, yet small, probability. The cases in which probabilities vanish, under given approximations, are called selection rules.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The dipole transitions described above are called dipolar electric transitions. The detailed computation of the key term $\scriptsize \left| \overrightarrow \epsilon \cdot \overrightarrow \mu_{if} \right|^2$ allows to determine the cases for which a&#xA;  transition is allowed:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;$\scriptsize{\overrightarrow \mu}\!_{i f}$ does not vanish: dipolar electric transition between states $i$ and $f$ is allowed;&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;$\scriptsize{\overrightarrow \mu}\!_{i f}$ vanishes but there exists non-zero higher-order terms in the development of $e^{i \overrightarrow k \cdot \overrightarrow r}\!\!$: dipolar electric transition is said to be semi-forbidden, but e.g. electric quadrupolar transition are allowed, though with much smaller rates;&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;$\scriptsize{\overrightarrow \mu}\!_{i f}$ and all higher-order terms of the development also vanish: the transition between $i$ and $f$ is forbidden. Other higher-order terms may still allow the emission/absorption of photons.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;The selection rules that apply to one-electron systems are:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;energy level: ∆n any&lt;/li&gt;&#xA;  &lt;li&gt;orbital angular momentum: ∆l = ±1&lt;/li&gt;&#xA;  &lt;li&gt;parity must change between i and f&lt;/li&gt;&#xA;  &lt;li&gt;magnetic quantum number: ∆m = 0,±1&lt;/li&gt;&#xA;  &lt;li&gt;spin does not change: ∆s = 0 (always true for the H atom)&lt;/li&gt;&#xA;  &lt;li&gt;total angular momentum: ∆j = 0,±1&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;In a multi-electron atom, these rules apply to the jumping electron. These rules completely determine the spectra of one-electron atoms, such as HI and HeII, and also the alkali metals. Diagrams which show the allowed transitions are called &lt;a href=&quot;https://en.wikipedia.org/wiki/Grotrian_diagram&quot; rel=&quot;noreferrer&quot;&gt;Grotrian diagrams&lt;/a&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;2.4 Selection rules for many-electrons systems&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We have thus far discussed the interaction of single-electron atoms with a radiation field. The results can be generalized to many-electron systems, by considering the total electric dipole moment, $\scriptsize\overrightarrow D = \sum{_i} −e\overrightarrow {r_i}$. For multi-electron atoms, the selection rules for dipolar electric transitions are:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;total angular momentum: ∆J = 0,±1 but J = 0 → 0 is strictly forbidden&lt;/li&gt;&#xA;  &lt;li&gt;magnetic quantum number: ∆M$_J$ = 0,±1&lt;/li&gt;&#xA;  &lt;li&gt;parity must change between i and f&lt;/li&gt;&#xA;  &lt;li&gt;spin does not change: ∆S = 0&lt;/li&gt;&#xA;  &lt;li&gt;one electron jumps, ∆n any, ∆l = ±1&lt;/li&gt;&#xA;  &lt;li&gt;orbital angular momentum: ∆L = 0,±1, 0 − 0 forbidden&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;The first three rules are rigorous and must be satisfied by all dipolar electric transitions. The last three are not necessarily satisfied in complex atoms, and are called propensity rules. Transitions which fulfill rules 1–3 are allowed. Transitions that violate 4 change the total spin and are called intercombination lines (e.g.CIII]). Transitions which violate the propensity rule 5 and/or 6 are strictly forbidden, and are labelled by square brackets (e.g. [CIII]). Table 1 gives examples of the 3 types of transitions.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\begin{array}{ccccccrrr} \\&#xA;\text{Species} &amp;amp; f \qquad \leftarrow \qquad i &amp;amp; λ (\unicode{x00C5}) &amp;amp; A_{ul}(s^{−1}) &amp;amp; \Delta J &amp;amp; \text{Parity}^\dagger &amp;amp; \Delta S &amp;amp; \Delta l &amp;amp; \Delta L \\&#xA;\hline &#xA;\,\text{NII} &amp;amp; 2{p ^{2}} \; {^{3}\!P} ^e_0 \leftarrow 2p3s \; {^{3}\!D} ^o_1&#xA;&amp;amp; 1084.0 &amp;amp; 2.18×10^8 &amp;amp; -1 &amp;amp; o \rightarrow e &amp;amp; 0 &amp;amp; -1 &amp;amp; -1 \\&#xA;\;\text{CIII]} &amp;amp; 2{s ^{2}} \; {^{1}\!S} ^e_0 \leftarrow 2s2p \; {^{3}\!P} ^o_1&#xA;&amp;amp; 1908.7 &amp;amp; 114 &amp;amp; +1 &amp;amp; o \rightarrow e &amp;amp; -1 &amp;amp; -1 &amp;amp; -1 \\&#xA;\text{[CIII]} &amp;amp; 2{s ^{2}} \; {^{1}\!S} ^e_0 \leftarrow 2s2p \; {^{3}\!P} ^o_2&#xA;&amp;amp; 1906.7 &amp;amp; 0.0052 &amp;amp; +2 &amp;amp; o \rightarrow e &amp;amp; -1 &amp;amp; -1 &amp;amp; -1 \\&#xA;\end{array}&#xA;$$&#xA;  $$\text{Table 1: Examples of allowed, semi-forbidden, and forbidden transitions.&quot;}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18054" LastActivityDate="2018-08-01T01:57:47.540" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27186" PostTypeId="1" CreationDate="2018-08-01T02:42:59.537" Score="0" ViewCount="1390" Body="&lt;p&gt;I have been watching the sky all night for a week now I see flying stars everywhere well things that look like stars, what are they? And the moon has been gone for two nights now? Why?&lt;/p&gt;&#xA;" OwnerUserId="23875" LastEditorUserId="7472" LastEditDate="2018-08-01T07:09:37.157" LastActivityDate="2020-06-13T07:39:18.537" Title="Why has there been no moon for 2 nights straight" Tags="&lt;solar-system&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27188" PostTypeId="5" CreationDate="2018-08-01T10:16:43.120" Score="0" Body="" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-08-01T10:16:43.120" LastActivityDate="2018-08-01T10:16:43.120" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27189" PostTypeId="4" CreationDate="2018-08-01T10:16:43.120" Score="0" Body="for questions regarding the dimming of light from distant objects when it passes through interstellar dust and gas clouds. &#xD;&#xA;&#xD;&#xA;This is not for questions about extinctions of species on Earth or other planets." OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-08-01T10:16:43.120" LastActivityDate="2018-08-01T10:16:43.120" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27191" PostTypeId="2" ParentId="27186" CreationDate="2018-08-01T18:20:15.160" Score="2" Body="&lt;p&gt;The Moon is out of sight for a few nights every four weeks, more or less. It moves around the sky slowly. Sometimes it's far from the Sun (as seen from our planet), and that's when you see it at night, up in the sky. Other times it seems close to the Sun, and then of course you won't see it at night. Then the cycle repeats, over and over again. Takes a bit over 4 weeks to repeat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The star-like things moving in the sky can be many things. Usually they are satellites in orbit around Earth, or airplanes flying at high altitude.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2018-08-01T18:20:15.160" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27192" PostTypeId="2" ParentId="27180" CreationDate="2018-08-01T18:43:37.893" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The question is if 1st and 2nd stars generation burned hydrogen to helium and more heavier elements, then should it be like 90% of all universe hydrogen already converted to helium and something else?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only a tiny portion of the primordial hydrogen has been converted to helium or something else. The explanation is fourfold.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Most of the universe's primordial hydrogen lies between galaxies. Some of this intergalactic gas might become captured by a galaxy, but much of it probably never will.&lt;/li&gt;&#xA;&lt;li&gt;Most of the hydrogen within a galaxy is in the form of the warm to hot interstellar medium. Some of this interstellar gas might condense to form an interstellar gas cloud, but as is the case with the intergalactic medium, much of this interstellar medium probably never will be incorporated into a star.&lt;/li&gt;&#xA;&lt;li&gt;While some of the gas in an interstellar gas cloud does collapse to form stars and planets, this process is incredibly inefficient. Well over 90% of the gas in a gas cloud gets ejected into the interstellar medium during the star formation process.&lt;/li&gt;&#xA;&lt;li&gt;While some of the hydrogen in a star is indeed converted to helium or more massive elements, this burning is incomplete. Stars between roughly 1/2 to 5 solar masses eject a lot of hydrogen during their death throes.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;That said, star formation in our galaxy is now drastically reduced compared to what it was at its peak. The reason isn't that hydrogen has been converted to helium and more massive elements. The reason is instead that a lot of the hydrogen is now locked up in low mass stars. The lifetime of a half solar mass star is several times the current age of the universe, and this lifetime grows as star mass decreases. All of the low mass stars that have ever formed are still stars, and that makes for a lot of locked-up hydrogen.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2018-08-01T19:02:17.383" LastActivityDate="2018-08-01T19:02:17.383" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27194" PostTypeId="1" CreationDate="2018-08-02T02:15:58.753" Score="7" ViewCount="125" Body="&lt;p&gt;I read from &lt;a href=&quot;https://en.wikipedia.org/wiki/Eclipse_cycle&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia$-$Eclipse cycle&lt;/a&gt; that the eclipse year, which is the period of Earth's hitting a certain node of its orbit around the sun, that is, the ecliptic, and the moon's orbit around Earth, can be calculated as the beat period of the synodic month and draconic month. The recent occurrence of planet transits across the sun lets me wonder if this principle can be applied to the case of a planet's transit across the sun as seen by an observer on another planet in an outer orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Venus transited across the sun on June 6, 2012, I read it takes place in pattern of the 8-year-interval pairs with each pair at the interval of 105.5 years and 121.5 years alternately, with one in each pair being when Venus crosses the ecliptic plane at its ascending node and the other being when Venus crosses the ecliptic plane at its descending node. This made me contemplate whether the period of the Venus transit across the sun when it crosses the ecliptic plane at its certain node can be calculated in the way the eclipse year is, that is, whether the following holds: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The period of the Venus transit across the sun as it crosses the&#xA;  ecliptic plane at its certain node = The beat period of the period of Venus'&#xA;  being in inferior conjunction with the sun and the period of Venus'&#xA;  hitting that node.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Early this July I read from &lt;a href=&quot;http://earthsky.org/astronomy-essentials/earth-transit-from-pluto-july-12-2018&quot; rel=&quot;noreferrer&quot;&gt;EarthSky&lt;/a&gt; an observer on Pluto may, through some optical aid, be able to see Earth transiting the sun on July 12, when Pluto was at opposition, which, together with the upcoming event, that on October 24, 2018 Pluto will cross the ecliptic plane at its descending node, is responsible for the Earth transit as seen from Pluto. Last time this kind of transit occurred on January 11, 1931 on account of both that Pluto was at opposition that day and that Pluto crossed the ecliptic plane at its ascending node on September 9, 1930. And next time this kind of transit will occur when Pluto is at opposition 161 years later, around when Pluto will cross the ecliptic plane at the same node it crossed in 1930. This makes me again contemplate whether what follows holds: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The period of the Earth transit across the sun as seen from Pluto as&#xA;  Pluto crosses the ecliptic plane at its certain node = The beat period of the&#xA;  period of Pluto's being at opposition and the period of Earth's&#xA;  hitting that node.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think the principal difference between the eclipse year and planet transit across the sun is that in the former case both the moon and the sun are in a closed orbit around Earth while in the latter case only the sun is in a closed orbit around the observer-sited planet (the transiting planet is in a tortuous path from the perspective of the observer). I don't know if this feature would affect the applicability of the beat-period principle. If the principle of beat period as it used to calculate the eclipse year can be applied to the above two examples, I think it should be able to be generally applied to calculate the period of any planet's transit across the sun as seen by an observer on an outer planet, right? That is, should what follows generally hold? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The period of a planet's transit across the sun as seen from an outer&#xA;  planet as it crosses the outer planet's orbit plane at its certain node =&#xA;  The beat period of the period of the transiting planet's being in&#xA;  inferior conjunction with the sun from the outer planet's perspective&#xA;  and the period of the transiting planet’s hitting that node&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="22933" LastEditorUserId="22933" LastEditDate="2018-08-02T03:25:32.873" LastActivityDate="2020-07-21T07:55:30.367" Title="Can the period of the planet transit across the sun be calculated in the same way the eclipse year is?" Tags="&lt;planetary-transits&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27195" PostTypeId="1" CreationDate="2018-08-02T02:42:31.827" Score="4" ViewCount="46" Body="&lt;p&gt;Why do tidal tails result from only disk galaxies but not ellipticals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that tidal tails are results of gravitational interactions not strong enough to completely plunge out the material violently. How is that they form only in organized motions (disk) rather than random motion (ellipticals). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any references to a detailed explanation is highly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="16900" LastEditorUserId="11264" LastEditDate="2018-08-02T06:00:55.527" LastActivityDate="2018-08-02T06:00:55.527" Title="Why Tidal Tails in Disk Galaxies, but not in Ellipticals?" Tags="&lt;galaxy&gt;&lt;tidal-forces&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27200" PostTypeId="1" CreationDate="2018-08-02T11:50:32.983" Score="3" ViewCount="151" Body="&lt;p&gt;If our universe is not infinite, what happens with the light beam (or photon) when it will travel through the whole universe?&#xA;For example, &lt;a href=&quot;https://en.wikipedia.org/wiki/Observable_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;observable universe&lt;/a&gt; according to wikipedia has diameter 93 billion light-years. What will happen with photon after 93 billion years if I generate it now? Will it return back to me? Or it will never reach the &quot;edge&quot; of the universe because the universe is expanding faster?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;UPDATE 1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Of course, I assume that photon will no be absorbed :) because it is just a theoretical thinking. And return back I mean not exactly to the same point, but approximately within some range (let say 5% of the universe size).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I correctly understand that there are a few options for the case when the universe is finite (means &quot;closed geometry&quot;): &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Photon will never return back because the universe will collapse before it (in case if there are no dark energy).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Photon will never return back because universe expanding faster than light speed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Photon will come back because universe expanding slower than light. And what &#xA;of these 3 options is most probable from a current scientists point of view?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15004" LastEditorUserId="15004" LastEditDate="2018-08-03T13:51:17.980" LastActivityDate="2018-08-03T13:51:17.980" Title="Light beam (1 photon) in the limited universe?" Tags="&lt;universe&gt;&lt;expansion&gt;&lt;observable-universe&gt;&lt;fate-of-universe&gt;&lt;hubble-constant&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27201" PostTypeId="2" ParentId="27200" CreationDate="2018-08-02T12:10:58.357" Score="2" Body="&lt;p&gt;No one can say for sure because, while we have theories which fit our observations quite well, those theories require that the universe be considerably bigger than the part we can see -- light simply not having had enough time to get to us yet.  But since we don't have observations of this extended universe, we run a significant risk that we're wrong in some important way.  Additionally, we know our current theories are incomplete: One of the things our theories do not tell us is whether the universe is finite or infinite.  (Note: Some theories do make a prediction, but they are not supported by better evidence than theories which make different predictions. The jury -- comprised of &lt;em&gt;experimentalists&lt;/em&gt; -- is still out.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we &lt;em&gt;can&lt;/em&gt; answer your question!  If the universe if infinite, the photon will never return.  That's obvious.  But even if it is finite, our theories show that widely-enough separated points in the universe separating faster than the speed of light due to the expansion of space, so in this case, too, the photon will never return.  There is strong (but not iron-clad) evidence in favor of this picture being true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even if space was &lt;em&gt;not&lt;/em&gt; expanding FTL, the photon would never come home again because space is not perfectly flat. Mass curves space and large masses (like giant galaxies and clusters of galaxies) curve space enough that we can measure it. So space acts like a very slightly ripply pane of glass, and even if it was finite and non-expanding, photons would not return home -- but they &lt;em&gt;would&lt;/em&gt; eventually get back to within maybe ten million light-years of home as they passed by on their way around the universe a second time!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course &lt;em&gt;this&lt;/em&gt; is true only if your photon doesn't get absorbed. There is a lot of &lt;em&gt;stuff&lt;/em&gt; in the universe which absorbs photons and if a photon traveled long enough, it would eventually be absorbed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, say good-bye now because this is the last time you'll ever see your photon.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-08-02T12:10:58.357" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27202" PostTypeId="2" ParentId="6625" CreationDate="2018-08-02T12:12:32.790" Score="1" Body="&lt;p&gt;The only known relationship between the &lt;a href=&quot;http://en.wikipedia.org/wiki/Epoch_(reference_date)&quot; rel=&quot;nofollow noreferrer&quot;&gt;epoch&lt;/a&gt; of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Julian_calendar&quot; rel=&quot;nofollow noreferrer&quot;&gt;Julian calendar&lt;/a&gt; (i.e., January 1&lt;sup&gt;st&lt;/sup&gt;, 45 BC) and any significant solar moment was its synchronization with the first &lt;a href=&quot;http://en.wikipedia.org/wiki/New_moon&quot; rel=&quot;nofollow noreferrer&quot;&gt;new moon&lt;/a&gt; following the previous year's &lt;a href=&quot;http://en.wikipedia.org/wiki/Winter_solstice&quot; rel=&quot;nofollow noreferrer&quot;&gt;winter solstice&lt;/a&gt; :&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;It was probably the original intention of &lt;a href=&quot;http://en.wikipedia.org/wiki/Julius_Caesar&quot; rel=&quot;nofollow noreferrer&quot;&gt;Caesar&lt;/a&gt; to commence the year with the shortest day. The winter solstice at Rome, in the year 46 B.C., occurred on the 24&lt;sup&gt;th&lt;/sup&gt; of December of the Julian calendar. His motive for delaying the commencement for seven days longer, instead of taking the following day, was probably the desire to gratify the superstition of the Romans, by causing the first year of the reformed calendar to fall on the day of the new moon. Accordingly, it is found that the mean new moon occurred at Rome on the 1&lt;sup&gt;st&lt;/sup&gt; of January, 45 B.C., at 6h. 16′ P. M. In this way alone can be explained the &lt;a href=&quot;http://penelope.uchicago.edu/Thayer/L/Roman/Texts/Macrobius/Saturnalia/1*.html#14&quot; rel=&quot;nofollow noreferrer&quot;&gt;phrase&lt;/a&gt; used by &lt;a href=&quot;http://en.wikipedia.org/wiki/Macrobius&quot; rel=&quot;nofollow noreferrer&quot;&gt;Macrobius&lt;/a&gt;: &lt;em&gt;Annum civilem Caesar, habitis ad lunam dimensionibus constitutum, edicto palam proposito publicavit&lt;/em&gt;. This edict is also &lt;a href=&quot;http://penelope.uchicago.edu/Thayer/E/Roman/Texts/Plutarch/Lives/Caesar*.html#59&quot; rel=&quot;nofollow noreferrer&quot;&gt;mentioned&lt;/a&gt; by &lt;a href=&quot;http://en.wikipedia.org/wiki/Plutarch&quot; rel=&quot;nofollow noreferrer&quot;&gt;Plutarch&lt;/a&gt; where he gives the anecdote of &lt;a href=&quot;http://en.wikipedia.org/wiki/Cicero&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cicero&lt;/a&gt;, who, on being told by some one that the constellation &lt;a href=&quot;http://en.wikipedia.org/wiki/Lyra&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lyra&lt;/a&gt; would rise the next morning, observed, &lt;em&gt;Yes, no doubt, in obedience to the edict.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;— &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/William_Smith_(lexicographer)&quot; rel=&quot;nofollow noreferrer&quot;&gt;William Smith&lt;/a&gt;&lt;/strong&gt;, &lt;em&gt;&lt;a href=&quot;http://penelope.uchicago.edu/Thayer/E/Roman/Texts/Secondary/SMIGRA&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Dictionary of Greek and Roman Antiquities&lt;/a&gt;&lt;/em&gt;, page &lt;a href=&quot;http://penelope.uchicago.edu/Thayer/E/Roman/Texts/Secondary/SMIGRA*/Calendarium.html#p231&quot; rel=&quot;nofollow noreferrer&quot;&gt;231&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/John_Murray_(publisher)&quot; rel=&quot;nofollow noreferrer&quot;&gt;John Murray&lt;/a&gt;, London, 1875.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14152" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-08-02T12:12:32.790" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27203" PostTypeId="1" CreationDate="2018-08-02T15:33:07.183" Score="2" ViewCount="189" Body="&lt;p&gt;I'm looking for (&lt;em&gt;'simplified'&lt;/em&gt;) equations governing stellar evolution. Especially how mass, luminosity, temperature and radius of a star change during it's lifetime. As well as equations which tell you how long a star stays in a certain stellar stage (PMS, MS, ...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extra clarification by example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that a star with a mass between &lt;em&gt;x&lt;/em&gt; and &lt;em&gt;y&lt;/em&gt; enters the main sequence for a duration of $\Delta T = formula$. During this phase it's mass changes according to $M = formula(t)$. after that it enters the &lt;strong&gt;xxx&lt;/strong&gt;  phase&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;For those who think this question is to broad, please give some suggestions, 'google terms', ... from which I can move forward.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="23870" LastEditorUserId="10196" LastEditDate="2018-08-03T02:34:16.653" LastActivityDate="2018-08-03T13:14:12.393" Title="What are the equations governing stellar evolution (Luminosity, Mass, Temperature, Radius)" Tags="&lt;astrophysics&gt;&lt;stellar-evolution&gt;&lt;resource-request&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27204" PostTypeId="1" AcceptedAnswerId="27209" CreationDate="2018-08-02T19:01:23.793" Score="2" ViewCount="352" Body="&lt;p&gt;Are there any standards or established practices on how to write / represent Right Ascension (RA) and Declination (DEC) values in the sexagesimal format?&#xA;I have seen all of the following. E.g. for RA:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;02 03 04&lt;/code&gt; (assumed to mean 2 hours, 3 hourly minutes and 4 hourly seconds)&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;2 3 4&lt;/code&gt; (same as above)&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;2h 3m 4s&lt;/code&gt; (same as above)&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;2d 3' 6&quot;&lt;/code&gt; (assumed to mean 2 degrees, 3 degree minutes and 4 degree seconds)&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;2d 3m 6s&lt;/code&gt; (same as above)&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;15.33&lt;/code&gt; (assumed to mean 15.33 degrees)&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;9.5H&lt;/code&gt; (assumed to mean 9 hours, 30 hourly minutes and zero hourly seconds)&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;8d 3.5&lt;/code&gt; (assumed to mean 8 degrees, 3 degree minutes and 30 degree seconds)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Likewise for DEC (even though for DEC when the first component does not have a suffix it is assumed by default to be in degrees, not hours). Does any astronomical association publish guidelines for the printing / presentation of such values or on how software should accept user's input of such values?&lt;/p&gt;&#xA;" OwnerUserId="522" LastActivityDate="2018-08-03T08:09:39.800" Title="Presenting / printing Right Ascension and Declination values" Tags="&lt;declination&gt;&lt;right-ascension&gt;&lt;standards&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27205" PostTypeId="2" ParentId="27200" CreationDate="2018-08-02T22:19:20.000" Score="5" Body="&lt;p&gt;A finite universe is said to have a &quot;closed geometry&quot;, or to be &quot;positively curved&quot;, meaning that, in principle, you may travel in a straight line and eventually return back to your starting point. In the 2D analogy, the surface of Earth is positively curved, and if you travel 40,000 km straight, you're back where you started. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A finite universe that does &lt;em&gt;not&lt;/em&gt; contain dark energy of some sort, however, is bound to collapse, and you can show mathematically that even a photon will not make it back to its starting point before the universe has collapsed. Only if you consider a photon emitted exactly when the universe is created (Big Bang), it will make it back exactly when the universe ends (Big Crunch).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the universe contains a sufficient amount of dark energy, this may prevent collapse. This may cause it to expand too fast for the photon to &quot;get back&quot;, but you could fine-tune its components such that the expansion/collapse is slow enough for the photon to get back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One such model is &quot;the loitering universe&quot;, which gained some interest in the 1960's, as an explanation for an excess of quasars around redshift $z\simeq2$ (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/1967ApJ...147.1222P&quot; rel=&quot;noreferrer&quot;&gt;Petrosian et al. 1967&lt;/a&gt;). If the deceleration of matter is almost exactly balanced by the acceleration of a cosmological constant, you can have an (arbitrarily long) period of time where the universe is static.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, observations indicate that &lt;em&gt;our&lt;/em&gt; universe is not such a universe (&lt;a href=&quot;https://arxiv.org/abs/1807.06209&quot; rel=&quot;noreferrer&quot;&gt;Planck Collaboration et al. 2018&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that since the universe is not empty, a photon may be absorbed on its way, not returning even if the expansion allows it. However, in our universe, photons that were emitted shortly after Big Bang (the so-called CMB photons) have traveled mostly freely through space ever since, with only 5% interacting with matter along its way.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2018-08-02T22:19:20.000" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="27206" PostTypeId="1" AcceptedAnswerId="27208" CreationDate="2018-08-03T04:47:47.823" Score="2" ViewCount="134" Body="&lt;p&gt;I have trouble picturing eclipses and whether the moon can be directly overhead.&lt;/p&gt;&#xA;" OwnerUserId="23904" LastActivityDate="2018-08-03T14:39:53.530" Title="Can a solar eclipse be directly overhead?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27208" PostTypeId="2" ParentId="27206" CreationDate="2018-08-03T06:22:06.477" Score="10" Body="&lt;p&gt;They could be directly overhead if you live in the tropics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tropics are the regions near the equator, between 23 degrees North and 23 degrees south. In the tropics, the sun and the moon can pass directly overhead. If both pass overhead at the same time there will be a solar eclipse directly overhead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The eclipse of &lt;a href=&quot;https://www.timeanddate.com/eclipse/in/mexico/tepic?iso=19910711&quot; rel=&quot;noreferrer&quot;&gt;July 11, 1991 was visible directly overhead&lt;/a&gt;, (to within a degree) and in totality, for observers in the Nayarit province in Mexico&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-08-03T14:39:53.530" LastActivityDate="2018-08-03T14:39:53.530" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27209" PostTypeId="2" ParentId="27204" CreationDate="2018-08-03T08:09:39.800" Score="2" Body="&lt;p&gt;There are no standards for representing right ascension and declination apart from the abbreviations RA and Dec, and the symbols $\alpha$ and $\delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The use of decimal degrees, dms, °'&quot;, hours and decimals, hms etc. depend on the [author's] style and intent. For example, determining transit times from local sidereal time and right ascension is easier when RA is expressed in hours, whereas doing calculations of angular separation is easier to calculate in software when decimal degrees are used. &lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2018-08-03T08:09:39.800" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27210" PostTypeId="1" AcceptedAnswerId="27215" CreationDate="2018-08-03T12:07:31.023" Score="2" ViewCount="212" Body="&lt;p&gt;I have been star gazing for awhile with a pair of binoculars and now I am ready to step it up a bit and purchase a telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been looking at the SkyWatcher Heritage 5&quot; Tabletop Dobsonian Telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this be a suitable first telescope , I hope to be able to make out the rings of Saturn using stock the Eyepiece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or would this be considered overkill and something like the Celestron 21035 70mm Travel Scope be more suitable.﻿&lt;/p&gt;&#xA;" OwnerUserId="23907" LastActivityDate="2018-08-03T18:14:41.290" Title="First Telescope SkyWatcher Heritage 5&quot; Tabletop Dobsonian Telescope" Tags="&lt;telescope&gt;&lt;star-gazing&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27211" PostTypeId="2" ParentId="27203" CreationDate="2018-08-03T13:14:12.393" Score="3" Body="&lt;p&gt;What you want is hydrodynamic stellar evolution. The following starts with a brief review of hydrostatic case to give the fundamentals. The last paragraph discusses the hydrodynamic case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For hydrostatic case, see &lt;a href=&quot;http://www.astro.caltech.edu/~george/ay20/Ay20-Lec7x.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;, or &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/0805304029&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;. Simplified stellar evolution assumes hydrostatic equilibrium (i.e., time-independent fashion). The system of equations compose of i) mass conservation, ii) mass transportation (i.e., hydrostatic equilibrium), iii) energy conservation, and iv) energy transportation. The system also requires i) the equation of state, ii) opacity, and iii) nuclear energy generation for being solved in a close form. &#xA;The close-form solution is typically in the form of polytrophs (i.e., &lt;a href=&quot;https://en.wikipedia.org/wiki/Lane%E2%80%93Emden_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lane-Emden equation&lt;/a&gt;). More complicated model, which is still hydrostatic equilibrium, will include other factors such as composition, rotation, and binary interaction. Since it is static and equilibrium case, the time-dependent terms are zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For hydrodynamic case (i.e., time dependent) as you wanted, there is no close-form solution, therefore only hydrodynamic simulation is the method for the study. The simulation model uses the set of equations mentioned above without dropping the time-dependent terms. See, e.g., &lt;a href=&quot;http://mesa.sourceforge.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;MESA&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-08-03T13:14:12.393" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27212" PostTypeId="1" CreationDate="2018-08-03T13:51:39.010" Score="4" ViewCount="695" Body="&lt;p&gt;If time slows down the closer you get to the speed of light, does it follow that time does not exist for photons. Do photons - even if they travel across billions of light years of space - arrive at the same time as they are created?&#xA;And if this is so, does this not explain the peculiarities of the double-slit experiment?&lt;/p&gt;&#xA;" OwnerUserId="19703" LastActivityDate="2018-08-03T13:58:32.170" Title="Does Time Exist for Photons?" Tags="&lt;light&gt;&lt;space-time&gt;&lt;photons&gt;&lt;time&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2018-08-03T18:09:06.880" ContentLicense="CC BY-SA 4.0" />
  <row Id="27213" PostTypeId="2" ParentId="27212" CreationDate="2018-08-03T13:58:32.170" Score="4" Body="&lt;p&gt;Time does not exist for photons.  Quoting &lt;a href=&quot;https://en.wikipedia.org/wiki/Proper_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;'s somewhat unnecessarily technical page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In relativity, proper time along a timelike world line is defined as the time as measured by a clock following that line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The proper time between any two points along any photon's path in empty space is zero.  So the photon &quot;experiences&quot; no passage of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not very clear to me how this could explain the double-slit experiment. Proper Time is an entirely relativistic concept, whereas the double slit results are &lt;em&gt;not&lt;/em&gt; explained by relativity, but are explained by Quantum Mechanics.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-08-03T13:58:32.170" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27214" PostTypeId="2" ParentId="25865" CreationDate="2018-08-03T17:10:42.380" Score="2" Body="&lt;p&gt;You are correct that the main way to determine planetary surface ages is through crater dating. This comes in two versions, relative and absolute dating. Both require counting the number of craters per area and so you need good resolution imaging since the number of impactors and craters follow a power-law (more small impactors/craters than large). It then basically boils down to older surfaces have had a longer time to be hit by more impactors and so there will be more craters per unit area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With relative dating, all you can do is put different surfaces in order of amount of craters and say &quot;surface A is older than surface B as it has more craters&quot;. If you have another method of dating, then you may be able to say where on the timeline a surface may be but you would still only be able to &quot;surface A is older than surface B (dated to be 250 Million years)&quot; but you don't know much older.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have an idea of the rate of cratering bodies and the rate of crater formation in the system you are trying to study, then you can perform absolute dating. This allows the cratering rates to be compared across different bodies. This has been done for the inner Solar System (Mercury, Moon, Mars) but is tricky where the surface can be altered (weather, lava flows etc). There are also problems with crater saturation with very old surfaces where new craters overlap and overwrite older ones and issues with understanding the lifetimes of the craters and what processes can fill-in and erase craters and how long this process takes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the Saturn system, determining the number of potential impactors and performing absolute dating is likely to be pretty tricky as the environment is dynamically complex. There are a lot of large Moons and you also have Jupiter and the other gas giants planets affecting things. In addition, the positions of the giant planets are thought to have changed dramatically around 500 Mya after formation (the &lt;a href=&quot;http://www.boulder.swri.edu/~kwalsh/GrandTack.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Grand Tack model and hypothesis&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The images of the Pluto and Charon system returned by New Horizons have also shown some of the difficulties with crater counting (&lt;a href=&quot;https://blogs.nasa.gov/pluto/2015/10/13/the-impact-of-craters/&quot; rel=&quot;nofollow noreferrer&quot;&gt;blog article&lt;/a&gt; on this by a member of the New Horizons team and crater dating expert). Some parts of the system such as Charon appear to be very old and have large numbers of craters; other areas such as Sputnik Planitia appear to be very young and no-one can find any craters of any size anywhere on that region.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-08-03T17:10:42.380" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27215" PostTypeId="2" ParentId="27210" CreationDate="2018-08-03T18:14:41.290" Score="1" Body="&lt;p&gt;No telescope that fits your budget and observing habits is overkill.&#xA;As you &lt;a href=&quot;https://www.skyandtelescope.com/astronomy-equipment/how-to-choose-a-telescope/&quot; rel=&quot;nofollow noreferrer&quot;&gt;may have read&lt;/a&gt;, the bigger the aperture, the more light it collects, and the finer detail it can resolve in steady air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 5-inch f/5 reflector (650mm focal length) is good for wide-field views of deep-sky objects and easy to transport to a dark site.&#xA;It should show the rings of Saturn clearly with a 10mm eyepiece (65x magnification); a shorter eyepiece could double that without losing clarity.&#xA;However, with such a short optical tube assembly, you might either operate mostly on your knees or put the mount on something less sturdy than the ground.&#xA;Also if the secondary mirror is exposed as &lt;a href=&quot;http://skywatcher.com/product/heritage-p130/&quot; rel=&quot;nofollow noreferrer&quot;&gt;shown here&lt;/a&gt;, it's vulnerable to dew, and stray light can enter the eyepiece and reduce contrast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're primarily interested in planets, a 6-inch f/8 reflector (1200mm focal length) could be an even better choice.&#xA;A 10mm eyepiece gives 120x magnification, the primary mirror is less sensitive to collimation error, and the secondary mirror can be smaller for slightly better contrast.&#xA;On the other hand, you might want an additional eyepiece longer than 25mm for better views of certain faint objects.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-08-03T18:14:41.290" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27216" PostTypeId="1" CreationDate="2018-08-03T18:18:28.620" Score="3" ViewCount="49" Body="&lt;p&gt;I seem to remember a TV science programme which re-examined the slides that led to the discovery of Pluto and then took a fresh look at earlier slides taken of the space which Pluto was passing through. These slides did indeed feature Pluto but were probably not part of Lowell’s search for the planet. Maybe the old data/images(if still on record) could be revisited with the advantage of later discoveries to reveal things of interest like say orphan planets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently there was this case:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.extremetech.com/extreme/234037-astronomers-observed-a-nova-explosion-for-the-first-time-by-accident&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.extremetech.com/extreme/234037-astronomers-observed-a-nova-explosion-for-the-first-time-by-accident&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which inadvertently did this but are there any programmes/projects retrospectively doing this intentionally?’&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it is the systematic review of past observations in the hope of finding things that may have been missed or even unknown at the time.&lt;/p&gt;&#xA;" OwnerUserId="23805" LastActivityDate="2018-08-03T18:43:59.613" Title="The systematic review of past observations in the hope of finding things that may have been missed" Tags="&lt;observational-astronomy&gt;&lt;history&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27217" PostTypeId="2" ParentId="27216" CreationDate="2018-08-03T18:38:37.780" Score="3" Body="&lt;p&gt;The first step for most of these projects is to make the old data available. Just about all of the old data is on photographic plates, so there are projects at many of the great observatories to scan their archives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good example is the &lt;a href=&quot;http://tdc-www.harvard.edu/plates/&quot; rel=&quot;nofollow noreferrer&quot;&gt;project at Smithsonian Astrophysical Observatory&lt;/a&gt; (located at Harvard) to digitize the entire plate collection, ranging from ultra-early photographs to sky surveys that Harvard Observatory did during a large part of the 1900s. The plates are scanned, and they are matched with star atlases to determine their exact coordinates and their limiting magnitude and the filters used, if any. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This data -- and lots more from all over, including the epochal Palomar Sky Survey -- is available (or soon will be) using special software such as &lt;a href=&quot;https://en.wikipedia.org/wiki/WorldWide_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Microsoft's Worldwide Telescope&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the old data is being collected and curated and being made available to everyone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that the old data, being photographic, is limited in how deep they go, so many objects are simply not visible in them, including most minor planets.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And &lt;em&gt;lots&lt;/em&gt; of projects make use of it.  The ones I'm familiar with tend to use the data to look for prediscovery images or to extend light curves into the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;New surveys like the &lt;a href=&quot;https://en.wikipedia.org/wiki/Large_Synoptic_Survey_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Large Synoptic Telescope&lt;/a&gt; will generate cubic shittons of data and are being &lt;em&gt;deliberately&lt;/em&gt; designed to separate the collection of data from the use -- it'll collect deep full-sky images every few days and then people will mine that data for things that the people who planned the survey hadn't thought of.  It will revolutionize astronomy.  Key point: All this old data is sure to be made available in the same way and the very bright people developing ways to find interesting stuff in the LSST will also find good stuff in old photographic plates from the 20s.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2018-08-03T18:43:59.613" LastActivityDate="2018-08-03T18:43:59.613" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27218" PostTypeId="1" AcceptedAnswerId="27244" CreationDate="2018-08-03T19:20:29.353" Score="5" ViewCount="92" Body="&lt;p&gt;I am currently working on determining the nth nearest neighbors of a large survey of galaxies. I am not an astronomer by trade, so I am unsure of interpreting some of my results. I have discovered for the sample of galaxies their 9th nearest neighbor distances. It turns out the mode of this distribution by far is between 2mpc and 3mpc. Since I am unsure of the scale of these implications I would prefer if a well versed astrophysicist can tell me if the distance to the 9th nearest neighbor galaxy of 2-3mpc would be considered densely clustered. These galaxies are at intermediate redshift.&lt;/p&gt;&#xA;" OwnerUserId="23909" LastActivityDate="2018-08-06T07:16:35.520" Title="What distance is considered 'close' for the 9th nearest neighbor of a galaxy?" Tags="&lt;galaxy&gt;&lt;galaxy-cluster&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27219" PostTypeId="1" AcceptedAnswerId="36581" CreationDate="2018-08-04T06:24:29.887" Score="7" ViewCount="444" Body="&lt;p&gt;I've been trying to build a model of the solar system in a game.  Thus far I've succeeded in placing each of the planets in position using Keplerian elements and formulae from &lt;a href=&quot;https://ssd.jpl.nasa.gov/?planet_pos&quot; rel=&quot;noreferrer&quot;&gt;https://ssd.jpl.nasa.gov/?planet_pos&lt;/a&gt;  An example for Jupiter is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;               a              e               I                L            long.peri.      long.node.&#xA;           AU, AU/Cy     rad, rad/Cy     deg, deg/Cy      deg, deg/Cy      deg, deg/Cy     deg, deg/Cy&#xA;-----------------------------------------------------------------------------------------------------------&#xA;Jupiter   5.20288700      0.04838624      1.30439695       34.39644051     14.72847983    100.47390909&#xA;         -0.00011607     -0.00013253     -0.00183714     3034.74612775      0.21252668      0.20469106&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;At this point, I want to add several of the Jovian moons, however the same site doesn't seem to provide the same type of data for these moons, perhaps because the moons behave differently.  Unfortunately, my math is not great; I can usually implement something that's written down, but I don't have the wherewithal to work it out myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Jovian moons, the data is listed at: &lt;a href=&quot;https://ssd.jpl.nasa.gov/?sat_elem#jupiter&quot; rel=&quot;noreferrer&quot;&gt;https://ssd.jpl.nasa.gov/?sat_elem#jupiter&lt;/a&gt;  Here is an example for Ganymede:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; Sat.           a       e          w           M        i            node      n             P       Pw      Pnode    R.A.        Dec.     Tilt&#xA; (km)        (deg)     (deg)      (deg)       (deg)    (deg/day)    (days)    (yr)          (yr)    (deg)    (deg)    (deg)&#xA; Ganymede  1070400.    0.0013    192.417    317.540    0.177        63.552    50.3176072    7.155   63.549  132.654   268.168    64.543    0.068&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here they state that the data is &quot;&lt;em&gt;Mean orbital elements referred to the local &lt;strong&gt;Laplace planes&lt;/em&gt;&lt;/strong&gt;&quot;, as opposed to &quot;&lt;strong&gt;&lt;em&gt;Keplerian Elements&lt;/strong&gt; for Approximate Positions of the Major Planets&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whilst the first 6 parameters appear to be the same (a, e, w, M, i, node), the Keplerian elements also have &quot;rates&quot;, or &lt;em&gt;time derivatives&lt;/em&gt; (according to this post: &lt;a href=&quot;https://space.stackexchange.com/questions/8911/determining-orbital-position-at-a-future-point-in-time&quot;&gt;https://space.stackexchange.com/questions/8911/determining-orbital-position-at-a-future-point-in-time&lt;/a&gt;).  The forumlae kindly provided by NASA for the Keplerian elements relies on these rates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I could understand how to determine the &quot;rates&quot; for the Jovian moons, then I might be able to use the same formula (is this the right thing to do?)  It looks to me like those rates are something I should be able to compute from the other data provided for the moons, however I haven't been able to determine how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My suspicion is that I either need a different formula for the computation using the Laplace Planes, or I need to compute those 'rate' values and use the same formula I already have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to generate cartesian (x, y, z) coordinates for each Moon where (0,0,0) is the centre of Jupiter itself, for a given date/time (typically &quot;now&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm beginning to wonder if I'm barking up the wrong tree on this.  Can anyone shed any light&lt;/p&gt;&#xA;" OwnerUserId="23911" LastActivityDate="2020-06-17T13:35:04.887" Title="Computing orbit positions of Jovian Satellites / Moons using JPL data" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;jupiter&gt;&lt;satellite&gt;&lt;algorithm&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27220" PostTypeId="1" AcceptedAnswerId="27222" CreationDate="2018-08-04T07:11:49.977" Score="2" ViewCount="98" Body="&lt;p&gt;The question &lt;a href=&quot;https://space.stackexchange.com/q/29879/12102&quot;&gt;What are the chances that the ISS transits the moon as seen from your house (or headquarters)?&lt;/a&gt; links to the SpaceX Instagram-posted photos of the ISS transiting the Moon's disk as seen from SpaceX headquarters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a bright object next to the Moon, but I'm not sure exactly when this happened. Instagram says &quot;last weekend&quot; which would appear to be the last weekend of July in 2018.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What bright object is this, next to the Moon? Planet? Star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.instagram.com/p/Bl8xjJFFrHM/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.instagram.com/p/Bl8xjJFFrHM/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BQQRB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BQQRBm.jpg&quot; alt=&quot;ISS lunar transit Spacex&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/DqX9U.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DqX9Um.jpg&quot; alt=&quot;ISS lunar transit Spacex&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-08-04T10:51:07.367" Title="What bright body is next to the Moon in these photos of the ISS transiting the lunar disk?" Tags="&lt;the-moon&gt;&lt;identify-this-object&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27221" PostTypeId="1" AcceptedAnswerId="27223" CreationDate="2018-08-04T10:20:52.107" Score="5" ViewCount="225" Body="&lt;p&gt;Below &lt;a href=&quot;https://space.stackexchange.com/q/29879/12102&quot;&gt;this question&lt;/a&gt; was left a comment linking to the Smarter Every Day video &lt;a href=&quot;https://youtu.be/lepQoU4oek4&quot; rel=&quot;nofollow noreferrer&quot;&gt;Space Station Transiting 2017 ECLIPSE, My Brain Stopped Working - Smarter Every Day 175&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At about &lt;a href=&quot;https://youtu.be/lepQoU4oek4?t=264&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;code&gt;04:30&lt;/code&gt;&lt;/a&gt; they show a short clip of a video recording of what looks like a white bed sheet hanging vertically and a shimmering pattern of light on it, I assume related to the total solar eclipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's going on? What causes this? What's it called? &lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-08-04T10:40:34.440" LastActivityDate="2018-08-04T12:04:07.780" Title="What is causing this strange shimmering pattern of light during a total solar eclipse?" Tags="&lt;solar-eclipse&gt;&lt;atmospheric-effects&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27222" PostTypeId="2" ParentId="27220" CreationDate="2018-08-04T10:51:07.367" Score="2" Body="&lt;p&gt;Given the appearance of the moon, that must have been photographed on the weekend of the 21-22 (the moon was nearly full on the 28-29)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The object is Jupiter, and from the position I'd say the photograph was taken at about 0400 UTC on the 21st July 2018.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think Antares can be dimmly seen towards the left of the photo, and perhaps alpha and beta Libra area visible above and below the moon, but I can't see any other stars.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-08-04T10:51:07.367" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27223" PostTypeId="2" ParentId="27221" CreationDate="2018-08-04T12:04:07.780" Score="6" Body="&lt;p&gt;These are called &lt;a href=&quot;https://en.wikipedia.org/wiki/Shadow_bands&quot; rel=&quot;noreferrer&quot;&gt;shadow bands&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the sunlight is reduced to a very narrow strip, in the last few moments before totality, turbulence and refraction in the atmosphere will cause shimmering bands of light and shadow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are seeing here is essentially the same as the twinkle that you see from stars.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-08-04T12:04:07.780" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27224" PostTypeId="1" CreationDate="2018-08-04T13:30:46.363" Score="2" ViewCount="78" Body="&lt;p&gt;I'm going to be in Chicago during the weekend of Aug 11th, 2018. Can you suggest any viewing parties that are not too far away? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm very very amateur. I don't really know too much about astronomy, but do like being around people that do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please post any events that you know of like this one that was one about an hour away at Indiana Dunes State Park. &lt;a href=&quot;https://www.trippy.com/a/5b65a86ae4b0e99fd1765e1b/507c527fe4b0c31db879944f&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.trippy.com/a/5b65a86ae4b0e99fd1765e1b/507c527fe4b0c31db879944f&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thank you&lt;/p&gt;&#xA;" OwnerUserId="23915" LastEditorUserId="23915" LastEditDate="2018-08-04T22:16:12.093" LastActivityDate="2018-08-04T22:16:12.093" Title="Perseid Meteor Shower Viewing events near Chicago" Tags="&lt;amateur-observing&gt;&lt;star-gazing&gt;&lt;meteor-shower&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27226" PostTypeId="1" AcceptedAnswerId="27280" CreationDate="2018-08-04T16:18:27.913" Score="3" ViewCount="63" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/27223/7982&quot;&gt;This answer&lt;/a&gt; explains that the shimmering patterns seen on surfaces at the moment before totality of a solar eclipse are called shadow bands, and that the phenomenon is closely related to the twinkling appearance of stars (but not angularly larger planets) through the Earth's atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If for some reason there happened to be a bright star close enough to Earth that its nighttime illumination visible on surfaces, but small enough in angular width that it twinkled, would we see shadow bands at night, every night when the star was out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As there's a good chance the answer is &quot;yes&quot; to the hypothetical, I'd like to ask if this situation is in fact possible, or if a star were close enough to be bright enough to visibly illuminate a white surface at night, would it then be large enough in angular extent that it wouldn't twinkle any more?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The pattern from the Eclipse can be seen in the Smarter Every Day video &lt;a href=&quot;https://youtu.be/lepQoU4oek4&quot; rel=&quot;nofollow noreferrer&quot;&gt;Space Station Transiting 2017 ECLIPSE, My Brain Stopped Working - Smarter Every Day 175&lt;/a&gt; at about &lt;a href=&quot;https://youtu.be/lepQoU4oek4?t=264&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;code&gt;04:30&lt;/code&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-08-04T16:25:28.873" LastActivityDate="2018-08-08T12:02:49.303" Title="Would an extremely bright star produce same kind of shadow band effect as seen in solar eclipses just before totality?" Tags="&lt;solar-eclipse&gt;&lt;atmospheric-effects&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27227" PostTypeId="2" ParentId="27224" CreationDate="2018-08-04T16:42:22.500" Score="1" Body="&lt;p&gt;The Astronomical League's &lt;a href=&quot;https://www.astroleague.org/societies/IL&quot; rel=&quot;nofollow noreferrer&quot;&gt;list of astronomy clubs&lt;/a&gt; leads to these opportunities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;a href=&quot;http://www.skokievalleyastronomers.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Skokie Valley Astronomers&lt;/a&gt; have a meeting at the &lt;a href=&quot;https://www.lcfpd.org/ryerson/welcome-center/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ryerson Forest Preserve&lt;/a&gt; in Riverwoods on Friday the 10th.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Some &lt;a href=&quot;http://www.stargazing.net/naa/index.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;Naperville Astronomical Association&lt;/a&gt; members will be at a &lt;a href=&quot;https://cantigny.org/event/french-picnic-stars/&quot; rel=&quot;nofollow noreferrer&quot;&gt;stargazing picnic at Cantigny Park&lt;/a&gt; in Wheaton on Saturday the 11th.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Like most meteor showers, the Perseids produce more meteors in the early morning, but there should be some in the late evening.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-08-04T16:42:22.500" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27228" PostTypeId="1" CreationDate="2018-08-04T23:16:38.463" Score="2" ViewCount="73" Body="&lt;p&gt;Today I could roughly see Saturn with ring structure for the first time with my Seben 114-1000 telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then later I was dealing with Mars and found out that I must do something wrong, as with no eyepiece I could really see this planet closer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, whether I take 20, 12.5, or 6 mm eyepieces, the size as I set sharp focus remains a small light orange circle of solid uniform colour.  The app sky view shows for today 64% clear view, maybe an explanation why I can't see more details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is this possible to see Mars filling in the view field? And why there is no recognizable difference between 20 and 6 mm - do I need them in combination with something else? &#xA;Here it says, even the &lt;a href=&quot;https://www.levenhuk.com/reviews/things_you_can_see/&quot; rel=&quot;nofollow noreferrer&quot;&gt;polar caps&lt;/a&gt; of Mars should be recognizable with this telescope model scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPD&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, it looks like that with &lt;strong&gt;up to 6 mm&lt;/strong&gt; I can zoom in fairly well to such details as:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;large moon craters&lt;/li&gt;&#xA;&lt;li&gt;Jupiter (seen as small white illuminated blob) with the Galilean moons (seen as  stars)&lt;/li&gt;&#xA;&lt;li&gt;Solid white-coloured Saturn with its ring and some space between planet and ring visible&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Mars keeps being small to medium yellow-orange disk of solid color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using 4 mm gives little bit more zoom but less detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I can see with 6 mm that Moon contours are slightly jittering, maybe through light refraction in the night air (all observation done from the ground on a field). The temperature in the night has been 21°C.&lt;/p&gt;&#xA;" OwnerUserId="22664" LastEditorUserId="22664" LastEditDate="2018-08-06T08:32:17.997" LastActivityDate="2018-08-06T08:32:17.997" Title="Mars and 115 mm - which eyepieces?" Tags="&lt;mars&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27229" PostTypeId="2" ParentId="27228" CreationDate="2018-08-05T02:21:40.627" Score="1" Body="&lt;p&gt;Mars is currently experiencing a global dust storm. There are virtually no features visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, some dark markings and the polar cap would be visible when conditions are favorable (meaning when it is close to us as it is now.)&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2018-08-05T02:21:40.627" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27230" PostTypeId="1" AcceptedAnswerId="27231" CreationDate="2018-08-05T06:07:12.450" Score="1" ViewCount="79" Body="&lt;p&gt;I would imagine that during a sunset, for a brief moment of time, there would be a shadow of the earth seen in the clouds but I have never seen this. Why does it not happen?&lt;/p&gt;&#xA;" OwnerUserId="23917" LastActivityDate="2018-08-05T08:26:28.450" Title="Why is the shadow of the earth not visible during sunsets?" Tags="&lt;the-sun&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27231" PostTypeId="2" ParentId="27230" CreationDate="2018-08-05T08:26:28.450" Score="0" Body="&lt;p&gt;To have a sharp shadow, you need a point source of light.  But the Sun is extended enough that it does not set all at once, so there is not a sharp shadow.  Also, there is refraction of the redder light, so instead of seeing a shadow, you'd see reddish clouds-- which I wager you see quite often (though much of that effect is from the way the atmosphere scatters away the blue light-- those effects are hard to disentangle but you know refraction matters because you see how the Sun's shape is distorted upon setting).&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-08-05T08:26:28.450" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27232" PostTypeId="1" AcceptedAnswerId="27234" CreationDate="2018-08-05T09:16:37.460" Score="5" ViewCount="231" Body="&lt;p&gt;The tropics is region of the Earth in which the Sun could be directly overhead.  What is the corresponding region for the Moon?  Is it the same or slightly larger?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I live in the UK and hence have to travel quite far to see the Sun directly overhead.  If I am content to see the Moon directly overhead, would my trip be much shorter?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My experience suggests so.  Once long ago, I travelled to India without flying.  I &quot;saw&quot; the Moon directly overhead while in Iran and hence not yet in the Tropics.  However, I had no way to accurately measure this and I may have mistaken the usually high (for me) Moon for being exactly overhead.&lt;/p&gt;&#xA;" OwnerUserId="21104" LastActivityDate="2018-08-05T10:47:46.077" Title="&quot;Lunar tropics&quot;" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27233" PostTypeId="1" AcceptedAnswerId="27238" CreationDate="2018-08-05T10:27:51.743" Score="6" ViewCount="530" Body="&lt;p&gt;The Next Big Future article &lt;a href=&quot;https://www.nextbigfuture.com/2018/08/rogue-exoplanet-12-7-times-bigger-than-jupiter-is-20-light-years-away.html&quot; rel=&quot;noreferrer&quot;&gt;Rogue Exoplanet 12.7 times bigger than Jupiter is 20 light years away&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://public.nrao.edu/news/planetary-mass-powerhouse/&quot; rel=&quot;noreferrer&quot;&gt;Astronomers using the National Science Foundation’s Karl G. Jansky Very Large Array (VLA) have made the first radio-telescope detection of a planetary-mass object beyond our Solar System.&lt;/a&gt; They found a rogue planet 12.7 times the mass of Jupiter twenty light years from earth.&lt;/p&gt;&#xA;&lt;p&gt;It is at the boundary of Jovian objects and Brown dwarf stars.&lt;/p&gt;&#xA;&lt;p&gt;[...]&lt;/p&gt;&#xA;&lt;p&gt;Last year, an independent team of scientists discovered that SIMP J01365663+0933473 was part of a very young group of stars. Its young age meant that it was in fact so much less massive that it could be a free-floating planet — only 12.7 times more massive than Jupiter, with a radius 1.22 times that of Jupiter. At 200 million years old and 20 light-years from Earth, the object has a surface temperature of about 825 degrees Celsius, or more than 1500 degrees Fahrenheit. By comparison, the Sun’s surface temperature is about 5,500 degrees Celsius.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;The difference between a gas giant planet and a brown dwarf remains hotly debated among astronomers&lt;/strong&gt;, but one rule of thumb that astronomers use is the mass below which deuterium fusion ceases, known as the “deuterium-burning limit”, around 13 Jupiter masses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In the spirit of&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/q/19963/7982&quot;&gt;Do astronomers generally agree that the distinction between comets and astroids is not so clear?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/q/19044/7982&quot;&gt;Is this object an asteroid or comet, and how can it produce so many tails?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;and the relatively recent reclassification of Pluto as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Dwarf_planet&quot; rel=&quot;noreferrer&quot;&gt;dwarf planet&lt;/a&gt; and not a planet proper, I'd like to ask what are the ways that have are currently proposed and given consideration to distinguish between a really big planet and a really small star?&lt;/p&gt;&#xA;&lt;p&gt;Is the &amp;quot;rule of thumb&amp;quot; proposed in the block quote the leading contender or are there other definitions that receive a similar amount of support&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-08-05T12:37:33.550" Title="Currently proposed ways to distinguish between a really big planet and a really small star" Tags="&lt;star&gt;&lt;planet&gt;&lt;exoplanet&gt;&lt;classification&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27234" PostTypeId="2" ParentId="27232" CreationDate="2018-08-05T10:47:46.077" Score="6" Body="&lt;p&gt;The orbital inclination of the Moon is 5.1°, which means it can be that amount above or below the ecliptic. That means it can theoretically reach a declination of 23.4° + 5.1° = 28.5° (North and South), so &lt;a href=&quot;https://en.wikipedia.org/wiki/Chabahar&quot; rel=&quot;noreferrer&quot;&gt;Chabahar&lt;/a&gt;, in the south of Iran, lying 25° North of the equator, is within the &quot;lunar tropics&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will not happen every month or even year; sometimes, the orbital inclination of the Moon (around the Earth) and the Earth (around the Sun) cancel out and the maximum declination will be 23.4° - 5.1° = 18.3°. This is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_standstill&quot; rel=&quot;noreferrer&quot;&gt;minor lunar standstill&lt;/a&gt;. One could say that at that moment, the &quot;lunar tropics&quot; are smaller than the &quot;solar tropics&quot;. The last minor lunar standstill happened in October 2015; we're slowly moving towards a major standstill which will occur in April 2025.&lt;/p&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2018-08-05T10:47:46.077" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27235" PostTypeId="1" AcceptedAnswerId="27236" CreationDate="2018-08-05T10:48:48.393" Score="1" ViewCount="89" Body="&lt;p&gt;The Next Big Future article &lt;a href=&quot;https://www.nextbigfuture.com/2018/08/rogue-exoplanet-12-7-times-bigger-than-jupiter-is-20-light-years-away.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rogue Exoplanet 12.7 times bigger than Jupiter is 20 light years away&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;SIMP J01365663+0933473 has a magnetic field more than 200 times stronger than Jupiter’s. The object was originally detected in 2016 as one of five brown dwarfs the scientists studied with the VLA to gain new knowledge about magnetic fields and the mechanisms by which some of the coolest such objects can produce strong radio emission. &lt;strong&gt;Brown dwarf masses are notoriously difficult to measure, and at the time, the object was thought to be an old and much more massive brown dwarf&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is this true? Are brown dwarf masses notoriously difficult to measure? Naively, it seems to me if I had brightness at a few wavelengths I could deduce a temperature, and if I had a distance, at least for a blackbody I'd know the area as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are one or both of these in fact difficult to get, or is there something else I'm missing? &lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-08-05T12:11:28.437" Title="Are brown dwarf masses notoriously difficult to measure?" Tags="&lt;temperature&gt;&lt;brown-dwarf&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27236" PostTypeId="2" ParentId="27235" CreationDate="2018-08-05T11:58:45.523" Score="4" Body="&lt;p&gt;Yes, they are difficult to measure. The only way to &lt;em&gt;measure&lt;/em&gt; a brown dwarf mass is if it is in a binary system. The binary frequency of brown dwarfs is low - perhaps 10-15%. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having found your binary system, then it has to be wide enough to resolve with your telescope, but not so wide that Kepler's third law means it takes centuries to do an orbit. The low mass of brown dwarfs does not help here. To have an orbit of say a decade or less means they have to orbit close to the other object. This in turn means adaptive optics are required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said all that, the masses of quite a few brown dwarfs &lt;em&gt;have&lt;/em&gt; been measured, so it is not extraordinarily difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The object talked about in the article is &lt;em&gt;not&lt;/em&gt; in a close binary, so it's mass cannot be &lt;em&gt;measured&lt;/em&gt;, it can only be estimated indirectly. But here come the difficulties. To estimate the mass requires accurate calculations of the evolutionary tracks of brown dwarfs. The &lt;em&gt;measurements&lt;/em&gt; of brown dwarf masses in binaries indicates that we do not have these. Secondly, to use the evolutionary tracks requires a luminosity (and therefore a distance), but because brown dwarfs are cooling objects, you also need &lt;em&gt;an age&lt;/em&gt;. In general, unless the brown dwarf can be associated with a group of other stars with known age then this isn't possible.  Measuring the temperature of a brown dwarf isn't helpful because the cooling tracks are very close together in the Hertzsprung-Russell diagram, and in any case, it is &lt;em&gt;very&lt;/em&gt; difficult because brown dwarfs are nowhere near blackbodies and their atmospheres are still poorly understood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, it seems that the brown dwarf has been newly associated with an (assumed to be) coeval group of stars at 200 Myr. Thus its luminosity can give a (highly) model-dependent mass. I cannot examine the details as the claimed published research cannot be found! &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-08-05T12:11:28.437" LastActivityDate="2018-08-05T12:11:28.437" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27237" PostTypeId="2" ParentId="27233" CreationDate="2018-08-05T12:14:08.433" Score="3" Body="&lt;p&gt;Like in each of the other cases you mention, when dividing really big planets from really small stars you're drawing a bright line across a continuous range of objects.  &lt;em&gt;Whenever&lt;/em&gt; you do that, you're going to get legitimate and reasonable disagreement, because different people with different purposes will see a different place to draw the (ultimately arbitrary) line as most useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Emphasis: This is not to say that drawing such lines is a useless business. Human beings need categories to think clearly and to communicate productively.  We need those subdivisions.  We just don't always agree on exactly what they ought to be.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at stars vs. planets, there are basically two dimensions to consider: Mass and metallicity. If you look at bodies which are nothing but primordial H and He, as you go down in mass at some point you read a mass where the H didn't ignite and at an even lower mass, you reach a point where the D didn't ignite. In bodies less massive than that, no fusion ever took place.  Since fusion (and the resultant high temperature and emission of radiation) is the outstanding characteristic of stars, it's very natural to say &quot;stars are massive bodies which sustained fusion at some point in their life; planets are those that never did.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you study the theories of star and planet formation, you see that ignition depends pretty much entirely on the object's mass. Unless the mass is large enough, conditions for fusion don't happen. (We're talking natural objects condensing out of nebulae here -- smaller fusing masses could be constructed with sufficient effort.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you add the second dimension of metallicity (basically how much of elements heavier than He are present) you discover that it doesn't make a huge difference. There doesn't seem to be a natural way for planets to get to the ignition mass without being mostly H and He, and these planets behave almost as if they had no heavier elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while have found a continuum of objects out there ranging from pebbles to blue-white super-giants, the deuterium ignition temperature is for most people a handy place to use as the border between categories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this is not always going to be right.  We can, for instance, imagine objects that are very old and have fused deuterium, but whose surfaces have cooled to the point of being solid and are made of ices. This would be a star by the above definition, but would be much more like a planet for anyone interested in studying its surface. (Kinda high gravity to visit, though!)  Here the D-fusion dividing line would be ill-chosen, and a dividing line based on the nature of the surface would be better -- perhaps dividing large bodies in space into categories with surfaces that are solid or liquid from those with surfaces that are gaseous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that being hot and radiating isn't a good dividing line, since the heat of formation -- gravitational energy released during formation -- makes every massive object start hot.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2018-08-05T12:34:11.703" LastActivityDate="2018-08-05T12:34:11.703" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27238" PostTypeId="2" ParentId="27233" CreationDate="2018-08-05T12:27:30.520" Score="5" Body="&lt;p&gt;The dividing line between star and brown dwarf is the mass at which hydrogen fusion via the pp-chain occurs at its core. A brown dwarf below this mass limit will never reach a sustainable equilibrium due to this reaction and will continue to cool as it gets older. A star will reach the long-lived main sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dividing line between planet and brown dwarf is much less well defined. Some want the definition to be mass-based, and there is a mass-based dividing line at about 13 Jupiter masses between brown dwarfs that fuse deuterium in their interiors (which provides only a brief hiatus in their cooling) and &quot;planets&quot; that do not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Others argue that any definition of a planets should be based on whether they formed around a star, but this is problematic, because such objects may then be disrupted or escape and also it is not clear where a binary brown dwarf becomes a star plus planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that some campaign for a different definition is that the composition and structure of a planet formed around a star could be different to that of a planet that collapses from a small gas cloud. According to the &quot;core accretion model&quot; for gas giant formation, the core of such a planet would consist of solid ice or rocky material, whereas it would be gas all the way down in a less massive version of a brown dwarf. However, the waters are further muddied by ideas that large planets can also form by direct collapse  in the disc of a forming planetary system.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-08-05T12:37:33.550" LastActivityDate="2018-08-05T12:37:33.550" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27240" PostTypeId="1" AcceptedAnswerId="27241" CreationDate="2018-08-05T18:45:28.777" Score="3" ViewCount="189" Body="&lt;p&gt;The below photos were taken yesterday (August 4, 2018) at about 10 p.m. from the plane over the Black Sea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We were seeing the light dot on the left for at least 10-20 minutes. It looked like a bright star, but there were no other stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How likely that it was Mars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GIB9D.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GIB9D.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/eriFN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eriFN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23923" LastActivityDate="2018-08-05T19:25:34.710" Title="Is the dot on the photo - Mars?" Tags="&lt;planet&gt;&lt;photography&gt;&lt;mars&gt;&lt;night-sky&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27241" PostTypeId="2" ParentId="27240" CreationDate="2018-08-05T19:25:34.710" Score="6" Body="&lt;p&gt;This was not Mars.  Mars is close to opposition now, so it almost opposite the sun, and visible as an obviously red &quot;star&quot; in the Southern sky at midnight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your photo shows that the sun has just set, and you note that other stars were not visible.  &lt;em&gt;This is the planet Venus&lt;/em&gt;. Currently shining at magnitude -5 it is 100 times brighter than the most bright stars. It is common, shortly after sunset for it to be the only visible &quot;star&quot;, and as it orbits closer to the sun than the Earth, it is always seen in the sky close to the sun, as your photo shows. The when the Planet Venus is an &quot;evening star&quot; it is seen in the west. This is consistent with your plane travelling in a roughly northerly direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is hard to be sure what time zone, you are in, or exactly where you are, but Venus will be visible shortly after sunset in most parts of the world.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-08-05T19:25:34.710" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27242" PostTypeId="1" AcceptedAnswerId="27243" CreationDate="2018-08-06T02:02:20.913" Score="2" ViewCount="185" Body="&lt;p&gt;I dug up my old 6&quot; Dobsonian telescope from storage and the mirror looks like the following:&lt;a href=&quot;https://i.stack.imgur.com/A6sCM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/A6sCM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture is down the tube, so you can see the spider holding the secondary. To my eye, the spots look like corrosion and not cleanable, but does anyone actually know what they are?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was stored relatively carefully - not totally sealed but in a house, with the tube covered with a plastic bag. So huge temperature changes are not likely, but it does quite humid here. It was last in action &gt; 10 years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only other thing I know about the scope is that it was made by Criterion, sometime before the 1980s.&lt;/p&gt;&#xA;" OwnerUserId="2010" LastActivityDate="2018-08-06T04:08:20.163" Title="Can this mirror be cleaned?" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;mirror&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27243" PostTypeId="2" ParentId="27242" CreationDate="2018-08-06T04:08:20.163" Score="3" Body="&lt;p&gt;The spots could be mold or fungus.&#xA;If a &lt;a href=&quot;https://www.skyandtelescope.com/astronomy-resources/caring-for-your-optics/&quot; rel=&quot;nofollow noreferrer&quot;&gt;gentle cleaning&lt;/a&gt; with distilled water, a little detergent, and sterile cotton (no abrasives!) fails to remove them, there are several vendors who can strip the old aluminum coating and apply a new one.&#xA;If you don't want to go that far, you can probably find a buyer who would.&#xA;Criterion scopes have a considerable fan base.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-08-06T04:08:20.163" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27244" PostTypeId="2" ParentId="27218" CreationDate="2018-08-06T07:16:35.520" Score="3" Body="&lt;p&gt;If galaxies were randomly distributed (a spatial Poisson process), then the probability of having $N$ galaxies inside a radius $r$ sphere is $\Pr[N]=\lambda^N e^{-\lambda} / N!$ where $\lambda = (4\pi /3)\rho r^3$. So the cumulative distribution function of the distance to the $N$th neighbour is $$\Pr[R_N&amp;lt;r]=1-e^{-\alpha r^3}\sum_{n=0}^{N-1} \frac{\alpha^n r^{3n}}{n!}$$ ($\alpha=4\pi\rho/3$ for brevity). Taking the derivative to get the PDF gives $$f(r) = e^{-\alpha r^3}\left [(3\alpha r^2) + 3\sum_{n=1}^{N-2} \frac{\alpha^n r^{3n-1}}{(n-1)!}\right ].$$ While one can try taking the derivative of this to get the mode (or setting the CDF to 1/2 to find the median) this becomes algebraically messy so a numerical solution is likely best. I get that the mode for the 9th neighbour is 2.44 times the mode of the first neighbour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5zn1B.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5zn1B.png&quot; alt=&quot;CDF and PDF of the closest neighbor distances in 3d space&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem here is of course that actual galaxies are highly clumped. That will cause a (potentially large) reduction in the median distance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, most galaxies have satellite galaxies, so unless one has a definition in the survey excluding them the answer will be that the 9th nearest neighbour is very close (in the case of the Milky Way the distance would be about 30 kpc). So any answer will depend on the criteria on inclusion in this particular catalogue. &lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2018-08-06T07:16:35.520" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27245" PostTypeId="1" AcceptedAnswerId="27248" CreationDate="2018-08-06T07:53:06.277" Score="-1" ViewCount="43" Body="&lt;p&gt;How at the beginning of this universe the fields as proposed in quantum fields theory get excited??&#xA;How exactly these fields originated?&lt;/p&gt;&#xA;" OwnerUserId="18093" LastActivityDate="2018-08-06T11:26:30.270" Title="Field excitement in quantum field theory?" Tags="&lt;magnetic-field&gt;&lt;early-universe&gt;&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2018-08-06T18:49:43.660" ContentLicense="CC BY-SA 4.0" />
  <row Id="27247" PostTypeId="1" AcceptedAnswerId="27581" CreationDate="2018-08-06T08:31:28.853" Score="1" ViewCount="252" Body="&lt;p&gt;Recent nights where I try to make my first observations with the entry levell telescope Seben with 114 mm mirror it seems that for lenses less than 6 mm details go lost due to warm air refraction (as observed with the Moon contour).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everywhere they say, it shouldn't be too warm: is it possible &lt;strong&gt;to recommend a rough temperature range&lt;/strong&gt; where you could go for say 4 mm and still see details?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Side note: the overall visibility index was about 77%).&lt;/p&gt;&#xA;" OwnerUserId="22664" LastActivityDate="2018-09-07T21:12:45.570" Title="Air temperature range recommendation for observations in a summer night with an entry level telescope?" Tags="&lt;telescope&gt;&lt;temperature&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27248" PostTypeId="2" ParentId="27245" CreationDate="2018-08-06T11:26:30.270" Score="0" Body="&lt;p&gt;Unfortunately, this question is one which is under active -- and so far fruitless -- investigation by the theoretical physics community.  Anyone who could answer it would have the choice of publishing here and getting +450 reputation or so, and a couple of badges. Or they could publish in &lt;em&gt;Nature&lt;/em&gt; of &lt;em&gt;Phys Rev&lt;/em&gt; and get at least one Nobel Prize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line: No one knows yet.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-08-06T11:26:30.270" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27249" PostTypeId="1" AcceptedAnswerId="27254" CreationDate="2018-08-06T11:35:35.930" Score="19" ViewCount="3809" Body="&lt;p&gt;Last night I was experimenting with my mobile phone shooting the night sky adjusting some settings like shutter time and ISO. One of the first pictures I thought was a complete failure, I only adjusted the shutter time to 30 seconds. But when I zoomed in on what I believed to be just some bright stars, I was amazed about what I saw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the original picture&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Q4ykY.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q4ykY.jpg&quot; alt=&quot;Original photo&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I zoomed in on the center most bright star and saw this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yJ3YB.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yJ3YB.jpg&quot; alt=&quot;cutout center bright star&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I zoomed in on that star at the bottom just a bit right of the center and saw this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZkowR.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZkowR.jpg&quot; alt=&quot;cutout bottom bright star&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm wondering what these are? Did I shoot some deep space objects? Or is this just static from the long exposure? Or maybe just pixalating from the zooming? I'm really curious about your answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My camera was just pointing at zenit and my location was 50°57'35.5&quot;N 5°05'10.3&quot;E the time was 1:06am on august the 6th 2018.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a screen with some more information about the photo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1NBh9.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1NBh9.png&quot; alt=&quot;Screen shot photo details&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23927" LastActivityDate="2018-08-07T10:39:52.263" Title="What did I photograph?" Tags="&lt;photography&gt;&lt;deep-sky-observing&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27250" PostTypeId="1" AcceptedAnswerId="27393" CreationDate="2018-08-06T11:44:20.070" Score="4" ViewCount="743" Body="&lt;p&gt;Given you take a (non-amateur) refractor telescope, what are aperture and magnification required to recognize phases of the four Galilean moons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What did Galileo see:&#xA;&lt;a href=&quot;http://www.astro.umontreal.ca/~paulchar/grps/site/images/galileo.4.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.astro.umontreal.ca/~paulchar/grps/site/images/galileo.4.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22664" LastEditorUserId="22664" LastEditDate="2018-08-10T13:28:45.067" LastActivityDate="2018-08-19T09:28:15.543" Title="What is minimal aperture/magnification to recognize phases of Galilean moons?" Tags="&lt;jupiter&gt;&lt;natural-satellites&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27251" PostTypeId="2" ParentId="27249" CreationDate="2018-08-06T12:32:35.377" Score="3" Body="&lt;p&gt;At the f-number you have used, you'd already expect to get weird images. In astronomy we usually use $f=\infty$, but this is of course only relevant for extended objects, not point like sources.&#xA;But with $f/2$ it's impossible to say whether it's extended or not.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still I'd say this is probably starlight, as also the fainter objects in your field seem to have some artifacts. Those can easily be generated by remnant vibrations, which degrade the quality of any long exposure without proper mounting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also without a &lt;a href=&quot;http://www.cfht.hawaii.edu/~baril/Pyxis/Help/flatdarkfield.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;dark field&lt;/a&gt; it is impossible to say what stems from imperfect pixelation in the camera (which every CCD has) and what part of the signal you see comes from the source object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In total, the question &quot;what did I see&quot; is hard to answer for faint objects without proper preparation of the photograph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point, your pupil entrance is very small, for nebulae/galaxies you usually need larger (amateur-sized) telescopes to get an image on a ~30 seconds timeframe.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditDate="2018-08-07T10:39:52.263" LastActivityDate="2018-08-07T10:39:52.263" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27252" PostTypeId="1" CreationDate="2018-08-06T13:41:21.613" Score="2" ViewCount="333" Body="&lt;p&gt;I would like to know, what is the average temperature of Earth as a whole planet, not just its surface temperature. Hope the following scenario will make it more clear what I mean:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine, that we isolated/sealed Earth (up to 30 km above sea level) as it is right now, so that there is no mass or energy being added or escaping and waited several millions of years till the temperature is everywhere the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would that temperature be? When I tried to google it, I only could find references to surface temperature, and not to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Based on remarks in comment section:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not an expert, that is why asked this question, however IMHO of a layman in this area, I presumed that radioactive decay wont play that much of a role (like a +-1%). And the role of a gravity is well beyond me. But the most important thing is, I pretty much wanted to know the average temperature by heat capacity of the Earth at this moment (so basically, all decays and all tidal forces excluded, even if they play significant role).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would just like to have (a fairly confident) idea about the potential temperature, and no need for highly exact numbers and calculations, +-15% is absolutely fine. &lt;/p&gt;&#xA;" OwnerUserId="18206" LastEditorUserId="18206" LastEditDate="2018-08-06T22:17:48.213" LastActivityDate="2018-08-06T22:17:48.213" Title="What is Earth's average temperature as a whole?" Tags="&lt;earth&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="7" ClosedDate="2018-08-07T07:07:27.117" ContentLicense="CC BY-SA 4.0" />
  <row Id="27254" PostTypeId="2" ParentId="27249" CreationDate="2018-08-06T15:03:06.943" Score="18" Body="&lt;p&gt;A contrast stretch reveals stars down to magnitude 4 or 5.&#xA;The stars you asked about are Deneb (center) and Vega (bottom).&#xA;The constellation in the center and below is Cygnus; we also see Cepheus at upper right and part of Draco at lower right.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/sJcKl.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sJcKl.jpg&quot; alt=&quot;OP&amp;#39;s image modified&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stars appear as blobs 10-12 arcminutes wide due to several factors:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Rotation around the north celestial pole, to the right of the upper right corner of the image. In your 30 second exposure, Deneb and Vega drifted 5-6 arcminutes in the 5 o'clock direction.&lt;/li&gt;&#xA;&lt;li&gt;Diffraction due to finite aperture. The width of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Airy_disk&quot; rel=&quot;noreferrer&quot;&gt;ring pattern&lt;/a&gt; is inversely related to the width of the aperture. These rings are consistent with a 2-3 mm aperture.&lt;/li&gt;&#xA;&lt;li&gt;Optical aberrations such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Coma_(optics)&quot; rel=&quot;noreferrer&quot;&gt;coma&lt;/a&gt;, increasing toward the edge of the image.&lt;/li&gt;&#xA;&lt;li&gt;Vibration as noted in another answer.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-08-06T15:19:57.623" LastActivityDate="2018-08-06T15:19:57.623" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27255" PostTypeId="2" ParentId="27252" CreationDate="2018-08-06T16:47:24.917" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is Earth's average temperature as a whole?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sealing the planet or using $E = mc^2$ isn't a practical means to measure the temperature. Here's how the &lt;em&gt;average&lt;/em&gt; is usually determined:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://www.tufts.edu/about/tufts-at-a-glance&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tufts University&lt;/a&gt; - Global Warming - &lt;a href=&quot;http://ase.tufts.edu/cosmos/view_chapter.asp?id=21&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heating by the greenhouse effect&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;... the average surface temperature of the planet is 288 degrees kelvin (15 degrees Celsius or 59 degrees Fahrenheit).&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: Space.com - &lt;a href=&quot;https://www.space.com/17816-earth-temperature.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;What Is Earth's Average Temperature?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;According to &lt;a href=&quot;https://data.giss.nasa.gov/gistemp/&quot; rel=&quot;nofollow noreferrer&quot;&gt;GISS&lt;/a&gt;, the global mean surface air temperature for that period was estimated to be 57 F (14 C). That would put the planet's average surface temperature in 2017 at 58.62 F (14.9 C).&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18054" LastActivityDate="2018-08-06T16:47:24.917" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27256" PostTypeId="1" CreationDate="2018-08-06T17:17:37.107" Score="4" ViewCount="2400" Body="&lt;p&gt;I really want to know why I can’t see detail on planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This summer my Dad and I went on a trip to Utah I went outside to look at the planets and couldn’t see detail on Mars, Saturn, and Jupiter. Saturn's rings were visible but very faint and Jupiter was a white ball and then it’s four moons, this was all out in the middle of nowhere with 0 light pollution at 6000ft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then went into Bryce Canyon at 9100ft and still the same outcome the only difference was a very very faint strip on Jupiter, and it’s so dry in Utah and at an altitude of 9100ft where the atmosphere is not as heavy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there still dew? Is my telescope not powerful enough I have an Astromaster 130 reflector and a 60az explorerscope refractor. I’m very confused&lt;/p&gt;&#xA;" OwnerUserId="23931" LastEditorUserId="2365" LastEditDate="2018-08-06T17:25:25.133" LastActivityDate="2020-07-10T04:53:52.007" Title="Why can’t I see any detail on planets?" Tags="&lt;telescope&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27257" PostTypeId="2" ParentId="27256" CreationDate="2018-08-06T17:35:53.580" Score="13" Body="&lt;p&gt;Planets are really small!  (when seen from Earth) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.deepskywatch.com/Articles/what-can-i-see-through-telescope.html&quot; rel=&quot;noreferrer&quot;&gt;Deepskywatch has an example&lt;/a&gt; of what can been seen with a 150mm reflector and a magnification of x180 (ie slightly more than your Astromaster which has a 130mm main mirror, and comes with a x65 eyepiece)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/d9gHn.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/d9gHn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to see the 4 moons, and one or two bands on Jupiter. This assumes good &quot;seeing&quot; (the amount of turbulence in the air, not the same as the clarity of the air), and a properly collimated and aligned telescope lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of other good deep-sky targets that benefit more from a dark sky than the planets.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-08-06T17:35:53.580" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27258" PostTypeId="2" ParentId="27247" CreationDate="2018-08-06T20:12:24.290" Score="0" Body="&lt;p&gt;The temperature is not important, at least not in the way that you are asking. The temperature of the telescope, the air around it, and the miles (kilometers) of air that you are looking through are what makes a difference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously you cannot control the miles (kilometers) of air, so the only control you have is the temperature of the scope compared to the local temperature. You want that to be as equal as possible, so let the scope acclimate to the air temperature (usually meaning to let the scope cool down). Of course, avoid looking over terrain that had uneven temperatures such as hot rooftops, roads, concrete, and do forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is likely that the 4 mm eyepiece will never give a good view (depending on the quality) or may give an acceptable view on rare occasions,&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2018-08-06T20:12:24.290" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27260" PostTypeId="2" ParentId="27252" CreationDate="2018-08-06T22:14:23.367" Score="0" Body="&lt;p&gt;Another way of saying this is that you're looking for the average temperature, weighted by heat capacity; you're looking for the total heat of the earth divided by the total heat capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at it weighted by mass, I found &lt;a href=&quot;https://www.colorado.edu/physics/phys2900/homepages/Marianne.Hogan/inside.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this site&lt;/a&gt; that gives the mass of the mantle as 67.3% and outer core as 30.8%. The temperature of the mantle is given as ranging from 1000 to 3700 C, while the outer core is given as ranging from 3700 to 4300. This gives a range of 1900 to 3900 C. The calculation would be a bit different if done in terms of heat capacity rather than mass. Almost half of the mantle is oxygen, but it wouldn't be in gaseous form, so it would be difficult to find what the appropriate heat capacity is.&lt;/p&gt;&#xA;" OwnerUserId="19010" LastActivityDate="2018-08-06T22:14:23.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27261" PostTypeId="1" AcceptedAnswerId="27290" CreationDate="2018-08-07T02:58:11.340" Score="6" ViewCount="199" Body="&lt;p&gt;How deep in Mars' surface would one have to go to both not need a pressurized suit and be warm enough to just wear clothes? in other words is there a dept Goldilocks where you would only need an oxygen supply? &lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-08-18T04:18:18.303" LastActivityDate="2018-08-18T04:18:18.303" Title="How deep in Mars would a person not need insulation and pressure?" Tags="&lt;planetary-atmosphere&gt;&lt;mars&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2018-08-20T07:30:46.063" ContentLicense="CC BY-SA 4.0" />
  <row Id="27263" PostTypeId="1" AcceptedAnswerId="27272" CreationDate="2018-08-07T03:39:15.067" Score="1" ViewCount="206" Body="&lt;p&gt;The LiveScience article &lt;a href=&quot;https://www.livescience.com/63268-incredible-hulk-comet.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Incredible Hulk? Nah, This Glowing, Green Light in the Night Sky Is a Comet&lt;/a&gt; published just hours ago discusses Comet PANSTARRS C/2017 S3 and mentions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;bright bursts exploding from its surface twice in close succession — &lt;a href=&quot;https://www.skyandtelescope.com/observing/panstarrs-comet-rocked-by-outburst-now-binocular-bright/&quot; rel=&quot;nofollow noreferrer&quot;&gt;first on June 30, and then again about two weeks later&lt;/a&gt;, Sky and Telescope reported. As Hulk himself might say, &quot;Comet flash!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The linked Sky and Telescope article &lt;a href=&quot;https://www.skyandtelescope.com/observing/panstarrs-comet-rocked-by-outburst-now-binocular-bright/&quot; rel=&quot;nofollow noreferrer&quot;&gt;PanSTARRS Comet, Rocked by Outburst, Goes Green&lt;/a&gt; is dated July 17, 2018 and offers the tantallizing information that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Comet PanSTARRS (C/2017 S3) has erupted again! Now bright enough to see in binoculars, it might become a naked-eye object if it survives until perihelion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How bright is the comet now, and how is it predicted to behave in the next week or two (baring further outbursts)?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; &quot;On July 15, Comet PanSTARRS (C/2017 S3) appeared bright and well condensed with a narrow ion tail.&quot; Photo Credit: Michael Jäger. From &lt;a href=&quot;https://www.skyandtelescope.com/observing/panstarrs-comet-rocked-by-outburst-now-binocular-bright/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Q3hyx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q3hyx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;update:&lt;/strong&gt; The Universe Today's article &lt;a href=&quot;https://www.universetoday.com/139573/catch-comet-c-2017-s3-panstarrs-in-outburst/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Catch Comet C/2017 S3 PANSTARRS In Outburst&lt;/a&gt; links to S Seiichi Yoshida's &lt;a href=&quot;http://www.aerith.net/comet/catalog/2017S3/2017S3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;page&lt;/a&gt; for the comet, but there's no recent data!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lriRk.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lriRk.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-08-07T05:37:02.520" LastActivityDate="2018-08-07T15:20:03.840" Title="How bright is Comet PANSTARRS C/2017 S3 now?" Tags="&lt;amateur-observing&gt;&lt;comets&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27264" PostTypeId="1" AcceptedAnswerId="27268" CreationDate="2018-08-07T03:40:41.763" Score="1" ViewCount="96" Body="&lt;p&gt;The WISE craft (Widefield Infrared Survey Explorer) surveys the sky in 4 wavelength bands; 3.4, 4.6, 12, and 22 $\mu$m. When it's observing an object with an estimated redshift (given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z = \frac{\lambda_{obs} - \lambda_{emit}}{\lambda_{emit}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from &lt;a href=&quot;https://en.wikipedia.org/wiki/Redshift&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;) of, say, 0.373 (the lowest redshift for one of the objects I'm studying), would the wavelength it's seen in be its observed wavelength? I'm interested in calculating the wavelength actually emitted by the object.&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2018-08-07T08:25:07.377" Title="Wavelengths for observed objects -- emitted or observed?" Tags="&lt;redshift&gt;&lt;wavelength&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27265" PostTypeId="1" AcceptedAnswerId="27269" CreationDate="2018-08-07T03:47:19.480" Score="3" ViewCount="97" Body="&lt;p&gt;In a text file I have a table of star data with Bayer/Flamsteed designations. I need to convert these into Hipparchos numbers. How can I automatically do this? &lt;/p&gt;&#xA;" OwnerUserId="23940" LastActivityDate="2018-08-07T10:58:25.610" Title="How can I convert Bayer/Flamsteed designations to Hipparchos numbers?" Tags="&lt;star-catalogues&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27267" PostTypeId="2" ParentId="27264" CreationDate="2018-08-07T05:59:50.677" Score="1" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;The WISE craft (Widefield Infrared Survey Explorer) surveys the sky in 4 wavelength bands; 3.4, 4.6, 12, and 22 &lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt;m.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;They are only using 3.4 and 4.6 since the coolant ran out. Source: FAQ &amp;quot;&lt;a href=&quot;http://wise2.ipac.caltech.edu/docs/release/neowise/neowise_2018_release_intro.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Near-Earth Object Wide-field Infrared Survey Explorer at IPAC&lt;/a&gt;&amp;quot;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;When it's observing an object with an estimated &lt;strong&gt;redshift of, say, 0.373&lt;/strong&gt; (the lowest redshift for one of the objects I'm studying), would the wavelength it sees be in its observed wavelength?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Yes, as long as it's within the very wide bandwidth of the imager filters it will be observable at such a small redshift (as long as you don't need the 12 and 22 &lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt;m bands, see spectrum below).&lt;/p&gt;&#xA;&lt;p&gt;Sources/Proof:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&amp;quot;&lt;a href=&quot;https://arxiv.org/abs/1801.04464&quot; rel=&quot;nofollow noreferrer&quot;&gt;Optical identifications of high-redshift galaxy clusters from Planck Sunyaev-Zeldovich survey&lt;/a&gt;&amp;quot; (Jan 13 2018), by R. A. Burenin, I. F. Bikmaev, I. M. Khamitov, I. A. Zaznobin, G. A. Khorunzhev, M. V. Eselevich, V. L. Afanasyev, S. N. Dodonov, J. A. Rubiño-Martín, N. Aghanim, and R. A. Sunyaev, page 2:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;Clusters located at redshifts below z ≈ 0.6 can be identified using the data of SDSS survey (Rykoff et al., 2014), using additional data from &lt;strong&gt;WISE&lt;/strong&gt; all-sky survey it is possible to identify galaxy clusters at higher &lt;strong&gt;redshifts, up to z ≈ 0.7 (Burenin, 2017)&lt;/strong&gt;. To identify clusters at even higher redshifts, deeper direct imaging data in red and near IR bands are required.&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&amp;quot;&lt;a href=&quot;https://arxiv.org/abs/1703.05597&quot; rel=&quot;nofollow noreferrer&quot;&gt;An extension of the Planck galaxy cluster catalogue&lt;/a&gt;&amp;quot; (Mar 20 2017), by R. A. Burenin, page 3:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;For galaxy cluster observations, the 3.4 µm photometric band is most useful. In this band, &lt;strong&gt;distant galaxy clusters are well detected at redshifts up to z ≈ 1–2&lt;/strong&gt; (e.g., &lt;a href=&quot;https://link.springer.com/article/10.1134%2FS1063773715050011&quot; rel=&quot;nofollow noreferrer&quot;&gt;Burenin, 2015&lt;/a&gt;).&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Source: &amp;quot;&lt;a href=&quot;http://wise2.ipac.caltech.edu/docs/release/allsky/expsup/sec4_4h.html#WISEZMA&quot; rel=&quot;nofollow noreferrer&quot;&gt;IV. WISE Data Processing - 4. Pipeline Science Modules&lt;/a&gt;&amp;quot;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/g6Ja8.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g6Ja8.gif&quot; alt=&quot;RSR Curves&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Figure 5a - The QE-based (response per photon) relative system response (RSR) curves normalized to a peak value of unity, on a logarithmic scale.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-08-07T05:59:50.677" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27268" PostTypeId="2" ParentId="27264" CreationDate="2018-08-07T08:25:07.377" Score="2" Body="&lt;p&gt;If the redshift is $z$, then the wavelength of light you observe is at a wavelength that is $1+z$ times longer than the wavelength emitted in the frame of reference of the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case $z=0.373$, so when you observe light centered on say the 3.4 micron band, then the light was emitted at a wavelength of 3.4/1.373 microns.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-08-07T08:25:07.377" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27269" PostTypeId="2" ParentId="27265" CreationDate="2018-08-07T10:58:25.610" Score="3" Body="&lt;p&gt;As the names and numbers are essentially arbitrary, the Hipparcos designation cannot be calculated from the Bayer or Flamsteed designation.  Instead you have to look up the star in a database. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a &lt;a href=&quot;http://cdsarc.u-strasbg.fr/viz-bin/Cat?IV/27A&quot; rel=&quot;nofollow noreferrer&quot;&gt;HD-DM-GC-HR-HIP-Bayer-Flamsteed Cross Index&lt;/a&gt; on line.  (&lt;a href=&quot;http://cdsarc.u-strasbg.fr/viz-bin/nph-Cat/txt?IV%2F27A/catalog.dat&quot; rel=&quot;nofollow noreferrer&quot;&gt;Direct link&lt;/a&gt;)You can get a computer to do the look up for you. Or if you only have a small number of stars, you can use SIMBAD and do the look-up by hand.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-08-07T10:58:25.610" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27272" PostTypeId="2" ParentId="27263" CreationDate="2018-08-07T15:20:03.840" Score="2" Body="&lt;p&gt;The solar elongation (angle from the Sun to the object) is about 15 degrees right now and decreasing as the comet moves in towards perihelion (closest approach to the Sun) on August 15. As &lt;a href=&quot;https://astronomy.stackexchange.com/users/2365/james-k&quot;&gt;James K&lt;/a&gt; says this means that comet is lost in the Sun's glare. Anything below 30 degrees solar elongation is tricky to impossible to see. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming it survives the close perihelion passage of 0.21 AU, it is predicted to enter the field of view of the SOHO LASCO Sun-observing coronographs on August 24:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Assuming there is anything left of C/2017 S3, perihelion will occur this month on the 15th at a rather small perihelion distance of 0.21 AU. Being only 2 weeks from perihelion, it is already at a small elongation of 31 degrees on the 1st. Most observers will lose sight of it during the first week of August as its elongation drops below 25 degrees on the 4th and 20 degrees on the 6th. If the comet is still a going concern it will be visible in the SOHO LASCO C3 FOV between August 24 and September 13. The comet appear around the 7:30 position, passes to the east (left) of the Sun (halfway between the Sun and edge of FOV) and exits the FOV around the 12:30 position. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The above info is from the &lt;a href=&quot;http://www.alpo-astronomy.org/cometblog/?page_id=23&quot; rel=&quot;nofollow noreferrer&quot;&gt;ALPO Comet Section&lt;/a&gt;. Latest LASCO image and movies are available &lt;a href=&quot;https://www.spaceweatherlive.com/en/solar-activity/solar-images/soho&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and many other places.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-08-07T15:20:03.840" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27273" PostTypeId="1" AcceptedAnswerId="27274" CreationDate="2018-08-07T15:32:41.983" Score="8" ViewCount="419" Body="&lt;p&gt;I understand factually that H&lt;sub&gt;$\delta$&lt;/sub&gt; lines are most prominent in type A stars and less so in more extreme types of stars on the H-R diagram. However I was wondering the reason for why they are not prominent in more extreme type stars such as type O or type F. For type F stars I believe (this may not be correct) none of the hydrogen is in the 4th excited state (or very few are) so there aren't many H&lt;sub&gt;$\delta$&lt;/sub&gt; balmer lines. However I am very unsure of why the EW H&lt;sub&gt;$\delta$&lt;/sub&gt; is low in type O stars. Can someone please explain this, Thanks!&lt;/p&gt;&#xA;" OwnerUserId="23909" LastEditorUserId="31410" LastEditDate="2020-11-29T00:27:51.713" LastActivityDate="2020-11-29T00:27:51.713" Title="Why is $H_\delta$ prominent in type A stars?" Tags="&lt;star&gt;&lt;observational-astronomy&gt;&lt;spectroscopy&gt;&lt;stellar-evolution&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27274" PostTypeId="2" ParentId="27273" CreationDate="2018-08-07T16:12:08.547" Score="12" Body="&lt;p&gt;H$\delta$ absorption is formed when hydrogen in the level $n=2$ is excited to $n=6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get strong H$\delta$ absorption lines you need large amounts of hydrogen in the first excited state $n=2$ and a radiation field that contains large numbers of photons with an energy equal to the difference between the $n=6$ and $n=2$ states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These requirements are satisfied in stars with photospheres between about 8,000K and 15,000K, and we call these early-F and A-type stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At cooler temperatures most of the hydrogen is in the ground state, there are few photons with sufficient energy (3eV) to be absorbed and the H$\delta$ line weakens. At much higher temperatures (in O-stars for example) then none of the hydrogen is in the $n=2$ level, and in fact most of it is ionised.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-08-07T22:59:50.830" LastActivityDate="2018-08-07T22:59:50.830" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27276" PostTypeId="1" AcceptedAnswerId="27277" CreationDate="2018-08-08T00:33:21.443" Score="2" ViewCount="86" Body="&lt;p&gt;From a search, I found a list of some &lt;a href=&quot;https://archive.stsci.edu/proposal_search.php?id=15450&amp;amp;mission=hst&quot; rel=&quot;nofollow noreferrer&quot;&gt;preparatory HST observations&lt;/a&gt; made for the August 4, 2018 occultation of a star by the asteroid 2014 MU69&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We propose observations in support of the upcoming New Horizons encounter for 2014MU69. One visit is for support observations of a star to be occulted by 2014MU69 on 2018 August 4 from which ultra-precise astrometry and shape information about the target can be derived. The second visit is for a backup epoch of astrometry on 2014MU69 itself at the end of the 2018 observing window at a time when New Horizons will also be observing the target. (minor typographical edits made)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Five lines are shown, for measurements between &lt;code&gt;2018-04-30 01:03&lt;/code&gt; and &lt;code&gt;01:48&lt;/code&gt;. The target is listed as &lt;code&gt;MU20180804&lt;/code&gt; but that name seems to be the &lt;a href=&quot;https://www.boulder.swri.edu/~buie/occ/MU20180804.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;identification of an occultation&lt;/a&gt;, not the name of a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I have not figured out the actual identity of the star that was predicted to be occulted by 2014 MU69 on &lt;code&gt;2018-08-04&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;http://www.asteroidoccultation.com/observations/NA/MU69Binzel.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;very nice prediction&lt;/a&gt; for three occultations by 2014 MU69 last year (2017). However when I go to the &lt;a href=&quot;http://www.asteroidoccultation.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;main page&lt;/a&gt; for that asteroid occultation site and look at &lt;code&gt;2018-08-04&lt;/code&gt; I don't see any events with MU69 as the asteroid!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;https://space.stackexchange.com/q/29881/12102&quot;&gt;Will the upcoming observations of occultation by “Ultima Thule” (2014 MU69) be of a single object, or two?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zhsyJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zhsyJ.jpg&quot; alt=&quot;Occultation by MU69 on 2018-08-04&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-08-08T00:39:38.457" LastActivityDate="2018-08-08T02:48:29.390" Title="What star was the asteroid 2014 MU69 predicted to occult on 2018-08-04?" Tags="&lt;asteroids&gt;&lt;hubble-telescope&gt;&lt;occultation&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27277" PostTypeId="2" ParentId="27276" CreationDate="2018-08-08T02:48:29.390" Score="2" Body="&lt;p&gt;A &lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR&quot; rel=&quot;nofollow noreferrer&quot;&gt;VizieR query&lt;/a&gt; within 15 arcseconds of &lt;a href=&quot;https://www.boulder.swri.edu/~buie/occ/MU20180804.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Buie's coordinates&lt;/a&gt; returns a 13th magnitude star, variously identified as:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2MASS &lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR-5?-ref=VIZ5b6a57cb8ba7&amp;amp;-out.add=.&amp;amp;-source=II/246/out&amp;amp;2MASS===19042146-2035362&amp;amp;-out.orig=o&quot; rel=&quot;nofollow noreferrer&quot;&gt;19042146-2035362&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;UCAC4 &lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR-5?-ref=VIZ5b6a57e88eea&amp;amp;-out.add=.&amp;amp;-source=I/322A/out&amp;amp;UCAC4===348-170642&amp;amp;-out.orig=o&quot; rel=&quot;nofollow noreferrer&quot;&gt;348-170642&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Gaia DR2 &lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR-5?-ref=VIZ5b6a4af18782&amp;amp;-out.add=.&amp;amp;-source=I/345/gaia2&amp;amp;-c=286.08948619284%20-20.59347479448,eq=ICRS,rs=0.005&amp;amp;-out.orig=o&quot; rel=&quot;nofollow noreferrer&quot;&gt;4082342058110504064&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-08-08T02:48:29.390" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27278" PostTypeId="1" AcceptedAnswerId="27279" CreationDate="2018-08-08T03:50:54.057" Score="2" ViewCount="51" Body="&lt;p&gt;@MikeG's &lt;a href=&quot;https://astronomy.stackexchange.com/a/27277/7982&quot;&gt;speedy answer&lt;/a&gt; to &lt;em&gt;&quot;What star was the asteroid 2014 MU69 predicted to occult on 2018-08-04?&quot;&lt;/em&gt; lists three names for one 13th magnitude star, from three catalogues; 2MASS, UCAC4, and Gaia DR2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question could be slightly subjective, but I believe that someone familliar with the topic will be able to provide an authoritative answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For stars around 13th magnitude that do appear in all three, are they electronically linked between catalogs in some automated way, or do you pretty much check each of them, decide the objects are the same, and then make note of your conclusions if you are writing something up?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2MASS All-Sky Catalog of Point Sources (Cutri+ 2003) &lt;/li&gt;&#xA;&lt;li&gt;UCAC4 Catalogue (Zacharias+, 2012) &lt;/li&gt;&#xA;&lt;li&gt;Gaia DR2 (Gaia Collaboration, 2018) &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-08-08T07:41:24.847" Title="For stars around 13th magnitude, are their entries in various catalogues usually linked, or does one do this manually?" Tags="&lt;star-catalogues&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27279" PostTypeId="2" ParentId="27278" CreationDate="2018-08-08T07:41:24.847" Score="3" Body="&lt;p&gt;If you search by position in the &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/&quot; rel=&quot;nofollow noreferrer&quot;&gt;CDS Simbad&lt;/a&gt; database by position it will give you entries in all three of these catalogues. If you search by name, then it will look for matched objects in these three catalogues. I am unsure what the matching algorithm is - it must cross-correlation on the position and look for objects within an appropriate matching radius, taking account of the proper motion of the object and epochs of the various catalogues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will not be foolproof, but the density of 13th magnitude objects on the sky is low enough that false matches are highly unlikely.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-08-08T07:41:24.847" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27280" PostTypeId="2" ParentId="27226" CreationDate="2018-08-08T12:02:49.303" Score="2" Body="&lt;p&gt;&lt;strong&gt;Yes, the nonuniform illumination pattern would be visible for extremely bright stars.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus (mag -4.6) is bright enough to cast visible shadows.&#xA;Let's move Sirius, the brightest star in the northern hemisphere closer, to match the Venus brightness:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;currently: mag -1.46, 8.6 light years away, 0.006&quot; angular diameter&lt;/li&gt;&#xA;&lt;li&gt;relative brightness difference: $\sqrt[5]{100}^{4.6-1.46} \approx 18$&lt;/li&gt;&#xA;&lt;li&gt;required distance factor: $\sqrt{18} \approx 4.24$&lt;/li&gt;&#xA;&lt;li&gt;at $8.6/4.24 \approx 2$ light years away, Sirius apparent diameter would be $0.006'' \cdot 4.24 \approx 0.026''$ which is still much smaller than planets&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Moreover, we can even &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10455/why-does-venus-flick&quot;&gt;see planets twinkling sometimes&lt;/a&gt; (depending on the seeing), so the &quot;point light source&quot; is not a strict requirement (well, strictly speaking, point light sources do not even exist ;))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, there are no mentions of shadowband-like phenomena in the &lt;a href=&quot;https://en.wikipedia.org/wiki/History_of_supernova_observation&quot; rel=&quot;nofollow noreferrer&quot;&gt;supernova observation history&lt;/a&gt;, even though some of them were bright enough to be visible in daylight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess we have to wait for the next Milky Way supernova to be sure...&lt;/p&gt;&#xA;" OwnerUserId="7654" LastActivityDate="2018-08-08T12:02:49.303" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27281" PostTypeId="1" CreationDate="2018-08-08T12:28:59.917" Score="5" ViewCount="727" Body="&lt;p&gt;I am living in the northern hemisphere in Greece and in class 4 area (light pollution scale). Is it possible to see the Andromeda galaxy with a naked eye?&lt;/p&gt;&#xA;" OwnerUserId="23951" LastEditorUserId="7982" LastEditDate="2018-08-10T11:59:42.567" LastActivityDate="2018-08-10T16:47:32.327" Title="Is it possible to view the Andromeda galaxy with the naked eye in Greece?" Tags="&lt;galaxy&gt;&lt;naked-eye&gt;&lt;light-pollution&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27283" PostTypeId="2" ParentId="27281" CreationDate="2018-08-08T14:12:33.540" Score="3" Body="&lt;p&gt;Assuming you mean Bortle class 4, then the Andromeda galaxy (M31) should be a fairly easy object with the naked eye. It won’t be particularly impressive, but rather will look like a hazy blob. On the other hand it is more than 2 million light years distant, so it’s impressive to be able to see it at all. &lt;/p&gt;&#xA;" OwnerUserId="11822" LastEditorUserId="11822" LastEditDate="2018-08-10T16:47:32.327" LastActivityDate="2018-08-10T16:47:32.327" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27284" PostTypeId="1" AcceptedAnswerId="27285" CreationDate="2018-08-08T15:44:29.473" Score="6" ViewCount="729" Body="&lt;p&gt;In many articles talking about telescopes, they talk often of &quot;bright, unresolved sources&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;bright&quot; usually mean that the magnitude should be less than a certain threshold, (or that the signal to noise ratio should bigger than a given value) but what I don't catch the meaning of &quot;unresolved&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;unresolved sources&quot; in &lt;a href=&quot;https://docushare.lsst.org/docushare/dsweb/Get/LPM-17/LSSTSRD_v5%202%203.pdf&quot; rel=&quot;noreferrer&quot;&gt;https://docushare.lsst.org/docushare/dsweb/Get/LPM-17/LSSTSRD_v5%202%203.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;bright unresolved point sources&quot; in the introduction of this paper : &lt;a href=&quot;https://docushare.lsstcorp.org/docushare/dsweb/Get/Publication-86/&quot; rel=&quot;noreferrer&quot;&gt;https://docushare.lsstcorp.org/docushare/dsweb/Get/Publication-86/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea ?&lt;/p&gt;&#xA;" OwnerUserId="23047" LastActivityDate="2018-08-08T16:40:20.273" Title="What is the meaning of &quot;bright unresolved sources&quot;" Tags="&lt;telescope&gt;&lt;photography&gt;&lt;observatory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27285" PostTypeId="2" ParentId="27284" CreationDate="2018-08-08T16:40:20.273" Score="7" Body="&lt;p&gt;&quot;Unresolved&quot; here means a source of light that appears as a point, not as a disc or a cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the planets can all be resolved to discs by even quite moderate telescopes. Galaxies also can be resolved, they don't appear as point sources. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, stars, quasars, and most asteroids cannot be resolved by amateur equipment. In the context of the LSST paper, resolved sources are &quot;galaxies&quot; and unresolved sources are &quot;stars and quasars&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The photometric measurements for resolved sources (galaxies) have to include several standard magnitudes, such as Petrosian magnitudes, as well as appropriate model magnitudes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-08-08T16:40:20.273" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27286" PostTypeId="1" AcceptedAnswerId="27293" CreationDate="2018-08-08T17:17:05.227" Score="5" ViewCount="213" Body="&lt;p&gt;Can anyone calculate the distance the moon travels in a single year? I can calculate it as it is close to 2.4 million km, but I need adjustments as the earth is moving and so is the Sun. So, I need the absolute distance rather than distance from earth's perspective.&lt;/p&gt;&#xA;" OwnerUserId="23954" LastEditorUserId="1891" LastEditDate="2018-08-09T07:04:26.910" LastActivityDate="2018-08-10T20:20:12.630" Title="Lunar orbit distance" Tags="&lt;orbit&gt;&lt;the-moon&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27287" PostTypeId="1" CreationDate="2018-08-08T17:46:27.040" Score="4" ViewCount="49" Body="&lt;p&gt;I'm still relatively new to amateur astronomy, and I would like to get into astrophotography. I've looked at different equatorial mounts, and I think I like the benefits of fork mounts. I'd like to design one that can be adjusted to fit a few different scopes and probably a camera with large lens. The way I would do this is by making various adapters, and possibly allowing the fork width to be changed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I have never done this before, what issues should I expect to encounter, especially with regard to tracking accuracy?&lt;/p&gt;&#xA;" OwnerUserId="23256" LastActivityDate="2018-08-08T17:46:27.040" Title="Things to be aware of when designing a custom equatorial fork mount" Tags="&lt;telescope&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27288" PostTypeId="1" CreationDate="2018-08-09T09:40:09.277" Score="1" ViewCount="53" Body="&lt;p&gt;I want to calculate the total angular momentum of a system in which two brown dwarfs are orbiting one another (P ~ 10 years), and this system is stably orbiting a third primary ZAMS star (P ~ $10^4$ years). I was thinking of approaching this question from a Hamiltonian mechanics perspective, but the problem doesn't become easier in the rotating reference frame. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let A denote the primary star, and B,C denote the two components of the binary. I was thinking the angular momentum would simply be the sum of rotational (S) and orbital angular (L) momentums:&#xA;$$&#xA;L_{tot} = S_{A} + S_{B} + S_{C} + L_{BC} + L_{A+BC}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am both worried that I am partially &quot;double counting&quot; with the last term and I don't really know how to treat the last term. Do I treat BC like a single mass at their barycenter? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a Hamiltonian method to approach this problem, I am all ears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question I am actually concerned with is what would the rotation rate of the primary star be if the total angular momentum of the system was in the primary star? But I would really just like an accurate description of the angular momentum of the system to do this.&lt;/p&gt;&#xA;" OwnerUserId="2675" LastActivityDate="2018-08-09T09:40:09.277" Title="Total angular momentum in system of binary brown dwarfs on wide orbit from primary star" Tags="&lt;binary-star&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27289" PostTypeId="1" CreationDate="2018-08-09T10:55:11.680" Score="6" ViewCount="225" Body="&lt;p&gt;I've stumbled upon this curious passage addressing the formation of stars in Feynman's lectures on physics:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Whether we have ever seen a star form or not is still debatable.&#xA;  Figure 7–12 shows the one piece of evidence which suggests that we&#xA;  have. At the left is a picture of a region of gas with some stars in&#xA;  it taken in 1947, and at the right is another picture, taken only 7&#xA;  years later, which shows two new bright spots.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While I fully agree with Feynman's later note† that it's extremely unlikely that this picture shows stars forming, &lt;em&gt;what does it actually show&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no reference to it, and no mention of filter used, the physical scale or even of the celestial object shown, and I've gotten rather curious as to what the seven-year change represents. Variable stars? Or is the displacement due to proper motions? Film artifact? Something else entirely?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure in question (&lt;a href=&quot;http://www.feynmanlectures.caltech.edu/I_07.html#Ch7-F12&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rlyMd.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rlyMd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;† Feynman concludes the paragraph with &quot;[has the star formation somehow occurred?] &lt;em&gt;Perhaps, and perhaps not. It is unreasonable that in only seven years we should be so lucky as to see a star change itself into visible form; it is much less probable that we should see two!&quot;&lt;/em&gt; - indeed, the star formation rate of the whole Galaxy would be of this order!&lt;/p&gt;&#xA;" OwnerUserId="14385" LastEditorUserId="14385" LastEditDate="2018-08-09T13:32:30.167" LastActivityDate="2018-08-25T14:47:05.253" Title="The formation of new stars as depicted in Feynman's lectures" Tags="&lt;star&gt;&lt;star-formation&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27290" PostTypeId="2" ParentId="27261" CreationDate="2018-08-09T11:34:56.530" Score="4" Body="&lt;p&gt;According to &lt;a href=&quot;https://agupubs.onlinelibrary.wiley.com/doi/pdf/10.1029/2007JE002905&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;, there are various but enough good estimations for the temperature gradient of the Martian soil. Note, direct measurements will we have first with the next Martian lander, the &lt;a href=&quot;https://www.space.com/40141-mars-lander-insight-will-look-inside-red-planet.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;InSight&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rWddZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rWddZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Image from &lt;a href=&quot;https://www.nasa.gov/mission_pages/mars/news/pia16079.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The for us important part of this estimation is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QqFYW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QqFYW.png&quot; alt=&quot;Estimated Martian soil temperature gradient&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Depth penetration of the annual temperature wave at $120^\circ E,&#xA;&amp;gt; 20^\circ N$, using data from the NASA/MSFC Mars GRAM as the surface&#xA;  boundary condition and assuming a planetary heat flow of 20&#xA;  $\frac{mW}{m^2}$. The snapshots of the temperature as a function of&#xA;  depth are given for the models with&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;(a) $k_\infty = 0.02 \frac{W}{m\cdot K}$ and&lt;/li&gt;&#xA;  &lt;li&gt;(b) $k_\infty = 0.1 \frac{W}{m\cdot K}$.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;Heat flows derived from the soil temperatures given in Figures 3a and&#xA;  3b for models with&#xA;  * (c) $k_\infty = 0.02 \frac{W}{m\cdot K}$ and&#xA;  * (d) $k_\infty = 0.1 \frac{W}{m\cdot K}$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The snapshots (gray lines) show the local heat flow. The area inside&#xA;  the envelope (black lines) represents possible heat flow values that&#xA;  can be obtained by a single measurement. The heat flows derived from&#xA;  the annual mean temperatures are indicated by squares. Note that in&#xA;  order to calculate heat flows, the thermal conductivity was assumed to&#xA;  be known.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What we can see here, is similar to the Earth:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;there are significant temperature differences in the upper soil, depending on the geographical position, yearly and daily cycles&lt;/li&gt;&#xA;&lt;li&gt;however, around from 4m depth, the soil temperature is roughly constant.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Extrapolating the upper two graphs, we can expect a pleasant earth-like temperature not very deeply, roughly around 20-40 m.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;However, according to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/14871/at-what-depth-on-mars-would-the-atmosphere-have-equal-pressure-of-that-on-earth&quot;&gt;this answer&lt;/a&gt;, the Earth-like pressure happens much deeper, around 20-40 km below the surface. This answer uses the &lt;a href=&quot;https://en.wikipedia.org/wiki/Barometric_formula&quot; rel=&quot;nofollow noreferrer&quot;&gt;Barometric formula&lt;/a&gt; to calculate the required depth, which is a very good estimation. It seems probably unreachable with the current technology, although it is not very far away from it: the main problems of the digging of deep holes are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;cooling&lt;/li&gt;&#xA;&lt;li&gt;water incursions&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The first is much easier, the second is non-existant on the Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, the Martian atmosphere is mainly &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Mars#Composition&quot; rel=&quot;nofollow noreferrer&quot;&gt;carbon dioxide&lt;/a&gt;, thus spacesuits won't be needed, but some oxygen tanks still will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no such depth where both the temperature and the pressure would be comfortable for us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.s. actually, 16% pure oxygen is already enough for us to breathe, which makes the required depth to around 10-15 km.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="1891" LastEditDate="2018-08-09T13:28:46.377" LastActivityDate="2018-08-09T13:28:46.377" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27291" PostTypeId="2" ParentId="27281" CreationDate="2018-08-09T13:33:25.630" Score="1" Body="&lt;p&gt;Yes, you can see Andromeda with the naked eye, assuming you have a site with suitable dark skies. Under good seeing conditions, you should be able to see it as a bit of a small hazy area using averted vision even in suburban skies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going a bit further out to a site with a good dark sky, you should be able to see it when looking directly at it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Greece is (approximately) located around 40°N to 36°N in latitude. Andromeda galaxy is at +41° declination (approx).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this information, you can see how high it will rise in the sky (in the best scenario) which is important because if it is too low on the horizon, it will be extremely difficult to see. By using the formula (90°-latitude)+declination, we can see that Andromeda Galaxy will get to just over 90° from your horizon, (at the best case, being in the Northern part of Greece)) which means it will be almost directly overhead! Obviously, this is an approximation because the earth does have an axial tilt! Either way, this shows that it will get far clear of the horizon, meaning less atmospheric distortion. Of course, it would be best to do it when it is 90° above the horizon during the night time!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Download one of the many free apps you can get on your phone which will tell you the best times to view it. &lt;/p&gt;&#xA;" OwnerUserId="17096" LastActivityDate="2018-08-09T13:33:25.630" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27292" PostTypeId="1" CreationDate="2018-08-09T16:06:19.167" Score="1" ViewCount="86" Body="&lt;p&gt;From my understanding D4000 is a ratio between the 'flux densities between 4050 and 4250 Angstroms and that between 3750 and 3950 Angstrom', from &lt;a href=&quot;https://arxiv.org/abs/astro-ph/9703067&quot; rel=&quot;nofollow noreferrer&quot;&gt;Poggianti et al (1997)&lt;/a&gt;. However I do not understand why this is a direct relationship with stellar age. Did someone just notice that D4000 has a strong correlation to stellar age, if so what is the scientific explanation of this relationship.&lt;/p&gt;&#xA;" OwnerUserId="23909" LastActivityDate="2018-08-09T16:06:19.167" Title="Why does D4000 affect stellar age?" Tags="&lt;astrophysics&gt;&lt;spectroscopy&gt;&lt;stellar-evolution&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27293" PostTypeId="2" ParentId="27286" CreationDate="2018-08-09T16:08:18.837" Score="8" Body="&lt;p&gt;It depends on where we observe it from. There is no absolute velocity reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the most distant quasars are moving away from us at nearly the speed of the light. They &quot;see&quot; our Moon going away from them at nearly light speed. So, for them, the Moon travels around a light year each year (its other movements are negligible, compared to that).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If we observe it from the Earth, then we see a Moon on a circular orbit around us. Consider the following well-known facts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Its orbital radius is $\approx$ 385000 km,&lt;/li&gt;&#xA;&lt;li&gt;Its orbital time is $\approx$ 29 days,&lt;/li&gt;&#xA;&lt;li&gt;A year is $\approx$ 365.25 days,&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;, then we can easily calculate the distance: $385000km \cdot \pi \cdot \frac{365.25}{29} \approx \underline{\underline{3\cdot 10^7 km}}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If we observe it from the point of view of the Sun: as we can see detailed in &lt;a href=&quot;https://astronomy.stackexchange.com/a/24249/1891&quot;&gt;this answer&lt;/a&gt;, the orbit of the Moon &lt;em&gt;around the Sun&lt;/em&gt; is practically circular: the 400000km orbital radius around the Earth is just a minor perturbation on its $\approx$ 150million km orbit around the Sun. Thus, we can calculate the length of the orbit of the Earth, which is $2 \cdot \pi \cdot 150 000 000 km \approx \underline{\underline{9.4245\cdot 10^8 km}}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Well... it is not entirely true that there is no absolute velocity reference: the &lt;a href=&quot;https://map.gsfc.nasa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;WMAP&lt;/a&gt; probe, measuring the minor perturbations of the cosmic microwave background radiation, found also a relativistic Doppler effect, from which you can read more &lt;a href=&quot;https://telescoper.wordpress.com/tag/cosmic-microwave-background-dipole/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. In essence, this is what it measured:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5KptA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5KptA.jpg&quot; alt=&quot;Plank CMB&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the results actually measured were these:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4Ch2a.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4Ch2a.png&quot; alt=&quot;Dipole Map&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deviation is due to our galaxy's movement of around 160km/s in the direction of the constellation Hercules, &lt;em&gt;in an reference frame of the Cosmic Microwave Background&lt;/em&gt;, that is the closest thing there is to an &lt;em&gt;absolute reference frame&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer of your question in this frame: all other movements (of the Sun, the orbit of the Earth around the Sun, and the orbit of the Moon around the Earth) are negligible or cancel themselves out. The result can be gotten by multiplying:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the speed in the frame of the CMB (160km/s)&lt;/li&gt;&#xA;&lt;li&gt;the seconds of a year ($\approx$ 31.5million)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is $\approx 160\frac{km}{s} * 31.5\cdot 10^7 s = \underline{\underline{5\cdot 10^{10} km}}$.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="1891" LastEditDate="2018-08-10T20:20:12.630" LastActivityDate="2018-08-10T20:20:12.630" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27294" PostTypeId="1" CreationDate="2018-08-09T19:49:59.933" Score="4" ViewCount="121" Body="&lt;p&gt;I would like a earth-moon system like Pluto-Charon (but with earth instead of Pluto) where both will be tidally locked to each other within something like a billion years. As the moon would probably lock to earth before the opposite how big the moon has to be?&lt;/p&gt;&#xA;" OwnerUserId="23961" LastEditorUserId="7472" LastEditDate="2018-08-09T20:41:11.050" LastActivityDate="2018-09-09T17:01:18.490" Title="How big does the moon have to be to have the Earth tidally locked to it" Tags="&lt;the-moon&gt;&lt;tidal-locking&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27296" PostTypeId="2" ParentId="27294" CreationDate="2018-08-09T22:55:41.303" Score="1" Body="&lt;p&gt;There's a lot of variables for the timing, but according to the formula from Wikipedia, &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_locking#Timescale&quot; rel=&quot;nofollow noreferrer&quot;&gt;the time ratio is inversely proportional to the square of the mass&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If our Moon will tidally lock the Earth to it in 50 billion years, a moon 7 times the mass should tidally lock the Earth, everything else being equal, about 7^2 or 49 times faster.  About 1 billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's numerous variables, such as how large the oceans are on your planet, the density, the radius, how long the surface stays molten after the impact, the initial rotation rate and initial distance.  (that one is proportional to the 6th power).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If somebody wants to give a better and more numerical explanation, please feel free, but this should get you started.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-08-10T16:35:40.003" LastActivityDate="2018-08-10T16:35:40.003" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27297" PostTypeId="1" CreationDate="2018-08-10T05:46:27.787" Score="2" ViewCount="1730" Body="&lt;p&gt;It is 1:15 am, 8/10/18 in Indiana. I looked out my kitchen window and saw an orange light to the north with the color of a street light but clearly out in space.  It was about as high as a full moon at 10pm in July. It must be a planet... Anybody know which one? Thanks for your time.&lt;/p&gt;&#xA;" OwnerUserId="23964" LastEditorUserId="23940" LastEditDate="2018-08-31T07:21:17.897" LastActivityDate="2018-08-31T07:21:17.897" Title="Orange light in sky" Tags="&lt;planet&gt;&lt;night-sky&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27298" PostTypeId="1" CreationDate="2018-08-10T06:10:26.200" Score="7" ViewCount="119" Body="&lt;p&gt;This &lt;a href=&quot;https://astronomy.stackexchange.com/a/26996/7982&quot;&gt;answer to &lt;em&gt;What defines the plane of an accretion disk around a black hole?&lt;/em&gt;&lt;/a&gt; mentions the &lt;a href=&quot;https://en.wikipedia.org/wiki/Penrose_process&quot; rel=&quot;nofollow noreferrer&quot;&gt;Penrose process&lt;/a&gt; as a potential mechanism that could change the axis of the orbital angular momentum of an accretion disk around a black hole away from the original axis of the infaling matter's source (e.g. a companion star, a galaxy) and towards the the black hole's axis of rotation.&lt;/p&gt;&#xA;&lt;p&gt;Here I'm just asking if the theoretical Penrose process has ever been linked to an observation, or has been used as part of a potential explanation of an observed phenomenon.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2021-03-18T10:12:10.590" LastActivityDate="2021-03-18T10:12:10.590" Title="Has the Penrose process ever been linked to an observed phenomenon in some way?" Tags="&lt;black-hole&gt;&lt;astrophysics&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27299" PostTypeId="1" AcceptedAnswerId="27317" CreationDate="2018-08-10T09:31:11.147" Score="8" ViewCount="1064" Body="&lt;p&gt;Dark matter interacts with the gravitational force right? Well, unlike black holes, neutron stars are actually visible, and they're an enormous gravitational sink, so dark matter should collect to them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if all that is true, which it seems to be, why haven't astronomers detected or used neutron stars to detect dark matter?&lt;/p&gt;&#xA;" OwnerUserId="23965" LastActivityDate="2019-02-01T04:29:38.207" Title="Why aren't neutron stars full of dark matter?" Tags="&lt;dark-matter&gt;&lt;neutron-star&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27300" PostTypeId="2" ParentId="27299" CreationDate="2018-08-10T10:38:11.397" Score="2" Body="&lt;p&gt;I have never seen &lt;em&gt;any&lt;/em&gt; discussion of the interaction between any of the various Dark Matter candidates and neutron star matter. But we can still say something useful about the prospect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, remember that we don't know what Dark Matter (DM) is. We do have a number of theories that are reasonable extensions of the Standard Model which contains particles which behave sorta-kinda like we think DM behaves, but not only do we not have &lt;em&gt;any&lt;/em&gt; good evidence for any of them, we have looked for most of them and have failed to find anything. The negative evidence falls well short of certain, but also suggests that there is something important we don't know yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, you're correct that DM ought to be attracted by the neutron star's (NS's) gravity, and it seems plausible that the DM would react with the NS's dense matter. But the only interactions that I'm aware of would release a bit of heat and a bit of electromagnetic radiation at the point of interaction. (DM particles aren't hugely energetic, and DM isn't very dense.) This would be promptly absorbed and result in an ultra-minuscule heating of the NS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And neutron stars are far away.  It is very difficult to see how we could hope to observe any effects of whatever interaction may be taking place.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-08-10T10:38:11.397" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="27301" PostTypeId="1" AcceptedAnswerId="27308" CreationDate="2018-08-10T11:14:56.350" Score="13" ViewCount="209" Body="&lt;p&gt;Writing about the 2018 occurrence of the annual Perseids meteor shower, the EarthSky.org Astronomy Essentials article &lt;a href=&quot;http://earthsky.org/astronomy-essentials/everything-you-need-to-know-perseid-meteor-shower&quot; rel=&quot;noreferrer&quot;&gt;Start watching for Perseids now&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;People tend to focus on the peak mornings of the shower and that’s entirely appropriate. But meteors in annual showers – which come from streams of debris left behind in space by comets – typically last weeks, not days. Perseid meteors have been streaking across our skies since around July 17. We’ll see Perseids for 10 days or so after the peak mornings on August 11, 12 and 13. &lt;strong&gt;What’s more, the Perseids tend to build up gradually, yet fall off rapidly&lt;/strong&gt;. So it’s possible to catch Perseid meteors now.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Looking at images in @userTLK's &lt;a href=&quot;https://astronomy.stackexchange.com/a/17869/7982&quot;&gt;excellent answer&lt;/a&gt; to &quot;What (actually) is Jupiter doing to this year's Perseids meteor shower?&quot; one can imagine the Earth passing through a &quot;fuzzy tube&quot; around the orbit of the shower's associated comet's orbit, which is often inclined with respect to the ecliptic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I can't imagine a mechanism that would make the ramp-up and then ramp-down happen at very different rates. Why would the Perseids meteor rate fall off after maximum faster than the increase before maximum?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I've just found this artist's conception of the &quot;fuzzy tube&quot; of debris in Sky and Telescopes's &lt;a href=&quot;https://www.skyandtelescope.com/press-releases/2018-perseid-meteor-shower/&quot; rel=&quot;noreferrer&quot;&gt;“Great Show” Predicted for Perseid Meteor Peak on August 12–13&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4oyEu.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4oyEum.jpg&quot; alt=&quot;Perseid Meteor Shower from Sky and Telescope&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; &quot;Every year, in mid-August, Earth passes collides with particles spread along the orbit of Comet Swift-Tuttle.&quot; Credit Sky &amp;amp; Telescope, from &lt;a href=&quot;https://www.skyandtelescope.com/press-releases/2018-perseid-meteor-shower/&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-08-10T11:35:47.727" LastActivityDate="2018-08-19T03:57:26.110" Title="Why would the Perseids meteor rate fall off after maximum faster than the increase before maximum?" Tags="&lt;comets&gt;&lt;meteor-shower&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27302" PostTypeId="1" AcceptedAnswerId="27303" CreationDate="2018-08-10T11:59:30.113" Score="2" ViewCount="99" Body="&lt;p&gt;This &lt;a href=&quot;https://astronomy.stackexchange.com/q/27281/7982&quot;&gt;question&lt;/a&gt; currently asks&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am living in the northern hemisphere in Greece and in class 4 area (light pollution scale).Is it possible to see the Andromeda galaxy with a naked eye?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd never heard of a light pollution scale before. How does the scale work, and if I wanted to gauge or estimate my own area, can I do that by looking at objects in they sky visually?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-08-10T12:45:19.107" Title="What is &quot;class 4 light pollution&quot; and could I estimate my own area's class visually?" Tags="&lt;naked-eye&gt;&lt;light-pollution&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27303" PostTypeId="2" ParentId="27302" CreationDate="2018-08-10T12:07:24.523" Score="1" Body="&lt;p&gt;Only the original poster can answer what was meant, but one system for rating the darkness of the night sky is the Bortle Scale. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Bortle_scale&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Bortle_scale&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2018-08-10T12:07:24.523" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27304" PostTypeId="2" ParentId="27302" CreationDate="2018-08-10T12:45:19.107" Score="1" Body="&lt;p&gt;As JohnHoltz said, only the OP of the linked question can answer definitively. Chances are, they are referring to the Bortle scale, as that is the most common one you will hear of. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the NELM method (Naked Eye Limiting Magnitude) that classes the darkness of the sky by the faintest star you can see with the naked eye. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the Antoniadi Seeing Scale. Both this and the NELM method can be seen &lt;a href=&quot;http://www.fjastronomy.com/observing/observing-info/&quot; rel=&quot;nofollow noreferrer&quot;&gt;HERE&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the problem with these are that people have different eyesight. Some people can see much fainter stars than other with a naked eye. This means these methods can be good for a general idea only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, you could always purchase a SQM (Sky quality meter) which is a small, portable battery powered device that you take out to your dark sky site, point it at the zenith, and the screen will display the darkness of the sky. Do beware that the SQM measures in 'magnitudes per square arcsecond' which means the &lt;em&gt;higher&lt;/em&gt; the number, the darker the sky.&lt;/p&gt;&#xA;" OwnerUserId="17096" LastActivityDate="2018-08-10T12:45:19.107" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27305" PostTypeId="1" CreationDate="2018-08-10T13:28:55.817" Score="4" ViewCount="178" Body="&lt;p&gt;@MCG's &lt;a href=&quot;https://astronomy.stackexchange.com/a/27304/7982&quot;&gt;answer&lt;/a&gt; mentions several methods to classify night sky quality or brightness, and goes on to say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Additionally, you could always purchase a SQM (Sky quality meter) which is a small, portable battery powered device that you take out to your dark sky site, point it at the zenith, and the screen will display the darkness of the sky. Do beware that the SQM measures in 'magnitudes per square arcsecond' which means the higher the number, the darker the sky.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I wanted to try to make a diy SQM, I'd need to understand if there was some established standards or practices associated with a sky quality measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I might try to use a low dark current &lt;a href=&quot;https://en.wikipedia.org/wiki/Photodiode&quot; rel=&quot;nofollow noreferrer&quot;&gt;silicon photodiode&lt;/a&gt; in photoconductive or photovoltaic mode, low frequency couple it to a low noise op-amp, then do a cover-on measurement to zero out the leakage then do a over-off measurement to quantify the sky brightness.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What solid angle should it integrate over? Certainly more than one arcsecond, but would it be one square degree, or a 10 degree wide circle for example?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Silicon covers the visual spectrum and some near infrared as well. Depending on the protective materials over it, it might measure all the way out to a micron. Is that a problem, or is light pollution pretty much in the visible part of the spectrum (for obvious reasons) and I wouldn't need to hunt for an IR-blocking filter like the ones found on imagers for cameras.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;With a single broadband (rather than spectrally resolved) measurement of photocurrent, I could choose a representative wavelength, or assume a flat 400-700 nm emission spectrum, or even make an educated guess about what the spectrum of noise pollution looks like, then use that along with an estimate of say 0.8 e-h pair per photon (roughly the shape of the sensitivity curve below) I could make some estimate on how much light was reaching the photodetector. But how might I convert that to magnitude? I understand the logarithmic scale, but right now I just start with 1300 W/m^2 is magnitude -27 (roughly the Sun's characteristic) as a reference. Is there anything better?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Anything else I haven't thought of?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QbfKx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QbfKx.png&quot; alt=&quot;silicon photodiode spectral response&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Silicon photodetector typical spectral response, from &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Response_silicon_photodiode.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2021-04-22T09:47:26.147" LastActivityDate="2021-04-22T09:47:26.147" Title="What would go into the design of a simple sky quality meter, used to measure night sky brightness?" Tags="&lt;amateur-observing&gt;&lt;night-sky&gt;&lt;instruments&gt;" AnswerCount="0" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="27306" PostTypeId="1" AcceptedAnswerId="27327" CreationDate="2018-08-10T15:45:56.433" Score="2" ViewCount="70" Body="&lt;p&gt;The Live Science article &lt;a href=&quot;https://www.livescience.com/63297-hydrogen-wall-glowing-interstellar-space.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA Spotted a Vast, Glowing 'Hydrogen Wall' at the Edge of Our Solar System&lt;/a&gt; links to Geophysical Research Letters &lt;a href=&quot;https://agupubs.onlinelibrary.wiley.com/doi/abs/10.1029/2018GL078808&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Lyman‐α Sky Background as Observed by New Horizons&lt;/a&gt; &lt;a href=&quot;https://doi.org/10.1029/2018GL078808&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://doi.org/10.1029/2018GL078808&lt;/a&gt; the abstract of which is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Recent observations of interplanetary medium (IPM) atomic hydrogen Lyman‐α (Lyα) emission in the outer solar system, made with the Alice ultraviolet spectrograph on New Horizons (NH), are presented. The observations include regularly spaced great‐circle scans of the sky and pointed observations near the downstream and upstream flow directions of interstellar H atoms. The NH Alice data agree very well with the much earlier Voyager UVS results, after these are reduced by a factor of 2.4 in brightness, in accordance with recent re‐analyses. &lt;strong&gt;In particular, the falloff of IPM Lyα brightness in the upstream‐looking direction as a function of spacecraft distance from the Sun is well‐matched by an expected 1/r dependence, but with an added constant brightness of ~40 Rayleighs. This additional brightness is a possible signature of the hydrogen wall at the heliopause or of a more distant background&lt;/strong&gt;. Ongoing observations are planned at a cadence of roughly twice per year.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The constant brightness (indicator of &quot;the wall&quot;) makes sense to me, a large resolved surface of constant emission intensity (W/m^2) will have a constant brightness, independent of distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, a point source's brightness will obey a $\frac{1}{r^2}$ (&lt;a href=&quot;https://en.wikipedia.org/wiki/Inverse-square_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;inverse square&lt;/a&gt;) law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If absorption were the reason for the fall off, it would be &lt;a href=&quot;https://en.wikipedia.org/wiki/Beer%E2%80%93Lambert_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Beer-Lambert&lt;/a&gt; as $e^{-\alpha r}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the abstract describes a 1/r fall off of the interplanetary medium (IPM) atomic hydrogen Lyman‐α (Lyα) emission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How does a $\frac{1}{r}$ falloff arise in this case? &lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-08-12T14:44:06.753" LastActivityDate="2018-08-13T13:25:16.717" Title="Why would the interplanetary medium atomic hydrogen Lyman‐α emission fall as 1/r rather than1/r^2?" Tags="&lt;interplanetary-medium&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27307" PostTypeId="1" AcceptedAnswerId="27309" CreationDate="2018-08-10T17:19:17.427" Score="2" ViewCount="340" Body="&lt;p&gt;I'd like to know how bright the Andromeda galaxy's core is as an extended object, not the total integrated brightness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally a contour map or scan across the center might show brightness in magnitudes per square arcminute or square arcsecond. Or a low power digital photo with some kind of brightness scale per pixel. Older images such as the one below are from emulsion and so if they are just scanned and uploaded they will not be linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't even know where to begin to look for such a thing. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Below:&lt;/strong&gt; &quot;This picture shows M31 (NGC 224) and its small companions M32 (NGC 221), lower center, and NGC 205 (sometimes designated M110), to the upper right. The image was made by combining three separate frames derived from photographic plates taken in 1979 at the Burrell Schmidt telescope of the Warner and Swasey Observatory of Case Western Reserve University (CWRU).&quot; From &lt;a href=&quot;https://www.noao.edu/image_gallery/html/im0424.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xxLeB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xxLeB.jpg&quot; alt=&quot;Andromeda Galaxy&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="616" LastEditDate="2018-08-11T12:27:19.397" LastActivityDate="2018-08-11T12:27:19.397" Title="How bright is Andromeda's center, in magnitudes per square arcminute or something similar?" Tags="&lt;galaxy&gt;&lt;photometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27308" PostTypeId="2" ParentId="27301" CreationDate="2018-08-10T19:32:22.603" Score="13" Body="&lt;p&gt;A very interesting question. Neither Bruce McClure at EarthSky nor I knew the answer, despite the fact that the observed gradual rise in numbers for the Perseids - and their quick drop-off - is something we've both observed for decades. So we asked a true expert on meteor showers, Robert Lunsford of the American Meteor Society. He also said: &quot;Good question!&quot; And, he said: &quot;It seems that the particle density of the meteor stream [what you're calling here a 'fuzzy tube'] is greater prior to maximum than after. Perhaps the core of the stream, which produces maximum activity, is not quite centered in the stream. This is true for some other major showers such as the Geminids. The core of the Geminids must be really off-center as the display nearly disappears 48 hours after maximum. Gravitational effects from the sun may also play a role in this shift.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here, as so often in astronomy, the answer suggests our knowledge of nature is - at present - inexact. Are any of you old enough to remember when the Voyager spacecraft first encountered Jupiter? Nowadays, we're used to seeing the very detailed images of Jupiter's cloud bands, but when we first saw that detail - via the Voyagers in the late 1970s - we were blown away! The situation with meteor streams in space is likely exactly the same. We picture them as symmetrical only because we don't have instruments refined enough to reveal their structure. All we have are the long-observed rates of meteors in annual showers - such as the Perseids and the Geminids - suggesting that meteor streams do indeed have a structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="23969" LastActivityDate="2018-08-10T19:32:22.603" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27309" PostTypeId="2" ParentId="27307" CreationDate="2018-08-10T23:59:37.893" Score="2" Body="&lt;p&gt;The surface brightness number depends on a few things: how large an area of the image is averaged, how close to the galactic nucleus it is, and in what wavelength band.&#xA;Some papers give a brightness profile along a line bisecting the galaxy image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1969ApJ...157...55S&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sandage, Becklin, and Neugebauer 1969&lt;/a&gt; report, for a 7.6 arcsec diameter circle centered on the nucleus and measured in the $V$ band, a total magnitude of 11.3 and an average surface brightness of 15.5 mag/arcsec$^2$.&#xA;The same aperture 30 arcsec east or west of the nucleus was 1.5 magnitude fainter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1983ApJ...266..562K&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kent 1983&lt;/a&gt; images the central bulge in the $r$ band and finds a surface brightness of 14 mag/arcsec$^2$ at the nucleus and 16 mag/arcsec$^2$ 10 arcsec away along the major axis of the galaxy image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1987A%26AS...69..311W&quot; rel=&quot;nofollow noreferrer&quot;&gt;Walterbos and Kennicutt 1987&lt;/a&gt;&#xA;measure a 60 arcsecond circle and find a surface brightness,&#xA;1 arcminute away from the nucleus along the major axis,&#xA;of 18.5 mag/arcsec$^2$ in the $B$ band.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-08-10T23:59:37.893" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27310" PostTypeId="1" CreationDate="2018-08-11T07:50:32.820" Score="4" ViewCount="76" Body="&lt;p&gt;The Magellanic clouds do not seem to move much relative to the other background stars, which of course move a good deal. Is there some explanation for this or am I observing poorly?&lt;/p&gt;&#xA;" OwnerUserId="23973" LastEditorUserId="9527" LastEditDate="2020-09-25T08:09:06.673" LastActivityDate="2020-09-25T08:09:06.673" Title="Magellanic Clouds" Tags="&lt;observational-astronomy&gt;&lt;magellanic-cloud&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27311" PostTypeId="1" AcceptedAnswerId="27312" CreationDate="2018-08-11T08:44:39.977" Score="-1" ViewCount="926" Body="&lt;p&gt;I am beginner and eager to buy telescope for viewing planets and deep sky objects. I have gone though several posts here and found that only magnification is not so important but along magnification image should be crisp and sharp. So that one can be amazed. &#xA;Only source of purchasing telescope i have is aliexpress, please could any one be kind enough for me to find a good telescope there.price range MAX USD300. &lt;/p&gt;&#xA;" OwnerUserId="23974" LastActivityDate="2018-08-12T15:17:58.930" Title="I am beginner here.Could any one please suggest telescope on AliExpress" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2018-08-12T16:24:23.630" ContentLicense="CC BY-SA 4.0" />
  <row Id="27312" PostTypeId="2" ParentId="27311" CreationDate="2018-08-11T10:22:21.407" Score="0" Body="&lt;p&gt;AliExpress advertise Celestron astronomical telescopes. Celestron is a reputable manufacturer, and these scopes should be good quality, providing that they are &lt;strong&gt;genuine&lt;/strong&gt; Celestron scopes and not Chinese rip-offs. Unfortunately, most of the none-computerised scopes have equatorial mounts, which are fiddly for beginners to use. An alt-azimuth mount will be easier to handle. Get the biggest aperture that you can afford, and check reviews on YouTube specifically for scopes sourced through AliExpress. Think twice before you sacrifice aperture for computerised control. You already know to ignore magnification claims.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is about the best advice that I can provide, given the nature of your post.&lt;/p&gt;&#xA;" OwnerUserId="13887" LastEditorUserId="13887" LastEditDate="2018-08-12T15:17:58.930" LastActivityDate="2018-08-12T15:17:58.930" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27313" PostTypeId="2" ParentId="27310" CreationDate="2018-08-11T11:02:38.893" Score="4" Body="&lt;p&gt;The Magellanic clouds are nearby galaxies made up of millions of stars. They are much further away than the visible stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stars, galaxies and other deep space objects do not move relative to each other (at least not enough to be seen by a casual observer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stars and Magellanic clouds will appear to move during the night, due to the rotation of the Earth. They do not move otherwise (at least not noticeably to a casual observer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The planets will move slowly, but this will only be noticed if you observe them over several weeks, and of course, nearby objects (the moon, man-made satellites, meteors) will move more quickly relative to the stars.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-08-11T11:02:38.893" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27314" PostTypeId="1" CreationDate="2018-08-11T14:11:06.620" Score="2" ViewCount="72" Body="&lt;p&gt;Why are mechanisms such as the Kozai-Lidov effect more prominent at at large semi-major axes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we had a binary system of a primary and secondary body, with the third perturber as the Sun. Why is it that the secondary feels more perturbations from the gravitational influence of the sun the further away it is from the primary?&lt;/p&gt;&#xA;" OwnerUserId="23978" LastEditorUserId="23977" LastEditDate="2018-08-12T01:32:44.517" LastActivityDate="2018-08-12T22:30:17.547" Title="Why are gravitational perturbations stronger at larger semi-major axes?" Tags="&lt;planet&gt;&lt;astrophysics&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27315" PostTypeId="1" AcceptedAnswerId="27316" CreationDate="2018-08-11T18:30:50.383" Score="1" ViewCount="192" Body="&lt;p&gt;I'm talking about the Light Pollution ratings on this page: &lt;a href=&quot;http://www.cleardarksky.com/csk/faq/2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.cleardarksky.com/csk/faq/2.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I live in an area which is a 7 on the Bortle Scale (&quot;red&quot; on clearsky.com), but within 2 hours of an area with a rating of 3. I am trying to determine how far I need to drive in order to get a proper viewing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the maximum rating that will enable me to view the Perseid Meteor Shower tonight (2018)?&lt;/p&gt;&#xA;" OwnerUserId="23979" LastActivityDate="2018-08-11T18:44:26.530" Title="What is the maximum Clear Sky Light Pollution rating to see the 2018 Perseid Meteor Shower?" Tags="&lt;light-pollution&gt;&lt;meteor-shower&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27316" PostTypeId="2" ParentId="27315" CreationDate="2018-08-11T18:44:26.530" Score="2" Body="&lt;p&gt;The darker the better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are meteors of different brightnesses in the Perseid shower, including some fireballs that can be seen in any (cloudless) sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But many of the meteors are dimmer, and the quoted rate of 60 per hour assumes dark skies.  As there is no moon tonight, you would benefit if you can get to a dark sky location.  If you can see the milky way, you can see the Perseids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am under a blanket of cloud, but have seen a few while out running on Wednesday.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-08-11T18:44:26.530" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27317" PostTypeId="2" ParentId="27299" CreationDate="2018-08-11T21:55:33.883" Score="13" Body="&lt;p&gt;Yes, neutron stars might actually accumulate weakly interacting dark matter and this allows some observational constraints on its nature. Basically, the temperature and continued existence of neutron stars places bounds on the density and interaction cross-section of dark matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A dark matter particle that does not interact with matter will just have its trajectory bent by the gravity field of a heavy object, so most unbound particles will just swoop past on a hyperbolic trajectory. But as discussed in &lt;a href=&quot;https://arxiv.org/abs/astro-ph/9701131&quot; rel=&quot;noreferrer&quot;&gt;(Adams &amp;amp; Laughlin 1997)&lt;/a&gt;, if there is some matter-dark matter interaction then the particle may scatter from a matter particle and now have less than escape velocity. This way white dwarfs and neutron stars would indeed accumulate dark matter in their cores. The rate of accumulation is proportional to $\rho v \sigma$ where $\rho$ is the dark matter density, $v$ the average relative velocity and $\sigma$ the cross section. Adams &amp;amp; Laughlin estimate that a white dwarf star would accumulate its own mass in $10^{25}$ years, but this is going to be depend on the cross section (if it is too small the dark matter will pass through) which is at present unknown. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Were this accumulation the only thing happening it would eventually make white dwarfs and later neutron stars implode. However, dark matter is plausibly a mix of particles and antiparticles that annihilate each other at a rate $\sim \rho^2$; in an enriched environment like a white dwarf core this would produce energy from emitted photons heating things up. Adams &amp;amp; Laughlin estimate the luminosity as to about $10^{-12}L_\odot$, which is imperceptible in the current era but would eventually keep white dwarfs at 63 K in the far future (until the dark matter halo runs out). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other, more elaborate, calculations lead to accretion estimates that are higher. If the rate were high enough, then we would not see any cool dense objects - &lt;a href=&quot;https://arxiv.org/abs/1002.0005&quot; rel=&quot;noreferrer&quot;&gt;so white dwarf and neutron star cooling gives some bounds on the possible density and cross sections, albeit not very strict ones&lt;/a&gt;. For example, &lt;a href=&quot;https://arxiv.org/abs/0708.2362&quot; rel=&quot;noreferrer&quot;&gt;one model suggests that neutron stars would level out at 10,000 K&lt;/a&gt;. &lt;a href=&quot;https://arxiv.org/abs/1001.2737&quot; rel=&quot;noreferrer&quot;&gt;Cool star observations can also already rule out some dark matter models&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are even &lt;a href=&quot;https://arxiv.org/abs/1103.5086&quot; rel=&quot;noreferrer&quot;&gt;some arguments that super-earth planets in dense dark matter halos might be heated significantly&lt;/a&gt;, although this may require unrealistically dense halos and big cross sections. &lt;a href=&quot;https://arxiv.org/abs/0705.4298&quot; rel=&quot;noreferrer&quot;&gt;The current heath flow of Earth does give some constraints on how strongly it can interact&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So neutron stars are not directly giving us dark matter detection, but they (and planets and white dwarfs) are giving us some information. &lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2018-08-11T21:55:33.883" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27318" PostTypeId="1" CreationDate="2018-08-11T23:26:52.037" Score="1" ViewCount="56" Body="&lt;p&gt;To give some context - I am trying to create a simple program that outputs the stellar properties of a star when given its initial mass and current age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Input&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Initial Mass = 2e30kg&lt;/li&gt;&#xA;&lt;li&gt;Age = 4.6 Billion years&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Output&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Type = Main Sequence&lt;/li&gt;&#xA;&lt;li&gt;Class = G2V&lt;/li&gt;&#xA;&lt;li&gt;Radius = 7e5km&lt;/li&gt;&#xA;&lt;li&gt;Surface Temp = 5778K&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Input&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Initial Mass = 2e30kg&lt;/li&gt;&#xA;&lt;li&gt;Age = 10.2 Billion years&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Output&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Type = Red Giant&lt;/li&gt;&#xA;&lt;li&gt;Class = K4V&lt;/li&gt;&#xA;&lt;li&gt;Radius = 4e7km&lt;/li&gt;&#xA;&lt;li&gt;Surface Temp = 4000K&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My issue is that the equations I have used to calculate these output properties are only applicable to the main sequence so cannot be used for other types such as red giants, super giants, or white dwarfs (e.g. The Mass-Luminosity relationship).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not had any luck finding the equivalent equations after many hours of googling so I would really appreciate any help or even a google keyword I can search for to put me on track.&lt;/p&gt;&#xA;" OwnerUserId="23981" LastActivityDate="2018-08-11T23:26:52.037" Title="Are there simplified M-L, M-R, and lifetime equations for non main sequence stars?" Tags="&lt;star&gt;&lt;stellar-evolution&gt;&lt;main-sequence&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27319" PostTypeId="1" AcceptedAnswerId="27321" CreationDate="2018-08-12T15:00:52.657" Score="8" ViewCount="755" Body="&lt;p&gt;In the Economist video &lt;a href=&quot;https://www.youtube.com/watch?v=OKKbkS_LOu4&quot; rel=&quot;noreferrer&quot;&gt;Why does time pass?&lt;/a&gt; there is some file footage of Albert Einstein attending some event where a model of a large telescope is on display, and being described to Prof. Einstein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which telescopes model is being shown to Prof. Einstein in this file footage? Is it possible to tell what event this might be, or who some of the individuals might be (any famous Astronomers)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pdN73.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pdN73.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/V1D6r.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V1D6r.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-04-25T23:55:20.550" LastActivityDate="2020-02-24T19:23:42.047" Title="Which telescope's model is being shown to Albert Einstein in this file footage? What event might this be?" Tags="&lt;telescope&gt;&lt;history&gt;&lt;observatory&gt;&lt;identify-this-object&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27321" PostTypeId="2" ParentId="27319" CreationDate="2018-08-12T17:59:52.380" Score="14" Body="&lt;p&gt;I'm pretty certain it is the 200inch Hale telescope at Palomar Observatory. There is an image galley of the &lt;a href=&quot;http://www.astro.caltech.edu/palomar/media/halemedia.html&quot; rel=&quot;noreferrer&quot;&gt;200 inch telescope&lt;/a&gt;. According to Richard Preston's book &lt;em&gt;First Light: The Search For The Edge Of The Universe&lt;/em&gt;, Einstein attended the completion of the telescope truss, yoke and horseshoe bearing at Westinghouse Electric and Manufacturing Company in South Philadelphia (page 41) so it may have been this event that the footage is from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Searching for &quot;celluloid model of Hale telescope&quot; in the &lt;a href=&quot;https://digital.hagley.org/islandora/object/islandora%3Aroot&quot; rel=&quot;noreferrer&quot;&gt;Hagley Digital Archive&lt;/a&gt; will yield information such as &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This celluloid mount was part of a 1/32 scale model of Hale Telescope that was constructed by Westinghouse Electric Corporation engineers to test design utility and points of stress. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and images such as these:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RRLOa.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RRLOa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/L9IX0.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L9IX0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="23748" LastEditorUserId="7982" LastEditDate="2018-08-12T19:42:27.827" LastActivityDate="2018-08-12T19:42:27.827" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27322" PostTypeId="2" ParentId="27314" CreationDate="2018-08-12T22:30:17.547" Score="1" Body="&lt;p&gt;&quot;Why is it that the secondary feels more perturbations from the gravitational influence of the sun the further away it is from the primary?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In brief, it is because the net perturbing acceleration on the secondary is just the (vector) difference between (a) the accelerative attraction towards the perturbing body experienced by the secondary, and &#xA;(b) the accelerative attraction towards the perturbing body experienced by the primary.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the closer the secondary is to the primary, the more nearly equal in size and direction are those two attractions towards the perturber, and the closer to zero is their vector difference.  Another result is that the the more similar are the changes in velocity in size and direction produced by the perturbing accelerations in the primary and the secondary, and the closer to zero is the resulting perturbation in their motions relative to each other.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has been known for a long time as a consequence of Newton's 6th corollary to the laws of motion: &quot;If bodies, anyhow moved among themselves, are urged in the direction of parallel lines by equal accelerative forces, they will all continue to move among themselves after the same manner as if they had been urged by no such forces.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because there can be immense variety in the possible trajectories of the primary and secondary apart from the perturbations, any detailed illustrations can speedily develop massive and intricate trigonometrical expressions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in all cases, including that of the Kozai-Lidov effect, the scale of the effect, however intricate its form, depends on the size of the net perturbing accelerations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A highly simplified configuration can at least show by example how a net perturbing force, affecting the relative motion of primary and secondary, is almost directly proportional to the first power of the distance between primary and secondary -- though it also depends of course on further factors due to changes in angular configuration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diagram below indicates some highly simplified configurations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Vs6kl.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vs6kl.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose first that the primary (E) and secondary (M) are for one instant in line with the perturbing body (S), with M at M1 between E and S.  Let s stand for distance ES, and d for distance EM (with d &amp;lt;&amp;lt; s).  Suppose also that the mass of E and M are negligibly small relative to the mass of S (though not negligible relative to each other). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these approximations, the accelerative attractions of S on M and S on E are respectively $ k/(s-d)^2 $ and $ k/s^2 ,$ and the net perturbing acceleration on M is the difference $ ( k/(s-d)^2 - k/s^2 ) .$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting $ s(1-d/s) $ for (s-d), and using the binomial expansion of $ 1/(1-d/s)^2 $ , one sees that the terms in $ k/s^2 $ cancel, leaving the net perturbing force as $ k/s^2 * (2d/s) $ , plus terms in higher powers of d/s, i.e in $ k d^2/s^4 $ and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where d is very much smaller than s, the higher-power terms in d/s can be neglected, and then the net perturbing acceleration on M at M1 in the chosen example-configuration is closely approximated by  $ +2 k d / s^3 $ , away from E and towards S.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If instead the configuration has M at M2 so that E is in line between M and S, then the net perturbing acceleration on M clearly becomes $ ~ -2 k d / s^3 $ , i.e. away from E and away from S.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If instead M is at M3, with the line EM3 at right-angles to ES, and if also the angle ESM3 can be treated as small enough that its cosine can be approximated as 1 and its sine as d/s , then it is easily found that the net perturbing acceleration on M at M3 is approximately $ k d/s^3 $ towards E, again neglecting higher powers of d/s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If M is at an intermediate position M4, and D represents the angle ESM4, it can be seen by using a little more trigonometry that the net perturbing acceleration on M at M4, under the assumptions already made, has a component parallel to line ES of approximately $ +2 k d cos D / s^3 $, and a component perpendicular to line ES (acting always towards line ES) of approximately $ k d sin D / s^3 $. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the components are proportional to the E-M separation d, to the extent that the omitted series of terms in higher powers of d/s can be treated as negligible as has been done here.   &lt;/p&gt;&#xA;" OwnerUserId="17202" LastActivityDate="2018-08-12T22:30:17.547" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27323" PostTypeId="2" ParentId="27319" CreationDate="2018-08-13T01:40:51.577" Score="7" Body="&lt;p&gt;The event was in 1937; the telescope began operating in 1948. If &lt;a href=&quot;https://www.gettyimages.ae/detail/news-photo/dr-albert-einstein-points-to-the-model-of-the-worlds-news-photo/517726740&quot; rel=&quot;noreferrer&quot;&gt;this Getty photo&lt;/a&gt; is the same scene from a different angle,&#xA;then those with Einstein in the lower frame are, from left to right,&#xA;Nobel laureate physicist and Caltech president &lt;a href=&quot;https://en.wikipedia.org/wiki/Robert_Andrews_Millikan&quot; rel=&quot;noreferrer&quot;&gt;Robert Millikan&lt;/a&gt;,&#xA;and Westinghouse engineers &lt;a href=&quot;http://digital.hagley.org/69170_P27704&quot; rel=&quot;noreferrer&quot;&gt;Guenther Froebel&lt;/a&gt;&#xA;and &lt;a href=&quot;http://digital.hagley.org/69170_P27706&quot; rel=&quot;noreferrer&quot;&gt;Jesse Ormondroyd&lt;/a&gt;.&#xA;In the upper frame, between Froebel (center, facing right) and Ormondroyd (right, seated)&#xA;may be Westinghouse chairman &lt;a href=&quot;https://outlet.historicimages.com/products/rsd48471&quot; rel=&quot;noreferrer&quot;&gt;A. W. Robertson&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-08-13T01:40:51.577" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27324" PostTypeId="1" CreationDate="2018-08-13T03:31:32.207" Score="3" ViewCount="112" Body="&lt;p&gt;I have a “sky” app on my ipad and expected it to be Mars, but my app shows Mars just below horizon and closer to East. Oh, I should add that we saw this about 10:15 tonight. Hanging out watching meteor shower, orange object didn’t move much, if at all. I’ve never seen something so orange (or perhaps it is more red).&lt;/p&gt;&#xA;" OwnerUserId="23990" LastActivityDate="2018-08-13T04:41:02.337" Title="We are in Green Lake, WI, 8/12/18, seeing bright orange “planet”/“star” almost due South, perhaps 20 degrees above horizon, what is it?" Tags="&lt;star&gt;&lt;planet&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27325" PostTypeId="2" ParentId="27324" CreationDate="2018-08-13T04:41:02.337" Score="8" Body="&lt;p&gt;Not all sky apps get DST right or compute planetary positions accurately.&#xA;&lt;a href=&quot;https://stellarium.org/&quot; rel=&quot;noreferrer&quot;&gt;Stellarium&lt;/a&gt; at your location shows Mars 17 degrees above the SSE horizon at 22:15 CDT, or 20 degrees above due south at 23:45.&#xA;&lt;a href=&quot;http://skymaps.com/downloads.html&quot; rel=&quot;noreferrer&quot;&gt;Skymaps.com&lt;/a&gt; northern edition also shows Mars above the horizon before 10 PM in mid-August.&#xA;As Mars is unusually close to Earth this summer, it appears brighter than usual--and orange as usual.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-08-13T04:41:02.337" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27327" PostTypeId="2" ParentId="27306" CreationDate="2018-08-13T13:25:16.717" Score="5" Body="&lt;p&gt;The brightness considered here is the brightness of scattered Lyman $\alpha$ as viewed in the &lt;em&gt;upstream&lt;/em&gt; direction, i.e. away from the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors assume that the density of the atoms responsible for scattering the Ly$\alpha$ — the neutral hydrogen atoms — is 1) roughly constant throughout the Solar System and 2) is optically thin, meaning that the radiation field is not attenuated considerably with distance from the Sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the intensity of a point source falls off as $1/r^2$, the scattering emissivity &lt;em&gt;in a given point&lt;/em&gt; should then also fall off roughly as $1/r^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when viewed from inside the Solar System and in the upstream direction, the brightness is the integral of all scattering events from that point to infinity, and since $\int 1/r^2 \propto 1/r$, you have your quoted brightness.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2018-08-13T13:25:16.717" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27328" PostTypeId="2" ParentId="27301" CreationDate="2018-08-14T12:33:22.740" Score="8" Body="&lt;p&gt;Interesting question indeed, especially since it's still a matter of investigation in meteor research area! Here are some elements of answer (but definitely not all elements...):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This observed asymetry is especially true for the Geminids, and less for the Perseids. Best is to check the activity profile, available e.g. at imo.net: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.imo.net/members/imo_live_shower?shower=PER&amp;amp;year=2018&quot; rel=&quot;noreferrer&quot;&gt;https://www.imo.net/members/imo_live_shower?shower=PER&amp;amp;year=2018&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.imo.net/members/imo_live_shower?shower=PER&amp;amp;year=2017&quot; rel=&quot;noreferrer&quot;&gt;https://www.imo.net/members/imo_live_shower?shower=PER&amp;amp;year=2017&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.imo.net/members/imo_live_shower?shower=GEM&amp;amp;year=2017&quot; rel=&quot;noreferrer&quot;&gt;https://www.imo.net/members/imo_live_shower?shower=GEM&amp;amp;year=2017&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It most probably has to do with orbital mechanics and the generation of the meteoroid stream (and not &quot;meteor stream&quot;...). A possible element of explanation is a differential precession, due e.g. to particle size, but this alone cannot explain all the observations. Another possibility is a variation of the amount of dust ejected by the parent body as a function of time, knowing that between two different perihelion passages (and hence 2 different ejection epochs), the parent orbit has changed. The net result is a region of the stream highly populated and another one less populated. Though this is hard to prove without direct observation of the dust production as a function of time, it is a reasonable explanation, but surely not the only one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More hypothesis welcome! :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jeremie&lt;/p&gt;&#xA;" OwnerUserId="24006" LastEditorUserId="7982" LastEditDate="2018-08-14T15:32:46.017" LastActivityDate="2018-08-14T15:32:46.017" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27329" PostTypeId="1" AcceptedAnswerId="27331" CreationDate="2018-08-14T12:54:39.157" Score="3" ViewCount="420" Body="&lt;p&gt;Does anybody know if there is an existing dataset that consists of all known star clusters, their name, and their apparent dimensions or similar? Is it possible to make a query to, or download all SIMBAD objects that are of type &quot;globular cluster&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="24007" LastActivityDate="2018-08-14T15:26:10.003" Title="Dataset containing list of known globular clusters" Tags="&lt;star&gt;&lt;data-analysis&gt;&lt;star-cluster&gt;&lt;globular-clusters&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27330" PostTypeId="2" ParentId="27329" CreationDate="2018-08-14T13:06:18.777" Score="2" Body="&lt;p&gt;So far we have about 150ish known globular clusters in the milky way. A list of all the known ones can be found &lt;a href=&quot;http://www.messier.seds.org/xtra/supp/mw_gc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;HERE&lt;/a&gt;. This includes their locations (may be slightly shifted by now as they were correct as of 2011, the time of writing) and where applicable, the Messier/NGC/IC numbers are included for ease.&lt;/p&gt;&#xA;" OwnerUserId="17096" LastActivityDate="2018-08-14T13:06:18.777" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27331" PostTypeId="2" ParentId="27329" CreationDate="2018-08-14T15:26:10.003" Score="0" Body="&lt;p&gt;Welcome to astro-informatics!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In SIMBAD with &quot;by identifier&quot;, using M13 as a working starting point, you find the type contains &quot;Glc&quot;, and there is a Glc catalog number, where the catalog designation is a hyperlink -- this takes you to a page where you can get enough info to track down catalogs in Vizier. In this case, an older catalog VII/44 is listed. It has 137 rows. (There is something like 250 GC for the Milky Way).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing the otype, using SIMBAD under &quot;Criteria query&quot; otype=&quot;Glc&quot; returns a count of 22831 globulars, not so great. These include those in other galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TOPCAT &lt;a href=&quot;http://www.star.bris.ac.uk/~mbt/topcat/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.star.bris.ac.uk/~mbt/topcat/&lt;/a&gt; is about the easiest way to get a workable table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With TOPCAT knowing the catalog is VII/44 ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...you can File-&gt;Load tables ... under the Vizier button;&#xA;in the &quot;Row Selection&quot; frame, check &quot;all rows&quot; and in the &quot;Catalog&#xA;Selection&quot; frame choose the &quot;By Keyword&quot; tab. Enter the VII/44 then the &quot;Search Catalog&quot; button. Up pops the catalog; double click the catalog name and you get the catalog downloaded into a sheet-like dialog box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general you may get a few things in the Table List frame. &#xA;Pick the one with the data. (Some have biblio references etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the table data displayed you can re-order the columns&#xA;by dragging them left/right and put the ra/dec into cols 1 and&#xA;2 for ds9 for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the proper table chosen in the main Table List frame --&#xA;use File -&gt; Save Tables(s)/Session and save the data.&#xA;The &quot;Output Format&quot; has CSV -- good enough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The little icon looks like 4 stacked coins (database) will let&#xA;you export directly to a PostgreSQL database! that is a different&#xA;can of worms, but once it works very handy. call the table rawxxxx&#xA;then select from that to make your internal working table, rename columns, add HEALPix indices as needed (Q3C) etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this properly, use the SIMBAD ADQL query and join on a table where you can choose GC's out to a distance in kpc. This trims away the galactic ones from the local(ish) ones. You may include the Magellanic clouds for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summary: The SIMBAD web interface is good; Vizier can find catalogs by a general description like globular clusters (which offers you  922 catalogs). TOPCAT is a program you have to install. The SIMBAD ADQL works, but requires a working knowledge of databases using a very restrictive SQL. The tables in TOPCAT differ a bit from the Vizier output, that differ too from ADQL queries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Older tables have brighter objects suited to actually observing them from the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: SIMBAD is not exactly complete in a general sense, check the results.&lt;/p&gt;&#xA;" OwnerUserId="24010" LastActivityDate="2018-08-14T15:26:10.003" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27333" PostTypeId="1" CreationDate="2018-08-14T21:44:56.360" Score="3" ViewCount="75" Body="&lt;p&gt;Have there been studies to compare structures that are 26 billion light years distant (i.e. from one side of the observable universe to the other) to search for faint differences?&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2018-08-15T16:35:44.973" LastActivityDate="2018-08-16T18:38:21.017" Title="Studies of isotropy from different ends of the visible universe?" Tags="&lt;galaxy&gt;&lt;cmb&gt;&lt;deep-sky-observing&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27334" PostTypeId="1" CreationDate="2018-08-14T22:07:40.220" Score="3" ViewCount="84" Body="&lt;p&gt;I have recently noticed an orange &quot;star&quot; (or what I presume to be a star), rise above the hill line behind my home. It rises later than every other star and rises within a few hours while the other stars are stationary! It moves off to the right when half way in the night sky this happens every night when clear for the past few months. I also believe I'm looking south. It's really bugging me. Could anyone identify this object from my description?&lt;/p&gt;&#xA;" OwnerUserId="24011" LastEditorUserId="1891" LastEditDate="2018-08-14T23:32:06.347" LastActivityDate="2018-08-14T23:32:06.347" Title="Orange light that rises quickly and moves of to the righ" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2018-08-20T07:29:56.797" ContentLicense="CC BY-SA 4.0" />
  <row Id="27335" PostTypeId="2" ParentId="27334" CreationDate="2018-08-14T22:55:55.877" Score="3" Body="&lt;p&gt;The object is (most likely) Mars! It's been rising in the south as of late. When it will rise of course depends on your location...for me it rises pretty close to sunset. Planets appear in an essentially constant position relative to the background stars over the course of the night, and as with the stars, the movement you see is due to the rotation of the Earth. Over the course of weeks, however, you can see the planets change position relative to the stars due to their motion in their orbit around the Sun. We don't see this wandering of position with stars - at least, not in an amount or on a timescale we can appreciate - because they're so far away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep an eye out for other bright objects in the sky that appear to move faster and (unless conditions are poor) don't really twinkle, especially near sunset before the stars properly come out - Mars probably isn't the only planet you can see! Jupiter, Saturn and Venus are often visible as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out this site if you want to see what planets are visible from your location and when.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.timeanddate.com/astronomy/night/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.timeanddate.com/astronomy/night/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ETA: Corrected explanation of planetary motion vs. motion of stars based on James' comment. &lt;/p&gt;&#xA;" OwnerUserId="7590" LastEditorUserId="7590" LastEditDate="2018-08-14T23:21:17.023" LastActivityDate="2018-08-14T23:21:17.023" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27336" PostTypeId="2" ParentId="27333" CreationDate="2018-08-15T00:17:32.197" Score="2" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_the_most_distant_astronomical_objects&quot; rel=&quot;nofollow noreferrer&quot;&gt;most distant objects&lt;/a&gt; which can be seen with telescopes are around 13.39 ly away, with a redshift $z=11$. Telescopes see a roughly uniform Universe, although some very large structures were already &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_largest_cosmic_structures&quot; rel=&quot;nofollow noreferrer&quot;&gt;identified&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://www.cv.nrao.edu/course/astr534/CMB.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cosmical Microwave Background&lt;/a&gt; has a $z \approx 1000$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was measured for its possible minor anisotropies by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_Background_Explorer&quot; rel=&quot;nofollow noreferrer&quot;&gt;COBE&lt;/a&gt;, and the &lt;a href=&quot;https://map.gsfc.nasa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wilkinson Microwave Anisotropy Probe (WMAP)&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Planck_(spacecraft)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck&lt;/a&gt; satellites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It measured the temperature of the 2.7K Cosmical Microwave Background with $\approx 10^{-5} K$ precision. The technology was not easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There were two results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, there is a global dipole anisotropy showing that we are &lt;a href=&quot;http://www.astro.ucla.edu/~wright/CMB-DT.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;moving with&lt;/a&gt; $\approx 368 \frac{km}{s}$ to the Hercules constellation. It is the nearest thing what could be seen, as if we had found an absolute reference frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4Ch2a.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4Ch2a.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, substracting this from the results, another, lesser anisotropy were found:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5KptA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5KptA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distribution of the anisotropy, combined with the result of other experiments, helped to get&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the most precise measurement of the age of the Universe (13.7 Gy).&lt;/li&gt;&#xA;&lt;li&gt;the most precise measurement of the curvature of the Universe (0, below measurement precision).&lt;/li&gt;&#xA;&lt;li&gt;that the expansion of the Universe is accelerating.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Also the polarity was measured, which has shown that the re-inozation happened earlier as it was previously believed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the best result what we have until now.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="1891" LastEditDate="2018-08-16T18:38:21.017" LastActivityDate="2018-08-16T18:38:21.017" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="27337" PostTypeId="1" AcceptedAnswerId="27350" CreationDate="2018-08-15T00:31:29.357" Score="1" ViewCount="78" Body="&lt;p&gt;I'm investigating the relationship between the magnitude of quasars in various wavelengths, and their redshift $z$. I've found that if I take the $log$-$mod$-transform ($L_{mag}$) of the magnitudes, where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L(y) = sgn(y) * log(|y| + 1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which I got from &lt;a href=&quot;https://blogs.sas.com/content/iml/2014/07/14/log-transformation-of-pos-neg.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://juluribk.com/2016/03/30/dealing-with-plotting-negative-zero-and-positive-values-in-log-scale/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, I can get a slightly tighter fit for my data, which is the aim of the study. Here's a plot for the W1 magnitude and $L(W1)$ as above, both against $log(z)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/549t2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/549t2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;showing the regression coefficient for each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question, before I take this to my supervisor, is: &lt;strong&gt;is there any reason why I shouldn't be taking the $log$ of an already $log$ quantity?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2018-08-16T13:28:55.733" Title="Can I take logarithms of magnitudes (which are already log)" Tags="&lt;redshift&gt;&lt;quasars&gt;&lt;magnitude&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2018-08-21T13:36:34.183" ContentLicense="CC BY-SA 4.0" />
  <row Id="27338" PostTypeId="2" ParentId="8227" CreationDate="2018-08-15T01:12:46.553" Score="2" Body="&lt;p&gt;I don't know of any sources for the data you're looking for, however, I did something like this a year ago in one of my classes. An option would be (keeping in mind that you don't want to be too rigorous) to get an image of a &lt;em&gt;good enough&lt;/em&gt; galaxy, i.e a fits file released to the public like &lt;a href=&quot;https://www.eso.org/public/images/eso9949a/?lang&quot; rel=&quot;nofollow noreferrer&quot;&gt;M83&lt;/a&gt;. Under the assumption that wherever there's light there's mass (mass = $kL$, where $k$ is a constant and $L$ luminosity), you can equate forces and solve for velocity. This would get you values for &quot;expected virtual velocities&quot; at each radii, so you could plot the data and get a curve with a certain shape.  What you expect to find is that as you increase radii, the values for velocity drop under the initial assumption. Then you go and &lt;em&gt;compare&lt;/em&gt; with different and more rigorous measurements, like &lt;a href=&quot;http://www.astro.ucla.edu/~lucian/papers/m83.co.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt; (page 18). Here you notice that the curve doesn't quite agree with the shape of the curve you got from the image, since it peaks and stays approximately constant (as opposed to dropping). Why is that? Maybe there's &lt;em&gt;something&lt;/em&gt; that light can't show you.&#xA;Note: as I said, this isn't rigorous. I've skipped conversion from counts per pixel, relationship between flux and luminosity and some other things. I've also assumed that you can work with something like astropy to process the data. Hope it helps&lt;/p&gt;&#xA;" OwnerUserId="24012" LastActivityDate="2018-08-15T01:12:46.553" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27339" PostTypeId="1" CreationDate="2018-08-15T06:30:40.593" Score="3" ViewCount="71" Body="&lt;p&gt;I wrote a code that follows the Barnes-Hut algorithm for gravitational dynamics. Everything looks pretty good, except that when I plot the total energy $$E_{tot}\equiv E_k+E_u=const$$ I get this result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4C1rP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4C1rP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you can see that although it seems constant for most of the time, it has some increasing transient in the beginning. I couldn't find a bug in the code that could be responsible for this, is there may be other explanation that can make sense of this?&lt;/p&gt;&#xA;" OwnerUserId="22030" LastEditorUserId="1891" LastEditDate="2018-08-15T21:13:40.003" LastActivityDate="2018-08-15T21:13:40.003" Title="Energy conservation in Barnes-Hut algorithm" Tags="&lt;galaxy-cluster&gt;&lt;n-body-simulations&gt;" AnswerCount="0" CommentCount="7" ClosedDate="2018-08-17T05:41:14.967" ContentLicense="CC BY-SA 4.0" />
  <row Id="27340" PostTypeId="2" ParentId="27337" CreationDate="2018-08-15T07:14:56.830" Score="1" Body="&lt;p&gt;You &lt;em&gt;can&lt;/em&gt;, the question is whether you should.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a theoretical reason why you think that magnitude should vary exponentially with redshift? If not then you may be falling into a fallacy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see with a polynomial with sufficiently many terms, I can get a curve that fits your data perfectly, &lt;em&gt;r&lt;/em&gt;=1, Does this my polynomial is a better model? Probably not.  Just throwing functions at your data and choosing the one that gives the biggest &lt;em&gt;r&lt;/em&gt; isn't mathematical modelling of data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would question why you are using magnitude at all? You seem to be finding that &quot;close objects appear brighter&quot; which isn't a new discovery.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-08-15T07:14:56.830" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27341" PostTypeId="1" AcceptedAnswerId="27342" CreationDate="2018-08-15T08:11:47.970" Score="2" ViewCount="117" Body="&lt;p&gt;I was solving a task that said (paraphrasing):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a flat mirror on the Moon, whose reflection coefficient is 100%, and observers on the Earth observe it as a star whose apparent magnitude is $3^m$. Knowing the magnitude of the Sun is $-27^m$, and its angular diameter is $0.5 {^\circ}$, calculate the diameter of the mirror.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution used Pogson's formula to state that the Sun is $1e12$ times brighter than the mirror ... and then the author said it is &quot;obvious&quot; that implies the surface of the mirror's disk in the sky, in $({^\circ})^2$, is $1e-12$ times that of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is it obvious, how do we derive this relation?&lt;/p&gt;&#xA;" OwnerUserId="24014" LastEditorDisplayName="user1569" LastEditDate="2018-08-15T08:37:17.333" LastActivityDate="2018-08-15T09:03:38.070" Title="How would a flat mirror on the Moon reflect sunlight" Tags="&lt;the-sun&gt;&lt;the-moon&gt;&lt;apparent-magnitude&gt;&lt;angular-diameter&gt;&lt;mirror&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27342" PostTypeId="2" ParentId="27341" CreationDate="2018-08-15T09:03:38.070" Score="3" Body="&lt;p&gt;The (weird for historical reasons) defintion of magnitude is that a difference of 5 magnitudes corresponds to a factor of 100 in the brightness of the source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a difference of 3-(-27) = 30 magnitudes is a difference of $10^{12}$ in brightness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface brightness of the mirror and the sun is effectively equal, since the mirror reflects 100% of the sun's light, so the only reason for the sun to be brighter is that it is larger. The brightness is in direct proportion to the apparent area of the source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because brightness is a simple linear scale. Two suns of the same size and brightness would be twice as bright as one. So if the sun is $10^{12}$ times brighter it must cover an area $10^{12}$ times larger.&lt;/p&gt;&#xA;&#xA;&#xA;" OwnerUserId="2365" LastActivityDate="2018-08-15T09:03:38.070" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27343" PostTypeId="1" AcceptedAnswerId="27363" CreationDate="2018-08-15T14:39:49.837" Score="6" ViewCount="11379" Body="&lt;p&gt;So, recently I was shopping for telescopes and I decided on the following telescope:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B0007UQNKY&quot; rel=&quot;noreferrer&quot;&gt;https://www.amazon.com/Celestron-21049-127EQ-PowerSeeker-Telescope/dp/B0007UQNKY/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had exactly $180 to spend, and that telescope fit the bill. However, I also noticed that it had a variety of options in sizes, mainly the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;60MM EQ Refractor &lt;/li&gt;&#xA;&lt;li&gt;70MM EQ Refractor &lt;/li&gt;&#xA;&lt;li&gt;80MM EQ Refractor&lt;/li&gt;&#xA;&lt;li&gt;114MM EQ Reflector&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;127EQ Newtonian Reflector&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is the difference between all these sizes/types, how does a &quot;Newtonian Reflector&quot; act differently than a &quot;regular&quot; (for lack of a better term) reflector and what are the benefits/drawbacks? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus Points:&lt;/strong&gt; Did I choose the &quot;highest-quality&quot; option for the price?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Side-by-side comparison of all sizes:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TxYiA.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TxYiA.jpg&quot; alt=&quot;Side-by-side comparison of all sizes&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23729" LastEditorUserId="23729" LastEditDate="2018-08-15T14:56:02.263" LastActivityDate="2020-10-22T01:45:58.443" Title="What's the difference between a Newtonian telescope and a regular telescope" Tags="&lt;telescope&gt;&lt;newtonian-telescope&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27344" PostTypeId="2" ParentId="27343" CreationDate="2018-08-15T15:35:46.713" Score="3" Body="&lt;p&gt;A &quot;Newtonian&quot; reflector has a main mirror which is curved and focuses the light back up the tube. It then has a secondary flat mirror at an angle which reflects the image sideways. A Newtonian reflector will have an eyepiece at the top, near the opening, and at 90 degrees to the tube. Newtonian reflectors are one of the simplest designs of reflector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other major design of reflecting telescope is the Cassegrain, which has a main mirror with a hole in it. It reflects light up the tube to a secondary which reflects the light back down the tube to an eyepiece at the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the image posted I'd guess both the reflectors are Newtonian. I don't know why only one is marketed as a Newtonian reflector. The main difference, apart from the larger main mirror of the 127, seems to be the focal length. A shorter tube allows for a wider field of view, but optical defects tend to be more significant. A longer tube is more tolerant of such defects. Short tube telescopes are good for looking at larger and diffuse objects like galaxies. Long tube is better for bright but small or detailed objects like the planets or the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other smaller telescopes seem to be refractor telescopes with a lens, not a mirror. &quot;Newtonian&quot; applies only to a design of reflector telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since &quot;Aperture Rules&quot;, normally one wants the biggest possible mirror. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-08-15T15:46:50.783" LastActivityDate="2018-08-15T15:46:50.783" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27345" PostTypeId="2" ParentId="27343" CreationDate="2018-08-15T18:05:23.363" Score="1" Body="&lt;p&gt;Some telescopes are refractors (convex lens at upper end), and some are reflectors (concave mirror at lower end).&#xA;This &lt;a href=&quot;https://www.skyandtelescope.com/astronomy-equipment/how-to-choose-a-telescope/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sky &amp;amp; Telescope article&lt;/a&gt; discusses the pros and cons of these and other optical designs.&#xA;Briefly, refractors are low-maintenance but require a multi-element lens to minimize &lt;a href=&quot;https://en.wikipedia.org/wiki/Chromatic_aberration&quot; rel=&quot;nofollow noreferrer&quot;&gt;chromatic aberration&lt;/a&gt;;&#xA;reflectors can provide larger aperture at lower cost but may need a little work to &lt;a href=&quot;https://garyseronik.com/a-beginners-guide-to-collimation/&quot; rel=&quot;nofollow noreferrer&quot;&gt;maintain the alignment&lt;/a&gt; of the mirrors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The essential feature of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Newtonian_telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Newtonian reflector&lt;/a&gt; is a flat diagonal secondary mirror near the upper end.&#xA;It folds the light path at a right angle, relocating the prime focus from the upper end of the main tube to a smaller tube on the side.&#xA;Besides the classical Newtonian design using a parabolic primary mirror, there are Maksutov-Newtonian and Schmidt-Newtonian designs using a spherical primary mirror and adding a refractive element at the upper end to correct &lt;a href=&quot;https://en.wikipedia.org/wiki/Spherical_aberration&quot; rel=&quot;nofollow noreferrer&quot;&gt;spherical aberration&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Cassegrain_reflector&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cassegrain reflectors&lt;/a&gt; instead use a convex secondary mirror at the upper end to put the focus behind a hole in the primary mirror.&#xA;As with the Newtonian family, there are Schmidt-Cassegrain, Maksutov-Cassegrain, and classical Cassegrain designs.&#xA;Other reflector designs, e.g. Gregorian, exist but are less common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an important difference between the two Newtonian reflectors in question.&#xA;The 114mm, whose focal length is the same as its tube length, is of the classical design Newton originated in 1668.&#xA;The 127mm, with a tube only half as long as its focal length, is probably a &lt;a href=&quot;https://en.wikipedia.org/wiki/Newtonian_telescope#Jones-Bird&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jones-Bird variant&lt;/a&gt; with a spherical primary mirror and a small corrective lens near the secondary mirror.&#xA;Aperture wins in theory, but in practice some users of the latter type have been unhappy with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this lineup, the 80mm refractor and the 114mm reflector may be better choices.&#xA;In addition to the 20mm and 4mm eyepieces that come with them, you might want an 8-10mm eyepiece for medium magnification.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-08-17T13:49:41.073" LastActivityDate="2018-08-17T13:49:41.073" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27346" PostTypeId="2" ParentId="27100" CreationDate="2018-08-15T19:43:11.847" Score="2" Body="&lt;p&gt;At the equinox this is quite easy to calculate, at least approximately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A geostationary orbit is 265000km long, but the Earth's shadow is only 12700km wide (since the Earth-Sun distance is much larger than the orbit). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At an equinox, the sun is in the plane of the equator, and since GEO satellites orbit on the equator, the satellite passes through the middle of the shadow. So it spends 12700/265000 of its orbit in shadow. Each orbit takes 1 day, so that fraction of one day is about 1 hour 10 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a solstice, the shadow deviates by 23.5 degrees = 0.41 radians from the equator. This means that at solstice the centre of the shadow will pass about 0.41\times 36000km =15000km from the satellite, and the satellite won't pass through the shadow at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At dates in between the calculation is harder. Approximations can be calculated with trigonometry if required.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-08-15T19:43:11.847" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27349" PostTypeId="1" CreationDate="2018-08-16T13:18:44.133" Score="2" ViewCount="60" Body="&lt;p&gt;I would like to use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gauss%27_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gauss' method&lt;/a&gt; to determine the orbit of planets of solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to calculate their orbits with personal observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the methodology/procedure/protocol to follow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What data do I need to measure during my observations?&lt;/p&gt;&#xA;" OwnerUserId="23812" LastEditorUserId="24027" LastEditDate="2018-08-16T20:28:45.053" LastActivityDate="2018-08-16T20:28:45.053" Title="What are the procedures and protocols to use with Gauss'method and personal observations?" Tags="&lt;astrophysics&gt;&lt;observational-astronomy&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27350" PostTypeId="2" ParentId="27337" CreationDate="2018-08-16T13:28:55.733" Score="1" Body="&lt;p&gt;You can do any mathematical operation to any number you have, but you should have well justifications for doing so. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the original transformation from flux to magnitude using -2.5*log transformation is for compressing the range of value presentation (i.e., instead of 10^n to just n), and the minus sign is to make it easily implemented and interpreted because of the ancient brightness system ranking brighter for lower number. (I am not sure if there is any good reason for making -2.5 instead of just -1. haha).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your transformation seems more like modelling reasons. There are a lot of cases in the history (including not only in physics but also other fields) that mathematical relations were ansatz, and purely phenomenological/observational rather than theorey driven. Some might be still used without underlying theory supporting, but we just keep using them because they provide good implications. Yours is one of the former case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you cannot make any sensible justification of why taking double-log to fluxes is something awesome, it is hard to make peer reviewers (who judge your works) to accept it. (This is for sure subjective to personal opinion). And by the meaning of sensible justification, since you are dealing with &quot;physics,&quot; it means the connection from the physical process to the functional form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the reason you should not take double-log might be because there is no physcial explanation of why doing so is physically meaningful.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-08-16T13:28:55.733" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27351" PostTypeId="1" CreationDate="2018-08-16T13:59:24.160" Score="3" ViewCount="2337" Body="&lt;p&gt;I'm an amateur astronomy lover and I have an Orion StarSeeker IV 150 mm, together with a bunch of lenses, which are the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1x 23mm (32.6x) and 1x 10mm (75x) wide-field 60-degree eyepieces 1x Plössl 32mm 1x Plössl 12,5mm 1x Plössl 6mm 1x Plössl 4mm 1x Barlow lens 2x 1x green filter No.56 1x red filter No. 23A 1x yellow filter No.12 1x orange filter No.21 1x blue filter No.82A 1x polarizer filter 1x lunar filter Crystal View.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been able to see the moon using the Barlow Lens and a 10mm lens, but I can't see other planets, star clusters and other stellar objects. I've searched on the internet, but haven't been able to find information that is understandable for an amateur, so here I am. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which lens is best to see which planets and star clusters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to whoever will help me :)&lt;/p&gt;&#xA;" OwnerUserId="24024" LastActivityDate="2018-08-16T17:27:24.497" Title="Which lens is right to see which planets with an Orion StarSeeker 150 mm?" Tags="&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27352" PostTypeId="1" AcceptedAnswerId="27359" CreationDate="2018-08-16T14:09:08.403" Score="2" ViewCount="111" Body="&lt;p&gt;Since these bits of debris crossing Earth's orbit have the same relative speed, why aren't they drawing parallel paths in the upper atmosphere? (consider one hour of observation to minimise Earth's rotation influence) &lt;/p&gt;&#xA;" OwnerUserId="16500" LastEditorUserId="2365" LastEditDate="2018-08-16T18:33:49.820" LastActivityDate="2018-08-17T14:36:21.833" Title="Why don't all perseid meteors follow the same trajectory?" Tags="&lt;orbit&gt;&lt;amateur-observing&gt;&lt;meteor-shower&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27353" PostTypeId="2" ParentId="27351" CreationDate="2018-08-16T15:17:26.017" Score="0" Body="&lt;p&gt;I believe this is the GoTo version, correct? First of all, make sure that it is set up correctly first of all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next you will need to find a target. You can do this via the handset. If the setup went well then you should find yourself looking at whatever target you wanted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this is a reflector type telescope, you will need to make sure it is collimated. If the collimation is off, then everything will appear blurry, no matter how much you play with the focus! Of course, make sure that your focus is good also! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your max power (zoom) is approximately 2*aperture (2*150), and the power is your objective focal length/eyepiece focal length. Say you have a 6mm eyepiece in, your power (zoom) will be 750/6 = 125.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To look at planets and stars etc you want quite a high power eyepiece. This will allow you to (hopefully) see some detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To look at open star clusters/nebula etc, then using a low power is better, as you open up your FOV (Field of view).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try looking at things without any filters first of all, then add them to experiment, to see if you can see things better (some say they work, some say they don't do much!). &lt;/p&gt;&#xA;" OwnerUserId="17096" LastActivityDate="2018-08-16T15:17:26.017" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27355" PostTypeId="2" ParentId="27351" CreationDate="2018-08-16T17:27:24.497" Score="2" Body="&lt;p&gt;You SHOULD be able to see Mars, Jupiter, and Saturn easily with your scope with any eyepiece.  In fact, You should be able to see them easily naked-eye.  But, I assume you want to see more detail, not just a bright dot in the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your telescope is a 150 mm Newtonian reflector with a focal length of 750 mm (giving you a focal ratio of f/5).  It looks like you have six eyepieces, each with different focal lengths, and a 2x Barlow.  Based on that, you should get the following magnifications:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Eyepiece        Mag         With Barlow&#xA;----------      -------     ---------------&#xA;32 mm           23x         47x&#xA;23 mm           33x         65x&#xA;12.5 mm         60x         120x&#xA;10 mm           75x         150x&#xA;6 mm            125x        250x&#xA;4 mm            188x        375x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Realistically speaking, anything over 200 or 250x is going to be a waste in this telescope.  The maximum &lt;em&gt;useful&lt;/em&gt; magnification of any telescope depends on a fe factors, such as aperture, eyepiece and telescope optics quality, atmospheric steadiness (seeing), humidity, air pollution, etc...  While you may often hear people say the maximum useful magnification is about 50 or 60x per inch of aperture, in reality, on an average night, it's more like 25 to 30x.  Based on this, you shouldn't expect anything over 200x to be useable very often, maybe 150x, unless you live in a place with unusually clear, steady air (like living at high altitude in the mountains).  Actually, I just looked, and Orion says the maximum useful is about 177x, which would be based on 30x per inch of aperture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, with even 177x, you can get some pretty good views of planets.  The 4mm will give you 188x, which should be a nice view, though probably a little fuzzy.  dropping to 6mm will give you 125x, and that shouldn't be too bad.  For Jupiter, you should be able to make out cloud bands and possibly the Great Red Spot, if it's in view.  And, of course, you can see the four Galilean moons.  With Saturn, you'll be able to see the rings, but I wouldn't expect to see the Cassini division - but you might get a hint that it's there.  With Mars you can make out hints of the polar caps, and just maybe a large feature like Syrtis Major.  The filters may help with planetary viewing, helping you bring out the details by increasing contrast.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a couple links to useful discussions of filters for planetary viewing:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://sas-sky.org/wp-content/uploads/2011/09/SAS-The-Use-of-Astronomical-Filters1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sangamon Astronomical Society Filter Advice&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://agenaastro.com/choosing-a-color-planetary-filter.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Agena Astro Planetary Filter Info&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You will find it easiest to start out with a wider view eyepiece, like your 32mm.  Get the planet centered, then switch out eyepieces. Trying to get right-on a target with a high-power eyepiece is much more difficult, even with GoTo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For other objects, things are a bit different.  Deep Sky Objects (DSOs: star clusters, galaxies, and nebulae, basically anything outside of our solar system), are many, varied, and can be quite difficult.  Even through larger telescopes, they are often nothing more than fuzzy patches.  They will NOT look like you see in pictures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your scope has an alt-az GoTo mount.  This is not always a good thing.  It must be properly aligned, and even then, a lot of alt-az GoTo mounts just aren't all that accurate. When you do your startup routine, it should ask you for at least two alignment stars.  You need to get these aligned as exactly as possible.  If it gives you the option to add more, do so. The more you add, and the more accurately you align them (meaning centered in the field of view, not just somewhere in the eyepiece), the better the GoTo and tracking performance will be.  I've had observing sessions with GoTo scopes where they were off by several degrees.  If you don't know your way around the sky to begin with, you have no idea where you're looking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For DSO's, aperture is important, but so is getting the right exit pupil.  A guy I know on Reddit wrote up a great &lt;a href=&quot;https://medium.com/@phpdevster/how-telescope-aperture-affects-your-view-24507147d7fc&quot; rel=&quot;nofollow noreferrer&quot;&gt;discussion of exit pupil you can read here&lt;/a&gt;.  Simply put, you should tailor your exit pupil to the need.  This partly will involve measuring your own pupil (&lt;a href=&quot;https://www.skyandtelescope.com/astronomy-equipment/a-pupil-primer/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Sky and telescope article includes how to measure your own pupil size&lt;/a&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding them, then is the issue.  Even with GoTo, it's not always easy. And just because your GoTo scope points at the object (if it's properly aligned) doesn't mean you can see it.  The description on Orion's Website of your telescope's GoTo capability says: &quot;Huge GoTo database of over 42,000 objects, and handy built-in Tour mode lets you enjoy great views on any clear night.&quot;  Unfortunately, a lot of these objects simply aren't going to be viewable in your scope, or from where you're observing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And observing site is an important consideration.  If you're in an area with a lot of light pollution, the sky background will be so bright that you can't get enough contrast between it and the object you're trying to view to make the object visible.  Still, that said, there are a lot of things you should be able to see from even moderate light pollution.  I would start with the Messier objects, as they are some of the brightest DSOs in the night sky and fairly readily found (a couple are even naked-eye objects).  There should be a menu option in your scope's hand control for these.  If you're properly set up and aligned, it should be able to find them for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mentioned above, if you don't know your way around the sky, it makes it harder.  You should definitely take the time to learn the constellations, at least the more common ones, as well as the names of some of the bright stars.  Most GoTo alignment systems ask you to locate 2 or 3 bright stars by name, and they mostly use the same stars.  There's between 150 and 200 commonly used bright stars for this purpose.  I wouldn't say you have to memorize them all, but if you can find a good chunk of them, say maybe 50-75 of the most common, it makes it easier to find your way around the sky.  If someone says &quot;it's 5 degrees south of Betelgeuse&quot; or &quot;Find Arcturus and move East a few degrees,&quot; you'll have a better understanding of what you're looking for.  Also, learning to estimate angular measurements (there's some great tricks out there for estimating degree measurements using fingers on an outstretched hand) can make a difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the best resources for the beginner is the book Turn Left At Orion by Guy Consolmagno and Dan M. Davis. It's written to help people new to the hobby learn their way around.  You can find it on Amazon, Barnes &amp;amp; Noble (bn.com), or order it from any good bookstore (if they don't stock it already).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the BEST resource, however, is to join an astronomy club or society.  In the US, membership is usually $50 (USD) or less per year.  I suspect they are similarly priced in other countries.  They are a great place to meet people with the same interest.  The best part is, most such groups have people who enjoy teaching newcomers the tricks to finding things and how to really observe them.  They also usually have people who can teach you how to properly collimate, maintain, and even repair your telescope.  And a lot of clubs have their own observing sites, or at least have scouted out good observing sites in the area.  It's really a worth it to join a club.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My club, the Houston Astronomical Society, only costs $36 per year, and we have some great benefits for members (including loaner telescopes, a dark site for observing, and even small observatory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck and clear skies!&lt;/p&gt;&#xA;" OwnerUserId="10546" LastActivityDate="2018-08-16T17:27:24.497" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27356" PostTypeId="2" ParentId="11375" CreationDate="2018-08-16T18:12:01.433" Score="3" Body="&lt;p&gt;It is exactly the rotation of a space object (satellite or space debris) combined with unequal light reflection ability of its surface parts which makes its brightness rapidly vary over time no matter how short it is while observed either visually or with a camera. Rotation periods can be even smaller than a second, and there is a plenty of rotating objects, especially defunct satellites out of control do rotate. Rather constant brightness speaks either for non-rotating objects or for objects with homogeneous surface in terms of light reflection or simply due to low amplitude of brightness fluctuation unperceivable by the detector (your eye in this case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phase angle (Sun-object-observer angle) effect mentioned by Rob Jeffries causes only a gradual raise or fall of brightness which takes place within a rather longer period of time depending on the apparent angular velocity which again depends on object's height. The greater the angle the fainter the object. Thus, even apparently non-rotating objects would at least demonstrate this effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And bright flashes are caused only at times of particular geometric orientation, when a certain highly reflective surface part reflects sunlight more or less exactly along the line of your sight.&lt;/p&gt;&#xA;" OwnerUserId="24027" LastActivityDate="2018-08-16T18:12:01.433" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27357" PostTypeId="2" ParentId="27352" CreationDate="2018-08-16T18:32:09.637" Score="1" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Why don't all perseid meteors follow the same trajectory?&lt;/p&gt;&#xA;&lt;p&gt;Since these debris crossing Earth's orbit have the same relative speed, why aren't they drawing parallel paths in the upper atmosphere ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Every 133 years the comet &lt;a href=&quot;https://en.wikipedia.org/wiki/Comet_Swift%E2%80%93Tuttle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Swift–Tuttle&lt;/a&gt; orbits the Sun, it last passed by Earth during November/December 1992 (26 years ago) and is due for another pass on August 14, 2126 (&lt;a href=&quot;http://www.cbat.eps.harvard.edu/iauc/05600/05636.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IAUC 5636&lt;/a&gt;: 1992t).&lt;/p&gt;&#xA;&lt;p&gt;When a comet travels through space it disintegrates leaving a shower of debris, every year the Earth passes through the tail's remnants (between July 14 - August 24 this year). The &lt;a href=&quot;https://en.wikipedia.org/wiki/Comet_nucleus&quot; rel=&quot;nofollow noreferrer&quot;&gt;comet nucleus&lt;/a&gt; is 26 km in diameter, thus it produces a wide tail of debris that can fall to Earth from various angles, appearing in the direction of the constellation Perseus (but not actually coming from that constellation).&lt;/p&gt;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Meteor_shower#Radiant_point&quot; rel=&quot;nofollow noreferrer&quot;&gt;meteor shower&lt;/a&gt; appears to radiate from a single point.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:PSM_V18_D201_Shower_of_perseids_sept_6_and_7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/d2shj.jpg&quot; alt=&quot;Perseid Meteor Shower&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;That is a result of passing through the trail of debris.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cnyo.org/2013/08/03/about-the-perseid-meteor-shower-perseids-for-short/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/73G3x.jpg&quot; alt=&quot;Comet Swift-Tuttle Path&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;See NASA's webpage: &amp;quot;&lt;a href=&quot;https://solarsystem.nasa.gov/small-bodies/meteors-and-meteorites/perseids/in-depth/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Perseids&lt;/a&gt;&amp;quot;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;&lt;strong&gt;Where Do Meteors Come From?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Meteors come from leftover comet particles and bits from broken asteroids. When comets come around the sun, they leave a dusty trail behind them. Every year Earth passes through these debris trails, which allows the bits to collide with our atmosphere and disintegrate to create fiery and colorful streaks in the sky.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;The Radiant&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Their radiant — the point in the sky from which the Perseids appear to come from — is the constellation Perseus. This is also where we get the name for the shower: Perseids. However, the constellation for which a meteor shower is named only serves to aid viewers in determining which shower they are viewing on a given night. &lt;strong&gt;The constellation is not the source of the meteors&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-08-16T18:32:09.637" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27358" PostTypeId="2" ParentId="27352" CreationDate="2018-08-16T18:32:18.567" Score="2" Body="&lt;p&gt;They are all following parallel paths. They are not on exactly the same trajectory but they are all travelling parallel to each to other in their orbit around the sun, in a stream of particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you see is then multiple meteors on parallel paths. When you look in the direction that the meteors are coming from you see them appearing to radiate from a point&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These meteors are all travelling parallel to each other. It is only perspective that makes them appear to be radiating from a point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TRmjc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TRmjc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is similar to the &quot;flying through space&quot; screensaver that used to be popular on windows. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-08-16T18:55:55.167" LastActivityDate="2018-08-16T18:55:55.167" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27359" PostTypeId="2" ParentId="27352" CreationDate="2018-08-16T18:51:12.917" Score="2" Body="&lt;p&gt;Meteoroids belonging to the same stream (so, this is valid not only for the Perseids stream) actually do follow the same trajectory (orbit) in space but meteors caused by them apparently appear to originate from a particular spot in the sky (the so called radiant) trailing in all possible directions away from it only due to our perspective of view. In fact, you could even spot a so called stationary meteor which would appear point-like because it appears exactly in the radiant and is directed along your line of sight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, to explain this it is compared to the same effect when you'd drive a car at a high velocity through a snow-fall: the snowflakes would apparently appear only to you to come from some frontal area radiating then in all directions away from it, although they all have the same trajectory in space for an external non-moving observer.&lt;/p&gt;&#xA;" OwnerUserId="24027" LastActivityDate="2018-08-16T18:51:12.917" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27360" PostTypeId="2" ParentId="14125" CreationDate="2018-08-16T19:25:05.973" Score="2" Body="&lt;p&gt;It is not a question of image coloring because there is a problem with the evolution of galaxies termed &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Faint_blue_galaxy&quot; rel=&quot;nofollow noreferrer&quot;&gt;FBG - Faint Blue Galaxies excess&lt;/a&gt;&quot; that has no satisfactory explanation. &#xA;There are important references not mentioned in the WP article. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A one page explanation of the problem (excess of blue galaxies and irregulars at z&gt;2 ) and the current state of affairs can be read at &lt;a href=&quot;https://guaix.fis.ucm.es/node/928&quot; rel=&quot;nofollow noreferrer&quot;&gt;GUAIX-The faint blue galaxies problem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from &lt;a href=&quot;http://10.1146/annurev-astro-081913-040037&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Evolution of Galaxy Structure Over Cosmic Time&lt;/a&gt;&#xA;Christopher J. Conselice (2014)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In addition to a faint blue excess, it was clear that there was also a&#xA;  peculiar excess, often for the faintest galaxies&#xA;  ..A deeper analysis showed that, using the rest-frame optical&#xA;  structures of galaxies, the Hubble sequence was nearly completely&#xA;  absent at z &gt; 2, and only at z ∼ 1.5 did spirals and ellipticals&#xA;  become as common as peculiar galaxies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;from &lt;a href=&quot;https://www.annualreviews.org/doi/abs/10.1146/annurev.astro.35.1.389&quot; rel=&quot;nofollow noreferrer&quot;&gt;FAINT BLUE GALAXIES, Richard S. Ellis, 1997&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;for a rapid decline in the volume-averaged star-formation rate of&#xA;  field galaxies since a redshift z ≈ 1. Together with the small angular&#xA;  sizes and modest mean redshift of the faintest detectable sources,&#xA;  these results can be understood in hierarchical models where the bulk&#xA;  of the star formation occurred at redshifts between z ≈ 1−2. The&#xA;  physical processes responsible for the subsequent demise of the faint&#xA;  blue galaxy population remain unclear&#xA;  ..Several puzzling features emerged that have collectively been&#xA;  referred to as the faint blue galaxy problem (Kron 1978). In its&#xA;  simplest manifestation, an apparent excess of faint blue galaxies is&#xA;  seen in the source counts over the number expected on the basis of&#xA;  local galaxy properties. A more specific version of the problem that&#xA;  attracted much attention followed the results of the first faint&#xA;  redshift surveys (Broadhurst et al 1988, Colless et al 1990). The&#xA;  count-redshift data from these surveys did not solve the number&#xA;  problem by revealing a redshift range (at either low or high redshift)&#xA;  where this additional population could be logically placed. &#xA;  Relatively complex evolutionary hypotheses were then proposed to&#xA;  reconcile these results, including luminosity-dependent evolution,&#xA;  galaxy merging, and the existence of a new population of source&#xA;  present at modest redshift but, mysteriously, absent locally&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11725" LastActivityDate="2018-08-16T19:25:05.973" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27361" PostTypeId="2" ParentId="21116" CreationDate="2018-08-16T22:41:10.450" Score="0" Body="&lt;p&gt;In photometry, we determine the &lt;strong&gt;color index&lt;/strong&gt; (or &lt;em&gt;colour&lt;/em&gt; index if you're from my part of the world) as the difference in magnitudes between two wavelength filters. The reason that we use the difference, is because the magnitude system is logarithmic, so using the difference tells us the  ratio of the intensities of the two wavelength bands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine the B-V index, you'd need the magnitudes in the B band (which is centered on $\lambda = 445$nm) and in the V band, centered on $\lambda = 551$nm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the U-B index, you'd also need the U-magnitude, at $\lambda = 365$nm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as @zephyr said, there's no way to convert between the two color indices.&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2018-08-16T22:41:10.450" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27362" PostTypeId="1" AcceptedAnswerId="27370" CreationDate="2018-08-16T22:55:54.967" Score="1" ViewCount="139" Body="&lt;p&gt;The &lt;strong&gt;color index&lt;/strong&gt; in the photometric system is defined by the differences in magnitudes between two wavelength filters, which gives the ratio of intensities. For example, to determine the B-V index, you'd need the magnitudes in the B band (which is centered on $\lambda = 445$nm) and in the V band, centered on $\lambda = 551$nm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about other operations? For instance, would there be any significance in taking, say, the ratio $B/V$ or $B\times V$?&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2018-08-17T11:14:39.897" Title="Ratios of magnitudes instead of differences" Tags="&lt;star&gt;&lt;observational-astronomy&gt;&lt;magnitude&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27363" PostTypeId="2" ParentId="27343" CreationDate="2018-08-16T23:02:22.480" Score="11" Body="&lt;p&gt;First off, I would not recommend any of these telescopes.  Of the ones you listed, the 114EQ is the least bad choice.  But, due to the poor quality of its mount, it's the better of the bunch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, before I get into the nuts and bolts, let me first offer a couple pieces of advice:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) Join an astronomy club or society.  In the US, membership usually runs $50 or less per year.  Most clubs offer guidance and advice to new members.  They also often conduct frequent star parties.  These are events where those who have them bring out telescopes and everyone spends the evening observing together.  At such events, you can see a lot of different types of telescopes of all different designs, prices, manufacturers, etc... and talk to owners about them.  Clubs also have other programs that can help those new to the hobby learn and better enjoy their observing time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) The best telescopes I can recommend in the under $200 price range (which is where these seem to be) are the &lt;a href=&quot;https://shop.astronomerswithoutborders.org/products/awb-onesky-reflector-telescope&quot; rel=&quot;noreferrer&quot;&gt;AWB OneSky 130&lt;/a&gt; (also known as the SkyWatcher Heritage 130P outside of the US) or the &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B01C5TG53Y&quot; rel=&quot;noreferrer&quot;&gt;Meade Lightbridge Mini 130&lt;/a&gt;.  Both are 130 mm Newtonian telescopes with fairly good optics for their price range and stable mounts.  Meade makes a 114 mm version, and so does Orion.  I would not recommend any telescopes smaller than that, as there's little they can show you that a set of binoculars in the price range wouldn't do just as well or better.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, your initial question was about the differences between them.  Please bear with me, I tend to be verbose and try to be thorough (in other words, grab a snack and settle in for the read).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, let me clear up a common misconception many people who are new to amateur astronomy tend to have.  The primary purpose of an astronomical telescope is not to magnify, but to gather light.  The problem with astronomical objects is less one of apparent size (this is a secondary concern, and not unimportant, but not as critical) and more one of brightness.  For example, the Andromeda Galaxy, also known as object 31 in the Messier catalog (M31), has an an angular size of about 3° by 1.5°.  The full moon is about 1/2° wide - so M31 is a lot bigger in apparent size.  But it's very faint in comparison.  If you know where to look and have fairly dark skies, you can actually see it with your naked eye at times, but not very clearly.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the light you see comes into your eye through your pupil.  The pupil of an average adult eye, when fully dark adapted, has a diameter of about 6 or so mm. That's about 28 or so square mm of light gathering area.   A modest 80mm telescope (80mm being the aperture) has an area of over 5,000 square mm - over 150 times as much light gathering area.  All that light is then funneled down and concentrated on your eye.  This makes dim objects much brighter.  Magnification helps, but it doesn't matter how big an object appears if you can't see it at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The light gathering ability of the telescope is dictated by its aperture - the size of the main optical element collecting light (either an objective lens or primary mirror).   As such, it's not really adjustable (well, you can adjust it downward by blocking the aperture, but you can't increase it).   Magnification, however, IS adjustable.  The magnification power of a telescope is calculated by dividing the focal length of the telescope by that of the eyepiece.  For example, if you have a telescope with a focal length of 1,000 mm and an eyepiece of 25 mm, you get 40x magnification. If you swap out the eyepiece for a 10 mm eyepiece, you get 100x.  Theoretically, you can magnify as much as you like, though you can only magnify so much before the image is a blurry mess (this is also a matter of aperture - larger apertures allow for higher magnification before blurriness sets in).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the most important consideration when choosing an astronomical telescope is usually the aperture of the telescope.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When most people who are inexperienced with telescopes hear the word telescope, they think of a refractor.  A refractor, or refracting telescope, uses a lens to collect light.  This lens is known as the objective lens (though sometimes referred to, erroneously, as the primary lens).  When light is passed through a substance, such as glass, it is slowed slightly by the substance.  An objective lens is shaped with a convex (outward) curvature.  When light passes through the thicker, center part of the lens it is slowed more than that in the thinner, outer portion of the lens,  This causes the light to refract, or bend. The symmetry of the lens causes the light to all bend toward a point along an axis through the center of the lens.  The point where the light rays meet is known as the focal point.  All the light it collects is concentrated here.  The eyepiece sits here to gather the light and reconstruct the image for viewing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refractors were the first telescopes made, invented in the early 1600's in the Netherlands.  People often mistakenly consider Galileo the inventor of the telescope.  In fact, the man most frequently credited with the invention is Hans Lippershey (though two other men sometimes are given credit).  Galileo was only the first to use it to examine the night sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in his day, however, there was a problem with the design.  When you refract light, you cause the colors to separate.  This is exactly how a prism works.  In fact, in a real sense, a lens IS a prism.  This is because different wavelengths of light, what we percieve as colors, are slowed at different rates.  Shorter wavelengths, which are bluer light, are slowed more than longer, redder, wavelengths.  What this means in a telescope is that the actual focal points of different colors of light are separated with bluer light coming to a focus a bit closer in than redder light.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We call this phenomenon chromatic aberration, or CA, and it can be seen in telescopes as ablueish blur or fringe around objects you're viewing.  It degrades the quality of the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two main methods for reducing its effects.  The first is to increase the focal length of the telescope. A lens with a longer focal length has less pronounced curve: the difference between the thinnest and thickest parts of the lens are less dramatic than in a lens with a shorter focal length. This means there's less dramatic bending, which leads to the focal points of the colors being closer together.  If you look at early telescopes, like those used by Galileo, they tend to be very long indeed with high focal ratios.  The focal ratio is the ratio of the focal length to the aperture.  My 8&quot; (203 mm) Schmidt-Cassegrain Telescope (SCT) has a focal length of 2,030 mm, making its focal ratio f/10.  My 8&quot; imaging Newtonian has a focal length of only 800 mm, making its focal ratio f/3.9.  I believe some of Galileo's early scopes had ratios like f/20 - the tube was 20 times longer than it was wide.  Mostly this was to reduce chromatic aberration (though it was also to increase magnification)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other method of reducing CA is by using multiple lenses with different shapes and made of different glass materials.  The most common option is to use two complementary lenses that work together to shift the focal points of bluer and redder light to bring them closer together.  This is called a doublet or achromat.  Using three lenses makes a telescope a triplet and and is often called an apochromatic refractor.  Sometimes even more lenses are used.  But this comes at a price. Each lens modifies the light, and while correcting, can also introduce distortion.  It also requires 2 surfaces (front and back) per lens to be precisely configured with a specific curvature.  This is costly and time consuming.  Lastly, lenses are heavy, and with a refractor, they are mounted at the front, which puts the weight at the front, and makes larger and longer such scopes much more difficult to properly maneuver. The largest refractor ever made was only about 49 inches, and the largest used for research was only 40 inches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galileo and his contemporaries understood the problem and that it could be solved by using a mirror instead of a lens.  Unfortunately, in those days, mirrors were much more difficult to make.  They were made from solid metal, which is much more difficult to precisely form and polish.  Most often,they were made of a metal known as speculum metal, which tarnishes very easy.  It wasn't until Sir Isaac Newton came along about 60 years later that a reflecting telescope was first created.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton used a mirror with a parabolic curvature to gather light.  It sat at the end of an open-ended tube. Light gathered by this mirror was bent and directed toward a secondary mirror, which, mounted at a 45 degree angle, would send the light out the side of the tube to the waiting eyepiece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because a mirror of this kind is all metal, there's no glass to pass through, so the glass does not separate the colors of light.  This eliminated chromatic aberration (well, there was still some in the eyepiece, but not much).  But, again, mirror technology was a problem.  Still, a few people adopted the design, and even built very large reflecting telescopes.  William Herschel's telescope was about 48 inches in diameter.  It was much easier to maneuver such a telescope, as the weight was all at the rear of the instrument (not to say it was easy, but putting a 4 foot glass lens at the end of the 40 foot tube would have been exceedingly difficult to work with - very, very top-heavy).  Herschel's mirrors, however, tarnished easily.  He actually had multiple mirrors, so one was being polished while another was in use, that he could just swap out as needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the mid 19th century, the process of depositing a thin layer of metal, originally silver, but now usually aluminum, onto glass was developed.  Glass, while heavy, is much easier to configure into a precise shape.  Once done, the reflective coating could be deposited onto the glass mirror blank.  Because the coating is on the front-side of the mirror, light doesn't pass through the glass, just bounces off the coating.  This made Newtonian reflecting telescopes much easier to produce.  Only the one surface, the primary mirror, has to be precisely configured, and this made them more cost-effective.  And the weight and balance issues made making larger and larger telescopes easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton's design wasn't the only.  Laurent Cassegrain and James Gregory both designed, but probably never constructed, reflecting telescopes relying on mirrors with a hole in the center.  The light is collected by the primary, sent to the secondary, and sent back toward the primary to pass through the hole.  In Cassegrain's design, the secondary is flat.  In Gregory's, it was curved.  Both allowed for a shortening of the overall physical length of the instrument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more recent times, the use of mirrors and lenses together lead to the development of the Catadioptric telescope.  The most common such designs are the Schmidt-Cassegrain Telescope (SCT) and Maksutov-Cassegrain Telescope (Mak).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as refractors suffer from Chromatic Aberration, Newtonian telescopes suffer from an aberration called Coma.  Coma appears as distortions of the stars around the edge of the field of view.  They appear to lengthen and have comet-like tails (hense the name, coma).  As with Refractors, the effect is reduced by increasing focal length.  In modern telescopes, corrective lenses are often used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Catadioptric telescopes resolve this partly by using a different kind of mirror shape, a spherical mirror.  The curvature of the parabolic mirror used in a Newtonian design varies along its profile.  The curvature of a spherical mirror is a constant curve.  This makes them easier to manufacture and reduces or eliminates coma, but it introduces another problem: spherical aberration.  Catadioptrics overcome this by using the combination of lenses and mirrors to correct the distortions.  However, the numerous surfaces then require precise configuring, which increases the cost to manufacture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of these designs has its pros and cons. There isn't one &quot;all-purpose&quot; option that's best for everything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as amateur telescopes go, however, the Newtonian design tends to be the most popular.  This is largely due to the average cost per inch of aperture.  For example, an 8&quot; Newtonian can be found in the neighborhood fo $300 USD for the telescope tube alone.  An 8&quot; SCT tube alone is likely to cost about three or four times that.  Very few 8&quot; refractors are available, and those that are would likely cost several thousand dollars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as to the telescope you chose.  The Celestron PowerSeeker 127EQ is actually not a true Newtonian.  It is what we call a Bird-Jones (or Jones-Bird, depending on who you talk to) Newtonian.  A true Newtonian has a parabolic primary mirror and a flat secondary mirror.  The only other optical surfaces are in any eyepieces used.  The Bird-Jones Newtonian uses a spherical primary mirror.  To counter the spherical aberration inherent in the design, a small correcting lens, which is not much more than a Barlow lens, is included in the optical path (usually mounted inside the focuser mechanism).  This allows for a more compact design (the PowerSeeker 127 is only 508 mm (20 inches) long, but has a focal length of 1,000 mm (almost 40 inches).  But it also decreases the quality of the view.  The optics on this telescope are pretty poor.  Worse still, it is extremely difficult to collimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All telescopes require the optical elements to be properly aligned to produce the best image.  This is collmation.  In the case of refractors, it rarely needs adjustment.  Most SCT's require it only occasionally, and Maksutovs rarely.  But Newtonians require it frequently.  Newtonians will have a set of adjustment screws or knobs on the back, behind the primary mirror, and on the front as part of the support that holds the secondary mirror (usually referred to as the spider).  Most telescopes use a 3-screw system for adjusting the tilt of the mirrors.  It can be a chore at first, but with a little practice, it becomes fast and easy to do.  It also requires a tool to insert into the focuser.  Traditionally, this is a collimation cap or Cheshire, which work like an eyepiece, but do not have lenses.  They help you visualize the alignment of the optics so that you can adjust it properly.  You can also get a laser collimator which uses a low-power laser to guide you.  But none of these work with a Bird-Jones scope.  The correcting lens makes them very, very difficult to collimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the mount on the PowerSeeker line, particularly the 127EQ, is not very stable.  It wobbles and shakes and makes for difficult observing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sadly, the reviews on sites like Amazon make it out to be a great instrument.  But these reviews themselves are flawed.  Amazon's servers use review rankings to determine the display order of search results.  If an item has higher ratings in its reviews, it appears toward the top of the list of results when you search.  Most buyers don't look past the first page, and rarely much further than 2 pages.  So if a seller wants to sell his or her product, it needs to be on the first page, preferably right at the top.  This leads a lot of companies and individual sellers to push customers to add reviews.  Some companies also falsify reviews (though it would be difficult to prove it) in order to game the system.  As such, the reviews on Amazon (and many other sites) are nearly useless to the consumer.  And the PowerSeeker telescopes are a prime example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On multiple other sites I spend time on, this telescope is outright hated.  It is sold by Celestron (a chinese-owned company) to unsuspecting buyers.  In the long run, I think it causes more harm to amateur astronomy than good - because users often end up discoraged by its poor peformance and cheap construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you've already purchased it, I'd strongly recommend returning it and getting something different (e.g. the two scopes I listed at the beginning).  If you haven't yet, then please do not.  The two telescopes I listed are significantly better in nearly every way for a similar price.  Even the slightly smaller 114 mm versions are better.  But the PowerSeeker line is nearly entirely composed of low-quality equipment that is not worth purchasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even before purchasing, joining a club is the best thing you can do.  Some clubs even have loaner telescopes that members can borrow to learn before they buy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish you luck with whatever you end up purchasing.  Clear skies!&lt;/p&gt;&#xA;" OwnerUserId="10546" LastActivityDate="2018-08-16T23:02:22.480" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="27364" PostTypeId="2" ParentId="27362" CreationDate="2018-08-16T23:25:59.983" Score="-1" Body="&lt;p&gt;The magnitude is logarithmic, so taking the difference between two values is effectively the same as taking the ratio between the two unlogged values.&lt;/p&gt;&#xA;" OwnerUserId="17655" LastActivityDate="2018-08-16T23:25:59.983" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27365" PostTypeId="1" AcceptedAnswerId="27366" CreationDate="2018-08-17T01:59:00.613" Score="2" ViewCount="171" Body="&lt;p&gt;I was reading an article by David O'Brien et al. where it stated&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;in meteorites only the water bound to the silicates can be found, all the water ice having been lost, whereas on asteroids water ice itself has been detected&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/S0019103514002620&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.sciencedirect.com/science/article/pii/S0019103514002620&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now this was published in 2014, so I'm wondering what is the validity of this statement today. Have we, since the publication of this article, discovered water ice on meteorites?&lt;/p&gt;&#xA;" OwnerUserId="24020" LastEditorUserId="24020" LastEditDate="2018-08-24T11:22:00.563" LastActivityDate="2018-08-24T11:22:00.563" Title="Water Ice on Meteorites" Tags="&lt;earth&gt;&lt;asteroids&gt;&lt;meteorite&gt;&lt;water&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27366" PostTypeId="2" ParentId="27365" CreationDate="2018-08-17T03:41:41.000" Score="5" Body="&lt;p&gt;It looks to me that you probably have some misunderstanding of the term meteorite. A meteorite is a remnant body which has reached the surface of a major or minor body (a planet, a moon, an asteroid, etc.). The meteorite's parent body (a meteoroid) would definitely lose its entire water ice (if it had any at all) no later than during its passage through the Earth's atmosphere. That's why no water ice was or will ever be found in meteorites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, unless you compare common meteorite impacts with hazardous asteroid impacts, these two body types are incomparable as water ice carriers. And thus, the statement by O'Brien et al still stands firm and will do so ever.&lt;/p&gt;&#xA;" OwnerUserId="24027" LastActivityDate="2018-08-17T03:41:41.000" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="27367" PostTypeId="2" ParentId="27362" CreationDate="2018-08-17T05:44:22.070" Score="3" Body="&lt;p&gt;Consider for example these bright stars (V and B-V values from &lt;a href=&quot;http://cdsarc.u-strasbg.fr/viz-bin/Cat?V/50&quot; rel=&quot;nofollow noreferrer&quot;&gt;Yale Bright Star Catalog&lt;/a&gt;).&#xA;Near V=0, B/V is ill-behaved, and B*V is near 0.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Name       B     V     B-V    B/V   B*V&#xA;Arcturus  1.19 -0.04  +1.23 -29.7  -0.05&#xA;Vega      0.03  0.03   0.00   1.00  0.00&#xA;Capella   0.88  0.08  +0.80  11.0   0.07&#xA;Rigel     0.09  0.12  -0.03   0.75  0.01&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now imagine that each of those stars were 10 times as far away.&#xA;Neglecting &lt;a href=&quot;https://en.wikipedia.org/wiki/Extinction_(astronomy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;interstellar extinction&lt;/a&gt;, this would decrease the apparent brightness by a factor of 100 and increase the apparent magnitude by 5.0.&#xA;As B and V increase together, B/V approaches 1 and B*V approaches V^2, telling us nothing meaningful about a given star.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Name       B     V     B-V   B/V    B*V&#xA;Arcturus  6.19  4.96  +1.23  1.25  30.7&#xA;Vega      5.03  5.03   0.00  1.00  25.3&#xA;Capella   5.88  5.08  +0.80  1.16  29.9&#xA;Rigel     5.09  5.12  -0.03  0.99  26.1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;B-V, on the other hand, has stood the test of time as a distance-independent metric of a star's color and surface temperature.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-08-17T05:44:22.070" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27368" PostTypeId="2" ParentId="21116" CreationDate="2018-08-17T07:24:22.903" Score="2" Body="&lt;p&gt;Both $B-V$ and $U-B$ are a reflection of how hot the surface of a star is. To first order (inhomogeneous surface temperatures do exist) this has a single value for a given stellar photosphere and so there ought to be a direct relationship between $B-V$ and $U-B$. Here it is, shown below as a graph appropriate for main sequence stars with a composition similar to the Sun. You can reproduce this yourself using data found (for example) in the &lt;a href=&quot;http://ads.harvard.edu/cgi-bin/bbrowse?book=hsaa&amp;amp;chap=2&amp;amp;page=0068&quot; rel=&quot;nofollow noreferrer&quot;&gt;Handbook of space Astronomy and Astrophysics (pp68-69)&lt;/a&gt; compiled by M.V. Zombeck.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/32k8z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/32k8z.jpg&quot; alt=&quot;colour-colour plot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that the relationship is complicated; this is because stellar surfaces are not blackbodies and the absorption by various chemical species cuts out chunks of the spectrum, particularly in the blue and ultraviolet. However, given a $B-V$ one could look up the appropriate $U-B$. Note that this is not straightforward the other way around, where there may be two values of $B-V$ possible for a given $U-B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice things are much more complicated. The plot above assumes the stars are on the main sequence. Stellar atmospheres change a bit as stars evolve and get bigger, with lower surface gravity. Thus changes the relationship a little. The relationship also changes if the composition is not similar to the Sun, especially around the bump at $B-V \simeq 0.4$. Finally, this relationship is &lt;em&gt;intrinsic&lt;/em&gt;, it assumes you are not viewing the star through extinction caused by dust. If that is not the case then the whole relationship slides redward in both colours (the vector direction of which is shown  in the plot above for about 2 magnitudes of visible extinction) but also changes shape slightly, depending on how much extinction there is.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-08-17T07:24:22.903" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27370" PostTypeId="2" ParentId="27362" CreationDate="2018-08-17T11:14:39.897" Score="2" Body="&lt;p&gt;The definition of a magnitude is something like&#xA;$$ B = -2.5\log_{10} f_B + Z_B,$$&#xA;where $f_B$ is a physical flux in whatever unit system you are using and $Z_B$ is a zeropoint for the magnitude system and the minus sign is there to ensure small magnitudes are brighter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A colour index is therefore something like&#xA;$$ B-V = -2.5\log_{10} f_B + 2.5\log_{10}f_V + Z_B - Z_V = -2.5\log_{10} \frac{f_B}{f_V} + Z_{BV}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's think about what you are asking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$BV = (-2.5\log_{10} f_B + Z_B)(-2.5\log_{10} f_V + Z_V)$$&#xA;$$BV= -2.5Z_v\log_{10} f_B -2.5Z_B\log_{10} f_V +Z_{B}Z_V -2.5 \log_{10} f_B^{-2.5\log_{10}f_V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This numerology has no physical significance at all and, because the fluxes themselves are distance-dependent, then like $B$ and $V$ individually, $BV$ and $B/V$ would be distance-dependent and therefore have no relationship to anything physical that is intrinsic to the star.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-08-17T11:14:39.897" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27371" PostTypeId="2" ParentId="21116" CreationDate="2018-08-17T14:00:28.850" Score="1" Body="&lt;p&gt;You can convert from B-V to U-B by assuming spectral energy distribution (SED).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you assume blackbody SED (BBSED). The BBSED is determined by only the temperature (+ scaling constant which is irrelevant here). Therefore, there is a mapping from B-V to the certain temperature. Once you get the temperature, you can find U-B by applying the temperature to the BBSED (again, there is a mapping).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Becareful in the situation where the object's SED is not BB (e.g., galaxy). Or, the object's SED in the range from U to V has issues such as strong line emission or absorption (especially 4000-A break). In these cases, you cannot assume BBSED.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People might provide you more informative idea how to deal with the problem if you specify what kind of object you are dealing with.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-08-17T14:00:28.850" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27372" PostTypeId="2" ParentId="27352" CreationDate="2018-08-17T14:36:21.833" Score="1" Body="&lt;p&gt;The answer has been given, it is the effect called &quot;perspective.&quot;  I would simply add an example that should make the situation quite clear: sunbeams.  When you see sunbeams come through clouds, it is natural to think they are not parallel rays, because they look very far from parallel.  Your eye tells you that the Sun is sending out those rays in all directions, from its apparent location in the center of the clouds.  But your eye cannot tell that the Sun is a trillion times farther than the clouds, so it cannot tell the divergence of the rays is a complete illusion.  Actually those rays are almost perfectly parallel, because they all point to that incredibly distant Sun.  But look at them carefully the next time you see sunbeams through the clouds-- they seem as far from parallel as they could possibly be.  So it is with the streaks of meteors, it's because your eye cannot pick out that third dimension so tends to imagine it is not important.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-08-17T14:36:21.833" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27373" PostTypeId="1" CreationDate="2018-08-17T18:18:33.983" Score="2" ViewCount="224" Body="&lt;p&gt;Let's say I have an picture of the sky or of an landscape and I want to measure an angle in the celestial sphere. Is there any software for that? In real life I would use the &lt;a href=&quot;https://i.stack.imgur.com/FahXj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;empirical hand gestures&lt;/a&gt; or a theodolite.&lt;/p&gt;&#xA;" OwnerUserId="24036" LastActivityDate="2018-08-18T14:36:58.603" Title="Measure an angle in the sky from a photo?" Tags="&lt;photography&gt;&lt;software&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27374" PostTypeId="2" ParentId="27373" CreationDate="2018-08-17T19:53:36.103" Score="1" Body="&lt;p&gt;A plate solver matches the stars in an image to a catalogue of star positions.&#xA;&lt;a href=&quot;http://nova.astrometry.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astrometry.net&lt;/a&gt; provides a web service and a software API for this.&#xA;Given only &lt;a href=&quot;https://astronomy.stackexchange.com/questions/27249/what-did-i-photograph&quot;&gt;another user's cell phone photo&lt;/a&gt;, Astrometry.net determined that the image scale was &lt;a href=&quot;http://nova.astrometry.net/user_images/2236311#annotated&quot; rel=&quot;nofollow noreferrer&quot;&gt;112 arcseconds per pixel&lt;/a&gt;.&#xA;It also estimated an image size of 72 by 54 degrees, apparently using an approximation which is only valid for small angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming low distortion (not a fisheye lens), the image is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Gnomonic_projection&quot; rel=&quot;nofollow noreferrer&quot;&gt;gnomonic projection&lt;/a&gt; of part of the celestial sphere.&#xA;If $r$ is the distance from the image center in pixels, $p$ is the pixel scale in radians per pixel, and $\theta$ is the angular separation from the center in radians, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$rp = \tan \theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I used the measurement tools in GIMP and Stellarium on two bright stars near the center to verify the central image scale of 112 arcsec/pixel or 0.54 mrad/pixel.&#xA;The edges of the 2320x1740 image are 1160 and 870 pixels away from the center.&#xA;Solving for $\theta$, the edges are 0.56 and 0.44 radian away from the center, so the image's angular dimensions are 64 by 50 degrees from edge to edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Daytime images would of course have the same pixel scale and angular dimensions as astrophotos taken with the same equipment.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-08-18T14:36:58.603" LastActivityDate="2018-08-18T14:36:58.603" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27375" PostTypeId="2" ParentId="27289" CreationDate="2018-08-17T22:10:19.807" Score="1" Body="&lt;p&gt;It could be caused by 2 rotating binary star systems. If the rotation rates are less than say 30 years then 7 years would show a significant movement of the bodies. I don't know how rare 2 rotating systems are so seeing both of these may be too rare to be practical.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastActivityDate="2018-08-17T22:10:19.807" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27376" PostTypeId="2" ParentId="27289" CreationDate="2018-08-18T03:37:01.457" Score="3" Body="&lt;p&gt;imo, all the 5 stars are already present and with a little separation in the left image.&lt;br&gt;&#xA;Their distance to Earth should have been investigated on the image of the right. The ones nearer could travel, proper motion, much faster than the ones on the background.&lt;br&gt;&#xA;In short: they could be at large distances one from the others and it is a coincidence the fact that in the left image they are aligned. With zillions of stars out there the probability is not null.&lt;br&gt;&#xA;There are very fast stars (VFS) for ex. &lt;a href=&quot;https://en.wikipedia.org/wiki/US_708&quot; rel=&quot;nofollow noreferrer&quot;&gt;US 708&lt;/a&gt; (1200 km/s on WP e 708 km/s heliocentric radial velocity - Hirsch et al. 2005) not known at Feynman time.&lt;br&gt;&#xA;The time of discovery of VFS was 2005 and 10 years later, Warren R, Brown et al, reported on the PROPER MOTIONS AND TRAJECTORIES FOR 16 EXTREME RUNAWAY AND HYPERVELOCITY STARS (2015).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT ADD:&lt;br&gt;&#xA;One thing that can be done is to use software enabled with &lt;a href=&quot;https://en.wikipedia.org/wiki/Point_spread_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;Point Spread Function (PSF)&lt;/a&gt; to discriminate the number of sources and decide how many stars are in each of the images.&lt;br&gt;&#xA;I will not do it because it takes time that I do not have.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several packages available, fi, DS9 (&lt;a href=&quot;https://www.youtube.com/watch?v=63umWDsHcEY&quot; rel=&quot;nofollow noreferrer&quot;&gt;video: find a moving star&lt;/a&gt;), MAXIM DL, &lt;a href=&quot;http://www.astromatic.net/software/sextractor&quot; rel=&quot;nofollow noreferrer&quot;&gt;SExtractor&lt;/a&gt;. The images have to be scaned, eventually formated as FITS (&lt;a href=&quot;http://docs.astropy.org/en/stable/generated/examples/io/split-jpeg-to-fits.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;python code&lt;/a&gt;) and if using SExtractor look at the procedure used by &lt;a href=&quot;https://ui.adsabs.harvard.edu//#abs/1997A&amp;amp;A...317...43B/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;E. Bertin and M. Dennefeld (1997)&lt;/a&gt;.&lt;br&gt;&#xA;Search on youtube &lt;a href=&quot;https://www.youtube.com/results?search_query=Detecting%20asteroids&quot; rel=&quot;nofollow noreferrer&quot;&gt;'Detecting asteroids'&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11725" LastEditorUserId="11725" LastEditDate="2018-08-25T14:47:05.253" LastActivityDate="2018-08-25T14:47:05.253" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27377" PostTypeId="1" CreationDate="2018-08-18T05:14:25.500" Score="1" ViewCount="140" Body="&lt;p&gt;These two statements seem to conflict, but since we're here they can't:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Electrically charged particles cannot be massless.&lt;/li&gt;&#xA;&lt;li&gt;Hydrogen was created at the moment of the Big Bang.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, the Higgs field theory states that the field was only created AFTER the Big Bang. This implies that all particles from which hydrogen could be formed were massless. Hydrogen consists of 2 charged particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could provide an answer without complex math I would really appreciate it.&lt;/p&gt;&#xA;" OwnerUserId="24039" LastEditorUserId="1891" LastEditDate="2018-08-18T07:14:55.207" LastActivityDate="2018-08-18T12:50:34.780" Title="If the Higgs field only formed after the Big Bang, how was hydrogen formed?" Tags="&lt;big-bang-theory&gt;&lt;hydrogen&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27378" PostTypeId="2" ParentId="27377" CreationDate="2018-08-18T07:14:17.613" Score="3" Body="&lt;p&gt;After the Big Bang, matter has formed in many steps. These are calles epoches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TsMGK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TsMGK.png&quot; alt=&quot;Big Bang epoches&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their length grows typically exponentially. A more detailed description can you read &lt;a href=&quot;https://en.wikipedia.org/wiki/Chronology_of_the_universe#Very_early_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Higgs field always existed, but its value were different. In the very early times, around between $10^{-43}$ and $10^{-36}$ seconds (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Grand_unification_epoch&quot; rel=&quot;nofollow noreferrer&quot;&gt;Grand unification epoch&lt;/a&gt;), its value could change, just like like electromagnetic field can change. Since then, its value is its current one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature of the Universe was higher than $10^{27}$ K at the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For atomic hydrogen, we need&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;protons&lt;/li&gt;&#xA;&lt;li&gt;electrons&lt;/li&gt;&#xA;&lt;li&gt;enough low mean energy for their bound state to be formed, it is around 3000K.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It happened much-much later, around 300000 year after the Big Bang. Before that, atomic hydrogen couldn't exist, it was too warm for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A remark: the visible hydrogen of the stars is still not atomic hydrogen, because the stars are still too hot for it to exist. It is in plasma state, a mixture of protons of electrons, without atomic hydrogen.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastActivityDate="2018-08-18T07:14:17.613" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27379" PostTypeId="2" ParentId="27377" CreationDate="2018-08-18T08:39:27.013" Score="7" Body="&lt;p&gt;Hydrogen was &lt;strong&gt;not&lt;/strong&gt; &quot;created at the moment of the big bang&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Particles (leptons and quarks) can attain mass via the Higgs field after the epoch of electroweak symmetry breaking, that occurred about a picosecond $(10^{-12}$s) after the big bang. Only after this was it possible to form the building blocks of hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact the formation of stable protons comes &quot;much&quot; later at about a millionth of a second. Most of the mass of a proton does not come about from the Higgs field, but from the binding energy of its constituent quarks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen (the atom, rather than the proton nucleus) forms much, much later; about 400,000 years after the big bang.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2365" LastEditDate="2018-08-18T11:20:03.027" LastActivityDate="2018-08-18T11:20:03.027" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27380" PostTypeId="1" AcceptedAnswerId="27386" CreationDate="2018-08-18T09:16:38.710" Score="4" ViewCount="101" Body="&lt;p&gt;According to Google, radial velocity is &quot;the velocity of a star or other body along the line of sight of an observer&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radial velocity of Sirius according to Google is    −5.50 km/s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the number is positive, does that mean it's moving away or towards us?&lt;/p&gt;&#xA;" OwnerUserId="5401" LastEditorUserId="24027" LastEditDate="2018-08-19T06:40:49.250" LastActivityDate="2018-08-19T15:04:18.667" Title="Radial Velocity Clarification Sought" Tags="&lt;radial-velocity&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27381" PostTypeId="1" AcceptedAnswerId="27387" CreationDate="2018-08-18T12:41:47.290" Score="1" ViewCount="107" Body="&lt;p&gt;Where is Voyager relative to me right now in realtime?&lt;/p&gt;&#xA;" OwnerUserId="24044" LastEditorUserId="6" LastEditDate="2018-08-19T16:43:56.903" LastActivityDate="2018-08-19T16:43:56.903" Title="Site to see direction to the Voyager by entering time and coordinates of my position on the Earth?" Tags="&lt;earth&gt;&lt;space&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2018-08-21T13:36:55.413" ContentLicense="CC BY-SA 4.0" />
  <row Id="27382" PostTypeId="2" ParentId="27377" CreationDate="2018-08-18T12:50:34.780" Score="1" Body="&lt;p&gt;Your basic assumption that Hydrogen was created at the Big Bang is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first protons did not appear until the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hadron_epoch&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hadron Epoch&lt;/a&gt; when the temperature of the universe was &quot;low&quot; enough to allow quarks to bind together into, among other things, protons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen is an atom formed from an electron and a proton and optional neutrons (for Deuterium and Tritium isotopes of Hydrogen).  These did not form until the &lt;a href=&quot;https://en.wikipedia.org/wiki/Recombination_(cosmology)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Recombination Epoch&lt;/a&gt; when temperatures had cooled enough to allow electrons to bind with protons to form Hydrogen atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the binding energies are also the energy required to break up particles that are bound, so at higher energies (= higher average temperatures) the atoms and hadrons would not remain in their bound states.  The binding energy of a proton to an electron is orders of magnitude lower than the binding energy of quarks to quarks, hence the Hadron epoch happens at a much early phase (a much higher temperature).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Electrically charged particles cannot be massless.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's an interesting point and there is an interesting question on &lt;a href=&quot;https://physics.stackexchange.com/questions/7905/massless-charged-particles#7909&quot;&gt;Physics SE&lt;/a&gt; about this.  There are arguments for saying it is not impossible on theoretical grounds, but also arguments for saying that we haven't seen them so they don't exist.  Suffice to say that the statement you make is subject to ongoing debate and probably a good one to start an argument between particle physicists at parties (assuming you ever see a particle physicist in the wild, let alone at a party :-)).&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2018-08-18T12:50:34.780" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27384" PostTypeId="2" ParentId="27380" CreationDate="2018-08-18T15:35:09.930" Score="4" Body="&lt;p&gt;If you are referring to the Earth...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the number is positive, the star is moving away from us. When the number is negative, the star is moving towards us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source with more info: &lt;a href=&quot;https://cseligman.com/text/stars/radialvelocity.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://cseligman.com/text/stars/radialvelocity.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24048" LastEditorUserId="24048" LastEditDate="2018-08-19T15:04:18.667" LastActivityDate="2018-08-19T15:04:18.667" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27386" PostTypeId="2" ParentId="27380" CreationDate="2018-08-18T16:39:50.690" Score="2" Body="&lt;p&gt;The number -5.5 km/s in this context means that the relative line of sight velocity between Sirius and the Sun is 5.5 km/s in the sense that Sirius and the Sun are getting closer to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the line of sight velocity of Sirius as observed from the Earth will vary throughout the year, since the Earth moves with an orbital speed of 30 km/s around the Sun. Astronomical radial velocities that are quoted in catalogues will almost certainly be corrected for this and will be referred to as &quot;heliocentric radial velocities&quot; if corrected to the Sun;s reference frame, or &quot;barycentric radial velocities&quot; if corrected to the centre of mass of the solar system (which is almost the same to within 10 m/s or so).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-08-18T16:39:50.690" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27387" PostTypeId="2" ParentId="27381" CreationDate="2018-08-18T18:27:44.510" Score="2" Body="&lt;p&gt;TheSkyLive has charts showing the &lt;a href=&quot;https://theskylive.com/spacecraft&quot; rel=&quot;nofollow noreferrer&quot;&gt;positions of a few interplanetary spacecraft&lt;/a&gt; in the sky for a given time and place on Earth.&#xA;Their positions for Voyagers 1 and 2 agree with the authoritative &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL HORIZONS&lt;/a&gt;, which can generate a table of &lt;a href=&quot;https://en.wikipedia.org/wiki/Equatorial_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;equatorial coordinates&lt;/a&gt; for a range of dates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://nssdc.gsfc.nasa.gov/nmc/spacecraftDisplay.do?id=1977-084A&quot; rel=&quot;nofollow noreferrer&quot;&gt;Voyager 1&lt;/a&gt; is in &lt;a href=&quot;https://theskylive.com/planetarium?obj=voyager1&quot; rel=&quot;nofollow noreferrer&quot;&gt;northern Ophiuchus&lt;/a&gt;.&#xA;This is in the August evening sky for most of the Earth.&#xA;As of August 2018, it is 143 &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_unit&quot; rel=&quot;nofollow noreferrer&quot;&gt;AU&lt;/a&gt; away from the Sun and receding 3.6 AU/year.&#xA;&lt;a href=&quot;https://nssdc.gsfc.nasa.gov/nmc/spacecraftDisplay.do?id=1977-076A&quot; rel=&quot;nofollow noreferrer&quot;&gt;Voyager 2&lt;/a&gt;&#xA;is in &lt;a href=&quot;https://theskylive.com/planetarium?obj=voyager2&quot; rel=&quot;nofollow noreferrer&quot;&gt;northern Pavo&lt;/a&gt;, in the September evening sky for the tropics and southern hemisphere.&#xA;Its current distance from the Sun is 118 AU, receding 3.2 AU/year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since both probes are so far away (compare Neptune at 30 AU) and moving away from the Sun, their apparent positions among the stars will not change much in the next few years.&#xA;Of course they are not visible in any telescope, and a very large dish is needed to receive their radio transmissions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you like using desktop planetarium software, you can try importing &lt;a href=&quot;https://voyager.jpl.nasa.gov/mission/science/hyperbolic-orbital-elements/&quot; rel=&quot;nofollow noreferrer&quot;&gt;hyperbolic orbital elements&lt;/a&gt; for the Voyagers.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-08-18T18:27:44.510" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27388" PostTypeId="2" ParentId="26870" CreationDate="2018-08-18T20:01:57.593" Score="-1" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Influences from outside the observable universe explaining dark energy and expansion?&lt;/p&gt;&#xA;&lt;p&gt;Could some influence from outside the observable universe be causing the expansion?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Outside of the &amp;quot;observable universe&amp;quot; is the portion of the universe which is unobservable.&lt;/p&gt;&#xA;&lt;p&gt;The &lt;em&gt;whole&lt;/em&gt; universe (observable or not) is all that there exists within our &lt;a href=&quot;https://en.wikipedia.org/wiki/Dimension#Additional_dimensions&quot; rel=&quot;nofollow noreferrer&quot;&gt;dimensions&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Outside of our universe is nothing, our universe is the full extent of existence in our dimensions.&lt;/p&gt;&#xA;&lt;p&gt;There is no proof that something outside of our universe can influence something within our universe, it's not as though we are contained in something.&lt;/p&gt;&#xA;&lt;p&gt;Even the rate of expansion is disagreed upon.&lt;/p&gt;&#xA;&lt;p&gt;In the paper: &amp;quot;&lt;a href=&quot;http://iopscience.iop.org/article/10.3847/1538-4357/aac82e&quot; rel=&quot;nofollow noreferrer&quot;&gt;Milky Way Cepheid Standards for Measuring Cosmic Distances and Application to Gaia DR2: Implications for the Hubble Constant&lt;/a&gt;&amp;quot; published in the journal of The American Astronomical Society (July 12 2018), by Adam G. Riess, Stefano Casertano, Wenlong Yuan, Lucas Macri, Beatrice Bucciarelli, Mario G. Lattanzi, John W. MacKenty, J. Bradley Bowers, WeiKang Zheng, Alexei V. Filippenko, Caroline Huang, and Richard I. Anderson they wrote:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;... the value of the Hubble constant has been determined to be &lt;span class=&quot;math-container&quot;&gt;$H_0 = 73.24 ±1.7\, $&lt;/span&gt;km s&lt;span class=&quot;math-container&quot;&gt;$^{−1}\, $&lt;/span&gt;Mpc&lt;span class=&quot;math-container&quot;&gt;$^{−1}$&lt;/span&gt;, from R16.&amp;quot;. [Note: R16 is Riess' paper: &amp;quot;&lt;a href=&quot;https://arxiv.org/abs/1604.01424&quot; rel=&quot;nofollow noreferrer&quot;&gt;A 2.4% Determination of the Local Value of the Hubble Constant&lt;/a&gt;&amp;quot;.]&lt;/p&gt;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&lt;p&gt;...  is inconsistent with the scale needed to match the Planck 2016&#xA;cosmic microwave background data combined with ΛCDM at the 2.9σ confidence level (99.6%). At 96.5% confidence we find that the formal DR2 errors may be underestimated as indicated. We identify additional errors associated with the use of augmented Cepheid samples utilizing ground-based photometry and discuss their likely origins. Including the DR2 parallaxes with all prior distance-ladder data raises the current tension between the late&#xA;and early universe route to the Hubble constant to 3.8σ (99.99%).&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In the &amp;quot;&lt;a href=&quot;https://www.cosmos.esa.int/web/planck/publications#Planck2018&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck 2018 Results&lt;/a&gt;&amp;quot; published as: &amp;quot;&lt;a href=&quot;https://arxiv.org/abs/1807.06209&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck 2018 results. VI. Cosmological parameters&lt;/a&gt;&amp;quot; (Jul 17 2018), by the&#xA;Planck Collaboration: N. Aghanim, Y. Akrami, M. Ashdown, J. Aumont, C. Baccigalupi, M. Ballardini, A. J. Banday, R. B. Barreiro, N. Bartolo, S. Basak, R. Battye, K. Benabed, J.-P. Bernard, M. Bersanelli, P. Bielewicz, J. J. Bock, J. R. Bond, J. Borrill, F. R. Bouchet, F. Boulanger, M. Bucher, C. Burigana, R. C. Butler, E. Calabrese, J.-F. Cardoso, J. Carron, A. Challinor, H. C. Chiang, J. Chluba, L. P. L. Colombo, C. Combet, D. Contreras, B. P. Crill, F. Cuttaia, P. de Bernardis, G. de Zotti, J. Delabrouille, J.-M. Delouis, E. Di Valentino, J. M. Diego, O. Doré, M. Douspis, A. Ducout, X. Dupac, S. Dusini, G. Efstathiou, F. Elsner, T. A. Enßlin, H. K. Eriksen, Y. Fantaye, M. Farhang, J. Fergusson, R. Fernandez-Cobos, F. Finelli, F. Forastieri, M. Frailis, E. Franceschi, A. Frolov, et al. (120 additional authors not shown) they wrote:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;The Planck base-ΛCDM cosmology requires a Hubble constant &lt;span class=&quot;math-container&quot;&gt;$H_0 = 67.4±0.5\, $&lt;/span&gt;km s&lt;span class=&quot;math-container&quot;&gt;$^{−1}\,$&lt;/span&gt;Mpc&lt;span class=&quot;math-container&quot;&gt;$^{−1}$&lt;/span&gt;, in substantial 3.6σ tension with the latest local determination by Riess et al. (2018b). The Planck measurement is in excellent agreement with independent inverse-distance ladder measurements using BAO, supernovae, and element abundance results. None of the extended models that we have studied in this paper convincingly resolves the tension with the Riess et al. (2018b) value of &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt;.&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Wikipedia's webpage &amp;quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Expansion_of_the_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;Expansion of the universe&lt;/a&gt;&amp;quot; explains:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;The expansion of the universe is the increase of the distance between two distant parts of the universe with time. It is an intrinsic expansion whereby the scale of space itself changes. &lt;strong&gt;The universe does not expand &amp;quot;into&amp;quot; anything and does not require space to exist &amp;quot;outside&amp;quot; it.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Technically neither space, nor objects in space, move. Instead &lt;strong&gt;it is the metric governing the size and geometry of spacetime itself that changes in scale&lt;/strong&gt;. Although light and objects within spacetime cannot travel faster than the speed of light, this limitation does not restrict the metric itself. To an observer it appears that space is expanding and all but the nearest galaxies are receding into the distance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-08-18T20:01:57.593" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27389" PostTypeId="2" ParentId="27250" CreationDate="2018-08-19T02:14:21.873" Score="1" Body="&lt;p&gt;Galileo saw Jupiter and its moons, now called the Galilean moons in his honor. They include Io, Europa, Ganymede, and Callisto. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When he took those &quot;images&quot;, that period of time was around early January in 1610. During that time, Galileo was testing his 30x magnification telescope (&lt;a href=&quot;https://www.universetoday.com/15763/galileos-telescope/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in terms of the &lt;em&gt;phases&lt;/em&gt; of the Galilean moons, it would definitely require a stronger telescope. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about some math...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a really complicated formula for this here - &lt;a href=&quot;https://rechneronline.de/sehwinkel/angular-diameter.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Website&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to see the moons themselves, you can put in the diamter of one of the moons in as R for a rough estimate. I'm going to use Io.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Io's diameter is 3642.58921 km. Its distance is 628287897.6 km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We insert that into the website/calculator... Getting 0.0003321805004 degrees or 1.19585 arcsecs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since you are referring to a refracting telescope, there definitely is a limit on how much resolution you can get due to blurring as the telescope gets larger and larger lenses are bound to break easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest refracting telescopes are less than a meter long. We can use diffraction limit to see the limit of their size and viewing before they start diffracting - &lt;a href=&quot;http://hosting.astro.cornell.edu/academics/courses/astro201/diff_limit.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Diffraction limit&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can use the wavelength of green, or approximately 530 nm or 5.3e-5 cm as a rough estimate. The largest telescopes are 1 m or 100 cm. We then get 0.000000647 or 6.466e-7 rad. We convert this to arcsecs and get 0.13337082 arcsecs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that Io is clearly visible, but a large telescope is definitely necessary. I would somewhat recommend not doing so, since a singular decent sized telescope probably wouldn't be strong enough...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit - 8/19/18 - made a mistake on the calculation, it is actually possible to see Io's phases, but would be rather difficult. Credit to Mike G for pointing it out.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="24048" LastEditorUserId="24048" LastEditDate="2018-08-19T04:01:07.097" LastActivityDate="2018-08-19T04:01:07.097" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27392" PostTypeId="1" CreationDate="2018-08-19T07:30:07.290" Score="0" ViewCount="263" Body="&lt;p&gt;Check out &lt;a href=&quot;https://twitter.com/natalie9280/status/1028400052924903424&quot; rel=&quot;nofollow noreferrer&quot;&gt;@natalie9280’s Tweet&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;long story short, friday morning at around 6:40am i looked outside and i saw three suns.&#xA;  i checked to see if there was an eclipse and i could[n't] find anything.&#xA;  twitter help pls&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;[VIDEO: three progressively brighter &quot;suns,&quot; left to right in a closely spaced row]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18432" LastEditorUserId="616" LastEditDate="2018-08-19T15:14:32.673" LastActivityDate="2018-08-19T16:20:20.290" Title="Why does the sky look like it has three suns in this video?" Tags="&lt;the-sun&gt;&lt;optics&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2018-08-20T07:29:33.220" ContentLicense="CC BY-SA 4.0" />
  <row Id="27393" PostTypeId="2" ParentId="27250" CreationDate="2018-08-19T09:28:15.543" Score="5" Body="&lt;p&gt;From Earth the resolution of is limited by the atmosphere.  Turbulence in the atmosphere make it impossible to resolve the moons of Jupiter to more than a &lt;a href=&quot;http://www.educationalwebadventures.com/portfolio/io/study2.html&quot; rel=&quot;noreferrer&quot;&gt;few pixels wide&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amateur refractor telescopes tend to be small and less powerful than reflector telescopes. I'm not aware of any amateur refactor telescope that could resolve a moon of Jupiter to more than a point of light. The eye is unable to resolve an object that is less than about 60 arcseconds and it would need to be closer to 1000 arcseconds before many surface details became visible. The Moon, for comparison, is 1800 arcseconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Professional equipment can partially overcome the atmosphere by using adaptive optics, but this is not a feature that is generally available to amateurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter and its moons orbit outside the orbit of Earth, so from our point of view Jupiter is never &lt;a href=&quot;http://www.astronomy.com/magazine/ask-astro/2017/02/jupiters-illumination&quot; rel=&quot;noreferrer&quot;&gt;partially lit less than 99%&lt;/a&gt;. The moons are sometimes eclipsed by Jupiter, but they too are never partially lit less than 99%.  &lt;em&gt;With amateur equipment, Jupiter and its moons are never anything other than &quot;full&quot;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even with the Hubble telescope, the resolution is not sufficient to identify phases of the moons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What Galileo saw was Jupiter, resolved as a disc and 4 moons orbiting it which appeared to him as &quot;stars&quot; (he called them the Medicean stars). The four moons were not always visible, sometimes a moon was hidden in the shadow of Jupiter, or behind Jupiter from our point of view. At other times a moon would be too close to Jupiter to be visible, or outside the field of view of his telescope. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What he realised was that these were bodies in orbit around Jupiter, and that they obeyed the same rule for orbits that Kepler had proposed for the planets.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-08-19T09:28:15.543" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27394" PostTypeId="1" AcceptedAnswerId="27396" CreationDate="2018-08-19T11:55:13.450" Score="1" ViewCount="124" Body="&lt;p&gt;I understand Planetary Eccentricity and Inclination.  However when it applies to a star, does the centre or ecliptic plane refer to the Sun or the Galactic Centre?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The star Sirius for example, 136.336 ± 0.040° inclination, the angle at which it is inclined to Sun, Earth or Galactic Centre?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The star Sirius has an Eccentricity of  0.59142 ± 0.00037.  The only thing the star orbits is the Galactic Centre (Sagittarius A*) so I guess that is the centre?&lt;/p&gt;&#xA;" OwnerUserId="5401" LastActivityDate="2018-08-20T00:33:32.040" Title="Eccentricity and Inclination of a Star" Tags="&lt;star&gt;&lt;inclination&gt;&lt;eccentricity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27395" PostTypeId="2" ParentId="27392" CreationDate="2018-08-19T14:46:25.237" Score="5" Body="&lt;p&gt;The ghost images in the video resemble the Sun and are separated by only 2 or 3 degrees (4-6 Sun widths), so we can rule out atmospheric phenomena.&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sun_dog&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sun dogs&lt;/a&gt; or parhelia have shapes different from the Sun and appear 22 degrees away from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect is more likely due to looking through a double-paned window at an oblique angle.&#xA;The bright image on the right is the Sun viewed directly.&#xA;The image in the middle is a reflection by the inner glass surfaces facing each other, offset due to the gap width and the viewing angle.&#xA;The faint image on the left is a further reflection of the middle image.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-08-19T16:20:20.290" LastActivityDate="2018-08-19T16:20:20.290" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27396" PostTypeId="2" ParentId="27394" CreationDate="2018-08-19T15:24:04.630" Score="5" Body="&lt;p&gt;Sirius is a binary system, composed of a main sequence star (Sirius A) and a white dwarf (Sirius B). The two orbit each other with a period of about 50 years, and the eccentricity given on Wikipedia is the eccentricity of the stars' orbits around one another. The inclination, according to &lt;a href=&quot;http://adsabs.harvard.edu/abs/2017ApJ...840...70B&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bond et al. 2017&lt;/a&gt; - the paper Wikipedia references for both numbers - is described as the inclination of the &quot;relative visual orbit&quot;, leading me to believe that the inclination is defined based on a plane relative to the point of view of an observer on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something it's important to realize is that &lt;a href=&quot;https://physics.stackexchange.com/a/148670/56299&quot;&gt;stars don't follow Keplerian orbits&lt;/a&gt;, because of the non-point-like distribution of matter in the galaxy; the motion of a star around a galaxy is therefore somewhat more complicated than the motion of a star in a binary system.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2018-08-20T00:33:32.040" LastActivityDate="2018-08-20T00:33:32.040" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27397" PostTypeId="1" CreationDate="2018-08-19T16:25:05.147" Score="2" ViewCount="110" Body="&lt;p&gt;This morning at 6:12 am looking south from Georgetown, Texas, there was a brief very bright stationary object what appeared to be a star that once I saw it within the next few seconds quickly vanished.  Did not appear to be other than a stationary object.  What was it?&lt;/p&gt;&#xA;" OwnerUserId="24054" LastEditorUserId="616" LastEditDate="2018-08-19T23:29:57.863" LastActivityDate="2018-08-30T14:34:56.810" Title="Bright star or flare near orion?" Tags="&lt;artificial-satellite&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27398" PostTypeId="2" ParentId="12935" CreationDate="2018-08-19T17:35:49.627" Score="3" Body="&lt;p&gt;The first star having rings seems to have been discovered in the late '90 by You-Hua Chu (University of Illinois). In &lt;em&gt;Cosmic Catastrophes: Exploding Stars, Black Holes, and Mapping the Universe&lt;/em&gt;, &lt;a href=&quot;https://www.as.utexas.edu/~wheel/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Craig Wheeler&lt;/a&gt; wrote (in 2007):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[You-Hua Chu] scored another coup a decade later, at a meeting in Chile to&#xA;  celebrate the tenth anniversary of the discovery of the supernova,&#xA;  when she reported that she had discovered the first star to have rings&#xA;  around it, like the progenitor of SN 1987A.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;SN 1987A was a supernovae that exploded in 1987. The progenitor (the star that actually exploded), Sk-69 20, was a blue supergiant. &lt;a href=&quot;http://chem.tufts.edu/science/astronomy/sn1987a.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pictures taken from Hubble showed three rings&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Yl79z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Yl79z.jpg&quot; alt=&quot;SN 1987A&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This video shows the lightning effect of the explosion on the inner ring from 1994 to 2016, that is, 7 to 29 years &lt;em&gt;after&lt;/em&gt; the explosion (the ring is a few light years across while the most rapidly moving outer portions of the exploded star are moving at least at 10 percent of the speed of light - source: same book, p. 136).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hubblesite.org/videos/news/release/2017-08&quot; rel=&quot;nofollow noreferrer&quot;&gt;This time-lapse video sequence of Hubble Space Telescope images reveals dramatic changes in a ring of material around the exploded star Supernova 1987A&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="24056" LastEditorUserId="24056" LastEditDate="2018-08-22T15:09:59.600" LastActivityDate="2018-08-22T15:09:59.600" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27399" PostTypeId="1" AcceptedAnswerId="27400" CreationDate="2018-08-19T22:30:29.550" Score="2" ViewCount="62" Body="&lt;p&gt;Night sky objects always interested me, nowadays Im trying to observe and photograph some of these objects. My phone is iPhone 6S, i used SkyGuide app to find positions, NightCap to photograph. Im sure brightest star is Mirach according to SkyGuide app. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's photographed at 00:30 at my local time (Istanbul, London + 3), SkyGuide says Mirach's azimuth was 77° 11’ 28” .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZXdYh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZXdYh.jpg&quot; alt=&quot;What are those?&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24059" LastEditorUserId="24059" LastEditDate="2018-08-19T22:46:45.567" LastActivityDate="2018-08-19T23:06:51.830" Title="What are these objects near Mirach" Tags="&lt;star&gt;&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27400" PostTypeId="2" ParentId="27399" CreationDate="2018-08-19T23:00:30.710" Score="2" Body="&lt;p&gt;If this is really Mirach, then I would say the point-like object above and a little to the left of it is the star mu And (another designation: 37 And) in the same constellation of Andromeda. Then the next point-like object slightly further above and to the left from mu And would be the star nu And (35 And). And a barely visible fuzzy object above nu And would be the Andromeda galaxy (M31). Then in the left lower corner we see another star Nembus (51 And), and above it the star phi And (42 And). And at the same height with nu And farther to the right from it at the edge of the building there is the star pi And (29 And).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would strongly suggest you to use Stellarium (www.stellarium.org) for identification of nearly any night sky object visible at your or any other location on Earth at any time.&lt;/p&gt;&#xA;" OwnerUserId="24027" LastEditorUserId="24027" LastEditDate="2018-08-19T23:06:51.830" LastActivityDate="2018-08-19T23:06:51.830" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27401" PostTypeId="2" ParentId="27397" CreationDate="2018-08-20T00:51:58.080" Score="4" Body="&lt;p&gt;Iridium satellites are known for predictable flare events but are gradually &lt;a href=&quot;https://www.heavens-above.com/IridiumDemise.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;being de-orbited&lt;/a&gt; and replaced.&#xA;In particular, Iridium 66 would have &lt;a href=&quot;https://www.heavens-above.com/flaredetails.aspx?fid=0&amp;amp;lat=30.6333&amp;amp;lng=-97.678&amp;amp;loc=Georgetown&amp;amp;alt=230&amp;amp;tz=CST&quot; rel=&quot;nofollow noreferrer&quot;&gt;flared at 6:04&lt;/a&gt; several degrees east of Orion,&#xA;but its orbit has almost certainly changed since its de-orbiting process &lt;a href=&quot;http://www.rod.sladen.org.uk/iridium.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;started on August 2&lt;/a&gt;.&#xA;Other Iridium satellites have taken 3 to 5 weeks to finish this process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Satellite_flare#Non-Iridium_flares&quot; rel=&quot;nofollow noreferrer&quot;&gt;Other satellites&lt;/a&gt; can flare too if a shiny surface catches the Sun just right.&#xA;Heavens-Above shows two satellites passing south of Orion at 6:12 on the morning of August 19:&#xA;&lt;a href=&quot;https://www.heavens-above.com/passdetails.aspx?lat=30.6333&amp;amp;lng=-97.678&amp;amp;loc=Georgetown&amp;amp;alt=230&amp;amp;tz=CST&amp;amp;satid=37162&amp;amp;mjd=58349.4700950893&quot; rel=&quot;nofollow noreferrer&quot;&gt;USA 215&lt;/a&gt;, an American spy satellite;&#xA;and &lt;a href=&quot;https://www.heavens-above.com/passdetails.aspx?lat=30.6333&amp;amp;lng=-97.678&amp;amp;loc=Georgetown&amp;amp;alt=230&amp;amp;tz=CST&amp;amp;satid=20663&amp;amp;mjd=58349.4651892621&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cosmos 2084&lt;/a&gt;, a failed Russian early warning satellite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is a head-on sporadic meteor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Update:&lt;/em&gt; Iridium 66 finished de-orbiting on August 23.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-08-30T14:34:56.810" LastActivityDate="2018-08-30T14:34:56.810" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27402" PostTypeId="2" ParentId="26870" CreationDate="2018-08-20T04:55:07.173" Score="-2" Body="&lt;p&gt;I think the original question is valid and I have been pondering that question myself for years&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I am having difficulty understanding the logic in stating there is nothing outside the observable universe. The term 'observable universe' by definition limits knowledge to whatever is observable and does not exclude any thing outside of it that is not observable.  This has been mentioned in earlier posts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there could be observable effects within our universe that may give a clue to the existence of an outside influence upon the known universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of these is the fact that distant galaxies are accelerating, apparently, away from us.  One phenomenon that has this ability to cause acceleration is gravity. One hypothesis could be that the distant galaxies are caught in a gravitational field outside our known universe. This would place our known universe within, for want of a better word, a 'bubble' within this gravitational influence causing it to expand in all directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this would change the idea of a big bang theory, not in its effect but its origins.  One has difficulty in imagining the whole of our universe being crammed into a single point as unstable as that may be.  However one could imagine it coming into existence by, for want of a better word, 'squirted' into a void within an outside existence of a meta-universe.  This also does away with the need for dark energy as we once thought we needed the ether to carry radio waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Karl Popper says that any hypothesis/theory/etc must be falsifiable.  The above  is certainly one of those things. It only needs to be proven non-falsifiable, possibly, with physics not yet known. &lt;/p&gt;&#xA;" OwnerUserId="24061" LastActivityDate="2018-08-20T04:55:07.173" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27403" PostTypeId="1" CreationDate="2018-08-20T10:52:59.243" Score="2" ViewCount="335" Body="&lt;p&gt;Imagine a neutron star could be broke into pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How big would be the piece matching Earth's mass?&lt;/p&gt;&#xA;" OwnerUserId="22664" LastActivityDate="2018-08-21T16:12:22.750" Title="How big should be a piece of neutron star to reach Earth's mass?" Tags="&lt;earth&gt;&lt;neutron-star&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27404" PostTypeId="2" ParentId="27403" CreationDate="2018-08-20T11:34:18.590" Score="2" Body="&lt;p&gt;Volume is mass divided by density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth's mass = 5.972x10²⁴ kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutron star's density = 10¹⁷ kg/m³.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sought volume of a neutron star's piece with Earth's mass is then Earth's mass divided by neutron star's density, which makes 59720000 m³. This is a cube with the side length of 390.87 m or a sphere with radius of 242.48 m.&lt;/p&gt;&#xA;" OwnerUserId="24027" LastEditorUserId="24027" LastEditDate="2018-08-20T11:45:34.367" LastActivityDate="2018-08-20T11:45:34.367" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27405" PostTypeId="2" ParentId="20973" CreationDate="2018-08-20T14:57:57.727" Score="0" Body="&lt;p&gt;Divide  the year by 19 and multiply the remainder by 11.&#xA;Add the number of complete months since March1&#xA;Add the day of the month.  Reduce this mod 30.&#xA;That is the approximate (+/- 1) number of days since the last new moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For January and February use the previous year.&lt;/p&gt;&#xA;" OwnerUserId="24067" LastActivityDate="2018-08-20T14:57:57.727" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27406" PostTypeId="2" ParentId="27403" CreationDate="2018-08-20T19:06:10.900" Score="6" Body="&lt;p&gt;The question has no answer. The &lt;strong&gt;minimum&lt;/strong&gt; mass for stable neutron matter is about 0.1 solar masses (&lt;a href=&quot;https://physics.stackexchange.com/questions/143166/what-is-the-theoretical-lower-mass-limit-for-a-gravitationally-stable-neutron-st/143174#143174&quot;&gt;see my answer here&lt;/a&gt;) and so is much more massive than the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As neutron stars become less massive they also become less dense, so your question is unanswerable at that level too, since we cannot put a figure on the density of something that cannot even exist in principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your question is merely what radius is a spherical object that has a similar  density to a typical neutron star, can you not do&#xA;$$\frac{4\pi r^3}{3} \rho = M_{\rm Earth}$$&#xA;to find $r$?&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-08-20T19:06:10.900" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27407" PostTypeId="2" ParentId="10222" CreationDate="2018-08-20T22:14:47.640" Score="2" Body="&lt;p&gt;There wouldn't be a direct correlation between the equinox and the moon's orbit, so there really isn't any special - it's just a coincidence.&lt;/p&gt;&#xA;" OwnerUserId="24048" LastActivityDate="2018-08-20T22:14:47.640" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27408" PostTypeId="2" ParentId="21869" CreationDate="2018-08-20T22:46:17.130" Score="2" Body="&lt;p&gt;One thing to keep in mind is that Earth isn't really all that special from the rest of the universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't believe the phenomenon exists, since that would really put Earth in a &quot;special&quot; position, and that's not really correct. If it did exist, then it'll probably be famous or well known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've never watched these &quot;alpha centauri&quot; videos before, and they seem to not be in english, but based off of my background knowledge, I do not recall any phenomenon of the sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, there are some exceptions to the &quot;every galaxy has a red shift&quot; rule, such as...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Andromeda Galaxy has a blue shift because it is moving towards us, and on a collision course.&lt;/li&gt;&#xA;&lt;li&gt;I saw this article (&lt;a href=&quot;http://www.slate.com/blogs/bad_astronomy/2016/10/18/m98_is_a_rare_blueshifted_galaxy.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;) that stated that a galaxy named M98 that is nearly traveling as fast as the Andromeda Galaxy, even though it isn't even close to us (50 million ly) and isn't on a collision course.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Besides these few exceptions I doubt that there really isn't any phenomenon. Not to mention that only galaxies are affected by the redshift-blueshift thing - not individual stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps!&lt;/p&gt;&#xA;" OwnerUserId="24048" LastActivityDate="2018-08-20T22:46:17.130" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27409" PostTypeId="2" ParentId="10131" CreationDate="2018-08-20T22:49:34.850" Score="2" Body="&lt;p&gt;Source: &lt;a href=&quot;https://www.jpl.nasa.gov/spaceimages/details.php?id=pia18821&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL Article&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The gravity anomalies along the border structures are interpreted as ancient, solidified, lava-flooded rifts that are now buried beneath the surface of the dark volcanic plains, or maria, on the near side of the moon.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the lava under Oceanus Procellarum, it's heavier/massive so it has a stronger gravitational pull than the area surrounding it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps!&lt;/p&gt;&#xA;" OwnerUserId="24048" LastActivityDate="2018-08-20T22:49:34.850" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27411" PostTypeId="1" AcceptedAnswerId="27412" CreationDate="2018-08-21T07:12:26.263" Score="6" ViewCount="91" Body="&lt;p&gt;If the density of the matter from which the first generation stars were made was homogeneous, so during the star formation nuclear fusion would have started around when they all reached to same critical mass and core temperature, once fusion started the outward pressure would have counteracted on inflow of matter so all stars should have been of similar size &lt;/p&gt;&#xA;" OwnerUserId="16136" LastEditorUserId="24027" LastEditDate="2018-08-21T12:54:57.937" LastActivityDate="2018-08-21T12:54:57.937" Title="Were all first generation stars of same mass and size?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27412" PostTypeId="2" ParentId="27411" CreationDate="2018-08-21T07:35:09.557" Score="5" Body="&lt;p&gt;If the universe had been entirely homogeneous then no stars would have formed. Star formation requires regions of over-density to collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are plenty of ideas of what determines the &lt;em&gt;spectrum&lt;/em&gt; of masses that stars are born with (known as the initial mass function) - mainly concerning turbulence, fragmentation and accretion. The commencement of nuclear fusion is not a factor, since nothing really changes at the surface of a star when this happens (the luminosity is the same).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reasons for the observed initial mass function in the present-day universe (which is remarkably similar from place to place) are not fully understood, so unsurprisingly, neither is it a settled matter in the early universe where it cannot even be observed. However, it seems likely on theoretical grounds that stars formed from pristine, primordial gas could be much more massive and less susceptible to fragmentation into smaller objects. Even if this is the case, simulations still suggest that stars would be born with masses that vary by orders of magnitude (e.g. &lt;a href=&quot;https://arxiv.org/abs/1603.09475&quot; rel=&quot;noreferrer&quot;&gt;Stacy (2016)&lt;/a&gt; and recent references therein). &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-08-21T07:35:09.557" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27413" PostTypeId="1" AcceptedAnswerId="27414" CreationDate="2018-08-21T08:37:08.583" Score="9" ViewCount="2032" Body="&lt;p&gt;As I understand, the evidence for dark matter is the observation of the gravitational dynamics of objects around the size of galaxies and above.  They move in a way such that the equations imply there is more mass than what we can see.  There has never been any direct observation of dark matter, and our theories tell us it cannot be composed of any elementary particles from the standard model.  Furthermore, there is no observation which suggests that dark matter is unevenly distributed, that some galaxies consisting of roughly the same amount of normal matter have varying amounts of dark matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if therefore an alternative explanation for the phenomena is that we have the wrong dynamical equations for gravity.  Just as Newtonian gravitation is a good approximation to general relativity at low masses/energies, perhaps general relativity is a good approximation to the true gravitational theory at sub-galactic distances.  Maybe when distances become very large we start seeing that gravatational force is described by slightly different curves than what general relativity predicts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any good evidence speaking against this alternative approach?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this question is different from the &quot;possible duplicate&quot; is that the other question asks if the source of the effect attributed to dark matter is parallel universes, rather than different dynamical laws.  However, the answer there does mention these alternative gravitational theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize now that this is not a new idea, and I thank the commenters and answerers for pointing me to existing research.  However, this part of StackExchange does not bill itself as an &quot;experts only&quot; forum, so I don't think the downvotes are fair.  In mathematics for example, there is one forum for experts, and another for everyone.  I'm an expert in something, but not astronomy.  Thanks.&lt;/p&gt;&#xA;" OwnerUserId="24077" LastEditorUserId="24077" LastEditDate="2018-08-21T13:27:37.043" LastActivityDate="2018-08-21T19:53:18.823" Title="Could dark matter be wrong?" Tags="&lt;gravity&gt;&lt;dark-matter&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="2" ClosedDate="2018-08-21T16:27:12.947" ContentLicense="CC BY-SA 4.0" />
  <row Id="27414" PostTypeId="2" ParentId="27413" CreationDate="2018-08-21T10:25:32.250" Score="23" Body="&lt;p&gt;Multiple theories and hypotheses have been proposed as an alternative to dark matter (DM). The most popular are, arguably,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Modified_Newtonian_dynamics&quot; rel=&quot;noreferrer&quot;&gt;MOND&lt;/a&gt; (MOdified Newtonian Dynamics):&lt;br&gt;A term for various theories where the gravitational force falls off less steeply than $1/r^2$ at large distances.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Tensor%E2%80%93vector%E2%80%93scalar_gravity&quot; rel=&quot;noreferrer&quot;&gt;TeVeS&lt;/a&gt; (Tensor–vector–scalar):&lt;br&gt;A relativistic generalization of MOND.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/F(R)_gravity&quot; rel=&quot;noreferrer&quot;&gt;f(R) gravity&lt;/a&gt;:&lt;br&gt;A generalization of general relativity, where the so-called Ricci scalar — which describes the curvature of space(time) — may take different forms than just a number.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Entropic_gravity&quot; rel=&quot;noreferrer&quot;&gt;Entropic gravity&lt;/a&gt;:&lt;br&gt;A string-theory-based theory describing gravity not as a fundamental interaction, but as a so-called entropic force.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note, however, that DM is &lt;em&gt;not&lt;/em&gt; just &quot;a fudge factor to explain weird dynamics&quot;. What makes most (but not all) physicists believe that DM exists is, I think, that several &lt;em&gt;independent&lt;/em&gt; observations yield consistent results, not only qualitatively, but quantitatively. In addition to &lt;a href=&quot;https://en.wikipedia.org/wiki/Galaxy_rotation_curve&quot; rel=&quot;noreferrer&quot;&gt;rotation curves&lt;/a&gt; of spiral galaxies, or &lt;a href=&quot;https://en.wikipedia.org/wiki/Velocity_dispersion&quot; rel=&quot;noreferrer&quot;&gt;velocity dispersions&lt;/a&gt; in stellar clusters, elliptical galaxies, and galaxy clusters, you have for instance&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_lens&quot; rel=&quot;noreferrer&quot;&gt;Gravitational lensing&lt;/a&gt;:&lt;br&gt;Massive clusters of galaxies &quot;bend&quot; space around them, causing the light from background sources to diverge from a straight path. Measuring this divergence tells us the mass of the foreground object, which can then be compared to the visible mass..&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;noreferrer&quot;&gt;Cosmic microwave background&lt;/a&gt;:&lt;br&gt;The patterns in the inhomogeneities of the background radiation is sensitive to the amounts of both dark and &quot;normal&quot; matter.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Structure_formation&quot; rel=&quot;noreferrer&quot;&gt;Structure formation&lt;/a&gt;:&lt;br&gt;DM explains how matter may collapse and form galaxies a few 100 million years after Big Bang, before the Universe expanded too much.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Type_Ia_supernova&quot; rel=&quot;noreferrer&quot;&gt;Supernova observations&lt;/a&gt;:&lt;br&gt;The way that the brightness of distant supernovae decreases with redshift depends on the expansion history of the Universe, which in turn depends on the densities of the various components of the Universe. A certain amount of matter is needed to explain the expansion, and this amount is larger than the observed amount by the same factor that the other experiments yield.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some of the above phenomena are explained by the alternative theories as well, but in general several ad hoc hypotheses have had to be made. In my view (but I am no expert on this!), DM is simply the simplest explanation, and so, by Occam's Razor, is favored by most physicists. I think the theory, unfairly, has a reputation for being &quot;some magic that astronomers pulled out of their hat to explain what they don't understand, and which we made sure can't be observed so you can't prove us wrong&quot;. But firstly, postulating an unknown object to explain some phenomenon is quite normal in physics — this is how e.g. Neptune and several elementary particles were found. And secondly, there are several candidates for what DM might be, and at least some of these make falsifiable predictions. If DM is a particle, then that particle might produce a signal that can be observed. There are many efforts to search for a DM annihilation signal, which is expected to be in the (tens of) GeV region (e.g. &lt;a href=&quot;https://arxiv.org/abs/1103.3266&quot; rel=&quot;noreferrer&quot;&gt;H.E.S.S Collaboration 2011&lt;/a&gt; and &lt;a href=&quot;https://arxiv.org/abs/1610.03840&quot; rel=&quot;noreferrer&quot;&gt;Cui et al. 2017&lt;/a&gt;). There are also efforts for direct detections — see e.g. &lt;a href=&quot;https://arxiv.org/abs/1709.00688&quot; rel=&quot;noreferrer&quot;&gt;Liu et al. (2017)&lt;/a&gt; for a review.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the most compelling pieces of evidence yet for DM came, I think, with the observation of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Bullet_Cluster&quot; rel=&quot;noreferrer&quot;&gt;Bullet Cluster&lt;/a&gt; (and subsequently many other clusters), where lensing clearly shows how unaccounted-for mass follows the collisionless galaxies, but not the collisional gas, as seen in this beautiful image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BTTDK.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BTTDK.jpg&quot; alt=&quot;Bullet&quot;&gt;&lt;/a&gt;&#xA;&lt;sup&gt;The Bullet Cluster. &lt;em&gt;Pink&lt;/em&gt; shows X-rays from the gas, whereas &lt;em&gt;blue&lt;/em&gt; shows where the mass is concentrated. Credit: &lt;a href=&quot;https://svs.gsfc.nasa.gov/30094&quot; rel=&quot;noreferrer&quot;&gt;NASA/CHANDRA&lt;/a&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-08-21T19:53:18.823" LastActivityDate="2018-08-21T19:53:18.823" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="27415" PostTypeId="2" ParentId="20973" CreationDate="2018-08-21T13:20:28.853" Score="0" Body="&lt;p&gt;The procedure for finding the phase of the Moon is outlined in &lt;em&gt;The Explanatory Supplement to the Astronomical Almanac&lt;/em&gt;, University Science Books, 3rd ed., p. 507. The annual &quot;Astronomical Almanac&quot; tabulates the phases to the nearest minute of UT. A phase occurs when the excess of the apparent geocentric ecliptic longitude of the Moon over the apparent geocentric ecliptic longitude of the Sun is 0, 90, 180, or 270 degrees. That section says &quot;there is no simple prediction formula with which to make a comparison&quot;. But that statement would apply to a desired accuracy of 1 minute. If your accuracy need is looser, maybe a usable formula exists, I don't know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your need is within a few hundred years of the present, there is a wide range of software to choose from. But the expressions used can deteriorate rapidly outside the intended range of validity. So if you are examining the distant past or future be sure to use software intended for that purpose.&lt;/p&gt;&#xA;" OwnerUserId="7385" LastActivityDate="2018-08-21T13:20:28.853" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27419" PostTypeId="1" CreationDate="2018-08-22T13:51:27.217" Score="4" ViewCount="1654" Body="&lt;p&gt;Gravity is most accurately described by the general theory of relativity (proposed by Albert Einstein in 1915) which describes gravity not as a force, but as a consequence of the curvature of spacetime caused by the uneven distribution of mass.(source -wikipedia)&lt;img src=&quot;https://i.stack.imgur.com/cTxY3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why it is moving on the boundary of that curvature, but not fall into the middle of that curvature and collide with sun?&lt;/p&gt;&#xA;" OwnerUserId="24088" LastEditorUserId="616" LastEditDate="2018-08-22T15:20:34.243" LastActivityDate="2018-08-22T15:53:31.913" Title="Why does earth not collide with the sun?" Tags="&lt;orbit&gt;&lt;gravity&gt;&lt;cosmology&gt;&lt;general-relativity&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2018-08-28T18:47:48.017" ContentLicense="CC BY-SA 4.0" />
  <row Id="27420" PostTypeId="2" ParentId="27419" CreationDate="2018-08-22T15:21:50.467" Score="8" Body="&lt;p&gt;The primary reason the Earth doesn't fall into the Sun is that it has a very large &lt;em&gt;tangential velocity&lt;/em&gt; with which it is able to maintain an orbit. The physics is the same for describing satellites which we launch into orbit around the Earth. I've heard it described as such: &quot;The ISS is falling towards the earth, but is moving forward so quickly that it continually misses.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, if you were to take a mass and insert it with zero velocity somewhere in the solar system, it would inevitably fall in towards the sun. But because almost all bodies in our solar system were formed from the same large clump of &lt;em&gt;rotating&lt;/em&gt; gas and dust which due to conservation of angular momentum during its collapse both sped up and flattened into a disk, all of the solar system bodies (Earth, other planets, asteroids, etc), have enough forward velocity to continue orbiting the Sun without being pulled to the center. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a physics perspective, you can see that &lt;a href=&quot;https://en.wikipedia.org/wiki/Centripetal_force&quot; rel=&quot;noreferrer&quot;&gt;centripetal acceleration&lt;/a&gt; is governed by the inward force of gravity, tangential to the direction of motion of moving objects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the &lt;a href=&quot;https://www.nasa.gov/content/goddard/parker-solar-probe&quot; rel=&quot;noreferrer&quot;&gt;Solar Parker Probe&lt;/a&gt; which just launched is attempting to do exactly as you describe --- fall into the Sun (or well, get close, to take lots of measurements). To do so is actually difficult, because the probe, launching from Earth, has to &lt;em&gt;lose&lt;/em&gt; roughly 30 km/s in velocity it already has (because that's the Earth's orbital velocity). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How it plans to do this is a cool story in and of itself, which you might be interested in, here's a good summary: &lt;a href=&quot;https://www.theatlantic.com/science/archive/2018/08/parker-solar-probe-launch-nasa/567197/&quot; rel=&quot;noreferrer&quot;&gt;https://www.theatlantic.com/science/archive/2018/08/parker-solar-probe-launch-nasa/567197/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24089" LastActivityDate="2018-08-22T15:21:50.467" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27421" PostTypeId="2" ParentId="27419" CreationDate="2018-08-22T15:53:31.913" Score="1" Body="&lt;p&gt;If you make a physical model of that curvature as you see in &lt;a href=&quot;https://www.youtube.com/watch?v=MTY1Kje0yLg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video&lt;/a&gt;, for instance, you see the model planet going around the model sun quite a few times before it finally hits it. The reason it hits at all is friction with the surface and the surrounding air, which doesn't apply to real planets, so they can orbit around and around for a long time without spiralling in. They are also helped by the fact they are very massive, so that what tiny amount of friction there is from interplanetary dust and such doesn't affect them much. &lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-08-22T15:53:31.913" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27422" PostTypeId="2" ParentId="21869" CreationDate="2018-08-22T18:32:02.743" Score="4" Body="&lt;p&gt;I have not heard of any idea that the nearest object is exceptional in any significant way, and in the case of the nearest pulsar, nearest star, nearest (large) galaxy etc. the properties are usually typical of the underlying population, though perhaps not the &lt;em&gt;observed&lt;/em&gt; population because of biases and selection effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take Proxima Centauri as an example. It is unlike &lt;em&gt;any&lt;/em&gt; star that can be observed with the naked eye. However, it is a very good example of a &quot;modal&quot; star - much less massive than the Sun, about half the age of the Galaxy and a member of a multiple system. The reason for the discrepancy is of course that stars have a wide range of luminosities and distances; in a &quot;flux-limited&quot; survey, we always pick out the brightest sources, which tend to be the most luminous sources at large distances, where our survey volume is largest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it &lt;em&gt;is&lt;/em&gt; true that nearby objects are different to the &lt;em&gt;observed&lt;/em&gt; population.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same may apply whenever some unusual chacteristic makes a type of object more likely to be detected. There is certainly a phenomenon in astronomy where the prototype of a class of objects (usually the first discovered) turns out to be atypical of the class as a whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example we could consider &lt;a href=&quot;https://en.wikipedia.org/wiki/RS_Canum_Venaticorum_variable&quot; rel=&quot;nofollow noreferrer&quot;&gt;RS CVn binaries&lt;/a&gt;, named after the object RS CVn, which is a close binary system containing an evolved (subgiant) component, and exhibiting extreme levels of magnetic activity (star spots, coronal X-ray emission).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was initially believed that the presence of an evolved component was a defining feature of the class. However, over the years it has become apparent that there are many more examples of close, magnetically active binaries with main sequence components. The prototype is atypical in having an evolved component - the explanation is that, although short-lived and rarer, evolved magnetically active stars are much more luminous and easier to find/study.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a final word on this, there &lt;em&gt;is&lt;/em&gt; a set of nearby/nearest objects, including the Sun and Moon, that are somewhat unusual. The Sun is more massive than 90+% of stars in the Galaxy; no other planets in our solar system have a moon that is so (relatively) large. An explanation for these things may lie with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Anthropic_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;anthropic principle&lt;/a&gt;, which argues that we would not observe any circumstance that would preclude the development of intelligent life on Earth or, to turn that around, we are more likely to find ourselves in a situation that is favourable for the development of intelligent life on Earth.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-08-22T18:32:02.743" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27423" PostTypeId="1" AcceptedAnswerId="27425" CreationDate="2018-08-22T20:36:15.063" Score="13" ViewCount="565" Body="&lt;p&gt;As far as I understand, both the Yarkovsky effect and the Yarkovsky–O'Keefe–Radzievskii–Paddack (YORP) effect exert forces upon bodies as a result of thermal photon radiation. How do both effects differ from one another?&lt;/p&gt;&#xA;" OwnerUserId="24091" LastActivityDate="2020-06-06T20:07:09.167" Title="What is the difference between the Yarkovsky effect and YORP effect?" Tags="&lt;orbit&gt;&lt;asteroids&gt;&lt;rotation&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27425" PostTypeId="2" ParentId="27423" CreationDate="2018-08-22T21:06:08.687" Score="15" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Yarkovsky_effect&quot; rel=&quot;noreferrer&quot;&gt;Yarkovsky effect&lt;/a&gt; is the thrust on a small object in space that has been heated by sunlight, created from radiant energy. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Yarkovsky%E2%80%93O%27Keefe%E2%80%93Radzievskii%E2%80%93Paddack_effect&quot; rel=&quot;noreferrer&quot;&gt;YORP effect&lt;/a&gt; is when the thrust produces rotation, likely due to uneven heating or differences in surface projections (IE: A mountain acts as a better solar collector than a flatter portion of a spherical object on the opposite side).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See: &quot;&lt;a href=&quot;https://arxiv.org/abs/1502.01249&quot; rel=&quot;noreferrer&quot;&gt;The Yarkovsky and YORP Effects&lt;/a&gt;&quot; (Feb 4 2015), by &#xA;D. Vokrouhlicky, W. F. Bottke, S. R. Chesley, D. J. Scheeres, and T. S. Statler:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;strong&gt;The Yarkovsky effect&lt;/strong&gt; describes a small but significant force that affects the orbital motion of meteoroids and asteroids smaller than 30 − 40 kilometers in diameter. It is caused by sunlight; when these bodies heat&#xA;  up in the Sun, they eventually re-radiate the energy away in the thermal waveband, which in turn &lt;strong&gt;creates a tiny thrust&lt;/strong&gt;. This recoil acceleration is much weaker than solar and planetary gravitational forces, but it can&#xA;  produce measurable orbital changes over decades and substantial orbital effects over millions to billions of years.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;The same physical phenomenon also creates a thermal torque that, complemented by a torque produced by scattered sunlight, can modify the rotation rates and obliquities of small bodies as well. This&lt;/strong&gt; &lt;strong&gt;&lt;em&gt;rotational variant&lt;/em&gt;&lt;/strong&gt; &lt;strong&gt;has been coined the Yarkovsky-O’Keefe-Radzievskii-Paddack (YORP) effect&lt;/strong&gt;. During the past decade or so, the Yarkovsky and YORP effects have been used to explore and potentially resolve a number of unsolved mysteries in planetary science dealing with small bodies.&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18054" LastActivityDate="2018-08-22T21:06:08.687" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27426" PostTypeId="1" AcceptedAnswerId="27431" CreationDate="2018-08-23T07:16:11.040" Score="2" ViewCount="228" Body="&lt;p&gt;I was just wondering, what is the relationship between C-type asteroids and carbonaceous chondrite (CC) meteorites? Do CC meteorites come from C-type asteroids?&lt;/p&gt;&#xA;" OwnerUserId="24020" LastEditorUserId="24027" LastEditDate="2018-08-23T16:17:27.027" LastActivityDate="2018-08-23T16:17:27.027" Title="C-Type Asteroids and Carbonaceous Chondrite Meteorites" Tags="&lt;asteroids&gt;&lt;space&gt;&lt;meteor&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27428" PostTypeId="2" ParentId="25343" CreationDate="2018-08-23T12:32:35.850" Score="1" Body="&lt;p&gt;I assume that the Moon was near the horizon. &#xA;This is just my assumption, since there kind of needs to be more detail, but the light scattering at the horizon probably was the cause. &#xA;It's the same light scattering that makes sunset that pinkish color.&#xA;Hope that helps!&lt;/p&gt;&#xA;" OwnerUserId="24048" LastActivityDate="2018-08-23T12:32:35.850" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27429" PostTypeId="1" AcceptedAnswerId="27435" CreationDate="2018-08-23T12:36:38.927" Score="1" ViewCount="188" Body="&lt;p&gt;What I want to know is how thick the observable universe is from the point of the cosmic microwave background and beyond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears the thickness of the cosmic microwave background itself (the part we can see) is above 100,000 light years, per the following article:&#xA;&lt;a href=&quot;http://scienceblogs.com/startswithabang/2013/06/19/5-facts-you-probably-dont-know-about-the-cosmic-microwave-background/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://scienceblogs.com/startswithabang/2013/06/19/5-facts-you-probably-dont-know-about-the-cosmic-microwave-background/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I want to know the thickness of that, plus what lies beyond that we cannot see, another way of looking at it would be the distance between the surface of last scattering (cosmic microwave background end) to the beginning (e.g. Big Bang).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the following article, it appears that this time from the beginning to the surface of last scattering is about 300,000 years:&#xA;&lt;a href=&quot;https://ned.ipac.caltech.edu/level5/Glossary/Essay_lss.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ned.ipac.caltech.edu/level5/Glossary/Essay_lss.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That would imply that the thickness should be about 300,000 light years, but that doesn't take inflation into account. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the thickness (in the observable universe) between the beginning (e.g. Big Bang) to the surface of last scattering (Cosmic Microwave Background), including inflation?&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2018-08-24T10:28:14.230" Title="How thick is the cosmic microwave background, including the part we cannot see within the observable universe?" Tags="&lt;distances&gt;&lt;big-bang-theory&gt;&lt;observable-universe&gt;&lt;cosmic-microwave-background&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27430" PostTypeId="2" ParentId="12707" CreationDate="2018-08-23T12:46:43.010" Score="2" Body="&lt;p&gt;Electron capture supernovae occur in a narrow regime where electron captures on to products such as Mg and Ne are happening at a rate that removes support of the core against collapse &quot;faster&quot; (or more efficiently than) the white dwarf will be able to explode via thermonuclear runaway as the white dwarf approaches the Chandrasekhar mass limit. Several factors are at play, but these electron capture collapses are thought to happen in white dwarfs that are rich in oxygen and neon, rather than carbon and oxygen (which have different critical densities for burning). These stars were still not massive enough to form &quot;regular&quot; neutron stars via the iron core collapse channel. &lt;/p&gt;&#xA;" OwnerUserId="24097" LastActivityDate="2018-08-23T12:46:43.010" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27431" PostTypeId="2" ParentId="27426" CreationDate="2018-08-23T12:50:33.727" Score="2" Body="&lt;p&gt;The difference between asteroids, comets and meteors is discussed in &lt;a href=&quot;https://astronomy.stackexchange.com/questions/213/what-is-the-difference-between-asteroids-comets-and-meteors?rq=1&quot;&gt;What is the difference between asteroids, comets and meteors?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some &lt;a href=&quot;https://en.wikipedia.org/wiki/Asteroid_spectral_types&quot; rel=&quot;nofollow noreferrer&quot;&gt;groupings of asteroids&lt;/a&gt; have been correlated with meteorite types:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;C-type – Carbonaceous chondrite meteorites&lt;br&gt;&#xA;  S-type – Stony meteorites&lt;br&gt;&#xA;  M-type – Iron meteorites&lt;br&gt;&#xA;  V-type – HED meteorites&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The S- and C-type asteroid classifications originate from the taxonomy proposed by &lt;a href=&quot;https://en.wikipedia.org/wiki/David_J._Tholen&quot; rel=&quot;nofollow noreferrer&quot;&gt;David Tholen&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meteors are generally considered to be asteroids which enter Earth's atmosphere. &#xA;Meteorites are meteor remnants which reach the ground. So essentially, carbonaceous chondrite meteor(ite)s start out as C-type asteroids; they're made of the same stuff.&lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2018-08-23T12:50:33.727" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27432" PostTypeId="1" AcceptedAnswerId="27433" CreationDate="2018-08-23T14:16:40.443" Score="24" ViewCount="4728" Body="&lt;p&gt;The title basically says it all. As seen from the Earth, is it possible for the Moon to eclipse Venus (or any other planet) or are the orbits inclined such that this never happens?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If such an eclipse is possible, is it a frequent or infrequent event? How would I find out when the next one occurs?&lt;/p&gt;&#xA;" OwnerUserId="24096" LastActivityDate="2020-02-28T23:31:24.533" Title="Can the Moon eclipse Venus?" Tags="&lt;the-moon&gt;&lt;venus&gt;&lt;eclipse-non-earth-related&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27433" PostTypeId="2" ParentId="27432" CreationDate="2018-08-23T15:01:59.123" Score="35" Body="&lt;p&gt;As @Donald.McLean said in comments, the answer is yes, the Moon can and does occult the other planets in the Solar System. When something apparently big (like the Moon) passes in front of something apparently small (another planet) it's called an &lt;em&gt;occultation&lt;/em&gt;. (I say &lt;em&gt;apparently&lt;/em&gt; because from our perspective the Moon &lt;em&gt;appears&lt;/em&gt; larger than the planets.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The planets lie* in the ecliptic plane whereas the Moon's orbit is inclined at about 5° to the ecliptic plane, so the Moon crosses the ecliptic plane twice each orbit, so it's not as frequent as it would be if the Moon's orbit were in the ecliptic plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Occultations of the planets by the Moon can be found &lt;a href=&quot;http://www.lunar-occultations.com/iota/planets/planets.htm&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. There is also software (&lt;a href=&quot;http://www.lunar-occultations.com/iota/occult4.htm&quot; rel=&quot;noreferrer&quot;&gt;Occult&lt;/a&gt;) that you can use to generate your own predictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* Although the planets are said to lie in the ecliptic plane, this is a generalisation - they actually have orbits that are within a few degrees of the ecliptic &quot;plane&quot;.&lt;/p&gt;&#xA;" OwnerUserId="18643" LastEditorUserId="18643" LastEditDate="2018-08-24T12:43:20.877" LastActivityDate="2018-08-24T12:43:20.877" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="27434" PostTypeId="2" ParentId="25343" CreationDate="2018-08-23T15:29:25.540" Score="3" Body="&lt;p&gt;What causes the moon to be red is our own atmosphere.  It's the same effect that causes a sunset to be red.  As the sun nears the horizon, the amount of air that we're looking through increases - and with it everything in the air such as dust, pollution, humidity, etc....  If you look at the moon near the horizon, you'll notice it is significantly yellower/oranger than when it's high in the sky, where it is normally whiter and bluer.  Again, this is all caused by the air and the stuff in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the moon turns read during an eclipse, it is because it is passing through the shadow of the atmosphere, which filters the light passing through and on to the moon, leaving more red light to make its way to the moon and, then, to reflect off of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar action happens when smoke from fires gets into the air - it filters out the shorter wavelength, bluer light and lets the red pass.  Other pollutants besides smoke particles can have the same effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the only way for this to happen is for something to pass between the sun and the moon, or the moon and the earth - between the source of light and where it is received.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the moon appearing larger: this is an optical illusion.  No one is entirely sure why it happens, but the moon does APPEAR larger on the horizon than when it is high in the sky, but this is merely an illusion.  To prove it, next full moon, go outside with a ruler and hold it up at arm's length and try to measure the apparent size of the moon (this will vary from person to person due to the length of your arm).  Then, wait until it's high in the sky and do the same thing.  It will measure the same size (give or take a small amount due to the change in position affecting the distance from your eye to the ruler).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have the right kind of telescope and mount, you could connect a camera, take a picture on the horizon, wait for it to pass close to its highest point and take another picture, then compare.  Assuming no changes were made to the camera or telescope, the images should show the same apparent size.&lt;/p&gt;&#xA;" OwnerUserId="10546" LastActivityDate="2018-08-23T15:29:25.540" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27435" PostTypeId="2" ParentId="27429" CreationDate="2018-08-23T15:34:41.110" Score="7" Body="&lt;p&gt;If I understand you correctly, you want to know the distance from the point from which we observe the CMB, to the edge of the observable Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During inflation, the observable Universe expanded from ridiculously small to some ten meters in radius, so that part can be safely ignored compared to the distances now$^1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance$^2$ to the &quot;CMB shell&quot;$^3$ is 45.4 billion lightyears (&quot;Glyr&quot;), and the distance to the edge of the observable Universe is 46.3 Glyr. Hence, the shell of the observable Universe that lies beyond the CMB shell has a thickness of only 0.9 Glyr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a sketch of how I interpret your question (not to scale):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QJNB0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QJNB0.png&quot; alt=&quot;cmb&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want, I can add details on how to calculate these numbers.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^1$&lt;sub&gt;The &lt;em&gt;relative&lt;/em&gt; expansion during inflation was huge, however: The Universe expanded roughly by the same factor that it has expanded afterwards, namely a factor of $\sim10^{26}$.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^2$&lt;sub&gt;Here, &quot;distance&quot; corresponds to the &lt;em&gt;comoving&lt;/em&gt; distance, which is what you would measure if you froze the Universe right now, and started laying out meter sticks.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^3$&lt;sub&gt;This shell is not infinitely thin, but has a thickness of some 60 &lt;em&gt;million&lt;/em&gt; lightyears, so let's ignore that.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-08-24T10:28:14.230" LastActivityDate="2018-08-24T10:28:14.230" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27436" PostTypeId="2" ParentId="27432" CreationDate="2018-08-23T15:40:56.920" Score="13" Body="&lt;p&gt;Yes, such events are called occultations.&#xA;The Moon can occult anything within about 5 degrees of the ecliptic.&#xA;The visibility of any lunar occultation is limited to a path as wide as the Moon is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://earthsky.org/sky-archive/occultations-galore-on-september-18&quot; rel=&quot;noreferrer&quot;&gt;EarthSky article&lt;/a&gt; discusses a series of lunar occultations of planets in September 2017, including Venus as seen from Australia.&#xA;The article includes a photo of a February 2014 occultation of Venus&#xA;and an excerpt of Guy Ottewell's &lt;a href=&quot;http://universalworkshop.com/guysblog/wp-content/uploads/2017/02/occ17.qxd_-1.pdf&quot; rel=&quot;noreferrer&quot;&gt;lunar occultation charts for 2017&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://www.skyandtelescope.com/observing/moon-flys-by-catalina-occults-venus-on-dec-7th120220150212/&quot; rel=&quot;noreferrer&quot;&gt;Sky &amp;amp; Telescope article&lt;/a&gt; which barrycarter found describes a lunar occultation of Venus visible from North America in December 2015 and includes a photo of an April 2009 occultation of Venus.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-08-23T15:40:56.920" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27437" PostTypeId="1" AcceptedAnswerId="27438" CreationDate="2018-08-23T20:30:42.923" Score="1" ViewCount="48" Body="&lt;p&gt;Imagine an earth getting warmer. Humans decide to make their homes ready for a voluntary ice age. All the space agencies work together to get an icy planet/moon into our orbit between us and the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My theory is that because the sun is heating up the ice and blowing it in the direction of earth, it'll make the Earth colder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That final part is my question: Could ice molten in space eventually reach earth and perhaps be pushed in that direction by its heat?? Not taking into account whether or not it would benefit the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm an amateur with no knowledge of physics, chemistry and astrology and want to learn answers to my questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="24098" LastEditorUserId="24048" LastEditDate="2018-08-24T02:08:14.163" LastActivityDate="2018-08-24T02:08:14.163" Title="Would this effect occur if an ice planet was put into orbit or trajectory?" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2018-08-24T07:27:02.850" ContentLicense="CC BY-SA 4.0" />
  <row Id="27438" PostTypeId="2" ParentId="27437" CreationDate="2018-08-23T20:59:38.090" Score="0" Body="&lt;p&gt;Welcome to Stack Exchange! I'm a little confused about your question, but I'll answer it the best to my ability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, I would like to state that such scenario is near impossible because getting the planet there would be extremely difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second of all, when the ice is being &quot;blown&quot;, it would have to be perfectly aligned with the Earth not just in its position in orbit, but also in its angle of orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third of all, the ice would have to travel probably hundreds of kilometres, meaning that it would either 1) not make it by getting in orbit around the Sun / melting or 2) get lost in space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat of the Sun can barely move things at all - I assume you're sort of thinking of a Light Sail, but that's light, and ice would be too heavy anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting the ice off of there would also be challenging because its connected to the planet, and can't just be &quot;lifted&quot; by heat. That's why the Earth's surface doesn't float / scrape away by the Sun's heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, I do not think it would be possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps!&lt;/p&gt;&#xA;" OwnerUserId="24048" LastEditorUserId="24048" LastEditDate="2018-08-23T21:17:53.973" LastActivityDate="2018-08-23T21:17:53.973" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27439" PostTypeId="1" CreationDate="2018-08-23T21:32:01.253" Score="4" ViewCount="2507" Body="&lt;p&gt;I've been thinking that everything has a relative escape velocity, if it has Mass, seemingly. You can escape orbit around the earth by accelerating enough. You can escape the orbit of the sun by accelerating enough. Heck, this is kind of a stretch, but even electrons can escape an atom relative to how much electric charge the nucleus holds. What would it be like to escape the orbital velocity of our Galaxy-- and what exactly would it be relative to-- a black hole at the center of the Galaxy? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quote was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Solar system's orbital velocity is estimated at roughly 220 km/s, and galactic escape velocity for our vicinity at about 537 km/s. So in the direction of Solar system's velocity vector, velocity required to escape Milky Way is ~ 317 km/s. And much more, if this Solar system's own orbital momentum cannot be used to full extent and a launch in other directions is required. This is of course assuming you can launch on a trajectory that avoids getting too close to gravitational influence of other solar systems.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How in the world do you calculate the escape velocity of other galaxies let alone our own-- do we know what the solar systems in the Milky Way orbit and how they orbit it? &lt;/p&gt;&#xA;" OwnerUserId="23729" LastEditorUserId="2365" LastEditDate="2018-08-23T23:04:06.807" LastActivityDate="2020-09-10T21:34:45.190" Title="What is the escape velocity of our Galaxy relative to?" Tags="&lt;galaxy&gt;&lt;galaxy-center&gt;&lt;escape-velocity&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27440" PostTypeId="2" ParentId="27439" CreationDate="2018-08-23T23:02:42.670" Score="10" Body="&lt;p&gt;One way to think of the Escape velocity is to imagine it backwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of a rocket being fired into space, think of the same rocket, starting at rest relative to the Earth at a great distance (for this thought experiment pretend the sun, doesn't exist and the rest of the universe is empty), then let it fall to the ground. No matter how far away you start, the speed when the rocket hits the Earth will always be 11km/s or less. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you start with 11km/s or more at the Earth's surface, you have enough energy to coast forever into space. The escape velocity &lt;em&gt;at the surface&lt;/em&gt; is 11km/s, the escape velocity would be different, and lower, if you were already in a high orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can do the same thought experiment with a galaxy. We know, roughly the distribution of mass in the galaxy (most is in a large blob of dark matter  with the visible disc inside it) Now take a rocket ship a very great distance from the galaxy, and allow it to fall.  As it passes the sun, the rocket will be moving at 537km/s relative to the centre of the galaxy. (there is a black hole there, but it is tiny compared to the galaxy, so has very little effect on the gravity of the galaxy)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know how the mass of a galaxy is distributed we can get a good idea of how fast something would be travelling at any point, if allowed to fall. And by the same argument as above, that gives the escape velocity &lt;em&gt;at that point&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The escape velocity around the sun is about 1/600 of the speed of light. The journey to leave the Milky way is at least 30000 light years long, This means it is going to take about 20 million years to reach the edge of the galaxy, since you will be getting slower as the galaxy's gravity pulls you back. Space (as has been observed) is big.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-08-23T23:02:42.670" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27441" PostTypeId="1" CreationDate="2018-08-24T01:42:47.417" Score="8" ViewCount="1486" Body="&lt;p&gt;A comment in response to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/27432/can-the-moon-eclipse-venus&quot;&gt;this question&lt;/a&gt; suggests that an eclipse differs from an occultation in that the former casts a shadow while the latter doesn't. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't particularly satisfactory since blocking the light from one object with a second will always cast a shadow even if it's not visible to the human eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, an annular eclipse of the sun by the moon could be referred to as a transit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a more rigorous definition of these three terms that distinguishes an eclipse from an occultation or a transit? &lt;/p&gt;&#xA;" OwnerUserId="24096" LastEditorUserId="2896" LastEditDate="2018-08-24T07:26:15.407" LastActivityDate="2018-08-24T12:45:30.137" Title="How does an eclipse differ from an occultation?" Tags="&lt;solar-eclipse&gt;&lt;terminology&gt;&lt;planetary-transits&gt;&lt;occultation&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27442" PostTypeId="2" ParentId="27441" CreationDate="2018-08-24T07:43:59.357" Score="4" Body="&lt;p&gt;There are a number of terms for these things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;conjunction&lt;/em&gt;&lt;/strong&gt;:&lt;br&gt;&#xA;When two objects have either the same right ascension or the same ecliptic longitude (usually as observed from Earth). That is, they lie in (almost) a straight line from the point of view of the observer. In the case of the Solar System, such as when the Sun and Venus line up from the point of view of an observer on Earth, this is a conjunction. (It is also a &lt;em&gt;conjunction&lt;/em&gt; when Earth lines up between e.g. Mars and the Sun - it is an &lt;em&gt;inferior conjunction&lt;/em&gt; from the point of view of Mars (but we can't observe from Mars)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, when Venus lines up between the Earth and the Sun this is an &lt;em&gt;inferior conjunction&lt;/em&gt;, and when Venus lines up with the Sun but on the opposite side of the Sun then this is a &lt;em&gt;superior conjunction&lt;/em&gt;. Thus the inner planets (Mercury, Venus) can have both &lt;em&gt;inferior conjunctions&lt;/em&gt; and &lt;em&gt;superior conjunctions&lt;/em&gt; with Earth while the outer planets (Mars, Jupiter, etc.) can only have &lt;em&gt;superior conjunctions&lt;/em&gt; with Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;&lt;em&gt;almost&lt;/em&gt;&quot; in &quot;&lt;em&gt;almost a straight line&lt;/em&gt;&quot; is important because the definition is by longitude. Although the planets are said to lie in the ecliptic plane, this is not completely flat - they do not all have the exact same orbit inclinations. Thus not every &lt;em&gt;conjunction&lt;/em&gt; is a &lt;em&gt;transit&lt;/em&gt;, &lt;em&gt;occultation&lt;/em&gt; or &lt;em&gt;eclipse&lt;/em&gt; - they do not pass &lt;em&gt;directly&lt;/em&gt; in front or behind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;transit&lt;/em&gt;&lt;/strong&gt;:&lt;br&gt;&#xA;When an object passes in front of a larger object*, this is a &lt;em&gt;transit&lt;/em&gt;. These are rarer than &lt;em&gt;conjunctions&lt;/em&gt; because the objects need to be directly in line so that one can be seen crossing the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of recent transit of Venus, when Venus passed in front of the Sun, the sunlight passing through Venus' atmosphere gave us the opportunity to collect data on its composition. When one of the moons of Jupiter pass in front of Jupiter itself* this is also a transit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;occultation&lt;/em&gt;&lt;/strong&gt;:&lt;br&gt;&#xA;When an object passes in front of a smaller object*, this is an &lt;em&gt;occultation&lt;/em&gt;. The farther object is hidden by the nearer object. These are also rare** because, like &lt;em&gt;transits&lt;/em&gt;, they also need to be directly in line so that the nearer object can be seen crossing in front of, and blocking out, the farther object. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asteroids &lt;em&gt;occulting&lt;/em&gt; stars are studied because while we can calculate the asteroid's orbit and its distance from us using its speed, knowing how gravity and orbits work, etc., we can't see clearly enough to determine the size (think few kilometres in diameter but millions of kilometres away). But when an asteroid passes in front of (&lt;em&gt;occults&lt;/em&gt;) a distant star, the star is effectively a point source (looks tiny), from the amount of time the &lt;em&gt;occultation&lt;/em&gt; lasts and the asteroid's speed we can calculate its size***.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;eclipse&lt;/em&gt;&lt;/strong&gt;:&lt;br&gt;&#xA;The Moon passing in front of the Sun is, depending on the conditions, a &lt;em&gt;partial eclipse&lt;/em&gt;, &lt;em&gt;total eclipse&lt;/em&gt; or &lt;em&gt;annular eclipse&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, by the definition, when the Moon fully blocks the Sun it is an &lt;em&gt;occultation&lt;/em&gt; and when it only partially blocks the Sun it is a &lt;em&gt;transit&lt;/em&gt;, but &#xA;this is a special case and is called a &lt;em&gt;solar eclipse&lt;/em&gt; rather than a &lt;em&gt;transit&lt;/em&gt; or &lt;em&gt;occultation&lt;/em&gt;. (Similarly when the Moon passes through the Earth's shadow - when the Earth is directly between the Sun and the Moon - it is called a &lt;em&gt;lunar eclipse&lt;/em&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The word &lt;em&gt;eclipse&lt;/em&gt; comes to us via Latin and French from ancient Greek &lt;em&gt;ekleipsis&lt;/em&gt;, from &lt;em&gt;ekleipein&lt;/em&gt; 'fail to appear, be eclipsed', from &lt;em&gt;ek&lt;/em&gt; 'out' + &lt;em&gt;leipein&lt;/em&gt; 'to leave'. (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Eclipse#Etymology&quot; rel=&quot;nofollow noreferrer&quot;&gt;eclipse on Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This alignment is also known as a &lt;em&gt;syzygy&lt;/em&gt;****.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* Smaller or larger as they appear to the observer, an asteroid is most definitely smaller than a star but because it is so much nearer it looks bigger.&lt;br&gt;&#xA;** A lookup of &lt;em&gt;occultations&lt;/em&gt; can return a long list so that it may not seem rare, but when you compare how many asteroids there are, and how many stars, in relative terms it actually does not happen very often.&lt;br&gt;&#xA;*** We can set limits on its size, given that the &quot;average&quot; asteroid looks like a lumpy potato rather than being some nice, regular shape that would make it easy to work these things out.&lt;br&gt;&#xA;**** We could keep adding terms and googling them all day... ;-)&lt;br&gt;&#xA;***** There is no footnote with five *&lt;/p&gt;&#xA;" OwnerUserId="18643" LastEditorUserId="18643" LastEditDate="2018-08-24T12:45:30.137" LastActivityDate="2018-08-24T12:45:30.137" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27443" PostTypeId="2" ParentId="23468" CreationDate="2018-08-24T08:54:50.000" Score="2" Body="&lt;p&gt;In general terms we have had the observational capability to discover interstellar objects like this for quite some time - before digital imaging, photographic plates were used, which allowed accurate and verifiable measurements to be made. The issue is not so much in the observational and recording capability, but in collecting sufficient data to be able to compute an orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a few hours of observation you can compute a path that allows further study on subsequent days. This usually assumes a circular orbit as a first approximation, but with increasing elapsed time after the observations the computed path increasingly diverges from the actual orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With observations over a span of days you can compute a better orbit, but again with increasing elapsed time after the observations the computed path increasingly diverges from the actual orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With observations over a span of weeks you can compute an orbit that allows you to find the object again in subsequent years... and so on. Increasing elapsed time correlates to increasing divergence from computed orbits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, with a moderate span of observations, we would find that an interstellar object is diverging from its predicted path (orbit around the Sun) and that it is passing through the Solar System instead of being part of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But with the greater number of automated surveys (especially compared to pre-digital imaging) that routinely flag asteroid and comet detections, the likelihood of interstellar objects being detected is greater now than in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;But&lt;/em&gt;&lt;/strong&gt; someone has to be looking for it. The &lt;em&gt;Minor Planet Center&lt;/em&gt; likely has a large number of objects in its database which are actually interstellar objects. But it's set up to correlate detections based on Solar System orbits and the interstellar objects will exist as unconnected short-path observations that don't fit to an orbit. Maybe someone could find more by mining the data, but unless they are still near enough to follow up with further observations these will be unverifiable. But it would be an interesting project!&lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2018-08-24T08:54:50.000" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27444" PostTypeId="2" ParentId="27441" CreationDate="2018-08-24T10:13:02.203" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a more rigorous definition of these three terms that distinguishes an eclipse from an occultation or a transit?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not quite rigorous, but&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Occultations refer to events where the intervening object's angular size is significant larger than is that of the more remote object,&lt;/li&gt;&#xA;&lt;li&gt;Transits refer to events where the intervening object's angular size is significantly smaller than is that of the more remote object, and&lt;/li&gt;&#xA;&lt;li&gt;Eclipses cover the fuzzy region where the angular sizes of the remote and intervening object are roughly equal.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The boundary is fuzzy. The Earth's angular diameter is 33 times larger than is that of the Sun from geosynchronous altitude, or 1.5 orders of magnitude larger. Yet the periods of time where geosynchronous satellites can be in the Earth's shadow are called &quot;eclipse season&quot; rather than &quot;occultation season&quot;. On the flip side, Venus's angular diameter is 33 times smaller than is that of the Sun when Venus crosses in front of the Sun, and those events are most definitely called transits.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2018-08-24T10:13:02.203" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27445" PostTypeId="1" CreationDate="2018-08-24T11:46:42.300" Score="4" ViewCount="135" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In general, are carbonaceous chondrites (CC) or comets more water rich?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that evidence has suggested that both CC and comets are partly composed of water, but which body is generally more water rich?&lt;/p&gt;&#xA;" OwnerUserId="24020" LastActivityDate="2018-10-27T12:02:37.413" Title="Water Content of Space Bodies" Tags="&lt;comets&gt;&lt;meteor&gt;&lt;water&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27446" PostTypeId="2" ParentId="27445" CreationDate="2018-08-24T11:52:52.953" Score="1" Body="&lt;p&gt;I compared C type asteroids and comets for this question. C type asteroids have around 30% of empty space, meaning that that much space can be for water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comets have around 40% of empty space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, just because they have more empty space does not mean they have more water. However, evidence for water on Earth originally comes from comets. Not only that, but comets are occasionally balls of ice with some rock. Asteroids are known for being just rocks, and only have a limited amount of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, comets are more water rich.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps...?&lt;/p&gt;&#xA;" OwnerUserId="24048" LastActivityDate="2018-08-24T11:52:52.953" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27447" PostTypeId="1" AcceptedAnswerId="27480" CreationDate="2018-08-24T16:35:55.987" Score="3" ViewCount="551" Body="&lt;p&gt;I've just started to learn &lt;a href=&quot;http://www.astropy.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astropy&lt;/a&gt;. It's a nifty tool. Using some tutorials, I wrote and then made modifications to a little script that finds the Alt/Az of an object at a certain place and time. I then checked &lt;a href=&quot;https://stellarium.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellarium's&lt;/a&gt; calculation for the Alt/Az of the same object (at exactly the same place and time) and found it to be slightly different. I suspected this was due to slight differences in the underlying coordinates, so I tried manually hard coding the coordinate data for the star from Stellarium; it still came out different. Here is the data without the hard coding:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Object: Rasalhague (alf Oph)&#xA;Time: 2018-08-28 23:00 (converted to universal in the script)&#xA;Latitude: 35&#xA;Longitude: -79&#xA;Stellarium reported Az/Alt: +247 35 12.3 / +48 00 57.8&#xA;Astropy reported Az/Alt: +247 35 37.4939s / +48 00 08.7586&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Manually coding:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;RA/Dec on date from Stellarium: 17 35 48.82, +12 32 55.9&#xA;Astropy reported Az/Alt: +247 22 42.0379 / +48 09 43.0755&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm guessing that the difference stems from the conversion algorithm to alt/az.&lt;/p&gt;&#xA;" OwnerUserId="23940" LastEditorUserId="23940" LastEditDate="2018-08-29T03:29:40.360" LastActivityDate="2018-08-29T03:29:40.360" Title="Differences between alt/az in Stellarium and Astropy" Tags="&lt;coordinate&gt;&lt;astropy&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27448" PostTypeId="1" CreationDate="2018-08-24T18:05:33.183" Score="5" ViewCount="202" Body="&lt;p&gt;I'm a fairly experienced photographer and have enjoyed taking lots of photos of the night sky with my existing gear - wide angle starscapes over pretty hills, mostly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to move on, having seen what's possible with telescopes and mounted cameras. I've read around, and am now confused....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plan sees me getting a telescope, tripod and mount of some sort (equatorial I believe I need?), and periodically putting that carefully into my car, driving into a rural area with dark skies and clear horizons, and pointing it at interesting things - my urban gardenless flat really isn't a good location for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have Nikon full and crop frame and Olympus m4/3 cameras available - I believe I'd need a T mount adaptor to fit them? Is one likely to be hugely better than the other for this in a way I've not foreseen? I know the Nikon has much better low light performance but the Olympus puts more pixels on the subject with its small chip and has the very useful live time mode. I'm used to raw processing, but I'm expecting there'll be more specialist software I'll need to learn.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Am I anywhere near right or is this all mad?&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What sort of things do I need to watch for on telescopes?&lt;/li&gt;&#xA;&lt;li&gt;Am I going to trash a telescope by moving it around like this? I'm used to moving big lenses which I'm hoping is tolerably similar (accepting the very different optical designs)?&lt;/li&gt;&#xA;&lt;li&gt;What have I completely missed?&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="24104" LastActivityDate="2018-08-24T22:11:09.753" Title="Telescopes for beginner astrophotography" Tags="&lt;telescope&gt;&lt;photography&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27449" PostTypeId="2" ParentId="27447" CreationDate="2018-08-24T18:11:00.910" Score="0" Body="&lt;p&gt;The alt/az coordinates will change as the object moves, and are relative to the observing position.  Both programs will need to be configured with the exact same time, latitude, longitude, and altitude.  If you want to compare the two, you have to get get readings at precisely the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even then, it's possible for variation if they are not using the same epoch.  If one is using J2000 and another J1950 or J2050, then there will be differences. It might be possible for one or the other to calculate precessional differences to provide something closer to exact coordinates.&lt;/p&gt;&#xA;" OwnerUserId="10546" LastActivityDate="2018-08-24T18:11:00.910" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27450" PostTypeId="2" ParentId="27448" CreationDate="2018-08-24T19:12:27.090" Score="2" Body="&lt;p&gt;Everything you've said is pretty much on the money. You're definitely going to be wanting an equatorial mount (mount and tripod are pretty much interchangeable here), because equatorial mounts don't have the issues with field rotation and zenith deadzones that alt-az mounts use. You're also going to want a mount with the ability to add automatic tracking, otherwise your objects will streak across the field during your longer exposures. Don't worry about damaging the telescope just by moving it around and transporting it; just treat it gently and make sure it doesn't roll around and hit things that can damage it. Store it appropriately, too, in a dry place with dust covers on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your cameras, it can end up being completely preferential whichever one is better. The Nikon's low light performance could make it so you can take fewer or shorter exposures without worrying about noise; however, you might decide that you prefer the live time mode. Whatever the case, consider investing in a remote switch for either so that you can trigger the shutter without physically touching the camera in order to avoid vibrating or moving the camera/telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Software wise, you're going to be wanting to find something that can do image stacking, bias subtraction, flat fielding, and other special processings. &lt;a href=&quot;http://deepskystacker.free.fr/english/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Deep Sky Stacker&lt;/a&gt; and &lt;a href=&quot;http://www.pixinsight.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;PixInsight&lt;/a&gt; are two popular options, and there are even more alternatives around. You'll also need to take those special exposures: biases, darks, and flat fields in order to reduce noise and other distortions even further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you're just gonna need to practice. There's a lot that goes into just figuring out how your personal workflow is best, and you can only get better at it with time.&lt;/p&gt;&#xA;" OwnerUserId="24091" LastActivityDate="2018-08-24T19:12:27.090" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27451" PostTypeId="2" ParentId="27447" CreationDate="2018-08-24T20:08:03.593" Score="0" Body="&lt;p&gt;The apparent alt/az position of a star in the night sky depends not only on its underlying equatorial coordinates and position/time of your interest, but also from other conditions which may or may not be accounted for by the one or the other software, depending on how you have configured the software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One simple example for those conditions would be the atmospheric refraction. Its effect on the apparent alt/az position of a start is: the lower the object is above the horizon, the higher altitude it seems to have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we don't know your script exactly and since you're saying that you've just started learning AstroPy, it could well be that you haven't configured both software with the same conditions for the alt/az calculation algorithm.&lt;/p&gt;&#xA;" OwnerUserId="24027" LastEditorUserId="24027" LastEditDate="2018-08-25T06:06:44.643" LastActivityDate="2018-08-25T06:06:44.643" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27452" PostTypeId="2" ParentId="27448" CreationDate="2018-08-24T22:11:09.753" Score="8" Body="&lt;p&gt;Ah, welcome to the world of AP... check your sanity at the door, and that loud sucking sound you hear is your bank account being emptied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first and best piece of advice for you is to NOT buy anything at all right off.  Find your local astronomy club(s) and join.  Most clubs have at least a few people doing imaging, and you would be well-advised to meet these people and learn from them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AP is not like conventional photography at all, for the most part. It's closer to digital signal processing.  It is a blend of science, engineering, and art.  What you know from conventional photography is not useless information, but you have a whole slew of new things to learn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we talk about telescopes and equipment for AP, we don't talk about &quot;tripods&quot; all that often.  A tripod is one kind of support structure for a mount head, which does the actual work.  The mount's role is to carry the telescope, camera, and associated equipment and point it at the object you wish to capture.  While the concept is quite simple, the execution is extremely complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two fundamental problems that we must overcome to capture good images of astronomical bodies.  The first is the fact that the majority of these objects are very faint.  While people often think the purpose of a telescope is to magnify, this is a misnomer: an astronomical telescope's primary purpose is to collect more light and concentrate it into a smaller area (the eye or the image sensor).  Magnification, while not irrelevant, is a secondary concern (and comes with its own issues).  Because these objects are so dim, we must resort to long-exposure photography.  While typical daylight photography uses exposure times measured in tenths and hundredths (even thousandths) of a second, exposure times for astrophotography often are measured in whole minutes.  Actually, total exposure is often measured in hours, with individual sub-exposures done in minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is where the second key problem emerges.  While our shutter is open to capture an exposure of 1 or more minutes, we have to contend with the fact that the objects we are capturing are moving relative to our instruments.  Due to the rotation of the earth, a given star, planet, galaxy, nebula, etc... in the sky above moves across the sky from east to west in an arc with a radius depending on the object's position relative to the celestial north or south pole.  In a full day, an object moves 365 degrees around the pole (well, we rotate, but for practical purposes, the object moves).  In a single hour, it moves 15 degrees.  In one minute, 15 arcminutes.  And in one second, 15 arcseconds.  This may seem like a small amount of motion, and when you look up at the stars in the sky, they do seem to be fixed in place.  But, when you start to reduce your field of view by using a telescope or camera lens, the motion starts to become more and more obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a conventional photographer, you are probably familiar with the standard alt-az style mount motion.  Alt-az is short for altitude-azimuth, and a typical camera tripod works this way.  The mount angles up/down (altitude) and left/right (azimuth) to point at an object.  In order to follow an object across the night sky, then, you have to move the mount in both axes.  The rates of motion between the axes will vary from one another as well as depending on the location of the object in the sky above.  To automate this process requires a great deal of computerization, and even then it is difficult to do so accurately enough to handle the demands of long exposure imaging.  And even if you do overcome those issues, you will find that the object you are following appears to rotate in the field of view as it crosses the night sky, a phenomenon we call frame-rotation.  This is because the frame of reference of an alt-az mount is the horizon, not the sky itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution to this issue is the Equatorial mount.  An equatorial mount has four axes.  The first two, altitude and azimuth, are only used for alignment purposes, not for motion.  They are used to align the third axis, known as Right Ascension (RA), with the celestial pole.  Once that is accomplished, motion in the RA axis shifts the telescope's aiming point east and west.  The fourth axis, Declination (or Dec), handles north and south.  When properly aligned, the mount then needs only to move in the RA axis and at a single rate of speed to follow an object across the sky.  This doesn't even require a computer, and thousands of mounts have been manufactured with clock-drive mechanisms to handle this motion.  Computerization only enhances the process and makes it more accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But not all equatorial mounts are created equally.  Many just do not have the accuracy needed to handle long exposure imaging, and the level of accuracy needed increases as you increase focal length, focal ratio, and exposure time.  A mount that could handle a 300mm camera lens or 80 mm short-tube refractor and produce 60 to 120 second exposures might not be able to handle an 8&quot; f/4 Newtonian telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weight is a common issue.  Mounts like this are all rated for a specific payload weight amount.  Payload here includes not only the telescope, but also any cameras, adapters, finders, guide scopes/cameras, and mounting hardware - everything attached to the top of the mount.  But these payload ratings are typically given for total payload maximums. When doing AP, the general wisdom is that you should keep your imaging payload to 1/2 or less the mount's rated maximum.  So if your mount can handle, say, 40 lbs. of weight, for imaging you shouldn't load it with more than about 20 lbs.  The rule isn't a hard limit, and not all manufacturers rate their mounts the same way, but it is definitely a strong suggestion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said, the tolerances become tighter as you increase focal length and ratio.  First, let me discuss ratio. The focal ratio of a telescope is calculated by dividing the focal length by the aperture.  Thus, if you have a telescope with a 200 mm (roughly 8&quot;) aperture and a focal length of 1,000 mm, you have a focal ratio of f/5.  If you have a focal length of 2,000 mm with the same aperture, it's f/10.  In general, the lower the focal ratio, the better for imaging.  The reason is the concentration of light.  Imagine you have the two telescopes I just mentioned, both 200 mm apertures, one f/10 the other f/5.  Now imagine you have identical cameras in each and are tandem-mounted on a GEM that can handle the payload without issue.  Now, let's say you are imaging an object that appears roughly circular and evenly illuminated.  Now, let's say in the shorter scope, f/5, the object you are imaging covers a spot on your camera's image sensor about 100 pixels across.  The total area of pixels covered is then around 7,850 pixels.  Let's say each pixel receives about 20 photons of light per second and you run a 1 minute exposure.  Thus each pixel gets a total of 1,200 photons of light in the exposure, for a total of about 9.4 million photons collected.  Now, if we look at the f/10 telescope, we find the image is larger.  A longer focal length increases magnification.  Double the focal length, you double the magnification.  So now the object we are imaging has a diameter of about 200 pixels on the image sensor.  This means that it has an area of about 31,400 pixels - double the diameter means 4 times the area.  Now, because the aperture is the same, we do NOT get more light, we get the same number of photons, they are just spread out over the larger area.  Instead of 20 photons per second, each pixel gets only 5.  In one minute, each of them gets only 300 photons.  The total number is the same, about 9.4 million, but they are less concentrated.  This means the image is only 1/4 as bright as that in the f/5 scope.  To get the same level of exposure, you would need to increase your exposure time by a factor of 4 to 240 seconds.  Yes, you get a larger, more detailed image, but it takes you four times as long, and this can be a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No mount is perfect and all will show drift over time, even with autoguiding technology.  Let's say that the mount you are using has an error that is equal to about 1 pixel per minute in the f/5 telescope.  In that 1 minute exposure, everything is stretched by 1 pixel.  This is hardly noticeable and not likely to be a problem.  Let's say you had a star near the center of the image that has a diameter of 5 pixels.  In the exposure, you would see it as 5 pixels wide by 6 long.  Hardly noticeable.  But now let's look at the f/10 telescope.  Because we double magnification, we now have 2 pixels of stretching.  Now the star is stretched 2 pixels.  Still, this isn't a huge problem, but we also need to increase exposure, so we quadruple the stretching to 8 pixels.  Now we have significant stretching in our image and blurring of our object (and that's assuming the mount is THAT accurate, which is not a given).  This is why astrophotographers refer to shorter focal ratios as &quot;faster&quot; telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aperture is typically the critical measure for a telescope.  But for AP, this isn't always the case.  A lot of very high quality imaging is done using short-tube refractors with fairly small apertures.  This also gives you a much wider field of view.  And this, in turn, reduces the exacting tolerances necessary for the mount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this isn't ideal for all targets.  And this is another key problem with AP: there is no one-size-fits-all solution.  What is ideal for planetary imaging is not ideal for imaging large nebulae, which is not ideal for imaging most galaxies... you need to consider multiple factors in equipment selection. Personally, I use an 8&quot; f/10 SCT (with a Barlow lens giving me f/20) for planetary imaging, an 8&quot; f/3.9 Newtonian for globular clusters, medium-sized nebulae, and most galaxies, and an 80mm f/3.75 refractor for wide-field nebulae and the Andromeda Galaxy (which is 3° across, so very large). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the mount is the key to all of it.  The conventional wisdom in AP is to focus on the mount first.  You can have a million dollar top-of-the-line camera and a million dollar top-of-the-line telescope, but if you don't have them on a mount that can provide accurate motion, you won't get good images.  Alternatively, you could put a mediocre camera and telescope on a good mount and get pretty decent images - the mount is THAT critical to the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prices on these kinds of mounts varies, and you can easily spend into the tens of thousands.  The cheapest mounts in this range are designed to handle DSLR Cameras and lenses, usually not more than 300 mm in length (perhaps a little shorter).   The price on a mount like this is around \$350 to \$600 USD, depending on quality and capability.  And, again, these are limited in what they can handle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stepping up from that are basic German Equatorial mounts with powered tracking ability.  While computerization (What we call GoTo) is not required, it is usually found on these kinds of mounts and allows for the addition of autoguiding technology.  Autoguiding is the use of a second camera and, usually, telescope that is harnessed to software to lock onto a guide star and make minor adjustments to the mount's motion to keep it on-target. Autoguiding will not make a bad mount into a good one, it will simply allow an already good mount to capture longer exposures.  The lowest-price mounts in this range usually have a payload weight maximum of 30 lbs. or less and a price tag between \$800 and \$1,200.  However, I wouldn't actually recommend ANY of them.  The two most popular here are the Celestron AVX, running about \$900, and the SkyWatcher HEQ5, at about \$1,100.  The SkyWatcher might actually be usable, but the 30lb limit means only 15lbs for imaging, and that's very low indeed.  You can still probably fit a short-tube refractor on here with a DSLR and lightweight guiding package.  The Celestron AVX promised to provide entry-level AP capability, but the reviews of the mount show that it hasn't lived up to its promises.  This mostly seems to be due to poor quality control in the production line, resulting in some that work just fine, and a lot that do not provide the accuracy needed for imaging.  They are still usually fine for visual observing, just not for imaging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step up finds mounts capable of handling between 40 and 45 lbs. of payload with prices between \$1,500 and \$2,000 USD or so.  This is where I STRONGLY recommend people start.  While they are not cheap (though you can often find them used for about 30% or so less), they often have the quality and accuracy necessary.  There are four models in this range I recommend: The Celestron CGEM II, Orion Atlas EQ-G, SkyWatcher EQ6-R, and the iOptron iEQ45 Pro.  My personal preference is the iOptron (I have the early version of this mount and love it), but all should be competent.  My mount is capable of 45lbs and I run it at around 26 (yes, over the recommendation, but hot hugely so, and it handles it fairly well) with an 8&quot; f/3.9 Imaging Newtonian and CCD Imaging camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You CAN go with a cheaper mount.  But you rapidly lose capability and accuracy.  The lower mounts aren't really intended for imaging use, the heavier ones are generally engineered to stricter tolerances needed for imaging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One nice thing about these four mounts, and most others in this range, is that they all use a common interchangeable modular mounting system for attaching the payload.  I personally have three sets of payloads and can switch them out in a matter of minutes. It also means that if you buy the mount now with a low-end telescope option, you can later upgrade the telescope itself without having to get a totally different mount.  It also means you can have different payloads for imaging and for visual observing.  I've never heard of anyone having too much mount for imaging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as for cameras....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DSLRs are not the best tool for the job, but they are widely used all the same.  There's a few considerations here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, DSLRs tend to have a great deal of signal noise - noise which must be dealt with.  The preference for AP is generally thermoelectrically cooled CCD imaging cameras.  These use Pelletier cooling systems, heat sinks, and fans to cool the image sensor and dramatically reduce signal noise.  Remember, we're dealing with faint objects, and the noise can often drown out the image signal in the exposure.  These kinds of systems also typically better isolate noisy components and reduce their effect on the image. Also, the nature of CCD technology makes them inherently less noisy than CMOS sensor technology prevalent in DSLR cameras. However, CMOS cameras are improving and many manufacturers are now offering cooled CMOS cameras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With or without a cooled system, you still will need to capture calibration frames. There are three common types: Dark frames, Bias frames, and Flat frames. Dark frames are images taken with the shutter closed.  They are done with the same exposure time/gain/ISO settings and, preferably, the same temperature range.  These are then digitally subtracted from the &quot;light&quot; frame image you capture.  Some cameras actually have built-in technology for this (my Nikon D5000 does).  But for AP, it is usually done through capturing separate dark frames and processing them in bulk after the fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Bias frame is also done with a closed shutter, but for the shortest exposure time possible. This maps the noise created when the image data is read off the image sensor itself.  It is also digitally subtracted in processing.  I don't believe any DSLRs have technology to do this automatically, though I could be wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flat frames are a different beast altogether.  They are taken with the camera still attached to the telescope or camera lens.  To produce flat frames, you want even illumination entering the optical train (e.g. the front of the lens/scope) and an exposure length that produces an average of 50% illumination.  These images are used to map the optical system's light and dark spots and even out the illumination (compensating for problems like vignetting).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said, noise is a major problem in DSLR cameras. In well-lit images such as daylight, well lit interior, or flash photography, the noise is nearly irrelevant.  But in low-light long-exposure imaging, it can be dramatic.  This is especially true if you use higher ISO levels to increase sensitivity.  CCD imaging systems are not immune, but they are designed specifically to handle it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CCD and CMOS systems designed for AP are also typically more sensitive.  The term Quantum Efficiency (Qe) describes the sensitivity level of an image sensor to various wavelengths of light.  Essentially, it indicates what percentage of photons that hit the image sensor are detected.  A typical DSLR has a peak Qe around 30-40%, with some newer ones moving up closer to 50% and a handful of newer ones in the 70% range.  CCD and CMOS systems for AP typically have peak Qe starting around 50% and some have peak Qe around 95%.  This means much more efficient imaging and shorter exposure times.  The actual Qe varies across the spectrum, with the peak being the highest sensitivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These issues, so far, are fairly simple to deal with, though often require longer exposure and/or more frames in the stack (I'll come back to stacking).  There's one problem with DSLR cameras, however, that isn't as easy to deal with.  With the exception of the Canon 20Da, Canon 60Da, and Nikon D810A, all DSLRs, and, in fact, nearly all digital cameras in general including cell phone cams, webcams, etc..., have a built-in filter for color-balance improvement.  In most cases, it's mounted directly over the imaging sensor &quot;chip&quot; and also provides some degree of protection from dust and damage.  This filter is designed to reduce the amount of deep red and near IR energy that reaches the sensor.  Human eyesight is not very sensitive to these longer wavelengths, and the filter helps produce a more &quot;life-like&quot; color balance to match human vision. Without the filter, images tend to be shifted toward the red.  Software color balance profiles can reduce this significantly, but for conventional imaging, the filter is the best option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for astronomy, this is a problem.  There is a lot of astronomically interesting light in the deeper red part of the spectrum.  When an electron in a Hydrogen atom falls from its second-highest to third-highest energy state, it emits a photon at a wavelength of 656.28 nanometers.  We call this Hydrogen Alpha, and it's a major component of emission nebulae and star-forming regions in galaxies.  When you see red in a photo of a nebula, it's often H-Alpha.  Some nebulae, such as the Rosette Nebula, are extremely heavy in H-alpha.  Unfortunately, the filter in a DSLR blocks the majority of this light from reaching the sensor - something like 90% if I remember correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution for this problem is to not have the filter in place.  To date, only three factory-issued DSLR models have been released with this filter removed (actually replaced with one that is transparent to these wavelengths).   For other DSLR cameras, you would need to have the camera modified. There are several services you can find on the internet that will do this for you, but there's a cost.  It is not something I recommend people do themselves unless they have experience with camera repair and/or can afford to run the (very large) risk of ruining the camera.  Doing so also makes the camera ill-suited for continued conventional imaging, unless you get a clip-in or screw-on replacement filter (more cost).  You can do AP without a modified camera, but you will be significantly handicapped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, for best results, a monochrome CCD and filter wheel is the best option of all, but this significantly increases the budget.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mentioned one of your cameras has more/smaller pixels.  In most cases, this is not actually desirable.  Larger pixels tend to be more sensitive (more collection area to gather photons), and more pixels may provide higher resolution, but also means larger file sizes and MUCH slower processing.   The Hubble Space Telescope's primary cameras have a resolution around 2048 x 2048 pixels (about 4 megapixels). The issue is not resolution as much as sensitivity.  I've worked with some very high end astronomical imaging CCD's with resolutions of only around 1 megapixel: but they produce FAR better images than any DSLR on the market.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to brands, Canon and Nikon are far and away the leaders, with Sony gaining some ground.  But overall, Canon is really the brand of preference among astrophotographers using DSLRs.  Canon has been very welcoming and accommodating to the AP community. They produced the first two pre-modified cameras well before Nikon produced their first.  Canon has made a great deal of information available, including detailed specifications and even firmware source code. Nikon is slowly coming around, but has been a LOT slower to do so and still has not offered anywhere near as much support as Canon.  Because of this, you will find a lot more resources and, especially, software for camera control of Canon cameras than Nikon or anyone else.  For people just getting started in AP and on a budget, I generally recommend looking for a used pre-modified Canon DSLR as an imaging camera.  Sites like CloudyNights and Astromart frequently have these in their classified ad sections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now... software...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A major difference between conventional photography and AP is the processing.  I already mentioned calibration images.  But the big deal here is image stacking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most Digital AP, whether DSLR or CCD based, uses stacking processes to produce better images.  Stacking is, essentially, a signal processing method using statistical analysis.  You capture a set of images of the same subject and exposure time.  You then run each through a calibration routine which removes the dark and bias signals and then applies the flat frame to even out the illumination.  The images are then aligned (nearly always through lining up stars in each frame).  At the end of the alignment phase, any pixel all the way through the stack represents the same point in space.  Then, each pixel in the stack is statistically analyzed.  For example, if you have 100 images in a a stack, and a particular pixel in the stack has brightness values between, say, 100 and 105, you average them out to come to what the most likely &quot;true&quot; value is, say 103.   In many cases, a weighted average is used in which pixels outside of 1 or 2 standard deviations above or below the average are discarded entirely, and only values in a consistent range are finally averaged together to produce the most likely &quot;true&quot; value.  This has the effect of reducing noise and, using drizzle techniques, can actually increase resolution.  Even without drizzle, you end up with a much better overall image, with greater clarity and less noise.  The larger the sample size, the better.   My most recent image of M51, the Whirlpool Galaxy, uses 3.5 hours of 120 second (2 minute) exposures (so over 100 individual exposures). The more exposures I use, the better the final result and the more faint detail I can gather.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this processing takes time.  The larger your image files, the longer it takes.  My main imaging camera produces image files in FITS format (the most common format for AP) that are about 50mb in size each.  100 of those comes to about 5 gigabytes, not including the flat, dark, and bias files (usually 10-20 of each).  This ends up taking a lot of storage space and a lot of processing time.  There are other file format options, but it is critical that you use a lossless format (like FITS or TIFF), and the only way to reduce those files is through compression, which works against you in processing (you have to decompress each frame in processing, so you save space, but dramatically increase processing time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is good processing software available, some for free.  I do most of my calibrating and stacking currently using DeepSky Stacker, which is a free product.  I do most of my image capture in Nebulosity, which is not free, but works well for what I need and I got before the last price increase.  While live view on a camera is helpful, it's not necessary with a program like Nebulosity which will loop exposures and provide metrics to help achieve accurate focus.  You can get MUCH better focus this way than through the live-view on a camera.  It will then also allow you to automate the capture, setting the number of exposures, length of exposures, what to name the files, where to store them on your hard drive (typically these store directly to hard drive, not on the camera), and in what format (usually FITS or TIFF, not the camera's proprietary RAW/NEF/etc... or JPG - and you really do NOT want to use JPG, it's FAR too lossy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After you've captured, calibrated, and stacked, you will then want to do post-processing.  Some people use GIMP, which is an open-source option.  I mostly use Photoshop, which has a lot of capability and there's a wealth of information and tutorial advice.  I am also trying to learn to use PixInsight, which is probably the best astrophotography processing software on the market, but is neither cheap nor easy to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, you probably are starting to see how very difficult Astrophotography can be.  Doing &quot;pretty decent&quot; AP is neither cheap nor easy.  I'm about 10 years into it and have spent about \$10,000 total (that includes several bad choices in equipment and also includes a backyard roll-off-roof observatory), and I'm just now feeling relatively competent.  My main issue now is time and weather.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For someone just starting, my recommended starting budget is at least \$2,000 and preferably closer to \$4,000 or more.  At \$2,000 you can get a competent mount and low-end telescope to get you started (and, depending on deals and what used equipment is available, a pre-modified DSLR or lower-end CCD camera).  You CAN get started cheaper than that... but you rapidly start losing capability and flexibility.  A \$1,500 or so mount can handle a camera/lens combo for very wide field imaging, a smaller scope for normal wide field, a medium sized scope for narrower field deep-sky (e.g. galaxies and smaller nebulae) and longer focal length scope like an SCT for planetary imaging.  If you go with a lesser mount, you're not likely to have as much flexibility there.  And, again, most of these mounts allow for interchangeable payloads.  So you can start with something simple like a short-tube refractor, and add to your equipment options as you have the ability to do so (my annual tax refunds have often ended up going to the hobby).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, I return to my original advice: joint a club.  Most clubs have people doing this you can learn from.  A lot of clubs have people regularly selling well-maintained used equipment to finance their next upgrade.  Some clubs have loaner equipment to borrow.  Most either have a dedicated observing site or have scouted out good locations away from light pollution. And many also have their own small observatories with higher-end equipment, and, sometimes, even imaging equipment.  Most clubs in the US have annual dues less than \$50 per year.  It becomes a very cost-effective way to pursue the hobby.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I've given you a lot here.  And I've honestly barely scratched the surface.  AP is not for the weak!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck and clear skies!&lt;/p&gt;&#xA;" OwnerUserId="10546" LastActivityDate="2018-08-24T22:11:09.753" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27453" PostTypeId="1" CreationDate="2018-08-25T02:11:07.247" Score="6" ViewCount="1834" Body="&lt;p&gt;I read that the CMB were all the photons left free when the universe was cold enough for electrons got bound with protons (recombination) and the photons free in that moment continued traveling in the directions they were.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I imagine:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the CMB as a sphere (with incredibly thick walls) increasing its radius continuously away from the 'center' of the universe. When the Milky Way formed, hadn't this sphere of photons already passed the space in which the Milky Way formed? How can we still see them if they are moving away from us.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the rate at which the space grew was bigger than the speed of these photons, so the Milky Way and our solar system had time to be formed before being reached by this photons lagged behind the speed in which the space increased.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;On the other hand, how can we continuously measure the CMB if is not being generated continuously as a stream of photons like looking at a distant star. We should be able to see them only once when they passed by earth.&lt;/p&gt;&#xA;" OwnerUserId="24108" LastEditorUserId="23115" LastEditDate="2018-08-25T10:32:04.287" LastActivityDate="2018-08-27T13:05:42.947" Title="Why can we still 'see' the CMB" Tags="&lt;cmb&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27456" PostTypeId="2" ParentId="27453" CreationDate="2018-08-25T12:46:19.877" Score="14" Body="&lt;p&gt;Firstly, there is no centre to the universe and the universe seems to continue indefinitely in all directions. It is best to imagine the universe as infinite in size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need to explain why the CMB appears to be a sphere surrounding us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's imagine, for a moment that I have filled the entire infinite universe with (magic) mud. This mud glows, and because its magic mud, it doesn't change the orbits of the Moon, Earth or anything else in the universe.  But if I hold up my my hand in front of my face I can't see it, because of the mud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now then, at one moment all the mud vanishes, everywhere, throughout the whole universe at the same time. One second later, if I hold my hand up I can see it. But if I look in the sky I can't see the moon. The moon is still hidden in mud.  In fact if I look at the sky, all I see is mud at 1 light-second (300,000km) distant. And because it is glowing mud, the sky appears to glow.  It looks like I am in a middle of sphere with radius 300000, surrounded by glowing mud. But this is an effect of the fact that light travels at 300000km/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After 1.2 seconds, the moon becomes visible, after about 8 minutes, the sun becomes visible too. But it takes 4½ years for the first stars to be visible. At this point it would look as if I'm in the middle of sphere with radius 4.5 light years surrounded by mud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what really happened is that 13.5 billion years ago the whole infinite universe was filled with hot gas that blocked light. As this cooled this became transparent. The whole universe became transparent at the same time, and while it wasn't instantaneous, it happened pretty quickly. So what we see now is that we seem to be in the middle of a sphere of hot gas 13.5 billion light-years across. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Except that in those 13.5 billion years, space has stretched. This has stretched the sphere from 13.5 billion light-years to 45 billion light years, and it has stretched the light from that hot gas, so it appears to be very cold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The photons that we see in the CMB now did not come from the hot gas that once filled this part of the universe, but from hot gas that was in part of the universe that is now 45 billion light years away. That being so, The photons that we saw last year came from gas that was in part of the universe that is now 44,999,999,999 light years away. Since we can't detect structures as small as few light years at that distance, the CMB appears as a constant even pattern in the sky, changing very little if at all.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-08-25T12:46:19.877" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27457" PostTypeId="1" AcceptedAnswerId="28617" CreationDate="2018-08-25T12:51:10.883" Score="3" ViewCount="276" Body="&lt;p&gt;Procrastinating, I clicked on a suggested video (&lt;a href=&quot;https://www.youtube.com/watch?v=3wOFAkggSiU&quot; rel=&quot;nofollow noreferrer&quot;&gt;How Large Can a Telescope Be?&lt;/a&gt;) and after about &lt;code&gt;01:20&lt;/code&gt; I saw a clip of a telescope shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was confused as I saw the reflections in the mirror until I decided that there are actually two small objects in front of the primary mirror, one far from the primary where you'd expect to see a secondary, which I've labeled &quot;A&quot;, and one very close to the primary, which I have labeled &quot;B&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both are supported by four vanes, and there does not seem to be a hole in the primary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've labeled the reflected image of the back side of B as B' (B prime), and I've indicated the reflections of an &quot;A vane&quot; and a &quot;B vane&quot; with arrows.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &quot;A&quot; looks very small compared to what I'd expect, only perhaps a few percent of the primary's diameter, and it looks like the telescope is not in a conventional dome, but perhaps has only a sliding or otherwise openable roof (currently off screen) to protect it when not in use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How does this telescope work, and what is it for? Is it for imaging, or performing some other function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oIuG5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oIuG5.jpg&quot; alt=&quot;screen shot of a telescope&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=3wOFAkggSiU&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=3wOFAkggSiU&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-08-25T14:03:25.627" LastActivityDate="2018-12-04T12:03:46.997" Title="How does this telescope work, and what is it for?" Tags="&lt;telescope&gt;&lt;observatory&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27458" PostTypeId="1" AcceptedAnswerId="27459" CreationDate="2018-08-25T04:32:52.337" Score="8" ViewCount="125" Body="&lt;p&gt;I was reading about how Mercury's core makes up most of the volume of the planet. I'm guessing this is both because of the small size and the distance to the Sun. If Mercury got too hot could it completely &quot;liquidify&quot;? Would it join the Sun and completely merge with it, gradually crack apart and spread, or something else entirely? Would it cause other problems in terms of life on Earth? &lt;/p&gt;&#xA;" OwnerUserId="23729" OwnerDisplayName="Magic Octopus Urn" LastActivityDate="2018-08-26T08:07:28.540" Title="Can Mercury's core &quot;boil over&quot; if it gets to close to the Sun?" Tags="&lt;mercury&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27459" PostTypeId="2" ParentId="27458" CreationDate="2018-08-25T07:42:50.217" Score="7" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;I was reading about how mercurys core makes up most of the volume of the planet. Im guessing this is both because of the small size and the distance to the sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;There are at least &lt;a href=&quot;https://en.wikipedia.org/wiki/Mercury_(planet)#Internal_structure&quot; rel=&quot;noreferrer&quot;&gt;two hypotheses&lt;/a&gt; about the causes of Mercury's composition:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Early in the Solar System's history, Mercury may have been struck by a planetesimal of approximately 1/6 that mass and several thousand kilometers across. The impact would have stripped away much of the original crust and mantle, leaving the core behind as a relatively major component.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Alternatively, Mercury may have formed from the solar nebula before the Sun's energy output had stabilized. It would initially have had twice its present mass, but as the protosun contracted, temperatures near Mercury could have been between 2,500 and 3,500 K and possibly even as high as 10,000 K. Much of Mercury's surface rock could have been vaporized at such temperatures, forming an atmosphere of &amp;quot;rock vapor&amp;quot; that could have been carried away by the solar wind.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;But if the second hypotheses is right Mercury should have been relocated to wider orbit later. &lt;a href=&quot;https://en.wikipedia.org/wiki/Planetary_migration&quot; rel=&quot;noreferrer&quot;&gt;Planetary migration&lt;/a&gt; is very interesting topic and I would say it's not solved completely yet.&lt;/p&gt;&#xA;&lt;p&gt;At current orbit Mercury is not hot enough to evaporate silicates at any significant rate.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If mercury got too hot could it completely &amp;quot;liquidify&amp;quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A lot of &amp;quot;hot jupiter&amp;quot; exoplanets have benn observed in recent years and results are very interesting. For example &lt;a href=&quot;http://blogs.discovermagazine.com/80beats/2010/07/16/osiris-the-scorched-exoplanet-with-a-comet-like-tail/#.W4Dz9M4zbIU&quot; rel=&quot;noreferrer&quot;&gt;HD 209458b &amp;quot;Oziris&amp;quot;&lt;/a&gt; has a comet-like tail of evaporating atnosphere. &lt;a href=&quot;http://astronomy.com/news/2018/08/ultra-hot-jupiters&quot; rel=&quot;noreferrer&quot;&gt;Clouds of iron and titanium dioxide&lt;/a&gt; were observed too. Iron rain (literally) could be one of exciting effects.&lt;/p&gt;&#xA;&lt;p&gt;Compared with &amp;quot;hot jupiters&amp;quot; Mercury is much smaller with less gravity. If closer to Sun it undoubtedly would have comet-like tail of evaporating elements, mostly silicates and metals. If intensive enough the tail could be seen by eye with black glass during &lt;a href=&quot;https://en.wikipedia.org/wiki/Transit_of_Mercury&quot; rel=&quot;noreferrer&quot;&gt;Mercury transits&lt;/a&gt; (also the transits would be much more often at closer orbit).&lt;/p&gt;&#xA;&lt;p&gt;At closer orbit Mercury should be tidally locked. The Sun-faced surface would be completely melted lava, but dark side would probably be solid because of lack of stable atmosphere. Evaporated rocks would cool down by thermal radiation and dowfall behind the terminator line.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Would it join the sun and completely merge with it, gradually crack&#xA;apart and spread, or something else entirely?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The planet should be affected by some force to migrate even close and close to Sun. I can't tell surely will evaporation slow down the planet and move it closer to Sun but I suppose it will not. I suppose Mercury will not fall to Sun but will be evaporating. I can't do math now to estimate how close to Sun it should be to evaporate completely in less than 5 billion years (after that Sun will become a red giant and will envelope Mercury by its atmosphere in &amp;quot;coup de grace&amp;quot; act :) ).&lt;/p&gt;&#xA;&lt;p&gt;I don't see reasons for Mercury to crack apart if it is not close to &lt;a href=&quot;https://en.wikipedia.org/wiki/Roche_limit&quot; rel=&quot;noreferrer&quot;&gt;Roche limit&lt;/a&gt; (again with assumption that evaporation only is not enough to lower its orbit).&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Would it cause other problems in terms of life on earth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I don't think so. Maybe Solar system space would be more dusty because of Mercury's evaporating tail. Even cracking of Mercury would not affect Earth because the debris would stay close to Mercury's orbit (much closer than current Mercury's orbit, don't forget!). Theoretically some of debris pieces could develop more eccentric orbits in millions of years  to reach Earth (because of collisions, uneven evaporation, orbital resonanses) but I bet all them would evaporate before that.&lt;/p&gt;&#xA;&lt;p&gt;P.S. Most of this answer consists of educated (or not so educated) assuptions. So objections, amendments, corrections are welcome.&lt;/p&gt;&#xA;" OwnerUserId="23872" OwnerDisplayName="Heopps" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-08-26T08:07:28.540" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27460" PostTypeId="1" CreationDate="2018-08-25T17:28:36.850" Score="4" ViewCount="127" Body="&lt;p&gt;I was fortunate enough to see the solar eclipse in southern Illinois last year. The clouds were not friendly, but we did have one spectacular moment of perfect eclipse. It is an eerie sensation, and I cannot imagine how utterly terrifying it must have been for our ancient ancestors to see the moon eating the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it got me wondering. A solar eclipse can occur in the spectacular fashion it does because the angular dimension on the moon and the Sun are almost exactly the same -- this despite the fact that they are dramatically different in actual dimension. This seems a remarkable coincidence to me -- the ratio of the distances being almost exactly the same as the ratio of their diameters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if there are any other known examples of such a phenomenon occurring anywhere else in the universe? Can we make any estimate of how often something like that might occur in a galaxy?&lt;/p&gt;&#xA;" OwnerUserId="24113" LastActivityDate="2018-09-25T01:01:05.697" Title="Solar and lunar eclipses -- how rare" Tags="&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27462" PostTypeId="2" ParentId="27460" CreationDate="2018-08-26T00:33:28.357" Score="1" Body="&lt;p&gt;TOTAL solar eclipses and TOTAL lunar eclipses are actually not that common in places outside of the solar system. For these TOTAL eclipses to occur, and I'm emphasizing total bc partial eclipses happen all the time, there needs to be the perfect distance so that an object perfectly covers the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, you are correct - it is such a remarkable coincidence that this occurs in our Solar System. I'm not entirely sure about how many exoplanetary systems also experience this phenomenon, but I assume that it won't be a lot. I do know that it DOES HAPPEN in other places, but it just isn't common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for a fun fact - the Moon is actually moving away from us, meaning that in a few million years, total solar eclipses won't be possible anymore - and we'll only see partial eclipses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps! :)&lt;/p&gt;&#xA;" OwnerUserId="24048" LastActivityDate="2018-08-26T00:33:28.357" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27463" PostTypeId="1" AcceptedAnswerId="27464" CreationDate="2018-08-26T04:54:02.967" Score="1" ViewCount="51" Body="&lt;p&gt;Just my curiosity. This is not my assignment etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let planet A take m Earth days to complete one full round around the Sun, and planet B take n Earth days for one full round around the Sun. Assuming that both are orbiting in the same direction, what is the duration recorded by an observer on the planet A, as the number of Earth days the planet B took to complete one full round around planet A?&lt;/p&gt;&#xA;" OwnerUserId="24121" LastEditorUserId="1891" LastEditDate="2018-08-26T07:32:54.297" LastActivityDate="2018-08-26T07:32:54.297" Title="What is the apparent rotation time for a planet relative to an other planet?" Tags="&lt;planet&gt;&lt;apparent-motion&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27464" PostTypeId="2" ParentId="27463" CreationDate="2018-08-26T07:23:57.967" Score="1" Body="&lt;p&gt;Note, an observer on Planet A can't see a rotation of Planet B. He can see a the Sun orbiting around him, and Planet B orbiting around the Sun. Thus&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If Planet B has an orbit of smaller radius than Planet A, then the observer can see Planet B sometimes before, sometimes after its Sun.&lt;/li&gt;&#xA;&lt;li&gt;If Planet B has an orbit of larger radius than Planet A, then the observer can see Planet B to cycle on the sky, roughly on the plane of the Ecliptics,  but a little bit faster as the Sun.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The calculation is this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Planet A takes $\frac{1}{m}$ full circle in a day.&lt;/li&gt;&#xA;&lt;li&gt;Planet B takes $\frac{1}{n}$ full circle in a day.&lt;/li&gt;&#xA;&lt;li&gt;Thus, the observer on Planet A can see an $|\frac{1}{m}-\frac{1}{n}|$ full circle of rotation of Planet B in a day.&lt;/li&gt;&#xA;&lt;li&gt;Thus, he can see a period of Planet B of $\frac{1}{|\frac{1}{m}-\frac{1}{n}|}$ Earth days, which is in simplified form&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\underline{\underline{|\frac{mn}{m-n}|}}.$$&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="1891" LastEditDate="2018-08-26T07:30:40.087" LastActivityDate="2018-08-26T07:30:40.087" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27466" PostTypeId="1" AcceptedAnswerId="27502" CreationDate="2018-08-26T19:31:35.763" Score="4" ViewCount="476" Body="&lt;p&gt;I'm having some trouble grasping the idea of planetary resonances due to what seems to me like an ambiguity. It is known that the asteroid main belt is sculpted by the Kirkwood gaps, corresponding to orbits that are in mean motion resonance with the Jupiter. As such, the asteroids reaching such orbits get ejected (unstable) from such resonance orbits. Nevertheless, objects like Pluto are dynamically stable due to their resonance with Neptune. Mercury's spin-orbital resonance makes its elliptical orbit stable around the Sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas how to explain the two-fold nature of the outcomes of planetary resonances ? Any references to some literature / mathematical proof would be highly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance,&lt;/p&gt;&#xA;" OwnerUserId="22174" LastActivityDate="2018-09-03T02:04:47.643" Title="Explanation of resonance stability and instability" Tags="&lt;solar-system&gt;&lt;jupiter&gt;&lt;mercury&gt;&lt;neptune&gt;&lt;orbital-resonance&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27467" PostTypeId="2" ParentId="27445" CreationDate="2018-08-26T19:56:02.497" Score="1" Body="&lt;p&gt;Well, in response to the response of @MystaryPi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;..However, evidence for water on Earth originally comes from comets..&quot;&#xA;The above statement is not 100% true, despite that fact that was the common belief how the Earth got its water until the recent Rosetta mission. This question can be partially answered by the ratio between the Deuterium and Hydrogen atoms of observed comets. The D/H isotope ratios of water measured on comet 67P/Churyumov-Gerasimenko by Rosetta and on other comets by other in-situ spacecraft have obtained accurate values, bringing up the prelude to the idea that water on terrestrial oceans may emanate from objects that were of non-cometary nature, simply because those D/H values do not correspond to the D/H values measured on terrestrial oceans. As such, it is thought now that the origin of water-brining objects to Earth to be more asteroid-like, which likely formed much closer to the Sun than the comets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information please refer to Altwegg et al., 2014 &#xA;(DOI: &lt;a href=&quot;http://science.sciencemag.org/content/347/6220/1261952.full&quot; rel=&quot;nofollow noreferrer&quot;&gt;10.1126/science.1261952&lt;/a&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rEHtf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rEHtf.jpg&quot; alt=&quot;D/H ratio of Solar bodies relative to Earth&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="22174" LastEditorUserId="11039" LastEditDate="2018-10-27T12:02:37.413" LastActivityDate="2018-10-27T12:02:37.413" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27468" PostTypeId="1" CreationDate="2018-08-26T20:45:09.267" Score="0" ViewCount="236" Body="&lt;p&gt;Is there a table which shows the dates of the planets retrograde motion? I want to know at which date a certain planet will start the retrograde motion.&lt;/p&gt;&#xA;" OwnerUserId="24127" LastEditorDisplayName="user1569" LastEditDate="2018-08-28T08:40:42.073" LastActivityDate="2018-08-28T08:40:42.073" Title="Table of dates for planet retrograde motion" Tags="&lt;orbit&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27469" PostTypeId="2" ParentId="16333" CreationDate="2018-08-26T21:57:27.763" Score="1" Body="&lt;p&gt;The fact is, the gravity and the distortion of the rate of time are the same thing. A free-falling object follows a spacetime geodesic into the future, which just means a path that is as straight as possible subject to the curvature of spacetime. But &quot;as straight as possible&quot; translates to &quot;the path in spacetime where the traveler experiences the most time passing over all possible paths&quot;. That's analogous to the way a geodesic on the Earth is the path with the &lt;em&gt;shortest&lt;/em&gt; distance. (The geometry of spacetime is not just the natural 4D generalization of 3D geometry, but instead it's more of an analogue with special rules based on the way physics actually works.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that the curve with the most &quot;proper time&quot; (time experienced by the traveler) is one that curves towards wherever time passes more slowly, and time passes more slowly near a massive object. That means that, as you move into the future, your path through spacetime curves towards the massive object. In other words, you move faster and faster towards the massive object. You fall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course it makes sense that a slowing down of time will reduce the frequency of a clock or a beam of light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, what's more, the slowing down of time is a part of the curvature. Any time you have the scale of measurement changing as you move from place to place, if it does so in a way that causes geodesics to change, then we call that curvature. And curvature is arguably the fundamental phenomenon. And so I would say that it really is the slowing down of time (along with the deformation of space) that is the basic cause of light not being able to escape. &lt;/p&gt;&#xA;" OwnerUserId="7752" LastActivityDate="2018-08-26T21:57:27.763" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27470" PostTypeId="2" ParentId="27468" CreationDate="2018-08-26T22:00:43.527" Score="0" Body="&lt;p&gt;Check this website out - it has a table that might help you: &lt;a href=&quot;https://www.wikiwand.com/en/Apparent_retrograde_motion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Website&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZauMz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZauMz.png&quot; alt=&quot;Or you can look here.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This site is also a little sketchy (it's for horoscopes and stuff) - but it still has some accurate information on the exact dates of retrograde: &lt;a href=&quot;https://www.findyourfate.com/astrology/year2018/2018-planetretrogrades.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Website&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps!&lt;/p&gt;&#xA;" OwnerUserId="24048" LastActivityDate="2018-08-26T22:00:43.527" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27471" PostTypeId="1" AcceptedAnswerId="27485" CreationDate="2018-08-26T23:44:46.787" Score="4" ViewCount="250" Body="&lt;p&gt;Does the major axis of the moon's orbit point in the same direction (with small deviations) throughout the year? or does it migrate along a path so that after a million years it might be 90° away from where it started?&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2018-08-27T22:54:38.057" Title="Direction of major axis of moon's orbit" Tags="&lt;the-moon&gt;&lt;orbital-migration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27472" PostTypeId="1" CreationDate="2018-08-27T01:14:24.233" Score="1" ViewCount="101" Body="&lt;p&gt;This is really a math question, but it's for use in astronomy, so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because reasons, I'm writing my own sky-viewing code.  I've found a few references for converting right ascension and declination to local azimuth and altitude; so far, so good, now I need to plot that coordinate in 2D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's trivial to turn az/alt into an azimuthal equidistant projection; just map azimuth to $\theta$ and altitude to $r$ where +90° is zero and -90° is 2.  However, what I really want is an azimuthal equidistant projection where the center is at a point of my choosing — for example, NW toward the horizon — ideally specified using two angles, $\theta$ and $\phi$.  With no modifications, az/alt is using a focal point directly overhead — how do I rotate that to my point?  (Continuing the example, I'd be rotating by $\theta$ 45°, $\phi$ 90°.)  I believe this is a spherical trigonometry question, but I'm having trouble grokking the references I've found well enough to do what I want.  Any guidance?&lt;/p&gt;&#xA;" OwnerUserId="24129" LastActivityDate="2018-08-27T01:14:24.233" Title="Rotating azimuth/altitude to a focal point?" Tags="&lt;coordinate&gt;&lt;mathematics&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27473" PostTypeId="1" AcceptedAnswerId="27477" CreationDate="2018-08-27T01:51:55.790" Score="6" ViewCount="835" Body="&lt;p&gt;I have a 12&quot; dobsonian with a 1500mm focal length and want to do some higher magnification planetary viewing. Currently my highest magnification is ~80x via my 19mm eyepiece (calculated &lt;a href=&quot;https://www.skyandtelescope.com/astronomy-equipment/telescope-eyepiece-guide/&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;), but I would like to do some viewing at higher magnification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My telescope is fitted for 2&quot; diameter eyepieces, but I have an adapter for  1¼&quot; eyepieces as well. I'm seeing a &lt;a href=&quot;https://smile.amazon.com/s?ie=UTF8&amp;amp;field-keywords=9mm%20telescope%20eyepiece&amp;amp;index=blended&amp;amp;link_code=qs&amp;amp;sourceid=Mozilla-search&amp;amp;tag=mozilla-20&quot; rel=&quot;noreferrer&quot;&gt;plethora of affordable 1¼-inch 9mm eyepieces&lt;/a&gt;, but &lt;a href=&quot;https://smile.amazon.com/s?url=search-alias%3Daps&amp;amp;field-keywords=telescope+eyepiece+2+inch+9mm&quot; rel=&quot;noreferrer&quot;&gt;very few 2&quot; diameter options&lt;/a&gt; and very pricey at that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, I see the following options, by increasing price:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1¼&quot; diameter 9mm eyepiece&lt;/li&gt;&#xA;&lt;li&gt;3x Barlow lens&lt;/li&gt;&#xA;&lt;li&gt;2&quot; diameter 9mm eyepiece&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I've read that the 2&quot; eyepieces yield greater field of view advantages. Is that still true when the goal is a narrow focal length on solar system objects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would using a Barlow lens on my existing 2&quot; eyepieces retain this advantage in a meaningful way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, is there any significant value in keeping to the 2&quot; diameter eyepiece capability of my telescope when doing planetary viewing with low focal length given the magnitude of the cost difference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;I've referenced a &quot;9mm&quot; eyepiece here merely as an example, I'm sure most anything in the 5mm-12mm range would be fine.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="15920" LastEditorUserId="15920" LastEditDate="2018-08-27T12:37:43.487" LastActivityDate="2018-08-30T23:37:56.380" Title="Eyepiece diameter for small focal length planetary viewing on a large dobsonian?" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;observational-astronomy&gt;&lt;dobsonian-telescope&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27475" PostTypeId="1" CreationDate="2018-08-27T05:50:20.257" Score="1" ViewCount="60" Body="&lt;p&gt;If the semi major and semi minor axis of a galaxy are 3&quot; and 1&quot; respectively, how would you find the radius of the galaxy?&lt;/p&gt;&#xA;" OwnerUserId="24132" LastEditorUserId="1891" LastEditDate="2018-08-27T06:24:40.893" LastActivityDate="2018-08-27T06:24:40.893" Title="How to the find radius of a galaxy given its inclination?" Tags="&lt;astrophysics&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2018-08-29T08:46:02.823" ContentLicense="CC BY-SA 4.0" />
  <row Id="27476" PostTypeId="2" ParentId="27453" CreationDate="2018-08-27T13:05:42.947" Score="1" Body="&lt;p&gt;&lt;strong&gt;TL;DR:&lt;/strong&gt;&lt;br&gt;&#xA;The light we now see from the Sun was emitted eight minutes prior, from the closest star, about 4.22 years prior. Observational astronomy inherently looks back in time, and the further away something is, the further we're looking back in time. We can still see the cosmic microwave background radiation because we can (at least conceptually) still see all the way back in time to the big bang itself. The time of last scattering is close to but not quite that far back in time.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So I imagine the CMB as a sphere (with incredibly thick walls) increasing its radius continuously away from the 'center' of the universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This represents a very common misunderstanding regarding big bang theory. The big bang was not an explosion &lt;strong&gt;in&lt;/strong&gt; space. A better way of looking at it  is that the big bang was instead an explosion &lt;strong&gt;of&lt;/strong&gt; space that took place everywhere. While the observable universe is finite and was very small moments after the big bang, the universe as a whole may well be infinite, and if that is the case, it was infinite moments after the big bang. (Science doesn't know what happened at the very instance the big bang started.) What this means is that every point in the universe appears to be the center of the universe, or better put, there is no center of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The early universe was very hot and very dense. The expansion of the universe made the universe cool from this state. Neutrinos decoupled from matter a second after the big bang. Future scientists might be able to detect the resulting cosmic neutrino background radiation, just as they can now detect the cosmic microwave background radiation that formed 380,000 years later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite the universe having cleared optically about 380,000 years after the big bang, the universe of 370,000 years after the big bang versus 390,000 years after the Big Bang wouldn't have looked all that much different. The universe of 370,000 years after the big bang was a rather thin and almost uniform warmish glowing gas that was slightly ionized. The universe of 20,000 years later (390,000 years after the big bang) was locally clear, but the only thing that was visible was a warmish glowing gas a bit over 20,000 light years away, in every direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That warmish glowing gas was still visible a few hundred million years later, but now in the deep infrared thanks to the continued expansion of the universe. The intervening space contained the first stars and perhaps the first galaxies. That warmish glowing gas remains visible to this day, but now in the microwave thanks to the continued expansion of the universe. The intervening space contains lots and lots of stars and galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether the surface of last scattering will always remain observable depends on the nature of the future expansion of the universe and dark energy. One possible far future outcome is that the expansion continues such that the distance to furthest into the past that is possible to &quot;see&quot; by any means (the particle horizon) is shorter than the distance to the surface of last scattering. The cosmic microwave background radiation (by then, the cosmic radio background radiation) will wink out of observability if that outcome is the case.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2018-08-27T13:05:42.947" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27477" PostTypeId="2" ParentId="27473" CreationDate="2018-08-27T13:14:11.610" Score="6" Body="&lt;p&gt;I would recommend a quality 1¼&quot; 9 mm or 12 mm eyepiece as the optimal solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You already have a 19 mm so something close to that such as 15 mm would bring small difference, so take a step or two down to a 9 mm or 12 mm Plössl lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While a 2&quot; eyepiece usually offers better eye relief (from bigger lens) the cost is also substantially more, and if you compromise on quality by buying a cheaper eyepiece the results will be less than satisfactory - lack of clarity, distortions from imperfections in the glass, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extra magnification can also be obtained by using a Barlow, but you are also adding extra lenses to the optical path which will slightly reduce the amount of light you are getting at the eyepiece, but also any imperfections will be magnified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, with Dobsonians (usually) not having any tracking mechanisms you will be manually tracking your object of interest, and extra magnification means that it will be moving more quickly across your field of view as well as making it a little more difficult to keep it smoothly in the eyepiece. These are also arguments in favour of a 2&quot; eyepiece over a 1¼&quot; eyepiece, but either way, good eyepieces can be kept and re-used when/if you upgrade your telescope.&lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2018-08-27T13:14:11.610" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27478" PostTypeId="2" ParentId="27468" CreationDate="2018-08-27T16:52:32.637" Score="0" Body="&lt;p&gt;There is an annual publication &quot;Astronomical Phenomena for the Year YYYY&quot; published by the United States Naval Observatory and United Kingdom Hydrogrphic Office. Page 6 of the 2018 edition gives a table of conjunctions, stationary, oppositions, etc. for the planets. See &lt;a href=&quot;http://aa.usno.navy.mil/publications/reports/ap18_for_web.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://aa.usno.navy.mil/publications/reports/ap18_for_web.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the superior (outer) planets, the earlier stationary date is when it starts its retrograde motion. The second stationary date is when it ends its retrograde motion.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2018-08-27T16:52:32.637" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27479" PostTypeId="2" ParentId="27468" CreationDate="2018-08-27T16:56:16.040" Score="1" Body="&lt;p&gt;EDIT: It turns out calsky.com will do these calculations for several years out, including other calculations commonly found in almanacs. Sample page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.calsky.com/cs.cgi?cha=7&amp;amp;sec=6&amp;amp;sub=3&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.calsky.com/cs.cgi?cha=7&amp;amp;sec=6&amp;amp;sub=3&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best place to start on this site isn't the home page, but rather the TOC:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.calsky.com/cs.cgi/TOC?&amp;amp;f=&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.calsky.com/cs.cgi/TOC?&amp;amp;f=&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I now consider my answer to be better than @JohnHoltz's :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I saw @JohnHoltz's answer while I was typing out my answer, so some of the below may be redundant. Upvote/checkmark his answer, not mine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't really an answer, but too long for a comment:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I was disappointed how difficult it was to find this information, given how easy it is to calculate and how many other astronomical phenomena are listed in various places on the Internet.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When a planet starts or ends retrograde motion, it is temporarily stationary before it &quot;turns around&quot;. Thus, some astronomical sources will say &quot;Jupiter stationary&quot; instead of &quot;Jupiter starts retrograde motion&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Based on the above I google'd &lt;code&gt;&quot;jupiter stationary&quot; &quot;mars stationary&quot;&lt;/code&gt; (as quoted; google tries to correct one of the stationary's to stationery, so I had to click one more time to do the search correctly), and finally found something: &lt;a href=&quot;http://aa.usno.navy.mil/publications/docs/ap.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://aa.usno.navy.mil/publications/docs/ap.php&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The publication doesn't mention retrograde, but it does mention when a given planet is stationary and provides a brief description of each planet's position in the sky for the given year.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Because this seems like a fairly easy calculation to make, I may do it myself and update this answer.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2018-08-27T18:48:52.330" LastActivityDate="2018-08-27T18:48:52.330" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27480" PostTypeId="2" ParentId="27447" CreationDate="2018-08-27T20:30:51.120" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Perhaps the difference stems from the conversion algorithm?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Almost certainly this is the answer.  It's not clear to me exactly what Stellarium assumes for is refraction calculation, but I know Astropy's algorithm.  Given that the altitude you've listed here is below the horizon (i.e., alt is negative), it's not really clear what the correct interpretation is: in particular, &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_refraction&quot; rel=&quot;nofollow noreferrer&quot;&gt;atmospheric refraction&lt;/a&gt; is not really well-defined below the horizon (since it should be invisible anyway!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, even well above the horizon where refraction isn't that important, there are several finicky details that you have to account for to properly do the Az/Alt conversions.  In addition to the precession that the other answers mention, if you want to get &lt;em&gt;exactly&lt;/em&gt; the same answer on two tools, you have to worry about things like how much the axis of the Earth shifts due to, say, major earthquakes (&quot;true polar wander&quot;), the effects of General Relativity on the apparent direction of the incoming light ,etc.  Most of these are pretty small, but together they mean it's very hard to ensure all of the software gives &lt;em&gt;exactly&lt;/em&gt; the same answer unless they carefully follow the definitions set by the International Astronomical Union.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, the differences you see are only ~20 arcsec.  The human eye isn't even capable of telling that small of a difference, and almost all telescopes that people would use Stellarium with probably wouldn't either (at least in an absolute sense).  So for &lt;em&gt;this particular application&lt;/em&gt; they essentially do match.&lt;/p&gt;&#xA;" OwnerUserId="24140" LastEditorUserId="23940" LastEditDate="2018-08-27T22:42:38.357" LastActivityDate="2018-08-27T22:42:38.357" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27481" PostTypeId="1" AcceptedAnswerId="27488" CreationDate="2018-08-27T21:04:34.057" Score="3" ViewCount="548" Body="&lt;p&gt;I'm trying to figure out at what right ascension (RA) — not celestial longitude — the sun enters the various traditional (Western) astrological signs of the Zodiac; but I can't figure out how to calculate this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the right ascensions of the boundaries between traditional zodiacal signs along the ecliptic?&lt;/p&gt;&#xA;" OwnerUserId="2369" LastActivityDate="2018-08-28T05:30:11.273" Title="What are the RAs of the boundaries between traditional zodiacal signs?" Tags="&lt;coordinate&gt;&lt;history&gt;&lt;zodiac&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27482" PostTypeId="2" ParentId="27473" CreationDate="2018-08-27T21:29:55.070" Score="2" Body="&lt;p&gt;Magnification is always a tricky subject. A lot of people don't understand it or its limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you already seem to know, the magnification factor is calculated by dividing the focal length of the telescope (in your case, 1,500 mm) by that of the eyepiece.  In theory, then, you can magnify as much as you like to infinity, assuming you use shorter and shorter focal length eyepieces and/or extend the focal length using a Barlow lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are hard limits to magnification.  When a wave of light enters the telescope, it begins to diffract.  If you want to understand the underlying physics, there's a great Khan Academy video &lt;a href=&quot;https://www.khanacademy.org/science/physics/light-waves/interference-of-light-waves/v/single-slit-interference&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;here&lt;/strong&gt;&lt;/a&gt; you can watch to understand it better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize the problem: the light waves essentially break down and create interference patterns which start to degrade the image. This results in a limit of the information (detail) carried in the light entering the telescope tube.  The larger the aperture, the more detail can be produced.  While the exact process is different, the result is not unlike trying to blow up a picture.  The more you try to blow it up, the blurrier it will get.  I put together a photographic example &lt;a href=&quot;http://astro.jmhastronomy.com/random/resolutionexample/index.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;here&lt;/strong&gt;&lt;/a&gt; using an image I captured of the moon.  Each frame in the sequence is effectively a doubling of magnification.  Because imaging and visual astronomy aren't quite equivalents, I can't really give you a magnification power here, but you can see the net effect as I &quot;zoom-in&quot; on the crater Herschel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the mid 1800's, William R. Dawes devised a forumla for determining the angular separation required between two stars before one can reasonably state that they are, in fact, two stars.  His formula provides a measurement in arcseconds we call the Dawes' limit that is frequently used as an indication of the resolution a telescope can provide.  His formula divides 4.56 by the aperture diameter in inches to produce the limit in arcseconds (or 116 divided by the aperture in millimeters).  For your 12&quot; aperture, this yields a Dawes' Limit of about 0.38 arcseconds.  Later in the 1800's, Lord Rayleigh devised another formula for determining resolution, which gives us a number, also in arcseconds, known as the Rayleigh Criterion.  It is a bit more conservative, producing a lower-resolution limit.  For your telescope, the Rayleigh Criterion is 0.45 arcseconds.  A further calculation can be done to ascertain the theoretic limitations of the optical system based on the diffraction limit. For this telescope, that would be about 0.35 arcseconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially what these numbers all tell us is how small a detail your telescope can resolve.  This measure is, of course, an angular measurement as opposed to a linear measurement.  The linear measurement depends on the object's distance.  A detail that appears as 1 arcminute on the moon would be a few hundredths of an arsecond at Jupiter's distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from the limitations imposed by diffraction/aperture, you also have the simple fact that the atmosphere limits your magnification.  This is going to vary depending on your altitude (e.g. how much atmosphere is between you and the object), your light pollution levels, air pollutants, humidity, and air currents.  For a telescope like yours, you can probably expect a maximum usable magnification of around 250x to 300x, depending on those factors.  In fact, on some nights, 200x might be pushing it, while on others, 350x might not be particularly bad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, back to your eyepiece question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you've found, there are two common formats for eyepiece barrel flange size: 1.25&quot; and 2&quot;.  2&quot; eyepieces are typically only needed for long-focal length eyepieces for wide fields.  In fact, for anything higher than about 25mm, it's entirely unnecessary.  And I've never seen any true 2&quot; eyepieces that have shorter focal lengths.  The 2&quot; compatible eyepieces I've seen with shorter focal lengths have always been combination 2&quot; and 1.25&quot;, they accomplish this by using nested barrel flanges.  The 1.25&quot; flange is inside the 2&quot; and slightly longer, so that it will fit into a 1.25&quot; focuser recepticle.  &lt;a href=&quot;http://televue.com/engine/TV3b_page.asp?ID=20&amp;amp;Tab=EP_ETH-10.0&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Here is an example&lt;/strong&gt;&lt;/a&gt;.  Notice how the 1.25&quot; flange protrudes below the 2&quot;?  This allows you to use it in either size focuser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what should you get?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is partly a matter of personal preference.  I use a set of Meade 5000 Series eyepieces (the earlier versions with the 60° AFOV). In that set, my shortest focal length is a 5.5m which on my 10&quot; Dob (1200mm focal length) gives me about 220x.  For the seeing conditions where I am outside of Houston, that's usually about as good as I can expect.  I have other eyepieces I occasionally try out.  I have a 3.6mm (not sure who made it), which gives mea round 333x.  It's rarely worth using - it just gives me a bigger blur.  Something around 5.5 or 6mm is about as high as is worth going.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to do it on the cheap, the Orion Expanse series isn't too bad, and you can find an un-branded version online cheaper.  The Expanse 6 mm has a blue line around the barrel toward the eyepiece.  If you do a search on sites like Amazon you can find the same eyepiece with a gold line instead of the blue for a lower price.  It's a dirty little secret in the telescope market that the majority of consumer telescopes and accessories are manufactured by only a handful of factories in China and re-branded for sale.  I have a Meade f/6.3 focal reducer and a Celestron f/6.3 reducer... they're identical other thant he labelling (and I've seen the same thing sold by Antares). The Orion SkyLine 8&quot; Dob is manufactured by GSO out of China and the exact same scope has been sold as the Apertura AD8 and Zhumell X8 (and by a few other brands as well).  In this case, the only difference in the eyepiece between the Orion Expanse and the gold line versions is the name and color of the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the question of &quot;to Barlow, or not to Barlow.&quot; Barlow lenses put more glass in the optical train, this is true.  But these days, that's not a huge problem, as long as you get a fairly good quality Barlow.  Generally speaking, however, it's better to go without than to use one, if you can avoid it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, overall, you're not really getting anything out of a 2&quot; eyepiece that you can't get out of a 1.25&quot; when you're talking higher end eyepieces.  For proof of that, look at the TeleVue line.  TeleVue arguably makes some of the best, if not THE best, eyepieces on the market.  Their high magnification eyepieces are either 1.25&quot; or dual format.  If there was a reason to make the high magnification eyepieces in true 2&quot; format, they'd definitely do so.   So there's no loss in going with the 1.25&quot; and saving the money.&lt;/p&gt;&#xA;" OwnerUserId="10546" LastActivityDate="2018-08-27T21:29:55.070" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27483" PostTypeId="2" ParentId="27481" CreationDate="2018-08-27T21:41:50.927" Score="-2" Body="&lt;p&gt;It is going to depend on where you draw the lines, but they're not equally spaced... and it also depends on whether or not you include Ophiuchus.  Since they're not evenly spaced, it's not like you can start with a point on the ecliptic and calculate an even spacing for them.&lt;/p&gt;&#xA;" OwnerUserId="10546" LastActivityDate="2018-08-27T21:41:50.927" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="27484" PostTypeId="2" ParentId="27471" CreationDate="2018-08-27T22:44:59.223" Score="3" Body="&lt;p&gt;As &lt;em&gt;pm-2ring&lt;/em&gt; already commented, the direction of the axis has a period of about 8.85 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth noting, though, that in reality the moon's orbit does not correspond very closely to an ellipse, because it is rather heavily perturbed by the Sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relative to this reality, the concept that the moon's orbit around the earth has a 'major axis' results from treating the real orbit approximately as if it were really an ellipse.  The approximations can be made in several different ways, and so in principle there can be a number of different approximating ellipses.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In effect the question presupposes the (idealised) rotating mean ellipse which results (in concept) when nearly all of the perturbation effects are treated as if they were averaged out -- except for the mean rate of axial rotation and the current position of the mean axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This idea of the mean ellipse is the one that underlies the 'mean elements' given in tables that express the fundamentals of the moon's motion, such as the 1954 &lt;em&gt;Improved Lunar Ephemeris&lt;/em&gt;, which was slightly adapted from Brown's theory and &lt;a href=&quot;https://archive.org/details/tablesofmotionof12browrich&quot; rel=&quot;nofollow noreferrer&quot;&gt;1919 &lt;em&gt;Tables of the Motion of the Moon&lt;/em&gt;&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more recent estimation of the position and rotation of the idealised mean ellipse, derived from lunar laser ranging via the JPL ephemeris DE200 (official basis of the &lt;em&gt;Astronomical Almanac&lt;/em&gt; from 1984-2002), is found in J L Simon et al. (1994) &quot;Numerical expressions for precession formulae and mean elements for the Moon and the planets&quot;, &lt;a href=&quot;http://adsabs.harvard.edu/abs/1994A%26A...282..663S&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Astronomy &amp;amp; Astrophysics&lt;/em&gt; 282, 663-683&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to this source, the mean longitude of the moon's perigee in arc-seconds (counted from the mean equinox of date) (see Simon 1994, p.670 at b.3) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$    =  300071&quot;.6752   +14648449&quot;.0869 t   -37&quot;.1582 t^2 $&#xA;$           -0&quot;.044970 t^3   +1&quot;.8948*10^-4 t^4 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at any time defined by t, where t is the time-interval counted from the J2000 epoch of 2000 Jan 1 at 12h TT (= JD 2451545.0), in units of Julian centuries (36525 days) of Terrestrial Time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The negative quadratic term in the expression above shows that the rate of rotation of the mean axis is very slightly slowing down; in 20 years it now rotates by about 1 minute of arc less than it used to do in 20 years around AD 1600 in the age of Tycho and Kepler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides the mean approximation to the moon's orbital ellipse just described, &#xA;there is also the (variable) &lt;a href=&quot;https://en.wikipedia.org/wiki/Osculating_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;osculating ellipse&lt;/a&gt;. The largest periodic variations in the osculating longitude of perigee, relative to the mean perigee, are given in the same 1994 paper by Simon et al. at p.671, Table 4, 5th column. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last but perhaps not least is the librating-and-rotating ellipse approximation which results when the rotating ellipse is taken to include the effect of the largest of the solar perturbations, now known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Evection&quot; rel=&quot;nofollow noreferrer&quot;&gt;evection&lt;/a&gt;.  This is described in older textbooks such as Godfray's &lt;em&gt;Elementary Treatise on the Lunar Theory&lt;/em&gt;, &lt;a href=&quot;https://archive.org/download/elementarytreati00godfuoft/elementarytreati00godfuoft.djvu&quot; rel=&quot;nofollow noreferrer&quot;&gt;see esp. pp.69-71 in the 1871 edition here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17202" LastActivityDate="2018-08-27T22:44:59.223" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27485" PostTypeId="2" ParentId="27471" CreationDate="2018-08-27T22:46:58.000" Score="3" Body="&lt;p&gt;The direction of the semimajor axis of the Moon's orbit actually changes quite quickly. In fact it does a cycle around the ecliptic in approximately 8.85 years. This is known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_precession#Apsidal_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;apsidal precession&lt;/a&gt;. That Wikipedia article also explains the Moon's nodal precession,  the precession of the Moon's orbital plane, which has a period around 18.6 years (retrograde).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These precessions have such short periods because of the strong effect of the Sun on the Moon's orbit (in fact, the Sun's gravitational force on the Moon is stronger than the Earth's). And because these cycles are so short and have a significant effect on the timing of eclipses, they have been known to astronomers since ancient times, although of course ancient astronomers didn't have a good model to explain the cycles they'd discovered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the Sun's influence, and the fact that the Moon is very large relative to its primary (Earth), the Moon's orbit is actually quite difficult to model accurately. It's certainly much harder to predict the lunar motion than the movement of the visible planets. The ancients managed to do an ok job on the Moon with the Ptolemaic system, although in that model the apparent size of the Moon varied by &lt;em&gt;much&lt;/em&gt; more than it actually does. I'm sure this was very annoying at the time, but in the long run it was a good thing, because it made it obvious that the Ptolemaic system was not correct, but it took many centuries before a better system was found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to know more about the complexity of lunar motion, please take a look at Wikipedia's article on &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;lunar theory&lt;/a&gt;, which is an excellent introduction to this fascinating topic.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="16685" LastEditDate="2018-08-27T22:54:38.057" LastActivityDate="2018-08-27T22:54:38.057" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27486" PostTypeId="1" CreationDate="2018-08-28T00:28:27.260" Score="0" ViewCount="180" Body="&lt;p&gt;Are there any known major events around the 50-60 million years preceding the cambrian explosion that could explain the sudden increase in biodiversity?  Specifically a supernova, or active galactic center either in the milky way or some other nearby galaxy?&lt;/p&gt;&#xA;" OwnerUserId="24144" LastActivityDate="2020-09-13T17:04:55.513" Title="Candidate major astronomical events circa 600-541 mya?" Tags="&lt;astrophysics&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="27488" PostTypeId="2" ParentId="27481" CreationDate="2018-08-28T01:56:52.593" Score="5" Body="&lt;p&gt;&lt;strong&gt;TL;DR&lt;/strong&gt;: Calculating the right ascension (i.e. converting from ecliptic to equatorial coordinates) requires some spherical trigonometry, but fortunately there are internet tools that will do this, such as &lt;a href=&quot;https://lambda.gsfc.nasa.gov/toolbox/tb_coordconv.cfm&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. The resulting RA coordinates are:&#xA;Aries 0h 0m 0s; Taurus 1h 51m 39s; Gemini 3h 51m 16s; Cancer 6h 0m 0s; Leo 8h 8m 44s; Virgo 10h 8m 21s; Libra 12h 0m 0s; Scorpio 13h 51m 39s; Sagittarius 15h 51m 16s; Capricorn 18h 0m 0s; Aquarius 20h 8m 44s; Pisces 22h 8m 21s.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation and background&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since traditional (Western) astrology uses the same &lt;a href=&quot;https://en.wikipedia.org/wiki/Tropical_year&quot; rel=&quot;noreferrer&quot;&gt;tropical year&lt;/a&gt; as astronomical celestial coordinates, the tropical zodiac commences with Aries at 0˚ longitude on the ecliptic, and each subsequent zodiac sign (NB &lt;em&gt;not&lt;/em&gt; the constellation, which is different) commences exactly 30˚ further along the ecliptic. So, the zodiac division of Taurus is from 30.00˚ to 59.99˚, Gemini from 60.00˚ to 89.99˚, and so on. It's then just a matter of converting ecliptic longitude to celestial coordinates (right ascension).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further useful information on the relationship between the ecliptic and right ascension can be found in &lt;a href=&quot;https://astronomy.stackexchange.com/a/25570/14094&quot;&gt;Mike G's answer&lt;/a&gt; to &quot;Effect of the obliquity of the ecliptic / tilt of the Earth on the equation of time&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both the traditional tropical zodiac and modern astronomical coordinates are based on the northern hemisphere vernal equinox. This starting point, also known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/First_Point_of_Aries&quot; rel=&quot;noreferrer&quot;&gt;First Point of Aries&lt;/a&gt;, is one of the two points at which the celestial equator crosses the ecliptic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Hipparchus defined the First Point of Aries in 130 BCE, it actually aligned with the border between the constellations of Pisces and Aries. However, due to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Precession_(astronomy)&quot; rel=&quot;noreferrer&quot;&gt;precession of the equinoxes&lt;/a&gt;, the tropical zodiac's starting point has a retrograde movement along the ecliptic at the rate of one degree every 71.5 years. As a result, the First Point of Aries is currently deep in Pisces! Using the current official IAU constellation boundaries, the point of 0h RA will cross into Aquarius in the year 2597, so we have quite some time to wait for the &quot;Age of Aquarius&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that not all astrological traditions are based on the tropical year. Hindu astrology, for instance, is based on the sidereal zodiac - i.e. it aligns permanently with the visible stars, regardless of the Earth's axial precession. Some modern Western astrologers also use a sidereal zodiac, in which sidereal Aries currently begins on 15 April.&lt;/p&gt;&#xA;" OwnerUserId="14094" LastEditorUserId="14094" LastEditDate="2018-08-28T05:30:11.273" LastActivityDate="2018-08-28T05:30:11.273" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27489" PostTypeId="2" ParentId="27486" CreationDate="2018-08-28T12:53:04.843" Score="1" Body="&lt;p&gt;There was japanese research into nebula winters, to find correlation of weather records with starbusts, dark bok clouds and supernovae:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/M2JMp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/M2JMp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/S1342937X13001597&quot; rel=&quot;nofollow noreferrer&quot;&gt;research reference&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A global winter caused by a proto-stellar clouds, called bok globules, is possible too, because there may be 100 times more of them than we can see, they are abot 1ly diameter and 2-50 solar masses, completely dark. Perhaps they would dim the sun by at least 10-40 percent. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/F1myA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F1myA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also find a document that regroups current research for Nebula Winters, corresponding to image 1. the geology evidence is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bodiselitsch et al. (2005) found iridium anomalies at the base of all cap carbonates both after&#xA;the Marinoan and Sturtian glaciations; iridium and other platinum-group elements are typical&#xA;proxies for extraterrestrial (including exosolar) materials and are much more abundant than in&#xA;the Earth’s upper mantle and crust. These authors interpreted the anomalies to be the product of&#xA;the rapid accumulation of extraterrestrial material once trapped in the surface ice and later&#xA;transported to the ocean floor via the melting of the ice at the end of the glaciation.&#xA;Nevertheless, the excursion of iridium may also reflect the enhanced flux of extraterrestrial&#xA;material due to an encounter with a nebula.&#xA;Moreover, 244Pu is expected to remain in the Fe/Mn crusts of interest. Fields et al. (2005)&#xA;estimated the 244Pu production by a supernova explosion to approximately 0.9 ×10-7&#xA;solar&#xA;masses. The expected flux of the 244Pu at the Earth is approximately 10-16 kg m-2&#xA;yr-1&#xA;. A&#xA;significant amount (0.1 Bq m-2&#xA;) of the accreted 244Pu remains in the sedimentary rocks, even&#xA;considering the half-life of 244Pu (80 Myr).&#xA;The existence of abundant 100-μm spherical iron balls in pelagic sediments is also possible&#xA;direct evidence. For example, Miono et al. (1993) argued that the spherules accumulate in&#xA;Paleozoic-Mesozoic bedded cherts. These spherules are generally believed to originate from &#xA;25&#xA;micro-meteorites. The nebula winter model can be tested by similar but further advanced&#xA;investigations, including the analysis of the abundance of the isotopes and platinum family&#xA;elements of the minerals separated from spherules in the pelagic sediments, detailing their&#xA;exosolar nature. In addition, mineralogical investigations on the ages, the abundances of&#xA;platinum family elements, and the isotope ratios of the minerals of the exosolar grains in the&#xA;pelagic sediments at that time, will clarify the influence of the galactic environment on the&#xA;natural history of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory is supported by time correlation, which is fairly close to starburts, which occur every 100Ma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specific evidence of historical astronomical events require the presence of supernovae elements, i.e. Iridium, in rocks from snowball times. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other reasons for the Cambrian explosion are the edvent of bilateral animals, 600Ma, the last snowball event from 630Ma, PH and oxygen instability in the ocean, algea blooms. I found this pic which gives an idea of the oxygen concentrations from the cambrian: &lt;a href=&quot;https://i.stack.imgur.com/tB1TL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tB1TL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2018-10-02T08:41:58.527" LastActivityDate="2018-10-02T08:41:58.527" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27490" PostTypeId="1" CreationDate="2018-08-28T23:30:31.553" Score="4" ViewCount="131" Body="&lt;p&gt;I’ve never purchased a telescope and I was considering buying one. However, I wasn’t sure what to look for. I love to tinker and so a telescope that I can program with libraries such as SPICE would be great. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="23722" LastActivityDate="2018-08-29T02:57:00.187" Title="What are some good telescopes for a programmer?" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2018-08-29T15:03:06.453" ContentLicense="CC BY-SA 4.0" />
  <row Id="27491" PostTypeId="2" ParentId="27490" CreationDate="2018-08-29T02:41:50.327" Score="2" Body="&lt;p&gt;One project you could work on is an Arduino-based Go-To telescope system. You could use, in theory, any scope that can accommodate the addition for this purpose. Also, it's my understanding that any Go-To system that can be connected to a computer can receive custom scripts you could write. At least, I know it's true for the NexStar system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, it's best to start with a simple Dobsonian telescope from a reputable dealer, without any programmable components. It will force you to learn the sky well before you start observing. And it will serve as an entry-level introduction to the world of telescopes.&lt;/p&gt;&#xA;" OwnerUserId="23940" LastActivityDate="2018-08-29T02:41:50.327" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27492" PostTypeId="2" ParentId="27490" CreationDate="2018-08-29T02:57:00.187" Score="2" Body="&lt;p&gt;What do you want to be able to see with it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The previous answer recommending a simple Dobsonian is my usual recommendation.  An 8&quot; or larger Dob is very easy to use and offers the best price to aperture ratio.  And aperture is the number one consideration in an telescope for visual observing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you start looking at computerized mounts, you find that you lose out on capability and quality unless you're willing to pay a lot more.  While you can get a decent 8&quot; Dob for under $500 USD, any computerized (what we call GoTo) telescope in that price range will give you about half the aperture, and all around mediocre to poor quality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before buying any telescope, my first recommendation is always to find your local astronomy club and join up.  In the US, memberships typically cost $50 per month or less (usually less).  By going to star parties, you can get a look at a wide range of telescopes and learn their capabilities before you buy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If and when you do get a GoTo telescope, I'd look into ASCOM (AStronomy Common Object  Model).  The ASCOM group is an open group of developers that work to provide a common interface for telescope control.  Each vendor has their own standards, but ASCOM seeks to bridge the gap.  There are drivers and plugins available to handle most of the popular vendors and manufacturers' hardware.  This goes not only for the telescope mounts themselves, but for some imaging systems, observatory control systems, planetarium software, motorized focusers, etc...  You can learn more about it &lt;a href=&quot;https://ascom-standards.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10546" LastActivityDate="2018-08-29T02:57:00.187" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27493" PostTypeId="2" ParentId="27299" CreationDate="2018-08-29T03:29:08.600" Score="7" Body="&lt;p&gt;I want to clarify a part of this question that some people may not understand. If dark matter literally were &lt;em&gt;only&lt;/em&gt; affected by gravity, then you would not expect to see it collecting at the center of neutron stars. As a dark matter falls towards the center of a neutron star, it picks up speed until it passes on through the neutron star and starts to slow down. But when it leaves the vicinity of the neutron star it will have the same speed that it had when entering the vicinity. In order to collect dark matter, the neutron star has to slow it down somehow. This is what Anders Sandberg meant when he mentioned the interaction cross-section of dark matter. That refers to the probability of interactions that might slow down the dark matter particles enough for them to be trapped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it a little differently, suppose a particle were far enough away from the neutron star so that the neutron star's gravity is can basically be ignored. Imagine that the particle is drifting towards the neutron star, so that eventually it will pass through it. Then, by the time the particle reaches the neutron star, it is mathematically guaranteed to be above escape velocity. The definition of escape velocity is the velocity that a particle will have if it falls from infinitely far away. The neutron star cannot trap anything because anything that drifts by is guaranteed to pick up precisely enough speed falling in so that, by the time it gets back out to the same distance it was before, it's moving away at the same speed it was moving in before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be fair, there is one caveat here. If the particles had significant gravity, so they could strongly influence each other, then they could interact in such a way that one particle flies out even faster, and the other one is trapped in orbit around (maybe also passing through) the neutron star. Some of Jupiter's moons may have been captured that way. But dark matter particles are thought to have negligible mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not saying anything different from Anders Sandberg here, but I just want to emphasize the importance of his statement that &quot;most particles will just swoop on past.&quot; &lt;/p&gt;&#xA;" OwnerUserId="7752" LastEditorUserId="7752" LastEditDate="2019-02-01T04:29:38.207" LastActivityDate="2019-02-01T04:29:38.207" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27494" PostTypeId="2" ParentId="13711" CreationDate="2018-08-29T09:10:07.683" Score="10" Body="&lt;p&gt;There's quite a simple answer to this that no-one seems to have focused on; The photos showing a starwheel effect are taken by pointing the camera at the celestial North Pole (roughly in the direction of Polaris in &lt;em&gt;Ursa Minor&lt;/em&gt;). The stars in that part of the sky appear to rotate around the pole and never set (viewed from Texas). This makes a good photo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun and Moon are never anywhere near the pole. They are found on the &lt;em&gt;ecliptic&lt;/em&gt;, which is roughly above the Earth's equator. Viewed from Texas, that's south of you, so as the Earth turns, the Sun and Moon follow an arc across the sky that reaches its highest point due South.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the image searches, there are a few photos showing Sun/Moon like paths (i.e., low arcs) for stars as well. Those are stars near the ecliptic. The photos aren't so striking as the circumpolar stars, which is why there are more of those.&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="2938" LastEditDate="2018-08-30T06:10:42.547" LastActivityDate="2018-08-30T06:10:42.547" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27495" PostTypeId="2" ParentId="27299" CreationDate="2018-08-29T13:30:12.930" Score="-12" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why aren't neutron stars full of dark matter?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because dark matter doesn't consist of particles. There's something of a myth  that it does, which I think comes from particle physicists who have never actually read Einstein's original material. I also think science is something of a competitive business, and there's a tendency for advocates to promote their own theory (eg WIMPs) and claim that a competitor theory (eg MOND) is flawed.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dark matter interacts with the gravitational force right? Well, unlike black holes, neutron stars are actually visible, and they're an enormous gravitational sink, so dark matter should collect to them. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Remember that we have good scientific evidence for flat galactic rotation curves and other phenomena. These suggest that either a) there's some unseen &quot;dark matter&quot; around somewhere, or b) that gravity doesn't work quite the way that people think. However the evidence does not actually say dark matter is made out of particles and falls down.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But if all that is true, which it seems to be, why haven't astronomers detected or used neutron stars to detect dark matter?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because &lt;em&gt;we do not live in some Chicken Little world where the sky is falling in&lt;/em&gt;. I'm referring to &lt;a href=&quot;https://en.wikipedia.org/w/index.php?title=Gullstrand%E2%80%93Painlev%C3%A9_coordinates&amp;amp;oldid=850190402#History&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gullstrand-Painlevé coordinates&lt;/a&gt; which model a gravitational field as a place where space is falling down. Einstein rejected the idea, but some contemporary physicists take it seriously, see &lt;a href=&quot;http://jila.colorado.edu/~ajsh/insidebh/waterfall.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eYY9M.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eYY9M.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Image credit Andrew Hamilton&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this relevant? Because in his 1916 &lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol6-trans/197?highlightText=gravitatively&amp;amp;backToSearch=https%3A%2F%2Feinsteinpapers.press.princeton.edu%2F~searchResults%3FsearchMode%3Dapp%26searchParam-titles%3Dfull%26searchParam-versions%3Dall%26searchParam-years%3D%257B%2522operator%2522%253A%2522%2522%257D%26searchParam-sort%3Ddate%26searchParam-terms%3Dgravitatively%26searchParam-start%3D0%26searchParam-page%3D20&quot; rel=&quot;nofollow noreferrer&quot;&gt;Foundation of General Relativity&lt;/a&gt; Einstein said &lt;em&gt;“the energy of the gravitational field shall act gravitatively in the same way as any other kind of energy”&lt;/em&gt;. This is spatial energy, and it isn't made of particles. The energy density of space near the Earth is greater than the energy density of space further away from the Earth. Because of this, there's a gravitational effect. This is why &quot;&lt;a href=&quot;https://www.google.co.uk/search?q=gravity%20gravitates&amp;amp;source=lnt&amp;amp;tbs=li:1&amp;amp;sa=X&amp;amp;ved=0ahUKEwivkqDgp5LdAhXILMAKHfyBD6UQpwUIHw&amp;amp;biw=1280&amp;amp;bih=866&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravity gravitates&lt;/a&gt;&quot;. Einstein also described a gravitational field as a place where space is &quot;&lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol7-trans/192?highlightText=%22neither%20homogeneous%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;neither homogeneous nor isotropic&lt;/a&gt;&quot;. So dark matter might simply be inhomogeneous space. Don't forget that as per the raisin-cake analogy, the space between the galaxies expands whilst the space between the galaxies does not. Conservation of energy tells me that this will surely lead to an inhomogeneous spatial energy density. And that an older galaxy will be surrounded by a bigger/steeper halo of inhomogeneous space than a younger galaxy, so it will look like there's more dark matter present. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What Einstein said means that there's &quot;dark matter&quot; of sorts in the room you're in, right in front of your face. Only it isn't made of particles, and it isn't falling down. Instead it's made of space. Don't forget that &lt;em&gt;space is dark, and there's a lot of about&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="9772" LastEditorUserId="9772" LastEditDate="2018-08-29T13:48:37.603" LastActivityDate="2018-08-29T13:48:37.603" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="27496" PostTypeId="1" CreationDate="2018-08-29T21:05:28.787" Score="1" ViewCount="246" Body="&lt;p&gt;It's somewhat of a no-brainer the &lt;a href=&quot;https://en.wikipedia.org/wiki/Revised_Julian_calendar&quot; rel=&quot;nofollow noreferrer&quot;&gt;revised Julian calendar&lt;/a&gt; (attibuted to Milankovitch) is more accurate than the Gregorian one just because it started with a more accurate estimate of the duration of the tropical year. From (the English translation) of &lt;a href=&quot;http://myweb.ecu.edu/mccartyr/orthodox-reform.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Milankovitch's 1923 paper&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A complete adoption of the Gregorian calendar was, however, not advisable either from a religious or scientific standpoint because the astronomical data concerning the length of the tropical year which lay at the base of the Gregorian reform are now replaced by others.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fact one (unpublished?) &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.527.7023&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper by Lichtenberg and Richter (1998)&lt;/a&gt; explains the difference between the two calendars as a better continued fraction approximation of the number 0.781 = (36 525 − 36 524.219), i.e. the of difference between the Julian year and the tropical year (as approximately measured more recently), with the Gregorian rational approximation of that 0.781 being 3/4 and the Milankovitch one being 7/9. (It's actually a bit more complicated as there's a second parameter to approximate, but I don't care to elaborate on the other one here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what I want to ask: &lt;strong&gt;is this superiority of the Milankovitch calendar still true considering (more realistic) models in which the tropical year is not constant?&lt;/strong&gt; I actually break this down into two sub-questions...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could easily find &lt;a href=&quot;https://www.jstor.org/stable/24966597&quot; rel=&quot;nofollow noreferrer&quot;&gt;a comparison (Moyer, 1982)&lt;/a&gt; of the Julian and Gregorian calendars using the Newcomb model for the tropical year:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G6ZVR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G6ZVR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What would the Milankovitch calendar look like on that graph?&lt;/strong&gt; (Or e.g. using a more sophisticated &lt;a href=&quot;https://en.wikipedia.org/wiki/Tropical_year#Mean_tropical_year_current_value&quot; rel=&quot;nofollow noreferrer&quot;&gt;cubic model of the tropical year&lt;/a&gt;, what would they look like).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And further, considering what we don't know about the uncertainties of the tropical year, e.g. &lt;a href=&quot;https://en.wikipedia.org/wiki/Tropical_year#Calendar_year&quot; rel=&quot;nofollow noreferrer&quot;&gt;according to Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If society in the future still attaches importance to the synchronization between the civil calendar and the seasons, another reform of the calendar will eventually be necessary. According to Blackburn and Holford-Strevens (who used Newcomb's value for the tropical year) if the tropical year remained at its 1900 value of 365.24219878125 days the Gregorian calendar would be 3 days, 17 min, 33 s behind the Sun after 10,000 years. Aggravating this error, the length of the tropical year (measured in Terrestrial Time) is decreasing at a rate of approximately 0.53 s per century. Also, the mean solar day is getting longer at a rate of about 1.5 ms per century. These effects will cause the calendar to be nearly a day behind in 3200. The number of solar days in a &quot;tropical millennium&quot; is decreasing by about 0.06 per millennium (neglecting the oscillatory changes in the real length of the tropical year).[3] This means there should be fewer and fewer leap days as time goes on. A possible reform would be to omit the leap day in 3200, keep 3600 and 4000 as leap years, and thereafter make all centennial years common except 4500, 5000, 5500, 6000, etc. But the quantity &lt;a href=&quot;https://en.wikipedia.org/wiki/%CE%94T&quot; rel=&quot;nofollow noreferrer&quot;&gt;ΔT&lt;/a&gt; is not sufficiently predictable to form more precise proposals (Blackburn &amp;amp; Holford-Strevens 2003, p. 692).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is the error with which we can estimate the duration of future tropical years &lt;em&gt;greater&lt;/em&gt; than the difference between the Milankovitch and Gregorian calendars?&lt;/strong&gt; The year 2800 is leap in Gregorian but not in Milankovitch's calendar. That's &quot;pretty close&quot; to the suggestion in the last quote to make 3200 not leap, which suggests that the difference between the two calendars is greater than the error currently incurred in estimating tropical year duration. So it looks like one could still answer the question whether Milankovitch's calendar is provably better than the (unmodified) Gregorian one even considering estimation errors for the tropical year duration. Is that a correct assessment though?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast to that model cited in Wikipedia, &lt;a href=&quot;http://adsabs.harvard.edu/full/1991JRASC..85..121B&quot; rel=&quot;nofollow noreferrer&quot;&gt;Borkowski (1991)&lt;/a&gt; (using a quadratic model for the tropical year, which he finds is a good approximation of the VSOP92 cubic model for the next 4000 years) concludes that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;our Gregorian calendar will not deviate from the exact solar calendar by significantly more than 1 day for some 2000 years to come.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So it looks like that would make the year 2800 (non-leap in Milankovitch's) a rather premature correction...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moyer also contains this passage:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 1582 the&#xA;  length of the tropical year was nearly&#xA;  365.24222 days, which differed from the&#xA;  Gregorian year by only a little more&#xA;  than 24 seconds. If the length of the year&#xA;  itself were not gradually diminishing,&#xA;  the calendar under the plan devised by&#xA;  Lilius would retain the date of the vernal&#xA;  equinox at or about March 21 for&#xA;  more than 3,550 years.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Clavius himself estimated that the&#xA;  calendar would deviate by one day in&#xA;  the year 5084. Considering the decline&#xA;  in the length of the year, which was quite&#xA;  unknown to 16th-century astronomers,&#xA;  the calendar will remain accurate to&#xA;  within one mean solar day for a much&#xA;  shorter time, some 2,417 years, or until&#xA;  about the year 4317. (Extrapolations are&#xA;  based on the determination of the tropical&#xA;  year at epoch 1900 by the American&#xA;  astronomer Simon Newcomb; his figure&#xA;  was 365.24219879 mean solar days.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which again is concluding (using the linear model of Newcomb) that Gregorian  calendar is better than  Milankovitch claims.&lt;/p&gt;&#xA;" OwnerUserId="9700" LastEditorUserId="9700" LastEditDate="2018-08-29T23:58:59.680" LastActivityDate="2018-08-29T23:58:59.680" Title="Is the Milankovitch calendar provably better than the Gregorian one considering a non-constant tropical year?" Tags="&lt;celestial-mechanics&gt;&lt;time&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27497" PostTypeId="2" ParentId="26410" CreationDate="2018-08-29T22:59:02.170" Score="1" Body="&lt;p&gt;You can easily use something like Astropy. It already has the capacity to accept time and location as inputs for an Alt/Az calculation. You can just output a stream of data in a table with a varying variable: time or location. It's been well-tested and is the recommended solution for this kind of thing.&lt;/p&gt;&#xA;" OwnerUserId="23940" LastActivityDate="2018-08-29T22:59:02.170" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27500" PostTypeId="1" CreationDate="2018-08-30T02:26:09.360" Score="3" ViewCount="136" Body="&lt;p&gt;&lt;a href=&quot;http://astrogeo.oxfordjournals.org/content/43/2/2.9.1.full.pdf+html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Duncan Steel (2002)&lt;/a&gt; makes an interesting argument that the notion of year is (1) somewhat arbitrary and, more interestingly (2) one notion of year can increase while another decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His examples are &lt;a href=&quot;https://en.wikipedia.org/wiki/Tropical_year#18th_and_19th_century&quot; rel=&quot;nofollow noreferrer&quot;&gt;tropical year&lt;/a&gt; (which decreases over time, at least in the linear Newcomb model) and the vernal equinox year which seems to be currently increasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually looking at the data that his point is based on, namely &lt;a href=&quot;http://adsabs.harvard.edu/abs/1992JBAA..102...40M&quot; rel=&quot;nofollow noreferrer&quot;&gt;Meeus and Savoie (1992)&lt;/a&gt;, we can find another measure that is decreasing, namely the solstice year(s), either June or December; the autumnal equinox year is also decreasing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sVbrf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sVbrf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steel actually uses the fact that the  vernal equinox year is increasing as an argument that the Gregorian calendar is &quot;getting better&quot; in a sense that it's closer to hitting its target year length. (The opposite is the case if one consider say either of the solstice years... or the tropical year, which is the point that a lot of other people make, e.g. &lt;a href=&quot;https://www.jstor.org/stable/24966597?seq=1#page_scan_tab_contents&quot; rel=&quot;nofollow noreferrer&quot;&gt;Moyer (1982)&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I really want to ask here: assuming a simple linear decrease model for the tropical year (e.g. Newcomb), will the trends of the equinox and solstice years reverse at some point in the future?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And if it's not too much extra to ask:) What are the implications of the more complex models for the tropical year (e.g. cubic VSOP) on the evolution of the equinox and solstice years?&lt;/p&gt;&#xA;" OwnerUserId="9700" LastActivityDate="2020-05-20T01:58:02.463" Title="What's a good model for the vernal equinox year? Can it increase forever?" Tags="&lt;earth&gt;&lt;time&gt;&lt;equinox&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27502" PostTypeId="2" ParentId="27466" CreationDate="2018-08-30T13:06:40.700" Score="3" Body="&lt;p&gt;Mercury and Spin-orbital resonance is pretty straight forward.  Planets and Moons are gravitationally lumpy and large bodies are somewhat fluid, even rocky bodies.   Both aspects are prone to tidal forces and that can lead to spin-orbital resonance if the tidal forces are strong enough.   Mercury's somewhat high eccentric orbit balances out with a 3:2 spin-orbital resonance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resonance in 3 body orbital system is much more complicated and the math is above my pay-grade.  A proper answer to this question would have lots of math where orbital stability can be measured over long time periods and thousands of orbits.  Something precision is probably lost in an intuitive explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard, 2 paragraph answer on Kirkwood gaps is that, if the Kirkwood object is in resonance with Jupiter, then Jupiter gives it a tug whenever it passes closest to the planet and those tugs over time destabilize the 2:1, 5:2, 7:3, 3:1 and less populated 4:1 orbital resonances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.daviddarling.info/encyclopedia/K/Kirkwood_gaps.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;See this summary&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Kirkwood gaps are regions in the main asteroid belt that have been&#xA;  cleared of asteroids by the perturbing effects of Jupiter, named for&#xA;  Daniel Kirkwood who discovered them. The Kirkwood gaps are due to&#xA;  resonances with Jupiter's orbital period. For example, an asteroid&#xA;  with a semimajor axis of 3.3 AU makes two circuits around the Sun in&#xA;  the time it takes Jupiter to make one and is thus said to be in a 2:1&#xA;  resonance orbit with Jupiter. Once every two orbits, Jupiter and such&#xA;  an asteroid would be in the same relative positions, so that the&#xA;  asteroid would experience a force in a fixed direction. Repeated&#xA;  applications of this force would eventually change the semimajor axes&#xA;  of asteroids in such orbits, creating gaps at that distance. Gaps&#xA;  occur at 4:1, 7:2, 3:1, 5:2, 7:3, and 2:1 resonances&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is only a partial answer to your question, because 3:2 (&lt;a href=&quot;https://en.wikipedia.org/wiki/Hilda_asteroid&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hildas&lt;/a&gt;) and the much less populated 4:3 (&lt;a href=&quot;https://en.wikipedia.org/wiki/279_Thule&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thules&lt;/a&gt;) collect around the resonance, they don't move away from it, and as you noted, Pluto and other &lt;a href=&quot;https://en.wikipedia.org/wiki/Resonant_trans-Neptunian_object&quot; rel=&quot;nofollow noreferrer&quot;&gt;Resonant Trans Neptunian Objects&lt;/a&gt; (TNOs) or Plutionos collect around the resonance as well.   They also get tiny tugs from the planet each orbit, but those tugs tend to keep them around the resonance rather than pulled away from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a perhaps obvious sidebar, Resonant objects, while they orbit the sun, the need to be &quot;herded&quot; so to speak, by a single planet.   That's why there are relatively few resonant objects between the gas giants, and the vast, vast majority of them are between Mars and Jupiter, because Mars is small enough to not have much gravitational influence, or outside Neptune, because there are no large planetary bodies in the vicinity outside Neptune.   That's why Kirkwoods and Hildas and Resonant TNOs dominate the known orbital resonance objects.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wiktionary.org/wiki/cis-Jovian&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cis-Jovian&lt;/a&gt;, for inside Jupiter's orbit or Trans-Neptunian, for outside Neptune's.   In time, we might discover Planet Nine and it might have a collection of resonant Trans and Cis-Ninian objects, but that remains to be seen, and I only mention it because I wanted to be say the word Cis-Ninian.  Obviously that won't stick if/when the planet is found and given an official name, but for now, there are theoretical resonant Trans and Cis-Ninian objects.   In fact, it's that theoretical resonance that lead to the hypothesis that Planet Nine was out there and gave a clue where to look.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So . . . &lt;strong&gt;meat of the answer&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter's 3/2 Hilda's, for the most part, have their aphelions passing close to one of Jupiter's Lagrange points, L3, L4 or L5.  That's why the Hilda's form a kind of triangle.   They don't orbit in a triangle, they orbit in slightly perturbed elipses, but together, they form a triangle because there are 3 main categories of Hildas, those with aphelions in Jupiter's L3, L4 and L5.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RbwX0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RbwX0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is &lt;a href=&quot;https://en.wikipedia.org/wiki/Hilda_asteroid&quot; rel=&quot;nofollow noreferrer&quot;&gt;explained in more detail here, under Dynamics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Hildas taken together constitute a dynamic triangular figure with&#xA;  slightly convex sides and trimmed apices in the triangular libration&#xA;  points of Jupiter—the &quot;Hildas Triangle&quot;.&lt;a href=&quot;https://en.wikipedia.org/wiki/Hilda_asteroid&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt; The &quot;asteroidal stream&quot;&#xA;  within the sides of the triangle is about 1 AU wide, and in the apexes&#xA;  this value is 20-40% greater. Figure 1 shows the positions of the&#xA;  Hildas (black) against a background of all known asteroids (gray) up&#xA;  to Jupiter's orbit at January 1, 2005.&lt;a href=&quot;https://astronomy.stackexchange.com/questions/22304/when-and-where-are-pluto-and-neptune-closest-to-each-other?rq=1&quot;&gt;7&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Each of the Hilda objects moves along its own elliptic orbit. However,&#xA;  at any moment the Hildas together constitute this triangular&#xA;  configuration, and all the orbits together form a predictable ring.&#xA;  Figure 2 illustrates this with the Hildas positions (black) against a&#xA;  background of their orbits (gray). For the majority of these&#xA;  asteroids, their position in orbit may be arbitrary, except for the&#xA;  external parts of the apexes (the objects near aphelion) and the&#xA;  middles of the sides (the objects near perihelion). The Hildas&#xA;  Triangle has proven to be dynamically stable over a long time span.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This resonance where the Aphelion is about 60 degrees or 180 degrees away from Jupiter keeps the Hilda objects from getting too close to Jupiter, which assists their long term stability.  Too close for a resonant object is generally bad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pluto, may have a similar resonant pattern with Neptune (note, I need to check my sources, I misread the linked answer) - will try to follow up. &#xA; some details in the &lt;a href=&quot;https://astronomy.stackexchange.com/questions/22304/when-and-where-are-pluto-and-neptune-closest-to-each-other?rq=1&quot;&gt;accepted answer to this question&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Kirkwood objects are mostly too close to the sun to be gravitationally tied to one of the 3 Trojan points.   Passing close to Jupiter's orbit would make a 2:1 resonance impossible.   As a result, there isn't a stable area for them to settle in, and they are pushed by Jupiter away from resonance, unlike the 3:2 resonant Hilda's which orbit through one of the Trojan regions.  That's the best I can work this out at least.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure I explained that as well as I should have, but that's the gist of it, as I understand it.  The requirements for resonance are very specific and the Kirkwood objects don't have a stable zone to gather around, so the periodic tugs of Jupiter sheppard them between the resonances rather than hold them to the resonances.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-09-03T02:04:47.643" LastActivityDate="2018-09-03T02:04:47.643" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27503" PostTypeId="2" ParentId="22522" CreationDate="2018-08-30T15:09:44.207" Score="1" Body="&lt;p&gt;In sailing, you might see another boat on a heading that will cross yours (i.e., it is coming in from port or starboard). You obviously want to know if there's any risk of collision. The trick is that you take a bearing on the boat and note it. Then you wait a while (long enough for you to close the range but short enough that you're still quite far apart) and take its bearing again.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If it has decreased, the boat will pass safely in front of you&lt;/li&gt;&#xA;&lt;li&gt;If it has increased, you will pass it front of it and it will be safely astern&lt;/li&gt;&#xA;&lt;li&gt;If it is the &lt;strong&gt;same&lt;/strong&gt;, you are on a collision course! So make an immediate turn...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This works for objects moving at a constant speed across a flat surface. Not sure if it also applies to accelerating objects on curved paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it does turn out to be true, the object would get steadily brighter night by night, while holding the same position against the fixed stars.&lt;/p&gt;&#xA;" OwnerUserId="2938" LastActivityDate="2018-08-30T15:09:44.207" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27504" PostTypeId="1" CreationDate="2018-08-30T19:02:53.267" Score="3" ViewCount="110" Body="&lt;p&gt;Good People,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am writing a short story, and I wanted to include a section wherein a family is told that an asteroid is going to strike their home. The thing is, I would like to set the story in 1987, but I'm not sure if there was the capability of that kind of prediction back then. To be honest, I'm not even sure that kind of accuracy is possible now! But I will allow myself some creative freedom in that area, but I don't want to stretch it so much that someone would keep thinking to themselves, &quot;Really, in 1987 they could predict &lt;em&gt;that&lt;/em&gt;?&quot; Would such a thing have been possible then? If not then, what about sometime in the 90s?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I have your eyes, is there a specific size that an asteroid would have to be in order to just destroy one home (or a couple on the side of the home).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for any insight and info! I appreciate the resources here. I did some digging around before asking this question: nothing seemed to be particular to my frame of reference. If it exists, please let me know.&lt;/p&gt;&#xA;" OwnerUserId="24165" LastActivityDate="2018-08-31T09:34:49.200" Title="Accuracy of predicting an asteroid strike in 1987" Tags="&lt;asteroids&gt;&lt;impact-probability&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27506" PostTypeId="2" ParentId="27504" CreationDate="2018-08-30T20:30:48.723" Score="5" Body="&lt;p&gt;There are problems with this:&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;What happens when an asteroid hits the atmosphere&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The asteroid will be travelling fast, more than 10 &lt;em&gt;km per second&lt;/em&gt;. In front of it the atmosphere will be compressed and heat up, this causes the &quot;meteor&quot; (a streak of light in the sky). It also causes puts a lot of stress on the asteroid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the asteroid is very small (sand grain to pebble sized), it gets destroyed completely.  If it is larger it can explosively break up. But several chunks can continue at terminal velocity (ie no faster than a stone dropped from a high building) and don't form craters. These chunks can later be picked up as meteorites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the body is very large it can survive the atmosphere, hit the Earth at hypervelocity (several km/s) and form a larger crater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the body to survive, but only form a small crater you need either a very strong body (a solid iron asteroid) or an odd set of events, such as at the &lt;a href=&quot;https://journals.uair.arizona.edu/index.php/maps/article/view/15749/15737&quot; rel=&quot;noreferrer&quot;&gt;Carancas impact&lt;/a&gt;. In these cases a small crater or crater field can form. The size of the impactor is about 50cm.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Detecting and forecasting impact&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;A 50cm asteroid is not currently detectable until very close to the Earth.  We have seen asteroids that are a couple of metres across, but they need to be close to the Earth (and we need to be lucky). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 2008 we did get lucky and saw a &lt;a href=&quot;https://en.wikipedia.org/wiki/2008_&quot; rel=&quot;noreferrer&quot;&gt;small&lt;/a&gt; object before it hit Earth. In that case we had a few hours of observations and by about an hour before impact we were able to narrow the impact zone down to &quot;Northern Sudan&quot;.  Now (North) Sudan is huge, an area about 3 times the size of Texas. And in 2008 we already had an active Near Earth Object search program, something that we did not have in 1987.  Even if the asteroid was noticed by lucky amateur, before the internet it would have been very hard to coordinate the observations needed even to predict &quot;Northern Sudan&quot;. This meteorite exploded at altitude and did not form a crater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To conclude, most meteorites will not take out a single house and leave neighbours standing.  Such meteorites that could are not currently detectable before impact.  Even with current technology and lots of luck we could not predict the impact location better than &quot;somewhere between New Mexico and Florida&quot; but not a single address.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other things that can destroy a house, perhaps a sinkhole? Perhaps these are more predictable (with luck) than an asteroid impact.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-08-30T20:51:44.567" LastActivityDate="2018-08-30T20:51:44.567" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27507" PostTypeId="1" CreationDate="2018-08-30T22:10:44.813" Score="3" ViewCount="93" Body="&lt;p&gt;What is the mass of the Kuiper belt? Obviously estimates will be highly uncertain, but for my application I would like to find a decent &lt;em&gt;current&lt;/em&gt; estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The papers I found so far: &lt;a href=&quot;http://adsabs.harvard.edu/abs/2001AJ....122.1051G&quot; rel=&quot;nofollow noreferrer&quot;&gt;(Gladman et al. 2001)&lt;/a&gt; estimates it to 0.1 Earth-masses, &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/321117/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;(Trujillo, Lewitt &amp;amp; Luu 2001)&lt;/a&gt; gets the mass of bodies between 100 km and 2000 km to 0.03 Earth-masses, &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/423697/pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;(Kenyon &amp;amp; Bromley 2004)&lt;/a&gt; estimate on the order of 0.5 Earth-masses (assuming a starting mass of 10 and then losing 95%), &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0004-637X/782/2/100&quot; rel=&quot;nofollow noreferrer&quot;&gt;(Fraser et al 2014)&lt;/a&gt; estimate just $\sim$0.01 Earth-masses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These papers are a bit old, and I suspect there have been more recent observations improving on the estimates. There is a fair number of papers discussing how to resolve this discrepancy between disk models predicting 10-30 Earth-masses and these low numbers, but I am more interested in empirical estimates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what is the current consensus on the Kuiper belt mass and its uncertainty?&lt;/p&gt;&#xA;" OwnerUserId="19816" LastEditorUserId="2531" LastEditDate="2018-09-02T08:09:29.580" LastActivityDate="2018-10-02T09:00:01.417" Title="Are there recent estimates of Kuiper belt mass?" Tags="&lt;kuiper-belt&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27508" PostTypeId="2" ParentId="27473" CreationDate="2018-08-30T23:37:56.380" Score="3" Body="&lt;p&gt;The only optical benefit from 2&quot; eyepieces is that they allow for a wider field of view at low magnifications, since you can fit more of the image formed by the scope into a 2&quot; barrel than you can with a 1.25&quot; one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At high magnifications, you're only looking at the central part of the image formed by the scope, and since that fits easily into a 1.25&quot; (or smaller) barrel, &lt;strong&gt;&lt;em&gt;there is no optical benefit&lt;/em&gt;&lt;/strong&gt; from having a 2&quot; barrel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;That's why many short focal length eyepieces only come in 1.25&quot; versions.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some (like most of Vixen's LVW range, or Baader's Hyperion range) have a&#xA;1.25&quot; barrel with a 2&quot; lower body, so you can use them without an adapter in either a 1.25&quot; or 2&quot; focuser. Some of the Televue Naglers use a dual barrel approach instead, with a longer 1.25&quot; barrel inside a shorter 2&quot; one. But optically, they're still 1.25&quot; designs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only time you need a 2&quot; barrel eyepiece is when the eyepiece is designed to show you more of the image formed by the scope than will fit in a 1.25&quot; barrel - for example, longer focal length ultrawide apparent field of view eyepieces, such as 82 degree AFOV eyepieces longer than around 17mm focal length, or most standard or wider view 40mm eyepieces - a standard Plossl 40mm in a 1.25&quot; barrel is limited by the barrel size, and will have a narrower AFOV than other (shorter) focal lengths as a result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for high magnification views of planets, you're just enlarging the middle of the image formed by the scope. And all the light rays that contribute to that part of the image fit easily inside a 1.25&quot; barrel; you're not losing anything by not having a 2&quot; barrel.&lt;/p&gt;&#xA;" OwnerUserId="17016" LastActivityDate="2018-08-30T23:37:56.380" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27509" PostTypeId="2" ParentId="27297" CreationDate="2018-08-31T01:02:53.587" Score="1" Body="&lt;p&gt;Most likely it was Mars, which was in the southern Indiana sky at the time you indicated.  I'm not aware of anything bright and orange in the northern sky that would have had a high altitude. &lt;/p&gt;&#xA;" OwnerUserId="23940" LastActivityDate="2018-08-31T01:02:53.587" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27512" PostTypeId="2" ParentId="27504" CreationDate="2018-08-31T09:34:49.200" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://www.nature.com/news/2009/090325/full/458401a.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Truth is stranger than fiction: an asteroid was observed less than 24 hours before striking Earth in 2008&lt;/a&gt;. JPL predicted the time of the collision to within 15 seconds, but the meteor exploded in the upper atmosphere, and the fragments were scattered over many kilometers of Sudanese desert.&lt;/p&gt;&#xA;" OwnerUserId="1498" LastActivityDate="2018-08-31T09:34:49.200" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27514" PostTypeId="2" ParentId="27507" CreationDate="2018-09-01T00:47:56.280" Score="1" Body="&lt;p&gt;Surprisingly, I also have not been able to find any generally recent (5 years or less ago) information about this topic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did find a research paper from 2007 (&lt;a href=&quot;https://web.archive.org/web/20070925203400/http://www.ifa.hawaii.edu/faculty/jewitt/papers/2006/DJ06.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt;). You may want to check out page 11 in Section 4.3. It states that the maximum mass of the Kuiper Belt is 0.1 Earth masses. They also talk about how it was estimated and such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't been able to find anything else besides Wikipedia articles (which you probably aren't looking for). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very strange how there hasn't been a lot of research on this. An intriguing topic, I must say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's all I could find... Hope that helps...?&lt;/p&gt;&#xA;" OwnerUserId="24048" LastActivityDate="2018-09-01T00:47:56.280" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27515" PostTypeId="1" CreationDate="2018-09-01T05:07:11.190" Score="0" ViewCount="34" Body="&lt;p&gt;I have two simulation snapshots in my hand at redshift 0 and 1. I know x, y, z coordinates of the galaxies in both redshifts 0 and 1, however there is no way for me to identify a single galaxy in both redshifts, i.e., there is no tracker/tracer for the galaxies that will trace the position of galaxies from redshift 1 to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have halo catalogue for both $z =0$ and $z=1$ and I am interested to trace a certain galaxy from z=0 to 1. I know that galaxy's exact x,y,z at $z =0$. My initial plan is to create a search with $\sqrt{x^2+y^2+z^2}\le r$ in kpc. So any galaxy at redshift 1 with coordinates less than r maybe a potential candidate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, is there any intelligible guess of what a potential value of r would be in kpc? In other words, I am interested to know with reduced Hubble parameter $h = 0.7$, how much far a galaxy might move from $z = 1$ to $z =0$?&lt;/p&gt;&#xA;" OwnerUserId="21282" LastEditorUserId="2365" LastEditDate="2018-09-01T05:54:49.167" LastActivityDate="2018-09-01T06:54:35.310" Title="How to calculate how much a galaxy moves from its coordinate at redshift 0 to redshift 1?" Tags="&lt;expansion&gt;&lt;hubble-constant&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27516" PostTypeId="2" ParentId="27515" CreationDate="2018-09-01T06:54:35.310" Score="1" Body="&lt;p&gt;The galaxies don't &quot;move&quot; (unless you have given them a peculiar velocity); space expands, such that the meaning of $x,y,z$ will change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hubble parameter is simply defined as the rate of change of the scale parameter divided by the scale parameter $\dot{a}/a$. But $z$ is also related to the Hubble parameter, so if your epochs are defined by $z$, the value of the Hubble parameter at any particular epoch doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am assuming that you have a coordinate system based on our Galaxy being at the &quot;origin&quot;. In which case, the &quot;initial&quot; position of a galaxy (at $z=1$) is related to where it is &quot;now&quot; at $z=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Redshift and scale factor are related by $ a = (1+z)^{-1}$. So taking your example of a galaxy at $x_0, y_0, z_0$ at $z=0$, then at $z=1$ all galaxies were closer together by a factor of two, and (ignoring any peculiar velocity) $x_1 = x_0/2$, $y_1 = y_0/2$ and $z_1 = z_0/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what you mean by &quot;halo catalogue&quot;. The simple relationship above completely breaks down on scales smaller than tens of Mpc, because galaxies then become influenced by their local gravitational potential. The relationship between galaxy separation and time then has little to do with cosmology and more to do with the dynamics of their local groups and clusters and no general answer can be given, other than to point out that $z=1$ corresponds to more than half the age of the universe - trying to predict the position of a galaxy within a cluster or group based on a snapshot 7 billion years ago is not possible!&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-09-01T06:54:35.310" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27518" PostTypeId="1" CreationDate="2018-09-01T10:27:20.163" Score="3" ViewCount="767" Body="&lt;p&gt;What factor determines the selection of star mass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And how we know that the number of massive stars is less than the number of low mass stars?&lt;/p&gt;&#xA;" OwnerUserId="23550" LastEditorUserId="2270" LastEditDate="2018-09-12T12:25:45.210" LastActivityDate="2018-09-12T12:25:45.210" Title="Why there are less massive stars than low mass stars?" Tags="&lt;star&gt;&lt;observable-universe&gt;&lt;stellar-astrophysics&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27519" PostTypeId="2" ParentId="27518" CreationDate="2018-09-01T15:51:51.093" Score="0" Body="&lt;p&gt;A combination of size and temperature of the protostar determines whether the actual star will be massive or small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can know that there are more small stars because we usually find the average amount of small stars and large stars in an area. As Rob Jeffries said, we usually just count them... Nothing too exciting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps!&lt;/p&gt;&#xA;" OwnerUserId="24048" LastActivityDate="2018-09-01T15:51:51.093" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27520" PostTypeId="2" ParentId="27518" CreationDate="2018-09-01T17:27:22.543" Score="2" Body="&lt;p&gt;A star has its life span: more massive, less life span.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are living at the point when the universe is about 13 billion years, when most of the massive stars have died.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, if you look back into the universe at younger ages, it is believed that massive stars are more common. For example, Population III stars, which are the most ancient type of stars with zero metallicity, it is believed that they are all very massive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the fraction of no. of stars with different initial mass actually depends on the age of the universe that you are considering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole study about the fraction is called `initial mass function.' There are many factors determining that, e.g., age of the universe as mentioned previous, metallicity, rotation, and binary.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-09-01T17:27:22.543" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27521" PostTypeId="1" AcceptedAnswerId="27542" CreationDate="2018-09-01T19:23:52.440" Score="10" ViewCount="168" Body="&lt;p&gt;The end of the relativistic jet that emerges from M87 looks curved like a hook. See here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WQvir.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WQvir.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Data from the VLA array show more of the jet: at the far end, the jet makes a complete 180 degree turn backwards toward the galaxy it emerges from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What forces the jet to make such a drastic turn around? &lt;/p&gt;&#xA;" OwnerUserId="2689" LastEditorUserId="23940" LastEditDate="2018-09-05T06:28:08.913" LastActivityDate="2018-09-05T06:28:08.913" Title="Why is M87's relativistic jet bent?" Tags="&lt;galaxy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27522" PostTypeId="1" AcceptedAnswerId="29694" CreationDate="2018-09-02T03:42:01.130" Score="5" ViewCount="618" Body="&lt;p&gt;How much magnetism is there between the Earth and Sun? Would the distance from the Sun differ for not the Earth iron core and how much?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastActivityDate="2021-03-06T20:30:24.810" Title="How much magnetic attraction is there between the Sun and Earth?" Tags="&lt;the-sun&gt;&lt;orbit&gt;&lt;earth&gt;&lt;magnetic-field&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27523" PostTypeId="1" AcceptedAnswerId="27524" CreationDate="2018-09-02T04:27:38.517" Score="3" ViewCount="115" Body="&lt;p&gt;I don’t think that black holes are really black as they are the only ones who have the capacity to bend light. At the centre of the black hole the laws of physics are totally defied, black holes appear black as space is black in colour. &lt;/p&gt;&#xA;" OwnerUserId="24177" LastEditorUserId="2365" LastEditDate="2018-09-02T06:07:36.773" LastActivityDate="2018-09-02T06:27:17.137" Title="Are black holes really black?" Tags="&lt;black-hole&gt;&lt;cosmology&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2018-09-04T08:22:15.967" ContentLicense="CC BY-SA 4.0" />
  <row Id="27524" PostTypeId="2" ParentId="27523" CreationDate="2018-09-02T06:27:17.137" Score="5" Body="&lt;p&gt;Black holes really look black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things look black because they don't reflect or emit light. A black hole is a region of spacetime which light can enter, but can't leave. Black holes don't reflect or emit light, so they look black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes are not the only things that bend light. Every object with mass bends light. Black holes bend light more than other objects because there is so much mass in such a small region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes don't have a surface, so they aren't black for the same reason that soot is black.  They are black because they don't reflect or emit light. Space also appears black because empty space does not refect or emit light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;At the centre&quot; is a difficult idea for black holes, because space and time are bent so much near the black hole that the centre becomes a &quot;when&quot; not a &quot;where&quot;.  The laws of physics aren't defied, but we don't know what the laws are at the time in the centre of a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Hawking radiation is predicted to be emitted by black holes. This makes black holes emit light, but for a large black hole Hawking radiation is very low energy and very dim.  It is theoretical, it has never been detected, but the theory is very strong and well supported)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Around the outside of the black hole matter may be falling in. This matter gets very very hot, and so shines very brightly. This means we can see where a black hole is, because we can see the matter that is falling into the black hole)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-09-02T06:27:17.137" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27527" PostTypeId="1" AcceptedAnswerId="27528" CreationDate="2018-09-02T14:18:07.173" Score="6" ViewCount="519" Body="&lt;p&gt;It is a well known fact that in stars, there exists a temperature gradiënt. The &#xA;observational reason is because we perceived spectral lines in the otherwise continuous spectrum of a star. If this wasn't the case, and thus if the temperature were uniform, then the absorption and the emission would cancel each other out and we would observe a net effect that results in a continuous spectrum. Now my question is: what is the physical explanation for this temperature gradient?&lt;/p&gt;&#xA;" OwnerUserId="24180" LastEditorUserId="1891" LastEditDate="2018-09-02T15:30:06.347" LastActivityDate="2018-09-03T04:22:39.033" Title="Temperature gradient in stars" Tags="&lt;star&gt;&lt;temperature&gt;&lt;spectra&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27528" PostTypeId="2" ParentId="27527" CreationDate="2018-09-02T15:23:02.037" Score="3" Body="&lt;p&gt;You're talking about the temperature gradients in the atmosphere of a star, that's where the emission and absorption lines are originating. This is a much more complicated matter than the star itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature of a star must increase from its surface downwards.&lt;br&gt;&#xA;This is a simple result of the fact that the stellar structure is given by pressure gradients balancing the force of gravity. We can express this via the law of hydrostatic equilibrium&lt;br&gt;&#xA;$$\frac{\partial P}{\partial r} = -g(r)\rho  $$&#xA;which is a pretty good approximation to reality in most cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure gradient then translates into a temperature gradient, as the pressure originates in the thermal motion of particles, formalized as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ideal_gas_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;law of ideal gases&lt;/a&gt;&lt;br&gt;&#xA;$$ P = \frac{\rho k_B T}{\mu} $$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So wherever there's non-zero gravity and no other force in a gaseous medium, there will be temperature gradients. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an atmosphere, things depend on how efficient the gas is at cooling. In very thin media, like in Earths or any planets / stars &lt;a href=&quot;https://en.wikipedia.org/wiki/Exosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;exosphere&lt;/a&gt; cooling is inefficient enough to make temperature gradients vanish.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2018-09-02T15:23:02.037" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27530" PostTypeId="1" AcceptedAnswerId="27533" CreationDate="2018-09-02T16:09:00.070" Score="3" ViewCount="135" Body="&lt;p&gt;Is there any software which will tell me accurately at what moment any part of the solar disk will become from a specific point on the ground &lt;strong&gt;taking account of terrain&lt;/strong&gt;. I feel like Google maps should have the terrain data I need, and the rest is geometry, but it looks like quite a bit of work, so I wondered if it already existed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, I am interested in when a viewer 1.5m above local ground level at latitude 46.497300, longitude 7.714651 will first see direct sunlight on the morning of 1 August 2023, taking account of the surrounding mountains. Ideally I'd like to know how that varies with height above the ground (1-2m) and for locations within 10-20m around that location (in other words, how and when the shadow of the mountains moves off that area).&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="21928" LastEditDate="2018-09-02T16:26:20.137" LastActivityDate="2018-09-03T05:30:10.643" Title="Predicting accurate time of local sunrise in mountainous area?" Tags="&lt;ephemeris&gt;&lt;sunrise&gt;&lt;prediction&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27533" PostTypeId="2" ParentId="27530" CreationDate="2018-09-03T03:07:38.620" Score="5" Body="&lt;p&gt;The &lt;a href=&quot;https://www.peakfinder.org/&quot; rel=&quot;noreferrer&quot;&gt;PeakFinder&lt;/a&gt; app&lt;sup&gt;1&lt;/sup&gt; can do most of what you're looking for.  It's available for Android and IOS as well as through a web browser.  The app shows the paths across the sky for both the sun and moon&lt;sup&gt;2&lt;/sup&gt; and displays rise/set times taking into account the local topography and viewpoint elevation&lt;sup&gt;3&lt;/sup&gt;.  At least the Android version works without a network connection (nice for hiking).  IIRC, it uses SRTM data&lt;sup&gt;4&lt;/sup&gt; for elevations so it's not super-accurate but does a good job over all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I plugged in the location, elevation (ground + 1m), and date you gave and got sunrise at 05:09 local time, reaching the ridgeline at 08:37 local time.  Here's a &lt;a href=&quot;https://www.peakfinder.org/?lat=46.4973&amp;amp;lng=7.7147&amp;amp;azi=79&amp;amp;zoom=4&amp;amp;ele=591&amp;amp;cfg=sm&amp;amp;date=2023-08-01T06:40Z&quot; rel=&quot;noreferrer&quot;&gt;link&lt;/a&gt; to the page; screenshot is below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; I'm not affiliated with PeakFinder, just a satisfied user.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2&lt;/sup&gt; If using a web browser, click on the &quot;sun&quot; and &quot;moon&quot; icons on the page's toolbar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;3&lt;/sup&gt; It also shows what the surrounding terrain looks like and the names of visible peaks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;4&lt;/sup&gt; As does OpenStreetMap, IIRC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/noUAb.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/noUAb.png&quot; alt=&quot;Cropped screenshot of PeakFinder site&quot;&gt;&lt;/a&gt;&#xA;Source: Cropped screenshot of PeakFinder site&lt;/p&gt;&#xA;" OwnerUserId="24184" LastEditorUserId="24184" LastEditDate="2018-09-03T05:30:10.643" LastActivityDate="2018-09-03T05:30:10.643" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27534" PostTypeId="2" ParentId="27527" CreationDate="2018-09-03T04:22:39.033" Score="2" Body="&lt;p&gt;Force balance only tells you that the &lt;em&gt;pressure&lt;/em&gt; must decrease outward, the cause of the decreasing &lt;em&gt;temperature&lt;/em&gt; is the nature of heat transport and the requirement that it increase entropy.  Hence heat is always transported from higher T to lower T, and a star must transport heat &lt;em&gt;outward&lt;/em&gt; because its surface is losing heat to the coldness of space.  That requires the temperature must drop as you go out, as long as you are just passing the same heat from layer to layer and the thermodynamics of heat transport rules the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was pointed out that the low-density atmospheres of stars can have their temperature rise with height, as the Earth's stratosphere also does, yet the net energy transport is still outward.  To avoid entropy decrease, this requires that more heat be deposited in the hot layers than is extracted from the cooler ones below, so the heat dumped high up in the atmosphere has to come from somewhere else, other than the cool layers.  That's what can't happen deep in the star-- the heat gets no &quot;Free ride&quot; to pass certain layers, it is passed from layer to layer, the same heat.  But when there are ways to dump heat into higher layers without extracting it from cooler ones below, then the temperature can rise.  (In the Earth's stratosphere, the extra heat comes from sunlight absorbed by ozone, and in the Sun's chromosphere, the extra heat comes from magnetic fields and plasma motions, which pass through the lower layers without being absorbed.)&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-09-03T04:22:39.033" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27535" PostTypeId="2" ParentId="10297" CreationDate="2018-09-03T10:47:44.387" Score="0" Body="&lt;p&gt;In 2007, Nobel laureate Steven Weinberg suggested that if the multiverse existed, &quot;the hope of finding a rational explanation for the precise values of quark masses and other constants of the standard model that we observe in our Big Bang is doomed, for their values would be an accident of the particular part of the multiverse in which we live.&lt;/p&gt;&#xA;" OwnerUserId="24177" LastActivityDate="2018-09-03T10:47:44.387" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27536" PostTypeId="1" CreationDate="2018-09-03T11:06:02.213" Score="1" ViewCount="30" Body="&lt;p&gt;Is it really possible for bending of space time to travel to another distant place in universe. Wormhole has two mouths which one is a black hole and another is a white hole which could let us out. Which is commonly called as Einstein-Rosen bridge, is there any pictorial proof of wormholes other than the theoretical proof by Einstein. Is there a possibility of having exotic matter which helps the wormhole to keep open up always??&lt;/p&gt;&#xA;" OwnerUserId="24177" LastActivityDate="2018-09-03T11:06:02.213" Title="Existence of wormholes and exotic matter" Tags="&lt;general-relativity&gt;&lt;wormhole&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2018-09-04T09:12:34.987" ContentLicense="CC BY-SA 4.0" />
  <row Id="27538" PostTypeId="2" ParentId="10538" CreationDate="2018-09-03T13:59:37.400" Score="2" Body="&lt;p&gt;The expansion of spacetime does not involve the creation of anything. What is happening in the process of expansion is the enlargement of the metric itself of the spacetime. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine this as any arbitrary coordinate system with geometric points a constant distance apart. As spacetime expands, the distance between these geometric points expand for the same distance for every point. Basically every region of spacetime is expanding away from every other region at the same rate and distance no matter where you happen to be situated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why thinking of new matter being &quot;created&quot; as the method of expansion is not just incorrect, but not helpful at all at it introduces the idea that spacetime has a concrete edge where new stuff is always coming into existence that could conceivably be thought of as a certain distance from us. It isn't so. Spacetime expansion is happening everywhere as the distances between every point on our coordinate system expand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At our scale (and indeed on any scale smaller than galaxy clusters) we don't see this isotropic expansion as gravitational forces between objects overcome following this coordinate expansion. Remember that Andromeda is currently heading towards us as a reminder. Galaxy clusters are following the expansion and are distancing themselves from each other presently. Care should be taken here though. It isn't the distances between the galaxies in the cluster that are expanding, it is the clusters themselves distancing from each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the 'big bang', the distances between coordinates was zero. This is where the idea of the initial singularity comes from, as the density of spacetime instantaneously becomes infinite with the claim of no geometric space existing at all (look up the definition of singularity for further clarification). Most will currently tell you that the notion of the singularity existing is doubtful and that General Relativity's descriptions of spacetime don't adhere to what is happening on the quantum level, but that is outside the scope of this question. &lt;/p&gt;&#xA;" OwnerUserId="23212" LastActivityDate="2018-09-03T13:59:37.400" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27539" PostTypeId="2" ParentId="9921" CreationDate="2018-09-03T14:26:01.867" Score="0" Body="&lt;p&gt;The universe isn’t expanding into anything. It is just the distance between the galaxies is increasing continuously, but the galaxies are relatively still but space between them is moving. However the universe is infinitely big enough to begin with and remain infinite, it’s size doesn’t change. For e.g. consider 1,2,3... upto infinity. If you multiply it with 2, then 2,4,6.... upto infinity. Here we can say that distance between adjacent has increased, but total extent of numbers remain the same.&lt;/p&gt;&#xA;" OwnerUserId="24177" LastActivityDate="2018-09-03T14:26:01.867" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27541" PostTypeId="1" AcceptedAnswerId="27543" CreationDate="2018-09-03T17:27:14.203" Score="2" ViewCount="241" Body="&lt;p&gt;I’ve heard that black hole will be rotating continuously which stores a huge amount of energy in its rotation. Will this energy be accessible, since the rotation is imposed on the space outside the hole.&lt;/p&gt;&#xA;" OwnerUserId="24177" LastActivityDate="2018-09-03T20:50:59.210" Title="Can black holes be used as an energy source" Tags="&lt;black-hole&gt;&lt;cosmology&gt;&lt;general-relativity&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27542" PostTypeId="2" ParentId="27521" CreationDate="2018-09-03T17:58:19.013" Score="7" Body="&lt;p&gt;I think the most common explanation for this kind of structure is ram pressure from the galaxy's motion through the surrounding &lt;a href=&quot;https://en.wikipedia.org/wiki/Intracluster_medium&quot; rel=&quot;noreferrer&quot;&gt;intracluster medium (ICM)&lt;/a&gt;. In the rest frame of the galaxy and jet, the ICM is blowing past with a velocity of several hundred km/sec, and this is enough to divert the flow of the jet. In this picture, the jet isn't moving back &lt;em&gt;toward&lt;/em&gt; the galaxy; this is a projection effect (the inner jet is thought to be pointed toward us, only 30 degrees or so away from our line-of-sight to the galaxy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second factor may be that the point where the jet seems to bend is where its smooth, relativistic flow breaks up due to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kelvin%E2%80%93Helmholtz_instability&quot; rel=&quot;noreferrer&quot;&gt;Kelvin-Helmholtz instability&lt;/a&gt;. The initial small kink visible at the right-hand edge of the HST-WFPC2 image (upper right) is probably a sign of the instability starting to disrupt the jet. Once the smooth flow of the jet breaks up, it is likely to slow down, making it more easily diverted by the ram pressure of the ICM.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2018-09-03T17:58:19.013" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27543" PostTypeId="2" ParentId="27541" CreationDate="2018-09-03T20:11:30.323" Score="6" Body="&lt;p&gt;What you are thinking of is called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Penrose_process&quot; rel=&quot;noreferrer&quot;&gt;Penrose process&lt;/a&gt; - a means of extracting rotational energy from a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic idea is that you allow a composite body to fall towards a rotating black hole. Once it reaches the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ergosphere&quot; rel=&quot;noreferrer&quot;&gt;ergosphere&lt;/a&gt; - that region, outside the event horizon, where the &quot;frame dragging&quot; due to the rotating spacetime is so severe that no body can remain stationary from the point of view of a distant observer - the body should split. One half falls into the black hole and the other orbits and flies out again. If things are arranged correctly then the returning part of the body can have a kinetic energy that exceeds the rest mass energy of the half that has fallen into the black hole. The energy comes at the expense of slowing the rotation of the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The Penrose Process (from &lt;a href=&quot;http://inspirehep.net/record/1341250/plots&quot; rel=&quot;noreferrer&quot;&gt;Brito et al. 2015&lt;/a&gt;)&lt;/em&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/o4gte.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o4gte.png&quot; alt=&quot;Penrose Process&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the rotating black hole has an accretion disc (as many do if they are in binary systems - that's how they are identified) then another method is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Blandford%E2%80%93Znajek_process&quot; rel=&quot;noreferrer&quot;&gt;Blandford-Znajek process&lt;/a&gt;. This may actually be the way that quasars and many of the energetic phenomena in gamma ray bursts and black hole binary systems are powered. In simple terms, a magnetic field that threads the accretion disc can be twisted up by the ergosphere. Material can then be propelled in jets along the rotation axis, but is given a spin by the field lines, which extracts angular momentum (and rotational energy) from the black hole. This energy can then be released when these jets run into other things.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-09-03T20:23:09.247" LastActivityDate="2018-09-03T20:23:09.247" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27544" PostTypeId="2" ParentId="27541" CreationDate="2018-09-03T20:50:59.210" Score="0" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While a rotating black hole will store a great deal of energy in its rotation, and there is even a way of converting that that energy into a form that can do useful work (the Penrose process), this is of no practical use in energy generation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nearest know black hole is more than 1000 light years away, and while there are probably quiet inactive black holes that are closer, they are still much too distant to be useful in energy generation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a stellar mass black hole approached the solar system close enough for it to be a potential source of energy, it's gravitational mass would start to disrupt the solar system (as it would have a mass greater than the sun, the planets would tend to move into orbits around the black hole, not the sun.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is like suggesting that Earth's core could be a source of Iron. Its not practical as it is too far away and if it were closer it would be too dangerous.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-09-03T20:50:59.210" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27545" PostTypeId="2" ParentId="2045" CreationDate="2018-09-03T22:27:50.767" Score="5" Body="&lt;p&gt;The formula :-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}\tag{A}&#xA;\frac{1}{P_{sid}^{-1}-P_{syn}^{-1}} = P_{e}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $P_{e}$ is the sidereal period of the Earth in its orbit around the Sun)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be derived with the aid of Fig. 1 below, using the different axes frames, and making the following simplifying assumptions about the motions of the Sun, Earth and Moon :-&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Earth has a circular orbit around the Sun of constant speed and a fixed period $P_{e}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Moon has a circular orbit around the Earth of constant speed and a fixed period $P_{sid}$. (Thus the moon's path wrt the fixed Sun frame is an &lt;em&gt;epicycle&lt;/em&gt;, ie a small circle whose center moves along the circumference of a larger circle, in this case the Earth's orbit).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The above two orbits are confined to a common plane (the &lt;em&gt;Ecliptic Plane&lt;/em&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;From the direction we are viewing (ie `North') both Earth and Moon orbit in the anti-clockwise direction&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In the diagram the angles $\theta$, $\phi$, $\psi$, and $\pi + \theta$ are all `polar coordinate angles' wrt the relevant $x$-axis. The axes-frame $x'y'$ always points towards the Sun, and it moves with the Earth in its orbit. The $y'$ axis is the 'noon line' always pointing towards the Sun. The phases of the Moon are determined by the phase angle $\psi$ which the Moon makes wrt the $x'$ axis :-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sYTPi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sYTPi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conjunction means in line with and in same direction as the Sun, opposition means in line with but in opposite direction as the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the diagram it is readily seen that :-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi = \pi/2 + (\phi - \theta)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we could write :-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{eqnarray*}&#xA;\theta &amp;amp; = &amp;amp; \omega_{e}t + \theta_{0} \\&#xA;\phi &amp;amp; = &amp;amp; \omega_{m}t + \phi_{0} \\&#xA;\end{eqnarray*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\omega_{e}$ = &lt;em&gt;constant&lt;/em&gt; angular velocity of the Earth (radians per second), $\omega_{m}$ = &lt;em&gt;constant&lt;/em&gt; angular velocity of the Moon, and $\theta_{0}$, $\phi_{0}$ are the angles at time $t$ = $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi = (\omega_{m} - \omega_{e})t + (\phi_{0} - \theta_{0}) + \pi/2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so change $\Delta \psi$ in $\psi$ after a time $\Delta t$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta\psi = (\omega_{m} - \omega_{e}) \Delta t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus moon phase angle $\psi$ varies linearly with time at the constant rate of $\omega_{m} - \omega_{e}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\omega_{m} = 2\pi / P_{sid}$ and $\omega_{e} = 2\pi / P_{e}$ we then have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta\psi = 2\pi (1 / P_{sid} - 1 / P_{e}) \Delta t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The synodic period of the moon is simply the time it takes for moon phase angle $\psi$ to make a full cycle, ie $2\pi$ radians (= $360^{\circ}$) so putting $\Delta \psi$ = $2\pi$ in the latter formula gives $P_{syn}$ = $\Delta t$ ie&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{syn} = \frac{1}{1 / P_{sid} - 1 / P_{e}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{P_{syn}} = \frac{1}{P_{sid}} - \frac{1}{P_{e}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is easily rearranged to obtain the required formula (A). In short we are just subtracting two angular velocities -- but these go by the &lt;em&gt;reciprocal&lt;/em&gt; of the relevant orbital period $P$, hence the reciprocals in the formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the values $P_{sid} = 27.321661$ days (from &lt;a href=&quot;https://en.wikipedia.org/wiki/Moon&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Moon)&lt;/a&gt;)) and $P_{e} = 365.256363$ days (from &lt;a href=&quot;https://en.wikipedia.org/wiki/Sidereal_year&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Sidereal_year&lt;/a&gt;) this gives as the moon's synodic period :-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{syn} = 29.530588 \mbox{ days}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which differs by only 0.000001 from the Wikipedia value of 29.530589 days, where 'day' is SI day = 86,400 SI seconds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the figures given on Wikipedia for these three orbital periods $P_{syn}$, $P_{sid}$, and $P_{e}$ are &lt;em&gt;very closely&lt;/em&gt; related by the formula (A), so that one of them has been calculated from the other two. Thus these are not entirely actual experimentally observed figures that have been quoted on Wikipedia (nor on WolframAlpha). For we could NOT expect the actual figures to be related &lt;em&gt;so exactly&lt;/em&gt; by the formula (A) because of the simplifying assumptions (1) - (4) above that are required to derive that formula (the actual orbits are not perfectly circular and they are not exactly in the same plane).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(NOTE: The figure I have for $P_{e}$ is 365.256363 days, and for the tropical year 365.242189 days, from Wikipedia (&lt;a href=&quot;https://en.wikipedia.org/wiki/Sidereal_year&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Sidereal_year&lt;/a&gt;), where 'day' is SI day = 86,400 SI seconds. This differs slightly from the values you have quoted.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reciprocal type formula is also applicable to the synodic period between two &lt;em&gt;planets&lt;/em&gt; orbiting the Sun (see Fig. 2 below), again with the simplifying assumptions of circular orbits in the same plane :-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{P_{syn}} = \frac{1}{P_{1}} - \frac{1}{P_{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $P_{1}$ and $P_{2}$ are the orbital periods of the planets, and $P_{1} &amp;lt; P_{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This same type of formula is derived because :-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi = \phi - \theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so the above derivation applies very similarly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/W28ZI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W28ZI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ny9Lv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ny9Lv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&lt;/p&gt;&#xA;" OwnerUserId="24192" LastEditorUserId="24192" LastEditDate="2018-09-04T08:41:51.070" LastActivityDate="2018-09-04T08:41:51.070" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27546" PostTypeId="2" ParentId="18380" CreationDate="2018-09-03T23:12:27.230" Score="2" Body="&lt;p&gt;The formula :-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}\tag{A}&#xA;\frac{1}{T_{syn}} = \frac{1}{T_{1}} - \frac{1}{T_{2}}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the synodic period between two planets orbiting the Sun is derived in a similar way to the derivation of the Moon's synodic period  :-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}\tag{B}&#xA;\frac{1}{T_{syn}} = \frac{1}{T_{sid}} - \frac{1}{T_{e}}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $T_{sid}$ is the sidereal perod of the Moon and $T_{e}$ is the sidereal period of the Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formula (B) is derived by making the following simplifying assumptions about the motions of the Sun, Earth and Moon :-&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Earth has a circular orbit around the Sun of constant speed and a fixed period $T_{e}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Moon has a circular orbit around the Earth of constant speed and a fixed period $T_{sid}$. (Thus the moon's path wrt the fixed Sun frame is an &lt;em&gt;epicycle&lt;/em&gt;, ie a small circle whose center moves along the circumference of a larger circle, in this case the Earth's orbit).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The above two orbits are confined to a common plane (the &lt;em&gt;Ecliptic Plane&lt;/em&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;From the direction we are viewing (ie `North') both Earth and Moon orbit in the anti-clockwise direction&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;and (A) is derived with similar assumptions about the planetary orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(See &lt;a href=&quot;https://astronomy.stackexchange.com/a/27545/24192&quot;&gt;this answer&lt;/a&gt; for the derivation of formula (B) for the Moon - for the planets the derivation of (A) would be virtually the same).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality the orbits are not perfectly circular and they are not in the exact same plane. The circular orbit approximation means the speed of the planet is constant, but in the case of an elliptical orbit the speed varies.&lt;/p&gt;&#xA;" OwnerUserId="24192" LastEditorUserId="24192" LastEditDate="2018-09-04T08:20:35.997" LastActivityDate="2018-09-04T08:20:35.997" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27547" PostTypeId="2" ParentId="2555" CreationDate="2018-09-04T04:01:03.030" Score="7" Body="&lt;p&gt;I saw a zig zag meteorite in Bukuru, Northern Nigeria, when I was about 13 in 1959. That started from small to increasing swings, then went out near the horizon. For many years I puzzled about this, but rarely mentioned it, because no one would believe it. Years later I attended an astronomy lecture at college in London. I cornered the astronomer with this sighting and he said we know what this is. A flat dish shaped meteorite enters the atmosphere shallowly &amp;amp; flies in a circular motion that gets wider as it approaches the earth then extinguishes. Seeing this from the horizon, it appears to zigzag down. He told me that I was very very lucky to witness this meteor show, and I am glad that others have seen it too. Adios, UFO folks.&lt;/p&gt;&#xA;" OwnerUserId="24194" LastEditorUserId="31410" LastEditDate="2020-12-24T02:59:18.747" LastActivityDate="2020-12-24T02:59:18.747" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27550" PostTypeId="1" AcceptedAnswerId="27686" CreationDate="2018-09-04T09:32:25.483" Score="2" ViewCount="114" Body="&lt;p&gt;I'm developing a telescope controller open-source application. I started this project with a very little knowledge of Astronomy.&#xA;Basically, the app is going to send data to telescope over a wireless connection.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Data of the celestial object is as follow:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[&quot;SAO#&quot;: 308,&#xA;&quot;HD&quot;: 8890,&#xA;&quot;Con&quot;: &quot;Alpha Ursae Minoris&quot;,&#xA;&quot;StarName&quot;: &quot;Polaris&quot;,&#xA;&quot;RAH&quot;: 2,&#xA;&quot;RAM&quot;: 31.812,&#xA;&quot;DED&quot;: 89,&#xA;&quot;DEM&quot;: 15.85,&#xA;&quot;Mag&quot;: 2.02,&#xA;&quot;PRA&quot;: 0.038,&#xA;&quot;PDec&quot;: -0.015]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;RAH = RA hours, RAM = RA minutes,&lt;br&gt;&#xA;DED = Dec degrees, DEM = Dec minutes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(PRA and PDec are the drift of the star in arc-seconds per year (from the catalog Epoch J2000.0). I don't need to use these in the app, that I'm told.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The controller server (data receiver) accepts in this format:&lt;br&gt;&#xA;Set target RA- HH:MM:SS,&lt;br&gt;&#xA;Set target Dec- DD:MM:SS,&lt;br&gt;&#xA;Set target Azm- DDD:MM:SS,&lt;br&gt;&#xA;Set target Alt- DD:MM:SS&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I would want to convert the given celestial object data to Ra, Dec, Azm, Alt.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm told that:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;RA = ((RAH + (RAM / 60.0)) * 15.0); //in degrees, RA is an earthly longitude projected onto the sky&#xA;DEC = (DED + (DEM / 60.0)); //in degrees, Dec is an earthly latitude projected onto the sky&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For a star or other celestial object from the catalogs they have a fixed Epoch (J2000.)  I need to apply a correction for precession/nutation (wobble of the Earth's axis of rotation) to get a half decent estimate of the star's RA/Dec &quot;now&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using a &lt;a href=&quot;http://onekiloparsec.github.io/SwiftAA/&quot; rel=&quot;nofollow noreferrer&quot;&gt;code library&lt;/a&gt; that seems to have support for this and does the calculation on it own.&lt;br&gt;&#xA;I would want to know the procedure of the data conversion from equatorial coordinates to horizontal including. &#xA;For an instance this is how the code works:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    let jd = Date().julianDay&#xA;    let RAH = 2.0&#xA;    let RAM = 31.812&#xA;    let DED = 89.0&#xA;    let DEM = 15.85&#xA;&#xA;    let eqCoor = EquatorialCoordinates.init(rightAscension: Hour((RAH+(RAM/60.0))*15.0), declination: (Degree(DED+(DEM/60.0))), epoch: Epoch.J2000, equinox: Equinox.standardJ2000)&#xA;    let annualAbb = eqCoor.correctedForAnnualAberration(julianDay: jd, highPrecision: true)&#xA;    let initEQ = AstronomicalObject.init(name: &quot;Polaris&quot;, coordinates: annualAbb, julianDay: jd, highPrecision: true)&#xA;    print(&quot;EquatorialCoordinates -&amp;gt; RA(α):&quot;, initEQ.equatorialCoordinates.alpha, &quot;DEC(δ):&quot;, initEQ.equatorialCoordinates.delta) &#xA;&#xA;    let userLocation = GeographicCoordinates(positivelyWestwardLongitude: Degree(.plus, 75, 51, 13.65), latitude: Degree(.plus, 30, 54, 43.55), altitude: 256)&#xA;    print(&quot;User Location -&amp;gt; Latitude:&quot;, userLocation.latitude, &quot;Longitude:&quot;, userLocation.longitude)&#xA;&#xA;    let preccCoor = annualAbb.precessedCoordinates(to: Equinox.standardJ2000)&#xA;    let horizontalCoor = preccCoor.makeHorizontalCoordinates(for: userLocation, at: jd)&#xA;    print(&quot;Horizontal Coordinates -&amp;gt; Azimuth:&quot;, horizontalCoor.azimuth.inHours, &quot;Altitude:&quot;, horizontalCoor.altitude.inRadians.inDegrees)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is the data that code returns:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;EquatorialCoordinates -&amp;gt; RA(α): +37h56m06.009s DEC(δ): +89°16'05.165&quot;&#xA;User Location -&amp;gt; Latitude: +30°54'43.550&quot; Longitude: +75°51'13.650&quot;&#xA;Horizontal Coordinates -&amp;gt; Azimuth: +12h1m07.396s Altitude: +30°13'16.650&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here's the data from &lt;a href=&quot;https://stellarium.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellarium&lt;/a&gt; according to the same location as my code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zfYAw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zfYAw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data calculated from my code is not even close to the one from Stellarium except apparent altitude.&lt;/p&gt;&#xA;" OwnerUserId="23083" LastEditorUserId="23083" LastEditDate="2018-09-04T09:38:15.330" LastActivityDate="2018-09-18T19:38:10.057" Title="Need help with the calculations/conversion of a celestial object" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;fundamental-astronomy&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27552" PostTypeId="1" CreationDate="2018-09-04T11:40:10.443" Score="2" ViewCount="175" Body="&lt;p&gt;I’m working on a project of which one of its functions is determining the XY pixel coordinates of an object given the object RA DEC coordinates and a platesolved image (this assumes that the target is near the image center).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn’t base this code on any algorithm, I thought it would just be basic vector manipulation. I’m aware that this basic approach doesn’t correct field flatness and stuff, I need it to be only roughly accurate.&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;'PS_DEC,PS_RA,PS_NORTH_ANGLE,PS_PIXEL_SCALE are retrieved from a platesolve&#xA;'Target_dec and Target_ra are the coordinates of the target&#xA;'VectorRotate(vector,angle) is a function that rotates a vector&#xA;&#xA;'PS_DEC = 3.3026014754&#xA;'PS_RA = 278.7921661190&#xA;'Target_dec = 3.136138&#xA;'Target_ra = 278.65&#xA;'PS_NORTH_ANGLE = 99.89&#xA;'PS_PIXEL_SCALE = 3.15&#xA;&#xA;Dim lengthTotal As Double = (90 - PS_DEC) * 3600 / PS_PIXEL_SCALE ' gets the length of the vector towards the celestial north pole in pixel&#xA;&#xA;Dim InitVector As PointF&#xA;InitVector.X = 0&#xA;InitVector.Y = -CSng(lengthTotal) 'Casts the length to a vector&#xA;&#xA;Dim FirstRotation As PointF = VectorRotate(InitVector, PS_NORTH_ANGLE * -1) 'rotates the vector towards the celestial north pole using north angle from the Plate Solve&#xA;Dim FlipVector As PointF&#xA;FlipVector.X = FirstRotation.X * -1&#xA;FlipVector.Y = FirstRotation.Y * -1 'flips the vector so that it points from the north pole to the image center&#xA;&#xA;Dim Multiplier As Single = CSng((90 - Target_dec) / (90 - PS_DEC)) 'multiplier for applying the DEC difference&#xA;Dim FlipVectorTrim As PointF&#xA;FlipVectorTrim.X = FlipVector.X * Multiplier &#xA;FlipVectorTrim.Y = FlipVector.Y * Multiplier 'after this the vector should have the length of the Target dec&#xA;&#xA;Dim Finalvector As PointF = VectorRotate(FlipVectorTrim, (target_Ra - PS_RA)) 'rotates the vector by the difference of the target ra and the ra from the plate solve&#xA;Dim Pixeloffset As Point&#xA;Pixeloffset.X = CInt(Finalvector.X - FlipVector.X)&#xA;Pixeloffset.Y = CInt(Finalvector.Y - FlipVector.Y) ' by subtracting the vectors I should get the relative offset from the image center in pixels&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;When I use the coordinates from the star marked green I get the yellow circle as a result, to the bottom left of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/acRpvH0h.png&quot; alt=&quot;Image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fiddling around with the code I found out that if I multiply the RA correction angle by 0.67 the target is correctly found.&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Dim Finalvector As PointF = VectorRotate(FlipVectorTrim, (target_Ra - PS_RA) * 0.67)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/nJI68C4h.png&quot; alt=&quot;Image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’ve tried this on another star where I got the same results. What I figure from this is that DEC is correctly handled, but the RA rotation is for some reason too much, by about one third.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’ve looked at other possible error sources(placement of the circle in the image, star coordinates, the plate solve itself) and I’m confident that they are all correct, so the error has to be somewhere in this code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can use the fix I found, but I’d much rather know what I’m doing wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://nova.astrometry.net/user_images/2303227#annotated&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is a link to the platesolved image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="24190" LastActivityDate="2018-09-04T11:40:10.443" Title="Code for getting the pixel coordinates of an object with known RA DEC not working" Tags="&lt;coordinate&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27553" PostTypeId="1" AcceptedAnswerId="27555" CreationDate="2018-09-04T16:54:35.973" Score="11" ViewCount="582" Body="&lt;p&gt;From my research, I found out that &lt;a href=&quot;https://en.wikipedia.org/wiki/Quasi-star&quot; rel=&quot;nofollow noreferrer&quot;&gt;quasistars&lt;/a&gt; theoretically existed because of a black hole core whose radiation pressure counteracted gravity within the star. However, a few websites stated that quasistars do not exist now because there are metals contaminating hydrogen and helium.&lt;/p&gt;&#xA;&lt;p&gt;Could somebody explain to me why metals (or a small displacement of hydrogen and helium) would influence a black hole's radiation pressure, or is it just that there's no where that that much mass (1000+ solar masses) can exist anywhere in one place now?&lt;/p&gt;&#xA;&lt;p&gt;I looked on Wikipedia (I usually do this and then research on different websites if I find something interesting there) and here&lt;a href=&quot;https://briankoberlein.com/2014/10/23/black-hole-sun/&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="24185" LastEditorUserId="37191" LastEditDate="2020-12-30T20:05:39.760" LastActivityDate="2020-12-30T20:05:39.760" Title="Why can't a quasistar exist now?" Tags="&lt;black-hole&gt;&lt;radiation&gt;&lt;metal&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27554" PostTypeId="2" ParentId="27553" CreationDate="2018-09-04T18:56:13.723" Score="1" Body="&lt;p&gt;The addition of metals (i.e., elements heavier than helium) to a stellar mixture makes it less transparent to radiation.  Basically, hydrogen and helium have relatively simple and uncrowded spectra, but the &quot;metals&quot; add many new spectral lines and the mixture absorbs much more light and is more efficiently heated by it and picks up more momentum from it, also..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gas in the early universe then had less metals, and was thus less affected by the radiation of a condensing new star. Consequently, the star could grow to higher mass before its radiation cut off the inflow of gasses allowing it to grow.  (Today, the upper limit for star formation is around 100 Ssolar masses; with a mixture of H and He only, it appears to be as high as 250 solar masses.)  See the &lt;a href=&quot;https://en.wikipedia.org/wiki/Quasi-star&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; for a good explanation.  These super-large stars are needed to form a quasistar, and can only form from unsullied H/He. So quasistars (if they exist at all) could only form very early in the evolution of the universe.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-09-04T18:56:13.723" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27555" PostTypeId="2" ParentId="27553" CreationDate="2018-09-04T18:56:46.027" Score="10" Body="&lt;p&gt;Gas clouds with masses much higher than $10^3\,M_\odot$ are plentiful in galaxies; the typical star-forming cloud (the so-called &lt;em&gt;molecular clouds&lt;/em&gt;) have masses of $10^3\,M_\odot$ to $10^7\,M_\odot$. When quasistars (hypothetical stars powered not by nuclear fusion, but by accretion onto a central black hole) cannot exist today, it is because all gas in the Universe has become polluted with metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars form from collapsing gas clouds. In order for a region of a cloud to collapse, it must be sufficiently dense, and sufficiently cool; if it's too dilute, there's not enough gravity, and if it's too hot, the energy of the individual atoms counteracts the collapse, making the atoms escape.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Jeans' mass&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This criterion is captured in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Jeans_instability&quot; rel=&quot;noreferrer&quot;&gt;Jeans instability equation&lt;/a&gt;. The relation can be expressed in several ways; one way is to say that the mass of the cloud — or a small region of it — must exceed the &quot;Jeans mass&quot;:&#xA;$$&#xA;M_\mathrm{cloud} \gtrsim M_J \simeq 3\times10^4 \frac{T^{3/2}}{n^{1/2}}\,M_\odot,&#xA;$$&#xA;where $T$ (in $K$) and $n$ (in $\mathrm{cm}^{-3}$) are the temperature and the number density of the gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this equation you see that the cooler the gas is, the smaller the threshold. In other words, the smaller stars you can form. If the gas is not able to cool, only the largest clumps will collapse, and hence such stars will be very massive.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Gas cooling&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;So, how does the gas cool? Hot gas means that the particles have large velocities. If the particles collide, they may excite each other, bringing an electron to a higher state at the expense of slowing down — i.e. cooling. When the electron de-excites, a photon is emitted, which may leave the system. Thus, the kinetic energy of the atoms is converted into electromagnetic energy which escapes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, an electron is only excited if the energy of the collision matches closely the energy needed for the excitation. If the collisional energy is too high, or too low, the atoms simply bounce off of each other, maintaining their total energy (although one may transfer some energy to the other).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The effect of metals&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;If the gas consists only of hydrogen and helium, there are only a few available energies for excitation. Hydrogen happens to be able to cool efficiently around $T\sim10^4\,\mathrm{K}$, while helium cools efficiently around $T\sim10^5\,\mathrm{K}$, but at other temperatures, the gas tends to stay at its given temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as soon as there a some metals, the many electrons of these metals, with their many possible transitions, allow for atoms with many possible energies to be excited. Thus, before a gas cloud of $M\sim10^3\,M_\odot$ collapses to form a $10^3\,M_\odot$ star, it will fragment into smaller pieces, forming smaller stars.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2018-09-04T18:56:46.027" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27556" PostTypeId="2" ParentId="27553" CreationDate="2018-09-04T19:10:47.993" Score="8" Body="&lt;p&gt;If you build a very massive protostar, more than a thousand solar masses, then it is possible for the core of the protostar to collapse directly to a black hole whilst it is still surrounded by a massive envelope. The collapse will happen &quot;inside out&quot;, so that the envelope collapses at a slower rate. However, there is a maximum rate at which black holes can grow, because the compressing material gets very hot and emits lots of radiation and the radiation pressure can stall the collapse (temporarily). This is a quasi star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key to a quasi star is it's large initial mass, which prevents the envelope being &quot;blown away&quot; by the initial release of energy during the black hole formation. Such massive protostars can only be built in the early universe from pristine material. If the material is polluted with heavier elements then it can cool more readily - the heavier atoms can form molecules and radiate away energy. This cooling allows a large cloud to fragment into much smaller pieces, so that in the present day universe, the collapse of such a large cloud would not lead to one massive protostar, but a cluster of smaller protostars.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-09-04T19:10:47.993" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27558" PostTypeId="1" CreationDate="2018-09-05T08:17:54.393" Score="3" ViewCount="25" Body="&lt;p&gt;Consider a semiregular variable with amplitude of 1.5 mag. Is it possible to explain brightness variations by purely dipole pulsations?&lt;/p&gt;&#xA;" OwnerUserId="24202" LastActivityDate="2018-09-05T08:17:54.393" Title="Large amplitude dipole pulsator" Tags="&lt;variable-star&gt;&lt;asteroseismology&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27560" PostTypeId="2" ParentId="13202" CreationDate="2018-09-05T09:37:45.827" Score="2" Body="&lt;p&gt;One can imagine turbulent eddies in atmosphere as very weak optical lenses which focus and defocus stellar radiation. This leads to image degradation (seeing) and fluctuations of flux being registered through some aperture. The latter effect is called scintillations. It is very prominent for naked-eye observations. For telescopes the averaging by large aperture reduces the magnitude of the scintillations. Nevertheless it is main limiting factor for high-precision photometry at telescopes larger than ca. 2 m. See e.g&#xA;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2012MNRAS.426..647K&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2012MNRAS.426..647K&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for spectrscopy, atmospheric turbulence does not affect the wavelength of radiation as long as in modern high-precision spectrographs (e.g. HARPS) actual detection takes place in vaccuum.&lt;/p&gt;&#xA;" OwnerUserId="24202" LastActivityDate="2018-09-05T09:37:45.827" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27563" PostTypeId="1" CreationDate="2018-09-06T12:52:15.457" Score="3" ViewCount="88" Body="&lt;p&gt;According to (german) wikipedia, Eta Carinae has a cloud shaped like two cones with the points at the poles of the star as well as a disk around the equator. These matter clouds are mostly or only condensed stellar wind from Eta Carinae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my understanding, to create the dual-cone chape, the stellar wind provding the matter originates at the poles and this gives the dual cone shape (Ok?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why the poles? Stellar wind would, I guess, be mostly stellar matter so hot it reaches escape velocity, further pushed by the radiation. This should be isotropic. I can only guess that maybe Eta Carina has a magnetic field shaped similarly to earth with a north and south pole, nearer the equator the stellar wind is perpendicular to the field lines so the flightpaths of the charged particles are bent back inward. Near the poles the flightpaths and field lines are mostly paralell, so the magnetic field does not really act on the stellar wind. I included this guesswork on my part so you know where to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ETA:&lt;/strong&gt; Turns out my guessing about stellar wind was wrong or mostly wrong, &lt;a href=&quot;https://en.wikipedia.org/wiki/Eta_Carinae#Homunculus&quot; rel=&quot;nofollow noreferrer&quot;&gt;the cloud appears to be mostly shaped by eruptions&lt;/a&gt; in recentish history. But the question why they are shaped that way still stands.&lt;/p&gt;&#xA;" OwnerUserId="24209" LastEditorUserId="24209" LastEditDate="2018-09-07T06:34:16.463" LastActivityDate="2018-09-07T13:36:23.370" Title="Why is the cloud around Eta Carinae shaped the way it is?" Tags="&lt;stellar-astrophysics&gt;&lt;solar-wind&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27564" PostTypeId="2" ParentId="27518" CreationDate="2018-09-06T14:56:51.210" Score="3" Body="&lt;p&gt;For this, you need to look at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Initial_mass_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;Initial Mass Function&lt;/a&gt; which gives you an estimate of the population distribution of stars in a galaxy. As you can see massive stars are formed exponentially less than low mass ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reason why we see a higher number of less massive stars &lt;em&gt;right now&lt;/em&gt; (present mass function) is because more massive stars die out much much quicker than say some &lt;a href=&quot;https://en.wikipedia.org/wiki/Red_dwarf&quot; rel=&quot;nofollow noreferrer&quot;&gt;red dwarfs&lt;/a&gt;, which haven't had time yet to die (as their lifespans are longer than the age of the universe).&lt;/p&gt;&#xA;" OwnerUserId="15901" LastEditorUserId="15901" LastEditDate="2018-09-12T11:41:57.533" LastActivityDate="2018-09-12T11:41:57.533" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27565" PostTypeId="2" ParentId="10538" CreationDate="2018-09-06T15:00:46.237" Score="2" Body="&lt;p&gt;This question also occurred in &lt;a href=&quot;https://www.youtube.com/watch?v=KdBrQMkkDz8#t=15m43s&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lecture 4 of Leonard Susskind's Modern Physics&lt;/a&gt; in relation to a model that he showed to explain expansion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Little bits of space, little bits of newly formed space, that's &lt;em&gt;one way&lt;/em&gt; to think about it. The other way to think about it is just trough the equations of general relativity...&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe that it would be wrong to think of space being 'created'. Or at least, that would be more like a metaphysical concept. It raises the question what &lt;em&gt;is&lt;/em&gt; space and what &lt;em&gt;is&lt;/em&gt; creation? I do not think that it has much place in general relativity (at least not without considering deeper models that try to unify general relativity with quantum field theory). Is space created or is it just stretched (compare with a rubber band, do we create rubber band when we stretch a rubber band)? Those concepts may be ways to think of it but they do not precisely describe what they &quot;are&quot;, or at least they go beyond the physics and mathematical description. The current equations (at least to the equations of general relativity) neither tell whether space is stretched or whether space is created. Those concepts are irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There isn't some physical mechanism by which we can say 'this or that is creating space'. An interesting contrasting view that puts the expansion and &quot;creation&quot; of space in a different light is the viewpoint that the expanding universe and the relativistic equations can already be derived using only Newtonian dynamics (&lt;a href=&quot;https://doi.org/10.1093/qmath/os-5.1.73&quot; rel=&quot;nofollow noreferrer&quot;&gt;first described by McCrea and Milne mid 30s&lt;/a&gt;). That framework, albeit leading to the same equations, is a Newtonian notion and has a different interpretation. Good to mention is that in Newtonian mechanics space is &lt;em&gt;absolute&lt;/em&gt; (so when a volume expands then one could say there must be something like creation of space), but in general relativity this is &lt;em&gt;not&lt;/em&gt; the case. What looks big to you can look small to a different observer. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In General Relativity the expansion of the universe can be described by the Einstein Field equations like any other motion. You could compare it to the Newtonian cosmological model and see the expansion of the universe as described by equations of motion (ie. there is no magical entity that is causing expansion by &quot;creating&quot; space, it is just ordinary motion as described by the Einstein field equations). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These equations can be very briefly stated as: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;sup&gt;Note: the following equations can be written in many different ways. I have used the notation from Jerzy Plebanski Andrzej Krasinski, in &lt;a href=&quot;http://www.cambridge.org/ch/academic/subjects/physics/astrophysics/introduction-general-relativity-and-cosmology?format=HB&amp;amp;isbn=9780521856232&quot; rel=&quot;nofollow noreferrer&quot;&gt;An Introduction to General Relativity and Cosmology&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expanding universe can be described by the Einstein field equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ R^{\mu\nu} - \frac{1}{2}g^{\mu\nu}R+\Lambda g^{\mu \nu} = G^{\mu \nu} + \Lambda g^{\mu\nu} = \kappa T^{\mu \nu} \\$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\kappa = \frac{8\pi G}{c^4}$ is Einstein's constant and $\Lambda$ the (unknown) cosmological constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of many specific solutions are the Friedman equations (for a homogeneous isotropic universe) that use the Robertson-Walker geometry&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{array}{rcl}  d s^2 &amp;amp;=&amp;amp; g_{\mu \nu} dx^{\mu} dx^{\nu} \\&#xA;&amp;amp;=&amp;amp; d t^2 - R(t) \left( \frac{d r^2}{1-kr} + r^2(d \theta^2 + \sin^2\theta d \psi^2) \right) \end{array}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $R(z)$ a time dependent scale factor and $k$ the curvature index.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will lead to a solution for the scale factor $R$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{array}{rcrcl} G_{00} &amp;amp;=&amp;amp; \frac{3k}{R^2} + 3 \frac{\dot R^2}{R^2} &amp;amp;=&amp;amp; \kappa \epsilon - \Lambda \\&#xA; -G_{11} = -G_{22} = -G_{33} &amp;amp;=&amp;amp; \frac{k}{R^2} + \frac{\dot R^2}{R^2} + 2 \frac{\ddot R}{R} &amp;amp;=&amp;amp; -\kappa p + \Lambda \end{array} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p$ is the pressure and $\epsilon$ is the energy density.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The expansion can be seen as a kinematic effect (like with the expansion in Newtonian Cosmology) and is when described by the equations of motion in the framework of general relativity equivalent to a change, in time, of the &lt;em&gt;'metric'&lt;/em&gt;. This change of the metric happens to be the way that we can describe motion, gravitation and electromagnetism in a relativistic way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wish you could call that effect something like 'creation of space', but then you would, imo, need to be consistent and call any other relativistic length contraction/stretch as 'space being created/eliminated' (e.g. space is removed from Michelson's and Morley's interferometer when it is moving, or when gravitational waves pass by then we experience a repeated removal and addition of space). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the next time you see an apple fall from a tree you should imagine that, from the point of view of that apple, space is being contracted, thus space is being removed.&lt;/p&gt;&#xA;" OwnerUserId="20819" LastEditorUserId="20819" LastEditDate="2018-09-06T15:11:48.413" LastActivityDate="2018-09-06T15:11:48.413" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27566" PostTypeId="2" ParentId="27518" CreationDate="2018-09-06T21:18:39.777" Score="4" Body="&lt;p&gt;One popular argument for the Initial Mass Function is that the mass spectrum -- i.e., the fact that there are more low-mass than high-mass stars -- is inherited from the density spectrum of molecular clouds. In other words, stars form from the collapsing cores of molecular clouds, and the distribution of cores is already biased towards there being more low-mass cores than high-mass cores (this is in fact observed in molecular clouds). So you naturally end up with more low-mass stars than high-mass stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the natural next question is: why is the distribution of clouds cores biased towards low-mass cores? The usual suggestion is that it is the result of supersonic turbulence, and simulations of supersonic turbulence for conditions similar to those of a molecular cloud tend to produce density distributions similar the observed distributions.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2018-09-06T21:18:39.777" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27567" PostTypeId="1" AcceptedAnswerId="27577" CreationDate="2018-09-06T22:01:09.507" Score="-2" ViewCount="157" Body="&lt;p&gt;Could a small curved mirror and/or lens in Earth's orbit be aimed at a different hemisphere from where you are be be viewed with a powerful telescope to see the ground through the mirror? Would a more powerful telescope than the Hubble see more or less because of the atmosphere diffracting light twice?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus: Would sending a mirror to space cost less this way then a huge camera/telescope?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="7982" LastEditDate="2021-02-28T11:40:34.320" LastActivityDate="2021-02-28T11:40:34.320" Title="Looking at ourselves with a powerful telescope?" Tags="&lt;telescope&gt;&lt;light&gt;&lt;hubble-telescope&gt;&lt;atmospheric-effects&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27568" PostTypeId="2" ParentId="27247" CreationDate="2018-09-07T00:40:20.690" Score="0" Body="&lt;p&gt;What really matters is that the telescope and the ambient air are at the same temperature. A simple way to accomplish this is to take the scope outside one hour before you begin observing.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;warm air refraction (as observed with the Moon contour)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is called seeing, and it's within the instrument, ambiental and environmental.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Environmental seeing is simply called seeing. It's basically air turbulence through the whole atmosphere. It varies in time; when it's really bad, the image is blurry and there's nothing you can do about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_seeing&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Astronomical_seeing&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Within the instrument it's just air convection in a hot telescope used outside in the cool night air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-1/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-1/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-2/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-2/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ambiental seeing is convection from a hot rooftop or concrete slab. Don't observe stars that are just above a house or tall building - convection tends to blur the image.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2018-09-07T00:40:20.690" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27569" PostTypeId="1" CreationDate="2018-09-07T06:27:51.333" Score="3" ViewCount="120" Body="&lt;p&gt;As we know, the total solar irradiance normally received by the Moon is about 1361 $W/m^2$ (the same as the top of the Earth's atmosphere). Of course, this changes during lunar eclipses, &lt;em&gt;but the Moon doesn't go completely dark&lt;/em&gt; even during total eclipses (it's usually just a dim red). It still receives light through the Earth's atmosphere. My question is how much?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried googling this, but all the search terms I could think of just gave me a bunch of unrelated astrology results. &lt;em&gt;Surely people have measured this.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The worst case irradiance for any part of the Moon during a total eclipse is fine for my purposes, but I wouldn't mind a paper with an intensity curve over time if there are any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;br&gt;&#xA;This is not a duplicate of &quot;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/2166&quot;&gt;Red Moon a Characteristic of all Total Lunar Eclipses?&lt;/a&gt;&quot; That question is asking if all lunar eclipses are red and/or if the amount of redness varies. I'm asking how much light hits the Moon (per second per area) during a total lunar eclipse. Furthermore, none of the answers there even mention irradiance.&lt;/p&gt;&#xA;" OwnerUserId="773" LastEditorUserId="7472" LastEditDate="2018-09-09T09:41:59.023" LastActivityDate="2018-10-09T10:00:21.750" Title="What irradiance does the Moon receive during a total lunar eclipse?" Tags="&lt;the-moon&gt;&lt;astrophysics&gt;&lt;lunar-eclipse&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27571" PostTypeId="1" AcceptedAnswerId="27575" CreationDate="2018-09-07T11:49:04.047" Score="1" ViewCount="113" Body="&lt;p&gt;How do they know that the pillars have been destroyed if the light will reach us in 1000 years time?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The &quot;Pillars of Creation&quot; from the Eagle Nebula. Evidence from the&#xA;  Spitzer Telescope suggests that the pillars may already have been&#xA;  destroyed by a supernova explosion, but the light showing us the&#xA;  destruction will not reach the Earth for another millennium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.spitzer.caltech.edu/news/249-ssc2007-01-Famous-Space-Pillars-Feel-the-Heat-of-Star-s-Explosion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Article explaining that we won't see the effects in another 1000 years&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24214" LastActivityDate="2018-09-07T13:44:20.740" Title="Eagle Nebula (Pillar of Creation) destroyed?" Tags="&lt;nebula&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27574" PostTypeId="2" ParentId="27563" CreationDate="2018-09-07T13:36:23.370" Score="1" Body="&lt;p&gt;It seems likely due to rotation.  The disklike &quot;skirt&quot; could be due to material that is orbiting the star that gets thrown out by the eruption, and the bipolar nature of the eruption can be because material is preferentially ejected from the poles of the star.  It may seem odd that the poles, where there is less angular momentum per gram, would be better at ejecting material, but this might be due to a greater role for radiation pressure in causing the eruption.  Radiation pressure produces forces that are proportional to the radiative flux, which can come out preferentially in the polar directions because this becomes the path of least resistance when rotation makes the shape of the star highly oblate.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-09-07T13:36:23.370" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27575" PostTypeId="2" ParentId="27571" CreationDate="2018-09-07T13:44:20.740" Score="2" Body="&lt;p&gt;It's a bit hard to say for sure as the article isn't terribly clear, but what I think they're saying is that there is a dust cloud to the upper right side of the Pillars (as viewed in the photo.) They further say that the dust cloud shows evidence of a supernova shock wave passing through it, presumably from the upper right and presumably from a supernova.  (Note that the supernova was not observed, and they don't report on a remnant being visible.) Then they say that when that shockwave reaches the Pillars, the Pillars will be destroyed.  They estimate that that will be observed from Earth in the next thousand years or so.  (Note that they don't report measuring of the speed of the shockwave -- I'm guessing they assume it's moving as typical speeds.) Since the Pillars are about 7000 light years away, than means that this destruction &quot;already&quot; has happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's nothing impossible about all this, but it looks like speculation based on limited actual facts.  It's also worth mentioning that what you see on the web went through a press office, and press offices are paid for doing publicity, not for doing science.  The quality of what a press office posts (already undermined by being a popularization, which is hard to do well even when you're trying to be accurate) is frequently biased by a publicity-at-any-cost attitude. &lt;em&gt;Always&lt;/em&gt; read press releases and the like with your sceptometer turned up to High.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I'd like to see stronger evidence for the existence of the shockwave, and evidence that the shockwave will actually disrupt the Pillars. (The Pillars are considerably denser than the medium around them, and so may be resistant.)&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-09-07T13:44:20.740" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27576" PostTypeId="1" AcceptedAnswerId="27578" CreationDate="2018-09-07T17:45:19.210" Score="0" ViewCount="187" Body="&lt;p&gt;Is it really possible for material such as dust, gas and other stellar debris that come near the black hole, doesn't fall into it, instead forms a flattened band of spinning material?&lt;/p&gt;&#xA;" OwnerUserId="24177" LastEditorUserId="1891" LastEditDate="2018-09-08T01:04:06.563" LastActivityDate="2018-09-08T01:04:06.563" Title="What exactly is the accretion disk?" Tags="&lt;black-hole&gt;&lt;cosmology&gt;&lt;general-relativity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27577" PostTypeId="2" ParentId="27567" CreationDate="2018-09-07T18:16:23.403" Score="3" Body="&lt;p&gt;If I interpret your system concept, you want to view over the horizon (OTH) by adding a distant mirror to your overall telescope optical prescription.&lt;br&gt;&#xA;First: it's not necessary to go into orbit, depending on how far OTH you want to view.&lt;br&gt;&#xA;Second, whether you try to place such a mirror on, e.g., a blimp or in orbit, the difficulties involved in launching, aiming, stabilizing, etc. a mirror of sufficient size to allow enough light-gathering at your ground location are greater than the difficulty in launching a spy satellite (of which there are many, each with a telescope comparable to Hubble's main scope).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third,  for optical reasons (try googling &quot;shower-curtain effect adaptive optics&quot;)  viewing the ground from orbit suffers far less from atmospheric distortion effects than vice versa.  Far better to put the telescope in orbit and send digitized image data to the ground. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-09-07T18:16:23.403" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27578" PostTypeId="2" ParentId="27576" CreationDate="2018-09-07T19:54:13.737" Score="3" Body="&lt;p&gt;A basic answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To us on Earth, it looks like everything falls mostly straight down and hitting the Earth.   We almost never see something falling, trapped in Earth's gravity and missing the Earth but in space that happens a LOT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we lived in space our observation would be very different.   Most of the stuff we see flies around, in orbits, around the Sun or Earth without colliding into them.  The reason for this is tangential velocity.  It's actually hard for something to hit something else in space, even with gravity, because space is (shockingly) mostly empty space.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets, the Sun, black holes, etc, are very small relative to their gravitationally dominant regions or hill spheres.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=LHvR1fRTW8g&quot; rel=&quot;nofollow noreferrer&quot;&gt;Why is it hard to hit the sun?&lt;/a&gt;  (this applies to black holes too, even more so)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=tmNXKqeUtJM&quot; rel=&quot;nofollow noreferrer&quot;&gt;Why is the solar system flat?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mostly, objects caught in the gravitationally dominant region around a black hole will just orbit the black hole.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When an object passes inside a black hole's Roche limit, it will begin to break apart into smaller parts, and those smaller parts into smaller parts, and close enough, pretty much all matter just becomes a bunch of debris, kind of similar to Saturn's rings but much faster moving and more dense.   Ring systems can form inside the Roche limit around any massive body, large moon, planet, sun or star, though stars which emit lots of light and heat aren't good for the stability of ring systems.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ring systems, kind of like Saturn's, can form around black holes when there is a collision or when an object passes too close and ring systems tend to flatten out over time due to collisions.  The angular momentum remains constant but the up or down motion of objects cancels out.  The flat accretion disk is actually conservation of momentum, it forms naturally, even if it seems like a lot of wasted space, the matter doesn't know where else to go but into a flattened, quickly rotating disk.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the laws of relativity and gravitational waves, inside about 3 Schwarzschild radii, the ring material much more quickly spirals in towards the black hole, moving faster into a smaller region, getting very hot and extremely magnetic and that's generally what's called the accretion disk.  (Ring systems are generally further out and not hot or spiraling in, though they are loosely similar)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All happens outside the black hole.   Nothing that happens inside the black hole gets out, but what happens outside the black hole in the accretion disk can be surprisingly bright and energetic.  &lt;a href=&quot;http://www.thephysicsmill.com/2013/11/09/accretion-disk/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brighter than the brightest stars&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might seem like stuff should just fall into a black hole like a vacuum sucking things in, but in gravity in space doesn't work like that.  Anything with zero tangential velocity will fall straight in (kinda-sorta, there's also the dragging space around it problem for rotating black holes), but for simple non rotating black holes, Only an object with close to zero tangential velocity will fall straight, but almost nothing that moves through space has zero tangential velocity relative to it's nearest large gravitational body.  This applies to dust, gas, comets, meteors, planets, stars, anything that passes close to or orbits a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the overall gist of it.  Corrections welcome.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-09-07T21:22:23.343" LastActivityDate="2018-09-07T21:22:23.343" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27579" PostTypeId="1" CreationDate="2018-09-07T20:25:27.737" Score="6" ViewCount="833" Body="&lt;p&gt;Titan is the only moon in our solar system with a thick atmosphere, similar to what our Earth's atmosphere was before life &quot;appeared&quot;. It also orbits Saturn close enough to actually see it beautifully in the sky... or can it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Titan has a very thick and hazy atmosphere. Tholin seems to be the main factor of this &quot;haziness&quot;. Also we can't see Titan's surface from space without infrared instruments. So in visible light, Titan looks like an almost uniform ball of beige, yellowish, golden brown because of its atmosphere, hiding the surface below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this should probably be the same case the other way around, right? Can we see Saturn from the surface of Titan, or does the haziness hide the parent planet from view during day time and night time (since I believe the atmosphere's opacity changes in day time and night time)?&lt;/p&gt;&#xA;" OwnerUserId="24220" LastEditorUserId="7982" LastEditDate="2021-02-28T11:39:51.187" LastActivityDate="2021-02-28T11:39:51.187" Title="Is it possible to see Saturn from Titan's surface, day and night?" Tags="&lt;planetary-atmosphere&gt;&lt;saturn&gt;&lt;atmospheric-effects&gt;&lt;titan&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27580" PostTypeId="2" ParentId="27579" CreationDate="2018-09-07T20:40:18.257" Score="3" Body="&lt;p&gt;&quot;since I believe the atmosphere's opacity changes in day time and night time&quot; maybe to a very small factor, somewhere in the upper atmosphere, in the UV-wavelengths where plasma opacities play a role.&lt;br&gt;&#xA;But not in the optical, where the bulk density of the atmosphere plays a role, wich at 1.6 bar doesn't change much between day and night. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your assertion &quot;So in visible light, Titan looks like an almost uniform ball of beige, yellowish, golden brown.&quot; is incorrect, where do you take that from? The optical depth integral is symmetrical both ways (ignoring scattering), so if we by far cannot see the surface, then one will certainly not be able to see Saturn from the surface.&lt;br&gt;&#xA;There is also &lt;a href=&quot;https://astronomy.stackexchange.com/questions/19085/are-there-other-pictures-of-titan-surface-from-huygens&quot;&gt;only one existing photo taken on Titans surface&lt;/a&gt;, so not sure where you get this idea from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from the other statement &quot;Titan is the only moon in our solar system with a thick atmosphere, similar to what our Earth's atmosphere was before life 'appeared'&quot;, which is a questionable statement at best.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2018-09-07T20:40:18.257" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27581" PostTypeId="2" ParentId="27247" CreationDate="2018-09-07T21:12:45.570" Score="1" Body="&lt;p&gt;The air is only part of your problem here.  There are several factors involved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is absolutely true that the air can cause a loss of detail in the image you see.  But even if there's no atmosphere between you and the moon or anything else you're looking at, there is only so much detail you can see.  Technically speaking, you can magnify as much as you want ad infinitim.  But there is only so much USABLE magnification  you can get out of a telescope.  This is due to the diffraction of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the light from the moon or any other object enters your telescope tube, it begins diffraction.  If you want to better understand the physics behind it, there's a &lt;a href=&quot;https://www.khanacademy.org/science/physics/light-waves/interference-of-light-waves/v/single-slit-interference&quot; rel=&quot;nofollow noreferrer&quot;&gt;fairly easy-to-undersatand Khan Academy video about it&lt;/a&gt;.   Essentially, what is happening is interference patterns form among the light waves and the result is a blurring of detail.  This blurring is less noticable at lower magnifications, and gets worse the more you magnify.  While the specifics are a little different from the result of blowing up a picture, the effect is similar.  I put together &lt;a href=&quot;http://astro.jmhastronomy.com/random/resolutionexample/index.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;this example&lt;/a&gt; of the effect.  In my example, I started with a picture I took of the moon in May or June and each image in the series is doubled in size then cropped.  I go from a full image of a waxing gibbous moon to zoom in on the crater Herschel.  As the magnification increases, you start to see the image getting fuzzier and fuzzier.  In the case of a picture, this is because there is only so much data to work with, depending on the geometry of the image sensor in the camera. Technically, it is also the case with the light that there is a fininte amount of information/data to work with, but the mechanics are a bit different.  In the end, the result is the image blurs past a certain point and you can't get a clearer image out of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Technically speaking, using speckle interferometry, a camera CAN increase detail when multiple images are captured and analyzed together, but there's still a finite limit, and this doesn't apply at all to visual observing.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your telescope has an aperture of 114mm, as you indicated.  I did a quick search for Seben 114mm telescopes and the only ones I found showed a focal length of 1,000mm.  As it turns out, it appears this is not a true Newtonian telescope, but a Bird-Jones Newtonian (or Jones-Bird, depending on who you ask), which is actually a type of catadioptric design and has some inherent issues, which I'll circle back around to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They eyepieces I found listed that come with this telescope include an SR4mm and a H6mm. The SR4mm is a symmetrical Ramsden design eyepiece with a 4 mm focal length.  The H6mm is a Huygens design with a  6mm focal length.  The Huygens eyepiece design was invented in about 1662 by Christian Huygens, and was an improvement upon the original basic single-lens eyepiece design of the first Lippershey and Galilean telescopes, and Kepler's minor improvements in 1611.  Huygens added a second element and produced an eyepiece that provides better eye relief (the distance from the eyepiece to your eye, and less chromatic aberration (the separation of colors of light caused by refraction.  These eyepieces, however, have fairly narrow apparent fields of view and perform poorly in telescopes with focal ratios under f/10.  The next major upgrade in eyepiece design was Jesse Ramsden's invention in 1782.  Ramsden's eyepiece provided better apparent fields of view, but decreased eye relief and increased chromatic aberration.  When used in longer focal lengths, this is less of a problem,b ut still a problem.  The Ramsden eyepiece is also susceptible to &quot;ghosting&quot; which is caused by reflections within the eyepiece.  These eyepieces aer fairly inexpensive to manufacture, however, which makes them common in lower-cost telescopes such as yours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;laying aside, for the moment, the inherent issues with these eyepieces, the sizes you specified are 4 mm and 6 mm.  Based on the design types, I am going to assume an apparent field of view (AFOV) of about 40° for both.  WhenI do the calculations, I find that the 6 mm gives you a magnification of about 167X with a TFOV (true field of view) of about 14 arcminutes. The 4 mm gives you 250X magnification with a TFOV around 10 arcmin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I mentioned above, however, aperture limits usable magnification.  In particular, it limits the angular size of details you can resolve.  A 114 mm aperture can provide details no larger than 0.93 arcseconds in size.  At the distance to the moon (about 384,400 km) this means the smallest details you can see are about 1.73 km in size.  But this is the theoretical limitation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the mid 1800's, William R. Dawes produced a method of calculating how far apart two stars must be to be able to determine that they are, in fact, two separate stars. His formula, which provides a measure in arcseconds we call the Dawes' Limit, is one of the most commonly referenced ways to describe how small a detail you can see with a given telescope aperture.  For your telescope, that would be about 1.02 arcseconds.  By this measure, you could not see details any smaller than about 1.9 km in size on the moon.  A third measure, known as the Raleigh Criterion, which is also measured in arcseconds, suggests the smallest details visible in a 114 mm telescope would be about 1.21 arcseconds in size or 2.25 km at the distance of the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on these formulae, then, if there was a particular crater on the moon that was about 2km in size, you could probably just make it out.  You could magnify this as much as you like, but you won't see more detail, just a blurrier and blurrier hint of a crater.  Eyepiece size/design won't change that, nor will your focus ability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on who you ask, for a given inch of aperture, you can only get about 25 to 60x of usable magnification - meaning that anything over that level is going to be so blurry as to not be worth viewing.  On a typical night, that would be toward the lower side, and on a night with very good air and sky conditions, toward the higher end.  Based on this, I'd say on a typical night you shouldn't expect more than about 135X magnification, and on a really nice night, maybe 225X.  Technically, the 4mm gets you much higher than that, and with the Barlow lens, double that (to around 500X) but there's absoloutely nothing you can do to improve the sharpness of the image - the limits of physics dictate that the telescope simply cannot do it.  Anything over 135X on an average night is likely to be fuzzy and get fuzzier still as you try to increase magnification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But wait, there's more!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I mentioned, your telescope isn't a true Newtonian.  A true Newtonian uses a mirror with a curvature defined by the formula of a parabola.  These kinds of mirrors are a bit harder to manufacture, so many manufacturers of lower-priced telescopes cheat a little and use a mirror with a spherical curvature.  The spherical curvature is a constant-radius curve, and this makes it MUCH easier to manufacture.  However, it introduces a problem known as spherical aberration  Luckilly, spherical aberration and most other aberrations are decreased by increasing focal length.  The Bird-Jones design uses a shorter-focal length spherical mirror and adds a correcting lens (almost identical to a Barlow lens) in the path of light to correct the errors. If you look into your focuser without an eyepiece, you'll probably notice a lens between your eye and the secondary mirror - this is the Bird-Jones corrector.  It adds more glass to the optical path. The more glass, the more chance for distortion and further aberration.  Additionally, this is not likely to be a very high quality optical component, making it more likely to produce a poorer image.  But worse, it makes the process of collimation very difficult indeed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing, though I may be wrong, you have little or no understanding of collimation.  Simply put, collimation is the proper alignment of the optical components in the telescope.  If the mirrors and lenses are not properly aligned, it introduces further errors in the path of light, which manifest themselves in image aberration and distortion.  When collimation is off, the image quality degrades and sharpness is lost.  Unfortunately, this built-in lens makes collimation more difficult, which means it's likely that your collimation is off somewhat, which will FURTHER degrade your image quality at higher magnifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what can you do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Honeslty, not much.  While warmer air will cause issues inside the tube of the telescope, this can be reduced through acclimatization - letting the telescope come to ambient temperature.  If you can collimate your telescpe accurately, that will improve your image.  And replacing your eyepieces with better designs (even lower-price Plossl eyepieces are significantly better) will help. Even then, you're not going to get sharp detail at higher magnification.  It is simply not possible. To get sharper detail, you need a larger aperture and better air conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My personal recommendation for a beginner telescope is an 8&quot; Dobsonian. Though somewhat more expensive, the 8&quot; aperture provides a lot more detail.  And a Dobsonian with a true Newtonian OTA is far easier to collimate and get a clearer image out of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For now, keep using your 114, but accept its limitations.  And start looking to your next scope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also recommend finding a local astronomy club you can join and learn more about how scopes work, how to maintain them, and ways to better enjoy your observing sessions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck and clear skies!&lt;/p&gt;&#xA;" OwnerUserId="10546" LastActivityDate="2018-09-07T21:12:45.570" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27582" PostTypeId="1" AcceptedAnswerId="27583" CreationDate="2018-09-07T22:09:48.807" Score="1" ViewCount="178" Body="&lt;p&gt;I've both fascinated and frustrated after learning about the existence of Dark Matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we find evidence of Dark Matter within our Solar system? What is currently the closest candidate for Dark Matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And please let me know what are good online resources where I can stay up to date with information about this?&lt;/p&gt;&#xA;" OwnerUserId="24221" LastEditorUserId="1891" LastEditDate="2018-09-08T01:00:09.350" LastActivityDate="2018-09-08T02:16:04.683" Title="What is the closest candidate for Dark Matter currently?" Tags="&lt;solar-system&gt;&lt;dark-matter&gt;&lt;dark-energy&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27583" PostTypeId="2" ParentId="27582" CreationDate="2018-09-08T00:58:41.953" Score="4" Body="&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/1306.5534&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; arxiv post says if there is dark matter in the solar systems, its density should be below $\approx 1.4\cdot 10^{-20} \frac{g}{cm^3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They tried to find the dark matter in the solar systems by measuring its possible gravitational effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, it is a negative result, i.e. they didn't find anything, but they are suspecting, if the dark matter density would be at least $\approx 1.4\cdot 10^{-20} \frac{g}{cm^3}$, they had found it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;a href=&quot;https://www.sciencealert.com/from-machos-to-wimps-meet-the-top-five-candidates-for-dark-matter&quot; rel=&quot;nofollow noreferrer&quot;&gt;various candidates&lt;/a&gt; what could it be, the currently considered most probable are the &lt;a href=&quot;https://en.wikipedia.org/wiki/Weakly_interacting_massive_particles&quot; rel=&quot;nofollow noreferrer&quot;&gt;WIMP&lt;/a&gt;s (particles interacting only gravitationally and maybe weakly).&lt;/p&gt;&#xA;" OwnerUserId="1891" LastActivityDate="2018-09-08T00:58:41.953" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27589" PostTypeId="1" CreationDate="2018-09-08T16:35:08.430" Score="0" ViewCount="212" Body="&lt;p&gt;Is there any absolute evidence of star formation inside of nebulae (AKA stellar nurseries) as is suggested and even stated as fact by many textbooks?&lt;/p&gt;&#xA;" OwnerUserId="24225" LastEditorUserId="2270" LastEditDate="2018-09-11T17:06:36.630" LastActivityDate="2018-09-11T17:06:36.630" Title="Is there any definitive proof of stars being &quot;born&quot; in nebula formations?" Tags="&lt;star&gt;&lt;star-formation&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27590" PostTypeId="2" ParentId="27589" CreationDate="2018-09-08T17:13:43.013" Score="0" Body="&lt;p&gt;Good question but what would you call absolute evidence? There is evidence that these nebula regions contain a large number of very young stars. A larger number than one would expect to find in a normal or regular region of space. I don't know if you consider that absolute evidence but absolute evidence may require us to view things that cannot be seen from afar. So we have to make assumptions. Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastActivityDate="2018-09-08T17:13:43.013" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27592" PostTypeId="1" AcceptedAnswerId="27593" CreationDate="2018-09-09T05:27:46.493" Score="1" ViewCount="142" Body="&lt;p&gt;We frequently observe lunar eclipse throughout the year. Have we ever observed Mars Eclipse or Jupiter Eclipse or Saturn eclipse?&lt;/p&gt;&#xA;" OwnerUserId="22552" LastEditorUserId="2365" LastEditDate="2018-09-09T06:58:14.663" LastActivityDate="2018-09-09T15:38:44.830" Title="Do we observe Mars Eclipse or Jupiter Eclipse?" Tags="&lt;cosmology&gt;&lt;observational-astronomy&gt;&lt;eclipse-non-earth-related&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27593" PostTypeId="2" ParentId="27592" CreationDate="2018-09-09T06:19:41.547" Score="3" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two types of eclipse: A lunar eclipse is when the moon goes into the Earth's shadow. We can see a lunar eclipse because the moon is so close to the Earth. If you were standing on the moon, The Earth would look bigger in the sky than the sun. If the Sun, Earth and moon are aligned, the Earth will block out the light from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars and Jupiter are further away. If you were standing on Mars, The Earth would look like a bright star, it would be too far away to block all the light from the sun, even if the Sun, Earth and Mars were perfectly aligned.  So there can be no eclipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second type of eclipse is a solar eclipse, when the moon blocks of light from the sun. Since Mars and Jupiter orbit further from the sun than the Earth, they can never block of its light. Venus and Mercury sometimes do cross the face of the sun. This is called a &quot;Transit&quot; of Venus. But all the planets are too far away to block of all the light, so there is no eclipse.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-09-09T06:19:41.547" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27594" PostTypeId="2" ParentId="27569" CreationDate="2018-09-09T09:59:41.977" Score="2" Body="&lt;p&gt;The following is more a series of observations than a full answer, but I hope other people can fill in the details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnitude_(astronomy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;apparent magnitude&lt;/a&gt; of the full moon is -13. The apparent magnitude of the moon during a lunar eclipse will differ from eclipse to eclipse (correlated to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Danjon_scale&quot; rel=&quot;nofollow noreferrer&quot;&gt;Danjon scale&lt;/a&gt; value), but let's suppose it is 0 for a certain eclipse. (Note that this magnitude is different from the (pen)umbral magnitudes mentioned on the &lt;a href=&quot;https://eclipse.gsfc.nasa.gov/LEcat5/LE2001-2100.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA website&lt;/a&gt;; these magnitudes can be computed and do not depend on the Earth's atmosphere.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means the full Moon is normally reflecting $100^{13/5}$ times more light than during this eclipse. I'm not sure if the &lt;a href=&quot;https://en.wikipedia.org/wiki/Albedo&quot; rel=&quot;nofollow noreferrer&quot;&gt;albedo&lt;/a&gt; of the Moon is different for red light than it is for normal sunlight, but if not, that means the answer is $\frac{1361}{100^{13/5}} \frac{W}{m^2} \approx 8.6 \frac{\text{m}W}{m^2}$.&lt;/p&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2018-09-09T09:59:41.977" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27595" PostTypeId="1" CreationDate="2018-09-09T15:23:50.390" Score="4" ViewCount="144" Body="&lt;p&gt;Newbie learning Astronomy and was curious how to locate a star giving a Latitude and LST.  For example if my LST is 4:00 and my latitude is 20*N, how would I find where the Horsehead Nebula appears in the sky?  The only formula I can think of is the Hour Angle formula.. &lt;/p&gt;&#xA;" OwnerUserId="24231" LastActivityDate="2018-12-09T19:00:35.103" Title="How to find star location knowing latitude and LST" Tags="&lt;star&gt;&lt;distances&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27596" PostTypeId="2" ParentId="27592" CreationDate="2018-09-09T15:38:44.830" Score="3" Body="&lt;p&gt;I agree with James K's answer, but just to play devil's advocate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth is much too small to cause what we think of as an eclipse when viewed from Mars or Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in a sense, a transit is like an &lt;a href=&quot;https://www.timeanddate.com/eclipse/solar-eclipse.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;annular eclipse&lt;/a&gt;.   Nobody would ever call a transit an annular eclipse, because it sounds stupid, but it's kind of the same thing.   Phobos is smaller than the Sun from the surface of Mars but &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_eclipses_on_Mars&quot; rel=&quot;nofollow noreferrer&quot;&gt;that's sometimes called an eclipse&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An eclipse is a transit, but most transits are never called eclipses, because the object in transit is much too small, but there's no formal percentage of the sun that must be blocked to earn the title eclipse.  Annular Eclipse vs Total is precise, but transit vs annular eclipse is undefined.  It's mostly common sense, but it's still undefined.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fun thought: . . . I wonder if they'll call the planet to planet transits &quot;eclipses&quot; on &lt;a href=&quot;https://en.wikipedia.org/wiki/TRAPPIST-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Trappist-1&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-09-09T15:38:44.830" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27597" PostTypeId="1" CreationDate="2018-09-09T15:47:14.567" Score="2" ViewCount="4346" Body="&lt;p&gt;In multiple video clips of the moon but also one of Jupiter, some kind of waves can be observed. Is this just the effect of the object moving behind layers of different refractive indexes (similar to what we observe at sunset) or is there a wave propagating through the atmosphere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://m.youtube.com/watch?v=xUGxysKSGEM&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://m.youtube.com/watch?v=xUGxysKSGEM&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://m.youtube.com/watch?v=EMSkd4Xhp0g&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://m.youtube.com/watch?v=EMSkd4Xhp0g&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an &lt;a href=&quot;https://www.quora.com/What-is-the-lunar-wave-from-a-non-conspiracy-anti-government-anti-science-standpoint&quot; rel=&quot;nofollow noreferrer&quot;&gt;answer on Quora&lt;/a&gt;, but I would love to see a more detailed answer, why we can see exactly these two rather sharp lines moving.&lt;/p&gt;&#xA;" OwnerUserId="24232" LastEditorUserId="7472" LastEditDate="2018-09-09T18:48:50.070" LastActivityDate="2019-04-18T20:31:09.150" Title="What phenomenon causes &quot;lunar waves&quot;?" Tags="&lt;the-moon&gt;&lt;observational-astronomy&gt;" AnswerCount="3" CommentCount="14" ClosedDate="2019-04-22T01:05:22.510" ContentLicense="CC BY-SA 4.0" />
  <row Id="27598" PostTypeId="2" ParentId="12178" CreationDate="2018-09-09T16:15:58.500" Score="1" Body="&lt;p&gt;The iron on Earth is being taken from the upper crust in well known processes of subduction.  Earth has less iron on its surface because it is being depleted through the process of subduction which has been going on for billions of years on Earth.  This process is yet to occur on Mars which is why there is very little granite on Mars surface as opposed to Earth where the continents cover a good portion of the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subduction cannot take place on Mars, because it sits outside the habitable zone. The good thing for Mars is that it will find itself in the habitable zone in the distant future because our star is growing in size and temperature. This will cause the habitable zone to move outward, and Earth will find itself unable to maintain water at this juncture and the viable planet will be Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key here is that Mars is inside out where the iron is on its surface causing the planet to appear smaller than Earth, because its silicate is in the inside while the iron is on the outside. Basically Earth through subduction has managed to turn the planet inside out allowing for Earth to grow from decompression. The iron is migrating toward the core in subduction processes because it does not melt and rise with the rest of the subducted basaltic plate to form continents it sinks to increase centripetal force and increase gravity because gravity stems from the core, the denser the core the deeper the indention on space time. Mars is compressed and its core is a silicate where Earths is now iron giving earth the greater magnetic field yet the magnetic field on Mars surface is greater than earths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars' rotation period is 24 hours and its axis are tilted like Earth's and it will sit in the habitable zone where the iron will have a chance to work its way to the center of the planet as it has on Earth.&lt;/p&gt;&#xA;" OwnerUserId="24233" LastEditorUserId="7472" LastEditDate="2018-09-09T18:49:20.360" LastActivityDate="2018-09-09T18:49:20.360" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27599" PostTypeId="2" ParentId="4755" CreationDate="2018-09-10T00:22:44.090" Score="3" Body="&lt;p&gt;I have been watching for almost 30 years on rosh ha shanah.  I have been able to see it when aprox. 3 percent is technically visible.&lt;/p&gt;&#xA;" OwnerUserId="24235" LastActivityDate="2018-09-10T00:22:44.090" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27600" PostTypeId="1" CreationDate="2018-09-10T00:44:00.280" Score="0" ViewCount="113" Body="&lt;p&gt;The key to finding the answer seems to involve processing data from astronomical databases, which I'm not sure how to do in this case. I'd prefer to limit the answer to the original list of approximately 1500 stars, but answers that rely on a list that contains stars with more recently acquired designations could be interesting to consider as well. &lt;/p&gt;&#xA;" OwnerUserId="23940" LastActivityDate="2018-09-11T20:12:41.063" Title="Which star with a Bayer designation is the farthest from the Sun?" Tags="&lt;distances&gt;&lt;data-analysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27601" PostTypeId="1" AcceptedAnswerId="27616" CreationDate="2018-09-10T01:29:33.063" Score="4" ViewCount="162" Body="&lt;p&gt;Let's say, for the sake of speculation, we acquire the technology necessary to travel to the outer edge of one of the spiral arms of our galaxy. We go to a star in this &quot;outer rim&quot; and land on a planet orbiting a star there. What would we see when our side of the planet is facing outward from the galaxy? (Assume no atmospheric interference.) The answer will, of course, depend on the galactic coordinates of the star we choose to visit. Provided a particular set of coordinates, how much do we know about what we could see with the naked eye or through some form of optical instrument? I'm more interested in our ability to answer the question, rather than the specific objects we'd see (although they're pretty interesting too). You're welcome to select an arbitrary but still useful and appropriate boundary as the edge of the spiral arm. &lt;/p&gt;&#xA;" OwnerUserId="23940" LastActivityDate="2018-09-11T00:03:13.470" Title="What would stargazing be like at the edge of our galaxy?" Tags="&lt;galaxy&gt;&lt;observational-astronomy&gt;&lt;future&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27602" PostTypeId="1" CreationDate="2018-09-10T06:38:23.423" Score="0" ViewCount="90" Body="&lt;p&gt;As I learn, the gravity is caused by the spacetime curvature. Is there any gravitational/magnetic field that really exists around the Earth, or any other planets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the planets/stars rotate, does this cause the space time curvature to twist in the direction of their rotation? If it does then the gravitational pull should also twist the the path of the objects falling(though we may not perceive the curvature of this path, we should be &#xA;able to calculate at-least theoretically)&lt;/p&gt;&#xA;" OwnerUserId="24237" LastEditorUserId="24237" LastEditDate="2018-10-08T06:01:06.347" LastActivityDate="2018-10-08T06:01:06.347" Title="Space curvature and its rotation" Tags="&lt;gravity&gt;&lt;space-time&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27603" PostTypeId="1" AcceptedAnswerId="27613" CreationDate="2018-09-10T08:14:18.233" Score="1" ViewCount="95" Body="&lt;p&gt;I remember reading an old astronomy book mentioning a certain planet and the main point of interest was that this planet has or may have polar ice caps but of acid. I can't remember the name though, what planet or planets fit this description? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there some very strange worlds out there though I'm unsure whether this one may or may not be all that strange. Is this something of an oddity or is a planet containing frozen acid not that uncommon?&lt;/p&gt;&#xA;" OwnerUserId="24149" LastEditorUserId="7472" LastEditDate="2018-09-10T11:18:28.637" LastActivityDate="2018-09-10T20:09:17.687" Title="Planet with acid ice caps" Tags="&lt;planet&gt;&lt;identify-this-object&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27604" PostTypeId="2" ParentId="27602" CreationDate="2018-09-10T13:56:48.043" Score="1" Body="&lt;p&gt;Unlike gravitation, which is entirely a result of space-time curvature, magnetic fields do exist and there is a magnetic field around the Earth and many other planets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There is a deeper level of understanding of electromagnetism called Quantum Electrodynamics which explains how fields work in terms of more fundamental concepts, but unlike gravity, the E-M field does exist. (And there is hope -- so far unrealized -- that eventually all of the fields of nature plus gravity will be explained as aspects of one yet deeper thing.  But not today.))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frame dragging -- the distortion of space-time by a rotating body -- does exist and the link suggested by @PM (&lt;a href=&quot;http://en.wikipedia.org/wiki/Frame-dragging&quot; rel=&quot;nofollow noreferrer&quot;&gt;en.wikipedia.org/wiki/Frame-dragging&lt;/a&gt;) is a good place to start.  Frame dragging has been measured by careful observation of satellite orbits, and possibly by astronomical observations.  The Gravity Probe B satellite measure a closely related effect.  As far as I know, the effect is too small to have been measured in the lab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frame dragging does indeed affect the path of a falling body, though for the Earth the effect is really, really small.  It would be large near a millisecond pulsar or a rapidly rotating black hole.  It's also large for merging black holes and is an essential part of the calculations of the gravitational waves emitted during the final rotations before merger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect can be precisely calculated and agrees perfectly with the (still rather crude, alas!) experiments and observations.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-09-10T13:56:48.043" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27605" PostTypeId="1" AcceptedAnswerId="27974" CreationDate="2018-09-10T13:57:05.850" Score="3" ViewCount="133" Body="&lt;p&gt;This is just a matter of curiosity; I was thinking about a scifi story. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;Isolated&quot;, I mean that if for every star (or binary) we found the distance to its closest neighboring star, then which one would be the &lt;strong&gt;furthest&lt;/strong&gt; from its closest neighbor?&lt;/p&gt;&#xA;" OwnerUserId="24241" LastEditorUserId="24241" LastEditDate="2018-09-11T03:19:16.953" LastActivityDate="2018-11-10T19:01:55.387" Title="Of the stars in the Gaia / Hipparchos catalogs, which one is the most isolated?" Tags="&lt;distances&gt;&lt;milky-way&gt;" AnswerCount="2" CommentCount="17" ContentLicense="CC BY-SA 4.0" />
  <row Id="27606" PostTypeId="1" CreationDate="2018-09-10T14:32:59.547" Score="2" ViewCount="221" Body="&lt;p&gt;I find the original reference seems to be &lt;a href=&quot;http://adsabs.harvard.edu/abs/1975A%26A....39..185I&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. I still do not understand the mechanism. It is necessarily occur in neutron binaries? Anyone can introduce it in detail? This paper is old.&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2018-09-10T17:27:07.790" Title="What is Propeller Effect exactly?" Tags="&lt;black-hole&gt;&lt;astrophysics&gt;&lt;observational-astronomy&gt;&lt;neutron-star&gt;&lt;high-energy-astrophysics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27607" PostTypeId="1" AcceptedAnswerId="27617" CreationDate="2018-09-10T16:04:03.043" Score="6" ViewCount="373" Body="&lt;p&gt;This may sound like a strange question, but say that two stars are relatively next to one another in a binary star system, what would happen to one of the stars if the other went supernova? Would it explode as well, or would it survive the blast?&lt;/p&gt;&#xA;" OwnerUserId="24242" LastEditorUserId="7472" LastEditDate="2018-09-10T19:42:34.343" LastActivityDate="2018-09-11T00:57:28.943" Title="What happens when a star gets hit by a supernova?" Tags="&lt;supernova&gt;&lt;binary-star&gt;&lt;star-systems&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27608" PostTypeId="2" ParentId="27597" CreationDate="2018-09-10T16:59:41.167" Score="-1" Body="&lt;p&gt;The guy's a wack job.   It's atmospheric turbulence that just happens to have a long regular structure in one direction.   Heck, he might even be deliberately blowing a fan across his telescope to create those patterns.  Ignore him.&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-09-10T16:59:41.167" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="27609" PostTypeId="2" ParentId="27606" CreationDate="2018-09-10T17:27:07.790" Score="3" Body="&lt;p&gt;The magnetic field of the rapidly rotating neutron star interacts with the material coming from the other star in the binary. This results in a transfer of angular momentum, spinning the neutron star down but accelerating the material out of the system. The effect is somewhat like a garden sprinkler seen from above, with material flung out (&quot;propelled&quot;) of the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there needs to be a quite delicate balance between the binary period (or separation) in the system and the strength of the magnetic field and rotation rate of the neutron star to form the propeller effect. It seems to happen most in the early stages of the X-ray binary when the neutron star is rotating rapidly. After the neutron star spins down by the torques exerted on it, the material from the secondary star in the binary forms either an accretion disk or is directly accreted onto the neutron star down the magnetic  field lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect has also been seen, possibly only the once, in Cataclysmic Variable  (CV) systems, which consist of a low-mass star (typically a K or M dwarf) and a magnetized white dwarf. The &quot;poster child&quot; for this type of propeller system in CVs is AE Aquarii &lt;a href=&quot;http://adsabs.harvard.edu/abs/1997MNRAS.286..436W&quot; rel=&quot;nofollow noreferrer&quot;&gt;A magnetic propeller in the cataclysmic variable AE Aquarii; Wynn et al 1997&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The propeller theory seems to be alive and well as there over 800 citations to the original 1975 paper. A recent highly cited review discussing propellers and accreting neutron stars seems to be &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/321393/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;Alipar 2001&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-09-10T17:27:07.790" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27610" PostTypeId="2" ParentId="27595" CreationDate="2018-09-10T17:41:14.390" Score="3" Body="&lt;p&gt;Welcome !. I think you still need the co-ordinates of the star that you are interested in, its Right Ascension (the sky equivalent of longitude on the Earth) and the Declination (equivalent to latitude). Then as you say, you can use the hour angle (HA) formula:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HA = LST - RA&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The Hour Angle tells you how many hours west or east of the meridian your star is (and can be further converted into an azimuth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the altitude of the star, you need a bit of trig:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;altitude = arcsin(sin(latitude) * sin(declination) + cos(latitude) * cos(latitude) * cos(HA))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-09-10T17:41:14.390" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27611" PostTypeId="1" AcceptedAnswerId="27614" CreationDate="2018-09-10T18:36:10.840" Score="2" ViewCount="150" Body="&lt;p&gt;Mars has been undergoing a planet-wide dust storm for a couple of months now.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/gsS6H.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gsS6H.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By my eye, while Mars currently looks a bright and handsome orange, there's been no change that I've been able to notice from this dust storm. Is it possible for an experienced naked-eye (or small telescope) astronomer to have noticed the change in appearance brought on by this dust storm?&lt;/p&gt;&#xA;" OwnerUserId="17655" LastEditorUserId="6" LastEditDate="2018-12-10T14:24:52.020" LastActivityDate="2018-12-10T14:24:52.020" Title="Has the Mars dust storm affected its naked-eye appearance?" Tags="&lt;mars&gt;&lt;dust-storm&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27612" PostTypeId="2" ParentId="27607" CreationDate="2018-09-10T18:58:17.123" Score="2" Body="&lt;p&gt;There are certain cases where the remaining star survives; this happened to &lt;a href=&quot;https://en.wikipedia.org/wiki/X-ray_binary#High-mass_X-ray_binary&quot; rel=&quot;nofollow noreferrer&quot;&gt;high-mass X-ray binary star systems&lt;/a&gt;, pairs which consist of a neutron star or black hole (the remnant of a supernova) and a massive star. The X-rays are produced when some of the material ejected by the star is captured by the supernova remnant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most famous of these is &lt;a href=&quot;https://en.wikipedia.org/wiki/Cygnus_X-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cygnus X-1&lt;/a&gt;, the first object widely accepted to be a black hole. The massive star in this pair has the same name as &lt;a href=&quot;https://astronomy.stackexchange.com/users/2153/hde-226868&quot;&gt;one of our moderators&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2018-09-10T18:58:17.123" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27613" PostTypeId="2" ParentId="27603" CreationDate="2018-09-10T20:09:17.687" Score="1" Body="&lt;p&gt;Of the planets, two have ice caps: Earth and Mars.  The Earth's ice caps are water ice, and Mars has ice caps of Carbon dioxide, and some Water ice. The CO2 on Mars would give a weakly acidic solution if melted but it would not be a caustic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Titan has hydrocarbon lakes at its poles, and other planets are too hot, or are uniformly frozen.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-09-10T20:09:17.687" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27614" PostTypeId="2" ParentId="27611" CreationDate="2018-09-10T20:26:48.127" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it possible for an experienced naked-eye (or small telescope)&#xA;  astronomer to have noticed the change in appearance brought on by this&#xA;  dust storm?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, it is definitely possible. Without dust storms it looks redder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://alpo-j.asahikawa-med.ac.jp/kk01/JBAA119123-143McKim.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;The opposition of Mars, 2001: Part I - Richard McKim&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="23940" LastEditorUserId="6" LastEditDate="2018-12-03T14:57:24.177" LastActivityDate="2018-12-03T14:57:24.177" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27615" PostTypeId="1" AcceptedAnswerId="40953" CreationDate="2018-09-10T21:03:40.547" Score="5" ViewCount="482" Body="&lt;p&gt;How stable is a donut-shaped star or planet configuration and under what conditions such object may form?&#xA;Is there any evidence suggesting that such objects might exist?&lt;/p&gt;&#xA;" OwnerUserId="2571" LastEditorUserId="31410" LastEditDate="2021-01-20T13:46:59.200" LastActivityDate="2021-01-20T13:46:59.200" Title="Can a donut-shape planet or star be formed?" Tags="&lt;star-formation&gt;&lt;planetary-formation&gt;" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27616" PostTypeId="2" ParentId="27601" CreationDate="2018-09-11T00:03:13.470" Score="2" Body="&lt;p&gt;Here is a render of what an observer might see near the edge of our galaxy. The &quot;edge&quot; itself being rather fuzzy, let's assume that we are in an area of the edge dense enough where a few occasional stars are still visible. (As far as we know, intergalactic space contains objects and stars, but so few that seeing even one naked-eye star in the sky there would be uncommon)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the sky towards the opposite side of the Milky Way, one could see the Magellanic Clouds, and a small quantity of naked-eye stars:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gbiXU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gbiXU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Milky Way side of the sky would be interesting to look at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KnFjf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KnFjf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as interesting as these artistic renditions may be, there is very little that we know about actual objects in these regions. Seeing individual stars there is quasi-impossible from here; we have a general idea of how dense some regions of the galaxy are, but for the rendition part here, it is mostly procedural based on some general facts and statistics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The software used for these renders is called Space Engine and contains all catalogued objects, and it procedurally generates objects to fill in the gaps based on what we know about space. &lt;/p&gt;&#xA;" OwnerUserId="19814" LastActivityDate="2018-09-11T00:03:13.470" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27617" PostTypeId="2" ParentId="27607" CreationDate="2018-09-11T00:27:42.867" Score="2" Body="&lt;p&gt;Remember that although a supernova expels a huge amount of matter, this matter travels outwards in all directions and only a tiny fraction of it impacts on the binary partner. In the case of a core collapse supernova, the vast majority of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Supernova#Energy_output&quot; rel=&quot;nofollow noreferrer&quot;&gt;energy released&lt;/a&gt; is in the form of neutrinos, which will generally fly straight through the binary partner without impact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a &lt;a href=&quot;https://en.wikipedia.org/wiki/Type_Ia_supernova&quot; rel=&quot;nofollow noreferrer&quot;&gt;type 1a supernova&lt;/a&gt;, most of the released energy is kinetic. However, as the exploding star is completely disrupted, this &quot;kicks&quot; the other star out of the system (think of swinging a ball on a string in a circle around you, and you suddenly relax your grip on the string), so it's not hanging around for the shock wave. Since this is travelling at 5,000-20,000 km/s, it will eventually (within a few hours) catch up with the escaping star, but will have become substantially attenuated by then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to &lt;a href=&quot;https://astronomy.stackexchange.com/a/27612&quot;&gt;Glorfindel's answer&lt;/a&gt;, it's worth noting that where the remnant partner of a high-mass X-ray binary is a black hole, it's likely that the progenitor star was also high-mass and collapsed directly to singularity, without a significant explosion. As Felix Mirabel and Irapuan Rodrigues note in &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0305205&quot; rel=&quot;nofollow noreferrer&quot;&gt;Formation of a Black Hole in the Dark&lt;/a&gt;: &quot;The observations suggest that high-mass stellar black holes may form promptly, when massive stars disappear silently.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also be interested in the answers to these questions on our sibling site [Physics.SE]:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/q/25357&quot;&gt;What happens to the neighboring star of a type Ia supernova?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/q/30082&quot;&gt;Regarding binary systems (with pulsars)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14094" LastEditorUserId="14094" LastEditDate="2018-09-11T00:57:28.943" LastActivityDate="2018-09-11T00:57:28.943" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27618" PostTypeId="2" ParentId="27615" CreationDate="2018-09-11T08:21:37.043" Score="5" Body="&lt;p&gt;No, such a configuration would not be stable. Matter tends to clump together and as the mass increases the rotation slows. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the rotation were too fast for the material at any given size then the stresses would be too great for it to remain as a cohesive body. In these cases you would get something more like a &lt;a href=&quot;https://en.wikipedia.org/wiki/Rubble_pile&quot; rel=&quot;noreferrer&quot;&gt;rubble pile&lt;/a&gt;. If the rotation were somehow sufficiently high enough to theoretically form a donut shape, I think the gravitational force of the matter would not be enough for it to coalesce into any sort of cohesive body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider, for conservation of angular momentum the angular velocity decreases as the distance from the centre of rotation increases. That is, stuff closer to the centre rotates faster than stuff farther away. But for a solid body the angular rotation is the same at all distances from the centre. This means that the angular velocity (in degrees per unit of time) is the same at all distances, and in this case the stuff closer to the centre rotates more slowly than the stuff farther away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to form a solid body the stuff in the centre needs to slow down as it accretes matter - it rotates more slowly as it gets bigger. This would prevent forming a donut because the centripetal forces don't outweigh the gravitational forces. Fluid bodies like stars or planets with liquid/molten cores are more complex but the same idea applies.&lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2018-09-11T08:21:37.043" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27619" PostTypeId="1" AcceptedAnswerId="27620" CreationDate="2018-09-11T12:24:58.907" Score="1" ViewCount="192" Body="&lt;p&gt;For jets, I found a pdf file &lt;a href=&quot;http://www.mpia.de/homes/fendt/Lehre/Lecture_JETS/lect_jets3.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Jet: collimated beam of matter of high velocity&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;sources: young stars, active galactic nuclei, $\mu$-quasars, gamma­ ray bursts, planetary nebulae, pulsars&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;jet sources host accretion disks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How about winds and outflows?&lt;/strong&gt; Can somebody make some clarifications for the &lt;strong&gt;three&lt;/strong&gt; concepts， their &lt;strong&gt;correlation and difference&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="2270" LastEditDate="2018-09-11T14:50:11.420" LastActivityDate="2018-09-11T14:50:11.420" Title="What is the difference between wind, jet and outflow?" Tags="&lt;astrophysics&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27620" PostTypeId="2" ParentId="27619" CreationDate="2018-09-11T13:57:29.937" Score="1" Body="&lt;p&gt;I'm not sure there are very clear official definitions, but here is my take on accepted usage in the star formation community:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Outflows&lt;/strong&gt; - general term describing both winds and jets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Jets&lt;/strong&gt; - collimated outflows along magnetic poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Winds&lt;/strong&gt; - outflow from the stellar surface in all directions, not just along the poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stellar winds are what comes from individual bright stars. Jets can be protostellar jets from young stars, or on a larger scale from galactic nuclei.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2018-09-11T13:57:29.937" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27621" PostTypeId="2" ParentId="27615" CreationDate="2018-09-11T14:03:07.853" Score="0" Body="&lt;p&gt;Simple answer: no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shapes like a doughnut (this shape is called a torus) can exist in space, but not in dense material. Dense material becomes spherical because gravity drags it together and a sphere is the smallest it can get (size depends on what is opposing gravity - turbulence or radiation etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of a planet actually involves spherical shape because this is related to size and mass. Asteroids can be made of similar material to planets, and be orbiting the sun, but one of the reasons we don't call them planets because they are not spherical.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2018-09-11T14:03:07.853" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27622" PostTypeId="2" ParentId="27605" CreationDate="2018-09-11T14:09:55.347" Score="0" Body="&lt;p&gt;I don't think anyone knows the answer to this! People have only recently started looking at the Gaia catalogues and they contain so much data that most of it won't get looked at unless it is in an interesting area. Astronomers have started combing through the data seeing what we can find in our favourite regions but most people focus on the more densely populated areas (clusters and places where stars are forming) rather than the mainly empty areas (probably older stars not interacting with each other).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think this is something you could find in the catalogue, the information does exist, but no one is particularly interested in spending a lot of time looking for it because we won't learn much from those regions.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2018-09-11T14:09:55.347" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27623" PostTypeId="2" ParentId="27589" CreationDate="2018-09-11T14:18:32.857" Score="4" Body="&lt;p&gt;In science there is no such thing as an irrefutable fact, there is just the best theory we currently have that explains all the data and makes sense. That is very different from an assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;theory&lt;/strong&gt; is something that you can test - make models and predictions and check that they match up to real life. An &lt;strong&gt;assumption&lt;/strong&gt; is something that you don't have evidence for but you think it should make sense, so you choose to believe it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our current understanding of star formation and stellar nurseries is an excellent theory - it has been around for a long time, built up of many parts that have each been tested and confirmed, it matches observations in many ways. It's not perfect and there are still things we don't know (otherwise scientists would have no jobs) but in high school level textbooks we can reasonably present this as a 'fact' because we are pretty certain it won't change any time in the next 50 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An assumption would be like me assuming you are reading a high school textbook. It fits the facts (the level of detail in your question) but I haven't tested it (asked you about the book).&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2018-09-11T14:18:32.857" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27624" PostTypeId="2" ParentId="27589" CreationDate="2018-09-11T15:08:34.907" Score="2" Body="&lt;p&gt;Nothing in science is &quot;irrefutable&quot;, but some things are well enough established that only a fool bets against them.  Theories of stellar birth in nebulae are close to being in that category.  (But the devil can still be found lurking in the details.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the things that give us confidence that a theory correctly describes the world is when it is large, covering a wide range of phenomena with only a small set of moving parts.  The idea that the galaxy started as hot gas which condensed into stars and planets explains a very great many facts without &lt;em&gt;any&lt;/em&gt; special assumptions (meaning without assuming anything we can't verify in the lab) except the assumption that there was a large cloud of relatively cool gas and dust to begin with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that, we can see how the laws of physics discovered in the lab cause the cloud to break into smaller, denser clouds each of which then collapses in multiple places forming a cluster of new stars of the same age embedded in the remains of the nebula.  And those same laws of physics show that some of the gas and dust that doesn't make it into the new star will condense into planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a very parsimonious explanation.  I know of no alternative explanations for stars which don't assume a bunch of new stuff we haven't seen in the lab or, worse, assume stuff which contradicts what we've seen in the lab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On top of all the physics, &lt;em&gt;astronomy&lt;/em&gt; observes clouds of gas and dust and clusters of stars and all the things in between.  The process is too slow for us to have watched a specific star start as a nebula, condense out, and end up shining in empty space, but we see all the stages. (It's like seeing a crowd of people from little babies to young adults to the elderly -- even though you don't have time to watch a baby grow up, it's much more reasonable to conclude that you're seeing individuals of the same species at different stages of their life cycle than that you're seeing many different species.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory is also strong because it is easily falsifiable. Strong theories make predictions which can be tested and which might fail the test. Weak theories are so vague or hand-waving that they can accommodate &lt;em&gt;any&lt;/em&gt; observation without failing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that stars condense out of clouds of gas and dust is rock-solid science.  As with any science, we need to always be prepared to change our theories when confronted with new observations...but not before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There is still a &lt;em&gt;lot&lt;/em&gt; we don't know about the details of the whole process that turns a cloud of gas and dust into a planetary system -- it's a hot topic of current research.  But so far, anyway, there's nothing visible that suggests the basic theory is wrong; just that the details are complicated enough to keep us working to understand them.)&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-09-11T15:08:34.907" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27626" PostTypeId="2" ParentId="27600" CreationDate="2018-09-11T20:12:41.063" Score="2" Body="&lt;p&gt;&lt;strong&gt;See important caveat to this answer below&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a distance of 2992 light years, Tau Canis Major is the furthest of the 1513 stars that have a Bayer designation, with several others almost equally distant. I computed the full list of Bayer-Sun distances: &lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-bayer-distance.csv&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-bayer-distance.csv&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Important Caveat&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used the HYG catalog at &lt;a href=&quot;https://github.com/astronexus/HYG-Database&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/astronexus/HYG-Database&lt;/a&gt; but it may be inaccurate. In particular, Wikipedia gives Tau Canis Major's distance as 5,000 light years, although I couldn't find this number in the source provided (&lt;a href=&quot;https://doi.org/10.1088/0067-0049/199/1/8&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://doi.org/10.1088/0067-0049/199/1/8&lt;/a&gt;), although I only skimmed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The far from canonical &lt;a href=&quot;https://www.universeguide.com/star/taucanismajoris&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.universeguide.com/star/taucanismajoris&lt;/a&gt; notes that 2992.32 is a revised figure, down from 3197.68, but, of course, the figure may have been revised again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Methodology&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After downloading the HYG database file, this one liner creates bc-bayer-distance.csv:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;zcat hygdata_v3.csv.gz | perl -F, -anle 'if ($F[9] &amp;lt; 100000 &amp;amp;&amp;amp; length($F[27])&amp;gt;0) {$F[9] *= 3.26156; print &quot;$F[9], $F[27] $F[29]&quot;}' | sort -nr | tee bc-bayer-distance.csv&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A brief explanation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The 9th field (starting our count with 0 as Perl does) is the distance, but there's a note on the HYG git page: &quot;A value &gt;= 10000000 indicates missing or dubious (e.g., negative) parallax data in Hipparcos&quot;. It turns out the actual number is 100,000 (I've submitted a correction), so we ignore distances greater than 100,000.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The distance is in parsecs, so I multiply by 3.26156 to convert to light years.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The 27th field is the stars Bayer letter designation without the constellation. If it's empty, we ignore the star.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The 29th field is the stars constellation name, which, combined with 27th field, yields the Bayer designation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2018-09-11T20:12:41.063" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27627" PostTypeId="1" AcceptedAnswerId="27628" CreationDate="2018-09-11T20:24:33.140" Score="15" ViewCount="3027" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WTVic.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a general question and more in specific regarding the image above: Which part of the milky way, planets, stars, clusters are included in the image?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I know which entities are included in different pictures?&lt;/p&gt;&#xA;" OwnerDisplayName="nick" LastEditorUserId="11822" LastEditDate="2018-09-11T21:29:33.047" LastActivityDate="2018-09-12T12:16:50.153" Title="Which part of the milky way does the image show?" Tags="&lt;photography&gt;&lt;milky-way&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27628" PostTypeId="2" ParentId="27627" CreationDate="2018-09-11T20:32:59.953" Score="30" Body="&lt;p&gt;I wholeheartedly recommend &lt;a href=&quot;http://nova.astrometry.net/&quot; rel=&quot;noreferrer&quot;&gt;astrometry.net&lt;/a&gt; for this sort of thing. It Just Works(tm); running on your image produced &lt;a href=&quot;http://nova.astrometry.net/user_images/2316280#annotated&quot; rel=&quot;noreferrer&quot;&gt;this output&lt;/a&gt; with absolutely no hints or guidance from me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bs86b.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bs86b.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For avoidance of doubt, I have no association with astrometry.net.&lt;/p&gt;&#xA;" OwnerUserId="2210" OwnerDisplayName="Philip Kendall" LastActivityDate="2018-09-11T20:32:59.953" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27629" PostTypeId="1" AcceptedAnswerId="27630" CreationDate="2018-09-11T22:44:22.873" Score="2" ViewCount="369" Body="&lt;p&gt;I currently have Celestron Nexstar 130STL.  I am trying to upgrade my telescope to next one.  I examined Meade 8&quot;ACF Goto telescope, but I was not too thrilled about it.  It gave me better view, but the details were not too different from my 130mm. &#xA;Maybe my expectation was too big.  I have an option of getting Orion 10&quot; dob motorized alternatively.  I am slightly concerned about the size and the portability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate any recommendation, thanks!&lt;/p&gt;&#xA;" OwnerUserId="24100" LastActivityDate="2018-09-14T03:00:04.493" Title="Upgrading telescope from 5&quot; to 8&quot; or 10&quot;?" Tags="&lt;telescope&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27630" PostTypeId="2" ParentId="27629" CreationDate="2018-09-12T01:35:28.170" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am trying to upgrade my telescope to next one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To compare the light-gathering power of two telescopes, all other properties being equal, you can calculate the ratio of the squares of their radii. In this sense, a telescope with a larger aperture will provide a &quot;better&quot; view than one with a smaller one. Keep in mind, though, that there are other factors to consider: the impact of different optical systems on light gathering power, the use of filters and eyepieces, the unpredictable effects of atmospheric interference, and skill in using whatever telescope you buy. You'll also have to consider specifically what you want to do with the scope (visual observation, astrophotography, citizen science, etc).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I would appreciate any recommendation, thanks!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This forum tends to be for scientific questions in astronomy, not for equipment recommendations. There's a forum primarily for equipment discussion on a separate site called &lt;a href=&quot;https://www.cloudynights.com/forum/57-beginners-forum/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cloudy Nights&lt;/a&gt; that you might want to visit. You can certainly post here, but you may not find people who are able to advise you on specific equipment issues.&lt;/p&gt;&#xA;" OwnerUserId="23940" LastEditorUserId="23940" LastEditDate="2018-09-13T00:24:56.323" LastActivityDate="2018-09-13T00:24:56.323" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27631" PostTypeId="2" ParentId="27627" CreationDate="2018-09-12T01:39:57.637" Score="6" Body="&lt;p&gt;Continuing on Philip Kendall's answer, Saturn is the bright object to the lower left of center (on the line from 14 Sgr to Kaus Borealis in the astrometry.net version of the image).&lt;/p&gt;&#xA;" OwnerUserId="17355" LastEditorUserId="2210" LastEditDate="2018-09-12T12:16:50.153" LastActivityDate="2018-09-12T12:16:50.153" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27632" PostTypeId="2" ParentId="27627" CreationDate="2018-09-12T02:18:30.837" Score="6" Body="&lt;p&gt;In addition to Saturn and the stars astrometry.net circled, the frame includes several &lt;a href=&quot;https://freestarcharts.com/messier&quot; rel=&quot;noreferrer&quot;&gt;Messier objects&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;open clusters M16, M18, M21, M23, M24, M25&lt;/li&gt;&#xA;&lt;li&gt;globular clusters M22, M28&lt;/li&gt;&#xA;&lt;li&gt;bright nebulae M8, M17, M20&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;https://freestarcharts.com/images/Articles/Messier/M16_M17_M18_M23_M24_M25_Finder_Chart.pdf&quot; rel=&quot;noreferrer&quot;&gt;chart from freestarcharts.com&lt;/a&gt; indicates which is which.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-09-12T02:25:52.427" LastActivityDate="2018-09-12T02:25:52.427" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27633" PostTypeId="1" AcceptedAnswerId="27634" CreationDate="2018-09-12T14:22:44.813" Score="29" ViewCount="7539" Body="&lt;p&gt;I was wondering if any planes had mounted telescopes with the intent to observe the stars. I understand that the atmosphere itself can warp and hinder incoming light and even completely obscure views on cloudy days. Would it be possible to have a telescope which could account for the speed of the airplane and remove motion blur? Would it even be worth it, seeing as the size of the lens would also be limited? Could we see anything interesting that a terrestrial scope of the same size would miss?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hobbes has mentioned SOFIA as a great example of this being done. Does anyone have more information on the benefits of a plane mounted telescope over one on the ground for noninfrared observation?&lt;/p&gt;&#xA;" OwnerUserId="23729" OwnerDisplayName="Magic Octopus Urn" LastEditorUserId="37191" LastEditDate="2021-02-12T10:18:48.247" LastActivityDate="2021-02-12T10:18:48.247" Title="Is it possible to mount a telescope on a plane? Is it beneficial?" Tags="&lt;telescope&gt;&lt;sofia&gt;" AnswerCount="5" CommentCount="10" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27634" PostTypeId="2" ParentId="27633" CreationDate="2018-09-12T14:32:41.027" Score="53" Body="&lt;p&gt;This has been done.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.nasa.gov/mission_pages/SOFIA/overview/index.html&quot; rel=&quot;noreferrer&quot;&gt;SOFIA is an infrared observatory&lt;/a&gt; built into a Boeing 747 SP:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/D3BLt.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D3BLt.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;SOFIA takes advantage of the fact that some infrared bands are visible at atltitude, these are attenuated by water in the atmosphere so they're less visible on the ground.&lt;/p&gt;&#xA;&lt;p&gt;There have been infrared observatories before SOFIA:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The first use of an aircraft for performing infrared observations was in 1965 when Gerard P. Kuiper used the NASA Convair 990 to study Venus. Three years later, Frank Low used the Ames Learjet for observations of Jupiter and nebulae.[20] In 1969, planning began for mounting a 910 mm (36 in) telescope on an airborne platform. The goal was to perform astronomy from the stratosphere, where there was a much lower optical depth from water-vapor-absorbed infrared radiation. This project, named the Kuiper Airborne Observatory, was dedicated on May 21, 1975. The telescope was instrumental in numerous scientific studies, including the discovery of the ring system around the planet Uranus.[21]&lt;/p&gt;&#xA;&lt;p&gt;The proposal for a larger aircraft-mounted telescope was officially presented in 1984 and called for a Boeing 747 to carry a three-meter telescope. The preliminary system concept was published in 1987 in a Red Book. It was agreed that Germany would contribute 20% of the total cost and provide the telescope.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Other airborne astronomy is more incidental. Around solar eclipses, you'll often see some astronomy flights. Some of these are tourism, others perform science. These are usually passenger aircraft temporarily modified (scopes installed that look through the existing windows). These take advantage of the fact you can lengthen the eclipse by flying along its path, and you can reach eclipses in places otherwise inaccessible.&lt;/p&gt;&#xA;" OwnerUserId="361" OwnerDisplayName="Hobbes" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-09-12T14:32:41.027" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27635" PostTypeId="2" ParentId="27615" CreationDate="2018-09-12T15:27:10.230" Score="1" Body="&lt;p&gt;COnsider this counterexample: the asteroid belt.  Taken as a whole it's toroidal, but only because all the components are revolving about the sun.  They wouldn't even form a torus in the absence of their orbital parameters. Larger clumps (i.e. protoplanets) can only form due to gravitational attraction, which produces straight-line attractive forces (as opposed to, e.g., the path of a free electron in a magnetic field).  All material is drawn towards a common center. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-09-12T15:27:10.230" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27636" PostTypeId="2" ParentId="27633" CreationDate="2018-09-12T16:20:11.247" Score="8" Body="&lt;p&gt;SOFIA (as described in @Hobbes answer) is currently the only astronomical telescope mounted on a plane. It does overcome some problems with the Earth's atmosphere and circumvents the huge costs and extreme environmental conditions of a space telescope, but it comes with a whole host of other difficulties!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plane can only take off and land in certain places, and it's better if it lands where it came from so the observations (which only come from one side of the plane) are looking at different places on the way out and the return. Coupled with the fact that the plane has a limited amount of fuel means you don't get very long integration times (how long the telescope is looking at one patch of sky) so you can't see a lot of detail or very faint objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plane shakes, moves and changes temperature, and all of that affects the telescope and reduces the quality of your observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, while this might be cheaper than a space mission, and give more opportunity for upgrades and maintenance, it's not super cheap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, SOFIA is cool, but probably no one else is going to build one and I'm not sure how much longer SOFIA will be funded. It is a good alternative in some ways but not good enough to make it viable.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2018-09-12T16:20:11.247" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27637" PostTypeId="2" ParentId="27633" CreationDate="2018-09-12T16:50:25.037" Score="11" Body="&lt;p&gt;As others have pointed out, SOFIA does exactly this.  The USAF has done some work with airborne telescopes.  I believe the US Navy has as well. I can only assume the Russians did during the years of the Soviet Union and perhaps continue to do so.  Not sure about anyone else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why the military? For imaging satellites and the like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the most part, airborne telescopes aren't worth the cost or trouble.  That's not to say they never are, but there's some big hurdles to overcome.  You have to provide an ultra-stable platform that compensates for all motion and vibration (I'm guessing they use strong gyroscopes for this), and cut a hole in the roof of the plane for it to stick out of (because using a window will screw with your image quality).  Then you have to have a plane that can fly high enough for long enough to make it worthwhile.  And THEN, you only get a limited amount of use out of it.  Add to the cost of building and maintaining the telescope the cost of modifying and maintaining a jet aircraft, costs which are non-trivial, and then take into account the fact you can't realistically carry all that large a telescope this way... it's just very impractical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SOFIA does some amazing things, but it is also very limited and specialized for those things.  Even then, it doesn't get anywhere near high enough to avoid all of the atmospheric filtering of IR energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For non-infrared observing, just dealing with the atmosphere, it's MUCH more cost-effective to construct a large telescope on top of a mountain where the air is much thinner and steadier, and then using active optics to enhance image quality.  And then, because these kinds of telescopes are used for photographic observation as opposed to active visual, you can further make use of speckle interferometry to further enhance the detail captured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, yes, it's possible, just not effective, to mount telescopes in aircraft.&lt;/p&gt;&#xA;" OwnerUserId="10546" LastActivityDate="2018-09-12T16:50:25.037" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="27638" PostTypeId="1" AcceptedAnswerId="27640" CreationDate="2018-09-12T18:40:49.120" Score="4" ViewCount="686" Body="&lt;p&gt;A while ago I saw something in the sky early in the evening. At first, I thought it was a plane since it seemed far too bright to be a star - no other stars could be seen. However, it didn't move, so I ruled out that possibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seemed to be relatively close to the horizon so I considered if it was a very high pole with some sort of light at the top, but after I walked for a few kilometres it was still at the same location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the few hours I looked at it, it moved a bit across the sky. &#xA;I considered a satellite, but it didn't seem to move quite enough of what I anticipated a satellite would move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since then, I have seen it every evening when I remember to look for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from this object only the moon could be seen. To me, it seemed like the moon moved in the same arched way across the sky as the object, but I could be wrong on the.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen several questions on this site where someone asks &quot;what was it?&quot; which has been closed with &quot;too broad&quot;. So I am not going to ask what it was.&#xA;Instead I am curious about what I can do to figure out what it is? Which observations is relevant to make when looking at such objects? As someone who is curious about the sky, but does not have any instruments for doing any actual measurements can I actually do anything?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first picture shows the object in the sky.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/iqVWa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iqVWa.png&quot; alt=&quot;Object in the sky&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second picture shows the same object, but where I have indicated the arch the object moved in as well as the moon and how that moved.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/8XEjk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8XEjk.png&quot; alt=&quot;Object in the sky and arched it moved in.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24256" LastActivityDate="2018-09-15T07:37:42.710" Title="How to figure out what something in the sky is?" Tags="&lt;identify-this-object&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27639" PostTypeId="2" ParentId="27638" CreationDate="2018-09-12T19:25:05.557" Score="4" Body="&lt;p&gt;It's hard to say 100% for sure without more info.  What is your location and what date/time was this?  How fast did it seem to move?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, it looks like Venus.  If it looked like a bright star, but much bigger and brighter than other stars, in the west around sunset, and moving slowly toward the horizon, then it was most likely Venus.&lt;/p&gt;&#xA;" OwnerUserId="10546" LastActivityDate="2018-09-12T19:25:05.557" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27640" PostTypeId="2" ParentId="27638" CreationDate="2018-09-12T20:04:05.570" Score="4" Body="&lt;p&gt;It could be Venus or Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;To help you identify the object, you could install &lt;a href=&quot;http://stellarium.org/en/&quot; rel=&quot;noreferrer&quot;&gt;Stellarium&lt;/a&gt;, an open source planetarium.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You could also try to record the azimuth with a compass as well as the exact time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A compact camera with a good zoom should be able to show the phase of Venus. If it looks like a very small waxing moon, it is Venus. If it's a full disk, it's probably Jupiter. These guidelines are only valid now, and will change over time. Stellarium will help.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="18474" LastActivityDate="2018-09-12T20:04:05.570" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27641" PostTypeId="1" AcceptedAnswerId="27656" CreationDate="2018-09-12T20:18:16.287" Score="2" ViewCount="304" Body="&lt;p&gt;I was solving a task that said (paraphrasing):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What percentage of the night sky can the Moon cover during the entire&#xA;  year, when observed from all points on Earth (what percentage of the&#xA;  night sky can be studied with the occultation method)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;This is a task I read from a book, and the solution states that the Moon covers a &quot;ribbon&quot; of width $\alpha$ where $\alpha = 0.5 ^{\circ} + 2*1^{\circ} + 2*5.3 ^ {\circ} = $ angular diameter of the Moon + 2*the Moon's daily parallax + 2*the inclination of the Moon's orbit to the ecliptic; and length $360 ^{\circ}$.&lt;br&gt;&#xA;Why is this like this, I understand the diameter and parallax part, but I couldn't visualize the problem very well. Why is the third number the inclination to the ecliptic and not to the equator?&lt;/p&gt;&#xA;" OwnerUserId="24014" LastActivityDate="2018-09-14T16:54:48.157" Title="What percentage of the celestial sphere can the Moon cover?" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;apparent-motion&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27643" PostTypeId="2" ParentId="27633" CreationDate="2018-09-13T02:47:38.817" Score="9" Body="&lt;p&gt;The other answers don't seem to stress one advantage of &lt;s&gt;snakes&lt;/s&gt; telescopes on a plane; that of &lt;em&gt;rapid movement and portability!&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here are examples of&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;IR viewed by a telescope on a plane other than SOFIA, and&lt;/li&gt;&#xA;&lt;li&gt;observation using SOFIA's plane but not it's large IR telescope!&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The questions &lt;a href=&quot;https://space.stackexchange.com/q/22411/12102&quot;&gt;Why the thermal imaging of Mercury's surface requires a telescope on a jet flying through an eclipse?&lt;/a&gt; as well as the linked NASA page &lt;a href=&quot;https://www.nasa.gov/feature/goddard/2017/chasing-the-total-solar-eclipse-from-nasa-s-wb-57f-jets&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chasing the Total Solar Eclipse from NASA’s WB-57F Jets&lt;/a&gt; describes two telescope-equipped jets:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Amir Caspi of the Southwest Research Institute in Boulder, Colorado, and his team will use two of NASA’s WB-57F research jets to chase the darkness across America on Aug. 21. Taking observations from twin telescopes mounted on the noses of the planes, Caspi will ­­­­­capture the clearest images of the Sun’s outer atmosphere — the corona — to date and the first-ever thermal images of Mercury, revealing how temperature varies across the planet’s surface.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;While IR is involved, the jets are necessary because of the speed at which the solar eclipse umbra moves relative to the surface of the Earth, or in this case, the speed relative to the atmosphere which due to friction mostly moves with the Earth's rotation.&lt;/p&gt;&#xA;&lt;p&gt;So &lt;strong&gt;SOFIA is not the only telescope on a plane!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Watch the video &lt;a href=&quot;https://www.youtube.com/watch?v=R0GNqlGNZkI&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA Jets Chase The Total Solar Eclipse&lt;/a&gt;! Here is a screenshot:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3lEHu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3lEHu.jpg&quot; alt=&quot;Screenshot from NASA Jets Chase The Total Solar Eclipse&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The question &lt;a href=&quot;https://space.stackexchange.com/q/21773/12102&quot;&gt;Timing shadows from the Kuiper belt! Any news? Did it work?&lt;/a&gt; describes the use of the SOFIA observatory, &lt;em&gt;but not it's primary infrared telescope&lt;/em&gt; to observe the occultation of a star:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;update 24-Jun-2017:&lt;/strong&gt; &lt;a href=&quot;https://www.nasa.gov/feature/sofia-arrives-in-new-zealand-to-observe-southern-skies&quot; rel=&quot;nofollow noreferrer&quot;&gt;SOFIA Arrives in New Zealand to Observe Southern Skies&lt;/a&gt; There are plans to fly SOFIA through another &lt;em&gt;predicted&lt;/em&gt; occultation path on July 17. &lt;a href=&quot;https://en.wikipedia.org/wiki/Stratospheric_Observatory_for_Infrared_Astronomy&quot; rel=&quot;nofollow noreferrer&quot;&gt;SOFIA&lt;/a&gt; is a huge (2.5 meter dia.) infrared telescope (1 ~ 250 um) with various cryogenic focal plane arrays that is flown above most of the water in Earth's atmosphere (&lt;a href=&quot;https://youtu.be/g5z6fZKOtP4&quot; rel=&quot;nofollow noreferrer&quot;&gt;YouTube&lt;/a&gt;). &lt;strong&gt;It &lt;a href=&quot;https://www.nasa.gov/ames/nisv-podcast-William-Reach&quot; rel=&quot;nofollow noreferrer&quot;&gt;sound like&lt;/a&gt; they will just use the visible light guide camera (behind the &lt;a href=&quot;https://en.wikipedia.org/wiki/Nasmyth_telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nasmyth&lt;/a&gt; mirror?), rather than the infrared capability, but the portability is certainly handy.&lt;/strong&gt; This mission might be &amp;quot;An Airplane Hunting for Shadows from the Kuiper Belt&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So it seems that in this case they are &lt;em&gt;using SOFIA for portability&lt;/em&gt;, but not necessarily it's infrared capability, and not even for the large telescope!&lt;/p&gt;&#xA;&lt;p&gt;See the NASA page &lt;a href=&quot;https://www.nasa.gov/feature/sofia-arrives-in-new-zealand-to-observe-southern-skies&quot; rel=&quot;nofollow noreferrer&quot;&gt;SOFIA Arrives in New Zealand to Observe Southern Skies&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NSp4X.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NSp4X.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; From AsteroidOccultations.com's &lt;a href=&quot;http://www.asteroidoccultation.com/observations/NA/MU69Binzel.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;News &amp;amp; Announcements&lt;/a&gt; for 2014 MU69&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;https://space.stackexchange.com/a/22095/12102&quot;&gt;this answer&lt;/a&gt; (this particular image sequence is from a ground telescope):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;h2&gt;Update July 20&lt;/h2&gt;&#xA;&lt;p&gt;At least &lt;a href=&quot;https://www.sciencenews.org/blog/science-ticker/new-horizons-next-target-caught-making-star-blink?tgt=nr&quot; rel=&quot;nofollow noreferrer&quot;&gt;5 telescopes&lt;/a&gt; have captured the occultation during the &lt;a href=&quot;https://www.nasa.gov/feature/nasa-s-new-horizons-team-strikes-gold-in-argentina&quot; rel=&quot;nofollow noreferrer&quot;&gt;July 17 observation campaign&lt;/a&gt; (animated GIF):&lt;br /&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Cpg4r.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cpg4r.gif&quot; alt=&quot;blink&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-09-14T09:49:21.400" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27644" PostTypeId="1" AcceptedAnswerId="27647" CreationDate="2018-09-13T08:16:29.240" Score="4" ViewCount="95" Body="&lt;p&gt;How do astronomers find the difference between a 'Cephid Variable Star' and distinct normal stars with opaque objects blocking it's light, like stars with planets revolving around them while both create a dip in the brightness-time graph at regular intervals of time, when this is how new stars are discovered?&lt;/p&gt;&#xA;" OwnerUserId="24255" LastActivityDate="2018-09-13T12:59:45.453" Title="Classification and discovery of stars" Tags="&lt;star&gt;&lt;classification&gt;&lt;cepheids&gt;&lt;light-curve&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27645" PostTypeId="2" ParentId="7780" CreationDate="2018-09-13T10:30:43.883" Score="2" Body="&lt;p&gt;First a warning, then a hack:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If you didn't run this simulation, if you are not a trained theoretical astronomer of undergraduate level or above, and if this simulation doesn't have the kind of data you need to answer the question you are asking &lt;strong&gt;you should not trust the results at all&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It takes years of study to understand the weird numerical effects (things that happen in the simulation because of the way it is set up that would never happen in reality), background physics and loads of assumptions and approximations that go into this kind of stuff. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We often say 'garbage in, garbage out'. The answers you get from this will probably not be very illuminating because they say more about the simulation than real space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) That said, if you are just doing this for fun or to learn the skills, GREAT! Keep playing! And a very basic assumption you can make in order to get some numbers out (that will be quite wrong) is that all gas in space is ~10K.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2018-09-13T10:30:43.883" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27646" PostTypeId="2" ParentId="27644" CreationDate="2018-09-13T11:12:53.690" Score="0" Body="&lt;p&gt;A Cepheid Variable is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a type of star that pulsates radially, varying in both diameter and temperature and producing changes in brightness with a well-defined stable period and amplitude. There is a strong direct relationship between a Cepheid variable's luminosity and pulsation period.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(From Wikipedia)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A star with planets does not give a regular dip in the same way, because orbits are not going to be exactly in line with our view, and the dip due to a planet is incredibly small. From &lt;a href=&quot;http://spiff.rit.edu/classes/resceu/lectures/transit_i/transit_i.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;spiff.rit.edu:&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;micromags: one micromag is one-millionth of a magnitude, or 0.000 001 magnitudes. People who look for transits by terrestrial planets sometimes use this unit: Earth produces a dip about 100 micromags deep.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no direct relationship between luminosity and period.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2018-09-13T11:12:53.690" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27647" PostTypeId="2" ParentId="27644" CreationDate="2018-09-13T12:43:50.733" Score="4" Body="&lt;p&gt;Cepheid variable stars and exoplanets transiting stars have very different light curves (the relation between brightness and time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exoplanet light curve from &lt;a href=&quot;https://www.nasa.gov/mission_pages/kepler/multimedia/images/transit-light-curve.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eCfNg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eCfNg.jpg&quot; alt=&quot;Exoplanet light curve from NASA&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cepheid variable light curve from &lt;a href=&quot;http://www.astronomynotes.com/ismnotes/s5.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;astronomynotes.com&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AO5Yy.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AO5Yy.gif&quot; alt=&quot;Cepheid variable light curve from astronomynotes.com&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as Rory Allsop points out, the scale of the change in brightness is very different. Cepheid variables can be seen at great distances, which allows us to use them to measure distances by their regular period-brighness relation, whereas exoplanets can only be found more locally by this method.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastEditorUserId="2270" LastEditDate="2018-09-13T12:59:45.453" LastActivityDate="2018-09-13T12:59:45.453" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27648" PostTypeId="2" ParentId="21623" CreationDate="2018-09-13T13:15:09.193" Score="3" Body="&lt;p&gt;There are &lt;em&gt;two&lt;/em&gt; Friedman equations. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One related to the energy-density component in the Einstein field equations.&lt;/li&gt;&#xA;&lt;li&gt;One related to the pressure component in the Einstein field equations.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The Einstein field equations are the following ten equations (16 equations for 4x4 tensor components but only ten of them are independent due to symmetry)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G_{\mu\nu} + \Lambda g_{\mu\nu} = \kappa T_{\mu\nu} = \frac{8 \pi G}{c^4} T_{\mu\nu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2} g_{\mu\nu}R$ is the Einstein tensor, $g_{\mu\nu}$ the metric tensor, $\Lambda$ is the cosmological constant, $\kappa = \frac{8 \pi G}{c^4}$ is the Einstein constant, $T_{\mu\nu}$ is the energy-momentum tensor&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Friedman was actually assuming that only the matter/energy density component is non-zero $T_{44}= \epsilon = \rho c^2$. He assumed &lt;em&gt;&quot;Die Materie ist inkohärent&quot;&lt;/em&gt; (ie non interacting dust with $p=0$) by which the components $T_{ii}$ are non-zero for $i \neq 4$. The other components (energy flux and momentum) are set equal to zero based on the assumption that the motion of energy and matter is considered to be negligibly small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LeMaître kept the pressure in place arguing that the pressure associated with the kinetic energy of matter might be negligible but this may not be the case for radiation pressure. So he used also $T_{11} = T_{22} = T_{33} = -p$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann%E2%80%93Lema%C3%AEtre%E2%80%93Robertson%E2%80%93Walker_metric&quot; rel=&quot;nofollow noreferrer&quot;&gt;Robertson Walker metric&lt;/a&gt; metric $d s^2 =  dt^2 - a(t)^2 \left[ \frac{dr^2}{1-k r^2}+r^2d\Omega^2\right]$ the equations become expressions in terms or the scale factor $a(t)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{rcrcl} G_{00} &amp;amp;=&amp;amp; c^{-2}\frac{3k}{a^2} + 3c^{-2} \frac{\dot a^2}{a^2} &amp;amp;=&amp;amp; \frac{8 \pi G}{c^4} \rho c^2 - \Lambda  \\ -G_{11} = -G_{22} = -G_{33} &amp;amp;=&amp;amp; c^{-2}\frac{k}{a^2} + c^{-2}\frac{\dot a^2}{a^2} + 2 c^{-2} \frac{\ddot a}{a} &amp;amp;=&amp;amp; -\frac{8 \pi G}{c^4} p + \Lambda \end{array} $$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you subtract the second equations from the first equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G_{00}+G_{11}+G_{22}+G_{33} = -6 c^{-2} \frac{\ddot a}{a} =&#xA;  \frac{8 \pi G}{c^4} (3p+ \rho c^2) + 2\Lambda   $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and rearrange some terms, then you could also write &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA; \frac{\ddot a}{a} = -\frac{4 \pi G}{3}  \left( \rho + \frac{3p}{c^2} \right) + \frac{ \Lambda c^2}{3}   $$&lt;/p&gt;&#xA;" OwnerUserId="20819" LastActivityDate="2018-09-13T13:15:09.193" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27649" PostTypeId="2" ParentId="21623" CreationDate="2018-09-13T15:29:54.920" Score="1" Body="&lt;p&gt;That was a lovely answer.  Since in a comment, titansarus asked for the connection between 3p (I'll take c=1, it's just a constant) and the radiation energy density, I will add that it is easy to show that pressure is always E/3 for any isotropic relativistic gas (where E is kinetic energy density and &quot;relativistic&quot; means rest energy is unimportant, so each particle has its kinetic energy equal to the magnitude of its momentum).  Pressure is momentum flux per area, so form a ratio of that to the kinetic energy density, for all particles in a given energy bin, integrating over all directions.  If that comes out 1/3, you have your answer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numerator of that ratio is the integral over direction cosine of the particle density times its momentum times its speed (here 1) times the square of the direction cosine (that's what gives momentum flux, one cosine comes from the normal component of the momentum, and the other from the need to transport it across the boundary via the speed normal to the boundary), and the denominator is exactly the same integral except without the square of the direction cosine (because it's energy density, so has neither of those two reasons to include the cosine).  Hence, the ratio is just the integral over the direction cosine of the cosine squared, divided by the integral over the direction cosine.  That's 1/3, simple as that.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-09-13T15:29:54.920" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27650" PostTypeId="2" ParentId="27629" CreationDate="2018-09-13T17:06:45.657" Score="1" Body="&lt;p&gt;The first thing I would recommend is to join an astronomy club or society.  There's most likely one near you, and membership is generally very affordable (in the US, prices usually are $50 or less per year).  Most clubs conduct freqeunt star parties, which would give you a chance to see a wide variety of telescopes in action and talk to their owners to find out more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Alphecca replied, the light gathering power of a telescope is proportional to the square of the radius of the aperture.  For example, your NexStar 130SLT has an aperture of 130 mm (just over 5&quot;).  If you use the formula for area (pi * r^2), this gives you an aperture area of about 13,273 mm^2 (square millimeters).  If you were to upgrade to an 8&quot; telescope (203 mm aperture), you'd get an aperture area of about 32,365 mm^2, which is about 240% the light gathering area of your 5&quot; scope.  If you go to a 10&quot; (254 mm), you end up with about 50,671 mm^2, or about 380% as much as the 5&quot; or 157% as much as the 8&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this actually mean?  That's a little more complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, there is more light gathering.  Contrary to the common opinion of those not experienced in using telescopes, magnification is not the key concern with a telescope, light gathering is.  But there's more to it than just that, and I'll circle back around shortly.   In the case of light gathering, larger apertures are better.  The telescope, in a sense, acts as a funnel for light, collecting from a larger area and concentrating on a smaller area.  In this sense, a larger telescope can, quite simply, gather more light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there's a catch: it's gathering ALL of the light that hits it, not just the light from the target object.  It will gather the light of your target object and any stray light, such as light pollution.  To deal with this, we need to increase contrast to make the dim deep sky object stand out from the sky.  This is heavily influenced, then, by things such as focal ratio and your eyepiece selection.   I spend a fair amount of time on Reddit r/telescopes, and we've had some discussions about this. One of the other regulars there wrote a &lt;a href=&quot;https://medium.com/@phpdevster/how-telescope-aperture-affects-your-view-24507147d7fc&quot; rel=&quot;nofollow noreferrer&quot;&gt;pretty good article on medium.com about this&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to put it mildly, there's more invovled than just getting a bigger aperture.  Though, in fairness, having the biggest aperture you can afford and handle as far as weight and size goes is always the best way to go - you will never have too LITTLE aperture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real questions then become what you want to do with it.  If you want to do astrophotography, you're going in entirely the wrong direction.  For visual observation, however, an 8 or 10 inch telescope is a good option.  I personally recommend an 8 inch for any beginner that can afford one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, based on what you said about being disappointed with the Meade 8&quot;, I think there is a question of expectation.  If you expect to see what you see in photographs, then you're just plain out of luck. Even a very large scope (say a 24&quot; Dob) won't do that for you. Our eyes are simply not capable of long exposure imaging as are cameras.  But a good observer who has learned to train their eye properly can still see some pretty amazing sights.  And this is another good reason to join a club: to spend time observing with other, more experienced observers and learning how to truly see through the eyepiece.  Visual observation is not just a matter of pointing the telescope in the right spot and taking a look.  Observing requires spending time letting your eye and brain work together to really see what's out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I jsut said our eyes aren't capable of long exposures.  This is true, but it's also true that extended observing does do something that's not entirely unlike long exposure imaging.  As you spend time at the eyepiece, your eyes collect more and more light and your brain does start to synthesize a more complete image.  It starts to fill in gaps and show you more and more.  This is especially the case when observing from darker skies where there is less light pollution to muddle the view.  A great example in my experience is M51.  From my club's dark site, when it's up in the sky, I can almost alawys make out the two galactic cores, but frequently the spiral structure is hard to see.  But if I have a comfortable seated viewing position and spend a few minutes, my eyes pick up hints of the spiral arms, which my brain starts to fill in.  A good technique for developing this is sketching.  If you sketch the object, your brain process it differently and you start to catch hints and glimpses of more detail which end up in your sketch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your current telescope is a GoTo, as is the Meade you mentioned, and you talked about an 10&quot; motorized Dob.  While GoTo technology and tracking can be very helpful, they are also often more of a hindrance than expected.  Most particuarly, the cost of the GoTo mounting inflates the overall cost of the instrument.  An 8&quot; standard Dob can be bought or under \$500 (occasionally under \$400).  An 8&quot; GoTo telescope is likely to run over \$1,000.  The GoTo equipment typically will at least double the price.  For the price of an 8&quot; GoTo dob you could get a 12&quot; manual Trus dob with almost 550% the light gathering area of your 5&quot;.  (Not to mention a larger aperture offers greater detail resolution capability for planetary observation). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, my recommendation is to start by joining a club and spending some time learning more.  Then you can make a more informed decision on a purchase (additionally, many clubs have a thriving buy/sell/trade culture with members constantly selling off something to fund their next purchse - I've bought most of my equipment at very good prices this way).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck and clear skies!&lt;/p&gt;&#xA;" OwnerUserId="10546" LastEditorUserId="10546" LastEditDate="2018-09-14T03:00:04.493" LastActivityDate="2018-09-14T03:00:04.493" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27651" PostTypeId="2" ParentId="1621" CreationDate="2018-09-14T00:07:41.983" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Gerald quoted Wikipedia: &quot;At high altitudes, Neptune's atmosphere is 80% hydrogen and 19% helium&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Interesting to read that what we see of Neptune's atmosphere is only the top part of the atmosphere. Thus, Neptune's always stated atmospheric composition of 80% hydrogen and 19% helium would only apply to this very outer layer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;www.Quora.com: &quot;In their diatomic forms, which is how we commonly encounter them, an oxygen molecule weighs 32 and a hydrogen molecule weighs 2. So oxygen is 16 times heavier than hydrogen&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since oxygen, which we breath, is one of the heavier gases, there could exist quite a lot of it near Neptune's surface.&lt;/p&gt;&#xA;" OwnerUserId="24270" LastActivityDate="2018-09-14T00:07:41.983" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27652" PostTypeId="2" ParentId="27633" CreationDate="2018-09-14T04:00:03.393" Score="6" Body="&lt;p&gt;In addition to the above examples, a telescope was once mounted in a Concorde for viewing a total eclipse. The speed of the aircraft allowed it to stay in the zone of totality for an extended period of time. The airplane's cruising altitude of around 60,000 ft. (half again as high as subsonic commercial jets) also reduced the absorption of light by the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://aviation.stackexchange.com/questions/22083/how-were-astronomical-observations-performed-from-concorde&quot;&gt;https://aviation.stackexchange.com/questions/22083/how-were-astronomical-observations-performed-from-concorde&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17298" LastEditorUserId="17298" LastEditDate="2018-09-14T04:08:03.567" LastActivityDate="2018-09-14T04:08:03.567" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27653" PostTypeId="1" CreationDate="2018-09-14T11:35:42.067" Score="1" ViewCount="59" Body="&lt;p&gt;I am an amateur so sorry for any misunderstanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to find a deep-sky object located 1° north east from the &lt;code&gt;δ Cas&lt;/code&gt; star. I am using the Stellarium program to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have toggled on the equatorial grid and Cardinal points are displayed as well, however, all of the cardinal points appear to be on circumference and I cannot really tell which direction is the wanted object at. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="24273" LastActivityDate="2018-09-14T16:27:48.827" Title="How to find a deep sky object 1° NE from a given star" Tags="&lt;deep-sky-observing&gt;&lt;navigation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27654" PostTypeId="2" ParentId="27653" CreationDate="2018-09-14T16:23:01.770" Score="2" Body="&lt;p&gt;Usually when the direction to an object is given relative to a star, the directions are related to equatorial coordinates (the coordinates on the sky). They are not related to the compass or cardinal points. It is best to avoid directions relative to the horizon/compass/cardinal points since those depend on the time and location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example, 1 degree northeast of $\delta$ Cas means this direction:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;north is toward Polaris along a line of right ascension, so you would go in the direction of increasing declination from $\delta$ Cas in the direction of Polaris.&lt;/li&gt;&#xA;&lt;li&gt;east is along a line of declination in the direction of increasing right ascension.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When looking at the sky, and if north (the direction toward Polaris) is thought of as the 12 o'clock position, the direction north east would be approximately 10:30. (Likewise, east would be the 9 o'clock position, south would be the 6 o'clock position, and so on.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a view of Cassiopeia for an arbitrary time and date with the directions shown. Are you trying to locate M103?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9Am89.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9Am89.png&quot; alt=&quot;Equatorial directions from delta Cas&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2018-09-14T16:23:01.770" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27655" PostTypeId="2" ParentId="27653" CreationDate="2018-09-14T16:27:48.827" Score="0" Body="&lt;p&gt;In the sky, north is the direction toward the north celestial pole, and east is 90 degrees counterclockwise of that.&#xA;On the equatorial grid, declination increases northward, and right ascension increases eastward.&#xA;If $\delta$ Cas is at J2000 1h26m +60.2$^\circ$, then 1 degree northeast of that would be about 1h32m +60.9$^\circ$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Terrestrial directions only help you find objects near the horizon, e.g. &quot;look for Mercury 10 degrees above the eastern horizon half an hour before sunrise.&quot;&#xA;For objects near the celestial pole, depending on the month and hour, north could be up, left, down, or right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you ask Stellarium to show deep-sky objects via the spiral galaxy icon in the lower toolbar, it should show a dotted circle symbol where M103 is.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-09-14T16:27:48.827" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27656" PostTypeId="2" ParentId="27641" CreationDate="2018-09-14T16:54:48.157" Score="4" Body="&lt;p&gt;If the Moon's orbit were fixed, the area covered by the Moon would just be $360^{\circ}*(0.5^{\circ}+2*1^{\circ})$. The 5.3 degree inclination relative to the ecliptic (or some other inclination relative to the equator) would not be a factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is important to this question about the coverage is that the orbit changes over time. In particular, the position of the ascending node (where the Moon's orbit crosses the ecliptic) moves &lt;em&gt;along the ecliptic&lt;/em&gt; with a period of 18.6 years (Wikipedia article on &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_precession&quot; rel=&quot;noreferrer&quot;&gt;Lunar precession&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a plot of the sky using ecliptic coordinates (so that the ecliptic is a straight line), the 5.3 degree inclination of the Moon's orbit to the ecliptic is shown as &quot;year X&quot; in my figure. The precession changes the orbit to positions such as year Y and year Z in my figure, and during the entire 18.6 year cycle, the entire band within 5.3 degrees of the ecliptic is covered by the Moon. In other words, the ascending node moves through all 360 degrees of ecliptic longitude. This results in the term 2*5.3 in your equation. (Note: I do not know if the precession is to the &quot;right&quot; or &quot;left&quot;, so do not interpret my figure to mean that year Y is later than year X. Year Y could be earlier than X. I'm sure someone will correct me!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XTqeu.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XTqeu.png&quot; alt=&quot;ecliptic projection&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the same plot were made in equatorial coordinates (so that the equator is a straight line), the band covered by the Moon is of course the same on the sky. However, the precession along the ecliptic limits the ascending node of the Moon's orbit on the equator to a small range or right ascension; the ascending node does not travel around the entire equator. Representing that band using equatorial coordinates and determining the area inside the bands is much more difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Md9RL.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Md9RL.png&quot; alt=&quot;equatorial projection&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2018-09-14T16:54:48.157" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27657" PostTypeId="2" ParentId="27638" CreationDate="2018-09-14T23:03:30.283" Score="0" Body="&lt;p&gt;Stellarium on your laptop, or SkySafari on your phone, could help you identify the object.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2018-09-14T23:03:30.283" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27658" PostTypeId="2" ParentId="27638" CreationDate="2018-09-15T07:20:37.363" Score="3" Body="&lt;p&gt;First, to figure out at what exact direction you were looking at, use Google Maps and set your location where you took the photo and then find for example that building to the left of the star on the map as a reference point and see on the map at what direction you were pointed (it doesn't have to be super precise, just whether it was south, south-west, west etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I recommend for a start use Stellarium online, which I think will be perfect for your needs because you don't need to install anything:&lt;br&gt;&#xA;&lt;a href=&quot;https://stellarium-web.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://stellarium-web.org/&lt;/a&gt;&lt;br&gt;&#xA;Set the location, date and time of when you took the picture and then look in the direction you determined earlier. Based on your picture, I'd say the altitude (that is, the angle between the ground plane and the star) is not more than 15 degrees. So then look at the brightest star or planet in that general direction and below that altitude line and I'm sure you'll figure it out pretty easily.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's at least how I do it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please let me know in the comment if that worked for you. :)&lt;/p&gt;&#xA;" OwnerUserId="24234" LastEditorUserId="24234" LastEditDate="2018-09-15T07:37:42.710" LastActivityDate="2018-09-15T07:37:42.710" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27659" PostTypeId="1" CreationDate="2018-09-15T09:49:19.073" Score="1" ViewCount="39" Body="&lt;p&gt;What's the time the sun takes in order to rise? (The time in which the solar disk crosses the horizon)&#xA;How does this time depend on local latitude and the sun's declination?&#xA;Assume that we're at sea level, and we ignore refraction and other atmospheric processes. (solar disk's altitude approximated with 0)&lt;/p&gt;&#xA;" OwnerUserId="24281" LastActivityDate="2018-09-15T09:49:19.073" Title="What's the time in which the sun rises?" Tags="&lt;the-sun&gt;&lt;time&gt;&lt;sunrise&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2018-09-15T17:55:25.737" ContentLicense="CC BY-SA 4.0" />
  <row Id="27661" PostTypeId="1" CreationDate="2018-09-15T12:21:49.857" Score="3" ViewCount="65" Body="&lt;p&gt;I recently purchased a Celestron Astromaster EQ130 telescope and have absolutely no clue about telescopes but would love to learn.&#xA;I have set this up but cannot even get a decent view of the moon, any advice of help would be very much appreciated.&#xA;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="24283" LastActivityDate="2018-09-15T17:24:33.833" Title="Celestron Astromaster EQ130" Tags="&lt;the-moon&gt;&lt;telescope&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27663" PostTypeId="1" CreationDate="2018-09-15T14:25:12.307" Score="1" ViewCount="227" Body="&lt;p&gt;I wrote a program based on this tutorial: &lt;a href=&quot;http://www.stjarnhimlen.se/comp/ppcomp.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.stjarnhimlen.se/comp/ppcomp.html&lt;/a&gt; to calculate altitude and azimuth of a celestial object. When I compare the calculated RA/Dec values to the ones from Stellarium, they are very accurate. But when I compare the Alt/Az values to the ones from Stellarium, there is an error of about 2 degrees! I use the following method to calculate Alt, Az:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;GMST0 = Ls + 180_degrees # Ls = Sun's longitude&#xA;GMST = GMST0 + UT&#xA;LST  = GMST + local_longitude&#xA;HA = LST - RA&#xA;&#xA;x = cos(HA) * cos(Decl)&#xA;y = sin(HA) * cos(Decl)&#xA;z = sin(Decl)&#xA;&#xA;xhor = x * sin(lat) - z * cos(lat)&#xA;yhor = y&#xA;zhor = x * cos(lat) + z * sin(lat)&#xA;&#xA;az  = atan2( yhor, xhor ) + 180_degrees&#xA;alt = asin( zhor ) = atan2( zhor, sqrt(xhor*xhor+yhor*yhor) )&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Some specific example:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Test date 15.09.2018, time 15:00 UT&#xA;Planet: Mercury&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coordinates: +47.55777777° +8.89888888&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What stellarium says:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RA = 11h 18m 13.26s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dec = +6°25'08.5&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Az = +250°21'13.2&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alt = +25°25'00.1&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What my program says:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RA = 11h 18m 14s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dec = 6° 25' 6.59&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Az = +248° 49' 6.9&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alt = +26° 33' 16.43&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="24285" LastEditorUserId="24285" LastEditDate="2018-09-15T14:56:58.643" LastActivityDate="2019-08-19T05:17:19.030" Title="Error when calculating Alt/Az from Ra/Dec" Tags="&lt;positional-astronomy&gt;&lt;azimuth&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27664" PostTypeId="2" ParentId="27661" CreationDate="2018-09-15T17:24:33.833" Score="2" Body="&lt;p&gt;Whatever difficulty you're having, the surest way to get past it is probably an in-person tutorial with an experienced telescope user.&#xA;Find a local astronomy club via&#xA;the &lt;a href=&quot;https://www.astroleague.org/societies&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomical League&lt;/a&gt;,&#xA;the &lt;a href=&quot;https://nightsky.jpl.nasa.gov/clubs-and-events.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA Night Sky Network&lt;/a&gt;,&#xA;&lt;a href=&quot;https://www.skyandtelescope.com/clubs-organizations/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sky and Telescope&lt;/a&gt;,&#xA;or &lt;a href=&quot;http://www.go-astronomy.com/astronomy-orgs.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Go Astronomy&lt;/a&gt;.&#xA;Then bring your scope to a public observing session, and most likely someone there will be willing and able to help.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-09-15T17:24:33.833" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27665" PostTypeId="1" AcceptedAnswerId="27699" CreationDate="2018-09-15T17:54:46.563" Score="3" ViewCount="90" Body="&lt;p&gt;I've implemented equations from Hurley's &lt;a href=&quot;http://adsabs.harvard.edu/abs/2000MNRAS.315..543H&quot; rel=&quot;nofollow noreferrer&quot;&gt;2000 paper&lt;/a&gt; on Single Stellar Evolution (SSE) to model the stages of stellar evolution without having to use the full SSE code (or MESA for that matter). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a more updated version of such a paper, where numerical formulae are given to model stages of stellar evolution? Has the knowledge base of the field of stellar evolution changed enough in the last 18 years to warrant an updated numerical formula recipe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0001295&quot; rel=&quot;nofollow noreferrer&quot;&gt;arxiv&lt;/a&gt; of the paper.&lt;/p&gt;&#xA;" OwnerUserId="23762" LastActivityDate="2018-09-19T18:17:51.007" Title="Is there a more modern/updated paper than Hurley's 2000 SSE?" Tags="&lt;astrophysics&gt;&lt;stellar-evolution&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27666" PostTypeId="1" CreationDate="2018-09-15T20:06:35.840" Score="3" ViewCount="44" Body="&lt;p&gt;Through trial and error (see script below that summarizes what I've found), I've learned Celestia's coordinate system appears to be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The x axis points to the vernal equinox&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The y axis points to the ecliptic north pole&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The z axis points to 18 hours right ascension, -23.4393 declination (ie, the obliquity of the ecliptic)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;regardless of whether you choose the &quot;equatorial&quot;, &quot;ecliptical&quot;, or &quot;universal&quot; frames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote the script below to confirm this: regardless of which setframe command you uncomment (or if you leave them all commented), the results are the same.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&#xA;&#xA;{&#xA;&#xA;# using Celestia 1.6.1&#xA;&#xA;# set to equatorial frame&#xA;# setframe {ref &quot;Sol/Earth&quot; coordsys &quot;equatorial&quot;}&#xA;# setframe {ref &quot;Sol/Earth&quot; coordsys &quot;ecliptical&quot;}&#xA;# setframe {ref &quot;Sol/Earth&quot; coordsys &quot;universal&quot;}&#xA;&#xA;# 27 Psc is near the vernal equinox at a distance of 234.48 ly&#xA;# this brings me within 14 ly&#xA;setposition {base [234.48 0 0]}&#xA;select {object &quot;27 Psc&quot;}&#xA;center {time 0.01}&#xA;wait {duration 3.0}&#xA;&#xA;# 42 Dra is near the north ecliptic pole at 18h, 66.56 declination&#xA;# this brings me within 15 ly&#xA;setposition {base [0 314.82 0]}&#xA;select {object &quot;42 Dra&quot;}&#xA;center {time 0.01}&#xA;wait {duration 3.0}&#xA;&#xA;# 7 Sgr is near the 18h, -23.4393 declination&#xA;# this brings me within 20 ly&#xA;setposition {base [0 0 1079.99]}&#xA;select {object &quot;7 Sgr&quot;}&#xA;center {time 0.01}&#xA;wait {duration 3.0}&#xA;&#xA;}&#xA;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I am aware that &lt;a href=&quot;https://en.wikibooks.org/wiki/Celestia/Binary_Star_File&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikibooks.org/wiki/Celestia/Binary_Star_File&lt;/a&gt; suggests the coordinates are stored internally as some gyration of ecliptical coordinates, but assumed &quot;setframe&quot; would change this.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2018-09-15T20:06:35.840" Title="Does Celestia respect 'setframe' coordinate systems?" Tags="&lt;frame-of-reference&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27667" PostTypeId="1" AcceptedAnswerId="27678" CreationDate="2018-09-16T03:30:51.090" Score="2" ViewCount="97" Body="&lt;p&gt;I have noticed that the units used in spectral data cubes coming from optical IFU instruments are usually in $erg/s/cm^2/Ang/pixel$. To my understanding this is a measure of Flux. Right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in radio astronomy, spectral data cubes coming from interferometers is usually in $Jy/beam$. To my understanding, this is a measure of surface brightness. Right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why optical and radio spectral data cubes don't use the same kind of unit (either flux or brightness)?&lt;/p&gt;&#xA;" OwnerUserId="1986" LastEditorUserId="14094" LastEditDate="2018-09-18T06:59:17.923" LastActivityDate="2018-09-18T06:59:17.923" Title="Units in optical and radio spectral data cubes: Flux vs Brightness" Tags="&lt;radio-astronomy&gt;&lt;units&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27668" PostTypeId="1" AcceptedAnswerId="27670" CreationDate="2018-09-16T18:37:06.527" Score="3" ViewCount="799" Body="&lt;p&gt;Yesterday evening, the moon from my place looked like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yk2jq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yk2jqm.jpg&quot; width=&quot;300&quot; alt=&quot;moon on september 2018, the 15th&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, instead of the crescent shape I am used to it looked like there was three tips: one at each extremity (as usual) and one in the middle between the two first ones. This picture shows the rough shape of the visible moon, with arrows pointing at all three tips:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Y9uWW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y9uWW.png&quot; width=&quot;300&quot; alt=&quot;annotated three tipped moon&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the explanation of this?&lt;/p&gt;&#xA;" OwnerUserId="24291" LastEditorUserId="-1" LastEditDate="2018-09-18T12:04:41.040" LastActivityDate="2018-09-18T12:04:41.040" Title="Why does the half moon appear to have a &quot;nose&quot;?" Tags="&lt;the-moon&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27670" PostTypeId="2" ParentId="27668" CreationDate="2018-09-16T23:22:20.473" Score="4" Body="&lt;p&gt;Here's an image I captured from a Moon globe app on my iPad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YnPN5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YnPN5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From your description, I think you may be referring to the southern rim of &lt;a href=&quot;https://en.wikipedia.org/wiki/Mare_Tranquillitatis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mare Tranquillitatis&lt;/a&gt;, the Sea of Tranquility, perhaps extending a little into the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hypatia_(crater)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rimae Hypatia&lt;/a&gt;. This is an area which is lower than the surrounding mountains and craters, into which the shadow is encroaching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, Mare Tranquillitatis is the region where Neil Armstrong and Buzz Aldrin landed on the Apollo 11 mission!&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2018-09-16T23:22:20.473" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27671" PostTypeId="2" ParentId="27668" CreationDate="2018-09-17T00:15:12.930" Score="2" Body="&lt;p&gt;The lunar phase is based on the &quot;age&quot; of the moon since the last new moon. Any illumination up until a first quarter moon will look crescent-shaped, without any &quot;nose&quot;. This crescent shape is a result of the positions of the Sun, Earth, and Moon in space. So, we can both observe and calculate the percent illumination and the shape of that illumination. In this particular case, the lunar terminator was running through a lunar mare, making it seem like the terminator was distorted. &lt;/p&gt;&#xA;" OwnerUserId="23940" LastActivityDate="2018-09-17T00:15:12.930" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27673" PostTypeId="1" CreationDate="2018-09-17T12:06:25.967" Score="6" ViewCount="517" Body="&lt;p&gt;The abstract of a paper currently in-press in Icarus: &lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/S0019103518303063&quot; rel=&quot;noreferrer&quot;&gt;The Reclassification of Asteroids from Planets to Non-Planets&lt;/a&gt; (Metzger et al. 2018, &lt;a href=&quot;https://doi.org/10.1016/j.icarus.2018.08.026&quot; rel=&quot;noreferrer&quot;&gt;https://doi.org/10.1016/j.icarus.2018.08.026&lt;/a&gt;) says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;It is often claimed that asteroids‘ sharing of orbits is the reason they were re-classified from planets to non-planets. A critical review of the literature from the 19th Century to the present shows this is factually incorrect. The literature shows the term asteroid was broadly recognized as a subset of planet for 150 years. On-going discovery of asteroids resulted in a de facto stretching of the concept of planet to include the ever-smaller bodies. Scientists found utility in this taxonomic identification as it provided categories needed to argue for the leading hypothesis of planet formation, Laplace‘s nebular hypothesis. In the 1950s, developments in planet formation theory found it no longer useful to maintain taxonomic identification between asteroids and planets, Ceres being the primary exception. At approximately the same time, there was a flood of publications on the geophysical nature of asteroids showing them to be geophysically different than the large planets. This is when the terminology in asteroid publications calling them planets abruptly plunged from a high level of usage where it had hovered during the period 1801 – 1957 to a low level that held constant thereafter. This marks the point where the community effectively formed consensus that asteroids should be taxonomically distinct from planets. The evidence demonstrates this consensus formed on the basis of geophysical differences between asteroids and planets, not the sharing of orbits. &lt;strong&gt;We suggest attempts to build consensus around planetary taxonomy not rely on the non-scientific process of voting, but rather through precedent set in scientific literature and discourse&lt;/strong&gt;, by which perspectives evolve with additional observations and information, just as they did in the case of asteroids. (emphasis added)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;On the website there is also:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Highlights&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The literature on asteroid classification is reviewed from 1801 to the present.&lt;/li&gt;&#xA;&lt;li&gt;Scientists considered asteroids to be planets until the 1950s.&lt;/li&gt;&#xA;&lt;li&gt;Thousands of asteroids sharing orbits was not debarring from planet status.&lt;/li&gt;&#xA;&lt;li&gt;They were reclassified as non-planets on the basis of geophysical characteristics.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Taxonomical terms like planet are determined by the scientific process, not voting&lt;/strong&gt;. (emphasis added)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It sounds like they are saying that Pluto's status as a planet was removed by a vote, more than by scientific consensus, and that while the title suggests that the paper is about &amp;quot;Asteroids&amp;quot; it seems the scenario might actually be proxy for Pluto's demotion.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Am I misunderstanding, or is that really the main point of the last few sentences of the abstract, if not the entire paper?&lt;/p&gt;&#xA;&lt;p&gt;Here's an example of another situation of classification being in flux: &lt;a href=&quot;https://astronomy.stackexchange.com/q/19963/7982&quot;&gt;Do astronomers generally agree that the distinction between comets and astroids is not so clear?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-04-28T18:47:59.477" Title="Is the point of this paper to say that Pluto's status as a planet should have been handled scientifically, rather than by a vote?" Tags="&lt;planet&gt;&lt;pluto&gt;&lt;classification&gt;" AnswerCount="5" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="27675" PostTypeId="2" ParentId="27673" CreationDate="2018-09-17T15:02:37.910" Score="3" Body="&lt;p&gt;If they wanted to make this paper about Pluto, why would they not just discuss Pluto?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may be right that the comment about voting could be linked to the Pluto debate, but I don't see any reason why they would need to conceal their intentions. I think it is more likely that they are actually researching and discussing asteroids, and they have linked their work briefly to the issues around Pluto because that gets more attention than asteroids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen many other examples of this: authors with throw in a paragraph at the end of a paper to link their work to a current hot topic (such as exoplanets or gravitational waves) because it boosts their readership, but it is not the main point of the paper of the focus of their research.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2018-09-17T15:02:37.910" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27676" PostTypeId="1" AcceptedAnswerId="27679" CreationDate="2018-09-17T15:21:48.963" Score="2" ViewCount="228" Body="&lt;p&gt;I have read that the Big Bang created space-time, but how could this be possible? The Big Bang was itself a singularity, and a singularity is a point in space-time where the curvature becomes infinite. So how could the Big Bang create something without which its existence would not have been possible?&lt;/p&gt;&#xA;" OwnerUserId="24297" LastEditorUserId="23940" LastEditDate="2018-10-01T12:06:31.540" LastActivityDate="2018-10-01T12:06:31.540" Title="Beginning of space-time" Tags="&lt;space-time&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27677" PostTypeId="2" ParentId="27676" CreationDate="2018-09-17T15:44:16.213" Score="4" Body="&lt;p&gt;There are questions on this topic on Physics.SE, but they require a certain level of understanding of the subject matter. Your question has some incorrect assumptions around the Big Bang. One being the definition of singularity - in reality this singularity is a point where certain variables may tend toward infinity, meaning our mathematics breaks down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the creation of space-time doesn't require anything to cause it - in fact it specifically implies that in our universe nothing could cause it, as there was nothing before it (there was no before)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were you I'd visit Physics.SE and read one of the &lt;a href=&quot;https://physics.stackexchange.com/questions/tagged/big-bang&quot;&gt;700 questions over there on this topic.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2018-09-17T15:44:16.213" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27678" PostTypeId="2" ParentId="27667" CreationDate="2018-09-17T15:50:54.997" Score="4" Body="&lt;p&gt;The answer is partially historically and partially scientifically. Flux, i.e. energy of radiation integrated across a solid angle (steradian) is basically counting all photons coming from a solid angle, while specific intensity (units of specific intensity are $erg/s/cm^2/Hz/sr$), is the energy per frequency along a line of sight! A surface brightness is flux divided by solid angle, i.e. the mean intensity within this solid angle. This evenly smears out all photons in this solid angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, flux is the preferred unit if one observes a point source, i.e. a source with a solid angle smaller than the angular resolution of the detector. For point sources, the beam size does not matter. If we observe the same point size with two instruments with different beam sizes, they both detect the same flux because they just count all photons coming from the source. A larger beam does not change this quantity. On the other hand, the brightness (mean intensity) of the object goes down with a larger beam size because we average the flux over a larger area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you observe an extended (larger than the beam!) object, e.g. a molecular cloud in radio, then surface brightness is the preferred quantity. Assume it is equally bright across its surface, then the surface brightness becomes independent of the beam size because the flux is proportional to the beam size. A telescope with a larger beam (smaller dish) observes a larger flux (always assuming the object is even larger) compared to a larger telescope with a smaller beam. Comparing the two measurements becomes difficult. The surface brightness of both observations is the same though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since optical astronomy was primarily observing stars, i.e. point sources, they used flux. Radio astronomers observed the interstellar medium, i.e. extended clouds of gas and dust, so they used surface brightness.   &lt;/p&gt;&#xA;" OwnerUserId="1988" LastEditorUserId="23748" LastEditDate="2018-09-18T00:55:05.870" LastActivityDate="2018-09-18T00:55:05.870" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27679" PostTypeId="2" ParentId="27676" CreationDate="2018-09-17T15:51:08.503" Score="6" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It is impossible to measure anything that happened in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Planck_units#Cosmology&quot; rel=&quot;noreferrer&quot;&gt;Planck Era&lt;/a&gt;, so physicists are unsure whether the universe actually began from a singularity or whether the singularity in the Big Bang theory represents an inability to describe the universe at that time. In fact, &quot;&lt;a href=&quot;https://physics.stackexchange.com/a/137497/28199&quot;&gt;Most physicists believe that this is a mathematical artefact and does not describe what actually happened.&lt;/a&gt;&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It would be more accurate to say that the Big Bang is the beginning of space-time, than that it &quot;created&quot; space-time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The Big Bang is a description of how the universe evolved in its earliest stages, not exactly a resolution to why the universe came to be. The latter is really a question for philosophy, not something that can be addressed by science &lt;em&gt;per se&lt;/em&gt;. Science informs our philosophy, but does not make philosophical claims.&lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorUserId="6" LastEditDate="2018-09-17T17:09:52.973" LastActivityDate="2018-09-17T17:09:52.973" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27680" PostTypeId="1" AcceptedAnswerId="27681" CreationDate="2018-09-17T19:43:56.760" Score="1" ViewCount="152" Body="&lt;p&gt;Is it theoretically possible that the Universe is expanding due to additional Big Bang events that have occurred at the same location that the original Big Bang occurred at?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps there has been an ongoing series of smaller Big Bang events over the past 13.7 billion years, which although were not as massive in scale as the original Big Bang, have created a large amount of new space-time and new matter that has resulted in the Universe expanding.&lt;/p&gt;&#xA;" OwnerDisplayName="user23327" LastActivityDate="2018-09-17T20:04:59.063" Title="Is it possible that the Universe is expanding due to additional Big Bang events?" Tags="&lt;universe&gt;&lt;astrophysics&gt;&lt;cosmology&gt;&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27681" PostTypeId="2" ParentId="27680" CreationDate="2018-09-17T20:04:59.063" Score="2" Body="&lt;p&gt;You may need to revisit the theory behind the &lt;a href=&quot;https://simple.wikipedia.org/wiki/Big_Bang&quot; rel=&quot;nofollow noreferrer&quot;&gt;Big Bang&lt;/a&gt;. The Big Bang is more or less a description of what the early universe looks like if you view expansion in reverse. The single &quot;Big Bang&quot; at the beginning of our universe is the only &quot;Big Bang&quot; event predicted by expansion. Also, the Big Bang did not happen in a single place, it happened everywhere in the universe. Finally, the Big Bang did not create space-time, and the expansion of the universe is not the creation of new matter. The Big Bang describes a time when the metric of space-time was extremely dense, and expansion is the metric of space-time increasing as below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/y3vMl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y3vMl.png&quot; alt=&quot;Expansion of space&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance between points in space increases, but the same mass-energy is there as before.&lt;/p&gt;&#xA;" OwnerUserId="6" LastActivityDate="2018-09-17T20:04:59.063" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27682" PostTypeId="1" AcceptedAnswerId="27702" CreationDate="2018-09-18T00:16:40.540" Score="4" ViewCount="102" Body="&lt;p&gt;Using Hurley's &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0001295&quot; rel=&quot;nofollow noreferrer&quot;&gt;2000&lt;/a&gt; paper on Single Stellar Evolution, I have graphed the core mass of a star at three stages as a function of metallicity, Z, for a chosen ZAMS mass: at the beginning of the HG (MS), at the end of the Hertzprung-Gap (HG), and at the Base of the Asymptotic Giant Branch (BAGB). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All three functions have a cusp at Z = 0.0005 which I 1) do not understand physically and 2) wonder if it is incorrect? Is this expected? I would have expected that the core mass is monotonic in metallicity, Z. The cusp is present for any ZAMS mass chosen. &lt;a href=&quot;https://i.stack.imgur.com/NqmiR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NqmiR.jpg&quot; alt=&quot;Core mass of Star at three stages as function of metallicity&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The equations I use are...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Core mass at beginning of HG:&#xA;$Mc_{MS}$ is the product of equation 29 and the core mass at the end of the HG described below ($\rho$ is always $&amp;lt; 1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Core mass at end of HG:&#xA;$Mc_{HG}$ is, using equation 28 over the relevant mass range the HG ends at Helium Ignition, given by equation 44, where I solved for the time, $t$, in equation 35 and substituted that into equation 34 to find that $M_{c} = \Big(\frac{L}{D}\Big)^{1/5}$, and following the procedure given in the paragraph below equation 65 (where it references equation 44). I used equation 49b for the Luminosity here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Core mass at BAGB:&#xA;$Mc_{BAGB}$ is equation 66.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm only interested in stars above $10 \odot{M}$, so I do not consider the possibility of a Blue Loop phase for simplicity. &lt;/p&gt;&#xA;" OwnerUserId="23762" LastEditorUserId="23762" LastEditDate="2018-09-19T02:54:12.930" LastActivityDate="2018-09-19T23:12:50.837" Title="Stellar core mass as a function of metallicity" Tags="&lt;star&gt;&lt;stellar-evolution&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27683" PostTypeId="1" AcceptedAnswerId="27730" CreationDate="2018-09-18T09:16:30.813" Score="-1" ViewCount="369" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I hope a meteor falls on your head.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I hope a meteorite falls on your head.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.merriam-webster.com/dictionary/meteor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Definition of meteor&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.merriam-webster.com/dictionary/meteorite&quot; rel=&quot;nofollow noreferrer&quot;&gt;Definition of meteorite&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24303" LastActivityDate="2018-09-23T07:30:05.223" Title="Which of the following is the correct usage: meteor / meteorite?" Tags="&lt;meteor&gt;&lt;meteorite&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27684" PostTypeId="2" ParentId="27683" CreationDate="2018-09-18T09:34:24.067" Score="1" Body="&lt;p&gt;As far as I know a meteor is the light phenomenon that occurs when an object enters the Earth's atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The meteorite is the object causing the light phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this the second sentence would be correct.&lt;/p&gt;&#xA;" OwnerUserId="22927" LastActivityDate="2018-09-18T09:34:24.067" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27685" PostTypeId="2" ParentId="27683" CreationDate="2018-09-18T16:29:39.823" Score="2" Body="&lt;p&gt;I think the easiest way to remember the distinction is to do with where you see them. You will see a meteor in the sky, for example, in a meteor shower. A meteorite is an object you find on the ground (many &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_rock_types&quot; rel=&quot;nofollow noreferrer&quot;&gt;types of rock&lt;/a&gt; have -ite endings) so a meteorite is the bit of a meteor that does not completely burn up and lands on Earth, (or on your head).&lt;/p&gt;&#xA;" OwnerUserId="2270" LastEditorUserId="2365" LastEditDate="2018-09-23T06:23:02.777" LastActivityDate="2018-09-23T06:23:02.777" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27686" PostTypeId="2" ParentId="27550" CreationDate="2018-09-18T19:38:10.057" Score="0" Body="&lt;p&gt;I couldn’t really figure out the way to get RA Dec from the data mentioned. I ended up using a catalog that already has RA Dec of objects. &lt;/p&gt;&#xA;" OwnerUserId="23083" LastActivityDate="2018-09-18T19:38:10.057" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27688" PostTypeId="1" AcceptedAnswerId="27704" CreationDate="2018-09-18T23:27:07.350" Score="5" ViewCount="216" Body="&lt;p&gt;I have a technical question regarding the Shack-Hartmann sensor in deformable mirrors.  &lt;strong&gt;What algorithm is used to take the Shack-Hartmann measurements of local tilt and turn them into a full phase estimate that can be passed to the  deformable mirror?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/g1g48.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g1g48.png&quot; alt=&quot;Adaptive Optics Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Source: Wikipedia&lt;sup&gt;1&lt;/sup&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that the the sensor itself can generate an estimate of the &lt;em&gt;derivative&lt;/em&gt; of the wavefront $w(x,y)$ by observing the offset of the centroid of the guidestar. If the delta in from center of the guidestar is $\delta_x$ and $\delta_y$ then the derivative of the wavefront in that region is proportional $\delta$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dw}{dx} \propto \delta_x $ and $\frac{dw}{dy} \propto \delta_y $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geary's Introduction to wavefront sensors&lt;sup&gt;2&lt;/sup&gt; has the following diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zGLbe.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zGLbe.png&quot; alt=&quot;Geary Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with the comment: &quot;To reconstruct the wavefront, the local tilts must be&#xA;stitched together&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it enough to connect them end-to-end, perhaps make them non-negative, and pass them on? The values transmitted are usually Zernike Polynomial modes, so do we we desire to make them positive or zero mean? What algorithm is actually used to take the Shack-Hartmann measurements of local tilt and turn them into a full phase estimate that can be passed to the deformable mirror?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; By 2pem - Own work, CC BY-SA 3.0, &lt;a href=&quot;https://commons.wikimedia.org/w/index.php?curid=15279624&quot; rel=&quot;noreferrer&quot;&gt;https://commons.wikimedia.org/w/index.php?curid=15279624&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2&lt;/sup&gt; Geary, Joseph M.. (1995). Introduction to Wavefront Sensors.&lt;/p&gt;&#xA;" OwnerUserId="10548" LastActivityDate="2018-09-30T21:09:28.607" Title="Adaptive Optics Shack-Hartmann Sensor and Deformable Mirror" Tags="&lt;adaptive-optics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27689" PostTypeId="1" CreationDate="2018-09-19T05:08:50.050" Score="0" ViewCount="122" Body="&lt;p&gt;When the magnetic poles of the earth get reversed, does the earth keep on revolving in the same direction or does it start revolving in the opposite direction?&lt;/p&gt;&#xA;" OwnerUserId="13871" LastEditorDisplayName="user1569" LastEditDate="2018-09-19T15:00:04.907" LastActivityDate="2018-09-27T02:31:59.200" Title="Reversal of magnetic poles of earth" Tags="&lt;earth&gt;&lt;magnetic-field&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="27691" PostTypeId="1" AcceptedAnswerId="27692" CreationDate="2018-09-19T07:08:25.907" Score="0" ViewCount="80" Body="&lt;p&gt;I am curious to know If dust extinction were to vary as a power-law so that $A_{\lambda}$ proportionate to $\lambda^{\beta}$ , what value of beta&#xA;would give rise to Rv =3.1 which is the average in diffuse ISM. I can see a plot but I wanted to see how to derive Rv=3.1 mathematically&lt;/p&gt;&#xA;" OwnerUserId="21282" LastActivityDate="2018-09-19T07:43:14.223" Title="Quesions about extinction power law" Tags="&lt;extinction&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27692" PostTypeId="2" ParentId="27691" CreationDate="2018-09-19T07:43:14.223" Score="1" Body="&lt;p&gt;The value is $\beta \simeq -1$ (e.g. see  &lt;a href=&quot;http://adsabs.harvard.edu/abs/1988ApJ...329L..33C&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cardelli et al. 1988&lt;/a&gt;). The extinction would not be well fitted by a power law, except over a very narrow wavelength range.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-09-19T07:43:14.223" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27693" PostTypeId="2" ParentId="27689" CreationDate="2018-09-19T07:49:15.677" Score="4" Body="&lt;p&gt;The Earth will keep revolving in the same direction during a &lt;a href=&quot;https://en.wikipedia.org/wiki/Geomagnetic_reversal&quot; rel=&quot;nofollow noreferrer&quot;&gt;geomagnetic reversal&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been a few articles about this phenomenon published during 2018:    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://news.nationalgeographic.com/2018/01/earth-magnetic-field-flip-north-south-poles-science/&quot; rel=&quot;nofollow noreferrer&quot;&gt;National Geographic - No, We're Not All Doomed by Earth's Magnetic Field Flip&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.sciencedaily.com/releases/2018/04/180430160437.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Science Daily - Earth's magnetic field is not about to reverse&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://theconversation.com/the-earths-magnetic-field-reverses-more-often-now-we-know-why-96957&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Conversation - The Earth’s magnetic field reverses more often – now we know why&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2018-09-19T07:49:15.677" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27694" PostTypeId="1" AcceptedAnswerId="27706" CreationDate="2018-09-19T09:54:31.413" Score="5" ViewCount="377" Body="&lt;p&gt;Is there a way to calculate the expected number density of Dark Matter Halos above a given mass, in a certain redshift range, and in a certain area?&lt;/p&gt;&#xA;" OwnerDisplayName="user19333" LastActivityDate="2018-09-21T05:03:07.243" Title="Number Density of Dark Matter Halos" Tags="&lt;galaxy&gt;&lt;cosmology&gt;&lt;dark-matter&gt;&lt;redshift&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27697" PostTypeId="1" AcceptedAnswerId="27751" CreationDate="2018-09-19T17:28:41.350" Score="2" ViewCount="104" Body="&lt;p&gt;News of the discovery of a &lt;em&gt;potentially&lt;/em&gt; habitable planet around the star 40 Eridani A (about 16 light-years away) is interesting for reasons including a 1991 letter described below in Sky &amp;amp; Telescope's recent article &lt;a href=&quot;https://www.skyandtelescope.com/astronomy-news/super-earth-vulcan/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Super-Earth Discovered in (Fictional) Vulcan System&lt;/a&gt;, part of which is shown below.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is there any way to estimate a low mass cut-off for the Dharma survey for this particular star? Could there also be a 1 or 2 Earth mass planet there as well, within a habitable zone, as-yet undetected? Or does data there exclude this possibility?&lt;/p&gt;&#xA;&lt;p&gt;The paper about the recent discovery: &lt;a href=&quot;https://arxiv.org/abs/1807.07098&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/1807.07098&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Almost three decades ago, Gene Roddenberry (producer of the Star Trek universe) wrote a letter to Sky &amp;amp; Telescope, &lt;strong&gt;along with Harvard-Smithsonian Center for Astrophysics astronomers Sallie Baliunas, Robert Donahue, and George Nassiopoulos&lt;/strong&gt;. In their Letter to the Editor, they argued that 40 Eridani A — an orange-ish star 16 light-years away — would make the ideal home for Vulcan, the home planet of Science Officer Mr. Spock.&lt;/p&gt;&#xA;&lt;p&gt;Now, a new discovery puts a little more science into that science fiction assertion.&lt;/p&gt;&#xA;&lt;p&gt;A Long-ago Letter&lt;/p&gt;&#xA;&lt;p&gt;In the July 1991 issue, the three astronomers and one movie-maker made the case for what star should be considered Vulcan's home:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The star around which Vulcan orbits was never identified in the original series or in any of the feature films based on it and so has never been officially established. But two candidates have been suggested in related literature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Two Star Trek books named the star 40 Eridani A as Vulcan's sun, while another publication named Epsilon Eridani instead. However, Roddenberry and the astronomers made an argument for 40 Eridani A:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;We prefer the identification of 40 Eridani as Vulcan's Sun because of what we have learned about both stars at Mount Wilson. . . . The HK [Project] observations suggest that 40 Eridani is 4 billion years old, about the same age as the Sun. In contrast Epsilon Eridani is barely 1 billion years old.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Based on the history of life on Earth, life on any planet around Epsilon Eridani would not have had time to evolve beyond the level of bacteria. On the other hand, an intelligent civilization could have evolved over the aeons on a planet circling 40 Eridani. So the latter is the more likely Vulcan sun. . . . Presumably Vulcan orbits the primary star, an orange main-sequence dwarf of spectral type K1. . . . Two companion stars — a 9th magnitude white dwarf and an 11th magnitude red dwarf — orbit each other about 400 astronomical units from the primary. They would gleam brilliantly in the Vulcan sky.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-09-24T18:56:15.460" Title="Low mass cut-off for the Dharma Planet Survey's detection of habitable planets around 40 Eridani A?" Tags="&lt;exoplanet&gt;&lt;astrobiology&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27698" PostTypeId="1" CreationDate="2018-09-19T18:01:13.733" Score="4" ViewCount="92" Body="&lt;p&gt;In several papers about Recombination, (for instance Peebles, Astrophysical Journal, vol. 153, eq. 39) it is claimed that the evolution of matter temperature follows the equation&#xA;\begin{equation}&#xA;\frac{dT}{dt} = -2\frac{\dot{a}}{a}T + \frac{2}{3 K_B n} \epsilon&#xA;\end{equation}&#xA;where the first term in the right hand side accounts for the expansion of the universe and $\epsilon$ is the energy injected into the gas per second per unit physical volume. How can I derive this equation?&#xA;I suppose that this equation can be derived applying the First Principle of Thermodynamics to a comoving volume in the expanding universe or maybe taking the second moment of a Boltzmann equation. Any help or reference in which this equation is discussed would be very helpful. &lt;/p&gt;&#xA;" OwnerUserId="24314" LastActivityDate="2018-09-30T06:53:25.733" Title="Matter Temperature during Recombination" Tags="&lt;thermodynamics&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27699" PostTypeId="2" ParentId="27665" CreationDate="2018-09-19T18:17:51.007" Score="1" Body="&lt;p&gt;My guess would be no because, as of this writing, from the paper &lt;a href=&quot;http://adsabs.harvard.edu/abs/2000MNRAS.315..543H&quot; rel=&quot;nofollow noreferrer&quot;&gt;link you provided&lt;/a&gt; you can see &lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/basic_connect?qsearch=hurley%20single%20star%20evolution&amp;amp;version=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;there are 949 citations&lt;/a&gt; to that paper with a good chunk of them in this year. You could always &lt;a href=&quot;http://astronomy.swin.edu.au/~jhurley/&quot; rel=&quot;nofollow noreferrer&quot;&gt;contact the author(s)&lt;/a&gt; to discuss it. Perhaps your effort could lead to an update.&lt;/p&gt;&#xA;" OwnerUserId="9749" LastActivityDate="2018-09-19T18:17:51.007" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27700" PostTypeId="2" ParentId="27688" CreationDate="2018-09-19T18:18:21.427" Score="3" Body="&lt;p&gt;Speaking as one who has dealt with those equations (and those problems) since 1984.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, the simplest approach is to use matrix methods to force the phases at the edges of every subaperture to be the same.  The assumption is that there is no phase tearing , and if there were, adaptive optics pretty much would be unable to handle such disruptions.   Once the full &quot;map&quot; is constructed, the mean value is removed (aka the &quot;piston&quot; term, zeroth Zernike) to reduce the overall stroke requirement on the deformable mirror (DM).  In most systems, the mean tip &amp;amp; tilt values (first Zernike terms) are removed as well and passed off to a full aperture tip/tilt correction mirror.  Again, that reduces overall stroke requirements on the DM.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's lots more in practice, because the subaperture actuators on the DM (other that segmented ones, which have a host of other problems) have what's known as the &quot;influence function&quot;.  That is, moving an actuator essentially creates a &quot;hill&quot; or &quot;valley&quot; over a decent region of the mirror.  These influence functions may or may not overlap neighboring actuators, depending on the mirror design, but in any case the final correction matrix has to take into account the difference between an idealized subaperture tip/tilt correction and what the mirror actually can do.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Follow that up with spatial and temporal (history) noise reduction filters, and off you go.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, back to the start: the constant of proportionality between Hartmann spot motion and phase tilt (or tip) is a simple function of the lens focal length. Draw the triangle consisting of the optic axis distance from lenslet to image plane (i.e. lens focal length), the distance from the origin to the centroid of the spot image, and the hypotenuse from the center of the lens to the centroid of the spot.  You can thus easily see that the tilt is the difference between the length of the hypotenuse and the focal length.  Convert that from meters (or microns) to wavelengths of the sensing light, and there you are. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-09-19T18:18:21.427" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27701" PostTypeId="1" CreationDate="2018-09-19T19:37:37.120" Score="2" ViewCount="152" Body="&lt;p&gt;I created this solar system model using POV-Ray:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fdXpK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fdXpK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by following these instructions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ssd.jpl.nasa.gov/txt/aprx_pos_planets.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ssd.jpl.nasa.gov/txt/aprx_pos_planets.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next thing I want to do is orient/rotate each globe correctly according to the time of day, time of year, obliqueness, position in its orbit, etc. but cannot find any instructions on the JPL website. How can I go about completing this task? Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part Two&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[deleted]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part Three&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &lt;em&gt;&amp;alpha;&lt;/em&gt; (right ascension), &lt;em&gt;&amp;delta;&lt;/em&gt; (declination) and &lt;em&gt;W&lt;/em&gt; are Euler rotation angles. But what are the starting conditions before applying these rotations? Should the North Pole of the planet be aligned with the z-axis? Should the plane formed by the prime meridian be perpendicular to the y-axis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, which axes do I rotate around, and in what order? I am guessing &lt;em&gt;&amp;delta;&lt;/em&gt; should be rotated first around the y-axis, then &lt;em&gt;&amp;alpha;&lt;/em&gt; around the z-axis. But &lt;em&gt;W&lt;/em&gt; also looks like it should go around the z-axis. I am confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part Four&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found another PDF document which on page 27 gives an equation for &lt;em&gt;&amp;delta;&lt;/em&gt;, &lt;em&gt;&amp;alpha;&lt;/em&gt; and &lt;em&gt;W&lt;/em&gt;. (Not sure how to reproduce it here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://depositonce.tu-berlin.de/bitstream/11303/6237/4/burmeister_steffi.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://depositonce.tu-berlin.de/bitstream/11303/6237/4/burmeister_steffi.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;em&gt;think&lt;/em&gt; that according to that PDF file I need to rotate around the z-axis by W, around the x-axis by (90-&amp;delta;), and around the z-axis again by (90+&amp;alpha;), in that order. Lastly, an additional rotation (around the x-axis?) by 23.43928 degrees must be done to get the body out of the ICRF frame and into the ecliptic frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I still don't know the starting conditions. I.e. before applying the transformations, should the globe's North Pole point upward? Should the intersection of the Prime Meridian and Equator lie along the x-axis? Or, am I totally wrong, and should I perform the inverse matrix calculations instead? So far I have not gotten results that match what I see in Celestia for the same Julian Date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part Five&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some instances I am able to get &quot;close&quot; results using the above method. In others not. For instance, Mars, Saturn, Uranus and Neptune are &quot;close&quot; to correct. (At least, the poles - the blue arrows - are pointing in the same directions.) The Earth is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Zk1M6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zk1M6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/4oFkZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4oFkZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/oDgzK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oDgzK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/zVepR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zVepR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/TN8xc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TN8xc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;correct&lt;/em&gt; results are the ones in the larger Celestia program window. My current POV-Ray results are in the smaller window to the left. The dark arrows indicate the parent (ecliptic) reference frame. The bright arrows indicate the body reference frame. The view in all images is from the top down on to the ecliptic plane.&lt;/p&gt;&#xA;" OwnerUserId="24284" LastEditorUserId="24284" LastEditDate="2018-09-28T09:23:22.023" LastActivityDate="2018-09-28T09:23:22.023" Title="Modeling planet rotations (time of day, obliqueness, etc.)" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;rotation&gt;" AnswerCount="0" CommentCount="11" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27702" PostTypeId="2" ParentId="27682" CreationDate="2018-09-19T23:12:50.837" Score="3" Body="&lt;p&gt;I feel silly, feel free to laugh at this, but I found my mistake. In the Hurley paper, he uses &quot;log&quot; for the logarithm of base 10. I had been using the natural logarithm. With the fix, the graph looks more realistic:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/coCJR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/coCJR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23762" LastActivityDate="2018-09-19T23:12:50.837" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27703" PostTypeId="2" ParentId="27694" CreationDate="2018-09-19T23:34:47.367" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a way to calculate the expected number density of Dark Matter Halos above a given mass, in a certain redshift range, and in a certain area?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are ways of counting mass distributions of dark matter halos. I'm not sure what you mean by &quot;expected number density,&quot; perhaps you could elaborate if my answer is not what you wanted? Do you mean the number density of the background galaxies if one is using lensing data? That is a very different question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with all astrophyiscs, limitations on computational power are limits on &quot;experimental&quot; power, since we use simulations as our laboratories, and the modeling of Dark Matter Halos is no exception. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, a standard technique is to run N-body simulations of a dark matter universe and try to see how everything arranges itself by &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_matter_halo#Density_profiles&quot; rel=&quot;nofollow noreferrer&quot;&gt;fitting functional forms&lt;/a&gt; to local regions (hence the density of a halo out to a certain mass and radius). Thus, the density you get depends on the functional form you choose! The standard profile is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Navarro%E2%80%93Frenk%E2%80%93White_profile&quot; rel=&quot;nofollow noreferrer&quot;&gt;NFW&lt;/a&gt; profile which treats the dark matter halo spherically, but this profile has many limitations that one must be careful of, for instance it is only valid approximately out to the virial radius of the galaxy. But, nowadays there are &lt;a href=&quot;https://arxiv.org/abs/1005.0411&quot; rel=&quot;nofollow noreferrer&quot;&gt;numerous&lt;/a&gt; available profiles (each with its own peculiarities) that try to include ellipticity of the dark matter halo. It is an area of active research to compare these profiles and improve them further, for instance see &lt;a href=&quot;https://arxiv.org/abs/1805.12226&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; recent paper, where the Deimer and NFW profiles are compared - the Deimer profile is capable of going beyond the virial radius because it's composed of an &quot;inner density&quot; that is similar to the NFW, and transitions at approximately the virial radius to an &quot;outer&quot; density profile. &lt;/p&gt;&#xA;" OwnerUserId="23762" LastEditorUserId="23762" LastEditDate="2018-09-19T23:56:02.730" LastActivityDate="2018-09-19T23:56:02.730" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27704" PostTypeId="2" ParentId="27688" CreationDate="2018-09-20T02:36:44.127" Score="3" Body="&lt;p&gt;Just adding some math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dKnAw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dKnAw.jpg&quot; alt=&quot;Shack-Hartmann wavefront sensor&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say your wavefront is &lt;span class=&quot;math-container&quot;&gt;$w(x,y)$&lt;/span&gt;, based on the diagram above you have sub-aperture local slope movements &lt;span class=&quot;math-container&quot;&gt;$\delta(x,y)$&lt;/span&gt;, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\nabla w = \delta(x,y) / f$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since for Shack-Hartmann the spatial resolution is usually low, so people fit wavefront &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt; to a Zernike polynomial. Denote the Zernike basis be matrix &lt;span class=&quot;math-container&quot;&gt;$Z$&lt;/span&gt; and coefficients be &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt;, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$w = Za$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle you solve a least-squares to estimate &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; first&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ a = \arg\min_a \| \nabla Za - \delta/f \|^2 = \left((\nabla Z)^T\nabla Z\right)^{-1} (\nabla Z)^{T}\delta/f$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you got the Zernike coefficients. And you would like to pass it to the deformable mirrors, which as @Carl mentioned contains many actuators. Say the commands to each actuator is a vector &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;, there is a command matrix &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; that maps the commands to actual shape on the deformable mirror. In math&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$d = Cc$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; the shape of the deformable mirror. Also since Shack-Hartmann and deformable mirror are in conjugate, there is another matrix describing this linear relationship, denoted as &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt;. And&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ w = M d $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to interpret matrix &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is, say your Zernike-fitted wavefront (i.e. &lt;span class=&quot;math-container&quot;&gt;$w = Za$&lt;/span&gt;) covers a square Shack-Hartmann sensor size of &lt;span class=&quot;math-container&quot;&gt;$\sqrt{n} \times \sqrt{n}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\text{mm}^2$&lt;/span&gt;. Your deformable mirror, at the conjugate plane, covers another square area of &lt;span class=&quot;math-container&quot;&gt;$\sqrt{m} \times \sqrt{m}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\text{mm}^2$&lt;/span&gt;. Assuming same sampling rate (e.g. 1 discretized pixel per &lt;span class=&quot;math-container&quot;&gt;$\text{mm}^2$&lt;/span&gt;), then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ w \in \mathbb{R}^n \quad d \in \mathbb{R}^m \quad  \text{and} \quad M \in \mathbb{R}^{n \times m}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the actual number of &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; depends on your manual specification on the resolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to interpret &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is, &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; models aperture occlusion as well. For simplicity say &lt;span class=&quot;math-container&quot;&gt;$n = m$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is a square matrix. If you wavefront sensor covers area outside of the deformable mirror, you can just let those areas be zeros in corresponding position of &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt;. Now &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is a diagonal matrix whose elements are either 1 or 0 (in ideal case). Diffraction or any setup misalignment will make &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; not diagonal anymore. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now go back to the formulation. Put it together&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$w = M d = MC c$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you got &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt; but want &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;, so you need to invert the above system matrix &lt;span class=&quot;math-container&quot;&gt;$MC$&lt;/span&gt;, that is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$c = (MC)^{-1} w = (MC)^{-1} Z a$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is the coefficients you solved from the least squares above. Usually &lt;span class=&quot;math-container&quot;&gt;$(MC)^{-1}$&lt;/span&gt; is calibrated before the whole adaptive optics loop. So in real time correction you need only to solve &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; and then pass it to the calibrated system to yield the reversal mirror shape.&lt;/p&gt;&#xA;" OwnerUserId="16553" LastEditorUserId="16553" LastEditDate="2018-09-30T21:09:28.607" LastActivityDate="2018-09-30T21:09:28.607" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27706" PostTypeId="2" ParentId="27694" CreationDate="2018-09-20T15:18:15.223" Score="5" Body="&lt;p&gt;The function you're requesting — i.e. the number density &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; of DM halos above a given mass &lt;span class=&quot;math-container&quot;&gt;$M_\mathrm{h}$&lt;/span&gt; — is called the &lt;em&gt;cumulative halo mass function&lt;/em&gt; (cHMF). It is obtained by integrating the &lt;em&gt;halo mass function&lt;/em&gt; (HMF) from a given mass to infinity. The HMF, in turn, is thus the function that describes the (differential) number density of DM halos of a given mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\boxed{N(&amp;gt;\!\!M_\mathrm{h})&#xA;= \int_{M_\mathrm{h}}^\infty \!\!\!dM_\mathrm{h}'\, \frac{dN}{dM_\mathrm{h}'}.}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Halo mass function&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;So, the problem is to determine the HMF, i.e. &lt;span class=&quot;math-container&quot;&gt;$dN/dM_\mathrm{h}$&lt;/span&gt;. This was first calculated, analytically, by &lt;a href=&quot;http://adsabs.harvard.edu/abs/1974ApJ...187..425P&quot; rel=&quot;noreferrer&quot;&gt;Press &amp;amp; Schechter (1974)&lt;/a&gt; assuming spherical collapse of structures from an initial, smoothed density field. It may be written as&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\frac{dN}{dM_\mathrm{h}} = \frac{\rho_{\mathrm{m,0}}}{M_\mathrm{h}}&#xA;                    \left| \frac{d\ln\sigma}{dM_\mathrm{h}} \right|&#xA;                    f(\sigma),&#xA;$$&lt;/span&gt;&#xA;where&#xA;&lt;span class=&quot;math-container&quot;&gt;$\rho_{\mathrm{m,0}}$&lt;/span&gt; is the present-day average mass density of the Universe,&#xA;&lt;span class=&quot;math-container&quot;&gt;$\sigma = \sigma(M_\mathrm{h},z)$&lt;/span&gt; is the rms fluctuations of the (smoothed) density field, and &#xA;&lt;span class=&quot;math-container&quot;&gt;$f(\sigma)$&lt;/span&gt; is the &quot;multiplicity function&quot; (note that because the number density decreases so fast with halo mass, for computational purposes often the HMF is expressed as &lt;span class=&quot;math-container&quot;&gt;$dN/d \ln M_\mathrm{h}$&lt;/span&gt; rather than &lt;span class=&quot;math-container&quot;&gt;$dN/dM_\mathrm{h}$&lt;/span&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want, I can give you more details about how to calculate &lt;span class=&quot;math-container&quot;&gt;$\sigma(M_\mathrm{h},z)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$f(\sigma)$&lt;/span&gt;. You can also find details on this in sec. 2.1 of &lt;a href=&quot;https://arxiv.org/abs/1806.07392&quot; rel=&quot;noreferrer&quot;&gt;Laursen et al. (2018)&lt;/a&gt;, but note that there's an error in Eq. 1. If you're happy with using a &quot;black box&quot;, you can get both the HMF and cHMF for your favorite cosmological parameters using &lt;a href=&quot;http://hmf.icrar.org&quot; rel=&quot;noreferrer&quot;&gt;this online HMF calculator&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only in the Press-Schechter formalism can an analytical form of &lt;span class=&quot;math-container&quot;&gt;$f(\sigma)$&lt;/span&gt; be derived, and it was subsequently found that it over(under)predicts the collapsed fraction at the low-(high-)mass end (see e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/1999MNRAS.307..949G&quot; rel=&quot;noreferrer&quot;&gt;Governato et al. 1999&lt;/a&gt;); in general one must obtain it by fitting halo abundances in cosmological &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt;-body simulations.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Defining a halo&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;This is what N. Steinle describes in his answer, but actually you don't need to assume a density profile of the halos. So, how do you count halos in a simulation? There are several ways, arguably the two most popular ones being the &lt;em&gt;spherical overdensity&lt;/em&gt; (SO) and the &lt;em&gt;friends-of-friends&lt;/em&gt; (FoF) method.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Spherical overdensity&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;In the SO method, you first calculate the center of mass (CoM) of a clump of particles in the simulation, and then you calculate the average density &lt;span class=&quot;math-container&quot;&gt;$\bar{\rho}$&lt;/span&gt; of particles in successively larger spheres centered on the CoM. As you increase the radius of the sphere, &lt;span class=&quot;math-container&quot;&gt;$\bar{\rho}$&lt;/span&gt; drops because you include less and less dense regions. When the density has reached a certain factor &lt;span class=&quot;math-container&quot;&gt;$\Delta$&lt;/span&gt; times the average density &lt;span class=&quot;math-container&quot;&gt;$\rho_\mathrm{m}(z)$&lt;/span&gt; in the Universe you stop, and the total mass of all particles inside that sphere is then the halo mass. The overdensity factor &lt;span class=&quot;math-container&quot;&gt;$\Delta$&lt;/span&gt; is usually chosen to be around 200, but other choices exist, e.g. 500 which then gives slightly smaller masses (because you stop counting earlier).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A variant of this method is using ellipsoids rather than spheres, allowing for more realistic results for elongated structures.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Friends-of-friends&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;In the FoF method, you start at an overdensity and count all particles that are &quot;linked&quot; together, meaning they are within some chosen distance (the &quot;linking length&quot;) of each other. This method may give more realistic results for very non-spherical structures, but also tends to include particles in filaments streaming into the galaxies, which perhaps shouldn't be considered a part of a galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure below (from &lt;a href=&quot;https://arxiv.org/abs/1002.3660&quot; rel=&quot;noreferrer&quot;&gt;Klypin et al. 2011&lt;/a&gt;) shows halos from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Bolshoi_Cosmological_Simulation&quot; rel=&quot;noreferrer&quot;&gt;Bolshoi simulation&lt;/a&gt; identified with the two methods; SO (&lt;em&gt;red points&lt;/em&gt;) and FoF (&lt;em&gt;blue points&lt;/em&gt;), compared to an extension of the P-S HMF (&lt;em&gt;solid black line&lt;/em&gt;) to allow for ellipsoidal structures (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1999MNRAS.308..119S&quot; rel=&quot;noreferrer&quot;&gt;Sheth &amp;amp; Tormen (1999&lt;/a&gt;, &lt;a href=&quot;http://adsabs.harvard.edu/abs/2002MNRAS.329...61S&quot; rel=&quot;noreferrer&quot;&gt;2002)&lt;/a&gt;. At all masses, FoF halos are seen to be more massive than SO halos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lniH4.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lniH4.png&quot; alt=&quot;HMF&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Redshift range and area&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Re-reading your question, I think maybe you're interested not in the number &lt;em&gt;density&lt;/em&gt;, but in the absolute number in a volume spanned by a redshift range &lt;span class=&quot;math-container&quot;&gt;$dz$&lt;/span&gt; and area &lt;span class=&quot;math-container&quot;&gt;$dA$&lt;/span&gt;. If that is the case, then you simply multiply your &lt;span class=&quot;math-container&quot;&gt;$N(&amp;gt;\!\!M_\mathrm{h})$&lt;/span&gt; by the cosmological volume given by &lt;span class=&quot;math-container&quot;&gt;$dz$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$dA$&lt;/span&gt;. Let me know if you also want to know how to calculate that.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-09-21T05:03:07.243" LastActivityDate="2018-09-21T05:03:07.243" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27707" PostTypeId="1" AcceptedAnswerId="27709" CreationDate="2018-09-20T16:07:17.810" Score="2" ViewCount="111" Body="&lt;p&gt;There is no cosmic definition of now because of special relativity, because of rate at which time flows for a particular object depends upon your point of view and your velocity. For a rocket traveling at speed of light, journey of alpha centauri might take a week. For us it would seem like 4 years. As a result of special relativity even the meaning of ‘simultaneous’ is related to your point of view&lt;/p&gt;&#xA;" OwnerUserId="24177" LastActivityDate="2018-09-20T16:31:45.803" Title="What exactly is the cosmic definition of ‘now’ is?" Tags="&lt;cosmology&gt;&lt;space-time&gt;&lt;special-relativity&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27708" PostTypeId="1" AcceptedAnswerId="27720" CreationDate="2018-09-20T16:21:08.407" Score="2" ViewCount="698" Body="&lt;p&gt;I am building some device as a gift for my girlfriend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this device to work, I need to be able to calculate the angle to which I should look towards (3D angle), in order to see the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This angle is affected by (and maybe not only):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The time of year.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The hour.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The location on earth from which I look.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm sure there's some formula that approximates the angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't need an exact value, I can live with +-1 degree.&lt;/p&gt;&#xA;" OwnerUserId="24321" LastEditorUserId="616" LastEditDate="2018-09-21T17:44:11.233" LastActivityDate="2018-09-24T11:41:37.620" Title="How to calculate exact angle to the moon?" Tags="&lt;the-moon&gt;&lt;amateur-observing&gt;&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="27709" PostTypeId="2" ParentId="27707" CreationDate="2018-09-20T16:31:45.803" Score="6" Body="&lt;p&gt;The cosmic &quot;now&quot; is well-defined: It is the time for an observer that has always been at rest in the Universe's &lt;em&gt;comoving&lt;/em&gt; coordinates, i.e. the coordinates that expand along with the Universe. Although this reference frame is no more special than any other frame, it makes sense to use this. For instance, it is the only frame in which the cosmic microwave background is (statistically) isotropic, and it is the frame in which you'll see the same number density of galaxies in all directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could (rightfully) define &quot;now&quot; in another frame, but it wouldn't be of much use. For instance, for a neutrino released 1 s after Big Bang, the age of the Universe is some ten thousand years, but it doesn't really tell you anything about the Universe.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2018-09-20T16:31:45.803" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27710" PostTypeId="1" CreationDate="2018-09-20T18:05:27.597" Score="2" ViewCount="1475" Body="&lt;p&gt;I have an image of a single unsaturated source (asteroid) that I'd like to fit guesses for various model PSFs to, so that I can then compare the total flux in each case. I've only recently begun studying PSFs, so I am a bit confused. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that astropy offers a package called &lt;code&gt;photutils.psf&lt;/code&gt; which does the PSF fitting for you, but I would like to better understand what's going on behind the fitting, and how the flux of the source is resolved from it. Furthermore, when I tried using this package, I wasn't getting any outputs anyway (it said a source was found, but that it didn't meet the sharpness or roundness criteria, despite me changing those parameters around quite a bit). I have tried something simpler below (where &lt;code&gt;model_data&lt;/code&gt; is the 2d numpy array representation of the PSF, and &lt;code&gt;img&lt;/code&gt; is the asteroid image data):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;from astropy.modeling import fitting&#xA;from photutils import FittableImageModel&#xA;&#xA;y, x = np.mgrid[:np.shape(img)[0], :np.shape(img)[1]]&#xA;&#xA;fitter = fitting.LevMarLSQFitter()&#xA;model2d = FittableImageModel(model_data)&#xA;result = fitter(model2d, x, y, img)&#xA;print(result)&#xA;print(np.median(img), np.max(img), np.sum(img))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The output I get is then:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;flux       x_0       y_0&#xA;&#xA;-----     -----     -----&#xA;&#xA;9282.784  466.44    531.69&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;2705.82 25301.56 980812434.98&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, these are not the correct centroid coordinates of the source in the image. The correct coordinates are around (300, 300). Also, to get this result, the PSF data was normalized so that the max intensity point is 1. But when I change this normalization so that the total flux (sum of all the pixels) is 1, the result for the flux after fitting I get is way higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what's happening behind the scenes here (I got lost in the documentation), and I'm not sure if I can trust any of these flux values, considering the &lt;code&gt;x_0&lt;/code&gt; and &lt;code&gt;y_0&lt;/code&gt; are always off. So I'd like to better understand -- how exactly does one find the flux of an object by fitting it to a PSF? And how should that PSF be normalized?&lt;/p&gt;&#xA;" OwnerUserId="24076" LastActivityDate="2018-09-23T15:20:24.017" Title="Fitting PSF to an image to get total flux (Python)" Tags="&lt;data-analysis&gt;&lt;python&gt;&lt;astropy&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27711" PostTypeId="2" ParentId="27710" CreationDate="2018-09-20T19:35:59.080" Score="2" Body="&lt;p&gt;You can't really get a PSF for an extended object.  The Point-Spread-Function is a parameter of your imaging system which describes the shape of the image of a putative point source.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have any knowledge of what your python package is intended to do, so I can't comment directly on the results.  I will point (sorry) out that the geometric centroid of the image is not generally the same as the energy(brightness) centroid of the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, the total flux is just the sum of all the pixels covered by the asteroid's image (less background noise and whatnot).  It's not clear why you need the PSF if you are just doing energy/power calculations. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-09-20T19:35:59.080" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27714" PostTypeId="1" CreationDate="2018-09-21T04:23:19.077" Score="4" ViewCount="149" Body="&lt;p&gt;I am wondering about what is the estimated size of astrophysical jet that shoots out from a black hole, I don’t know if it is determined by the size of a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would classify a black hole that is able to have at least 10 planets orbit it? Is it considered stellar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, what would the estimated astrophysical jet or quasar be?&lt;/p&gt;&#xA;" OwnerUserId="24242" LastEditorUserId="37191" LastEditDate="2021-02-17T21:13:37.837" LastActivityDate="2021-05-19T11:00:30.043" Title="Astrophysical Jet Size?" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;general-relativity&gt;&lt;accretion-discs&gt;&lt;astronomical-jet&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27715" PostTypeId="1" AcceptedAnswerId="27755" CreationDate="2018-09-21T08:08:56.500" Score="2" ViewCount="133" Body="&lt;p&gt;Could anyone tell me what the names of the Greek letters in the screenshot are? Is the first one alpha? They are used in calculating ICRF rotations. Wikipedia uses a different font for their list of characters, so it's hard to tell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Rxd4T.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rxd4T.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24284" LastEditorUserId="24284" LastEditDate="2018-09-26T07:04:30.217" LastActivityDate="2018-09-26T07:04:30.217" Title="Identify Greek letters used in formulas" Tags="&lt;rotation&gt;" AnswerCount="1" CommentCount="11" ClosedDate="2018-09-26T16:21:04.037" ContentLicense="CC BY-SA 4.0" />
  <row Id="27717" PostTypeId="2" ParentId="27043" CreationDate="2018-09-21T11:55:17.143" Score="2" Body="&lt;p&gt;It's very disappointing that you have received only one reply to this so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But @Ingolifs is correct, the image is clearly out of focus since you can see the spider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as he has indicated, it also clearly needs to be collimated.  The mirrors are very badly out of alignment.  Just follow the manual.  If you're part of a club, most of the members can show you how to do it.  You probably didn't get a collimation cap with your 'scope so you'll have to go back to the store and get one.  Or you can make one.  Or, if you're part of a club, there will probably be members who have uber-cool collimation tools to help you out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first time you do it by yourself you'll want to have good, daytime-level light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing at the lines:  they're probably a product of the poor alignment and may just go away after proper collimation.  They may be an indication that your primary is being &quot;pinched&quot;.  There may be something twisting or squeezing your mirror.  Make sure it's not mounted too tightly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.  Good luck with it.&lt;/p&gt;&#xA;" OwnerUserId="24065" LastActivityDate="2018-09-21T11:55:17.143" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27718" PostTypeId="1" AcceptedAnswerId="27719" CreationDate="2018-09-21T17:03:36.013" Score="4" ViewCount="87" Body="&lt;p&gt;This answer (here in the forum) provides background to my question: &lt;a href=&quot;https://astronomy.stackexchange.com/a/5953&quot;&gt;https://astronomy.stackexchange.com/a/5953&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to understand the process in which a stellar population synthesis (SPS) is created, in order to calculate the stellar mass of a galaxy. (This is my main goal: understand how the stellar mass is calculated via SPS and broadband photometry)(Section 2.5:  &lt;a href=&quot;https://ned.ipac.caltech.edu/level5/Sept14/Conroy/Conroy2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ned.ipac.caltech.edu/level5/Sept14/Conroy/Conroy2.html&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok.&#xA;I have already understood the process to obtain a stellar population library (SPL). Such SPL will be fitted to observational data in order to create a SED (Spectral Energy Distribution)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok. Now I have the fitted SED on a galaxy, here are my questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How can I estimate the M/L ratio of that galaxy?&lt;/li&gt;&#xA;&lt;li&gt;How can I estimate the L?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I feel it is an easy and direct answer, but I don't get it. &lt;/p&gt;&#xA;" OwnerUserId="24330" LastEditorUserId="14094" LastEditDate="2018-09-25T07:34:37.460" LastActivityDate="2018-09-25T07:34:37.460" Title="How can I estimate the M/L ratio from a fitted SED on a galaxy?" Tags="&lt;galaxy&gt;&lt;mass&gt;&lt;luminosity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27719" PostTypeId="2" ParentId="27718" CreationDate="2018-09-21T17:14:22.443" Score="3" Body="&lt;p&gt;The luminosity L isn't estimated, it's observed. You make models of stellar populations with many different parameters and those give you a set of SEDs. For each set of parameters, you will have an SED covering a range of wavelengths (a continuous line).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From your observations, you will have an SED of a few points depending on how many wavelengths you observed. You use &lt;span class=&quot;math-container&quot;&gt;$\chi^{2}$&lt;/span&gt; fitting or some other technique to compare all your model lines with those few points and see which fits best. This gives you the best fit value for the parameters in your model, one of which is M/L.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have the actual luminosity of the galaxy from the observations (at some wavelength) so you multiply the estimated M/L by the measured L to get an estimate of M.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2018-09-21T17:14:22.443" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27720" PostTypeId="2" ParentId="27708" CreationDate="2018-09-21T17:18:02.840" Score="5" Body="&lt;p&gt;For a one-off calculation, the easiest solution would be to use planetarium software, such as stellarium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a programming solution, a python package such as pyephem is an efficient way to calculate the position of the moon at multiple times, in a way that can be imported into a spreadsheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The apparent motion of the moon is more complex than you might think.  The moon moves about 15 degrees per hour due to the rotation of the Earth. But on top of this fairly simple motion is the Moon's actual orbit around the Earth, which is elliptical (and so this motion is not even, the moon moves faster when it is close to the Earth) and the position and inclination of the orbit is perturbed by the Sun. The perturbations are fairly regular, and can be calculated, but these effects (rotation of the earth, the orbit of the moon, the eccentricity of the orbit, perturbation of orbit) combine to make the actual calculation more complex  than &quot;a formula&quot; that can fit easily in a spreadsheet, which is why a package like pyephem is recommended.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-09-21T17:18:02.840" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27721" PostTypeId="2" ParentId="27673" CreationDate="2018-09-21T19:59:54.187" Score="4" Body="&lt;p&gt;To respond to the question as is: yes the paper is trying to point out that the concensus was not scientific.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this claim is not true at all. In order to understand how classifications within the solar system are made, one can look at the history of the planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short answer is, the demotion of a planet is  a result of much scientific consensus than the classification of a body as a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For thousands of years, we had 7 planets: Sun, Moon, Mercury, Venus, Mars, Jupiter and Saturn. These were the seven wanderers in the night sky. However, as we understood more about our system, we dropped the moon and the sun and added Earth. That process happened through a lot of bloodshed, since back then, voting upon an idea seemed not very logical(!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the time came that we discovered a planet for the very first time: Uranus. Without doubt it was classified as a planet because its place perfectly fit the Titus-Bode law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uranus's discovery did not stop the astronomers because, according to Titus-Bode law, there was a missing planet between Mars and Jupiter. And so the day came and in the year 1801, Ceres was discovered and announced as the new planet (again because it fit the law). However, as time passed by and we understood more and more about the Asteroid belt, 46 years later, Ceres (and others that were considered a planet) was demoted from the planet status.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same story (more or less) happened with Pluto, and after much consideration (and a lot of published papers about Pluto, Kuiper Belt objects &amp;amp; planetary taxonomy) 76 years later Pluto was demoted as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the difference between 1847 and 2006 is IAU, which is the organisation to decide astronomical classifications and facts by looking at scientific evidence and reaching a consensus by vote! Through all these information, it is a blatant ignorance to tell that planetary taxonomy relies on non-scientific process of voting. &lt;/p&gt;&#xA;" OwnerUserId="20818" LastActivityDate="2018-09-21T19:59:54.187" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27722" PostTypeId="1" AcceptedAnswerId="27726" CreationDate="2018-09-22T02:02:59.167" Score="6" ViewCount="198" Body="&lt;p&gt;Pretty much every definition I've read about asteroids is something like &quot;Asteroids are rocky worlds revolving around the sun that are too small to be called planets.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would an asteroid be called if it was in deep space, not orbiting any sun? &lt;/p&gt;&#xA;" OwnerUserId="24331" LastEditorUserId="37191" LastEditDate="2021-02-25T21:39:34.570" LastActivityDate="2021-02-25T22:28:02.890" Title="What is an asteroid called if it is in deep space?" Tags="&lt;asteroids&gt;&lt;definition&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27725" PostTypeId="1" CreationDate="2018-09-22T05:12:46.007" Score="1" ViewCount="42" Body="&lt;p&gt;What are recent research techniques to find Dark matter? As they don't interact with other matters, which properties of it can help us to detect it? &lt;/p&gt;&#xA;" OwnerUserId="23604" LastEditorUserId="1891" LastEditDate="2018-09-22T08:41:53.200" LastActivityDate="2018-09-22T08:41:53.200" Title="Techniques to find dark matter" Tags="&lt;dark-matter&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2018-09-23T09:45:01.987" ContentLicense="CC BY-SA 4.0" />
  <row Id="27726" PostTypeId="2" ParentId="27722" CreationDate="2018-09-22T13:38:31.650" Score="8" Body="&lt;p&gt;We have only discovered one such object so far: ʻOumuamua. (Although a second, 2I/Borisov, was discovered in 2019) The general practice has been to call it an &amp;quot;interstellar asteroid&amp;quot; or &amp;quot;interstellar comet&amp;quot;, even though it is not in orbit.  You could just call it an &amp;quot;interstellar object&amp;quot;.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2021-02-25T22:28:02.890" LastActivityDate="2021-02-25T22:28:02.890" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27727" PostTypeId="2" ParentId="12704" CreationDate="2018-09-22T18:15:06.953" Score="2" Body="&lt;p&gt;Don't forget that there are brown dwarfs between planets and stars in mass. The heaviest planets are around a dozen Jupiter masses, and the lightest stars are a little over 80 Jupiter masses (from the Wikipedia article on brown dwarfs). This means you would need some kind of system of seven super-Jupiters for one red dwarf to orbit around. As userLTK made clear in the accepted answer, there's really no plausible stable configuration where this works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can propose an implausible stable configuration, though. In an empty universe, or deep intergalactic space, two objects can be very far apart and still stably orbit each other. So, imagine two super-Jupiters orbiting their common center of gravity. Far away from them, a third super-Jupiter orbits the pair. More accurately, it orbits the common center of gravity of the 3-planet system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the system is so isolated, we can, unless I am mistaken, repeat the process indefinitely. You can always move far enough from a cluster so that it is nearly a point from your perspective, and then orbit the center of gravity between you and that point. Using that process, you build a system with seven or more super-Jupiters and one red dwarf orbiting very far away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If such a system existed, I would consider it evidence of intentional stellar engineering. Such a chain of perfect orbits is just too big a coincidence.&lt;/p&gt;&#xA;" OwnerUserId="7752" LastActivityDate="2018-09-22T18:15:06.953" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27730" PostTypeId="2" ParentId="27683" CreationDate="2018-09-23T07:30:05.223" Score="3" Body="&lt;p&gt;In ancient times people sometimes saw things in the sky, and they did not know what caused them.  They called these things (in Greek) &quot;ta meteora&quot; or &quot;things in heaven&quot;.  Some of these we know call &quot;stars&quot; others &quot;planets&quot; and some &quot;comets&quot;. We understand those to be bodies in space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other &quot;meteora&quot; we now understand to be atmospheric effects, for example &quot;clouds&quot; or &quot;lightning&quot; are other examples of &quot;meteora&quot;. By the 15 century &quot;meteor&quot; (in English) had come to mean any atmospheric phenomenon, including the &quot;streaks of light&quot; that we sometimes see at night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over time the word became specialised. Although we still talk of &quot;meteorology&quot; (which studies clouds, storms and weather in general) The word &quot;meteor&quot; came to be used mainly for those streaks of light in the sky at night. Though at the time we still didn't know what they were caused by.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been reports of &quot;stones falling from the sky&quot; since ancient times, but they were rare and could be regarded as fantasy or illusion; they could be dismissed as being like the reports of dragons or ghosts. But eventually, careful studies realised that &quot;stones falling from the sky&quot; were a &lt;em&gt;real&lt;/em&gt; thing. And these stones could be identified and collected. We wanted a name for this type of stone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On very rare occasions it was realised that the &quot;stone from the sky&quot; was associated with the &quot;streak in the sky&quot;. And this gave rise to a name for these stones: &quot;meteor-rock&quot; or &quot;meteorite&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So meteor means &quot;a light phenomenon visible as a streak of light in the sky&quot;. A meteor is mostly ionised air molecules. The meteor is caused by the passage at hypervelocity (several km/s) of a piece of rock from space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If any parts of that rock are found on Earth, they are called meteorites. Something that can hit you on the head is a rock, and hence a &lt;strong&gt;meteorite&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rock, before it passes through the atmosphere, is very different from the rock that lands.  If it is necessary to make the distinction, you can call the very space rock a &quot;meteoroid&quot;, by analogy with &quot;meteor-asteroid&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people care about the naming of things, but if you just want to insult someone, I doubt they care much about the correctness of your nomenclature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obligitory xkcd &lt;a href=&quot;https://xkcd.com/1405/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://xkcd.com/1405/&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-09-23T07:30:05.223" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27731" PostTypeId="1" CreationDate="2018-09-23T08:10:22.910" Score="1" ViewCount="173" Body="&lt;p&gt;Does the Sun undergo any cycles or revolutions approximately every 24,000 years or 8.6 million years? In Hinduism it is believed that the Sun revolves in an elliptical orbit around a large mass at these rates.&lt;/p&gt;&#xA;" OwnerUserId="24341" LastEditorUserId="2531" LastEditDate="2018-09-23T08:42:32.393" LastActivityDate="2018-09-23T16:06:52.620" Title="Movement of the Sun in space" Tags="&lt;the-sun&gt;&lt;black-hole&gt;&lt;galaxy&gt;&lt;milky-way&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27732" PostTypeId="1" AcceptedAnswerId="27734" CreationDate="2018-09-23T10:19:36.030" Score="1" ViewCount="125" Body="&lt;p&gt;Or are there points on Earth where certain parts of the 360° night sky are never visible throughout the whole year?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the answer to this question, are there points on Earth where parts of the celestial sphere as such are never in the sky, no matter if it's a day or night sky?&lt;/p&gt;&#xA;" OwnerUserId="24234" LastActivityDate="2018-09-23T11:26:11.540" Title="Throughout the year, are all parts of the night sky visible from any given point on Earth?" Tags="&lt;amateur-observing&gt;&lt;observational-astronomy&gt;&lt;night-sky&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27733" PostTypeId="2" ParentId="27731" CreationDate="2018-09-23T10:41:21.323" Score="1" Body="&lt;p&gt;The sun revolves around the galactic centre and completes one revolution in approximately 250 million years. The Sun's orbit is around 25,000 light years away from the galactic centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than these, it doesn't have any cycles or movement corresponding to the numbers you specified. &lt;/p&gt;&#xA;" OwnerUserId="20818" LastEditorUserId="20818" LastEditDate="2018-09-23T16:06:52.620" LastActivityDate="2018-09-23T16:06:52.620" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27734" PostTypeId="2" ParentId="27732" CreationDate="2018-09-23T11:26:11.540" Score="7" Body="&lt;p&gt;At any one time, only half the sky can be visible, and half is below the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the poles the sky appears to rotate about the zenith, and so no new parts of the sky ever become visible.  At the North pole, you can never see any stars in the Southern hemisphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the equator different parts of the sky become visible and over the year all the stars can be seen (but stars very close to the pole will be very hard to see, as they are too close to the horizon.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For someone who is at (for example) a latitude of 40 degrees North will not see any stars that are more than degrees south of the equator; 90-40=50; and so won't see the southern cross, the Magellanic clouds or other features of the Southern sky. Similarly an observer in Tasmania won't see Polaris, Cassiopeia, or the &quot;big dipper&quot; as these are features of the Northern sky.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-09-23T11:26:11.540" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27735" PostTypeId="1" CreationDate="2018-09-23T14:49:18.590" Score="1" ViewCount="799" Body="&lt;p&gt;I don't know much about astronomy, but my kids are asking for telescopes to view planets, stars etc. Could anyone tell me if &lt;a href=&quot;http://www.argos.co.uk/product/7523700&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Eduscience telescope&lt;/a&gt; is any good?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This kit contains a 40x with a 41cm body, 20x 30x 40x eyepiece, a diagonal mirror, 30mm diameter objective lens and tripod.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Size H51, W44cm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EvPP7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EvPP7.jpg&quot; alt=&quot;Edu-Science 30mm telescope&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24343" LastEditorDisplayName="user1569" LastEditDate="2018-09-26T09:38:49.577" LastActivityDate="2018-10-07T20:43:45.250" Title="Edu-Science telescope" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27736" PostTypeId="2" ParentId="27731" CreationDate="2018-09-23T14:50:21.800" Score="3" Body="&lt;p&gt;The &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Axial_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;precession of the Earth's spin axis orientation&lt;/a&gt; takes place on a 25,700 year cycle. But this isn't connected with the Sun's movement through space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun orbits the Galaxy about once every 220-240 million years. It also oscillates about the Galactic plane with a period of about 70 million years.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-09-23T15:01:52.990" LastActivityDate="2018-09-23T15:01:52.990" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27737" PostTypeId="2" ParentId="27710" CreationDate="2018-09-23T15:20:24.017" Score="1" Body="&lt;p&gt;The image is 2D pixels, the PSF fitting routine would do something like this. First, you pick the PSF function, which is a 2D gaussian in this case. Then, the routine will choose a centroid, mu, and variance (or mus and variances).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of things that will complicate the routine. i) Local peak due to a noise. This will mess up the centroid. Fixing is to do the centroid search by using averages given some sizes of the areas. Also, you might want to specify how large the search area should be given the initially specified spot. Or, if you are certain about the location of centroid, you can fix it and the centroid would not be a choice variable. ii) Aperture size. If the object is a perfect 2D gaussian point source, the aperture size does not matter. But with noise and other surrounding objects, you have to judge what should be the optimal size: too large = more noise and including other objects, too small = incorrect fit shape due to noise dominating. There might be something more to think about that I cannot recall at this moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speaking of this, the fitting routine can fit even an extended source, but you won't get a good result unless you use a proper kernel (i.e., not gaussian).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally don't know about astropy. I used IRAF, which has its equivalent as PyRAF in Python, I believe. You might want to check that as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your problem about incorrect centroid, I believe, roots from the local spikes from noise and the search routine not having enough size for a good average. This problem directly leads &gt;1 total flux as you mentioned as well.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-09-23T15:20:24.017" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27738" PostTypeId="2" ParentId="27735" CreationDate="2018-09-23T16:15:06.313" Score="1" Body="&lt;p&gt;I'm not familiar with this telescope, but if you only want a telescope, and not a microscope as well, then for the same money you could probably get a better telescope.  Just google &quot;beginner telescope astronomy&quot; or something similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't mention the age of your children, but young children seem to find looking through optical equipment quite difficult to do, especially using just one eye, so it would be worth discussing suitability with the retailer - assuming you choose a specialist company.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2018-09-23T16:15:06.313" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27744" PostTypeId="1" CreationDate="2018-09-24T02:34:50.867" Score="3" ViewCount="65" Body="&lt;p&gt;My sundial shows real time until 11am when it will drop behind an hour until sunset. This is in keeping with daylight saving but why does it show real time before 11am? Am I missing something elementary?&lt;/p&gt;&#xA;" OwnerUserId="24349" LastActivityDate="2018-09-24T02:34:50.867" Title="Erratic Sundial" Tags="&lt;the-sun&gt;" AnswerCount="0" CommentCount="6" FavoriteCount="1" ClosedDate="2020-02-20T23:26:57.523" ContentLicense="CC BY-SA 4.0" />
  <row Id="27745" PostTypeId="2" ParentId="14041" CreationDate="2018-09-24T07:15:37.193" Score="4" Body="&lt;p&gt;Various &lt;a href=&quot;https://www.google.com/search?q=meteors%20%20%20ecliptic%20longitude.&amp;amp;tbm=isch&amp;amp;ved=2ahUKEwitstej_MHvAhXK_4UKHQ1-D9IQ2-cCegQIABAA&amp;amp;oq=meteors%20%20%20ecliptic%20longitude.&amp;amp;gs_lcp=CgNpbWcQA1DkC1jkC2CaDGgAcAB4AIABAIgBAJIBAJgBAKABAaoBC2d3cy13aXotaW1nwAEB&amp;amp;sclient=img&amp;amp;ei=dYpXYO2jGMr_lwSN_L2QDQ&amp;amp;bih=711&amp;amp;biw=1267&quot; rel=&quot;nofollow noreferrer&quot;&gt;graphs illustrate the times and areas&lt;/a&gt; of the globe where meteorites are more likely to hit.&#xA;&lt;a href=&quot;https://ars.els-cdn.com/content/image/1-s2.0-S0019103516303761-gr4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://ars.els-cdn.com/content/image/1-s2.0-S0019103516303761-gr4.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;That's the best scientific indicator that we have regarding daytime: it was significantly more likely to occur either side of dawn.&lt;/p&gt;&#xA;&lt;p&gt;Specific scientific evidence would have to come in the form of biochemical markers from fossils, which is possible but practically impossible. One of the few fossils from the KT boundary was found 13cm below it, a Ceratopsian dinosaur horn, about a year previous, and we'd need multiple fossils from the same day.&lt;/p&gt;&#xA;&lt;p&gt;Difficult to recover data from a 10 billion Hiroshima explosions that happened 65,500,000 years ago.&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2021-03-21T18:02:24.773" LastActivityDate="2021-03-21T18:02:24.773" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27747" PostTypeId="1" AcceptedAnswerId="27750" CreationDate="2018-09-24T09:28:37.763" Score="9" ViewCount="1413" Body="&lt;p&gt;A photo in the NYTimes article &lt;a href=&quot;https://www.nytimes.com/2018/09/21/climate/greenland-arctic-ice.html&quot; rel=&quot;noreferrer&quot;&gt;Ice Surveys and Neckties at Dinner: Here’s Life at an Arctic Outpost&lt;/a&gt; has got me thinking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the winter months at either the north or south pole, when it is constantly dark, how does the Moon behave? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would a nearly-full moon sometimes circle for days without setting, or would it still rise and set, or would it be all together missing, or some combination of all of the above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/562uq.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/562uq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-09-24T15:09:32.007" Title="How does the Moon move in the &quot;night&quot; sky as seen from the poles?" Tags="&lt;the-moon&gt;&lt;night-sky&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27748" PostTypeId="1" CreationDate="2018-09-24T10:43:47.687" Score="1" ViewCount="155" Body="&lt;p&gt;I'm new to this could someone tell me if I'd be able to see planets with the&#xA;&lt;a href=&quot;https://www.celestron.com/products/cometron-firstscope-telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Celestron Cometron FirstScope&lt;/a&gt;&#xA;described as below on &lt;a href=&quot;http://www.argos.co.uk/product/8339876&quot; rel=&quot;nofollow noreferrer&quot;&gt;argos.co.uk&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A compact, easy to use and powerful 76mm reflector telescope which will give fantastic views of the Moon and stars.&#xA;  The Dobsonian mount design is best used on a table or window ledge.&#xA;  The included eyepieces allow for 30x and 15x magnification.&#xA;  The included finderscope makes locating objects a snap!&#xA;  No tools are required to assemble.&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Optimum magnification 30x.&lt;/li&gt;&#xA;  &lt;li&gt;Maximum magnification 30x.&lt;/li&gt;&#xA;  &lt;li&gt;Portable design perfect for use in the garden, from a window ledge or camping.&lt;/li&gt;&#xA;  &lt;li&gt;Two eyepieces included for 15x &amp;amp; 30x magnification.&lt;/li&gt;&#xA;  &lt;li&gt;Weight 1.95kg.&lt;/li&gt;&#xA;  &lt;li&gt;Size H39.4, W24.1, D23.8cm.&lt;/li&gt;&#xA;  &lt;li&gt;Manufacturer's 2 year guarantee.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="24343" LastEditorUserId="616" LastEditDate="2018-09-24T19:22:15.507" LastActivityDate="2018-09-24T22:26:17.540" Title="Amateur observing" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27749" PostTypeId="2" ParentId="27708" CreationDate="2018-09-24T11:22:49.853" Score="1" Body="&lt;p&gt;In &lt;a href=&quot;https://astronomy.stackexchange.com/a/18029/7982&quot;&gt;this answer&lt;/a&gt; I mention a source where handy formula for approximate behavior of celestial objects can be found.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Astronomical Algorithms&lt;/em&gt; by Jean Meeus&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't have a copy personally so I can't verify if there is exactly what you need can be found there or not, but this &lt;a href=&quot;http://www.math.harvard.edu/~knill/3dprinter/exhibits/bc/Meeus1988.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;pdf&lt;/a&gt; of a small subset of pages from the &lt;em&gt;fourth edition&lt;/em&gt; includes the table of contents, which includes:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;30. Position of the moon....................... 147&#xA;31. Illuminated Fraction of the Moon’s Disk.... 155&#xA;32. Phases of the Moon......................... 159&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;so there's a good chance in those seven or eight pages for chapter 30 there may be just what you are looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the &quot;3D angles&quot; you mention would be just altitude (or elevation angle) and azimuth. The first is the angle above (or below) the horizon, and the second is like the &quot;compass direction&quot; except starting at true north, not magnetic north, so it's just 2D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are comfortable using Python, one of the easiest computer languages, there is a package called &lt;a href=&quot;http://rhodesmill.org/skyfield/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Skyfield&lt;/a&gt;. I believe you should be able to get it to run on a Raspberry Pi for example, if your creation is electromechanical. The package is very precise so it generally needs a connection to the internet to &quot;check stuff&quot; which is a little frustrating sometimes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also calculate a table for your location, then save it and use it to run a small computer or controller, if you are using one.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;GREEN  = Moon (being made of green Cheese of course)&#xA;YELLOW = Sun&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/O9evh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O9evh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/R0dZe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R0dZe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xTzSn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xTzSn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;from skyfield.api import Loader, Topos&#xA;&#xA;halfpi, pi, twopi = [f*np.pi for f in 0.5, 1, 2]&#xA;degs, rads        = 180/pi, pi/180&#xA;&#xA;load   = Loader('~/Documents/fishing/SkyData')&#xA;data   = load('de421.bsp')&#xA;ts     = load.timescale()&#xA;&#xA;de421  = load('de421.bsp')&#xA;earth  = de421['earth']&#xA;sun    = de421['sun']&#xA;moon   = de421['moon']&#xA;&#xA;Haifa   = earth + Topos(32.82, 34.98)&#xA;minutes = np.arange(60*24)&#xA;times   = ts.utc(2018, 9, 24, 0, minutes)&#xA;&#xA;salt, saz, sd = Haifa.at(times).observe(sun).apparent().altaz()&#xA;malt, maz, md = Haifa.at(times).observe(moon).apparent().altaz()&#xA;&#xA;saltr, sazr, maltr, mazr = [x.radians for x in (salt, saz, malt, maz)]&#xA;&#xA;sr, mr = [1. - x/halfpi for x in (saltr, maltr)]&#xA;&#xA;sx, sy = [sr * f(sazr) for f in (np.sin, np.cos)]&#xA;mx, my = [mr * f(mazr) for f in (np.sin, np.cos)]&#xA;&#xA;theta = np.linspace(0, twopi, 201)&#xA;xc, yc = [f(theta) for f in (np.cos, np.sin)]&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;    plt.plot(sx, sy, '-y')&#xA;    plt.plot(mx, my, '-g')&#xA;    plt.plot(xc, yc, '-k', linewidth=2)&#xA;    plt.plot([0], [0], 'ok')&#xA;    plt.xlim(-1.8, 1.8)&#xA;    plt.ylim(-1.8, 1.8)&#xA;    plt.text(-0.15, 0.1, 'Zenith', fontsize=16)&#xA;    plt.text(-0.02, 1.04, 'N', fontsize=16)&#xA;    #plt.text(-1.6, 1.0, 'North is up', fontsize=16)&#xA;    plt.show()&#xA;&#xA;&#xA;saltd, sazd, maltd, mazd = [x.degrees for x in (salt, saz, malt, maz)]&#xA;mbreaks = np.where(np.abs(mazd[1:]-mazd[:-1] &amp;lt; 0))&#xA;sbreaks = np.where(np.abs(sazd[1:]-sazd[:-1] &amp;lt; 0))&#xA;&#xA;saltd, sazd, maltd, mazd = [x[:-1] for x in (saltd, sazd, maltd, mazd)]&#xA;sazd[sbreaks] = np.nan&#xA;mazd[mbreaks] = np.nan&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;    plt.plot(sazd, saltd, '-y')&#xA;    plt.plot(mazd, maltd, '-g')&#xA;    plt.plot([0, 360], [0, 0], '-k')&#xA;    plt.xlabel('azimuth (degs)', fontsize=18)&#xA;    plt.ylabel('altitude/elevation (degs)', fontsize=18)&#xA;    plt.xlim(0, 360)&#xA;    plt.show()&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;    hours = minutes[:-1]/60.&#xA;&#xA;    plt.subplot(2, 1, 1)&#xA;&#xA;    plt.plot(hours, sazd, '-y')&#xA;    plt.plot(hours, mazd, '-g')&#xA;    # plt.xlabel('UTC hours', fontsize=16)&#xA;    plt.ylabel('azimuth (degs)', fontsize=16)&#xA;&#xA;    plt.subplot(2, 1, 2)&#xA;&#xA;    plt.plot(hours, saltd, '-y')&#xA;    plt.plot(hours, maltd, '-g')&#xA;    plt.xlabel('UTC hours', fontsize=16)&#xA;    plt.ylabel('altitude/elevation (degs)', fontsize=16)&#xA;&#xA;    plt.suptitle(&quot;Haifa 24/9/2018&quot;, fontsize=16)&#xA;&#xA;    plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-09-24T11:41:37.620" LastActivityDate="2018-09-24T11:41:37.620" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27750" PostTypeId="2" ParentId="27747" CreationDate="2018-09-24T14:19:29.427" Score="10" Body="&lt;p&gt;The moon is quite interesting as you get near the poles. It still moves round due to the Earth's rotation once a day, but it also orbits with a period of a month so what you get is two weeks of moon above the horizon followed by two weeks where you cannot see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What changes through the year is what phase you see when the moon is visible, and in the middle of winter you have two weeks around full moon, whereas in summer you have two weeks around the new moon - not so exciting. Spring and autumn obviously have two weeks around first and last quarter, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=Toc6JAUWv_M&quot; rel=&quot;noreferrer&quot;&gt;This simulation using Stellarium&lt;/a&gt; shows the effect quite well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://curious.astro.cornell.edu/about-us/46-our-solar-system/the-moon/observing-the-moon/127-is-the-moon-always-visible-during-winter-on-the-north-pole-intermediate&quot; rel=&quot;noreferrer&quot;&gt;astro.cornell.edu&lt;/a&gt; - this wording may help: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The second movement caused by the Moon's orbit around the Earth is analogous to the movement of the Sun over the course of a year only it repeats over the course of a lunar month. Near the new Moon phase, the Moon is near the Sun and therefore never rises during the winter. As the Moon approaches full, it will start to pop up above the horizon. Eventually near the full Moon phase it will be high enough in the sky to stay up all day and circle like the Sun in the video above. The elevation of the circle will rise as the Moon becomes completely full and then start to decrease until it begins to dip below the horizon. Eventually the Moon will stop rising at all as it gets close enough to the new phase. The cycle then repeats.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And another useful reference at &lt;a href=&quot;https://planetarium.madison.k12.wi.us/mooncal/north-pole-moon.htm&quot; rel=&quot;noreferrer&quot;&gt;planetarium.madison.k12.wi.us&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At the beginning of Winter, when it's nighttime all of the time, the moon would be in the sky for the 2 weeks closest to Full Moon, and then below the horizon for the next 2 weeks. And at the beginning of Spring &lt;a href=&quot;https://planetarium.madison.k12.wi.us/mooncal/moon2wks-march&quot; rel=&quot;noreferrer&quot;&gt;(click on the graphic)&lt;/a&gt;, when the sun is at sunrise all of the time, the moon would be up in the sky for the 2 weeks closest to First Quarter (waxing), and then below the horizon for the next 2 weeks.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The animated graphic above shows what we would see from the North Pole if we went out every day at noon, for 14 days in a row, from March 1st to March 14, 2006. We start with a thin crescent moon near the horizon, and end with a full moon near the horizon. Halfway through, the First Quarter moon would be when the moon is highest above the horizon.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Keep in mind, that if you were observing the moon constantly, throughout a 24 hour period, the moon would seem to move to the right in the sky along with the sun, stars, and planets due to the Earth's rotation. Nothing would seem to rise and set: they would just seem to circle around you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="43" LastEditorUserId="43" LastEditDate="2018-09-24T15:09:32.007" LastActivityDate="2018-09-24T15:09:32.007" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="27751" PostTypeId="2" ParentId="27697" CreationDate="2018-09-24T18:56:15.460" Score="2" Body="&lt;p&gt;In general, finding the low mass cutoff of a particular RV survey is normally quite complicated as there are a lot of selection effects on many different variables (such as planet period, planet eccentricity, host star type, signal-to-noise, stellar activity level etc) that determine whether you can recover a planet signal. In the case of this paper, this would be particularly true as they are combining datasets from several instruments, each with different coverage, sensitivity and systematic noise properties. The way this is normally handled is through a &quot;bootstrap&quot; technique where you inject planet signals of known parameters and see what your detection code finds and study the recovery percentage as a function of the varied parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the online &lt;a href=&quot;http://depts.washington.edu/naivpl/sites/default/files/hz.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;Habitable Zone calculator&lt;/a&gt; for a Teff=5072 K, Stellar luminosity=0.47 * L_Sun (luminosity of the Sun) matching the host star, produces 0.678 AU as the inner edge of a conservative habitable zone. If we use this, the star's mass from the paper (0.78 M_Sun) and the 1 M_Earth mass you would like to find, then making use of the formulas given on the &lt;a href=&quot;https://exoplanetarchive.ipac.caltech.edu/docs/poet_calculations.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Exoplanet Archive website&lt;/a&gt;, gives us a period of 230.9 days. Plugging this into an equation for the radial velocity semi-amplitude (also on the Exoplanet Archive), produces an amplitude of about 12 cm/sec. This is assuming an edge-on orbit, so the &lt;em&gt;sin i&lt;/em&gt; term from the inclination is 1; given the lack of detected transits in the paper, the inclination is likely lower than 90 degrees, which would lower the RV amplitude further. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This RV amplitude is far too small to be detected with current instruments. The best instrument we have right now is HARPS on the ESO 3.6m telescope, whose data is included in the paper. If I am understanding their modeling correctly, they get error bars of 181 cm/s on the HARPS data for this star in their analysis, 15x the expected signal. Also note that the calculated RV signal is somewhat best case (unless it is more massive); decreasing the inclination from edge-on (likely) and moving the planet further out in the habitable zone from the inner edge I calculated (also likely) will both decrease the RV amplitude further. Even for future instruments such as &lt;a href=&quot;https://www.eso.org/sci/facilities/paranal/instruments/espresso.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ESO's ESPRESSO&lt;/a&gt; which will go the 8.4m VLT and reach 10 cm/s precision, this would be a challenging measurement.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-09-24T18:56:15.460" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27752" PostTypeId="2" ParentId="27748" CreationDate="2018-09-24T19:27:10.863" Score="1" Body="&lt;p&gt;Visually without a telescope or binoculars you should be able to see Venus, Mars, Jupiter and Saturn, even in most light polluted cites.(Possibly Mercury). So with the telescope you describe you should be able to see some planets, the Moon, stars, etc. The rings of Saturn and the Galilean moons of Jupiter and Lunar details should also be visible. Planetary details will probably not be visible because of the small light gathering ability of the telescope.&lt;/p&gt;&#xA;" OwnerUserId="24358" LastActivityDate="2018-09-24T19:27:10.863" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27754" PostTypeId="2" ParentId="27748" CreationDate="2018-09-24T22:20:32.297" Score="2" Body="&lt;p&gt;For your telescope I'm reading different statistics on Celestron's website than the ones you posted in your question: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Highest Useful Magnification 180x  &#xA;Lowest Useful Magnification 11x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm new to this. Could someone tell me if I'd be able to see planets&#xA;  with the Celestron Cometron FirstScope?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can see the planets with your eyes in a clear sky (even with some light pollution). They look like colorful stars that don't twinkle. In a telescope of low magnification and poor build quality you can also see the planets: they just look like bigger disks of light without recognizable details. The question you're not asking is &quot;What do I need in order to see the details of the planets?&quot; or &quot;What do I need in order to get spectacular views of the planets?&quot;. I'm including an excellent answer to this follow-up question &lt;a href=&quot;https://astronomy.stackexchange.com/questions/8788/how-much-magnification-is-needed-to-see-planets-of-solar-system&quot;&gt;from a separate post to this site&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As a general rule, there isn't much point in pushing the magnification&#xA;  above 2x the diameter of the instrument, measured in mm. 3 inch,&#xA;  that's 75mm, that's 150x max. Beyond that limit, even under ideal&#xA;  skies the image is large but blurry.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;After that, seeing (or air turbulence) pushes that limit further down.&#xA;  Your aperture is small enough that it almost never suffers from&#xA;  seeing, but larger instruments are often affected. It varies greatly&#xA;  with time, place and season. There are times when a 12&quot; dobsonian,&#xA;  that in theory could do 600x, is clamped down by seeing to 150 ...&#xA;  180x. There are times when you could take a 20&quot; dobsonian all the way&#xA;  up to 1000x - but that's very, VERY rare, it's the stuff of legends.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Assuming average seeing conditions and instruments of usual size&#xA;  (refractors of 3...4&quot; aperture, reflectors 6&quot; or larger), here are&#xA;  some rules of thumb:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Jupiter is seen best under mid-high magnification. It's rare that more&#xA;  than 200x is beneficial. This is because it's a very low contrast&#xA;  object, and additional magnification comes at the cost of less&#xA;  contrast, which makes things worse.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Saturn works best at high-ish magnification, bit more than Jupiter but&#xA;  maybe not much more. Around 200 ... 250x usually works. It depends on&#xA;  what you do - if you're trying to see the ring divisions, push it a&#xA;  bit higher.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Mars can use the highest magnification that you could generate, given&#xA;  the instrument and the conditions. It's a very small object, contrast&#xA;  is not bad, so crank it all the way up. Most instruments are limited&#xA;  by seeing when observing Mars.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Moon is the same as Mars.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;As you can see, magnification is never an issue for you. More&#xA;  magnification will not make it better. In fact, more magnification&#xA;  always means the image is more blurry, not more crisp - it's always a&#xA;  compromise between size and blurriness that decides the optimal&#xA;  magnification.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming this reasoning is correct in your case, it looks like you'd be able to see Jupiter with this telescope in some detail, but perhaps not the other planets with any helpful level of detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, I'd strongly recommend replacing those Kellner eyepieces with something better (Plossls or better). And quite frankly, if planetary viewing is your passion, I'd look into saving for a refracting telescope and eyepieces of high quality rather than a budget starter kit.&lt;/p&gt;&#xA;" OwnerUserId="23940" LastEditorUserId="23940" LastEditDate="2018-09-24T22:26:17.540" LastActivityDate="2018-09-24T22:26:17.540" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27755" PostTypeId="2" ParentId="27715" CreationDate="2018-09-24T22:37:10.397" Score="2" Body="&lt;p&gt;The first one is alpha (α); the second one is delta (δ). You can study the shapes of the letters here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Greek_alphabet&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Greek_alphabet&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what they mean (information from a comment to your question):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Orientation models use three Euler angles to describe body&#xA;  orientation. The first two angles are the right ascension and&#xA;  declination of the north pole of a body as a function of time. The&#xA;  third angle is the prime meridian location (represented by &quot;W&quot;), which&#xA;  is expressed as a rotation about the north pole, and is also a&#xA;  function of time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="23940" LastActivityDate="2018-09-24T22:37:10.397" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27756" PostTypeId="1" CreationDate="2018-09-24T23:31:52.983" Score="1" ViewCount="58" Body="&lt;p&gt;A colour-colour diagram (CCD) like this&#xA;&lt;a href=&quot;https://i.stack.imgur.com/lSwev.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lSwev.jpg&quot; alt=&quot;http://www.brucegary.net/All-Sky/&quot;&gt;&lt;/a&gt;&#xA;compares the difference in magnitudes at different wavelengths. The trend in the pattern of colour-colour comparisons tells us about what type of source the light comes from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any significance in the best-fit line of the trend? As in, if I did a linear regression of the left plot above, I'd get a gradient of (V-R)/(B-V). I realise that this ratio doesn't have any physical significance in terms of a single source, but &lt;strong&gt;is there any value in calculating the best fit line and its slope or regression coefficient? Does it tell us anything about the group of sources it's associated with, or is it just used to calibrate filters&lt;/strong&gt;, as it states on &lt;a href=&quot;https://en.wikipedia.org/wiki/Color%E2%80%93color_diagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;? The answer to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/25813/what-is-equation-of-straight-line-for-a-ideal-blackbody-on-the-color-color-diagr&quot;&gt;this question&lt;/a&gt; doesn't really answer mine.&lt;/p&gt;&#xA;" OwnerUserId="17535" LastEditorUserId="17535" LastEditDate="2018-09-25T22:30:33.550" LastActivityDate="2018-09-25T22:30:33.550" Title="Regression line in a colour-colour diagram" Tags="&lt;photometry&gt;&lt;magnitude&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27758" PostTypeId="1" AcceptedAnswerId="27766" CreationDate="2018-09-25T01:50:05.220" Score="4" ViewCount="292" Body="&lt;p&gt;This &lt;a href=&quot;https://astronomy.stackexchange.com/a/27751/7982&quot;&gt;excellent answer&lt;/a&gt; points to &lt;a href=&quot;https://www.eso.org/sci/facilities/paranal/instruments/espresso.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ESPRESSO&lt;/a&gt;, - Echelle SPectrograph for Rocky Exoplanets and Stable Spectroscopic Observations. From there I looked at the &lt;a href=&quot;https://www.eso.org/sci/facilities/paranal/instruments/espresso/inst.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Instrument Description and Performance&lt;/a&gt; page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The page describes three parts or th esystem:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Coudé Trains (transporting light from each of the VLT's four telescopes)&lt;/li&gt;&#xA;&lt;li&gt;Front Ends (corrections, stabilizations, fiber pick-up) &lt;/li&gt;&#xA;&lt;li&gt;Spectrograph (fiber output, pupil slicer, Echelle grating, collimator, dichroic splitter, Red and Blue cameras)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm wondering what the pupil slicer is, and how it works exactly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At the entrance of the spectrograph, an anamorphic pupil slicing unit (APSU) shapes the beam in order to compress the beam in cross-dispersion direction but not in main-dispersion direction, where high resolving power needs to be achieved. In the latter direction, however, the pupil is sliced and superimposed on the echelle grating to minimize its size. The rectangular white-pupil is then re-imaged and compressed by the anamorphic VPH grism.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm familliar with the basics of how an Echelle spectrograph works, and as far as anamorphic optics, I understand this to mean that not all of the elements are not cylindrically symmetric so that the size and divergence changes in one plane is different than in the perpendicular plane. (Classic example is &lt;a href=&quot;https://en.wikipedia.org/wiki/Anamorphic_format&quot; rel=&quot;nofollow noreferrer&quot;&gt;Anamorphic format&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is a pupil slicer, and how does it work with anamorphic optics in VLT's ESPRESSO Echelle spectrograph?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; The Spectrograph itself. Light enters from the right from an optical fiber, and the anamorphic pupil slicer (whatever that is) forms a diverging beam to illuminate the entire Echelle grating (beam shown in gray) via the collimator. Everything else in this drawing is post-grating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SpN8N.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SpN8N.png&quot; alt=&quot;ESPRESSO Echelle spectrograph&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-09-25T05:16:08.533" LastActivityDate="2018-09-25T16:20:57.560" Title="What is a pupil slicer, and how does it work with anamorphic optics in VLT's ESPRESSO Echelle spectrograph?" Tags="&lt;spectroscopy&gt;&lt;optics&gt;&lt;very-large-telescope&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27759" PostTypeId="2" ParentId="21487" CreationDate="2018-09-25T03:35:36.080" Score="2" Body="&lt;p&gt;I don't think the question was clearly phrased. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question that I will answer is how did the ancients see the invisible line known as the ecliptic.  The problem is similar in that it is easy to tell how far north and south you are but difficult to tell how far east and west you are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can just keep track of where the heart of the sun rises on the horizon and then match that point up with an equivalent point on the celestial sphere. You have to have a fairly good map of the heavens either drawn in your head or on paper. It would be easy to tell where the sun is North vs South but a little bit more difficult to see exactly where it is east to west.  My guess is that they understood that the sun moves very steadily and they figured out its daily rate forward. This would have been confirmed during total eclipses hence the name ecliptic.&lt;/p&gt;&#xA;" OwnerUserId="24360" LastActivityDate="2018-09-25T03:35:36.080" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27760" PostTypeId="1" AcceptedAnswerId="28152" CreationDate="2018-09-25T05:15:45.500" Score="14" ViewCount="1108" Body="&lt;p&gt;I grew up with long cold winters, and saw a lot of remarkably transparent ice formed by refreezing meltwater, both in puddles and ponds, and in large icicles. I'd always thought about making optical elements but never tried, as I knew getting a good surface figure would be quite a lot of work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After clicking through links in &lt;a href=&quot;https://outdoors.stackexchange.com/a/20596/12619&quot;&gt;this answer&lt;/a&gt; to the question &lt;em&gt;Are there any natural materials that can be used as a magnifying lens (or to craft one)?&lt;/em&gt; I came upon the images below, which have rekindled my interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I'm currently near the Tropic of Cancer and far from any &quot;Winter Wondernalds&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'd like to ask if anyone has ever tried to build a simple, low magnification telescope from ice or if there are any references to such an attempt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't care if it's a front-surface reflector for the Sun or a refractor for astronomical or terrestrial viewing. I'd just enjoy seeing any, even slightly successful attempt at getting some optical magnification from surfaces produced in ice. Conventional eyepieces would be fine, as would projection on a piece of paper.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Some related imagery to &quot;get the ball rolling&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://wildwoodsurvival.com/survival/fire/ice/ist.html&quot; rel=&quot;noreferrer&quot;&gt;http://wildwoodsurvival.com/survival/fire/ice/ist.html&lt;/a&gt; also see &lt;a href=&quot;https://www.icesculptingtools.com/home/knowledge-center/misc-knowledge/fire-from-ice/&quot; rel=&quot;noreferrer&quot;&gt;Fire from Ice&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/C1PYc.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/C1PYc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/eK2zm.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eK2zm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://wildwoodsurvival.com/survival/fire/ice/istmake.html&quot; rel=&quot;noreferrer&quot;&gt;http://wildwoodsurvival.com/survival/fire/ice/istmake.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UC682.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UC682.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/kVpl4.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kVpl4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/zIEKR.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zIEKR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/sEpox.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sEpox.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/BVB7Z.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BVB7Z.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/Of88m.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Of88m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.primitiveways.com/fire_from_ice.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.primitiveways.com/fire_from_ice.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6OVQZ.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6OVQZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-09-27T20:10:40.913" LastActivityDate="2021-05-17T21:37:13.910" Title="Has anyone ever tried to make a simple telescope using ice?" Tags="&lt;history&gt;&lt;optics&gt;&lt;telescope-making&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27761" PostTypeId="2" ParentId="27760" CreationDate="2018-09-25T11:57:07.453" Score="10" Body="&lt;p&gt;I'm not sure it counts as &quot;simple&quot; but there is the &lt;a href=&quot;https://en.wikipedia.org/wiki/IceCube_Neutrino_Observatory&quot; rel=&quot;noreferrer&quot;&gt;ice cube neutrino observatory&lt;/a&gt; whose detector consists of a cubic kilometer of very clear ice a mile or so down in the antarctic icecap.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-09-25T11:57:07.453" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27762" PostTypeId="2" ParentId="27760" CreationDate="2018-09-25T12:43:10.027" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Has anyone ever tried to make a simple telescope using ice?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have done some research and not found any evidence of an attempt to build one. Another poster has added information on a neutrino telescope that relies on ice. If that's the kind of telescope you're looking for, then there's an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it theoretically possible to build an optical one? Yes, but I'm not sure what kind of optics you'd get even in the best of (non-melting) cases. Try and see what happens. &lt;/p&gt;&#xA;" OwnerUserId="23940" LastEditorUserId="23940" LastEditDate="2018-09-25T23:17:32.653" LastActivityDate="2018-09-25T23:17:32.653" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="27763" PostTypeId="2" ParentId="27735" CreationDate="2018-09-25T13:17:45.560" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Could anyone tell me if this telescope is any good&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Telescopes of this kind (toy telescopes) have always been, in my experience, of quite poor quality. Maybe this one is great, but I don't see any information to suggest it would be. Again in my experience, a good telescope manufacturer will provide helpful information on the optics of the system (eyepiece design, f-number, highest useful magnification, types of object that can be profitably viewed); this company hasn't. It's just giving magnification numbers. My assumption is that the optics aren't manufactured to a high level of quality, and the eyepieces are of the kind typically provided with this kind of scope, which makes the whole experience even worse; not to mention the typically shaky mount these come with, and the extremely narrow field of view these often provide. I'd recommend checking out AWB's reflector or something like it, which is a great entry-level option:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://shop.astronomerswithoutborders.org/products/awb-onesky-reflector-telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://shop.astronomerswithoutborders.org/products/awb-onesky-reflector-telescope&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This kind of telescope is relatively simple to use and can actually produce decent views of certain objects. &lt;/p&gt;&#xA;" OwnerUserId="23940" LastEditorUserId="23940" LastEditDate="2018-09-25T23:00:51.303" LastActivityDate="2018-09-25T23:00:51.303" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27765" PostTypeId="1" CreationDate="2018-09-25T16:08:26.373" Score="1" ViewCount="264" Body="&lt;p&gt;I know that LST = HA + RA.&lt;br&gt;&#xA;And GST is LST + longitude(L).&lt;br&gt;&#xA;If a star culminates, it's HA=0. Therefore, GST = RA + L.&lt;br&gt;&#xA;Since RA and L are constants, GST must also be a constant. So, how can GST be different on different days of the year for the culmination of the same object for the same observer? Also if I'm given the L and RA, why am I given the GST in this question?&lt;a href=&quot;https://i.stack.imgur.com/nQ93A.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nQ93A.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16506" LastActivityDate="2018-09-25T17:58:07.997" Title="Clarification on Greenwich Sidereal Time (GST)" Tags="&lt;coordinate&gt;&lt;time&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27766" PostTypeId="2" ParentId="27758" CreationDate="2018-09-25T16:20:57.560" Score="4" Body="&lt;p&gt;In order to reach the stability required for planet detection, the majority of high resolution echelle spectrographs are mounted off the telescope in a stable environment. This involves pressure and temperature stabilization, often by housing the spectrograph in a vacuum vessel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to feed the spectrograph with light from the telescopes, and to reduce the effect of image motion causing variations in the slit illumination and corresponding variations in the spectral line shape which would mask the tiny radial velocity changes     we are trying to measure, optical fibers are used to feed the light. However there are two disadvantages to this: 1) the fiber is now round (although octagonal fibers are the new hotness for their better image scrambling properties) and not a long slit, and 2) due to &lt;em&gt;focal ratio degradation&lt;/em&gt;, the output of the fiber is wider (lower f/ratio) than the fiber input which decreases the spectrograph resolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to combat these effects and increase the resolution, which is needed to measure the tiny 1/1000th pixel shifts caused by low mass planets, image slicers are typically used to reformat the output fiber into a new more slit-like shape. The classic example is the Bowen-Walraven image slicer which uses a glass plate and prisms to slice the circular fiber image into 2 semi-circles and stack them on top of each other. Higher efficiency versions using mirrors or glass slabs have been made more recently; &lt;a href=&quot;https://www.eso.org/sci/libraries/SPIE2012/8446-370.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this ESO SPIE publication&lt;/a&gt; has some examples of alternative constructions and the resulting image shapes. There is also &lt;a href=&quot;https://astrospectroscopy.wordpress.com/image-slicer/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt; which has more details, including construction of a prototype.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not an expert on optical design or anamorphic optics but I suspect this is done because the beam is not the same size and shape in the spatial/cross-dispersion direction compared to the dispersion/wavelength direction and so equal power in the optical elements in each direction is not needed. Given the size of the input beam from the 8.4m VLT that will feed ESPRESSO, and the need to get the whole instrument into an affordable vacuum vessel that will keep vacuum for a long time, I would imagine the designers employed as many tricks as they could in order to keep the size of the instrument down; the Optical Design section of the &lt;a href=&quot;https://arxiv.org/pdf/1711.05250.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;ESPRESSO Overview paper&lt;/a&gt; talks a little more about this.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-09-25T16:20:57.560" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27767" PostTypeId="2" ParentId="27765" CreationDate="2018-09-25T17:58:07.997" Score="2" Body="&lt;p&gt;The flaw in your logic is that &quot;RA and L are constants&quot;. The right ascension (RA) of the object on the meridian is not constant. It changes continuously, otherwise we would not say that the sky is rising in the east and setting in the west.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the Sun moves about 1 degree per day along the ecliptic, so the RA of the Sun is constantly changing. If the Sun is on the meridian at 12:00:00 every day, and if the RA of the Sun is changing every day, then the same RA is not on the meridian at the same time on every day. (Note: Because of the equation of time, the sun does not cross the meridian at the same time each day. For purposes of this thought experiment, pretend that it is true.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two things taken together lead to the fact that 24 hours of right ascension cross the meridian every 23 hours 56 minutes 4 seconds. And this is why the question asks for a date when the LMC culminates (crosses the meridian) at a specific time (9 pm) and a specific location.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2018-09-25T17:58:07.997" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27768" PostTypeId="1" CreationDate="2018-09-25T19:52:30.747" Score="1" ViewCount="32" Body="&lt;p&gt;I am interested in seeing a plot of dark to baryonic matter ratios in terms of the galactic disk size (when it applies, as not all galaxies fit into disk shapes, and probably analyzing disk galaxies with amorphous ones would only contribute to murky the curve with noise) and how the galactic populations are distributed over that range of sizes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does something like this exist?&lt;/p&gt;&#xA;" OwnerUserId="7400" LastActivityDate="2018-09-25T19:52:30.747" Title="Population histograms and relationship between size of galactic disk and amount of dark matter" Tags="&lt;dark-matter&gt;&lt;galactic-dynamics&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27769" PostTypeId="1" AcceptedAnswerId="27808" CreationDate="2018-09-25T19:53:11.177" Score="4" ViewCount="257" Body="&lt;p&gt;The context is the ionization zones in stellar atmospheres or interiors, the Sun, for instance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The adiabatic exponent is the heat capacity ratio:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\gamma = \frac{c_P}{c_V} = \frac{C_P}{C_V}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, for an ideal gas, it can be shown that  &lt;span class=&quot;math-container&quot;&gt;$\gamma=(f+2)/f$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt; is the degrees of freedom. Resulting in &lt;span class=&quot;math-container&quot;&gt;$\gamma\approx1.66$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most of the extent of the solar interior, &lt;span class=&quot;math-container&quot;&gt;$\gamma\sim1.66$&lt;/span&gt; but it presents dips when ionization is present, like the H, HeI and HeII ionization zones. &lt;strong&gt;My question is why does this happen?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought it could be because when ionization, the heat is used to ionize atoms instead of increasing the temperature and, therefore, the heat capacity would be larger in those regions. But, because the &lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt; is the ratio between the heat capacities, the argument does not work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestion?&lt;/p&gt;&#xA;" OwnerUserId="17043" LastEditorUserId="7982" LastEditDate="2021-02-28T11:25:54.643" LastActivityDate="2021-02-28T11:25:54.643" Title="Why does the adiabatic exponent decrease at the ionization zones?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;astrophysics&gt;&lt;stellar-astrophysics&gt;&lt;stellar-atmospheres&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27771" PostTypeId="2" ParentId="27673" CreationDate="2018-09-25T20:47:10.893" Score="6" Body="&lt;p&gt;Part of the problem here is semantics.  The meaning of &quot;planet&quot; was really poorly defined prior to 2006 (and some argue that it still is).  In the early 2000's, this started to become an issue. You can blame Mike Brown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In January 2005, Mike Brown's team at CalTech in Pasadena, California, discovered an object we now know as Eris (they nicknamed it Xena early on).  The early observations of Eris suggested it was larger than Pluto (which was later conclusively proven incorrect with the New Horizons flyby of Pluto in 2015). But larger or not, it was close in size to Pluto.  Did this make it the tenth planet?  If so, then what about the other bodies like Sedna? Haumea? MakeMake? For that mater, what about Ceres, or Chiron?  By some counts, you could say we had upwards of 23 planets.  Or do we only have 9? Or 8?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For that matter, what makes a planet a planet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The word planet comes to us from the ancient Greek &quot;planetes asteres&quot; which meant &quot;wandering stars.&quot;  To the ancients, even well before the Greeks, a few of the lights in the night sky were definitely different from the others.  The stars remained in fixed patterns.  Sure, they rose and set, and they rose and set at different times through the year, but their relative positions were always the same - as far as the ancients could tell.  But five of them were different.  Their positions changed.  Two of them would rise before the sun in the morning for a while, then disappear and set after the sun in the evening for a while.  These are Mercury and Venus.  The other three, Mars, Jupiter, and Saturn, all seemed to move in the same direction, interestingly through the constellations they knew as the zodiac, but now and then they'd back-track a little bit, then move forward again.  Very odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ancients had no idea that they were, in fact, other worlds, completely different in nature from the burning stars.  Of course, for that matter, they had no idea what the stars really were either.  They just knew the planets, though they looked pretty much the same as stars, moved differently. They were wandering stars.  It wasn't until Galileo turned his scope on the heavens that we learned that these aren't just bright points of light like the other stars - but whole different worlds.  The ancients barely could see Uranus, and gave it little heed, and couldn't see Neptune at all.  Other, smaller bodies like Pluto, Ceres, and others were far to small to see without a telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Herschel discovered Uranus in 1781, he first thought it was a comet.  In 1783, observations and calculations had lead to the discovery that Uranus' orbit was nearly circular, and it behaved like the other planets, and so it was accepted as one (and Herschel wanted to name it George).  When the predicted movements of Uranus didn't match the observed motions, astronomers and mathematicians suggested the reason was due to another planet, further out, pulling on Uranus and affecting its orbit.  In 1846 it was finally discovered, largely based on mathematical predictions.  At the time, no telescope could show either Uranus or Neptune as more than a very small disk.  But they were in nearly circular orbits, and that seemed to be good enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ceres was discovered in 1801, and for several years was considered a planet. German Astronomer Johann Elerte Bode had previously observed what he thought was a mathematical relationship between the planetary orbital radii.  According to his calculations, there should be a planet between Jupiter and Mars, right where Ceres was discovered by Guiseppe Piazzi.  Bode's observation was actually erroneous - there was no such mathematical realationship, but it was convenient that an object was found there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it turned out, Neptune's orbit didn't quite fit the mathematical predictions either.  This lead to a belief that there might be a tenth planet causing the orbital discrepancies.  In the late 1800's and into the 1900's, the hunt was on.  Wealthy eccentric and astronomer Percival Lowell founded the Lowell Observatory in Fagstaff, Arizona, and set out to find &quot;Planet X.&quot;  In 1929, the director of the Lowell Observatory hired a 23 year old farmer from Kansas who'd been constructing his own telescopes, mostly out of salvaged farm equipment, and doing his own observations. Clyde Tombaugh was brought in to work on the hunt for Planet X, a rather arduous task involving taking pictures of the same patch of sky a few days apart and then comparing them using a device called a &quot;blink comparator&quot; and looking or something different between the two images.  On February 18, 1930, Tombaugh noticed a tiny dot moving between images taken a few weeks before. Looking back at a previous image a few days before, he added to the evidence.  After capturing a few more images still and comparing, he finally reported the discovery on March 13, 1930.  He had discovered Pluto - the ninth planet!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, at least, that's what the hype said.  It didn't take too long to notice that Pluto was a bit different.  While no two planets orbit in the same precise plane, the previously-discovered eight are very close to co-planar.  They also have orbits that are very close to circular.  Pluto's orbit is tilted dramatically with respect to the other planets, and its orbit is significantly more elliptical.  In fact - Pluto's orbit crosses inside that of Neptune (though they'll never hit each other).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pluto was also tiny. Our own moon is larger, as are several other moons of other planets. In 1978, Pluto's moon Charon was discovered. With further research, it was found that Charon was so large, that the point of orbit between them, what is known as the barycenter of the orbit, is actually well outside Pluto's radius.  Many astronomers were starting to talk of Pluto-Charon as a &quot;co-planet.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, in the late 1990's and into the 2000's, a bunch of other objects started turning up out past Pluto.  Some of them were spherical, like a planet.  So, were they planets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take a look back at Pluto's discovery, you can make an argument that the only reason Pluto was called a planet from the start was because Tombaugh was trying to find a planet.  He found something, there's no doubt there.  But would it have been called a planet if they weren't specifically looking for a planet?  As it turns out, Pluto's mass (including the whole Pluto/Charon system which includes four other &quot;moons&quot;) isn't nearly enough to explain the discrepancies between the observed and calculated positions of Neptune (which, as it happens, were later explained away by calculations based on Einstein's discovery of special relativity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Mike Brown discovered something that was potentially larger than Pluto, but much further out, and just as weird, people started questioning what it meant.  If Eris was a planet, then what made it a planet and not Sedna, or Orcus, Or MakeMake, or the others.  And then, if those were planets, why not Ceres, which had been downgraded to an asteroid.  What does &quot;planet&quot; mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here come the semantic police. The problem is, who decides what a word means.  We run into this in language all the time.  But for science, we need precise definitions so we can make and categorize observations.  If I simply define a star as a bright light in the sky, that includes planets, the moon, and even airplanes.  For science to work, we have to have agreed upon definitions - operational definitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, in many cases, there's no single body to define things.  There is no &quot;official&quot; body to make these decisions.  The purpose of the International Astronomical Union is to provide an agreed-upon group to set standards like this.  When they met in 2006, they had a working group to formulate a definition of what should define a planet and differentiate it from another body. To some greater or lesser extent, it IS arbitrary - as is all language.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I say fruit, you might think &quot;banana&quot; or &quot;apple.&quot; But not everyone would think of &quot;tomato&quot; as a fruit, even though it is, by definition.  How about a cucumber?  Both a tomato and a cucumber are the fruiting bodies of their plants.  They are, by definition, fruits.  Yet how many people do you know that talk of a cucumber as a fruit?  Most people consider fruits to be sweet while vegetables are savory.  So, to an extent, it will depend on your point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for science, we need a standardized set of definitions.  And this is what the IAU did in 2006.  Their working group formulated a set of criteria to define a planet, and put it to a vote of the general assembly, which passed it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors of the cited article state that &quot;taxonomical terms like planet are determined by the scientific process, not voting.&quot;  However, when you are arguing semantics, then you are talking about consensus.  Looking at another taxonomic system, scientists agree that a &quot;cat&quot; is a cat because it has certain characteristics.  More recently, those characteristics include genetics and assumed evolution, but historically taxonomy was more based on observed characteristics.  In defining a planet, we have the same process at work - only the defining characteristics were put to a vote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless of what we call Pluto, no one argues that it is not a definite body of some substantial size orbiting the star Sol.  The argument does not deny Pluto's existence, only its classification.  Pluto, along with the eight other fully-accepted planets, along with the thousands of asteroids, along with the myriad other trans-Neptunian objects, Plutinos, Kuiper Belt objects, and everything else orbiting the sun, are all planet-like in some ways, and not in others.  At some point, a definition must be agreed upon, and this is exactly what the IAU did in 2006.  Not everyone agreed, but that wasn't necessary.  We weren't stating a scientific fact, we were assigning a label.  It's not a FACT that anything is a planet, it is simply a label of convenience.  In fact, if you wanted to be nit-picky, you could argue that Jupiter and Earth are not at all the same type of object, and that they shouldn't both be called planets but should have different classifications.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By reclassifying Pluto, the IAU basically played a game of &quot;one of these things is not like the others&quot; and drew a line.  And whether or not you agree with that line, the majority of the assembled members of the convention did, and adopted the definition.  Sure, it could be changed, but why bother? It really makes no difference other than in the minds of people who don't understand otherwise.  It's PR, and nothing more.&lt;/p&gt;&#xA;" OwnerUserId="10546" LastActivityDate="2018-09-25T20:47:10.893" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27772" PostTypeId="1" AcceptedAnswerId="27779" CreationDate="2018-09-25T21:09:39.937" Score="1" ViewCount="70" Body="&lt;p&gt;Can an image of a person using a radio telescope see a person from far away? Could a radio telescope be made to see someone walking on Mars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://space.stackexchange.com/questions/30477/could-we-see-someone-walking-on-mars-from-earth&quot;&gt;https://space.stackexchange.com/questions/30477/could-we-see-someone-walking-on-mars-from-earth&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-09-26T03:29:12.470" LastActivityDate="2018-09-26T17:01:44.753" Title="Is it possible to see an image with radio telescope of something not so far away in detail?" Tags="&lt;telescope&gt;&lt;photography&gt;&lt;radio-telescope&gt;&lt;post-processing&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="27773" PostTypeId="2" ParentId="27735" CreationDate="2018-09-25T21:16:38.210" Score="6" Body="&lt;p&gt;Someone else posted a question about this very scope not long ago. Simply put, no, it's little more than a cheap toy and a waste of money.  The optical quality is going to be extremely poor.  In my opinion, it is more likely to dis-interest someone in astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You didn't mention how old your kids are.  Not knowing, it's hard to make a clear recommendation.  Still, I wouldn't recommend this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your kids were interested in music and wanted to learn to play guitar, would you buy them the cheapest, lowest-quality instrument available, one that has poor sound, is difficult to tune and then doesn't stay in tune, and with such poor intonation that even when the strings are open-tuned, you can't produce proper chords?  You wouldn't necessarily want to rush out and get them a professional-quality recording-ready instrument, but you wouldn't get them a toy either (at least, I hope not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd say the same thing about a telescope.  You want a real telescope, with quality optics.  I'm sure you don't want to pay a lot, but you don't want to get something that will be so frustrating as to make them lose interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, like playing a guitar, unless you know what to do with a telescope, it will be frustrating.  Too many people buy telescopes with the belief they'll see things like they see in pictures on the internet, only to be badly disappointed.  I've seen so many telescopes put up for sale on Craigslist and eBay or thrown in a closet or garage and ignored because the user had no idea what they were doing and were completely disappointed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on your kids age, my first recommendation would be to find and join a local astronomy club. The club I belong to, the Houston Astronomical Society, does regular star parties for schools and children.  Many of our members join because of their kids - because they want to be able to teach their kids.  We also have many adolescent and teen members who are active in our club.  Membership is inexpensive (most clubs are $50 USD or less per year) and you can learn a lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for a telescope itself,  I would ignore ANY AND ALL telescopes under \$1,000 USD that are computerized, what we call GoTo technology.  While some people might disagree, my experience, which is fairly extensive, with such telescopes is that they are poorly manufactured and prone to breaking down - at which point they usually are not usable in a &quot;manual&quot; mode.  You also end up paying a lot for the electronics and mechanical parts, and getting less telescope.  On top of that, the majority of those telescopes still require you know your way around the sky to be able to find key landmarks (usually a number of bright, named stars) to set up the telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My general recommendation is an 8&quot; Dobsonian telescope.  A Dobsonian telescope is a Newtonian OTA (Optical Tube Assembly, the telescope-proper portion of the full instrument) with a simple altitude-azimuth rocker-box mount (meaning the motion is up/down and left/right, a natural motion for most people).  Dobsonians or Dobs, offer the best ratio of price to aperture, and are very easy to use.  However, an 8&quot; Dob may be too much for smaller children, may be more than you're willing to spend, and may be larger than you're comfortable with. However, if you can, an 8&quot; aperture is a very good starting place for an amateur telescope as it offers a good balance between cost, physical size, and optical capability.  Most 8&quot; Dobs weigh a total of about 50 lbs, evenly divided between the OTA and base, making the heaviest single piece around 25 lbs, with an overall total height/length of about 55 inches, give or take a couple.  If you live in a light polluted area, they typically fit comfortably on the back seat of a midsize sedan or larger, and often will fit in a compact (though not something like a Smart car).  Brand new, you can sometimes find them under \$400, but usually you can expect between \$400 and \$500 with a decent starting accessory package.  I personally recommend the Apertura AD8, though the Orion 8&quot; Skyline and XT8 are also decent models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If price and/or Size is an issue, the next step down would be a 6&quot; Dobsonian.  The length/height is about the same, but they're somewhat thinner and a little lighter.  The smaller aperture reduces light gathering ability, which reduces overall capability. Brand new, the price runs under \$300 for a decent instrument, but they can often be found used or refurbished for closer to \$200.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smaller still, I'd consider the AWB OneSky 130 (known outside of the US as the SkyWatcher Heritage 130p).  This is a 130 mm (5&quot;) Newtonian telescope on a simple alt-az mount (often referred to as a Dobsonian, but not a true Dob).  The price is around \$200.  Another option at the same size is the Meade LightBridge Mini 130, which is similar in size and capability.  I've seen this closer to \$175 lately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stepping down further, there are a handful of 114 mm (4.5&quot;) scopes, including the Meade LightBridge 114 and the Orion StarBlast 4.5&quot;  Again, you are stepping downward in capability, but also price, and these smaller scopes are much lighter and easy to transport, though you Won't see as much with them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would not recommend ANY telescopes under this.  There are many available, but most are of poor quality or design.  Under \$150-ish, I'd recommend a decent pair of 7x50 or 10x70 binolculars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="10546" LastActivityDate="2018-09-25T21:16:38.210" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27775" PostTypeId="1" CreationDate="2018-09-26T08:46:34.980" Score="3" ViewCount="106" Body="&lt;p&gt;Is the acceleration of expansion we observe with supernovae direct or is this an interpretation according &lt;span class=&quot;math-container&quot;&gt;$\Lambda$&lt;/span&gt;CDM model ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean what we observe is luminosity distance versus redshift, so to see acceleration we should have scale factor vs time, which need a model, right (assuming Friendmann-Lemaître equations in which we need to consider homogenous universe for example) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to say &quot;our universe is in accelerated expansion&quot; without any  interpretation with a model ?&lt;/p&gt;&#xA;" OwnerUserId="23047" LastActivityDate="2018-09-26T08:46:34.980" Title="Is the acceleration of the expansion of the universe a direct observation or an interpretation?" Tags="&lt;universe&gt;&lt;supernova&gt;&lt;expansion&gt;&lt;observable-universe&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27776" PostTypeId="2" ParentId="27735" CreationDate="2018-09-26T09:30:22.937" Score="0" Body="&lt;p&gt;I bought a very similar one in a supermarket for 30 euros when I was an undergrad.&lt;br&gt;&#xA;My experience goes along with what the other answers expect: It's a mostly useless toy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'd have a better experience with binoculars and a proper mount. The problem with my cheap scope was not that you wouldn't see anything (you could, Moon, Jupiter's moons, resolve binary stars like Albireo), but that it was extremely shaky.&lt;br&gt;&#xA;If your cheap telescope doesn't have a mount that absorbs the force of your eye bumping into it, you're in for a hard time seeing anything, because that happens all the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After tweaking the mount a bit do-it-yourself style, it was good enough to impress some of my fellow students however. So my recommendation would be: As long as it's cheap and you don't mind possibly some handiwork, go for it.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2018-09-26T09:30:22.937" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27778" PostTypeId="1" CreationDate="2018-09-26T15:03:44.307" Score="2" ViewCount="146" Body="&lt;p&gt;Let an universe with no content and positive curvature.&#xA;Friedmann-Lemaître equation&#xA;&lt;span class=&quot;math-container&quot;&gt;$$H^2=\frac{8\pi G}{3}\left(\rho_m+\rho_r+\rho_{\Lambda}\right)-\frac{k\, c^2}{a^2},$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; corresponds to the scale factor of Friedmann-Lemaître-Robertson-Walker Metric, and the &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; to the density of the contents, &#xA;will become &lt;span class=&quot;math-container&quot;&gt;$$H^2=-\frac{\, c^2}{a^2},$$&lt;/span&gt; as positive curvature (closed universe) corresponds to &lt;span class=&quot;math-container&quot;&gt;$k=+1$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as &lt;span class=&quot;math-container&quot;&gt;$H=\frac{\dot{a}}{a}$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$\dot{a}=\frac{da}{dt}$&lt;/span&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$da=\pm i \, c \, dt$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$a(t)=\pm i \, c \, t + cst$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what am I doing wrong ?&lt;/p&gt;&#xA;" OwnerUserId="23047" LastEditorUserId="23047" LastEditDate="2018-09-26T15:18:42.047" LastActivityDate="2019-01-07T19:52:55.140" Title="What happens for a &quot;closed&quot; universe without any content?" Tags="&lt;universe&gt;&lt;cosmology&gt;&lt;fate-of-universe&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27779" PostTypeId="2" ParentId="27772" CreationDate="2018-09-26T17:01:44.753" Score="3" Body="&lt;p&gt;In brief: you cannot resolve details of an object on a scale less than roughly a wavelength or two.  The distance to the object only marginally affects this problem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So take a look at the wavelengths your radiotelescope is capable of receiving and compare with dimensions of interesting &quot;features&quot; on a human body.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, as the comments pointed out, unless the human is either emitting radio-wavelength energy in significant amounts (unlikely) or is being &quot;painted&quot; by a large emitting source (as active radar systems do), you won't get any signal in the first place. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-09-26T17:01:44.753" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27780" PostTypeId="1" AcceptedAnswerId="27781" CreationDate="2018-09-26T17:31:54.673" Score="1" ViewCount="1585" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/u9Tga.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the Celestron astromaster 130 EQ Model able to see nebulas and galaxies?&#xA;If this telescope is not able to see nebulae, please suggest me the best one telescopes for nebulas.&lt;/p&gt;&#xA;" OwnerUserId="24370" LastEditorUserId="7472" LastEditDate="2018-09-28T06:42:50.167" LastActivityDate="2018-09-28T06:42:50.167" Title="Is the Celestron astromaster 130 EQ Model able to see nebulas and galaxies?" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27781" PostTypeId="2" ParentId="27780" CreationDate="2018-09-26T18:18:46.643" Score="2" Body="&lt;p&gt;The question isn't whether you can see deep-sky objects but how many.&#xA;Given a reasonably dark observing site, the Orion nebula and the Andromeda galaxy are visible in 7x50 binoculars.&#xA;A 130mm objective should be enough to see most of the nebulae and galaxies in the &lt;a href=&quot;http://www.messier.seds.org/xtra/similar/dataRASC.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Messier catalogue&lt;/a&gt;.&#xA;Once you get the hang of it, you may be able to see a few &lt;a href=&quot;http://www.messier.seds.org/xtra/similar/rasc-ngc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NGC objects&lt;/a&gt; too, but a bigger scope will show more of them.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-09-26T18:18:46.643" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27782" PostTypeId="2" ParentId="23634" CreationDate="2018-09-26T18:24:30.440" Score="9" Body="&lt;p&gt;A recently released paper on arXiv (&lt;a href=&quot;https://arxiv.org/abs/1809.09009&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bailer-Jones et al., AJ&lt;/a&gt;) has made use of the recently released GAIA-DR2 data set, and specifically the 7 million stars that have full 6D information, to try and find plausible home stars for 'Oumuamua. They found 4 potential systems of which the closest encounter was at 0.6 parsecs with the M2.5 dwarf HIP 3757 at a relative velocity of 24.7 km/s, 1 million years ago.  However all of the encounter stars have quite high encounter velocities.  A giant planet or a binary system could provide the needed velocity but none of the 4 systems are known to be binaries or to harbor planets. However 7 million stars with full information is still a pretty small fraction of the total, a situation which will improve with later data releases.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastEditorUserId="28167" LastEditDate="2021-04-25T21:54:32.720" LastActivityDate="2021-04-25T21:54:32.720" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27783" PostTypeId="1" AcceptedAnswerId="27786" CreationDate="2018-09-26T20:29:57.857" Score="1" ViewCount="26" Body="&lt;p&gt;I'm more in the CS based stack exchange, but this question prompted me to come in here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been compiling a large set of ephemerides for solar system, stars with mag &gt; 6.8, exo planets etc, and I am now working on the DSO catalog.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The app actually holds celestial objects in 3D space, and I made conversions from RA/DEC/dist(parsec) for various objects, but I am having trouble with DSO objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know of a red shift time base formula, but heard the conversion is not very good. Also, I understand that most estimations cannot be counted as precise, but I would like to have these objects within 3D space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone tell me what features I should look for in a catalog and what formula I can base myself on for this conversion, if any?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks &lt;/p&gt;&#xA;" OwnerUserId="24372" LastEditorUserId="616" LastEditDate="2018-09-26T21:08:27.677" LastActivityDate="2018-09-27T00:59:24.387" Title="DSO carthesian coordinate estimations" Tags="&lt;galaxy&gt;&lt;distances&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27784" PostTypeId="1" AcceptedAnswerId="27797" CreationDate="2018-09-26T21:47:58.967" Score="0" ViewCount="29" Body="&lt;p&gt;I'm planning an observation for a proposal and I need to determine which dates the moon is within 30 degrees of an object so I can ask to not be scheduled for those dates.  This object is near the ecliptic so it is guaranteed the moon will pass close by at times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the past I've used skycalc to manually figure out such dates, but now I need to figure out dates over a several month period, and I was hoping there was some more automated tool I could use for this purpose.&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2018-09-27T17:01:24.087" Title="Planning observation: determine dates where moon is within 30 degrees of my object?" Tags="&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27785" PostTypeId="2" ParentId="27784" CreationDate="2018-09-26T23:32:28.700" Score="0" Body="&lt;p&gt;Not a tool, but you could try looking at an Astronomical Almanac - for example, the&#xA;2019-2023 Astronomical Almanac by Richard Bartlett lists daily moon RA and Dec values in the PDF version, which may meet your needs - here's a link to the free early PDF:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.dropbox.com/s/smm2eprodpod0cf/2019-2023%20Astronomical%20Almanac%20%28Free%20Early%20PDF%29.pdf?dl=0&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.dropbox.com/s/smm2eprodpod0cf/2019-2023%20Astronomical%20Almanac%20%28Free%20Early%20PDF%29.pdf?dl=0&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Free to share, but if you find it useful the author asks you to consider buying a copy and/or leaving a review.&lt;/p&gt;&#xA;" OwnerUserId="17016" LastEditorUserId="17016" LastEditDate="2018-09-26T23:39:46.990" LastActivityDate="2018-09-26T23:39:46.990" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27786" PostTypeId="2" ParentId="27783" CreationDate="2018-09-27T00:59:24.387" Score="0" Body="&lt;p&gt;There are many cosmological &lt;a href=&quot;https://en.wikipedia.org/wiki/Distance_measures_(cosmology)&quot; rel=&quot;nofollow noreferrer&quot;&gt;distance measures&lt;/a&gt; and what you need depends on what you want to do. If you want to position the object in 3D space where it is today, you need its proper distance. If you want to e.g. display the object with a certain angular size on the sky, and you want to figure that size out from how far away the object is, you need the angular diameter distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate any of the measures, you need the object's redshift &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;. You will probably need to store the comoving distance (= proper distance today) and redshift so that you can convert to other distance measures quickly without redoing the integral.&lt;/p&gt;&#xA;" OwnerDisplayName="user18746" LastActivityDate="2018-09-27T00:59:24.387" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27787" PostTypeId="2" ParentId="27689" CreationDate="2018-09-27T02:31:59.200" Score="2" Body="&lt;p&gt;I had sought an answer from Dr. Christopher S. Baird and am forwarding his answer.                       That is incorrect. At present, the north geographic pole points toward the North Star, and the &lt;em&gt;south&lt;/em&gt; magnetic pole points approximately toward the North Star. After the next reversal, the &lt;em&gt;south&lt;/em&gt; magnetic pole will point away from Antarctica.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the magnetic poles flip, it will have zero effect on the physical orientation of the earth or the location of its geographic poles. The north geographic pole will still be in the arctic near Canada and point towards the North Star, and the south geographic pole will still be in Antarctica. The rotation of the earth (which determines the geographic poles) has a slight, indirect, complicated influence on the spiraling flow of earth's liquid outer core (which determines the magnetic poles). That is why the geographic poles and magnetic poles are close to being aligned. However, the spiraling flow of earth's liquid core has no effect on earth's overall rotation. This means that changes to the earth's magnetic field do not affect the earth's overall rotation. The sun will still rise in the East.&lt;/p&gt;&#xA;" OwnerUserId="13871" LastActivityDate="2018-09-27T02:31:59.200" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27788" PostTypeId="1" CreationDate="2018-09-27T07:06:41.313" Score="6" ViewCount="4255" Body="&lt;p&gt;So as a Sailor I have experienced some times where sunset and moonrise are very close together (this happens when it's full moon) which is particularly beautiful. I assumed that this was always the case but looking more closely at time difference between sunset and moonrise they can change quite dramatically. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2019 (Porto-Vecchio France)&#xA;Date        sunset   moonrise  diff&#xA;04/20       20:08    21:24     1:16&#xA;05/19       20:39    21:20     0:41&#xA;06/18       20:59    22:05     1:06&#xA;07/17       20:54    21:31     0:37&#xA;08/16       20:21    21:12     0:51&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As you can see there is some degree of variation, can someone explain why this happens and how I can find out where and when they are closest together?&lt;/p&gt;&#xA;" OwnerUserId="24375" LastEditorUserId="7472" LastEditDate="2018-09-27T16:48:08.917" LastActivityDate="2020-05-08T22:56:29.927" Title="Difference in time between sunset and moonrise on a full moon" Tags="&lt;the-moon&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27789" PostTypeId="1" CreationDate="2018-09-27T08:00:11.533" Score="0" ViewCount="64" Body="&lt;p&gt;If the universe is expanding how come Andromeda is heading for us? &#xA;In isolated spatial areas does gravity overcome universal expansion? &#xA;What is the gravitational force between us and Andromeda?&lt;/p&gt;&#xA;" OwnerUserId="24376" LastActivityDate="2018-09-27T13:10:56.433" Title="Expansion ofthe Universe" Tags="&lt;gravity&gt;&lt;expansion&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2018-09-27T16:46:18.373" ContentLicense="CC BY-SA 4.0" />
  <row Id="27791" PostTypeId="2" ParentId="27789" CreationDate="2018-09-27T09:56:22.620" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the gravitational force between us and Andromeda?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The masses of the Milky Way and Andromeda (depending on exactly whay you consider &quot;part of&quot; each) are &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014MNRAS.443.2204P&quot; rel=&quot;nofollow noreferrer&quot;&gt;estimated&lt;/a&gt; as about 850 Billion solar masses and 1.7 Trillion solar masses respectively. Their mean distance is about 2.54 million light years, so it's easy to calculate the force using &lt;span class=&quot;math-container&quot;&gt;$$F = GMM'/R^2$$&lt;/span&gt; We get a force of &lt;span class=&quot;math-container&quot;&gt;$$6.6\times 10^{29} N$$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-09-27T09:56:22.620" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27792" PostTypeId="2" ParentId="27778" CreationDate="2018-09-27T11:24:12.920" Score="1" Body="&lt;p&gt;I see that you are using the convention that a has units of distance and is like the radius of curvature of the universe, and you are getting that either a or t would need to be imaginary.  So you can interpret that in two ways, either you say that is impossible, and conclude that it is impossible for an empty universe to be closed, or you attribute some physical meaning to imaginary radius of curvature or imaginary time.  I'm not sure what that significance would be, but what seems clear is there is something bizarre, or even bogus, about claiming that an empty universe can have positive curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to see this is to interpret your equation as an equation for k via H=Kc/a, where K^2 = -k.  If you further demand that nothing be imaginary, this requires that k be either 0 or -1.  Empty universes are flat or open, or else they are imaginary in some way.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-09-27T11:24:12.920" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27794" PostTypeId="2" ParentId="27788" CreationDate="2018-09-27T15:16:14.633" Score="0" Body="&lt;p&gt;The reason for this is quite simple. While we orbit the sun, the moon is orbiting us. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AFA7I.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AFA7I.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the direction of the sunlight (this will show where the Sun is), then look at the position of the Moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us take the Full Moon as an example. Imagine, in this 2D image, the Earth is rotating anti-clockwise. As the Sun is on one side, the moon is on the other. So Sunset and Moonrise will occur at close to the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let us look at the first quarter. At sunset, the moon is going to be almost directly overhead, which means moonrise occurred hours previously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a 2D visual representation makes it easy to visualise. Of course, it is not always so black and white, as the moons orbit is not a perfect circle same as ours is not perfect around the sun (hence different rise times throughout the year during the same phases), then there is axial tilt etc, etc. But this will give you a pretty good idea. The closer to the full moon it is, the closer Moonrise and Sunset will be.&lt;/p&gt;&#xA;" OwnerUserId="17096" LastEditorUserId="17096" LastEditDate="2018-09-28T07:52:56.913" LastActivityDate="2018-09-28T07:52:56.913" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="27795" PostTypeId="1" AcceptedAnswerId="27803" CreationDate="2018-09-27T15:57:37.473" Score="1" ViewCount="97" Body="&lt;p&gt;I downloaded the info from NASA exoplanet archive at: &lt;a href=&quot;https://exoplanetarchive.ipac.caltech.edu/cgi-bin/TblView/nph-tblView?app=ExoTbls&amp;amp;config=compositepars&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://exoplanetarchive.ipac.caltech.edu/cgi-bin/TblView/nph-tblView?app=ExoTbls&amp;amp;config=compositepars&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I named the file &quot;NASA.Composite.Planet.Data.csv&quot;. I had to edit the file to remove the header which was preventing the program below from running. (I put the header at the top of the program below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then plotted it with Matplotlib using this python (anaconda) program: &lt;a href=&quot;https://planet.wikia.com/wiki/User:Granpa/Python_program&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://planet.wikia.com/wiki/User:Granpa/Python_program&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got this image:&#xA;&lt;a href=&quot;https://i.imgur.com/1057YvQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.imgur.com/1057YvQ.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain what I am seeing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ngl5G.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ngl5G.png&quot; alt=&quot;Exoplanet radius density relation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24374" LastEditorUserId="24374" LastEditDate="2018-09-27T16:49:36.050" LastActivityDate="2018-09-28T10:12:43.940" Title="Exoplanet radius density relation according to NASA exoplanet archive" Tags="&lt;exoplanet&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27796" PostTypeId="2" ParentId="27788" CreationDate="2018-09-27T16:25:11.883" Score="1" Body="&lt;p&gt;I'm wondering where you got your list.  It appears that other days might be better choices.  As an example, for Porto-Vecchio, April 20,2019 is a day late:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apr 19, 2018 sunset: 20:07, moonrise: 20:13, diff: 0:06&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon rises about 50 minutes later on consecutive days.  In addition, the sunset time might be changing by a couple of minutes a day as well.  Since sunset only happens once a day, then we'd expect the rise after a full moon to be anywhere in that ~50 minute period late.  On average then, I'd expect the closest sunset/moonrise of a month to usually be a quarter of that range (cut in half because closest could be before or after, and another half for variation) or about 15 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a lot of sites with accurate sunset/moonrise times, but I don't know of any that directly give the delta between them.  Could probably create one with some python scripts, but not something I've done.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2018-09-27T16:25:11.883" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27797" PostTypeId="2" ParentId="27784" CreationDate="2018-09-27T17:01:24.087" Score="1" Body="&lt;p&gt;I would use &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL Horizons&lt;/a&gt; to calculate the position of the Moon and save the results to a text file. Then use a spreadsheet to calculate the separation between the Moon and your object. &lt;span class=&quot;math-container&quot;&gt;$$\cos \theta = \sin \delta_1 \sin \delta_2 + \cos \delta_1 \cos \delta_2 \cos (\alpha_1 - \alpha_2)$$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; is the separation, &lt;span class=&quot;math-container&quot;&gt;$\delta$&lt;/span&gt; is the declination, and &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; is the right ascension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Moon moves about 1 degree in 2 hours, you can use that to determine what time interval to use in the output, depending on how accurately you want to find the 30 degree separation. &lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2018-09-27T17:01:24.087" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27798" PostTypeId="2" ParentId="27795" CreationDate="2018-09-27T17:49:42.760" Score="3" Body="&lt;p&gt;What you are seeing is a combination of several things as you move across the diagram from left to right with increasing planet radius. As this diagram &lt;a href=&quot;https://i.stack.imgur.com/CJ3Zx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CJ3Zx.png&quot; alt=&quot;this version of your diagram&quot;&gt;&lt;/a&gt; from a &lt;a href=&quot;http://nexsci.caltech.edu/conferences/2017/knowthystar/Unterborn_KnowThy.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;talk by Cayman Unterborn&lt;/a&gt; at the recent &lt;a href=&quot;http://nexsci.caltech.edu/conferences/2017/knowthystar/agenda.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Know Thy Star - Know Thy Planet&quot; conference&lt;/a&gt; shows, is a transition from rocky (like Mars, Venus, Earth; the 'Ma', 'V' and 'E' on the diagram) exoplanets with increasing density as ~1.5 Earth radii is reached to &quot;Neptune-like&quot; gas giants with increasing amount of a gas envelope, possibly with a denser core. (The size and type of Jupiter's core is not a totally settled question for example and so the wider range of exoplanets could have a wide variety of core properties). There is a lot more information in &lt;a href=&quot;https://rockyworlds.wordpress.com/2017/08/05/which-small-exoplanets-are-rocky/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this blog post on 'Which small exoplanets are rocky'&lt;/a&gt; which talks about what we can learn about planets in the 0.5 to 4 Earth radius regime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At higher radii (approximately 11 Earth radii/1 Jupiter radius) what you are seeing is the effect of electron-degeneracy pressure. This causes the radii of exoplanets with a mass of about Jupiter, brown dwarfs up to 80 times the mass of Jupiter and low mass M dwarf stars (with a mass of hundreds of Jupiters) all to have very similar radii but with widely varying densities. This causes the near-vertical line on the right hand side of your plot. This is also what makes it very difficult to know whether you have a star, brown dwarf or a planet if you only a measure of a transiting object's radius (from e.g. Kepler) and don't have a mass measurement (usually from radial velocities) to go with it.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-09-27T17:49:42.760" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27799" PostTypeId="2" ParentId="27788" CreationDate="2018-09-27T19:12:14.387" Score="0" Body="&lt;p&gt;Going out on a limb a bit here -- depending on the month, the compass location where the sun sets varies as well as the length of time the sun is above the horizon (note also there are different definitions of &quot;sunrise&quot; - visible, half up, etc).  The moon's compass location varies with a different cycle because the moon's orbit is not perfectly over the equator.&lt;br&gt;&#xA;Combine those two variables, keeping in mind that a &quot;perfect&quot; full moon happens when the sun and the moon are exactly &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; radians apart from our point of view, and it would appear quite reasonable to have significant variation in their relative rise/set times. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-09-27T19:12:14.387" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27801" PostTypeId="1" CreationDate="2018-09-28T09:58:52.533" Score="0" ViewCount="137" Body="&lt;p&gt;We know &lt;a href=&quot;https://www.physicsoftheuniverse.com/topics_blackholes_singularities.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;that&lt;/a&gt; &lt;bR&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;In the center of a black hole is a gravitational singularity, a&#xA;  one-dimensional point which contains a huge mass in an infinitely&#xA;  small space, where density and gravity become infinite and space-time&#xA;  curves infinitely, and where the laws of physics as we know them cease&#xA;  to operate.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So where does this mass go? A measure of how much matter is there in a object is called mass right? so how does that matter stays in a 1-D space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that matter occupy space so how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: My question is about how can a singularity exists not about singularity itself.&lt;/p&gt;&#xA;" OwnerUserId="24386" LastEditorUserId="1891" LastEditDate="2018-09-28T21:47:06.280" LastActivityDate="2018-09-28T21:47:06.280" Title="How does the center of a black hole can contain so much mass?" Tags="&lt;black-hole&gt;&lt;singularity&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2018-09-28T23:09:01.210" ContentLicense="CC BY-SA 4.0" />
  <row Id="27803" PostTypeId="2" ParentId="27795" CreationDate="2018-09-28T10:12:43.940" Score="0" Body="&lt;p&gt;The straight lines you see in the diagram must be coming from extrapolated radius or mass values. Keep in mind that most planets don't have both values simultaneously measured, because their discovery technique is only sensitive to radius (transits) or mass (doppler).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what people in those archives often do is assume some mass-radius relationship in order to fill those gaps. I'm 99.9% sure that this is where those perfectly straight lines come from. The other, scattered points are what @astrosnapper's answer refers to.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2018-09-28T10:12:43.940" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27804" PostTypeId="1" CreationDate="2018-09-28T14:37:01.383" Score="1" ViewCount="79" Body="&lt;p&gt;My friend showed me a video of a CME that looked as if it was swirling into a gigantic circular &quot;spaceship&quot; or &quot;UFO&quot; and he explained it as &quot;an alien ship potentially refueling&quot;. I did a bit of research and I &lt;s&gt;know&lt;/s&gt; assumed the images he was showing me were from &lt;s&gt;SETI&lt;/s&gt; SOHO and they were yellow, meaning it was obviously only a single or a couple wavelengths of light shown (possibly UV?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/YnzZLNsSGyY&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/YnzZLNsSGyY&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;disclaimer: this clip is overly-liberal in its subtitles...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yoy habe the original source that would be awesome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making a total BLIND guess- I'd say the plasma was swirling into the filament of the CME before it reached a critical mass to be pushed away from the sun? Once departed from the energy of the sun under certain conditions it would no longer show because it wouldn't have enough energy to emit the correct wavelength of light?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I even remotely correct? Obviously the space ship explanation has no merit- it's just from some dude. Hoping to get a clarification on the video, I do not believe it is a &quot;UFO refueling&quot; heh. Im not a conspiracy nut, and neither is my bud- so we're looking for a science answer :).&lt;/p&gt;&#xA;" OwnerUserId="23729" LastEditorDisplayName="user1569" LastEditDate="2018-09-30T16:52:10.230" LastActivityDate="2018-09-30T16:52:10.230" Title="What phenomena was this video from SOHO actually showing?" Tags="&lt;the-sun&gt;&lt;coronal-mass-ejection&gt;&lt;soho&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27805" PostTypeId="2" ParentId="27804" CreationDate="2018-09-28T15:35:34.607" Score="6" Body="&lt;p&gt;It's a &lt;em&gt;polar crown cavity&lt;/em&gt; or &lt;em&gt;coronal prominence cavity&lt;/em&gt;. &lt;a href=&quot;https://www.youtube.com/watch?v=QNMjb17uIOM&quot; rel=&quot;noreferrer&quot;&gt;Here is another Youtube video&lt;/a&gt; explaining it. It references &lt;a href=&quot;https://arxiv.org/pdf/1107.3451v1.pdf&quot; rel=&quot;noreferrer&quot;&gt;A new look at a polar crown cavity as observed by SDO/AIA&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Conclusions.&#xA;  We propose to define a polar crown cavity as a density depletion sitting above denser polar crown filament plasma drained down the cavity due to gravity. As part of the polar crown filament,&#xA;  plasma at different temperatures (ranging from 50000K to 0.6 MK) is observed at&#xA;  the same location on the cavity dips and sustained by a competition between the gravity and the curvature of magnetic field&#xA;  lines. The eruption of the polar crown cavity as a solid body can be decomposed into two phases: a slow rise at a speed of 0.6 km/s,&#xA;  and an acceleration phase at a mean speed of 25 km/s&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.uksolphys.org/uksp-nugget/15-structure-and-dynamics-of-a-polar-crown-cavity-as-observed-by-sdoaia/&quot; rel=&quot;noreferrer&quot;&gt;video on this page&lt;/a&gt; shows one forming nicely.&lt;/p&gt;&#xA;" OwnerDisplayName="user1569" LastActivityDate="2018-09-28T15:35:34.607" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27806" PostTypeId="1" AcceptedAnswerId="27807" CreationDate="2018-09-28T16:05:23.567" Score="1" ViewCount="111" Body="&lt;p&gt;If a spaceship was to travel to the barycenter of a binary star (the stars having identical size and mass) and come to a stop there, would the people on board this spaceship find that time is moving slower there (relative to the passage of time experienced outside of the gravitational fields of this binary star) due to the two stars overlapping gravitational fields at this point in space?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does time dilation increase within overlapping gravitational fields?&lt;/p&gt;&#xA;" OwnerDisplayName="user23327" LastActivityDate="2018-09-28T16:40:01.850" Title="Does time dilation increase within overlapping gravitational fields?" Tags="&lt;gravity&gt;&lt;astrophysics&gt;&lt;space-time&gt;&lt;time-dilation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27807" PostTypeId="2" ParentId="27806" CreationDate="2018-09-28T16:40:01.850" Score="2" Body="&lt;p&gt;Gravitational time dilation depends on the observers' relative positions in the gravitational potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For relatively weak gravitational fields the rate at which a clock ticks for observer A compared to the clock rate for observer B is given by&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{d \tau_{\rm A}}{d\tau_{\rm B}} \simeq \sqrt{1 + \frac{2\Delta \Phi_{\rm AB}}{c^2}} \sim 1 + \frac{\Delta \Phi_{\rm AB}}{c^2},$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\Delta \Phi_{\rm AB}$&lt;/span&gt; is the difference in gravitational potential between observers A and B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational potential is a scalar quantity, so the basic answer to your question is yes - the effects of the two neutron stars would be summative. However, I would hesitate to use the approximation above when dealing with the strong gravity near neutron stars.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-09-28T16:40:01.850" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27808" PostTypeId="2" ParentId="27769" CreationDate="2018-09-28T17:00:56.137" Score="3" Body="&lt;p&gt;There are two things going on. (1) When you add heat to a gas that is on the threshold of ionisation or partially ionised, some of that heat goes into ionisation. This means that it takes a much larger amount of energy to produce a rise in temperature. (2) However, as the gas becomes ionised the number of particles per unit mass also increases and so the pressure increases at a given temperature. The net effect is that &lt;span class=&quot;math-container&quot;&gt;$C_V$&lt;/span&gt; (heat capacity for a fixed volume) increases by more than &lt;span class=&quot;math-container&quot;&gt;$C_P$&lt;/span&gt; (heat capacity for a fixed pressure) and so the adiabatic index decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a tricky and detailed computation to work out how &lt;span class=&quot;math-container&quot;&gt;$C_V$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_P$&lt;/span&gt; change as ionisation proceeds. An example calculation for pure hydrogen gas is give on pp 123-126 of &quot;Principles of stellar evolution and nucleosynthesis&quot; (1983, D. Clayton). Both &lt;span class=&quot;math-container&quot;&gt;$C_V$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_P$&lt;/span&gt; increase drastically (by factor of 30-40) as ionisation increases, both peaking at around 50% ionisation before falling back to exactly &lt;em&gt;twice&lt;/em&gt; their unionised values once the gas is completely ionised (when it just behaves like a monatomic perfect gas but with twice the number of particles per unit mass). However, the behaviour of &lt;span class=&quot;math-container&quot;&gt;$C_V$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_P$&lt;/span&gt; are slightly different, &lt;span class=&quot;math-container&quot;&gt;$C_P$&lt;/span&gt; increases by less than &lt;span class=&quot;math-container&quot;&gt;$C_V$&lt;/span&gt;, and so the ratio of &lt;span class=&quot;math-container&quot;&gt;$C_P/C_V$&lt;/span&gt; changes and goes through a minimum at around 50% ionisation before regaining the standard monatomic gas value once completely ionised.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-09-29T10:29:09.883" LastActivityDate="2018-09-29T10:29:09.883" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27809" PostTypeId="2" ParentId="27801" CreationDate="2018-09-28T18:01:29.903" Score="1" Body="&lt;p&gt;The notion of &quot;centre of the black hole&quot; is misleading. It treats the spacetime in and around the black hole as if it were normal flat spacetime, with no gravitational distortion.  In fact the spacetime is highly curved. So curved, in fact, that &quot;space&quot; behaves more like &quot;time&quot;.  Instead of &quot;where is the singularity&quot; you should ask &quot;when is the singularity.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to this question is always &quot;in the future&quot;.  There are no singularities in our past.  This is significant, since when we say &quot;there is a pen on the table&quot; we mean that the pen was there a few nano-seconds ago, because of the time light takes to reach us. Everything that we say &quot;exists&quot; actually exists in our past.  Singularities are a &quot;when&quot; that have a &quot;before&quot; but don't have an &quot;after&quot;. Nothing can have a singularity in its past, because there is no future after the singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this sense, singularities don't exist. We can't study them because they are always in the future (due to the extreme bending of spacetime). If we fall into a black hole we will (rapidly) reach the singularity but there is no way to report back about it, since the singularity is surrounded by an event horizon. Some of us may (if we are unlucky) fall into a black hole. For these people the singularity is in their future. But it is not possible for a singularity to be observed, as it can never be in the past of any timeline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a black hole behaves &quot;as if&quot; all the mass was concentrated in one point. But don't think of the singularity as a &quot;thing&quot; in the centre of the black hole, but think of it as a &quot;when&quot;, and always remember &quot;this is weird, and analogies and metaphor are never a complete explanation&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-09-28T18:49:44.863" LastActivityDate="2018-09-28T18:49:44.863" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27810" PostTypeId="2" ParentId="27788" CreationDate="2018-09-28T21:32:13.983" Score="5" Body="&lt;p&gt;Technically, the Moon is Full at a specific time. Therefore you need to check the time of sunset/moonrise OR sunrise/moonset that is closest to the time of Full Moon. Then you can more accurately compare the variation. Here is the list for 2019 created using the &lt;a href=&quot;http://aa.usno.navy.mil/data/docs/RS_OneDay.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;U.S. Naval Observatory&lt;/a&gt; website.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;Longitude E9° 8', Latitude N41° 35'&#xA;All times in Universal Time (UT)&#xA;Date of Full Moon   -------- Sun ----------     -------- Moon --------  Moon-Sun&#xA;     Date    Time              Date    Time              Date    Time     Diff&#xA; 1/21/2019    5:16   Rise  1/21/2019    6:45     Set 1/21/2019    6:54   +0:09&#xA; 2/19/2019   15:53    Set  2/19/2019   17:01    Rise 2/19/2019   16:55   -0:06&#xA; 3/21/2019    1:43   Rise  3/21/2019    5:26     Set 3/21/2019    5:58   +0:32&#xA; 4/19/2019   11:12    Set  4/19/2019   18:08    Rise 4/19/2019   18:14   +0:06&#xA; 5/18/2019   21:11    Set  5/18/2019   18:39    Rise 5/18/2019   18:14   -0:25&#xA; 6/17/2019    8:31   Rise  6/17/2019    3:49     Set 6/17/2019    3:54   +0:05&#xA; 7/16/2019   21:38    Set  7/16/2019   18:56    Rise 7/16/2019   18:49   -0:07&#xA; 8/15/2019   12:29   Rise  8/15/2019    4:32     Set 8/15/2019    4:09   -0:23&#xA; 9/14/2019    4:33   Rise  9/14/2019    5:02     Set 9/14/2019    4:59   -0:03&#xA;10/13/2019   21:08    Set 10/13/2019   16:46    Rise10/13/2019   17:02   +0:16&#xA;11/12/2019   13:34    Set 11/12/2019   16:07    Rise11/12/2019   16:25   +0:18&#xA;12/12/2019    5:12   Rise 12/12/2019    6:40     Set12/12/2019    6:42   +0:02&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;The difference is much smaller than your original list.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Why this happens&lt;/strong&gt;&#xA;MCG and Carl Witthoft explained why it happens, namely because:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If the moonrise/set is not the same time as the exact Full Moon, then moonrise/set  occurs earlier or later than the sunset/rise.&lt;/li&gt;&#xA;&lt;li&gt;Because the Moon passes above and below the ecliptic (the path of the Sun in the sky), the Full Moon is rarely directly opposite from the Sun. Therefore, it can rise earlier or later than the Sun even when exactly &amp;quot;Full&amp;quot;.&lt;/li&gt;&#xA;&lt;li&gt;Even if the Full Moon were exactly opposite from the Sun (180 degrees apart, and in a total lunar eclipse), there would still be a difference between the time of sunset/moonrise. Those times are when the limb of the sun/moon is visible, and the limbs are closer together than 180 degrees apart. Also, atmospheric refraction shrinks the separation a small amount.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;strong&gt;Where and when are they closest?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;This is more involved but can be reasoned using the following figure. The Sun is in the direction of 1, so the ring of sunset-sunrise is the circle A-A. The Full Moon is in the direction of 2, so the ring of moonrise-moonset is the circle B-B. Where these two circles cross at C is where the sunset and moonrise (or sunrise and moonset) occur at the same time. The highlighted area is where the Moon and Sun are both above the horizon at the same time. Note that this figure is highly exaggerated for illustrative purposes.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dKyjf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dKyjf.png&quot; alt=&quot;illustration of moonrise and sunset planes&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17355" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-09-28T21:32:13.983" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27811" PostTypeId="1" CreationDate="2018-09-29T03:29:41.193" Score="0" ViewCount="345" Body="&lt;p&gt;In the first episode of &lt;em&gt;Star Trek: Discovery&lt;/em&gt;, we learn that Captain Georgiou has an old-fashioned telescope in her ready room. Michael Burnham inherits it a bit later in the season. I wonder if anyone can identify exactly which telescope it is. What's the model and manufacturer? It seems to be an equatorial-mounted refractor with a finder scope. &lt;/p&gt;&#xA;" OwnerUserId="23940" LastActivityDate="2018-10-29T05:00:07.443" Title="Type of telescope on Star Trek" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27812" PostTypeId="2" ParentId="27811" CreationDate="2018-09-29T04:47:29.910" Score="1" Body="&lt;p&gt;I'm not sure this fits this site, maybe better for TV &amp;amp; Movies, but it doesn't appear to be a historic or current telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.spotern.com/en/spot/tv/star-trek-discovery/90656/le-telescope-du-capitaine-han-bo-michelle-yeoh-dans-star-trek-discovery&quot; rel=&quot;nofollow noreferrer&quot;&gt;This website advertises that they sell that model&lt;/a&gt; Based on the look and it is similar.   It may be a  nice telescope but it's nothing special.  Certainly not worthy of being a family heirloom passed down for generations, so that doesn't really make sense and it's not a perfect match either, though it's similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/j6P9W.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j6P9W.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://media.agonybooth.com/wp-content/uploads/2017/09/25202409/00002247.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Larger image here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It not being a perfect match and there being virtually no references to that telescope representing something real, it's likely not a historical telescope, which is fine.   It could be a future telescope and still be in Captain Georgiou's family for generations.  That would work because they used it in the show to look at things the ship couldn't sense due to some kind of obscuring.  A historic telescope probably wouldn't be able to do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, I think Signourney Weaver as Lt. Madison said it best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qEFxA.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qEFxA.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-09-29T04:47:29.910" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27813" PostTypeId="1" CreationDate="2018-09-29T05:21:27.993" Score="1" ViewCount="51" Body="&lt;p&gt;I've been reading the book Welcome to The Universe by Niel De Grasse Tyson. In the tenth chapter he has mentioned &quot;Forty light-years is 12 parsecs. A planet at 1 AU from its star, 40 light-years away, will be 1/12 of a second of arc away from the star in the sky.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain how he got the figure 1/12 arc seconds?&lt;/p&gt;&#xA;" OwnerUserId="24392" LastActivityDate="2018-09-29T08:25:16.197" Title="How far in arc seconds is a planet from its star? The star is 12 parsecs away from earth. The planet is 1 AU from its star" Tags="&lt;parsec&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27815" PostTypeId="2" ParentId="27813" CreationDate="2018-09-29T08:25:16.197" Score="6" Body="&lt;p&gt;Because a separation of 1 au subtends an angle of 1 arcsecond at 1 parsec. That is the definition of a parsec. As you move further away, the angle gets smaller by a factor of the increase in distance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just draw a right-angled triangle where the base length is the distance and the height is the star-planet separation. Now increase the base length and ask yourself what happens to the angle subtended at the Earth? NB All angles are small so &lt;span class=&quot;math-container&quot;&gt;$\theta \simeq \tan \theta$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-09-29T08:25:16.197" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27816" PostTypeId="1" AcceptedAnswerId="27824" CreationDate="2018-09-29T15:53:23.147" Score="3" ViewCount="327" Body="&lt;p&gt;Was taking some extremely amatuer pictures of the moon from my cheap telescope with my camera phone...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MK4rt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see the crater Im talking about? Top right.The one with a little nub in the middle- which Im sure is actually massive. How was this formed differently than the other craters? What couldve caused the hill? &lt;/p&gt;&#xA;" OwnerUserId="23729" LastActivityDate="2018-09-30T00:29:50.910" Title="What are the craters on the moon with peaks in the middle of them?" Tags="&lt;the-moon&gt;&lt;crater&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27817" PostTypeId="1" AcceptedAnswerId="27820" CreationDate="2018-09-29T16:06:53.577" Score="0" ViewCount="183" Body="&lt;p&gt;How it is possible for astronomers to find out when the end of the Sun's life will be?&lt;/p&gt;&#xA;" OwnerUserId="18651" LastEditorUserId="7472" LastEditDate="2018-09-30T09:27:10.427" LastActivityDate="2018-10-01T13:02:21.140" Title="How do astronomers know when the Sun will die?" Tags="&lt;the-sun&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27820" PostTypeId="2" ParentId="27817" CreationDate="2018-09-29T17:35:00.987" Score="7" Body="&lt;p&gt;Yes, astronomers could be wrong.  Part of being a scientist is always having to be ready to admit that you were wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomers have developed models of how the sun and other stars work. We understand them to be nuclear furnaces. These models may not be perfect, but they are well supported by the evidence and can predict much of what we observe from the sun and other stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these models we can expect the sun to exist for about 10 billion years, and it is currently about 4.7 billion years old. While the model could be wrong, it is probably not very wrong, and the sun will go on shining for a long time yet.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-09-29T19:34:04.173" LastActivityDate="2018-09-29T19:34:04.173" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27821" PostTypeId="1" AcceptedAnswerId="27822" CreationDate="2018-09-29T20:37:05.300" Score="2" ViewCount="720" Body="&lt;p&gt;Does the Moon have a night and day like Earth? When sitting on the Moon and viewing the Earth what does it look like when the Earth goes through phases like the Lunar Phase? How does the sun travel through the Moon's sky facing the Earth like this &lt;a href=&quot;https://i.stack.imgur.com/f6zAI.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;gif&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of my question was a duplicate so I deleted it and posted the other link below.&#xA;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/8133/how-bright-is-the-full-earth-during-the-lunar-midnight&quot;&gt;How bright is the full Earth during the lunar midnight?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-09-30T23:46:09.143" LastActivityDate="2018-09-30T23:46:09.143" Title="What is the Terra phase?" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;earth&gt;&lt;moon-phases&gt;&lt;visualization&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27822" PostTypeId="2" ParentId="27821" CreationDate="2018-09-29T22:03:45.723" Score="13" Body="&lt;p&gt;The Earth-Moon System is a system where the moon is tidally locked to the planet, but the planet is not tidally locked to the moon. Earth has a 24 hour day, whereas the Moon has a 29.5 Earth day long day. It takes the Moon 29.5 Earth days to orbit around the Earth once as well, which means the Moon is tidally locked in a 1:1 ratio. This means that if you were to stand on a particular point on the Moon, the Earth would not appear to move because the Moon is spinning on its axis at a rate such that it matches how much the Earth should appear to move from orbiting around it. Basically, the same side of the moon is always facing the Earth, which is why there is a concept of a &quot;dark&quot; side of the moon. The &quot;dark&quot; side is not always actually dark, it's just that no one had ever seen it before until we sent some satellites up there to look at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a gif showing tidal locking. The tidal locked moon is on the left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SHSn9.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SHSn9.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine you are back on Earth. Throughout the course of the day you can see the Moon move across the sky. That's because 24 hours is drastically shorter than 29.5 days. However, if you were standing on the Moon, the Earth would appear to stand still. Now, it would actually appear to wobble a bit because of the Moon's orbit's eccentricity, but it more or less stays in the same place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an artistic yet scientifically accurate rendition of what it would look like. It's from Space Engine. Notice how the gas giant (the planet) wobbles a bit but stays in roughly the same place. It would look similar if you were on the Moon, but without the atmosphere, and the Earth would obviously be smaller and further away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f6zAI.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f6zAI.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, the Earth would go through the same phases as the Moon. It would complete a cycle every 29.5 days because that is how long it takes for the Moon to orbit around the Earth. However, something different is that you would be able to see the Earth rotate, since it takes only 24 hours to rotate. When we look up at the Moon from Earth, as I said before, we cannot see it rotate because it is tidally locked, and thus only one side gets shown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brightness depends on the phase, but generally the Earth is brighter from the Moon than the Moon is from the Earth because the Earth has a higher albedo (reflects more light) than the Moon, and the Earth appears larger from the Moon than the Moon appears from the Earth. The Moon takes up roughly 30 arcminutes of the sky, whereas the Earth takes up roughly 1 degree and 50 arcminutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At full moon, the Moon has an apparent magnitude of roughly -12.75. At full earth, the Earth has an apparent magnitude of -16.75. That's about 40x brighter! If we were to qualify phases as a % from 0% being a new moon to 100% being a full moon, then every 10% you go below 100%, the apparent magnitude of each phase decreases by about the same amount. For example: 90% Earth would be roughly -16.5 (+.25 magnitude), and 90% Moon would be roughly 15.5 (also +.25 magnitude). The differences in magnitude only get larger when you go below 10% phase, where each little bit of the celestial body that shows adds a lot more to the magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a rough chart:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7eLI2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7eLI2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have more questions feel free to ask.&lt;/p&gt;&#xA;" OwnerUserId="24373" LastEditorUserId="24373" LastEditDate="2018-09-30T12:47:44.110" LastActivityDate="2018-09-30T12:47:44.110" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27823" PostTypeId="2" ParentId="27816" CreationDate="2018-09-29T23:43:05.890" Score="4" Body="&lt;p&gt;That is the rebound effect. Just like a rock falling into a pond causes a big ripple and a big drop to jump out in the center, the same happens with a rock striking the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refer to &lt;a href=&quot;https://physics.stackexchange.com/questions/26091/how-does-the-central-peak-in-moon-craters-form&quot;&gt;this&lt;/a&gt; astrophysics stackexchange for more information.&lt;/p&gt;&#xA;" OwnerUserId="24373" LastActivityDate="2018-09-29T23:43:05.890" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27824" PostTypeId="2" ParentId="27816" CreationDate="2018-09-30T00:29:50.910" Score="5" Body="&lt;p&gt;Nice picture! Those peaks are actually a very common feature in craters formed by large impacts. Craters with these features are known as &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Complex_crater&quot; rel=&quot;noreferrer&quot;&gt;complex craters&lt;/a&gt;&quot;. Basically what happens is that when the meteor hits, it pushes in the surface. Then the underlying mantle produces an &quot;elastic rebound&quot; and pushes outwards. This is a similar effect to what happens when you hit a drum. Your drumstick rebounds off the drum, and the drum begins to vibrate. Since rock (even molten rock) is much more dense than a drumskin, it does not vibrate very much. The vibrations solidify, resulting in a crater with one central peak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How exactly this happens is still being investigated. &lt;a href=&quot;http://www.planetary.brown.edu/pdfs/4779.pdf&quot; rel=&quot;noreferrer&quot;&gt;This paper&lt;/a&gt; provides a pretty good explanation of what happens, but it's quite technical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kj4cR.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kj4cR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even larger impacts will produce &lt;a href=&quot;https://en.wikipedia.org/wiki/Peak_ring_(crater)&quot; rel=&quot;noreferrer&quot;&gt;ring structures&lt;/a&gt;. These are relatively rare on the Earth, since such structures are quickly weathered away. A good example of this is the Chicxulub crater off the Yucatan peninsula (the impact that killed the dinosaurs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AjUDf.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AjUDf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13426" LastActivityDate="2018-09-30T00:29:50.910" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27825" PostTypeId="1" CreationDate="2018-09-30T05:21:55.293" Score="1" ViewCount="2853" Body="&lt;p&gt;I have a 125mm Jones Bird newtonian telescope (very shoddy) with 6x lens (2× + 3×) essentially we were able to see a very small O (yeah that's to scale) with varied shades of red. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is Mars far away from earth right now relative to the average distance or is it just my telescope not powerful enough?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I wanted to seriously have a chance at picking up significant features on Mars or perhaps pickup the rings of saturn or the red spot of jupiter, whats the minimum setup that I am going to need? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This setup would run me around 3,000 dollars:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y5TOk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G5ZTb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its named...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Celestron CPC 1100 StarBright XLT GPS Schmidt-Cassegrain 2800mm Telescope with Tripo... &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B000ARFND2&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amazon.com/dp/B000ARFND2/ref=cm_sw_r_sms_c_api_x7mSBb39EAS9H&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this be overkill? For $3,000 is this one of the better choices for local planetary observation?&lt;/p&gt;&#xA;" OwnerUserId="23729" LastEditorUserId="23729" LastEditDate="2018-09-30T14:23:25.033" LastActivityDate="2018-10-04T00:09:44.777" Title="What magnification is required to see detail on Mars / other local bodies?" Tags="&lt;telescope&gt;&lt;mars&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27826" PostTypeId="1" AcceptedAnswerId="27828" CreationDate="2018-09-30T08:17:15.247" Score="2" ViewCount="98" Body="&lt;p&gt;Hawking radiation is supposed to very slowly evaporate a black hole (terms and conditions apply :] ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from Hawking radiation, is there any mechanism or effect that can make a black hole cease to exist? Or once they are formed are they expected to exist in this form &quot;forever&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I posted the same question to Physics but I think it applies well to both SEs. So I hope it is okay to try here as wwell. &lt;/p&gt;&#xA;" OwnerUserId="16462" LastActivityDate="2018-09-30T13:04:04.020" Title="Once a black hole is formed, is there anything other than Hawking radiation which shortens its life?" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27827" PostTypeId="1" CreationDate="2018-09-30T10:24:52.260" Score="0" ViewCount="96" Body="&lt;p&gt;Hypothetically,if our sun was yellow would we still be able to see white colour on objects like snow,cotton etc.? Would we have the phenomenon of rainbow?&lt;/p&gt;&#xA;" OwnerUserId="24392" LastActivityDate="2018-10-01T18:13:55.870" Title="Hypothetically, if our sun was yellow would we still see &quot;white&quot; colours?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;astrophysics&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27828" PostTypeId="2" ParentId="27826" CreationDate="2018-09-30T10:26:24.397" Score="3" Body="&lt;p&gt;It can merge with another black hole, if that counts. A simulation of that is illustrated &lt;a href=&quot;https://www.youtube.com/watch?v=jEs6yHJZz5I&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That aside, no black hole with decay &lt;em&gt;more&lt;/em&gt; quickly, as far as I can find out, than the &quot;normal&quot; decay of a non-rotating, uncharged black hole into which nothing is falling. Let's call this the &quot;standard rate of decay&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Feeding it&quot; (allowing any matter or energy to fall into it), will reduce the decay below the standard rate, or even reverse it. Since the CMB pervades space and will fall into the black hole, blocking this (for instance by surrounding the hole by an opaque refrigerated shell) will be necessary to allow it to decay at the standard rate. Indeed for a black hole of mass over roughly &lt;span class=&quot;math-container&quot;&gt;$10^{22} kg$&lt;/span&gt; the CMB would feed it faster than it decays, if not blocked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy from the spin of the  black hole can be extracted using &lt;a href=&quot;https://en.wikipedia.org/wiki/Penrose_process&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Penrose process&lt;/a&gt;. Since &lt;a href=&quot;https://physics.stackexchange.com/questions/44323/at-what-rate-does-a-rotating-black-hole-lose-mass-via-hawking-radiation&quot;&gt;this answer&lt;/a&gt; shows that rotation also reduces the emission of Hawking radiation, you can clearly speed up the decay by extracting energy from as much of the spin as possible, but the fastest decay you get is, once again, the standard rate.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="21928" LastEditDate="2018-09-30T13:04:04.020" LastActivityDate="2018-09-30T13:04:04.020" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27829" PostTypeId="2" ParentId="27827" CreationDate="2018-09-30T10:30:02.763" Score="0" Body="&lt;p&gt;We see  white objects as white under a very wide range of lighting conditions.  Photography resources will be quite helpful.  Look at a white piece of paper at dawn, at noon, and under a variety of artificial light sources.  The spectrum of the light hitting your eye will vary a lot but, except in extreme cases, it will st6il look white.  The answer is not in astronomy but the visual processing in our brains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some extra material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I just mentioned in a comment, digital cameras do something similar and call it automatic white balance.  Search for that and you will loads of discussions in photography websites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It fails in some extreme cases.  E.g. sodium street lamps which are close to monochromatic and a combination of UV lamps and fluorescent substances.&lt;/p&gt;&#xA;" OwnerUserId="21104" LastEditorUserId="21104" LastEditDate="2018-09-30T12:42:22.233" LastActivityDate="2018-09-30T12:42:22.233" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27830" PostTypeId="2" ParentId="27827" CreationDate="2018-09-30T10:47:27.530" Score="1" Body="&lt;p&gt;Every &quot;normal&quot; star, produces a more or less &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Black-body_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;black-body&lt;/a&gt;&quot; spectrum, at some temperature (more-or-less because there are brighter and darker lines caused by specific elements in the upper layers of the star), so there will always be a range of colours of light (or electromagnetic radiation at least) in its light, which can be spread to give a rainbow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A white object (like ice crystals, or bleached cotton) reflects all colours of light more or less equally (over some range of wavelengths of interest, anyway). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if the Sun suddenly cooled, so that it was emitting a 5000K black body spectrum instead of its current 6000K, with more red light and less blue light, that change would be reflected in the light reflected from a white object, which would look more yellow or orange to our eyes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/07b4r.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/07b4r.png&quot; alt=&quot;this chart attempts, within the limitations of the web to illustrate the colours&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a couple of caveats though:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;As @badjohn pointed out in a simultaneous answer, our brain processes visual information in odd ways. If the colour of sunlight changed, the brain might try and compensate for the overall shift and you might see everything more as you expect it, and less as it now is.&lt;/li&gt;&#xA;&lt;li&gt;We evolved to perceive sunlight as white. If we had evolved around a cooler star, we would probably perceive a different colour as white.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-09-30T10:47:27.530" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27831" PostTypeId="1" CreationDate="2018-09-30T12:47:33.433" Score="6" ViewCount="5746" Body="&lt;p&gt;&lt;strong&gt;Observations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The heaviest elements known in abundance in nature are forged deep within stars.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;These elements are made possible by the high densities/pressures within the stars.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Black holes are known to have a much higher density/pressure than any known star.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Black holes are also known to be a phase of stellar evolution - &lt;strong&gt;this suggests that the original star's internal process of forging metals would persist within the resultant black hole.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Scientists have forged synthetic/ephemeral heavy metals under conditions which could hypothetically be sustained within a black hole.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hypothesis:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes forge heavier elements that have not been observed on earth. The conditions needed to sustain these elements are unique to the black hole, due to its high density/pressures. These conditions can be glimpsed, but not sustained in any experimental context &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Follow up questions&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Has this been hypothesized?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Where can I find research on this topic?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="24405" LastEditorUserId="24405" LastEditDate="2018-09-30T21:39:06.280" LastActivityDate="2018-10-01T18:34:51.307" Title="Could Black holes forge heavier elements that have yet to be discovered?" Tags="&lt;black-hole&gt;&lt;stellar-evolution&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27832" PostTypeId="2" ParentId="27831" CreationDate="2018-09-30T13:26:08.280" Score="4" Body="&lt;p&gt;The problem of the superheavy elements is not that we can't forge them. Their problem is that they decay very quickly. For example, &lt;a href=&quot;https://en.wikipedia.org/wiki/Oganesson&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oganesson&lt;/a&gt;, the heaviest element synthetised until now, has a half life of 181 ms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory, even much heavier elements could be created in particle accelerators, but there is no way to even detect them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In neutron stars, or in exploding supernovas, all the elements are created, but there is no way to even detect them. We can consider a neutron star as a large nucleus with &lt;span class=&quot;math-container&quot;&gt;$\approx 10^{56}$&lt;/span&gt; neutrons.&lt;sup&gt;1&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In black holes, the fact is that no one knows, what is in them. They don't radiate anything (with a very little &lt;a href=&quot;https://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;exception&lt;/a&gt;), and nothing leaves the singularity in them. To understand what is in them, would require currently unrealistic &lt;a href=&quot;https://en.wikipedia.org/wiki/Theory_of_everything&quot; rel=&quot;nofollow noreferrer&quot;&gt;advances&lt;/a&gt; in Physics. The singularity in their center is probably not &lt;a href=&quot;https://en.wikipedia.org/wiki/Baryon#Baryonic_matter&quot; rel=&quot;nofollow noreferrer&quot;&gt;baryonic matter&lt;/a&gt;, though, thus we could hardly say that it would be any chemical element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;&lt;sub&gt;As @PM2Ring's excellent &lt;a href=&quot;https://astronomy.stackexchange.com/questions/27831/could-black-holes-forge-heavier-elements-that-have-yet-to-be-discovered/27832#comment48784_27832&quot;&gt;comment&lt;/a&gt; says, neutron stars also have a significant number of other particles, too, not only neutrons. I also extend it that they are bound gravitationally and not by the strong interaction, which makes them in this aspect essentially different from nuclei.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="1501" LastEditDate="2018-10-01T10:12:27.490" LastActivityDate="2018-10-01T10:12:27.490" CommentCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="27833" PostTypeId="2" ParentId="27831" CreationDate="2018-09-30T15:45:25.140" Score="23" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The heaviest elements known in nature are forged deep within stars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, the heaviest elements are made on Earth in scientific laboratories, or in the extreme gravity of a neutron star's crust. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;These elements are made possible by the high densities/temperature/pressures within the stars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many of the larger elements can be made in supernovae and neutron star collisions, not in stars. It requires extreme conditions for these elements to form.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Black holes are known to have a much higher density/temperature/pressure than any known star.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Black holes are actually very cold, they &quot;absorb&quot; any radiation that passes their event horizon.  Outside the event horizon may be some very hot material, but it is not actually so hot compared to the core of a star.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Black holes are also known to be a phase of stellar evolution - this suggests that the original star's internal process of forging metals would persist within the resultant black hole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, inside the black hole everything falls, and reaches a singularity in a short amount of time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Scientists have forged synthetic/ephemeral heavy metals under conditions which could hypothetically be sustained within a black hole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As above, the conditions beyond the event horizon are unlike anything we have on Earth, because there is the unavoidable singularity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some matter has crossed the event horizon it will certainly come to the singularity. (in the same way as you will certainly reach tomorrow) And as it gets closer the tidal effects get greater, eventually ripping the atoms apart.  The extreme gravity in a black hole will tend to pull matter apart not fusing it to larger atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be &lt;a href=&quot;http://lup.lub.lu.se/luur/download?func=downloadFile&amp;amp;recordOId=8912003&amp;amp;fileOId=8912097&quot; rel=&quot;nofollow noreferrer&quot;&gt;nucleosynthesis in the accretion disc&lt;/a&gt; around a black hole. While the amount of high mass atoms made here is relatively small, it may be useful for the sake of detecting and distinguishing black holes from neutron stars or white dwarfs.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-10-01T18:34:51.307" LastActivityDate="2018-10-01T18:34:51.307" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="27834" PostTypeId="5" CreationDate="2018-09-30T16:57:10.223" Score="0" Body="&lt;p&gt;The Solar and Heliospheric Observatory (SOHO) is a spacecraft built by a European industrial consortium led by Matra Marconi Space (now Astrium) that was launched on a Lockheed Martin Atlas II AS launch vehicle on December 2, 1995 to study the Sun. SOHO has also discovered over 3,000 comets. It began normal operations in May 1996. It is a joint project of international cooperation between the European Space Agency (ESA) and NASA.&lt;/p&gt;&#xA;" OwnerDisplayName="user1569" LastEditorDisplayName="user1569" LastEditDate="2018-09-30T19:32:57.297" LastActivityDate="2018-09-30T19:32:57.297" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27835" PostTypeId="4" CreationDate="2018-09-30T16:57:10.223" Score="0" Body="Use for questions about observations of the Sun done with The Solar and Heliospheric Observatory (SOHO) spacecraft." OwnerDisplayName="user1569" LastEditorDisplayName="user1569" LastEditDate="2018-09-30T19:32:40.827" LastActivityDate="2018-09-30T19:32:40.827" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27837" PostTypeId="1" AcceptedAnswerId="27838" CreationDate="2018-09-30T18:17:24.603" Score="8" ViewCount="371" Body="&lt;p&gt;A recent &lt;a href=&quot;https://space.stackexchange.com/questions/27906/can-a-solar-cell-sail-powered-satellite-orbit-both-the-moon-and-earth-rendezvous#comment93046_27906&quot;&gt;comment&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An object far enough away can certainly orbit the Moon and the Earth (and the Sun) -- Mars, for instance does this. An object in the Earth-Moon L2 is also orbiting both the Earth and the Moon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A recent &lt;a href=&quot;https://space.stackexchange.com/questions/27906/can-a-solar-cell-sail-powered-satellite-orbit-both-the-moon-and-earth-rendezvous#comment93117_27906&quot;&gt;comment&lt;/a&gt; about that comment:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;according to your way of thinking all solar system bodies orbit Mercury, except the Sun, which is just silly. Or does the Sun in fact orbit Mercury as well? I think your use of the word orbit is not workable, nor shared by almost anyone else.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I'd like to ask if astronomers have an established, systematic way for saying what does or doesn't orbit what within the solar system, and if so, a link to point to in the future would be greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-09-30T18:23:04.130" LastActivityDate="2020-03-27T21:12:33.080" Title="Do astronomers have an established, systematic way for saying what does or doesn't orbit what? (e.g. &quot;Mars orbits Earth&quot;)" Tags="&lt;solar-system&gt;&lt;history&gt;&lt;terminology&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27838" PostTypeId="2" ParentId="27837" CreationDate="2018-09-30T19:23:44.963" Score="11" Body="&lt;p&gt;It is possible that there is nothing &quot;official&quot;, there is just the technical use of language. For example Phil Plait notes that &lt;a href=&quot;https://www.syfy.com/syfywire/mea-culpa-circling-ryugu&quot; rel=&quot;noreferrer&quot;&gt;He incorrectly used the word &quot;orbit&quot; for the motion of the Hayabusa probe&lt;/a&gt;, which does not orbit the asteroid Ryugu. But hovers over the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what does Phil mean when he talks about orbiting?  I summarize the meaning thus: Object A is in orbit around object B if A moves around B (ie the True Anomaly of A increases from 0 to 360 degrees in fairly regular fashion) principally as a result of the gravitational field of B.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So although Mars does move around the Earth, its motion is not principally due to the gravitation field of the Earth, so it is not in orbit.  However the moon's motion is principally due to the gravitational field of the Earth. It is in orbit around the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon is also in orbit around the sun: It moves around the sun and the reason for this motion is principally the graviational field of the sun.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A body at the Earth Sun L2 point is not in orbit around the Earth (it is fixed relative to the Earth) It is in orbit around the sun. A body in the Earth Moon L2 point (which is probably very unstable) is orbiting the Earth and orbiting the sun. Its motion relative to the Earth is primarily due to the Earth (and secondarily due to the moon). Its motion relative to the sun is due to the gravity of the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cruithne is in a 1:1 resonance with the Earth, but its postion is not primarily due to the gravity of the Earth. It is in orbit around the sun, not the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hayabusa is not in orbit around an asteroid, it is maintaining position by thrusters. Similarly, A Jumbo jet is flying not orbiting, it's motion is not mainly due to gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This idea can be sharpened by the notion of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;noreferrer&quot;&gt;Hill sphere&lt;/a&gt;.&#xA;Inside the Hill sphere, a body will orbit. Outside it, the body will not orbit.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="616" LastEditDate="2018-09-30T22:03:10.540" LastActivityDate="2018-09-30T22:03:10.540" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27839" PostTypeId="2" ParentId="27837" CreationDate="2018-09-30T19:34:45.897" Score="-1" Body="&lt;p&gt;What if I told you that technically speaking, the Moon does not orbit the Earth, and the Earth doesn't orbit the Sun, and that the Sun orbits something that is nothing? Welcome to the world of barycenters, which seem to be skipped over in elementary school science, though for good reason. It would be a mouthful for the students to learn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Earth-Moon system, the Earth does not orbit the Moon, and the Moon does not orbit the Earth. Rather, they orbit a common center of mass, referred to as the Earth-Moon Barycenter, or just barycenter if the situation is unambiguous (as you will see later, you can have multiple barycenters). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon tugs on the Earth with the same force that the Earth tugs on the Moon. Therefore, the orbit looks more like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Sor1t.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sor1t.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon has roughly 1.2% the mass of the Earth. Therefore, the Moon should orbit roughly 83x further away (it's just &lt;span class=&quot;math-container&quot;&gt;$\frac{100}{1.2}$&lt;/span&gt;) from the barycenter than the Earth does. The Moon's semi major axis is 385000 km. Therefore, the Earth's semi major axis around the Earth-Moon barycenter (not the Sun) is roughly 4500 km. And that is the case, as shown below. The diamond symbol is what Space Engine uses for barycenter symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4SXlZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4SXlZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we zoom in we see the Earth's orbit around the barycenter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZE3DA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZE3DA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the Moon has such a low mass compared to the Earth, the barycenter is actually located inside the Earth. As the Moon slowly travels away from the Earth in the next few hundred million years, the barycenter will slowly leave the Earth's surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the barycenter be located outside of the planet? It sure can. The Pluto-Charon system looks a bit like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ckdD4.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ckdD4.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Space Engine it looks like this (brightness increased to you can actually see it):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oDaEd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oDaEd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the Sun-Earth system? Well the problem is that the Sun is also affected by all the other planets in the solar system. The sum of all these interactions is where the Solar Systemic Barycenter is located. It doesn't travel in a nice circle though, but it is mostly affected by Jupiter since Jupiter is the most massive planet in the Solar System. It looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ChsOk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ChsOk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about binary stars? Well it's pretty much the same thing. Here's &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; Centauri.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lpLMx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lpLMx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, last thing, how about systems with 3 or more stars? Well here's the thing. barycenters can orbit other barycenters. The Castor system is a sextuple (6) system. Here's what it looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wtZoa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wtZoa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those three points of light are actually two stars orbiting very closely together, so that they look like one star. So to answer you question, no they don't orbit anything really, but they instead orbit a common barycenter, and we determine what orbits what by seeing if it goes around that barycenter.&lt;/p&gt;&#xA;" OwnerUserId="24373" LastActivityDate="2018-09-30T19:34:45.897" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27841" PostTypeId="2" ParentId="27831" CreationDate="2018-10-01T02:04:55.213" Score="8" Body="&lt;p&gt;Superheavy elements have short half lives because of their extreme instability with respect to alpha decay and fission. This is a result of their high electric charge, which results in strong forces of electrical repulsion. Although theorists have predicted an &quot;island of stability&quot; due to quantum mechanical shell effects, this stability is a relative thing. We're still talking about half-lives on the order of seconds or less. So any such element created by astrophysical processes will not survive for very long, even if it doesn't fall into the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So conceivably in the accretion disk, outside the event horizon, you could get some fusion events resulting in the formation of superheavy elements, but those elements would not survive for very long, even if they were somehow ejected rather than infalling past the horizon. And the normal methods for detecting and characterizing superheavy elements would not work here. Normally we look for things like alpha-decay chains with characteristic alpha energies. Those would not be detectable from outside the accretion disk, since charged particles interact strongly with matter and are stopped.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The conditions needed to sustain these elements are unique to the black hole, due to its high density/pressures.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most of the interior of a black hole (inside the event horizon) is probably an extremely good vacuum. The only high densities and pressures would be near the singularity.  So any exotic matter formed at high densities and pressures would not be observable from earth or have any consequences for the outside universe, because nothing can escape from inside the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were to send a space probe inside a black hole to look for exotic matter being formed near the singularity, the probe would not be able to report its results back. Also, the region of high-density and high-pressure infalling matter would exist near the singularity, which would probably not be detectable to the probe until the probe itself had been destroyed by the same processes. (On the inside of a black hole, if general relativity is correct, you can't see the singularity. You only see infalling photons from the outside.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If exotic matter is formed near the singularity, it will only exist for a very short time before accreting onto the singularity. (IIRC the maximum infall time for a 10-solar-mass black hole is on the order of milliseconds from horizon to singularity.) We don't really know what happens at the singularity, but we certainly can't have atomic nuclei under those conditions.&lt;/p&gt;&#xA;" OwnerUserId="15381" LastActivityDate="2018-10-01T02:04:55.213" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27842" PostTypeId="1" CreationDate="2018-10-01T10:38:55.867" Score="3" ViewCount="357" Body="&lt;p&gt;I understand that before the Big Bang time (as well as dimensions) didn't exist. But how could this be? If there was no time then nothing could change, and so time itself couldn't come into existence. &lt;/p&gt;&#xA;" OwnerUserId="24410" LastEditorUserId="23940" LastEditDate="2018-10-01T12:07:05.440" LastActivityDate="2020-11-02T08:03:19.360" Title="How could time only have started with the Big Bang?" Tags="&lt;big-bang-theory&gt;&lt;time&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27844" PostTypeId="1" AcceptedAnswerId="27847" CreationDate="2018-10-01T11:29:22.387" Score="0" ViewCount="107" Body="&lt;p&gt;Wasn't even sure about what the title should be, but I had a thought and I wonder whether it may be possible or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible that there is a star which acts as a mirror and reflects back the light almost as it was received?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My though is that if there is something like that and let's say that the star is far from earth 50 light years, than we could see 100 years back to the past.&lt;/p&gt;&#xA;" OwnerUserId="24413" LastActivityDate="2018-10-02T11:04:08.833" Title="Existence of a &quot;reflecting&quot; star" Tags="&lt;light&gt;&lt;time&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2018-10-02T09:14:10.050" ContentLicense="CC BY-SA 4.0" />
  <row Id="27845" PostTypeId="2" ParentId="27844" CreationDate="2018-10-01T12:25:21.230" Score="2" Body="&lt;p&gt;See &lt;a href=&quot;https://astronomy.stackexchange.com/a/26173&quot;&gt;this&lt;/a&gt; post about binary stars, which contains relevant information: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Stars are far from perfect blackbodies due to scattering/reflection.&#xA;  This is especially true for hotter stars, because of all the free&#xA;  electrons, but even cooler stars can reflect a significant amount. For&#xA;  example, in aanda.org/articles/aa/pdf/2001/19/aa1009.pdf you will see&#xA;  that they use a reflection albedo of 0.30 for the K star and 1.00 for&#xA;  the F star, but the latter number is not meant to be taken seriously,&#xA;  they simply don't care if the light is reflected or absorbed and&#xA;  re-emitted because it isn't an important term. But the value of 0.30&#xA;  for the K star might be meant more seriously, though it is still not&#xA;  regarded as a critical parameter because it only affects the color of&#xA;  the light that is reflected, not the total amount of light (given that&#xA;  stars are in radiative equilibrium, so must ultimately return all the&#xA;  incident light, whether it happens by reflection or heating).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Indeed, stellar emissions are often characterized by &quot;effective&#xA;  temperature,&quot; to connect the surface flux of a star to the&#xA;  Stefan-Boltzmann formula for the emission of a blackbody by using a T&#xA;  parameter that is not necessarily the actual temperature. When using&#xA;  this notion, as is quite common for dealing with stars, there is no&#xA;  essential difference between heating of and reflection from the&#xA;  surface of the star in question. The details of the difference have to&#xA;  do with the shape of the spectrum, but that shape is generally not a&#xA;  Planck function anyway, so as soon as one is using the &quot;effective&#xA;  temperature&quot; concept one has already parted company from a detailed&#xA;  understanding of the shape of the spectrum. (When you do want the&#xA;  details of the spectrum, you will have to model the situation with&#xA;  some care.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Is it possible that there is a star which acts as a mirror and reflects back the light almost as it was received?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That depends on what you mean by &quot;as it was received&quot;. That post shows that the concept of albedo applies to stars as well; they can be reflective.&lt;/p&gt;&#xA;" OwnerUserId="23940" LastEditorUserId="14094" LastEditDate="2018-10-02T11:04:08.833" LastActivityDate="2018-10-02T11:04:08.833" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27846" PostTypeId="2" ParentId="27842" CreationDate="2018-10-01T12:49:22.537" Score="2" Body="&lt;p&gt;This is both a physical and philosophical question because it depends what you mean by &quot;time&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a physical perspective, time doesn't really exist. We know that things change, and we measure a rate of change in relation to other things that also change (like clocks) but how we define time as an absolute concept is not something we have a firm answer to. It may not exist at all. To understand the details fo this you need to learn a whole lot of quantum mechanics, string theory and theoretical physics in general, which will take about a decade, two degrees and a doctorate. Alternatively, I've been reading a great book called &quot;The Order of Time&quot; by Carlo Rovelli. He is a theoretical physicist who has worked for decades to understand this stuff and then written all about it with hardly any maths and some drawings of Smurfs. I would highly recommend it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, from a more philosophical perspective, if the Big Bang started with a singularity in which time and space and all matter were compressed into one single point with no dimensions, then that point contained no information about what was before it. If we can never know what was before the singularity (or even if there was anything before), then there is very little point in discussing it, so we can choose to set that as the zero point of time and call it the beginning because anything that happened before that is irrelevant. Like starting a stopwatch to time a race when the cars set off - there was time before, but that's not important for the race so we call the start of the race a time of zero.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2018-10-01T12:49:22.537" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27847" PostTypeId="2" ParentId="27844" CreationDate="2018-10-01T12:51:32.360" Score="2" Body="&lt;p&gt;I have never heard of any naturally occurring object acting as an optical mirror on a scale large enough to form an image at a distance of light-years, and it seems deeply implausible that anything of the kind would exist. Natural objects are not that smooth or uniform. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Independently of that question though, let's look at how large a mirror you would need. If we assume light with a wavelength of 500 nm, and an optical path 50 light years (&lt;span class=&quot;math-container&quot;&gt;$5\times 10^{17}m$&lt;/span&gt;) long, and we want to see Earth with &lt;span class=&quot;math-container&quot;&gt;$1m$&lt;/span&gt; resolution, we need a resolution of &lt;span class=&quot;math-container&quot;&gt;$2\times 10^{-18} radians$&lt;/span&gt; which requires an aperture &lt;span class=&quot;math-container&quot;&gt;$2\times 10^{18}$&lt;/span&gt; wavelengths wide, which is just about &lt;span class=&quot;math-container&quot;&gt;$10^{12}m$&lt;/span&gt; or about a billion kilometers -- most of the diameter of the orbit of Jupiter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even if they were perfectly flat optical mirrors, most stars are simply too small.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="21928" LastEditDate="2018-10-01T12:58:06.600" LastActivityDate="2018-10-01T12:58:06.600" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27848" PostTypeId="2" ParentId="27817" CreationDate="2018-10-01T13:02:21.140" Score="1" Body="&lt;p&gt;To understand stellar evolution (the life cycle of all stars, including our Sun), we start by simply looking at all the stars. We can find out what they are made of and what temperature they are by looking at the spectrum - different elements emit and absorb different wavelengths of light, and the brightness gives us information about temperature. Over hundreds of years, we have then developed models of stars and tried to fit them to explain how many of different types of stars we can see. The models that fit well we keep and refine, the ones that don't fit reality we discard. There are still things we don't understand about stellar structure and evolution, but our Sun is a rather well understood type - it is low to medium mass, and in about the middle of its life. We can't predict to the day or year or hundred years exactly what will happen, but we can predict to around a billion years accuracy when it will run out of hydrogen fuel, swell up to a red giant, and then explode as a supernova and become a white dwarf.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2018-10-01T13:02:21.140" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27850" PostTypeId="2" ParentId="27837" CreationDate="2018-10-01T13:45:15.297" Score="1" Body="&lt;p&gt;Take binary star systems as an example.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://what-if.xkcd.com/150/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Circumbinary, the planet orbits two stars&lt;/a&gt;, or, you could say it orbits their center of mass or barycenter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gSX15.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gSX15.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also imagine a star with a brown dwarf/heavy jupiter orbiting it and a planet orbiting a bit further out.   There's no precise point where the brown dwarf/heavy Jupiter stops being a binary system and becomes another planet orbiting the star.  Because there's no precise place where the system changes, I'd be very surprised if there's a verifiable, official answer.  I'd bet money that there isn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomy is full of situations like this.  Sometimes definitions can be set up, like the qualifications for being a planet.  Sometimes it's harder to set up a precise definition, for example, what is the smallest size a moon can be?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no precise switching point where a planet orbiting two stars becomes two planets orbiting one star and there doesn't need to be.  It's OK to say that a brown dwarf of low enough mass could be a heavy Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easier to say that Mars orbits the Sun, but in a sense, Mars orbits the center of mass of the Sun + Venus + Earth &amp;amp; Moon (and all the small stuff in there too).  Both statements have truth to them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus and Earth together, when lined up, create a barycenter that's only about 70,000 km from the center of the Sun, or about 0.03% of the distance Mars is from the Sun (a bit less than 1 part in 3,000).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe (but can't do the math), that it's more accurate to say that Mars orbits the Venus/Earth/Sun barycenter more closely than it orbits the Sun.  Both likely deviate from a perfect Kepler orbit to some degree.   Mars' mass also plays a role in deviating from a pure Kepler orbit at least in terms of orbital velocity, and there's the perturbations from the outer planets, Jupiter and Saturn primarily and relativistic dilation has a tiny effect too (more noticeable with Mercury).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When NASA wants to land a craft on Mars, they need to take into account the gravitational influence from several planets, not just Mars orbiting the Sun, though I gather the early lunar landings, they were able to ignore relativitistic effects.  With Mars, they likely need to factor in relativity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also important to &lt;a href=&quot;http://articles.latimes.com/1999/oct/01/news/mn-17288&quot; rel=&quot;nofollow noreferrer&quot;&gt;remember to convert to the metric system&lt;/a&gt; when needed, but, I digress.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-10-01T13:45:15.297" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27854" PostTypeId="2" ParentId="27827" CreationDate="2018-10-01T18:13:55.870" Score="0" Body="&lt;p&gt;You have confused solar spectral output with our neurological processing of collected photons.  While it's true that a careful mix of red,blue, green photons may be perceived as &quot;white&quot; or &quot;off-white,&quot;  whiteness in general occurs when either (1) the cones, which discriminate on the basis of wavelength, saturate and all &quot;output&quot; signals are peak, or (2) the light levels are so low that only the rods, which do not discriminate by wavelength but have higher sensitivity than the cones, respond.  For example, in a dim night locale, everything is black, grey or white.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It gets worse, as our brain reprocesses the incoming signals to make the colors we see &quot;fit expectations.&quot;  That's why we tend to observe everything in our backyard garden as having the same coloration all day, but if we take a photo at sunset, the image will show a skew towards the red (due to extreme loss of blue, green thru the atmosphere).&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-10-01T18:13:55.870" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27855" PostTypeId="1" CreationDate="2018-10-01T18:27:08.037" Score="2" ViewCount="165" Body="&lt;p&gt;I was reading about heavy elements recently, and I took note of the theorized islands of stability, and how some the elements involved are considerably heavier than anything we've been able to synthesize so far. Is there any known astronomical process that could create elements well above the 120-proton mark?&lt;/p&gt;&#xA;" OwnerUserId="24420" LastActivityDate="2018-10-02T15:51:40.137" Title="Could an astronomical process create elements in the islands of stability?" Tags="&lt;supernova&gt;&lt;neutron-star&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27856" PostTypeId="2" ParentId="27855" CreationDate="2018-10-01T18:42:13.897" Score="2" Body="&lt;p&gt;To this point, this is purely theoretical.  We have no solid evidence of islands of stability.  They may or may not exist to begin with.  It's hard to speculate on the methods of creating something we don't even know exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, currently, the heaviest elements found in nature appear to be synthesized in the mergers of neutron stars.  It might be possible that a similar process could create even heavier elements.  It's possible that some of the discovered transuranic elements we've discovered might be formed in such mergers, however their half-lives are so short that they break down before they're detectable.  If there are stable transuranic elements, then they might be formed in this manner, but we've never seen any evidence of them.  Me might not know what to look for, but so far we've detected no suggestions of such elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option is the accretion disks of black holes.  It is believed that nucleosynthesis occurs in accretion disks, and one could speculate that there may be conditions in accretion disks that could make it possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But without any evidence of the islands of instability, this is entirely conjecture.&lt;/p&gt;&#xA;" OwnerUserId="10546" LastActivityDate="2018-10-01T18:42:13.897" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27857" PostTypeId="2" ParentId="27825" CreationDate="2018-10-01T21:31:21.370" Score="2" Body="&lt;p&gt;You have several questions here:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Mars is relatively close to Earth right now (October 2018) - about 90 million km.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is your telescope powerful enough for you to see detail on Mars?  From what you have said, it's difficult to tell.  You say you have magnification of 6x.  This is almost certainly wrong.  If you state the focal lengths of your telescope and its eyepieces we can work out the correct magnification available.  To see much detail on Mars you would need in excess of 100x magnification (same for Saturn, less for Jupiter), and ideally a lot more.  Your scope should be able to manage 100x.  However the Jones Bird design is known to be difficult to align (collimate) properly, so even with that magnification, you may just see a blur.  How much detail you can see will also be heavily affected by the state of the atmosphere - ideally you need a night when the air is calm. Sometimes a slight mist can indicate a steady atmosphere.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Would the CPC 1100 be overkill etc?  This is much more opinion based, but in short, yes - for a beginner anyway.  For an experienced observer, the answer is much more nuanced, and I'm not going to go into any of that.  The scope is generally regarded as pretty good for what it is.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2018-10-01T21:31:21.370" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27860" PostTypeId="1" AcceptedAnswerId="27863" CreationDate="2018-10-02T07:09:51.450" Score="0" ViewCount="58" Body="&lt;p&gt;I'm sorry if this sounds like a silly question but I don't understand how measuring the location of a star would help in sea navigation. &lt;/p&gt;&#xA;" OwnerUserId="24332" LastActivityDate="2018-10-02T10:11:22.757" Title="How was calculating the celestial latitude using an astrolable of any help to sailors?" Tags="&lt;amateur-observing&gt;&lt;observational-astronomy&gt;&lt;history&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27862" PostTypeId="1" AcceptedAnswerId="27871" CreationDate="2018-10-02T08:49:19.830" Score="2" ViewCount="138" Body="&lt;p&gt;Bart Bok gave his name to dust clouds of 2-50 solar masses, which by their darkness may be 100 times more frequent than telescopes can detect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the solar system traversed a &lt;a href=&quot;https://en.wikipedia.org/wiki/Bok_globule&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bok Globule&lt;/a&gt;, how dark would the sun become towards the surface of the Earth, as visible by humans? &lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2018-10-03T07:15:43.483" LastActivityDate="2018-10-03T07:15:43.483" Title="How dark would the sun become if we traversed a Bok globule?" Tags="&lt;radiation&gt;&lt;nebula&gt;&lt;interstellar-medium&gt;&lt;dust&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27863" PostTypeId="2" ParentId="27860" CreationDate="2018-10-02T10:11:22.757" Score="3" Body="&lt;p&gt;Measuring the angle to a known star allows the navigator to calculate the latitude of the ship in the absence of any visible (known) landmarks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since they would know (or refer to an almanac) the celestial latitude (declination) or altitude of the star at their home port or some other known port, they can measure the position of the star and take the difference between the measurement and the reference to calculate their latitude at sea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if their home port were located at 54°N then the angle from the horizon to Polaris would be 54°. If they take a measurement of Polaris while at sea and find it to be 30° above the horizon then they know that they are 24° farther south, i.e. latitude 30°. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if there were a star that is directly overhead (90° from the horizon) when it transits the celestial meridian at home, then they know that that star has a declination of 54°. If they take a measurement of it while at sea and find it to be 6° north of zenith (84° from the horizon) then they know that they are 6° south of home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From all of this, knowing the date lets them calculate the sunrise and sunset times, and if they have an accurate clock they can work out their longitude. In the past, longitude was a bit more guesswork as clocks were not all that accurate at sea.&lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2018-10-02T10:11:22.757" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27865" PostTypeId="1" AcceptedAnswerId="27873" CreationDate="2018-10-02T14:45:40.580" Score="1" ViewCount="212" Body="&lt;p&gt;&lt;a href=&quot;https://www.theguardian.com/science/2018/oct/02/dwarf-planet-the-goblin-discovery-planet-nine-oort-cloud&quot; rel=&quot;nofollow noreferrer&quot;&gt;The recently discovered Planet Goblin&lt;/a&gt; is just 300 kms across. While I perfectly understand what it literally means, I'm having trouble wrapping this in my head. Does it mean I can imagine the planet as (approximately) a sphere for 300 kms diameter? That gives a radius of 150 kms, and thus a circumference of 942 kms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A circumference of 942 kms means that the longest distance between any two points on the surface of the planet is just 471 kms -- That's just over 1.5 times the distance between &lt;a href=&quot;https://www.google.com/search?q=london%20to%20pairs&amp;amp;ie=utf-8&amp;amp;oe=utf-8&amp;amp;client=firefox-b-ab&quot; rel=&quot;nofollow noreferrer&quot;&gt;London to Paris&lt;/a&gt; or 100 kms more than the distance between New &lt;a href=&quot;https://www.google.com/search?q=new%20york%20to%20washington%20dc&amp;amp;ie=utf-8&amp;amp;oe=utf-8&amp;amp;client=firefox-b-ab&quot; rel=&quot;nofollow noreferrer&quot;&gt;York and Washington DC&lt;/a&gt; or about half the distance between &lt;a href=&quot;https://www.google.com/search?q=st%20petersburg%20to%20moscow&amp;amp;ie=utf-8&amp;amp;oe=utf-8&amp;amp;client=firefox-b-ab&quot; rel=&quot;nofollow noreferrer&quot;&gt;St. Petersburg and Moscow&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the planet &lt;em&gt;that small&lt;/em&gt;, really? Or am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="24425" LastActivityDate="2018-10-03T12:24:47.053" Title="What does it mean for a planet to be just 300 kms across?" Tags="&lt;dwarf-planets&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27866" PostTypeId="2" ParentId="27855" CreationDate="2018-10-02T15:51:40.137" Score="1" Body="&lt;p&gt;The relatively well known process that is believed to produce most heavy atoms in the universe is the &lt;a href=&quot;https://en.wikipedia.org/wiki/R-process&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;r-process&quot;&lt;/a&gt;. The wikipedia page mentions s&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;. It has been suggested that multiple nuclear explosions would make it possible to reach the island of stability, as the affected nuclides (starting with uranium-238 as seed nuclei) would not have time to beta decay all the way to the quickly spontaneously fissioning nuclides at the line of beta stability before they absorbed more neutrons in the next explosion, thus providing a chance to reach neutron-rich superheavy nuclides like copernicium-291 and -293 which should have half-lives of centuries or millennia.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not clear (at least from this source) whether this could also happen in a supernova or a neutron star collision, or whether other processes like photodissociation might destroy the large nuclei before they reach this putative island of stability. If they did reach it, it might be quite hard to detect:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Presumably only a small proportion of the nuclei would become that heavy&lt;/li&gt;&#xA;&lt;li&gt;No one really expects the isotopes on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Island_of_stability&quot; rel=&quot;nofollow noreferrer&quot;&gt;island to actually&lt;/a&gt; be stable, just to be more stable than &quot;nearby&quot; isotopes.&lt;/li&gt;&#xA;&lt;li&gt;Since we don't have samples of these elements on Earth, we don't have accurate spectra that might enable us to detect them in space.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-10-02T15:51:40.137" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27868" PostTypeId="1" AcceptedAnswerId="27874" CreationDate="2018-10-02T19:40:08.640" Score="2" ViewCount="109" Body="&lt;p&gt;How would the Earth be affected if the Moon was destroyed as far as our orbit around the Sun? What if Mars or Jupiter was destroyed?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-10-02T21:54:49.323" LastActivityDate="2018-10-03T15:10:40.943" Title="How dependent is Earth is with the other Planets' gravity?" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;gravity&gt;&lt;solar-system&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ClosedDate="2018-10-03T17:36:08.317" ContentLicense="CC BY-SA 4.0" />
  <row Id="27869" PostTypeId="1" CreationDate="2018-10-02T21:26:13.543" Score="3" ViewCount="60" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/21P/Giacobini%E2%80%93Zinner&quot; rel=&quot;nofollow noreferrer&quot;&gt;Comet 21P/Giaccobini-Zinner&lt;/a&gt; just made its closest pass to Earth and also made its Solar perihelion on the same day (Sept 10) which is roughly one month ahead of the annual &lt;strong&gt;Draconids&lt;/strong&gt; meteor shower (Oct 9).  In both &lt;strong&gt;1933&lt;/strong&gt; and &lt;strong&gt;1946&lt;/strong&gt; these same circumstances also occurred and there were reportedly 1000s of meteors per hour in those years - &lt;strong&gt;meteor storms&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt; - I know the expert forecasts are not calling for more than 20 or so per hour based on some reference data about debris streams, but is there any chance that they may be wrong this year in their forecasts? Could the favorable circumstances this year plus a new debris stream result in a &lt;strong&gt;meteor storm surprise next week&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="22542" LastEditorUserId="14094" LastEditDate="2018-10-02T23:56:21.093" LastActivityDate="2018-10-03T20:01:13.597" Title="Will comet 21P that just passed Earth on Sept 10 produce Draconid storm on Oct 9?" Tags="&lt;star&gt;&lt;solar-system&gt;&lt;amateur-observing&gt;&lt;comets&gt;&lt;meteor&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27871" PostTypeId="2" ParentId="27862" CreationDate="2018-10-02T22:04:22.720" Score="4" Body="&lt;p&gt;Not noticeably darker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming such a globule has a mass of 50 solar masses and a diameter of 1 light year, that would make it's average density about &lt;span class=&quot;math-container&quot;&gt;$2.2\times 10^{-16}\,kg\,m^{-3}$&lt;/span&gt; which is fairly close to not being there at all in human terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An imaginary tube of this stuff 1 AU tall and of area 1 meter squared would contain about &lt;span class=&quot;math-container&quot;&gt;$3\times 10^{-5}\,kg$&lt;/span&gt; of mass.  That is an imaginary tube from Earth to the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That much matter won't block much light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparison the density of air at ground level is about &lt;span class=&quot;math-container&quot;&gt;$1.2\,kg\,m^{-3}$&lt;/span&gt;.  So less than a millimeter of air at ground level blocks more sunlight than the dust cloud would.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also &lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2018-10-02T22:04:22.720" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27872" PostTypeId="2" ParentId="18474" CreationDate="2018-10-03T03:53:45.057" Score="2" Body="&lt;p&gt;Your methodology seems to be based on an algorithm that uses successive approximations to obtain the desired values. I was not able to arrive quite at the same values as you mentioned, but I noticed an error in the posted script, &lt;code&gt;zeta = sqrt(ro^2 + x^2 + y^2)&lt;/code&gt; clearly should be &lt;code&gt;zeta = sqrt(ro^2 - x^2 - y^2)&lt;/code&gt;, so perhaps slightly different values or operations were used somewhere else too, but I couldn't find out where. I also noticed that you did not take into account the difference between Ephemeris Longitude and True Longitude in your calculation, True Longitude shifts towards the East by &lt;code&gt;1.002738 * At (Delta t) / 240&lt;/code&gt; degrees. Moreover, executing more iterations could possibly converge to a better result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, to compute the intersection of the Moon's shadow axis from Besselian Elements, I suggest using the following closed-form solution, which is derived from solving the intersection of a parametric line (Moon's shadow axis) and the Earth ellipsoid.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f = 1 / 298.257223563&#xA;b = 1 - f&#xA;sin_d = Sin(d * PI / 180)&#xA;cos_d = Cos(d * PI / 180)&#xA;y0_r = y * cos_d&#xA;z0_r = -y * sin_d&#xA;y1_r = y0_r + sin_d&#xA;z1_r = z0_r + cos_d&#xA;v = y1_r - y0_r&#xA;w = z1_r - z0_r&#xA;a_quadratic = v ^ 2 + b ^ 2 * w ^ 2&#xA;b_quadratic = 2 * z0_r * w * b ^ 2 + 2 * y0_r * v&#xA;c_quadratic = z0_r ^ 2 * b ^ 2 + y0_r ^ 2 + b ^ 2 * (x ^ 2 - 1)&#xA;p = (-b_quadratic + Sqrt(b_quadratic ^ 2 - 4 * a_quadratic * c_quadratic)) / (2 * a_quadratic)&#xA;y_i = y0_r + v * p&#xA;z_i = z0_r + w * p&#xA;dist_axis = Sqrt(x ^ 2 + z_i ^ 2)&#xA;lat = Atan(y_i / (dist_axis * (1 - f) ^ 2)) * 180 / PI&#xA;lon = Atan2(x, z_i) * 180 / PI - u + 1.002738 * At / 240&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here is the explanation of the formulas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use the flattening value to define the WGS84 ellipsoid, and calculate the semiminor-axis:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f = 1 / 298.257223563&#xA;b = 1 - f&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Convert declination angle d to radians and execute sine and cosine functions:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;sin_d = Sin(d * PI / 180)&#xA;cos_d = Cos(d * PI / 180)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Select 2 points (points 0 and 1 in the code) on the shadow axis, at z = 0 and z = 1, and apply the declination rotation (_r) on them:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;y0_r = y * cos_d&#xA;z0_r = -y * sin_d&#xA;y1_r = y0_r + sin_d&#xA;z1_r = z0_r + cos_d&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The x value always remains the same, so we don't need to recalculate it. &#xA;Next, evaluate the parametric line's vectors v,w (delta y, z) from point 0 as a base:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;v = y1_r - y0_r&#xA;w = z1_r - z0_r&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;By substituting the parametric line's equations in the ellipsoid equation, and rearranging to quadratic form and solve:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a_quadratic = v ^ 2 + b ^ 2 * w ^ 2&#xA;b_quadratic = 2 * z0_r * w * b ^ 2 + 2 * y0_r * v&#xA;c_quadratic = z0_r ^ 2 * b ^ 2 + y0_r ^ 2 + b ^ 2 * (x ^ 2 - 1)&#xA;p = (-b_quadratic + Sqrt(b_quadratic ^ 2 - 4 * a_quadratic * c_quadratic)) / (2 * a_quadratic)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;'p' here represents the distance from point 0 along the parametric line, expressed in Earth's equatorial radius.&#xA;The contents of the square root must ideally be tested beforehand for negative values to avoid exceptions if the line does not intersect the ellipsoid.&#xA;With this, we find the coordinates of our points of intersection:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;y_i = y0_r + v * p&#xA;z_i = z0_r + w * p&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then we convert this to geographic latitude and longitude:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;dist_axis = Sqrt(x ^ 2 + z_i ^ 2)&#xA;lat = Atan(y_i / (dist_axis * (1 - f) ^ 2)) * 180 / PI&#xA;lon = Atan2(x, z_i) * 180 / PI - u + 1.002738 * At / 240&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Direct conversion for the Latitude is possible here because our point is on the surface of the ellipsoid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end, you can use the following logic to capture out-of-bounds longitude values:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;If lon &amp;lt;= -180 Then&#xA;    lon += 360&#xA;ElseIf lon &amp;gt; 180 Then&#xA;    lon -= 360&#xA;End If&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In terms of accuracy, you can expect pretty good results with this method, but it does not take into account some factors, like local topography, geoid height, refraction, etc. Topography can lead to differences of a few kilometers, the geoid, +-100 meters, and refraction can bend the shadow by a few meters to a few hundred meters depending on the incident angle. Also, there is no correction for the moon's center of figure, that can shift the umbral center by several hundred meters as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the above formulas for UT1 = 18h and delta t = 72.3s, the computed intersection point is at 58.26828°N, 21.57331°W, which matches F. Espenak's values. &lt;/p&gt;&#xA;" OwnerUserId="19814" LastEditorUserId="19814" LastEditDate="2018-10-03T19:06:06.707" LastActivityDate="2018-10-03T19:06:06.707" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27873" PostTypeId="2" ParentId="27865" CreationDate="2018-10-03T12:24:47.053" Score="6" Body="&lt;p&gt;The reports that &lt;em&gt;Goblin&lt;/em&gt; is a planet are incorrect. It is a dwarf planet, at best. Dwarf planet and planet are mutually exclusive terms. Planets are objects whose masses are so large that they have collided with, ejected, or captured the vast majority of the smaller stuff that share the same orbit as the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, &lt;em&gt;Goblin&lt;/em&gt; might well not even be a dwarf planet. The distinguishing characteristics of a dwarf planet is an object that is large enough to have pulled itself into a more or less spherical shape via self gravitation but is not anywhere near large enough to have the potential to clear the neighborhood of its orbit about the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The threshold that distinguishes dwarf planets from small solar system bodies is the &lt;a href=&quot;https://arxiv.org/pdf/1004.1091.pdf&quot; rel=&quot;noreferrer&quot;&gt;potato radius&lt;/a&gt;. Objects somewhat smaller than the potato radius look like lumpy potatoes, while objects larger than that have sufficient mass to more or less pull themselves into hydrostatic equilibrium. The potato radius depends a bit on composition, and probably also depends on distance from the Sun. (Water ice is quite rocky at trans-Neptunian distances from the Sun.) The potato radius for rocky objects is about 300 km (diameter = 600 km), while for icy objects it's about 200 km (diameter = 400 km).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With an estimated diameter of 300 km, &lt;em&gt;Goblin&lt;/em&gt; is probably too small to qualify as a dwarf planet, let alone as a planet.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2018-10-03T12:24:47.053" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27874" PostTypeId="2" ParentId="27868" CreationDate="2018-10-03T15:10:40.943" Score="1" Body="&lt;p&gt;The orbital speed of the Moon around the Earth is 1022 m/s. The Earth is 81 times more massive, than the Moon. Thus, a sudden disappearance of the Moon would cause an &lt;span class=&quot;math-container&quot;&gt;$\frac{1022}{81}=12 \frac{m}{s}$&lt;/span&gt; change of the Earth's orbit, which would be negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The disappearance of the Mars or Jupiter would have yet lesser effect.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastActivityDate="2018-10-03T15:10:40.943" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27875" PostTypeId="1" AcceptedAnswerId="27935" CreationDate="2018-10-03T16:32:52.207" Score="2" ViewCount="246" Body="&lt;p&gt;@DavidHammen's &lt;a href=&quot;https://astronomy.stackexchange.com/a/27873/7982&quot;&gt;answer&lt;/a&gt; explains that Goblin has roughly a potato radius and I can't believe I just wrote Goblin and potato in a serious question. That happened.&lt;/p&gt;&#xA;&lt;p&gt;The linked paper; &lt;a href=&quot;https://arxiv.org/abs/1004.1091&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Potato Radius: a Lower Minimum Size for Dwarf Planets&lt;/a&gt; (Lineweaver &amp;amp; Norman 2010) says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Derivation of the Potato Radius&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;One could begin, as the author did initially, by setting the gravitational acceleration at the surface of a body equal to the acceleration needed to undo the electronic bonds. But this is not what we want since it is equating surface gravity with the force needed to crush rocks. The Earth is spherical but its gravitational pull is not so strong that it crushes rock at the surface. One has to go down ~10 or ~20 kilometers before the overburden pressure can do that.&lt;/p&gt;&#xA;&lt;p&gt;To our knowledge, the published derivation of something most analogous to the potato radius is in Chap. 1.2.1 of Stevenson (2009). &lt;strong&gt;Focusing on energy (not force or pressure) the electronic bond energy of ~ 1 eV is set equal to the gravitational energy ~ GMµ/R where µ is the mass of the particle in question and R is the radius of the object&lt;/strong&gt;. The result is a radius of a few thousand kilometers for a rocky body and ~ 1000 km for an icy body. This was meant to be only an order of magnitude calculation. We try to improve on it below.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;We have there&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$1 \text{eV} \approx \frac{GM\mu}{R} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;1 eV is about 1.6E-19 Joules and &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; is the gravitational constant, but how to I use this to arrive at a potato radius &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; of about 200 to 300 km? I can read the words in the quote, but I don't understand how to proceed.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-10-07T20:43:02.383" Title="How did early estimates of a &quot;potato radius&quot; set 1 eV ~ GMμ/R and get 200 to 300 km?" Tags="&lt;planet&gt;&lt;planetary-science&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27876" PostTypeId="1" AcceptedAnswerId="27883" CreationDate="2018-10-03T18:23:19.323" Score="1" ViewCount="357" Body="&lt;p&gt;Let say that the Earth's Moon was knocked out of orbit and landed on the Earth with some damage but not world ending. Would Earth's atmosphere envelop to include the Moon? Would the moon sink into the Earth? What would happen?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastActivityDate="2018-10-04T18:59:57.880" Title="Could the Moon sit on the Earth?" Tags="&lt;the-moon&gt;&lt;gravity&gt;&lt;earth&gt;&lt;planetary-atmosphere&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27877" PostTypeId="2" ParentId="27876" CreationDate="2018-10-03T18:47:24.137" Score="4" Body="&lt;p&gt;It can't happen. The moon is quite big and can't get knocked out of orbit easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were somehow knocked out of orbit and collided with the Earth, the energy released would melt the entire crust. Huge amounts of matter would be knocked into space. After everything had settled down, you would get a lifeless world, still spherical but now slightly larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon can't get knocked out of orbit.  If it did, It can't &quot;land&quot;. If it did &quot;land&quot; it would be crushed by its own weight, and all life would end in one way or another.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-10-03T18:47:24.137" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27878" PostTypeId="2" ParentId="27837" CreationDate="2018-10-03T19:21:57.047" Score="5" Body="&lt;p&gt;This is a neat question in that it made me think deeper about the word &quot;orbit&quot;. The answer that mentions barycenters has a point, but in the end, it is a bit about references frames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take a relatively uncontroversial statement and make it controversial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The Moon orbits the Earth&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would have never challenged this statement, except that once I plotted the path of the Moon and the Earth together relative to the SSB, I realized that the Moon is actually orbiting the sun (SSB) and is merely being heavily perturbed because it happens to be in close proximity to the Earth's gravitational influence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we zoom out we would next say that we are all merely orbiting the galactic core and simply because we are in proximity to the Sun, our galactic orbit is heavily perturbed by the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, depending on the relative sizes of the influences we can make practical (sensible) inferences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, does &quot;Mars orbit the Earth&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, because if we place a reference frame at Mars and try and explain the motion of the Earth and the Sun we get nonsensical results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we place a reference frame on the Sun, or at the SSB and then try to explain the motion of the Earth and Sun it becomes much clearer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does &quot;the Moon orbit the Earth&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, yes, if we place a reference at the Earth we can explain the motion of the Moon pretty sensibly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an article that is decently written about such frames of reference questions: &lt;a href=&quot;https://www.wired.com/2012/12/does-the-moon-orbit-the-sun-or-the-earth/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.wired.com/2012/12/does-the-moon-orbit-the-sun-or-the-earth/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the video suggested by @uhoh:&lt;br&gt;&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=z52WWLE8bBo&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=z52WWLE8bBo&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19292" LastEditorUserId="19292" LastEditDate="2020-03-26T16:47:07.060" LastActivityDate="2020-03-26T16:47:07.060" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27880" PostTypeId="2" ParentId="27869" CreationDate="2018-10-03T20:01:13.597" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://earthsky.org/?p=180611&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earthsky&lt;/a&gt; discusses this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There isn't a storm every perihelion, and predicting meteors is tricky, as we can't observe the meteoroids in space, we don't know if the Earth will pass through a dense patch or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Key quote &quot;How many Draconid meteors will you be able to catch? No one expects a Draconid storm this year, but one can always hope!&quot;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-10-03T20:01:13.597" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27881" PostTypeId="1" AcceptedAnswerId="27882" CreationDate="2018-10-03T22:27:19.307" Score="1" ViewCount="169" Body="&lt;p&gt;I'm not talking about tomorrow but in the far far future could we say our Sun could collide with another star or otherwise within our current detectable range? &lt;/p&gt;&#xA;" OwnerUserId="15415" LastActivityDate="2018-10-04T09:13:31.240" Title="Are there any stars that one day we could collide with?" Tags="&lt;star&gt;&lt;galaxy&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27882" PostTypeId="2" ParentId="27881" CreationDate="2018-10-03T22:54:01.263" Score="2" Body="&lt;p&gt;Statistically speaking, the Sun will probably not crash into any other stars in the near future. Currently, we do not believe the Sun is on a collision course with any other star either, but we can only predict so far due to Chaos Theory and what I believe is called Lyapunov Time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this is that solar systemic distances are minuscule compared to interstellar distances. The distance from the Earth to the Sun is 1 AU, or 150,000,000 km. Our sun is 0.004 AU in radius. The nearest star (which isn't on a collision course) is approximately 275,000 AU from the Sun. As you can see stars are generally far apart, and thus they rarely interact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The locations of the stars do, in fact, change over hundreds of thousands of years, but again, stars generally stay far apart. That is not to say that stars never collide. However, it only happens in very dense globular clusters, young star systems (where they may form close), and in contact stars, as well as the occasional exception, which may be 1-2 in the entire galaxy when Andromeda collides with our Milky Way Galaxy in a few billion years (the Sun might already be dead though).&lt;/p&gt;&#xA;" OwnerUserId="24373" LastActivityDate="2018-10-03T22:54:01.263" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27883" PostTypeId="2" ParentId="27876" CreationDate="2018-10-03T23:00:51.800" Score="4" Body="&lt;p&gt;This is a highly improbable scenario, since the Moon has a lot of momentum. However, let's say you teleport the Moon onto the surface of the Earth. Then Moon can't just sit on the Earth, and it will collapse under its own weight, probably killing everyone on Earth. The two would probably merge into a spherical shape (as most massive objects do).&lt;/p&gt;&#xA;" OwnerUserId="24373" LastEditorUserId="907" LastEditDate="2018-10-04T18:59:57.880" LastActivityDate="2018-10-04T18:59:57.880" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27884" PostTypeId="2" ParentId="27876" CreationDate="2018-10-03T23:11:24.183" Score="4" Body="&lt;p&gt;At the scale of the Earth or the Moon there are no solid objects. The Earth does not behave like a solid, nor does the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to think of these objects as more like piles of sand, made round by their own gravity. So if one round pile of sand touches another round pile of sand, they just merge and make a bigger pile of sand, also round.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the Earth's Moon was knocked out of orbit and landed on the Earth with&#xA;  some damage but not world ending&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's essentially impossible. I can't think of any mechanism that would allow the Moon to gently touch the Earth. Regardless, both the Earth and the Moon would be severely bent out of shape before they even begin to touch - they would look more like eggs, or a teardrop shape, then they merge, and that's the end of everyone and everything on this planet.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2018-10-03T23:11:24.183" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27885" PostTypeId="2" ParentId="27825" CreationDate="2018-10-03T23:12:02.540" Score="2" Body="&lt;p&gt;It's a lot easier to see detail on Jupiter, or pick out the rings of Saturn than it is on Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although Mars is closer to us, Jupiter and Saturn are a LOT bigger - and so appear larger. And because Mars is closer to us, its apparent size changes a lot berween closest approach (opposition) and when it's further away. When it's close to opposition, it's significantly larger than when its further away; From memory, I think there's a least a 3X difference in apparent size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even when it's close, it's still pretty small. And unlike Jupiter and Saturn, where the details are easy to see, Mars tends to need more patience and practice. Look at jupiter or Saturn and the details are immediately obvious. Look at Mars, and you really want to keep looking for several minutes. Most of the time you may not see much, but every so often you'll get a bit of good seeing and be able to glimpse some detail. (planets are also best seen when higher above the horizon; when it's low, you're looking through much more atmospheric crud and turbulence).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should certainly be able to see the rihgs of Saturn and details on Jupiter with your current scope - but even at high magnifications the planets are still going to be small.  We're spoilt by the full page images we see in magazine photos - through the eyepiece things aren't going to be anything like that big. Expect to see a small planet in the middle of the eyepiece - your O is typical :) .it may not be big, but can be nicely detailed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bigger aperture scope has more resolution, allowing you to see finer details, and allows more magnification before you get to the point where increasing magnification just makes things bigger and blurrier without showing more detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical maximum everyday magnification is usually around your aperture in mm, which you get with an eyepiece focal length equal to your scopes f-ratio (so 8mm for an f8 scope, and so on).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With very good seeing (steady atmosphere), which can be rare, you can go to around twice the magnification. This is what manufacturers usually quote as the maximum magnification; what they don't say is that it requires near perfect conditions which may only happen a few times a year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a N11GPS, the model before the CPCs. Views are nice, and it's a good light bucket but it's a bulky and heavy beast that's a pain to move any great distance. Not a problem if you can leave it set up in an observatory, but if you have to take it in and out then you might want to consider some sort of trolley or balance better views with portability - I think the 8&quot; fork mounted SCTS are probably the sweet spot between performance and portability.  I can lift an 8&quot; on its tripod, but I have to move the 11&quot; OTA+Mount and tripod separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the specs quote 661x as the maximum magnification, I've found that to be optimistic in practice - while I can use a 12mm eyepiece all the time (233x) It's very rare that I've been able to use an 8mm (350x) and I don't remember ever being able to use a 5mm eyepiece (576x). I think the quoted 661x is being wildly optimistic, at least for the seeing conditions where I am,  but those can vary significantly with location.  Just remember that a lot of the time, the atmosphere will limit how much magnification you can use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But also remember, that for a lot of deep sky objects, it's more about light gathering than it is about high magnifications. Many objects work better at lower magnifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By the way, magnification = focal length of scope / focal length of eyepiece, so a 10 mm eyepiece in a 700mm focal length scope will give you 700/10 = 70x magnification. Adding a 2x barlow doubles that, a 3x triples it, and using both multiplies by six - which is likely to be way over the useful magnification for the scope and won't do wonders for the image quality...)&lt;/p&gt;&#xA;" OwnerUserId="17016" LastEditorUserId="17016" LastEditDate="2018-10-04T00:09:44.777" LastActivityDate="2018-10-04T00:09:44.777" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27886" PostTypeId="1" CreationDate="2018-10-04T00:51:02.563" Score="2" ViewCount="190" Body="&lt;p&gt;I was looking at the sun during a sunset earlier today and wondered how much atmosphere you need to be able to look directly at the sun without damaging your eyes. The thought crossed my mind because I was continuously driving into an extremely oppressive sunset, my eyes were battered by the sun continuously for about a half an hour. Then, suddenly, the sun became more and more dull as it lowered into the sky. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the approximate ratio of atmosphere between the sun and me, when setting as compared to when its positioned at a peak point in the sky? If you have the math behind an approximation like this from a topological observer, I would love to see that worked out.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="23729" LastActivityDate="2018-10-05T07:04:23.843" Title="How much more atmosphere is the sun passing through at sunset as opposed to at its peak in the sky?" Tags="&lt;the-sun&gt;&lt;horizon&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27888" PostTypeId="1" AcceptedAnswerId="35129" CreationDate="2018-10-04T01:43:00.727" Score="8" ViewCount="135" Body="&lt;p&gt;A while back NASA made a site called &lt;a href=&quot;https://solarsystem.nasa.gov/planets/dwarf-planets/pluto/plutotime/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pluto Time&lt;/a&gt;, where you can enter your address and find that &quot;...moment near dawn and dusk each day, where the illumination on Earth matches that of high noon on Pluto.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a generic formula where we can calculate this for any body in the solar system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about any body around any star?&lt;/p&gt;&#xA;" OwnerUserId="19880" LastEditorUserId="9527" LastEditDate="2020-02-20T10:12:48.973" LastActivityDate="2020-02-20T10:12:48.973" Title="&quot;Pluto Time&quot; for any solar system body" Tags="&lt;the-sun&gt;&lt;planet&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27890" PostTypeId="1" CreationDate="2018-10-04T04:02:23.210" Score="1" ViewCount="55" Body="&lt;p&gt;According to wikipedia (&lt;a href=&quot;https://en.wikipedia.org/wiki/Coma_Cluster&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Coma_Cluster&lt;/a&gt;), Coma cluster contains over 1000 galaxies. Is there any upto date catalogue of these galaxies with the following given parameters &lt;code&gt;(Ra, Dec, z, magnitude, position angle, major diameter, minor diameter)&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="24447" LastActivityDate="2018-10-04T04:02:23.210" Title="Coma cluster of galaxies" Tags="&lt;galaxy-cluster&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27891" PostTypeId="1" AcceptedAnswerId="27896" CreationDate="2018-10-04T04:15:23.890" Score="2" ViewCount="106" Body="&lt;p&gt;The new paper in Science Advances &lt;a href=&quot;http://advances.sciencemag.org/content/4/10/eaav1784&quot; rel=&quot;nofollow noreferrer&quot;&gt;Evidence for a large exomoon orbiting Kepler-1625b&lt;/a&gt; (open access!) uses a sophisticated combination of occultation light curve simulation, signal processing and statistical analysis to show that there is a significant chance that they have identified an exomoon, a moon orbiting an exoplanet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's too sophisticated for me to understand how this Hubble data indicates a potential moon. Terms like &quot;detrending&quot;,  &quot;Markov chain Monte Carlo&quot;, and &quot;Bayesian evidences&quot; are beyond me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to choose one of those plots and add an arrow to say &quot;this blip here, this is Kepler-1625b potential moon&quot;? If not, is it possible to at least explain in a simple way what it is from this analysis that has led them to believe there may be a moon there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Abstract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Exomoons are the natural satellites of planets orbiting stars outside our solar system, of which there are currently no confirmed examples. We present new observations of a candidate exomoon associated with Kepler-1625b using the Hubble Space Telescope to validate or refute the moon’s presence. We find evidence in favor of the moon hypothesis, based on timing deviations and a flux decrement from the star consistent with a large transiting exomoon. Self-consistent photodynamical modeling suggests that the planet is likely several Jupiter masses, while the exomoon has a mass and radius similar to Neptune. Since our inference is dominated by a single but highly precise Hubble epoch, we advocate for future monitoring of the system to check model predictions and confirm repetition of the moon-like signal. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-10-04T04:20:40.623" LastActivityDate="2018-12-05T08:45:10.250" Title="Where is the exomoon? (Possible discovery of a companion to Kepler-1625b)" Tags="&lt;exoplanet&gt;&lt;hubble-telescope&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27892" PostTypeId="2" ParentId="27891" CreationDate="2018-10-04T07:16:00.400" Score="0" Body="&lt;p&gt;Monte Carlo refers to a technique of running multiple simulations, each with the input data very slightly changed, usually one variable at a time. After teens or hundreds of thousands of simulations you end up with a probability plot - if these trend towards a particular distribution you can gain confidence that the data is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bayesian analysis looks more at similarities (which is why it is often used for identifying spam: if it looks more like these thousand examples of spam than those thousand valid emails then it probably is spam)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this work uses probability theory, and the specific tools mentioned to provide a likelihood measure. And in this case, the balance of probability from the data is that it is an exomoon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there will not be a blip that looks like a moon - there will be a vast cloud of data that happens to be more similar to possible models of data expected from an exomoon than from other alternatives considered.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2018-10-04T07:16:00.400" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27893" PostTypeId="2" ParentId="27881" CreationDate="2018-10-04T07:56:22.787" Score="1" Body="&lt;p&gt;Currently there are no stars in our vicinity that will enter the solar system or even collide with the sun in the, astronomically speaking, near future. Calculations can only be made for a limited time as they become more and more uncertain to more we look into the future &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Near-stars-past-future-en.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;but maybe this diagram is helpful for you&lt;/a&gt; which depicts the distance of some of our closests stellar neighbors over the next 80000 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coincidentially I stumbled over a Wikipedia article about &lt;a href=&quot;https://en.wikipedia.org/wiki/Gliese_710&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gliese 710&lt;/a&gt;.&#xA;In 1.5 million years, this red dwarf will come very near to the Solar System, possibly even cross the Kuiper Belt. It is of course still very uncertain how close it will come but it is possible that it disturbs the orbits of object in the Oort Cloud or the Kuiper Belt and thus sends them flying towards the inner Solar System.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although it is no collision it is as close to one as we currently know.&lt;/p&gt;&#xA;" OwnerUserId="22927" LastEditorUserId="22927" LastEditDate="2018-10-04T09:13:31.240" LastActivityDate="2018-10-04T09:13:31.240" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27894" PostTypeId="2" ParentId="18753" CreationDate="2018-10-04T09:07:40.600" Score="-2" Body="&lt;p&gt;The reason the night sky is dark is our distance from the nearest stars (over 4 light years) which would be as bright as our sun (93 million miles) if they were not over 2500 times further away and only the brightness of a star, not the sun.&lt;/p&gt;&#xA;" OwnerUserId="24448" LastEditorDisplayName="user1569" LastEditDate="2018-10-04T11:29:21.343" LastActivityDate="2018-10-04T11:29:21.343" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27895" PostTypeId="2" ParentId="27886" CreationDate="2018-10-04T10:32:13.787" Score="4" Body="&lt;p&gt;The amount of atmosphere that light from outside Earth has to pass through before it reaches the ground is given by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Air_mass_(astronomy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;air mass&lt;/a&gt;, &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt;. It is normalized to unity for a source directly overhead (at &quot;zenith&quot;), and increases as the source &quot;sets&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't look too close to the horizon, you can use the formula &lt;span class=&quot;math-container&quot;&gt;$X = \sec z$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\sec \equiv 1/\cos$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; is the angle from zenith. For instance, if the Sun is &lt;span class=&quot;math-container&quot;&gt;$30^\circ$&lt;/span&gt; above the horizon, so that &lt;span class=&quot;math-container&quot;&gt;$z=60^\circ$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$X = 2$&lt;/span&gt;, i.e. its light passes through twice the amount of atmosphere compared to when its at zenith.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula above makes (at least) three approximations, however. Firstly, it assumes a &lt;em&gt;plane-parallel&lt;/em&gt; atmosphere (an infinite, flat Earth), and so &lt;span class=&quot;math-container&quot;&gt;$X\rightarrow\infty$&lt;/span&gt; as the Sun sets. Secondly, it assumes that the atmosphere is homogeneous, whereas in fact its density decreases with distance from Earth. This not only causes a non-uniform absorption&lt;span class=&quot;math-container&quot;&gt;$^\dagger$&lt;/span&gt; along the ray, but also causes refraction which increases the path length for high &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;. For astronomers, these approximations don't matter much, since they're quite accurate out to &lt;span class=&quot;math-container&quot;&gt;$z\sim70^\circ \text{–}\,80^\circ$&lt;/span&gt; and we usually don't observe at greater angles anyway, precisely because of the large air mass. But &lt;a href=&quot;https://en.wikipedia.org/wiki/Air_mass_(astronomy)#Interpolative_formulas&quot; rel=&quot;nofollow noreferrer&quot;&gt;several more exact formulas&lt;/a&gt; based on &lt;a href=&quot;https://en.wikipedia.org/wiki/Air_mass_(astronomy)#Atmospheric_models&quot; rel=&quot;nofollow noreferrer&quot;&gt;various models of the atmosphere&lt;/a&gt; have been created to fit observed air masses. These models converge roughly to &lt;span class=&quot;math-container&quot;&gt;$X \simeq 38$&lt;/span&gt; for &lt;span class=&quot;math-container&quot;&gt;$z=90^\circ$&lt;/span&gt;. In other words, &lt;strong&gt;the amount of atmosphere that light goes through for a source at the horizon, compared to a source at zenith, is roughly 38&lt;/strong&gt;. However, note that just &lt;span class=&quot;math-container&quot;&gt;$10^\circ$&lt;/span&gt; above the horizon, &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; is already &lt;span class=&quot;math-container&quot;&gt;$&amp;lt;6$&lt;/span&gt;, so it's the last few degrees that really matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact value depends on many factors, including temperature (and its change with height), position on Earth, and the altitude of the observer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^\dagger$&lt;/span&gt;&lt;sub&gt;Most photons are actually not &lt;em&gt;absorbed&lt;/em&gt;, but rather scattered out of the line of sight. Effectively, the result is the same; photons are removed from the ray. The scattered photons then become part of the background (and since blue photons scatter more than red, the sky is blue, and the Sun becomes redder as it sets).&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-10-05T07:04:23.843" LastActivityDate="2018-10-05T07:04:23.843" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27896" PostTypeId="2" ParentId="27891" CreationDate="2018-10-04T17:31:00.570" Score="3" Body="&lt;p&gt;&quot;Detrending&quot; refers to removing systematic effects that are caused by the spacecraft, detector or the environment and which are not part of the particular star. The systematic trends can be caused by a large variety of things; the main ones affecting Kepler are drifts in position and focus, thruster firings to compensate for angular momentum build up in the reaction wheels and the roll of the spacecraft every quarter. Because these trends are systematic and affect all the stars that Kepler sees in a similar way, the effects can be modeled and the trends removed - detrending. There are some examples in slide 15 and 16 of &lt;a href=&quot;http://nexsci.caltech.edu/workshop/2012/talks/MartinStill.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Kepler workshop presentation&lt;/a&gt; which shows the &quot;raw&quot; brightness measurements (labeled 'SAP') and the detrended data on the right (labeled 'PDC').&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see examples of this in the &lt;a href=&quot;http://advances.sciencemag.org/content/advances/suppl/2018/10/01/4.10.eaav1784.DC1/aav1784_SM.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Supplementary Materials&lt;/a&gt;. On page 47 in Figure S2, you can see the data before detrending and you can see the dips caused by the transiting exoplanet (Kepler-1625b) but these are superimposed on long-term upwards or downwards slopes or curves. Detrending removes these effects (which will be common to all the stars, not just to Kepler-1625b's star) allowing you to get corrected flat light curves, enabling the search for the much smaller exomoon signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The evidence for the exomoon is in two parts. The timing of the planet transit that was observed by HST in October 2017 was 77.8 minutes early compared to the predicted time from the Kepler transits. This is shown (but not very well) in Figure S12 in the Supplementary Materials. If there was no exomoon or anything else tugging on Kepler-1625b, we would have expected the HST transit (the data point at transit epoch 7) to be on the extended black prediction line at an O-C value (y axis) of +77.8 minutes, ( off the top of their plot as it scaled). The second piece of evidence is that is if the transit is early, then they should expect to see the shallower/smaller transit of the exomoon after the transit as it would be on the other side of the orbit with the exoplanet. This is what they show in the bottom half of Figure 4; no matter which detrending method they use for the HST data, they see a small shallow dip after the main transit is over, centered around a time (x value) of 3056.25. This gives more confidence that the exomoon interpretation is real and not an artifact of the detrending process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Rnl7U.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rnl7U.png&quot; alt=&quot;Modified Figure 4 of paper show proposed exomoon transit&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What they are saying after Table 1 is that the Δχ2 (the measure of the change in the goodness of fit of the model as you add extra things to the model) is better with the moon models (models Z and M) over one that just accounts for the transit timing variations (model T). This is because the presence of the exomoon is predicted to cause changes in the length and depth of the exoplanet transit but these are harder to measure than the change in timing.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastEditorUserId="7982" LastEditDate="2018-12-05T08:45:10.250" LastActivityDate="2018-12-05T08:45:10.250" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27897" PostTypeId="1" CreationDate="2018-10-04T18:59:28.527" Score="2" ViewCount="1099" Body="&lt;p&gt;I'm making a very basic orbit simulator in C# using the Helix Toolkit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to position the Helix lighting object at the sun, but also be able to calculate solar eclipse by the moon and earth for a satellite (position defined in ECEF).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon some brief research, I'm reading that getting an accurate position calculation for the sun in ECEF is a lot of work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an algorithm that can perform this calculation for me that I can include into an UpdateSunPosition() method so that I can have a reasonably accurate position calculation of the sun in ECEF?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps I have to scratch the eclipse calculation ambitions for now, instead calculate Geographical Position of the sun to get the Declination and Greenwich Hour Angle, Call those latitude and longitude, and then just give it a satisfactory altitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've noticed that people are referring to SOFA a lot when talking about this, but I'm not sure what SOFA is, or whether I can use it in C#. I'm pretty sure it doesn't operate in a .NET environment like visual studio, but please correct me if I'm wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I appreciate any and all suggestions or insights anyone may be able to offer here. &lt;/p&gt;&#xA;" OwnerUserId="24452" LastActivityDate="2019-09-02T17:37:07.827" Title="Calculate position of the Sun in ECEF" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;coordinate&gt;&lt;frame-of-reference&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27898" PostTypeId="1" AcceptedAnswerId="27899" CreationDate="2018-10-04T19:07:22.700" Score="1" ViewCount="165" Body="&lt;p&gt;Of all the different ways that the Earth can be destroyed what ways is the most likely that could happen?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastActivityDate="2018-10-04T20:50:20.940" Title="What cosmic disaster is most likely to happen?" Tags="&lt;black-hole&gt;&lt;earth&gt;&lt;asteroids&gt;&lt;comets&gt;&lt;flare&gt;" AnswerCount="1" CommentCount="9" ClosedDate="2018-10-05T07:47:08.667" ContentLicense="CC BY-SA 4.0" />
  <row Id="27899" PostTypeId="2" ParentId="27898" CreationDate="2018-10-04T20:24:55.417" Score="7" Body="&lt;p&gt;Asteroid impacts are a possibility, but a killer asteroid typically affects the Earth every few dozen million years. This would probably destroy all life as we know it, but life will probably nonetheless persist. Statistically, by 500,000 years from now, we should have been hit by an asteroid 1 km in diameter, and 100,000,000 years from now, we should have been hit by one that is the size of the one that killed the dinosaurs. Statistically it's very unlikely we will get one that completely obliterates the Earth, causing it to be completely and utterly destroyed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gamma-ray bursts may irradiate the Earth to the point that complex life will no longer be able to survive, and only very resilient organisms like waterbears may live. At some point in the next 300,000 years, WR 104 is expected to go supernova, and if the poles are somewhere around 12 degrees or lower  we could be in the path of the supernova. We aren't 100% sure what its axis is angled at though. Either way, it won't actually destroy the Earth, but just irradiate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In roughly 7.5 billion years, the Earth will probably have been destroyed by falling into the Sun, as it engulfs the inner planets. This is nearly certain, but if somehow, the Earth-Moon system were to survive, then 50 billion years from now, the Earth and Moon will have become tidally locked. Then the Sun would extract angular momentum from the system as a white dwarf, and will cause the Earth's spin in accelerate and the Moon's orbit to decay. If this happens, then roughly 65 billion years from now the Moon will crash into the Earth, which would probably destroy it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that there are many more ways that life as we know it could end, but these are not cosmic disasters, but are geologic and biological disasters, so they are not included. Generally speaking, it's very hard to completely destroy the Earth.&lt;/p&gt;&#xA;" OwnerUserId="24373" LastEditorUserId="24373" LastEditDate="2018-10-04T20:50:20.940" LastActivityDate="2018-10-04T20:50:20.940" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27900" PostTypeId="1" AcceptedAnswerId="27902" CreationDate="2018-10-04T20:33:23.777" Score="3" ViewCount="131" Body="&lt;p&gt;What I'm asking is if the tilt of a planet can &quot;follow&quot; the rotation around its star so that, for example, its north pole would always lean towards the star, and thus always be under daylight. The planet could have a tilt similar to earth's 23.5°. And can that planet have a rotation around its axis, allowing for day/night cycles closer to the equator? Hopefully this image will explain what I mean. The tilt (red line) stays the same, but the pole is leaning toward the star throughout the whole year, while the planet rotates around its axis.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/hdO4E.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hdO4E.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24454" LastEditorUserId="23940" LastEditDate="2018-10-06T17:10:00.907" LastActivityDate="2018-10-06T17:10:00.907" Title="Can a planet's changing tilt be aligned with the rotation around its star?" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;rotation&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2018-10-07T10:14:30.323" ContentLicense="CC BY-SA 4.0" />
  <row Id="27901" PostTypeId="2" ParentId="27900" CreationDate="2018-10-04T20:47:22.180" Score="-3" Body="&lt;p&gt;Kinda. Uranus has an axial tilt of 98&lt;span class=&quot;math-container&quot;&gt;$^\circ$&lt;/span&gt;, which means that it is only 8&lt;span class=&quot;math-container&quot;&gt;$^\circ$&lt;/span&gt; off of having a pole that faces exactly the Sun. However, it wasn't formed that way. We predict that large objects, probably protoplanets, struck Uranus early in its life, which caused its axial tilt to be shifted almost sideways. Uranus definitely has day night cycles (since the 8&lt;span class=&quot;math-container&quot;&gt;$^\circ$&lt;/span&gt; give some leeway), but, just like the Earth has midnight days and polar nights close to the poles, Uranus will have the same, but with a much larger range. In fact, only the equator and places near the equator would be spared from this, just as you described.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the pole is not locked to face the star, so I guess the answer is no because of angular torque, there's more to this &lt;a href=&quot;https://astronomy.stackexchange.com/questions/18399/are-there-any-known-planets-whose-axis-is-oriented-such-that-one-pole-always-fac&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="24373" LastEditorUserId="24373" LastEditDate="2018-10-04T21:54:35.363" LastActivityDate="2018-10-04T21:54:35.363" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27902" PostTypeId="2" ParentId="27900" CreationDate="2018-10-04T21:47:25.747" Score="9" Body="&lt;p&gt;No, or at least not for a planet of any size. To change the absolute direction of the axis of rotation of a planet needs a large force acting in some way asymmetrically. In the case of the Earth this comes from the gravity of the sun and the moon acting on the Earth's equatorial bulge and it takes 26000 years to rotate the Earth's axis around a cone 46 degrees wide. To rotate the axis in a full circle (a cone 180 degrees wide) in a single year would only be imaginable for a very irregular planet (which means a small one, since big ones collapse into spheres under their own gravity) probably with a very large and close moon to pull on it.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-10-04T21:47:25.747" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27904" PostTypeId="1" AcceptedAnswerId="27905" CreationDate="2018-10-05T14:19:35.383" Score="1" ViewCount="687" Body="&lt;p&gt;If in the future a spaceship can accelerate to nearly the speed of light, will it encounter any kind of resistance/drag from the spacetime in front of it not being able to bend/warp fast enough around it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is my understanding that as an object accelerates, its mass will increase and this increase in mass should result in the object having more gravity, and this gravity will continue to increase as the object's speed increases. The creation of more and more gravity should result in an increased area of spacetime surrounding the object being bent/warped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if there is a limit to how fast spacetime can bend/warp around an accelerating object, along with an increased area of spacetime being affected by an accelerating object's increased gravity, will this spaceship start to heat up and begin to melt the faster it accelerates? &lt;/p&gt;&#xA;" OwnerDisplayName="user23327" LastActivityDate="2018-10-05T17:51:50.640" Title="Is there a limit to how fast spacetime can bend/warp around an accelerating object?" Tags="&lt;gravity&gt;&lt;astrophysics&gt;&lt;space-time&gt;&lt;general-relativity&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27905" PostTypeId="2" ParentId="27904" CreationDate="2018-10-05T14:28:19.747" Score="3" Body="&lt;p&gt;Changes in the bending of spacetime propagate through spacetime at the speed of light. If I wave my hand &lt;em&gt;thus&lt;/em&gt; the (very, very, very tiny) changes to spacetime will propagate outwards as gravitational waves at the speed of light and won't reach the Sun for 8 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No matter how massive an object or how great the acceleration, spacetime elsewhere only &quot;feels&quot; that acceleration after it has passed into the acceleration's lightcone.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-10-05T14:28:19.747" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27906" PostTypeId="2" ParentId="27897" CreationDate="2018-10-05T16:13:51.287" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://www.iausofa.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;SOFA&lt;/a&gt; is the Standards Of Fundamental Astronomy, approved by the International Astronomical Union (IAU) as being the canonical standard for calculating positions and time in astronomy. It is available in C and F77 versions as standard, and wrappers have been written in other languages, most notably Python as part of AstroPy. For your purpose, the thirdparty part of the website mentions the &lt;a href=&quot;https://github.com/abrudana/wwa&quot; rel=&quot;nofollow noreferrer&quot;&gt;World Wide Astronomy (WWA) library&lt;/a&gt; which is C# so that would be worth a look for your application. I would recommend reading the &lt;a href=&quot;http://www.iausofa.org/cookbooks.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;SOFA Cookbooks&lt;/a&gt;, particularly the ones on Earth Attitude and Time Scales to get a better idea of how the (many) various time and co-ordinate systems fit together.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-10-05T16:13:51.287" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27907" PostTypeId="2" ParentId="27904" CreationDate="2018-10-05T17:51:50.640" Score="1" Body="&lt;p&gt;I have nothing to add to Mark Olson's answer except to point out that there are far more severe problems to contend with if you want to have a spaceship travelling at highly relativistic speeds &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cosmic microwave background will be blue shifted, boosted and beamed such that it might become a hazardous source of X-rays/gamma rays. Of course these problems would be much, much more severe should you fly towards anything brighter - like another star!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, cosmic dust and debris is going to become a serious problem. Any relativistic mass increase your ship manages will be mirrored by an increase in the kinetic energy of anything that runs into you (from the point of view of the rest frame of the ship). &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-10-05T17:51:50.640" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27908" PostTypeId="1" CreationDate="2018-10-05T18:28:25.077" Score="1" ViewCount="107" Body="&lt;p&gt;&lt;strong&gt;Overview:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking a planet Ganymede for example (moon, technically I suppose). Ganymede is huge (well, in contrast to other bodies) and I've noticed a pattern between larger bodies and atmospheres. Usually it seems that the larger a body is, the more likely it will have an atmosphere (I know this isn't true, but it definitely seems true, with obvious exceptions). Even when playing Kerbal space program I came up on Vall expecting to be slowed due to its huge size. Big mistake, slammed into the surface. So, if size isn't the end-all-be-all of why a planet would have an atmosphere, what really causes it? Why wouldn't something like Ganymede, doubly protected by it's own magnetic field and Jupiter's field not have an atmosphere? What causes the planet to lose it once an atmosphere is created? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;You can ignore my overview rambling, I suppose my one-liner &lt;em&gt;focused&lt;/em&gt; question is:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the real conditions for the creation of an atmosphere/loss of an existing atmosphere?&lt;/p&gt;&#xA;" OwnerUserId="23729" LastEditorUserId="7982" LastEditDate="2021-04-09T23:20:08.150" LastActivityDate="2021-04-11T02:03:24.713" Title="What are the real conditions for the creation of an atmosphere/loss of an existing atmosphere?" Tags="&lt;planetary-atmosphere&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27909" PostTypeId="2" ParentId="27908" CreationDate="2018-10-05T20:34:14.427" Score="2" Body="&lt;p&gt;Technically speaking, every planet has an atmosphere (though most drop &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Mercury&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mercury&lt;/a&gt; from the list because it's so thin). Let's start with the big guys, which are pretty straightforward. Gas giants just keep their atmosphere from when they gathered the remnants of gas from the nebula that their parent star formed from. If you're like the gas giants in our solar system, you usually keep this atmosphere for a very long time because you have a strong magnetosphere and you also are fairly far from your star. Your star also isn't particularly powerful. Some exoplanets are categorized as hot Jupiters because they are similar to jupiter in composition but are orbiting very closely to their star. If this is the case, despite the strong magnetosphere, their star may quite literally boil their atmosphere off, either adding it to the star's mass, or blowing it away, sort of like a comet. Eventually they may become what we call &lt;a href=&quot;https://en.wikipedia.org/wiki/Chthonian_planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;chthonian planets&lt;/a&gt;, which are hypothetical planets that are the rocky cores left over from boiled gas giants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about rocky planets like Earth? They get their atmosphere from a process called outgassing, which basically means that volcanoes burp out gases. Earth is unique in that we believe our atmosphere was at least partially created from comets and wet asteroids. Venus burped out too much greenhouse gases and got a hellhole for an atmosphere. At least it rains there...rains sulfuric acid...and even that boils before it reaches the surface! Mars probably had a thicker atmosphere than today, as evidenced by big volcanoes like Olympus Mons. What happened to it? Well first, Mars doesn't have a very strong magnetosphere because it is simply smaller than the Earth. Because it's smaller its core cooled down faster, and its mantle no longer supported a good magnetic field. More charged particles means a higher chance of atmospheric stripping, which Mars suffered from at least partially. However, we think that a large impact probably knocked a lot of Mars's atmosphere off. Finally, that leaves us with Mercury. Mercury is pretty small and very close to the Sun. It just can't retain an atmosphere because its gravity is weak. The same can be said for most moons in our solar system, as they are typically smaller than Mercury. They are simply too small to retain an atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wait, then how come some moons have atmospheres and others don't? Well the ones that don't, as I said before, are too small to retain one, and they also have to deal with interactions with the solar wind, atmospheric stripping, and interactions with their planet. Some moons have volcanic activity, so their thin atmosphere can be replenished with burped up gases. In fact, some may have weather! I believe Io and Triton both have weather systems, despite having very thin atmospheres. Titan has the thickest atmosphere of the bunch, with one that is about 50% denser than Earth's. Thus, it also has complex weather systems. But why? Titan is similar to Ganymede in structure. Why does Titan get to have an atmosphere? Short answer: we don't fully know why. We think it might've been comets, but we really aren't 100% sure on anything at this point. Go check out the &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Titan&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia article&lt;/a&gt; for more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer your question, atmospheres are usually formed from volcanic activity or, for gas giants, retained gas from nebula remnants. Atmospheres are lost due to low gravity, high temperatures, solar wind, atmospheric stripping, large impacts, and planet-moon interactions.&lt;/p&gt;&#xA;" OwnerUserId="24373" LastActivityDate="2018-10-05T20:34:14.427" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27910" PostTypeId="1" CreationDate="2018-10-06T00:13:50.137" Score="3" ViewCount="85" Body="&lt;p&gt;Two days ago the New York Times &lt;a href=&quot;https://nyti.ms/2O2Frou&quot; rel=&quot;nofollow noreferrer&quot;&gt;featured the Event Horizon Telescope (E.H.T.)&lt;/a&gt;, a huge virtual radio telescope emerging from combining data from several dishes scattered around the world. While the E.H.T. is designed and built to observe black holes, specifically Sagittarius A*, the article claimed that the E.H.T. could &quot;spot a doughnut on the moon&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was probably just meant as an illustration of its enormous resolution, but: Could it? I'm aware that the moon is not a radio source but it was apparently possible to make &lt;a href=&quot;https://sservi.nasa.gov/articles/new-radar-images-uncover-remarkable-features-below-the-surface-of-the-moon/&quot; rel=&quot;nofollow noreferrer&quot;&gt;radar images of its surface&lt;/a&gt; using reflected radar radiation from one of the involved dishes. Would something like that be possible with the E.H.T.? Could we &quot;see&quot; the landing sites from earth? Left behind rovers, descent and ascent stages?&lt;/p&gt;&#xA;" OwnerUserId="11190" LastActivityDate="2018-10-06T08:04:15.933" Title="Could the E.H.T. produce an image of the human artifacts on the moon?" Tags="&lt;the-moon&gt;&lt;radio-astronomy&gt;&lt;interferometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27911" PostTypeId="2" ParentId="27910" CreationDate="2018-10-06T05:13:39.773" Score="4" Body="&lt;p&gt;This is a really interesting question!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;tl;dr:&lt;/strong&gt; A definite maybe, but you would have to engineer a clever way to focus the transmitted power to a much smaller spot first; &lt;em&gt;possibly several orders of magnitude smaller&lt;/em&gt; than what any one dish can do. Phasing several widely-spaced dishes alone would not be enough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's see what can be checked easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Resolution:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using round numbers of &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt; = 300 GHz, &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; = 3E+08 m/s gives a wavelength &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; of 1 millimeter. With a baseline &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; = 10,000 km, the angular resolution &lt;span class=&quot;math-container&quot;&gt;$\lambda / D$&lt;/span&gt; is 1E-10. At a lunar distance &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; of 360,000 km that's a resolution of 36 millimeters. Not only could you spot the donut, but &lt;em&gt;the donut's hole would be resolved (barely)!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Radar at lunar distance:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second image in your linked article &lt;a href=&quot;https://sservi.nasa.gov/articles/new-radar-images-uncover-remarkable-features-below-the-surface-of-the-moon/&quot; rel=&quot;nofollow noreferrer&quot;&gt;New Radar Images Uncover Remarkable Features below the Surface of the Moon&lt;/a&gt; include images of &lt;a href=&quot;https://en.wikipedia.org/wiki/Luther_(crater)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Luther&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Aristillus_(crater)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aristillus&lt;/a&gt; craters which are one and five million centimeters in size, respectively. The images already look noisy at this scale, so going a factor of 1E+05 smaller in resolution is likely to require a stronger transmitted radar signal, as measured in focused power per unit area on the lunar surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might think that that can be fixed by making many of the EHT dishes (distributed across the globe) into transceivers, using very long baseline to focus the radar spot's power, and that &lt;em&gt;could work to a limited extent&lt;/em&gt; but the problem is that it would not work the same way a filled aperture works. A sparsely populated arrayed aperture will still put most of the power into a plethora of ugly sidelobes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a name for exactly this problem. I can't remember it now but I'll look for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Something similar, but different:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the question &lt;a href=&quot;https://en.wikipedia.org/wiki/Aristillus_(crater)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Why was the 100m Green Bank dish needed together with DSN's 70m Goldstone dish to detect Chandrayaan-1 in lunar orbit?&lt;/a&gt; I talk about the use of two very large dish antennas to &lt;s&gt;image&lt;/s&gt; &lt;em&gt;detect the existence of&lt;/em&gt; a small object at the lunar distance. The experiment is described &lt;a href=&quot;https://www.jpl.nasa.gov/news/news.php?feature=6769&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://phys.org/news/2017-03-nasa-radar-technique-lost-lunar.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They used two &quot;tricks&quot; to make this feat possible with only two dishes.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The DSN's Goldstone 70 meter dish transmitted the signal, and the large diameter of the dish made it possible to put most of the transmitted power into a circle which intercepted the spacecraft at its maximum elongation from the Moon (shown below) so that very little of the power would intercept the much larger Moon. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;They took advantage of the Doppler shift of the reflected signal caused by the spacecraft's orbital velocity to further separate reflected signals from the Moon from the reflected signals from the spacecraft.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cLYeF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cLYeF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; &quot;This computer-generated image depicts the Chandrayaan-1's location at time it was detected by the Goldstone Solar System radar on July 2, 2016. The 120-mile (200-kilometer) wide purple circle represents the width of the Goldstone radar beam at lunar distance. The white box in the upper-right corner of the animation depicts the strength of echo. Inside the radar beam (purple circle), the echo from the spacecraft alternated between being very strong and very weak, as the radar beam scattered from the flat metal surfaces.&quot; Credit: NASA/JPL-Caltech.  From &lt;a href=&quot;https://phys.org/news/2017-03-nasa-radar-technique-lost-lunar.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QxaEk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QxaEk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; &quot;Radar imagery acquired of the Chandrayaan-1 spacecraft as it flew over the moon's south pole on July 3, 2016. The imagery was acquired using NASA's 70-meter (230-foot) antenna at the Goldstone Deep Space Communications Complex in California. This is one of four detections of Chandrayaan-1 from that day.&quot; Credit: NASA/JPL-Caltech.  From &lt;a href=&quot;https://phys.org/news/2017-03-nasa-radar-technique-lost-lunar.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-10-06T08:04:15.933" LastActivityDate="2018-10-06T08:04:15.933" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27912" PostTypeId="1" CreationDate="2018-10-06T05:36:15.207" Score="7" ViewCount="498" Body="&lt;p&gt;The April 2015 CBS News article &lt;a href=&quot;https://www.cbsnews.com/news/an-ingenius-fix-for-hubbles-famously-flawed-vision/&quot; rel=&quot;noreferrer&quot;&gt;How NASA fixed Hubble's flawed vision - and reputation&lt;/a&gt; describes the modification taken to the Hubble Space Telescopes by the Space Shuttle that was used to correct for the strong spherical aberration of the primary mirror due a highly accurate but incorrect figure.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dFH0b.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dFH0b.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; &amp;quot;Due to an oversight during fabrication, the concave shape of Hubble's primary mirror was too shallow toward its outer edges by 2 microns, a tiny fraction of the width of a human hair. As a result, starlight was not brought to a focus at the same point, resulting in blurry images. These three images show the same star as viewed from a ground-based telescope, left, and from Hubble's uncorrected mirror, center. The corrected image, after a space shuttle repair mission in 1993, is seen on the right&amp;quot; from &lt;a href=&quot;https://www.cbsnews.com/news/an-ingenius-fix-for-hubbles-famously-flawed-vision/&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. credit: NASA&lt;/p&gt;&#xA;&lt;p&gt;From the article:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;But the mirror's biggest impact was on the Wide Field Planetary Camera, the instrument expected to provide the dazzling visible and near-infrared images most easily understandable by the public and of tremendous value to astronomers.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;The real killer was Jim Westphal (the WFPC principal investigator at the Jet Propulsion Laboratory in Pasadena, California), who tended to be very negative,&amp;quot; Weiler recalled. &amp;quot;Basically, he said he couldn't do anything. So that's what I was preparing to walk into the press conference with.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;But as it turned out, Weiler had two aces up his sleeve. One he knew about, one he didn't.&lt;/p&gt;&#xA;&lt;p&gt;The ace he knew about was drawn seven years earlier, when Weiler kicked off work to build a backup camera, the Wide Field Planetary Camera 2, a carbon copy of the original. John Trauger was the principal investigator, the man in charge of the instrument at JPL.&lt;/p&gt;&#xA;&lt;p&gt;The ace Weiler didn't yet realize was built into the design of Trauger's camera.&lt;/p&gt;&#xA;&lt;p&gt;Shortly after the first light image came down, &amp;quot;one of the grand old men of optics here at JPL, Aden Meinel, he and his wife Marjorie wanted to see what the image looked like,&amp;quot; Trauger said in an interview. &amp;quot;We brought it up on the screen. ... And he looked at that for five minutes and he said 'that looks like spherical aberration.'&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;That was the first time I heard spherical aberration,&amp;quot; Trauger recalled. &amp;quot;The thing was, he also said 'well, if it is spherical aberration, you could fix it with Wide Field Planetary Camera 2.' He knew that we had these nickel-sized mirrors in our optical system, which received a sharp image of the primary mirror itself. ... The very last mirror before the image is created on our CCD was an opportunity to straighten the wavefront out. He recognized that, too.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;By giving that mirror a prescription that exactly countered the spherical aberration in the primary mirror, the WFPC 2 would be able to achieve a perfect focus and do all of the science that was intended.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Where exactly is the modification that first corrected the spherical aberration in Hubble's primary mirror? Can it be identified within the image below, or is it located elsewhere? Was the modification only a change in the figure of one of the small mirrors, or were other optical elements modified as well as part of the correction?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bnVuq.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bnVuq.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; &amp;quot;The business end of COSTAR: Small mirrors on motorized arms directed corrected light into the Hubble Space Telescope's spectrographs and faint object camera. COSTAR and the Wide Field Planetary Camera 2 were installed during a 1993 shuttle mission, correcting Hubble's flawed vision.&amp;quot; From &lt;a href=&quot;https://www.cbsnews.com/news/an-ingenius-fix-for-hubbles-famously-flawed-vision/&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. Eric Long/Smithsonian Institution&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-01-24T13:19:09.483" Title="Where exactly is the modification that first corrected the spherical aberration in Hubble's primary mirror?" Tags="&lt;telescope&gt;&lt;hubble-telescope&gt;&lt;space-telescope&gt;&lt;optics&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27914" PostTypeId="1" AcceptedAnswerId="27955" CreationDate="2018-10-06T07:41:59.747" Score="0" ViewCount="1055" Body="&lt;p&gt;It's relatively easy to find tables for Earth aphelion and perihelion dates and times, such as this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astropixels.com/ephemeris/perap2001.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.astropixels.com/ephemeris/perap2001.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about the other planets in the solar system? Are there tables or lists of aphelion and perihelion dates for them too? I checked Wikipedia, but it only lists perihelion distances and not dates, so it is insufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the error I get on the SPICE web interface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G5ZiR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G5ZiR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24284" LastEditorUserId="24284" LastEditDate="2018-10-08T22:02:07.950" LastActivityDate="2018-10-10T05:16:09.403" Title="Tables of aphelion/perihelion dates for other planets?" Tags="&lt;planet&gt;&lt;orbital-elements&gt;&lt;resource-request&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="27915" PostTypeId="2" ParentId="27914" CreationDate="2018-10-06T08:35:57.620" Score="0" Body="&lt;p&gt;If you search for &quot;Astronomical Ephemeris&quot; you will find plenty of resources.  For example:  &lt;a href=&quot;http://astropixels.com/skycal/skycalmain.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://astropixels.com/skycal/skycalmain.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2018-10-06T08:35:57.620" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27916" PostTypeId="1" AcceptedAnswerId="28286" CreationDate="2018-10-06T08:37:51.550" Score="2" ViewCount="662" Body="&lt;p&gt;The NPR News item &lt;a href=&quot;https://www.npr.org/2018/10/04/654518484/macarthur-fellow-and-planetary-scientist-sarah-stewart-discusses-how-the-moon-wa&quot; rel=&quot;nofollow noreferrer&quot;&gt;MacArthur Fellow And Planetary Scientist Sarah Stewart Discusses How The Moon Was Formed&lt;/a&gt; and audio podcast begins:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Ari Shapiro, Host:&lt;/strong&gt; Sarah Stewart likes to think about what happens when planets collide. She uses two actual cannons to simulate those massive impacts. Here's one firing in her lab at UC Davis.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Unidentified Person:&lt;/strong&gt;  Firing in three, two, one.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Soundbite of Cannon FIring&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Ari Shapiro, Host:&lt;/strong&gt; Her work earned her a spot in this year's class of MacArthur Fellows. Many of us call it the genius grant. For years, experts have thought that Earth's moon formed after a large collision knocked off a bunch of rock. Stewart told me her research suggests a different story.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Planetary Scientist Sarah Stewart:&lt;/strong&gt; During planet formation, when two bodies collide, there's so much energy released that most of these bodies are vaporized. That means that a rocky planet like Earth is mostly rock vapor.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Shapiro:&lt;/strong&gt; What is rock vapor, and what does that have to do with our moon?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Stewart:&lt;/strong&gt; Rock vapor is taking the rocks that we stand on and heating it up to the point where it becomes a gas. And when that occurs, the Earth becomes much larger because vapor is much less dense. And it extends out into this enormous object hundreds of times larger than the Earth today. And we proposed that our moon grows within the rock vapor of the Earth after a giant impact.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Shapiro:&lt;/strong&gt; So the moon actually came from the Earth.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Stewart:&lt;/strong&gt; The moon grows within the rock vapor of the Earth. And that gives the moon the same chemistry as the Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;We don't learn about &amp;quot;rock vapor&amp;quot; in Earth Science class, but I know it's got to be a lot hotter than the lava we see in the news. A significant fraction of Earth's crust is SiO₂ based and it's boiling point is roughly 3,000°C, and I have a hunch the temperatures involved here are much much higher than that. The kinetic energy associated with say a relative velocity of 40 km/s is roughly 8 eV per AMU, over 130 eV for every oxygen atom for example.&lt;/p&gt;&#xA;&lt;p&gt;So does &amp;quot;rock vapor&amp;quot; start out as a highly ionized &amp;quot;rock plasma&amp;quot; with almost no covalent bonds remaining, or does most of the energy of the original impactor get transferred to a much greater mass of Earth?&lt;/p&gt;&#xA;&lt;p&gt;Is there a good place to read about her and her students' MacArthur grant-getting research described in the podcast?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2020-10-15T07:56:57.047" Title="What is the nature of &quot;rock vapor&quot; in this description of the formation of the Moon?" Tags="&lt;the-moon&gt;&lt;planetary-formation&gt;&lt;formation&gt;&lt;collision&gt;" AnswerCount="2" CommentCount="28" ContentLicense="CC BY-SA 4.0" />
  <row Id="27917" PostTypeId="2" ParentId="27916" CreationDate="2018-10-06T11:47:20.513" Score="1" Body="&lt;p&gt;In short:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Not every gas is a plasma. Covalent bonds can be absent in a neutral gas. Rock vapour is just vapour, silicate atoms in their gasous state. And just as  oxygen can freeze, so can silicates evaporate. Of course they can thermally ionize as well at even higher temperatures, but I don't see that this is implied in the text.&lt;/li&gt;&#xA;&lt;li&gt;They're not the only ones investigating scenarios like that: &lt;a href=&quot;https://www.nature.com/articles/nature19341&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article&lt;/a&gt; for example argues that the isotopic differentiation in Moon rocks can be explained by an extended, Earth-moon spanning hot rock-vapour atmosphere just after the impact.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2018-10-06T11:47:20.513" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27918" PostTypeId="1" CreationDate="2018-10-06T16:58:12.797" Score="2" ViewCount="292" Body="&lt;p&gt;What is the planet relative to earth that shows the greatest change in apparent brightness?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the answer is Mars, and a quick google search reveals it is Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have my reasoning here, but I am not sure if I am correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to do this problem is to get all the planets and collect their perihelion and aphelion distances from the sun, and see the different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I want to be able to explain this without the extensive use of numbers. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is my explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest planet to us is Mars, both in terms of its aphelion and its perihelion distances&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth-Mars distance varies between 2.66 AU and .38 AU&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury is almost always blinded by the sun, so we rarely see it in the sky&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus is frequently viewed as a crescent, and thus its apparent brightness also dimishes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter and rest of the superior planets are simply too far away for us to notice a big differene&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Are all my points correct here?&lt;/p&gt;&#xA;" OwnerUserId="17327" LastActivityDate="2018-10-10T17:08:51.180" Title="What is the planet relative to earth that shows the greatest change in apparent brightness?" Tags="&lt;the-sun&gt;&lt;planet&gt;&lt;apparent-magnitude&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27919" PostTypeId="1" AcceptedAnswerId="27962" CreationDate="2018-10-06T17:10:56.970" Score="1" ViewCount="81" Body="&lt;p&gt;If we find asteroid flying towards us, can we change it's direction with white painting? Will the ray from sun do something?&lt;/p&gt;&#xA;" OwnerUserId="20727" LastActivityDate="2018-10-10T13:32:23.350" Title="Move asteroid away with white color?" Tags="&lt;asteroids&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27920" PostTypeId="1" AcceptedAnswerId="27923" CreationDate="2018-10-06T20:50:20.280" Score="2" ViewCount="207" Body="&lt;p&gt;Voyager 1 (&lt;a href=&quot;https://en.wikipedia.org/wiki/Voyager_1&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Voyager_1&lt;/a&gt;) is approximately 140 AU away, and is considered to have left the solar system. Now, the layman's definition of solar system provided by Google is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...the collection of eight planets and their moons in orbit around the&#xA;  sun, together with smaller bodies in the form of asteroids,&#xA;  meteoroids, and comets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Solar System[a] is the gravitationally bound system of the Sun and&#xA;  the objects that orbit it, either directly or indirectly,[b] including&#xA;  the eight planets and five dwarf planets as defined by the&#xA;  International Astronomical Union (IAU).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At this point, Voyager 1 is well beyond the aphelion of all known planets, so according to the definitions above, so at least colloquially, saying that it has left the Solar System seems to make sense. However, according to Wikipedia, it was only officially considered to have left the Solar System once it left the heliopause. This is not really in conflict with the above definitions, as, to my knowledge, the heliopause extends beyond the orbit of all the original 8/9 planets at all times. The Encyclopedia Britannica (&lt;a href=&quot;https://www.britannica.com/science/heliopause&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.britannica.com/science/heliopause&lt;/a&gt;) states that the heliopause&#xA;extends for 123AU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the hypothesised Planet 9/X is conjectured to have an aphelion of approximately 1,200 AU and a Perihelion of 200 AU (&lt;a href=&quot;https://en.wikipedia.org/wiki/Planet_Nine&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Planet_Nine&lt;/a&gt;). This would seem to bring the two competing definitions of &quot;solar system&quot; discussed here into sharp conflict.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which definition would likely win out? Could the confirmation of such a planet force a reckoning on whether Voyager 1 has truly left the solar system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;I wanted to add a more authoritative source that says Voyager 1 has indeed left the solar system.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Voyager 1 is the first man-made object to leave our solar system and pass into interstellar space. Scientists confirmed this finding a year later after studying Voyager’s data, which showed clear changes in the plasma or ionized gas right outside of the solar bubble.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.nasa.gov/directorates/heo/scan/images/history/August2012_2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.nasa.gov/directorates/heo/scan/images/history/August2012_2.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24465" LastEditorUserId="24465" LastEditDate="2019-01-16T04:12:18.373" LastActivityDate="2019-01-16T04:12:18.373" Title="How is the term &quot;solar system&quot; defined? Could confirmation of a new planet lead to a change in this definition?" Tags="&lt;solar-system&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27921" PostTypeId="1" CreationDate="2018-10-06T21:38:48.820" Score="2" ViewCount="38" Body="&lt;p&gt;I've readen &quot;When photographing faint objects monochrome photos will resolve more than RGB pics&quot;&#xA;Is this really a thing? If yes, how?&lt;/p&gt;&#xA;" OwnerUserId="24059" LastActivityDate="2018-10-06T21:38:48.820" Title="What is the advantage of taking monochrome photographs in standard lenses" Tags="&lt;amateur-observing&gt;&lt;light&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27922" PostTypeId="1" CreationDate="2018-10-06T23:46:54.873" Score="1" ViewCount="15" Body="&lt;p&gt;Around 13 years ago I saw fast moving bright star with eliptic course which crossed half a sky in 1-2 sec and stopped instantly like it was a smashed like a bug, and didn't move for a while few minutes, then I went my way. I saw many falling stars and sattelits mouvements, planes on night, but never something similar.  Thank you in advance!&lt;/p&gt;&#xA;" OwnerUserId="24468" LastActivityDate="2018-10-06T23:46:54.873" Title="Fast moving bright star with eliptic course, instant stop" Tags="&lt;star&gt;&lt;movement&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2018-10-08T07:38:52.420" ContentLicense="CC BY-SA 4.0" />
  <row Id="27923" PostTypeId="2" ParentId="27920" CreationDate="2018-10-06T23:57:28.873" Score="8" Body="&lt;p&gt;The definition of the edge of the Solar System is not one of particular scientific importance, so it is not particularly well-defined. There are a number of plausible definitions, all reasonable.  I've listed one which occur to me below in order of increasing size:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The pancake-shaped volume which contains the orbits of the major planets.&lt;/li&gt;&#xA;&lt;li&gt;The sphere big enough to contain the major planets.&lt;/li&gt;&#xA;&lt;li&gt;The volume inside the solar bubble.&lt;/li&gt;&#xA;&lt;li&gt;The thickish disk which contains the Kupier Belt.&lt;/li&gt;&#xA;&lt;li&gt;The sphere which contains the Oort Cloud.&lt;/li&gt;&#xA;&lt;li&gt;The Sun's Hill sphere.&lt;/li&gt;&#xA;&lt;li&gt;The irregularly-shaped volume of space where the Sun is the closest star.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As you note, the discovery of a Planet X would change some of them, but not others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How choose to define the edge of the Solar System depends a lot on what you intend to use it for.  I very much doubt that any one will win out, at least not until the definition of the Solar System makes an important difference somewhere: government, commerce, science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think there's no clear answer to your question possible. &lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-10-06T23:57:28.873" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27925" PostTypeId="1" AcceptedAnswerId="27944" CreationDate="2018-10-07T00:38:35.880" Score="6" ViewCount="475" Body="&lt;p&gt;Imagine a stellar-mass black hole passing through the Oort cloud, perhaps a quarter  to half a light-year from the Sun. How black would the black hole be? That is, any matter that fell into the black hole would release energy on the way in, and we know there are comets out there, and presumably smaller objects down to microscopic dust. But we also know they are very few and far between. Has anyone done a serious analysis of whether there would be enough infalling matter for the resulting radiation to be detectable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As two commenters already recognized, this really is about radiation from infalling matter, and not the broader question of how we could detect such a black hole. Answers to that broader question in the comments are interesting, though.&lt;/p&gt;&#xA;" OwnerUserId="7752" LastEditorUserId="7752" LastEditDate="2018-10-08T00:31:34.330" LastActivityDate="2018-10-08T07:51:07.427" Title="Could a black hole pass quiescently through the Oort cloud?" Tags="&lt;black-hole&gt;&lt;oort-cloud&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27926" PostTypeId="1" AcceptedAnswerId="27928" CreationDate="2018-10-07T02:19:11.910" Score="2" ViewCount="432" Body="&lt;p&gt;What is true purpose of polar aligning? Last night, I looked at Mars. It was moving fast.  I still need to adjust RA axis to track fast-moving Mars. Why the telescope didn't  track it by itself?&lt;/p&gt;&#xA;" OwnerUserId="24470" LastEditorUserId="11822" LastEditDate="2018-10-07T09:34:35.893" LastActivityDate="2018-10-07T18:44:02.577" Title="How exactly 'polar aligning' a Telescope with equatorial mount helps?" Tags="&lt;alignment&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27927" PostTypeId="2" ParentId="27912" CreationDate="2018-10-07T03:00:31.313" Score="3" Body="&lt;p&gt;The image provided is of COSTAR which replaced the High Speed Photometer and fed corrected light to the Faint Object Camera (FOC), Faint Object Spectrograph (FOS) and the Goddard High Resolution Spectrograph (GHRS). Since the WFPC2 was already going to replace WFPC1, the correction for the primary mirror's spherical aberration could be designed in (as it has been for all subsequent instruments).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall layout of the HST optics (the Optical Tube Assembly (OTA)) and the WFPC camera optics is shown below in a figure taken from the &lt;a href=&quot;https://www.spiedigitallibrary.org/conference-proceedings-of-spie/1920/0000/Development-of-the-second-generation-Wide-Field-Planetary-Camera-for/10.1117/12.152673.short&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rodgers and Vaughan 1993 SPIE paper&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light enters from the top left and is focused as normal by the HST primary and secondary mirrors and then part of the beam is picked-off into the WFPC2 camera. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oTDxp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oTDxp.png&quot; alt=&quot;Optical layout of the HST and WFPC optics&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The WFPC2 optical configuration is shown in more detail below (taken from the instrument description section of the &lt;a href=&quot;http://documents.stsci.edu/hst/wfpc2/documents/handbooks/cycle17/ch2_instrument4.html#441147&quot; rel=&quot;nofollow noreferrer&quot;&gt;WFPC2 Instrument Handbook&lt;/a&gt;) &lt;a href=&quot;https://i.stack.imgur.com/hhHYV.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hhHYV.gif&quot; alt=&quot;WFPC2 optical configuration&quot;&gt;&lt;/a&gt;. The text just above that states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The aberrated HST wavefront is corrected by introducing an equal but&#xA;  opposite error in each of the four Cassegrain relays. An image of the HST&#xA;  primary mirror is formed on the secondary mirrors in the Cassegrain relays.&#xA;  (The fold mirror in the PC channel has a small curvature to ensure this.)&#xA;  The spherical aberration from the telescope's primary mirror is corrected&#xA;  on these secondary mirrors, which are extremely aspheric.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;indicating that the correction for HST's primary is done on the part labeled 'Secondary Mirror' in the diagram. Note that is the secondary mirror of the mini-Cassegrain telescope that is inside each of the 3 widefield and 1 planetary camera channels inside WFPC2 itself and not the main telescope secondary mirror which is off the diagram and part of the 'OTA' (Optical Telescope Assembly)&lt;/p&gt;&#xA;" OwnerUserId="23748" LastEditorUserId="23748" LastEditDate="2019-01-23T21:40:05.100" LastActivityDate="2019-01-23T21:40:05.100" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27928" PostTypeId="2" ParentId="27926" CreationDate="2018-10-07T04:36:38.847" Score="1" Body="&lt;p&gt;There are two main coordinate systems used in astronomy: equatorial and horizontal. The equatorial system imagines a sphere around the earth on which stars occupy (for our purposes) fixed positions. (In fact, the stars do move slightly, but not with a speed that matters for casual amateur observation.) The point is that this system works well for objects (particularly, stars) with negligible amounts of motion. We use declination and right ascension to describe positions on the sphere, just as we use latitude and longitude to describe a position on the sphere of the earth. The point of polar aligning an equatorial mount is to find a convenient starting point, which is the North Celestial Pole: it's how your telescope gets aligned with this grid on the sky. Once your telescope &quot;knows&quot; where the pole is, you can manually move it using calibrated mechamisms to a particular position in RA &amp;amp; Dec. The second coordinate system is the horizontal one, which uses the observer's position as its starting point (it is location-dependent), and measures objects using altitude (from the observer's horizon) &amp;amp; azimuth (true bearing from the observer's position). An alt-az telescope mount can move freely up and down, left and right. Sometimes it has markings for horizontal coordinates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An equatorial mount was not designed to track planets, but it can do so. The advantage of using an equatorial mount is that you can finely adjust the scope to account for the movement of the planet during the (limited) time you're observing. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why didn't the telescope track it by itself?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Do you have a GoTo system? &lt;/p&gt;&#xA;" OwnerUserId="23940" LastEditorUserId="23940" LastEditDate="2018-10-07T04:55:53.157" LastActivityDate="2018-10-07T04:55:53.157" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27929" PostTypeId="1" CreationDate="2018-10-07T13:49:04.963" Score="2" ViewCount="87" Body="&lt;p&gt;Are there printable 2D star maps that are distorted such that, when printed, they can be cut (along printed lines) to be applied to the outside of a sphere to represent the celestial sphere, as seen from the center of the substrate sphere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a term for this type of map?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they don't exist, how can one be created?&lt;/p&gt;&#xA;" OwnerUserId="24126" LastEditorUserId="7472" LastEditDate="2018-10-07T14:50:20.020" LastActivityDate="2018-10-07T14:50:20.020" Title="2D Distorted Star Map for Pasting on Sphere Surface" Tags="&lt;star-maps&gt;" AnswerCount="0" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27930" PostTypeId="1" AcceptedAnswerId="27931" CreationDate="2018-10-07T13:55:35.133" Score="1" ViewCount="137" Body="&lt;p&gt;In this &lt;a href=&quot;https://space.stackexchange.com/a/31156/12102&quot;&gt;answer&lt;/a&gt; I mention that for very low orbits around spherical bodies, the period &lt;em&gt;tends to&lt;/em&gt; scale only as the inverse square root of the density, and not the diameter.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;For a low orbits where the semi-major axis is close to the radius of the central body, the period is related to the average density of the body and unrelated to it's size.&lt;/p&gt;&#xA;&lt;p&gt;So a low orbit around a spherical asteroid (which there usually aren't) made of a mixture of rock and iron (which there usually aren't) will be roughly 90 minutes just like LEO (low Earth orbit), even if it's only 1 km in diameter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I'd like to use an example, but I don't know where to find an asteroid density table, nor even a histogram, that includes something close to 5 g/cm^3.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Are there any known asteroids with average density similar to that of Earth's?&lt;/p&gt;&#xA;&lt;p&gt;Some example mean densities from Wikipedia (some rocky planets included for good measure):&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;body      mean density (g/cm^3)&#xA;&#xA;Earth            5.5&#xA;Mercury          5.4&#xA;Venus            5.2&#xA;&#xA;Mars             3.9&#xA;Io               3.5&#xA;Vesta            3.4&#xA;Moon             3.3&#xA;Hygiea           2.8&#xA;Pallas           2.7&#xA;Ceres            2.1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="34459" LastEditDate="2020-07-17T08:25:47.327" LastActivityDate="2020-07-17T08:25:47.327" Title="Are there any known asteroids with average density similar to that of Earth's?" Tags="&lt;asteroids&gt;&lt;density&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27931" PostTypeId="2" ParentId="27930" CreationDate="2018-10-07T15:55:30.957" Score="3" Body="&lt;p&gt;Apparently &lt;a href=&quot;https://en.wikipedia.org/wiki/21_Lutetia&quot; rel=&quot;nofollow noreferrer&quot;&gt;21 Lutetia&lt;/a&gt; (~80x120 km) is one of the most dense asteroids known at about 3.4 g/cm&lt;span class=&quot;math-container&quot;&gt;$^3$&lt;/span&gt;. Unfortunately that is barely over 60% of Earth's density. 21 Lutetia is unique in that scientists believe that it has a primordial core, that is, it has a core from the formation of the solar system (and it hasn't been destroyed after all these years). We believe it is hidden under the surface. It would be rich in metals, explaining the density, and because it's hidden under the surface, 21 Lutetia would look no different from any other asteroid, which is the case. This asteroid is a prime target for mining because of its metal core. I suppose the answer to your question would be no, unless 3.4 is sufficient for your definition of &quot;close&quot;.&lt;/p&gt;&#xA;" OwnerUserId="24373" LastEditorUserId="24373" LastEditDate="2018-10-07T16:06:56.760" LastActivityDate="2018-10-07T16:06:56.760" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27932" PostTypeId="2" ParentId="27926" CreationDate="2018-10-07T17:19:25.667" Score="4" Body="&lt;p&gt;The Earth's rotation makes the stars and planets appear to move westward across the eyepiece field of a stationary telescope at about 900 arcseconds per minute.&#xA;An equatorial mount can counter this by moving only on its polar axis at a constant rate.&#xA;An altitude-azimuth mount must move on both axes at variable rates in order to track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In early October 2018, Mars appeared to move about 1.1 arcseconds per minute relative to the stars.&#xA;In an equatorially mounted telescope, properly aligned and tracking at the sidereal rate, Mars would take about 13 minutes to move through its own apparent diameter of 15 arcseconds.&#xA;Without a tracking motor, you'd have to move the RA axis by hand, but you could leave the declination axis alone for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the mount has a tracking motor designed to be usable in either hemisphere, it may have a switch to reverse direction.&#xA;If that switch is in the wrong position, stars appear to move westward twice as fast as if the motor were switched off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an equatorial mount is misaligned, stars appear to drift in declination, depending on their position in the sky, up to 15 arcseconds per minute per degree of polar alignment error.&#xA;For example, if you're in the northern hemisphere and the mount's polar axis is 5 degrees too far to the east, objects near the meridian appear to drift southward at 75 arcseconds per minute, whereas objects 4 hours off the meridian drift only half that fast.&#xA;If the polar axis is more than 30 degrees off the celestial pole, it might as well be an alt-az mount.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-10-07T18:44:02.577" LastActivityDate="2018-10-07T18:44:02.577" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27935" PostTypeId="2" ParentId="27875" CreationDate="2018-10-07T20:43:02.383" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How do I use this to arrive at a potato radius &lt;em&gt;R&lt;/em&gt; of about 200 to 300 km?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can't, for multiple reasons. One reason is that the authors themselves disclaim this as a valid approach:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The result is a radius of a few thousand kilometers for a rocky body and ~1000 km for an icy body. This was meant to be only an order of magnitude calculation. We try to improve on it below.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another is that the authors don't show their math. On trying to replicate it, I get rather different results. The mass of a more or less spherical object is &lt;span class=&quot;math-container&quot;&gt;$M=\frac43\pi\rho R^3$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; is the mean density of the object. Plugging this into their expression yields &lt;span class=&quot;math-container&quot;&gt;$1\,\text{eV}\approx \frac43\pi G\rho\mu R^2$&lt;/span&gt;, or&#xA;&lt;span class=&quot;math-container&quot;&gt;$$R \approx \sqrt{\frac{1\,\text{eV}}{\frac43\pi G\rho\mu}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This suggests the potato radius is inversely proportional to the square root of density and inversely proportional to the square root of particle mass. This obviously is not the case; the observed potato radius of rocky objects is about 50% more than that of ice objects, even though rocky objects have the greater density and are made of more massive molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took this as a throwaway paragraph. The key results are further on in the article.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2018-10-07T20:43:02.383" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27936" PostTypeId="2" ParentId="27735" CreationDate="2018-10-07T20:43:45.250" Score="1" Body="&lt;p&gt;The polite term we use for these department store fixtures is &quot;Hobby killers&quot;.  The detailed answers you have received are spot on. I would only temper them by saying &#xA;it is not the absolute amount of cash you need to spend. &#xA;One can come by even greater scopes for much less.  And that if you are somewhere dark, even a six inch reflector &#xA;could be a good first scope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then of course there is building your own...&lt;/p&gt;&#xA;" OwnerUserId="24480" LastActivityDate="2018-10-07T20:43:45.250" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27937" PostTypeId="1" AcceptedAnswerId="27939" CreationDate="2018-10-07T20:53:13.613" Score="-1" ViewCount="92" Body="&lt;p&gt;Can a black hole suck in time? People claim that if you were to put a bowling ball on a trampoline and rolled much smaller balls through the deppression, some would stay in &quot;orbit&quot;, some would roll through (if they were fast enough) and the others would just get stuck at the bottom. And if you put something that was heavy enough, the trampoline would snap. So could a black hole suck in the matrix??&lt;/p&gt;&#xA;" OwnerUserId="23317" LastEditorUserId="23317" LastEditDate="2018-10-08T00:23:44.927" LastActivityDate="2018-10-08T00:23:44.927" Title="Since gravity is like an indent on a piece of paper (the space/time matrix) to a marble, can the black hole eventually suck in the matrix?" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;space-time&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27938" PostTypeId="2" ParentId="27937" CreationDate="2018-10-07T21:19:43.020" Score="0" Body="&lt;p&gt;The ball on a trampoline is a neat visualization of spacetime, but it is only an analogy. The more massive an object is, the greater its gravity. A person has very little gravity compared to our Earth, so the Earth would make the trampoline sink further than a person. The same applies to a comparison between our Sun and the Earth. The Sun has much greater gravity than the Earth, so the Sun makes the trampoline sink further than the Earth. The singularity at the center of a black hole has infinite gravity, and thus, will stretch that trampoline infinitely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ste0l.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ste0l.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24373" LastActivityDate="2018-10-07T21:19:43.020" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27939" PostTypeId="2" ParentId="27937" CreationDate="2018-10-07T21:25:59.540" Score="4" Body="&lt;p&gt;There's a lot going on in the body of the question, so I'll try and be as specific as I can in addressing the separate points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can a black hole suck in time? No. It changes the passage of time locally due to its immense gravity crushing frequency (our observation of time), but it can't &quot;eat&quot; time. Time is an observational property, and has no physical feature to be eaten.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can a black hole lead to a snap in space-time such that there's a hole for stuff to pass through into some unknown &quot;layer&quot;? Possibly, but the answer isn't beyond theory at this point as we can't see past the event horizon. Such a &quot;rip&quot; into a new &quot;layer&quot; is one theory on how this universe was created, with the big bang being the white hole end of a black hole in another universe (layer). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could the black hole suck in the matrix (space-time) itself? No, gravity only &lt;em&gt;distorts&lt;/em&gt; space-time, so the matrix is only stretched and not consumed. &lt;/p&gt;&#xA;" OwnerUserId="24476" LastActivityDate="2018-10-07T21:25:59.540" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27940" PostTypeId="1" AcceptedAnswerId="27950" CreationDate="2018-10-08T00:16:40.237" Score="3" ViewCount="173" Body="&lt;p&gt;I'm not sure why type Ia supernova reaches its peak magnitude 15-20 days after the explosion. Is this luminosity generated only by the radioactive decay of nickel-56 to cobalt-56?(considering a half-life, I don't think the decay to iron-56 can be involved with it) Or is there any combination that produces the peak luminosity?&lt;/p&gt;&#xA;" OwnerUserId="24460" LastActivityDate="2018-10-12T14:10:59.873" Title="Why type Ia supernova reaches its peak absolute magnitude(~ -19.3mag) at around 15-20 days after the explosion?" Tags="&lt;cosmology&gt;&lt;supernova&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27941" PostTypeId="2" ParentId="27940" CreationDate="2018-10-08T01:35:15.240" Score="-3" Body="&lt;p&gt;It has to do with the Chandrasekhar limit, and how all type 1a supernova are created from the destruction of a white dwarf. It takes the same amount of time to hit absolute magnitude because they convert the same amount of mass.&lt;/p&gt;&#xA;" OwnerUserId="24476" LastActivityDate="2018-10-08T01:35:15.240" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27942" PostTypeId="1" AcceptedAnswerId="27943" CreationDate="2018-10-08T05:24:01.453" Score="2" ViewCount="189" Body="&lt;p&gt;Does the solar system also have cardinal directions? Which one is North and in what direction does it point?&lt;/p&gt;&#xA;" OwnerUserId="24284" LastActivityDate="2018-10-08T20:46:49.503" Title="Cardinal directions in solar system?" Tags="&lt;solar-system&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27943" PostTypeId="2" ParentId="27942" CreationDate="2018-10-08T05:42:58.507" Score="4" Body="&lt;p&gt;There are a bunch of different measurements for celestial coordinates, but in regards to the solar system - or specifically the Sun as its &quot;master&quot; - there's the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ecliptic_coordinate_system#Primary_direction&quot; rel=&quot;nofollow noreferrer&quot;&gt;ecliptic celestial coordinate system&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that system, centred around the sun and our solar system by extension, the North pole is in the constellation Draco. According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_pole&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, the coordinates from Earth are right ascension 18h 0m 0.0s (exact), declination +66° 33′ 38.55″.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WbrK7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WbrK7.png&quot; alt=&quot;Ecliptic&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/File:Ecliptic_grid_globe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24476" LastEditorUserId="7982" LastEditDate="2018-10-08T15:18:42.373" LastActivityDate="2018-10-08T15:18:42.373" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27944" PostTypeId="2" ParentId="27925" CreationDate="2018-10-08T06:38:24.433" Score="9" Body="&lt;p&gt;An approach would be to use the Bondi-Hoyle (B-H) accretion rate and then assume that some large fraction of the gravitational potential energy was radiated from the innermost stable circular orbit (ISCO) as the material spiralled into the black hole via an accretion disc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the B-H accretion rate we need how fast the black hole traverses the Oort cloud and the density of matter there.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\dot{M} \simeq \lambda\frac{4\pi G^2 M^2 \rho}{v^3},$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is the black hole mass, and &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; is an accretion efficiency parameter that is judged to be &lt;span class=&quot;math-container&quot;&gt;$\sim 0.01$&lt;/span&gt; on the basis of observed compact objects (see &lt;a href=&quot;https://arxiv.org/abs/1301.1341&quot; rel=&quot;noreferrer&quot;&gt;Fender et al. 2013&lt;/a&gt;).&#xA;The luminosity will be&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ L \simeq \eta \frac{GM \dot{M}}{R_{\rm ISCO}},$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$R_{\rm ISCO}= 6GM/c^2$&lt;/span&gt; for a Schwarzschild black hole. Thus&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ L \sim \eta \lambda \frac{4\pi G^2 M^2 \rho c^2 }{6v^3}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting all my information from &lt;a href=&quot;https://en.wikipedia.org/wiki/Oort_cloud&quot; rel=&quot;noreferrer&quot;&gt;wikipedia&lt;/a&gt;, I assume there are 5 Earth masses in the Oort cloud in a spherical shell from 20,000 to 50,000 au. This gives an average density &lt;span class=&quot;math-container&quot;&gt;$\rho \sim 2\times 10^{-23}$&lt;/span&gt; kg/m&lt;span class=&quot;math-container&quot;&gt;$^3$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is actually &lt;em&gt;lower&lt;/em&gt; by a factor of a thousand or so than the density of the interstellar medium. Thus accretion from the ISM is far more important on average. Putting the numbers in for a 1 solar mass black hole and &lt;span class=&quot;math-container&quot;&gt;$\lambda=0.01$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\eta =1$&lt;/span&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$L = 3\times 10^{23}\left(\frac{\rho}{10^{-20}\ {\rm kg m}^{-3}}\right)\left(\frac{v}{10\ {\rm km s}^{-1}}\right)^{-3}\ {\rm W.}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is quite an appreciable luminosity and much would emerge at UV and X-ray wavelengths. It would be readily detectable. The flux at a distance of 50,000 au would be &lt;span class=&quot;math-container&quot;&gt;$5\times 10^{-10}$&lt;/span&gt; W/m&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt;, which is &lt;strong&gt;4 orders of magnitude&lt;/strong&gt; above the faint flux limit of the &lt;a href=&quot;https://heasarc.gsfc.nasa.gov/docs/rosat/rass.html&quot; rel=&quot;noreferrer&quot;&gt;ROSAT all sky X-ray survey&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-10-08T07:51:07.427" LastActivityDate="2018-10-08T07:51:07.427" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27945" PostTypeId="2" ParentId="27940" CreationDate="2018-10-08T12:05:33.030" Score="1" Body="&lt;p&gt;SN Ia is a thermonuclear explosion. At very early time, Ni-56 is the main source of energy. Since the hard photons from the Ni-56 decay have to travel from the inside out, they interact with ejecta. The diffusion timescale due mainly to the mass of ejecta determines the peak timescale of a light curve. Check Arnett &lt;a href=&quot;http://adsabs.harvard.edu/abs/1980ApJ...237..541A&quot; rel=&quot;nofollow noreferrer&quot;&gt;1980&lt;/a&gt;, &lt;a href=&quot;http://adsabs.harvard.edu/abs/1982ApJ...253..785A&quot; rel=&quot;nofollow noreferrer&quot;&gt;1982&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fe-56 is not radioactive. Co-56 supplies energy at later times after the peak.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastEditorUserId="22468" LastEditDate="2018-10-08T18:53:42.383" LastActivityDate="2018-10-08T18:53:42.383" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27948" PostTypeId="2" ParentId="27942" CreationDate="2018-10-08T20:46:49.503" Score="2" Body="&lt;p&gt;Generally you use celestial coordinates, which are analogous to longitude and latitude, but instead use right ascension and declination. Right Ascension uses hours, minutes, and seconds, and Declination uses degrees, minutes, and seconds. Longitude and latitude are the same no matter where you are on the Earth. If you are at the North Pole, then the coordinates of New York City are the same as if you were standing in Beijing. However, with celestial coordinates it changes based on where you are, but the difference is very very small if you stay within the solar system and nearby stars. The difference gets larger when you are far away from our Sun, so generally you would need to declare a reference frame. For example, the location of Sirius from the Earth would be different than the location of Sirius from...say...Polaris. Also, due to the proper motion of the stars and the Milankovitch cycles (notably axial tilt), the coordinates would need to be updated every 50 or so years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the ISS, they use North, South, East, and West (they predefined where those pointed), and they also use Zenith and Nadir for the up and down directions, since you can move in three dimensions.&lt;/p&gt;&#xA;" OwnerUserId="24373" LastActivityDate="2018-10-08T20:46:49.503" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27950" PostTypeId="2" ParentId="27940" CreationDate="2018-10-09T08:05:22.100" Score="3" Body="&lt;p&gt;The early light curve of a Type Ia SN (i.e., the fact that the brightness increases to a peak, then declines) is the result of the combination of two things, both of which are decreasing over time: energy from the decay of Ni-56 to Co-56 (and Co-56 to Fe-56), and the opacity of the explosion as it expands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just after the actual explosion, the expanding fireball is dense and opaque. Only light from the very outermost layer escapes to the outside.[*] Since the Ni-56 created in the thermonuclear burning that caused the explosion is in the interior regions, the initially visible part of the expanding fireball has little or no Ni-56, and thus no energy from its decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the fireball expands, it gets less dense, and thus less opaque, to the point where light from the layers heated by radioactive decay can actually escape. Over time, more and more of the inner part of the fireball becomes visible, and so we see more and more of the regions where Ni-56 (and its daughter product Co-56) is decaying. Roughly speaking, the peak of the light curve is when the opacity is low enough that all the light from the Ni-56 + Co-56 decay can escape.[**]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since there are fewer and fewer radioactive nuclei as time goes by, the amount of energy produced by the decay decreases with time. This is why the light curve turns over: now we're seeing all the light produced by Ni-56 and Co-56 decay, but &lt;em&gt;that&lt;/em&gt; just keeps getting fainter over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[*] At this point, the SN is too faint to be seen. Even the initial thermonuclear explosion is too faint to be seen under most circumstances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[**] Ni-56 and Co-56 actually produce gamma rays (and energetic positrons) when they decay, so what &quot;the light from Ni-56 + Co-56 decay&quot; means is thermal radiation from the gas heated by the gamma rays. Also, as I understand it, what we're seeing is really the combination of energy from current decay &lt;em&gt;and&lt;/em&gt; residual heat from previous radioactive decay.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2018-10-12T14:10:59.873" LastActivityDate="2018-10-12T14:10:59.873" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27952" PostTypeId="2" ParentId="27912" CreationDate="2018-10-09T14:20:22.137" Score="2" Body="&lt;p&gt;The small mirror inside WFPC2 which provided the correction for that instrument is not visible in that photograph, for the simple reason that the photograph shows &lt;a href=&quot;https://en.wikipedia.org/wiki/Corrective_Optics_Space_Telescope_Axial_Replacement&quot; rel=&quot;nofollow noreferrer&quot;&gt;COSTAR&lt;/a&gt;, a difference device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For what it's worth, three of the four projecting arms in that picture carry the corrective mirrors that COSTAR used to fix the optical input to the Faint Object Camera, the Goddard High Resolution Spectrograph, and the Faint Object Spectrograph. You can see that the far-left and the second-from-right arms have two mirrors each (the small, dark, circular objects), which means they provide correction for the FOS and the FOC (I think the FOS corrector is the second-from-right arm); both of these instruments have two modes, requiring separate correction mirrors for each mode. One of the other arms has the correction mirror for GHRS. (I'll admit I don't know why there are four arms.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2018-10-09T14:20:22.137" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27953" PostTypeId="2" ParentId="6589" CreationDate="2018-10-10T03:47:34.453" Score="2" Body="&lt;p&gt;The only way we can detect black holes is through the effect that their gravitating mass has on objects outside the event horizon. These objects can gain energy from the black hole, as they fall into the potential well they heat up and radiate (often in the X-ray region of the spectrum). So black holes have gravitational energy from their mass. A black hole could also have charge, but this is rare in the universe which is largely neutral. Many black holes probably rotate, and these also have rotational energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://curious.astro.cornell.edu/disclaimer/86-the-universe/black-holes-and-quasars/general-questions/436-what-type-of-energy-does-a-black-hole-have-intermediate&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://curious.astro.cornell.edu/disclaimer/86-the-universe/black-holes-and-quasars/general-questions/436-what-type-of-energy-does-a-black-hole-have-intermediate&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24497" LastActivityDate="2018-10-10T03:47:34.453" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27954" PostTypeId="2" ParentId="27918" CreationDate="2018-10-10T04:53:53.860" Score="7" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$&#xA;\begin{array}{cccc}&#xA;                   \text{Body} &amp;amp; \text{Max} &amp;amp; \text{Min} &amp;amp; \text{Delta} \\&#xA;                   \text{Sun} &amp;amp; -26.78 &amp;amp; -26.71 &amp;amp; 0.07 \\&#xA;                   \text{Mercury} &amp;amp; -2.45 &amp;amp; 5.58 &amp;amp; 8.03 \\&#xA;                   \text{Venus} &amp;amp; -4.89 &amp;amp; -3.82 &amp;amp; 1.07 \\&#xA;                   \text{Moon} &amp;amp; -12.87 &amp;amp; -3.76 &amp;amp; 9.11 \\&#xA;                   \text{Mars} &amp;amp; -2.88 &amp;amp; 1.84 &amp;amp; 4.72 \\&#xA;                   \text{Jupiter} &amp;amp; -2.94 &amp;amp; -1.66 &amp;amp; 1.28 \\&#xA;                   \text{Saturn} &amp;amp; 0.42 &amp;amp; 1.47 &amp;amp; 1.05 \\&#xA;                   \text{Uranus} &amp;amp; 5.31 &amp;amp; 5.95 &amp;amp; 0.64 \\&#xA;                   \text{Neptune} &amp;amp; 7.8 &amp;amp; 8. &amp;amp; 0.2 \\&#xA;                   \text{Pluto} &amp;amp; 13.75 &amp;amp; 15.96 &amp;amp; 2.21 \\&#xA;                   \text{Comet Halley} &amp;amp; 2 &amp;amp; 25.66 &amp;amp; 23.66 \\&#xA;                   \text{Tesla Roadster} &amp;amp; 6.66 &amp;amp; 29.29 &amp;amp; 22.63 \\&#xA;                  \end{array}&#xA;$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The answer is Mercury (as above), subject to the proecedure/caveats below:&lt;/p&gt;&#xA;&lt;h3&gt;Procedure&lt;/h3&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I used HORIZONS to generate daily brightness data for a century for:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;all the planets as viewed from Earth (except Earth itself)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the Sun, the Moon, Pluto, Comet Halley, and the Tesla Roadster&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I then noted the minimum and maximum brigtness, along with the magnitude difference of these brightnesses, in the table above.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can use HORIZONS to compute the results yourself, or view the results in the *-brightness.txt.bz2 files in &lt;a href=&quot;https://github.com/barrycarter/bcapps/blob/master/ASTRO/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/blob/master/ASTRO/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h3&gt;Caveats&lt;/h3&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Although Mercury's brightness changes more than Mars, the Sun's glare makes it impossible to see Mercury though Earth's atmosphere when Mercury's angular distance from the Sun is small. Therefore, Mars may be a better practical answer.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Because I used daily brightnesses, it's theoretically possible I missed absolute (intraday) minimums or maximums, especially for the Moon, whose brightness changes rapidly. However:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I took 100 years worth of data, and the moon's brightness doesn't have a period that's a multiple of one day. Since the moon's synodic period is approximately 29.5 days, its brightness is almost periodic in 59 days (2 synodic periods), but it's far enough from 29.5 days that this shouldn't be too much of a problem.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Moon isn't a planet: I just added it for reference&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Unless the Moon's actual brightness difference was higher than 22.64 magnitudes (which is probably unlikely[?]), it would remain in 3rd place in terms of brightness change, so the exact value isn't as important.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Because I used only a 100 year period, I did not include a complete orbit for either Neptune or Pluto. This shouldn't be an issue because:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The synodic period of both planets is just over a year, and much of the brightness change comes from Earth's own orbit, not Neptune's or Pluto's.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Even if the maximum magnitude change were slightly higher than in the table, it wouldn't make much of a difference.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Note that &amp;quot;Max&amp;quot; and &amp;quot;Min&amp;quot; refer to brightness, which is ordered the opposite of magnitude: lower magnitude means greater brightness.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Because I used daily data, I missed rare events such as transits and eclipses. The table above is for an &amp;quot;average&amp;quot; orbit, excluding special cases.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In some cases, HORIZONS gives &amp;quot;n.a.&amp;quot; for magnitude data. I ignore these &amp;quot;n.a.&amp;quot; values.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Data for the Tesla Roadster is only available from 2018-Feb-07 03:00 UTC to 2090-Jan-01 23:00 UTC, not the entire century.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h3&gt;Complexity&lt;/h3&gt;&#xA;&lt;p&gt;The problem is nontrivial. As you correctly note, the planet's geocentric and heliocentric distance play into the formula, but there's more to it.&lt;/p&gt;&#xA;&lt;p&gt;Quoting Oliver Montenbruck and Thomas Pfleger's &amp;quot;Astronomy on the Personal Computer&amp;quot; (&lt;a href=&quot;https://books.google.com/books?id=nHUqBAAAQBAJ&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://books.google.com/books?id=nHUqBAAAQBAJ&lt;/a&gt;):&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CzYLg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CzYLg.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/wPjLz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wPjLz.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Paul Schlyter's &lt;a href=&quot;http://www.stjarnhimlen.se/comp/ppcomp.html#15&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.stjarnhimlen.se/comp/ppcomp.html#15&lt;/a&gt; provides similar nontrivial formulas.&lt;/p&gt;&#xA;&lt;p&gt;EDIT: &lt;a href=&quot;https://en.wikipedia.org/wiki/Phase_curve_(astronomy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Phase_curve_(astronomy)&lt;/a&gt; provides more information on how planetary brightness varies nontrivially with planetary phase.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-10-10T17:08:51.180" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27955" PostTypeId="2" ParentId="27914" CreationDate="2018-10-10T05:16:09.403" Score="2" Body="&lt;p&gt;(too long for a comment, this answers why barycenter data is available for a much longer time span than planet center data)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When NASA computes planetary positions, they treat each planet and its moons as a single point mass (the barycenter of the planet and its moons). They compute these positions +-15000 years from now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For most planet systems, the barycenter is very close to the planet center (Mars' &lt;em&gt;planet&lt;/em&gt; barycenter is only a few inches from Mars' &lt;em&gt;system&lt;/em&gt; barycenter because Phobos and Deimos have such little mass-- for the moonless Mercury and Venus, the system barycenters and the planet barycenters are the same).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They then compute the locations of the moons and the planet's center itself. These calculations are done for a much shorter period of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, you can compute the location of the barycenter of Jupiter and its moons for +-15000 years, but the actual barycenter of Jupiter itself for a much shorter period of time, even though Jupiter is massive enough that its own barycenter is very close to the barycenter of Jupiter and its moons.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastActivityDate="2018-10-10T05:16:09.403" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27957" PostTypeId="1" AcceptedAnswerId="27959" CreationDate="2018-10-10T07:25:00.543" Score="5" ViewCount="306" Body="&lt;p&gt;I want to know the explanation of Phillips relation, the relation seen in type Ia supernovae(SN Ia with broader light curve is brighter). More Ni-56 implies a larger luminosity at peak but what about broader light curve?&lt;/p&gt;&#xA;" OwnerUserId="24460" LastEditorUserId="2365" LastEditDate="2018-10-10T09:03:31.297" LastActivityDate="2018-10-10T11:15:16.553" Title="Why is there a Phillips relation?" Tags="&lt;supernova&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27959" PostTypeId="2" ParentId="27957" CreationDate="2018-10-10T11:06:36.827" Score="5" Body="&lt;p&gt;Check this paper: &lt;a href=&quot;http://adsabs.harvard.edu/abs/2001ASPC..229..287M&quot; rel=&quot;noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2001ASPC..229..287M&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LargerNi-56 directly implies brighter at peak, since it is the dominant source of radiation energy at early times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The broadness of a light curve depends on the light curve timescale, which depends on ejecta mass, opacity, and kinetic energy. Since SNe Ia are from WD progenitors, there are not much different in ejecta mass. From simulations, kinetic energy is roughly the same across broad parameters. This leaves opacity to be the origin of the broadness. Since Fe-group elements have higher opacity than intermediate mass elements (IME), and more Ni-56 implies more Fe-group/IME, it is to be expected that high Ni-56 implies high opacity that implies broader light curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slide, that someone provided in your comment, explains the same idea here (with more references).&lt;/p&gt;&#xA;" OwnerUserId="22468" LastEditorUserId="22468" LastEditDate="2018-10-10T11:15:16.553" LastActivityDate="2018-10-10T11:15:16.553" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27960" PostTypeId="1" CreationDate="2018-10-10T11:29:56.580" Score="2" ViewCount="424" Body="&lt;p&gt;From what I understand, &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Starry night (&lt;a href=&quot;https://starrynight.com/starry-night-7-professional-astronomy-telescope-control-software.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://starrynight.com/starry-night-7-professional-astronomy-telescope-control-software.html&lt;/a&gt;) is more focussed on simulating the night sky whereas &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Alcyone (&lt;a href=&quot;http://www.alcyone.de/aat_list_of_available_tables.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.alcyone.de/aat_list_of_available_tables.html&lt;/a&gt;) is more focussed on creating data tables.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Both programs provide simulation of the past (Starry Night going farther into the past while Alcyone stops at -3000), which is actually what I am looking for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that, 2 years after buying Starry Night and being confirmed a bug makes it impossible to export ephemerides prior to 1600 AD, I'm told it was never part of the features at all - contrary to their marketing material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'm stuck trying to find another way of getting the information I need (outside of pointing to each object and writing the information by hand). Does anyone know whether the Alcyone software is reliable? It does not go so back in time as I would need it to, but -3000 is far better than 1600AD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Further clarification:&lt;/strong&gt; I have found the following webpage (&lt;a href=&quot;http://www.archaeocosmology.org/eng/skyprog.htm#Evaluation&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.archaeocosmology.org/eng/skyprog.htm#Evaluation&lt;/a&gt;) and I quote:&#xA;&quot;This page will provide some insight on who to make a choice in planetarium computer programs. This will be done along the following lines:&quot;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a specification is made on the features needed in such a program.&#xA;an evaluation will be given of the accuracy reachable with present theories.&#xA;a sensitivity analysis will be made with regard to:&#xA;    azimuth/altitude accuracies of normal celestial events&#xA;    local time/location/declination accuracy with regard to solar eclipses/occultations&#xA;    set/rise events&#xA;a benchmark will be proposed to check planetarium computer programs&#xA;an evaluation of some 20 computer programs&#xA;and of course I want to thank people who helped me with this!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, Alcyone is not mentioned here, and I have found no review whatsoever or comment about it's accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At Alcyone's website they say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The ephemeris calculation is based upon Steve Moshier's analytical ephemeris using trigonometric expansions for the earth and planets and the lunar ephemeris ELP2000-85 of Chapront-Touzé and Chapront for the moon, both adjusted to Jet Propulsion Laboratory's DE404 (see www.moshier.net). There are further adjustments in Alcyone Ephemeris, some optional, to JPL's more recent DE406, the most accurate long-term ephemeris.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;All eclipse predictions were performed by Fred Espenak and Jean Meeus (NASA's GSFC).&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The calculation of the local circumstances of solar eclipses is based upon the data from the Five Millenium Canon of Solar Eclipses -1999 to 3000 (Fred Espenak, NASA's GSFC) with Besselian Elements provided by Jean Meuus. The calculation of the local circumstances of lunar eclipses is based upon the data from the Five Millenium Canon of Lunar Eclipses -1999 to 3000 (Fred Espenak, NASA's GSFC).&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for Starry Night, which I have found reviews concerning its accuracy, they say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The most precise planetary positions available (using JPL ephemerides) plus an expanded asteroid catalog with highlighted families and groups.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Checkout the simple demonstration video showcasing the Starry Night Pro 7 Time Travel function. In this video, we travel through 200,000 years, from 99999 BC to 99999 AD, at 100 year intervals. Intervals can be set for seconds, minutes, hours, days, weeks, months, years, etc.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;You’ll always know where and when to look with Starry Night’s built-in ephemeris generator. It’s a handy observational aid that creates a table of positional data for any object over the time span and at the interval you specify. You can then export the generated values to a text file and print it out. Ephemeris values are handy for knowing where an object will be in the sky at a particular time. If you are using manual or digital setting circles, for instance, you can dial in the generated positional data to locate an object quickly. For example, if you are tracking the path of a fast-moving asteroid as it makes its closest approach to Earth, your ephemeris table will let you know exactly where the asteroid will be during the time period you’ll be observing it.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as the ephemerides don't work before 1600 AD, I'm stuck at watching the night sky in the hopes of finding a past event instead of getting its date out of the ephemerides search - I was hoping I could still make use of Starry Night (It was very expensive!) by using some other program to find event dates so I know when to look for them in Starry Night. I know they'll probably have slightly different dates for the distant past, but it'll give me a pointer in the right direction - unless the level of accuracy is very different. Then there'll be no hope to use both programs in tandem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further Clarifications&lt;br&gt;&#xA;&lt;strong&gt;1) The exact date for which you're looking for ephemerides,&lt;/strong&gt;&lt;br&gt;&#xA;From about -10,000 to about 2,500.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2) The objects you're interested in,&lt;/strong&gt;&lt;br&gt;&#xA;Mostly the longitude of the Sun as referring to Earth; longitude and distance of the planets (Mercury to Saturn)  referring to the Sun; Moon phases; solar and lunar eclipses&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3) How you intend to use the data.&lt;/strong&gt;&lt;br&gt;&#xA;I'll be collecting the data in excel so as to create a couple of calendars based on astronomical events for a fictional story. I have already plotted out a basic one without astronomic connections (it's juxtaposed to Julian day and Julian/Gregorian years to allow easy identification of dates), but there are two others which I need to juxtapose that are purely solar and purely lunar, with the planets being important to create sort of decades. It is only important to note where the planets are in the ecliptic (or zodiac houses, that is, 30º areas in the sky). The stars themselves aren't important, only the course of planets, sun and moon in the sky.&lt;/p&gt;&#xA;" OwnerUserId="24500" LastEditorUserId="24500" LastEditDate="2018-10-12T19:09:22.197" LastActivityDate="2019-06-19T13:03:25.433" Title="Comparing accuracy in Starry Night and Alcyone Software" Tags="&lt;night-sky&gt;&lt;software&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27961" PostTypeId="2" ParentId="27897" CreationDate="2018-10-10T12:37:56.643" Score="0" Body="&lt;p&gt;@astrosnapper's &lt;a href=&quot;https://astronomy.stackexchange.com/a/27906/7982&quot;&gt;answer&lt;/a&gt; is probably the one you want. But if you'd like something you can implement yourself, have a look at &lt;strong&gt;Astronomical Algorithms&lt;/strong&gt; in &lt;a href=&quot;https://astronomy.stackexchange.com/a/18029/7982&quot;&gt;this answer&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;This is something you'll have to dig into a bit, but if you like to program, it may be exactly what you're looking for. The &lt;a href=&quot;http://www.gaisma.com/en/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gaisma website&lt;/a&gt; is one of my favorites on the internet - easy to use and presents a bunch of information in easy-to-understand graphics. Click around!&lt;/p&gt;&#xA;&lt;p&gt;I believe that this site uses algorithms from the collection found at &lt;a href=&quot;http://www.esrl.noaa.gov/gmd/grad/solcalc/calcdetails.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this NOAA site&lt;/a&gt;. Click around there as well. They provide Excel spreadsheets which contain the algorithms and other resources. The &amp;quot;main&amp;quot; resource is a collection of algorithms published in the book &lt;a href=&quot;http://www.willbell.com/math/mc1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomical Algorithms - Jean Meeus&lt;/a&gt;. Search the books title and you can find that there are many similarly titled books. I'd recommend going to a library if possible, because it's (in my opinion) always good to go to libraries. However parts of these can be found on-line. For example, a few pages shown from the book &lt;a href=&quot;http://www.math.harvard.edu/%7Eknill/3dprinter/exhibits/bc/Meeus1988.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomical Formulae for Calculators (1988)&lt;/a&gt; include an interesting table of contents.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-10-10T12:37:56.643" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27962" PostTypeId="2" ParentId="27919" CreationDate="2018-10-10T13:32:23.350" Score="3" Body="&lt;p&gt;@called2voyage's &lt;a href=&quot;https://astronomy.stackexchange.com/questions/27919/move-asteroid-away-with-white-color#comment48966_27919&quot;&gt;comment&lt;/a&gt; points you to an in depth discussion. I'll add to that a simple summary of some things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you imagine that the astroid is in a simple, circular orbit around the Sun that's defined only by gravity, let's call that the nominal orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the effects of sunlight. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;If the asteroid were white or reflective&lt;/em&gt;, then a lot of the incident sunlight will reflect back towards the Sun, pushing the asteroid away from the Sun. Since both gravitational attraction and the Sun's brightness vary as &lt;span class=&quot;math-container&quot;&gt;$1/r^2$&lt;/span&gt;, so to the asteroid it feels like gravity is just a little weaker or the Sun is a little lighter, but otherwise still feels like gravity. If the asteroid became reflective slowly, it would just move to a slightly larger radius and remain in circular orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that the incident photons, and the reflected photons both contribute to the push!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;If the asteroid were black and absorbing&lt;/em&gt;, then the absorption of the photons would give a push, and then they would warm the asteroid. Warm things radiate heat, and once equilibrium were reached, the radiation would also be a source of pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the astroid didn't rotate fast, but instead turned slowly so that it had a warm side always facing the Sun and a cold side always facing away, then the radiated infrared photons &lt;em&gt;would also result in a push away from the Sun!&lt;/em&gt; So a dark asteroid that turned so slow that it had a permanent hot and permanent cold side would also be at a larger radius than nominal. The difference between the two depends on the functional form of the diffusivity and roughness of the surfaces.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt; if the dark asteroid were rotating fast, say every few hours or tens of hours, then the radiation would be a little stronger in the forward or backward direction with respect to the orbit. If it was rotating the same sense as its orbit, the hot side would turn backwards, radiate backwards, and push the asteroid forward. That would slowly, but continuously raise the orbit of the asteroid. The longer this happened, the larger and larger its radius would become. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Yarkovsky_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Yarkovsky effect&lt;/a&gt; and over long periods of time (millions of years) can have a big impact. For an asteroid to just miss the Earth, maybe a few years would be enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that @ BobJacobsen's &lt;a href=&quot;https://space.stackexchange.com/a/28202/12102&quot;&gt;answer&lt;/a&gt; to &lt;a href=&quot;https://space.stackexchange.com/q/26131/12102&quot;&gt;Why paint only one-half of Bennu?&lt;/a&gt; is, in a highly condensed and stylized form; &quot;because the Yarkovsky effect is strong enough that there's no point in wasting good 'space paint' on the other half.&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;If you are a tiny dust particle&lt;/em&gt; near the Sun, there is yet another thing that happens,  called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Poynting%E2%80%93Robertson_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Poynting–Robertson effect&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tiny bits of dust have a higher area to mass ratio, so even though they may be in thermal equilibrium and not have a hot and cold side, tiny forces make a big difference. And if you are close to the Sun, then the forces are &lt;em&gt;less tiny&lt;/em&gt; i.e. larger. @honeste_vivere's &lt;a href=&quot;https://astronomy.stackexchange.com/a/16803/7982&quot;&gt;excellent answer&lt;/a&gt; points out that the Poynting–Robertson effect can cause the radius of the dust particles' orbits to decrease over time. An effect similar to (identical to) &lt;a href=&quot;https://en.wikipedia.org/wiki/Aberration_of_light&quot; rel=&quot;nofollow noreferrer&quot;&gt;astronomical aberration&lt;/a&gt; will cause more pressure at the front of the dust particle, pushing it backwards and causing it to spiral closer to the Sun over time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-10-10T13:32:23.350" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27963" PostTypeId="1" CreationDate="2018-10-10T14:30:30.193" Score="0" ViewCount="111" Body="&lt;p&gt;This is a somewhat simplified, and drawn / formulated in a manner more targeted to astronomy, version of a diagram and a question that I've also posted in &lt;a href=&quot;https://math.stackexchange.com/q/2949028/50256&quot;&gt;math.stackexchange.com&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the diagram below I depict:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the celestial sphere with the North Celestial Pole (NCP), the South Celestial Pole (SCP) and the celestial equator&lt;/li&gt;&#xA;&lt;li&gt;a latitude circle at latitude &lt;span class=&quot;math-container&quot;&gt;$\delta$&lt;/span&gt; (declination)&lt;/li&gt;&#xA;&lt;li&gt;two points &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; on that latitude circle&lt;/li&gt;&#xA;&lt;li&gt;the center of the celestial sphere &lt;span class=&quot;math-container&quot;&gt;$O$&lt;/span&gt; and the center of the latitude circle &lt;span class=&quot;math-container&quot;&gt;$O'$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x03mQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x03mQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given angle &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt; between the rays &lt;span class=&quot;math-container&quot;&gt;$OA$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$OB$&lt;/span&gt;, what is the angle &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; between the rays &lt;span class=&quot;math-container&quot;&gt;$O'A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$O'B$&lt;/span&gt; ? I am trying to derive &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; given &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\delta$&lt;/span&gt;. Observe that if &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is the first point of Aries, then the angle &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; is the Right Ascension of a star located at point &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My assumption is that the following formula holds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\phi = \theta*cos(\delta)$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="522" LastEditorUserId="522" LastEditDate="2018-10-11T13:50:26.630" LastActivityDate="2020-06-20T06:01:10.103" Title="angles on the celestial sphere" Tags="&lt;coordinate&gt;&lt;declination&gt;&lt;mathematics&gt;&lt;right-ascension&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27964" PostTypeId="1" AcceptedAnswerId="27973" CreationDate="2018-10-10T21:42:19.227" Score="1" ViewCount="399" Body="&lt;p&gt;I have downloaded a star database from this page (The HYG Database):&lt;br&gt;&#xA;&lt;a href=&quot;http://astronexus.com/node/34&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://astronexus.com/node/34&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And all the stars have (among other parameters) their XYZ coordinates in parsecs according to this definition:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;X,Y,Z&lt;/strong&gt;: The Cartesian coordinates of the star, in a system based on the&#xA;  equatorial coordinates as seen from Earth. +X is in the direction of&#xA;  the vernal equinox (at epoch 2000), +Z towards the north celestial&#xA;  pole, and +Y in the direction of R.A. 6 hours, declination 0 degrees&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The problem is: I cannot figure out how to calculate the XYZ coordinates of the center of the galaxy (the Milky Way). So the question is: What are the coordinates of the Milky Way center? (Or is it not possible to calculate it just with that information?). This is an example of the data:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/q4PJB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q4PJB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24506" LastEditorUserId="23940" LastEditDate="2018-10-12T16:01:57.107" LastActivityDate="2018-10-12T16:01:57.107" Title="Where is the center of the Milky Way located?" Tags="&lt;distances&gt;&lt;milky-way&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27965" PostTypeId="2" ParentId="27963" CreationDate="2018-10-10T22:13:59.217" Score="1" Body="&lt;p&gt;This answer in &lt;a href=&quot;https://math.stackexchange.com/a/2949040/50256&quot;&gt;math.stackexchange&lt;/a&gt; provides the correct relationship between &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt;. And &lt;a href=&quot;https://math.stackexchange.com/a/2950640/50256&quot;&gt;this answer&lt;/a&gt; supplies the proof.&lt;/p&gt;&#xA;" OwnerUserId="522" LastActivityDate="2018-10-10T22:13:59.217" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27966" PostTypeId="1" CreationDate="2018-10-10T22:50:11.450" Score="7" ViewCount="375" Body="&lt;p&gt;How could this be estimated, or how accurately could it be extrapolated from a present-day measurement of the Earth's spin-down rate?&lt;/p&gt;&#xA;" OwnerUserId="24486" LastEditorUserId="2531" LastEditDate="2018-10-11T15:45:48.243" LastActivityDate="2018-10-12T01:12:13.517" Title="What was the length of the solar day 73 million years ago?" Tags="&lt;astrophysics&gt;&lt;fundamental-astronomy&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27967" PostTypeId="2" ParentId="27966" CreationDate="2018-10-11T05:05:18.773" Score="13" Body="&lt;p&gt;There is evidence from &lt;a href=&quot;https://www.theatlantic.com/science/archive/2016/02/fossilized-coral-calendar-changes-leap-day/471180/&quot; rel=&quot;noreferrer&quot;&gt;Silurian coral fossils&lt;/a&gt; that, 430 My ago, there were 420 daily rotations in each annual season, compared to 365 days per year now.  If we assume that the year (fixed by mean Earth-Sun distance and Kepler's Laws) hasn't changed much in that interval, but that instead the day has gotten longer due to the Earth-Moon tidal interaction, that means that 430 My ago there were about 20.9 hours in each day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Linear extrapolation in between says that only about one-sixth of that change has taken place in the past 73 My, which would give a day length of about 23.5 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a comment, &lt;a href=&quot;https://astronomy.stackexchange.com/users/17355/johnholtz&quot;&gt;JohnHoltz&lt;/a&gt; cautiously extrapolates your currently-known slowing of 0.0016 seconds/century over that interval, which gives an ancient day length of closer to 23.7 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;JohnHoltz's estimate is probably better. Tidal slowing is occurring as the mean distance between the Earth and Moon increases, which makes all tidal effects (including tidal slowing) weaker. So a linear extrapolation from today's well-measured rate of rotational slowing is probably more accurate for recent times than a linear extrapolation over a much longer period that probably includes more rapid slowing in the past. It inspires confidence that the two methods come out close to each other, though.&lt;/p&gt;&#xA;" OwnerUserId="1498" LastEditorUserId="2531" LastEditDate="2018-10-11T06:48:36.097" LastActivityDate="2018-10-11T06:48:36.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27969" PostTypeId="1" AcceptedAnswerId="27970" CreationDate="2018-10-11T13:08:34.293" Score="3" ViewCount="270" Body="&lt;p&gt;We are not able to quantize spacetime and the theory of  &lt;strong&gt;loop quantum gravity&lt;/strong&gt; has been proved wrong but we still use the Planck length in astrophysics. Why?&lt;/p&gt;&#xA;" OwnerUserId="24509" LastActivityDate="2018-10-12T07:34:11.310" Title="Why do we use the Planck length?" Tags="&lt;space-time&gt;&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27970" PostTypeId="2" ParentId="27969" CreationDate="2018-10-11T14:18:08.373" Score="5" Body="&lt;p&gt;First off, you're much more likely to see parsecs, light years, or astronomical units than Planck lengths in astrophysics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for those who like to work in a &lt;a href=&quot;https://en.wikipedia.org/wiki/Natural_units&quot; rel=&quot;nofollow noreferrer&quot;&gt;system of natural units&lt;/a&gt;, the Planck length is &lt;strong&gt;the&lt;/strong&gt; obvious choice for the unit of length. This choice has nothing to do with quantizing spacetime. It instead results from a desire to make conversion factors such as the speed of light &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;, the reduced Planck constant &lt;span class=&quot;math-container&quot;&gt;$\hbar$&lt;/span&gt;, the Boltzmann constant &lt;span class=&quot;math-container&quot;&gt;$k_b$&lt;/span&gt;, and the gravitational constant &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; all have a numerical value of one or some other nice number such as &lt;span class=&quot;math-container&quot;&gt;$&#xA;\frac{1}{4\pi}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to look at this: Compare the conventional units used in the US and a few other backward countries. These countries have to use &lt;span class=&quot;math-container&quot;&gt;$F=kma$&lt;/span&gt; rather than the &lt;span class=&quot;math-container&quot;&gt;$F=ma$&lt;/span&gt; used in the metric system. The &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; is a conversion factor that results from having those conventional units being an inherently inconsistent system. From a modern perspective, force &lt;em&gt;is&lt;/em&gt; the product of mass and acceleration; no conversion should be needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, that the mass-energy equivalence equation expressed in metric units is &lt;span class=&quot;math-container&quot;&gt;$E^2=\bigl(m c^2\bigr)^2 + \bigl(pc\bigr)^2$&lt;/span&gt; is a sign that the metric system, too, is inconsistent. With natural units, this expression simplifies to &lt;span class=&quot;math-container&quot;&gt;$E^2=m^2+p^2$&lt;/span&gt;. Energy, mass, and momentum all have the same units; no conversion is needed.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="14152" LastEditDate="2018-10-12T07:34:11.310" LastActivityDate="2018-10-12T07:34:11.310" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="27971" PostTypeId="2" ParentId="27969" CreationDate="2018-10-11T14:48:59.000" Score="3" Body="&lt;p&gt;The Planck length also still has meaning. &lt;a href=&quot;http://www.physlink.com/Education/AskExperts/ae281.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;This source&lt;/a&gt; describes it as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Planck length is the scale at which classical ideas about gravity and space-time cease to be valid, and quantum effects dominate. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we may not be sure what happens on such small length scales, but we can be sure that it is &lt;em&gt;not&lt;/em&gt; even approximately described by &quot;classical&quot; space-time. Conversely, for larger distances, and absent very large concentrations of energy of some kind, it is reasonable to expect that classical space-time might be a reasonable approximation to the universe.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-10-11T14:48:59.000" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27973" PostTypeId="2" ParentId="27964" CreationDate="2018-10-11T17:49:09.220" Score="6" Body="&lt;p&gt;Distance estimates to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_Center&quot; rel=&quot;noreferrer&quot;&gt;center of our galaxy&lt;/a&gt; vary around &lt;span class=&quot;math-container&quot;&gt;$r \approx$&lt;/span&gt; 8000 &lt;span class=&quot;math-container&quot;&gt;$\pm$&lt;/span&gt; 200 parsecs.&#xA;Its spherical &lt;a href=&quot;https://en.wikipedia.org/wiki/Equatorial_coordinate_system&quot; rel=&quot;noreferrer&quot;&gt;equatorial coordinates&lt;/a&gt; are &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; = 17h45m40s, &lt;span class=&quot;math-container&quot;&gt;$\delta$&lt;/span&gt; = -29&lt;span class=&quot;math-container&quot;&gt;$^\circ$&lt;/span&gt;00'28&quot; (J2000).&#xA;Converting those to rectangular form, we have:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;X &amp;amp;= r \cos \delta \cos \alpha \approx -437 \pm 11 \ \mathrm{pc} \\&#xA;Y &amp;amp;= r \cos \delta \sin \alpha \approx -6983 \pm 175 \ \mathrm{pc} \\&#xA;Z &amp;amp;= r \sin \delta \approx -3879 \pm 97 \ \mathrm{pc}&#xA;\end{align}$$&lt;/span&gt;&#xA;As with most celestial objects, its direction is known much more precisely than its distance.&#xA;We can't pinpoint its 3D position, but we can narrow it down to a long, fuzzy ellipsoid pointing away from the Sun.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-10-11T17:49:09.220" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27974" PostTypeId="2" ParentId="27605" CreationDate="2018-10-11T18:28:53.780" Score="1" Body="&lt;p&gt;From a comment by &lt;a href=&quot;https://astronomy.stackexchange.com/users/21/barrycarter&quot;&gt;barrycarter&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In our galaxy, it might be &lt;a href=&quot;https://www.jpl.nasa.gov/news/news.php?feature=6578&quot; rel=&quot;nofollow noreferrer&quot;&gt;CX330&lt;/a&gt;, which is 1000 light years from any star-forming region, much less any star. However, the article describes it as &quot;Loneliest Young Star&quot;, so there may be older stars that are even more isolated. &lt;a href=&quot;https://www.smh.com.au/technology/cx330-galaxys-loneliest-baby-star-discovered-wandering-far-from-stellar-nurseries-20160728-gqfg1i.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This source&lt;/a&gt; contains the phrase &quot;CX330 [...] has been described as the galaxy's loneliest star.&quot;, but this may just be bad reporting.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1498" LastActivityDate="2018-10-11T18:28:53.780" CommentCount="0" CommunityOwnedDate="2018-10-11T18:28:53.780" ContentLicense="CC BY-SA 4.0" />
  <row Id="27975" PostTypeId="1" AcceptedAnswerId="27987" CreationDate="2018-10-11T20:15:40.853" Score="6" ViewCount="174" Body="&lt;p&gt;I've noticed that sunrises are not reverse sunsets regarding the sky hues and lighting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sunrises always start pinkish and end with a glorious yellow sunshine.&#xA;On the other hand as the sun sets lighting gets orange and the sky gets a hellish red colour. I've noticed this many times regardless the weather or temperture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this contrast between colors (pastel-like in the morning vs vivid in the evening) in a fenomenon that should be same in both instances in a day ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I the only one that perceives this difference ?&lt;/p&gt;&#xA;" OwnerUserId="22022" LastActivityDate="2018-10-12T15:40:52.863" Title="Why are there optical differences between sunrise and sunset?" Tags="&lt;optics&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27976" PostTypeId="2" ParentId="27966" CreationDate="2018-10-12T01:12:13.517" Score="2" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$~\quad~$&lt;/span&gt; &lt;em&gt;A quick Google search after &lt;a href=&quot;http://www.google.com/search?q=daylength+dinosaurs&quot; rel=&quot;nofollow noreferrer&quot;&gt;daylength dinosaurs&lt;/a&gt; reveals the following relevant results :&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Period : Upper Cretaceous &amp;ndash; Age (years) : 70 million &amp;ndash; Days per &#xA;  year : 370 &amp;ndash; Hours per day : 23.7&lt;/em&gt; &amp;mdash; &lt;a href=&quot;http://spacemath.gsfc.nasa.gov/earth/6Page58.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;If we say the &quot;time of the dinosaurs&quot; started about 200 million years ago, then the years has lengthened by 1.2 million seconds in that time - or just under an hour a day less than today. To get all the way to a day of 17 hours, you need to go back a little before the dinosaurs - about 2 billion years.&lt;/em&gt; &amp;mdash; &lt;a href=&quot;http://physics.stackexchange.com/questions/179649&quot;&gt;Physics Stack Exchange&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;This means, for instance, that 100m years ago, when dinosaurs ruled Earth, a day was nearer 23 than 24 hours.&lt;/em&gt; &amp;mdash; &lt;a href=&quot;http://www.economist.com/science-and-technology/2016/12/10/ancient-eclipses-show-how-days-are-getting-longer&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Economist&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Fossil evidence informs us about how the day-length has changed over geological time. When the dinosaurs first walked the Earth there were about 375 days in a year, 10 more than at present, each day more than 30 minutes shorter than now.&lt;/em&gt; &amp;mdash; &lt;a href=&quot;http://www.independent.co.uk/arts-entertainment/historical-notes-total-eclipses-old-hat-for-dinosaurs-1108786.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Independent&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;For Jurassic-era stegosauruses 200 million years ago, the day was perhaps 23 hours long and each year had about 385 days. Two hundred million years from now, the daily dramas for whatever we evolve into will unfold during 25-hour days and 335-day years.&lt;/em&gt; &amp;mdash; &lt;a href=&quot;http://www.washingtonpost.com/2013/09/23/a75c548a-f2dc-11e2-ae43-b31dc363c3bf_story.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Washington Post&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Meyers and a colleague recently showed that 1.4 billion years ago, a single “day” lasted only about 18 hours. Back in the Proterozoic, a few million years after multicellular life appeared on the planet, the moon was about 211,825 miles away from Earth—about 21,000 miles closer than it is today. This, they figured out, compressed the length of a day down to 18.68 hours. By the time land plants evolved about 470 million years ago, the days had stretched to 21.46 hours. And when the dinosaurs showed up 225 million years ago, the days lasted 22.68 hours. The roughly 24-hour day we have now was almost in place as the first ice sheets bloomed near the Northern Hemisphere’s poles about 3.5 million years ago.&lt;/em&gt; &amp;mdash; &lt;a href=&quot;http://news.nationalgeographic.com/2018/06/summer-solstice-2018--days-are-getting-longer&quot; rel=&quot;nofollow noreferrer&quot;&gt;National Geographic&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;So would the day length during the age of the dinosaurs have been 21 hours? &quot;The dinosaurs were around 100 million years ago, which at the current rate [ of day lengthening ] adds up to 2000 seconds, which is less than an hour.&quot; Geological evidence for increasing day length can help us pin this time down more accurately. Tidal records laid down in ancient estuaries can show daily, monthly and seasonal cycles in alternating deposits of sand and silt. They indicate that 620 million years ago the day was 21 hours, says Mardling. Since the dinosaurs lived during the Mesozoic era, from 250 million years ago to 65 million years ago, day length would have been longer than 21 hours and probably closer to 23 hours.&lt;/em&gt; &amp;mdash; &lt;a href=&quot;http://www.abc.net.au/science/articles/2012/11/28/3642932.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;ABC Science&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14152" LastActivityDate="2018-10-12T01:12:13.517" CommentCount="0" CommunityOwnedDate="2018-10-12T01:12:13.517" ContentLicense="CC BY-SA 4.0" />
  <row Id="27977" PostTypeId="1" AcceptedAnswerId="27978" CreationDate="2018-10-12T02:22:07.527" Score="19" ViewCount="3026" Body="&lt;p&gt;If I connect an eight foot &lt;a href=&quot;https://en.wikipedia.org/wiki/Yagi%E2%80%93Uda_antenna&quot; rel=&quot;noreferrer&quot;&gt;Yagi&lt;/a&gt; or other comparable sized antenna to my oscilloscope and point the antenna at a bright star will I see a voltage on my oscilloscope?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not interested in turning the voltage into an image just wondering if I would see a voltage increase when it is on a bright star. I’d like to know your thoughts before I take the time to build the antenna. I’m thinking about in the 25cm range. I’ve heard that’s an active area. My oscilloscope will read down to about 20 millivolts.&lt;/p&gt;&#xA;" OwnerUserId="19597" LastEditorDisplayName="user1569" LastEditDate="2018-10-12T12:24:29.630" LastActivityDate="2018-10-12T17:19:47.923" Title="Can I sense a bright star pointing an eight foot antenna towards it?" Tags="&lt;star&gt;&lt;radio-astronomy&gt;&lt;radio-telescope&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27978" PostTypeId="2" ParentId="27977" CreationDate="2018-10-12T03:44:32.750" Score="30" Body="&lt;p&gt;Stars are too dim for amateur radio equipment. There are two possible radio sources that you can detect: the sun and Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.spacetoday.org/SolSys/Jupiter/JupiterRadio.html&quot; rel=&quot;noreferrer&quot;&gt;Jupiter is particularly interesting&lt;/a&gt; as interactions between Io and its magnetic field produce beams of radio waves that sweep past earth every 10 hours.  These are detectable in the amateur range, at about 20 MHz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://radiojove.gsfc.nasa.gov/&quot; rel=&quot;noreferrer&quot;&gt;Nasa make a kit&lt;/a&gt; for detecting these radio signals, or it &lt;a href=&quot;https://radiojove.gsfc.nasa.gov/help/faq1.htm&quot; rel=&quot;noreferrer&quot;&gt;is possible to use a ham antenna&lt;/a&gt;, but of course it must be cut for the frequency of operation. The Nasa kit uses a phased dipole antenna which must be set up in a field or similar as the antenna is about 7m long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars are not very good radio sources. Supernovae remnants such as Cassiopeia A or the Crab nebula are much brighter at radio wavelengths. Most supernovae are too distant to be powerful radio sources; &lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/R/Radio+Supernova&quot; rel=&quot;noreferrer&quot;&gt;radio supernovae are rare&lt;/a&gt;. A local supernova would be a radio source but we haven't observed a supernova in the milky way for several hundred years.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-10-12T17:19:47.923" LastActivityDate="2018-10-12T17:19:47.923" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27979" PostTypeId="2" ParentId="27964" CreationDate="2018-10-12T05:52:14.347" Score="3" Body="&lt;p&gt;@MikeG's &lt;a href=&quot;https://astronomy.stackexchange.com/a/27973/7982&quot;&gt;answer&lt;/a&gt; is excellent. This is a supplementary answer only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of the problem finding the coordinates of the &quot;center&quot; is defining the center. One example might be the center of mass, but finding all the mass is quite a challenge as we can't even see all of the galaxy's mass due to dust clouds, as discussed in &lt;a href=&quot;https://astronomy.stackexchange.com/q/23927/7982&quot;&gt;How was the galactic plane established?&lt;/a&gt;, links therein and @RobJeffries &lt;a href=&quot;https://astronomy.stackexchange.com/a/27981/7982&quot;&gt;excellent answer&lt;/a&gt;, and of course dark matter being invisible and all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even though you can't see and identify all of the mass, you can &lt;em&gt;try to have a look&lt;/em&gt; at the way that &lt;em&gt;all of the stars that you can see&lt;/em&gt; seem to be moving via their radial Doppler shift, and at least try to estimate it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for the purposes of your question, I'm not sure if that's the only definition of &quot;center&quot; that could work for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If instead you'd like to consider something more catchy and visual, how about the &lt;a href=&quot;https://astronomy.stackexchange.com/q/23927/7982&quot;&gt;supermassive black hole&lt;/a&gt; that's almost for sure at the &quot;center&quot; of our galaxy, affectionately known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Sagittarius_A*&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sagittarius A*&lt;/a&gt;? That's just Sgr A* to her friends (see also &lt;a href=&quot;https://astronomy.stackexchange.com/q/23692/7982&quot;&gt;Why is Sagittarius A* called so?&lt;/a&gt;). According to that article, the distance from our Sun to Sgr A* is 7,860 ±140 ±40 pc, which is a remarkable thing in itself because it has &lt;em&gt;error bars on the error bars!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That number is linked to the ArXiv preprint &lt;a href=&quot;https://arxiv.org/abs/1607.05726&quot; rel=&quot;nofollow noreferrer&quot;&gt;An Improved Distance and Mass Estimate for Sgr A* from a Multistar Orbit Analysis&lt;/a&gt;, which is also an analysis of how some very special stars move. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a GIF, found in &lt;a href=&quot;https://physics.stackexchange.com/a/406609/83380&quot;&gt;this excellent answer&lt;/a&gt; to &lt;em&gt;What is the evidence for a supermassive black hole at the center of Milky Way?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HLjZZ.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HLjZZ.gif&quot; alt=&quot;stars orbiting Sgr A*&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that I call this a supplementary answer is that you end up with very nearly the same answer as @MikeG. Here I've used the same procedure:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;               X                     Y                    Z&#xA;parsecs:     -430±8,              -6861±122,           -3812±68&#xA;lightyears:  -1401±30,            -22,379±399,         -12433±221&#xA;AU:          -8.86(0.16)E+07,     -1.415(0.025)E+09,   -7.86(0.14)E+08&#xA;km:          -1.326(0.024)E+16,   -2.117(0.038)E+17,   -1.176(0.021)E+17&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-10-12T08:05:34.943" LastActivityDate="2018-10-12T08:05:34.943" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27980" PostTypeId="1" AcceptedAnswerId="27982" CreationDate="2018-10-12T06:09:28.720" Score="2" ViewCount="61" Body="&lt;p&gt;Articles summarizing a recent result about certain heavy element concentrations in stars near the galactic center all say that earlier reports about high levels in stellar atmospheres were the product of an &quot;optical illusion&quot;. Examples include&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;UPI.com &lt;a href=&quot;https://www.upi.com/Study-explains-optical-illusion-at-the-center-of-the-Milky-Way/3741539184868/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Study explains optical illusion at the center of the Milky Way&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Science Daily &lt;a href=&quot;https://www.sciencedaily.com/releases/2018/10/181010093625.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mystery at the center of the Milky Way solved&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Phys.org &lt;a href=&quot;https://phys.org/news/2018-10-high-scandium-galaxy-giant-black.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;High levels of scandium near the galaxy's giant black hole were illusory, astronomers find&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;At least Phys.org goes on to explain with real &quot;science words&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;According to the new study, the lower temperatures of the giant stars helped to create the &lt;strong&gt;optical illusion&lt;/strong&gt; that appeared in the measurements of spectral lines. Specifically, it means that the electrons in the elements behave differently at different temperatures, which in turn can be misleading when measuring the spectral lines of elements in different stars. The conclusion is the result of a close collaboration between astronomers and atomic physicists. (emphasis added)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What was the &quot;optical illusion&quot; that led to erroneous metal concentrations in stellar atmospheres in the galactic center? Beyond just &quot;electrons in the elements behave differently at different temperatures&quot; (because how electrons behave in atoms at different temperatures should be at least fairly well understood by astronomers) what specifically is the source of the misunderstanding? Is it atomic physics, incorrect assumption just a mistake, something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mention of &quot;close collaboration between astronomers and atomic physicists&quot; suggests that the answer will turn out to be fairly interesting!&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-10-12T06:45:31.897" Title="What was the &quot;optical illusion&quot; that led to erroneous metal concentrations in stellar atmospheres in the galactic center?" Tags="&lt;spectroscopy&gt;&lt;metallicity&gt;&lt;stellar-atmospheres&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27981" PostTypeId="2" ParentId="23927" CreationDate="2018-10-12T06:23:51.900" Score="8" Body="&lt;p&gt;The zero of Galactic latitude (i.e. the Galactic plane) was defined by a working group of the International Astronomical Union. The observations used to do this were 21 cm radio observations of atomic hydrogen, which are unaffected by dust absorption near the plane. The details of the observations are found in &lt;a href=&quot;http://adsabs.harvard.edu/abs/1960MNRAS.121..132G&quot; rel=&quot;noreferrer&quot;&gt;Gum, Kerr &amp;amp; Westerhout (1960)&lt;/a&gt;. This paper uses least squares fits to the 21cm data &lt;em&gt;assuming&lt;/em&gt; that the greatest concentration of neutral hydrogen is found in the Galactic plane itself. Whether this &lt;em&gt;is&lt;/em&gt; the dynamical plane was considered by &lt;a href=&quot;http://adsabs.harvard.edu/abs/1960MNRAS.121..150G&quot; rel=&quot;noreferrer&quot;&gt;Gum &amp;amp; Pawsey (1960)&lt;/a&gt;, as part of the same IAU study. They concluded that it was (bar the location of the Sun with respect to the plane), citing the flatness and relative narrowness of the neutral hydrogen and the coincidence of the source Sgr A*, which even then was thought to be at the Galactic centre, with zero latitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark matter was not well explored or widely considered in the dynamics of individual galaxies in the late 1950s, however it is not unexpected that neutral hydrogen &lt;em&gt;should&lt;/em&gt; lie in the Galactic plane, as defined by the symmetry of the Galactic potential. A rotating mass of gas, capable of dissipative interactions, &lt;em&gt;should&lt;/em&gt; settle to a plane perpendicular to its angular momentum vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 21cm definition of Galactic coordinates cannot itself define the position of the Galactic plane in 3d space unless the Sun is exactly in the plane. Detailed observations of star counts and the dynamics of stars around the Sun and perpendicular to the plane have revealed this isn't quite the case. The details are given in &lt;a href=&quot;https://astronomy.stackexchange.com/questions/822/how-far-is-the-earth-sun-above-below-the-galactic-plane-and-is-it-heading-towar&quot;&gt;How far is the Earth/Sun above/below the galactic plane, and is it heading toward/away from it?&lt;/a&gt; ; the Sun is currently around 17 pc above the Galactic plane and heading upwards. This is close enough that it makes very little difference to the apparent flatness and narrowness of the 21cm data.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-10-12T06:23:51.900" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27982" PostTypeId="2" ParentId="27980" CreationDate="2018-10-12T06:45:31.897" Score="4" Body="&lt;p&gt;The journal paper is &lt;a href=&quot;https://arxiv.org/abs/1808.07489&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thorsbro et al. (2018)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The facts are somewhat mundane. The atmospheres of cool M-giants are not well understood in detail. The infrared lines of neutral Scandium that had previously been used to claim massive overabundances in stars near the Galactic centre are problematic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper by Thorsbro et al. shows that the same &quot;anomalies&quot; are present in other M giants nowhere near the Galactic centre. Instead they argue that the details of line formation, especially the formation low excitation lines formed high in the atmosphere, like the IR transitions of neutral Scandium, are particularly vulnerable to poorly modelled NLTE (non local thermodynamic equilibrium) effects. If analysed with atmospheric codes that assume LTE, then you get the &quot;illusion&quot; of a much higher abundance.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-10-12T06:45:31.897" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27983" PostTypeId="1" CreationDate="2018-10-12T09:58:42.037" Score="3" ViewCount="612" Body="&lt;p&gt;Planet Nine (a variant of the old X-Planet and Tyche-Thelistos hypothesis) proposes a super-earth planet at the edges of the solar system with a mass of around ten Earths. However, given the interest in exotic objects and gravitational variants, maybe it's interesting to study &quot;exotic&quot; scenarios: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Tiny black hole. Using the famous online calculator &lt;a href=&quot;http://xaonon.dyndns.org/hawking/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://xaonon.dyndns.org/hawking/&lt;/a&gt;, for ten earth masses, we get a 89 mm black hole. Temperature about 2 mK! How can we put bounds on this scenario?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An exotic compact object. Of course, we have no evidence so far of general relativity modifications of gravity, but if not a black hole but another &quot;thing&quot; (a gravastar, a quark star, a preon star,...), I find hard to estimate how could we be sure to give up experimentally these examples. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Of course, the real thing is to think something less exotic (I am not sure how to differentiate the P9 hypothesis from a swarm of thousands of ETNO dwarf planet-like or planetesimal-like, but the real issue is how to test these alternative models and if these could affect what we know about the origin and formation of the solar system and its actual current structure, any hint to this in the answer to above exotic cases would also welcome).&lt;/p&gt;&#xA;" OwnerUserId="2391" LastEditorUserId="43" LastEditDate="2018-10-12T14:22:57.650" LastActivityDate="2018-10-12T14:22:57.650" Title="Could Planet Nine be a tiny black hole or an exotic compact object such that it escaped detection? What would its bounds be?" Tags="&lt;black-hole&gt;&lt;solar-system&gt;&lt;earth-like-planet&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27985" PostTypeId="2" ParentId="27977" CreationDate="2018-10-12T11:29:10.253" Score="7" Body="&lt;p&gt;Connecting an antenna directly to oscilloscope will not give reception, even with a strong radio source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First problem is the power level. Typical received power from antenna would be around -100 dBm, i.e. &lt;span class=&quot;math-container&quot;&gt;$10^{-10}\,\textrm{mW}$&lt;/span&gt;. A typical oscilloscope has an input impedance of 1 Mohm, which means if all the received power went there, it would give a voltage of &lt;span class=&quot;math-container&quot;&gt;$\sqrt{10^{-10}\,\textrm{mW}\cdot1\,\textrm{Mohm}} \approx 0.3\,\textrm{mV}$&lt;/span&gt;. With the minimum scale of 20 mV, you wouldn't see much anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second problem is &lt;a href=&quot;https://en.wikipedia.org/wiki/Mismatch_loss&quot; rel=&quot;noreferrer&quot;&gt;mismatch loss&lt;/a&gt;. Most antennas are matched to 50 ohm impedance instead of the 1 Mohm. The mismatch means that only about 0.01% of the power would actually go into the oscilloscope, the rest would reflect back.&lt;/p&gt;&#xA;" OwnerUserId="14071" LastActivityDate="2018-10-12T11:29:10.253" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="27986" PostTypeId="2" ParentId="27983" CreationDate="2018-10-12T11:31:32.433" Score="3" Body="&lt;p&gt;I think the answer is included in my answer to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/27925/could-a-black-hole-pass-quiescently-through-the-oort-cloud&quot;&gt;Could a black hole pass quiescently through the Oort cloud?&lt;/a&gt; The mini black hole would be detected via its accretion from the interplanetary medium, probably at X-ray wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ten Earth mass black hole is a ten thousandth (or so) less massive than the Sun, but if it were in a distant orbit, it &lt;em&gt;could&lt;/em&gt; be moving slower with respect to the accreting material than the 10 km/s I assumed in the answer cited above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net outcome is that this &lt;em&gt;might&lt;/em&gt; still be an X-ray source well above the detection limits of all-sky surveys. What's more, it would move with time, so if followed up would show appreciable proper motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variables here are the distance to the black hole and the speed at which is travels with respect to any medium it is travelling through.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-10-12T12:32:15.787" LastActivityDate="2018-10-12T12:32:15.787" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27987" PostTypeId="2" ParentId="27975" CreationDate="2018-10-12T14:15:05.940" Score="5" Body="&lt;p&gt;You are not the only one whom sees a difference but remember that the statistics are skewed. Fewer people are awake early to see, or photograph, the sunrise while more people are awake to see the sunset.&lt;/p&gt;&#xA;&lt;p&gt;People and heating of the Earth are factors explaining the difference in the atmosphere through which the sunlight is filtered, that affects the appearance of the light.&lt;/p&gt;&#xA;&lt;p&gt;It's well &lt;a href=&quot;https://photo.stackexchange.com/a/26782/37074&quot;&gt;explained at our Photo.SE site&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Most of the physical reasons (I can think of) probably do favor sunsets as well. During the day, the sun is heating the air and the ground. This leads to air movement that tends to pick up particulates (e.g., dust, smoke, etc.) and keep it in the air. It also leads to water evaporating (relatively) quickly from anything like lakes and rivers in the area. In addition, quite a few things that people do (e.g., driving) tends to happen primarily during the day, so it &amp;quot;pushes&amp;quot; more pollution into the air during the day. Thus, at the end of the day, the light is shining through air that carries more particles, which almost inevitably does more to diffuse the light.&lt;/p&gt;&#xA;&lt;p&gt;During the night, there's a lot less pollution being put into the air, and the lack of heating means the air tends to be calmer as well, so by morning many of those particulates will have simply fallen out of the air, and the humidity either formed into clouds or precipitated. The clearer air does less to diffuse the light, so the sunrise isn't as spectacular.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A more scientific, but not complicated, explanation is offered at Wikipedia's &amp;quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sky&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sky&lt;/a&gt;&amp;quot; webpage. The heat affects the &lt;a href=&quot;https://en.wikipedia.org/wiki/Rayleigh_scattering&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rayleigh scattering&lt;/a&gt; and the Lorenz–Mie–Debye solution or &lt;a href=&quot;https://en.wikipedia.org/wiki/Mie_scattering&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mie scattering&lt;/a&gt;, that explains the differences between &lt;a href=&quot;https://en.wikipedia.org/wiki/Sunrise&quot; rel=&quot;nofollow noreferrer&quot;&gt;sunrise&lt;/a&gt; and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Sunset&quot; rel=&quot;nofollow noreferrer&quot;&gt;sunset&lt;/a&gt; - a difference in the number and movement of particles in the sky.&lt;/p&gt;&#xA;&lt;p&gt;A somewhat related factor is explained in a Q&amp;amp;A over at Space.SE: &amp;quot;&lt;a href=&quot;https://space.stackexchange.com/a/29916/20766&quot;&gt;Where does space begin on planets without atmospheres?&lt;/a&gt;&amp;quot;. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere#Earth&quot; rel=&quot;nofollow noreferrer&quot;&gt;height of the atmosphere and its layers&lt;/a&gt; along with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Turbopause&quot; rel=&quot;nofollow noreferrer&quot;&gt;distribution of its component gases&lt;/a&gt; is altered by the heating of the planet and &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_circulation&quot; rel=&quot;nofollow noreferrer&quot;&gt;atmospheric circulation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-10-12T15:40:52.863" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27988" PostTypeId="2" ParentId="18853" CreationDate="2018-10-12T14:38:10.193" Score="-1" Body="&lt;p&gt;The discovery of TON 618 have created a new black hole species (already fingerprinted by M87 or even IC1101 cores): the ultramassive black holes with masses greater than &lt;span class=&quot;math-container&quot;&gt;$10^{10}M_\odot$&lt;/span&gt;. As said in the previous answer, in classical settings, there is no upper limit of the mass of black holes (I am not so sure if you get a theory beyond General Relativity even in classical settings). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe, one day we will learn that quantum gravity says something about that. Interestingly, any supermassive, stellar, intermediate and ultramassive black hole has a mass much much greater than Planck mass, about a microgram. The issue is that we think quantum gravity applies only to VERY MASSIVE TINY (very dense) objects, not to very massive only. Indeed, any person has a mass much greater than Planck mass, but it is not &quot;concentrated&quot;. When you have concentrated mass in very tiny regions, we have no idea of how to handle quantum fluctuations and amplitudes excepting with superstring theory. Another related question, is if you can have black holes of any DENSITY. Again, as said, you need to consider quantum processes like Hawking radiation, ... However, there is a subtle point, called the transplanckian problem. In principle, as the black holes evaporates it gets smaller and smaller, such as at certain size the wavelength would be lesser than the Planck length. We have to expect for a definitive theory of quantum gravity before to answer the ultimate fate of black holes and thus, the destiny of both: black holes and the whole universe (even the spacetime could be metastable and provisional/transitional state).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How large can a black hole formed from the collapse of a massive star grow&#xA; in 1 Gyr? Suppose the black hole can grow as fast as it can. Suppose, by the moment, it satisfies the Eddington limit. Then, an exponential law follows up:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\dot{M}=kM=M/\tau$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$k=4\cdot 10^{-16}s^{-1}$&lt;/span&gt; for a ten solar masses inicial mass function accordingtly to the Eddington limit. Then, as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M=M_0\exp(kt)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugin into this formula &lt;span class=&quot;math-container&quot;&gt;$M_0=10M_\odot$&lt;/span&gt; and the value of k, you get that the maximum mass it yields is in the range of ultramassive BH, i.e., &lt;span class=&quot;math-container&quot;&gt;$M_f\sim 10^{10}M_\odot$&lt;/span&gt; for a timescale about 1 Gyr (be aware, the numbers are tricky). Of course, transEddington limit is tricky, but there are some reasons to believe black holes bigger than &lt;span class=&quot;math-container&quot;&gt;$10^{10}M_\odot$&lt;/span&gt; are unstable and eject material. Of course, in the absence of any other argument, the above argument does NOT provide an upper limit in principle. Only other considerations relativo to quasars and jets seem to apply. But the issue is a hot topic of debate in astrophysics.&#xA;By the other hand, the minimal (or tiniest) black hole mass is also a mystery. In macroscale, we have NOT found black holes tinier than 3-5 solar masses (stellar black holes). However, primordial black holes or microblack holes could made some bits of the dark matter hidden in clusters and other parts of the galaxies. Again, the only hint are inflationary ideas, astronomical measures and experimental bounds (recently, it has been analyzed the probability of dark matter being totally black holes, but some evidence seems to say that that is not the case: black holes can not be all the dark matter).  &lt;/p&gt;&#xA;" OwnerUserId="2391" LastEditorUserId="2391" LastEditDate="2018-10-15T14:41:27.293" LastActivityDate="2018-10-15T14:41:27.293" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27989" PostTypeId="1" CreationDate="2018-10-12T14:48:05.687" Score="1" ViewCount="97" Body="&lt;p&gt;I'm currently looking into writing a story or two set on twin worlds, which I came to find out in my research is apparently not a new idea (Double planets? &#xA; Rocheworlds?  Am I close?).  But I want it to be as scientifically plausible as possible, so I'm here fact-checking.  I guess my main question going forward is: &lt;strong&gt;is the description of my worlds, as they are, scientifically plausible, and if not, what should I change?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, these two worlds are very similar to our own Earth.  To simplify things, let's say they have the exact same or relatively similar atmosphere and living conditions, so that humans and Earth-based life can live on it without much or noticeable acclimation.  If you were to hop a ship from Earth and landed on one or the other, getting off the ship you would notice little to no difference than you would on Earth.  Similarly, the short trip between each world shows little to no difference, although one has slightly more gravity than the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only real difference between the two worlds themselves is that one is slightly smaller than the other, but by an almost negligible amount (a few thousand miles); the difference isn't as dramatic as, say, Earth and Mars orbiting each other, maybe Earth and Venus being the better comparison.&lt;br&gt;&#xA;Let's also suppose that they are also far enough away from one another that they aren't dramatically effecting one another's gravity or tides, etc., but are still visible in one another's skies, prominent enough to be important to the culture of the other.  I initially wanted them to be very prominent in one another's skies, but I suppose being as close as the moon is to our world would be too close: I'm curious as to how far away one would have to be and still be caught in the other's gravity (I suppose the smaller of the two would have to be the one caught in the bigger one's gravity?  I plead ignorance on that one).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm in love with the idea that they broke off from one another in the same way the Earth and the moon supposedly did billions of years ago, in an impact event, and they slowly came to orbit one another (although this idea has recently been challenged, is my understanding).  From what I've read, though, it might be more plausible that one formed naturally in the disk formation of that planet's solar system, and a rogue planet got caught in its gravitational pull.  I like that idea less, but I'm not so proud I can't just get over it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my head, they have their own day and night cycles, just like Earth.  They may not have identical hours in a day or days in a year, etc., but its similar.  They same face of the planet is never quite facing the face of the other: it's not like the moon, which is locked to one side always facing us and the other always away.  From what I've researched, though, it's probably more likely that the planets would be facing one another on one side constantly, and their day-night systems would be greatly affected by that.  Again, not in love with that idea, but if it has to go, it has to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's all I have for now.  I would try to explain how they rotate one another better, but I can't: I'm terrible in that area, and I've forgotten a lot of the terms you use in describing this.  Please let me know what your thoughts are.&lt;/p&gt;&#xA;" OwnerUserId="21850" LastActivityDate="2018-10-12T15:48:19.417" Title="The specifics of two Earth-like worlds orbiting each other?" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;gravity&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ClosedDate="2018-10-13T09:38:41.663" ContentLicense="CC BY-SA 4.0" />
  <row Id="27990" PostTypeId="2" ParentId="27989" CreationDate="2018-10-12T15:48:19.417" Score="1" Body="&lt;p&gt;A couple thoughts:&lt;br&gt;&#xA;Tidal lock depends on the age of the system, so you can have it or not as you wish, depending on how long they came to orbit each other -- which is to say, of course, orbiting a common epicenter very nearly halfway between the pair.&lt;br&gt;&#xA;Depending on your whim, you might want the orbit plane to be in or 90 degrees out of the ecliptic, the latter leading to interesting effects on tumbling.  I'd recommend staying in the ecliptic, which is more stable and allows each of the planets to rotate about an axis which allows night/day similar to Earth.  Notice that there will be a dramatic bunch of total solar eclipses of decent length, on both planets in turn. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to their spatial separation, take a look at Newton's Laws.  Nearly any separation is allowed, though if they were very close they'd have suffered severe gravitational distortions (of shape) , messed up each others' atmospheres, and almost certainly have crashed together with horrific results. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-10-12T15:48:19.417" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27991" PostTypeId="2" ParentId="27977" CreationDate="2018-10-12T16:44:26.657" Score="10" Body="&lt;p&gt;As others have noted, you will not be able to detect a star using an oscilloscope and an antenna.  The received signal level is too low, and the oscilloscope not nearly sensitive enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A radio telescope consists of an antenna, an amplifier, and a receiver (that incorporates other amplifiers and other stuff besides - like filters and mixers to select the desired frequency range.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An antenna by itself wouldn't pick up enough signal to be directly useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oscilloscope lacks amplification and filtering needed to make the antenna signal useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As others have said, you can use commercial antennas and receivers to pick up the signals.  There are kits you can buy with everything you need, or you can get the components piece at a time from various sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an alternative, you might consider building a small radio telescope using standard satellite TV components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have one, and besides the sun and the TV satellites, it can detect the moon.  I haven't gotten around to trying to detect smaller or less intense things.  I do have it mounted on servos, though, and have made pictures of ambient RF signals.  Houses and trees are surprisingly &quot;bright&quot; sources of 13GHz RF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.stargazing.net/david/radio/itty_bitty_radio_telescope.html&quot; rel=&quot;noreferrer&quot;&gt;The folks here have instructions for building one, as well as examples of what you can do with it.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.aoc.nrao.edu/epo/teachers/ittybitty/procedure.html&quot; rel=&quot;noreferrer&quot;&gt;Here is another example of making such a small radio telescope.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think both projects link back to the same original source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can usually get all of the needed parts at any store that sells satellite TV receivers.  I bought my stuff on Amazon, but most of the hardware stores here stock those things as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All you need is a dish, an LNB, (both can be bought in a set) and one of the little gadgets that helps you aim the dish properly.  And a few feet of cable and connectors, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dish has high gain.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The LNB contains amplifiers and filters to make the signal strong enough to be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alignment device is the final bit.  It has yet more amplification, and converts the received radio signal to a (somewhat noisy) voltage that represents the strength of the received signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The signal strength indication is shown on a small meter.  You can also open the box, and add a couple of wires - you can then connect that to your oscilloscope and see how strong the signal is that you are picking up from the sun or whatever.  The two wires driving the meter are the correct place to connect to.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My profile picture is an image I made in my garage using my servo aimed satellite dish.  Not terribly impressive, but that was made without any kind of additional &quot;lighting.&quot;  All just ambient RF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a fluorescent light, you can pick up 60Hz modulated RF by pointing just the LNB at the light.  Fluorescent lights cause broadband RF interference, and the LNB can pick it up at 13GHz.  The signal strength meter demodulates it, and you can see a nice 60Hz signal if you connect an oscilloscope to the meter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My detector is a little more advanced than just the little meter.  I built a controller out of an Arduino.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It uses a MAX2015 as a signal strength detector, and has a 24 bit analog to digital converter.  It also has a chip to generate control signals for the LNB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The LNBs can actually receive two bands, and can use horizontal or vertical polarisation.  My controller lets me switch among the various combinations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Arduino operates the hardware (it also drives the servos,) makes measurements, and delivers results to my PC over the serial port.  It also takes commands as to what to do.  The smarts are all in the PC - an Arduino just hasn't got what it takes to build an image out of a bunch of measurements.&lt;/p&gt;&#xA;" OwnerUserId="24518" LastActivityDate="2018-10-12T16:44:26.657" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27992" PostTypeId="1" CreationDate="2018-10-12T18:50:18.287" Score="0" ViewCount="79" Body="&lt;p&gt;I assume expansion implies an increase in the volume of the universe. But if the volume of the universe is increasing, does that not imply the area it is expanding into's volume is being reduced?&lt;/p&gt;&#xA;" OwnerUserId="21145" LastActivityDate="2018-10-12T20:37:01.627" Title="How can the universe be expanding?" Tags="&lt;universe&gt;&lt;space-time&gt;&lt;expansion&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2018-10-12T23:27:17.257" ContentLicense="CC BY-SA 4.0" />
  <row Id="27997" PostTypeId="2" ParentId="27992" CreationDate="2018-10-12T20:37:01.627" Score="0" Body="&lt;p&gt;In 1929, Hubble published his discovery that the sizes of galaxy redshifts and proportional to galaxy distances. That is, nearby galaxies are less redshifted than further away galaxies. This implies that further away galaxies are travelling away from us faster than nearby galaxies, which implies that we live in an expanding universe. We do not expand into anything, but rather, space is stretching. The Earth is not expanding, nor are you, me, or any other matter in the universe. It is space itself that is stretching.&lt;/p&gt;&#xA;" OwnerUserId="24373" LastActivityDate="2018-10-12T20:37:01.627" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27998" PostTypeId="1" CreationDate="2018-10-12T20:54:16.337" Score="2" ViewCount="127" Body="&lt;p&gt;Might such a thing give better images than we could get from other virtual telescopes with multiple collectors? Or is having multiple collectors not a limiting factor? Could a single collector compensate for atmospheric distortion in a way multiple collectors cannot?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We sometimes use arrays of radio telescopes spread over a wide distance to increase the effective aperture. I think that array is called a virtual telescope, but I might be wrong. A normal radio telescope has a collector in the same way that a satellite dish has a collector. The reflector of a radio telescope is parabolic so as to focus parallel waves to a single point. That works because dish and collector are in a fixed geometry. The reflectors of my imagined telescope would need to turn so that they reflected waves from whichever part of the sky was being observed onto the collector. That turning would change the required curvature for perfect parabolic-ness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The telescope I'm imaging would have a (perhaps geostationary) satellite as the collector and an array of flat-ish reflectors spread over hundreds of kilometers working together to collect waves from the same part of the sky. Being non-parabolic, the reflectors wouldn't be very efficient, but they could compensate by being numerous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The collector wouldn't &lt;em&gt;have&lt;/em&gt; to be geostationary, because the reflectors already need to compensate for the earth turning; adding the satellite's movement into the equation wouldn't be terribly complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/y0J7L.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y0J7L.png&quot; alt=&quot;crude diagram of satellite collector&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24520" LastEditorUserId="24520" LastEditDate="2018-10-14T08:45:09.623" LastActivityDate="2018-10-14T16:11:46.643" Title="Array of telescopes with single orbiting collector?" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27999" PostTypeId="2" ParentId="16124" CreationDate="2018-10-13T13:11:09.620" Score="2" Body="&lt;p&gt;This is not an answer, but it is too long to put in the comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are formulas for cooling timescales of different radiative processes. Cannot recall right now about sources for reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing about nebulae. Some of them might have compact objects such as pulsars or black holes feeding energy from the centers. In those cases, you have to consider the energy input from the central sources and radiative cooling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing about X-ray emission processes. There are thermal X-ray emission (i.e., bremmstrahlung) and nonthermal ones (i.e., synchrotron and inverse Compton; IC). Each process dominates in different environment. Regarding to SN nebulae, nonthermal processes are the dominant ones, the electrons involving in the processes are relativistic and thermal process requires dense electron medium (for collision) to maintain the thermal equilibrium. IC normally dominates at early ages of nebulae due to the compactness of the system, while synchrotron self-absorption is also strong at these ages. At older ages, synchrotron dominates, while the IC can interact with the background photons (i.e., cosmic background radiation) and results in X/gamma-ray photons as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last about IR emission. Since IR emission is mostly from thermal process while X-rays are from nonthermal processes, X-rays tend to last longer.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-10-13T13:11:09.620" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28000" PostTypeId="1" CreationDate="2018-10-13T21:34:04.547" Score="2" ViewCount="47" Body="&lt;p&gt;When the two main galaxies of the local group will collide, it is very unlikely for stars to collide with another, but how unlikely is it in the initial collision? And over the whole merging process?&lt;/p&gt;&#xA;" OwnerUserId="24528" LastActivityDate="2018-10-13T21:34:04.547" Title="What are the odds for any two bodies to collide during the milky way-Andromeda collision?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;collision&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2018-10-14T10:57:11.540" ContentLicense="CC BY-SA 4.0" />
  <row Id="28005" PostTypeId="2" ParentId="25850" CreationDate="2018-10-14T01:29:48.797" Score="-2" Body="&lt;p&gt;The Hubble constant discrepancy with Planck lambda CDM results has been reconciled by the Quantum Mass-Energy theory (QME)= 67.76 (km/s)/Mpc ... See research paper &lt;a href=&quot;http://www.gsjournal.net/Science-Journals/Research%20Papers/View/7439&quot; rel=&quot;nofollow noreferrer&quot;&gt;QME Theory Universal Hubble Constant Ho = 67.76 (km/s)/Mpc Allows Full Reconciliation of Space Telescopes with Satellites Results&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Abstract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;By applying the principles of Quantum Mass-Energy (QME) theory, the vastly different Hubble constant (Ho) values from various sources such as satellite measurements, space telescope observations, QME theory, LIGO, BOA, WMAP, and sky surveys are scientifically reconciled into a single true universal value of Ho(Universal)=67.76 (km/s)/Mpc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="24531" LastEditorDisplayName="user1569" LastEditDate="2018-10-15T09:23:21.880" LastActivityDate="2018-10-15T09:23:21.880" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28006" PostTypeId="1" CreationDate="2018-10-14T01:56:20.010" Score="0" ViewCount="65" Body="&lt;p&gt;Let’s say that I had some device capable of outputting insane amounts of radiation, across the entire EM spectrum. If I wanted to affect the CMB to the point that, were Earth-based scientists to know that the device existed, they would be unable to gather any useful data from the CMB regarding the early universe, how much radiation would this theoretical device have to output? Peaking in what frequency? Where would be the optimal place to put it? Or multiple copies of it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone who actually knows stuff about the CMB is scratching their head thinking I’m nuts, I probably am. I don’t know much about the CMB, except that it’s random EM waves that are leftovers from the Big Bang, peaking in the microwave frequency range. So if what I’m describing is absolutely implausible across all frequencies and amplitudes, I would appreciate if you could explain why, given my low level of knowledge on the topic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the event that for certain waves what I’m describing is possible, I would be curious to know if there’s actually anything in the known universe that could do it.&lt;/p&gt;&#xA;" OwnerUserId="17533" LastEditorUserId="17533" LastEditDate="2018-10-14T21:22:18.357" LastActivityDate="2018-10-14T22:48:56.070" Title="How much radiation would something need to output to sufficiently affect the CMB?" Tags="&lt;radiation&gt;&lt;cosmic-microwave-background&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2018-10-22T15:54:27.507" ContentLicense="CC BY-SA 4.0" />
  <row Id="28007" PostTypeId="2" ParentId="28006" CreationDate="2018-10-14T02:25:39.910" Score="3" Body="&lt;p&gt;The CMB is basically the cooling of the high energy radiation of the early universe to achieve a nearly uniform dispersion of microwave shifted energy. It came to be via the decoupling, an early universe expansion-related event (~380,000 years after the big bang) that allowed particles to fall out of thermal equilibrium for the first time. This decoupling phase allowed hydrogen to form for the first time, and photons to roam free; the photons being what you're seeing cooled to the microwave spectrum in CMB pictures. Basically, the CMB is a snapshot of the first time photons could move freely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to effect the CMB as we see it you'd have to have something major happen at the same time the photons first escaped their &quot;confinement,&quot; altering the energy/mass of the universe right before that, or changing one of its already intrinsic properties at the time. Nobody can do that as matter (and therefore a machine or an intelligence to use it) couldn't exist at that time, so let's call that impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, though nearly impossible and entirely improbable, you'd need something along the lines of another big bang event to occur inside our already formed universe, pouring enough energy/mass into the universe to both destroy what's there and cause another event where photons were confined and then let loose - making an entirely new CMB image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, it's not possible without destroying the universe and then allowing the universe to reform; so both the machine and the intelligence using it would be destroyed along with it. &lt;/p&gt;&#xA;" OwnerUserId="24476" LastActivityDate="2018-10-14T02:25:39.910" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28008" PostTypeId="1" AcceptedAnswerId="28009" CreationDate="2018-10-14T02:56:20.150" Score="0" ViewCount="100" Body="&lt;p&gt;Strangely, there isn't a lot of information on Ceres on the internet. I want to know the specific percentages of its composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="24048" LastActivityDate="2018-10-14T06:29:03.227" Title="What's the composition of Ceres?" Tags="&lt;asteroids&gt;&lt;ceres&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28009" PostTypeId="2" ParentId="28008" CreationDate="2018-10-14T06:29:03.227" Score="1" Body="&lt;p&gt;The details are not known for certain, however it is generally thought to have a rocky core, formed of silicates (aluminium and magnesium silicate for example) and a large mantle of water ice, perhaps 50% by volume, 25% by mass. The surface crust is formed of various carbonaceous and iron minerals: siderite, dolomite, cronstedtite. There may be a small iron core. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-10-14T06:29:03.227" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28010" PostTypeId="2" ParentId="28006" CreationDate="2018-10-14T08:27:43.290" Score="6" Body="&lt;p&gt;It is not very clear what you mean. There are many cosmic sources of microwave radiation that emit so much energy and/or are so close to us that they confuse measurements of the cosmic microwave &lt;em&gt;background&lt;/em&gt;. For example &quot;compact sources&quot; such as giant molecular clouds, supernova remnants, H II regions; &quot;diffuse&quot; sources due to dust in the galaxy and in our solar system. In order to get a clean signal, these &lt;em&gt;foreground&lt;/em&gt; sources have to be removed from the data. There are also more subtle cosmological effects caused by lensing and gravitational time dilation caused by the large scale structure in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many ways and techniques that are used to characterise and remove these backgrounds (e.g. see &lt;a href=&quot;https://arxiv.org/abs/1303.5072&quot; rel=&quot;noreferrer&quot;&gt;Ade et al. 2013&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most confusing background source would be one that is diffuse, spread over the whole sky almost uniformly and with a spectrum that is very similar to the cosmic microwave background. The source that most closely matches these properties is dust in our own solar system. An imperfect subtraction of this may have led to controversy over &lt;a href=&quot;https://www.newscientist.com/article/dn23301-planck-shows-almost-perfect-cosmos-plus-axis-of-evil/&quot; rel=&quot;noreferrer&quot;&gt;possible large scale anisotropies in the CMB&lt;/a&gt;- for example see &lt;a href=&quot;https://arxiv.org/abs/1510.07929&quot; rel=&quot;noreferrer&quot;&gt;Schwarz et al. (2015)&lt;/a&gt;; who discuss the problems of removing the signatures of solar system dust. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-10-14T08:27:43.290" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28011" PostTypeId="2" ParentId="27998" CreationDate="2018-10-14T16:11:46.643" Score="4" Body="&lt;p&gt;If I understand your question, what you're talking about is one of the things that has been on the astronomical wish list for ever.  There are two ways to combine separated telescopes.  One of them (the easy way) is to just overlay the images from the two telescopes.  In this way you get to double the light you collect, but you have to take care to overlay the two images very well, such that they have the same magnification, rotation, &lt;em&gt;etc&lt;/em&gt;.  The resolution you get doing it this way is only the resolution you get from either individual telescope, so the main reason for doing this is to increase you signal-to-noise (it doesn't make sense to do this if you're looking at a bright scene).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other way to combine multiple telescopes, which is along the lines of what you are thinking, is to phase the two telescopes, meaning that not only are you laying the images on top of each other, you are also doing it so that the two images have the same phase, meaning they have traveled exactly the same distance from source to collector.  What you gain by doing this is that the resolution you get this way is what you would get if you had a telescope aperture that circumscribed the two telescopes.  For instance, if you had two 10 cm telescopes separated by 10 meters, you'd end up with an image that had a resolution of a 10 meter diameter telescope; however it would only have the light collection capability of the two individual telescopes, so though you have the resolution of a 10 meter telescope, you have the light collection capability that is only &lt;span class=&quot;math-container&quot;&gt;$\frac{2*{0.05}^2}{5^2} = 0.02\%$&lt;/span&gt;, meaning that you'd need to take an exposure of 5000 seconds to get the same signal you'd get with a 1-second exposure with a 10 meter telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason you can easily make these virtual telescopes with radio telescopes is that the level you need to match path length goes as a fraction of the wavelength you're looking in.  Radio waves are many many meters long, so if you need to match the path length to, say, 1/50th of that, you're talking distances that are on the order of many centimeters or even meters.  What this means is that the radio telescopes can measure the incoming radio waves and time-tag them with atomic clocks; the ability to time-tag and the level that we can synchronize separate clocks results in phasing errors that easily fit within the phasing requirements.  Radio astronomers can record the data at each telescope then after the fact combine their signals in a computer.  However, for visible light, a fraction of a wavelength is now on the order of nanometers.  We currently don't have the ability to time-tag optical data to be able to go back later and do the phasing, which means the phasing has to be done real-time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the challenge to do what you are suggesting is that we need to phase the separate telescopes in real-time.  This means that we need to hold the optics in the separate telescopes so that they are not only matched to the nanometer level, but you need to hold them to that level as you are imaging.  We can easily do that on an optics table in the lab, which is basically what you are doing when you make a Michelson interferometer, but doing it on separated satellites that are moving around, vibrating, etc., it is a bear of a challenge.&lt;/p&gt;&#xA;" OwnerUserId="9749" LastActivityDate="2018-10-14T16:11:46.643" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28012" PostTypeId="1" AcceptedAnswerId="28022" CreationDate="2018-10-14T20:53:58.677" Score="13" ViewCount="4103" Body="&lt;p&gt;Would I be correct if I stated that all galaxies have a black hole in the center? Since other galaxies orbit around a center point in a galaxy - the barycenter, right? - I would assume at this point there must be some insane gravitational pull. Which only black holes can create, right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If this is a bit vague, please tell me so I can clarify!)&lt;br&gt;&#xA;I am a complete newbie to astronomy in general, so sorry if this question makes no sense whatsoever.&lt;/p&gt;&#xA;" OwnerUserId="20235" LastEditorUserId="-1" LastEditDate="2018-10-15T13:18:18.650" LastActivityDate="2018-10-15T18:47:41.843" Title="Do all galaxies have a black hole in the center?" Tags="&lt;black-hole&gt;&lt;galaxy&gt;&lt;galaxy-center&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28013" PostTypeId="2" ParentId="28012" CreationDate="2018-10-14T21:03:05.863" Score="6" Body="&lt;p&gt;It's generally assumed that &lt;em&gt;most&lt;/em&gt; galaxies have a black hole at their centre. There are galaxies we've detected that don't seem to have a black hole though, so it's not a necessary requirement or a guaranteed find in every case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In cases without a black hole, it has been postulated that the gravity of the stars and gases combined is enough to hold the galaxy together.&lt;/p&gt;&#xA;" OwnerUserId="24476" LastActivityDate="2018-10-14T21:03:05.863" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="28016" PostTypeId="2" ParentId="4769" CreationDate="2018-10-14T23:35:36.043" Score="0" Body="&lt;p&gt;Cameras are sold separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fork mounted SCTs like the CPC run in alt-az mode (up/down left/right movements). Although the controlling software can track objects fine for visual use, because neither axis points at the celestial pole, tracking alt-az mounts suffer from an effect called field rotation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of watching a point on the rim of a wheel as it rotates - you can track the point, but the direction to the centre of the wheel changes as it rotates. in images, that means that the field of view appears to gradually rotate around the object you're tracking, which causes trailing that increases as you move away from the object.&#xA;This limits your maximum exposure time before the traiing becomes noticeable. Exactly how long varies with your latitude and where in the sky you're pointing, but around 30 seconds or so is a rough ballpark figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For longer exposures, you either need a field derotator (Meade used to do one, but as far as I know Celestron never have), or (more commonly) an equatorial mount. With one axis pointed at the celestial pole, tracking an object doesn't cause field rotation from the point of view of an attached camera - for the wheel example, think of a camera attached to an extension of the axle - now, the camera rotates with the wheel, so from the camera's point of view the wheel stays in the same relative position - so no trailing. This means that exposures are now only limited by tracking accuracy (and how long it takes for light pollution to saturate the image).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a fork mounted SCT, the usual way to achieve this is something called an equatorial wedge. This bolts between the fork mount and the tripod, tilting it so that the the azimuth axis points at the celestial pole, and making the mount work as an equatorial mount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DSLR cameras have the advantage of being relatively cheap (especially secondhand), and can be modified to be more sensitive to h-alpha light (the standard infrared cut filter in front of the sensor also blocks most h-alpha light - replacing this with&#xA;something that passes h-alpha makes a big difference for images of many emission nebulae that have h-alpha.  The other advantage of DSLRs is that the sensors are big.&#xA;Downside is that they're not cooled cameras, which makes the images more noisy - especially in warmer weather.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomical cameras (CCD or CMOS) for long exposure imaging usually include peltier effect coolers - this reduces the sensor temperature by a lot - often tens of degrees, to below freezing point in some cases.  Dropping the temperature significantly reduces thermal noise, giving you less noisy images. You can also get monochrome cameras, which make working with narrowband filters easier. Downside is that cooled CCD/CMOS cameras can be very expensive, especially at sensor sizes comparable to DSLRs. Many of the more affordable ones have sensors significantly smaller, and with fewer pixels than DSLRs.&lt;/p&gt;&#xA;" OwnerUserId="17016" LastActivityDate="2018-10-14T23:35:36.043" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28019" PostTypeId="1" AcceptedAnswerId="28029" CreationDate="2018-10-15T12:03:54.310" Score="3" ViewCount="300" Body="&lt;p&gt;I've been trying to find a clean definition what people mean by when they talk about &lt;em&gt;diffuse ionized gas in the interstellar medium&lt;/em&gt;, but I couldn't find anything so far.&#xA;Apparently it's supposed to be trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it simply ionized gas, mixed with non-ionized gas in the ISM through the process of diffusion?&lt;/p&gt;&#xA;" OwnerUserId="20403" LastActivityDate="2018-10-16T10:30:15.467" Title="What is the diffuse ionized gas?" Tags="&lt;interstellar-medium&gt;&lt;gas&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28021" PostTypeId="1" CreationDate="2018-10-15T13:27:40.563" Score="3" ViewCount="106" Body="&lt;p&gt;first post here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found 1 related Question, which has not solved my Problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SHORT QUESTIONs VERSION:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suns axis is 7,25° tilted from the ecliptic. But in which direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LONG QUESTION:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nasa defines the Suns Rotation axis as described in the title:&#xA;&lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/sunfact.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://nssdc.gsfc.nasa.gov/planetary/factsheet/sunfact.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0mkXW.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0mkXW.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it described with (Ra, Declination)? Is this a (Earth) Aequatorial System placed in the middle of the Sun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If true, then this should work (&lt;a href=&quot;https://astronomy.stackexchange.com/questions/11849/what-is-the-angle-between-the-equator-of-the-sun-and-the-plane-of-the-earths-o&quot;&gt;What is the angle between the equator of the sun, and the plane of the Earth&amp;#39;s orbit&lt;/a&gt;). As the best anwer says, the angle between ecliptical pole and sun Rotation axis should be 7.25°. I get 4°, what is going on here? Have I messed up the angle calculation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot, this is making my Brain burn : D&lt;/p&gt;&#xA;" OwnerUserId="24544" LastEditorUserId="24544" LastEditDate="2018-10-15T14:32:00.687" LastActivityDate="2018-10-15T22:54:30.863" Title="What does it mean that the Sun`s rotation axis is: RA 286.13° , Dekl: 63.87°?" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28022" PostTypeId="2" ParentId="28012" CreationDate="2018-10-15T14:22:36.730" Score="12" Body="&lt;p&gt;The answer here is certainly no. Many dwarf galaxies (e.g. the &lt;a href=&quot;https://en.wikipedia.org/wiki/Magellanic_Clouds&quot; rel=&quot;noreferrer&quot;&gt;Magellanic Clouds&lt;/a&gt;) don't contain supermassive black holes, although &lt;a href=&quot;https://www.nasa.gov/mission_pages/chandra/images/cosmos-legacy-survey-shows-evidence-of-intermediate-mass-black-holes.html&quot; rel=&quot;noreferrer&quot;&gt;some may contain less massive intermediate-mass black holes&lt;/a&gt; (&lt;a href=&quot;https://arxiv.org/abs/1802.01567&quot; rel=&quot;noreferrer&quot;&gt;Mezcua et al. 2018&lt;/a&gt;). As they are the most numerous type of galaxy in the universe, it's quite possible that &lt;em&gt;most&lt;/em&gt; galaxies, in fact, do not contain supermassive black holes. This shouldn't be too surprising; if the bottom-up theories of supermassive  black hole formation are correct, many intermediate-mass black holes must coalesce to form a massive one. Dwarf galaxies simply wouldn't have (or be able to form) the requisite population, in most cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, a number of dwarf galaxies have been observed to host supermassive black holes (notably including &lt;a href=&quot;https://en.wikipedia.org/wiki/M60-UCD1&quot; rel=&quot;noreferrer&quot;&gt;M60-UCD1&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Hen_2-10&quot; rel=&quot;noreferrer&quot;&gt;Hen 2-10&lt;/a&gt;, among others). These have provided, and continue to provide, insights into the formation and evolution of low-mass galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some users here have also identified some massive galaxies without supermassive black holes, including &lt;a href=&quot;https://en.wikipedia.org/wiki/Triangulum_Galaxy&quot; rel=&quot;noreferrer&quot;&gt;M33&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/A2261-BCG&quot; rel=&quot;noreferrer&quot;&gt;A2261-BCG&lt;/a&gt;. The former has a bulge, but the central stellar velocity dispersion is low, implying there is no black hole at its core (&lt;a href=&quot;https://arxiv.org/abs/astro-ph/0107359&quot; rel=&quot;noreferrer&quot;&gt;Merritt et al. 2001&lt;/a&gt;), assuming the required parameters are known well enough.  In A2261-BCG, it's possible that &lt;a href=&quot;https://www.nasa.gov/mission_pages/hubble/science/a2261-bcg.html&quot; rel=&quot;noreferrer&quot;&gt;one or more central supermassive black holes were ejected&lt;/a&gt;, leaving a &quot;puffy&quot; core without a black hole. Keep in mind, of course, that this is only one possible explanation for the observed distribution of stars in A2261-BCG.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, a central supermassive black hole does &lt;em&gt;not&lt;/em&gt; influence the large-scale kinematics of stars and gas within a galaxy. Most supermassive black holes are much less massive than their host galaxy; I believe Sag A*, at the center of the Milky Way, is on the order of about 0.001% of the total mass of the galaxy. Even massive ellipticals like &lt;a href=&quot;https://en.wikipedia.org/wiki/NGC_4889&quot; rel=&quot;noreferrer&quot;&gt;NGC 4889&lt;/a&gt; have black holes about 0.1% of their total mass. As far as I know, ultra-compact dwarfs have the highest black-hole-to-galaxy mass ratios, with an extreme example being the previously-mentioned M60-UCD1.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2018-10-15T18:47:24.337" LastActivityDate="2018-10-15T18:47:24.337" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28024" PostTypeId="2" ParentId="28021" CreationDate="2018-10-15T16:53:01.070" Score="5" Body="&lt;p&gt;You can think of the position given in your title as the coordinates of the star that is directly over the Sun's north pole. (Similar to how Polaris is &lt;em&gt;almost&lt;/em&gt; directly over the Earth's north pole.) Likewise, the ecliptic plane has a north pole which can be located by right ascension and declination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7.25° to 7.26° appears to be the correct calculation, based on the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;North pole of sun: 286.13° RA, 63.87° declination&lt;/li&gt;&#xA;&lt;li&gt;North pole of ecliptic: 270° RA, 66.5738° declination (per &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_pole&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$cos(angle)=sin(dec1)sin(dec2)+cos(dec1)cos(dec2)cos(RA1-RA2)$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$cos(angle)=sin(63.87)sin(66.5738)+cos(63.87)cos(66.5738)cos(286.13-270)$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$angle=7.26$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2018-10-15T16:53:01.070" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28025" PostTypeId="2" ParentId="28019" CreationDate="2018-10-15T17:28:58.507" Score="0" Body="&lt;p&gt;I don't think it means anything special.  Take, for example, the claim  at  &lt;a href=&quot;https://en.wikipedia.org/wiki/Interstellar_cloud&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An interstellar cloud is generally an accumulation of gas, plasma, and&#xA;  dust in our and other galaxies. Put differently, an interstellar cloud&#xA;  is a denser-than-average region of the interstellar medium, (ISM), the&#xA;  matter and radiation that exists in the space between the star systems&#xA;  in a galaxy. Depending on the density, size, and temperature of a&#xA;  given cloud, its hydrogen can be neutral, making an H I region;&#xA;  ionized, or plasma making it an H II region; or molecular, which are&#xA;  referred to simply as molecular clouds, or sometimetimes dense clouds.&#xA;  Neutral and ionized clouds are sometimes also called diffuse clouds.&#xA;  An interstellar cloud is formed by the gas and dust particles from a&#xA;  red giant in its later life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As well as another &lt;a href=&quot;https://en.wikipedia.org/wiki/Interstellar_medium&quot; rel=&quot;nofollow noreferrer&quot;&gt;statement&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In all phases, the interstellar medium is extremely tenuous by&#xA;  terrestrial standards. In cool, dense regions of the ISM, matter is&#xA;  primarily in molecular form, and reaches number densities of &lt;span class=&quot;math-container&quot;&gt;$10^6$&lt;/span&gt;&#xA;  molecules per &lt;span class=&quot;math-container&quot;&gt;$cm^3$&lt;/span&gt; (1 million molecules per cm3). In hot, diffuse&#xA;  regions of the ISM, matter is primarily ionized, and the density may&#xA;  be as low as &lt;span class=&quot;math-container&quot;&gt;$10^{−4}$&lt;/span&gt; ions per &lt;span class=&quot;math-container&quot;&gt;$cm^3$&lt;/span&gt;. Compare this with a number density of&#xA;  roughly &lt;span class=&quot;math-container&quot;&gt;$10^{19}$&lt;/span&gt; molecules per cm3 for air at sea level, and &lt;span class=&quot;math-container&quot;&gt;$10^{10}$&lt;/span&gt;&#xA;  molecules per &lt;span class=&quot;math-container&quot;&gt;$cm^3$&lt;/span&gt; (10 billion molecules per &lt;span class=&quot;math-container&quot;&gt;$cm^3$&lt;/span&gt;) for a laboratory&#xA;  high-vacuum chamber.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="907" LastEditorUserId="907" LastEditDate="2018-10-16T10:30:15.467" LastActivityDate="2018-10-16T10:30:15.467" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28028" PostTypeId="1" AcceptedAnswerId="28030" CreationDate="2018-10-15T21:55:04.020" Score="2" ViewCount="104" Body="&lt;p&gt;I just saw in the advertisement-laden tabloid &lt;a href=&quot;https://www.express.co.uk/news/science/1031721/NASA-asteroid-warning-three-asteroids-Earth-Close-Approach&quot; rel=&quot;nofollow noreferrer&quot;&gt;Express&lt;/a&gt; that &lt;em&gt;three near-Earth astroids will pass the Earth today&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;The first of the three, dubbed Asteroid 2018 TU4, will come as close as 0.03637 astronomical units (au) or 14.16 lunar distances. around 3.13pm BST (2.13pm2 UTC)&lt;/li&gt;&#xA;  &lt;li&gt;The next asteroid to pass the Earth is Asteroid SL3 at around 3.36pm BST (2.36pm BST).&lt;/li&gt;&#xA;  &lt;li&gt;Lastly, Asteroid 2018 TS1 will come closest to Earth at around 3.45pm BST (2.45pm UTC) today.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is this just coincidental, or is there some relationship between these three particular asteroids, perhaps in the past, that would make their near-coincident arrival more likely?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-10-15T22:44:09.147" Title="Is the passage of three asteroids near Earth today just coincidental?" Tags="&lt;asteroids&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28029" PostTypeId="2" ParentId="28019" CreationDate="2018-10-15T22:16:48.460" Score="4" Body="&lt;p&gt;The &quot;diffuse ionized gas&quot; (DIG) is another term for the phase of the interstellar medium (ISM) usually called the &lt;strong&gt;warm ionized medium (WIM)&lt;/strong&gt;. With a temperature of the order &lt;span class=&quot;math-container&quot;&gt;$10^4\,\mathrm{K}$&lt;/span&gt;, but extenting to lower and higher temperatures, it is hot enough to keep hydrogen ionized, and various metals exist as low-ionization species, such as S II, N II, and O II, and even (weak) O III (e.g. &lt;a href=&quot;https://arxiv.org/abs/1211.0369&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hill et al. 2012&lt;/a&gt;; &lt;a href=&quot;https://arxiv.org/abs/1612.02000&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zang et al. 2017&lt;/a&gt;; &lt;a href=&quot;https://arxiv.org/abs/1712.04450&quot; rel=&quot;nofollow noreferrer&quot;&gt;Weilbacher et al. 2018&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rough pressure equilibrium with the other phases of the ISM results in characteristic densities of &lt;span class=&quot;math-container&quot;&gt;$0.1\,\mathrm{cm}^{-1}$&lt;/span&gt;, but typical ranges are an order of magnitude to both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A review of the DIG/WNM can be found in &lt;a href=&quot;https://arxiv.org/abs/0901.0941&quot; rel=&quot;nofollow noreferrer&quot;&gt;Haffner et al. (2009)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2018-10-15T22:16:48.460" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28030" PostTypeId="2" ParentId="28028" CreationDate="2018-10-15T22:40:02.483" Score="5" Body="&lt;p&gt;It is just a coincidence, and isn't that uncommon of an occurrence. You can see a list of all such occurrences on &lt;a href=&quot;https://cneos.jpl.nasa.gov/ca/&quot; rel=&quot;noreferrer&quot;&gt;this page&lt;/a&gt;. Looking at the specific orbital parameters, the orbits are very different. They range from an eccentricity of .2-.5, periods of 349-725 days, and otherwise are just plain different. There are 7 listed in the next week of a similar closeness, so having 3 on a single day isn't particularly uncommon. Much further then a week doesn't really matter because these were only spotted recently, they are all 3 quite small objects. The oldest detection of the 3 was only September 30, and one was only spotted on October 10th.&lt;/p&gt;&#xA;" OwnerUserId="182" LastActivityDate="2018-10-15T22:40:02.483" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28031" PostTypeId="2" ParentId="28028" CreationDate="2018-10-15T22:44:09.147" Score="3" Body="&lt;p&gt;Nothing to be very excited about. There are usually multiple rocks in the 0-50m size range passing by Earth.  The &quot;&lt;a href=&quot;http://www.hohmanntransfer.com/news4.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Traffic report&lt;/a&gt;&quot; states&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Traffic Report  on 15 October '18&lt;br&gt;&#xA;  Eight objects reported inside ten LD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that 2018 TU4 is so far away it doesn't even get mentioned on this list. So this is not newsworthy. Even the Express states &quot;dozens of objects get this close on a daily basis&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-10-15T22:44:09.147" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28035" PostTypeId="1" CreationDate="2018-10-16T00:19:26.843" Score="3" ViewCount="1142" Body="&lt;p&gt;In years, what is the time it takes our sun to orbit around our solar system's barycenter?&lt;/p&gt;&#xA;" OwnerUserId="24553" LastEditorUserId="7472" LastEditDate="2018-10-16T12:37:34.693" LastActivityDate="2021-05-03T13:10:05.520" Title="What is the orbital period of our sun around its barycenter?" Tags="&lt;the-sun&gt;&lt;solar-system&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28036" PostTypeId="2" ParentId="28035" CreationDate="2018-10-16T00:56:14.093" Score="13" Body="&lt;p&gt;As you can see from the diagrams below of the motion of the Solar System's barycenter relative to the Sun (courtesy of Wikipedia), the Sun's motion with respect to the system's barycenter is &lt;em&gt;not&lt;/em&gt; a simple closed curve, and it doesn't have a simple period.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/0/0c/Solar_system_barycenter.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QeG73.png&quot; alt=&quot;Solar system barycentre 1945-1995&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;(Click the above image for a SVG version).&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VjyLz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VjyLz.png&quot; alt=&quot;Solar System barycenter 2000-2051&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;It is primarily dominated by Jupiter's period (11.862 years) because Jupiter has most of the non-solar mass of the Solar System. (On a related note, Jupiter has most of the angular momentum of the solar system). Jupiter's mass is approximately 0.0009543 solar masses, so the Sun is almost 1048 times more massive than Jupiter.&lt;/p&gt;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Barycenter#Inside_or_outside_the_Sun?&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;To calculate the actual motion of the Sun, you only need to consider&#xA;the motions of the four giant planets (Jupiter, Saturn, Uranus,&#xA;Neptune). The contributions of all other planets, dwarf planets, etc.&#xA;are negligible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;https://sagecell.sagemath.org/?z=eJyNVmtv4jgU_Z5f4aaVklBIAx1mt5WiHUpTbWd4qWRG6qZp5BID2Q1JxnaYwmr_-17bgdBW013UCOf63ONzX6bJqsgpR5Q04fleEsaZNqf5CiWcUJ7nKUOJgiQsTWZE00qaIhfpS84Ldnl2xlhs_1mkdoYZthf5-myZ02SbZyx6wny2tGeLRNe0T0kGfHjGtZjMEWNPJuOYcldvX3zotpw2_OlNxHheCNNFbcIlz6OyiDEn7g1OGbEuNQSf2SoGFXNd17Wjk5Pp-Ebrj4fD3ujaRUbbMbS-N_K9O3jpOs4neB_2vniRN_ndG4Lt3psamt-7GniRfz_xwPLN6_vjO7BO_d6dH_m3Q2H9W4r8R1jHkwNjXkibN4mmt38IWxvdG9r4qx99Hd36UzB8GbauDe3Ou4kmg95IQLz-4Hbi3_aVdXoP7kLK547jgDxxftQf3wnFo_HIk5Zo0LvyBlNpUoZrb-D3Ir-y9Kffopvx3bDn72IaX32Ornt-78BFhgnvHUMkyoNEiZzJHDLIICU2K59Mqh8Fjw9Z2ICU6_BAei2JSSDzGHBxMuMmgOdJFuM0BQczeHTB49R6YA0XHtMIHo2w8ZthCRIm-skeWgcsgSE7wgiBri3N0HFSguo7e0G4Cd3VRAWmeMVc6aUIjgsKHQQCvtucPHNLBbCqAiCYzpYg6eHkAVrBM-2GJZbe2NNlW0sfKWha6ZmDb8LQKM-I6ifxUWfooxyRYklWhAJChj_Pyyw-0q09khJe0ky-VvlZ2Qual4XZtoL2ZWizIk14mmSEmZVWNsMpAWCXtLqHjgFTYNNoGsK31Q7hPArFSTKJCSWaR0XOojURsxcok0DR_McOV8exkbMBW4FTKTGBOazjrJlaSlYDzdMcc5NZ9dnCH0Kp3cRZ4ALsITrdkex39wJtXBQki02Aq4TFcJQQqPLAkxUUmxRAdf6xa3e66Bh2N0wlie17Q9wTcWyun5tovYFna9UBrgUMsI2a7Qydv6oOWj8DYM2E_26xlYtKSLEGFnWrmXv1TeQ0ZVtACzOlHwJuIgQ6QMYWVrUk8M-gs0ygqkKFrRi2YrH1Ur26tUq6JkJ8EJgVodVE5jPkU7huxLcwiu-tBbnfF_odEVArEFBn579E1EICaDYbAhDlAkGoValoVSpaQoXQV0kNXxPsav2_oqm7oT5dBQi_GAuIwkWmI_PvKCgrM4pFK3-86P7iONAof63UhoisTTryZQLrIofJPY9NRQSlS7bEPXcaDFKVF3iW8I3r2F241_I0p66eU5wtSDXQE3QK3QQTT0lNIMaioQQcUrT3FMfzuaNXsy0ZdhoCNTgfwsMB3fdX2FQRCIFdqa_ie0pLsuM7cHynP7vWi5Fv7rqjmn1QsN-WAsUlCPoAGnRgs64r1MexHaHGeitA1ao-6fhFsOqSqRL-60_zPdvg7OD6fMsh771d3fZuK7wgGcdvPMXFWjledsI9nJL4PWjn8mdQiXwi24TQqoCvkarMNlvmP8w5_DgR16cl1AA_E1b9X_Ivg16XcQ==&amp;amp;lang=sage&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sage / Python script which displays an interactive 3D plot&lt;/a&gt; of the above diagrams, using data from&#xA;the &lt;a href=&quot;https://ssd.jpl.nasa.gov/?horizons&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL Horizons&lt;/a&gt; on-line solar system data and ephemeris computation service. Horizons can produce data covering a wide range of dates, please see the Horizons documentation for details.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="16685" LastEditDate="2021-05-03T13:10:05.520" LastActivityDate="2021-05-03T13:10:05.520" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="28040" PostTypeId="1" CreationDate="2018-10-16T11:12:28.793" Score="2" ViewCount="40" Body="&lt;p&gt;On the Wikipedia Article on “Geodesics in general relativity”, it says the following: “&lt;strong&gt;&lt;em&gt;Thus, for example, the path of a planet orbiting a star is the projection of a geodesic of the curved 4-D spacetime geometry around the star onto 3-D space.&lt;/em&gt;&lt;/strong&gt;” &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let’s suppose that the planet is, for example, the Mercury far from us. In this case, I think that, for the above “projection onto 3-D space”, it is necessary to &lt;strong&gt;&lt;em&gt;extend a Euclidean coordinate system&lt;/em&gt;&lt;/strong&gt;, which is constructed by us living on the Earth, to a region near the Mercury &lt;strong&gt;&lt;em&gt;in a simple and continuous manner&lt;/em&gt;&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, &quot;the extension in the simple and continuous manner&quot; means that three orthogonal basis vectors, defined by an observer on Earth, are used to span the entire universe beyond the earth. That is, &lt;strong&gt;in order to compare the observation with the theoretically predicted result, the trajectory of Mercury should be observed and described in terms of coordinates based on the basis vectors.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, it seems that the extended Euclidean coordinate system is also used in describing galaxies and so forth.  Specifically, if a distance to a star is expressed based on a traveling time of light traveling from the star to us, a distance to a star, which is located beyond the massive black hole at the center of our Milky Way, will be infinity. However, the distance to the star is not expressed in such a manner.  Rather, it seems that the distance to the star is expressed based on the extended Euclidean coordinate system. &lt;strong&gt;Is this my understanding correct?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am a self-taught person who is not good in English. Thus, any comment on technical contents in this article or any improved edition on English expressions would be very welcome.&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="24502" LastActivityDate="2018-10-16T11:12:28.793" Title="How to compare the observation with the theoretically predicted result?" Tags="&lt;observational-astronomy&gt;&lt;space-time&gt;&lt;general-relativity&gt;&lt;frame-of-reference&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28042" PostTypeId="1" CreationDate="2018-10-16T14:01:21.047" Score="2" ViewCount="52" Body="&lt;p&gt;If the surface temperature of a young neutron star is hot enough, say, several KeV, its blackbody radiation could be super-&lt;a href=&quot;https://en.wikipedia.org/wiki/Eddington_luminosity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eddington&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did any paper discuss this scenario before?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditDate="2018-10-17T07:21:56.500" LastActivityDate="2018-10-17T07:21:56.500" Title="A high-temperature isolated neutron star could be super-Eddington?" Tags="&lt;astrophysics&gt;&lt;neutron-star&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28043" PostTypeId="1" CreationDate="2018-10-16T22:58:37.653" Score="1" ViewCount="156" Body="&lt;p&gt;I know that Sirius A and B orbit each other. I'd like to ask how large is their relative apparent motion in arc seconds, seen from Earth?&lt;/p&gt;&#xA;" OwnerUserId="24569" LastEditorUserId="7982" LastEditDate="2018-10-19T09:35:17.550" LastActivityDate="2018-10-19T09:35:17.550" Title="How many arc seconds is Sirius &quot;spiralling motion&quot; around Sirius B?" Tags="&lt;star-systems&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28044" PostTypeId="1" AcceptedAnswerId="28045" CreationDate="2018-10-17T00:03:11.787" Score="0" ViewCount="735" Body="&lt;p&gt;The website &lt;a href=&quot;https://astronomy.swin.edu.au/cosmos/S/Semi-major+Axis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Swinburne University of Technology (AU)&lt;/a&gt; gives the definition of semi-major axis as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...half of the longest diameter of an ellipse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://astro.wsu.edu/worthey/astro/html/lec-ellipse.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This website&lt;/a&gt; give the definition of semi-major axis as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Half of the major axis is called the semi-major axis, and the semi-major axis is also the average sun-planet distance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But the average sun-planet distance should be the average of the semi-major and semi-minor axes. I've also seen the mean orbital distance described as the semi-major axis in many other places, not just the one cited, so that seems to be a fairly consistent definition, too. But both seem to be inconsistent; how could it be both?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it also seems to be that the semi-major axis ought to be the same as the aphelion distance, and the semi-minor axis ought to be the same perihelion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, the confusion must be mine, but I'm not sure where I'm getting things wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; It occurred to me shortly after posting this and shutting down my computer for the day that (as I'm sure 90% of this SE community knows) the semi-major and semi-minor axes are measured to the center of the ellipse, whereas the aphelion and perihelion are measured to the Sun. I think I'm starting to understand how the semi-major axis can be both the mean orbital radius with respect to the sun and the largest distance from the center of the ellipse.&lt;/p&gt;&#xA;" OwnerUserId="7108" LastEditorUserId="7108" LastEditDate="2018-10-17T18:36:00.273" LastActivityDate="2018-10-17T18:36:00.273" Title="Semi-major axis, vs aphelion, vs mean orbital distance?" Tags="&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28045" PostTypeId="2" ParentId="28044" CreationDate="2018-10-17T00:15:44.717" Score="2" Body="&lt;p&gt;I &lt;em&gt;think&lt;/em&gt; your confusion is just an incorrect (though not uncommon!) visualization of a planetary orbit. The parent star (usually) doesn't lie at the center of the ellipse, but at one of the &lt;em&gt;foci&lt;/em&gt;; only for circular orbits are the foci at the same location as the center of the ellipse. Therefore, for non-circular orbits, the aphelion distance is greater than the semi-major axis, and the average of the semi-major and semi-minor axes is the average distance to the center of the ellipse - not the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, note that the perihelion and aphelion distances are related to the semi-major axis &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; and eccentricity &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt; by&#xA;&lt;span class=&quot;math-container&quot;&gt;$$R_p=a(1-e),\quad R_a=a(1+e)$$&lt;/span&gt;&#xA;As a quick check, adding these together and dividing by 2 yields &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt;. In the case of a circular orbit, where &lt;span class=&quot;math-container&quot;&gt;$e=0$&lt;/span&gt;, we have &lt;span class=&quot;math-container&quot;&gt;$R_p=R_a=a$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2018-10-17T00:32:59.067" LastActivityDate="2018-10-17T00:32:59.067" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28047" PostTypeId="2" ParentId="28043" CreationDate="2018-10-17T00:36:17.957" Score="1" Body="&lt;p&gt;From Sirius A to Sirius B on the sky is around 3 arcsec on average.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The separation between Sirius A and Sirius B actually varies from 3 to 11 arcseconds b/c of their orbital motion over a long period of time (&lt;a href=&quot;https://florin.myip.org/blog/how-see-sirius-b&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe you're referring to that. There's a lot of information (and even some graphs and diagrams) here: &lt;a href=&quot;http://www.as.utexas.edu/white_dwarfs/wkshp_05/SiriusAB.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt;. The information on that website is sort of vague, but I think it'll be helpful in visualization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps!&lt;/p&gt;&#xA;" OwnerUserId="24048" LastActivityDate="2018-10-17T00:36:17.957" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28049" PostTypeId="1" AcceptedAnswerId="28052" CreationDate="2018-10-17T02:34:57.817" Score="0" ViewCount="47" Body="&lt;p&gt;I've recently noticed that in Wikipedia articles such as Pluto (&lt;a href=&quot;https://www.wikiwand.com/en/Pluto&quot; rel=&quot;nofollow noreferrer&quot;&gt;Link&lt;/a&gt;), there's this number that says how much flattened it is (&lt;a href=&quot;https://www.wikiwand.com/en/Haumea&quot; rel=&quot;nofollow noreferrer&quot;&gt;Definition Here&lt;/a&gt;). In Pluto, it's &gt;1%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Haumea, which is a very flattened dwarf planet, actually doesn't have this information on Wikipedia, and I haven't been able to find it online. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be great if someone could find this info (and it's source as well), because I can't seem to find it anywhere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="24048" LastActivityDate="2018-10-17T04:41:53.500" Title="What is the percent flattening in Haumea?" Tags="&lt;dwarf-planets&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28050" PostTypeId="1" CreationDate="2018-10-17T02:55:26.083" Score="2" ViewCount="581" Body="&lt;p&gt;It's commonly known that the center of gravity of our solar system is not the suns axis, and, that the same goes for other orbital systems like Pluto and Charon, and so on. When was it discovered that our sun has an orbit around a barycenter?&lt;/p&gt;&#xA;" OwnerUserId="24569" LastActivityDate="2018-10-17T06:25:12.087" Title="When did we discover that our sun does not orbit its own axis but a barycenter?" Tags="&lt;solar-system&gt;" AnswerCount="1" CommentCount="16" FavoriteCount="1" ClosedDate="2018-10-19T09:48:52.717" ContentLicense="CC BY-SA 4.0" />
  <row Id="28051" PostTypeId="1" CreationDate="2018-10-17T03:32:08.463" Score="4" ViewCount="137" Body="&lt;p&gt;This is for a novel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rogue planet, is, by a one-in-a-zillion freak chance, on its way to collide with Earth. It has been drifting in interstellar space since its original formation (billions of years ago), and thus will be about as cold as it can get. It is approaching at a sharp angle to the ecliptic, and at its nearest point it will pass within the orbit of the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently envisioning its mass to be roughly equal to Neptune, but the only real constraint is that even if it does not collide with the Earth, its passage will end life on Earth with near certainty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: How close can such a planet come and still have a reasonable chance of not being detected from Earth by modern astronomy? Assume that the astronomical community is at its usual level of vigilance. I don't want the failure to detect it earlier to be due to some fluke; the oncoming collision is flukey enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that this will depend on its composition, whether predominantly rocky or predominantly gaseous. I'd prefer the answer to be based on whichever type can get closer without being detected.&lt;/p&gt;&#xA;" OwnerUserId="24570" LastEditorUserId="24571" LastEditDate="2018-10-17T07:35:22.427" LastActivityDate="2018-10-17T07:35:22.427" Title="Detecting a Rogue Planet from Earth" Tags="&lt;rogue-planet&gt;" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="28052" PostTypeId="2" ParentId="28049" CreationDate="2018-10-17T04:41:53.500" Score="2" Body="&lt;p&gt;That would be because Haumea is a Triaxial ellipsoid. It's dimensions are different in every cardinal direction. The term 'flattening' only makes sense for the biaxial ellipsoids, because two of the axes are the same and one is different (in the case of Pluto, the equatorial axis is 1% larger, hence 1% flattened). The concept of 'flattened' doesn't make terribly much sense when all three axes are different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Haumea's axis ratios are approximately 1 : 0.75 : 0.5, if that's the information you're after.&lt;/p&gt;&#xA;" OwnerUserId="17655" LastActivityDate="2018-10-17T04:41:53.500" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28053" PostTypeId="2" ParentId="28050" CreationDate="2018-10-17T06:14:50.303" Score="5" Body="&lt;p&gt;Newton explained the motions of planets in the solar system in Principia in 1687.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This amongst many other things &quot;defines the very slow motion of the Sun relative to the solar-system barycenter;&quot; (&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Philosophi%C3%A6_Naturalis_Principia_Mathematica&quot; rel=&quot;noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Philosophiæ_Naturalis_Principia_Mathematica&lt;/a&gt; ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Newton came up with the ideas of gravity and barycentres, surely this is the answer? (Though, I note there is some debate that the inverse square law may have been around and discussed during the previous decade).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideas that the Sun &lt;em&gt;moved&lt;/em&gt; were of course commonplace before then.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-10-17T06:25:12.087" LastActivityDate="2018-10-17T06:25:12.087" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28054" PostTypeId="1" AcceptedAnswerId="28057" CreationDate="2018-10-17T07:33:29.710" Score="5" ViewCount="162" Body="&lt;p&gt;I cannot understand the meaning of &quot;shock cooling&quot; which occurred in supernova. &#xA;And I saw the sentence like &quot;Early photometry and spectroscopy reveal evidence of shock cooling of an extended helium-rich envelope&quot;. Could you explain the mechanism of shock cooling in supernova?&lt;/p&gt;&#xA;" OwnerUserId="24460" LastActivityDate="2018-10-17T12:28:51.630" Title="What does the word &quot;shock cooling&quot; mean in supernova physics?" Tags="&lt;supernova&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28055" PostTypeId="1" AcceptedAnswerId="39857" CreationDate="2018-10-17T08:10:09.630" Score="20" ViewCount="598" Body="&lt;p&gt;There are of course many, many sources that quote Kepler's laws of planetary motion. This is preventing me from finding out what I really want to know: which is - what are Kepler's laws &lt;em&gt;as he wrote them&lt;/em&gt;? As opposed to how they may have been later interpreted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very specifically I am interested in whether &quot;Kepler's First Law&quot;, as proposed by Kepler, which is usually stated in the form that orbiting objects in the solar system execute elliptical orbits with the Sun at one focus, actually recognised, or had a caveat, that in a 2-body system it is the barycentre that is at the focus of an ellipse for both bodies, and that it is even more complex in the case of 3-bodies or more.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="5506" LastEditDate="2018-10-19T09:45:19.050" LastActivityDate="2020-11-15T20:12:26.567" Title="What are Kepler's laws (as he wrote them)?" Tags="&lt;planet&gt;&lt;history&gt;&lt;orbital-mechanics&gt;&lt;johannes-kepler&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28057" PostTypeId="2" ParentId="28054" CreationDate="2018-10-17T12:28:51.630" Score="6" Body="&lt;p&gt;Shock cooling in a SN is the scenario where kinetic energy carried by the shock transforms into other energy forms: adiabatic expansion, radiation, acceleration. A shock is actually a term meaning discontinuity. For example, a density profile of steady-wind medium is &lt;span class=&quot;math-container&quot;&gt;$r^{-2}$&lt;/span&gt; while that of external part of ejecta is &lt;span class=&quot;math-container&quot;&gt;$r^{-n}$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$7 \le n \le 12$&lt;/span&gt; as typical values in core-collapse supernovae, but at the interface between the medium and the external ejecta, where they interacts, its density profile spikes before asymptotically decaying to the characteristic profile of each region. The interface is the shock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism of shock cooling in SN involves processes which are not completely understood. What we know is that the shock is formed as a double-shock structure: forward and reverse. The forward shock propagates outward, while the reverse one goes inward in mass coordinate. (Note that in spatial coordinate both shocks propagate outward). Kinetic energy in each shock is transformed by interactions. At a forward shock, typically the energy is transformed into acceleration of relativistic particles, which in turn radiate via synchrotron or inverse Compton processes, and collisionally interacts with the medium. At a reverse shock, the shock energy thermalizes ejecta, which in turn thermally radiates.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-10-17T12:28:51.630" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28058" PostTypeId="1" AcceptedAnswerId="39021" CreationDate="2018-10-17T13:58:40.300" Score="10" ViewCount="451" Body="&lt;p&gt;The path of the Earth around the sun is described as being an ellipse, with the Sun being at one of the foci of the ellipse.&lt;/p&gt;&#xA;&lt;p&gt;Since an ellipse, by definition, is symmetrical with respect to the foci, I would expect the time for the earth to transit any two halves of the ellipse to be the same, so long as it moves at the same speed around the ellipse, or at least the speed characteristics are symmetrical as well.&lt;/p&gt;&#xA;&lt;p&gt;However, this does not appear to be the case according to the following rounded figures for the time in days between equinoxes and solstices:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;Summer Solstice to Winter Solstice    184.3&#xA;Winter Solstice to Summer Solstice    181.0&#xA;Vernal Equinox to Autumnal Equinox    186.4&#xA;Autumnal Equinox to Vernal Equinox    178.9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;These values can be found, for example, in astronomer Robert Newton's book of 1976.&lt;/p&gt;&#xA;&lt;p&gt;Why are the values different?&lt;/p&gt;&#xA;" OwnerUserId="2800" LastEditorUserId="2800" LastEditDate="2020-09-22T08:45:09.313" LastActivityDate="2020-09-22T14:08:20.847" Title="Why is the time between equinoxes different?" Tags="&lt;orbit&gt;&lt;ephemeris&gt;&lt;equinox&gt;&lt;solstice&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28059" PostTypeId="1" CreationDate="2018-10-17T13:59:43.450" Score="1" ViewCount="34" Body="&lt;p&gt;Long story short, what I am doing is to split the spherical space around my device into equal size pixels (49,152 pixels) with HealPix algorithm. If the power that my device can receive at each pixel is equal or higher than a certain threshold, I will color that pixel, if not, the pixel remains white. At the end, I want to find the angular correlations of the colored pixels (can be seen as galaxies). I started with the simplest estimator Davis &amp;amp; Peebles (&lt;span class=&quot;math-container&quot;&gt;$w(\theta) = DD/RR$&lt;/span&gt;) and it works fine. However what I achieve with LS estimator is hard to interpret since I get &lt;span class=&quot;math-container&quot;&gt;$w(\theta) = 1$&lt;/span&gt; for large angular separations between my pixels (solid angle elements) which are separated around 170-180 degrees (I express it in degrees since I measure the arc length between 2 solid angle elements) apart and that is because I have no pairs in my DD and DR sets to be separated by 170-180 degrees. For instance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$w(\theta) = (DD-2DR+RR)/RR$&lt;/span&gt;    ---------&gt;    &lt;span class=&quot;math-container&quot;&gt;$w(170) = (0-2\cdot0+RR)/RR = 1$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is what does this &lt;span class=&quot;math-container&quot;&gt;$w(\theta) = 1$&lt;/span&gt; mean while I have no pairs in my data set separated by the mentioned large angles. In case if needed, attached is a set of plots to highlight the problem plus 2D representation of my data and random sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be greatly appreciated if you can help me how to interpret my angular correlation results from LS estimator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6JJgl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6JJgl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/VQQ0G.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VQQ0G.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Angular Correlation results for LS estimator and Peebles estimator:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/hZpdj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hZpdj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24576" LastEditorUserId="7472" LastEditDate="2018-10-17T14:03:28.327" LastActivityDate="2018-10-17T14:03:28.327" Title="Interpretation of Angular Correlation results" Tags="&lt;galaxy-cluster&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28060" PostTypeId="2" ParentId="28058" CreationDate="2018-10-17T14:12:15.517" Score="13" Body="&lt;p&gt;This is because the summer and winter solstices (approx. June 21st and December 21st) do not correspond to the aphelion and perihelion (approx. July 5th and January 4th). Therefore, the average distance from the Sun is longer in the period from the Summer Solstice to Winter Solstice than vice versa, so the Earth is moving slower (on average) and it takes longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the values you mentioned are incorrect (switched around); 181.0 is a more likely value for Winter to Summer (February is a few days shorter than August).&lt;/p&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2018-10-17T14:12:15.517" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28061" PostTypeId="1" AcceptedAnswerId="28064" CreationDate="2018-10-17T15:00:15.973" Score="1" ViewCount="82" Body="&lt;p&gt;In the recent Acta Astronautica article &lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/S0094576518308221&quot; rel=&quot;nofollow noreferrer&quot;&gt;The edge of space: Revisiting the Karman Line&lt;/a&gt;, Harvard-Smithsonian Center for Astrophysics astronomer, &lt;a href=&quot;https://space.stackexchange.com/a/21168/12102&quot;&gt;Space SE contributor&lt;/a&gt; and &quot;inverse namesake&quot; of &lt;em&gt;asteroid (4589) McDowell&lt;/em&gt; Johnathan C. McDowell lists a number of different boundaries in the solar system in section 3.2. One of them is described as the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The ν6 secular Sun-Jupiter-Saturn resonance which marks the conventional inner edge of the asteroid belt at 2.06 astronomical units (308 million km) from the Sun (26); it coincides with the 1:4 Sun-Jupiter resonance (27) and asteroid orbits near this resonance are unstable, soon perturbed to enter the inner solar system. Although there is no generally agreed definition, this location is a reasonable place to mark as the boundary between the inner and outer solar system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What (the heck) are the &lt;em&gt;ν6 secular Sun-Jupiter-Saturn&lt;/em&gt; and the &lt;em&gt;1:4 Sun-Jupiter&lt;/em&gt; resonances exactly?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-10-17T15:06:29.923" LastActivityDate="2018-10-19T15:02:41.927" Title="What exactly are the &quot;ν6 secular Sun-Jupiter-Saturn&quot; and the &quot;1:4 Sun-Jupiter&quot; resonances?" Tags="&lt;asteroids&gt;&lt;orbital-mechanics&gt;&lt;terminology&gt;&lt;orbital-resonance&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28063" PostTypeId="1" CreationDate="2018-10-17T17:56:34.963" Score="0" ViewCount="1034" Body="&lt;p&gt;I'm thinking about getting a telescope and i'm debating between these 2 telescopes: &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B00LY8JWAQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;meade&lt;/a&gt; and &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B00D12U1IK&quot; rel=&quot;nofollow noreferrer&quot;&gt;orion&lt;/a&gt;. Theyre both the same price so this question is not about the money, but about getting the most out of your money. I like the Meade 102 mm (or 4 inches) because it is a refractor, which means it is more lightweight and compact. The orion starblast 4.5 on the other hand is a reflector and therefore more clumsy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, is a reflector necessarily better than a refractor in terms of magnification and sharpness of image? The orion does have a slightly bigger aperture (half inch bigger) but I don't think that makes a huge difference. The question really is, a reflector vs a refractor, which one is more desirable in terms of magnification, and how much of a difference does it make? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just judging from the reviews it would seem like the reflector (orion) wins the fight. For example, Saturn's rings according to some reviewers are &quot;clearly visible&quot; in the reflector, but not so with the refractor which most reviewers agree, Saturn will be very tiny and the rings barely visible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So will I settle for magnification if I get the refractor instead of the reflector, or are they basically the same and in that case I would choose the refractor (meade) over the reflector since it is more compact, and is also easier for beginners like me?&lt;/p&gt;&#xA;" OwnerUserId="18226" LastActivityDate="2018-10-20T23:14:17.010" Title="Reflector vs refractor in terms of magnification, which one wins?" Tags="&lt;telescope&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28064" PostTypeId="2" ParentId="28061" CreationDate="2018-10-17T19:06:07.893" Score="4" Body="&lt;p&gt;The 1:4 (or 4:1) Jupiter resonance is a mean-motion resonance: an asteroid there takes 1/4 as long to orbit the Sun as Jupiter does.&#xA;Perturbations by Jupiter at recurring ecliptic longitudes alter the asteroid's orbital period, so asteroids do not remain long in this state.&#xA;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Kirkwood_gap&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kirkwood gaps&lt;/a&gt; in the asteroid belt include this and other mean-motion resonances with Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;span class=&quot;math-container&quot;&gt;$\nu_6$&lt;/span&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Secular_resonance&quot; rel=&quot;nofollow noreferrer&quot;&gt;secular resonance&lt;/a&gt; (also called &lt;span class=&quot;math-container&quot;&gt;$g = g_6$&lt;/span&gt;) is more complex: in certain combinations of orbital period and inclination, an asteroid experiences &lt;a href=&quot;https://en.wikipedia.org/wiki/Apsidal_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;perihelion precession&lt;/a&gt; at the same rate as Saturn (the 6th planet) does, resulting in a gradual increase of the asteroid's orbital eccentricity.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-10-19T15:02:41.927" LastActivityDate="2018-10-19T15:02:41.927" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="28066" PostTypeId="2" ParentId="28063" CreationDate="2018-10-17T19:58:18.580" Score="7" Body="&lt;p&gt;Magnification is (mostly) done by the eyepiece. You can change the magnification by using a different eyepiece. Whether it is a reflector or refractor doesn't make much difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a limit to the useful magnification that can be done. At a certain point, you are just magnifying the blur. Larger apertures can resolve better, and so can allow greater magnification. Using more magnification also makes the view dimmer. A bigger aperture collects more light, so also allows for more magnification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The choice between reflector or refractor is not about magnification. Lenses work by refraction, and this can create coloured fringes (which can be compensated for, but this increases the cost). Thicker lenses also absorb light, but so do mirrors to some degree.  Generally, bigger is better, but other factors (stability of mount, ease of use, ease of transport, computer control etc.) may be significant. &quot;Magnification&quot; is often not a good measure of quality.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-10-17T19:58:18.580" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28067" PostTypeId="1" CreationDate="2018-10-17T23:09:32.193" Score="0" ViewCount="330" Body="&lt;p&gt;If an observer travels away from our sun, at what distance would our sun occupy 7 arc seconds of space?&lt;/p&gt;&#xA;" OwnerUserId="24569" LastEditorUserId="7472" LastEditDate="2018-10-25T14:48:00.167" LastActivityDate="2018-10-25T14:48:00.167" Title="From what distance would our sun have an angular diameter of 7 arc seconds?" Tags="&lt;the-sun&gt;&lt;angular-diameter&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28068" PostTypeId="1" CreationDate="2018-10-18T00:00:06.980" Score="2" ViewCount="194" Body="&lt;p&gt;I am trying to find a source of mirror reflectivity versus wavelength for the typical protected overcoat Aluminum mirrors used in astronomy. I need this to cover the whole visible range (300-1200nm) and ideally machine-readable. I've only been able to find part coverage of the wavelength range or single spot measurements at a specific wavelength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best I've managed is &lt;a href=&quot;https://www.edmundoptics.com/resources/application-notes/optics/metallic-mirror-coatings/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page at Edmund Optics&lt;/a&gt; which has (small) plots but no table or downloadable file.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2019-01-28T19:05:17.723" Title="Source of mirror reflectivity data" Tags="&lt;telescope&gt;&lt;light&gt;&lt;mirror&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28069" PostTypeId="2" ParentId="28067" CreationDate="2018-10-18T00:42:24.847" Score="3" Body="&lt;p&gt;From us, the Sun is visible around a half grad, which is &lt;span class=&quot;math-container&quot;&gt;$\approx$&lt;/span&gt; 30 arc min = 1800 arc seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get to 7 arc seconds, you need to go 1800/7 times farther away, so the result is around 250 AU.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorDisplayName="user1569" LastEditDate="2018-10-18T08:55:33.310" LastActivityDate="2018-10-18T08:55:33.310" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28071" PostTypeId="1" CreationDate="2018-10-18T16:32:49.527" Score="7" ViewCount="1376" Body="&lt;p&gt;In which direction does the ecliptic plane make an angle of 60 degrees with galactic plane? Is it toward the galactic center? (It can have any 360 degree orientation.)&lt;/p&gt;&#xA;" OwnerUserId="24584" LastEditorUserId="2531" LastEditDate="2019-10-29T07:08:21.610" LastActivityDate="2020-12-11T04:59:27.343" Title="In which direction does the ecliptic plane make an angle of 60 degrees with galactic plane?" Tags="&lt;milky-way&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28072" PostTypeId="2" ParentId="28068" CreationDate="2018-10-18T17:34:20.647" Score="3" Body="&lt;p&gt;The spectral reflectance curve depends on the coatings applied.  For 'scopes used by people there is no point in maintaining reflectivity below 450 or above 700 nm.  You idea of &amp;quot;visible&amp;quot; range is, I fear somewhat optimistic.  You may be better off with an uncoated Al mirror (spectral curves easily found on the net) and giving up some signal in the photopic range.  -- or with a nice Ag overcoat, perhaps.&lt;/p&gt;&#xA;&lt;h2&gt;edit&lt;/h2&gt;&#xA;&lt;p&gt;Per request, some data on uncoated Al.&#xA;From   &lt;a href=&quot;https://laserbeamproducts.wordpress.com/2014/06/19/reflectivity-of-aluminium-uv-visible-and-infrared/&quot; rel=&quot;nofollow noreferrer&quot;&gt;laserbeamproducts&lt;/a&gt;,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;REFLECTIVITY OF PURE UNCOATED ALUMINIUM AT NORMAL INCIDENCE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;Wavelength            % Reflectivity &#xA;248nm                      92.6 &#xA;400nm                      92.0 &#xA;532nm                      91.6 &#xA;633nm                      90.7 &#xA;800nm                      86.8 &#xA;900nm                      89.0 &#xA;1um                          94.0 &#xA;3um                          98.0&#xA;10.6um                     98.7 &#xA;20um                        99.0 &#xA;100um                      99.4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;An image published (by the author) at &lt;a href=&quot;https://www.researchgate.net/figure/Spectral-reflectance-of-aluminum-and-various-other-metals-O-Elsevier-Reprinted-with_fig1_283437763&quot; rel=&quot;nofollow noreferrer&quot;&gt;Researchgate&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Nu3Rc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nu3Rc.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="907" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-10-19T12:18:00.247" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28073" PostTypeId="2" ParentId="28063" CreationDate="2018-10-18T17:58:42.980" Score="3" Body="&lt;p&gt;In this price range some telescopes are stripped-down &quot;real&quot; telescopes, and some telescopes are &quot;toys&quot;.  Meade is generally a reputable brand, but I suspect a 4&quot; f/5.9 refractor for $200 is essentially going to be a &quot;toy&quot; (bad color correction, shaky mount) or at best a spotting scope designed for daytime use (where bad color correction is less noticeable). High quality refractors of the same size are apochromats and go for 5-10x the price (or more...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Orion telescope seems to be more of a &quot;real&quot; telescope stripped down (very small dobsonian reflector, standard size eyepieces).&lt;/p&gt;&#xA;" OwnerUserId="14434" LastActivityDate="2018-10-18T17:58:42.980" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28075" PostTypeId="2" ParentId="28071" CreationDate="2018-10-18T22:46:57.267" Score="4" Body="&lt;p&gt;The line defining where the ecliptic plane and the Galactic plane coincide lies (approximately) between the points defined by right ascensions  and declinations  of 6h +23 and 18h -23.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be seen below - the plot is in celestial coordinates with the ecliptic plane marked (thicker solid line) and the Galactic plane clearly visible in the background map of dust extinction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4HjKQ.png&quot; alt=&quot;Galactic and ecliptic planes&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second of these coordinates is close to, but not coincident, with the Galactic centre (marked Sgr A*). So the relative tilt is sort of at right angles to the Galactic centre at present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this will change over tens of millions of years as the solar system orbits the Galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Another way of specifying the direction is just to ask where the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Orbital_pole&quot; rel=&quot;nofollow noreferrer&quot;&gt;ecliptic north pole&lt;/a&gt; is. It is at RA=18h, Dec=66d 34m, in the constellation of Draco.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A further way of thinking about it is to consider at which points in the sky do the ecliptic and Galactic planes have their maximum angular separation. These are 90 degrees away from the two points I defined above and are almost on the celestial equator at about 0h and 12h.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-10-29T07:25:53.923" LastActivityDate="2019-10-29T07:25:53.923" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28076" PostTypeId="1" CreationDate="2018-10-19T05:20:35.597" Score="2" ViewCount="50" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Are there any observations that more distant galaxies appear to rotate more slowly than closer galaxies? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What methods are used to measure the rotational speed of galaxies? (both orbital speed profiles and/or bulk average rotation speed) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the current limit for how far away a galaxy can be to make such measurements? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I greatly appreciate any answer or links to info on this topic.  I am not an astronomer by profession or education, just deeply fascinated with cosmology. &lt;/p&gt;&#xA;" OwnerUserId="24592" LastActivityDate="2018-10-19T05:20:35.597" Title="Galactic rotation speeds - questions from a non-astronomer" Tags="&lt;galaxy&gt;&lt;galactic-dynamics&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28077" PostTypeId="1" AcceptedAnswerId="28092" CreationDate="2018-10-19T08:17:23.997" Score="2" ViewCount="389" Body="&lt;p&gt;It has been argued that it is not possible to trace back the remains of the progenitor star of the Sun (sometimes called Coatlicue), which, being estimated in the mass range of 30 &lt;span class=&quot;math-container&quot;&gt;$M_{\odot}$&lt;/span&gt;, should have left a black hole after its explosion as a supernova, nor is it possible to find the sister stars formed from the compression of the same nebula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/10729&quot;&gt;https://astronomy.stackexchange.com/a/10729&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://phys.org/news/2012-08-solar-genealogy-revealed-meteorites.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://phys.org/news/2012-08-solar-genealogy-revealed-meteorites.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because &quot;[The Sun] has completed many Galactic orbits [since its formation]. Not only that, but it could have migrated in orbital radius as well.&quot; and &quot;These twin stars did not influence its development nor that of its planets [and] are now dispersed in the Galaxy and cannot be identified&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why did the supernova's remains and the sister stars not remain gravitationally bound together? Why did they not share the same angular momentum of revolution around the Galaxy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why did these massive bodies formed from the same nebula not become a bound system among them and with the supernova's remains, like the Sun and its planets in the Solar system?&lt;/p&gt;&#xA;" OwnerUserId="24594" LastEditorUserId="24594" LastEditDate="2018-10-21T06:40:02.457" LastActivityDate="2018-10-21T08:46:37.330" Title="Remains of the progenitor of the Sun" Tags="&lt;the-sun&gt;&lt;star-formation&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28078" PostTypeId="2" ParentId="28067" CreationDate="2018-10-19T12:06:31.527" Score="3" Body="&lt;p&gt;While this might seem to be strictly a math problem, it's really loaded with Astronomy, let's see what we can learn!&lt;/p&gt;&#xA;&lt;p&gt;If you're far enough away that you can see essentially a full hemisphere (which you can't if you're close), the apparent angular width is twice the half-width, and that's given by&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\theta_{HW} = \arcsin\frac{r}{R} \approx \frac{r}{R}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;for small angles, where &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; is the physical radius of the object and &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; is the distance from the viewer to its center. This is also similar to or the same as &lt;a href=&quot;https://en.wikipedia.org/wiki/Moon&quot; rel=&quot;nofollow noreferrer&quot;&gt;@peterh's method&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;There's (at least) two problems with trying to talk about the radius of the Sun;&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The Sun is oblate&lt;/li&gt;&#xA;&lt;li&gt;The Sun is diffuse; it doesn't have a well defined edge&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Let's do #2 first. There is a formal working definition for the optical edge of the Sun, and &lt;a href=&quot;https://astronomy.stackexchange.com/a/2106/7982&quot;&gt;this excellent answer&lt;/a&gt; to the question &lt;em&gt;How do you define the diameter of the Sun&lt;/em&gt; states:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Most literature will define the diameter of the Sun up to the photosphere, the layer of the solar atmosphere you would see if you were to observe the Sun in white light.&lt;/p&gt;&#xA;&lt;p&gt;The base of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Photosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;Photosphere&lt;/a&gt; is defined as the region where the &lt;a href=&quot;https://en.wikipedia.org/wiki/Optical_depth&quot; rel=&quot;nofollow noreferrer&quot;&gt;optical depth&lt;/a&gt; is around 2/3, or the region where the plasma becomes transparent to most optical light wavelengths.&lt;/p&gt;&#xA;&lt;p&gt;Of course the true edge of the solar atmosphere could be considered as the heliopause, where the direct influence of the Sun's magnetic field and solar wind end and interstellar space begins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;With that definition of the edge, let's look at the shape of the Sun.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Sun&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; gives both 695,700 and 696,392 km for the equatorial radius of the Sun, sourcing from &lt;a href=&quot;https://arxiv.org/abs/1510.07674&quot; rel=&quot;nofollow noreferrer&quot;&gt;IAU 2015 Resolution B3...&lt;/a&gt; and &lt;a href=&quot;https://arxiv.org/abs/1203.4898&quot; rel=&quot;nofollow noreferrer&quot;&gt;Measuring the Solar Radius from Space during the 2003 and 2006 Mercury Transits&lt;/a&gt; respectively.&lt;/p&gt;&#xA;&lt;p&gt;Let's use 695,700 km because it's the one that I see most often and &amp;quot;because IAU&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;The Wikipedia article gives a &lt;a href=&quot;https://astronomy.stackexchange.com/a/28069/7982&quot;&gt;flattening&lt;/a&gt; of 9E-06, which makes the polar radius only about 10 parts per million smaller, which is a &lt;em&gt;much smaller equator-to-pole difference&lt;/em&gt; than I remembered it to be. I guess we can ignore it after all!&lt;/p&gt;&#xA;&lt;p&gt;Seen from the Earth then, who's orbit takes it from 152.1 million to 147.1 million km from the Sun, the angular width of the Sun would vary from about 1887 to 1951 arcseconds (31.4 to 32.5 arcminutes).&lt;/p&gt;&#xA;&lt;p&gt;At what distance would an object with a radius of 695,700 km have an apparent width of 7 arcseconds (which is 3.39E-05 radians)? Flipping the equation around gives 2×695700/3.39E-05 or 4.105E+10 kilometers about or 274 AU.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;What is it like 274 AU from the Sun?&lt;/h2&gt;&#xA;&lt;p&gt;In addition to appearing to be roughly 274 times smaller from here than from Earth, the Sun is 274×274 times dimmer. Using &lt;span class=&quot;math-container&quot;&gt;$2.5 \times \log_{10}(274×274)$&lt;/span&gt; we get that the Sun would appear about 12.2 magnitudes dimmer than it's -26.8 magnitude brightness at 1 AU, or -14.6 magnitude, which is still 2 magnitudes brighter than a typical full moon.&lt;/p&gt;&#xA;&lt;p&gt;Your orbital period around the Sun would be over 4,500 years, and you'd be moving at 1.8 km/sec in that orbit, as opposed to 30 km/sec for the Earth.&lt;/p&gt;&#xA;&lt;p&gt;You'd be well past the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kuiper_belt&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kuiper belt&lt;/a&gt; but nowhere near the &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_Solar_System_objects_most_distant_from_the_Sun_in_2018&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oort cloud&lt;/a&gt; (this is a simplification, but it will do for now), and you'd be way past the &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_Solar_System_objects_most_distant_from_the_Sun_in_2018&quot; rel=&quot;nofollow noreferrer&quot;&gt;farthest known solar system body&lt;/a&gt; in 2018! &lt;a href=&quot;https://en.wikipedia.org/wiki/V774104&quot; rel=&quot;nofollow noreferrer&quot;&gt;V774104&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZmQGg.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZmQGg.gif&quot; alt=&quot;V774104&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; from EarthSky.org's &lt;a href=&quot;http://earthsky.org/space/new-most-distant-object-in-solar-system-v774104&quot; rel=&quot;nofollow noreferrer&quot;&gt;New most distant object in solar system&lt;/a&gt; Image via S. Sheppard / C. Trujillo / D. Tholen / Subaru Telescope/ &lt;a href=&quot;http://www.skyandtelescope.com/astronomy-news/v774104-most-distant-solar-system-object-11212015&quot; rel=&quot;nofollow noreferrer&quot;&gt;skyandtelescope.com&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;And at 274 AU, you'd be much farther from the Sun than any deep space probe as well. Voyagers 1 and 2 are &amp;quot;only&amp;quot; 118 and 142 AU from the Sun now, and New Horizons is only about 42 AU.&lt;/p&gt;&#xA;&lt;p&gt;I don't know the name of the place you'd be, it looks pretty lonely though!&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Rg9nC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rg9nC.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:PIA17046_-_Voyager_1_Goes_Interstellar.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-10-19T12:29:10.503" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28079" PostTypeId="1" CreationDate="2018-10-19T12:43:54.390" Score="4" ViewCount="172" Body="&lt;p&gt;I had the following question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A comet enters the solar system. Assume that it is influenced only by the Sun, not by any of the planets. Which statement about its orbit is correct? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I chose as correct statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Its orbit might be an ellipse, or is might be a parabola or a hyperbola (because a comet is not necessarily confined to the solar system forever).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I want to know what it means by &lt;em&gt;a comet is not necessarily confined to the solar system forever&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it mean that eventually the comet will &quot;die&quot; out, with the sun vaporizing all of its ice, so it will basically cease to exist while it is in the solar system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or does it mean that the comet can leave the solar system? If the comet enters the solar system it is subject to the mighty force of the suns gravity, then how can it escape?&lt;/p&gt;&#xA;" OwnerUserId="17327" LastEditorDisplayName="user1569" LastEditDate="2018-10-19T13:25:12.037" LastActivityDate="2018-10-22T16:10:17.943" Title="Is a comet necessarily confined to the solar system?" Tags="&lt;comets&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28080" PostTypeId="2" ParentId="28079" CreationDate="2018-10-19T13:25:21.030" Score="4" Body="&lt;p&gt;It means that the sum of the kinetic and gravitational potential energies can be greater than zero. Such objects are not bound to the solar system. They fall towards the Sun on hyperbolic trajectories, picking up speed and kinetic energy as they do so. After rounding the Sun they then have enough speed to escape from the solar system again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The origin of such comets is not known for certain - some may have come from other star systems. Some comets can enter the solar system on &lt;em&gt;parabolic&lt;/em&gt; orbits - these have come from a long way out - perhaps the Oort cloud - and are &lt;em&gt;just&lt;/em&gt; bound to the solar system. But interactions with other planets and the Sun can result in them gaining energy and leaving the solar system on unbound hyperbolic orbits. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-10-19T13:25:21.030" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28082" PostTypeId="1" AcceptedAnswerId="28085" CreationDate="2018-10-19T13:51:36.850" Score="0" ViewCount="176" Body="&lt;p&gt;Yet again, I just stumbled across another baffling eclipse question (actually, two now). I was interested in two eclipses (Saros series 147 and 152) that will both be occurring in 2021. Each of these eclipses will occur relatively near the opposite poles of the earth. &lt;strong&gt;The Wikipedia link for saros 147 provided the following additional information:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Solar eclipses of 1964-1967[edit]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;This eclipse is a member of a semester series&lt;/strong&gt;. &lt;em&gt;An eclipse in a&#xA;  semester series of solar eclipses repeats approximately every 177 days&#xA;  and 4 hours (a semester)&lt;/em&gt; at alternating nodes of the Moon's orbit.[1]&#xA;  &quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These are both fairly young saros series. Saros 152 is comprised of 70 eclipses, while saros 147 contains a total of 80 eclipses. &lt;strong&gt;Observations:&lt;/strong&gt; &lt;strong&gt;They are 10 eclipses difference in length - They each began 180 years apart.&lt;/strong&gt; So, following on from that, I checked out their ending dates and found, much to my amazement, that &lt;strong&gt;&lt;em&gt;both series end during the exact same year&lt;/em&gt;&lt;/strong&gt; - &lt;strong&gt;exactly 177 day apart - in the year 3049&lt;/strong&gt; !!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, since this additional information, my new questions concerning these eclipses have become: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) &lt;strong&gt;What exactly is a &quot;semester series&quot; for eclipses (any detail)?&lt;/strong&gt; &lt;em&gt;The day spacing definition is obvious but doesn't explain much else.&lt;/em&gt; &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2) &lt;strong&gt;Does this mean that &lt;em&gt;some or all&lt;/em&gt;&lt;/strong&gt; eclipses in series 147 and 152 are in&#xA;  the same semester series?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="22542" LastActivityDate="2018-10-19T21:38:30.783" Title="What is a &quot;semester series&quot; for eclipse prediction? - Solar saros series 147 and 152" Tags="&lt;the-sun&gt;&lt;solar-system&gt;&lt;observational-astronomy&gt;&lt;solar-eclipse&gt;&lt;lunar-eclipse&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28083" PostTypeId="1" AcceptedAnswerId="28084" CreationDate="2018-10-19T14:34:39.347" Score="9" ViewCount="1486" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Jan_Oort#A_few_of_Oort&amp;#39;s_discoveries&quot; rel=&quot;noreferrer&quot;&gt;Jan Oort&lt;/a&gt; was a pioneer in radio astronomy. Wikipedia says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It has been written that “Oort was probably the first astronomer to realize the importance” of radio astronomy. “In the days before radio telescopes,” one source notes, “Oort was one of the few scientists to realise the potential significance of using radio waves to search the heavens. His theoretical research suggested that vast clouds of hydrogen lingered in the spiral arms of the Galaxy. These molecular clouds, he predicted, were the birthplaces of stars.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The article includes the image shown below, which is a 21 cm radio map of the Milky Way galaxy. (for more on the transparency of dust at 21 cm see &lt;a href=&quot;https://astronomy.stackexchange.com/a/27981/7982&quot;&gt;this excellent answer&lt;/a&gt; to &lt;em&gt;How was the galactic plane established?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also read more about Oort's work during this time in this &lt;a href=&quot;https://www.aip.org/history-programs/niels-bohr-library/oral-histories/4806&quot; rel=&quot;noreferrer&quot;&gt;1976 AIP Oral History Interview interview&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing that the center of the plot is the galactic center and the point at 8 kpc above it is the Earth. What is the reason for the two blank wedges, projecting downward and upward from the Earth? Are they geometrical, blind spots from the few early radio telescopes in the 1950's perhaps, or do they reflect real phenomenon in the galaxy? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bjDoE.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bjDoE.jpg&quot; alt=&quot;Oort et al. 1958 21 cm map of Milky Way&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original source is &lt;a href=&quot;http://adsabs.harvard.edu/abs/1958MNRAS.118..379O&quot; rel=&quot;noreferrer&quot;&gt;The galactic system as a spiral nebula&lt;/a&gt; Oort, J. H.; Kerr, F. J.; Westerhout, G. MNRAS 118, (1958) p. 379&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8dGvl.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8dGvl.jpg&quot; alt=&quot;Oort et al. 1958 21 cm map of Milky Way&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-10-19T15:00:44.537" LastActivityDate="2018-10-19T15:42:08.390" Title="Why the blank wedges in this very early 21 cm map of the Milky Way? (Oort et al. 1958)" Tags="&lt;observational-astronomy&gt;&lt;milky-way&gt;&lt;radio-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28084" PostTypeId="2" ParentId="28083" CreationDate="2018-10-19T15:42:08.390" Score="9" Body="&lt;p&gt;TLDR: these wedges are bits where things are moving around the centre of the galaxy at about the same speed as us, so we can't understand what is there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it states on page 4 of the paper you have linked,   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the great gap between 315 and 340, where, except at small R, the differential rotation is too small to separate the various arms&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The method used to deduce this structure is described on page 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Observations in the decimetre continuum can only give the integrated radiation over the line of sight. The 21-cm line gives discrimination in distance. But although the 21-cm observations give discrimination in distance they cannot by themselves provide actual distances. The distance distribution in a given direction can only be inferred from radial velocities. For this we have to suppose that in each part of the Galactic System the average motion of the gas coincides with the circular velocity at the corresponding distance from the centre.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The map is given by combining several observations at different wavelengths - the 'decimetre' gives the total amount of stuff along the line of sight (because it isn't absorbed much along the way), the 21-cm line gives a measure of distance (because it is absorbed so predictably wakened by the amount of stuff it has passed through), and then to we have to assume a model of Keplerian rotation as a guess at how things are moving around the centre of the galaxy. Combining these three things allows you to map out the material even though you are seeing different bits stacked together along your line of sight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't work in the blank wedges because the speed of rotation is too similar to ours, so we can't use the rotation velocity to disentangle distances.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2018-10-19T15:42:08.390" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28085" PostTypeId="2" ParentId="28082" CreationDate="2018-10-19T17:45:36.707" Score="2" Body="&lt;p&gt;Any eclipse is a coincidence of a new or full moon (repeating every 29.53 day lunation or &lt;a href=&quot;https://moonblink.info/Eclipse/why/months#Synodic&quot; rel=&quot;nofollow noreferrer&quot;&gt;synodic month&lt;/a&gt;)&#xA;and a lunar node (moon crossing the ecliptic, repeating every 27.21 day &lt;a href=&quot;https://moonblink.info/Eclipse/why/months#Draconic&quot; rel=&quot;nofollow noreferrer&quot;&gt;draconic month&lt;/a&gt;).&#xA;A &lt;a href=&quot;http://www.staff.science.uu.nl/~gent0113/eclipse/eclipsecycles_cycles.htm#semester&quot; rel=&quot;nofollow noreferrer&quot;&gt;semester&lt;/a&gt;&#xA;is 6 synodic months, close enough to 6.5 draconic months that eclipses can occur at that interval in a series 3 or 4 years long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interval between any two solar or lunar eclipses &lt;a href=&quot;http://users.skynet.be/felixverbelen/cycles.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;can be expressed&lt;/a&gt; as &lt;span class=&quot;math-container&quot;&gt;$aI + bS$&lt;/span&gt;,&#xA;where &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; is the saros interval of 223 lunations,&#xA;&lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Inex&quot; rel=&quot;nofollow noreferrer&quot;&gt;inex&lt;/a&gt; interval of 358 lunations between adjacent saros series,&#xA;and &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; are integers.&#xA;The semester interval of 6 lunations is &lt;span class=&quot;math-container&quot;&gt;$5I - 8S$&lt;/span&gt;; for each eclipse in saros &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;, there may be an eclipse in saros &lt;span class=&quot;math-container&quot;&gt;$n-5$&lt;/span&gt; one semester earlier, an eclipse in saros &lt;span class=&quot;math-container&quot;&gt;$n+5$&lt;/span&gt; one semester later, or both.&#xA;As the eclipses in a given saros series are about 18 years apart, each is in a different semester series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, one semester series runs from 2015 to 2018 and includes one solar eclipse in each of saros series 120, 125, 130, 135, 140, 145, 150, and 155.&#xA;Another runs from 2018 to 2021 and includes one solar eclipse in each of saros series 117, 122, 127, 132, 137, 142, 147, and 152.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2018-10-19T21:38:30.783" LastActivityDate="2018-10-19T21:38:30.783" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="28086" PostTypeId="1" CreationDate="2018-10-19T18:20:17.077" Score="2" ViewCount="101" Body="&lt;p&gt;I have a Dobson Omegon with a 640mm focal length and a 102mm aperture.&lt;br /&gt;&#xA;My finderscope is broken but I can do without.&lt;br /&gt;&#xA;I have located the planets with the naked eye but I do  I do not see them through my eyepiece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can focus perfectly on the moon, using a 20mm eyepiece. (I have a 6mm eyepiece as well.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I proceed?&lt;/p&gt;&#xA;" OwnerUserId="24597" LastEditorDisplayName="user1569" LastEditDate="2018-10-19T21:07:48.227" LastActivityDate="2018-10-19T21:11:26.243" Title="Finding Jupiter, Saturn" Tags="&lt;telescope&gt;&lt;jupiter&gt;&lt;saturn&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28087" PostTypeId="2" ParentId="28086" CreationDate="2018-10-19T21:11:26.243" Score="3" Body="&lt;p&gt;Always start with the low magnification (20 mm) eyepiece.&#xA;If the target is not in the eyepiece field at first, methodically sweep the surrounding area.&#xA;Any sweep pattern that makes sense to you is fine.&#xA;When the telescope is pointed near a bright planet, stray light may brighten the background slightly.&#xA;After you find and center the target, switch to the high magnification (6 mm) eyepiece if needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course it's much easier to aim a telescope with a working finder.&#xA;If your finder is not repairable, most modern consumer telescopes have a dovetail fitting which makes it easy to install a replacement.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2018-10-19T21:11:26.243" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28088" PostTypeId="1" CreationDate="2018-10-20T02:33:20.237" Score="1" ViewCount="78" Body="&lt;p&gt;I am exploring the outcome of a problem assigned in a quantum mechanics homework where the mass of of some beam of &quot;something&quot; is insignificant to the result. So much so that something weighing approximately &lt;span class=&quot;math-container&quot;&gt;$1.2285 * 10^{17} kg$&lt;/span&gt; would be needed to make a difference in respect to the problem. (that is, assuming I've solved the problem correctly). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been googling for quite a bit and I can't find anything that is in the &lt;span class=&quot;math-container&quot;&gt;$10^{17} kg$&lt;/span&gt; range. Does anything come to mind?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I can't seem to find any good tags for this question)..&lt;/p&gt;&#xA;" OwnerUserId="18174" LastEditorUserId="18174" LastEditDate="2018-10-20T02:58:51.417" LastActivityDate="2018-10-20T02:58:51.417" Title="Looking for a celestial body that weights ~$10^{17} kg$?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28090" PostTypeId="1" CreationDate="2018-10-20T20:17:52.527" Score="3" ViewCount="138" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;This question was left over from the last question I had, so I am asking it separately here:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have been looking into two solar eclipses &lt;strong&gt;(Saros series 147 and 152) that will both be occurring in 2021&lt;/strong&gt;. Each of these eclipses will be happening near the opposite poles of the earth.  Both are part of young saros series. Saros 152 is comprised of 70 eclipses, while saros 147 contains a total of 80 eclipses. These series are 10 eclipses difference in length and began 180 years apart. So, I happened to check out their ending dates and found that both series are projected to end during the exact same year - 177 day apart – together in the year 3049. &lt;strong&gt;I thought that both of these series ending together one semester apart in the very same year was a pretty amazing coincidence.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, checking further – it seems new saros series start from as few as 7 years apart to as many as 100 years apart (this is extremely variable spacing, but it averages to about 31 years apart over the usual 1260+ year active series lifespan). &lt;strong&gt;I wanted to know why series 147 and 152 would both be ENDING at the same time.&lt;/strong&gt;  I went back to the NASA website, and checked the 5000 year listing of all solar saros series numbering 0 through 180. I only found one other such instance of two solar saros series ENDING in the same year – series 49 and 54 both ENDED together in 32 CE – so it seems like a pretty rare event.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But then, I checked to see if there were any occasions of two saros series BEGINNING together in the same year.&lt;/strong&gt; Going back to the NASA website again with this question, I checked the 5000 year listing of all solar saros series numbering 0 through 180. I only found one single example of two saros series both BEGINNING in the same year. This time series 142 and 147 both BEGAN together in the year 1624 CE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Not only is solar saros 147 involved in BOTH of these &quot;same year&quot; cases, but all three series eclipses will be presenting themselves during a single upcoming semester series in 2021 – solar series 142&lt;/em&gt;&lt;/strong&gt; (Dec 14, 2020), &lt;strong&gt;&lt;em&gt;147&lt;/em&gt;&lt;/strong&gt; (June 10, 2021),&#xA;&lt;strong&gt;&lt;em&gt;and 152&lt;/em&gt;&lt;/strong&gt; (Dec 4, 2021). &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Can anyone explain to me the actual mechanics behind what causes saros series to begin and end?&lt;/strong&gt; I am familiar with Saros/Inex&#xA;  Panorama – but unfortunately they do not answer this question, and I&#xA;  can’t even read them online anyway.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Does anyone know why solar saros 147 may have STARTED in the same year as saros 142 or why saros 147 will be ENDING together in the same&#xA;  year as saros 152?&lt;/strong&gt; Are these two occasions both very rare&#xA;  coincidences, while at the same time, both involve the same series 147?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="22542" LastActivityDate="2021-05-17T15:37:18.863" Title="Solar series 142 and 147 started same year - series 147 and 152 will be ending same year. Why? Coincidence?" Tags="&lt;solar-system&gt;&lt;observational-astronomy&gt;&lt;solar-eclipse&gt;&lt;lunar-eclipse&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28091" PostTypeId="2" ParentId="14540" CreationDate="2018-10-20T20:18:43.693" Score="1" Body="&lt;p&gt;@Hackworth has also asked this question on the Geosciences SX site (&lt;a href=&quot;https://earthscience.stackexchange.com/q/7855/4439&quot;&gt;link to question&lt;/a&gt;), where I provided an answer (&lt;a href=&quot;https://earthscience.stackexchange.com/a/15399/4439&quot;&gt;link to my answer&lt;/a&gt;). I'm including a link to my answer there from here in case it proves useful to somebody. &lt;/p&gt;&#xA;" OwnerUserId="7965" LastActivityDate="2018-10-20T20:18:43.693" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28092" PostTypeId="2" ParentId="28077" CreationDate="2018-10-20T20:33:15.850" Score="4" Body="&lt;p&gt;In the article you link to, it suggests that the sun formed in a cluster of stars.  There are many clusters, the most famous is the Pleiades, or 7-sisters. Clusters are formed of stars that were born at the same time, and share similar motions in space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the motions of the stars in a cluster are similar, not the same. Over time the stars in a cluster spread out, and over much more time can end up far far away. The stars in a cluster are not close enough to hold on to each other by gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is impossible to tell for certain which of the 100 billion stars in the galaxy were born at the same time as the sun. Nor is it possible to identify a particular black hole. Again the black hole may have had a similar, but not identical motion to the sun and would have drifted very far away over the course of 4.6 billion years.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-10-20T20:33:15.850" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28093" PostTypeId="2" ParentId="28077" CreationDate="2018-10-20T21:32:51.720" Score="6" Body="&lt;p&gt;Most stars are born in groups or clusters of anywhere between 100 and millions of stars. Most of these groups and clusters do not survive more than about 10 million years or so and the fraction that survive decreases further with time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two basic reasons. First, it seems that most young clusters and associations may be born in an unbound state. Second, those that are gravitationally bound can become unbound through the expulsion of gas or through tidal interactions with the Galaxy, spiral arms and giant molecular fields. A supernova is one event that &lt;em&gt;might&lt;/em&gt; be a culpable agent in expelling mass from a young cluster and consequently unbinding it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is therefore highly unlikely that a 4.5 billion year old star will be found in a cluster. There are very few examples of clusters at this age or older.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, the theory that the Sun was born from the debris of just one or even a few stars is totally false and easily disproved by looking at the isotopic composition of dust grains formed in other stars before the solar system was born. These can be found inside meteorites and demonstrate that the solar system contains debris from many, many stars; of both high and relatively low masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://astronomy.stackexchange.com/questions/16311/how-can-there-be-1-000-stellar-ancestors-before-our-sun&quot;&gt;How can there be 1,000 stellar ancestors before our Sun?&lt;/a&gt; and &lt;a href=&quot;https://astronomy.stackexchange.com/questions/22694/parent-stars-of-our-sun-where-are-its-remains&quot;&gt;Parent stars of our Sun - Where are its remains?&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: It seems that you are just asking why we can't see &lt;em&gt;siblings&lt;/em&gt; of the Sun, including this massive star that may have been involved in the formation of the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First - it isn't clear that the Sun was born in a cluster from material seeded by a supernova from another star &lt;em&gt;in that cluster&lt;/em&gt;. There is a timing problem. The massive star that exploded would have to form more than 10 million years before the Sun. Clusters we see today do not seem to have such an extended star formation history.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second - even if that weren't the case, a supernova very frequently gives rise to a &quot;kick&quot;. The velocity dispersion of pulsars (recent supernova remnants) is exceedingly large (100s of km/s). Many remnants would escape the Galaxy and there is no chance of predicting where they would be in billions of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thirdly - even if the first and second problems were not applicable, as I mentioned above, even bound clusters do disperse. Once free of any binding potential stars can undergo a process called &quot;disc heating&quot; - the action of encounters with spiral density waves and giant molecular clouds gradually increases their velocity dispersions (changing their kinetic energies and exerting torques that can change their angular momentum wrt the Galactic centre). This allows the solar siblings to go their separate ways. A Galactic orbit at the solar radius is more than 200 million years. A velocity difference of just 1 km/s (typical in a cluster even before it unbinds or is heated) could result in a separation by 200 pc in that time. The Sun has executed more than 20 such orbits.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-10-21T08:46:37.330" LastActivityDate="2018-10-21T08:46:37.330" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28094" PostTypeId="2" ParentId="28063" CreationDate="2018-10-20T23:14:17.010" Score="5" Body="&lt;p&gt;Magnification is calculated as scope focal length / eyepiece focal length (so shorter focal length eyepieces give you a more magnified result).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the more you magnify, the more the captured light is spread out, so the image gets darker the more you magnify it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also - and more importantly - you eventually get to a point where magnifying the image more doesn't show you any more detail, it just gets bigger, dimmer, and fuzzier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretical maximum magnification is detemined by aperture - the size of the objective lens (for refractors) or primary mirror (for reflectors). The bigger the scope, the higher the maximum useful magnification. For everyday conditions, the maximum useful magnification is around the scopes aperture in mm (and you get it with an eyepiece with focal around the focal ratio of the scope - so around 6mm for an f/6 scope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given good seeing conditions (steady atmosphere) you can go to around double that magnification, which is closer to what what manufacturers usually quote. What they fail to say is that seeing conditions good enough to allow it can be rare. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, other factors also come into play. Inexpensive refractors are usually what's known as &quot;achromatic&quot; - they don't focus all colours of light at the same point, which causes what's called &quot;chromatic aberration&quot;. This usually shows up as a violet halo around bright objects, which makes the high magnification performance suffer (slower long tube refractors - with larger f-ratio numbers are less affected than fast, short tube refractors). By contrast, reflectors reflect all colours of light the same, so don't suffer from this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get more expensive refractors - ED or Apochromatic (&quot;Apo&quot;) designs that have much better colour correction than achromats, but the cost goes up significantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, don't get too hung up on magnification - it's handy for looking at planets or the moon, but many deep sky objects actually look better at lower magnification; a bigger scope helps here because it captures more light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reflectors have  the advantage when you're looking for a large light bucket; you can get reflectors with much bigger apertures (8 to 16&quot; or larger) than you can with practical amateur refractors (around 6&quot; or so).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, refractors don't have a central obstruction, and tend to give higher contrast views than an equivalent sized reflector - but reflectors don't suffer from chromatic aberration (but fast reflectors suffer from coma, which refractors don't).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that newtonian reflectors aren't really suited to non-astronomical uses, while refractors usually do better for terrestrial use (depending on the diagonal, you may end up with a left/right reversed view, but at least it won't be upside down).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With entry level scopes, you'll probably find that a dobsonian style mount (as used by many reflectors) is more stable than the tripod for a refractor; entry level scopes are built down to a price point, and for scopes with a tripod, that usually means that the tripod is fairly flimsy and vibration prone - so it may take a little while to settle down after adjusting something (either pointing or focus).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the two scopes you mention, the refractor is likely to do better at low power wide field views than at high magnifications. With no chromatic aberration, the reflector is likely to do a better job on planets. &lt;/p&gt;&#xA;" OwnerUserId="17016" LastActivityDate="2018-10-20T23:14:17.010" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28096" PostTypeId="1" CreationDate="2018-10-21T01:08:54.153" Score="2" ViewCount="6015" Body="&lt;p&gt;I was out with a couple friends a night and we looked up to the sky just watching the stars. I noticed what looked like a star moving then told my friends and they saw it too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After we saw the one moving star we noticed 2 more, it seriously looked like 3 stars just decided to have a meeting. It was the weirdest thing I've ever seen&lt;/p&gt;&#xA;" OwnerUserId="24607" LastActivityDate="2018-10-21T02:26:17.887" Title="Moving lights in the sky looks like they were moving coordinated" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2018-10-22T09:14:11.590" ContentLicense="CC BY-SA 4.0" />
  <row Id="28097" PostTypeId="2" ParentId="28096" CreationDate="2018-10-21T01:19:44.653" Score="3" Body="&lt;p&gt;These are almost certainly satellites. Keep in mind that satellites look like slowly moving stars; well, they move more slowly than meteors do at least! If they were moving somewhat slowly and constantly, and if they had no additional flashing lights and no bright colors, they were definitely satellites. NOSS satellites used to be known for flying in groups of three, but it looks like none of the generations of those triplets are still flying in formation. It looks like the Chinese Yaogan satellites are the only ones flying in triplet formation these days; you probably saw one of them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.satobs.org/noss.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="23940" LastEditorUserId="23940" LastEditDate="2018-10-21T02:26:17.887" LastActivityDate="2018-10-21T02:26:17.887" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28098" PostTypeId="1" AcceptedAnswerId="28110" CreationDate="2018-10-21T04:00:37.420" Score="0" ViewCount="75" Body="&lt;p&gt;I was looking at a very bright moon tonight through light cloud cover and saw a halo of light far bigger than the moon imposed on the clouds. Almost in a conic shape by comparison. I was wondering if early astronomers could use this difference to calculate the distance of the moon given the height of the cloud and the dimensions of earth.&lt;/p&gt;&#xA;" OwnerUserId="23729" LastActivityDate="2018-10-22T09:21:53.507" Title="How could original astronomers have used the clouds to find distances?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2018-10-22T12:37:03.443" ContentLicense="CC BY-SA 4.0" />
  <row Id="28099" PostTypeId="2" ParentId="28090" CreationDate="2018-10-21T09:12:41.980" Score="-1" Body="&lt;p&gt;The reason Saros Cycles begin and end is down to the positions of the orbits of the Sun, the Moon and the Earth. From the &lt;a href=&quot;https://en.wikipedia.org/wiki/Saros_(astronomy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;main Wikipedia page on Saros&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;p&gt;For an eclipse you need the 3 bodies to line up at a full or new moon,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;and repeat occurrences of these lunar phases result from solar and lunar orbits producing the Moon's synodic period of 29.53059 days...however for the majority of these, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Saros_(astronomy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;shadow of the Earth or Moon fall to the north or south of the other body&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It can only happen when a full or new moon is around the ecliptic.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The period of time for two successive lunar passes through the ecliptic plane (returning to the same node) is termed the draconic month, a 27.21222 day period. The three-dimensional geometry of an eclipse, when the new or full moon is near one of the nodes, occurs every 5 or 6 months when the Sun is in conjunction or opposition to the Moon and coincidentally also near a node of the Moon's orbit at that time, or twice per eclipse year.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Also:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Two Saros eclipses have the same appearance and duration due to the distance between the Earth and Moon being nearly the same for each event because the Saros eclipse cycle is also an integer of the anomalistic month, the period of the eccentricity of lunar orbit, 27.5545 days.&lt;/p&gt;&#xA;&lt;p&gt;After one saros, the Moon will have completed roughly an integer number of synodic, draconic, and anomalistic periods (223, 242, and 239) and the Earth-Sun-Moon geometry will be nearly identical&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The reason we get varying start and ends is because of this. Those roughly integer periods vary.&lt;/p&gt;&#xA;&lt;p&gt;This diagram from that Wikipedia page illustrates perfectly why they can be different lengths, stop and start at changing intervals etc.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/M6EhF.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/M6EhF.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-10-22T15:03:07.253" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="28100" PostTypeId="1" AcceptedAnswerId="28102" CreationDate="2018-10-21T10:39:29.607" Score="8" ViewCount="702" Body="&lt;p&gt;I mean that the angle between these angular momenta is acute as opposed to obtuse (23.5 and not 180-23.5), so that the rotations are &quot;roughly in the same direction&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related but distinct question -- when was this actually discovered? &lt;/p&gt;&#xA;" OwnerUserId="24608" LastEditorUserId="24608" LastEditDate="2018-10-21T10:45:47.650" LastActivityDate="2018-10-21T12:52:31.813" Title="Can one detect from earth that angular momenta of its axial rotation and its orbital rotation are &quot;almost parallel&quot;?" Tags="&lt;the-sun&gt;&lt;orbit&gt;&lt;earth&gt;&lt;rotation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28101" PostTypeId="1" AcceptedAnswerId="28103" CreationDate="2018-10-21T12:24:13.647" Score="3" ViewCount="163" Body="&lt;p&gt;Recently on the discovery channel I heard that a black hole burps matter out of it when it consumes more than it could process at a time. Given that even light cannot escape from a black hole, how does this &quot;burp&quot; matter escape?&lt;/p&gt;&#xA;" OwnerUserId="24609" LastActivityDate="2018-10-22T07:12:42.230" Title="How does a blackhole burp?" Tags="&lt;star&gt;&lt;black-hole&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28102" PostTypeId="2" ParentId="28100" CreationDate="2018-10-21T12:28:37.997" Score="13" Body="&lt;p&gt;Certainly! Ancient astronomers, eg in Babylon ~3000 years ago were sufficiently familiar with the geometry of the celestial equator and the ecliptic to be able to predict eclipses, and discover the eclipse series now known as the Saros series. But of course, in those days astronomers (mostly) didn't conceive of them in terms of the motion of the Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;OTOH, there were a few ancient astronomers in ancient Greece (notably Aristarchus of Samos), and even in Babylon (Seleucus of Seleucia), who did believe in a heliocentric system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The celestial pole is easy to discover, (especially if you're in the northern hemisphere), and it doesn't take a lot of observation to notice that the ecliptic differs from the celestial equator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although it's not possible to observe directly that the Sun's path is the ecliptic, it was known that all of the other celestial wanderers stay fairly close to the ecliptic, and that eclipses only occur when the Moon is very close to the ecliptic. And you can get a pretty accurate idea of the Sun's path by observing the rising &amp;amp; setting of stars and planets that are near the Sun's path. So astronomers across the ancient world recognised that the ecliptic is the Sun's path. And by the 2nd century BC, knowledge of these movements was sufficiently detailed that Hipparchus of Nicea was able to estimate the precession of the equinoxes.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="16685" LastEditDate="2018-10-21T12:52:31.813" LastActivityDate="2018-10-21T12:52:31.813" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28103" PostTypeId="2" ParentId="28101" CreationDate="2018-10-21T14:05:30.677" Score="6" Body="&lt;p&gt;The matter that the black hole is ingesting begins with some initial conditions outside of the black hole's event horizon, and due to numerous processes this surrounding matter loses energy and angular momentum and so falls into the black hole. The physics of this infall process can be quite complicated - especially if an accretion disk forms - but essentially, viscous friction between the particles that compose the infalling matter will cause it to lose momentum, among other possible processes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a theoretical limit of how much matter a black hole (or any astrophysical object) may accrete: the &lt;a href=&quot;https://en.wikipedia.org/wiki/Eddington_luminosity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eddington&lt;/a&gt; luminosity limit. Basically, as the black hole swallows the matter it illuminates electromagnetic radiation due to the intense collisions the infalling matter is experiencing near the event horizon. Eddington found that once the luminosity of this emitted radiation reaches a certain limit, the accretor (in this case a black hole) will actually begin to expel the surrounding matter rather than continue swallowing it, due the fact that once the limit is reached the surrounding matter will be blown away by the radiation. See &lt;a href=&quot;https://ned.ipac.caltech.edu/level5/ESSAYS/Blandford/blandford.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://arstechnica.com/science/2017/02/a-black-hole-has-been-devouring-a-star-for-a-decade/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for deeper explanations of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the surrounding matter that is expelled (once the Eddington limit has been reached) did not ever reach the event horizon - it's expelled by radiation before it makes it that close to the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And &lt;a href=&quot;https://www.nasa.gov/mission_pages/chandra/images/researchers-catch-supermassive-black-hole-burping-twice.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;'s a lovely image of a double burp! :)&lt;/p&gt;&#xA;" OwnerUserId="23762" LastEditorUserId="23762" LastEditDate="2018-10-21T23:57:47.063" LastActivityDate="2018-10-21T23:57:47.063" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28104" PostTypeId="2" ParentId="12122" CreationDate="2018-10-21T16:05:43.540" Score="3" Body="&lt;p&gt;Projected separation is the minimum physical separation of two astronomical objects, as determined from their angular separation and estimated distance.&lt;a href=&quot;https://en.wikipedia.org/wiki/File:Stellarparallax_parsec1.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Parallax shift of a star at a distance of one parsec as seen from the Earth. Not to scale&lt;/a&gt; For planets and double stars, this distance is usually given in Astronomical Units. The actual separation of the two objects depends on the angle of the line between the two objects to the line-of-sight of the observer.&lt;/p&gt;&#xA;" OwnerUserId="24610" LastEditorUserId="24610" LastEditDate="2019-07-07T16:17:00.507" LastActivityDate="2019-07-07T16:17:00.507" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28105" PostTypeId="1" AcceptedAnswerId="28114" CreationDate="2018-10-21T17:20:26.257" Score="0" ViewCount="206" Body="&lt;p&gt;Time for the sun rays to reach earth is approximately 8 mins. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If time of the sun rise is 7:30 - which time is this? Time when sun rays reaches earth or time when earth and sun comes into correct position where sun is visible again after night?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is time when sun rays reaching earth - that means sun is already 8 mins away from that first position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is second option - it means it's still 8 mins to wait until we actually see 'sun rising' - sun rays reaching earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So which time sun rise really is?&lt;/p&gt;&#xA;" OwnerUserId="24612" LastActivityDate="2018-10-22T18:04:17.020" Title="Sunrise - from earth - sun position or when sun rays reaches earth?" Tags="&lt;the-sun&gt;&lt;sun-rays&gt;&lt;sunrise&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28107" PostTypeId="1" CreationDate="2018-10-22T06:33:30.377" Score="0" ViewCount="86" Body="&lt;p&gt;Let there be a planet with Earth-like life. However, no trash or monuments or other visible structures, just organisms. How far can that planet be from Earth before our equipment just cannot observe the life anymore? Furthermore, let's say that our equipment is perfected and limited only by natural laws: how far can the planet be now?&lt;/p&gt;&#xA;" OwnerUserId="24617" LastActivityDate="2018-10-22T07:43:00.147" Title="How far can a planet with living beings get before we cannot see the living beings?" Tags="&lt;observational-astronomy&gt;&lt;life&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2018-10-22T15:53:43.913" ContentLicense="CC BY-SA 4.0" />
  <row Id="28108" PostTypeId="2" ParentId="28101" CreationDate="2018-10-22T07:06:31.000" Score="0" Body="&lt;p&gt;Burp is a misleading word because the explosion energy of compressed hydrogen disk causes the burp energy. The black hole force/energy only draws inwards. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black hole compression of matter is most powerful energy source in the visible universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accretion disks and quasars happen especially in the early universe when free hydrogen was burped into projectile vomits going travelling nearly at light speed higher than the width of our galaxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's like 1 supernova happening every minute. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Artist impressions and transverse sections of the &quot;burp&quot;.&#xA;&lt;a href=&quot;https://www.google.fr/search?q=quasar+accretion+disk&amp;amp;num=100&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ved=0ahUKEwiF0tnFvpneAhVnyoUKHYkQCMwQ_AUIDigB&amp;amp;biw=1257&amp;amp;bih=646&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.google.fr/search?q=quasar+accretion+disk&amp;amp;num=100&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ved=0ahUKEwiF0tnFvpneAhVnyoUKHYkQCMwQ_AUIDigB&amp;amp;biw=1257&amp;amp;bih=646&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2018-10-22T07:12:42.230" LastActivityDate="2018-10-22T07:12:42.230" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28109" PostTypeId="2" ParentId="28107" CreationDate="2018-10-22T07:43:00.147" Score="2" Body="&lt;p&gt;The simple answer to this would be anywhere that is physically unreachable. If there happened to be Earth-like life out there in our solar system, we would likely be able to detect it, then land a probe and confirm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to planets outside our solar system, then it becomes difficult. If there was no trash orbiting (and thus no way to notice slight dips in light etc) then we would likely be limited to studying the atmosphere using the transit spectroscopy method (more on this &lt;a href=&quot;http://spiff.rit.edu/classes/extrasol/lectures/spectra/spectra.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;HERE&lt;/a&gt;) where we can detect the gasses and chemical makeup of an exoplanets atmosphere. There may possibly be other things to look for in xray or infrared spectrums, such as localised hot-spots if there is intelligent life, although depending on distance and size of the planet, this could be difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, no matter what we detected, if it was to far away for us to physically be there, then we would only be theorising. We could say 'This planet is a great candidate for life' and things along those lines and present papers on findings. We would, however never be able to confirm it. &lt;/p&gt;&#xA;" OwnerUserId="17096" LastActivityDate="2018-10-22T07:43:00.147" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28110" PostTypeId="2" ParentId="28098" CreationDate="2018-10-22T09:21:53.507" Score="3" Body="&lt;p&gt;No, because the size of the halo is determined by the properties of ice crystals, the distance to the Moon is irrelevant. So you get a &lt;a href=&quot;https://en.m.wikipedia.org/wiki/22%C2%B0_halo&quot; rel=&quot;nofollow noreferrer&quot;&gt;22° halo&lt;/a&gt;&#xA; with both the Sun &amp;amp; the Moon, even though the Sun's much further away.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2018-10-22T09:21:53.507" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28112" PostTypeId="1" CreationDate="2018-10-22T16:00:29.573" Score="1" ViewCount="48" Body="&lt;p&gt;My family and I saw small dots moving rapidly from north to south that flew in a V formation (the kind that ducks fly in). There aren't any ducks in my country. This phenomenon appeared twice. They're not stars or satellites. Could they be any type of UFO?&lt;/p&gt;&#xA;" OwnerUserId="24622" LastEditorDisplayName="user1569" LastEditDate="2018-10-23T06:22:39.693" LastActivityDate="2018-10-23T06:22:39.693" Title="My family saw small dots in the sky moving rapidly at night" Tags="&lt;observational-astronomy&gt;" AnswerCount="0" CommentCount="6" ClosedDate="2018-10-23T02:10:21.147" ContentLicense="CC BY-SA 4.0" />
  <row Id="28113" PostTypeId="2" ParentId="28079" CreationDate="2018-10-22T16:10:17.943" Score="2" Body="&lt;p&gt;Since you have not accepted Rob's answer let's try a simpler version of it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short answer is Conservation of Energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe below is Newtonian.  So, not quite exact but very close except in extreme cases.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, a simple thought experiment.  Go to the Moon or some other quite large and airless body.  Take a rock and measure its mass e.g. 1kg.  Throw it directly upwards.  Measure the height at which you let go and its velocity at that time.  We will use the height of release as the baseline for potential energy so the rock starts with zero potential energy.  You can calculate the initial kinetic energy from the mass and release velocity.  As the rock ascends, it will gain potential energy from its increasing height but lose kinetic energy as it slows.  Since the body is airless, no energy is lost due to friction etc.  The total of potential and kinetic energy will be constant.  Since I said that the body that you are on is large, I will assume that the rock is released below escape velocity.  So, eventually it will stop and begin to fall.  It will begin to lose potential energy and gain kinetic energy.  Since no energy is lost to friction, when it gets back down to the release height, its potential energy is will be zero again and its kinetic energy will be as when it started.  So, it will come down as fast as it went up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now head out into deep space to a point where the Sun's gravity is negligible.  Clear away the Earth, the other planets, asteroids and other junk.  So, there is just you, a handy small rock, and the Sun.  Throw the rock towards the Sun but not quite on a a collision course.  Use its initial distance from the Sun as the zero point for potential energy.  So, it starts with no potential energy and a small kinetic energy.  As it approaches the Sun, it will lose potential energy (so it goes negative, it can because the zero point is arbitrary) and gain kinetic energy.  If your aim was good then it will pass close to the Sun but not so close that it hits it or encounters any substantial drag from solar wind, etc.  It will curve around the Sun due to the Sun's gravity but eventually start heading away again.  As it heads away, it will start to gain potential energy (less negative) and lose kinetic energy.  Since no energy is being lost to drag, it will eventually reach its initial distance from the Sun (where the Sun's gravity is negligible) but be heading away.  Its kinetic energy will be the same as you gave it at the beginning of the experiment so it will escape.  &lt;/p&gt;&#xA;" OwnerUserId="21104" LastActivityDate="2018-10-22T16:10:17.943" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28114" PostTypeId="2" ParentId="28105" CreationDate="2018-10-22T18:04:17.020" Score="3" Body="&lt;p&gt;It's when we see the Sun's rays and includes refraction. It also is the moment we see the first glimpse of the sun. Think of it this way: we humans need to know sunrise and sunset for our human activities; the fact that the sun is about eight light-minutes away doesn't concern us, doesn't affect the things we do. BTW, this is also true of solstices and equinoxes.&lt;/p&gt;&#xA;" OwnerUserId="7108" LastActivityDate="2018-10-22T18:04:17.020" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28115" PostTypeId="1" CreationDate="2018-10-23T02:47:04.143" Score="1" ViewCount="111" Body="&lt;p&gt;I've seen most flat earthers mention if earth was a sphere the earth should curve at x value per unit length. Is that even a proper logic? Radius of curvature is constant for sphere and a circle of a particular radius. If that logic exist can anyone share a simple diagram? I just can't comprehend how they got to that logic. Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="24392" LastActivityDate="2018-10-23T02:47:04.143" Title="Just out of curiosity. What's this curvature per unit length for a sphere? I see a lot of flat earthers using it." Tags="&lt;earth&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28118" PostTypeId="1" CreationDate="2018-10-23T22:22:56.257" Score="2" ViewCount="728" Body="&lt;p&gt;Where are all the nebulae located?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, a simple question that should be obvious but apparently is not to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more explicit, how far are all these objects? Are they mostly in our galaxy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not suggesting that the Milky Way galaxy is the only galaxy that has nebulae, but that the ones we have pictures of are only within the Milky Way...if that is indeed the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may also apply to stars themselves. What is the radius in ly of discernibility of (most) stars? Is it beyond the distance to the center of our galaxy (where even in principle individual stars might be discernible but there's too much of a mess to differentiate)?&lt;/p&gt;&#xA;" OwnerUserId="673" LastEditorUserId="673" LastEditDate="2018-10-24T13:50:02.980" LastActivityDate="2018-10-24T21:07:20.953" Title="Are the viewable nebulae only in our galaxy?" Tags="&lt;distances&gt;&lt;milky-way&gt;&lt;nebula&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28119" PostTypeId="1" CreationDate="2018-10-23T23:49:15.200" Score="2" ViewCount="26" Body="&lt;p&gt;A very severe corolnal mass ejection in the lines of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_storm_of_1859&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carrington event&lt;/a&gt; could potentially disrupt power grids and communications,  but would it affect the earth evenly or would the part of the earth facing away at the time be less affected? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question &lt;a href=&quot;https://astronomy.stackexchange.com/questions/87/what-are-the-main-differences-between-solar-flares-and-coronal-mass-ejections&quot;&gt;What are the main differences between solar flares and coronal mass ejections?&lt;/a&gt; seems to explain a coronal mass ejection but it does not answer the question about difference&lt;/p&gt;&#xA;" OwnerUserId="969" LastActivityDate="2018-10-23T23:49:15.200" Title="Would a large coronal mass ejection affect the earth evenley" Tags="&lt;solar-wind&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28120" PostTypeId="2" ParentId="28118" CreationDate="2018-10-24T00:02:54.197" Score="3" Body="&lt;p&gt;1) The nebulae are spread out over the universe in many different galaxies. Nebulae called extra-galactic nebulae are nebulae outside of our galaxy. They aren't common, but we can see them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Most of the ones we can see, such as the Helix nebula or the Orion nebula, are in our galaxy, but the ones outside of our galaxy are just more harder to see - there is a same dispersion of nebulae in our galaxy than in other galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Credit to @AtmosphericPrisonEscape (thanks!) - many nebulae outside of our galaxy are too dim for us to see. I assume that since the Tarantula Nebula is so massive and prominent, it is able to be visible from such a far distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, NGC 2070, or the Tarantula nebula, located in the LMC. (Link: &lt;a href=&quot;https://www.wikiwand.com/en/Tarantula_Nebula&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt;). The LMC is rather close to us, and the nebula is around 160,000 light years away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really can't find other major nebulae that are outside of our galaxy, but there are a lot of research papers on them, so they are visible - just not as prominent as the ones in our galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some research papers with extra galactic nebulae include: &lt;a href=&quot;http://adsabs.harvard.edu/abs/1926ApJ....64..321H&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt;, &lt;a href=&quot;https://ned.ipac.caltech.edu/level5/March17/Zwicky/frames.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt;, &lt;a href=&quot;http://www.pnas.org/content/15/3/168&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt;, and there are many more online.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand the 2nd part of your question with the coordinates, but I hope this at least helps!&lt;/p&gt;&#xA;" OwnerUserId="24048" LastEditorUserId="24048" LastEditDate="2018-10-24T21:07:20.953" LastActivityDate="2018-10-24T21:07:20.953" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28122" PostTypeId="2" ParentId="28118" CreationDate="2018-10-24T01:51:51.833" Score="0" Body="&lt;p&gt;Firstly you have to specify what you mean by nebulae.&lt;br&gt;&#xA;There is a historical term that some of @MystaryPi's links refer to, &quot;extragalactic nebulae&quot;. An extragalactic nebula only existed until around the 1930s, as it was then realised, that extragalactic nebulae are not nebulae but galaxies like our  milky way.&lt;br&gt;&#xA;The classical papers by Hubble and Zwicky that were linked in the other answer were part of establishing this fact. This is why until today, misleadingly, galaxies are sometimes referred to as nebulae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A planetary nebula (which is an ejected shell of a late-type star) wouldn't be visible in another galaxy neither with the eye nor large telescopes and long-time exposure. They simply posses a much too low surface brightness.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditDate="2018-10-24T02:00:11.523" LastActivityDate="2018-10-24T02:00:11.523" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28123" PostTypeId="1" AcceptedAnswerId="28132" CreationDate="2018-10-24T01:55:51.673" Score="5" ViewCount="464" Body="&lt;p&gt;Different sources online say that the Chandrasekhar Limit is either 1.40 or 1.39, or 1.44 solar masses. Why the discrepancy? I heard it might have to do with the composition of the white dwarf, but, correct me if I'm wrong, don't pretty much all white dwarf stars have the same composition?&lt;/p&gt;&#xA;" OwnerUserId="24633" LastActivityDate="2018-10-24T16:44:36.443" Title="Which Chandrasekhar Limit do I use? 1.39 or 1.44?" Tags="&lt;white-dwarf&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28125" PostTypeId="1" CreationDate="2018-10-24T02:52:43.500" Score="0" ViewCount="110" Body="&lt;p&gt;Sirius B has a distance to Sirius A of 11 arc seconds at apastron; are there any good images of it at that point in its orbit?&lt;/p&gt;&#xA;" OwnerUserId="24634" LastEditorUserId="7472" LastEditDate="2018-10-24T17:09:42.680" LastActivityDate="2018-10-24T20:12:10.227" Title="Are there any good images of Sirius B at apastron in the Sirius system?" Tags="&lt;star&gt;&lt;binary-star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28126" PostTypeId="1" CreationDate="2018-10-24T03:13:56.400" Score="4" ViewCount="98" Body="&lt;p&gt;In &lt;a href=&quot;https://en.wikipedia.org/wiki/Numerical_model_of_the_Solar_System&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Wikipedia article on numerical integration of planetary motion&lt;/a&gt; it states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...the flattening of the Earth causes precession, which causes the&#xA;  axial tilt to change, which affects the long-term movements of all&#xA;  planets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why would precession affect the motion of the other planets?&lt;/p&gt;&#xA;" OwnerUserId="2800" LastActivityDate="2018-10-24T05:50:01.763" Title="Why does the precession of the earth affect other planets?" Tags="&lt;celestial-mechanics&gt;&lt;precession&gt;&lt;modeling&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28127" PostTypeId="1" CreationDate="2018-10-24T03:32:44.663" Score="-2" ViewCount="156" Body="&lt;p&gt;It is common to base models on deduction rather than empirical observation. The orbit of Sirius A is easily seen as variation in proper motion. The orbit of Sirius B, has it been observed, or is it derived from the orbit of Sirius A (which has empirical data underlying it. )&lt;/p&gt;&#xA;" OwnerUserId="24634" LastActivityDate="2018-10-24T06:23:38.413" Title="Is the orbit of Sirius B based on observation, or derived from the variation in proper motion of Sirius A?" Tags="&lt;star&gt;&lt;data-analysis&gt;&lt;orbital-mechanics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28128" PostTypeId="2" ParentId="28126" CreationDate="2018-10-24T05:50:01.763" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why would precession affect the motion of the other planets?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First things first: That's an unreferenced portion of a wikipedia article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pcWZp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pcWZp.png&quot; alt=&quot;Citation needed, xkcd 285.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, a perfectly spherical body acts exactly like a point mass in Newtonian mechanics. A non-spherical body does not. The Earth's equatorial bulge has a significant effect on satellites in low Earth orbit. Sun synchronous orbits would be impossible without that bulge. For example, a sun synchronous orbit about Venus with its very slow rotation rate and hence very small equatorial bulge is not possible (&lt;a href=&quot;https://space.stackexchange.com/questions/5312/why-would-sun-synchronous-orbits-be-impossible-around-perfectly-spherical-bodies&quot;&gt;reference&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth's equatorial bulge adds a quadrupole (inverse quartic, or &lt;span class=&quot;math-container&quot;&gt;$1/r^4$&lt;/span&gt;) term to the inverse square law relation for point masses, and this inverse quartic term depends not only on distance but also on latitude. The added term is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$3 J_2 \frac {GM_\oplus a^2}{r^4}\left(\frac 3 2 \cos^2 \lambda - 1\right)$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$J_2$&lt;/span&gt; is the Earth's second dynamic form and &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; is the geocentric latitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this term drops off as &lt;span class=&quot;math-container&quot;&gt;$1/r^4$&lt;/span&gt;, it becomes very small beyond the Moon's orbit. But it still exists beyond the Moon's orbit. However, I've yet to see a paper that does account for (or says we need to account for) Earth's &lt;span class=&quot;math-container&quot;&gt;$J_2$&lt;/span&gt; effect on the other planets.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2018-10-24T05:50:01.763" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28129" PostTypeId="2" ParentId="28127" CreationDate="2018-10-24T06:23:38.413" Score="1" Body="&lt;p&gt;Observation. There are many pictures on the internet showing both stars, but the pair have been &quot;visually&quot; observed for decades. For example, see below from &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/1538-4357/aa6af8/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bond et al. (2017)&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the accuracy of visual observers was poor, especially near minimum separation where the separation was over-estimated, but HST observations have nailed the orbit, allowing a precise mass determination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if only one star is observed then precise mass estimates cannot be obtained (think about Kepler's third law).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image also answers your other question(s) about Sirius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Y3C2z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y3C2z.jpg&quot; alt=&quot;The orbit of Sirius AB&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-10-24T06:23:38.413" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28130" PostTypeId="2" ParentId="28125" CreationDate="2018-10-24T06:40:36.053" Score="2" Body="&lt;p&gt;HST took pictures close to periastron in 1997, when the separation of the stars was about 3.7 arcsecs. See &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/1538-4357/aa6af8/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bond et al. (2017)&lt;/a&gt;. There are no pictures exactly at periastron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HST took pictures close to apastron in August 2016, when the separation was 10.7 arcsecs (see the same reference). There are no pictures from HST closer to apastron but some may be taken over the next year or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a raw HST image from August 2016, taken from the &lt;a href=&quot;https://hla.stsci.edu/hlaview.html#Inventory%7CfilterText%3D%24filterTypes%3D%7Cquery_string=Sirius%20&amp;amp;posfilename=&amp;amp;poslocalname=&amp;amp;posfilecount=&amp;amp;listdelimiter=whitespace&amp;amp;listformat=degrees&amp;amp;RA=101.287155&amp;amp;Dec=-16.716116&amp;amp;Radius=0.200000&amp;amp;inst-control=all&amp;amp;inst=ACS&amp;amp;inst=ACSGrism&amp;amp;inst=WFC3&amp;amp;inst=WFPC2&amp;amp;inst=NICMOS&amp;amp;inst=NICGRISM&amp;amp;inst=COS&amp;amp;inst=WFPC2-PC&amp;amp;inst=STIS&amp;amp;inst=FOS&amp;amp;inst=GHRS&amp;amp;imagetype=best&amp;amp;prop_id=&amp;amp;spectral_elt=&amp;amp;proprietary=both&amp;amp;preview=1&amp;amp;output_size=256&amp;amp;cutout_size=12.8%7Cra=&amp;amp;dec=&amp;amp;sr=&amp;amp;level=&amp;amp;image=&amp;amp;inst=ACS%2CACSGrism%2CWFC3%2CWFPC2%2CNICMOS%2CNICGRISM%2CCOS%2CWFPC2-PC%2CSTIS%2CFOS%2CGHRS&amp;amp;ds=&quot; rel=&quot;nofollow noreferrer&quot;&gt;HST archive&lt;/a&gt;  (freely available and easy to find).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/j9AOo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j9AOo.jpg&quot; alt=&quot;HST image of Sirius AB&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many photographs of the binary taken at around 11 arcseconds separation. Table~3 of the same reference above lists a stack of observations taken between 1970 and 1975 when the separation was around 11 arcseconds. One of these images is shown below; Sirius B is the faint object below and to the right of Sirius A. The other symmetrically spaced blobs are image artefacts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IvKmq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IvKmq.png&quot; alt=&quot;Sirius AB&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-10-24T20:12:10.227" LastActivityDate="2018-10-24T20:12:10.227" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28131" PostTypeId="2" ParentId="28123" CreationDate="2018-10-24T07:48:50.463" Score="4" Body="&lt;p&gt;A white dwarf is a stellar remnant composed almost entirely of electron degenerate matter. Chandrasekhar determined that there is a &lt;em&gt;mass limit&lt;/em&gt; at which electron degeneracy pressure is insufficient to prevent gravitational collapse, and the equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M_{\mathrm{limit}} = \frac{\omega_3^0 \sqrt{3\pi}}{2} \left(\frac{\hbar c}{G}\right)^{3/2} \frac{1}{(\mu_e m_\mathrm{H})^2},$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\mu_e$&lt;/span&gt; is the average molecular weight per electron (depending on stellar composition), &lt;span class=&quot;math-container&quot;&gt;$m_\mathrm{H}$&lt;/span&gt; is the mass of hydrogen, and &lt;span class=&quot;math-container&quot;&gt;$\omega_3^0$&lt;/span&gt; is a numerical constant (approx. 2.018236).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Chandrasekhar limit &lt;span class=&quot;math-container&quot;&gt;$M_{CH}$&lt;/span&gt; is therefore the maximum mass of a stable white dwarf. For a typical white dwarf composed mostly of carbon and oxygen, &lt;a href=&quot;https://en.wikipedia.org/wiki/Chandrasekhar_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; gives the limit as &quot;about 1.4 &lt;span class=&quot;math-container&quot;&gt;$M_{\odot}$&lt;/span&gt;&quot;, but as the question notes, other sources give quite a broad range of values. For example, &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0702351v1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mazzali, P. A.; Röpke, F. K.; Benetti, S.; Hillebrandt, W. (2007)&lt;/a&gt; use &lt;span class=&quot;math-container&quot;&gt;$M_{CH}$&lt;/span&gt; = 1.38 &lt;span class=&quot;math-container&quot;&gt;$M_{\odot}$&lt;/span&gt;; and in &lt;a href=&quot;https://astronomy.stackexchange.com/a/10880&quot;&gt;this answer&lt;/a&gt;, Rob Jeffries gives &lt;span class=&quot;math-container&quot;&gt;$M_{CH}$&lt;/span&gt; = 1.45 &lt;span class=&quot;math-container&quot;&gt;$M_{\odot}$&lt;/span&gt;. Rob goes on to note that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In more recent years, the Chandrasekhar limit has evolved to&#xA;  colloquially mean the maximum possible mass for a white dwarf. The&#xA;  main corrections to the ideal case considered by Chandrasekhar are&#xA;  Coulomb corrections, the possible onset of inverse beta decay&#xA;  (electron capture) and the instability at a finite density predicted&#xA;  by using GR rather than Newtonian gravity. The latter probably sets&#xA;  the &quot;Chandrasekhar mass&quot; for carbon white dwarfs to be 1.38&#xA;  &lt;span class=&quot;math-container&quot;&gt;$M_{\odot}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A critical factor is &lt;span class=&quot;math-container&quot;&gt;$\mu_e$&lt;/span&gt;, which depends upon the chemical composition of the star. This is important, as not all white dwarfs are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a low to medium mass main-sequence star (like the Sun) has fused all the hydrogen to helium in its core, it evolves into a red giant and starts fusing the helium into carbon and oxygen. If the star isn't massive enough to reach the billion degrees required for carbon to start fusing, the C and O accumulate in the core, and when the giant eventually sheds its outer envelope as a planetary nebula, the remnant is a &quot;standard&quot; white-hot C-O white dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a star of 8-10.5 &lt;span class=&quot;math-container&quot;&gt;$M_{\odot}$&lt;/span&gt; is big enough to reach the core temperature required to fuse carbon. On the other hand, it won't be hot enough to fuse heavier nucleosynthesis products such as oxygen, neon and magnesium. The end result is not a C-O white dwarf but an O-Ne-Mg white dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add to this that accreting white dwarfs are adding hydrogen and/or helium to their outer envelope, and it's clear that the composition of a white dwarf can vary considerably, and this is no doubt one of the reasons for the range of &lt;span class=&quot;math-container&quot;&gt;$M_{CH}$&lt;/span&gt; values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To complicate things, a surprising number of physicists continue to assume that a type Ia supernova occurs when an accreting white dwarf (for example, with a red giant binary companion) &lt;em&gt;reaches the relevant Chandrasekhar limit for its chemical composition&lt;/em&gt;. However, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Type_Ia_supernova#Consensus_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;current view of astronomers&lt;/a&gt; is that &lt;em&gt;just before&lt;/em&gt; the white dwarf reaches &lt;span class=&quot;math-container&quot;&gt;$M_{CH}$&lt;/span&gt;, the increased pressure and density raise the core temperature to the point where carbon can fuse. Exactly what happens next is an area of active research, but the result is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Carbon_detonation&quot; rel=&quot;nofollow noreferrer&quot;&gt;carbon detonation&lt;/a&gt; and runaway thermonuclear reaction that spreads through the core within seconds, releasing a massive amount of energy that completely disrupts the white dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Rob's answer to &lt;a href=&quot;https://astronomy.stackexchange.com/a/12208&quot;&gt;How massive does a main sequence star need to be to&#xA;go type 1 supernova&lt;/a&gt;,&#xA;and&lt;/li&gt;&#xA;&lt;li&gt;both answers to &lt;a href=&quot;https://astronomy.stackexchange.com/q/14740/14094&quot;&gt;Why does the Chandrasekhar limit affect white dwarfs&#xA;differently?&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14094" LastActivityDate="2018-10-24T07:48:50.463" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28132" PostTypeId="2" ParentId="28123" CreationDate="2018-10-24T08:16:56.953" Score="5" Body="&lt;p&gt;The &quot;classic&quot; Chandrasekhar mass is given by&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ M_{\rm Ch} = 1.445 \left(\frac{\mu_e}{2}\right)^{-2}\ M_{\odot},$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\mu_e$&lt;/span&gt; is the number of mass units per electron (&lt;span class=&quot;math-container&quot;&gt;$\mu_e =2$&lt;/span&gt; for ionised carbon, oxygen or helium; &lt;span class=&quot;math-container&quot;&gt;$\mu_e = 1$&lt;/span&gt; for hydrogen, &lt;span class=&quot;math-container&quot;&gt;$\mu_e= 56/26$&lt;/span&gt; for iron (56)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This assumes a white dwarf star of uniform, ionised composition and an equation of state given by that for ideal fermions (electrons in this case). It also ignores rotation and it uses Newtonian gravity. When defined in this way, the Chandrasekhar mass is obviously composition dependent - but most white dwarfs are expected to be formed from something with &lt;span class=&quot;math-container&quot;&gt;$\mu_e =2$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, a white dwarf can never become this massive. There are (at least) 4 effects that can mean the de facto Chandrasekhar mass (if it is taken to mean the maximum mass of a stable white dwarf) and at least one effect that can increase it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effects that reduce it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Electrostatic interactions&lt;/em&gt; The electrons and ions do not form an ideal Fermi gas because of Coulomb interactions. The net result is to make a white dwarf slightly more compressible and the maximum mass about 2% lower. The correction &lt;em&gt;is&lt;/em&gt; composition dependent; it is stronger for white dwarfs made of material with a larger atomic number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;General Relativity&lt;/em&gt; Massive white dwarfs are strongly affected by General Relativity. White dwarfs will become unstable at a finite density when the hydrostatic equilibrium is treated with GR. This finite density is reached at about &lt;span class=&quot;math-container&quot;&gt;$1.38-1.39 M_{\odot}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Inverse beta decay&lt;/em&gt; At high densities the electron Fermi energy becomes high enough to initiate inverse beta decay reactions. Electrons are captured by protons in the nuclei and this renders the star unstable. For a Carbon white dwarfs this will also happen at about &lt;span class=&quot;math-container&quot;&gt;$1.39 M_{\odot}$&lt;/span&gt;, for oxygen the threshold is lower at &lt;span class=&quot;math-container&quot;&gt;$\sim 1.37M_{\odot}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Pyconuclear reactions&lt;/em&gt; At high densities, even at low temperatures, fusion reactions can be initiated by quantum tunnelling. This changes the composition of the white dwarf and can change &lt;span class=&quot;math-container&quot;&gt;$\mu_e$&lt;/span&gt; or lower the density threshold for inverse beta decay making the star unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect that can increase the maximum mass for stability is rotation. Some authors have claimed that the limit can be increased to as high as &lt;span class=&quot;math-container&quot;&gt;$2.6M_{\odot}$&lt;/span&gt; in certain circumstances, though these are usually referred to as &lt;em&gt;Super-Chandrasekhar mass&lt;/em&gt; (e.g. &lt;a href=&quot;https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.110.071102&quot; rel=&quot;nofollow noreferrer&quot;&gt;Das &amp;amp; Mukhopadhyay 2013&lt;/a&gt;) and are &lt;em&gt;not stable&lt;/em&gt;. A stability analysis of rotating white dwarfs in GR found that the maximum &lt;em&gt;stable&lt;/em&gt; mass was only increased to around &lt;span class=&quot;math-container&quot;&gt;$1.47 M_{\odot}$&lt;/span&gt; for a Carbon white dwarf (&lt;a href=&quot;https://arxiv.org/abs/1204.2070&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boshkayev et al. 2012&lt;/a&gt;), but these would be rotating faster than any observed white dwarfs.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-10-24T16:44:36.443" LastActivityDate="2018-10-24T16:44:36.443" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28135" PostTypeId="1" CreationDate="2018-10-24T17:45:31.580" Score="1" ViewCount="184" Body="&lt;p&gt;I have flux time series of quasars in the observed frame and want to convert it in the rest or quasar frame. Any one who can help? Thanks&lt;/p&gt;&#xA;" OwnerUserId="24642" LastActivityDate="2019-08-21T07:02:50.327" Title="How to convert quasar flux time series from observed frame to rest frame?" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28136" PostTypeId="1" AcceptedAnswerId="28137" CreationDate="2018-10-24T17:56:39.093" Score="3" ViewCount="113" Body="&lt;p&gt;Astronomical telescopes observe particular points in the sky for hours or even days, and they have to remain particularly steady for the duration of the observation, as far as I know. What happens when an earthquake shakes the ground? Does it ruin the observation? If so, how frequently does that happen?&lt;/p&gt;&#xA;" OwnerUserId="24588" LastActivityDate="2018-10-25T09:42:05.530" Title="How do ground-based telescopes deal with earthquakes?" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28137" PostTypeId="2" ParentId="28136" CreationDate="2018-10-24T19:09:11.140" Score="10" Body="&lt;p&gt;Earthquakes are not uncommon in Hawaii and Chile, where several observatories are built. I only have personal experience with the &lt;a href=&quot;https://en.wikipedia.org/wiki/La_Silla_Observatory&quot; rel=&quot;noreferrer&quot;&gt;observatory in La Silla&lt;/a&gt;, where a minor earthquake once occurred while I was observing. This was an old telescope, and the image was ruined. Modern telescopes may be able to withstand small seismic activities, but since &lt;em&gt;we&lt;/em&gt; felt the ground shaking, I think the image would've been ruined on &lt;em&gt;any&lt;/em&gt; telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luckily, images are not taken in one, long exposure. Rather, they're composed of multiple exposures that are added subsequently. You typically don't expose for more than 30 minutes. A faint object that requires, say, 5 hours of exposure, is then given by the &lt;em&gt;median&lt;/em&gt; of 10 half-hour exposures. Why the median and not the mean? Although earthquakes are, after all, rare, &lt;em&gt;cosmic rays&lt;/em&gt; quite often ruin a small region on the image, but when you take the median, such outliers are removed&lt;span class=&quot;math-container&quot;&gt;$^\dagger \!\!\!$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the whole night isn't ruined, only the one exposure that was taken when the earthquake occurred.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of greater concern are more violent earthquakes, which may in principle damage the telescope. Modern telescopes are prepared to deal with such seismic activity. For instance,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;…ESO's &lt;a href=&quot;https://en.wikipedia.org/wiki/Very_Large_Telescope&quot; rel=&quot;noreferrer&quot;&gt;VLT&lt;/a&gt; has a complex mechanical system that is automatically activated when&#xA;  it experiences a strong earthquake of magnitude 7 or higher. There are&#xA;  a series of clamps around the edges of the mirror, which can — at a&#xA;  moment's notice — lift the entire 23-tonne mirror off the actuators&#xA;  and secure it to the telescope’s support structure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In 2006, an earthquake close to Hawaii shifted slightly the position of various things on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Subaru_Telescope&quot; rel=&quot;noreferrer&quot;&gt;Subaru telescope&lt;/a&gt;, apparently resulting in people there working with pre- and post-2006 numbers, according to former telescope operator &lt;a href=&quot;https://www.quora.com/Were-the-observatories-on-Maunakea-and-Mauna-Loa-affected-by-the-magnitude-5-0-5-4-and-6-9-earthquakes-on-May-4-2018&quot; rel=&quot;noreferrer&quot;&gt;Dan Birchall&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Source: &lt;a href=&quot;https://www.eso.org/public/blog/protecting-elt-from-earthquakes/&quot; rel=&quot;noreferrer&quot;&gt;ESO&lt;/a&gt;, where you can also find more about how to prevent damage in case of earthquakes.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^\dagger$&lt;/span&gt;&lt;sub&gt;For instance, if the values in a given pixel of 10 images are &lt;span class=&quot;math-container&quot;&gt;$[21,25,22,19,22,27,23,65536,18,24]$&lt;/span&gt;, then obviously something went wrong in the 8th exposure, and the median &lt;span class=&quot;math-container&quot;&gt;$(22.5)$&lt;/span&gt; is more characteristic of the &quot;true&quot; value than the mean &lt;span class=&quot;math-container&quot;&gt;$(6573.7)$&lt;/span&gt;.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-10-25T09:42:05.530" LastActivityDate="2018-10-25T09:42:05.530" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28139" PostTypeId="1" CreationDate="2018-10-24T21:58:36.397" Score="0" ViewCount="167" Body="&lt;p&gt;Stars appear larger than they are, could that be the stellar corona here around Sirius, making it appear n-times larger than the stellar body? NASA recently mapped the solar corona to extending 12 solar radii above the surface &lt;sup&gt;&lt;a href=&quot;https://www.nasa.gov/content/goddard/nasas-stereo-maps-much-larger-solar-atmosphere-than-previously-observed/&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;&lt;/sup&gt;, the corona as a whole occupying 13x the suns diameter if viewed from a distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=EWOkkZnsCR4&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kQ8kT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/kQ8kT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Refences:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;a href=&quot;https://www.nasa.gov/content/goddard/nasas-stereo-maps-much-larger-solar-atmosphere-than-previously-observed/&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA's STEREO Maps Much Larger Solar Atmosphere Than Previously Observed - nasa.gov&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24634" LastActivityDate="2018-10-24T22:37:48.897" Title="In the closest stars, can you see the stellar corona?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="28140" PostTypeId="1" CreationDate="2018-10-24T22:00:45.087" Score="4" ViewCount="1505" Body="&lt;p&gt;The Sun currently has a stellar class of G2V. However, since its birth, the Sun has brightened significantly. What would the Sun's class have been when it was born ~4.6 gyr ago?&lt;/p&gt;&#xA;" OwnerUserId="24644" LastActivityDate="2018-10-24T23:20:37.953" Title="Sun's Stellar Class at Birth?" Tags="&lt;the-sun&gt;&lt;stellar-evolution&gt;&lt;classification&gt;" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28141" PostTypeId="1" AcceptedAnswerId="28146" CreationDate="2018-10-24T22:10:39.013" Score="4" ViewCount="520" Body="&lt;p&gt;I am trying to use Kepler's second law to find the duration of Venus's orbit. I am assuming circular orbits (using Earth and Venus, so low eccentricity). Here is my process:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the radius of Earth's orbit is 150 million km, then the swept area in one day is &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{365.25}\times\pi\times 150^2 \approx 194 \text{ million km}^2$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus must sweep the same area in the same time. Assuming a orbital radius of 108 million km for Venus, and using &lt;span class=&quot;math-container&quot;&gt;$A = \frac{\theta}{360}\pi r^2$&lt;/span&gt;, we can find the central angle for the swept sector, that is, the angle traveled in one Earth day:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$194 = \frac{\theta}{360}\pi \times108^2 \implies \theta = 1.90 ^{\circ}$&lt;/span&gt; per Earth day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the orbital period should be &lt;span class=&quot;math-container&quot;&gt;$\frac{360}{1.90}\approx 189$&lt;/span&gt; Earth days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the orbital period of Venus is &lt;span class=&quot;math-container&quot;&gt;$224.7$&lt;/span&gt; Earth days. The difference between 189 and 224.7 appears to be well beyond the error introduced by my assumption of circular orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this is perhaps a circuitous way of doing this calculation. My goal is to write a mathematics exercise that uses the area of sectors in a meaningful way.&lt;/p&gt;&#xA;" OwnerUserId="24645" LastActivityDate="2018-10-24T23:34:43.720" Title="What is wrong with my calculations of Venus' orbital period?" Tags="&lt;orbit&gt;&lt;solar-system&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28142" PostTypeId="2" ParentId="28140" CreationDate="2018-10-24T22:24:37.467" Score="1" Body="&lt;p&gt;Mass is one of the major factors that determine the spectral class of a star when it is born. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;During a star's protostar stages, a spectral class determines the length of its protostar stage and other factors such as its lifetime. Even though a spectral class is not physically assigned to them, the life stages that they undergo are essentially determined by their spectral class at birth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even though protostars don't have a spectral class determined by scientists, the star that it ends up becoming in its main sequence is really what its spectral class is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this case, the Sun in it's protostar stages still was on the G2V route - it was taking the same (or at least similar) life cycles as stars in the G2V spectral class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To the 2nd part:&#xA;Check out this website, especially from page 18: &lt;a href=&quot;http://www.astro.umass.edu/~myun/teaching/a100_old/Astro100Mar25.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.astro.umass.edu/~myun/teaching/a100_old/Astro100Mar25.pdf&lt;/a&gt;. There's a lot of equations on there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also more information here on the luminosity-mass relationship: &lt;a href=&quot;https://www.e-education.psu.edu/astro801/content/l7_p3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.e-education.psu.edu/astro801/content/l7_p3.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a side note, looks like someone asked about the luminosity+temperature and mass relationship on here as well: &lt;a href=&quot;https://physics.stackexchange.com/questions/12589/are-there-formulae-for-calculating-stellar-luminosity-and-effective-temperature&quot;&gt;https://physics.stackexchange.com/questions/12589/are-there-formulae-for-calculating-stellar-luminosity-and-effective-temperature&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I find more information, I'll comment it or edit it onto my answer. Hope that helps!&lt;/p&gt;&#xA;" OwnerUserId="24048" LastEditorUserId="24048" LastEditDate="2018-10-24T22:35:08.587" LastActivityDate="2018-10-24T22:35:08.587" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28143" PostTypeId="2" ParentId="28139" CreationDate="2018-10-24T22:30:26.693" Score="3" Body="&lt;p&gt;Stars other than the Sun do not have photospheres or coronae that can be directly imaged (the exception would be the red giant Betelgeuse that can just be resolved from the HST). They &quot;appear larger than they are&quot; as you put it because of the limitations of physical optics or turbulence in the atmosphere. There is a minimum-sized &quot;point spread function&quot; image that will be convolved with the intrinsic image of a star, whatever its size. Thus even a point-like source will appear to have a finite size when viewed with a telescope with a finite aperture. Images of very bright objects may also be affected by &quot;saturation&quot; of the detector - the counts in saturated pixels can &quot;bleed&quot; into surrounding pixels. All stars (bar Betelgeuse) are too small and far away to be anything but intrinsically &quot;point-like&quot;, even when observed with space telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image of Sirius in your question is the saturated image of a very bright, &lt;strong&gt;point-like&lt;/strong&gt; object blurred by the imperfect optics (and perhaps a turbulent atmosphere) on a telescope of finite aperture. Any &quot;corona&quot; you see on this picture is the extended wings of the telescope point spread function. The size of Sirius &lt;em&gt;cannot&lt;/em&gt; be directly inferred from this picture. (It is also the case that Sirius, an A-type star, does not have a corona like the Sun's, because it lacks the magnetic dynamo and strong magnetic fields that are responsible for heating the solar corona, and it lacks the very strong winds that shock heat coronae in much hotter stars).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, there are clever, interferometric techniques to synthesise a large telescope using extended baselines between multiple telescopes. These are capable of directly measuring the angular diameters of stars.  The assumption is that what is being measured is the angular diameter of the photosphere, since this is what dominates the light at optical and near-infrared wavelengths from most stars. In particular, the corona - a regions of tenuous, optically thin, ionised gas - emits hardly any visible/infrared light (it is of order &lt;strong&gt;a billion times less bright&lt;/strong&gt; than the solar disk and due mainly to Thomson scattering of photospheric light from free electrons and dust scattering) and can safely be neglected. That is why the optical/infrared corona of the Sun can only be seen during total solar eclipses on Earth or from spacecraft with occulting disks that can totally block out the dominant light from the photosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distant stars are too far away for this occultation technique to work. The presence of any coronae around them can only be inferred from the X-ray emission due to the hot coronal gas. In some cases it has been &lt;em&gt;indirectly&lt;/em&gt; argued that the coronal gas can extend several radii or more above the stellar surfaces (e.g. by analysing the fall and rise times of stellar X-ray flares or from the lack of X-ray eclipses in some eclipsing binary systems).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-10-24T22:37:48.897" LastActivityDate="2018-10-24T22:37:48.897" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28144" PostTypeId="2" ParentId="28140" CreationDate="2018-10-24T23:13:21.890" Score="8" Body="&lt;p&gt;Spectral type is determined by (and is an indicator of) photospheric temperature and surface gravity. The classification of G2V applies to a star with a photospheric temperature of about 5800K and a surface gravity of about 100-1000 m/s&lt;span class=&quot;math-container&quot;&gt;$^{2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun's temperature and gravity have not changed greatly since it first appeared on the &quot;zero age main sequence&quot; - it has got a little bigger and more luminous and a little bit hotter (by about 170 K- &lt;a href=&quot;https://academic.oup.com/mnras/article/386/1/155/977315&quot; rel=&quot;noreferrer&quot;&gt;Schroder &amp;amp; Smith 2008&lt;/a&gt;) - so maybe it was a G3V-G4V star at the zero age main sequence (&lt;a href=&quot;http://www.pas.rochester.edu/~emamajek/EEM_dwarf_UBVIJHK_colors_Teff.txt&quot; rel=&quot;noreferrer&quot;&gt;Mamajek 2018&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there were much bigger changes during the 20 million years or so before the Sun reached the ZAMS. It was cooler and briefly much larger. This would be reflected in a spectral classification that was something like K5IV- M0IV; indicating a subgiant-like gravity and a temperature of about 4000K. As it aged, it would then shrink and become hotter, passing through the G5V-K5V spectral types before reaching the main sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can synthesise these two diagrams in your head, you can see what I mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Pre main sequence evolution in the Hertzsprung-Russell diagram&lt;/em&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/cWQNt.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cWQNt.png&quot; alt=&quot;PMS evolution&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Spectral types on the Hertzsprung-Russell diagram&lt;/em&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/IZVtC.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IZVtC.jpg&quot; alt=&quot;Spectral type HR&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-10-24T23:13:21.890" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28145" PostTypeId="2" ParentId="28140" CreationDate="2018-10-24T23:20:37.953" Score="2" Body="&lt;p&gt;It is true that on the main sequence, where most stars spend most of their lifetimes, mass is the main determining factor of all stellar parameters, with metallicity and angular momentum distant second-most important factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However when a star is born, from the moment it becomes detectable, it moves through several stages of structural evolution, which surely do not classify as the stellar class that it later has on the main sequence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The evolutionary stages of this pre-main-sequence (PMS) star can be described in the astronomer's favourite diagram - the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hertzsprung%E2%80%93Russell_diagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;HR-diagram&lt;/a&gt; - via the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hayashi_track&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hayashi track&lt;/a&gt;.&#xA;The Hayashi track is for any star of given mass the sequence of colour-luminosity values that it attains during its PMS evolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As colour gives the stellar class, there is an evolution in stellar class for all stars, even on the main sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;since its birth, the Sun has brightened significantly.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is incorrect. Once the sun arrived at the main sequence, it still brightened, yes, but that means there was an increase in Luminosity. As the &lt;a href=&quot;https://en.wikipedia.org/wiki/Sun#Life_phases&quot; rel=&quot;nofollow noreferrer&quot;&gt;wiki page on the sun's evolution&lt;/a&gt; clarifies in the figure seen below, the temperature however didn't change much during that time, therefore also not the stellar class.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pB4Ac.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pB4Ac.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summarizing&lt;/strong&gt;&#xA;During the pre-main-sequence of a star the stellar class evolves significantly, on the main sequence it doesn't.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2018-10-24T23:20:37.953" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28146" PostTypeId="2" ParentId="28141" CreationDate="2018-10-24T23:34:43.720" Score="9" Body="&lt;p&gt;Kepler's laws state that a planet sweeps equal areas in equal times as it moves in its elliptical orbit. It doesn't state that different planets will sweep the same area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;equal areas&quot; law can be derived from the &quot;conservation of angular momentum&quot;. In fact dA/dt = L/(2m)  (where A is the area, L is the angular momentum and m is the (reduced) mass). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different planets will sweep out different areas. To calculate the period you used Kepler's third Law: &lt;span class=&quot;math-container&quot;&gt;$T^2 = k a^3$&lt;/span&gt; (T= orbital period,a = semi-major axis). If, for convenience you take a in AU and T in Earth Years, then the constant &lt;span class=&quot;math-container&quot;&gt;$k=1$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Venus, a = 0.72. so &lt;span class=&quot;math-container&quot;&gt;$T=\sqrt{0.72^3}=0.61$&lt;/span&gt; or about 223 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hyperphysics has a section on &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/kepler.html&quot; rel=&quot;noreferrer&quot;&gt;Kepler's Laws&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-10-24T23:34:43.720" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28147" PostTypeId="1" AcceptedAnswerId="28150" CreationDate="2018-10-25T03:27:13.203" Score="0" ViewCount="721" Body="&lt;p&gt;So, I know that the standard for conversion from bolometric magnitude to luminosity established in 2015 equates to:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;L = 10^(-0.4 * magnitude) * 3.0128e28&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Easy enough. But what does one do about luminosities in specific bands? I've seen some information that makes it look like it should be something like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;U: 10^(0.4*(5.61 - U-band magnitude)) * 3.828e26&#xA;B: 10^(0.4*(5.48 - B-band magnitude)) * 3.828e26&#xA;V: 10^(0.4*(4.83 - V-band magnitude)) * 3.828e26&#xA;R: 10^(0.4*(4.42 - R-band magnitude)) * 3.828e26&#xA;I: 10^(0.4*(4.08 - I-band magnitude)) * 3.828e26&#xA;J: 10^(0.4*(3.64 - J-band magnitude)) * 3.828e26&#xA;H: 10^(0.4*(3.32 - H-band magnitude)) * 3.828e26&#xA;K: 10^(0.4*(3.28 - K-band magnitude)) * 3.828e26&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But I'm not sure about these, because I've seen varying figures from varying sources for the magnitude of the sun in various bands.  I'm trying to extract the luminosity values from PARSEC data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speaking of PARSEC data, the documentation is rather lacking, so any clue about the following fields?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;McoreTP: Sounds like it should be something about core temperature, but the value is usually either 0, or something small like &quot;0.546&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mloss: Presumably mass loss, but there's no units!  But I can probably work out the units just by comparing how star masses change over time....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tau1m: ??? Usually 0.00, but sometimes something small like &quot;0.06&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cexcess: ??? Generally &quot;-1.000&quot;, but sometimes things like &quot;7.356&quot;.  Note that there's an entirely separate field for Xc, which is clearly carbon mass fraction.&lt;/p&gt;&#xA;" OwnerUserId="15606" LastEditorUserId="37191" LastEditDate="2021-02-11T09:23:12.840" LastActivityDate="2021-02-11T09:23:12.840" Title="Conversion of UBVRIJHK magnitudes to luminosity?" Tags="&lt;units&gt;&lt;absolute-magnitude&gt;&lt;parsec&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28148" PostTypeId="2" ParentId="28147" CreationDate="2018-10-25T04:26:44.823" Score="0" Body="&lt;p&gt;No comment on PARSEC data, but here is how to do bolometric luminosity from UBVRIJHK magnitudes:&lt;br&gt;&#xA; 1. Map filter to its effective wavelength (e.g., U = 3650 A).&lt;br&gt;&#xA; 2. Convert magnitudes to specific fluxes + extinction correction and other. Make sure that magnitudes, which are expressed in Vega, AB, or other reference systems, are converted accordingly. Extinction correction and other corrections may or may not what you want to do depending on what you want to get in the end.&lt;br&gt;&#xA; 3. Integrate {specific flux (i.e., in erg/s/A/cm^2) * effective wavelength (i.e., in A)}. Normally, trapezoid rule is enough.&lt;br&gt;&#xA; 4. Consider how to deal with wavelength outside the observed bandpass (i.e., wavelength shorter than U, and longer than K). Normally, people have the edges to be zero and quote the luminosity as &quot;pseudo.&quot; If you really want &quot;bolometric,&quot; you can assume blackbody spectral energy distribution (BB-SED) to reconstruct the edges. But, this depends on whether the BB-SED is suitable for your object. If this is the case, then you fit the observation to BB function to get temperature and scale factor. Then, bolometric flux = flux calculated in step 3 + correction from the edges estimated by the BB-SED.&lt;br&gt;&#xA; 5. L = flux * area. If you assume spherical symmetry, area = &lt;span class=&quot;math-container&quot;&gt;$4 \pi r^2$&lt;/span&gt;, where r = luminosity distance in this case. Note that you get the units right.&lt;br&gt;&#xA; 6. If you assume BB-SED, an easier way to get the bolometric luminosity, but less preferred in most cases, is to simply do &lt;span class=&quot;math-container&quot;&gt;$L = 4 \pi r^2 \sigma T^4$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to your formula up there, U: 10^(0.4*(5.61 - bolometric magnitude)), this is the 2nd step with magnitude system reference to Vega system (I guess). '* 3.828e26' is probably the area.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastEditorUserId="907" LastEditDate="2018-10-25T17:54:40.640" LastActivityDate="2018-10-25T17:54:40.640" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28149" PostTypeId="2" ParentId="28135" CreationDate="2018-10-25T04:39:34.390" Score="1" Body="&lt;p&gt;This is not an answer, but it is too long to put in the comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not familiar with quasars. In general, converting fluxes from an observed frame to a rest frame requires following consideration: extinction correction, cosmological expansion, k-correction. I think the simplest form is &lt;span class=&quot;math-container&quot;&gt;$f_{rest} = (1 + z)^n f_{obs}$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; is redshift and &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is some integer depending on how to express the unit of fluxes. This considers only the cosmological expansion. But, I am not sure if this transformation is valid with your quasars because they might be too far that &lt;span class=&quot;math-container&quot;&gt;$(1+z)^n$&lt;/span&gt; is no longer a good approximation.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-10-25T04:39:34.390" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28150" PostTypeId="2" ParentId="28147" CreationDate="2018-10-25T08:15:15.040" Score="1" Body="&lt;p&gt;Luminosities &lt;em&gt;cannot&lt;/em&gt; be calculated from magnitudes without knowing the distance to the source (and perhaps something about extinction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume therefore that you are talking about &lt;em&gt;absolute&lt;/em&gt; magnitudes in those filter band passes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luminosity can also &lt;em&gt;not&lt;/em&gt; be arrived at from a single absolute magnitude measurement in the way you suggest in your post, since for a source of a given luminosity, how much of that luminosity emerges in a specific band is entirely dependent on the spectrum of the source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, but now I see what you want - you want to divide up the bolometric luminosity of the Sun between the individual bands using the absolute magnitude of the Sun in those bands. But the luminosity doesn't divide up neatly like that, partly because the bands overlap and partly because the bandpasses are not top-hat functions so it's difficult to see what a &quot;luminosity in the U-band&quot; actually means other than an integration of the intrinsic stellar spectrum (which you don't have) over the (normalised) bandpass sensitivity profile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ L_{\rm band} = \int f_{\nu} S_{\nu} d\nu$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formulae you quote contain the absolute magnitudes of the Sun in those bands. That is the first number in the brackets. If you subtract the bolometric magnitude from this, then that gives you (minus) the bolometric correction. The problem is that the UBV etc. magnitude system and the bolometric magnitude system do not share the same zeropoint. The bolometric correction in the V band is about zero, so your formulae would have the entire luminosity of the Sun emerging in the V band. The zeropoints of the UBV etc. system are set by defining an A0 star (like Vega) to have an absolute magnitude of zero in all pass bands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can do is convert the magnitudes into some sort of mean flux density over the bandpass and then just multiply this by the &quot;effective width&quot; of the filter. This is very approximate and essentially assumes the spectrum is flat across the filter bandpass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Page 100 of &lt;a href=&quot;http://ads.harvard.edu/cgi-bin/bbrowse?book=hsaa&amp;amp;chap=2&amp;amp;page=0100&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zombeck (1992)&lt;/a&gt; gives the flux &lt;em&gt;density&lt;/em&gt; zeropoints in UBVRIJHK corresponding to magnitude 0.0. So take your magnitude and scale the zeropoint flux accordingly (by &lt;span class=&quot;math-container&quot;&gt;$10^{-0.4m}$&lt;/span&gt;) and then multiply the flux density by the effective bandpass width. Finally to convert an observed flux to a luminosity, multiply by &lt;span class=&quot;math-container&quot;&gt;$4\pi d^2$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$d=10$&lt;/span&gt; pc, if you are dealing with absolute magnitudes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;PARSEC&quot; I assume you are talking about the stellar evolution models produced by the &lt;a href=&quot;http://stev.oapd.inaf.it/cgi-bin/cmd&quot; rel=&quot;nofollow noreferrer&quot;&gt;Padova group&lt;/a&gt;. In which case a basic output of the model is the stellar luminosity. Any magnitudes that are provided by the models &lt;em&gt;are calculated afterwards&lt;/em&gt; by using a particular set of &lt;a href=&quot;https://en.wikipedia.org/wiki/Bolometric_correction&quot; rel=&quot;nofollow noreferrer&quot;&gt;bolometric corrections&lt;/a&gt; (the relationship between absolute bolometric magniude and absolute magnitude in a particular passband).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As regards the specific columns in the tables, you will likely have to go to the original papers that describe the models. I cannot comment further without knowing exactly which models they are or where you got them.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-10-25T21:34:55.100" LastActivityDate="2018-10-25T21:34:55.100" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="28151" PostTypeId="1" CreationDate="2018-10-25T09:10:01.027" Score="1" ViewCount="39" Body="&lt;p&gt;&lt;strong&gt;Can NASA's sound recordings, and photos of, astronomical bodies be used in for-profit/commercial projects (movies, video games)?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read NASA's Media Usage Guidelines, but am still not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(here's the link &lt;a href=&quot;https://www.nasa.gov/multimedia/guidelines/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.nasa.gov/multimedia/guidelines/index.html&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="24511" LastEditorUserId="24511" LastEditDate="2018-11-01T03:35:32.243" LastActivityDate="2018-11-01T03:35:32.243" Title="Can NASA's sound recordings of planets, and photos of astronomical bodies, be used freely in for-profit projects?" Tags="&lt;nasa&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2018-10-25T23:23:36.280" ContentLicense="CC BY-SA 4.0" />
  <row Id="28152" PostTypeId="2" ParentId="27760" CreationDate="2018-10-25T10:21:44.833" Score="2" Body="&lt;p&gt;This is a coincidence, but it seems that the photographer &lt;a href=&quot;https://www.mathieustern.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mathieu Stern&lt;/a&gt; has been thinking about making and using an ice lens for photography for years. The video linked below was just recently released.&lt;/p&gt;&#xA;&lt;p&gt;See &lt;a href=&quot;https://www.mathieustern.com/blog/2018/10/22/l437fjpq58g619vlkm6t1iwhk8s6dr&quot; rel=&quot;nofollow noreferrer&quot;&gt;their blogpost&lt;/a&gt; and their YouTube video &lt;a href=&quot;https://youtu.be/jWvwt3mCu68&quot; rel=&quot;nofollow noreferrer&quot;&gt;I made a CAMERA LENS with an ICEBERG&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;I'm going to call this an &amp;quot;almost&amp;quot; telescope in that a large refracting telescope exposing directly on to a photographic plate, certainly called a telescope by astronomers, is no different than a camera except that it tends to be much larger. (e.g. &lt;a href=&quot;https://astronomy.stackexchange.com/q/18074/7982&quot;&gt;What (actually) is the &amp;quot; deprojected half-light radius&amp;quot; of this almost-all-dark-matter Galaxy?&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;p&gt;Here are some of the points listed in the description of the YouTube video&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First of all the life-span of a lens made of ice is very short, so you can't move very far from where you create the lens.&lt;/li&gt;&#xA;&lt;li&gt;Focusing is extremely hard because of the water melting on the lens itslef.&lt;/li&gt;&#xA;&lt;li&gt;Shaping one &amp;quot; lens&amp;quot; takes 45 min because of the ambiant cold.&lt;/li&gt;&#xA;&lt;li&gt;If your lens breaks in the mold, you have to start all over again&lt;/li&gt;&#xA;&lt;li&gt;I hacked a Japanese cocktail ice ball maker to create half spheres.&lt;/li&gt;&#xA;&lt;li&gt;The housing of the lens was made using a home made 3D print model.&lt;/li&gt;&#xA;&lt;li&gt;All photos where shot during the 1 minute of the last ice lens.&lt;/li&gt;&#xA;&lt;li&gt;The video at the beginning of the video was shot in the last usable seconds of the last ice lens.&lt;/li&gt;&#xA;&lt;li&gt;No I didn't ruined my camera with water damages, even if it was pretty dangerous, I managed to keep the inside dry.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jcNl7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jcNl7.jpg&quot; alt=&quot;I made a CAMERA LENS with an ICEBERG, Mathieu Stern&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5wvEc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5wvEc.jpg&quot; alt=&quot;I made a CAMERA LENS with an ICEBERG, Mathieu Stern&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/K8RTu.jpg&quot; width=&quot;320&quot; alt=&quot;I made a CAMERA LENS with an ICEBERG, Mathieu Stern&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/oOdLQ.jpg&quot; width=&quot;320&quot; alt=&quot;I made a CAMERA LENS with an ICEBERG, Mathieu Stern&quot;&gt;&lt;/p&gt;&#xA;&lt;p&gt;Source:  &lt;a href=&quot;https://www.mathieustern.com/blog/2018/10/22/l437fjpq58g619vlkm6t1iwhk8s6dr&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mathieu Stern&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2021-05-17T21:37:13.910" LastActivityDate="2021-05-17T21:37:13.910" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28154" PostTypeId="1" AcceptedAnswerId="28157" CreationDate="2018-10-25T13:27:05.183" Score="0" ViewCount="359" Body="&lt;p&gt;I am planning to buy celestron nexstar 127SLT and I want to know what can I see using my telescope.&lt;/p&gt;&#xA;" OwnerUserId="24649" LastEditorUserId="1891" LastEditDate="2020-12-23T22:53:41.833" LastActivityDate="2020-12-23T22:53:41.833" Title="What can I see using a celestron nexstar 127SLT?" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2018-10-26T08:06:43.773" ContentLicense="CC BY-SA 4.0" />
  <row Id="28155" PostTypeId="1" CreationDate="2018-10-25T14:13:52.123" Score="-2" ViewCount="84" Body="&lt;p&gt;In the 1700s, Edmond Halley and others discovered that stars appeared larger than they are, and attributed it to diffraction in the lens of an observer (telescope, eye) and/or in the Earth's atmosphere. Are there any ways to view stars without that effect?&lt;/p&gt;&#xA;" OwnerUserId="24634" LastActivityDate="2018-10-25T22:14:24.940" Title="Are there any optical method of viewing stars without &quot;diffraction&quot; making them appear larger?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="8" ClosedDate="2018-10-25T23:27:42.440" ContentLicense="CC BY-SA 4.0" />
  <row Id="28156" PostTypeId="2" ParentId="28155" CreationDate="2018-10-25T15:35:56.457" Score="3" Body="&lt;p&gt;There is a physical limit to the angular resolution of any imaging system. For a circular aperture it is often quoted that the angular resolution corresponds to around &lt;span class=&quot;math-container&quot;&gt;$\theta_{\rm min} = 1.22 \lambda/D$&lt;/span&gt; (in radians), where &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; is the wavelength of observation and &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; is the diameter of the telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar limit applies in interferometry but now &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; might represent the separation between your telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To translate an angular resolution to an actual physical resolution of size you would have to multiply by the distance to the source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the ways in which you can improve matters are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Use a bigger telescope or use an interferometer with a longer baseline.&lt;/li&gt;&#xA;&lt;li&gt;Work at shorter wavelengths.&lt;/li&gt;&#xA;&lt;li&gt;Get closer to the source.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are of course technical challenges with each of these. In particular you might have thought just increasing the baseline would be easy enough - but you need to know the separation of your telescope to a fraction of the wavelength you are using and that turns out to be tough for larger separations. Shorter wavelengths are problematic for optics - normal lenses and mirrors don't work at X-ray wavelengths for example. Getting closer to the source requires interstellar travel...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The OP wants to argue with me about whether the atmosphere is a significant factor. Of course the atmosphere degrades stellar images (it's not diffraction, but refraction). Going into space overcomes this and &lt;strong&gt;in the past&lt;/strong&gt; this &lt;strong&gt;would&lt;/strong&gt; have been the limiting issue. However, now it is commonplace for ground-based telescopes to achieve close to their diffraction-limited performance using adaptive optics techniques or &quot;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Lucky_imaging&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lucky Imaging&lt;/a&gt;&quot;, both of which can and have got better images than the Hubble Space Telescope (examples below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 1: &lt;em&gt;The middle picture is from HST, the right hand image is Lucky Imaging from the ground based Palomar 5m telescope. The larger aperture of the Palomar produces better resolution despite looking through the atmosphere.&lt;/em&gt; See &lt;a href=&quot;https://www.ast.cam.ac.uk/research/lucky&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.ast.cam.ac.uk/research/lucky&lt;/a&gt;&#xA;for further details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GRkWi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GRkWi.jpg&quot; alt=&quot;HST Vs Lucky Imaging&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 2: &lt;em&gt;HST image of Titan vs an image taken using adaptive optics with the 10m ground-based Keck telescope. Larger aperture wins again because the effects of the atmosphere can be avoided.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3uFOb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3uFOb.jpg&quot; alt=&quot;HST Vs Keck&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-10-25T22:14:24.940" LastActivityDate="2018-10-25T22:14:24.940" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="28157" PostTypeId="2" ParentId="28154" CreationDate="2018-10-25T17:03:10.293" Score="4" Body="&lt;p&gt;First of all, the high end of magnification is generally not as important as several other factors. However, considering that no additional information was given in the question beyond &quot;reflecting&quot;, I will make some assumptions and only give you a very general estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book &quot;Fundamental Astronomy: H. Karttunen et al.&quot;, page 51, formula (3.5) gives a relationship between the diameter of a telescope's aperture and it's &lt;em&gt;useful magnification&lt;/em&gt;.&lt;br&gt;This relation is as follows (&lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt; is the resolving capacity of the human eye, &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; is the aperture diameter, &lt;span class=&quot;math-container&quot;&gt;$ω$&lt;/span&gt; is the maximum magnification and &lt;span class=&quot;math-container&quot;&gt;$λ$&lt;/span&gt; is the wavelength of visible light, used in the formula for the telescope's resolving capacity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$θ ≈ 1.22×λ/D ≈ λ/D,$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ω_{max} = \frac{e}{θ} ≈ \frac{eD}{λ} = \frac{(5.8 × 10^{-4} × D)}{(5.5 × 10^{-7} m)} ≈ \frac{D}{1 mm}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a key concept for you, because any additional magnification above that can be thought of as &quot;empty magnification&quot;. It exists mathematically, but the image is so blurry that you will not see any additional detail, and probably be quite disappointed in your purchase if you paid a lot of money for that magnification!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's try backtracking from 500x. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the true &lt;em&gt;maximum useful magnification&lt;/em&gt; this telescope can achieve, then its aperture diameter would be about 50cm. This aperture allows for &lt;span class=&quot;math-container&quot;&gt;$D^2/d^2$&lt;/span&gt; brighter stars to be observed (&lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; is the diameter of the human pupil). When this is calculated and converted to magnitudes, it turns out you should be able to see stars up to magnitude &lt;span class=&quot;math-container&quot;&gt;$9+6=15$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this is a very rough estimation and you certainly will not be able to see magnitude 15 stars due to atmospheric extinction, seeing and other factors ... my estimate is you'll see stars about the 12th magnitude on a good night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the 500x telescope you are looking at &lt;strong&gt;has a much smaller aperture than 50 cm&lt;/strong&gt; (and probably that's true) then that 500x is not going to be so useful in reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take some time and read further about telescopes. This site has many existing questions and answers on the subject.&lt;/p&gt;&#xA;" OwnerUserId="24014" LastEditorUserId="7982" LastEditDate="2018-10-26T13:03:24.917" LastActivityDate="2018-10-26T13:03:24.917" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28158" PostTypeId="2" ParentId="18737" CreationDate="2018-10-25T18:21:42.570" Score="0" Body="&lt;p&gt;It was Kashgar in China back in 1991 when there was DST.  Noon occured at 4:10 pm and sunrise and 8:30 am and sunset at 11:45 pm. Civil twilight continued even after midnight.&lt;/p&gt;&#xA;" OwnerUserId="24652" LastActivityDate="2018-10-25T18:21:42.570" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28159" PostTypeId="1" CreationDate="2018-10-25T23:39:20.340" Score="12" ViewCount="332" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_seeing&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomical seeing&lt;/a&gt; is the limiting factor for the resolution of all but the smallest Earthbound telescopes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e0NFD.gif&quot; alt=&quot;Astronomical Seeing&quot;&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Seeing_Moon.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stunning advances in &lt;a href=&quot;https://en.wikipedia.org/wiki/Adaptive_optics&quot; rel=&quot;nofollow noreferrer&quot;&gt;adaptive optics&lt;/a&gt; (along with it's predecessor &lt;a href=&quot;https://en.wikipedia.org/wiki/Speckle_imaging&quot; rel=&quot;nofollow noreferrer&quot;&gt;speckle interferometry&lt;/a&gt; and it's budget-minded cousin &lt;a href=&quot;https://en.wikipedia.org/wiki/Lucky_imaging&quot; rel=&quot;nofollow noreferrer&quot;&gt;lucky imaging&lt;/a&gt;) get around this but only with substantial compromises in (some combination of) throughput, cost, complexity and wavelength range (see answers to &lt;a href=&quot;https://astronomy.stackexchange.com/q/14795/7982&quot;&gt;Why aren't ground-based observatories using adaptive optics for visible wavelengths?&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If identical telescopes sat on the surface of Earth and Mars and looked at a distant and &lt;em&gt;equi&lt;/em&gt;-distant  body, how would the following differ between the Mars telescope and the Earth telescope?:&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-12-12T11:05:24.383" LastActivityDate="2020-01-29T23:29:10.677" Title="How would astronomical seeing on Mars differ from that on Earth?" Tags="&lt;observational-astronomy&gt;&lt;mars&gt;&lt;angular-resolution&gt;&lt;atmospheric-effects&gt;&lt;planetary-atmosphere&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28161" PostTypeId="1" CreationDate="2018-10-26T14:56:29.247" Score="1" ViewCount="294" Body="&lt;p&gt;I need a bit of advice. I purchased a Orion xt8 and a lens kit with a 2x Barlow 6mm 20 mm eyepieces and every time I try pulling things into focus / magnification or focus goes out the window recently purchased a 5x Barlow to try to get the images bigger. But I’m starting to get a bit discouraged as the only thing I can see in the sky is the moon. All others. Clusters. Andromeda. Mars  I can not pull into focus to save my life. &lt;img src=&quot;https://i.stack.imgur.com/qOj1H.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iMABy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don’t know what I’m doing wrong. I see all these pictures of Andromeda from the same telescope and on mine I see dots. (Above picture is the most dots I could see near a cluster. But it’s not the cluster. I guess my question would be. How can I see objects in higher mag / detail. Andromeda is a dot like a star. So are all the planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes I know what to expect. I know things will look very blurry or out of focus. But only seeing star dots (when they are planets) concerns me.&lt;/p&gt;&#xA;" OwnerUserId="24658" LastEditorUserId="7472" LastEditDate="2018-10-26T17:32:55.863" LastActivityDate="2018-10-26T17:32:55.863" Title="Help new Dobsonian owner" Tags="&lt;telescope&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28162" PostTypeId="1" AcceptedAnswerId="28167" CreationDate="2018-10-26T15:43:15.990" Score="3" ViewCount="190" Body="&lt;p&gt;Many books explain the difference between synodic and sideral periods. The synodic period of planets are tabulated for centuries and it is defined as the time interval between two consecutive repetition of phase, phase in this case meaning the terrestrial measure of its angular position and speed relative to the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how do astronomers measure the angular position of a planet relative to the Sun if the major observations are made during the night?&lt;/p&gt;&#xA;" OwnerUserId="24650" LastActivityDate="2018-10-29T11:26:37.737" Title="Synodic period of planets" Tags="&lt;positional-astronomy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28163" PostTypeId="2" ParentId="28161" CreationDate="2018-10-26T16:36:14.160" Score="4" Body="&lt;p&gt;I have experience using this telescope.  It looks like you can focus correctly if you get a good view of the moon.  You probably don't want more magnification (the Andromeda galaxy is comparable in size to the moon).  It is easier to get and keep targets in the field of view with a low power eyepiece, and I suspect that is your problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with planets.  Use your low-power eyepiece, 20 mm.  Jupiter is the easiest to see as a disk, but it is unfortunately close to the sun right now so it is low in the west in the evenings.  With your 20mm eyepiece, it should be quite obvious as a disk with satellites and you may be able to see bands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars is quite prominent right now in the evening sky but it is smaller than Jupiter.  Still, with your 20mm eyepiece, you should be able to at least tell it is a disk (and a bright orange one) if you have it in the view of the eyepiece.  Once you have it centered in view of your 20 mm eyepiece, you can try switching to your high-power (6mm) and see if that reveals any more detail.  Markings on Mars or quite subtle; often it looks like a featureless orange ball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your night sky isn't that dark because you are close to a city with its outdoor lights, it can be a challenge to get a good view of even spectacular &quot;deep sky&quot; objects like the Andromeda galaxy.  Again, use your low power eyepiece and no Barlow lens.  The core of the galaxy is much brighter than the extended glow around it, so it can appear star-like.  Try looking at with &quot;averted vision&quot; (don't look directly at it, to use the more sensitive light-detectors towards the edge of your field of vision) to pick out the faint glow that surrounds the core.  If you go to a dark sky location (one where you can really see the Milky Way) the XT8 can give a pretty impressive view of the galaxy where the elliptical outer glow will fill the FOV of your low-power eyepiece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is hard for you to travel to a dark-sky location, you can try buying a deep-sky filter.  This filter tries to block the specific wavelengths thrown into the sky by outdoor lighting.  It makes everything you see through the scope dimmer and kind of green, but it can improve contrast on galaxies and nebula by making the background sky darker.&lt;/p&gt;&#xA;" OwnerUserId="14434" LastActivityDate="2018-10-26T16:36:14.160" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28164" PostTypeId="2" ParentId="28161" CreationDate="2018-10-26T17:02:02.183" Score="0" Body="&lt;p&gt;What is your location? If you're in a high-humidity or high-wind area, atmospheric turbulence will affect image quality as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are you attaching the camera to the 'scope, and have you set up a remote shutter release?  Any small vibrations will, again, affect image quality, and these effects are much more noticeable on small objects vs. the Moon.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start, if you have them available, by sighting the planets with 8x or 10x binoculars. If you get clear images with them, then it's time to look at your 'scope's alignment -- it could be in focus but the elements not on-axis, leading to aberrations as well.  &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-10-26T17:02:02.183" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28165" PostTypeId="2" ParentId="4753" CreationDate="2018-10-26T17:25:42.977" Score="-4" Body="&lt;p&gt;It is definitely possible, but in a very hot and active part of the universe. In order to discover these elements, there would have to be a lot of waiting for these elements to form. Our solar system isn't active enough, and a nebula would be the best place to look.&lt;/p&gt;&#xA;" OwnerUserId="24660" LastEditorUserId="7472" LastEditDate="2018-10-26T17:33:14.360" LastActivityDate="2018-10-26T17:33:14.360" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28166" PostTypeId="1" AcceptedAnswerId="28169" CreationDate="2018-10-26T18:00:29.547" Score="0" ViewCount="99" Body="&lt;p&gt;I'm trying to create a little program in Unity3D that positions the planets in 3D space at a given time. I've followed this document: &lt;a href=&quot;http://stjarnhimlen.se/comp/ppcomp.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://stjarnhimlen.se/comp/ppcomp.html&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the data from section 4 and the code from section 6 and 7 I should have the correct algorithm right? I did skip on the changes in the orbital attributes (eg + 4.70935E-5 * d) because I just need to have them move in a set orbit. I don;t really need them to change over time. For now I'm just increasing the mean anomaly over time without taking the period in account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that the planets do not seem to get the correct offset to the ecliptic, and their movement is really weird: they speed up, then slow down and reverse in their paths. Here is a video of their movement: &lt;a href=&quot;https://www.youtube.com/watch?v=zHOBChFpan0&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=zHOBChFpan0&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like the eccentricity is correct on their ellipses and the distances to the sun are also correct, but the rest is just wrong. I'm not that great at math or astronomy to fully understand what's going on in this algorithm, but I'm hoping someone here might spot the error.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;using UnityEngine;&#xA;using System;&#xA;&#xA;public class Orbit : MonoBehaviour &#xA;{&#xA;public float longitudeOfTheAscendingNode;&#xA;public float inclination;&#xA;public float argumentOfPerihelion;&#xA;public float semimajorAxis;&#xA;public float eccentricity;&#xA;&#xA;public float meanAnomalySpeed = 0.1f;&#xA;&#xA;float M = 0; //mean anomaly&#xA;&#xA;void Update()&#xA;{&#xA;    SetPosition();&#xA;}&#xA;&#xA;void SetPosition()&#xA;{&#xA;    float N = longitudeOfTheAscendingNode;&#xA;    float i = inclination;&#xA;    float w = argumentOfPerihelion;&#xA;    float a = semimajorAxis;&#xA;    float e = eccentricity;&#xA;&#xA;    float E = M + e * (180 / Mathf.PI) * Mathf.Sin(M) * (1 + e * Mathf.Cos(M));&#xA;&#xA;    float xv = a * (Mathf.Cos(E) - e);&#xA;    float yv = a * (Mathf.Sqrt(1 - e * e) * Mathf.Sin(E));&#xA;&#xA;    float v = Mathf.Atan2(yv, xv);&#xA;    float r = Mathf.Sqrt(xv * xv + yv * yv);&#xA;&#xA;    float xx = r * (Mathf.Cos(N) * Mathf.Cos(v + w) - Mathf.Sin(N) * Mathf.Sin(v + w) * Mathf.Cos(i));&#xA;    float yy = r * (Mathf.Sin(N) * Mathf.Cos(v + w) + Mathf.Cos(N) * Mathf.Sin(v + w) * Mathf.Cos(i));&#xA;    float zz = r * (Mathf.Sin(v + w) * Mathf.Sin(i));&#xA;&#xA;    transform.position = new Vector3(xx, yy, zz);&#xA;&#xA;    M += meanAnomalySpeed * Time.deltaTime;&#xA;    if (M &amp;gt; 360)&#xA;        M -= 360;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;}&lt;/p&gt;&#xA;" OwnerUserId="24661" LastActivityDate="2018-10-27T11:52:34.553" Title="Problems with computing X,Y,Z positions of planets at a certain time in Unity3D (C#)" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;orbital-elements&gt;&lt;orbital-mechanics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28167" PostTypeId="2" ParentId="28162" CreationDate="2018-10-26T18:43:11.717" Score="3" Body="&lt;p&gt;I'm pretty far from an expert at ancient astronomy, but as I understand it, the ancient astronomers looked for the date when the planet was highest in the sky.   This is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Opposition_(planets)&quot; rel=&quot;nofollow noreferrer&quot;&gt;being in opposition&lt;/a&gt; or opposite the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EiWS6.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EiWS6.gif&quot; alt=&quot;https://www.heavens-above.com/images/opposition.gif&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.heavens-above.com/glossary.aspx?term=opposition&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source of image&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.polaris.iastate.edu/EveningStar/Unit1/unit1_sub3.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;See here for further explanation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also read that the synodic period can be calculated between &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_retrograde_motion#Apparent_motion&quot; rel=&quot;nofollow noreferrer&quot;&gt;periods of retrograde&lt;/a&gt;.   In neither case, the position of the sun is necessary.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The synodic period, because it was initially based entirely on observation, works for both Geocentric and Heliocentric models.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Position of the sun&quot; infers our current understanding of the heliocentric system.  The triangle, between the 3 points, Earth, Sun and observed planet and that only works for the heliocentric model.  Most of the ancient astronomers didn't use that model.  Their synodic period was based entirely on the motion of the planet relative to the fixed background stars.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-10-29T05:55:01.207" LastActivityDate="2018-10-29T05:55:01.207" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28168" PostTypeId="1" CreationDate="2018-10-26T22:28:03.280" Score="0" ViewCount="148" Body="&lt;p&gt;The sun in the Voyager 1 photo here from 1990 appears 13x bigger than its diameter, diffraction of course. On Earth, photographs of the sun do not make it appear 13x bigger. Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A description of the photo from &lt;a href=&quot;https://www.jpl.nasa.gov/spaceimages/details.php?id=PIA00450&quot; rel=&quot;nofollow noreferrer&quot;&gt;jpl.nasa.gov&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The wide-angle was taken with the camera's darkest filter (a methane&#xA;  absorption band), and the shortest possible exposure (5 thousandths of&#xA;  a second) to avoid saturating the camera's vidicon tube with scattered&#xA;  sunlight. &quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JF9xZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JF9xZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user24662" LastEditorDisplayName="user24662" LastEditDate="2018-10-27T16:13:27.783" LastActivityDate="2018-10-28T12:18:38.893" Title="Why does the sun appear 13x bigger through diffraction in Voyager 1 image from 1990, but not from Earth?" Tags="&lt;star&gt;&lt;optics&gt;" AnswerCount="1" CommentCount="8" ClosedDate="2018-10-28T17:37:48.257" ContentLicense="CC BY-SA 4.0" />
  <row Id="28169" PostTypeId="2" ParentId="28166" CreationDate="2018-10-26T22:33:29.990" Score="5" Body="&lt;p&gt;I think E and M are in degrees in your code, but the trig functions in the &lt;code&gt;math&lt;/code&gt; library in C# expect an argument in radians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to handle this is to define constants  &lt;code&gt;radians = Pi/180&lt;/code&gt; and &lt;code&gt;degrees = 180/pi&lt;/code&gt; then always do &lt;code&gt;Mathf.Sin(M*radians)&lt;/code&gt;. You will also need to do this for the lines &lt;code&gt;float xx=&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you could work entirely in radians. Convert all angles to radians at the start and only use radians in the code. You would need to convert &lt;code&gt;n, i, w&lt;/code&gt; and &lt;code&gt;M&lt;/code&gt; to radians and remember that &lt;code&gt;e&lt;/code&gt; and &lt;code&gt;v&lt;/code&gt; will also be in radians.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-10-27T11:52:34.553" LastActivityDate="2018-10-27T11:52:34.553" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28170" PostTypeId="1" CreationDate="2018-10-26T23:15:10.493" Score="5" ViewCount="74" Body="&lt;p&gt;Objects which are temporarily captured into an Earth-Moon system orbit (temporarily captured orbiters or TCOs) by definition will leave this orbit. While the term suggest an Earth orbit, it's really an n-body orbit involving the gravitational effects of the Moon and Sun as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are also sometimes referred to as &quot;mini-moons&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon exiting, can TCOs go as far out as the asteroid belt, or do they then just orbit the Sun in a heliocentric orbit very similar to the Earth's 1 AU?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone tracked or calculated the orbit of a known TCO to this extent?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side question (it can be split this off separately if necessary) have any of these been larger than say 10 meters in diameter?&lt;/p&gt;&#xA;" OwnerUserId="24663" LastEditorUserId="7982" LastEditDate="2018-10-27T04:41:46.717" LastActivityDate="2018-11-02T12:29:47.930" Title="How far from the Sun can an Earth TCO (temporarily captured orbiter) go after leaving?" Tags="&lt;orbit&gt;&lt;natural-satellites&gt;&lt;orbital-mechanics&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28173" PostTypeId="2" ParentId="28168" CreationDate="2018-10-27T14:46:55.540" Score="5" Body="&lt;p&gt;Don't base the size of the Sun on what you see in that image because the Sun is over-exposed.  The big spot in the center of the diffraction spikes are saturated pixels because they all appear to have the same value (the spot looks uniform) so you don't know what their real values are; you would need to take a shorter-exposure image (or put filters in front of the lens) if you wanted to take a proper picture of the Sun.  Since Venus is a lot dimmer than the Sun, they set the exposure long enough to see it and accepted the fact that the Sun would be over-exposed.  You see the same thing in &lt;a href=&quot;http://hubblesite.org/image/4229/gallery&quot; rel=&quot;noreferrer&quot;&gt;some of the Hubble images&lt;/a&gt; where to bring out the background stars, the brighter ones are over-exposed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of looking at it, if you see an image where you can very clearly see the diffraction spikes, that part of the image is over-exposed.&lt;/p&gt;&#xA;" OwnerUserId="9749" LastActivityDate="2018-10-27T14:46:55.540" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="28174" PostTypeId="1" AcceptedAnswerId="34583" CreationDate="2018-10-27T16:24:22.887" Score="3" ViewCount="93" Body="&lt;p&gt;I understand how to construct South-facing vertical an horizontal sundials and the geometry behind their construction by projecting an equatorial sundial into a plane, but I can't quite get how to do that for a declined plane.&lt;/p&gt;&#xA;" OwnerUserId="24667" LastActivityDate="2020-01-03T14:55:27.747" Title="How does the geometry for constructing a declining vertical sundial work?" Tags="&lt;the-sun&gt;&lt;time&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28175" PostTypeId="1" AcceptedAnswerId="28176" CreationDate="2018-10-27T17:17:33.580" Score="1" ViewCount="364" Body="&lt;p&gt;Assuming circular orbit, the evaluation of the angular speed of a planet around the Sun, just requires its orbital period. Similarly, to calculate its angular speed relative to the Earth (also seen from Sun and considering the approximate circular orbit), its synodic period is needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: &lt;strong&gt;what is the angular speed of an interior planet across the sky in the conjunction with Sun and how can I deduce its formula?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="24650" LastEditorUserId="24650" LastEditDate="2018-10-29T20:06:33.370" LastActivityDate="2018-10-29T20:06:33.370" Title="Angular speed of a planet across the sky" Tags="&lt;positional-astronomy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28176" PostTypeId="2" ParentId="28175" CreationDate="2018-10-27T17:17:33.580" Score="2" Body="&lt;p&gt;Let me use the following image as reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vyDLV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vyDLV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the planet moves, its &lt;strong&gt;linear displacement on its orbit relative to Earth&lt;/strong&gt; is &lt;span class=&quot;math-container&quot;&gt;$s = d \cdot \psi$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; is its distance to the sun and &lt;span class=&quot;math-container&quot;&gt;$\psi = \phi - \tau$&lt;/span&gt; is its &lt;strong&gt;orbital angular displacement &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt; minus the Earth's orbital angular displacement&lt;/strong&gt;, both seen from Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Earth, we see the same linear displacement, but given by &lt;span class=&quot;math-container&quot;&gt;$s = D \cdot \theta$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; is our distance to the Sun and &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; is the angular distance we watched the planet moving. Therefore &lt;span class=&quot;math-container&quot;&gt;$D \cdot \theta = d \cdot \psi$&lt;/span&gt;, and therefore, the angular speed &lt;span class=&quot;math-container&quot;&gt;$\dot{\theta}$&lt;/span&gt; across the sky is simply &lt;span class=&quot;math-container&quot;&gt;$\dot{\theta} = d \cdot \dot{\psi}/D$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\dot{\psi}$&lt;/span&gt; is the angular speed of the planet relative to Earth, &lt;em&gt;i.e.&lt;/em&gt;, &lt;span class=&quot;math-container&quot;&gt;$\dot{\psi} = 2\pi/T_\mathrm{s}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$T_\mathrm{s}$&lt;/span&gt; is the synodic period of the planet.&lt;/p&gt;&#xA;" OwnerUserId="24650" LastEditorUserId="24650" LastEditDate="2018-10-29T13:31:00.253" LastActivityDate="2018-10-29T13:31:00.253" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28177" PostTypeId="2" ParentId="28175" CreationDate="2018-10-27T20:21:36.567" Score="4" Body="&lt;p&gt;There is no simple formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, it's not true that the angular speed of a planet relative to the sun can be found from its orbital period.  This is because planets don't move in circles, but in eclipses and the speed of the planet changes, as described by Kepler's second Law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When describing the motion of a planet in the sky you need to consider the elliptical motion of both the Earth and the planet, the superposition of these motions causes the planets to make looping motions in the sky, sometimes slowing down and moving backwards&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BTIRK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BTIRK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image shows how Mars would have appeared to loop and move backwards in 2003 and 2005. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motion in the sky can be calculated, but not by a simple formula. You start with the orbital elements of the planets, solve Kepler's equation to determine the position on their elliptical orbit. Solve some trigonometry to determine the position of each planet relative to the sun, then some vector calculations and spherical geometry to find the position of the planet relative to Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a &lt;a href=&quot;http://stjarnhimlen.se/comp/ppcomp.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;guide online&lt;/a&gt; for writing a computer program to do this. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-10-27T20:21:36.567" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28181" PostTypeId="1" AcceptedAnswerId="28184" CreationDate="2018-10-28T01:00:04.247" Score="22" ViewCount="7470" Body="&lt;p&gt;I have no doubt that we have been to the moon. This question has nothing to do with a moon landing hoax.  But, there are two quotes from two different astronauts regarding the size of the earth as viewed from the moon that are puzzling to me. Both quotes talk about how small the Earth looked. Shouldn’t the Earth look very large when viewed from the moon&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;It suddenly struck me that that tiny pea, pretty and blue was the Earth. I put up my thumb, shut one eye and my thumb blotted out the planet Earth. I didn't feel like a giant. I felt very, very small. — Neil Armstong&lt;/p&gt;&#xA;&lt;p&gt;As we got further and further away it [the Earth] diminished in size. Finally it shrank to the size of a marble, the most beautiful you can imagine. That beautiful, warm living object looked so fragile, so delicate that if you touched it with a finger it would crumble and fall apart. Seeing this has to change a man. — James Irwin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I know the term large is subjective, but still, the comments seem off. Please let me know what I’m missing.&lt;/p&gt;&#xA;" OwnerUserId="19597" LastEditorUserId="-1" LastEditorDisplayName="user1569" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-10-30T07:12:48.177" Title="Puzzling quotes from astronauts about earth size" Tags="&lt;the-moon&gt;&lt;observational-astronomy&gt;&lt;moonlanding&gt;" AnswerCount="6" CommentCount="8" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28182" PostTypeId="2" ParentId="28181" CreationDate="2018-10-28T01:47:27.577" Score="8" Body="&lt;p&gt;From the moon, the earth would appear to span across about 2 degree of sky.  If you hold your thumb a few inches in front of your face (or your facemask if you are on the moon), your thumb will block about two degrees of your field of view (Armstrong mentions closing one eye because his thumb is close enough that the different perspective between his two eyes is significant).  So there doesn't seem to be anything too surprising about these statements.&lt;/p&gt;&#xA;" OwnerUserId="14434" LastActivityDate="2018-10-28T01:47:27.577" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28183" PostTypeId="2" ParentId="28181" CreationDate="2018-10-28T08:06:26.993" Score="11" Body="&lt;p&gt;If you reduce the scale of the Earth to make it the size of a tennis ball, then the moon would be about the size of a nerf ball (20mm or under an inch), and the moon would be about 7 feet from the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can get a good idea of what the Earth would look like from the Moon: Get a tennis ball and nerf ball set them up 7 ft (2 m) apart put your head next to the nerf ball, and the tennis ball will look about the same size as the Earth would look on the moon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=GCTuirkcRwo&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=GCTuirkcRwo&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now hold up your thumb, cover the tennis ball and imagine &quot;on it [is] everyone you love, everyone you know, everyone you ever heard of, every human being who ever was, lived out their lives&quot;.   &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-10-29T16:47:14.377" LastActivityDate="2018-10-29T16:47:14.377" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28184" PostTypeId="2" ParentId="28181" CreationDate="2018-10-28T08:13:32.263" Score="45" Body="&lt;p&gt;The Earth is 4 times the diameter of the Moon. The Earth viewed from the Moon will therefore appear to have 4 times the angular diameter of the Moon viewed from the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon is &lt;em&gt;easily&lt;/em&gt; obscured by a thumb at arm's length (by a factor of 3-4).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now bring your thumb closer (because you can't fully extend your arm in a bulky space suit) and put on the biggest pair of ski gloves you can find.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not a great stretch of the imagination to think your thumb, in a spacesuit, would easily obscure something 4 times the size of the Moon.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-10-29T18:59:53.000" LastActivityDate="2018-10-29T18:59:53.000" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="28185" PostTypeId="2" ParentId="28181" CreationDate="2018-10-28T10:58:02.697" Score="11" Body="&lt;p&gt;You can get a rough idea of how large the Earth appears from these two photos:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/OS83y.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OS83y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/696OK.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/696OK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not quite that large, is it?&lt;/p&gt;&#xA;" OwnerUserId="24234" LastActivityDate="2018-10-28T10:58:02.697" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28186" PostTypeId="2" ParentId="28181" CreationDate="2018-10-28T13:34:12.673" Score="1" Body="&lt;p&gt;Watching the moon from earth, it seems to be larger at the horizon than at the sky. This means that optical illusions may play a role here, too. The lack of  reference objects or maybe the different shape of the horizon on moon may also contribute to the apparent size of earth.&lt;/p&gt;&#xA;" OwnerUserId="24669" LastActivityDate="2018-10-28T13:34:12.673" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28188" PostTypeId="1" AcceptedAnswerId="28216" CreationDate="2018-10-28T15:18:23.927" Score="6" ViewCount="233" Body="&lt;p&gt;In his appearance on Television contained in the video &lt;a href=&quot;https://youtu.be/TgA2y-Bgi3c?t=427&quot; rel=&quot;noreferrer&quot;&gt;The Mystery That Keeps Neil deGrasse Tyson Up At Night&lt;/a&gt;, Astronomer andeducator &lt;a href=&quot;https://en.wikipedia.org/wiki/Neil_deGrasse_Tyson&quot; rel=&quot;noreferrer&quot;&gt;Neil deGrasse Tyson&lt;/a&gt; says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I don’t want to be blamed if you can not get to sleep tonight.&lt;/p&gt;&#xA;&lt;p&gt;This dark energy, in the future, will render the universe so large, having accelerated so significantly, that all the galaxies of the night sky will have accelerated beyond our horizon.&lt;/p&gt;&#xA;&lt;p&gt;All the galaxies are the source of our knowledge of the big bang.&lt;/p&gt;&#xA;&lt;p&gt;Everything we know about the history of the universe comes to us from these galaxies.&lt;/p&gt;&#xA;&lt;p&gt;If they accelerate beyond our horizon, the next generation of cosmic explorers will only have the stars of the milky way to think about.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What does this mean, in simple terms if possible. How will nearby galaxies some day expand beyond our horizon (if that's what he's saying and if it is correct)?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-10-30T21:50:31.697" Title="How will nearby galaxies some day expand beyond our horizon? (Neil deGrasse Tyson)" Tags="&lt;cosmology&gt;&lt;local-group&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="28189" PostTypeId="1" AcceptedAnswerId="28190" CreationDate="2018-10-28T15:29:31.170" Score="2" ViewCount="104" Body="&lt;p&gt;In his appearance on television contained in the video &lt;a href=&quot;https://youtu.be/TgA2y-Bgi3c?t=465&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Mystery That Keeps Neil deGrasse Tyson Up At Night&lt;/a&gt;, in the backdrop behind astronomer and educator &lt;a href=&quot;https://en.wikipedia.org/wiki/Neil_deGrasse_Tyson&quot; rel=&quot;nofollow noreferrer&quot;&gt;Neil deGrasse Tyson&lt;/a&gt; and the talk-show host, what latitude on Earth would have Orion's belt and Aldebaran at equal elevation (about 10 degrees) above the horizon? Could that be Sunrise/Sunset, or are those just city lights?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He's known to be a stickler about such things (the story about the stars in the movie &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Titanic_(1997_film)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Titanic&lt;/a&gt;&lt;/em&gt; for example). This background must be calculated from a particular latitude at least (if not longitude and/or season).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dJASQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dJASQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7411" LastEditDate="2018-10-30T13:21:39.753" LastActivityDate="2018-10-30T13:21:39.753" Title="At what latitude could Orion's belt and Aldebaran be ~10 degrees above the horizon at the same time?" Tags="&lt;constellations&gt;&lt;astrometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28190" PostTypeId="2" ParentId="28189" CreationDate="2018-10-28T15:58:43.530" Score="4" Body="&lt;p&gt;This sky configuration is consistent with 40 degrees N, at about 6.30am (local time) at the end of November.  If you went to mid November, this would look the same at about 7.30am (ie 1 hour per half month).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latitude of New York is about 40N, which makes sense for &lt;s&gt;a&lt;/s&gt; this particular US TV show,  which is recorded live in New York City, &quot;just down the street&quot; from where Dr. Tyson works as directory of the &lt;a href=&quot;https://www.haydenplanetarium.org/tyson/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hayden Planetarium&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I guessed was from a combination of things. From where I live I'm familiar with the eastern sky, so it was obvious this wasn't looking east. So I played around with different times of day and directions, which confirmed this was a view from looking west, and south of where I live (50N).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be quickly confirmed with any on-line &quot;planetarium&quot; or sky viewer. For example here's a shot from &lt;a href=&quot;https://in-the-sky.org/skymap.php?year=2018&amp;amp;month=10&amp;amp;day=29&amp;amp;town=5128581&quot; rel=&quot;nofollow noreferrer&quot;&gt;in-the-sky.org&lt;/a&gt; in planetarium mode, set to New York City, 6:15 AM November 20. It puts Aldebaran and Orion's belt both at about 10 degrees above the Western horizon, with only the beginning of twilight. Obviously other combinations of nearby dates and times can work as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that the light in the image in the question is not the sunset or sunrise, so it must be a distant city in the West.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/I8kGj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I8kGj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WGxp6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WGxp6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11822" LastEditorUserId="7982" LastEditDate="2018-10-28T17:07:03.113" LastActivityDate="2018-10-28T17:07:03.113" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28191" PostTypeId="1" CreationDate="2018-10-28T18:12:01.830" Score="2" ViewCount="63" Body="&lt;p&gt;I'm trying to do photometry on my stacked images of the night sky, but when I use the program IRIS and try to do automatic photometry, it always says it's a wrong image number and in every example of photometry the numbers have only 3 digits while my numbers are always 4 digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've selected 3 objects that have the coordinates (360,3377) (591,2197) and (746,3437). The input generic name is the name of the folder with the pictures (FITS). I've also selected the aperture photometry option with the radius 8, 12 and 20. The coordinates VX and VY are both 0. The boxes &quot;no matching&quot; and &quot;magnitude output&quot; are selected. When I try to process it the program tells me &quot;incorrect image number&quot;.&lt;/p&gt;&#xA;" OwnerUserId="24671" LastEditorUserId="14094" LastEditDate="2018-10-31T12:59:27.950" LastActivityDate="2018-10-31T12:59:27.950" Title="Why can't I use automatic photometry on IRIS?" Tags="&lt;photography&gt;&lt;photometry&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28192" PostTypeId="1" AcceptedAnswerId="28194" CreationDate="2018-10-28T20:36:13.110" Score="-6" ViewCount="330" Body="&lt;p&gt;Most natural elliptical orbits are one sided as shown below from wiki.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/zkHZe.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zkHZe.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is an elliptical orbit where perigee and apogee are closer to being equal in distance than not? The object would have to skip off the atmosphere to maintain this orbit.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/QaFKt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QaFKt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have learned that this orbit would not be possible without spending energy and is not a stable orbit. Which makes this question off-topic no matter how I word it. I have flagged it for being off-topic.&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2019-05-04T19:33:38.540" LastActivityDate="2019-05-04T19:33:38.540" Title="What is this elliptical orbit?" Tags="&lt;orbit&gt;&lt;identify-this-object&gt;" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="28194" PostTypeId="2" ParentId="28192" CreationDate="2018-10-29T05:16:43.020" Score="8" Body="&lt;p&gt;That second picture is not an elliptical orbit (at least, not as elliptical as depicted). It is a roughly circular orbit viewed from the side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any orbit that appears as such (elliptical, but with the main body in the middle rather than at one of the two foci) is just a circular orbit viewed from just above or below the plane of the orbit.&lt;/p&gt;&#xA;" OwnerUserId="17655" LastActivityDate="2018-10-29T05:16:43.020" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28196" PostTypeId="2" ParentId="28162" CreationDate="2018-10-29T11:26:37.737" Score="2" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;how do astronomers measure the angular position of a planet relative&#xA;to the Sun if the major observations are made during the night?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Modern astronomers don't need to &amp;quot;see&amp;quot; the Sun or the planets, as their positions are precisely known and constantly verified. It's therefore a fairly simple matter to determine when a planet returns to the same relative position in relation to the Earth and Sun (i.e. the synodic period). However, for ancient astronomers (or anyone without access to planetary ephemerides), other methods are necessary.&lt;/p&gt;&#xA;&lt;p&gt;The simplest method is to define the synodic period as the time from one Sun-planet &amp;quot;opposition&amp;quot; to the next, an opposition occurring when the Sun and the planet form an angle of 180º as seen from the Earth. [See the image in &lt;a href=&quot;https://astronomy.stackexchange.com/a/28167&quot;&gt;this answer&lt;/a&gt; for what such an opposition looks like]. However, the challenge remains: when the planet is visible and approaching opposition, the Sun is likely to be below the horizon. How can you tell when they are at exactly 180º?&lt;/p&gt;&#xA;&lt;p&gt;One solution is to wait until midnight. By definition, this is when the Sun is at its nadir, so any celestial object at its upper &lt;a href=&quot;https://en.wikipedia.org/wiki/Culmination&quot; rel=&quot;nofollow noreferrer&quot;&gt;culmination&lt;/a&gt; (i.e. its highest point in the sky in a given 24-hour period) &lt;em&gt;at midnight&lt;/em&gt; must be exactly on the other side of the Earth from the Sun (i.e. in opposition to the Sun).&lt;/p&gt;&#xA;&lt;p&gt;The two problems with this method are that (a) you need a reasonably accurate way to measure the time so that you know when it's actually midnight, and (b) you need to be up and about at midnight. Well, not insuperable difficulties, even for the ancients....&lt;/p&gt;&#xA;&lt;p&gt;If you're able to measure time relatively well, there's a more complex solution based on &lt;em&gt;rising or setting&lt;/em&gt; rather than culmination. It involves plotting the movements of the celestial bodies along the ecliptic (in simple terms, the path the Sun takes across the sky). The ecliptic intersects &lt;em&gt;exactly&lt;/em&gt; with due East and due West twice a year - at the equinoxes - but the ancient astronomers had a sophisticated understanding of its movement north or south at other times of the year. By noting how long after sunset it takes for a planet to rise, it's possible to work out the corresponding location of the Sun on the &amp;quot;great circle&amp;quot;, and therefore what angle the two bodies form. If it's 180º, then they are in opposition.&lt;/p&gt;&#xA;&lt;p&gt;I should add that the most obvious opposition is the &lt;strong&gt;Full Moon&lt;/strong&gt;, the synodic period of which is 29.5 days (i.e. from one Full Moon to the next), as opposed to the sidereal period of 27.3 days (i.e. how long it takes for the Moon to return to the same position relative to the background stars).&lt;/p&gt;&#xA;&lt;p&gt;This image shows how it works with the New Moon, but the principle is the same:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ynaxN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ynaxN.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://community.dur.ac.uk/john.lucey/users/lunar_sid_syn.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source (and explanation)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14094" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-10-29T11:26:37.737" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28197" PostTypeId="1" CreationDate="2018-10-29T14:04:26.893" Score="2" ViewCount="86" Body="&lt;p&gt;Suppose there is a planet made of pure tungsten (a quite common and heavy element), with perfect spherical shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which should be its radius so that the air pressure and ground/sea level is the same as on earth ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if it is sufficient to have a 1G gravity, because the smaller radius could affect the density of the atmosphere as the gravity will decrease differently that from earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For sake of simplificy, say that this planet is on the same orbit as Earth, orbiting around a twin of our Sun. Also magnetic field and atmosphere composition should be comparable to the Earth's one (altough this is not realistic, due to the lack of body of waters).&lt;/p&gt;&#xA;" OwnerUserId="24675" LastActivityDate="2018-10-29T15:32:03.340" Title="What is the smallest possible planet with same sea level atmospheric pressure as Earth?" Tags="&lt;gravity&gt;&lt;planetary-atmosphere&gt;&lt;earth-like-planet&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2018-10-31T11:19:00.153" ContentLicense="CC BY-SA 4.0" />
  <row Id="28199" PostTypeId="1" CreationDate="2018-10-29T14:46:10.487" Score="2" ViewCount="108" Body="&lt;p&gt;Dust along the ecliptic scatters light, causing “light pollution” throughout the orbital plane of the solar system. If an object like, say, the sun's corona, had its brightness measured from above the ecliptic, would the measurement reflect that?&lt;/p&gt;&#xA;" OwnerDisplayName="user24677" LastEditorUserId="7982" LastEditDate="2018-12-15T15:10:41.413" LastActivityDate="2018-12-15T15:10:41.413" Title="Light pollution and apparent magnitude of objects in solar system" Tags="&lt;star&gt;&lt;light&gt;&lt;apparent-magnitude&gt;&lt;light-pollution&gt;&lt;zodiacal-light&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28200" PostTypeId="2" ParentId="28197" CreationDate="2018-10-29T15:32:03.340" Score="4" Body="&lt;p&gt;In the textbook treatment the pressure decreases exponentially with height as &lt;span class=&quot;math-container&quot;&gt;$P(z)=P(0)\exp(-z/H)$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$H=kT/mg$&lt;/span&gt; is the scale height of the atmosphere. This assumes that the gravity &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; is constant with altitude, which is a decent approximation for planets with an atmospheric height much smaller than their radius. Since &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; is about 8 km for Earth, this makes sense to use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an atmosphere with Earth-like temperature &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; and mean molecular mass &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; it would hence seem that for the tungsten planet that if we want &lt;span class=&quot;math-container&quot;&gt;$P(0)$&lt;/span&gt; to be like on Earth &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; must be equal to Earth's 9.82 m/s&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt;. So if we have a planet with &lt;span class=&quot;math-container&quot;&gt;$g=G\rho (4\pi r^3/3)/r^2$&lt;/span&gt;, we get &lt;span class=&quot;math-container&quot;&gt;$r=3g/4\pi G\rho$&lt;/span&gt;. For &lt;span class=&quot;math-container&quot;&gt;$\rho=19300$&lt;/span&gt; kg/m&lt;span class=&quot;math-container&quot;&gt;$^3$&lt;/span&gt; this gives &lt;span class=&quot;math-container&quot;&gt;$r=1820$&lt;/span&gt; km. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limit is met somewhere before white dwarf density. They have a density of about a billion kg/m&lt;span class=&quot;math-container&quot;&gt;$^3$&lt;/span&gt; and a 35 meter radius white dwarf would have terrestrial surface gravity. At this point the above approximation has broken down (the scale height is bigger than the radius), and besides the low gravity would not be enough to hold it together. The real determinant of the limit will be the highest density material you can get that is stable at (near) zero pressure. For molecular matter tungsten is likely close to the limit, and we have reason to doubt degenerate white dwarf matter is up to the job. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that small but less dense bodies can have quite dense atmospheres. Titan has radius about 2550 km and a surface pressure 1.48 times Earth, despite lower surface gravity. The trick here is that the atmosphere is colder and slightly denser. &lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2018-10-29T15:32:03.340" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28201" PostTypeId="1" AcceptedAnswerId="28203" CreationDate="2018-10-29T18:07:25.527" Score="-2" ViewCount="388" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Rtzgh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rtzgh.jpg&quot; alt=&quot;Blood Moon Jul 2018&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't undertand why the moon can be that big a few hundred miles sout of where I am in London...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And when I went to Greece, even Africa, the moon wasn't so large...&lt;/p&gt;&#xA;" OwnerUserId="21644" LastActivityDate="2018-10-29T20:17:43.240" Title="Why is the moon so large in this picture from Athens" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28202" PostTypeId="1" AcceptedAnswerId="28204" CreationDate="2018-10-29T19:09:00.790" Score="2" ViewCount="354" Body="&lt;p&gt;I am experimenting with the Python library PyEphem for astronomy, and I am trying to recreate/understand some of the basic calculations there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a function called separation that, given two planets and a date/time, it calculates the separation (angle) between those 2 planets with respect to their &quot;x&quot; projection in a plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for 2018/1/1 for planets Mercury and Mars, we have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import ephem&#xA;import math&#xA;&#xA;mercury = ephem.Mercury('2018/1/1')&#xA;mars = ephem.Mars('2018/1/1')&#xA;&#xA;s1 = ephem.separation(mercury, mars)&#xA;print(math.degrees(s1))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which returns&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;33.792384499568264&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But if I wanted to calculate this without the separation function, then I think that the calculation would be as simple as the &quot;right ascension&quot; of 1 minus the &quot;right ascension&quot; of the other:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;math.degrees(mercury.ra) - math.degrees(mars.ra)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which returns&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;35.114532008671574&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Why are the angles different? Since I am not including a latitude and longitude of the observer, all the calculations are supposed to be geo-centric, according to PyEphem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is anybody familiar with the calculations going on behind PyEphem, or another library with built-in ephemeris that can produce consistent results for separation?&lt;/p&gt;&#xA;" OwnerUserId="24682" LastEditorUserId="24682" LastEditDate="2018-10-29T20:12:43.623" LastActivityDate="2018-10-29T20:35:36.363" Title="Calculating right ascension for two planets and their separation using python" Tags="&lt;python&gt;&lt;ephemeris&gt;&lt;right-ascension&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28203" PostTypeId="2" ParentId="28201" CreationDate="2018-10-29T19:49:44.357" Score="4" Body="&lt;p&gt;The moon is the same size in Athens as in London.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To take a photo like that you go a long way from the Greek temple, so the temple appears to be smaller than your little-fingernail with your arm stretched out.  You then get a powerful telephoto lens to zoom in and wait for the moon to rise. If you have done your calculations correctly you have positioned yourself so that the moon appears right behind the temple. If you were standing there, the moon and the temple would both be small and the temple would be in the distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is the temple of Poseidon, which is about 10m along the side, we can estimate that the photographer must have been standing about 1km away to get this shot, perhaps &lt;a href=&quot;https://www.google.com/maps/@37.6559701,24.0125648,3a,60y,136.04h,90.87t/data=!3m6!1e1!3m4!1s7OBMAOaEAPrV8So1GwlAZw!2e0!7i13312!8i6656?hl=en&quot; rel=&quot;nofollow noreferrer&quot;&gt;standing by the road across the bay&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get similar photos in London. You just need a viewpoint from which you can see the buildings of the city and is in the right position (direction and distance) for the moon to rise (or set) in just the right place. &lt;a href=&quot;https://fstoppers.com/spotlight/london-cityscape-photographer-who-caught-super-blue-blood-moon-219149&quot; rel=&quot;nofollow noreferrer&quot;&gt;Michael Tomas captured such a picture.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-10-29T20:17:43.240" LastActivityDate="2018-10-29T20:17:43.240" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28204" PostTypeId="2" ParentId="28202" CreationDate="2018-10-29T20:35:36.363" Score="2" Body="&lt;p&gt;The issue appears to be in your understanding of the &lt;code&gt;separation&lt;/code&gt; function. You state that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;it calculates the separation (angle) between those 2 planets with respect to their &quot;x&quot; projection in a plane.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;however, &lt;a href=&quot;https://rhodesmill.org/pyephem/tutorial.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;the documentation states&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The &lt;code&gt;separation()&lt;/code&gt; function computes the angle in degrees between two bodies, &lt;strong&gt;as measured by their right ascension and declination.&lt;/strong&gt; &lt;br/&gt;&lt;br/&gt;In general &lt;code&gt;separation()&lt;/code&gt; &lt;strong&gt;can measure the angle between any pair of spherical coordinates&lt;/strong&gt;, so long as the elements of each coordinate are spherical longitude (angle around the sphere) followed by spherical latitude (angle above or below its equator). Each pair should be provided as a two-item sequence like a tuple or list. Appropriate coordinate pairs include right ascension and declination; heliocentric longitude and latitude; azimuth and altitude; and even the geographic longitude and latitude of two locations on earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This means that this function is calculating the angular separation between the two objects, not just the separation in a particular plane. If you want to perform the same calculation (as the documentation says they're performing), it looks like you actually want the general &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_distance#Equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;angular distance formula&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\theta = \cos^{-1}\left[\sin(\delta_1)sin(\delta_2)+\cos(\delta_1)\cos(\delta_2)\cos(\alpha_1-\alpha_2)\right]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\delta\in[-\pi/2,\pi/2]$&lt;/span&gt; is the declination, and &lt;span class=&quot;math-container&quot;&gt;$\alpha\in[0,2\pi]$&lt;/span&gt; is the Right Ascension.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2018-10-29T20:35:36.363" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28205" PostTypeId="2" ParentId="2247" CreationDate="2018-10-29T22:24:12.080" Score="2" Body="&lt;p&gt;I'd argue it isn't a total coincidence, but it isn't artificial either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out of possible arrangements allowing for stable orbits, what positions of the moon on the Sun-Moon-Earth line offer near equal angular measures?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$l_s=150*10^6 km=1AU$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$r_s=695,508km$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$r_m=1,737km$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$l_m=384,400km$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$l_s$&lt;/span&gt; is distance from earth to the sun, &lt;span class=&quot;math-container&quot;&gt;$r_s$&lt;/span&gt; is radius of the sun, &lt;span class=&quot;math-container&quot;&gt;$r_m$&lt;/span&gt; is radius of moon. &lt;span class=&quot;math-container&quot;&gt;$l_m$&lt;/span&gt; is distance from earth to moon, hear assumed to be along the straight line running from earth to sun. &lt;span class=&quot;math-container&quot;&gt;$l_m$&lt;/span&gt; will be variable, but empirically on average has the value given above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For what values of &lt;span class=&quot;math-container&quot;&gt;$l_m$&lt;/span&gt; does &lt;span class=&quot;math-container&quot;&gt;$r_m/l_m$&lt;/span&gt; , the tangent of the angular radius of the moon, fall within 10% of &lt;span class=&quot;math-container&quot;&gt;$r_s/l_s$&lt;/span&gt;, the tangent of the angular radius of the sun? i.e.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$(0.9)(r_s/l_s)&amp;lt;(r_m/l_m)&amp;lt;(1.1)(r_s/l_s)$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$0&amp;lt;\Delta(r_m/l_m)&amp;lt;(0.2)(r_s/l_s)$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$0&amp;lt;\frac{r_m}{l_m^2}\Delta l_m&amp;lt;(0.2)(r_s/l_s)$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$0&amp;lt;\Delta l_m&amp;lt;(0.2)\frac{r_sl_m^2}{(r_ml_s)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if &lt;span class=&quot;math-container&quot;&gt;$(r_m/l_m\approx r_s/l_s)$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$l_m$&lt;/span&gt; can vary by &lt;span class=&quot;math-container&quot;&gt;$20\%$&lt;/span&gt;  of its current value and still maintain an approximately equal angular radius- a Goldilock's Zone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few back of the envelope methods to gauge how far the moon can get from earth to remain in a stable orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How far can the moon get without escaping from the earth? The &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hill Sphere&lt;/a&gt; (h/t uhoh!)of the Earth has a radius of about &lt;span class=&quot;math-container&quot;&gt;$(0.01)l_s$&lt;/span&gt;. Further away from there and the sun perturbs the moon away from earth, establishing a maximum distance for our allowable range. The Roche Limit is a lower bound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas the probability of the angular radii being about equal if allowing the moon to be any place on the straight line joining the earth and the moon is approximately &lt;span class=&quot;math-container&quot;&gt;$\frac{(0.2)l_m}{l_s}=0.051253\%$&lt;/span&gt;, factoring in that the moon can only be just so far from the earth without decoupling, that probability goes up to at least &lt;span class=&quot;math-container&quot;&gt;$\frac{(20.0)l_m}{l_s}=5.1253\%$&lt;/span&gt; by the above arguments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more thorough approach uses tighter bounds. There's a Goldilock's Zone for the moon and sun to have the same angular radius. The probability of falling within that Goldilock's Zone is the ratio of the Goldilock's Zone &lt;span class=&quot;math-container&quot;&gt;$(0.2l_m)$&lt;/span&gt; to the interval  between the minimum and maximum distances from the earth to the moon factoring in orbital stability, the moon's recession,  and eventual return closer to earth. From answer above, the moon is 60.3 Earth Radii from Earth and will not get further away than 75 Earth Radii. The Goldilock's Zone spans from about 38 Earth Radii to about 72 earth radii.  So between now and maximum recession of the moon, and assuming uniform time spent at each distance, the angular radii will match to within 10%  about 75% of the time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall its a low probability that the angular radii match so well. Given the distance the Earth lies from the sun and the radius of the moon which are fairly arbitrary, gravity allows the above mentioned Goldilock's Zone to have significant overlap with realistic lunar trajectories. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put another way, if the Equal Angular Goldilock's Zone has sufficient overlap with allowable lunar orbits, then the angular match becomes about as likely as having a moon in the first place which itself is a coincidence. &lt;/p&gt;&#xA;" OwnerUserId="24685" LastEditorUserId="24685" LastEditDate="2018-10-31T18:25:38.360" LastActivityDate="2018-10-31T18:25:38.360" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28206" PostTypeId="2" ParentId="28181" CreationDate="2018-10-29T22:45:49.073" Score="4" Body="&lt;p&gt;The average distance &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; between the Earth and the Moon is about 385000 km. The Earth's mean radius &lt;span class=&quot;math-container&quot;&gt;$r_E$&lt;/span&gt; is 6371 km while the Moon's mean radius &lt;span class=&quot;math-container&quot;&gt;$r_M$&lt;/span&gt; is 1737 km. This means that when the Earth is seen directly overhead from the Moon, it subtends an angle of &lt;span class=&quot;math-container&quot;&gt;$2\arcsin\left(\frac{r_E}{D-r_M}\right)$&lt;/span&gt;, or 1.9°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good rule of thumb is that the &quot;&lt;a href=&quot;https://www.ncbi.nlm.nih.gov/pubmed/1762884&quot; rel=&quot;nofollow noreferrer&quot;&gt;visual angle of the width of the thumb held at arm's length is about 2 degrees&lt;/a&gt;.&quot; (Specifically, it's 2.12°±0.26° at the joint per the referenced paper.) A typical person could just barely blot out the Earth as seen from the Moon with a naked (not gloved) thumb held at arms length. The angular size of a thumb at arm's length varies from person to person; from the referenced paper the visual size of a thumb held at arm's length varied between 1.23° and 3.20° for the 112 tested subjects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thumb at arm's length is 2.45° wide at the middle of the thumbnail, so I could easily blot out the Earth from the surface of the Moon with my thumb at arm's length.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2018-10-29T22:45:49.073" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28207" PostTypeId="1" AcceptedAnswerId="28209" CreationDate="2018-10-30T03:16:42.687" Score="3" ViewCount="294" Body="&lt;p&gt;When I look at the Alnitak,the left most star of Orion's belt, it is 736 light years away. How could I convert this distance to an estimate of how long ago what I am seeing happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be simply be 736 years ago?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there instances of when this isn't as straight forward?&lt;/p&gt;&#xA;" OwnerUserId="4208" LastActivityDate="2018-10-30T12:02:28.317" Title="How is distance in light-years converted to years back in time?" Tags="&lt;distances&gt;&lt;space-time&gt;&lt;time&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28209" PostTypeId="2" ParentId="28207" CreationDate="2018-10-30T04:39:01.480" Score="7" Body="&lt;p&gt;Yes, for stars in our galaxy, if a star is 736 light years away, the light took 736 years to reach here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For very very distant objects, you need to account for the expansion of space. The light might have been travelling for 10 billion years, but in that time space has expanded and so the distance that you would find if you froze time and measured would be greater (to get the proper distance at the current time) it would be larger.  This effect is not significant at distances less than about a billion light years.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-10-30T04:39:01.480" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28212" PostTypeId="1" CreationDate="2018-10-30T07:21:06.210" Score="1" ViewCount="148" Body="&lt;p&gt;What is the reason of Magnetic Braking in contact binary stars?&lt;br&gt;&#xA;Just dark spots?&lt;br&gt;&#xA;I mean hot spots can have this effect on binaries.&lt;/p&gt;&#xA;" OwnerUserId="24692" LastEditorUserId="7472" LastEditDate="2018-11-04T18:42:29.457" LastActivityDate="2018-11-04T18:42:29.457" Title="What is the reason for magnetic braking?" Tags="&lt;astrophysics&gt;&lt;binary-star&gt;&lt;magnetic-field&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28214" PostTypeId="1" AcceptedAnswerId="28225" CreationDate="2018-10-30T09:51:58.003" Score="10" ViewCount="288" Body="&lt;p&gt;When we look at photo's of the centre of the Milky Way, it looks extremely bright due to the concentraion of star systems...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If our solar system was only 100 light years from the centre of the Galaxy, would we still have a dark night sky?&lt;/p&gt;&#xA;" OwnerUserId="21644" LastActivityDate="2018-11-06T01:43:11.720" Title="What would our night sky look like if the solar system was 100 light years from the centre of the Milky Way" Tags="&lt;milky-way&gt;&lt;night-sky&gt;&lt;globular-clusters&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28215" PostTypeId="2" ParentId="28207" CreationDate="2018-10-30T12:02:28.317" Score="3" Body="&lt;p&gt;Actually, the common standard is that whenever distances are reported in light years, they are actually lookback times and not distances at all.  There is no difference in our galaxy, but at much larger distances where the metric expansion of the universe comes into play, it is almost always lookback times that are reported in the popular press, rarely comoving distances (even though &quot;light year&quot; is a distance, not a time).  If they don't say &quot;comoving distance,&quot; and they generally don't, they mean lookback time!  Apparently a decision was made long ago that this distinction would only confuse people, so ingrained are we in the idea of a static universe.  Here's a random example, at &lt;a href=&quot;https://www.eso.org/public/usa/news/eso9304/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.eso.org/public/usa/news/eso9304/&lt;/a&gt;  : &quot; This corresponds to a look-back time of 83% of the age of the Universe, i.e. we see them as they were when the Universe was less than one-fifth as old as it is now. Assuming the age of the Universe to be 20,000 million years, the distance to this quasar would be about 16,000 million light-years.&quot;  Don't ask me why they also decided to &quot;dumb down&quot; the numbers themselves!  It's pretty awful what article writers think the public can understand.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-10-30T12:02:28.317" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28216" PostTypeId="2" ParentId="28188" CreationDate="2018-10-30T14:31:29.963" Score="7" Body="&lt;p&gt;Assuming that Neil deGrasse Tyson is referring to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Event_horizon#Cosmic_event_horizon&quot; rel=&quot;nofollow noreferrer&quot;&gt;cosmic event horizon&lt;/a&gt;, he's right. But that doesn't really mean that galaxies &quot;fall out of sight&quot;. He's right that they will disappear, but not because they leave the horizon.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Nearby galaxies don't disappear&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;First of all, there's a small thing to note that I'm sure he just left out to simplify matter a bit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Universe expands and drags galaxies along, but for galaxies that are sufficiently close, their mutual gravitational attraction keeps them &quot;locked&quot;. Hence, we will always be able to see the galaxies in our &lt;a href=&quot;https://en.wikipedia.org/wiki/Local_Group&quot; rel=&quot;nofollow noreferrer&quot;&gt;Local Group&lt;/a&gt;. In the case of Andromeda, we will even end up merging, and I suppose that after a sufficiently long time, we'll merge with the other galaxies in our group, simply because we will keep orbiting around each other, so at some point we'll come so close so as to merge.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Galaxies don't leave the observable Universe…&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Second, and more importantly, &lt;em&gt;galaxies do not leave our observable Universe&lt;/em&gt;. The observable Universe consists of the region of the Universe from which light has had the time to reach us since Big Bang, and the &quot;boundary&quot; between the observable and the un-observable Universe is called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Particle_horizon&quot; rel=&quot;nofollow noreferrer&quot;&gt;particle horizon&lt;/a&gt;. As time goes, the light from an ever-increasing number of galaxies will have had the time to reach us. This number will never &lt;em&gt;de&lt;/em&gt;crease. Once inside our observable Universe, a galaxy will stay inside. &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;…but fewer and fewer galaxies will be able to send a light signal to us…&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;However, the distance at which galaxies may emit a light signal &lt;em&gt;today&lt;/em&gt;, that we may receive some day in the future, only asymptotically increases to a finite value (of some 17 billion lightyears; see e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2004PASA...21...97D&quot; rel=&quot;nofollow noreferrer&quot;&gt;Davis &amp;amp; Lineweaver 2004&lt;/a&gt;), while the galaxies themselve continue to increase their distance. I suppose this is what Neil deGrasse Tyson refers to: Even though the galaxies we see don't vanish from sight, we will never see what's going on in them today.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;…and in practice, even the ones we see now &lt;em&gt;will&lt;/em&gt; disappear…&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Moreover, even the galaxies that we do see will become increasingly redshifted until eventually all light from them will be in the radio regime. Furthermore, photons will arrive at a smaller and smaller rate, until for all practical purposes, they will be invisible to us.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;…especially if &lt;span class=&quot;math-container&quot;&gt;$w&amp;gt;-1$&lt;/span&gt;&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The above considerations hold true if the Universe expands, and accelerates, as we think it does. However, there is a possibility that the acceleration itself will increase in magnitude. If the equation of state &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt; of the dark energy thought to be responsible for the acceleration is smaller than &lt;span class=&quot;math-container&quot;&gt;$-1$&lt;/span&gt;, the distance to the point where the Universe expands faster than the speed of light may decrease indefinitely, such that galaxies &lt;em&gt;will&lt;/em&gt; indeed leave the horizon. Eventually, even gravitationally bound galaxies will be separated, stars will be torn out of the galaxies, planets will be pulled from the suns and finally disrupted, culminating in a &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Rip&quot; rel=&quot;nofollow noreferrer&quot;&gt;Big Rip&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-10-30T21:50:31.697" LastActivityDate="2018-10-30T21:50:31.697" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28218" PostTypeId="2" ParentId="28199" CreationDate="2018-10-30T17:45:45.960" Score="3" Body="&lt;p&gt;The answer is yes: dust is highly constrained to the ecliptic plane and so viewing from above the plane would result in less light that needs to be subtracted from observations of the solar corona.  The zodiacal light is quite strong in the direction of the sun because the dust strongly scatters in the forward direction. Near the Sun, its surface brightness is somewhere around &lt;a href=&quot;https://en.wikipedia.org/wiki/Sky_brightness&quot; rel=&quot;nofollow noreferrer&quot;&gt;21 - 23 mag/arcsec&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt;&lt;/a&gt; from our viewpoint in the ecliptic, and this would be an additive contamination to any measurement of the Sun's corona.  From the ecliptic pole, this light foreground would be much less.  Of course, a careful observer would know to also measure the zodiacal light contribution and to subtract it off, so both observers would get the same brightness for the corona roughly.  The observer at the pole would have lower noise for a comparable instrument and observing time.  &lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2018-10-30T17:45:45.960" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28219" PostTypeId="2" ParentId="28212" CreationDate="2018-10-30T18:10:52.430" Score="4" Body="&lt;p&gt;The magnetic braking in contact binary stars AFAIK is assumed to be similar to the magnetic braking mechanism that spins down the rotation of single stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A stellar wind couples to the large scale magnetic field and maintains this coupling out to the Alfven radius (where the thermal pressure and magnetic pressure are roughly equal). At this point the wind decouples, stops co-rotating with the star and hence removes a certain amount of specific angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation will be more complicated in a binary system (entangled, asymmetric fields etc.) but the basic principle will be the same. Many contact binaries have fast rotation and strong, dynamo-generated magnetic fields.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-10-30T18:10:52.430" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28221" PostTypeId="1" CreationDate="2018-10-30T20:06:55.433" Score="1" ViewCount="68" Body="&lt;p&gt;I have just read on this site that there is a Newtonian solution to Einstein's field equation but that it is spherically symmetrical and therefore not isotropic (having a centre for example) and so considered not allowed. But if the universe is a 3-sphere it could be both spherically symmetric and isotropic. Would the Newtonian solution then be a possibility?&lt;/p&gt;&#xA;" OwnerUserId="24703" LastActivityDate="2018-10-30T20:06:55.433" Title="3-sphere universe" Tags="&lt;universe&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2018-11-02T11:28:43.270" ContentLicense="CC BY-SA 4.0" />
  <row Id="28222" PostTypeId="1" AcceptedAnswerId="28236" CreationDate="2018-10-31T02:31:47.593" Score="6" ViewCount="90" Body="&lt;p&gt;I was trying to compose an answer to &lt;a href=&quot;https://space.stackexchange.com/questions/22115/is-it-possible-to-touch-stars-with-our-bare-hands&quot;&gt;this question here&lt;/a&gt;, about touching the surface of a star, and I was going to mention that the popular depiction of the sun's surface as 'basically lava' (as depicted in &lt;a href=&quot;https://www.youtube.com/watch?v=TBzVPXr8SNY&quot; rel=&quot;noreferrer&quot;&gt;Sunshine&lt;/a&gt;, for example) is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual definition of a star's radius is where &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_radius&quot; rel=&quot;noreferrer&quot;&gt;the optical depth is 2/3.&lt;/a&gt; At this radius, the density of the sun is about 1/10th of air at sea level and can hardly be described as solid or liquid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to get a sense of what this diffuse plasma would look like if you were in it and somehow impervious to the heat and radiation. Would it appear as a thin, glowing fog, through which you could still discern distant objects? Or would it appear as opaque as, say, a bonfire?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rephrasing in a little more technical way, &lt;strong&gt;What is the extinction coefficient of solar plasma at ~1 Solar radius in the visible spectrum? &#xA;What is the path length at which all light is effectively attenuated?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Since it varies by wavelength, I mean specifically the region of the visible spectrum where the plasma is at its most transparent).&lt;/p&gt;&#xA;" OwnerUserId="17655" LastActivityDate="2018-11-01T09:33:37.647" Title="How far can you see in solar plasma at just under 1 solar radius?" Tags="&lt;the-sun&gt;&lt;plasma-physics&gt;&lt;opacity&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28223" PostTypeId="1" AcceptedAnswerId="28227" CreationDate="2018-10-31T03:02:17.583" Score="35" ViewCount="6610" Body="&lt;p&gt;I think it's probably unlikely that there are more planets between Mercury and Mars, but out from Jupiter, there's lots of empty space between the planets. Could there be some small planet hidden out there?&lt;/p&gt;&#xA;" OwnerUserId="21521" LastActivityDate="2018-11-01T04:01:14.293" Title="Considering our methods of exploration, how likely is it that there are unfound planets (not dwarf planets) in our solar system?" Tags="&lt;planet&gt;&lt;solar-system&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28224" PostTypeId="2" ParentId="26883" CreationDate="2018-10-31T03:42:53.103" Score="2" Body="&lt;p&gt;The simplest answer might be that mass/energy have always existed.  For field-based inflation, their past eternality is prohibited by the Borde-Guth-Vilenkin theorem, but, as that theorem only prohibits it in a universe that is expanding &quot;on average&quot;, they appear compatible with some of the bouncing cosmologies developed more recently, of which the simplest that's compatible with CMB data may be Poplawski's entirely relativistic version (using the fermions of the Einstein-Cartan version of GR, which have spatial extent) of the &quot;baby universe&quot; scenario, in which the gravitational collapse of any large rotating star includes tidal effects which initially separate the particles and antiparticles of virtual pairs, that are subsequently blown outward by energy until then tied up in particulate spin, when the collapsing star's fermions lock up against each other at extremely high densities.  The &quot;parenting&quot; universe (which is on vastly larger scales of space and time) is analogous to the 3-D skin of a basketball, and is causally separated by the inertia of its own continuing expansion, as it would've begun in an earlier (and larger-scaled) version of the same process.  Poplawski feels this cosmology to require a prevalent direction of rotation, so it is falsifiable and consequently scientific, however marginal that prevalence may be.  (My reading of physics notation is very limited, so I've relied on his English verbiage for some of the inferences I've made here.)&lt;/p&gt;&#xA;" OwnerUserId="22115" LastActivityDate="2018-10-31T03:42:53.103" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28225" PostTypeId="2" ParentId="28214" CreationDate="2018-10-31T03:54:17.867" Score="8" Body="&lt;p&gt;I have tried to collate some info from various other answers.&lt;/p&gt;&#xA;&lt;p&gt;Density of stars where we are:&lt;/p&gt;&#xA;&lt;h1&gt;About 0.15 per cubic parsec&lt;/h1&gt;&#xA;&lt;p&gt;Density of stars &amp;quot;at the middle of the galaxy&amp;quot;.  Unfortunately I don't know how &amp;quot;100 ly&amp;quot; affects this. It's probably about this figure.&lt;/p&gt;&#xA;&lt;h1&gt;About 50-100 stars per cubic parsec&lt;/h1&gt;&#xA;&lt;p&gt;So in this QA it's &amp;quot;about 500x as dense&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;It would seem that this is in fact:&lt;/p&gt;&#xA;&lt;h1&gt;&lt;em&gt;Not very bright.&lt;/em&gt; About the same amount of light overall as when there's a full moon.&lt;/h1&gt;&#xA;&lt;p&gt;However the overwhelming factor is&lt;/p&gt;&#xA;&lt;h1&gt;If one particular star was &amp;quot;very close&amp;quot; to us?&lt;/h1&gt;&#xA;&lt;p&gt;How &lt;em&gt;&lt;strong&gt;likely would it be&lt;/strong&gt;&lt;/em&gt;, in the region under discussion, that another star is &amp;quot;very&amp;quot; close to us - close enough that that one particular star is extremely bright?&lt;/p&gt;&#xA;&lt;p&gt;The answer seems to be&lt;/p&gt;&#xA;&lt;h1&gt;&lt;em&gt;No&lt;/em&gt;, the stars there are not so close that you'd typically be close enough to a really bright star that it would be bright.&lt;/h1&gt;&#xA;&lt;p&gt;That seems to be the bottom line.&lt;/p&gt;&#xA;&lt;p&gt;(There's a completely pointless, if pretty, image here showing what it's like &amp;quot;inside a star cluster&amp;quot; {where? At what depth?} &lt;a href=&quot;https://astronomy.stackexchange.com/a/3681/13071&quot;&gt;here&lt;/a&gt;  Unfortunately that image does not simply state the star density they are simulating - so it is of no help at all to this excellent QA!)&lt;/p&gt;&#xA;&lt;p&gt;the bottom line then:&lt;/p&gt;&#xA;&lt;h1&gt;&lt;em&gt;&amp;quot;If our solar system was only 100 light years from the centre of the Galaxy, would we still have a dark night sky?&amp;quot;&lt;/em&gt;  Answer, yes, there would just be &amp;quot;more stars&amp;quot;. It would be hardly any brighter.&lt;/h1&gt;&#xA;&lt;p&gt;To improve on this answer, the next thing we need to quantify is one piece of information:&lt;/p&gt;&#xA;&lt;h1&gt;What actually is the star density at &amp;quot;100 ly from the center&amp;quot;?&lt;/h1&gt;&#xA;&lt;p&gt;Is it 50, 100, 25, 300, 72.275 stars per cubic parsec?&lt;/p&gt;&#xA;&lt;p&gt;I don't know.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A fun point, would it be different (slightly? a lot?) if inside the galactic bar?  IDK.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-11-06T01:43:11.720" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28226" PostTypeId="1" AcceptedAnswerId="28823" CreationDate="2018-10-31T04:00:08.033" Score="14" ViewCount="214" Body="&lt;p&gt;How much denser is it in the galactic bar than the &quot;normal&quot; density at the same radius?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it just a few percent? or is it, say, &quot;three times&quot; as dense?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or are other factors at play: star brightness, gasses?&lt;br /&gt;&#xA;Or do we really not know?&lt;/p&gt;&#xA;" OwnerUserId="13071" LastEditorDisplayName="user1569" LastEditDate="2018-12-20T16:02:01.343" LastActivityDate="2018-12-21T10:26:13.693" Title="How much greater is the star density in our galactic bar?" Tags="&lt;galaxy&gt;&lt;galactic-center&gt;" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="28227" PostTypeId="2" ParentId="28223" CreationDate="2018-10-31T04:16:38.523" Score="72" Body="&lt;h1&gt;There are no undiscovered planets between the sun and Neptune.&lt;/h1&gt;&#xA;&lt;p&gt;Objects closer (to Sun) than Neptune that are large enough to be considered planets (and not dwarf planets) can't remain 'hidden'. If it's there, the light from the sun will bounce off it and we will see it. As it moves in its orbit, we will notice the position in the sky change, so we will know it isn't a star.&lt;/p&gt;&#xA;&lt;p&gt;I would like to give a more broad answer to this question though:&#xA;&lt;strong&gt;What is the maximum size of an undiscovered solar system object and how does it change as you get further from the sun?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The further out a solar system object is, the harder it is to detect. The rate at which it gets harder is severe; the light we receive from an object scales roughly as &lt;span class=&quot;math-container&quot;&gt;$1/r^4 $&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;(&lt;span class=&quot;math-container&quot;&gt;$1/r^2$&lt;/span&gt; for the light travelling from the Sun to the object, and again, &lt;span class=&quot;math-container&quot;&gt;$1/r^2$&lt;/span&gt; for the light travelling from the object to us on earth).&lt;/p&gt;&#xA;&lt;p&gt;We are able to detect some very small earth-crossing asteroids. Some as small as ~50 metres across. At a guess (and this is just based on my intuition, not any calculations), there are probably no undiscovered objects larger than 1 km close to earth.&lt;/p&gt;&#xA;&lt;p&gt;As you travel to the outer solar system (Jupiter to Neptune), the number of bodies increases dramatically. There are currently ~700,000 known solar system bodies and most of them occur in this area. It is believed that all asteroids larger than 10 km have been found.&lt;/p&gt;&#xA;&lt;p&gt;In the area immediately beyond Neptune (30 AU to 100 AU), we have been finding many pluto-like objects in the past two decades, objects with diameters of 500 km or more. For reference, Pluto is about 2200 km across. It is entirely plausible that there are some, if not many, similar objects at this distance range that have not yet been found. Some of these would classify as dwarf planets, in that they are round, orbit the sun, but have not cleared 95% of their orbit of other matter.&lt;/p&gt;&#xA;&lt;p&gt;And finally &lt;em&gt;- and this is where it gets exciting -&lt;/em&gt; where it gets really far out, &lt;a href=&quot;https://en.wikipedia.org/wiki/Planet_Nine&quot; rel=&quot;noreferrer&quot;&gt;there may be an undiscovered large planet&lt;/a&gt;, out at ~700 AU. Referred to as Planet Nine, this is a hypothetical object that some (noteworthy) astronomers believe exist because of patterns they see in the orbits of other distant dwarf planets. From calculations, they estimate that it would be as heavy as 10 earths and travel between 200 and 1200 AU. However, these distances are so large and the sunlight out there is so dim, that even with the best telescopes and two years of looking, they haven't found it.&lt;/p&gt;&#xA;&lt;p&gt;Finally, I'd like to share two graphs. The first one is of distance (of closest approach, or perihelion) vs diameter of various outer solar system bodies. In the bottom right corner there is a noticeable lack of dots, indicating roughly where we don't yet have the capability of seeing objects that small and distant.&lt;/p&gt;&#xA;&lt;p&gt;The second is an expanded but slightly less accurate diagram of the same.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/MInnH.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MInnH.gif&quot; alt=&quot;TNO size versus perihelion distance graph&quot; /&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/r7VLh.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/r7VLh.png&quot; alt=&quot;xkcd comic about undiscovered planets&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Image 1 from &lt;a href=&quot;http://www.johnstonsarchive.net/astro/tnos.html&quot; rel=&quot;noreferrer&quot;&gt;Johnston's trans-Neptunian objects page&lt;/a&gt;. Image 2 from &lt;a href=&quot;https://xkcd.com/1633/&quot; rel=&quot;noreferrer&quot;&gt;xkcd&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17655" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-11-01T04:01:14.293" CommentCount="21" ContentLicense="CC BY-SA 4.0" />
  <row Id="28229" PostTypeId="2" ParentId="28222" CreationDate="2018-10-31T17:45:48.057" Score="-1" Body="&lt;p&gt;I suspect part of the problem here is that cosmological optical depth is defined vertically (so far as I can discern from the wikipedia article), while you seem to be asking for the extinction coefficient for a horizontal path at the specified star-radius.  So you'll need to find out what the &lt;span class=&quot;math-container&quot;&gt;$\alpha(z)$&lt;/span&gt; is at that radius, play some games with integrated radiance due to the plasma in the horizontal line of site to estimate the noise factor, and compare that with the putative radiance from the distant object you wish to see.  &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-10-31T17:45:48.057" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28230" PostTypeId="2" ParentId="19687" CreationDate="2018-10-31T19:16:06.480" Score="2" Body="&lt;p&gt;To see the difference between forward and inverse models, consider our understanding that an atom can absorb and emit only certain discrete wavelengths of light. This is what we &lt;em&gt;observe&lt;/em&gt;; we can build a simple (inverse) model of atomic structure based on these observations. But only after we had a well-developed  model of the atom, such as quantum theory, were we able to predict the absorption and emission of &lt;em&gt;any&lt;/em&gt; atom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forward modeling is based on these well-developed understandings and is generally the most useful form of modeling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, inverse models are important when we don't yet have a good understanding of a system; in that case, ad hoc models may ultimately lead us to develop entirely new models and understandings -- as was the case in understanding atoms and molecules before quantum theory was fully developed.&lt;/p&gt;&#xA;" OwnerUserId="24719" LastEditorUserId="14094" LastEditDate="2018-11-01T16:59:01.553" LastActivityDate="2018-11-01T16:59:01.553" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28231" PostTypeId="1" AcceptedAnswerId="28233" CreationDate="2018-10-31T20:23:23.180" Score="1" ViewCount="137" Body="&lt;p&gt;So there's a boatload of planets being discovered around stars in our solar neighborhood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler-22&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kepler 22&lt;/a&gt;, 638 ly away, was discovered to have a planet &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler-22b&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kepler 22b&lt;/a&gt;, calculated to be twice the size of Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The observations that make these measurements aren't high resolution images but periodic changes in luminosity (the transit method).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: would a similar situation with similar telescope properties (Kepler) detect our own Earth around our Sun from a similar distance? If not, then how close would you have to be to see Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably there is some correspondence among: distance, luminosity/size of a star, size of a planet. Any other factors that make Kepler-22 different from us?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/8605/if-alpha-centauri-as-solar-system-exactly-mirrored-our-own-what-would-we-be-ab&quot;&gt;this question asks about Alpha Centauri&lt;/a&gt;. I'm asking specifically about distances like Kepler-22.&lt;/p&gt;&#xA;" OwnerUserId="673" LastEditorUserId="673" LastEditDate="2018-11-01T12:45:02.387" LastActivityDate="2018-11-02T06:35:24.580" Title="Would you get the same planet-discovering data for our own system at the same distance we're getting from Earth?" Tags="&lt;exoplanet&gt;&lt;distances&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28232" PostTypeId="1" AcceptedAnswerId="28234" CreationDate="2018-10-31T20:44:36.333" Score="2" ViewCount="75" Body="&lt;p&gt;On Mars is there an area in which all the &lt;span class=&quot;math-container&quot;&gt;$CO_2$&lt;/span&gt; evaporates exposing dirt near the pole? How deep can solid &lt;span class=&quot;math-container&quot;&gt;$CO_2$&lt;/span&gt; accumulate during winter in that area during the Martian Winter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://space.stackexchange.com/questions/31724/solar-carbon-dioxide-collector-on-mars&quot;&gt;https://space.stackexchange.com/questions/31724/solar-carbon-dioxide-collector-on-mars&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastActivityDate="2018-10-31T21:41:07.237" Title="How thick or deep does solid carbon dioxide accumulate on Mars?" Tags="&lt;planetary-atmosphere&gt;&lt;mars&gt;&lt;weather&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28233" PostTypeId="2" ParentId="28231" CreationDate="2018-10-31T20:46:40.973" Score="6" Body="&lt;p&gt;There is a lot of detail in my answer to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/8605/if-alpha-centauri-as-solar-system-exactly-mirrored-our-own-what-would-we-be-ab&quot;&gt;If Alpha Centauri A&amp;#39;s solar system exactly mirrored our own, what would we be able to detect?&lt;/a&gt; I showed that current Doppler techniques and transit techniques would be unlikely to detect the Earth at the distance of Alpha Cen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not get any easier if you are further away. Stars are point sources at 1pc or 200pc, so these spatially unresolved techniques just care what signal level is being received - further away is worse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of Doppler techniques we need an improvement in radial velocity precision and stability of a factor of a few.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With transits, Kepler was &lt;em&gt;almost&lt;/em&gt; there. So another Kepler with a slightly bigger telescope that observed the Sun for say 5 years would do the trick (if the orbital inclination was about 90 degrees). The reason Kepler 22b was just detected is because it has twice the Earth's radius so eclipses 4 times the area and gives a deeper transit signal for a Sun-like star. It is this contrast that is the key to transit detectability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe the missing piece of information here is that there is a floor level to photometric precision that can't be bettered even if the star is close and bright. So long as you can reach this level in a time shorter than any transit, then the distance is irrelevant, all that matters is the fractional change in the brightness during transit and the number of transits you observe. The first of these grows as the square of the planet radius (for a fixed stellar radius), the second just depends on the length of your mission divided by the orbital period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line here is that the Earth in an Earth-like orbit around a Sun-like star would &lt;em&gt;not quite&lt;/em&gt; be detectable by Kepler in a Kepler-length mission, even if it were very close and nor if it were further away.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-11-02T06:35:24.580" LastActivityDate="2018-11-02T06:35:24.580" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28234" PostTypeId="2" ParentId="28232" CreationDate="2018-10-31T21:41:07.237" Score="2" Body="&lt;p&gt;1-2 m in the north and 8m in the south. &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Mars#Physical_characteristics&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Mars#Physical_characteristics&lt;/a&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Martian_polar_ice_caps#North_polar_cap&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Martian_polar_ice_caps#North_polar_cap&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the northern cap that disappears in summer.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-10-31T21:41:07.237" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28236" PostTypeId="2" ParentId="28222" CreationDate="2018-11-01T09:27:25.610" Score="4" Body="&lt;p&gt;If you have a look at the top-left panel of Fig.11 in &lt;a href=&quot;https://www.staff.science.uu.nl/~rutte101/rrweb/rjr-edu/coursenotes/rutten_issf_notes_1993.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;these lecture notes&lt;/a&gt; by Rob Rutten, you will see that the continuum opacity at optical wavelengths at the photosphere is about &lt;span class=&quot;math-container&quot;&gt;$10^{-6.7}$&lt;/span&gt; cm&lt;span class=&quot;math-container&quot;&gt;$^{-1}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inverse of this is the optical depth. You can see stuff through about 2-3 optical depths, so your &quot;horizon&quot;, looking horizontally in the solar atmosphere is about 100 km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the same plot shows that, because of the temperature and density variations with height, the photosphere becomes an order of magnitude more or less opaque at only 50km below or above the &quot;photosphere&quot; respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means is that you can see about 100 km horizontally, but less than this (guesstimate 30km) looking downwards and more than this (guesstimate infinity - that is how the photosphere is defined!) looking upwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These numbers would all be revised downwards at the wavelengths of prominent absorption lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could look all around you, you would see a very asymmetric brightness/colour distribution. Overhead would be dimmer and cooler (redder); underneath, much brighter and hotter (more or less the same colour we see the photosphere).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-11-01T09:33:37.647" LastActivityDate="2018-11-01T09:33:37.647" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28237" PostTypeId="2" ParentId="19687" CreationDate="2018-11-01T23:47:55.797" Score="4" Body="&lt;p&gt;Inverse modelling is where you use features of your data to estimate a set of underlying parameters of your physical model of what is going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forward modelling is where you use your model to predict what you would observe and use a comparison of these predictions to your data to infer your model parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple exoplanet example. Consider a sparsely sampled radial velocity curve. You could fit a sinusoid (or an elliptical orbit solution) to these data and estimate the period, radial velocity amplitude and then deduce a minimum mass for the orbiting exoplanet by plugging these numbers, along with an estimate of the stellar mass into the mass function formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A forward modelling approach would start with the mass of the star and planet, specify an orbital period and inclination and then predict what would be observed - including if necessary, functions that allow for imperfections and uncertainties in the measurements. Many such models are produced and compared with the observations until one can estimate probability functions for each of the model parameters.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-11-01T23:47:55.797" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28238" PostTypeId="1" AcceptedAnswerId="28244" CreationDate="2018-11-02T00:42:29.060" Score="3" ViewCount="300" Body="&lt;p&gt;ESA's HST page &lt;a href=&quot;https://www.spacetelescope.org/news/heic1819/?lang&quot; rel=&quot;nofollow noreferrer&quot;&gt;heic1819 — Photo Release; Hubble reveals cosmic Bat Shadow in the Serpent’s Tail&lt;/a&gt; is of course beautiful and stunning, but my eyes are drawn to the diffraction artifacts of the bright stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming the crosses are due to four vanes supporting the secondary mirror, but are the tight concentric rings due to Hubble's large aperture, or it's smaller secondary mirror blocking that aperture, or something else, perhaps image processing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without a scale for reference, it's hard to get an angular frequency and compare to an Airy-like diffraction pattern to get a diameter, and that's where it gets more puzzling, because you need a narrow wavelength range to get a coherent oscillation for so many cycles (I think I can see perhaps 15 or more sharp, distinct rings), and to zeroth order stars are mostly black-body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a bit of a puzzle, or am I missing something obvious (e.g. filters)? Or both?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a cropped, monochromed, ROI:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Sour1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sour1.png&quot; alt=&quot;heic1819 — Photo Release; Hubble cropped, monochrome&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further stretched in contrast and size:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Y2OGQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y2OGQ.png&quot; alt=&quot;heic1819 — Photo Release; Hubble cropped, monochrome&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bcyLJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bcyLJ.jpg&quot; alt=&quot;heic1819 — Photo Release; Hubble&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-11-21T00:58:02.283" LastActivityDate="2018-11-21T00:58:02.283" Title="What is the cause of all of these sharp, concentric rings around bright stars in this HST image?" Tags="&lt;hubble-telescope&gt;&lt;optics&gt;&lt;angular-resolution&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28239" PostTypeId="1" AcceptedAnswerId="28296" CreationDate="2018-11-02T00:51:11.920" Score="6" ViewCount="250" Body="&lt;p&gt;This &lt;a href=&quot;https://astronomy.stackexchange.com/a/14780/7982&quot;&gt;answer&lt;/a&gt; to the question &lt;a href=&quot;https://astronomy.stackexchange.com/q/14660/7982&quot;&gt;V838 Monocerotis “light-echo” images morphed into nice video, but why so few original images?&lt;/a&gt; contains the following information below. There are other answers and more images there, so it's recommended to enjoy there as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What does V838 Monocerotis look like today? Have there been subsequent light echo image sequences, or is the party over?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;https://astronomy.stackexchange.com/q/14781/7982&quot;&gt;Why weren't the Hubble light-echo images of V838 Monocerotis supplemented by ground telescopes?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The V838 Monocerotis &lt;em&gt;expansion&lt;/em&gt; (not a supernova) and the observation of the subsequent &quot;spectacular&quot; light echo was quite a notable event! From &lt;a href=&quot;http://www.nature.com/nature/journal/v422/n6930/abs/nature01508.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nature &lt;strong&gt;422&lt;/strong&gt;, 405-408 (27 March 2003)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xU4aE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xU4aE.png&quot; alt=&quot;Nature Cover&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/TSrZZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TSrZZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0004-6256/135/2/605&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronom. J. &lt;strong&gt;135&lt;/strong&gt;, 2, 2008&lt;/a&gt; or &lt;a href=&quot;http://arxiv.org/abs/0711.1495&quot; rel=&quot;nofollow noreferrer&quot;&gt;ArXiv&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;em&gt;.&lt;strong&gt;Galactic light echoes are extremely rare. The only other known example of extent similar to that of V838 Mon was the echo produced by Nova GK Persei 1901&lt;/strong&gt; (Kapteyn 1902; Perrine 1902; Ritchey 1902). Following early misunderstandings, light-echo geometry was properly described by Couderc (1939), and more recent discussions are given by many authors, including Chevalier (1986), Felten (1991), Sparks (1994), Sugerman (2003), and references therein&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It was the sole topic of an international conference &lt;a href=&quot;http://www.ing.iac.es/conferences/v838mon/&quot; rel=&quot;nofollow noreferrer&quot;&gt;photo from here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BgHSZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BgHSZ.jpg&quot; alt=&quot;Conference&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.nature.com/nature/journal/v422/n6930/fig_tab/nature01508_F2.html#figure-title&quot; rel=&quot;nofollow noreferrer&quot;&gt;Figure 2&lt;/a&gt; of the Nature paper describes the preservation of the actual light curve (history) within the structure of the light-echo shell:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YztnU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YztnU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;strong&gt;FIGURE 2. HST images of the light echoes&lt;/strong&gt;&#xA;  The apparently superluminal expansion of the echoes as light from the outburst propagates outward into surrounding dust is shown dramatically. Images were taken in 2002 on 30 April (a), 20 May (b), 2 September (c) and 28 October (d). Each frame is 83&quot; times 83&quot;; north is up and east to the left. Imaging on 30 April was obtained only in the B filter, but B, V and I were used on the other three dates, allowing us to make full-colour renditions. The time evolution of the stellar outburst (Fig. 1) is reflected by structures visible in these colour images. In b, for example, note the series of rings and filamentary structures, especially in the upper right quadrant. &lt;strong&gt;Close examination shows that each set of rings has a sharp, blue outer edge, a dip in intensity nearer the star, and then a rebrightening to a redder plateau. Similar replicas of the outburst light curve are seen propagating outwards throughout all of the colour images&lt;/strong&gt;.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;also see:&lt;/strong&gt; This is an artificially constructed animation, morphing eight existing images to suggest what a higher cadence imaging campaign might have seen: &lt;a href=&quot;https://www.spacetelescope.org/videos/heic0617a/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.spacetelescope.org/videos/heic0617a/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-11-07T02:47:28.160" LastActivityDate="2020-01-09T13:36:13.907" Title="How does V838 Monocerotis look today? Any subsequent light echo images, or is the party over?" Tags="&lt;light&gt;&lt;dust&gt;&lt;hubble-telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28240" PostTypeId="1" AcceptedAnswerId="28241" CreationDate="2018-11-02T01:05:38.313" Score="15" ViewCount="2222" Body="&lt;p&gt;CNET.com's &lt;a href=&quot;https://www.cnet.com/news/scientists-confirm-theres-a-supermassive-black-hole-at-the-centre-of-our-galaxy/&quot; rel=&quot;noreferrer&quot;&gt;SCI-TECH&#xA;Scientists &lt;strong&gt;confirm&lt;/strong&gt; a 'supermassive black hole' at the heart of our galaxy&#xA;It's &quot;mind-boggling&quot;, they say.&lt;/a&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This links to Astronomy.com's &lt;a href=&quot;http://www.astronomy.com/news/2018/10/scientists-confirm-the-milky-way-has-a-supermassive-black-hole&quot; rel=&quot;noreferrer&quot;&gt;Scientists finally &lt;strong&gt;confirm&lt;/strong&gt; the Milky Way has a supermassive black hole&lt;/a&gt;&lt;br /&gt;&#xA;This links to ESO's eso1835 — Science Release &lt;a href=&quot;https://www.eso.org/public/news/eso1835/?lang&quot; rel=&quot;noreferrer&quot;&gt;Most Detailed Observations of Material Orbiting close to a Black Hole; ESO’s GRAVITY instrument &lt;strong&gt;confirms&lt;/strong&gt; black hole status of the Milky Way centre&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FYI, the GRAVITY instrument measures infrared light, not gravity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What is it exactly&lt;/em&gt; about these flares of infrared light from Sgr A* that &lt;strong&gt;confirms&lt;/strong&gt; it is a supermassive black hole?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly there is plenty of evidence, the orbits of stars nearby for one. But this use of the word &lt;strong&gt;confirm&lt;/strong&gt; seems strong and frequent enough to suggest the deal has been done, the ink is dry, and it's a black hole without question and that until now that wasn't so.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I've shamelessly absconded the following images from &lt;a href=&quot;https://physics.stackexchange.com/a/406609/83380&quot;&gt;this answer&lt;/a&gt; to the question &lt;em&gt;What is the evidence for a supermassive black hole at the center of Milky Way?&lt;/em&gt;, which may need a new answer!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, the size of these stellar orbits are compared to those of Sedna, Eris, Pluto and Neptune in the bottom right corner (same scale)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/m186c.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/m186c.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.m.wikimedia.org/wiki/File:Galactic_centre_orbits.svg&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HLjZZ.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HLjZZ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="16685" LastEditDate="2018-11-02T11:06:52.717" LastActivityDate="2018-11-02T11:06:52.717" Title="What is it exactly about these flares of infrared light from Sgr A* that &quot;confirms&quot; it is a supermassive black hole?" Tags="&lt;observational-astronomy&gt;&lt;supermassive-black-hole&gt;&lt;accretion-discs&gt;&lt;sagittarius-a&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28241" PostTypeId="2" ParentId="28240" CreationDate="2018-11-02T02:06:05.083" Score="28" Body="&lt;p&gt;We have reasonably good measurements of the mass of Sagittarius A*, thanks to measurements of the movements of stars like S0-2 over several decades. It's been well-established that the mass of the central object is &lt;span class=&quot;math-container&quot;&gt;$M\approx4\times10^6M_{\odot}$&lt;/span&gt;; this alone is fairly good evidence for a supermassive black hole, and we can constrain the size of the object with the measurements (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2008ApJ...689.1044G&quot; rel=&quot;noreferrer&quot;&gt;Ghez et al. 2008&lt;/a&gt;). Models of the orbits of other stars have further improved these results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The recently-published paper (&lt;a href=&quot;https://www.aanda.org/component/article?access=doi&amp;amp;doi=10.1051/0004-6361/201834294&quot; rel=&quot;noreferrer&quot;&gt;Abuter et al. 2018&lt;/a&gt;) uses a similar technique. The motion of the flares is well-described by the rotation of a &quot;hot spot&quot; of gas in the inner reaches of the accretion disk, with the flares arising from magnetic recombination or some similar event. In particular, it lies near the innermost stable prograde circular orbit (ISCO) - just outside it, actually (the orbital radius can be found because we know the mass of the object and the period of the orbit, with the best-fit model of the latter being &lt;span class=&quot;math-container&quot;&gt;$40\pm 8$&lt;/span&gt; minutes). This both constrains the object's size and even qualitatively provides support for the black hole model, as we might expect to see such events around a supermassive black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at those articles, I think the use of the word &quot;confirm&quot; is inaccurate. As far as I can tell, the term is only used by the project leader, Reinhard Genzel, in &lt;a href=&quot;https://www.eso.org/public/news/eso1835/?lang&quot; rel=&quot;noreferrer&quot;&gt;that ESO statement&lt;/a&gt; - and it's not a claim repeated in the paper. The team describes their results as &quot;strong support&quot; for the supermassive black hole model, and say that their findings are &quot;consistent with&quot; that theory. As scientists should be, they're cautious. The results don't definitely confirm that Sagittarius A* corresponds to a supermassive black hole; they're simply additional (excellent) evidence for it.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2018-11-02T02:33:38.503" LastActivityDate="2018-11-02T02:33:38.503" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28243" PostTypeId="2" ParentId="28170" CreationDate="2018-11-02T12:08:56.287" Score="2" Body="&lt;p&gt;No answers after a week, so I will put a toe into the water an make a rough estimate. I think a thorough answer would require both a significant 3D orbital dynamics simulation with a million particles, and a way to define dynamically a threshold for what passing bodies would be considered &amp;quot;just barely temporarily captured&amp;quot; and which *not quite temporarily captured&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;There should in fact be some good papers to cite that report this kind of simulation. I'll try to look around; I think there is a good opportunity for one or two more answers here.&lt;/p&gt;&#xA;&lt;p&gt;Also see &lt;a href=&quot;https://space.stackexchange.com/q/16738/12102&quot;&gt;Have there been any documented mini-moons since 2006 RH120?&lt;/a&gt; and all of the sources linked there!&lt;/p&gt;&#xA;&lt;p&gt;An example of an object in a medium-term stable orbit in the Earth-Moon system might be found in a halo orbit around the Earth-Moon L1, or L2 points. A spherical envelope-back calculation using spherical cow milk for ink suggests that in the rotating frame, it cycles in a 100,000 km orbit every two weeks around the Lagrange point, or about 0.5 km/sec in that frame. In a non-rotating Earth-centered frame, the Moon itself is moving at about 1 km/sec.&lt;/p&gt;&#xA;&lt;p&gt;Since these temporary orbits are weakly bound, I'm going to say that an object entering the Earth-Moon system at roughly 1.5 km/sec can hang around for a while, then exit with roughly that velocity.&lt;/p&gt;&#xA;&lt;p&gt;Looking inthe Sun's inertial frame now, The Earth-Moon system is moving at about 30 km/sec. That 1.5 km/sec is now looking fairly small, only about 5%. So if it exited prograde or retrograde, it would end up in a heliocentric orbit with a semi-major axis no more than 10% different than the Earth's. TCO's are NEOs that take the time to flirt with the Earth-Moon system, then continue on as NEOs to perhaps flirt again another day.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I just visited my mini-moon question, found Gravnik 2012 and realized the plots are there! Here are some of them. It turns out everything I said is right. That's not a surprise, because I believe I read this paper a few years ago.&lt;/p&gt;&#xA;&lt;p&gt;1 billion initial conditions, 10 million particles actually integrated.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/S0019103511004684&quot; rel=&quot;nofollow noreferrer&quot;&gt;The population of natural Earth satellites&lt;/a&gt; Mikael Granvik, Jeremie Vaubaillon, and Robert Jedicke, Icarus, Volume 218, Issue 1, March 2012, Pages 262-277  &lt;a href=&quot;https://doi.org/10.1016/j.icarus.2011.12.003&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://doi.org/10.1016/j.icarus.2011.12.003&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Abstract&lt;/p&gt;&#xA;&lt;p&gt;We have for the first time calculated the population characteristics of the Earth’s irregular natural satellites (NESs) that are temporarily captured from the near-Earth-object (NEO) population. The steady-state NES size–frequency and residence-time distributions were determined under the dynamical influence of all the massive bodies in the Solar System (but mainly the Sun, Earth, and Moon) for NEOs of negligible mass. To this end, we compute the NES capture probability from the NEO population as a function of the latter’s heliocentric orbital elements and combine those results with the current best estimates for the NEO size–frequency and orbital distribution. At any given time there should be at least one NES of 1-m diameter orbiting the Earth. The average temporarily-captured orbiter (TCO; an object that makes at least one revolution around the Earth in a co-rotating coordinate system) completes (2.88 ± 0.82) rev around the Earth during a capture event that lasts (286 ± 18) d. We find a small preference for capture events starting in either January or July. Our results are consistent with the single known natural TCO, 2006 RH120, a few-meter diameter object that was captured for about a year starting in June 2006. We estimate that about 0.1% of all meteors impacting the Earth were TCOs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6X6hR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6X6hR.png&quot; alt=&quot;Fig. 1 Gravnik et al. 2012 Icarus, 218, 1 262-277&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o77gr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o77gr.jpg&quot; alt=&quot;Fig. 2 Gravnik et al. 2012 Icarus, 218, 1 262-277&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tLTul.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tLTul.png&quot; alt=&quot;Fig. 3 Gravnik et al. 2012 Icarus, 218, 1 262-277&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J2QlU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J2QlU.png&quot; alt=&quot;Fig. 5 Gravnik et al. 2012 Icarus, 218, 1 262-277&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-11-02T12:29:47.930" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28244" PostTypeId="2" ParentId="28238" CreationDate="2018-11-02T18:01:27.190" Score="7" Body="&lt;p&gt;The diffraction pattern at the focal plane created by a circular aperture is called an &lt;a href=&quot;https://en.wikipedia.org/wiki/Airy_disk&quot; rel=&quot;noreferrer&quot;&gt;Airy Disk or Airy Pattern&lt;/a&gt;.  Both the outer opening and the inner hole plus secondary contribute to the exact function.  This is usually not easily observed with ground based telescopes because the seeing fluctuations due to air turbulence smears it out.  These images use filters that are a few 100 Angstroms wide, so although not monochromatic, they are narrow enough to clearly see the pattern.  The spectral shape of the star's light within the filter makes the effective width of frequencies more narrow. &lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2018-11-02T18:12:16.123" LastActivityDate="2018-11-02T18:12:16.123" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="28246" PostTypeId="1" CreationDate="2018-11-03T09:45:51.497" Score="2" ViewCount="65" Body="&lt;p&gt;The polar circle is the limit for how far north or south the sun will shine 24h. But is there a similar limit for the moon?&lt;/p&gt;&#xA;" OwnerUserId="24673" LastActivityDate="2018-11-03T09:45:51.497" Title="Polar circle for the moon" Tags="&lt;the-moon&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28247" PostTypeId="1" AcceptedAnswerId="28263" CreationDate="2018-11-03T19:05:56.460" Score="5" ViewCount="225" Body="&lt;p&gt;There are numerous lists available of past and future &lt;a href=&quot;https://en.wikipedia.org/wiki/Planetary_transits_and_occultations&quot; rel=&quot;noreferrer&quot;&gt;planetary transits and occultations&lt;/a&gt;, but I can't find any that list transits behind the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do the planets in our solar system ever pass directly behind the sun, from Earth's perspective? Is there a special name for this kind of occultation, and is there a list of these events?&lt;/p&gt;&#xA;" OwnerUserId="14302" LastActivityDate="2018-11-05T03:22:43.247" Title="When are the planets occulted by the sun?" Tags="&lt;the-sun&gt;&lt;planetary-transits&gt;&lt;occultation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28248" PostTypeId="1" CreationDate="2018-11-03T19:07:11.190" Score="3" ViewCount="306" Body="&lt;p&gt;I know there is a gioscopic explanation but I was under the impression that Earth is traveling in a straight line through space that has been curved by our sun's gravity ,so my question is why doesn't one hemisphere always point away from the sun ? It just seems like a bit of a contradiction that gravity can curve the fabric of space but has no effect on earth's axis...&lt;/p&gt;&#xA;" OwnerUserId="24757" LastActivityDate="2018-11-04T00:11:10.643" Title="Why does earth's north pole roughly point in the same direction throughout the year?" Tags="&lt;the-sun&gt;&lt;earth&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28249" PostTypeId="2" ParentId="28248" CreationDate="2018-11-03T20:06:13.043" Score="2" Body="&lt;p&gt;It is just &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_momentum#Conservation_of_angular_momentum&quot; rel=&quot;nofollow noreferrer&quot;&gt;conservation of angular momentum&lt;/a&gt;. In order to change the direction of the spin axis of the Earth (which would change the &lt;em&gt;direction&lt;/em&gt; of its angular momentum) you would have to subject it to an external torque. The curvature of space has no significant bearing on the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact the direction in which the Earth's spin axis points &lt;em&gt;does&lt;/em&gt; change, but on much longer timescales. The &quot;&lt;a href=&quot;https://www.britannica.com/science/precession-of-the-equinoxes&quot; rel=&quot;nofollow noreferrer&quot;&gt;precession of the equinoxes&lt;/a&gt;&quot; (as it is called) is cyclical with a 26,000 year period.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-11-03T20:06:13.043" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28250" PostTypeId="2" ParentId="28248" CreationDate="2018-11-03T21:46:24.500" Score="2" Body="&lt;p&gt;The path of the Earth in space is curved (by gravity).  The explanation of gravity in Relativity is that the Earth follows a geodesic in space&lt;em&gt;time&lt;/em&gt;, which is not usually a straight line in &lt;em&gt;space&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ball that is thrown will follow a geodesic, a ball that is thrown harder will also follow a geodesic. The reason that the two balls don't follow the same path, is that they are following geodesics in curved spacetime, not straight lines in curved space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This being the case, the direction of the Earth's axis has very little to do with curved spacetime. The direction of the axis in space remains fixed, unless some force acts on the Earth to change it. The sun and moon can provide such torque (and cause precession and nutation), but this can be accurately described by the Newtonian approximation of gravity. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-11-03T21:46:24.500" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28251" PostTypeId="2" ParentId="28248" CreationDate="2018-11-04T00:11:10.643" Score="0" Body="&lt;p&gt;The main effect of the Sun's gravity on the Earth is to keep it in a (very nearly) steady elliptical orbit. So from one perspective, the Sun pulls the Earth towards it according to the inverse square law (with some rather tiny variations given by general relativity). From another perspective, the Sun's mass distorts space-time so that the natural geodesic path that the Earth follows returns to the same place (relative to the Sun and oriented by distant stars) every year. These are two descriptions of the same physics, and both are equally true (or false).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Newtonian physics (without GR) (and assuming that both the Earth and the Sun were perfectly spherical and that the other planets did not exist) the Earth's axis of rotation would not change at all. General relativity does predict some very small changes: &lt;a href=&quot;https://en.wikipedia.org/wiki/Frame-dragging&quot; rel=&quot;nofollow noreferrer&quot;&gt;frame dragging&lt;/a&gt; and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Geodetic_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;geodetic effect&lt;/a&gt;. Their effect on the Earth, though is incredibly tiny. The effect of the Moon's gravity on the Earth's equatorial bulge is much greater.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-11-04T00:11:10.643" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28252" PostTypeId="1" CreationDate="2018-11-04T00:34:54.173" Score="4" ViewCount="403" Body="&lt;p&gt;I'm looking to make a custom pulsar map like the one on the Voyager Records, but for an arbitrary location within the Milky Way Galaxy. I'm not even sure where to begin.&lt;/p&gt;&#xA;" OwnerUserId="24760" LastEditorUserId="2153" LastEditDate="2018-11-04T00:59:26.467" LastActivityDate="2018-11-04T12:42:31.693" Title="How do I make a custom pulsar map?" Tags="&lt;positional-astronomy&gt;&lt;pulsar&gt;&lt;star-maps&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28253" PostTypeId="2" ParentId="28252" CreationDate="2018-11-04T00:47:59.430" Score="5" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Pioneer_plaque#Relative_position_of_the_Sun_to_the_center_of_the_Galaxy_and_14_pulsars&quot; rel=&quot;noreferrer&quot;&gt;The pulsar map&lt;/a&gt; used by &lt;em&gt;Pioneer&lt;/em&gt; and &lt;em&gt;Voyager&lt;/em&gt; is a diagram of the position of the Sun relative to 14 galactic pulsars. It encodes their positions, distances, and pulse periods, which in theory makes it a decent tool for identifying where the Solar System is (although &lt;a href=&quot;https://www.forbes.com/sites/startswithabang/2017/08/17/voyagers-cosmic-map-of-earths-location-is-hopelessly-wrong/#732ea8e269d5&quot; rel=&quot;noreferrer&quot;&gt;there are some issues&lt;/a&gt; with it). To make a similar map for an arbitrary point in the galaxy, you'd need to find a selection of pulsars near that location, and encode the same amount of information. To do that, you need a database of pulsars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.atnf.csiro.au/&quot; rel=&quot;noreferrer&quot;&gt;Australian Telescope National Facility (ATNF)&lt;/a&gt; has a really excellent &lt;a href=&quot;http://www.atnf.csiro.au/research/pulsar/psrcat/&quot; rel=&quot;noreferrer&quot;&gt;pulsar catalogue&lt;/a&gt; that I like to use, with over 2600 entries. You can &lt;a href=&quot;http://www.atnf.csiro.au/research/pulsar/psrcat/tutorial/intro.html&quot; rel=&quot;noreferrer&quot;&gt;query it fairly easily&lt;/a&gt; using a number of parameters, including distance, dispersion measure, coordinates, pulse period, etc. This makes it a handy tool to match observations with already known pulsars, but it should also be well-suited to your needs. You can get a list of pulsar names, coordinates, periods and distances by checking the &lt;code&gt;JName&lt;/code&gt;, &lt;code&gt;RaJ&lt;/code&gt;, &lt;code&gt;DecJ&lt;/code&gt;, &lt;code&gt;P0&lt;/code&gt; and &lt;code&gt;Dist&lt;/code&gt; parameters in the search interface, then click the button to get &lt;a href=&quot;http://www.atnf.csiro.au/research/pulsar/psrcat/proc_form.php?version=1.59&amp;amp;table_top.x=75&amp;amp;table_top.y=18&amp;amp;Name=Name&amp;amp;RaJ=RaJ&amp;amp;DecJ=DecJ&amp;amp;P0=P0&amp;amp;Dist=Dist&amp;amp;startUserDefined=true&amp;amp;c1_val=&amp;amp;c2_val=&amp;amp;c3_val=&amp;amp;c4_val=&amp;amp;sort_attr=jname&amp;amp;sort_order=asc&amp;amp;condition=&amp;amp;pulsar_names=&amp;amp;ephemeris=short&amp;amp;coords_unit=raj%2Fdecj&amp;amp;radius=&amp;amp;coords_1=&amp;amp;coords_2=&amp;amp;style=Long+with+last+digit+error&amp;amp;no_value=*&amp;amp;fsize=3&amp;amp;x_axis=&amp;amp;x_scale=linear&amp;amp;y_axis=&amp;amp;y_scale=linear&amp;amp;state=query&quot; rel=&quot;noreferrer&quot;&gt;a table&lt;/a&gt; of the results. You should then be able to compute the positions of all of the pulsars in the catalogue in your coordinate system of choice, then re-compute their positions with respect to a chosen location in space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The output of the first hundred rows looks like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;---------------------------------------------------------------------------------------------------------------------------------------------------&#xA;#     NAME                     RAJ                                DECJ                               P0                                     DIST&#xA;                               (hms)                              (dms)                              (s)                                    (kpc)&#xA;---------------------------------------------------------------------------------------------------------------------------------------------------&#xA;1     J0002+6216    cwp+17     00:02:58.17          2  cwp+17     +62:16:09.4          1  cwp+17     0.1153635682680         14  cwp+17     *&#xA;2     J0006+1834    cnt96      00:06:04.8           2  cn95       +18:34:59            4  cn95       0.69374767047           14  cn95       0.86&#xA;3     J0007+7303    aaa+09c    00:07:01.7           2  awd+12     +73:03:07.4          8  awd+12     0.3158731909             3  awd+12     1.40&#xA;4     J0011+08      dsm+16     00:11:34           114  dsm+16     +08:10               8  dsm+16     2.55287                  0  dsm+16     5.40&#xA;5     B0011+47      dth78      00:14:17.75          4  hlk+04     +47:46:33.4          3  hlk+04     1.240699038946          11  hlk+04     1.78&#xA;&#xA;6     J0023+0923    hrm+11     00:23:16.878141      9  abb+18     +09:23:23.8652       3  abb+18     0.003050203104480002     7  abb+18     1.11&#xA;7     J0024-7201Z              00:24:06.041         2  frk+17     -72:05:01.480        6  frk+17     0.004554447383906        3  frk+17     2.56&#xA;8     J0024-7204aa  phl+16     00:24:05.67          0  mam+06     -72:04:52.62         0  mam+06     0.00184                  0  phl+16     2.69&#xA;9     J0024-7204ab  phl+16     00:24:08.1615        5  frk+17     -72:04:47.602        2  frk+17     0.0037046394947985       6  frk+17     2.54&#xA;10    B0021-72C     mld+90     00:23:50.3546        1  frk+17     -72:04:31.5048       4  frk+17     0.00575677999551635     14  frk+17     4.69&#xA;&#xA;11    B0021-72D     mlr+91     00:24:13.88092       6  frk+17     -72:04:43.8524       2  frk+17     0.00535757328486573      9  frk+17     4.69&#xA;12    B0021-72E     mlr+91     00:24:11.10528       5  frk+17     -72:05:20.1492       2  frk+17     0.00353632915276244      4  frk+17     4.69&#xA;13    B0021-72F     mlr+91     00:24:03.85547      10  frk+17     -72:04:42.8183       2  frk+17     0.00262357935251262      4  frk+17     4.69&#xA;14    B0021-72G     rlm+95     00:24:07.9603        1  frk+17     -72:04:39.7030       5  frk+17     0.00404037914356515     14  frk+17     4.69&#xA;15    B0021-72H     mlr+91     00:24:06.7032        2  frk+17     -72:04:06.8067       6  frk+17     0.00321034070935032     11  frk+17     4.69&#xA;&#xA;16    B0021-72I     mlr+91     00:24:07.9347        2  frk+17     -72:04:39.6815       7  frk+17     0.00348499206166289     13  frk+17     4.69&#xA;17    B0021-72J     mlr+91     00:23:59.4077        1  frk+17     -72:03:58.7908       5  frk+17     0.00210063354535246      5  frk+17     4.69&#xA;18    B0021-72L     rlm+95     00:24:03.7721        3  frk+17     -72:04:56.923        2  frk+17     0.0043461679994616       4  frk+17     4.69&#xA;19    B0021-72M     mlr+91     00:23:54.4899        3  frk+17     -72:05:30.756        2  frk+17     0.0036766432176002       3  frk+17     4.69&#xA;20    B0021-72N     rlm+95     00:24:09.1880        2  frk+17     -72:04:28.8907       7  frk+17     0.00305395434626085     10  frk+17     4.69&#xA;&#xA;21    J0024-7204O   clf+00     00:24:04.65254       6  frk+17     -72:04:53.7670       2  frk+17     0.00264334329724357      5  frk+17     4.69&#xA;22    J0024-7204P   clf+00     00:24:20            29  rft+16     -72:04:10           62  rft+16     0.0036430207             3  rft+16     4.69&#xA;23    J0024-7204Q   clf+00     00:24:16.4909        1  frk+17     -72:04:25.1644       6  frk+17     0.00403318118457258     15  frk+17     4.69&#xA;24    J0024-7204R   clf+00     00:24:07.5851        2  frk+17     -72:04:50.3954       5  frk+17     0.00348046270749329     13  frk+17     4.69&#xA;25    J0024-7204S   clf+00     00:24:03.9794        1  frk+17     -72:04:42.3530       4  frk+17     0.00283040595787912      8  frk+17     4.69&#xA;&#xA;26    J0024-7204T   clf+00     00:24:08.5491        5  frk+17     -72:04:38.932        3  frk+17     0.0075884798073671      12  frk+17     4.69&#xA;27    J0024-7204U   clf+00     00:24:09.8366        1  frk+17     -72:03:59.6882       4  frk+17     0.00434282669639233     12  frk+17     4.69&#xA;28    J0024-7204V   clf+00     00:24:05.67          0  mam+06     -72:04:52.62         0  mam+06     0.00481016762            7  rft+16     4.69&#xA;29    J0024-7204W   clf+00     00:24:06.058         1  rft+16     -72:04:49.088        2  rft+16     0.0023523445319370       3  rft+16     4.69&#xA;30    J0024-7204X   fre08      00:24:22.38565       9  rft+16     -72:01:17.4414       7  rft+16     0.00477152291069356      5  rft+16     4.69&#xA;&#xA;31    J0024-7204Y   frk+17     00:24:01.4026        1  frk+17     -72:04:41.8363       4  frk+17     0.00219665714352123      5  frk+17     4.69&#xA;32    J0024-7204Z   fre08      00:24:05.67          0  mam+06     -72:04:52.62         0  mam+06     0.004554                 0  fre08      4.69&#xA;33    J0026+6320    jml+09     00:26:50.561         8  jml+09     +63:20:00.87         5  jml+09     0.318357731235           1  jml+09     6.42&#xA;34    J0030+0451    lzb+00     00:30:27.42781       4  abb+18     +04:51:39.7113      14  abb+18     0.0048654532112551       3  abb+18     0.32&#xA;35    J0033+57      hrk+08     00:33                0  hrk+08     +57                  0  hrk+08     0.315                    0  hrk+08     2.25&#xA;&#xA;36    J0033+61      hrk+08     00:33                0  hrk+08     +61                  0  hrk+08     0.912                    0  hrk+08     1.64&#xA;37    J0034-0534    bhl+94     00:34:21.83422       8  dcl+16     -05:34:36.722        3  dcl+16     0.0018771818845850       2  aaa+10b    1.35&#xA;38    B0031-07      lvw69a     00:34:08.8703        1  cbv+09     -07:21:53.409        2  cbv+09     0.9429509945598         17  hlk+04     1.03&#xA;39    J0034+69      slr+14     00:34                3  slr+14     +69:43              18  slr+14     0.03680412              16  slr+14     2.28&#xA;40    B0037+56      dtws85     00:40:32.362         6  hlk+04     +57:16:24.91         6  hlk+04     1.1182253452412         13  hlk+04     2.42&#xA;&#xA;41    J0045-7042    mfl+06     00:45:25.69         17  mfl+06     -70:42:07.1         13  mfl+06     0.63233580002            6  mfl+06     59.70&#xA;42    J0045-7319    mmh+91     00:45:35.16          7  kbm+96     -73:19:03.0          2  kbm+96     0.92627590497            3  kbm+96     59.70&#xA;43    B0045+33      dtws85     00:48:33.98          3  hlk+04     +34:12:08.0          4  hlk+04     1.217094296058           8  hlk+04     4.50&#xA;44    J0050+03      dsm+16     00:50:31           113  dsm+16     +03:48               8  dsm+16     1.36656                  0  dsm+16     25.00&#xA;45    J0051+0423    snt97      00:51:30.1           2  snt97      +04:22:49            6  snt97      0.35473179890            1  snt97      1.25&#xA;&#xA;46    J0053+69      slr+14     00:53                3  slr+14     +69:39              18  slr+14     0.832938                28  slr+14     2.75&#xA;47    J0054+66      hrk+08     00:54                0  hrk+08     +66                  0  hrk+08     1.39                     0  hrk+08     0.90&#xA;48    J0054+69      kkl+15     00:54:28            25  kkl+15     +69:26               2  kkl+15     *                        0  *          2.35&#xA;49    B0052+51      dtws85     00:55:45.378        10  hlk+04     +51:17:24.97        11  hlk+04     2.115171148865           4  hlk+04     1.94&#xA;50    B0053+47      dtws85     00:56:25.51          2  hlk+04     +47:56:10.5          3  hlk+04     0.472036662076           7  hlk+04     1.12&#xA;&#xA;51    J0057-7201    ckm+01     00:57:44.0           4  ckm+01     -72:01:19            2  ckm+01     0.7380624426             2  ckm+01     3.48&#xA;52    J0058+6125    hrk+08     00:58                0  hrk+08     +61:25               0  hrk+08     0.637                    0  hrk+08     2.61&#xA;53    J0059+50      slr+14     00:59                2  slr+14     +50:02              18  slr+14     0.996009                61  slr+14     2.61&#xA;54    J0100-7211    lfmp02     01:00:43.03         11  mgr+05     -72:11:33.6          5  mgr+05     8.020392                 9  mgr+05     59.70&#xA;55    J0101-6422    kcj+12     01:01:11.1163        2  kcj+12     -64:22:30.171        2  kcj+12     0.0025731519721683       2  kcj+12     1.00&#xA;&#xA;56    J0102+4839    hrm+11     01:03                0  hrm+11     +48:00               0  *          0.00296                  0  hrm+11     2.38&#xA;57    B0059+65      dbtb82     01:02:32.958        16  hlk+04     +65:37:13.40        11  hlk+04     1.679164229064          12  hlk+04     1.98&#xA;58    J0103+54      kkl+15     01:03:37            20  kkl+15     +54:02               5  kkl+15     0.354304                 6  kkl+15     2.01&#xA;59    J0106+4855    pga+12     01:06:25.06          1  pga+12     +48:55:51.8          2  pga+12     0.0831573050050          6  pga+12     3.06&#xA;60    J0108-1431    tnj+94     01:08:08.347016     88  dtbr09     -14:31:50.1871      11  dtbr09     0.807564614019          20  hlk+04     0.21&#xA;&#xA;61    B0105+65      dls72      01:08:22.64          5  hlk+04     +66:08:34.0          3  hlk+04     1.2836598429             5  hlk+04     1.46&#xA;62    B0105+68      dtws85     01:08:29.50          2  hlk+04     +69:05:52.63        11  hlk+04     1.071118109859           5  hlk+04     1.98&#xA;63    J0111-7131    mfl+06     01:11:28.77          9  mfl+06     -71:31:46.8          6  mfl+06     0.68854151164            5  mfl+06     59.70&#xA;64    J0112+66      slr+14     01:12                3  slr+14     +66:22              18  slr+14     4.30124                 70  slr+14     2.42&#xA;65    J0113-7220    ckm+01     01:13:11.09          3  ckm+01     -72:20:32.20        15  ckm+01     0.32588301613            1  ckm+01     59.70&#xA;&#xA;66    B0114+58      stwd85     01:17:38.661         3  hlk+04     +59:14:38.39         3  hlk+04     0.1014390664824         20  hlk+04     1.77&#xA;67    J0125+62      slr+14     01:26                3  slr+14     +62:35              18  slr+14     1.708233                98  slr+14     2.40&#xA;68    J0131-7310    mfl+06     01:31:28.51          3  mfl+06     -73:10:09.34        13  mfl+06     0.348124045581           7  mfl+06     59.70&#xA;69    J0133-6957    lml+98     01:33:32.46          1  dsb+98     -69:57:29.68         3  dsb+98     0.463473741703           3  dsb+98     2.44&#xA;70    J0134-2937    mld+96     01:34:18.6824        3  lwy+16     -29:37:17.042        7  lwy+16     0.136961610513651       14  lwy+16     25.00&#xA;&#xA;71    J0136+63      slr+14     01:36                3  slr+14     +63:42              18  slr+14     0.717895                27  slr+14     9.12&#xA;72    J0137+1654    lxf+05     01:37:23.88          1  lxf+05     +16:54:42.1          4  lxf+05     0.4147630265082          5  lxf+05     3.06&#xA;73    J0139+33      ttm+18     01:39:10           120  ttm+18     +33:10              15  ttm+18     *                        0  *          1.47&#xA;74    J0139+5621    cvh+14     01:39:38.56          2  cvh+14     +56:21:36.8          4  cvh+14     1.77534382596           10  cvh+14     2.41&#xA;75    B0136+57      dth78      01:39:19.7401       12  cbv+09     +58:14:31.819       17  cbv+09     0.272450630950          15  hlk+04     2.60&#xA;&#xA;76    B0138+59      mth72      01:41:39.938         8  hlk+04     +60:09:32.30         7  hlk+04     1.2229485205457         18  hlk+04     2.30&#xA;77    J0141+63      slr+14     01:41                2  slr+14     +63:08               8  slr+14     0.04668588              49  slr+14     8.77&#xA;78    J0146+31      ttol16     01:46:15            40  ttol16     +31:04              10  ttol16     0.9381                   0  ttol16     1.83&#xA;79    J0146+6145    ims94      01:46:22.42         10  pkw+03     +61:45:02.8          7  pkw+03     8.6889941                4  dk14       3.60&#xA;80    B0144+59      stwd85     01:47:44.6457       17  ywml10     +59:22:03.293       15  ywml10     0.19632137543003        16  ywml10     1.58&#xA;&#xA;81    B0148-06      mlt+78     01:51:22.701        11  hlk+04     -06:35:02.8          4  hlk+04     1.464664549334           7  hlk+04     25.00&#xA;82    J0152+0948    clm+05     01:52:23.73          5  clm+05     +09:48:10            2  clm+05     2.74664729014           10  clm+05     2.48&#xA;83    B0149-16      mlt+78     01:52:10.8536       17  hlk+04     -16:37:52.99         4  hlk+04     0.8327416126878         13  hlk+04     0.92&#xA;84    J0156+04      dsm+16     01:56:01           114  dsm+16     +04:02               8  dsm+16     *                        0  *          25.00&#xA;85    B0153+39      dtws85     01:56:55.2           3  hlk+04     +39:49:29            5  hlk+04     1.81156061131           10  hlk+04     4.91&#xA;&#xA;86    B0154+61      dls72      01:57:49.93          3  hlk+04     +62:12:25.90        17  hlk+04     2.35174493646            7  hlk+04     1.39&#xA;87    J0158+21      dsm+13     01:58:29            32  dsm+13     +21:08               8  dsm+13     0.5053                   0  dsm+13     1.57&#xA;88    J0201+7005    kkl+15     02:01:41.344         7  kkl+15     +70:05:18.11         6  kkl+15     1.349184471846          10  kkl+15     1.17&#xA;89    J0205+6449    mss+02     02:05:37.92247      15  bkr+13     +64:49:41.3343      10  bkr+13     0.06571592849324         9  lrc+09     3.20&#xA;90    B0203-40      mlt+78     02:06:01.268         9  smd93      -40:28:04.33        14  smd93      0.630550140847           6  smd93      1.26&#xA;&#xA;91    J0211-8159    lml+98     02:11:59.0           3  dsb+98     -81:59:36.2          3  dsb+98     1.0773327013             8  dsb+98     1.52&#xA;92    J0212+5222    bck+13     02:12:52.2           6  bck+13     +52:22:45           13  bck+13     0.376386292              1  bck+13     1.56&#xA;93    J0214+5222    slr+14     02:14:55.2713       34  slr+14     +52:22:40.95        19  slr+14     0.02457529479071        22  slr+14     1.16&#xA;94    J0215+6218    llc98      02:15:56.626        10  hlk+04     +62:18:33.37        10  hlk+04     0.548879819155           3  hlk+04     2.00&#xA;95    J0218+4232    nbf+95     02:18:06.358569      4  dyc+14     +42:32:17.37515      8  dyc+14     0.00232309053151224      8  dcl+16     3.15&#xA;&#xA;96    J0220+36      ttol16     02:20:50            60  ttol16     +36:22              15  ttol16     1.0297                   0  ttol16     2.25&#xA;97    J0229+20      dsm+13     02:29:03            32  dsm+13     +20:58               8  dsm+13     0.8069                   0  dsm+13     2.09&#xA;98    B0226+70      dtws85     02:31:13.99          3  hlk+04     +70:26:33.88        17  hlk+04     1.466820306152           7  hlk+04     1.76&#xA;99    J0241+16      dsm+13     02:41:46            31  dsm+13     +16:04               8  dsm+13     1.5454                   0  dsm+13     0.94&#xA;100   J0242+62      hrk+08     02:42:35             3  cvh+14     +62:56:50           36  cvh+14     0.592                    0  hrk+08     0.22&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The peculiar numbers are links in the table to the referenced publications that are the sources of the data. Ignoring those, the output is pretty simple to parse (manually or via a script, if you want).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some things to keep in mind:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Not all of the pulsars have all five parameters, so not all of the entries will be useful for you.&lt;/li&gt;&#xA;&lt;li&gt;This isn't a complete list of all pulsars in the galaxy; it's just the most comprehensive one I know (and have used). We only know a small fraction of the total galactic pulsar population.&lt;/li&gt;&#xA;&lt;li&gt;Keep your units in mind. Right ascension is given in hours-minutes-seconds format, declination is give in degrees-minutes-seconds format, and distances are in kiloparsecs.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This information should allow you to create &lt;em&gt;a&lt;/em&gt; pulsar map for use within the Milky Way. It might not be terribly helpful, given our small (and incomplete) sample, but it's a start.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2018-11-04T00:57:52.610" LastActivityDate="2018-11-04T00:57:52.610" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28254" PostTypeId="1" CreationDate="2018-11-04T03:24:04.543" Score="3" ViewCount="141" Body="&lt;p&gt;Viewing &lt;a href=&quot;https://youtu.be/zDUn_AV0HS0?t=267&quot; rel=&quot;nofollow noreferrer&quot;&gt;the animations of Oumuamua passing through the solar system&lt;/a&gt;, it is apparent that the plane of its hyperbola departs from the plane of the ecliptic by a large amount.  The plane of the ecliptic departs from that of the Milky Way Galaxy by a large amount.  Now that there is so much speculation about a potential artificial origin, one might be forgiven for speculating that there might be a coincidence between the Milky Way Galaxy's plane and that of Oumuama.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the angle between the planes of Oumuamua's hyperbola and the Milky Way Galaxy?&lt;/p&gt;&#xA;" OwnerUserId="24762" LastActivityDate="2018-11-04T15:19:34.030" Title="What is the angle between the planes of Oumuamua's hyperbola and the Milky Way Galaxy?" Tags="&lt;milky-way&gt;&lt;oumuamua&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28255" PostTypeId="1" AcceptedAnswerId="28291" CreationDate="2018-11-04T07:33:50.157" Score="4" ViewCount="4587" Body="&lt;p&gt;I took my telescope out tonight (I am new, just got the telescope a few days ago, it's my first one) and I was observing Mars, however all I see is just an bright orange circle, with no surface detail, regardless of magnification. The telescope I'm using is a 8in Newtonian, 1200mm focal length. Here's what I saw with a 8mm eyepiece and 2x Barlow:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/dGhVC.jpg&quot; alt=&quot;Mars 8mm 2x&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars was about 32° above the horizon when I took this image. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did I just do a poor job with collimation? Is the the observation angle? Please let me know!&lt;/p&gt;&#xA;" OwnerUserId="24763" LastActivityDate="2020-11-13T16:32:57.577" Title="Mars looks like a blur" Tags="&lt;telescope&gt;&lt;mars&gt;&lt;newtonian-telescope&gt;" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28256" PostTypeId="2" ParentId="28252" CreationDate="2018-11-04T09:09:36.480" Score="3" Body="&lt;p&gt;The idea of a pulsar map that can be used anywhere in the Milky Way is probably a non-starter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pulsars emit beamed radiation. That is, the radiation is restricted to a relatively narrow cone that sweeps around like a light house. If you aren't within the cone, then you don't see the pulsar. We likely only see 10% of the existing pulsars in our neighborhood (e.g. &lt;a href=&quot;https://academic.oup.com/mnras/article/414/2/1314/977538&quot; rel=&quot;nofollow noreferrer&quot;&gt;Maciesiak et al. 2011&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if you move any appreciable distance in the Milky Way you will see a quite different set of pulsars, &lt;span class=&quot;math-container&quot;&gt;$&amp;gt;90$&lt;/span&gt;% of which we don't even know about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you might be able to recognise the small subset of known pulsars that are seen at some different point, except there is a further problem. These maps are not especially useful for posterity; they are only valid over a limited cosmic epoch as well as over a limited volume. The pulsar phenomenon switches off within a million years or so and pulsar periods and pulse shapes will change over a considerably shorter time than this. The few pulsars that can be seen here and at another point in the Milky Way might be unrecognisable in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend the use of globular clusters for Milky Way mapping. Each one has a unique chemical composition mixture and HR diagram; they are more or less symmetrically distributed, can be observed over long distances, their distances can be estimated without a parallax, they will look the same tens of millions of years into the future and our catalogues are reasonably complete except on the other side of the Galactic centre.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-11-04T12:42:31.693" LastActivityDate="2018-11-04T12:42:31.693" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28257" PostTypeId="1" AcceptedAnswerId="28258" CreationDate="2018-11-04T10:43:12.570" Score="1" ViewCount="258" Body="&lt;p&gt;How should I approach this problem?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A star has a motion on the sky of 0.78 arcsec/yr in declination and 1.33 arcsec/yr in right ascension. &#xA;Calculate the velocity of the star in the plane of the sky in km/sec.&lt;/li&gt;&#xA;&lt;li&gt;The star also has a radial component of its velocity, coming towards earth&#xA;at 4.3 km/s. What is the time and distance of closest approach?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I know how to do 1. but I am not sure how to approach 2.&lt;/p&gt;&#xA;" OwnerUserId="24764" LastEditorUserId="7472" LastEditDate="2018-11-04T18:41:46.870" LastActivityDate="2018-11-04T19:38:24.057" Title="How to calculate the time and distance of closest approach of a star?" Tags="&lt;distances&gt;&lt;declination&gt;&lt;right-ascension&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28258" PostTypeId="2" ParentId="28257" CreationDate="2018-11-04T12:56:54.357" Score="1" Body="&lt;p&gt;This is not really an astronomy so much a basic Maths problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put the Earth at the origin and the star at &lt;span class=&quot;math-container&quot;&gt;$\vec{r}(t)$&lt;/span&gt; with a position &lt;span class=&quot;math-container&quot;&gt;$\vec{r_0}$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$t=0$&lt;/span&gt;. Define your axes such that the star is in the z direction and its tangential motion components are along the x and y axes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k},$$&lt;/span&gt;&#xA;where these components are the two proper motion velocities and the radial&#xA; velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume no acceleration, so that&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\vec{r} = \vec{r_0} + \vec{v} t $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use Pythagoras to get the distance to the Earth using the components of &lt;span class=&quot;math-container&quot;&gt;$\vec{r}$&lt;/span&gt; at any time &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; and then find the time at which this is a minimum (and if you are unsure how to do that &lt;span class=&quot;math-container&quot;&gt;$\rightarrow$&lt;/span&gt; Maths SE).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-11-04T19:38:24.057" LastActivityDate="2018-11-04T19:38:24.057" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28259" PostTypeId="2" ParentId="28254" CreationDate="2018-11-04T14:27:42.733" Score="2" Body="&lt;p&gt;The inclination of the orbit is 122.7 degrees (with respect to the ecliptic), whilst the Galactic plane is inclined at 60 degrees to the ecliptic. So there is 63 degrees between the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a high inclination by no means rules out that the object is orbiting within the Galactic disk. A study by &lt;a href=&quot;https://arxiv.org/abs/1710.11364&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mamajek (2017)&lt;/a&gt; concluded that it's approach trajectory was entirely consistent with an object travelling at the median velocity of stars around us. It had insignificant vertical or radial motion with respect to the local standard of rest, but is travelling about 10 km/s too slowly for a circular orbit at the Sun's Galactocentric radius (whereas the Sun moves about 10 km/s too fast).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This appears to be discrepant with the inclination angles discussed above until you recall that the sun is moving up and out of the Galactic plane at around 7 km/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at the animation at &lt;a href=&quot;https://www.cosmos.esa.int/web/gaia/iow_20180925&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.cosmos.esa.int/web/gaia/iow_20180925&lt;/a&gt; , it says a thousand words. It shows time running &lt;strong&gt;backwards&lt;/strong&gt; from the time of encounter and shows how Oumuamua comes in from inside the Sun's Galactic orbit and from &quot;above&quot;, with respect to the Galactic plane.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-11-04T15:19:34.030" LastActivityDate="2018-11-04T15:19:34.030" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28260" PostTypeId="2" ParentId="28255" CreationDate="2018-11-04T17:16:06.330" Score="2" Body="&lt;p&gt;Surface details on Mars are very low contrast (except for ice caps).  Sometimes you just can't see anything.  Poor collimation may be a problem, or bad seeing (turbulence in the atmosphere, especially a problem when Mars is low in the sky); sometimes you have to look for a while for a moment where you can glimpse detail on the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A newtonian's central obstruction will always reduce contrast on planetary details.  One answer to this is to use an &lt;a href=&quot;https://www.bbastrodesigns.com/OffAxisMasks.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;off-axis aperture mask&lt;/a&gt; for planetary and lunar observation; they are not that hard to make (sometimes they will come with your scope).  This will essentially convert your 8&quot; newtonian to a 2.5&quot; (or so) obstruction-free off-axis reflector.  This will make the overall image dimmer, but can improve contrast and lessen the impact of bad seeing.&lt;/p&gt;&#xA;" OwnerUserId="14434" LastActivityDate="2018-11-04T17:16:06.330" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28261" PostTypeId="2" ParentId="19687" CreationDate="2018-11-04T17:59:43.400" Score="6" Body="&lt;p&gt;From math's point of view it is simple. In linear algebra, for both, the model is the same, says &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;. Then:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$y = Ax$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; the observation, and &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; the physical parameters.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Forward modeling: Given &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;, calculate &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;. This is straightforward.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Inverse modeling: Given &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;, estimate &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;. Usually it is considered hard, because &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; might be a fat matrix (more cols than rows; that said, more unknowns than number of equations), and hence hard for inversion. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The reason why forward modeling is important, is that if you solve the inverse problem using, say iterative solvers, then for each step you need at least to calculate the primal matrix-vector product (&lt;span class=&quot;math-container&quot;&gt;$Ax$&lt;/span&gt;). So when it comes to inverse modeling, forward modeling is always important (so that you know how to forward modeling for &lt;span class=&quot;math-container&quot;&gt;$Ax$&lt;/span&gt;).&lt;/p&gt;&#xA;" OwnerUserId="16553" LastActivityDate="2018-11-04T17:59:43.400" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28262" PostTypeId="1" AcceptedAnswerId="28360" CreationDate="2018-11-05T00:12:38.417" Score="2" ViewCount="77" Body="&lt;p&gt;Are Vampire Stars and Symbiotic Stars the same type of star or is there a difference?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vampire Stars siphon off hydrogen from its neighbour.  According to &lt;a href=&quot;https://www.cfa.harvard.edu/~kenyon/ss/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.cfa.harvard.edu/~kenyon/ss/&lt;/a&gt;, Symbiotic Stars seem like the same description.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In all systems, the hot component - a main sequence star, a white dwarf, or a neutron star - accretes material lost by the red giant.&quot; (Harvard)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SS Leporis has been described as Symbiotic Star and in some places as Vampire Stars.  &lt;/p&gt;&#xA;" OwnerUserId="5401" LastActivityDate="2018-11-13T05:12:26.200" Title="Are Symbiotic and Vampire Stars the same type of star or different?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28263" PostTypeId="2" ParentId="28247" CreationDate="2018-11-05T01:08:49.863" Score="5" Body="&lt;p&gt;The search for sungrazing comets requires looking very close to the Sun for faint objects. This requires a chronograph and for faint objects is better done above the atmosphere. The SOHO spacecraft near the Sun-Earth L1 point does this nicely.&lt;/p&gt;&#xA;&lt;p&gt;Here is a GIF I made for &lt;a href=&quot;https://space.stackexchange.com/a/15429/12102&quot;&gt;this answer&lt;/a&gt;. In addition to some comets you can see Venus on its approach to occultation by the Sun in 2016. Seeing the Pleides so close to the Sun is exciting as well!&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;These &lt;a href=&quot;http://lasco-www.nrl.navy.mil/index.php?p=content/handbook/hndbk_3&quot; rel=&quot;nofollow noreferrer&quot;&gt;LASCO C3&lt;/a&gt; images from &lt;a href=&quot;http://sohowww.nascom.nasa.gov/home.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;SOHO&lt;/a&gt; were downloaded &lt;a href=&quot;http://sohodata.nascom.nasa.gov/cgi-bin/data_query&quot; rel=&quot;nofollow noreferrer&quot;&gt;sohodata.nascom.nasa.gov/cgi-bin/data_query&lt;/a&gt;. &lt;em&gt;The square frame is about 15.9 degrees wide&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LgrIs.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LgrIs.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You can see many trajectories for 2016 in &lt;a href=&quot;http://sungrazer.nrl.navy.mil/index.php?p=transits/transits_2016&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://sungrazer.nrl.navy.mil/index.php?p=transits/transits_2016&lt;/a&gt; (shown below) and replacing 2016 with other years works as well.&lt;/p&gt;&#xA;&lt;p&gt;The problem with this page is that while it may flag likely occultations, it does not unambiguously predict them, so you may need to use another source for accurate predictions.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/01OYE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/01OYE.jpg&quot; alt=&quot;from sungrazer(dot)nrl(dot)navy(dot)mil&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here's a quick calculation for 2016 using the Python package &lt;a href=&quot;https://rhodesmill.org/skyfield/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Skyfield&lt;/a&gt;. It seems both Mercury and Venus were able to hide behind the Sun.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kNFZ2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kNFZ2.png&quot; alt=&quot;2016 Solar occultation of planets&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h2&gt;Jackpot!  Mercury, Jupiter, Saturn, Mercury&lt;/h2&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vQZiw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vQZiw.png&quot; alt=&quot;2019.5 to 2020.5 Solar occultation of planets&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4KPVp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4KPVp.png&quot; alt=&quot;2000 to 2030 Solar occultation of planets&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;from skyfield.api import Loader, Topos, EarthSatellite&#xA;&#xA;halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)] &#xA;degs, rads        = 180/pi, pi/180&#xA;Rsun              = 696392. # https://en.wikipedia.org/wiki/Sun&#xA;&#xA;load    = Loader('~/Documents/fishing/SkyData') &#xA;ts      = load.timescale()&#xA;data    = load('de421.bsp')&#xA;&#xA;names   = ('Mercury', 'Venus', 'Mars', 'Jupiter', 'Saturn')&#xA;objects = [data[name + ' barycenter'] for name in names]&#xA;earth   = data['Earth']&#xA;sun     = data['Sun']&#xA;&#xA;ts      = load.timescale()&#xA;days    = np.arange(1, 366, 0.1)&#xA;times   = ts.utc(2016, 1, days)&#xA;&#xA;observations = [earth.at(times).observe(obj) for obj in objects]&#xA;sunobs       =  earth.at(times).observe(sun)&#xA;sundist      = sunobs.distance().km&#xA;separations  = [obs.separation_from(sunobs) for obs in observations]&#xA;&#xA;sunhalfangledegs     = degs*np.arctan2(Rsun, sundist)&#xA;sunhalfangledegsmin  = sunhalfangledegs.min()&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;    for i, (name, sep) in enumerate(zip(names, separations)):&#xA;        plt.subplot(5, 1, i+1)&#xA;        angle = degs*sep.radians&#xA;        minangle = angle.min()&#xA;        plt.plot(days, angle)&#xA;        plt.ylim(0, None)&#xA;        ymin, ymax = plt.ylim()&#xA;        ymax = 5*max(ymin, minangle, sunhalfangledegsmin)&#xA;        plt.ylim(0, ymax)&#xA;        plt.plot(days, sunhalfangledegs, '-r', linewidth=0.5)&#xA;        plt.title('observed Sun-Earth-' + name + ' angle (degs)')&#xA;    plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-11-05T03:22:43.247" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28264" PostTypeId="2" ParentId="14769" CreationDate="2018-11-05T07:27:35.803" Score="0" Body="&lt;p&gt;Pixel size is usually (beamsize/3). Based on the axes and number of pixels, you can find out pixel size by doing&#xA;pixel_size = image size(in arcsec) / number_of_pixels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, the image is done by first finding the beamwidth of the telescope and then putting 3-5 pixels inside a beamwidth in any axis.&lt;/p&gt;&#xA;" OwnerUserId="24768" LastActivityDate="2018-11-05T07:27:35.803" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28266" PostTypeId="1" CreationDate="2018-11-05T13:34:21.767" Score="2" ViewCount="170" Body="&lt;p&gt;According to Hubble more distant galaxies are moving away faster but since the light took billions of years to reach us we should say they WERE moving away faster billions of years ago. How do we know that those galaxies where light is just reaching us know are STILL moving away at all.... let alone faster&lt;/p&gt;&#xA;" OwnerUserId="24773" LastActivityDate="2018-11-05T16:56:47.083" Title="how do we know the universe is STILL expanding?" Tags="&lt;expansion&gt;&lt;observable-universe&gt;&lt;hubble-constant&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28268" PostTypeId="2" ParentId="28266" CreationDate="2018-11-05T16:28:56.743" Score="2" Body="&lt;p&gt;We don't know. This is a depressing thing that makes me, at least, feel so small. The only thing we might assume, that if they once were accelerating that fast, they probably will do it still (even faster), but of course we can't be sure if anything have happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's like this with all the long distances in the universe. How do we know if Proxima Centauri (4.2 light years away) have died/exploded? We wait, 4.2 years.&lt;/p&gt;&#xA;" OwnerUserId="24775" LastActivityDate="2018-11-05T16:28:56.743" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28269" PostTypeId="1" CreationDate="2018-11-05T16:38:04.173" Score="2" ViewCount="387" Body="&lt;p&gt;I was debating with someone online (I know, great way to get nowhere fast) regarding the size of Earth as it appears in the famous 'Earthrise' photos from Apollo 8.&#xA;Below is an accurate (almost pixel-perfect) calculation of the apparent size Earth should appear. In this instance it checks out, however if there is a better approach in calculating this and similar images I would be interested to know : &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The photograph was shot on 70mm film using a 250mm lens. We can calculate the what the Field of View should be for the image, using the following formula :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         FOV (rectilinear) = 2 * arctan (frame size/(focal length * 2))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;i.e.&#xA;             FOV (rectilinear) = 2 * arctan (70/(250 * 2))&#xA;             = a FOV of 15.93 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From our point of view on Earth, the average angular diameter of the moon is 0.5 degrees. From the Moon, the Earth has an angular diameter of about 1.9 .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this it can be calculated how many pixels in size the Earth should appear in the photograph :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diameter of Earth in Pixels = Photo's Diagonal Resolution * (Earth's Angular Diameter / FOV)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diameter of Earth in Pixels = 3841.87 * (1.9 / 15.93)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= 458.23 pixels&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For reference, The full, un-cropped version of the 'Earthrise' photo that the above calculation is basing the image resolution upon can be found at this location :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.nasa.gov/images/content/297755main_GPN-2001-000009_full.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.nasa.gov/images/content/297755main_GPN-2001-000009_full.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine that this could be further refined by deriving the correct angular diameter of Earth from the exact distance of the moon at the time the photo was taken, as well as the altitude of the Apollo module above the moon.&lt;/p&gt;&#xA;" OwnerUserId="20623" LastActivityDate="2018-11-06T10:10:33.803" Title="Calculating the Size of Earth in the Apollo 8 Earthrise Photo" Tags="&lt;the-moon&gt;&lt;photography&gt;&lt;nasa&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28270" PostTypeId="2" ParentId="28266" CreationDate="2018-11-05T16:56:47.083" Score="0" Body="&lt;p&gt;I think you raise an interesting point that is usually glossed over, and it's really a cosmology question (since it concerns the large-scale doings of the universe). And I think the best answer is that we assume the universe to be 1) isotropic and 2) homogeneous, meaning that if the universe itself has some property somewhere, then it has it essentially &lt;strong&gt;everywhere&lt;/strong&gt;. These cosmological assumptions, known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmological_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cosmological&lt;/a&gt; Principle, are very well motivated: see for instance &lt;a href=&quot;http://www.ir.isas.ac.jp/~cpp/teaching/cosmology/documents/cosmology01-02.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, which explains that so far large scale experiments justify the assumptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in truth, we don't &lt;strong&gt;know&lt;/strong&gt; in the sense that I know what time it is on my wrist watch; rather, we &lt;strong&gt;know&lt;/strong&gt; it in the sense that I know the sun will rise again tomorrow - it is a well verified assumption about nature's doings.&lt;/p&gt;&#xA;" OwnerUserId="23762" LastActivityDate="2018-11-05T16:56:47.083" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28271" PostTypeId="1" AcceptedAnswerId="28272" CreationDate="2018-11-05T17:43:16.080" Score="5" ViewCount="919" Body="&lt;p&gt;Many people say that stars twinkle because of Earth's atmosphere. However, the atmosphere is still there when we look at planets and don't see they twinkling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a clear scientific consensual explanation for the twinkling of some objects but not for others?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; &lt;em&gt;just as a metter of distinction, my question is different from &lt;a href=&quot;https://astronomy.stackexchange.com/questions/14082/why-doesnt-the-moon-twinkle&quot;&gt;this one&lt;/a&gt;. In fact, the moon is obviously different from stars, while my question is based on the fact that the only actual difference between stars and planets is that planets move across the sky, differently from stars (since they are the sky's reference). Proof of my words is the late distinction of Uranus as a planet due its extremely slow motion, although it also doesn't twinkle.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="24650" LastEditorUserId="24650" LastEditDate="2018-11-06T19:10:06.450" LastActivityDate="2018-11-06T19:35:53.113" Title="Why do stars twinkle but planets don't" Tags="&lt;star&gt;&lt;planet&gt;&lt;observational-astronomy&gt;&lt;planetary-atmosphere&gt;" AnswerCount="3" CommentCount="8" ClosedDate="2018-11-07T08:35:45.680" ContentLicense="CC BY-SA 4.0" />
  <row Id="28272" PostTypeId="2" ParentId="28271" CreationDate="2018-11-05T18:01:21.973" Score="7" Body="&lt;p&gt;The answer to this is that stars are effectively point sources, but planets have disks.  Borrowing the &lt;a href=&quot;http://skyandtelescope.com/astronomy-resources/why-do-stars-twinkle&quot; rel=&quot;noreferrer&quot;&gt;link to Sky and Telescope's explanation&lt;/a&gt; from @N. Steinle's comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I envision it is to think of the light from a celestial object as a fiber optic cable, continually being snipped at with scissors (which stand in for atmospheric turbulence changing the path of light traveling through it).  The cable from a star is a single glass fiber with a very bright beam shining down it.  The cable from a planet is a thick bundle of fibers with a much dimmer beam shining down each fiber in the bundle (compare the surface brightness of Mars with the surface brightness of the Sun)-- the total amount of light coming through the cables is comparable.  The scissors can cut through the single-fiber cable completely, but they can only take out a few strands of the thick bundle.  So atmospheric turbulence can cause the light from a star to be drastically interrupted, while the change for a planet is much less.  Therefore, there is visible twinkling from stars, but not from planets.&lt;/p&gt;&#xA;" OwnerUserId="14434" LastActivityDate="2018-11-05T18:01:21.973" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28273" PostTypeId="1" AcceptedAnswerId="28275" CreationDate="2018-11-05T19:14:35.107" Score="3" ViewCount="143" Body="&lt;p&gt;Do all stars go from Protostar stage to T Tauri and then to Main Sequence or is T-Tauri stage only for certain stars?  If only certain stars, is that those that are less than 3 solar masses which i see mentioned?  (&lt;a href=&quot;https://en.wikipedia.org/wiki/T_Tauri_star&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/T_Tauri_star&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="5401" LastActivityDate="2018-11-05T19:52:30.840" Title="Is T Tauri a compulsory stage for stellar evolution?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28274" PostTypeId="1" AcceptedAnswerId="28276" CreationDate="2018-11-05T19:49:20.893" Score="15" ViewCount="3962" Body="&lt;p&gt;When a star has finished fusing all its hydrogen into helium, it will then start fusing helium into beryllium and so on and so forth up until iron.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the star is fusing to beryllium, will the star still be in the main sequence phase and will it at that point start to grow into the red giant phase, or is there no given rule for when it will start growing?&lt;/p&gt;&#xA;" OwnerUserId="5401" LastEditorUserId="2153" LastEditDate="2018-11-05T20:14:34.703" LastActivityDate="2018-11-06T16:59:09.580" Title="Does a star fuse helium to beryllium on the main sequence?" Tags="&lt;star&gt;&lt;stellar-evolution&gt;&lt;nucleosynthesis&gt;&lt;main-sequence&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28275" PostTypeId="2" ParentId="28273" CreationDate="2018-11-05T19:52:30.840" Score="1" Body="&lt;p&gt;Yes, only certain stars that are less than about 3 solar masses were T Tauri stars. Here are some reference and quotes that I saw to answer this question. There are all review papers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Our planetary system was formed in early stages of the suns evolution. What the sun was like at that time can be learned by observing young solar-mass stars. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://link.springer.com/article/10.1023/B:ASYS.0000003267.35552.f7&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://link.springer.com/article/10.1023/B:ASYS.0000003267.35552.f7&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;T Tauri stars are newly formed low-mass stars that have recently become visible in the optical range,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.annualreviews.org/doi/abs/10.1146/annurev.aa.27.090189.002031&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.annualreviews.org/doi/abs/10.1146/annurev.aa.27.090189.002031&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This review presents the main aspects of the body of knowledge on T Tauri variable stars, which are presumed to be young, gravitationally contracting low-mass stars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0034-4885/47/2/001&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://iopscience.iop.org/article/10.1088/0034-4885/47/2/001&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24683" LastActivityDate="2018-11-05T19:52:30.840" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28276" PostTypeId="2" ParentId="28274" CreationDate="2018-11-05T20:05:57.783" Score="10" Body="&lt;h2&gt;What defines the main sequence?&lt;/h2&gt;&#xA;&lt;p&gt;Main sequence stars are characterized by hydrogen fusion in their cores, either through the proton-proton chain (for lower-mass stars) or the CNO cycle (for stars more than about 1.5 times the Sun's mass). Outside the core, no significant fusion takes place; the outer layers are involved in radiative or convective energy transport, but not energy generation. In general, if hydrogen fusion is occurring in the core, we say that a star is still on the main sequence.&lt;/p&gt;&#xA;&lt;p&gt;This changes in stars that evolve off the main sequence. Some low-mass red giants may fuse hydrogen to helium via the CNO cycle in a layer outside a largely non-reactive helium core; this is referred to as &lt;em&gt;shell burning&lt;/em&gt;. In more massive stars, heavier elements (e.g. helium, carbon, etc.) are fused inside the core, and shell burning continues in the outer layers. For instance, in a fairly high-mass star that is far into the post-main sequence phase of its life, you might see oxygen, neon, carbon, helium and hydrogen being fused in successive layers farther and farther from the core.&lt;/p&gt;&#xA;&lt;p&gt;A common misconception is that a star uses up all its hydrogen before leaving the main sequence; this is not true. It merely uses up the majority of the hydrogen in its core; there is still plenty in the outer layers, which is what makes shell fusion possible.&lt;/p&gt;&#xA;&lt;h2&gt;Post-main sequence evolution&lt;/h2&gt;&#xA;&lt;p&gt;Let's consider stars of around one solar mass. As hydrogen fusion stops in the (now degenerate) core, the source of pressure keeping the star in hydrostatic equilibrium vanishes. Hydrogen burning starts in a shell around the core. After some time, the core begins to contract, the outer envelope expands, and the star is said to be on the red giant branch. Eventually, temperatures rise to the point where the triple-alpha process can occur, and a helium flash occurs, marking the beginning of the horizontal branch and helium fusion via the triple-alpha process. Hydrogen shell burning continues.&lt;/p&gt;&#xA;&lt;p&gt;As you'll notice - and as others have said - stars &lt;em&gt;don't&lt;/em&gt; fuse helium to beryllium to any significant degree during any part of this process, or post-main sequence evolution in general. It's endothermic; the triple-alpha process is exothermic.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-11-06T16:59:09.580" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28277" PostTypeId="1" AcceptedAnswerId="28282" CreationDate="2018-11-05T22:16:42.947" Score="2" ViewCount="315" Body="&lt;p&gt;Recently, I've watched &lt;a href=&quot;https://youtu.be/y9c8oZ49pFc&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; video by Veritasium describing Poisson's Spot, or the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Arago_spot&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arago Spot&lt;/a&gt;. It is explained in the video that near circular (or spherical) objects can produce this optical effect, given that there aren't many obstructions about the circumference of the object to diffuse and scatter the light and the light producing the shadow is more or less in phase.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, &lt;strong&gt;Are Arago Spots possible/Have Arago Spots been observed in the solar shadows of any celestial bodies?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="19454" LastActivityDate="2018-11-06T03:01:32.657" Title="Arago Spot in the Shadows of Celestial Bodies" Tags="&lt;optics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28278" PostTypeId="2" ParentId="28274" CreationDate="2018-11-05T22:48:55.280" Score="19" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does a star fuse helium to beryllium on the main sequence?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stars &lt;strong&gt;don't&lt;/strong&gt; fuse helium to beryllium except as a very, very short intermediate step toward carbon. Helium-helium fusion to form beryllium is endothermic: It consumes energy. To make matters worse, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Beryllium-8&quot; rel=&quot;noreferrer&quot;&gt;beryllium-8&lt;/a&gt; that results has an extremely short half-life, less than &lt;span class=&quot;math-container&quot;&gt;$10^{-16}$&lt;/span&gt; seconds. Helium would be the end of fusion in stars (and there would be no us) if not for a fluke: The beryllium-8 formed by helium-helium fusion has almost exactly the same energy as does an excited state of carbon-12.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This greatly increases the probability of a third helium-4 nucleus combining with a short-lived beryllium-8 nucleus to form carbon-12. This is stable. The next stage after hydrogen burning is thus triple helium burning (the &lt;a href=&quot;https://en.wikipedia.org/wiki/Triple-alpha_process&quot; rel=&quot;noreferrer&quot;&gt;triple alpha process&lt;/a&gt;), essentially bypassing beryllium except as an intermediary.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When the star is fusing to beryllium, will the star still be in the main sequence phase and will it at that point start to grow into the red giant phase, or is there no given rule for when it will start growing?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A star leaves the main sequence well before it starts fusing helium. It leaves the main sequence when the star can no longer sustain hydrogen fusion in the core. This happens when the core becomes void of hydrogen. At this point, the helium left behind by hydrogen fusion is essentially ash. Hydrogen fusion proceeds on the edge of the core (shell burning), but the hydrogen-depleted core at this point is far too cold to fuse helium to carbon (not beryllium). So it collapses, and gradually becomes hotter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The star starts fusing helium to carbon (and also oxygen) if the post main sequence star's mass is large enough. At this point, the red giant collapses and behaves almost like a main sequence star with a second life. That second life doesn't last very long, however.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="332" LastEditDate="2018-11-06T03:16:40.220" LastActivityDate="2018-11-06T03:16:40.220" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28279" PostTypeId="2" ParentId="28277" CreationDate="2018-11-05T22:49:41.497" Score="2" Body="&lt;p&gt;The Arago spot phenomenon requires the light source to be point-like and very bright. The sun is not a point-like light source, as it has an apparent diameter of about 0.5 degrees&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only body large enough to cast a full solar shadow is the moon.  Arago's spot has not been observed during a total solar eclipse.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-11-05T22:49:41.497" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28280" PostTypeId="2" ParentId="28271" CreationDate="2018-11-06T01:20:34.437" Score="2" Body="&lt;p&gt;Thought I'd add on the above. Starlight is like a very narrow laser beam pointed at you. A planet is like a bunch of those arranged in a circle all pointed at you. Any one of those lasers can be deflected by the atmosphere. There is only one to deflect in the case of star light. There's several in the case of a planet. If any few lasers deflect, there's still plenty that are undisturbed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light from both stars andcplanets deflect. The planet is less like a point source. Still deflects, but the spread of source points compensates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently both planets and stars can twinkle when closer to the horizon. There is more deflection. This works to offset  the spread effects. Can't say I've  witnessed this personally though.&lt;/p&gt;&#xA;" OwnerUserId="24685" LastEditorUserId="24685" LastEditDate="2018-11-06T13:10:13.353" LastActivityDate="2018-11-06T13:10:13.353" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28281" PostTypeId="2" ParentId="11882" CreationDate="2018-11-06T02:37:39.853" Score="-1" Body="&lt;p&gt;The bulges are caused by the fact that the earth's center is free falling  and the rest of the planet is not properly.  The vector force aplied by the Moon in a certain point minus the vector force aplied by the Moon on the Earth's center is not equal to zero. This difference causes the bulges. The Earth and Moon are orbiting the center of mass of the system. Consider someone in a box free falling. The person will not feel force acting. The diference between what the Moon's force on an Earth's point is and what the Moon's force on this same point should be is the cause of the bulges. When I said should be I meant that this point falls with the Earth but only the Earth's center of mass is free falling &quot;correctly&quot; with F =Mm/r^2&lt;/p&gt;&#xA;" OwnerUserId="24396" LastEditorUserId="24396" LastEditDate="2018-11-08T21:49:17.273" LastActivityDate="2018-11-08T21:49:17.273" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28282" PostTypeId="2" ParentId="28277" CreationDate="2018-11-06T03:01:32.657" Score="2" Body="&lt;p&gt;@JamesK's &lt;a href=&quot;https://astronomy.stackexchange.com/a/28279/7982&quot;&gt;answer&lt;/a&gt; is basically correct. I'd change &quot;be point-like&quot; to &quot;be parallel&quot; or &quot;be collimated&quot;, but then I'd change that to &quot;have a high degree of partial coherence&quot; or maybe &quot;have high &lt;a href=&quot;https://en.wikipedia.org/wiki/Etendue&quot; rel=&quot;nofollow noreferrer&quot;&gt;etendue&lt;/a&gt; because a slightly converging or diverging wave front could produce a similar spot-like structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important thing is that the spread in the directions of rays from the source does not blur the spot out so much that it becomes so weak and spread out that it is not noticeable, or isn't really a spot anymore. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several other problems as well, mostly related to scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For visible light, a planet's roughness is so huge compared to a wavelength that it will not produce a small well-defined spot. See how &lt;a href=&quot;https://en.wikipedia.org/wiki/Arago_spot#Deviation_from_circularity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Deviation from circularity&lt;/a&gt; points out that the spot's existence and size &lt;em&gt;can be though of in terms of&lt;/em&gt; a point-spread function of the source. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You wouldn't compare the planet's roughness with a wavelength directly, but perhaps to something like &lt;span class=&quot;math-container&quot;&gt;$\sqrt{\lambda \ R_{planet}}$&lt;/span&gt;. For the Earth that's a few meters, so the Earth and other planets are too rough. For the &lt;a href=&quot;https://en.wikipedia.org/wiki/Arago_spot#The_circular_object&amp;#39;s_surface_roughness&quot; rel=&quot;nofollow noreferrer&quot;&gt;roughness example&lt;/a&gt; in Wikipedia using a 4 mm sphere, &lt;strong&gt;that's 45 microns&lt;/strong&gt;, and that nicely explains why the simulated edge corrugations of 10 microns do not remove the central peak, but they are weaker at 50 microns and nearly gone at 100 microns!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And while much of the contribution comes from rays that pass fairly near the obstruction, a planet's atmosphere's refraction will muddle that and have more &lt;em&gt;refractive&lt;/em&gt; than &lt;em&gt;diffractive&lt;/em&gt; effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why most demonstrations use the collimated light from a laser. It's not for the wavelength purity (the detailed fine structure depends on wavelength, but the spot is an enhancement extended along the axis for any wavelength) but for the extremely high partial coherence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that shouldn't completely stop you from looking for an Arago spot in the radio wavelengths, especially at low frequency where all the problems related to scale are less severe. Long wavelength signals from the Earth (e.g. AM radio) are blocked from reaching space by the ionosphere unfortunately, so you can't sneak around behind the Moon to look for your favorite radio station's Arago spot. That's too bad because the Moon would likely be smooth enough for those half-km wavelengths, and it's quite round (rather than oblate) as well. The nice thing about that would be the narrow frequency of the carrier in standard AM transmissions; detection of that would be far simpler than a broadband radio source.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-11-06T03:01:32.657" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28283" PostTypeId="2" ParentId="28269" CreationDate="2018-11-06T04:02:41.097" Score="1" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;What is your actual question? Are you asking others to confirm your process/formulas?&lt;/p&gt;&#xA;&lt;p&gt;Correct, specifically whether the diagonal fov is always the correct method?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;No, an incorrect approximation is never &amp;quot;always the correct method&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;Even with the basic &lt;a href=&quot;https://en.wikipedia.org/wiki/Pinhole_camera_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;pinhole camera model&lt;/a&gt; (the basis for almost all of the better approximations) position on the focal plane is never linear with incident angle. Instead, you draw similar triangles.&lt;/p&gt;&#xA;&lt;p&gt;Here is a simple 2D analogy, usually people use 3D, &lt;a href=&quot;https://en.wikipedia.org/wiki/Projective_geometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Projective Geometry&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Homography_(computer_vision)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Homography&lt;/a&gt; and the matrix math it requires.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{w}{d}=\frac{x}{f}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$x=f\frac{w}{d}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt; are the distances along the optical axis from the source and image plane to the lens pupil, and &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; are distances from the axis to the points on those planes which are perpendicular to the axis.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o8ueC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o8ueC.png&quot; alt=&quot;pinhole camera approximation&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Most higher order methods that incorporate lens distortions star with the pinhole camera approximation, and quantify the imaging system's deviations from it.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;no arctangents were harmed in the making of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Projective_plane&quot; rel=&quot;nofollow noreferrer&quot;&gt;projective plane&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-11-06T10:10:33.803" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28285" PostTypeId="1" AcceptedAnswerId="28382" CreationDate="2018-11-06T05:32:01.397" Score="3" ViewCount="517" Body="&lt;p&gt;I am trying to understand the topic of &quot;Metallicity estimates&quot;, either HII regions or planetary nebula. For this latter, the electron temperature (and density) must be known. I am studying the formalism of Osterbrock. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know, for instance, that if you use the lines [OIII]4363 and [OIII]5007 the electron temperature can be computed. In the same way, the ratio between [OII]3729 and [OII]3726 provides an estimation of electronic density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, I also know that, for the case of [OIII], its metastable excited states are relatively separated. For the case of [OII], such metastable excited states are very close. Such separation, in the metastable excited states, is crucial to compute the temperature and density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My point is that I do not understand why this separation is important, from a physical point of view. Why the ratio between these lines can estimate such properties?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can check examples of what I am talking about in &quot;Electronic Density&quot; and &quot;Electronic Temperature&quot; sections of this link: &lt;a href=&quot;http://www.astronomie-amateur.fr/feuilles/Spectroscopie/NGC2392.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.astronomie-amateur.fr/feuilles/Spectroscopie/NGC2392.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope to make me understood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much for your help.&lt;/p&gt;&#xA;" OwnerUserId="24330" LastEditorUserId="24330" LastEditDate="2018-11-13T23:40:41.107" LastActivityDate="2018-12-14T15:00:28.897" Title="Why the forbidden lines of [OIII],[NII] and [SII],[OII] are sensitive to electron temperature and electron density, respectively?" Tags="&lt;temperature&gt;&lt;spectra&gt;&lt;density&gt;&lt;metallicity&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28286" PostTypeId="2" ParentId="27916" CreationDate="2018-11-06T06:06:49.690" Score="2" Body="&lt;p&gt;Planetary Scientist Sarah Stewart's research is on the formation of the moon, not, as far as I can tell, as much on the chemical composition and precise temperature of the atmosphere after impact, so I don't know if plasma is all that relevant to her work, but I think she'd have to model and account for total energy and temperature, similar to what you did in your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand you correctly, you want to know what Earth's atmosphere was like, lets say a weeks or a few weeks or maybe a year after the giant impact.   Dr. Stewart's team has a word for this type of planet, &lt;a href=&quot;https://en.wikipedia.org/wiki/Synestia&quot; rel=&quot;nofollow noreferrer&quot;&gt;a Synestia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plasma temperature is tricky for 2 reasons.  One, there's not a specific temperature where gas becomes plasma.  Unlike melting points of boiling points, which happen at specific temperatures (and specific pressure for boiling points), the plasma phase of matter is closer to a dimmer switch that turns on gradually than a specific plasma point.  Similar to the temperature where fusion happens, individual electrons are unpredictable, so heating a gas, it will turn into a plasma gradually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also possible for a rock vapor, take SiO2 as a baseline, to retain it's double bonds as a gas and at the same time, be a low level plasma, emitting some electrons, so it can be both rock vapor and a plasma.   That's not possible for water, for example, because those bonds are too weak.  Water splits into Hydrogen and Oxygen a couple thousand degrees lower than when the individual molecules begin to enter the plasma state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem is pressure.  The center of the Earth is plasma temperature (low level Plasma but it's in that temperature range), but people generally don't call that state of matter a plasma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think your 40 km/s estimate is too high, because Theia was thought to be a Trojan object before it collided with Earth, so the collision rate should be not much more than escape velocity, maybe 12 or 13 km/s because they shared the same orbit.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;40 km/s meteor collisions on Earth happen because they approach at a different inclination, where the orbital directions are much less lined up, that's how you get 11 to 70 km/s for meteors, but Theia was probably on the low side of that, perhaps 14 or 15 km/s tops depending on it's eccentricity - if I can make a bad guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure that plasma temperature happens during giant impacts.  But the temperature is highest where the two objects collide, so initially, the highest temperature corresponds with the highest pressure.   After the impact you have the explosive rebound, because collisions of this magnitude are more like large explosions than anything else and after that you can model where the temperature goes as the Earth begins to settle.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Models would have to account for how the heat moves around and through the planet, how much heat ends up burred vs goes into rock vapor, heat of vaporization, heat lost due to expansion of rock vapor, how quickly heat radiates away (I would think it would be highly opaque, so radiation would be somewhat slow), &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bigger factors would be how much is lost in rebound and ejected material and how much is transferred to angular momentum.  There's also the uncertainty on how massive Theia was.  I think later estimates put it at 1/2 to 1/3rd the mass of Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to look at this question is, after formation, the synestia would have layers, similar to any gas giant, though gas giant layers aren't well understood, we could use the sun as an example.  There might be convection, conduction and condensation layers, and perhaps lapse rate could be applied, and obviously gravity would be lower with the material more spread out.   You might also have layers where the pressure was sufficient that different types of matter would form, like, hot enough to be liquid but enough pressure to be a solid, similar to Earth's core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that said, trying to calculate the lower atmospheric temperature of this theoretical, recently formed synestia is a little bit more math than I want to do, and I'd probably get it wrong anyway even if I did the math.  But it seems entirely reasonable that the lower atmosphere was at plasma temperatures if much of the upper atmosphere was at rock-vapor temperature.  But if you can get a temperature model, that would be a step in the direction of a plasma model.   I'd guess the low level plasma temperature for rock vapor would begin somewhere in the 5,000 or 6,000 C range, but it's a hard thing to look up as different compounds have different plasma temperatures.  There's even some cold plasma, like florescent bulbs work on that property, but they require an electric field.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if my long &quot;I don't know if there was plasma&quot; counts as an answer, but it's a fun question and I thought I'd give it a shot.   I like her idea a lot and I've read a few articles that indicated problems with the more &lt;a href=&quot;https://www.extremetech.com/extreme/253830-giant-impact-hypothesis-moon-formation-synestia&quot; rel=&quot;nofollow noreferrer&quot;&gt;traditional giant impact models&lt;/a&gt;, so she may end up being right.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-11-06T11:25:19.870" LastActivityDate="2018-11-06T11:25:19.870" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28288" PostTypeId="1" CreationDate="2018-11-06T16:54:22.190" Score="1" ViewCount="207" Body="&lt;p&gt;Why can't we travel faster than light? And is the theory of relativity absolutely flawless?&lt;/p&gt;&#xA;" OwnerUserId="24317" LastEditorUserId="7472" LastEditDate="2018-11-06T19:03:00.257" LastActivityDate="2018-11-06T22:19:44.643" Title="Why can't we travel faster than light? And is the theory of relativity absolutely flawless?" Tags="&lt;general-relativity&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2018-11-08T13:23:41.010" ContentLicense="CC BY-SA 4.0" />
  <row Id="28289" PostTypeId="1" CreationDate="2018-11-06T19:25:50.533" Score="1" ViewCount="108" Body="&lt;p&gt;Every once in a while, we have a celestial body that passes through our solar system. What would be needed to reel one of these in? Not crashing down to Earth, that would be bad, but say into a high earth orbit (above the satellite belt)?&lt;/p&gt;&#xA;" OwnerUserId="22637" LastEditorUserId="17096" LastEditDate="2018-11-07T16:37:15.090" LastActivityDate="2019-01-06T23:39:14.323" Title="What would it take to capture a passing asteroid or other celestial body into Earth's orbit?" Tags="&lt;asteroids&gt;&lt;meteor&gt;" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="28290" PostTypeId="2" ParentId="28271" CreationDate="2018-11-06T19:35:53.113" Score="7" Body="&lt;p&gt;Planets do definitely twinkle sometimes, and stars don't always twinkle as much. It's not a binary yes-or-no, it's a continuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even Venus can be seen twinkling if it's close to the horizon and the atmospheric turbulence (a.k.a. &quot;seeing&quot;) is bad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for why stars twinkle more easily than planets, you've received some answers already.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2018-11-06T19:35:53.113" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28291" PostTypeId="2" ParentId="28255" CreationDate="2018-11-06T19:59:26.683" Score="9" Body="&lt;p&gt;There's another answer here claiming that an off-axis aperture mask will improve things. That is a fairly popular meme, but it's entirely incorrect, and makes misleading claims about the effects of the central obstruction. All that does is it hides existing problems with the instrument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the off-axis hole appears to improve things, something else is very wrong and you need to fix that first; it's not a solution, it's a symptom pointing at serious issues. More importantly, it reduces precious aperture, therefore capping the resolving power of the instrument in a situation (observing Mars) when you need all the resolving power you can get. Please don't do that. Ignore the off-axis mask altogether.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what to do to get better results with Mars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Collimation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure your telescope is in perfect collimation. There is no one technique to rule them all. There are many ways to collimate a telescope. Whichever method you choose, make sure you actually use it before observing Mars. Small deviations from perfect collimation will affect results on this difficult target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your instrument's manual, or the manufacturer's website should provide some info. Here are some links to get you started:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://garyseronik.com/a-beginners-guide-to-collimation/&quot; rel=&quot;noreferrer&quot;&gt;https://garyseronik.com/a-beginners-guide-to-collimation/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.skyandtelescope.com/astronomy-resources/how-to-align-your-newtonian-reflector-telescope/&quot; rel=&quot;noreferrer&quot;&gt;https://www.skyandtelescope.com/astronomy-resources/how-to-align-your-newtonian-reflector-telescope/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some info about the importance of collimation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astrophoto.fr/collim.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.astrophoto.fr/collim.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding laser collimators, which are very popular: A laser collimator is very quick and can be very good, but provided the laser itself is perfectly centered. Plug the laser into the focuser, watch the laser spot on the primary mirror, and slowly turn the laser like a knob - if the spot moves in a circle as you turn the laser, then it's not centered. Some lasers can be adjusted until they are centered (2 or 3 adjustment screws on the barrel), others can't. A laser that's not centered will not provide good collimation for your telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many other collimation techniques, keep learning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't need to be in perfect collimation every single time - a lot of faint fuzzies (galaxies, etc) don't require any careful collimation. But for Mars and the Moon you better do it right. Don't obsess over it, but try and do a good job. Personally, I at least check collimation every time no matter what, but with practice now it only takes me 5 minutes to fix any deviation, so it's not a chore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thermal equilibrium&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your telescope is not at the same temperature as the air, convection vortices will appear inside which will affect performance. Take the instrument outside at least 1 hour before the observation and let it vent out all the heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some reading material:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-1/&quot; rel=&quot;noreferrer&quot;&gt;https://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-1/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-2/&quot; rel=&quot;noreferrer&quot;&gt;https://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-2/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Seeing&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is basically atmospheric turbulence. When it's bad, the image is distorted and fuzzy. When it's good, the image is clear and you can see small details. Basically, Mars is impossible to watch if seeing is not good or very good. Sometimes, spells of bad seeing can last for days or weeks. There's not much you can do except put the scope in the car and drive someplace else where seeing is good at that time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_seeing&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Astronomical_seeing&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing is basically weather-related, so it can be predicted to some extent. Go on the Clear Dark Sky website, search for a location close to your home, and pull up the chart:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cleardarksky.com/csk/&quot; rel=&quot;noreferrer&quot;&gt;http://www.cleardarksky.com/csk/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KpdiB.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KpdiB.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the row called Seeing. If it's dark blue for the time when you plan your observation, chances are good that seeing will not get in the way. If it's light blue or white, the forecast is not good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opposition&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars is only worth trying around oppositions, when it's closest to Earth. This happens roughly every two years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The previous opposition was in July 2018 when Mars had an apparent size of 24 arcsec.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently (November 2018) the apparent size of Mars is 11 arcsec, less than half the size at opposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next opposition is in October 2020, when it will be about 22 arcsec.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few weeks before and after opposition Mars is worth trying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.wolframalpha.com/input/?i=mars+apparent+size+in+october+13+2020&quot; rel=&quot;noreferrer&quot;&gt;https://www.wolframalpha.com/input/?i=mars+apparent+size+in+october+13+2020&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Focus&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure the scope is perfectly focused. Point it at a fixed star (Polaris is quite convenient since it does not move), and adjust the focuser until the image of the star is as small as it can be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's pointless to try fine focusing if the scope is not collimated (see above). Collimate first, then you'll be able to achieve a sharper focus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if there are seeing issues, or thermal issues, a sharp focus will be hard to achieve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other factors&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars is a difficult object to observe. Good quality optics are important. Aperture is important (more is better - please disregard the memes about limiting the aperture). Experience is important - you'll get better in time.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2018-11-07T05:40:18.480" LastActivityDate="2018-11-07T05:40:18.480" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28292" PostTypeId="2" ParentId="28288" CreationDate="2018-11-06T20:44:01.267" Score="3" Body="&lt;p&gt;No scientist really considers a theory flawless, but as theories put together by mankind go, it is one of the best ones put out there.  One of the things that makes for a great theory is not only being able to describe what's already happened, but to make predictions about new stuff, and general relativity did that with predictions of things like gravitational waves that took 80 years for technology to get good enough to see.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;General relativity also passes an aesthetics test, where some people feel that the most fundamental laws are beautifully simple and elegant, and though as far as we know, there is no scientific reason that says the Universe is simple and elegant, most of the best theories turn out to be that way, including general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you acknowledge that relativity handles all the many many tests that have been thrown at it, the answer to your first question has to do with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lorentz_factor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lorentz factor&lt;/a&gt;, which goes as &lt;span class=&quot;math-container&quot;&gt;$ \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} $&lt;/span&gt;, so you see that as your velocity &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; gets closer and closer to the speed of light &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;, the bottom of the fraction gets closer to zero and the Lorentz factor blows up.  Since the kinetic energy of something moving really fast involves the Lorentz factor, that means as you approach the speed of light, our energy blows up, and to actually travel at the speed of light, you'd need an infinite amount of energy.&lt;/p&gt;&#xA;" OwnerUserId="9749" LastActivityDate="2018-11-06T20:44:01.267" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28293" PostTypeId="2" ParentId="28288" CreationDate="2018-11-06T22:19:44.643" Score="1" Body="&lt;p&gt;One of the predictions of Special Relativity is that the relativistic mass of an object increases toward infinity as the object approaches the speed of light. The formula is given by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;m&lt;sub&gt;r&lt;/sub&gt; = m&lt;sub&gt;0&lt;/sub&gt; /sqrt(1 - v&lt;sup&gt;2&lt;/sup&gt;/c&lt;sup&gt;2&lt;/sup&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where m&lt;sub&gt;r&lt;/sub&gt; = relativistic mass, m&lt;sub&gt;0&lt;/sub&gt; is rest or invariant mass, v = velocity, and c = the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the mass of an object approaches infinity near the speed of light, it takes more and more energy to cause acceleration, and at the speed of light, it would take infinite energy to cause any further acceleration, since the object has infinite mass. Of course, you could never get to the speed of light because there isn't enough energy represented as mass in the universe to make that happen.&lt;/p&gt;&#xA;" OwnerUserId="7108" LastActivityDate="2018-11-06T22:19:44.643" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28294" PostTypeId="1" CreationDate="2018-11-07T02:10:09.057" Score="1" ViewCount="91" Body="&lt;p&gt;How old could population I stars theoretically be? What's the earliest they could have formed?&lt;/p&gt;&#xA;" OwnerUserId="6753" LastEditorUserId="6753" LastEditDate="2018-11-14T01:21:52.703" LastActivityDate="2018-11-14T01:21:52.703" Title="What is the theoretical ages for Pop I stars" Tags="&lt;stellar-evolution&gt;" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="28296" PostTypeId="2" ParentId="28239" CreationDate="2018-11-07T03:36:41.383" Score="5" Body="&lt;p&gt;I'm not an observational expert, but here's what I found. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this object (V838 Monocerotis) on the HST data archive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://archive.stsci.edu/hst/search.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://archive.stsci.edu/hst/search.php&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I input the 'Target Name' as &quot;&lt;strong&gt;V838 Monocerotis&lt;/strong&gt;&quot;, clicked all 'Imagers', 'Spectrographs', and 'Other,' and searched it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You can also do the exactly the same thing I did because this site is opened for everyone.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there are 199 images of this object. And the last observation was made in 2012. Unfortunately, I couldn't see any echoes in the picture taken in 2012. So I looked from the beginning. And I found this!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lrFVj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lrFVj.jpg&quot; alt=&quot;J9SF01020&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This picture was taken from 2006-09-09 20:04:57 to 2006-09-10 01:14:57. Dataset number is J9SF01020. You can see that the shape of the echo. There are other pictures in the archive, but I couldn't see significant echo shape anymore after that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I'm sorry to say that the party is over, the echo was faded away.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also love this picture, it means a lot to me because this picture made me study Astronomy. Thanks for making me remind this memory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: If you want other pictures, especially color pictures, check out this link too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hla.stsci.edu/hlaview.html#Images|filterText%3D%24filterTypes%3D|query_string=V838%20Monocerotis&amp;amp;posfilename=&amp;amp;poslocalname=&amp;amp;posfilecount=&amp;amp;listdelimiter=whitespace&amp;amp;listformat=degrees&amp;amp;RA=106.020093&amp;amp;Dec=-3.847399&amp;amp;Radius=0.200000&amp;amp;inst-control=all&amp;amp;inst=ACS&amp;amp;inst=ACSGrism&amp;amp;inst=WFC3&amp;amp;inst=WFPC2&amp;amp;inst=NICMOS&amp;amp;inst=NICGRISM&amp;amp;inst=COS&amp;amp;inst=WFPC2-PC&amp;amp;inst=STIS&amp;amp;inst=FOS&amp;amp;inst=GHRS&amp;amp;imagetype=best&amp;amp;prop_id=&amp;amp;spectral_elt=&amp;amp;proprietary=both&amp;amp;preview=1&amp;amp;output_size=256&amp;amp;cutout_size=12.8|ra=&amp;amp;dec=&amp;amp;sr=&amp;amp;level=&amp;amp;image=&amp;amp;inst=ACS%2CACSGrism%2CWFC3%2CWFPC2%2CNICMOS%2CNICGRISM%2CCOS%2CWFPC2-PC%2CSTIS%2CFOS%2CGHRS&amp;amp;ds=&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://hla.stsci.edu/hlaview.html#Images|filterText%3D%24filterTypes%3D|query_string=V838%20Monocerotis&amp;amp;posfilename=&amp;amp;poslocalname=&amp;amp;posfilecount=&amp;amp;listdelimiter=whitespace&amp;amp;listformat=degrees&amp;amp;RA=106.020093&amp;amp;Dec=-3.847399&amp;amp;Radius=0.200000&amp;amp;inst-control=all&amp;amp;inst=ACS&amp;amp;inst=ACSGrism&amp;amp;inst=WFC3&amp;amp;inst=WFPC2&amp;amp;inst=NICMOS&amp;amp;inst=NICGRISM&amp;amp;inst=COS&amp;amp;inst=WFPC2-PC&amp;amp;inst=STIS&amp;amp;inst=FOS&amp;amp;inst=GHRS&amp;amp;imagetype=best&amp;amp;prop_id=&amp;amp;spectral_elt=&amp;amp;proprietary=both&amp;amp;preview=1&amp;amp;output_size=256&amp;amp;cutout_size=12.8|ra=&amp;amp;dec=&amp;amp;sr=&amp;amp;level=&amp;amp;image=&amp;amp;inst=ACS%2CACSGrism%2CWFC3%2CWFPC2%2CNICMOS%2CNICGRISM%2CCOS%2CWFPC2-PC%2CSTIS%2CFOS%2CGHRS&amp;amp;ds=&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24683" LastEditorUserId="24683" LastEditDate="2018-11-07T03:57:44.913" LastActivityDate="2018-11-07T03:57:44.913" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28297" PostTypeId="2" ParentId="28294" CreationDate="2018-11-07T04:37:15.663" Score="-1" Body="&lt;p&gt;&lt;a href=&quot;https://www.scientificamerican.com/article/the-first-stars-in-the-un/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Since&lt;/a&gt; Pop III stars are the oldest stars, then they could have theoretically began to form in the Cosmic &lt;a href=&quot;https://en.wikipedia.org/wiki/Chronology_of_the_universe#The_Dark_Ages_and_large-scale_structure_emergence&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dark Age&lt;/a&gt; (CDA), see section 5, since the &lt;a href=&quot;https://en.wikipedia.org/wiki/UDFj-39546284&quot; rel=&quot;nofollow noreferrer&quot;&gt;oldest&lt;/a&gt; galaxy we've observed so far (in infrared) is about half way through the CDA (~&lt;span class=&quot;math-container&quot;&gt;$480$&lt;/span&gt; million years). As the article states,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Stars and galaxies are formed when dense regions of gas form due to the action of gravity, and this takes a long time within a near-uniform density of gas and on the scale required, so it is estimated that stars did not exist for perhaps hundreds of millions of years after recombination. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So not only must the temperature of the universe be sufficiently cool, gravity must have acted for a long enough amount of time for baryonic structure to begin to form. &lt;/p&gt;&#xA;" OwnerUserId="23762" LastEditorUserId="23762" LastEditDate="2018-11-07T04:46:07.510" LastActivityDate="2018-11-07T04:46:07.510" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28298" PostTypeId="1" AcceptedAnswerId="28306" CreationDate="2018-11-07T08:47:42.703" Score="12" ViewCount="229" Body="&lt;p&gt;During the rapid passing of ʻOumuamua through our solar system, it was &lt;a href=&quot;https://www.youtube.com/watch?v=rfi3w9Bzwik&amp;amp;t=3m59s&quot; rel=&quot;noreferrer&quot;&gt;difficult&lt;/a&gt; to get instant telescope time. Usually, formal requests and a review process are necessary before telescope time is provided to a particular project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now everybody is excited and eager to observe more interstellar objects. Did &quot;they&quot; set up an appropriate request procedure for immediate telescope usage for that matter? Or can one expect that similar hurdles and paperwork are required when the next such object is discovered?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any kind of international standard, or are such requests managed individually by each observatory?&lt;/p&gt;&#xA;" OwnerUserId="24789" LastEditorUserId="24789" LastEditDate="2018-11-10T15:34:54.750" LastActivityDate="2018-11-10T15:34:54.750" Title="After Oumuamua passage, has an &quot;urgency&quot; telescope time request procedure been set up?" Tags="&lt;telescope&gt;&lt;observational-astronomy&gt;&lt;observatory&gt;&lt;oumuamua&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28299" PostTypeId="1" CreationDate="2018-11-07T11:09:58.520" Score="2" ViewCount="248" Body="&lt;p&gt;I have read that if we image without a filter we get no information about the color or SED of objects.  Can anyone elaborate the reasons for using filters for imaging/photometry?  What happens if we image without a filter?&lt;/p&gt;&#xA;" OwnerUserId="24188" LastActivityDate="2018-11-07T22:36:08.620" Title="Why do we use filters in telescopes for astronomical imaging?" Tags="&lt;telescope&gt;&lt;observational-astronomy&gt;&lt;photography&gt;&lt;photometry&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28300" PostTypeId="2" ParentId="28299" CreationDate="2018-11-07T12:14:20.550" Score="3" Body="&lt;p&gt;In general, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Charge-coupled_device&quot; rel=&quot;nofollow noreferrer&quot;&gt;CCDs&lt;/a&gt; used to capture images do not register the energy (therefore, colour) of the incident photons on them - they just count the number of photons observed by each pixel (or a value proportional to the number of photons, as the are not 100% efficient). So, they essentially just show overall brightness variations across the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to capture colour information you therefore have to use filters. I.e., if you want to get information on the number of red photons (e.g., the intensity of red light incident on each pixel of your image), you use a filter to block out all other light. You can do this with multiple different filters to build up a full colour image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most commercial digital colour cameras the &lt;a href=&quot;https://en.wikipedia.org/wiki/Charge-coupled_device#Color_cameras&quot; rel=&quot;nofollow noreferrer&quot;&gt;CCDs&lt;/a&gt; have a &lt;a href=&quot;https://en.wikipedia.org/wiki/Bayer_filter&quot; rel=&quot;nofollow noreferrer&quot;&gt;filter mask&lt;/a&gt; over four pixel patches: two filtering for green light, one for blue light, and one for red light. The outputs from these pixel is used to build the full colour image.&lt;/p&gt;&#xA;" OwnerUserId="16795" LastActivityDate="2018-11-07T12:14:20.550" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28301" PostTypeId="2" ParentId="28299" CreationDate="2018-11-07T18:45:27.510" Score="3" Body="&lt;p&gt;As &lt;a href=&quot;https://astronomy.stackexchange.com/questions/28299/why-do-we-use-filters-in-telescopes-for-astronomical-imaging#comment50135_28299&quot;&gt;WDC pointed out in his comment&lt;/a&gt;, without filters, you simply get a recording of the received irradiance as a function of the sensor's spectral response function. In other words, a normal CCD that detects the light in a camera isn't capable of picking up every wavelength of light perfectly and the response function tells you how good that CCD is at picking up every wavelength of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes though, you don't want to take a picture and record every possible photon the CCD is capable of recording. Sometimes you want to record specific wavelengths. You do this by applying a filter, before the CCD, that let's in only specific wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has all sorts of uses. A simple example would be taking three pictures, one with a red filter to primarily let in red light, another with a green filter to primarily let in green light, and a third with a blue filter to primarily let in blue light. When you look at the individual images on your screen, the computer doesn't know what colors (i.e., wavelengths) of light  the CCD saw, it just knows how many photons were observed so it can only show you a greyscale. However, you can then, in post-processing, tint your image with the red filter red, tint your green image green, etc. and then combine you red, green, and blue images into a single, colored picture to get a close-to-true, color image of your object. In fact, &lt;a href=&quot;https://electronics.howstuffworks.com/cameras-photography/digital/digital-camera4.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;that's actually how digital cameras actually work to take colored pictures&lt;/a&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides using filters to get colored pictures, Astronomers use filters for a wide variety of science goals. It is very possible to create a special filter that only lets in a single wavelength (or as near to a single wavelength as one can get). Often single wavelengths of light are tied to specific physical processes. By that I mean, only specific physical processes can create that exact wavelength. So by looking at something with a filter for a specific wavelength, you're looking at the components of that object that created that wavelength of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common single-wavelength filter people like to use is the &lt;a href=&quot;https://en.wikipedia.org/wiki/H-alpha&quot; rel=&quot;nofollow noreferrer&quot;&gt;H-alpha&lt;/a&gt; filter and a common target for observation is the Sun. Shown below, and &lt;a href=&quot;https://apod.nasa.gov/apod/ap111115.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;taken from APOD&lt;/a&gt;, is a picture of the Sun using the H-alpha filter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cXqlq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cXqlq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or simlarly, the &lt;a href=&quot;https://www.nasa.gov/content/goddard/how-sdo-sees-the-sun&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solar Dynamics Observatory (SDO) spacecraft is constantly observing the sun in all sorts of filters&lt;/a&gt;. Note how different the sun looks at the different wavelengths!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YNZGt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YNZGt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;sup&gt; Note: these are false color images for effect! &lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="11143" LastEditorUserId="11143" LastEditDate="2018-11-07T22:36:08.620" LastActivityDate="2018-11-07T22:36:08.620" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28302" PostTypeId="2" ParentId="28289" CreationDate="2018-11-07T20:17:20.133" Score="1" Body="&lt;p&gt;You could look it up :-) .  The basic problem is that it takes a [bleep]-load of energy to force a distant object to &lt;strong&gt;stop&lt;/strong&gt; orbiting way out there and fall towards the sun, or to be exact, towards Earth.  It then takes another [bleep]-load of energy to kick it back up to the needed velocity relative to Earth to be able to orbit Earth.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To some extent, you can reduce the energy required by sending it on a slow spiral, as is more or less done when sending probes to Mars, but that leads to tens of years (at best) waiting for it to arrive.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're thinking of comets in their extreme elliptical orbits, well, same problem, since their path is nowhere near tangent to Earth's orbit. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-11-07T20:17:20.133" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28303" PostTypeId="2" ParentId="28298" CreationDate="2018-11-07T21:23:20.453" Score="2" Body="&lt;p&gt;Most observatories have a pot of Director's Discretionary Time (DDT) available to respond to these sort of unexpected events that were not part of the regular (normally) 6-month proposal cycle. The procedure and how quickly they respond varies from observatory to observatory but generally you send an abbreviated proposal of what you want to do and how much telescope time it will need and it gets considered in a short time and granted time if appropriate. The kilonova last year and SN1987A would be other examples were these this has happened in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For queue-scheduled telescopes, the new observations can be fitted into the queue fairly easily; for classically scheduled telescope where the proposer is at the telescope, then things get more complicated as that observer and observing will need to moved to a different later time if possible. For very rapid turnaround, particularly on classically scheduled telescopes, the &quot;normal&quot; procedure is to find out who's on the telescope that night from the schedule and email that observer or phone the control room and see if they will observe your object for you while you work on the DDT proposal...&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-11-07T21:23:20.453" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28304" PostTypeId="1" CreationDate="2018-11-08T01:15:45.567" Score="13" ViewCount="258" Body="&lt;p&gt;Does the axis of a galaxy as a whole &quot;wobble&quot; the same way that the Earth does?  If so do we have any idea how much or how fast the Milky Way is doing so?&lt;/p&gt;&#xA;" OwnerUserId="24791" LastActivityDate="2018-11-10T14:27:27.720" Title="Does the rotation of galaxies precess?" Tags="&lt;galaxy&gt;&lt;rotation&gt;&lt;precession&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28305" PostTypeId="1" CreationDate="2018-11-08T05:50:35.123" Score="6" ViewCount="178" Body="&lt;p&gt;I'm a math undergrad, but I'm really interested in learning positional astronomy; However, the only well-referenced textbook I've come up with is &lt;em&gt;Spherical Astronomy&lt;/em&gt; by W. M. Smart. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to understand how suitable it is today. I have the 6th edition which seems to be the latest, but which was issued &lt;em&gt;back in 1977&lt;/em&gt;. I know that a relatively old math text book, for example Apostol's Calculus (published in the 60's, if I'm not wrong), won't present too much trouble, but I'm a newcomer to astronomy, so I don't know if a book printed in 1977 is &quot;up to date&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the case that there are better or more up-to-date books, are there other recommended texts for me to read as well or is this one a good choice?&lt;/p&gt;&#xA;" OwnerUserId="24792" LastEditorUserId="7472" LastEditDate="2018-11-09T16:06:10.660" LastActivityDate="2018-11-09T16:06:10.660" Title="Is Spherical Astronomy by W. M. Smart a good book on positional/spherical astronomy?" Tags="&lt;positional-astronomy&gt;&lt;resource-request&gt;" AnswerCount="0" CommentCount="9" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28306" PostTypeId="2" ParentId="28298" CreationDate="2018-11-08T11:12:22.557" Score="6" Body="&lt;p&gt;For observing &quot;predictable&quot; transient events (e.g. Gamma Ray burst or gravitational wave events), observers &lt;em&gt;apply&lt;/em&gt; (i.e. write a proposal for peer review) for &quot;target of opportunity&quot; (ToO)time in advance. Typically, you might say we expect LIGO to give us 2 reasonably located, strong GW events dutring the next semester and we would like 3 hours of telescope time to follow each one of these up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the proposal is approved, then the ToO time can be &quot;triggered&quot; by the proposers at any time (up to their maximum of 2 events per semester, or whatever they got approved for), by contacting the observers who are currently at the telescope and asking them to execute the ToO observing that they asked for. The current observers programme is interrupted (and they may &lt;em&gt;not&lt;/em&gt; be compensated for this, other than perhaps co-authorship on a paper arising from the ToO).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Totally unpredictable events like Oumuamua or a new Galactic Supernova have to rely on director's discretionary time (DDT). This is a fast track, way of getting &lt;em&gt;some&lt;/em&gt; observing time. It is still peer-reviewed to some extent and only meritorious proposals will get any time allocated. Often a DDT proposal could be approved within 48 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If something truly extraordinary happened (and I would not classify Oumuamua in this way) that required observing &lt;strong&gt;right now&lt;/strong&gt;, such as for example a local supernova, then I imagine that any observer at a telescope would be happy to break off what they are doing and point whatever instruments they had at it. It should be noted though that there is a risk here. At many observatories (for example ESO) you are only supposed to observe what you said you were going to observe. Departures from this can get you into trouble. A wise observer would seek permission from observatory staff/astronomers before chasing up some object somebody phoned up and asked them to observe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://www.eso.org/sci/facilities/paranal/sciops/ToO_policies.html&quot; rel=&quot;noreferrer&quot;&gt;https://www.eso.org/sci/facilities/paranal/sciops/ToO_policies.html&lt;/a&gt; for some more information on how these policies and procedures work at ESO.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-11-08T11:12:22.557" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28307" PostTypeId="2" ParentId="28304" CreationDate="2018-11-08T11:49:56.223" Score="1" Body="&lt;p&gt;Precession is a general property of rotating systems under forces from external sources. For instance, Earth's precession is due to gravitational force from Sun and Moon. Since the galaxy is very far away from any other gravitational source, I don't see a reason for its precession.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; as pointed out in the comment below (@RobJeffries), if it is not sufficiently far from massive system, than it will possibly precess...&lt;/p&gt;&#xA;" OwnerUserId="24650" LastEditorUserId="24650" LastEditDate="2018-11-08T14:18:57.550" LastActivityDate="2018-11-08T14:18:57.550" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28308" PostTypeId="2" ParentId="28304" CreationDate="2018-11-08T14:05:08.177" Score="2" Body="&lt;p&gt;The whole galactic disk is made up of many individual stars and the interstellar gas between them. The whole thing doesn't tilt or wobble as a solid object would.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other parts of the Galaxy do precess though. The central supermassive black hole might and therefore the jest coming out of it might move around like the axis of the Earth's rotation does. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, galactic disks do wobble but not like a solid disk, more like a series of rings - the orbits are different at different distances from the centre. These can change very slowly and parts of the disk can be as a different angle than other parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is generally not well understood because we can't see the disks very well, being so far away, and they would change very slowly.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2018-11-08T14:05:08.177" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28309" PostTypeId="2" ParentId="14769" CreationDate="2018-11-08T14:10:44.093" Score="0" Body="&lt;p&gt;You can estimate the pixel size just from the number of pixels if you know the scale of your image or certain object in the image, but really you should already know this and the beamsize (which you could take a guess at but can't be sure). This information should be given with the image by whoever actually made it. If you have a data file it may be in the headers, or if you got the image from a website check the caption for details or an original source.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2018-11-08T14:10:44.093" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28310" PostTypeId="1" CreationDate="2018-11-08T16:29:47.303" Score="1" ViewCount="66" Body="&lt;p&gt;Does anyone know where I could obtain a program called Moonkeps? It was an old MsDos based program that produced, I believe, the Keplerian elements for tracking the Moon and also the Sun. I wish to control a rotator for tracking the Sun for a project I am messing with. Dave.&lt;/p&gt;&#xA;" OwnerUserId="24796" LastActivityDate="2018-11-08T21:51:03.860" Title="Moonkeps PC program" Tags="&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28311" PostTypeId="2" ParentId="28310" CreationDate="2018-11-08T21:51:03.860" Score="1" Body="&lt;p&gt;It can be obtained from the internet archive of amsat ftp server.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://web.archive.org/web/20120412030910/https://www.amsat.org/amsat/ftp/software/PC/tracking/moonkeps.zip&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://web.archive.org/web/20120412030910/https://www.amsat.org/amsat/ftp/software/PC/tracking/moonkeps.zip&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-11-08T21:51:03.860" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28312" PostTypeId="1" CreationDate="2018-11-08T22:37:03.110" Score="7" ViewCount="70" Body="&lt;p&gt;I am trying to detect tidal forces influence on a pendulum. I use a laser interferometer that is able to detect very tiny space changes. The pendulum never stops even after long times quiet and protected from wind by a kind of plastic box. The pendulum is 2,60 m,  long, 10Kg and one of the mirrors of the interferometer is placed in the Iron cable near to the pendulum hanging point. I can see that the top energy of the pendulum changes but I am not sure of the causes.&lt;/p&gt;&#xA;" OwnerUserId="24396" LastActivityDate="2018-11-09T00:29:20.863" Title="Does Earth changes angular velocity in 24 hours evaluation?" Tags="&lt;tidal-forces&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28313" PostTypeId="2" ParentId="28312" CreationDate="2018-11-09T00:29:20.863" Score="6" Body="&lt;p&gt;There are a series of resources on the &lt;a href=&quot;http://www.leapsecond.com/hsn2006/&quot; rel=&quot;noreferrer&quot;&gt;leapsecond.com&lt;/a&gt; site which specifically talk about detecting tides with pendulum clocks. It has been done by e.g. the &lt;a href=&quot;https://en.wikipedia.org/wiki/Shortt%E2%80%93Synchronome_clock&quot; rel=&quot;noreferrer&quot;&gt;Shortt&lt;/a&gt; pendulum clocks which had a master pendulum in a vacuum to remove disturbances synchronized to a slave pendulum. From the first link in the resources at &lt;a href=&quot;http://www.leapsecond.com/hsn2006/&quot; rel=&quot;noreferrer&quot;&gt;1&lt;/a&gt;, the order of magnitude of the change in pendulum rate caused by the change in gravity is ~1e-7 so your pendulum clock needs to be stable to a few milliseconds per day (few x10^-8) if you hope to see any tidal effects. You will need a more stable source of time such as a GPS Disciplined Oscillator (GPSDO) and normally a time interval counter to measure the stability with.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-11-09T00:29:20.863" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28314" PostTypeId="1" AcceptedAnswerId="28318" CreationDate="2018-11-09T03:16:13.220" Score="6" ViewCount="218" Body="&lt;p&gt;The question &lt;a href=&quot;https://astronomy.stackexchange.com/questions/1715/how-are-rogue-planets-discovered&quot;&gt;How are rogue planets discovered?&lt;/a&gt; describes the difficulties in finding planets (or planet-sized objects) that are floating through space without being under the influence of any star or Galaxy system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would the possible surface conditions of these planets be? Could chemotrophic life arise on such planets? &lt;/p&gt;&#xA;" OwnerUserId="24797" LastEditorDisplayName="user1569" LastEditDate="2018-11-09T15:10:02.773" LastActivityDate="2018-11-10T09:00:34.123" Title="Surface conditions of lonely (rogue) planets" Tags="&lt;terrestrial-planets&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28315" PostTypeId="1" AcceptedAnswerId="28316" CreationDate="2018-11-09T04:09:30.510" Score="0" ViewCount="160" Body="&lt;p&gt;Why was Uranus named what it was, and who came up with the name?&lt;/p&gt;&#xA;" OwnerUserId="24758" LastEditorDisplayName="user1569" LastEditDate="2018-11-09T08:36:05.500" LastActivityDate="2018-11-09T08:36:05.500" Title="Why was Uranus named what it was?" Tags="&lt;naming&gt;&lt;uranus&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2018-11-09T14:12:02.013" ContentLicense="CC BY-SA 4.0" />
  <row Id="28316" PostTypeId="2" ParentId="28315" CreationDate="2018-11-09T07:10:20.437" Score="5" Body="&lt;p&gt;Herschel, the discoverer of the planet named it Georgium Sidus, &quot;George's star&quot; after his patron, King George III of Great Britain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The name was not popular outside of Great Britain, and there were various other proposals. It was Johann Elert Bode who proposed Uranus, the Latin form of the Greek god of the sky.  It fits with the existing planets having the names of Roman Gods, and just as Saturn was the father of Zeus, so Uranus was the father of Saturn.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-11-09T07:10:20.437" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28317" PostTypeId="1" CreationDate="2018-11-09T09:55:53.150" Score="7" ViewCount="354" Body="&lt;p&gt;Guys what is that moon like thing inside the halo?&lt;img src=&quot;https://i.stack.imgur.com/Tthdf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="24804" LastEditorUserId="7982" LastEditDate="2018-11-09T10:32:27.887" LastActivityDate="2018-11-09T16:12:28.437" Title="22° Sun halo today in Zimbabwe. But there's a moon like thing inside the halo beneath the sun that cannot be seen with the naked eye, what is that?" Tags="&lt;the-sun&gt;&lt;solar-system&gt;&lt;photography&gt;" AnswerCount="1" CommentCount="8" ClosedDate="2018-11-09T18:55:31.680" ContentLicense="CC BY-SA 4.0" />
  <row Id="28318" PostTypeId="2" ParentId="28314" CreationDate="2018-11-09T14:21:49.137" Score="2" Body="&lt;p&gt;Rogue planets have two formation mechanisms: Independent formation and ejection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An independently formed rogue planet would have condensed out of the nebular material by itself and not formed from a young star's protoplanetary disk. We understand how individual stellar-mass objects condense and which have been tested by observation, but I'm not aware of any tested studies which predict rate at which Earth-massed objects independently form.  (We certainly have not observed a large population of them, and the gravitational micro-lensing studies which have been done would have detected them if they existed.)  But since Brown Dwarfs are fairly common, it seems safe to assume that there is a population of super-Jupiters that formed by direct condensation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The planets formed by ejection should range in size from meteors right up to super-Jupiters and should be broadly similar to the size distribution of bound planets. (If small planets tend to be closer to their star, there may be some bias against them being ejected.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets are mostly ejected early in a system's life, but it can happen at any point in a star's lifetime -- planetary system dynamics &lt;em&gt;never&lt;/em&gt; becomes perfectly stable.  And if there's a near passage of another star or a large rogue planet, ejections can occur even after billions of years of stability.  (See Fritz Leiber's &quot;A Pail of Air&quot;!)  But the population of ejected planets is probably very heavily weighted towards planets ejected soon after formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distinction between independently formed and ejected planets is important because when a planet forms it is very hot from the gravitational energy released in formation, and most likely initially inimical to life forming.  But it cools (the surface much faster than the interior) and eventually life can form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the planet remains in orbit around a star, the surface temperature drops asymptotically towards an equilibrium temperature where the sum of the star's radiational input and the heat leakage from the still-hot interior balances the IR radiation leaked by the planet into space. Frequently, that will be in a temperature range where life can form on the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once a planet is ejected, the equilibrium temperature will be much lower. For example, for Earth at 4.5 GY old, the Sun's radiation dumps 3000 times as much energy onto the Earth's surface than does leakage from the interior, (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth%27s_internal_heat_budget&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Earth%27s_internal_heat_budget&lt;/a&gt; for details.)  The time it takes for an Earth-sized rogue planet to cool to Earth's current temperature is on the order of 10 MY and after that it would just keep on cooling down to a surface temperature on the order of 30K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question of life &lt;em&gt;forming&lt;/em&gt; on an Earth-like planet is one of whether or not there is time before the surface layers freeze solid, since that would appear to halt the formation of life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All bets are off for larger planets which (a) cool more slowly and (b) probably have enough more water to have subsurface liquid oceans even after the surface freezes solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For planetary bodies where life does have time to form, the only source of energy would be the heat leaking from the interior either directly due to the heat gradient (a very diffuse source), or indirectly from the equivalent of Earth's deep sea vents.  The latter seems more likely since the steeper heat and chemical potential gradients are a lot easier to exploit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hal Clement wrote two scientifically excellent stories set on such planets.  One, &lt;em&gt;Star Light&lt;/em&gt;, has high-gravity aliens working with humans to explore the surface of Dhrawn, a Brown Dwarf.  The other was short fiction and had humans meeting intelligent life living on an closer to Earth-sized rogue planet. (I think it was &quot;Sortie&quot; and sequels, but I'm not sure.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any event, it seems likely that there would be a great variety of different kinds of rogue planet that might potentially support life.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorDisplayName="user1569" LastEditDate="2018-11-09T15:12:24.117" LastActivityDate="2018-11-09T15:12:24.117" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="28319" PostTypeId="2" ParentId="28317" CreationDate="2018-11-09T16:12:28.437" Score="4" Body="&lt;p&gt;It's a &lt;a href=&quot;https://en.wikipedia.org/wiki/Lens_flare&quot; rel=&quot;nofollow noreferrer&quot;&gt;lens flare&lt;/a&gt;, caused by scattered sunlight within the camera's lens system. It looks very much like one of the examples on the Wikipedia page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Qhddj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qhddj.jpg&quot; width=&quot;300&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;a href=&quot;https://commons.wikimedia.org/w/index.php?curid=3323354&quot; rel=&quot;nofollow noreferrer&quot;&gt;By Hustvedt - Own work&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2018-11-09T16:12:28.437" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28320" PostTypeId="2" ParentId="10213" CreationDate="2018-11-09T22:07:09.220" Score="5" Body="&lt;p&gt;Mercury, Venus, Mars, Jupiter and Saturn had been named as such in ancient times after various gods. We can speculate that Mercury (named after the speedy messenger god) is fast moving, Venus (a god of love and beauty) appears white and beautiful.  Mars looks red as blood (appropriate for a god of war). Jupiter is the brightest planet that can reach opposition to the sun, and so appear at midnight. Saturn is slow moving and was a &quot;grandfather&quot; god. The use of Latin names reflect the influence that Roman civilisation had on the development of European culture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Herschel, the discoverer of the planet Uranus named it Georgium Sidus, &quot;George's star&quot; after his patron, King George III of Great Britain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The name was not popular outside of Great Britain, and there were various other proposals. It was Johann Elert Bode who proposed Uranus, the Latin form of the Greek god of the sky. It fits with the existing planets having the names of Roman Gods, and just as Saturn was the father of Zeus, so Uranus was the father of Saturn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neptune was named, apparently after as a major God that did not have a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these names were well established by the time of the founding of the IAU in 1919&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="6" LastEditDate="2018-11-09T22:10:42.877" LastActivityDate="2018-11-09T22:10:42.877" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28321" PostTypeId="1" CreationDate="2018-11-09T22:56:14.380" Score="3" ViewCount="94" Body="&lt;p&gt;&lt;img src=&quot;https://airandspace.si.edu/sites/default/files/images/5317h.jpg&quot; alt=&quot;Illustration of the Apollo 11 Flight Path with labels and stages.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Howdy, this might be an odd question, but I'm using whatever information I gather here to help write an alternate universe version of the 1950s through 1980s. Due to some convoluted plot contrivance involving several delays to the moon landing, the very last pass behind the moon the Apollo crew takes place during a lunar eclipse. Now, I unfortunately don't have access to any simulation software, so I'm not sure what that would look like to them! How much of the lunar surface would the crew be able to see in passes that occur beforehand, and what does the phenomenon look like from space? Do we know what it might look like?&#xA;Time is not of the essence, but I still need to know these details before writing this particular section of the story.&lt;/p&gt;&#xA;" OwnerUserId="24807" LastEditorUserId="24807" LastEditDate="2018-11-09T23:08:55.860" LastActivityDate="2018-11-10T20:40:51.287" Title="Apollo 11 Under Different conditions" Tags="&lt;lunar-eclipse&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28322" PostTypeId="2" ParentId="28321" CreationDate="2018-11-09T23:53:06.407" Score="2" Body="&lt;p&gt;Lunar eclipses occur when the moon is full.  When the moon is full, the far side of the moon is unlit. With neither the Sun nor the Earth in the sky, the only ambient light would come from stars, and this is not bright enough to see anything by.  The moon would look completely black. The fact that a lunar eclipse is happening is largely irrelevant. The far side would be unlit during any full moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The near side of the moon would be lit by light refracted around the Earth. This would be bright enough to see the moon surface. It may be bright enough to read by. The exact brightness would depend on lots of unpredictable factors such as the amount of volcanic dust in the stratosphere. It would certainly be red just like we can see the moon during a lunar eclipse, so would the astronauts orbiting it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In previous passes the Earth would at most partially cover the moon and the nearside surface would be well lit. The far side would be unlit and dark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact the landing happened when the moon was just less than half full and so much of the far side would have been well lit.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-11-10T20:40:51.287" LastActivityDate="2018-11-10T20:40:51.287" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28324" PostTypeId="2" ParentId="28314" CreationDate="2018-11-10T08:55:31.840" Score="0" Body="&lt;p&gt;David Stevenson has theorized that a rouge planet could be ejected with considerable hydrogen in the atmosphere that would lead to a high-pressure hydrogen atmosphere.  This is highly opaque to infrared and could conceivably hold the little bit of internally-generated heat well enough that water could exist on the surface.  Here's an abstract of that paper:  &lt;a href=&quot;https://www.nature.com/articles/21811&quot; rel=&quot;nofollow noreferrer&quot;&gt;Life-sustaining Planets in Interstellar Space&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;During planet formation, rock and ice embryos of the order of Earth's&#xA;  mass may be formed, some of which may be ejected from the Solar System&#xA;  as they scatter gravitationally from proto-giant planets. These bodies&#xA;  can retain atmospheres rich in molecular hydrogen which, upon cooling,&#xA;  can have basal pressures of 102 to 104 bars. Pressure-induced&#xA;  far-infrared opacity of H2 may prevent these bodies from eliminating&#xA;  internal radioactive heat except by developing an extensive adiabatic&#xA;  (with no loss or gain of heat) convective atmosphere. This means that,&#xA;  although the effective temperature of the body is around 30 K, its&#xA;  surface temperature can exceed the melting point of water. Such bodies&#xA;  may therefore have water oceans whose surface pressure and temperature&#xA;  are like those found at the base of Earth's oceans. Such potential&#xA;  homes for life will be difficult to detect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another such paper is:  &lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/S009457651100172X&quot; rel=&quot;nofollow noreferrer&quot;&gt;Constraints on the free-floating planets supporting aqueous life&lt;/a&gt;, by Viorel Badescu.&lt;/p&gt;&#xA;" OwnerUserId="7937" LastEditorUserId="7937" LastEditDate="2018-11-10T09:00:34.123" LastActivityDate="2018-11-10T09:00:34.123" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28325" PostTypeId="2" ParentId="28304" CreationDate="2018-11-10T14:27:27.720" Score="3" Body="&lt;p&gt;To expand a bit on FJC's answer: galactic disks are not rigid bodies, and so we should not expect them to rigidly precess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most disk galaxies &lt;em&gt;do&lt;/em&gt; show warps at large radii (including our own), in both the interstellar gas and the stars. From observations of warps in gas disks, we know that the azimuthal angle of the warp often changes with radius: i.e., the tilt of the disk changes as a function of radius.[*] Since the &lt;em&gt;angular&lt;/em&gt; rotation speed changes as a function of radius,[**] this suggests the tilt could precess at different rates for different radii.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, since there isn't really a consensus on what causes or maintains warps, it's very hard to go from observing a warp to predicting what the underlying precession rates might or might not be. (Although one can argue that the precession rate can't vary &lt;em&gt;too&lt;/em&gt; much with radius, otherwise the warp would rapidly &quot;wind up&quot; and lose its coherent warped shape, and we wouldn't see warps as often as we do.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://ned.ipac.caltech.edu/level5/March11/Sellwood/Sellwood8.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for some discussion of the theoretical issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[*] This is a &lt;em&gt;kind&lt;/em&gt; of precession, although it's a change in the orientation of the rotational axis with radius rather than a change with time (as in the Earth's precession).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[**] Even if the circular &lt;em&gt;velocity&lt;/em&gt; is approximately constant, as it often is in the outer regions of galaxies, the increase in the circumference as a function of increasing radius means the angular velocity goes down.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2018-11-10T14:27:27.720" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28326" PostTypeId="1" CreationDate="2018-11-10T14:27:52.757" Score="1" ViewCount="20" Body="&lt;p&gt;We know that the Galaxies are moving very fast, not only the Galaxies other known and unknown objects in the universe, are moving. But when it was started where is the starting point, and how it move with so much speed. Is there any theory on that  ? Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="19385" LastActivityDate="2018-11-10T14:27:52.757" Title="How the Galactic objects are moving?" Tags="&lt;galaxy&gt;&lt;speed&gt;&lt;movement&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2018-11-10T22:01:26.603" ContentLicense="CC BY-SA 4.0" />
  <row Id="28327" PostTypeId="1" CreationDate="2018-11-10T16:35:43.103" Score="3" ViewCount="155" Body="&lt;p&gt;I am asking myself this question: if a giant celestial object would approach the Earth, would Earth start to turn faster (then day duration would be diminished) or slower (day duration increased) while the object starts to become nearer and nearer?&#xA;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="24816" LastActivityDate="2018-12-12T09:05:33.510" Title="If a giant object would approach the Earth, would Earth start to turn faster or slower while the object starts to become nearer and nearer?" Tags="&lt;earth&gt;&lt;rotation&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28328" PostTypeId="1" CreationDate="2018-11-10T16:55:19.837" Score="6" ViewCount="523" Body="&lt;p&gt;Correct me if I'm wrong, but the factors than enable a planet to retain or lose an atmosphere seem to be: 1. Magnetic Field, 2. Solar Winds, 3.Weight of Gases, 4. Planetary Temperatures. 5. Distance from the Sun In regards to solar winds and planetary temperatures.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A strong magnetic field usually helps to retain all the gases, even the lightest as on Earth.  The magnetic field prevents solar winds and radiation from reaching the atmosphere and exciting the gases enough to escape the gravitation planet.&lt;/li&gt;&#xA;&lt;li&gt;The Strength of the Solar Winds have the ability to excite gas molecules enough to escape the gravity of the planet. Their strength is determined by the planets proximity to the Sun. Their effects are mitigated by the strength of the magnetic fields and molecular weight of the different gases.&lt;/li&gt;&#xA;&lt;li&gt;Molecular Weight of the Gases and the gravity of the planet. While Venus has a much stronger solar wind than Mars and almost no magnetic fields, it's not a question of maintaining an atmosphere, as much as maintaining what kind of atmosphere.  The lighter gases like Hydrogen, Oxygen and Nitrogen are not present on Venus, but CO2 and heavier gases are.  The heavier gases are much less easily excited due to their heavier molecular weight and thus remain, unable to escape the gravity of the planet.  That is not to say that the heavier gases aren't aren't blown away to one degree or another, but aren't blown away in more significant amounts.  Thus Venus has an atmosphere of these gases.&lt;/li&gt;&#xA;&lt;li&gt;Planetary Temperature as well as the lack of a molecular field has a great deal to do with the lack of an atmosphere on Mars.  The lighter gases still escape but the heavier ones, more specifically CO2, remain, but mostly in frozen form.  The lower temperatures are also thought to be responsible for some remaining water, stored in the form of ice below the surface.  &lt;/li&gt;&#xA;&lt;li&gt;Distance from the sun determines the strength of the solar winds and planetary temperature.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;From what I see, these factors, interacting, determine the ability of the planets to retain an atmosphere, and of course, what kind of atmosphere.  While my writing may have some issues, I'm not so much concerned with that this point(I can work on it), as I am to know if I have an accurate and comprehensive understanding of all the known factors that determine a planets atmosphere.  Thank you.&lt;/p&gt;&#xA;" OwnerUserId="24813" LastActivityDate="2018-11-11T21:37:43.930" Title="Mars vs Venus: the retention of atmospheres in relationship to Earth" Tags="&lt;the-sun&gt;&lt;planetary-atmosphere&gt;&lt;mars&gt;&lt;venus&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28329" PostTypeId="1" AcceptedAnswerId="28337" CreationDate="2018-11-10T17:22:37.900" Score="5" ViewCount="1516" Body="&lt;p&gt;This is probably more of a math question than astronomy, but I'd like to know the distance the Earth travels in one revolution around the sun (i.e. a year).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth&quot; rel=&quot;nofollow noreferrer&quot;&gt;data in Wikipedia&lt;/a&gt;, I think there may be several ways for me to calculate this but I'm not sure whether they are correct or how to do them:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Calculate by velocity and time: the average velocity is given as &lt;span class=&quot;math-container&quot;&gt;$107200\ \mathrm{km/h}$&lt;/span&gt; (a suspiciously round number; imprecise?) and one revolution takes &lt;span class=&quot;math-container&quot;&gt;$365.256363004$&lt;/span&gt; days. That gives &lt;span class=&quot;math-container&quot;&gt;$\text{distance} = 107\,200\ \mathrm{km/h} \times (365.256363004 \times 24\ \mathrm h) = 939\,731\,570.736691\ \mathrm{km} \approx 939.731\ \mathrm{Gm}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Calculate by semi-major axis and eccentricity: my math is too weak for that. I searched the web and didn't find a solution for that (calculating the circumference of an ellipse &lt;em&gt;exactly&lt;/em&gt; seems to be a problem?)&lt;/li&gt;&#xA;&lt;li&gt;Any other way using the orbital data given in the Wikipedia article?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In essence, this can be made into a pretty generic question: &lt;em&gt;How do I calculate the distance an object travels in one revolution using the usual orbital parameters?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="16093" LastEditorDisplayName="user8021" LastEditDate="2018-11-11T11:14:15.157" LastActivityDate="2019-02-06T07:58:56.520" Title="Calculate Earth's distance travelled in a year" Tags="&lt;orbital-mechanics&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28330" PostTypeId="2" ParentId="28329" CreationDate="2018-11-10T18:23:56.770" Score="10" Body="&lt;p&gt;You need to find the perimeter of an ellipse. This might be accurate enough for you (recalling that an ellipse is only an accurate description of an orbit for a two-body system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An ellipse has no simple formula to find the perimeter. There appear to be several commonly used approximations based on infinite series (see for example &lt;a href=&quot;https://www.mathsisfun.com/geometry/ellipse-perimeter.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Maths is Fun&quot;&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. Define&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ h = \frac{(a-b)^2}{(a+b)^2},$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; are the semi-major and semi-minor axes and &lt;span class=&quot;math-container&quot;&gt;$b = a\sqrt{1-e^2}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt; is the eccentricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the result you want is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ p = \pi(a+b)\left[1 + \frac{h}{4} + \frac{h^2}{64} + \frac{h^3}{256} ...\right]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this is general, but might be overkill for your problem involving the Earth. If my calculator skills are working, then for the Earth's orbit around the Sun, &lt;span class=&quot;math-container&quot;&gt;$a = 1.00000011$&lt;/span&gt; au, &lt;span class=&quot;math-container&quot;&gt;$e = 0.01671022$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$b = 0.99986048$&lt;/span&gt; au and &lt;span class=&quot;math-container&quot;&gt;$h = 7\times 10^{-5}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this and just the first 2 terms in the series, I get a perimeter of &lt;span class=&quot;math-container&quot;&gt;$1.999895\pi$&lt;/span&gt; au. So hardly a correction from &lt;span class=&quot;math-container&quot;&gt;$2\pi a$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you were serious about accuracy then you would have to add to this about 12.5 orbits around the barycentre of the Earth-Moon system and a further correction for the motion of the Sun around the solar system barycentre.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-11-11T18:03:08.437" LastActivityDate="2018-11-11T18:03:08.437" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="28331" PostTypeId="2" ParentId="28329" CreationDate="2018-11-10T20:41:24.827" Score="3" Body="&lt;p&gt;I pondered this myself a couple of weeks ago. For Earth's orbital circumference Ramanujan's (&lt;em&gt;&lt;a href=&quot;https://www.johndcook.com/blog/2013/05/05/ramanujan-circumference-ellipse/&quot; rel=&quot;nofollow noreferrer&quot;&gt;q.v.&lt;/a&gt;&lt;/em&gt;) formula for the perimeter of an ellipse is good enough. (Actually, Jean Meeus says so in his book, but I can't find the reference. Sorry.) The Excel formula I used is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;=PI() * (F4 + F6) * (1 + 3 * ((F4 - F6)^2/(F4 + F6)^2)/(10 + (4 - 3 * ((F4 - F6)^2/(F4 + F6)^2))^1/2))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where F4 and F6 are the semi-major and semi-minor axes, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this formula I get a circumference of 939.8855 X 10&lt;sup&gt;6&lt;/sup&gt; km.&lt;/p&gt;&#xA;" OwnerUserId="7108" LastEditorUserId="7108" LastEditDate="2018-11-10T22:28:01.953" LastActivityDate="2018-11-10T22:28:01.953" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28332" PostTypeId="1" CreationDate="2018-11-11T04:37:55.533" Score="-1" ViewCount="502" Body="&lt;p&gt;I’ve heard about them and know that they are used as standard candles, but what exactly are they? What makes them differ from other supernovae? Also, are they only special because of their use as standard candles, or are the use for something else?&lt;/p&gt;&#xA;" OwnerUserId="24818" LastActivityDate="2018-11-11T05:56:57.103" Title="What is a type la supernova?" Tags="&lt;star&gt;&lt;supernova&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2018-11-18T07:22:19.333" ContentLicense="CC BY-SA 4.0" />
  <row Id="28333" PostTypeId="2" ParentId="28332" CreationDate="2018-11-11T05:53:00.267" Score="0" Body="&lt;p&gt;Type Ia is a thermonuclear explosion of a white dwarf. The other type of supernovae is core-collapse supernovae, which happen in massive &lt;span class=&quot;math-container&quot;&gt;$&amp;gt;$&lt;/span&gt;8 M&lt;span class=&quot;math-container&quot;&gt;$_\odot$&lt;/span&gt; at zero age main sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SNe Ia are used for measuring distances (as standard candles) and measuring cosmological parameters. SNe Ia are also well known to significantly produce Fe group compared to other types of SNe, therefore they are natural sites for studying related nucleosynthesis. There are some more reasons why SNe Ia are interesting.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-11-11T05:53:00.267" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28334" PostTypeId="2" ParentId="28332" CreationDate="2018-11-11T05:56:57.103" Score="2" Body="&lt;p&gt;I think these links are highly useful for you.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Standard Candle and other applications&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Type_Ia_supernova&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Type_Ia_supernova&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;This type Ia category of supernovae produces consistent peak luminosity because of the uniform mass of white dwarfs that explode via the accretion mechanism. The stability of this value allows these explosions to be used as standard candles to measure the distance to their host galaxies because the visual magnitude of the supernovae depends primarily on the distance. (...)&lt;/p&gt;&#xA;&lt;p&gt;The similarity in the absolute luminosity profiles of nearly all known Type Ia supernovae has led to their use as a secondary standard candle in extragalactic astronomy. Improved calibrations of the Cepheid variable distance scale and direct geometric distance measurements to NGC 4258 from the dynamics of maser emission, when combined with the Hubble diagram of the Type Ia supernova distances, have led to an improved value of the Hubble constant.&lt;/p&gt;&#xA;&lt;p&gt;In 1998, observations of distant Type Ia supernovae indicated the unexpected result that the universe seems to undergo an accelerating expansion. Three members from two teams were subsequently awarded Nobel Prizes for this discovery. (...)&lt;/p&gt;&#xA;&lt;p&gt;It has been discovered that Type Ia supernovae that were considered the same are in fact different; moreover, a form of the Type Ia supernova that is relatively infrequent today was far more common earlier in the history of the universe. This could have far-reaching cosmological significance and could lead to a revision of estimation of the rate of expansion of the universe and the prevalence of dark energy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Difference between other Supernovae&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Supernova#Classification&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Supernova#Classification&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;This table in the link shows the differences between them very precisely.&lt;/p&gt;&#xA;&lt;p&gt;Type Ia supernovae occur a white dwarf accretes a material from the companion in a binary system. When the white dwarf excess the Chandrasekhar limit, which is about 1.4 solar mass, it explodes. Only the explosion happens when they exceed that limit, that's why the luminosity of these explosions are nearly constant. So we can use this as a 'standard candle.'&lt;/p&gt;&#xA;&lt;p&gt;Type II supernovae, in short, are the massive star explosion. A star must have at least 8 times, but no more than 40 to 50 times, the mass of the Sun. So their luminosities are quite different from each other, therefore we cannot use these as a standard candle like SN Type Ia.&lt;/p&gt;&#xA;" OwnerUserId="24683" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-11-11T05:56:57.103" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28335" PostTypeId="2" ParentId="28327" CreationDate="2018-11-11T07:38:23.267" Score="1" Body="&lt;p&gt;Could be, you'd be better to ask this on the Physics site, by the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, the &lt;strong&gt;&lt;em&gt;spinning&lt;/em&gt;&lt;/strong&gt; of a perfectly round object is not affected at all, it just &lt;strong&gt;&lt;em&gt;keeps spinning&lt;/em&gt;&lt;/strong&gt; identically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say ..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Jls9u.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jls9u.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;say another Sun appeared, just like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, the &lt;strong&gt;&lt;em&gt;path&lt;/em&gt;&lt;/strong&gt; that Earth is taking would go crazy.  We might fly off, head towards this new Sun, starting making a figure of 8, or some crazy path!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the &lt;strong&gt;&lt;em&gt;spinning&lt;/em&gt;&lt;/strong&gt; as such of the Earth would be totally unchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;way we point&quot; (&quot;North up&quot; so to speak) would be completely unchanged, and the speed we turn at (ie, 24 hours a turn) would be unchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might seem surprising but, well, you can learn all about it in Newtonian physics!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once again, no change to the spin as such of the Earth.  the &quot;direction we point&quot; (ie the axis we spin around) and the time taken is totally unchanged if (for some bizarre reason!) we are &quot;moved around in the heavens&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note that there would be some &lt;strong&gt;&lt;em&gt;absolutely tiny&lt;/em&gt;&lt;/strong&gt; effects due to things like relativity (&lt;a href=&quot;https://www.sciencenews.org/article/einstein-general-relativity-mercury-orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;have a read!&lt;/a&gt;), the water on our planet sloshing around, and after all our Moon could be moved, bashed, etc.  These effects on the speed we are spinning would be extremely small over the short period of time you're considering, in the &quot;sci fi&quot; scenario you outline, as seen in the diagram.&lt;/p&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="13071" LastEditDate="2018-11-11T10:57:58.297" LastActivityDate="2018-11-11T10:57:58.297" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28336" PostTypeId="1" AcceptedAnswerId="28341" CreationDate="2018-11-11T09:32:39.103" Score="6" ViewCount="344" Body="&lt;p&gt;I've been experimenting with Binary Star System calculations but came across some issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can correctly calculate the Semi-Major Axis of the Earth's orbit around the Sun by doing the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/acbQf.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/acbQf.png&quot; alt=&quot;T=2pi*SQRT(a^3/GM)&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using data from Wikipedia. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a = (1.521e11 + 1.47095e11) / 2.   &amp;lt;- average radius of Earth's orbit (meters)&#xA;μ = G * M = 6.67408e(-11) * 1.98847e30 gravitational constant * Mass of Sun (kg)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I got ≈ 31557884 seconds which equates to ≈ a year&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my issue is that I've been searching around for a formula that calculates the period and etc of a Binary System where two masses (such as stars) orbit around each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So then I try essentially the same formula, &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_period&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dIaZ6.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dIaZ6.png&quot; alt=&quot;T = 2pi * sqrt(a^3 / (G(M1 + M2)))&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used data from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Procyon&quot; rel=&quot;noreferrer&quot;&gt;Procyon A and B&lt;/a&gt; (binary system)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a = Semi-Major Axis = 4.3 AU&#xA;M1 = Procyon A Mass = 1.499 Solar Masses&#xA;M2 = Procyon B Mass = 0.602 Solar Masses&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;these give me a T (period) of 4731227. It says 40.82 years on Wiki.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong? Is it the data? is it the formula? My comprehension of the question? Thank you for the help!&lt;/p&gt;&#xA;" OwnerUserId="24819" LastActivityDate="2018-11-11T17:23:08.333" Title="Doing Correct Calculations with Binary Star Systems" Tags="&lt;orbit&gt;&lt;binary-star&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28337" PostTypeId="2" ParentId="28329" CreationDate="2018-11-11T09:48:32.210" Score="4" Body="&lt;p&gt;The distance travelled by Earth in one revolution around the Sun as calculated by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth&quot; rel=&quot;nofollow noreferrer&quot;&gt;data from Wikipedia's entry about Earth&lt;/a&gt; is about &lt;span class=&quot;math-container&quot;&gt;$939.886Gm$&lt;/span&gt;, or &lt;span class=&quot;math-container&quot;&gt;$939.801Gm$&lt;/span&gt;, depending on the approach used for calculation. Since my math-fu is not very strong please point out any mistakes I've made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wolfram Alpha's answer to &lt;a href=&quot;https://www.wolframalpha.com/input/?i=circumference+of+earth+orbit+around+the+sun&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;circumference of earth orbit around the sun&quot;&lt;/a&gt; is &lt;span class=&quot;math-container&quot;&gt;$9.399 \times 10^8km$&lt;/span&gt; which is the rounded solution of the first four approaches I've arrived at and I'm thus assuming that &lt;span class=&quot;math-container&quot;&gt;$939.886Gm$&lt;/span&gt; is the most accurate solution.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Integral approach&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It seems the &lt;em&gt;exact&lt;/em&gt; way to calculate the perimeter/circumference is via the formula &lt;span class=&quot;math-container&quot;&gt;$C = 4aE(e^2)$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Complete_elliptic_integral_of_the_second_kind&quot; rel=&quot;nofollow noreferrer&quot;&gt;complete elliptic integral of the second kind&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found a &lt;a href=&quot;https://scipython.com/book/chapter-8-scipy/examples/the-circumference-of-an-ellipse/&quot; rel=&quot;nofollow noreferrer&quot;&gt;solution using Python&lt;/a&gt;. I tried using the numbers from Wikipedia here for consistency:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;from scipy.special import ellipe&#xA;a = 149598023000 # semi-major in meter&#xA;e = 0.0167086 # eccentricity&#xA;pe = 4 * a * ellipe(e * e)&#xA;print(pe) # 939886493337.0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So about &lt;span class=&quot;math-container&quot;&gt;$939.886 Gm$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Ramanujan's first approximation&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Srinivasa_Ramanujan&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ramanujan&lt;/a&gt; has given two approximations. The first is &lt;span class=&quot;math-container&quot;&gt;$C \approx \pi[3(a+b) - \sqrt{10ab + 3(a^2+b^2)} ]$&lt;/span&gt;. In Python:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import math&#xA;a = 149598023000&#xA;e = 0.0167086&#xA;b = a * math.sqrt(1 - e**2) # derive semi-minor&#xA;pe = math.pi * ( 3*(a+b) - math.sqrt( (3*a + b) * (a + 3*b) ) )&#xA;print(pe) # 9.39886493337e+11&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This gives &lt;span class=&quot;math-container&quot;&gt;$939.886Gm$&lt;/span&gt; again.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Ramanujan's second approximation&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The formula used by @BillDOe is Ramanujan's second approximation: &lt;span class=&quot;math-container&quot;&gt;$C = \pi(a+b)(1 + \frac{3h}{10 + \sqrt{4-3h}})$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$h = \frac{(a-b)^2}{(a+b)^2}$&lt;/span&gt;. In Python:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a = 149598023000&#xA;e = 0.0167086&#xA;b = a * math.sqrt(1 - e**2) # derive semi-minor&#xA;h = (a-b)**2 / (a+b)**2&#xA;pe = math.pi * (a+b) * (1 + ((3*h) / (10 + math.sqrt(4 - 3*h))))&#xA;print(pe) # 9.39886493337e+11&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;That's the same result as the first approximation. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Ellipse#Circumference&quot; rel=&quot;nofollow noreferrer&quot;&gt;error for Ramanujan's approximations&lt;/a&gt; is stated as &lt;span class=&quot;math-container&quot;&gt;$h^3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$h^5$&lt;/span&gt; respectively. That's about &lt;span class=&quot;math-container&quot;&gt;$1.157^{-25}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$2.747^{-42}$&lt;/span&gt;. I'm not a math geek, so these sound pretty low for me and I wonder why the solution is so different than the integral approach.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;@RobJeffries's &lt;a href=&quot;https://astronomy.stackexchange.com/a/28330/7982&quot;&gt;approach&lt;/a&gt;&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I don't know how this is called. The formula is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ C = \pi(a+b)\left[1 + \frac{h}{4} + \frac{h^2}{64} + \frac{h^3}{256} ...\right]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Python:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a = 149598023000&#xA;e = 0.0167086&#xA;b = a * math.sqrt(1 - e**2) # derive semi-minor&#xA;h = (a-b)**2 / (a+b)**2&#xA;pe = math.pi * (a+b) * (1 + (h/4) + (h**2/64) + (h**3/256))&#xA;print(pe) # 9.39886493337e+11&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Again, &lt;span class=&quot;math-container&quot;&gt;$939.886Gm$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Calculate from average speed&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The average speed is given as &lt;span class=&quot;math-container&quot;&gt;$29.78km$&lt;/span&gt; and a sidereal day is 365.256. Plugging in the numbers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;s = 29780&#xA;y = 365.256&#xA;pe = y * 24 * 60 * 60 * s&#xA;print(pe) # 9.39800765952e+11&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;That gives &lt;span class=&quot;math-container&quot;&gt;$939.801Gm$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16093" LastEditorUserId="16093" LastEditDate="2019-02-06T07:58:56.520" LastActivityDate="2019-02-06T07:58:56.520" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28339" PostTypeId="1" CreationDate="2018-11-11T14:10:22.417" Score="6" ViewCount="484" Body="&lt;p&gt;Like there are names for various events such as sunset, sunrise, solar noon, solar midnight, astro, nautical and civil twilight, what exactly is name for time between sunset and solar noon. Afternoon it is but what exactly between these two points? Is golden hour the only subcategory for the afternoon? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did come across the word midafternoon. similarly, midnight, midmorning and midevening. but then again midafternoon would mean 3 o clock or does it mean point between solar noon and sunset? because midnight in general is more acceptable as 12 o clock. is it?&lt;/p&gt;&#xA;" OwnerUserId="24820" LastEditorUserId="2365" LastEditDate="2018-11-12T22:45:35.267" LastActivityDate="2019-06-07T19:06:49.407" Title="In astronomy Is there a name for the exact time between solar noon and sunset?" Tags="&lt;earth&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28340" PostTypeId="2" ParentId="28328" CreationDate="2018-11-11T16:43:44.877" Score="3" Body="&lt;p&gt;One thing we have to keep in mind:&lt;br&gt;&#xA;&lt;strong&gt;All atmospheres escape, always.&lt;/strong&gt; Only the degree to which this happens is different on different planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do appreciate your question a lot, as somehow people always forget about the existence of Venus when asking &quot;what keeps an atmosphere in place?&quot;. As we will see, the classic story of &quot;a magnetic field protects, that's all&quot; is more of an urban myth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also all factors you have mentioned play a certain role. But let's pick them apart one-by-one, starting from the back in order to tell the story of escaping particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. Distance&lt;/strong&gt;&lt;br&gt;&#xA;As you've correctly noted, distance plays a role indirectly by determining solar wind and irradiation at a given distance, so this factor is out as a direct cause.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all, when modeling escape from planetary atmospheres one would rather take solar wind strength and UV-irradiation as free parameters rather than distance. This would make certain that one can compare planets around different stellar types, or stellar ages with each other.  Thus, distance is usually not considered a free parameter in the literature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Planetary Temperature&lt;/strong&gt;&lt;br&gt;&#xA;Temperature is not equal to temperature. The surface of a planet has a variation in temperature from equator to poles &lt;span class=&quot;math-container&quot;&gt;$\Delta T_{\rm EP}$&lt;/span&gt;.&lt;br&gt;&#xA;Then again, the temperature varies with altitude above the planet. Similar to the sun, the temperature difference from surface to high up in the atmosphere/corona is &lt;span class=&quot;math-container&quot;&gt;$\Delta T_{SC}$&lt;/span&gt; and one finds &lt;span class=&quot;math-container&quot;&gt;$\Delta T_{SC} \gg \Delta T_{\rm EP}$&lt;/span&gt;. &#xA;On Earth temperature variations go (very roughly) from &lt;span class=&quot;math-container&quot;&gt;$+40^{\circ}C$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$-60^{\circ}C$&lt;/span&gt;, so &lt;span class=&quot;math-container&quot;&gt;$\Delta T_{\rm EP} \approx 100 K$&lt;/span&gt;, while the temperatures in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Exosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;exosphere&lt;/a&gt; at 500km height go up to &lt;span class=&quot;math-container&quot;&gt;$T_{exo} = 1000K$&lt;/span&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modern measurments and calculations for Mars and Venus give low exospheric temperatures of around &lt;span class=&quot;math-container&quot;&gt;$300K$&lt;/span&gt;, due to efficient cooling to space of &lt;span class=&quot;math-container&quot;&gt;$\rm CO_2$&lt;/span&gt; (see this presentation by &lt;a href=&quot;http://sci.esa.int/science-e/www/object/doc.cfm?fobjectid=50647&quot; rel=&quot;nofollow noreferrer&quot;&gt;Coates et al.&lt;/a&gt; for Venus and a recent review by &lt;a href=&quot;http://www.issibern.ch/teams/evolexoplanet/Articles/Lammer-et-al_SpaceSciRev_2008.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lammer et al. (2008)&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the escape of particles into space, it is the lower boundary of the exopshere, the exobase, that determines escape rates. Particularly, it is the temperature and the density at the exobase that determines escape rates, as from here the fast tail of the Maxwell-Boltzmann distribution can directly head off into space. This is a particular escape process called &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_escape&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jeans-escape&lt;/a&gt;. Below the exobase particles that travel upwards at escape speeds will encounter on average more than one particle and thus be scattered without escape. This is btw. also the definition of where to find the exobase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is to illustrate that neither the surface temperature of planets, nor the variations thereof play any role for atmospheric escape. Very strictly this is not true, as the atmosphere of a star and possible young rocky planet can get hot enough to launch a &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrodynamic_escape&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrodynamic wind&lt;/a&gt; into space, greatly increasing escape rates. For the sun we call this phenomenon &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_wind&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Parker solar wind&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's take a look how particles can escape from the exobase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3. Particle mass&lt;/strong&gt;&lt;br&gt;&#xA;High up in the atmosphere, at around 100km height, there lies another atmospheric boundary, the homopause. Here mixing of particles due to large-scale motions of the air becomes so inefficient, that particles start forming layers according to their masses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heavier atomic and molecular species will thus, once they reach the exosphere &lt;span class=&quot;math-container&quot;&gt;$\sim$&lt;/span&gt;500km be there in much smaller numbers than, for example Hydrogen. Thus, hydrogen escape rates will always be higher than other species.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the 'classical' picture, which cannot explain some mysteries involving the noble gases. Noble gases are chemically inert, meaning they don't form molecules and once in the atmosphere, they should just hang around there for billions of years. Except for those who escape. This is why noble gases are giving important datapoints when constructing theories of what happened to Earth's atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important problem is that Xenon is missing, relative to its lighter counterpart Krypton and the other noble gases. This shouldn't be if the classical picture is correct. I recommend reading the introduction in the article of &lt;a href=&quot;https://arxiv.org/abs/1809.06960v1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zahnle et al. (2018)&lt;/a&gt; for a very detailed picture of this &quot;missing Xenon problem&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to address this problem and this article, let me introduce another particle parameter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4. Ionization energies&lt;/strong&gt;&lt;br&gt;&#xA;So particle mass only doesn't explain escape rates. the article of &lt;a href=&quot;https://arxiv.org/abs/1809.06960v1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zahnle et al. (2018)&lt;/a&gt; (and references therein, this is not a new idea) however proposes that if escape rates are mostly given by ions, then a different picture emerges, that could reconcile escape via mass fractionation and missing Xenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general idea is that because the ionization energies differ as we move through the periodic system of elements, &lt;a href=&quot;http://butane.chem.uiuc.edu/pshapley/GenChem1/L6/1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;it is much easier to ionize Krypton and Xenon&lt;/a&gt;. A mass of ionized hydrogen, or protons will be pre-existent in the high atmosphere. Ion-ion collisions are much more efficient in coupling species together than ion-neutrals.&lt;br&gt;&#xA;So if we now assume that instead of the neutral hydrogen, the protons mostly escape, they'd have dragged Krypton and Xenon ions along.&lt;br&gt;&#xA;Except that under Earth atmospheric conditions the Krypton ions quickly recombine to neutral Krypton, while Xenon doesn't. So the Zahnle article concludes that the missing Xenon is a signpost of past, dominant ion escape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if ions are this important for atmospheric escape, as opposed to the neutral species, we probably have to think more about the magnetic field lines that they follow. Finally we get to discuss the planetary magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5. Magnetic field and solar wind&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some magnetic field lines connect to the field in interplanetary space. Ions travelling along those will be inevitabely lost to space and picked up by the solar wind. The strength of this effect is dominated by the geometry of field lines intersecting with the solar wind, and can lead to net protection or net erosion of the atmosphere.&lt;br&gt;&#xA;This, among with other effects like pick-up and sputtering have been summarized in a comparative article on Earth, Mars and Venus in &lt;a href=&quot;http://adsabs.harvard.edu/abs/2018A%26A...614L...3G&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gunell et al. (2018)&lt;/a&gt;. Their key finding is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;While a planetary magnetic field protects the atmosphere&#xA;  from sputtering and ion pickup, it enables polar cap and cusp escape,&#xA;  which increases the escape rate. Furthermore, the induced&#xA;  magnetospheres of the unmagnetised planets also provide protection&#xA;  from sputtering and ion pickup in the same way as the&#xA;  magnetospheres of the magnetised planets. Therefore, contrary&#xA;  to what has been believed and reported in the press (Achenbach&#xA;  2017), the presence of a strong planetary magnetic field does not&#xA;  necessarily protect a planet from losing its atmosphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They find, that with all those complications of different escape processes and different ionized and neutral species, still for any single planet the ion escape rates can outcompete the neutral ones by a factor of &lt;span class=&quot;math-container&quot;&gt;$\sim 4$&lt;/span&gt;. Is this enough to explain the retention of the Venusian atmosphere and the escape of the Martian one? I think not. There is another factor that we've ignored so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;6. Planetary mass&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To close off this escape-story, I want to come back to the beginning of my answer, where I was happy that you've mentioned Venus. This is because Mars and Venus form a small set of a nearly identical case-study on atmospheric escape. Only two factors are different, while it is much harder to compare any other two planets in the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sake of comparing Mars and Venus to zeroth order, one could say that their atmospheres have identical composition, which is mostly &lt;span class=&quot;math-container&quot;&gt;$\rm CO_2$&lt;/span&gt;, as you have already stated. Both have no intrinsic magnetic field. Then, Venus has a hotter exosphere, stronger solar wind conditions, but still is somehow able to retain an atmosphere that has several thousand times more mass than the martian one.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we now add to our perspective that a magnetic field doesn't play such a huge role for the retention of atmospheres, then the only remaining parameter that is different between Mars and Venus is the mass, which differs by a factor of about 10.&lt;br&gt;&#xA;A factor of 10 is significant here, because if we go to planets now with 10 Earth-masses we already get into the regime of the Ice-giants Uranus and Neptune, which are able to hold on to their neutral hydrogen.&lt;br&gt;&#xA;Understanding this is comparatively simple, as the totel escaping flux &lt;span class=&quot;math-container&quot;&gt;$\Phi_0$&lt;/span&gt; is the integral of the Maxwell-Boltzmann-tail which goes very roughly as&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \Phi_0 \sim n(z_{\rm exo}) \cdot v_{\rm rms} \cdot \left( \frac{v_{\rm esc}^2}{v_{\rm rms}^2} + 1  \right) \cdot \exp(-\frac{v_{\rm esc}^2}{v_{\rm rms}^2}) $$&lt;/span&gt;&#xA;(source: &lt;a href=&quot;http://sci.esa.int/science-e/www/object/doc.cfm?fobjectid=50647&quot; rel=&quot;nofollow noreferrer&quot;&gt;Coates&lt;/a&gt;, or homework problems...)&#xA;where &lt;span class=&quot;math-container&quot;&gt;$n(z_{\rm exo})$&lt;/span&gt; is the number density of a species at the height of the exosphere, &lt;span class=&quot;math-container&quot;&gt;$v_{\rm esc} = \sqrt{2GM/R}$&lt;/span&gt; is the escape velocity of a planet with &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; the mass and &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; the radius and &lt;span class=&quot;math-container&quot;&gt;$v_{\rm rms} = \sqrt{2 kT_{\rm exo}/m}$&lt;/span&gt; is the root-mean-square velocity of the MB-distribution at a given exospheric temperature &lt;span class=&quot;math-container&quot;&gt;$T_{\rm exo}$&lt;/span&gt; for a given mean molecular mass &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt;. So as there is an exponential factor involved, the Jeans-escape rate must go up quickly as planetary mass goes down.&#xA;So as ionic escape rates are a factor of a few times the neutrals, even if they dominate the scape, they're still bound to the exponential in the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More information can also be found &lt;a href=&quot;https://geosci.uchicago.edu/~kite/doc/Catling2009.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;in this&lt;/a&gt;, a bit older article by the same authors as the Zahnle paper, where they speculate also a little bit about escape from exoplanets, and the role of atmospheric chemistry for this effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The escape rate of ionic species can dominate over the escape rates of neutral species.&#xA;Ionic escape rates however don't react as strongly as one could think to the presence of a magnetic field. This leads to the interesting picture that the presence of a magnetic field is only a second order effect for determining escape rates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dominant parameter is then simply how deep the gravitational well of a planet is, that ions need to escape from. A escape rates go exponentially with planetary mass vs. molecular weight and available thermal energy at the exobase this is what determines escape rates most strongly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noble gases and their depletion can possibly resolve some mysteries of Earth's past, all the while telling us how much hydrogen has escaped from Earth.&lt;br&gt;&#xA;As technology progresses, we might be able to address the same questions for Mars and Venus one day, but we're not there yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was a long rant, please tell me if something is unclear.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditDate="2018-11-11T21:37:43.930" LastActivityDate="2018-11-11T21:37:43.930" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28341" PostTypeId="2" ParentId="28336" CreationDate="2018-11-11T17:23:08.333" Score="5" Body="&lt;p&gt;The semi-major axis is 4.3 &lt;em&gt;arcseconds&lt;/em&gt;. At a distance of 3.51 parsecs, this corresponds to 15.09 au. Using your formula I get 40.3 years, which is close enough given the 2 significant figure precision of the semi-major axis.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-11-11T17:23:08.333" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28342" PostTypeId="2" ParentId="28339" CreationDate="2018-11-11T17:35:31.273" Score="7" Body="&lt;p&gt;No, there is no such word better than &quot;afternoon&quot;. I can't think of a situation that would require technical language for this period of time, the exact point halfway. When astronomers want to be technical about time they use UTC or UT1 or AT1 or any of the other &quot;times&quot;. When they want to talk about more loosely defined periods they use &quot;morning&quot;, &quot;afternoon&quot;, &quot;evening&quot; just like everybody else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Mid-afternoon&quot; is an ill-defined period, from about 2pm to about 4pm&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-11-12T22:44:34.570" LastActivityDate="2018-11-12T22:44:34.570" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28343" PostTypeId="2" ParentId="20868" CreationDate="2018-11-11T18:05:53.380" Score="5" Body="&lt;p&gt;This is kind of old, but I've been reading a bit about CMB and stumbled on this question. If you go back to the original discovery of the CMB, they basically pointed their antenna at the sky and measured the microwave signal. They found one that was &quot;isotropic, unpolarized, and free from seasonal variation&quot; having run the antenna from July 1964 to April 1965. To calibrate the instrument they had a liquid helium cooled reference that they could switch into the receiving circuit. Blocking the CMB signal would be difficult because it's at 2.7 K, and most anything you would block it with would be a larger source (i.e. warmer). Instead, they used a liquid He reference source as a calibration, and then accounted for all other sources of microwave signal. The biggest &quot;other&quot; source was from looking through the atmosphere. You can read about this here:&#xA;&lt;a href=&quot;http://adsbit.harvard.edu/full/seri/ApJ../0142//0000420.000.html&quot; rel=&quot;noreferrer&quot;&gt;http://adsbit.harvard.edu/full/seri/ApJ../0142//0000420.000.html&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.wmrc.edu/resources/docs/ch1.pdf&quot; rel=&quot;noreferrer&quot;&gt;https://www.wmrc.edu/resources/docs/ch1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To really see what they did, you'd have to read more about that antenna, and there are references in the paper. The original publication is just a letter and so it is terse and refers to previous results instead of offering a detailed description.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...but if you're in orbit, it's easier to make something cold. For FIRAS on COBE, they actually had a device called XCAL (External Calibrator, I guess) that fit into the CMB receiving antenna like a trumpet mute. It looks like they held it at 2.7K and used it to calibrate by moving it into and out of the microwave receiving horn. That's pretty close to blocking the measurement signal. Here's the reference:&#xA;&lt;a href=&quot;https://www.researchgate.net/publication/2669262_Calibration_of_the_COBE_FIRAS_instrument&quot; rel=&quot;noreferrer&quot;&gt;https://www.researchgate.net/publication/2669262_Calibration_of_the_COBE_FIRAS_instrument&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24822" LastEditorUserId="24822" LastEditDate="2018-11-11T19:02:33.663" LastActivityDate="2018-11-11T19:02:33.663" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28344" PostTypeId="2" ParentId="28255" CreationDate="2018-11-11T18:29:13.347" Score="1" Body="&lt;p&gt;You wrote that your telescope comes with 1200mm focus. Your diameter is 200mm. Check out the length of your telescope. If it's less then 600mm, then you have troubles, because your telescope could use &quot;corrector&quot; lense in front of your ocular lenses. Such &quot;corrector&quot; is weak part, because it most likely corrects the aspheric error of main mirror and making such correctors is very costly. Producers could use cheaper parts which destroy your telescope, but increase their profits. If you find out that there is corrector, then you got solution: buy proper corrector. But such task is not for beginners. Consult with other peers!&lt;/p&gt;&#xA;" OwnerUserId="24823" LastActivityDate="2018-11-11T18:29:13.347" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28345" PostTypeId="2" ParentId="20868" CreationDate="2018-11-11T19:08:34.047" Score="0" Body="&lt;p&gt;COBE, WMAP and Planck are all spacecraft that have measured the same cosmic microwave background (in increasing detail) from &lt;em&gt;outside&lt;/em&gt; the Earth's atmosphere. The same background is found, with similar anisotropies, from different spacecraft using different telescopes, detectors and analysis techniques to remove non-cosmological sources.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-11-11T19:08:34.047" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28347" PostTypeId="1" AcceptedAnswerId="28359" CreationDate="2018-11-12T01:30:32.730" Score="3" ViewCount="590" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;GMST = 6.697374558 + 0.06570982441908 D&lt;sub&gt;0&lt;/sub&gt; + 1.00273790935 H + 0.000026 T&lt;sup&gt;2&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the &lt;a href=&quot;http://aa.usno.navy.mil/faq/docs/GAST.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;GMST calculation&lt;/a&gt; on the USNO website (and many other places) we encounter these constants:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;6.697374558&lt;/li&gt;&#xA;&lt;li&gt;0.06570982441908&lt;/li&gt;&#xA;&lt;li&gt;1.00273790935&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Can someone enlighten me as to what they represent?&lt;/p&gt;&#xA;" OwnerUserId="24826" LastEditorUserId="2618" LastEditDate="2018-11-12T09:33:32.337" LastActivityDate="2018-11-12T22:44:14.777" Title="In the calculation of GMST, what are these constants?" Tags="&lt;time&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2018-11-12T23:59:09.090" ContentLicense="CC BY-SA 4.0" />
  <row Id="28349" PostTypeId="2" ParentId="23449" CreationDate="2018-11-12T03:37:17.617" Score="4" Body="&lt;p&gt;Yes the Alpha Centauri system is moving closer and it's not the only one. Barnard's Star, the fastest moving star in our skies, will get close sooner. The major concern is whether or not they will get close enough to disrupt our Oort cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3Bqxg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3Bqxg.png&quot; alt=&quot;Near stars, past &amp;amp; future&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Near-stars-past-future-en.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;diagram&lt;/a&gt; shows the relative distance of various nearby stars over time. So Alpha Centauri was close to 6LY away 20,000 years ago and will be at its closest position at around 3LY about 30,000 years from now.&lt;/p&gt;&#xA;" OwnerUserId="24827" LastEditorUserId="16685" LastEditorDisplayName="user1569" LastEditDate="2020-03-19T09:46:10.593" LastActivityDate="2020-03-19T09:46:10.593" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28350" PostTypeId="2" ParentId="28327" CreationDate="2018-11-12T04:09:04.027" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If a giant object would approach the Earth, would Earth start to turn faster or slower while the object starts to become nearer and nearer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;tl;dr:&lt;/strong&gt; Probably will slow it down a small amount, but may induce a wobble that make a simple answer impossible without further details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is complicated. I'll leave a short answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is carefully written in (at least two) ways. It is agnostic to the size of the effect, instead asking only about its sign. So arguments that the effects are too tiny to matter are moot. It also does not specify the nature, distance, or duration of the approach. That makes the question quite difficult to answer definitively. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;induced tidal distortion&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The most straight-forward effect we know about is slowing due to tidal forces. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_acceleration#Theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Theory&lt;/a&gt; subsection of Wikipedia's article &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_acceleration&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tidal acceleration&lt;/a&gt; outlines this nicely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the Earth is slightly stretchy, the Moon's or another close body's gravity induces a tidal bulge. Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth_tide&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth tide&lt;/a&gt; gives several components of a thorough decomposition, but the top two are about 38 cm due to the Moon and 17 cm due to the Sun. These exist on both the near and far side of each body. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the Earth responds slowly, these have a &quot;lag&quot;. The lumps exist offset from the line between the Earth and the other body, and this allows for a torque. In the case of the Earth-Moon system that torque then tends to slow the Earth's rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the body is double in mass, the induced bulge's mass will be roughly linearly proportional and since the torque proportional to both it will be quadruple, proportional to &lt;span class=&quot;math-container&quot;&gt;$m^2$&lt;/span&gt;. As shown in that subsection it is also proportional to $r^{-6}. So just for example, a planet about the size of the Earth (~80 Moons) and half the distance of the moon would have a torque 80 x 80 x 64 times larger. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it's hard to imagine a scenario where that could last for a thousand years to have a huge effect on Earth's rotation speed. This would be small, but definitely negative.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;interaction with equatorial bulge&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The body could pass perpendicular to the Earth's equator as well, and produce a really huge torque on Earth's equatorial bulge (expressed often as the &lt;span class=&quot;math-container&quot;&gt;$J_2$&lt;/span&gt; term). That bulge is more like &lt;em&gt;30 kilometers&lt;/em&gt;, compared to the 30 centimeters of the induced tidal bulge by the Moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A perpendicular pass like that would induce a substantial wobble in the Earth's rotation during the time of passage, but I don't think it is easy to say with certainty the sign of the effect on the rotation rate with such complex motion and no further details.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-11-12T04:09:04.027" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="28353" PostTypeId="2" ParentId="28339" CreationDate="2018-11-12T14:31:17.827" Score="2" Body="&lt;p&gt;If you are writing technical documentation, you could define, within your document, that &quot;mid-afternoon&quot; means the instant midway between noon (mean or apparent?) and sunset.  You will need to specifically define what that instant is when sunset does not occur (polar day or polar night).  Or you can ignore the seasons and declare that this moment is defined to be always 15:00 apparent solar time ;-)&lt;/p&gt;&#xA;" OwnerUserId="1462" LastActivityDate="2018-11-12T14:31:17.827" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28354" PostTypeId="1" AcceptedAnswerId="28357" CreationDate="2018-11-12T14:57:08.853" Score="7" ViewCount="91" Body="&lt;p&gt;Gravitational wave detectors &lt;a href=&quot;https://www.aps.org/publications/apsnews/201710/virgo.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;VIRGO and LIGO&lt;/a&gt; are currently down for some updates, slated to be completed sometime early next year (they hope). What exactly are they doing? I hadn't heard it was down again after the last set of updates awhile ago. What are they updating now-- is that knowledge public domain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt; I went to an astronomy club meeting and it was stated in the presentation.&lt;/p&gt;&#xA;" OwnerUserId="23729" LastEditorUserId="7982" LastEditDate="2018-11-12T23:55:36.630" LastActivityDate="2018-11-12T23:55:36.630" Title="What changes are being made to VIRGO and LIGO (if any)?" Tags="&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28355" PostTypeId="1" CreationDate="2018-11-12T17:44:06.433" Score="4" ViewCount="72" Body="&lt;p&gt;So there's stuff very far away that is, due to expansion, accelerating away faster than the speed of light, so fast that its light will never reach us even with infinite time - and we can't reach that point even if we travel at the speed of light for all eternity. The comoving distance of where the universe becomes fundamentally unexploreable is, afaik, some 46.5 billion light years or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretical Einstein-Rosen bridges could connect any two points of space-time. What I wonder though is whether this could also apply to points that are farther than these 46.5 billion light years apart, or whether this is a theoretical maximum distance for wormholes as well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether the answer is yes or no - an explanation would be great :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: maybe I'm mixing up observable universe and hubble volume here - I mean the one relevant to the question of course :)&lt;/p&gt;&#xA;" OwnerUserId="24834" LastEditorUserId="24834" LastEditDate="2018-11-12T17:49:28.383" LastActivityDate="2018-11-12T17:49:28.383" Title="Can Einstein-Rosen bridges - if they exist - link two points of which one lies outside of the observable universe of the other?" Tags="&lt;observable-universe&gt;&lt;wormhole&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28356" PostTypeId="1" CreationDate="2018-11-12T19:44:16.450" Score="0" ViewCount="539" Body="&lt;p&gt;I'm looking for a mathematical justification of the &lt;strong&gt;longitude of periapsis&lt;/strong&gt; as it makes no mathematical sense to me as a notion. By definition, it is the sum of the argument of periapsis and the longitude of the ascending node, each of which makes sense on their own.  I understand these latter two notions geometrically. But I don't know what the longitude of periapsis is supposed to represent.  The longitude of the ascending node is an actual angle measured within the ecliptic, while the argument of periapsis is an actual angle measured within the given bodies orbital plane.  These two planes are skewed with respect to one another by an inclination angle.  Thus, you cannot mathematically just add the two angles together; it geometrically has no representation.  At least not without making certain unmentioned assumptions which cannot possibly be true for all bodies.  I cannot make sense of why it seems to be the preferred orbital parameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is not a duplicate of &lt;a href=&quot;https://astronomy.stackexchange.com/questions/11477/what-is-exactly-the-longitude-of-the-perigee&quot;&gt;What is exactly the &amp;quot;longitude of the perigee&amp;quot;&lt;/a&gt;. The two-sentence afterthought regarding this topic therein doesn't adequately respond to the spirit of my question. I'm looking for a mathematical justification for its use, including why its regarded as more accurate than splitting up the angles into their separate components.&lt;/p&gt;&#xA;" OwnerUserId="10832" LastEditorDisplayName="user1569" LastEditDate="2018-11-13T08:56:17.193" LastActivityDate="2019-04-01T20:55:39.777" Title="Explanation of Longitude of Periapsis" Tags="&lt;orbital-elements&gt;&lt;mathematics&gt;" AnswerCount="2" CommentCount="11" ClosedDate="2018-11-13T11:06:06.390" ContentLicense="CC BY-SA 4.0" />
  <row Id="28357" PostTypeId="2" ParentId="28354" CreationDate="2018-11-12T19:47:16.523" Score="6" Body="&lt;p&gt;Currently, LIGO and Virgo are in what's known as a &lt;a href=&quot;https://www.ligo.caltech.edu/news/ligo20180921&quot; rel=&quot;noreferrer&quot;&gt;commissioning phase&lt;/a&gt;, a scheduled stage where the teams work on improving the sensitivity of the detectors. A big part of this is noise reduction, but there are other technical updates (&lt;a href=&quot;https://arxiv.org/abs/1304.0670&quot; rel=&quot;noreferrer&quot;&gt;an evolving plan&lt;/a&gt; also mentions increasing the power of the lasers and adding a &quot;squeezed vacuum source&quot;). Virgo in particular is &lt;a href=&quot;https://cplberry.com/2018/06/16/1304-0670-2-the-explosioning/&quot; rel=&quot;noreferrer&quot;&gt;replacing the steel fibers that hold up their mirrors with silica ones&lt;/a&gt;. The original silica fibers failed and were replaced with steel ones before last year's detection of GW170817, but now new silica ones are being added.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As of October, &lt;a href=&quot;https://www.ligo.org/scientists/GWEMalerts.php&quot; rel=&quot;noreferrer&quot;&gt;the Livingston site has reached low noise levels, while Hanford is still a bit away&lt;/a&gt;. Virgo isn't quite there yet. Soon, though, the three detectors should begin a couple of joint &lt;a href=&quot;https://www.ligo.caltech.edu/news/ligo20180921&quot; rel=&quot;noreferrer&quot;&gt;engineering runs&lt;/a&gt;, testing the hardware updates before O3 (the third, year-long, observing session) begins in February.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2018-11-12T19:47:16.523" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28358" PostTypeId="2" ParentId="28356" CreationDate="2018-11-12T21:21:07.707" Score="2" Body="&lt;p&gt;If you are dealing with planets, or bodies with a pretty low inclination, it might make sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the inclination of all the planets was exactly zero the longitude of perihelion would be an actual physical angle in space. You would be able to compare the relative position of perihelion of all the planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now actually the planets are slightly inclined but you can still a general idea of the relative position of perihelion from the longitude of perihelion, whereas the argument of perihelion can't be compared between planets, as it is not measured from the same direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a second slight convenience in that the longitude of perihelion can be defined for a planet with zero inclination, whereas the argument of perihelion isn't.  For an orbiter with a very small inclination, it may be possible to determine the longitude of perihelion with much greater accuracy than the argument of perihelion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take, for example an asteroid with an inclination of 0.1 degrees +-0.05. The very small inclination means that the longitude of the ascending node is not well known. A slight error in inclination could make a big error in the ascending node, and consequently the argument of periapsis is also not known accurately.  But the longitude of periapsis can be measured, and a small error in the inclination won't significantly affect it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when dealing with bodies that orbit close to the reference plane, it may make sense to use the longitude. But for objects that are far from the reference plane it may be better to use the argument.  &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-11-12T21:26:19.770" LastActivityDate="2018-11-12T21:26:19.770" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28359" PostTypeId="2" ParentId="28347" CreationDate="2018-11-12T22:44:14.777" Score="4" Body="&lt;p&gt;As the comments have said, the second and third numbers are the ratio of sidereal to solar days but it is split into two parts from the original formula which gives GMST at 0h UT1 by including the hours part directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first number comes from the definition of sidereal time as &quot;&lt;em&gt;in general terms, sidereal time is the hour angle of the vernal equinox&lt;/em&gt;&quot; (Explanatory Supplement to the Astronomical Almanac, 1992, p48 (referred to on p77 of the 3rd Edition of the Explanatory Supplement)). This is bound up with the original definition of the right ascension of the &lt;em&gt;fictitious mean sun&lt;/em&gt; which moves around the celestial equator at a fixed rate which was due to Simon Newcomb in 1895 (&lt;a href=&quot;https://archive.org/details/06AstronomicalPapersPreparedForTheUse/page/n17&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomical Papers Prepared For The Use Of The American Ephemeris And Nautical Almanac Volume 6 (1898)&lt;/a&gt;. This was based on noon and shifting it by 12 hours to the definition used now at converts the 18.64606 to 6.646065. Differences in the measured values of the precession constants and changes in the star catalog reference systems from FK4 to FK5 have resulted in changes in the values from the original.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://articles.adsabs.harvard.edu/pdf/1982A%26A...105..359A&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aoki et al. 1982&lt;/a&gt; discuss this in more detail and added a &lt;span class=&quot;math-container&quot;&gt;$T^3$&lt;/span&gt; term (their equation 13, although note the values there are in seconds not hours as in the original USNO version above) for what became the IAU 1984 system. &lt;a href=&quot;http://articles.adsabs.harvard.edu/pdf/2000A%26A...355..398C&quot; rel=&quot;nofollow noreferrer&quot;&gt;Capitaine et al 2000&lt;/a&gt; added a &lt;span class=&quot;math-container&quot;&gt;$T^4$&lt;/span&gt; term along with another 32 more terms to the so-called &lt;em&gt;equation of the equinoxes&lt;/em&gt; which is added onto the formula for GMST to get GAST (Apparent Sidereal Time). The equation of the equinoxes takes account of the accumulated precession and nutation as the fictitious mean sun is not a very good reflection of reality for precise work. Because of these increased complexities and coupling between the Earth's rotation and the background reference frame of the stars needed for precise work, the IAU did away with sidereal time in 2006 and replaced it with Earth Rotation Angle (ERA) which is a simple linear function of UT1 time.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-11-12T22:44:14.777" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28360" PostTypeId="2" ParentId="28262" CreationDate="2018-11-13T05:12:26.200" Score="0" Body="&lt;p&gt;As the title of the &lt;a href=&quot;https://www.eso.org/public/archives/releases/sciencepapers/eso1148/eso1148.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;research paper&lt;/a&gt; linked to in the ESO press release is &quot;&lt;em&gt;An incisive look at the symbiotic star SS Leporis&lt;/em&gt;&quot;, it is safe to conclude that they are the same type of star. My suspicion is that, since symbiotic stars transfer and accrete mass via either Roche-lobe overflow or stellar wind overflow, the term &quot;vampire star&quot; was coined as part of the press release as a &quot;cute&quot; way to imply that one star is &quot;feeding&quot; on another.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-11-13T05:12:26.200" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28361" PostTypeId="2" ParentId="28356" CreationDate="2018-11-13T05:51:20.747" Score="5" Body="&lt;p&gt;I agree that the common explanation of the longitude of periapsis, as the question gave it, appears at first sight to make no sense -- because it constructs a sum of two angles in two different planes, and that seems at first sight to lack geometrical meaning. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact the fault, if any, is in the explanation rather than in the geometry. It is not so difficult to show a geometrically sensible construction for the longitude of periapsis, without adding any angles in different planes. But before showing this, it may be worth answering the question 'Why do it at all, what's the point?'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point the present answer links up with the answer given by &lt;em&gt;James K&lt;/em&gt;.  One of the uses of such a construction came to attention in the early days of astrodynamics. When programmed orbit calculations are to be done for artificial space objects, the orbits to be calculated can in principle have any values whatever of the inclination and eccentricity, in principle down to zero.  When either or both of these orbital quantities approaches zero, then the position of the node and/or the periapsis, or else computations involving them, can become in practice numerically ill-defined, or anyhow show some undesired numerical behavior. As a result of the difficulty, searches were made a long time ago now, for alternative systems of orbital elements that do not generate undesired numerical behavior at zero or near-zero values of inclination or eccentricity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting formulae for orbital equations, usable for low-inclination, low-eccentricity orbits, typically made use of devices that included the equivalent of replacing the anomaly of an orbiting object by its longitude from a defined initial point, and replacing the argument of periapsis by a corresponding longitude from the initial point.  The development of orbital equations in terms of such quantities can usefully be generated, or at least illustrated, by the following construction that also provides an in-plane account of the longitude of periapsis, the subject of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources for the geometry and formulae connected with this kind of development include R M L Baker (1967) '&lt;em&gt;Astrodynamics&lt;/em&gt;', especially Appendix C, and applications are shown for example in &lt;a href=&quot;https://books.google.co.uk/books?id=Hzv7k2vH6PgC&quot; rel=&quot;nofollow noreferrer&quot;&gt;A E Roy, '&lt;em&gt;Orbital Motion&lt;/em&gt;' (especially 4th ed., 2005)&lt;/a&gt; (or google for other sources of the same title), especially chapter 8, Special Perturbations, section 8.5, for perturbational equations and their derivation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Baker's 'Astrodynamics' illustrates the definition for these purposes of an alternative ('F G W') coordinate system.  See the figure below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zTGHR.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zTGHR.gif&quot; alt=&quot;F G W coordinate diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One starts with ordinary rectangular coordinate axes defined by points X, Y, Z on a unit sphere with center O, where the xy plane is for example the ecliptic plane, and OX is in a chosen initial direction, e.g. towards the equinox of date.  Define an orbital plane through F N G, intersecting the xy-plane at node N with inclination angle &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; .  Angle XON is then the longitude &lt;span class=&quot;math-container&quot;&gt;$\Omega$&lt;/span&gt; of node N. Define axis OW perpendicular to orbital plane F N G. Let angle FON = XON, and let angle FOG = 90°. Then make F the initial point of coordinate system FGW. In this coordinate system, the longitude of the orbital periapsis starting from the initial point is FOP, which is the sum, as usual, of &lt;span class=&quot;math-container&quot;&gt;$\Omega$&lt;/span&gt; the longitude of the node FON and &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; the argument NOP of periapsis, but now they are measured in the same plane -- which was the thing to be illustrated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This FGW system can be conveniently oriented by the coordinates of point W, the pole of the orbital plane, &#xA;which has coordinates (in terms of the x,y,z axes and the quantities &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\Omega$&lt;/span&gt;) of  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ W_x =  + \sin (\Omega) . \sin (i) $&lt;/span&gt; ,&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$ W_y =  - \cos (\Omega) . \sin (i) $&lt;/span&gt; ,&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$ W_z =  \cos (i) $&lt;/span&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then initial point F can be shown (with no difficulty but at a little length) to have coordinates relative to initial axes x, y, z, of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ F_x = 1 - W_x^2 / (1 + W_z) $&lt;/span&gt; ,&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$ F_y =  - W_x W_y / (1 + W_z) $&lt;/span&gt; ,&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$ F_z =  - W_x $&lt;/span&gt; . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantities Wx, Wy, Wz are sometimes identified as &lt;span class=&quot;math-container&quot;&gt;$h_x/h_m, h_y/h_m, h_z/h_m,$&lt;/span&gt; and can be called direction cosines made by the orbital angular momentum vector (in direction OW) with axes x, y, z. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also find some of the formulae useful in &lt;a href=&quot;http://adsabs.harvard.edu/abs/1977CeMec..15..137B&quot; rel=&quot;nofollow noreferrer&quot;&gt;Preliminary orbit-determination method having no co-planar singularity&lt;/a&gt; by R M L Baker and N H Jacoby, &lt;em&gt;Celestial Mechanics&lt;/em&gt; 15 (1977) 137-160.&lt;/p&gt;&#xA;" OwnerUserId="17202" LastEditorUserId="17202" LastEditDate="2019-04-01T20:55:39.777" LastActivityDate="2019-04-01T20:55:39.777" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28362" PostTypeId="1" AcceptedAnswerId="28608" CreationDate="2018-11-13T10:46:59.557" Score="3" ViewCount="91" Body="&lt;p&gt;There are many articles like this one &lt;a href=&quot;https://arxiv.org/pdf/1504.03428.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Testing the isotropy of the Universe by using the JLA compilation of type-Ia supernovae&lt;/a&gt; (PDF, arxiv.org) trying to search for a dipole effect in cosmology with type Ia supernovae (used as standard candles). The idea of this kind of search is to do a test of the cosmological principle on a phenomenological approach independantly of some given models (e.g. Tolman-Bondi anistropic model). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They consider a dipole effect on the distance modulus &lt;span class=&quot;math-container&quot;&gt;$ \mu $&lt;/span&gt;: &lt;span class=&quot;math-container&quot;&gt;$$ \mu \leftarrow \mu \times \left(1+A_D (\hat{\textbf{n}}\cdot \hat{\textbf{p}})\right)$$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$\hat{\textbf{n}}$&lt;/span&gt; is unitary vector pointing in the direction &lt;span class=&quot;math-container&quot;&gt;$(l,b)$&lt;/span&gt; of the dipole with an amplitude of &lt;span class=&quot;math-container&quot;&gt;$A_D$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\hat{\textbf{p}}$&lt;/span&gt; points the direction of each Type Ia supernova.&#xA;The distance modulus is related to luminosity distance &lt;span class=&quot;math-container&quot;&gt;$d_l$&lt;/span&gt; as &lt;span class=&quot;math-container&quot;&gt;$ \mu =5 \log \left(\frac{d_l}{10 pc}\right)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is something that bothers me a bit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our peculiar velocity to the universe is usually estimated with the CMB dipole measuments, right? And so the reshifts of SNIa are computed in the CMB frame using this correction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, supposing we have that kind of cosmological ansisotropy, how it is not mistakenly corrected with the dipole anisotropy of the CMB (and could also be why all searches of this kind give us an amplitude of anisotropy compatible to 0)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean, the CMB dipole implies a velocity of &lt;span class=&quot;math-container&quot;&gt;$369.5\pm3.0$&lt;/span&gt; km/s in the direction of &lt;span class=&quot;math-container&quot;&gt;$l=264.4^{\circ}\pm0.3^{\circ}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$b= 48.4^{\circ}\pm0,5^{\circ}$&lt;/span&gt; according to &lt;a href=&quot;http://adsabs.harvard.edu/abs/1993ApJ...419....1K&quot; rel=&quot;nofollow noreferrer&quot;&gt;COBE measurments&lt;/a&gt;. But what if the CMB dipole is not just due to our peculiar velocity but also to an unknown &quot;cosmological effect&quot;? This effect would be corrected and nothing will be see when try to make a dipole fit on SNIa measurement, right?&lt;/p&gt;&#xA;" OwnerUserId="23047" LastEditorUserId="23047" LastEditDate="2018-12-03T21:48:41.153" LastActivityDate="2018-12-03T21:48:41.153" Title="Searches for dipole form of anisotropies (on luminosity distances) in the universe with type Ia supernovae" Tags="&lt;cosmology&gt;&lt;supernova&gt;&lt;cosmological-principle&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28363" PostTypeId="1" AcceptedAnswerId="28367" CreationDate="2018-11-13T13:11:43.060" Score="1" ViewCount="93" Body="&lt;p&gt;I have a SN spectrum here (already reduced) that seems to have some artifacts. Professor tells me that I may have to normalize it using photometry.&#xA;As far as I understand, I just have to multiply it by some factor to make the integral over all frequencies equal to bolometric luminosity. Am I right, or is there some more complicated thing here?&lt;/p&gt;&#xA;" OwnerUserId="22098" LastActivityDate="2018-11-13T15:08:10.533" Title="Normalizing the supernova spectrum by photometry" Tags="&lt;supernova&gt;&lt;photometry&gt;&lt;spectra&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28364" PostTypeId="1" AcceptedAnswerId="28376" CreationDate="2018-11-13T13:39:39.980" Score="9" ViewCount="323" Body="&lt;p&gt;The eccentricity of Venus is almost 14 times smaller than that of Mars, and 2.5 times smaller than that of Earth, and 30 times smaller than that of Mercury. Is it just a pure fluke? Or is there some “physical reasons”? I realize those are the ratios as of now, and that they change over time: will the evolution of those eccentricities dramatically change those ratios? If not, then the question of a “physical reason” emerges again.&lt;/p&gt;&#xA;" OwnerUserId="24843" LastEditorUserId="1056" LastEditDate="2018-11-23T16:30:18.950" LastActivityDate="2018-11-23T16:30:18.950" Title="Why has the orbit of Venus such a low eccentricity?" Tags="&lt;venus&gt;&lt;eccentricity&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28365" PostTypeId="1" AcceptedAnswerId="28366" CreationDate="2018-11-13T13:57:03.640" Score="0" ViewCount="30" Body="&lt;p&gt;In a couple of papers I've read that it's usual practice to fit early rise of supernova luminosity curve by &lt;span class=&quot;math-container&quot;&gt;$t^2$&lt;/span&gt;. Let me describe how I see it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I take the early fast rising part of luminosity curve and then fit them to a curve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$L = a\cdot\left( t - t_0 \right) ^ 2 + L_0,$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$t_0$&lt;/span&gt; is the moment of explosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this approach right?&lt;/p&gt;&#xA;" OwnerUserId="22098" LastActivityDate="2018-11-13T14:44:02.190" Title="Finding the SN explosion time by t-square fitting" Tags="&lt;supernova&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28366" PostTypeId="2" ParentId="28365" CreationDate="2018-11-13T14:44:02.190" Score="0" Body="&lt;p&gt;I think you misunderstood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$(y-k) = a (x-h)^2$&lt;/span&gt; is a parabola with its apex at &lt;span class=&quot;math-container&quot;&gt;$(h,k)$&lt;/span&gt;. Therefore, the formula you mentioned above has &lt;span class=&quot;math-container&quot;&gt;$t_0$&lt;/span&gt; as the time at peak, not at the explosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the case of SN, we normally do not know when the explosion happened. Mostly, we have an image before SN detection that can set a constraint for the explosion happening in between the last non-detection to the first detection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, most analytical models require the explosion time as the reference, so the parabola fit is applied as an approximation.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-11-13T14:44:02.190" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28367" PostTypeId="2" ParentId="28363" CreationDate="2018-11-13T15:08:10.533" Score="1" Body="&lt;p&gt;I think you understood correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to be clear. Your photometric data have enough images including bias, flat, and standard stars, and are reduced + calibrated with standard stars. And, your issue is that you can reduce spectrographs (with bias, flat, and arc lamp) but you don't have standard stars for the calibration. Hence, you want to calibrate your spectra with the photometric data. Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is for more detailed example how to proceed this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, note that a photometric data point is the accumulation of photons within a bandpass. See the response curve shown &lt;a href=&quot;http://www.aip.de/en/research/facilities/stella/instruments/data/johnson-ubvri-filter-curves&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for example. In the example, B band has its effective wavelength 4353 A, FWHM 781 A, and with its own shape of the response function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity, the following demonstration will assume that the response function of the B band is delta function, i.e., zero elsewhere except at 4353 A. We will extend this example to other response function later. Suppose that you have a spectrum containing 4353 A, and you want to calibrate it with the known B photometry. So you can find a ratio = flux(B; photometry) / flux(B; spectrum), and apply this ratio elsewhere in your spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the spectrum and photometry must be from the same epoch ideally. Otherwise, use the nearest available epoch as an approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have more than one photometric data points to calibrate the one spectrum, ideally the ratios elsewhere will be the same. Realistically, they are not. Hence, you average the ratios by somehow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, if you would like to consider the dispersion of the response function, instead of using the flux at a point, you accumulate the fluxes by considering the shape of the response function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity, let's say the B band response function is a Heaviside step function = 1 within [4353-781, 4353+781] A and = 0 elsewhere. So, you accumulate the fluxes weighted by the response function from your spectrum. Then find the ratio, and follow the rest of the process.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-11-13T15:08:10.533" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28368" PostTypeId="1" CreationDate="2018-11-13T00:34:07.973" Score="8" ViewCount="143" Body="&lt;p&gt;I was reading about how the planet Mercury changes its orbit. I was wondering if farther away planets, particularly ones with elongated orbits, could do that as well? Planets and objects such Halley's comet, Pluto, Sedna, the dwarf planets, or even planet Nine (if there is one). Could the same happen to those objects, and if so how great are those changes? Can other things change as well like the speed of the orbit, or the size?&lt;/p&gt;&#xA;" OwnerDisplayName="Nomad" LastEditorUserId="2153" LastEditDate="2018-11-13T15:58:01.630" LastActivityDate="2018-11-13T15:58:01.630" Title="Can orbits change?" Tags="&lt;orbit&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28369" PostTypeId="2" ParentId="28368" CreationDate="2018-11-13T00:48:44.773" Score="1" Body="&lt;p&gt;Yes, especially early in a system's life-cycle, when things are less sorted.  Resonances between neighboring planets' orbits can arise, inducing eccentricities which can enhance their mutual interactions, resulting in altered orbits, sometimes catastrophically of course, as with &lt;a href=&quot;https://en.wikipedia.org/wiki/Theia_(planet)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Theia&lt;/a&gt; and the proto-earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a theory that the &lt;a href=&quot;https://en.wikipedia.org/wiki/Late_Heavy_Bombardment&quot; rel=&quot;nofollow noreferrer&quot;&gt;Late Heavy Bombardment&lt;/a&gt; of the earth (4.1 - 3.8Ga ago) was caused by disruptions in the orbits of Saturn and Jupiter, which further disrupted asteroids in the asteroid and/or Kuiper belts.  This is called the &lt;strong&gt;giant-planet migration&lt;/strong&gt; hypothesis.&lt;/p&gt;&#xA;" OwnerUserId="24705" OwnerDisplayName="theRiley" LastActivityDate="2018-11-13T00:59:54.963" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28370" PostTypeId="2" ParentId="28368" CreationDate="2018-11-13T15:57:21.123" Score="5" Body="&lt;h1&gt;Precession&lt;/h1&gt;&#xA;&lt;p&gt;I assume you're talking about the &lt;a href=&quot;https://en.wikipedia.org/wiki/Tests_of_general_relativity#Perihelion_precession_of_Mercury&quot; rel=&quot;nofollow noreferrer&quot;&gt;precession of Mercury&lt;/a&gt;, one of the famous tests of general relativity. One reason Mercury is a good planet to test the relativistic predictions of precession is that its precession is more dramatic than the other planets. &lt;a href=&quot;https://astronomy.stackexchange.com/a/7945/2153&quot;&gt;We can calculate the shift in the longitude of perihelion&lt;/a&gt; by&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\mathrm{d}\varpi}{\mathrm{d}t} \propto n^3a^2$$&lt;/span&gt;&#xA;with some other (important!) factors multiplied in. &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is the mean motion (where &lt;span class=&quot;math-container&quot;&gt;$n\propto 1/T$&lt;/span&gt;) and &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is the semi-major axis. Over one orbit, the net change is approximately&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\Delta\varpi=\frac{\Delta\varpi}{\Delta t}\Delta t\approx\frac{\mathrm{d}\varpi}{\mathrm{d}t}T\propto n^3a^2T=\frac{a^2}{T^2}\propto a^{-1}$$&lt;/span&gt;&#xA;Orbits with larger semi-major axes should have smaller &lt;span class=&quot;math-container&quot;&gt;$\Delta\varpi$&lt;/span&gt;, meaning that Mercury's shift is the largest in the Solar System; therefore, the other planets' orbits do precess, but by smaller amounts.&lt;/p&gt;&#xA;&lt;h1&gt;Planetary migration&lt;/h1&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/28369/2153&quot;&gt;theRiley mentioned this in their answer.&lt;/a&gt; Planets' orbits can change over time through a process called &lt;a href=&quot;https://en.wikipedia.org/wiki/Planetary_migration&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;planetary migration&lt;/em&gt;&lt;/a&gt;. This often occurs by one of two mechanisms:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Disk-driven migration, where interactions with a protoplanetary disk change a planet's semi-major axis. This is the main theory for the formation of &lt;a href=&quot;https://en.wikipedia.org/wiki/Hot_Jupiter&quot; rel=&quot;nofollow noreferrer&quot;&gt;hot Jupiters&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Scattering, where interactions with other planets or planetesimals causes drastic orbital changes. This is thought to have happened in the Solar System several billion years ago (see the &lt;a href=&quot;https://en.wikipedia.org/wiki/Nice_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nice model&lt;/a&gt;, the Nice II model, and variations).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-11-13T15:57:21.123" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28371" PostTypeId="1" CreationDate="2018-11-13T17:04:32.303" Score="0" ViewCount="71" Body="&lt;p&gt;This math states, that by multiplying the gravitational pull (we assume square of density of the planet), by its inclination (being how high and out it goes) and subtracting that loop from the velocity in a slope formula that measures to the borderline of being horizontal on the sun's x-axis, most parallel with the solar system's other planets, relegating the offset in order as to consign the preferred balance of the planet with the Sun. We can take then from the furthest point out once the Sun can lesser not take from the product and sum and difference of the clouds on the planet so that we may therefore continue on and calculate the priori aft dividing by 100 to delineate in a reciprocity to the true count, the atmospheric makeup of which the count of surface bars is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This site is were to get the constants for the planetary facts you will need to use the formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://nssdc.gsfc.nasa.gov/planetary/factsheet/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n = Orbital Velocity - root of density * inclination to sun&lt;br&gt;&#xA;round((Apogee - n)/100) = Planet's Atmospheric layer count;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an easy formula to figure out how many layers of atmosphere your favorite planets have on them. It's preliminary because I don't have the satellite data from each planet to synthesize with my data from this formula. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is wrong, please tell me. If it is not, I suppose tell me. But I am only in search criticism and laud.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;NASA Deg.   Real Deg.   Planet + Rank   Density Orbital Vel.    Apogee  Bars (Atmospheres)&#xA;&#xA;17.2        15.3        Pluto+4         2095    4.7             7375.9  6.6 Bars&#xA;&#xA;7           5.1         Mercury+3       5427    47.4            69.8    2.5 Bars&#xA;&#xA;1.8         3.7         Neptune+2       1638    5.4             4545.7  44.01 Bars&#xA;&#xA;1.3         3.2         Jupiter+1       1326    13.1            816.6   34.13 Bars&#xA;&#xA;1.9         -           Mars -0         3933    24.1            2492.2  2.1 Bars&#xA;&#xA;2.5         -0.6        Saturn -1       687     9.7             1514.5  15.07 Bars&#xA;&#xA;3.4         -1.5        Venus -2        5243    35.0            108.9   2.52 Bars&#xA;&#xA;0           -1.9        Earth -3        5514    29.8            152.1   3.22 Bars&#xA;&#xA;0.8         -2.7        Uranus -4       1271    6.8             3003.6  31.06 bars&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="24845" LastEditorUserId="2365" LastEditDate="2018-11-14T21:42:55.600" LastActivityDate="2018-11-14T23:26:07.313" Title="Solution for counting atmospheric layers of planets according to relational data" Tags="&lt;amateur-observing&gt;&lt;planetary-atmosphere&gt;" AnswerCount="1" CommentCount="13" ClosedDate="2018-11-18T07:21:26.107" ContentLicense="CC BY-SA 4.0" />
  <row Id="28373" PostTypeId="2" ParentId="28255" CreationDate="2018-11-13T18:20:45.763" Score="1" Body="&lt;p&gt;As both Earth and Mars move in their orbits, the actual distance between us varies.  We past the close-approach point in this orbit in July, and now we are further ahead of Mars in the orbit. As distance increases, the angular size of Mars decreases.  Right now, Mars is about 11 arcseconds in diameter from Earth.  The size of details you can make out depends on the aperture of the telescope.  As light enters the telescope, the wave-fronts begin to break down in a process known as diffraction.  The practical result of this is that the interference patterns between the wavelets limits the amount of detail that can be resolved. This is a bit different from, but analogous to, the way a photographic sensor can only capture so much data and as you enlarge the image you lose detail through pixellation.  In theory, magnification is unlimited.... but USABLE magnification is limited primarily due to the aperture of the instrument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diffraction effects are also somewhat dependent upon the wavelength of the light - shorter wavelengths are less affected than longer ones.  For deep-blue light around 425 nanometers, the diffraction limit of an 8&quot; (203 mm) telescope is about 0.57 arcseconds.  This means that the smallest details it can resolve will be about that size or larger.  For redder light, say around 675 nm, the diffraction limit is about 0.84 arcseconds.  At 425 nanometers, and at the current distance to Mars (as I write this, about 131,443,000 km), .57 arcsec equates to about 363 kilometers.  Thus, the smallest details you should be able to see on Mars are about 363 km in size - if they're blue.  For redder objects (most of Mars), something would have to be about 510 km in size to be visible.  Anything smaller is just lost in the blur, and even object that size are going to be a significant challenge.  Of course, these numbers are based on the theoretic limits of the instrument. This doesn't take into account the quality of the optics or the effects of the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The generally accepted rule of thumb is that the highest usable magnification of a telescope can be calculated by multiplying the aperture in inches by 50x or the aperture in mm by 2x.  For an 8&quot;/203 mm scope, this gives a value of 400x.  However, this is maximum and assumes ideal conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For average conditions with humidity in the air, turbulence in the atmosphere, dust, pollen, and other pollutants, etc... as well as the air density itself, the practical limit for most observers under normally experienced conditions is probably closer to 20 to 30x per inch or 1x or so per mm.  This would mean a maximum normal usable magnification between about 180x and 240x.  Beyond that, the image gets too blurry to be worth viewing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of this, a Barlow lens doesn't necessarily help maters.  Further, most Barlow lenses suffer from some amount of chromatic aberration, worsening the view.  A good quality Barlow is usually apochromatic, but the more common ones aren't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, looking at the image you posted, it's hard to get a good idea what the issues are for sure.  Camera sensors and human eyes work differently, and don't equate well.  But, most likely your primary issue is simply that of distance and resolution.  This gets worse as Mars gets closer to the horizon.  If it's under 40° above the horizon, you're starting to get into substantially more atmosphere between you and the object.  Like looking through a glass of water, the atmosphere will distort the view more and more.  If Mars wasn't high in the night sky when you were observing, this could be part of your problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, Mars won't be a good target again until October 2020, at which time it should pass within 62.12 million miles, at which time its angular size should be more than twice what it currently is, and you should be able to resolve details half the size you currently can see.&lt;/p&gt;&#xA;" OwnerUserId="10546" LastActivityDate="2018-11-13T18:20:45.763" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28374" PostTypeId="1" AcceptedAnswerId="28375" CreationDate="2018-11-13T19:13:36.910" Score="-8" ViewCount="133" Body="&lt;p&gt;Describe how this instrument works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6l2IY.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6l2IY.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24847" LastActivityDate="2018-11-13T21:26:54.707" Title="Please dsescribe how this instrument works" Tags="&lt;radio-telescope&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2018-11-13T21:36:37.653" ContentLicense="CC BY-SA 4.0" />
  <row Id="28375" PostTypeId="2" ParentId="28374" CreationDate="2018-11-13T20:55:29.577" Score="5" Body="&lt;p&gt;That's the Very Large Array.  The very large array is a Y shaped array of massive Radio astronomy observatory. It has the computing power to function as an interferometer.  Each radio is mounted on double parallel railroad tracks, so the radius and density of the array can be transformed to adjust the balance between its angular resolution and its surface brightness sensitivity.  If you want to read more, read &lt;a href=&quot;https://en.wikipedia.org/wiki/Very_Large_Array&quot; rel=&quot;noreferrer&quot;&gt;this Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="24849" LastEditorUserId="7472" LastEditDate="2018-11-13T21:26:54.707" LastActivityDate="2018-11-13T21:26:54.707" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28376" PostTypeId="2" ParentId="28364" CreationDate="2018-11-13T21:33:24.843" Score="10" Body="&lt;p&gt;It has a low eccentricity, but there may not be a particular reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JUYMI.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JUYMI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image by &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Eccentricity_rocky_planets.jpg&quot; rel=&quot;noreferrer&quot;&gt;Kheider on wikipeda using Gravity Simulator by Trevor Dunn&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a simulation of the solar system, both Earth and Venus had orbital eccentricities that were much below those of Mars and Mercury (note the two axes. Mars and Mercury are on a scale that is 10 times bigger). But in 25000 years, Earth will actually have a smaller eccentricity, and 20000 years ago Venus's eccentricity was much larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there may not need to be a special explanation for this. Planetary eccentricities do change over 1000s of years due mostly to perturbation of other planets.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-11-13T21:33:24.843" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28377" PostTypeId="2" ParentId="28371" CreationDate="2018-11-13T23:28:18.307" Score="3" Body="&lt;p&gt;While the question is rather hard to follow, the formula simply doesn't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Mercury the formula using a naive application of the numbers predicts:&#xA;four layers&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Orbital Velocity (km/s)        47.4&#xA;Density (kg/m3)              5427&#xA;Orbital Inclination (degrees)   5.1&#xA;n                            -328.65368&#xA;Aphelion (10**6 km)            69.8&#xA;Layers                          4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;That is clearly not true, since Mercury has no atmosphere. You can vary the formula, by choosing different units, or taking the inclination relative to a different plane, but no matter what I try, can't get the formula to give a sensible value for all the planets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula is inconsistent, since you generally can't add velocity (km/s) to sqrt(density)*inclination (kg^0.5 degrees/m^(1.5)) as these quantities have different units. It rarely makes sense to add quantities with different units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is, moreover, no reason to suppose that the orbital characteristics of a planet, and its density would have any effect on the number of layers in the atmosphere, which is not a very well defined quantity anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry but this formula is entirely wrong. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-11-14T23:26:07.313" LastActivityDate="2018-11-14T23:26:07.313" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="28378" PostTypeId="1" CreationDate="2018-11-14T02:03:21.783" Score="5" ViewCount="118" Body="&lt;p&gt;When I look into the history of the Milky Way, it seems like the Milky Way has been the pinnacle of stability. It seems highly likely that our galaxy has eaten several smaller galaxies/dwarf galaxies in the past. Also, our galaxy seems to be one of the first to form. I am wondering if our seeming stability is the result of our galaxy's age. So, was the Milky Way more active (as far as stellar formation, supernovae, etc.) in its early years? How do we/ could we know? &lt;/p&gt;&#xA;" OwnerUserId="6753" LastActivityDate="2021-02-17T22:36:47.243" Title="How active was the Milky Way in its first 5 billion years?" Tags="&lt;galaxy&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28379" PostTypeId="1" CreationDate="2018-11-14T07:12:11.693" Score="3" ViewCount="61" Body="&lt;p&gt;CMB anisotropy was measured in 1992. I assume that astronomers, then, like now, would have been able to deduce the cosmological constant and things like that from the CMB anisotropy data. Then, from the cosmological constant and things like that, they would have been able to find that the current universe was accelerating. Did they indeed know that the current universe should be accelerating before its observation in 1998? Or, was the discovery in 1998 a great surprise? What was the exact situation between 1992 and 1998?&lt;/p&gt;&#xA;" OwnerUserId="24850" LastActivityDate="2018-11-14T14:48:01.233" Title="From CMB anisotropy data observed in 1992, did astronomers figure out that the universe should be accelerating before its discovery in 1998?" Tags="&lt;expansion&gt;&lt;history&gt;&lt;cmb&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28380" PostTypeId="1" CreationDate="2018-11-14T09:25:42.643" Score="1" ViewCount="62" Body="&lt;p&gt;For cluster of galaxies we can observe velocity distributions and so we can deduce mass from multiwavelength observations. I suppose total mass of the cluster could be deduced by observing velocities of galaxies and applying virial theorem. Is it be done using general relativity or just with classical mechanics ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An other way could be whith looking at gravitational lensing it causes on far galaxies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these two methods giving same results ?&lt;/p&gt;&#xA;" OwnerUserId="23047" LastActivityDate="2018-11-14T09:25:42.643" Title="Questions on dark matter observations in galaxy clusters" Tags="&lt;general-relativity&gt;&lt;dark-matter&gt;&lt;galaxy-cluster&gt;&lt;gravitational-lensing&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28381" PostTypeId="2" ParentId="28379" CreationDate="2018-11-14T13:58:59.397" Score="3" Body="&lt;p&gt;The only CMB anisotropies that were well determined in 1992 (and by the date, I assume you refer to the results from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_Background_Explorer&quot; rel=&quot;nofollow noreferrer&quot;&gt;COBE satellite&lt;/a&gt;) were those at small angular frequencies. This revealed the overall scale of the CMB structure (which tells you about a need for dark matter in order to get the large scale structure seen at the present epoch) and the dipole anisotropy, which tells you about the Galactic motion with respect to the CMB rest frame. There were no constraints on any possible dark energy - or rather, the effect of dark energy on the low frequency anisotropies is degenerate (i.e. cannot be distinguished from) the effects of the overall curvature of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To measure the curvature of the universe and constrain the mix of dark matter and dark energy (and constrain other cosmological parameters) needs information from higher angular frequencies - in particular information from the location and amplitudes of the first three &quot;acoustic peaks&quot; at angular scales of . This really only became available with the WMAP data (launched in 2001) and with improvements in ground based CMB measurements that took place at the end of the 1990s. The results and conclusions from this emerged over several years from 2000 onwards (e.g. &lt;a href=&quot;http://iopscience.iop.org/article/10.1086/345079&quot; rel=&quot;nofollow noreferrer&quot;&gt;Durrer et al. 2003&lt;/a&gt;; &lt;a href=&quot;http://adsabs.harvard.edu/abs/2003ApJS..148..175S&quot; rel=&quot;nofollow noreferrer&quot;&gt;Spergel et al. 2003&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In actual fact, the evidence for &lt;span class=&quot;math-container&quot;&gt;$\Lambda$&lt;/span&gt; from the CMB &lt;em&gt;alone&lt;/em&gt; is rather weak. As explained in one of &lt;a href=&quot;http://background.uchicago.edu/~whu/araa/node15.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wayne Hu's cosmology tutorials&lt;/a&gt;, the strongest evidence from the CMB is that the total density of the universe &lt;span class=&quot;math-container&quot;&gt;$\Omega = 1$&lt;/span&gt; from the CMB and this can then be combined with other data (e.g. primordial abundances of D, He; galaxy cluster dynamics; gravitational lensing) to suggest that the matter contribution to this is only &lt;span class=&quot;math-container&quot;&gt;$\simeq 0.3$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-11-14T14:48:01.233" LastActivityDate="2018-11-14T14:48:01.233" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28382" PostTypeId="2" ParentId="28285" CreationDate="2018-11-14T14:16:41.630" Score="2" Body="&lt;p&gt;Forbidden lines like these arise in thin gases, where collisional de-excitation is unlikely. Forbidden lines become &quot;quenched&quot; by collisional de-excitation when the densities reach levels that depend on the detailed quantum physics of each transition (i.e. different for every line/transition).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you study the ratio of a pair of lines at densities well below their &quot;quenching thresholds&quot;, the ratio will depend only on the relative occupation of their upper levels, which in turn depends on the temperature, but is insensitive to density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If on the other hand you choose a pair of lines with very similar upper energy levels, the ratio will be insensitive to temperature. However, if one transition is around or even above its &quot;quenching threshold&quot;, then the ratio becomes very sensitive to density.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-11-14T14:16:41.630" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28383" PostTypeId="1" CreationDate="2018-11-14T16:32:47.797" Score="8" ViewCount="228" Body="&lt;p&gt;As far as I've been able to determine, it appears there are some significant  limitations on our ability to detect exo-planets that are below certain sizes or beyond certain distances from their stars.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that Uranus and Neptune are about as small as we can detect unless the planet is relatively close to its star, and Earth is near the lower size limit we can detect, regardless of proximity to the star.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, our bias toward detecting large planets, and planets close to their stars, means I have been able to find reasonably good data regarding how many of those kinds of planets are generally present in a system.  But since we can't seem to tell (by direct observation) how many small planets are generally present in a system, or how many planets are present that are just too far from their star for us to detect, I'm struggling to find data regarding how many plants of those types are common in planetary systems.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: our system has 8 planets, 4 rocky, 2 gas giants, 2 ice giants.  But chances are that if we were observing (with current technology) our own Solar system from some other solar system, we'd probably only see 2 to 6 of them.  We'd almost certainly see Jupiter and Saturn, and we'd almost certainly NOT see Mercury or Mars (they're just too small), Venus and Earth &quot;might be&quot; close enough to the sun for us to see them even though they are on the small end of what we can detect, and Uranus and Neptune also &quot;might be&quot; detectable, even though they are small-ish for their distance from the sun.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My specific question:  &lt;strong&gt;How many planets are in a 'normal' or 'average' planetary system, according to humanity's best scientific theories available so far?&lt;/strong&gt;(beyond what we can simply detect.)  In other words, where can I find &lt;strong&gt;reasonable scientific research data sets, models, estimates, theories, evidence, etc.&lt;/strong&gt;, that describe the likelihood of &quot;undetectable&quot; (or very hard to detect) planets being present in a planetary system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my first question on Astronomy stackexchange, so please be gentle, but don't hesitate to provide constructive criticism if I'm doing anything wrong.&lt;/p&gt;&#xA;" OwnerUserId="24853" LastEditorUserId="24853" LastEditDate="2018-11-14T22:28:30.873" LastActivityDate="2018-12-21T05:20:30.280" Title="Estimates for &quot;undetectable&quot; planets in extra-solar systems" Tags="&lt;astrophysics&gt;&lt;observational-astronomy&gt;&lt;exoplanet&gt;&lt;planetary-formation&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="28384" PostTypeId="1" CreationDate="2018-11-14T18:28:30.623" Score="2" ViewCount="617" Body="&lt;p&gt;I understand stars fuse atoms into heavier and heavier elements. When the star starts to fuse iron, the amount of energy released is less than the amount of energy it to takes to fuse the atoms. This creates an energy deficiency and the core is unable to maintain its outward pressure against gravity's inward collapse. The star collapses, rebounds against electron degeneracy pressure (I believe?) and goes BOOM. But how much iron is actually fused? And, for how long is the star fusing iron before the collapse? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as a little bonus, I read somewhere recently that the presence of nickel is possible prior to going supernova. I can't find where though. Was that person wrong? &lt;/p&gt;&#xA;" OwnerUserId="6753" LastActivityDate="2018-11-16T16:51:39.567" Title="How much iron does a star fuse before going supernova?" Tags="&lt;supernova&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ClosedDate="2018-11-18T07:20:27.030" ContentLicense="CC BY-SA 4.0" />
  <row Id="28385" PostTypeId="1" AcceptedAnswerId="28387" CreationDate="2018-11-14T18:52:42.117" Score="2" ViewCount="138" Body="&lt;p&gt;Hundreds of millions of years after the Big Bang, the very first stars began to form consisting of mostly hydrogen, a bit of helium, and maybe some lithium. These stars lacked any &quot;metals&quot; (elements heavier than helium) and are categorized as Population III stars. We have yet to observe any of these Pop III stars, and observation remains unlikely. The reason for this is because Pop III stars are thought to have been incredibly massive, so they died out so long ago that we can no longer see their light. Pop I stars are metal-rich; their metallicity is 1/10 to 3 times that of our Sun's. In my research, everyone seems to agree that that the first generation of stars had no metal, second generation stars had very little, and third generation stars have some metal.But, no one ever addresses the possibilities. So, is it possible that some Pop III stars were so massive that they were able to skip Pop II and go straight to Pop I? Maybe if several are close by when they go supernova?   &lt;/p&gt;&#xA;" OwnerUserId="6753" LastEditorUserId="2531" LastEditDate="2018-11-17T07:52:33.680" LastActivityDate="2018-11-17T07:52:33.680" Title="Could population III stars develop directly into population I stars?" Tags="&lt;star&gt;&lt;supernova&gt;&lt;stellar-evolution&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="28386" PostTypeId="1" CreationDate="2018-11-14T19:25:41.850" Score="4" ViewCount="143" Body="&lt;p&gt;Are there any scenarios in which two stars could collide, and simply fuse without triggering a supernova. Maybe a mid-sized star slowly spiraling in together with a smaller star? It seems like any time stars collide it's incredibly violent. It seems that one star always gets torn to pieces and accretes matter onto the other, or ends in a red nova. Is there any way this process could be calmer?&lt;/p&gt;&#xA;" OwnerUserId="6753" LastActivityDate="2018-11-22T17:10:30.923" Title="Could two stars collide without a resultant explosion?" Tags="&lt;supernova&gt;&lt;collision&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28387" PostTypeId="2" ParentId="28385" CreationDate="2018-11-14T20:25:03.130" Score="5" Body="&lt;p&gt;No they couldn't. Population I stars contain elements like strontium, barium, gold, lead etc. that are &lt;em&gt;not&lt;/em&gt; formed (much) in type II (core collapse) supernovae. They are also much richer in iron, nickel, manganese etc. than are stars formed from gas enriched only by supernovae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Population I stars (like the Sun) can be termed third generation stars because they contain material that has been through at least two stars - but not just two high mass stars that exploded as supernovae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron-peak elements are mainly formed and disseminated by type Ia supernova that are exploding white dwarfs. White dwarfs are the remnants of &lt;em&gt;low-mass&lt;/em&gt; stars with long lives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many heavy elements (strontium, barium, lead etc) are formed by the s-process in intermediate mass stars that also have (relatively) long lives and do not explode as supernovae. Furthermore, this neutron capture requires iron-peak nuclei to act as seeds, so these stars must in turn have formed out of material already enriched with iron-peak nuclei.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also now thought that other elements (silver, gold, osmium etc) are formed by the merger of neutron stars, where there must be a significant delay between the initial supernovae that produced the neutron stars and the subsequent merger by orbital decay. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-11-14T20:25:03.130" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28388" PostTypeId="1" CreationDate="2018-11-15T03:39:00.800" Score="2" ViewCount="81" Body="&lt;p&gt;I recently read a post on a discussion board about the creation of our current universe that stated, and I quote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[T]he shrinkage activity of the old universe and the new Big Bang expansion movement still made a tremendous collision, resulting in the diffusion and spiral of fine particles.The diffusion of fine particles was integrated into a swirl to reduce the collision and the thermal activity became mature, but the expansion of the fine particle spiral group by the Big Bang is still progressing, expanding while colliding with the shrinking old matters of the old universe There. New matters cooperate solidly with a spiral net to the whole universe as a gear that rotates at the same speed. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am just an amateur interested in astronomy, but I have never heard of a theory regarding such an event before. Is this a known theory that has been researched?&lt;/p&gt;&#xA;" OwnerUserId="24856" LastActivityDate="2019-02-27T18:41:00.523" Title="Is there a theory about a collision between a shrinking universe and a new expanding universe?" Tags="&lt;big-bang-theory&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28389" PostTypeId="2" ParentId="28386" CreationDate="2018-11-15T06:31:24.233" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there any scenarios in which two stars could collide, and simply fuse without triggering a supernova&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As said in the comments, binary evolution of stars is very uncertain, but there exist numerous codes (i.e. MESA, STARTRACK, BSE, etc...) which incorporate LOTS of physical processes over the course of millions of years (and if they evolve to black holes billions of years) of binary evolution. Ono Pols has a nice textbook free on his website &lt;a href=&quot;http://www.astro.ru.nl/~onnop/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stellar binaries usually experience at least a couple mass-transfer (MT) phases after the binary components fill their Roche lobes. To put it simply, MT is either stable or it is unstable, but either way it results in the donor star losing it's envelope so that it's core emerges (which will later turn into a compact object). In stable MT, the donor star transfers some mass to its companion and thusly loses rotational angular momentum (it's spin decreases) while the accretor's spin increases. In unstable mass transfer, the mass transfer rate is too much for the accretor to handle so that the binary is engulfed by the envelope, this is also called a Common Envelope (CE). The CE results in either 1) the envelope is ejected from the binary resulting in a drastic decrease in the binary separation, or 2) the binary merges leaving a single star which accretes the leftover envelope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I think the case 2) is an answer to your question. Keep in mind that CE is very very uncertain, as Postnov explains, and it is a relatively quick process (less than 1 million years). It's worth noting that case 1) of CE is crucial for forming compact binaries that have small separations so that they merge within the lifetime of the universe (so we can measure they're gravitational waves).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your nutrition, &lt;a href=&quot;https://arxiv.org/abs/1403.4754&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; in Postnov's review on compact binary evolution the section on CE is lovely (section 3.6 on page 35).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you're willing to include black hole binaries in your question, then their merging due to the emission of gravitational waves is certainly a calmer scenario. &lt;/p&gt;&#xA;" OwnerUserId="23762" LastEditorUserId="23762" LastEditDate="2018-11-22T17:10:30.923" LastActivityDate="2018-11-22T17:10:30.923" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28390" PostTypeId="2" ParentId="28386" CreationDate="2018-11-15T07:39:51.323" Score="2" Body="&lt;p&gt;The merger of two stars may only result in a supernova if the merging stars are white dwarfs, or possibly a white dwarf with a neutron star, and even in these scenarios a supernova is not certain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other mergers do not cause supernovae. Examples are the formation of contact binaries (&lt;a href=&quot;https://en.m.wikipedia.org/wiki/W_Ursae_Majoris_variable&quot; rel=&quot;nofollow noreferrer&quot;&gt;W UMa binaries&lt;/a&gt;), which may then fully merge to produce fast-rotating FK Com stars, or more generally, the merger of main sequence stars is thought to produce blue stragglers. It is possible that some of these events may result in some sort of outburst, but nothing like a supernova - e.g &lt;a href=&quot;https://arxiv.org/abs/1012.0163&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tylenda et al. (2011)&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-11-15T14:07:47.230" LastActivityDate="2018-11-15T14:07:47.230" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28391" PostTypeId="1" AcceptedAnswerId="28397" CreationDate="2018-11-15T07:54:48.470" Score="5" ViewCount="136" Body="&lt;p&gt;ʻOumuamua passed its closest point at about 0.25 UA from the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone researched what the odds were that the first ever spotted interstellar object passed so close to the Sun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being the interstellar space almost totally empty, I would intuitively suppose that such odds would be extremely low.&lt;/p&gt;&#xA;" OwnerUserId="24857" LastActivityDate="2018-11-15T19:37:09.697" Title="What were the odds that ʻOumuamua passed so close to the Sun?" Tags="&lt;asteroids&gt;&lt;oumuamua&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28392" PostTypeId="2" ParentId="28384" CreationDate="2018-11-15T08:31:55.520" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Type_II_supernova&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia says&lt;/a&gt; the iron-nickel core is inert, and doesn't mention any endothermic iron fusion reactions, just the core collapse, which converts the core to neutrons (and a not insignificant proportion of protons &amp;amp; electrons), and releases a lot of neutrinos. The nickel is radioactive, decaying to cobalt &amp;amp; iron. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Silicon-burning_process&quot; rel=&quot;nofollow noreferrer&quot;&gt;silicon burning&lt;/a&gt; phase leading to nickel only lasts about 5 days. The inert core collapses when it exceeds the Chandrasekhar limit of about 1.2 M&lt;span class=&quot;math-container&quot;&gt;$\odot$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="16685" LastEditDate="2018-11-15T14:24:52.223" LastActivityDate="2018-11-15T14:24:52.223" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28395" PostTypeId="1" CreationDate="2018-11-15T11:01:31.443" Score="0" ViewCount="149" Body="&lt;p&gt;Why is the expansion of the universe allowed to faster than light if everything that moves within a vacuum can't reach the speed limit of light?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read that astronomers looked at the distance between galaxies but I'm still not understanding what's going on. If objects in the universe can't go faster than the speed of light in a vacuum, how are galaxies (which objects within the universe) moving away from each other faster than light?&lt;/p&gt;&#xA;" OwnerUserId="24859" LastActivityDate="2018-11-15T11:45:58.047" Title="Why Is The Expansion of The Universe Allowed To Move Faster Than Light?" Tags="&lt;universe&gt;&lt;expansion&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2018-11-15T13:25:48.183" ContentLicense="CC BY-SA 4.0" />
  <row Id="28396" PostTypeId="2" ParentId="28395" CreationDate="2018-11-15T11:45:58.047" Score="0" Body="&lt;p&gt;You are not allowed to move faster than c in sense of classical motion that is velocity can't be greater than  c but ,universe is expanding not because objects are moving away but new space time is being created between them which increases distance between them and that can happen such that expansion is faster than c , there is no violation of theory of relativity.&#xA;I hope that's clear&lt;/p&gt;&#xA;" OwnerUserId="24861" LastActivityDate="2018-11-15T11:45:58.047" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28397" PostTypeId="2" ParentId="28391" CreationDate="2018-11-15T14:20:31.840" Score="6" Body="&lt;p&gt;How could it be found otherwise? It is currently (November 2018) only at about the orbital distance of Mars and cannot be detected by the world's largest telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such objects are small and only seen in reflected light - they have to get close to the Sun (and Earth) to be seen. There may well be a substantial population of interstellar rocks passing through the solar system unseen at greater distances.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-11-15T14:27:32.930" LastActivityDate="2018-11-15T14:27:32.930" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28398" PostTypeId="1" CreationDate="2018-11-15T14:37:35.907" Score="4" ViewCount="116" Body="&lt;p&gt;Like if you had two planets that had a close orbital path could the moon be pulled away from the first planet?&lt;/p&gt;&#xA;" OwnerUserId="24864" LastActivityDate="2018-11-16T10:50:02.863" Title="Could a moon be passed from one planet's orbit to another?" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;gravity&gt;&lt;natural-satellites&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28399" PostTypeId="2" ParentId="28398" CreationDate="2018-11-15T17:49:48.647" Score="0" Body="&lt;p&gt;For this to happen randomly?  That would be nearly impossible.  I'm guessing it would be possible to construct a situation where this could be accomplished for an object under some control (like a spacecraft with minimal thrust).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if the situation isn't precisely controlled, it is much more likely that during any interaction the satellite will either be driven closer to the first planet or that it will be ejected away from both planets (probably to a solar orbit, becoming a planetoid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's hard enough for a random object to get captured by a planet.  To do so immediately from orbiting a nearby planet makes the odds near zero.  Also two planets (each large enough to capture a moon) on close orbital paths would be unstable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's quite possible for (small) satellites to move from one planet to another.  Just there would normally be a long period in between when it's not orbiting either one.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2018-11-15T17:49:48.647" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28400" PostTypeId="2" ParentId="28391" CreationDate="2018-11-15T19:37:09.697" Score="0" Body="&lt;p&gt;The odds are actually very good, for the simple fact that the object could be observed. If it was much further away than that, it could not be observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's like eating a rare food, then asking yourself, what are the odds that that tasty morsel was ever so close to my mouth? Well, if you ate it already, the odds are actually 100%.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2018-11-15T19:37:09.697" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28402" PostTypeId="2" ParentId="28398" CreationDate="2018-11-15T22:04:08.243" Score="4" Body="&lt;p&gt;Given the nature of 3 body orbital mechanics, it's possible, just unlikely and I hope you'll forgive me if I skip the math.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our solar-system, it's pretty much not possible because our planets are too far from each other.   Speaking in solar-system distances, Moons have to be close to planets, within the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere#True_region_of_stability&quot; rel=&quot;nofollow noreferrer&quot;&gt;stable regions of their planet's hill spheres&lt;/a&gt; in order to have long-lasting orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to assume that I can ignore &lt;a href=&quot;https://phys.org/news/2011-12-earth-temporary-moon.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;temporary moons&lt;/a&gt;, which in my opinion, aren't real moons anyway.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty with one planet giving another planet it's moon is that they'd have to pass quite close to each other, and one of the definitions of planets is that they have their own orbital region around their star, but orbits can move over time, so it's possible for two planets to pass close to each other but there would be other effects as well because, as they got close, the planets would interfere with each other as well.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a just right scenario a moon could theoretically be passed from one planet to another during a close pass.   This is similar to the theory about how an asteroid in a &lt;a href=&quot;https://en.wikipedia.org/wiki/Binary_asteroid&quot; rel=&quot;nofollow noreferrer&quot;&gt;binary-asteroid pair&lt;/a&gt; could get captured if it passes close to a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A planet could also lose a moon and maybe in a few million years that moon could find it's way to being captured by another planet, but I don't think that's what you're asking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might think that around a red-dwarf where planets are quite close to each other and much closer to their star, that a transfer could happen due to proximity, but the difficulty there is that planets that close to their star have a hard time having moons at all because they have very short Hill radii.   It's probably more likely, but still unlikely, to happen between two distant planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the planet Neptune, as an example.  Neptune has a Hill radius of a bit over 100 million km, about the distance between Venus and the Sun, and it's true region of stability, between 33% and 50% of that, so Neptune could in theory have a moon in a stable orbit as distant as perhaps up to 50 million km.  At that distance, it's moon would orbit it very slowly, about .37 km/s with an orbital period of about 25 years.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A fun mathematical quirk, which has nothing to do with your question, but if you calculate the maximum orbital period a moon can have inside the stable orbital region around it's planet, it works out to about 1/6th the period the planet has around it's star, it doesn't matter the mass of the planet or the distance to the star that 1/6th rule roughly applies to every scenario).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in theory, if Uranus passed close to Neptune, say within 100 million km, the theoretical distant moon, especially if it had a somewhat elliptical orbit, could get transferred from Neptune to Uranus and as is passed.  In the process, it would become a wrong-way moon around it's new planet.  (insert wacky theory about Triton having once orbited Uranus here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uranus and Neptune never get that close, though perhaps they did when our solar-system was young and Jupiter was migrating.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's more likely, if two planets got that close that they'd alter each other's orbits rather than trade a moon.  The moon could also be ejected from one planet and not captured by the other.  That's probably a more likely scenario.   If one of the planets was considerably smaller than the other, the entire planet could be tossed around significantly, perhaps tossed out of the solar-system.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What probably does happen, is binary systems like Pluto-Charon, when they pass close enough to a planet, one of them can be captured.  &lt;a href=&quot;https://www.nature.com/articles/nature04792&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article suggests that Triton&lt;/a&gt;, Neptune's wrong way moon, was captured by that method.  A capture involving 3 gravitational bodies isn't that strange, but the criteria that two of those bodies are planets makes it more rare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 body orbits allow for a variety of scenarios, and trading a moon is probably one of the least likely options but theoretically possible.  This just might happen when solar systems are young and chaotic and full of planetesimals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a related note, I've read an article a while back that stars in the center of our galaxy can trade planets, but I looked, I couldn't find that article.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-11-16T10:50:02.863" LastActivityDate="2018-11-16T10:50:02.863" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28403" PostTypeId="1" CreationDate="2018-11-16T03:03:02.397" Score="4" ViewCount="393" Body="&lt;p&gt;For example, if astronomers photograph a &quot;planet&quot;, which is in close proximity to a given star, then this &quot;planet&quot;, could actually just be another &quot;star&quot; in the background, many light years away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might just be in the photograph by chance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can astronomers determine if this is, in fact, a planet orbiting that star or just another star many light years away?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For example, a common method to find planets orbiting stars is the transit method. ie, we can observe how the star appears to slowly decrease brightness, then increase back to normal again. This basically tells astronomers that this must be a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do astronomers know that this isn't just the &quot;front star&quot; transitting a star &quot;in the background&quot;?&lt;/p&gt;&#xA;" OwnerUserId="17327" LastEditorUserId="2270" LastEditDate="2018-11-19T14:59:58.720" LastActivityDate="2018-11-20T04:33:08.717" Title="How do astronomers know if a planet is orbiting a star, or it's simply another star in the background?" Tags="&lt;star&gt;&lt;planet&gt;&lt;planetary-transits&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28404" PostTypeId="1" CreationDate="2018-11-16T03:13:51.353" Score="1" ViewCount="74" Body="&lt;p&gt;A new analysis of gravitational wave (and other data) from GW170817 on 2017-Aug-17 has been published, strongly suggesting that the merger of two neutron stars resulted in a large, rapidly rotating neutron star, rather than a black hole. From the abstract of the recent &lt;em&gt;open access Letter in MNRAS&lt;/em&gt; &lt;a href=&quot;https://academic.oup.com/mnrasl/article/482/1/L46/5090425&quot; rel=&quot;nofollow noreferrer&quot;&gt;Observational evidence for extended emission to GW170817&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;[...]Here, we report on a possible detection of extended emission (EE) in gravitational radiation during GRB170817A: a descending chirp with characteristic time-scale τs = 3.01 ± 0.2 s in a (H1,L1)-spectrogram up to 700 Hz with Gaussian equivalent level of confidence greater than 3.3σ based on causality alone following edge detection applied to (H1,L1)-spectrograms merged by frequency coincidences. Additional confidence derives from the strength of this EE. The observed frequencies below 1 kHz indicate a hypermassive magnetar rather than a black hole, spinning down by magnetic winds and interactions with dynamical mass ejecta.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Maurice H P M van Putten and Massimo Della Valle,&#xA;Monthly Notices of the Royal Astronomical Society: Letters, Volume 482, Issue 1, 1 January 2019, Pages L46–L49, &lt;a href=&quot;https://doi.org/10.1093/mnrasl/sly166&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://doi.org/10.1093/mnrasl/sly166&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Discussion in paper points to &lt;a href=&quot;https://academic.oup.com/mnrasl/article/482/1/L46/5090425#supplementary-data&quot; rel=&quot;nofollow noreferrer&quot;&gt;supplementary data&lt;/a&gt; and in the introduction to that document, it says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;For GW170817A/GRB170817A, we perform a model-independent deep search for broadband extended gravitational-wave emission in 2048 s (LIGO 2017) of data at 4096 Hz according to Fig. A1 comprising&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pre-processing: cleaning and glitch removal (Abbott et al. 2017a) followed by whitening of H1, L1 and V1 data;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Singe detector spectrograms by GPU-accelerated butterfly filtering of H1, L1 and V1 by matched filtering over a dense bank of time-symmetric chirp-like templates&lt;/strong&gt; (van Putten et al. 2014; van Putten 2017);&lt;/li&gt;&#xA;&lt;li&gt;Merging spectrograms by coincidences in frequency or amplitude, producing merged spectrograms as input to image analysis (van Putten 2018).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I am having trouble understanding what &lt;em&gt;GPU-accelerated butterfly filtering of H1, L1 and V1 by matched filtering over a dense bank of time-symmetric chirp-like templates&lt;/em&gt; means. &lt;code&gt;van Putten et al. 2014&lt;/code&gt; in the supplementary data document refers to both &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0004-637X/786/2/146&quot; rel=&quot;nofollow noreferrer&quot;&gt;BROADBAND TURBULENT SPECTRA IN GAMMA-RAY BURST LIGHT CURVES&lt;/a&gt; and to &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/0004-637X/819/2/169&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;/a&gt; as well. These appear to be thorough explanations, but pretty in-depth.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is it possible to explain the basics of what &amp;quot;GPU-accelerated butterfly filtering of H1, L1 and V1 by matched filtering over a dense bank of time-symmetric chirp-like templates&amp;quot; means? A dense bank of chirp-like templates sounds like it could be analogous to a wavelet-type analysis, but with basis waveforms tailored to this specific problem.&lt;/p&gt;&#xA;&lt;p&gt;First image is a cropped and annotated version of the second which ia from &lt;a href=&quot;https://academic.oup.com/mnrasl/article/482/1/L46/5090425&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JF6wh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JF6wh.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TKBhp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TKBhp.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Figure2. Ascending–descending chirp in the (H1,L1)-spectrogram produced by the double neutron star merger GW170817 concurrent with GRB170817A (Goldstein et al. 2017) past coalescence (tc = 1842.43 s). Minor accompanying features around 100 Hz ( 1840-1852 s) are conceivably due to dynamical mass ejecta. Colour coding (blue-to-yellow) is proportional to amplitude defined by butterfly output ρ of time-symmetric chirp-like template correlations to data.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-11-16T03:13:51.353" Title="What does &quot;GPU-accelerated butterfly matched filtering over dense bank of time-symmetric chirp-like templates&quot; mean? (GW170817)" Tags="&lt;gravitational-waves&gt;&lt;neutron-star&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28405" PostTypeId="2" ParentId="28403" CreationDate="2018-11-16T05:49:03.797" Score="9" Body="&lt;p&gt;I think there are two parts to this question: How do we know that the transiting object is a planet, and how do we know that it's gravitationally bound to the parent star, rather than an interloper?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the object is indeed orbiting a star, we should see periodic transits. If a candidate transit is observed, follow-up observations should also be able to observe it. A lack of further confirmation would be a strike against the idea that the object is indeed bound. An additional tool that's useful here is spectroscopy. If the body is orbiting, it should influence the motion of the parent star, because they'll orbit the shared center of mass. This means the star's motion with respect to an observer will change over time, causing a Doppler shift in its light. This shift can be measured throughout the body's orbit. If a periodic shift is observed - with the same period as the transits - then we have another bit of evidence for an exoplanet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the light curves of transiting exoplanets can look similar to those of &lt;a href=&quot;https://en.wikipedia.org/wiki/Binary_star#Eclipsing_binaries&quot; rel=&quot;nofollow noreferrer&quot;&gt;eclipsing binaries&lt;/a&gt; - binary stars where the orbital plane of the components is aligned with the line of sight of the observer. A couple key elements can help astronomers differentiate between the two:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;An eclipsing binary should display a secondary eclipse when the less luminous component passes behind the first. This is not likely to be prominent in the case of a transiting exoplanet.&lt;/li&gt;&#xA;&lt;li&gt;The mass of the orbiting object can be determined by measuring the radial velocity of the star, as determined by looking at the Doppler shift. This is an easy way to distinguish an exoplanet from a star.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In short, an occultation of a background star by an interloper would not be periodic and would not be associated with Doppler shifts in the other star's spectrum. The mass of the orbiting body - as well as the presence or absence of a significant secondary eclipse - can help us determine its nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to note that &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_directly_imaged_exoplanets&quot; rel=&quot;nofollow noreferrer&quot;&gt;we've directly imaged a number of exoplanets&lt;/a&gt; over the past decade. Beta Pictoris and HR 8799 are well-known examples. There are some difficulties with this, of course; planets are much less luminous than their parent stars, and so the light from the central star must be blocked during observations. Direct imaging is useful for planets in face-on orbits, where we're looking nearly perpendicular to the orbital plane. These exoplanets will, naturally, never transit their parent stars from our perspective, and so transiting planets aren't usually directly imaged planets. Additionally, direct imaging is also good for planets in orbits with large semi-major axes - which are also planets that are unlikely to yield strong transits from any position.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2018-11-20T04:33:08.717" LastActivityDate="2018-11-20T04:33:08.717" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28408" PostTypeId="1" AcceptedAnswerId="28411" CreationDate="2018-11-16T12:23:49.147" Score="4" ViewCount="227" Body="&lt;p&gt;In &lt;a href=&quot;http://aoi.com.au/OortSoup/index.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; article (and references therein) the author claims that the Cosmic Background Radiation can be explained as black-body radiation originating from the cold bodies filling the interstellar space (which the author refers to as “Oort soup”).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, the mainstream point of view is that CMB is an echo of the Big Bang which dates back to the electroweak symmetry breaking,and has cooled down since then due to the expansion of Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the first time I learned about the alternative “Oort soup” hypothesis, which, according to the Author, also favors the Dark Matter model made of cosmic dust. I don’t have enough evidence to judge for myself whether it is pseudoscience or a valid alternative explanation of the CMB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know what the position of the scientific community is — is the hypothesis considered pseudoscientific, or a valid (though maybe unlikely) alternative? What scientific evidence against the author’s hypothesis do we have?&lt;/p&gt;&#xA;" OwnerUserId="24871" LastActivityDate="2020-10-10T23:19:27.667" Title="Scientific evidence against CMB originating from “Oort Soup”?" Tags="&lt;dark-matter&gt;&lt;oort-cloud&gt;&lt;cosmic-microwave-background&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28409" PostTypeId="1" AcceptedAnswerId="28414" CreationDate="2018-11-16T15:42:11.620" Score="2" ViewCount="124" Body="&lt;p&gt;How much does subatomic particles such as photons, tachyons, and electron being the biggest particles etc. occupy space? If put in a ball how big of a ball would that be?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-11-16T16:39:19.277" LastActivityDate="2018-11-19T20:18:56.300" Title="How much of the Universe is subatomic?" Tags="&lt;astrophysics&gt;&lt;universe&gt;&lt;space&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2018-11-22T17:09:13.910" ContentLicense="CC BY-SA 4.0" />
  <row Id="28410" PostTypeId="2" ParentId="28384" CreationDate="2018-11-16T16:30:35.800" Score="-1" Body="&lt;p&gt;The collapse of the stellar core has actually 5 steps, and only the first is well known.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;1. exotherm fusion&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The first is the fusion of the nuclei until iron/nickel. Also the normal life of the star belongs here.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;2. endotherm fusion&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;While the iron-nickel core is inert on the Wikipedia, it is true only in &quot;normal&quot; circumstances. In fact, also iron nuclei can fuse, but it is already an endotherm process: they get away energy, thus they cool the core, accelerating the contraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until iron (and nickel) is created due to nuclear fusion, the nuclear reactions are working against the collapse, because they generate energy (photons) and their pressure tries to avoid the collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, this obstacle stops and the core can contract. From a point, also the iron (and nickel) nuclei start to fuse, to create all known elements of the periodical system, and probably even much higher, which can't exist in our low-pressure environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is already an endotherm process, thus it takes away energy from the core. Thus, instead of decelerating the contraction, it accelerates.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;3. exotherm fission&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;But it can't balance the temperature elevation due to the gravitational contraction, and over some billions kelvins, the nuclei start to decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the reverse process what the star did in the previous millions of years. It is exotherm for nuclei higher than iron/nickel and endotherm for nuclei below it.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;4. endotherm fission&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Thus, first the fission of the nuclei decelerate the contraction, then (after iron nuclei decayed), they accelerate again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The produced energy radiates away (more exactly, it hits the upper layers of the star, causing the supernova explosion what we can see), and the contraction goes, this time with elevating temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over 60billion degrees, even the helium nuclei decay, which is still a highly endotherm process. From that point, there is no more way to get away energy from the core by nuclear processes: there is too hot to fuse the nuclei and there is no more way to decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is a mess of protons, neutrons and electrons. Essentially what the star did millions (or hundred millions) years long, is reversed in some seconds. The difference is that this mess is compressed into a size somewhere between a neutron star and a white dwarf, and it is 60billion degrees hot (normal stellar core temperature is in the order of some ten millions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the point where there is no more obstacle before the further contraction.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;5. proton-electron fusion&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;More exactly, the next obstacle will be that the beta decay of protons becomes energetically favorable. It can happen either by the fusion of a proton and an electron to a neutron and an anti-neutrino, or by the fission of a proton to a neutron, a positron and an anti-neutrino. The positron finds an electron quickly and annihilates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although neutrinos can permeate normal matter very quickly, their mean free path in such dense matter is only some hundred meters. They explode away the upper layers of the forming neutron star, dispersing all the elements of the periodical system in the Universe.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This all happens quicker than a minute, the steps 2-5 happen in seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the answer is, that in a supernova explosion, practically all matter in the stellar core will be iron. Not even once, but twice. Although both is happening only for a very short time, they might be iron nuclei maybe some tenth of seconds long.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="1891" LastEditDate="2018-11-16T16:51:39.567" LastActivityDate="2018-11-16T16:51:39.567" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28411" PostTypeId="2" ParentId="28408" CreationDate="2018-11-16T16:39:18.387" Score="14" Body="&lt;p&gt;The idea that belts or spheres of dust might be responsible for (some) microwave emission is not crazy. Indeed we know that dust does emit microwaves and indeed the contribution of such dust has to be &lt;em&gt;removed&lt;/em&gt; from the CMB signal before it can be interpreted cosmologically.&lt;/p&gt;&#xA;&lt;p&gt;There is some debate about some of the larger scale anisotropies (particularly the &amp;quot;quadrupole anisotropy&amp;quot; (of about 1 part in &lt;span class=&quot;math-container&quot;&gt;$10^5$&lt;/span&gt;) might be the result of badly-subtracted emission from within the solar system because the anisotropy is roughly aligned with the ecliptic plane.&lt;/p&gt;&#xA;&lt;p&gt;However, the bulk of the article you refer to is nonsense and betrays a basic lack of understanding of physics and a lack of knowledge about the microwave background.&lt;/p&gt;&#xA;&lt;p&gt;Just to address a couple of things.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The CMB is an almost perfect black body spectrum. A blackbody spectrum is emitted by perfect blackbodies. The thermal emission from dust is &lt;em&gt;not&lt;/em&gt; blackbody radiation. Thermal radiation &lt;span class=&quot;math-container&quot;&gt;$\neq$&lt;/span&gt; blackbody radiation. To become blackbody radiation the emission has to come from something that is &amp;quot;optically thick&amp;quot; to that radiation - i.e. it should have a surface that we can see. In other words if we had an &lt;strong&gt;opaque&lt;/strong&gt; sphere of dust around us that had somehow come into perfect thermal equilibrium at 2.7K then this would produce something that looked like the CMB. Unfortunately we would then be unable to see anything else beyond that! For a cosmological interpretation that isn't a problem. We cannot see (at any wavelength) &amp;quot;beyond&amp;quot; the CMB at a redshift of &lt;span class=&quot;math-container&quot;&gt;$\sim 1100$&lt;/span&gt;. It is a fatal problem for any model where the CMB is optically thick emission from a sphere surrounding the solar system, since we wouldn't be able to see anything beyond the solar system!&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The CMB &lt;em&gt;is&lt;/em&gt; anisotropic. The most obvious anisotropy is the large (0.1%) dipole anisotropy caused by the motion of the solar system with respect to the rest frame of the CMB. Why should the Sun be moving at about 370 km/s with respect to a spherically symmetric Oort cloud (in the direction with Galactic coordinates &lt;span class=&quot;math-container&quot;&gt;$l=264$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$b=48$&lt;/span&gt;? Even if the Oort cloud were at a distance of 10,000 au, the Sun would travel beyond it in only &lt;span class=&quot;math-container&quot;&gt;$\sim 100$&lt;/span&gt; years! If the sphere is supposed to be further away, what is preventing gravity disturbing its almost perfect uniformity and how does it maintain a perfect thermal equilibrium?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2020-10-10T23:19:27.667" LastActivityDate="2020-10-10T23:19:27.667" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28412" PostTypeId="1" CreationDate="2018-11-16T18:42:05.063" Score="1" ViewCount="139" Body="&lt;p&gt;Planet A and Planet B orbit a sun-like star.  Planet A has a semi-major axis of 1 AU from the star.  Planet B has an albedo of .8 and has the same radius as Jupiter.  When Planet B is observed from Planet A when the planets are at opposition (Planet B is directly opposite from the star from the perspective of Planet A), Planet B has an apparent magnitude of 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How far is Planet A from Planet B?  (alternatively, what is the semi major axis of Planet B?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume negligible eccentricity in the orbits.  Assume co-planar orbits.  &lt;/p&gt;&#xA;" OwnerUserId="24853" LastEditorUserId="24853" LastEditDate="2018-11-16T19:15:39.517" LastActivityDate="2018-11-16T19:15:39.517" Title="Distance between two planets" Tags="&lt;planet&gt;&lt;observational-astronomy&gt;&lt;exoplanet&gt;&lt;distances&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2018-11-18T07:21:55.343" ContentLicense="CC BY-SA 4.0" />
  <row Id="28413" PostTypeId="1" CreationDate="2018-11-16T19:58:08.697" Score="5" ViewCount="184" Body="&lt;p&gt;Some of the ultra-hot Jupiters (WASP-33b, KELT-9b) have dayside surface temperatures similar to low-mass stars. Would they be capable of producing flares? To be specific, I am asking about the possibility of flaring on the planet itself, rather than &lt;a href=&quot;https://astronomy.stackexchange.com/questions/25188/can-hot-jupiters-cause-solar-flares&quot;&gt;the planet inducing flares on the star&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastActivityDate="2021-04-17T05:22:54.863" Title="Could an ultra-hot Jupiter flare?" Tags="&lt;exoplanet&gt;&lt;hot-jupiter&gt;&lt;flare&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28414" PostTypeId="2" ParentId="28409" CreationDate="2018-11-16T21:10:20.300" Score="8" Body="&lt;p&gt;First things first ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Tachyons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These are &lt;em&gt;entirely&lt;/em&gt; hypothetical and no evidence for them exists.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How much does subatomic particles such as photons, &lt;del&gt;tachyons&lt;/del&gt;, and electron being the biggest particles etc. occupy space?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How long is a piece of string ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no simple answer to what size an elementary particle is.  You can view them, depending on context, as having no size (point like) or being spread all over space (wave distribution).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can't say how many free particles there are (i.e. particles not bound to other particles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a lot more complicated by the current mainstream thinking that dark matter and dark energy make up most of the mass/energy content of the universe.  Whatever dark matter may be, if it turns out to be a particle or particles we don't know about yet (quite likely on existing evidence) then those particles will almost certainly exist as free particles (only gravitationally bound together) as current thinking is that dark matter does not interact much at all with other matter or itself.  And we have no idea what dark energy could be.  So with most of the mass/energy is in forms we cannot identify we can't really say anything much about what you're asking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very, very, very rough estimate on the amount of dark matter is about 25% of the mass-energy of the universe.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If put in a ball how big of a ball would that be?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In a ball it would be (possibly) a huge black hole comprising anywhere from 25% to 95% of the mass-energy of the universe.  I won't even bother estimating the size of the event horizon as it's just a useless number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However &quot;how big a ball&quot; could equally mean it was, as it is now, spread out over the entire universe.  So a perfectly correct answer is - universe sized.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2018-11-16T21:10:20.300" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28415" PostTypeId="1" CreationDate="2018-11-16T22:13:50.013" Score="4" ViewCount="88" Body="&lt;p&gt;I live in a big city: São Paulo, Brazil. I would like to build a simple telescope dome to do some amateur sky obseravtions. But the whole project lies on the viability to see something interresting more than just some stars. Due to high luminosity of cities this is a good question to ask.&#xA;So, which physical parameters are relevant to decide if a backyard observatory is a good holiday DIY project?&lt;/p&gt;&#xA;" OwnerUserId="24876" LastActivityDate="2018-11-21T19:39:10.920" Title="Question about the relevant physical parameters to build a backyard observatory" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;observatory&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28416" PostTypeId="2" ParentId="28415" CreationDate="2018-11-17T01:09:31.003" Score="1" Body="&lt;p&gt;You've identified sky brightness as a potential concern for your observatory project. In that case, you should start by trying to measure it. Unihedron manufacturers sky quality meters that you can use to get scientific measurements in mag/arcsec2. This could give you a baseline for your considerations. You could roughly convert this measurement into a class on the Bortle scale. (But keep in mind that the conversion is approximate.) You could research whether the objects you might be interested in could be seen under that class of sky. More simply, you can use &lt;a href=&quot;http://darksitefinder.com/maps/world.html#4/39.00/-98.00&quot; rel=&quot;nofollow noreferrer&quot;&gt;this site&lt;/a&gt; to see a map of light pollution in your area. Keep in mind that this map is generally accurate, but may be somewhat inaccurate at your specific location (as there are a huge number of factors that might affect sky brightness). You could, of course, buy a telescope and try to observe objects of interest with it. This would be the most direct way, obviously, to determine this project's feasibility in terms of what you're interested in observing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure you realize that such an observatory should not be directly in the path of any artificial lights. It should be (obviously) on flat, sturdy land with an unobstructed view of the horizon. If you're asking specifically about building the dome itself, I have no experience with it. Someone else would have to add information form their experience. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="23940" LastActivityDate="2018-11-17T01:09:31.003" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28418" PostTypeId="1" CreationDate="2018-11-17T05:59:55.990" Score="2" ViewCount="35" Body="&lt;p&gt;I have noticed in my experience that the coordinate solution within the headers of default reduced images from radio interferometers are commonly trustworthy to the arcsecond or less. From IR and optical observatories it is never so simple: the header solution from the default images is at most useful as a very rough guidance, off by an arcminute or more. Sometimes there is no useful header whatsoever, and the user needs to compare with, e.g., DSS of 2MASS to get good coordinates for her images, if she cares about that at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has other consequences which in my opinion affect the idiosyncrasy of optical and radio astronomers, for example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;No coordinates in published figures is much more common within the optical-IR community than in the radio community&lt;/li&gt;&#xA;&lt;li&gt;Finding charts. Radio astronomers do not know what are they, because they are not needed for radio observations.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Radio telescopes (this depends on the frequency also) sometimes require doing pointing calibrations during the observations. But in optical telescopes they also often require a guiding star, whose coordinates I imagine are properly tabulated. Regardless, the delivered images do not end up with a proper header.&lt;/p&gt;&#xA;" OwnerUserId="24878" LastActivityDate="2018-11-17T05:59:55.990" Title="Why do default coordinates from IR-optical observatories are usually not trustworthy, while from radio interferometers they usually are?" Tags="&lt;observational-astronomy&gt;&lt;coordinate&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28419" PostTypeId="1" CreationDate="2018-11-17T06:30:12.547" Score="2" ViewCount="79" Body="&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/5/59/Diurnal_motion_by_latitude.svg/500px-Diurnal_motion_by_latitude.svg.png&quot; alt=&quot;Diurnal motion by latitude&quot; /&gt;&#xA;Somewhat related to this my picture showing only diurnal motion but Moon—seen from Earth—has a rapid motion with respect to the inertial celestial sphere, unlike stars and planets. Its motion relative to stars partially cancels the diurnal motion, and as the Moon’s orbit is inclined wrt Earth’s equator, Moon’s declination also changes and hence it can rise in 180°, 181°, 182°… or 360°, 359°, 358°… where objects moving &lt;em&gt;only&lt;/em&gt; diurnally (such as stars and planets) cannot rise.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;How far outside the polar circles can Moon rise in the Western celestial hemisphere or in north/south exactly?&lt;/strong&gt; How is it best to see &lt;em&gt;practically?&lt;/em&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Also a meta request: where is a tag on astronomy.stackexchange about astronomy at high latitudes?&lt;/p&gt;&#xA;" OwnerUserId="2287" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-11-17T06:30:12.547" Title="Where on Earth can moon rise on the meridian?" Tags="&lt;the-moon&gt;&lt;positional-astronomy&gt;&lt;declination&gt;&lt;azimuth&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28420" PostTypeId="1" CreationDate="2018-11-17T11:02:42.083" Score="2" ViewCount="116" Body="&lt;p&gt;Could anyone point me to some manual or tutorial, possibly step-by-step one, on reconstruction of (quasi-) bolometric light curve using UBVRI-photometry data?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both links and names are fine for me.&lt;/p&gt;&#xA;" OwnerUserId="22098" LastActivityDate="2021-04-05T21:06:50.770" Title="Comprehensible full manual for bolometric light curve reconstruction" Tags="&lt;supernova&gt;&lt;photometry&gt;&lt;light-curve&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28421" PostTypeId="2" ParentId="28420" CreationDate="2018-11-17T15:38:51.677" Score="0" Body="&lt;p&gt;If you want a bolometric light curve from photometric points, you don't need to reconstruct its SED inside the range you already have data points. Use effective central wavelength of each band with its magnitude. Do trapezoid. This is an approximation, but it is better than reconstructing the SED that makes you rely more on assumptions for most circumstances, especially that you have UBVRI.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, under some circumstances, you really need some more data points in between, or you really want a SED, blackbody is the first-order approximation, and is commonly used in SNe (at young ages). Less known issue in optical VBRI. There might be some issues with U like UV excess/suppressed. In this case, just fit the data to a blackbody.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-11-17T15:38:51.677" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28422" PostTypeId="1" CreationDate="2018-11-17T18:44:22.947" Score="1" ViewCount="104" Body="&lt;p&gt;In apsidal lunar precession, the moon's elliptic major axis precesses eastward and completes one revolution eastward in 8.85 years.  Because the orbit is inclined relative to the ecliptic, basic geometry dictates that the ascending and descending nodes must also rotate about the ecliptic at the same rate (all else constant.)  But then there's nodal precession, rotating the nodes in the opposite (west) direction, one revolution every 18.6 years.  So what is the net effect of these two precessions on the positions of the nodes? Subtract the two?&lt;/p&gt;&#xA;" OwnerUserId="18331" LastActivityDate="2021-04-09T12:42:00.130" Title="what is the net effect of nodal and apsidal lunar precession?" Tags="&lt;the-moon&gt;&lt;precession&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28423" PostTypeId="1" CreationDate="2018-11-17T21:16:34.630" Score="3" ViewCount="197" Body="&lt;p&gt;Here we will be concerning ourselves with the transit of a planet with it's parent sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a few questions that I am not sure about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;strong&gt;Do all planets transit their parent sun?&lt;/strong&gt; It would seem to me that if all planets &quot;cross&quot; their parent star in some form, then depending on our point of view, we should be able to see the transits. For example, we never see a Mars transit, but if we alter our point of view to be farther then mars, then we would.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;strong&gt;If all the planets are in the same orbital plane, would the transit of one, imply the transit of all the others&lt;/strong&gt;? For example, our solar system is relevatily in the same plane, and it seems like Mercury/Venus transit, would also imply earth and beyond transits. We just have to be in right &lt;strong&gt;perspective&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3) If a planet transits its star, does the probability of a transit increase/decrease/same if planet semi major axis changes&lt;/strong&gt;? I am not sure about this one. I still think that it would be 100% &lt;strong&gt;the fact&lt;/strong&gt; that a &lt;strong&gt;transit&lt;/strong&gt; will occur, but its more unlikely that we would ever see it, because we need our angle to be just right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) &lt;strong&gt;If a planetary system has multiple planets, would we detect transits from several planets if they are all in the same orbital plane, or if they have different orbital planes&lt;/strong&gt;? This kind of builds off of #2, and I think the answer is the same orbital plane, because the &quot;different&quot; orbital planes could be such that are 90 degrees from our point of view, meaning we would never see the transit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know if my understanding of transits is correct by these questions. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="17327" LastActivityDate="2018-11-19T14:50:10.087" Title="A few questions regarding the transit of planets" Tags="&lt;the-sun&gt;&lt;planet&gt;&lt;star-formation&gt;&lt;planetary-transits&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28424" PostTypeId="2" ParentId="28423" CreationDate="2018-11-17T23:08:13.290" Score="3" Body="&lt;p&gt;When we talk about a transit, we are normally talking about one that is visible from Earth, or at least from some planet in the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you are willing to go to a point is space, it is always the case that every planet is transiting the sun from somewhere. So the Earth is right now transiting the  sun from any position that is directly opposite the sun and more than about 1600000km from Earth (otherwise the sun won't be transited, it will be eclipsed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transits of all the planets can be seen from exterior planets, though the transits of (say) Saturn from Neptune are exceedingly rare&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are in the orbital plane of the planet, you will eventually see a transit.&#xA;But you may be waiting a long time. The semi major axis doesn't have any effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the planet in a planetary system are in the same plane, then you may see more than one transit. But you are less likely to see the outer planets as they are further from the star.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="1462" LastEditDate="2018-11-19T14:50:10.087" LastActivityDate="2018-11-19T14:50:10.087" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28425" PostTypeId="2" ParentId="28423" CreationDate="2018-11-18T00:35:52.337" Score="3" Body="&lt;p&gt;You need to draw yourself a diagram. A right angle triangle with one side equal to the radius of the star &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; and another, the distance from the planet to the star &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless your line of sight is in the angle defined by &lt;span class=&quot;math-container&quot;&gt;$\theta = \pm \tan^{-1} (r/a)$&lt;/span&gt; then you won't see a transit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$a \gg r$&lt;/span&gt; then we can say the probability of a transit given a random orbital orientation is about &lt;span class=&quot;math-container&quot;&gt;$r/a$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://ay20class.blogspot.com/2011/11/transit-probability.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ay20class.blogspot.com/2011/11/transit-probability.html&lt;/a&gt; for the details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For multiple transits then all planetary orbits would have to be confined within a range of angles defined by &lt;span class=&quot;math-container&quot;&gt;$\pm \tan^{-1} (r/a)$&lt;/span&gt;, using the &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; appropriate for each planet. Thus more distant planets are less likely to transit and there are tighter constrains on their orbital inclinations if they are to be one of multiple transiting planets &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-11-18T00:35:52.337" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28426" PostTypeId="1" CreationDate="2018-11-18T02:15:34.913" Score="2" ViewCount="79" Body="&lt;p&gt;Do you have in mind any database/sites that I can search for stars that matches a given surface temperature and an absolute magnitude?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, when I enter &lt;code&gt;Temperature=9,500K &amp;amp; AbsoluteMagnitude=0.58&lt;/code&gt; then the search engine returns the result &lt;code&gt;Vega&lt;/code&gt; and some other stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions are appreciated!&lt;/p&gt;&#xA;" OwnerUserId="24885" LastActivityDate="2018-11-18T10:55:36.783" Title="Searching for stars with a given surface temperature and an absolute magnitude?" Tags="&lt;star&gt;&lt;data-analysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28428" PostTypeId="1" AcceptedAnswerId="28432" CreationDate="2018-11-18T03:14:15.463" Score="2" ViewCount="979" Body="&lt;p&gt;I've been thinking about getting a new telescope. From what I've read online, reflector telescopes are newer and better then refractor telescopes. Is there any reason to get a refractor telescope instead of a reflector?&lt;/p&gt;&#xA;" OwnerUserId="24758" LastActivityDate="2018-11-19T23:15:54.917" Title="What are the advantages of refractor telescopes over reflector telescopes?" Tags="&lt;telescope&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28429" PostTypeId="1" AcceptedAnswerId="28431" CreationDate="2018-11-18T10:00:03.990" Score="5" ViewCount="341" Body="&lt;p&gt;Are there any astronomical examples where the transverse doppler effect(Horizontal doppler effect) is applied (Derives a meaningful result)?&lt;/p&gt;&#xA;" OwnerUserId="24885" LastEditorUserId="24885" LastEditDate="2018-11-18T10:25:43.457" LastActivityDate="2018-11-18T11:36:31.947" Title="Are there any examples where the transverse doppler effect is applied in astronomy?" Tags="&lt;light&gt;&lt;special-relativity&gt;&lt;doppler-effect&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28430" PostTypeId="2" ParentId="28426" CreationDate="2018-11-18T10:55:36.783" Score="1" Body="&lt;p&gt;Though not much convenient, I found one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/astronexus/HYG-Database&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/astronexus/HYG-Database&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a database that contains information of a lot of stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may use excel to use this database and derive any needed results.&lt;/p&gt;&#xA;" OwnerUserId="24885" LastActivityDate="2018-11-18T10:55:36.783" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28431" PostTypeId="2" ParentId="28429" CreationDate="2018-11-18T11:36:31.947" Score="7" Body="&lt;p&gt;Accounting for the transverse Doppler effect (and other relativistic effects) is essential in modelling the X-ray spectral emission lines from the accretion discs around black holes (e.g. &lt;a href=&quot;https://academic.oup.com/mnras/article/363/1/177/1299829&quot; rel=&quot;noreferrer&quot;&gt;Cadaz &amp;amp; Calvani 2005&lt;/a&gt;). In this case the transverse Doppler effect is &quot;mixed up&quot; with gravitational redshift and it is treated holistically in the Schwarzschild or Kerr metrics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transverse Doppler effect is also essential in interpreting the signals from binary pulsars and even in single pulsars because of the rotation of the Earth and its motion around the Sun.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-11-18T11:36:31.947" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28432" PostTypeId="2" ParentId="28428" CreationDate="2018-11-18T14:35:35.770" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;reflector telescopes are newer and better then refractor telescopes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's true that the first examples of reflectors we know of were made shortly after the first refractors we know of. But it's not correct to say that they're (always) better. First, each telescope model you might buy has its own specifications and build quality, which have to be carefully considered in terms of what you're trying to view. Second, certain refractors can outperform certain reflectors for certain tasks: for instance, many refractors can offer better views of planetary/lunar/cluster details than similarly priced reflectors could. Reflectors designed such that eyepieces have to be used at a standing height have the disadvantage of being difficult to use for things (e.g. planets) near the horizon. Refractors could also have the advantage of being potentially smaller and more portable than, for instance, something like a 10&quot; dob. Reflectors have their advantages too: lack of chromatic aberration, for instance (although they do suffer from other forms of aberration). I hope this gives you a better idea of the pros and cons of different kinds of telescopes.  &lt;/p&gt;&#xA;" OwnerUserId="23940" LastActivityDate="2018-11-18T14:35:35.770" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28435" PostTypeId="2" ParentId="28428" CreationDate="2018-11-18T23:59:24.040" Score="3" Body="&lt;p&gt;Every type of scope has its own strengths and weaknesses. That's one reason why it's fairly common for enthusiastic amateur astronomers to end up with several scopes :) .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reflectors have the advantage of being available in large apertures, and when coupled with a Dobsonian style mount, are usually the cheapest way to get a given aperture - because the Dobsonian mounts are cheap to make, more of the purchase price can go towards the optics - and the dobsonian mounts are usually nice and stable, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refractors have the advantage of having no central obstruction, a basically fixed setup (they practically never need collimating, unlike reflectors), freedom from diffraction spikes (caused by the &quot;spider&quot; secondary support in newtonian reflectors),&#xA;and low maintenance - &#xA;and they are also good for terrestrial use, unlike newtonian reflectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smaller refractors make great grab and go widefield scopes, and make a nice complement to a larger reflector &quot;light bucket&quot;, and a good ED or APO refractor is an excellent choice for wide field astrophotography - and complements something like a longer focal length SCT or similar for imaging smaller objects (which needs better tracking/guiding/seeing). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For long exposure deep sky astrophotography, the most important thing is a good motorised equatorial mount - whatever type of scope you use, you need to be able to track your target accurately as it moves across the sky. Alt-az goto mounts are fine for visual use, but suffer from an effect called field rotation - although they can keep the object centered, the field of view appears to gradually rotate, causing trailing in long exposures. Equatorial mounts have one axis pointing at the celestial pole, and don't suffer from field rotation.)&lt;/p&gt;&#xA;" OwnerUserId="17016" LastEditorUserId="17016" LastEditDate="2018-11-19T23:15:54.917" LastActivityDate="2018-11-19T23:15:54.917" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28437" PostTypeId="1" CreationDate="2018-11-19T03:38:48.557" Score="2" ViewCount="69" Body="&lt;p&gt;I mean a total number of stars, not the only ones that we can see.&lt;/p&gt;&#xA;" OwnerUserId="24895" LastActivityDate="2018-11-19T03:38:48.557" Title="What is the total number of stars in the universe?" Tags="&lt;star&gt;&lt;universe&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2018-11-19T21:11:33.783" ContentLicense="CC BY-SA 4.0" />
  <row Id="28438" PostTypeId="1" CreationDate="2018-11-19T04:19:56.003" Score="2" ViewCount="77" Body="&lt;p&gt;I woild like to build a &lt;strong&gt;Radio Jove receiver&lt;/strong&gt; whose &lt;a href=&quot;https://radiojove.gsfc.nasa.gov/telescope/&quot; rel=&quot;nofollow noreferrer&quot;&gt;design is available on NASA website&lt;/a&gt;. However, I don't want to purchase the receiver, I would like to make it, but but I don't have any knowledge about the &lt;a href=&quot;https://en.wikipedia.org/wiki/Printed_circuit_board&quot; rel=&quot;nofollow noreferrer&quot;&gt;PCB design&lt;/a&gt;. &lt;strong&gt;May&#xA;someone please help me to get the PCB manufacturing file?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="24896" LastEditorUserId="37191" LastEditDate="2021-05-28T07:47:29.523" LastActivityDate="2021-05-28T07:58:16.837" Title="Printed Circuit Board manufacturing file for Radio Jove?" Tags="&lt;amateur-observing&gt;&lt;radio-astronomy&gt;&lt;jupiter&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28439" PostTypeId="1" CreationDate="2018-11-19T14:42:10.230" Score="5" ViewCount="701" Body="&lt;p&gt;I got the Orbital Elements of Jupiter around the Sun, which describe how it orbits relative to the &quot;fix&quot; Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter&#xA;Semi Major Axis (AU): 5.20336301&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eccentricity: 0.04839266&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inclination to Ecliptic Plane (deg): 1.30530&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longitude of the Ascending Node (deg): 100.55615&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longitude of Perihelion (deg): 14.75385&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mean Anomaly at J2000.0 (deg): 19.65053&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to convert this One-Body Problem back into the Two-Body Problem to find out where the Sun is gonna be at a given time (i couldnt find the Orbital Elemens of Sun around Sun-Jupiter Barycenter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eWTLz.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eWTLz.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the Semi-Major Axis depends on the Mass Ratio. The Eccentricity and Inclination should be equal for both, just like Longitude of the Ascending Node. The Longitude of Perihelion should be the one from Jupiter+180°, doesnt it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main Problem is the Mean Anomaly at J2000.0. The two Ellipses are equal, besides of the Size. Thus, the True Anomaly at a given time should be the same for Each Coordinate System. Then the eccentric Anomaly will be the same too. Does that mean the Mean Anomaly for J2000.0 will be also the same for both?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for reading and sunny greetings!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GkEKK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GkEKK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24544" LastEditorUserId="24544" LastEditDate="2018-11-20T09:56:54.467" LastActivityDate="2018-11-22T13:12:24.370" Title="How does the Sun orbit the Sun-Jupiter Barycenter?" Tags="&lt;the-sun&gt;&lt;jupiter&gt;&lt;celestial-mechanics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28440" PostTypeId="2" ParentId="28409" CreationDate="2018-11-19T20:18:56.300" Score="3" Body="&lt;p&gt;Sadly, your question has no meaning unless you both posit a finite size for the particles (unlikely) and some magical way to defeat all 3 of the E-M, strong, and weak nuclear forces.  Otherwise you don't even define how &quot;closely&quot; the particles could be placed.  That's pretty much what StephenG is alluding to when he suggests a black hole -- the black hole itself has undefined dimension; the event horizon would pointless.&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-11-19T20:18:56.300" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28441" PostTypeId="1" CreationDate="2018-11-19T20:39:40.747" Score="2" ViewCount="122" Body="&lt;p&gt;I have real data of “Direct Normal Irradiation” and “Diffuse Horizontal Irradiation” and I should calculate “Global Horizontal Irradiation”. I have also longitude, latitude and elevation. &#xA;Could you advise me how I can get zenith from these data, then to calculate the GHI?&#xA;Many thanks,&lt;/p&gt;&#xA;" OwnerUserId="24899" LastEditorUserId="7472" LastEditDate="2018-11-19T21:12:12.660" LastActivityDate="2018-12-21T09:00:55.160" Title="How to get the solar zenith from the longitude and latitude?" Tags="&lt;the-sun&gt;&lt;positional-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28442" PostTypeId="1" CreationDate="2018-11-19T22:26:35.500" Score="0" ViewCount="183" Body="&lt;p&gt;Are Trans-Neptunian Objects (TNOs) asteroids?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please, base your answer. If possible, give sources. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="24900" LastActivityDate="2018-11-20T07:33:36.520" Title="Are Trans-Neptunian Objects (TNOs) asteroids?" Tags="&lt;asteroids&gt;&lt;tno&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28443" PostTypeId="2" ParentId="28442" CreationDate="2018-11-19T23:02:12.067" Score="1" Body="&lt;p&gt;The IAU does not give a definition of &quot;asteroid&quot; (in contrast to its defintion of &quot;planet&quot;) The definition of &quot;asteroid&quot; is therefore determined by actual use, rather than a specific authoritative source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The IAU in its &lt;a href=&quot;https://www.iau.org/public/themes/naming/#minorplanets&quot; rel=&quot;nofollow noreferrer&quot;&gt;guidelines on &quot;naming&quot;&lt;/a&gt; doesn't use the term &quot;asteroid&quot; (except qualified as &quot;near earth asteroid&quot;. The term is somewhat ambiguous.  Instead it considers &quot;planets&quot;, &quot;dwarf planets&quot; and &quot;minor planets&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://minorplanetcenter.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Minor planets&lt;/a&gt; can be dwarf planets, asteroids, trojans, centaurs, Kuiper belt objects, and other trans-Neptunian objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially solar system bodies were classed as &quot;comet&quot; (appears fuzzy due to a coma of dust and gas) &quot;asteroid (appears star like, a single point of light with no fuzziness). The planets appear disc-like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TNOs have no coma, but their surfaces are dominated by ices (and would start to evaporate if ever brought to the inner solar system).  Thus they are rather different from the traditional rocky asteroids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you should avoid using an unqualified &quot;asteroid&quot; as it may be ambiguous. You can say &quot;minor planet&quot; if you want to include both TNOs KBOs and classical asteroids.  Or &quot;Main belt asteroid&quot; for those between Mars and Jupiter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you read &quot;asteroid&quot; it would normally exclude the outer solar system bodies. But as it may be ambiguous, so read the context.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-11-20T07:33:36.520" LastActivityDate="2018-11-20T07:33:36.520" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28444" PostTypeId="1" CreationDate="2018-11-20T01:16:19.707" Score="1" ViewCount="81" Body="&lt;p&gt;On this post,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/10467/time-according-to-the-gravity-of-sagittarius-a/10470#10470&quot;&gt;Time according to the gravity of Sagittarius A?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They found the factor of time on the varying distances from the SMBH, I am curious on how to turn those factors into relatable terms such as:&#xA;At 1 km from the event horizon ____ time passes on earth in an hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just not sure how to figure this out.&lt;/p&gt;&#xA;" OwnerUserId="24897" LastEditorUserId="24897" LastEditDate="2018-11-20T05:43:21.543" LastActivityDate="2018-11-20T22:35:18.647" Title="How to convert factors of time dilation into something relatable?" Tags="&lt;black-hole&gt;&lt;space-time&gt;&lt;space&gt;&lt;supermassive-black-hole&gt;&lt;time-dilation&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2018-11-21T13:12:40.647" ContentLicense="CC BY-SA 4.0" />
  <row Id="28449" PostTypeId="1" CreationDate="2018-11-20T12:32:00.967" Score="2" ViewCount="2645" Body="&lt;p&gt;I am the owner of a small 4&quot;, f/6.3 &lt;strong&gt;tabletop&lt;/strong&gt; Dobsonian telescope with a 102mm aperture and 640mm focal length. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is based on a Dobsonian mount but I would like to use a tripod instead because of the complications of &lt;em&gt;having to cary a table with me everywhere I would like to use the telescope.&lt;/em&gt; The alternative is to but the telescope on the ground which makes viewing uncomfortable and is not very stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good tripod could be foldable or collapsible and make my rig much more portable and bring the eyepiece up to a more convenient viewing height, without having to bring along a &quot;telescope table&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this possible? If so, are there any specific tripods I should use or other pointers on how best to proceed?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is not me or my telescope or mount, but it gives an idea of viewing with a similarly-sized tabletop telescope. I'd like to mount the telescope on a tripod instead. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/joizs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/joizs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://www.skyandtelescope.com/astronomy-equipment/coda-binoculars-versus-starblast/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24597" LastEditorUserId="7982" LastEditDate="2018-11-25T02:38:41.347" LastActivityDate="2018-11-27T00:45:57.547" Title="How to adapt small Dobsonian to use with a tripod?" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;dobsonian-telescope&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28451" PostTypeId="1" AcceptedAnswerId="28452" CreationDate="2018-11-20T17:45:46.420" Score="1" ViewCount="203" Body="&lt;p&gt;I do know nothing can escape a black hole, and I understand popsci explanation of geodesics of space-time around and inside the event horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I however have some trouble with following thought experiment, given there are many references that crossing event horizon of SMBH should be totally unnoticeable for the subject falling through: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am at gym - a regular one, on Earth. Escape velocity of Earth is around 11.2 km/s. Yet I can lift weights up say one meter even at slower speed, say one meter per second. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, consider me hovering one meter above event horizon. As far as I know, for non-rotating BH this is totally possible. Similar to earth, I could lift weight up one meter at speed less than escape velocity (which at this point is speed of light &lt;em&gt;minus&lt;/em&gt; x). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I’m hovering one meter below the event horizon. Would I not be able to lift the weight up one meter at speed less than escape velocity (which at this point is speed of light &lt;em&gt;plus&lt;/em&gt; y; not sure if x and y are equal), or would I not be able to hover at this position?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming answer is no to either or both, how this would appear to me as the subject?&lt;/p&gt;&#xA;" OwnerUserId="10813" LastActivityDate="2021-04-11T21:19:21.357" Title="Black hole, escape velocity, going up?" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;supermassive-black-hole&gt;&lt;escape-velocity&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28452" PostTypeId="2" ParentId="28451" CreationDate="2018-11-20T20:57:40.947" Score="0" Body="&lt;p&gt;This is more of a pure physics question and I'm not sure it would do well there either, but I can touch on some basics.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am at gym - a regular one, on Earth. Escape velocity of Earth is&#xA;  around 11.2 km/s. Yet I can lift weights up say one meter even at&#xA;  slower speed, say one meter per second.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can lift a weight at the gym and hold it there, but only by doing constant work.   If you let go, it'll fall to the floor and make a noise, and perhaps get your membership revoked. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could, in a thought experiment anyway, climb a ladder up into Earth's orbit, past the satellites and keep going out past Earth's Hill Sphere, where the gravity of the Sun would take over and you could let go and part ways with the Earth.   It makes no difference how fast you climb the ladder but that would require applying continued work.    Applying work has nothing to do with escape velocity.    Escape velocity is like throwing a ball up into the air, once you let go of the ball, it has an initial velocity, and the &quot;what goes up must come down&quot; law takes over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What comes up must come down, assuming we ignore air resistance, is a good way to explain escape velocity.   If you throw something faster than 11.2 km/s, it won't ever come down, it'll keep going up until it escapes the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Escape velocity doesn't apply to rockets which apply several minutes of thrust to escape Earth's gravity and enter orbit.  (technically in orbit is still in Earth's gravity well, but it's escaping a significant part of Earth's gravity it in order to remain in orbit) - &lt;a href=&quot;https://what-if.xkcd.com/58/&quot; rel=&quot;nofollow noreferrer&quot;&gt;see xkcd&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Escape velocity only applies to initial velocity and it ignores air resistance.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Next, consider me hovering one meter above event horizon. As far as I&#xA;  know, for non-rotating BH this is totally possible. Similar to earth,&#xA;  I could lift weight up one meter at speed less than escape velocity&#xA;  (which at this point is speed of light minus x).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can lift a weight 1 meter in your gym because the Earth's surface is holding you up.   An event horizon has no surface, so there's nothing for you to stand upon.   Also, your gym-weight would weigh (I don't want to calculate how much), but near the surface of a black hole gravity is so high that a feather would weigh probably thousands maybe hundreds of thousands of tons in your scenario.   Lifting with no surface below you and feathers weighing tons poses several problems with your scenario.   I'm not sure any of that applies to your question in your thought experiment, I'm just pointing out some of the difficulties that arise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@PM2Ring points out that if the black hole is large enough the weight and surface gravity does decrease, which sounds odd, but seems correct, but no black holes are that large, and there's still the no surface to stand on.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now, I’m hovering one meter below the event horizon. Would I not be&#xA;  able to lift the weight up one meter at speed less than escape&#xA;  velocity (which at this point is speed of light plus y; not sure if x&#xA;  and y are equal), or would I not be able to hover at this position?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can't hover below the surface of a black hole unless you have an energy source that defies the laws of physics.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps it's easier to think of in terms of space-time, which I'm not an expert with, but the definition of space-time inside a black hole is such that no amount of acceleration can move any thing further away from the singularity.  No fuel, no manual force can move any part of anything in a direction away from the center of the black hole.   Every direction points to the center.  Space inside a black hole becomes like like time.  You can slow time down but you can't reverse it.   Lifting something upwards is no more possible than stepping backwards in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if that helps.   Maybe someone can explain it better than me.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-11-20T20:57:40.947" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28453" PostTypeId="2" ParentId="28444" CreationDate="2018-11-20T22:35:18.647" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am curious on how to turn those factors into relatable terms such as: At 1 km from the event horizon ____ time passes on earth in an hour.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The key here is to understand what the formula relates to.  The (all important) &lt;em&gt;observer&lt;/em&gt; is in this case a notional observer so distant that curvature (and hence gravitational effects) are negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That observer is the &quot;base clock&quot; we reference.  Everything is dilated relative to &lt;em&gt;that&lt;/em&gt; clock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that the dilation on Earth (not strictly speaking an ideal distant observer) has to be taken into account to get an answer in the form you mean.  So starting with the formula you had :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$t = t_\infty \sqrt{1 - \frac{r_s}{r}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want this relative to Earth so we need :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$t_e = t_\infty \sqrt{1 - \frac{r_s}{r_e}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And combining youy get :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$t = t_e \sqrt{ \frac {1 - \frac {r_s} r} {1 - \frac {r_s} {r_e}} }&#xA;= t_e \sqrt { \frac { r_e(r-r_s) } { r(r_e-r_s) } }$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in the specific case of Earth and Sagitarius A*, &lt;span class=&quot;math-container&quot;&gt;$r_e$&lt;/span&gt; is &lt;em&gt;huge&lt;/em&gt; so the dilation factor for Earth is as close to one as makes no difference in practice, so you don't &lt;em&gt;really&lt;/em&gt; need to go to all that trouble and you can use &lt;span class=&quot;math-container&quot;&gt;$t_\infty \approx t_e$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2018-11-20T22:35:18.647" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28454" PostTypeId="1" AcceptedAnswerId="28494" CreationDate="2018-11-21T04:15:45.013" Score="1" ViewCount="118" Body="&lt;p&gt;Wondering what primary astronomical cycles there are that are greater than a year (which is a cycle around the sun), but less than 1000 years-ish (on that scale). There is the cycle of the moon phases too, and the cycle of the earth making a full rotation (a day). But I wonder if there are any cycles at a bigger-than-a-year scale, but not too big. I guess there's also the cycle of a stars life (but that is kind of a big scale, on the order of billions of years), but I am more wondering in terms of &lt;em&gt;rotation&lt;/em&gt; what sort of cycles there are. I guess there is another one, &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;Axial Precession&lt;/a&gt;, but that is a bit large such as on the order of tens of thousands of years. Maybe there are some cycles of the planets in which they are visible from earth, or of their moons, or of comets, or of nearby stars or that kind of thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there were ones around each of these points that would be nice: 2, 5, 10, 20, 50, 100, 500, or 1000 years sort of thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically I am looking for &quot;interesting&quot; or &quot;popular&quot; cycles rather than just picking any of them (since every planet and moon has an orbital period). I am looking for ones that (a) we can observe most easily either directly without a telescope, or with a simple telescope/binoculars, &lt;em&gt;and/or&lt;/em&gt; (b) have significance of some kind historically perhaps. If there is one that is historically significant or historically popular (say something Newton or Kepler talked about for example) but is difficult to see, that would be okay too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I vaguely remember hearing people mention &quot;hey these x and y planets just aligned, it only happens every x years&quot;, so did a quick search and found &lt;a href=&quot;https://www.quora.com/How-often-does-Jupiter-align-with-Mars-while-the-moon-is-in-the-seventh-house&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; saying Jupiter and Mars align every 2-3 years. That is pretty interesting, since it is a small timeframe, bigger than a year, and you can see it pretty easily. I would like to find a few others though, spread over the scale (as mentioned above) from 1-1000 years. &lt;a href=&quot;https://www.forbes.com/sites/startswithabang/2018/01/24/how-rare-is-the-all-in-one-supermoon-blue-moon-and-lunar-eclipse-really/#2237862a3cf2&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is another interesting one, a &quot;Super Blood Blue Moon&quot; as NASA is calling it, which hasn't happened since 1866 (so let's say ~150 years). Not sure if that will cycle every 150 years like that, but something that does cycle similar to this is also exactly the sort of &quot;cycle&quot; I am looking for.&lt;/p&gt;&#xA;" OwnerUserId="7620" LastEditorUserId="7620" LastEditDate="2018-11-21T07:08:24.033" LastActivityDate="2018-11-22T10:00:44.510" Title="Cycles that are longer than the year, but less than 1000 years" Tags="&lt;rotation&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28455" PostTypeId="1" CreationDate="2018-11-21T07:50:31.620" Score="-2" ViewCount="82" Body="&lt;p&gt;I  know that there is very little O2 on mars, very little pressure, and very cold temperature, but what would kill her first-suffocation, exploding, boiling body fluids, or exploding?&lt;/p&gt;&#xA;" OwnerUserId="24914" LastEditorUserId="2365" LastEditDate="2018-11-21T07:55:43.333" LastActivityDate="2018-11-21T09:13:55.777" Title="What would happen to a woman who was teleported to mars while wearing a dress and skirt" Tags="&lt;mars&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2018-11-21T13:06:53.227" ContentLicense="CC BY-SA 4.0" />
  <row Id="28456" PostTypeId="1" CreationDate="2018-11-21T08:05:30.903" Score="0" ViewCount="83" Body="&lt;p&gt;A galaxy at the boundary of the Hubble sphere is receeding from us at the speed of light, right? If it emits a photon now, how long will it take to reach us? &lt;/p&gt;&#xA;" OwnerUserId="16506" LastActivityDate="2018-11-22T09:14:06.377" Title="How long would a photon take to reach us if it's emitted from a galaxy receeding at c?" Tags="&lt;cosmology&gt;&lt;cosmological-horizon&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28459" PostTypeId="1" AcceptedAnswerId="28478" CreationDate="2018-11-21T08:22:23.960" Score="2" ViewCount="271" Body="&lt;p&gt;I have seen the Gaia project from &lt;a href=&quot;https://astronomy.stackexchange.com/questions/3668/where-can-i-find-a-catalog-of-all-stars-in-the-milky-way&quot;&gt;here&lt;/a&gt; which is about mapping the &lt;em&gt;stars&lt;/em&gt; in the Milky Way galaxy. But I'm wondering if there is any open data on the positions (or something like &quot;positions&quot; since it probably not calculable now), or &quot;coordinates&quot;, of the &lt;em&gt;galaxies&lt;/em&gt; in the visible universe, as opposed to the stars. I wonder how animations such as &lt;a href=&quot;https://www.youtube.com/watch?v=HiN6Ag5-DrU&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; (a time=1m30s point it shows the galaxies zooming out from the Milky Way) get their data on the positions of the galaxies.&lt;/p&gt;&#xA;" OwnerUserId="7620" LastActivityDate="2018-11-22T09:43:18.120" Title="Catalog of Galaxies in the Visible Universe" Tags="&lt;galaxy&gt;&lt;data-analysis&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28460" PostTypeId="2" ParentId="28441" CreationDate="2018-11-21T08:29:50.293" Score="1" Body="&lt;p&gt;Direct Normal Irradiance (DNI) is the the amount of solar radiation from the direction of the Sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diffuse Horizontal Irradiance (DHI) is the amount of solar radiation that does not arrive on a direct path from the Sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Global Horizontal Irradiance (GHI) is the total amount of solar radiation. It is the sum of Direct Normal Irradiance (DNI), Diffuse Horizontal Irradiance (DHI) and ground-reflected radiation. Ground-reflected radiation is usually insignificant (compared to DNI and DHI) so for all practical purposes GHI is the sum of DNI and DHI.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus &lt;span class=&quot;math-container&quot;&gt;$GHI = DHI + DNI \times cos (Z)$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$Z$&lt;/span&gt; is the solar zenith angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solar zenith angle is derived from latitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solar zenith occurs at local solar noon, which is derived from longitude.&lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2018-11-21T08:29:50.293" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28463" PostTypeId="2" ParentId="28459" CreationDate="2018-11-21T08:49:11.847" Score="1" Body="&lt;p&gt;There is the &lt;a href=&quot;https://heasarc.gsfc.nasa.gov/db-perl/W3Browse/w3table.pl?tablehead=name%3Dugc&amp;amp;Action=More+Options&quot; rel=&quot;nofollow noreferrer&quot;&gt;Uppsala General Catalog of Galaxies&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has galaxies down to 19 magnitude. It doesn't have distances, but it does have radial velocity.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-11-21T08:49:11.847" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28465" PostTypeId="2" ParentId="28459" CreationDate="2018-11-21T09:03:13.563" Score="1" Body="&lt;p&gt;There is the HyperLeda catalog &lt;a href=&quot;http://leda.univ-lyon1.fr/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://leda.univ-lyon1.fr/&lt;/a&gt;, which has about 5 million galaxies, with redshift for about 3 million (if I understand the summary).&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2018-11-21T09:03:13.563" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28466" PostTypeId="2" ParentId="28455" CreationDate="2018-11-21T09:08:28.070" Score="1" Body="&lt;p&gt;I don't think this is astronomy, so it'll probably be closed, but the vacuum is the biggest immediate stress on the body and would very likely be the cause of death, with suffocation, which takes a few minutes, coming in a distant 2nd.   The temperature of Mars won't kill somebody that quickly and body parts don't explode and most bodily fluids won't boil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The body will swell, but the skin will hold the body together and only exposed liquids like saliva will quickly boil and &lt;a href=&quot;https://www.youtube.com/watch?v=ECOMAbLI1V4&quot; rel=&quot;nofollow noreferrer&quot;&gt;boiling in a vacuum will also lead to freezing&lt;/a&gt; so your person's mouth will fill with ice and some ice will likely form in their lungs and throat, not by freezing, but by the temperature change due to phase transition of evaporation and they'll probably explosively defecate, but their blood won't boil.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you ever are tossed out of an airlock, or teleported to Mars without your permission (where does that fall &lt;a href=&quot;https://what-if.xkcd.com/146/&quot; rel=&quot;nofollow noreferrer&quot;&gt;in the things to worry about list&lt;/a&gt;, not very high I'd imagine), but if it does happen to you, the first thing to do is exhale.  You might think &quot;no air, I better hold my breath&quot;, but that's the worst thing you can do, you want as little air as possible inside you're body in that situation.  Any air inside your body will want to expand at 14.4 lbs per square inch and the inside of your lungs are delicate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Animal lovers shouldn't read, but NASA did studies on this before sending the astronauts into space.    &lt;a href=&quot;https://www.scientificamerican.com/article/survival-in-space-unprotected-possible/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.scientificamerican.com/article/survival-in-space-unprotected-possible/&lt;/a&gt;   Human studies have never been done, but with animals, 2 minutes of vacuum exposure usually lead to the heart stopping and death.  90 seconds was often survivable without significant long term negative effects.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-11-21T09:13:55.777" LastActivityDate="2018-11-21T09:13:55.777" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28467" PostTypeId="1" CreationDate="2018-11-21T10:12:53.523" Score="4" ViewCount="75" Body="&lt;p&gt;I am looking for the information on the percentage illumination of the moon during the different stages of a total lunar eclipse. Is there a database from where I can acquire information of lunar illumination, distance and angle for the July 27th 2018 lunar eclipse for a given location? &lt;/p&gt;&#xA;" OwnerUserId="24918" LastEditorUserId="24918" LastEditDate="2018-11-21T11:19:32.383" LastActivityDate="2020-12-04T15:10:23.563" Title="Lunar illumination (percentage) data on an eclipse day needed" Tags="&lt;the-moon&gt;&lt;distances&gt;&lt;lunar-eclipse&gt;&lt;moon-phases&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28471" PostTypeId="2" ParentId="28439" CreationDate="2018-11-21T12:37:39.410" Score="2" Body="&lt;p&gt;The barycenter of two point masses M1 and m2 -- or their 'center of gravity' (which to avoid any doubt is not in itself a center of gravitational attraction) -- lies on the straight line joining them, and divides the distance between them in the proportion m2:M1 -- in classical/Newtonian terms. That is, the barycenter is separated from center M1 by the fraction m2/(M1+m2) of the total distance M1 to m2, and by the complementary fraction M1/(M1+m2) from the center of m2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a two-body system, the positions and motions of each body relative to their barycenter are similar in the way that two triangles can be similar: the corresponding angles for each are the same, the distances are in a constant proportion. (See also &lt;a href=&quot;https://en.wikipedia.org/wiki/Barycenter&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Barycenter&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Two-body_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Two-body_problem&lt;/a&gt; .) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at least part of the answer to the question about the Sun and Jupiter -- imagined as a 2-body system -- is that the angles including the mean anomaly are indeed the same for each. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this only applies approximately to the figures quoted in the question. These are not 'one-body' or 'two-body' orbit figures, they are heliocentric approximate mean elements for Jupiter in the n-body solar system as at J2000 (i.e. 2000 Jan 1.5), from a derivation applied to data over the interval 1800-2050. The figures appear in a &lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/jupiterfact.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;GSFC/NASA Jupiter fact-sheet&lt;/a&gt;, and also on p.316 of 'Explanatory Supplement to the Astronomical Almanac' (1992, ed. K P Seidelmann et al.), where the error in Jupiter's position computed from the elements is stated as up to 300&quot; in right ascension.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mean elements such as those quoted in the question can be derived from the time-varying &lt;a href=&quot;https://en.wikipedia.org/wiki/Osculating_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;osculating elements&lt;/a&gt; obtained over a long period. Such elements may be obtained for the chosen period in different ways, for example &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) by deriving them from results of a long-term combined numerical integration of the orbits of a set of solar-system bodies -- usually involving all of the major planets along with either some of the minor planets as further individual bodies and/or an approximate allowance made for the effects of those not included individually;  or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) by deriving them from an analytical theory (a method used for example to obtain the results in J L Simon et al., 'Precession formulae and mean elements for the Moon and the planets', Astronomy &amp;amp; Astrophysics 282 (1994) 663-683). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In either of these cases, subtraction of the periodic terms leaves the mean elements as non-periodic polynomials as a function of time-interval from a chosen epoch associated with the data.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A third and approximate approach, which seems to have been the one adopted for the figures quoted in the question, involves fitting the osculating elements to a chosen polynomial function over a chosen interval. The website &lt;a href=&quot;http://www.met.rdg.ac.uk/~ross/Astronomy/Planets.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.met.rdg.ac.uk/~ross/Astronomy/Planets.html&lt;/a&gt; suggests that the figures here resulted from a 250-year least squares fit of data from the JPL DE200 numerically integrated planetary ephemeris to a Keplerian orbit where each element was allowed to vary linearly with time.  The mean elements for a date such as J2000 would then represent the value of the linear function at the chosen date.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is wished to take the approximate mean elements for Jupiter at J2000 to represent 1-body elements on that date, then that is perhaps about as close an approximation of the kind as can be made: and the angles will then be similar as already described, and the barycentric distances will be obtained by the fractional split already mentioned. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more precise way of tracking the relation between the Sun and the solar-system barycenter is outlined in &lt;a href=&quot;http://adsabs.harvard.edu/abs/1983A%26A...125..150N&quot; rel=&quot;nofollow noreferrer&quot;&gt;a 1983 paper from JPL on solar-system numerical integration by XX Newhall et al.&lt;/a&gt;). This takes into account all of the planets and their masses, and the result is a complex path that has perhaps never been represented by any trigonometric series, but only numerically and diagrammatically -- e.g. at &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Solar_system_barycenter.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/File:Solar_system_barycenter.svg&lt;/a&gt;, thus: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fM7hn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fM7hn.png&quot; alt=&quot;from Wikimedia&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17202" LastEditorUserId="17202" LastEditDate="2018-11-21T12:55:57.777" LastActivityDate="2018-11-21T12:55:57.777" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28473" PostTypeId="1" AcceptedAnswerId="28475" CreationDate="2018-11-21T14:57:02.550" Score="4" ViewCount="703" Body="&lt;p&gt;I know that we don't see green stars because in blackbody radiation star doesn't produce one spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stars that have peaks in the green spectrum produce other spectrum in nearly same amout. This make them appear white.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If stars have peaks in the infrarad spectrum that has wavelength little longer than red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see those stars red because they produce red spectrum more than other visible spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if stars have peaks in the ultraviolet spectrum that has wavelength little shorter than violet. Why don't we see those stars purple ?&lt;/p&gt;&#xA;" OwnerUserId="24925" LastActivityDate="2018-11-21T16:06:10.413" Title="Why don't we see purple stars" Tags="&lt;star&gt;&lt;spectra&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28474" PostTypeId="2" ParentId="28473" CreationDate="2018-11-21T15:04:06.700" Score="3" Body="&lt;p&gt;Short answer: most stars emit a very wide range of wavelengths, from MHz (radio) to gamma ray, &lt;span class=&quot;math-container&quot;&gt;$10^{19} Hz$&lt;/span&gt;  -- albeit rather little of the latter. But what we see is the combined wavelengths that our retina responds to, and due to the blackbody curve, by the time there's enough energy/temperature to produce a lot of blue or violet photons, the number of yellow/red photons swamps them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coldest stars appear red, as you mention, because the black body spectral peak is low enough that there isn't much at all in the green-to-violet range. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you take the light from the hottest stars and run it through a prism or grating spectrometer, you can see that there is blue/purple light included. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-11-21T15:04:06.700" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28475" PostTypeId="2" ParentId="28473" CreationDate="2018-11-21T16:06:10.413" Score="4" Body="&lt;p&gt;Purple and green star exists. For example, a blue supergiant star can have its surface temperature to 50000 K, which peaks at purple in the blackbody spectrum. However, our eyes are more sensitive to blue than purple, and the contrast between blue and purple in the spectrum is insignificant, therefore we see blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, check this article: &lt;a href=&quot;https://parade.com/295384/marilynvossavant/can-stars-be-green-or-purple/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://parade.com/295384/marilynvossavant/can-stars-be-green-or-purple/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-11-21T16:06:10.413" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28476" PostTypeId="2" ParentId="28449" CreationDate="2018-11-21T16:32:16.573" Score="1" Body="&lt;p&gt;First, a tripod won't help you point to a particular location in the sky.  For that you would need a correctly-aligned equatorial mount with setting circles, or a mount with digital encoders (you can get these for dobsonians) that when correctly configured can convert the telescopes position to a position on the celestial sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is &lt;em&gt;possible&lt;/em&gt; to convert a dobsonian to a different kind of mounting.  If your telescope doesn't already have a tripod bracket, you would probably need to buy &lt;a href=&quot;https://www.bhphotovideo.com/c/product/386033-USA/Vixen_Optics_2664_Tube_Rings_with_90mm.html/?ap=y&amp;amp;gclid=EAIaIQobChMI46ya467j3gIVAoezCh17rwwZEAQYASABEgL0B_D_BwE&amp;amp;lsft=BI%3A514&amp;amp;smp=Y&quot; rel=&quot;nofollow noreferrer&quot;&gt;tube rings such as these&lt;/a&gt;, and put one on either side of the circles on the sides of the tube that sit in the rocker box, so that the center of gravity of the scope was between the rings.  Then you'd probably need a plate connecting the rings and providing a mount point between them that you could attach to a tripod or a more complex mount (like an equatorial).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My suspicion is that rather than trying to adapt the dobsonian to a different mount, you'd be better served either by buying after-market digital encoders for your dobsonian, or just getting a whole new telescope with the kind of mounting you want.&lt;/p&gt;&#xA;" OwnerUserId="14434" LastActivityDate="2018-11-21T16:32:16.573" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28478" PostTypeId="2" ParentId="28459" CreationDate="2018-11-21T19:37:54.613" Score="2" Body="&lt;p&gt;There are &lt;em&gt;lots&lt;/em&gt; of galaxy catalogs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're only interested in nearby galaxies, there are things like the &lt;a href=&quot;https://heasarc.gsfc.nasa.gov/w3browse/all/neargalcat.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Updated Nearby Galaxy Catalog&lt;/a&gt;; downloadable versions can be found &lt;a href=&quot;http://cdsarc.u-strasbg.fr/viz-bin/cat/J/AJ/145/101&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A classic all-sky galaxy catalog is the Third Reference Catalogue of Bright Galaxies (RC3, published in the early 1990s), which you can query &lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR?-source=VII/155&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;; if you follow&#xA;the &quot;FTP&quot; links, you can find links to download the actual catalog.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The HyperLEDA catalog that Dr. Chuck mentioned started as the RC3, but has various&#xA;bits of data and re-classifications added more or less continuously since.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more recent catalog, based on the near-infrared observations of the 2 Micron All-Sky&#xA;Survey (2MASS) is the &lt;a href=&quot;http://wise2.ipac.caltech.edu/staff/jarrett/2mass/XSCz/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;2MASS Galaxy Redshift Catalog&lt;/a&gt;, with about one million galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even larger catalogs (though not covering the whole sky) can be found among the data&#xA;products of the Sloan Digital Sky Survey, e.g. &lt;a href=&quot;https://www.sdss.org/dr14/data_access/value-added-catalogs/?vac_id=large-scale-structure-galaxy-catalogs&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many more catalogs exist, though some do not have accurate redshift information (meaning their distance from us is unknown or uncertain), and many are only for parts of the sky, sometimes very small regions (e.g., catalogs from &lt;em&gt;Hubble Space Telescope&lt;/em&gt; observations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As for that video, I suspect very little is based on real data once you get beyond the Andromeda Galaxy...)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2018-11-22T09:43:18.120" LastActivityDate="2018-11-22T09:43:18.120" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="28479" PostTypeId="2" ParentId="28415" CreationDate="2018-11-21T19:39:10.920" Score="2" Body="&lt;p&gt;Many astronomers are using electronic eyepieces to overcome light pollution. You may want to see what these devices can do in your area'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use several of the Mallincam cameras in my observatory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.mallincam.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.mallincam.net/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://groups.io/g/MallinCam/topics&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://groups.io/g/MallinCam/topics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some photos of my observatory and assistants&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.flickr.com/photos/142615862@N04/sets/72157672437929982&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.flickr.com/photos/142615862@N04/sets/72157672437929982&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More home observatory sites&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://obs.nineplanets.org/obs/obslist.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://obs.nineplanets.org/obs/obslist.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24927" LastActivityDate="2018-11-21T19:39:10.920" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28480" PostTypeId="1" AcceptedAnswerId="28517" CreationDate="2018-11-21T20:43:14.417" Score="-2" ViewCount="107" Body="&lt;p&gt;How big would a ball of frozen water need to be to last one orbit? How would atmospheric drag effect the ball of water? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://space.stackexchange.com/questions/32188/what-is-the-darkest-orbit-around-earth&quot;&gt;https://space.stackexchange.com/questions/32188/what-is-the-darkest-orbit-around-earth&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-11-22T20:20:30.943" LastActivityDate="2018-11-23T21:44:48.890" Title="How long will a ball of ice stay in orbit around Earth?" Tags="&lt;orbit&gt;&lt;earth&gt;&lt;astrophysics&gt;&lt;water&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2018-11-24T07:07:31.557" ContentLicense="CC BY-SA 4.0" />
  <row Id="28481" PostTypeId="1" AcceptedAnswerId="28492" CreationDate="2018-11-21T22:26:45.283" Score="4" ViewCount="635" Body="&lt;p&gt;As far as I understand, metallicities of old stars are usually low and those of new stars are usually high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If elliptical galaxies form after merger of galaxies, then surely that means the elliptical galaxy passed through a stage where it was an irregular galaxy. Now, irregular galaxies are said to be sites of rapid, new star formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why then do we say that elliptical galaxies have low metallicity, if the star formation went on for a long time there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[New to astronomy, so a detailed explanation would help]&lt;/p&gt;&#xA;" OwnerUserId="24929" LastActivityDate="2018-11-22T14:41:39.427" Title="Why is the metallicity of elliptical galaxies low?" Tags="&lt;star&gt;&lt;galaxy&gt;&lt;metallicity&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28482" PostTypeId="1" AcceptedAnswerId="28485" CreationDate="2018-11-21T23:10:27.107" Score="1" ViewCount="195" Body="&lt;p&gt;If all living things like animals require oxygen to live, how can people think that there could be life on other planets such as mars for example ? Living things require oxygen and carbon dioxide, and these planets do not have plants and such to provide this. So why do scientists try to still look for life on these planets ? &lt;/p&gt;&#xA;" OwnerUserId="24930" LastActivityDate="2018-11-22T08:50:04.267" Title="Life on other planets" Tags="&lt;life&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28483" PostTypeId="1" CreationDate="2018-11-21T23:14:43.070" Score="6" ViewCount="118" Body="&lt;p&gt;I am not quite sure if it would, considering i'v looked through countless websites and found nothing and would like to know if it could clear its path and join the other planets? &lt;/p&gt;&#xA;" OwnerUserId="24931" LastActivityDate="2018-11-22T06:55:48.863" Title="Will Pluto ever clear its path?" Tags="&lt;pluto&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28484" PostTypeId="1" CreationDate="2018-11-22T00:20:42.673" Score="0" ViewCount="775" Body="&lt;p&gt;Whether the Moon (Earth's natural satellite) is a planet is an old debate in planetary science.&lt;br /&gt;&#xA;Though I do not agree that Moon is a planet, I found something difficult to clearly understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the &lt;a href=&quot;https://en.wikipedia.org/wiki/IAU_definition_of_planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;2006 IAU definition of planet&lt;/a&gt;, a planet must satisfy &lt;em&gt;Clearing the neighbourhood&lt;/em&gt; condition.&#xA;And some criteria for the condition are introduced in the Wikipedia article about &lt;a href=&quot;https://en.wikipedia.org/wiki/Clearing_the_neighbourhood&quot; rel=&quot;nofollow noreferrer&quot;&gt;Clearing the neighbourhood&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Stern–Levison's &lt;span class=&quot;math-container&quot;&gt;$\Lambda$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Lambda = \frac {m^2}{a^{\frac 3 2}} k$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Margot's &lt;span class=&quot;math-container&quot;&gt;$\Pi$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Pi=\frac {m}{m^{\frac 5 2}a^{\frac 9 8}} k $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above equations, &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; is the mass of the body, &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is the body's semi-major axis, and other factors are described in the Wikipedia article. Both criteria are said to be satisfied if the calculated value is greater than 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My calculations are simple. Let &lt;span class=&quot;math-container&quot;&gt;$m_{Moon} = 0.0123 \times m_{Earth}$&lt;/span&gt; and Moon's other factors (such as &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt;) have the same values as Earth's. Then I can get...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Lambda_{Moon} = 0.0123^2 \times \Lambda_{Earth} = 23.147$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Pi_{Moon} = 0.0123 \times \Pi_{Earth} = 9.963$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both &lt;span class=&quot;math-container&quot;&gt;$\Lambda_{Moon}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\Pi_{Moon}$&lt;/span&gt; thus satisfy the &lt;em&gt;Clearing the neighbourhood&lt;/em&gt; condition. Though I do not calculate the Soter's &lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt;, it will not change my results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is wrong with my calculations and results? Do I miss something about the condition and the criteria? Or is there something wrong with the &lt;em&gt;Clearing the neighbourhood&lt;/em&gt; condition?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; in response to comments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question can be related to the idea that the &lt;em&gt;Clearing the neighborhood&lt;/em&gt; condition is inadequate and ambiguous for the planet definition.&#xA;More deeply, this question can be related to the idea that 2006 IAU definition of planet is logically ambiguous and clumsy.&lt;br /&gt;&#xA;For example, the 2006 definition of planet defines the Moon as a satellite and not a planet.&#xA;If one celestial body can be defined as a planet or not without rules and conditions, why other conditions are needed? All we need is just a planet list, not other planet conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frankly speaking, the &lt;em&gt;Clearing the neighborhood&lt;/em&gt; condition is like comedy. &#xA;The condition says &quot;Planet must sweep out other bodies&quot; ...&#xA;Then says &quot;Except satellites&quot; ...&#xA;Then again says &quot;Except bodies having no significant size&quot; ...&#xA;Then again &quot;Except bodies having resonant orbit&quot; ...&#xA;Then again &quot;It may take billions of years&quot; ...&#xA;Then again &quot;Sweep-out process and time are influenced by other planets and bodies, actually&quot; .......&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is obvious that sometimes the &lt;em&gt;Clearing the neighborhood&lt;/em&gt; theory is useful. But it is inappropriate for the rule and definition requiring strictness.&lt;/p&gt;&#xA;" OwnerUserId="23440" LastEditorUserId="23440" LastEditDate="2018-11-25T07:04:11.400" LastActivityDate="2018-11-26T18:44:11.590" Title="Moon is a planet by &quot;Clearing the neighborhood&quot; condition?" Tags="&lt;planet&gt;&lt;the-moon&gt;" AnswerCount="4" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="28485" PostTypeId="2" ParentId="28482" CreationDate="2018-11-22T00:31:38.157" Score="4" Body="&lt;p&gt;&lt;strong&gt;TL;DR: Not all life requires atmospheric oxygen or carbon dioxide. There are many organisms here on Earth that use &lt;a href=&quot;https://en.wikipedia.org/wiki/Chemosynthesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;chemosynthesis&lt;/a&gt; to process chemicals like methane or hydrogen sulphide for energy.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The search for life on Mars has concentrated on finding evidence of &lt;a href=&quot;https://en.wikipedia.org/wiki/Autotroph&quot; rel=&quot;nofollow noreferrer&quot;&gt;autotrophic&lt;/a&gt; microorganisms – organisms that convert simple chemicals in the environment into complex organic compounds. Most autotrophs use water as the reducing agent, but some can use other chemicals like hydrogen sulphide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy source for these autotrophs can either be sunlight, in which case we call them &lt;a href=&quot;https://en.wikipedia.org/wiki/Phototroph&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;phototrophs&lt;/em&gt;&lt;/a&gt; (which includes algae and green plants here on Earth), or obtained through the oxidation of electron donors in their environment, in which case they're called &lt;a href=&quot;https://en.wikipedia.org/wiki/Chemotroph&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;chemotrophs&lt;/em&gt;&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a wide variety of chemotrophs here on Earth, typically living in completely dark and often hostile environments such as next to deep-sea volcanic vents. For example, iron-oxidizing bacteria even colonise new lava beds in the deep ocean, and in effect use the oxidation of available ferrous iron as their respiratory process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no reason to think that autotrophs couldn't find niche environments on Mars in which to exist, but the biggest hurdles are the significantly lower temperature and pressure, and the significantly greater radiation. The latter can be overcome by colonising &lt;em&gt;sub-surface&lt;/em&gt; environments, which eliminates the phototrophs from contention. While UV radiation penetrates soils only in the millimetre range, &lt;a href=&quot;http://discovery.ucl.ac.uk/134609/1/2006GL027494.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;one study&lt;/a&gt; found that viable dormant cells would have to be at least 7.5m below the planet's surface to survive the cumulative damage from cosmic radiation.&lt;/p&gt;&#xA;" OwnerUserId="14094" LastActivityDate="2018-11-22T00:31:38.157" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28486" PostTypeId="2" ParentId="28483" CreationDate="2018-11-22T01:18:07.597" Score="-2" Body="&lt;p&gt;Space is super empty, and Pluto has a very small mass. There's a extremely small chance that it could attract any objects in its &quot;neighborhood&quot;, so if it ever does happen, it would be a very, very long time from now. Perhaps, it would take so long that by the time that happens, Pluto would have been ejected from the Solar System after the Sun dies.&lt;/p&gt;&#xA;" OwnerUserId="24373" LastActivityDate="2018-11-22T01:18:07.597" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28487" PostTypeId="2" ParentId="28484" CreationDate="2018-11-22T02:14:32.827" Score="5" Body="&lt;p&gt;According to the &lt;a href=&quot;https://en.wikipedia.org/wiki/IAU_definition_of_planet&quot; rel=&quot;noreferrer&quot;&gt;2006 IAU definition of a planet&lt;/a&gt;, a celestial body must meet all 3 criteria to be considered a planet.  The Moon fails the first criteria &quot;1. Is in orbit around the Sun&quot;.  The Moon orbits Earth, not the Sun, and thus it doesn't have a legitimate &quot;neighbourhood&quot; (solar orbit) to clear in that context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also to note, according to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Clearing_the_neighbourhood#Disagreement&quot; rel=&quot;noreferrer&quot;&gt;Clearing the neighbourhood&lt;/a&gt; article, this criteria is contentious as it is vaguely written and Earth, Mars, Jupiter and Neptune have not &quot;cleared the neighbourhood&quot; of their respective solar orbits.&lt;/p&gt;&#xA;" OwnerUserId="24934" LastActivityDate="2018-11-22T02:14:32.827" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="28489" PostTypeId="2" ParentId="28483" CreationDate="2018-11-22T06:55:48.863" Score="5" Body="&lt;p&gt;In addition to the answer above, Pluto is a Neptune crossing &lt;a href=&quot;https://en.wikipedia.org/wiki/Trans-Neptunian_object&quot; rel=&quot;noreferrer&quot;&gt;trans neptunium object&lt;/a&gt; or TNO as it's often abreviated.  Being a TNO doesn't exclude the possibility of an object being a planet but crossing the orbit of Neptune does.   So, even if Pluto cleared out it's entire orbital neighborhood, it would still cross Neptune and naming it a planet would be problematic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's orbit changed to being entire outside of Neptune, it would still face the problem of being one of several minor planets in that general region of similar size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LffKR.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LffKR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really wanted to push the theoretical to the limit, if Pluto's orbit moved so it stopped crossing Neptune's and if it merged with all the other similarly sized TNOs, then you could make an argument that it stands out as unique for it's region and a strong argument could be made that it should under those conditions, be called a planet, but that's a stack of improbabilities so high that winning next weeks mega millions top prize seems likely by comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if all the large known TNOs were somehow combined, they would still have a mass less than Mercury, though perhaps a slightly larger diameter, but that would be large enough where it just might get planet status.   It won't happen though.  Space is too big and object's orbits are too far from each other.   Planet formation happened within the proto-planetary disk at distances much closer to the Sun.   Uranus and Neptune likely formed much closer to the Sun than they are now.   There's not enough material and too much space in between for a planet to coalesce at Pluto distance from our Sun.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-11-22T06:55:48.863" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28490" PostTypeId="2" ParentId="28481" CreationDate="2018-11-22T07:36:13.770" Score="1" Body="&lt;p&gt;Elliptical galaxies have no regions of high density, so they are not forming stars any more.  So all the stars in an elliptical galaxy are old stars, and as you say, tend to have low metallicity (actually, as pointed out, that doesn't seem to be true-- they have higher metallicity, presumably from old star formation no longer occurring).  You are right that they might have had active star formation in the distant past, but it ended long ago.  Spiral galaxies like the Milky Way have been forming new stars for their entire history, so have more opportunity for current metallicity increase, but seem to be starting out behind, as pointed out above.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2018-11-22T11:09:39.500" LastActivityDate="2018-11-22T11:09:39.500" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28491" PostTypeId="2" ParentId="28482" CreationDate="2018-11-22T08:50:04.267" Score="1" Body="&lt;p&gt;What's more, in addition to finding potential evidence for life now, there is also the equally important possibility of past life that has gone extinct but left biosignatures like fossils or other geological byproducts.  It is important to realize that there is considerable evidence that the Martian climate used to be very different, and there may have been a much denser atmosphere, providing the carbon dioxide that plant life would need to live on the surface and use photosynthesis.  The fact that there is not a lot of oxygen in the atmosphere now suggests that there was not a multibillion year history of plant life like the Earth has (a lot of it without much oxygen), but there could have been a short period where life did thrive, and it might be hard to tell now that this ever happened-- without looking closely.  But yes, it does seem likely that there are places in the galaxy more conducive to life than Mars, Mars is just the closest-- and you always start looking for your lost keys under the streetlights, going on only after you've eliminated that possibility.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-11-22T08:50:04.267" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28492" PostTypeId="2" ParentId="28481" CreationDate="2018-11-22T08:50:37.353" Score="7" Body="&lt;p&gt;It isn't true that elliptical (or early-type) galaxies are metal-poor. This sounds like a pop-sci myth being propagated. The first line of the abstract of &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0610831&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pipino &amp;amp; Matteucci (2006)&lt;/a&gt; states that &quot;Elliptical galaxies probably host the most metal rich stellar populations in the Universe&quot;.  Thus your suspicion of the idea that &quot;ellipticals are metal-poor&quot; is well-founded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measurements of galaxy metallicities and metallicity gradients have been around for decades and show that ellipticals obey fairly simple scaling relationships between mass, luminosity, velocity dispersion and metallicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of metallicity, massive ellipticals with bigger velocity dispersions are a bit more metal-rich than the Sun, whilst smaller galaxies may get down half the solar metallicity. The range is &lt;em&gt;much&lt;/em&gt; larger for spiral galaxies. In addition, there is a tendency for ellipticals that are &lt;em&gt;older&lt;/em&gt; to have higher metallicities. Examples of these trends can be seen in &lt;a href=&quot;https://arxiv.org/abs/1005.0014&quot; rel=&quot;nofollow noreferrer&quot;&gt;Graves &amp;amp; Faber (2010)&lt;/a&gt; and &lt;a href=&quot;https://arxiv.org/abs/1802.01819&quot; rel=&quot;nofollow noreferrer&quot;&gt;Li et al. (2018)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The central plot here (taken from Graves &amp;amp; Faber 2010) shows how the metallicity of galaxies varies with their stellar velocity dispersion (an indicator of their mass). [Z/H] is a logarithmic base 10 scale where 0.0 means the metallicity of the Sun and -1 would a tenth the metallicity of the Sun. You can see that the elliptical galaxies (red) have a relatively narrow range of metallicities, that most are as metal-rich as the Sun, that less massive ellipticals have lower metallicity and that spirals (blue) have a much bigger range of metallicity.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RWXX6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RWXX6.png&quot; alt=&quot;The fundamental plane from Graves &amp;amp; Faber (2010)&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reasons behind these trends are still debated. But as you rightly point out, ellipticals likely form from mergers of objects that have already been actively forming stars. A low metallicity is therefore not necessarily expected. Much depends on how much enriched gas from the ejecta of previous (higher mass) stellar populations is &lt;em&gt;retained&lt;/em&gt; within the galaxy. Higher mass galaxies with deeper gravitational potentials are able to retain more of that enriched gas.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-11-22T14:41:39.427" LastActivityDate="2018-11-22T14:41:39.427" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28493" PostTypeId="2" ParentId="28456" CreationDate="2018-11-22T09:14:06.377" Score="3" Body="&lt;p&gt;Actually, this is a perfectly good question if interpreted in a reasonable way, i.e., interpreted as asking how long will light take to get to us if emitted at comoving age 13.8 billion years from a galaxy whose comoving distance is currently increasing from us at rate c (and the Hubble law with H=70 tells us that comoving distance is 4.3 Gpc or 14 billion LY).  That is not an infinite time, as that is not the definition of the Hubble sphere.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to how long it would take, less than infinity, depends on the cosmology model, and is a little hard to calculate because cosmology calculators (such as &lt;a href=&quot;http://www.astro.ucla.edu/~wright/DlttCalc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.astro.ucla.edu/~wright/DlttCalc.html&lt;/a&gt;) tend to focus on light reaching us now-- as that's what we see.  For example, that calculator gives that a galaxy that is receding from us at rate c now would have had to emit light 9.25 billion years ago for us to be seeing it now, but that's not what was asked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned, it is not easy to calculate the time it would take light emitted at the Hubble sphere now to get to us, but it is not infinite.  The Hubble sphere would only be the edge of what we could ever see if the Hubble constant were constant with time (as in an accelerating universe completely dominated by a cosmological constant), but even though the expansion does seem to be accelerating, it is not yet completely dominated by a cosmological constant (assuming that is what is causing the acceleration).  Hence, we would indeed eventually see the light emitted at the Hubble sphere now, but it might take a very long time, perhaps 50-100 billion years but that's just a guess without doing a real calculation with a full cosmological model.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-11-22T09:14:06.377" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28494" PostTypeId="2" ParentId="28454" CreationDate="2018-11-22T10:00:44.510" Score="1" Body="&lt;p&gt;The Moon does not follow the ecliptic precisely like the Sun does; rather, it follows a path inclined of about 5.14&amp;deg;. The two points where the ecliptic and the Moon's path coincide are called &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_node&quot; rel=&quot;nofollow noreferrer&quot;&gt;'nodes'&lt;/a&gt;, and they travel around the ecliptic in a period of &lt;strong&gt;18.66 years&lt;/strong&gt;. This is a combination of the orbital period of the Moon, also called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_month#Sidereal_month&quot; rel=&quot;nofollow noreferrer&quot;&gt;sidereal month&lt;/a&gt;, about 27.32 days, and the slightly shorter &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_month#Draconic_month&quot; rel=&quot;nofollow noreferrer&quot;&gt;draconic month&lt;/a&gt;, the average period between the moment where the Moon traverses a node, which is 27.21 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once every 18.66 years, the ascending node coincides with the &lt;a href=&quot;https://en.wikipedia.org/wiki/First_Point_of_Aries&quot; rel=&quot;nofollow noreferrer&quot;&gt;vernal equinox&lt;/a&gt;, and the Moon will reach declinations of about +28&amp;deg; and -28&amp;deg; (much more than the Sun, which reaches ±23&amp;deg;); this is called a 'major lunar standstill'. 9.33 years later, the opposite is true, and the Moon will only reach ±18&amp;deg; during a 'minor lunar standstill'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_standstill&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; these moments have had historical significance, all the way back to the Bronze Age:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This time appears to have had special significance for the Bronze Age societies, who built the megalithic monuments in Britain and Ireland. It also has significance for some neopagan religions. Evidence also exists that alignments to the moonrise or moonset on the days of lunar standstills can be found in ancient sites of other ancient cultures, such as at Chimney Rock in Colorado and Hopewell Sites in Ohio. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2018-11-22T10:00:44.510" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28495" PostTypeId="2" ParentId="28439" CreationDate="2018-11-22T10:56:30.730" Score="0" Body="&lt;p&gt;Thanks for the good explanation. Tomorrow there is Sun over here and i am gonna test if it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay so this is the mean orbital elements won from measuring over a long period. Still i can take this Elements and make a Modell. I think getting the &quot;counter-Orbital-Elements&quot; for the Sun should at least be a good first Approximation into this complex Problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_elements&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Orbital_elements&lt;/a&gt;: &quot;...When viewed from a non-inertial frame centred on one of the bodies, only the trajectory of the opposite body is apparent; &lt;strong&gt;Keplerian elements describe these non-inertial trajectories&lt;/strong&gt;. &quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means the Orbital Elements describe an Object orbiting relative to a second, fixed Object. So it should be correct to get the &quot;absolute&quot; Orbit Elements like i did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways Jupiter seems to raise the accuracy of the Sun position in my algorithm, now i would like to build Saturn, Neptun... Do u have any idea how to start with it? I dont think i can add the Sun-Saturn Orbit geometrically to the first one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will see, i will report Tomorrow.&lt;/p&gt;&#xA;" OwnerUserId="24544" LastEditorUserId="24544" LastEditDate="2018-11-22T13:12:24.370" LastActivityDate="2018-11-22T13:12:24.370" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28496" PostTypeId="1" AcceptedAnswerId="28498" CreationDate="2018-11-22T10:58:05.210" Score="4" ViewCount="443" Body="&lt;p&gt;It seems very logical, to call the star, closest to earth, &quot;Proxima&quot; (&quot;Proximus&quot; is Latin for &quot;close&quot; and the word for star, &quot;stella&quot;, is feminin, hence &quot;Proxim&lt;strong&gt;a&lt;/strong&gt;&quot;), but there's one thing I don't understand: in order to call a star ±&quot;The closest one&quot;, you must be sure that in future, no closer star will be found.&lt;br&gt;&#xA;This is not so simple: lots of very close stars, like Barnard's star, Kapteyn's, ..., are very weak and even not visible to the naked eye.&lt;br&gt;&#xA;So, how can we be that sure that Proxima is indeed the nearest one? Have astronomers indeed examined the complete sky in order to prove this?&lt;/p&gt;&#xA;" OwnerUserId="14381" LastActivityDate="2018-11-22T20:24:38.360" Title="Why is Proxima Centauri called Proxima Centauri?" Tags="&lt;star&gt;&lt;distances&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28498" PostTypeId="2" ParentId="28496" CreationDate="2018-11-22T12:49:35.390" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So, how can we be that sure that Proxima is indeed the nearest one? Have astronomers indeed examined the complete sky in order to prove this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The minimum absolute magnitude for a star is, according to &lt;a href=&quot;https://en.wikipedia.org/wiki/Hertzsprung%E2%80%93Russell_diagram&quot; rel=&quot;noreferrer&quot;&gt;this well-known diagram&lt;/a&gt; about +16. &lt;a href=&quot;https://en.wikipedia.org/wiki/Proxima_Centauri&quot; rel=&quot;noreferrer&quot;&gt;Proxima Centuari&lt;/a&gt; is at the very low end of this scale with an absolute magnitude of +15.60. Still, it has an apparent magnitude of +11.13. That means that if a star is closer to us, it needs to be +11.5 or brighter. Every object of that magnitude or brighter has been categorized, and a hypothetical close star would've stood out due to its large &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_parallax&quot; rel=&quot;noreferrer&quot;&gt;parallax&lt;/a&gt;. So it's safe to say there are no closer stars ... at this moment.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;you must be sure that in future, no closer star will be found.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stars move (they rotate around the galaxy center), so in the (far) future, there might be a star closer to use than Proxima Centauri, just like sometimes Venus is the planet closest to Earth, and sometimes Mars. So it's possible that Proxima might not be the nearest star in the far future. This is not something we have to worry in our lifetimes, though :)&lt;/p&gt;&#xA;" OwnerUserId="7472" LastEditorUserId="7472" LastEditDate="2018-11-22T13:03:51.530" LastActivityDate="2018-11-22T13:03:51.530" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28499" PostTypeId="1" CreationDate="2018-11-22T20:24:30.923" Score="1" ViewCount="103" Body="&lt;p&gt;Does anybody know of a good tutorial on how to use python and astropy to generate plots of GAIA, RAVE or any other database of barycentric (heliocentric) radial velocity data of stars in our galaxy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to generate various plots of the heliocentric velocities of stars in our galaxy, making different restrictions on distances from the sun, and I'd like to see what the data looks like with correction for the proper motion of the sun and without. Now, I know I could find plots done by others, but I really want to learn how to generate my own plots. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty new to python, so please target answers accordingly...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(added later:) There's probably a simple astropy function that if you call it with the right variables, say, a certain slice of the sky in RA and Dec, and a certain distance range in kpc, it will spit back a list of all the stars in that region along with their radial velocities.  I just need to learn the name of that function and how to call it.&lt;/p&gt;&#xA;" OwnerUserId="24950" LastEditorUserId="24950" LastEditDate="2018-11-23T13:42:36.400" LastActivityDate="2018-11-23T13:42:36.400" Title="Plotting GAIA barycentric radial velocity data with python" Tags="&lt;data-analysis&gt;&lt;python&gt;&lt;gaia&gt;&lt;astropy&gt;&lt;radial-velocity&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28500" PostTypeId="2" ParentId="28496" CreationDate="2018-11-22T20:24:38.360" Score="5" Body="&lt;p&gt;A &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1928AJ.....39...20A&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;noreferrer&quot;&gt;1928 paper mentions this very point:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Iɴɴᴇs has&#xA;  suggested the name &lt;em&gt;Proxima Centauri&lt;/em&gt; for this star.&#xA;  Since it is the nearest of the three components of the&#xA;  system, the name is most appropriate and may well&#xA;  be retained even though some star nearer the Sun&#xA;  be subsequently discovered in the same constellation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are a couple of points to note here.  &quot;Proxima Centauri&quot; can be taken to mean &quot;close one of the Centaur&quot;.  So a more close star in a &lt;em&gt;different&lt;/em&gt; constellation would not invalidate the name. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Innes, the discover of the star, was hopeful that his discovery was important. He attempted to make a distance measurement by parallax (seeing how far the star appears to move as the Earth orbits the sun.) The value for the distance he published was slightly smaller than the accepted distance of Alpha Centauri, but the error bars in he measurement meant that in truth he did not know if it was slightly nearer, or slightly further from the Sun. Nevertheless he suggested:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If this small star had a name it would be convenient—it is therefore suggested that it should be referred to as &lt;em&gt;Proxima Centaurus&lt;/em&gt;. (&lt;a href=&quot;http://adsabs.harvard.edu/full/2007AfrSk..11...39G&quot; rel=&quot;noreferrer&quot;&gt;quoted&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is basically grandstanding.  Innes hoped that his star might turn out to be the very closest star and suggested that name. In 1928, it was established to be the closest star in the Alpha centauri system, the name was set and seen to be appropriate, even if some other star was later discovered to be closer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the future other stars will come closer. A star called Ross 248 will briefly be the closest star in about 30000 years time. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-11-22T20:24:38.360" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28501" PostTypeId="1" AcceptedAnswerId="28504" CreationDate="2018-11-22T20:53:22.597" Score="0" ViewCount="109" Body="&lt;p&gt;Can a liquid be used to slow a satellite to loose orbit by adding or taking momentum from a satellite, a type of devise that releases a liquid that expands and solidified adding surface area making more drag and or by adding weight?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-11-25T21:14:04.647" LastActivityDate="2018-11-25T21:14:04.647" Title="Can a liquid be used to de orbit debris?" Tags="&lt;orbital-mechanics&gt;&lt;space-debris&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28502" PostTypeId="1" AcceptedAnswerId="28699" CreationDate="2018-11-22T21:14:41.973" Score="3" ViewCount="376" Body="&lt;p&gt;Would a smaller satellite be more stable around a smaller planet where the same satellite would not be stable or last around a larger planet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LffKR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LffKR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;I added the picture to show how moon size planets can have even smaller moons.&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-12-12T04:43:14.760" LastActivityDate="2018-12-12T04:43:14.760" Title="What is the minimum size to be a natural satellite?" Tags="&lt;orbit&gt;&lt;astrophysics&gt;&lt;natural-satellites&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28503" PostTypeId="1" AcceptedAnswerId="28541" CreationDate="2018-11-22T21:43:36.547" Score="6" ViewCount="399" Body="&lt;p&gt;If Europa has an oxygen atmosphere with a water mantle, is it possible that there could be life under the crust of Europa ice on the ocean where it is warmer? What could be the temperature under the ice? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x5h6C.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x5h6C.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Image of all the water of Earth beside it in a sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bScsI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bScsI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-12-09T16:45:51.327" LastActivityDate="2018-12-11T01:26:38.403" Title="What is the underwater temperature of Europa?" Tags="&lt;light&gt;&lt;planetary-atmosphere&gt;&lt;life&gt;&lt;core&gt;&lt;europa&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28504" PostTypeId="2" ParentId="28501" CreationDate="2018-11-22T22:15:07.700" Score="3" Body="&lt;p&gt;Temperature is not the only factor for matter phases, pressure does make a difference too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As space is practically a vacuum, thus pressure is zero. As far as I know, there can be no liquids in the vacuum of space. No element nor compound is liquid in zero pressure no matter what the temperature might be. The best candidate for this would be Helium, but even in this case the vacuum of space cannot keep equilibrium between liquid and gas phases, and thus all liquid helium will evaporate into gas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Liquid could exist in space, if it would be able to collapse on its own gravity and maintain gaseous atmosphere bound to the liquid body. This would require way more matter than is feasible for your purpose, however. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Also adding mass to an orbiting object does not in general cause its orbit to decay, as mass does not significant role in orbit equation. Gravitational force IS proportional to product of masses, but then again acceleration is force divided by mass - ie it gets cancelled out; applies only if adding mass does not significantly change the barycenter of said two objects (ignoring relativistic effects and gravitational waves since obviously far outside the scope of this question)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding surface area does not affect orbit in vacuum since there is no drag nor friction. Also shape of the object does not matter for the very same reason. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 2: As noted in comments, liquids may remain liquids in vacuum long enough for this scenario. I couldn’t find very good references how this might look like in the vacuum of space, but seemingly water can exist as a liquid for extended periods of time in very low pressures, see eg &lt;a href=&quot;https://www.hou.usra.edu/meetings/lpsc2014/pdf/2060.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.hou.usra.edu/meetings/lpsc2014/pdf/2060.pdf&lt;/a&gt; which states that 250ml of liquid water did fully evaporate in about 2.5 hours in 0.003 arm (Personally I find this very surprising and contradicting to my prior knowledge but I have been wrong many times before)&lt;/p&gt;&#xA;" OwnerUserId="10813" LastEditorUserId="10813" LastEditDate="2018-11-25T15:48:15.280" LastActivityDate="2018-11-25T15:48:15.280" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28505" PostTypeId="2" ParentId="28502" CreationDate="2018-11-22T22:21:39.970" Score="2" Body="&lt;p&gt;As long as the satellite is way smaller than the object it is orbiting, the mass does not make any difference to its orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because even while gravitational force depends on product of masses of both objects, the acceleration of orbiting object is this force divided by its mass. Therefore any change to its mass will keep its acceleration and thus velocity and thus orbit about the same as before (as long as there are no significant changes to barycenter the two objects are orbiting; gravitational waves do change things, but only for very extreme cases such as close black hole and/or neutron star binaries)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will get more complicated if there are more than one satellite orbiting the body, and stability in these cases is often hard to determine (aspects like orbital resonances and phase do matter in these cases)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: added explanation why mass does not affect orbit&lt;/p&gt;&#xA;" OwnerUserId="10813" LastEditorUserId="10813" LastEditDate="2018-11-22T22:42:10.893" LastActivityDate="2018-11-22T22:42:10.893" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28506" PostTypeId="1" CreationDate="2018-11-23T00:26:39.790" Score="2" ViewCount="123" Body="&lt;p&gt;Can someone please tell me what the mass (or size) of the star at the center of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Helix_Nebula&quot; rel=&quot;nofollow noreferrer&quot;&gt;Helix Nebula&lt;/a&gt; is, and how much larger it is than our own star (we call the Sun or Sol).  Also, what is the name of this star?  I cannot find a website that addresses these questions.  I believe I have heard, however, that the star is roughly 100 times larger than our sun.&lt;/p&gt;&#xA;" OwnerUserId="24951" LastEditorUserId="14094" LastEditDate="2018-11-23T08:35:23.900" LastActivityDate="2018-11-23T20:53:34.407" Title="Information on the star at the centre of the Helix &quot;Eye Of God&quot; Nebula" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28507" PostTypeId="2" ParentId="28484" CreationDate="2018-11-23T01:02:27.663" Score="3" Body="&lt;p&gt;What you have shown is that the Moon is large enough so that it would be a planet by the IAU definition if the Earth were not present. But, because the Moon does orbit the Earth, it is not a planet by the IAU definition. If the &quot;clearing the neighborhood&quot; criterion were sufficient by itself to define a planet, then the other criteria would not be part of the definition.&lt;/p&gt;&#xA;" OwnerUserId="7752" LastActivityDate="2018-11-23T01:02:27.663" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28508" PostTypeId="1" CreationDate="2018-11-23T04:32:53.707" Score="3" ViewCount="61" Body="&lt;p&gt;In the beginning of the short YouTube video &lt;a href=&quot;https://www.youtube.com/watch?v=JjDJLNubKKw&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arecibo Uncut: Under the Dish with Phil Perillat&lt;/a&gt; it is mentioned that while dish antennas are usually a parabola, the problem is that a static parabola would only be useful to see objects for a few minutes as they passed overhead.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;But then some guy at DARPA says &amp;quot;Hey why don't you guys make it a sphere?&amp;quot; A sphere has the unique property that from the center of curvature, no matter where you look, it's the same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;That is indeed how the Arecibo dish works; a static primary sphere and a big, funny-looking secondary mirror and associated optics/electronics suspended above on cables that can be moved around in order to &amp;quot;point&amp;quot; in different directions.&lt;/p&gt;&#xA;&lt;p&gt;Two questions:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;While the statement in the block quote about the sphere is correct as far as it goes, was the shape of the correcting optics (secondary, etc) above also independent of where you look? Or ideally would you like a different secondary shape depending on how far off-vertical you look? If so, is this what happens?&lt;/li&gt;&#xA;&lt;li&gt;Any ideas who &amp;quot;the guy at DARPA&amp;quot; might be?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2021-02-09T01:32:32.423" LastActivityDate="2021-02-09T01:32:32.423" Title="Did Arecibo's secondary optics compensate aberrations when viewing farther away from vertical?" Tags="&lt;radio-astronomy&gt;&lt;optics&gt;&lt;radio-telescope&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28510" PostTypeId="1" CreationDate="2018-11-23T09:56:15.170" Score="2" ViewCount="108" Body="&lt;p&gt;The Science News article &lt;a href=&quot;https://www.sciencenews.org/article/mars-reconnaissance-orbiter-glitch-liquid-water&quot; rel=&quot;nofollow noreferrer&quot;&gt;An orbiter glitch may mean some signs of liquid water on Mars aren’t real&lt;/a&gt; begins:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Some signs of water on Mars may have just dried up.&lt;/p&gt;&#xA;&lt;p&gt;Thanks to the way data from NASA’s Mars Reconnaissance Orbiter are handled, &lt;strong&gt;the spacecraft may be seeing &lt;a href=&quot;https://agupubs.onlinelibrary.wiley.com/doi/abs/10.1029/2018GL080077&quot; rel=&quot;nofollow noreferrer&quot;&gt;signs of hydrated salts that aren’t really there&lt;/a&gt;&lt;/strong&gt;, planetary scientists report online November 9 in Geophysical Research Letters.&lt;/p&gt;&#xA;&lt;p&gt;That lack of salts could mean that certain sites proposed as places where life could exist on Mars today, including purported streaks of liquid water on the walls of Martian craters, are probably dry and lifeless.&lt;/p&gt;&#xA;&lt;p&gt;“People think these environments might be inhabitable by microbes,” says planetary scientist Ellen Leask of Caltech. But “there might not actually be any real evidence for it,” at least not from orbit.&lt;/p&gt;&#xA;&lt;p&gt;Leask and her colleagues found the problem while searching for hydrated salts called perchlorates in maps of Mars taken by the orbiter’s &lt;a href=&quot;http://crism.jhuapl.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Compact Reconnaissance Imaging Spectrometer for Mars&lt;/a&gt;, or CRISM. Perchlorates can lower the freezing point of water by up to 80 degrees Celsius, which could be enough to melt ice in the frigid Martian climate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The report is &lt;em&gt;Open Access&lt;/em&gt; in Geophysical Research Letters: &lt;em&gt;Challenges in the Search for Perchlorate and Other Hydrated Minerals with 2.1‐μm Absorptions on Mars&lt;/em&gt;&#xA;Leask, E. K., Ehlmann, B. L., Dundar, M. Murchie, S. L. Seelos, F. P., &lt;a href=&quot;https://agupubs.onlinelibrary.wiley.com/doi/abs/10.1029/2018GL080077&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://agupubs.onlinelibrary.wiley.com/doi/abs/10.1029/2018GL080077&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The data in question comes from Mars Reconnaissance Orbiter's CRISM (&lt;a href=&quot;https://mars.nasa.gov/mro/mission/instruments/crismcompactreconnaissanceimagingspectrometerformars/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Compact Reconnaissance Imaging Spectrometer for Mars&lt;/a&gt;), a Vis-NIR push broom hyperspectral imager, where a 1D line image is cross-dispersed in wavelength and recorded on a 2D array as the spacecraft moves over Mars' surface.&lt;/p&gt;&#xA;&lt;p&gt;If I understand correctly, during later image analysis and , some of the corrections to the data meant to compensate for &amp;quot;spikes&amp;quot; &lt;em&gt;in some observables&lt;/em&gt; (&lt;strong&gt;despiking&lt;/strong&gt;) occurring during large changes in brightness resulted in spectral features that suggested presence of hydrated minerals that didn't exist.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; But I find the description of the step in the analysis responsible for the error a little bit complicated. Is it something simple like division by very small values leading to the spike in a derived quantity, followed by heavy-handed interpolation to smooth it over? What exactly does it mean when they talk about a few pixels, are there some physical locations on the detector that behave differently, or a spectral hot spot from artifacts in the grating? (e.g. Woods anlomalies)? “But for 0.05 percent of the pixels, it smooths out in a way that looks like perchlorate.”&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Potentially helpful resource:&lt;/strong&gt; &lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/S0019103516301579&quot; rel=&quot;nofollow noreferrer&quot;&gt;Regularization of Mars Reconnaissance Orbiter CRISM along‐track oversampled hyperspectral imaging observations of Mars&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Any help translating the &amp;quot;processingese&amp;quot; language to something easier to understand for a non-astronomer would be a great help!&lt;/p&gt;&#xA;&lt;p&gt;More from the &lt;a href=&quot;https://www.sciencenews.org/article/mars-reconnaissance-orbiter-glitch-liquid-water&quot; rel=&quot;nofollow noreferrer&quot;&gt;Science News article&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;CRISM’s camera doesn’t work perfectly, though. It can be thrown off by boundaries between light and dark, like a shadowed region at the edge of a cliff. &lt;strong&gt;Some pixels in the orbiter’s camera take a fraction of a millisecond to realize the surface color changed, so they record an extra spot of light or dark where it shouldn’t be. Planetary scientists have software to correct for these “spikes” in the spectra and make the data more reliable and easier to read.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;But the correction sometimes introduces dips in the spectra at the same wavelengths as perchlorates, Leask and Ehlmann and their colleagues found. “We cleverly created a way to get rid of the spiky noise,” Ehlmann says. “But for 0.05 percent of the pixels, it smooths out in a way that looks like perchlorate.”&lt;/p&gt;&#xA;&lt;p&gt;The researchers found the glitch while looking for small signs of the salts in CRISM images. Assuming that the orbiter would have already spotted large deposits of perchlorates if they existed, the team wrote an algorithm to find smaller traces that covered fewer than 10 pixels in a CRISM image. And the scientists started seeing perchlorates everywhere, including Jezero crater, which NASA announced was selected as the landing site for the Mars 2020 rover on November 19 (SN Online: 11/19/18).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/F0ter.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F0ter.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;© 2018 American Geophysical Union. All rights reserved.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Figure 1.Example of the (a) spatial and (b) spectral appearance of the artifact creating the spurious absorptions, shown in ratioed I/F and ratioed radiance. The threshold for an artifact is defined to be where ratioed radiance differs from ratioed I/F by more than 1% (c).  Laboratory spectra of minerals with a 2.1-μm absorption are shown (d) for comparison (from USGS spectral library (Clark et al. (2007)) and Hanley et al. (2015)).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2020-11-24T01:01:16.340" LastActivityDate="2020-11-25T05:24:41.717" Title="Help understanding the false MRO observations of hydrated minerals on Mars" Tags="&lt;mars&gt;&lt;spectroscopy&gt;&lt;artificial-satellite&gt;&lt;water&gt;&lt;spectrometry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28512" PostTypeId="2" ParentId="28506" CreationDate="2018-11-23T10:47:39.587" Score="5" Body="&lt;p&gt;The central star is a hot DAO white dwarf. As far as I can tell, the numbers that are the most reliable are those from &lt;a href=&quot;http://cdsads.u-strasbg.fr/abs/1999A%26A...350..101N&quot; rel=&quot;nofollow noreferrer&quot;&gt;Napiwotzki et al. (1999)&lt;/a&gt;. They use a direct spectral fitting technique to estimate the surface gravity and effective temperature of the central white dwarf star. These are &lt;span class=&quot;math-container&quot;&gt;$g = 10^{(7.00 \pm 0.22)}$&lt;/span&gt; cm/s&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$T_{\rm eff} = 103600 \pm 5500$&lt;/span&gt; K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They use these numbers and theoretical white dwarf evolutionary models (there is little alternative for a star that isn't part of a binary system) to estimate a mass of &lt;span class=&quot;math-container&quot;&gt;$0.57 \pm 0.02 M_{\odot}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a distance now known to be around  &lt;span class=&quot;math-container&quot;&gt;$219 \pm 24$&lt;/span&gt; pc (&lt;a href=&quot;https://arxiv.org/abs/astro-ph/0611543&quot; rel=&quot;nofollow noreferrer&quot;&gt;Harris et al. 2007&lt;/a&gt;), this makes the &lt;strong&gt;luminosity&lt;/strong&gt; of the white dwarf around &lt;strong&gt;100 times that of the Sun&lt;/strong&gt;. A bit of calculation shows that for this gravity and mass, then although the star is small (about &lt;span class=&quot;math-container&quot;&gt;$0.04 R_{\odot}$&lt;/span&gt;), it is much bigger than most white dwarfs. That is because it is recently formed (planetary nebulae are recently formed objects) and hot; the white dwarf is still contracting towards its final equilibrium size of around &lt;span class=&quot;math-container&quot;&gt;$0.015 R_{\odot}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CDS SIMBAD entry for the star is &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=WD+2226-210&amp;amp;NbIdent=1&amp;amp;Radius=2&amp;amp;Radius.unit=arcmin&amp;amp;submit=submit+id&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=WD+2226-210&amp;amp;NbIdent=1&amp;amp;Radius=2&amp;amp;Radius.unit=arcmin&amp;amp;submit=submit+id&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is listed in several catalogues, so goes by various names e.g. 2MASS J22293854-2050136, WD 2226-210.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most recent (and probably most accurate) distance measurement is from the Gaia DR2 parallax survey, giving &lt;span class=&quot;math-container&quot;&gt;$201 \pm 4$&lt;/span&gt; pc (&lt;span class=&quot;math-container&quot;&gt;$655 \pm 13$&lt;/span&gt; light years).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-11-23T20:53:34.407" LastActivityDate="2018-11-23T20:53:34.407" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="28513" PostTypeId="1" AcceptedAnswerId="29702" CreationDate="2018-11-23T14:18:22.627" Score="10" ViewCount="181" Body="&lt;p&gt;Shouldn't captured moons have the same size distribution as asteroids? And asteroids are more common the smaller size they are. Moons are likely captured if they are in highly inclined orbits, and those moons should've originated as asteroids or Kuiper Belt objects. But there are only &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_natural_satellites#List&quot; rel=&quot;noreferrer&quot;&gt;two moons&lt;/a&gt;, out of 194 known, smaller than 500 meters in radius. Aegaeon and S/2009 S1 both of Saturn. Although it is expected to exist over a million asteroids smaller than that. And all of Pluto's moons were large enough to be found before New Horizons arrived (sorry Alan Stern, no moon for you!)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Is this purely an observational bias? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is it expected that the planets are surrounded by a myriad of moons too small to have been detected yet, from hundreds of meters down to dust grains?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do they all somehow gather into planar rings below a certain size? (I think the main line is that a ring system forms as a result of a single collision or tidal crunching event).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Or what mechanism makes them lack small to tiny moons and dust?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/udjxz.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/udjxz.png&quot; alt=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/6/63/Asteroids_by_size_and_number.svg/1024px-Asteroids_by_size_and_number.svg.png&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;How would this asteroid size frequency plot compare to a moon size frequency plot?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2019-02-22T13:21:44.050" Title="Why haven't more captured small moons been found?" Tags="&lt;asteroids&gt;&lt;natural-satellites&gt;&lt;size&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28514" PostTypeId="1" AcceptedAnswerId="28515" CreationDate="2018-11-23T17:58:16.810" Score="-3" ViewCount="2223" Body="&lt;p&gt;I saw a shooting star as I was standing on the roof of my home. My home is double storey. And the height of that shooting star that I saw was maybe 10 to 13 meters. It was burned down at that height. My question is why haven't the shooting stars burnt completely when they enter the earth's atmosphere where the air is thick? Why sometimes a small piece gets to come very close to the ground and then burns?&lt;/p&gt;&#xA;" OwnerUserId="24464" LastEditorUserId="7472" LastEditDate="2018-11-25T21:29:08.560" LastActivityDate="2019-10-13T05:17:34.607" Title="Why do shooting stars burn close to the ground sometimes?" Tags="&lt;meteor&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="28515" PostTypeId="2" ParentId="28514" CreationDate="2018-11-23T20:05:22.023" Score="5" Body="&lt;p&gt;What you described is simply impossible. Meteors glow because its surface is being vaporized. When they enter the upper layers of the Earth's atmosphere they are typically travelling several kilometers per second (that's thousands of mph). They are going so fast that the air in front of it doesn't have time to get out of the way of the hypersonic meteor, so the air gets compressed. Compressed air naturally heats up, and because the air can't get out of the way, it stays very hot. This heat will vaporize the leading surface of the meteor (the &quot;front&quot;, so to speak), making it glow very brightly. It also leaves a trail of plasma, which is the streak you see behind it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the meteor travels lower and lower into the atmosphere, it will slow down. Once it gets to about 15 km (9 miles up), the meteor will have slowed down to below 2 km/s (4000 mph). This speed is too slow to vaporize the surface. Thus, the meteor stops glowing red hot, and is essentially invisible from then on. Eventually it will slow down to about 150 m/s (300 mph). That's only a bit faster than terminal velocity of the average human, and slower than a jet plane. Eventually it will hit the ground at roughly the same as ambient temperature (this vaporization process is very heat efficient).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You claimed that you saw the meteor 10-13 m above your head, but this would not be possible. In order to be glowing, it would need to be travelling at least 2 km/s (4000 mph), which is not possible for a meteor as it would have already slowed down in its flight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you may have been mistaken. Distance is generally hard to measure. In fact, the ancients did not know if the Sun was close by and fairly small, or far away and fairly large. The same went for the Moon. It was only until the Greeks figured out how to measure the distance using some neat trigonometry that we got the Moon's size distance and size. They also got the Sun's size and distance but it was pretty far off, though the way of measuring was correct.&lt;/p&gt;&#xA;" OwnerUserId="24373" LastActivityDate="2018-11-23T20:05:22.023" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28516" PostTypeId="1" CreationDate="2018-11-23T21:07:15.420" Score="17" ViewCount="5575" Body="&lt;p&gt;I’m not sure all of the details in the experiment but basically it went as follows. Someone marked their location of take off in a hot air balloon and went straight up in the air. They stayed suspended for some time before coming straight back down. Apparently the result was that they had not drifted from the original take-off point underneath them. Maybe I got all of that incorrect, but I’m sure it’s close. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if in that experiment the Earth had no effect on the landing point, why does the “rotation of the Earth” need to be factored into making the shot with a long range rifle for example? Now I know it isn’t the best comparison and that there are many different factors in both scenarios. Such as “the bullet is traveling and in the experiment the balloon wasn't”, but I’d just like to hear different theories and opinions as to why this is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I know a popular answer is going to be that the Earth is too massive for us to feel its spin, but I’m thinking deeper than that! &lt;/p&gt;&#xA;" OwnerUserId="24961" LastEditorUserId="2701" LastEditDate="2018-11-25T00:21:20.510" LastActivityDate="2018-12-02T10:06:52.027" Title="Why don’t we feel the Earth rotating beneath us?" Tags="&lt;gravity&gt;&lt;rotation&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28517" PostTypeId="2" ParentId="28480" CreationDate="2018-11-23T21:44:48.890" Score="2" Body="&lt;p&gt;Simply put. It all depends on the balance of an objects straight line speed orbiting the planet and gravity. The further away the object is from earth the slower the speed requirement for this balance is. Great question by the way. &lt;/p&gt;&#xA;" OwnerUserId="24961" LastActivityDate="2018-11-23T21:44:48.890" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28518" PostTypeId="2" ParentId="28516" CreationDate="2018-11-23T21:46:11.233" Score="17" Body="&lt;p&gt;There are two different things going on, and they aren't too related (from what I can see). For the hot air balloon situation, you would think that if you hovered above where you are for 6 hours, the Earth could spin under you, and you would land back in a completely different place. Unfortunately, because the hot air balloon was on the Earth to begin with, it was already moving with the Earth. There are many reference frames in play, even though we feel like we are standing still. Everyone on Earth is standing still on the surface. However, the surface is rotating around the Earth's axis. The Earth's axis (and the Earth itself) orbits the Sun. The Sun is goes around our galaxy and our galaxy is travelling through intergalactic space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how does this relate to the hot air balloon situation? Well because the balloon was on the surface, it was already moving with the surface of the Earth. Remember how I said that the surface of the Earth rotates around the Earth's axis? Well, since the balloon was on the surface to begin with, it will also rotate with Earth's axis, just like the surface! What if we wanted to achieve that hover effect? I said earlier that the Earth orbits the Sun. In order to achieve that hover effect, we would need our hot air balloon to orbit the Sun &lt;em&gt;without&lt;/em&gt; rotating around the Earth. You can't do this with a hot air balloon, since the atmosphere also happens to move with the Earth, and hot air balloons can't go into space. To achieve that hover effect, we would need some sort of spaceship with lots of fuel. If I had a hovering spaceship that didn't orbit with the Earth (this would, again, cost lots and lots of fuel), then yes, I could hover in the same spot and have the Earth rotate under me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you're probably wondering how I could achieve this with a spacecraft that was originally on the surface of the Earth. There wouldn't be much point in this hovering effect since you may as well just fly to your destination (like the theoretical Big Falcon Rocket), but if you wanted to do it, you would need to go above the Earth's atmosphere, use your boosters to go opposite of the rotation of the Earth (to cancel out your velocity), hover for some time, then use your boosters to go along with the rotation of the Earth (to return to rotation velocity), and land on Earth. Of course, you could skip the 2nd part by using heat shields and ramming into Earth's atmosphere like every other spacecraft we have, and no rocket would ever do this as it's just much more practical to orbit around the Earth (hovering costs fuel, orbits don't).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, what about the sniper situation? The Earth rotate on its axis. Since a day is 24 hours long, a location at the equator goes around once in the same time it takes for someone in the arctic to go around once. However, the further north or south you are from the equator, the slower that part of the Earth has to rotate to complete 1 rotation in 24 hours. Think about spinning a ball. The equator of the ball whips around, but the top and bottom move much slower. It's the same thing. Lets say my sniper is at the equator. When the sniper shoots east or west, they do not need to correct for the Earth's rotation because everywhere along that latitude, the Earth is rotating at the same speed. However, if the sniper shoots north, the bullet will go to the east. That's because when the bullet was shot at the latitude closer to the equator (shot from the south), that spot of the Earth was moving faster than the spot of the Earth that the target was standing on. It's like saying that the spot I'm standing rotates at a speed of say... 1000 mph, but my target's spot rotates at say...995 mph. Since my bullet came from the 1000 mph part, it will obviously go out of the barrel at high speeds because of the gun, but it will also be rotating around the Earth's axis at 1000 mph. However, once my bullet starts approaching my target's latitude, since the Earth rotates slower, it will appear to deviate in the direction of the rotation (which is east, or to the right). Why? Well since my target's speed is 995 mph from rotation, you do 1000 - 995 = 5. That means that my bullet will have a net velocity of 5 mph to the right relative to my target. That means my bullet will miss if the target is far away enough. If you shoot from the equator to the south, the bullet will also go east, so as a result, it will deviate to the left. It's the same effect, but &quot;upside down&quot;. This effect is called the Coriolis effect, and it's what gives hurricanes their power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you asked &quot;why don't we feel the Earth rotating beneath us?&quot; This is because we are moving with the Earth. When you are on a train that is going at a constant speed of 50 mph, you do not feel it move (you may feel some bumps due to bumpy tracks). It is only when you accelerate or decelerate do you feel things move. When you are on the train, your speed is constant, so you do not feel anything. It's the same for the Earth, except you are rotating around the Earth's axis at a constant 1000 mph. Nothing has changed, except the speed is faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I probably explained this horribly so feel free to ask anything.&lt;/p&gt;&#xA;" OwnerUserId="24373" LastActivityDate="2018-11-23T21:46:11.233" CommentCount="16" ContentLicense="CC BY-SA 4.0" />
  <row Id="28520" PostTypeId="2" ParentId="28516" CreationDate="2018-11-24T03:41:32.443" Score="16" Body="&lt;p&gt;It's hard to feel the Earth's rotation personally, but only because we are not very sensitive to very gradual changes and very weak &quot;forces&quot;. &lt;em&gt;However, there are everyday objects that can show us the rotation of the Earth.&lt;/em&gt; The balloon may have been a bad choice because even a slight wind over time will dominate the problem. (also &lt;a href=&quot;https://space.stackexchange.com/a/22614/12102&quot;&gt;see&lt;/a&gt;]&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;A pendulum&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;One object is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Foucault_pendulum&quot; rel=&quot;noreferrer&quot;&gt;Foucault pendulum&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Wq7tH.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wq7tH.gif&quot; alt=&quot;Foucault pendulum&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Foucault-rotz.gif&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;A gyroscope&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Another is an (expensive) camera with inertial image stabilization. Some cameras have small gyroscope chips that sense very tiny rotations of the camera body and compensate with image processing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gp7Vp.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gp7Vp.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Gyroscope_wheel_animation.gif&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the following questions and their answers for more on this. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://physics.stackexchange.com/q/288958/83380&quot;&gt;SLR camera gyroscopic stabilization - limited by rotation of the earth?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://photo.stackexchange.com/q/83007/50110&quot;&gt;Why does Image Stabilization have a Limit?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-11-24T03:59:01.477" LastActivityDate="2018-11-24T03:59:01.477" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28522" PostTypeId="2" ParentId="28516" CreationDate="2018-11-24T06:39:19.583" Score="8" Body="&lt;p&gt;That's three questions in one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why don’t we feel the earth rotating beneath us?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you're asking what one feels when standing still on the surface of the Earth, it's because&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Static friction and the normal force are constraint forces,&lt;/li&gt;&#xA;&lt;li&gt;The difference between what you would feel on a nonrotating planet with the same mass and size of the Earth and our rotating Earth us very small, and&lt;/li&gt;&#xA;&lt;li&gt;You've never experienced anything but standing on our rotating Earth.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Constraint forces take on whatever value is needed to satisfy the constraint if doing so doesn't violate some other condition. For example, the normal force pushes normal to the surface so as to keep the constrained object on the surface (the constraint). The &quot;other condition&quot; in the case of the normal force is that the normal force can only be directed outward. Friction comes into play if the surface on which one is standing isn't level. The other condition in the case of static friction is that it can't exceed the coefficient of static friction times the normal force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what values do these forces need to take on to keep you standing still on the surface of the Earth? From the perspective of an inertial frame, an object at rest on the Earth's surface is undergoing uniform rotation about the Earth's axis. This means the net force on the object must be inward, toward the Earth's axis of rotation (note well: not toward the center of the Earth, except at the equator), and have a magnitude of &lt;span class=&quot;math-container&quot;&gt;$F_{\text{net}}= m r \Omega^2$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; is the object's mass, &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; is the distance to the axis of rotation, and &lt;span class=&quot;math-container&quot;&gt;$\Omega$&lt;/span&gt; is the Earth's sidereal rotation rate. The other forces acting on the object are gravitation and buoyancy, which is very small. Ignoring buoyancy, we must have &lt;span class=&quot;math-container&quot;&gt;$\vec F_\text{surface} + \vec F_\text{gravity} = \vec F_\text{net}$&lt;/span&gt;. The net force is largest at the equator, where it's very small compared to gravity, about 0.3% of gravitational force. In other words, the upward force exerted by the Earth's surface and the downward force exerted by the Earth as a whole are nearly equal to one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same result arises if one looks at things from the perspective of a rotating frame. From this perspective, the apparent net force is zero. Rotating frames involve fictitious forces such as the fictitious centrifugal force. This has the exact same magnitude as but the opposite direction of the net force calculated above. The end result is the same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Someone marked their location of take off in a hot air balloon and went straight up in the air. They stayed suspended for some time before coming straight back down. Apparently the result was that they had not drifted from the original take off point underneath them. Maybe I got all of that incorrect but I’m sure it’s close.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's the second part of the question. Unless tethered, hot air balloons typically do not return to the spot from which they take off. Hot air balloons go wherever the wind carries them. That's why hot air balloons need chase teams. A hot air balloon will return to its take off point if the wind is nonexistent or if it reverses direction during the course of the balloon's flight. The buoyancy that I ignored above is anything but negligible for a hot air balloon. Here it's the normal force that is negligible (it's nonexistent). For a balloon to remain stationary, the wind needs to be nonexistent and the buoyancy force must take on the same form as the normal force takes for an object sitting still on the surface of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does the “rotation of the earth” need to be factored into making the shot in a long range shot from a rifle for example?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because the bullet is moving with respect to the surface of the Earth. Note well: It's only the very long range shots, more than a kilometer, where the Earth's rotation needs to be taken into account. The motion of the bullet brings another fictitious force into account, the Coriolis effect. The Coriolis effect is zero for a stationary person and for a stationary balloon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Coriolis effect has two primary effects on the flight of a bullet, a horizontal deflection and a vertical deflection. The horizontal deflection depends on latitude, the effect being strongest at the poles and zero at the equator. The Coriolis effect makes moving objects turn to the right in the northern hemisphere but to the left in the southern hemisphere. The vertical deflection depends on latitude and direction. This vertical deflection is strongest at the equator, zero at the poles or if the direction of motion is to the north or south. The deflections are small, even for a bullet fired toward a target a kilometer away. Centimeters however can be important.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2018-11-24T06:39:19.583" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28525" PostTypeId="2" ParentId="28484" CreationDate="2018-11-25T01:39:41.700" Score="2" Body="&lt;p&gt;(Please excuse my English)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above question arose during my investigation on the double planet/dwarf planet system. As we know, the definition of satellite and the definition of double planet/dwarf planet system are closely related. And currently there is no clear definition of the satellite, and so is the double planet/dwarf planet system.&lt;br /&gt;&#xA;I thought a simple and obvious criterion is better. However, the simple size criterion is not sufficient for complex combinations of celestial bodies. In other hand, the position of barycenter is often said to be a criterion for satellite, but it can not be an appropriate criterion as described &lt;a href=&quot;https://astronomy.stackexchange.com/questions/28484/moon-is-a-planet-by-clearing-the-neighbourhood-condition?noredirect=1#comment50692_28487&quot;&gt;here&lt;/a&gt;.&lt;br /&gt;&#xA;What I thought as a criterion is : If two celestial bodies bound by gravity are classified as same class, we should classify the two bodies as a double body system. For example, if two stars are bound by gravity, we classify it as double star system. And if two black holes are bound, we classify it as a double black hole system.&lt;br /&gt;&#xA;So if two planet/dwarf planet bodies are bound, it is natural to classify it as a double planet/dwarf planet system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For test, I applied the criterion to Pluto-Charon system. Pluto is classified as a dwarf planet by 2006 IAU definition. If we consider Charon as a seperated body, Charon also satisfies the conditions for dwarf planet. So we can classify Pluto-Charon system as a double dwarf planet system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, I applied the criterion to Earth-Moon system. As I mentioned in above question, I do not agree that Moon is a planet and Earth-Moon system is a double planet system. I thought Earth and Moon should be classified to different classes, because there are too many differences between two bodies.&lt;br /&gt;&#xA;However, Moon as a seperated body satisfies IAU planet conditions(especially &lt;em&gt;Clearing the neighborhood&lt;/em&gt; condition) as I described in above question. I was embarrassed. So I asked my calculation is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have thought for a long time that there are many problems with the &lt;em&gt;Clearing the neighborhood&lt;/em&gt; condition for the planet definition. Above situation increased my distrust of the &lt;em&gt;Clearing the neighborhood&lt;/em&gt; condition. Thus, as I replied the comments and answers, I became over-expressing my opinion on the condition.&lt;/p&gt;&#xA;" OwnerUserId="23440" LastEditorUserId="23440" LastEditDate="2018-11-25T09:47:40.450" LastActivityDate="2018-11-25T09:47:40.450" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28526" PostTypeId="1" CreationDate="2018-11-25T08:05:04.053" Score="4" ViewCount="195" Body="&lt;p&gt;How do we know that ʻOumuamua is interstellar and not otherwise? I fail to understand the logic that the extraordinary speed of the object proves it to be of interstellar origin, and that our sun's gravity failed to capture it.&lt;/p&gt;&#xA;" OwnerUserId="24974" LastEditorDisplayName="user1569" LastEditDate="2018-11-25T13:43:58.100" LastActivityDate="2018-11-26T15:26:18.563" Title="How do we know about the origin of ʻOumuamua?" Tags="&lt;interstellar-travel&gt;&lt;interstellar&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="28527" PostTypeId="1" CreationDate="2018-11-25T12:06:24.620" Score="1" ViewCount="35" Body="&lt;p&gt;How to estimate upper limits on flux values for a radio source if it is not seen in NVSS (1.4 GHz survey) but is observed in lower radio frequency maps?&lt;/p&gt;&#xA;" OwnerUserId="24976" LastEditorUserId="1891" LastEditDate="2018-11-25T18:47:10.703" LastActivityDate="2018-11-25T18:47:10.703" Title="Estimating upper limits on flux values (Radio)" Tags="&lt;radio-astronomy&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28528" PostTypeId="2" ParentId="28526" CreationDate="2018-11-25T13:27:02.923" Score="8" Body="&lt;p&gt;I'll start with the second question first:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Also, do we exactly know about each and every object inside our solar system?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Each and every object? Of course not. That however is irrelevant to the question at hand. The vast majority of objects in our solar system are very, very small. Quite literally as small as dust. But because they're so small, they have essentially zero impact on the orbits of objects that are not as small as dust.&lt;/p&gt;&#xA;&lt;br&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;How do we know that ʻOumuamua is interstellar and not otherwise?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Because it's moving so quickly. ʻOumuamua's huge 26 kilometers per second excess means that is not possible that ʻOumuamua originated within the solar system.&lt;/p&gt;&#xA;&lt;p&gt;I'll start with the two body problem, e.g., the Sun and some other object interacting gravitationally. The Newtonian gravitational interaction between two objects is a conic section: Either a circle, an ellipse, a parabola, or a hyperbola. The total mechanical energy (kinetic energy plus gravitational energy) for objects following a circular or elliptical trajectory is negative. Such objects are gravitationally bound to one another. Hyperbolic trajectories on the other hand have a positive mechanical energy. Parabolic trajectories form the boundary cases, where the total mechanical energy is exactly zero. Objects on parabolic or hyperbolic trajectories are not bound. They visit once and then they're gone. The excess velocity of an object on a parabolic trajectory (zero excess velocity) or a hyperbolic trajectory (positive excess velocity) is a constant of motion in the two body problem.&lt;/p&gt;&#xA;&lt;p&gt;The N-body problem makes things a bit dicier. Long-period comets are believed to have originated from within the solar system. These comets appear because something perturbed their very long-period orbits. The perturbation make those long-period comets dive well inside Neptune's orbit. A few of those long-period comets appear to be following an unbound trajectory (non-negative excess velocity). A chance close encounter with a planet can make that happen.&lt;/p&gt;&#xA;&lt;p&gt;This raises the question: Could ʻOumuamua be yet another long-period comet whose orbit has been perturbed to an extent that puts it on a hyperbolic trajectory? This would not be the first case of an object with an over-unity eccentricity. The answer to this question is no. A close encounter with any of the four known giant planets, or even with the conjectured Planet IX, or even multiple encounters, could not have given ʻOumuamua the excess velocity that has been observed for it.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-11-25T13:27:02.923" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28531" PostTypeId="1" AcceptedAnswerId="28532" CreationDate="2018-11-25T18:53:48.763" Score="4" ViewCount="2097" Body="&lt;p&gt;Is there any material natural or otherwise a free floating liquid that can exist in space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://space.stackexchange.com/questions/32274/what-liquids-last-the-longest-in-space&quot;&gt;https://space.stackexchange.com/questions/32274/what-liquids-last-the-longest-in-space&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-11-26T16:28:13.657" LastActivityDate="2018-11-27T08:13:22.390" Title="What liquids can be found in the void space?" Tags="&lt;astrophysics&gt;&lt;amateur-observing&gt;&lt;fundamental-astronomy&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28532" PostTypeId="2" ParentId="28531" CreationDate="2018-11-25T20:48:27.750" Score="11" Body="&lt;p&gt;No liquid can be completely stable in a vacuum, since all liquids have some non-zero vapour pressure, and so will evaporate at some rate. However some liquids have an exceptionally low vapour pressure, and so can be used in a vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vapour pressure of &lt;a href=&quot;http://www.kayelaby.npl.co.uk/chemistry/3_4/3_4_3.html&quot; rel=&quot;noreferrer&quot;&gt;silcone fluid DC705&lt;/a&gt;, which is &lt;a href=&quot;http://www.testbourne.com/invoil-silicone-diffusion-pump-oil-702-704-705&quot; rel=&quot;noreferrer&quot;&gt;used in diffusion pumps&lt;/a&gt; is 2.6e-8, and it is designed to function in a high vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a location could be found at which it was warmed sufficiently to remain liquid (in deep space it would just freeze, whereas too close to the sun and its vapour pressure would rise) it could remain in a liquid state for some time.  Not indefinitely but it could be stable for a while.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-11-25T20:48:27.750" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28537" PostTypeId="2" ParentId="28526" CreationDate="2018-11-26T15:26:18.563" Score="2" Body="&lt;p&gt;It's not that much about the speed, but about the orbital eccentricity. Wikipedia gives a good explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Based on observations spanning 34 days, ʻOumuamua's orbital eccentricity is 1.20, the highest ever observed. An eccentricity exceeding 1.0 means an object exceeds the Sun's escape velocity, is not bound to the Solar System and may escape to interstellar space. While an eccentricity slightly above 1.0 can be obtained by encounters with planets, as happened with the previous record holder, C/1980 E1, ʻOumuamua's eccentricity is so high that it could not have been obtained through an encounter with any of the planets in the Solar System. Even undiscovered planets in the Solar System, if any should exist, could not account for ʻOumuamua's trajectory nor boost its speed to the observed value. For these reasons, ʻOumuamua can only be of interstellar origin.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/%CA%BBOumuamua#Trajectory&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/%CA%BBOumuamua#Trajectory&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24234" LastActivityDate="2018-11-26T15:26:18.563" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28538" PostTypeId="2" ParentId="28484" CreationDate="2018-11-26T18:44:11.590" Score="0" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;At present, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Moon&quot; rel=&quot;nofollow noreferrer&quot;&gt;Moon&lt;/a&gt; is defined as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Natural_satellite&quot; rel=&quot;nofollow noreferrer&quot;&gt;satellite&lt;/a&gt; of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Earth&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth&lt;/a&gt;, but over time the Earth-Moon &lt;a href=&quot;http://en.wikipedia.org/wiki/Barycenter&quot; rel=&quot;nofollow noreferrer&quot;&gt;barycenter&lt;/a&gt; will drift outwards (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Tidal_acceleration&quot; rel=&quot;nofollow noreferrer&quot;&gt;tidal acceleration&lt;/a&gt;) and could eventually become situated outside of both bodies.&lt;sup&gt;&lt;strong&gt;25&lt;/strong&gt;&lt;/sup&gt; This development would then upgrade the Moon to planetary status at that time, according to the definition. The time taken for this to occur, however, would be billions of years, long after many &lt;a href=&quot;http://en.wikipedia.org/wiki/Astronomy&quot; rel=&quot;nofollow noreferrer&quot;&gt;astronomers&lt;/a&gt; expect the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sun&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sun&lt;/a&gt; to expand into a &lt;a href=&quot;http://en.wikipedia.org/wiki/Red_giant&quot; rel=&quot;nofollow noreferrer&quot;&gt;red giant&lt;/a&gt; and destroy both Earth and Moon.&lt;sup&gt;&lt;strong&gt;26&lt;/strong&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;&lt;strong&gt;25&lt;/strong&gt;&lt;/sup&gt; &lt;sub&gt;Robert Roy Britt (2006). &lt;em&gt;&lt;a href=&quot;http://web.archive.org/web/20100822072302/http://www.space.com/scienceastronomy/moon_mechanics_0303018.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Moon Mechanics: What Really Makes Our World Go 'Round'&lt;/a&gt;&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;&lt;strong&gt;26&lt;/strong&gt;&lt;/sup&gt; &lt;sub&gt;Robert Roy Britt (2006). &lt;em&gt;&lt;a href=&quot;http://edition.cnn.com/2006/TECH/space/08/18/moon.planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth's moon could become a planet&lt;/a&gt;&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$&lt;/span&gt; — &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/IAU_definition_of_planet#Criticism&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Wikipedia :&lt;/strong&gt; Criticism of the IAU definition of 'planet'&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="14152" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-11-26T18:44:11.590" CommentCount="1" CommunityOwnedDate="2018-11-26T18:44:11.590" ContentLicense="CC BY-SA 4.0" />
  <row Id="28539" PostTypeId="2" ParentId="28449" CreationDate="2018-11-27T00:20:35.370" Score="0" Body="&lt;p&gt;There are a couple of simple approaches that would work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a system tripod that lets you swap the tripod head, Manfrotto do a 183 aluminium projector table attachment (&lt;a href=&quot;https://www.amazon.co.uk/Manfrotto-183-Aluminium-Attachment-Projectors/dp/B001HS697C&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amazon.co.uk/Manfrotto-183-Aluminium-Attachment-Projectors/dp/B001HS697C&lt;/a&gt;) that attaches to the 3/8&quot; thread usually used for the tripod head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, if you have access to the appropriate tools, you can tap a 3/8&quot; or 3/4&quot; threaded hole into a metal block, and screw that to the bottom of a piece of wood for a DIY equivalent - that's what an uncle of mine did years ago when I was a youngster :).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the design of your mini dob, you might be able to attach a similar block to the bottom of your baseplate. (As a side note, I think there are a few mini reflector scopes around that already have a tripod socket in the base - Though tripod sockets are a lot more common with small refractors, maks, and spotting scopes, since they're usually designed for use with a tripod in the first place)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to go the photo tripod plus platform route, and you want to keep the weight down, then if your budget is big enough, it may be worth looking at one of the carbon fibre tripods. They're more expensive, but noticeably lighter - which can feel like a bargain if you're lugging one around all day... They also have the advantage of not freezing your fingers off after a session outdoors in cold weather.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you're looking for a more compact tripod, it's fairly common to find similar models with different numbers of leg sections - for example, 3 or 4 section legs. The 3 section ones are longer, compared to the 4 section versions, so they're bulkier when closed - but tend to be a bit more stable than the 4 section ones since they have a thinner fourth section (to fit inside the other three).&lt;/p&gt;&#xA;" OwnerUserId="17016" LastEditorUserId="17016" LastEditDate="2018-11-27T00:45:57.547" LastActivityDate="2018-11-27T00:45:57.547" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28541" PostTypeId="2" ParentId="28503" CreationDate="2018-11-27T05:38:26.240" Score="9" Body="&lt;p&gt;No, nothing on Europa could possibly be photosynthesizing as we know it. Jupiter doesn't emit light, and what it reflects from the sun is not enough, plus there's no significant amount of carbon dioxide there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the first life on Earth was not plants, it was &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Chemotroph&quot; rel=&quot;nofollow noreferrer&quot;&gt;chemotrophs&lt;/a&gt;. These bacteria gained their energy from the heat of hydrothermal vents, and if there's life on Europa, it would almost certainly take this form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the temperature: Europa's heat comes from friction due to tidal forces, so it comes from within. The water temperature increases as you go down, but the pressure is crushing down there and it's unknown whether life could form under those conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tenuous oxygen atmosphere comes from charged particles hitting it's cold surface and creating vapor. It does not indicate biological processes.&lt;/p&gt;&#xA;" OwnerUserId="18617" LastEditorUserId="18617" LastEditDate="2018-12-11T01:26:38.403" LastActivityDate="2018-12-11T01:26:38.403" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28542" PostTypeId="2" ParentId="28531" CreationDate="2018-11-27T08:13:22.390" Score="1" Body="&lt;p&gt;My answer is: &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Superionic_water&quot; rel=&quot;nofollow noreferrer&quot;&gt;superionic water&lt;/a&gt;, at high pressure and low temperature. During the superionic phase, the hydrogen and oxygen within water molecules behave bizarrely; hydrogen ions move like a liquid, inside of a solid crystal lattice of oxygen.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Derivation of answer, and references:&lt;/p&gt;&#xA;&lt;p&gt;Any element or molecule will evaporate (essentially &lt;em&gt;boil&lt;/em&gt;) in low enough pressure, even at extremely low temperatures; albeit slowly (by human standards).&lt;/p&gt;&#xA;&lt;p&gt;Both extreme pressure and low temperature is necessary to prolong the lifespan (duration) of an element or molecule's liquidity.&lt;/p&gt;&#xA;&lt;p&gt;Something &lt;em&gt;contained&lt;/em&gt; must also be &amp;quot;free floating&amp;quot; (to comply with the terms of your question) - thus suggesting liquid hydrogen (or deuterium) seems a stretch, but it's a clue for the direction to look.&lt;/p&gt;&#xA;&lt;p&gt;From the article: &amp;quot;&lt;a href=&quot;https://www.nytimes.com/2018/08/16/science/metallic-hydrogen-lasers.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Settling Arguments About Hydrogen With 168 Giant Lasers&lt;/a&gt;&amp;quot;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;At ultracold temperatures, below -423 degrees Fahrenheit, hydrogen condenses into a liquid. It also turns into a liquid at higher temperatures when squeezed under immense pressure. The molecules remain intact, and this state of liquid hydrogen is an insulator — a poor conductor of electricity.&lt;/p&gt;&#xA;&lt;p&gt;Under even higher pressures, the molecules break apart into individual atoms, and the electrons in the atoms are then able to flow freely and readily conduct electricity — the definition of a metal.&lt;/p&gt;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&lt;p&gt;Liquid metallic hydrogen does not naturally occur on Earth — except possibly at the core. But at Jupiter, the solar system’s largest planet, most of the hydrogen could be flowing as a liquid metal and generating the planet’s powerful magnetic fields.&lt;/p&gt;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Preventing &lt;em&gt;freezing&lt;/em&gt; (solidification) of the hydrogen is the key, and the use of pressure will alter the phase change diagram.&lt;/p&gt;&#xA;&lt;p&gt;The article: &amp;quot;&lt;a href=&quot;https://projects.iq.harvard.edu/silvera&quot; rel=&quot;nofollow noreferrer&quot;&gt;Two Pathways to Metallic Hydrogen and Deuterium&lt;/a&gt;&amp;quot; (by the Silvera Group, Department of Physics, Harvard) has some photos of liquid hydrogen:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gg9h5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gg9h5.png&quot; alt=&quot;Gaseous, Liquid, and Metallic Hydrogen&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;sub&gt;&amp;quot;Images of hydrogen at different pressures and low temperature showing the progression from a transparent molecular solid to a black semiconducting solid to a brilliant shiny metal of hydrogen.&amp;quot;&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;It was also realized that at a lower, but still high pressure and very high temperatures, there is a temperature driven transition to liquid atomic metallic hydrogen (LMH).  This liquid-liquid phase transition is sometimes called the Plasma Phase Transition or the PPT. We have determined the phase line for LMH and liquid atomic deuterium for several values of P and T, as well as optical properties and optical conductivity. We observe isotopic differences in the phase lines.&lt;/p&gt;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&lt;p&gt;LMH is the principle component of giant outer planets such as Jupiter and gives rise to its magnetic field via the dynamo.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;BUT&lt;/strong&gt;, there are the terms of your question (which the answerer ought not to &lt;a href=&quot;https://www.merriam-webster.com/dictionary/fudge&quot; rel=&quot;nofollow noreferrer&quot;&gt;fudge&lt;/a&gt; on): &amp;quot;Is there any material, natural or otherwise, a &lt;strong&gt;free floating liquid&lt;/strong&gt; that can exist in space?&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;So simply pressurizing hydrogen didn't seem to meet the requirements of &amp;quot;free flowing&amp;quot; (unless you'll accept it as &lt;em&gt;&lt;strong&gt;one&lt;/strong&gt;&lt;/em&gt; answer) - but it hinted at the direction I should look.&lt;/p&gt;&#xA;&lt;p&gt;The clue led me to superionic water, at high pressure and low temperature.&lt;/p&gt;&#xA;&lt;p&gt;From the Nature article: &amp;quot;&lt;a href=&quot;https://www.nature.com/articles/s41567-017-0017-4&quot; rel=&quot;nofollow noreferrer&quot;&gt;Experimental evidence for superionic water ice using shock compression&lt;/a&gt;&amp;quot; (Nature Physicsvolume 14, pages297–302 (2018)) and also written up at LiveScience: &amp;quot;&lt;a href=&quot;https://www.livescience.com/62373-superionic-ice-lab-created.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This Ice Is Nearly As Hot As the Sun. Scientists Have Now Made It on Earth&lt;/a&gt;&amp;quot;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;It's both solid and liquid, it's 60 times denser than ordinary water ice, and it forms at temperatures almost as hot as the sun's surface.&lt;/p&gt;&#xA;&lt;p&gt;It's superionic ice — and for the first time, scientists have made it in the lab.&lt;/p&gt;&#xA;&lt;p&gt;This high-pressure form of water ice has long been thought to exist in the interiors of Uranus and Neptune. But until now, its existence was only theoretical.&lt;/p&gt;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&lt;p&gt;Scientists first predicted the existence of a weird water phase that makes the substance both solid and liquid at the same time 30 years ago. It's also way denser than ordinary water ice because it forms only under extreme heat and pressure, such as those found inside giant planets. &lt;strong&gt;During the superionic phase, the hydrogen and oxygen within water molecules behave bizarrely; hydrogen ions move like a liquid, inside of a solid crystal lattice of oxygen&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Making the ice was complicated. First, the team compressed water into an ultrastrong cubic crystalline ice, in a different crystal form than what you see in ordinary ice cubes. To do that, the researchers used diamond anvil cells to apply 360,000 pounds per square inch (2.5 gigapascals (GPa) of pressure; that's about 25,000 times the atmospheric pressure on Earth). Next, the researchers heated and compressed the cells even further, using laser-driven shocks. Each crystal ice structure received up to six laser beams of more than 100 times that high pressure.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Because we pre-compressed the water, there is less shock-heating than if we shock-compressed ambient liquid water,&amp;quot; Millot said. The new method lets researchers &amp;quot;access much colder states at high pressure than in previous shock-compression studies.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Once the superionic ice was ready, the team moved quickly to analyze its optical and thermodynamic properties. They had only 10 to 20 nanoseconds to perform the work, before pressure waves released the compression, and the water dissolved. And the results were bizarre. They found that the ice melts at an extraordinary 8,540 degrees Fahrenheit (4,725 degrees Celsius ) at 29 million pounds per square inch (200 GPa) of pressure. That pressure is about 2 million times the atmospheric pressure on Earth&lt;/p&gt;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&lt;p&gt;The new findings could provide a peek inside the interiors of planets such as Uranus and Neptune. Planetary scientists suggest these worlds' innards are composed of up to 65 percent water by mass, plus some ammonia and methane.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Thus, &lt;strong&gt;hydrogen&lt;/strong&gt; can be &lt;strong&gt;liquid&lt;/strong&gt; &lt;em&gt;&lt;strong&gt;and&lt;/strong&gt;&lt;/em&gt; &amp;quot;float freely&amp;quot; in space - but only in a lattice of oxygen, under the extreme pressures, at the core of one of our gas giants (or other similar objects in space).&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-11-27T08:13:22.390" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28545" PostTypeId="1" CreationDate="2018-11-27T22:31:00.137" Score="5" ViewCount="592" Body="&lt;p&gt;How does a gravitational wave travel over vast distances without losing its energy? Could they go on forever as long as the Space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="24221" LastActivityDate="2018-11-28T07:41:42.990" Title="Does a gravitational wave loses energy over distance?" Tags="&lt;black-hole&gt;&lt;gravitational-waves&gt;&lt;wavelength&gt;&lt;high-energy-astrophysics&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28546" PostTypeId="2" ParentId="28545" CreationDate="2018-11-27T22:35:56.070" Score="2" Body="&lt;p&gt;In empty space, just like a light wave, they spread out, becoming less intense as they get further from their source, but never vanishing completely. At some stage the waves from a distant event might become undetectable in local noise, or so weak that quantum effects might become relevant, but essentially they never die out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @uhoh points out, they do lose some energy when they interact with matter.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="21928" LastEditDate="2018-11-28T07:41:42.990" LastActivityDate="2018-11-28T07:41:42.990" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28547" PostTypeId="2" ParentId="28545" CreationDate="2018-11-27T22:39:51.730" Score="1" Body="&lt;p&gt;It's a wave, and like any other wave, it loses flux (power over area) with the square of the distance. That is, if you double the distance it travels, your wave will have a quarter of the flux it had. If you triple the distance, your wave will have a ninth of the flux it had. This is a property of waves and can be observed in light intensity, sound, and any other wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically, the wave will never fully dissipate. If the wave travels a million times the distance, the wave will weaken to one trillionth of what it was. Still not zero, but very tiny nonetheless.&lt;/p&gt;&#xA;" OwnerUserId="24373" LastEditorUserId="24373" LastEditDate="2018-11-27T23:34:33.267" LastActivityDate="2018-11-27T23:34:33.267" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28548" PostTypeId="1" CreationDate="2018-11-28T01:07:47.167" Score="3" ViewCount="92" Body="&lt;p&gt;Probably this is a extremely basic question, but after a lot of time searching about it without results (in all the sites/publications i've found they only use this term but there is no specific definition)i came here with the hope of get an answer to my question: in the context of chemical abundance in stars, what is a differential abundance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already know what is a chemical abundance, the question is simply about the implication of the &quot;differential&quot; word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, my doubt emerged when i was reading &lt;a href=&quot;https://arxiv.org/abs/1610.09067&quot; rel=&quot;nofollow noreferrer&quot;&gt;this publication&lt;/a&gt;.&#xA;In the second paragraph of the section 3.2 &quot;Stellar Parameters&quot; is the first encounter with this concept:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We adopted the same line list as in Ramírez et al. (2014). The&#xA;  EW  measurements  were  used  to  obtain  elemental  abundances&#xA;  with the&#xA;  &lt;em&gt;abfind&lt;/em&gt;&#xA;  driver of MOOG (Sneden 1973), using Kurucz&#xA;  model atmospheres (Castelli &amp;amp; Kurucz 2004). Then, line-by-line&#xA;  differential abundances were obtained.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The stellar parameters&#xA;  &lt;span class=&quot;math-container&quot;&gt;$(T_\text{eff}, \log g, {\rm[Fe/H]}, v_t)$&lt;/span&gt;&#xA;  of HIP 68468&#xA;  were determined by imposing a differential spectroscopic equilibrium  of  iron  lines  relative  to  the  Sun  (e.g.  Meléndez  et  al.&#xA;  2014a), using as initial parameters the values given in Ramírez&#xA;  et al. (2014). The solar values were kept fixed at&#xA;  &lt;span class=&quot;math-container&quot;&gt;$(T_\text{eff}, \log g, v_t)&#xA;= \rm&#xA;(5777\,K, 4.44\,dex, 1.0\,km\,s^{-1})$&lt;/span&gt;. Our solution for HIP 68468&#xA;  gives consistent differential abundances for neutral and ionised&#xA;  species of different elements, and also for atomic and molecular&#xA;  lines, as discussed in more detail in the next section.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="25004" LastEditorUserId="1498" LastEditDate="2018-11-28T03:24:20.037" LastActivityDate="2018-11-28T10:08:43.020" Title="What is a &quot;differential chemical abundance&quot;?" Tags="&lt;stellar-astrophysics&gt;&lt;elemental-abundances&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28549" PostTypeId="1" CreationDate="2018-11-28T01:59:52.820" Score="5" ViewCount="67" Body="&lt;p&gt;I would like to ask; how does Water Vapour in the atmosphere interfere with astronomical observations? Is it in the same manner as Carbon dioxide?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What brought this to mind&lt;/em&gt; is the following: A member of Monash University staff claimed recently in a newspaper in Australia that both CO2 and H2O absorbed heat and hence their presence needed to be reduced for astronimical observations. Hence the siting of optical observatories on mountain tops where the absolute amount of water vapour in the atmosphere is reduced by the relatively low temperatures there.&lt;/p&gt;&#xA;" OwnerUserId="25005" LastEditorUserId="7982" LastEditDate="2018-11-28T04:04:26.373" LastActivityDate="2018-12-28T10:01:35.313" Title="Impact of Atmospheric Water vapour on Optical Observations" Tags="&lt;observational-astronomy&gt;&lt;observatory&gt;&lt;weather&gt;&lt;atmospheric-effects&gt;&lt;climate&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28550" PostTypeId="2" ParentId="28545" CreationDate="2018-11-28T04:01:40.573" Score="6" Body="&lt;h2&gt;tl;dr: Yes. Feynman's beads on a string argument&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The other answers skirt what I think is the issue that the OP is asking about. In a &lt;strong&gt;lossless medium&lt;/strong&gt; a spherical wave packet itself, caused by a disturbance, will not &quot;loose energy&quot; itself. If you integrate over a large volume you get a constant energy versus time. Of course the flux per unit area will decrease by the same rate that the area increases, but that's a trivial result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But&lt;/strong&gt;, when gravitational waves pass matter, they really do deposit energy in the matter, and thereby become attenuated. The effect is very very tiny for a given amount of matter, but it is non-zero and the universe is pretty big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;https://physics.stackexchange.com/a/262962/83380&quot;&gt;this answer&lt;/a&gt; to the question &lt;em&gt;Transfer of energy from gravity back to other “more familiar” forms of energy?&lt;/em&gt; I've said:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This is from the Wikipedia article &lt;a href=&quot;https://physics.stackexchange.com/questions/262926/theoretically-possible-to-extract-energy-from-gravity-waves-how-might-it-be-imp/262947#comment582545_262947&quot;&gt;mentioned&lt;/a&gt;. &#xA;  I think this is really worth reading, as it not only describes some of the physics associated with the &lt;a href=&quot;https://physics.stackexchange.com/a/262947/83380&quot;&gt;accepted answer&lt;/a&gt; and it's citation of &lt;a href=&quot;https://en.wikipedia.org/wiki/Sticky_bead_argument#Feynman.27s_argument&quot; rel=&quot;noreferrer&quot;&gt;Feynman's argument&lt;/a&gt; about a bead a stick, but perhaps gives a little insight into the development of the field as well.&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&lt;strong&gt;Feynman's argument&lt;/strong&gt;&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;Later in the Chapel Hill conference, Richard Feynman — who had insisted on registering under a pseudonym to express his disdain for the contemporary state of gravitational physics — used Pirani's description to point out that a passing gravitational wave should in principle cause a bead on a stick (oriented transversely to the direction of propagation of the wave) to slide back and forth, thus heating the bead and the stick by friction. This heating, said Feynman, showed that the wave did indeed impart energy to the bead and stick system, so it must indeed transport energy, contrary to the view expressed in 1955 by Rosen.&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;In two 1957 papers, Bondi and (separately) Joseph Weber and John Archibald Wheeler used this bead argument to present detailed refutations of Rosen's argument.(5)(6)&lt;/p&gt;&#xA;    &#xA;    &lt;hr&gt;&#xA;    &#xA;    &lt;p&gt;(5) Bondi, Hermann (1957). &quot;Plane gravitational waves in general relativity&quot;. Nature 179 (4569): 1072–1073. Bibcode:1957Natur.179.1072B. doi:10.1038/1791072a0.&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;(6) Weber, Joseph &amp;amp; Wheeler, John Archibald (1957). &quot;Reality of the cylindrical gravitational waves of Einstein and Rosen&quot;. Rev. Mod. Phys. 29 (3): 509–515. Bibcode:1957RvMP...29..509W. doi:10.1103/RevModPhys.29.509.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-11-28T04:08:12.570" LastActivityDate="2018-11-28T04:08:12.570" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28551" PostTypeId="2" ParentId="28549" CreationDate="2018-11-28T09:59:53.563" Score="2" Body="&lt;p&gt;&lt;strong&gt;Yes, water vapour interferes even stronger with astronomical observations than &lt;span class=&quot;math-container&quot;&gt;$\rm CO_2$&lt;/span&gt;.&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the optical, water easily forms droplets which scatter light. But also in the gaseous phase water has strong absorption bands, some of those reach into the optical, while most of those however lie in the infrared, making water a greenhouse gas at the same time.&lt;br&gt;&#xA;Let's have a look at the absorption spectrum of various gaseous species in the atmosphere:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/S8A8C.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S8A8C.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we see with data what would otherwise take too many words to explain. Take note that the observable spectrum for humans is at the wavelengths &lt;span class=&quot;math-container&quot;&gt;$\rm \lambda = 0.4-0.8 \mu m$&lt;/span&gt;. The many absorbtion bands of water make clear why this interferes with astronomical observations:&lt;br&gt;&#xA;In the same way that outgoing thermal radiation from Earth is stopped at going into space, also infrared coming from space is absorbed in the atmosphere and thus not reaching telescopes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To deal with this there is a number of ways to cope:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Go high up above all the water vapour, to a dry place like Chile. Or a dry place like space. Then the absorption spectrum looks much more benevolent, and that's part of the reason we have all the major observatories in Chile, Mauna Kea, etc.&lt;/li&gt;&#xA;&lt;li&gt;Try to observe in one of the absorption bands, but &lt;em&gt;fly&lt;/em&gt; above all the water vapour. This is what the two airborne-based observatories &lt;a href=&quot;https://en.wikipedia.org/wiki/Stratospheric_Observatory_for_Infrared_Astronomy&quot; rel=&quot;nofollow noreferrer&quot;&gt;SOFIA&lt;/a&gt; and its predecessor do/did.&lt;/li&gt;&#xA;&lt;li&gt;Observe through the infrared-'windows' that let light pass through, from the ground. This is a bit exotic science wise nowadays - AFAIK this is a method mostly used in history, as today it is hard to justify building a massive telescope when most of your prospective signal gets absorbed. This was employed at &lt;a href=&quot;https://en.wikipedia.org/wiki/Mount_Wilson_Observatory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mt. Wilson Observatory&lt;/a&gt;, but also many others.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2018-11-28T09:59:53.563" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28552" PostTypeId="2" ParentId="28548" CreationDate="2018-11-28T10:08:43.020" Score="3" Body="&lt;p&gt;If the line emission is optically thin, then the measured line flux is proportional to the abundance &lt;span class=&quot;math-container&quot;&gt;$F_{\nu} \sim n$&lt;/span&gt; of a species along the line of sight (if it's only one species).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem here is that &lt;span class=&quot;math-container&quot;&gt;$F_{\nu}$&lt;/span&gt; needs to be measured relative to a baseline. If that fails, e.e. when the stellar signal is too noisy, one doesn't know what this should be. But a measurment at two different wavelengths &lt;span class=&quot;math-container&quot;&gt;$\nu$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\nu'$&lt;/span&gt; that lie reasonably close together, can at least yield a difference in abundances &lt;span class=&quot;math-container&quot;&gt;$F_{\nu}-F_{\nu'} \approx n-n'$&lt;/span&gt; relative to this unknown base level.&lt;br&gt;&#xA;This allows then to construct an atmospheric model, up to the unknown base level, which one can try to fix with other physical arguments.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2018-11-28T10:08:43.020" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28553" PostTypeId="2" ParentId="3668" CreationDate="2018-11-28T10:42:42.387" Score="1" Body="&lt;p&gt;You might already have your answer at this point, but never mind, here is an up to date (2018) answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gaia Data release is out (1 &amp;amp; 2).&#xA;Here is a link to the download page of the Gaia archive data release 2: &lt;a href=&quot;http://cdn.gea.esac.esa.int/Gaia/gdr2/gaia_source_with_rv/csv/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://cdn.gea.esac.esa.int/Gaia/gdr2/gaia_source_with_rv/csv/&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the link is dead or you want more, you can just type Gaia archive on a search engine and then go to downloads and you should find what you are searching for. You can also select only the ones that also have radial velocities if you want to include time evolution.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more precise requests, you can register on the Gaia archive (just need email address + name and takes 5 minutes) and you can then do queries to the database allowing you to filter the sources as you please. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note:&lt;/em&gt;&lt;br&gt;&#xA;1. The distance measurements have about a 10% margin of error, but it should be the state of the art for the time being.&lt;br&gt;&#xA;2. There are plenty of parameters more than just the astronomic parameters (position and velocity), that you can use for rendering your application as accurately and interestingly as you like.&lt;/p&gt;&#xA;" OwnerUserId="25009" LastEditorUserId="31391" LastEditDate="2020-05-07T07:49:31.547" LastActivityDate="2020-05-07T07:49:31.547" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28554" PostTypeId="1" CreationDate="2018-11-28T11:53:42.450" Score="2" ViewCount="33" Body="&lt;p&gt;I'm aware that our solar system is largely/roughly on a single orbital plane, but haven't other star system's orbital planes been scrambled by billions of years of interaction? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so - how much chance is there that an exoplanetary system's orbital plane is inclined favourably enough to witness the dimming of a transiting planet across its sun? &lt;/p&gt;&#xA;" OwnerUserId="4255" LastEditorDisplayName="user1569" LastEditDate="2018-11-28T13:11:21.240" LastActivityDate="2018-11-28T13:11:21.240" Title="Orbital plane evolution and planet discovery via transit method" Tags="&lt;exoplanet&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2018-12-03T11:46:43.103" ContentLicense="CC BY-SA 4.0" />
  <row Id="28555" PostTypeId="1" AcceptedAnswerId="28593" CreationDate="2018-11-28T15:10:07.363" Score="1" ViewCount="90" Body="&lt;p&gt;I was reading about &lt;a href=&quot;https://en.wikipedia.org/wiki/NGC_5866&quot; rel=&quot;nofollow noreferrer&quot;&gt;NGC 5866&lt;/a&gt;, sometimes referred to as the Spindle Galaxy. It's notable because of its inclination relative to Earth; we see the disk from the side, rather than face-on. The difference is akin to viewing a person's head in profile, instead of looking at their face from the front.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do lose some information about the morphology of NGC 5866 because of the unusual orientation, as it's difficult to see any spiral arms, bars, or other structures. That said, I'm wondering whether we can still learn more about the galaxy than if we viewed it face-on. For instance, can we gain information about the density distribution of the galaxy at various distances removed from the galactic plane, or perhaps about the structure of its central bulge (if there is one)?&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2018-12-01T18:49:23.287" Title="Can we learn anything from observing galaxies edge-on, like the Spindle Galaxy?" Tags="&lt;galaxy&gt;&lt;observational-astronomy&gt;&lt;inclination&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28556" PostTypeId="1" AcceptedAnswerId="28559" CreationDate="2018-11-28T15:23:05.223" Score="3" ViewCount="339" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/28503/what-is-the-underwater-temperature-of-europa&quot;&gt;What is the underwater temperature of Europa?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know Jupiter emits infrared light but does it emit any light able to support plant life of any type on near by Moons? Can under water volcanic activity of Europa produce enough light to support plant life?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TaenC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TaenC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-11-28T21:12:56.277" LastActivityDate="2018-11-28T21:12:56.277" Title="Is there enough light to grow plants in the ocean of Europa?" Tags="&lt;light&gt;&lt;jupiter&gt;&lt;natural-satellites&gt;&lt;life&gt;&lt;europa&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28557" PostTypeId="1" AcceptedAnswerId="28558" CreationDate="2018-11-28T16:46:05.503" Score="2" ViewCount="132" Body="&lt;p&gt;If there were an observer on a planet in the centre of a ring galaxy (omitting the galactic core) what could they expect to see in their night sky? I am currently under the impression that their night sky would be divided by a dense band of stars stretching across the celestial sphere, similar to the milky band of our sky - though to a greater extent. But is there any reason this wouldn't be the case? &lt;/p&gt;&#xA;" OwnerUserId="25010" LastActivityDate="2018-11-28T18:02:28.253" Title="What would the night sky look like from a planet at the centre of a ring galaxy?" Tags="&lt;galaxy&gt;&lt;observational-astronomy&gt;&lt;night-sky&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28558" PostTypeId="2" ParentId="28557" CreationDate="2018-11-28T18:02:28.253" Score="2" Body="&lt;p&gt;It would perhaps be less impressive than you imagine, and not much different from how any spiral galaxy would look from near the centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The many stars near the centre would mean that there were lots of foreground stars and dust near the centre would obscure parts of the ring. The stars of the bulge would mean the whole sky was filled with star fields.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ring itself would be distant and so quite dim, it would look like a &quot;milky way&quot;, but without a central bulge. So it might look something like the milky way in Auriga. Nice, but not especially dramatic.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-11-28T18:02:28.253" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28559" PostTypeId="2" ParentId="28556" CreationDate="2018-11-28T21:01:31.543" Score="3" Body="&lt;p&gt;The surface of Europa gets some sunlight direct from the sun and some long wave radiation from Jupiter so it's not in darkness.  It's also in &lt;a href=&quot;https://www.astrobio.net/news-exclusive/hiding-from-jupiters-radiation/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jupiter's radiation belt&lt;/a&gt; so it gets hit with high energy charged particles and the surface of Europa, despite being bombarded by particles is still a vacuum.  The surface of Europa is a terrible place to look for life, with the exception being that the liquid ocean of Europa explodes onto it's surface regularly, so evidence of life, if it exists, may well be found on the surface, but it's unlikely that anything can live there for long.  It's very inhospitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inside of Europa is more interesting.  That's why Neil deGrass Tyson says he wants to go &quot;&lt;a href=&quot;https://www.businessinsider.com/neil-degrasse-tyson-on-colbert-late-show-2015-10&quot; rel=&quot;nofollow noreferrer&quot;&gt;ice fishing on Europa&lt;/a&gt;&quot; to see what &quot;licks the camera&quot; and if we find life there we'll &quot;Call them Europeans&quot; . . . funny comments aside, the inside of Europa is probably dark, outside of some infra-red rays due to temperature.   Photosynthesis doesn't work well in darkness and with just long wave light, though maybe it's possible, it wouldn't nearly as much growth due to far lower energy levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted &lt;a href=&quot;https://astronomy.stackexchange.com/questions/28503/what-is-the-underwater-temperature-of-europa&quot;&gt;in the answer&lt;/a&gt; to your other question, if there is life in Europa's ocean, it's more likely to be something like &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Chemotroph&quot; rel=&quot;nofollow noreferrer&quot;&gt;chemotrophs&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes Europa interesting is the energy it gets from tidal forces being stretched between Jupiter and Ganymede, so it gets a regular influx of heat and perhaps internal volcanism, so it may have undersea vents similar to our oceans.   The energy influx and possibility of underwater volcanoes makes it more interesting than most other moons or dwarf planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If life is found in Europa's oceans, even if it's very primitive life by most standards, it would still be a very cool discovery because it could tell us something about how life starts and that's a big unknown right now.  The discovery of life outside the Earth would be a very big story.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll probably have to &lt;a href=&quot;https://en.wikipedia.org/wiki/Europa_Lander_(NASA)&quot; rel=&quot;nofollow noreferrer&quot;&gt;wait till 2030&lt;/a&gt; before we begin to get some answers.  Sorry for the long wait, but it is what it is.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-11-28T21:01:31.543" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28560" PostTypeId="1" AcceptedAnswerId="28561" CreationDate="2018-11-28T21:18:47.413" Score="5" ViewCount="1066" Body="&lt;p&gt;How different would being under water in Europa differ from being under water on Earth? Would it be possible to dive deeper on Europa with a submarine than on Earth and by how much?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mLYOy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mLYOy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-11-29T16:15:07.800" LastActivityDate="2018-12-01T07:28:03.117" Title="How does gravity affect pressure under water?" Tags="&lt;gravity&gt;&lt;astrophysics&gt;&lt;planetary-atmosphere&gt;&lt;europa&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28561" PostTypeId="2" ParentId="28560" CreationDate="2018-11-28T21:44:32.433" Score="11" Body="&lt;p&gt;For a liquid, hydrostatic pressure is &lt;span class=&quot;math-container&quot;&gt;$\rho g h$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; is density (this is always the same for all water) g is gravitational acceleration and h is depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational acceleration on Europa is 1.3 &lt;span class=&quot;math-container&quot;&gt;$\text{m/s}{}^2$&lt;/span&gt; (compared with 9.8 &lt;span class=&quot;math-container&quot;&gt;$\text{m/s}{}^2$&lt;/span&gt; on Earth). But on Europa there is 20km of ice floating on the water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a rough estimate, the gravity in Europa is 1/10 that on Earth, so the pressure at the bottom of the ice/top of the water would be comparable to the pressure at 2km in the Earth's Oceans. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we can build submarines that can operate at 10km, we could reasonably suppose that we could operate submarines down to about 100km on Europa, which takes us to the rocky surface. However we have no way to pass through 20km of ice crust to reach the water layer. Building the submarine may be the easy part!&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="7982" LastEditDate="2018-12-01T07:28:03.117" LastActivityDate="2018-12-01T07:28:03.117" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="28562" PostTypeId="1" AcceptedAnswerId="28563" CreationDate="2018-11-28T22:16:36.300" Score="4" ViewCount="422" Body="&lt;p&gt;I've heard of a few moons in the solar system that have water, like europa and enceladus. However, I can't find information about whether there are other moons as well. I'm guessing there are more, but I'm not totally sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many moons in the solar system have water (either liquid or frozen)?&lt;/p&gt;&#xA;" OwnerUserId="24758" LastActivityDate="2018-12-04T13:45:00.870" Title="How many moons in the solar system have water?" Tags="&lt;solar-system&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28563" PostTypeId="2" ParentId="28562" CreationDate="2018-11-28T23:10:18.817" Score="9" Body="&lt;p&gt;It's probably easier to consider the reverse question: which ones don't have water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Talking about the major satellites of the planets, there are only two which are not ice moons: our own Moon, which has &lt;a href=&quot;https://www.nasa.gov/feature/ames/ice-confirmed-at-the-moon-s-poles&quot; rel=&quot;nofollow noreferrer&quot;&gt;ice deposits at the poles&lt;/a&gt;, and Io. As far as I am aware, the evidence for water on Io is tentative at best (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2004Icar..169..175D&quot; rel=&quot;nofollow noreferrer&quot;&gt;Douté et al. 2004&lt;/a&gt;), so of the major planetary satellites, the answer is all of them have water (usually in the form of ice, but some of them also have liquid subsurface oceans) with the possible exception of Io.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The small satellites of the giant planets are probably icy but I'm not sure the possibility that one or two of them are dry rocky asteroids that have been captured has been fully excluded. Phobos and Deimos have rocky surfaces but the possibility of ice in the interior of the Martian moons has not been ruled out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are probably lots of ice-free asteroid moons but finding out which ones they are would require going there and checking whether there's any ice in the subsurface, so confirming this is not currently possible. Small solar system bodies in the outer system are probably going to be icy, so dwarf planet satellites (Charon, Vanth, etc.) all likely contain ice.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2018-12-02T18:31:58.410" LastActivityDate="2018-12-02T18:31:58.410" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28564" PostTypeId="2" ParentId="1013" CreationDate="2018-11-29T00:40:27.810" Score="0" Body="&lt;p&gt;There are many different ways to measure temperature of an astronomical object. Typically, effective temperature means just simply a blackbody temperature. However, the blackbody model is just the first-order approximation that we know it is inaccurate in many circumstances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a nice spectrum from a wide wavelength, your effective temperature might be better to be defined as the excitation temperature. However, what definition you should use really depends on what context you are in. Check this for a short summary: &lt;a href=&quot;https://www.physics.byu.edu/faculty/christensen/Physics%20427/FTI/Measures%20of%20Temperature.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.physics.byu.edu/faculty/christensen/Physics%20427/FTI/Measures%20of%20Temperature.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-11-29T00:40:27.810" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28566" PostTypeId="1" AcceptedAnswerId="28567" CreationDate="2018-11-29T07:27:43.517" Score="16" ViewCount="2881" Body="&lt;p&gt;I am not completely sure whether the Parallax is a &quot;technique to measure the star distances&quot; or a &quot;tiny shift in star's position&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The a book says two points about Parallax:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Astronomers have developed a variety of clever &lt;strong&gt;techniques&lt;/strong&gt; for measuring the vast distances to stars, &lt;strong&gt;one is called Parallax&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Astronomers can measure a star's position once and then again 6 months later and calculate the apparent change in position. &lt;strong&gt;This tiny shift in stars position is called its parallax&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Apart from the book source, &lt;a href=&quot;https://whatis.techtarget.com/definition/parallax&quot; rel=&quot;noreferrer&quot;&gt;whatis.techtarget.com&lt;/a&gt; defines Parallax as &lt;strong&gt;the way an object’s position or direction seems to change depending on viewing angle.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="23422" LastEditorUserId="7472" LastEditDate="2018-11-29T21:16:19.670" LastActivityDate="2018-11-29T21:16:19.670" Title="Is parallax a &quot;technique to measure the star distances&quot; or a &quot;tiny shift in star's position&quot;?" Tags="&lt;star&gt;&lt;distances&gt;" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28567" PostTypeId="2" ParentId="28566" CreationDate="2018-11-29T07:48:32.563" Score="20" Body="&lt;p&gt;It is both - a small shift of the position of a star on the sky &lt;em&gt;as we see it&lt;/em&gt;, and a means of estimating the distance to the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The apparent position (with respect to very distant objects like quasars) changes because our viewing point changes as the Earth moves around the Sun in its orbit. The amount by which the position changes is inversely proportional to is distance.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-11-29T07:48:32.563" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28568" PostTypeId="2" ParentId="28566" CreationDate="2018-11-29T07:53:37.630" Score="8" Body="&lt;p&gt;It's all of that! If you look at an object with just your left, and then with just your right eye, you'll see its position slightly differently. The angle that your lines of sight form (from one eye to the object, and from the other), divided by two, is called &quot;parallax angle&quot;. With trigonometry, one can calculate the distance to the object by knowing this angle and the distance between your eyes. Instead of the eyes, astronomers typically use the distance from the Earth to the Sun, by observing when the Earth is at diametrically opposite points during its orbit (approximately).&#xA;See: &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Parallax&quot; rel=&quot;noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Parallax&lt;/a&gt; under &quot;distance measurement&quot; for more details.&lt;/p&gt;&#xA;" OwnerUserId="24014" LastActivityDate="2018-11-29T07:53:37.630" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28569" PostTypeId="1" AcceptedAnswerId="28579" CreationDate="2018-11-29T10:09:05.177" Score="0" ViewCount="90" Body="&lt;p&gt;One of ideas of how &lt;a href=&quot;https://en.wikipedia.org/wiki/Breakthrough_Starshot&quot; rel=&quot;nofollow noreferrer&quot;&gt;Breakthrough Starshot&lt;/a&gt; could communicate data from Proxima Centauri is to have a massive swarm of the solar sail powered craft to pivot (with their sails) synchronously, taking either 'sideways' or 'head on' orientation to the star - in effect either obstructing it or revealing for an observer 'behind' - on Earth. If the swarm is big enough, it would cause an observable dimming of the star - and timing the pivots right it could encode data in the star's visible luminosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary question of viability of this idea lies in 'how big a swarm?'. What percentage, or what absolute area or a star like Proxima would need to be occluded for the effect to be observable? We can notice transition of exoplanets quite reliably so that would be the upper bound, but what's the smallest object area that would make a detectable difference?&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2018-11-30T08:30:14.247" Title="What area of Proxima Centauri would need to be occluded for the effect to be observable on Earth?" Tags="&lt;star&gt;&lt;observational-astronomy&gt;&lt;luminosity&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28570" PostTypeId="1" AcceptedAnswerId="28572" CreationDate="2018-11-29T11:17:56.167" Score="1" ViewCount="533" Body="&lt;p&gt;I want to know how to calculate the rising time of a star, if &#xA;culmination and setting times are available with me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say Culmination time is 03:00 on 28-Nov-2018&#xA;and Setting time is 05:30 on 28-Nov-2018.&lt;/p&gt;&#xA;" OwnerUserId="25018" LastActivityDate="2018-11-30T07:03:18.403" Title="calculating rising time from culmination time" Tags="&lt;time&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28571" PostTypeId="2" ParentId="28570" CreationDate="2018-11-29T12:05:19.350" Score="2" Body="&lt;p&gt;From the book &quot;Fundamental Astronomy&quot; by H. Karttunen et al. we have the following expression (to calculate the hour angle when the altitude is &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\cos(h) = -\tan(\delta)\tan(\phi)$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt; is the latitude, &lt;span class=&quot;math-container&quot;&gt;$\delta$&lt;/span&gt; the&#xA;  declination)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;Since &lt;span class=&quot;math-container&quot;&gt;$\Theta = \alpha + h$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$\Theta$&lt;/span&gt; is the Mean Sidereal Time, &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; the right ascension), we can use this formula to calculate rising and setting times. &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\Theta$&lt;/span&gt; matches GMT at the time of the Autumnal Equinox (roughly, and we're talking about Mean Sidereal Time), and every sidereal day is roughly 23h 56min so if you know the time of the observation, you can approximate the rising/setting time.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This of course does not take refraction, precession and other effects into account, which I assumed you do not need since you gave rather vague culmination times (stars rarely culminate around such &quot;exact&quot; times).&lt;/p&gt;&#xA;" OwnerUserId="24014" LastEditorUserId="7472" LastEditDate="2018-11-30T07:03:18.403" LastActivityDate="2018-11-30T07:03:18.403" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28572" PostTypeId="2" ParentId="28570" CreationDate="2018-11-29T17:11:26.457" Score="4" Body="&lt;p&gt;For a star, the time from rising to culmination is the same length as the time from culmination to setting. From knowing the time of culmination and setting, you can calculate that difference and subtract it from the time of culmination to obtain the time of rising.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method is generally accurate for the planets and Sun, but it is not always accurate for the Moon. Culmination refers to when an object is highest in the sky. For a star, this is when the object is on the meridian. For the Moon, there can be a significant difference between the time the Moon crosses the meridian (transits) and when it culminates. See &lt;a href=&quot;http://www.geoastro.de/TransitCulm/index.html&quot; rel=&quot;noreferrer&quot;&gt;Transit and Culmination&lt;/a&gt; (although the images on that site were not working today).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I have never seen tables that give the time of culmination for the Sun and Moon. Times are generally given for transits (when the object is on the meridian).&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2018-11-29T17:11:26.457" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28573" PostTypeId="1" CreationDate="2018-11-29T18:15:34.077" Score="9" ViewCount="319" Body="&lt;p&gt;Would it be possible that Mercury had a natural satellite but the gravitational pull from the Sun became so overwhelming from it growing over the years that it just sucked in Mercury's moon?&lt;/p&gt;&#xA;" OwnerUserId="25024" LastEditorUserId="7472" LastEditDate="2018-11-29T21:16:47.610" LastActivityDate="2018-11-30T20:25:02.940" Title="Natural satellites for Mercury" Tags="&lt;natural-satellites&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28574" PostTypeId="2" ParentId="28573" CreationDate="2018-11-29T18:48:28.047" Score="6" Body="&lt;p&gt;The larger an object is and the further away it is from the Sun, the greater its &lt;a href=&quot;https://wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hill Sphere&lt;/a&gt; is. Mercury's is very small (only about &lt;span class=&quot;math-container&quot;&gt;$175000km$&lt;/span&gt; in radius from what @userLTK said) and its &quot;moon&quot; would therefore be yanked away by the Sun relatively quickly. See &lt;a href=&quot;https://planetpailly.com/2015/02/13/why-doesnt-mercury-have-a-moon-sciency-words-hill-sphere/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; for more details. &lt;br&gt; The orbit of this moon would be highly unstable and therefore unpredictable, so it may hit the Sun.&lt;/p&gt;&#xA;" OwnerUserId="24014" LastEditorUserId="24014" LastEditDate="2018-11-30T20:25:02.940" LastActivityDate="2018-11-30T20:25:02.940" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28575" PostTypeId="2" ParentId="28566" CreationDate="2018-11-29T19:12:19.027" Score="4" Body="&lt;h1&gt;It is both&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;But, specifically, the small motion is what we observe of the star at opposing points of Earth's orbit around the Sun. Then, because we know the radius of that orbit (1 AU) we can use the &lt;em&gt;parallax&lt;/em&gt;, the observed small angular change in the location of a star against a distant background between opposite sides of the Earth's orbit, to make a &lt;a href=&quot;https://en.wikipedia.org/wiki/Parallax#Distance_measurement&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;parallax&lt;/em&gt; estimate&lt;/a&gt; of the star's distance, which is the inverse of small angular change (in radians) in units of &lt;a href=&quot;https://en.wikipedia.org/wiki/Parsec&quot; rel=&quot;nofollow noreferrer&quot;&gt;parsecs&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fLoXB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fLoXB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14989" LastActivityDate="2018-11-29T19:12:19.027" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28576" PostTypeId="2" ParentId="28566" CreationDate="2018-11-29T20:46:46.530" Score="3" Body="&lt;p&gt;It is really just imprecise language in the first example.&lt;/p&gt;&#xA;&lt;p&gt;Parallax is the change in apparent position. To be formally correct, you &lt;em&gt;measure&lt;/em&gt; or &lt;em&gt;observe&lt;/em&gt; parallax and use it, with trigonometry, to determine the distance.&lt;/p&gt;&#xA;&lt;p&gt;You can call this the parallax &lt;em&gt;method&lt;/em&gt; of distance measurement. What you have found is that people often omit the word &amp;quot;method&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;More precise language would be either:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Astronomers have developed a variety of clever techniques for measuring the vast distances to stars. One uses an effect called parallax.&lt;/p&gt;&#xA;&lt;p&gt;Astronomers have developed a variety of clever techniques for measuring the vast distances to stars, one is by measuring parallax.&lt;/p&gt;&#xA;&lt;p&gt;Astronomers have developed a variety of clever techniques for measuring the vast distances to stars. The parallax method is one of them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A similar issue comes up with &amp;quot;Doppler&amp;quot;.&lt;/p&gt;&#xA;" OwnerDisplayName="user23052" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-11-29T20:46:46.530" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28577" PostTypeId="1" AcceptedAnswerId="28578" CreationDate="2018-11-29T21:54:46.613" Score="1" ViewCount="112" Body="&lt;p&gt;Just a theory, from &lt;a href=&quot;https://www.youtube.com/watch?v=qsN1LglrX9s&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video&lt;/a&gt;. At last, only the black holes are gonna exist and even they'll decay because of hawkings radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As black holes emit hawkings radiation, then if one emits it, another should absorb it, this will create infinite loop kind of thing &amp;amp; universe is never gonna be empty as blackholes will keep on refilling!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; &lt;em&gt;Can a black hole absorb hawkings radiation from another black hole?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(this question is quite obvious if we think that black holes can absorb anything but how's that gonna make the universe empty)&lt;/p&gt;&#xA;" OwnerUserId="20655" LastActivityDate="2018-11-29T22:39:46.220" Title="Can hawking radiation be eaten by another blackhole?" Tags="&lt;black-hole&gt;&lt;hawking-radiation&gt;&lt;black-body&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28578" PostTypeId="2" ParentId="28577" CreationDate="2018-11-29T22:14:08.567" Score="4" Body="&lt;p&gt;Black holes can &quot;eat&quot; Hawking radiation, but the radiation has to be directed towards the black hole before it can be &quot;eaten&quot;. Black holes are pretty small, and space is very very very big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as a black hole evaporates, and gives off&quot; Hawking radiation, some may eventually fall into another black hole, but most will not. It will just travel out into space, travelling forever (or until it is red-shifted to such low energy that it loses an individual identity - the so-called &quot;heat death&quot;).  We can't picture this because our brains are not equipped to deal with the massive amounts of time. But on these time scales there is nothing very mysterious about a black hole evaporating and so vanishing. It is no more mysterious than a candle flame going out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a metaphor:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some children in a field, each child is carrying some balls (the children represent black holes, the balls are the mass of the black holes and the field is the universe). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The children are walking away from each other (the represents the expansion of the universe*) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly every minute each child throws a ball in a random direction. (This is Hawking radiation) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the ball passes another child, he catches it and adds it to his collection (Radiation falling into the black hole). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a child runs out of balls, he sits down, he is out of the game and can't catch any more balls (this is the black hole running out of mass, and so evaporating). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the game end?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the balls will never be caught by other children because they are not thrown in the right direction. Eventually, all the children will be sitting down and all the balls will be lost (This is the &quot;heat death&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;sub&gt;This isn't quite right because it is an expansion, not walking from a centre, there is no centre to the universe.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-11-29T22:39:46.220" LastActivityDate="2018-11-29T22:39:46.220" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28579" PostTypeId="2" ParentId="28569" CreationDate="2018-11-30T07:11:19.870" Score="2" Body="&lt;p&gt;The photometric precision achieved by satellites like TESS is around 100 parts per million. The original Kepler mission achieved a bit better on bright stars - maybe 20 ppm, but the question of precision may be moot since stars themselves are quite variable;  McQuillan et al. (2012) &lt;a href=&quot;https://arxiv.org/abs/1111.5580&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/1111.5580&lt;/a&gt; suggest that the floor level of intrinsic variability in M-dwarfs may be as high as 1000 ppm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, let's assume that a rapidly encoded signal at 100 ppm could easily be distinguished from stellar noise and maybe even be generous and say 10 ppm might be possible. That would mean that this fraction of the stellar surface would need to be obscured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radius of the star is 0.15 that of the Sun. So the physical area to be obscured is &lt;span class=&quot;math-container&quot;&gt;$\sim 10^{-5} \pi R^2 = 3.4\times 10^{11}$&lt;/span&gt; m&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The web page referred to in the question suggests that Starshot consists of a thousand space craft with light sail areas of 16 m&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt;. It thus seems to fall short of providing detectable modulation by 7 orders of magnitude (unless I have misunderstood the proposal, which isn't mentioned on that web page). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also going to be the question of the low bit rate and what limited information could be transmitted using such a crude technique. Recall that the spacecraft will be travelling at an appreciable fraction of the speed of light, so won't have long to send their message.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A further point is that I don't think it can be assumed that the trajectory of the spacecraft will be such that they are travelling in a direct straight line &#xA;between the Earth and Proxima Centauri, which are of course in relative motion. &#xA;Thus the spacecraft would approach Proxima Cen &quot;at an angle&quot; as viewed from the position of the Earth some 30 years after the spacecraft were launched, unless the trajectories can be tuned quite carefully en-route.&#xA;Even if they were, the parallax motion of the Earth around the Sun would mean that the spacecraft would not necessarily be aligned be when they reached Proxima Cen, although I suppose the launch time and speed could be arranged so that was the case. However, from what I have read, the idea is just to pepper the inner au of the Proxima Cen system with these 1000 spacecraft and so only a few of them would actually be lined up so that they could obscure the star as seen from Earth.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-11-30T08:30:14.247" LastActivityDate="2018-11-30T08:30:14.247" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28580" PostTypeId="2" ParentId="28573" CreationDate="2018-11-30T09:39:17.400" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Would it be possible that Mercury had a natural satellite but the&#xA;  gravitational pull from the Sun became so overwhelming from it growing&#xA;  over the years that it just sucked in Mercury's moon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let me clarify a point, that the gravitational pull of a star doesn't &quot;suck things in&quot;.   One definition of an orbit is a balance between gravitational pull and tangential velocity.   Mercury orbits the Sun at an average (tangential) velocity of about 47 km/s.   If the Sun did pull a theoretical moon away from Mercury, which is absolutely possible, that Moon would exit Mercury's grasp with a similar tangential velocity to Mercury and it would enter a near-Mercury orbit.  It wouldn't fall into the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without getting too off-topic, a near-mercury orbit would be precarious because every time it flew past Mercury, which it would from time to time because the orbits would either cross or pass very close to one another, then with each near pass, it would get a push from Mercury, perhaps seeing an increase in eccentricity, or moving closer to the sun at points in it's orbit where it might begin to vaporize.  There are no known objects &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_Mercury-crossing_minor_planets&quot; rel=&quot;noreferrer&quot;&gt;with a smaller semi-major axis than Mercury&lt;/a&gt;, which implies that near-mercury orbits wouldn't be long-term stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted in the other answer, the region where a moon can continue to orbit a planet is the planet's Hill Sphere, but for long-term orbital stability, it's the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere#True_region_of_stability&quot; rel=&quot;noreferrer&quot;&gt;true region of stability&lt;/a&gt; that matters, which is about 1/2 to 1/3rd of the Hill Radius.  In the case of Mercury, using Mercury's closest point to the Sun of about 46,000 km, it's true region of stability is in the 60,000 km to 85,000 km distance.   It's not an exact number, but inside that distance, a moon could theoretically orbit Mercury and be safe from the Sun's pull.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's other potential problems though.   Mercury isn't a perfect sphere, it's somewhat lumpy and a lumpy planet is less good for stable satellites.  This lumpyness is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Mass_concentration_(astronomy)&quot; rel=&quot;noreferrer&quot;&gt;Mass Concentrations or Mascons&lt;/a&gt;    NASA discovered that when they &lt;a href=&quot;https://www.space.com/21364-moon-gravity-mascons-mystery.html&quot; rel=&quot;noreferrer&quot;&gt;tried to orbit two satellites around the Moon&lt;/a&gt;.   Our moon is unusually unballanced, so it's especially bad for orbits.   Mercury is much more symmetrical than the Moon, so that's less of a problem, but it does have a heavy side and a lighter side which keeps it tidally locked in a 3/2 spin/orbit resonance, so it's not ideal for close orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another, potentially even bigger problem is that with a close orbiting Moon around a planet tends to be drawn in towards the planet due to creating a tidal bulge and tidal forces.   Mercury has a very slow rotation, so any moon it might have had would almost certainly orbit ahead of it's tidal bulge and that interaction would draw the Moon closer to the planet over time.   This is happening &lt;a href=&quot;https://www.space.com/20346-phobos-moon.html&quot; rel=&quot;noreferrer&quot;&gt;with Mars and Phobos&lt;/a&gt;.  Phobos is also very small.  A moon of more size would create a larger bulge in the planet which would cause it to spiral down faster.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth noting that Mercury may, when it was much younger, have had a fast rotation that was slowed down over time by the sun, so there's some wiggle room on this point if you allow for a faster rotation in the distant past, but today, Mercury would have some trouble holding a Moon because it doesn't have a good window between too close and too far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nobody knows if Mercury ever had a moon, but there's some reason to believe it didn't.   The 3 methods from which a planet can get a moon, Formation, Impact or Capture are all somewhat problematic for Mercury.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formation moons require a fair amount of space where the planet forms, a lot of material and the most important factor, a considerable amount of angular momentum.  There are no known rock planet formation moons and they would probably require a more massive planet to have a chance of forming at all.   Mercury is too small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Impact moons are also rare because to create an impact moon, the impact has to be very large and perhaps also hit the planet at the right angle.   There's no such thing (to my knowledge) of a small impact moon.   Individual bits of debris blown off a planet by impact would follow an elliptical orbit back to the planet.   You need a critical mass of debris before the debris has enough of it's own gravitational field to coalesce.  For that reason, impact of sufficient size and proper angle of impact are rare and the Moons, quite large relative to their planets.  Our Moon and Charon are the only two known impact moons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Captured moons tend to be small asteroids but could in theory be larger bodies.   The difficulty with capture is that it requires a 3 body interaction.   Mercury is deep in the Sun's gravity well, where objects orbit very fast and there's no significant 3rd body to assist with capture, and on top of that, Mercury's Hill sphere is small, so Mercury capturing a moon is unlikely.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth occasionally and temporarily &lt;a href=&quot;https://www.space.com/15151-earth-multiple-moons-asteroids.html&quot; rel=&quot;noreferrer&quot;&gt;captures a Moon&lt;/a&gt;, but Earth's Moon can act as a 3rd body, assisting in the capture, and then assisting in the later ejection of the same body and I use the word Moon pretty loosely in this case, as these are mostly small and very temporary bodies.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars is thought to have two captured asteroids/moons, Deimos and Phobos, but Mars is a fair bit further from the Sun and closer to the Asteroid belt.  The point is that capture is rare.   Sure, Jupiter has dozens of them, but Jupiter is a gravitational monster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it's not just keeping a moon, but getting a moon in the first place can be a rare event for a rocky planet close to the sun.   That's not your question, but I thought I'd put it out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't want to say never, because when the solar-system was very young, during the proto-planet stage or the late heavy bombardment . . . Mercury might have, theoretically had a moon by capture or impact, and if that was the case early enough, there might be no evidence of it today, especially since Mercury &lt;a href=&quot;https://www.sciencedaily.com/releases/2018/04/180427100306.htm&quot; rel=&quot;noreferrer&quot;&gt;may have lost a significant percentage of it's crust&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if Mercury had a moon today and if that Moon was in the right orbit, close to the outer edge of it's true region of stability, such a moon might orbit the planet for some time, perhaps millions of years, and it wouldn't get pulled away by the Sun, because it would more likely slowly spiral into Mercury eventually breaking up into a rocky ring system inside the Roche Limit, and over time, falling into the planet with some of the smaller bits of debris being deflected away from Mercury by the solar wind, radiative pressure and/or the &lt;a href=&quot;https://en.wikipedia.org/wiki/Poynting%E2%80%93Robertson_effect&quot; rel=&quot;noreferrer&quot;&gt;Poynting-Robertson effect&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury window to have a stable or somewhat long term moon is the smallest of the 8 known planets in our solar-system.  Further away from the Sun it becomes much easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope I didn't drift too off course in my answer.  Pun intended.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-11-30T09:39:17.400" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28582" PostTypeId="1" AcceptedAnswerId="28583" CreationDate="2018-11-30T12:26:06.160" Score="2" ViewCount="73" Body="&lt;p&gt;All I know is its main mirror diameter; 2.40 m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know that the eye, which its pupil is 5mm in diameter can see up to an apparent magnitude of 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate the apparent magnitude that the telescope can see?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have used the equation of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$5⋅log(\frac{D_1}{D_0})$&lt;/span&gt;&#xA;However, it gave the answer of 13.40, while the answer should be 19.40&lt;/p&gt;&#xA;" OwnerUserId="25034" LastEditorUserId="25034" LastEditDate="2018-11-30T13:35:29.203" LastActivityDate="2018-11-30T13:48:29.610" Title="What is the apparent magnitude of the weakest stars we can see in one telescope on the ground with a diameter of 2.4m?" Tags="&lt;astrophysics&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2018-12-09T20:25:26.483" ContentLicense="CC BY-SA 4.0" />
  <row Id="28583" PostTypeId="2" ParentId="28582" CreationDate="2018-11-30T13:48:29.610" Score="4" Body="&lt;p&gt;You just need to add the 13.4  that you got to the 6 that you can see without a telescope and that's it.  The formula is telling you the difference in magnitude that arises from the different apertures of the eye and the telescope.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2018-11-30T13:48:29.610" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28584" PostTypeId="1" AcceptedAnswerId="28586" CreationDate="2018-11-30T14:35:46.600" Score="16" ViewCount="4726" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bN6Op.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bN6Op.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relative to this simple drawing, can someone explain to me,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;in which direction is the sun and its solar system travelling,&lt;/li&gt;&#xA;&lt;li&gt;which star (irrespective of distance) is closest to that direction,&lt;/li&gt;&#xA;&lt;li&gt;and in which direction the milky way center is.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="25038" OwnerDisplayName="Alonda" LastEditorUserId="19049" LastEditDate="2018-11-30T23:22:47.783" LastActivityDate="2020-03-28T05:04:52.300" Title="In which direction is the sun travelling?" Tags="&lt;the-sun&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28585" PostTypeId="2" ParentId="28584" CreationDate="2018-11-30T15:52:52.383" Score="4" Body="&lt;p&gt;1) and 2) The direction that the Sun is moving is known as the Solar Apex. This Wikipedia article shows a star map, so you can decide which star you want to say is closest. &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_apex&quot; rel=&quot;noreferrer&quot;&gt;Solar Apex&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The center of the Milky Way is toward the constellation of Sagittarius.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2018-11-30T15:52:52.383" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28586" PostTypeId="2" ParentId="28584" CreationDate="2018-11-30T16:04:23.813" Score="11" Body="&lt;p&gt;The direction of solar motion is referred to as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_apex&quot; rel=&quot;noreferrer&quot;&gt;solar apex&lt;/a&gt;. It is at around RA=18h28m and Dec=+30d in the constellation of Hercules (you can see a map on the page linked to). Note that this is the direction in which you would see the &quot;star wars&quot; effect of onrushing stars diverging from the apex point. i.e. It demonstrates the sun's motion &lt;em&gt;with respect to the local stellar population&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On your 2-dimensional cartoon of the Earth's, the circle is the ecliptic plane and the Earth's northern hemisphere (at winter solstice) is tilted away from the Sun at an angle of  23 degrees. The celestial coordinate system defined above is geocentric and the Sun is at 18h -23d at the winter solstice. The apex point is therefore towards  the Sun but 57 degrees above it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the map below (produced by Christian Ready), you can judge where the solar apex is with respect to the Sun and ecliptic plane (green line), which is roughly the path (drawn in your diagram) that the Earth takes around the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZvgGt.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZvgGt.png&quot; alt=&quot;Celestial sphere&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The very bright star Vega is reasonably close to the solar apex direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Milky way centre is at RA=17h45m, Dec=-29d in the constellation of Sagittarius - i.e. almost where the Sun is in December. In Galactic coordinates (see &lt;a href=&quot;https://astronomy.stackexchange.com/questions/12310/best-approximation-for-suns-trajectory-around-galactic-center/12311#12311&quot;&gt;Best approximation for Sun&amp;#39;s trajectory around galactic center?&lt;/a&gt; ) the Sun moves at about 10 km/s towards the Galactic centre, about 5 km/s faster than the average circular speed (which is itself about 220 km/s) of local stars in the tangential direction and 7 km/s upwards out of the Galactic plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the motion of the Sun &lt;em&gt;with respect to the Galactic centre&lt;/em&gt;, as opposed to the local stellar population, is basically at &lt;span class=&quot;math-container&quot;&gt;$\sim 230$&lt;/span&gt; km/s towards a point that is slightly less than 90 degrees around the Galactic plane and slightly above it. This would be at 9h30m -50d in the constellation of Vela. I think that is into the page and down about 30 degrees in your cartoon.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-12-01T07:43:28.583" LastActivityDate="2018-12-01T07:43:28.583" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="28588" PostTypeId="2" ParentId="28584" CreationDate="2018-11-30T23:31:23.103" Score="9" Body="&lt;p&gt;Relative to your drawing, the sun is moving south-ish:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ly8dr.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ly8dr.gif&quot; alt=&quot;solar system moving through space&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solar system is tilted about 60 degrees of the milky way plane, on which it's moving around like everyone else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=IJhgZBn-LHg&amp;amp;t=1015&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is another example/explanation.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To know which star is closest, you'd have to define what &quot;closest&quot; mean. Are we setting a straight line and seeing if it hits any star? If a start is 50 light years away from that straight line and 100 light years away from the sun, is that closer than a start that's 5 light years away from line but 800 light years away from the sun? The video above says we're moving towards Vega, I don't know how satisfactory this information is to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The milky way center is to the right of the path the sun is moving on. But I don't know the location when earth is at winter solstice, and you didn't specify northern hemisphere's winter or southern hemisphere's. ;-)&lt;/p&gt;&#xA;" OwnerUserId="25041" LastEditorUserId="25041" LastEditDate="2020-03-28T05:04:52.300" LastActivityDate="2020-03-28T05:04:52.300" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28589" PostTypeId="1" CreationDate="2018-12-01T11:44:13.940" Score="1" ViewCount="84" Body="&lt;p&gt;Currently we observe that the universe is expanding. And this expansion even goes faster and faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do we know (or presume) that this expansion started slowly at a singularity (i.e. big bang) and is getting faster and faster?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did a serious scientist ever presumed that the universe might be oscillating, i.e. like a breathing bubble? &lt;strong&gt;Currently&lt;/strong&gt; the universe is expanding but who knows, perhaps in 1 Million year we will observe a shrinking universe and 3 Million year later again expanding - provided the humans will still exist at this time which I doubt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this would open many new questions like: How and when did it start? What is the frequency and amplitude? But let's keep this question off side.&lt;/p&gt;&#xA;" OwnerUserId="1838" LastEditorUserId="1838" LastEditDate="2018-12-01T19:27:30.133" LastActivityDate="2018-12-01T19:27:30.133" Title="Is the universe always expanding" Tags="&lt;universe&gt;&lt;expansion&gt;&lt;fate-of-universe&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28590" PostTypeId="2" ParentId="28589" CreationDate="2018-12-01T14:10:15.213" Score="1" Body="&lt;p&gt;It will always expand (probably).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass and energy in the universe curves space, and it seems that there is exactly the right amount of mass and energy to keep the universe expanding forever. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Had you asked this question 20 years ago the answer would have been &quot;we don't know&quot;. Recent discoveries of &quot;Dark Energy&quot; suggest that the expansion of the universe is not slowing down, but speeding up. This would mean that the universe is destined to expand forever, at a faster and faster rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The notion that there will be a &quot;Big Crunch&quot; followed by another &quot;Big Bang&quot; was seriously considered by multiple scientists: among them Willem de Sitter and George Gamow. There are problems with this hypothesis: It is not well supported by observations of the rate of expansion. Moreover, there would need to be way to reduce the entropy of the big crush, as entropy can only increase. And moreover any &quot;pre-big bang state&quot; may be hidden behind the singularity at the start of the (current) universe and so fundamentally hidden from us.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-12-01T14:10:15.213" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28592" PostTypeId="1" AcceptedAnswerId="28598" CreationDate="2018-12-01T16:33:10.113" Score="1" ViewCount="153" Body="&lt;p&gt;Question implies all, why is that? Why we exist even after equally creation of matter and anti-matter, any possible exception?&lt;/p&gt;&#xA;" OwnerUserId="23217" LastActivityDate="2018-12-02T17:01:36.080" Title="Why more matter still left after collision with anti-matter?" Tags="&lt;big-bang-theory&gt;&lt;matter&gt;&lt;antimatter&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28593" PostTypeId="2" ParentId="28555" CreationDate="2018-12-01T18:49:23.287" Score="2" Body="&lt;p&gt;Yes, we can certainly learn things from edge-on galaxies. The easiest things to learn, as you suggest, have to to with the vertical structure of the stars (and gas). For example, our knowledge of how thick galaxy stellar and gas disks are -- and whether this thickness varies with distance from the galaxy center -- comes almost exclusively from studying edge-on galaxies. It is also in many cases easier to study faint stellar and gaseous halos in edge-on galaxies, because in face-on galaxies, these things would be superimposed on -- and largely lost in the glare of -- the face-on disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traditionally, it was also easier to study the &lt;em&gt;radial&lt;/em&gt; structure of disks in edge-on systems, because the fact that the line of sight through a disk intersects more stars when the galaxy is edge-on than when it is face-on meant you could make observations at higher signal/noise at large radii. Thus, the detection that many disk are &quot;truncated&quot; (that is, that there is a radius beyond which their brightness falls off more rapidly than it does inside that radius) was first made using photographic data of edge-on spirals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some elements of bulge structure have been traditionally easier to study in edge-on galaxies. The realization that some galaxies had &quot;boxy&quot; or &quot;peanut-shaped&quot; bulges -- along with the existence of &quot;cylindrical&quot; off-plane stellar rotation in such bulges and their likely nature as the vertically thickened inner parts of bars -- came about from observations of edge-on galaxies. (Although we now know how to identify such structures in moderately inclined and even face-on galaxies as well.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that our view of the Milky Way is essentially that of an edge-on galaxy, albeit one seen from the inside.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2018-12-01T18:49:23.287" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28596" PostTypeId="2" ParentId="11467" CreationDate="2018-12-02T14:24:04.740" Score="2" Body="&lt;p&gt;An astronomical telescope can view objects on Earth if they’re far enough away to be focused, and if the telescope can be aimed at them. Most often, the targets will appear to be inverted, but you can get prisms to correct that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space satellites cannot place on Earth they are around the Earth but they can see things on the ground less than a foot across.&lt;/p&gt;&#xA;" OwnerUserId="24526" LastEditorUserId="14094" LastEditDate="2018-12-03T09:10:46.610" LastActivityDate="2018-12-03T09:10:46.610" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28598" PostTypeId="2" ParentId="28592" CreationDate="2018-12-02T17:01:36.080" Score="-3" Body="&lt;p&gt;I presume you're asking about the &lt;a href=&quot;https://www.google.co.uk/search?q=mystery+antimatter&amp;amp;*&amp;amp;cad=h&quot; rel=&quot;nofollow noreferrer&quot;&gt;mystery of the missing antimatter&lt;/a&gt;. This features extensively in the popular and scientific media. For example the CERN webpage on &lt;a href=&quot;https://home.cern/science/physics/matter-antimatter-asymmetry-problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;the matter-antimatter asymmetry problem&lt;/a&gt; is fairly typical. It says the Big Bang should have created equal amounts of matter and antimatter, but everything we see today is made almost entirely of matter. It goes on to say &lt;em&gt;“something must have happened to tip the balance. One of the greatest challenges in physics is to figure out what happened to the antimatter”&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's something of a &quot;clickbait&quot; mystery I'm afraid, to try to curry favour with the paying public. Why do I think that? Because there’s baryon asymmetry, where we see lots of protons but hardly any antiprotons, and there’s also lepton asymmetry where we see lots of electrons but hardly any positrons. And they &lt;em&gt;match&lt;/em&gt;. Take a look at positronium. It’s &lt;em&gt;“a system consisting of an electron and its anti-particle, a positron, bound together into an exotic atom”.&lt;/em&gt; The electron is matter, and the positron is antimatter. Positronium however is neither matter nor antimatter. It’s &lt;em&gt;both&lt;/em&gt;. It’s an exotic atom. Now see &lt;a href=&quot;https://books.google.co.uk/books?id=Q74GCAAAQBAJ&amp;amp;pg=PA314&amp;amp;lpg=PA314&amp;amp;dq=%22light+hydrogen%22&amp;amp;#v=onepage&amp;amp;q=%22light%20hydrogen%22&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;page 314&lt;/a&gt; of &lt;em&gt;photon and electron collisions with atoms and molecules&lt;/em&gt; edited by Philip Burke and Charles Joachain. It says &lt;em&gt;“the positronium atom thus created is in essence a very light hydrogen atom”&lt;/em&gt;. So positronium is both matter and antimatter, and it’s like light hydrogen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clear inference from that is &lt;em&gt;hydrogen is also a combination of matter and antimatter&lt;/em&gt;. We say the electron is matter, and the proton too. But why? Surely since the proton is more like the positron than the electron, we ought to say the proton is the antimatter. Like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RdUNN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RdUNN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That would mean antimatter is a lot more common than people say. The proton mass is 1836 times the electron mass. Which would mean that weight for weight, 99.95% of everything around you is made of antimatter. Where’s all the antimatter gone? Nowhere. Because it’s hiding in plain sight. Because truth be told and by and large, &lt;em&gt;you are made of it&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2018-12-02T17:01:36.080" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="28599" PostTypeId="2" ParentId="11835" CreationDate="2018-12-02T18:31:11.333" Score="2" Body="&lt;p&gt;Going by multiple online exoplanet catalogues and a quick scan through the papers listed on the SIMBAD pages for the host stars:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Gliese 581&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;At the time of writing, the Extrasolar Planets Encyclopaedia (&lt;a href=&quot;http://exoplanet.eu&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://exoplanet.eu&lt;/a&gt;), the NASA Exoplanet Archive (&lt;a href=&quot;https://exoplanetarchive.ipac.caltech.edu&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://exoplanetarchive.ipac.caltech.edu&lt;/a&gt;) and the Exoplanet Orbit Database (&lt;a href=&quot;http://exoplanets.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://exoplanets.org&lt;/a&gt;) are in agreement and list only three confirmed planets for this system: in order of increasing orbital period these are &lt;em&gt;e&lt;/em&gt;, &lt;em&gt;b&lt;/em&gt; and &lt;em&gt;c&lt;/em&gt;. It's not just Robertson et al. claiming that &lt;em&gt;d&lt;/em&gt; is an activity signal, see for example &lt;a href=&quot;https://ui.adsabs.harvard.edu/?#abs/2016A%26A...585A.144H&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hatzes (2016)&lt;/a&gt;. The most recent RV study of the system I have seen is &lt;a href=&quot;https://ui.adsabs.harvard.edu/?#abs/2018A&amp;amp;A...609A.117T/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Trifonov et al. (2018)&lt;/a&gt; who give a 3-planet solution, not sure whether they searched for evidence of a fourth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gliese 581 is known to host a debris disc with an inner edge around 25 AU (&lt;a href=&quot;https://ui.adsabs.harvard.edu/?#abs/2012A%26A...548A..86L&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lestrade et al., 2012&lt;/a&gt;), much further out than &lt;em&gt;c&lt;/em&gt; (or indeed any of the claimed planets beyond &lt;em&gt;c&lt;/em&gt;), so I wouldn't be too surprised if there are additional planets to be found.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Gliese 667 C&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This one's a bit less settled because it doesn't seem to have attracted so many studies. Extrasolar Planets Encyclopaedia lists 6 confirmed planets, the NASA Exoplanet Archive lists 5 (they don't have planet &lt;em&gt;d&lt;/em&gt;), while the Exoplanet Orbit Database only lists 2 (&lt;em&gt;b&lt;/em&gt; and &lt;em&gt;c&lt;/em&gt;). Again it is not just Robertson et al. claiming that there are only two planets, there's also the study by &lt;a href=&quot;https://ui.adsabs.harvard.edu/?#abs/2014MNRAS.437.3540F&quot; rel=&quot;nofollow noreferrer&quot;&gt;Feroz &amp;amp; Hobson (2014)&lt;/a&gt;. The identification of planet &lt;em&gt;d&lt;/em&gt; as being caused by the stellar rotation is supported by &lt;a href=&quot;https://ui.adsabs.harvard.edu/?#abs/2017MNRAS.468.4772S/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Suárez Mascareño et al. (2017)&lt;/a&gt; which is probably why this one was removed from the NASA list. Planet &lt;em&gt;e&lt;/em&gt; is at roughly 2/3 of the rotation period and so is also considered suspicious. There doesn't seem to be nearly as much discussion about the outer candidates &lt;em&gt;f&lt;/em&gt; and &lt;em&gt;g&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps as a summary. More data would definitely be useful for these systems if there is going to be any consensus, particularly for Gliese 667 C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update (June 2019)&lt;/strong&gt;: the &lt;a href=&quot;https://arxiv.org/abs/1906.04644&quot; rel=&quot;nofollow noreferrer&quot;&gt;recent study posted on the arXiv by Tuomi et al.&lt;/a&gt; brings back planet &lt;em&gt;d&lt;/em&gt;, unfortunately they don't discuss this case in detail. Confusingly they also decided to swap the designations of planets &lt;em&gt;d&lt;/em&gt; and &lt;em&gt;e&lt;/em&gt; (at least they did in their tables, they kept the old name elsewhere in their text). They also include the 91-day planet &lt;em&gt;d&lt;/em&gt; at Gliese 667C, also unfortunately without a detailed discussion.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-06-13T19:27:26.060" LastActivityDate="2019-06-13T19:27:26.060" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28600" PostTypeId="1" AcceptedAnswerId="28601" CreationDate="2018-12-02T20:04:59.140" Score="3" ViewCount="170" Body="&lt;p&gt;In the last few weeks, I read a few news that presented new advances in our understanding of the solar system, which implied we don’t really know our neighboorhood that well. I was wondering, how likely is it that we make a significant discovery around us in the next decade? Like discovering a new massive object (moon or even planet etc.), or that a planet area is habitable &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was assuming it’s very unlikely because we have good measures and models (gravity based or other means I guess) but those few papers (sorry I didn’t bookmark them) have me wonder. How much do we really know of our nearest environment, before looking at interstellar stuff? &lt;/p&gt;&#xA;" OwnerUserId="25054" LastActivityDate="2018-12-02T21:45:17.123" Title="Is there likely other significant objects we haven’t discovered in the solar system?" Tags="&lt;planet&gt;&lt;solar-system&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2018-12-11T01:30:27.647" ContentLicense="CC BY-SA 4.0" />
  <row Id="28601" PostTypeId="2" ParentId="28600" CreationDate="2018-12-02T21:45:17.123" Score="5" Body="&lt;p&gt;&lt;strong&gt;Moons&lt;/strong&gt;.  We've explored the planets up to Saturn with probes that remained in orbit for long periods and found moons down to quite small sizes.  While it is certain that there are more small undiscovered moons, these won't be large spherical objects.  Uranus and Neptune have not been visited by orbiting probes, and so there are surely lots of undiscovered moons. Nothing very large, or we could see it from Earth but relatively substantial moons could be orbiting Neptune but be undetected. However there are no planned missions to either Uranus or Neptune in the next 10 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dwarf Planets&lt;/strong&gt;.  There are probably spherical bodies (and hence &quot;dwarf planets&quot;) in the Kuiper belt, in the region of space around Pluto. There could well be further discoveries of dwarf planets here. We are certain there are no more dwarf planets in the asteroid belt. On the other hand we discover new asteroids all the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Major Planets&lt;/strong&gt;.  There is the intriguing suggestion that the orbits of the known Kuiper belt objects are correlated with each other in a way that could be due to a large, Neptune sized planet in orbit in the outer part of the solar system. It is far from certain if such a body exists, but if it does, it could be discovered within a decade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Habitable zones.&lt;/strong&gt;  The only Habitable zone is the one which is approximately 150,000,000 km from the sun.  If you go closer, water boils. If you go further it freezes. Liquid water surely exists under the ice of some of the outer solar system moons, but I would not call them &quot;habitable&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Unknown Unknowns&lt;/strong&gt;.  If you look back 10 years most of the new discoveries could not have been predicted. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-12-02T21:45:17.123" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28603" PostTypeId="1" CreationDate="2018-12-03T01:41:09.950" Score="1" ViewCount="1114" Body="&lt;p&gt;Mars looks like a small white blob/circle and the same with all other planets. This is very frustrating. I've looked all over for a solution jumping from site to site and nothing. Please be different on this site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Many people have marked this post as duplicate. Yes, I know, but the other post did not answer my question.&lt;/p&gt;&#xA;" OwnerUserId="25055" LastEditorUserId="25055" LastEditDate="2018-12-04T02:59:54.870" LastActivityDate="2018-12-04T02:59:54.870" Title="Can't see any detail on planets with my new Celestron NexStar 130 SLT Computerized Telescope" Tags="&lt;planet&gt;&lt;telescope&gt;&lt;amateur-observing&gt;&lt;mars&gt;&lt;saturn&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="0" ClosedDate="2018-12-03T12:34:21.203" ContentLicense="CC BY-SA 4.0" />
  <row Id="28604" PostTypeId="1" CreationDate="2018-12-03T02:20:29.393" Score="2" ViewCount="100" Body="&lt;p&gt;My understanding is that gas and ice giants form outside the &quot;snow line&quot; of their protoplanetary disk, then sometimes through gravitational interactions they are able to migrate into the inner star system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this situation, if the giant planet had moons, what would they be like?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Would the planet be able to bring initial moons with it? Or would they be lost?&lt;/li&gt;&#xA;&lt;li&gt;If they are kept, are they likely to retain an icy composition due to forming outside the snowline?&lt;/li&gt;&#xA;&lt;li&gt;If not, can it form or capture new moons post-migration?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="24644" LastActivityDate="2018-12-03T02:20:29.393" Title="Moons of migrated giant planets: Icy, Rocky, or none at all?" Tags="&lt;natural-satellites&gt;&lt;planetary-formation&gt;&lt;orbital-migration&gt;" AnswerCount="0" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="28605" PostTypeId="2" ParentId="28603" CreationDate="2018-12-03T04:46:54.430" Score="2" Body="&lt;p&gt;Mars is a small planet and currently far from Earth (just over 1 Astronomical Unit). Therefore, it currently appears to be very small -- smaller than 9 arcseconds. At that size, it will look like a small disk. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on this one observation, I think that no conclusion can be made about any possible problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, a 130 mm f/5 scope is rather short for planetary observing where you want high magnification. This scope is better for wide fields of view, such as the Pleiades or Double Cluster.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2018-12-03T04:46:54.430" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28607" PostTypeId="1" CreationDate="2018-12-03T12:48:11.683" Score="3" ViewCount="65" Body="&lt;p&gt;Doing research for a game and trying to understand the implications of the stellar flares from &lt;a href=&quot;https://en.wikipedia.org/wiki/YZ_Ceti&quot; rel=&quot;nofollow noreferrer&quot;&gt;YZ Ceti&lt;/a&gt;. In short (ignoring the planets for now), if people were to build large space stations of various types (O'Neil cylinders, hollowed out asteroids, etc.), do the YZ Ceti flares pose a significant danger to them? If so, what would be the frequency and scale of such danger?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Better still, what reliable sources can I consult and how would I interpret them so I can figure this out on my own? I'm getting tired of searching and finding the &lt;a href=&quot;https://forum.kerbalspaceprogram.com/index.php?/topic/164868-three-planets-for-yz-ceti/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kerbal Space Program forums&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="25058" LastEditorUserId="25058" LastEditDate="2018-12-03T12:57:33.490" LastActivityDate="2018-12-03T12:57:33.490" Title="How strong are the stellar flares on YZ Ceti?" Tags="&lt;fundamental-astronomy&gt;&lt;life&gt;&lt;flare&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28608" PostTypeId="2" ParentId="28362" CreationDate="2018-12-03T20:18:10.903" Score="1" Body="&lt;p&gt;Isn't the point of using type Ia supernovae that they have an estimated distance that is independent of redshift?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;${\bf \hat{n}}$&lt;/span&gt; here is not the direction of the CMB dipole, it is a free parameter in their model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If distances were being estimated using redshift I guess you would have a point.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-12-03T20:18:10.903" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28609" PostTypeId="1" AcceptedAnswerId="28627" CreationDate="2018-12-03T23:39:38.007" Score="1" ViewCount="389" Body="&lt;p&gt;If I know the semi-major axis, the density of the planet, and orbital period.. how can I find the radius of the satellite?  I can't think of any formula to use.&lt;/p&gt;&#xA;" OwnerUserId="24231" LastActivityDate="2018-12-05T04:41:54.847" Title="How to find the radius of a satellite not knowing the mass" Tags="&lt;astrophysics&gt;&lt;fundamental-astronomy&gt;&lt;satellite&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28610" PostTypeId="1" AcceptedAnswerId="30199" CreationDate="2018-12-04T00:20:08.040" Score="2" ViewCount="317" Body="&lt;p&gt;The Ars Technica article &lt;a href=&quot;https://arstechnica.com/science/2018/12/physicists-detected-gravitational-waves-from-four-new-black-hole-mergers/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Physicists detected gravitational waves from four new black-hole mergers&lt;/a&gt; notes that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This brings the total number of events detected by LIGO and Virgo to 11.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It also has a link to the SXS Collaboration' YouTube video &lt;a href=&quot;https://youtu.be/gmmD72cFOU4&quot; rel=&quot;nofollow noreferrer&quot;&gt;LIGO Orrery&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SXS stands for Simulating eXtreme Spacetimes &lt;a href=&quot;https://www.black-holes.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.black-holes.org/&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The SXS project is a collaborative research effort involving multiple institutions. Our goal is the simulation of black holes and other extreme spacetimes to gain a better understanding of Relativity, and the physics of exotic objects in the distant cosmos.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so quite a substantial amount of math and physics went into this work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But as an outsider I don't really understand what I'm looking at.&lt;/strong&gt; The video's comments just say that the surface is the event horizon:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Computer calculations modeling the gravitational waves LIGO has observed to date and the black holes that emitted the waves. The image shows the horizons of the black holes above the corresponding gravitational wave. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What does the color represent, could it be some kind of &quot;temperature&quot; or energy density of distorted space on the event horizon surface? Why can I see some evidence of rotation in the resulting merged object for a few rotations, then it disappears? What else (if anything) can be learned from this video?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Screen shot from the YouTube video &lt;a href=&quot;https://youtu.be/gmmD72cFOU4&quot; rel=&quot;nofollow noreferrer&quot;&gt;LIGO Orrery&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LQzqF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LQzqF.jpg&quot; alt=&quot;LIGO Orrery video screenshot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-04-01T14:07:09.167" LastActivityDate="2019-04-02T06:24:28.343" Title="What can be learned from, or noted in this LIGO Orrery video?" Tags="&lt;black-hole&gt;&lt;general-relativity&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28612" PostTypeId="1" AcceptedAnswerId="28614" CreationDate="2018-12-04T05:15:11.397" Score="1" ViewCount="96" Body="&lt;p&gt;The BBC article &lt;a href=&quot;https://www.bbc.com/news/science-environment-45967655&quot; rel=&quot;nofollow noreferrer&quot;&gt;Archaeopteryx: The day the fossil feathers flew&lt;/a&gt; has nothing to do with Astronomy but it shows a file photo of the famous astronomer &lt;a href=&quot;https://en.wikipedia.org/wiki/Fred_Hoyle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sir Fred Hoyle&lt;/a&gt; with a model of what looks like some kind of telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if the &quot;dish&quot; is supposed to be fixed metal to reflect radio waves, or a rotating dish of mercury to reflect light, but I'm guessing the former is more likely. It looks like an off-axis, tilted secondary mirror reflects radiation back to ground level, but it's just a model so I can't be sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any ideas what real instrument (if any) or at least what design that this fixed-&quot;dish&quot; model might be a model of?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's scary is that at the bottom right it looks like a building several stories tall, which means that the telescope is huge, and the size of Sir Hoyle is on par with Godzilla. The time frame seems right for that too! ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RWALz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RWALz.jpg&quot; alt=&quot;Sir Fred Hoyle with model of some kind of telescope&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-12-04T07:41:38.613" Title="What telescope might this be a model of, with Sir Fred Hoyle?" Tags="&lt;radio-astronomy&gt;&lt;radio-telescope&gt;&lt;identify-this-object&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28614" PostTypeId="2" ParentId="28612" CreationDate="2018-12-04T07:35:51.870" Score="2" Body="&lt;p&gt;The photo looks like it could be from the late 1950s/early 1960s. Jodrell Bank's Lovell telescope had been commissioned and people were looking for the next big thing in radio astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That next big thing was the fixed Arecibo dish, with a diameter of 305m, commissioned in 1963.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear though that the picture is not a (very accurate) model of Arecibo and the location is wrong. I would guess instead that this picture purports to be the design for a radio telescope on the Moon or is an upscaled model of the &lt;a href=&quot;https://csiropedia.csiro.au/radio-astronomy-at-dover-heights/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dover Heights&lt;/a&gt; &quot;sandpit&quot; telescope (see below) built in the 1950s in Australia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zGeXj.jpg&quot; alt=&quot;Dover Heights sandpit telescope&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-12-04T07:41:38.613" LastActivityDate="2018-12-04T07:41:38.613" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28615" PostTypeId="1" CreationDate="2018-12-04T09:56:48.783" Score="1" ViewCount="115" Body="&lt;p&gt;CHIME, the Canadian Hydrogen Intensity Mapping Experiment uses four parabolic cylinder reflectors and a dense linear array of 256 x 2 polarization feeds along each focal line. Each feed is a dual polarization cloverleaf. The axes extend north-south (NS) and the cloverleaves appear to have their resolved polarizations EW and NS as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is the radiation reaching the cloverleaves from a source at the zenith, reflected from one tip of the parabola to the other is spread out over a 180 degree arc since the parabola is f/D of 0.25. The focal plane is the same height as the edges of the parabola.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;imagining the electric field in the NS direction and I imagine a significant cancellation of the electric fields scattered from the east and west tips of the parabolas. If the incident wave is coming down vertically and it's electric field is pointed E, the ray from the east tip may be pointed up, while that from the west tip would be pointed down, so these would tend to cancel leaving the effective aperture smaller for this polarization. (An NS polarized ray would not experience this interference.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would call this a kind of polarization-dependent apodization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why not simply rotate the cloverleaves by 45 degrees so that the two polarizations were matched and there was no polarization-dependent apodization?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yf3Nu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yf3Nu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't found any details about this in &lt;a href=&quot;https://arxiv.org/abs/1803.11235&quot; rel=&quot;nofollow noreferrer&quot;&gt;The CHIME Fast Radio Burst Project: System Overview&lt;/a&gt; nor in the reference below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below x2:&lt;/strong&gt; from &lt;a href=&quot;https://arxiv.org/abs/1708.08521&quot; rel=&quot;nofollow noreferrer&quot;&gt;The cloverleaf antenna: A compact wide-bandwidth dual-polarization feed for CHIME&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/u6J0Q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/u6J0Q.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fig. 1. At left, the shape of each petal consists of two perpendicular straight&#xA;lines, two 45 degree circular arcs with radius R and one half an ellipse. W is&#xA;the major axis of the ellipse and L is the length from the intersection of the&#xA;straight sides to the outer edge of the ellipse. The shape is illustrated here for&#xA;the adopted values of gap, R, L, and W. Each of the four tabs shown at the&#xA;centre is connected to one side of a vertical microstrip transmission line and&#xA;in each case the full width of the adjacent petal is connected to the other lead.&#xA;At right, CST simulated currents for one linear polarization at 600MHz are&#xA;shown. Note the small asymmetry in the curent distribution near the centre&#xA;becasue of the tab geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1Sb4J.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1Sb4J.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fig. 3. A linear array of eight cloverleaf antennas installed at the focal line of&#xA;the CHIME Pathfinder at the Dominion Radio Astrophysical Observatory in Penticton, BC, Canada. The picture is taken through the wire mesh reflective surface (mesh spacing 19 mm) illustrating a photonsview of the antennas and ground plane. Notice that each feed has an image-feed in the ground plane, 1/2 λ away at the passband centre frequency. Notice also the four slots cut to remove dielectric material from the gaps between the petals.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; From &lt;a href=&quot;https://chime-experiment.ca/instrument&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://chime-experiment.ca/instrument&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;CHIME consists of four adjacent 20m x 100m cylindrical reflectors oriented north-south. The focal axis of each cylinder is lined with 256 dual-polarization antennas, each of which receives radiation from a large swath of sky that nearly stretches from the northern horizon to the southern horizon. This gives CHIME its enormous field of view. The CHIME antennas are custom-designed to have good sensitivity from 400 to 800 MHz, in both linear polarizations. This gives CHIME its large frequency coverage. Signals from the antennas are amplified in two stages, using low-noise amplifiers developed by the cell-phone industry. This novel application of consumer technology makes CHIME affordable. The 2048 signals from these inputs (256 antennas x 2 polarizations x 4 cylinders) are fed to the F-Engine for the initial stage of digital processing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NSO9C.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NSO9C.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; From SKA_telescope &lt;a href=&quot;https://twitter.com/SKA_telescope/status/1065277565429010432&quot; rel=&quot;nofollow noreferrer&quot;&gt;tweet&lt;/a&gt; I believe this is looking north.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1cVDT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1cVDT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2021-01-23T23:49:59.713" LastActivityDate="2021-01-23T23:49:59.713" Title="How does Canada's radiotelescope CHIME manage polarization-dependent apodization of the fast f/0.25 optics?" Tags="&lt;radio-astronomy&gt;&lt;radio-telescope&gt;&lt;angular-resolution&gt;" AnswerCount="0" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28617" PostTypeId="2" ParentId="27457" CreationDate="2018-12-04T12:03:46.997" Score="3" Body="&lt;p&gt;I'm pretty sure this is one of the four auxiliary (or &quot;outrigger&quot;) telescopes for the European Southern Observatory's Very Large Telescope Interferometer, part of the VLT complex at Paranal Observatory in Chile. It's used in conjunction with the four main 8 m (&quot;unit&quot;) telescopes in an optical/infrared interferometric system, so its purpose is to send images of the sky in the optical or near-IR to be combined with images from the unit telescopes and the other auxiliary telescopes (via underground tunnels) in the interferometric system. (The octagonal gray frame is kind of a giveaway, and the fact that this video is ESO-based helps confirm this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.eso.org/public/teles-instr/paranal-observatory/vlt/auxiliarytelescopes/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.eso.org/public/teles-instr/paranal-observatory/vlt/auxiliarytelescopes/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that, per the link above, the primary mirror in each auxiliary telescope has a diameter of 1.8 m, while the secondary mirror is only 0.14 m -- very small, as you guessed.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The object you identify as &quot;B&quot; is the tertiary mirror, which is flat, elliptical, and tilted at a 45-degree angle so that light from the secondary mirror is redirected off to the side. It sits in a mount (supported by the &quot;B vanes&quot;) located just &lt;em&gt;in front of&lt;/em&gt; the primary mirror, so there is no need for a hole in the primary mirror. You're probably used to versions of this (Nasmyth) design where the tertiary mirror is located &lt;em&gt;behind&lt;/em&gt; the primary mirror, requiring a hole in the primary. Since the tertiary mirror in a Nasmyth design needs to sit along the elevation rotation axis of the telescope, whether it goes in front of or behind the primary depends on the overall telescope balance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sometimes there &lt;em&gt;is&lt;/em&gt; a hole in the primary, and the tertiary sits on top of a small &quot;tower&quot; that protrudes up through the hole from behind the primary; this would help disguise the existence of the hole. That's how the tertiary mirrors in the 8 m unit telescopes of the VLT work, though I don't think they have support vanes the way this one does. The existence of the support vanes and the nature of the B' reflection suggests there really isn't a hole in the primary here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;dome&quot; is an example of a &quot;clamshell&quot; design, in which two halves fold down out of the way; you can see one of the two hinges in the front, near the bottom right of your images. If you google for images of these telescopes, you can see them in the daytime, folded-up configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an image of one of the telescopes when it was still in the factory (before the mirrors were installed), which may help you understand the design better:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OHyt5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OHyt5.jpg&quot; alt=&quot;VLTI auxiliary telescope in the AMOS factory (credit:ESO)&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2018-12-04T12:03:46.997" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28619" PostTypeId="1" CreationDate="2018-12-04T12:36:07.230" Score="3" ViewCount="254" Body="&lt;p&gt;How do they know that this is a spherical shell of gas, and not just something like an Airy pattern-like artifact produced by the VLT's large interferometric aperture?&lt;/p&gt;&#xA;&lt;p&gt;Image from: &lt;a href=&quot;https://www.eso.org/public/images/eso0906b/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.eso.org/public/images/eso0906b/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oYu00.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oYu00.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;I get 180 pixels from center to the first minimum, and 72 pixels = 4 mas. That makes the first minimum at 10 mas or 4.8E-08 radians. The wavelengths are around 1.6 microns so that would correspond to a circular aperture of 40 meters using 1.22 λ/d.&lt;/p&gt;&#xA;&lt;p&gt;Are they really sure this is a real shell, and not an Airy disk? How was that proven?&lt;/p&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;https://www.eso.org/public/news/eso0906/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.eso.org/public/news/eso0906/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;“We were able to construct an amazing image, and reveal the onion-like structure of the atmosphere of a giant star at a late stage of its life for the first time,” says Antoine Mérand, member of the team. “Numerical models and indirect data have allowed us to imagine the appearance of the star before, but it is quite astounding that we can now see it, and in colour.”&lt;/p&gt;&#xA;&lt;p&gt;Although it is only 15 by 15 pixel across, the reconstructed image shows an extreme close-up of a star 100 times larger than the Sun, a diameter corresponding roughly to the distance between the Earth and the Sun. This star is, in turn, surrounded by a sphere of molecular gas, which is about three times as large again.&lt;/p&gt;&#xA;&lt;p&gt;T Leporis, in the constellation of Lepus (the Hare), is located 500 light-years away. It belongs to the family of Mira stars, well known to amateur astronomers. These are giant variable stars that have almost extinguished their nuclear fuel and are losing mass. They are nearing the end of their lives as stars, and will soon die, becoming white dwarfs. The Sun will become a Mira star in a few billion years, engulfing the Earth in the dust and gas expelled in its final throes.&lt;/p&gt;&#xA;&lt;p&gt;Mira stars are among the biggest factories of molecules and dust in the Universe, and T Leporis is no exception. It pulsates with a period of 380 days and loses the equivalent of the Earth’s mass every year. Since the molecules and dust are formed in the layers of atmosphere surrounding the central star, astronomers would like to be able to see these layers. But this is no easy task, given that the stars themselves are so far away — despite their huge intrinsic size, their apparent radius on the sky can be just half a millionth that of the Sun.&lt;/p&gt;&#xA;&lt;p&gt;“T Leporis looks so small from the Earth that only an interferometric facility, such as the VLTI at Paranal, can take an image of it. VLTI can resolve stars 15 times smaller than those resolved by the Hubble Space Telescope,” says Le Bouquin.&lt;/p&gt;&#xA;&lt;p&gt;To create this image with the VLTI astronomers had to observe the star for several consecutive nights, using all the four movable 1.8-metre VLT Auxiliary Telescopes (ATs). The ATs were combined in different groups of three, and were also moved to different positions, creating more new interferometric configurations, so that astronomers could emulate a virtual telescope approximately 100 metres across and build up an image.&lt;/p&gt;&#xA;&lt;p&gt;“Obtaining images like these was one of the main motivations for building the Very Large Telescope Interferometer. We have now truly entered the era of stellar imaging,” says Mérand.&lt;/p&gt;&#xA;&lt;p&gt;A perfect illustration of this is another VLTI image showing the double star system Theta1 Orionis C in the Orion Nebula Trapezium. This image, which was the first ever constructed from VLTI data, separates clearly the two young, massive stars from this system. The observations themselves have a spatial resolution of about 2 milli-arcseconds. From these, and several other observations, the team of astronomers, led by Stefan Kraus and Gerd Weigelt from the Max-Planck Institute in Bonn, could derive the properties of the orbit of this binary system, including the total mass of the two stars (47 solar masses) and their distance from us (1350 light-years).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-04-28T02:08:44.350" Title="Are they really sure this isn't an Airy disk? How was that ruled out?" Tags="&lt;optics&gt;&lt;angular-resolution&gt;&lt;interferometry&gt;" AnswerCount="0" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="28620" PostTypeId="1" CreationDate="2018-12-04T12:37:48.423" Score="1" ViewCount="165" Body="&lt;p&gt;One serious theory advanced for the creation of Earth's moon is that it resulted from a collision. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However matter would be thrown off in all directions and some will fall back. How could a mass of ejecta as large as the Moon all end up travelling together in a single direction and therefore a single orbit? For a spacecraft to attain orbit requires some form of propulsion. Even if the impacting object 'bounced off' I'm still not clear that it could end up where the Moon is and travelling in the right direction. This would require it to be 238,900 miles away and travelling roughly at right angles to the Earth's gravitational field.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The giant-impact hypothesis, sometimes called the Big Splash, or the&#xA;  Theia Impact suggests that the Moon formed out of the debris left over&#xA;  from a collision between Earth and an astronomical body the size of&#xA;  Mars&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Orbit_of_the_Moon&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Orbit_of_the_Moon&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7828" LastEditorUserId="7828" LastEditDate="2018-12-04T13:20:06.587" LastActivityDate="2018-12-04T23:58:33.547" Title="How could the Moon have arisen from a collision - what would make for an elliptical orbit?" Tags="&lt;orbit&gt;&lt;the-moon&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="28624" PostTypeId="2" ParentId="28609" CreationDate="2018-12-04T16:15:53.950" Score="0" Body="&lt;p&gt;As the comments suggest, in general you cannot.   So long as the satellite's mass is a small fraction of the planet's mass (or a planet vs. the Sun's mass),  the difference between a circular orbit and the elliptical orbit about the system center of mass is too small to allow accurate calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you do know the mass of the planet (or sun), then knowing the shape of the elliptical orbit will tell you the mass of the satellite, and since you state you know the density -- I'm assuming that was a typo in your original question, finding the radius (of a perfect sphere) is trivial. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2018-12-04T16:15:53.950" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28625" PostTypeId="2" ParentId="28620" CreationDate="2018-12-04T23:53:02.087" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;This would require it to be 238,900 miles away and travelling roughly&#xA;  at right angles to the Earth's gravitational field.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your thinking is correct that an impact would have a very hard time forming the Moon where it is, but the Moon wasn't 238,000 miles away when it formed.   The Moon was much closer, &lt;a href=&quot;http://curious.astro.cornell.edu/our-solar-system/37-our-solar-system/the-moon/the-moon-and-the-earth/31-how-close-was-the-moon-to-the-earth-when-it-formed-intermediate&quot; rel=&quot;noreferrer&quot;&gt;about 3-5 Earth radii distant&lt;/a&gt;.  It's impossible to say precisely how close because there's no way to model it precisely.   The interactive tidal forces push the Moon slowly away, and that's still happening today.  Working backwards, it's not hard to deduce that 4.4 billion years ago, the Moon was over 10 times closer to the Earth.    &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Even if the impacting object 'bounced off'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Terminology isn't perfect but a giant impact with incoming velocity of at least 12 or 13 km/s doesn't exactly bounce off.   It plows into it, kind of merges into then explodes or rebounds off.   Meteor impacts are most accurately represented as explosions and the size of an impact crater tends to be about 10 times the size of the object that impacts.    Think about what that means.   If the impactor was the size of Mars, or even the size of mercury (estimates vary).   A crater 10 times the size of the planet Mercury would be 3 times bigger than the Earth.   An impact that large at the very least liquifies the Earth and sends it spinning very fast and perhaps, according to a new theory, doesn't only liquify the Earth but turns it into a kind of temporary gas giant made of vaporized rock &lt;a href=&quot;https://www.space.com/36959-planet-giant-impact-vaporized-rock-synestia.html&quot; rel=&quot;noreferrer&quot;&gt;called a synestia&lt;/a&gt;.   Whichever theory you like, it's a lot more involved than just hitting and bounding off.   It's a messy splash that turns the Earth into a ball of lava, at the very least.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another hypothesis is that &lt;a href=&quot;https://www.space.com/35258-earth-moon-formed-in-moonlet-mashup.html&quot; rel=&quot;noreferrer&quot;&gt;it took multiple impacts to form the Moon&lt;/a&gt;, not a single impact.  We don't have any photographs of the event, so all anyone can do is try to find a good model that explains the Moon's formation and the material it's made of.   What everyone agrees on, as I said above, is that the Moon was over 10 times closer when it formed.   If the Moon is a golfball in the sky to our eyes, then after it formed, it was a volleyball.  That would have been quite something to see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How could a mass of ejecta as large as the Moon all end up travelling&#xA;  together in a single direction and therefore a single orbit?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using the synestia model, then what basically happens is the moon forms out of the rotating donut of gas.   Collapsing clouds of dust can't form into a single ball in the center because angular momentum needs to be conserved.   When the gas that formed our solar-system collapsed there was enough debris with angular momentum to form the planets.   Likewise, when Jupiter formed, not all the material fell into Jupiter, where the angular momentum was too high, Jupiter's 4 Galilean moons formed, and, perhaps more than 4 formed initially.   Binary stars often form that way.   Angular momentum must be conserved, so moons forming out of a rapidly rotating ball/donut of gas is pretty common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the more traditional impact model, which is what you address in your question, it's important to think of it from both frames of reference.   There's the surface of the Earth where the impact happens frame of reference and there's the gravitational field or non-rotating frame of reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're standing on the Earth and you witness a giant meteor impact.   Ignoring the fact that you'd die, what you'd see is the surface basically liquify and contract from the impact, then explode outwards.   From your frame of reference, material would be blown out in all directions equally, in a simple explosion model.  The angle of impact wouldn't change the explosion model from that frame of reference all that much.  But what would actually happen is more complicated.  The impacting object would have not only straight down velocity but sideways velocity, accelerating the entire crust of the Earth in the direction it impacts.  The actual plowing into the Earth would take about 8 minutes before the rebound outwards began, and in those 8 minutes, the Earth's rotation would accelerate significantly.   The rotational acceleration would be perhaps 1/2 g force for those 8 minutes and, initially the impact spot and tangential velocity of impact would be shoving it's way across the Earth as the impact continued and the surface of the Earth would rotate faster than the inner part and the axis of rotation would change too, initially just on the surface, and the core would be slower to catch up.  It would be an absolutely crazy event to witness, but ignoring all that, it's probably accurate enough to say that the explosion model would shoot debris equally in all directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you're observing from space, the other frame of reference, lets say, a geostationary orbit (which OK, as the rotation of the planet changes, so does the geostationary distance, but lets say you're just observing from above) for optimal impact viewing, you'd see the Earth's rotation speed increase from the impact.  Earth is now rotating about once every 4 hours.  Blazingly fast for planetary standards, and as a result, it bulges around the equator and flattens at the poles, visibly flattened, like Jupiter.   At the equator relative to the axis of rotation, which, as a result of the impact, the impact spot would be close to the newly positioned equator, in that scenario equator is spinning around at 6 times it's current speed or about 6,000 mph.   We usually use km/s for asteroid and escape velocity, so 6,000 mph = about 2.7 km/s.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From your point of view from space, when debris is blown off the Earth from a rotating surface moving at 2.7 km/s, that means, material blown off in the direction of the rotation has a plus 2.7 km/s and material blown off in the opposite direction has a -2.7 km/s.   Material blown off perpendicularly, then it would be a vector addition, but the result, material blown roughly off in the direction of rotation is far more likely to make it into orbit and material blown off against the rotation, even if the material is blown away equally in all directions, the material that makes it into orbit would be much more likely to be in line with the Earth's rotation and that's where the tangential velocity comes from.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We usually just call this conservation of angular momentum, but I think, explaining it from the 2 frames of reference can make it easier to visualize.   An explosion on a rotating sphere, effects the direction of the ejected orbiting material, where most of the material would orbit with some alignment to the direction to the rotation of the sphere.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So, can I make a moon by simply throwing stuff off the Earth at very high speed?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to add this point, even if it's not quite what you asked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbits move in ellipses, so, If we ignore air resistance and if you're standing on the surface of the Earth, anything you throw, or, lets say you have a space cannon.  Anything you cannon into space, would follow an elliptical orbit and potentially spiral around the Earth and hit you in the back.   Of course, the Earth rotates, so that wouldn't really happen, and it wouldn't happen for other reasons as well, like there's not just 2 bodies in space so all orbits are perturbed, and Earth has mascons that would affect the neatness of the orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, like &lt;a href=&quot;https://en.wikipedia.org/wiki/Newton%27s_cannonball&quot; rel=&quot;noreferrer&quot;&gt;Newton's cannonball&lt;/a&gt;, which in theory, when fired from a high mountain at just the right velocity, would circle the Earth and hit the back of the cannon.  Newton focused on the circle or orbit around the Earth, but I always thought that the Cannon shooting itself in the back was a fun aspect too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An object, shot into orbit with an initial velocity, would either escape, if the escape velocity was high enough (hyperbolic orbit) or it would circle back in an ellipse and return to where it was launched from, because orbits, once set in motion, repeat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you launch heavy enough cannonball into space, lets say, 1/81th the mass of the Earth, then something else happens.   You're not just launching a new moon into an orbit that reaches 3-5 Earth radii at it's furthest distance before circling back to the Earth, but you're pushing the Earth away at the same time.  Every action has an equal and opposite reaction and 1/81th the Mass of the Earth is enough mass where you have to consider the push on both objects, not just the theoretical cannonball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this scenario, when your new moon spiraled back to where you launched it, the Earth would have moved about 150 miles and it wouldn't be where it was.  Earth would be lighter than it was when the object was launched.   The two objects might still collide as each would in effect orbit the other, but you'd no longer have a neat Kepler orbit, but something a little more mathematically complicated when both masses need to be considered.  (That's why I said 12 or 13 km/s above, not Earth's escape velocity, because both Earth and Theia would have escape velocities, so the minimum impact speed is slightly higher than for a standard space rock).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kepler's orbits and Newton's calculated orbits assume the mass ratio is 100 to 0.   The math changes when you consider that both objects have mass.   In the case of the giant impact, enough mass was blown off the Earth that the Kepler eclipse model doesn't quite apply.   But that's not the primary factor that made it possible for the Moon to form with a close to circular orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The newly formed debris shot into orbit has enough mass to have a gravitational effect on itself, and that you have collisions, so, individual bits of debris entered orbit with elliptical orbits the combination of debris, after collisions and gravitational interaction doesn't retain high eccentricity orbits.   A lot of the ejecta, perhaps most of it, probably did fall back to the Earth, but enough entered a temporary ring-like system of orbiting cloud of debris that the Moon was able to form, and it likely formed with a reasonably circular orbit too, because the debris, by collision and gravitational interaction, corrected for a lot of individual bits' eccentric orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar to how the rings of Saturn which are neat and circular.   Non circular orbits in a crowded debris field would be more likely to collide, so over time, individual bits of debris that remained in orbit would move towards more circular orbits in a similar way that enough material tends to spiral into a disk, because variations in inclination or eccentricity would tend to cancel out over time.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be remiss if I didn't point out that &lt;a href=&quot;https://www.extremetech.com/extreme/253830-giant-impact-hypothesis-moon-formation-synestia&quot; rel=&quot;noreferrer&quot;&gt;the giant impact theory has some issues&lt;/a&gt;, it was still a brilliant theory and it would be more accurate to say that it needed some adjustments, so, I don't like saying it was wrong.  I think it was brilliant, but there are problems working out the details with the debris field formation of the Moon, or so I've read, but the Moon having a circular orbit from formation isn't one of those problems.  That part of the model works.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't show you the numbers because you need a super-computer to run those numbers and that's above my pay-grade, so I tried to explain it in a way that can be visualized.   Hope that helps. &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-12-04T23:53:02.087" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28626" PostTypeId="2" ParentId="28620" CreationDate="2018-12-04T23:58:33.547" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How could a mass of ejecta as large as the Moon all end up travelling together in a single direction and therefore a single orbit?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Moon did not form this way. The giant impact hypothesis instead proposes a collision between the Earth and an object that was significantly larger than the Moon, roughly ten times the mass of the Moon. Per this hypothesis, large amounts of material ejected from the surface of the Earth. The energy released from the collision would have heated the impactor and the surface of the Earth immensely. The proto-Earth would have been enshrouded in an atmosphere of vaporized rock. Meanwhile, a handful of Earth radii out, there was an orbiting collection of bits, a few somewhat large, but most rather small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over the course of decades to centuries, these bits coalesced to form what would become our Moon. The Earth cooled and formed oceans. The close-by Moon raised huge tides on the Earth. Like today, but amped up by orders of magnitude by those huge tides, the tidal interactions between the Earth and the Moon made the Moon's orbit recess from the Earth. Now we see our Moon at about sixty Earth radii from the Earth, compared to the roughly one tenth of that when the Moon first formed.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2018-12-04T23:58:33.547" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28627" PostTypeId="2" ParentId="28609" CreationDate="2018-12-05T00:36:09.540" Score="1" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;How to find the radius of a satellite not knowing the mass:&lt;/p&gt;&#xA;&lt;p&gt;If I know the&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;semi-major axis,&lt;/li&gt;&#xA;&lt;li&gt;density of the planet, and&lt;/li&gt;&#xA;&lt;li&gt;orbital period..&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;how can I find the radius of the satellite? I can't think of any formula to use.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I have a hunch that you've misunderstood the question, and that it asks for the &lt;strong&gt;radius of the planet&lt;/strong&gt; not the artificial satellite, and that's certainly doable. Otherwise, why would the planet's density be a given?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The equation that links semi-major axis, &lt;em&gt;mass&lt;/em&gt; of the planet, and orbital period is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T=2 \pi \sqrt{\frac{a^3}{GM}} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_constant&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravitational constant&lt;/a&gt; 6.674E-11 m^3 s^-2  kg^-1.&lt;/p&gt;&#xA;&lt;p&gt;Moving things around we get&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{4 \pi^2 a^3}{T^2G} = M $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Density is mass per volume, or:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\rho = \frac{M}{\frac{4}{3} \pi R^3} = \frac{3M}{4 \pi R^3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Moving that around we get&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$R = \left(\frac{3M}{4 \rho \pi}\right)^{1/3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Put in the earlier result for &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt;, we get&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$R = a \left(\frac{3 \pi}{\rho T^2G}\right)^{1/3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;For Earth's average density, Wikipedia says 5514 kg/^3, for its radius, try 6,371,000. meters. For the ISS at 400 km (400,000 meters) above the Earth, use 92.5 minutes (5550 sec). It should all work out nicely!&lt;/p&gt;&#xA;&lt;p&gt;The reason this works with average density, even though Earth's density is higher in the core than the surface is the application of Newton's &lt;a href=&quot;https://en.wikipedia.org/wiki/Shell_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;shell theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-12-05T04:41:54.847" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28628" PostTypeId="1" CreationDate="2018-12-05T04:19:28.000" Score="1" ViewCount="212" Body="&lt;p&gt;I am trying to use a Celestron NexStar 130 SLT Computerized Telescope, but I am not able to use the GoTo function.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I do the three start alignment, it confirms matches, &lt;/li&gt;&#xA;&lt;li&gt;I make sure the level is exactly level before and after,&lt;/li&gt;&#xA;&lt;li&gt;I finally punch in a planet to slew to&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;...and once the telescope has slewed to the planet I see nothing in the eyepiece except pin point dots spread out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the words on the hand control are very hard to read because the letters are overlapping each other while the words are panning off the screen. I already tried factory resting which didn't work. The words are in English.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would my next step be? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1ZVDQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1ZVDQ.jpg&quot; alt=&quot;Example of Celestron NexStar 130 SLT Computerized Telescope&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example image, &lt;a href=&quot;https://www.telescopesplus.com/products/celestron-nexstar-130-slt-computerized-telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25055" LastEditorUserId="25055" LastEditDate="2018-12-05T21:11:49.707" LastActivityDate="2018-12-05T21:11:49.707" Title="Celestron NexStar 130 SLT Computerized Telescope GoTo Not Accurate" Tags="&lt;star&gt;&lt;planet&gt;&lt;telescope&gt;&lt;amateur-observing&gt;" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="28629" PostTypeId="1" AcceptedAnswerId="28653" CreationDate="2018-12-05T06:03:39.747" Score="2" ViewCount="260" Body="&lt;p&gt;&lt;strong&gt;Where can I find the &#xA;dataset(s) for a globular star cluster?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The datasets should have the coordinates, color index (for different wavelengths) , magnitude etc. of every star in the cluster. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to do data analysis on it just for fun.&lt;/p&gt;&#xA;" OwnerUserId="25078" LastActivityDate="2018-12-18T10:02:13.177" Title="Dataset for globular star cluster" Tags="&lt;data-analysis&gt;&lt;globular-clusters&gt;&lt;star-cluster&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28630" PostTypeId="1" CreationDate="2018-12-05T09:31:40.137" Score="7" ViewCount="110" Body="&lt;p&gt;I am a data scientist and curious about their roles in modern Astronomy, both now, and in the near future. What are type of roles and projects where data scientist are most active in Astronomy? What are the basic solutions provided by data scientists in astronomy. Are there data scientists as principle investigators, or are they at least team members in research efforts?&lt;/p&gt;&#xA;" OwnerUserId="25082" LastEditorUserId="6" LastEditDate="2019-01-29T14:42:38.370" LastActivityDate="2019-01-29T14:42:38.370" Title="What are the present, and potentially the future roles of data scientists in Astronomy?" Tags="&lt;data-analysis&gt;&lt;mathematics&gt;&lt;future&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="0" ClosedDate="2018-12-06T14:19:10.180" ContentLicense="CC BY-SA 4.0" />
  <row Id="28632" PostTypeId="1" AcceptedAnswerId="28633" CreationDate="2018-12-05T15:09:28.540" Score="0" ViewCount="114" Body="&lt;p&gt;In the visible universe, we have a lot of matter to energy conversion going on in stars.  Is there an equivalent structure that is converting dark matter to dark energy?&lt;/p&gt;&#xA;" OwnerUserId="14085" LastActivityDate="2018-12-05T15:28:01.347" Title="Are there Dark Stars" Tags="&lt;dark-matter&gt;&lt;dark-energy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28633" PostTypeId="2" ParentId="28632" CreationDate="2018-12-05T15:28:01.347" Score="3" Body="&lt;p&gt;Dark matter has different properties than ordinary matter. It cannot collide, and thus it alone cannot coalesce to form a star. Furthermore, dark matter and dark energy are unassociated properties. There is not currently any reason to assume dark matter can be converted to dark energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, there is a &lt;em&gt;strictly hypothetical&lt;/em&gt; entity referred to as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_star_(dark_matter)&quot; rel=&quot;nofollow noreferrer&quot;&gt;dark star&lt;/a&gt;, which does involve dark matter. While composed mostly of ordinary matter, this entity is hypothesized to include a high concentration of dark matter, generating heat and preventing fusion of the ordinary matter. If dark stars existed, it is likely they existed in the early universe and did not survive to the present. They are modeled to be large clouds--4 to 2000 astronomical units in diameter--of hydrogen and helium, with a surface temperature low enough to make them invisible to the naked eye. &lt;/p&gt;&#xA;" OwnerUserId="6" LastActivityDate="2018-12-05T15:28:01.347" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28634" PostTypeId="1" CreationDate="2018-12-05T16:03:06.090" Score="1" ViewCount="37" Body="&lt;p&gt;Is there a database with cartesian coordinates (x, y, z) of galaxies? I only need positional data and calculating that using degrees and redshift for all galaxies could take a while and would be complicated.&lt;/p&gt;&#xA;" OwnerUserId="25085" LastEditorUserId="25085" LastEditDate="2018-12-05T16:11:38.583" LastActivityDate="2018-12-05T16:11:38.583" Title="What's the easiest way to get 3D-positional data of galaxies" Tags="&lt;galaxy&gt;&lt;data-analysis&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28635" PostTypeId="2" ParentId="28630" CreationDate="2018-12-05T16:57:00.990" Score="5" Body="&lt;p&gt;Astronomer is moving towards the big(ger) data era because of many sky survey technologies. The coming ones include, e.g., LSST, JWST, and WFIRST.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the meaning of survey, it normally means observing the whole sky over a few days, and keep repeating over and over. Also, since most of the surveys are imaging technologies, every pixel in an image is important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you can imagine how much data is incoming per day, and how much data science is important in many fields of astronomy (and cosmology). So, data science is, kind of, taking over when it comes to analyzing data from surveys.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-12-05T16:57:00.990" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28637" PostTypeId="1" CreationDate="2018-12-06T01:13:55.257" Score="3" ViewCount="122" Body="&lt;p&gt;This &lt;a href=&quot;https://astronomy.stackexchange.com/a/22996/7982&quot;&gt;excellent answer&lt;/a&gt; to the question &lt;em&gt;What is the physics of the “spinning dust” contribution to Cosmic Microwave Background measurements?&lt;/em&gt; Is well sourced. The abstract of one of the papers there, Purcell 1975 &lt;a href=&quot;http://adsabs.harvard.edu/abs/1975duun.book..155P&quot; rel=&quot;nofollow noreferrer&quot;&gt;Interstellar grains as pinwheels&lt;/a&gt; begins with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Contrary to the assumption usually made in theories of grain alignment, the rotational energy of an interstellar grain is likely to be very much greater than 3/s kT, where T is the gas temperature, or the grain temperature, or any other temperature in the system. Any more or less permanent irregularity of the grain’s surface with respect to accommodation coefficient, distribution of H-H recombination sites, or photoelectric emissivity will result in an unbalanced torque capable of spinning the grain up to high angular velocity. Such a grain is, in effect, the rotor of a heat engine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the body of the paper, the first example given is of a sort-of &lt;a href=&quot;https://en.wikipedia.org/wiki/Thought_experiment&quot; rel=&quot;nofollow noreferrer&quot;&gt;gedankenexperiment&lt;/a&gt; involving a hypothetical &lt;a href=&quot;https://en.wikipedia.org/wiki/Crookes_radiometer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Crookes radiometer&lt;/a&gt; with the gas removed. The author explains that it would spin up to high energy (without saying how):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] The kinetic energy of the rotating paddle wheel is now enormously greater than it would be if it were immersed in, and en equilibrium with, a 6000 K radiation field, for then it would exhibit only a residual “Brownian rotation” appropriate to that temperature. So the wheel is really a heat engine, depending on the difference between the temperature of the vanes and that of the radiation field to which they are exposed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without the gas to cool them, I suppose that after the whole thing reaches equilibrium temperature the black sides radiate more strongly than the white sides and so there should always be a torque, and the final speed and kinetic energy would be limited only by friction and other practical losses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/11577/7982&quot;&gt;This excellent answer&lt;/a&gt; to the question &lt;em&gt;What is the difference between gas and dust in astronomy?&lt;/em&gt; points out that there is no real distinction between the two, but that they are limits in size, gas particles being small and dust particles being large(r).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if dust can become pinwheels, then I have a hunch that very very large molecules can do this too, and if so, then so can smaller molecules. All you need is some mechanism for torque. Molecules can certainly have a non-uniform distribution of  H-H recombination probability just as dust can, and they can have some regions that are concave and some that aren't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes me wonder if these gas molecules no longer obey the &lt;a href=&quot;https://en.wikipedia.org/wiki/Equipartition_theorem#Basic_concept_and_simple_examples&quot; rel=&quot;nofollow noreferrer&quot;&gt;Equipartition theorem&lt;/a&gt; of thermodynamics. The spinning dust certainly doesn't. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How can spinning dust and by induction, my hypothetical spinning molecules avoid conforming to the equipartition theorem? Why doesn't the equipartition theorem disallow spinning dust?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6bDyB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6bDyB.png&quot; alt=&quot;turbine&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1975duun.book..155P&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt; borrowed from &lt;a href=&quot;https://astronomy.stackexchange.com/a/22996/7982&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus points:&lt;/strong&gt; Is that little arrow (&lt;span class=&quot;math-container&quot;&gt;$\omega_{rot}$&lt;/span&gt;) pointing the wrong way?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="6" LastEditDate="2018-12-06T13:54:32.747" LastActivityDate="2018-12-06T14:41:08.867" Title="Why doesn't the equipartition theorem disallow spinning dust?" Tags="&lt;temperature&gt;&lt;dust&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28639" PostTypeId="1" AcceptedAnswerId="28674" CreationDate="2018-12-06T04:18:44.767" Score="5" ViewCount="233" Body="&lt;p&gt;Results from the IBEX satellite indicate the nose of the solar system is pointing to the constellation Scorpius.  Various sources say the solar apex is pointing to the constellation Hercules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the two different, or is simply the information about the location in conflict?&lt;/p&gt;&#xA;" OwnerUserId="25088" LastActivityDate="2019-01-25T01:53:18.947" Title="Is the nose of the solar system and the solar apex the same thing?" Tags="&lt;solar-system&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28640" PostTypeId="1" AcceptedAnswerId="28645" CreationDate="2018-12-06T07:24:43.723" Score="1" ViewCount="125" Body="&lt;p&gt;In science fiction, when a starship captain commands &quot;all stop,&quot; the audience simply assumes that, like a terrestrial naval ship anchored to the ocean floor, the ship literally comes to a complete stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does our current understanding of celestial mechanics provide a meaningful definition of a complete stop in space &amp;mdash; for any object?&lt;/p&gt;&#xA;" OwnerUserId="18065" LastActivityDate="2018-12-06T15:00:45.637" Title="What does &quot;all stop&quot; mean in reality?" Tags="&lt;celestial-mechanics&gt;&lt;apparent-motion&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28641" PostTypeId="2" ParentId="28630" CreationDate="2018-12-06T12:00:07.050" Score="3" Body="&lt;p&gt;I'm an astronomer and I get lots of job offers to retrain as a data scientist, but it might be more tricky to go the other way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomy is definitely a field in which 'big data' is important, and the analysis and visualisation techniques we use every day are probably decades behind what is taught to computer scientists. However, most astronomy software is very niche, only used by a handful of people around the world, and usually edited to fit their particular purpose. I have seen attempts to make more 'generic' astronomical software that can be standardised and widely used, but it is difficult to find a task that can be automated and standardised because we are normally working with slightly different constraints each time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few years ago I met a PhD student who had studied computer science at undergraduate and then moved into astronomy, and he was amazed at how old the techniques we were just getting to grips with were, but he also had real trouble understanding what he was trying to analyse because he didn't have the astronomical knowledge to know all the sources of error and noise in his data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe there are data scientists working on more abstract problems, like telescope pipelines, but unfortunately, I can't give much advice in this area as that is an observational problem and my work is theoretical.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2018-12-06T12:00:07.050" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28642" PostTypeId="1" AcceptedAnswerId="28652" CreationDate="2018-12-06T12:25:08.020" Score="0" ViewCount="374" Body="&lt;p&gt;The moon follows 5.14 deg inclined path. the ecliptic and the Moon's path coincide are called 'nodes', and they travel around the ecliptic in a period of 18.66 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once every 18.66 years, the ascending node coincides with the vernal equinox, and the Moon will reach declinations of about +28° and -28° (much more than the Sun, which reaches ±23°); this is called a 'major lunar standstill'. 9.33 years later, the opposite is true, and the Moon will only reach ±18° during a 'minor lunar standstill'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read &lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20121018091531AArHgNS&amp;amp;guccounter=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://answers.yahoo.com/question/index?qid=20121018091531AArHgNS&amp;amp;guccounter=1&lt;/a&gt; here, the precise declination of moon is at +28.88 deg, over the earth. Where can I find the exact, value from other sources? I don't know how he found this value? &quot;  I get plus or minus 28 degrees 53', over a 18.6 year cycle. Eventually it will cover the whole region, but it will take many thousands of years (depending on how close you consider it has to get to a point to cover it.) &quot; 28 degrees and 53' = 28.88 deg &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 0.88 deg, to time is 53 min, I think it seems to coincide with the fact the moon rises, 53 min sooner, over each day. Is this related to this? 13.2 deg * 4 min/deg = 53 min &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum inclination of the moon, for standstill, can it be specified by the eclipse point? On, august 21 2017, an eclipse occurs in 28.88 deg on latitude point. Is this angle the same as the inclination angle specified here: &lt;a href=&quot;http://www.umass.edu/sunwheel/pages/moonteaching.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.umass.edu/sunwheel/pages/moonteaching.html&lt;/a&gt; - 28.8 deg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just want 28.88 deg = 28 deg 53' confirmed. &lt;/p&gt;&#xA;" OwnerUserId="25090" LastActivityDate="2018-12-07T18:10:12.930" Title="Precise declination of Moon" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;declination&gt;&lt;inclination&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28643" PostTypeId="2" ParentId="28637" CreationDate="2018-12-06T14:16:04.140" Score="3" Body="&lt;p&gt;The first relevant point is that the equipartition theorem only formally applies in thermodynamic equilibrium, which requires that all temperatures be the same.  Since the spinning dust only happens when there are temperature differences, holding to the analogy with a heat engine mentioned in the source, we should not expect that theorem to be the whole story.  Granted, we often apply that theorem even when there are temperature differences, so the issue boils down to, when is the behavior more or less the same when all temperatures are the same, and when is the behavior fundamentally a heat engine?  I will assume the source is correct that this is heat engine-like behavior for spinning grains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have a heat engine in operation, the sky is the limit on how much energy you can put in any given mode (as long as you can maintain the temperature difference), because a heat engine does work, and work energy never needs to be equipartitioned, you can partition it any way you like by setting up an appropriate apparatus.  The dust particle is therefore a kind of apparatus for doing work, in the presence of temperature differences, and channeling the work energy into rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How and why that heat engine operates and picks rotation to channel the work is a much more subtle issue.  For example, do we need temperature differences across the dust grain itself, or just between the grain and the surroundings?  If the &quot;hypothetical&quot; Crooke's radiometer (meaning, one that works quite differently from actual Crooke's radiometers and actually spins the opposite direction) is a good analogy, then the spin is from temperature differences across the dust grain itself, which warm the gas in the surroundings of the dust grain, producing gas flows that maintain constant pressure.  Those flows require the gas to receive torques, and conservation of angular momentum requires an opposite torque on the dust grain.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get torque like that, it seems that you not only need a temperature difference across the dust grain, but it also has to have a kind of &quot;quadrupole&quot; character-- hot-cold on one side of the grain, cold-hot on the other, producing a &quot;handedness&quot; to the temperature structure in the grain.  So we immediately see that one cannot get spin until one has a large enough system to be able to support the concept of &quot;temperature difference&quot; across the object, and molecules are generally not thought of that way-- though maybe in the case of very large molecules, they could be.  So the question then comes down to, how large does a molecule need to get before it can support a concept of spatially nonuniform temperature? I don't know how big that needs to be, but my guess is, very big indeed-- there are certainly polymers that can get very large, but the molecules talked about in astronomy are generally not polymers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the sign of the rotation, that can be very difficult to figure out!  It seems to depend on how the temperature differences get created in response to absorption of a radiation field (or other interactions with the environment, perhaps thermal contact with warmer gas).  Generally speaking, concave sides warm up more quickly than convex, so cause the gas in the concave part to be warmer than in the convex, so a flow sets up from concave to convex.  That acts like a jet engine that pushes the concave side forward.  That this is subtle can be seen from the fact that the higher pressure gas would seem to push the other way on the dust, but it's like the way a sail on a sailboat works-- follow the deflection of the air and conclude that the boat deflects the opposite way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is all correct (see the Wiki on Crooke's radiometers for background information), then I would say the shape of the dust grain you drew produces spin in the direction that you indicated, but that no astrophysical molecules are large enough to produce the same effect.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2018-12-06T14:41:08.867" LastActivityDate="2018-12-06T14:41:08.867" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28644" PostTypeId="2" ParentId="26892" CreationDate="2018-12-06T14:37:23.537" Score="1" Body="&lt;p&gt;From the introduction of the source paper, we can see that what is described is a particular bump in the extinction curve of the galaxy centered at 2175 Ångström (a unit of length used for wavelength). The strength of this bump is large compared to the Large Magellanic Cloud, and is rarer toward the Small Magellanic Cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned in the comments, the main candidates for the origin of the feature are non-graphitic carbon or polycyclic aromatic hydrocarbons (PAHs). The absence in other environments is attributed to low metallicities or different radiative environments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, this is a case where we lose information in science reporting. &quot;Very high content of carbonaceous dust&quot; was a translation of a specific strength at a specific wavelength, and &quot;other galaxies&quot; was a translation of surrounding galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://arxiv.org/abs/1806.00293&quot; rel=&quot;nofollow noreferrer&quot;&gt;The 2175 Å extinction feature in the optical afterglow spectrum of GRB 180325A at z=2.25 - Zafar, et al.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6" LastActivityDate="2018-12-06T14:37:23.537" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28645" PostTypeId="2" ParentId="28640" CreationDate="2018-12-06T14:58:01.357" Score="2" Body="&lt;p&gt;As mentioned in the comments, in relativity, &quot;all stop&quot; can only be in reference to some other object, or set of objects.  Still, I would argue there remains some interesting potential for the &quot;all stop&quot; concept in physics theory, because Einstein was motivated by Mach's principle when creating general relativity (regardless of whatever version of that principle is formally supported in that theory), and so we might suspect that &quot;all stop&quot; can have the theoretical local meaning of &quot;at rest with respect to the center of mass of everything around it that is affecting it.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That notion is trickier in an expanding universe, where the concept of a center of mass gets replaced with the concept of &quot;at rest with respect to the cosmic microwave background&quot;, which is normally treated as a kind of accident of history, rather than a theoretical truth.  Still, since the Big Bang model is often taken as asserting that the accidental history of our universe is actually relevant to the origin of our universe and its theoretical properties, as well as to the origin of the laws that our universe obeys, it is not a great stretch to imagine that an &quot;accident of history&quot; kind of preferred frame could still, nevertheless, have a formal status in the laws as a kind of true &quot;all stop&quot; frame-- albeit only locally.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2018-12-06T15:00:45.637" LastActivityDate="2018-12-06T15:00:45.637" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28646" PostTypeId="1" CreationDate="2018-12-06T16:31:47.733" Score="-1" ViewCount="107" Body="&lt;p&gt;A non rotating asteroid of huge mass collides with earth.&#xA;After the collision, would the length of the day on earth increase or decrease? Why?&lt;/p&gt;&#xA;" OwnerUserId="25092" LastEditorUserId="1498" LastEditDate="2018-12-06T23:40:05.423" LastActivityDate="2018-12-06T23:40:05.423" Title="A non rotating asteroid of huge mass collides with earth. After the collision, does the length of the day on earth change?" Tags="&lt;asteroids&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28649" PostTypeId="2" ParentId="28646" CreationDate="2018-12-06T17:35:43.853" Score="7" Body="&lt;p&gt;There is insufficient information to give an answer.  There are two effects in play here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bigger one depends on the impact parameter -- basically how off-center the impact is.  If you imagine the impactor coming in in the equatorial plane, due to the Earth's rotation, one side of the Earth (as viewed from the impactor) is approaching it and one side receding. Unless it hits straight-on, it has angular momentum relative to the Earth and the impactor's angular momentum will remain with the combined body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if it hits on the &lt;em&gt;approaching&lt;/em&gt; side it's angular momentum will be oppositely-directed to the Earth's and the final body will have less angular momentum and the day will be longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it his on the &lt;em&gt;receding&lt;/em&gt; side, it's angular momentum will be directed the same as the Earth's and the final body will have more angular momentum than the Earth does now and the day will be shorter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it his dead-on-center, the combined body will have the same angular momentum as the Earth now does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second effect is due to the changed moment of inertia of the final body compared to the Earth's.  The moment of inertia of a sphere depends on its mass (all else being equal, the moment of inertia is proportional to the mass), the distribution of mass (a body with a denser core has a lower moment of inertia than one of the same mass which is uniform in density), and the size (for the same mass, the larger body has the higher moment of inertia.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of that, if the collision leaves the combined body's angular momentum the same as the Earth's, the result will depend on how the resultant body compares with Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will have a larger mass, which will increase the moment of inertia.  It may be larger (which also increases the moment of inertia), but it also may be smaller if the added mass actually winds up compacting the Earth's material to higher density.  (I think this is a long-shot, but it's certainly possible.)  If the impactor is extremely dense, it will sink to the center of the combined body and will have little effect on the final moment of inertia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The combination of these moment of inertia effects will probably more often be a higher moment of inertia and thus a slower rotation rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this effect is probably dominant only for a head-on collision. If the collusion is it all glancing, the final body will have substantially different angular momentum than the Earth presently has and the rotation rate change will be mostly controlled by that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line: To answer this -- even to give you the &lt;em&gt;sign&lt;/em&gt; of the change -- you need to specify exactly how the asteroid is approaching the Earth (what direction, what speed) its mass, and its composition (though this last only matters if it's pretty exotic.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;On average&lt;/em&gt; for &quot;normal&quot; asteroids the effect is probably a small lengthening of the day since the first effect should average out to close to zero, while for normal materials the second effect will probably mostly increase the day's length.  But that's the average of a big range in results from a significant shortening of the day to a significant lengthening of it.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2018-12-06T17:50:00.973" LastActivityDate="2018-12-06T17:50:00.973" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28650" PostTypeId="1" AcceptedAnswerId="28651" CreationDate="2018-12-06T17:56:02.400" Score="3" ViewCount="217" Body="&lt;p&gt;I've been meaning to post it on the day when InSight landed but just got around to it now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When they broadcasted landing in real-time, it really wasn't in real-time because radio signals from Mars can take from 3 to 24 minutes to reach the Earth (depending on the distance between Mars and Earth) if I am not mistaken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what was the delay on the date InSight landed? Each time someone mentioned &quot;real-time&quot; I cringed a little bit because everything already happened some time ago and we were just only learning about it now.&lt;/p&gt;&#xA;" OwnerUserId="25093" LastActivityDate="2018-12-06T22:23:06.107" Title="Current delay for radio signals from the Mars InSight mission" Tags="&lt;mars&gt;&lt;radio&gt;&lt;communication&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28651" PostTypeId="2" ParentId="28650" CreationDate="2018-12-06T18:35:15.157" Score="3" Body="&lt;p&gt;&lt;strong&gt;Current delay for radio signals from the Mars InSight mission.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The delay at the time of the landing was 8.07 minutes (8 minutes 4+ seconds), as Mars was at the time at a relative distance from Earth of 0.9703 Au.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Current delay as at 8th December 2018, we're at a distance of 8.85 (8 minutes 51 seconds) light minutes or 1.064 Au.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Using WolframAlpha's tool &lt;a href=&quot;https://www.wolframalpha.com/input/?i=Mars%2026%20Nov%202018&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, you should be able to satisfy your curiosity without effort.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Times taken are at midnight on those dates and vary minimally throughout the day (eg. the delay on the day following launch at midnight was approx four seconds longer).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This does not take into account the time delay between the signal being sent to the satellite and it being decoded by your box and displayed by your tv or the time it may have taken to be distributed over the net if you watched it that way. I'm not able to determine if any Broadcast-Delay was deliberately introduced, which is a common practice and commonly is about 7 seconds.&lt;/p&gt;&#xA;" OwnerUserId="768" LastEditorUserId="2365" LastEditDate="2018-12-06T22:23:06.107" LastActivityDate="2018-12-06T22:23:06.107" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28652" PostTypeId="2" ParentId="28642" CreationDate="2018-12-06T22:53:23.863" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I just want 28.88 deg = 28 deg 53' confirmed. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, 0.88 degrees is equal to 53 arc minutes, so 28.88 deg = 28 deg 53'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the Yahoo source is off in its value. According to this reference, the maximum declination is closer to 28° 44' (28.73°).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Title: Extreme declinations of the moon.&#xA;Authors: Können, G. P. &amp;amp; Meeus, J.&#xA;Journal: Journal of the British Astronomical Association, Vol. 82, p. 192 - 193&#xA;Bibliographic Code: 1972JBAA...82..192K&#xA;Available on &lt;a href=&quot;http://adsbit.harvard.edu//full/1972JBAA...82..192K/0000193.000.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA Astrophysics Data System&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That reference gives a declination of 28° 43' 21&quot; on Sep 15 2006. Using the Lunar Polynomials from the &lt;a href=&quot;http://asa.usno.navy.mil/SecD/LunarPoly.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomical Almanac Online&lt;/a&gt;, which permit you to calculate the declination to a far greater precision than you can know the time, I calculate a value of 28° 43' 22&quot;. Thus, I trust the article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;28° 43' 22&quot; = 28.723 degrees (Sep 15, 2006). The maximum between 1920 and 2050 is 28° 44' 11&quot; = 28.736 degrees (March 15, 1932).&lt;/p&gt;&#xA;" OwnerUserId="17355" LastEditorUserId="17355" LastEditDate="2018-12-07T18:10:12.930" LastActivityDate="2018-12-07T18:10:12.930" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28653" PostTypeId="2" ParentId="28629" CreationDate="2018-12-06T22:55:54.953" Score="5" Body="&lt;p&gt;I would start by doing a search on &lt;a href=&quot;https://vizier.u-strasbg.fr/viz-bin/VizieR&quot; rel=&quot;noreferrer&quot;&gt;VizieR&lt;/a&gt;, select &quot;Globular_Clusters&quot; from the right-hand list box (this tells me there are around 900 catalogs in this category) and click &quot;Find Catalogs&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the returned results will be catalogues of stars within a given cluster, the density map on the right-hand side of the list will show these as a single point. Other catalogues will be lists of multiple globular clusters, the density map will show there is data from all over the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that's useful for you!&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastActivityDate="2018-12-06T22:55:54.953" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28654" PostTypeId="2" ParentId="387" CreationDate="2018-12-07T04:24:30.850" Score="-5" Body="&lt;p&gt;Newton’s Law of universal gravitation and Kepler’s Law of planetary motion describes the movement of planets around the sun. Equations are derived from these laws and hence the birth of Celestial Mechanics. But nowhere in the literature can we find an accepted law of Planetary Rotation because everyone is convinced that there is nothing special about the rotation of the planets. Except for the “overused” explanation…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“A long time ago in a galaxy far, far away...spinning gas and dust flattened into a protoplanetary disk and due conservation of angular momentum the planets are now rotating with RANDOM velocities”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It’s the same as saying we don’t really know how it works. We have a concept but not enough to express it in numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a quote from Lord Kelvin (William Thomson):“I often say that when you can measure what you are speaking about, and express it in numbers, you know something about it; but when you cannot measure it, when you cannot express it in numbers, your knowledge is of a meagre and unsatisfactory kind; it may be the beginning of knowledge, but you have scarcely, in your thoughts, advanced to the stage of science, whatever the matter may be.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Follow the below link for the Planetary Spin Equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.quora.com/What-determines-the-rotation-period-of-planets/answer/Randy-Evangelista-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.quora.com/What-determines-the-rotation-period-of-planets/answer/Randy-Evangelista-1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25095" LastEditorUserId="25095" LastEditDate="2018-12-07T05:05:45.270" LastActivityDate="2018-12-07T05:05:45.270" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28655" PostTypeId="1" AcceptedAnswerId="28658" CreationDate="2018-12-07T06:28:33.713" Score="2" ViewCount="142" Body="&lt;p&gt;What could this square thing be? Roughly how big is it? From BGR's &lt;a href=&quot;https://bgr.com/2018/12/06/bennu-photo-asteroid-osiris-rex/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Our first up-close look at the Bennu asteroid is a messy one&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/h30FN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h30FN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nc7ri.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nc7ri.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;rotated, 2x zoomed, and sharpened. unscientifically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Gcn8w.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gcn8w.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-12-07T09:32:05.230" LastActivityDate="2018-12-07T11:02:10.510" Title="What is this thin, square-ish object on asteroid Bennu? How big is it?" Tags="&lt;asteroids&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28656" PostTypeId="1" CreationDate="2018-12-07T06:38:50.420" Score="2" ViewCount="96" Body="&lt;p&gt;The solar system and space in general has plenty of dust, and dust has major impacts on observing both visually and in radio astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I wanted to &quot;see dust&quot; in space, or see its effects, using my eyes or at most a pair of regular binoculars, where could I look and what would I &lt;em&gt;look at&lt;/em&gt; in order to say to myself &quot;yep, that's dust!&quot; or a direct result of dust's properties in space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; It doesn't necessarily have to be interstellar dust!&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-12-15T15:05:50.333" LastActivityDate="2018-12-16T00:59:07.893" Title="What direct or indirect observations of dust can one make by eye or with binoculars?" Tags="&lt;dust&gt;&lt;zodiacal-light&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28657" PostTypeId="2" ParentId="28656" CreationDate="2018-12-07T07:04:48.590" Score="3" Body="&lt;p&gt;If you can see the Milky Way, then in Summer through early Autumn you should be able to see the &quot;dark rift&quot; which runs through the middle of the milky way from Saggitarius to Deneb in Cygnus. This is a dust band in the Milky Way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you live far enough South, you can see the Coalsack, a quite dense dust cloud near the Southern Cross, and also clearly visible without optics.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-12-07T07:04:48.590" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28658" PostTypeId="2" ParentId="28655" CreationDate="2018-12-07T10:47:27.313" Score="8" Body="&lt;p&gt;In &lt;a href=&quot;https://svs.gsfc.nasa.gov/12318&quot; rel=&quot;noreferrer&quot;&gt;this composite&lt;/a&gt; of the rotating asteroid you can see that it is not thin, but actually protruding:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SNGut.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SNGut.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BxvLc.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BxvLc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sup&gt;This series of images taken by the OSIRIS-REx spacecraft shows Bennu in one full rotation from&lt;br /&gt;&#xA;a distance of around 50 miles (80 km). The spacecraft’s PolyCam camera obtained&lt;br /&gt;the thirty-six 2.2-millisecond frames over a period of four hours and 18 minutes.&lt;br /&gt;&#xA;Credit: NASA/Goddard/University of Arizona&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like an 'ordinary' boulder.&lt;br /&gt;&#xA;(You can see several more in that sequence)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply comparing the 'height' from the second picture with Bennu's diameter of 492m shows this 'boulder' to be 28 meters high.&lt;/p&gt;&#xA;" OwnerDisplayName="user1569" LastEditorDisplayName="user1569" LastEditDate="2018-12-07T11:02:10.510" LastActivityDate="2018-12-07T11:02:10.510" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28659" PostTypeId="1" AcceptedAnswerId="28663" CreationDate="2018-12-07T14:13:43.967" Score="0" ViewCount="61" Body="&lt;p&gt;The Phys.org article &lt;a href=&quot;https://phys.org/news/2018-12-unknown-treasure-trove-planets.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Unknown treasure trove of planets found hiding in dust&lt;/a&gt; reports on recently published radio measurements (~1.3mm) of planetary systems forming around stars in Taurus:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Scientists base this scenario of how our solar system came to be on observations of protoplanetary disks around other stars that are young enough to currently be in the process of birthing planets. Using the Atacama Large Millimeter Array, or ALMA, comprising 45 radio antennas in Chile's Atacama Desert, the team performed a survey of young stars in the Taurus star-forming region, a vast cloud of gas and dust located a modest 450 light-years from Earth. When the researchers imaged 32 stars surrounded by protoplanetary disks, they found that 12 of them—40 percent—have rings and gaps, structures that according to the team's measurements and calculations can be best explained by the presence of nascent planets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The article ends by saying:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Going forward, &lt;strong&gt;the research group plans to move ALMA's antennas farther apart&lt;/strong&gt;, which should increase the array's resolution to around five astronomical units (one AU equals the average distance between the Earth and the sun), and to make the antennas sensitive to other frequencies that are sensitive to other types of dust.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Our results are an exciting step in understanding this key phase of planet formation,&amp;quot; Long said, &amp;quot;and by making these adjustments, we are hoping to better understand the origins of the rings and gaps.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Changing from 15AU to 5 AU suggests a roughly factor of 3 increase in the overall scale of the pattern of antennas. The &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/1538-4357/aae8e1/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;linked paper in ApJ&lt;/a&gt; Feng Long et al. has a preprint in &lt;a href=&quot;https://arxiv.org/abs/1810.06044&quot; rel=&quot;nofollow noreferrer&quot;&gt;ArXiv&lt;/a&gt; as well, which says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;All observations were obtained from late August to early September 2017 &lt;strong&gt;using 45-47 12-m antennas on baselines of 21∼3697 m (15∼2780 kλ)&lt;/strong&gt;, with slight differences in each group (see Table 1).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Several years ago I remember reading about the design of ALMA and there are a lot more &lt;em&gt;places to put dishes&lt;/em&gt; than there are dishes. I think there were five optimized configurations that all looked to have a similar spiral shape, but each one was maybe a factor of 2 or 3 larger than the previous.&lt;/p&gt;&#xA;&lt;p&gt;Since the dishes take time to move and setup and calibrate, this probably doesn't happen that often.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How often does it happen? How often do they move most of ALMA's dishes to change resolution?&lt;/p&gt;&#xA;&lt;p&gt;If there's a site that shows how ALMA is set up now, and what the next setup will be and when, that would be great to know about as well.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SUVLI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SUVLI.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; &amp;quot;Until recently, protoplanetary disks were believed to be smooth, like pancake-like objects. The results from this study show that some disks are more like doughnuts with holes, but even more often appear as a series of rings. The rings are likely carved by planets that are otherwise invisible to us.&amp;quot; &lt;a href=&quot;https://phys.org/news/2018-12-unknown-treasure-trove-planets.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt; Credit: Feng Long/ALMA&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-12-08T02:57:22.893" Title="How often do they move most/all of ALMA's dishes?" Tags="&lt;radio-astronomy&gt;&lt;radio-telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28660" PostTypeId="2" ParentId="10660" CreationDate="2018-12-07T17:33:58.373" Score="3" Body="&lt;p&gt;The latest adaptive optics systems now allows diffraction limited images from the 8 meter Very Large Telescope in the visible spectrum.  So it seems reasonable that with continued improvements this will be feasible with larger telescopes in the future, including the 24-39 meter hug telescopes currently planned:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.eso.org/public/usa/news/eso1824/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.eso.org/public/usa/news/eso1824/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25099" LastActivityDate="2018-12-07T17:33:58.373" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28661" PostTypeId="5" CreationDate="2018-12-07T17:50:27.027" Score="0" Body="&lt;p&gt;Interstellar Dust is comprised of tiny particulate matter that has undergone the stronger intermolecular bonds to create substances like ice, silicates, and carbon compounds that float around in varying densities between the stars and between the galaxies. Since these particles are still extremely small (typically a fraction of a micron across), they can appear to be a gas, but these tiny, irregularly-shaped objects still exist individually in a solid or liquid state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dust clouds absorb light that passes through them, causing extinction of that light, and so appear as dark shadows against the Milky Way or bright nebulae.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="6" LastEditDate="2018-12-10T14:27:58.210" LastActivityDate="2018-12-10T14:27:58.210" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28662" PostTypeId="4" CreationDate="2018-12-07T17:50:27.027" Score="0" Body="Questions about particulate matter found between stars, not to be confused with gas or ionised plasma. For planetary dust storms, use [dust-storm]." OwnerUserId="2365" LastEditorUserId="37191" LastEditDate="2021-02-16T14:11:08.947" LastActivityDate="2021-02-16T14:11:08.947" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28663" PostTypeId="2" ParentId="28659" CreationDate="2018-12-07T18:30:40.963" Score="2" Body="&lt;p&gt;The list of array configurations is given in the &lt;a href=&quot;https://almascience.nrao.edu/observing/proposing/proposers-guide#section-53&quot; rel=&quot;nofollow noreferrer&quot;&gt;Proposers Guide&lt;/a&gt; and ten configurations have been defined for the 43 12-meter antennas. They have the form &lt;code&gt;C43-x&lt;/code&gt; where &lt;code&gt;x&lt;/code&gt; goes from 1 (most compact; 15-161 meter separation) to 10 (least compact; 244-16200 meter separation). The changes between configurations are shown in the &lt;a href=&quot;https://almascience.nrao.edu/observing/observing-configuration-schedule&quot; rel=&quot;nofollow noreferrer&quot;&gt;Configuration Schedule&lt;/a&gt;; over the ~12 months of Cycle 6, it looks like they change configuration 14 times. There is a month long &quot;Antennae Relocation Shutdown&quot; in May 2019 as they transition from fairly compact to least compact - presumably because it takes a longer time to move the 43 dishes out multiple kilometers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, here is a snapshot of the planned schedule for a year starting 01-Oct-2018 as it stands now:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0j4ra.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0j4ra.png&quot; alt=&quot;ALMA configuration schedule&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23748" LastEditorUserId="7982" LastEditDate="2018-12-08T02:57:22.893" LastActivityDate="2018-12-08T02:57:22.893" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28664" PostTypeId="1" AcceptedAnswerId="28665" CreationDate="2018-12-07T19:36:30.947" Score="4" ViewCount="157" Body="&lt;p&gt;Near Earth Objects (NEO's) have been mapped by NASA and JPL since 1990's. The graph below shows how many of these potential planet-killers have been found and classified over the years. I am curious as to why there have been so many more 140 meter to 1 km NEO's discovered, as opposed to NEO's with a diameter of 1 km or more. I think it is as simple as there are just more &quot;smaller&quot; objects left over from the formation of the solar system? Am I correct in saying this, or are there deeper astronomical processes at play here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G5AKg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G5AKg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24911" LastActivityDate="2018-12-07T23:49:02.317" Title="Discovery of Near Earth Objects (NEO's)" Tags="&lt;asteroids&gt;&lt;near-earth-object&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28665" PostTypeId="2" ParentId="28664" CreationDate="2018-12-07T23:49:02.317" Score="3" Body="&lt;p&gt;This is because there are many more small NEOs than there are large ones and the larger ones are easier to find and so were found first. The numbers of NEOs follow what's know as a power law distribution with an exponent of ~1.75. This means that given the roughly factor 7 difference between 140m and 1000m NEOs, there will be 7^1.75 = ~31 times as many 140m NEOs as 1000m NEOs. This is shown in this plot below. &lt;a href=&quot;https://i.stack.imgur.com/Z8KoS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z8KoS.png&quot; alt=&quot;NEO size-frequency distribution&quot;&gt;&lt;/a&gt; We estimate there are about 950 NEOs 1km or larger and about 25,000 NEOs larger than 140m. We have found over 93% of the 1+km NEOs and so there aren't many left to find. In comparison, over 2/3rds of the 140-1000m NEOs still remain to be found, most in the 140-300m size range. However this improved considerably since even 2010 when about 82% of the population was still to be found.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-12-07T23:49:02.317" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28666" PostTypeId="1" AcceptedAnswerId="39656" CreationDate="2018-12-08T03:18:32.123" Score="6" ViewCount="178" Body="&lt;p&gt;The ALMA (Atacama Large Millimetre/submillimetre Array) radio telescope's band-10 capabaility is now operational, per &lt;a href=&quot;https://astronomy.stackexchange.com/a/28663/7982&quot;&gt;this answer&lt;/a&gt;. That's confirmed by NRAO's &lt;a href=&quot;https://public.nrao.edu/news/2018-alma-band10/&quot; rel=&quot;noreferrer&quot;&gt;First Science with ALMA’s Highest-Frequency Capabilities&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://www.eso.org/public/teles-instr/alma/receiver-bands/&quot; rel=&quot;noreferrer&quot;&gt;this site&lt;/a&gt; ALMA's bands and frequencies are as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;band  wavelength   noise   frequency&#xA;        (mm)        (K)      (GHz)&#xA; 1    6.0 - 8.5     26     35 -  50&#xA; 2    3.3 - 4.5     47     65 -  90&#xA; 3    2.6 - 3.6     60     84 - 116&#xA; 4    1.8 - 2.4     82    125 - 163&#xA; 5    1.4 - 1.8    105    163 - 211&#xA; 6    1.1 - 1.4    136    211 - 275&#xA; 7    0.8 - 1.1    219    275 - 373&#xA; 8    0.6 - 0.8    292    385 - 500&#xA; 9    0.4 - 0.5    261    602 - 720&#xA;10    0.3 - 0.4    344    787 - 950&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;900 GHz (0.9 THz) is quite a high frequency for a radio receiver! Each ALMA dish down-converts received frequency to a baseband of a few GHz before they get digitized and sent to the correlator for digital interferometry, but you still need an ultra-stable local oscillator (LO) for downconversion, and all of the LO's of all of the dishes need to be mutually coherent. That's quite a feat considering they can be tens of kilometers apart!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How does ALMA produce stable, mutually coherent ~THz local oscillators for all of their dishes?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As suggested in this &lt;a href=&quot;https://astronomy.stackexchange.com/a/20086/7982&quot;&gt;partial answer&lt;/a&gt; to the question &lt;em&gt;How does the Event Horizon Telescope implement the interferometry?&lt;/em&gt; hard drives collect digital data down-converted by local atomic clocks, probably using GPS as a secondary reference, then brought to a central location for post analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am guessing that they spend a significant amount of time trying to reconstruct coherence at the millimetre (picosecond) level, but that option is not available for ALMA as data is continuously recorded at some large fraction of 24/7, and the huge volume of data at a location where magnetic hard drives don't work is overwhelming. So they need to get it right the first time.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-12-08T03:28:21.763" LastActivityDate="2020-11-13T00:48:31.420" Title="How does ALMA produce stable, mutually coherent ~THz local oscillators for all of their dishes?" Tags="&lt;radio-astronomy&gt;&lt;radio-telescope&gt;&lt;instruments&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28667" PostTypeId="1" AcceptedAnswerId="28670" CreationDate="2018-12-08T03:28:04.337" Score="-1" ViewCount="550" Body="&lt;p&gt;The Moon does not rotate on its own axis. If I am right, for something to gain a spherical shape it has to rotate on its own axis. We know that the Moon does not rotate on its own axis. How can this be?&lt;/p&gt;&#xA;" OwnerUserId="23759" LastEditorUserId="23940" LastEditDate="2018-12-10T08:02:13.123" LastActivityDate="2018-12-10T08:02:13.123" Title="Why is the shape of the Moon spherical?" Tags="&lt;the-moon&gt;&lt;rotation&gt;" AnswerCount="1" CommentCount="8" ClosedDate="2018-12-08T17:19:01.173" ContentLicense="CC BY-SA 4.0" />
  <row Id="28669" PostTypeId="2" ParentId="14808" CreationDate="2018-12-08T03:54:49.663" Score="5" Body="&lt;p&gt;It is interesting, it looks like you can combine a &quot;normal&quot; lens with a fresnel to correct for chromatic abberation and have a lighter lens. I don't think it helps with making a bigger refractor though, but maybe a more portable one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Nikon Phase Fresnel, and Canon Diffractive Optics are lenses you can buy now with fresnel lenses. They seem to have extra flaring, which you may not want in a astrophotograph.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.cloudynights.com/topic/598354-astrophotography-with-do-pf-lenses/&quot; rel=&quot;noreferrer&quot;&gt;https://www.cloudynights.com/topic/598354-astrophotography-with-do-pf-lenses/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Phase Frenel lenses:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.nikonusa.com/en/learn-and-explore/a/ideas-and-inspiration/phase-fresnel-from-wildlife-photography-to-portraiture.html&quot; rel=&quot;noreferrer&quot;&gt;https://www.nikonusa.com/en/learn-and-explore/a/ideas-and-inspiration/phase-fresnel-from-wildlife-photography-to-portraiture.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://petapixel.com/2015/01/10/closer-look-nikons-new-phase-fresnel-pf-lens-technology/&quot; rel=&quot;noreferrer&quot;&gt;https://petapixel.com/2015/01/10/closer-look-nikons-new-phase-fresnel-pf-lens-technology/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Canon DO:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://cpn.canon-europe.com/content/education/infobank/lenses/multi_layer_diffractive_optical_element.do&quot; rel=&quot;noreferrer&quot;&gt;https://cpn.canon-europe.com/content/education/infobank/lenses/multi_layer_diffractive_optical_element.do&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="25103" LastEditorUserId="25103" LastEditDate="2018-12-08T05:14:43.817" LastActivityDate="2018-12-08T05:14:43.817" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28670" PostTypeId="2" ParentId="28667" CreationDate="2018-12-08T07:16:27.390" Score="4" Body="&lt;p&gt;The question is based on two false premises, that the Moon does not rotate about its own axis (it does), and that for something to gain spherical shape it has to rotate own its own axis (this is irrelevant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addressing the first premise, the translation and rotation behaviors of any rigid body can be described in terms of the translational behavior of the body's center of mass and the rotational behavior of the body about its center of mass. In the special case of an orbiting object that is tidally locked to a larger object, it appears to observers on the larger object that the smaller object is not rotating. For example, people on the Earth looking at the Moon. From the perspective of an object orbiting Venus, it would be quite clear that the Earth's Moon does indeed rotate about its own axis. Every object rotate about its own axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addressing the second premise, rotation about its own axis has nothing to do with whether an object of sufficient size becomes more or less spherical over time. What matters is the object's mass and composition. A key concept in this regard is the potato radius, about 300 kilometers for icy objects, 400 kilometers for rocky objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Self gravitation is overwhelming for objects significantly larger than the potato radius. Sufficiently large objects are inevitably be drawn into a spherical shape over time because self gravitation overwhelms the comparatively puny chemical forces that would otherwise maintain the object's non-spherical shape. For small objects, it's self gravitation that is the puny force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The forces that work toward making a body take on a more or less spherical shape scale roughly with the cube of linear size. The size of our Moon is over four times that of the potato radius, over eighty times larger when cubed. Except for objects made of &lt;em&gt;unobtanium&lt;/em&gt;, any object that is ~3500 kilometers across will inevitably be drawn into a spherical shape over time.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2018-12-08T07:16:27.390" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28671" PostTypeId="1" CreationDate="2018-12-08T10:06:42.367" Score="1" ViewCount="70" Body="&lt;p&gt;What frequency distribution law applies to a random point in deep space that is assumed to be far away from any particular source of radiation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can it be expressed by some form of Plancks/Bose-Einsteins/Fermi-Diracs laws or a mix of all of them? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the mean frequency/wawelength?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the mysterious &quot;dark energy&quot; be a major contributor to the distribution?&lt;/p&gt;&#xA;" OwnerUserId="17223" LastActivityDate="2018-12-12T21:24:57.287" Title="Deep space radiation distribution" Tags="&lt;radiation&gt;&lt;interstellar-medium&gt;&lt;wavelength&gt;&lt;cosmic-microwave-background&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28672" PostTypeId="1" AcceptedAnswerId="28673" CreationDate="2018-12-08T15:29:55.417" Score="5" ViewCount="100" Body="&lt;p&gt;The radial velocity technique has apparently been used to discover exoplanets with extremely long orbital periods of ~10,000 days. For example, see this link:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://exoplanetarchive.ipac.caltech.edu/cgi-bin/TblView/nph-tblView?app=ExoTbls&amp;amp;config=planets&quot; rel=&quot;noreferrer&quot;&gt;https://exoplanetarchive.ipac.caltech.edu/cgi-bin/TblView/nph-tblView?app=ExoTbls&amp;amp;config=planets&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the period of a planet is ~10,000 days, then the period of the star must be the same. However, this would mean that observations would need to be made over ~30 years to get a complete picture of the radial velocity variation of the star, a time which is far too long. How then are radial velocity measurements used to discover such planets?&lt;/p&gt;&#xA;" OwnerUserId="16092" LastActivityDate="2018-12-08T17:18:44.550" Title="How are exoplanets with 10,000 day periods discovered with the radial velocity method?" Tags="&lt;observational-astronomy&gt;&lt;exoplanet&gt;&lt;doppler-effect&gt;&lt;radial-velocity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28673" PostTypeId="2" ParentId="28672" CreationDate="2018-12-08T17:18:44.550" Score="1" Body="&lt;p&gt;You don't need to collect data for one cycle in order to determine the period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Newtonian's physics, we know precisely how a trajectory of one particle orbitting around another one would be. Since the instantaneous velocity (v_inst) at any point on the trajectory is tangential to the point on the trajectory, we can decompose the v_inst into the radial velocity which we can match with the observation from, e.g., redshift. With assumption of constant v_inst and trajectory, we can determine the period from just a couple of observations (depending on how accurate the redshift can be measured, and the number of variables in the equation).&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-12-08T17:18:44.550" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28674" PostTypeId="2" ParentId="28639" CreationDate="2018-12-08T17:43:53.577" Score="4" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Is the nose of the solar system and the solar apex the same thing?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;No, they're roughly 60 degrees apart.&lt;/p&gt;&#xA;&lt;h1&gt;The Solar Apex.&lt;/h1&gt;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_apex&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar apex&lt;/a&gt; does indeed indicate &lt;em&gt;a component&lt;/em&gt; of direction of travel within our corner of the galaxy:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The coordinates as obtained by visual observation of the&#xA;apparent motion is right ascension 18h 28m 0s and declination&#xA;of 30° North in galactic coordinates: 56.24° longitude, 22.54°&#xA;latitude. The radioastronomical position is 18h 03m 50.2s and dec&#xA;30° 00′ 16.8″ galactic coordinates: 58.87° longitude, 17.72°&#xA;latitude.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;There are two different precise measurements here which conflict, I won't speculate as to why Wikipedia is unreferenced on the subject. The galaxy is a place containing a great deal of complex movement and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Local_standard_of_rest&quot; rel=&quot;nofollow noreferrer&quot;&gt;Local Standard of Rest&lt;/a&gt; is moving relative to the galactic plane, the centre of the galaxy and the other local stars:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In astronomy, the local standard of rest or LSR follows the mean&#xA;motion of material in the Milky Way in the neighborhood of the Sun.&#xA;The path of this material is not precisely circular. The Sun follows&#xA;the solar circle (eccentricity e &amp;lt; 0.1 ) at a speed of about 255 km/s&#xA;in a clockwise direction when viewed from the galactic north pole at a&#xA;radius of ≈ 8.34 kpc about the center of the galaxy, and&#xA;has only a slight motion, towards the solar apex, relative to the LSR.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;We're traveling relative to an ever shifting cloud of stars towards Hercules (also shifting), southwest of the star Vega.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The sun's motion in the Milky Way is not confined to the galactic&#xA;plane; it also shifts (&amp;quot;bobs&amp;quot;) up and down with respect to the plane&#xA;over millions of years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So you see, there are many components of direction of travel to consider.&lt;/p&gt;&#xA;&lt;p&gt;There's an unquotable abstract from a Harvard article &lt;a href=&quot;http://adsabs.harvard.edu/full/1981ApJ...248..408D&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, which may give you an idea of how much needs to be taken into account in determining the relative vectors.&lt;/p&gt;&#xA;&lt;h1&gt;The Nose.&lt;/h1&gt;&#xA;&lt;p&gt;The &amp;quot;&amp;quot;Nose&amp;quot; of the solar system is sometimes used to refer to the direction of travel of the local galactic wind. Just as a comet's tail points away from the sun, the tail of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Heliosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heliopause&lt;/a&gt; points away from the wind, the nose points in the direction it's coming from.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o6YUR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o6YUR.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The nose is as you have stated is pointing towards Scorpius.&lt;/p&gt;&#xA;" OwnerUserId="768" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-12-08T19:53:32.653" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28676" PostTypeId="2" ParentId="14808" CreationDate="2018-12-08T21:11:20.073" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If we were to make telescopes with Fresnel lenses instead of regular lenses, would it be more practical?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is a Wikipedia article about using &lt;a href=&quot;https://en.wikipedia.org/wiki/Fresnel_Imager&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fresnel zone plates&lt;/a&gt; successfully in the visible spectrum. A paper about it called: &quot;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2012ExA....33..129K&quot; rel=&quot;nofollow noreferrer&quot;&gt;First high dynamic range and high resolution images of the sky obtained with a diffractive Fresnel array telescope&lt;/a&gt;&quot; was written by Laurent Koechlin et al. in 2011:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;... this concept is well fitted for space missions in any spectral domain, from far UV (100 nm) to IR (25 μm). However, the advantages of Fresnel arrays become outstanding in the UV domain, for which a 6 m size aperture yields milliarcseconds angular resolutions with a very rough manufacturing precision compared to standard optics: typically 50 μm compared to 10 nm.&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From Laurent Koechlin's paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f61XN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f61XN.jpg&quot; alt=&quot;Figure 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;strong&gt;Fig. 1&lt;/strong&gt; Top: Fresnel array, close view on the central zones. Bottom: sketch of our prototype, not to scale for clarity. On the real prototype, the Fresnel lens and the achromat doublet are much&#xA;  smaller than the entrance Fresnel array. The distance between the Fresnel array and the field&#xA;  optics is 18 m. The rest of the light path is short (2 m). The zero-order mask blocks the light that has not been focused by the Fresnel array: all diffraction orders are blocked, except one. The achromat forms the final image after chromatic correction by the secondary Fresnel lens.&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another flat lens is the &lt;a href=&quot;https://www.seas.harvard.edu/news/2018/01/single-metalens-focuses-all-colors-of-rainbow-in-one-point&quot; rel=&quot;nofollow noreferrer&quot;&gt;metalens&lt;/a&gt; which uses structures on a flat  surface to focus each frequency individually, but &quot;Blade Optics&quot; (prisms) might prove to be the most practical, more below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Would this also allow for much bigger telescopes to be made?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, they will be bigger; but not in a &lt;em&gt;good&lt;/em&gt; way:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Fresnel arrays of 6m and larger have focal lengths of a few kilometers in the UV; they will require two satellites flying in formation around &lt;a href=&quot;https://en.wikipedia.org/wiki/Lagrangian_point&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lagrangian point L2&lt;/a&gt;, but with tolerant positioning in translation (a few centimeters).&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With modern technology that isn't considered completely impractical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Blade Optics&quot; have both supporters and naysayers. They have been &lt;a href=&quot;http://calgary.rasc.ca/nexoptic.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;tested by RASC&lt;/a&gt; and a few patents (&lt;a href=&quot;https://patents.google.com/patent/US20170307864A1/en&quot; rel=&quot;nofollow noreferrer&quot;&gt;US 2017 / 0307864 A1&lt;/a&gt;) have been issued.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principle is simple: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fR5c9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fR5c9.png&quot; alt=&quot;Image from Spectrum Optix&amp;#39;s patent&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://nexoptic.com/project/proof-of-concept-prototype/&quot; rel=&quot;nofollow noreferrer&quot;&gt;NexOptix&lt;/a&gt; has a working prototype which is far shorter than an equivalent focal length refractor (straight tube), Newtonian (right angle eyepiece) or even Schmidt-Cassegrain (&lt;em&gt;folded&lt;/em&gt;) design:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2EOja.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2EOja.png&quot; alt=&quot;NexOptic BladeOptic Telescope&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That squeezes 146 cm (57.48 inches) of focal &lt;strong&gt;length&lt;/strong&gt; into just 12.7 cm (5 inches), though the other dimensions increase (but not proportionally).&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="18054" LastEditDate="2018-12-08T21:17:42.147" LastActivityDate="2018-12-08T21:17:42.147" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28677" PostTypeId="1" AcceptedAnswerId="28679" CreationDate="2018-12-08T21:56:36.803" Score="26" ViewCount="7372" Body="&lt;p&gt;Would that person go blind?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutrino detectors and the abundance of Neutrinos would detect the upcoming visible show about 3 hours before any visible signs, so there would be time to point certain telescopes that could handle the brightness towards it.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious if an individual with a telescope pointed in that direction would have an unpleasant surprise.    Would the scientific community be wise to not announce the massive stellar explosion until after it's visible to avoid potential negative effects from over-eager amateur astronomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize this is a kind of silly question and it might depend too much on the telescope, but I'm curious.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2020-01-11T19:13:52.020" Title="What would happen if someone had a telescope and watched Betelgeuse when it goes supernova?" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;supernova&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28678" PostTypeId="2" ParentId="28677" CreationDate="2018-12-08T22:52:22.877" Score="4" Body="&lt;p&gt;Brightness varies inversely with the square of the distance. Betelgeuse is about 642.5 light years away and has an apparent magnitude of 0.42. My grasp of apparent magnitude concepts is a bit wobbly, but I believe if it grew a million times as bright, it might have an apparent magnitude of -14.5 or so, which is a lot more like the brightness of the moon than the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the great distance, the decrease in brightness due to distance, and the countless amounts of dust &amp;amp; gas between earth and Betelgeuse, I think you'd probably be fine. You might be dazzled by its brightness -- a bit like looking at a light bulb, I imagine -- but I doubt it would cause any physical harm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I hope a real astronomer sounds off here. I'm not sure what kind of supernova we might expect from Betelgeuse, but apparently supernovas (supernovae?) can achieve a theoretical brightness equal to 5 trillion suns!&lt;/p&gt;&#xA;" OwnerUserId="16705" LastActivityDate="2018-12-08T22:52:22.877" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28679" PostTypeId="2" ParentId="28677" CreationDate="2018-12-08T23:22:56.307" Score="39" Body="&lt;p&gt;No, it would not be a problem. Supernovae are not at all like flashbulbs – they brighten over a period of many days and dim again even more slowly. Here are a number of different light curves taken from Wikipedia:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/GH7QI.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GH7QI.png&quot; alt=&quot;Luminous output of different types of Supernovae&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rise is fast on an astronomical scale – several orders of magnitude over a period of roughly ten days – but &lt;em&gt;very&lt;/em&gt; slow on a human scale. An amateur looking at it would not notice any significant change in brightness, but if the same person came back a few hours later or the next night, the change would be very evident.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as we can tell, the reason is that the light at peak brightness is caused by emissions from material blown off by the explosion. For example, in Type 1a SNe, most of the light is from the radioactive decay of the &lt;em&gt;huge&lt;/em&gt; mass of ejected nickel-56 (half life 6 days).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Supernova&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article on supernovae&lt;/a&gt; is quite good and covers this all in more detail.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="1435" LastEditDate="2018-12-10T01:51:56.673" LastActivityDate="2018-12-10T01:51:56.673" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28680" PostTypeId="2" ParentId="28677" CreationDate="2018-12-08T23:57:24.797" Score="24" Body="&lt;p&gt;If you insist on observing the exploding Betelgeuse at peak brightness, you could potentially damage your eye. The complete answer enters the realm of physiology. Here I'll discuss the astronomical parts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Betelgeuse will explode as a type II supernova, the typical brightness of which is around &lt;span class=&quot;math-container&quot;&gt;$M \sim -17$&lt;/span&gt;. With a distance of &lt;span class=&quot;math-container&quot;&gt;$d\simeq200\,\mathrm{pc}$&lt;/span&gt;, its distance modulus is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\mu = 5\log(d/\mathrm{pc}) - 5 \simeq 6.5,&#xA;$$&lt;/span&gt;&#xA;so its apparent magnitude will be&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;m = M + \mu \simeq -10.5.&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For these calculations I assume that the Sun is the threshold for damaging your eye (a brief look at the Sun is okay, a longer look will cause permanent damage. But… physiology…). The Sun has an apparent magnitude of &lt;span class=&quot;math-container&quot;&gt;$m_\odot = -26.7$&lt;/span&gt;, i.e. it is &lt;span class=&quot;math-container&quot;&gt;$\Delta m = 16.2$&lt;/span&gt; magnitudes brighter. In other words, Betelgeuse will be&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;f = 10^{\Delta m/2.5} \simeq 3\times10^6&#xA;$$&lt;/span&gt;&#xA;times dimmer than the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the Sun is an extended source, spanning an angle of roughly &lt;span class=&quot;math-container&quot;&gt;$\theta_\mathrm{Sun} = 32$&lt;/span&gt; arcminutes across. In contrast, Betelgeuse is a point source, which when transferred through the atmosphere and the telescope, is spread out over &lt;span class=&quot;math-container&quot;&gt;$\theta_\mathrm{Bet} \sim$&lt;/span&gt; a few square arcseconds. Thus its light will be more concentrated; i.e. it will be much brighter, but it will hit a much smaller area of your retina. However, your eye will also move around, smearing out the light. Not being a physiologist, for the sake of this calculation I assume that the light is smeared out over a disk 1 arcminute across (about the size of a planet seen from Earth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the factor &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt; will itself be a factor &lt;span class=&quot;math-container&quot;&gt;$(\theta_\mathrm{Sun} / \theta_\mathrm{Bet})^2 \simeq 1000$&lt;/span&gt; times larger — that is, Betelgeuse is only &lt;span class=&quot;math-container&quot;&gt;$\sim 3\,000$&lt;/span&gt; times dimmer than the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, for our assumptions your eye will be damaged if you observe exploding Betelgeuse through a telescope with an area &lt;span class=&quot;math-container&quot;&gt;$\sim 3\,000$&lt;/span&gt; larger — or roughly 55 times wider —  than your pupil. In bright light, the pupil contracts to roughly 3 mm in diameter, so if observing through a telescope of 16 cm or larger, you could damage your eye. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on evolutionary models of Betelgeuse, &lt;a href=&quot;https://arxiv.org/abs/1406.3143v2&quot; rel=&quot;noreferrer&quot;&gt;Dolan et al. (2016)&lt;/a&gt; estimate an apparent magnitude of &lt;span class=&quot;math-container&quot;&gt;$m=-12.4$&lt;/span&gt;, i.e. roughly 6 times brighter than our estimate. This would mean that you only need a 7 cm telescope to damage your eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as Mark writes in his answer, supernovae don't increase to their peak brightness in matters of seconds, but rather in matters days (roughly half a mag per day), so you have plenty of time to look away.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2018-12-11T09:55:26.417" LastActivityDate="2018-12-11T09:55:26.417" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28681" PostTypeId="1" AcceptedAnswerId="28690" CreationDate="2018-12-09T01:17:26.890" Score="1" ViewCount="201" Body="&lt;p&gt;I heard about the radial speed method to discover exoplanets a few years ago, however, these questions kept confusing me. If we know a planet star of mess &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; is orbiting a star by interpreting the  radial speed of the central mass, how can we tell wether this is a even more massive object whose orbiting plane has a larger inclination? Further more, even if we know the angle of our sight and the axis of orbit, can we determine the direction of orbit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastEditorUserId="24351" LastEditDate="2018-12-09T13:29:20.883" LastActivityDate="2018-12-09T13:29:20.883" Title="How can we learn the inclination of the orbiting plane of the exoplanets using radial speed method?" Tags="&lt;orbit&gt;&lt;exoplanet&gt;&lt;inclination&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28682" PostTypeId="1" CreationDate="2018-12-09T02:53:14.823" Score="7" ViewCount="137" Body="&lt;p&gt;I don't fully understand why clouds of particles which turn into galaxies and solar systems usually result in an (approximately) 2D shape, but whatever. Would they still result in an almost 2D shape if our space was n dimensional with &lt;span class=&quot;math-container&quot;&gt;$ n &amp;gt; 3 $&lt;/span&gt; and the laws of physics were the same (adapted for that &lt;span class=&quot;math-container&quot;&gt;$ n $&lt;/span&gt;)? Or would it be an (approximately) &lt;span class=&quot;math-container&quot;&gt;$ n-1 $&lt;/span&gt; dimensional shape?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be of interest to have an answer to this question under the assumption that the attractive force was still proportional to &lt;span class=&quot;math-container&quot;&gt;$1/r^2$&lt;/span&gt; even when the dimensionality is greater than 3. &lt;/p&gt;&#xA;" OwnerUserId="25107" LastEditorUserId="7752" LastEditDate="2018-12-11T01:03:27.417" LastActivityDate="2018-12-11T01:03:27.417" Title="If space was n&gt;3 dimensional, would clouds of particles still turn into 2D disks due to gravity?" Tags="&lt;accretion-discs&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28683" PostTypeId="1" AcceptedAnswerId="28792" CreationDate="2018-12-09T03:11:48.330" Score="8" ViewCount="316" Body="&lt;p&gt;In very long baseline interferometry, it is typical to use a hydrogen maser atomic clock to sample data. However, given how expensive these atomic clocks are, it seems surprising that more observatories aren't using non-local atomic clocks, such as GPS clocks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Very-long-baseline_interferometry&quot; rel=&quot;noreferrer&quot;&gt;This&lt;/a&gt; Wikipedia article says &quot;The antenna signal is sampled with an extremely precise and stable atomic clock (usually a hydrogen maser) that is additionally locked onto a GPS time standard.&quot; My question is, why not do away with the hydrogen maser entirely?&lt;/p&gt;&#xA;" OwnerUserId="25108" LastActivityDate="2018-12-18T20:36:18.183" Title="Why aren't GPS clocks sufficient to synchronize VLBI observations?" Tags="&lt;time&gt;&lt;vlbi&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28684" PostTypeId="2" ParentId="28682" CreationDate="2018-12-09T03:15:43.237" Score="6" Body="&lt;p&gt;It just so happens that this problem has been rigorously analyzed.  Max Tegmark wrote an interesting paper on it which can be &lt;a href=&quot;https://arxiv.org/abs/gr-qc/9702052&quot; rel=&quot;noreferrer&quot;&gt;gotten from the arXiv&lt;/a&gt;.  It is not super-technical, so I recommend you look at it.  Basically, with more than three space dimensions there are no stable orbits. Particles either disperse to infinity or collapse to a point (presumably a black hole).  (Tegmark also analyzes the effect of extra time dimensions -- weird.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: A comment asks if a &lt;em&gt;cloud&lt;/em&gt; of particles behaves differently and might still be stable. The answer is &quot;no&quot;.  Here's why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two cases here &quot;gas&quot; and &quot;dust&quot;. The difference is that gasses have internal pressure which does affect the cloud's dynamics, while the particles in dust are big enough that collisions between them become infrequent and can be ignored for their effect on the dynamics. (Note: the only important difference between &quot;gas&quot; and &quot;dust&quot; is if the particles interact often enough to produce enough pressure to significantly affect dynamics.  It's not their composition. The terms &quot;gas&quot; and &quot;dust&quot; are simply terms of art used to describe the two cases.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dust case is trivial: The dust particles spend nearly all of the time in their individual orbits without colliding and if the orbits are not stable, the dust cloud isn't either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gas is more complicated, but it's probably easiest to look at what happens in three dimensions.  A non-rotating gas cloud collapses -- that's how galaxies and stars form!  A gas cloud that's rotating enough to support itself against collapse is relying on the centrifugal force to support it against gravity -- which doesn't work in four or more dimensions.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2018-12-09T14:37:20.983" LastActivityDate="2018-12-09T14:37:20.983" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28685" PostTypeId="1" AcceptedAnswerId="28688" CreationDate="2018-12-09T06:03:48.797" Score="0" ViewCount="210" Body="&lt;p&gt;Isn't Titan better to colonize than Mars? It has a thick atmosphere which is made up of nitrogen like our planet's. It doesn't have a magnetic field, but it does get a type of protection from Saturn. Maybe the only downside I see is that it is too far away. While Mars has neither a thick atmosphere nor a magnetic field, it is bombarded with solar radiation every day.&lt;/p&gt;&#xA;" OwnerUserId="25109" LastEditorUserId="23940" LastEditDate="2018-12-10T08:02:31.477" LastActivityDate="2018-12-10T08:02:31.477" Title="Is Titan Better or Mars?" Tags="&lt;planetary-atmosphere&gt;&lt;mars&gt;&lt;magnetic-field&gt;&lt;radiation&gt;&lt;titan&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2018-12-09T10:24:57.157" ContentLicense="CC BY-SA 4.0" />
  <row Id="28686" PostTypeId="2" ParentId="28685" CreationDate="2018-12-09T06:51:21.383" Score="0" Body="&lt;p&gt;I don't know if this question will make it past opinion close votes, but my favorite line about colonizing Mars came from &lt;a href=&quot;https://www.youtube.com/watch?v=TDYDdr4HVdQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bill Nye in an interview&lt;/a&gt;.   He said (and I'm paraphrasing) that we should think of Mars as like Antarctica.   At any given time there's 700 or so scientists living on Antarctica doing various studies.   But nobody wants to live there.   Nobody wants to move to Antarctica and stay there.   It's too barren.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He said he'd love to go to Mars if he gets the opportunity, but he'd want a round trip ticket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think either is a good colonization option.   I don't disagree with a scientist who might want to go to Mars or Titan for the rest of their life, to dedicate their work to that - that's fine, and it's a choice.   I'm not sure it's call it colonization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A vacuum is one of the more difficult and expensive things to build technology for.   Because Titan has an atmosphere, in many ways it's easier to set up a human base on.   It's also a lot farther, so you have to either build faster spacecraft or deal with a several year journey and that's a huge disadvantage for Titan.   A 6 month journey to Mars is hard enough to plan for.    But once people get to titan, air tight is much easier to build than vacuum resistant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both have advantages and disadvantages.    I don't think there's a true answer to which is better.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars is much better known, and that appeal could make it the next body in space that people land on, unless we land on an asteroid first or return to the Moon next.   But Mars is being discussed and ideas are being explored.   Titan, to my knowledge, isn't in that kind of stage of discussion, so Mars, may or may not be better but it's much more likely to be next in line.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-12-09T06:51:21.383" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28687" PostTypeId="2" ParentId="28681" CreationDate="2018-12-09T08:39:20.783" Score="1" Body="&lt;p&gt;I assume you are talking about the &lt;em&gt;radial&lt;/em&gt; velocity method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analysis of a radial velocity curve yields the &lt;em&gt;minimum&lt;/em&gt;  mass of a planet &lt;span class=&quot;math-container&quot;&gt;$M \sin i$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is the true planetary mass and &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; is its orbital inclination (90 degrees would mean the orbital plane is in our line of sight).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, with no other information that's all you can say. In principle &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; could have any value &lt;span class=&quot;math-container&quot;&gt;$\geq M\sin i$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course increasingly small values of &lt;span class=&quot;math-container&quot;&gt;$\sin i$&lt;/span&gt; are increasingly unlikely and the average &lt;span class=&quot;math-container&quot;&gt;$\sin i$&lt;/span&gt; of a randomly oriented population is &lt;span class=&quot;math-container&quot;&gt;$\pi/4$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you see a transit then you know that &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; is close to 90 degrees and can in fact estimate &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; from the transit duration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you measure the &lt;em&gt;star's&lt;/em&gt; rotation period and can estimate it's projected equatorial velocity from spectral line broadening, then you can estimate it's inclination to the line of sight. One could then assume that the planet orbits in the equatorial plane of the star and use the same inclination for its orbit.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-12-09T08:39:20.783" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28688" PostTypeId="2" ParentId="28685" CreationDate="2018-12-09T09:21:44.313" Score="2" Body="&lt;p&gt;We have no practical means to travel to either planet and set up even a base, let alone a colony.  As such neither is better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that we have no reason to do it and we would need an extremely good reason because it would almost certainly involved the expenditure of vast resources to even set up a base, let alone a viable colony.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a practical sense neither have any real human habitability options other that entirely sealed bases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However current knowledge suggests that the geology on Titan would be really challenging.  We'd be operating a base at significantly higher temperatures than the local norms and as such we'd always be subject to problems in a atmosphere with active weather.  The only attraction on Titan is studying the locale, but we'd have done that in enormous depth before we could even consider a colony, so it's a mute point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars is closer, less hostile and more familiar.  We'd go there before Titan for the simple reasons it's easier, cheaper and more likely to succeed.  But again, why ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we'd be unlikely to do either before we'd establish a viable self-sustaining Moon base of at least &quot;colony&quot; status.  Again that's not even remotely possible and would not be possible without a unified global effort.  Well this particular globe can't even agree the wording of one sentence in a document about it's own report on climate change.  &lt;em&gt;They can't even agree about one lousy word !&lt;/em&gt;  So no one is going to even start talking about the possibility of setting up a committee to discuss the likelihood of setting  up another committee to start planing for a colonization effort research program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, even if we solved the technical problems to make it worth deciding which place to colonize, we have no political chance in Hell of actually doing it.  Humanity will die squabbling in a committee room.  No colonists required until we replace humans with something that works.  Skynet anyone ? :-)&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2018-12-09T09:21:44.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28690" PostTypeId="2" ParentId="28681" CreationDate="2018-12-09T13:13:30.793" Score="1" Body="&lt;p&gt;Most of the time you can't determine the inclination of a planet detected by the radial velocity method, but in certain favourable cases the dynamical interactions between two planets in the same system will allow you to constrain the inclination. This has been used to determine the inclination of the resonant planets in the Gliese 876 system, e.g. &lt;a href=&quot;https://ui.adsabs.harvard.edu/?#abs/2005ApJ...634..625R&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rivera et al. (2005)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise you would need to combine radial velocities with another detection method, for example transits or astrometry.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastActivityDate="2018-12-09T13:13:30.793" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28691" PostTypeId="1" AcceptedAnswerId="28704" CreationDate="2018-12-09T14:31:24.083" Score="8" ViewCount="351" Body="&lt;p&gt;So I've been having fun with trying to determine where the perihelia of different planets are (using JPL Horizons).&lt;/p&gt;&#xA;&lt;p&gt;And by the time I got to Saturn, something weird started happening. I've noticed that at some moments true anomaly (that I've used to track perihelion time) started falling rather than rising. I've also noticed that it seems to be only a slight effect, which also demonstrated ~ 17 days period. So I though: perhaps that's the effect of Titan, right? Problem solved.&lt;/p&gt;&#xA;&lt;p&gt;Yet now I look at Neptune's true anomaly (see the last column on the picture) and &lt;em&gt;have absolutely no idea what's going on&lt;/em&gt;.&#xA;Can somebody explain, please? How is something like that even possible? It's like the planet decided to go backwards for a change.&lt;/p&gt;&#xA;&lt;p&gt;My settings are:&lt;/p&gt;&#xA;&lt;p&gt;Ephemeris Type [change] :   OBSERVER&lt;/p&gt;&#xA;&lt;p&gt;Target Body [change] :  Neptune [899]&lt;/p&gt;&#xA;&lt;p&gt;Observer Location [change] :    Sun (body center) [500@10]&lt;/p&gt;&#xA;&lt;p&gt;Time Span [change] :    Start=1900-01-01, Stop=2100-12-01, Step=1 Y&lt;/p&gt;&#xA;&lt;p&gt;Table Settings [change] :   QUANTITIES=18,41&lt;/p&gt;&#xA;&lt;p&gt;Display/Output [change] :   default (formatted HTML)&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zQ4JG.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zQ4JG.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25113" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-12-27T21:11:35.410" Title="Why is Neptune true anomaly decreasing?" Tags="&lt;orbit&gt;&lt;orbital-mechanics&gt;&lt;orbital-elements&gt;&lt;neptune&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28692" PostTypeId="1" AcceptedAnswerId="28696" CreationDate="2018-12-09T16:03:00.130" Score="0" ViewCount="154" Body="&lt;p&gt;Is there a name for the orbit of a natural satellite of a planet who's period is the same as the rotational period of the planet, but the orbital motion &lt;em&gt;has the opposite sense&lt;/em&gt; than the planet's rotation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there isn't an existing term or name, how would it at least be described, technically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://space.stackexchange.com/questions/32817/how-much-does-the-tidal-effect-affect-satellites-in-a-retrograde-orbit&quot;&gt;How much does the tidal effect affect satellites in a retrograde orbit?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-12-13T22:00:23.637" LastActivityDate="2018-12-13T22:00:23.637" Title="Is there a name for this type of orbit of a natural satellite?" Tags="&lt;orbit&gt;&lt;orbital-mechanics&gt;&lt;identify-this-object&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28693" PostTypeId="1" AcceptedAnswerId="28694" CreationDate="2018-12-09T16:08:52.997" Score="0" ViewCount="97" Body="&lt;p&gt;Is it possible for a natural moon to have a natural moon? What is the limit to when satellites on satellites becomes to unnatural for it to be possible?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-12-09T16:46:54.360" LastActivityDate="2018-12-09T16:46:54.360" Title="Can a moon have a moon?" Tags="&lt;natural-satellites&gt;&lt;orbital-mechanics&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2018-12-09T17:14:43.513" ContentLicense="CC BY-SA 4.0" />
  <row Id="28694" PostTypeId="2" ParentId="28693" CreationDate="2018-12-09T16:44:52.110" Score="1" Body="&lt;p&gt;Of course it is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the bigger moon (A) is far enough from a parent body so that the latter one cannot pull the smaller moon (a) away from it.&#xA;So the factors at play here are the mass of the planet, the mass of the moon (A) and the distance between them. These determine what is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Sphere_of_influence_(astrodynamics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;sphere of influence (SOI)&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the moon (a) orbits the moon (A) inside its SOI, it would be possible for (a) to exist there for a long time.&#xA;How long that time would be - it's a slightly more difficult question. As the tidal forces and various disturbances might eventually make the system unstable, and eventually (a) might be pulled away from (A) or even fall onto it etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that by &quot;unnatural&quot; you mean &quot;impossible&quot;? Because usually when people talk about &quot;non-natural&quot; satellites, they mean &quot;artificial&quot;, that is, &quot;made by humans&quot;. So it has nothing to do with the type of the orbit. Natural satellites are formed, well, naturally, and artificial satellites are launched by us humans.&lt;/p&gt;&#xA;" OwnerUserId="25113" LastActivityDate="2018-12-09T16:44:52.110" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28695" PostTypeId="2" ParentId="28692" CreationDate="2018-12-09T17:20:28.563" Score="5" Body="&lt;p&gt;There is probably no specific term, as I don't think such orbits occur naturally anywhere, nor there is any sense to launch an artificial satellite into such an orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in general terms if an orbital period of a satellite is equal to the period of rotation of a body it orbits, such orbits are called synchronous (e.g. geosynchronous if the satellite orbits the Earth). And if the orbit is in the opposite sense w/respect to the rotation of a body, the orbit is called retrograde.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would say that &quot;&lt;em&gt;synchronous retrograde orbit&lt;/em&gt;&quot; would pretty unambiguosly point to a type of orbit you're describing.&lt;/p&gt;&#xA;" OwnerUserId="25113" LastActivityDate="2018-12-09T17:20:28.563" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28696" PostTypeId="2" ParentId="28692" CreationDate="2018-12-09T17:22:21.537" Score="6" Body="&lt;p&gt;No.  Such an orbit would be co-incidental and dynamically uninteresting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a planet of given mass and rotation period, there is a class of orbits, whose period is equal to the rotation period of the planet. If the orbit is prograde this gives an interesting effect: the satellite would remain in the same position in the sky (or if it had an elliptical orbit, it would return to the same position each day)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the orbit is retrograde, the orbit would not have any particularly interesting features. While it is possible for tidal locking to produce a synchronous orbit, there is no mechanism to stabilise a retrograde orbit like this, and tidal effects will slowly disrupt it.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-12-09T17:22:21.537" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28697" PostTypeId="1" AcceptedAnswerId="28698" CreationDate="2018-12-09T19:32:00.897" Score="1" ViewCount="578" Body="&lt;p&gt;I am reading about the interior temperature structure of Jupiter. It says in various texts that there are regions in Jupiter's atmosphere which are well-approximated by a dry adiabatic lapse rate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that in these regions there is convection happening? Is adiabatic just a general synonym for convection?&lt;/p&gt;&#xA;" OwnerUserId="16092" LastActivityDate="2020-07-26T08:34:51.630" Title="Adiabatic versus convection" Tags="&lt;planetary-atmosphere&gt;&lt;jupiter&gt;&lt;atmospheric-effects&gt;&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28698" PostTypeId="2" ParentId="28697" CreationDate="2018-12-09T21:43:50.650" Score="2" Body="&lt;p&gt;An adiabatic process in physics is at first a process where a system of limited size does not exchange any heat with the outside world, or where the energy exchange timescale is very long compared to the interesting timescale under which the adiabatic process occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also loosely used as a term meaning &lt;a href=&quot;https://en.wikipedia.org/wiki/Adiabatic_invariant&quot; rel=&quot;nofollow noreferrer&quot;&gt;'constant in the framework of a certain approximation'&lt;/a&gt;. One sees this often used in quantum mechanics where it essentially means 'quasi-static', or in plasma physics where particles can gyrate in a field gradient. If the field gradient is much smaller than the gyration radius, such that every individual particle gyration experiences a near uniform field, then the particle experiences an adiabatic change of the underlying field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coming back to planetary/stellar atmospheres, here the two concepts are used interchangeably. Should a piece of atmosphere experience a super-adiabatic temperature gradient, then this situation is usually unstable towards buoyant instabilities. As soon as convective motion sets in, the temperature gradient will be brought down towards the adiabatic one.&lt;br&gt;&#xA;This is a very general process and happens in stars as well as planets, if no other physics is present. Thus it is used interchangeably, while people using it are aware that this is a bit sloppy.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2018-12-09T21:43:50.650" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28699" PostTypeId="2" ParentId="28502" CreationDate="2018-12-10T05:06:25.043" Score="2" Body="&lt;p&gt;There are two questions here, a terminological question and one about orbital dynamics and satellite mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The International Astronomical Union sometimes establishes criteria for whether astronomical bodies fit into certain classes. Famously, they formalized the definition of &quot;planet&quot; in such a way that Pluto was excluded. I am not aware that they have specified a lower bound for the mass of a natural satellite (so I don't get the bounty since I haven't found documentation either way). But for the rest of this answer, I will define &quot;natural satellite&quot; to be any naturally occurring body in a stable orbit around another body that isn't a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that definition, there is no clear lower bound on the mass of a natural satellite, for the reason given by tuomas. However, the vacuum of space is imperfect and the Earth's atmosphere really just gradually fades away into the interplanetary medium. The smaller an object is, the more it is affected by drag, and this is still true even when the drag is really tiny as is experienced by satellites in orbit. So dust could be expected to last fewer orbits than pebbles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On an even smaller scale, individual atoms and molecules are considered to be part of the  extended atmosphere, either the thermosphere or exosphere, and not treated as satellites. In the thermosphere, there are still enough particle-to-particle collisions to treat it as a gas, so those particles aren't even in orbit themselves since they tend to collide before they make it all the way around. This is true even though the ISS orbits right in the middle of the thermosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, higher up is the exosphere, where individual atoms hardly ever collide. Some of those are presumably in orbits that last at least a few periods before they collide with another atom, but many others are in the process of falling back toward the thicker atmosphere, or heading outward after getting an especially large random kick from other particles in the thermosphere below. The particles that manage to orbit for a while meet the definition I gave for a natural satellite, but it is certainly true that the word is not generally used for them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there is no lower bound for something to be a natural satellite in the broadest sense. It is possible for a particle to be so small people wouldn't normally refer to it as a satellite, but I don't think there is an exact threshold for that.&lt;/p&gt;&#xA;" OwnerUserId="7752" LastActivityDate="2018-12-10T05:06:25.043" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28700" PostTypeId="1" AcceptedAnswerId="28703" CreationDate="2018-12-10T07:42:38.400" Score="0" ViewCount="76" Body="&lt;p&gt;In &lt;a href=&quot;http://wirtanen.astro.umd.edu/46P/46P_status.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;this post&lt;/a&gt; on the &lt;a href=&quot;http://wirtanen.astro.umd.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;website for the Comet Wirtanen Observing Campaign&lt;/a&gt; at wirtanen.astro.umd.edu, there's mention of a CN filter being used to observe the comet, and this cool GIF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exactly is a CN filter? What wavelengths does it pass? Where can I read more about it? Why is it showing a spiral shape of stuff being emitted from comet 46P? Is this comet dust, gas, something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GIF:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vkGYB.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vkGYB.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UDiW3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UDiW3.png&quot; alt=&quot;http://wirtanen.astro.umd.edu/46P/46P_status.shtml#brightness_plot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-12-10T15:19:15.187" Title="What might a CN filter be in the context of comet watching? Is it showing dust, or gas, or something else?" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28701" PostTypeId="5" CreationDate="2018-12-10T14:25:33.593" Score="0" Body="" OwnerUserId="6" LastEditorUserId="6" LastEditDate="2018-12-10T14:25:33.593" LastActivityDate="2018-12-10T14:25:33.593" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28702" PostTypeId="4" CreationDate="2018-12-10T14:25:33.593" Score="0" Body="Questions about planetary dust storms, such as those on Mars and Titan" OwnerUserId="6" LastEditorUserId="6" LastEditDate="2018-12-10T14:25:33.593" LastActivityDate="2018-12-10T14:25:33.593" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28703" PostTypeId="2" ParentId="28700" CreationDate="2018-12-10T15:19:15.187" Score="2" Body="&lt;p&gt;A CN filter &lt;a href=&quot;http://wirtanen.astro.umd.edu/46P/CN_filter_test.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;shows abundances of cyanogen&lt;/a&gt;, a molecule with the structure &lt;span class=&quot;math-container&quot;&gt;$(\mathrm{CN})_2$&lt;/span&gt;. Cyanogen emission lines at 387 nm allow us to study gas in the coma of a comet, telling us more about its structure, properties and outgassing. Other gases, like &lt;span class=&quot;math-container&quot;&gt;$\mathrm{OH}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\mathrm{NH}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{C}_2$&lt;/span&gt;, also give away a lot of information about a comet, but the often-strong cyanogen lines are a good tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism behind cyanogen production from parent gases isn't entirely clear, although &lt;a href=&quot;https://www2.jpl.nasa.gov/comet/news65.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;a number of theories have been proposed&lt;/a&gt;. For instance, cyanogen could be released by icy grains undergoing sublimation, or by the decomposition of more complex molecules.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2018-12-10T15:19:15.187" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28704" PostTypeId="2" ParentId="28691" CreationDate="2018-12-10T17:10:11.407" Score="8" Body="&lt;p&gt;Short answer: because Neptune's argument of the perifocus, the reference point for the true and mean anomalies jumps around a lot because Neptune's orbit is nearly circular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you run HORIZONS w/ the following parameters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0VEb2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0VEb2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you'll notice Neptune's mean anomaly also decreases at times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of these occur because Neptune's argument of the perifocus (from which both anomalies are measured) is itself jumping around quite a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Except when Neptune is actually at perihelion, the argument of the perifocus is a theoretically computed point, not representing any actual physical quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function NASA uses to calculate the argument of the perifocus is &lt;a href=&quot;https://naif.jpl.nasa.gov/pub/naif/toolkit_docs/C/cspice/oscelt_c.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;oscelt&lt;/a&gt;, and specifically notes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;   6) If the eccentricy of the orbit is very close to but not&#xA;      equal to zero, the argument of periapse may not be accurately&#xA;      determined.&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To summarize, using the true anomaly to determine's Neptune's position is a bad idea, unless you also account for the changing argument of the perifocus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you note, Neptune's ecliptic longitude behaves much better, because the zero point of the ecliptic, which, while based on the Earth's precession and thus not completely fixed, nonetheless moves very slowly, much slower than Neptune's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the &quot;best&quot; coordinate system to use is something like ICRF, which is unchanging with respect to the fixed stars.&lt;/p&gt;&#xA;" OwnerDisplayName="user21" LastEditorUserId="7472" LastEditDate="2018-12-27T21:11:35.410" LastActivityDate="2018-12-27T21:11:35.410" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28705" PostTypeId="2" ParentId="28671" CreationDate="2018-12-10T22:41:26.757" Score="3" Body="&lt;p&gt;If  you were in deep space, i,e., a spot somewhere between galaxies, then the night sky would look a lot like the sky seen by astronauts in the ISS, when they look away from both the ecliptic plane and the galactic plane, minus the individual Galactic stars.  That is, it is like our deep sky photos in those directions that are dominated by galaxies, but no atmospheric distortions, no night sky lines, and no zodiacal light.  The spectrum is dominated by different things at different wavelengths, but basically it is the sum of the light of galaxies and perhaps intergalactic stars that escaped their galaxies.  At long radio waves, it is dominated by supernova remnants and AGN as self-absorbed synchrotron radiation, peaking between 1 and 10 Mhz (&lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/bib_query?1977MNRAS.180..429S&quot; rel=&quot;nofollow noreferrer&quot;&gt;Simon 1977&lt;/a&gt;).  The mm band is dominated by the microwave background radiation at 2.7 C. The far-ir is dominated by cold galaxy dust at 60 microns. The optical is dominated by galaxy thermal starlight at 1 micron.  The UV is dominated by hot stars at 0.2 microns (&lt;a href=&quot;http://www.isas.jaxa.jp/en/feature/forefront/151030.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;see summary&lt;/a&gt;).  The Xray spectrum from 3 to about 45 keV (&lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/bib_query?1980ApJ...235....4M&quot; rel=&quot;nofollow noreferrer&quot;&gt;Marshall et al 1980&lt;/a&gt;, &lt;a href=&quot;https://ned.ipac.caltech.edu/level5/Fabian/Fabian2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fabian 1992&lt;/a&gt;) is well-fitted by a ~40 keV thermal bremsstrahlung model. Above 45 kev the spectrum falls and is thought to be the sum of many AGN at all redshifts.  The spectral energy density (SED) from radio to X-ray is shown in this diagram from the paper by &lt;a href=&quot;https://arxiv.org/abs/1802.03694&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hill et al. (2018)&lt;/a&gt; &lt;em&gt;The Spectrum of the Universe&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kMqXU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kMqXU.png&quot; alt=&quot;Spectrum of the Universe from Hill et al. (2018)&quot;&gt;&lt;/a&gt;&#xA;The width of the line in the diagram shows the present uncertainties in the measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark energy does not shine, it has an effect on the redshift as a function of time and thereby indirectly affects the shape of the background spectrum. &lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2018-12-12T21:24:57.287" LastActivityDate="2018-12-12T21:24:57.287" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28706" PostTypeId="2" ParentId="28697" CreationDate="2018-12-10T23:56:29.367" Score="2" Body="&lt;p&gt;I would say that an adiabatic lapse rate is indeed synonymous with convection, though what is not known when you just say there is convection is the turnover timescale.  The timescale could be so long that it wouldn't look much like convection, but then, that can happen in convective atmospheres like that of the Sun-- in some layers, the convection can be so slow you wouldn't notice it on a sound crossing time across the scale of gradients.  But the reason I'm claiming they are synonymous is that convection is about the only reason you can think of why there would be an adiabatic lapse rate for the overall gradient in the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way adiabatic convection works is, you can think of every parcel of gas in the convective region as if it was the same parcel, just found at a different time.  In short, every parcel goes everywhere eventually.  What's more, as the parcel goes everywhere, it remains insulated, so it's just the same parcel expanding and contracting with no heat exchange.  That's what sets the adiabatic gradient everywhere.  Another way to say that is, the entropy per gram is the same everywhere, because entropy doesn't change if there is no heat transport, and if the entropy of each parcel stays the same, and if each parcel goes everywhere, then the entropy per gram is constant everywhere.  That's what an adiabatic lapse rate means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if there is not convection, so parcels just stay put, how would they know to have the same entropy as each other?  So it is generally convection that induces an adiabatic lapse rate, though the convection could still be rather slow if the heat transport rate is also very slow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No doubt there are complications I am leaving out, like what happens when there are changes in ionization or phase which alter what we mean by the adiabatic lapse rate, so I'm just talking about the simplest situation where we are not considering such details.  We are basically assuming the parcel is always in thermodynamic equilibrium, even though its environment is slowly changing.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2018-12-11T00:01:48.580" LastActivityDate="2018-12-11T00:01:48.580" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28707" PostTypeId="1" CreationDate="2018-12-11T09:07:39.957" Score="4" ViewCount="262" Body="&lt;p&gt;I am getting confused by some &quot;popular science&quot; about time travel, such as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XPnQd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XPnQd.png&quot; alt=&quot;sensational...&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;a href=&quot;https://www.wired.com/2014/11/time-dilation/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Time Travel Is Real. Here Are the People and Spacecraft Who Have Done It&quot;&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ggi0h.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ggi0h.png&quot; alt=&quot;more sensational news&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is indeed an interesting property of physics as we know it, and a great gift from the theory of relativity. It leads to useful perspectives on the nature of time as exemplified in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Twin_paradox&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Twin Paradox&quot;&gt;Twin Paradox&lt;/a&gt; example: When you travel for instance at a Lorentz factor of γ = 100 (99.995% the speed of light) for 10 years, time &quot;dilates&quot; to thousand years. For instance it has been proven experimentally that muons would &quot;tick&quot; slower, and therefore for a &lt;em&gt;measure&lt;/em&gt; of time in seconds. Similarly, I would assume that extending this property at a macroscopic scale, your whole body would slow down... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is to know if will you &quot;live for a thousand years&quot; (internet abounds of such statements as in &lt;a href=&quot;https://www.quora.com/If-I-traveled-close-to-light-speed-for-one-year-then-returned-to-earth-how-many-years-would-my-friends-have-aged&quot; rel=&quot;nofollow noreferrer&quot;&gt;this example forum&lt;/a&gt; and may be interpreted as such in the headlines above), or rather that during the 10 years of your travel (in your spaceship) and which would be measured as 1000 years in another frame of reference (the  earth), you would have aged by &quot;only&quot; 10 years in your spaceship - time &quot;froze&quot; during the travel.&lt;/p&gt;&#xA;" OwnerUserId="10877" LastEditorUserId="10877" LastEditDate="2018-12-20T11:48:51.853" LastActivityDate="2018-12-28T22:39:00.953" Title="What is this &quot;time travel&quot; these popular blogs refer to?" Tags="&lt;time-dilation&gt;&lt;time&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28708" PostTypeId="2" ParentId="28707" CreationDate="2018-12-11T10:46:28.957" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My question is to know if will you &quot;live for a thousand years&quot; (...) or &quot;only&quot; 10 years in your spaceship - time &quot;froze&quot; during the travel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you were to travel to close the speed of light (not doing the math, plainly speaking), you would appear to live for a thousand years to the people that are outside of your frame of reference. Time would actually pass the same to you and you would observe the time outside your frame of reference to pass significantly faster.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it a scientific fact?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Human time travel is a very real scientific fact, but the implications are too complex to be presented to the public sphere since it has no practical applications &lt;strong&gt;yet&lt;/strong&gt;. If we were to sent people to another star system (Alpha Centaury A/B - at 4.3 light-years from Earth), accelerating their ship close to the speed of light would make the time in their frame of reference pass slower relative to the outside of their frame of reference, thus allowing them to reach the destination in a shorter time from their perspective. It would be useful when managing supplies on this vast journey.&lt;/p&gt;&#xA;" OwnerUserId="25127" LastEditorUserId="10877" LastEditDate="2018-12-26T17:05:24.997" LastActivityDate="2018-12-26T17:05:24.997" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28709" PostTypeId="1" AcceptedAnswerId="28710" CreationDate="2018-12-11T21:24:42.883" Score="2" ViewCount="528" Body="&lt;p&gt;Thanks to reflected starlight, many planets and comets in the Solar System have been visible to humans since long before the development of modern astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the starlight from outside the Solar System should be reflected light as well. That is, light emitted by one star in the Milky Way may encounter another star, and be reflected toward Earth. Normally, stars are assumed to be perfect blackbodies, but in reality they must bounce off &lt;em&gt;some&lt;/em&gt; radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, starlight should be reflected off exoplanets, moons, asteroids, meteors, comets, brown dwarfs, etc. What fraction of starlight, seen by the naked eye from Earth, is actually reflected light?&lt;/p&gt;&#xA;" OwnerUserId="23100" LastActivityDate="2018-12-12T03:19:30.840" Title="What fraction of starlight, seen from Earth, is actually reflected light?" Tags="&lt;star&gt;&lt;light&gt;&lt;naked-eye&gt;&lt;interstellar&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28710" PostTypeId="2" ParentId="28709" CreationDate="2018-12-11T21:42:47.433" Score="3" Body="&lt;p&gt;A truly negligible amount. You only need compare the brightness of the planets as viewed from the Earth with the brightness of the sun. A very rough calculation (considering the relative magnitudes of the planets and the Sun) suggests that light emitted directly by the sun is 100 million times brighter than light reflected off planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You note that the sun is not a perfect black body, and so does reflect some light. However it only reflects light reflected on to it from the stars, which are a very weak light source. And most starlight that does fall on the sun is absorbed. I haven't tried to estimate how much of the sun's light is reflections of starlight, but it is a very very small proportion. I'd guess much less than a billionth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one situation where reflected light is significant, and that is when very bright stars illuminate dust and gas, to form a reflection nebula. These are common and at least part of the orion nebula is reflected starlight. Other reflection nebulae are too dim to see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While a I don't have an exact figure, it is clear that nearly all starlight is directly emitted from the star, and it takes special equipment to see any reflected light at all.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-12-11T21:42:47.433" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28711" PostTypeId="2" ParentId="28709" CreationDate="2018-12-12T03:19:30.840" Score="0" Body="&lt;p&gt;@James K's answer, above, is correct (and I've upvoted it), but I'd like to expand on it more than comments allow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't forget the inverse square law! The starlight light available to be reflected drops off with the square of the distance from the star to the reflector. Except for very close binaries the inverse square law dramatically diminishes the total amount of reflected light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a simple way to work it out: &lt;em&gt;All&lt;/em&gt; the light from a star passes through the sphere centered on the Sun with a radius of the Earth's orbital distance. (The Earth's orbit can be thought of as the equator of that sphere.)  The &lt;em&gt;only&lt;/em&gt; light Earth can reflect is the tiny part its surface intercepts.  The rest streams past it into space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;em&gt;r&lt;/em&gt; is the Earth's radius and &lt;em&gt;R&lt;/em&gt; is the radius of its orbit, the fraction intercepted and available to be reflected is (&lt;sup&gt;r&lt;/sup&gt;/&lt;sub&gt;2R&lt;/sub&gt;)&lt;sup&gt;2&lt;/sup&gt;.  With 4=4000 miles and R=93,000,000 miles, that's just a bit over 4*10&lt;sup&gt;-10&lt;/sup&gt; of the Sun's light is available to be reflected by the Earth. (Earth reflects most of it.) If you do the arithmetic, Mercury, Venus, Jupiter and Saturn roughly the same and the rest of the planets make a pretty minor contribution.  The total reflected (&lt;em&gt;before&lt;/em&gt; albedo losses!) is around 4*10&lt;sup&gt;-9&lt;/sup&gt; of the Sun's light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A closer-in planetary system will reflect a bigger proportion of the central star's light -- it scales as the inverse square, of course.  If the solar system were ten times as compact but the planets' sizes unchanged, the planets would reflect around 4*10&lt;sup&gt;-7&lt;/sup&gt; of the Sun's light.  Still tiny.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only time you get anything that isn't tiny is in the case of a close binary. If another star the same size as the Sun orbited the Sun just one solar diameter away, it would intercept (and thus have the possibility of reflecting) around 6% of the sun's light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This supports James K's conclusion pretty strongly.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2018-12-12T03:19:30.840" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28714" PostTypeId="1" CreationDate="2018-12-12T13:48:26.590" Score="3" ViewCount="242" Body="&lt;p&gt;This post comes from physics exchange forum, I have transfered it here, maybe I will be luckier since I have not received answers on the other forum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the issue :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asked to give the formal expression of the &lt;strong&gt;total number density of galaxies&lt;/strong&gt; and explain &lt;strong&gt;why is this expression problematic in practice&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I saw from my research and into my lectures, I have found the follwing relation which gives the number of galaxies &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; with magnitude &lt;span class=&quot;math-container&quot;&gt;$(m &amp;lt; M)$&lt;/span&gt; (number counts) :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\text{log}\,N(m&amp;lt;M)\,\propto\,0.6 M + \text{constant}\quad(1)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$N(m&amp;lt;M)\,\propto\,\text{exp}(0.6 M)\,\text{exp(constant)}\quad(2)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what the teacher wants to highlight, i.e by saying that the formal expression of the total number density of galaxies is problematic in practice, given that I even haven't the kind of expression or equation which gives the estimation of galaxy density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1) Concerning this distribution to use, do you advise me to use a distribution as a function of magnitude, mass or a distribution as function of luminosity (like Schechter) ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2) I have seen on web the HMF (Halo Mass function) but it seems to be about the dark matter halos : can we count one dark matter halo per galaxy ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3) But the problem is here is that a dark matter halo is more massive than the galaxy hosted in this halo : how can we deal with this ?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone could help me, this would be kind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PS:&lt;/strong&gt; &lt;strong&gt;I have also to explain why the estimation of galaxies density is difficult and problematic&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4) What are the issues when we want to estimate density of galaxies ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning the distribution as a function of luminosity, I have found the &lt;strong&gt;Schechter luminosity function&lt;/strong&gt; :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$N(L)\ \mathrm {d} L=\phi^{*}\left({\frac{L}{L^{*}}}\right)^{\alpha}\mathrm {e}^{-L/L^{*}}{\frac{\mathrm {d} L}{L^{*}}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE 1:&lt;/strong&gt; my apologies, &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; represent apparent magnitude and not masses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the questions remain about the counting of galaxies and the computation of total density at large scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In practise, what are the major problems when we want to estimate this density ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE 2:&lt;/strong&gt; people here not really motivated by this bounty ... feel free to ask for precisions about this post .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the moment, I am faced to know at which scale I can compute a mean density : have I got to take into account this scale since it seems to have a dependance between density and volume considered, doesn't it ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there a general formula to express this dependance ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards&lt;/p&gt;&#xA;" OwnerUserId="16492" LastEditorUserId="16492" LastEditDate="2018-12-23T15:09:16.933" LastActivityDate="2018-12-23T15:09:16.933" Title="Total number density of galaxies and problematic expression in practise" Tags="&lt;galaxy&gt;&lt;astrophysics&gt;&lt;cosmology&gt;&lt;galaxy-cluster&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28715" PostTypeId="1" AcceptedAnswerId="28753" CreationDate="2018-12-12T20:18:48.253" Score="0" ViewCount="56" Body="&lt;p&gt;&lt;strong&gt;Can we count only one dark matter halo for each Galaxy ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, I am trying to estimate the total density of galaxies in given volume and I have often seen on physics exchange and other forums the estimation of dark matter halos distribution (as a function of the mass of halo or as a function of Luminosity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I could count one galaxy per halo, this would make the things simplier but I think that things are more complicated, especially since a halo and the galaxy associated to this halo have not the same mass, so distribution mass is different (mass of halo is not linear towards the mass of galaxy, isn't it ?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone could help me on this issue, regards&lt;/p&gt;&#xA;" OwnerUserId="16492" LastActivityDate="2018-12-16T21:34:21.153" Title="number density of Dark Matter Halos Versus number density of Galaxies" Tags="&lt;galaxy&gt;&lt;dark-matter&gt;&lt;galaxy-cluster&gt;&lt;density&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28716" PostTypeId="1" CreationDate="2018-12-12T23:00:07.830" Score="0" ViewCount="94" Body="&lt;p&gt;Viewed from Stonehenge, when did the Sun and Regulus rise together at the summer solstice? Also when did Regulus rise before the Sun and would have been seen before the dawn glare?&lt;/p&gt;&#xA;" OwnerUserId="25143" LastEditorUserId="14094" LastEditDate="2018-12-13T01:10:05.260" LastActivityDate="2019-01-12T12:01:05.230" Title="The Sun and Regulus together at the summer solstice" Tags="&lt;positional-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28719" PostTypeId="2" ParentId="28716" CreationDate="2018-12-13T11:07:44.567" Score="1" Body="&lt;p&gt;Using sky simulator software (in my case Guide 9.1), the sun and regulus rose at the same time in about 2500 bc.  I think you would need to go back another 1500 years to be able to see regulus in the pre dawn glow. At that time, the sun would be about 7 degrees below the horizon as regulus was rising.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the surrounding woodland wasn't cleared until about 5500 years ago (per Wikipedia), so it may not have been possible to make this earlier observation.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2018-12-13T11:07:44.567" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28720" PostTypeId="1" CreationDate="2018-12-14T02:17:49.123" Score="0" ViewCount="81" Body="&lt;p&gt;Brevard Co. FL. At first I thought it was a comet it appeared to have a tail. Starts out low in the east and gains elevation as the sun rises. As the sun is just above the horizon it is still visible at about an 80 degree elevation. What am I seeing?&lt;/p&gt;&#xA;" OwnerUserId="25152" LastEditorUserId="11822" LastEditDate="2018-12-14T10:35:28.177" LastActivityDate="2018-12-14T10:35:28.177" Title="very bright object (visible even after sun risen)" Tags="&lt;planet&gt;&lt;comets&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28721" PostTypeId="1" CreationDate="2018-12-14T07:23:28.057" Score="3" ViewCount="187" Body="&lt;p&gt;Taking into account the physical characteristics of Titan (e.g. surface gravity, atmospheric pressure), what would be the most efficient method of motion for an astronaut to navigate the surface of Titan on foot?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be a leaping or hopping motion? Or would it make more sense to walk somehow? I'm unsure how the friction of the ground would accommodate for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the purpose of this thought experiment, lets assume that humanity has the necessary equipment to put an astronaut on Titan and to have protective clothing to enable reasonable safety from environmental hazards on the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="25129" LastActivityDate="2019-05-30T08:58:29.270" Title="How would humans with appropriate equipment navigate the surface of Saturn's moon Titan on foot?" Tags="&lt;gravity&gt;&lt;astrophysics&gt;&lt;saturn&gt;&lt;titan&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2019-05-29T16:01:48.733" ContentLicense="CC BY-SA 4.0" />
  <row Id="28723" PostTypeId="2" ParentId="28720" CreationDate="2018-12-14T09:51:17.697" Score="5" Body="&lt;p&gt;This is the planet Venus.  It is always bright (when visible) but exceptionally so at the moment.  Venus shows phases (similar to the moon) and is currently nearly half illuminated.  That, combined with its relative closeness (about 75million km) is why it is particularly bright at the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is often visible in broad daylight, but can be difficult to spot because of the bright sky background.  &lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2018-12-14T09:51:17.697" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28724" PostTypeId="2" ParentId="28721" CreationDate="2018-12-14T11:17:02.987" Score="5" Body="&lt;p&gt;I wonder if this would be better for worldbuilding or space travel, but I can touch on the basics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'd need oxygen and protection from the cold, but a space suit on Titan would be less restrictive, not more restrictive than the ones used on the Moon.   The pressure isn't too much, so that actually helps and cold temperature is much easier to protect from than a vacuum.    They'd probably need some kind of full body wrap to avoid exposure and frostbite, and goggles but that's not terrible hard.   Electric heating or just insulation would both be options.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's safe to assume that even if they carry an electric heating device, their suits would still be less bulky and less restrictive than the ones the astronauts wore on the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hop or Walk?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very likely hop.   I'm going to &lt;a href=&quot;https://www.quora.com/Why-did-astronauts-hop-on-the-moon&quot; rel=&quot;noreferrer&quot;&gt;borrow from this Quora answer&lt;/a&gt;, because I agree with it and it makes logical sense.   Walking is kind of a controlled fall.  Walking in 0.14 g, means your body doesn't fall forward in pace with every step, so walking would be weird and hoping would be easier.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even on Mars, astronauts might hop due to the lower gravity.  If I was to make a wild guess, for comfortable walking I think it would begin somewhere around 50% or 60% Earth gravity.   Less then that, the slower rate of falling with each step and the lighter body weight, I think hopping is the way to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The human brain and inner ear begins to stop working somewhere around lunar gravity.   There are videos of astronauts on the moon falling forward and looking clumsy, but they were very fit individuals.  In low enough gravity the human brain loses a sense of which way is down.  (I read that, I'll see if I can find a link).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would think, if it was for an extended time there, astronauts on Titan would wear lead boots, to help with their sense of which way is down and also for helping maintain musculature, but that's just speculation on my part.  There are other ways to maintain musculature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Titan's surface might/probably gets soggy as well and humans aren't buoyant in methane lakes the way they are in water, (so maybe the lead boots aren't a good idea).   Falling into a lake or falling through ice into a like might happen on titan, but they'd be wearing full body wraps and have oxygen, so falling into a methane lake wouldn't necessarily mean death, they could probably get fished out and rescued.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long story short, hopping is likely the way to go, and perhaps, either some kind of snow-shoe feature or perhaps an ice/jagged edges feature if parts of Titan's surface are slippery.   I'm far from an expert in such things, this is just my best guess.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-12-14T11:17:02.987" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28725" PostTypeId="1" CreationDate="2018-12-14T21:51:26.427" Score="0" ViewCount="136" Body="&lt;p&gt;How do I find some of the constellations of the zodiac without remembering their locations on the sky?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it should be quite simple if you are located in the northern hemisphere and you know &#xA;1) which astrological sign it is right now (this month)&#xA;2) where the sun set&#xA;3) the number of hours left until sunrise&#xA;4) the order of the constellations in the circle of the zodiac: Aries, Taurus, Gemini, Cancer, Leo, Virgo, Libra, Scorpio, Sagittarius, Capricorn, Aquarius and Pisces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Straight forward logic says that it is enough to make calculations in order to understand where they should be, but I can't find any such rule on the internet. Maybe it is not so simple. Have you seen such a rule?&lt;/p&gt;&#xA;" OwnerUserId="25157" LastEditorUserId="23940" LastEditDate="2018-12-17T03:15:49.303" LastActivityDate="2018-12-17T03:15:49.303" Title="Simple rule to find some of the zodiac constellations after sunset?" Tags="&lt;zodiac&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="28726" PostTypeId="2" ParentId="28725" CreationDate="2018-12-14T22:24:12.173" Score="4" Body="&lt;p&gt;This is far too complicated. You can find any asterism in the sky far more simply by remembering its shape and its position relative to other (perhaps brighter) constellations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Zodiac constellations, they are all in a band that (in the Northern hemisphere) goes from East through South, to West. In the winter, that band will be high is the sky. In summer it will be low. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then you learn the sky. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In winter you first learn Orion and from there you can find your way to Gemini, the V of the Hyades in Taurus.  Cancer is dim (but has that nice cluster), but next in line, Leo, is nice and bright.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to find out what is actually visible right now, go outside! If you want to check on the location or visibility of a particular constellation right now, use planetarium software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't need to calculate anything.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-12-14T22:59:17.660" LastActivityDate="2018-12-14T22:59:17.660" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28730" PostTypeId="2" ParentId="1767" CreationDate="2018-12-15T08:15:14.973" Score="-2" Body="&lt;p&gt;Yes all of the atoms in any organism, and in fact the whole Earth, have been around for about 13.8 bn years (hydrogen, helium, lithium), since the beginning of the Universe, or have been made in various stars. Some elements were formed in supernovae, but others have been formed in other types of stellar element foundries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since about 4.5 bn years ago the Earth is being formed. I use 'being' since, although most already is in place, the process is still on going, meteorites and other types of material is continuing to fall down on Earth but there are also atoms leaving Earth. Most atoms that form the Earth have been here for at last 4 bn years though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now any organism literally builds itself from these atoms. Not only during growth, but in fact all atoms are continually being replaced in an upkeep process, which is one reason why adults still need to eat, we are continually losing atoms and thus continue to replace them with new ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since some of the atoms are recent arrivals, some of our atoms might actually have been in outer space as recently as last week, but most have been part of Earth for a long time, and the atoms in our bodies have been in many more organisms. They have been in bacteria, in dinosaurs, trees, mushrooms, fish, potatoes, in trilobites, lettuce, and countless others. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of our atoms have been in the oceans ever since they were formed; others have been locked in rock for billions of years before they ended up in our bodies. And since they are just temporarily in us, they might be in a plant next week and in a turtle the next. &lt;/p&gt;&#xA;" OwnerUserId="25161" LastEditorUserId="14094" LastEditDate="2018-12-15T16:53:02.717" LastActivityDate="2018-12-15T16:53:02.717" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28734" PostTypeId="2" ParentId="28656" CreationDate="2018-12-15T14:42:50.470" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Zodiacal_light&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zodiacal light&lt;/a&gt;!&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Zodiacal light (also called false dawn when seen before sunrise) is a faint, diffuse, and roughly triangular white glow that is visible in the night sky and appears to extend from the Sun's direction and along the zodiac, straddling the ecliptic. &lt;strong&gt;Sunlight scattered by interplanetary dust causes this phenomenon&lt;/strong&gt;. Zodiacal light is best seen during twilight after sunset in spring and before sunrise in autumn, when the zodiac is at a steep angle to the horizon. However, the glow is so faint that moonlight and/or light pollution outshine it, rendering it invisible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pL2on.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pL2on.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/rT2OS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rT2OS.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/File:Cantin1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Zodiacal_Light_Seen_from_Paranal.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Brian_May&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brian May&lt;/a&gt; (CBE):&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;In 2007, Brian May, lead guitarist with the band Queen, completed his thesis, A Survey of Radial Velocities in the Zodiacal Dust Cloud, thirty-six years after abandoning it to pursue a career in music.&lt;/strong&gt; He was able to submit it only because of the minimal amount of research on the topic undertaken during the intervening years. May describes the subject as being one that became &amp;quot;trendy&amp;quot; again in the 2000s.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Brian May's Thesis: &lt;a href=&quot;https://spiral.imperial.ac.uk:8443/handle/10044/1/1333&quot; rel=&quot;nofollow noreferrer&quot;&gt;A survey of radial velocities in the zodiacal dust cloud&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;And for good measure, Brian May's announcement of IAU's naming of the asteroid &lt;a href=&quot;https://en.wikipedia.org/wiki/17473_Freddiemercury&quot; rel=&quot;nofollow noreferrer&quot;&gt;17473 Freddiemercury&lt;/a&gt; YouTube: &lt;a href=&quot;https://youtu.be/o2vo6VR51eA&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/o2vo6VR51eA&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Zodiacal light is sunlight scattered by &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_dust&quot; rel=&quot;nofollow noreferrer&quot;&gt;interplanetary dust&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Cosmic dust, also called extraterrestrial dust or space dust, is dust which exists in outer space, or has fallen on Earth. Most cosmic dust particles are between a few molecules to 0.1 µm in size. Cosmic dust can be further distinguished by its astronomical location: intergalactic dust, interstellar dust, interplanetary dust (such as in the zodiacal cloud) and circumplanetary dust (such as in a planetary ring).&lt;/p&gt;&#xA;&lt;p&gt;In the Solar System, interplanetary dust causes the zodiacal light. Solar System dust includes comet dust, asteroidal dust, dust from the Kuiper belt, and interstellar dust passing through the Solar System. Thousands of tons of cosmic dust are estimated to reach the Earth's surface every year, with each grain having a mass between 10^−16 kg (0.1 pg) and 10^−4 kg (100 mg). The density of the dust cloud through which the Earth is traveling is approximately 10^−6/m3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MMdm4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MMdm4.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/File:Porous_chondriteIDP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-12-16T00:59:07.893" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28736" PostTypeId="1" AcceptedAnswerId="34127" CreationDate="2018-12-15T15:17:36.807" Score="8" ViewCount="141" Body="&lt;p&gt;This &lt;a href=&quot;https://astronomy.stackexchange.com/a/28734/7982&quot;&gt;answer&lt;/a&gt; to the question &lt;em&gt;What direct or indirect observations of dust can one make by eye or with binoculars?&lt;/em&gt; mentions Brian May's thesis and it's connection to &lt;a href=&quot;https://en.wikipedia.org/wiki/Zodiacal_light&quot; rel=&quot;noreferrer&quot;&gt;Zodical Light&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Brian_May&quot; rel=&quot;noreferrer&quot;&gt;Brian May&lt;/a&gt; (CBE):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;In 2007, Brian May, lead guitarist with the band Queen, completed his thesis, A Survey of Radial Velocities in the Zodiacal Dust Cloud, thirty-six years after abandoning it to pursue a career in music.&lt;/strong&gt; He was able to submit it only because of the minimal amount of research on the topic undertaken during the intervening years. May describes the subject as being one that became &amp;quot;trendy&amp;quot; again in the 2000s.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Brian May's Thesis: &lt;a href=&quot;https://spiral.imperial.ac.uk:8443/handle/10044/1/1333&quot; rel=&quot;noreferrer&quot;&gt;A survey of radial velocities in the zodiacal dust cloud&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;And for good measure, Brian May's announcement of IAU's naming of the asteroid &lt;a href=&quot;https://en.wikipedia.org/wiki/17473_Freddiemercury&quot; rel=&quot;noreferrer&quot;&gt;17473 Freddiemercury&lt;/a&gt; YouTube: &lt;a href=&quot;https://youtu.be/o2vo6VR51eA&quot; rel=&quot;noreferrer&quot;&gt;https://youtu.be/o2vo6VR51eA&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I believe that MgI stands for neutral, atomic magnesium, and since this is optical (not X-rays) this would have to be free atoms, or perhaps molecular, but not bound magnesium in compounds or metals.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Why is there free magnesium in the ecliptic (if there is) and why would some be orbiting within the ecliptic plane but retrograde?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Here is the abstract of the thesis in full:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;This thesis documents the building of a pressure-scanned Fabry-Perot Spectrometer, equipped with a photomultiplier and pulse-counting electronics, and its deployment at the Observatorio del Teide at Izaña in Tenerife, at an altitude of 7,700 feet (2567 m), for the purpose of recording high-resolution spectra of the Zodiacal Light. The aim was to achieve the first systematic mapping of the MgI absorption line in the Night Sky, as a function of position in heliocentric coordinates, covering especially the plane of the ecliptic, for a wide variety of elongations from the Sun. More than 250 scans of both morning and evening Zodiacal Light were obtained, in two observing periods – September-October 1971, and April 1972. The scans, as expected, showed profiles modified by components variously Doppler-shifted with respect to the unshifted shape seen in daylight. Unexpectedly, MgI emission was also discovered. These observations covered for the first time a span of elongations from 25º East, through 180º (the Gegenschein), to 27º West, and recorded average shifts of up to six tenths of an angstrom, corresponding to a maximum radial velocity relative to the Earth of about 40 km/s. The set of spectra obtained is in this thesis compared with predictions made from a number of different models of a dust cloud, assuming various distributions of dust density as a function of position and particle size, and differing assumptions about their speed and direction. The observations fit predominantly prograde models fairly well, but show a morning-evening asymmetry, different in the two observing periods. Models are investigated containing various components, including prograde and retrograde orbiting dust around the Sun, a drift of interstellar material though the Solar System, and light from distant emitting matter. The implications for possible asymmetries of the Zodiacal Cloud are discussed. Other researches on the Zodiacal Dust Cloud, before, during, and after my observations, are reviewed, including recent insights into its structure, orientation, and evolution, up to the present day, and my observations are evaluated in this context. Period of study, 1970-2007.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/303wc.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/303wc.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://spiral.imperial.ac.uk:8443/handle/10044/1/1333&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-11-30T17:22:17.417" Title="Why is there free magnesium in the ecliptic (if there is) and why would some be orbiting within the ecliptic plane but retrograde?" Tags="&lt;dust&gt;&lt;ecliptic&gt;&lt;zodiacal-light&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28738" PostTypeId="1" AcceptedAnswerId="28893" CreationDate="2018-12-15T20:51:12.467" Score="2" ViewCount="98" Body="&lt;p&gt;When carbon dioxide forms on Mars is it like ice or snow? How dense is newly formed solid carbon dioxide? Would more carbon dioxide form inside a pit vs a flat plain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/28232/how-thick-or-deep-does-solid-carbon-dioxide-accumulate-on-mars&quot;&gt;How thick or deep does solid carbon dioxide accumulate on Mars?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://space.stackexchange.com/questions/31724/martian-carbon-dioxide-turbine-oxygen-generator&quot;&gt;https://space.stackexchange.com/questions/31724/martian-carbon-dioxide-turbine-oxygen-generator&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2019-01-26T17:20:25.663" LastActivityDate="2019-01-26T17:20:25.663" Title="What are the characteristics of solid carbon dioxide on the Martian poles?" Tags="&lt;planetary-atmosphere&gt;&lt;mars&gt;&lt;weather&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28740" PostTypeId="1" CreationDate="2018-12-15T22:03:56.093" Score="2" ViewCount="88" Body="&lt;p&gt;I've got a Celestron Astromaster 114 telescope. I live in Los Angeles and would like to get a look at 46P/Wirtanen. Can anyone tell me &lt;em&gt;when&lt;/em&gt; this comet is going to brightest and how best to locate it in the sky? It's my (perhaps incorrect) understanding that it is not especially bright. Any tips/hints/information would be much appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aOH2E.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aOH2E.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://telescopereviewer.com/celestron-astromaster-114-eq-reflector-telescope-review/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16705" LastEditorUserId="7982" LastEditDate="2018-12-16T01:28:11.830" LastActivityDate="2018-12-16T01:28:11.830" Title="How to see Comet 46P/Wirtanen?" Tags="&lt;telescope&gt;&lt;comets&gt;&lt;night-sky&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28741" PostTypeId="2" ParentId="28740" CreationDate="2018-12-15T22:36:20.300" Score="1" Body="&lt;p&gt;Peak magnitude is &quot;now&quot;, or a few days either side  of the 17th of December. The peak magnitude is about 4.0 (so this is not like Hale-Bopp), (assuming nothing unexpected happens) it can be found passing the Pleiades over the next few days. The waxing moon will affect the evenings, which you should consider when planning observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In-the-sky have an &lt;a href=&quot;https://in-the-sky.org/ephemeris.php?ird=1&amp;amp;irs=1&amp;amp;ima=1&amp;amp;iob=1&amp;amp;objtype=3&amp;amp;objtxt=46P%2FWirtanen&amp;amp;tz=0&amp;amp;startday=1&amp;amp;startmonth=12&amp;amp;startyear=2018&amp;amp;interval=0&quot; rel=&quot;nofollow noreferrer&quot;&gt;ephemeris for the comet:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space.com have a &lt;a href=&quot;https://www.space.com/42751-see-comet-46p-wirtanen-earth-flyby-december-2018.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;finder chart&lt;/a&gt; and general discussion&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-12-15T22:36:20.300" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28742" PostTypeId="1" AcceptedAnswerId="28747" CreationDate="2018-12-16T01:21:23.603" Score="1" ViewCount="51" Body="&lt;p&gt;Suppose that all stars in this galaxy were born in a single major-merger burst event about 10&#xA;Gyr ago. From this original burst, I want to compute the fraction of stellar mass still surviving as stars in the&#xA;main sequence ? For this, I have got to use a Salpeter IMF, and a star formation range between 0.1 and 120 solar&#xA;masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have done is starting from Salpeter IMF : &lt;span class=&quot;math-container&quot;&gt;$$\Phi(m)\text{d}m=\Phi_{0}\,m^{-2.35}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with &lt;span class=&quot;math-container&quot;&gt;$$\Phi_{0}$$&lt;/span&gt; a constant normalization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, I integrate from &lt;span class=&quot;math-container&quot;&gt;$$m_{1}=0.1\,\text{M}_{\odot}$$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$$m_{2}=120\,\text{M}_{\odot}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$N(0.1&amp;lt;m&amp;lt;120) = \int_{0.1}^{120}\,\Phi(m)\,\text{d}m = \Phi_{0}\,\bigg[\dfrac{0.1^{-1.35}-120^{-1.35}}{1.35}\bigg]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This result depends on the valeur of &lt;span class=&quot;math-container&quot;&gt;$$\Phi_{0}$$&lt;/span&gt; and I don't know how to deal with it in order to get &lt;span class=&quot;math-container&quot;&gt;$$N(0.1&amp;lt;m&amp;lt;120)$$&lt;/span&gt; ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, it seems that I have to take into account of the age of the major-merger burst event (10 Gyr).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From these 2 principles, how could I calculate the fraction of stars surviving in the main sequence ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is wlecome, Regards&lt;/p&gt;&#xA;" OwnerUserId="16492" LastActivityDate="2018-12-16T08:28:08.673" Title="From original burst, fraction of stellar mass still surviving on Main sequence" Tags="&lt;galaxy&gt;&lt;stellar-evolution&gt;&lt;density&gt;&lt;initial-mass-function&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28743" PostTypeId="2" ParentId="28740" CreationDate="2018-12-16T01:27:09.633" Score="2" Body="&lt;p&gt;If you have, or can borrow binoculars, that is probably a &lt;em&gt;much better way&lt;/em&gt; to try to find the comet than your telescope. If it's set up like the one in the image in your question (that I've added) you don't have a computer read-out for RA/Dec or any way to position your mount accurately so that the polar axis points at polaris. The bright skies in LA will make doing so manually a bit of a challenge, and lack of previous experience doing so will make it even more of one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go to in-the-sky.org and then to &lt;strong&gt;Planetarium view&lt;/strong&gt;. Go to the bottom right and type &lt;strong&gt;46P&lt;/strong&gt; into the Find box and it will find your object's name and then display it for you. Turn on the Alt/Az grid instead of the RA/Dec grid and you can see that 46P is about 60 degrees up at 8 PM your time. Lay on your back and take some time to look around with your binoculars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should set your location to Los Angeles automatically, but double check that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that there are some very bright stars lined up to make your viewing easier. First is &lt;a href=&quot;https://en.wikipedia.org/wiki/Betelgeuse&quot; rel=&quot;nofollow noreferrer&quot;&gt;Betelgeuse&lt;/a&gt; the bright corner star in Orion (&lt;em&gt;do not&lt;/em&gt; say Betelgeuse three times in a row!) then &lt;a href=&quot;https://en.wikipedia.org/wiki/Aldebaran&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aldebaran&lt;/a&gt; the eye of the bull, then the comet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At about 4th magnitude you should be able to see it as a green fuzzy star, unlike anything else. You can also zoom in using in-the-sky.org to see the local stars. The comet is going to be moving relative to the stars, so don't make zoomed-in print-outs and try to use them six hours or a day later. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set up anything you have around the spot where you'll be laying down, to block lights from neighbors or the street, turn your own lights off. You are going to be looking &lt;strong&gt;East-Southeast about 2/3 of the way up to the zenith&lt;/strong&gt; in the evening, and 46P will remain high in the sky most of the night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have fun!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Oh, the telescope!&lt;/strong&gt; Once you've gotten good at finding 46P with binoculars and have some feel for the other stars nearby, then try the absolute lowest magnification you have on your telescope and just look around slowly. Loosen up your mount so you can move the tube by hand, and scan around until you see the blue fuzzy blob again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bhMh9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bhMh9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ix8zH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ix8zH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-12-16T01:27:09.633" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28744" PostTypeId="1" AcceptedAnswerId="28765" CreationDate="2018-12-16T01:51:16.827" Score="9" ViewCount="1274" Body="&lt;p&gt;I'm writing a program that involves calculating the apparent magnitude of satellites from a ground location. I currently have the intrinsic magnitude of the satellites and the solar phase angle in degrees. I can't seem to find a formula that works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;magnitude = intrinsicMagnitude - 15 + 5 * Math.Log(distanceToSatellite) - 2.5 * Math.Log(Math.Sin(B) + (Math.PI - B) * Math.Cos(B));&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(B is the phase angle)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...but it does not work (it's returning numbers like +30). I know it's wrong because I'm comparing it to heavens-above.com satellite passes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;intrinsicMagnitude = Visual magnitude at 1000km away (Use -1.3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;distanceToSatellite = Observer distance to the satellite in km (Use 483)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B = This is what I am trying to figure out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the paper it says what this is but it says some other things that I do not understand. The phase angle you use to get this should be 113.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The target output of this equation should be around -3.&lt;/p&gt;&#xA;" OwnerUserId="25168" LastEditorUserId="1082" LastEditDate="2018-12-17T18:54:30.967" LastActivityDate="2018-12-21T18:24:04.137" Title="Calculating the apparent magnitude of a satellite" Tags="&lt;satellite&gt;&lt;artificial-satellite&gt;&lt;mathematics&gt;&lt;iss&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28745" PostTypeId="1" CreationDate="2018-12-16T03:39:55.390" Score="9" ViewCount="120" Body="&lt;p&gt;Is there any indication of how much dust is deposited, especially in areas that are not open where wind might produce a cleaning event?&lt;/p&gt;&#xA;" OwnerUserId="25088" LastEditorUserId="7982" LastEditDate="2021-02-13T01:41:44.097" LastActivityDate="2021-04-07T09:58:10.223" Title="How much is usually deposited in a Mars regional dust storm?" Tags="&lt;observational-astronomy&gt;&lt;mars&gt;&lt;dust-storm&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28747" PostTypeId="2" ParentId="28742" CreationDate="2018-12-16T08:28:08.673" Score="1" Body="&lt;p&gt;When you are calculating &lt;em&gt;fractions&lt;/em&gt;, rather than absolute numbers, the value of &lt;span class=&quot;math-container&quot;&gt;$\Phi_0$&lt;/span&gt; does &lt;em&gt;not&lt;/em&gt; matter, since it will be a multiplying factor in both the numerator and denominator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have (almost, see below)  successfully got an expression for the &lt;em&gt;denominator&lt;/em&gt; of your fraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numerator is found by calculating an equivalent integral from your lower limit to an upper limit that is instead defined by the most long-lived main sequence stars that are still &quot;alive&quot; - i.e. those with a lifetime equal to the age of our Galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you were asked to find the fraction of stellar mass surviving, not the fraction of stars. The stellar mass existing between two mass intervals is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$M_* = \int_{m_1}^{m_2} m\Phi(m)\ dm$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-12-16T08:28:08.673" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28748" PostTypeId="1" CreationDate="2018-12-16T09:27:08.630" Score="3" ViewCount="77" Body="&lt;p&gt;Star forming molecular clouds and especially Bok globules are low temperature &lt;span class=&quot;math-container&quot;&gt;$\sim 10 $&lt;/span&gt; K environments with density on the order of &lt;span class=&quot;math-container&quot;&gt;$10^{-18}$&lt;/span&gt; kg/m&lt;span class=&quot;math-container&quot;&gt;$^3$&lt;/span&gt;, mostly hydroogen with about 1% dust. What is their heat capacity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to find good data for low-temperature heat capacities for hydrogen (which would be nice to have anyway), but most experiments also use dense hydrogen rather than the by Earth standard very low-pressure molecular gas of the clouds. Are there any decent estimates of how much a given increase in energy increases the cloud temperature? Or should one just use the ideal gas approximation for a diatomic gas and assume &lt;span class=&quot;math-container&quot;&gt;$C_v\approx (5/2)R$&lt;/span&gt; per mole?&lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2018-12-16T09:27:08.630" Title="What is the specific heat capacity of interstellar clouds?" Tags="&lt;interstellar-medium&gt;&lt;heat&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28749" PostTypeId="1" CreationDate="2018-12-16T11:25:14.577" Score="1" ViewCount="63" Body="&lt;p&gt;Suppose that all stars in this galaxy were born in a single major-merger burst event about 10 Gyr ago&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the luminosity in the B band (absolute magnitude in B-band is equal to -21.22) is dominated by stars of in the RG branch, with masses &lt;span class=&quot;math-container&quot;&gt;$m \sim 1\,\text{L}_{\odot}$&lt;/span&gt; (within &lt;span class=&quot;math-container&quot;&gt;$\sim 10\%$&lt;/span&gt;) and average luminosities &lt;span class=&quot;math-container&quot;&gt;$\sim 1000\,\text{L}_{\odot}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I estimate the total stellar mass of this galaxy using the Schechter relation ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that I have to use the definition of &lt;strong&gt;Schechter law&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$N(L)\ \mathrm {d} L=\phi^{*}\left({\frac{L}{L^{*}}}\right)^{\alpha}\mathrm {e}^{-L/L^{*}}{\frac{\mathrm {d} L}{L^{*}}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or maybe &lt;strong&gt;Salpeter relation&lt;/strong&gt; : &lt;span class=&quot;math-container&quot;&gt;$$\text{d}N=0.06\,\times\,M^{-2.35}\,\text{d}M$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how to introduce the parameters of Red-Giants of &lt;span class=&quot;math-container&quot;&gt;$1\,\text{M}_{\odot}$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$L=1000\,\text{L}_{\odot}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially, I calculate the fraction of masse between &lt;span class=&quot;math-container&quot;&gt;$m_{1}=0.9$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$m_{2}=1.1\,\text{M}_{\odot}$&lt;/span&gt; :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text{d}N(m_{1}&amp;lt;m&amp;lt;m_{2})=\int_{m_{1}}^{m_{2}}\,\Phi(m)\,\text{d}m=0.06\,\dfrac{(0.9^{-1.35}-1.1^{-1.35})}{1.35} = 1.22 \%$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Anyone could see the trick to compute total stellar mass from these parameters with above laws ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards&lt;/p&gt;&#xA;" OwnerUserId="16492" LastEditorUserId="16492" LastEditDate="2018-12-16T15:25:27.597" LastActivityDate="2018-12-16T15:25:27.597" Title="Compute the total stellar mass of galaxy with major influence of Red Giants luminosity" Tags="&lt;galaxy&gt;&lt;stellar-evolution&gt;&lt;mass&gt;&lt;stellar-astrophysics&gt;&lt;initial-mass-function&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28750" PostTypeId="1" CreationDate="2018-12-16T13:34:16.787" Score="4" ViewCount="240" Body="&lt;p&gt;I've done the 'sparkly curtains behind landscapes' form of astrophotography for years with DSLRs, lately full frame Nikon. I know what I'm doing with the camera, and want to take the next step into incompetence with telescopes to point at stars with a camera attached, but I'm getting conflicting advice on what kit I need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One group are saying 'basic motorised Newtonian will be fine' - e.g.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.harrisontelescopes.co.uk/acatalog/skywatcher-explorer-130m-telescope.html#SID=34&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.harrisontelescopes.co.uk/acatalog/skywatcher-explorer-130m-telescope.html#SID=34&lt;/a&gt;, &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.firstlightoptics.com/celestron-astromaster-series/celestron-astromaster-130eq-telescope.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.firstlightoptics.com/celestron-astromaster-series/celestron-astromaster-130eq-telescope.html&lt;/a&gt;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Another are saying 'that's not powerful enough, you need a cadiotropic goto like &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.picstop.co.uk/goto-telescopes/celestron-nexstar-evolution-6-schmidt-cassegrain-goto-telescope.html?wgu=217_60210_15449648240362_d6add8be16&amp;amp;wgexpiry=1547556824&amp;amp;source=webgains&amp;amp;siteid=60210&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.picstop.co.uk/goto-telescopes/celestron-nexstar-evolution-6-schmidt-cassegrain-goto-telescope.html?wgu=217_60210_15449648240362_d6add8be16&amp;amp;wgexpiry=1547556824&amp;amp;source=webgains&amp;amp;siteid=60210&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They're obviously very different bits of kit...! And I can see both arguments TBH. Either one is trying to waste my time with inadequate kit, or another to waste my money into his pocket with overspecified kit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help....?&lt;/p&gt;&#xA;" OwnerUserId="24104" LastEditorUserId="7982" LastEditDate="2018-12-18T03:42:30.827" LastActivityDate="2018-12-21T19:34:05.413" Title="How much telescope do I actually need for astrophotography?" Tags="&lt;photography&gt;" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28751" PostTypeId="2" ParentId="28750" CreationDate="2018-12-16T18:56:20.707" Score="2" Body="&lt;p&gt;I would start with at least an equatorial mount sth as the HEQ5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are many factors to take into account such as what kind of targets you want, mostly, to shoot. Also, you you want to take photos with a wide of narrow field of view, just like Stephen asked in his comment. &#xA;Then you also have t think about guiding the mount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I wouldn't choose any of these set ups. But we need more information in order to narrow down your options within your budget's restrictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers &lt;/p&gt;&#xA;" OwnerUserId="5355" LastActivityDate="2018-12-16T18:56:20.707" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28752" PostTypeId="2" ParentId="1767" CreationDate="2018-12-16T21:26:27.727" Score="-2" Body="&lt;p&gt;Because atoms are neither created nor destroyed we could be made of anything. Not only particles that use to be stars, but just particles in the air, the oxygen we breathe, the atmosphere. Because we are created from a sperm and egg cell from two people, we contain molecules from their bodies. This is a very interesting discussion. &lt;/p&gt;&#xA;" OwnerUserId="25189" LastActivityDate="2018-12-16T21:26:27.727" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28753" PostTypeId="2" ParentId="28715" CreationDate="2018-12-16T21:34:21.153" Score="0" Body="&lt;p&gt;No. You need to specify structure formation and galaxy evolution for details, but in general, you will expect many galaxies in a dark matter halo. If you need a relation, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Halo_occupation_distribution&quot; rel=&quot;nofollow noreferrer&quot;&gt;halo occupation distribution&lt;/a&gt; might be a start. The Wikipedia page references three good texts on the matter.&lt;/p&gt;&#xA;" OwnerDisplayName="user18746" LastActivityDate="2018-12-16T21:34:21.153" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28755" PostTypeId="1" CreationDate="2018-12-17T01:27:10.610" Score="0" ViewCount="186" Body="&lt;p&gt;Suppose the amount of mass in a black hole doubles. Does the event horizon change? If so, how much does it change?&lt;/p&gt;&#xA;" OwnerUserId="25192" LastEditorUserId="6" LastEditDate="2019-01-29T14:42:53.193" LastActivityDate="2019-01-29T14:42:53.193" Title="Need help with a simple Blackhole question" Tags="&lt;black-hole&gt;&lt;mass&gt;&lt;event-horizon&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28759" PostTypeId="2" ParentId="28755" CreationDate="2018-12-17T08:54:14.387" Score="1" Body="&lt;p&gt;Let's have a look at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Black_hole#Physical_properties&quot; rel=&quot;nofollow noreferrer&quot;&gt;Physical_properties&lt;/a&gt; subsection of the Wikipedia article on Black holes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schwarzschild radius&lt;/a&gt; which is defined only for a non-rotating black hole is given there as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$r_S=\frac{2GM}{c^2}\approx 2.95 \frac{M}{M_{Sun}} \text{km}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Schwarzschild radius was named after the German astronomer Karl Schwarzschild, who calculated this exact solution for the theory of general relativity in 1916.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the Schwarzschild radius increases linearly with mass, by 2.95 kilometers for every solar mass you add, provided you add it carefully and don't spill any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to express the &quot;size&quot; of the event horizon for an arbitrary black hole is going to be more complicated, so I'm not going to adventure into that territory. However, @Chappo's &lt;a href=&quot;https://astronomy.stackexchange.com/questions/28755/need-help-with-blackhole-question-asap#comment51238_28755&quot;&gt;comment&lt;/a&gt; suggests the dependence on mass will also be approximately linear as well.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-12-17T08:54:14.387" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28760" PostTypeId="1" AcceptedAnswerId="28769" CreationDate="2018-12-17T09:50:40.630" Score="0" ViewCount="91" Body="&lt;p&gt;This &lt;a href=&quot;https://astronomy.stackexchange.com/a/22996/7982&quot;&gt;excellent answer&lt;/a&gt; to the question &lt;em&gt;What is the physics of the “spinning dust” contribution to Cosmic Microwave Background measurements?&lt;/em&gt; includes the figure below which helps show what it was the triggered the existence-hypothesizing of cosmic spinning dust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eHR9q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eHR9q.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question &lt;a href=&quot;https://astronomy.stackexchange.com/q/28637/7982&quot;&gt;Why doesn't the equipartition theorem disallow spinning dust?&lt;/a&gt; is my way of starting to wonder if spinning dust is real, or if it really happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that would make me more comfortable than an &lt;s&gt;argument&lt;/s&gt; gedankenexperiment involving a modified Crookes radiometer with the gas removed would be some kind of experiment showing that simply being exposed to some very low pressure hydrogen, or photons, can make dust spin at anything close to 20 GHz!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These days optical trapping of particles in very high vacuum is readily accessible for example, so I think if someone wanted to try to make GHz-spinning-dust they could certainly have a go. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Has this been tried? Has &quot;GHz-spinning dust&quot; ever been demonstrated in the laboratory? &lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-12-17T23:05:45.697" Title="Has &quot;GHz-spinning dust&quot; ever been demonstrated in the laboratory?" Tags="&lt;astrophysics&gt;&lt;interstellar-medium&gt;&lt;dust&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28762" PostTypeId="1" CreationDate="2018-12-17T15:25:41.533" Score="1" ViewCount="175" Body="&lt;p&gt;According to the Helical model, as the Sun orbits around the galaxy, the planets that orbit the sun are travelling perpendicular to the path of the Sun.  This is better shown in this video:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=0jHsq36_NTU&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=0jHsq36_NTU&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, which pole on Earth faces in the direction that the Sun moves?  The South Pole or the North Pole?&lt;/p&gt;&#xA;" OwnerUserId="2844" LastActivityDate="2020-02-29T00:55:07.910" Title="Which direction does the Earth face as it travels through the Galaxy" Tags="&lt;galaxy&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="9" ClosedDate="2020-02-29T05:27:07.617" ContentLicense="CC BY-SA 4.0" />
  <row Id="28763" PostTypeId="1" CreationDate="2018-12-17T15:43:09.823" Score="5" ViewCount="52" Body="&lt;p&gt;I am wondering. I am sure that they are related. How can we derive one from the other?&lt;/p&gt;&#xA;" OwnerUserId="24850" LastActivityDate="2018-12-17T15:43:09.823" Title="What is the relation between the saros (18 years, 11 days, 8 hours) and the period of lunar nodal precession (18.6 years)?" Tags="&lt;the-moon&gt;&lt;solar-eclipse&gt;&lt;lunar-eclipse&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28765" PostTypeId="2" ParentId="28744" CreationDate="2018-12-17T17:57:40.707" Score="3" Body="&lt;p&gt;This is for satellites with unknown size and orientation but known standard magnitude (Standard magnitude can be found on the satellite info page of heavens above, the number is called intrinsic magnitude)&#xA;The proper formula is &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;            double distanceToSatellite = 485; //This is in KM&#xA;            double phaseAngleDegrees = 113.1; //Angle from sun-&amp;gt;satellite-&amp;gt;observer&#xA;            double pa = phaseAngleDegrees * 0.0174533; //Convert the phase angle to radians&#xA;            double intrinsicMagnitude = -1.8; //-1.8 is std. mag for iss&#xA;&#xA;&#xA;            double term_1 = intrinsicMagnitude;&#xA;            double term_2 = 5.0 * Math.Log10(distanceToSatellite / 1000.0);&#xA;&#xA;            double arg = Math.Sin(pa) + (Math.PI - pa) * Math.Cos(pa);&#xA;            double term_3 = -2.5 * Math.Log10(arg);&#xA;&#xA;            double apparentMagnitude = term_1 + term_2 + term_3;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This will give the apparent magnitude of the satellite.&#xA;Note: I gave the formula in C#&lt;/p&gt;&#xA;" OwnerUserId="25168" LastEditorUserId="25168" LastEditDate="2018-12-21T18:24:04.137" LastActivityDate="2018-12-21T18:24:04.137" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28766" PostTypeId="1" CreationDate="2018-12-17T18:35:12.167" Score="1" ViewCount="332" Body="&lt;p&gt;I read somewhere that sun's daily course and complete cycle is visible in the ecliptic belt by zodiac constellations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't understand what that means closest I could find online is: &lt;a href=&quot;https://www.youtube.com/watch?v=x1DqwC7lRpk&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=x1DqwC7lRpk&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the sun's route visible during the night in some areas? Can someone explain?&lt;/p&gt;&#xA;" OwnerUserId="25200" LastEditorUserId="2365" LastEditDate="2018-12-17T18:39:29.867" LastActivityDate="2018-12-18T03:34:49.953" Title="What is meant by sun's path through zodiac" Tags="&lt;the-sun&gt;&lt;rotation&gt;&lt;ecliptic&gt;&lt;zodiac&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28767" PostTypeId="2" ParentId="28766" CreationDate="2018-12-17T18:48:00.283" Score="2" Body="&lt;p&gt;The stars are not visible during the day, but they are still there (we just can't see them because of the brightness of the sun). We can't see the stars, but we can calculate where they would be if we could see them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over the course of the year the sun appears to move relative to the stars. This is actually due to the orbit of the Earth around the sun. The apparent motion of the sun is quite simple: it moves around the sky in a great circle, always following the same path, and taking exactly one year to complete. The route it takes is called the ecliptic. And most of the constellations that lie on the ecliptic are called the zodiac (they are very important to astrologers, but not especially important to astronomers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun's route is not visible to the naked eye from Earth, but it can be calculated with great precision.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-12-17T18:48:00.283" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28768" PostTypeId="2" ParentId="28750" CreationDate="2018-12-17T19:54:06.160" Score="1" Body="&lt;p&gt;The mount is the most important thing. Then the camera. Then the telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Get the best mount you could possibly afford. I'm not saying you should start directly with a Paramount MYT (although that would be great), but I suggest you take a good long look at Losmandy and figure out what works for you. It's going to be expensive, yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.bisque.com/sc/pages/ParamountLandingPage.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.bisque.com/sc/pages/ParamountLandingPage.aspx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.losmandy.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.losmandy.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mounts need to be designed for the carrying capacity that will take all your gear - telescope, camera, filters, autoguider, etc. Always allow some headroom in terms of carrying capacity - don't max out the mount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If those are too expensive, the bare minimum I would recommend is something like a Celestron CGEM or equivalent, but these are less worry- and fiddle-free than the higher end mounts. You could get good results with the CGEM (I have), but more work is required. Be prepared for some learning curve there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.celestron.com/products/cgem-computerized-mount&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.celestron.com/products/cgem-computerized-mount&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With mounts in this class, it's actually best to stay below 1/2 of their declared capacity - if the mount claims it can carry X lb or kg of gear, stay below X/2. More expensive mounts can carry closer to their stated max load. For planetary shots you could increase the load a little bit and get closer to the limits, but for DSO (galaxies, nebulae) definitely follow the guidelines I've told you. An oversized mount (designed to carry way more than your gear's weight) will always work better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of camera, some sort of cooled, large sensor camera for deep space objects (DSO), and a small sensor non-cooled camera with a fast chip (a.k.a. planetary camera) for planets. There are many good camera makers out there. I use ZWO products with good results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy-imaging-camera.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://astronomy-imaging-camera.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big sensor camera will be quite similar to a DSLR (some actually use the same sensor as some popular DSLR cameras), except it's cooled and it's optimized for astronomy. Cooling helps quite a bit with the noise. You could use a DSLR instead, it's just going to be more noisy so you probably need to collect a bit more data to remove noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The planetary camera could be used for autoguiding when taking DSO shots with the big cooled camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Telescopes: you will probably end up with two instruments - a refractor for wide angle DSOs, and an SCT for zooming into smaller targets (like small DSOs and planets). These don't need to be very fancy. Even a doublet refractor will do well (like an Orion ED80 with a focal reducer). A Celestron C8 (or the better corrected EdgeHD series) would work well as an SCT, but more aperture is better (assuming your mount can carry it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.telescope.com/Orion/Orion-ED80-80mm-f75-Apochromatic-Refractor-Telescope/rc/2160/p/9895.uts&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.telescope.com/Orion/Orion-ED80-80mm-f75-Apochromatic-Refractor-Telescope/rc/2160/p/9895.uts&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.celestron.com/products/edgehd-8-optical-tube-assembly&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.celestron.com/products/edgehd-8-optical-tube-assembly&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;High resolution planetary shots can be taken with a large dobsonian on an equatorial platform. This way you could get way more aperture than any SCT, and the dobsonian could do double duty as a visual instrument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's really no aperture limit for planetary shots. More is always better. Yay dobs!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For DSOs, start with a short focal length instrument, like an ED80 with a focal reducer. It's a heck of a lot easier to figure out how the mount works, and autoguiding and all that, when the scope has a focal length in the range of 400 ... 600 mm. Pick easy targets first, like M31 or the Orion nebula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Learn how that works, and then move up to bigger focal lengths, like with an SCT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For planetary shots you could use a huge focal length from the beginning (SCT with barlow), since you don't need high precision in tracking (also, no autoguiding either).&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2018-12-17T21:15:08.150" LastActivityDate="2018-12-17T21:15:08.150" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28769" PostTypeId="2" ParentId="28760" CreationDate="2018-12-17T23:05:45.697" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://spacephysics.msfc.nasa.gov/science/dusty/posters/rot_exp.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This poster&lt;/a&gt; suggests an answer of &lt;em&gt;yes and no&lt;/em&gt;. Essentially the poster reports a demonstration of some of the proposed mechanisms, but the rotation speeds produced are in the &lt;span class=&quot;math-container&quot;&gt;$kHz$&lt;/span&gt; not &lt;span class=&quot;math-container&quot;&gt;$GHz$&lt;/span&gt; range because the dust partcles are bigger, the vacuum less good and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To quote the summary:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;First laboratory measurements of rotation of the analogs of interstellar&#xA;  dust grains induced by radiative torques have been made&lt;/li&gt;&#xA;  &lt;li&gt;The measurements were made on irregularly shaped SiC particles&#xA;  particles of ~ 0.17 to 8.2 &lt;span class=&quot;math-container&quot;&gt;$\mu m$&lt;/span&gt; effective radii levitated in an electrodynamic&#xA;  balance and illuminated with laser light at a wavelength of 5320 Å, at&#xA;  pressures of &lt;span class=&quot;math-container&quot;&gt;$10^{-3}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$10^{-5}$&lt;/span&gt; torr.&lt;/li&gt;&#xA;  &lt;li&gt;The rotation rates with radiation intensities of ~ &lt;span class=&quot;math-container&quot;&gt;$4-30 Wcm^{-2}$&lt;/span&gt; were&#xA;  observed in the range of 1,000-22,000 rot/sec.&lt;/li&gt;&#xA;  &lt;li&gt;The grain rotation rates were observed to follow the expected functional&#xA;  relationships, being directly proportional to the incident radiation&#xA;  intensities and inversely proportional to the drag represented by the&#xA;  ambient pressure.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2018-12-17T23:05:45.697" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28770" PostTypeId="1" CreationDate="2018-12-17T23:08:28.070" Score="2" ViewCount="259" Body="&lt;p&gt;How could a &lt;em&gt;great year&lt;/em&gt; be explained in &lt;em&gt;laymen's terms&lt;/em&gt;?&lt;/p&gt;&#xA;&lt;p&gt;According to the definition in NASA's &lt;a href=&quot;https://www.hq.nasa.gov/office/hqlibrary/aerospacedictionary/aerodictall/g.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aerospace Dictionary&lt;/a&gt;  it is a complete cycle of the equinoxes around the ecliptic that completes in 26000 years:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;great year&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;(NASA SP-7, 1965)&#xA;The period of one complete cycle of the equinoxes around the ecliptic, about 25,800 years. Also called platonic year. See &lt;a href=&quot;https://www.hq.nasa.gov/office/hqlibrary/aerospacedictionary/aerodictall/p.html#precession%20of%20the%20equinoxes&quot; rel=&quot;nofollow noreferrer&quot;&gt;precession of the equinoxes&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Does this mean that every X years the March 21st and September 22nd will be in a different part of the zodiac and it will take 26000 years to make a full circle?&lt;/p&gt;&#xA;" OwnerUserId="25200" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-12-18T02:59:46.990" Title="What is the great (platonic?) year? How could it be explained in laymen's terms?" Tags="&lt;the-sun&gt;&lt;ecliptic&gt;&lt;equinox&gt;&lt;zodiac&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28771" PostTypeId="1" AcceptedAnswerId="28780" CreationDate="2018-12-18T02:25:25.317" Score="1" ViewCount="265" Body="&lt;p&gt;The Carnegie Science article &lt;a href=&quot;https://carnegiescience.edu/news/discovered-most-distant-solar-system-object-ever-observed&quot; rel=&quot;nofollow noreferrer&quot;&gt;Discovered: The Most-Distant Solar System Object Ever Observed&lt;/a&gt; begins:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Washington, DC— A team of astronomers has discovered the most-distant body ever observed in our Solar System.  It is the first known Solar System object that has been detected at a distance that is more than 100 times farther than Earth is from the Sun.&lt;/p&gt;&#xA;&lt;p&gt;The new object was announced on Monday, December 17, 2018, by the International Astronomical Union’s Minor Planet Center and has been given the provisional designation 2018 VG18. The discovery was made by Carnegie’s Scott S. Sheppard, the University of Hawaii’s David Tholen, and Northern Arizona University’s Chad Trujillo.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;2018 VG18, nicknamed “Farout”&lt;/strong&gt; by the discovery team for its extremely distant location, is at about 120 astronomical units (AU), where 1 AU is defined as the distance between the Earth and the Sun.  The second-most-distant observed Solar System object is Eris, at about 96 AU.  Pluto is currently at about 34 AU, making 2018 VG18 more than three-and-a-half times more distant than the Solar System’s most-famous dwarf planet.&lt;/p&gt;&#xA;&lt;p&gt;2018 VG18 was discovered as part of the team’s continuing search for extremely distant Solar System objects, including the suspected Planet X, which is sometimes also called Planet 9.  In October, the same group of researchers &lt;a href=&quot;https://carnegiescience.edu/news/new-extremely-distant-solar-system-object-found-during-hunt-planet-x&quot; rel=&quot;nofollow noreferrer&quot;&gt;announced the discovery&lt;/a&gt; of another distant Solar System object, called 2015 TG387 and nicknamed “The Goblin,” because it was first seen near Halloween.  The Goblin was discovered at about 80 AU and has an orbit that is consistent with it being influenced by an unseen Super-Earth-sized Planet X on the Solar System’s very distant fringes.&lt;/p&gt;&#xA;&lt;p&gt;The existence of a ninth major planet at the fringes of the Solar System was first proposed by this same research team in 2014 when they &lt;a href=&quot;https://carnegiescience.edu/news/solar-systems-edge-redefined&quot; rel=&quot;nofollow noreferrer&quot;&gt;discovered 2012 VP113&lt;/a&gt;, nicknamed Biden, which is currently near 84 AU.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;That's great! But the next bit confuses me.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The discovery images of 2018 VG18 were taken at the Japanese Subaru 8-meter telescope located atop Mauna Kea in Hawaii on &lt;strong&gt;November 10, 2018&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Once 2018 VG18 was found, it needed to be re-observed to confirm its very distant nature. (&lt;strong&gt;It takes multiple nights of observing to accurately determine an object’s distance.&lt;/strong&gt;)  2018 VG18 was seen for the second time in &lt;strong&gt;early December&lt;/strong&gt; at the Magellan telescope at Carnegie’s Las Campanas Observatory in Chile.  These recovery observations were performed by the team with the addition of graduate student Will Oldroyd of Northern Arizona University.  &lt;strong&gt;Over the next week, they monitored 2018 VG18 with the Magellan telescope to secure its path across the sky and obtain its basic physical properties such as brightness and color.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;The Magellan observations confirmed that 2018 VG18 is around 120 AU&lt;/strong&gt;, making it the first Solar System object observed beyond 100 AU.  Its brightness suggests that it is about 500 km in diameter, likely making it spherical in shape and a dwarf planet. It has a pinkish hue, a color generally associated with ice-rich objects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question(s):&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How was the value of 120 AU &lt;em&gt;actually determined?&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;or at least, how &lt;em&gt;could it be determined&lt;/em&gt; if the explanation isn't readily available?&lt;/li&gt;&#xA;&lt;li&gt;Is this measurement &lt;em&gt;more of&lt;/em&gt; an orbit reconstruction, or a parallax measurement of current distance?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; &amp;quot;Discovery images of 2018 VG18, nicknamed 'Farout,' from the Subaru Telescope on November 10, 2018. Farout moves between the two discovery images while the background stars and galaxies do not move over the one hour between images. Image is courtesy of Scott S. Sheppard and David Tholen.&amp;quot; &lt;a href=&quot;https://carnegiescience.edu/news/discovered-most-distant-solar-system-object-ever-observed&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jpb3a.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jpb3a.jpg&quot; alt=&quot;2018 VG18, nicknamed &amp;quot;Farout,&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-12-18T07:24:32.677" Title="How do they know how far out Farout is? (2018 VG18)" Tags="&lt;asteroids&gt;&lt;astrometry&gt;&lt;dwarf-planets&gt;&lt;trans-neptunian-object&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28772" PostTypeId="1" CreationDate="2018-12-18T02:33:19.253" Score="5" ViewCount="179" Body="&lt;p&gt;I live in New England, Cape Cod to be exact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw a very bright, twinkling object. I watched it for several minutes and it never moved. It was in the east just above the tree line, what might it be? &lt;/p&gt;&#xA;" OwnerUserId="25203" LastEditorUserId="7982" LastEditDate="2018-12-18T02:54:16.710" LastActivityDate="2018-12-18T02:54:16.710" Title="South-eastern sky above the tree line just now, NE Cape Cod, I saw a very bright, twinkling &quot;star&quot;" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="28773" PostTypeId="1" AcceptedAnswerId="28775" CreationDate="2018-12-18T02:34:15.813" Score="4" ViewCount="148" Body="&lt;p&gt;The Carnegie Science article &lt;a href=&quot;https://carnegiescience.edu/news/solar-systems-edge-redefined&quot; rel=&quot;noreferrer&quot;&gt;Solar System's Edge Redefined&lt;/a&gt; describes the discovery of the object 2012 VP113.&lt;/p&gt;&#xA;&lt;p&gt;Wikipedia's article &lt;a href=&quot;https://en.wikipedia.org/wiki/2012_VP113&quot; rel=&quot;noreferrer&quot;&gt;2012 VP113&lt;/a&gt; says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Nickname:&lt;/p&gt;&#xA;&lt;p&gt;2012 VP113 was abbreviated &amp;quot;VP&amp;quot; and nicknamed &amp;quot;Biden&amp;quot; by the discovery team, after Joe Biden, who at the time of discovery, was Vice President (VP) of the United States.&lt;sup&gt;11&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;11&lt;/sup&gt; &lt;a href=&quot;https://www.nature.com/news/dwarf-planet-stretches-solar-system-s-edge-1.14921&quot; rel=&quot;noreferrer&quot;&gt;Dwarf planet stretches Solar System's edge&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;and the referenced Nature News item says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The newfound object's official name is 2012 VP113, but the discovery team calls it VP for short, or just 'Biden' — after US Vice-President Joe Biden. In several years time, after observations have pinned down its orbit, the scientists will submit a name for consideration by the International Astronomical Union (IAU), the organization in charge of celestial nomenclature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; I understand that VP is short for Vice President, but where do those two letter codes come from? Is the VP in the object's name short for something literal (did the team choose these two letters?), or just a random computer-generated code, or something else?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-12-18T02:56:56.737" Title="How did Biden become VP? (naming of 2012 VP113)" Tags="&lt;asteroids&gt;&lt;naming&gt;&lt;dwarf-planets&gt;&lt;iau&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28774" PostTypeId="2" ParentId="28772" CreationDate="2018-12-18T02:50:11.660" Score="3" Body="&lt;p&gt;I just did the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;opened &lt;a href=&quot;https://in-the-sky.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;in-the-sky.org&lt;/a&gt;, &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://in-the-sky.org/location.php?&quot; rel=&quot;nofollow noreferrer&quot;&gt;manually chose the location&lt;/a&gt; New Bedford, MA (close to Cape Cod) because I'm somewhere else (you wouldn't have to do that) &lt;/li&gt;&#xA;&lt;li&gt;selected &lt;a href=&quot;https://in-the-sky.org/skymap.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;planetarium&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and looked in the southeast and found that the brightest star in the sky &lt;a href=&quot;https://en.wikipedia.org/wiki/Sirius&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sirius&lt;/a&gt; is currently only 20 degrees above the horizon.  At that low altitude, and being a star (not a planet) it would definitely be bright and twinkle, and not appear to move like a plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is rising, and moving diagonally to the right and up. Check back in an hour or so to confirm that if you are interested, or check in the next several days to confirm it's in the &lt;em&gt;almost&lt;/em&gt; the same place each night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ov825.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ov825.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yOxNF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yOxNF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-12-18T02:50:11.660" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28775" PostTypeId="2" ParentId="28773" CreationDate="2018-12-18T02:56:56.737" Score="6" Body="&lt;p&gt;This has to do with &lt;a href=&quot;https://www.minorplanetcenter.net/iau/info/OldDesDoc.html&quot; rel=&quot;noreferrer&quot;&gt;how minor planets obtain provisional names&lt;/a&gt; (see also &lt;a href=&quot;https://en.wikipedia.org/wiki/Provisional_designation_in_astronomy#Minor_planets&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;). The year is divided into 24 half months, with a particular letter associated with each one. Each of these half-months has a number of cycles of length 25 depending on how many minor planets are discovered; a minor planet is assigned a letter corresponding to its order in the cycle, and the number of the cycle it is in. This means that the provisional name for a minor planet is, in general,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;year&lt;/code&gt;+&lt;code&gt;half-month&lt;/code&gt;+&lt;code&gt;position in cycle&lt;/code&gt;+&lt;code&gt;cycle number&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2012 VP&lt;sub&gt;113&lt;/sub&gt; was discovered in 2012, in the first half-month of November (corresponding to the letter V) and was the 15th minor planet discovered in the 113th cycle that half-month.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2018-12-18T02:56:56.737" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28776" PostTypeId="2" ParentId="28766" CreationDate="2018-12-18T03:34:49.953" Score="1" Body="&lt;p&gt;As the question is pointing out, because of Earth's thick atmosphere we have quite a bright sky, and so we can not easily see stars during the day. So how can we talk about the position of the Sun with respect to the stars if the stars around the Sun can't be seen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Basic answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By using star charts, and the very steady and well known rotation speed of the Earth (360 degrees in about 23 hours, 56 minutes, 4 seconds) ancient (and modern) astronomers can &lt;em&gt;calculate&lt;/em&gt; where the Sun is with respect to the stars.  Before the invention of telescopes, astronomers had a huge amount of time on their hand with nothing better to do than these kinds of calculations, so they got quite good at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Practical answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a century of measuring the positions of all the planets using both optical telescopes and radar reflection timings and round-trip signal delays from distant spacecraft, orbiters, and landers, and calculating masses and orbits to extremely high precision, the locations of the Sun and all the planets are known to a &lt;a href=&quot;https://space.stackexchange.com/q/23408/12102&quot;&gt;handful of kilometer accuracy&lt;/a&gt;. The &lt;a href=&quot;https://physics.stackexchange.com/q/251700/83380&quot;&gt;Moon's position is known to &lt;em&gt;centimeters!&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this, one can predict with computers where the Sun is with respect to the stars.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From space, in orbit above the Earth or in orbits around the Sun, spacecraft can indeed see the stars and the Sun at the same time! Even telescopes on balloons or aircraft high in the atmosphere can do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a cool GIF I made of images taken by the SOHO spacecraft. You can see stars, comets, the planet Venus (the &lt;em&gt;really bright&lt;/em&gt; object) very close to the Sun, and even the &lt;a href=&quot;https://en.wikipedia.org/wiki/Pleiades&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pleiades&lt;/a&gt; pass over the top of the Sun!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://astronomy.stackexchange.com/a/28263/7982&quot;&gt;this answer&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;These &lt;a href=&quot;http://lasco-www.nrl.navy.mil/index.php?p=content/handbook/hndbk_3&quot; rel=&quot;nofollow noreferrer&quot;&gt;LASCO C3&lt;/a&gt; images from &lt;a href=&quot;http://sohowww.nascom.nasa.gov/home.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;SOHO&lt;/a&gt; were downloaded &lt;a href=&quot;http://sohodata.nascom.nasa.gov/cgi-bin/data_query&quot; rel=&quot;nofollow noreferrer&quot;&gt;sohodata.nascom.nasa.gov/cgi-bin/data_query&lt;/a&gt;. &lt;em&gt;The square frame is about 15.9 degrees wide&lt;/em&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LgrIs.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LgrIs.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-12-18T03:34:49.953" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28778" PostTypeId="2" ParentId="28744" CreationDate="2018-12-18T05:30:11.273" Score="2" Body="&lt;p&gt;Congratulations to @NickBrown for his &lt;a href=&quot;https://astronomy.stackexchange.com/a/28765/7982&quot;&gt;solution&lt;/a&gt;! Based on that equation and some additional references I'll just add a little more.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://apps.dtic.mil/dtic/tr/fulltext/u2/785380.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://apps.dtic.mil/dtic/tr/fulltext/u2/785380.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.researchgate.net/publication/268194552_Large_phase_angle_observations_of_GEO_satellites&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.researchgate.net/publication/268194552_Large_phase_angle_observations_of_GEO_satellites&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Calculating visual magnitude takes three input parameters&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;how good of a reflector the object is&lt;/li&gt;&#xA;&lt;li&gt;the angle between the illumination and the viewing&lt;/li&gt;&#xA;&lt;li&gt;the distances from illuminator and viewer are from the object&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For astronomical objects we use absolute magnitude for item #1, for satellite viewing both absolute magnitude and &lt;em&gt;intrinsic magnitude&lt;/em&gt; are used. Absolute magnitude is the visual magnitude of the object at 1 AU from the Sun and 1 AU from you, viewed full-on (phase angle = 0) which means you are sitting right next to the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intrinsic magnitude is similar, but you are now only 1,000 km from the object with the Sun over your shoulder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, all of the albedo, size and shape information is lumped into the absolute or intrinsic magnitude, leaving only distances and angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle between the direction of illumination and the direction of viewing is called the &lt;em&gt;phase angle&lt;/em&gt;. Think &lt;em&gt;phases of the Moon&lt;/em&gt; for example. If the phase angle of the Moon were 90 degrees, it would be a half-moon. Zero would be full Moon and 180 degrees would be new Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The modulation of the brightness as a function of phase angle was proposed by Vallerie, E.M. III, &lt;em&gt;Investigation of Photometric Data Received from an Artificial Earth Satellite&lt;/em&gt;, AD #419069, Air Force Institute of Technology, Defense Documentation Center, Alexandria, Virginia, 1963, which I found in &lt;a href=&quot;https://amostech.com/TechnicalPapers/2013/POSTER/COGNION.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Observations and Modeling of GEO Satellites at Large Phase Angles&lt;/a&gt; by Rita L. Cognion, also in &lt;a href=&quot;https://www.researchgate.net/publication/268194552_Large_phase_angle_observations_of_GEO_satellites&quot; rel=&quot;nofollow noreferrer&quot;&gt;Researchgate&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dependence is given by the term&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{1}{\pi}(\sin(\phi) + (\pi-\phi) \cos(\phi))$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and looks like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Fjah9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fjah9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the satellite in the question at a distance of 483 kilometers and an intrinsic magnitude of -1.3, the the apparent magnitude seems to be about -2.0 and its dependence on phase angle is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/g0jBN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g0jBN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Not all spacecraft are spherical with diffuse white surfaces nor spherical-cow-shaped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MaJzM.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MaJzM.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For phase angle dependence of some more familliar shapes, see Figure 2 in &lt;a href=&quot;https://apps.dtic.mil/dtic/tr/fulltext/u2/785380.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Visible Magnitude of Typical Satellites in Synchronous Orbits&lt;/a&gt; William E. Krag, MIT, 1974 AD-785 380, which describes the problem nicely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/upWil.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/upWil.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def Mapparent_from_Mintrinsic(Mint, d_km, pa):&#xA;    term_1 = Mint&#xA;    term_2 = +5.0 * np.log10(d_km/1000.)&#xA;    arg    = np.sin(pa) + (pi - pa) * np.cos(pa)&#xA;    term_3 = -2.5 * np.log10(arg)&#xA;    return term_1 + term_2 + term_3&#xA;&#xA;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;&#xA;halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]&#xA;degs, rads = 180/pi, pi/180&#xA;&#xA;Mintrinsic   = -1.3&#xA;d_kilometers = 483.&#xA;&#xA;phase_angles = np.linspace(0, pi, 181)&#xA;&#xA;Mapp = Mapparent_from_Mintrinsic(Mintrinsic, d_kilometers, phase_angles)&#xA;&#xA;# https://astronomy.stackexchange.com/q/28744/7982&#xA;# https://www.researchgate.net/publication/268194552_Large_phase_angle_observations_of_GEO_satellites&#xA;# https://amostech.com/TechnicalPapers/2013/POSTER/COGNION.pdf&#xA;# https://apps.dtic.mil/dtic/tr/fulltext/u2/785380.pdf&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;&#xA;    F = (1./pi)*(np.sin(phase_angles) + (pi-phase_angles)*np.cos(phase_angles))&#xA;&#xA;    plt.suptitle('F = (1/pi)(sin(phi) + (pi-phi)cos(phi))', fontsize=16)&#xA;&#xA;    plt.subplot(2, 1, 1)&#xA;    plt.plot(degs*phase_angles, F)&#xA;    plt.ylabel('F', fontsize=16)&#xA;&#xA;    plt.subplot(2, 1, 2)&#xA;    plt.plot(degs*phase_angles, -2.5*np.log10(F))&#xA;    plt.xlabel('phase angle (degs)', fontsize=16)&#xA;    plt.ylabel('-2.5log10(F)', fontsize=16)&#xA;    plt.ylim(-1, 11)&#xA;&#xA;    plt.show()&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;    plt.plot(degs*phase_angles, Mapp)&#xA;    plt.plot(degs*phase_angles[113], Mapp[113], 'ok')&#xA;    plt.text(90, -5, '{:0.2f} at {:0.1f} deg'.format(Mapp[113], 113), fontsize=16)&#xA;    plt.xlabel('phase angle (degs)', fontsize=16)&#xA;    plt.ylabel('mag', fontsize=16)&#xA;    plt.title('apparent mag of intrinsic mag=-1.3 at 483 km', fontsize=16)&#xA;    plt.ylim(-10, 15)&#xA;    plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-12-18T05:54:06.707" LastActivityDate="2018-12-18T05:54:06.707" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28779" PostTypeId="2" ParentId="28771" CreationDate="2018-12-18T07:12:19.073" Score="1" Body="&lt;p&gt;It is a process of orbit determination. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The apparent position of any solar system body will change from night to night as a result of the combination of its actual motion around the sun and the motion of the Earth. A very distant body such as &quot;Farout&quot; will be moving very slowly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine an orbit, a minimum of three observations is needed. From three observations it should be possible to fit a Keplerian elliptical orbit to the observations and from the orbit one can tell the object's position at any time. The method takes into account the effect of parallax as the position of the Earth is used in the determination of the orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there will be observational errors; the exact location of the body can't be seen, only a white blob on a photograph. This means that there can be errors in the orbit determined from the minimum of three observations. To counter this, further observations over a longer period of time can be taken. The process of orbit determination then becomes a process of finding the best fit orbit to the observations. The more spread out (in time) the observations are, the better is the orbit determined.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-12-18T07:12:19.073" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28780" PostTypeId="2" ParentId="28771" CreationDate="2018-12-18T07:24:32.677" Score="1" Body="&lt;p&gt;It's parallax. An object this far from the Sun is essentially &quot;stationary&quot; compared with how fast the Earth goes around the Sun. Thus, observing it over even a short duration will see it move in a retrograde way against the background stars, that is almost entirely due to parallax rather than orbital motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting a good orbit will need observations over a longer time, since the period will be around 1300 years.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-12-18T07:24:32.677" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28783" PostTypeId="1" CreationDate="2018-12-18T11:21:40.223" Score="4" ViewCount="56" Body="&lt;p&gt;What is the bright star or planet in eastern sky on Dec 18 2018 in Laurens SC. It is very bright and beautiful on a clear night. I have been seeing it for about a month or more. It seems to follow the same path as the moon. This object is sighted at sundown to sunset. Is it a star or a planet? It is very bright and does not twinkle. &lt;/p&gt;&#xA;" OwnerUserId="25211" LastEditorUserId="1891" LastEditDate="2018-12-18T22:29:03.127" LastActivityDate="2018-12-18T22:29:03.127" Title="What is the bright star or planet in eastern sky on dec 18 2018?" Tags="&lt;galaxy&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2018-12-18T22:28:53.860" ContentLicense="CC BY-SA 4.0" />
  <row Id="28784" PostTypeId="1" CreationDate="2018-12-18T12:37:30.143" Score="6" ViewCount="77" Body="&lt;p&gt;Being a daytime software engineer and a sparetime photographer, I was looking for a nice DIY project for the next weeks. I finally decided to build a star tracker with a camera mount, so that I can take deep sky pictures with my DSLR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While planning and layouting the components for this, I had to decide whether I want to manually adjust the rotation axis of the device so that it aligns with the rotation axis of the earth - or go the nerd way and let the system auto-arrange itself by attaching a suitable fixed USB camera, evaluate the video/images and write some software to move the device with stepper motors and screw drives until the correct orientation is reached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I do not want to reinvent the wheel, is there any star recognition framework that you've heard of? Preferrably open source, the language is of marginal importance as long as I can get the relevant data in and out. I'm thinking of something that I can feed with an image file and get back an array with the recognized objects with their x/y locations within the image. I did some research on that but could not find anything promising. What I need is an automatic identification of the brightest stars, so that I can check the current orientation against a reference image and get some vectors for the stepper motors. I hope I got the idea across.&lt;/p&gt;&#xA;" OwnerUserId="25213" LastActivityDate="2018-12-18T18:32:36.847" Title="Is there any star recognition framework/library available?" Tags="&lt;star&gt;&lt;photography&gt;&lt;positional-astronomy&gt;&lt;software&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28785" PostTypeId="2" ParentId="25133" CreationDate="2018-12-18T14:40:51.267" Score="2" Body="&lt;p&gt;I'm pretty far from an expert, but there are a lot of moving parts to this question and it's not easily answered, partially because we don't have a good definition of precisely what a black hole is.   Special relativity tells us that it's a point singularity surrounded by an event horizon from which noting can escape.  Quantum physics tells us that we don't know what it is, but hawking radiation can escape the event horizon and information is conserved on the outside . . . er, I think.   A recent gravity wave discovery (see special relativity), indicates that they lose a pretty fair portion of mass when they merge.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To ship 1 it would appear that ship 2 would be moving slower to the&#xA;  point it would seem like ship 2 stopped moving the closer to the black&#xA;  hole ship 2 gets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the common explanation, but I'd like to add some detail to it.   This only works if the ship falls direct into the black hole with zero tangential velocity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we imagine a ship that's both reflecting visible light and radiating infrared light as it falls into a black hole, it speeds up very fast as it falls into the black hole, until the time dilation takes over and it appears to slow down, at which point it's very close to the black hole.  It's velocity is a significant fraction of the speed of light - somebody else can do the math if they like, but If memory from another question is correct, about 1/3rd the speed of light when the slowing down takes over - to the observer.  To the ship it continues to accelerate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ship continues to red-shift however, because the light that escapes the ship is red shifted both by the &lt;strong&gt;SHIP&lt;/strong&gt;'s increasing velocity combined with the gravitational red shift, so as a visual, you'd lose sight of the ship pretty quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something else happens because of a quirk with relativistic gravity.   An object reflects and emits light in all directions, but as it gets close enough to the black hole, the only light that can escape is almost entirely perpendicular.   Light can only escape the &lt;a href=&quot;https://en.wikipedia.org/wiki/Photon_sphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;photonsphere of a black hole&lt;/a&gt; if it's close to perpendicular, and maybe my wording could use improvement there.  This question has a very nice answer on &lt;a href=&quot;https://physics.stackexchange.com/questions/207816/what-is-the-relation-between-orbital-velocity-and-escape-velocity-in-strongly-re&quot;&gt;orbital velocity around a black hole&lt;/a&gt;.  Tangential light gets considerably bent or trapped entirely.  What this means is you pretty quickly will lose sight of anything that isn't directly perpendicular between the ship, the center of the black hole and where you are viewing from.   To actually continue to see the ship you'd need numerous cameras spread around the black hole and the view would need to be reconstructed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And even if not for the photonsphere, the fact that parts of the ship would be closer to the event horizon than other parts, and that light would travel from the ship to your eyes at slightly different directions, you'd likely see some distortion, similar to the high velocity simulations you might have seen on relativity tv shows, or &lt;a href=&quot;https://www.spacetimetravel.org/ueberblick/ueberblick1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this website gives some examples&lt;/a&gt;, or when Randy Johnson throws a baseball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And lets not forget, any object would likely break apart or spaghettify due to tidal forces, unless you had a really big black hole where &lt;a href=&quot;https://www.spacetimetravel.org/ueberblick/ueberblick1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;spagetification&lt;/a&gt; wouldn't be an issue and the warping would be less but the photonsphere would still be a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My point is, the story of the ship remaining visible is a nice simplification, in reality it's oversimplified.  It ignores tangential velocity and it's frequently not mentioned that it would both spread out and quickly become so red shifted that it couldn't be recognized.     The particle nature of light would suggest that in not too much time, the ship would be reduced to individual photons flickering and you'd need an enormous number of telescopes (assuming you could build a telescope to capture light that far red shifted), and an enormous amount of time to capture enough photons, to even track the ship at all.   In practical terms it disappears pretty quickly.   You might see it slow down and red shift and stretch for a while, but not very long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if the ship falls into the black hole with tangential velocity, well, that's even worse for keeping a visual track on it, and almost everything in space that falls into something else, has tangential velocity.  The idea that the ship appears to freeze kind of in place and fade is probably not accurate at all but if any experts here want to correct me on that, I invite correction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to point out this complexity before giving a layman's answer to your question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way I can personally imagine what it would look like when two black holes fall into each other is to think of the photons outside the event horizon and what happens to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dxMmz.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dxMmz.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is probably a pretty good estimate of what it looks like from ship 2.    I think images like this are computer generated and not bad.  One thing I disagree with is that these images are always drawn with a kind of ring around the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black circle that you see is actually the photonshpere, not the event horizon and it's not completely black because some light with a tangential velocity can escape the photonsphere, but it's probably pretty dark, unless something very energetic is going on inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, those black holes above are drawn with a kind of light-circle around them.  That's probably not accurate either.   The entire space is warped, so there's no light circle.   We imagine seeing a circle around an object that blocks a light because light can wrap around an object and we think of things as having edges, but that's not accurate.   There would be warping around the photonsphere, extending outwards, but not a circle of light.  I suspect there's a good chance you wouldn't see one black hole pass infront of the other, you'd just see it disappear, like black on black, then reappear as it warped space behind it when it was no longer infront of the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't object to the liberties taken to make videos like that more visible, but if you're asking what it would look like, I felt those corrections were worth pointing out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Onto your questions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;because of the time dilatation is this gif picture below a good&#xA;  rendering of how the merger would appear to ship 1?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(give posted above as well).    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's an OK rendering to give amateurs an idea of what it would look like.  Time dilation is largely irrelevant because the black holes would orbit each other very very fast, unless they were very large black holes.   That's another difficulty with your question.  The answer changes somewhat with 2 supermassive black holes, like is expected to happen when Andromeda and the Milky Way's black holes are expected to merge, and with stellar mass black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stellar mass black holes would orbit each other very quickly and the time dilation would slow it down a little bit, but not enough where it would appear slow.   The interesting stuff happens when they get very close.  &lt;a href=&quot;https://www.scientificamerican.com/article/gravitational-waves-discovered-from-colliding-black-holes1/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Take the collision that was detected by Ligo&lt;/a&gt;, when two roughly 30 solar mass black holes merged.    A 30 solar mass black hole has a radius of about 90 km and if the two objects are orbiting each other close enough where you'd want to park a space ship and watch them collide, figure about 5 or 10 times that distance, say 900 km apart.   Objects travelling at relativistic speeds around a (3.14 x 2 x 900), 5,652 km track, at relativistic speeds, you're looking at several orbits per second, give or take, that would likely speed up as they got closer.   Time dilation doesn't catch up fast enough.   You might get some red shifted slowed down bits and pieces, but like the space ship, the idea of watching the black holes kind of freeze on each other is probably not what it would actually look like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few other things to keep in mind.   As the black holes spiral towards each other, they lose mass, so their event horizons actually shrink, which would be hard to observe until the very last moment where most of the gravitational wave energy is lost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, the black part is actually the photonsphere, not the event horizon.   It could look to the viewer that the black holes are touching when it's just their photonspheres shaking hands, so to speak.   If it's a small black hole and a bigger one, the small one could actually disappear into the larger one's photonsphere before the merger actually happens and the enormous pulse of gravitational wave energy is released.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the video probably isn't a very accurate representation of what it looks like, even allowing for the slow-down, but it's not a bad one either.   I don't think it's meant to be accurate as much as it's meant to try to show what happens. It's like when they give Jupiter all these colors that it doesn't really have.  It helps people like us &quot;see&quot; Jupiter, but an actual picture would be less colorful.   It's OK to sacrifice some accuracy in order to make something more visual.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I imagine that 2 naked black hole(an infinity dense point with out a&#xA;  surrounding black hole) merging would look like a blur from the speed&#xA;  of them,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;OK, here's where I'm in over my head, cause this is hard.   I imagine that you're doing a thought experiment, not actually proposing &quot;naked black holes&quot;, that is, if you could see the two singularities spiral towards each other, what would it look like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First thing is, you can't see a singularity.   Imagine you want to see one, so you sail your ship into a black hole.  Sure it's a suicide mission but you will be the first person to see a &quot;naked singularity&quot; - WRONG.   No light can travel from the theoretical point singularity outwards, so as you're inside the event horizon, not much changes.  You see stuff that fell in with you, with different variations of tangential velocity.  You see things stretched and warped by the curvature of space but you can't see the singularity because it doesn't send any photons in your direction.   You could maybe observe some hawking radiation but even that is iffy, because unlike the border of the event horizon where one particle can escape and one can leave, inside, both particles would be drawn to the singularity and probably annihilate each other like good virtual particles are supposed to do.   You'd see space, star light behind you, and stuff around you but you can never see the singularities (assuming that's what happens anyway, we don't have a good answer for what happens inside a black hole).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While you couldn't see it, you could model it.  Two theoretical point singularities would, in a decaying orbit, spiral into each other.   The gravitational waves released inside the black hole would be enormous, but, somehow, all that energy inside the black hole would likely remain trapped inside the black hole.   The gravitational wave energy that's lost probably all happens before the theoretical singularity crosses the other black hole's event horizon, or before the event horizons touch and merge, so to speak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you imagine there are two point singularities in this scenario, I see no reason why they wouldn't just very quickly spiral into each other.  To get a better picture of this, we'd need an accepted model for quantum gravity, which we don't have.  There are also models of black holes where the &lt;a href=&quot;https://www.newscientist.com/article/dn28159-black-holes-may-be-brick-walls-that-bounce-information-back-out/&quot; rel=&quot;nofollow noreferrer&quot;&gt;matter accrues on the outer edge of the event horizon&lt;/a&gt;, where time stops and there is no central singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So the real answer to your question&lt;/strong&gt; is that nobody knows what it would happen, but if we imagine two point singularities when two black holes merge, the laws of special relativity suggest that they would fly into each other very very fast, perhaps a tiny fraction of a second.  Gravitational energy would be emitted as they spiral into each other, as defined by special relativity, but I don't imagine that energy would escape, it would remain inside the event horizon, but on that, I'm just speculating.   There's many reasons why this question probably can't be neatly answered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also a kerr black hole possibility where the point singularity is more of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Ring_singularity&quot; rel=&quot;nofollow noreferrer&quot;&gt;rotating hoop or ring&lt;/a&gt; in order to retain it's angular momentum, but I've always had a hard time seeing how that meshes with losing energy via gravitational waves . . . but I'm not going to try to explain that one, cause it's above my pay grade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apologies if I waxed poetic a little too much, but I think this question almost works as a thought experiment, though it's probably impossible to answer.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2018-12-18T14:40:51.267" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28786" PostTypeId="1" AcceptedAnswerId="28787" CreationDate="2018-12-18T14:46:03.810" Score="9" ViewCount="1001" Body="&lt;p&gt;I was amazed to find out about the discovery of The Goblin &lt;a href=&quot;https://en.wikipedia.org/wiki/2015_TG387&quot; rel=&quot;noreferrer&quot;&gt;2015 TG387&lt;/a&gt; with an orbital period of more than 32 thousand years and a aphelion of more than 1955 AU! That is an astonishing distance. What would the sun look like from aphelion on The Goblin? Could a naked eye observer distinguish it from the rest of the sky?&lt;/p&gt;&#xA;" OwnerUserId="6870" LastActivityDate="2018-12-18T21:01:21.490" Title="What would the sun look like from aphelion of The Goblin's orbit?" Tags="&lt;the-sun&gt;&lt;dwarf-planets&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28787" PostTypeId="2" ParentId="28786" CreationDate="2018-12-18T15:49:25.587" Score="13" Body="&lt;p&gt;Yes, easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1955 AU is a long way, but it is only 0.03 of a light year. The sun would still be less than 1/100 of the distance to the next star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The brightness of the sun would be much less: about 4000000 times less bright, but that is about 16 magnitudes. The sun from Earth is -27 magnitude, so the sun would appear as a magnitude -10 star. It would be much brighter than any other star, and would easily outshine Venus as seen from Earth. In fact it would be 100 times brighter than Venus. It would be as bright as a half-moon (strong enough to cast shadows, but not to read by)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-12-18T21:01:21.490" LastActivityDate="2018-12-18T21:01:21.490" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28788" PostTypeId="2" ParentId="28784" CreationDate="2018-12-18T18:32:36.847" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;As I do not want to reinvent the wheel, is there any star recognition&#xA;  framework that you've heard of? Preferably open source, the language&#xA;  is of marginal importance as long as I can get the relevant data in&#xA;  and out. I'm thinking of something that I can feed with an image file&#xA;  and get back an array with the recognized objects with their x/y&#xA;  locations within the image.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd take a look at the &lt;a href=&quot;http://nova.astrometry.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;astrometry&lt;/a&gt; service. You can upload a picture to their site and receive an astrometric labeled overlay. They do have an &lt;a href=&quot;http://nova.astrometry.net/api_help&quot; rel=&quot;nofollow noreferrer&quot;&gt;API&lt;/a&gt;. I'm not sure if an array output option is builtin, but I have a feeling it's doable within what they're offering. &lt;/p&gt;&#xA;" OwnerUserId="23940" LastActivityDate="2018-12-18T18:32:36.847" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28790" PostTypeId="1" CreationDate="2018-12-18T20:10:39.460" Score="0" ViewCount="71" Body="&lt;p&gt;Are stellar spectra unique enough that they could be used to identify particular stars? If so, has any attempt been made to use them for navigation or equipment alignment?&lt;/p&gt;&#xA;" OwnerUserId="23940" LastActivityDate="2018-12-18T23:23:21.800" Title="Spectra and star identification" Tags="&lt;spectroscopy&gt;&lt;spectra&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2018-12-19T11:31:09.370" ContentLicense="CC BY-SA 4.0" />
  <row Id="28791" PostTypeId="1" AcceptedAnswerId="28793" CreationDate="2018-12-18T20:15:14.567" Score="2" ViewCount="156" Body="&lt;p&gt;When we read that, for instance, M31 is 2.54 ± 0.11 Mly (778 ± 33 kpc) away from us, does this distance estimate take into account the travel time of the light we observe? In other words, what is the &quot;timestamp&quot; on the distance estimate? Does it actually reflect how far away M31 is &lt;em&gt;now&lt;/em&gt;?  &lt;/p&gt;&#xA;" OwnerUserId="23940" LastActivityDate="2019-12-10T21:07:22.937" Title="Distance to objects" Tags="&lt;astrophysics&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28792" PostTypeId="2" ParentId="28683" CreationDate="2018-12-18T20:36:18.183" Score="7" Body="&lt;p&gt;There is a presentation that covers this subject which is a regular at the International VLBI Service workshops. The most recent version is at &lt;a href=&quot;https://www.haystack.mit.edu/workshop/TOW2017/files/Seminars/tow-time2017.pdf&quot; rel=&quot;noreferrer&quot;&gt;High-accuracy Time and&#xA;Frequency in VLBI&lt;/a&gt; and the relevant derivation of needed accuracy is on slide 5. Basically for interferometery, you are trying to match up the phase of the waveforms (position along a sine wave in the simplistic sense) of the two signals from each of the dishes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we use an observing frequency of 10 GHz (10e9 Hz) and want to keep the phase coherent to ~10 degrees (out of the 360 degrees) after 1000 seconds integration then you need 10 / (360 * 10 * 1e9 * 1e3) or 10 / 3.6e16 or a clock stability of about 2.8e-15. Higher frequencies such as the 230 GHz mentioned in the comment will be even more demanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at Allan Deviation/Variance plots, which measure the stability of clocks, such as the plot on slide 4 (and copied below): &lt;a href=&quot;https://i.stack.imgur.com/dxwgb.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dxwgb.png&quot; alt=&quot;Allan Deviation plot&quot;&gt;&lt;/a&gt; this level of performance at 1000 seconds is only reached by hydrogen masers. The typical GPS Disciplined Oscillator will be at about 1e-12@1000seconds, see &lt;a href=&quot;http://leapsecond.com/u/sims/gpsdo17/&quot; rel=&quot;noreferrer&quot;&gt;this recent comparison of 17 GPSDOs&lt;/a&gt;. With careful correction of the sawtooth error caused by the internal oscillator of the GPS (discussed at the end of the presentation) you can improve the typical 50 ns jitter of GPS to ~5 ns. Beyond this atmospheric/ionospheric effects come in and you need to use a dual-frequency GPS to take those out.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2018-12-18T20:36:18.183" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28793" PostTypeId="2" ParentId="28791" CreationDate="2018-12-18T20:55:18.973" Score="3" Body="&lt;p&gt;It represents the distance when light measured was emitted. That is the conventional way that the distance to anything is defined isn't it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do anything else requires a very accurate knowledge of what will happen to an object &lt;em&gt;in its future&lt;/em&gt;. You could in principle do that, if you have very good measurements of an object's velocity and  know all the forces acting upon it) but it doesn't make much difference unless something is travelling at a fair fraction of the speed of light (and M31 isn't).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-12-18T21:27:00.473" LastActivityDate="2018-12-18T21:27:00.473" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="28794" PostTypeId="1" CreationDate="2018-12-18T22:43:36.993" Score="6" ViewCount="143" Body="&lt;p&gt;If space is undergoing accelerated expansion then regions of space are becoming inaccessible to us even at light speed. If gravity is bounded by light speed then inaccessible regions of space no longer contribute gravitational influence on us. If true wouldn't gravity waves... tiny variations in gravity given sensitive enough equipment... be all around us as matter crosses the threshold and becomes inaccessible to us? Does this violate conservation?&lt;/p&gt;&#xA;" OwnerUserId="25219" LastActivityDate="2018-12-21T07:28:17.260" Title="Expansion and Gravity" Tags="&lt;gravitational-waves&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28795" PostTypeId="2" ParentId="28790" CreationDate="2018-12-18T23:23:21.800" Score="2" Body="&lt;p&gt;No, you can't use them to identify your star. Indeed, you have too many stars in the universe. Even in our galaxy you can find two stars which have pretty much the same spectrum. You can classify the star though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another problem with redshift and blueshift. It change the spectrum of the star depending its movement. &lt;/p&gt;&#xA;" OwnerUserId="25214" LastActivityDate="2018-12-18T23:23:21.800" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28796" PostTypeId="2" ParentId="28794" CreationDate="2018-12-19T04:00:15.427" Score="5" Body="&lt;p&gt;Joshua, your question combines a number of complex but related astrophysics issues, so let me try to untangle them as best I can in layperson's terms, with links where you can find more detailed information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you're describing is known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmological_horizon&quot; rel=&quot;nofollow noreferrer&quot;&gt;cosmological horizon&lt;/a&gt;, which defines the limit beyond which a particle's light can never reach us because of the expansion of the Universe. While gravitational waves (&quot;GWs&quot;) are unrelated to electromagnetic waves (what we call &lt;em&gt;light&lt;/em&gt;), it's believed GWs are propagated at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Speed_of_light&quot; rel=&quot;nofollow noreferrer&quot;&gt;speed of light&lt;/a&gt; (the universal limit for how fast information can travel), so the cosmological horizon would apply to gravitational waves as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, &lt;em&gt;gravitational waves&lt;/em&gt; are different from &lt;em&gt;gravity&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it overly simply, the general theory of relativity says that mass distorts both space and time, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravity&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravity&lt;/a&gt; is what we observe of this curvature of spacetime. &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_wave#Properties_and_behaviour&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gravitational waves&lt;/a&gt;, on the other hand, are &quot;ripples&quot; in that spacetime curvature caused by non-symmetric &lt;em&gt;acceleration&lt;/em&gt; of mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;GWs are indeed &quot;all around us&quot;, but they are extraordinarily tiny - Albert Einstein himself doubted they could ever be detected. Our most sensitive equipment, &lt;a href=&quot;https://en.wikipedia.org/wiki/LIGO&quot; rel=&quot;nofollow noreferrer&quot;&gt;LIGO&lt;/a&gt;, looked for the very largest GWs for eight years (2002-2010) and found nothing. It wasn't until a significant enhancement came on line just three years ago (Sept 2015) that the first GW was detected - from the collision of two black holes with a combined mass equivalent to 65 Suns! - and since then we've only detected 11. GWs are &lt;em&gt;really&lt;/em&gt; hard to detect!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since gravity is an &lt;em&gt;intrinsic geometric property&lt;/em&gt; of spacetime, it exists even beyond the cosmological horizon: it's inseparable from the Universe as a whole. All parts of the Universe contribute to the curvature of the Universe, no matter how far away one point is from another. There's no &quot;threshold&quot; for gravity.&lt;/p&gt;&#xA;" OwnerUserId="14094" LastEditorUserId="14094" LastEditDate="2018-12-20T21:00:06.983" LastActivityDate="2018-12-20T21:00:06.983" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="28798" PostTypeId="1" AcceptedAnswerId="28799" CreationDate="2018-12-19T06:50:21.387" Score="2" ViewCount="93" Body="&lt;p&gt;I am a beginner in astronomy. In fact, I just bought my first refractor telescope with 60 mm aperture and 600 focal lengths with 4mm, 12.5mm and 20 mm eyepiece and 3x barlow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could I get info on android app or book where I can see real-time position of planets and other amazing celestial bodies, I am in the northern hemisphere.&lt;/p&gt;&#xA;" OwnerUserId="25212" LastEditorUserId="43" LastEditDate="2018-12-19T09:19:07.773" LastActivityDate="2018-12-19T09:19:07.773" Title="Locate planets and stars for beginners" Tags="&lt;star&gt;&lt;planet&gt;&lt;telescope&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28799" PostTypeId="2" ParentId="28798" CreationDate="2018-12-19T07:58:24.643" Score="3" Body="&lt;p&gt;There are lots of apps you can download on your laptop / smartphone : &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Stellarium : the one i use on my computer. Useful for beginners. You will see planets, some of the best known object, ... &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Skymap for android : used it before, quite nice.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;SkySafari : don't use it but some people told me it was great.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All of them are free. However, you can just search on google, there are plenty of online solutions.&lt;/p&gt;&#xA;" OwnerUserId="25214" LastActivityDate="2018-12-19T07:58:24.643" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28800" PostTypeId="1" AcceptedAnswerId="28801" CreationDate="2018-12-19T09:14:36.130" Score="7" ViewCount="121" Body="&lt;p&gt;On New Year's Eve 2018, the New Horizons spacecraft will fly past Ultima Thule (&lt;a href=&quot;https://en.wikipedia.org/wiki/(486958)_2014_MU69&quot; rel=&quot;noreferrer&quot;&gt;2014 MU69&lt;/a&gt;) at about a third of the distance it flew by Pluto some years ago. So it should be able to get some detailed photographs, but the closing speed is rather high. What scientific observations will be done during the flyby? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a case of a couple of snaps, assuming the camera is pointed correctly? Or is there more that can be done? Any other instruments that will get useful data?&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorDisplayName="user1569" LastEditDate="2018-12-19T11:16:41.090" LastActivityDate="2018-12-19T17:16:39.597" Title="What observations will be done during the New Horizons flyby of Ultima Thule?" Tags="&lt;kuiper-belt&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28801" PostTypeId="2" ParentId="28800" CreationDate="2018-12-19T11:00:14.193" Score="12" Body="&lt;p&gt;Emily Lakdawalla in her &lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/2018/what-to-expect-new-horizons-mu69-ultima-thule.html&quot; rel=&quot;noreferrer&quot;&gt;Planetary Society blog post of 17 December&lt;/a&gt; describes fairly well what we can expect:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A few precious data downlinks in the days before and after the flyby will contain some small photos from New Horizons’ Long Range Reconnaissance Imager (LORRI), which takes grayscale snapshots. Lower-resolution color comes from the Multicolor Visible Imaging Camera (MVIC), a component of the Ralph instrument. The first images that we get back from New Horizons will not be the best images that New Horizons has taken.&lt;br /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Because of the uncertainty in the position of MU69...&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;... the real position of the body may be completely outside the field of view of a photo focused on the predicted position. In order to make sure it doesn’t miss the target, the spacecraft will have to scan its instruments across a region of space in which it is most likely to find the tiny world, taking a lot of data on empty space to make sure to get data on the object.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;She shows what to expect in the days around closest approach, using comet 67P images as an example:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xwNS1.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xwNS1.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The Core phase will last 10 days, from 24 December to 2 January. Nearly all of the science will happen in just a 2-day period around closest approach, with intense imaging and spectroscopy to map MU69, study its composition, figure out its rotation rate and pole, and search for dust, rings, and satellites&lt;/p&gt;&#xA;&lt;p&gt;Several things will make the MU69 flyby significantly more difficult than the Pluto flyby. MU69 is small, dark, and faint. New Horizons won’t see it as more than a dot until 3 days before closest approach. MU69 is darker, and the sunlight dimmer, than the LORRI camera was designed for. Also, because MU69 is so small, New Horizons has to pass much closer to it than it did at Pluto in order to see interesting detail on the surface, resulting in a rapidly changing viewpoint near closest approach.&lt;br /&gt;&#xA;As a result of all of these factors, LORRI’s images may be a little smeared and will be noisier than we saw at Pluto. MVIC is less affected by these challenges; the way that MVIC works, it’s possible for New Horizons to compensate for the low light and high speed by sweeping more slowly across MU69 to build up a stronger signal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Other measurements to be made are:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Toward the end of 29 December, science data gathering will really begin. A dense set of observations will gather a lightcurve on MU69, a measure of how its surface appears to change with rotation, over a period of about 6 hours.&lt;/p&gt;&#xA;&lt;p&gt;30 December will see another 6-hour block of science, a mix of color imaging, deep satellite searches, and lightcurve data. The same happens early on 31 December, after which New Horizons enters the most intense, Inner Core phase of the flyby.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You should read the blog post for all information, because she gives much more detail than can reasonably be quoted here.&lt;br /&gt;&#xA;She also quotes more resources.&lt;/p&gt;&#xA;&lt;p&gt;Here is another simulated image of Ultima Thule, showing how the primary camera aboard NASA’s New Horizons spacecraft will “see” the Kuiper Belt object as it approaches for a 1 January flyby:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YCPMm.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YCPMm.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&lt;sup&gt;Source: Johns Hopkins University Applied Physics Laboratory&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here is the operations time line overview for the approach:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oFZXF.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oFZXF.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sup&gt;(&lt;a href=&quot;https://theskylive.com/pages/new-horizons-2014-mu69/&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;That &lt;em&gt;flyby distance decision&lt;/em&gt; &lt;a href=&quot;https://solarsystem.nasa.gov/news/795/nasas-new-horizons-spacecraft-takes-the-inside-course-to-ultima-thule/&quot; rel=&quot;noreferrer&quot;&gt;has now been made&lt;/a&gt;: the spacecraft will remain on course for a 3500 km flyby.&lt;/p&gt;&#xA;" OwnerDisplayName="user1569" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-12-19T17:16:39.597" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28802" PostTypeId="1" AcceptedAnswerId="28806" CreationDate="2018-12-19T17:11:33.720" Score="1" ViewCount="1081" Body="&lt;p&gt;I’m trying to calculate visible ISS passes and I need to calculate if the ISS is behind the earth by “drawing” a line between the ISS and the sun and seeing if it crosses earth. I have the ECI position of the ISS, but I don’t have the position of the sun. How do I calculate the suns ECI position given a date and time? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I need to use c# for this so I can’t use python libraries &lt;/p&gt;&#xA;" OwnerUserId="25168" LastEditorUserId="25168" LastEditDate="2018-12-19T17:55:02.360" LastActivityDate="2020-07-29T19:00:33.740" Title="Calculating the sun’s position in ECI" Tags="&lt;the-sun&gt;&lt;astrophysics&gt;&lt;mathematics&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28803" PostTypeId="2" ParentId="28802" CreationDate="2018-12-19T17:22:26.467" Score="2" Body="&lt;p&gt;If you know how to use python, you can do something like : &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import astropy as ap&#xA;ap.coordinates.get_sun(time)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It will return  a position in geocentric coordinates for the given time.&lt;/p&gt;&#xA;" OwnerUserId="25214" LastActivityDate="2018-12-19T17:22:26.467" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28804" PostTypeId="2" ParentId="28750" CreationDate="2018-12-19T20:26:59.227" Score="3" Body="&lt;p&gt;Electronic viewing is filling the area between traditional eyepiece viewing and Astro Photography. Electronic images may be viewed live, near live and with digital processing on the fly as the image is viewed or processed after the images have been collected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cameras have lower resolution but higher sensitivity than DSLR's and astro photo cameras. Short exposure time are the norm. Imaging/processing time is generally measured in seconds and minutes instead of hours. Stacking algorithms have been integrated into the viewing software. While tracking is still necessary the software is much more forgiving. People are viewing/imaging DSO's with Dobsonians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use Mallincams in my observatory. They are used with a 10&quot; SCT, a 5&quot; SCT and several Canon DSLR lenses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One has the choice of going big and beautiful or small and good enough is good enough.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at the &lt;a href=&quot;https://groups.io/g/MallinCam&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mallincam io forum&lt;/a&gt;&lt;sup&gt;*&lt;/sup&gt; and judge for yourself. They have a photo section with thousands of images and answers to the questions those interested might have. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;* You'll need to request membership first&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="24928" LastEditorDisplayName="user1569" LastEditDate="2018-12-20T11:11:06.563" LastActivityDate="2018-12-20T11:11:06.563" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28805" PostTypeId="1" AcceptedAnswerId="28809" CreationDate="2018-12-20T03:43:06.460" Score="5" ViewCount="80" Body="&lt;p&gt;Why are M dwarf stars the most active (in terms of starspots and faculae) of all stars? &lt;/p&gt;&#xA;" OwnerUserId="16092" LastActivityDate="2018-12-20T05:55:50.760" Title="Activity of M dwarf stars" Tags="&lt;star&gt;&lt;stellar-evolution&gt;&lt;star-formation&gt;&lt;stellar-astrophysics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28806" PostTypeId="2" ParentId="28802" CreationDate="2018-12-20T03:55:22.577" Score="0" Body="&lt;pre&gt;&lt;code&gt;public static Vector3 GetSunDirection(DateTime time)&#xA;        {&#xA;            time = time.ToUniversalTime();   &#xA;            double JD = 367*time.Year-Math.Floor(7.0*(time.Year+Math.Floor((time.Month+9.0)/12.0))/4.0)+Math.Floor(275.0*time.Month/9.0)+time.Day+1721013.5+time.Hour/24.0+time.Minute/1440.0+time.Second/86400.0;&#xA;            double pi = 3.14159265359;&#xA;            double UT1 = (JD-2451545)/36525;&#xA;            double longMSUN = 280.4606184+36000.77005361*UT1;&#xA;            double mSUN = 357.5277233+35999.05034*UT1;&#xA;            double ecliptic = longMSUN+1.914666471*Math.Sin(mSUN*pi/180)+0.918994643*Math.Sin(2*mSUN*pi/180);&#xA;            double eccen = 23.439291-0.0130042*UT1;&#xA;&#xA;            double x = Math.Cos(ecliptic*pi/180);&#xA;            double y = Math.Cos(eccen*pi/180)*Math.Sin(ecliptic*pi/180);&#xA;            double z = Math.Sin(eccen*pi/180)*Math.Sin(ecliptic*pi/180);&#xA;&#xA;            return new Vector3(x, y, z);&#xA;        }&#xA;&#xA;        public static Vector3 GetSun(DateTime time)&#xA;        {&#xA;            double sunDistance = 0.989 * 1.496E+8;&#xA;            var sunPosition = GetSunDirection(time.ToUniversalTime());&#xA;&#xA;            sunPosition = sunPosition * sunDistance;&#xA;&#xA;            return sunPosition;&#xA;        }&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="25168" LastActivityDate="2018-12-20T03:55:22.577" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28807" PostTypeId="1" CreationDate="2018-12-20T04:24:44.180" Score="1" ViewCount="84" Body="&lt;p&gt;I have a plot in Log(g) vs T_eff coordinates from my stellar evolution model. The parameters of the model are M=0.8 M_solar, [Fe/H]=-1.19, Y=0.245. I want to convert it to the y vs (v-y) coordinates (&lt;a href=&quot;https://en.wikipedia.org/wiki/Str%C3%B6mgren_photometric_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Strömgren photometric system&lt;/a&gt;) in order to compare my model with observations of NGC 6752 globular cluster. How do I make this conversion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://arxiv.org/pdf/astro-ph/0311458v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This paper&lt;/a&gt; provides a &lt;a href=&quot;http://vizier.cfa.harvard.edu/viz-bin/VizieR?-source=J/AJ/127/1227&quot; rel=&quot;nofollow noreferrer&quot;&gt;table&lt;/a&gt; that includes the following columns: Log(g), T_eff, V, (B-V), (b-y), m1, c1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I could use this if I can figure out how to calculate y and (v-y) from these data. Is it possible?&lt;/p&gt;&#xA;" OwnerUserId="21367" LastActivityDate="2018-12-20T04:24:44.180" Title="How to convert Log(g) vs T_eff to y vs (v-y)?" Tags="&lt;astrophysics&gt;&lt;stellar-evolution&gt;&lt;photometry&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28809" PostTypeId="2" ParentId="28805" CreationDate="2018-12-20T05:55:50.760" Score="4" Body="&lt;p&gt;The magnetic activity is driven by a dynamo that relies on a combination of rotation and convection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M dwarfs have large convection zones or are fully convective beyond spectral type M4. They also have much longer spin-down timescales than G- and K-dwarfs; billions of years rather than tens of millions of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that a large fraction of cool M dwarfs are rotating fast and have deep convection zones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the basic reason, but it is still a very busy research area.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-12-20T05:55:50.760" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28810" PostTypeId="1" AcceptedAnswerId="28833" CreationDate="2018-12-20T06:17:35.387" Score="3" ViewCount="909" Body="&lt;p&gt;I am planning on buying my first telescope. I am getting an 80mm refractor and a 114mm reflector for the same price range. Which would be a better option? Further, would a Barlow lens solve the chromatic aberration issues with a f/5 refractor?&lt;/p&gt;&#xA;" OwnerUserId="25234" LastEditorUserId="23940" LastEditDate="2018-12-20T08:31:41.370" LastActivityDate="2018-12-22T12:47:41.063" Title="Advice on first Telescope (refractor vs reflector), Barlow lens" Tags="&lt;telescope&gt;&lt;observational-astronomy&gt;&lt;optics&gt;&lt;newtonian-telescope&gt;&lt;refractor-telescope&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28811" PostTypeId="2" ParentId="16583" CreationDate="2018-12-20T06:48:47.740" Score="3" Body="&lt;p&gt;It's been quite a while with no activity, so I'll post an answer based on my &lt;a href=&quot;https://astronomy.stackexchange.com/questions/16583/when-did-resilvering-large-telescope-mirrors-actually-refer-to-aluminization#comment25644_16595&quot;&gt;comment&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;tl;dr: It started in the 1930's but even the Kepler space telescope still used silver!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.cloudynights.com/topic/277592-when-did-aluminum-replace-silver-on-mirrors/?p=3533758&quot; rel=&quot;nofollow noreferrer&quot;&gt;This post&lt;/a&gt; in cloudynights.com says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Aluminizing mirrors was developed by John Strong, in the early 1930's. One can read about it in his book &lt;em&gt;Procedures in Experimental Physics&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;From Archive.org's &lt;a href=&quot;https://archive.org/details/ProceduresInExperimentalPhysics/page/n181&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://archive.org/details/ProceduresInExperimentalPhysics/page/n181&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GxrJa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GxrJa.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The 40-inch tank, Figure 13, shows the type of equipment used at the California Institude of Technology for larger mirrors. Still larger systems have been used&lt;sup&gt;26&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;sup&gt;26&lt;/sup&gt; &lt;a href=&quot;http://articles.adsabs.harvard.edu/full/1936ApJ....83..401S&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://articles.adsabs.harvard.edu/full/1936ApJ....83..401S&lt;/a&gt;&#xA;&lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1936ApJ....83..401S&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Evaporation Process and its Application to the Aluminizing of Large Telescope Mirrors&lt;/a&gt; also &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0004-637X/728/2/102/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;From that paper:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;OXIDE FILMS ON ALUMINUM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;The aluminum film is automatically protected from tarnishing by an oxide film (presumably corundum, Al203, or bauxite, Al2O3*2H20) which starts to form as soon as the aluminum comes in contact with the air. This oxide becomes thicker with time for about sixty days, when it is very hard and tough, forming a surface not easily scratched when it is being dusted and cleaned.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The formation of the aluminum oxide film on an exposed metallic aluminum surface has been studied by Vernon,31 who weighed the sample to a precision of 1/100 mg and plotted the weight increment due to oxidation as a function of time. The period of time required for the oxide film to attain its natural thickness was seven to fourteen days. After this the thickness of the film remained nearly constant at 100 A.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;This oxide layer is, of course, too thin to give interference effects. It could be much thicker, however, and still not give interference, because of its transparency and the high reflectivity of the under- lying aluminum.32&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The article shows why shows an additional reason why aluminum was more attractive than silver for astronomers, even when aluminumization was a challenging and difficult process, spectral range!&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/X32AJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X32AJ.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Silvering&quot; rel=&quot;nofollow noreferrer&quot;&gt;Silvering&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;An aluminum vacuum-deposition process &lt;strong&gt;invented in 1930 by Caltech physicist and astronomer John Strong, led to most reflecting telescopes shifting to aluminum.&lt;/strong&gt;&lt;sup&gt;9&lt;/sup&gt; Nevertheless, some modern telescopes use silver, such as the Kepler space observatory. The Kepler mirror's silver was deposited using ion assisted evaporation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;sup&gt;9&lt;/sup&gt;&lt;a href=&quot;https://web.archive.org/web/20091011082949/http://www.pfonline.com/articles/030805.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;From that archived source, a post by Jim Destefani: &lt;a href=&quot;https://web.archive.org/web/20091011082949/http://www.pfonline.com/articles/030805.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mirror, Mirror - Keeping the Hale Telescope optically sharp&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Why Clean?&#xA;You might think the Hale mirror, nestled deep inside the telescope workings in the Observatory dome, would be relatively sheltered from dirt and soils. But telescope mirrors get a lot of use, and the Hale Telescope is no exception. According to Palomar Observatory’s W. Scott Kardel, the 200-inch Hale Telescope at the Observatory is used an average of 300 nights per year. Even with weekly cleanings, mirrors periodically need to be given a new coat of reflective aluminum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;“Soils that may accumulate on the mirror include oils from the machinery,” Kardel explains. “We have motors driving the telescope workings that are more than 60 years old. Some of them are suspended above the primary mirror, because we also use smaller mirrors that have to be moved in and out of position. Occasionally, one of the motors might leak. Oil may come from elsewhere, too.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Other possible soils include occasional drops of rain or condensation, which leave water spots. “A water spot with a bit of dust or ash from a fire, for example, can actually form an acid eventually,” Kardel says. Obviously, that’s not acceptable on a surface that needs to be as optically perfect as possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;To combat routine soil accumulations, observatory staff perform a weekly carbon dioxide dusting of the mirror surface. But even this will not remove water spots.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Eventually, the mirror surface degrades to a point where full refurbishment, including cleaning, stripping of the old vacuum metallized aluminum coating, and recoating becomes necessary. “On a historical basis, we’ve done the full stripping, cleaning and re-metallizing about every 18 months to two years,” says Bruce Baker, skilled trades senior at the Observatory. “Of course, some of that is contingent on scheduling of the telescope and how it’s being used.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Aluminum  quickly forms a native oxide when exposed to atmosphere. Aluminum is incredibly exothermically reactive with oxygen. The resulting thin Al2O3 or alumina film is only tens of angstroms thick but the reaction is self-limiting because the resulting layer can not be penetrated by any further oxygen once it gets this thick. It's a bit similar to the formation of native oxides on silicon for similar reasons.&lt;/p&gt;&#xA;&lt;p&gt;Here is an example of what the &amp;quot;resilvering&amp;quot; of the Hale 200 inch mirror looks like. Between &lt;code&gt;01:00&lt;/code&gt; and &lt;code&gt;01:30&lt;/code&gt; you can see what an involved project simply washing the mirror can be. At this point it's probably a better idea to resurface than to try to do this without damaging an existing aluminum surface.&lt;/p&gt;&#xA;&lt;p&gt;&lt;div class=&quot;youtube-embed&quot;&gt;&lt;div&gt;&#xD;&#xA;                &lt;iframe width=&quot;640px&quot; height=&quot;395px&quot; src=&quot;https://www.youtube.com/embed/lkBgQaJVKjc?start=0&quot;&gt;&lt;/iframe&gt;&#xD;&#xA;            &lt;/div&gt;&lt;/div&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The  paper &lt;a href=&quot;http://articles.adsabs.harvard.edu//full/1997PASP..109..303M/0000304.000.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reflectivity Degradation Rates of Aluminum Coatings at the CFHT&lt;/a&gt; Publications of the Astronomical Society of the Pacific, v.109, p.303-306 shows that aluminum mirrors that are washed regularly are just fine without resurfacing.&lt;/p&gt;&#xA;&lt;p&gt;The experimental data in the paper supports the basic science; native aluminum oxide forms quickly, stays very very thin (tens of angstroms) and becomes a permanent barrier to further reaction of the aluminum. &lt;em&gt;Alumina&lt;/em&gt; is un-reactive as well; alumina ceramics are used in a wide variety of chemically reactive and corrosive situation because of their non-reactivity.&lt;/p&gt;&#xA;&lt;p&gt;Resurfacing of aluminized mirrors is necessary from time to time when cleaning with soap and water would cause more damage than it would benefit because dirt is stuck to the mirror's surface. It is not done because the integrity of the aluminum is compromised, but because it's a better choice than cleaning for practical reasons.&lt;/p&gt;&#xA;&lt;p&gt;GIF (&amp;quot;blink comparator&amp;quot;)&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XIvTX.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XIvTX.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;http://global.kyocera.com/fcworld/charact/chemistry/chemiresist.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://global.kyocera.com/fcworld/charact/chemistry/chemiresist.html&lt;/a&gt; - alumina can potentially be remarkably resistant to chemical attack compared to other materials - note logarithmic scale.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RftTf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RftTf.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The individual figures:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/abvnq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/abvnq.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Abn42.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Abn42.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9yDgk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9yDgk.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2021-03-10T00:05:48.567" LastActivityDate="2021-03-10T00:05:48.567" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28812" PostTypeId="2" ParentId="28810" CreationDate="2018-12-20T07:17:03.193" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Further would a Barlow lens solve the chromatic aberration issues with&#xA;  a f/5 refractor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I suppose in theory they could, but they would have to be designed to do so. In practice, they vary widely in quality, and either don't have an effect on chromatic aberration, or actually make it worse. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which would be a better option?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no objective way of answering this question. Based simply on price per square unit of aperture, a reflector is more economical, and therefore better. But if you're looking for a telescope that doesn't require collimation, a refractor might be best. You have to consider the portability of the specific models you're considering as well. Finally, what would you like to see with your telescope? A high-end refractor with correcting optics is, in theory, better than a similar reflector for high-quality imaging of planets. Ultimately, the best telescope is the one you're comfortable using regularly without feeling overwhelmed. &lt;/p&gt;&#xA;" OwnerUserId="23940" LastEditorUserId="23940" LastEditDate="2018-12-20T07:25:35.973" LastActivityDate="2018-12-20T07:25:35.973" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28813" PostTypeId="2" ParentId="24143" CreationDate="2018-12-20T08:52:43.107" Score="2" Body="&lt;p&gt;Even the optical spectrum alone has lots of goodies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_spectroscopy#Temperature_and_size&quot; rel=&quot;nofollow noreferrer&quot;&gt;chemical abundance, size, and other outer properties from emission and absorption lines&lt;/a&gt; you can get the rotational rate from a narrow line's doppler profile, and very careful spectroscopy &lt;a href=&quot;https://arxiv.org/abs/0811.2908&quot; rel=&quot;nofollow noreferrer&quot;&gt;information on vibrational modes&lt;/a&gt; of the star through spectral &lt;a href=&quot;https://en.wikipedia.org/wiki/asteroseismology&quot; rel=&quot;nofollow noreferrer&quot;&gt;asteroseismology&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also most stars are part of multiple-star systems, so spectroscopy from a distance will yield information on &lt;em&gt;each star&lt;/em&gt; and so even if two different stars are barely differentiable, chances are that their companions won't be also! This also gives some potential additional differentiation from the relative velocities, but that depends on how long you observe and if you know the orientation.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-12-20T08:52:43.107" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28814" PostTypeId="1" CreationDate="2018-12-20T09:27:23.067" Score="4" ViewCount="88" Body="&lt;p&gt;The YouTube video &lt;a href=&quot;https://youtu.be/V-v0DwNa194&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tracing the History of Starlight with NASA's Fermi Mission&lt;/a&gt; describes observation by the Fermi gamma ray telescope  of EBL (extragalactic background light) gamma rays from 2,000 &lt;a href=&quot;https://en.wikipedia.org/wiki/Blazar&quot; rel=&quot;nofollow noreferrer&quot;&gt;Blazars&lt;/a&gt;. It seems to suggest that they used an attenuation measurement of the gamma rays to determine... I'm not exactly sure exactly what was determined by the investigation discussed in the video. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The attenuation is due to reverse pair-production, two photons in, and an electron-positron pair out. I am not sure if they measured 511 keV gamma rays from subsequent annihilation of the positrons, or attenuation of the primary gammas, and I am also not sure if the experiment was used to measure the primary gamma ray rates, or the total density of UV-Vis-IR light photons with which the gamma rays annihilated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is there a simple but physically clear way to explain all of this; the experiment the video is supposed to be explaining?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do they differentiate attenuation due to revere pair production from attenuation by say Compton scattering considering dust and gas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MsV8L.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MsV8L.jpg&quot; alt=&quot;screenshot: Tracing the History of Starlight with NASA&amp;#39;s Fermi Mission&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-12-25T01:50:24.887" LastActivityDate="2019-07-24T01:17:31.210" Title="What experiment does this &quot;Tracing the History of Starlight&quot; video try to explain?" Tags="&lt;astrophysics&gt;&lt;dust&gt;&lt;active-galaxy&gt;&lt;gamma-rays&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28815" PostTypeId="2" ParentId="28794" CreationDate="2018-12-20T09:51:54.107" Score="2" Body="&lt;p&gt;In extremely layman terms and I can be wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually such a far away matter has no direct newtonian gravitational influence on us and as such, its departure (crossing the horizon) has not effect on us.&#xA;Moreover, on the event &quot;crossing the horizon&quot; nothing  change and oscillates the texture of spacetime and so we should not expect GW.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that moment on, it will disappear (to us) also for, eventually, GW created by it (in case it undergos event such a merger etc.). Those GW won't be able to reach us anymore. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps my answer is a duplicate of the one above. I just offer this very simple picture - I would not able to do more, by the way. &lt;/p&gt;&#xA;" OwnerUserId="17437" LastEditorUserId="17437" LastEditDate="2018-12-21T07:28:17.260" LastActivityDate="2018-12-21T07:28:17.260" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28816" PostTypeId="1" AcceptedAnswerId="28817" CreationDate="2018-12-20T11:24:12.217" Score="3" ViewCount="72" Body="&lt;p&gt;Michael Stevens from V-Sauce has a video called &lt;a href=&quot;https://youtu.be/MwjsO6aniig?t=776&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Mandelbrot Set&lt;/a&gt; on a different channel. At &lt;code&gt;12:56&lt;/code&gt; he says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There’s a moon of Saturn that has liquid hydrocarbon lakes that literally have waves that are only like a fraction of a centimeter large, but, you could surf that, right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While Michael occasionally does indeed &lt;a href=&quot;https://earthscience.stackexchange.com/q/9978/6031&quot;&gt;say things that appear to be wrong&lt;/a&gt; I'm guessing he got this correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are waves a fraction of a centimeter tall measured on a moon of Saturn? How is the existence of these waves either measured or deduced?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-12-20T18:49:43.483" Title="How do they measure sub-centimeter wave height on a moon of Saturn?" Tags="&lt;natural-satellites&gt;&lt;saturn&gt;&lt;planetary-science&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28817" PostTypeId="2" ParentId="28816" CreationDate="2018-12-20T18:49:43.483" Score="3" Body="&lt;p&gt;According to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Titan_(moon)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page on Titan&lt;/a&gt;, that data was deduced from Cassini.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On December 21, 2008, Cassini passed directly over Ontario Lacus and observed specular reflection in radar. The strength of the reflection saturated the probe's receiver, indicating that the lake level did not vary by more than 3 mm (implying either that surface winds were minimal, or the lake's hydrocarbon fluid is viscous).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That claim cites two sources:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2009GeoRL..3616201W&quot; rel=&quot;nofollow noreferrer&quot;&gt;newscientist.com&lt;/a&gt; (definitely not a primary source)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2009GeoRL..3616201W&quot; rel=&quot;nofollow noreferrer&quot;&gt;Harvard ADSABS&lt;/a&gt; which does appear to be a primary source analysis of Cassini data. However, I don't have time right now to see what peer-review made of this analysis.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cassini RADAR altimetry data collected on the 49th flyby of Titan (2008 December 21) over Ontario Lacus in Titan's south polar region provides strong evidence for an extremely smooth surface, with less than 3 mm rms surface height variation over the 100m-wide Fresnel zone. Histograms of the raw radar echoes imply a mirror-like specular reflection of the transmitted signal. Such an echo is possible only if the surface is extremely flat relative to our 2.2-cm wavelength. The 3 mm upper bound follows from analyzing the strength of the specular return, which declines exponentially with increasing surface height variance. In this experiment, the strength of the echo was larger than expected, severely saturating the receiver. We developed a method to partially correct the echoes for the distortion incurred. While the implied mm-scale smoothness is not proof that the surface is liquid, it is unlikely that a solid surface is so smooth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the abstract, it seems they are deriving this conclusion indirectly, but I'm not educated enough in this area to personally vouch for the methods.&lt;/p&gt;&#xA;" OwnerUserId="14255" LastActivityDate="2018-12-20T18:49:43.483" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28818" PostTypeId="1" CreationDate="2018-12-20T21:29:09.740" Score="2" ViewCount="130" Body="&lt;p&gt;First of all, I have absolutely no background in astronomy and physics. I just watch a lot of &quot;i-hope-scientific&quot; youtube channels talking about physics and astrophysics and I can kind of follow the maths and concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I was thinking, with absolutely no physics to support this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REAL QUESTION STARTS HERE:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say, the big bang happened everywhere, followed by inflation that never stopped. This is wat eternal inflation described (if i am correct)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, as eternal inflation goes on and more and more universe outside of the observable universe is created. Then there would be an unlimited amount of mass outside our observable universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this mass create an gravitational effect on the galaxies in our universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are at the center of our observable universe, so the gravitational effect of all the mass outside our observable universe would never affects us. But from the perspective of another galaxy, it would be affected by a bit of universe that is outside our observable universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you might explain the effect that galaxies that are farther away from us move faster away from us. And that galaxies closer to use move slower away. Then you can explain this by saying that the galaxies farther away are closer to the &quot;unlimited&quot; mass outside our observable universe. And thus are more affected by this. And that galaxies close by are less affected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory with dark energy claims that dark energy causes an outward push from the inside of our universe. Wouldn't this be indistinguishable from a gravitational pull from the outside?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, could this explain the red-shift we see in light from the far end of the observable universe? If I remember correct, gravity can red-shift light if the light moves away from the source of the gravitational pull (as seen near black holes). Light form the far end of the universe has moved trough the gravitational pull away from the &quot;infinite mass&quot; of the outside universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was my idea, and I would like to see where I am thinking wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing where I obviously might be thinking wrong is that this effect would be homogeneous throughout the universe and it should cancel out. But is this also true when you throw some relativity into the mix?&lt;/p&gt;&#xA;" OwnerUserId="25237" LastActivityDate="2021-03-15T17:00:45.133" Title="Can eternal inflation and gravity explain dark energy?" Tags="&lt;gravity&gt;&lt;cosmological-inflation&gt;&lt;dark-energy&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28819" PostTypeId="1" AcceptedAnswerId="28841" CreationDate="2018-12-20T21:43:39.730" Score="8" ViewCount="373" Body="&lt;p&gt;I am learning some machine learning (I have previous background in statistics) on my own, and I am interested in getting some hands-on practice by doing some ML analysis in R and Python. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given how mad I am about astronomy, I would love to get my hands dirty on doing some ML using open data. I am aware that organisations like NASA have lots of open datasets. What I am less clear about is the kind of analysis that needs to be done. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'd like to know if there is any website where I can get a repository of space data (images, signals, whatever) and which asks what analysis is required; kind of like Kaggle where one gets the dataset and has the required question which one can attempt.&lt;/p&gt;&#xA;" OwnerUserId="25238" LastEditorUserId="6" LastEditDate="2019-01-29T14:41:49.973" LastActivityDate="2021-03-08T10:30:41.737" Title="How do I get started in astronomy data analysis?" Tags="&lt;data-analysis&gt;&lt;raw-data&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28820" PostTypeId="1" AcceptedAnswerId="28822" CreationDate="2018-12-21T00:03:18.520" Score="6" ViewCount="116" Body="&lt;p&gt;This NASA article talks about how there's a dearth of hot neptune exoplanets that have been discovered so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.nasa.gov/feature/goddard/2018/in-search-of-missing-worlds-hubble-finds-a-fast-evaporating-exoplanet&quot; rel=&quot;noreferrer&quot;&gt;https://www.nasa.gov/feature/goddard/2018/in-search-of-missing-worlds-hubble-finds-a-fast-evaporating-exoplanet&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/467da.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/467da.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article makes no attempt at postulating why this gap may exist. One possible explanation I have is that the silicate material of a terrestrial planet is difficult to vapourise and the surface gravity of a hot jupiter is too great to drive off hydrogen to any great extent, whereas a Hot Neptune lacks both the solid surface and the gravity to maintain its size.&lt;/p&gt;&#xA;" OwnerUserId="17655" LastActivityDate="2018-12-21T10:22:13.397" Title="Plausible reason for the Hot Neptune gap" Tags="&lt;exoplanet&gt;&lt;hot-jupiter&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28821" PostTypeId="2" ParentId="28383" CreationDate="2018-12-21T05:20:30.280" Score="3" Body="&lt;p&gt;This is a partial answer. It is partial because this is an active area of research, with this very topic as one of the prime questions, second I've only chased down the estimates from one method; microlensing, because this method is not biased toward large (in size and mass) planets in close orbits around small stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The microlensing method is biased toward more massive planets in Jupiter like orbits i.e. cold planets. It is biased toward distant planets about stars much closer toward the galactic center. This is important because stars there have lower concentrations of heavy elements need to make planets. It is equally sensitive to planets round all sorts of stars, and even free floating planets. It can detect down to Earth mass planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link to a technical summary of the method. The part you are most interested in is section 4.1.3 pages 23. &lt;a href=&quot;https://www.mdpi.com/2076-3263/8/10/365/pdf-vor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Microlensing Searches for Exoplanets Yiannis Tsapras&lt;/a&gt;&#xA;This poster paper has a summary plot Fig 3 of planet occurrence rates from various methods &lt;a href=&quot;http://www.mpia.de/homes/ppvi/posters/2K018.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Poster: Planet frequency from microlensing observations Arnaund Caasan&lt;/a&gt;&#xA;There is also a link to an archive of papers on the subject: &lt;a href=&quot;https://exoplanetarchive.ipac.caltech.edu/docs/occurrence_rate_papers.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA archive: Planet Occurrence Rate Papers&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do a quick summary of the main results: (microlensing perspective).&#xA;Estimating the biases, combined with actual detections it is possible to estimate (or put limits on) the numbers of planets and how theoccurence depends on mass, distance from star.&#xA;Snow line == distance from star where water remains frozen. In our solar system it is between Mars and Jupiter. Jupiter's moons have lots of ice. Dimmer stars this is closer in.&#xA;1 AU = Earth-Sun distance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of planets beyond snow line (Jupiter and beyond in our solar system) is 7 times greater than close in hot planets. In the distance range 0.5 to 10 Au of 20% of stars have Jupiters. 50% have Neptunes, 60% have super earths. There are many stars with more than one planet. So there is about one planet for every star in the galaxy. At least!. About 1/6th of stars with planets have a solar analog (massive planets like Jupiter and Saturn with room for small rocky planets close in). &lt;/p&gt;&#xA;" OwnerUserId="10761" LastActivityDate="2018-12-21T05:20:30.280" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28822" PostTypeId="2" ParentId="28820" CreationDate="2018-12-21T07:29:53.073" Score="6" Body="&lt;p&gt;An explanation could be provided by irradiation and evaporation of planetary envelopes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The introduction in the article by &lt;a href=&quot;https://arxiv.org/abs/1812.05119&quot; rel=&quot;noreferrer&quot;&gt;Bourrier et al. 2018&lt;/a&gt;, which is the source of the NASA press release, explains that as exoplanets migrate inwards, they are increasingly challenged to hang on to their envelopes due to irradiation and heating (insolation) by their star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Neptune-sized planets are less capable of retaining their envelopes because of lower densities and lower &quot;surface&quot; gravities. The idea is that many of them get stripped of their outer envelopes and end up as &quot;Cthonian&quot; planets that are of &quot;super-Earth&quot; size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hypothesis is that many hot Jupiters have enough mass and density to maintain a large size for billions of years and hence it is only the hot Neptunes that are depleted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The recent work on GJ 3470b (cited in the NASA press release) supports this hypothesis by finding a very large atmospheric loss rate from a Neptune-sized object at the edge of the &quot;hot Neptune desert&quot;. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-12-21T10:22:13.397" LastActivityDate="2018-12-21T10:22:13.397" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28823" PostTypeId="2" ParentId="28226" CreationDate="2018-12-21T09:52:05.247" Score="3" Body="&lt;p&gt;In other barred galaxies which are vaguely similar to the Milky Way, the contrast in (projected) stellar surface density between the bar and the inter-bar region at the same radius (e.g., along the bar's minor axis, perpendicular to the bar) is typically a factor of at least two; in particularly strong bars it can be as high as six (see, e.g., Figure 5 in &lt;a href=&quot;http://adsabs.harvard.edu/abs/1990ApJ...357...71O&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ohta et al. 1990&lt;/a&gt;). Similar contrasts are seen in N-body models of disk galaxies that form bars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's much harder to figure this out for the Milky Way, because we're not looking down it from above. The best attempt to derive a model of the bar's 3D stellar density from star counts and distance estimates that I know of is &lt;a href=&quot;http://adsabs.harvard.edu/abs/2015MNRAS.450.4050W&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wegg et al. (2015)&lt;/a&gt;. From the face-on projected view of their model (their Figure 14), I would guesstimate the maximum contrast as a factor of 4 or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9QSSC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9QSSC.jpg&quot; alt=&quot;Face-on view of Milky Way from Wegg et al. (2015)&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure 14 of Wegg et al.: projected face-on view of stellar density for the Milky Way (full model in right-hand panel).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 3D density (which is maybe what you're really asking about) in the inner part of the bar is not quite as great as this suggests, because the inner part of the bar is vertically thick, forming a &quot;boxy/peanut-shaped&quot; bulge (this would correspond to the red region in the figure above). So the contrast would be a little less compared to the (less thickened) inter-bar region. But the outer part of the bar is roughly as thin as the rest of the disk, so the projected surface density contrast would mean a similar contrast in 3D stellar density.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2018-12-21T09:52:05.247" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28824" PostTypeId="2" ParentId="28226" CreationDate="2018-12-21T10:26:13.693" Score="3" Body="&lt;p&gt;In the Milky Way, the density in the bar seems to roughly 5 times larger than &quot;next to the bar&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most recent model of the Galactic bar I could find is &lt;a href=&quot;https://arxiv.org/abs/1608.07954&quot; rel=&quot;nofollow noreferrer&quot;&gt;Portail et al. (2017)&lt;/a&gt;, whose model is constructed to match a range of observational surveys &#xA;(&lt;a href=&quot;https://en.wikipedia.org/wiki/VVV_Survey&quot; rel=&quot;nofollow noreferrer&quot;&gt;VVV&lt;/a&gt;,&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/UKIRT_Infrared_Deep_Sky_Survey&quot; rel=&quot;nofollow noreferrer&quot;&gt;UKIDSS&lt;/a&gt;,&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/2MASS&quot; rel=&quot;nofollow noreferrer&quot;&gt;2MASS&lt;/a&gt;,&#xA;&lt;a href=&quot;http://brava.astro.ucla.edu&quot; rel=&quot;nofollow noreferrer&quot;&gt;BRAVA&lt;/a&gt;,&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Optical_Gravitational_Lensing_Experiment&quot; rel=&quot;nofollow noreferrer&quot;&gt;OGLE&lt;/a&gt;, and&#xA;&lt;a href=&quot;https://arxiv.org/abs/1212.1541&quot; rel=&quot;nofollow noreferrer&quot;&gt;ARGOS&lt;/a&gt;). The figure below from this paper shows the density profile of the bar/bulge (left panel), the disk (middle panel), and the combined mass (right panel).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red curve shows the density along the bar (i.e. the major axis), and the blue curve shows it perpendicular hereto (the minor axis). The central bump in the blue curve thus is inside the bar, but after roughly 2 kpc (i.e. 6-7000 lightyears), it flattens out. Here the mass (surface) density is roughly &lt;span class=&quot;math-container&quot;&gt;$10^9\,M_\odot\,\mathrm{kpc}^{-2}$&lt;/span&gt;. Off-bar, however, the blue line shows that the density is only &lt;span class=&quot;math-container&quot;&gt;$2\times10^8\,M_\odot\,\mathrm{kpc}^{-2}$&lt;/span&gt;, i.e. &lt;strong&gt;5 times lower&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/moOVo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/moOVo.png&quot; alt=&quot;bar&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2018-12-21T10:26:13.693" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28825" PostTypeId="1" CreationDate="2018-12-21T12:38:30.850" Score="4" ViewCount="369" Body="&lt;p&gt;How do we calculate the maximum angular separation between sun and moon during:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Solar Eclipse&lt;/li&gt;&#xA;&lt;li&gt;Lunar Eclipse&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="23302" LastEditorUserId="17296" LastEditDate="2018-12-29T20:04:15.470" LastActivityDate="2018-12-29T20:04:15.470" Title="What is the maximum possible separation between sun and moon in the earth sky for an eclipse to occur?" Tags="&lt;solar-eclipse&gt;&lt;lunar-eclipse&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28826" PostTypeId="2" ParentId="28825" CreationDate="2018-12-21T13:08:05.890" Score="5" Body="&lt;p&gt;We'll need the angular diameter of the Sun: 31.6-32.7 arcminutes; the Moon's is between 29.3 and 34.1 arcminutes (found on Wikipedia).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solar eclipse case is 0 degrees; it can only be an eclipse if the Moon is touching the Sun. In that case, their centers are at most (32.7 + 34.1) / 2 = 33.4 arcminutes apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lunar eclipse would occur if the Sun and Moon are diametrically opposite in the sky; the separation of their centers is then 180 degrees. The actual separation is at least (31.6 + 29.3) / 2 = 30.45 arcminutes &lt;em&gt;less&lt;/em&gt; than 180 degrees.&lt;/p&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2018-12-21T13:08:05.890" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28827" PostTypeId="1" AcceptedAnswerId="28828" CreationDate="2018-12-20T15:02:57.963" Score="4" ViewCount="106" Body="&lt;p&gt;When the black hole is only detected due to its gravitational effects on other stars and their light, could it be possible that it is in reality a working Dyson sphere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would still have at least the mass of the star but also blocks all the emitting light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got the idea from watching &lt;a href=&quot;https://www.youtube.com/watch?v=pP44EPBMb8A&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Kurzgesagt video about Dyson spheres&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="Sebus" LastActivityDate="2018-12-21T13:09:40.480" Title="Could a distant black hole (detected due to its mass) be in reality a Dyson sphere?" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28828" PostTypeId="2" ParentId="28827" CreationDate="2018-12-20T15:13:30.370" Score="7" Body="&lt;p&gt;The thing here is that a Dyson Sphere doesn't &quot;block all the emitting light&quot; -- that would violate the conservation of energy.  It does absorb the radiation from whatever body it surrounds and then re-radiates it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, it's likely to radiate something like a relatively normal-looking blackbody spectrum, rather than the plasma X-rays from matter falling into a black hole, or the almost-nothing (thus &quot;black&quot;) from a black hole with no infall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed out in @SteveLinton's &lt;a href=&quot;https://space.stackexchange.com/questions/32995/could-a-distant-black-hole-detected-due-to-its-mass-be-in-reality-a-dyson-sphe#comment101889_32996&quot;&gt;comment&lt;/a&gt; the Dyson sphere would be much larger than the star or a black hole of the same mass in order to not melt, so at that distance the deflection would be far less, since it &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_lens#Explanation_in_terms_of_space%E2%80%93time_curvature&quot; rel=&quot;noreferrer&quot;&gt;scales as inverse radius&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\approx \frac{m}{r}$$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="25911" OwnerDisplayName="Roger" LastActivityDate="2018-12-21T01:10:09.550" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28831" PostTypeId="2" ParentId="28750" CreationDate="2018-12-21T19:34:05.413" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;Strong mount. EQ5. Quality bronze parts.&lt;/li&gt;&#xA;&lt;li&gt;Wide angle telescope. One with high &lt;em&gt;Diameter / Focus Length&lt;/em&gt; ratio. Such telescopes are not popular, because astronomers do not need wide angle normally. And making angle high will significantly grow the costs. Refractors are better, because you will have less trouble with probable &quot;correctors&quot; which could blur your image. You will pay also for good lenses without chromatic aberrations.&lt;/li&gt;&#xA;&lt;li&gt;Good motors and CCD camera. CMOS cameras are not going to work, you need CCD one with possibility to cool it to eliminate dark current.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="24823" LastActivityDate="2018-12-21T19:34:05.413" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28833" PostTypeId="2" ParentId="28810" CreationDate="2018-12-21T21:13:33.327" Score="2" Body="&lt;p&gt;The refractor will be better for wide-field observing of starfields and big nebula and galaxies. For planets and small to medium (also larger with 2&quot; eyepieces or just 1.25&quot; eyepieces with a low magnification and big apparent field of view) sized DSOs (nebulae, galaxies and star clusters) the reflector will have an advantage due to its aperture. The only downside of the reflector is collimation. The reflector will be equal to a 90-95m refractor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fast reflectors (low f number) will suffer from coma (f5 will be ok) and fast refractors suffer from chromatic aberration (at f5 this will be a problem for planets, the moon and brighter stars). Some find the chromatic aberration a problem, and some don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your goal is to observe very large nebulae, open clusters and star fields, then I would advise the refractor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For planets, deep sky objects including planetary nebulae, globular clusters, all the small-medium sized deep sky objects and larger nebulae and galaxies (if the reflector is f6 or lower). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the reflector is f8 or more it will be harder to find objects and to observe the bigger deep-sky objects.&lt;/p&gt;&#xA;" OwnerUserId="25250" LastEditorUserId="14094" LastEditDate="2018-12-22T12:47:41.063" LastActivityDate="2018-12-22T12:47:41.063" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28834" PostTypeId="2" ParentId="28810" CreationDate="2018-12-22T01:21:24.543" Score="0" Body="&lt;p&gt;The optical scheme doesn't matter for beginner. All you need is design which is well used by community, so you don't run into &quot;blurry image&quot; problem with &quot;noname&quot; scope. I recommend you to use the so-called &quot;white lists&quot; of telescopes. Plenty of those exists, many of those are connected to live communities. I do not advocate for any in particular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Barlow lens does not solve anything except of elongating focal length of condensing light ray. You loose view angle but gain &quot;dirty magnification&quot; and loose more brightness because of two extra surfaces for light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do not fall for price. Fall for proven design with emphasis on &quot;plug-n-view&quot;.&lt;/p&gt;&#xA;" OwnerUserId="24823" LastEditorUserId="24823" LastEditDate="2018-12-22T01:31:43.123" LastActivityDate="2018-12-22T01:31:43.123" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28837" PostTypeId="1" CreationDate="2018-12-22T04:24:06.497" Score="8" ViewCount="2446" Body="&lt;p&gt;Someone asked me this question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To see the color of the Great Nebula in Orion, how large the aperture of the telescope should be.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don’t even know whether this is possible. I understand that the colorful patterns of the nebulae are from the cameras. Besides, my own experience shows that when I target the telescope at them, I see nothing but some white blur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is this possible when we watch the deep sky objects with a really huge telescope, under the finest conditions?&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2021-01-15T12:49:10.537" Title="Can we see the color of nebulae?" Tags="&lt;telescope&gt;&lt;nebula&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28838" PostTypeId="2" ParentId="28837" CreationDate="2018-12-22T12:00:53.723" Score="-5" Body="&lt;p&gt;Yes, it is, and that is the reason why telescope has been getting larger and larger. You need a good telescope to be able to see a distant object bigger in an image. When the object is bigger in the image, the light will spread out following the real shape of the object, and instead of seeing a white blur spot you will see detailed colour. Think of a car approaching you from a far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be precise, what you need is a telescope with good &quot;angular resolution,&quot; which is equivalent to the &quot;aperture&quot; size in this context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you know the physical size of an object and its distant, you can work on some math to estimate the minimum aperture that your object will be spatially resolved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_resolution&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Angular_resolution&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-12-22T12:00:53.723" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28839" PostTypeId="1" CreationDate="2018-12-22T15:30:14.623" Score="2" ViewCount="120" Body="&lt;p&gt;While not exactly the most exciting question, I'm wondering: is there any real, semantic difference between a &lt;strong&gt;gas planet&lt;/strong&gt; and &lt;strong&gt;gas giant&lt;/strong&gt;, or are the two terms used interchangeably by most in popular usage (i.e. popular-scientific texts)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;now, a &lt;strong&gt;gas planet&lt;/strong&gt; is obviously the umbrella term for all planets composed mainly of gas, but I often see &quot;gas planet&quot; and &quot;gas giant&quot; used interchangeably without any real change in meaning. I'd class Neptune &amp;amp; Uranus as &quot;&lt;strong&gt;ice giants&lt;/strong&gt;&quot; and Jupiter &amp;amp; Saturnus as &quot;&lt;strong&gt;gas giants&lt;/strong&gt;&quot;, but (popular) dictionaries like &lt;a href=&quot;https://www.thefreedictionary.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.thefreedictionary.com/&lt;/a&gt; ; &lt;a href=&quot;https://en.oxforddictionaries.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.oxforddictionaries.com/&lt;/a&gt; still seem to group them all into &lt;strong&gt;gas giants&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="25259" LastActivityDate="2018-12-24T02:40:08.450" Title="Terminology question: gas giant vs gas planet" Tags="&lt;gas-giants&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28840" PostTypeId="1" CreationDate="2018-12-22T16:15:10.287" Score="3" ViewCount="184" Body="&lt;p&gt;The weather tonight here is great and I decided to shoot some hand-held (too bad I don't have any tripod) night mode pictures shot with P10. I have no experience in stars observation therefore I can't tell what was I looking at. But I'd definitely love to know what are those. Forgive me if the quality of pictures are bad; I've tried my best. At some point I found some of the stars look clear on my phone but fainter over my laptop monitor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Picture 1 - Taken under hand-held night mode: ISO/400 S/0.5s F/2.2&lt;/sup&gt;&lt;a href=&quot;https://i.stack.imgur.com/dn4Zx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dn4Zx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Picture 2 - Taken under hand-held night mode: ISO/800 S/0.5s F/2.2&lt;/sup&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/zd9QJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zd9QJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;strong&gt;Latitude, Longitude &amp;amp; Compass details:&lt;/strong&gt; Right is for P1; left belongs to P2. Keep in mind that it may not 100% accurate.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;strong&gt;Time:&lt;/strong&gt; 22 Dec 2018 / 22:52 P.M. UTC+8&lt;/sup&gt;&#xA;&lt;sup&gt;&lt;strong&gt;Location:&lt;/strong&gt; Ampang, Malaysia.&lt;/sup&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/mlShS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mlShS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Star labelled 4 is the only brightest among the others. It shines and flickers so bright that visible very well with naked eyes. The rest are barely to be seen if not focus well.&lt;/p&gt;&#xA;" OwnerUserId="25125" LastEditorUserId="7472" LastEditDate="2018-12-22T20:58:34.393" LastActivityDate="2018-12-23T12:56:44.670" Title="I've seen some stars tonight. What could I possibly have been looking at with these pictures?" Tags="&lt;amateur-observing&gt;&lt;photography&gt;&lt;deep-sky-observing&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28841" PostTypeId="2" ParentId="28819" CreationDate="2018-12-22T18:39:46.547" Score="3" Body="&lt;p&gt;A possibility might be a course at a university, youtube, or a self-study-course via a book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we've used in our astrostatistics class was &quot;Wall &amp;amp; Jenkins: Practical Statistics for Astronomers&quot; which presents theoretical sections with exercises as follow-ups. No answer sections exist, as you need to do some programing for some of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you can maybe lend this or similar books at a local university library, without the need to spend money.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2018-12-22T18:39:46.547" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28842" PostTypeId="1" CreationDate="2018-12-22T19:51:34.580" Score="3" ViewCount="556" Body="&lt;p&gt;If the moon is moving away at a distance of 1.48&quot; per year (BBC), than wouldn't that mean the gravitational pull by the moon would weaken over time causing the tidal bulge to shrink?  Thus causing sea levels to rise?&lt;/p&gt;&#xA;" OwnerUserId="25261" LastActivityDate="2020-11-18T03:27:42.273" Title="Are rising sea levels caused by the moon?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28843" PostTypeId="2" ParentId="28840" CreationDate="2018-12-22T20:57:49.740" Score="1" Body="&lt;p&gt;You could try using an online planetarium to identify the stars. Right now, it's hard to tell because of the scale, but given &lt;a href=&quot;https://in-the-sky.org/skymap.php?no_cookie=1&amp;amp;latitude=3.14&amp;amp;longitude=101.69&amp;amp;timezone=8.00&amp;amp;year=2018&amp;amp;month=12&amp;amp;day=22&amp;amp;hour=22&amp;amp;min=52&amp;amp;PLlimitmag=2&amp;amp;zoom=200&amp;amp;ra=6.51140&amp;amp;dec=30.90950&quot; rel=&quot;nofollow noreferrer&quot;&gt;this view&lt;/a&gt;, stars 1-3 in the first picture belong to the Orion constellation (1 is Betelgeuze, 3 is Rigel, and 2 is one of the three belt stars) and 4 is Sirius, the brightest star in the night sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second compass must be wrong; going to the East should move the Moon to the left in the picture, not to the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd recommend shooting somewhere else than the Moon; your phone will probably increase exposure time, making more stars visible than right now in the pictures. Phones aren't really suitable for astronomical photos because of their small sensors and lenses, though.&lt;/p&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2018-12-22T20:57:49.740" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28844" PostTypeId="1" CreationDate="2018-12-22T20:58:53.583" Score="1" ViewCount="51" Body="&lt;p&gt;The original post was on physics exchange but I prefer to move it here :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to estimate the distance of closest galaxy neighbor knowing the expression of number of neighbors &lt;span class=&quot;math-container&quot;&gt;$\text{d}N$&lt;/span&gt; into a volume &lt;span class=&quot;math-container&quot;&gt;$\text{d}V$&lt;/span&gt;, the mean density &lt;span class=&quot;math-container&quot;&gt;$n_\text{gal}$&lt;/span&gt; and the correlation function, i.e with this expression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text{d}N=n_{\text{gal}}\,\text{d}V\,(1+\xi(r))$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with &lt;span class=&quot;math-container&quot;&gt;$\xi(r)=\bigg(\dfrac{r}{r_{0}}\bigg)^{-\gamma}\quad\text{with}\quad\gamma\,\sim\,1.77 \quad\text{and}\,r_{0}\,\sim\,5\,\text{Mpc}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must precise that expression of &lt;span class=&quot;math-container&quot;&gt;$\xi(r)$&lt;/span&gt; is valid for &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; between &lt;span class=&quot;math-container&quot;&gt;$0.5\,\text{and}\,10\,\text{Mpc}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I take into my calculation the following value for &lt;span class=&quot;math-container&quot;&gt;$n_{\text{gal}}=0.0420\,\text{h}^{-3}\,\text{Mpc}^{-3}$&lt;/span&gt; : I tried to choose a typical value for this density but it depends on which scale we consider (galaxy cluster, super cluster, very large scales ...), so maybe don't focus on this value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What it interests me is to have elements to answer to the following question :&#xA;&lt;strong&gt;How could I infer the distance of the closest galaxy from us?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE 1:&lt;/strong&gt; Maybe I should take the lower limit of validity for the definition of &lt;span class=&quot;math-container&quot;&gt;$\xi(r)$&lt;/span&gt;, i.e &lt;span class=&quot;math-container&quot;&gt;$r=0.5\,\text{Mpc}$&lt;/span&gt; and so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text{d}N/\text{d}V=n_{\text{gal}}\,(1+\xi(r=0.5))$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text{D}_{\text{closest}}=\bigg(n_{\text{gal}}(1+(0.5/5)^{-1.77})\bigg)^{-1/3}=0.7353\,\text{Mpc}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But by taking &lt;span class=&quot;math-container&quot;&gt;$r=0.5\,\text{Mpc}$&lt;/span&gt;, I am already located myself like the closest neighbor was at &lt;span class=&quot;math-container&quot;&gt;$r=0.5\,\text{Mpc}$&lt;/span&gt;, am I not ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this reasoning correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is welcome &lt;/p&gt;&#xA;" OwnerUserId="16492" LastActivityDate="2019-01-08T20:01:42.153" Title="Deducing distance of closest neighbor galaxy from expression of correlation and mean galaxy density" Tags="&lt;galaxy&gt;&lt;cosmology&gt;&lt;distances&gt;&lt;galaxy-cluster&gt;" AnswerCount="0" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="28845" PostTypeId="2" ParentId="28842" CreationDate="2018-12-22T21:14:31.840" Score="5" Body="&lt;p&gt;You are correct in saying that the moon is receding from the Earth. However, as you can see, this rate is very very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.48 inches = 0.037592 meters&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon is 384400 kilometers away (on average), or 384400000 meters. That means every year, the Moon's distance increases by 0.00000000009779%. That's such a small difference, it is negligible on human timeframes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, although the tides would shrink, this would not mean the sea level would rise. When we refer to rising sea levels, we refer to an average rise in sea levels that have been seen across the world, mostly due to climate change. During high tide, a beach may experience 4 meter higher sea level, but at low tide, it will experience 4 meter lower sea level. If the moon doubled its distance, then gravity would be a quarter its strength (gravity decreases with the square of the distance), so you would have 1 meter above during high tide and 1 meter below during low tide. As you can see, there is still no net change in sea level, it's just that the changes between high and low tide would be smaller. Again, this is negligible on human timeframes.&lt;/p&gt;&#xA;" OwnerUserId="24373" LastActivityDate="2018-12-22T21:14:31.840" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28847" PostTypeId="2" ParentId="28840" CreationDate="2018-12-23T09:53:03.150" Score="2" Body="&lt;p&gt;in-the-sky.org planetarium mode is your friend. It will pick up your location and time zone from your ip address, but you can manually set it to anywhere else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've added screenshots at both &lt;code&gt;20:52&lt;/code&gt; where the Moon is about 26 degrees above the horizon, and at the stated time of&lt;code&gt;22:52&lt;/code&gt; where the Moon is more like 53 degrees.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;20:52:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NwBxd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NwBxd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;22:52:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AZuf4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AZuf4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's a plot of altitude and azimuth of the Moon on that night. Python script as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z8AGp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z8AGp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;from skyfield.api import Topos, load, Loader&#xA;&#xA;halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]&#xA;degs, rads = 180/pi, pi/180&#xA;&#xA;load = Loader('~/Documents/fishing/SkyData')  # avoids multiple copies of large files&#xA;&#xA;&#xA;data    = load('de421.bsp')&#xA;earth   = data['earth']&#xA;moon    = data['moon']&#xA;ts      = load.timescale()&#xA;&#xA;print earth.at(ts.now()).position.km   # for no reason at all&#xA;&#xA;local_hours = np.linspace(18, 24, 13)&#xA;UTC_hours   = [h-8 for h in local_hours]&#xA;times       = ts.utc(2018, 12, 22, UTC_hours, 0, 0)&#xA;&#xA;KL     = earth + Topos(3.1333, 101.6833, elevation_m=66)&#xA;Ampang = earth + Topos(3.1255, 101.7588, elevation_m=66)&#xA;&#xA;alt, az, dist = KL.at(times).observe(moon).apparent().altaz()&#xA;KLalt, KLaz   = alt.degrees, az.degrees&#xA;&#xA;alt, az, dist        = Ampang.at(times).observe(moon).apparent().altaz()&#xA;Ampangalt, Ampangaz  = alt.degrees, az.degrees&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;&#xA;    pts = [6, 8, 10]&#xA;&#xA;    plt.suptitle('Moon  22-Dec-2018 KL &amp;amp; Ampang', fontsize=16)&#xA;&#xA;    plt.subplot(4, 1, 1)&#xA;    plt.plot(local_hours, KLalt)&#xA;    plt.plot(local_hours, Ampangalt)&#xA;    plt.plot(local_hours[pts], Ampangalt[pts], 'ok')&#xA;    plt.ylabel('Altitude (degs)', fontsize=16)&#xA;    plt.xlabel('hours (local)', fontsize=16)&#xA;    plt.ylim(0, 90)&#xA;&#xA;    plt.subplot(4, 1, 2)&#xA;    plt.plot(local_hours, KLaz)&#xA;    plt.plot(local_hours[pts], Ampangaz[pts], 'ok')&#xA;    plt.ylabel('Azimuth (degs)', fontsize=16)&#xA;    plt.xlabel('hours (local)', fontsize=16)&#xA;    plt.ylim(0, 90)&#xA;&#xA;    plt.subplot(2, 1, 2)&#xA;    r = (1. - Ampangalt/90.)&#xA;    r[r&amp;gt;1] = np.nan&#xA;    theta = rads*Ampangaz&#xA;    x, y = r*np.sin(theta), r*np.cos(theta)&#xA;    plt.plot(x, y)&#xA;    plt.plot(x[pts], y[pts], 'ok')&#xA;    th = np.linspace(0, twopi, 361)&#xA;    xc, yc = np.sin(th), np.cos(th)&#xA;    plt.plot(xc, yc, '-k', linewidth=2)&#xA;    d = 0.05&#xA;    plt.plot([-d, d], [ 0, 0], '-k')&#xA;    plt.plot([ 0, 0], [-d, d], '-k')&#xA;    plt.xlim(-1.1, 1.1)&#xA;    plt.ylim(-1.1, 1.1)&#xA;&#xA;    plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2018-12-23T12:56:44.670" LastActivityDate="2018-12-23T12:56:44.670" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28848" PostTypeId="1" CreationDate="2018-12-23T14:28:55.187" Score="3" ViewCount="288" Body="&lt;p&gt;I wish I could attach a good quality photograph of this red color moon I'm looking at right now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I searched on internet but there was no lunar eclipse scheduled to occur on 23rd December 2018&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in  HYDERABAD, INDIAN SUBCONTINENT around 7:20pm&#xA;Please tell me why the moon is red tonight:) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/k5sFb.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k5sFb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tmMuH.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tmMuH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25263" LastEditorUserId="907" LastEditDate="2018-12-23T20:19:07.580" LastActivityDate="2018-12-26T04:41:09.797" Title="Why is this moon red?" Tags="&lt;the-moon&gt;" AnswerCount="5" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28849" PostTypeId="2" ParentId="28848" CreationDate="2018-12-23T14:47:01.763" Score="5" Body="&lt;p&gt;First of all there could not be an eclipse on December 23rd because that is one day later than the full moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can be a couple of reasons for the moon looking red or orange:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It is near the horizon, and being viewed through the atmosphere over a large city.  Air pollution (for example from cars, trucks, fires etc) push small sooty particles in to the air.  These particles scatter blue light more than red, so there is less blue light coming towards you directly from the moon.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Sometimes sand from deserts can get blown up into the atmosphere, and it can get carried a long way.  The sand has a similar effect to the soot.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2018-12-23T14:47:01.763" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28850" PostTypeId="1" AcceptedAnswerId="28861" CreationDate="2018-12-23T21:13:06.740" Score="1" ViewCount="122" Body="&lt;p&gt;Every time astronomers look &quot;13 billion&quot; years into the past, not only is there more space than there normally would be at that instant which increases from observation to observation, but there was also expanding space for eons prior to that which must offset photons over those billions of years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't seen astronomers ever actually state they've reconciled that difference whenever they talk about their observations, so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: Do they actually?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: How do they?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3: How does that change the reality of their public statements?&lt;/p&gt;&#xA;" OwnerUserId="25265" LastActivityDate="2018-12-24T23:43:12.790" Title="When astronomers &quot;look into the past&quot; are they accounting for the expansion of space?" Tags="&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28852" PostTypeId="2" ParentId="28819" CreationDate="2018-12-24T00:29:57.193" Score="3" Body="&lt;p&gt;The Python &lt;a href=&quot;http:///http://www.astroml.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;AstroML&lt;/a&gt; module might be interesting to you. It is accompanied by a textbook about machine learning applied to astronomical datasets, but if you don't have the cash to spend on the book, there are some exercises and datasets included in the module itself that should give you a lot to dig into. &lt;/p&gt;&#xA;" OwnerUserId="2182" LastActivityDate="2018-12-24T00:29:57.193" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28853" PostTypeId="2" ParentId="28839" CreationDate="2018-12-24T02:40:08.450" Score="2" Body="&lt;p&gt;&quot;Gas giant&quot; is the most standard term for planets in the Jupiter/Saturn class, and, as you said, &quot;ice giant&quot; (since the 90's) is the term for planets with compositions like Uranus and Neptune. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Gas planet&quot;, on the other hand, is just not a commonly used term. Judging by Google n-grams, it is less than 1/12 as frequent. Also, if you do a Google book search, of the times it is used, it is very often as a part of the phrase &quot;giant gas planet.&quot; I would consider &quot;gas planet&quot; simply to be a description rather than a technical term, at this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of likely materials for Earth-sized non-rocky planets would be more likely to be classified as ices in the outer solar system. So planets that deserve to be called &quot;gas planets but not gas giants&quot; may be quite rare, so it's possible a distinction may not be needed.&lt;/p&gt;&#xA;" OwnerUserId="7752" LastActivityDate="2018-12-24T02:40:08.450" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28855" PostTypeId="2" ParentId="28848" CreationDate="2018-12-24T06:36:29.453" Score="-3" Body="&lt;p&gt;During a total lunar eclipse, Earth completely blocks direct sunlight from reaching the Moon. The only light reflected from the lunar surface has been refracted by Earth's atmosphere. This light appears reddish for the same reason that a sunset or sunrise does: the Rayleigh scattering of bluer light.&lt;/p&gt;&#xA;" OwnerUserId="25268" LastActivityDate="2018-12-24T06:36:29.453" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28857" PostTypeId="2" ParentId="28850" CreationDate="2018-12-24T16:39:04.687" Score="0" Body="&lt;p&gt;Welcome new user,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simple answer is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &quot;yes&quot; scientists are absolutely aware of precisely the difficulty you bring up. Google lots about &quot;comving distance&quot;, for instance, to see this issue addressed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) and YES in caps, you are totally 100% correct that the usual language used in such popular science explanations, is, nonsensical / a total simplification  :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what can we do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note carefully that the phrase&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;space is expanding&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(aka &quot;the expanding universe&quot; &quot;cosmological inflation&quot; etc etc) is basically - an utter, total mess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many scientists suggest we should never again say &quot;space expanding&quot; and instead talk about the ......... &quot;metric expansion&quot;.  (Unfortunately - absolutely nobody, including those who propose it, has a clue what that means.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note for example that in the article referenced here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/a/223697/10319&quot;&gt;https://physics.stackexchange.com/a/223697/10319&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the authors conclude that all the common phrases we hear in discussing this .. are total claptrap.  :O&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So just to repeat,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;your broad point (1) yes, scientists are 1000% aware of this and deal with it all the time (read about &quot;comoving distances&quot;) and your broad point (2) you are spectacularly correct, the language/meaning used is a shambles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Merry Xmas!&lt;/p&gt;&#xA;" OwnerUserId="13071" LastActivityDate="2018-12-24T16:39:04.687" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28858" PostTypeId="1" AcceptedAnswerId="28860" CreationDate="2018-12-24T20:06:56.590" Score="4" ViewCount="299" Body="&lt;p&gt;If Jupiter was orbiting at 1AU, replacing Earth, but everything else in the solar system remained as it is currently, how much would the increased heat from the sun increase Jupiter's radius?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, how much would Jupiter fluff up if it was closer to the sun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would that be close enough to classify it as a warm jupiter, or hot jupiter?  or would it still just be a standard jovian planet?&lt;/p&gt;&#xA;" OwnerUserId="24853" LastActivityDate="2018-12-24T23:02:36.893" Title="Warm jupiter vs hot jupiter, fluffyness" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;astrophysics&gt;&lt;gas-giants&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28859" PostTypeId="2" ParentId="28837" CreationDate="2018-12-24T20:51:43.347" Score="4" Body="&lt;p&gt;Very large (for amateur) scopes in very dark places, with healthy eyes on a good night have credibly reported seeing color in nebula. The Orion is about as good as it gets,&#xA;lots of photons, my eyes are not particularly good, but have been out with others who claim noticing color in the Trapezium in a thirty-ish inch reflector &lt;/p&gt;&#xA;" OwnerUserId="24480" LastActivityDate="2018-12-24T20:51:43.347" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28860" PostTypeId="2" ParentId="28858" CreationDate="2018-12-24T23:02:36.893" Score="8" Body="&lt;p&gt;Radius inflation seems to mainly occur in planets with irradiation in the range 10&lt;sup&gt;5&lt;/sup&gt;–10&lt;sup&gt;6&lt;/sup&gt; W/m&lt;sup&gt;2&lt;/sup&gt;, see &lt;a href=&quot;https://ui.adsabs.harvard.edu/#abs/2018A&amp;amp;A...616A..76S/abstract&quot; rel=&quot;noreferrer&quot;&gt;Sestovic et al. (2018)&lt;/a&gt;. In contrast, at 1 AU from the Sun, &lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html&quot; rel=&quot;noreferrer&quot;&gt;the irradiation is 1361 W/m&lt;sup&gt;2&lt;/sup&gt;&lt;/a&gt;, which is well below the threshold at which the inflated gas giant radii are observed. It seems unlikely to me that there would be a significant increase in the radius of Jupiter if it were moved to a 1 AU orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As regards the hot/warm Jupiter classification, the term &quot;hot Jupiter&quot; seems to mainly be used to describe the population of giant planets with orbital periods below about 10 days or so. Your Jupiter at 1 AU would not fall into this category.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastActivityDate="2018-12-24T23:02:36.893" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28861" PostTypeId="2" ParentId="28850" CreationDate="2018-12-24T23:43:12.790" Score="3" Body="&lt;p&gt;When astronomers say something is seen &quot;13 billion years in the past&quot;, this is (in most cases) a model-dependent statement, but a perfectly valid statement that tells us how long the light from an object has been travelling to us and which &lt;em&gt;does&lt;/em&gt; take account of universal expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What has actually been measured is a redshift, and to convert that into a &quot;look-back&quot; time requires an understanding (and a model) of the expansion history of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the actual scientific research publication, as opposed to a press release, the details of what assumptions have been made will be stated. The current uncertainties in the cosmological parameters are small - they would not for example lead to uncertainties in the look-back time of even a few percent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is always a possibility that our understanding of cosmological expansion is wrong. In particular, the assumption that the cosmological constant is constant might be wrong, in which case there will be systematic shifts in estimated look-back times.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-12-24T23:43:12.790" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28862" PostTypeId="2" ParentId="28848" CreationDate="2018-12-25T01:42:04.687" Score="0" Body="&lt;p&gt;Judging color from snapshots can be a little tricky sometimes but I'm sure in this case the moon was quite reddened by atmospheric effects (normal Rayleigh scattering plus pollution, both natural and artificial). Reddening of the Moon is no different than reddening of the Sun. We may not notice it happening to the Sun until it becomes extreme near the horizon because the same sunlight is also illuminating everything else around us which re-adjusts our internal &quot;white balance&quot; mechanisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon isn't actually white or gray in color, it's a bit orange in reality. Here's a plot of the Moon's spectral albedo; it's average reflectivity as a function of wavelength. Our visual sensitivity is strongly peaked in the middle (yellow/green) so the linear slope has less of an impact than we'd expect, but it's still there!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(most of the rest is borrowed from &lt;a href=&quot;https://space.stackexchange.com/q/21589/12102&quot;&gt;Why doesn't a full/gibbous moon high in the sky ever seem to look orange? Shouldn't it?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4wtWq.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4wtWq.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; &quot;Figure 8: Averaged geometrical moon albedos measured by GOME from July 1995, November 1995, and September 1996.&quot; From ESA's &lt;a href=&quot;https://earth.esa.int/workshops/ers97/papers/dobber/index-2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;GOME moon measurements, including instrument characterisation and moon albedo&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FNXzZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FNXzZ.jpg&quot; alt=&quot;Moon photobombs Earth&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; &quot;Moon photobombs Earth&quot; photo from DSCOVR, widely circulated NASA public domain image, from &lt;a href=&quot;https://astronomy.stackexchange.com/q/16704/7982&quot;&gt;Why does the Moon appear gray when passing between the Sun and the Earth?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As a non-scientific enquiry, I took the photobomb image as-is and used my computer to color analyze the moon. I got this, which does show a definite progression, blue darker than green darker than red:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/biELX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/biELX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1ulZJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1ulZJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I isolate the moon with Python and calculate the average pixel values in the R, G, and B channels, I get [0.33, 0.31, 0,28] which is a pretty good match to the albedo plot if I just look at the plot around 450, 550, and 650nm, (which is what normal people do after spending hours trying and failing to understand the intricacies of human color perception).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MPCKG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MPCKG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2018-12-25T01:42:04.687" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28864" PostTypeId="2" ParentId="28848" CreationDate="2018-12-25T05:04:46.987" Score="1" Body="&lt;p&gt;@Human Interesting question. I am editing my answer to remove the old post that may be misleading for new readers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Human, @uhoh makes good points. Moonrise was 6:34pm on December 23th in Hyderabad. Azimuth 70.5 degrees (i.e. approx ENE), and altitude approx 9.2 degrees above the horizon.  Therefore the moon was likely too far due north from the Krakatoa volcanic plume to be a reddening factor. The reddish moon in the picture that @Human posted is likely due to the effect of the thick atmospheric layer that the moon's light had to travel through at that low angle, over the city of Hyderabad - I don't know much about the condition of air pollution/dust in that area, but @Wayfaring Stranger noted that air pollution was high on Monday 24th Dec. Thank you @uhoh for your helpful comments. &lt;a href=&quot;https://i.stack.imgur.com/k3Lql.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k3Lql.jpg&quot; alt=&quot;Position of moon over Hyderabad on 23 December 2018&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25275" LastEditorUserId="25275" LastEditDate="2018-12-26T04:41:09.797" LastActivityDate="2018-12-26T04:41:09.797" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28865" PostTypeId="1" CreationDate="2018-12-25T11:12:25.333" Score="2" ViewCount="212" Body="&lt;p&gt;I'm coding on a space simulator and I want to draw a predictable orbit for a satellite. I have read about orbits on &lt;a href=&quot;http://www.braeunig.us/space/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.braeunig.us/space/&lt;/a&gt; and under the section &quot;&lt;a href=&quot;http://www.braeunig.us/space/orbmech.htm#launch&quot; rel=&quot;nofollow noreferrer&quot;&gt;Launch of a Space Vehicle&lt;/a&gt;&quot; they describe how I can obtain the orbital elements like eccentricity, true anomaly etc. given that you have the position vector, velocity vector and the zenith angle between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two questions that I need to understand:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the range of the zenith angle (0-180 degrees or 0-360 degrees etc.) ?&lt;/li&gt;&#xA;&lt;li&gt;Can the angle tell which &quot;half side&quot; of the orbit the satellite is? I.e. if it's onwards to periapsis or apoapsis?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="25276" LastEditorUserId="7982" LastEditDate="2018-12-25T15:24:22.293" LastActivityDate="2020-01-19T20:01:25.163" Title="What is the range of the zenith angle between velocity and position vectors?" Tags="&lt;orbital-mechanics&gt;&lt;orbital-elements&gt;&lt;artificial-satellite&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28866" PostTypeId="2" ParentId="28865" CreationDate="2018-12-25T15:54:52.707" Score="2" Body="&lt;p&gt;Partial answer:&lt;/p&gt;&#xA;&lt;p&gt;What Braeunig calls gamma or zenith angle is actually beta!&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.braeunig.us/space/orbmech.htm#launch&quot; rel=&quot;nofollow noreferrer&quot;&gt;...and γ is the angle between the position and the velocity vectors, called the zenith angle (see Figure 4.7)&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G1vLw.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G1vLw.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;According to @TomSpilker's &lt;a href=&quot;https://space.stackexchange.com/a/29437/12102&quot;&gt;answer&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;This is a problem that has plagued groups of people very knowledgeable about orbital dynamics but who learned using different textbooks: &lt;em&gt;there are two different definitions of &amp;quot;flight path angle&amp;quot;!!&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;In addition to &lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt;, the angle between the &lt;em&gt;tangential direction&lt;/em&gt; and the velocity vector, there is &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt;, the angle between the &lt;em&gt;radial direction&lt;/em&gt; and the velocity vector. People often say &amp;quot;flight path angle&amp;quot; without saying &lt;em&gt;which definition they're using&lt;/em&gt;. Confusing! (I just noticed that the diagram in Julio's answer also shows &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt;)&lt;/p&gt;&#xA;&lt;p&gt;If you work with &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; instead of &lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; is given by&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\arccos\left(\frac{\mathbf{r \centerdot v}}{|\mathbf{r}| \ |\mathbf{v}|} \right) \tag{1} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;So Braeunig's gamma is NASA's beta. It's positive from peri to apo, and negative from apo to peri. Because I'm not smart enough to do it with math, I'll do it with Python.&lt;/p&gt;&#xA;&lt;p&gt;I'm doing this in dimensionless units, so the period is 2π and the semi-major axis is unity.  You can see beta's behavior over one orbit, positive moving towards apoapsis, then negative moving towards periapsis.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/73K84.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/73K84.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eruMw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eruMw.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;def deriv (X, t):&#xA;    x, v = X.reshape(2, -1)&#xA;    acc = -x * ((x**2).sum())**-1.5&#xA;    return np.hstack((v, acc))&#xA;&#xA;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;from scipy.integrate import odeint as ODEint&#xA;&#xA;halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]&#xA;degs, rads = 180/pi, pi/180&#xA;&#xA;time = np.linspace(0, twopi, 361)&#xA;&#xA;r0 = 1.8&#xA;v0 = np.sqrt(2./r0 - 1)  # vis-viva&#xA;&#xA;X0 = np.hstack([r0, 0, 0, v0])&#xA;&#xA;answer, info = ODEint(deriv, X0, time, full_output=True)&#xA;&#xA;x,    v      = answer.T.reshape(2, 2, -1)&#xA;rn = x/np.sqrt((x**2).sum(axis=0))&#xA;vn = v/np.sqrt((v**2).sum(axis=0))&#xA;&#xA;x, y, vx, vy = answer.T&#xA;&#xA;beta = np.arccos((rn*vn).sum(axis=0))&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;    plt.plot(x, y)&#xA;    plt.plot([0], [0], 'ok', markersize=8)&#xA;    x0, y0 = 0.5*(x.min() + x.max()), 0.5*(y.min() + y.max())&#xA;    hwx, hwy = (np.abs(x-x0)).max(), (np.abs(y-y0)).max()&#xA;    hw = 1.05*max(hwx, hwy)&#xA;    plt.xlim(x0-hw, x0+hw)&#xA;    plt.ylim(y0-hw, y0+hw)&#xA;    plt.show()&#xA;    &#xA;if True:&#xA;    plt.figure()&#xA;    plt.subplot(3, 1, 1)&#xA;    plt.plot(time, x)&#xA;    plt.plot(time, y)&#xA;    plt.plot(time, np.zeros_like(time), '-k')&#xA;    plt.title('x and y vs time')&#xA;    plt.subplot(3, 1, 2)&#xA;    plt.plot(time, vx)&#xA;    plt.plot(time, vy)&#xA;    plt.plot(time, np.zeros_like(time), '-k')&#xA;    plt.title('vx and vy vs time')&#xA;    plt.subplot(3, 1, 3)&#xA;    plt.plot(time, beta)&#xA;    plt.plot(time, halfpi*np.ones_like(time), '-k')&#xA;    plt.title('beta vs time')&#xA;    plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2018-12-25T16:01:57.370" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28867" PostTypeId="2" ParentId="28848" CreationDate="2018-12-25T18:02:30.147" Score="0" Body="&lt;p&gt;Particulate air pollution was &lt;a href=&quot;https://air.plumelabs.com/en/week/hyderabad&quot; rel=&quot;nofollow noreferrer&quot;&gt;quite high in Hyderabad&lt;/a&gt; on Monday.&#xA;181 particulate Monday vs the usual 98. The link doesn't seem to say what the units are, but it does say it's all particulate, and that it is &quot;excessive pollution&quot;. Note that the contents of this link probably change on a daily or hourly basis.&#xA;Is this the time of year when the rice farmers burn the stubble in their fields? If so, it's probably smoke making the moon so red.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2018-12-25T18:02:30.147" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28868" PostTypeId="1" AcceptedAnswerId="28872" CreationDate="2018-12-26T00:12:02.293" Score="2" ViewCount="457" Body="&lt;p&gt;I am faced to a problem of interpretation illustrated on figure below :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/guO20.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/guO20.png&quot; alt=&quot;Two-line element set&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From wikipedia : &lt;a href=&quot;https://en.wikipedia.org/wiki/Two-line_element_set&quot; rel=&quot;nofollow noreferrer&quot;&gt;Two-line element definition&lt;/a&gt;, &lt;strong&gt;I can extract the &#xA;following meaning anomaly  : 308.7017 ° (in degrees)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the 2 questions, I am asked to find : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1) at which anomaly is the ISS at the two lines epcoh&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2) at wchich anomaly is the ISS on December 1st 2018.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the answer to question 1) is maybe what I have extracted above &lt;strong&gt;(308.7017 °)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But how to compute the anomaly of ISS on December 1st 2018 ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't undertand very well the difference between these two anomalies requested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone could help me, this would be nice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE 1:&lt;/strong&gt; Sorry, I must precise that I talk about &lt;strong&gt;mean anomaly&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16492" LastEditorUserId="16492" LastEditDate="2018-12-26T14:24:26.717" LastActivityDate="2018-12-26T14:24:26.717" Title="Extract anomalies from two-line element set (TLE) on ISS (International Space Station)" Tags="&lt;orbit&gt;&lt;orbital-mechanics&gt;&lt;iss&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28869" PostTypeId="1" CreationDate="2018-12-26T07:06:32.017" Score="4" ViewCount="158" Body="&lt;p&gt;Can we estimate the order of the total mass of any galaxy from the apparent magnitude and its distance to us? Just like &#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{equation}&#xA;M_{galaxy}=c\times 10^{8}M_{\odot },&#xA;\end{equation}&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; is a constant (can be arbitrary or ignored, the order is important).&#xA;For example mass of the Tucana Dwarf of the Local group isn't known. But its apparent magnitude is 15.7 and average distance is &lt;span class=&quot;math-container&quot;&gt;$8.8\times10^{5}pc$&lt;/span&gt;. How can I estimate its approximate total mass?&lt;/p&gt;&#xA;" OwnerUserId="25284" LastActivityDate="2018-12-27T19:27:03.033" Title="Galactic mass estimation from apparent magnitude and distance" Tags="&lt;galaxy&gt;&lt;distances&gt;&lt;mass&gt;&lt;apparent-magnitude&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28872" PostTypeId="2" ParentId="28868" CreationDate="2018-12-26T09:29:09.077" Score="3" Body="&lt;p&gt;First you should get an idea of what &quot;Anomaly&quot; means.  An anomaly is an angle which can describe the position of the satellite in its orbit. The true anomaly is the actual angle made between the position of the satellite and the perigee. The true anomaly changes quickly when the satellite is close to the Earth and slowly when it is far from the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mean anomaly is much simpler: it just increases linearly with from 0 to 360. It is zero when the satellite is at perigee. So if a satellite takes 100 minutes to orbit and it is perigee at time t=0, then it will be at 90degrees at t=25, 180degrees at t=50.  The mean anomaly is just proportional to time.  &quot;Mean&quot; here means &quot;average&quot;, not &quot;meaning&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbital elements will tell you the mean anomaly at a particular time (given as &quot;days in the year&quot; In your data, at day 320.81254693 of 2018 (the &quot;epoch&quot;) the mean anomaly was 308.7017.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also know that the mean motion is 15.53998777. That is the number of orbits the satellite makes each day. One orbit is 360 degrees, so in 'n' days the mean anomaly will increase by 15.53998777 × 360 × n&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can now find the mean anomaly at any future time. For example 2.5 days after the epoch, at day 323.31254693, the anomaly will be 308.7017 + (15.53998777 × 360 × 2.5). You may need to reduce this angle to the range 0 to 360.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To add accuracy you can also use the fact that the orbital period is slowly changing (It is getting shorter as the ISS falls towards Earth due to drag). But this change is not quick and can be neglected over short periods of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can work out how many days pass between the epoch and Dec 1st and you can work out the mean anomaly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is worded strangely. It is an odd use of &quot;which&quot; and is unclear about the time. It should ask &quot;What is the mean anomaly of the  ISS at the epoch&quot; and &quot;What is the mean anomaly of the ISS at 00:00 on Dec 18th 2018&quot;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-12-26T09:29:09.077" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28874" PostTypeId="2" ParentId="28869" CreationDate="2018-12-26T10:00:36.840" Score="4" Body="&lt;p&gt;First you must estimate the luminosity. That may be difficult unless the magnitude you have is a bolometric magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you assume some &lt;a href=&quot;https://en.wikipedia.org/wiki/Mass-to-light_ratio&quot; rel=&quot;nofollow noreferrer&quot;&gt;mass to light ratio&lt;/a&gt; (actually mass to luminosity ratio, where in solar units the M/L of the Sun is 1). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The M/L ratios of galaxies depend on galaxy type (and age, metallicity, star forming history) and vary from about 2-10. However, this is just the stellar mass. In most cases these ratios would need to be increased by a factor of 10 to give the total mass of a galaxy, which may be dominated by dark matter. The total M/L ratio of dwarf spheroidals like Tucana is likely to be &lt;span class=&quot;math-container&quot;&gt;$\sim 100$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, there is no single value of &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; that can convert from a magnitude (luminosity) to a mass. Further reading: &lt;a href=&quot;https://courses.lumenlearning.com/astronomy/chapter/properties-of-galaxies/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://courses.lumenlearning.com/astronomy/chapter/properties-of-galaxies/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-12-26T10:00:36.840" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28875" PostTypeId="1" AcceptedAnswerId="28877" CreationDate="2018-12-26T11:56:22.020" Score="3" ViewCount="858" Body="&lt;p&gt;I'm not quite sure how I find the orbital phase  (&lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt;) of a transitting exoplanet. For example, I have this RV data &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/2041-8205/713/2/L140/pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;in this article&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/W5uw6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W5uw6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, I don't know how that phase or &quot;cycles&quot; is derived. But that's not the phase &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt; I'm looking for but this (in the same article):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ccHuM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ccHuM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you see, that phase goes from 0 to 1, and I know it has to do with the time of passage from the periastron and the data time (the HJD column), but I'm not able to derive this number in the interval [0,1]. So, if we assume the orbital parameters are known, how can we derive that number?&lt;/p&gt;&#xA;" OwnerUserId="9949" LastActivityDate="2018-12-26T13:31:27.950" Title="Calculate the orbital phase when plotting the Radial Velocity data of a transitting exoplanet" Tags="&lt;exoplanet&gt;&lt;orbital-elements&gt;&lt;radial-velocity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28877" PostTypeId="2" ParentId="28875" CreationDate="2018-12-26T13:14:05.440" Score="3" Body="&lt;p&gt;The phase as plotted on the graph is constrained in the range [0,1) to plot everything in one orbital cycle (this makes it easier to see the shape of the RV curve). The phase can be obtained by taking the fractional part of the total number of cycles. So the first row in the table with the cycles number 28.677 corresponds to phase 0.677.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: As to what the cycles number is, it is a measurement of time since a reference time (usually the transit time), counted in units of the orbital period of the planet. In the case of a transiting planet, the orbital period will be known from the transit interval, so it doesn't need to be derived from the RV curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the first row in Table 1 has time &lt;span class=&quot;math-container&quot;&gt;$t = \mathrm{HJD}\ 2455107.37937$&lt;/span&gt;, while in Table 2 the transit time is given as &lt;span class=&quot;math-container&quot;&gt;$E = \mathrm{HJD}\ 2454967.27571$&lt;/span&gt; and the orbital period as &lt;span class=&quot;math-container&quot;&gt;$P = 4.885525\ \mathrm{days}$&lt;/span&gt;. So the number of cycles can be calculated as &lt;span class=&quot;math-container&quot;&gt;$\mathrm{cycles} = (t-E)/P$&lt;/span&gt; giving 28.677 (to 3 decimal places).&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2018-12-26T13:31:27.950" LastActivityDate="2018-12-26T13:31:27.950" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28879" PostTypeId="2" ParentId="526" CreationDate="2018-12-26T16:35:36.367" Score="2" Body="&lt;p&gt;What's more, the Neupert effect is often stated that the hard X-rays look like the time derivative of the soft X-rays.  The physical significance of that is it suggests that the presence of hard X-rays relates directly to the energization of soft X-ray gas.  It supports the common model that hard X-rays are a byproduct of particle acceleration, and the accelerated particles deposit heat that gradually leaks out as soft X-rays.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-12-26T16:35:36.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28880" PostTypeId="1" AcceptedAnswerId="28883" CreationDate="2018-12-26T16:55:19.017" Score="27" ViewCount="3228" Body="&lt;p&gt;I recently got an app that let me track the ISS. I noticed that during the time it's visible, it disappears &lt;strong&gt;before&lt;/strong&gt; reaching the horizon, and sometimes reappears soon after for a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody explain to me how the ISS and other satellites orbit? Specifically, what makes their visible period such a narrow one, that doesn't span all the way down to the horizon? (which is what my layman brain expected)&lt;/p&gt;&#xA;" OwnerUserId="22136" LastActivityDate="2018-12-26T18:47:21.170" Title="Why do satellites arcs end abruptly when observed from Earth" Tags="&lt;orbit&gt;&lt;artificial-satellite&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28881" PostTypeId="1" CreationDate="2018-12-26T17:38:00.157" Score="1" ViewCount="68" Body="&lt;p&gt;I'm located in NYC. I know it's not a planet, although it is very spherical and looks like one. Perhaps it's a satellite, because it doesn't move much and I don't recall ever seeing a star that appears so round.&lt;/p&gt;&#xA;" OwnerUserId="25288" LastActivityDate="2018-12-26T17:38:00.157" Title="Can someone please tell me what the very bright star is that I see throughout the evening in the northern sky?" Tags="&lt;star&gt;" AnswerCount="0" CommentCount="6" ClosedDate="2018-12-27T10:24:55.253" ContentLicense="CC BY-SA 4.0" />
  <row Id="28882" PostTypeId="1" AcceptedAnswerId="28885" CreationDate="2018-12-26T18:08:33.347" Score="8" ViewCount="1362" Body="&lt;p&gt;The gravitational acceleration, g, inside the Earth generally decreases with decreasing distance to the center:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QzqEi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QzqEi.jpg&quot; alt=&quot;Gravity Acceleration of Earth&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, apparently for Jupiter, the gravitational acceleration only increases with decreasing distance to its center. Why is this?&lt;/p&gt;&#xA;" OwnerUserId="16092" LastEditorUserId="10031" LastEditDate="2018-12-27T10:23:39.103" LastActivityDate="2018-12-27T10:23:39.103" Title="Gravitational acceleration inside a planet" Tags="&lt;gravity&gt;&lt;earth&gt;&lt;jupiter&gt;&lt;newtonian-gravity&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28883" PostTypeId="2" ParentId="28880" CreationDate="2018-12-26T18:47:21.170" Score="41" Body="&lt;p&gt;Because satellites are only visible when they are in sunlight, they are not visible when they go into the Earth's shadow. The app most likely predicts where this occurs and ends the arc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, it does not make sense for an observer to look for a satellite when it is not visible, so there is no need to draw the path when it is in the shadow.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2018-12-26T18:47:21.170" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28884" PostTypeId="2" ParentId="28882" CreationDate="2018-12-26T20:10:40.863" Score="2" Body="&lt;p&gt;No idea whether that's true or not for Jupiter. However, it is theoretically possible if the outer layers of the planet are much less dense than the stuff underneath. In that case the shell theorem applied to the outer layers does not remove much gravity, while the gravity from the inner parts increases fast due to the inverse square law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even for the image you've posted, gravity increases in the lower half of the lower mantle, close to the outer core - and for the same reason. Stuff in the outer layers is &quot;fluffy&quot;, stuff in the core is dense.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2018-12-26T20:10:40.863" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28885" PostTypeId="2" ParentId="28882" CreationDate="2018-12-26T20:18:21.560" Score="10" Body="&lt;p&gt;You can use &lt;a href=&quot;https://en.wikipedia.org/wiki/Gauss%27s_law_for_gravity#Qualitative_statement_of_the_law&quot; rel=&quot;noreferrer&quot;&gt;Gauss's law for gravitation&lt;/a&gt; to work out the gravity as a function of (interior) radius.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \oint \vec{g} \cdot d\vec{A} = -4\pi G \int \rho\ dV\ .$$&lt;/span&gt;&#xA;What this means is that the flux of gravitational field &lt;span class=&quot;math-container&quot;&gt;$\vec{g}$&lt;/span&gt; out of a closed surface is proportional to the mass enclosed within that surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us assume that the density is a function of (interior) radius, such that &lt;span class=&quot;math-container&quot;&gt;$\rho = Ar^{\alpha}$&lt;/span&gt; and that the planet is spherically sysmmetric, so that the left hand side becomes &lt;span class=&quot;math-container&quot;&gt;$4 \pi r^2 g(r)$&lt;/span&gt; and the volume element &lt;span class=&quot;math-container&quot;&gt;$dV = 4\pi r^2\ dr$&lt;/span&gt;. Then&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ 4 \pi r^2 g(r) = -4\pi G \int_{r=0}^{r} Ar^{\alpha} 4\pi r^2\ dr$$&lt;/span&gt;&#xA;as long as &lt;span class=&quot;math-container&quot;&gt;$\alpha \neq -3$&lt;/span&gt;.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ r^2 g(r) = - 4\pi G \frac{Ar^{3+\alpha}}{3+\alpha}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ g(r) = -\frac{4\pi GA }{3+\alpha} r^{1+\alpha}$$&lt;/span&gt;&#xA;What this shows, is that if &lt;span class=&quot;math-container&quot;&gt;$\alpha &amp;lt; -1$&lt;/span&gt;, then gravity will increase with decreasing radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we are in a position to answer your question. In the Earth's deep interior &lt;span class=&quot;math-container&quot;&gt;$\alpha \simeq 0$&lt;/span&gt; and the density does not increase greatly as we head towards the centre. This means that &lt;span class=&quot;math-container&quot;&gt;$g(r) \propto r$&lt;/span&gt; as is seen in the plot you reference. In the outer part of the crust &lt;span class=&quot;math-container&quot;&gt;$\alpha \simeq -1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$g(r)$&lt;/span&gt; is roughly constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a model density profile for Jupiter (solid line).&#xA;&lt;a href=&quot;https://i.stack.imgur.com/M7kRG.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/M7kRG.jpg&quot; alt=&quot;Jupiter density profile&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just looking at this I would say that &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; is just a bit &lt;span class=&quot;math-container&quot;&gt;$&amp;lt;-1$&lt;/span&gt; for most of the interior and so the gravity should be &lt;em&gt;increasing&lt;/em&gt; slowly as &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; decreases, but there is a sharp uptick in density at the boundary to the core which would see &lt;span class=&quot;math-container&quot;&gt;$\alpha \ll -1$&lt;/span&gt; and the gravity becoming proportional to &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; to the power of a negative number (i.e. increasing strongly with decreasing &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt;), but then &lt;em&gt;must&lt;/em&gt; drop to &lt;span class=&quot;math-container&quot;&gt;$g=0$&lt;/span&gt; when &lt;span class=&quot;math-container&quot;&gt;$r=0$&lt;/span&gt; shortly afterwards.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-12-26T20:53:02.047" LastActivityDate="2018-12-26T20:53:02.047" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28886" PostTypeId="2" ParentId="28882" CreationDate="2018-12-26T20:37:53.110" Score="5" Body="&lt;p&gt;The short answer because Jupiter is a gas giant, so it's kind of got a very large atmosphere and atmosphere's aren't very dense.   Also, if you look at your chart, the gravitation inside the Earth increases until you get to the outer core.  This is likely to be much more pronounced on gaseous bodies like gas giants and stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longer answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In layman's terms, if we use &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell2.html&quot; rel=&quot;noreferrer&quot;&gt;Newton's shell theorem&lt;/a&gt;, you can ignore the mass &quot;above&quot; you because the shell of matter with a greater distance from the central point than you has a close to zero gravitational effect on you.   Technically that's a shell of mass around you but I'm going to call it &quot;above&quot; cause it's easier to say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, looking at the gravitational field, you can just take into account the mass below you and the radius and ignore the shell above you.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, for example, you tunnel 10% into a planet, using the law of cubes, 72.9% of the planet's volume is below you, but you're 11.1% closer to the center, using the inverse square law, that's 23.4% greater gravitational pull from the 72.9% of planet that remains below you than you'd get from that 72.9% if you were on the planet's surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the 72.9% of the planet below you weights more than 81% of the planet's mass, then gravity increases.   Lets say it weighs exactly 81%.    81% of the mass x 1.234 greater pull at 10% closer works out exactly the same.   In other words, if the mass above you is light enough, then gravity increases as you tunnel or fall inside a planet.    There's likely a pretty simple logarithmic ration between density ratio and where the gravity stops increasing.   If I can work it out, I'll post it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of Jupiter and starting near it's equator, with Jupiter's rapid rotation, that should be taken into account as well.  In the case of Earth, Earth's rotation is pretty negligible relative to it's gravity and can be ignored unless you want high accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With planetary bodies, density can play a bigger role than mass regarding surface gravity.   Mercury, for example, &lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/planet_table_ratio.html&quot; rel=&quot;noreferrer&quot;&gt;is about 52% the mass of Mars, but it 38% denser&lt;/a&gt;, which enables it to have a slightly higher gravitational pull on it's surface than Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two people have pointed out that they're not sure the gravity really does increase inside Jupiter.   I'm sure it does because the density deep inside Jupiter likely increases significantly.   We can't get a good look at the inside of Jupiter, so precise numbers are impossible, but it seems very probably to me that the gravitation increases much of the way inside Jupiter, only beginning to decrease when the core reaches considerable density.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The majority of Jupiter's external layers is hydrogen and hydrogen, even under very high pressure, is not very dense.   At 700 atm, for example, and Earth temperature, not hot temperature deep inside Jupiter, &lt;a href=&quot;https://energies.airliquide.com/resources-planet-hydrogen/how-hydrogen-stored&quot; rel=&quot;noreferrer&quot;&gt;Hydrogen still has a density of less than 1/10th the density of water&lt;/a&gt;.    The mass of Jupiter's outer layers are almost certainly too low to have as much gravitational effect as the denser inner parts when you take into account falling into the planet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets like Uranus or Neptune, which much less hydrogen and helium - probably not as much, but for gas giants and most stars, gravity very likely increases considerably for a considerable percentage of their radius, for an object falling inside of them.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2018-12-26T20:51:09.283" LastActivityDate="2018-12-26T20:51:09.283" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28887" PostTypeId="2" ParentId="28707" CreationDate="2018-12-26T20:38:01.487" Score="2" Body="&lt;p&gt;Saying that time &quot;froze&quot; is a poor way to put it. To &quot;freeze&quot; means to stop entirely, but time only slows down. Also, saying that you could &quot;live for a thousand years&quot; is a very poor way to put it, because that makes it sound like you could experience a thousand years of life, which is false. What is true is what you say here, assuming you travel at the correct speed:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;during the 10 years of your travel (in your spaceship) and which would be measured as 1000 years in another frame of reference (the earth), you would have aged by &quot;only&quot; 10 years in your spaceship&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you travel faster than that, even more time would pass on earth compared to ten years on the ship, and if you travel slower, less time would pass on earth. A trip like this would go far past Alpha Centauri, since you would be traveling at nearly the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No one knows if we will ever be able to travel at these speeds.&lt;/p&gt;&#xA;" OwnerUserId="7752" LastActivityDate="2018-12-26T20:38:01.487" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28888" PostTypeId="1" AcceptedAnswerId="28892" CreationDate="2018-12-26T22:15:43.117" Score="4" ViewCount="79" Body="&lt;p&gt;Excluding 'fluffy' or 'puffy' gas giants that are 'inflated by heat from their stars, what is the maximum radius of a gas giant planet.  I keep reading things like &quot;...Jupiter is 'about' as big as they get...&quot;, but I can't seem to find a specific number to go with that word &quot;about&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a gas giant was made up of the highest plausible ratio of Hydrogen, and the lowest plausible ratios of other elements (and not significantly heated by the star), just how big could a gas giant get before it started shrinking again under its own weight/gravity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it really just 1 Jupiter radius? 1.0001? 1.1? 1.3?&lt;/p&gt;&#xA;" OwnerUserId="24853" LastActivityDate="2018-12-26T23:16:54.380" Title="Largest non-hot gas giants" Tags="&lt;planet&gt;&lt;gravity&gt;&lt;astrophysics&gt;&lt;exoplanet&gt;&lt;gas-giants&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28889" PostTypeId="1" CreationDate="2018-12-26T22:29:16.320" Score="3" ViewCount="117" Body="&lt;p&gt;Sudarsky's &lt;a href=&quot;https://en.wikipedia.org/wiki/Sudarsky%27s_gas_giant_classification&quot; rel=&quot;nofollow noreferrer&quot;&gt;gas giant classification&lt;/a&gt; predicts the visual appearance of gas giant planets based on their temperatures.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what determines their temperatures in the first place?  Is it just the irradiation from their star, or is there more to it than that?  If there's more to it, what else is involved?&lt;/p&gt;&#xA;" OwnerUserId="24853" LastEditorUserId="2531" LastEditDate="2018-12-27T14:44:07.637" LastActivityDate="2018-12-27T14:44:07.637" Title="Gas Giant temperatures" Tags="&lt;planet&gt;&lt;exoplanet&gt;&lt;gas-giants&gt;&lt;planetary-formation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28890" PostTypeId="2" ParentId="28889" CreationDate="2018-12-26T22:56:13.627" Score="5" Body="&lt;p&gt;The other thing(s) involved are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Contraction and the release of gravitational potential energy. Gas giants are born bigger than they are now. They gradually contract towards a completely degenerate configuration, and as they do so half the released potential energy is radiated away, whilst the other half heats the planet.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Gravitational diffusion and separation also releases gravitational potential energy. Heavier elements tend to settle towards the centre, falling through the majority hydrogen.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Radioactive decay of heavy elements that may form part of the giant planet's core.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the &quot;planet&quot; is more than about 13 times the mass of Jupiter, then it will likely attain a central temperature sufficient to fuse its deuterium. Some would refer to such objects as brown dwarfs at that point.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-12-27T08:33:37.393" LastActivityDate="2018-12-27T08:33:37.393" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28891" PostTypeId="1" AcceptedAnswerId="28894" CreationDate="2018-12-26T23:01:47.813" Score="5" ViewCount="369" Body="&lt;p&gt;What factors determine the atmospheric density on various planets and moons? Some larger bodies have less dense atmospheres, while some smaller ones have a more dense atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="25289" LastActivityDate="2018-12-27T00:16:57.930" Title="what determines the density of atmospheres" Tags="&lt;planet&gt;&lt;planetary-atmosphere&gt;&lt;density&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28892" PostTypeId="2" ParentId="28888" CreationDate="2018-12-26T23:16:54.380" Score="6" Body="&lt;p&gt;According to the models of &lt;a href=&quot;https://www.aanda.org/articles/aa/full_html/2012/11/aa18464-11/aa18464-11.html&quot; rel=&quot;noreferrer&quot;&gt;Mordasini et al. 2012&lt;/a&gt; it is about 13 Earth radii (about 1.2 Jupiter radii) for gas giants older than a billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Younger gas giants that are still contracting significantly can be bigger. How much bigger is wrapped up in uncertainties in the formation mechanism and the core mass and composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Gas giant mass-radius relationships from Mordasini et al. 2012. The colour coding corresponds to the fraction of heavy elements in the planet. The larger planets have smaller fractions of heavy elements.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/twYLQ.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/twYLQ.jpg&quot; alt=&quot;Planet models&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-12-26T23:16:54.380" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28893" PostTypeId="2" ParentId="28738" CreationDate="2018-12-26T23:39:31.140" Score="2" Body="&lt;p&gt;On Earth we know of frozen carbon dioxide as &quot;dry ice.&quot; If shaved, it could exist as &quot;dry ice snow.&quot; On Mars the dry ice snow could precipitate, then become compressed into dry ice. It would probably be a stiffly packed granular solid.&lt;br&gt;&#xA;I had only heard of &quot;water ice&quot; on Mars. Check the article on kottke.org  Link:  &lt;a href=&quot;https://kottke.org/18/12/a-massive-ice-filled-crater-on-mars&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://kottke.org/18/12/a-massive-ice-filled-crater-on-mars&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25289" LastActivityDate="2018-12-26T23:39:31.140" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28894" PostTypeId="2" ParentId="28891" CreationDate="2018-12-27T00:16:57.930" Score="5" Body="&lt;p&gt;Most simply, the density is determined by the number of molecules, the surface area of the planet, the temperature, and the gravity.  But it sounds like what you really care about would be called the &quot;column density&quot;, which depends only on the number of molecules and the surface area, and you are wondering why there is not some direct correlation between the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say that the reason you don't see a clear correlation is that atmospheric gas generally comes through the surface of the planet, so all else being equal, you'd expect all planets to have the same column density.  But all else is not equal, because you have issues of how volcanic is the interior (so how much gas gets liberated), and how much escape into space happens.  The amount of volcanism is higher for bigger planets (they take longer to cool), and bigger planets also have stronger gravity and cling better to the gas, so one might generally expect larger planets to have higher column density (and indeed the Earth and Venus have higher column densities).  But a higher temperature will lead to escape, and a lower temperature will lead to freezing out many of the species, so temperature is a key factor as well.  Magnetic fields can also play a role, by protecting the atmosphere from being stripped by the stellar wind, and it's still not real clear what planets will have strong fields (Earth) and what planets won't (Venus, whose interior is almost the same as the Earth but doesn't spin much).  Finally, you can have even more complicated effects, such as the way Earth's oceans have dissolved and locked up carbon dioxide.  So when there are a lot of potential factors, you can expect a lot of variation.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2018-12-27T00:16:57.930" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28895" PostTypeId="1" CreationDate="2018-12-27T01:51:55.063" Score="6" ViewCount="151" Body="&lt;p&gt;Aren't there any rock or similar firm material on/in the gas giant planets? What happens if a rock asteroid hits one of these planets? Shouldn't the rock accumulate in the center of the planet due to its high density?&lt;/p&gt;&#xA;" OwnerUserId="25291" LastActivityDate="2018-12-27T09:29:51.797" Title="Aren't there any rock or similar firm material on/in the gas giant planets?" Tags="&lt;jupiter&gt;&lt;gas-giants&gt;&lt;saturn&gt;&lt;neptune&gt;&lt;uranus&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28897" PostTypeId="2" ParentId="28895" CreationDate="2018-12-27T09:29:51.797" Score="5" Body="&lt;p&gt;The inside of a giant planet is not like regular gas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, it is hot. You are away of how it gets hot inside the Earth (causing volcanos). It is also hot inside Jupiter, but since Jupiter is bigger, it is hotter. It would be hot enough to vaporise rock under &quot;normal conditions&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure is immense, and things stop behaving as you are used to when the pressure is very high. For example, all the atoms in a gas are pushed so close together that they are touching (like they would be in a liquid). This means that there is no boiling point and no real difference between the gas and liquid phases. At extreme pressures, hydrogen atoms are pushed so close together that their electrons can start to flow from atom to atom, forming a fluid metal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressure that can push atoms so close that they can turn hydrogen to a metal mean that you can't send a probe to this region. You can't swim in a metallic hydrogen sea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, heavier atoms like iron, silicon, carbon, oxygen, will tend to fall towards the centre of the planet, and so there may be a since rocks are made mostly of atoms like iron, silicon, carbon, oxygen, you could describe this as a &quot;rocky&quot; core.  But don't think this is a kind of solid surface that you could stand on. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-12-27T09:29:51.797" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="28898" PostTypeId="1" CreationDate="2018-12-27T13:02:35.657" Score="8" ViewCount="538" Body="&lt;p&gt;For Earth, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_month&quot; rel=&quot;noreferrer&quot;&gt;lunar month&lt;/a&gt; is well known. But Earth has only one moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But on Jupiter, how long does it takes for the same moon configuration to appear in the &quot;sky&quot; of the planet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sake of sanity, you can consider only the four Galileian Mooons.&lt;/p&gt;&#xA;" OwnerUserId="11366" LastEditorUserId="11366" LastEditDate="2018-12-28T11:06:58.347" LastActivityDate="2019-01-02T14:23:00.667" Title="How long is a &quot;lunar month&quot; in Jupiter?" Tags="&lt;jupiter&gt;&lt;natural-satellites&gt;&lt;orbital-mechanics&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28899" PostTypeId="2" ParentId="28707" CreationDate="2018-12-27T13:04:30.813" Score="0" Body="&lt;p&gt;These popular books refer to the moving of people from the present, past or future to any other willingly or intentionally. The world we are currently living in is the present. We move towards the future at an almost constant rate that is 1min/minute(say). If by some reason some people found out how to travel slower or faster than this normal rate, they are able to travel in time. &#xA;Physical laws of physics have no such limitation or rule that this cannot happen, its the engineering and machinery which lacks the ability to do so. Even so, there are some quite interesting stuff which might happen if you went in future or past. These movements through time introduce paradoxes which are hard to wrap our heads around. Like, the very famous of them all--Grandfather Paradox, where you travel back in time and somehow prevent your grandparents from the meeting. This generates a question, who was the person who prevented the meeting in the first place? &#xA;There is a very good explanation thought up by our scientists that, the time you go which you regard as past is actually a parallel universe and in no way affect the reality of the time from which you came from. Its said that we will not be able to travel to the past, but we can travel to the future quickly. &#xA;This can be achieved by living inside a high gravitational field where time slows down making you travel through time slowly with respect to others.&#xA;These ideas are thought upon and modified a bit according to the imagination of the authors to provide a new version to it, making it interesting to read.&lt;/p&gt;&#xA;" OwnerUserId="25297" LastActivityDate="2018-12-27T13:04:30.813" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28900" PostTypeId="1" AcceptedAnswerId="40097" CreationDate="2018-12-27T15:17:01.790" Score="14" ViewCount="343" Body="&lt;p&gt;Does anyone know a good reference for this?&lt;/p&gt;&#xA;&lt;p&gt;I once read how Jupiter's gravitation can be treated as a perturbation. I think they expanded Jupiter's gravitation as an infinite sum of one of the special functions (such as Bessel function, Legendre function, Hermite polynomial, I can't remember).&lt;/p&gt;&#xA;&lt;p&gt;I am not interested in how the motion of a planet at each time looks like, but just the overall effects on parameters such as period, semi-major axis, and eccentricity.&lt;/p&gt;&#xA;" OwnerUserId="24850" LastEditorUserId="15835" LastEditDate="2020-10-13T12:14:10.060" LastActivityDate="2020-11-29T21:57:40.287" Title="How does the gravitation of Jupiter or other big planets affect Kepler's third law?" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;orbital-mechanics&gt;&lt;celestial-mechanics&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28902" PostTypeId="2" ParentId="28837" CreationDate="2018-12-27T18:34:53.680" Score="2" Body="&lt;p&gt;I have been able to see color in the Orion nebula (M42) with my 16 inch dobsonian from a heavily light polluted city. This is only in the center (trapezium area), where I can see a reddish color. Interestingly, it's much more difficult to see this red color from a dark location. Several other people in my astronomy club have also noticed this, and one explanation that I got was that the cone cells are still active in a light polluted city, so I'm able to notice the red color ( I don't know whether it's true). From a good dark location, I've seen color on many planetary nebulae such as the Ghost of Jupiter. You usually see these are greenish in color, if you have a suitably large scope. With my 16 inch, I've seen color in a small number of bright objects (almost always limited to greens), but most objects don't show color. Having observed in instruments ranging from 2 inch to 18 inches, it certainly seems like a larger aperture allows you to see color in a larger number of objects. In my personal experience, I've only been able to notice color in the brighter objects in scopes 16 inches and above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disclaimer: I don't have any mathematical relation for aperture v/s color visibility; it appears to depend heavily on the observers' eye.&lt;/p&gt;&#xA;" OwnerUserId="1759" LastActivityDate="2018-12-27T18:34:53.680" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28903" PostTypeId="2" ParentId="28869" CreationDate="2018-12-27T19:27:03.033" Score="1" Body="&lt;p&gt;As Rob Jeffries said, you typically assume a value of &lt;span class=&quot;math-container&quot;&gt;$M/L$&lt;/span&gt; to find the mass from luminosity. I want to share a nice problem which guides you through the calculation for NGC1052 in the problem T8 &lt;a href=&quot;http://ioaa2018.bjp.org.cn/download/Theory/IOAA2018%20Theoretical%20Competition-final.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. Solutions are available &lt;a href=&quot;http://ioaa2018.bjp.org.cn/download/Theory/IOAA2018%20Theoretical%20Solution-final.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16744" LastActivityDate="2018-12-27T19:27:03.033" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28904" PostTypeId="2" ParentId="28898" CreationDate="2018-12-27T21:04:09.560" Score="5" Body="&lt;p&gt;The three inner Galilean moons are in orbital resonances. For each orbit of Ganymede, Europa makes two orbits and Io makes four orbits.  So after 1 orbit of Ganymede, the two inner moons are in the same configuration. That is 7.154 (Earth) days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Callisto is not in an orbital resonance, It won't be in the same position, but its orbit (of 16.689 days) is close to a ratio of 7:3 to Ganymede. So after 7 orbits of Ganymede (50.08 days) Callisto will have made just over 3 orbits (50.07 days)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want all known moons to be in a similar configuration you will have to wait a while, probably longer than the life of the sun (there are a lot of moons)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2018-12-28T16:14:53.570" LastActivityDate="2018-12-28T16:14:53.570" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28905" PostTypeId="1" AcceptedAnswerId="28989" CreationDate="2018-12-27T22:59:09.527" Score="8" ViewCount="841" Body="&lt;p&gt;I know star phenomena like solar flares can (at some degree) be predicted:&lt;br&gt;&#xA;&lt;a href=&quot;https://www.dailymail.co.uk/sciencetech/article-5371481/Scientists-predict-solar-flares.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.dailymail.co.uk/sciencetech/article-5371481/Scientists-predict-solar-flares.html&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is there any phenomena in a star that could be used to predict its supernova explosion &lt;em&gt;with years or decades&lt;/em&gt; of advance notice?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answers to &lt;a href=&quot;https://astronomy.stackexchange.com/q/18423/7982&quot;&gt;Are there observable changes in a star about to become supernova, minutes or hours before the explosion?&lt;/a&gt; address shorter timescales like hours or days, but I'm looking for times that are orders of magnitude longer than those discussed there.&lt;/p&gt;&#xA;" OwnerUserId="24506" LastEditorUserId="6" LastEditDate="2019-01-29T14:40:39.640" LastActivityDate="2021-01-11T21:01:09.743" Title="Is it possible to predict a supernova with years or decades of advanced notice?" Tags="&lt;star&gt;&lt;supernova&gt;&lt;prediction&gt;" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28906" PostTypeId="1" AcceptedAnswerId="28908" CreationDate="2018-12-27T23:50:33.377" Score="-2" ViewCount="85" Body="&lt;p&gt;Depending on the angle you view a black hole could it appear white, black or grey? Excluding how the &lt;a href=&quot;https://physics.stackexchange.com/questions/297890/rainbow-blackhole?r=SearchResults&amp;amp;s=1%7C94.6884&quot;&gt;gravitational lensing would make all the light emitted either red or blue shifted creating a rainbow effect&lt;/a&gt; from the accretion disk from and remove all the other light sources. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could a black hole appear black from the side where matter enters and white from the top where the jet is emitted? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/E0ztI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E0ztI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.physics.utah.edu/~bromley/blackhole/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.physics.utah.edu/~bromley/blackhole/index.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/207317/could-the-black-hole-in-the-center-of-the-galaxy-be-a-white-hole/207379#207379&quot;&gt;https://physics.stackexchange.com/questions/207317/could-the-black-hole-in-the-center-of-the-galaxy-be-a-white-hole/207379#207379&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2018-12-28T16:38:04.003" LastActivityDate="2018-12-28T16:38:04.003" Title="Can blackholes appear white and black?" Tags="&lt;black-hole&gt;&lt;white-hole&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2018-12-28T10:26:12.647" ContentLicense="CC BY-SA 4.0" />
  <row Id="28907" PostTypeId="1" CreationDate="2018-12-28T04:21:42.897" Score="3" ViewCount="404" Body="&lt;p&gt;I wish to do an astronomy related project which incorporates machine learning.Do you have any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="23302" LastActivityDate="2018-12-28T04:21:42.897" Title="Any ideas for an astronomy + ML project?" Tags="&lt;astrophysics&gt;&lt;cosmology&gt;&lt;fundamental-astronomy&gt;&lt;data-analysis&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ClosedDate="2018-12-28T15:55:41.813" ContentLicense="CC BY-SA 4.0" />
  <row Id="28908" PostTypeId="2" ParentId="28906" CreationDate="2018-12-28T05:43:31.307" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can blackholes be both &lt;a href=&quot;https://en.wikipedia.org/wiki/White_hole&quot; rel=&quot;nofollow noreferrer&quot;&gt;white&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Black_hole&quot; rel=&quot;nofollow noreferrer&quot;&gt;black&lt;/a&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The white hole is the end-life of a black hole, the (as you call it) &lt;em&gt;grey&lt;/em&gt; area is the tunnelling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many dispute the existence of white holes, described inaccurately (briefly) as a black hole partly reversed. A credible explanation from highly regarded sources is given in the paper: &quot;&lt;a href=&quot;https://arxiv.org/abs/1802.04264&quot; rel=&quot;nofollow noreferrer&quot;&gt;White Holes as Remnants: A Surprising Scenario for the End of a Black Hole&lt;/a&gt;&quot; by Eugenio Bianchi, Marios Christodoulou, Fabio D'Ambrosio, Hal M. Haggard, Carlo Rovelli, (Submitted on 12 Feb 2018 (v1), last revised 17 Mar 2018 (this version, v2)):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Briefly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Quantum tunneling of a black hole into a white hole provides a model for the full life cycle of a black hole. The white hole acts as a long-lived remnant, solving the black-hole information paradox. The remnant solution of the paradox has long been viewed with suspicion, mostly because remnants seemed to be such exotic objects. We point out that (i) established physics includes objects with precisely the required properties for remnants: white holes with small masses but large finite interiors; (ii) non-perturbative quantum-gravity indicates that a black hole tunnels precisely into such a white hole, at the end of its evaporation. We address the objections to the existence of white-hole remnants, discuss their stability, and show how the notions of entropy relevant in this context allow them to evade several no-go arguments. A black hole’s formation, evaporation,&#xA;  tunneling to a white hole, and final slow decay, form a unitary process that does not violate any known physics.&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Depending on the angle you view a black hole could it appear white, black or grey? I understand that matter goes in from the side and particles come out the top.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, that is something entirely different, nothing to do with a &quot;white hole&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you first divide black holes into two categories based on rotation you have a &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_metric#Singularities_and_black_holes&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schwarzschild black hole&lt;/a&gt; (non-rotating) and a &lt;a href=&quot;https://en.wikipedia.org/wiki/Frame-dragging#Astronomical_evidence&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kerr black hole&lt;/a&gt; (rotation speed &amp;gt; 0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rapidly rotating black hole has an &lt;a href=&quot;https://en.wikipedia.org/wiki/Ergosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;ergosphere&lt;/a&gt; that dips within the Schwarzchild radius:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The equatorial (maximum) radius of an ergosphere corresponds to the Schwarzschild radius of a non-rotating black hole; the polar (minimum) radius can be as little as half the Schwarzschild radius (the radius of a non-rotating black hole) in the case that the black hole is rotating maximally (at higher rotation rates the black hole could not have formed).&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is because of that reason that &lt;a href=&quot;https://en.wikipedia.org/wiki/Astrophysical_jet#Relativistic_jet&quot; rel=&quot;nofollow noreferrer&quot;&gt;relativistic jets&lt;/a&gt; eject matter from the poles. This is not visible light. It's not a white or grey hole, it's still a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to a number of close votes coming in I will post this brief answer before I'm locked out.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastActivityDate="2018-12-28T05:43:31.307" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28911" PostTypeId="1" CreationDate="2018-12-28T16:46:53.777" Score="2" ViewCount="392" Body="&lt;p&gt;Assume the &quot;Big Splash&quot; or &quot;Giant Impact&quot; (whichever name you prefer) theory is more or less correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; &lt;em&gt;With so many variations of this theory, with so many names and qualifiers on them, such as &quot;direct impact theory&quot; and &quot;glancing blow hypothesis&quot;, etc., let's assume that the impact was direct enough that nothing of Theia (other than Earth and the Moon) was left that was large enough to for its gravity to compress it into a hydrostatically equilibrious shape, and scatters too much to coalesce (on its own) in to anything large enough either.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume &quot;Theia&quot; was the size of Mars (some variations on the theory use other sizes, but for the sake of this question, let' stick with Mars)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How big was Earth before the impact?&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can't be as simple as &quot;Earth+moon-Theia(mars)=pre-impact size&quot;, can it?  Because there must have been some debris ejected completely from the earth+moon system, right? or was it such a small amount that it can be considered negligible? &lt;/p&gt;&#xA;" OwnerUserId="24853" LastEditorUserId="24853" LastEditDate="2018-12-28T22:39:36.227" LastActivityDate="2018-12-29T04:30:48.160" Title="What was earth's mass before the collision with theia?" Tags="&lt;planet&gt;&lt;solar-system&gt;&lt;earth&gt;&lt;planetary-formation&gt;&lt;theia&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28913" PostTypeId="1" AcceptedAnswerId="28916" CreationDate="2018-12-28T16:50:48.493" Score="16" ViewCount="3478" Body="&lt;p&gt;I was reading on white dwarfs, and I came across this sentence—&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Without energy sources, the white dwarf cools to a black dwarf in a few billion years.&lt;sup&gt;[1]&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, when I looked into the Wikipedia page on &lt;a href=&quot;https://en.wikipedia.org/wiki/White_dwarf&quot; rel=&quot;noreferrer&quot;&gt;White dwarf&lt;/a&gt;, it says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because the length of time it takes for a white dwarf to reach this state is calculated to be longer than the current age of the universe (approximately 13.8 billion years), it is thought that no black dwarfs yet exist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So which is true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what is the proper definition of a black dwarf?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Introductory Astronomy and Astrophysics. Zellik, M. Gregory, S. 4th edition. Brooks/Cole. 1998&lt;/p&gt;&#xA;" OwnerUserId="24929" LastEditorUserId="332" LastEditDate="2018-12-29T10:34:54.260" LastActivityDate="2019-05-10T06:55:41.997" Title="How long does it take for a white dwarf to cool to a black dwarf?" Tags="&lt;stellar-evolution&gt;&lt;white-dwarf&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28914" PostTypeId="2" ParentId="28913" CreationDate="2018-12-28T17:36:06.737" Score="20" Body="&lt;p&gt;I don't think there is an accepted definition of a &quot;black dwarf&quot; - it is not a term used in the scientific literature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A popular definition that appears to circulate on the internet is that it is a white dwarf that has cooled down to the extent that it no longer emits any radiation in the visible part of the spectrum. But this is an unworkable theoretical definition. Even the coolest objects emit &lt;em&gt;some&lt;/em&gt; radiation in the optical part of the spectrum; whether or not it is detectable will depend on the size of the object and how far away it is from us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a black dwarf is simply to be given the practical definition of being unobserved/unobservable at visible wavelengths then such objects exist already. &lt;a href=&quot;https://arxiv.org/abs/1406.0488&quot; rel=&quot;noreferrer&quot;&gt;Kaplan et al. (2014)&lt;/a&gt; have found evidence for a &lt;span class=&quot;math-container&quot;&gt;$\sim 1M_{\odot}$&lt;/span&gt; white dwarf companion to a pulsar that they are unable to detect at visible wavelengths even with the Keck telescope down to an &lt;em&gt;absolute&lt;/em&gt; (red) magnitude of &lt;span class=&quot;math-container&quot;&gt;$M_R&amp;gt;19.1$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They argue that this means the white dwarf has cooled below 3000 K. This is possible within the age of the universe because more massive white dwarfs cool more rapidly because their higher interior densities lead to more rapid crystallisation and a steep decrease in their heat capacity when they enter the Debye cooling regime. A more typical old white dwarf of mass &lt;span class=&quot;math-container&quot;&gt;$\sim 0.6M_{\odot}$&lt;/span&gt; would still be at a temperature of &lt;span class=&quot;math-container&quot;&gt;$\sim 4000$&lt;/span&gt;K and emit a reasonable amount of light at visible wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if you are happy to accept the definition above, then black dwarfs can result from the cooling of massive white dwarfs in about 10 billion years. The plot below, taken from the paper of Kaplan et al. (2014), shows some cooling models for massive white dwarfs with hydrogen atmospheres. They easily cool below 3000 K within 10 billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xeeXZ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xeeXZ.png&quot; alt=&quot;Massive DA white dwarf cooling models&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To address Mark Foskey's point: Once in the Debye cooling regime, the cooling rate hardly slows because the heat capacity is &lt;span class=&quot;math-container&quot;&gt;$\propto T^3$&lt;/span&gt;, whilst the luminosity is roughly &lt;span class=&quot;math-container&quot;&gt;$\propto T^{3.5}$&lt;/span&gt; (here, &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is the isothermal interior temperature). Luminosity is also &lt;span class=&quot;math-container&quot;&gt;$\propto T_s^4$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$T_s$&lt;/span&gt; is the surface temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relevant differential equation is that the rate of change of thermal energy equals (minus) the luminosity&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{d}{dt}\ \int T^3\ dT \propto -T^{7/2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting this together, we find that &lt;span class=&quot;math-container&quot;&gt;$dT_s/dt \propto -T_s^{3/7}$&lt;/span&gt; and so the cooling rate does not change much once Debye cooling starts. As a consequence the temperature plummets as &lt;span class=&quot;math-container&quot;&gt;$T_s \sim -t^{7/4}$&lt;/span&gt; and the time taken to cool well below 1000K from 3000K would only be a few times longer (maybe a billion years) than the time to cool from 4000K to 3000K (as suggested by a simple by-eye extrapolation of the models shown in the plot).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same treatment in the &quot;Mestel&quot; cooling regime, where the heat capacity is constant yields a much shallower &lt;span class=&quot;math-container&quot;&gt;$T_s \sim t^{-7/20}$&lt;/span&gt;, and accounts for the sharp &quot;knee&quot; at the onset of Debye cooling.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-01-01T09:53:57.977" LastActivityDate="2019-01-01T09:53:57.977" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28915" PostTypeId="2" ParentId="1965" CreationDate="2018-12-28T18:01:00.887" Score="2" Body="&lt;p&gt;It is possible that accretion onto a white dwarf companion causes type Ia supernovae. This is most likely to occur if the white dwarf is &lt;em&gt;massive&lt;/em&gt; to begin with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The timescale for &lt;em&gt;massive&lt;/em&gt; white dwarfs (&lt;span class=&quot;math-container&quot;&gt;$&amp;gt;1 M_{\odot}$&lt;/span&gt;) to cool below 3000 K and emit almost no radiation at visible wavelengths (which &lt;em&gt;you&lt;/em&gt; might call a &quot;black dwarf, though the term is not used extensively in the scientific literature AFAIK) is considerably shorter than for more typical white dwarfs of mass &lt;span class=&quot;math-container&quot;&gt;$0.6 M_{\odot}$&lt;/span&gt; and can be of order 10 billion years. Such white dwarfs &lt;em&gt;have&lt;/em&gt; been identified (&lt;a href=&quot;https://arxiv.org/abs/1406.0488&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kaplan et al. 2014&lt;/a&gt;). The plot below shows the cooling curves of massive white dwarfs in the form of either surface temperature or absolute red magnitude vs time. Clearly they can cool (according to these widely used models) on timescales considerably less than the age of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nZ6CD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nZ6CD.png&quot; alt=&quot;White dwarf cooling models&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since massive white dwarfs evolve from massive progenitors with short lives, it would seem eminently possible for a lower mass companion to have a much longer evolutionary timescale and for mass transfer to occur later on that will trigger a supernova explosion, and within the current lifetime of our Galaxy. I see no reason for such a supernova to look any different to other type Ia supernovae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the cooling time to a &quot;black dwarf&quot; would still be of order 10 billion years, so these events would not be occurring in high redshift galaxies that are considerably younger than the Milky Way.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2018-12-28T19:03:46.350" LastActivityDate="2018-12-28T19:03:46.350" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28916" PostTypeId="2" ParentId="28913" CreationDate="2018-12-28T18:09:39.547" Score="12" Body="&lt;p&gt;I think what you need is here on the &lt;a href=&quot;https://en.wikipedia.org/wiki/White_dwarf#Radiation_and_cooling&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;. In section &quot;Radiation and cooling,&quot; it says &quot;The rate of cooling has been estimated ... After initially taking approximately 1.5 billion years to cool to a surface temperature of 7140 K, cooling approximately 500 more K ... takes around 0.3 billion years, but the next two steps of around 500 K ... take first 0.4 and then 1.1 billion years.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One takeaway is that the rate of cooling (giving a fixed change in temperature, i.e., every 500 K) is increasing non-linearly. This is because the cooling is governed by diffusion process. So, at low temperature, to cool down 500 K more would take very long time than what it did in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As someone said in the comment, there is no precise definition of a black dwarf. So, I would not say who is right or wrong without understanding how they define the cutoff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you roughly define it to be at the level that its color temperature passes beyond the visible wavelength (i.e. &gt;7000 A or &amp;lt; 4000 K), and if you follow the info mentioned above by extrapolating from about 5500 K and assuming the rate of changing 500 K is constant as what it did in the previous step (i.e., from 6000 to 5500K taking 1.1 billion years), approximately we get the upper limit for cooling from 5500 to 4000 K as 3 billion years. By adding about the previous 2 billion years from the initial temperature down to 5500 K, we have &gt;5 billion years for a white dwarf from its initial state down to about 4000 K. Note that the 5 billion is a lower limit because we did not consider non-linearity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that you can also approximate the non-linearity effect by assuming an increment of 1 billion years in each step implying by the step 6000-5000 K. By doing this, the lower limit would be &gt;7 billion years.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the age of the universe is 13 billion years, whether you believe that a black dwarf exists or not depend on i) definition, ii) rate of cooling, and iii) variation (which means there might be a white dwarf that was born cool or living in the environment that supports better cooling than the typical population).&lt;/p&gt;&#xA;" OwnerUserId="22468" LastEditorUserId="25318" LastEditDate="2018-12-29T20:04:22.110" LastActivityDate="2018-12-29T20:04:22.110" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28917" PostTypeId="1" AcceptedAnswerId="28919" CreationDate="2018-12-28T18:51:08.987" Score="9" ViewCount="164" Body="&lt;p&gt;I searched and have found that the questions mainly focus on neutron stars, white dwarves and black holes. This was not what I looked for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the bigger the star's mass is, the more intense is its fusion reaction and the shorter is its lifespan in the main sequence. Now, imagine that a star spins much faster. There will be less surface gravity at the equator than at the poles. How does a fast rotation affect the star's fusion reaction?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Will the pressure of the star's mass on its core be lower at the equator and therefore reduce nuclear fusion rate?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Will there be a deeper convection because of the Coriolis effect?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What shall we notice in term of longevity in the main sequence, luminosity and emission spectrum of a fast rotating star compared to a slow rotating one &lt;strong&gt;of the same initial mass&lt;/strong&gt;?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="22874" LastActivityDate="2018-12-28T19:32:51.667" Title="How a star's rotation affects a star in the main sequence?" Tags="&lt;rotation&gt;&lt;main-sequence&gt;&lt;spectral-type&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28918" PostTypeId="2" ParentId="8782" CreationDate="2018-12-28T18:56:53.147" Score="2" Body="&lt;p&gt;&quot;Far away&quot; = &quot;very old&quot; in cosmology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All quasars are far away because all of them are old. These are objects that occurred when our bubble of Universe was young. So when our telescopes look far into space, they look back in time, and see a lot of quasars then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They're mostly giant black holes devouring gas and dust and cosmic junk, of which there was lots back then near these black holes. After they vacuum clean their environment, they calm down and the quasars basically turn off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are essentially no quasars forming in our time. Therefore there are no quasars visible nearby.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2018-12-28T18:56:53.147" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28919" PostTypeId="2" ParentId="28917" CreationDate="2018-12-28T19:32:51.667" Score="7" Body="&lt;p&gt;This is a well-studied problem. The effect of rotation on the structure of a low-ish mass star (like the Sun) is summarised by &lt;a href=&quot;https://www.epj-conferences.org/articles/epjconf/pdf/2013/04/epjconf_liac40_01005.pdf&quot; rel=&quot;noreferrer&quot;&gt;Eggenberger (2013)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such stars are never observed to rotate so fast that the rotation plays any significant role in their hydrostatic equilibrium, however rotation &lt;em&gt;does&lt;/em&gt; play a role by causing additional mixing in the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is important for two reasons: (i) it inhibits the gradual diffusion of helium towards the core, this slightly lowers the opacity in the core and raises it in the envelope (compared with a non-rotating star). This results in a slightly higher luminosity and a slightly higher surface temperature. (ii) More importantly, the additional mixing brings additional hydrogen into the core and this increases the lifetime on the main sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effects on stars of a solar mass are however unlikely to be very significant in practice, because these stars efficiently lose angular momentum via a magnetised wind during their early lives and the effects of rotation are unlikely to be significant even at several times the solar rotation rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effects on more massive stars can be more severe. These can spin at an appreciable fraction of their break-up rate for a large part of their lives and do not lose angular momentum as efficiently as lower mass stars (they do not have magnetised winds). The effects are described in the canonical paper of &lt;a href=&quot;http://adsabs.harvard.edu/abs/2000A%26A...361..101M&quot; rel=&quot;noreferrer&quot;&gt;Meynet &amp;amp; Maeder (2000)&lt;/a&gt;; they are more pronounced than for lower mass stars and more complicated because of the radiative envelope and uncertainties in the rotation-dependence of the significant mass-loss. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydrostatic effects of rotation &lt;em&gt;are&lt;/em&gt; expected to be important early in the main sequence and contribute to a slightly &lt;em&gt;lower&lt;/em&gt; surface temperature. At later times the dominant effects are caused by changes in the mixing and diffusion near the core and in the envelope as for the lower mass stars, resulting in higher luminosities and hotter temperatures. The main sequence lifetimes can be extended by 30% due to the additional mixing of fresh hydrogen fuel into the core.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2018-12-28T19:32:51.667" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28920" PostTypeId="1" CreationDate="2018-12-28T20:11:15.413" Score="1" ViewCount="54" Body="&lt;p&gt;We all know that one day the distance between the Earth and the sun will be zero. They will be bonded with gravity. My Question is which side of the earth will face toward the sun? I can't calculate that. Someone please calculate and don't say it's impossible.&lt;/p&gt;&#xA;" OwnerUserId="25311" LastActivityDate="2018-12-29T00:12:12.530" Title="Which side of earth will face toward sun when earth and sun will be get closer and be bonded by Gravitational Force?" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2019-01-07T06:40:20.840" ContentLicense="CC BY-SA 4.0" />
  <row Id="28921" PostTypeId="2" ParentId="28913" CreationDate="2018-12-28T20:16:17.620" Score="9" Body="&lt;p&gt;I think the most important part of any answer is that, as Rob Jeffries said, &quot;black dwarfs&quot; aren't really a thing in the astronomical literature, and I suspect that's the reason that you get different answers about how long it takes to become one. Different people come up with different thresholds for becoming one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would argue that 3000 K is too hot to call something black. The filament of a glowing light bulb is cooler than that. According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Incandescence&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, &quot;virtually all solid or liquid substances start to glow around 798 K (525 °C) (977 °F), with a mildly dull red color&quot;. I have not been able to find a reference for the theoretical time required to cool to that temperature.&lt;/p&gt;&#xA;" OwnerUserId="7752" LastEditorUserId="14094" LastEditDate="2019-01-02T05:23:17.787" LastActivityDate="2019-01-02T05:23:17.787" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28922" PostTypeId="2" ParentId="28707" CreationDate="2018-12-28T22:39:00.953" Score="1" Body="&lt;p&gt;They're referring to Time Dilation. The faster you go the slower time is moving for you. If you were to reach the speed of light for you time would come to a dead stop.&lt;/p&gt;&#xA;" OwnerUserId="25312" LastActivityDate="2018-12-28T22:39:00.953" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28923" PostTypeId="2" ParentId="28920" CreationDate="2018-12-29T00:12:12.530" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;We all know that one day the distance between the Earth and the sun will be zero.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This may be true, as the sun will get much bigger and engulf the Earth.  But this is irrelevant to the rest of the question, which seems to be about tidal locking.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They will be bonded with gravity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Earth is already gravitationally bound to the sun. But it isn't tidally locked. The Earth+moon has too much angular momentum to ever become tidally locked to the sun. It will be engulfed by the sun before it becomes tidally locked.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;which side of the earth will face toward the sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As noted above, the Earth will continue to rotate and will not become tidally locked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if it would become tidally locked it would be impossible to calculate. It is like putting a battery into a clock and asking exactly what time the battery will run out.  You can't measure the present nor predict the future with enough accuracy to forecast such things&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words. This won't happen, and if it does we can't calculate it.  It's impossible.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-12-29T00:12:12.530" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28924" PostTypeId="1" CreationDate="2018-12-29T03:58:15.053" Score="1" ViewCount="141" Body="&lt;p&gt;I noticed you replied to a question before about this topic and i came across your answer. Way back in the 80s when i was in my teens i saw a point of light moving between the stars at a steady rate across the sky. It was about the same size and brightnesz of those visible. It was moving in such a was as it approached a star ir veered off left or right then continued on its original path across the night sky. I showed my sister who was older than me at the time. It kind of reminded me of a whiplash efect using gravity to maintain spèed. Mabie too much Star Trek, but you get the idea. I am not an astronomer so forgive my ignorance, but a plane or a satalite would not move in this manor would it? I am convinced to this day that it was not one of ours. Can you clarify this as it was definately avoiding the stars?&lt;/p&gt;&#xA;" OwnerUserId="25314" LastActivityDate="2018-12-29T05:24:20.147" Title="A point of light moving between the stars" Tags="&lt;identify-this-object&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2019-01-03T15:15:24.137" ContentLicense="CC BY-SA 4.0" />
  <row Id="28925" PostTypeId="2" ParentId="28911" CreationDate="2018-12-29T04:30:48.160" Score="3" Body="&lt;p&gt;I'm not sure if there was a significant amount of debris, however the calculation you mentioned would at least give an upper bound on the value:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$&#xA;M_{piE}=\text{Greatest possible mass of pre-impact Earth}\\&#xA;M_E=\text{Mass of Earth today}=6.0\times10^{24}kg\\&#xA;M_M=\text{Mass of the Moon}=7.3\times10^{22}kg\\&#xA;M_T=\text{Mass of Theia}\approxeq6.4\times10^{23}kg\\&#xA;$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{eqnarray*}&#xA;M_{piE}&amp;amp;=&amp;amp;M_E+M_M-M_T\\&#xA;&amp;amp;=&amp;amp;(6.0\times10^{24})+(7.3\times10^{22})-(6.4\times10^{23})\\&#xA;&amp;amp;=&amp;amp;5.4\times10^{24}kg&#xA;\end{eqnarray*}&#xA;$$&lt;/span&gt;&#xA;So it wouldn't have been more than about &lt;span class=&quot;math-container&quot;&gt;$5.4\times10^{24}kg$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="25313" LastActivityDate="2018-12-29T04:30:48.160" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28926" PostTypeId="2" ParentId="28924" CreationDate="2018-12-29T05:24:20.147" Score="3" Body="&lt;p&gt;It's almost definitely not alien, if that's what you're asking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A satellite in low Earth orbit (around 200 km altitude) would have needed to move about 60m in the time you noticed it for the movement to have been visible. So, given that satellites at this altitude generally travel at around 7km/s or 7000 metres per second, it is not inconceivable that it could have been a man-made satellite correcting its orbit.&lt;/p&gt;&#xA;" OwnerUserId="25313" LastActivityDate="2018-12-29T05:24:20.147" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28928" PostTypeId="2" ParentId="12534" CreationDate="2018-12-29T06:23:02.383" Score="0" Body="&lt;p&gt;I think if you check out how to do spectrograph reduction, that will give you a better idea about how to identify line features from a real spectrograph. May be, you can start with this &lt;a href=&quot;https://slideplayer.com/slide/7477538/&quot; rel=&quot;nofollow noreferrer&quot;&gt;one&lt;/a&gt;, and go from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, after you clean the spectrograph with a bunch of calibration techniques, you will end up with a data set of (flux, pixel number). The pixel number is equivalent to the continuous wavelength, that is (pixel number, wavelength), but you have to find the way to map it. So, what you do is to have another spectrograph of an object with known spectral profile (i.e., arc lamps) occupying the same (pixel number, wavelength). Therefore, you can establish the (pixel number, wavelength) map. Then, you apply the same map back to your science object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moving away effect is called &quot;redshift.&quot; The redshift is easily predictable, and has been taken care of during the mapping process mentioned above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are blending effects as you mentioned. However, during the process since you identify the map by using an arc lamp, this is not a problem. This is how an arc lamp profile looks like (&lt;a href=&quot;https://telescope.livjm.ac.uk/TelInst/Inst/Meaburn/&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;). In your science object, you cannot separate the blending unless you perform further analysis such as simulation or fitting.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2018-12-29T06:23:02.383" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28929" PostTypeId="1" AcceptedAnswerId="30323" CreationDate="2018-12-29T12:39:09.590" Score="6" ViewCount="160" Body="&lt;p&gt;There has been no news on the obvious public channels from the EHT project since last (Northern) summer. Does anyone have more information on when any results from the April 2017 run might be released? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A secondary question is has the data from the April 2018 run been successfully gathered? I know they have to wait for Antarctic spring to ship the hard drives from the South pole, but that should have arrived a couple of months ago.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="7472" LastEditDate="2019-04-10T15:02:13.860" LastActivityDate="2019-04-10T15:02:13.860" Title="Is there any update on the Availability of Event Horizon Telescope Images?" Tags="&lt;radio-astronomy&gt;&lt;sagittarius-a&gt;&lt;event-horizon-telescope&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28931" PostTypeId="1" CreationDate="2018-12-29T13:55:16.817" Score="4" ViewCount="52" Body="&lt;p&gt;For a star with a given Zero-Age Main Sequence (ZAMS) mass, as a function of metallicity how strongly is the star's core coupled to its envelope?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the core-envelope boundary is only (semi-) well defined for Main Sequence stars that do have a core, and that in principle the final spin (angular velocity) of the core should depend on the spin of the star prior to losing its envelope in a mass transfer event. I've read, &lt;a href=&quot;http://adsabs.harvard.edu/abs/2000MNRAS.315..543H&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for instance, that the core should be nearly completely uncoupled from the envelope, however in &lt;a href=&quot;https://arxiv.org/abs/1706.07053&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; recent paper they assume that the coupling is sufficient enough to determine the natal spin of the stellar core (see figure 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a simple way to understand the strength of core-envelope coupling in terms of the ZAMS quantities of the star, without performing full stellar population synthesis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked this question at the Physics SE but it has received very little interest.&lt;/p&gt;&#xA;" OwnerUserId="23762" LastEditorUserId="23762" LastEditDate="2018-12-30T06:21:38.663" LastActivityDate="2018-12-30T06:21:38.663" Title="Strength of core-envelope coupling in stars" Tags="&lt;star&gt;&lt;astrophysics&gt;&lt;stellar-evolution&gt;&lt;stellar-astrophysics&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28932" PostTypeId="1" AcceptedAnswerId="28934" CreationDate="2018-12-29T15:57:39.473" Score="3" ViewCount="175" Body="&lt;p&gt;The formula of the radial velocity is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$v_i= V - K(\sin(f_i+\omega)\ + e\sin\omega),$$&lt;/span&gt; with&#xA;&lt;span class=&quot;math-container&quot;&gt;$$K = \frac{m_p}{m_s+m_p}\frac{na\sin i}{\sqrt{1-e^2}}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt;the systematic velocity of the system,&#xA;&lt;span class=&quot;math-container&quot;&gt;$m_p$&lt;/span&gt; the mass of the planet, &lt;span class=&quot;math-container&quot;&gt;$m_s$&lt;/span&gt; the mass of the star, &lt;span class=&quot;math-container&quot;&gt;$n = \frac{2\pi}{P}$&lt;/span&gt; the mean motion and &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; is orbital period of planet, &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; the length of the semi-major axis of the planet, &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; the inclination of the orbital plane with the ecliptic,&#xA;&lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt; the eccentricity of the planet, &lt;span class=&quot;math-container&quot;&gt;$f_i$&lt;/span&gt; the true anomaly at time &lt;span class=&quot;math-container&quot;&gt;$t_i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; the longitude of periastron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's well known the usual procedure. You have data about the instances &lt;span class=&quot;math-container&quot;&gt;$t_i$&lt;/span&gt;, the radial velocities &lt;span class=&quot;math-container&quot;&gt;$v_i$&lt;/span&gt; at time &lt;span class=&quot;math-container&quot;&gt;$t_i$&lt;/span&gt; and then you also have the measurements uncertainties &lt;span class=&quot;math-container&quot;&gt;$\sigma_i$&lt;/span&gt;. With that data, you can estimate 6 parameters: &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; (the periastron passage time) and &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;. After you estimate these parameters, you can calculate (if you know the mass of the star and you assume that &lt;span class=&quot;math-container&quot;&gt;$m_p&amp;lt;&amp;lt;m_s$&lt;/span&gt;), &lt;span class=&quot;math-container&quot;&gt;$m_p\sin i$&lt;/span&gt;, which is known as the &lt;em&gt;minimum mass&lt;/em&gt;. So, as far as this way shows, you got no way to estimate the mass of the planet and the inclination separately. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if instead of estimate this set of parameters, we introduce another one? Just one with all the parameters in the equation: &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$m_p$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt;, etc.? Couldn't that be possible? But nonetheless, it's always assumed that it's impossible to know the inclination with the RV data. Is there some reason why?&lt;/p&gt;&#xA;" OwnerUserId="9949" LastActivityDate="2018-12-31T11:46:18.657" Title="Why is it impossible to estimate the inclination of an exoplanet with the Radial Velocity Method?" Tags="&lt;orbital-elements&gt;&lt;radial-velocity&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28933" PostTypeId="1" AcceptedAnswerId="28937" CreationDate="2018-12-29T15:57:59.833" Score="1" ViewCount="154" Body="&lt;p&gt;According to wikipedia, Schmidt corrector plates must be aligned with the center of curvature to eliminate the spherical aberration. However, in compact SCT designs by Celestron and Meade, that is clearly not the case since the optical tube is usually shorter than the radius of curvature of the primary mirror. My question here is &lt;strong&gt;&lt;em&gt;WHY?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="25320" LastActivityDate="2018-12-29T19:35:57.950" Title="Why aren't corrector plates aligned with the center of curvature in an SCT?" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;optics&gt;&lt;telescope-lens&gt;&lt;mirror&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28934" PostTypeId="2" ParentId="28932" CreationDate="2018-12-29T17:36:06.410" Score="3" Body="&lt;p&gt;I am not sure what you meant by &quot;introduce another one.&quot; The system of equations is derived simply from Newtonian mechanics, and it is already completed in itself. So, my guess answer is, no that is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think Figure 2 &lt;a href=&quot;https://www.physics.ncsu.edu/classes/astron/orbits.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; is a good one to have in mind when thinking about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nature of RV data is that you have a bunch of data points specifying (vi,ti). From there, you can get P straightforwardly. There was a post I saw recently asking how to get P from RV data, you might want to check it out. Or, simply google.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;V is something that is observable or known because it is just a Hubble flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me express the equation like this: &lt;span class=&quot;math-container&quot;&gt;$v_i = V - K * f(t_i,P(\cdot))$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$P(\cdot)$&lt;/span&gt; is the period which is a function of some parameters that is irrelevant here. Since we know &lt;span class=&quot;math-container&quot;&gt;$(t_i,P)$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$f(\cdot)$&lt;/span&gt; is known, regardless of &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; because it is just a multiplicative constant. However, we know &lt;span class=&quot;math-container&quot;&gt;$(v_i,V,f(\cdot))$&lt;/span&gt;, we also know &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we want &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt;, we actually have to break the degeneracy in &lt;span class=&quot;math-container&quot;&gt;$K = K(m_p,m_s,e,P,a,i)$&lt;/span&gt;. We can get &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt; from &lt;span class=&quot;math-container&quot;&gt;$f(\cdot) = f(t_i,e,\omega) = g(t_i,\omega) + h(e,\omega)$&lt;/span&gt; by choosing &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; that fits the equation because we can treat &lt;span class=&quot;math-container&quot;&gt;$h(\cdot)$&lt;/span&gt; as a constant. Once we get &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt;, we resolve &lt;span class=&quot;math-container&quot;&gt;$h(\cdot)$&lt;/span&gt; and get &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far from &lt;span class=&quot;math-container&quot;&gt;$K(m_p,m_s,e,P,a,i)$&lt;/span&gt;, we know &lt;span class=&quot;math-container&quot;&gt;$e,P$&lt;/span&gt;. I am not sure how to estimate &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; but it does not matter because the leftover degeneracy between &lt;span class=&quot;math-container&quot;&gt;$m_p, i$&lt;/span&gt; at the nominator of &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; cannot be broken unless you can estimate one of the two (given that &lt;span class=&quot;math-container&quot;&gt;$m_s$&lt;/span&gt; is known as you mentioned).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the reason why you can only do the minimum mass from RV data.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastEditorUserId="22468" LastEditDate="2018-12-29T17:44:10.450" LastActivityDate="2018-12-29T17:44:10.450" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28935" PostTypeId="2" ParentId="28932" CreationDate="2018-12-29T17:41:09.450" Score="4" Body="&lt;p&gt;Essentially the problem comes down to geometry: radial velocity provides a 1-dimensional view of a 3-dimensional system. See the diagram below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zZQYJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zZQYJ.png&quot; alt=&quot;RV diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The observer sees only the component of the velocity vectors along the line from the observer to the star. That means that they cannot distinguish between the two blue velocity vectors: they will see only the purple (radial) component. This means a larger velocity vector that is tilted further away from the line-of-sight can masquerade as a smaller vector in the line of sight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No matter how you define your variables, you will always run into this issue because it is inherent to the geometry of the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does break down in the case of planet-planet interactions, which can break the degeneracy here (a good example of this is Gliese 876).&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastActivityDate="2018-12-29T17:41:09.450" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28936" PostTypeId="2" ParentId="26" CreationDate="2018-12-29T18:27:36.850" Score="1" Body="&lt;p&gt;There is also the theory which involves Mercury as an ex-moon of&#xA;Venus, largely based on calculations done by Van Flandern and Harrington (A Dynamic Investigation of the Conjecture that Mercury is an Escaped Satellite of Venus. Icarus 28: 435-40 (Abstract), 1976) and goes as follows (Van Flandern, Missing Planets, Dark Matter, and New Comets, 1999):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As Mercury tidally drifted outward it necessarily produced rotational&#xA;  drag on Venus, and it raised even bigger tides on the Venusian&#xA;  atmosphere causing it to circulate in retrograde direction. After&#xA;  billions of years this might impart retrograde motion on the whole&#xA;  planet.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Tides caused on Venus by Mercury while the latter was still spinning&#xA;  rapidly would have caused great interior heating and outgassing, and&#xA;  probably a great deal of surface upheaval (mountain building), too,&#xA;  causing the very dense atmosphere, the massive release of carbonate in&#xA;  the rocks as CO2 into the atmosphere, and the very high mountains.&#xA;  Mercury is massive enough to have taken much of Venus's spin in the&#xA;  1st half-billion years after formation and Venus's orbit is close&#xA;  enough to the Sun that complete escape occurs. The interchange of&#xA;  energy between Venus and Mercury would have been enormous, given&#xA;  Mercury's large mass (4 1/2 times more massive than the Moon).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Most of the iron (which eventually produces the magnetic field) in&#xA;  Venus would have been forced up into the crust by an excessively high&#xA;  spin rate, with Mercury getting most of the iron during fissioning,&#xA;  which would explain why Mercury has a stronger magnetic field than&#xA;  Venus. By contrast, the Earth's iron was not forced to the surface,&#xA;  perhaps because the Earth was not as hot and molten as Venus during&#xA;  that phase of its formation.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;During its lunar phase Mercury would have acquired a prolate shape&#xA;  (somewhat elongated towards Venus) because of tidal forces.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Both planets would have been melted by tidal heating in the early&#xA;  stages following escape. If this occured before Venus differentiated,&#xA;  it might have caused Mercury's high density and stronger magnetic&#xA;  field. Subsequently, both planets would have melted from mutual tidal&#xA;  heating.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;After escape, Mercury acquired greater tilt and eccentricity, and&#xA;  Venus would have lost more of its spin. Its prolate shape would have&#xA;  been reduced after escape but still maintained. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;At the point of escape Mercury would have had a period of revolution&#xA;  of about 40 days, and would have retained its spin period, which would&#xA;  also be 40 days since it was locked with Venus. But tides raised by&#xA;  the Sun would slow down its spin to its present 60 days, which gives&#xA;  it a 3-2 spin-revolution ratio (3 spins per 2 revolutions, in other&#xA;  words, its rotational period is 2/3 its period of revolution, which is&#xA;  88 days), because the next stable configuration for such a body&#xA;  (Mercury mass and diameter and degree of prolateness) is this ratio,&#xA;  so it is a predicted outcome of its having been a moon of Venus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This model, then, explains all the anomalies of both Venus and Mercury. Musser (2006) says it would require too much time for Venus to lose a moon but does not provide any reference for this, and the possibility has been corroborated by Kumar (1977) and Donnison (1978). This is the abstract from Donnison:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Kumar's (1977) suggestion that the slow rotations of Mercury and Venus are in part due to natural satellites that subsequently escaped is discussed. A more useful criterion for the escape of such satellites than that previously proposed is derived, and it is shown that this distance is sufficiently small for Mercury and Venus to make the escape of satellites a likely possibility.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And this is the abstract from Kumar:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is suggested that the slow rotations of Mercury and Venus may be connected with the absence of natural satellites around them. If Mercury or Venus possessed a satellite at the time of formation, the tidal evolution would have caused the satellite to recede. At a sufficiently large distance from the&#xA;  planet, the sun's gravitational influence makes the satellite orbit unstable. The natural satellites of Mercury and Venus might have escaped as a consequence of this instability.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They do not, however, specifically say that Mercury was once a moon of Venus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the abstract from Van Flandern and Harrington (gizidda.altervista.org):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The possibility that Mercury might once have been a satellite of Venus, suggested by a number of anomalies, is investigated by a series of numerical computer experiments. Tidal interaction between Mercury and Venus would result in the escape of Mercury into a solar orbit. Only two escape orbits are possible, one exterior and one interior to the Venus orbit. For the interior orbit, subsequent encounters are sufficiently distant to avoid recapture or large perturbations. The perihelion distance of Mercury tends to decrease, while the orientation of perihelion librates for the first few thousand revolutions. If dynamical evolution or nonconservative forces were large enough in the early solar system, the present semimajor axes could have resulted. The theoretical minimum quadrupole moment of the inclined rotating Sun would rotate the orbital planes out of coplanarity. Secular perturbations by the other planets would evolve the eccentricity and inclination of Mercury's orbit through a range of possible configurations, including the present orbit. Thus the conjecture that Mercury is an escaped satellite of Venus remains viable, and is rendered more attractive by our failure to disprove it dynamically.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="25322" LastEditorUserId="16685" LastEditDate="2019-08-29T06:49:35.013" LastActivityDate="2019-08-29T06:49:35.013" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28937" PostTypeId="2" ParentId="28933" CreationDate="2018-12-29T19:35:57.950" Score="2" Body="&lt;p&gt;In a schmidt camera, the corrector is placed at the centre of curvature, which is at a distance of twice the focal length from the primary mirror, and minimizes aberrations such as coma and astigmatism.  This makes for relatively long telescopes, particularly with large primary mirrors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sct design is similar, in that it uses a corrector plate, but this (along with the secondary mirror) is placed slightly closer to the primary mirror than its focal length.  This design allows much more compact telescopes, but at the expense of more aberrations.  As with most if not all optical designs, it is a compromise.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastActivityDate="2018-12-29T19:35:57.950" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28940" PostTypeId="1" AcceptedAnswerId="28942" CreationDate="2018-12-30T05:44:17.933" Score="11" ViewCount="772" Body="&lt;p&gt;The ESA video &lt;a href=&quot;https://www.youtube.com/watch?v=if2opecmev8&quot; rel=&quot;nofollow noreferrer&quot;&gt;ESOcast 173: First Successful Test of Einstein’s General Relativity Near Supermassive Black Hole&lt;/a&gt; includes a clip of images of stars at the center of our galaxy orbiting around SgrA*, a presumed supermassive black hole. This isn't visible light because it's obscured by dust, so it may be radio or long wave infrared, but I don't know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the middle, I can see something flashing at whatever wavelength this image has been produced from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How are these images obtained, and&lt;/li&gt;&#xA;&lt;li&gt;what process is it that is believed to be causing that flashing?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;GIF made from video at around &lt;code&gt;02:50&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eXQ0F.gif&quot; alt=&quot;flashing at the center of the Milky Way galaxy and SgrA*&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Six annotated frames from GIF highlighting the flashing that I'm seeing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RdyAG.gif&quot; alt=&quot;flares? from Sgr A*&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-07-26T04:50:54.160" LastActivityDate="2019-09-12T05:21:29.500" Title="How did they make a video of the center of the galaxy, and what is it exactly that's flashing there?" Tags="&lt;supermassive-black-hole&gt;&lt;dust&gt;&lt;accretion-discs&gt;&lt;identify-this-object&gt;&lt;sagittarius-a&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28941" PostTypeId="2" ParentId="11467" CreationDate="2018-12-30T06:00:07.477" Score="3" Body="&lt;p&gt;Yes, there is no hard distinction between a telescope and a camera with a long focal-length lens.&lt;/p&gt;&#xA;&lt;p&gt;When it's stuck directly to a camera body, it's often referred to as a &amp;quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Catadioptric_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;mirror lens&lt;/a&gt;&amp;quot; because they still want to call it a &amp;quot;camera lens&amp;quot; but it's made from a mirror:&lt;/p&gt;&#xA;&lt;p&gt;But when it's stuck directly to a &lt;em&gt;tripod&lt;/em&gt; and the camera body is stuck to it, we call it a telescope. You can point it at terrestrial or extraterrestrial things, it doesn't make a difference.&lt;/p&gt;&#xA;&lt;p&gt;Something to remember, no matter what you are looking at, &lt;em&gt;extended objects&lt;/em&gt; like wildlife, barns, the Moon, resolved nebula, etc. get dimmer the more you magnify them or spread them out, whereas stars remain at roughly the same brightness, independent of magnification, until you go too high and they go out of focus. So While a large aperture telescope can show you &lt;em&gt;stars&lt;/em&gt; that a thousand times dimmer or even ten thousand times dimmer than what you can see with your eye, if you try to use a telescope at high magnification to look at terrestrial objects at night, they can become really dim. This is why for a good line of binoculars, the higher the magnification, the larger the lenses become.&lt;/p&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;https://space.stackexchange.com/q/17008/12102&quot;&gt;What's the largest aperture telescope sent beyond the Earth-Moon system?&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;h2&gt;&amp;quot;Mirror Lens&amp;quot;&lt;/h2&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yGhsW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yGhsW.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NxvHK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NxvHK.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Sources: &lt;a href=&quot;https://en.wikipedia.org/wiki/File:500mmMirrorLens.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;top&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Sony_Alpha_55_with_Minolta_500_F8_Reflex.JPG&quot; rel=&quot;nofollow noreferrer&quot;&gt;bottom&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h2&gt;&amp;quot;Telescope&amp;quot;&lt;/h2&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9dXwN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9dXwN.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/File:Maksutov_150mm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h2&gt;&amp;quot;Telescope? Camera?&amp;quot;&lt;/h2&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1aNgV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1aNgV.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6RZ1e.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6RZ1e.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above x2:&lt;/strong&gt; image of a Dragonfly refractive array telescope [from dunlap.utoronto.ca][http://www.dunlap.utoronto.ca/instrumentation/dragonfly/]. &lt;em&gt;Image: P. Van Dokkum; R. Abraham; J. Brodie&lt;/em&gt; Images borrowed from &lt;a href=&quot;https://astronomy.stackexchange.com/q/18074/7982&quot;&gt;What (actually) is the “ deprojected half-light radius” of this almost-all-dark-matter Galaxy?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2021-05-24T21:14:18.057" LastActivityDate="2021-05-24T21:14:18.057" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28942" PostTypeId="2" ParentId="28940" CreationDate="2018-12-30T13:24:35.023" Score="12" Body="&lt;h2&gt;Question: How are these images obtained?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Later in the video the narrator says they took the images using ESO's VLT.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;03:40 [Narrator] 14.​ Making these measurements pushed the power of ESO’s Very Large Telescope to the limits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Source: &lt;a href=&quot;https://www.eso.org/public/archives/videos/script/eso1825a_en.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;ESO transscript&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over the whole observation period multiple telescopes and imaging instruments were used. Early observations were accomplished using the &lt;a href=&quot;https://www.eso.org/public/teles-instr/lasilla/ntt/&quot; rel=&quot;nofollow noreferrer&quot;&gt;NTT&lt;/a&gt;. Since around the year 2002 the VLT observed Sagittarius A* with the &lt;a href=&quot;https://www.eso.org/public/teles-instr/paranal-observatory/vlt/vlt-instr/naco/&quot; rel=&quot;nofollow noreferrer&quot;&gt;NACO&lt;/a&gt; and &lt;a href=&quot;https://www.eso.org/sci/facilities/paranal/instruments/gravity.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;GRAVITY&lt;/a&gt; instruments too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This graph puts the rotation period of Sagittarius A* in relation to the time and the observing instruments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/q2mKf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q2mKf.jpg&quot; alt=&quot;&amp;quot;S2&amp;quot; orbit with instruments and timescale&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;https://www.eso.org/public/images/eso1825c/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Question: What process is it that is believed to be causing that flashing?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;ESO provides older footage over a minutes-timescale showing a flare of Sagittarius A* in May 2003: &lt;a href=&quot;https://www.eso.org/public/videos/eso0330a/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Flashes of light from disappearing matter&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;ESO Press Video eso0330 shows the detection of a powerful flare from the centre of the Milky Way galaxy. These and other adaptive optics (AO) images (with resolution 0.040 arcsec in the near-infrared H-band at wavelength 1.65 µm) of the central region of the Milky Way were obtained with the NACO imager on the 8.2-m VLT YEPUN telescope at the ESO Paranal Observatory on May 9, 2003. [...] The position of the 15-year orbiting star S2 (cf. ESO Press Release eso0226) is marked by a cross and the astrometric location of the black hole is indicated by a circle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The cause of the flickering is probably the same for the video sequence you showed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A October 2018 ESO publication states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;ESO’s GRAVITY instrument on the Very Large Telescope (VLT) Interferometer has been used by scientists from a consortium of European institutions, including ESO [1], to observe flares of infrared radiation coming from the accretion disc around Sagittarius A*, [...]. The observed flares provide long-awaited confirmation that the object in the centre of our galaxy is, as has long been assumed, a supermassive black hole. The flares originate from material orbiting very close to the black hole’s event horizon — making these the most detailed observations yet of material orbiting this close to a black hole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;https://www.eso.org/public/news/eso1835/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ESO also has &lt;a href=&quot;https://www.youtube.com/watch?v=DRCD-zx5QFA&quot; rel=&quot;nofollow noreferrer&quot;&gt;another video&lt;/a&gt; showing the therory with more detail: skip to about 1:00.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flares associated with Sgr A* are also described in &lt;a href=&quot;https://astronomy.stackexchange.com/a/28241/7982&quot;&gt;this answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Artist's illustration of heated gas orbiting the black hole:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xpg0G.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xpg0G.png&quot; alt=&quot;Screengrab from Youtube DRCD-zx5QFA @ 1:25&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Screengrab from 1:25 of that video)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;See also &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Phys.org (2019): &lt;a href=&quot;https://phys.org/news/2019-09-black-hole-center-galaxy-hungrier.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Black hole at the center of our galaxy appears to be getting hungrier&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Do et al. 2019 &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/ab38c3&quot; rel=&quot;nofollow noreferrer&quot;&gt;Unprecedented Near-infrared Brightness and Variability of Sgr A*&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Chen et al. 2019 &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/ab3c68&quot; rel=&quot;nofollow noreferrer&quot;&gt;Consistency of the Infrared Variability of SGR A* over 22 yr&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AyNc5.jpg&quot; alt=&quot;Figure 1 from Do et al. 2019, Unprecedented Near-infrared Brightness and Variability of Sgr A*&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure 1 from Do et al. 2019&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Figure 1. Top row: a series of K' images taken on 2019 May 13 centered on Sgr A* showing the large variations in brightness throughout the night. The first image from the left is the brightest measurement ever made of Sgr A* in the near-infrared. Also labeled are nearby stars S0-2 (K' = 14 mag) and S0-17 (K' = 16 mag) for comparison. Bottom panel: K' (black) and H-band light curves of Sgr A* from 2019 May 13. On this night, we alternated between H and K' observations. The H-band magnitudes are offset using H − K' = 2.45 mag. There appear to be no significant color changes during the large change in brightness. Red circles show the location of the four images in the panels above.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16746" LastEditorUserId="7982" LastEditDate="2019-09-12T05:21:29.500" LastActivityDate="2019-09-12T05:21:29.500" CommentCount="5" CommunityOwnedDate="2018-12-30T21:05:08.220" ContentLicense="CC BY-SA 4.0" />
  <row Id="28944" PostTypeId="1" CreationDate="2018-12-30T16:50:48.967" Score="4" ViewCount="600" Body="&lt;p&gt;The Earth-Moon barycentre is said to be inside Earth. But Moon-Earth mass ratio (1.2%) is larger than Jupiter-Sun mass ratio (0.09%). Then how come the Jupiter-Sun barycentre is outside the surface of Sun?&lt;/p&gt;&#xA;" OwnerUserId="9735" LastActivityDate="2018-12-30T21:06:40.870" Title="Why the Sun-Jupiter barycentre is outside Sun?" Tags="&lt;solar-system&gt;&lt;mass&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28946" PostTypeId="2" ParentId="28944" CreationDate="2018-12-30T21:01:35.367" Score="8" Body="&lt;p&gt;There are three factors that all play into this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Jupiter/Sun mass ratio &lt;/li&gt;&#xA;&lt;li&gt;The Jupiter/Sun distance&lt;/li&gt;&#xA;&lt;li&gt;The Sun's radius&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The barycenter of any pair of orbiting masses lies on the line connecting their centers of mass, and its position depends on the masses of the two objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two objects have the same mass, their barycenter will lie halfway between them.  If one is twice as massive as the other, the center of mass will be 1/3 of the way from the larger mass to the smaller.  The formula is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D&lt;sub&gt;Sb&lt;/sub&gt;=D&lt;sub&gt;SP&lt;/sub&gt;*M&lt;sub&gt;P&lt;/sub&gt;/(M&lt;sub&gt;S&lt;/sub&gt;+M&lt;sub&gt;P&lt;/sub&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M&lt;sub&gt;S&lt;/sub&gt; and M&lt;sub&gt;P&lt;/sub&gt; are the masses of the Sun and planet, respectively,&#xA;D&lt;sub&gt;SP&lt;/sub&gt; is the distance between their centers of mass, and&#xA;D&lt;sub&gt;Sb&lt;/sub&gt; is the distance from the center of mass of the Sun to the barycenter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, whether or not the barycenter is inside the Sun depends on the radius of the Sun. How far it is from the center of mass of the Sun is unaffected by the Sun's radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Earth-Sun pair, M&lt;sub&gt;P&lt;/sub&gt;/(M&lt;sub&gt;S&lt;/sub&gt;+M&lt;sub&gt;P&lt;/sub&gt;) is 3.0×10&lt;sup&gt;-6&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Jupiter-Sun pair, M&lt;sub&gt;P&lt;/sub&gt;/(M&lt;sub&gt;S&lt;/sub&gt;+M&lt;sub&gt;P&lt;/sub&gt;) is 9.5×10&lt;sup&gt;-4&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the Earth-Sun barycenter is about 3.0×10&lt;sup&gt;-6&lt;/sup&gt; of the way from the Center of the Sun to the Earth, or about 279 miles.  The Jupiter-Sun barycenter is about 9.5×10&lt;sup&gt;-4&lt;/sup&gt; (a number 300 times bigger) times Jupiter's distance (a number 5.2 times bigger) and comes out at 459,000 miles, which is &lt;em&gt;just slightly&lt;/em&gt; larger than the Sun's radius of 430,000 miles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's outside the Sun because Jupiter is more massive than the Earth and further away from the Sun.  It's just how the numbers work out -- nothing deeper than that.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2018-12-30T21:06:40.870" LastActivityDate="2018-12-30T21:06:40.870" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28947" PostTypeId="1" AcceptedAnswerId="28948" CreationDate="2018-12-30T21:24:58.960" Score="0" ViewCount="1712" Body="&lt;p&gt;I know that blue stars are the hottest and youngest (according to &lt;a href=&quot;https://www.universetoday.com/24596/what-is-the-hottest-star/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Universe Today&lt;/a&gt;). But what makes them blue, instead of other colors like purple? &lt;/p&gt;&#xA;" OwnerUserId="25325" LastActivityDate="2018-12-30T21:41:24.817" Title="Why are blue stars the hottest?" Tags="&lt;star&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2018-12-31T17:14:55.757" ContentLicense="CC BY-SA 4.0" />
  <row Id="28948" PostTypeId="2" ParentId="28947" CreationDate="2018-12-30T21:41:24.817" Score="1" Body="&lt;p&gt;Our eyes and brains make them blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars are (close to) a black body, they emit a broad band of radiation. The hottest stars will emit most of their radiation in the ultraviolet but they will still be very bright in visible light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As they get hotter, they emit even more ultraviolet, and more visible light but the mixture of wavelengths in visible light doesn't change much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This mixture is interpreted by our brains as being a sort of light blue. No matter how hot the star gets, it will still be emitting lots of visible light in a mixture that looks &quot;light blue&quot;.  It will never look purple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Purple can only be seen if either a single short-wavelength visible light is being emitted, or a mixture of short and long wavelengths (red and blue). Neither of the mixtures can occur in light emitted from a black body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So stars can appear red, orange, yellow, white or light blue, but never green or purple.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-12-30T21:41:24.817" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28949" PostTypeId="2" ParentId="28837" CreationDate="2018-12-31T02:32:07.227" Score="3" Body="&lt;p&gt;YES and NO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;YES, you will get &quot;color&quot; of nebulae. Reason of this is that because of very low luminosity for small diameter telescope, your eye is using &quot;grey vision&quot;.&#xA;When luminosity will reach certain level, your eye will start adopt color receptors. Good luminosity levels begin from diameters above 200mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NO, that &quot;color&quot; which you may observe does NOT have anything in common with colors which you see everywhere. Because most nebulae emitting in near-infrared and ultra-blue regions, which when mixing, give &quot;greenish&quot; light. But this color is result of your brain mixing different wavelengths, not actual green wave. Some people see it green, others see &quot;pink&quot; etc...&lt;/p&gt;&#xA;" OwnerUserId="24823" LastActivityDate="2018-12-31T02:32:07.227" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28950" PostTypeId="1" AcceptedAnswerId="28951" CreationDate="2018-12-31T06:54:51.063" Score="2" ViewCount="149" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What would happen if we painted the whole moon with black, will it still reflect light, or it will become invisible?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What would happen if we painted the whole moon with silver paint or make its surface shiny?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="25331" LastEditorUserId="7472" LastEditDate="2018-12-31T13:11:59.480" LastActivityDate="2018-12-31T13:11:59.480" Title="Questions related to the reflection of the moon" Tags="&lt;the-moon&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28951" PostTypeId="2" ParentId="28950" CreationDate="2018-12-31T09:19:40.003" Score="8" Body="&lt;p&gt;The moon is already a pretty dark grey. It only reflects about 12% of the light that hits it. It is about the colour of (worn) tarmac on the road. You can see how grey it is in NASA's &lt;a href=&quot;https://www.nasa.gov/feature/goddard/from-a-million-miles-away-nasa-camera-shows-moon-crossing-face-of-earth&quot; rel=&quot;noreferrer&quot;&gt;gif from a million miles away&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Black&quot; paint reflects less light. Common household black paint reflects only about &lt;a href=&quot;https://corona-renderer.com/forum/index.php?topic=2359.0&quot; rel=&quot;noreferrer&quot;&gt;5% of light&lt;/a&gt;. So the moon would be rather less than half as bright as it is now. But you could still see it.  There are some &lt;a href=&quot;https://www.geek.com/news/carbon-nanotubes-used-to-create-the-blackest-black-ever-1599202/&quot; rel=&quot;noreferrer&quot;&gt;specialised black substances&lt;/a&gt; that reflect as little as 0.035%. A moon coated in these would actually be hard to see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;White paint and silver paint reflect about 80% of light, so if the moon were white it would be about 7 times brighter, but much less bright than the sun which is about 400000 times brighter than the full moon.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2018-12-31T09:19:40.003" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28952" PostTypeId="2" ParentId="28950" CreationDate="2018-12-31T09:29:24.720" Score="6" Body="&lt;p&gt;The albedo of the Moon (the proportion of the light that falls on it which is reflected) is about &lt;a href=&quot;http://www.asterism.org/tutorials/tut26-1.htm&quot; rel=&quot;noreferrer&quot;&gt;0.12&lt;/a&gt;. The darkest known material is &lt;a href=&quot;https://en.wikipedia.org/wiki/Vantablack&quot; rel=&quot;noreferrer&quot;&gt;Vantablack&lt;/a&gt; with an albedo of about 0.0004. So if you painted the Moon with that, the reflected light level would drop by a factor of roughly 300, taking the magnitude of a full moon from -13 to -6 or so. This is still brighter than the Moon appears &lt;a href=&quot;https://www.researchgate.net/publication/281146827_Apparent_magnitude_of_earthshine_A_simple_calculation&quot; rel=&quot;noreferrer&quot;&gt;when illuminated by Earthshine&lt;/a&gt; during a lunar eclipse (about -3), when it is still perfectly visible, so the Moon would not become invisible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the other direction, making the Moon perfectly white would increase its brightness 8 fold or about 2 magnitudes, taking it down to -15. This would make moonlight a bit brighter, but probably not change anything fundamental.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="7472" LastEditDate="2018-12-31T13:11:23.183" LastActivityDate="2018-12-31T13:11:23.183" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28953" PostTypeId="1" CreationDate="2018-12-31T11:40:07.517" Score="1" ViewCount="97" Body="&lt;p&gt;I need to convert the Keplerian orbital elements to the equinoctial elements and then use the converted elements&quot;equinoctial&quot; in the function &quot;d&quot; below. To do this, first, I converted the Keplerian elements to equinoctial elements (a,h,k,p,q,lamda)  using the formula in the attached picture, BUT I need to know what is the equinoctial elements that correspond to each of e,i,w,omega so that I can replace each keplerian elements in the function &quot;d&quot; with the equinoctial element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d= Sin[omega]* Cos[w]2 +[w] + Cos[omega] &lt;em&gt;Cos[i]&lt;/em&gt; Sin[w - Omega]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UCBzr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UCBzr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25334" LastEditorUserId="1891" LastEditDate="2019-01-07T06:43:32.420" LastActivityDate="2019-01-07T06:43:32.420" Title="How to convert a function of keplerian elements to a function of equinoctial elements?" Tags="&lt;orbit&gt;&lt;orbital-mechanics&gt;&lt;orbital-elements&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28954" PostTypeId="2" ParentId="28932" CreationDate="2018-12-31T11:46:18.657" Score="2" Body="&lt;p&gt;Let's assume we had a mathematical method that could separate the planet's mass from the inclination of the planet-star system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we could derive the mass of the planet and the inclination independently from each other.  In particular we would be able to determine the mass of the planet in the case of inclination &lt;span class=&quot;math-container&quot;&gt;$i = 0$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this cannot be true for the Radial Velocity Method since for inclination &lt;span class=&quot;math-container&quot;&gt;$i = 0$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$sin~i = 0$&lt;/span&gt;) the planet's motion in the line of sight is &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; and hence the radial velocity measurements give the same result (&lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt;) for all planet masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that we cannot separate the planet's mass from the inclination with the Radial Velocity Method. As a corollary it follows that it is impossible to determine the inclination of an exoplanet with the Radial Velocity Method.&lt;/p&gt;&#xA;" OwnerUserId="10568" LastActivityDate="2018-12-31T11:46:18.657" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28957" PostTypeId="1" AcceptedAnswerId="28959" CreationDate="2019-01-01T00:23:26.667" Score="2" ViewCount="72" Body="&lt;p&gt;The Kármán line is often cited as the beginning of outer space.  The Voyager probes have identified the boundary between the heliosphere and interstellar space.  Is there a agreed upon definition of the boundary of interplanetary space?  Is it the Earth's magnetopause?&lt;/p&gt;&#xA;" OwnerUserId="25088" LastEditorUserId="25088" LastEditDate="2019-01-01T03:13:01.817" LastActivityDate="2019-01-01T06:48:13.450" Title="How far above the Earth's surface does interplanetary space begin?" Tags="&lt;solar-system&gt;&lt;definition&gt;&lt;interplanetary-medium&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28958" PostTypeId="2" ParentId="28957" CreationDate="2019-01-01T02:18:53.583" Score="2" Body="&lt;p&gt;It it wherever is convenient for the article you are writing. There is no single consenus definintion of &quot;interplanetary space&quot;.  If you are in orbit around a planet then you are not in interplanetary space, if you are between planets, then you are.  But there is no substantial change in the matter filling space as you move from planetary orbit to interplanetary trajectory&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can use any reasonable point that fits the article you happen now to be writing&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-01-01T02:18:53.583" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28959" PostTypeId="2" ParentId="28957" CreationDate="2019-01-01T06:19:49.257" Score="2" Body="&lt;p&gt;I think that @JamesK's &lt;a href=&quot;https://astronomy.stackexchange.com/a/28958/7982&quot;&gt;answer&lt;/a&gt; summarizes the situation nicely. Here I can summarize the way that I see space commonly sliced and diced by people in &lt;a href=&quot;https://space.stackexchange.com&quot;&gt;Space Exploration SE&lt;/a&gt;. I can't defend these as &quot;right&quot;, but recollect seeing them typically used this way.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Earth orbit&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;LEO or low Earth orbit&lt;/li&gt;&#xA;&lt;li&gt;MEO or medium Earth orbit&lt;/li&gt;&#xA;&lt;li&gt;GEO or geostationary orbit (though it is sometimes mixed with the broader geosynchronous)&lt;/li&gt;&#xA;&lt;li&gt;HEO or high Earth orbit&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;cis-Lunar space&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;out to about 400,000 km from the Earth, still gravitationally bound to Earth (as the Moon is as well) but usually associated with gravitational effects of the Moon as well. Since the Moon is fairly massive (&gt; 1% of the Earth) it has a lot of &quot;pull&quot; in cis-lunar space. Anything a half-million km from the Earth has to deal with the Moon.&lt;/li&gt;&#xA;&lt;li&gt;includes orbits associated with the Earth-Moon Lagrange points&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Heliocentric orbit = &quot;Deep Space&quot;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;orbits that are bound to the Sun but not strongly bound to the Earth or other bodies.&lt;/li&gt;&#xA;&lt;li&gt;includes orbits associated with the Sun-Earth Lagrange points (we don't call it &quot;cis-Earth space&quot;). These are heliocentric orbits that are in 1:1 resonance with the Earth's orbit.&lt;/li&gt;&#xA;&lt;li&gt;exception in &lt;a href=&quot;https://space.stackexchange.com/q/23810/12102&quot;&gt;some specialized cases&lt;/a&gt; where HEO is called &quot;deep space&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So to answer your question, the distance from Earth where we start calling the space &quot;interplanetary&quot; or not associated with the Earth or another planet begins at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hill sphere&lt;/a&gt; which is about the same distance as the Sun-Earth Lagrange L1 and L2 points are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's about 1.5 million kilometers from Earth, and a little less from Mars, and several million miles for the &quot;big planets&quot;!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://space.stackexchange.com/a/26037/12102&quot;&gt;this answer&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For L1, find the smallest value of &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{M_2}{r^2} + \frac{M_1}{R^2} - \frac{r(M_1+M_2)}{R^3} - \frac{M_1}{(R-r)^2} = 0 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For L2, find the smallest value of &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{M_1}{R^2} + \frac{r(M_1+M_2)}{R^3}  - \frac{M_1}{(R+r)^2} - \frac{M_2}{R^2}  = 0 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though Mars is 50% farther from the Sun than the Earth, &lt;strong&gt;it's mass is only 11% that of Earth's&lt;/strong&gt;, so while the distances to Earth's Lagrange point are about 1% of that to the Sun for Earth, those of Mars are only about 0.5% of the distance to the Sun for Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In either case, a diagram would show two dots very close to each planet. The diagrams on the internet usually exaggerate this &lt;em&gt;greatly&lt;/em&gt; to make it easier to see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The values for the distance from the planets to the Sun and to their Sun-associated L1 and L2 points look like this. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a_Earth:     149598023  km&#xA;Sun-Earth L1:  1491524  km&#xA;Sun-Earth L2:  1501504  km&#xA;Earth r_Hill:  1496531  km&#xA;&#xA;a_Mars:      227939200  km&#xA;Sun-Mars L1:   1082311  km&#xA;Sun-Mars L2:   1085748  km&#xA;Mars r_Hill:   1084032  km&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IXzBf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IXzBf.png&quot; alt=&quot;find Earth and Mars L1 and L2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Python script based on scipy.optimize's &lt;a href=&quot;https://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.brentq.html#scipy.optimize.brentq&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brentq&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def solve_L1 (r, R, M1, M2):&#xA;    return M2/r**2 + M1/R**2 - r*(M1 + M2)/R**3 - M1/(R-r)**2&#xA;&#xA;def solve_L2 (r, R, M1, M2):&#xA;    return M1/R**2 + r*(M1 + M2)/R**3 - M1/(R+r)**2 - M2/r**2&#xA;&#xA;def r_Hill(R, M1, M2):&#xA;    return R * (M2 / (3.*M1))**(1./3.)&#xA;&#xA;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;from scipy.optimize import brentq&#xA;&#xA;a_Earth  = 149598023.   # Earth's semi-major axis (km)&#xA;a_Mars   = 227939200.   # Mars'   semi-major axis (km)&#xA;r_low    =   1000000.   # 1.0  million km  (lower guess)&#xA;r_high   =   1600000.   # 1.6  million km  (upper guess)&#xA;&#xA;M_Sun    = 1.9886E+30  # approximate mass (kg)&#xA;M_Earth  = 5.9724E+24  # approximate mass (kg)&#xA;M_Mars   = 6.4171E+23  # approximate mass (kg)&#xA;&#xA;r_Hill_Earth = r_Hill(a_Earth, M_Sun, M_Earth)&#xA;r_Hill_Mars  = r_Hill(a_Mars,  M_Sun, M_Mars)&#xA;&#xA;r = np.linspace(r_low, r_high)&#xA;if True:&#xA;    plt.figure()&#xA;    plt.plot(r, solve_L1(r, a_Earth, M_Sun, M_Earth), '-g')&#xA;    plt.plot(r, solve_L1(r, a_Mars,  M_Sun, M_Mars), '-r')&#xA;&#xA;    plt.plot(r, solve_L2(r, a_Earth, M_Sun, M_Earth), '--g')&#xA;    plt.plot(r, solve_L2(r, a_Mars,  M_Sun, M_Mars), '--r')&#xA;&#xA;    plt.plot([r_Hill_Earth], [0], 'ok')&#xA;    plt.plot([r_Hill_Mars ], [0], 'ok')&#xA;&#xA;    plt.text(1040000, 1.1E+11, 'L1 Mars L2', fontsize=14)&#xA;    plt.text(1450000, 3.0E+11, 'L1 Earth L2', fontsize=14)&#xA;&#xA;    plt.plot(r, np.zeros_like(r), '-k')&#xA;    plt.ylim(-4E+11, 4E+11)&#xA;&#xA;    plt.show()&#xA;&#xA;# for Mars:&#xA;r_L1_Mars = brentq(solve_L1, r_low, r_high, args=(a_Mars, M_Sun, M_Mars))&#xA;r_L2_Mars = brentq(solve_L2, r_low, r_high, args=(a_Mars, M_Sun, M_Mars))&#xA;&#xA;# for Earth:&#xA;r_L1_Earth = brentq(solve_L1, r_low, r_high, args=(a_Earth, M_Sun, M_Earth))&#xA;r_L2_Earth = brentq(solve_L2, r_low, r_high, args=(a_Earth, M_Sun, M_Earth))&#xA;&#xA;print &quot;a_Earth:    &quot;, int(a_Earth), &quot; km&quot;&#xA;print &quot;Sun-Earth L1: &quot;, int(r_L1_Earth), &quot; km&quot;&#xA;print &quot;Sun-Earth L2: &quot;, int(r_L2_Earth), &quot; km&quot;&#xA;print &quot;Earth r_Hill: &quot;, int(r_Hill_Earth), &quot; km&quot;&#xA;print ''&#xA;print &quot;a_Mars:     &quot;, int(a_Mars), &quot; km&quot;&#xA;print &quot;Sun-Mars L1:  &quot;, int(r_L1_Mars), &quot; km&quot;&#xA;print &quot;Sun-Mars L2:  &quot;, int(r_L2_Mars), &quot; km&quot;&#xA;print &quot;Mars r_Hill:  &quot;, int(r_Hill_Mars), &quot; km&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-01-01T06:48:13.450" LastActivityDate="2019-01-01T06:48:13.450" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28962" PostTypeId="1" AcceptedAnswerId="35127" CreationDate="2019-01-01T15:45:45.993" Score="8" ViewCount="1025" Body="&lt;p&gt;The moon has an elliptical orbit (of course), but as it inches away from the Earth each year, does its orbit become more circular and how come? Io's orbit is becoming more circular due to tidal heating effects, but why?&lt;/p&gt;&#xA;" OwnerUserId="16048" LastEditorUserId="16048" LastEditDate="2019-01-01T16:08:11.263" LastActivityDate="2020-02-18T23:54:24.683" Title="Is the moon's orbit circularizing? Why does tidal heating circularize orbits?" Tags="&lt;the-moon&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28963" PostTypeId="2" ParentId="28962" CreationDate="2019-01-01T17:41:54.403" Score="8" Body="&lt;p&gt;Short answer: YES. I will assume you mean the eccentricity in the orbit of the Moon around the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, tidal forces in binary systems (like the Earth-Moon system, or a binary star, etc...) effect the binary in three main ways: in order of longest timescale to shortest timescale&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) circularization of the orbit (eccentricity goes to zero, binary separation goes to minimum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) alignment of the binary components' spin angular momenta with the orbital angular momentum (the directions of &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; are the same).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) synchronization of the binary components' rotational frequency with the orbital frequency.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;but why?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's different ways of answering &quot;why,&quot; and &lt;a href=&quot;https://arxiv.org/abs/0807.4870&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; is a great conceptual answer given by the God Father of astrophysical tides himself, Z. P. Zahn:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A fundamental property of closed mechanical systems is that they conserve their&#xA;  total momentum. This is true in particular for binary stars, star-planet(s) systems,&#xA;  whether they possess or not a circumstellar disc, if one can ignore the angular&#xA;  momentum that is carried away by winds and by gravitational waves. Through&#xA;  tidal interaction, kinetic energy and angular momentum are exchanged between&#xA;  the rotation of the components their orbital motion and the disc. In the absence&#xA;  of such a disc, which is the case that we shall consider here, they evolve due&#xA;  to viscous and radiative dissipation to the state of minimum kinetic energy, in&#xA;  which the orbit is circular, the rotation of both stars is synchronized with the&#xA;  orbital motion, and their spin axis are perpendicular to the orbital plane. How&#xA;  rapidly the system tends to that state is determined by the strength of the tidal&#xA;  interaction, and thus by the separation of the two components...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, basically, the tidal torque drives dissipation, and this dissipation brings the binary to a minimum kinetic energy state, i.e. circular orbit, synchronized spins with the orbit, aligned spins with the orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a contour plot of the timescales provided by equations #&lt;span class=&quot;math-container&quot;&gt;$9 - 13$&lt;/span&gt; from Hut's 1981 seminal &lt;a href=&quot;http://adsabs.harvard.edu/abs/1981A&amp;amp;A....99..126H&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper&lt;/a&gt;, assuming the separation does not change much relative to the other quantities, for a binary composed of a black hole and a Wolf-Rayet star, which is a similar system to a planet-satellite system, where the timescales are parameterized in terms of the mass of the WR star and the separation of the binary:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0kZol.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0kZol.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dotted black line is the merger timescale for the binary due to gravitational waves, meaning below that line you are binary that mergers within the lifetime of the universe. The synchronization timescale is independent of the initial spin of the WR star, which is why there is only one synchronization line in the plot, but the alignemnt timescale does depend on the initial spin of the component feeling the tides. Points below the contours achieve that process (below the red dashed line are synchronized). The quantity &lt;span class=&quot;math-container&quot;&gt;$f_{B}$&lt;/span&gt; is the break-up fraction parameter, is between &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;, and the choice of which determines the initial WR star spin as a fraction of its break-up spin. (the tidal timescale contours were made by equating the WR lifetime with the tidal timescale). I did not include the circularization timescale because it is essentially always less than the sync. timescale (seen simply from the fact that generally, &lt;span class=&quot;math-container&quot;&gt;$S &amp;lt;&amp;lt; L$&lt;/span&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, it must be said that a WR star is quite different from a moon/satellite, however the effects of tides on each is similar enough to make a conceptual comparison (since a WR star is much less massive than the BH, as is the moon than the Earth, the hierarchy of tidal timescales is the same, but you must keep in mind that the WR's apsidal response to the tidal gradient is very different than a rocky moon's). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you pointed out, the Moon's orbit is slightly eccentric, but the moon is also tidally synchronized with the Earth's rotation (which are both sync'ed to the orbital rotation). This should make sense since it takes longer for tides to circularize than to synchronize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure somebody can post an example of using tides in an actual planet-moon system, rather than my black hole-star system. :)&lt;/p&gt;&#xA;" OwnerUserId="23762" LastEditorUserId="23762" LastEditDate="2019-01-03T15:21:59.520" LastActivityDate="2019-01-03T15:21:59.520" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28964" PostTypeId="1" AcceptedAnswerId="28966" CreationDate="2019-01-01T18:21:05.373" Score="2" ViewCount="184" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$2MG/c^2$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the expression for both, but some authors use one name and it has confused me. Is there any distinction?&lt;/p&gt;&#xA;" OwnerUserId="25199" LastEditorUserId="2365" LastEditDate="2019-01-01T18:33:43.380" LastActivityDate="2019-01-01T18:38:21.597" Title="Is the Bondi Radius the Same as the Schwarzschild radius?" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28965" PostTypeId="1" CreationDate="2019-01-01T18:34:53.423" Score="0" ViewCount="105" Body="&lt;p&gt;Maybe just maybe it is possible for a particle to travel FLT inside of black hole  because of the singularity gravitational force&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could be due to the fabric of space it self falling towards the singularity FTL beyond the event horizon. So technically &#xA;A particle can go FTL moving with space itself not not moving through it, and this would break any laws of relativity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if I'm wrong please tell me.&lt;/p&gt;&#xA;" OwnerUserId="25346" LastEditorUserId="25346" LastEditDate="2019-01-02T17:55:17.893" LastActivityDate="2019-01-02T17:55:17.893" Title="Is it possible for particles to be accelerated FTL inside BH travelling towards the singularity?" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ClosedDate="2019-01-02T03:10:05.767" ContentLicense="CC BY-SA 4.0" />
  <row Id="28966" PostTypeId="2" ParentId="28964" CreationDate="2019-01-01T18:38:21.597" Score="7" Body="&lt;p&gt;The Bondi radius is &lt;span class=&quot;math-container&quot;&gt;$2MG/s^2$&lt;/span&gt; where M is the mass, G is the gravitational constant and &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; is the speed of &lt;em&gt;sound&lt;/em&gt; in the medium of the accretion disk.  It represents the boundary between sub-sonic and super-sonic infall of matter into the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Schwartzchild radius is &lt;span class=&quot;math-container&quot;&gt;$2MG/c^2$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; is the speed of light and is the radius of the event horizon.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-01-01T18:38:21.597" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28967" PostTypeId="2" ParentId="28965" CreationDate="2019-01-01T19:01:30.440" Score="2" Body="&lt;p&gt;It's a little hard to follow the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short answer is: No it doesn't move faster than light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off from the perspective of the outside observer, the particle doesn't reach the Schwartzchild radius, due to gravitational time dilation. From the viewpoint of the distant observer (the best kind to be when black holes are involved) the particle's clock slows to a stop as it nears the event horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your velocity is a relative and a local property, if you want to talk about the velocity you need to say what it is relative, and when black holes are involved you need to be careful about which metric you are measuring the velocity in. From the point of view of the infalling particle, it enters and crosses the event horizon. It may then be falling faster than the speed of light relative to people on the outside, but it is behind the event horizon, and can not return to them. It is not moving faster than light relative to any point which it can return to. So this doesn't break causality, nor does it allow for high-speed transport.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that is certain is that in a short time (as the particle sees it) it will reach a singularity and then time and space will cease to exist for it, and all you are left with is mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hadrons might become some kind of quark gluon plasma on their way in, but this doesn't make it lose mass, nor does it have any effect on the whether it moves faster or slower than light.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-01-01T19:01:30.440" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28968" PostTypeId="2" ParentId="11467" CreationDate="2019-01-01T22:51:33.520" Score="1" Body="&lt;p&gt;For terrestrial use, avoid Newtonian reflectors. The eyepiece goes directly into the focuser, and you'll get an inverted (upside down) image.  Fine for astronomy, a pain for terrestrial use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other scope types - like refractors, Schmidt cassegrain telescopes (SCTs), and Maksutov cassegrain telescopes, use a diagonal mirror or prism between the focuser and the eyepiece.  This flips the image vertically, so you end up with an image that's the right way up, but left/right reversed. This can often be OK for wildlife, but if you want correct reading text (with no reversal) you can get specialised alternative diagonals (&quot;Amici prisms&quot;) for terrestrial use. They're not quite as good for astronomical use, but do give you a correct image for terrestrial use.&lt;/p&gt;&#xA;" OwnerUserId="17016" LastActivityDate="2019-01-01T22:51:33.520" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28969" PostTypeId="1" CreationDate="2019-01-02T02:31:06.520" Score="50" ViewCount="11093" Body="&lt;p&gt;I just got the news that the &lt;a href=&quot;https://en.wikipedia.org/wiki/New_Horizons&quot; rel=&quot;noreferrer&quot;&gt;New Horizons&lt;/a&gt; space probe has passed by some remote planet on the edge of the solar system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was surprised that the guy from NASA says that it might take 24 months from us to get the photo of that planet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solar system is not that big, right? It is slow because the signal transmission is slow, right? But why is the transmission so slow?&lt;/p&gt;&#xA;" OwnerUserId="11754" LastEditorUserId="23748" LastEditDate="2019-01-07T02:11:57.347" LastActivityDate="2019-01-07T02:11:57.347" Title="Why does it take so long to transmit an image from New Horizons to Earth?" Tags="&lt;solar-system&gt;&lt;data-analysis&gt;&lt;nasa&gt;" AnswerCount="4" CommentCount="14" FavoriteCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28970" PostTypeId="2" ParentId="28969" CreationDate="2019-01-02T03:50:41.627" Score="69" Body="&lt;p&gt;New Horizons has just passed the &lt;a href=&quot;https://en.wikipedia.org/wiki/Classical_Kuiper_belt_object&quot; rel=&quot;noreferrer&quot;&gt;Kuiper Belt Object&lt;/a&gt; (KBO) &lt;a href=&quot;https://en.wikipedia.org/wiki/(486958)_2014_MU69&quot; rel=&quot;noreferrer&quot;&gt;2014 MU69&lt;/a&gt; also known as Ultima Thule. KBOs form a belt of asteroids (the Kuiper Belt) from Neptune's orbit outwards and of which Pluto is the largest member of the Belt. During the encounter with Ultima Thule, all of the 7 instruments on New Horizons were gathering data (although not all at the same time) and the total data collected is expected to be about 50 gigabits of data (compared to 55 gigabits of data taken during the Pluto encounter in 2015).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &lt;a href=&quot;https://en.wikipedia.org/wiki/New_Horizons&quot; rel=&quot;noreferrer&quot;&gt;New Horizons&lt;/a&gt; is about another billion miles further out than Pluto was and 3 more years have elapsed, there is less power for the (tiny) transmitter and the signals are much weaker. The bit rate is about 1000 bits per second and so the 50 gigabits to transmit this will take 50e9 bits / 1000 bits per second = 50,000,000 seconds or about 579 days. Converting (roughly) to months by dividing by 365.25 and multiplying by 12 shows that it will indeed take about 19-20 months to transmit everything back. The first image at about 300 meters per pixel resolution and so about 100 pixels across the 30 km KBO, should be received on 2019 Jan 1. A second higher resolution image with about 300 pixels across the KBO is expected to be downloaded by 2019 Jan 2. There will be a press conference on 2019 Jan 2 when these images are due to be released and shown. (more details on what to expect when at &lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/2018/what-to-expect-new-horizons-mu69-ultima-thule.html&quot; rel=&quot;noreferrer&quot;&gt;Emily Lakdawalla's Planetary Society blog entry&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the initial data download, they expect to perform some analysis to see which images have the best data with 2014 MU69 in the frame. Given the uncertainty in the position of 2014 MU69 and the high speed of the encounter, they had to shoot strips of images and not all will contain the target. These data will be prioritized in the downlink so they arrive on the ground first and can be analyzed first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned by @luis-g there is also the Solar conjunction which will cause a 5 day period (according to PI Alan Stern in the 2019 Jan 3 press briefing) when reception of the data won't be possible. We would expect this to re-occur in January 2020 but these approx. 10 days don't make a large difference to the time taken which is dominated by the weakness of the received signal after the 15W transmission travels the ~4 billion miles and falls off due to the inverse square law, the corresponding low bitrate allowed by the need to have the transmitted data decodeable and the amount of data to transfer.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastEditorUserId="13772" LastEditDate="2019-01-07T02:10:58.937" LastActivityDate="2019-01-07T02:10:58.937" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="28973" PostTypeId="1" CreationDate="2019-01-02T09:25:14.813" Score="0" ViewCount="97" Body="&lt;p&gt;I'd like to buy a telescope with a focal length of 700mm and an aperture of 60mm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would I be able to use a telescope of this power to study the closer planets (for example Mars and Venus) and nebulae?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xQNpB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="25354" LastEditorDisplayName="user20278" LastEditDate="2019-01-02T16:13:57.423" LastActivityDate="2019-01-03T20:24:19.367" Title="Seeking advice regarding buying a telescope" Tags="&lt;telescope&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28974" PostTypeId="1" CreationDate="2019-01-02T11:22:39.310" Score="1" ViewCount="125" Body="&lt;p&gt;I was curious to find out that as to why, the moon looks more sharper during the day in winter, whereas it's almost not visible during summer.&lt;/p&gt;&#xA;" OwnerUserId="24938" LastEditorUserId="7472" LastEditDate="2019-01-02T13:42:21.753" LastActivityDate="2019-01-02T13:42:21.753" Title="Why is the moon more visible during the winter?" Tags="&lt;the-moon&gt;&lt;earth&gt;" AnswerCount="0" CommentCount="6" ClosedDate="2019-01-07T14:13:54.517" ContentLicense="CC BY-SA 4.0" />
  <row Id="28975" PostTypeId="2" ParentId="28969" CreationDate="2019-01-02T11:43:40.617" Score="16" Body="&lt;p&gt;On top of the slow data transmission rate (explained in &lt;a href=&quot;https://astronomy.stackexchange.com/a/28970&quot;&gt;astrosnapper's answer&lt;/a&gt;), I think it is worth pointing out that New Horizons will enter &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_conjunction&quot; rel=&quot;noreferrer&quot;&gt;solar conjunction&lt;/a&gt; next week, meaning that we won't be able to receive any transmissions from it due to the Sun blocking them.&lt;br&gt;&#xA;I don't know how many times this will happen over those 24 months, but it is an additional reason for the long(er) wait.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: NASA News Conference [&lt;a href=&quot;https://www.youtube.com/watch?v=hW1rc-D3A5I&amp;amp;t=42m18s&quot; rel=&quot;noreferrer&quot;&gt;42:18&lt;/a&gt;]&lt;/p&gt;&#xA;" OwnerUserId="25357" LastActivityDate="2019-01-02T11:43:40.617" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28976" PostTypeId="2" ParentId="28969" CreationDate="2019-01-02T12:00:10.350" Score="46" Body="&lt;p&gt;The other answer mentions it, but this gives a bit more theory as to the &lt;em&gt;why&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's effectively for the same reason that your phone or Wi-Fi don't work as well and slow down when that they are far from the hotspot or cannot get a clear line of access to the cell tower, more commonly known as having &quot;few bars&quot;: the signal gets weaker and as a result the signal-to-noise ratio (SNR) goes down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the error rate - failure to successfully transmit a bit and have it received correctly at the sender - goes up, because there is a greater probability that some fluctuation, like other sources of radio waves such as the stars and astrophysical phenomena, or even thermal fluctuation within the receiving devices themselves, can be taken as representing data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, to ensure that the bits successfully make it through, they have to be transmitted for a longer time so that they can be more clearly distinguished over that noisy background and won't be spuriously flipped. The poorer the SNR, the longer you need to transmit to make it clear. Another way to say it is that when you have a noisy background, and you turn on the transmitter, it creates a statistical bias in the noise fluctuations as its transmissions become superimposed upon them, e.g. putting a sinusoidal variation on top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At very low levels, this statistical bias is very small and thus requires a long sampling time to collect enough data to tease it out with high probability and since you don't know what data is coming at you by definition, you want the thing you're trying to tease to be as predictable as possible over the teasing time, and thus you must be sending only a single specific type of signal over that time and not switching between bits, limiting the bit rate to exactly that time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A mathematical theorem called the &lt;em&gt;Shannon-Hartley Theorem&lt;/em&gt; analyses this precisely and gives the exact bounds on just how fast you can transmit data and still have it reliably heard over a given level of noise relative to the strength of the transmitting signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an understanding of the spatial scales involved here and thus exactly what one is up against: your phone has to deal with a cell tower maybe 10 km away ... but here the probes are easily over 6000 Gm away (that's 6000 &lt;em&gt;billion&lt;/em&gt; meters and so 600 &lt;em&gt;million&lt;/em&gt; times further), and naturally we need a very large antenna, and because of the concerns just mentioned, the transmission rate is limited to, as said, about 1 kbit/s, taking a full millisecond for every bit transmitted, versus your phone at several Mbit/s or more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To downlink an uncompressed 8-bit (greyscale) 640x480 picture at that rate of 1 kbit/s takes 640*480*8/1000 ~ 2500 s or 2.5 ks (kiloseconds). A 4K UHD image would take 3840*2160*8/1000 ~ 66 ks to downlink, or the better part of a day (86.4 ks). Compare that to your broadband domestic Internet connection where &lt;em&gt;streaming&lt;/em&gt; 4K video (up to 60 frames per second so four million times faster) comes down with ease. (ADD NOTE: as mentioned in the comments, this last comparison may not be entirely accurate as there is also a significant amount of (lossy) compression on &quot;real&quot; 4K streams, or any Internet video streams for that matter, which is unacceptable for high-fidelity scientific data that can at best use purely lossless compression only so as not to introduce unnecessary errors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even with no compression, however, your typically decent 100 Mbit/s Internet connection would still be able to downlink maybe about 1-2 frames of video per second which is still enough to perceive something understandable as motion, albeit greatly slowed and incremental, and far higher than the data rates achieved here of a bit more than one frame per day.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also one of the reasons that Martian exploration would be significantly aided by, and it has been proposed to use, telepresence robotics controlled from a human base near, but in orbit of, the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADD: More accurately, the distance to &lt;a href=&quot;https://en.wikipedia.org/wiki/(486958)_2014_MU69y&quot; rel=&quot;noreferrer&quot;&gt;2014 MU&lt;sub&gt;69&lt;/sub&gt;&lt;/a&gt; is around 6600&amp;nbsp;Gm.&lt;/p&gt;&#xA;" OwnerUserId="9807" LastEditorUserId="332" LastEditDate="2019-01-05T01:32:39.073" LastActivityDate="2019-01-05T01:32:39.073" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28977" PostTypeId="1" CreationDate="2019-01-02T12:03:51.153" Score="9" ViewCount="2137" Body="&lt;p&gt;There is a bright object above the moon in Sudbury, highly visible to the eye, nothing else as bright in the sky except for the moon. I'm looking around 6 and 7am.  What is it ? &lt;/p&gt;&#xA;" OwnerUserId="25359" LastEditorUserId="7472" LastEditDate="2019-01-02T13:43:07.523" LastActivityDate="2019-01-09T01:00:09.173" Title="What is the bright object above the moon in Sudbury, Ontario?" Tags="&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28978" PostTypeId="2" ParentId="28977" CreationDate="2019-01-02T13:48:35.033" Score="14" Body="&lt;p&gt;It's probably the planet Venus. You can use an online planetarium (or one of the many mobile apps) to identify objects. &lt;a href=&quot;https://in-the-sky.org/skymap.php?no_cookie=1&amp;amp;latitude=46.49&amp;amp;longitude=-80.99&amp;amp;timezone=-5.00&amp;amp;year=2019&amp;amp;month=1&amp;amp;day=2&amp;amp;hour=6&amp;amp;min=45&amp;amp;PLlimitmag=2&amp;amp;zoom=152&amp;amp;ra=16.04509&amp;amp;dec=-22.22574&quot; rel=&quot;noreferrer&quot;&gt;This link&lt;/a&gt; shows the sky this morning at 6:45am at your location:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vsCiV.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vsCiV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2019-01-02T13:48:35.033" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28980" PostTypeId="2" ParentId="28898" CreationDate="2019-01-02T14:23:00.667" Score="2" Body="&lt;p&gt;This is basically a question of asking about the time of simultaneous recurrence of periodic events happening with some given period, and thus is fundamentally mathematical, and it is a type of phenomenon that turns up in many, many, many different places in the study of the universe at all scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; cycles of lengths &lt;span class=&quot;math-container&quot;&gt;$P_1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$P_2$&lt;/span&gt;, ..., &lt;span class=&quot;math-container&quot;&gt;$P_N$&lt;/span&gt;, then the shortest time for all of them to complete at once must be the &lt;em&gt;least common multiple&lt;/em&gt; of all the times, since that will ensure that at the end, a whole number of every period will have elapsed and thus total recurrence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P_\mathrm{recur} = \mathrm{lcm}(P_1, P_2, \cdots, P_N)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this lcm is only finite if the periods are all integral or at best rational multiples of some common unit to begin with. If not, and they are related by irrational numbers, then the period will be infinite, i.e. there will be no recurrence, though they may come within arbitrary closeness to their original configuration, but the interval for this will be very sensitive to the amount of proximity to the original configuration desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, in fact, actually not only is it good to restrict it to the Galilean moons, and in fact only the inner three, for practicality, it is &lt;em&gt;necessary&lt;/em&gt; if you want a finite periodicity because those moons, thanks to their orbital resonances, bear just such a relationship to each other. Even then though, it will have to be approximate, because the resonance is not perfect. Nonetheless, using the 1:2:4 relationship, since each number is a divisor of the next, that means the lcm and thus the period of the system are set by its longest member, the 4x one, which is the period of the futhest moon, Ganymede, so if you want the closest analogue should be Ganymede's orbital period, which is interestingly only a bit longer than our week, at 7.14 days (617 kiloseconds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that would be the best &quot;lunar month&quot; that I believe one can do, without too much arbitrary choosing, and you have to leave Callisto out of it. And if one wants, as a close week analogue, and embracing all the weirdness that a month is effectively a week, it could be a useful time unit for the Jupiter system, though I think that it makes more sense to get away with oddly wedding clocks and times to planetary movements as much as possible since they can simply be mapped by computers onto simpler time measures anyways as &quot;events&quot;, should they be of concern. Instead I think for astronautical use in both measuring and also organizing we should be moving to just use SI units - seconds, kiloseconds, megaseconds, and gigaseconds - as much as possible use megaseconds for weeks, and for the day analogue, the only non-SI unit still necessary to deal with thanks to humans' biological planetary baggage in the form of the circadian rhythm system and really an organizer and not to be used as a measurer, you can use actually alternatively either 86 or 87 ks (actually 87 ks is almost exactly the natural period of humans' biological clocks), and if you balance both right 23 such days (87 ks each with the last being 86) will equal to exactly two megaseconds, regularizing the system as much as possible. These units will then be understood everywhere.&lt;/p&gt;&#xA;" OwnerUserId="9807" LastActivityDate="2019-01-02T14:23:00.667" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28982" PostTypeId="2" ParentId="28973" CreationDate="2019-01-02T16:49:32.090" Score="1" Body="&lt;p&gt;All the magnifications listed above 84x will be useless on a 60mm telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, such a telescope can be useful if it is well made (particularly a steady tripod/mount) and you have reasonable expectations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will show lots of detail on the moon.  You will be able to see the phases of Venus and Mercury (which is pretty much all anyone can see of those from Earth).  You will be able to see Mars as a disk and maybe glimpse an ice cap or a little surface detail under excellent seeing with a favorable apparition.  You will be able to see Saturn's rings and Jupiter's moons and maybe with good conditions a band or two on Jupiter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will only show a few of the very brightest nebulae, and only as a faint smudge.&lt;/p&gt;&#xA;" OwnerUserId="14434" LastActivityDate="2019-01-02T16:49:32.090" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28983" PostTypeId="1" AcceptedAnswerId="29025" CreationDate="2019-01-02T16:52:09.433" Score="2" ViewCount="963" Body="&lt;p&gt;The Schmidt-Cassegrain telescope is amazing! It is compact, suitable for astrophotography ,and it requires little to no collimation. However, the SCT has some inherent optical issues. What are they? What are their causes? and how can they be eliminated?&lt;/p&gt;&#xA;" OwnerUserId="25320" LastActivityDate="2019-01-04T09:04:22.537" Title="What are the aberrations of an SCT? And how can they be eliminated?" Tags="&lt;telescope&gt;&lt;optics&gt;&lt;telescope-lens&gt;&lt;mirror&gt;&lt;collimation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28984" PostTypeId="1" AcceptedAnswerId="29086" CreationDate="2019-01-02T17:40:50.310" Score="0" ViewCount="146" Body="&lt;p&gt;Due to accelerating cosmic inflation, other galaxies and galaxy clusters will eventually race toward the Hubble horizon apparently away from us, placing them effectively out of our sphere of influence. Due to Virgocentric flow, nearby galaxies, including our own, may become gravitationally bound to the Virgo Supercluster, resisting cosmic expansion (for at least some time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, &lt;strong&gt;Is there any hypothetical time-frame for when this event happens?&lt;/strong&gt; That is, in what supposed number of billions or trillions of years will the Virgo Supercluster become the only region within the Hubble sphere with which we may interact with?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Is such an estimate currently impossible? I know I've read one in the past, I just don't remember it well. Somewhen in the ballpark of a trillion years. All I'm looking for is an educated guess.&lt;/p&gt;&#xA;" OwnerUserId="19454" LastEditorUserId="19454" LastEditDate="2019-01-04T19:32:45.153" LastActivityDate="2019-01-10T14:48:20.573" Title="Time-frame regarding cosmic expansion and the bound state of the Virgo Cluster" Tags="&lt;cosmology&gt;&lt;expansion&gt;&lt;cosmological-inflation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28985" PostTypeId="1" AcceptedAnswerId="28991" CreationDate="2019-01-02T18:04:42.607" Score="6" ViewCount="112" Body="&lt;p&gt;I imagine New Horizons data from its encounter with Ultima Thule (2014 MU69) has started arriving. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know if this data is public?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, if so, how you can access it?&lt;/p&gt;&#xA;" OwnerUserId="25366" LastEditorUserId="24184" LastEditDate="2019-01-03T13:36:04.653" LastActivityDate="2019-01-03T23:56:25.567" Title="Will there be public access to the New Horizons data from Ultima Thule?" Tags="&lt;data-analysis&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28987" PostTypeId="2" ParentId="28973" CreationDate="2019-01-02T19:18:04.470" Score="1" Body="&lt;p&gt;In my opinion, 75mm is the minimal requirement to say that device belongs to &quot;telescope&quot; class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnification itself is useless. For every doubling of magnification you pay with dividing luminosity of your image by 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also check out this black list of telescopes ref.2. All of these have defects which you, as beginner, will have real pain to cope with. As you can see, many of those telescopes are from serious firms. It means that even those are buying cheap parts and save their expenses when producing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From lower end of telescopes (of refractors), which I consider less or equal of 90mm, I think that most beginners should choose between computerized versus non-computerized model. If you pay for computerized model, you will loose significantly in optical quality of your scope. Example of such choice: Meade Polaris 90 VERSUS Celestron NexStar 90 GT. The most &quot;bang for the buck&quot; choices, from trusted makers. (I do not advise in favor of any manufacturer! If you fall deeply into astronomy, build your own telescope!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't fall for chineese ultra-low-end scopes. You cannot check their optics by web.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. If you are &lt;em&gt;strongly&lt;/em&gt; on budget, I would even advise to NOT buy any telescope. Make your own. Buy 200mm mirror from trusted source (approx 100-200 bucks) and work your way to telescope by yourself and garage tools (such telescope costs over 500 bucks in stores). Check out ref.1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS2. For the planets, your choice of telescope falls more into category of large focal ratio. F/10 works good. Downside is that nebulae are large and could not fit into such narrow angles. Upside is that telescopes with F/10 are moderate at cost. Parabolic mirror could be swapped with spheric and so on. Another downside is bigger weight, because of longer tube (newtonian and galilelian telescopes). Modern schemes like Maksutov/Schmidt-Cassegrains are somewhat &quot;more universal&quot;, combining good elimination of spheric aberration with wide angle. But weight is &lt;em&gt;very large&lt;/em&gt;. Only buy with EQ5 and EQ6. Or with custom computerized mount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At last. Best telescope is the one which is looked through everynight. Best telescope manufacturer is your own hands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References.&#xA;[1] Number one site for ATM &lt;a href=&quot;https://stellafane.org/tm/atm/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://stellafane.org/tm/atm/&lt;/a&gt;&#xA;[2] Black list of telescopes &lt;a href=&quot;http://www.star-hunter.ru/en/black-list/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.star-hunter.ru/en/black-list/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24823" LastEditorUserId="24823" LastEditDate="2019-01-02T22:09:31.897" LastActivityDate="2019-01-02T22:09:31.897" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28988" PostTypeId="1" CreationDate="2019-01-02T19:36:29.953" Score="1" ViewCount="93" Body="&lt;p&gt;Everyine knows its dangerous to look directly at the sun, but can you look safely at sunrises and sunsets without going blind? Because the uv radiaton and the light is much weaker. Or is it still very dangerous? How longcould you look at the sunrise or sunset without damaging your eyes? Will you be ok if you look at the sunrise, or sunset for a few seconds? Or is sungazing unsafe, no matter the time?&lt;/p&gt;&#xA;" OwnerUserId="25367" LastActivityDate="2019-01-02T19:36:29.953" Title="Can you go blind if you look at the sunrise and sunset?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;light&gt;&lt;time&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2019-01-03T15:15:53.960" ContentLicense="CC BY-SA 4.0" />
  <row Id="28989" PostTypeId="2" ParentId="28905" CreationDate="2019-01-02T19:48:33.520" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;..phenomena like solar flares can be predicted&quot;&#xA;It depends on weather. Solar weather. And timescale.&#xA;BUT you cannot predict it's motion. Solar dot measure, any of this form - is unpredictable. It may go down, it may go up - both ways 50/50. Stochastical process. The process itself is non-linear. You can have solar weather in some bounded range, but you cannot say what it will be tomorrow. &quot;Soft physics&quot; works on such things. Still some unexplained things, same as in meteorology.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;could be used to predict its supernova explosion&quot;&#xA;Yes. Direct observation of star could be useful. From close range. With magnetometric, X-ray etc. Which none of us have. On the other hand, supernova relies on changes in chemical constituents in star material. It could be defined.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;but I'm looking for times that are orders of magnitude longer than those discussed there&quot;. Yes, there is. Chain of events starts with changes in spectrum, which corresponds to changes of chemical structure in star material. Knowing all star parameters (from close range), you can have the time until event. Ofc you cannot definitely say &lt;em&gt;what follows&lt;/em&gt;, because small changes in parameters are varying outcome --- either there will be white dwarf or neutron star etc. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Ofc you cannot compute exact second of event, because (first) there is no such thing. Event occurs continiously. And, second, your prediction will rely on existing star models, and equipment accuracy. So if you measure with accuracy of 2%, do not expect prediction be more precise then 2%. Common practice shows that model makes more miscalculations almost always (99% of events miscalculated are from wrong models and/or miscalculated model range).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider weather prediction. It fails. When it fails, nobody notices. Because it fails when model fails. It fails less often then it works, but it happens. It doesn't fail catastrophically, that's why nobody notices. &quot;Oh, it's more colder at 4 o'clock, but prediction said it will get colder at 2 o'clock&quot; - nobody says that. It happened anyway, but model was slightly out of range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. Also categorize exactly what &lt;em&gt;prediction&lt;/em&gt; means. Because there is still no such thing as &quot;prediction&quot;. In science, we say that &quot;time period until Moon falls on Earth is 200-300 billion years&quot;. Always in range. According to model. And according to data.&lt;/p&gt;&#xA;" OwnerUserId="24823" LastEditorUserId="24823" LastEditDate="2019-01-02T20:06:11.870" LastActivityDate="2019-01-02T20:06:11.870" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28990" PostTypeId="2" ParentId="28969" CreationDate="2019-01-03T03:29:19.933" Score="16" Body="&lt;p&gt;Just to put some perspective on things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. New Horizons is really far away from the Earth.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment of closest approach, New Horizons was over 6,600,000,000 kilometers away from Earth. This is about 6 light-hours. And the spacecraft is continuing to get farther by about 14 kilometres per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pluto.jhuapl.edu/Mission/Where-is-New-Horizons.php&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cikbX.png&quot; alt=&quot;Solar system diagram of the location of NH at around the time of the closest approach&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Transmissions from farther away are weaker.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Inverse-square_law&quot; rel=&quot;noreferrer&quot;&gt;inverse square law&lt;/a&gt; states that the intensity of things like radio signals and sources of light (energy per unit of area perpendicular to the source) is inversely proportional to the square of the distance. That means doubling the distance results in us receiving only a quarter of the energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Inverse-square_law#/media/File:Inverse_square_law.svg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/2/28/Inverse_square_law.svg/800px-Inverse_square_law.svg.png&quot; width=&quot;479&quot; alt=&quot;Inverse square law diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3. New Horizons only has &lt;a href=&quot;http://pluto.jhuapl.edu/Mission/Spacecraft.php#Systems-and-Components&quot; rel=&quot;noreferrer&quot;&gt;so much power to work with&lt;/a&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spacecraft is powered by a single RTG (radioisotope thermoelectric generator) that contains ~11 kg of Plutonium-238. At launch, this produced 245 watts (at 30 volts of direct current) of power, but due to radioactive decay, this decreased to 200 watts by the time of the July 2015 Pluto flyby, and further to 190 watts by the time of the January 2019 MU69 flyby. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For data transmission, it has a 2.1-meter diameter high gain dish antenna, a 30-centimeter diameter medium-gain dish antenna, and two broad-beam, low-gain antennas. The high-gain beam is 0.3 degrees wide, and the medium-gain beam is 4 degrees wide (used in situations when the pointing might not be as accurate). New Horizon's &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.472.8206&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;noreferrer&quot;&gt;radio system&lt;/a&gt; is powered by a TWTA (Traveling Wave Tube Amplifier), which consumes 12 watts. (That's about the same as a modern &lt;a href=&quot;https://en.wikipedia.org/wiki/Compact_fluorescent_lamp&quot; rel=&quot;noreferrer&quot;&gt;CFL light bulb&lt;/a&gt;!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.planetary.org/multimedia/space-images/spacecraft/20130726_new_horizons_diagram.html&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RpV83.jpg&quot; alt=&quot;New Horizons labelled diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are actually two TWTAs for redundancy; one with left-hand circular polarization, and one with right-hand circular polarization. After launch, they figured out a trick to use both TWTAs at the same time, which increased the data transfer rate by 1.9 times. They used this two-TWTA mode to &lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/2015/01300800-talking-to-pluto-is-hard.html&quot; rel=&quot;noreferrer&quot;&gt;get all the data back from the Pluto flyby more quickly&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4. There's a limit how sensitive the antennas on Earth can be.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though we listen for New Horizon's transmissions using enormous 70-meter dish antennas from the &lt;a href=&quot;https://en.wikipedia.org/wiki/NASA_Deep_Space_Network&quot; rel=&quot;noreferrer&quot;&gt;Deep Space Network&lt;/a&gt;, there comes a point where it starts getting difficult to discern the signal amongst a sea of white noise and other interference, because the signal is so weak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the 70-meter dish from Madrid. It's hard to do much better than this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://spaceflightnow.com/2018/12/30/new-horizons-ultima-thule-mission-status-center/&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dUh2q.jpg&quot; alt=&quot;absurdly large dish antenna&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5. So, the downlink speed has to be restricted because of the very weak signal.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As elaborated upon in &lt;a href=&quot;https://astronomy.stackexchange.com/a/28976&quot;&gt;The_Sympathizer's answer&lt;/a&gt;, the signal-to-noise ratio gets lower when the signal gets fainter, and so you have to transmit data more slowly in order to make sure that the data you receive is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NASA has a neat &lt;a href=&quot;https://eyes.nasa.gov/dsn/dsn.html&quot; rel=&quot;noreferrer&quot;&gt;interactive page&lt;/a&gt; that shows what each antenna in the DSN is doing right now. Here's a screenshot from January 3, 2019, 01:11 UTC:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AMTsE.png&quot; alt=&quot;Canberra 70-m dish is talking to new Horizons, with data rate 1.06 kb/s, frequency 8.44 GHz, and power received -148.90 dBm&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the signal that this dish is receiving from New Horizons is only 1.29E-18 W in strength. That's 1.29 attowatts. That's &lt;em&gt;extremely&lt;/em&gt; weak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as a result of the faint signal, it looks like the people at NASA decided to restrict the downlink rate at about 1000 bits per second (125 bytes per second), as an optimal balance between data integrity and downlink speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a point of comparison, the &lt;a href=&quot;https://google.ca&quot; rel=&quot;noreferrer&quot;&gt;https://google.ca&lt;/a&gt; homepage (when you're not logged in) comes out to about 1 MB. So, if you tried to open the Google homepage at the speed of the New Horizons downlink, it'd take &lt;em&gt;over 2 hours&lt;/em&gt; for the page to fully load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;6. There is a lot of data.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;New Horizons was busy during the flyby. It collected about 50 gigabits of data (6 GB). So at 1,000 bits per second, on-and-off (the &lt;a href=&quot;https://astronomy.stackexchange.com/a/28975&quot;&gt;solar conjunction that Luis G. pointed out&lt;/a&gt; will also briefly delay the data transfer), it'll take about &lt;a href=&quot;http://pluto.jhuapl.edu/Mission/Spacecraft.php#Data-Collection&quot; rel=&quot;noreferrer&quot;&gt;20 months&lt;/a&gt; for the full set of the Ultima flyby data to be sent back to Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparison:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;During the Pluto flyby in July 2015, downlink speed was at about 2,000 bits per second, and it took about 15 months to download all 55 gigabits (7 GB) of Pluto data.&lt;/li&gt;&#xA;&lt;li&gt;During the Jupiter flyby in February 2007, downlink speed was at about 38,000 bits per second.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Further reading:&lt;/strong&gt; Here's an interesting related question: &lt;a href=&quot;https://space.stackexchange.com/questions/24338/how-to-calculate-data-rate-of-voyager-1&quot;&gt;How to calculate data rate of Voyager 1?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13784" LastEditorUserId="13784" LastEditDate="2019-01-03T18:15:18.243" LastActivityDate="2019-01-03T18:15:18.243" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28991" PostTypeId="2" ParentId="28985" CreationDate="2019-01-03T04:28:54.730" Score="11" Body="&lt;p&gt;The &lt;a href=&quot;http://pluto.jhuapl.edu/soc/UltimaThule-Encounter/&quot; rel=&quot;nofollow noreferrer&quot;&gt;raw images&lt;/a&gt;&lt;sup&gt;1&lt;/sup&gt; from LORRI (high resolution greyscale) are available from JHU APL but the rest of the data such MVIC (wide angle in greyscale plus four color bands: near IR, methane, red, and blue), LEISA (infrared spectroscopic imagery), Alice (ultraviolet spectroscopic imagery), as well as other instruments seems not to be available yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some &lt;a href=&quot;http://pluto.jhuapl.edu/Galleries/Featured-Images/index.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;processed imagery&lt;/a&gt; is also available elsewhere on APL's site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually all the data will be released via the &lt;a href=&quot;https://pds-smallbodies.astro.umd.edu/data_sb/missions/newhorizons/index.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA PDS&lt;/a&gt;&lt;sup&gt;2&lt;/sup&gt; but this may take some time since it first has to be transmitted to Earth (this will take 20 months) then processed and calibrated (several more months) or (for some images) converted to a public-friendly JPEG.  Once this is done the data will be released since, per &lt;a href=&quot;https://pds-smallbodies.astro.umd.edu/data_sb/missions/newhorizons/05310-DMAP-01_R0_C0.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this document&lt;/a&gt; (p. 15), the New Horizons data is not subject to embargo by the research teams:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are no proprietary data rights for the New Horizons mission. Selected uncalibrated (CODMAC Level 2) data, particularly image data, will be publicly released by the New Horizons project over the Internet in close to real time. ... Fully reduced, calibrated and corrected data products .. will be [released as it becomes available].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Quoting further (from Appendix D):    &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Selected uncalibrated data, including image data, will be publicly released over the INTERNET in close to real time.&lt;/li&gt;&#xA;  &lt;li&gt;Fully reduced, calibrated and corrected data products will be published and forwarded to the Planetary Data System as soon as they&#xA;  have been generated and validated.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I've asked &lt;a href=&quot;https://space.stackexchange.com/questions/33273/where-to-find-raw-images-of-ultima-thule&quot;&gt;a question&lt;/a&gt; on Space.SE regarding whether there are currently any archives of raw(ish) New Horizons/Ultima Thule imagery besides the one linked above.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; Converted to JPEG files.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2&lt;/sup&gt; Thanks to @astrosnapper for that link.&lt;/p&gt;&#xA;" OwnerUserId="24184" LastEditorUserId="24184" LastEditDate="2019-01-03T23:56:25.567" LastActivityDate="2019-01-03T23:56:25.567" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28992" PostTypeId="1" AcceptedAnswerId="28995" CreationDate="2019-01-03T05:48:28.647" Score="4" ViewCount="126" Body="&lt;p&gt;Is this correct?  &lt;a href=&quot;http://blogs.discovermagazine.com/badastronomy/2012/03/15/bafact-math-how-bright-is-the-sun-from-pluto/#.XC2clS-ZNTY&quot; rel=&quot;nofollow noreferrer&quot;&gt;If the Sun is 400,000 times brighter than the full moon&lt;/a&gt; am I correct in figuring that one would have to be about 632AU from the Sun for it to appear as bright as the full moon.  &lt;a href=&quot;https://solarsystem.nasa.gov/solar-system/oort-cloud/overview/&quot; rel=&quot;nofollow noreferrer&quot;&gt;If the Oort Cloud begins about 1000AU from the Sun&lt;/a&gt;, one would have to be more than 60% the way to the Oort Cloud for the Sun to appear as bright as the full moon.&lt;/p&gt;&#xA;" OwnerUserId="25088" LastActivityDate="2019-01-03T08:32:54.173" Title="Would one have to be more than halfway to the Oort Cloud for the Sun to appear as bright as the full moon?" Tags="&lt;the-sun&gt;&lt;the-moon&gt;&lt;apparent-magnitude&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28993" PostTypeId="2" ParentId="17956" CreationDate="2019-01-03T06:37:33.337" Score="5" Body="&lt;p&gt;This is called &quot;physical libration&quot;. &lt;a href=&quot;https://hal.archives-ouvertes.fr/hal-00588671/document&quot; rel=&quot;noreferrer&quot;&gt;The Moon’s physical librations and determination of its free modes&lt;/a&gt; (2011) estimates it using the Apollo mission retroreflectors. Their result is very small. They note that the &quot;damping times for these two modes are estimated as &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;x&lt;span class=&quot;math-container&quot;&gt;$10^4$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;x&lt;span class=&quot;math-container&quot;&gt;$10^6$&lt;/span&gt; years&quot;, so it is a mystery why it is even as large as their small result. They say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some possible excitation mechanisms have been explored in the past without satisfactory explanation. It has been shown that a recent meteoroid impact is an unlikely source of such excitation Peale (1975). Eckhardt (1993) proposed an excitation process related to a resonance crossing of the longitude normal mode (of 2.9 years) and a close forced frequency. During the evolution of the lunar orbit, the free and forced frequencies change slowly and can cross. However, the mechanism excites only the libration in longitude mode. Yoder (1981) proposed an alternative mechanism, based on turbulent fluid core interaction, to excite the wobble mode. The new determination of the amplitudes of the free librations invites new investigation of their excitation mechanisms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I did not search to see whether anyone has taken up their invitation.&lt;/p&gt;&#xA;" OwnerUserId="17743" LastActivityDate="2019-01-03T06:37:33.337" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28994" PostTypeId="1" AcceptedAnswerId="29018" CreationDate="2019-01-03T08:30:45.220" Score="3" ViewCount="186" Body="&lt;p&gt;When I see this picture, I'm trying to understand how these two elements merged together and have two hypothesis :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;two balls of lava in fusion together &lt;/li&gt;&#xA;&lt;li&gt;two solid rocks fused at their contact point &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But it appears that the collision speed was &lt;a href=&quot;http://pluto.jhuapl.edu/News-Center/News-Article.php?page=20190102&quot; rel=&quot;nofollow noreferrer&quot;&gt;not particularly fast&lt;/a&gt; so I'm not sure in the end. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rsDVW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What process likely formed Ultima Thule? &lt;/p&gt;&#xA;" OwnerUserId="1630" LastActivityDate="2019-01-04T06:18:22.067" Title="What process shaped Ultima Thule?" Tags="&lt;planetary-formation&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28995" PostTypeId="2" ParentId="28992" CreationDate="2019-01-03T08:32:54.173" Score="4" Body="&lt;p&gt;Yes, brightness is inversely proportional to the square of the distance, and the square of 632 is about 400,000. It's probably appropriate to round the value to 1s.f. to give &quot;600AU&quot;.  Of course, at that distance the sun appears point-like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance to the Oort cloud is rather hard to define. It's not like you reach 1000 AU and see a &lt;em&gt;Star Wars&lt;/em&gt; like &quot;Asteroid field&quot; of comet nuclei. For example Wikipedia gives a distance of 2000AU. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-01-03T08:32:54.173" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28996" PostTypeId="2" ParentId="28994" CreationDate="2019-01-03T08:46:03.720" Score="3" Body="&lt;p&gt;It was a slow collision between two lumps of ices, but at these temperatures, ice is a kind of rock. Energy would be released in such a collision which resulted in the partial fusion of the two bodies, but not enough energy to significantly disrupt either one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting body does not have sufficient mass for its gravity to pull it into a spherical shape, and hence after the two bodies had joined, they remained in the &quot;snowman&quot; shape.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-01-03T08:46:03.720" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28997" PostTypeId="2" ParentId="28994" CreationDate="2019-01-03T08:46:11.077" Score="4" Body="&lt;p&gt;Ultima Thule is a contact binary comprised of two formerly separate solid objects that gently&lt;sup&gt;1&lt;/sup&gt; (probably &amp;lt;5 km per hour) came together.  If the objects had been molten when they came together they would have formed a spherical object.  Since the lobes are distinct the objects must have been solid.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As for &quot;understand[ing] how these two elements merged together&quot;, Ultima and Thule are believed to be have been two objects that formed separately through accretion.  At some point the two (solid) objects collided at a very low velocity of &quot;one or a few miles per hour&quot;&lt;sup&gt;1&lt;/sup&gt;.  Since the escape velocity at point of impact is &lt;a href=&quot;https://news.ycombinator.com/item?id=18811511&amp;amp;goto=item%3Fid%3D18809772&quot; rel=&quot;nofollow noreferrer&quot;&gt;about 16 miles per hour&lt;/a&gt;&lt;sup&gt;2&lt;/sup&gt;, gravity alone would be sufficient to keep the two lobes together&lt;sup&gt;1&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; This topic was discussed in the 2019-01-02 New Horizons press conference.&#xA;The presentation talking about formation mechanism &lt;a href=&quot;https://youtu.be/5ZnFiN1cXS0?t=626&quot; rel=&quot;nofollow noreferrer&quot;&gt;starts here&lt;/a&gt; (the portion discussing approach speed is &lt;a href=&quot;https://youtu.be/5ZnFiN1cXS0?t=708&quot; rel=&quot;nofollow noreferrer&quot;&gt;~80 seconds later&lt;/a&gt;).  A discussion of how strongly bound the lobes are &lt;a href=&quot;https://youtu.be/5ZnFiN1cXS0?t=971&quot; rel=&quot;nofollow noreferrer&quot;&gt;starts here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A note about the video links above: The official JHU APL video linked to above is missing the first few minutes.  If the video is updated to include the complete press conference then the links above will point to the wrong times.  The first two links are to Jeff Moore's presentation and the third is him fielding a question from Bill Horowitz (CBS News).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2&lt;/sup&gt; To quote &lt;a href=&quot;https://news.ycombinator.com/item?id=18811511&amp;amp;goto=item%3Fid%3D18809772&quot; rel=&quot;nofollow noreferrer&quot;&gt;the source&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Modeling it as two spheres, one 16 km across the other 12 km, with a density the same as water, ... [a]t a distance of 8 km from the center of mass, the escape velocity ... would be about ... 16 miles/hour.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="24184" LastEditorUserId="24184" LastEditDate="2019-01-04T06:18:22.067" LastActivityDate="2019-01-04T06:18:22.067" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28998" PostTypeId="2" ParentId="28983" CreationDate="2019-01-03T09:04:54.520" Score="2" Body="&lt;p&gt;The primary aberration present in a &lt;a href=&quot;https://en.wikipedia.org/wiki/Schmidt%E2%80%93Cassegrain_telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schmidt–Cassegrain telescope&lt;/a&gt; is spherical, due to the primary mirror being spherical in shape; that causes light at the edges to have a different path length than light at the center, and thus a different focal point.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/8EXTp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8EXTp.png&quot; alt=&quot;Exaggerated cross section of a Schmidt corrector plate&quot;&gt;&lt;/a&gt;&#xA;The image above, from Wikipedia's &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Schmidt_corrector_plate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schmidt corrector plate&lt;/a&gt;&quot; webpage, shows an exaggerated corrector; they are actually much flatter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The secondary aberration is coma, though this is present to a much lesser extent. Coma is a spherical aberration from rays that come in off-axis. Coma shows up as little off-axis comet-shaped blobs that point inwards towards the center of the field and get bigger as you look towards the edge of the field of view. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Persons wanting a lengthy technical and mathematical explanation will enjoy reading sections: &quot;&lt;a href=&quot;https://www.telescope-optics.net/Schmidt-camera.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;10.2.2. Full-aperture Schmidt corrector: Schmidt camera&lt;/a&gt;&quot; &lt;strong&gt;and&lt;/strong&gt; &quot;&lt;a href=&quot;https://www.telescope-optics.net/SCT.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;10.2.2.4. Schmidt-Cassegrain telescope (SCT)&lt;/a&gt;&quot; of the &lt;a href=&quot;https://www.telescope-optics.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.telescope-optics.net/&lt;/a&gt; website.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recent improvements in the design are offered using &lt;a href=&quot;https://en.wikipedia.org/wiki/Aplanatic_lens&quot; rel=&quot;nofollow noreferrer&quot;&gt;aplanatic&lt;/a&gt; flat field Schmidt-Cassegrain optics. Here's a screenshot from &lt;a href=&quot;https://www.celestron.com/products/edgehd-8-optical-tube-assembly-cg5&quot; rel=&quot;nofollow noreferrer&quot;&gt;Celestron's video&lt;/a&gt; showing the difference between a standard versus aplanatic Schmidt telescope design:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZrFhd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZrFhd.png&quot; alt=&quot;Standard versus aplanatic Schmidt telescope&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The competing system from &lt;a href=&quot;https://www.meade.com/lx65-acf-6.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Meade is called Advanced Coma-Free&lt;/a&gt; (ACF™) optical design. Claiming also to offer coma-free pinpoint star images and flatter image fields while reducing astigmatism and eliminating diffraction spikes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The website forum &quot;&lt;a href=&quot;https://www.cloudynights.com/topic/462785-meade-vs-celestron/page-3&quot; rel=&quot;nofollow noreferrer&quot;&gt;Starry Nights - Cats &amp;amp; Casses&lt;/a&gt;&quot; compares Celestron and Meade.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastActivityDate="2019-01-03T09:04:54.520" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29001" PostTypeId="1" CreationDate="2019-01-03T12:00:32.653" Score="2" ViewCount="106" Body="&lt;p&gt;I wrote software which uses GMT real time to calculate Earth's mean, eccentric, and true anomaly. I encountered a bug where after reaching 360 degrees, instead of flipping back to zero, it subtracts from 360.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I checked Wolfram Alpha with search for Earth's true anomaly, just to see if we had passed through 0 degrees triggering the bug. As of a few days ago Wolfram and I were in agreement, approaching 360 degrees. But now Wolfram reads 179 degrees approx. I remember passing 180 in July or August.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it appears Wolfram is also experiencing an error. To double check, true anomaly is the angle between Earth and perihelion side of major axis. We know Earth reaches its closest point to the Sun in the northern hemisphere winter, January roughly. And aphelion in summer months.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally the j2000 epoch I uses 358 degrees roughly, putting the perihelion or closest approach right around early january roughly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus I can only conclude Wolfram is in error as well?&lt;/p&gt;&#xA;" OwnerUserId="25378" LastEditorUserId="14094" LastEditDate="2019-01-03T14:17:47.510" LastActivityDate="2019-01-05T01:32:20.350" Title="Is Earth's true anomaly roughly 1 degree currently?" Tags="&lt;orbit&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29002" PostTypeId="1" CreationDate="2019-01-03T12:52:09.267" Score="4" ViewCount="41" Body="&lt;p&gt;I just read on &lt;a href=&quot;https://en.wikipedia.org/wiki/Wow!_signal&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; that...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;By 1977, Ehman was working at the SETI project as a volunteer; his job involved analyzing &lt;strong&gt;by hand&lt;/strong&gt; large amounts of data processed by an IBM 1130 computer and recorded on line printer paper. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Emphasis mine.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understand, the program had been running for four years before this happened, (1973–1995), and considering the amount of data that must have been generated and how bad humans are at doing boring repetitive tasks, it seems to me that the risk is very high that a similar event was overlooked. Here is a small example of the type of printouts he was looking at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/d/d3/Wow_signal.jpg/320px-Wow_signal.jpg&quot; alt=&quot;Printout of the Wow!-signal&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can easily imagine a stream of &lt;code&gt;5IJSI&lt;/code&gt; in one of the 50 columns looking the same as &lt;code&gt;21121&lt;/code&gt; to a tired human operator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I have two related questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Was manual human inspection the only analysis done on the data during the seventies, or did he just do it out of interest? Was the program set up to alert on any &quot;suspicious&quot; signals, for example anything above a certain threshold for at least two samples?&lt;/li&gt;&#xA;&lt;li&gt;Is all data digitally preserved? I assume that by the eighties, everything became trivial to store, but I'm not so sure everything was stored in the seventies when they just started. If the data is preserved, has further automated analysis been done on the data set before and after the Wow!-event to rule out anything similar?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16328" LastActivityDate="2019-01-03T12:52:09.267" Title="Is all of the data from the same program that found the Wow!-signal analyzed?" Tags="&lt;seti&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29004" PostTypeId="1" CreationDate="2019-01-03T14:18:46.770" Score="5" ViewCount="107" Body="&lt;p&gt;Just a quick question regarding Neptune. Because Phobos is spiraling in towards Mars, Mars' rotation speeds up. Does the same thing happen with Neptune's rotation?&lt;/p&gt;&#xA;" OwnerUserId="16048" LastActivityDate="2019-01-03T16:33:55.117" Title="Does the orbital decay of Triton affect Neptune's rotation?" Tags="&lt;tidal-forces&gt;&lt;neptune&gt;&lt;tidal-locking&gt;&lt;triton&gt;" AnswerCount="0" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29005" PostTypeId="1" AcceptedAnswerId="29008" CreationDate="2019-01-03T15:10:39.623" Score="1" ViewCount="1534" Body="&lt;p&gt;For a school project I am making a gravitational computer model. As I want to Milankovitch cycles, I need to calculate the eccentricity of an orbit after the model has completed its simulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have thought about calculating the eccentricity using the aphelion and parohelion height, but these are not available as it is a simulation and as it therefore stores the data for one point on the ellipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data I have after the model has completed:&lt;br&gt;&#xA;X,Y,Z coordinates of points that it has saved&lt;br&gt;&#xA;X,Y,Z velocities of points that it has saved  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been able to calculate the semi-major axis using the vis-viva equation. I have tried to get the value using the vis-viva equation but I got stuck as I think I need at least the Aphelion or Parohelion to calculate the eccentricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question: Does anyone know how  to calculate the eccentricity at any given point in the orbit using the velocity and coordinates (without simulating all points around it)?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If not, does anyone maybe have a better idea/suggestion that I can try to gain the eccentricity values?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="25382" LastEditorUserId="7472" LastEditDate="2019-01-03T18:26:21.220" LastActivityDate="2019-01-03T18:34:19.377" Title="Calculation of Eccentricity of orbit from velocity and radius" Tags="&lt;orbital-mechanics&gt;&lt;eccentricity&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29006" PostTypeId="2" ParentId="29001" CreationDate="2019-01-03T15:18:31.177" Score="1" Body="&lt;p&gt;I've found an almanac for Earth perihelion up to the year 2100. 2019 perihelion occurred at roughly 5:00 January 3, 2019. Thus the angle now of true anomaly is some extent past 0 degrees. And Wolfram Alpha is wrong by negative 180 degrees. By Wikipedia and other standard definitions of mean anomaly which have persisted since Kepler's time.&lt;/p&gt;&#xA;" OwnerUserId="25378" LastEditorUserId="14094" LastEditDate="2019-01-05T01:32:20.350" LastActivityDate="2019-01-05T01:32:20.350" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29008" PostTypeId="2" ParentId="29005" CreationDate="2019-01-03T18:04:16.810" Score="5" Body="&lt;p&gt;If you have the relative position &lt;span class=&quot;math-container&quot;&gt;$\mathbf{r}$&lt;/span&gt; and velocity &lt;span class=&quot;math-container&quot;&gt;$\mathbf{v}$&lt;/span&gt; at the same point in time, compute the &lt;a href=&quot;https://en.wikipedia.org/wiki/Eccentricity_vector&quot; rel=&quot;nofollow noreferrer&quot;&gt;eccentricity vector&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\mathbf{e} = \frac{\mathbf{v} \times \mathbf{h}}{\mu} - \frac{\mathbf{r}}{\|\mathbf{r}\|}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$\mathbf{h} = \mathbf{r} \times \mathbf{v}$&lt;/span&gt; is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Specific_relative_angular_momentum&quot; rel=&quot;nofollow noreferrer&quot;&gt;specific relative angular momentum&lt;/a&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mu = GM$&lt;/span&gt; is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Standard_gravitational_parameter&quot; rel=&quot;nofollow noreferrer&quot;&gt;standard gravitational parameter&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The eccentricity is then given by the norm of the eccentricity vector &lt;span class=&quot;math-container&quot;&gt;$\left\| \mathbf{e} \right\|$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-01-03T18:34:19.377" LastActivityDate="2019-01-03T18:34:19.377" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29009" PostTypeId="2" ParentId="29005" CreationDate="2019-01-03T18:31:36.943" Score="1" Body="&lt;p&gt;One uses the so called eccentricity vector, also called (up to a factor) Laplace-Runge-Lenz vector. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the position &lt;span class=&quot;math-container&quot;&gt;$$\vec{r} = \begin{bmatrix}x\\ y\\ z\end{bmatrix}$$&lt;/span&gt; of the object at given time &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; and its velocity &lt;span class=&quot;math-container&quot;&gt;$$\vec{v} = \begin{bmatrix}u\\ v\\ w\end{bmatrix}$$&lt;/span&gt; both with respect to an inertial coordinate system cantered at the origin of the gravitational field, the eccentricity of the orbit (Keplerian case) can be calculated by first computing the eccentricity vector:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1: Calculate the angular momentum &lt;span class=&quot;math-container&quot;&gt;$$\vec{L} = \vec{r} \times \vec{v}$$&lt;/span&gt;&#xA;Step 2: Calculate the eccentricity vector &lt;span class=&quot;math-container&quot;&gt;$$\vec{e} = \frac{1}{\mu}\, \big(\,\vec{v} \times \vec{L}\,\big) - \frac{\vec{r}}{|r|}$$&lt;/span&gt;&#xA;Step 3: Calculate eccentricity&#xA;&lt;span class=&quot;math-container&quot;&gt;$$e = |\vec{e}| = \sqrt{\vec{e} \circ \vec{e}\,}$$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$\cdot \circ \cdot$&lt;/span&gt; is the dot product, e.g. &#xA;&lt;span class=&quot;math-container&quot;&gt;$$|r| = \sqrt{\vec{r} \circ \vec{r}\,} = \sqrt{x^2 + y^2 + z^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="24019" LastActivityDate="2019-01-03T18:31:36.943" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29010" PostTypeId="2" ParentId="29001" CreationDate="2019-01-03T18:46:58.627" Score="1" Body="&lt;p&gt;To fix the bug, maybe try something like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume you have managed to calculate &lt;span class=&quot;math-container&quot;&gt;$\sin(f)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\cos(f)$&lt;/span&gt; of the true anomaly &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt;. Then the true anomaly can be expressed in code as &lt;span class=&quot;math-container&quot;&gt;$$f = (\sin(f) &amp;gt; = 0)\arccos\big(\cos(f)\big)\, + \, &#xA;(\sin(f) &amp;lt; 0) \Big(\, 360^{\circ} - \arccos\big(\cos(f)\big)\,\Big)$$&lt;/span&gt; In this expression the logical operation &lt;span class=&quot;math-container&quot;&gt;$(\sin(f) &amp;gt; = 0)$&lt;/span&gt; produces as output either &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; and so does &lt;span class=&quot;math-container&quot;&gt;$(\sin(f) &amp;lt; 0)$&lt;/span&gt;. Also make sure the your &lt;span class=&quot;math-container&quot;&gt;$\arccos$&lt;/span&gt; function produces output in degrees and not radians, because that's what &lt;span class=&quot;math-container&quot;&gt;$\arccos$&lt;/span&gt; does mathematically. If it produces radians, then multiply the output by &lt;span class=&quot;math-container&quot;&gt;$\frac{180^{\circ}}{\pi}$&lt;/span&gt; i.e. &#xA; &lt;span class=&quot;math-container&quot;&gt;$$f = (\sin(f) &amp;gt; = 0)\frac{180^{\circ}}{\pi}\,\arccos\big(\cos(f)\big)\, + \, &#xA;(\sin(f) &amp;lt; 0) \Big(\, 360^{\circ} - \frac{180^{\circ}}{\pi}\,\arccos\big(\cos(f)\big)\,\Big)$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="24019" LastActivityDate="2019-01-03T18:46:58.627" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29011" PostTypeId="1" AcceptedAnswerId="29016" CreationDate="2019-01-03T19:03:57.953" Score="2" ViewCount="109" Body="&lt;p&gt;I want to fully understand how an SCT works. Thus, I figured out I can use some formulas that describe how an incident ray that is parallel with the axis would act after refracting off a Schmidt plate. I presume that the formulas would depend on the vertical and horizontal distances from the center of the plate. My goal now is to make a ray tracing model of an SCT with certain specifications. &lt;strong&gt;Please make your answer as simple as possible&lt;/strong&gt;.I will be thankful if a nontechnical source is provided at the end of the answer.&lt;/p&gt;&#xA;" OwnerUserId="25320" LastActivityDate="2019-01-06T09:38:26.083" Title="What are some formulas that are associated with the Schmidt corrector?" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;optics&gt;&lt;telescope-lens&gt;&lt;mirror&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29012" PostTypeId="1" AcceptedAnswerId="29014" CreationDate="2019-01-03T19:12:35.640" Score="8" ViewCount="202" Body="&lt;p&gt;Chang-E has just sent a shot after landing on the far side of the moon. It looks tricky, how  can a device send information from a place that is oriented opposed to Earth?&lt;/p&gt;&#xA;" OwnerUserId="25387" LastEditorUserId="1891" LastEditDate="2020-10-01T14:51:53.737" LastActivityDate="2020-10-01T14:51:53.737" Title="How to send information from the far side of the Moon to the Earth?" Tags="&lt;the-moon&gt;&lt;communication&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29013" PostTypeId="1" CreationDate="2019-01-03T19:13:50.163" Score="3" ViewCount="110" Body="&lt;p&gt;I have not been able to find an answer to my question through google, and therefore ask it on this medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do brown dwarfs have spots and faculae similar to other heavier stars? &lt;/p&gt;&#xA;" OwnerUserId="16092" LastActivityDate="2019-01-03T19:54:38.800" Title="Brown dwarf magnetic activity" Tags="&lt;magnetic-field&gt;&lt;brown-dwarf&gt;&lt;variable-star&gt;&lt;dwarf-star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29014" PostTypeId="2" ParentId="29012" CreationDate="2019-01-03T19:16:44.233" Score="11" Body="&lt;p&gt;You need a satellite that has a view of the landing site and the Earth. There are several possibilities for doing this. According to the &lt;a href=&quot;https://nssdc.gsfc.nasa.gov/nmc/spacecraft/display.action?id=QUEQIAO&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA Space Science Data Coordinated Archive&lt;/a&gt;, the communications are relayed through the satellite Queqiao which is in a halo orbit around the Earth-Moon L&lt;sub&gt;2&lt;/sub&gt; point.&lt;/p&gt;&#xA;&lt;p&gt;You could also have a satellite in lunar orbit, transmit data from the lander while it's over the far side, transmit data to Earth while it's over the near side.&lt;/p&gt;&#xA;&lt;p&gt;Or relay via a satellite in interplanetary space, etc.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2020-09-30T18:23:53.560" LastActivityDate="2020-09-30T18:23:53.560" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29015" PostTypeId="2" ParentId="29013" CreationDate="2019-01-03T19:54:38.800" Score="3" Body="&lt;p&gt;So far, opinion is divided, as evidence has not been found to prove either way. Back in 2015, a study published in Nature, and l&lt;a href=&quot;https://www.theverge.com/2015/7/29/9067697/powerful-aurorae-observed-brown-dwarf-first-time&quot; rel=&quot;nofollow noreferrer&quot;&gt;inked via the Verge, here&lt;/a&gt; suggested that no, brown dwarfs do not exhibit star-like behaviours, but are in fact much closer to planets (due to them being below critical mass for hydrogen fusion in their cores)&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;All planets that have magnetic fields within our Solar System have aurorae, so Hallinan and other researchers believed that if brown dwarfs were more like planets, they too would exhibit aurorae.&lt;/p&gt;&#xA;&lt;p&gt;...each time LSRJ 1835+3259 rotated — every 2.84 hours — it pulsed a radio emission and an aurora appeared at a location on the brown dwarf's surface.&lt;/p&gt;&#xA;&lt;p&gt;...&amp;quot;In the case of the brown dwarf, the aurora is so much more powerful than anything we've seen in our Solar System,&amp;quot; says Hallinan. &amp;quot;Jupiter's aurorae are 1,000 times more powerful than the Earth's aurorae. Well this brown dwarf has aurorae that are at least 10,000 times more powerful than Jupiter's aurora.&amp;quot; The aurorae's colors are also a bit more diverse, emitting red along with blue and green.&lt;/p&gt;&#xA;&lt;p&gt;...it strengthens the argument that brown dwarfs are more closely related to planets, kind of like &amp;quot;beefed-up Jupiters.&amp;quot; &amp;quot;Typically, people tend to think of these things as failed stars,&amp;quot; says Nichols. &amp;quot;My preferred explanation is they’re more like very successful planets, and this study demonstrates more planet behavior than stellar behavior.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;However this 2017 &lt;a href=&quot;http://iopscience.iop.org/article/10.3847/1538-4357/aa866b/meta;jsessionid=7DBF62727C46034DEFAFA5AF54962DD3.ip-10-40-1-105&quot; rel=&quot;nofollow noreferrer&quot;&gt;article in the Astrophysical Journal&lt;/a&gt; which looked at the same data from LSRJ1835+3259 suggests otherwise, identifying the magentic data not as aurorae, but as effects more like that of our own sun. A nice summary of the article on &lt;a href=&quot;https://science.howstuffworks.com/brown-dwarf-isnt-failed-star-magnetic-powerhouse.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Howstuffworks.com:&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;As the object rotated, a powerful magnetic region came into view, more powerful than the magnetic fields associated with sunspots that we observe on the sun. Sunspots are magnetically active regions that can trigger coronal mass ejections, solar flares and produce powerful streams of solar wind — all of which can generate powerful geomagnetic storms on Earth.&lt;/p&gt;&#xA;&lt;p&gt;In an interview with New Scientist, the researchers point out that LSR J1835+3259 is very young (approximately 20 million years old) and the powerful magnetic field could be interacting with the object's protoplanetary disk (if it has one). But if this active magnetic region is long-lasting and representative of its global magnetic field, LSR J1835+3259 is way more &amp;quot;star-like&amp;quot; than we give brown dwarfs credit for.&lt;/p&gt;&#xA;&lt;p&gt;So rather than calling brown dwarfs &amp;quot;failed stars,&amp;quot; perhaps we should call them overachieving planets or magneto-dwarfs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="43" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-01-03T19:54:38.800" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29016" PostTypeId="2" ParentId="29011" CreationDate="2019-01-03T20:04:42.090" Score="3" Body="&lt;p&gt;The specific purpose of a Schmidt Corrector plate is to have an equal but opposite spherical aberration to the primary mirror they compensate for. So any formula you try to come up with will depend on the primary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact the information on SCT's shows how it is made using physical forms, not using formulae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://www.revolvy.com/page/Schmidt-corrector-plate&quot; rel=&quot;nofollow noreferrer&quot;&gt;revolvy.com&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A thin glass disk with a perfectly polished accurate flat form was placed on a heavy metal pan. The upper edge of the pan was ground at a precise angle or bevel based on the coefficient of elasticity of the particular type of glass plate that was being used. The glass plate was sealed to the ground edge of the pan, then a vacuum pump was used to exhaust the air until a particular negative pressure had been achieved. This caused the glass plate to warp slightly. The exposed side was then ground and polished to a perfect flat. When the vacuum was released, the plate sprang back until its bottom surface was again plane, while the upper surface had the correct figure. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If, however, you do wish to work out the SCT parameters from formulae, &lt;a href=&quot;https://www.telescope-optics.net/Schmidt-camera.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page on telescope-optics.net&lt;/a&gt; has the fairly complex info you need to use this formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$z=\frac{(\rho^4-\Lambda\rho^2)d^4}{4(n^\prime-n)R^3}=\frac{(\rho^4-\Lambda\rho^2)D}{512(n^\prime-n)F^3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not an easy page to work through - but for someone who hasn't needed complicated maths for some years I could just about handle it.&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="10196" LastEditDate="2019-01-06T09:38:26.083" LastActivityDate="2019-01-06T09:38:26.083" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29017" PostTypeId="2" ParentId="28973" CreationDate="2019-01-03T20:24:19.367" Score="0" Body="&lt;p&gt;Have you ever looked through a telescope of that size? Astronomy is a hobby enjoyed worldwide. There are astronomical clubs just about everywhere there are people. I would suggest joining a club nearby and getting familiar with the hobby before buying anything. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electronics has found it's way into Astronomical viewing in recent years. What was no more than a faint smudge in an eyepiece is now clearly visible with modest equipment. Find out what others are doing before making any decisions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As already said the best telescope is the one which is looked through every night. The only way to determine what is best for you is to look through a few telescopes first then think about what sort of equipment will make you want to use it whenever the skies are clear. &lt;/p&gt;&#xA;" OwnerUserId="24928" LastActivityDate="2019-01-03T20:24:19.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29018" PostTypeId="2" ParentId="28994" CreationDate="2019-01-04T01:49:55.113" Score="2" Body="&lt;p&gt;This graphic in Wired.com's &lt;a href=&quot;https://www.wired.com/story/new-horizons-first-photos-ultima-thule/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Finally, New Horizons' First Photos of Ultima Thule&lt;/a&gt; suggest the contact process could have been extremely slow. Rather than two bodies on independent but similar heliocentric orbits approach slowly, touch, and stick, it proposes that they were gravitationally bound and very slowly approached each other as angular momentum was lost somehow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LVLUg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LVLUg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;NASA/JHUAPL/SWRI/JAMES TUTTLE KEANE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are no fast processes for two bound bodies so far from the Sun to loose angular momentum quickly, so it would have had to be very slow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astrobob.areavoices.com/2019/01/02/ultima-thule-more-clearly-now-two-spheres-touching/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astro Bob says&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In this step-by-step view we see how Ultima Thule probably formed when smaller chunks stuck together to form two objects. The twin-lobed structure of Ultima Thule vindicates accepted theory — called the accretion process — of how the solar system built itself up. NASA / JHUAPL / SwRI&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The reason the two objects still appear so neatly intact is that they approached each other at about the same speed you back into a parking spot — 1 to 2 miles an hour — and slowly fused together. Take a look again at color photo and focus on the “neck” of the asteroid. The rest of Ultima Thule is reddish-brown, probably from methane or other organic compounds exposed to sunlight, but the neck is pale, almost white and indicates lots of fine-grained material. During today’s press conference, mission scientists suggested it could be material that rolled downhill from the steep slopes where the two bodies make contact. Much more detailed photos are on the way along with compositional measurements that will help answer that question in more detail.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Ultima Thule formed at the beginning of it all from small, icy chunks that coalesced into two planetesimals. The two spiraled into toward one another and joined to form what astronomers call a “contact binary.” If you want to know what things looked like shortly after the first solid materials coalesced from the solar nebula, Ultima Thule appears to be as close as we’ve come so far. Like taking a ride back in a time machine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hov6w.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hov6w.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Here you can see the difference in the neck area vs. the rest of Ultima Thule. NASA / JHUAPL / SwRI&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6whcU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6whcU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's calculate the gravitational attraction between a 10 km and a 20 km diameter snowball. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll assume the density is 1. The could have both voids (many &quot;cosmic snowballs&quot; do) which would lower it, but rocks as well which would raise it, but this is just an order of magnitude calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance between centers is the sum of the radii, or 15,000 meters, and we can approximate them as spheres, then use Newton's &lt;a href=&quot;https://en.wikipedia.org/wiki/Shell_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;shell theorem&lt;/a&gt; to treat each one as a point at their center. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ F=\frac{G m_1 m_2}{r^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$m = \rho \frac{4 \pi}{3} r^3$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With rho of 1 g/cm^3 or 1000 kg/m^3 I get m1, m2 = 5.2E+11, 4.2E+12 kilograms, the force pushing each body into the other (using G=6.7E-11 N m^2 kg^-2) is about 650,000 Newtons, or &lt;strong&gt;roughly 66 tons&lt;/strong&gt;. That's 2/3 of the thrust of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Merlin_(rocket_engine_family)#Merlin_1D&quot; rel=&quot;nofollow noreferrer&quot;&gt;Merlin-1D engine&lt;/a&gt; (~900 kN) on each asteroid pushing it towards the other 24/7!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a while, the touching similar ices will start to fuse, adding some mechanical strength as well. That's over a long time, but things this far out have plenty of undisturbed time on their hands. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZCnDC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZCnDC.jpg&quot; alt=&quot;Merlin 1D engine&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/File:SpaceX_Testing_Merlin_1D_Engine_In_Texas.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-01-04T02:23:24.480" LastActivityDate="2019-01-04T02:23:24.480" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="29021" PostTypeId="1" CreationDate="2019-01-04T05:37:12.890" Score="1" ViewCount="48" Body="&lt;p&gt;The author in this article: &lt;a href=&quot;https://news.nationalgeographic.com/news/2010/07/100707-aphelion-heat-wave-earth-sun-science-weather-nation/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://news.nationalgeographic.com/news/2010/07/100707-aphelion-heat-wave-earth-sun-science-weather-nation/&lt;/a&gt; has explained that positional differences on the orbit is a negligible factor - about 7% more sunlight at the perihelion compared to the aphelion. I suppose the explanation is relatable to people in the Northern/Southern hemispheres, where temperatures are driven by changing seasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for people in the equator, where daily temperature swings measure only in a matter of few degrees all year round, it would have definitely made some difference. Does anyone know if this temperature change has been investigated before? I would like to read up more about it. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="24558" LastActivityDate="2019-01-04T05:37:12.890" Title="How much does being at the perihelion heat Earth up?" Tags="&lt;orbit&gt;&lt;temperature&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29022" PostTypeId="1" CreationDate="2019-01-04T06:16:00.550" Score="2" ViewCount="63" Body="&lt;p&gt;I have two fits images from the WFC3 that I am trying to combine with a C# program.  When I try to combine multiple images I don't think the values I am getting for my x/y coordinates (calculated from Right Ascension/Declination) are correct.  I am expecting a final image that is about the same width as the two images combined, but it turns out to be about the same width and about twice the height.  I know the final image should be about double the width of a single image because I manually combined the images in photoshop and the final image was about twice as wide as either of the two original images.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE:  when I say &quot;image&quot; they are fits images, so they are just a bunch of single values in a file, so to combine them I create a new file and initialize the correct number of single values (width * height) to zero, and then fill in the values from the images I am using to combine.  They are not jpg or tif or png.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using the following formula to change from world coordinate system to cartesian:&#xA;formula is (since distance is the same for everything):&#xA;x = cos(dec) * cos(ra)&#xA;y = cos(dec) * sin(ra)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get the right ascension and declination from the header in the fits file.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the final image dimensions, I calculate the distance between x1 and x2 and create a new image that is 1/2 image 1 width + distance + 1/2 image 2 width.  For the final height I do a similar calculation with y and image heights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The images do also have a rotational component, but I am ignoring that as both images share the same rotation.  This could be part of my problem.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;public const double PixelsPerArcSecond = .039; // per WFC3 spec from Nasa&#xA;&#xA;public static ImageDataModel Combine(List&amp;lt;ImageDataModel&amp;gt; inputImages)&#xA;{&#xA;    //  Right ascension is CRVAL1&#xA;    //  Declination is CRVAL2&#xA;&#xA;    //  formula is (since distance is the same for everything):&#xA;    //     x = cos(dec) * cos(ra)&#xA;    //     y = cos(dec) * sin(ra)&#xA;&#xA;    ImageDataModel returnImage = new ImageDataModel();&#xA;    ImageDataModel bm = inputImages[0];&#xA;&#xA;    double x1, y1, x2, y2;&#xA;&#xA;    x1 = Math.Cos(bm.CRVAL2) * Math.Cos(bm.CRVAL1);&#xA;    y1 = Math.Cos(bm.CRVAL2) * Math.Sin(bm.CRVAL1);&#xA;&#xA;    int mult = 4; // todo: set this based off of the bitpix of the incoming images.&#xA;&#xA;    for (int i = 1; i &amp;lt; inputImages.Count; i++)&#xA;    {&#xA;        ImageDataModel cm = inputImages[i];&#xA;&#xA;        x2 = Math.Cos(cm.CRVAL2) * Math.Cos(cm.CRVAL1);&#xA;        y2 = Math.Cos(cm.CRVAL2) * Math.Sin(cm.CRVAL1);&#xA;&#xA;        double dx = x1 - x2;&#xA;        double dy = y1 - y2;&#xA;&#xA;        int distX = (int)((dx * 3600) / PixelsPerArcSecond);&#xA;        int distY = (int)((dy * 3600) / PixelsPerArcSecond);&#xA;&#xA;        // This is what I expect to be wider than tall, but the converse is true.&#xA;        int w = Math.Abs(distX) + (bm.ImageWidth / 2) + (cm.ImageWidth / 2);&#xA;        int h = Math.Abs(distY) + (bm.ImageHeight / 2) + (cm.ImageHeight / 2);&#xA;        // This is where the two images are combined into the final image.&#xA;        ImageDataModel imd = CombineTwoImages(bm, cm, i, w, h, mult);&#xA;        bm = imd;&#xA;    }&#xA;&#xA;    return returnImage;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I am expecting an image that turns out like this:&lt;br&gt;&#xA;&lt;a href=&quot;http://wierdling.net/stack-overflow-images/ManuallyCombined.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://wierdling.net/stack-overflow-images/ManuallyCombined.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But getting this:&lt;br&gt;&#xA;&lt;a href=&quot;http://wierdling.net/stack-overflow-images/CombinedTest.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://wierdling.net/stack-overflow-images/CombinedTest.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stats for the first image are: Width = 4139, Height = 4535, RA = 350.1584456860353 (CRVAL1), DEC = 61.16155335032816 (CRVAL2), ORIENTAT = -125&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stats for the second image are:Width = 4139, Height = 4535, RA = 350.1159150008405 (CRVAL1), DEC = 61.19543100394401 (CRVAL2), ORIENTAT = -125&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final expected width is close to 7733 with a height near 4773.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final actual width is 4284, and the height is 7662.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any insight into what I am doing wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The full source code for the program can be downloaded from &lt;a href=&quot;https://bitbucket.org/wierdling/fitscombiner/src/master/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://bitbucket.org/wierdling/fitscombiner/src/master/&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It currently only works with WFC3 data, and the program is very much a work in progress.&lt;/p&gt;&#xA;" OwnerUserId="25394" LastEditorUserId="6" LastEditDate="2020-03-11T19:50:23.813" LastActivityDate="2020-03-11T19:50:23.813" Title="Issue translating World Coordinate system to cartesian coordinate" Tags="&lt;coordinate&gt;&lt;hubble-telescope&gt;&lt;image-processing&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29025" PostTypeId="2" ParentId="28983" CreationDate="2019-01-04T09:04:22.537" Score="1" Body="&lt;p&gt;Astigmatism. No way to eliminate in SCT. Reduce it with proper fixing of every optical element. Astigmatism is only eliminated with 3-mirror scheme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Curvature of field. No way to eliminate in SCT.&#xA;You are forced to use so-called &quot;field flatteners&quot;.&#xA;If kept in good range, does not hurt, if you are not going to do astrophotography (which with SCT nobody does).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chromatism. No way to eliminate residual chromatism from corrector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corrector blur. No way to eliminate low-quality corrector aberrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, SCT is good consumer design (together with MCT), providing nice compromise for all aberrations for astronomical observations (not astrophotography).&lt;/p&gt;&#xA;" OwnerUserId="24823" LastActivityDate="2019-01-04T09:04:22.537" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29026" PostTypeId="1" CreationDate="2019-01-04T12:22:32.123" Score="8" ViewCount="90" Body="&lt;p&gt;Meteor showers occur when Earth passes through debris shed by a comet (if I get that right). Does this mean that the entire orbit of the comet is strewn with debris? I suppose that the debris follows the same orbit as the comet core it was released from as it passed close enough to the Sun for ices of different kinds to sublimate. Is the debris that cause meteor showers somewhat evenly distributed along a huge ellipse crossing the orbits of Earth and (often) Jupiter? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are such cometary debris streams, and not only those crossing Earth's orbit, somehow cataloged and characterized?&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="7982" LastEditDate="2019-01-05T04:57:34.373" LastActivityDate="2019-06-06T17:59:46.313" Title="How uniformly is cometary debris distributed along the entire orbit of the comet?" Tags="&lt;orbit&gt;&lt;comets&gt;&lt;meteor-shower&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29027" PostTypeId="1" AcceptedAnswerId="29028" CreationDate="2019-01-04T12:25:01.537" Score="0" ViewCount="869" Body="&lt;p&gt;I'm starting my astronomy hobby and looking for a beginner telescopes to start with. I've searched couple websites and found these 2 within my budget. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know for a fact that refractors gives you sharper and clearer images but its original price is actually cheaper than the reflector telescope. I will list the specifications below. It would be really appreciated if you guys can tell me the relative benefits and drawbacks of each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Refractor&lt;/em&gt;&lt;/strong&gt;: &lt;strong&gt;&lt;em&gt;Saxon 70mm Telescope&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OPTICAL DESIGN: Refractor&lt;/p&gt;&#xA;&#xA;&lt;p&gt;APERTURE: 70mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LOWEST PRACTICAL POWER: No&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HIGHEST PRACTICAL POWER: 165x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FOCAL LENGTH: 400mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FOCAL RATIO: F/10.0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EYEPIECES: 20mm &amp;amp; 10mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FINDERSCOPE: 5x24&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DIAGONAL: 1.25&quot; 45° Erect Image Diagonal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MOUNT TYPE: Alt-Azimuth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TRIPOD: Aluminium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This telescope includes a pre-assembled tripod, two 1.25” eyepieces, a cleaning cloth and a bonus padded backpack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Reflector&lt;/em&gt;&lt;/strong&gt;: &lt;strong&gt;&lt;em&gt;Saxon 767AZ Reflector Telescope&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OPTICAL DESIGN:Newtonian Reflector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;APERTURE: 76mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LOWEST PRACTICAL POWER: No&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HIGHEST PRACTICAL POWER: 152x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FOCAL LENGTH: 700mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FOCAL RATIO: F/9.2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EYEPIECES: H25, H12.5 and SR4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FINDER-SCOPE: 5x24&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BARLOW LENS: 3x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DIAGONAL: No&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MOUNT TYPE: Alt-Azimuth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TRIPOD: Aluminium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OPTICAL TUBE DIMENSIONS: 12.5 cm x 68.5cm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OPTICAL TUBE WEIGHT: 1.76kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SHIPPING DIMENSIONS: 74 x 25 x 23cm3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SHIPPING WEIGHT: 5.4kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The accessory pack includes three 1.25&quot; eyepieces,​ a finder-scope, an erecting eyepiece for terrestrial viewing, an adjustable aluminium tripod with accessory tray and a moon filter, which reduces the brightness of the moon and makes surface details sharper and clearer.&lt;/p&gt;&#xA;" OwnerUserId="25400" LastEditorUserId="43" LastEditDate="2019-01-06T09:24:38.953" LastActivityDate="2019-01-06T09:24:38.953" Title="76mm reflector vs 70mm refractor? Both are at sale at $200" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29028" PostTypeId="2" ParentId="29027" CreationDate="2019-01-04T13:31:39.957" Score="0" Body="&lt;p&gt;Personally, I would go for neither of those. $200 is a bit steep for such a small refractor or reflector telescope (I am assuming this is USD?) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a budget of $200, I would be looking at some second hand telescopes, as you will likely be able to pick up a 90mm refractor or 130mm reflector second hand with that budget. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also depends what you want to do with it. For simple observations of planets and the moon, it will be fine, but don't expect to see much detail with such a small aperture. The optics on these aren't fantastic either so DSO's will not look great, although some of the brighter ones will be viewable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my answer would be don't get either of them, shop around second hand places and most importantly, read reviews on telescopes before buying them. If you can, join an astronomy group in your area and have a look through their scopes and see how you get on.  &lt;/p&gt;&#xA;" OwnerUserId="17096" LastActivityDate="2019-01-04T13:31:39.957" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29029" PostTypeId="1" CreationDate="2019-01-04T13:50:56.203" Score="8" ViewCount="123" Body="&lt;p&gt;It's really good to see that man (well, machine) has gone back to the moon again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chang'e 4 is studying potential mantle material from the lunar depths and will be conducting radio astronomy in frequencies that simply aren't possible from Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a huge amount to be learned that will fill some gaps in our knowledge of the early Moon, Earth, and the Solar system as well as the wider galaxy and beyond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will this information be open for the greater scientific community, or will we have to rely on the Chinese scientists to collate and publish the data as and when they can do?&lt;/p&gt;&#xA;" OwnerDisplayName="user20278" LastActivityDate="2021-03-12T02:31:24.497" Title="How open is China in sharing data from the Chang'e 4 mission?" Tags="&lt;the-moon&gt;&lt;radio-astronomy&gt;&lt;data-analysis&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29030" PostTypeId="1" CreationDate="2019-01-04T15:09:51.783" Score="1" ViewCount="132" Body="&lt;p&gt;What is the exact meaning of vicinity of Blackhole or a disk? Is this the same as inside the horizon for black holes? If so, why people don't say inside of BH, simply. What is the difference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use this phrase actually! and it may seem obvious but not to me, please somebody make it clear. I searched but I couldn't find an exact answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited: Until which radius from Blackhole I can use this word meaningfully? It's not clear to me its astronomical dimension. &quot;near&quot; is not enough to me. If it could be inside a horizon and also outside of it, it makes a huge difference from point of view of the observer at infinity, as you know, how I can call them by the same name and sometimes I mean one sometimes the other one. &lt;/p&gt;&#xA;" OwnerUserId="25404" LastEditorUserId="25404" LastEditDate="2019-01-07T12:11:32.487" LastActivityDate="2019-01-07T12:11:32.487" Title="What is the exact meaning of vicinity of Black hole or a disk? the same as inside?" Tags="&lt;black-hole&gt;&lt;space-time&gt;&lt;general-relativity&gt;&lt;accretion-discs&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="29031" PostTypeId="2" ParentId="29030" CreationDate="2019-01-04T15:21:15.720" Score="3" Body="&lt;p&gt;From the Oxford English Dictionary:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;vicinity&quot; just means &quot;the area near to or surrounding a place&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so &quot;in the vicinity of a black hole&quot; just means &quot;near to a black hole.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other dictionaries say &quot;immediately next to&quot; so in response to your comment about vagueness, treat vicinity as the immediate neighbourhood of the black hole. You will not find an exact meaning - it is used as it is in English. There is no specific distance measurement. You don't have a distance from radius which would stop being in the vicinity.&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="43" LastEditDate="2019-01-04T15:32:45.343" LastActivityDate="2019-01-04T15:32:45.343" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29033" PostTypeId="1" CreationDate="2019-01-04T17:46:57.550" Score="5" ViewCount="141" Body="&lt;p&gt;The first hint we had as to the shape of Ultima Thule was via an occultation study using an array of small telescopes. Has anyone investigated the feasibility and effectiveness of conducting a survey campaign using the same technique?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that this was managed for an arbitrarily selected object in a window of few years, it would seem that, if you just wanted to do the same thing for as many Kuiper belt objects as you can manage, that you should be able to get similar data for a significant number of objects a year at relatively low cost.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: the kind of project I'm wondering about would be targeted at getting detailed shape (e.g. what fraction of KBOs are contact binaries?) and size data. For example a line of a few hundred 0.5m class interments at 100m to km spacing.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit 2: I guess I kind of didn't actually say what the target of the survey would be. To clarify, the objective would be &lt;strong&gt;to collect fine detail on the &lt;em&gt;shape&lt;/em&gt; of KBOs&lt;/strong&gt;. E.g. how common are contact binaries?&lt;/p&gt;&#xA;" OwnerUserId="11011" LastEditorUserId="11011" LastEditDate="2019-01-24T06:36:17.593" LastActivityDate="2019-01-24T06:36:17.593" Title="Could stellar occultations be used to survey Kuiper belt objects?" Tags="&lt;asteroids&gt;&lt;kuiper-belt&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29034" PostTypeId="2" ParentId="29033" CreationDate="2019-01-04T18:44:54.107" Score="8" Body="&lt;p&gt;Yes this is possible and has been tested by the TAOS (Transneptunian Automated Occultation Survey) group among others. That group is in the process of building the &lt;a href=&quot;https://taos2.asiaa.sinica.edu.tw/&quot; rel=&quot;nofollow noreferrer&quot;&gt;TAOS-II&lt;/a&gt; project which will use several telescopes and fast readout CMOS cameras (since the occultations only last a few seconds) to search for occultations caused by unknown Kuiper Belt Objects. Multiple telescopes are very useful to build confidence that you have a real event and not atmospheric fluctuations or other instrumental effect. A detailed workthrough and calculations of how many objects you would see, what is needed to detect them and resolve the Fresnel diffraction and other issues is given in a section by Roques et al. in the &lt;em&gt;Solar System Beyond Neptune&lt;/em&gt; book and which is available &lt;a href=&quot;https://www.lpi.usra.edu/books/ssbn2008/7022.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general for surveys such as occultations by unknown KBOs, it is &lt;em&gt;etendue&lt;/em&gt; (product of telescope collecting area and amount of sky covered in a single exposure) that matters (but it is often more complicated as survey depth doesn't just scale with telescope area) in that higher etendue value equals &quot;better&quot; survey. So for the camera(s) in a survey, it is sensor &lt;em&gt;area&lt;/em&gt; that matters, which is a product of pixel size and no. of pixels (and number of sensors and size of gaps between if you are tiling them). The telescope area/diameter tends to be set by how bright your targets of your survey are and practicalities such as cost and manufacturability tend to influence the f/ratio and therefore focal length and pixel scale at your detector. Rarity of your targets and how many you want to find, drive how much area you need to cover. (The telescope area also influences this since in general you get more targets per area if you use a bigger telescope and go fainter). So you end up trading all of these things off to get your best survey for your particular science and type of targets, normally constrained by available or potential budget.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of trying to time-resolve occultations by KBOs, your exposure time is set by the event timescale which depends on the velocity of the KBO across the line of sight and the diameter of the KBO. For typical KBO velocities of 20 km/s, this means that a Pluto (approx 2000km diameter) occultation lasts &lt;span class=&quot;math-container&quot;&gt;$2000\,km/20\,  kms^{-1}=100\,s$&lt;/span&gt;. If you wanted to find small 20 km KBOs similar to 2014 MU69/Ultima Thule, then the events are &lt;span class=&quot;math-container&quot;&gt;$20\,km/20\,  kms^{-1}=1\,s$&lt;/span&gt; long (both of these calculations assume the occultation goes over the largest part of the KBO and isn't grazing). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that you will want several points during the occultation, both to convince yourself it's real and to measure diameters etc, you will need at least 5 points across the event. For the 20km KBOs/1s events, this means at least 5 frames per second, ideally more. This means that you need to get enough signal-to-noise on your target stars in a 0.2 or 0.1s exposure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As discussed in the reply comments to the question author, scientific CMOS sensors such as the &lt;a href=&quot;https://andor.oxinst.com/products/scmos-camera-series/marana-scmos&quot; rel=&quot;nofollow noreferrer&quot;&gt;Andor Marana&lt;/a&gt; (a commercially available closeish match to the custom detectors used by TAOS-II with ~$31k pricetag; Jan 2019) will be better at both the 'signal' part (due to higher quantum efficiency) and the 'noise' parts (larger, less noisy pixels, cooling and temperature stabilization) than a consumer DSLR sensor. Looking at a study of &lt;a href=&quot;https://www.photonics.com/Articles/Dark_Current_in_Consumer_Cameras_Not_What_you/a44298&quot; rel=&quot;nofollow noreferrer&quot;&gt;DSLR dark current&lt;/a&gt; by Photonics.com, the dark current is ~50 counts/pixel/second (the Photonics.com figure is for 30s exposures and assuming 1500DN) rising to ~100 counts/pixel/second (averaging at ~3000DN for 3 of the pixels) after 100 frames as the sensor heats. For the Marana, the figure is 0.26 counts/pixel/second (taking datasheet value of 0.2 electrons/pix/s and assuming a gain of 1.3 to match the full well capacity of 85k electrons to the 16 bit ADC (65536 levels)), an almost 400x lower value. (The ADC data range is another area of difference; the Marana can do 16bit digitization whereas DSLRs are typically 12 or 14bit, giving only 4096 or 16,384 levels to record all the signal levels in the image)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other use of occultations is to monitor predicted occultations by &lt;em&gt;known&lt;/em&gt; asteroids and Kuiper Belt Objects. This can be used to get a better idea on the size and shape of the object (as you note for Ultima Thule/2014 MU69), search for additional satellites or rings around the object (as has been done for the KBO Haumea (&lt;a href=&quot;https://news.nationalgeographic.com/2017/10/rings-found-dwarf-planet-haumea-space-science/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nat Geo article&lt;/a&gt;) and the Centaur Chariklo (&lt;a href=&quot;https://www.space.com/39261-ringed-asteroid-chariklo-giant-planets-secrets.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Space.com article&lt;/a&gt;)) and look for and monitor changes in the atmosphere (as has been done for Pluto since the first occultation in 1988).&lt;/p&gt;&#xA;" OwnerUserId="23748" LastEditorUserId="23748" LastEditDate="2019-01-18T22:48:03.563" LastActivityDate="2019-01-18T22:48:03.563" CommentCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="29036" PostTypeId="1" AcceptedAnswerId="29042" CreationDate="2019-01-04T20:37:41.017" Score="2" ViewCount="118" Body="&lt;p&gt;If the photosphere was blocked, before the sun rises on an airless moon in the outer solar system, would either the inner or outer corona be visible to the naked eye?  &lt;/p&gt;&#xA;" OwnerUserId="25088" LastActivityDate="2019-01-06T15:22:29.747" Title="Would the solar corona be visible from the distance of the outer planets?" Tags="&lt;the-sun&gt;&lt;naked-eye&gt;&lt;corona&gt;&lt;outer-solar-system&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29037" PostTypeId="5" CreationDate="2019-01-04T21:23:58.147" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2019-01-04T21:23:58.147" LastActivityDate="2019-01-04T21:23:58.147" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29038" PostTypeId="4" CreationDate="2019-01-04T21:23:58.147" Score="0" Body="For questions about the observation of the sky through binoculars." OwnerUserId="24758" LastEditorUserId="24758" LastEditDate="2019-01-05T01:32:09.620" LastActivityDate="2019-01-05T01:32:09.620" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29040" PostTypeId="5" CreationDate="2019-01-05T02:23:03.570" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2019-01-05T02:23:03.570" LastActivityDate="2019-01-05T02:23:03.570" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29041" PostTypeId="4" CreationDate="2019-01-05T02:23:03.570" Score="0" Body="The corona is the large field of plasma surrounding a star, normally only visible when the star is fully eclipsed by another heavenly object. For questions about the sun or any other star's corona. " OwnerUserId="24758" LastEditorUserId="24758" LastEditDate="2019-01-07T06:45:45.760" LastActivityDate="2019-01-07T06:45:45.760" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29042" PostTypeId="2" ParentId="29036" CreationDate="2019-01-05T02:32:15.807" Score="2" Body="&lt;p&gt;If the sun is ever perfectly eclipsed as it is on Earth, then yes, the corona would be visible. The only reason we can't usually see the corona is because it is &lt;em&gt;way&lt;/em&gt; dimmer than the rest of the sun. If the entire sun other than the corona is ever covered fully by another object, then the corona will be visible.&lt;/p&gt;&#xA;" OwnerUserId="24758" LastActivityDate="2019-01-05T02:32:15.807" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29043" PostTypeId="1" CreationDate="2019-01-05T04:53:39.650" Score="1" ViewCount="37" Body="&lt;p&gt;@astrosnapper's &lt;a href=&quot;https://astronomy.stackexchange.com/a/29034/7982&quot;&gt;answer&lt;/a&gt; to the question &lt;em&gt;Could stellar occultations be used to survey Kuiper belt objects?&lt;/em&gt; mentions the TAOS-II project and asteroid occultation measurements (which occur on the timescale of seconds) and that got me thinking about the term Time Domain Astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick search led me to an Astrobites post with a catchy title &lt;a href=&quot;https://astrobites.org/2012/11/12/a-big-step-backward-for-time-domain-astronomy/&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Big Step Backward for Time Domain Astronomy&lt;/a&gt; that (among other things) describes the DASCH project (Digital Access to a Sky Century @ Harvard).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The post is dated 2012 and says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The project is working to (1) digitally scan all of the plates, (2) detect sources on the plates, (3) assign physical coordinates to the plates, and (4) measure/calibrate the magnitudes of all sources.  The end product will be a publicly available database which can be queried to produce the 100 year light curve for individual objects.  As a part of the project the team had to develop a specialized scanner, as commercially available units would have been too slow to finish the project in anything resembling a reasonable amount of time. As it is, when operating at full efficiency, the project can now scan and analyze 400 plates a day.  As of the writing of this article, 38,346 plates have been scanned (~8% completion; see Figure 2 for a map of their current coverage) and over 2.7 billion magnitudes have already been measured.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The article ends with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...Given that we’re only 8% of the way through analyzing the plates, &lt;strong&gt;we can look forward to further interesting discoveries from this survey in the next few years.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is now six years later. Presumably the scan rate has accelerated a bit and a much larger fraction of the plates have been scanned, and techniques for reducing and analyzing the huge amounts of data have improved as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since 2012, what new discoveries have been made from this effort? There may be too many to list, so a few highlights and a link to a summary might be the best way to answer.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-01-05T05:02:18.847" LastActivityDate="2019-01-05T12:22:33.050" Title="What discoveries have come from the DASCH project (Digital Access to a Sky Century @ Harvard)" Tags="&lt;data-analysis&gt;&lt;sky-survey&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29044" PostTypeId="1" AcceptedAnswerId="29045" CreationDate="2019-01-05T07:05:31.890" Score="0" ViewCount="69" Body="&lt;p&gt;Some days ago i read about &lt;a href=&quot;http://www.astronomy.com/news/2018/10/china-artificial-moons&quot; rel=&quot;nofollow noreferrer&quot;&gt;China's artificial moon mission for lighting cities&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my junior classes i was taught the effect of Moon's gravity on tidal waves.&#xA;&lt;br&gt;According to me this is a very dangerous mission this can cause floods due to additional gravitational force.&#xA;&lt;br&gt;Am I wrong?Please explain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I had searched on Internet about this defect of China's artificial moon for lighting but i didn't found any result.Many astronomers are against this mission as this will disturb natural timing of flora and fauna.But no one has raised the point ,that it will disturb natural motion of tidal waves.&#xA;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="25331" LastEditorUserId="25331" LastEditDate="2019-01-05T11:51:03.233" LastActivityDate="2019-01-05T13:33:46.117" Title="Objection on artificial moon for lighting" Tags="&lt;the-moon&gt;&lt;artificial-satellite&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29045" PostTypeId="2" ParentId="29044" CreationDate="2019-01-05T08:13:31.287" Score="3" Body="&lt;p&gt;This has been called &lt;a href=&quot;https://www.linkedin.com/pulse/chengdu-fake-moon-news-eric-bretschneider/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;fake news&quot;&lt;/a&gt; However tides are caused by gravity, and the proposal doesn't seem to require very massive satellites, hence no tidal waves.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-01-05T08:13:31.287" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29047" PostTypeId="2" ParentId="10447" CreationDate="2019-01-05T10:19:21.680" Score="5" Body="&lt;p&gt;The detection of Alpha Centauri Bb unfortunately did not survive follow-up analysis and it is no longer thought to exist, see &lt;a href=&quot;http://adsabs.harvard.edu/abs/2016MNRAS.456L...6R&quot; rel=&quot;noreferrer&quot;&gt;Rajpaul et al. (2016)&lt;/a&gt;. This planet candidate was in any case too close to the star to support liquid water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There has been a claim of a single transit of a longer-period planet orbiting Alpha Centauri B, although this has not been confirmed. The relevant paper is &lt;a href=&quot;http://adsabs.harvard.edu/abs/2015MNRAS.450.2043D&quot; rel=&quot;noreferrer&quot;&gt;Demory et al. (2015)&lt;/a&gt;. Once again, this planet candidate was inferred to be too close to the star for liquid water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently it has been confirmed that Proxima Centauri is a bound member of the Alpha Centauri system (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2017A%26A...598L...7K&quot; rel=&quot;noreferrer&quot;&gt;Kervella et al., 2017&lt;/a&gt;). The planet Proxima b has been found in the liquid water zone (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2016Natur.536..437A&quot; rel=&quot;noreferrer&quot;&gt;Anglada-Escudé et al., 2016&lt;/a&gt;) and has a minimum mass of 1.27 Earth masses, so if the orbit is close to being edge-on it is likely to be a terrestrial planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, just being in the liquid water zone is not sufficient for habitability (obvious counterexample: our Moon). A major issue is that Proxima Centauri is a flare star and the space weather environment experienced by Proxima b is extremely hostile. This brings into question whether or not Proxima b can maintain an atmosphere under such conditions, for example see &lt;a href=&quot;http://adsabs.harvard.edu/abs/2017ApJ...844L..13G&quot; rel=&quot;noreferrer&quot;&gt;Garcia-Sage et al. (2017)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the issue that low-mass stars like Proxima Centauri spend billions of years in a luminous pre-main sequence stage of their evolution. This would likely trigger a runaway greenhouse on planets in the main sequence liquid water zone and result in the loss of significant amounts of water from the planet, as discussed in &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014ApJ...797L..25R&quot; rel=&quot;noreferrer&quot;&gt;Ramirez &amp;amp; Kaltenegger (2014)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while Proxima b appears to meet some requirements for being able to support life, in several other ways it appears to be far less than ideal and we should not be too surprised if it turns out to be non-habitable.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastActivityDate="2019-01-05T10:19:21.680" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29049" PostTypeId="2" ParentId="29043" CreationDate="2019-01-05T11:58:31.867" Score="1" Body="&lt;p&gt;The most notable one I'm aware of is the detection of the secular dimming of Boyajian's Star (KIC 8462852), as described in &lt;a href=&quot;https://ui.adsabs.harvard.edu/#abs/2016ApJ...822L..34S/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schaefer (2016)&lt;/a&gt;. Follow-up by &lt;a href=&quot;https://ui.adsabs.harvard.edu/#abs/2018MNRAS.481.2235S/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schaefer et al. (2018)&lt;/a&gt; suggest that the ratios of the dimming across the BVRI bands are consistent with dust, though the reason for the lack of detectable infrared excess remains unknown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DASCH data has also been used to constrain the orbital period of the ringed object transiting J1407, see &lt;a href=&quot;https://ui.adsabs.harvard.edu/#abs/2018A&amp;amp;A...619A.157M/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mentel et al. (2018)&lt;/a&gt;. The DASCH plates did not show any previous transit in the data, ruling out a large fraction of orbital periods and opening the possibility that the ringed object J1407b may not be gravitationally bound to J1407.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-01-05T12:22:33.050" LastActivityDate="2019-01-05T12:22:33.050" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29050" PostTypeId="1" CreationDate="2019-01-05T12:43:00.440" Score="0" ViewCount="135" Body="&lt;p&gt;An astronomer, T observed the spectrum of a distant star X. The spectral lines were observed to have tiny blueshifts and redshifts regularly with a period of 1.27 years. T concluded that the star is wobbling slightly due to a planet orbiting it. Let the estimated mass of the planet when                                                     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) eccentricity of the orbit e = 0 be k*10^23 kg&lt;br&gt;&#xA;(b) and for e = 0.6 be x*10^22 kg.&lt;br&gt;&#xA;Assume that there is only one major planet and the axis of rotation of the system is perpendicular to the line of sight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass of Star = 23.86163 x 10^30 kg &#xA;Time Period = 463.55 days &#xA;Shift in wavelength = 1.7627 x 10^-5 nm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to find out 'k' and 'x'                       &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I studied about the Doppler Spectroscopy method.&#xA;I know that the relation&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;K = \left(\dfrac{2\pi G}{P_{orb}}\right) ^{1/3}\dfrac{M_p}{(M_{*}+M_p)^{2/3}}\dfrac{1}{\sqrt{1-e^2}}&#xA;$$&lt;/span&gt;&#xA;holds where K = radial velocity half amplitude.&#xA;How can I use the doppler shift given into this?&#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="25418" LastEditorUserId="25418" LastEditDate="2019-01-05T12:51:22.153" LastActivityDate="2020-03-26T14:02:10.227" Title="Doppler Spectroscopy - Finding mass of an exoplanet" Tags="&lt;exoplanet&gt;&lt;mass&gt;&lt;spectroscopy&gt;&lt;doppler-effect&gt;&lt;radial-velocity&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29052" PostTypeId="1" AcceptedAnswerId="29059" CreationDate="2019-01-05T12:57:39.287" Score="2" ViewCount="700" Body="&lt;p&gt;A friend has a vintage telescope with a broken lens, I asked him to send me its catalog so I can see if I can buy him a new one, he gave me what looks like a Russian catalog, it seems a soviet telescope, I can't read it, I hope you could tell me what it is. I want to search for its lens using English characters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8yQQH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8yQQH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/QOBh8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QOBh8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/YQb9L.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YQb9L.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/wY8Wz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wY8Wz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/8Vw7B.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8Vw7B.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5485" LastEditorUserId="5485" LastEditDate="2019-01-06T14:39:19.500" LastActivityDate="2019-09-16T15:35:08.693" Title="What soviet telescope is this?" Tags="&lt;telescope&gt;&lt;telescope-lens&gt;&lt;identify-this-object&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29053" PostTypeId="2" ParentId="29044" CreationDate="2019-01-05T13:33:46.117" Score="3" Body="&lt;p&gt;The &quot;Artificial Moon&quot; which is being proposed is not nearly as close a copy of the real moon as your concerns suggest. It will be smaller, closer and probably hollow, whereas the real moon is a rather large solid ball of rock quite a long way away. In particular the artificial moon would not be massive enough to exert significant gravity.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-01-05T13:33:46.117" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29056" PostTypeId="2" ParentId="29036" CreationDate="2019-01-05T16:15:06.193" Score="2" Body="&lt;p&gt;When the sun is about to rise for a given location is disk or photosphere is totally covered by the planet. The situation is similar to that of a total eclipse but we can't see the corona rising first because of two and at least overlapping reasons :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the corona image undergo multiple diffusion and reflection and is not bright enough to contrast with the skydome which is already illuminated by the same; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;when the sun further approach a &quot;geometrical sunrise&quot; refraction makes it already visible so its light is overwhelming the corona anyway&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7fcza.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7fcza.jpg&quot; alt=&quot;Picture from Wikipedia&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Atmospheric_refraction&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Atmospheric_refraction&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corona is somehow hidden in the twilight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume the star not too far or its corona bright enough then on a planet without atmosphere the latter can be indeed seen rising. The conditions for its observation would be even better that those on earth during a total eclipse as for the skydome in this terrestrial case still receive light from the penumbra and illuminated nearby regions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt; after I had a look at this pic that I have taken just yesterday and I realised it is related to the topic here. It makes me wonder about the nature of the observation &lt;em&gt;without altering the answer above&lt;/em&gt; when a planet without atmosphere is concerned. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/196dA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/196dA.jpg&quot; alt=&quot;Bright spot before the appearance of the photosphere&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bright spot was deep red to the naked eyes. And it was there minutes before the photosphere made its appearance!  Its red tint is well explained as usual by diffusion but its brightness confuses me a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Might it be indeed a corona picture &lt;em&gt;superimposed&lt;/em&gt; to the twilight of the morning?&#xA;Or is just an ill-defined image of the disk originating as in the scheme above but with an even lower (as compared to horizon) Sun? Probably this is the explanation but I wonder to observe always something new. It is not the first time that I see a sunrise :) but I never observed a sun-like bright spot lasting so long before the actual photosphere rising. &lt;/p&gt;&#xA;" OwnerUserId="17437" LastEditorUserId="17437" LastEditDate="2019-01-06T15:22:29.747" LastActivityDate="2019-01-06T15:22:29.747" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29057" PostTypeId="1" CreationDate="2019-01-05T20:11:24.850" Score="9" ViewCount="203" Body="&lt;p&gt;The two component parts of the Ultima Thule contact binary are approximately spherical. Why have the same forces that made the two parts spherical when they were separate not made the whole thing spherical after they joined?&lt;/p&gt;&#xA;" OwnerUserId="13125" LastActivityDate="2019-01-09T01:24:44.567" Title="Why isn't Ultima Thule spherical, while its two component parts are" Tags="&lt;planetary-formation&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29059" PostTypeId="2" ParentId="29052" CreationDate="2019-01-05T23:37:25.367" Score="3" Body="&lt;p&gt;Possible search terms to use:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Vintage Russian CCCP Spotting Scope 3PT-460 20x&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Komz Russian Military Telescope - 3PT-460 20x&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;ZRT 460 Russian Scope&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Images of item:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/66IBH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/66IBH.jpg&quot; alt=&quot;Picture of 3PT-460 - Image 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ytNls.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ytNls.jpg&quot; alt=&quot;Picture of 3PT-460 - Image 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LwaER.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LwaER.jpg&quot; alt=&quot;Picture of 3PT-460 - Image 3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prices I found range from \$50 to less than \$100.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastActivityDate="2019-01-05T23:37:25.367" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29060" PostTypeId="2" ParentId="29057" CreationDate="2019-01-06T01:43:04.697" Score="4" Body="&lt;p&gt;The explanation is that as Ultima and Thule were forming, they collected out of small hunks of debris and piles of accumulated rubble.  The key thing is that once U and T became the biggest piles of rubble around, any new rock or pile of rubble that gently approached them would be much smaller then U and T and hit and gently bounce a bit and, on average, settles down in a low spot. The collisions are mostly pretty gentle and even rubble heaps probably mostly stay stuck together because the force of gravity isn't enough to overcome the cohesion of the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;U and T grow and are -- roughly -- spherical since the things colliding with them and sticking to them are much smaller than they are, there are lots and lots of them, and they have a small tendency to end up in low spots. So a rough sphere emerges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When U and T finally collide to produce Ultima Thule, it's another gentle collision, and the same thing happens: Gravitational forces are too small to overcome cohesion, so U and T stay pretty much the shape they were in and you have a body which is two spheres stuck together at the point of collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;U and T formed spherical because they formed out of many much smaller pieces which are probably still pretty distinct, but which are too small to see -- yet.  UT formed out of two big pieces which are also still pretty distinct, but much easier to see..&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2019-01-06T01:43:04.697" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29061" PostTypeId="1" AcceptedAnswerId="29069" CreationDate="2019-01-06T04:29:41.800" Score="9" ViewCount="1025" Body="&lt;p&gt;The theory of the expanding universe is so widely accepted, that the redshift is sometimes used as a measure of distance to far away galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is it still possible that the redshift is caused by some unknown phenomena and not by galaxies moving away from each other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any other proof (apart from the redshift) that the universe is indeed expanding and far away galaxies are moving away from us?&lt;/p&gt;&#xA;" OwnerUserId="23720" LastEditorUserId="24184" LastEditDate="2019-01-07T07:08:05.153" LastActivityDate="2019-01-11T08:16:49.533" Title="Are there other proofs of the expanding universe apart from the redshift?" Tags="&lt;cosmology&gt;&lt;expansion&gt;&lt;redshift&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29062" PostTypeId="2" ParentId="29061" CreationDate="2019-01-06T08:28:34.617" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But is it still possible that the redshift is caused by some unknown phenomena and not by galaxies moving away from each other?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In history some alternative theories were proposed, like the &lt;a href=&quot;https://en.wikipedia.org/wiki/Tired_light&quot; rel=&quot;nofollow noreferrer&quot;&gt;tired light hypothesis&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Steady_state_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;the steady state universe&lt;/a&gt; etc. But the observation &lt;a href=&quot;http://www.sciencemag.org/news/2001/06/tired-light-hypothesis-gets-re-tired&quot; rel=&quot;nofollow noreferrer&quot;&gt;ruled these and other theories out&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;https://rationalwiki.org/wiki/Alternative_cosmology&quot; rel=&quot;nofollow noreferrer&quot;&gt;Alternative cosmology&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25424" LastEditorDisplayName="user1569" LastEditDate="2019-01-08T12:57:14.857" LastActivityDate="2019-01-08T12:57:14.857" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29064" PostTypeId="1" AcceptedAnswerId="31859" CreationDate="2019-01-06T23:30:08.297" Score="3" ViewCount="206" Body="&lt;p&gt;If one was orbiting in the Jovian system, as depicted in this artist conception,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2FufF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2FufF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;could the aurora be seen unassisted? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hubblesite.org/image/3501/news_release/2015-09&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubblesite-Illustration Credit: NASA, ESA, and G. Bacon (STScI)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25088" LastEditorUserId="25088" LastEditDate="2019-01-07T02:22:17.587" LastActivityDate="2019-05-14T15:51:02.387" Title="Could Ganymede's aurora be seen with the naked eye?" Tags="&lt;hubble-telescope&gt;&lt;naked-eye&gt;&lt;aurora&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29065" PostTypeId="2" ParentId="28289" CreationDate="2019-01-06T23:39:14.323" Score="2" Body="&lt;p&gt;There have been proposals to capture a small (boulder-sized) asteroid into Earth orbit (more or less) as several other answers have discussed. This needs a lot of planning, a big rocket and a great deal of patience, but all of the asteroids being considered already orbit the Sun, and generally are among the relatively small number of asteroids that periodically come close to Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Capturing an extra-solar object like Oumuamua would be &lt;em&gt;much&lt;/em&gt; harder. Firstly we'd need to detect it a long time  (decades, ideally) before it gets anywhere near the Sun, which means spotting something that is far away, very dimly illuminated, and coming from a completely unpredictable direction.  We'd want to get to it as quickly as possible, break a lump off if it was too big to redirect as a whole, and then use something, probably nuclear explosives to change the course of the lump enough for it to come close to one of the giant planets, ideally Jupiter. That encounter would have to be arranged so that the planet's gravity deflects our (now slightly radioactive) lump of rock into a closed orbit around the Sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once that is done, the time pressure is off, and we can use less drastic measures (rockets, light-sails,...), over possibly many decades, to steer our target lump, via whatever gravitational slingshots seem useful until it encounters the Earth and Moon at a fairly low relative velocity. Then a combination of more rockets and the Moon's gravity and maybe aero-braking in the upper atmosphere can be used (over several passes if necessary) to actually capture it.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-01-06T23:39:14.323" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29067" PostTypeId="1" CreationDate="2019-01-07T03:31:33.210" Score="4" ViewCount="231" Body="&lt;p&gt;&lt;em&gt;In the early 2010's&lt;/em&gt; there were many popular articles that said that the total bandwidth of the the Square Kilometer Array's optical fiber network for raw data from each dish would be larger than the Earth's internet. Here is just one example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Square_Kilometre_Array&quot; rel=&quot;noreferrer&quot;&gt;Square Kilometre Array&#xA;&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It would require very high performance central computing engines and long-haul links with a capacity greater than the global Internet traffic as of 2013.&lt;a href=&quot;https://www.skatelescope.org/wp-content/uploads/2011/03/SKA-Brochure.pdf&quot; rel=&quot;noreferrer&quot;&gt;3&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you can see that link is broken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The SKA project has several &lt;a href=&quot;https://en.wikipedia.org/wiki/Square_Kilometre_Array#Precursors,_pathfinders_and_design_studies&quot; rel=&quot;noreferrer&quot;&gt;Precursors, pathfinders and design studies&lt;/a&gt; but the SKA proper has not been built yet, and perhaps is not even finalized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, with available data, by the time it might be built (2025?) is it likely to still exceed the Earth's internet's bandwidth?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-01-15T19:55:09.990" Title="Will the SKA's total bandwidth still exceed the Earth's internet's bandwidth?" Tags="&lt;data-analysis&gt;&lt;radio-telescope&gt;&lt;raw-data&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29069" PostTypeId="2" ParentId="29061" CreationDate="2019-01-07T06:34:50.157" Score="9" Body="&lt;p&gt;Yes, there is direct, non-red-shift evidence of expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The past temperature of the Cosmic Microwave Background Radiation (CMBR) has been directly measured and found to be substantially higher than it is today.  Its reduction in temperature over time is direct evidence of expansion.  Here are the details:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://arxiv.org/pdf/astro-ph/0012222.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;, the CMBR was measurably hotter in the past (&lt;a href=&quot;https://www.eso.org/public/news/eso0043/&quot; rel=&quot;nofollow noreferrer&quot;&gt;less-technical synopsis here&lt;/a&gt;).  The researchers observed absorption lines in a gas cloud located in a distant galaxy and found that the pattern of lines seen could only be explained if the CMBR temperature at time of absorption was between 6&amp;nbsp;K  and 14&amp;nbsp;K (it's 3&amp;nbsp;K now).  This temperature is consistent with the expected temperature for that galaxy's redshift (9&amp;nbsp;K).  Note that the temperature was measured from the specific pattern of lines seen and not from how much the lines had been red-shifted; this measurement would yield the same temperature even if there were no red-shifting.  Since a hotter temperature implies a higher density, this cooling of the CMBR over time is direct evidence for expansion of the universe.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Additional comments&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;What is the relationship between red-shift and absorption lines?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inspired by a conversation with uhoh in the comments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my answer I refer to a &quot;pattern&quot; of &quot;absorption lines&quot;.  For those not versed in the topic allow me to explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a light shines through a cloud of gas specific frequencies of light get absorbed.  When this light is then shone through a prism the blocked frequencies will appear as black lines in the spectrum (see illustration below).  The exact lines that appear and their positions in the spectrum (the &quot;pattern&quot; of &quot;absorption lines&quot;) depends on the elements present in the gas and the gas's environment.  The effect is most clearly seen with a light that emits photons at all frequencies; this kind of light is known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Black-body_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;black-body radiation&lt;/a&gt;.  Although emitting light at all frequencies, a black-body radiator will emit the most light at a particular wavelength; the location of this peak is referred to as the black-body's temperature.  The CMBR discussed in the question is an example of black-body radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/p5mCk.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p5mCk.gif&quot; alt=&quot;Red shift over cosmological distances&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sup&gt;Source: &lt;a href=&quot;http://www.astro.ucla.edu/~wright/doppler.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Doppler Shift&lt;/a&gt;, Edward L. Wright&lt;br&gt;&#xA;(Excellent site BTW, the &lt;a href=&quot;http://www.astro.ucla.edu/~wright/cosmology_faq.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;FAQ&lt;/a&gt; is worth a look for more info on red shifts and cosmology in general)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As light is travels through (expanding) space it's wavelength &lt;em&gt;and the wavelengths of the absorption lines&lt;/em&gt; stretches at a fixed rate for all frequencies.  Let's say that at time of emission/absorption a spectrum shows lines at wavelengths of 1, 3, and 5&amp;nbsp;nm&lt;sup&gt;1&lt;/sup&gt;.  After the photons have traveled for a certain amount of time, all of the spectrum's wavelengths will appear to have doubled&lt;sup&gt;2&lt;/sup&gt;.  The line formerly at 1&amp;nbsp;nm is now seen at 2&amp;nbsp;nm, the one formerly at 3&amp;nbsp;nm is now seen at 6&amp;nbsp;nm, and the one originally at 5&amp;nbsp;nm is now seen at 10&amp;nbsp;nm.  Though their absolute frequencies change over time, the ratio of the lines' wavelengths (and frequencies) relative to one-another remains constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The precise amount that a given object's spectrum is shifted directly correlates with its distance.  As seen in the diagram above, close objects (like the Sun) show no red-shifting.  As one looks at objects further and further away one sees increasing amounts of red shift&lt;sup&gt;3&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the discussion in the answer above, it's this &lt;em&gt;pattern&lt;/em&gt; of relative positions in the lines that is affected by the CMBR temperature at time of absorption and not the degree to which the lines have been shifted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;sup&gt;1&lt;/sup&gt; To put it technically, this point is at &lt;span class=&quot;math-container&quot;&gt;$z=0$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; indicates the magnitude of the shift, positive for red shifts (moving away) and negative for blue shifts (approaching).  A more in-depth discussion of this topic (including the precise definition of &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;) can be &lt;a href=&quot;http://www.astro.ucla.edu/~wright/doppler.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;found here&lt;/a&gt;.&lt;/sup&gt;&lt;br&gt;&#xA;&lt;sup&gt;&lt;sup&gt;2&lt;/sup&gt; The wavelength-doubling (frequency-halving) point is at &lt;span class=&quot;math-container&quot;&gt;$z=1$&lt;/span&gt;&lt;/sup&gt;&lt;br&gt;&#xA;&lt;sup&gt;&lt;sup&gt;3&lt;/sup&gt; It should be noted that since there is some uncertainty in the rate that the Universe is expanding at, red-shifts do not refer to precisely-known distances.  Thus astronomers and cosmologists rarely refer to the distances to distant objects in absolute terms of, say, light-years or &lt;a href=&quot;https://en.wikipedia.org/wiki/Parsec&quot; rel=&quot;nofollow noreferrer&quot;&gt;parsecs&lt;/a&gt;, preferring, rather, to use the amount of red-shift observed (the &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; mentioned above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism behind red shifting isn't that the photons themselves are changing rather it is that the very space that the electromagnetic waves are moving through is expanding.  (Photons are both particles and waves; no, it's not exactly intuitive.)  This constant stretching of space stretches the light's wavelength giving rise both to the effect of red-shifting and the increase in a given photon's red shift over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KMtAF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CTGYS.png&quot; alt=&quot;Light is a PwARaTIvCLeE!&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;Douglas Hofstadter, CC A-SA 3.0&lt;/sup&gt;&lt;br&gt;&#xA;&amp;nbsp;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;How does red-shift relate to the CMBR?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the comments Alchimista asked &quot;Isn't CMBR actually the quintessence of red-shift anyway?&quot;&lt;br&gt;&#xA;(I'm assuming you're using the common, and not &lt;a href=&quot;https://en.wikipedia.org/wiki/Quintessence_(physics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;cosmological&lt;/a&gt;, meaning of &quot;quintessence&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, the current CMBR temperature (3&amp;nbsp;K) is generally agreed to be the result of relatively high-energy photons (3000&amp;nbsp;K) emitted about 380,000 years after the Big Bang that have had their wavelengths stretched over time by the expansion of the Universe towards the red (i.e. cooler or lower-energy) end of the spectrum.  This expansion was inferred by Hubble &lt;em&gt;et al.&lt;/em&gt; from the observation that smaller and dimmer galaxies (as seen from Earth) have a greater shift in their spectra.  The farther the apparent distance, the greater the observed shift.  Using this apparent distance-correlated red shift we can &lt;em&gt;infer&lt;/em&gt; that the Universe was smaller in the past and thus denser with a higher temperature for the CMBR.  Based on observed red shifts of distant galaxies, we can then deduce, but not directly measure, what the CMBR temperature was at each distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the authors of the above paper did was make a &lt;strong&gt;direct&lt;/strong&gt; measurement of the CMBR's temperature at a specific time in the past.  The measured temperature is higher than it is today which implies a denser and thus smaller Universe.  The researchers further found that the directly-measured temperature fit neatly with that inferred from the observed red shift of the galaxy being studied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a nutshell, the chain of inference is swapped:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For reasoning based on red-shifting:&lt;br&gt;&#xA;Increasing red shifts with apparent distance (directly measured) ⇒ Expansion ⇒ Denser Universe in the past ⇒ Higher CMBR temperature in the past.&lt;/li&gt;&#xA;&lt;li&gt;For a direct measurement of past temperature (as with this paper):&lt;br&gt;&#xA;Higher CMBR temperature in the past (directly measured) ⇒ Denser Universe in the past ⇒ Expansion ⇒ Observed red shift.&lt;br&gt;&#xA;&amp;nbsp;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These two inference chains based on different sets of evidence neatly complement and support one-another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing to note is that the CMBR wasn't &lt;em&gt;created&lt;/em&gt; by expansion (at least not directly) rather it's expansion that explains its current temperature and uniformity.  Per the Big Bang theory, the early universe was very dense; so dense and hot that all matter was a plasma of subatomic particles, opaque to photons. At about 380,000 years after the Big Bang the Universe had cooled (through expansion) enough that protons and electrons could combine to form neutral Hydrogen gas (which is transparent). The CMBR is the light that was set free at this time and has been cooling ever since. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="24184" LastEditorUserId="24184" LastEditDate="2019-01-09T04:01:28.507" LastActivityDate="2019-01-09T04:01:28.507" CommentCount="26" ContentLicense="CC BY-SA 4.0" />
  <row Id="29071" PostTypeId="2" ParentId="29061" CreationDate="2019-01-07T07:15:15.387" Score="1" Body="&lt;p&gt;Yes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Distribution of 1a supernova data&lt;/li&gt;&#xA;&lt;li&gt;WMAP measurements of the CMB&lt;/li&gt;&#xA;&lt;li&gt;Sloan galactic sky survey (catalog of the galaxies)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The important thing is that these results not only say the same, but &lt;em&gt;they are also corresponding eachother&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastActivityDate="2019-01-07T07:15:15.387" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29072" PostTypeId="1" AcceptedAnswerId="29075" CreationDate="2019-01-07T08:39:46.280" Score="7" ViewCount="255" Body="&lt;p&gt;In the picture sequence which was taken by New Horizons during the approach to Ultima Thule on the 3rd of Dec there can be seen a speck of light in the lower left corner, not as bright as 2014 MU69 but clearly visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WxKQ0.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WxKQ0.gif&quot; alt=&quot;Animation of LORRI images taken during UT approach&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;Source: &lt;a href=&quot;https://www.heise.de/newsticker/meldung/NASA-Sonde-New-Horizons-2014-MU69-gibt-vor-historischem-Vorbeiflug-Raetsel-auf-4257960.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here (German)&lt;/a&gt; or &lt;a href=&quot;https://blogs.nasa.gov/pluto/2018/12/20/the-pis-perspective-on-final-approach-to-ultima/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here (English)&lt;/a&gt;, NASA / Johns Hopkins Applied Physics Laboratory / Southwest Research Laboratory / Henry Throop&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any theory what may have caused this signal? A particle from the probe itself, coincidentally passing in front of the lens? A variable star? A small object in the right distance and angle to be glimpsed upon by New Horizons? I think the latter would be an astronomically improbable event...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some single frames of that image sequence, the speck is clearly visible on the 3rd of December.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OsZ3E.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OsZ3E.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="844" LastEditorUserId="24184" LastEditDate="2019-01-09T03:56:10.760" LastActivityDate="2019-01-09T03:56:10.760" Title="Strange speck of light in the picture sequence of the approach to 2014 Mu69 (Ultima Thule)" Tags="&lt;kuiper-belt&gt;&lt;space-probe&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29073" PostTypeId="2" ParentId="16621" CreationDate="2019-01-07T11:44:04.650" Score="1" Body="&lt;p&gt;Lorenzo Raynard, spokesperson for SKA Africa was gracious enough to send me the map with the sixteen sites below, and I've duplicated the dots on a larger format map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's been quite a while, probably the array has grown since this &quot;first-light&quot; image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next step (strictly for fun) will be to try to make a &lt;a href=&quot;https://astronomy.stackexchange.com/q/26413/7982&quot;&gt;UV map&lt;/a&gt; for this configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/W4FGQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W4FGQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0Gv1F.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0Gv1F.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-01-07T11:44:04.650" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29074" PostTypeId="2" ParentId="29067" CreationDate="2019-01-07T13:54:31.463" Score="-2" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Will the SKA's total bandwidth still exceed the Earth's Internet's bandwidth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;No.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;First some conversions:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1024 Gigabytes = 1 Terabyte&lt;/li&gt;&#xA;&lt;li&gt;1024 Terabytes = 1 Petabyte&lt;/li&gt;&#xA;&lt;li&gt;1024 Petabytes = 1 Exabyte&lt;/li&gt;&#xA;&lt;li&gt;1024 Exabytes = 1 Zettabyte&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;From the SKA Telescope website: &amp;quot;&lt;a href=&quot;http://www.skatelescope.org/wp-content/uploads/2018/08/16231-Factsheets-operational-model-v4.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Operational Model&lt;/a&gt;&amp;quot; (.PDF), on the webpage: &amp;quot;&lt;a href=&quot;https://www.skatelescope.org/technical/info-sheets/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Info Sheets&lt;/a&gt;&amp;quot;, this explanation and graphic is offered:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;The data journey&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Basic data products will be produced and stored in Cape Town for SKA1-mid and Perth for SKA1-low. From there, they will be delivered to a global alliance of SKA Regional Centres for further processing and archiving and access by the user community.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;New data models will need to be developed as current software packages for radio astronomy data reduction don’t have the capabilities to handle the SKA’s large bandwidths and Field of View dataset.&amp;quot;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/updgR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/updgR.jpg&quot; alt=&quot;SKA Data Path and Bandwidth&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;Major dates&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;SKA1 OPERATIONAL CONCEPT DOCUMENT - Operational Principles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;late 2027 - Commencement of Key Science Projects (KSPs)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;late 2026 - Commencement of PI-led programmes&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;2024–2027 - Key Science Project (KSP) planning &amp;amp; proposals&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;2022 - Start of Observatory &amp;amp; Science commissioning&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;2020 - Start of construction activities&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;2017–2018 - Prototypes deployed at the telescope sites&amp;quot;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;See also: &amp;quot;&lt;a href=&quot;https://www.skatelescope.org/ska-community-briefing-18jan2017/&quot; rel=&quot;nofollow noreferrer&quot;&gt;SKA Community Briefing – 18 January 2017&lt;/a&gt;&amp;quot;, page 12:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uGhnM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uGhnM.jpg&quot; alt=&quot;Data Flow through the SKA&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Nothing will flow to &lt;a href=&quot;https://en.wikipedia.org/wiki/Null_device&quot; rel=&quot;nofollow noreferrer&quot;&gt;/dev/null&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Source: Cisco - Global - IP Traffic: &lt;a href=&quot;https://www.cisco.com/c/m/en_us/solutions/service-provider/vni-forecast-highlights.html#&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.cisco.com/c/m/en_us/solutions/service-provider/vni-forecast-highlights.html#&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Globally, IP traffic will grow 3-fold from 2017 to 2022, a compound annual growth rate of 26%.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Globally, IP traffic will reach 396.0 EB per month by 2022, up from 122.4 EB per month in 2017.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Global IP networks will carry 13.0 EB per day by 2022, up from 4.0 EB per day in 2017.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&lt;p&gt;Internet Traffic&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Globally, Internet traffic will grow 3.7-fold from 2017 to 2022, a compound annual growth rate of 30%.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Globally, busy hour Internet traffic will grow 4.8-fold from 2017 to 2022, a compound annual growth rate of 37%.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Globally, Internet traffic will reach 350.8 EB per month by 2022, up from 96.0 EB per month in 2017.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Global Internet traffic will be 11.5 EB per day by 2022, up from 3.2 EB per day in 2017.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Global Internet traffic by 2022 will be equivalent to 1 trillion DVDs per year, 88 billion DVDs per month, or 120 million DVDs per hour.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By 2022, the gigabyte equivalent of all movies ever made will cross the Internet every 1 minutes.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Global Internet traffic by 2022 will be equivalent to 193x the volume of the entire Global Internet in 2005.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;No, it's not even close.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-01-15T19:55:09.990" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="29075" PostTypeId="2" ParentId="29072" CreationDate="2019-01-07T14:31:22.097" Score="11" Body="&lt;p&gt;It's probably a cosmic ray strike, a common artifact on New Horizons images.  The image, however, is of very poor quality which makes it difficult to analyse the exact nature of the spot.  I'll go into this more  at the end of the answer.&lt;/p&gt;&#xA;&lt;p&gt;Possibilities:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Cosmic ray hit&lt;br /&gt;&#xA;A common occurance (see below); the most likely cause without evidence to the contrary.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Background star (e.g. a nova)&lt;br /&gt;&#xA;Novæ (and variable stars) brighten and dim over several days.  Were this a nova it would have shown in images from previous and later days.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Kuiper belt object&lt;br /&gt;&#xA;To not show in images from adjacent days it would need large relative motion and thus a sizable trail would be expected (none is evident).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Aliens&lt;br /&gt;&#xA;Possible?  Yes.  Likely?  Far, far, from it.  Not worth serious consideration.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Here's the New Horizons team's standard disclaimer:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;This image contains one or more objects whose brightness exceeds the detector's saturation level. This sometimes produces a &amp;quot;tail&amp;quot; of bright and/or dark pixels to the right of the object. You may also notice a faint vertical white stripe passing through the saturated object; this is an artifact called &amp;quot;frame transfer smear&amp;quot; and is associated with the incomplete removal of signal produced when the image is transferred from the optically active region of the detector to the storage region of the detector. If the target is badly saturated, you may also notice a faint, X-shaped feature nearly centered on the object; these are optical diffraction spikes.&lt;/p&gt;&#xA;&lt;p&gt;This image contains one or more streaks associated with cosmic rays passing through the detector. &lt;strong&gt;Nearly every LORRI image has at least one cosmic ray strike, but most are &amp;quot;single pixel&amp;quot; events (i.e., they only appear to be in single pixel and can easily be mistaken for stars). But sometimes a cosmic ray is energetic enough that it leaves a &amp;quot;trail&amp;quot; as it passes through the LORRI detector.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Example of cosmic ray hit and saturation trails:&lt;/strong&gt;&lt;br /&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/x55IL.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x55IL.png&quot; alt=&quot;Example of cosmic ray hit and saturation trails&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sup&gt;Source: &lt;a href=&quot;http://pluto.jhuapl.edu/soc/UltimaThule-Encounter/&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; (via &amp;quot;Learn more...&amp;quot; link), NASA/JHUAPL/SwRI&lt;br /&gt;&#xA;Note that despite being a PNG, JPEG compression artifacts can be seen in this image&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;The frame in question:&lt;/strong&gt;&lt;br /&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/vUxUd.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vUxUd.png&quot; alt=&quot;Extract from GIF of frame in question&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sup&gt;Source: Frame K-29 of &lt;a href=&quot;https://blogs.nasa.gov/pluto/wp-content/uploads/sites/253/2018/12/movie_w150_d50_Greys_r_n42_f8.gif&quot; rel=&quot;noreferrer&quot;&gt;this image&lt;/a&gt; taken from &lt;a href=&quot;https://blogs.nasa.gov/pluto/2018/12/20/the-pis-perspective-on-final-approach-to-ultima/&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;, NASA/JHUAPL/SwRI/Henry Throop&lt;br /&gt;&#xA;Enlargement below is from frame K-29 of this image&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;The image in question is almost certainly a stacked composite of ~10 source images and appears to have been 2×2 binned (compare &lt;a href=&quot;http://pluto.jhuapl.edu/Galleries/Featured-Images/image.php?page=1&amp;amp;gallery_id=2&amp;amp;image_id=555&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;).  To make faint objects more visible, the image was also likely &lt;a href=&quot;http://www.boulder.swri.edu/pkb/ssr/ssr-lorri.pdf&quot; rel=&quot;noreferrer&quot;&gt;contrast-enhanced&lt;/a&gt;.  The image also shows clear signs of lossy compression and quantization (from 12 bits per pixel to ~5), see enlargement below.  While these are (mostly) acceptable choices for the image's intended use it tends to obscure detail from the original images.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Crop showing artifacts:&lt;/strong&gt;&lt;br /&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/dlsME.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dlsME.png&quot; alt=&quot;Crop showing artifacts&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Looking at the spot in question (enlargement below) a couple of things are noticable:&lt;/p&gt;&#xA;&lt;p&gt;It is compact and relatively dim.  Its overall size in pixels is about half that of the stars and of MU69.  In addition, the drop-off in brightness moving away from its center is much more rapid than for the other objects.  This is consistent with a bright but compact object appearing in only one image in the K-29 stack.&lt;/p&gt;&#xA;&lt;p&gt;This object also exhibits a less-blurry structure with two distinctly brighter spots or lines.  This is consistent with spray from a cosmic ray hit.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Crop showing mystery spot:&lt;/strong&gt;&lt;br /&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/va1uw.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/va1uw.png&quot; alt=&quot;Crop showing mystery spot&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;All in all, I think a cosmic ray hit is the most likely explanation.  Given the very poor quality of the image in question though it's difficult to give a definitive answer.  To make a proper determination one would need access to the yet-to-be-released raw imagery.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;I've seen a paper with a more thorough discussion of New Horizons image artifiacts but don't seem to have saved a copy.  I'll add references from it when I track it down.&lt;/p&gt;&#xA;" OwnerUserId="24184" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-01-08T04:47:09.037" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="29076" PostTypeId="2" ParentId="26099" CreationDate="2019-01-07T19:37:04.183" Score="5" Body="&lt;p&gt;In the book &lt;a href=&quot;https://press.princeton.edu/titles/8697.html&quot; rel=&quot;noreferrer&quot;&gt;Galactic Dynamics by Binney and Tremaine (second edition)&lt;/a&gt; there is a whole section explaining the difference between the Jacobi radius and the tidal radius (page 677-chapter 8).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, &lt;span class=&quot;math-container&quot;&gt;$r_J$&lt;/span&gt; is defined as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$r_J=  R_0\left(\frac{m}{3M}\right)^{1/3}$$&lt;/span&gt; &#xA;  The Jacobi radius &lt;span class=&quot;math-container&quot;&gt;$r_J$&lt;/span&gt; (also, Roche or Hill radius) of an orbiting stellar system is expected to correspond to the observational tidal radius, the maximum extent of the satellite system. However this correspondence is only approximate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Five reasons are subsequently given to explain why the correspondence is only approximate. &lt;/p&gt;&#xA;" OwnerUserId="25448" LastEditorUserId="354" LastEditDate="2019-01-08T08:25:31.867" LastActivityDate="2019-01-08T08:25:31.867" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29077" PostTypeId="2" ParentId="27778" CreationDate="2019-01-07T19:52:55.140" Score="1" Body="&lt;p&gt;You are not doing anything wrong. Positively curved empty universes are forbidden. Since as you figure it out the scale factor would be imaginary.&lt;/p&gt;&#xA;" OwnerUserId="25424" LastActivityDate="2019-01-07T19:52:55.140" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29078" PostTypeId="2" ParentId="29072" CreationDate="2019-01-07T22:38:05.543" Score="-2" Body="&lt;p&gt;Looking closely, the spot has roughly same slightly noncircular, diffuse, extended shape and size as 2014 MU69 (optical or kinematic artifact) which would not happen from a cosmic ray or hot pixel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could be a real object, and simply much closer to New Horizons than MU69 is. That could certainly account for the object being visible on one day only. If its 10x closer than MU69 it can be 100x smaller than MU69 and still have a similar apparent magnitude, and the number of asteroids increases rapidly with decreasing size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Left is object in question, right is MU69.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/E0GiQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E0GiQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/Wrywh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wrywh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Lsxy4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lsxy4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; purely unscientific processing in order to perform &quot;blobology&quot;: OP's image, blown up and then cropped twice (to get rid of the red arrow) then just plotted in python &lt;a href=&quot;https://pastebin.com/8KWBHqnf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://pastebin.com/8KWBHqnf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-01-08T11:03:03.323" LastActivityDate="2019-01-08T11:03:03.323" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29080" PostTypeId="1" AcceptedAnswerId="29088" CreationDate="2019-01-08T05:11:40.133" Score="6" ViewCount="308" Body="&lt;p&gt;If, for instance, Hubble's camera was a &lt;span class=&quot;math-container&quot;&gt;$10^{300}$&lt;/span&gt; times better, could it see a distant star in details, or is there a limit to the amount of light that reaches the Earth's orbit or perhaps does the light get distorted on its way so you must come closer?&lt;/p&gt;&#xA;&lt;p&gt;Is there a limit to the number of light beams that hit a telescope camera's &amp;quot;eye&amp;quot; or if you built a huge telescope would you then be able to see that distant star in high detail (for instance, see an earth size satellite around Betelgeuse's orbit).&lt;/p&gt;&#xA;" OwnerUserId="25452" LastEditorUserId="37191" LastEditDate="2021-02-12T21:05:49.263" LastActivityDate="2021-02-12T21:05:49.263" Title="If we had the right technology could we see a distant star in detail?" Tags="&lt;telescope&gt;&lt;photography&gt;&lt;hubble-telescope&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="29082" PostTypeId="1" AcceptedAnswerId="35216" CreationDate="2019-01-08T11:18:26.850" Score="15" ViewCount="412" Body="&lt;p&gt;There are plenty of optical interferometers in use with baselines of up to maybe 1km. As far as I can find out, they all work by directly collecting the light at all the telescopes, using mirrors to bring them together and then actually interfering them and observing the fringes with photodetectors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, radio astronomers do interferometry over distances of thousands of km, but they do it quite differently. Essentially, they record the baseband electromagnetic waveform at each telescope, with an accurate timestamp, transport them all to a central location and combine them computationally. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is whether &lt;em&gt;in principle&lt;/em&gt; you could use the radio astronomers methods with an optical (or near IR) signal. Obviously the engineering and computational challenges would be formidable -- you would need to record data at speeds close to 1 Pb/s, timestamp it with femtosecond accuracy and do a ludicrously huge computation to recover an image, but that's not my question here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the one hand, it seems like the answer should be &quot;yes&quot; -- electromagnetic waves are electromagnetic waves. On the other hand, why should the phase of a photon observed in one telescope have anything to do with the phase of one observed in another? If you are digitising the data at each telescope, you are ruling out interference  fringes on a quantum level, just as happens if you find a way to measure which slit a photon went through in the double-slit interferometer. This suggests that the answer should be &quot;no&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the answer is &quot;no&quot; then what wavelength does the cutoff happen at? We have radio-style VLBI at 1mm now and optical-style interferometry at 1 &lt;span class=&quot;math-container&quot;&gt;$\small\mathsf{\mu m}$&lt;/span&gt;, what is possible in between?&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="7982" LastEditDate="2020-01-09T02:22:12.550" LastActivityDate="2020-06-21T21:56:47.997" Title="Is Optical VLBI theoretically feasible? If not why not?" Tags="&lt;interferometry&gt;&lt;vlbi&gt;" AnswerCount="1" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="29084" PostTypeId="1" CreationDate="2019-01-08T14:37:33.857" Score="1" ViewCount="42" Body="&lt;p&gt;I need to get the X-ray light curve of Classical Nova V1494 Aquilae (1999 No. 2) in Outburst. It is done long time ago, but I need to this by myself. While reading the papaer about the V1494 Aquilae, they have analyzed the data belonging three seperate time intervals, then they combined the three graph to get final light curve. The final one shows a clear pick corresponding to the outburst.&lt;/p&gt;&#xA;" OwnerUserId="25405" LastActivityDate="2019-01-08T14:37:33.857" Title="How to decide which part of the data is analyzed?" Tags="&lt;data-analysis&gt;&lt;light-curve&gt;&lt;nova&gt;&lt;raw-data&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2019-01-09T16:38:22.387" ContentLicense="CC BY-SA 4.0" />
  <row Id="29085" PostTypeId="2" ParentId="29061" CreationDate="2019-01-08T15:30:17.743" Score="3" Body="&lt;p&gt;There are no other reasonably direct methods, but there are definitely indirect methods.  One, in @Alex Hajnal's answer, the higher CMB temperatures measured further out is a very nice indirect measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another indirect piece of evidence, which no one has noted yet, is that as we look further and further out, the universe looks younger and younger, and less and less like what we see in our neighborhood.  You are pretty much forced to explain that scientifically by saying that the universe had a beginning on the rough order of 10 billion years ago, and that stars and galaxies only started forming then. (This isn't proof of a Big Bang specifically, but it does eliminate most alternatives to it. The Steady State model, for example, is falsified.)  It's very very hard to explain what we see except as being due to a universe expanding from a hot dense state ca. 10&lt;sup&gt;10&lt;/sup&gt; years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More indirect evidence comes from General Relativity, a theory of space, time and gravity which is &lt;em&gt;very&lt;/em&gt; well verified -- it's been tested for a century now and challenged by countless other theories, and only GR has passed all experimental tests.  GR robustly predicts that a static universe is impossible and that it must either be expanding or contracting.  This is indirect evidence from mostly local experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet more indirect evidence comes from nucelosynthesis calculations which show that the H/He/Li ratios that we observe in the oldest and least evolved stars is exactly what we predict based on applying the measured properties of nuclei to a Big Ban fireball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's &lt;em&gt;so&lt;/em&gt; much science other than the red shifts which point to the universe expanding from an initial very hot, dense state that even without the observation of red shifts, we'd eventually be forced to that conclusion.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2019-01-08T15:30:17.743" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29086" PostTypeId="2" ParentId="28984" CreationDate="2019-01-08T15:57:17.380" Score="2" Body="&lt;p&gt;I don't know how to calculate but Wikipedia has a page about it:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Future_of_an_expanding_universe#Timeline&quot; rel=&quot;nofollow noreferrer&quot;&gt;Timeline of the Universe&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The galaxies in the Local Group, the cluster of galaxies which includes the Milky Way and the Andromeda Galaxy, are gravitationally bound to each other. It is expected that between 10^11 (100 billion) and 10^12 (1 trillion) years from now, their orbits will decay and the entire Local Group will merge into one large galaxy.[4]&lt;/p&gt;&#xA;&lt;p&gt;Assuming that dark energy continues to make the universe expand at an accelerating rate, in about 150 billion years all galaxies outside the Local Supercluster will pass behind the cosmological horizon. It will then be impossible for events in the Local Group to affect other galaxies. Similarly it will be impossible for events after 150 billion years, as seen by observers in distant galaxies, to affect events in the Local Group.[3] However, an observer in the Local Supercluster will continue to see distant galaxies, but events they observe will become exponentially more red shifted as the galaxy approaches the horizon until time in the distant galaxy seems to stop. The observer in the Local Supercluster never observes events after 150 billion years in their local time, and eventually all light and background radiation lying outside the local supercluster will appear to blink out as light becomes so redshifted that its wavelength has become longer than the physical diameter of the horizon.&lt;/p&gt;&#xA;&lt;p&gt;Technically, it will take an infinitely long time for all casual interaction between our local supercluster and this light; however, due to a the redshifting explained above, the light will not necessarily be observed fit an infinite amount of time, and after 150 billion years, no new causal interaction will be observed.&lt;/p&gt;&#xA;&lt;p&gt;Therefore, after 150 billion years intergalactic transportation and communication beyond the Local Supercluster becomes causally impossible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This wikipedia data, comes from &lt;a href=&quot;https://arxiv.org/abs/astro-ph/9902189&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, which is used in the text as reference (4)&lt;/p&gt;&#xA;" OwnerUserId="25424" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-01-10T14:48:20.573" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29087" PostTypeId="2" ParentId="29061" CreationDate="2019-01-08T16:11:13.217" Score="3" Body="&lt;p&gt;In addition to the circumstantial evidence provided by the other answers, a strong verification of galaxies moving away from each other is given by the fact that we see physical processes — such as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Light_curve#Supernovae&quot; rel=&quot;nofollow noreferrer&quot;&gt;decline time for the brightness of supernovae&lt;/a&gt; — increase, the farther away it is. For a source with a redshift of &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;, the amount of this &lt;em&gt;time dilation&lt;/em&gt; is observed to be &lt;span class=&quot;math-container&quot;&gt;$(1+z)$&lt;/span&gt;, exactly in accordance with what is expected from general relativity in an expanding universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, a supernova observed with a redshift of &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; takes twice the time to decline as a local supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note though that this is not a verification of the expanding Universe, only of galaxies moving away from each other. If the Universe were static, but the galaxies moved &lt;em&gt;through&lt;/em&gt; space, you would observe the processes dilated by the same factor, as predicted by &lt;em&gt;special&lt;/em&gt; relativity. There are, however, other evidence that the galaxies do not move through a static space, but instead lie more or less still in an expanding space.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2019-01-08T16:11:13.217" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="29088" PostTypeId="2" ParentId="29080" CreationDate="2019-01-08T19:35:51.477" Score="8" Body="&lt;p&gt;The only limitations would be related to building an instrument large enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a limit to the size of the finest detail a telescope can see. &quot;Size&quot; here is angular size, the angle that the detail is covering in the visual field. For a given wavelength of light, the smallest angular size depends on the aperture (diameter) of the telescope in a linear fashion: double the aperture, and you cut the angular size in half.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rule of thumb: 100 mm aperture has a resolving power of 1 arcsecond, for visible light (not exactly, close enough). Therefore 200 mm aperture has 0.5 arcsec resolving power, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By building larger and larger telescopes, the angular size of the smallest details becomes smaller and smaller. But are there any limits here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There aren't any cosmologic mechanisms that I'm aware of that would prevent telescopes from operating at arbitrarily large resolving powers (arbitrarily small angular sizes). But we can't build arbitrarily large instruments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest monolithic (single mirror) telescope has an aperture of 8.4 meters. Using segmented mirrors, the aperture aim of the current largest project is 30 meters - the Thirty Meter Telescope, still in a very early stage, completion date unknown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond that size, large scale interferometers are the only option, with synthetic aperture varying from dozens of meters to kilometers, and larger projects are being discussed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory, very large scale interferometers could be built in space. But there might be inherent limits for the aperture of such systems. Anyway, this is hypothetical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Giant telescopes in general do not require a continuous surface of glass. Interferometers almost by definition do not use a single reflector. You could build a gigantic telescope or interferometer where the active reflective surfaces are just these tiny chunks at the periphery of a huge perimeter. Most of the area in between is empty / unused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT2: The performance of the &quot;tiny chunks&quot; reflector is not the same as the performance of the full size mirror, as mentioned in the comments.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2019-01-10T01:33:39.887" LastActivityDate="2019-01-10T01:33:39.887" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="29090" PostTypeId="1" CreationDate="2019-01-09T00:41:05.893" Score="5" ViewCount="125" Body="&lt;p&gt;This is a theoretical question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/29088/7982&quot;&gt;This answer&lt;/a&gt; to the question &lt;em&gt;If we had the right technology could we see a distant star in detail?&lt;/em&gt; (presumably space-based) primarily addresses the scaling of diffraction-limited resolution with aperture &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt;. For existing and near-future telescopes this is assumed to follow &lt;span class=&quot;math-container&quot;&gt;$\lambda/D$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a future optical or radio interferometric telescope of absurdly large &lt;em&gt;effective aperture&lt;/em&gt; or baseline, say AU or even larger, operated similarly to the Event Horizon Telescope (local downconversion and timestamping, &lt;a href=&quot;https://astronomy.stackexchange.com/q/20082/7982&quot;&gt;interferometry performed offline&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could &lt;a href=&quot;https://en.wikipedia.org/wiki/Weak_gravitational_lensing&quot; rel=&quot;nofollow noreferrer&quot;&gt;weak gravitational lensing&lt;/a&gt; effects or &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_microlensing&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravitational microlensing&lt;/a&gt; result in wavefront distortion that if not corrected, would result in &lt;em&gt;degraded resolution&lt;/em&gt;? Note I'm not asking about distortion of the field, but instead a limitation to the resolution; degraded point spread function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, would this problem be amenable to correction in the same way that atmospheric and ionospheric effects (visible and radio) are dealt with? Would it generally be static on the timescale of the observation?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-04-04T03:40:29.170" LastActivityDate="2019-04-04T03:41:09.273" Title="Can weak gravitational lensing or microlensing-induced wavefront distortion limit resolution of absurdly large aperture telescopes?" Tags="&lt;radio-astronomy&gt;&lt;gravitational-lensing&gt;&lt;interferometry&gt;&lt;adaptive-optics&gt;&lt;weak-lensing&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29092" PostTypeId="2" ParentId="28977" CreationDate="2019-01-09T01:00:09.173" Score="0" Body="&lt;p&gt;I, too, saw a very bright object in the eastern sky before dawn a few days ago (Saturday-5-Jan-2019).  Based on previous experience viewing astronomical objects, I reckoned that it was probably Venus.  Jupiter would not be that bright unless it were opposite the sun, i.e., closer to us.  Saturn, Mars, and Mercury are NEVER that bright.&#xA;With kind regards,&lt;br&gt;&#xA;Mark Creek-water Dorazio,  long-time astronomy enthusiast&lt;/p&gt;&#xA;" OwnerUserId="13290" LastActivityDate="2019-01-09T01:00:09.173" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29094" PostTypeId="2" ParentId="26099" CreationDate="2019-01-09T04:57:51.763" Score="1" Body="&lt;p&gt;thanx Gabriel for adding the detailed definition to my reply.. an additional definition of the tidal radius hereafter&#xA;&lt;a href=&quot;https://i.stack.imgur.com/tdZFQ.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tdZFQ.gif&quot; alt=&quot;tidal radius as defined in (Tan 2000-equation 9-http://adsabs.harvard.edu/abs/2000ApJ...536..173T&quot;&gt;&lt;/a&gt;&#xA;in this definition it describes molecular clouds rather than stellar systems..again notice the difference (2 is in the numerator).. there is no reference on where he got this formalism from..this is from Tan 2000 eq (9) check also eq (8)(&lt;a href=&quot;http://adsabs.harvard.edu/abs/2000ApJ...536..173T&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2000ApJ...536..173T&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25448" LastActivityDate="2019-01-09T04:57:51.763" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29097" PostTypeId="1" AcceptedAnswerId="29099" CreationDate="2019-01-09T14:55:40.923" Score="2" ViewCount="91" Body="&lt;p&gt;I understand this is an icky subject, but I recently got interested in units for large distances for applications in cosmology and what not (after hearing about the redefinition of the kilo and kelvin), and I was surprised to see many very weird units which seem difficult to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I was thinking in astronomy distances only get larger and larger, so I looked to see if I could find any logarithmic scale for distances and I couldn't find any. Then again, I'm not an astronomer and not suited to argue anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any logarithmic scales for larger distances, and if so why aren't they being used? And if non exist, is there a reason why(is it simply a bad idea)? Or is it simply a matter of keeping historical background?&lt;/p&gt;&#xA;" OwnerUserId="25465" LastEditorUserId="25465" LastEditDate="2019-01-09T15:15:36.490" LastActivityDate="2019-01-09T15:20:51.897" Title="Logarithmic scale for large distances?" Tags="&lt;distances&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29098" PostTypeId="1" AcceptedAnswerId="29101" CreationDate="2019-01-09T15:07:32.700" Score="0" ViewCount="207" Body="&lt;p&gt;The way I usually see spacetime explained, it says that if you just sit still, you are moving at 0kph and so you are moving at maximum speed through time (ie. time for you is flowing at maximum speed).  However, sitting still we're moving at the speed of our galaxy through space, which is about 600km/s.  The speed of light is about 300,000km/s.  So for simplicity's sake, let's say the galaxy is actually moving through space at 1,000km/s.  Does this in fact mean that if I sit still on Earth, time for me is actually flowing 1/300th slower than if I were sitting still with respect to the universe?  And does it also mean that if the Milky Way were travelling at 99% the speed of light, time for me would be flowing at 1% the rate of the flow of time for someone that was sitting still with respect to the universe?&lt;/p&gt;&#xA;" OwnerUserId="7717" LastActivityDate="2019-01-09T23:33:29.780" Title="Does the Milky Way's movement through space have spacetime implications?" Tags="&lt;space-time&gt;&lt;general-relativity&gt;&lt;time&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29099" PostTypeId="2" ParentId="29097" CreationDate="2019-01-09T15:20:51.897" Score="5" Body="&lt;p&gt;Having some experience with the only logarithmic unit in regular use (the dB, for sound intensity), I have to say: please, no. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A logarithmic &lt;em&gt;representation&lt;/em&gt; can be &lt;a href=&quot;https://www.xkcd.com/482/&quot; rel=&quot;noreferrer&quot;&gt;handy in some cases&lt;/a&gt;. This chart is an excellent example of what can be done if you choose the scale on your image correctly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fIy22.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fIy22.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this has been done without introducing a new unit for distance, it just plots log(x) instead of x on the vertical scale. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you were to use a logarithm in the fundamental unit for distance, you'd introduce some problems: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;cosmological distances no longer match other distance units and any conversion between them involves using a log function. &lt;/li&gt;&#xA;&lt;li&gt;calculations involving cosmological distances get more complicated because you've now got to use log() everywhere. &lt;/li&gt;&#xA;&lt;li&gt;you lose the linear relationship between numbers: object X is twice as far away as object Y, but the difference in your unit becomes log(2) making distances far less intuitive. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="361" LastActivityDate="2019-01-09T15:20:51.897" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29100" PostTypeId="2" ParentId="29098" CreationDate="2019-01-09T16:58:43.530" Score="-1" Body="&lt;p&gt;As far as you are considering the proper motion of the galaxy towards the Great Attractor the speed is less than a percent of the speed of light c.&lt;/p&gt;&#xA;&lt;p&gt;Although astonishing this speed doesn't require a relativistic correction.&lt;/p&gt;&#xA;&lt;p&gt;Of course if we were to sit on a galaxy moving at, say, 0.9 × c, then we would be like the travelling twin of the famous &amp;quot;paradox&amp;quot; as compared to someone still &lt;em&gt;nearby&lt;/em&gt; and watching us.&lt;/p&gt;&#xA;&lt;p&gt;However for calculating the time dilation  you can't go by % proportion as in your question but apply the relativistic formula&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta t^\prime = \frac {\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;which expresses the fact that the moving observer's period of the clock &lt;span class=&quot;math-container&quot;&gt;$\Delta t^\prime$&lt;/span&gt; is longer that the period &lt;span class=&quot;math-container&quot;&gt;$\Delta t$&lt;/span&gt; in the frame of the clock itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;that I copied from &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Time_dilation&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Time_dilation&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;It might be worth pointing out that the formula is valid for both observers and thus they will see each other's clock ticking slower.&lt;/p&gt;&#xA;&lt;p&gt;Related but more general questions and discussion by following &lt;a href=&quot;https://astronomy.stackexchange.com/questions/25239/how-does-space-time-differ-between-galaxies&quot;&gt;How does space time differ between galaxies?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;in which the effect (lack of) of galaxies on spacetime is attacked from a general relativity point of view (how does galaxy's mass influence the spacetime curvature).&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt; after receiving comments I am not mirroring anymore the wording &amp;quot;moving through space&amp;quot; as it leads to think, or that the authors of the Q and the A think, of absolute motion in absolute space. I also add that the conclusions for the observer on the galaxy and the one still coincides and both are right (unless the galaxy for some reason turns back ;)&lt;/p&gt;&#xA;&lt;p&gt;Nevertheless&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;I would like to point out that moving through space is a perfectly sensible sentence in both common speaking and in physics&lt;/strong&gt;. Through what something should move when its spatial coordinates are changing if not space? :)&lt;/p&gt;&#xA;" OwnerUserId="17437" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-01-09T23:33:29.780" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29101" PostTypeId="2" ParentId="29098" CreationDate="2019-01-09T19:19:47.020" Score="2" Body="&lt;p&gt;Your question has a misunderstanding at its root:  You say, &quot;So for simplicity's sake, let's say the galaxy is actually moving through space at 1,000km/s&quot;. The problem here is that the whole point of Relativity is that the &lt;em&gt;idea&lt;/em&gt; of movement with respect to space is meaningless -- the only thing that has meaning is movement with respect to other bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time dilation is also relative. If you're moving at 1000kps relative to Fred, Fred sees your clocks as running slower, and in his &quot;reference frame&quot; they are.  But &lt;em&gt;you&lt;/em&gt;, in your reference fame, see &lt;em&gt;Fred's&lt;/em&gt; clocks as running slower! (This is the essence of the famous &lt;a href=&quot;https://en.wikipedia.org/wiki/Twin_paradox&quot; rel=&quot;nofollow noreferrer&quot;&gt;Twin Paradox&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're being led astray by your intuitions about space and time. (And don't be ashamed about that. Newton, who was probably the greatest physicist ever, even greater than Einstein, and Kant (one of the great philosophers) and many, many others had the same intuitions. Yet experiment has taught us that they were wrong.)  Neither space nor time is absolute, and an understanding of &lt;a href=&quot;https://en.wikipedia.org/wiki/Special_relativity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Special Relativity&lt;/a&gt; (which is much easier to understand than it was to discover in the first place!) is required to ask as well as to answer your question.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2019-01-09T21:13:18.003" LastActivityDate="2019-01-09T21:13:18.003" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29102" PostTypeId="1" AcceptedAnswerId="29109" CreationDate="2019-01-10T04:48:13.937" Score="4" ViewCount="306" Body="&lt;p&gt;What are the factors that prevent planets from also turning into disks, like the stellar dust in a nebula does?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth is not a perfect sphere, but rather it's squished at the poles and it bulges at the Equator, owing to its rotation working to squish it into a disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm tempted to say that the Earth couldn't become a disk because of the density of the material in it fighting back against being pressured into a disk in a way that in stellar gas you can't? But then, why isn't the gas around the gas giants a disk? Is it because of the solid nucleus of those planets fighting against that?&lt;/p&gt;&#xA;" OwnerUserId="25469" LastActivityDate="2019-01-10T20:17:44.933" Title="Why do nebulae turn into accretion disks but planets do not?" Tags="&lt;gravity&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29103" PostTypeId="2" ParentId="29102" CreationDate="2019-01-10T06:32:49.723" Score="0" Body="&lt;p&gt;First point: planet VS nebula&#xA;Your intuition is correct. Nebula is less dense and is mainly is gaseous state, so it has more fluidity than a solid state like a planet. Think of water can change shape regarding to its container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second point: why some gas does not turn into a disk shape?&#xA;This has to do with the force field that the gas is under the influence, and the initial velocity of the gas. In case of having a solid core, the force field is just gravity pulling everything inward radially to the center. If the pulled stuff has the perpendicular (to the radial direction) component of the initial velocity not equal zero, instead of falling inward as a straight line radially, the stuff spirals into the center. The accumulation of stuff spiralling to the center gives the shape of disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the giant gas cloud, it does not have a disk shape because the influence of gravity of a core (if exists) is negligible due to the outer part of the cloud (that we see) is very far from the core.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2019-01-10T06:32:49.723" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29108" PostTypeId="2" ParentId="29102" CreationDate="2019-01-10T16:02:02.213" Score="3" Body="&lt;p&gt;The circumstellar disk is perhaps about 2%-3% the mass of the central star.   Large planets may form in similar ways.   The mass ratio of the disk to what it orbits depends a lot on the angular momentum and I would think that only large planets that clear out a considerably larger volume of space are likely to have enough angular momentum to form disks around them.    Jupiter's Galilean moons and most formation moons probably form planetary debris disks during their planet's formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.rochester.edu/newscenter/gigantic-ring-system-around-j1407b/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This ring system around a brown dwarf&lt;/a&gt; that orbits a larger star could be compared to a circumstellar disk.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disks of this nature, both stellar or theoretical-planetary are temporary, so only very young solar systems might have them and we probably don't have telescopes good enough to get a good look at young planets during their formation period.   A large part of the reason we don't hear about planetary disks is that we might not have good enough telescopes to observe them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note, if someone with better knowledge than me answers this, I can remove my answer).&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2019-01-10T16:02:02.213" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29109" PostTypeId="2" ParentId="29102" CreationDate="2019-01-10T18:39:27.610" Score="3" Body="&lt;p&gt;They both form disks, both disks are transient, and both disks are small compared with the central body.  The parallels between the Sun and a planet are closer than one may think!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A collapsing nebula forms a star and a circumstellar disk. &lt;em&gt;Nearly all of the angular momentum in the system is in the disk.&lt;/em&gt; The disk is unstable and much of the material condenses into small icy or rocky bodies which mostly combine to form planets or are expelled from the system by close encounters with larger bodies.  After a few tens of millions of years, a planetary system is all that remains of the disk. (Note that even today, the planets contain 96% of the angular momentum in the Solar System.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When planets form it seems likely that they also have disks of material around them. Because the planets are mainly forming from planetoids and bits of rubble, the disks don't start out mostly as gas.  (The best current theory of the Moon's formation is that it condensed out of a disk formed around Earth from a large, late collision.)  These disks also dissipate, falling onto the planet through drag or being ejected from orbit by perturbing bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very unlikely that Saturn's (or Jupiter's or any of the others') rings are primordial, but are probably the result of the disruption of a icy body in the (comparatively) recent past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that neither the Sun nor the planets are rotating fast enough for their shape to be much affected: They're all pretty much spherical.  In each case, there are mechanisms which carry off angular momentum from the central body and deposits it in the disk or into the bodies which condense from the disk. &lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2019-01-10T18:39:27.610" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29110" PostTypeId="2" ParentId="29102" CreationDate="2019-01-10T20:17:44.933" Score="0" Body="&lt;p&gt;I like the answers provided by @userLTK and @MarkOlson but let me add my own phrasing here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape in each case is given by the particular characteristics of the scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An irregular shaped nebula is a bunch of gas and dust only very loosely bound by gravity, or unbound. There's not much overall angular momentum (the whole thing is not spinning, typically). It's just a sparse pile of random stuff thrown out by a stellar explosion or something. The shape can be anything, depending on how exactly it came out to be. It could be slowly expanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An accretion disk is different. The stuff is bound together by gravity - perhaps the overall density of the cloud is greater, perhaps there's a dense chunk in the middle (protoplanet or protostar), perhaps both. Each bit of cosmic junk is orbiting the common center of mass - bound by gravity. You would think the whole collection could be spherical, but that only lasts for a blink of an eye: particles in the cloud collide with each other all the time, and whichever direction has the most angular momentum ends up winning in the end, so all particles will rotate more or less in the same plane; heretics will slam into the disk and join it in the end; within the disk there are far fewer collisions, so the disk configuration is more stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A planet is basically a giant pile of stuff that behaves, at the macroscopic level, like an incompressible fluid. Even &quot;rocky&quot; planets behave like fluids at the scale of the whole body. Of course, as with any fluid, gravity will eventually pull it into a spherical shape. If there's any angular momentum, the sphere will be squished a little, but most planets are close to a perfect sphere (extreme angular momentums are rare).&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2019-01-10T20:17:44.933" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29111" PostTypeId="1" AcceptedAnswerId="29115" CreationDate="2019-01-10T20:30:10.173" Score="3" ViewCount="125" Body="&lt;p&gt;Optical fibers are well-known to observe astronomical objects, say, galaxies, to generate massive spectroscopic surveys. The galaxies are often very far from the optical fibers of a telescope. However, I'm thinking about the possibility of using optical fibers to observe much smaller objects (compared to galaxies) like satellites in much shorter ranges like from earth to GEO orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, &lt;strong&gt;given an object at the distance &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; from an optical fiber mounted at the focal plane of a telescope, how big (in terms of dimensions) that object should be to be detectable by the signal coming from the optical fiber?&lt;/strong&gt;  (&lt;em&gt;I couldn't find any related specification in the web. Since I need to use the answer as a part of my research, any cited reference is extremely appreciated.&lt;/em&gt;) &lt;/p&gt;&#xA;" OwnerUserId="7843" LastActivityDate="2019-01-11T04:05:54.490" Title="What is the dimensions of the smallest object detectable by an optical fiber from a specific distance?" Tags="&lt;observational-astronomy&gt;&lt;spectroscopy&gt;&lt;sky-survey&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29113" PostTypeId="1" CreationDate="2019-01-10T23:55:26.060" Score="2" ViewCount="77" Body="&lt;p&gt;Sometimes it does not let me delete text in an xgterm terminal and outputs ^H instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Run IRAF:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;xgterm -e cl&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It opens the xgterm, there I type:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;splot black_hole.fits[0]&#xA;Image line/aperture to plot (0:) (12): 192&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then I realise I made a mistake and press the Delete key on Mac, and it writes the ^H text instead:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Image line/aperture to plot (0:) (12): 192^H^H&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is very annoying. Is there a way to fix it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that delete key works in xgterm initially, but breaks and writes ^H in the case above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've installed IRAF 2.16 through astroconda iraf-all package on Mac.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Update&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The only solution I found was to use PyRAF instead. PyRAF uses the normal Mac's terminal, which does not have those Delete key issues. I can still run the same IRAF commands in PyRAF terminal, so it works fine for me.&lt;/p&gt;&#xA;" OwnerUserId="21367" LastEditorUserId="21367" LastEditDate="2019-01-11T10:03:51.053" LastActivityDate="2019-01-11T10:03:51.053" Title="Delete key does not always work in IRAF's xgterm on Mac" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29114" PostTypeId="2" ParentId="29113" CreationDate="2019-01-11T03:08:31.687" Score="2" Body="&lt;p&gt;This sounds like a problem with your terminal settings.  Quoting the &lt;strong&gt;Setting up IRAF&lt;/strong&gt; section &lt;a href=&quot;http://mips.as.arizona.edu/~khainline/intro_to_IRAF.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;, &quot;We assume the user has defined this correctly when issuing the MKIRAF and no longer key off the unix TERM to set a default.&quot;  The page goes on to say that one should add &lt;code&gt;stty xgterm&lt;/code&gt; the to your &lt;code&gt;login.cl&lt;/code&gt; file (directly above the final &lt;code&gt;keep&lt;/code&gt; line might be a good place).  The page linked above goes into more detail.&lt;/p&gt;&#xA;" OwnerUserId="24184" LastEditorUserId="24184" LastEditDate="2019-01-11T03:39:02.163" LastActivityDate="2019-01-11T03:39:02.163" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29115" PostTypeId="2" ParentId="29111" CreationDate="2019-01-11T03:28:54.307" Score="4" Body="&lt;p&gt;Assuming you are using &lt;a href=&quot;https://en.wikipedia.org/wiki/Single-mode_optical_fiber&quot; rel=&quot;nofollow noreferrer&quot;&gt;single mode fiber&lt;/a&gt;, the entrance (mode diameter) is small, usually a few microns (extremes can range from 1 to say 30 microns, but 2-8 microns is common for visible SMF).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The only constraint:&lt;/strong&gt; you have is to get enough light into that small spot. You can do that with a single lens or convex mirror, a microscope, or a telescope, or even by proximity (hold the fiber so very close to a tiny object that enough light reaches the core, works for certain nanoparticle or biological experiments).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Size of target object:&lt;/strong&gt; So far &lt;em&gt;I can not think of any constraint on the size of the object&lt;/em&gt;. Really the question is how bright the object is and how sensitive the instrument is at the other end. If you have a high-dispersion spectrometer, you probably need a lot more light than if you are doing broadband photometry by counting photons with a photomultiplier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Object is moving:&lt;/strong&gt; It doesn't matter if the object is moving or fixed, as long as your optical system can move to track it so that the focused image falls on the core of the fiber. That's simply a mechanical and tracking issue not related to the fiber. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The big advantage&lt;/strong&gt; of using a fiber is now obvious, you can move your telescope, but keep your measurement instrument somewhere else and fixed, and the flexible fiber will guide the light to it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Further background&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The trick is to put a &lt;em&gt;very large telescope&lt;/em&gt; between the object and the fiber, and to use a robot to place hundreds or thousands of fibers at the focal plane, each at the exact location of a galaxy or a star you would like to collect. This way you can line up the other ends of the fibers along a single slit of a spectrometer and study the spectra of thousands or even millions of objects over time without having to point the telescope to each object one-at-a-time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fibers have some big advantages:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Multiplicity and configurability (as mentioned above)&lt;/li&gt;&#xA;&lt;li&gt;Flexibility (literally) it allows the telescope and the instrument to be mounted separately and not have to be mechanically aligned and stabilized.&lt;/li&gt;&#xA;&lt;li&gt;Mechanical stabilization of source for instruments: even if the telescope drifts and the object moves slightly with respect to the entrance of the fiber, the physical location of the &lt;em&gt;fiber exit&lt;/em&gt; can remain fixed. So tracking errors don't affect the distribution of intensity within the entrance slit (virtual or real) of the fiber and therefore don't cause instrumental spectral shifts.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2xmtX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2xmtX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; From the astrophysicist Ángel R. López-Sánchez's enjoyable &lt;a href=&quot;https://angelrls.wordpress.com/2014/05/06/a-2df-night-at-the-aat/&quot; rel=&quot;nofollow noreferrer&quot;&gt;A 2dF night at the Anglo-Australian Telescope&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uV0wo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uV0wo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above::&lt;/strong&gt; From AAO's &lt;a href=&quot;https://www.aao.gov.au/news-media/media-releases/the-stellar-cluster-behind-sirius&quot; rel=&quot;nofollow noreferrer&quot;&gt;The stellar cluster behind Sirius&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/X9igP.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X9igP.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; GIF from the video (with exciting music) &lt;a href=&quot;https://www.youtube.com/watch?v=qOtkI4HYzTk&quot; rel=&quot;nofollow noreferrer&quot;&gt;A 2dF night at the Anglo-Australian Telescope&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;And if you like &lt;em&gt;that robot&lt;/em&gt; then consider the &quot;magic fingers&quot; at the &lt;a href=&quot;https://www.desi.lbl.gov/focal-plane-system/&quot; rel=&quot;nofollow noreferrer&quot;&gt;focal plane of the Dark Energy Spectroscopic Instrument&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x89Xx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x89Xx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lZJJ3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lZJJ3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;DESI’s focal plane is designed to sit high atop the Mayall Telescope and to carry 5,000 robotic positioners, each holding a fiber-optic cable. Each of these fiber-toting robots is automatically positioned to fix on a preset sequence of individual galaxies and quasars so that the fibers can collect their light. The movements of these positioners must be carefully choreographed to avoid bumping into one another. The roundish focal plane, which measures nearly a meter in diameter, consists of 10 pie-shaped wedges that are fit snugly together. Each wedge holds 500 robotic positioners. The focal plane also contains sensors and light sources called field fiducials that help ensure the positioners are in proper alignment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And also these robotic fibers in Australia's &lt;a href=&quot;https://www.aao.gov.au/news-media/media-releases/TAIPAN-australian-robotic-eyes-to-observe-the-heavens&quot; rel=&quot;nofollow noreferrer&quot;&gt;TAIPAN instrument at Siding Spring Observatory in North-Western NSW&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DaTq8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DaTq8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-01-11T04:05:54.490" LastActivityDate="2019-01-11T04:05:54.490" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29116" PostTypeId="1" AcceptedAnswerId="29117" CreationDate="2019-01-11T04:18:27.373" Score="0" ViewCount="111" Body="&lt;p&gt;Many older or &quot;classic&quot; telescope domes have a horizon-to-zenith opening in the dome, and this helps speed up the thermal equilibration between the inside and outside air, decreasing turbulence and its effects upon resolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ncIsj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ncIsj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:P200_Dome_Open.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/joYQP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/joYQP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://gostargazing.co.uk/observatories-in-the-uk/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the Anglo-Australian Telescope has second mechanism that tightly constrains the vertical extent of the opening as well, leaving just a tiny hole barely big enough for the telescope to see the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the benefits of each of these, and how do the minimal aperture domes deal with temperature differences?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; Screen shot from the video &lt;a href=&quot;https://www.youtube.com/watch?v=qOtkI4HYzTk&quot; rel=&quot;nofollow noreferrer&quot;&gt;A 2dF night at the Anglo-Australian Telescope&lt;/a&gt; found &lt;a href=&quot;https://angelrls.wordpress.com/2014/05/06/a-2df-night-at-the-aat/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wKMix.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wKMix.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-01-11T06:13:56.757" Title="Why is the opening in the Anglo-Australian Telescope's dome so small?" Tags="&lt;observatory&gt;&lt;angular-resolution&gt;&lt;weather&gt;&lt;atmospheric-effects&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29117" PostTypeId="2" ParentId="29116" CreationDate="2019-01-11T05:30:08.390" Score="4" Body="&lt;p&gt;This is a two part windscreen designed to minimize the effects of windshake on the telescope and to avoid the deterioration in image quality that wind would cause. The AAT is in a tall 6 story dome on a pretty exposed part of Siding Spring Mountain and so is likely more affected by wind gusts. Initially there were issues with the mount being too flexible and having an lower resonant frequency than expected, which was in the range that can be excited by wind buffeting (&lt;a href=&quot;https://www.science.org.au/fellowship/fellows/biographical-memoirs/harry-clive-minnett-1917%E2%80%932003#8&quot; rel=&quot;nofollow noreferrer&quot;&gt;a mention&lt;/a&gt; in the bio of Harry Minnett, a CSIRO engineer who worked on the AAT). This may have also lead to a more extensive windscreens but windscreens, particularly upper ones, are fairly common and exist at several other observatories. As an example, the &lt;a href=&quot;http://www.ing.iac.es/Astronomy/telescopes/int/int_pointing_limits.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;pointing limits&lt;/a&gt; of the Isaac Newton Telescope page discusses the sky area that is available with and without lowering the upper windshield.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastEditorUserId="23748" LastEditDate="2019-01-11T06:13:56.757" LastActivityDate="2019-01-11T06:13:56.757" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29118" PostTypeId="1" AcceptedAnswerId="29135" CreationDate="2019-01-11T06:25:47.573" Score="0" ViewCount="65" Body="&lt;p&gt;There is news of an analysis of a large number of nearby white dwarfs revealing a &quot;pile-up&quot; at a certain point in their cooling which has been predicted to happen due to a phase change where nuclear material crystalizes. This is analogous to a mixture of ice and water remaining at 0C while it is in the process of freezing (or thawing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is this crystalline material like? Is it really a lattice of Carbon and Oxygen nuclei in some recognizable structure (cubic BCC... etc.)? Is it related in some way to &lt;a href=&quot;https://en.wikipedia.org/wiki/Nuclear_pasta&quot; rel=&quot;nofollow noreferrer&quot;&gt;nuclear pasta&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nature: &lt;a href=&quot;https://www.nature.com/articles/s41586-018-0791-x&quot; rel=&quot;nofollow noreferrer&quot;&gt;Core crystallization and pile-up in the cooling sequence of evolving white dwarfs&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-01-12T08:10:52.670" Title="Is the structure of crystalized white dwarf related to nuclear pasta?" Tags="&lt;astrophysics&gt;&lt;stellar-evolution&gt;&lt;white-dwarf&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29119" PostTypeId="2" ParentId="29061" CreationDate="2019-01-11T07:57:17.463" Score="-1" Body="&lt;p&gt;OK, this answer involves red-shifts but hear me out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under General Relativity, multiple mechanisms can create red-shifts: expansion of space, objects moving relative to an observer (i.e. us), and light moving out of gravity well.  The latter option is outside the scope of this question and the former is excluded from consideration at the questioner's request.  That leaves only the second option (relative motion, a.k.a. the relativistic Doppler effect) under consideration; this shifting can be (and has been) tested here on Earth and has been shown to exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Red-shift is observed in all apparently distant objects (dim, low metallicity, etc.).  From the red-shift of the spectra seen in any given object we can determine how quickly it is moving away from us.  For example, an object with a measured red-shift of &lt;span class=&quot;math-container&quot;&gt;$z=0.5$&lt;/span&gt; is moving away from us at about half the speed of light.  So far so good.  The trouble arises when we observe objects with &lt;span class=&quot;math-container&quot;&gt;$z&amp;gt;1$&lt;/span&gt;.  Many such objects have been found; the current record-holder is &lt;a href=&quot;https://en.wikipedia.org/wiki/GN-z11&quot; rel=&quot;nofollow noreferrer&quot;&gt;GN-z11&lt;/a&gt; with a red-shift of &lt;span class=&quot;math-container&quot;&gt;$z=11.09$&lt;/span&gt;.  Put another way, if only relativistic shifting were at play this object would be travelling away from us at over 11 times the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that no object having mass can reach light-speed it is clear that the observed red-shifts cannot be caused by relativistic motion.  Since there are no known mechanisms beyond the three listed above that can cause red-shifts &lt;em&gt;in the spectra&lt;/em&gt; (compare &lt;a href=&quot;https://en.wikipedia.org/wiki/Extinction_(astronomy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;extinction&lt;/a&gt;), the only explanation matching these observations is the expansion of space.  Put succinctly, the fact that superluminal red-shifts are observed &lt;strong&gt;at all&lt;/strong&gt; is evidence that space is expanding.&lt;/p&gt;&#xA;" OwnerUserId="24184" LastEditorUserId="24184" LastEditDate="2019-01-11T08:16:49.533" LastActivityDate="2019-01-11T08:16:49.533" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29120" PostTypeId="2" ParentId="28929" CreationDate="2019-01-11T08:35:32.040" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://www.theguardian.com/science/2019/jan/11/scientists-close-to-capturing-first-image-of-black-hole-at-the-centre-of-the-milky-way&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; neewpaper article appears to reflect new information from the project.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The observations, by the Event Horizon Telescope, are expected to be unveiled in March . . .&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The team is in the final phase of reviewing data that was gathered in 2017 . . .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nothing there about the 2018 data.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-01-11T08:35:32.040" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29122" PostTypeId="1" AcceptedAnswerId="29126" CreationDate="2019-01-11T10:21:25.150" Score="53" ViewCount="15569" Body="&lt;p&gt;Light doesn’t accelerate in a gravitational field, which things with mass would do, because light has a universally constant velocity. Why is that exception?&lt;/p&gt;&#xA;" OwnerUserId="23217" LastEditorUserId="-1" LastEditDate="2019-01-14T12:25:34.050" LastActivityDate="2021-04-08T09:46:18.757" Title="If light has no mass, why is it affected by gravity?" Tags="&lt;light&gt;&lt;general-relativity&gt;" AnswerCount="5" CommentCount="6" FavoriteCount="15" ContentLicense="CC BY-SA 4.0" />
  <row Id="29123" PostTypeId="2" ParentId="29122" CreationDate="2019-01-11T10:45:24.893" Score="31" Body="&lt;p&gt;There are a couple of ways one could approach your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes are regions of space that have been deformed by a sufficiently concentrated mass.  Light waves/particles always travel in a straight line at a constant velocity (&lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;).  Although a photon approaching a black hole will continue traveling in a straight line through space, space itself has curved so the photon's path will curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While photons don't speed up in the presence of a gravity well, they are affected by it in other ways.  In specific, photons entering a gravity well are blue-shifted while photons leaving one are red-shifted.  This red/blue-shifting happens because time passes slower within a gravity well than without.  In all frames of reference, though, the speed of light remains constant.  There's more info on this &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_redshift&quot; rel=&quot;noreferrer&quot;&gt;on the wiki&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Note: The question originally referred specifically to black holes.  The above hold for any concentration of matter (of which black holes are an extreme example).&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="24184" LastEditorUserId="24184" LastEditDate="2019-01-12T00:05:19.800" LastActivityDate="2019-01-12T00:05:19.800" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29126" PostTypeId="2" ParentId="29122" CreationDate="2019-01-11T14:24:40.713" Score="50" Body="&lt;p&gt;Another way to answer this question is to apply the Equivalence Principle, which Einstein called his &quot;happiest thought&quot; (so you know it has to be good).  The equivalence principle says that if you are in an enclosed box in the presence of what Newton would call a gravitational field, then everything that happens in that box must be the same as if the box was not in a gravitational field, but accelerating upward instead.  So when you release a ball, you can imagine the ball is accelerated downward by gravity, or you can imagine everything but the ball is accelerated upward, and the ball is simply being left behind (which, ironically, checks better with the stresses you can easily detect on every object around you that are not present on the ball, including the feeling you are receiving from your bottom right now).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that rule, it is easy to see how light would be affected by gravity-- simply imagine shining a laser horizontally.  In the &quot;left behind&quot; reference frame, we see what would happen-- the beam would start from a sequentially higher and higher point, and that raising effect is &lt;em&gt;accelerating.&lt;/em&gt;  So given the finite speed of light, the shape of the beam would appear to curve downward, and the beam would not strike the point on the wall of the box directly opposite the laser.  Therefore, this must also be what is perceived from inside the box-- the beam does not strike the point directly across from the laser (as that point is getting higher then the point across from it where the light was emitted), and its path appears to curve downward.  Ergo, light &quot;falls.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, this is the crucial simplification of the Equivalence Principle-- you never need to know what the substance is, all substances &quot;fall the same&quot; because it's nothing happening to the substance, it is just the consequences of being &quot;left behind&quot; by whatever actually does have forces on it and is actually accelerating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, it is interesting to note that even in Newtonian gravity, massless objects would &quot;fall the same&quot; as those with mass, but to see it requires taking a limit.  Simply drop a ball in a vacuum, then a lower mass ball, then a lower still mass.  All objects fall the same under Newtonian gravity.   So simply proceed to the limit of zero mass, you will not see any difference along the path of that limit.  Nevertheless, Newtonian gravity doesn't get the answer quite right for the trajectory of light in gravity, because Newtonian physics doesn't treat the speed of light correctly.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2019-01-11T14:30:01.227" LastActivityDate="2019-01-11T14:30:01.227" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="29127" PostTypeId="1" AcceptedAnswerId="29131" CreationDate="2019-01-11T14:33:08.980" Score="1" ViewCount="749" Body="&lt;p&gt;How can I obtain ephemeris data which fits to a specific location? Differently phrased: I need ephemeris data of satellites which orbited a specific location. Is there a document where I can look up references?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now I am looking in &lt;a href=&quot;ftp://cddis.gsfc.nasa.gov/gnss/data/daily/2019/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ftp://cddis.gsfc.nasa.gov/gnss/data/daily/2019/&lt;/a&gt; but the subfolders only contain weirdly named broadcast files which don't make much sense to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I need these ephemeris files is that I want to test &lt;code&gt;gps-sdr-sim&lt;/code&gt; with a USRP I got. I know that those ephemeris broadcast files contain data to every other satellite in orbit (but not in the same file as I understand). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried multiple ephemeris files with &lt;code&gt;gps-sdr-sim&lt;/code&gt; but none of the data transmitted worked for the location I am in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="25492" LastEditorUserId="7472" LastEditDate="2019-01-11T21:10:42.787" LastActivityDate="2019-01-11T21:45:32.193" Title="How to get ephemeris data for location?" Tags="&lt;satellite&gt;&lt;artificial-satellite&gt;&lt;ephemeris&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29128" PostTypeId="2" ParentId="29122" CreationDate="2019-01-11T14:39:45.570" Score="0" Body="&lt;p&gt;Acceleration is not relevant here. Any given gravity well has a definable escape velocity. Particles faster then that velocity escape the well, particles slower do not. The very definition of a black hole is a gravity well (hole) where the escape velocity exceeds 'c' the speed of light particles, so by definition, light can not escape from the hole, making it 'black'.  &lt;/p&gt;&#xA;" OwnerUserId="25494" LastActivityDate="2019-01-11T14:39:45.570" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29129" PostTypeId="1" CreationDate="2019-01-11T20:00:01.187" Score="2" ViewCount="49" Body="&lt;p&gt;In the past few days, journalists have been playing the usual &quot;X could be aliens&quot; game. This time, X is repeating FRBs from a galaxy 1.5 billion light years away. (Or is it 1.5 billion years ago? The proper/conformal time distinction isn't clear in pop sci.) I couldn't find the galaxy's &lt;em&gt;name&lt;/em&gt; anywhere, but then I suppose a galaxy that distant might not have been named, even if it's now of some interest. Still, the claim of &quot;repeated&quot; bursts (regardless of their cause) made me wonder: what is our tightest empirical upper bound on how far apart the detected bursts' sources are?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm prepared to believe we can see 1.5 billion light years away, clearly enough to pin two events to the same 100,000-light-year wide galaxy, with no others within 2 million light years of it (these are plausible example numbers based on comparing the Milky way to Andromeda). They're small fractions of 1.5 billion light years, but not so small we couldn't resolve an image that clearly. On the other hand, I sincerely doubt we can prove any two bursts that distant are from the same star. I presume what's notable about repeat FRBs is they're so close together in time we &lt;em&gt;expect&lt;/em&gt; the source to be common, but I'm curious as to how tightly observations constrain that.&lt;/p&gt;&#xA;" OwnerUserId="16921" LastActivityDate="2021-03-18T17:00:45.080" Title="How tightly clustered are the CHIME-detected FRB repeats?" Tags="&lt;galaxy&gt;&lt;distances&gt;&lt;fast-radio-bursts&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29130" PostTypeId="1" CreationDate="2019-01-11T21:38:50.473" Score="3" ViewCount="74" Body="&lt;p&gt;Since observers in different parts of Earth will be observing the moon at slightly different angles relative to its (choose reference vector: north pole, origin of latitude longitude position, whatever), they will be observing very slightly different faces of the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried looking online to see if anyone had examples of photos of the moon taken from different parts of Earth at the same time but got overwhelmed of photos showing that the moon appears &quot;upside down&quot; in the southern hemisphere relative to what a northern observer would expect and couldn't find any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have or know where to find an example of what I'm asking?  Please note that I am &lt;em&gt;not&lt;/em&gt; asking about the &quot;upside down&quot; moon effect I just described, and neither am I asking about lunar libration: I am specifically asking about being able to observe slightly different faces of the moon simply by standing somewhere else on Earth as someone else at the same time.&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2019-01-11T21:38:50.473" Title="Any pictures showing difference in moon between southern and northern observers?" Tags="&lt;the-moon&gt;" AnswerCount="0" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="29131" PostTypeId="2" ParentId="29127" CreationDate="2019-01-11T21:45:32.193" Score="1" Body="&lt;p&gt;The merged broadcast ephemeris is in the &lt;code&gt;YYn&lt;/code&gt; directory where (&lt;code&gt;YY&lt;/code&gt; is the 2 digit year, &lt;code&gt;n&lt;/code&gt; means navigation e.g. &lt;code&gt;19n&lt;/code&gt;) and you are looking for the &lt;code&gt;brdcDDD0.YYn.Z&lt;/code&gt; where &lt;code&gt;DDD&lt;/code&gt; is the day number (of the year) and &lt;code&gt;YY&lt;/code&gt; is the 2 digit year as before. These are compressed RINEX files; &lt;code&gt;gunzip&lt;/code&gt; etc will be able to uncompress them. The RINEX format is pretty well documented and I've written Python code to read them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other files in those directories are the received GPS signals from the various GNSS monitoring sites. You may be able to make progress by using the map on the &lt;a href=&quot;https://cddis.nasa.gov/Data_and_Derived_Products/CddisArchiveExplorer.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Archive Explorer&lt;/a&gt; to find a site near you, noting its code and then downloading its data file. For me the nearest is VNDP (Vandenberg AFB) and so for 2019-Jan-01 I would download &lt;code&gt;vndp0010.19n.Z&lt;/code&gt; from &lt;code&gt;ftp://cddis.gsfc.nasa.gov/gnss/data/daily/2019/001/19n/&lt;/code&gt;. Otherwise I think you would be looking at having to do your own visibility code to work out what satellites were visible from your location at the specific time (since the satellites have a ~12 hour orbital period and come and go)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may want to look at &lt;a href=&quot;https://github.com/tomojitakasu/RTKLIB/tree/rtklib_2.4.3&quot; rel=&quot;nofollow noreferrer&quot;&gt;RTKlib&lt;/a&gt; (binaries available there at github also) which has a lot of tools for processing these sorts of data and has &lt;code&gt;rtkget&lt;/code&gt; which can automate the download of the various types of files from CDDIS etc.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2019-01-11T21:45:32.193" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29132" PostTypeId="1" CreationDate="2019-01-11T22:02:41.807" Score="20" ViewCount="1208" Body="&lt;p&gt;&lt;a href=&quot;https://xkcd.com/2003/&quot; rel=&quot;noreferrer&quot;&gt;xkcd #2003&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6BqQi.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6BqQi.png&quot; alt=&quot;Ties are broken by whoever was closest to the surface of Europa when they were born.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gives a humorous line of succession to the United States presidency. The comic's title text, found by hovering over the comic, states that a tie is broken by whoever was closest to the surface of Europa when they were born.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only potential tie on the list occurs in item #19. It states the &quot;champion&quot; of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Nathan%27s_Hot_Dog_Eating_Contest&quot; rel=&quot;noreferrer&quot;&gt;Nathan's Hot Dog Eating contest&lt;/a&gt;, but does not specify the men's or women's champion. At the time of the comic's publication, the men's champion was &lt;a href=&quot;https://en.wikipedia.org/wiki/Joey_Chestnut&quot; rel=&quot;noreferrer&quot;&gt;Joey Chestnut&lt;/a&gt; (born 1983-11-25) and the women's champion was &lt;a href=&quot;https://en.wikipedia.org/wiki/Miki_Sudo&quot; rel=&quot;noreferrer&quot;&gt;Miki Sudo&lt;/a&gt; (Wikipedia gives only 1985 as her year of birth, but &lt;a href=&quot;https://www.famousbirthdays.com/people/miki-sudo.html&quot; rel=&quot;noreferrer&quot;&gt;famousbirthdays.com claims July 22&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both were ineligible to hold the presidency at the time of publication due to not meeting the minimum age requirement of 35. But supposing they both were, who would be first in line by the tiebreaker of distance from the surface of Europa at birth?&lt;/p&gt;&#xA;" OwnerUserId="25504" LastActivityDate="2019-01-12T10:39:24.683" Title="Was Earth closer to Europa on 1983-11-25 or 1985-07-22?" Tags="&lt;distances&gt;&lt;europa&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29133" PostTypeId="2" ParentId="29132" CreationDate="2019-01-11T22:15:58.780" Score="12" Body="&lt;p&gt;Any online planetarium or equivalent mobile app will tell you that on 1983-11-25 Jupiter was near to its &lt;a href=&quot;https://en.wikipedia.org/wiki/Conjunction_(astronomy)#Superior_and_inferior&quot; rel=&quot;nofollow noreferrer&quot;&gt;conjunction&lt;/a&gt; with the Sun:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aUdnS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aUdnS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while on 1985-07-22 it was close to &lt;a href=&quot;https://en.wikipedia.org/wiki/Opposition_(planets)&quot; rel=&quot;nofollow noreferrer&quot;&gt;opposition&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CRZRe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CRZRe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So on Miki Sudo's birthday, Europa was about 300 million kilometers closer to the Earth than on Joey Chestnut's birthday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;images taken from the Star Walk iOS app&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="7472" LastEditorUserId="7472" LastEditDate="2019-01-12T10:39:24.683" LastActivityDate="2019-01-12T10:39:24.683" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29134" PostTypeId="2" ParentId="29132" CreationDate="2019-01-12T05:51:35.183" Score="22" Body="&lt;p&gt;tl;dr &lt;strong&gt;Miki Sudo&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using JPL's &lt;a href=&quot;https://naif.jpl.nasa.gov/naif/toolkit.html&quot; rel=&quot;noreferrer&quot;&gt;SPICE toolkit&lt;/a&gt;, I computed the positions of Earth and Europa for the times in question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4zxYK.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4zxYK.png&quot; alt=&quot;Relative Positions of Earth and Europa on 1983-Nov-25&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/37dQu.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/37dQu.png&quot; alt=&quot;Relative Positions of Earth and Europa on 1985-Jul-22&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On 1983-Nov-25, Earth and Europa are 935.2 million km apart, while 1985-Jul-22, they are 612.5 million km apart. Miki Sudo wins by 323 million km, given the assumed date for her birthday. If we don't trust famousbirthdays.com for Sudo's birthday, but assume it was still sometime in 1985, she still wins.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/osp3Y.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/osp3Y.png&quot; alt=&quot;Earth-Europa Distance in 1985&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, it turns out Joey Chesnut was born at a time when Earth was close to its farthest point from Europa, so he's fairly unlikely to ever win in this tiebreaker:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/vR2RJ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vR2RJ.png&quot; alt=&quot;Earth-Europa Distance, 1980-220&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25505" LastActivityDate="2019-01-12T05:51:35.183" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29135" PostTypeId="2" ParentId="29118" CreationDate="2019-01-12T08:10:52.670" Score="2" Body="&lt;p&gt;It's thought to be crystalline in the same sense that diamonds are a crystal. The coulomb interactions between the positively charged ions becomes much larger than their thermal energies and they settle into a crystal lattice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This both releases latent heat &lt;em&gt;and&lt;/em&gt; increases the thermal heat capacity of the ions (roughly doubles it). This causes a &quot;pile up&quot; of white dwarfs with interior temperatures at around the crystallisation phase transition. However, the pile-up is not too steep because the freezing temperature increases with density and white dwarf interiors are roughly isothermal. This means they freeze from the inside out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crystalline structure is expected to be bcc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the only thing this has in common with nuclear pasta is that they are both phase transitions. Nuclear pasta is the aggregation of nucleons (neutrons and protons) into wacky shapes in order to minimise the total energy density of the gas. The temperature is not really an issue here and it is all to do with density.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-01-12T08:10:52.670" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29136" PostTypeId="1" CreationDate="2019-01-12T08:58:03.353" Score="27" ViewCount="3865" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Carl_Sagan&quot; rel=&quot;noreferrer&quot;&gt;Carl Sagan&lt;/a&gt; has said several times that we are &quot;star-stuff&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One instance can be found in Good Reads' &lt;a href=&quot;https://www.goodreads.com/quotes/144310-the-nitrogen-in-our-dna-the-calcium-in-our-teeth&quot; rel=&quot;noreferrer&quot;&gt;Carl Sagan &gt; Quotes &gt; Quotable Quote&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The nitrogen in our DNA, the calcium in our teeth, the iron in our blood, the carbon in our apple pies were made in the interiors of collapsing stars. We are made of starstuff.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Was most of my nitrogen really made in the interior of a star during its collapse? Was my calcium and iron made there as well, and not (for example) in an expanding shell after a supernova?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="2153" LastEditDate="2019-01-13T15:18:32.527" LastActivityDate="2019-01-13T15:18:32.527" Title="Are we really star-stuff from the interior of collapsing stars?" Tags="&lt;star&gt;&lt;supernova&gt;&lt;nucleosynthesis&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29137" PostTypeId="2" ParentId="29136" CreationDate="2019-01-12T12:53:37.830" Score="43" Body="&lt;p&gt;The straightforward answer is, &quot;Yes, we are made of star stuff.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of it will be from the interior of collapsing stars, some will be from supernovas, some from normal everyday fusion, and some from other processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers from @HDE226868 and @RobJeffries on &lt;a href=&quot;https://astronomy.stackexchange.com/q/8894/43&quot;&gt;this question&lt;/a&gt; on where heavier elements come from gives good background, including this nugget:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The split between r-process and s-process production of heavier than iron (peak) elements is about 50:50. ie They weren't mainly made in supernovae, which is a frequent, incorrect claim.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but of most importance is Rob's final point:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The relative contributions of various sites to the r-process remains an unsettled matter. You could also read my answers on this topic in Physics Stack Exchange.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On following Rob's links I think &lt;a href=&quot;https://physics.stackexchange.com/questions/7131/origin-of-elements-heavier-than-iron-fe&quot;&gt;this&lt;/a&gt; provides you with an excellent overall answer (and relative percentages)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A more up-to-date visualisation of what goes on (produced by &lt;a href=&quot;http://www.astronomy.ohio-state.edu/~jaj/nucleo/&quot; rel=&quot;noreferrer&quot;&gt;Jennifer Johnson&lt;/a&gt;) and which attempts to identify the sites (as a percentage) for each chemical element is shown below. It should be stressed that the details are still subject to a lot of model-dependent uncertainty.&#xA;  &lt;a href=&quot;https://i.stack.imgur.com/Gy0lF.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gy0lF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Looking at C and N - the majority seems to be from dying low mass stars, and Ca and Fe are from exploding stars, which indicates that Carl is not far off the mark.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2019-01-12T12:53:37.830" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29138" PostTypeId="2" ParentId="29136" CreationDate="2019-01-12T15:31:44.203" Score="19" Body="&lt;p&gt;Sagan's quote is half-correct. While some of these elements are created during or immediately prior to a supernova of some sort, others are either partially or entirely fused during normal stellar nucleosynthesis. Nitrogen falls into the latter category, whereas calcium and iron have one foot in each. On the whole, though, calling these elements &amp;quot;starstuff&amp;quot; is pretty accurate.&lt;/p&gt;&#xA;&lt;h1&gt;Nitrogen&lt;/h1&gt;&#xA;&lt;p&gt;My answer serves largely to complement &lt;a href=&quot;https://astronomy.stackexchange.com/a/29137/2153&quot;&gt;Rory's&lt;/a&gt;, and to address the issue of nitrogen production in particular, partly since &lt;a href=&quot;http://blog.sdss.org/2017/01/09/origin-of-the-elements-in-the-solar-system/&quot; rel=&quot;noreferrer&quot;&gt;there is some disagreement&lt;/a&gt; as to how much high-mass stars produce. It is thought that the majority of nitrogen is produced in the &lt;a href=&quot;https://en.wikipedia.org/wiki/CNO_cycle&quot; rel=&quot;noreferrer&quot;&gt;carbon-nitrogen-oxygen (CNO) cycle&lt;/a&gt;, which includes the subprocess originally known as the CN cycle. The CNO cycle is only dominant in stars more massive than the Sun, partly because the energy generation rate is &lt;a href=&quot;http://burro.case.edu/Academics/Astr221/StarPhys/nuclear2.html&quot; rel=&quot;noreferrer&quot;&gt;much more sensitive to temperature than the proton-proton chain&lt;/a&gt; (scaling as &lt;span class=&quot;math-container&quot;&gt;$\epsilon\sim T^{20}$&lt;/span&gt;, compared to &lt;span class=&quot;math-container&quot;&gt;$\epsilon\sim T^4$&lt;/span&gt;), largely because the Coulomb barrier is much higher for the CNO cycle.&lt;/p&gt;&#xA;&lt;p&gt;Intermediate-mass &lt;a href=&quot;https://en.wikipedia.org/wiki/Asymptotic_giant_branch&quot; rel=&quot;noreferrer&quot;&gt;AGB stars&lt;/a&gt;, with masses in the vicinity of &lt;span class=&quot;math-container&quot;&gt;$5M_{\odot}$&lt;/span&gt;, enrich the interstellar medium with nitrogen through strong stellar winds (&lt;a href=&quot;https://www.aanda.org/articles/aa/pdf/2002/01/aadj242.pdf&quot; rel=&quot;noreferrer&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;https://ned.ipac.caltech.edu/level5/Sept06/Tosi/Tosi4.html&quot; rel=&quot;noreferrer&quot;&gt;2&lt;/a&gt;), and are thought to be the most important contributors to nitrogen synthesis. AGB stars are post-main sequence stars that have ascended the red giant branch and are now large and luminous, undergoing shell burning of helium and hydrogen. Their high mass-loss rates are responsible for the enrichment, and most of this mass loss happens &lt;em&gt;before&lt;/em&gt; the planetary nebula phase; therefore, I'd be reluctant to characterize the sources of nitrogen as even dying stars. They're simply old, evolved intermediate-mass stars - still not massive enough to undergo supernovae, but also not true low-mass stars.&lt;/p&gt;&#xA;&lt;p&gt;In short, the answer to the nitrogen question is no, most nitrogen in the universe was &lt;em&gt;not&lt;/em&gt; made from supernova nucleosynthesis, but was indeed made by lower-mass stars, in particular intermediate-mass AGB stars. The contributions of supernovae are, as indicated above, not agreed upon.&lt;/p&gt;&#xA;&lt;h1&gt;Calcium&lt;/h1&gt;&#xA;&lt;p&gt;Calcium can indeed be produced via nucleosynthesis in massive stars, usually via &lt;a href=&quot;https://en.wikipedia.org/wiki/Silicon-burning_process&quot; rel=&quot;noreferrer&quot;&gt;silicon-&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Oxygen-burning_process&quot; rel=&quot;noreferrer&quot;&gt;oxygen-&lt;/a&gt; based pathways that synthesize &lt;span class=&quot;math-container&quot;&gt;$^{40}\text{Ca}$&lt;/span&gt;, a common calcium isotope. Recently, discoveries of &lt;a href=&quot;https://en.wikipedia.org/wiki/Calcium-rich_supernovae&quot; rel=&quot;noreferrer&quot;&gt;calcium-rich supernovae&lt;/a&gt; have indicated that those could be substantial contributors to calcium abundances. The characteristics of the progenitors are not yet known; they could be &lt;a href=&quot;http://adsabs.harvard.edu/abs/2010Natur.465..322P&quot; rel=&quot;noreferrer&quot;&gt;low-mass white dwarfs&lt;/a&gt; accreting matter from a companion, compact objects colliding, or higher-mass stars undergoing traditional core collapse supernovae. We don't have enough data to determine what the contribution of these supernovae is to calcium production, although &lt;a href=&quot;http://adsabs.harvard.edu/abs/2014ApJ...780L..34M&quot; rel=&quot;noreferrer&quot;&gt;it's being worked on&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;h1&gt;Iron&lt;/h1&gt;&#xA;&lt;p&gt;Much of the iron produced by stars is in the form of &lt;a href=&quot;https://en.wikipedia.org/wiki/Iron-56&quot; rel=&quot;noreferrer&quot;&gt;the isotope &lt;span class=&quot;math-container&quot;&gt;$^{56}\text{Fe}$&lt;/span&gt;&lt;/a&gt;, which is one of the end results of &lt;a href=&quot;https://en.wikipedia.org/wiki/Silicon-burning_process&quot; rel=&quot;noreferrer&quot;&gt;silicon burning&lt;/a&gt; in the extreme late stages (essentially the last day or so) of a high-mass star's life, as well as in supernovae. &lt;span class=&quot;math-container&quot;&gt;$^{56}\text{Ni}$&lt;/span&gt; is initially synthesized but decays to &lt;span class=&quot;math-container&quot;&gt;$^{56}\text{Co}$&lt;/span&gt; and eventually &lt;span class=&quot;math-container&quot;&gt;$^{56}\text{Fe}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-01-12T18:26:18.720" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29140" PostTypeId="2" ParentId="29122" CreationDate="2019-01-12T18:26:01.330" Score="-6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If light has no mass, why is it affected by gravity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because light has a wave nature, and a gravitational field is a place where the speed of light varies. So light curves downwards. It's rather like the way sonar waves tend to curve downwards in the sea:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LsfoC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LsfoC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Image from FAS and the US Navy, see course ES310 &lt;a href=&quot;https://fas.org/man/dod-101/navy/docs/es310/SNR_PROP/snr_prop.htm?&quot; rel=&quot;nofollow noreferrer&quot;&gt;chapter 20&lt;/a&gt;&lt;/sup&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Light doesn’t speed up its acceleration, which things with mass would do, because light has a universally constant velocity. Why is that exception?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's not right I'm afraid. See what Einstein said:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol4-trans/144&quot; rel=&quot;nofollow noreferrer&quot;&gt;1912&lt;/a&gt;: “On the other hand I am of the view that the principle of the constancy of the velocity of light can be maintained only insofar as one restricts oneself to spatio-temporal regions of constant gravitational potential”.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol4-trans/325&quot; rel=&quot;nofollow noreferrer&quot;&gt;1913&lt;/a&gt;: “I arrived at the result that the velocity of light is not to be regarded as independent of the gravitational potential. Thus the principle of the constancy of the velocity of light is incompatible with the equivalence hypothesis”.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol4-trans/295&quot; rel=&quot;nofollow noreferrer&quot;&gt;1914&lt;/a&gt;: “In the case where we drop the postulate of the constancy of the velocity of light, there exists, a priori, no privileged coordinate systems.”&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol4-trans/271?highlightText=%22abandoned%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;1915&lt;/a&gt;: “the writer of these lines is of the opinion that the theory of relativity is still in need of generalization, in the sense that the principle of the constancy of the velocity of light is to be abandoned”.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol6-trans/340?highlightText=%22cannot%20claim%20any%20unlimited%20validity%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;1916&lt;/a&gt;: “In the second place our result shows that, according to the general theory of relativity, the law of the constancy of the velocity of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity and to which we have already frequently referred, cannot claim any unlimited validity”.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol7-trans/156?highlightText=%22spatially%20variable%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;1920&lt;/a&gt;: “Second, this consequence shows that the law of the constancy of the speed of light no longer holds, according to the general theory of relativity, in spaces that have gravitational fields. As a simple geometric consideration shows, the curvature of light rays occurs only in spaces where the speed of light is spatially variable”.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Einstein also &lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol5-trans/230?highlightText=%22refraction%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;spoke of&lt;/a&gt; &lt;em&gt;“the refraction of light rays by the gravitational field”&lt;/em&gt;. So did Newton, see &lt;a href=&quot;http://homepages.wmich.edu/~mcgrew/nq.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Opticks&lt;/a&gt; query 20: &lt;em&gt;“Doth not this aethereal medium in passing out of water, glass, crystal, and other compact and dense bodies in empty spaces, grow denser and denser by degrees, and by that means refract the rays of light not in a point, but by bending them gradually in curve lines?”&lt;/em&gt; It really is a refraction, and gravitational lensing is an appropriate phrase. Also see the GR section of &lt;a href=&quot;http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/speed_of_light.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Is The Speed of Light Everywhere the Same?&lt;/a&gt; It’s a PhysicsFAQ article by editor Don Koks. He talks about Einstein and refraction, and says this: &lt;em&gt;&quot;light speeds up as it ascends from floor to ceiling, and it slows down as it descends from ceiling to floor; it's not like a ball that slows on the way up and goes faster on the way down&quot;.&lt;/em&gt; Isn't that interesting? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of people will tell you that light curves &lt;em&gt;&quot;because it follows the curvature of spacetime&quot;&lt;/em&gt;, but that isn't right either. Spacetime curvature is associated with the tidal force, not the force of gravity. See my &quot;physics detective&quot; articles on &lt;a href=&quot;http://physicsdetective.com/the-speed-of-light/&quot; rel=&quot;nofollow noreferrer&quot;&gt;the speed of light&lt;/a&gt; and &lt;a href=&quot;http://physicsdetective.com/how-gravity-works/&quot; rel=&quot;nofollow noreferrer&quot;&gt;how gravity works&lt;/a&gt; for details and references.     &lt;/p&gt;&#xA;" OwnerUserId="9772" LastEditorUserId="9772" LastEditDate="2019-01-13T16:47:08.870" LastActivityDate="2019-01-13T16:47:08.870" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="29141" PostTypeId="1" AcceptedAnswerId="29144" CreationDate="2019-01-12T18:31:28.480" Score="1" ViewCount="173" Body="&lt;p&gt;I've read about Hubble seeing the brightest quasar in the early universe. Question is, how does science knows what is early and what is late?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please correct me if I'm wondering, but after the big bang, everything went in all directions equally. How do we know what direction is the center of the big bang? And continuing on the line of thought, is &quot;early universe&quot; considered closer to the origin of the big bang or further away?&lt;/p&gt;&#xA;" OwnerUserId="1305" LastEditorUserId="1305" LastEditDate="2019-01-15T17:57:55.487" LastActivityDate="2019-01-16T10:20:29.700" Title="How does science knows what is the early universe?" Tags="&lt;big-bang-theory&gt;&lt;hubble-telescope&gt;&lt;early-universe&gt;&lt;quasars&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29142" PostTypeId="2" ParentId="27219" CreationDate="2019-01-12T19:08:08.287" Score="5" Body="&lt;p&gt;An excellent book by Jean Meeus, &quot;Astronomical Algorithms&quot;, provides the calculations for planetary positions and Jupiter's Galilean Moons (Io, Europa, Ganymede, and Callisto). Meeus does not really explain the math to a great degree, so it is relatively simple to follow it as a formulaic approach. He uses VSOP for planetary positions and Lieske's E5 theory for the moons (you can get an explanation of both on Wikipedia). As a bonus, Meeus shows how to make calculations for both parameters with a lesser degree of accuracy which might improve your calculation speed. Though this book can be a bit dense, you might want to read up on the difference between &quot;ephemeris time&quot; and &quot;terrestrial time&quot;, as this will affect your accuracy as well.&lt;/p&gt;&#xA;" OwnerUserId="15619" LastEditorUserId="15619" LastEditDate="2019-01-31T20:14:35.280" LastActivityDate="2019-01-31T20:14:35.280" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29143" PostTypeId="1" CreationDate="2019-01-12T20:38:48.677" Score="3" ViewCount="84" Body="&lt;p&gt;I am practising some exam questions for a radio interferometry exam, and I am struggling with this question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The physical size of a single LOFAR station, operating at 150MHz is about 50m.&#xA;Estimate the number of pixels which would be needed to map the full primary beam if one wished to include all LOFAR stations (maximum baseline ~ 800 km) at full resolution.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, so the resolution of the entire array is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wavelength / Baseline Length which is 2m/800km = 2.5 x 10^-6 radians, or 0.52 arcseconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how to relate this angular resolution to the image size which I think is a 50m square.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help with this?&lt;/p&gt;&#xA;" OwnerUserId="25513" LastEditorUserId="14094" LastEditDate="2019-01-13T09:09:41.170" LastActivityDate="2019-01-13T09:09:41.170" Title="Estimate the number of pixels required to map the full primary beam of a Radio Telescope station" Tags="&lt;angular-resolution&gt;&lt;interferometry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29144" PostTypeId="2" ParentId="29141" CreationDate="2019-01-12T21:09:57.597" Score="5" Body="&lt;p&gt;I assume you're referring to &lt;a href=&quot;https://www.spacetelescope.org/news/heic1902/?lang&quot; rel=&quot;noreferrer&quot;&gt;the recent press release about the quasar J043947.08+163415.7&lt;/a&gt;, observed recently using &lt;em&gt;Hubble&lt;/em&gt;. &lt;a href=&quot;http://imgsrc.hubblesite.org/hvi/uploads/science_paper/file_attachment/363/main.pdf&quot; rel=&quot;noreferrer&quot;&gt;The paper about the observations&lt;/a&gt; details how the authors measured the distance to the quasar, by calculating its &lt;a href=&quot;https://en.wikipedia.org/wiki/Redshift&quot; rel=&quot;noreferrer&quot;&gt;redshift&lt;/a&gt;, a quantity that describes how the wavelength of light appears to change based on whether the object is moving relative to an observer. On cosmological scales, &lt;a href=&quot;https://physics.stackexchange.com/q/294099/56299&quot;&gt;redshift can then be converted to distances&lt;/a&gt;. We usually see quasars at &lt;span class=&quot;math-container&quot;&gt;$z&amp;gt;0.1$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As is usually the case, the quasar's redshift was determined by looking at spectral lines - in this case, Mg II emission around 21000 angstroms (see in particular the inset):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UnLpN.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UnLpN.png&quot; alt=&quot;Quasar spectrum, showing Mg II line&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This emission line was observed to have a different wavelength than it would if the quasar was at rest, enabling the astronomers to calculate its redshift: &lt;span class=&quot;math-container&quot;&gt;$z=6.511\pm0.003$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the typical process used to determine redshift, and therefore distance - well, not necessarily using that particular Mg II line, of course, but nonetheless the use of spectroscopy to measure how the location of spectral lines changes. The difference between &quot;early&quot; and &quot;late&quot; isn't quite clear, and probably varies depending on who you're talking to. For instance, someone studying the Big Bang might refer to the period before &lt;a href=&quot;https://en.wikipedia.org/wiki/Recombination_(cosmology)&quot; rel=&quot;noreferrer&quot;&gt;recombination&lt;/a&gt; as &quot;early&quot; - although we're seeing the quasar as it existed much later than that; it wasn't around for recombination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to note that &lt;a href=&quot;https://www.forbes.com/sites/startswithabang/2019/01/11/sorry-astronomers-havent-found-the-brightest-quasar-in-the-entire-universe/&quot; rel=&quot;noreferrer&quot;&gt;it's somewhat misleading to call this quasar the brightest (or even brightest known) quasar in the universe&lt;/a&gt;, because - as that press release notes - the gravitational lensing by a foreground galaxy, which enabled the object to be discovered, also magnified its brightness. It appears to have a brightness of 600 trillion solar luminosities, but without the lensing, that number is reduced to 11 trillion solar luminosities - a much more modest figure.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2019-01-12T21:27:20.847" LastActivityDate="2019-01-12T21:27:20.847" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29148" PostTypeId="1" AcceptedAnswerId="29156" CreationDate="2019-01-13T05:41:20.933" Score="1" ViewCount="125" Body="&lt;p&gt;I'm wondering exactly in which situations forces between bodies are, and are not consider to be &lt;em&gt;tidal forces&lt;/em&gt; in the context of planetary and astronomical science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two rigid, non-deformable spherical masses &lt;em&gt;with &lt;a href=&quot;https://space.stackexchange.com/questions/33463/which-point-in-an-orbiting-body-most-closely-follows-its-keplerian-trajectory#comment104012_33474&quot;&gt;shperically symmetric mass distributions&lt;/a&gt; orbit each other&lt;/em&gt;,  can we say there are no tidal forces or tidal effects there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one of them has a permanent, static deformation (e.g. quadrupole moments and higher) are there tidal forces then? If both do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or are dynamic, induced deformations required before we invoke the &quot;T-word&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can tidal forces be radial, such that they do not tend to raise or lower an orbit over time, or must they have tangential components, like the one that is slowly raising the Moon's orbit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to draw a clear line here, at least within the scope of this SE site and &lt;a href=&quot;https://astronomy.meta.stackexchange.com/q/480/7982&quot;&gt;This month's focus tag&lt;/a&gt; which is &lt;a href=&quot;/questions/tagged/tidal-forces&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;tidal-forces&amp;#39;&quot; rel=&quot;tag&quot;&gt;tidal-forces&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to that tag's Tag Info:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is no usage guidance for this tag … yet!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Slightly related: &lt;a href=&quot;https://physics.stackexchange.com/q/453460/83380&quot;&gt;Constraints on the mass distribution within each body such that their mutual orbits are Keplerian?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-01-13T05:48:43.993" LastActivityDate="2021-05-23T23:41:52.323" Title="In planetary and astronomical science, what exactly is, or is not, a tidal force?" Tags="&lt;tidal-forces&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29149" PostTypeId="1" AcceptedAnswerId="29154" CreationDate="2019-01-13T08:33:17.140" Score="3" ViewCount="109" Body="&lt;p&gt;Is there an compression algorithm for storing the information?&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What if there is a uniformaly complex object dropped to the black hole? Then there will be no compression of information.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How is there enough space on the surface (2D) to encode something which is in 3D. The planck distance is the same in the black hole as in the universe. So how can the black hole encode information about an object in 3D that is both uniformly complex and as dense as the black hole? Or is there a law that says 3D objects cannot be more dense then some level - which is sufficiently sparse so that when collapsed to a black hole all its information can be encoded on the surface of a sphere? Has this limit been found - given that a sphere has the lowest surface to volume of any object: &lt;a href=&quot;https://en.wikipedia.org/wiki/Surface-area-to-volume_ratio&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Surface-area-to-volume_ratio&lt;/a&gt; ?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Edit: added '3D' to first sentence of 2. question.&lt;/p&gt;&#xA;" OwnerUserId="25522" LastEditorUserId="37191" LastEditDate="2021-01-29T23:29:15.717" LastActivityDate="2021-01-30T01:09:01.017" Title="How is information preserved on the surface of a Black Hole?" Tags="&lt;black-hole&gt;&lt;holographic-principle&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="29150" PostTypeId="2" ParentId="29122" CreationDate="2019-01-13T08:48:49.353" Score="3" Body="&lt;p&gt;TL;DR &lt;strong&gt;Light is affected by gravity because it travels along the space-time grid and its curvature which &lt;em&gt;IS&lt;/em&gt; gravity. This gets very visible in black holes. &lt;code&gt;also: Einstein &amp;gt; Newton&lt;/code&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Black holes are black because no light that crosses the &amp;quot;Event Horizon&amp;quot; can escape ever again. Mass bends the &amp;quot;grid&amp;quot; of space-time. Light - &lt;strong&gt;&lt;em&gt;2-dimensionally speaking&lt;/em&gt;&lt;/strong&gt; - travels along the floor of the space-time grid and follows its curvature i.e. it goes down a cone created by a presence of mass, and moves along the shortest path outwards again. &lt;strong&gt;This makes the journey of the light take longer&lt;/strong&gt;. Now for a black hole things are more extreme: A black hole forms, when a lot of matter is crammed into a space that is at or smaller than the Schwarzschild Radius. &lt;strong&gt;The Schwarzschild Radius&lt;/strong&gt; of any stellar object is determined solely by its mass. Any mass with a high enough denisty turns into a black hole:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;r&lt;sub&gt;s&lt;/sub&gt;&lt;/strong&gt; = 2 * G / 2 c &lt;br&gt;&#xA;Schwarzschild Radius = &lt;code&gt;2* the gravitational constant / 2 * the speed of light&lt;/code&gt;.&lt;br&gt;&#xA;Multiply that with &lt;code&gt;M&lt;/code&gt;, the mass of an object in kg and you got the &lt;strong&gt;r&lt;sub&gt;s&lt;/sub&gt;&lt;/strong&gt; for that mass.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;To understand however how black holes curve space so much that they let no light escape, we must look at only a small part of Schwarzschilds equation.&lt;br&gt;&#xA;&lt;img src=&quot;https://wikimedia.org/api/rest_v1/media/math/render/svg/66000e19e524c041d07557e91158c5001ba9280e&quot;/&gt;&lt;br&gt;&#xA;To paint an image for understanding black holes, we only need this middle section:&lt;br&gt;1)&lt;a href=&quot;https://i.stack.imgur.com/0Ce43.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0Ce43.png&quot; alt=&quot;Schwarzschild Radius equation rs/r&quot; /&gt;&lt;/a&gt; 2)&lt;a href=&quot;https://i.stack.imgur.com/M6ZsF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/M6ZsF.png&quot; alt=&quot;Schwarzschild Radius equation rs/rs&quot; /&gt;&lt;/a&gt; 3)&lt;a href=&quot;https://i.stack.imgur.com/v00eQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/v00eQ.png&quot; alt=&quot;Schwarzschild Radius equation 1/1&quot; /&gt;&lt;/a&gt; 4) &lt;a href=&quot;https://i.stack.imgur.com/f4Q4H.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f4Q4H.png&quot; alt=&quot;Schwarzschild Radius equation 0,singularity,doom&quot; /&gt;&lt;/a&gt;&lt;br&gt;&#xA;We've already established &lt;strong&gt;r&lt;sub&gt;s&lt;/sub&gt;&lt;/strong&gt; as being the Schwarzschild Radius of a particular object, &lt;strong&gt;r&lt;/strong&gt; is the radius of the stellar object. When &lt;strong&gt;r&lt;/strong&gt; becomes as small as &lt;strong&gt;r&lt;sub&gt;s&lt;/sub&gt;&lt;/strong&gt; you get a singularity&lt;sup&gt;1&lt;/sup&gt; and weird stuff starts happening, most importantly to OPs question, the &lt;strong&gt;space-time curvature&lt;/strong&gt; at the black hole &lt;strong&gt;becomes infinite(!)&lt;/strong&gt;, this means that any light that intersects the event horizon at any point will take an &lt;strong&gt;infinite amount of time&lt;/strong&gt; to travel across the black holes funnel. Even at a very flat angle relative to the event horizon, where its just ever so slightly poking it, it is lost because set theory teaches us: &lt;em&gt;any&lt;/em&gt; subset of infinity is also infinite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Here are some extra visalisiations:&lt;br&gt;&#xA;&lt;strong&gt;Gravity space-time cone of earth:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/VBsEV.jpg&quot; width=&quot;330&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Gravity space-time funnel of a black hole:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/z0EAR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z0EAR.jpg&quot; width=&quot;330&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;https://static.interestingengineering.com/images/import/2016/05/black-hole-space-time.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;interestingengineering.com&lt;/a&gt;)&lt;/sub&gt;&lt;br /&gt;&#xA;&lt;img src=&quot;https://i.imgur.com/sGubgg6.png&quot; width=&quot;330&quot; /&gt;&lt;/p&gt;&#xA;&lt;br&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Singularity: A singularity is basically, in calculus/algebra terms, just when you divide by zero (which you shall never do!). A 2D singularity might just look like this: &lt;code&gt;f(x) = 1/x&lt;/code&gt; (the singularity is there in the middle at x=0).&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/47gP9.png&quot; width=&quot;180&quot; /&gt; &lt;img src=&quot;https://i.imgur.com/37H9lZm.png&quot; width=&quot;280&quot;/&gt;&lt;br&gt;&#xA;A 3D singularity can look like this /, singularity at x=1 (this is Riemanns zeta function).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15223" LastEditorUserId="7472" LastEditDate="2021-01-01T19:03:08.443" LastActivityDate="2021-01-01T19:03:08.443" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29151" PostTypeId="2" ParentId="29148" CreationDate="2019-01-13T14:39:50.060" Score="5" Body="&lt;p&gt;The &quot;tidal force&quot; is simply a manifestation of good old inverse-square-law gravity.  It doesn't need a rigid body to be present to manifest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a giant circular ring of dozens of space stations established in interstellar space. They're far away from any star and just sit there happily doing whatever it was they were set up to do.  Suddenly, a star sweeps in from stage left (perhaps a fast-moving star created by a supernova-disrupted close binary system).  What happens to the ring of stations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are attracted towards the rogue star, of course and start moving noticeably towards it when it gets close enough.  They accelerate towards it pulled by the force of the star's gravity, which follows the inverse square law. &lt;em&gt;But they don't all accelerate at the same rate.&lt;/em&gt; The station on the edge of the circle closest to the star accelerates fastest, stations half-way around the circle accelerate, but not as much, and the station on the far side of the circle accelerates least.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soon the circle becomes an oval with its long axis pointing towards the star.  This is the tidal effect.  Let's say that the rogue star was dark -- a neutron star, perhaps, which had been stripped of its accretion disk by the explosion.  What would the people on the stations observe? They wouldn't notice the acceleration &lt;em&gt;per se&lt;/em&gt; because they're in free fall.  Unless they were doing doppler spectroscopy of distant stars, they wouldn't detect their velocity (at least not at first), but they &lt;em&gt;would&lt;/em&gt; see their neat ring of stations turning into an oval. From their point of view, some force was tugging on them and dragging them out of alignment. &lt;em&gt;That's&lt;/em&gt; the tidal force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the point of view of the omniscient observer (us!) there are no tides, there's just simple gravity.  But from the point of vie of the freely falling observers in the gravity field, most of the effects of gravity disappears and what's left looks like a force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia has an &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_force&quot; rel=&quot;noreferrer&quot;&gt;article on the tidal force&lt;/a&gt; which gives a &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_force#Mathematical_treatment&quot; rel=&quot;noreferrer&quot;&gt;mathematical treatment&lt;/a&gt;. (Personally, I don't think the article is terribly clear, though it appears to be correct.)&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="7982" LastEditDate="2019-01-13T15:48:41.903" LastActivityDate="2019-01-13T15:48:41.903" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29152" PostTypeId="1" AcceptedAnswerId="29153" CreationDate="2019-01-13T14:48:38.383" Score="2" ViewCount="69" Body="&lt;p&gt;Why the earth is rising and setting seen from the moon,&#xA;when the moon is tidally locked?&#xA;Shouldn't the earth be always on the same spot because of the tidal lock, if observed from the moon?&lt;/p&gt;&#xA;" OwnerUserId="25527" LastActivityDate="2019-01-13T15:34:58.247" Title="Earth rising and setting from moon's perspective" Tags="&lt;the-moon&gt;&lt;earth&gt;&lt;tidal-locking&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29153" PostTypeId="2" ParentId="29152" CreationDate="2019-01-13T15:34:58.247" Score="6" Body="&lt;p&gt;You're quite right.  The Earth is (nearly) stationary in the Moon's sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I say &quot;nearly&quot; because the Moon is in a slightly elliptical orbit around the Earth, but rotates perfectly smoothly. This means that the Earth's motion through the Lunar sky is &#xA; a bit faster when the Moon is at perigee and a bit slower when it's at apogee. Because the Moon is tide-locked, on average they match perfectly, but during the course of the sidereal month the Earth appears to move back and forth by a few degrees.  The effect isn't huge, but it's easily observable with a telescope.  See the Wikipedia article on &#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Libration&quot; rel=&quot;noreferrer&quot;&gt;&quot;libration&quot;&lt;/a&gt; for an OK discussion and a really great animated gif illustration of libration as viewed from Earth.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible that you're reacting to the amazing movie from Apollo 8 showing Earthrise over the Lunar horizon? It's been int he news a lot given that this is the 50th anniversary of the flight.  If so, what you saw there was not from the Moon's surface, but from orbit around the Moon and the Earthrise happened as Apollo 8 came around from behind the moon (where the Earth is not visible in the sky) and the Earth first became visible around the edge of the Moon.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2019-01-13T15:34:58.247" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29154" PostTypeId="2" ParentId="29149" CreationDate="2019-01-13T17:02:52.997" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How is information preserved on the surface of a black hole? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We have no evidence that it is. We have some claims of this nature, associated with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Holographic_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;holographic principle&lt;/a&gt;. But I'd say the physics concerned is hypothetical and speculative. It isn't in the same league as general relativity, which is one of the best tested theories we've got. See Clifford M Will's paper on &lt;a href=&quot;https://arxiv.org/abs/1403.7377&quot; rel=&quot;nofollow noreferrer&quot;&gt;the confrontation between general relativity and experiment&lt;/a&gt; for details of that.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there an compression algorithm for storing the information? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not that I know of. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What if there is a uniformly complex object dropped to the black hole? Then there will be no compression of information.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Instead there will be a gamma ray burst. Friedwardt Winterberg wrote about this. See the 2013 AMPS paper &lt;a href=&quot;https://arxiv.org/abs/1304.6483&quot; rel=&quot;nofollow noreferrer&quot;&gt;an apologia for firewalls&lt;/a&gt;. Tucked away in the conclusion is footnote 31, containing a reference 87 to Winterberg’s 2001 paper &lt;a href=&quot;https://www.degruyter.com/downloadpdf/j/zna.2001.56.issue-12/zna-2001-1216/zna-2001-1216.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;gamma ray bursters and Lorentzian relativity&lt;/a&gt;. I'm confident that Winterberg is essentially correct.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How is there enough space on the surface (2D) to encode something which is in. The planck distance is the same in the black hole as in the universe. So how can the black hole encode information about an object in 3D that is both uniformly complex and as dense as the black hole? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't think it can. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Or is there a law that says 3D objects cannot be more dense then some level - which is sufficiently sparse so that when collapsed to a black hole all its information can be encoded on the surface of a sphere? Has this limit been found - given that a sphere has the lowest surface to volume of any object: &lt;a href=&quot;https://en.wikipedia.org/wiki/Surface-area-to-volume_ratio&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Surface-area-to-volume_ratio&lt;/a&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As far as I know there is no law or limit. And as far as I know there is no evidence to support the holographic principle. I for one do not take it seriously. But please advise if you receive any useful answers about this subject on any other website. &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2019-01-13T17:02:52.997" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29156" PostTypeId="2" ParentId="29148" CreationDate="2019-01-13T17:23:43.330" Score="1" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;I'm wondering exactly in which situations forces between bodies are, and are not consider to be tidal forces in the context of planetary and astronomical science.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Don't overthink it. All a tidal force is, is the net force you get when two gravitational forces aren't quite the same. The force of gravity is the first derivative of potential (the &amp;quot;slope&amp;quot;), whilst the tidal force is the second derivative of potential (the change in &amp;quot;slope&amp;quot;). The classic example is spaghettification. You're falling into a black hole, and the force of gravity is greater at your feet than at your head. So you get stretched by a tidal force.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\phi(\mathbf{r}) = -\frac{GM}{|\mathbf{r}|}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathbf{F}(\mathbf{r}) = \nabla  \phi = -\frac{GMm}{r^2}\mathbf{\hat{r}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathbf{F_{Tidal}}(\mathbf{r'}-\mathbf{r}) =  -GMm \left( \frac{1}{r^2}\mathbf{\hat{r}} - \frac{1}{r'^2}\mathbf{\hat{r'}} \right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\mathbf{r})$&lt;/span&gt; is the position of the secondary body relative to the &amp;quot;main&amp;quot; or more massive body, and &lt;span class=&quot;math-container&quot;&gt;$\mathbf{r'}-\mathbf{r}$&lt;/span&gt; is the position of the particular point &lt;em&gt;relative&lt;/em&gt; to the secondary body's center of mass where the tidal force is experienced, and &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; is the mass of the test particle at that point experiencing the tidal force.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If two rigid, non-deformable spherical masses with spherically symmetric mass distributions orbit each other, can we say there are no tidal forces or tidal effects there?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;No. Like Mark said, the tidal force is independent of your non-deformable spherical masses.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If one of them has a permanent, static deformation (e.g. quadrupole moments and higher) are there tidal forces then? If both do?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;There are tidal forces because one side of each object is closer to the other object than the other side is.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Or are dynamic, induced deformations required before we invoke the &amp;quot;T-word&amp;quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;No deformations are required!&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Can tidal forces be radial, such that they do not tend to raise or lower an orbit over time, or must they have tangential components, like the one that is slowly raising the Moon's orbit?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;They're usually radial. The simplest case is where one object is falling straight towards another much larger object. Things get more complicated when you've got two large orbiting objects like the Earth and the Moon.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Is it possible to draw a clear line here, at least within the scope of this SE site and This month's focus tag which is tidal-forces?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Probably not, because to really understand the tidal force you have to understand the force of gravity. A lot of people don't, even though Einstein made it clear. I said something about that yesterday in &lt;a href=&quot;https://astronomy.stackexchange.com/a/29140/9772&quot;&gt;this answer&lt;/a&gt;, but I don't think it was appreciated.&lt;/p&gt;&#xA;" OwnerUserId="9772" LastEditorUserId="7982" LastEditDate="2021-05-23T23:41:52.323" LastActivityDate="2021-05-23T23:41:52.323" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29157" PostTypeId="1" CreationDate="2019-01-13T18:20:58.807" Score="2" ViewCount="118" Body="&lt;p&gt;Radiometric dating relies on accurate knowledge of the half-lives of radioactive elements. Skeptics have argued that since many elements' half-lives are too long for anyone to have observed them, we can't know what they were in the distant past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various counterarguments based on terrestrial evidence, like agreement between elements in the same sample, and corroboration from varves, tree rings, and ice layers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another counterargument I've seen is by looking at supernova, we get a window &quot;back in time&quot; to when their light began its long journey to us, and that much of this light was generated by radioactive decay, confirming that decay rates were the same then as now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't quite follow this explanation, and would like some help. Distant starlight doesn't give us a rock sample in which to analyze proportions of parent and daughter elements. Does it still somehow give evidence of specific radioactive decay rates? If so, how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know next to nothing about astronomy, so please explain this to me like I'm a child.&lt;/p&gt;&#xA;" OwnerUserId="25026" LastEditorUserId="25026" LastEditDate="2019-01-13T18:29:39.890" LastActivityDate="2020-05-26T02:40:07.637" Title="Can radioactive decay rates be confirmed by looking at supernova?" Tags="&lt;supernova&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29158" PostTypeId="1" CreationDate="2019-01-13T18:26:09.777" Score="0" ViewCount="171" Body="&lt;p&gt;In the context of the study of a cubesat (little satellite), I am asked to define the local orbital frame and compute the local orbital frame vector directions as a function of time :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the context :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LF06s.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LF06s.png&quot; alt=&quot;Attitude for a cubesat&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone could give me some tracks or links to find the answer of part 2.1 Local orbital frame ? this would be kind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards&lt;/p&gt;&#xA;" OwnerUserId="16492" LastActivityDate="2019-06-14T07:03:31.030" Title="Computing the local orbital frame vector directions in a geocentric reference frame" Tags="&lt;orbit&gt;&lt;orbital-mechanics&gt;&lt;satellite&gt;&lt;artificial-satellite&gt;&lt;geocentrism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29159" PostTypeId="2" ParentId="29157" CreationDate="2019-01-13T18:39:36.307" Score="-1" Body="&lt;p&gt;When an unstable atom decays, it releases energy. At an earlier time, there are more unstable atoms, so the released energy is larger. At different times, there are different atoms left, so the energy released is different. You can measure the decay rate by following the time-series of this energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ E(t) = E_0 \exp[-\lambda t]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; is the energy, &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; is time, &lt;span class=&quot;math-container&quot;&gt;$E_0$&lt;/span&gt; is energy at &lt;span class=&quot;math-container&quot;&gt;$t=0$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; is the decay constant. This is a very simple picture. In reality, it will be a bit more complicated because of other effects causing the deviation of the observed energy, such as scattering.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2019-01-13T18:39:36.307" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29161" PostTypeId="1" AcceptedAnswerId="33326" CreationDate="2019-01-13T21:56:49.963" Score="10" ViewCount="325" Body="&lt;p&gt;I had this thought, and my first guess was &quot;high density = lots of absorption, so I guess it's neutron stars&quot; but &lt;a href=&quot;https://physics.stackexchange.com/questions/280586/how-much-of-neutrinos-pass-through-a-neutron-star&quot;&gt;this physics.se question&lt;/a&gt; about that has a great answer which covers why that's incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what object will absorb the highest fraction of neutrinos passing through it, or at least be a good candidate? Feel free to assume a certain energy range of neutrinos. Exclude black holes because they just absorb everything and that's not as interesting.&lt;/p&gt;&#xA;" OwnerUserId="25532" LastActivityDate="2019-09-11T08:03:08.903" Title="What object in the universe is most opaque to neutrinos?" Tags="&lt;neutrinos&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29162" PostTypeId="1" CreationDate="2019-01-14T00:31:09.567" Score="2" ViewCount="64" Body="&lt;p&gt;GW170729 was gravitational wave signal's detected by the Virgo and LIGO detectors. I read the following in a news report:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Known as GW170729, in accordance with the date of its discovery (in US&#xA;  format), the gravitational wave that astronomers can listen in to was&#xA;  caused by the formation of a black hole more than 80 times as massive&#xA;  as our sun and about nine billion light years away. (&lt;a href=&quot;https://www.smh.com.au/national/astrophysics-will-explode-scientists-hear-10-black-holes-merging-20181203-p50jub.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am specifically interested in the claim that it comes from 9-billion light years away. I have been unable to find a good source to back this claim up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How accurate is this claim? Please provide a source for your answer and where applicable, a simple explanation of how the data backs this up. More information about this event can be found &lt;a href=&quot;https://cplberry.com/2018/12/07/o2-catalogue/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="25535" LastEditorUserId="1891" LastEditDate="2019-01-14T05:51:54.637" LastActivityDate="2019-01-14T05:51:54.637" Title="How far away was the binary black hole that produced GW170729?" Tags="&lt;black-hole&gt;&lt;observational-astronomy&gt;&lt;gravitational-waves&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29163" PostTypeId="1" CreationDate="2019-01-14T08:16:40.543" Score="1" ViewCount="316" Body="&lt;p&gt;I'm asking this question as a follow-up to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/29122/if-light-has-no-mass-why-is-it-affected-by-gravity&quot;&gt;if light has no mass why is it affected by gravity?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you’re standing on a gedanken planet, shining a laser beam straight up into space. The light goes straight up. It doesn’t curve, and it doesn’t fall back down. Now imagine it’s a denser more massive planet. The light still goes straight up. It still doesn’t curve, and it still doesn’t fall back down. Let’s make it a really massive planet. That light still goes straight up. It &lt;em&gt;still&lt;/em&gt; doesn’t curve, and it &lt;em&gt;still&lt;/em&gt; doesn’t fall back down:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when we make our gedanken planet so massive that it’s a black hole, all of a sudden light can’t escape. Why? Why doesn’t the light get out? &lt;em&gt;Why doesn't the vertical light beam get out of a black hole?&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2019-01-17T20:54:46.247" Title="Why doesn't the vertical light beam get out of a black hole?" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;general-relativity&gt;" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="29164" PostTypeId="2" ParentId="29163" CreationDate="2019-01-14T09:08:15.940" Score="3" Body="&lt;p&gt;Using the General Relativity and Schwarzschild metric we can define the redshift of the photon as,&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac {v_{\infty}} {v_e} = (1-r_s/R_e)^{1/2}$$&lt;/span&gt; in this equation &lt;span class=&quot;math-container&quot;&gt;$v_{\infty}$&lt;/span&gt; represents the frequency of the light measured by an observer at infinity, &lt;span class=&quot;math-container&quot;&gt;$v_e$&lt;/span&gt; is the frequency of the emitted wavelength, &lt;span class=&quot;math-container&quot;&gt;$r_s$&lt;/span&gt; is the schwarzschild radius, &lt;span class=&quot;math-container&quot;&gt;$r_S=2GM/c^2$&lt;/span&gt;, and finally &lt;span class=&quot;math-container&quot;&gt;$R_e$&lt;/span&gt; is the radius which photon is emitted.&lt;/p&gt;&#xA;&lt;p&gt;When we set &lt;span class=&quot;math-container&quot;&gt;$r_s=R_e$&lt;/span&gt; we can see that &lt;span class=&quot;math-container&quot;&gt;$\frac {v_{\infty}} {v_e}=0$&lt;/span&gt; which means that the redshift will be infinitely large and the photon cannot escape from the black hole.&lt;/p&gt;&#xA;&lt;p&gt;For more information, you can look here, &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_redshift#Prediction_by_the_equivalence_principle_and_general_relativity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gravitational redshift&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;For an object compact enough to have an event horizon, the redshift is not defined for photons emitted inside the Schwarzschild radius, both because signals cannot escape from inside the horizon and because an object such as the emitter cannot be stationary inside the horizon, as was assumed above. Therefore, this formula only applies when &lt;span class=&quot;math-container&quot;&gt;$R_{e}$&lt;/span&gt;  is larger than &lt;span class=&quot;math-container&quot;&gt;$r_{s}$&lt;/span&gt; . When the photon is emitted at a distance equal to the Schwarzschild radius, the redshift will be infinitely large, and it will not escape to any finite distance from the Schwarzschild sphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;For the energy change in the photon, there has been done many experiments which some of them are explained in the Wikipedia page and I have found another experiment which is, &lt;a href=&quot;https://en.wikipedia.org/wiki/Pound%E2%80%93Rebka_experiment&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pound–Rebka experiment&lt;/a&gt;. That perfectly explains the energy change and for the math part, you can look &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/gratim.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="25424" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-01-15T05:14:00.730" CommentCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="29165" PostTypeId="1" AcceptedAnswerId="29166" CreationDate="2019-01-14T09:14:56.370" Score="2" ViewCount="228" Body="&lt;p&gt;I want to add realistic noise to a simulated image but I am a little confused about the process. I want to have some Gaussian random noise representing the readout noise and to also add Poisson noise correlated with the signal in the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand how to add the Gaussian random noise; I draw samples from a Gaussian with fixed width and add the samples onto the pixel values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am confused about the Poisson noise though. Is this noise only correlated to the noiseless signal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e I should draw samples from the Poisson distribution with &lt;span class=&quot;math-container&quot;&gt;$k = \mathrm{pixel  value}$&lt;/span&gt; (before adding the random noise) and add it to the noiseless image.&lt;/p&gt;&#xA;" OwnerUserId="14536" LastActivityDate="2019-01-14T10:57:02.073" Title="Simulating Noise in Astronomical Images" Tags="&lt;observational-astronomy&gt;&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29166" PostTypeId="2" ParentId="29165" CreationDate="2019-01-14T10:30:08.707" Score="4" Body="&lt;p&gt;Yes, one can superpose these two noise sources: add a Gaussian readout noise (with a spatially constant sigma) to a Poisson noise (with spatially variable number count k).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, as the number counts are high for optical CCD images, one often approximates the Poisson noise by a spatially-variable Gaussian noise, using the square root of the photon number counts as sigma. If your images have some significant sky background level (e.g., hundreds of photons), this is fine even for faint sources that add only few photons.  A Poisson distribution gets very close to a Gaussian already for k &gt; 20. Note that despite this common approximation by a Gaussian, such a simulated noise contribution is still often called &quot;Poisson noise&quot;, to stress that it originates from the spatially variable signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One warning: using the square root of the signal as sigma of a Gaussian to approximate Poisson noise &lt;strong&gt;only holds for a signal in units of photons (or almost equivalently, in units of electrons)&lt;/strong&gt;. If an image is given in ADU (analog to digital unit), the gain is required to convert the pixel values into electrons. The same holds when using a true Poisson distribution: one has to be careful that the counts have not been rescaled.&lt;/p&gt;&#xA;" OwnerUserId="9150" LastEditorUserId="9150" LastEditDate="2019-01-14T10:57:02.073" LastActivityDate="2019-01-14T10:57:02.073" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29167" PostTypeId="1" AcceptedAnswerId="29168" CreationDate="2019-01-14T12:14:32.717" Score="3" ViewCount="66" Body="&lt;p&gt;The  Canadian Hydrogen Intensity Mapping Experiment (&lt;a href=&quot;https://en.wikipedia.org/wiki/Canadian_Hydrogen_Intensity_Mapping_Experiment&quot; rel=&quot;nofollow noreferrer&quot;&gt;CHIME&lt;/a&gt;) has &lt;a href=&quot;https://arstechnica.com/science/2019/01/new-canadian-telescope-spots-a-second-repeating-fast-radio-burst/&quot; rel=&quot;nofollow noreferrer&quot;&gt;discovered&lt;/a&gt; the second repeated fast radio burst recently. However, its structure is simple apparently. I wonder why we did not build a telescope before. What is the challenge? I just know that its data volume is large.&#xA;It is difficult to update the telescope?&lt;/p&gt;&#xA;&lt;p&gt;There are 4 cylinders and 1024 antennas. What determines the distribution of the 1024 antennas? I mean why is it not just one longer cylinder?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="37191" LastEditDate="2021-01-19T13:57:42.667" LastActivityDate="2021-01-19T13:57:42.667" Title="What are the challenges for the building and data analysis of the CHIME telescope?" Tags="&lt;observational-astronomy&gt;&lt;radio-astronomy&gt;&lt;data-analysis&gt;&lt;interferometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29168" PostTypeId="2" ParentId="29167" CreationDate="2019-01-14T13:52:47.553" Score="1" Body="&lt;p&gt;There are many radiotelescopes that consist of large arrays of antennas. We usually think of an array individual dishes and &lt;a href=&quot;https://en.wikipedia.org/wiki/Very_Large_Array&quot; rel=&quot;nofollow noreferrer&quot;&gt;VLA&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Atacama_Large_Millimeter_Array&quot; rel=&quot;nofollow noreferrer&quot;&gt;ALMA&lt;/a&gt; are two of the most commonly recognized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you want to keep an eye on a huge patch of sky in order to try to catch a very rare event, you want a wide field of view more than ultrahigh spatial resolution. And if what you are looking for is super-hot science, you want to build your device quickly which means a modest budget.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So wide field, lower resolution means a smaller baseline, and CHIME is of order 100 meters on a side. but building 1000 small say 3 meter dishes or 1000's of log-periodic dipole arrays and trying to steer them all would be a mess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So instead CHIME built four, simple, parabolic cylinders and told all those antennas to &lt;em&gt;be nice to each other and share!&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each antenna picks up signals from a huge area of the cylinder. If you include all directions it's like 25x25 meter section. The antennas are spaced only about a wavelength or so apart, but that's plenty far to do interferometry with the signals in a computer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The ArXiv preprint &lt;a href=&quot;https://arxiv.org/abs/1803.11235&quot; rel=&quot;nofollow noreferrer&quot;&gt;The CHIME Fast Radio Burst Project: System Overview&lt;/a&gt; goes into great detail about the design and optimization of the antenna, considering what it is supposed to do and the necessity to make it straightforward to build an have a reasonable budget.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kfSYm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kfSYm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Figure 1. Photograph of the CHIME telescope on 15 September, 2016, looking North-West. The shipping containers housing the X-Engine and CHIME/FRB backend can be seen adjacent to the right-most cylinder. The receiver huts containing the F-Engine are beneath the reflectors and cannot be seen here. The DRAO Synthesis Telescope (Kothes et al. 2010) can be seen in the background. See Table 1 for detailed properties&#xA;  of CHIME.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Field of View in the North-South direction (the axis of the parabolic cylinders) is about 110 degrees, but in the East-West direction it's of course very narrow, only 2.5 to 1.3 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z1lZH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z1lZH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I've borrowed the following text, images and links from my question &lt;a href=&quot;https://astronomy.stackexchange.com/q/28615/7982&quot;&gt;How does Canada's radiotelescope CHIME manage polarization-dependent apodization of the fast f/0.25 optics?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a picture of some of the antennas, looking down into the trough:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below x2:&lt;/strong&gt; from &lt;a href=&quot;https://arxiv.org/abs/1708.08521&quot; rel=&quot;nofollow noreferrer&quot;&gt;The cloverleaf antenna: A compact wide-bandwidth dual-polarization feed for CHIME&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/u6J0Q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/u6J0Q.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fig. 1. At left, the shape of each petal consists of two perpendicular straight&#xA;lines, two 45 degree circular arcs with radius R and one half an ellipse. W is&#xA;the major axis of the ellipse and L is the length from the intersection of the&#xA;straight sides to the outer edge of the ellipse. The shape is illustrated here for&#xA;the adopted values of gap, R, L, and W. Each of the four tabs shown at the&#xA;centre is connected to one side of a vertical microstrip transmission line and&#xA;in each case the full width of the adjacent petal is connected to the other lead.&#xA;At right, CST simulated currents for one linear polarization at 600MHz are&#xA;shown. Note the small asymmetry in the curent distribution near the centre&#xA;becasue of the tab geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1Sb4J.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1Sb4J.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fig. 3. A linear array of eight cloverleaf antennas installed at the focal line of&#xA;the CHIME Pathfinder at the Dominion Radio Astrophysical Observatory in Penticton, BC, Canada. The picture is taken through the wire mesh reflective surface (mesh spacing 19 mm) illustrating a photonsview of the antennas and ground plane. Notice that each feed has an image-feed in the ground plane, 1/2 λ away at the passband centre frequency. Notice also the four slots cut to remove dielectric material from the gaps between the petals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; From &lt;a href=&quot;https://chime-experiment.ca/instrument&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://chime-experiment.ca/instrument&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;CHIME consists of four adjacent 20m x 100m cylindrical reflectors oriented north-south. The focal axis of each cylinder is lined with 256 dual-polarization antennas, each of which receives radiation from a large swath of sky that nearly stretches from the northern horizon to the southern horizon. This gives CHIME its enormous field of view. The CHIME antennas are custom-designed to have good sensitivity from 400 to 800 MHz, in both linear polarizations. This gives CHIME its large frequency coverage. &lt;strong&gt;Signals from the antennas are amplified in two stages, using low-noise amplifiers developed by the cell-phone industry. This novel application of consumer technology makes CHIME affordable.&lt;/strong&gt; The 2048 signals from these inputs (256 antennas x 2 polarizations x 4 cylinders) are fed to the F-Engine for the initial stage of digital processing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NSO9C.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NSO9C.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-01-14T14:13:45.043" LastActivityDate="2019-01-14T14:13:45.043" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="29169" PostTypeId="1" CreationDate="2019-01-14T14:03:03.397" Score="3" ViewCount="155" Body="&lt;p&gt;I understand that we can calculate the gravitational force  other planets exert on Earth, but can it be measured?&lt;/p&gt;&#xA;" OwnerUserId="25548" LastEditorUserId="7472" LastEditDate="2019-01-14T20:52:50.473" LastActivityDate="2019-01-15T15:44:13.647" Title="Are other planets' gravity measurable on Earth?" Tags="&lt;planet&gt;&lt;gravity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29170" PostTypeId="1" AcceptedAnswerId="29172" CreationDate="2019-01-14T14:45:39.023" Score="1" ViewCount="65" Body="&lt;p&gt;I'm wondering why there is a shadow under the Yutu 2 rover, of the Chang'e 4 mission, since it is on the far/dark side of the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ryVFm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ryVFm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;sup&gt;The Chang'e 4 lander, as seen by the Yutu 2 rover. (Image credit: CNSA)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.space.com/42981-china-moon-far-side-panorama-chang-e-4.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;See the Far Side of the Moon in This Amazing Panorama from China's Chang'e 4 Lander&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4127" LastEditorDisplayName="user1569" LastEditDate="2019-12-18T12:02:54.557" LastActivityDate="2019-12-18T12:02:54.557" Title="Pekin Chang'e 4 Mission, Shadow?" Tags="&lt;the-moon&gt;&lt;moonlanding&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29171" PostTypeId="1" CreationDate="2019-01-14T14:47:47.580" Score="2" ViewCount="243" Body="&lt;p&gt;It would seem that if life were relatively common in the universe that there would be a cacophony of noise somewhere in the electromagnetic spectrum. Is life so rare and distant that the waves haven't reached us yet, or have they attenuate out, or are they unintelligible, or are we the first, or are we alone?&lt;/p&gt;&#xA;" OwnerUserId="25548" LastActivityDate="2019-01-15T23:22:58.843" Title="Why haven't detected life in the universe?" Tags="&lt;life&gt;&lt;origin-of-life&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29172" PostTypeId="2" ParentId="29170" CreationDate="2019-01-14T14:50:10.100" Score="3" Body="&lt;p&gt;Oh sorry, I just understand what is going on. Being the far side of the moon, did not mean that it's never exposed to the sun.&#xA;So the Expression &quot;Dark Side Of The Moon&quot; is not so true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has  written in Wikidepia :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Tidal forces from Earth have slowed down the Moon's rotation to the point where the same side is always facing the Earth—a phenomenon called tidal locking. The other face, most of which is never visible from the Earth, is therefore called the &quot;far side of the Moon&quot;. Over time, some parts of the far side can be seen due to libration.[7] In total, 59 percent of the Moon's surface is visible from Earth at one time or another. Useful observation of the parts of the far side of the Moon occasionally visible from Earth is difficult because of the low viewing angle from Earth (they cannot be observed &quot;full on&quot;).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The phrase &quot;dark side of the Moon&quot; does not refer to &quot;dark&quot; as in the absence of light, but rather &quot;dark&quot; as in unknown: until humans were able to send spacecraft around the Moon, this area had never been seen.[1][2][3][4] While many misconstrue this to think that the &quot;dark side&quot; receives little to no sunlight, in reality, both the near and far sides receive (on average) almost equal amounts of light directly from the Sun. However, the near side also receives sunlight reflected from the Earth, known as earthshine. Earthshine does not reach the area of the far side that cannot be seen from Earth. Only during a full Moon (as viewed from Earth) is the whole far side of the Moon dark. The word &quot;dark&quot; has expanded to refer also to the fact that communication with spacecraft can be blocked while the spacecraft is on the far side of the Moon, during Apollo space missions for example.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4127" LastActivityDate="2019-01-14T14:50:10.100" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29173" PostTypeId="2" ParentId="29171" CreationDate="2019-01-14T15:05:10.940" Score="3" Body="&lt;p&gt;To quote a favourite movie of mine&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pardon me sir, but it's a big-ass sky.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our current method of detecting life (electromagnetic signals) is by targeting and scanning stars indivdually using radio telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method is pretty costly because we don't have a huge amount of them and they can be better used for other (proper) research purposes.  Basically, it costs a lot to monopolise a telescope's resources by repeatedly pointing it at stars without a reasonable hope of a result.  Understandably, astronomers want to keep their funding by producing results rather than nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current estimates is that we can stumble upon signals by about 2040&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.space.com/24622-intelligent-alien-life-detection-2040.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;By 2040 or so, astronomers will have scanned enough star systems to give themselves a great shot of discovering alien-produced electromagnetic signals, said Seth Shostak of the SETI (Search for Extraterrestrial Intelligence) Institute in Mountain View, Calif.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But again, this is pure guesswork.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A related question here on SE asks this same question from another direction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/8146/how-far-away-could-we-detect-that-earth-has-life?rq=1&quot;&gt;how far away could we detect that Earth has life?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the answers, quoted in part:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A &quot;blind&quot; search could look for radio signatures and of course this is what SETI has been doing. If we are talking about detecting &quot;Earth&quot;, then we must assume that we are not talking about deliberate beamed attempts at communication, and so must rely on detecting random radio &quot;chatter&quot; and accidental signals generated by our civilisation. The SETI Phoenix project was the most advanced search for radio signals from other intelligent life. Quoting from Cullers et al. (2000): &quot;Typical signals, as opposed to our strongest signals, fall below the detection threshold of most surveys, even if the signal were to originate from the nearest star&quot;. Quoting from Tarter (2001): &quot;At current levels of sensitivity, targeted microwave searches could detect the equivalent power of strong TV transmitters at a distance of 1 light year (within which there are no other stars)...&quot;. The equivocation in these statements is due to the fact that we do emit stronger beamed signals in certain well-defined directions, for example to conduct metrology in the solar system using radar. Such signals have been calculated to be observable over a thousand light years or more. But these signals are brief, beamed into an extremely narrow angle and unlikely to be repeated. You would have to be very lucky to be observing in the right direction at the right time if you were performing targeted searches&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd really recommend reading the rest of this &lt;a href=&quot;https://astronomy.stackexchange.com/a/8658/20278&quot;&gt;answer&lt;/a&gt;, and other answers for this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, our &quot;noise&quot; isn't really strong enough to be detected over long distances above the greater (and stronger) chatter of the natural universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While we could engineer a stronger broadcast message and hope for a reply, it's a long, long project and unlikely to be funded.&lt;/p&gt;&#xA;" OwnerDisplayName="user20278" LastEditorDisplayName="user20278" LastEditDate="2019-01-15T07:57:34.577" LastActivityDate="2019-01-15T07:57:34.577" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29174" PostTypeId="2" ParentId="29171" CreationDate="2019-01-14T17:26:12.393" Score="1" Body="&lt;p&gt;There are a number of questions here on the &lt;a href=&quot;https://astronomy.stackexchange.com/search?q=fermi&quot;&gt;Fermi Paradox&lt;/a&gt; with answers ranging from an Infinite Universe, the parameters for the Drake Equation and others. Well worth having a read through them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my favourite is author Cixin Liu's theory outlined in The &lt;a href=&quot;https://www.amazon.co.uk/Three-Body-Problem-Cixin-Liu/dp/1784971553&quot; rel=&quot;nofollow noreferrer&quot;&gt;Three Body Problem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Civilisations which are noisy draw attention to themselves, and are extinguished by civilisations thinking long term who don't want to have competition for resources in the far distant future.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously that theory is not easily proven or disproven in any useful way, so currently it may be as valid as any other. Well worth a read in any case!&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2019-01-14T17:26:12.393" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29175" PostTypeId="1" CreationDate="2019-01-14T18:23:05.030" Score="3" ViewCount="396" Body="&lt;p&gt;From my understanding/been told, the color red has the ability to travel the farthest in our atmosphere, being the reason why we have red sunsets and such. But with this in mind, why is it that at night the moonlight we receive seems to have a slight blue tint to it, and since red can travel the best out of our visible spectrum, how come our moonlit nights don't have a red tint instead of a blue one?&lt;/p&gt;&#xA;" OwnerUserId="25551" LastEditorUserId="7982" LastEditDate="2021-02-19T23:26:54.910" LastActivityDate="2021-02-19T23:26:54.910" Title="Why is it that night has a almost a bluish tint to it instead of red?" Tags="&lt;the-sun&gt;&lt;the-moon&gt;&lt;light&gt;&lt;atmospheric-effects&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29176" PostTypeId="2" ParentId="29169" CreationDate="2019-01-14T18:36:20.583" Score="6" Body="&lt;p&gt;I think the questions you're asking is, &quot;All of the planets exert a gravitation force on everything else in the Solar System, including the Earth. Can we make measurements of that force in a laboratory on the surface of the earth?  (I.e., not by observing events on the other planets or their orbits.)&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm right, the answer is &quot;No, the effects are too small.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, calculate the gravitational acceleration we feel on Earth from, say, Jupiter.  As it happens, Earth is roughly 10,000 Jupiter radii away from Jupiter.  The acceleration due to gravity follows an inverse square law, so the acceleration due to Jupiter on Earth is about 10&lt;sup&gt;-10&lt;/sup&gt; that at the surface of Jupiter or about 3x10&lt;sup&gt;-10&lt;/sup&gt; g.  This is small, but perfectly measurable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that we can't measure gravity directly. We're sitting on a freely-falling Earth and (as far as Jupiter is concerned) are freely falling with it. Whatever acceleration Jupiter is imparting, it imparting the same acceleration to the Earth, us and our laboratories and everything in them.  So there's no force to measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We &lt;em&gt;can&lt;/em&gt; measure tidal forces directly, since they are in essence the differences in acceleration due to objects being a different distances from the gravitating object, but &lt;em&gt;they&lt;/em&gt; drop off as the inverse cube of distance and are undetectable at this distance using current technology.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2019-01-15T15:44:13.647" LastActivityDate="2019-01-15T15:44:13.647" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29177" PostTypeId="2" ParentId="29163" CreationDate="2019-01-14T19:59:46.983" Score="13" Body="&lt;p&gt;There is no &quot;up&quot; direction within the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most people get fixated on the speed of light, or energy or whatever. They're like, if light was faster, could it escape the black hole? If my rocket had bigger engines, could I escape? The problem is, all these questions make no sense. You can't get out because there is no way out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A black hole is formed when gravity is so strong, it ties spacetime into a giant knot. Space is tied into itself. It's not just a little bent; it gets curved until it closes onto itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside the event horizon, all paths lead to the center. No matter how you're turning, which direction you're looking, you're actually facing the center. It's hard to visualize, but that's how it is. This is not normal spacetime, it's something unlike anything you've thought about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting from point A inside the event horizon, there is no path you could draw that leads to point B outside. All paths lead to the center. Spacetime is really sick and broken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the real reason why nothing gets out of a black hole.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2019-01-14T19:59:46.983" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="29178" PostTypeId="2" ParentId="29175" CreationDate="2019-01-14T22:39:39.130" Score="3" Body="&lt;p&gt;There are a couple of slight misapprehensions here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Red light doesn't travel furthest, it just is scattered least. From math.ucl.edu:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A clear cloudless day-time sky is blue because molecules in the air scatter blue light from the sun more than they scatter red light. When we look towards the sun at sunset, we see red and orange colours because the blue light has been scattered out and away from the line of sight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) In low light human eyes pretty much lose any colour sensitivity. Our eyes just respond to brightness. So while there may be a hint of a blue tint as no red will make its way round the earth - really it is just black and white and grey.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2019-01-14T22:39:39.130" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29179" PostTypeId="1" AcceptedAnswerId="29192" CreationDate="2019-01-14T22:41:39.210" Score="5" ViewCount="366" Body="&lt;p&gt;I understand that the radiation of a body can be described using the curve for black-body radiation. In the sense that a hotter body will be blue shifted and a cooler body will be red shifted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The doppler effect gives similar results in the sense that a body is approaching us is blue shifted. While an object that is travelling away from us is red shifted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How is it possible to distinguish between temperature and doppler effect?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="25550" LastActivityDate="2019-09-18T10:06:22.843" Title="How to Distinguish between Temperature and Doppler effect using Black-body radiation?" Tags="&lt;star&gt;&lt;temperature&gt;&lt;radiation&gt;&lt;doppler-effect&gt;&lt;black-body&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29180" PostTypeId="2" ParentId="29175" CreationDate="2019-01-14T23:25:11.670" Score="6" Body="&lt;p&gt;The blue tint is an illusion caused by the wavelength sensitivity shift when switching from rods to cones as the light intensity decreases below certain level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Purkinje_effect&quot; rel=&quot;noreferrer&quot;&gt;Purkinje effect&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objectively, &lt;a href=&quot;https://physics.stackexchange.com/questions/244922/why-does-moonlight-have-a-lower-color-temperature/245275&quot;&gt;moonlight is not blue. In fact, it is even more yellow than sunlight!&lt;/a&gt; (comparing the same elevation, because obviously the true color can be distorted by the atmoshpere, too). The moonlight is not bright enough to fully enable the photopic vision and our red color sensitivity suffers, producing the blue moonlight illusion.&lt;/p&gt;&#xA;" OwnerUserId="7654" LastActivityDate="2019-01-14T23:25:11.670" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29181" PostTypeId="2" ParentId="29179" CreationDate="2019-01-15T00:58:57.137" Score="7" Body="&lt;p&gt;Without any other information, you cannot distinguish between the two effects.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ T = T_0 (1 + z) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A blackbody spectrum of temperature &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is identical to a blackbody spectrum of temperature &lt;span class=&quot;math-container&quot;&gt;$T_0$&lt;/span&gt; with redshift &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For stellar/galactic radiation, we can use the fact that the radiation is not a perfect blackbody.  For the CMB, we can use the fact that atomic combination happens only below a particular temperature.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2019-01-15T00:58:57.137" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29183" PostTypeId="2" ParentId="29171" CreationDate="2019-01-15T05:03:39.133" Score="2" Body="&lt;p&gt;The experimental data only shows that there is no life and no intelligent life in the parameter space where we could detect it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The experimental data doesn't explain the cause.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That we are alone in the Universe, it is still a possibility.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastActivityDate="2019-01-15T05:03:39.133" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29184" PostTypeId="2" ParentId="29158" CreationDate="2019-01-15T06:07:08.317" Score="1" Body="&lt;p&gt;I would interpret this as the Local Vertical, Local Horizontal (LVLH) frame (see, e.g., Vallado and McClain, Fundamentals of Astrodynamics and Applications, 4th ed., p. 156). This is also known as the RIC frame (radial, in-track, cross-track) or the RTN frame (radial, transverse, normal).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The radial or local vertical vector &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; is just the position vector from the center of the Earth to the spacecraft.&lt;/li&gt;&#xA;&lt;li&gt;The in-track, transverse, or local horizontal vector &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; is the vector in the orbital plane that is both prograde and perinduclar to the radial vector. &lt;strong&gt;It is not, in general, the velocity vector.&lt;/strong&gt; (The in-track and velocity vectors are coincident if and only if the flight path angle is zero, i.e., you have one of a) a circular orbit, b) the spacecraft is at apoapsis, or c) the spacecraft is at periapsis). You can construct it by applying the Graham-Schmidt Process to the position and velocity vectors; &lt;span class=&quot;math-container&quot;&gt;$I = v - \frac{R\cdot v}{R \cdot R}R$&lt;/span&gt;. &lt;/li&gt;&#xA;&lt;li&gt;The cross-track or normal vector &lt;span class=&quot;math-container&quot;&gt;$C = R \times I$&lt;/span&gt; completes the right-handed set.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="25505" LastActivityDate="2019-01-15T06:07:08.317" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29185" PostTypeId="2" ParentId="4800" CreationDate="2019-01-15T06:47:53.020" Score="1" Body="&lt;p&gt;Layman's explanation here, so take it for what it's worth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark matter has a localized gravitational effect (localized to galaxies, yes, but localized nonetheless); its existence is hypothesized on the basis of this observed effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If dark matter were equally distributed throughout the universe, it could not have a localized effect, but instead its effect would be observed in the overall rate of the universe's expansion (or contraction); and at any rate it could not remain equally distributed, but would begin to clump together with the ordinary, visible matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the dark matter has to clumped so that there's more of it in and around galaxies, and less of it in intergalactic space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light, however, could not clump in this way. While some of the light generated in a galaxy hits some of the matter in that galaxy and is absorbed, or hits a black hole in that galaxy and get stuck there until the end of time, for the most part the light emitted by a celestial body will make it to intergalactic space, and unless the universe is dense enough to eventually collapse, that light will never return to its home galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore&amp;mdash;as others have pointed out&amp;mdash;dark matter is postulated because the visible matter in galaxies is insufficient to explain its rotation. Since the visible matter has far and away more mass than the light emitted by that matter (by multiple orders of magnitude), but is insufficient to account for the rotation of galaxies, then surely the radically smaller mass of the emitted light must necessarily be even less sufficient, and thus could not be an explanation for what dark matter is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So no, dark matter is not light.&lt;/p&gt;&#xA;" OwnerUserId="24571" LastActivityDate="2019-01-15T06:47:53.020" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29186" PostTypeId="2" ParentId="12416" CreationDate="2019-01-15T16:20:05.950" Score="0" Body="&lt;p&gt;Because the gravitational pull of a non-rotating black hole in a void space is equal at all sides (as a result of it being a point) the event horizon should in theory be a perfect sphere. Because of the nature of a blackhole not letting light escape you would need a background for it to be seen. Also, in real world physics a blackhole  would need to interact with light to be &quot;seen&quot; which &quot;correct me if I'm wrong&quot; has it's own very small gravitational pull, it would slightly bend the sphere a bit, the observer will give the same effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If rotating it would (as pointed out before) form an ergosphere around the event horizon connecting to it at the poles an being the farthest over the equator, thus forming a more oval shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you know who to subscribe to right?&lt;/p&gt;&#xA;" OwnerUserId="25570" LastEditorUserId="7472" LastEditDate="2019-01-15T16:57:58.203" LastActivityDate="2019-01-15T16:57:58.203" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29189" PostTypeId="1" CreationDate="2019-01-15T19:44:33.593" Score="8" ViewCount="313" Body="&lt;p&gt;Uranus has an obliquity of 98° which means that the mutual inclination between a satellite orbiting in its equatorial plane and the orbit of Uranus around the Sun would exceed the critical angle for &lt;a href=&quot;https://en.wikipedia.org/wiki/Kozai_mechanism&quot; rel=&quot;noreferrer&quot;&gt;Kozai oscillations&lt;/a&gt;. This would drive the satellites to high eccentricities which would likely be extremely damaging to the stability of the system, yet Uranus does host a system of regular satellites on near-circular orbits in its equatorial plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This suggests something is suppressing the Kozai oscillations for the regular satellites, so what is doing this? I suspect it is linked either to the non-spherical shape of the planet or gravitational interactions between the moons, but I'm not sure which would be a more relevant factor.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-01-17T22:35:29.263" LastActivityDate="2019-01-17T22:35:29.263" Title="Why is Uranus able to support a regular satellite system?" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;natural-satellites&gt;&lt;celestial-mechanics&gt;&lt;uranus&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29190" PostTypeId="1" AcceptedAnswerId="29196" CreationDate="2019-01-15T19:50:55.617" Score="2" ViewCount="106" Body="&lt;p&gt;I am trying to figure out how fast an object will have to travel (at a height around 10,600 m) to never be in the presence of the moon. Meaning the object will always be on the opposite side of the moon to the Earth.&lt;/p&gt;&#xA;" OwnerUserId="25572" LastEditorUserId="1891" LastEditDate="2019-01-15T20:21:01.680" LastActivityDate="2019-01-16T06:37:25.903" Title="How fast would an object at cruising altitude(around 10,600 m) need to travel to stay on the opposite side of the Earth to the Moon?" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;earth&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29191" PostTypeId="2" ParentId="29190" CreationDate="2019-01-15T20:14:48.367" Score="1" Body="&lt;p&gt;Well, as a first approximation,  you know the Earth rotates once every 24 hours while the moon moves &quot;only a little&quot; in that time.  This means your object must appear to an observer on the Earth' surface to rise and fall 12 hours opposite to the moon.  First thing, then, is you better be moving East to West, just as the moon appears to do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the Earth's radius as approximately 6400 km,  your object is essentially at the Earth's surface to 2 significant digits.  The orbital circumference is thus &lt;span class=&quot;math-container&quot;&gt;$2*\pi*6400$&lt;/span&gt; km.  You want to cover that distance in  24 hours, or 1676 km/hr . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to get very accurate results, you'll need to take into account the moon's eccentricity with respect to the equator (so that you are always on the opposite end of a diameter thru the Earth's center from the moon) as well as the fact that it moves approximately 1/28 of a revolution per day.&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2019-01-15T20:14:48.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29192" PostTypeId="2" ParentId="29179" CreationDate="2019-01-15T20:19:24.650" Score="3" Body="&lt;p&gt;If the object you're observing is something good and hot, like a star, it's pretty easy to identify a set of &lt;a href=&quot;https://en.wikipedia.org/wiki/Balmer_series&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ballmer lines&lt;/a&gt;   by the relative spacing between the lines.  Then compare the absolute wavelengths to the known stationary values and &lt;em&gt;viola&lt;/em&gt; you've got the Doppler shift value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this would &quot;technically&quot; be using the blackbody radiation from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Photosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;photosphere&lt;/a&gt; of a star to illuminate its cooler atmosphere above, making its dark absorption lines visible.&lt;/p&gt;&#xA;" OwnerUserId="907" LastEditorUserId="907" LastEditDate="2019-01-16T12:57:53.293" LastActivityDate="2019-01-16T12:57:53.293" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29193" PostTypeId="2" ParentId="29171" CreationDate="2019-01-15T23:22:58.843" Score="0" Body="&lt;p&gt;I assume you mean &quot;civilization&quot;, not &quot;life&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a pure speculation that an advanced civilization would emit a &quot;cacophony of noises&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even our civilization on Earth is already working on reducing the amount of electromagnetic noise it produces. We are reducing light pollution, we are moving communications from radio waves into fiber optic cables and into low-power cellular networks, a lot of our equipment is shielded to avoid unwanted electromagnetic emissions. The general push to conserve energy means less and less will be emitted into space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So based on our own recent historical experience, a civilization is unlikely to generate an electromagnetic signal noticeable at interstellar distances.&lt;/p&gt;&#xA;" OwnerUserId="23720" LastActivityDate="2019-01-15T23:22:58.843" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29194" PostTypeId="2" ParentId="29190" CreationDate="2019-01-15T23:26:26.527" Score="1" Body="&lt;p&gt;There are 24.84 hours before the moon appears in the same position in the sky. So if you wanted to keep the moon apparently fixed on the opposite side of Earth, you would need to travel the circumference at 10,000m of the Earth (40,100km) in 24.84 hours, or 1610 km/hr.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At that altitude the speed of sound is about 1090 km/hr, so you would need to travel substantially above the speed of sound&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-01-15T23:26:26.527" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29195" PostTypeId="2" ParentId="29189" CreationDate="2019-01-16T01:37:46.133" Score="4" Body="&lt;p&gt;The solar perturbations on most of the satellites of Uranus are on a very small scale indeed, which may explain the absence of the instabilities noted in the question.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A perturbational effect depends on the scale of the solar perturbing accelerations relative to the ordinary inverse-square attraction of the primary body. The scale factor (often designated &lt;span class=&quot;math-container&quot;&gt;$ m^2 $&lt;/span&gt; in texts on the lunar theory such as Airy's 'Mathematical Tracts' and Godfray's 'Lunar Theory') increases with the separation of the satellite from the primary, it is also as the cube of the distance-ratio satellite-primary : primary-Sun (see detailed calculation below). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking as an example the most distant of the major moons of Uranus, Oberon at 583500 km: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scale factor for the solar perturbing accelerations on Oberon in its orbit relative to Uranus is only about 1/5,000,000 of the gravitational acceleration towards Uranus felt by Oberon (calculation below). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing the corresponding scale factor for Earth's moon and its solar perturbations, the factor is very much larger, close to  1/178, as is well-known.  The Moon has a fairly strongly sun-perturbed orbit relative to the Earth, but the solar perturbations on Oberon are more than four orders of magnitude smaller, really very tiny. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that the two tiny outermost satellites XVI and XVII of Uranus may be experiencing perturbations large enough to drive growth of their eccentricities, their eccentricities at 0.18 and 0.52  are very large compared with the major satellites, all with eccentricity &amp;lt; 0.004 (source, Astronomical Almanac 2016).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Detail of the approximate calculation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 'sampled' solar perturbing acceleration on a satellite is taken for present purposes to be represented by the perturbing acceleration on the satellite when it is close to quadrature with the Sun (as seen from the planet). In this configuration, the solar perturbation on the satellite is directed towards the planet, adding here to the ordinary inverse-square attraction of the satellite towards the planet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With mass-constants of the Sun, S, and of the planet, P ;  planetary semi-axis a ; and planet-satellite separation d , plus an assumption that the mass of the satellite is very small relative to the other two: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 ** the planet's accelerative attraction on the satellite is &lt;span class=&quot;math-container&quot;&gt;$P/d^2$&lt;/span&gt; ; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 ** the Sun's accelerative attraction on the planet is &lt;span class=&quot;math-container&quot;&gt;$S/a^2$&lt;/span&gt; ; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 ** and so also (to a quite close approximation) is the amplitude of the Sun's accelerative attraction on the satellite  &lt;span class=&quot;math-container&quot;&gt;$S/a^2$&lt;/span&gt; . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 ** When the satellite is in quadrature, the resolved part of the Sun's vector attraction on it that does not cancel with the Sun's attraction on the planet, i.e. the net perturbation on the satellite at that point, is given very nearly by the proportions of the triangle satellite-planet-sun:  the lengths of two of the sides are approximately a and the third d. In this configuration, the resolved part of acceleration #3 in the direction of the planet, to a close approximation, is thus  -- &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ (S/a^2) .  (d/a) $&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio of perturbing acceleration 4 to ordinary planetary attraction 1 is thus &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ (S/P) . (d/a)^3 $&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Sun and Uranus, S/P ~= 22902 , &lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the Sun and Earth+Moon, S/P ~= 328901 . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uranus is approximately 19 au from the Sun to the Earth's 1, the au is  149597871 km, the mean planetocentric distance of Oberon is 583500 km and of the Moon 385000 km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these figures, the ratio &lt;/p&gt;&#xA;&#xA;&lt;p&gt;'solar perturbation on satellite : planetary attraction on satellite'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;comes to ~ 1/178 for the Earth and Moon, and  ~ 1/5,000,000 for Uranus and Oberon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tiny outer satellites of Uranus are farther away from it than Oberon is, in a ratio of about 20.8 for the outer (XVII). Thus the perturbations on it, as a proportion of the ordinary attraction of Uranus at its distance, is &lt;span class=&quot;math-container&quot;&gt;$(20.8)^3$&lt;/span&gt; times larger than for Oberon, making the relevant perturbation scale factor as large as about 1/550 , still smaller than for the one for our Moon, but perhaps enough for the disturbing effects to be reflected in its higher eccentricity of about 0.52.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;{Update:} It turns out that the orbits of the outer Uranian moons have actually been studied: Brozovic, M.; Jacobson, R. A. (2009), &quot;The Orbits of the Outer Uranian Satellites&quot;, &lt;a href=&quot;http://adsabs.harvard.edu/abs/2009AJ....137.3834B&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Astronomical Journal, 137 (4): 3834-42&lt;/a&gt;. Just one of the (high-eccentricity) outer satellites was found to be disturbed by a Kozai resonance and may be in unstable orbit. It appears that the conditions for the effect are not fulfilled for the others.  &lt;/p&gt;&#xA;" OwnerUserId="17202" LastEditorUserId="17202" LastEditDate="2019-01-17T01:52:10.840" LastActivityDate="2019-01-17T01:52:10.840" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29196" PostTypeId="2" ParentId="29190" CreationDate="2019-01-16T06:29:49.767" Score="1" Body="&lt;p&gt;Other answers already note that the average period of the Moon in the sky (in hours) is about&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{1}{\frac{1}{1 \ \text{day}} - \frac{1}{27.32 \ \text{days}}} \approx 1.038 \ \text{days, or } 24.91 \ \text{hours}. $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the Earth's equator, where the circumference is about &lt;a href=&quot;https://space.stackexchange.com/a/23847/12102&quot;&gt;6378.137 km&lt;/a&gt; plus the OP's 10.6 km alitude = 6388.7 km, that would be about 1611 km/hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you can prorate that by &lt;span class=&quot;math-container&quot;&gt;$\cos(\text{latitude})$&lt;/span&gt; so at 60 degrees north or south that would be only half, or only about 806 km/s and that's only about 0.76 the &lt;a href=&quot;https://en.wikipedia.org/wiki/Mach_number#Overview&quot; rel=&quot;nofollow noreferrer&quot;&gt;speed of sound at cruise altitude&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what's really interesting is that if you go far enough towards one of the poles, you may be able to hide for the moon &lt;em&gt;for almost two weeks without moving at all!&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://astronomy.stackexchange.com/a/27750/7982&quot;&gt;this answer&lt;/a&gt; which makes for very interesting reading, the Moon can remain above the horizon at one pole for two weeks while being below the horizon and therefore invisible at the other for those two weeks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That can get you ready for your flight from one pole to the other, in a sort-of spiral shape to stay hidden from the Moon just long enough to reach the the other pole. Then you've got close to two more weeks before you make your return flight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will have quite a challenge finding a place to land near the North pole, and global warming means that's even more challenging for more months a year, but who knows maybe after a few decades of warming there will be floating airports in the arctic ocean and hotels and...&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-01-16T06:37:25.903" LastActivityDate="2019-01-16T06:37:25.903" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29197" PostTypeId="2" ParentId="29163" CreationDate="2019-01-16T08:02:14.023" Score="6" Body="&lt;p&gt;Florin Andrei's answer is correct in my opinion, and flows directly from the mathematics of GR. But here is an alternative way of thinking about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light always travels at the speed of light when measured locally. An observer inside the event horizon can emit light moving radially outward (according to them). The problem is that they and everything else is falling inwards. The OP's &quot;gedanken experiment&quot; is simply not possible; within the event horizon there can be no stationary observer that launches a light beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An oft-used analogy is drifting in a boat on a river. You release fish into the water that swim at constant speed, upstream or downstream &lt;em&gt;relative to your boat&lt;/em&gt;. However, if the river flows fast enough, the fish can never make their way upstream as far as an observer on the bank is concerned, and both boat and fish will end up going over the waterfall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The event horizon marks the point where the river flows too fast for the fish to escape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For anyone interested in exploring the so-called &quot;river model&quot; of thinking about dynamics in and around black holes, there is an excellent (though somewhat mathematical) introduction by &lt;a href=&quot;https://arxiv.org/abs/gr-qc/0411060&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hamilton &amp;amp; Lisle (2006)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-01-17T20:54:46.247" LastActivityDate="2019-01-17T20:54:46.247" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29198" PostTypeId="1" AcceptedAnswerId="29205" CreationDate="2019-01-16T08:53:47.163" Score="3" ViewCount="185" Body="&lt;p&gt;Reading about Uranus almost 90° tilt, I was wondering if some rocky planet with mass concentration at one pole could possibly spin around its own axis, while still being locked to its star? Which means the more massive pole always points toward the star?&lt;/p&gt;&#xA;" OwnerUserId="16500" LastActivityDate="2019-01-16T15:11:27.770" Title="Tidally locked, and yet spinning?" Tags="&lt;planet&gt;&lt;orbit&gt;&lt;tidal-locking&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29199" PostTypeId="2" ParentId="29141" CreationDate="2019-01-16T10:20:29.700" Score="3" Body="&lt;p&gt;To answer the question &quot;how does science knows what is early and what is late?&quot; in simple terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the &lt;a href=&quot;https://en.wikipedia.org/wiki/Expansion_of_the_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;Universe is expanding&lt;/a&gt;. Because of this, the light from things further away from us gets &quot;stretched&quot; more, and is &lt;a href=&quot;https://en.wikipedia.org/wiki/Redshift&quot; rel=&quot;nofollow noreferrer&quot;&gt;redshifted&lt;/a&gt;. The bigger the redshift, the further the light has travelled. We can use formulas to work out the distance based on the amount of redshift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The further away an object is, the longer it's taken for that light to reach us. So if we see an object with a very large redshift and we know this means it's very far away, the light must therefore have travelled a very long time to reach us. This is how we can tell whether the light comes from very long ago, i.e. earlier in the Universe's history. By examining the redshift of the quasar that you referred to, we can work out that the light we're now seeing was emitted 12.8 billion years ago, when the Universe was only 1 billion years old. That is &lt;em&gt;amazingly early&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;oldest&quot; light we can detect is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow noreferrer&quot;&gt;cosmic microwave background (CMB) radiation&lt;/a&gt;, which you can think of as the &quot;remnant heat&quot; from the original Big Bang. When we look at its redshift we can calculate that the CMB photons started travelling towards us when the Universe was about 380,000 years old. We can't detect photons from earlier than that because the Universe was still full of free electrons, which scattered the light. As the Universe cooled enough for hydrogen ions to trap electrons and form neutral atoms, the &quot;fog&quot; cleared and photons were free to travel uninterrupted. &lt;/p&gt;&#xA;" OwnerUserId="14094" LastActivityDate="2019-01-16T10:20:29.700" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29202" PostTypeId="1" AcceptedAnswerId="29203" CreationDate="2019-01-16T12:22:55.320" Score="0" ViewCount="42" Body="&lt;p&gt;I want to know if there can be any planet that can fulfill the following conditions and if any planet has been observed with such characteristics:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The planet must not be tidally locked.&lt;/li&gt;&#xA;&lt;li&gt;The day-night cycle must be similar to our planet or smaller.&lt;/li&gt;&#xA;&lt;li&gt;The planet must be so close to the star that the light emitted from the star 'burns' (can cause third degree burns to human skin within seconds after exposure).&lt;/li&gt;&#xA;&lt;li&gt;Temperatures in the shadow cannot exceed 40-50°C (whether prolonged shadow during daytime or conditions during nighttime).&lt;/li&gt;&#xA;&lt;li&gt;Atmosphere is not necessary.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="25128" LastEditorUserId="1891" LastEditDate="2019-01-16T13:08:13.290" LastActivityDate="2019-01-16T13:42:18.607" Title="Possibility of certain conditions on a planet" Tags="&lt;star&gt;&lt;planet&gt;&lt;orbit&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2019-01-17T16:52:29.627" ContentLicense="CC BY-SA 4.0" />
  <row Id="29203" PostTypeId="2" ParentId="29202" CreationDate="2019-01-16T13:42:18.607" Score="1" Body="&lt;p&gt;The basic condition is you'd need is a more massive, hotter star than our sun.  Not necessarily huge, but 1.5 - 2.5 times the mass of the sun should be enough because that corresponds to considerable extra luminosity and UV rays, so you could move the planet far enough away than Earth is from the sun and avoid any risk of tidal locking or slowing, so the planet could retain fast rotation period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you'd need a very thin atmosphere so the planet doesn't retain much heat under shadow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the star is bright enough in the sky that you'd get a burn in seconds under direct light, then shade probably wouldn't be cool unless it was permanent shade, as you noted, perhaps a cave. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there's no atmosphere, you'd probably get colder temperatures than what you mention in shade, so some atmosphere, but a thin one.   Maybe somewhere between 1/10th of an ATM give or take and very low greenhouse gas, so basically very little water, CO2 or CH4.    1/10th of an ATM is just a very ballpark guess. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only part I don't see happening is 3rd degree burns in seconds.   3rd degree burns by photon radiation from a star would probably require longer than that, unless the planet's surface was hot enough to melt rock.   Bad sunburn in minutes, sure.   3rd degree burns in seconds with your other criteria sounds unlikely to me.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2019-01-16T13:42:18.607" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29204" PostTypeId="2" ParentId="26082" CreationDate="2019-01-16T14:19:39.343" Score="1" Body="&lt;p&gt;First off blackholes can orbit eachother, we can and have proven that because of the amence energy in the form of gravitational waves they give off which we can measure here on earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's possible that the ergosphere would prolong the mergetion of the two blackholes if they have the correct spin corresponding to the (very unlikely) orbit of the other black hole, although it would almost inevitable fall into one of the event horizons eventually because of the gravitational waves it produces. Which gives off a masive amount of kenetic energy. For the blackhole to gain kenetic energy it would have to overcome this energy loss.&lt;/p&gt;&#xA;" OwnerUserId="25570" LastEditorUserId="25570" LastEditDate="2019-01-16T20:43:08.147" LastActivityDate="2019-01-16T20:43:08.147" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29205" PostTypeId="2" ParentId="29198" CreationDate="2019-01-16T15:11:27.770" Score="5" Body="&lt;p&gt;I think what you are envisioning is having one pole always pointed towards the primary star. So the planet would rotate about that pole, but then the direction of the pole would change over the course of a year to keep the pole pointing towards a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This phenomenon of the direction of the pole changing is called &quot;precession&quot;. The Earth does it over a period of 26,000 years, but the pole only traces out a fairly small circle in the sky, rather than twisting about a perpendicular line like you envision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think your scenario is improbable at best. Planets are like gyroscopes and resist having their axes of rotation be twisted about a perpendicular axis like that. Even a planet with mass concentrations at both poles (maybe a long cigar shape) to maximize tidal force would probably end up losing the perpendicular rotation to tidal friction, and just end up rotating slowly about an axis nearly parallel to the axis of the orbit.&lt;/p&gt;&#xA;" OwnerUserId="7752" LastActivityDate="2019-01-16T15:11:27.770" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29206" PostTypeId="1" CreationDate="2019-01-16T15:34:50.647" Score="6" ViewCount="207" Body="&lt;p&gt;Forgive me if any of my terminology is off I am no physicist but I have wondered about this question for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read in various books that if the universe was infinite, and uniformly created the whole sky would be white. This is because in every possible location in the sky there would be a star emitting light. Sort of like pixels, filling a screen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought about this and wondered, doesn't this therefore imply that light is never absorbed by another medium?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if a distant stars light is absorbed, therefore never reaching earth, giving the illusion that there are less stars?&lt;/p&gt;&#xA;" OwnerUserId="25579" LastActivityDate="2019-01-17T16:04:52.377" Title="Universe isn't infinite theory" Tags="&lt;astrophysics&gt;&lt;light&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29208" PostTypeId="2" ParentId="29206" CreationDate="2019-01-16T17:31:01.900" Score="-3" Body="&lt;p&gt;Those books have it wrong apparently, and are not up-to-date on astronomy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can't even see the stars from the other side of our galaxy. Why? There's just not enough photons coming in from them, they're too faint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one would want to place so many stars in every line of sight to fill the sky 'white', either the universe would collapse because those stars would represent so much mass that it couldn't be stable, or time-of-flight effects come into play, if the stars were farther apart.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2019-01-16T17:31:01.900" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29211" PostTypeId="2" ParentId="29206" CreationDate="2019-01-16T20:20:09.610" Score="4" Body="&lt;p&gt;Olber's paradox describes an infinite, static, eternal, and homogenous universe. If there were a cloud of opaque gas situated in some region of the sky, light coming from all conceivable directions via the paradox would bombard the cloud until either it begins emitting light itself via the radiation it absorbs or its particles scatter to allow light to pass through. &lt;/p&gt;&#xA;" OwnerUserId="19454" LastActivityDate="2019-01-16T20:20:09.610" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29212" PostTypeId="1" CreationDate="2019-01-16T20:41:55.020" Score="1" ViewCount="292" Body="&lt;p&gt;A question for a random reader passing through: because of the nature of things spinning faster when it gets smaller, wouldn't a singularity spin infinitely fast as it's infinitely small, if it has a spinn from the beginning that is? Btw I know what a &quot;ringularity&quot; but does it automaticly become one when rotational velocity is added? Is it just &quot;fake news&quot;?&lt;/p&gt;&#xA;" OwnerUserId="25570" LastEditorUserId="1891" LastEditDate="2019-01-21T07:02:43.927" LastActivityDate="2019-01-21T16:28:07.000" Title="Why doesn't a singularity spin infinitely fast?" Tags="&lt;singularity&gt;&lt;infinite&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29213" PostTypeId="2" ParentId="29206" CreationDate="2019-01-16T21:25:48.850" Score="0" Body="&lt;p&gt;The reason that there isn't stars at every single point of the outlook from the earths surface is because of something called dark energy which is expanding the universe (the main point being that the universe in not shrinking). The further something is from something else the more the expantion effect applies. We can for an example see this because of something called &quot;redshift&quot; which is basically light being stretched to lower wavelengths. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NHl4O.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NHl4O.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If something is far enough so the relative expation for us compared to a obj is faster then light we can't see it. The part inside of that range is called the &quot;observable universe&quot;. Only in the obeservable universe light can interact with you. This takes out a big chunk of stars making it not being obs-infinite. The stars left are too spread out to cover the entire &quot;sky&quot;. Also the look of stars in the observalbe universe varys a lot. The stars feathest away have a very low ratio of photons reaching the earth compered to stars closeby. this is because light disperches is squared. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other factors as well like the dispersion of light in the atmoasphear!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stein aus!&lt;/p&gt;&#xA;" OwnerUserId="25570" LastEditorUserId="25570" LastEditDate="2019-01-17T16:04:52.377" LastActivityDate="2019-01-17T16:04:52.377" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29214" PostTypeId="1" CreationDate="2019-01-17T01:30:38.763" Score="1" ViewCount="60" Body="&lt;p&gt;If Earth suddenly stops rotating, will be thrown at 465 m/s ?&lt;/p&gt;&#xA;" OwnerUserId="24552" LastActivityDate="2019-01-17T01:30:38.763" Title="If Earth suddenly stops rotating, will be thrown at 465 m/s?" Tags="&lt;earth&gt;&lt;rotation&gt;&lt;celestial-mechanics&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2019-01-17T15:21:45.240" ContentLicense="CC BY-SA 4.0" />
  <row Id="29219" PostTypeId="1" AcceptedAnswerId="29788" CreationDate="2019-01-17T15:47:16.850" Score="7" ViewCount="1559" Body="&lt;p&gt;I've been reading and thinking about this subject for some time, but I can't seem to find a clear answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading &lt;a href=&quot;https://motherboard.vice.com/en_us/article/3k48qv/the-42-biggest-questions-about-life-the-universe-and-everything&quot; rel=&quot;nofollow noreferrer&quot;&gt;Q14&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Aspects of the Higgs boson suggest that our Universe is only&#xA;  “marginally stable,” or is perhaps in a transition phase to a more&#xA;  stable state that would result in a universe with fundamentally&#xA;  different properties.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And &lt;a href=&quot;https://phys.org/news/2014-09-universe-stable-quantum.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;With the discovery of the Higgs field, one question we can ask is&#xA;  whether the Universe as a quantum system it is in a stable, low-energy&#xA;  state, or in a metastable higher energy state?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Standard_Model&quot; rel=&quot;nofollow noreferrer&quot;&gt;standard model&lt;/a&gt; of particle physics, the answer depends upon the masses of the top quark and the Higgs boson. The more massive the Higgs boson, the more likely the Universe is to be stable. The more massive the top quark, the more likely the Universe is to be metastable, or even unstable. From our current measurements of the top and Higgs masses, it seems that our Universe is metastable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe an answer already been found, but I just didn't find it, or am I not on the right level. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is our universe stable or metastable?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="25592" OwnerDisplayName="tom" LastEditorUserId="25592" LastEditDate="2019-04-22T11:47:34.197" LastActivityDate="2021-05-25T02:43:56.093" Title="Is our universe stable or metastable?" Tags="&lt;astrophysics&gt;&lt;universe&gt;&lt;cosmology&gt;&lt;observable-universe&gt;&lt;fate-of-universe&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29221" PostTypeId="1" CreationDate="2019-01-17T21:00:31.220" Score="2" ViewCount="121" Body="&lt;p&gt;Just wondering so I can see it because I don't know if  it is happening this month or not?&lt;/p&gt;&#xA;" OwnerUserId="25595" LastActivityDate="2019-01-22T05:38:12.950" Title="Is there really a blood wolf moon this month?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="29222" PostTypeId="1" AcceptedAnswerId="29230" CreationDate="2019-01-17T22:51:47.310" Score="4" ViewCount="156" Body="&lt;p&gt;What are the white reflections on Ceres?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://solarsystem.nasa.gov/planets/dwarf-planets/ceres/overview/&quot; rel=&quot;noreferrer&quot;&gt;https://solarsystem.nasa.gov/planets/dwarf-planets/ceres/overview/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yUBBd.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yUBBd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25597" LastEditorUserId="7472" LastEditDate="2019-01-18T12:15:38.130" LastActivityDate="2020-08-13T11:53:56.023" Title="What are the white reflections on Ceres?" Tags="&lt;ceres&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29223" PostTypeId="1" CreationDate="2019-01-17T22:59:59.547" Score="12" ViewCount="201" Body="&lt;p&gt;Since the actual position of stars is relative, and we see the light they emitted long time ago, and one galaxy has thousands of millions of stars, and those stars can be separated by hundred of thousands of light years, why don't we see galaxies as abstract forms or lines in the sky  as the light of every galaxy star arrives to us at different times instead of well formed spiral shapes? It's like if those millions stars were forming the galaxy as a single light spot, instead of millions of spots separated by light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has been in my head for decades, but I didn't find the question (not even the answer, but the question) anywhere, so either there must be an really easy answer to it or a really complex one or it's a glitch in the Matrix.&lt;/p&gt;&#xA;" OwnerUserId="25596" LastActivityDate="2019-01-19T11:56:54.723" Title="Why do we see galaxies with their actual spiral shapes?" Tags="&lt;galaxy&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29224" PostTypeId="2" ParentId="29223" CreationDate="2019-01-17T23:54:47.170" Score="11" Body="&lt;p&gt;Basically, because the relative speeds of stars within galaxies are &lt;em&gt;much&lt;/em&gt; less than the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a structure with dimension 100,000 light years (about the size of a galaxy). Now suppose the components of that structure move with speeds relative to each other of around 100 km/s (or 0.03% of the speed of light). In the time it takes light to cross the structure (100,000 years of course), the components within it will have moved by just 30 light years (0.03% of the size of the structure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There would therefore be no significant &quot;blurring&quot; of our snapshot of a galaxy caused by the differing light travel times from its structural components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thing you may have not understood, is that the lives of stars are generally &lt;em&gt;much&lt;/em&gt; longer than 100,000 years. So the stars that emit the light on the far side of a galaxy are still in  existence 100,000 years later, when their light passes stars that are present at the near side of a galaxy.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-01-18T07:30:10.233" LastActivityDate="2019-01-18T07:30:10.233" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29227" PostTypeId="1" CreationDate="2019-01-18T07:21:14.227" Score="1" ViewCount="101" Body="&lt;p&gt;Why do most of the &lt;a href=&quot;https://www.windows2universe.org/mythology/const_nav.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;myths regarding constellations&lt;/a&gt; happen to have begun from Greek civilization? Why not other civilizations? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most people who lived in the past era probably observed star patterns in the sky and created their own constellation myths. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Also, I would like to know more about some of the &lt;a href=&quot;http://www.tcoe.org/scicon/instructionalguide/constellations.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;stories regarding the myths of those constellations&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="25602" LastEditorDisplayName="user1569" LastEditDate="2019-01-18T10:24:19.207" LastActivityDate="2019-01-20T14:14:01.570" Title="Why do most of the myths behind constellations originate from Greek?" Tags="&lt;star&gt;&lt;fundamental-astronomy&gt;&lt;constellations&gt;" AnswerCount="1" CommentCount="7" ClosedDate="2019-01-20T17:33:11.327" ContentLicense="CC BY-SA 4.0" />
  <row Id="29229" PostTypeId="1" CreationDate="2019-01-18T12:10:13.103" Score="1" ViewCount="43" Body="&lt;p&gt;I am looking for an animation that not only shows the inclination of the solar system respective to the plane of the galaxy, but also shows the motion of the sun relative to and in conjunction with (i.e keeping the local frame of reference of rest intact) the local group of stars. If anyone has seen such a animation or if anyone has the skill set to make this animation then it would be a great help!&lt;/p&gt;&#xA;" OwnerUserId="25606" LastActivityDate="2019-01-18T12:10:13.103" Title="Looking for a 3D animation showing the motion of the sun relative to other stars and the Milky Way, any one seen anything like that?" Tags="&lt;the-sun&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29230" PostTypeId="2" ParentId="29222" CreationDate="2019-01-18T12:25:41.100" Score="8" Body="&lt;p&gt;According to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Bright_spots_on_Ceres&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia page&lt;/a&gt; linked by @BowlOfRed, they're ice and/or salt.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a material with a high level of reflection, and [the scientists] suggested ice and salt as possibilities.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;On 9 December 2015, scientists reported that the bright spots on Ceres may be related to a type of salt, particularly a form of brine containing hydrated magnesium sulfate (MgSO&lt;sub&gt;4&lt;/sub&gt;·6H&lt;sub&gt;2&lt;/sub&gt;O); the spots were also found to be associated with ammonia-rich clays. Arizona State University scientists have proposed that the bright spots are best explained as resulting from briny water erupted from Ceres's interior that subsequently sublimated, leaving behind only the salt deposits. Near-infrared spectra of these bright areas are consistent with a large amount of sodium carbonate, (Na&lt;sub&gt;2&lt;/sub&gt;CO&lt;sub&gt;3&lt;/sub&gt;) and smaller amounts of ammonium chloride (NH&lt;sub&gt;4&lt;/sub&gt;Cl) or ammonium bicarbonate (NH&lt;sub&gt;4&lt;/sub&gt;HCO&lt;sub&gt;3&lt;/sub&gt;). These materials have been suggested to originate from the recent crystallization of brines that reached the surface from below.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2019-01-18T12:25:41.100" CommentCount="0" CommunityOwnedDate="2019-01-18T12:25:41.100" ContentLicense="CC BY-SA 4.0" />
  <row Id="29232" PostTypeId="1" AcceptedAnswerId="29243" CreationDate="2019-01-18T23:35:37.057" Score="1" ViewCount="110" Body="&lt;p&gt;In 2015 the Planck satellite consortium released an analysis of data collected by the Planck satellite that showed this percentage break down of the universe’s mass/energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ordinary matter: 4.9%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark Matter: 26.8%,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark Energy: 68.3%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did they calculate the &quot;stuff&quot; of the observable universe and how did they come to this conclusion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume  this took some tough calculations and some work with a supercomputer but i wonder what the entire process was.&lt;/p&gt;&#xA;" OwnerUserId="25592" LastActivityDate="2019-01-28T16:33:59.247" Title="What was the process of calculating how much darkmatter/matter/dark energy the observable universe consists of and how did it come at this conclusion" Tags="&lt;universe&gt;&lt;dark-matter&gt;&lt;observable-universe&gt;&lt;dark-energy&gt;&lt;matter&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29234" PostTypeId="2" ParentId="27960" CreationDate="2019-01-19T00:59:43.317" Score="1" Body="&lt;p&gt;Although I have Starry Night I have used Alcyone Ephemeris, version 4.3, in conjunction with Solex120 to create a 7000 year Biblical Calendar from -4000 to 3000. The drawback with the Alcyone software is that it uses DE406 which is really only accurate for the specified 6000-year range. Also it does not seem to properly take into account the impact of altitude on sunrise and sunset times. Solex 120 uses DE431 which I have used in other projects to cover a range from -20000 to +20000 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have compared both Alcyone and Solex120 with my own spreadsheet-based sunrise/sunset calculator based on Meeus' algorithms and determined that Solex is correct. What I did was calculate the sunrise/set times in Solex and then imported them into Alcyone to calculate the position of the sun and moon around the time of sunset to search for the 1st visible crescent of the moon. Alcyone did this 16000 entries at a time and I checked a few of them manually with Solex and determined a good match.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although one can type any time into Solex to find the planets positions it doesn't allow one to import a list of slightly variable times, like sunset, as it only allows fixed interval steps to be specified. Solex can search for eclipses, oppositions, occultations etc. and you have to learn how to use it. Last I checked, and this may have changed, I used the User Manual from Solex110 to find out how to use Solex120, but that version only uses DE406. However the manual did apply to Solex120. Solex does allow you to export ephemeris date pretty much over any period, depending upon how big you can tolerate your files being.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My Excel files get up to nearly 200 MB. I had to get 16GB RAM to support that. You need to use a Windows command prompt to run Solex. You can use a companion Alcyone program called Planetry, Lunar and Stellar Visibility to created calendars back to about -3990, and output 100 years worth of data to XML at a time. But be warned you need to grab each file before you create the next one and you also need to find the folder it puts them in. This depends on where your browser stores stuff.&lt;/p&gt;&#xA;" OwnerUserId="25613" LastEditorUserId="1891" LastEditDate="2019-01-20T11:53:43.470" LastActivityDate="2019-01-20T11:53:43.470" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29236" PostTypeId="2" ParentId="29223" CreationDate="2019-01-19T09:09:17.917" Score="-2" Body="&lt;p&gt;I am not rewriting what nicely said in another answer. I just point out, as it is just implicit there, that even the closest stars in our galaxy are really far away!  Not too say other galaxies. If, as I suppose, you aren't surprised by the fact that the stars look as fixed over millennia, then considering distance should fix your question. &lt;/p&gt;&#xA;" OwnerUserId="17437" LastEditorUserId="17437" LastEditDate="2019-01-19T11:56:54.723" LastActivityDate="2019-01-19T11:56:54.723" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29242" PostTypeId="1" CreationDate="2019-01-19T11:42:31.463" Score="6" ViewCount="282" Body="&lt;p&gt;Because the moon distance from earth grows, how long will it take, that it is no longer possible to see a total eclipse but only circular eclipses? And does it make a difference if you are on the mount everest or on sea level?&lt;/p&gt;&#xA;" OwnerUserId="20334" LastActivityDate="2019-01-24T18:06:10.110" Title="When will the last total solar eclipse occur?" Tags="&lt;the-moon&gt;&lt;solar-eclipse&gt;" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="29243" PostTypeId="2" ParentId="29232" CreationDate="2019-01-19T11:56:19.680" Score="2" Body="&lt;p&gt;We are not actually measuring the &lt;span class=&quot;math-container&quot;&gt;$\Omega_r$&lt;/span&gt;,&lt;span class=&quot;math-container&quot;&gt;$\Omega_m$&lt;/span&gt;,&lt;span class=&quot;math-container&quot;&gt;$\Omega_{\Lambda}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;At first, we measure the &lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/D/Distance+Modulus&quot; rel=&quot;nofollow noreferrer&quot;&gt;distance modulus&lt;/a&gt; to certain celestial objects by using the &lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/S/Standard+Candle&quot; rel=&quot;nofollow noreferrer&quot;&gt;standart candles&lt;/a&gt; in the astronomy.&lt;/p&gt;&#xA;&lt;p&gt;This distance can be calculated from,&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$d_L=\frac {c} {H_0}z(1+\frac {1-q_0} {2}z)\,\,\, (Eqn.1)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here &lt;span class=&quot;math-container&quot;&gt;$q_0$&lt;/span&gt; represents the deceleration parameter. Now I am not going to go in mathematical details but you should now that, using the acceleration equation we can write &lt;span class=&quot;math-container&quot;&gt;$q_0$&lt;/span&gt; in terms of &lt;span class=&quot;math-container&quot;&gt;$\Omega_r$&lt;/span&gt;,&lt;span class=&quot;math-container&quot;&gt;$\Omega_m$&lt;/span&gt;,&lt;span class=&quot;math-container&quot;&gt;$\Omega_{\Lambda}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$q_0=1/2 \sum \Omega_{w,0}(1+3w)$$&lt;/span&gt;  and for matter ,&lt;span class=&quot;math-container&quot;&gt;$w=0$&lt;/span&gt;, for radiation &lt;span class=&quot;math-container&quot;&gt;$w=1/3$&lt;/span&gt; and for dark energt &lt;span class=&quot;math-container&quot;&gt;$w=-1$&lt;/span&gt; we get,&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$q_0=\Omega_r+1/2\Omega_m - \Omega_{\Lambda}\,\,\, (Eqn.2)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;And the relationship between distance modulus &lt;span class=&quot;math-container&quot;&gt;$(m-M)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$d_L$&lt;/span&gt; can be written as,&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$m-M=5log_{10}(d_L/1Mpc)+25\,\,\, (Eqn.3)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;If we combine Eqn.1 and Eqn.3 we get a relationship between distance modulus and z.&lt;/p&gt;&#xA;&lt;p&gt;We plot the &lt;span class=&quot;math-container&quot;&gt;$m-M$&lt;/span&gt; for distant objects with respect to &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;. From the slope of the graph we can determine the &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt; (for &lt;span class=&quot;math-container&quot;&gt;$z\rightarrow 0$&lt;/span&gt;) And for large z,  deviation of the plot from the straight line gives us to determine the &lt;span class=&quot;math-container&quot;&gt;$q_0$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Now we have experimental data. Then we put different values for &lt;span class=&quot;math-container&quot;&gt;$q_0$&lt;/span&gt; or equally &lt;span class=&quot;math-container&quot;&gt;$\Omega_m$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\Omega_{\Lambda}$&lt;/span&gt; (with respect to Eqn.2) to determine the best fit for the graph (since we can take &lt;span class=&quot;math-container&quot;&gt;$\Omega_r=0$&lt;/span&gt;)&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RIu9u.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RIu9u.png&quot; alt=&quot;Distance modulus versus redshift for type Ia supernovae from the Supernova Cosmology Project (Perlmutter et al. 1999, ApJ, 517, 565) and the High-z Supernova Search Team (Riess et al. 1998, AJ, 116, 1009). &quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Edit: I clearly missed the &amp;quot;The Planck data&amp;quot; in the OP's question but its the same general idea. For different density parameters, the CMBR would look different, like the graphs that I shared are different for different density parameters. So by looking at the CMBR, we can determine the density parameters and which one fits well to the CMBR.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Further Note: I want to also suggest that the &lt;span class=&quot;math-container&quot;&gt;$\Omega_b$&lt;/span&gt; can be calculated from the early universe by measuring the amount of deuterium and other elements present in primordial gas clouds.&lt;/p&gt;&#xA;&lt;p&gt;And observering the rotational curves on the galaxies we can have approximate values on the &lt;span class=&quot;math-container&quot;&gt;$\Omega_{dm}$&lt;/span&gt;. (finding the dark matter density)&lt;/p&gt;&#xA;&lt;p&gt;We cannot deduce the &lt;span class=&quot;math-container&quot;&gt;$\Omega_m$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\Omega_{\Lambda}$&lt;/span&gt; just by looking at the Supernova data, for example the supernova data is also consistent with &lt;span class=&quot;math-container&quot;&gt;$\Omega_m=0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\Omega_{\Lambda}=0.4$&lt;/span&gt; or with &lt;span class=&quot;math-container&quot;&gt;$\Omega_m=1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\Omega_{\Lambda}=1.7$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="25424" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-01-28T16:33:59.247" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29244" PostTypeId="2" ParentId="29221" CreationDate="2019-01-19T12:03:18.887" Score="4" Body="&lt;p&gt;On January 20th, 2019, there will be a total eclipse of the moon. Since Rayleigh scattering of sunlight passing through the earth's atmosphere and then hitting the moon during a total lunar eclipse makes the moon look reddish, total lunar eclipses are sometimes given the (portentous and slightly silly) name of &quot;blood moon&quot;. (This seems to be the result of vague New Testament references to lunar eclipses being promoted by a couple of apocalyptic fundamentalists in the last decade; see &lt;a href=&quot;https://en.wikipedia.org/wiki/Blood_moon_prophecy&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for more.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, there are various folk traditions which give special names to the (first) full moon in each month; it's apparently an old Anglo-Saxon tradition to call the first full moon in January the &quot;Wolf Moon&quot;. Since total lunar eclipses necessarily happen during full moons, you could thus (somewhat) legitimately call this a &quot;blood wolf moon&quot;. (Some discussion of other &quot;full moon names&quot;, including the &quot;harvest moon&quot;, can be found &lt;a href=&quot;https://www.timeanddate.com/astronomy/moon/full-moon-names.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, there is a recent trend of referring to a full moon that happens near to lunar perigee (when the moon makes its closest approach to the earth and appears somewhat larger in the sky) as a &quot;super moon&quot;; this term was apparently coined by the astrologer Richard Nolle in 1979. This sort of thing happens three or four times a year. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you &lt;em&gt;could&lt;/em&gt; call the upcoming lunar eclipse on January 20th a &quot;blood wolf super moon&quot; (or &quot;super blood wolf moon&quot;, or whatever), if you really wanted to....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Personally, I don't think that's nearly enough in the way of unnecessary and portentous adjectives. I'm holding out for &quot;blood wolf demon vampire super extreme power moon&quot;.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2019-01-22T05:38:12.950" LastActivityDate="2019-01-22T05:38:12.950" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29246" PostTypeId="2" ParentId="29242" CreationDate="2019-01-19T18:00:09.370" Score="9" Body="&lt;p&gt;I will take a stab at it.  Assuming &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the last eclipse happens when the moon is at perigee and the earth is at aphelion.&lt;/li&gt;&#xA;&lt;li&gt;the moon's perigee is increasing by 4 cm per year,&lt;/li&gt;&#xA;&lt;li&gt;the earth's aphelion does not change with time,&lt;/li&gt;&#xA;&lt;li&gt;the sun and the moon are perfect spheres,&lt;/li&gt;&#xA;&lt;li&gt;the radius of the sun and moon don't change, and&lt;/li&gt;&#xA;&lt;li&gt;a total eclipse occurs when &lt;span class=&quot;math-container&quot;&gt;$r_{\mathrm{sun}}/d_{\mathrm{sun}} &amp;lt; r_{\mathrm{moon}}/d_{\mathrm{moon}}$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; is radius and &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; is distance,&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I get the following results:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The last total eclipse at sea level will occur in 721,587,917 years at sea&#xA;level  (obviously too many significant digits).&lt;/li&gt;&#xA;&lt;li&gt;The last total eclipse at mount Everest will occur in 721,807,917 years.&lt;/li&gt;&#xA;&lt;li&gt;The difference is 220,000 years = (height_of_Everest) * 25000 years/km.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Mathematica Source Code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;dSun = 152097000; rSun = 695508; dMoonNow = 357347; rMoon = 1737; &#xA;rEarth = 6371; hEverest = 8 + 8/10;&#xA;sol2 = Solve[ rMoon/(d + t 4/100/1000) == rSun/(dSun - rEarth), t][[1]];&#xA;Print[ t /. sol2 /. d -&amp;gt; dMoonNow - rEarth  // Round];&#xA;Print[ t /. sol2 /. d -&amp;gt; dMoonNow - rEarth - hEverest // Round];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="25554" LastEditorUserId="25554" LastEditDate="2019-01-21T01:22:50.733" LastActivityDate="2019-01-21T01:22:50.733" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="29247" PostTypeId="1" CreationDate="2019-01-19T20:31:21.843" Score="4" ViewCount="568" Body="&lt;p&gt;On Earth, the sky is blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What color would the sky be on other planets in our solar system? What about outside the solar system? Are there planets with purple skies, or green skies? Or are the all just blue?&lt;/p&gt;&#xA;" OwnerUserId="25624" LastActivityDate="2021-05-09T03:39:29.330" Title="What color is the Sky on other Planets?" Tags="&lt;planet&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29249" PostTypeId="2" ParentId="29227" CreationDate="2019-01-20T14:14:01.570" Score="0" Body="&lt;p&gt;Ptolemy was Greek.  He wrote an important book called &lt;em&gt;the Almagest&lt;/em&gt;. In it, he lists and describes the constellations as used in Greece in the 4th century BC. Naturally, the Greeks used elements from Greek Mythology in their sky culture, just as Chinese sky culture uses elements from Chinese legend and myth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ptolemy's Almagest and Greek sky culture influenced the Hellenistic world, and through Rome, it influenced the whole of Western Europe.  Then through European expansion, it came to influence the whole world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several additions to Ptolemy's system: The Southern Sky was divided into constellations by Europeans in the 15th century, and there are several additions to the Northern sky, such as Coma Berencies (Hair of Berenice II of Egypt) and Camelopardalis, the giraffe, (added in the 15th century)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-01-20T14:14:01.570" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29250" PostTypeId="1" AcceptedAnswerId="29252" CreationDate="2019-01-20T15:30:48.397" Score="2" ViewCount="250" Body="&lt;p&gt;The smallest distance probed in any experiment so far conducted is &lt;span class=&quot;math-container&quot;&gt;$10^{-18}$&lt;/span&gt;m and the largest distance we can have is the radius of the universe, nearly &lt;span class=&quot;math-container&quot;&gt;$10^{26}$&lt;/span&gt;m. The ratio is &lt;span class=&quot;math-container&quot;&gt;$10^{44}$&lt;/span&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In another case, if we think about time, the shortest time studied is about &lt;span class=&quot;math-container&quot;&gt;$10^{-26}$&lt;/span&gt; sec and the longest time is &quot;&lt;em&gt;the lifetime of the universe&lt;/em&gt;&quot;, which is estimated to be &lt;span class=&quot;math-container&quot;&gt;$13.7$&lt;/span&gt; billion years, or nearly &lt;span class=&quot;math-container&quot;&gt;$10^{18}$&lt;/span&gt; sec. The ratio is again &lt;span class=&quot;math-container&quot;&gt;$10^{44}$&lt;/span&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this(the ratio belongs is of same order) a coincidence or there is some deeper explanation(probably space and time are related etc.)? I am an undergrad student, don't know much about astronomy. I wondering about this question while studying quantum mechanics from the book &lt;a href=&quot;https://www.goodreads.com/book/show/100023.The_Quantum_World&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Quantum World&lt;/a&gt;, which claims that it is &lt;em&gt;not&lt;/em&gt; a coincidence. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The book explains it in the following way - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;This is not a coincidence. The outermost regions of the universe are moving at a speed near the speed of light, and the particles flying about in the subatomic world are also moving at such speed. On both subatomic and cosmological frontiers, the speed of light is the natural link between distance and time measurements.&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't posted exactly what the book is saying because I wanted to hear what other people are saying about this and also because I haven't understood what the book is saying.&lt;/p&gt;&#xA;" OwnerUserId="23604" LastEditorUserId="23604" LastEditDate="2019-01-21T13:00:50.067" LastActivityDate="2021-02-01T21:05:22.890" Title="Is this a coincidence?" Tags="&lt;distances&gt;&lt;time&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29251" PostTypeId="2" ParentId="29250" CreationDate="2019-01-20T18:14:16.907" Score="4" Body="&lt;p&gt;It's indeed a coincidence. The human limits on observables cannot have some deeper explanation.&lt;/p&gt;&#xA;&lt;p&gt;In fact, we can claim many things like this, without having some sort of deep meaning.&lt;/p&gt;&#xA;&lt;p&gt;The other thing is they are not &amp;quot;exactly&amp;quot; equal. Their ratios just happens to be in the same order. There are also uncertainties in all these measurements and the ratios are never equal to each other. If they were &lt;strong&gt;exactly&lt;/strong&gt; equal in some sense then we may thought there is some deeper meaning.&lt;/p&gt;&#xA;" OwnerUserId="25424" LastEditorUserId="37191" LastEditDate="2021-02-01T21:05:22.890" LastActivityDate="2021-02-01T21:05:22.890" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29252" PostTypeId="2" ParentId="29250" CreationDate="2019-01-20T23:10:45.343" Score="2" Body="&lt;p&gt;Giving it some thought, I think this is a question worth some examination, so I voted your question up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest distance we can see and the amount of time we can see by looking as far in the past as we can are directly related.   We can see &lt;span class=&quot;math-container&quot;&gt;$13.8$&lt;/span&gt; billion light years away, though we can infer that those most distant objects are currently 44 billion light years away (a variation of 3.2), and we can also infer that the diameter of the observable universe is twice that, but let's ignore the expansion and just use the farthest distant object we can observe &lt;span class=&quot;math-container&quot;&gt;$- 13.8$&lt;/span&gt; billion light years and &lt;span class=&quot;math-container&quot;&gt;$13.8$&lt;/span&gt; billion years. Using those numbers doesn't significantly change your question and those two numbers are clearly related.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Translating to seconds and meters, we get &lt;span class=&quot;math-container&quot;&gt;$4.35 \times 10^{17}$&lt;/span&gt; seconds (let's not call that nearly &lt;span class=&quot;math-container&quot;&gt;$10^{18}$&lt;/span&gt;, it's less than half).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And translating light years to meters, we get &lt;span class=&quot;math-container&quot;&gt;$1.31 \times 10^{26}$&lt;/span&gt; meters.  About 300 million times the other, which makes sense cause that's how many meters light travels in 1 second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, let's look at small and see if we get the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Large Hadron Collider gives a pretty precise measurement of the smallest distance we can measure, &lt;a href=&quot;https://physics.stackexchange.com/questions/146791/what-is-the-smallest-length-scale-ever-measured&quot;&gt;see this question for details&lt;/a&gt;.  That number is what you said, about &lt;span class=&quot;math-container&quot;&gt;$10^{-18}$&lt;/span&gt; meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shortest time we can measure isn't quite as clear cut.   &lt;a href=&quot;https://www.scientificamerican.com/article/what-is-the-fastest-event/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article gives a time&lt;/a&gt; of 250 atto seconds (&lt;span class=&quot;math-container&quot;&gt;$250 \times 10^{-18}$&lt;/span&gt; seconds):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Most recently, the techniques of pulse generation and measurement have&#xA;  been extended to the ultraviolet and soft x-ray region of the spectrum&#xA;  where pulses on the order of 250 attoseconds have been created and&#xA;  used to study the motion of electrons around the nucleus of a neon&#xA;  atom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But it mentions that quantum models can infer time periods much shorter. The shortest is the estimated average lifespan of the top quark at about 0.4 yoctoseconds or about &lt;span class=&quot;math-container&quot;&gt;$4 \times 10^{-25}$&lt;/span&gt; seconds (40 times your estimate of the 1 to the power of -26).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Events as short as about &lt;span class=&quot;math-container&quot;&gt;$10^{-25}$&lt;/span&gt; second have been indirectly inferred in&#xA;  extremely energetic collisions in the largest particle accelerators.&#xA;  For example, the mean lifetime of the top quark, the most massive&#xA;  elementary particle so far observed, has been inferred to be about 0.4&#xA;  yoctosecond.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, when I run the numbers, I get the shortest estimated time and the age of the universe have a ratio of about &lt;span class=&quot;math-container&quot;&gt;$1.1 \times 10^{42}$&lt;/span&gt; and the shortest measured distance and the longest observable distance have a ratio of about &lt;span class=&quot;math-container&quot;&gt;$1.3 \times 10^{44}$&lt;/span&gt;. A variation of more than 100. That's pretty close, but nothing remarkable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I can't for the life of me see what the lifespan of the top quark and the energy they can generate per collision in the LHC would have to do with each other, so I would say that there is no relation at all, and at risk of taking on your author, I would disagree with his claim.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="23604" LastEditDate="2019-01-21T09:36:40.923" LastActivityDate="2019-01-21T09:36:40.923" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29253" PostTypeId="1" CreationDate="2019-01-21T01:59:48.613" Score="2" ViewCount="53" Body="&lt;p&gt;I'm a statistics student, currently working on &quot;quantum inference&quot;, which is the statistical problem of inference in the quantum model:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$y_i = \alpha + \beta x_i + \epsilon_i$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is similar to a simple linear model, except instead of &lt;span class=&quot;math-container&quot;&gt;$x_i$&lt;/span&gt; being a known constant, it is instead a random integer. The prototypical example is von Mises's analysis of atomic weights in 1918, trying to assess the significance of the observation that they tended to be integer multiples of the mass of hydrogen. For more information on the subject see &lt;a href=&quot;https://onlinelibrary.wiley.com/doi/abs/10.1002/0471667196.ess2129.pub2&quot; rel=&quot;nofollow noreferrer&quot;&gt;DG Kendall's overview&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure if I should continue working on this problem because the applications are so few, mostly in archaeology, where weights or lengths are often integer multiples of an ancient unit of measure. However, it occurred to me that there may be applications in astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, perhaps the brightness of a spot has been measured, and the spot could represent any number of objects which are not distinguishable at the present resolution. This could be a problem of quantum inference... but only if if were for some reason known &lt;em&gt;a priori&lt;/em&gt; that all objects had to have the same brightness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, while that specific scenario might be implausible, is there some problem in astronomy data analysis that involves quantum inference in this sense? In other words, are there ever cases where a histogram of the data shows multiple, &lt;em&gt;equally spaced&lt;/em&gt; modes?&lt;/p&gt;&#xA;" OwnerUserId="25635" LastActivityDate="2019-01-21T01:59:48.613" Title="Do integer multiples occur in astronomy data?" Tags="&lt;data-analysis&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29254" PostTypeId="2" ParentId="29247" CreationDate="2019-01-21T03:29:18.610" Score="2" Body="&lt;p&gt;The color you can see in the sky depends on two things, first the color (if you want to be more specific, the emission spectra of the light of the star) of the light shining, this is the Sun in our case (but may be another star in order to answer your question about outside the solar system), and second depends on the chemical components of the atmosphere of the planet in question, because this atmosphere is the reason for what is called the absorption lines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Absorption light is something like that, if you have white light and a gas with green absorption lines, then after crossing the gas the light will turn the opposite of green, I don't know what color that is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion if our atmosphere changes (for some unknown reason) here on planet Earth, then the color of the sky is going to change. &lt;/p&gt;&#xA;" OwnerUserId="25315" LastEditorUserId="25315" LastEditDate="2019-01-21T06:32:06.400" LastActivityDate="2019-01-21T06:32:06.400" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29255" PostTypeId="2" ParentId="18750" CreationDate="2019-01-21T05:13:07.167" Score="1" Body="&lt;p&gt;The answer from @JamesK is pretty much all you need to know on this topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you mention the word &quot;singularity&quot; in the title, let me add this - the singularity is not a thing. I know how pop sci articles talk about it as if it was a thing, but it's just a mathematical abstraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the center of the black hole - we don't really know what happens. The equations from GR (general relativity) basically blow up, we get divisions by zero and whatnot. This in mathematics is called a &quot;singularity&quot;. But in the physical reality that simply means our theory is inadequate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will have to wait for a future theory to properly understand the center of BH. For now, the GR equations are just hitting a wall. GR predicts infinite densities and zero diameters and so on - this is virtually certain to be wrong. But as of now there is no other theory we can use to figure this out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So don't say it's falling towards the &quot;singularity&quot;. Say it falls toward the center, whatever that may be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make no mistake, your helium atom is doomed. Its particles will get shredded down to whatever scale things exist at the center. Maybe down to the Planck length, but we don't know for sure.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2019-01-21T11:27:39.173" LastActivityDate="2019-01-21T11:27:39.173" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29256" PostTypeId="1" CreationDate="2019-01-21T07:13:21.573" Score="-1" ViewCount="31" Body="&lt;p&gt;I was watching the blood moon with a perfectly clear night- using only my naked eyes. At such a micro level, is it possible that all the shimmer I saw was an artefact of our atmosphere or just my poor eyes being the apeture? It basically had the look of a dark flame upon the moon- flickering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My assumption was that the suns rays could be being attenuated through varied atmospheric densities causing a flickering effect. What is really happening?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note... I wanted to observe it closer but snow and -9 degree weather prevented it. Bad to bring my telescope out in that. Also sorry for the shorthand. On mobile as well.&lt;/p&gt;&#xA;" OwnerUserId="23729" LastActivityDate="2019-01-21T08:29:10.333" Title="Do clouds on earth have an affect on a blood moon to the naked eye?" Tags="&lt;lunar-eclipse&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2019-01-23T01:11:04.527" ContentLicense="CC BY-SA 4.0" />
  <row Id="29258" PostTypeId="2" ParentId="29256" CreationDate="2019-01-21T08:23:58.237" Score="1" Body="&lt;p&gt;Probably the cold weather was playing tricks on your eyes. Or might have been faint clouds in the local sky (too faint to notice otherwise), or even your own breath. Cold weather astronomy is tricky. Sometimes you see your own breath as turbulence in the telescope. It's weird when you see it, and fun when you are able to explain it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;BTW, it's called a lunar eclipse, or a Moon eclipse. That's how astronomers refer to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Blood&quot;, &quot;wolf&quot;, &quot;super&quot; and other terms from vampire fantasy books for teenagers are just how bloggers and uncritical media like to talk about it. It's a fad - this too shall pass, although it may take years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's an eclipse. A Moon eclipse.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2019-01-21T08:29:10.333" LastActivityDate="2019-01-21T08:29:10.333" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29259" PostTypeId="1" AcceptedAnswerId="29272" CreationDate="2019-01-21T09:04:25.443" Score="6" ViewCount="173" Body="&lt;p&gt;The very beautiful Jupiter Venus conjunction of ~ 2019-01-20 (practically something out of &quot;2001 A Space Odyssey&quot;) is seen as &quot;vertically aligned&quot; in the dawn sky, Venus &quot;above&quot; Jupiter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See this picture by &lt;a href=&quot;https://earthsky.org/?p=296766&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jeff Majewski at EarthSky.com&lt;/a&gt; to see what I mean (not sure I can copy it in here directly)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should mean my &quot;vertical&quot; is aligned with the Ecliptic. I'm in northern latitudes (49° N), how is that possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And is this alignment geometrically linked with the &quot;Super Wolf Blood Moon&quot; which was very visible at the same time?&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Update&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The problem seems hard to explain so here is a little free-hand drawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/amOwi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/amOwi.jpg&quot; alt=&quot;free hand drawing of observation by an observer standing on the equator&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The observer sits on the equator and is just being moved into sunlight by Earth's rotation (the image shows Winter Time for the Northern Hemisphere). The observer's &quot;UP&quot; vector lies in the Ecliptic Plane. This being so, and Jupiter and Venus also being in the Ecliptic Plane, the observer can see Jupiter and Venus aligned &quot;vertically&quot; (same azimuth), separated by a couple of degrees in elevation. As the day progresses, that vertical alignment will evidently cease as the line connecting the planets rotates relative to the observer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works on the Equator, but what if the observer is at higher latitudes? I can't get my head around the idea that the planets can be vertically aligned while still being on the Ecliptic Plane.&lt;/p&gt;&#xA;" OwnerUserId="16289" LastEditorUserId="16289" LastEditDate="2019-01-21T22:25:48.070" LastActivityDate="2019-01-21T22:25:48.070" Title="Jupiter-Venus conjunction is vertically aligned. I'm at 49° N. How is this possible?" Tags="&lt;observational-astronomy&gt;&lt;alignment&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29261" PostTypeId="2" ParentId="29259" CreationDate="2019-01-21T10:30:22.920" Score="2" Body="&lt;p&gt;I'm even further north than you and it was vertically aligned for me at one point through that night as well, as the night sky turned&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your latitude is not really relevant. It only really matters at what time you observed it. I'm trying to find a useful picture sequence showing them moving overhead - I don't have one unfortunately, so perhaps try and imagine Orion, for example. It rotates through the sky overnight, so when Orion rises, his belt is almost vertical, with Mintaka being higher in the sky. By the time Orion is setting, Alnitak is uppermost in the belt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it has absolutely nothing to do with the lunar eclipse :-)&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="43" LastEditDate="2019-01-21T12:39:35.547" LastActivityDate="2019-01-21T12:39:35.547" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29262" PostTypeId="1" CreationDate="2019-01-21T11:06:18.943" Score="1" ViewCount="49" Body="&lt;p&gt;Shooting stars are actually meteors falling down on the surface of earth. How come some of them appear shooting up and not down?&lt;/p&gt;&#xA;" OwnerUserId="25643" LastActivityDate="2019-01-22T19:38:37.157" Title="Meteor shower shooting stars" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29263" PostTypeId="2" ParentId="29250" CreationDate="2019-01-21T11:11:25.003" Score="3" Body="&lt;p&gt;Actually, I favor the idea that it is not a coincidence, at least it's not a coincidence that those two ratios bear some strong resemblance.  The reason is because at the large scale, as seen already, the largest time is we can detect is the age of the universe T, and the largest distances L we can measure are going to be directly related to that age by L~cT, at least in rough order (the expansion of the universe makes distance L a rather nonunique issue).  On the smallest scales, the energies E we require to probe the shortest times t obey Et ~ h by the uncertainty relation, but if we assume those energies will be manifested in relativistic particles, the best we can localize those particles in a length scale l obeys E/c ~ h/l, also by the uncertainty principle.  Eliminating E gives l ~ ct, and so forming the ratio of the largest and smallest scales gives L/l ~ (cT)/(ct) ~ T/t.  So whatever that ratio comes out, the two should not be all that much different.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that those ratios are similar is not nearly as surprising as that they come out so incredibly huge. Why does our universe exhibit such incredible contrasts of times and distances? One might look to the anthropic principle there, in the sense that life requires both long timescales to appear and short timescales to support the required complexity, but then there's still the issue of whether or not one regards anthropic explanations as explanations at all.   Also, even if we take anthropic thinking, it seems likely the universe could have produced intelligent life even if it was much smaller, and could have supported such life using much slower processes than those we can detect in our supercolliders.  It appears our universe is built with a significant degree of &quot;overkill&quot; in regard to both the volume available for events to happen, and the smallest possible distance scales on which they do happen.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2019-01-21T11:18:18.783" LastActivityDate="2019-01-21T11:18:18.783" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29265" PostTypeId="2" ParentId="29212" CreationDate="2019-01-21T15:08:58.103" Score="1" Body="&lt;p&gt;Your intuition is correct, as far as classical dynamics is concerned. The main problem, I think, arises in that you are referring to a strictly quantum mechanical phenomenon - what is the physical singularity? In General Relativity (GR), we know these singularities exist, but we cannot really know more than that (aside from events that occur not too close to the physical singularity). But, it's deeper than that, because in GR we cannot get information out of a black hole, so we'd never really know. That's why people in the comments are referring to a &quot;theory of quantum gravity&quot; which would take over where GR breaks down (at the physical singularity). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a non-rotating (Schwarzschild) BH, the physical singularity is at &lt;span class=&quot;math-container&quot;&gt;$r = 0$&lt;/span&gt;, but the coordinate singularity is at the event horizon, &lt;span class=&quot;math-container&quot;&gt;$r = 2M$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a rotating (Kerr) BH, the physical singularity is now a ring with no thickness but non-zero radius - this is &lt;strong&gt;the&lt;/strong&gt; ringularity. We literally don't know what should happen here (need &quot;quantum gravity&quot;), but in GR we can't even know in principle, because if someone were to try to fall into the ringularity we would never receive any signals back from them (as far as GR tells us).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OKay, with all that out of the way, we can still make a similar comment like what you want to &lt;strong&gt;within&lt;/strong&gt; the framework of GR:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a Kerr BH (in Boyer-Lindquist coordinates) with dimensionless spin &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; whose symmetry axis is along the &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;, so that the ringularity lies in the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;-&lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; plane. One can show that at &lt;span class=&quot;math-container&quot;&gt;$z = 0$&lt;/span&gt; and at &lt;span class=&quot;math-container&quot;&gt;$r = 0$&lt;/span&gt; (where &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; is not the physical radial distance), the equation of the ringularity is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ x^2 + y^2 = a^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so the radius of the ringularity is the spin. Therefore, as the spin of the Kerr BH increases, the radius of the ringularity also increases. This is the opposite of what your classical intuition leads to, but then again, this is a very nontrivial spacetime metric, and it exists on the fringe between classical and quantum physics.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why doesn't a singularity spin infinitely fast?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the ringularity is not spinning itself, rather it is defined by the spin of the BH. Mathematically, we can construct some kind of &quot;spin&quot; of the ringularity, but it wouldn't make much physical sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the BH was spinning infinitely fast, i.e. &lt;span class=&quot;math-container&quot;&gt;$a \rightarrow \infty$&lt;/span&gt;, then the ringularity would have an infinite radius!!! And THAT's certainly problematic (but interesting!). Fortunately, as far as we know, BH's cannot have a spin greater than 1 - the maximum spin of a Kerr BH is capped by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Naked_singularity&quot; rel=&quot;nofollow noreferrer&quot;&gt;naked singularity&lt;/a&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_censorship_hypothesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;hypothesis&lt;/a&gt;. You see, if BH's could have a spin greater than 1, then the ringularity would extend &quot;outside&quot; the event horizon, which is problematic because it would mean that we could observe the collapse of an object into infinite density. ;)&lt;/p&gt;&#xA;" OwnerUserId="23762" LastEditorUserId="23762" LastEditDate="2019-01-21T16:28:07.000" LastActivityDate="2019-01-21T16:28:07.000" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29268" PostTypeId="1" CreationDate="2019-01-21T18:51:45.480" Score="8" ViewCount="844" Body="&lt;p&gt;We all know that our sun will become a red giant in a couple of billion years? But how fast will the process be when it starts? Would it be theoretically possible (disregarding the deadly effects of this event) for a human to experience the sun as its present size until it swallows earth?&lt;/p&gt;&#xA;" OwnerUserId="25646" LastActivityDate="2019-01-22T19:09:26.713" Title="How fast will the sun grow when it becomes a red giant?" Tags="&lt;the-sun&gt;&lt;red-giant&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29269" PostTypeId="2" ParentId="29268" CreationDate="2019-01-21T19:32:14.347" Score="1" Body="&lt;p&gt;According to a &lt;a href=&quot;https://phys.org/news/2016-05-earth-survive-sun-red-giant.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Physics.org&lt;/a&gt; article: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When the sun does begin to expand, it will do so quickly, sweeping through the inner solar system in just 5 million years. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, no. No human alive could possibly see the sun change from its more-or-less present size to a red giant in their life time, unless extreme advances in human longevity approaching science fiction were to occur before then.&lt;/p&gt;&#xA;" OwnerUserId="7108" LastEditorUserId="7108" LastEditDate="2019-01-22T19:09:26.713" LastActivityDate="2019-01-22T19:09:26.713" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29271" PostTypeId="1" AcceptedAnswerId="29275" CreationDate="2019-01-21T21:40:08.580" Score="8" ViewCount="681" Body="&lt;p&gt;When is Earth closest to the Sun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, you say, perihelion of course.  But perihelion is when the &lt;a href=&quot;https://en.wikipedia.org/wiki/Barycenter&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Earth-Moon barycentre&lt;/em&gt;&lt;/a&gt; is closest to the Sun. Not Earth itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The barycentre is on average 4,671 km from the Earth, which should mean that even if the orbit of the Earth-Moon barycentre around the Sun were perfectly circular, the Earth-Moon orbit around the barycentre would cause an oscillation in the Earth-Sun distance by twice that amount on a monthly basis.  That should be enough to move Earth's closest approach to the Sun away from perihelion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering this, how many days from perihelion may Earth's closest approach to the Sun be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question inspired by: &lt;a href=&quot;https://earthscience.stackexchange.com/a/16001/6&quot;&gt;&lt;em&gt;What place on Earth is closest to the Sun?&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1462" LastEditorUserId="1462" LastEditDate="2019-01-22T00:00:58.723" LastActivityDate="2019-01-22T03:58:35.380" Title="When is Earth closest to the Sun?" Tags="&lt;orbit&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29272" PostTypeId="2" ParentId="29259" CreationDate="2019-01-21T22:21:33.573" Score="6" Body="&lt;p&gt;A combination of several movements can lead two objects in the sky to appear vertically aligned. The planets are constantly moving relative to each other, they are orbiting on slightly different planes, the Earth is orbiting and rotating, so such configurations between planets can occur frequently. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An observer does not need to be in the tropics to see two planets being vertically aligned at some point. The planets orbit close to the ecliptic, but their orbital planes are slightly inclined relative to ours. Jupiter's orbit is about 1° from the ecliptic, and Venus is about 3°. This means that both planets can move above or below the ecliptic by a few degrees. When they pass each other as seen from the Earth in a conjunction, they are rarely on top of each other, most of the time, they are separated by a few degrees, and they are not exactly on the ecliptic line. (Such occurences of planets occulting or transiting each other do happen, but are rather uncommon: the next time Venus will transit in front of Jupiter will be on 22 November 2065)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll add a few pictures to help visualizing the positions, movements and alignments involved here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a view of the sky on 20 January at 49° North latitude:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IsYzf.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IsYzf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, Jupiter and Venus appear pretty much vertically aligned in the morning sky. The yellow line is the ecliptic. Note that the ecliptic does not pass through the planets here. Jupiter is pretty close, but Venus is a few degrees north of it. As both planets pass each other in the next days, at some point, their alignment will be perpendicular to the ecliptic (called the ecliptic conjunction), as we can see from this January 22 simulated image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Wjo3O.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wjo3O.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But they won't appear vertical anymore from your vantage point. (In fact, they would apprear vertical again later during that day, because of Earth's rotation, but the Sun will be up, so it wouldn't be observable directly)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few days later, on January 26, they will appear horizontally aligned!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aD4h4.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aD4h4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Images generated in &lt;a href=&quot;https://stellarium.org/&quot; rel=&quot;noreferrer&quot;&gt;Stellarium&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="19814" LastActivityDate="2019-01-21T22:21:33.573" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29273" PostTypeId="1" AcceptedAnswerId="29277" CreationDate="2019-01-22T03:03:19.453" Score="4" ViewCount="174" Body="&lt;p&gt;I am having a lot of trouble getting a sharp image in my telescope. It is a skywatcher with a focal length of 1200mm and a diameter of 254mm. I have a laser colimater that I got for Christmas, so I have been using that but when I try to look at mars I see three mars in an overlapping triangle. I am not sure what to do. I have made sure that the laser is straight by 1. spinning it in the eyepiece holder and the laser stayed in the same spot 2. zooming in and out and again the laser stayed in the same place. the laser seems to be centered so I put it in the middle of the doughnut on the mirror and the laser bounces back into the middle of the colimating lens, but when I look at mars there is still three and there is no detail its just a red fuzz ball. I also checked the collimation of the telescope by zooming all the  way out to see where the shadow was and it seemed to be right in the middle.hopefully this makes sense, but let me know if I should say something in a different way. thank you In advance.&lt;/p&gt;&#xA;" OwnerUserId="25650" LastActivityDate="2019-01-22T11:59:43.920" Title="I need help getting mars to look like a circle" Tags="&lt;planet&gt;&lt;solar-system&gt;&lt;mars&gt;&lt;newtonian-telescope&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29274" PostTypeId="2" ParentId="29268" CreationDate="2019-01-22T03:30:19.697" Score="5" Body="&lt;p&gt;Yes, the time is very long, but exactly how long depends on when you &quot;start the clock&quot; for the expansion to a red giant.  If you take the end of the &quot;main sequence&quot; (when the core hydrogen runs out) as the start of the expansion phase, it's more like a billion years to complete the process of becoming a maximally expanded red giant.  But the answer that gets 5 million years focuses only on the phase of most rapid expansion (which begins after the star is already quite a bit larger than the Sun).  Either way, our lives a the smallest blink of an eye to a star!&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2019-01-22T03:40:06.307" LastActivityDate="2019-01-22T03:40:06.307" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29275" PostTypeId="2" ParentId="29271" CreationDate="2019-01-22T03:42:17.993" Score="11" Body="&lt;p&gt;Earth's perihelion timings and distances are based on the distance between the center of the Earth and the center of the Sun. They are effectively the times and distances of the closest approach between the two bodies, not the barycenters. As we can see on Fred Espenak's &lt;a href=&quot;http://www.astropixels.com/ephemeris/perap2001.html&quot; rel=&quot;noreferrer&quot;&gt;astropixel page&lt;/a&gt; about perihelion and aphelion, the time intervals between two successive perihelions tend to vary between 363 and 368 days. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data on astropixels seems perfectly valid. With JPL Horizons, for example, the next perihelion is calculated to be on 2020 Jan 5 at 07:47:42, at a distance of 0.98324356482 AU (147,091,144 km) which matches astropixel's values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I analyzed the perihelions of the 21st century to see how much the actual timing can differ from an idealized &quot;average&quot; perihelion moment (based on the anomalistic year length). Perihelions this century remain within 1.4 days of that average. The differences from the average are mainly due to the gravitational effect of the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using all perihelion times in the 21st century, and the average perihelion interval (anomalistic year, 365.2596 days) I constructed the following graph showing these differences:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f3tkt.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f3tkt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19814" LastEditorUserId="19814" LastEditDate="2019-01-22T03:58:35.380" LastActivityDate="2019-01-22T03:58:35.380" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29277" PostTypeId="2" ParentId="29273" CreationDate="2019-01-22T11:53:34.653" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I have made sure that the laser is straight by 1. spinning it in the&#xA;  eyepiece holder and the laser stayed in the same spot 2. zooming in&#xA;  and out and again the laser stayed in the same place. the laser seems&#xA;  to be centered so I put it in the middle of the doughnut on the mirror&#xA;  and the laser bounces back into the middle of the colimating lens&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Very good. You're doing all the right things. Kudos for taking care of your telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, about Mars looking like a fuzzball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angular size of Mars right now is 6.5 arcsec. The maximum theoretical resolving power of your scope is 0.4 arcsec. In ideal conditions (assuming everything was perfect, which it never is), you'd get something like 16 &quot;pixels&quot; diameter for the image of Mars in your scope. That, I'm sure you realize, is a very low resolution. No wonder it looks like a fuzzball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars is a tricky target. The best times to observe it is around oppositions, which happen once every two years (plus a few months). Mars is nice and big a few weeks before and after opposition, and otherwise it's pretty small. The last opposition was in July 2018, when it measured 24 arcsec. The next one is in October 2020, and it will measure 22 arcsec.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nakedeyeplanets.com/mars-oppositions.htm&quot; rel=&quot;noreferrer&quot;&gt;http://www.nakedeyeplanets.com/mars-oppositions.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stick to a few weeks before and after opposition. In the rest of the time, Mars is just too small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another source of problems is seeing - or atmospheric turbulence. When seeing is bad, all the high resolution targets (like all planets and the Moon) look bad. Nothing you can do about it. When seeing is good, your scope can work closer to its theoretical resolving power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_seeing&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Astronomical_seeing&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To predict seeing, go on the Clear Dark Sky site, choose a location closest to where you live, and check the forecast they give you for that location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.cleardarksky.com/csk/&quot; rel=&quot;noreferrer&quot;&gt;https://www.cleardarksky.com/csk/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The row called Seeing is a seeing forecast. When it's dark blue, seeing is good. When it's white or light blue, seeing is bad. No point in looking at Mars when seeing is bad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, make sure collimation is actually good. You seem to be doing all the good stuff with the laser collimator. That's great. Now, when seeing is good, point the scope at the North Star, defocus a bit (move the focuser in or out of perfect focus a tiny bit) and look at the star image. It looks like a shooting target, a bunch of concentric rings. If those rings are perfectly round and concentric, your collimation is probably good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astrophoto.fr/collim.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.astrophoto.fr/collim.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is called a star test, and it's the most accurate test of collimation. It's also the most difficult. So use your laser collimator on a daily basis, which is much easier, and maybe do the star test once or twice, just to keep your laser honest. Once you're convinced your laser technique works well, there's no need for the star test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, one more thing. Make sure your scope is at thermal equilibrium with the environment. If you're storing your scope indoors, it's probably way hotter than the cold night air. That will create internal turbulence within the scope. It's basically like bad seeing, except it's all inside your scope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-1/&quot; rel=&quot;noreferrer&quot;&gt;https://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-1/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-2/&quot; rel=&quot;noreferrer&quot;&gt;https://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-2/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No need to install a fan on the primary (although that would definitely work if you're willing to make the extra effort). Just take the scope outside 1 hour before you begin using it. It will &quot;breathe out&quot; all the heat and it will be close to equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck and clear sky!&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2019-01-22T11:59:43.920" LastActivityDate="2019-01-22T11:59:43.920" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29278" PostTypeId="1" CreationDate="2019-01-22T16:21:29.723" Score="2" ViewCount="92" Body="&lt;p&gt;I don't know astronomy that well, but I am preparing for the second stage of &lt;a href=&quot;https://en.wikipedia.org/wiki/National_Standard_Examination_in_Junior_Science&quot; rel=&quot;nofollow noreferrer&quot;&gt;IJSO&lt;/a&gt;, and here in India they do add some questions like these ones.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A star is seen rising from Kolkata (23.5N 92 E) at 7 pm IST. What time can it be seen rising from Mumbai (19N 72E)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If someone could explain how &lt;em&gt;I might go about solving this&lt;/em&gt;, and some surrounding theory so I can grasp the concept, I'll be really grateful. &lt;/p&gt;&#xA;" OwnerUserId="25658" LastEditorUserId="7982" LastEditDate="2019-02-24T00:19:34.367" LastActivityDate="2019-02-24T00:22:58.193" Title="Star rising times in a different place given the latitude and time of one place" Tags="&lt;coordinate&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29279" PostTypeId="2" ParentId="29262" CreationDate="2019-01-22T19:38:37.157" Score="1" Body="&lt;p&gt;Shooting stars are indeed meteors falling down, but consider this: meteors which are part of a meteor shower actually appear to come from a point in the sky called the &lt;em&gt;radiant point&lt;/em&gt;:&#xA;&lt;a href=&quot;https://www.google.com/url?sa=i&amp;amp;source=imgres&amp;amp;cd=&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=2ahUKEwiItuubkYLgAhWZV30KHV-9Dn8QjRx6BAgBEAU&amp;amp;url=https%3A%2F%2Fen.wikipedia.org%2Fwiki%2FRadiant_(meteor_shower)&amp;amp;psig=AOvVaw0lP8KGKheVCSl9wC0P6qp_&amp;amp;ust=1548272112613916&quot; rel=&quot;nofollow noreferrer&quot;&gt;google image&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you're looking upwards, you may be looking straight at the radiant point, and therefore the meteors are 'radiating' away from that point in all directions; not only 'downwards' towards the horizon, but 'upwards' from your point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that, even for a meteor which is not part of a shower, it's just as likely for a meteor to be going 'upwards' across the sky, as it is to be going 'downwards'.&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2019-01-22T19:38:37.157" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29281" PostTypeId="2" ParentId="27912" CreationDate="2019-01-22T23:16:53.700" Score="2" Body="&lt;p&gt;The image you have is of COSTAR.  The modification to Hubble was to actually remove another instrument and place COSTAR inside.  Once inside, it deployed and the other instruments were corrected.  So the entire thing in the image is the modification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over time, new instruments were built and installed in Hubble.  Those instruments (like WFPC2) were built to have the correction inside the instrument.  When all of the original cameras were replaced, COSTAR was no longer needed and was also removed.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2019-01-22T23:16:53.700" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29284" PostTypeId="1" CreationDate="2019-01-23T07:28:58.347" Score="1" ViewCount="504" Body="&lt;p&gt;I'm trying to simulate Earth's orbit using 2nd Newton's Law and cartesian coordinates. The program seems ok, but when plotting graph, it's a parabola.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the shape of orbit changes according velocity. I'm using vy0 = sqrt(G*M/R) and it supposed to be an circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry the bad english)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;'-------------------Bibliotecas Importadas-------------------'&#xA;&#xA;import math&#xA;import matplotlib.pyplot as plt&#xA;&#xA;'-------------------Listas Criadas-------------------'&#xA;&#xA;pos_x = []&#xA;pos_y = []&#xA;vel_x = []&#xA;vel_y = []&#xA;teo_virial = []&#xA;tempo = []&#xA;&#xA;'-------------------------- Definição de Funções -----------------------------'&#xA;&#xA;def velocidade_atualizador(velocidade_inicial, aceleracao):&#xA;    velocidade = velocidade_inicial + (aceleracao * dt)&#xA;    return velocidade&#xA;&#xA;def tabela(nome, lista): &#xA;    arquivo = open(nome + '.txt', 'a')&#xA;    for i in range(len(lista)):&#xA;        arquivo.write(str(lista[i]) + '\n')&#xA;    arquivo.close&#xA;&#xA;def graph(x, y, eixo_x, eixo_y): &#xA;    plt.plot(x, y)&#xA;    plt.xlabel(eixo_x)&#xA;    plt.ylabel(eixo_y)&#xA;    plt.show()&#xA;&#xA;def aceleracao_equacao(pos_1, pos_2):   &#xA;    r_cubo = (pos_1**2 + pos_2**2)**1.5&#xA;    k = constgrav*massasol&#xA;    aceleracao = (k*pos_1)/r_cubo&#xA;    return aceleracao&#xA;&#xA;def posicao_equacao(posicao_inicial, velocidade_inicial, aceleracao):&#xA;    pos = posicao_inicial + (velocidade_inicial*dt) + ((aceleracao*(dt**2))/2)&#xA;    return pos&#xA;&#xA;'-------------------Condições Iniciais-------------------'&#xA;&#xA;constgrav = math.pow(6.67, -11)             #constante gravitacional&#xA;massasol = math.pow(1.98, 30)               #massa do sol&#xA;massaterra = math.pow(5.97, 24)             #massa da terra&#xA;xo = math.pow(1.5, 11)                      #posição inicial no eixo x&#xA;yo = 0                                    #posição inicial no eixo y&#xA;vxo = 0                                   #velocidade inicial no eixo x&#xA;vyo = math.sqrt((constgrav*massasol)/xo)                               #velocidade inicial no eixo y&#xA;npasso = 10000                                #numero de passos&#xA;t = math.pow(3.1536, 7)                               #tempo total = tempo de um ano&#xA;conttempo = 0                             #contador de tempo&#xA;dt = t/npasso                               #delta t &#xA;&#xA;'-------------------Laço de Repetição-------------------'&#xA;&#xA;while (t &amp;gt; conttempo):&#xA;&#xA;    ax = aceleracao_equacao(xo, yo)&#xA;    ay = aceleracao_equacao(yo, xo)&#xA;    a_modulo = math.sqrt(ax**2 + ay**2)&#xA;&#xA;    vx = velocidade_atualizador(vxo, ax)&#xA;    vy = velocidade_atualizador(vyo, ay)&#xA;    v_modulo = math.sqrt(vx**2 + vy**2)&#xA;&#xA;    x = posicao_equacao(xo, vxo, ax)&#xA;    y = posicao_equacao(yo, vyo, ay)&#xA;    r = math.sqrt(x**2 + y**2)&#xA;&#xA;    energia_pot = (- constgrav * massasol * massaterra)/math.sqrt(x**2 + y**2)&#xA;    energia_cin = (massaterra * (v_modulo**2))/2&#xA;    virial = energia_pot / energia_cin    &#xA;&#xA;    print('x =', x)&#xA;    print('y =', y)&#xA;    print('virial =', virial)&#xA;&#xA;    pos_x.append(x)&#xA;    pos_y.append(y)&#xA;    vel_x.append(vx)&#xA;    vel_y.append(vy)&#xA;    teo_virial.append(virial)&#xA;    tempo.append(conttempo)&#xA;&#xA;    xo = x&#xA;    yo = y&#xA;    vxo = vx&#xA;    vyo = vy&#xA;&#xA;    conttempo = conttempo + dt&#xA;&#xA;tabela('Lista x', pos_x)&#xA;tabela('Lista y', pos_y)&#xA;tabela('Lista vx', vel_x)&#xA;tabela('Lista vy', vel_y)&#xA;tabela('Lista tempo', tempo)&#xA;&#xA;'-------------------Gráfico-------------------'&#xA;&#xA;graph(pos_x, pos_y, 'x (m)', 'y (m)')&#xA;graph(vel_x, tempo, 'vx (m/s)', 't (s)')&#xA;graph(vel_y, tempo, 'vy (m/s)', 't (s)')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Thank you very much, guys!! I was thinking but don't know what to do anymore. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After changing the code, it works. I can rest in peace now. Hahaha.&lt;/p&gt;&#xA;" OwnerUserId="25667" LastEditorUserId="2153" LastEditDate="2019-01-23T16:09:48.770" LastActivityDate="2019-01-23T16:09:48.770" Title="Earth Orbit Simulator" Tags="&lt;orbit&gt;&lt;python&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="0" ClosedDate="2019-01-25T14:42:44.857" ContentLicense="CC BY-SA 4.0" />
  <row Id="29285" PostTypeId="2" ParentId="29284" CreationDate="2019-01-23T09:11:26.527" Score="5" Body="&lt;p&gt;There is no obvious astronomy error, but your use of math.pow is wrong&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You write &lt;code&gt;constgrav = math.pow(6.67,-11)&lt;/code&gt;.  That means &lt;span class=&quot;math-container&quot;&gt;$G = 6.67^{-11}$&lt;/span&gt;.  You mean &lt;code&gt;constgrav = 6.67e-11&lt;/code&gt; or &lt;span class=&quot;math-container&quot;&gt;$G=6.67\times 10^{-11}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With completely different initial conditions, it is not surprising that the Earth flies off into space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a suggestion, try not using SI units for this, as it makes for some very small and very large intermediate values, the computers have difficulty in rounding.  Instead measure distance in Astronomical Units, Mass in &quot;solar masses&quot; and time in years. You can convert back to SI units when printing or plotting.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-01-23T09:11:26.527" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29286" PostTypeId="2" ParentId="29284" CreationDate="2019-01-23T10:17:39.317" Score="1" Body="&lt;p&gt;I think you are using &lt;code&gt;math.pow(x, y)&lt;/code&gt; incorrectly. You should write &lt;code&gt;constgrav = 6.67E-11&lt;/code&gt; for example, standard scientific notation. Python's &lt;code&gt;math.pow&lt;/code&gt; returns &lt;span class=&quot;math-container&quot;&gt;$x^y$&lt;/span&gt; which is not the same thing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, &lt;code&gt;math.pow(x, y)&lt;/code&gt; is similar, perhaps the same as &lt;code&gt;x**y&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you forgot a negative sign in the acceleration: gravity is attractive! Changing it to a negative and fixing the scientific notation, no more hyperbola; &lt;strong&gt;your program works nicely, congratulations!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add more digits and it should close nicely into a circle.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def aceleracao_equacao(pos_1, pos_2):   &#xA;    r_cubo = (pos_1**2 + pos_2**2)**1.5&#xA;    k = constgrav*massasol&#xA;    aceleracao = -(k*pos_1)/r_cubo        # ¡negative sign!&#xA;    return aceleracao&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;also&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;constgrav     = 6.67E-11                #constante gravitacional&#xA;massasol      = 1.98E+30                #massa do sol&#xA;massaterra    = 5.97E+24                #massa da terra&#xA;xo            = 1.5E+11                 #posição inicial no eixo x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;t             = 3.1536E+07              #tempo total = tempo de um ano&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0pfFz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0pfFz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-01-23T10:34:56.773" LastActivityDate="2019-01-23T10:34:56.773" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29287" PostTypeId="2" ParentId="27912" CreationDate="2019-01-23T11:34:15.660" Score="1" Body="&lt;p&gt;The HST has space for 4 main instruments. Each of them consists of a box with electronics, with a CCD or other detector on one end, &lt;a href=&quot;http://documents.stsci.edu/hst/proposing/documents/cp_cy13/primer.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;in the focal plane of the telescope&lt;/a&gt;. The fifth instrument, WFPC is placed radially in front of the focal plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DyYYk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DyYYk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image is from later in Hubble's career. FOS, FOC and GHRS (see below) have been replaced by STIS, NICMOS and ACS. The principle remains the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During Servicing Mission 1, 2 major changes were made: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the WFPC was replaced with WFPC2. WFPC had correctional optics built in, as described in astrosnapper's answer. &lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the High Speed Photometer (HSP) was removed and COSTAR installed in its place. COSTAR did not have a CCD. In its place was a structure containing a series of mirrors, this structure protruded into the optics space in front of the focal plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.montana.edu/jshaw/documents/17c%20Hubble_correctors_AO1993.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This diagram&lt;/a&gt; shows the COSTAR mirrors in place: &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xlifd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xlifd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After installation, small arms with one mirror each unfolded from the structure and into the optics paths for the Faint Object Camera (FOC), the Faint Object Spectrograph (FOS) and the Goddard High Resolution Spectrograph (GHRS). There were two mirrors per instrument. These mirrors corrected the image. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the optical path for the FOC:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DAWFE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DAWFE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In COSTAR each corrector consists of a spherical field mirror Ml and an aspherical reimaging mirror M2 (Fig. 1). The function of Ml is to image the telescope primary mirror at M2. A spherical mirror suffices. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;M2 is a fourth-order anamorphic asphere. The fourth-order term compensates for the spherical aberration and must be slightly anamorphic because the corrector mirrors do not form a centered system. Control of astigmatism requires M2 to be anamorphic of the second order also.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In total five sets of correctors are required. They serve the two cameras in the FOC (f/48 and f/96 cameras), the short-wavelength spectrograph in the FOS (FOS blue), the long-wavelength spectrograph in the FOS (FOS red), and the GHRS. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In later service missions, the &lt;a href=&quot;https://space.stackexchange.com/questions/26840/hubble-gained-weight-while-in-space-which-components-account-for-this-extra-mas&quot;&gt;main instruments were replaced&lt;/a&gt; with instruments that had corrective optics built in, the last instrument that needed COSTAR was removed during Servicing Mission 3B.&#xA;So on the last servicing mission (4), COSTAR could be replaced with the Cosmic Origins Spectrograph (COS). COSTAR was taken back to Earth and can now be seen in the Smithsonian, which is where the photo in the question was taken. &lt;/p&gt;&#xA;" OwnerUserId="361" LastEditorUserId="361" LastEditDate="2019-01-24T13:19:09.483" LastActivityDate="2019-01-24T13:19:09.483" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29290" PostTypeId="1" AcceptedAnswerId="29291" CreationDate="2019-01-23T18:15:23.690" Score="1" ViewCount="130" Body="&lt;p&gt;I have been reading some literature on using the HST to take spectra of exoplanet atmospheres when exoplanets transit in front of their stars. The literature mention two modes of observations, 'spatial scanning' and 'staring', in order to obtain spectra of the atmospheres of these exoplanets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain conceptually what the difference is between these two modes?&lt;/p&gt;&#xA;" OwnerUserId="16092" LastActivityDate="2019-02-06T08:54:47.947" Title="Hubble Space Telescope (HST) observing modes" Tags="&lt;star&gt;&lt;exoplanet&gt;&lt;spectroscopy&gt;&lt;hubble-telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29291" PostTypeId="2" ParentId="29290" CreationDate="2019-01-23T18:49:56.917" Score="3" Body="&lt;p&gt;&quot;Staring mode&quot; is the traditional way of simply pointing the telescope and exposing the detector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &quot;spatial scanning mode&quot;, you slowly move the telescope, so that the image is slightly smeared out in the spatial direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure below shows the principle: On the left is how a regular image is taken; in the middle, a regular spectrum is taken in &lt;em&gt;staring mode&lt;/em&gt;, and in the right panel, the detector is moved during the exposure, resulting in the &lt;em&gt;spatial scanning mode&lt;/em&gt;. The farther you scan spatially, the brighter and object you may observe before over-exposing, but the downside is that more objects will start to overlap, ruining the spectra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nmzsL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nmzsL.jpg&quot; alt=&quot;spec&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;strong&gt;Left panel&lt;/strong&gt;: &quot;Imaging&quot;; each observed object shows up on the detector at its corresponding position on the sky.&lt;br&gt;&#xA;&lt;strong&gt;Middle panel&lt;/strong&gt;: &quot;Staring mode&quot;; when a &lt;em&gt;grism&lt;/em&gt; (sort of like a prism) is inserted, the light from each object is spread out according to its wavelength, producing a spectrum.&lt;br&gt;&#xA;&lt;strong&gt;Right panel&lt;/strong&gt;: &quot;Scanning mode&quot;; if the detector is moved slightly during exposure, the spectra are spread out &lt;em&gt;perpendicular&lt;/em&gt; to the wavelength direction, effective using a larger number of pixel to collect the photons in. However, if the detector is moved too much, the spectra begin to blend.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This picture, from &lt;a href=&quot;http://www.stsci.edu/hst/wfc3/documents/handbooks/currentIHB/c08_slitless7.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;HST's WFC3 handbook&lt;/a&gt; shows the difference in a real observation: It's quite small, but particularly for the &quot;&lt;span class=&quot;math-container&quot;&gt;$0^\mathrm{th}$&lt;/span&gt; object&quot; (on the left) you can see that in the staring mode it's over-exposed, creating diagonal spikes, which are reduced in spatial scanning mode:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wvuf4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wvuf4.jpg&quot; alt=&quot;WFC3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the picture, the objects are located in the vertical direction, while their spectra are dispersed in the horisontal direction. You can see that in the spatial scanning mode, the objects are slightly broader vertically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantage of the scanning mode is that you can expose for a longer time, without the detector getting saturated (which can be a problem for bright stars). When you expose for longer, you collect more photons and hence obtain a higher signal-to-noise ratio (S/N), meaning that your observation becomes more &quot;accurate&quot;. The figure below (from &lt;a href=&quot;https://terpconnect.umd.edu/~toh/spectrum/SignalsAndNoise.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;these notes&lt;/a&gt;) shows the advantage of a high S/N:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iKqVR.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iKqVR.gif&quot; alt=&quot;S/N&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Two spectra of the same object, with the right having been exposed 9 times longer than the left, resulting in a &lt;span class=&quot;math-container&quot;&gt;$\sqrt{9}=3$&lt;/span&gt; times higher S/N. Determining for instance the central wavelength of this spectral line becomes much more accurate with higher S/N.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that both modes refer to so-caled &lt;em&gt;slitless spectroscopy&lt;/em&gt;. A way to avoid the spectra to overlap is to insert a &quot;slit&quot; which allows only the light from a narrow (of the order of an arcsecond) part of the sky to be observed. In that way you will only obtain spectra of objects lying exactly along the slit, but in return you will have a &quot;cleaner&quot; spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that you can obtain the same effect by dividing your observation into multiple staring mode exposures and then add them together. Scanning mode is a more practical way of not having to do this.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2019-02-06T08:54:47.947" LastActivityDate="2019-02-06T08:54:47.947" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29292" PostTypeId="1" CreationDate="2019-01-23T21:05:35.760" Score="1" ViewCount="393" Body="&lt;p&gt;This is more a terminology question, since Mercury and Venus certainly &quot;move backwards&quot; in the sky. The question is whether we call this retrograde motion. I'm referring to apparent retrograde motion here, not retrograde rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm under the impression we don't, and reserve that name only for planets that are further away from the Sun than us; however conclusive evidence for whether this is (or isn't) the widely-accepted meaning for &quot;retrograde motion&quot; is hard to find.&lt;/p&gt;&#xA;" OwnerUserId="20348" LastEditorUserId="20348" LastEditDate="2019-01-23T21:16:37.783" LastActivityDate="2019-01-23T22:17:03.987" Title="Do Mercury and Venus exhibit retrograde motion?" Tags="&lt;orbit&gt;&lt;venus&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29293" PostTypeId="1" AcceptedAnswerId="29294" CreationDate="2019-01-23T21:19:03.060" Score="3" ViewCount="1432" Body="&lt;p&gt;I'm working on a program that generates the basics stats of a terrestrial planets in binary star systems. I'm not the best at this kind of math, so I'm having trouble calculating the planetary equilibrium temperature.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula I'm currently referencing was posted in a thread from almost 4 years ago: &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10113/how-to-calculate-the-expected-surface-temperature-of-a-planet&quot;&gt;How to calculate the expected surface temperature of a planet&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TJOp5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TJOp5.png&quot; alt=&quot;Formula&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems reasonably accurate since (if memory serves me) using it yields a good answer for Earth and Mars. However this is for a system with one radiating body not a system with two radiating bodies (binary stars). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My instinct is to treat the two stars as one big star and add the wattage &lt;span class=&quot;math-container&quot;&gt;$(w/m^2)$&lt;/span&gt; the planet receives from the various stars. So total watts = watts star 1 + watts star 2. That seems reasonable to me, at least for the purposes of calculating planetary temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've already done work to calculate the watts received by the planets from the various sources. I've used this equation, subbing in values for both stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;F(L/S^2)&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;L = Luminosity (in fraction of Solar), stars 1 &amp;amp; 2 (respectively): &lt;code&gt;0.0159&lt;/code&gt; &amp;amp; &lt;code&gt;0.7758&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;S = Semimajor Axis (in AU), stars 1 &amp;amp; 2 (respectively): &lt;code&gt;0.1228&lt;/code&gt; &amp;amp; &lt;code&gt;3.6000&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;F = Earth Mean Solar Flux &lt;span class=&quot;math-container&quot;&gt;$(w/m^2)$&lt;/span&gt; &lt;code&gt;1362&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the above calculation yields a Mean Solar Flux for the example planet of &lt;span class=&quot;math-container&quot;&gt;$1344w/m^2$&lt;/span&gt; from Star 1 and &lt;span class=&quot;math-container&quot;&gt;$92w/m^2$&lt;/span&gt; from Star 2. Adding those together I get a mean global wattage of &lt;span class=&quot;math-container&quot;&gt;$1436w/m^2$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm unsure how to modify the original equation to utilize the above calculate solar flux (ideally without having to reference the original luminosity and semimajor axis variables). What is the correct way to alter the formula? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be very much appreciated! &lt;/p&gt;&#xA;" OwnerUserId="20372" LastActivityDate="2020-09-30T18:29:53.823" Title="How to Calculate Planetary Equilibrium Temperature in Binary systems" Tags="&lt;planet&gt;&lt;temperature&gt;&lt;mathematics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29294" PostTypeId="2" ParentId="29293" CreationDate="2019-01-23T21:45:53.770" Score="3" Body="&lt;p&gt;The power radiated from the planet, assuming it is at the same temperature across its entire surface:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P_{\mathrm{rad}}=\epsilon\sigma T_{\mathrm{eq}}^4 \cdot 4\pi R_{\mathrm{pl}}^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$\epsilon$&lt;/span&gt; is the emissivity (to match the formula in the question, set this to 1), &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; is the &lt;a href=&quot;https://physics.nist.gov/cgi-bin/cuu/Value?sigma&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stefan-Boltzmann constant&lt;/a&gt;, &lt;span class=&quot;math-container&quot;&gt;$T_{\mathrm{eq}}$&lt;/span&gt; is the equilibrium temperature, and &lt;span class=&quot;math-container&quot;&gt;$R_{\mathrm{pl}}$&lt;/span&gt; is the radius of the planet.&lt;/p&gt;&#xA;&lt;p&gt;The power absorbed by the planet from the stars, assuming it is sufficiently far away from each star that there is negligible illumination from that star beyond 90° from the substellar point:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P_{\mathrm{abs}}=\sum_i{f_i\left(1 - A_i\right) \cdot \pi R_{\mathrm{pl}}^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$f_i$&lt;/span&gt; is the flux from the &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt;th star, and &lt;span class=&quot;math-container&quot;&gt;$A_i$&lt;/span&gt; is the planetary albedo with respect to the &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt;th star (this may be different because the reflectivity will be wavelength-dependent, and the stellar temperature affects which wavelengths are being emitted).&lt;/p&gt;&#xA;&lt;p&gt;So equate radiated and absorbed power, and rearrange:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T_{\mathrm{eq}} = \left[ \frac{\sum_i{f_i\left(1-A_i\right)}}{4\epsilon\sigma} \right]^{1/4}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The limitation here is that the fluxes are going to be time-variable due to the planet's distance changes from the stars (same applies to a planet in an eccentric orbit around a single star), and it is non-trivial to figure out how to average this out. The real planet will take time to heat up and cool down, and in that time the distances and fluxes will change.&lt;/p&gt;&#xA;&lt;p&gt;So take the numbers you get from this &amp;quot;equilibrium temperature&amp;quot; calculation with a pinch of salt.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;As requested, here are some example calculations. First off, the sanity check: Earth. Taking 1361 W/m&lt;sup&gt;2&lt;/sup&gt; as the value of the &lt;a href=&quot;https://www.britannica.com/science/solar-constant&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar constant&lt;/a&gt; and using an albedo of 0.3 and an emissivity of 1, the temperature works out as&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T_{\mathrm{eq}} = \left[\frac{1361\ \mathrm{W\!\cdot\!m^{-2}\,\times \left(1-0.3\right)}}{4 \times 1 \times \left(5.6704 \times 10^{-8}\ \mathrm{W\!\cdot m^{-2}\!\cdot\!K^{-4}} \right)}\right]^{1/4} \approx 255\ \mathrm{K} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;A bit chilly, but not too bad: the greenhouse effect keeps the actual Earth warmer, which could be taken into account by using an emissivity &lt;span class=&quot;math-container&quot;&gt;$\epsilon &amp;lt; 1$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Now for the example in your question, with fluxes of 1344 W/m&lt;sup&gt;2&lt;/sup&gt; and 92 W/m&lt;sup&gt;2&lt;/sup&gt;. Let's also suppose that while the first star is Sunlike (so I will keep the 0.3 value for the albedo), the second star is cooler than the Sun. This star emits more of its light in the red where the planet is less reflective. I will use an albedo of 0.25 for this star.&lt;/p&gt;&#xA;&lt;p&gt;The calculation works out as follows (dropping the units in the calculation step for the sake of space):&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T_\mathrm{eq}=\left[\frac{1344\times\left(1-0.3\right) + 92\times\left(1-0.25\right)}{4\times1\times\left(5.6704\times10^{-8}\right)}\right]^{1/4} \approx 258\ \mathrm{K}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Hope that helps!&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2020-09-30T18:29:53.823" LastActivityDate="2020-09-30T18:29:53.823" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29295" PostTypeId="2" ParentId="29292" CreationDate="2019-01-23T22:17:03.987" Score="1" Body="&lt;p&gt;As noted by PM2Ring, the notion of &quot;retrograde motion&quot; predates the &quot;Copernican revolution&quot; So you should think about the motions of the planets from a Ptolemaic view:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The planets have two major motions (in addition to the diurnal motion around the Earth) motion with a period of one year, and a second motion with a period that is specific to each planet. For Mars, Jupiter and Saturn, the main orbit has a long period, and the epicycle has a period of one year.  For Mercury and Venus, the main orbit has a period of a year (keeping them aligned with the sun) and the epicycle has a period of less than a year.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When the right ascension of the planet is decreasing, then we say it's motion is retrograde. Retrograde periods occur during inferior conjunctions of the planet with the sun, and so the planet cannot be easily observed while it is retrograde.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-01-23T22:17:03.987" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29300" PostTypeId="2" ParentId="16255" CreationDate="2019-01-24T10:53:53.473" Score="1" Body="&lt;p&gt;In order to pinpoint the original position of gravitational waves, you need more than  1 interferometer. These interferometers will detect gravitational waves at different timing. If we know the exact locations of our interferometers and the timming at which they pick up the waves, we can work out the exact location of the source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made a short demo video on interferometers and how to pinpoint waves with them. You can check it at here:&#xA;&lt;a href=&quot;https://youtu.be/4enCrprmLn8&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/4enCrprmLn8&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a nice day!&lt;/p&gt;&#xA;" OwnerUserId="25676" LastActivityDate="2019-01-24T10:53:53.473" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29302" PostTypeId="2" ParentId="2231" CreationDate="2019-01-24T12:51:45.807" Score="1" Body="&lt;p&gt;The most important reason that core collapse happens is that particles moving close to the speed of light are notoriously gravitationally unstable.  That's because when particles moving slower than that are compressed by gravity without energy escape, the gravitational energy released by gravity causes the particles to move faster, and this raises the pressure more than the increase in gravity (due to getting smaller), which allows the object to bounce back when compressed.  But if the particles are already moving close to c, then they do not speed up enough, and even though the pressure does increase, so does the gravity, and there is no tendency for the star to bounce back from the contraction.  Tack on escape of energy, and the pressure is incapable of rising as much as gravity does.  This leads to a runaway, where a little escape of energy leads to a large amount of contraction of the core.  This causes &quot;collapse.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cure for this is to prevent net loss of heat.  That can be done in two ways-- either by release of heat from fusion (self-regulated to occur at whatever rate heat is escaping), or the particles responsible for the pressure can be close enough to their quantum mechanical ground state that they are disallowed from giving up any more heat (even in the absence of any fusion).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So core collapse is a kind of race between whether the electrons (which provide the pressure as they approach their quantum mechanical ground state because that raises their specific heat and they become gluttons for kinetic energy and therefore pressure) can reach their ground state and be disallowed from giving up any more heat, or if they reach close to the speed of light and suffer from the abovementioned instability as they suffer net heat loss.  The winner of this race is determined by the amount of mass in the core-- if the mass is high, the speed of light is reached prior to the quantum mechanical ground state, and if the mass is low, the opposite occurs.  There is nothing about iron that leads to this, except that iron cannot be a source of nuclear energy so does not provide the other type of protection from core collapse.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2019-01-24T19:45:25.077" LastActivityDate="2019-01-24T19:45:25.077" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29303" PostTypeId="1" CreationDate="2019-01-24T12:53:43.617" Score="3" ViewCount="66" Body="&lt;p&gt;I have two Galileo 700mm x 60mm Astro/Terrestrial Refractor Telescopes. Can I take the lenses out and make one powerful telescope, something like an &lt;a href=&quot;https://en.wikipedia.org/wiki/Series_and_parallel_circuits#Series_circuits&quot; rel=&quot;nofollow noreferrer&quot;&gt;optical series circuit&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="25682" LastEditorUserId="37191" LastEditDate="2021-05-28T09:13:21.230" LastActivityDate="2021-05-28T09:13:21.230" Title="Combining lenses from two telescopes to build a more powerful one?" Tags="&lt;telescope&gt;&lt;telescope-lens&gt;&lt;diy&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29305" PostTypeId="1" AcceptedAnswerId="29552" CreationDate="2019-01-24T15:34:56.647" Score="5" ViewCount="117" Body="&lt;p&gt;A photon is travelling through space. As it travels, the space through which it travels is expanding&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this describe&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A - a photon travelling from a distant galaxy to earth&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;B - a photon travelling from a point of higher to a point of lower&#xA;gravity&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;C - either&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If like me you think the answer is C, can you explain why light should travel differently through a gravitational field to how it travels through an expanding universe? In other words, why in one case does the speed of the photon vary, while in the other case it doesn't? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively can you explain why you don't think it's C?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have asked a similar question on the Physics Stack, but I probably worded it badly, so I just got a load of down votes and no answers. I hope I haven't offended anyone here with this wording.&lt;/p&gt;&#xA;" OwnerUserId="25100" LastEditorUserId="25100" LastEditDate="2019-01-24T16:47:54.550" LastActivityDate="2019-02-11T21:50:20.637" Title="What is the difference between the way light travels in a gravitational field and the way it travels through an expanding universe?" Tags="&lt;light&gt;&lt;space-time&gt;&lt;cosmological-inflation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29306" PostTypeId="2" ParentId="29305" CreationDate="2019-01-24T17:15:51.657" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If like me you think the answer is C, can you explain why light should travel differently through a gravitational field to how it travels through an expanding universe? In other words, why in one case does the speed of the photon vary, while in the other case it doesn't?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think the reason is, we can take the Cosmological redshift as a Doppler redshift, where the speed of the photon does not change but only the wavelength changes due to the expansion of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, in Gravitational redshift, its the &quot;redshift&quot; is purely caused by GR and it's possible that the light loses its speed due to this effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the link why I think cosmological redshift can be thought as Doppler redshift. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.physicsforums.com/insights/coordinate-dependent-statements-expanding-universe/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25424" LastEditorUserId="25424" LastEditDate="2019-01-24T20:54:47.017" LastActivityDate="2019-01-24T20:54:47.017" CommentCount="19" ContentLicense="CC BY-SA 4.0" />
  <row Id="29307" PostTypeId="2" ParentId="29278" CreationDate="2019-01-24T19:32:53.140" Score="1" Body="&lt;p&gt;&quot;Time&quot; in astronomy has many meanings. In essence, let us look at something called Sidereal Time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a fixed day, the difference between local and sidereal time is roughly the same. Sidereal time is the sum of the star's right ascension (constant) and its hour angle. The rising hour angle is given by the expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\arccos(-\tan (p) \tan(d)),$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where p is the latitude and d is the declination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the latitude is &quot;roughly&quot; the same we only need to take longitude into account. This means that the time when an event happens, since it happens at the same sidereal time happens &lt;span class=&quot;math-container&quot;&gt;$S_2-S_1$&lt;/span&gt; (S are local sidereal times) later in the second city (the difference is roughly the same since the difference between local and sidereal times is constant for this day).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it happens 80 minutes later since 20 degrees to 360 is 80 min to 24h.&lt;/p&gt;&#xA;" OwnerUserId="24014" LastEditorUserId="7982" LastEditDate="2019-02-24T00:22:58.193" LastActivityDate="2019-02-24T00:22:58.193" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29308" PostTypeId="1" CreationDate="2019-01-24T20:04:00.877" Score="6" ViewCount="175" Body="&lt;p&gt;As far as I am aware, the latest spectral types that have been assigned are around Y2, for objects like &lt;a href=&quot;https://en.wikipedia.org/wiki/WISE_0855%E2%88%920714&quot; rel=&quot;nofollow noreferrer&quot;&gt;WISE 0855-0714&lt;/a&gt; that have temperatures around 250 K or so. I've also seen several directly-imaged exoplanets have had their spectra classified as L- or T-type (e.g. &lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2016A%26A...587A..58B/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bonnefoy et al. 2016&lt;/a&gt; who find that the giant planets HR 8799 d and e are good matches for L6–L8 dwarfs).&lt;/p&gt;&#xA;&lt;p&gt;Jupiter is colder and less massive than the observed Y dwarfs and differs from isolated brown dwarfs because it is being illuminated by the Sun, but there is thermal emission which could be observed from the nightside of the planet. Has Jupiter's nightside spectrum (i.e. the spectrum of Jupiter excluding contributions from reflected sunlight) been observed in sufficient detail to compare it to the spectra of Y dwarfs, and if so does it fit the trends observed in spectral type Y or is it so different that it cannot be spectrally classified as a Y dwarf?&lt;/p&gt;&#xA;&lt;p&gt;For clarity: I am asking purely in terms of spectral classification, i.e. the classification of the &lt;em&gt;spectrum&lt;/em&gt;. I am not asking about whether Jupiter is a brown dwarf making the Sun+Jupiter a binary system. That has been asked &lt;a href=&quot;https://astronomy.stackexchange.com/q/16129/24157&quot;&gt;elsewhere&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2020-07-25T07:51:36.503" LastActivityDate="2020-08-24T08:05:20.510" Title="Can Jupiter's nightside be classified as spectral type Y?" Tags="&lt;jupiter&gt;&lt;temperature&gt;&lt;classification&gt;&lt;spectral-type&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29309" PostTypeId="1" CreationDate="2019-01-24T20:45:06.633" Score="2" ViewCount="115" Body="&lt;p&gt;A search of the Internet is so clouded with discussions about the asteroid belt or asteroids &lt;em&gt;orbiting the planet Earth&lt;/em&gt; that I couldn't see an answer to my question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we know if there are asteroids orbiting the Sun &lt;em&gt;in Earth's orbit&lt;/em&gt; (NOT &lt;em&gt;orbiting the Earth).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be clear: these would be asteroids located somewhere around the Sun at the same distance as Earth, either following it or leading it in the Earth's own orbit around the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I hope that was clear.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="18065" LastEditorUserId="16685" LastEditDate="2019-01-24T21:55:17.273" LastActivityDate="2020-10-04T10:19:09.907" Title="Do we know if there are asteroids leading or following Earth in Earth's orbit around the Sun?" Tags="&lt;the-sun&gt;&lt;orbit&gt;&lt;earth&gt;&lt;asteroids&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29310" PostTypeId="1" AcceptedAnswerId="29330" CreationDate="2019-01-24T20:57:56.813" Score="2" ViewCount="201" Body="&lt;p&gt;Edit: &lt;a href=&quot;https://math.stackexchange.com/questions/3086906/circumpolar-equation-derivation/3087343#3087343&quot;&gt;Geometric answer from Math.stackexchange&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to understand where the circumpolar equation comes from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\delta \geq \frac{\pi}{2} - l$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can understand that the sort of &quot;rotation&quot; of the reference frame that one experiences in horizon coordinates. I'm not sure where to go from the fact that a star's lowest altitude must be greater than 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vMme1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vMme1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is pretty simple to show that &lt;span class=&quot;math-container&quot;&gt;$a = l$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$c = \frac{\pi}{2} - l$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; in this case is the altitude of the north celestial pole. What I need to do is show that the angle of the cone made by a circumpolar star with &lt;span class=&quot;math-container&quot;&gt;$\delta + l \geq \frac{\pi}{2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can deduce that &lt;span class=&quot;math-container&quot;&gt;$a_{min} \geq 0$&lt;/span&gt; for a star to be considered circumpolar. I just need to do this geometrically in terms of its declination from the celestial equator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I think my latest attempt has found a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/n4s7T.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/n4s7T.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a star to be circumpolar, the star must, at the very least, come above the horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$0 \leq a \leq \frac{\pi}{2}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the celestial sphere coordinate system, &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; can be found to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$c = \frac{\pi}{2} - \delta$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the horizon coordinate system, &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; can be found to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$c = l - a_{min}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$l - a_{min} = \frac{\pi}{2} - \delta$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$\\$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$l + \delta = \frac{\pi}{2} + a_{min}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$a_{min}$&lt;/span&gt; can be set to 0 because it is the lower bound for circumpolarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$l + \delta = \frac{\pi}{2}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\delta$&lt;/span&gt; must be in the cone made by the star's rotation about its declination angle. Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{\pi}{2} - l \leq \delta$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="25685" LastEditorUserId="25685" LastEditDate="2019-01-25T20:31:05.813" LastActivityDate="2019-01-25T20:31:05.813" Title="Circumpolar Equation Derivation" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29311" PostTypeId="2" ParentId="29309" CreationDate="2019-01-24T21:20:19.227" Score="4" Body="&lt;p&gt;There are several known &lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2002Icar..160....1M/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth co-orbital asteroids&lt;/a&gt;. The &lt;a href=&quot;https://doi.org/10.1038%2F42662&quot; rel=&quot;nofollow noreferrer&quot;&gt;first to be discovered&lt;/a&gt; was &lt;a href=&quot;https://ssd.jpl.nasa.gov/sbdb.cgi?sstr=3753&quot; rel=&quot;nofollow noreferrer&quot;&gt;(3753) Cruithne&lt;/a&gt;, which is often mistakenly described as &amp;quot;Earth's second moon&amp;quot;. Cruithne has a bean-shaped orbit when viewed in a reference frame rotating with the Earth.&lt;/p&gt;&#xA;&lt;p&gt;The only known &lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2011Natur.475..481C/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth Trojan&lt;/a&gt; is &lt;a href=&quot;https://ssd.jpl.nasa.gov/sbdb.cgi?sstr=3548081&quot; rel=&quot;nofollow noreferrer&quot;&gt;2010 TK&lt;sub&gt;7&lt;/sub&gt;&lt;/a&gt;, which orbits the L&lt;sub&gt;4&lt;/sub&gt; point located 60° ahead of the Earth.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2020-10-04T10:19:09.907" LastActivityDate="2020-10-04T10:19:09.907" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29312" PostTypeId="1" CreationDate="2019-01-24T21:55:20.197" Score="1" ViewCount="68" Body="&lt;p&gt;I have long considered astronomical observing as a form of remote sensing, though quite different than typical remote sensing in that many objects observed are not resolved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering, though, if the  formal definitions of remote sensing include or exclude astronomical observing (and specifically, observing things outside the solar system)?  And/or are there any resources that can be pointed to that either do or don't call astronomical observing remote sensing?&lt;/p&gt;&#xA;" OwnerUserId="128" LastEditorUserId="7982" LastEditDate="2019-01-25T07:57:40.367" LastActivityDate="2019-01-25T10:38:30.253" Title="Does astronomical observing &quot;count&quot; as remote sensing?" Tags="&lt;amateur-observing&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29313" PostTypeId="1" CreationDate="2019-01-24T22:19:32.037" Score="5" ViewCount="130" Body="&lt;p&gt;Wikipedia gives the following physical dimensions for &lt;a href=&quot;https://en.wikipedia.org/wiki/Moon&quot; rel=&quot;noreferrer&quot;&gt;the Moon&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h3&gt;Physical characteristics&lt;/h3&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Mean radius        1737.1 km  (0.273 of Earth's)&#xA;Equatorial radius  1738.1 km  (0.273 of Earth's)&#xA;Polar radius       1736.0 km  (0.273 of Earth's)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In all cases, the proportion of Earth's equivalent measurement is 0.273. If these figures are accurate, why aren't these numbers (0.273) different from each other at this scale? Obviously, if measured to enough decimal places, differences will appear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon has less gravity and rotates at a different speed from the Earth. Why don't these cause variability in the different radius measurements? Do they cancel out? It is a coincidence? Is there something about the Earth-Moon dynamics that means over millions of years these proportions gradually equalise?&lt;/p&gt;&#xA;" OwnerUserId="1435" LastEditorUserId="1435" LastEditDate="2019-01-24T22:40:22.943" LastActivityDate="2019-01-25T06:31:28.377" Title="Why is the difference from a perfect sphere the same for the Earth and the Moon?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29315" PostTypeId="1" AcceptedAnswerId="29323" CreationDate="2019-01-25T03:24:18.107" Score="2" ViewCount="55" Body="&lt;p&gt;There are several different classifications of gravitational lensing phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I am asking for any examples of strong lensing where the lensed image of an extended object is magnified and recognizable as an extended object. Of course it may be distorted somewhat, but if the lensed object is a galaxy, then the lensed image should appear larger and &quot;galaxy-shaped&quot; even though distorted. If it's a tight pattern of stars or other objects, then the pattern should be larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a strong memory of reading about such a case in Nature or Science within the last few years, but I can not find it. I believe that the nearer, &lt;em&gt;lensing&lt;/em&gt; object was a galaxy.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-01-25T03:31:00.503" LastActivityDate="2019-01-25T11:11:48.600" Title="Cases of gravitational lensing resulting in a recognizable image of an extended object?" Tags="&lt;observational-astronomy&gt;&lt;gravitational-lensing&gt;&lt;deep-sky-observing&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29316" PostTypeId="1" AcceptedAnswerId="29325" CreationDate="2019-01-25T04:48:08.220" Score="2" ViewCount="1016" Body="&lt;p&gt;I have read that it takes billions of years for stars of approximately the size of the Sun to get through the main sequence, after which it will enter the red giant phase. When that happens, how long does it take for the star to balloon outward when it becomes a red giant?&lt;/p&gt;&#xA;" OwnerUserId="21306" LastActivityDate="2019-01-25T14:50:44.890" Title="How long does it take a star to transition from main sequence to red giant?" Tags="&lt;star&gt;&lt;main-sequence&gt;&lt;red-giant&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29317" PostTypeId="1" CreationDate="2019-01-25T05:30:31.977" Score="2" ViewCount="283" Body="&lt;p&gt;The drawing below, found in Wikimedia and at &lt;a href=&quot;https://iiif.lib.harvard.edu/manifests/view/ext:aHR0cHM6Ly9paWlmLmxpYi5oYXJ2YXJkLmVkdS9tYW5pZmVzdHMvaWRzOjE2NDcwMzE1$1i&quot; rel=&quot;nofollow noreferrer&quot;&gt;lib.harvard.edu&lt;/a&gt; is of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Aerial_telescope#Very_long_%22tubed%22_telescopes&quot; rel=&quot;nofollow noreferrer&quot;&gt;very long tubed aerial telescope&lt;/a&gt;. I believe it is taken from his 1673 work &lt;a href=&quot;https://library.si.edu/digital-library/book/machinae-coelestis-pars-prior-posterior-hevelius-johannes&quot; rel=&quot;nofollow noreferrer&quot;&gt;Machinae coelestis&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've always wondered about the purposes of all those (20+) square blocks with round holes, spaced evenly from one end to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've included two cropped views from &lt;a href=&quot;https://iiif.lib.harvard.edu/manifests/view/ext:aHR0cHM6Ly9paWlmLmxpYi5oYXJ2YXJkLmVkdS9tYW5pZmVzdHMvaWRzOjE2NDcwMzE1$1i&quot; rel=&quot;nofollow noreferrer&quot;&gt;Harvard University, Houghton Library, pga_typ_620_73_451_fig_aa&lt;/a&gt; (found &lt;a href=&quot;https://hollis.harvard.edu/primo-explore/fulldisplay?context=L&amp;amp;vid=HVD2&amp;amp;search_scope=everything&amp;amp;tab=everything&amp;amp;lang=en_US&amp;amp;docid=01HVD_ALMA211764767520003941&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;) the first of which I've zoomed and enhanced the contrast both to better appreciate the cute bird, and to try to read the annotations; there seems to be a 772 or 722 next to one of the blocks near the top left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What was the function of all of these blocks? Is it optical, or mechanical, both, neither? When I was young I thought there were lenses in each one, but I'm assuming now that at least the central circular holes are completely open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is pretty easy to guess either one, but presumably with such detailed annotation there is text to go along with this image somewhere and it is likely that we can know Hevelius's intention. I just don't know where to read the text. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that &lt;a href=&quot;https://archive.org/details/bub_gb_TVtGAQAAIAAJ&quot; rel=&quot;nofollow noreferrer&quot;&gt;the text is here&lt;/a&gt; but it's 1,000 pages long and in a language that I can't understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iFTbX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iFTbX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1I0eh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1I0eh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-08-16T04:57:37.877" LastActivityDate="2019-08-16T11:23:20.633" Title="How did Johannes Hevelius' long telescope work? Why all the round holes?" Tags="&lt;telescope&gt;&lt;history&gt;&lt;optics&gt;&lt;telescope-lens&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29318" PostTypeId="2" ParentId="29313" CreationDate="2019-01-25T06:31:28.377" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is the difference from a perfect sphere the same for the Earth and the Moon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It isn't, as soon as you look one more digit further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From NASA's &lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html&quot; rel=&quot;noreferrer&quot;&gt;Earth Fact Sheet&lt;/a&gt; and &lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html&quot; rel=&quot;noreferrer&quot;&gt;Moon Fact Sheet&lt;/a&gt; which I found &lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/planet_table_ratio.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                             Moon (km)   Earth (km)      Ratio     (rounded)&#xA;Mean radius (volumetric)     1737.4       6371.000      0.27270      0.273&#xA;Equatorial radius            1738.1       6378.137      0.27251      0.273&#xA;Polar radius                 1736.0       6356.752      0.27310      0.273&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you would like to compare the Moon and Earth parameters further, see the &lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html&quot; rel=&quot;noreferrer&quot;&gt;Moon Fact Sheet&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HIPm7.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HIPm7.png&quot; alt=&quot;nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-01-25T06:31:28.377" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29319" PostTypeId="1" CreationDate="2019-01-25T06:48:29.243" Score="10" ViewCount="1018" Body="&lt;p&gt;How can gravitational lensing makes a quasar brighter than it would in the absence of a foreground galaxy?&lt;/p&gt;&#xA;" OwnerUserId="23550" LastActivityDate="2019-01-25T07:33:17.883" Title="How can gravitational lensing makes a quasar appear brighter?" Tags="&lt;cosmology&gt;&lt;observational-astronomy&gt;&lt;gravitational-lensing&gt;&lt;quasars&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29320" PostTypeId="2" ParentId="29319" CreationDate="2019-01-25T07:33:17.883" Score="16" Body="&lt;p&gt;The quasar gives out light in all directions. The light spreads out in space. Only a very small amount of that light would be pointed exactly in the direction of your telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if a large galaxy or galaxy cluster is between the quasar and us, it bends some of the light towards us, making the quasar brighter (it would also distort the shape, but quasars are too small for their shape to be seen) This makes it brighter for us (and dimmer in other parts of space which would have got the light that is now bent towards us.)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-01-25T07:33:17.883" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29322" PostTypeId="2" ParentId="29312" CreationDate="2019-01-25T10:38:30.253" Score="2" Body="&lt;p&gt;As suggested by, e.g., &lt;a href=&quot;https://en.wikipedia.org/wiki/Remote_sensing&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Wikipedia page&lt;/a&gt; (or &lt;a href=&quot;https://www.usgs.gov/faqs/what-remote-sensing-and-what-it-used?qt-news_science_products=7#qt-news_science_products&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page at the US Geological Survey&lt;/a&gt;), &quot;remote sensing&quot; is used pretty much exclusively for airborne or satellite-based observations of the surface of the Earth (including the ocean floor in some cases). &#xA;This makes sense if you consider that &quot;remote&quot; is used in contrast to observations made &lt;em&gt;locally&lt;/em&gt; (or &quot;on-site&quot;). E.g., if you walk around on the ground and make observations, those are &quot;local&quot; observations; if you fly a plane and make observations of the same place, those are &quot;remote&quot; observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since astronomical observations are overwhelming of places you simply can't make local observations of, it's unnecessary to specify them as &quot;remote&quot; observations, just as no one call looking up at high-altitude clouds or distant mountain peaks &quot;remote sensing&quot;. (And it seems more than a little silly to claim that humans have been doing &quot;remote sensing&quot; for millions of years, even though humans and proto-humans have indeed been looking at the Sun, the Moon, planets, and stars &quot;remotely&quot; for about that long.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I have occasionally seen satellite observations of planetary surfaces -- e.g., imaging by probes orbiting Mars -- referred to as a type of remote sensing; this is clearly based on the direct analogy with Earth remote-sensing satellites. But that's about as far as you can push the term.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're free to consider astronomy to involve &quot;remote sensing&quot;, based on an overly literal and overly logical interpretation of the term, if you like; in practice, no one working in either astronomy or remote sensing does.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2019-01-25T10:38:30.253" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29323" PostTypeId="2" ParentId="29315" CreationDate="2019-01-25T11:11:48.600" Score="4" Body="&lt;p&gt;There are actually quite a few examples of this. A particularly nice one, where you can see star-forming substructure within the distant, lensed galaxy (the extended, vertical &quot;snake&quot; structure just right of center), is &lt;a href=&quot;https://www.spacetelescope.org/images/potw1747a/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this HST image&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SoPaP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SoPaP.jpg&quot; alt=&quot;HST image of background galaxy lensed by cluster MACSJ1206.2-0847&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is another HST-based example, in which the same background galaxy is lensed multiple times (indicated by white ovals) by an intervening galaxy cluster, with a reconstruction of the background galaxy in undistorted form shown in the lower-left:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1rskq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1rskq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.space.com/14481-hubble-photo-brightest-galaxy-gravitational-lens.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.space.com/14481-hubble-photo-brightest-galaxy-gravitational-lens.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People have even obtained spatial resolved spectroscopy within the lensed galaxies, as discussed in &lt;a href=&quot;https://arxiv.org/abs/1503.07873&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt; (which includes spectroscopy of the first lensed galaxy pictured above).&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2019-01-25T11:11:48.600" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29324" PostTypeId="2" ParentId="29317" CreationDate="2019-01-25T11:21:39.120" Score="0" Body="&lt;p&gt;Machinae coelestis has two volumes. You've linked to volume 2, but this drawing is in &lt;a href=&quot;https://library.si.edu/digital-library/book/machinacoelesti1heve&quot; rel=&quot;nofollow noreferrer&quot;&gt;volume 1 on page 410.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The annotation isn't '772' but 'M'. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;hanc vero M cui lens objectiva inserta est&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;the objective lens has been inserted at M&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if there are lenses in all the other blocks. There's at least 10 pages of description. The text needs extensive editing before Google Translate can make sense of it (most of the 'f' in the text need to be replaced with 's'), and even then half of it is gibberish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hevelius' 150-foot telescope is called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Refracting_telescope#Keplerian_telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Keplerian telescope&lt;/a&gt; in various places, suggesting it used 2 or maybe 3 lenses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the blocks allowed Hevelius to easily move the lens around (by inserting it into different blocks)? You can often adjust the length of a refractor telescope (they are built with two or more tubes that can slide inside each other).&lt;/p&gt;&#xA;" OwnerUserId="361" LastEditorUserId="361" LastEditDate="2019-01-25T16:17:14.317" LastActivityDate="2019-01-25T16:17:14.317" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29325" PostTypeId="2" ParentId="29316" CreationDate="2019-01-25T14:50:44.890" Score="1" Body="&lt;p&gt;It takes a star like the Sun about a billion years to go from the end of core hydrogen burning to the beginning of helium core burning.  One might not call that entire phase a red giant phase, however, because the puffing out process takes quite a while to get going.  In the mean time, the star is what is called a &quot;subgiant,&quot; rather than a red giant, and physically the difference is that in a full-blown red giant, the electrons in the core have lost so much heat that they are approaching close to their quantum mechanical ground state.  This implies the core is very small, about the size of the Earth, so a lot of heat must be lost before the star gets to that state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, we can say the star is a red giant, but it won't be all that large until the core builds up more mass.  A ball of electrons near their ground state will shrink as more mass is added to it, and as the core shrinks, the electrons gain kinetic energy, and importantly, so do the helium ions.  When the core mass gets large enough, the helium gets enough kinetic energy to start fusion, and that's the end of the red giant phase.  The star ceases to expand and instead contracts to the &quot;horizontal branch.&quot;  Most of the huge expansion occurs near to the end of that process, so the Sun will look like a red giant for less than perhaps 100 million years, depending on how big and red you want it to be in order to count.  A reasonable source is &lt;a href=&quot;https://www.space.com/22471-red-giant-stars.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.space.com/22471-red-giant-stars.html&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2019-01-25T14:50:44.890" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29327" PostTypeId="1" AcceptedAnswerId="29333" CreationDate="2019-01-25T16:15:34.217" Score="16" ViewCount="1378" Body="&lt;p&gt;The Daily Galaxy article &lt;a href=&quot;https://dailygalaxy.com/2019/01/the-lost-hubble-new-deepest-image-of-the-universe-ever-taken/&quot; rel=&quot;noreferrer&quot;&gt;“The Lost Hubble” –New! Deepest Image of the Universe Ever Taken&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To produce the image a group of researchers from the Instituto de Astrofísica de Canarias (IAC) led by Alejandro S. Borlaff used original HUDF images from the Hubble Space Telescope. After improving the process of combining several images the group was able to recover a large quantity of light from the outer zones of the largest galaxies in the HUDF. Recovering this light, emitted by the stars in these outer zones, was equivalent to recovering the light from a complete galaxy (“smeared out” over the whole field) and for some galaxies this missing light shows that they have diameters almost twice as big as previously measured.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The image looks really strange, what is going on? Is there a technical article associated with this work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iPsaV.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iPsaV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-01-26T10:15:10.537" Title="What is the &quot;lost light&quot; in this unusual Hubble Deep Sky image?" Tags="&lt;data-analysis&gt;&lt;hubble-telescope&gt;&lt;deep-sky-observing&gt;&lt;image-processing&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29329" PostTypeId="2" ParentId="29327" CreationDate="2019-01-25T16:27:30.287" Score="8" Body="&lt;p&gt;When you plug the lead researcher's name into Arxiv, the first search result is &lt;a href=&quot;https://arxiv.org/abs/1810.00002&quot; rel=&quot;noreferrer&quot;&gt;The missing light of the Hubble Ultra Deep Field&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZendY.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZendY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 main steps:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Creation of sky flat fields for the four filters. This process is fully described in Sect. 2.4.   &lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;– Creation of a catalogue of all WFC3/IR datasets that may affect our mosaics (including calibration exposures) to generate a set of improved persistence models for each exposure of the HUDF. We detail this process in Sect. 2.5.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;– Download and reduction of all the WFC3/IR datasets that include observations using the F105W, F125W, F140W and F160W filters on the HUDF.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Flat sky field: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In order to measure the relative sensitivity of the pixels of a detector (flat field), the optimal process would be to observe a uniform external source of light. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Basically they're trying to remove all sources of noise from the image, in an attempt to make faint signals appear in places where that signal has been overwhelmed by noise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Persistence models: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A known effect that affects HgCdTe IR array detectors (as is the case of the WFC3/IR) is persistence. Persistence shows up as an afterglow on the pixels that were exposed to a bright source of light in a previous exposure. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The current method of persistence correction of WFC3/IR con- sists in modeling the number of electrons that would be created by persistence in each pixel by all the previous exposures (up to a certain time) that were taken before the one to correct (Long et al. 2012). &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;During long exposures, sky background can vary noticeably, introducing a non-linear component to the count rates calculated by calwf3. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We individually estimate and subtract the sky background emission from each readout of the intermediate ima.fits files.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In order to avoid systematic biases due to the presence of de- fects in some regions of the detector, we created a manual data quality mask to flag those regions were the flat field cannot fully correct the differences in sensitivity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More image processing to remove sky background: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In this Section we describe the methods used to remove the sky background from the individual exposures and the final mosaics of the HUDF. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Image alignment:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As a consequence, when comparing images from different visits, it is usual to see that they are not exactly aligned. In order to exploit the full resolving capabilities of WFC3, we need to carefully re-align the images of different visits to a single reference world coordinate system solution (WCS hereafter).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and as a final step, image combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Result: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The XDF version of the HUDF WFC3/IR mosaics is dominated by a systematic bias in the form of a significant oversubtraction of the sky background around the objects with large angular size. A similar result (to a lesser extent) is obtained for the HUDF12. We successfully recover a significant amount of over-subtracted diffuse light around the largest objects of the HUDF, not detected by the previous versions of the mosaics. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Summary: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They've processed the images to bring out details in the galaxies. In the space between the galaxies, the image processing gives garbage results (the white areas), but they've managed to bring out detail on the edge of the galaxies that was hidden before. &lt;/p&gt;&#xA;" OwnerUserId="361" LastEditorUserId="361" LastEditDate="2019-01-25T17:29:42.090" LastActivityDate="2019-01-25T17:29:42.090" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29330" PostTypeId="2" ParentId="29310" CreationDate="2019-01-25T18:14:40.050" Score="2" Body="&lt;p&gt;There is nothing wrong with your explanation.  You could add some explanatory notes like &lt;span class=&quot;math-container&quot;&gt;$a_{min} = l-c$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$a_{max}=l+c$&lt;/span&gt;. Also, I suggest that you put the equation &lt;span class=&quot;math-container&quot;&gt;$l+\delta = \frac{\pi}2-a_{min}$&lt;/span&gt; on a separate line.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very good to figure out these equations on your own. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an alternative explanation with less algebra:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you walk outside at night, you will notice that the north star has an elevation of &lt;span class=&quot;math-container&quot;&gt;$l$&lt;/span&gt; degrees if your latitude is &lt;span class=&quot;math-container&quot;&gt;$l$&lt;/span&gt;.  &lt;/li&gt;&#xA;&lt;li&gt;The north star does not move.  &lt;/li&gt;&#xA;&lt;li&gt;The angular distance between stars does not change  (in the short term!).&lt;/li&gt;&#xA;&lt;li&gt;In order for the a star to be circumpolar, it needs to be within &lt;span class=&quot;math-container&quot;&gt;$l$&lt;/span&gt; degrees of the north star (otherwise it will dip below the horizon when its &lt;a href=&quot;https://en.wikipedia.org/wiki/Hour_angle&quot; rel=&quot;nofollow noreferrer&quot;&gt;hour angle&lt;/a&gt; is 12 hours.)&lt;/li&gt;&#xA;&lt;li&gt;So, the angle &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; between the north star to a circumpolar star needs to be less than &lt;span class=&quot;math-container&quot;&gt;$l$&lt;/span&gt;.  That angle is &lt;span class=&quot;math-container&quot;&gt;$\theta=90^\circ-\delta = \pi/2 -\delta$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$\delta$&lt;/span&gt; is the declination of the star.  &lt;/li&gt;&#xA;&lt;li&gt;Finally, &#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{align} \theta &amp;amp;\leq l\\&#xA;\pi/2 -\delta &amp;amp;\leq l \\&#xA;\pi/2  &amp;amp;\leq  \delta + l \\&#xA;\pi/2  - l &amp;amp;\leq \delta. &#xA;\end{align}&#xA;$$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="25554" LastActivityDate="2019-01-25T18:14:40.050" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29331" PostTypeId="2" ParentId="29327" CreationDate="2019-01-25T19:11:58.890" Score="0" Body="&lt;p&gt;In response to a couple comments that Hobbes' answer is a bit thick,  how about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To reduce noise effects, the team did  flat-fielding adjustment and then summed multiple exposures, thus allowing weak signals to add while noise effects cancelled out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the TL;DR  which leaves out a lot of really cool methods of identifying &quot;true dark&quot; and noise patches vs. reliable signals (stars or galaxies or whatever). &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2019-01-25T19:11:58.890" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29332" PostTypeId="2" ParentId="29317" CreationDate="2019-01-25T19:18:47.623" Score="1" Body="&lt;p&gt;In general,  the less the support structure weighs, the easier it is to move the whole thing around.  Home-user telescopes usually have a solid body because it's by far the cheapest and simplest way to get a small rigid structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Large observatory 'scopes often have a minimal framework supporting the secondary (the far element in a reflector);  here the primary refractor is supported with a similar structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most certainly the support blocks do not contain lenses.  For one thing, just imagine trying to align and focus a 12 or 15-element system!  For another, producing the lenses for telescopes of that size was incredibly tedious back then (and now, really).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a couple great chapters in &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/The_Age_of_Wonder&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Age Of Wonder&lt;/a&gt;&quot; by Richard Holmes about a couple of the early astronomers &amp;amp; scope builders in Great Britain for those who are interested.&lt;/p&gt;&#xA;" OwnerUserId="907" LastEditorUserId="7982" LastEditDate="2019-01-26T00:01:22.047" LastActivityDate="2019-01-26T00:01:22.047" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29333" PostTypeId="2" ParentId="29327" CreationDate="2019-01-25T21:14:30.307" Score="9" Body="&lt;p&gt;Let me see if I can explain the main aim and accomplishment of this work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off: the picture you're puzzling over is a &quot;luminance RGB&quot; image, in which the bright regions are represented by color (a sort of pseudo-true color using near-infrared images), with the second faintest parts represented with black and the very faintest parts with white. The latter aren't qite &quot;garbage&quot;, as Hobbes suggests in their answer, but they are relatively speaking the noisiest parts of the image, so there's little real information be found there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This paper (Borlaff et al.; see the link in Hobbes's answer) is about re-processing near-infrared &lt;em&gt;HST&lt;/em&gt; images originally taken about ten years ago as part of the Ultra Deep Field. Previous processing of these images (e.g., Koekemoer et al. 2013 [&quot;HUDF12&quot;] and Illingworth et al. 2013 [&quot;XDF&quot;]) was focused on getting information about the smallest, faintest galaxies, which are mostly the really distant, high-redshift galaxies. Because of this, the crucial stage of sky subtraction had some biases: in particular, it tended to treat the faint outer regions of the large, nearer galaxies as part of the sky to be subtracted. This is actually fine for the analysis of the small, distant galaxies, but means that if you &lt;em&gt;do&lt;/em&gt; want to analyze the outer regions (outer disks, faint stellar halos, remnants of merger structures, etc.) of the larger, closer galaxies, you have the problem that their outer regions are &lt;em&gt;over-subtracted&lt;/em&gt; (hence the &quot;missing light&quot;) and thus unmeasurable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The &quot;sky&quot; being subtracted is a combination of emission from certain atoms in the tenuous outer atmosphere above &lt;em&gt;HST&lt;/em&gt;, sunlight scattered from dust grains in the inner solar system, and the so-called &quot;extragalactic background&quot; = the combined light from unresolved distant galaxies.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The abstract mentions four improvements the new study implemented when they re-processed the HST images: &quot;1) creation of new absolute sky flat fields, 2) extended persistence models, 3) dedicated sky background subtraction and 4) robust co-adding.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest that the third item is perhaps the most important: they implement a method which does &lt;em&gt;not&lt;/em&gt; subtract off the faint outer regions of the larger galaxies, and thus the resulting images still have information about the outer parts of these galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plot below (extracted from Figure 20 of the paper) illustrates the kind of improvement they were after. It shows the surface brightness (in the F105W near-infrared filter) of one of the largest galaxies (a giant elliptical -- I think it's the big, round, yellow galaxy in the lower middle of the color image) as a function of radius (measured in elliptical annuli). The red triangles were measured using the XDF-processed image, the blue squares used the HUDF12-processed image, and the black points use the newly re-processed image produced as part of this paper [ABYSS]. You can see that the XDF points drop off at a radius of about 55 kpc, the HUDF12 points fall off at around 90 kpc -- but the light from this galaxy can be traced out to 140 kpc in the ABYSS-reprocessed image. This also means you can potentially analyze structure in the outer part of the galaxy out to ~ 150 kpc, which wouldn't be possible with the previous versions of the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hXPcP.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hXPcP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;(I should point out that I'm friends with, and have co-authored papers with, a couple of the authors, so I may be a bit biased -- but I think this is really impressive work!)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2019-01-26T10:15:10.537" LastActivityDate="2019-01-26T10:15:10.537" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29335" PostTypeId="1" AcceptedAnswerId="29337" CreationDate="2019-01-26T00:23:51.320" Score="3" ViewCount="431" Body="&lt;p&gt;@antlersoft's &lt;a href=&quot;https://astronomy.stackexchange.com/a/28260/7982&quot;&gt;answer&lt;/a&gt; describes some of the challenges to seeing any details in the small disk of Mars in small amateur telescopes. In the case of reflecting telescopes, it mentions the use of either an off-axis parabolic mirror, or an off-axis aperture mask on a standard Newtonian to produce a clear unobstructed view. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer links to a short article on the amateur telescope site &lt;a href=&quot;https://www.bbastrodesigns.com/BBAstroDesigns.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;BBAstroDesigns&lt;/a&gt; which is titled &lt;a href=&quot;https://www.bbastrodesigns.com/OffAxisMasks.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Off-Axis Masks: Light Reduction and Resolution&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand the explanation correctly, besides the light reduction function, the off-axis masking excludes the secondary mirror and support, thereby removing diffracted light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a bit surprised that for amateur viewing on Earth that diffraction from a Newtonian or Cassegrain's secondary mirror and supports would really have any impact on resolution of small extended objects like Mars. The example in the article is a 16 inch telescope. I would have thought that &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_seeing&quot; rel=&quot;nofollow noreferrer&quot;&gt;seeing&lt;/a&gt; would be the limitation, but maybe for bright objects a smaller diameter (1/3 to 1/6 of the full aperture) is actually better? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to explain the physics/optic behind the use of off-axis apertures to improve resolution as discussed in these links?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If possible, can a bit of math be used to demonstrate this quantitatively and convincingly?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-01-26T02:01:15.123" Title="Under what situations can an aperture mask improve the resolution of a small/medium amateur telescope? Is this demonstrable mathematically?" Tags="&lt;amateur-observing&gt;&lt;newtonian-telescope&gt;&lt;angular-resolution&gt;&lt;atmospheric-effects&gt;&lt;cassegrain-telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29337" PostTypeId="2" ParentId="29335" CreationDate="2019-01-26T02:01:15.123" Score="4" Body="&lt;p&gt;The vast majority of people - almost everyone, really - should not rely on this on a daily basis. The chances are overwhelming that they will reduce the performance of their instruments. There are so many variables involved, and it depends on so many factors, a likely result is worse performance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The link is to a site maintained by Mel Bartels. Mel is a very experienced observer and telescope builder, with hundreds if not thousands of hours behind the eyepiece. He has enough practical experience so that, just by looking at the overall image, and the shape and size of the Airy disk and diffraction figure, he can tell exactly in what way the telescope's behavior is affected by changes.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The effects of the secondary mirror and spider are a topic which is much misunderstood. You will encounter many strong opinions on this topic with not much basis in reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spider vanes, if straight, will each create two spikes, diametrically opposed, perpendicular to the length of the vane. So a 3-vane spider makes 6 spikes. A 4-vane spider makes 8 spikes, but coincidental two by two (look like 4 brighter spikes). If the vanes are curved, then the diffraction figure is more like a faint, diffuse cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So using an aperture mask can remove the spikes due to the spider vanes. But will it make the image &quot;better&quot;? Spikes don't matter from a practical perspective. If a spike is blocking one of the components of a double star, just rotate the scope. And the theoretical resolving power is diminished. You are, after all, using a smaller aperture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The secondary mirror will increase the size of the diffraction figure. The Airy disk and the first minimum become larger in diameter - and so do the rest of the rings. This is a matter of much confusion, with some claiming that obstructed telescopes (scopes with a secondary mirror) are essentially inferior to unobstructed designs - this is demonstrably false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple rule of thumb: if D is the diameter of the primary, and d is the diameter of the secondary, then the performance of an obstructed scope is equivalent, from the p.o.v. of resolving power (high frequency components), to an unobstructed instrument of aperture D - d. This is not an exact formula but it's close enough. A 300mm reflector with a 60mm secondary has about the same fine detail performance on planetary observations like an unobstructed 240mm instrument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are cases when you're observing close double stars, and it would be great if one of the stars could fall into the diffraction minimum of the other, brighter star, just outside the Airy disk. Sometimes you can do that with an aperture mask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article by Mel Bartels makes an important point, and I wish he emphasized it more: the aperture mask should only be used at low or medium magnification. In other words, if you're planning anyway to use low or medium magnification, then in some rare cases the aperture mask might be useful. It &lt;strong&gt;NEVER&lt;/strong&gt; makes sense to use an aperture mask at high magnification - it's like pushing a telescope above its maximum useful magnification, and the result is just a blurry mess.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The linked article also mentions the issue of brightness. Some observers claim that the Moon is too bright in a telescope. Mel Bartels sees this as an opportunity to use an aperture mask (again, NOT at high magnification). Others use Moon filters - neutral density filters that reduce the incoming flux of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a third school of thought here. Do not become dark-adapted while observing the Moon. Do not observe the Moon from a dark place. Observe it with lights turned on around you, like from the sidewalk of a well-illumined street. This is what I, and others like me, do. Suddenly the Moon is not too bright anymore. The telescope is not performance-limited by the aperture mask, and you can zoom in at high magnification to observe small details of craters and mountains. And your eyes are working in their optimal regime, when the pupil is not fully dilated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surface brightness of extended objects never increases in a telescope, it just seems to increase. In fact, it always decreases. The Moon seems too bright because you're dark-adapted - but in reality it's quite pale.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There is another case when the image is demonstrably improved by an aperture mask. This is when the primary mirror has strong defects near the perimeter, like TDE (turned down edge), which is a defect both quite common (for mass-manufactured instruments) and very damaging to performance. A scope with strong TDE has poor resolving power and poor high frequency contrast. Sometimes it's so bad that even an aperture mask smaller than D/2 will improve performance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution here is not the aperture mask, but instead masking off the perimeter. Make a ring with an inner diameter somewhat smaller than the aperture and place it permanently on the mirror, so as to block off the defective edge. Aperture is somewhat diminished, but now the optics are good across the whole surface.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One more case: let's assume seeing is so bad, the image is shaking like a giant, blurry pile of jelly. There's no way you're going to use high magnification anyway. You operate at low-to-medium magnification. Now apply the aperture mask. Is the image better?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objectively the answer is no. If you're doing lucky imaging with a camera, it's still better to leave aperture alone, and by luck you will capture a few frames where detail resolution is pretty good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the human visual system is weird. Some people prefer to let go of the higher performance ceiling, instead opting for an image that's perceived as more smooth and stable. Personally I prefer to let low magnification do its thing, and leave aperture alone, but not everyone is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even visually, if you don't mask the aperture, sometimes seeing gets better briefly, and you can see small details for a while. With an aperture mask, you will miss those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is a subjective thing, it's not something that can be objectively measured, except as a study of the psychology of visual perception.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you want to experiment with the aperture mask, by all means go ahead and do it. But be aware of the many, &lt;strong&gt;many&lt;/strong&gt; pitfalls here. It's not a silver bullet. It's not the solution to all of humanity's problems. It's just a trick that sometimes works, in a fairly narrow slice of use cases.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2019-01-26T02:01:15.123" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29339" PostTypeId="1" CreationDate="2019-01-26T19:28:33.163" Score="0" ViewCount="96" Body="&lt;p&gt;Will Earth at some point stop rotating anti-clockwise, stay still, and then rotate clockwise?&lt;/p&gt;&#xA;" OwnerUserId="25704" LastEditorUserId="1891" LastEditDate="2019-01-27T10:13:46.847" LastActivityDate="2019-01-27T10:13:46.847" Title="Will Earth at some point stop rotating anti-clockwise, stay still, and then rotate clockwise?" Tags="&lt;earth&gt;&lt;rotation&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2019-01-27T00:22:23.903" ContentLicense="CC BY-SA 4.0" />
  <row Id="29340" PostTypeId="2" ParentId="29339" CreationDate="2019-01-26T21:29:28.923" Score="0" Body="&lt;p&gt;No. This would require some force imposed externally. Momentum is a wonderful thing.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2019-01-26T21:29:28.923" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29341" PostTypeId="1" AcceptedAnswerId="29345" CreationDate="2019-01-26T23:46:34.410" Score="3" ViewCount="133" Body="&lt;p&gt;If the Milky Way galaxy is as big as planet earth, how big is Earth?&lt;/p&gt;&#xA;" OwnerUserId="17024" LastEditorUserId="17024" LastEditDate="2019-01-27T00:35:45.730" LastActivityDate="2019-01-27T13:02:21.223" Title="If the Milky Way galaxy is as big as planet earth, how big is Earth?" Tags="&lt;milky-way&gt;&lt;size&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29342" PostTypeId="1" CreationDate="2019-01-27T00:10:18.750" Score="0" ViewCount="254" Body="&lt;p&gt;In Star Wars, the planet Naboo has a plasma core, instead of a molten one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can a planet really have a plasma core?&lt;/p&gt;&#xA;" OwnerUserId="18855" LastEditorUserId="18855" LastEditDate="2019-01-27T00:34:36.793" LastActivityDate="2019-01-27T04:39:22.183" Title="Is Naboo's core possible in real life?" Tags="&lt;planet&gt;&lt;exoplanet&gt;&lt;planetary-formation&gt;&lt;core&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29343" PostTypeId="1" AcceptedAnswerId="29352" CreationDate="2019-01-27T00:36:38.817" Score="1" ViewCount="172" Body="&lt;p&gt;If the whole observable Universe is as big as planet Earth, how big would Earth be?&lt;/p&gt;&#xA;" OwnerUserId="17024" LastActivityDate="2019-01-27T09:14:23.753" Title="If the whole observable Universe is as big as planet Earth, how big would Earth be?" Tags="&lt;observable-universe&gt;&lt;size&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29344" PostTypeId="2" ParentId="28413" CreationDate="2019-01-27T00:57:09.063" Score="-2" Body="&lt;p&gt;Flares in space are known for coming from stars. Sunlike stars, red dwarfs, etc, almost any kind of star has flares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hot Jupiters can be the size of Jupiter,(hence the name). They can be bigger; however, if they get too big, they become brown dwarfs. This means that hot Jupiters are too small to have flares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, many hot Jupiters have unstable atmospheres. They are so close to their parent stars, that they lose their atmospheres. No atmosphere, no flares.&lt;/p&gt;&#xA;" OwnerUserId="18855" LastActivityDate="2019-01-27T00:57:09.063" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29345" PostTypeId="2" ParentId="29341" CreationDate="2019-01-27T01:05:43.740" Score="6" Body="&lt;p&gt;Well, they are different shapes, but we can get a rough answer by doing a ratio calculation based on their radii.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; be the radius of the galaxy, about &lt;span class=&quot;math-container&quot;&gt;$5\times 10^{20}\ \mathrm m$&lt;/span&gt;.&lt;br&gt;&#xA;Let &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt; be the radius of the Earth, about &lt;span class=&quot;math-container&quot;&gt;$6371\ \mathrm{km}$&lt;/span&gt;.&lt;br&gt;&#xA;Let &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; be the radius of the scaled-down Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac e g = \frac s e$$&lt;/span&gt;&#xA;So &lt;span class=&quot;math-container&quot;&gt;$$s = \frac{e^2} g = \frac {(6371\ \mathrm{km})^2} {5\times 10^{20}\ \mathrm m}$$&lt;/span&gt;&#xA;which is approximately 80 nanometres. So it's about 1000 times the size of an atom.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorDisplayName="user8021" LastEditDate="2019-01-27T13:02:21.223" LastActivityDate="2019-01-27T13:02:21.223" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29346" PostTypeId="1" CreationDate="2019-01-27T03:49:51.557" Score="5" ViewCount="172" Body="&lt;p&gt;I am teaching myself astronomy and I'm working on a basic problem regarding time dilation around black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is basically this: if a neutron were ejected from a nucleus at a distance of 3km from a 1 solar mass black hole, how long would it appear to take for the neutron to decay (the assumption here is that the neutron would decay in 15 minutes... right on the half-life) to an outside observer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first thought was that we'd never see it decay, because 3km is the Schwarzschild radius of a BH of that mass, so the frequency would basically be &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; and we'd never observe the decay event.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The text notes the time dilation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\frac{\Delta t'}{\Delta t_\text{obs}} = \frac{\nu_f}{\nu_i}$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\Delta t'$&lt;/span&gt; is the the time difference between two events at the source and &lt;span class=&quot;math-container&quot;&gt;$\Delta t_{obs}$&lt;/span&gt; at the observer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can use the relativistic red shift equation to get the wavelength difference (9.2 in my case using 1 solar mass and a radius of 3km) so that wouldn't be infinite, just big. For a 15 min &lt;span class=&quot;math-container&quot;&gt;$\Delta t'$&lt;/span&gt; then, the &lt;span class=&quot;math-container&quot;&gt;$\Delta t_{obs}$&lt;/span&gt; would be 15 min * 9.2, yeah?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my two answers obviously contradict each other. What am I missing in my approach here?&lt;/p&gt;&#xA;" OwnerUserId="25705" LastActivityDate="2020-02-27T03:03:45.463" Title="Time Dilation and Particle Decay" Tags="&lt;black-hole&gt;&lt;general-relativity&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29348" PostTypeId="2" ParentId="29342" CreationDate="2019-01-27T04:39:22.183" Score="3" Body="&lt;p&gt;The mass will pull into the shape of the sphere if the body is massive enough. The body will then basically be in hydrostatic equilibrium. A planet like Naboo couldn't exist b/c the denser materials would move to the centre, and the lighter elements would then form an atmosphere around the solid core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plasmas require really high temps and a terestrial plantet couldn't compress gases to the temps required to completely ionize the gas. That and what I said about the way the planet would form. There's just no way that a shell of rocky material could be supported by gas pressure in that way. The whole thing would be really unstable.&lt;/p&gt;&#xA;" OwnerUserId="25705" LastActivityDate="2019-01-27T04:39:22.183" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29350" PostTypeId="1" CreationDate="2019-01-27T06:44:01.477" Score="0" ViewCount="98" Body="&lt;p&gt;There is no edge to the atmosphere, and therefor I expect there would be a gradual reduction of concentrations of bacteria as we ascend into space. Logically anaerobic bacteria should be able to obtain its energy from the sun and nutrients from whatever scarce but available organic floating materials. Cosmic rays and high energy particles could damage DNA, but there might be someway life could protect itself against them. Could it be that bacteria is evolving/ has already evolved at the edge of the atmosphere to live in space?&lt;/p&gt;&#xA;" OwnerUserId="25707" LastActivityDate="2019-01-28T19:07:49.487" Title="Could some bacteria theoretically evolve to survive in space?" Tags="&lt;life&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29351" PostTypeId="2" ParentId="29350" CreationDate="2019-01-27T08:32:14.320" Score="0" Body="&lt;p&gt;Some bacteria already have evolved to survive in space. They merged with other bacteria (to form eukaryotic cells), formed large structured colonies (or &quot;organisms&quot;) with developed &quot;brains&quot; which were able to design and build rockets and space stations. It took rather more than a billion years of evolution to go from a bacteria to space dwelling humans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space is very hostile environment, bacteria exposed directly to space will eventually die. They could remain viable for some time, but in the absence of liquid water, growth is not possible. It is conceivable that bacterial spores could survive for an extended period, if embedded in rock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We cannot answer this question, as nothing that we know of has evolved in a vacuum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Survive? Yes. Thrive? No.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2019-01-27T18:16:22.230" LastActivityDate="2019-01-27T18:16:22.230" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29352" PostTypeId="2" ParentId="29343" CreationDate="2019-01-27T08:38:07.840" Score="4" Body="&lt;ul&gt;&#xA;&lt;li&gt;The radius of the observable Universe is 41.5billion light years (&lt;a href=&quot;https://en.wikipedia.org/wiki/Observable_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;ref&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;li&gt;A light year is &lt;span class=&quot;math-container&quot;&gt;$9.5\cdot10^{12}$&lt;/span&gt; km.&lt;/li&gt;&#xA;&lt;li&gt;The radius of the Earth is 6371 km.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the answer is &lt;span class=&quot;math-container&quot;&gt;$6371 km \cdot \frac{6371 km}{41.5\cdot10^9\cdot9.5\cdot10^{12}km} = 1.03\cdot10^{-16}km = \underline{\underline{1.03\cdot10^{-13}m}}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of the atoms are &lt;span class=&quot;math-container&quot;&gt;$\approx 10^{-10}m$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="1891" LastEditDate="2019-01-27T09:14:23.753" LastActivityDate="2019-01-27T09:14:23.753" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29353" PostTypeId="1" CreationDate="2019-01-27T10:52:04.687" Score="1" ViewCount="86" Body="&lt;p&gt;Constellations are dominant features of night sky. But these are basically group of stars that forms particular shapes when viewed from Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much the shape of a constellation will change if we observe them from another planet, say, Mars? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no significant change be visible, then how much distance we need to travel to observe a change? &lt;/p&gt;&#xA;" OwnerUserId="9213" LastActivityDate="2019-01-27T15:12:10.487" Title="Will the shape of the constellations change if we observe from another planet, say Mars?" Tags="&lt;mars&gt;&lt;constellations&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2019-01-28T05:13:02.117" ContentLicense="CC BY-SA 4.0" />
  <row Id="29355" PostTypeId="2" ParentId="29350" CreationDate="2019-01-27T12:58:50.310" Score="2" Body="&lt;p&gt;Currently there is no life that we know of that evolved in, or can survive for prolonged time in a vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To say 'Life in space is possible because there is anaerobic life' is a completely wrong point, that ignores the effects of vacuum on life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sure, there are &lt;a href=&quot;https://www.newscientist.com/article/dn14690-water-bears-are-first-animal-to-survive-space-vacuum/&quot; rel=&quot;nofollow noreferrer&quot;&gt;extremophiles&lt;/a&gt; that can survive for a limited time in space (see also &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_microorganisms_tested_in_outer_space&quot; rel=&quot;nofollow noreferrer&quot;&gt;this list&lt;/a&gt;) when going into hibernation modes. However all life as we know it needs liquids and/or gases to survive. Those would evaporate once the cell/water bear would become active again. Activity is necessary to sustain and multiply life, so no, nothing that we know of can stay active and alive in space.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cosmic rays and high energy particles could damage DNA, but there might be someway life could protect itself against them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Until life comes up with such mechanisms, we just don't know. The problems with liquids and gases evaporating would need some tougher solutions even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bacteria are also pretty heavy 'particles' and therefore cannot float anywhere near the &lt;a href=&quot;https://en.wikipedia.org/wiki/K%C3%A1rm%C3%A1n_line&quot; rel=&quot;nofollow noreferrer&quot;&gt;'edge of space'.&lt;/a&gt; There are &lt;a href=&quot;https://www.sciencedaily.com/releases/2009/03/090318094642.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;reports&lt;/a&gt; of bacteria in the stratosphere, but that requires already strong upwinds to put them there.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2019-01-27T12:58:50.310" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29356" PostTypeId="2" ParentId="24600" CreationDate="2019-01-27T13:10:27.733" Score="2" Body="&lt;p&gt;We can see at least some of the stars on the edge of the Milky Way. With Gaia, we now have not only the position, but also motion data for 1.7 billion stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means we can see e.g. &lt;a href=&quot;https://www.cam.ac.uk/research/news/gaia-spots-a-ghost-galaxy-next-door&quot; rel=&quot;nofollow noreferrer&quot;&gt;small clusters like Ant 2 hiding behind the Milky Way&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Perhaps the most valuable of these is the wealth of high-quality all-sky proper motion (PM) information. Gaia’s astrometry makes it possible to filter out nearby contaminating populations, revealing the distant halo behind them. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is Ant 2, which can be separated from the foreground stars thanks to the proper motion data supplied by Gaia. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HkJum.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HkJum.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gaia was used to study the &lt;a href=&quot;http://sci.esa.int/gaia/60186-gaia-s-surprising-discoveries-scrutinising-the-milky-way/&quot; rel=&quot;nofollow noreferrer&quot;&gt;halo of stars around the Milky Way&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Amina Helmi and colleagues published a study titled “A box full of chocolates” in which they began to reveal the rich structure of the Galaxy’s stellar halo. The halo is the cloud of stars that surrounds the main disc of the Milky Way. Its structure reveals the formation processes of the Galaxy. The investigation of this structure was one of the prime motivations for building Gaia in the first place. An immediate surprise was the discovery of a large population of stars that orbit the Milky Way in the opposite direction to most stars. These `ghost rider’ stars probably came from a smaller galaxy that was cannibalized by the Milky Way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gaia was designed to &lt;a href=&quot;http://www.esa.int/Our_Activities/Space_Science/Gaia/Frequently_Asked_Questions_about_Gaia&quot; rel=&quot;nofollow noreferrer&quot;&gt;map all visible stars down to magnitude 20.5&lt;/a&gt;, with an initial estimate of 1 billion stars. M 20.5 is pretty dim, but an even more sensitive camera would be able to see more stars. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;During its anticipated lifetime of five years, Gaia will observe each of its one billion sources about 70 times, resulting in a record of the brightness and position of each source over time. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By Data Release 2, Gaia has already &lt;a href=&quot;http://sci.esa.int/gaia/60186-gaia-s-surprising-discoveries-scrutinising-the-milky-way/&quot; rel=&quot;nofollow noreferrer&quot;&gt;mapped 1.7 billion stars&lt;/a&gt; and has proper motion for 1.3 billion. &lt;/p&gt;&#xA;" OwnerUserId="361" LastActivityDate="2019-01-27T13:10:27.733" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29357" PostTypeId="1" AcceptedAnswerId="29364" CreationDate="2019-01-27T13:22:52.667" Score="0" ViewCount="752" Body="&lt;p&gt;I'm measuring equivalent widths of absorption lines using a spectrum of a star. I make two or three measurements of each line by making reasonable gaussian fits of the line with IRAF's splot tool. Then I calculate the mean of the measurements, which serves as my final equivalent width estimate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a good way of estimating the uncertainty of this measurement?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;My current method&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I'm currently using half of the range for the uncertainty. For example, if I made two measurements 10 and 16 mA (milliangstrom), then the mean is 13 mA and uncertainty is 3 mA. This gives the estimate of equivalent width to be 13±3 mA. Do you see any problems with this method of estimating uncertainty?&lt;/p&gt;&#xA;" OwnerUserId="21367" LastEditorUserId="21367" LastEditDate="2019-01-27T23:50:21.003" LastActivityDate="2019-01-27T23:50:21.003" Title="How to estimate uncertainty of measurements of equivalent widths?" Tags="&lt;spectroscopy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29358" PostTypeId="2" ParentId="29353" CreationDate="2019-01-27T14:59:00.090" Score="4" Body="&lt;p&gt;There will be no change observable by naked eye from any planet in the Solar System.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to estimate the distances for which change is visible.  There are two key points. First, changes in the constellations are due to parallax.  Second, by definition, a baseline of 2 AU (astronomical units) produces a parallax of 1 second of arc at a parsec's distance (wihc turns out to be about 3 light years) and half an arc second at two parsecs, and so forth.  Double the displacement to four AU and you double the parallax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nearest visible star to the Sun (Alpha Centauri) is just over a parsec away.  For it to move a barely noticeable distance (say a tenth of a degree or 1/5 the diameter of the Moon) in the night sky, the baseline would have to be 360 times as long, since 0.1 degree is 360 arc seconds.  So the closest star in the night sky would have moved a just-perceptible amount if observed from 500 AU away from Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The brighter stars which make up the constellations range from the distance of Alpha Centauri out to thousands of light years, but many of them are within 100 light years (30 parsecs).  While a movement of 0.1 degree is perceptible, for the constellations to be obviously distorted, you'd probably want many stars to move by more than a degree (two full Moons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: 30 parsecs and 1 degree requires a movement of 1AU * 30 * 3600 (arc second/degree) or about 100,000 AU.  That's about 1.5 light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you voyaged away from the Sun, a keen observer would notice an slight effect somewhere out past the Kuiper Belt, and a few stars would be massively displaced by the time you were out in the Oort cloud, but for wholesale disruption of the constellations, you'd have to go about a light year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two points: First, at a light year's distance from the Sun, the constellations would still be somewhat recognizable, though there'd be no doubt at all that things were scrambled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, the shifts would be largest in the direction &lt;em&gt;perpendicular&lt;/em&gt; to your line of travel (the baseline for parallax) and much smaller fore and aft.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2019-01-27T15:12:10.487" LastActivityDate="2019-01-27T15:12:10.487" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29359" PostTypeId="1" AcceptedAnswerId="29365" CreationDate="2019-01-27T18:39:20.057" Score="9" ViewCount="384" Body="&lt;p&gt;I came across post in quora where somebody said that &quot;most astronomers no longer think Planet 9 exists&quot; and being somewhat shocked by that, I did a google search and found a &lt;a href=&quot;https://www.forbes.com/sites/startswithabang/2018/09/14/this-is-why-scientists-think-planet-nine-doesnt-exist/#5ba235c76e77&quot; rel=&quot;noreferrer&quot;&gt;fairly recent article that claims exactly that&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can take several years to prove a negative within pretty good certainty, as opposed to a discovery which can be checked pretty quickly.   Because I don't want to wait several years, I wanted to ask if the general consensus is that Planet 9 probably doesn't exist?  Or is there still a respectable amount of belief among astronomers in the original evidence and any new evidence that's followed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it ends up not existing, I'll be sorry to hear it.   It was a fun story when it lasted.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2019-12-16T20:20:58.320" Title="3 years later, how strong is the evidence for planet 9?" Tags="&lt;planet&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29360" PostTypeId="2" ParentId="29357" CreationDate="2019-01-27T18:44:32.830" Score="1" Body="&lt;p&gt;Estimating uncertainty... the most preferred one would be Bayesian. However, if you follow the frequentist, MCMC is the most preferred, and it would be somewhat similar to what your advisor suggested with more complicated algorithm. The simplest method would be what your advisor suggested, but you can measure for a larger sample size and use simple statistics like mean and standard error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various variations along these lines that I would not make a list here.&lt;/p&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2019-01-27T18:44:32.830" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29361" PostTypeId="1" AcceptedAnswerId="29362" CreationDate="2019-01-27T19:51:07.533" Score="3" ViewCount="632" Body="&lt;p&gt;Does water on Mars form from the atmosphere as ice free from salt?&#xA;Does water desalinize through the evaporation and transpiration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the fresh top ice is placed in a container then pressurized to melt, would the water be carbonated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Particulates of salt and dust is less in the atmosphere during a Martian winter so would the ice that forms be also drinkably cleaner?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://space.stackexchange.com/questions/33884/how-to-drink-water-on-mars&quot;&gt;https://space.stackexchange.com/questions/33884/how-to-drink-water-on-mars&lt;/a&gt;&#xA;&lt;a href=&quot;https://earthscience.stackexchange.com/questions/16076/how-salty-can-ice-be&quot;&gt;https://earthscience.stackexchange.com/questions/16076/how-salty-can-ice-be&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gwI9m.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gwI9m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://kottke.org/18/12/a-massive-ice-filled-crater-on-mars&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://kottke.org/18/12/a-massive-ice-filled-crater-on-mars&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2019-01-27T20:45:00.247" LastActivityDate="2019-01-27T20:45:00.247" Title="Would drinking melted ice water from Mars taste like sparkling water?" Tags="&lt;mars&gt;&lt;water&gt;&lt;weather&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29362" PostTypeId="2" ParentId="29361" CreationDate="2019-01-27T20:02:41.847" Score="8" Body="&lt;p&gt;Any ice that forms from water will be free of salt -- the process of crystallization does an excellent job of removing impurities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But&lt;/em&gt; that's only when freezing out of liquid water, which is not a process that happens today at the Martian surface. When ice forms out of the air, it forms as tiny crystals -- &quot;hoar frost&quot; we call it on Earth -- and forms &lt;em&gt;on&lt;/em&gt; substrates, such as grains of sand.  Since the soil of Mars is laden with salts of various sorts (including things a lot nastier than NaCl, such as perchlorate salts), it's hard to believe that the Martian hoar frost wouldn't be contaminated with salty sand and dust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, of course, the Martian atmosphere has dust suspended in it which is continually precipitating out and dust which is blown up by the not-infrequent dust storms and dust devils and all of this falls on any newly-formed frost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very hard to see how there could be any macroscopic ice deposits on Mars that don't have dust and salt as impurities.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2019-01-27T20:02:41.847" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29363" PostTypeId="1" AcceptedAnswerId="29695" CreationDate="2019-01-27T21:08:18.067" Score="3" ViewCount="188" Body="&lt;p&gt;Answers to and comments on &lt;a href=&quot;https://space.stackexchange.com/q/33884/12102&quot;&gt;How to drink water on Mars?&lt;/a&gt; suggest that one would not want to drink melted water ice obtained from exposed surface deposits near the poles because it may contain various salts and other dangerous chemicals (perchlorate salts are one particularly nasty suspected component of mars soil and it's water content).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars' atmosphere contains dust even when there are no dust storms, and in &lt;a href=&quot;https://astronomy.stackexchange.com/a/10772/7982&quot;&gt;this answer&lt;/a&gt; it is explained to be the cause of the reddish color of the Martian sky as has been photographed by several Mars landers and rovers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any measurements or even models or predictions on how much of the particulate matter in Mars' atmosphere could be tiny particles of salt, or salt mixed with other minerals? Put more simply, &lt;em&gt;how salty might Mars' atmosphere be?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Hy7dC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hy7dC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Martian sunset (credit: &lt;a href=&quot;https://mars.nasa.gov/resources/5300/a-martian-sunset/?site=insight&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA/JPL-Caltech/Texas A&amp;amp;M/Cornell&lt;/a&gt;) borrowed from &lt;a href=&quot;https://astronomy.stackexchange.com/a/10772/7982&quot;&gt;this answer&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="7982" LastEditDate="2019-01-28T03:08:05.500" LastActivityDate="2019-03-02T17:57:04.540" Title="Any information on the salt content of Mars' atmospheric dust?" Tags="&lt;mars&gt;&lt;water&gt;&lt;weather&gt;&lt;planetary-atmosphere&gt;" AnswerCount="1" CommentCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="29364" PostTypeId="2" ParentId="29357" CreationDate="2019-01-27T21:39:01.550" Score="2" Body="&lt;p&gt;Yes there is a problem. You seem to be trying to derive an uncertainty in the measurement of EW by doing repeated measurements of the &lt;em&gt;same&lt;/em&gt; data?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can only give you the uncertainty associated with your measurement technique (i.e. where you define the limits of the line and how you set the continuum level) - the systematic error you might call it (although there can be other systematic errors inherent to EW measurements, like whether you subtracted the sky or scattered light in your spectrograph correctly for example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What it does not do is evaluate the uncertainty in the EW caused by the quality or signal-to-noise ratio of the data itself. You might assess this using some rule-of-thumb formulae for a Gaussian line, e.g.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\Delta EW \simeq 1.5 \frac{\sqrt{fp}}{{\rm SNR}},$$&lt;/span&gt;&#xA;(eqn 6 of &lt;a href=&quot;http://adsabs.harvard.edu/abs/1988IAUS..132..345C&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cayrel de Strobel 1988&lt;/a&gt;) where &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt; is the FWHM of the spectral line (in wavelength units), &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; is the size of one pixel in wavelength units and SNR is the signal-to-noise ratio of the data in an average pixel. Or you could take a synthetic spectrum and add some artificial noise to it with the appropriate properties and measure the EW of several randomisations of the same spectrum, taking the standard deviation of your EW measurements to indicate the EW uncertainty for a particular level of signal-to-noise ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this statistical uncertainty is not negligible, then you would then need to add it to any systematic uncertainties associated with your analysis of the spectrum. As far as the latter is concerned then your suggested method does give some indication of what that error might be, though I suspect it will overestimate the 1-sigma uncertainty. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-01-27T21:39:01.550" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29365" PostTypeId="2" ParentId="29359" CreationDate="2019-01-27T23:15:42.490" Score="9" Body="&lt;p&gt;It's worth noting that Siegel's article seems to extrapolate from &lt;em&gt;one paper&lt;/em&gt; (&lt;a href=&quot;https://arxiv.org/abs/1706.05348&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shankman et al., 2017&lt;/a&gt;) to &quot;most scientists&quot; without much justification, so it's not particularly clear what most scientists think anyway (and that's before we get into the question of whether the &quot;most scientists&quot; referred to includes scientists whose discipline is not astronomy-related). Usually in these kind of situations there's a lot of back-and-forth as new objects are found in the outer solar system and the analyses of systematic errors and biases in the various surveys are refined. If there is an actual detection of Planet&amp;nbsp;9, that would resolve things fairly definitively, the reverse situation is going to be rather more difficult to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In response to the OSSOS paper (Shankman et al., 2017) Michael Brown and Konstantin Batygin released a new analysis of the clustering in the outer solar system, taking into account the OSSOS results that are the subject of Ethan Siegel's article. The paper can be &lt;a href=&quot;https://arxiv.org/abs/1901.07115&quot; rel=&quot;nofollow noreferrer&quot;&gt;downloaded from the arXiv&lt;/a&gt;. Brown also has a &lt;a href=&quot;https://twitter.com/plutokiller/status/1087495838379651072&quot; rel=&quot;nofollow noreferrer&quot;&gt;Twitter thread summarising the result&lt;/a&gt;. From the paper, their conclusion on the apparent absence of clustering in OSSOS:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;That is,&#xA;  the uncertainties in the measurement of clustering from&#xA;  the OSSOS data are so large that OSSOS would not be&#xA;  capable of confidently detecting the clustering seen in&#xA;  the larger data set even if it were real and present in the&#xA;  OSSOS data. Because of the limited survey region and&#xA;  small number of detected objects, OSSOS observations&#xA;  are equally consistent with being drawn from a uniform&#xA;  distribution of longitudes of perihelion and with being&#xA;  clustered in longitude of perihelion as strongly as seen&#xA;  in the ensemble data. No conclusions on clustering of&#xA;  longitude of perihelion observed in the complete dataset&#xA;  can be drawn from the OSSOS data.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They do note that this does not necessarily prove the existence of Planet&amp;nbsp;9, but in the absence of Planet&amp;nbsp;9 there would still need to be an explanation for the clustering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Batygin et al. subsequently posted &lt;a href=&quot;https://arxiv.org/abs/1902.10103&quot; rel=&quot;nofollow noreferrer&quot;&gt;a review of the Planet Nine hypothesis&lt;/a&gt; on the arXiv, favouring a lower planetary mass and a less eccentric orbit than originally proposed. The new parameters do not favour Planet&amp;nbsp;9 as an explanation for the solar obliquity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the anti-Planet&amp;nbsp;9 side of things, it's worth looking into &lt;a href=&quot;https://arxiv.org/abs/1909.09673&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kavelaars et al. 2019&lt;/a&gt;, a subsequent paper from the same authors as the study based on the OSSOS data, who note that there seems to be an absence of objects with perihelia (&lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt;) between 50 and 75&amp;nbsp;AU that would be expected if a planet was causing the alignment, which so far doesn't seem to be explainable by detection bias.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A survey&#xA;  that might have detected a TNO at &lt;span class=&quot;math-container&quot;&gt;$a ∼ 500\ \mathrm{au}$&lt;/span&gt;&#xA;  and &lt;span class=&quot;math-container&quot;&gt;$q ∼ 75\ \mathrm{au}$&lt;/span&gt; is actually more likely to have detected objects with similar &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; but smaller values&#xA;  of &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt;. The same is true of the other &lt;span class=&quot;math-container&quot;&gt;$q &amp;gt; 75\mathrm{au}$&lt;/span&gt;&#xA;  detections: the lower-&lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; but similar &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; detections&#xA;  are always more likely. Thus, the lack of detections in the &lt;span class=&quot;math-container&quot;&gt;$50 &amp;lt; q &amp;lt; 75\ \mathrm{au}$&lt;/span&gt; range may be indicating that there really is an absence of TNOs&#xA;  on orbits in this range. This strongly contradicts models of orbital evolution that include&#xA;  an additional planet, as the gravitational action of such an object would cause TNOs to&#xA;  be distributed across a range of &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; values at any&#xA;  given moment (Figure 2; Shankman et al. 2017a;&#xA;  Lawler et al. 2017). Thus, if the lack of objects&#xA;  in the &lt;span class=&quot;math-container&quot;&gt;$50\ \mathrm{au} &amp;lt; q &amp;lt; 70\ \mathrm{au}$&lt;/span&gt; range is real, the&#xA;  hypothesized external planet can be excluded.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It still is an active area of research, so it remains to be seen how the various studies will hold up.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-12-16T20:20:58.320" LastActivityDate="2019-12-16T20:20:58.320" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29366" PostTypeId="1" AcceptedAnswerId="29394" CreationDate="2019-01-28T00:58:27.267" Score="3" ViewCount="214" Body="&lt;p&gt;I'm just curious about that the existence of non-rotating stars. Is it physically possible for a star not to rotate at all? Does magnetic braking eventually stop the stellar rotation?&lt;/p&gt;&#xA;" OwnerUserId="24683" LastEditorUserId="24683" LastEditDate="2019-01-28T01:05:10.670" LastActivityDate="2019-01-30T02:51:10.870" Title="Is it possible for stars not to rotate?" Tags="&lt;star&gt;&lt;rotation&gt;&lt;stellar-astrophysics&gt;&lt;stellar-dynamics&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29368" PostTypeId="1" CreationDate="2019-01-28T07:49:40.243" Score="0" ViewCount="129" Body="&lt;p&gt;What’s the closest earth like planet or moon to earth we have discovered in our galaxy? &lt;/p&gt;&#xA;" OwnerUserId="25714" LastEditorUserId="7472" LastEditDate="2019-01-30T10:56:17.717" LastActivityDate="2019-01-30T10:56:17.717" Title="What’s the closest earth like planet we have discovered in our galaxy?" Tags="&lt;exoplanet&gt;&lt;milky-way&gt;&lt;earth-like-planet&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2019-01-30T10:56:23.727" ContentLicense="CC BY-SA 4.0" />
  <row Id="29370" PostTypeId="1" CreationDate="2019-01-28T13:17:45.280" Score="1" ViewCount="85" Body="&lt;p&gt;I was trying to help my son with a project and thought it shouldn't be hard but turned out I'm not as smart as I thought I was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His project is about celestial navigation. The first step is, given the J2000 ra/dec of a specific star and a UTC time of observation first find the geographical lat/lon ground point (substellar point) for the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My assumption was lat = dec and lon was simply the ra - sidereal time from UTC time. So we knocked up a simple program to perform this. Then we used Stellarium (with refraction off for now) to test the output. And the star is indeed overhead but it is always off by a number of minutes. I suspected precession but that seems to be measured in arcseconds per year?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now I realise my simple assumption isn't correct, only approximate. Could someone more knowledgeable please fill the gap in my knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="25720" LastEditorUserId="25720" LastEditDate="2019-01-28T15:16:32.607" LastActivityDate="2019-01-28T15:16:32.607" Title="Geographical Substellar position calculation from RA/DEC" Tags="&lt;star&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29372" PostTypeId="2" ParentId="28068" CreationDate="2019-01-28T19:05:17.723" Score="2" Body="&lt;p&gt;I came across an article in the &lt;a href=&quot;https://www.noao.edu/noao/noaonews/oct18/118news.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;NOAO 2018 October Newsletter&lt;/a&gt; which discussed recoating of the 4-meter Blanco primary in Chile. This contained before, after and theoretical reflectivity data. I contacted one of the authors of the article for the source of the theoretical data who sent me a scan of the pages from R. N. Wilson's &quot;Reflecting Telescope Optics II&quot;, page 426. I transcribed the data from the table for Alumin(i)um (Al) and Silver (Ag) and below is a plot of the Al and Ag data in the visible range:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jYiIc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jYiIc.png&quot; alt=&quot;Reflectivity of Aluminum (Al) and Silver (Ag) mirrors&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I've made more plots and the data files available in a &lt;a href=&quot;https://adventuresinprecision.space/2019/01/28/mirror-reflectivity-data/&quot; rel=&quot;nofollow noreferrer&quot;&gt;post&lt;/a&gt; on my blog)&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2019-01-28T19:05:17.723" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29373" PostTypeId="2" ParentId="29350" CreationDate="2019-01-28T19:07:49.487" Score="-1" Body="&lt;p&gt;Given that we've found bacteria thriving inside &quot;solid&quot; rock a couple km straight down, living on local sulfur and heat, there's no overriding reason that something couldn't evolve on some rock (planet) of sufficient size to hold some heat in, but without any need for an atmosphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were specifically asking about free-floating life that survives entirely on solar irradiance, then no, but more because of a complete lack of various elements (let alone long organic chains) in most of space.   Veering off into world-building, imagine my intra-rock bacteria slowly growing to the surface, forming a protective layer of dead relatives which absorbs irradiance but seals against vaccuum,  and sometime later a volcanic eruption launches the entire magillah into the vacuum (and the protective layer manages to envelope the entire colony).  Highly improbable.  You'll need The Heart Of Gold here. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2019-01-28T19:07:49.487" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29374" PostTypeId="2" ParentId="1291" CreationDate="2019-01-28T20:04:57.733" Score="2" Body="&lt;p&gt;The fastest celestial body we know so far is neutron star XTE J1739-285. It rotates 1122 times per second. Source: &lt;a href=&quot;https://arxiv.org/abs/0712.4310?fbclid=IwAR2g9RYhybbBlxxzKOfquBa9C33H8mMwuZ_mHZbm1RbcsetrxO3vbvXq7gA&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/0712.4310?fbclid=IwAR2g9RYhybbBlxxzKOfquBa9C33H8mMwuZ_mHZbm1RbcsetrxO3vbvXq7gA&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25724" LastActivityDate="2019-01-28T20:04:57.733" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29376" PostTypeId="1" AcceptedAnswerId="29383" CreationDate="2019-01-29T03:02:20.180" Score="1" ViewCount="254" Body="&lt;p&gt;I have been wondering for a while now about a scenario of where if you combined Venus &amp;amp; Mercury would it be habitable if located in the goldy locks zone? I have found in my little research that Venus doesn't have an internally generated magnetic field. But Mercury does indeed have an internally generated magnetic field. Though weaker than Earths of course. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really know what other factors for making it to whether or not this new plant would be able to sustain life. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is, would the new planet be habitable, or are there more factors needed to determine this?&lt;/p&gt;&#xA;" OwnerUserId="25727" LastActivityDate="2019-01-29T13:17:38.383" Title="If Mercury and Venus where to collide would it make a habitable planet?" Tags="&lt;solar-system&gt;&lt;venus&gt;&lt;mercury&gt;&lt;habitable-zone&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2019-02-02T02:18:17.703" ContentLicense="CC BY-SA 4.0" />
  <row Id="29379" PostTypeId="1" CreationDate="2019-01-29T11:23:21.837" Score="1" ViewCount="29" Body="&lt;p&gt;Planets like Saturn &amp;amp; Jupiter are called &quot;gas giants&quot;--does that mean that there is no surface to walk upon? Though some source of gravity is required to hold the gas in place, isn't it?&#xA;Venus: we are told that it rains sulphuric acid, there. Given the reported surface-temp. of 450C., wouldn't these &quot;acid drops&quot; vaporise long before they reached the ground--producing vapour-clouds, in the upper atmosphere?&#xA;If a person could stand on the surface of Venus, what would he see--would it be just like our moon?&lt;/p&gt;&#xA;" OwnerUserId="25734" LastActivityDate="2019-01-29T11:23:21.837" Title="Surface Conditions" Tags="&lt;planet&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2020-02-26T21:24:43.060" ContentLicense="CC BY-SA 4.0" />
  <row Id="29380" PostTypeId="1" AcceptedAnswerId="29448" CreationDate="2019-01-29T12:41:30.097" Score="5" ViewCount="203" Body="&lt;p&gt;Is it theoretically possible for a  moon to orbit a planet (let us say Earth size, mass and solar distance) in such a way as to create a planet-wide solar eclipse lasting several weeks?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are my assumptions (some of them likely flawed) about the circumstances under which this might be possible:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;To simplify the problem, let us assume that there is no planetary axial tilt and that the moon has an equatorial orbit.&lt;/li&gt;&#xA;&lt;li&gt;The orbital speed of the moon would need to be quite slow in order to stay in position for long enough to create a graduating eclipse over several weeks.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In order to create a planet wide eclipse, the moon would need to be quite large and/or relatively close to the planet. Is this circumstance feasible if the mass of the object were low compared to its size?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How stable could the described relationship be?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thank you for your interest and answers.&lt;/p&gt;&#xA;" OwnerUserId="25735" LastEditorUserId="25735" LastEditDate="2019-01-31T07:28:48.253" LastActivityDate="2019-02-03T13:04:07.840" Title="Could a moon orbit an Earth size planet in such a way as to create a solar eclipse lasting for several weeks from start to finish?" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;solar-eclipse&gt;&lt;earth-like-planet&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29381" PostTypeId="1" CreationDate="2019-01-29T12:54:51.060" Score="16" ViewCount="1954" Body="&lt;p&gt;I have an interest in cosmology, but I would no way consider myself an expert. It would be great if the more knowledgeable contributors would be able to answer the question I have posed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intense gravitational field near a black hole causes time dilation. This means, to an observer close to its event horizon, time  runs more slowly. It can mathematical be expressed as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time dilation = &lt;span class=&quot;math-container&quot;&gt;$\frac{\delta t_f}{\delta t_0} = \frac{1}{\sqrt{1-\frac{R_s}{R}}}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$\delta t_f$&lt;/span&gt; is an interval of time a long distance from the black hole, &lt;span class=&quot;math-container&quot;&gt;$\delta t_0$&lt;/span&gt; is the time interval a distance &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; from the black hole’s centre and &lt;span class=&quot;math-container&quot;&gt;$R_s$&lt;/span&gt; is the radius of the event horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula means that time dilation becomes &lt;strong&gt;infinite&lt;/strong&gt; at the event horizon so to a distant observer a falling object would take an infinite time to reach the event horizon and would never cross it.  However, in the frame of an observer falling into the black hole it should them take a finite time to reach the event horizon. To me the complexity is that all black holes are believed to &lt;strong&gt;‘evapourate’ and eventually disappear in a finite time&lt;/strong&gt;, by emitting Hawking radiation. Although this an incredibly slow process. For example, a solar mass hole would have a lifetime of &lt;span class=&quot;math-container&quot;&gt;$10^{66}$&lt;/span&gt; years (in a time-frame measured a distance from the black hole).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Question&lt;/strong&gt;&#xA;Assuming Hawking radiation exists and  black hole evaporation occurs, does this mean that an infalling observer would never actually reach the event horizon, because the black hole has finite lifetime and would evaporate before an infalling observer would reach it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Further Note&lt;/strong&gt;&#xA;Since posing this question I have had one further thought, which may or may be relevant, since my knowledge of GR and quantum theory is pretty sketchy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way that the time dilation formula works is that when the observer is a distance of one Planck length ( ~ &lt;span class=&quot;math-container&quot;&gt;$10^{-35}$&lt;/span&gt; m ) away from the event horizon of a solar mass black hole (which has radius ~3km), the time dilation would be still only be  ~ &lt;span class=&quot;math-container&quot;&gt;$10^{19}$&lt;/span&gt;. However, the lifetime of a solar mass black hole is ~&lt;span class=&quot;math-container&quot;&gt;$10^{66}$&lt;/span&gt; years. Therefore, even when they are within 1 Planck length of the event horizon, the time dilation near the black hole is not sufficient for the observer to see the black hole evaporate. &lt;/p&gt;&#xA;" OwnerUserId="25718" LastEditorUserId="25718" LastEditDate="2019-02-01T14:27:48.953" LastActivityDate="2019-02-01T14:27:48.953" Title="Infalling observer could never cross Black Hole event horizon?" Tags="&lt;black-hole&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" ClosedDate="2019-01-31T06:22:19.130" ContentLicense="CC BY-SA 4.0" />
  <row Id="29383" PostTypeId="2" ParentId="29376" CreationDate="2019-01-29T13:17:38.383" Score="1" Body="&lt;p&gt;The answer is &quot;Probably, but not very quickly.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy of a collision comes from two sources, the kinetic energy of the collision and the gravitational energy. If Mercury were to impact Venus, even if it began by creeping up on it very, very slowly so as to have as little kinetic energy as possible, it would fall into Venus's gravitational well and release gravitational energy enough to melt both bodies.  The resulting mess would take &lt;em&gt;ca.&lt;/em&gt; 10-100 million years for a crust to resolidify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting body &lt;em&gt;could&lt;/em&gt; evolve towards habitability.  One theory of planetary development says that active plate tectonics are critical to habitability.  We know that new rock absorbs large amounts of CO&lt;sub&gt;2&lt;/sub&gt; -- new volcanic rock in Indonesia has been estimated to absorb a quarter of the CO&lt;sub&gt;2&lt;/sub&gt; released each year.  So the continuous exposure of new rock by plate tectonics may be a major force in preventing the runaway greenhouse effect that has made Venus uninhabitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't know for sure how plate tectonics starts or ends, but if the resulting planet had a particularly vigorous case of it, this might compensate for its closeness to the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the down side, Venus is already short of hydrogen and thus water and a collision with Mercury would eject Venusian volatiles into space and Mercury would not add any new, so the resulting planet might be rather barren.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any event, whatever happens, it's going to take a very long time for the resulting body to cool enough to be livable.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2019-01-29T13:17:38.383" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29385" PostTypeId="2" ParentId="29381" CreationDate="2019-01-29T13:52:59.747" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The intense gravitational field near a black hole causes time dilation. This means, to an observer close to its event horizon, time  runs more slowly. It can mathematical be expressed as:&lt;br&gt;&#xA;  Time dilation &lt;span class=&quot;math-container&quot;&gt;$= \frac{\Delta t_f}{\Delta t_0} = \frac{1}{\sqrt{1-\frac{R_s}{R}}}$&lt;/span&gt;... &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Actually, time dilation is associated with gravitational &lt;em&gt;potential&lt;/em&gt; rather than field. The strength of the field at some elevation is proportional to the gradient in the potential at that location. This is important. If there's a bigger difference in the gravitational time dilation between two elevations A and B, then when I drop you at A, you will be falling faster when you fall past B.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This formula means that time dilation becomes &lt;strong&gt;infinite&lt;/strong&gt; at the event horizon &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, the gravitational time dilation becomes infinite at the event horizon. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;so to a distant observer a falling object would take an infinite time to reach the event horizon and would never cross it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's what people say. That's what Einstein said too. He wrote a paper in 1939. It was &lt;a href=&quot;http://www.cscamm.umd.edu/tiglio/GR2012/Syllabus_files/EinsteinSchwarzschild.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;on a stationary system with spherical symmetry consisting of many gravitating masses&lt;/a&gt;. He said this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;“&lt;span class=&quot;math-container&quot;&gt;$g_{44}=\left(\frac{1-\frac{\mu}{2r}}{1+\frac{\mu}{2r}}\right)^2$&lt;/span&gt;&lt;br&gt;&#xA;vanishes for &lt;span class=&quot;math-container&quot;&gt;$r = \mu/2$&lt;/span&gt;. This means that a clock kept at this place would go at the rate zero. Further it is easy to show that both light rays and material particles take an infinitely long time (measured in “coordinate time”) in order to reach the point &lt;span class=&quot;math-container&quot;&gt;$r = \mu/2$&lt;/span&gt; when originating from a point &lt;span class=&quot;math-container&quot;&gt;$r &amp;gt; \mu/2$&lt;/span&gt;”&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm a big fan of Einstein. But there's two problems with this. One is that falling bodies don't slow down. The bigger the difference in gravitational time dilation between elevations A and B, the faster they fall past B. In fact, if there was no difference in gravitational time dilation, a body wouldn't fall down at all. The other problem is that Einstein concluded that black holes cannot form, and we have great evidence that there are black holes out there. So I have to sigh and say my hero Einstein got this one wrong, more's the pity.         &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, in the frame of an observer falling into the black hole it should them take a finite time to reach the event horizon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's right. There is said to be some &quot;paradox&quot; here. You can read about it in the New Scientist article &lt;a href=&quot;https://www.newscientist.com/article/mg19225751-200-the-elephant-and-the-event-horizon/&quot; rel=&quot;nofollow noreferrer&quot;&gt;the elephant and the event horizon&lt;/a&gt;. The article talks about Leonard Susskind, and says this: &lt;em&gt;&quot;If his calculations are correct, the elephant must be in more than one place at the same time&quot;&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To me the complexity is that all black holes are believed to &lt;strong&gt;‘evaporate’ and eventually disappear in a finite time&lt;/strong&gt;, by emitting Hawking radiation. Although this an incredibly slow process. For example, a solar mass hole would have a lifetime of &lt;span class=&quot;math-container&quot;&gt;$10^{66}$&lt;/span&gt; years ( in a time-frame measured a distance from the black hole). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think you should sideline the Hawking radiation actually. We have a major problem here even without it. How can the infalling body fall faster and faster, and yet take forever to reach the event horizon? And how can it be in two places at once? It can't. Something's wrong, so how do we resolve it? That's perhaps another question you could ask.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;My Question&lt;/strong&gt;&#xA;  Assuming Hawking radiation exists and  black hole evaporation occurs, does this mean that an infalling observer would never actually reach the event horizon, because the black hole has finite lifetime and would evaporate before an infalling observer would reach it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. Black holes exist. If the infalling observer never reached the event horizon, black holes wouldn't exist. And they do. &lt;/p&gt;&#xA;" OwnerUserId="9772" LastEditorUserId="25759" LastEditDate="2019-01-31T07:02:09.470" LastActivityDate="2019-01-31T07:02:09.470" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="29386" PostTypeId="2" ParentId="29381" CreationDate="2019-01-29T14:17:36.787" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Assuming Hawking radiation exists and black hole evaporation occurs,&#xA;  does this mean that an infalling observer would never actually reach&#xA;  the event horizon, because the black hole has finite lifetime and&#xA;  would evaporate before an infalling observer would reach it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The formulas I know which say that a black hole has a finite lifetime because of evaporation were done in a free-falling inertial frame. As an example see formula (11.108) from the book &lt;em&gt;A First Course in General Relativity&lt;/em&gt; by Bernard Schutz; it says,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\tau \sim M^3$$&lt;/span&gt;where &lt;span class=&quot;math-container&quot;&gt;$\tau $&lt;/span&gt; is the time in a freely falling system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in order to answer your question. If you want to compare two times you must be sure that both times were calculated in the same coordinate system. Then to know what happens first (the whole evaporation or the body crossing the Schwarzschild limit) you can take both times in a freely falling system, and compare if the proper time for the particle is more than &lt;span class=&quot;math-container&quot;&gt;$M^3$&lt;/span&gt; or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It makes no sense to compare times measured in different systems.&lt;/p&gt;&#xA;" OwnerUserId="25315" LastEditorUserId="14094" LastEditDate="2019-01-31T07:01:56.653" LastActivityDate="2019-01-31T07:01:56.653" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29387" PostTypeId="2" ParentId="29381" CreationDate="2019-01-29T16:39:26.427" Score="10" Body="&lt;p&gt;It is often pointed out that the infinite time dilation at the EH is what is known as a &quot;coordinate singularity&quot;, rather than an essential singularity that says something is physically singular there (like never reaching the EH for the infaller).  A coordinate singularity is essentially a hiccup in the language we are using, like the way a north-south-east-west coordinate system is of no use at the North pole.  The problem is that if you want to talk about &quot;whether something has crossed the EH yet&quot;, you must match up your own concept of &quot;now&quot; with some time at the EH, and that matching up is the job of a coordinate system.  So saying &quot;it hasn't reached the EH yet,&quot; and repeating that every day for infinity of time here at Earth, is not the same thing as a physical statement about what is happening at the EH, it's just a reflection of the way we are matching up our &quot;nows&quot; with theirs-- it is purely a matter of the chosen coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To know what is physically happening to the observer at the EH, the ideal solution  would be to simply ask them, and they would say &quot;yes I am crossing the EH now&quot;.  Here it gets tricky, because the normal coordinates we use to talk about the EH (the one you were using also), says that we could never receive that communication.  But those coordinates make physical assumptions that are violated by having something falling into the black hole.  At this point it would be best to have a GR expert, but I believe what happens is that the very act of having mass fall into the black hole in effect moves the EH out to swallow that mass, and we should therefore be able to receive the communication that the observer has crossed the EH in a finite time.  This must be true because, as pointed out, the very existence of the black hole is like information coming to us that stuff has crossed its EH, an EH that formed and moved out as the black hole was created by mass infall.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2019-01-29T16:39:26.427" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29388" PostTypeId="2" ParentId="29368" CreationDate="2019-01-29T18:45:25.547" Score="1" Body="&lt;p&gt;Well, this really does depend on what is considered an Earth-like planet. For instance, one could conclude that Venus is an Earth-like planet based off of them being similar in mass and also in regards to the similarities in radius. As for Earth-like moon, Titan would be a great candidate if located inside the so-called Goldilocks zone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Counting planets outside our solar system, the nearest planet I could find was about 4.243 light-years away. Astronomers found it on August 24, 2016. It is a rocky planet in the habitable zone of its parent star Proxima Centauri, the closest star to Earth. The name of the planet is Proxima b, this planet is approximately  1.3 times the mass of Earth and possesses an orbital period of roughly 11.2 Earth days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since Proxima Centauri is classified as a red dwarf this casts reasonable doubts on whether Proxima b is habitable. To be habitable Proxima b would have to deal with low stellar flux, which causes a high probability of tidal locking, and high stellar variation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/List_of_nearest_exoplanets&quot; rel=&quot;nofollow noreferrer&quot;&gt;List of nearest planets&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://solarsystem.nasa.gov/news/301/10-things-about-our-solar-systems-most-marvelous-moons/&quot; rel=&quot;nofollow noreferrer&quot;&gt;List of possible habitable moons&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25727" LastEditorUserId="2365" LastEditDate="2019-01-29T22:16:14.040" LastActivityDate="2019-01-29T22:16:14.040" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29389" PostTypeId="1" AcceptedAnswerId="29393" CreationDate="2019-01-29T19:12:07.797" Score="4" ViewCount="350" Body="&lt;p&gt;I'm wondering about Mars and Titan. If Mars is larger than Titan how come it lost most of its atmosphere? Mars also has a magnetic field while Titan doesn't. Yet Titans atmospheric pressure on the surface is a whopping 1.5 times that of Earth's. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: What causes Mars not to have much of an atmosphere, but Titan does?&lt;/p&gt;&#xA;" OwnerUserId="25727" LastEditorUserId="25727" LastEditDate="2019-01-29T19:39:38.663" LastActivityDate="2019-01-30T02:20:21.290" Title="What causes Titan a moon to have a denser atmosphere than that of a Mars?" Tags="&lt;earth&gt;&lt;planetary-atmosphere&gt;&lt;mars&gt;&lt;magnetic-field&gt;&lt;titan&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29392" PostTypeId="2" ParentId="10373" CreationDate="2019-01-29T23:13:58.247" Score="3" Body="&lt;p&gt;According to Opatrný et al. (2016) &quot;&lt;a href=&quot;https://arxiv.org/abs/1601.02897&quot; rel=&quot;nofollow noreferrer&quot;&gt;Life under a black sun&lt;/a&gt;&quot;, it might be possible to have warm temperatures around an isolated supermassive black hole thanks to the blue-shifted cosmic background radiation. Their calculations lead to an estimated equilibrium temperature of 890°C for Miller's planet (this is without the additional radiation coming from the accretion disc), which does not bode well for the watery environment shown in the movie. A planet orbiting further out might be able to support liquid water. In the past the planet would need to be located further away from the black hole because of the higher temperature of the background radiation in the early universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether such systems actually exist is another matter. The inhabitants of such a planet would have to hope that nothing came too close to the black hole, as accretion would make the environment rather hostile.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastActivityDate="2019-01-29T23:13:58.247" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29393" PostTypeId="2" ParentId="29389" CreationDate="2019-01-30T02:20:21.290" Score="7" Body="&lt;p&gt;Mars and Titan differ markedly in distance from the Sun, composition, and possibly geological activity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Titan is about 6.3 times as remote from the Sun as is Mars, which means Titan receives about 1/40th solar radiation that Mars does, with about the same reduction in solar wind. Mars' small size coupled with its much closer proximity to the Sun allowed the solar wind to rip much of Mars' atmosphere away. Despite the marked differences between the atmospheres of Venus, Earth, and Mars, the atmospheric loss rates (in terms of mass loss per unit time) from these three planets are estimated to be about the same. This rate represents a pittance for Venus' very large atmosphere of Venus, a small amount for Earth's somewhat large atmosphere, but a huge proportional loss rate for Mars. Titan is spared to some extent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mar's average density is 3.93 g/cm&lt;sup&gt;3&lt;/sup&gt;. Mars is a terrestrial planet. Titan's average density is less than half that, 1.88 g/cm&lt;sup&gt;3&lt;/sup&gt;. Titan is an icy moon: some rock, but a lot of volatiles. There's a lot of ammonia mixed in with the water ice (and possibly liquid water deep down, if Titan's core is warm). Some hypothesize that Titan regenerates its atmosphere via releases of ammonia. Sunlight splits ammonia near the top of Titan's atmosphere into hydrogen and nitrogen. The hydrogen escapes, but not the nitrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, Mars is most likely dead geologically. It's covered with craters that may date back to the Late Heavy Bombardment. Titan exhibits some signs of being somewhat active, geologically. It too has craters, but it's not as plastered with them as is Mars. It's mountains appear to be younger, at least according to some. The geological activity may help to regenerate Titan's atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2019-01-30T02:20:21.290" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29394" PostTypeId="2" ParentId="29366" CreationDate="2019-01-30T02:51:10.870" Score="2" Body="&lt;p&gt;No, this is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the stellar formation, some angular momentum will always be present. And any &quot;braking&quot; effects (magnetic, relativistic, tidal etc.) will become weaker as the rotation slows down. So the rotation will never completely stop, because any forces slowing down the rotation will weaken as well.&lt;/p&gt;&#xA;" OwnerUserId="23720" LastActivityDate="2019-01-30T02:51:10.870" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29397" PostTypeId="1" CreationDate="2019-01-30T10:15:16.077" Score="17" ViewCount="3938" Body="&lt;p&gt;Is it possible for there to be a dense enough nebula to form stars outside of any galaxy? Does a galaxy have a minimum size to produce stars? Or could you have a few dozen stars clustered together by themselves?&lt;/p&gt;&#xA;" OwnerUserId="15605" LastEditorUserId="2153" LastEditDate="2019-01-30T15:45:35.667" LastActivityDate="2019-01-30T15:45:35.667" Title="Could stars form outside of galaxies?" Tags="&lt;star&gt;&lt;star-formation&gt;&lt;intergalactic-space&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29399" PostTypeId="2" ParentId="29397" CreationDate="2019-01-30T12:18:22.563" Score="27" Body="&lt;p&gt;Yes, stars can form outside galaxies if the conditions are right. &lt;a href=&quot;http://www.sci-news.com/astronomy/spiral-galaxy-d100-ram-pressure-stripped-tail-06851.html&quot; rel=&quot;noreferrer&quot;&gt;An impressive example is D100&lt;/a&gt;, a galaxy that is moving through a cluster so fast that the ram pressure from the ambient gas forces galactic gas out of it leaving a long tail. &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/1538-4357/aaefff&quot; rel=&quot;noreferrer&quot;&gt;That tail is dense and cold enough to allow star formation, and there are newly formed clusters in it&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SmQ0s.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SmQ0s.jpg&quot; alt=&quot;D100&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle star formation can happen anywhere where there is dense, cold gas that can collapse gravitationally. Stars can sometimes help by forcing gas together by radiation pressure. There are intergalactic gas clouds, but they are usually too hot.  Dark matter halos are not great at capturing hot gas, but intra-cluster gas is losing energy from Bremsstrahlung so over time some of it would accumulate there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space is large, so no doubt there is some star formation happening in corners where gas clouds have cooled off remotely from galaxies. After all, that was how stars and galaxies once started.&lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2019-01-30T12:18:22.563" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29400" PostTypeId="1" CreationDate="2019-01-30T12:50:24.513" Score="1" ViewCount="26" Body="&lt;p&gt;I read that, if the moon would not be there or if it would dissapear or be destroyed it could cause the earth to rotate into a possible axis of 45° degrees. Because it would be missing the gravitational forces of the moon that keeps te earth in the current axis of +-23.5°. If that happend a part of the earth, for example Europe, could turn in a way that it would be permanently &amp;quot;night&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;If I understood that correctly and it is possible, would that transformation take long? Or would it be fast? And with fast I am talking about a couple of years or decades and long about thousands or millions of years.&lt;/p&gt;&#xA;&lt;p&gt;I am working on a story that would involve this and I am trying to see if I can get the logic to work with a real scientific explanation.&lt;/p&gt;&#xA;" OwnerUserId="25751" LastEditorUserId="37191" LastEditDate="2021-02-12T20:57:54.640" LastActivityDate="2021-02-12T20:57:54.640" Title="Change in axis of the Earth if the Moon would dissappear" Tags="&lt;the-moon&gt;&lt;gravity&gt;&lt;earth&gt;" AnswerCount="0" CommentCount="13" ClosedDate="2019-01-31T07:52:00.367" ContentLicense="CC BY-SA 4.0" />
  <row Id="29401" PostTypeId="1" CreationDate="2019-01-30T14:34:14.363" Score="3" ViewCount="179" Body="&lt;p&gt;As I understand it, our solar system travels through the milky way with the orbital plane tilted about 60 degrees so that we are sort of traveling in the direction of earth's north. My question is if our orbital plane rotates with our galactic orbit so that it maintains its orientation in relation to the center of the galaxy as I've show in this first picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/havkS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/havkS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is the orientation unaffected by the galactic orbit as I've shown in this second picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NCueh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NCueh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21307" LastActivityDate="2021-01-20T00:23:34.690" Title="does the orientation of orbital plane of our solar system rotate with the galaxy" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;milky-way&gt;&lt;orbital-mechanics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29402" PostTypeId="2" ParentId="29397" CreationDate="2019-01-30T15:21:12.893" Score="10" Body="&lt;p&gt;It's quite possible for stars to form outside of galaxies, typically in environments where large amounts of gas have been stripped from a galaxy. This usually involves either a tidal interaction with another galaxy or the intracluster medium (ICM). In the latter group are a set of peculiar galaxies sometimes dubbed &lt;a href=&quot;https://en.wikipedia.org/wiki/Jellyfish_galaxy&quot; rel=&quot;noreferrer&quot;&gt;&quot;jellyfish galaxies&quot;&lt;/a&gt;. Gas, dust and stars are being stripped away by ram pressure from the high-speed collision with the ICM, and some of this gas can then collapse to form star clusters in the gaseous tail behind the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some candidate intergalactic star formation regions, it's unclear whether the young, blue stars found in blue knots were ejected with the gas or formed from it, but it's still reasonable to assume that star formation will eventually take place in the cold gas clumps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notable examples of stars forming in gas stripped by ram pressure include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/29399/2153&quot;&gt;D100&lt;/a&gt;, as mentioned by Anders Sandberg (&lt;a href=&quot;https://iopscience.iop.org/article/10.3847/1538-4357/aaefff&quot; rel=&quot;noreferrer&quot;&gt;Cramer et al. 2019&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Comet_Galaxy&quot; rel=&quot;noreferrer&quot;&gt;The Comet Galaxy&lt;/a&gt; (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2007MNRAS.376..157C&quot; rel=&quot;noreferrer&quot;&gt;Cortese et al. 2011&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/ESO_137-001&quot; rel=&quot;noreferrer&quot;&gt;ESO 137-001&lt;/a&gt; (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2007ApJ...671..190S&quot; rel=&quot;noreferrer&quot;&gt;Sun et al. 2007&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the other group, gaseous structures ejected during galaxy interactions can also form stars, under the right condition. The one notable example I'm aware of is &lt;a href=&quot;https://en.wikipedia.org/wiki/Leo_Ring&quot; rel=&quot;noreferrer&quot;&gt;the Leo Ring&lt;/a&gt; (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2010ApJ...717L.143M&quot; rel=&quot;noreferrer&quot;&gt;Michel-Dansac et al. 2010&lt;/a&gt;), but I assume that other cases have been proposed.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2019-01-30T15:21:12.893" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29403" PostTypeId="2" ParentId="29401" CreationDate="2019-01-30T18:14:38.397" Score="3" Body="&lt;p&gt;The orbits in our solar system add up to a certain total angular momentum, which is a vector so points in some direction.   To get that vector to point some other way requires torque, it won't happen just because the solar system is orbiting the galaxy. So the &quot;precession&quot; rate of the plane of our solar system is controlled by very different things than the orbital rate, so there would never be any reason for picture #1.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture #2 is what you get if there is no torque, so at least there's a reason for it.  The kind of torque you'd need would be a force that points up from the plane on one side of the orbit and down from the plane on the other side, but a force like that would depend on distance from the Sun so it would be different for different planets.  That would tend to make their orbital planes undergo nodal precession at different rates, so they would not be in the same plane. So either such torques have had no effect over the lifetime of the solar system, or else something else shepherds the planets into the same plane.  But it can't be the latter, because Neptune is in the plane but Pluto isn't.  So I think it must be picture #2.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2019-01-30T18:14:38.397" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29404" PostTypeId="2" ParentId="22560" CreationDate="2019-01-30T19:41:41.570" Score="3" Body="&lt;p&gt;Wikipedia says that Francis Baily explained Baily's beads by different elevations on the Moon' surface in 1836.  However, it is possible that earlier astronomers also explained them by mountains and valleys on the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read somewhere that when Galileo observed the Moon with a telescope c. 1620 and saw mountains and craters and &quot;seas&quot; on the Moon, Johannes Kepler said that he already knew the Moon had different elevations, without a telescope.  As I remember, Kepler said that his astronomy teacher used a camera obscura to project an image of a solar eclipse showing the jagged edges of the Moon's surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if my memory is correct Francis Baily may have independently found an explanation that had earlier been found by a German astronomer, probably Michael Maestlin (1550-1631).&lt;/p&gt;&#xA;" OwnerUserId="25755" LastActivityDate="2019-01-30T19:41:41.570" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29405" PostTypeId="1" AcceptedAnswerId="30510" CreationDate="2019-01-30T20:13:50.133" Score="1" ViewCount="98" Body="&lt;p&gt;In 2012, the &lt;a href=&quot;https://ui.adsabs.harvard.edu/?#abs/2012Sci...337.1511O&quot; rel=&quot;nofollow noreferrer&quot;&gt;discovery paper&lt;/a&gt; for the multiplanet circumbinary system Kepler-47 mentioned a potential third transiting planet in the system:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A 0.2% deep transit-like event is present at time 2,455,977.363 (BJD) that is not caused by&#xA;  either of the two planets. A search for additional transits has revealed several more tentative&#xA;  transit events (12), but we caution that the star is faint (the Kepler magnitude is 15.178), there&#xA;  are large modulations due to star-spots, and the data contain correlated “red” noise, making&#xA;  small, non-periodic transit detection challenging. The marginal evidence at the present time is&#xA;  insufficient to place confidence on any additional candidate planet(s).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In 2013, Jerome Orosz gave a talk at the Second Kepler Science conference titled &quot;&lt;a href=&quot;http://nexsci.caltech.edu/conferences/KeplerII/abstracts_talks/Orosz.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Confirmation of a Third Planet in the Kepler-47 Circumbinary System&lt;/a&gt;&quot;, claiming the detection of two further transits attributed to the third planet and giving the orbital period as 187.3 days and noting that the dynamical fit indicates that transits would stop for four years starting in July 2013.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 2015, the &lt;a href=&quot;https://ui.adsabs.harvard.edu/?#abs/2015ApJ...809...26W/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;discovery paper for Kepler-453 b&lt;/a&gt; was published, describing this planet as the tenth circumbinary planet discovered by the Kepler mission, which lists Kepler-47 d as the ninth of the previously discovered planets, giving a reference of &quot;Orosz 2015, in preparation&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I have been able to tell, there has been no further update since then - the Orosz 2015 paper never appeared. Is anyone aware of any further updates about this system and whether or not the third planet is still considered to be present?&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastActivityDate="2019-04-17T17:41:53.417" Title="What happened to the third planet in the Kepler-47 system?" Tags="&lt;exoplanet&gt;&lt;kepler&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29406" PostTypeId="1" AcceptedAnswerId="29407" CreationDate="2019-01-30T22:40:51.447" Score="5" ViewCount="278" Body="&lt;p&gt;The Pleiades are 450 light years away. Thus when we see the twinkle in the night sky the light that hits our retina emanates from the year 1569 or so. On that proviso; how can we be sure that the star is still alive given the fact that we are seeing light from hundreds of years ago.&lt;/p&gt;&#xA;" OwnerUserId="25756" LastActivityDate="2019-01-31T07:44:53.737" Title="How do we know distant stars still exist" Tags="&lt;star&gt;&lt;observational-astronomy&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29407" PostTypeId="2" ParentId="29406" CreationDate="2019-01-30T23:03:34.993" Score="8" Body="&lt;p&gt;We don't know from sensory evidence.  As far as we can tell, no information &lt;em&gt;ever&lt;/em&gt; moves faster than the speed of light, so we are always at least 450 years out of date for anything happening in the Pleiades.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only evidence we have is inferential: As far as we can tell (and we've looked &lt;em&gt;real hard&lt;/em&gt; at this) the world works the same both here and at the Pleiades.  And as far as we can tell (and, once again, we've looked &lt;em&gt;real hard&lt;/em&gt; at this) stars like those in the Pleiades don't just disappear or go out in only 450 years -- it takes millions of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, based on what we observe of them as of 450 years ago, and based on what we know of how stars work, we conclude that they are pretty much the same &quot;today&quot; -- though, really, what we're concluding is that when we observe them 450 years from now, they'll still look pretty much like they do today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all inferential, but I'd bet just about anything on it.  (If only I were here in 450 years to collect!)&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2019-01-30T23:03:34.993" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29408" PostTypeId="2" ParentId="29406" CreationDate="2019-01-30T23:17:06.323" Score="6" Body="&lt;p&gt;We have a fairly good idea of how stars are born, how they evolve, and how they die. Numerical simulations allow us to the model the entire life of a star starting from its initial state (specified by mass and composition, mostly). Furthermore, &lt;a href=&quot;https://astronomy.stackexchange.com/q/942/2153&quot;&gt;we have various ways of determining the age of a star&lt;/a&gt;, and so we can therefore place it on a particular stellar evolutionary track and figure out where it goes from there. This means that we know if a star is likely to undergo some catastrophic transformation (say, a supernova) soon; if we don't observe it to be near the end of its life, we can be pretty sure that a relatively nearby star is still around (note also that 450 years is a tiny amount of time, even for the short lives of the most massive stars).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Age estimation is not an easy thing to do. &lt;a href=&quot;https://astronomy.stackexchange.com/a/8678/2153&quot;&gt;There might be certain options available&lt;/a&gt; - comparing the target star to evolutionary simulations, measuring its magnetic field and rotation, etc. In most cases, not all of them are possible. Stars in clusters might be easier than field stars to date because the characteristics of the other stars in the cluster (e.g. place on a Herztsprung-Russell diagram) can tell us something about the age of the cluster itself, and thus of its constituents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a boon for the Pleiades, and we have age estimates to within a factor of two. Observations of the amount of lithium in brown dwarfs (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1996ApJ...458..600B&quot; rel=&quot;noreferrer&quot;&gt;Basri et al. 1996&lt;/a&gt;) place it at &lt;span class=&quot;math-container&quot;&gt;$\sim$&lt;/span&gt;115 million years; comparisons to isochrones give &lt;span class=&quot;math-container&quot;&gt;$\sim$&lt;/span&gt;100 million years (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1993A%26AS...98..477M&quot; rel=&quot;noreferrer&quot;&gt;Meynet et al. 1993&lt;/a&gt;), and &lt;a href=&quot;http://www.messier.seds.org/m/m045.html&quot; rel=&quot;noreferrer&quot;&gt;earlier models gave values of &lt;span class=&quot;math-container&quot;&gt;$\sim$&lt;/span&gt;75 million years&lt;/a&gt;. That said, the stars in the cluster as a whole may not have formed &lt;em&gt;precisely&lt;/em&gt; at the same time. Nonetheless, most of the stars in the cluster have lifetimes &lt;span class=&quot;math-container&quot;&gt;$\gg$&lt;/span&gt;100 million years, and we can therefore say with near certainty that they are still around.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2019-01-30T23:17:06.323" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29409" PostTypeId="1" CreationDate="2019-01-31T00:20:06.003" Score="5" ViewCount="56" Body="&lt;p&gt;For my research purposes, I am wondering how many horizontal branch (HB) stars are within 2 kpc of the sun?  I tried to find some with the Gaia DR2 data with a magnitude cut that leaves HB stars out if they're beyond ~2 kpc, and there were surprisingly few unambiguous HB stars, so I'm just wondering how many there are and/or how to go about estimating that.&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2019-02-02T08:11:07.727" Title="How many horizontal branch stars are within 2 kpc of the sun?" Tags="&lt;star&gt;&lt;milky-way&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29410" PostTypeId="1" CreationDate="2019-01-31T01:31:12.107" Score="1" ViewCount="50" Body="&lt;p&gt;I want to build a Machine Learning model to approximate effective temperatures of stars based on their B-V index but there aren't that many catalogues that include that. I came across the Vizier catalogue (&lt;a href=&quot;http://vizier.u-strasbg.fr/viz-bin/VizieR-3?-source=V/19/data&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://vizier.u-strasbg.fr/viz-bin/VizieR-3?-source=V/19/data&lt;/a&gt;) and I wanted to be sure if that's a catalogue for individual stars that I can leverage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, do you know what other resources could help in that regard? Or suggestions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="25723" LastActivityDate="2019-01-31T01:31:12.107" Title="Regarding the VizierR catalogue" Tags="&lt;star&gt;&lt;temperature&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29411" PostTypeId="2" ParentId="29406" CreationDate="2019-01-31T07:44:53.737" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;...how can we be sure that the star is still alive given the fact that&#xA;  we are seeing light from hundreds of years ago.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answers by &lt;a href=&quot;https://astronomy.stackexchange.com/a/29407/14094&quot;&gt;Mark Olson&lt;/a&gt; and &lt;a href=&quot;https://astronomy.stackexchange.com/a/29408/14094&quot;&gt;HDE 226868&lt;/a&gt; both address the relative certainty that little will change in such a short (stellar) timeframe as 450 years. However, there's a deeper level that I thought deserves attention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fundamental nature of our Universe involves an information speed limit which is commonly referred to as the speed of light, with a value of exactly 299,792,458 metres per second. Our model of physics suggests that nothing can travel faster through spacetime than this, although expanding space, entanglement and hypothetical wormholes all make this more complicated than a mere paragraph can explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, a direct outcome of this information speed limit is that it doesn't really make sense to talk about &quot;now&quot; for any object at a distance. We can talk about that object &lt;em&gt;as we currently observe it&lt;/em&gt;, and we can make &lt;em&gt;predictions&lt;/em&gt; of what we expect it will be like when we observe it in the future (as the above answers have explained), but it makes no sense to treat it as though it's in the room next door. In reality, we live in an information bubble, where we can only ever observe objects in the past or events that have already happened. This is even true of the Sun: it was there eight minutes ago, but we can't say &lt;em&gt;for certain&lt;/em&gt; it's there &quot;now&quot;, we'll just have to wait another eight minutes to verify our extremely confident predictions.&lt;/p&gt;&#xA;" OwnerUserId="14094" LastActivityDate="2019-01-31T07:44:53.737" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29414" PostTypeId="2" ParentId="29380" CreationDate="2019-01-31T15:33:18.037" Score="2" Body="&lt;p&gt;The items 2 and 3 seems to be incompatible, I agree that you need it if you want an eclipse lasting for several weeks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You said &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The orbital speed of the moon would need to be quite slow. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and then &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the moon would need to be quite large and/or relatively close to the planet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In genera both thing can happen only &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;if the mass of the object were low compared to its size.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I agree with you at this point. The mass of the object should be tiny compared to its size, because if not its speed would be fast for conservation of angular momentum (which establis in the moment of the cretion of the system earth-moon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But It is so diffucult that a natural satelite has so low density. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the only way to avoid the comflict between 2 and 3, is a sistem which at the moment of its formation had a very very low angular momentum, but this is actually an uncommon situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;May be this last one is the arguement I would chosse  between both. I prefer low angular momentum before low density.&lt;/p&gt;&#xA;" OwnerUserId="25315" LastEditorUserId="25315" LastEditDate="2019-02-03T13:04:07.840" LastActivityDate="2019-02-03T13:04:07.840" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29415" PostTypeId="5" CreationDate="2019-01-31T18:25:14.663" Score="0" Body="" OwnerUserId="6" LastEditorUserId="6" LastEditDate="2019-01-31T18:25:14.663" LastActivityDate="2019-01-31T18:25:14.663" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29416" PostTypeId="4" CreationDate="2019-01-31T18:25:14.663" Score="0" Body="Questions about the apparent force stretching a body towards and away from the center of mass of another body due to the variance in acceleration due to gravity" OwnerUserId="6" LastEditorUserId="6" LastEditDate="2019-01-31T18:25:14.663" LastActivityDate="2019-01-31T18:25:14.663" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29417" PostTypeId="2" ParentId="29380" CreationDate="2019-01-31T18:45:29.963" Score="1" Body="&lt;p&gt;First question is what do we want for a long eclipse? The answer would be a large moon on a slow orbit. The following analysis assumes the system is exactly coplanar, and I neglect the angular size of the star (which would result in increased satellite radii) and the size of the planet itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a slow satellite orbit, we need to figure out what the outermost stable orbit. Neglecting the small eccentricity of the Earth's orbit, the radius of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hill sphere&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$R_\mathrm{H}$&lt;/span&gt; is given as follows, where &lt;span class=&quot;math-container&quot;&gt;$a_\oplus$&lt;/span&gt; is the radius of the Earth's orbit, &lt;span class=&quot;math-container&quot;&gt;$m_\oplus$&lt;/span&gt; is the mass of the Earth and &lt;span class=&quot;math-container&quot;&gt;$m_\odot$&lt;/span&gt; is the mass of the Sun:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$R_\mathrm{H} \approx a_\oplus\sqrt[3]{\frac{m_\oplus}{3m_\odot}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works out as about 1,946,400 km. Moons are stable out to a factor of about 1/3 for the Hill radius in the case of prograde moons, and about 1/2 of the Hill radius for retrograde moons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a prograde satellite, this puts the orbital radius &lt;span class=&quot;math-container&quot;&gt;$a_\mathrm{m}$&lt;/span&gt; at about 498,800 km with an orbital period &lt;span class=&quot;math-container&quot;&gt;$P_\mathrm{m}$&lt;/span&gt; of 40.5 days. We need to also take care of the Earth's orbit around the Sun, which has a period &lt;span class=&quot;math-container&quot;&gt;$P_\oplus$&lt;/span&gt; of 365.25 days. This gives the resultant angular velocity as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\omega = 2\pi\left(\frac{1}{P_\mathrm{m}}-\frac{1}{P_\oplus}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can then be multiplied by the time taken to figure out the angle the moon travels through in a given time &lt;span class=&quot;math-container&quot;&gt;$\theta = \omega t$&lt;/span&gt;. For a week-long eclipse, the angle works out as about 55°. The radius of the moon can then be obtained by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$r_\mathrm{m}=a_\mathrm{m} \arcsin\left(\frac{\theta}{2}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works out as 231,100 km - which is about the radius of an M3V red dwarf star. So a prograde satellite doesn't look like a good possibility for week-long eclipses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the retrograde case? In this case, the orbital radius and period of the outermost stable orbit end up being 748,200 km and 74.5 days. The angular velocities add up, so you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\omega = 2\pi\left(\frac{1}{P_\mathrm{m}}+\frac{1}{P_\oplus}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing the same calculation again gives a moon radius of 260,200 km - even worse! The increased distance wins out over the slower orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if we use a different type of host star? According to &lt;a href=&quot;http://www.pas.rochester.edu/~emamajek/EEM_dwarf_UBVIJHK_colors_Teff.txt&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eric Mamajek's list of the properties of main sequence stars&lt;/a&gt;, an A5V star as having a mass of 1.85 times that of the Sun and a luminosity of 10&lt;sup&gt;1.16&lt;/sup&gt; times that of the Sun, which puts the habitable zone at 3.80 AU with an orbital period of 1990 days. Doing the calculations again results in smaller satellite radii of 136,400 km (prograde) and 150,900 km (retrograde). These are still far too large: the prograde case is roughly the size of an M5V red dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So unfortunately it looks like the case of a week-long eclipse is not very plausible!&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-01-31T22:27:45.200" LastActivityDate="2019-01-31T22:27:45.200" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29420" PostTypeId="1" CreationDate="2019-01-31T22:11:53.893" Score="1" ViewCount="42" Body="&lt;p&gt;How could the lunar eclipse of January 20, 2019 begin from bottom to top and then finish from left to right?  Here's the video of it's progression's directions.&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=A2sTkuZht4s&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=A2sTkuZht4s&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25171" LastActivityDate="2019-01-31T22:19:49.673" Title="Eclipse Changes Directions?" Tags="&lt;lunar-eclipse&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29421" PostTypeId="1" CreationDate="2019-01-31T22:17:02.217" Score="5" ViewCount="223" Body="&lt;p&gt;I've had some discussions recently about black holes, and the issue of infalling bodies taking forever to reach the event horizon. That's essentially what Einstein said in his 1939 paper &lt;a href=&quot;http://www.cscamm.umd.edu/tiglio/GR2012/Syllabus_files/EinsteinSchwarzschild.pdf&quot; rel=&quot;noreferrer&quot;&gt;on a stationary system with spherical symmetry consisting of many gravitating masses&lt;/a&gt;. He said &lt;em&gt;&quot;it is easy to show that both light rays and material particles take an infinitely long time (measured in “coordinate time”) in order to reach the point r = μ/2 when originating from a point r &gt; μ/2”.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm a big fan of Einstein. But there seems to be two issues with this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;One is that Einstein concluded that black holes cannot form, but we&#xA;have good evidence that there are black holes out there. The obvious&#xA;example is &lt;a href=&quot;https://en.wikipedia.org/w/index.php?title=Sagittarius_A*&amp;amp;oldid=769794306#Observation_and_description&quot; rel=&quot;noreferrer&quot;&gt;Sagittarius A*&lt;/a&gt;. That's where's there’s something&#xA;with a mass 4.28 million times the Sun, with a diameter of less than&#xA;44 million kilometres, and we can't see it. Surely it's just got to be a black hole.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The other issue is that falling bodies don't slow down. Imagine you&#xA;drop a body at elevation A and it falls down to elevation B. The&#xA;&quot;force&quot; of gravity relates to the first derivative of gravitational&#xA;potential. Hence the bigger the difference in gravitational time&#xA;dilation between elevations A and B, the faster the body falls past&#xA;B. Then if you drop a body at elevation B it falls down to elevation&#xA;C. Again the bigger the difference in gravitational time dilation&#xA;between elevations B and C, the faster the body falls past C. In a&#xA;typical gravitational field the force of gravity at B is greater than&#xA;at A. Hence as the body descends, the acceleration increases as well as the falling speed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Imagine a gedanken spaceship from which we've suspended a cable. We have clocks at different elevations, so we can measure the gravitational time dilation at each elevation. We can also release test bodies at each elevation and record the  clock readings as they fall past other elevations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/21xOq.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/21xOq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of the experiment we can reel in the cable and upload the recorded measurements to ascertain how our test bodies behaved. My understanding is that we will always find that time dilation always increases as we descend, that the falling body always accelerate downwards, and that both the acceleration and the falling speed always increases as the body descends. Is this correct? Or do falling bodies somehow stop accelerating? And do falling bodies ever slow down?   &lt;/p&gt;&#xA;" OwnerUserId="9772" LastEditorUserId="9772" LastEditDate="2019-01-31T23:55:02.780" LastActivityDate="2019-02-01T13:59:26.223" Title="Do falling bodies ever slow down?" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;general-relativity&gt;" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="29422" PostTypeId="2" ParentId="29420" CreationDate="2019-01-31T22:19:49.673" Score="3" Body="&lt;p&gt;Because we are standing on a rotating globe.  We are changing direction as the Earth spins. This appears exaggerated in the video as the horizon is not visible, so there is no way of seeing how the camera was rotated to view the moon as the eclipse progressed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relative to the stars, the moon appears to enter the Earth's shadow from one side and leave it from the other,&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-01-31T22:19:49.673" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29423" PostTypeId="1" CreationDate="2019-01-31T22:28:21.443" Score="3" ViewCount="193" Body="&lt;p&gt;This is my first post here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am studying the following article: &lt;a href=&quot;https://arxiv.org/abs/1901.03574&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/1901.03574&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Page 5, Sec 3.1, the authors are presenting their results they obtained via Isochrone fitting to the CMD of the two GCs being discussed. My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How can I find out the distance modulus and the Total extinction &lt;span class=&quot;math-container&quot;&gt;$A_{F606W}$&lt;/span&gt; from just the fitting? I know that Isochrone fitting is done when we need to estimate the age. In order to fit, we need to move the Isochrone both vertically (Distance modulus?) and horizontally (Color excess/Reddening/Extinction?). Is my understanding correct?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How do they convert the Reddening in filters to reddening in B-V? What is the technical details/physics of it?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the role of the Extinction law in getting these numbers?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the end, they calculate the distance, which is simple once we get the distance modulus. Why is it the distance from the Sun? Did we assume a heliocentric coordinate system somewhere?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thank You. &lt;/p&gt;&#xA;" OwnerUserId="23967" LastEditorUserId="23967" LastEditDate="2019-02-01T00:13:32.143" LastActivityDate="2019-07-04T12:54:07.863" Title="How to obtain distance modulus and Total extinction from Isochrone fitting to CMD?" Tags="&lt;main-sequence&gt;&lt;extinction&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29424" PostTypeId="1" CreationDate="2019-02-01T02:04:44.537" Score="2" ViewCount="69" Body="&lt;p&gt;While reading about the discovery of quasars and the spectroscopic analysis of 3C 273 in &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1964ApJ...140....1G&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt; by Maarten Schmidt, I came across the following quote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At these distances, corrections to luminosities depend on the specific world model. For&#xA;  an approximate determination of the fluxes in the emission lines we use the formulae&#xA;  given by Sandage (1961) based on Mattig (1958). For the energy contained within an&#xA;  emission line, in which we integrate over the entire profile, the correction to the inverse square law is of the form &lt;span class=&quot;math-container&quot;&gt;$(1 + z/2)^2$&lt;/span&gt;&#xA;  if the cosmological constant and deceleration parameter are zero. It is &lt;span class=&quot;math-container&quot;&gt;$(1 + z)^2$&lt;/span&gt;&#xA;  for the steady-state model.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First, why is a correction to the inverse square law necessary? Is it to do with interstellar extinction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, why are the two models different? I understand what the cosmological constant and deceleration parameter are, but why is the redshift prediction for a steady state model twice as much as when &lt;span class=&quot;math-container&quot;&gt;$\Lambda=q=0$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2019-02-01T02:04:44.537" Title="Correction to inverse square law for cosmology" Tags="&lt;cosmology&gt;&lt;spectroscopy&gt;&lt;quasars&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29425" PostTypeId="1" CreationDate="2019-02-01T11:06:15.263" Score="0" ViewCount="53" Body="&lt;p&gt;If the axis of the Earth was through Greenland, what would the north celestial pole be?&lt;/p&gt;&#xA;" OwnerUserId="25778" LastEditorUserId="25778" LastEditDate="2019-02-01T11:57:59.510" LastActivityDate="2019-02-01T14:07:24.753" Title="If the axis of the Earth was through Greenland, what would the north celestial pole be?" Tags="&lt;star&gt;&lt;amateur-observing&gt;&lt;constellations&gt;&lt;pole&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2019-02-02T15:34:57.427" ContentLicense="CC BY-SA 4.0" />
  <row Id="29426" PostTypeId="1" CreationDate="2019-02-01T11:27:28.470" Score="1" ViewCount="452" Body="&lt;p&gt;I'm studying statistical properties of a cosmological density field and I have some doubts; in the following quote and pictures you can see what my problem is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A fundamental limitation arises in cosmology – because there is only one universe to observe, i.e. there is only one realization of the stochastic process that generates the fluctuations whose consequences we observe. Therefore we cannot measure ensemble averages&#xA;  or expectation values, as we would in a repeatable laboratory experiment. What we can do when observing a fluctuation on a given scale &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; is to average over many distinct regions of size &lt;span class=&quot;math-container&quot;&gt;$\sim\lambda$&lt;/span&gt;. An ergodic-type hypothesis allows us to replace the ensemble average by a spatial average over these regions. (see pag 307 from &lt;em&gt;Relativistic cosmology&lt;/em&gt;-George F.R. Ellis, R. Maartens, M.A.H MacCallum; Cambridge University Press (2012))&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So here we talk about an average over many distinct regions of the Universe (someone call them &quot;fair sample&quot; of the Universe if I understood correctly) and an ensemble average is replaced by a spatial average over these region&lt;strong&gt;S&lt;/strong&gt;. Thus I should average the field over region 1, region 2 and so on and I would get many mean values of the field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in this presentation things seem different:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/6eaOv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6eaOv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/hpyiz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hpyiz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here instead we talk about an average over an ensemble replaced by a spatial average over &lt;strong&gt;one&lt;/strong&gt; realization of the random field (one of the possible  universes I imagine). In this case I have only one realization: our Universe, so I should average my field spatially over the entire Universe. And what about the regions mentioned above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confused and I would appreciate an explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So theoretically if i wanted to get the average of a stochastic field or of a function, we say &lt;span class=&quot;math-container&quot;&gt;$\delta(\vec x)$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$f[\delta(\vec x)]$&lt;/span&gt;, in a point &lt;span class=&quot;math-container&quot;&gt;$\vec x_1$&lt;/span&gt; I should do the average over the ensemble of all the universes using (see slide): &lt;span class=&quot;math-container&quot;&gt;$\int{\delta}\cal P(\delta) d\delta$&lt;/span&gt; (it is obviously impossible because we have only one Universe). I have imagined this thing thinking about a stack of planes representing the different universes of the ensemble and drawing the point &lt;span class=&quot;math-container&quot;&gt;$\vec x_1$&lt;/span&gt; in each plane. Now i can take the value of &lt;span class=&quot;math-container&quot;&gt;$\delta(\vec x_1)$&lt;/span&gt; in the first universe, in the second, and so on and finally make the average of these values; if I repeat this procedure for all the points I'll get the average of &lt;span class=&quot;math-container&quot;&gt;$\delta (\vec x)$&lt;/span&gt;, i.e. &lt;span class=&quot;math-container&quot;&gt;$\langle\delta\rangle$&lt;/span&gt; over the ensemble of the universes. Ok it's fine...but it is not possible because I have only one Universe, so I have to appeal to Ergodic Hypothesis (really a &lt;em&gt;spatial&lt;/em&gt; version of the Ergodic Hypothesis).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An ergodic-type hypothesis allows us to replace the ensemble average by a spatial average over these regions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems that I have to divide the Universe in regions (we can imagine squares for simplicity in two dimensions, I thought the Universe like a chessboard) and I can replace the ensemble averege with a spatial average over any single region. Any square of the chessboard is a fair sample of the Universe, so I have many squares that can &quot;reproduce&quot; the Universe (I divided my sky in many regions and I can consider these squares like the realisations of the &quot;Universe experiment&quot;, the outcomes of the experiment). Thus I can do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\langle\delta\rangle=\frac{1}{\sum_i V_i}\sum_i \langle\delta\rangle_{V_i}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\langle\delta\rangle_{V_i}=\int_{V_i}\delta(\vec x) d\vec x^3$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="24355" LastEditorUserId="24355" LastEditDate="2019-02-02T17:24:27.837" LastActivityDate="2019-07-02T19:03:05.477" Title="Ensemble average VS spatial average for a cosmological density field" Tags="&lt;cosmology&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29427" PostTypeId="2" ParentId="29421" CreationDate="2019-02-01T12:43:02.007" Score="6" Body="&lt;p&gt;According to the standard interpretation of General Relativity (e.g. as presented in &quot;Exploring black holes&quot; by Taylor &amp;amp; Wheeler chapter 3, or &quot;Black holes, white dwarfs &amp;amp; neutron stars&quot; by Shapiro &amp;amp; Teukolsky, pp 343-345) then yes they do. But it depends on the frame of reference of the observer - there is no absolute answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to an observer far from the black hole, the rate of change of radial coordinate with time (for an object that started falling radially inwards far from a non-rotating black hole) is given by&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{dr}{dt} = -\left(1 - \frac{r_s}{r}\right)\left(\frac{r_s}{r}\right)^{1/2}$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$r_s$&lt;/span&gt; is the Schwarzschild radius and &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; are Schwarzschild coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we call this the infalling speed as measured by a distant observer then we can see by differentiation that it goes through a maximum at &lt;span class=&quot;math-container&quot;&gt;$r=3r_s$&lt;/span&gt; and that &lt;span class=&quot;math-container&quot;&gt;$dr/dt \rightarrow 0$&lt;/span&gt; as &lt;span class=&quot;math-container&quot;&gt;$r \rightarrow r_s$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, an observer accompanying the falling particle would totally disagree. To them, their velocity is given by &lt;span class=&quot;math-container&quot;&gt;$dr/d\tau$&lt;/span&gt;, the rate of change of &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; with respect to the time on their clock.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{dr}{d\tau} = -\left( \frac{r_s}{r}\right)^{1/2}$$&lt;/span&gt;&#xA;which continues to increase up to and below the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter appears to admit the possibility of faster than light travel, but no more so than me (correctly) saying that if you travel at close to the speed of light you can get to a star 10 light years away in much less than 10 years (as measured on your clock).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally we could have the point of view of stationary &quot;shell&quot; observers at fixed radial distances (outside the event horizon, because no stationary observer is possible below the event horizon). They would measure the speed of objects falling past them to be&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{dr_{\rm shell}}{d\tau_{\rm shell}} = -\left(\frac{r_s}{r}\right)^{1/2}.$$&lt;/span&gt;&#xA;This means that the reports of stationary observers (which is the gist of your question I think) at increasingly lower heights, is indeed that the velocity of the falling object is increasing as it falls. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no paradox to these apparently contradictory points of view. Measurements of non-local events and phenomena are not bound to agree in General Relativity, where there isn't even agreement on what is meant by &quot;now&quot; or whose coordinate system in what frame of reference should be used in any particular circumstance.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-02-01T13:59:26.223" LastActivityDate="2019-02-01T13:59:26.223" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="29428" PostTypeId="2" ParentId="29425" CreationDate="2019-02-01T14:07:24.753" Score="3" Body="&lt;p&gt;If you assume that the shift was due to natural process, it would still point more-or-less towards Polaris.  (If the shift is due to magic or advanced technology, all bets are off.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;em&gt;think&lt;/em&gt; your question may be based on a false premise, that the celestial sphere is fixed relative to the Earth.  It isn't at all.  Over the timescales we're dealing with it's basically just fixed and unchanging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rotating Earth acts like a gyroscope and its axis points in &lt;em&gt;some&lt;/em&gt; direction.  To a good approximation, that direction is fixed in space due to conservation of angular momentum.  So in the absence of external torques, the Earth just sits there and spins like a top.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(To a better approximation, there &lt;em&gt;are&lt;/em&gt; some external torques and the gravitational effects of the Sun and Moon causes the Earth's axis to slowly &lt;em&gt;precess&lt;/em&gt;. It wobbles (a bit like a top slowing down) once every 26,000 years, actually chaning where it points to in space.  I think this is probably an irrelevant detail though, for your question.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if the Earth's axis passed through Greenland, what would happen?  (Other than the climate changes due to the polar and equatorial regions moving around, of course.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have in fact seen something like this in the Earth's past, and it's called &lt;a href=&quot;https://en.wikipedia.org/wiki/True_polar_wander&quot; rel=&quot;nofollow noreferrer&quot;&gt;Polar Wandering&lt;/a&gt;. We observe that the Earth's north pole has moved relative to the Earth's surface and actually has moved quite a lot. But what happened was that the Earth's axis stayed fixed relative to the stars and the Earth's bulk shifted relative to its axis of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The True Polar Wander page referred to above has an excellent diagram: &lt;a href=&quot;https://i.stack.imgur.com/kQrrg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kQrrg.jpg&quot; alt=&quot;True polar wandering&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the answer is that if the shift occurred because of changes to the Earth's moment of inertia, the axis of rotation would still point towards Polaris.  If it happened due to outside influences (a giant impact, aliens, magic) it might wind up pointing most anywhere.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2019-02-01T14:07:24.753" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29429" PostTypeId="1" CreationDate="2019-02-01T16:21:25.653" Score="4" ViewCount="537" Body="&lt;p&gt;Whenever there is a picture of a black hole, (even though they are artist depictions, this is how they are described) most are shown kind of in a flat disk like this. &lt;a href=&quot;https://i.stack.imgur.com/gD6DQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gD6DQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, why do we see the black hole itself? If it is sucking in material from all directions, then shouldn't the material be kind of covering the black hole, not just spinning around it to be sucked in from only the sides. Especially in the case that a star would be sucked in, then wouldn't all of its light &amp;amp; matter in the accretion disk be encasing the black hole? &#xA;In other words, why is the accretion disk flat, and not a sphere surrounding the black hole? So if this were true, we shouldn't be able to see some black holes, only their accretion disks around them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;unrevised/old post:&lt;/strong&gt; If a black hole is spherical and has infinite density, then it would be sucking in objects from all directions because the gravity would be the same all around. In other words, the event horizon should be all around it! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand everything would orbit the black hole a little before 'falling in', but since the amount of matter and light is so large, shouldn't it encase the black hole, thus making it covered in a ball of light? The event horizon would be a shell so we could not see some black holes, which would mean there could be billions more out there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the other black holes which we have seen/proven, if the information above is true, then maybe they are the result of black holes with poles with low force/gravity/magnetivity/energy. Possibly they could be a disk of some sort or have an unequal distribution of density, either of which could lead to the flat event horizon, maybe even both. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Poles on a black hole would also help account for the quasars shooting out of some. Since the gravitational pull would not be as great from the poles of a black hole, then this would allow the quasars to beam out (otherwise they would have to be going faster than the escape velocity, which is the speed of light). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe the quasars are even the black holes that I was talking about earlier, and this is why you would see no event horizon, only a shell of bright light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am no expert in astronomy but if any of these theories sound plausible, please let me know! Just looking for some explanation, I thought of this earlier and couldn't figure it out, I could be completely wrong though. Thanks!)&lt;/p&gt;&#xA;" OwnerUserId="25782" LastEditorUserId="25782" LastEditDate="2019-02-02T02:08:49.783" LastActivityDate="2019-09-17T04:02:57.937" Title="Shouldn't we not be able to see some black holes?" Tags="&lt;black-hole&gt;&lt;event-horizon&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29430" PostTypeId="1" AcceptedAnswerId="29431" CreationDate="2019-02-01T18:59:15.947" Score="0" ViewCount="79" Body="&lt;p&gt;Stars A and B are both equally bright as seen from Earth, but A is 120 pc away while B is 24 pc away. Which star has the greater luminosity? How many times brighter is it?&lt;/p&gt;&#xA;" OwnerUserId="25774" LastActivityDate="2019-02-01T19:20:57.647" Title="The Luminosity of two stars with same brightness but at different PC away" Tags="&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29431" PostTypeId="2" ParentId="29430" CreationDate="2019-02-01T19:20:57.647" Score="1" Body="&lt;p&gt;The absolute luminosity of A is bigger than B. Like two lamps.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{120}{24}=5,$$&lt;/span&gt;&#xA;so A is five times farther B, but as the luminosity decays with the square of the distance, then  A is &lt;span class=&quot;math-container&quot;&gt;$5^{2}=25$&lt;/span&gt; times more luminous.&lt;/p&gt;&#xA;" OwnerUserId="25315" LastActivityDate="2019-02-01T19:20:57.647" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29432" PostTypeId="2" ParentId="29346" CreationDate="2019-02-01T20:10:23.437" Score="2" Body="&lt;p&gt;If you use the equation of the Schwarzschild metric.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ds^{2}=-(1-\frac{2M}{r})dt^{2}+(1-\frac{2M}{r})^{-1}dr^{2}+r^{2}d\Omega^{2}$$&lt;/span&gt; &#xA;(equation 11.1 from Schutz's book A first course in GR).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we think the neutron stay at rest in &lt;span class=&quot;math-container&quot;&gt;$r=r_{S}=2M$&lt;/span&gt; then between the born and dead of the neutron &lt;span class=&quot;math-container&quot;&gt;$dr=0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$d\Omega=0$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$ds$&lt;/span&gt; will be the proper time, which I will call &lt;span class=&quot;math-container&quot;&gt;$\tau$&lt;/span&gt; (I'm folowing the convention &lt;span class=&quot;math-container&quot;&gt;$c=1=G$&lt;/span&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting this on the metric we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$  d\tau^{2}=-(1-\frac{2M}{r})dt_{obs}^{2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$  \frac{d\tau^{2}}{dt_{obs}^{2}}=-(1-\frac{2M}{r})$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if we put &lt;span class=&quot;math-container&quot;&gt;$r=r_{s}$&lt;/span&gt; (where the neutros was), we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$  \frac{d\tau^{2}}{dt_{obs}^{2}}=0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then &lt;span class=&quot;math-container&quot;&gt;$\Delta t_{obs}=\infty$&lt;/span&gt; . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't write a comment that's  why I answered, but I think you had some mistake using the red shift equation to get the wavelength difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you show your calculus? May be I can help.&lt;/p&gt;&#xA;" OwnerUserId="25315" LastEditorUserId="25315" LastEditDate="2019-02-02T00:04:12.063" LastActivityDate="2019-02-02T00:04:12.063" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29433" PostTypeId="2" ParentId="29423" CreationDate="2019-02-01T21:03:01.723" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The isochrones are not straight lines in the CMD. Extinction moves the stars both redward (right) as well as fainter (down) in the CMD. The vertical shift (the extinction &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;) is &lt;em&gt;calculated&lt;/em&gt; from the horizontal shift (the reddening) using an extinction law. (e.g. &lt;span class=&quot;math-container&quot;&gt;$A_V = 3.1 E(B-V)$&lt;/span&gt;). The effects of extinction and distance (which only moves the stars vertically) can therefore be separated. &#xA;In the case of the paper referred to, the globular clusters have a horizontal branch and a giant branch. The horizontal branch is almost ...horizontal and the giant branch is almost vertical. This enables the authors to be independently sensitive to a vertical and horizontal shift of the model isochrones, as described above.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;+3. There are simple conversion factors published in several papers. Extinction is wavelength dependent, so the spectrum of a star is attenuated in a wavelength dependent way  The reddening in any colour (e.g. E(B-V)) is just the difference in extinction at the two wavelengths (well it needs to be integrated over the photometric bands) i.e. &lt;span class=&quot;math-container&quot;&gt;$A_B - A_V$&lt;/span&gt;. These differences in extinction coefficients can be calculated for any pair of bands. They depend on the wavelength dependence of extinction, aka the extinction law, but also on the intrinsic spectrum of the star (which is frequently ignored). Extinction laws  can be empirically determined by observing stars with a known intrinsic spectrum.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;'4. Does it matter? Have you compared a parsec with an astronomical unit?&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-07-04T12:54:07.863" LastActivityDate="2019-07-04T12:54:07.863" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29434" PostTypeId="1" CreationDate="2019-02-01T21:24:44.637" Score="1" ViewCount="137" Body="&lt;p&gt;Suppose two stars have the same apparent brightness, but one star is 8 times further away than the other. What is the ratio of their luminosities? Which one is more luminous, the closer or the further star?&lt;/p&gt;&#xA;" OwnerUserId="25774" LastActivityDate="2019-02-01T22:37:49.437" Title="Luminosities and Apparent Brightness of Two Stars of which one is 8 times the distance of the first star" Tags="&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2019-02-02T23:52:17.690" ContentLicense="CC BY-SA 4.0" />
  <row Id="29435" PostTypeId="1" AcceptedAnswerId="29446" CreationDate="2019-02-01T21:49:14.883" Score="2" ViewCount="152" Body="&lt;p&gt;I want to draw a Sun in a 3d scene, so I figured the easiest way would be to calculate the latitude and longitude of the point perpendicular to Sun position and then just move it &quot;up&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a starting point, I want to calculate it's equatorial coordinates: declination and right ascension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I followed instructions from &lt;a href=&quot;https://en.wikipedia.org/wiki/Position_of_the_Sun&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;, also cross-referencing with &lt;a href=&quot;http://njsas.org/projects/tidal_forces/altaz/pausch/ppcomp.html#6&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my code (in Swift):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import Foundation&#xA;&#xA;class Astronomy {&#xA;&#xA;  static func julianDays(since: Date) -&amp;gt; Double {&#xA;    return since.timeIntervalSince1970 / 86400 + 2440587.5&#xA;  }&#xA;&#xA;  static func daysSinceJ2000(for date: Date) -&amp;gt; Double {&#xA;    return julianDays(since: date) - 2451545&#xA;  }&#xA;&#xA;  static func sunAxialTilt(for daysSinceJ200: Double) -&amp;gt; Double {&#xA;    return 23.4393 - 3.563e-7 * daysSinceJ200&#xA;  }&#xA;&#xA;  static func sunEquatorialPosition(on date: Date) -&amp;gt; (declination: Double, RA: Double, distance: Double) {&#xA;    let daysSinceGreenwichNoon = daysSinceJ2000(for: date)&#xA;    var meanLongitude = 280.460 + 0.9856474*daysSinceGreenwichNoon&#xA;    var meanAnomaly = 357.528 + 0.9856003*daysSinceGreenwichNoon&#xA;    meanLongitude = meanLongitude.rangedUpTo360&#xA;    meanAnomaly = meanAnomaly.rangedUpTo360&#xA;    let eclipticLongitude = meanLongitude + 1.915*sin(meanAnomaly.radians) + 0.020*sin(2*meanAnomaly.radians)&#xA;&#xA;    let distanceToEarth = 1.00014 - 0.01671*cos(meanAnomaly.radians) - 0.00014*cos(2*meanAnomaly.radians)&#xA;&#xA;    let axialTilt = sunAxialTilt(for: daysSinceGreenwichNoon)&#xA;    let rightAscension = atan2(cos(axialTilt.radians)*sin(eclipticLongitude.radians), cos(eclipticLongitude.radians))&#xA;    let declination = asin(sin(axialTilt.radians)*sin(eclipticLongitude.radians))&#xA;&#xA;    return (declination: declination.degrees, RA: rightAscension.degrees, distance: distanceToEarth)&#xA;  }&#xA;}&#xA;&#xA;extension Double {&#xA;  var radians: Double {&#xA;    return self * .pi / 180&#xA;  }&#xA;&#xA;  var degrees: Double {&#xA;    return self  * 180 / .pi&#xA;  }&#xA;&#xA;  var rangedUpTo360: Double {&#xA;    let multiples = (self/360).rounded(.down)&#xA;    return self - 360*multiples&#xA;  }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However, it does not seem to be working correctly. As of the time I write this article, the result is &lt;code&gt;declination -16.98772292697088, RA -44.80674952684117&lt;/code&gt;. Compared with &lt;a href=&quot;https://theskylive.com/sun-info&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://theskylive.com/sun-info&lt;/a&gt; the declination is about right (-17° 03’ 38”), but right ascension is way off:  20h 59m 43s, which is 314.93 degrees according to &lt;a href=&quot;https://www.vercalendario.info/en/how/convert-ra-degrees-hours.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this site&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What went wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;As it turned out nothing was wrong, I can get to the second part: calculating the geographical point. (the title suggests it after all)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AFAIK for &lt;strong&gt;latitude&lt;/strong&gt;, I can use &lt;em&gt;declination&lt;/em&gt; as is and for &lt;strong&gt;longitude&lt;/strong&gt;, I &#xA;need to calculate &lt;em&gt;Greenwich Hour Angle(GHA)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After examining several articles, I found these equations:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;SHA[°] = 360 – 15 · RA[h] &#xA;GHAAries[°] = 15 · GST[h] &#xA;GHA = SHA + GHAAries &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For my case, I suppose I may use RA as is, without multiplying its hour component by 15. What keeps me a bit puzzled is GST:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;May I use Greenwich Mean Sidereal Time instead if I don't need the calculations to be super precise?&lt;/li&gt;&#xA;&lt;li&gt;Is there any &quot;GST for dummies&quot;? All the calculations I found rely on some floating second offset, I found &lt;a href=&quot;ftp://maia.usno.navy.mil/ser7/ser7.dat&quot; rel=&quot;nofollow noreferrer&quot;&gt;this table&lt;/a&gt; but have no idea how to use it.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="25784" LastEditorUserId="25784" LastEditDate="2019-02-02T14:57:45.113" LastActivityDate="2019-02-03T16:02:46.150" Title="Geographical Point of the Sun" Tags="&lt;the-sun&gt;&lt;coordinate&gt;&lt;positional-astronomy&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="29437" PostTypeId="2" ParentId="29434" CreationDate="2019-02-01T22:37:49.437" Score="1" Body="&lt;p&gt;Welcome to Astronomy Stack Exchange!&lt;br&gt;&#xA;Let the star which is farther away be A, and the fainter one be B.&lt;br&gt;Since this perceived brightness is something called flux density (sometimes referred to as just flux), which the following formula holds its own for: &lt;span class=&quot;math-container&quot;&gt;$F = \frac{L}{4\pi r^2}$&lt;/span&gt; &#xA;(&lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; is the luminosity, and &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; is the distance) it would mean the luminosity ratio can be calculated in the following manner:&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$F_A = F_B$&lt;/span&gt;&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$\frac{L_A}{4\pi r_A^2} = \frac{L_B}{4\pi r_B^2}$&lt;/span&gt;&lt;br&gt;Substituting for &lt;span class=&quot;math-container&quot;&gt;$r_A = 8r_B$&lt;/span&gt;:&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$\frac{L_A}{64r_B^2} = \frac{L_B}{r_B^2}$&lt;/span&gt;&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$L_A = 64L_B$&lt;/span&gt;&#xA;&lt;br&gt;The farther star is much more luminous than the closer one.&lt;br&gt;&#xA;P.S. This is only if there is no extinction, if it is present, I cannot make an estimate without more details.&lt;/p&gt;&#xA;" OwnerUserId="24014" LastActivityDate="2019-02-01T22:37:49.437" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29438" PostTypeId="2" ParentId="29429" CreationDate="2019-02-01T23:15:13.673" Score="5" Body="&lt;p&gt;Well it doesn't exactly work the way you describe it. Matter doesn't just fall into a black hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A black hole still has a finite mass, which means other matter orbits its just like it would a comparable object of the same mass, like a star. In fact the gravitational disturbance it causes in this way is usually how we determine that there are, for example, supermassive blackholes in the center of galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the light you speak of is an accretion disk. When a star orbits a black hole, the star can get torn apart due to gravitational variation in the star as a consequence of the nearby black hole. Matter from the star starts orbiting the black hole, but it cant just do that in any direction. Due to conservation of angular momentum (the star itself was rotating) it is forced to rotate in one specific circular path, thats what we call the accretion disk and is the reason it doesn't become a huge sphere of light. As for light itself, it can't orbit, it either gets deflected or it gets trapped inside.&lt;/p&gt;&#xA;" OwnerUserId="25786" LastActivityDate="2019-02-01T23:15:13.673" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29439" PostTypeId="5" CreationDate="2019-02-02T02:33:19.753" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2019-02-02T02:33:19.753" LastActivityDate="2019-02-02T02:33:19.753" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29440" PostTypeId="4" CreationDate="2019-02-02T02:33:19.753" Score="0" Body="An event horizon is the notion of a distance away from the gravitational center of mass of a black hole.  Observers outside of the event horizon can escape however any object including light which approaches a black hole and crosses the event horizon cannot escape the gravitational pull and are consumed. " OwnerUserId="25787" LastEditorUserId="25787" LastEditDate="2019-02-02T15:33:10.227" LastActivityDate="2019-02-02T15:33:10.227" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29441" PostTypeId="1" CreationDate="2019-02-02T04:18:12.153" Score="1" ViewCount="44" Body="&lt;p&gt;I have a problem with the collection of photometric data for star WW Aur. I can't find any file with HJD and mag column beside &lt;a href=&quot;http://cdsarc.u-strasbg.fr/viz-bin/cat/J/MNRAS/363/529&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one on Vizier&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need more than one source of data. I will use the data to create a light curve via Scidavis software.  &lt;/p&gt;&#xA;" OwnerUserId="25789" LastEditorUserId="14094" LastEditDate="2019-07-04T09:59:24.253" LastActivityDate="2019-07-04T09:59:24.253" Title="How to collect photometric data for WW Aur" Tags="&lt;photometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29442" PostTypeId="2" ParentId="29426" CreationDate="2019-02-02T08:01:51.360" Score="1" Body="&lt;p&gt;I can't see the contradiction. One version says average over the regions, the other says integrate over one realisation (containing many regions). I suppose you could argue that one is performing a summation,whilst the other is an integration, but if the regions are small compared with the size of your observable universe then they amount to the same thing.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{1}{\sum_i V_i} \sum_i \lambda_i  V_i \simeq \frac{1}{V}\int \lambda(r)\ dV$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-02-02T08:01:51.360" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29443" PostTypeId="2" ParentId="29409" CreationDate="2019-02-02T08:11:07.727" Score="1" Body="&lt;p&gt;You could try selecting them in an &lt;em&gt;absolute&lt;/em&gt; magnitude versus colour diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. Use the apparent G magnitude and parallax to find absolute G magnitude and plot this versus a colour (&lt;span class=&quot;math-container&quot;&gt;$B_p-R_p$&lt;/span&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The HB stars occupy a reasonably distinct region (box?) in this diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your problem is that this distinct box is blurred by differences in composition and by extinction/reddening. You could sidestep the latter to some extent by avoiding low Galactic latitudes.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-02-02T08:11:07.727" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29445" PostTypeId="1" CreationDate="2019-02-02T09:52:26.530" Score="4" ViewCount="127" Body="&lt;p&gt;I mean, we experience tides here on earth because of combined effect of gravitational field of moon and sun. So I thought that tides could also occur on other planets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Let us consider Neptune where the gas transits into a slushy ice and water layer. The water-ammonia ocean serves as the planet's mantle, and contains more than ten times the mass of Earth. Also it has &lt;strong&gt;14&lt;/strong&gt;  moons. What I was thinking was that more moons would mean more tidal effect would be seen even if distance from sun has increased.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then I thought that the moons of Neptune would all have different orbits and different time periods of rotation. So this will definitely impact the extent of tides. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-To what extent will this affect tides?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;And how will tides formed on Neptune be different than that  formed on earth? I mean, how will they vary in properties?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="25791" LastActivityDate="2019-02-02T15:56:04.220" Title="Do tides occur on planets like Neptune, having water?" Tags="&lt;neptune&gt;&lt;tidal-forces&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" ClosedDate="2019-08-15T13:45:33.097" ContentLicense="CC BY-SA 4.0" />
  <row Id="29446" PostTypeId="2" ParentId="29435" CreationDate="2019-02-02T10:05:10.667" Score="0" Body="&lt;p&gt;If we draw a circle and draw a 315 degree angle on that circle, in the counter clockwise direction, we would roughly get the same point as when we draw an angle of 44.8 degrees in the opposite direction (the equivalent of drawing -44.8 in the same direction) because 315+45=360. There therefore does not seem to be an error in your results... Visually the two angles are the same. As for the slight inaccuracy, it may be due to floating point precision errors or measurement errors. &lt;br&gt;&#xA;If you want positive right ascension, all you need to do is add 360 if it is negative. Before doing that (this is general advice, it may not be important to your code), you ought to subtract the decimal part, get the remainder on dividing by 360, and re-add the decimal part. &lt;br&gt; Edit: The difference between GST and GMST is called the equation of time (for Grinich), and it does not grow beyond 16 minutes, so I think you can use it for estimates, but then there would be no point of rounding your results to the tenth decimal or so. I honestly don't know if there is a &quot;GST for dummies&quot; (normally I wouldn't post an incomplete answer, but the question was edited), but using GMST should suffice if the 16 minute error margin is acceptable. &lt;/p&gt;&#xA;" OwnerUserId="24014" LastEditorUserId="24014" LastEditDate="2019-02-03T16:02:46.150" LastActivityDate="2019-02-03T16:02:46.150" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29447" PostTypeId="2" ParentId="29452" CreationDate="2019-02-02T15:56:04.220" Score="8" Body="&lt;p&gt;There will be tides, but they will be not very large.  It's pretty easy to get a good estimate of their size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two things that control tides on Earth and both will be present anywhere else: The tidal forces from celestial bodies, and the size of the body of liquid in which the tides are raised along with resonances int he body of liquid.  Taking them in turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tidal force is a bit funny, since the size of the tides raised on a planet by a celestial body is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) &lt;em&gt;proportional&lt;/em&gt; to the mass of the celestial body raising the tide, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) &lt;em&gt;inversely proportion to the &lt;strong&gt;cube&lt;/strong&gt; of the distance away&lt;/em&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) &lt;em&gt;inversely&lt;/em&gt; proportional to the surface gravity of the planet,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) and (roughly) &lt;em&gt;proportional&lt;/em&gt; to the diameter of the body of liquid in which the tides are being raised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That inverse cube of the distance means that the ability of a body to raise a tide drops off very quickly with distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing Earth and Neptune: The Sun is 30 times further from Neptune than Earth, so Solar tides will be roughly 27,000 times smaller on Neptune.  They can be completely ignored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Triton (Neptune's largest moon) is the jackpot, since it's a third as massive as Earth's Moon, and about the same distance away.  Neptune is 17 times more massive than Earth and four times the diameter.  So Triton tides of Neptune will be about fifteen times smaller than lunar tides on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the other moons are much smaller or further away or both and will raise completely negligible tides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So from gravitational effects alone, Triton will raise small tides and nothing else will be measurable without good instruments and a lot of care.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note, I'm talking about tides in oceans comparable in size to the planet -- for smaller bodies of liquid, the tides will scale down.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Earth, tides vary all over the place, and this is due to the configuration of the body of water and resonance effects.  Basically, if a body of water has a resonance -- a natural period for sloshing back and forth -- that matches the period of the tidal force, the tide can build up like repeated pushes to a kid in a swing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, there can be a funnel effect with two arms of land coming together, or shallowing water that can amplify the tides. (E.g., the Bay of Fundy.)  Since as far as we know, Neptune has no land, it's effectively an planetary ocean and will get the full effect of the gravitational tides, but no resonances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, bottom line: Small Triton tides and nothing else.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2019-02-02T15:56:04.220" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29448" PostTypeId="2" ParentId="29380" CreationDate="2019-02-02T22:07:22.890" Score="6" Body="&lt;p&gt;Short answer: No, it could not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon already has little enough mass relative to the earth so that its orbital period is not much shorter than a satellite with negligible mass. So, any object at about the distance of the moon would take around a month. If the object really had &lt;em&gt;negligible&lt;/em&gt; mass it would be moving a little slower than the moon, so let's be generous and say 5 weeks. That means the object would have to be large enough to extend 1/5 of the way around the sky to block the sun for just one week. This is not possible. Such an object would be much larger than the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you imagine an object closer to the earth, it would be orbiting faster so it would have to be even larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving farther away is a little better. The orbital period is proportional to the radius raised to the 3/2 power, so moving the moon four times farther away would make its period eight times as long. But it would have to be four times larger, or 64 times the volume, just to eclipse the sun at all. And at that point it's approaching the size of the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It &lt;em&gt;may&lt;/em&gt; be possible to construct a synthetic object that orbits the sun at earth's L1 point (between the earth and the sun at just the right distance to stay in line with them) and permanently blocks the sun. You could tweak its solar orbit so that it only blocked the sun part of the time. It could be something like a mylar disk that rotates to avoid collapsing into a ball. But this becomes more of an engineering question than an astronomy one. No plausible natural object would have this configuration.&lt;/p&gt;&#xA;" OwnerUserId="7752" LastActivityDate="2019-02-02T22:07:22.890" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29449" PostTypeId="1" CreationDate="2019-02-03T03:33:03.533" Score="3" ViewCount="633" Body="&lt;p&gt;What is the difference between albedo, absolute magnitude or aparent magnitude? I can’t really tell what’s the difference between them.&lt;/p&gt;&#xA;" OwnerUserId="25797" LastActivityDate="2019-02-04T03:43:13.680" Title="What is the difference between albedo, absolute magnitude or aparent magnitude?" Tags="&lt;light&gt;&lt;magnitude&gt;&lt;albedo&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29450" PostTypeId="2" ParentId="29441" CreationDate="2019-02-03T22:35:04.567" Score="2" Body="&lt;p&gt;I did a bibliography search on the object via &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?bibyear1=1850&amp;amp;bibyear2=%24currentYear&amp;amp;submit=Display&amp;amp;Ident=%40885136&amp;amp;Name=V*%20WW%20Aur&amp;amp;bibdisplay=refsum&amp;amp;bibyear1=1850&amp;amp;bibyear2=%24currentYear#lab_bib&quot; rel=&quot;nofollow noreferrer&quot;&gt;Simbad&lt;/a&gt; and there isn't a lot of data available since the &lt;a href=&quot;https://academic.oup.com/mnras/article/363/2/529/1127173&quot; rel=&quot;nofollow noreferrer&quot;&gt;Southworth et al. 2005&lt;/a&gt; article you linked to. Looking at the paper it seems the only other source of photometric data that is mentioned is &lt;em&gt;Kiyokawa and Kitamura, 1975&lt;/em&gt; in &lt;em&gt;Ann. Tokyo Astron. Obs., second series&lt;/em&gt; which seems to be a in-house publication of Tokyo Observatory and doesn't seem to be available online. The only recent references are a few times of eclipse, which would help any period change modelling, but not the general modelling of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This situation is not uncommon; there are a lot of binary stars out there and at a period of 2.5 days, this one is quite &quot;expensive&quot; in terms of observing time. This is even more the case if you want to get lightcurves in several filters, to aid the modelling by constraining temperatures etc.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2019-02-03T22:35:04.567" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29451" PostTypeId="2" ParentId="29449" CreationDate="2019-02-04T00:14:55.477" Score="4" Body="&lt;p&gt;&lt;strong&gt;Albedo&lt;/strong&gt; is a measurement of reflected light. It varies between 0 and 1, with 0 being completely black (reflecting no light) and 1 being completely white (reflecting all light). The albedo of coal is about 0.04, which is actually very similar to the albedo of our moon. The Albedo of ice is about 0.7. Only planets and other small objects are described using Albedo because they cannot make their own light, and so its apparent brightness is due to reflecting sunlight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Absolute magnitude&lt;/strong&gt; is a measure of how much light a star is emitting overall. Absolute magnitude is related to a bunch of other properties of the star, mainly how massive it is and at what stage of its lifetime it's in.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Apparent magnitude&lt;/strong&gt; can be used to describe both stars and planets. It is simply a measure of how bright the object appears to us. An illustrative example is Sirius and Canopus. Sirius is the brightest star in the sky, but it is bright because it is close, only being 8 light years away. Canopus is the second brightest star in the sky, and appears about half as bright as Sirius, but is about 300 light years away, which means the star itself must much brighter than Sirius. If we check the absolute magnitudes and do some maths (Sirius absolute magnitude = +1.42, Canopus = -5.71, the more negative the number the brighter the star, brightness difference = 2.5^(5.71+1.42)) we get a brightness difference of 680 times between Canopus and Sirius.   &lt;/p&gt;&#xA;" OwnerUserId="17655" LastActivityDate="2019-02-04T00:14:55.477" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29452" PostTypeId="1" CreationDate="2019-02-01T15:19:11.337" Score="6" ViewCount="546" Body="&lt;p&gt;I mean, we experience tides here on earth because of combined effect of gravitational field of moon and sun. So I thought that tides could also occur on other planets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Let us consider Neptune where the gas transits into a slushy ice and water layer. The water-ammonia ocean serves as the planet's mantle, and contains more than ten times the mass of Earth. Also it has &lt;strong&gt;14&lt;/strong&gt;  moons. What I was thinking was that more moons would mean more tidal effect would be seen even if distance from sun has increased.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then I thought that the moons of Neptune would all have different orbits and different time periods of rotation. So this will definitely impact the extent of tides. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-To what extent will this affect tides?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;And how will tides formed on Neptune be different than that             formed on earth? I mean, how will they vary in properties?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="25791" OwnerDisplayName="Garima" LastActivityDate="2019-02-04T01:40:45.037" Title="Do tides occur on planets like Neptune, having water?" Tags="&lt;tidal-forces&gt;&lt;neptune&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29453" PostTypeId="2" ParentId="29452" CreationDate="2019-02-01T17:23:05.207" Score="7" Body="&lt;p&gt;Neptune's largest moon, Triton, is about a third the mass of Earth's moon, and roughly the same distance away from its primary, so it will indeed have noticeable tidal effect on Neptune's atmosphere and oceans. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neptune's other moons are all tiny (all of them together being about 0.5% of Triton's mass), so while they do exert tidal forces, those effects would be dwarfed by Triton even when all the small moons are on the same side of Neptune. &lt;/p&gt;&#xA;" OwnerUserId="2573" OwnerDisplayName="Russell Borogove" LastActivityDate="2019-02-01T17:23:05.207" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29454" PostTypeId="2" ParentId="29452" CreationDate="2019-02-02T01:28:37.290" Score="5" Body="&lt;p&gt;Neptune’s moon Triton will cause tidal potential changes that are less than Earths, but still good fractions of a meter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But whether there are visible tides is a harder question. On Earth, we have visible tides because the land doesn’t respond as much as the water does. A planet completely covered by deep water would have no detectable ocean tides: everything would just move together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Neptune’s surface is slush &amp;amp; liquid, it may move together enough that Neptunians don’t notice any tides. &lt;/p&gt;&#xA;" OwnerDisplayName="Bob Jacobsen" LastActivityDate="2019-02-02T01:28:37.290" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29455" PostTypeId="2" ParentId="29452" CreationDate="2019-02-03T02:58:07.120" Score="4" Body="&lt;p&gt;As Russell Borogove and Bob Jacobsen write, indeed Triton wil raise tides on Neptune. But they won't be like the tides people envision here on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2005AGUFM.P11A0088A&quot; rel=&quot;nofollow noreferrer&quot;&gt;Complex modeling of the vertical structure of Neptune&lt;/a&gt; (behind a paywall, but the abstract says enough) by researchers like Sushil Atreya suggests there is no liquid water ocean at Neptune, but instead an &quot;ionic ocean&quot; of water with dissolved ammonia &lt;em&gt;very&lt;/em&gt; deep in the atmosphere, deep enough to have pressures in the thousands of bars. At those depths the &lt;a href=&quot;https://www.space.com/18920-neptune-composition.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;temperatures are so high (2,000-5,000 K)&lt;/a&gt; they are well above the &lt;a href=&quot;https://en.wikipedia.org/wiki/Critical_point_(thermodynamics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;647 K critical temperature of water&lt;/a&gt;, so liquid water doesn't exist. Instead you have a &lt;em&gt;supercritical fluid&lt;/em&gt; that is dense like a liquid but compressible like a gas, with other characteristics (like chemical reactivity) that you don't find in the liquid or gaseous state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notably, at Neptune (and at other giant planets) &lt;em&gt;there is no discontinuous surface&lt;/em&gt; where above it is one state and below it is another. Here on Earth we're quite accustomed to such a discontinuous (at least on human scales) surface, like the surface of our oceans: above that surface is a gas, our atmosphere, and below it is a liquid. Instead, at Neptune the water density and content of other species grades gradually as you go down. So you can't see a discontinuous surface rising and falling with respect to its surroundings, like an ocean level at a shoreline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here on Earth we see the tides at the boundaries of liquids and solids—shorelines of continents and islands. The response of the liquid in our oceans to tidal forces is larger than the response of the solid part of Earth. Seeing that difference with human eyes is only possible when you have solid Earth extending above the ocean's level. Were Earth's lithosphere (the solid part) completely smooth, i.e. no topography, and the oceans a uniformly deep layer above that, someone floating on that ocean wouldn't detect the tidal rise and fall without instruments that go far, far beyond human senses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, you couldn't detect the tidal response at Neptune. It's made even more difficult to detect by the absence of a discontinuous surface as a reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, &lt;a href=&quot;https://www.researchgate.net/publication/234211521_Water-Ammonia_Ionic_Ocean_on_Uranus_and_Neptune-Clue_from_Tropospheric_Hydrogen_Sulfide_Clouds&quot; rel=&quot;nofollow noreferrer&quot;&gt;this reference&lt;/a&gt; (that I found while searching for Sushil's) attributes the &lt;em&gt;same paper&lt;/em&gt; to the &lt;em&gt;wrong Atreya!&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="Tom Spilker" LastActivityDate="2019-02-03T02:58:07.120" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29456" PostTypeId="1" AcceptedAnswerId="29474" CreationDate="2019-02-01T21:45:58.480" Score="11" ViewCount="227" Body="&lt;p&gt;Often in discussing exoplanets we hear that they are close to the parent star (usually a red dwarf) but then there's a caveat that the planet is tidally locked to that star and that that fact severely decreases prospects for that planet to harbor life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do we know that an exoplanet is tidally locked to a star? I doubt that we can directly observe the rotation of that planet. Does that mean that we conclude that it's tidally locked because that's what our theoretical models tell us, and not that we observe it directly?&lt;/p&gt;&#xA;" OwnerUserId="24234" OwnerDisplayName="stackzebra" LastActivityDate="2019-02-05T19:26:42.543" Title="How do we know that an exoplanet is tidally locked to the parent star?" Tags="&lt;orbital-mechanics&gt;&lt;exoplanet&gt;&lt;rotation&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29457" PostTypeId="1" AcceptedAnswerId="29465" CreationDate="2019-02-04T01:57:40.787" Score="4" ViewCount="62" Body="&lt;p&gt;Captions in the Hubble Space Telescope YouTube channel video &lt;a href=&quot;https://youtu.be/M8XQFVyYfL4&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shedding New Light on the Whirlpool Galaxy&lt;/a&gt; say:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;(visible) Along the dark dust lanes, bright pink nebulae are the birthplace of new stars.&lt;/p&gt;&#xA;&lt;p&gt;(X-ray, lower energy) Before leaving these stellar nurseries, the most massive stars die in supernova explosions, which heat the surrounding gas to X-ray temperatures.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I'm not very familliar with star formation processes.&lt;/p&gt;&#xA;&lt;p&gt;These might be considered as separate questions, but I've put them here together because they all refer to the same two sentences in this video, and it's likely that those familliar with nurseries, or wanting to know more about them, will have something to say or learn about all three.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Do stars tend to leave &amp;quot;these stellar nurseries&amp;quot; after a while, and it's only the short lifetime of the most massive stars that keeps them from leaving before going supernova?&lt;/li&gt;&#xA;&lt;li&gt;What impact does the heating of surrounding gas to X-ray temperatures have on the formation and evolution of other stars within a given &amp;quot;nursery&amp;quot;?&lt;/li&gt;&#xA;&lt;li&gt;Are there really such things as spatially well defined &amp;quot;nurseries&amp;quot;? Are they distinct and roughly countable, or is this a subjective popular term? Can one point to something and say &amp;quot;this little blob here at x, y, z in the spiral arm of the Whirlpool Galaxy, it's a classic nursery&amp;quot;?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/12ul2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/12ul2.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DdMTl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DdMTl.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-02-04T16:51:06.593" Title="Stellar nurseries in the &quot;Shedding New Light on the Whirlpool Galaxy&quot; video" Tags="&lt;galaxy&gt;&lt;stellar-evolution&gt;&lt;star-formation&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29458" PostTypeId="2" ParentId="29449" CreationDate="2019-02-04T03:31:44.817" Score="1" Body="&lt;p&gt;Here is how I understand these terms.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Apparent magnitude&lt;/em&gt; is what you can see, or measure from where you really are and where the object really is.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Absolute magnitude&lt;/em&gt; is a mathematical prediction based on a standardized distance and configuration (see below) so that you don't have to specify the particular conditions. Venus always has roughly the same absolute magnitude because it's always roughly the same size and &lt;em&gt;albedo&lt;/em&gt; (see next), but varies greatly in apparent magnitude be cause of changes in distance and phase angle (sun-venus-earth angle).&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Albedo&lt;/em&gt; is a little tricky because different objects have different surface roughness and angular scattering properties. If it was a smooth ping-pong ball and had a well defined diffusivity profile, then it's easy to imagine a white, gray, or black ping-pong ball representing transitions from 1 to 0.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I've just asked &lt;a href=&quot;https://space.stackexchange.com/q/34009/12102&quot;&gt;Is albedo ever greater than unity? How is it calculated for various surface diffusivities and specular reflection?&lt;/a&gt; to get at some of those subtleties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For stars and other objects that make their own light:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Q6tyC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q6tyC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For planets, asteroids, comets, satellites (artificial and natural) &lt;strong&gt;and even electric cars&lt;/strong&gt; &lt;a href=&quot;https://space.stackexchange.com/a/32065/12102&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;https://space.stackexchange.com/a/25121/12102&quot;&gt;2&lt;/a&gt; that are seen in reflected light from the Sun:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uuA2L.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uuA2L.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-02-04T03:43:13.680" LastActivityDate="2019-02-04T03:43:13.680" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29459" PostTypeId="1" CreationDate="2019-02-04T03:45:21.330" Score="0" ViewCount="47" Body="&lt;p&gt;I'm trying to understand how albedo would be &lt;em&gt;calculated&lt;/em&gt; for a body with well-characterized surface reflectance properties, so that it's absolute magnitude could be defined. One could argue that this is going backwards (one should measure apparent magnitude, correct for distances and phase angle to extract absolute magnitude, correct for radius to extract albedo) but consider if the body were not smooth and diffusely reflecting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine two identically sized spherical bodies; one diffuse or matte white and the other somehow covered in liquid mercury and exhibiting specular rather than diffuse reflectivity, with no wind or waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would they both have albedo of unity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If these two reflective bodies were very irregularly-shaped and had an extremely large, flattened (much larger radius of curvature) face, would its albedo be larger than unity at times?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I'm looking for how astronomers and planetary scientists &lt;em&gt;use the term albedo&lt;/em&gt; and not just a mathematical hypothesis. There may be ways this is dealt with practically.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-02-04T03:45:21.330" Title="Is albedo ever greater than unity? How is it calculated for various surface diffusivities and specular reflection?" Tags="&lt;apparent-magnitude&gt;&lt;magnitude&gt;&lt;absolute-magnitude&gt;&lt;albedo&gt;" AnswerCount="0" CommentCount="11" ClosedDate="2019-02-04T07:11:00.093" ContentLicense="CC BY-SA 4.0" />
  <row Id="29462" PostTypeId="1" CreationDate="2019-02-04T08:05:18.270" Score="2" ViewCount="179" Body="&lt;p&gt;I am writing a piece of python code to convert right ascension and declination with a given radius to minimum and maximum right ascension and declination. I firstly can't understand why the min and max is not always just the given coordinates +- the radius, and secondly how I would write a piece of code to make this conversion. Cheers. &lt;/p&gt;&#xA;" OwnerUserId="25806" LastEditorUserId="1891" LastEditDate="2019-02-04T09:03:43.953" LastActivityDate="2019-11-08T02:02:05.303" Title="How to convert celestial Coordinates with a radius, to max and min celestial coordinates?" Tags="&lt;python&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29465" PostTypeId="2" ParentId="29457" CreationDate="2019-02-04T16:47:12.563" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Do stars tend to leave &quot;these stellar nurseries&quot; after a while, and it's only the short lifetime of the most massive stars that keeps them from leaving before going supernova?&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. The lifetimes of very massive stars (10-50 Myr) are comparable with the dispersion timescale for young clusters and associations. So massive stars tend to die near where they were born and they will be surrounded by their (more numerous) lower mass siblings. Lower mass stars have long lives and end up dispersing and making up the general population of a galaxy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;What impact does the heating of surrounding gas to X-ray temperatures have on the formation and evolution of other stars within a given &quot;nursery&quot;?&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is unclear. The injection of energy and heating of gas ought to make it less susceptible to collapse and fragmentation. On the other hand that doesn't matter isf most of the gas that was going to turn into stars has already turned into stars. On the other hand, it is possible that the shock waves and outflows from supernovae actually compress surrounding structures and trigger new star formation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Are there really such things as spatially well defined &quot;nurseries&quot;? Are they distinct and roughly countable, or is this a subjective popular term? Can one point to something and say &quot;this little blob here at x, y, z in the spiral arm of the Whirlpool Galaxy, it's a classic nursery&quot;?&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To some extent yes. There are some massive clusters of young stars that can be observed in other galaxies and can be certainly identified as the birthplaces of tens of thousands or even millions of stars. On the other hand, in our own galaxy it appears that although most stars are probably born in clusters and associations, there is no such thing as a &quot;typical&quot; cluster and that star formation takes place at a range of scales and at a range of densities. i.e. Much of the star formation taking place in other galaxies may take place away from some of the obvious blobs of star formation.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="7982" LastEditDate="2019-02-04T16:51:06.593" LastActivityDate="2019-02-04T16:51:06.593" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29466" PostTypeId="1" CreationDate="2019-02-04T21:12:52.417" Score="4" ViewCount="69" Body="&lt;p&gt;In the description of Eudoxus’s model for the sun he says one the three circles which describe solar motion is “the circle which runs along the middle of the zodiac”. Does this refer to the ecliptic plane?&#xA;What I thought is that it does mean the same thing as the ecliptic, which as seen from earth when looking up to the night sky, appears as if it cuts the zodiac in the middle. In a spherical model, the Sun’s apparent movement can be described as running along the constellations of the zodiac throughout the year on a “circle” which is known as the ecliptic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I getting this wrong?&lt;/p&gt;&#xA;" OwnerUserId="25814" LastActivityDate="2019-02-04T21:12:52.417" Title="Did Eudoxus describe the eclipitic on his sun motion model?" Tags="&lt;the-sun&gt;&lt;ecliptic&gt;&lt;apparent-motion&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29467" PostTypeId="1" CreationDate="2019-02-04T22:43:45.870" Score="10" ViewCount="136" Body="&lt;p&gt;Approximately 468 million years ago there were a large number of L-chondrite meteorites falling to Earth. This is called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ordovician_meteor_event&quot; rel=&quot;noreferrer&quot;&gt;Ordovician meteor event&lt;/a&gt; and was probably the result of a large collision in the asteroid belt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the lunar cratering record also show evidence of this event?&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-02-04T23:25:32.420" LastActivityDate="2021-05-21T04:08:26.453" Title="Is the Ordovician meteor event detectable in the lunar cratering record?" Tags="&lt;the-moon&gt;&lt;meteorite&gt;&lt;crater&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29468" PostTypeId="1" CreationDate="2019-02-05T05:22:59.233" Score="-1" ViewCount="109" Body="&lt;p&gt;Lets suppose there is a huge planet having large amount of gravitational pull, then will the light passing through the planet get attracted towards it.&lt;/p&gt;&#xA;" OwnerUserId="24938" LastEditorUserId="1891" LastEditDate="2019-02-05T16:33:52.827" LastActivityDate="2019-02-05T17:28:43.927" Title="Can light be curved due to heavy gravitational force?" Tags="&lt;planet&gt;&lt;gravity&gt;&lt;light&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="29469" PostTypeId="1" AcceptedAnswerId="29470" CreationDate="2019-01-31T07:33:17.557" Score="4" ViewCount="136" Body="&lt;p&gt;On this clear early morning over London, I was studying the moon with a pair of binoculars and noticed a tiny dot of light that appeared to trace a slow orbit around the moon. I reckon it moved roughly 30 degrees in as many minutes. I'm wondering whether I was seeing a star or if it's remotely possible that I was seeing sunlight reflecting off an orbiting craft? Although the sky was clear it certainly wasn't starry; it was about 100 minutes before sunrise.&lt;/p&gt;&#xA;" OwnerUserId="25820" OwnerDisplayName="Matt Jenkins" LastActivityDate="2019-02-05T19:06:35.827" Title="Is it possible to see a spacecraft orbiting the moon with a pair of binoculars?" Tags="&lt;the-moon&gt;&lt;orbit&gt;&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="29470" PostTypeId="2" ParentId="29469" CreationDate="2019-02-05T13:58:04.840" Score="4" Body="&lt;p&gt;No. There are currently 5 spacecraft orbit the Moon (Or the E-M L2 point). These are LRO, Artemis P1 and P2, Chang'e 5 T1, and the Queqiao relay satellite. Of these, the largest is probably &lt;a href=&quot;https://www.nasa.gov/pdf/360020main_LRO_LCROSS_presskit2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;LRO&lt;/a&gt;. The solar panels, which are the largest part of most spacecraft, &quot;he three panels together are 168 inches wide and extend out from the spacecraft 126 inches&quot;. Converted to meters, that is 4.25 x 3.2m. Just to give an idea of how visible that would be, Elon's Tesla Roadster, according to &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL Horizons&lt;/a&gt;, when it was at the distance of the Moon, had an apparent brightness of about 15.27. Using &lt;a href=&quot;http://www.icq.eps.harvard.edu/MagScale.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this chart,&lt;/a&gt; an aperture of about 50 cm is required. LRO should be a bit brighter, but not enough to make a huge difference. And none of this takes in to account the glare of the Moon itself. Whatever you saw, it probably wasn't a satellite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect it was a star of some kind. If you can provide the date and time we might be able to figure out a candidate star.&lt;/p&gt;&#xA;" OwnerUserId="182" LastActivityDate="2019-02-05T13:58:04.840" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29471" PostTypeId="1" CreationDate="2019-02-05T15:57:54.613" Score="3" ViewCount="2855" Body="&lt;p&gt;I came across something that won't let me go, but no matter how much I calculate (which I do not like), it doesn't seem to work.&#xA;Also: I've used so much apps and links already to convert, but it doesn't work in that direction. You see, local time converting to sidereal time = not that hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to know the local 'normal' time of my local sidereal time.&#xA;The sidereal time is exactly 13.30h, from that I would like to know the normal local time.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The time zone = Brussels (CET)&lt;/li&gt;&#xA;&lt;li&gt;Olson time zone Europe/Brussels&lt;/li&gt;&#xA;&lt;li&gt;Time difference: same as Amsterdam and Europe UTC+1&lt;/li&gt;&#xA;&lt;li&gt;no summer time at the moment (and what if that was the case, what would the local time be then?)&lt;/li&gt;&#xA;&lt;li&gt;Latitude: 50,929315°&lt;/li&gt;&#xA;&lt;li&gt;Longitude: 5,337367°&lt;/li&gt;&#xA;&lt;li&gt;Address: Grote Markt 18, 3500 Hasselt Belgium.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The reason I need to know this has to do with a sort of awareness I found in ancient writings and books. It has extra power than to meditate, and I would like very much to try this out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course, I need to know the local time then.&lt;/p&gt;&#xA;" OwnerUserId="25819" LastEditorUserId="616" LastEditDate="2019-02-05T20:06:35.680" LastActivityDate="2019-02-06T13:41:29.807" Title="How to convert sidereal time to local time?" Tags="&lt;time&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29472" PostTypeId="2" ParentId="29468" CreationDate="2019-02-05T17:28:43.927" Score="3" Body="&lt;p&gt;Light (in general relativity) follows paths in spacetime dictated by the curvature of spacetime.  That curvature is a result of gravitational distortion of spacetime by energy (and this includes mass as a form of energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So light follows curved paths through curved spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should not, however, think of light as feeling a gravitational force in a Newtonian sense.  This is &lt;em&gt;not&lt;/em&gt; what happens.  You need to forget Newtonian ideas like that in general relativity.   For example, it's tempting to say that light is attracted to a gravitational source, but this is not really the case and can lead to incorrect thinking (e.g. like imagining light was being accelerated, which it is not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The technical term for paths that light follows in spacetime is &lt;a href=&quot;https://physics.stackexchange.com/questions/188859/what-is-a-null-geodesic&quot;&gt;null geodesic&lt;/a&gt; (the link is to a Q&amp;amp;A on the Physics SE site).  I suspect that's possibly more that the OP wants to knows, but the link is there to investigate if they want to.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2019-02-05T17:28:43.927" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29473" PostTypeId="2" ParentId="29469" CreationDate="2019-02-05T19:06:35.827" Score="1" Body="&lt;p&gt;The other answer is conclusive (it probably wasn't a satellite). Let me just add more data w.r.t. time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the lowest altitude possible, a lunar orbit is completed in 1 hour 51 minutes. The higher the orbit, the longer it takes - say at 400 km altitude (similar to the ISS altitude around Earth), it would take 2 hours 20 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say it takes 2 hours sharp. That's 120 minutes for 360 degrees, or 3 degrees per minute. In 30 minutes you would expect the satellite to do about 1/4 of its orbit. But it's pretty complicated to translate that into what you see with a telescope from Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the above is for very low orbits, which in a small telescope would seem to graze the surface of the Moon. If there's any appreciable distance between the satellite and the Moon, the orbit would be much, much slower.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2019-02-05T19:06:35.827" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29474" PostTypeId="2" ParentId="29456" CreationDate="2019-02-05T19:16:48.577" Score="6" Body="&lt;p&gt;They are &lt;em&gt;probably&lt;/em&gt; tidally locked, given some assumptions about age of the planetary system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_locking#Moons&quot; rel=&quot;noreferrer&quot;&gt;formula&lt;/a&gt; to calculate how long it'll take for an orbiting object to tidally lock, and one of the outcomes of that formula is that large objects in close orbits don't take very long to tidally lock. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since many of the exoplanets we've found are large objects in close orbits of old stars, tidal locking is a likely outcome. &lt;/p&gt;&#xA;" OwnerUserId="361" LastActivityDate="2019-02-05T19:16:48.577" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29475" PostTypeId="2" ParentId="29456" CreationDate="2019-02-05T19:26:42.543" Score="6" Body="&lt;p&gt;Hobbes's answer is correct for planets assumed to be tidally-locked. There are nevertheless a couple of exoplanets we know definitely &lt;em&gt;aren't&lt;/em&gt; tidally locked because their rotations have been measured. The first rotation measurement of an exoplanet was for &lt;a href=&quot;https://www.eso.org/public/news/eso1414/&quot; rel=&quot;noreferrer&quot;&gt;Beta Pictoris b&lt;/a&gt;, which turns out to have an 8-hour rotation period. Another example is &lt;a href=&quot;https://scitechdaily.com/hubble-measures-rotation-of-super-jupiter-exoplanet-2m1207b/&quot; rel=&quot;noreferrer&quot;&gt;2M1207b&lt;/a&gt; with a 10-hour rotation. Admittedly these objects are so far away from their stars that we wouldn't expect them to be tidally locked: the rotation measurements rely on these objects being imaged, which requires a large star-planet separation.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastActivityDate="2019-02-05T19:26:42.543" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29476" PostTypeId="2" ParentId="19943" CreationDate="2019-02-05T21:20:32.340" Score="1" Body="&lt;p&gt;An educated guess: on Earth, the working fluid in a &quot;geyser&quot; is water, which remains a fluid more or less permanently after each eruption.  This means that geyser eruptions are not associated with buildup of new geological material around the geyser (apart from the much slower process of dissolved minerals accumulating as the water evaporates).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contrariwise, on Earth, the working fluid in a &quot;volcano&quot; is molten rock, which solidifies after reaching the surface.  Each new eruption permanently deposits new solid material in the vicinity of the volcano.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were writing definitions, I would define a &quot;cryovolcano&quot; as a geological vent system where the working fluid is water, but where the environment is cold enough that the ejecta from eruptions accumulates in a permanent mountain of ice around the vent.&lt;/p&gt;&#xA;" OwnerUserId="1498" LastActivityDate="2019-02-05T21:20:32.340" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29477" PostTypeId="2" ParentId="12967" CreationDate="2019-02-05T23:58:49.620" Score="2" Body="&lt;p&gt;There is another bottom to this. The question would be who were the astronomers voting in the IAU at that time and what were the geopolitical settings and resentments in Europe after 1918 end of war. Who were the gainers and losers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some answers can be found in historical documents of Polskie Towarzystwo Astronomiczne or Centrum Astronomiczne Mikołaja Kopernika w Toruniu. Happy hunting ;)&lt;/p&gt;&#xA;" OwnerUserId="25822" LastEditorUserId="7472" LastEditDate="2019-02-06T10:30:44.623" LastActivityDate="2019-02-06T10:30:44.623" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29479" PostTypeId="1" AcceptedAnswerId="29493" CreationDate="2019-02-06T00:56:02.403" Score="2" ViewCount="38" Body="&lt;p&gt;Firecrackers all night and all libraries being closed here are two good indicators to a nerd like me that &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_New_Year&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lunar New Year&lt;/a&gt; has come again, and that means that it's the beginning of a lunar month and new Moon as seen from Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That also means that the Chinese lander &lt;a href=&quot;https://en.wikipedia.org/wiki/Chang%27e_4&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chang'e-4&lt;/a&gt; and rover &lt;a href=&quot;https://en.wikipedia.org/wiki/Chang%27e_4#Chang&amp;#39;e_lander_and_Yutu-2_rover&quot; rel=&quot;nofollow noreferrer&quot;&gt;Yutu-2&lt;/a&gt; should be in sunlight since they are on the far side of the Moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What phases of the Moon &lt;em&gt;as seen from Earth&lt;/em&gt; correspond to times of sunrise and sunset for those two?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-02-07T01:04:58.203" Title="Happy lunar new year! When will &quot;sunset&quot; happen for Chang'e-4 and Yutu-2?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29480" PostTypeId="1" CreationDate="2019-02-06T01:30:41.283" Score="1" ViewCount="99" Body="&lt;p&gt;A fixed magnetic field will induce an electrical current in a conductive object moving through the field.  The strength of the eddy currents will depend on the shape and composition of the conductor and also how fast it is moving.  The electrical current produces ohmic heating in the conductor, and some of the kinetic energy of the moving conductor will be turned to heat. This is how induction brakes work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if this principle applies to fast moving conductors like asteroids or satellites that encounter planetary magnetic fields.  Are planetary magnetic fields strong enough to act as induction brakes on moving conductors?  A satellite experiencing induction braking would fall to a lower and lower orbit and eventually out of orbit.  Passing asteroids might have their course deviated more than would be expected just from gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My search turned up only this abstract, which attempted to explain why meteorites are hot via this principle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/full/1946PA.....54..482R&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/full/1946PA.....54..482R&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0aCFV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0aCFV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16293" LastActivityDate="2019-08-23T04:27:53.193" Title="Do planetary magnetic fields slow the movement of conductive extraplanetary moving objects via induction?" Tags="&lt;asteroids&gt;&lt;magnetic-field&gt;&lt;satellite&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="29482" PostTypeId="2" ParentId="29471" CreationDate="2019-02-06T04:54:42.703" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;Convert your current local date and time &quot;t&quot; to UTC if necessary.&lt;/li&gt;&#xA;&lt;li&gt;Use an app to calculate the local sidereal time (LST) for the current date and time t.&lt;/li&gt;&#xA;&lt;li&gt;Calculate the following equation:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$local\;time\;for\;13.30\;LST = t+\frac {13.30-LST\;at\;current\;date\;and\;time\;t}{1.0027379}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, at 3:00 local time (2:00 UTC) on Feb 6, 2019, the LST=11:25:02 (11.4172 hours) at 5 deg 20' 15&quot; east longitude (5.3375 deg). The above formula gives the time for 13.30 hours LST as&#xA;&lt;span class=&quot;math-container&quot;&gt;$$=3:00+\frac{13.30-11.4172}{1.0027379}$$&lt;/span&gt;&#xA;which equals 4.8777 hours local time (4:52:40). Plugging that time (converted to UTC if necessary) back into an app confirms the LST is 13:18:00 (13.30 hours).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After you know one date and time, you can calculate the next date and time approximately by adding 23:56:04 to your answer. For a date more than a few weeks in the future, I suggest you repeat the calculation for a the new date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was using this online site, but any site should give a similar answer: &lt;a href=&quot;http://www.csgnetwork.com/siderealjuliantimecalc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Calendar Date And Time To Julian Day And Sidereal Times&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17355" LastEditorUserId="17355" LastEditDate="2019-02-06T13:41:29.807" LastActivityDate="2019-02-06T13:41:29.807" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29483" PostTypeId="1" AcceptedAnswerId="29484" CreationDate="2019-02-06T07:10:56.513" Score="7" ViewCount="3113" Body="&lt;p&gt;Around 6.15am (GMT + 1) on the 6th of February 2019 I &#xA;was looking out of my window somewhat north, when I saw something one would probably call a shooting star. The features were:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A magnitude I haven't seen before; it even had a soft blinding effect, that's how bright it was.&lt;/li&gt;&#xA;&lt;li&gt;From my view frame it moved from my right to my left (so most likely from east to west). &lt;/li&gt;&#xA;&lt;li&gt;The size was a quarter to half of what the moon usually has from my view frame.&lt;/li&gt;&#xA;&lt;li&gt;It traveled almost through the sky half I was able to see within half a second before its glow was immediately gone again. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am assuming the object must have been either very large and fast or quite close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was there any meteor worthy mentioning passing by Earth at that given time?&lt;br /&gt;&#xA;Or are the features I just described way more common than I am assuming, and I just spotted an event that happens quite regularly?&lt;/p&gt;&#xA;" OwnerUserId="10102" LastEditorDisplayName="user1569" LastEditDate="2019-02-06T14:10:45.553" LastActivityDate="2019-02-09T14:57:19.127" Title="Did I see a mentionable event in Germany early morning on 6th of February 2019?" Tags="&lt;meteor&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29484" PostTypeId="2" ParentId="29483" CreationDate="2019-02-06T07:18:01.173" Score="11" Body="&lt;p&gt;Fireballs are real but rare events. Meteors are formed when small rocks enter the atomsphere from space. Larger rocks or small asteroids will produce brighter meteors. The brightest meteors are called fireballs or bolides. It is not uncommon for parts of the rock that made the fireball to survive and can later be found as meteorites.&lt;/p&gt;&#xA;&lt;p&gt;A couple of things can be confused with fireballs.  When satellites re-enter, they can burn up. Satellites move much more slowly than true fireballs, most fireballs last less than 10 seconds.&lt;/p&gt;&#xA;&lt;p&gt;A comet is an astronomical, not atmospheric, object. It looks like a fuzzy star with a tail. It remains almost motionless in the sky, sometimes for months. Really bright comets are rare, and even the brightest comets are much much dimmer than a fireball.&lt;/p&gt;&#xA;&lt;p&gt;There are several organisations that collect reports of fireballs:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://fireball.amsmeteors.org/members/imo/report_intro&quot; rel=&quot;noreferrer&quot;&gt;https://fireball.amsmeteors.org/members/imo/report_intro&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://ukmeteornetwork.co.uk/fireball-report/&quot; rel=&quot;noreferrer&quot;&gt;https://ukmeteornetwork.co.uk/fireball-report/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The advice given states:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Please, don't report sighting that lasted more than 30 seconds: the vast majority of fireballs are only visible for few seconds.&lt;/li&gt;&#xA;&lt;li&gt;Please, don't report recurring events: seeing a fireball is extremely rare and often a once in a lifetime event.&lt;/li&gt;&#xA;&lt;li&gt;Please, don't report slow blinking objects or lights crossing the sky going by 2 or 3: a fireball looks like a big shooting star.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-02-06T20:20:04.620" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29486" PostTypeId="1" AcceptedAnswerId="29947" CreationDate="2019-02-06T11:41:30.213" Score="2" ViewCount="100" Body="&lt;p&gt;In the ABC News video &lt;a href=&quot;https://www.youtube.com/watch?v=ubYhcVWmrbE&quot; rel=&quot;nofollow noreferrer&quot;&gt;New space discovery has astronomers buzzing&lt;/a&gt; at about 02:30 &lt;a href=&quot;https://en.wikipedia.org/wiki/Hakeem_M._Oluseyi&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dr. Hakeem M. Oluseyi&lt;/a&gt; explains that the shortness of &lt;a href=&quot;https://en.wikipedia.org/wiki/Fast_radio_burst&quot; rel=&quot;nofollow noreferrer&quot;&gt;FRB&lt;/a&gt;s (before dispersion) suggests a compact source size and illustrates this by using the Sun as an example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd never though about this until just now, but the radius of the Sun is indeed about 2.3 light-seconds. It takes light from the edge of the Sun's disk over 2 seconds longer to reach us than light from the central area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any ways in which this light time difference is noticeable, or needs to be taken into account?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-02-06T13:12:18.290" LastActivityDate="2019-03-11T10:45:51.363" Title="Are there any ways in which the light time difference between the center and the edge of the solar disk is noticeable?" Tags="&lt;the-sun&gt;&lt;light&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29487" PostTypeId="1" AcceptedAnswerId="29549" CreationDate="2019-02-06T13:25:11.537" Score="3" ViewCount="170" Body="&lt;p&gt;According to Wikipedia;s &lt;a href=&quot;https://en.wikipedia.org/wiki/Fast_radio_burst#Frequencies_and_dispersion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fast Radio Bursts; Frequencies and dispersion&lt;/a&gt; are recorded they &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The component frequencies of each burst are delayed by different amounts of time depending on the wavelength. This delay is described by a value referred to as a dispersion measure (DM). This results in a received signal that sweeps rapidly down in frequency, as longer wavelengths are delayed more.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The time between the arrival of the pulse at two different frequencies can be used to generate a &lt;em&gt;kind of&lt;/em&gt; measure of distance, based on a dispersion constant. The measure does not have units of length, but of integrated electron density over the path from source to observer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using some fancy Fourier tricks one could first undo the &lt;span class=&quot;math-container&quot;&gt;$1/\nu^2$&lt;/span&gt; delay and then try to reconstruct what the original pulse might have looked like before dispersion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has this been done? If so, how fast (narrow in time) might the original disturbance be? A millisecond? Less?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-02-07T00:48:31.250" LastActivityDate="2019-05-12T00:44:18.433" Title="Just how fast is a Fast Radio Burst thought to be?" Tags="&lt;radio-astronomy&gt;&lt;fast-radio-bursts&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29488" PostTypeId="1" CreationDate="2019-02-06T05:33:22.217" Score="4" ViewCount="125" Body="&lt;p&gt;If this is possible, would it then lead to an object undergoing varying time dilation (predictable variation with elliptical and random variation with three body)?&lt;/p&gt;&#xA;" OwnerDisplayName="Oscar Whalley" LastActivityDate="2019-02-06T14:09:05.727" Title="Is an elliptical orbit around a black hole, or an orbit consisting of three bodies (including the black hole) possible?" Tags="&lt;gravity&gt;&lt;black-hole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29489" PostTypeId="2" ParentId="29488" CreationDate="2019-02-06T06:08:03.230" Score="8" Body="&lt;p&gt;&lt;strong&gt;Elliptical Orbit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, an elliptical orbit around a black hole is possible.  &lt;a href=&quot;https://space.stackexchange.com/questions/1911/is-there-any-stable-orbit-around-a-black-hole&quot;&gt;Provided the orbit doesn't cross too near the event horizon&lt;/a&gt;, the orbital mechanics are the same as orbiting any massive object.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_time_dilation&quot; rel=&quot;noreferrer&quot;&gt;Gravitational time dilation&lt;/a&gt; is a function of distance only, so the time dilation experienced at any point in the orbit should be easily calculated and presumably predicted ahead of time or calculated for past times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Three Body Orbit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you mention an orbit consisting of three bodies, do you mean the black hole, another massive object, and a much smaller object?  Or do you mean three massive bodies?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If the orbit consists of the black hole, another massive object, and a smaller body, the system should be predictable as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Three-body_problem#Restricted_three-body_problem&quot; rel=&quot;noreferrer&quot;&gt;restricted three body&lt;/a&gt; system.  This is made even easier if the smaller body finds its way to a Lagrange point, or resides in a stable repeating orbit.  There are limits to the predictability of the restricted three body cases, which differ based on requisite constraints (e.g. circular vs. elliptical orbit for the smaller body).  If the constraints are not met, or near their limits (not hard limits), the predictability of the orbit will suffer and thus the predictability of the time dilation will suffer as well.  Constraints for the elliptical restricted three body problem include the mass of the small body being &quot;negligible&quot; relative to the two larger bodies, and the ability to approximate the co-orbit of the larger bodies as Keplerian between point masses.  Thus if the mass of the smaller body is some appreciable fraction of either larger body, or if the larger bodies orbit very closely to one another, this may not hold.  The circular restricted three body problem adds the constraint that the co-orbit between the larger bodies must be circular, so if it is near-circular that condition may or may not be said to be met.  Whether constraints are met would be most easily judged by setting a time over which predictability of the system may be maintained.  So if the goal is to be able to predict the system for 1000 years, and the smaller mass is small enough that the system is predictable under the restricted three body problem, the system can be said to meet the constraints for those purposes.&lt;/li&gt;&#xA;&lt;li&gt;If the orbit consists of the black hole and two other massive bodies, the system will not be perfectly predictable (this may not be totally true, apparently solutions exist in the form of &lt;a href=&quot;http://www.math.uvic.ca/faculty/diacu/diacuNbody.pdf&quot; rel=&quot;noreferrer&quot;&gt;convergent power series&lt;/a&gt; though they may not be practical due to their size).  If this is the case, time dilation will not be predictable, but will be calculable for the present and past (provided there is position data from the past).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="25831" OwnerDisplayName="ben" LastActivityDate="2019-02-06T06:08:03.230" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29490" PostTypeId="2" ParentId="2441" CreationDate="2019-02-06T17:41:12.867" Score="6" Body="&lt;p&gt;Several wonderful yet technical answers have been given, and I cannot add anything to those very nice answers that explain why it is not useful to think black holes get &quot;frozen&quot; at their event horizons.  But I can give an answer with a more essentially useful philosophical perspective, which is that the central lesson of relativity is that reality involves a bunch of things happening at various places and times, so reality is something &lt;em&gt;local.&lt;/em&gt;  As such, if you want to know what happened at some place and time (regardless of how you decide to give numbers to that place and time, that's like choosing how to coordinatize the surface of the Earth), then you should ask someone who was at that place and time!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to this simple rule, we should imagine asking someone falling past an event horizon if a black hole has formed yet or not.  They will say it has, and they will say they get to that central black hole in a finite time.  Whether or not you get to receive that message is a more difficult issue, but they will say it all the same because reality happens somewhere, and we can always imagine someone there to experience it-- and &lt;em&gt;ask them&lt;/em&gt;.  Or at least, &lt;em&gt;imagine&lt;/em&gt; what they would say in cases where communication gets difficult or impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you follow that one simple rule, then all these apparent coordinate paradoxes disappear immediately.  Coordinates are a useful language for making calculations, but they are not a useful language for making assertions about &quot;what is.&quot;  That is an issue for observation, and &lt;em&gt;all observations are local&lt;/em&gt;-- no one ever observes a coordinate, and way too much is made out of arbitrary coordinate choices.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2019-02-06T17:41:12.867" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29491" PostTypeId="1" AcceptedAnswerId="29492" CreationDate="2019-02-06T21:49:25.383" Score="2" ViewCount="140" Body="&lt;p&gt;I know it's extremely unlikely a rogue planet passing by there, and I know the size of rocky planets can vary a lot (IIRC there is a rocky planet 5 times the size of Earth) but in any hypothetical case, could a rogue planet be trapped in any part of the orbit between Earth and Mars? If so, which are the possible effects it could have over Earth and Mars? In which cases would they be expelled from the solar system?&lt;/p&gt;&#xA;" OwnerUserId="11562" LastActivityDate="2019-02-06T22:11:06.347" Title="Could a rocky rogue planet get trapped in the orbit between Earth and Mars?" Tags="&lt;solar-system&gt;&lt;earth&gt;&lt;mars&gt;&lt;rogue-planet&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29492" PostTypeId="2" ParentId="29491" CreationDate="2019-02-06T22:11:06.347" Score="3" Body="&lt;p&gt;It is exceptionally unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that there was a planet already there. What could cause it to suddenly escape from the solar system?  It would have to be a massive event, such as a second passing rogue planet catapulting it out of orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Running time backwards, for a rogue planet to be captured and neatly end up orbiting between Mars and Earth would require an unbelievably unlikely sequence of events, such as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;two rogue planets happen to pass between Earth and Mars at the same time and the interaction between the two results in a transfer of energy and momentum, which (by lucky chance) leaves one of the planets in a nearly circular orbit, (while the other escapes)&quot;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given how big space is, and in consequence, how rarely a rogue planet enters the inner solar system, this would make the above sequence of events practically impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this unbelievable scenario, the effects on Earth and Mars would depend on how closely the new planet passes. If we are close enough for significant effects on the planet, we are in a great deal of danger.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-02-06T22:11:06.347" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29493" PostTypeId="2" ParentId="29479" CreationDate="2019-02-07T00:37:10.790" Score="2" Body="&lt;p&gt;&lt;strong&gt;tl;dr:&lt;/strong&gt; To an accuracy of about a half of a day:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Third quarter is sunrise&lt;/strong&gt; for Chang'e-4 and Yutu-2&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;First quarter is sunset&lt;/strong&gt; for Chang'e-4 and Yutu-2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Moon seen from Earth:&lt;/h2&gt;&#xA;&lt;p&gt;From Wikipedia's article &lt;a href=&quot;https://en.wikipedia.org/wiki/Selenographic_coordinates&quot; rel=&quot;nofollow noreferrer&quot;&gt;Selenographic coordinates&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Longitude&lt;/p&gt;&#xA;&lt;p&gt;Longitude on the Moon is measured both east and west from its prime meridian. When no direction is specified, east is positive and west is negative.&lt;/p&gt;&#xA;&lt;p&gt;Roughly speaking, the Moon's prime meridian lies near the center of the Moon's disc as seen from Earth. For precise applications, many coordinate systems have been defined for the Moon, each with a slightly different prime meridian. The IAU recommends the &amp;quot;mean Earth/polar axis&amp;quot; system,&lt;sup&gt;2&lt;/sup&gt; in which the prime meridian is the average direction (from the Moon's center) of the Earth's center.&lt;sup&gt;3&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;2&lt;/sup&gt; &lt;a href=&quot;https://astrogeology.usgs.gov/Projects/WGCCRE/WGCCRE2003preprint.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://astrogeology.usgs.gov/Projects/WGCCRE/WGCCRE2003preprint.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;3&lt;/sup&gt; &lt;a href=&quot;http://lunar.gsfc.nasa.gov/library/451-SCI-000958.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://lunar.gsfc.nasa.gov/library/451-SCI-000958.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Simplifying further, &lt;em&gt;seen from Earth&lt;/em&gt;, &lt;strong&gt;0° longitude &lt;a href=&quot;https://en.wikipedia.org/wiki/Libration&quot; rel=&quot;nofollow noreferrer&quot;&gt;wobbles&lt;/a&gt; near the center of the lunar disk.&lt;/strong&gt; (see also &lt;a href=&quot;https://space.stackexchange.com/q/19936/12102&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;https://physics.stackexchange.com/q/251700/83380&quot;&gt;2&lt;/a&gt;, &lt;a href=&quot;https://space.stackexchange.com/q/14943/12102&quot;&gt;3&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;h2&gt;Chang'e-4:&lt;/h2&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Chang%27e_4&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chang'e-4&lt;/a&gt; is in &lt;a href=&quot;https://en.wikipedia.org/wiki/Von_K%C3%A1rm%C3%A1n_(lunar_crater)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Von Kármán crater&lt;/a&gt; (also &lt;a href=&quot;https://tools.wmflabs.org/geohack/geohack.php?pagename=Von_K%C3%A1rm%C3%A1n_(lunar_crater)&amp;amp;params=44.8_S_175.9_E_globe:moon_type:landmark&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;) at about 44.8° South, &lt;strong&gt;175.9° East.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;So &lt;strong&gt;Chang'e-4 and Yutu-2 sit almost diametrically opposed to the Earth&lt;/strong&gt; longitude-wise, and so sunset on the Moon's prime meridian is close to sunrise for the pair, and vice-verse.&lt;/p&gt;&#xA;&lt;h2&gt;Answer:&lt;/h2&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;What phases of the Moon as seen from Earth correspond to times of sunrise and sunset for those two?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First quarter is sunset for Chang'e-4 and Yutu-2&lt;/li&gt;&#xA;&lt;li&gt;Third quarter is sunrise for Chang'e-4 and Yutu-2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Location Confirmation:&lt;/h2&gt;&#xA;&lt;p&gt;We know where Chang'e-4 landed because it returned video of it's own landing.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;YouTube Video:&lt;/strong&gt; &lt;a href=&quot;https://www.youtube.com/watch?v=jyjyf4ANGIo&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=jyjyf4ANGIo&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;https://space.stackexchange.com/q/33495/12102&quot;&gt;How did Chang'e-4 hover, rotate, and then descend so gracefully? (Video)&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Additional verification that Chang'e-4 and Yutu-2 are in Von Kármán crater:&lt;/p&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;http://lroc.sese.asu.edu/about&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Lunar Reconnaissance Orbiter Camera&lt;/a&gt; recent post &lt;a href=&quot;http://lroc.sese.asu.edu/posts/1090&quot; rel=&quot;nofollow noreferrer&quot;&gt;First Look: Chang'e 4&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RmU9H.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RmU9H.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Arrows indicate position of Chang'e 4 lander on the floor of Von Kármán crater. The sharp crater behind and to the left of the landing site is 3900 meters across (12,800 feet) and 600 meters (1970 feet) deep. Image was shrunk by more than a factor of ten (full resolution available inside the post). LROC NAC M1303521387LR [NASA/GSFC/Arizona State University].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hZRiS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hZRiS.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Area around lander enlarged by a factor of two relative to the native pixel scale (see zoomify below), bright speck between two arrows is the lander. The large crater in the center (just right and below arrows) is about 440 meters (1440 feet) across. LROC NAC M1303521387LR [NASA/GSFC/Arizona State University].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-02-07T01:04:58.203" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29494" PostTypeId="1" AcceptedAnswerId="29495" CreationDate="2019-02-07T06:21:58.007" Score="8" ViewCount="1119" Body="&lt;p&gt;How does the Earth differ from a (low-mass) stellar remnant, which has stopped fusion and the outer layers of which have been blown away?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could a stellar remnant end up with a similar relative abundance of elements as the Earth?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;i.e. Is it possible that a star loses so much matter when it dies that the remnant doesn't turn into degenerate forms of matter?&lt;/p&gt;&#xA;" OwnerUserId="6968" LastEditorUserId="6968" LastEditDate="2019-02-07T13:45:07.263" LastActivityDate="2019-02-07T20:49:41.217" Title="Properties of low-mass stellar remnants vs the Earth" Tags="&lt;earth&gt;&lt;stellar-evolution&gt;&lt;stellar-remnants&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="29495" PostTypeId="2" ParentId="29494" CreationDate="2019-02-07T06:35:30.873" Score="4" Body="&lt;p&gt;By stellar remnant, it sounds like you mean a white dwarf.  These each have a composition that is determined by their history and how far into nuclear burning they've gone.   Often they have lots of carbon and oxygen, sometimes they get as far as iron.  But the Earth formed from dust orbiting the Sun, and since its formation mechanism was so different, it has a very different composition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the same, it should perhaps be noted that a planet whose metallic core has cooled to a solid is not so much different from a white dwarf.  The main difference is the mass is lower, so the kinetic energy of the free electrons is lower, so many of them get captured by the nuclei.  So it is like a white dwarf with many fewer free degenerate electrons.  That's mostly what the electrostatic attractions are doing-- removing degenerate free electrons.  So it has a smaller radius for its gravity, since there are fewer electrons producing the degeneracy pressure.  In condensed matter lingo, that population is called the &quot;conduction band.&quot;&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2019-02-07T20:49:41.217" LastActivityDate="2019-02-07T20:49:41.217" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29496" PostTypeId="2" ParentId="29494" CreationDate="2019-02-07T06:47:58.187" Score="13" Body="&lt;p&gt;Stellar remnants are completely different from planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth was never a star and fusion has never occurred in the Earth's core at any time in its history. When a small to medium sized star dies, and the outer layers are lost, what remains is a white dwarf. It is still much more massive than the Earth, and is very hot. It is crushed by its own gravity, causing the matter to become &quot;degenerate&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Earth, atoms are held together by chemical bonds. Lots of interesting patterns can be formed by the atoms, producing minerals, rocks, seas and life. Nothing like this can happen in degenerate matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In degenerate matter, the atoms are pushed together by gravity. Degenerate matter is unlike regular matter. It is much much more dense, and chemical bonds are not a significant force between atoms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;White dwarfs are formed of carbon, oxygen, hydrogen and helium (with the lighter elements on the surface). Even after a white dwarf has cooled, it would still be degenerate, and quite unlike the Earth, in composition, properties and in density.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-02-07T06:47:58.187" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29497" PostTypeId="1" AcceptedAnswerId="29514" CreationDate="2019-02-07T07:02:07.427" Score="1" ViewCount="180" Body="&lt;p&gt;I'm working with data from the &lt;a href=&quot;https://astro.troja.mff.cuni.cz/projects/asteroids3D/web.php?page=db_description&quot; rel=&quot;nofollow noreferrer&quot;&gt;DAMIT database&lt;/a&gt; of asteroid shape models.  I'm adding them to a visualization in which the sun is at [0, 0, 0] and the X, Y axes constitute the ecliptic plane of the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each asteroid model comes with some attributes that define its orientation and spin:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;λ (ecliptic longitude), &lt;/li&gt;&#xA;&lt;li&gt;β (ecliptic latitude)&lt;/li&gt;&#xA;&lt;li&gt;P (sidereal rotation period)&lt;/li&gt;&#xA;&lt;li&gt;φ0 (initial rotation angle)&lt;/li&gt;&#xA;&lt;li&gt;JD0 (initial date).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I've applied the matrix formulas suggested by the folks at DAMIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FQaHe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FQaHe.png&quot; alt=&quot;ecliptic rotation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in which&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xtuln.png&quot; alt=&quot;matrix definitions[2]&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where I get confused.  My understanding is that my visualization uses a heliocentric ecliptic coordinate system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I set &lt;code&gt;r_ast&lt;/code&gt; to the XYZ location of a vertex in my visualization.  I'm not sure this is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I compute &lt;code&gt;r_ecl&lt;/code&gt; from the equation above.  But I think variables λ and β are angles in a geocentric ecliptic coordinate system and I need to do some transformation into heliocentric ecliptic coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my approach correct and if so, how can I convert the asteroid's rotation to my visualization's heliocentric ecliptic coordinate system?&lt;/p&gt;&#xA;" OwnerUserId="2605" LastEditorUserId="2605" LastEditDate="2019-02-07T07:34:31.110" LastActivityDate="2019-02-08T18:55:47.447" Title="Transform asteroid rotation to heliocentric ecliptic coordinates" Tags="&lt;asteroids&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29498" PostTypeId="1" CreationDate="2019-02-07T12:16:42.503" Score="3" ViewCount="84" Body="&lt;p&gt;In this article &lt;a href=&quot;https://www.quora.com/What-is-the-shortest-distance-between-stars&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shortest distance between stars&lt;/a&gt; it is mentioned that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some double stars can actually be in contact with each other.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This occurs when a red giant star has a white dwarf star companion in&#xA;  a very close orbit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, this isnt the case&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://earthsky.org/space/hottest-heaviest-touching-double-star-vfts-352&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hottest, heaviest touching double stars&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the case of VFTS 352, however, both stars in the system are of&#xA;  almost identical size. Material is, therefore, not sucked from one to&#xA;  another, but instead may be shared.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question then is, can any type of stars be contact binaries, or are there some limitations in the type of stars for contact binaries to appear?&lt;/p&gt;&#xA;" OwnerUserId="11562" LastEditorUserId="11562" LastEditDate="2019-02-07T19:26:02.537" LastActivityDate="2019-02-07T19:26:02.537" Title="Can any type of stars be contact binaries?" Tags="&lt;star-systems&gt;&lt;contact-binaries&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29499" PostTypeId="1" AcceptedAnswerId="29500" CreationDate="2019-02-07T15:56:13.517" Score="2" ViewCount="75" Body="&lt;p&gt;Wikipedia's article &lt;a href=&quot;https://en.wikipedia.org/wiki/Helioseismology&quot; rel=&quot;nofollow noreferrer&quot;&gt;Helioseismology&lt;/a&gt; outlines several ways to analyze, data but is fairly thin on the nature of the measurements themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there hundreds or thousands of high resolution spectroscopic measurements of the Doppler shift of some emission or absorption line distributed across the disk, or can this be done using low resolution spectroscopy to generate some variation in temperature or brightness?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in a basic appreciation of the nature and granularity of the data (e.g. 0.1 angstrom in wavelength, 1 second of time, 1 minute of arc, etc.) and what kind of instruments collect said data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer doesn't necessarily have to be long, just something to get me started.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7472" LastEditDate="2019-09-01T09:50:24.820" LastActivityDate="2019-09-01T09:50:24.820" Title="Helioseismology of the Sun, what is actually measured?" Tags="&lt;the-sun&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29500" PostTypeId="2" ParentId="29499" CreationDate="2019-02-07T17:19:32.340" Score="5" Body="&lt;p&gt;The two main methods to detect the solar &lt;em&gt;p&lt;/em&gt;-mode oscillations, which have a period of about 5 minutes (so frequency of ~3 milliHertz) are line of sight Doppler shift and change in irradiance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Doppler method works by measuring the intensity of light in (normally) two narrow regions either side of a prominent spectral line in the Solar spectrum and using this change to measure the velocity shift caused by the oscillations (once the solar rotation is removed). The ground-based &lt;a href=&quot;https://gong.nso.edu/instrument/&quot; rel=&quot;noreferrer&quot;&gt;GONG&lt;/a&gt; instrument uses the Ni I at 676.8nm, &lt;a href=&quot;http://bison.ph.bham.ac.uk/index.php?page=bison,background&quot; rel=&quot;noreferrer&quot;&gt;BiSON&lt;/a&gt; network and the GOLF instrument on the SOHO spacecraft both use the sodium D lines. After isolating the light around a suitable spectral feature, then you can either use a Michelson interferometer to get a velocity image of the shifts across the surface (GONG and MDI on SOHO) or a disk-averaged velocity, treating the Sun as a star (BiSON and GOLF on SOHO). The difficulties are in keeping everything stable so that the measured shifts only come from the Sun and not the instrument. For example GONG note that their optical elements are stabilized in an oven to the order of 0.00001 K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The irradiance method simply involves measuring the disk-integrated intensity of the Sun and watching for changes caused by the oscillations. However fluctuations in the Earth's atmosphere make this very difficult to do from the ground and the most successful has been the &lt;a href=&quot;https://www.pmodwrc.ch/en/research-development/space/virgo-soho/&quot; rel=&quot;noreferrer&quot;&gt;VIRGO instrument&lt;/a&gt; on the SOHO spacecraft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The imaging instruments have a resolution of about 2&quot; (MDI) to 5&quot; (GONG) so return velocity data at approx. 800,000 to 100,000 locations across the Sun. This allow the study of oscillations with very high degree (up to 1000) which have very small scale on the surface. The disk-integrated &quot;Sun-as-a-star&quot; instruments by comparison are sensitive to the low order oscillations.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2019-02-07T17:19:32.340" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29501" PostTypeId="2" ParentId="29480" CreationDate="2019-02-07T18:10:40.453" Score="-1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A fixed magnetic field will induce an electrical current in a conductive object moving through the field.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. It will induce an EMF in one direction as described in &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/genwir2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;HyperPhysics&lt;/a&gt;. It will not induce a current since there is no return path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is different from induction braking, where you have a magnet with a non-uniform field moving near a conductor. It is the changing magnetic field at points within the conductor that induces the EMF and causes currents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, we can consider how electromagnetic fields transforn according to Relativity. If an object is moving at constant velocity through a uniform B field then, in its rest frame, it will experience a &lt;a href=&quot;https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity#The_E_and_B_fields&quot; rel=&quot;nofollow noreferrer&quot;&gt;combination of uniform E and B fields&lt;/a&gt; which would not produce a current.  (Any current would have to be due to the spacial rate of change of the field, which would be tiny on the scale of the Earth's field.)&lt;/p&gt;&#xA;" OwnerUserId="17743" LastEditorUserId="17743" LastEditDate="2019-08-23T04:27:53.193" LastActivityDate="2019-08-23T04:27:53.193" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29502" PostTypeId="2" ParentId="10901" CreationDate="2019-02-07T20:54:18.803" Score="1" Body="&lt;p&gt;I obtained a more recent (2002) book by Meeus, &lt;em&gt;More Mathematical Astronomy Morsels&lt;/em&gt;, published by Willmann-Bell. Chapter 63 is about the Gregorian calendar and the vernal equinox. It contains expressions from J. L. Simon et al. &quot;&lt;a href=&quot;https://ui.adsabs.harvard.edu/link_gateway/1994A&amp;amp;A...282..663S/ADS_PDF&quot; rel=&quot;nofollow noreferrer&quot;&gt;Numerical expressions for precession formulae and mean elements for the Moon and planets&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can use the expressions to find the orbital elements of the Earth for a given date near an equinox or solstice, and then solve Kepler's equation to find the &quot;true&quot; longitude of the Earth. The longitude is true in the sense that the elliptical orbit is used rather than the fictitious circular orbit that is used for mean mean longitude, but the periodic perturbations from the Moon and other planets are smoothed out. One iterates until dates are found where the longitude is 0, 90, 180 and 270 in years either side of the year of interest. The intervals between solstices or equinoxes are found, and divided by 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried this for year 0 and year 2000 and it agreed with the full precision that Meeus published in 2002, six decimal places, or about 0.1 second of time. On the other hand, around 2000, the results were different by about 8 minutes from the results from the book in the question. The 2002 book and 1992 paper were about finding the &lt;strong&gt;mean&lt;/strong&gt; interval between successive equinoxes or solstices, in a discussion of how &quot;tropical year&quot; has been defined throughout history. I suspect the methods in the book in the first question were not used for this purpose.&lt;/p&gt;&#xA;" OwnerUserId="7385" LastEditorUserId="7385" LastEditDate="2019-02-08T13:05:16.230" LastActivityDate="2019-02-08T13:05:16.230" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29503" PostTypeId="1" AcceptedAnswerId="29504" CreationDate="2019-02-07T22:12:04.203" Score="11" ViewCount="1623" Body="&lt;p&gt;I have some questions about redshift. I think that the first two hang on the last two, but feel free to point out anything you think I should know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we know if an object is actually redshifted, and that we aren't just seeing the regular spectrum of the star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we tell the extent to which a star is redshifted? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do astronomers have a full baseline spectrum that they compare the measured spectrum against? Or do they just use the hydrogen spectrum since all stars are mostly hydrogen? &lt;/p&gt;&#xA;" OwnerUserId="25843" LastActivityDate="2019-02-08T15:37:47.320" Title="How do we know if an object is redshifted?" Tags="&lt;the-sun&gt;&lt;distances&gt;&lt;redshift&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29504" PostTypeId="2" ParentId="29503" CreationDate="2019-02-07T22:32:29.120" Score="22" Body="&lt;p&gt;There are two methods, one more reliable than the other (though both are pretty good.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Key point: The brighter a star is, the more detail we can see in its spectrum -- you can think of it as being able to magnify the spectrum more so as to be able to see finer details.  This also allows us to see fainter lines (not all spectral lines are equally intense.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have very extensive reference spectra of all the elements under a wide variety of conditions, so when a star is bright enough we see many spectral lines and can match them up to reference spectra.  No two elements have similar spectra, so if you can see many lines, you can't easily confuse one for another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you can account for nearly all of the lines in a star's spectrum by a plausible set of elements in plausible abundances and all with the same red shift, you have a very, very reliable match and accurately know the object's red shift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GZXNs.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GZXNs.jpg&quot; alt=&quot;credit: https://en.wikipedia.org/wiki/Redshift&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But distant stars (and galaxies) are very faint and so the spectra we get are very low resolution and show only a few of the strongest spectral lines. They can frequently be matched because hydrogen is so common that the strongest lines seen will be hydrogen lines and even if you can only spot two or three of them, if they bear the same relative positions as bright hydrogen lines, you can safely identify them and read off the red shift.  This is less reliable, but still reliable enough that it rarely is a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the very dimmest cases you may only see a single line. You have a very rough idea of the redshift from the object's brightness and all you can do is assume it's the strongest hydrogen line and see if it's in roughly the right position given the distance deduced from the object's brightness and type.  This works too, but is still less reliable.  (Much of Hubble's original work relied on just a single line for the more distance objects, but proved to be correct when observed later with dramatically better equipment.)&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="-1" LastEditDate="2019-02-08T15:37:47.320" LastActivityDate="2019-02-08T15:37:47.320" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29505" PostTypeId="1" AcceptedAnswerId="29508" CreationDate="2019-02-08T03:01:57.267" Score="1" ViewCount="84" Body="&lt;p&gt;I am writing a novel about what it will be like on earth 500 or so years before the death of our Sun. The Sun is heating up gradually and one day that will negatively affect life on Earth. I know that we are a long ways away from the heat being to much to permit life though. That being said, the novel is more about the social/political/religious ramifications of living in such a time. I still am very interested in making it as relatable and understandable as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any causal mechanism (hypothetically or imaginatively possible) that could cause the sun to heat up faster than it is at present? This could be internal (something collapses and the increase in core pressure speeds up the reaction rate of hydrogen fusion) or external (something impacts the sun with enough extra mass to increase net g, thus increasing pressure in the core).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not a causal mechanism, maybe something you can think of that current science/ scientists could have missed/ aren't taking into account in their models that, if true, would mean the sun is going to be much hotter, much sooner than we thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you so much for your help! I am pretty passionate about writing and I am only an amateur in the subject of astronomy. I'm hoping some brilliant minds can help me flesh out this idea to make it more concrete. &lt;/p&gt;&#xA;" OwnerUserId="25846" LastEditorUserId="1891" LastEditDate="2019-02-08T09:48:19.000" LastActivityDate="2019-02-08T10:16:39.547" Title="Is there a plausible causal mechanism to explain why the Sun would be heating up faster/ sooner than we thought? Question for a novel" Tags="&lt;the-sun&gt;&lt;gas&gt;&lt;solar-flare&gt;&lt;solar-storm&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2019-02-08T19:57:33.370" ContentLicense="CC BY-SA 4.0" />
  <row Id="29506" PostTypeId="1" CreationDate="2019-02-08T03:37:58.190" Score="2" ViewCount="125" Body="&lt;p&gt;I've seen that a large block of Pan-Starrs data has been released and is available on-line and how the distributors of the data expect professionals and the general public to use the data in their research. I am part of the general public and was wondering how accessible the data actually is?  For instance if i were interested in searching a small section of sky for a moving object, i would need multiple &quot;photos&quot; &quot;data sets&quot;? over time in order to see the movement.  Would special software be required? How would one even start the process?&lt;/p&gt;&#xA;" OwnerUserId="25848" LastEditorUserId="616" LastEditDate="2020-10-10T02:52:06.860" LastActivityDate="2020-10-10T02:52:06.860" Title="How does one &quot;use&quot; the Pan-Starrs data?" Tags="&lt;observational-astronomy&gt;&lt;sky-survey&gt;&lt;raw-data&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29507" PostTypeId="1" CreationDate="2019-02-08T04:26:14.247" Score="6" ViewCount="226" Body="&lt;p&gt;Let's assume the moon was formed because Earth collided into Theia. (There are numerous other theories, but this one is the most widely accepted.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case, are there theories that address what the Earth surface looked like? E.g. would the Earth have had bodies of water at the time? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read some websites and a book!, and some say it did, some say it did not.&lt;/p&gt;&#xA;" OwnerUserId="25849" LastEditorUserId="7982" LastEditDate="2019-02-25T06:42:28.427" LastActivityDate="2019-02-25T06:42:28.427" Title="What is Earth believed to have looked like before the collision with Theia?" Tags="&lt;the-moon&gt;&lt;earth&gt;&lt;theia&gt;&lt;collision&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29508" PostTypeId="2" ParentId="29505" CreationDate="2019-02-08T10:16:39.547" Score="-1" Body="&lt;h3&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_neutrino_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Solar Neutrino Problem&lt;/a&gt;&lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The solar neutrino problem concerned a large discrepancy between the&#xA;  flux of solar neutrinos as predicted from the Sun's luminosity and&#xA;  measured directly. The discrepancy was first observed in the mid-1960s&#xA;  and finally resolved around 2002.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The flux of neutrinos at Earth is several tens of billions per square&#xA;  centimetre per second, mostly from the Sun's core. They are&#xA;  nevertheless hard to detect, because they interact very weakly with&#xA;  matter, traversing the whole Earth as light does thin air. Of the&#xA;  three types (flavors) of neutrinos known in the Standard Model of&#xA;  particle physics, the Sun produces only electron neutrinos. When&#xA;  neutrino detectors became sensitive enough to measure the flow of&#xA;  electron neutrinos from the Sun, the number detected was much lower&#xA;  than predicted. In various experiments, the number deficit was between&#xA;  one half and two thirds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The problem is practically solved today. It wasn't so until &lt;span class=&quot;math-container&quot;&gt;$\approx$&lt;/span&gt; 2002. However, a little chance that we know something not enough well, still exists, but it is little.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The currently accepted solution of the problem is this: yes, we know the Sun enough well, but the neutrinos can change flavor on the way they reach us. Thus, for example, if an electron-neutrino becomes muon-neutrino, then the detectors capable to see only electron-neutrinos, won't detect it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is today a hot research topic in the particle physics, thus some surprising result in the future isn't yet closed out. The main problem is that the neutrino practically don't interact with anything (a light year of lead would stop half of them), so there are only &lt;em&gt;very&lt;/em&gt; little possibilities to measure their properties.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/KATRIN&quot; rel=&quot;nofollow noreferrer&quot;&gt;The KATRIN neutrino experiment&lt;/a&gt;&lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;KATRIN is a German acronym (Karlsruhe Tritium Neutrino Experiment) for&#xA;  an undertaking to measure the mass of the electron antineutrino with&#xA;  sub-eV precision by examining the spectrum of electrons emitted from&#xA;  the beta decay of tritium. The core of the apparatus is a 200-ton&#xA;  spectrometer. In 2015, the commissioning measurements on this&#xA;  spectrometer were completed, successfully verifying its basic vacuum,&#xA;  transmission and background properties. The experiment began&#xA;  running tests in October 2016. Regular measurements started&#xA;  2018-06-11, with a projected duration of 5 years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As far I know, too much surprising results aren't expected from this - most likely, the electron neutrino mass of some tenths of eV will be found, or possible the result will be that the experiment is not enough precise to find anything. (In this case, we will still get an &quot;upper limit&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is a little chance that some surprising result will happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, such a neutrino mass could be found, which can be explained only by some hyphotetical effect (like &lt;a href=&quot;https://en.wikipedia.org/wiki/Sterile_neutrino&quot; rel=&quot;nofollow noreferrer&quot;&gt;sterile neutrinos&lt;/a&gt;, which might also contribute to the solution of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_matter&quot; rel=&quot;nofollow noreferrer&quot;&gt;dark matter&lt;/a&gt; problem).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The idea&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Go back to the 80ties, where a possible solution to the solar neutrino problem was a different working and internal structure of the Sun, which looks the same for us, but generates lesser neutrinos. By analogy, a similarly different Sun should be expected, which goes into a red giant phase soon. (Or, at least its luminosity will increase enough in the near future to make the Earth inhabitable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, we also know a lot of other stars today, there are very good models running on big computers, and these are enough sophisticated to close this possibility out with a good probability. Thus, the Sun should be somehow exceptional on these results.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastActivityDate="2019-02-08T10:16:39.547" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29509" PostTypeId="2" ParentId="29507" CreationDate="2019-02-08T12:21:44.057" Score="5" Body="&lt;p&gt;&lt;a href=&quot;http://advances.sciencemag.org/content/4/3/eaao5928&quot; rel=&quot;noreferrer&quot;&gt;Greenwood et al&lt;/a&gt; suggest that Earth had a lot of its water (maybe upto 70%) before Theia, but I can't find anything definite saying that it was liquid, although several sites reporting on the Greenwood paper assume that Earth was cool enough to have an ocean. It almost certainly had a crust, though. And immediately after the impact it had a magma ocean for quite a while. ;)&lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2019-02-08T12:21:44.057" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29510" PostTypeId="1" AcceptedAnswerId="29518" CreationDate="2019-02-08T12:39:44.657" Score="8" ViewCount="431" Body="&lt;p&gt;A few days ago, I took a photo of the night sky in the mountains of Georgia with my DSLR. There, I noticed that one single star (in the center of the part of the photo that I added below) looks very different than all the other stars on the photo. Why does it have this white glow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jhip7.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jhip7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25856" LastActivityDate="2019-02-09T12:34:50.573" Title="What's the white glow around this star?" Tags="&lt;star&gt;&lt;amateur-observing&gt;&lt;photography&gt;&lt;night-sky&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29513" PostTypeId="2" ParentId="29506" CreationDate="2019-02-08T18:15:46.220" Score="4" Body="&lt;p&gt;There are two main datasets available from the PanSTARRS survey which are available from the &lt;a href=&quot;https://outerspace.stsci.edu/display/PANSTARRS/Pan-STARRS1+data+archive+home+page&quot; rel=&quot;nofollow noreferrer&quot;&gt;archive at MAST/STScI&lt;/a&gt;. These are the Object Catalog, a (large) list of parameters such as position, brightness, shape etc (&lt;a href=&quot;https://outerspace.stsci.edu/display/PANSTARRS/PS1+MeanObjectView+table+fields&quot; rel=&quot;nofollow noreferrer&quot;&gt;full list of catalog fields&lt;/a&gt;) and the Image Cutouts service which, well, cuts out sections of the images. (The survey went over the same patch of sky several times in 5 filters so there are deeper stacks available)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking for moving objects then you will need to use the &quot;&lt;em&gt;warp&lt;/em&gt;&quot; single epoch images as a moving object would be averaged away in the stack (unless it was done on the object's motion). You will also need a source of positions in order to cutout the suitable postage stamps in the right places as the PanSTARRS archive doesn't support querying by moving objects or orbital elements. You would need to use something like the Minor Planet Center's &lt;a href=&quot;https://www.minorplanetcenter.net/iau/MPEph/MPEph.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ephemeris Service&lt;/a&gt; or JPL's &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;HORIZONS service&lt;/a&gt; to generate these positions (use 'F51' as the Observatory Code/Observers Location as this is the MPC site code for PanSTARRS1). Once you have a source of positions, you will probably want to investigate querying the Image Cutout Service via script/code; the &lt;a href=&quot;https://outerspace.stsci.edu/display/PANSTARRS/PS1+Image+Cutout+Service#PS1ImageCutoutService-PythonExampleScript&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image Cutout Service&lt;/a&gt; documentation has instructions and a Python Jupyter notebook showing how to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should also caution you that one of the goals of the PanSTARRS survey, which my institution was a partner in, was to find moving objects. A lot of effort was put into the Moving Object Processing System (MOPS; paper, which I'm on, available (for free) &lt;a href=&quot;https://iopscience.iop.org/article/10.1086/670337/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;here at IOPscience&lt;/a&gt;) and PanSTARRS was one of the two major sources of new Near Earth Objects (NEOs), see purpley bars in &lt;a href=&quot;https://cneos.jpl.nasa.gov/stats/site_all.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this plot&lt;/a&gt; at the Center for NEO Studies. I'm not saying there isn't new moving objects in there, we go back and search through the PanSTARRS data for precoveries of newly discovered NEOs that the automatic software missed on a roughly monthly basis, but bear this in mind before launching into a big project.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2019-02-08T18:15:46.220" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29514" PostTypeId="2" ParentId="29497" CreationDate="2019-02-08T18:55:47.447" Score="1" Body="&lt;p&gt;&lt;code&gt;r_ast&lt;/code&gt; is a vector that goes from the center of the asteroid to each vertex of the shape model as if you had stuck a set of rods in the center of a potato and pointing along the long axis, to the pole on top and 90 degrees to these to give X,Y,Z directions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matrices above that you give then transform these &lt;code&gt;r_ast&lt;/code&gt; vectors for the co-ordinates of the surface of the asteroid into the ecliptic plane as a function of time as the asteroid rotates, giving you &lt;code&gt;r_ecl&lt;/code&gt;. It says in the Kaasalainen paper (I think) that &lt;code&gt;r_ecl&lt;/code&gt; has an origin shifted to the center of the asteroid. If you are wanting a visualization of what the asteroid looks like relative to the Earth or Sun etc, then you will need to calculate the Sun-&gt;asteroid vector and the Sun-&gt;Earth vector using e.g. &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL HORIZONS&lt;/a&gt; and using the Vector Table ephemeris type or a programmatic wrapper to this (e.g. &lt;a href=&quot;https://astroquery.readthedocs.io/en/latest/jplhorizons/jplhorizons.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;astroquery&lt;/a&gt;) or the &lt;a href=&quot;https://naif.jpl.nasa.gov/naif/toolkit.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NAIF SPICE toolkit&lt;/a&gt; and getting or generating a SPK kernel file for the asteroid (the normal e.g. &lt;code&gt;de430.bsp&lt;/code&gt; JPL planetary ephemeris will handle the Sun-&gt;Earth part)&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2019-02-08T18:55:47.447" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29515" PostTypeId="1" CreationDate="2019-02-08T19:22:46.233" Score="2" ViewCount="101" Body="&lt;p&gt;This question is inspired by the &quot;Stargate SG-1&quot; episode &quot;The Road Not Taken&quot;. In that story a device is used to shift the entire Earth to another dimension to defend against an attack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not asking about the science of whether that is even theoretically possible but rather the effect on the rest of the objects in our solar system if it did happen. Since this was apparently a short shift (likely days at the most) I would imagine there might be little overall effect to most things but it occurred to me the Moon might immediately start drastically altering its position relative to the now massless Earth (there was no indication that the effect extended to the Moon but given that ships very close to Earth were unaffected it would seen that the Moon was also unaffected) so that when the Earth returns to &quot;normal&quot; it may no longer be in Earth's orbit or worse, Earth may rematerialize partially within or completely around it.&lt;/p&gt;&#xA;" OwnerUserId="25859" LastEditorUserId="23748" LastEditDate="2019-02-08T19:27:17.003" LastActivityDate="2019-02-09T00:21:34.113" Title="What would happen to the solar system if Earth's mass was removed?" Tags="&lt;the-moon&gt;&lt;gravity&gt;&lt;solar-system&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29516" PostTypeId="2" ParentId="29515" CreationDate="2019-02-08T20:20:28.917" Score="1" Body="&lt;p&gt;There would be three effects to consider, one very short-term, one medium term and one long-term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The medium-term effect is that the Moon has an orbital velocity around the Earth of about 1 km/s.  If the Earth suddenly disappeared, the Moon would just keep on going in a straight line at 1 km/s.  This would have the effect of putting it into a Solar orbit quite different from the Earth's. It would have a different inclination to the ecliptic (the Moon's orbit is inclined to the ecliptic by 5 degrees), different period and different eccentricity. (How different depends on exactly where in the Moon's monthly orbit the Earth disappeared.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect wouldn't be enough to cause the orbit to intersect that of Venus or Mars right away, but perturbations on the new Planet Luna would potentially be very different from those on Earth and it might evolve fairly quickly (centuries) to a Mars-crossing or Venus-crossing orbit with an eventual collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is &lt;em&gt;highly&lt;/em&gt; unlikely that it would be very near the spot where Earth reappears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The long-term effect is smaller: The Earth is part of the generally stable Solar System and contributes small perturbations to all the planets. It is &lt;em&gt;possible&lt;/em&gt; (but I suspect unlikely) that the disappearance of the Earth and the subsequent lack of the Earth's perturbations would result in significant changes to the Solar System over many centuries to tens of thousands of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The very short-term effect is that the disappearance of the Earth's mass would result in a speed-of-light propagation of an abrupt change in the curvature of spacetime.  I.e., gravitational waves.  I don't know enough to calculate if these waves would have any noticeable effect within the Solar System. My inexpert guess is that it's plausible that they would.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23748" LastEditDate="2019-02-09T00:21:34.113" LastActivityDate="2019-02-09T00:21:34.113" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29517" PostTypeId="1" CreationDate="2019-02-09T00:20:53.170" Score="5" ViewCount="320" Body="&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; This was NOT a duplicate of &lt;a href=&quot;https://astronomy.stackexchange.com/questions/29270/horseshoe-orbit-cycle-times&quot;&gt;Horseshoe orbit cycle times&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;strong&gt;that other question has been deleted, regardless&lt;/strong&gt;. My original question asked for the answer to the equation(s), and provides many of the variables.  But since no one was apparently willing/able to answer it, this new question asks only for the form of the equation(s) itself and a list of variables involved, so that I can either provide more information on a future question, or try and solve the equations myself (though I doubt my math ability to do that, hence the original question, if I know what equations and variables are involved I can at least make an attempt myself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What variables need to be known in order to calculate a horseshoe orbit cycle time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/d/d6/Lagrange_Horseshoe_Orbit.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words: What is the equation, and what do the variables in that equation represent (speed? mass? semi major axes? etc.), to calculate the time it takes for an object to go from Point A on that image, through Points B, C, D, and E and back to A, along the light blue line in the image above (image also available &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/d/d6/Lagrange_Horseshoe_Orbit.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, Earth and Cruithne complete their cycle in 770 years, while Janus and Epimetheus complete theirs in 8 years.  I'm confident that semi-major axes are key factors in the equation, but I don't know what other variables are included, or how the variables relate to each other in the calculation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I understand that these orbits are not stable, in the long term, and that my example of Cruithne is especially complex, as it will actually change orbit types periodically, and will likely be ejected from the solar system entirely, or impact the Sun or Jupiter, at some point.  But those complexities are for another time.  For this question, I just want to know the equation in its simplest form. &lt;/p&gt;&#xA;" OwnerUserId="24853" LastEditorUserId="24853" LastEditDate="2019-02-11T23:51:33.517" LastActivityDate="2019-02-14T14:32:46.093" Title="What variables are needed to calculate simple horseshoe orbit times?" Tags="&lt;orbit&gt;&lt;natural-satellites&gt;&lt;orbital-mechanics&gt;&lt;celestial-mechanics&gt;" AnswerCount="2" CommentCount="12" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29518" PostTypeId="2" ParentId="29510" CreationDate="2019-02-09T00:39:05.293" Score="16" Body="&lt;p&gt;&lt;a href=&quot;http://nova.astrometry.net/user_images/2654510#annotated&quot; rel=&quot;noreferrer&quot;&gt;Astrometry.net&lt;/a&gt; has identified your star field as being part of the Andromeda constellation. The diffuse object in the centre of your image is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Andromeda_Galaxy&quot; rel=&quot;noreferrer&quot;&gt;Andromeda Galaxy&lt;/a&gt; (Messier 31). The bright star to the left of it is Mirach (β Andromedae). Given a reasonably dark sky and averted vision, it is possible to see the core of M31 with the naked eye. Since M31 is approximately 2.5 million light years away from us, it is one of the most distant objects that can be seen with the naked eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diametrically opposite Mirach to M31, and equidistant from it, is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Triangulum_Galaxy&quot; rel=&quot;noreferrer&quot;&gt;Triangulum Galaxy&lt;/a&gt; (M33), although it is only just discernible in your photograph (even here, averted vision helps). If you can see M33 with your naked eye, then you have excellent vision and really dark skies. At a distance of 3 million light years, it is also one of the most distant objects that can be seen with the naked eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Msakq.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Msakq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astrometry.net&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that only the core of M31 is visible to the naked eye (as it is in your photograph). If M31 was brighter, the arms of the galaxy would be visible and it would appear to be several times larger than the full Moon, as shown in the composite image below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fNDWG.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fNDWG.jpg&quot; alt=&quot;M31 and Moon composite photo&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NASA&lt;/p&gt;&#xA;" OwnerUserId="13887" LastEditorUserId="13887" LastEditDate="2019-02-09T12:34:50.573" LastActivityDate="2019-02-09T12:34:50.573" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29519" PostTypeId="1" AcceptedAnswerId="29520" CreationDate="2019-02-09T12:08:44.177" Score="5" ViewCount="464" Body="&lt;p&gt;For this question assume that the entire universe is completely empty. The universe is not expanding or contracting, it is completely motionless since time immemorial and has never moved before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only two identical earths without moon are left in this entire universe and they are 5 billion light years apart. From their starting position the earths do not spin or move at all they are completely motionless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would these planets ever meet or at least be pulled together?  Or are the their respective gravities too small to have any meaningful effect?&lt;/p&gt;&#xA;" OwnerUserId="25592" LastEditorUserId="25592" LastEditDate="2019-03-14T20:07:34.947" LastActivityDate="2019-03-14T20:07:34.947" Title="Can two planets in an empty universe meet/be pulled together?" Tags="&lt;gravity&gt;&lt;universe&gt;&lt;observable-universe&gt;&lt;rogue-planet&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29520" PostTypeId="2" ParentId="29519" CreationDate="2019-02-09T12:36:12.867" Score="7" Body="&lt;p&gt;Yes, they would experience gravitational attraction. It would take a long time for them to collide... the formula (derived &lt;a href=&quot;https://physics.stackexchange.com/a/14718/83380&quot;&gt;here&lt;/a&gt; and shown &lt;a href=&quot;https://physics.stackexchange.com/a/90722/83380&quot;&gt;here&lt;/a&gt;) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$t = \frac{\pi}{2} \sqrt{\frac{d^3}{2G(m_1+m_2)}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; is the initial distance between the two planets of mass &lt;span class=&quot;math-container&quot;&gt;$m_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$m_2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_constant&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gravitational constant&lt;/a&gt;. This gives a time of about &lt;span class=&quot;math-container&quot;&gt;$10^{23}$&lt;/span&gt; years, much much longer than the universe has existed. This assumes Newtonian mechanics. Relativity would not change the conclusion much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no known upper limit to gravity, and plenty of indirect evidence that it has no upper limit.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="7982" LastEditDate="2019-02-28T08:21:15.387" LastActivityDate="2019-02-28T08:21:15.387" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29521" PostTypeId="2" ParentId="29483" CreationDate="2019-02-09T14:51:16.047" Score="1" Body="&lt;p&gt;Someone near Heidelberg also &lt;a href=&quot;https://fireball.imo.net/members/imo_view/report/160653&quot; rel=&quot;nofollow noreferrer&quot;&gt;observed a fireball&lt;/a&gt; at that time.&#xA;&lt;a href=&quot;https://www.amsmeteors.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;AMS&lt;/a&gt; and &lt;a href=&quot;https://www.imo.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;IMO&lt;/a&gt; designate the event &lt;a href=&quot;https://fireball.imo.net/members/imo_view/event/2019/635&quot; rel=&quot;nofollow noreferrer&quot;&gt;635-2019&lt;/a&gt;, the 635th fireball reported in 2019.&#xA;As there is only one other report so far and your description is fairly detailed, they would welcome &lt;a href=&quot;https://fireball.imo.net/members/imo/report_intro&quot; rel=&quot;nofollow noreferrer&quot;&gt;your report&lt;/a&gt; too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fireball is a space rock whose orbit around the Sun happens to intersect with Earth's orbit at the right time - bigger than an everyday meteor, but smaller than the &lt;a href=&quot;https://cneos.jpl.nasa.gov/about/basics.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;near-Earth objects&lt;/a&gt; (NEOs) which astronomers try to detect and track.&#xA;For comparison, the &lt;a href=&quot;https://earthsky.org/space/small-meteorite-asteroid-cuba-feb-2019&quot; rel=&quot;nofollow noreferrer&quot;&gt;February 1 meteorite in western Cuba&lt;/a&gt; was event &lt;a href=&quot;https://fireball.imo.net/members/imo_view/event/2019/513&quot; rel=&quot;nofollow noreferrer&quot;&gt;513-2019&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-02-09T14:57:19.127" LastActivityDate="2019-02-09T14:57:19.127" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29522" PostTypeId="1" CreationDate="2019-02-09T16:08:08.777" Score="3" ViewCount="205" Body="&lt;p&gt;It took some effort, but in the end, I successfully calculated the geographical point of the Sun. I assumed that I could do the same for the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My plan is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Find RA and declination of the Moon&lt;/li&gt;&#xA;&lt;li&gt;Find GHA of the Moon, using the same math as for the Sun&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, after following &lt;a href=&quot;http://njsas.org/projects/tidal_forces/altaz/pausch/ppcomp.html#6&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; almost to the letter (except for precession and perturbations), my calculated answer is much different from &lt;a href=&quot;https://theskylive.com/moon-info&quot; rel=&quot;nofollow noreferrer&quot;&gt;this site&lt;/a&gt;. I've examined the article and my code many times, but can't see an error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use Swift programming language, but here it's merely a way to represent my calculations and results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eNVEJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eNVEJ.png&quot; alt=&quot;my code and results&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone would like to try it for yourself, here's my code in text format:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import Foundation&#xA;&#xA;extension Double {&#xA;  var radians: Double {&#xA;    return self * .pi / 180&#xA;  }&#xA;&#xA;  var degrees: Double {&#xA;    return self  * 180 / .pi&#xA;  }&#xA;&#xA;  var rangedUpTo360: Double {&#xA;    let multiples = (self/360).rounded(.down)&#xA;    return self - 360*multiples&#xA;  }&#xA;&#xA;  var rangedUpTo86400: Double {&#xA;    let multiples = (self/86400).rounded(.down)&#xA;    return self - 86400*multiples&#xA;  }&#xA;}&#xA;&#xA;extension Date {&#xA;  var greenwich: Date {&#xA;    let greenwichDate = self.addingTimeInterval(TimeInterval(-TimeZone.current.secondsFromGMT()))&#xA;    return greenwichDate&#xA;  }&#xA;}&#xA;&#xA;class Astronomy {&#xA;&#xA;  static func julianDays(since: Date) -&amp;gt; Double {&#xA;    return since.timeIntervalSince1970 / 86400 + 2_440_587.5&#xA;  }&#xA;&#xA;  static func daysSinceJ2000(for date: Date) -&amp;gt; Double {&#xA;    return julianDays(since: date) - 2_451_545&#xA;  }&#xA;&#xA;  static func sunAxialTilt(for daysSinceJ200: Double) -&amp;gt; Double {&#xA;    return 23.4393 - 3.563e-7 * daysSinceJ200&#xA;  }&#xA;&#xA;  static func greenwichHourAngle(rightAscension: Double, date: Date) -&amp;gt; Double {&#xA;    let greenwichTimeZone = TimeZone(secondsFromGMT: 0)!&#xA;    var calendar = Calendar.current&#xA;    calendar.timeZone = greenwichTimeZone&#xA;    let components = calendar.dateComponents([.hour, .minute, .second], from: date)&#xA;    let secondsOfDay = Double(components.hour!)*3600 + Double(components.minute!)*60 + Double(components.second!)&#xA;&#xA;    let centuriesSinceJ200 = daysSinceJ2000(for: date.greenwich) / 36525&#xA;    let greenwichSiderealTime = 24110.54841 + 8640184.812866*centuriesSinceJ200 + 0.093104*pow(centuriesSinceJ200, 2) - 0.0000062*pow(centuriesSinceJ200, 3)&#xA;    let earthSiderealRotationRate = 1.00273790935 + 5.9e-11*centuriesSinceJ200&#xA;    let earthRotation = earthSiderealRotationRate*secondsOfDay&#xA;    let gmstSeconds = greenwichSiderealTime + earthRotation&#xA;    let gmstSecondsNormalized = gmstSeconds.rangedUpTo86400&#xA;    let gmst = gmstSecondsNormalized / 3600 * 15&#xA;    return gmst.rangedUpTo360 - rightAscension&#xA;  }&#xA;&#xA;  static func moonPosition(on date: Date) -&amp;gt; (declination: Double, RA: Double, distance: Double) {&#xA;    date&#xA;    // day number&#xA;    let d = daysSinceJ2000(for: date)&#xA;    // longitude of the ascending node&#xA;    var N = 125.1228 - 0.0529538083 * d // have to be normalized&#xA;    if N &amp;lt; 0 {&#xA;      N += 360&#xA;    }&#xA;    // inclination to the ecliptic (plane of the Earth's orbit)&#xA;    let i = 5.1454&#xA;    // argument of perihelion&#xA;    var w = 318.0634 + 0.1643573223 * d // have to be normalized&#xA;    w = w.rangedUpTo360&#xA;    // semi-major axis, or mean distance from Sun&#xA;    let a = 60.2666 // (Earth radii)&#xA;    // eccentricity (0=circle, 0-1=ellipse, 1=parabola)&#xA;    let e = 0.054900&#xA;    // mean anomaly (0 at perihelion; increases uniformly with time)&#xA;    var M = 115.3654 + 13.0649929509 * d // have to be normalized&#xA;    M = M.rangedUpTo360&#xA;    // eccentric anomaly&#xA;    let E = M + e * sin(M.radians) * (1.0 + e * cos(M.radians))&#xA;    let xv = a * (cos(E.radians) - e)&#xA;    let yv = a * (sqrt(1.0 - e*e) * sin(E.radians))&#xA;    // true anomaly (angle between position and perihelion)&#xA;    let v = atan2(yv, xv).degrees&#xA;    let r = sqrt(pow(xv, 2) + pow(yv, 2))&#xA;    // geocentric ecliptical position&#xA;    let xG = r * (cos(N.radians) * cos((v + w).radians) - sin(N.radians) * sin((v + w).radians) * cos(i.radians))&#xA;    let yG = r * (sin(N.radians) * cos((v + w).radians) + cos(N.radians) * sin((v + w).radians) * cos(i.radians))&#xA;    let zG = r * (sin((v + w).radians) * sin(i.radians))&#xA;    // equatorial coordinates&#xA;    let ecl = sunAxialTilt(for: d)&#xA;    let xE = xG&#xA;    let yE = yG * cos(ecl.radians) - zG * sin(ecl.radians)&#xA;    let zE = yG * sin(ecl.radians) + zG * cos(ecl.radians)&#xA;    //right scension&#xA;    var rA = atan2(yE, xE)&#xA;    rA = rA.degrees&#xA;    // declinatino&#xA;    var dec = atan2(zE, sqrt(xE * xE + yE * yE) )&#xA;    dec = dec.degrees&#xA;    // geocentric distance&#xA;    let rG = sqrt(xE * xE + yE * yE + zE * zE)&#xA;    return (declination: dec, RA: rA, distance: rG)&#xA;  }&#xA;}&#xA;&#xA;Astronomy.moonPosition(on: Date().greenwich)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="25784" LastActivityDate="2020-12-30T16:18:02.170" Title="Moon equatorial coordinates" Tags="&lt;the-moon&gt;&lt;declination&gt;&lt;right-ascension&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="29523" PostTypeId="1" CreationDate="2019-02-09T16:29:48.333" Score="9" ViewCount="805" Body="&lt;p&gt;I was wondering.. is there any habitable exoplanet around Tau Ceti?&lt;/p&gt;&#xA;" OwnerDisplayName="user22712" LastActivityDate="2020-12-07T17:40:29.787" Title="Is there any habitable exoplanet around Tau Ceti?" Tags="&lt;exoplanet&gt;&lt;space&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29525" PostTypeId="1" CreationDate="2019-02-09T18:00:55.970" Score="6" ViewCount="224" Body="&lt;p&gt;I was reading about the coolest stars, and was surprised to find about stars with a surface temperature lower than a candle, like 2MASS 0939-2448 A/B, of about 100 Celsius like CFBDSIR 1458+10B and even below 0 Celsius temperature where the water freezes like WISE J0855-0714. Then I thought, they might be stars where fusion only happens very deep in the core, and then I found out this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Brown_dwarf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brown dwarf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In addition, many brown dwarfs undergo no fusion; those at the low end&#xA;of the mass range (under 13 MJ) are never hot enough to fuse even&#xA;deuterium, and even those at the high end of the mass range (over 60&#xA;MJ) cool quickly enough that after 10 million years they no longer&#xA;undergo fusion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Are these brown dwarfs which don't sustain any type of fusion considered stars? If so, what's the reason?&lt;/p&gt;&#xA;" OwnerUserId="11562" LastEditorUserId="26112" LastEditDate="2020-10-25T15:00:17.333" LastActivityDate="2020-10-25T15:00:17.333" Title="Are brown dwarfs which don't sustain any fusion considered stars?" Tags="&lt;star&gt;&lt;brown-dwarf&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29526" PostTypeId="2" ParentId="29523" CreationDate="2019-02-09T18:27:58.313" Score="29" Body="&lt;p&gt;TLDR version: probably not, and claims for the habitability of planets in this system are on shaky ground.&lt;/p&gt;&#xA;&lt;p&gt;Long version follows.&lt;/p&gt;&#xA;&lt;h1&gt;Planets&lt;/h1&gt;&#xA;&lt;p&gt;So as of &lt;a href=&quot;https://arxiv.org/abs/1708.02051&quot; rel=&quot;nofollow noreferrer&quot;&gt;Feng et al. (2017)&lt;/a&gt;, there are four planet candidates around Tau Ceti:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;Planet&lt;/th&gt;&#xA;&lt;th&gt;Minimum mass (&lt;span class=&quot;math-container&quot;&gt;$M_\odot$&lt;/span&gt;)&lt;/th&gt;&#xA;&lt;th&gt;Semimajor axis (AU)&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Tau Ceti g&lt;/td&gt;&#xA;&lt;td&gt;&lt;span class=&quot;math-container&quot;&gt;$1.75^{+0.25}_{-0.40}$&lt;/span&gt;&lt;/td&gt;&#xA;&lt;td&gt;&lt;span class=&quot;math-container&quot;&gt;$0.133^{+0.001}_{-0.002}$&lt;/span&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Tau Ceti h&lt;/td&gt;&#xA;&lt;td&gt;&lt;span class=&quot;math-container&quot;&gt;$1.83^{+0.68}_{-0.26}$&lt;/span&gt;&lt;/td&gt;&#xA;&lt;td&gt;&lt;span class=&quot;math-container&quot;&gt;$0.243^{+0.003}_{-0.003}$&lt;/span&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Tau Ceti e&lt;/td&gt;&#xA;&lt;td&gt;&lt;span class=&quot;math-container&quot;&gt;$3.93^{+0.83}_{-0.64}$&lt;/span&gt;&lt;/td&gt;&#xA;&lt;td&gt;&lt;span class=&quot;math-container&quot;&gt;$0.538^{+0.006}_{-0.006}$&lt;/span&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Tau Ceti f&lt;/td&gt;&#xA;&lt;td&gt;&lt;span class=&quot;math-container&quot;&gt;$3.93^{+1.05}_{-1.37}$&lt;/span&gt;&lt;/td&gt;&#xA;&lt;td&gt;&lt;span class=&quot;math-container&quot;&gt;$1.334^{+0.017}_{-0.044}$&lt;/span&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;Note that the designations Tau Ceti b, c and d refer to planet candidates that are no longer thought to exist. The error bars refer to the 1% and 99% percentiles. &lt;span class=&quot;math-container&quot;&gt;$M_\oplus$&lt;/span&gt; is the mass of the Earth.&lt;/p&gt;&#xA;&lt;p&gt;The Feng et al. (2017) paper also notes that the system is dynamically packed, which does not bode well for the prospects for additional planets between the known planet candidates (note that their figure 17 shows the regions where the planets would interfere with &lt;em&gt;each other&lt;/em&gt;, not the regions of stability for an additional planet).&lt;/p&gt;&#xA;&lt;h1&gt;The habitable zone&lt;/h1&gt;&#xA;&lt;p&gt;The conclusion of the paper gives the luminosity of Tau Ceti as 0.52 times solar and the effective temperature as 5344 K. Using these values, the habitable zone limits can be calculated from &lt;a href=&quot;https://arxiv.org/abs/1301.6674&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kopparapu et al. (2013)&lt;/a&gt;, which assumes that habitable conditions are maintained by the carbonate-silicate cycle with carbon dioxide as the main (non-condensible) greenhouse gas.&lt;/p&gt;&#xA;&lt;h2&gt;Inner boundaries&lt;/h2&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;Boundary&lt;/th&gt;&#xA;&lt;th&gt;Distance (AU)&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Recent Venus&lt;/td&gt;&#xA;&lt;td&gt;0.551&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Runaway greenhouse&lt;/td&gt;&#xA;&lt;td&gt;0.723&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Moist greenhouse&lt;/td&gt;&#xA;&lt;td&gt;0.729&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;The moist greenhouse limit is the most conservative inner boundary, it occurs where sufficient water vapour enters the upper atmosphere that water loss begins to occur from the planet. In our solar system, Earth is located close to this limit in the inner part of the most conservative habitable zone.&lt;/p&gt;&#xA;&lt;p&gt;The runaway greenhouse limit occurs where the positive feedback from water vapour overwhelms the stabilising negative feedback from the silicate-carbonate cycle, driving further evaporation of the oceans and higher temperatures. This is thought to have occurred on Venus, leaving the planet in the state it is in today.&lt;/p&gt;&#xA;&lt;p&gt;The recent Venus limit is based on the possibility that Venus may have retained oceans for several billion years. This is not known for certain as our knowledge of Venus's evolution is rather incomplete and the conditions on the planet's surface are not favourable for driving rovers around investigating the geology.&lt;/p&gt;&#xA;&lt;p&gt;From this we see that Tau Ceti e is located close to the recent Venus limit and is closer to the star than the runaway greenhouse limit. This suggests that any oceans that may once have existed would likely have boiled off, leaving the planet in a Venus-like state.&lt;/p&gt;&#xA;&lt;p&gt;Planets g and h are too close to the star.&lt;/p&gt;&#xA;&lt;h2&gt;Outer boundaries&lt;/h2&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;Boundary&lt;/th&gt;&#xA;&lt;th&gt;Distance (AU)&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Maximum greenhouse&lt;/td&gt;&#xA;&lt;td&gt;1.279&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Early Mars&lt;/td&gt;&#xA;&lt;td&gt;1.330&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;The maximum greenhouse limit is the furthest distance from the star that a cloud-free carbon dioxide atmosphere can support conditions compatible with liquid water. Beyond this, the increased scattering leads to increased reflectivity of the planet and the CO&lt;sub&gt;2&lt;/sub&gt; would begin to condense, removing it from the atmosphere and leading to runaway cooling. This is the most conservative outer habitable zone boundary. Note that by this point, the planet would require several bars of CO&lt;sub&gt;2&lt;/sub&gt; which would make it toxic for humans.&lt;/p&gt;&#xA;&lt;p&gt;The early Mars limit is based on the observation that Mars managed to maintain surface water (e.g. various rivers, and a possible northern ocean) in the early solar system when the Sun was significantly fainter than it is today. Tau Ceti f is located right at this limit.&lt;/p&gt;&#xA;&lt;h2&gt;Extensions to the habitable zone&lt;/h2&gt;&#xA;&lt;p&gt;None of the planets fall into the most conservative habitable zone, and Tau Ceti e and f are at the boundaries of the most optimistic estimates for the habitable zone boundaries. There are nevertheless options for extending the habitable zone.&lt;/p&gt;&#xA;&lt;p&gt;At the inner boundary, a runaway greenhouse effect could be avoided on dry planets, where there simply isn't enough water to evaporate to drive the positive feedback, see &lt;a href=&quot;https://arxiv.org/abs/1304.3714&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zsom et al. (2013)&lt;/a&gt;. It isn't clear to me that such a planet can be described as habitable, since such planets may lack the hydrothermal systems that could act as the sites for abiogenesis. Their geological evolution would likely be substantially different to Earth's without water to lubricate plate tectonics.&lt;/p&gt;&#xA;&lt;p&gt;Another possibility is on slowly-rotating planets, where substantial cloud layers can build up on the day side of the planet and increase the reflectivity, as noted by &lt;a href=&quot;https://arxiv.org/abs/1404.4992&quot; rel=&quot;nofollow noreferrer&quot;&gt;Yang et al. (2014)&lt;/a&gt;. On the other hand, &lt;a href=&quot;https://arxiv.org/abs/1804.07380&quot; rel=&quot;nofollow noreferrer&quot;&gt;Scholz et al. (2018)&lt;/a&gt; have noted that there appears to be a universal mass-spin relationship that extends from planets to brown dwarfs. This predicts that super-Earths would likely spin too fast for this mechanism to work unless they have been spun down by stellar tides or a large moon.&lt;/p&gt;&#xA;&lt;p&gt;On the outer boundary, adding additional greenhouse gases such as methane may work to extend the outer habitable zone, see for example &lt;a href=&quot;https://arxiv.org/abs/1805.02801&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ramirez &amp;amp; Kaltenegger (2018)&lt;/a&gt;. This has been suggested as the mechanism for allowing surface water on Mars, which would suggest that the &amp;quot;Early Mars&amp;quot; limit is an observed data point within the methane habitable zone. Another possibility is that a dense hydrogen atmosphere could maintain liquid water, e.g. &lt;a href=&quot;https://arxiv.org/abs/1105.0021&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pierrehumbert &amp;amp; Gaidos (2011)&lt;/a&gt; though the pressure of such an atmosphere may well have implications for the geology of the planet and hence the potential for abiogenesis.&lt;/p&gt;&#xA;&lt;p&gt;Planets whose climates are stabilised by something other than the carbonate-silicate cycle, or have substantially different atmospheric compositions would have different habitable zone boundaries (if subsurface oceans on icy worlds are habitable, there may be interesting prospects for dwarf planets in the outer debris belt), but this is already getting speculative enough, besides there is another possible objection to the habitability of these planets...&lt;/p&gt;&#xA;&lt;h1&gt;Planetary masses&lt;/h1&gt;&#xA;&lt;p&gt;A limitation of the radial velocity method is that only the minimum masses can be derived. With Tau Ceti, we have a possible means to estimate the true masses: the star is surrounded by a debris disc (this would likely provide a source of impactors onto the planets, how bad the situation is depends on how much material is being perturbed into the inner system). Using Herschel observations, &lt;a href=&quot;https://arxiv.org/abs/1408.2791&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lawler et al. (2014)&lt;/a&gt; give an inclination of 35±10 degrees. Assuming that the planets lie in the same plane as the disc, the true masses would therefore be approximately 1.74 times greater than the minimum masses.&lt;/p&gt;&#xA;&lt;p&gt;Under this assumption, the true masses of the planets e and f both come out as about 6.85 Earth masses. Taking the 99% lower limit on the minimum mass error bars and a 45° orbital inclination as a low estimate, these would be 4.65 Earth masses for e and 3.62 Earth masses for f.&lt;/p&gt;&#xA;&lt;h1&gt;The nature of the planets&lt;/h1&gt;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://arxiv.org/abs/1407.4457&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rogers (2014)&lt;/a&gt;, the transition between rocky and Neptune-like planets is somewhere in the region of 1.4 to 1.6 Earth radii. Using the mass-radius relationship from &lt;a href=&quot;https://arxiv.org/abs/1512.08827&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zeng et al. (2016)&lt;/a&gt; and their core mass fraction of 0.26 for typical terrestrial planets, these radius limits correspond to terrestrial planets of roughly 3.3 to 5.4 Earth masses.&lt;/p&gt;&#xA;&lt;p&gt;This suggests that Tau Ceti e and f are fairly likely to be sub-Neptunes rather than rocky planets, although the caveats are that in the optimistic case they can have masses below the rocky/Neptune-like transition, and that there do seem to be a few cases of rocky planets above the transition (most of those are likely to be evaporated cores of Neptune-like planets, which wouldn't apply to Tau Ceti e and f as they have much lower levels of stellar irradiation).&lt;/p&gt;&#xA;&lt;h1&gt;Conclusion&lt;/h1&gt;&#xA;&lt;p&gt;Given the current state of knowledge, Tau Ceti does not look like a good prospect for habitable planets. Tau Ceti e and f are rather marginal in terms of their location within the habitable zone, and their masses are sufficiently high that there is a good chance that they are sub-Neptunes rather than rocky planets. The dynamical packing of the system makes it unlikely that there can be a smaller, temperate planet in the habitable zone between the known planets.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2020-12-07T17:40:29.787" LastActivityDate="2020-12-07T17:40:29.787" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29527" PostTypeId="2" ParentId="29519" CreationDate="2019-02-09T20:22:01.467" Score="1" Body="&lt;p&gt;Yes: Given that the universe isn't expanding at a faster rate than the speed of the gravity effect of those objects, then the gravity effect of them would be able to reach one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even then it would take a very long time for the gravity effects of the objects to reach one another, so the objects would just stay there motionless for ages(in this case 5 billion light years) before starting to move towards eachother.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nowykurier.com/toys/gravity/gravity.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.nowykurier.com/toys/gravity/gravity.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user25868" LastEditorDisplayName="user25868" LastEditDate="2019-02-10T18:45:40.143" LastActivityDate="2019-02-10T18:45:40.143" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29528" PostTypeId="1" AcceptedAnswerId="29529" CreationDate="2019-02-09T21:52:32.833" Score="1" ViewCount="257" Body="&lt;p&gt;While researching about galaxies in &lt;a href=&quot;https://www.google.com/sky/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google sky&lt;/a&gt;, I came upon a strange area.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Co-ordinates: 11h 6m 15.0 seconds | -77° 22' 9.8&amp;quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;A lot of space around this body is empty, is it possible that the body is a black hole?&lt;/p&gt;&#xA;&lt;p&gt;The reason I'm asking is because of this image - &lt;a href=&quot;https://i.stack.imgur.com/Stpbp.png%3Cbr%3E&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Stpbp.png%3Cbr%3E&quot; alt=&quot;Black hole&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://chandra.harvard.edu/photo/2003/perseus/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This is a photograph and readings are of a real black hole.&lt;/a&gt;.&#xA;It looks like the one on the co-ordinates which I observed with all the dust around it.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;The co-ordinates of the real black hole are - 03h 19m 47.60s | +41° 30' 37.00&amp;quot;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="24386" LastEditorUserId="35106" LastEditDate="2020-10-04T20:35:16.123" LastActivityDate="2020-10-04T20:35:16.123" Title="Is this a Black Hole?" Tags="&lt;black-hole&gt;&lt;observational-astronomy&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29529" PostTypeId="2" ParentId="29528" CreationDate="2019-02-09T22:08:53.170" Score="6" Body="&lt;p&gt;Going by those coordinates, the dark area you're seeing in Google Sky is the &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=Cha%20I&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chamaeleon I cloud&lt;/a&gt;. It's a nearby star-forming region that is part of the &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-basic?Ident=Cha%20Complex&amp;amp;submit=SIMBAD%20search&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chamaeleon complex&lt;/a&gt; (see also the &amp;quot;Notable features&amp;quot; section of &lt;a href=&quot;https://www.universetoday.com/20114/chamaeleon/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article on the Chamaeleon constellation&lt;/a&gt;). The star nearest to the coordinates you provided is the T Tauri star &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=%09TYC%209414-787-1&amp;amp;submit=submit%20id&quot; rel=&quot;nofollow noreferrer&quot;&gt;TYC 9414-787-1&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2020-10-04T11:06:15.043" LastActivityDate="2020-10-04T11:06:15.043" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29531" PostTypeId="2" ParentId="29462" CreationDate="2019-02-09T23:26:06.450" Score="3" Body="&lt;p&gt;If you were working with a rectangular coordinate system as shown in the figure below (that also had no bounds), then it is correct that the right ascension would be &lt;span class=&quot;math-container&quot;&gt;$RA=RA_p\pm d$&lt;/span&gt; (points 1 and 3) and the declination would be &lt;span class=&quot;math-container&quot;&gt;$Dec=Dec_p \pm d$&lt;/span&gt; (points 2 and 4) where d is the radius of the circle.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/OqdDZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OqdDZ.png&quot; alt=&quot;rectangular coordinate system&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sky is spherical, so the lines of RA are not a constant separation. (It also has limits such as declination less than &lt;span class=&quot;math-container&quot;&gt;$\pm 90$&lt;/span&gt;.) Therefore you need to use an equation for spherical trigonometry. The angle d between two points i and j is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\cos(d) = \sin(Dec_i)\sin(Dec_j) + \cos(Dec_i)\cos(Dec_j)\cos(RA_i - RA_j)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8alBn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8alBn.png&quot; alt=&quot;spherical coordinates used in the sky&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For points 1 and 3 on the circle, &lt;span class=&quot;math-container&quot;&gt;$Dec_i=Dec_j=Dec_p$&lt;/span&gt; in the above equation, and you can solve for the difference in right ascension &lt;span class=&quot;math-container&quot;&gt;$RA_i - RA_j$&lt;/span&gt; to be added to or subtracted from &lt;span class=&quot;math-container&quot;&gt;$RA_p$&lt;/span&gt;. For points 2 and 4, &lt;span class=&quot;math-container&quot;&gt;$RA_i=RA_j=RA_p$&lt;/span&gt; and the above equation simplifies to &lt;span class=&quot;math-container&quot;&gt;$Dec=Dec_p \pm d$&lt;/span&gt;. Of course, keep in mind that declination is between -90 and +90. For any other point on the circle, you can assume either an RA or Dec (within the bounds of the above 4 points) and solve for the other unknown. &lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2019-02-09T23:26:06.450" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29532" PostTypeId="1" CreationDate="2019-02-10T01:24:49.320" Score="6" ViewCount="439" Body="&lt;p&gt;I know that the Earth will never become tidally locked to the Moon, so this is purely hypothetical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand correctly, in this hypothetical scenario an Earth day is now a lot longer at around 29 current days, and you can only ever see the Moon from half of the Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So standing on the &quot;lucky&quot; side of the Earth you should be able to see &lt;em&gt;all the lunar phases&lt;/em&gt; during a single night. The Moon become gradually more illuminated, peak as a full Moon and then darken as the night ends. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="25870" LastEditorUserId="7982" LastEditDate="2019-02-10T02:08:35.880" LastActivityDate="2019-02-11T04:21:47.317" Title="Lunar phases if Earth was tidally locked to the Moon" Tags="&lt;the-moon&gt;&lt;moon-phases&gt;&lt;tidal-locking&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29533" PostTypeId="2" ParentId="29532" CreationDate="2019-02-10T02:47:27.987" Score="5" Body="&lt;p&gt;&lt;strong&gt;tl;dr:&lt;/strong&gt; An observer on the &quot;Moon side&quot; would see only &lt;em&gt;half&lt;/em&gt; the phases during the fortnight-long night: from waxing half-moon to waning half-moon. They would also see the Moon during the day (early morning, late afternoon), as we currently can, but they would never see the New Moon or crescent moons &lt;strong&gt;at night&lt;/strong&gt; since these would be only visible during the day.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;br&gt;&#xA;While the Earth being tidally locked to the Moon would indeed mean that the side of the Earth facing the Moon always sees the Moon, the &lt;em&gt;phases&lt;/em&gt; of the Moon result from the angle between the Earth, the Moon and &lt;strong&gt;the Sun&lt;/strong&gt;, and probably wouldn’t change significantly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon orbits around the Earth-Moon barycentre with an orbital period of 27.3 days and a synodic period (i.e. from Full Moon to Full Moon) of 29.5 days. An observer on the Moon-facing side of the Earth would certainly see the Moon go through its different phases, but it would still take 29.5 days. [For the purposes of this question I'm ignoring the increasing lunar period due to the Moon receding from the Earth.] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the question notes, one obvious outcome of Earth becoming tidally locked to the Moon is that Earth’s “day” becomes the same as the Moon’s synodic period: 29.5 days. Daylight will last for about 15 days, baking everything other than at the poles. The 360-hour night will be freezing, even in the tropics. Local weather will be utterly different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another notable outcome is there will only be really tiny “solar” tides spaced about 15 days apart (due to the Earth’s slow rotation in relation to the Sun). The loss of significant tidal flows would probably have a dramatic negative effect on marine life. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth tidally locked to the Moon is a nightmare scenario!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The core question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So standing on the &quot;lucky&quot; side of the Earth you should be able to see&#xA;  &lt;em&gt;all the lunar phases&lt;/em&gt; during a single night. The Moon become gradually&#xA;  more illuminated, peak as a full Moon and then darken as the night&#xA;  ends.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No: you would only see only &lt;em&gt;half&lt;/em&gt; the phases during the fortnight-long night: from waxing half-moon to waning half-moon. You would also see the Moon during the day (early morning, late afternoon), as we currently can, but you would never see the New Moon or crescent moons &lt;strong&gt;at night&lt;/strong&gt; since these would be only visible during the day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We currently see the New Moon because when it's roughly overhead for some at midday, it's not far above the horizon just before sunrise or just after sunset for others. Since the New Moon phase currently lasts about 3-5 days (depending on how you define &quot;new&quot;), but the Earth rotates in just one 24-hour day, most habitable places on Earth get to see this phase every lunar month. However, in the tidally-locked-Earth scenario, the phase still lasts 3-5 days but the Earth rotates once every 29.5 days. Those viewing the Moon just after sunset will only &lt;em&gt;ever&lt;/em&gt; see the Moon in that position above the horizon – it more or less &quot;sits&quot; there, apart from libration and nodal period effects – and they will miss some of the other phases because the Sun will be too bright at, say, the waning half-moon.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT - Two additional things to note:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(1) Earth's phases&lt;/strong&gt;&lt;br&gt;&#xA;While the Earth becoming tidally locked to the Moon is hypothetical, the Moon is already tidally locked to the Earth, which means the above descriptions can be applied in reverse for a real observer on the Moon: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a lunar &quot;day&quot; (sunrise to sunrise) is 29.5 Earth days;&lt;/li&gt;&#xA;&lt;li&gt;the Earth's position in the lunar sky is relatively constant over the lunar day, within about 15º (due to a slow but noticeable &lt;a href=&quot;https://en.wikipedia.org/wiki/Libration&quot; rel=&quot;nofollow noreferrer&quot;&gt;libration&lt;/a&gt;);&lt;/li&gt;&#xA;&lt;li&gt;the Moon's orbit is elliptical, so over much longer periods the Earth would alternately recede and grow larger (a &quot;Super Earth&quot;);&lt;/li&gt;&#xA;&lt;li&gt;unlike the Earth-locked scenario, all of the Earth's phases – from Full Earth at lunar midnight to New Earth at lunar noon - would be visible, because there's no atmosphere and hence no &quot;bright daytime sky&quot; to block out the Earthlight.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note also that the famous &lt;a href=&quot;https://en.wikipedia.org/wiki/Earthrise&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earthrise&lt;/a&gt; experience of the Apollo 8 astronauts would not be available to an observer on the Moon's surface. The Earth &quot;rose&quot; above the limb of the Moon because the astronauts were &lt;em&gt;in orbit&lt;/em&gt;. Nonetheless, someone located on the border of the &quot;dark side of the Moon&quot; would see the Earth on the lunar horizon, and libration could result in the Earth alternately setting below the horizon and rising again over a lunar day. For that observer, however, earthrise and earthset occur at roughly &lt;em&gt;the same point&lt;/em&gt; on the horizon!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(2) Will the Earth really end up tidally locked to the Moon?&lt;/strong&gt;&lt;br&gt;&#xA;Tidal braking is slowing the Earth's rotation by 23 milliseconds (and increasing the Moon's distance from us by 38 metres) per 1,000 years. At that rate, the Andromeda Galaxy will have &lt;a href=&quot;https://en.wikipedia.org/wiki/Andromeda_Galaxy#Collision_with_the_Milky_Way&quot; rel=&quot;nofollow noreferrer&quot;&gt;collided with the Milky Way&lt;/a&gt; long before the Earth's rotation has slowed down enough to approach tidal locking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, &lt;a href=&quot;https://astronomy.stackexchange.com/a/29533/14094&quot;&gt;this AstronomySE answer&lt;/a&gt; suggests that as the Earth's oceans &quot;are responsible for almost all of the tidal deceleration&quot; (&lt;a href=&quot;https://astronomy.stackexchange.com/questions/18794/will-the-earth-ever-be-tidally-locked-to-the-moon/18795#comment30069_18795&quot;&gt;David Hammen's comment&lt;/a&gt;), the evaporation of the oceans &lt;a href=&quot;https://en.wikipedia.org/wiki/Future_of_Earth#Loss_of_oceans&quot; rel=&quot;nofollow noreferrer&quot;&gt;in a billion years&lt;/a&gt; will eliminate the main drag on the Earth's spin and tidal locking may therefore never occur.&lt;/p&gt;&#xA;" OwnerUserId="14094" LastEditorUserId="14094" LastEditDate="2019-02-10T23:47:57.843" LastActivityDate="2019-02-10T23:47:57.843" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="29535" PostTypeId="1" CreationDate="2019-02-10T09:44:37.817" Score="2" ViewCount="277" Body="&lt;p&gt;How are the cardinal points (North-East-South-West) defined? On equatorial regions it is somewhat obvious. But what about the higher latitudes? If the pole star is up above the sky instead of horizon, how do we define North? What is the case on the geographic poles? &lt;/p&gt;&#xA;" OwnerUserId="9735" LastActivityDate="2019-03-12T17:00:42.107" Title="How are cardinal points on horizon defined?" Tags="&lt;earth&gt;&lt;pole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29537" PostTypeId="2" ParentId="29535" CreationDate="2019-02-10T16:31:13.450" Score="1" Body="&lt;h2&gt;Geographic Meridian&lt;/h2&gt;&#xA;&lt;p&gt;Every point on Earth is on some &lt;a href=&quot;https://en.wikipedia.org/wiki/Meridian_(geography)&quot; rel=&quot;nofollow noreferrer&quot;&gt;meridian&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A (geographical) meridian (or line of longitude) is the half of an imaginary great circle on the Earth's surface, terminated by the North Pole and the South Pole, connecting points of equal longitude, as measured in angular degrees east or west of the Prime Meridian.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;From where you are standing, North is the direction on the horizon along this meridian or &amp;quot;line&amp;quot; of longitude that extends to the Earth's geographic North pole, and South is the direction on the horizon 180 degrees opposite, towards the South pole.&lt;/p&gt;&#xA;&lt;p&gt;East and west are the perpendicular directions, which are also found extending lines of latitude from where you are standing towards the horizon.&lt;/p&gt;&#xA;&lt;p&gt;Locally you can think of the region under your feet as flat, and so North, South, East and West all form right angles even far from the equator.&lt;/p&gt;&#xA;&lt;p&gt;Close to the poles, none of that changes mathematically, and exactly on the poles those terms suddenly no longer have any meaning.&lt;/p&gt;&#xA;&lt;h2&gt;Celestial Meridian&lt;/h2&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Meridian_(astronomy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;meridian&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In astronomy, the meridian is the great circle passing through the celestial poles, as well as the zenith and nadir of an observer's location.&lt;/p&gt;&#xA;&lt;p&gt;Consequently, it contains also the north and south points on the horizon, and it is perpendicular to the celestial equator and horizon. &lt;strong&gt;A celestial meridian is coplanar with the analogous terrestrial meridian projected onto the celestial sphere.&lt;/strong&gt; Hence, the number of celestial meridians is also infinite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UVXbW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UVXbW.png&quot; alt=&quot;Meridian_on_celestial_sphere.png&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/File:Meridian_on_celestial_sphere.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-02-10T16:31:13.450" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29539" PostTypeId="1" AcceptedAnswerId="29540" CreationDate="2019-02-10T17:58:47.590" Score="2" ViewCount="88" Body="&lt;p&gt;The IAU has different classifications for objects in the solar system orbiting the sun, they can be according to their shape and if they clear their neighborhood of other material around its orbits , planets, dwarf planets or Small Solar System bodies. But for objects in the solar system orbiting a planet or a dwarf planet they are all satellites, no matter if they are rounded or if they have an irregular shape like an asteroid. Is there any known reason why this is so? Why is it needed to differenciate its shape when the object it's orbiting the sun, but it isnt needed when the object it's orbiting a planet? (or dwarf planet)?&lt;/p&gt;&#xA;" OwnerUserId="11562" LastActivityDate="2020-08-05T04:55:29.633" Title="Is there a reason why the IAU differenciates objects in the SS orbiting the sun into a particular classification but it doesnt for satellites?" Tags="&lt;planet&gt;&lt;satellite&gt;&lt;dwarf-planets&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29540" PostTypeId="2" ParentId="29539" CreationDate="2019-02-10T21:36:41.183" Score="3" Body="&lt;p&gt;History, and recent history at that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At one time (1980) things seemed clear enough: There were planets (Mercury-Pluto), Asteroids (mostly between Mars and Jupiter) and Satellites.  No one care much about whether a body was spherical or not, since it was pretty clear that this was all there was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.&lt;br&gt;&#xA;.&lt;br&gt;&#xA;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the Kuiper belt happened, and in particular it became clear that Eris was big enough to be rounded... Now why should not Eris be a planet? Could we have tens of rounded icy objects orbiting beyond Neptune?  How many would we expect grade-schoolers to memorise?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The decision was taken to classify solar system bodies into:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Planets (round and not shareing their orbit with anything comparible in size)&lt;/li&gt;&#xA;&lt;li&gt;Dwarf planets (round, but have not cleared their orbit)&lt;/li&gt;&#xA;&lt;li&gt;Minor planets (not round)&lt;/li&gt;&#xA;&lt;li&gt;Satellites (not orbiting the sun)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is a matter of convenience for humans. It does not reflect any fundamental reality.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For convenience, there is no need to distinguish between round moons and non-round ones, so no distinction is needed (for humans)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-02-10T21:36:41.183" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29541" PostTypeId="2" ParentId="29462" CreationDate="2019-02-11T00:28:46.723" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rkyz1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rkyz1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that you are at 80 degrees north and 0 degrees east (on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Prime_meridian&quot; rel=&quot;nofollow noreferrer&quot;&gt;Prime Meridian&lt;/a&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is only 10 latitude degrees from you to the north pole.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the distance from you to the 45th meridian (shown in green) is about 7 latitude degrees (&lt;span class=&quot;math-container&quot;&gt;$\sin{\pi/4}\cdot 10$&lt;/span&gt;) because if you walked to the closest point on the 45th meridian, the north pole would be directly to your left and the three points 1) the closest point on the 45th meridian, 2) the north pole, and 3) 80 degrees north on the prime meridian almost form an isosceles 45-45-90 triangle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you draw a circle around the position (80 North, 0 East) with radius of 7 latitude degrees, the maximum latitude would be 87, the minimum latitude would be 73, the maximum longitude would be about 45 degrees east, and the minimum longitude would be about -45 degrees (45 degrees west).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the point on the 45th meridian closest to your starting position (80N 0E) would have a latitude of about 83 degrees, not 80 degrees.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example above illustrates that the spacing between longitude lines gets closer near the poles.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if you are at the position (&lt;span class=&quot;math-container&quot;&gt;$\delta$&lt;/span&gt; north, &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; east) and you draw a circle around yourself with &quot;radius&quot; &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt;, then the min/max latitude/longitude of the enclosed points would be &lt;span class=&quot;math-container&quot;&gt;$\delta-\rho$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\delta+\rho$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\alpha - \gamma$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\alpha + \gamma$&lt;/span&gt; where&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\gamma = \mathrm{arcsin}\left(\frac{\sin\rho}{\cos\delta}\right).&#xA;$$&lt;/span&gt;&#xA;(where &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; is the angle between you and the edge of the circle as seen from the center of the Earth).&#xA;Notice that &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if &lt;span class=&quot;math-container&quot;&gt;$\delta=0$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$\gamma=\rho$&lt;/span&gt;,&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\gamma\geq\rho$&lt;/span&gt;, &lt;/li&gt;&#xA;&lt;li&gt;if &lt;span class=&quot;math-container&quot;&gt;$\pi/2-|\delta|=\rho$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$\sin\rho=\cos\delta$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\gamma=\mathrm{arcsin}(1)=\pi/2$&lt;/span&gt;, and&lt;/li&gt;&#xA;&lt;li&gt;if &lt;span class=&quot;math-container&quot;&gt;$\pi/2-|\delta|&amp;lt;\rho$&lt;/span&gt;, then there is no solution for &lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt; because a pole is enclosed by the circle.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The formula can be derived from dot products between your position vector and the vector perpendicular to the either extreme meridian.  I need to do a lot of visualization and to fix several false starts to get it right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the original example of a circle around 80N 0E with radius 7 degrees, the formula yields &#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\gamma = \mathrm{arcsin}( \sin(7\pi/180)/\cos( 80\pi/180))\cdot180/\pi = 44.573 \mathrm{\ degrees}&#xA;$$&lt;/span&gt; &#xA;which is close to the estimated 45 degrees.  &lt;/p&gt;&#xA;" OwnerUserId="25554" LastEditorUserId="25554" LastEditDate="2019-02-11T00:38:13.947" LastActivityDate="2019-02-11T00:38:13.947" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29542" PostTypeId="1" AcceptedAnswerId="29545" CreationDate="2019-02-11T00:35:37.573" Score="1" ViewCount="58" Body="&lt;p&gt;&lt;a href=&quot;https://space.stackexchange.com/a/34094/12102&quot;&gt;In this answer&lt;/a&gt; I show that the (now pretty famous) Kuiper Belt Object &lt;code&gt;2014-MU69&lt;/code&gt; has two entries in JPL's ephemeris generating Horizons site; Major Body &lt;code&gt;3713011&lt;/code&gt; and Small Body &lt;code&gt;486958&lt;/code&gt;. I'm pretty sure these represent the same body, their trajectories differ by about a half-million kilometers only, and that difference is probably only because for one of them the trajectory is only based on a 118 day arc of it's ~100,000 day orbit at ~43 AU back in 2015.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are both &lt;code&gt;3713011&lt;/code&gt; and &lt;code&gt;486958&lt;/code&gt; still official references to the same body, but in different catalogs or databases? Is there a disambiguation page linking the two numbers somewhere? How did 2014-MU69 (unofficially &lt;em&gt;Ultima Thule&lt;/em&gt;) get promoted from SB to MB?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OuX4K.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OuX4K.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="23748" LastEditDate="2019-02-11T17:37:34.790" LastActivityDate="2019-02-11T17:37:34.790" Title="Does the KBO 2014-MU69 have two numbers and entries in asteroid databases? How did it get promoted to Major Body designation?" Tags="&lt;asteroids&gt;&lt;terminology&gt;&lt;nasa&gt;&lt;kbo&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29543" PostTypeId="1" CreationDate="2019-02-11T01:17:18.090" Score="0" ViewCount="78" Body="&lt;p&gt;When the Earth passes by Mars they exert a slight gravitational pull on each other, but the force is so small that their orbits are not significantly disturbed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we imagined that the orbits of the Earth and Mars were coplanar and much closer to one another, then there would presumably be some critical distance at which they would move together and collide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is that critical distance? In other words, if we shrunk the radius of the orbit of Mars, how much would it have to shrink before a collision would occur by the natural force of gravity between the two bodies?&lt;/p&gt;&#xA;" OwnerUserId="2800" LastActivityDate="2019-02-14T09:48:50.980" Title="How close would the orbit of Mars have to be to that of the Earth to induce a collision?" Tags="&lt;orbital-mechanics&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="29544" PostTypeId="1" AcceptedAnswerId="29546" CreationDate="2019-02-11T02:24:32.790" Score="9" ViewCount="1681" Body="&lt;p&gt;I was using Stellarium to watch Saturn from its moon Pan and I saw the rings were very close to this moon. Now Saturn's rings extend for a large distance so several moons see them from close. I was wondering, is it know which Saturn moon passes closest to Saturn's rings and at what distance?&lt;/p&gt;&#xA;" OwnerUserId="11562" LastEditorUserId="7472" LastEditDate="2019-02-11T07:49:16.503" LastActivityDate="2019-02-11T07:49:16.503" Title="Which Saturn satellite passes closest to Saturn's rings and at what distance?" Tags="&lt;natural-satellites&gt;&lt;saturn&gt;&lt;satellite&gt;&lt;planetary-ring&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29545" PostTypeId="2" ParentId="29542" CreationDate="2019-02-11T03:06:22.327" Score="3" Body="&lt;p&gt;The &lt;code&gt;3713011&lt;/code&gt; object (that is the internal SPICE/SPK id) is a special trajectory computed by the New Horizons team at SwRI which included data which is not in the public domain. The other one for &lt;code&gt;(486958) 2014 MU69&lt;/code&gt; (its official asteroid number and temporary discovery designation) is for the publicly available data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generating an ephemeris in HORIZONS for the former produces:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; Revised: Dec 29, 2018            486958 (2014 MU69)                    2486958&#xA; *SPECIAL*&#xA;&#xA; This pre-computed trajectory is consistent with the New Horizons spacecraft&#xA; Kuiper-Belt extended mission, with the planned 3500 km flyby of Ultima Thule&#xA; on 2019-Jan-1 @ 05:33 UTC.&#xA;&#xA; Trajectories were provided by the New Horizons mission planning team at SWRI &#xA; and are fits to internal flight-project data that has not been made available &#xA; outside the flight project.&#xA;&#xA; To use this official mission trajectory to ...&#xA;&#xA;  A) Generate ephemerides with respect TO the mission target:&#xA;      Specify the coordinate center in Horizons as &quot;@2486958&quot; &#xA;&#xA;  B) Generate ephemerides OF the mission target&#xA;      Specify the target in Horizons as &quot;2486958&quot;. &#xA;&#xA; To instead use the latest JPL ground-based orbit solution (based on the &#xA; publicly available data) and numerically integrate over an arbitrary &#xA; time-span with statistical uncertainties of that public dataset, do a &#xA; look-up in Horizons with syntax &quot;2014 MU69;&quot; or &quot;486958;&quot; or &quot;DES= 2486958;&quot; &#xA; or &quot;DES= 2014 MU69&quot; or &quot;2486958;&quot; (i.e., with a semi-colon). &#xA;&#xA; Note: the object previously had a provisional SPK ID of 3713011.  It was &#xA; then assigned an IAU number of 486958 such that the primary SPK ID became&#xA; the present 2486958 (2000000 + IAU number).&#xA;&#xA; Trajectory name                            Start (TDB)         Stop (TDB)&#xA; ---------------------------------------  -----------------  -----------------&#xA; 20180601_20250101_od141_tcm30_burn_V0.1  1994-Jan-08 00:00  2033-Dec-21 00:00 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(486958) is 2014 MU69's official number and is used at e.g the &lt;a href=&quot;https://minorplanetcenter.net/db_search/show_object?utf8=%E2%9C%93&amp;amp;object_id=486958&quot; rel=&quot;nofollow noreferrer&quot;&gt;MPC&lt;/a&gt;. 3713011 and 2486958 are the JPL SPK id's for the kernels with and without the extra trajectory information respectively; they are not used outside JPL's software such as HORIZONS, Small Body Browser or the SPICE toolkit. &lt;a href=&quot;ftp://naif.jpl.nasa.gov/pub/naif/toolkit_docs/Tutorials/pdf/individual_docs/18_spk.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tutorial 18&lt;/a&gt; of the SPICE Tutorial covers SPK files if you want more details. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The JPL software system handles multiple versions of the trajectory and/or orbital elements by creating multiple &quot;bodies&quot; for the same object. This is most commonly seen with comets which have 1 or more orbital elements for each perihelion passage e.g. this screenshot for comet 46P (Wirtanen):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/p2y7T.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p2y7T.png&quot; alt=&quot;Screenshot of JPL HORIZONS showing multiple element sets/&amp;quot;bodies&amp;quot;&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="23748" LastEditorUserId="23748" LastEditDate="2019-02-11T17:26:33.487" LastActivityDate="2019-02-11T17:26:33.487" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29546" PostTypeId="2" ParentId="29544" CreationDate="2019-02-11T03:37:29.137" Score="23" Body="&lt;p&gt;Pan, Daphnis, and various other moonlets, I would argue, are inside the rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you explicitly discount the &lt;a href=&quot;https://en.wikipedia.org/wiki/Rings_of_Saturn#Encke_Gap&quot; rel=&quot;noreferrer&quot;&gt;Encke gap&lt;/a&gt; (which Pan orbits in) and the Keeler gap (which Daphnis orbits in) as being part of the ring system, Daphnis would be your answer, as it is a ~8 km object in a 42 km gap. (for comparison, Pan is a ~35 km object in a 325 km gap)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really, your answer depends on what you want to consider a moon. There are many objects small enough to be classed as moonlets (several hundred metres across) embedded in the rings, and presumably countless more of smaller size.  &lt;/p&gt;&#xA;" OwnerUserId="17655" LastActivityDate="2019-02-11T03:37:29.137" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29547" PostTypeId="2" ParentId="29532" CreationDate="2019-02-11T04:21:47.317" Score="-1" Body="&lt;p&gt;&lt;strong&gt;The moon will look as it normally does with regards to phases, but will be in a fixed position in the sky&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Practically speaking, that's all that will change. (At least in regards to the moon's appearance that is, having a 29 day-long day would wreak havoc on the earth's geology.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon will only be visible from one half of earth's hemisphere, as you speculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also worth noting, in response to Chappo's answer, that not being able to see a new moon at night is something that happens currently, not just in this scenario. You can't see it at night because it's in the same rough direction as the sun. Similarly, you can see a crescent moon at night, but only if you're near the edge of the moonward hemisphere and the earth is blocking out the sun.    &lt;/p&gt;&#xA;" OwnerUserId="17655" LastActivityDate="2019-02-11T04:21:47.317" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29549" PostTypeId="2" ParentId="29487" CreationDate="2019-02-11T15:15:49.727" Score="3" Body="&lt;p&gt;The publication describing the original detection of the first known FRB (Lorimer, D. R., Bailes, M., McLaughlin, M. A., Narkevic, D. J. &amp;amp; Crawford, F.: A Bright Millisecond Radio Burst of Extragalactic Origin. &lt;a href=&quot;https://arxiv.org/abs/0709.4301&quot; rel=&quot;nofollow noreferrer&quot;&gt;arXiv:0709.4301&lt;/a&gt;) has a plot of the measurement that makes the effects of dispersion on this particular FRB nicely visible. Take a look at Fig. 2 in the paper. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual signal is less than 10 ms, while dispersion delays the signal by around 200 ms over the 200 MHz frequency range between 1.3 and 1.5 GHz (note that this relationship is nonlinear).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your idea about algorithmically removing the effects of dispersion on the signal is regularly done in practice, search for &quot;dedispersion&quot;. At our (hobbyist) observatory, we are using D. Lorimer's own &lt;a href=&quot;http://sigproc.sourceforge.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;sigproc&lt;/a&gt; package to do this, and it seems to be in widespread use amongst professional observers as well. The basic idea is to simulate a classical filterbank arrangement and shift each filter channel according to the DM.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://arxiv.org/abs/0709.4301&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lorimer et al.&lt;/a&gt; (cited above):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Zz4IS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zz4IS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Figure 2: Frequency evolution and integrated pulse shape of the radio burst. The survey data, collected on 2001 August 24, are shown here as a two-dimensional ‘waterfall plot’ of intensity as a function of radio frequency versus time. The dispersion is clearly seen as a quadratic sweep across the frequency band, with broadening towards lower frequencies. From a measurement of the pulse delay across the receiver band using standard pulsar timing techniques, we determine the DM to be 375±1 cm−3 pc. The two white lines separated by 15 ms that bound the pulse show the expected behavior for the cold-plasma dispersion law assuming a DM of 375 cm−3 pc. The horizontal line at ∼ 1.34 GHz is an artifact in the data caused by a malfunctioning frequency channel. This plot is for one of the offset beams in which the digitizers were not saturated. By splitting the data into four frequency sub-bands we have measured both the half-power pulse width and flux density spectrum over the observing bandwidth. Accounting for pulse broadening due to known instrumental effects, we determine a frequency scaling relationship for the observed width W = 4.6 ms (f/1.4 GHz)−4.8±0.4 , where f is the observing frequency.&#xA;  A power-law fit to the mean flux densities obtained in each sub-band yields a spectral index of −4 ± 1. Inset: the total-power signal after a dispersive delay correction assuming a DM of 375 cm−3 pc and a reference frequency of 1.5165 GHz. The time axis on the inner figure also spans the range 0–500 ms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20361" LastEditorUserId="7982" LastEditDate="2019-02-11T15:32:25.793" LastActivityDate="2019-02-11T15:32:25.793" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29551" PostTypeId="1" AcceptedAnswerId="29557" CreationDate="2019-02-11T21:06:20.120" Score="4" ViewCount="101" Body="&lt;p&gt;I live in northern hemisphere around 44th latitude. Starting from Winter Solstice, December 21, 2018 the duration of daylight is naturally increasing. What I observe, however, is that the duration increase is happening more in the evening rather than in the morning. In other words as days pass, the sunset seems to move forward much faster to a later hour than the first morning light moving backward to an earlier hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is what I am observing a true phenomenon or is it merely a wrong observation? If so why is this happening in an uneven fashion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0Xnwy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0Xnwy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the sunrise/sunset times for my area. It in fact confirms the situation. In the first half of January sunrise moved back only 3 minutes while sunset moved forward 15 minutes with respect to civil time.&lt;/p&gt;&#xA;" OwnerUserId="25891" LastEditorUserId="25891" LastEditDate="2019-02-12T11:46:12.683" LastActivityDate="2019-02-12T11:46:12.683" Title="Daylight Time Expansion" Tags="&lt;sunrise&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29552" PostTypeId="2" ParentId="29305" CreationDate="2019-02-11T21:50:20.637" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the difference between the way light travels in a gravitational field and the way it travels through an expanding universe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the former situation spacetime is curved across space, in the latter situation spacetime is curved over time. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A photon is travelling through space. As it travels, the space through which it travels is expanding. Does this describe: &#xA;  A - a photon travelling from a distant galaxy to earth&lt;br&gt;&#xA;  B - a photon travelling from a point of higher to a point of lower gravity&#xA;  C - either&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It describes A. It can also be likened to a photon travelling from a point of low gravitational potential to a point of high gravitational potential. Hence we talk of gravitational redshift, and cosmological redshift. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If like me you think the answer is C, can you explain why light should travel differently through a gravitational field to how it travels through an expanding universe? In other words, why in one case does the speed of the photon vary, while in the other case it doesn't? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who says it doesn't? We can find Einstein saying the speed of light varies with gravitational potential year after year:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol4-trans/144&quot; rel=&quot;nofollow noreferrer&quot;&gt;1912&lt;/a&gt;: “On the other hand I am of the view that the principle of the constancy of the velocity of light can be maintained only insofar as one restricts oneself to spatio-temporal regions of constant gravitational potential”.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol4-trans/325?highlightText=%22incompatible%&quot; rel=&quot;nofollow noreferrer&quot;&gt;1913&lt;/a&gt;: “I arrived at the result that the velocity of light is not to be regarded as independent of the gravitational potential. Thus the principle of the constancy of the velocity of light is incompatible with the equivalence hypothesis”.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol4-trans/295?highlightText=%22postulate%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;1914&lt;/a&gt;: “In the case where we drop the postulate of the constancy of the velocity of light, there exists, a priori, no privileged coordinate systems”.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol4-trans/271?highlightText=%22abandoned%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;1915&lt;/a&gt;: “the writer of these lines is of the opinion that the theory of relativity is still in need of generalization, in the sense that the principle of the constancy of the velocity of light is to be abandoned”.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol6-trans/340?highlightText=%22cannot%20claim%20any%20unlimited%20validity%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;1916&lt;/a&gt;: “In the second place our result shows that, according to the general theory of relativity, the law of the constancy of the velocity of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity and to which we have already frequently referred, cannot claim any unlimited validity”.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol7-trans/156?highlightText=%22spatially%20variable%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;1920&lt;/a&gt;: “Second, this consequence shows that the law of the constancy of the speed of light no longer holds, according to the general theory of relativity, in spaces that have gravitational fields. As a simple geometric consideration shows, the curvature of light rays occurs only in spaces where the speed of light is spatially variable”.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also find Irwin Shapiro saying the same thing in 1964. See line 7 of his paper in the fair-use excerpt &lt;a href=&quot;https://i.stack.imgur.com/IlSrh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. It's even on &lt;a href=&quot;https://en.wikipedia.org/w/index.php?title=Shapiro_time_delay&amp;amp;oldid=875091556&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;: &lt;em&gt;“the speed of a light wave depends on the strength of the gravitational potential along its path”&lt;/em&gt;. For something more recent see &lt;a href=&quot;http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/speed_of_light.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Is The Speed of Light Everywhere the Same?&lt;/a&gt; by  PhysicsFAQ editor Don Koks. He says &lt;em&gt;“Einstein talked about the speed of light changing in his new theory”&lt;/em&gt; and &lt;em&gt;&quot;a &quot;global&quot; observer can say that ceiling light does travel faster than floor light&quot;&lt;/em&gt;. Also see Ned Wright’s &lt;a href=&quot;http://www.astro.ucla.edu/~wright/deflection-delay.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Deflection and Delay of Light&lt;/a&gt;, where you can read this: &lt;em&gt;“In a very real sense, the delay experienced by light passing a massive object is responsible for the deflection of the light&quot;&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, some people will claim that c=1 and the speed of light is truly constant, but they're contradicting Einstein and not understanding the tautology. John Moffat and João Magueijo explained that in their &lt;a href=&quot;https://arxiv.org/abs/0705.4507&quot; rel=&quot;nofollow noreferrer&quot;&gt;Comments on “Note on varying speed of light theories”&lt;/a&gt;. Contemporary physicists use the local motion of light to define the second and the metre, then use them to measure the local speed of light. Hence they'll say the local speed of light is 299,792,458 m/s, at both the floor and the ceiling. Moffat and Magueijo (and others) were the authors of variable speed of light theories as an alternative to inflation. See &lt;a href=&quot;https://en.wikipedia.org/w/index.php?title=Variable_speed_of_light&amp;amp;oldid=836079117#Einstein&amp;#39;s_updated_proposals_(1905%E2%80%931915)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;. They said the speed of light was much faster in the early universe. Sadly I think they got that back to front, but never mind.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Alternatively can you explain why you don't think it's C?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It isn't C because space isn't expanding in a gravitational field. Instead it's &lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol7-trans/192?highlightText=%22neither%20homogeneous%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;neither homogeneous nor isotropic&quot;&lt;/a&gt;. That's what Einstein said in 1920. Note though that when space is expanding, it's inhomogeneous over time. Note this paper: &lt;a href=&quot;https://iopscience.iop.org/article/10.1088/0256-307X/25/5/014/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;Inhomogeneous Vacuum: An Alternative Interpretation of Curved Spacetime&lt;/a&gt;. Einstein referred to inhomogeneous vacuum whilst most modern authors refer to curved spacetime, but they're talking about the same thing.  &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2019-02-11T21:50:20.637" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="29554" PostTypeId="1" CreationDate="2019-02-11T23:23:40.480" Score="6" ViewCount="61" Body="&lt;p&gt;In January, the Chinese probe lander Chang'e-4 was &lt;a href=&quot;https://phys.org/news/2019-01-chinese-rover-lunar-nights-colder.html&quot; rel=&quot;noreferrer&quot;&gt;was announced to have found temperatures dipping lower on the far side than expected (&quot;Chinese rover finds lunar nights 'colder than expected'&quot; by R. Jackson, Jan. 31, 2019. AFP via Phys.org)&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Temperatures on the moon's surface plummeted to minus 190 degrees celsius (-310 degrees Fahrenheit) during the probe's first lunar night, which &quot;was colder than scientists expected,&quot; the China National Space Administration (CNSA) said.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This was said to be surprising, because near-side temperature measurements by U.S. missions showed less substantial dips. It was suggested that this was probably due to a characteristic difference between regolith on the two sides:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They were lower than those recorded by previous US missions to the near side of the moon, Zhang He, executive director of the Chang'e-4 mission, told Xinhua news agency.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;That's probably due to the difference in lunar soil composition between the two sides of the moon,&quot; he said.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But the U.S. surface missions were not the last thermal readings taken of the Moon, and the Chinese mission isn't the first mission to take thermal readings of the far side. Unless the thermal variability measured by Chang'e-4 is particular to a very localized area of regolith on which it took those lunar dawn readings (in which case it can't be said to be representative of a generalizable difference from near-side regolith), how would that not be known already, from the &lt;a href=&quot;https://www.diviner.ucla.edu/&quot; rel=&quot;noreferrer&quot;&gt;Infrared measurements&lt;/a&gt; that have been taken by the Lunar Reconnaissance Orbiter &lt;a href=&quot;https://www.nasa.gov/topics/solarsystem/features/divinerb20090917.html&quot; rel=&quot;noreferrer&quot;&gt;for the past decade&lt;/a&gt;, &#xA; with &lt;a href=&quot;https://arstechnica.com/science/2019/01/the-earth-has-been-experiencing-more-frequent-asteroid-strikes/&quot; rel=&quot;noreferrer&quot;&gt;enough resolution and sensitivity to distinguish the thermal variation of craters from surrounding regolith&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="22118" LastActivityDate="2021-03-09T15:43:19.407" Title="Why would the Chang'e-4 lander find lunar far side temp. &quot;colder than scientists expected&quot;, when the LRO has already been taking thermal readings?" Tags="&lt;the-moon&gt;&lt;temperature&gt;&lt;space-probe&gt;&lt;infrared&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29555" PostTypeId="1" AcceptedAnswerId="29559" CreationDate="2019-02-12T00:53:38.570" Score="11" ViewCount="1101" Body="&lt;p&gt;In one of the episodes from Carl Sagan's show Cosmos he explains that Beta Andromedae is the second brightest star in the constellation Andromeda, and is 75 light years away. The link to the video is &lt;a href=&quot;https://youtu.be/kr7Cd23Vo0w?list=PLKSi40WEKtMxykDBP8_vrC6bKXotys8KJ&amp;amp;t=439&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've &lt;a href=&quot;https://en.wikipedia.org/wiki/Beta_Andromedae&quot; rel=&quot;noreferrer&quot;&gt;looked up this star&lt;/a&gt; on Wikipedia and it says it's the brightest star in the constellation, not the second brightest, and instead of 75 light years away it says it's about 197 light years away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that this show Cosmos is quite old. It aired in 1980. I understand our measurements are more accurate nowadays, but I was just wondering if anyone knew why the discrepancy with the brightest and second brightest star claims, and also that the Wikipedia article says it's 197 light years away, a difference of about 260% from the 75 light years mentioned by Sagan. That's quite a significant difference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;His information of the distance from us to the center of our galaxy is quite close: he says 30 thousand light years; Wikipedia says 27 thousand light years. As to our distance to Andromeda, he says: 2 million light years; Wikipedia says 2.5 million light years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have I completely got the wrong star that I'm looking up?&lt;/p&gt;&#xA;" OwnerUserId="25893" LastEditorUserId="1174" LastEditDate="2019-02-12T02:00:11.357" LastActivityDate="2019-02-13T06:12:38.750" Title="Beta Andromedae (Mirach) and distances mentioned in original Cosmos series from 1980" Tags="&lt;star&gt;&lt;distances&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29556" PostTypeId="2" ParentId="29555" CreationDate="2019-02-12T01:45:24.773" Score="3" Body="&lt;p&gt;I think I found the source of the 75 light year distance for Beta Andromeda as being from the 1978 edition of &lt;a href=&quot;https://en.wikipedia.org/wiki/Robert_Burnham_Jr.#Burnham&amp;#39;s_Celestial_Handbook&quot; rel=&quot;nofollow noreferrer&quot;&gt;Burnham's Celestial Handbook&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this excerpt out of &lt;a href=&quot;https://books.google.com/books?id=kxJwq3mtn8kC&amp;amp;pg=PA111&amp;amp;dq=beta+andromeda+distance&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwi10JDYhLXgAhWvuFkKHc45ApEQ6AEIKjAA#v=onepage&amp;amp;q=beta%20andromeda%20distance&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google Books&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Name - Mirach. Mag 2.03 ... The distance is about 75 light years according to parallaxes obtained at Mt. Wilson, Allegheny, and McCormick; the resulting luminosity is about 75 times that of the Sun, and the absolute magnitude about +0.2. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just reading the excerpt alone, I'm unclear if this is a misinterpretation of the results from the observatories or that at the time the understanding the characteristics of star type would result in the discrepancy.&lt;/p&gt;&#xA;" OwnerUserId="1174" LastActivityDate="2019-02-12T01:45:24.773" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29557" PostTypeId="2" ParentId="29551" CreationDate="2019-02-12T01:50:25.353" Score="8" Body="&lt;p&gt;This phenomenon is the equation of time, the annually varying difference between apparent solar time (what a sundial indicates) and mean solar time (related to civil time).&#xA;It has two components:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The eccentricity of the Earth's orbit around the Sun causes a cycle with 7.5-minute amplitude and 12-month period, with zeroes at perihelion (Earth moving fastest and closest to the Sun) and aphelion (Earth moving slowest and farthest from the Sun).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The oblique angle between the Earth's equator (rotation plane) and ecliptic (orbital plane) causes a cycle with 10-minute amplitude and 6-month period, with zeroes at equinoxes (Sun crossing Earth's equator) and solstices (Sun farthest north or south of the equator). &lt;a href=&quot;https://astronomy.stackexchange.com/a/25570/616&quot;&gt;This answer&lt;/a&gt; explains how that works.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These combine so that apparent solar time is 17 minutes ahead of mean solar time in early November and 14 minutes behind in mid February.&#xA;Between those times of year, the apparent Sun falls several seconds behind the mean Sun each day, affecting both sunset and sunrise times as you observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further explanation, see the &lt;a href=&quot;https://aa.usno.navy.mil/faq/docs/eqtime.php&quot; rel=&quot;noreferrer&quot;&gt;US Naval Observatory&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/Equation_of_time&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2019-02-12T01:50:25.353" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29558" PostTypeId="2" ParentId="29555" CreationDate="2019-02-12T03:11:46.940" Score="6" Body="&lt;p&gt;Distances of stars measured by parallax in general (and red giants in particular) can be notoriously inexact, and the distance values can vary markedly between different measurements. For example, &lt;a href=&quot;https://en.wikipedia.org/wiki/Betelgeuse#Distance_measurements&quot; rel=&quot;noreferrer&quot;&gt;Betelgeuse has been measured&lt;/a&gt; as being 520±73 ly by the Hipparcos mission, and  643±146 ly by the Very Large Array. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I often find distances to stars listed on wikipedia are different from what I remember them to be a decade ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mirach is a red giant star. It is entirely possible that the measurements may have changed considerably in that time.&lt;/p&gt;&#xA;" OwnerUserId="17655" LastActivityDate="2019-02-12T03:11:46.940" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29559" PostTypeId="2" ParentId="29555" CreationDate="2019-02-12T05:37:06.217" Score="17" Body="&lt;p&gt;Looking at the &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=Beta%20Andromedae&amp;amp;NbIdent=1&amp;amp;Radius=2&amp;amp;Radius.unit=arcmin&amp;amp;submit=submit%20id&quot; rel=&quot;nofollow noreferrer&quot;&gt;SIMBAD&lt;/a&gt; data page for Beta Andromeda shows the source of both the parallax (distance) and the magnitudes. In this case, as it will be for many bright stars, the source of the parallax is the reprocessed data from the Hipparcos satellite, described in &lt;a href=&quot;https://www.aanda.org/articles/aa/abs/2007/41/aa8357-07/aa8357-07.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;. Prior to the launch of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hipparcos&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hipparcos&lt;/a&gt; satellite by ESA in 1989, parallaxes were very difficult to obtain and only low precision values were available for the closest stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parallax given in SIMBAD for Beta Andromeda is &lt;span class=&quot;math-container&quot;&gt;$16.52\pm0.56$&lt;/span&gt; milliarcseconds which translates to a distance of &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{0.01652&quot;} = 60.5\pm2.1$&lt;/span&gt; parsecs or &lt;span class=&quot;math-container&quot;&gt;$60.5\times3.26=197$&lt;/span&gt; light years. This small parallax would have been extremely challenging or impossible to measure accurately with the pre-CCD technology before 1980. Errors of several hundred percent were not uncommon. The prior parallax measurement which probably was an improvement on what was available in 1980 at the time of &lt;em&gt;Cosmos&lt;/em&gt; is from &lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-ref?bibcode=1995GCTP..C......0V&quot; rel=&quot;nofollow noreferrer&quot;&gt;van Altena et al. 1995&lt;/a&gt;. This lists a parallax of &lt;span class=&quot;math-container&quot;&gt;$47.7\pm7.9$&lt;/span&gt; millarcsec, a nearly 5x larger error and which gives a distance of 68 light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly we can see that the &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt;-band magnitude comes from &lt;a href=&quot;https://ui.adsabs.harvard.edu/?#abs/2002yCat.2237....0D&quot; rel=&quot;nofollow noreferrer&quot;&gt;this collection&lt;/a&gt; and the difference between Beta Andromeda (&lt;span class=&quot;math-container&quot;&gt;$V=2.05$&lt;/span&gt;) and Alpha Andromeda (&lt;span class=&quot;math-container&quot;&gt;$V=2.06$&lt;/span&gt;), the in-theory brightest star in the constellation is only 0.01 magnitude. Measuring a star that bright to that precision was, and in fact still is, quite difficult as most detectors will saturate. So it's not particularly surprising given how close Alpha and Beta Andromeda are in brightness, that the early measurements got them reversed when Beta is in fact (very slightly) brighter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This high brightness probably means that we will not see a more accurate from the &lt;a href=&quot;http://sci.esa.int/gaia/28820-summary/&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;_Gaia_&quot;&gt;&lt;em&gt;Gaia&lt;/em&gt;&lt;/a&gt; satellite, the successor to Hipparcos. The Gaia DR2 data release lists a brightness limit of &lt;span class=&quot;math-container&quot;&gt;$G\sim3$&lt;/span&gt; which may be improved slightly in later data releases with more sophisticated data processing and treatment of saturated stars.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastEditorUserId="23748" LastEditDate="2019-02-13T06:12:38.750" LastActivityDate="2019-02-13T06:12:38.750" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29561" PostTypeId="2" ParentId="29517" CreationDate="2019-02-12T07:56:43.810" Score="4" Body="&lt;p&gt;That depends on the accuracy you want to work with. To zeroth order, as outlined in Murray &amp;amp; Dermott, &quot;Solar System dynamics&quot;, Chapt 3., you can do the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The zero-velocity contours that are plotted in your image will not be coinciding with particle orbits to infinite precision, but they're a good zeroth order approximation for objects with low eccentricity w.r.t the star (&lt;span class=&quot;math-container&quot;&gt;$v_{r}/v_{\theta} \ll 1$&lt;/span&gt;)&lt;/li&gt;&#xA;&lt;li&gt;A particle on those orbits is on a regular Keplerian orbit of radius &lt;span class=&quot;math-container&quot;&gt;$r_{\rm H}$&lt;/span&gt;, out of reach of the gravitational influence of the planet at &lt;span class=&quot;math-container&quot;&gt;$r_{\rm P}$&lt;/span&gt;. Thus, to get the largest &quot;chunk&quot; of traveltime of one part of the horseshoe orbit, either inner or outer part, you can work out with relative velocities and the assumption of keplerian velocities.&lt;/li&gt;&#xA;&lt;li&gt;Caution is required as to which orbital time you're interested in: If &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is Earth's distance to the sun, and the horseshoe is &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; away from a perfectly circular orbit, thus alternating between distances &lt;span class=&quot;math-container&quot;&gt;$a\pm d$&lt;/span&gt;, then the orbital time in the rest-frame of the sun will be &lt;span class=&quot;math-container&quot;&gt;$v_{-}=\sqrt{\frac{GM}{a-d}}$&lt;/span&gt;, for a large number of orbits until close encounter, and after that &lt;span class=&quot;math-container&quot;&gt;$v_{+}=\sqrt{\frac{GM}{a+d}}$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Therefore, the simplified case that the satellite is on its own Keplerian orbit is true for most of the time. Armed with this knowledge, we can approximate the Planetocentric recurrence time as &lt;span class=&quot;math-container&quot;&gt;$t_{\rm rec}=\frac{2\pi a}{v_{K} - v_{-}}$&lt;/span&gt; as a simple catch-up time between objects on different orbits.&lt;/li&gt;&#xA;&lt;li&gt;The relative velocity &lt;span class=&quot;math-container&quot;&gt;$v_{\rm rel} = v_{K} - v_{-} = \sqrt{\frac{GM}{a}}-\sqrt{\frac{GM}{a-d}}$&lt;/span&gt; can be expanded in the limit of &lt;span class=&quot;math-container&quot;&gt;$d/a\ll1$&lt;/span&gt; into &lt;span class=&quot;math-container&quot;&gt;$v_{\rm rel} \approx \frac{1}{2} \sqrt\frac{GM}{a} \frac{d}{a}$&lt;/span&gt; and thus I derive from this &lt;span class=&quot;math-container&quot;&gt;$t_{rec}\approx \frac{4\pi}{\sqrt{GM}} \frac{a^{5/2}}{d}$&lt;/span&gt;. As expected, the cycle time diverges for &lt;span class=&quot;math-container&quot;&gt;$d=0$&lt;/span&gt;, as in the co-orbiting case with Earth this time must be infinite. An upper limit on &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; cannot be derived from this, one would need to turn for the full solution for that.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Just out of curiosity, I've plugged in a few values into this formula and written something quick in python:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;&#xA;#Basic physics quantities&#xA;G      = 6.678e-8 #cgs units&#xA;pi     = 3.141592&#xA;navo   = 6e23 # particles per mole&#xA;sigma  = 5.67e-5   #erg cm-2 s-1 K-4&#xA;kb     = 1.38e-16  #erg/K&#xA;km     = 1e5 #kilometers in cm&#xA;mearth = 5.98e27  #g&#xA;msun   = 2.0e33   #g&#xA;au     = 1.49e13  #cm&#xA;yr   = 365*24*3600&#xA;rearth   = 6370e5&#xA;rjupiter = 74000*km&#xA;&#xA;#&#xA;# Returns the approximate horseshoe-cycle time in the reduced 3body problem&#xA;# Masses of bodies: m0&amp;gt;&amp;gt;m1&amp;gt;&amp;gt;m2&#xA;# Semimajor axis distance is from m0 to m1, radial distance is a(m0-&amp;gt;m1)-a(m0-&amp;gt;m2)&#xA;#&#xA;def hs_cycle(mcentral, semimajor_axis, radial_distance):&#xA;    return 4*pi/np.sqrt(G*mcentral)*semimajor_axis**(5./2.)/radial_distance/yr&#xA;&#xA;#&#xA;# https://en.wikipedia.org/wiki/(419624)_2010_SO16 around the Sun&#xA;#&#xA;# Quoted cycle time ~350 years, with d=0.004 AU&#xA;#&#xA;print(&quot;Predicted 2010_SO16 cycletime [years] = &quot;, hs_cycle(msun, 1*au,0.004*au), &quot; predicted = 350 yrs&quot;)&#xA;&#xA;#&#xA;# Janus/Epimetheus around Saturn&#xA;#&#xA;# a = 151410 km, d = 25 km, as stated in https://en.wikipedia.org/wiki/Epimetheus_(moon)&#xA;# Quoted cycle time = 8 years (from comments)&#xA;#&#xA;print(&quot;Predicted Janus/Epimetheus cycletime [years] = &quot;, hs_cycle(95*mearth, 151410*km,50*km), &quot; predicted = 4 yrs&quot;)&#xA;&#xA;&#xA;#&#xA;# 3753 Cruithne&#xA;#&#xA;# a = 1 AU and semimajor axis difference from https://en.wikipedia.org/wiki/3753_Cruithne&#xA;# Quoted cycle time = 770 years&#xA;#&#xA;print(&quot;Predicted 3753 Cruithne cycletime [years] = &quot;,hs_cycle(msun, 1*au, (1.0-0.99774)*au), &quot; predicted = 770 yrs&quot;)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and the results I get are&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Predicted 2010_SO16 cycletime [years] =  495.7747141830971  predicted = 350 yrs&#xA;Predicted Janus/Epimetheus cycletime [years] =  11.542076781209305  predicted = 8 yrs&#xA;Predicted 3753 Cruithne cycletime [years] =  877.4773702355546  predicted = 770 yrs&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So the formula may be off by up to a factor of ~2. This is for sure simply because reality is more complex than a simple circular-orbit approximation, but also due to the quality of the values used.&#xA;Wikipedia isn't well known for researching particular values well. I took those that I found there. For SO16 that was particularly confusing to select, so I took those two that were mentioned in the same line of text, hoping they would come from the same source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone finding more consistent values, is free to comment.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditDate="2019-02-13T23:34:24.243" LastActivityDate="2019-02-13T23:34:24.243" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29562" PostTypeId="1" AcceptedAnswerId="29581" CreationDate="2019-02-12T10:02:20.610" Score="1" ViewCount="142" Body="&lt;p&gt;If you live on the equator and look up at the night's sky at, say, midnight, would you see a completely different sky on the 1st of January than you would on the 1st of July?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume you would as the night side would be facing away from the sun by 180 degrees and necessarily looking outward from the solar system 180 degrees from each of the two points at those times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that correct?&#xA;Why?&lt;/p&gt;&#xA;" OwnerUserId="25216" LastActivityDate="2019-02-13T19:06:18.807" Title="Are the stars seen at the equator different throughout the year?" Tags="&lt;amateur-observing&gt;&lt;equinox&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29563" PostTypeId="2" ParentId="29543" CreationDate="2019-02-12T13:30:40.463" Score="1" Body="&lt;p&gt;It's more likely that perturbations by another planet might induce a collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other systems with bodies in almost the same orbit, what happens is like a game of orbital catch-up: one overtakes the other, then they swap places, so in your (rather vague) scenario, that might well be the outcome instead of an outright collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the Saturn ring moonlets interchange positions, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solar system is stable over a timeline of millions of years (as far as we know), but there have been instabilities in the past, and due to &quot;butterfly effects&quot; instabilities could occur in the future.&#xA;(See also &lt;a href=&quot;https://plus.maths.org/content/rocks-suggest-theres-chaos-solar-system&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://plus.maths.org/content/rocks-suggest-theres-chaos-solar-system&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a play with a simulator such as orbitsimulator.com !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://orbitsimulator.com/gravitySimulatorCloud/simulations/1433364162018_Solar%20System.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://orbitsimulator.com/gravitySimulatorCloud/simulations/1433364162018_Solar%20System.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25889" LastEditorUserId="25889" LastEditDate="2019-02-14T09:48:50.980" LastActivityDate="2019-02-14T09:48:50.980" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29566" PostTypeId="1" CreationDate="2019-02-12T15:13:57.093" Score="6" ViewCount="100" Body="&lt;p&gt;Some neutrinos are fermions. There are quite of few of them out there. Could their degeneracy pressure give rise to an expansion force to explain Dark Energy?&lt;/p&gt;&#xA;" OwnerUserId="25899" LastEditorUserId="1891" LastEditDate="2019-02-12T15:36:02.250" LastActivityDate="2019-02-12T18:03:16.003" Title="Can the Dark Energy be Pauli pressure?" Tags="&lt;dark-energy&gt;&lt;degenerate-matter&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29568" PostTypeId="2" ParentId="29566" CreationDate="2019-02-12T17:57:37.017" Score="6" Body="&lt;p&gt;There are several reasons dark energy cannot be pressure due to the Pauli exclusion principle.  First of all, pressure does not cause expansion of the universe, because pressure is not a force-- pressure &lt;em&gt;gradients&lt;/em&gt; are a force (per unit volume), and the cosmological principle precludes them.  Indeed, the pressure that exists everywhere only appears as a force when it is put into gravity, and there it is an &lt;em&gt;attraction&lt;/em&gt; if it is positive, as it is for what you mean by Pauli pressure.  So what you'd need for dark energy is &lt;em&gt;negative&lt;/em&gt; pressure, appearing in the gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second reason is that not only is what you mean by Pauli pressure positive, not negative, it is not any different from normal gas pressure, in the sense that it appears as a consequence of kinetic energy density in exactly the same way as any gas pressure does.  This is widely misunderstood, many people think there is some kind of extra pressure that comes from the Pauli exclusion principle, but this is not the case if you simply track the kinetic energy involved.  If you are tracking the kinetic energy, then gas pressure is 2/3 the kinetic energy density for nonrelativistic particles, and 1/3 for relativistic, and that's all-- it's just the same for &quot;Pauli pressure,&quot; there's nothing at all different about it when you track kinetic energy density-- which is generally exactly what you want to do when you have an adiabatic expansion such as for the universe.  Hence, in order for us to understand the pressure from neutrinos, we only need to understand their kinetic energy density, which we think we do because it's pretty easy to track given some assumptions about how neutrinos couple to matter and what their mass is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be wondering, if there's nothing different or extra about &quot;Pauli pressure,&quot; then when can it be treated as an extra source of pressure?  Only when you are talking about the temperature, rather than the energy.   If for some reason it made sense to think of temperature as the fundamental constraint, rather than energy, then it would make sense to think of the Pauli exclusion principle as adding pressure.  But even then it's kind of a strange way to think about it, because all the Pauli principle does is constrain how the energy gets partitioned, which in turn affects the temperature.  The chain of causality in any gas is energy + partition -&gt; temperature  and energy -&gt; pressure independently of temperature, so it's kind of weird to do what is often done, and pretend that temperature + partition -&gt; pressure.  It's only that last way that makes degeneracy look like an added pressure, and it would only make sense in a system whose temperature is, for some reason, being controlled.  That is rarely the case in an expanding universe, where the temperature is just whatever it needs to be given the energy and the contraints on its partitioning.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2019-02-12T18:03:16.003" LastActivityDate="2019-02-12T18:03:16.003" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29569" PostTypeId="1" AcceptedAnswerId="29586" CreationDate="2019-02-12T19:56:01.223" Score="2" ViewCount="260" Body="&lt;p&gt;Earth's orbit is about 99% circular. How circular is Earth around the equator? I know it bulges around the equator and is a spheroid. I also know it is not smooth and has oceans and mountains and all that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a similar measure of eccentricity to describe how circular it is around the equator? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it stretched into an ellipse by the sun/moon, and if so by how much?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for comparison, does the Earth's equator deviate from a circle more than the Earths' orbit around the Sun deviates from a circle?&lt;/p&gt;&#xA;" OwnerUserId="25498" LastEditorUserId="25498" LastEditDate="2019-02-13T13:58:19.003" LastActivityDate="2019-02-13T23:31:57.490" Title="How close to circular is the Earth's equator" Tags="&lt;orbit&gt;&lt;earth&gt;&lt;eccentric-orbit&gt;&lt;eccentricity&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29570" PostTypeId="1" CreationDate="2019-02-12T21:03:53.513" Score="4" ViewCount="128" Body="&lt;p&gt;I am looking for examples where amateur astronomers in these days contributed sigificantly to important scientifically discoveries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware of one example: &lt;a href=&quot;https://www.nature.com/articles/d41586-018-02331-4?utm_source=twt_nnc&amp;amp;utm_medium=social&amp;amp;utm_campaign=naturenews&amp;amp;sf182688358=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Victor Buso&lt;/a&gt; from Argentina was lucky enough to observe a Supernova just when it formed, which has led to follow-up observations by large scientific observatories, and in the end resulted in a &lt;a href=&quot;https://www.nature.com/articles/nature25151&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nature article&lt;/a&gt; which has Buso as a co-author. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What might some other similar examples be?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; &quot;Astronomer Victor Buso poses in front of the telescope with which he witnessed the birth of the supernova 2016gkg. (Victor Buso)&quot; From &lt;a href=&quot;https://www.washingtonpost.com/news/speaking-of-science/wp/2018/02/21/a-self-taught-astronomer-spotted-something-no-scientist-had-ever-seen/?utm_term=.48aef0106509&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, borrowed from &lt;a href=&quot;https://astronomy.stackexchange.com/q/25229/7982&quot;&gt;Does this amateur observation hold a record even including robotic searches for supernovae?&lt;/a&gt;. Click/open for full size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H3fPU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H3fPUm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2662" LastEditorUserId="7982" LastEditDate="2019-02-13T06:38:12.533" LastActivityDate="2019-02-14T00:27:08.943" Title="Scientifically important discoveries with the help of amateurs" Tags="&lt;amateur-observing&gt;&lt;supernova&gt;&lt;history&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29572" PostTypeId="1" CreationDate="2019-02-13T01:53:23.633" Score="4" ViewCount="179" Body="&lt;p&gt;Can anyone please give me an explanation on how to calculate photon's path by using the null geodesic equation? N.B. I know all of the non-zero values of Christoffel symbols. &lt;/p&gt;&#xA;" OwnerUserId="25901" LastActivityDate="2019-04-18T14:02:35.477" Title="Necessary steps to calculate photon's path by using null geodesic equation" Tags="&lt;astrophysics&gt;&lt;cosmology&gt;&lt;fundamental-astronomy&gt;&lt;general-relativity&gt;&lt;observable-universe&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29573" PostTypeId="2" ParentId="29519" CreationDate="2019-02-13T05:30:53.563" Score="0" Body="&lt;p&gt;YES!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity will eventually pull them together. &#xA;The thing is that gravity exerts itself over very vast distances, although the force decreases exponentially over distance. BUT, two motion less massive objects will attract each other, without any other influence whatsoever.&lt;/p&gt;&#xA;" OwnerUserId="25606" LastActivityDate="2019-02-13T05:30:53.563" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29574" PostTypeId="1" CreationDate="2019-02-13T06:56:32.583" Score="2" ViewCount="32" Body="&lt;p&gt;As part of my worldbuilding for a book, I’d like to have a planet with extreme variance in its tides. I want lowest tide  reveal about 30 miles of floodplain, and occasional massive tidal waves. I understand that, in theory, this might be achieved by giving the planet several relatively large moons with different orbital periods, such that sometimes they are spaced out so they balance out the tides (a kind of medium tide), and sometimes they are all together on one side of the planet so that they create a highest / lowest tide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How stable can such a system be? (The star and other solar system features can be whatever to improve stability, if that is a thing)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any spitballed numbers for viable orbital periods and tidal patterns?&lt;/p&gt;&#xA;" OwnerUserId="25904" LastActivityDate="2019-02-13T06:56:32.583" Title="Tides with 3 Largish Moons" Tags="&lt;orbital-mechanics&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2019-02-13T21:23:20.910" ContentLicense="CC BY-SA 4.0" />
  <row Id="29575" PostTypeId="1" AcceptedAnswerId="29579" CreationDate="2019-02-13T08:26:12.837" Score="2" ViewCount="85" Body="&lt;p&gt;I am writing a program to render an asteroid as seen from various parts of the solar system.  My goal is to roughly reproduce the &lt;a href=&quot;http://astro.troja.mff.cuni.cz/projects/asteroids3D/php.php?script=db_sky_projection&amp;amp;model_id=1863&amp;amp;jd=2443568.0&quot; rel=&quot;nofollow noreferrer&quot;&gt;image of Cacus&lt;/a&gt; according to information on its DAMIT &lt;a href=&quot;http://astro.troja.mff.cuni.cz/projects/asteroids3D/web.php?page=db_asteroid_detail&amp;amp;asteroid_id=1046&quot; rel=&quot;nofollow noreferrer&quot;&gt;detail page&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my approach:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Create a solar system at JD0=2443568.0.  The axes of the visualization are arranged according to the ecliptic with positive X axis in the direction of the vernal equinox and other axes according to the right-hand rule.  Earth is positioned at t=JD0.  (&lt;a href=&quot;https://i.imgur.com/Wer2pb6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;image&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Place the asteroid at (0, 0, 0) and begin to rotate the object according to the matrix multiplication described by the &lt;a href=&quot;http://astro.troja.mff.cuni.cz/projects/asteroids3D/web.php?page=db_description&quot; rel=&quot;nofollow noreferrer&quot;&gt;Description page&lt;/a&gt; and &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.546.7648&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kaasalainen &amp;amp; Torppa 2001&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;φ0=0 and t=JD0, so there is no initial Z axis rotation&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Latitude β is -63, so rotate 90 - (-63) = 153 degrees on the Y axis (&lt;a href=&quot;https://i.imgur.com/2VCJA4u.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;image&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Longitude λ is 251, so rotate the object 251 degrees anticlockwise around the Z axis of the ecliptic (&lt;a href=&quot;https://i.imgur.com/fTwbKVk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;image&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Using ephemeris from the NASA/JPL Small Body Database, move Cacus to its position in orbit at JD0. (&lt;a href=&quot;https://i.imgur.com/ULb5NgH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;image&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Move the camera to Earth's position at JD0 and point it at the asteroid. (&lt;a href=&quot;https://i.imgur.com/3884O5X.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;image&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My final rendering does not match the &lt;a href=&quot;https://i.imgur.com/1ubeYRz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;projection from DAMIT&lt;/a&gt;.  I understand that there are factors like light-time correction that I'm not including, but it looks completely different.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is wrong with my approach to rotating the asteroid on its spin axis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RHCLw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RHCLw.png&quot; alt=&quot;My rendering of Cacus&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The work-in-progress visualization is online &lt;a href=&quot;http://www.ianww.com/asteroid-spin/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="2605" LastEditorUserId="2605" LastEditDate="2019-02-13T20:13:52.240" LastActivityDate="2019-02-13T20:13:52.240" Title="Rendering asteroid orientation as seen from Earth" Tags="&lt;asteroids&gt;&lt;rotation&gt;&lt;orbital-mechanics&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29576" PostTypeId="2" ParentId="29570" CreationDate="2019-02-13T11:29:03.737" Score="4" Body="&lt;p&gt;A quite recent example is &lt;a href=&quot;https://en.wikipedia.org/wiki/Hanny%27s_Voorwerp&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hanny's Voorwerp&lt;/a&gt;, not done though direct observations, but discovered in 2007 by Dutch schoolteacher Hanny van Arkel while she was participating as a volunteer in the &lt;a href=&quot;http://zoo1.galaxyzoo.org/Project.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;Galaxy Zoo project&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vNFtY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vNFtY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sup&gt;(&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Hannys_voorwerp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://commons.wikimedia.org/wiki/File:Hannys_voorwerp.jpg&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user1569" LastActivityDate="2019-02-13T11:29:03.737" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29577" PostTypeId="1" CreationDate="2019-02-13T12:47:15.177" Score="2" ViewCount="103" Body="&lt;p&gt;I was listening to a scientific podcast about CPT-symmetry and mirror-universes, and when speaking about matter-antimatter annihilation, they stated that several emissions would occur at this time, notably gamma rays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this raised that question to me: I know that FRB (Fast Radio Bursts) are theorized differently today (lifecycle of some stars, etc.) but, could they actually be the result of matter-antimatter annihilation occurring locally (i.e. a bunch of matter colliding another bunch of matter)? After all, matter cohabiting with antimatter has been one theory, but we failed to find evidence of it... unless it is FRB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this is a newbie question and I'm not convinced of that, but would rather know &lt;em&gt;why&lt;/em&gt; this cannot be the case (like, &quot;we have solid proof of what FRB are&quot;, or &quot;the matter-antimatter emissions would not be the same as the ones in FRBs&quot;, etc.). &lt;/p&gt;&#xA;" OwnerUserId="14962" LastEditorUserId="14962" LastEditDate="2019-02-13T12:52:44.160" LastActivityDate="2019-02-13T12:52:44.160" Title="Could FRB be the result of matter-antimatter annihilation?" Tags="&lt;gamma-ray-bursts&gt;&lt;antimatter&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29579" PostTypeId="2" ParentId="29575" CreationDate="2019-02-13T17:28:32.190" Score="1" Body="&lt;p&gt;From the paper by &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.546.7648&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;M. Kaasalainen, J. Torppa, and K. Muinonen&lt;/a&gt;, equation (1) is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \vec r_{ast} = R_z(\omega(t-t_0)+\phi_0) R_y(\hat\beta) R_z(\lambda) \vec r_{ecl} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;To reverse this equation you must reverse the order of the rotations &lt;em&gt;and&lt;/em&gt; the rotation angles:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \vec r_{ecl} = R_z(-\lambda)R_y(-\hat\beta)R_z(-\omega(t-t_0)-\phi_0) \vec r_{ast} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;From your description, you reversed the order of the rotations but not the rotation angles.&lt;/p&gt;&#xA;" OwnerUserId="22897" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-02-13T17:28:32.190" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29581" PostTypeId="2" ParentId="29562" CreationDate="2019-02-13T19:06:18.807" Score="3" Body="&lt;p&gt;Yes, the constellations that are visible at midnight on Jan 1 are different than those visible at midnight on July 1 (6 months later). The website &lt;a href=&quot;https://www.stelvision.com/en/sky-map/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stelvision&lt;/a&gt; has a star chart that you can customize for date, time, and location. This is due to the Earth's revolution around the Sun and the direction that an observer is facing at midnight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a given location, a given star or constellation rises, transits (when an object is due south), and sets 2 hours earlier each month. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In generally, the above is true for all latitudes. Of course, stars/constellations that are circumpolar never rise and set, and from the Earth's pole all objects are circumpolar. But even in this case, if you would pick a direction to look at midnight on Jan 1, those constellations would be behind you (180 degrees away) at midnight on July 1 (assuming you could see them in the daylight on one of those dates ;-).&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2019-02-13T19:06:18.807" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29582" PostTypeId="1" CreationDate="2019-02-13T19:49:18.843" Score="4" ViewCount="166" Body="&lt;p&gt;I'm starting to learn Kepler's &lt;strong&gt;&lt;em&gt;Astronomia Nova&lt;/em&gt;&lt;/strong&gt; using William Donahue's translation . In the translator's introduction, it is demonstrated how Kepler did his parallax computation. I failed to understand the following paragraph:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eHYa3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eHYa3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;specifically: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) nonagesimal's definition: there are 2 points of intersection between the ecliptic and the horizon: one in the East, the other in West. another ambiguity is the direction of the &lt;span class=&quot;math-container&quot;&gt;$90^\circ$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I can't see why the altitude of the nonagesimal is &quot;the same as the angle between the ecliptic and the horizon&quot;. It does make sense to me, but I'm missing a kind of &quot;proof&quot; for that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Actually I also failed to understand what follows in that paragraph - but maybe your explanation for the points above, will clarify that for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;your help is very much appreciated&lt;/p&gt;&#xA;" OwnerUserId="25913" LastActivityDate="2019-03-06T19:24:18.703" Title="nonagesimal in Kepler's parallax computation" Tags="&lt;orbit&gt;&lt;solar-system&gt;&lt;astrophysics&gt;&lt;kepler&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29583" PostTypeId="1" CreationDate="2019-02-13T21:38:35.360" Score="3" ViewCount="153" Body="&lt;p&gt;How common or rare is it to find stars outside of a galaxy? Can such a star still have planets that contain life? Recently, I think that I heard of a star with a planet that entered our Galaxy. Did this star come here from another galaxy or was it originally from the Milky Way?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastEditorUserId="2270" LastEditDate="2019-02-14T14:21:53.187" LastActivityDate="2020-10-04T10:52:48.967" Title="Stars located outside of a galaxy" Tags="&lt;star&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29584" PostTypeId="2" ParentId="29583" CreationDate="2019-02-13T21:57:26.823" Score="8" Body="&lt;p&gt;I don't think we have a good estimate of the number of stars not in galaxies -- they're actually pretty hard to find unless they're really bright, since even if you spot it, how do you know that it's extragalactic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But galaxy mergers (which are &lt;em&gt;very&lt;/em&gt; common) tend to throw out large numbers of stars from the merging galaxies, so it seems likely there are many stars in space between the galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no reason why such stars can't have planets, but it's likely that the planets formed along with the star inside a galaxy and got ejected from the galaxy along with the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars (and planets) form when Giant Molecular Clouds collapse under their own gravity, the collapse frequently being triggered by tidal forces from a passing star or a nearby supernova.  It seems unlikely that there are many GMCs in intergalactic space, and there are certainly fewer encounters which might trigger collapse. So it seems likely that relatively few stars (or planets) actually form in intergalactic space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Life does not seem to have evolved because of the existence of stars in the sky, so once a planet exists in a habitable zone, life should form regardless of where it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gaia mission is providing us with superb position and velocity data for billions of stars, and it has identified a handful (maybe more by now) of stars which seem clearly to have come from intergalactic space.  &lt;a href=&quot;https://www.sciencedaily.com/releases/2018/10/181002103000.htm&quot; rel=&quot;noreferrer&quot;&gt;Here's an article&lt;/a&gt; which covers the report last year.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2019-02-14T14:35:41.970" LastActivityDate="2019-02-14T14:35:41.970" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29585" PostTypeId="2" ParentId="29582" CreationDate="2019-02-13T22:11:21.160" Score="1" Body="&lt;p&gt;It looks like the assumption regarding the 90° point is that the amount of the ecliptic visible is 180°. So the 90° direction is the halfway point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nonagesimal is on the ecliptic, so the altitude approximates the arc. The wording in the initial part has that correct, but then it goes on to say &quot;the same as the angle&quot; - this seems to be a mistake. The arc is directly related to the angle, yes, but it is not the angle.&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="43" LastEditDate="2019-02-14T20:18:27.340" LastActivityDate="2019-02-14T20:18:27.340" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29586" PostTypeId="2" ParentId="29569" CreationDate="2019-02-13T22:29:07.713" Score="2" Body="&lt;p&gt;My answer won't be complete, from lack of time and resources, but I still wanted to share some interesting aspects here that could be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty in aswering this question revolves around the complex and irregular shapes involved here. Also, finding a &quot;best-fit&quot; ellipse for comparison is not as easy as it seems, because it depends on what and how you want to model. The actual shape of the equator is rather complex and irregular. Generally speaking, it is pretty much circular, but indeed topography and the geoid complicate matters. The Earth's movement in the solar system is not a perfect ellipse either, because of the gravitational interactions with other celestial bodies. Let's review the different irregularities involved here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth is usually modeled as an ellipsoid of revolution (oblate spheroid), and the equator as a circle. A good example of that is the WGS 1984 geodetic reference system used by the Global Satellite Navigation System. Of course, the equator is not a perfect circle, it has irregularities mainly because of topography, and even sea level itself is a little irregular too. We can approximate sea level with a &lt;a href=&quot;https://en.wikipedia.org/wiki/Geoid&quot; rel=&quot;nofollow noreferrer&quot;&gt;geoid&lt;/a&gt;, for example, here is a map of the EGM2008, a geoid used with WGS 1984 to transform ellipsoidal heights to geoid heights:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vflxU.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vflxU.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, this map shows the height of the geoid (the idealized sea level without the effects of tides and currents) with respect to the WGS84 reference ellipsoid of revolution (semi-major axis 6,378,137 m, semi-minor axis 6,356,752.314 m). The differences are mostly less than 100 meters, and are caused by the irregular distribution of mass inside the Earth itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, some studies show that the Earth's shape could be slightly better modeled by a triaxial ellipsoid, and one could try to model the equator as an ellipse, and the Earth as a triaxial ellipsoid, however, even with a best-fit triaxial, we would still need geoid corrections for the irregular mean sea level, let alone topography, and geodetic computations would be more complex on a triaxial. Other funny models and names have come up over time, like a pear-shaped (because of a slight bulge in mid southern latitudes) model as a best-fit shape. But if you look at the map above, good luck visually finding the pear shape in there, or other mathematically modelizable aspects of these bumps. We are talking about very subtle differences here, that do not necessarily need to be taken into account for most purposes when describing the general shape of the Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So depending on how you consider the shape of the equator (i.e. by topography and ocean floor, or sea level) you will arrive at a shape that is mostly circular with irregular bumps along the way. There is no authoritative agreement that I know of about an &lt;em&gt;eccentricity&lt;/em&gt; of the equator. For instance, &lt;a href=&quot;https://www.arcjournals.org/pdfs/ijms/v1-i1/4.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This study&lt;/a&gt; proposes a flattening of about 70 meters for the equator's ellipse, &lt;a href=&quot;https://www.britannica.com/science/geoid&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article&lt;/a&gt; on Encyclopaedia Britannica proposes 80 meters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Earth's orbit, for the sake of this comparison, we can use an best-fit ellipse of 149.598 million km by 149.577 million km. Of course, that is only a idealized ellipse, the real movement of the Earth in the Solar System is more complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, say we scale down and superimpose the Earth's orbit's ellipse on the equator to compare. The eccentricity of the Earth's orbit is 0.0167, the semi-major axis is 149.598 million km and semi-minor axis is 149.577 million km. Scaled down by a factor of 23,455 to the equator's size, this corresponds to a difference of about 900 meters in both orbit axes. So I think we can agree that a best-fit ellipse of sea level along the equator is more circular than the Earth's orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, topography-wise, there are bumps up to over 4,000 meters in the Andes, and the sea floor reaches 5000 meters below sea level in several places. So the &quot;topographic surface&quot; equator would, for one thing, definitely appear more bumpy than the Earth's orbit. Unfortunately, I haven't found a example or study showing what a best-fit ellipse of the equator (including topography and bathymetry) could look like, mainly because we tend to approximate sea level, not topography itself, but with more time, tools and data, it might be possible to work out the answer with a &quot;topographic best-fit&quot; ellipse. &lt;/p&gt;&#xA;" OwnerUserId="19814" LastEditorUserId="19814" LastEditDate="2019-02-13T23:31:57.490" LastActivityDate="2019-02-13T23:31:57.490" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29587" PostTypeId="2" ParentId="29570" CreationDate="2019-02-14T00:27:08.943" Score="6" Body="&lt;p&gt;I guess it depends on what you consider &lt;em&gt;scientifically important&lt;/em&gt;, but users at &lt;a href=&quot;https://www.zooniverse.org/projects/nora-dot-eisner/planet-hunters-tess/&quot; rel=&quot;noreferrer&quot;&gt;Planet Hunters&lt;/a&gt; have helped with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Planet_Hunters#Discoveries&quot; rel=&quot;noreferrer&quot;&gt;discovery of several exoplanets&lt;/a&gt; by analyzing data from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler_(spacecraft)&quot; rel=&quot;noreferrer&quot;&gt;Kepler Space Telescope&lt;/a&gt;, as well as the famous &lt;a href=&quot;https://en.wikipedia.org/wiki/KIC_8462852&quot; rel=&quot;noreferrer&quot;&gt;Tabby's star (KIC 8462852)&lt;/a&gt; which has led to a &lt;a href=&quot;https://arxiv.org/abs/1509.03622&quot; rel=&quot;noreferrer&quot;&gt;paper&lt;/a&gt; by Tabetha Boyajian co-written by the users who participated in the analysis of the light curve of the star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The star is currently still the subject of numerous studies and publications.&lt;/p&gt;&#xA;" OwnerUserId="25153" LastActivityDate="2019-02-14T00:27:08.943" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29589" PostTypeId="1" CreationDate="2019-02-14T02:09:53.620" Score="1" ViewCount="245" Body="&lt;p&gt;Why does each phase of the moon stay up a different number of hours per phase, so that, for example, January’s full moon was up twice as long as the new moon? Plus every month this bulge shifts and at some point the opposite phase is doubled. Three June full moon nights takes six March 1st quarter nights. Why does this happen?&lt;/p&gt;&#xA;" OwnerUserId="25916" LastEditorUserId="25916" LastEditDate="2019-02-15T19:35:58.037" LastActivityDate="2019-02-16T04:27:23.157" Title="What causes the duration of moon phases (the time the moon is visible above the horizon) to vary for an observer in the northern hemisphere?" Tags="&lt;the-moon&gt;&lt;moon-phases&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="29590" PostTypeId="1" AcceptedAnswerId="29601" CreationDate="2019-02-14T02:32:27.477" Score="3" ViewCount="115" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EI9eE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EI9eE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Aristarchus estimated the relative distance of the Sun and Moon by&#xA;  observing the angle between the Sun and the Moon (α in the diagram)&#xA;  when the Moon is exactly half lit. Angle β must be 90° for the Moon to&#xA;  be half lit. By observing the angle α, he could then set the scale of&#xA;  the triangle and thus the relative lengths of the sides. (Sizes and&#xA;  distances are not to scale.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does half-lit mean exactly 45 degrees?&lt;/p&gt;&#xA;" OwnerUserId="25918" LastActivityDate="2019-02-14T17:35:52.613" Title="What does half-lit means?" Tags="&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29591" PostTypeId="1" CreationDate="2019-02-14T02:43:27.267" Score="4" ViewCount="121" Body="&lt;p&gt;There are many TV shows such as &quot;How the Universe Works&quot; that explain theories about the origin of the universe such as the Big-Bang Theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What resources can I access to verify the scientific thinking on the origin of the universe theories presented on such TV shows?&lt;/p&gt;&#xA;" OwnerUserId="25917" LastEditorUserId="25988" LastEditDate="2019-02-24T23:07:53.093" LastActivityDate="2019-02-24T23:07:53.093" Title="How can you verify the theories presented on TV shows like &quot;How the Universe Works&quot; is current and/or valid?" Tags="&lt;universe&gt;&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2019-02-14T16:10:46.867" ContentLicense="CC BY-SA 4.0" />
  <row Id="29592" PostTypeId="1" AcceptedAnswerId="29594" CreationDate="2019-02-14T06:19:56.200" Score="3" ViewCount="109" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Rogue_planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nomad planets&lt;/a&gt; (also called planemos or rogues planets) which drift in space without  a star or substellar object to orbit. The majority of such planets we have discovered are gas planets with thick atmospheres and no landmass, although some of them could be frozen deserts with rocky surfaces like Earth's. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most nomad planets we have found are alone/solitary, and do not orbit any other companion objects. We have only discovered &lt;strong&gt;1&lt;/strong&gt; (as in 1.0) binary planemo called &lt;a href=&quot;https://www.centauri-dreams.org/2017/07/12/a-binary-rogue-planet/&quot; rel=&quot;nofollow noreferrer&quot;&gt;2MASS J11193254–1137466&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm asking here is that can a large system of 6 or 7 rogue planets and other planetary bodies orbit each other and remain stable; basically like a tiny open cluster or huge multiple star system, just comprised of planets instead of stars?&lt;/p&gt;&#xA;" OwnerUserId="25486" LastActivityDate="2019-02-16T00:28:50.503" Title="Can rogue/nomad planets and other starless planetary bodies orbit each other in a stable system?" Tags="&lt;rogue-planet&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29593" PostTypeId="2" ParentId="29592" CreationDate="2019-02-14T06:38:53.660" Score="2" Body="&lt;p&gt;One of the main hypotheses for the formation of rogue planets is ejection from a preexisting planetary system. This is an unlikely scenario for the formation of a &lt;em&gt;cluster&lt;/em&gt; of rogue planets, because it doesn't seem feasible for a whole bunch of bodies to be ejected together and remain in a stable system. If multiple planets were ejected from a planetary system, I would expect them to exit randomly and separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, a couple sub-brown dwarfs have been observed to host what may be protoplanetary disks (although that term may not be applicable in these cases). The important objects here are &lt;a href=&quot;https://en.wikipedia.org/wiki/Cha_110913-773444&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cha 110913-773444&lt;/a&gt; (&lt;span class=&quot;math-container&quot;&gt;$\sim8M_J$&lt;/span&gt;) and &lt;a href=&quot;https://en.wikipedia.org/wiki/OTS_44&quot; rel=&quot;nofollow noreferrer&quot;&gt;OTS 44&lt;/a&gt; (&lt;span class=&quot;math-container&quot;&gt;$\sim11M_J$&lt;/span&gt;). These objects have not been classified yet; they could be considered rogue planets or sub-brown dwarfs (or even brown dwarfs, honestly). Assuming the normal processes active in protoplanetary disks take place - collisions, accretion and the like - then in the future, additional bodies could form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OTS 44 is thought to have a disk of around 30 Earth masses (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2013A&amp;amp;A...558L...7J&quot; rel=&quot;nofollow noreferrer&quot;&gt;Joergens et al. 2013&lt;/a&gt;); depending on the efficiency of planet formation, we might see bodies similar in mass to terrestrial planets (note that this also depends on disk composition; much of it is likely to be hydrogen and helium, not silicates). Giant planets seem unlikely, as the disks mass is comparable to those of Uranus and Neptune, within a factor of two, and it is highly unlikely that the entire disk would coalesce into one planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a rogue planet formed &lt;em&gt;in situ&lt;/em&gt;, rather than being ejected from a system, we do have a different picture to consider. In this case, it seems more likely for binary rogue planets to form. At this point, of course, the label &quot;sub-brown dwarf&quot; is more appropriate than &quot;rogue planet&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2019-02-14T06:38:53.660" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29594" PostTypeId="2" ParentId="29592" CreationDate="2019-02-14T06:58:56.820" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What I'm asking here is that can a large system of 6 or 7 rogue planets and other planetary bodies orbit each other and remain stable; basically like a tiny open cluster or huge multiple star system, just comprised of planets instead of stars?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Sure! (but...)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;If you have a half-dozen bodies in a group, and the total energy (kinetic plus potential) is less than the energy it would take to move any of them to infinity, then they are a gravitationally bound system and they'll stay together forever.&lt;/s&gt;†&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;†&lt;/sup&gt;&lt;strong&gt;This is probably not true&lt;/strong&gt;; @RobJeffries' comments (&lt;a href=&quot;https://astronomy.stackexchange.com/questions/29592/can-rogue-nomad-planets-and-other-starless-planetary-bodies-orbit-each-other-in/29594?noredirect=1#comment53527_29594&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;https://astronomy.stackexchange.com/questions/29592/can-rogue-nomad-planets-and-other-starless-planetary-bodies-orbit-each-other-in/29594?noredirect=1#comment53539_29594&quot;&gt;2&lt;/a&gt;) bring up &lt;em&gt;pairings&lt;/em&gt;, and so even in a situation where moving &quot;any one of them&quot; was energetically possible, it could might for example be possible for a bound pair. I'll update this soon...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An extreme example would be a cluster that are touching and at rest, a second extreme example would be that cluster given a bit of a &quot;kick&quot; to start it jostling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good question might be &quot;Given a half-dozen bodies of equal mass &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; what is the farthest one can be from it's nearest neighbor while the system is still energetically bound?&quot; A follow-up might be the same except remove the &quot;equal mass &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt;&quot; part. I don't know the answer to that question, perhaps someone else will post an answer here with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without a &quot;Sun&quot; or particularly large central body, their motion will be chaotic, and any apparent organization (pairings or higher) will sooner or later be disturbed by another body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually there is chaos in (probably) all real-work n-body system, but in cases like our solar system this would take quite a long time to show up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a particular configuration in mind, you can calculate if the system is bound or not with a little math. Here the total energy &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; is the sum of potential energy &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt; and kinetic energy &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$E = U + T = \sum_{i\neq j} -G\frac{m_i m_j}{|\mathbf{r_i}-\mathbf{r_j}|} + \sum_{i} \frac{1}{2}m_i v_i^2 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$|\mathbf{r_i}-\mathbf{r_j}|$&lt;/span&gt; is the distance between each pair of objects with masses &lt;span class=&quot;math-container&quot;&gt;$m_i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$m_j$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$v_i^2$&lt;/span&gt; is the square of the speed, and &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_constant&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gravitatinal constant&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;but...&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Systems are never isolated in the real world, so for example&#xA; - another body could pass by and disturb the system and eject one or more members. &#xA; - two bodies could collied and exchange mass, momentum, and energy, allowing one or more &lt;em&gt;fragment&lt;/em&gt; to escape.&#xA; - other things...&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-02-16T00:28:50.503" LastActivityDate="2019-02-16T00:28:50.503" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29595" PostTypeId="2" ParentId="29591" CreationDate="2019-02-14T08:00:38.773" Score="3" Body="&lt;p&gt;Use other, more scientific sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have no way to test them by some home measurements - science was tuned to pass even more sophisticated tests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Popular media is often not very exact. They try to open the results to the public. For example, to find the geometry of the Kerr Black Holes was a decade long work for the best mathematicians of the time. The result is a relative simple formula, which is understable with not far more than high school math. Most in the popular media about the wormholes and white holes is based on it, or at least strongly related to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to check it, best if you try to find the formula and try to learn, what does it mean. It still doesn't need to understand the decade long work behind that, and also this is amazing - but only after you've understood it.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastActivityDate="2019-02-14T08:00:38.773" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29596" PostTypeId="2" ParentId="29517" CreationDate="2019-02-14T08:42:00.963" Score="2" Body="&lt;p&gt;Ernest W. Brown's &lt;a href=&quot;https://academic.oup.com/mnras/article/71/5/438/972133&quot; rel=&quot;nofollow noreferrer&quot;&gt;On a New Family of Periodic Orbits in the Problem of Three Bodies: (Plates 6, 7.)&lt;/a&gt; in MNRAS, 71, (5), pp 438–454 &lt;strong&gt;published on 10 March 1911&lt;/strong&gt; seems to be where horseshoe orbits were first proposed. (Available &lt;a href=&quot;http://adsabs.harvard.edu/abs/1911MNRAS..71..438B&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; as well). It begins:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;There are four known asteroids which appear to oscillate about one or other of the vertices of the two equilateral triangles which have the line joining Jupiter and the Sun as base. These vertices are the well-known positions of relative equilibrium. The heliocentric vector of one of these asteroids can apparently move as far as 17° away from its equilibrium position.* The oscillations cannot therefore be considered very small. One naturally asks whether oscillations of this kind in arcs of still greater extent are possible ; and if so, in what manner the orbits may he most conveniently obtained.&lt;/p&gt;&#xA;&lt;p&gt;*L. J. Linders, Arhivfor Mat., Ast. och Fys., So. Vet. Ak. i Stockholm, Bd. 4, No. 20.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;I'll make some horseshoe orbits in the Circular Restricted Three-Body Problem formalism and plot them in Python, then compare to the synodic period estimation described in @AtmosphericPrisonEscape's&#xA;&lt;a href=&quot;https://astronomy.stackexchange.com/a/29561/7982&quot;&gt;answer&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;tl;dr:&lt;/strong&gt; There's good qualitative agreement, no surprises!&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;A brief summary of CR3BP math in dimensionless units. The distance between the two bodies is equal to 1, as is the gravitational constant. They orbit around a common center of mass in circular orbits, with a period of &lt;span class=&quot;math-container&quot;&gt;$2 \pi$&lt;/span&gt;. It's easier to visualize and calculate if you do it in a rotating frame, so the two masses are fixed. The third body at position &lt;span class=&quot;math-container&quot;&gt;$x, y, z$&lt;/span&gt; is considered to have no gravitational effect on the first two,&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mu = \frac{m_2}{m_1 + m_2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$x_1 = -\mu $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$x_2 = 1-\mu $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$r_1 = \sqrt{(x-x1)^2 + y^2 + z^2}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$r_2 = \sqrt{(x-x2)^2 + y^2 + z^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The Jacobi Energy &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; is a conserved quantity in this rotating frame:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$C = x^2 + y^2 + 2\frac{1-\mu}{r_1} + 2\frac{\mu}{r_2} - (\dot{x}^2 + \dot{y}^2 + \dot{z}^2)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where the &lt;span class=&quot;math-container&quot;&gt;$x^2 + y^2$&lt;/span&gt; is the pseudopotential. If you set the velocity dependent terms &lt;span class=&quot;math-container&quot;&gt;$(\dot{x}^2 + \dot{y}^2 + \dot{z}^2)$&lt;/span&gt; to zero, you get a &lt;em&gt;zero velocity surface&lt;/em&gt;, that surface that's pasted into many/most questions about three body orbits. These plots do not apply when an object is moving, and so you can't superimpose orbits on top of them!&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fDq7l.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fDq7l.png&quot; alt=&quot;random zero-velocity curve&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The acceleration felt by the third body in this rotating frame has both the expected &lt;span class=&quot;math-container&quot;&gt;$1/r^2$&lt;/span&gt; forces and a velocity-dependent &lt;em&gt;pseudoforce&lt;/em&gt; which is not real, but accounts for the fact that the frame is rotating and not inertial.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ddot{x} = x + 2\dot{y} - \frac{(1-\mu)(x+\mu)}{r_1^3} - \frac{\mu(x-1+\mu)}{r_2^3}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ddot{y} = y - 2\dot{x} - \frac{(1-\mu)y}{r_1^3} - \frac{\mu y}{r_2^3}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ddot{z} =  -\frac{(1-\mu) z}{r_1^3} - \frac{\mu z}{r_2^3} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Here are some calculations. I chose &lt;span class=&quot;math-container&quot;&gt;$\mu = 0.001$&lt;/span&gt; which is pretty close to the situation Jupiter and the Sun. I chose an array of starting points at the opposite point from &lt;span class=&quot;math-container&quot;&gt;$m_2$&lt;/span&gt; at about &lt;span class=&quot;math-container&quot;&gt;$x=-1$&lt;/span&gt; &lt;em&gt;but that's not what I really did.&lt;/em&gt; What I really did is choose a bunch of starting velocities &lt;span class=&quot;math-container&quot;&gt;$-0.08 &amp;lt; \dot{y} &amp;lt; 0.08$&lt;/span&gt; and for each I calculated the position on the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; axis near &lt;span class=&quot;math-container&quot;&gt;$x=-1$&lt;/span&gt; where the acceleration in the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; direction was zero.&lt;/p&gt;&#xA;&lt;p&gt;That gives the solutions a tiny bit of starting symmetry, but halo orbits are bumpy and wiggly and not always so stable, so this effort isn't really necessary.&lt;/p&gt;&#xA;&lt;p&gt;I propagated each orbit until it came back around to the same area and stopped it when it crossed the x-axis, producing a family of half-cycles.&lt;/p&gt;&#xA;&lt;p&gt;To make a long story short, the method shown in @AtmosphericPrisonEscape's&#xA;&lt;a href=&quot;https://astronomy.stackexchange.com/a/29561/7982&quot;&gt;answer&lt;/a&gt; of estimating the cycle time by calculating the synodic period in the inertial frame is in pretty good agreement with these halo orbits, and that shouldn't be very surprising!&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OkuSe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OkuSe.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; half-cycles of some wobbly horseshoe orbits&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rWKMP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rWKMP.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; times to first x-axis crossings of the same wobbly horseshoe orbits, used to calculate half-cycle times.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cHG8s.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cHG8s.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; cycle times from this calculation (black dots) versus from the synodic period estimation method (red dots). Good qualitative agreement. Also the starting y velocities at each starting point in x.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; Python script for these plots.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;def x_acc(x, ydot):&#xA;    r1    = np.abs(x-x1)&#xA;    r2    = np.abs(x-x2)&#xA;    xddot = x + 2*ydot  -  ((1-mu)/r1**3)*(x+mu) - (mu/r2**3)*(x-(1-mu))&#xA;    return xddot&#xA;&#xA;def C_calc(x, y, z, xdot, ydot, zdot):&#xA;    r1 = np.sqrt((x-x1)**2 + y**2 + z**2)&#xA;    r2 = np.sqrt((x-x2)**2 + y**2 + z**2)&#xA;    C = (x**2 + y**2 + 2.*(1-mu)/r1 + 2.*mu/r2 - (xdot**2 + ydot**2 + zdot**2))&#xA;    return C&#xA;&#xA;def deriv(X, t): &#xA;    x, y, z, xdot, ydot, zdot = X&#xA;    r1 = np.sqrt((x-x1)**2 + y**2 + z**2)&#xA;    r2 = np.sqrt((x-x2)**2 + y**2 + z**2)&#xA;    xddot = x + 2*ydot  -  ((1-mu)/r1**3)*(x+mu) - (mu/r2**3)*(x-(1-mu))&#xA;    yddot = y - 2*xdot  -  ((1-mu)/r1**3)*y      - (mu/r2**3)*y&#xA;    zddot =             -  ((1-mu)/r1**3)*z      - (mu/r2**3)*z&#xA;    return np.hstack((xdot, ydot, zdot, xddot, yddot, zddot))&#xA;&#xA;# http://cosweb1.fau.edu/~jmirelesjames/hw4Notes.pdf&#xA;&#xA;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;from scipy.integrate import odeint as ODEint&#xA;from scipy.optimize import brentq&#xA;&#xA;halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]&#xA;&#xA;mu = 0.001&#xA;&#xA;x1 = -mu&#xA;x2 = 1. - mu&#xA;&#xA;x = np.linspace(-1.4, 1.4, 1201)&#xA;y = np.linspace(-1.4, 1.4, 1201)&#xA;&#xA;Y, X = np.meshgrid(y, x, indexing='ij')&#xA;Z    = np.zeros_like(X)&#xA;&#xA;xdot, ydot, zdot = [np.zeros_like(X) for i in range(3)]&#xA;&#xA;C = C_calc(X, Y, Z, xdot, ydot, zdot)&#xA;C[C&amp;gt;8] = np.nan&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;    plt.imshow(C)&#xA;    plt.colorbar()&#xA;    levels = np.arange(2.9, 3.2, 0.04) &#xA;    CS = plt.contour(C, levels,&#xA;                 origin='lower',&#xA;                 linewidths=2) &#xA;    plt.show()&#xA;&#xA;ydot0s   = np.linspace(-0.08, 0.08, 20)&#xA;x0ydot0s = []&#xA;for ydot0 in ydot0s:&#xA;    x0, infob =  brentq(x_acc, -1.5, -0.5, args=(ydot0), xtol=1E-11, rtol=1E-11,&#xA;                           maxiter=100, full_output=True, disp=True)&#xA;    x0ydot0s.append((x0, ydot0))&#xA;&#xA;states = [np.array([x0, 0, 0, 0, ydot0, 0]) for (x0, ydot0) in x0ydot0s]&#xA;&#xA;times  = np.arange(0, 150, 0.01)&#xA;&#xA;results = []&#xA;for X0 in states:&#xA;    answer, info = ODEint(deriv, X0, times, atol = 1E-11, full_output=True)&#xA;    results.append(answer.T.copy())&#xA;&#xA;resultz = []&#xA;for x0ydot0, thing in zip(x0ydot0s, results):&#xA;    y     = thing[1]&#xA;    check = y[2:]*y[1:-1] &amp;lt; 0&#xA;    zc    = np.argmax(y[2:]*y[1:-1] &amp;lt; 0) + 1&#xA;    if zc &amp;gt; 10:&#xA;        resultz.append((thing, zc, x0ydot0))&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;    hw = 1.6&#xA;    for j, (thing, zc, x0ydot0) in enumerate(resultz):&#xA;        x, y = thing[:2,:zc]&#xA;        plt.plot(x, y)&#xA;    plt.xlim(-hw, hw)&#xA;    plt.ylim(-hw, hw)&#xA;    plt.plot([x1], [0], 'ok')&#xA;    plt.plot([x2], [0], 'ok')&#xA;    plt.show()&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;    for j, (thing, zc, x0ydot0) in enumerate(resultz):&#xA;        x, y = thing[:2]&#xA;        plt.plot(times[:zc], y[:zc])&#xA;    plt.show()&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;    for j, (thing, zc, x0ydot0) in enumerate(resultz):&#xA;        x0, ydot0 = x0ydot0&#xA;        cycle_time = 2. * times[zc] / twopi&#xA;        ratio = abs(x0/x2)&#xA;        T_simple_model = twopi * abs(x0/x2)**1.5&#xA;        T_synodic_simple_model = 1. / (1. - twopi/T_simple_model) # https://astronomy.stackexchange.com/a/25002/7982&#xA;        plt.subplot(2, 1, 1)&#xA;        plt.plot(x0, cycle_time, 'ok')&#xA;        plt.plot(x0, abs(T_synodic_simple_model), 'or')&#xA;        plt.subplot(2, 1, 2)&#xA;        plt.plot(x0, ydot0, 'ok')&#xA;    plt.subplot(2, 1, 1)&#xA;    plt.xlabel('x0', fontsize=16)&#xA;    plt.ylabel('cycle times (periods)', fontsize=16)&#xA;    plt.subplot(2, 1, 2)&#xA;    plt.xlabel('x0', fontsize=16)&#xA;    plt.ylabel('ydot0', fontsize=16)&#xA;    plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-02-14T14:32:46.093" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29597" PostTypeId="1" CreationDate="2019-02-14T11:06:02.373" Score="3" ViewCount="989" Body="&lt;p&gt;Below are two cropped views of &amp;quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Aerial_telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Johannes Hevelius's 8 inch telescope with an open work wood and wire &amp;quot;tube&amp;quot; that had a focal length of 150 feet to limit chromatic aberration.&lt;/a&gt;&amp;quot; from &lt;a href=&quot;https://iiif.lib.harvard.edu/manifests/view/ext:aHR0cHM6Ly9paWlmLmxpYi5oYXJ2YXJkLmVkdS9tYW5pZmVzdHMvaWRzOjE2NDcwMzE1$1i&quot; rel=&quot;nofollow noreferrer&quot;&gt;Harvard University, Houghton Library, pga_typ_620_73_451_fig_aa&lt;/a&gt; (found &lt;a href=&quot;https://hollis.harvard.edu/primo-explore/fulldisplay?context=L&amp;amp;vid=HVD2&amp;amp;search_scope=everything&amp;amp;tab=everything&amp;amp;lang=en_US&amp;amp;docid=01HVD_ALMA211764767520003941&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;) the first of which I've zoomed and enhanced the contrast.&lt;/p&gt;&#xA;&lt;p&gt;According to Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Aerial_telescope#Very_long_%22tubed%22_telescopes&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aerial telescope; Very long &amp;quot;tubed&amp;quot; telescopes&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Very long &amp;quot;tubed&amp;quot; telescopes&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Telescopes built in the 17th and early 18th century used single element non-achromatic objective lenses that suffered from interfering rainbow halos (chromatic aberration) introduced by the non-uniform refractive properties of single glass lenses. This degraded the quality of the images they produced. Telescope makers from that era found that very long focal length objectives had no appreciable chromatic aberration (the uncorrected chromatic aberration fell within the large diffraction pattern at focus). They also realized that when they doubled the diameter of their objectives they had to make the objective's focal length 4 times as long (focal length had to be squared) to achieve the same amount of minimal chromatic aberration. As the objective diameter of these refracting telescopes was increased to gather more light and resolve finer detail they began to have focal lengths as long as 150 feet. Besides having very long tubes, these telescopes needed scaffolding or long masts and cranes to hold them up. Their value as research tools was minimal since the telescope's support frame and tube flexed and vibrated in the slightest breeze and sometimes collapsed altogether.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A refractor's primary lens produces an image because the angle it deflects is proportional to the distance &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; from the center of the lens of focal length &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt; as&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\theta \approx x/f$$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;For a dispersive material the deflection at a given point on the lens varies with wavelength, and therefore the focal length of the lens varies with wavelength.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KBEYI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KBEYI.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Chromatic_aberration_lens_diagram.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;This is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Chromatic_aberration&quot; rel=&quot;nofollow noreferrer&quot;&gt;chromatic aberration&lt;/a&gt; and its severity for a given type of glass is often expressed by its &lt;a href=&quot;https://en.wikipedia.org/wiki/Abbe_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;abbe number&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is there a simple mathematical way to explain how making the focal length longer could possibly &lt;em&gt;improve&lt;/em&gt; a telescope's performance by decreasing the impact of chromatic aberration on its performance?&lt;/p&gt;&#xA;&lt;p&gt;Or do I misunderstand what the benefits are to making these very long telescopes?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iFTbX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iFTbX.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1I0eh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1I0eh.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-02-15T01:24:54.877" Title="How does making a refracting telescope very long reduce the chromatic aberration of an uncorrected lens?" Tags="&lt;telescope&gt;&lt;history&gt;&lt;optics&gt;&lt;angular-resolution&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29598" PostTypeId="2" ParentId="29583" CreationDate="2019-02-14T14:20:53.147" Score="1" Body="&lt;p&gt;I don't think it is common at all to &lt;em&gt;find&lt;/em&gt; stars outside of galaxies, but most astronomers would agree that they must exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that individual stars are not very bright. We can see stars in our own Galaxy, but beyond that (in other galaxies or in the space between galaxies) we wouldn't be able to detect a single star on its own. We would generally observe star clusters (hundreds or thousands or tens of thousands of stars) in other galaxies as one bright blob, we can't see each star inside it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you give us a source for the star with a planet that entered our Galaxy? I haven't heard of this so would need to see the details to explain where it came from.&lt;/p&gt;&#xA;" OwnerUserId="2270" LastActivityDate="2019-02-14T14:20:53.147" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29599" PostTypeId="1" CreationDate="2019-02-14T16:23:32.957" Score="2" ViewCount="92" Body="&lt;p&gt;I read in a book that the earth's rotation sometimes shows unpredictable irregularities(not caused by precession, nutation, tidal forces) which are accounted for using dynamical time. What would be the cause of these?&lt;/p&gt;&#xA;" OwnerUserId="25925" LastActivityDate="2019-02-14T22:37:22.720" Title="What causes random irregularities in the earth's rotation accounted for by dynamical time" Tags="&lt;observational-astronomy&gt;&lt;rotation&gt;&lt;time&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29600" PostTypeId="1" AcceptedAnswerId="29608" CreationDate="2019-02-14T17:16:54.707" Score="2" ViewCount="95" Body="&lt;p&gt;How was it determined that Cruithne takes 770 years to complete its cycle with Earth? Janus and Epimetheus were disovered long enough ago, and have short enough cycle times, that we've had time to directly observe their 8 year cycle more than 6 times over. But Cruithne was only discovered 30-ish years ago, so certainly it's cycle time wasn't determined by simple observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what method was used to arrive at this 770 year number?&lt;/p&gt;&#xA;" OwnerUserId="24853" LastActivityDate="2019-02-15T00:21:09.277" Title="How was Cruithne's orbital time calculated?" Tags="&lt;orbit&gt;&lt;natural-satellites&gt;&lt;orbital-mechanics&gt;&lt;celestial-mechanics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29601" PostTypeId="2" ParentId="29590" CreationDate="2019-02-14T17:35:52.613" Score="5" Body="&lt;p&gt;Half lit means the Moon is at a quarter phase, either First Quarter or Third Quarter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original question asks &quot;Does half-lit mean exactly 45 degrees?&quot;.  I assume this is asking if the angle &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; is 45 degrees. The answer is no. The angle &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; is found from trigonometry to be &lt;span class=&quot;math-container&quot;&gt;$cos(\alpha)=EM/ES$&lt;/span&gt; where EM is the distance from the Earth to the Moon and ES is the distance from the Earth to the Sun. EM/ES is approximately 1/390 (=238000 miles/93 million miles), so &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; works out to be 89.85 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aristarchus was trying to do the reverse calculation to find the distances. If the angle &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; could be measured accurately, the relative distances EM/ES could be calculated using the same formula. The problem is the difference between 89.85 degrees (the Sun is 390 times farther than the Moon) and 90 degrees (the Sun is infinitely farther than the Moon) is a very small angle and difficult to measure exactly. If the Sun were a lot closer, then &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; would be a smaller value, and the error in the measured angle of &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; would be restrict the value of EM/ES to a more meaningful range than &quot;infinity&quot;.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2019-02-14T17:35:52.613" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29602" PostTypeId="2" ParentId="29583" CreationDate="2019-02-14T19:45:57.387" Score="1" Body="&lt;p&gt;The other answers address the question of whether or not interstellar stars can have planets. You also mentioned the star with a planet that entered our galaxy, so I will try to answer that.&lt;/p&gt;&#xA;&lt;p&gt;I suspect the planetary system you are referring to is the one at &lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2014MNRAS.443L..89A/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kapteyn's Star&lt;/a&gt;, which was &lt;a href=&quot;https://ph.qmul.ac.uk/ancient-worlds-around-kapteyns-star&quot; rel=&quot;nofollow noreferrer&quot;&gt;described in the press release&lt;/a&gt; as being a former member of the star cluster &lt;a href=&quot;https://www.eso.org/public/news/eso0844/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Omega Centauri&lt;/a&gt;. Omega Centauri is traditionally categorised as a globular cluster but there is evidence to suggest that it is actually the remnant core of a dwarf galaxy that merged with the Milky Way. The origin of Kapteyn's Star in Omega Centauri has since been challenged by &lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2015ApJ...808..103N/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Navarrete et al. (2015)&lt;/a&gt;, who find that the abundances of various elements in Kapteyn's Star and other members of the Kapteyn Moving Group do not match the abundances in Omega Centauri and are indistinguishable from Milky Way halo stars.&lt;/p&gt;&#xA;&lt;p&gt;Another possibility is that it is &lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2011IAUS..276..121K/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;HIP 13044&lt;/a&gt;, a star in the Helmi stream. The Helmi stream is another remnant of a destroyed dwarf galaxy that merged with the Milky Way. Unfortunately the evidence for the planet was found to be a result of errors in the data analysis, see &lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2014MNRAS.443L..89A/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jones &amp;amp; Jenkins (2014)&lt;/a&gt; for details.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2020-10-04T10:52:48.967" LastActivityDate="2020-10-04T10:52:48.967" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29603" PostTypeId="1" AcceptedAnswerId="29604" CreationDate="2019-02-14T20:07:09.017" Score="2" ViewCount="399" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Circumstellar_habitable_zone&quot; rel=&quot;nofollow noreferrer&quot;&gt;Circumstellar habitable zone&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In astronomy and astrobiology, the circumstellar habitable zone (CHZ), or simply the habitable zone, is the range of orbits around a star within which a planetary surface can support liquid water given sufficient atmospheric pressure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But then you have moons in the solar system such as Titan, Europe, Enceladus where some scientists believe life could be possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.theguardian.com/science/2018/jun/17/the-most-likely-cradles-for-life-our-solar-system-mars&quot; rel=&quot;nofollow noreferrer&quot;&gt;The most likely cradles for life inside our solar system&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Life_on_Titan&quot; rel=&quot;nofollow noreferrer&quot;&gt;Life on Titan&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why the CHZ is defined as it is, if life could be possible outside of it? Is it pourly defined or scientists believing there could be life outside the CHZ in the solar system are having too much wishful thinking?&lt;/p&gt;&#xA;" OwnerUserId="11562" LastActivityDate="2019-02-14T20:38:39.520" Title="Why the Circumstellar Habitable Zone is defined as it is, if life could be possible outside of it?" Tags="&lt;solar-system&gt;&lt;habitable-zone&gt;&lt;titan&gt;&lt;europa&gt;&lt;enceladus&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29604" PostTypeId="2" ParentId="29603" CreationDate="2019-02-14T20:27:55.627" Score="9" Body="&lt;p&gt;From your first link, the definition is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The circumstellar habitable zone (CHZ), or simply the habitable zone, is the range of orbits around a star within which a planetary surface can support liquid water given sufficient atmospheric pressure.[1][2][3][4][5] The bounds of the CHZ are based on Earth's position in the Solar System and the amount of radiant energy it receives from the Sun.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but as you will see further down in that same article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;In subsequent decades, the CHZ concept began to be challenged as a primary criterion for life, so &lt;strong&gt;the concept is still evolving&lt;/strong&gt;. Since the discovery of evidence for extraterrestrial liquid water, substantial quantities of it are now thought to occur outside the circumstellar habitable zone. The concept of deep biospheres, like Earth's, that exist independently of stellar energy, are &lt;strong&gt;now generally accepted&lt;/strong&gt; in astrobiology given the large amount of liquid water known to exist within in lithospheres and asthenospheres of the Solar System. Sustained by other energy sources, such as tidal heating or radioactive decay or pressurized by non-atmospheric means, &lt;strong&gt;liquid water may be found even on rogue planets&lt;/strong&gt;, or their moons.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So really, it's just historical - it is accepted now that it is a limited view, and there is just as likely to be liquid water elsewhere. That doesn't necessarily mean life is as likely, as with lower solar radiation, there may be insufficient energy to support life even with liquid water.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2019-02-14T20:27:55.627" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29606" PostTypeId="2" ParentId="29597" CreationDate="2019-02-14T21:08:08.457" Score="7" Body="&lt;p&gt;The actual math is a bit complicated, but there's a simple intuitive explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longitudinal chromatic aberration happens because, when you cut a convergent lens in two, and you look at the cross-section, the edge of the lens looks a bit like a prism, doesn't it? (look at the diagram that you've posted above, the top of the lens) And it does exactly what a prism does - it makes a &quot;pretty&quot; little rainbow. Not so pretty when you're trying to do astronomy, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short, fat, triangular prisms make widespread rainbows. Thin, elongated, thin-wedge shaped prisms make tight, narrow rainbows. The spread of the colors increases as the two sides of the prism are separated by a wider angle. It decreases as the two sides get closer together at a narrow angle. At the extreme, if the two sides of a &quot;prism&quot; were parallel, there would be no rainbow - but then it's not a prism anymore, it's just a glass plate, completely flat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All else being equal, a single-lens-objective refractor has a &quot;fatter&quot; lens when focal length gets shorter (the two sides of the lens are more strongly curved, and/or less parallel); it has a &quot;thinner&quot; lens when the focal length gets bigger (the two sides of the lens are less strongly curved, and/or closer to parallel). A &quot;fat&quot; lens at the edge is like a fat, blunt prism. A &quot;thin&quot; lens at the edge is like a sharp, slender prism. The former makes a big spread of color. The latter makes a much more tight rainbow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that's what the old telescope makers did, back in the day (Hevelius, late 1600s). They just pumped up the focal ratio, made these super-long refractors with weakly curved lenses to combat the rainbow edge images which are so annoying and ultimately can falsify your observations.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, if you study the properties of lenses, you notice that divergent lenses (the kind that make the image smaller) have a chromatic aberration opposite to convergent lenses (lenses that magnify, and that were used as primary, or objective lenses in telescopes back then). It could be argued that by combining a convergent and a divergent lens, chromatic aberration might be greatly reduced; as long as the convergent lens is stronger than the divergent companion, the combination should remain convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, different glasses need to be used, so that the convergent lens is strongly convergent, but weakly dispersing, while the divergent lens is weakly divergent, but strongly dispersing. That's how you get near zero dispersion, but still pretty decent convergence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ty6XV.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ty6XV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the early 1700s this was accomplished by British opticians. Chester Moore Hall is credited with the making of the first achromatic objective lens using the scheme shown above. But the actual work of grinding/polishing the glass was contracted to George Bass, who noticed that the two lenses he was making, when put together, had near-zero chroma. Hall failed to recognize the practical applications of the achromats, neglected his own invention, and years later Bass mentioned the achromat scheme to John Dollond, who understood its value and started to popularize the new design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/History_of_the_telescope#Achromatic_refracting_telescopes&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/History_of_the_telescope#Achromatic_refracting_telescopes&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Achromatic_lens&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Achromatic_lens&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The achromatic combo guarantees zero chromatic aberration at two wavelengths. So you basically correct the aberration in the red and blue, or something like that, and the rest is &quot;good enough&quot;. Things can be improved further by adding another lens - the apochromatic objective, or the &quot;apo&quot;, or the &quot;triplet&quot;. This was invented by Peter Dollond, John's son, in the mid-1700s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/E6feo.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E6feo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The apochromat guarantees zero chromatic aberration at three wavelengths - usually in the middle and then closer to the two extremes of the visual spectrum. So that's overall better than the previous scheme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Apochromat&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Apochromat&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best apo refractors have extremely low chromatic aberration, basically invisible to the eye in most situations. But they are expensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further improvements can be made. The most expensive amateur refractors are &quot;quad&quot; schemes, whereby using 4 lenses zeroes out chroma at 4 wavelengths. Takahashi is a brand nowadays known for excellent quality quad refractors, used in top-level astrophotography rigs.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;BTW, not just chroma, but basically all aberrations can be reduced by increasing the focal ratios of single lens objectives. Pretty much every single aberration gets better if you put a piece of paper in front of it, with a hole smaller than the diameter of the lens. But the aperture gets smaller, resolving power gets worse, there's less light captured by the instrument, etc. Tradeoffs...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;W.r.t detailed math of single lens chromatic aberrations, see chapter 5.3.1. Chromatic Aberration of a Singlet in 'Telescopes, Eyepieces, Astrographs' by G.H. Smith, R. Ceragioli, R. Berry. I believe that chapter is exactly the answer to your question, complete with working out the math for the aberration as a function of lens parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.willbell.com/tm/TelescopesEyepiecesAstrographs.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.willbell.com/tm/TelescopesEyepiecesAstrographs.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given your interest in the theory of optical systems, you should read that book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also worth mentioning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Introduction to Lens Design, by J.M. Geary&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.willbell.com/tm/lens_design/index.htm&quot; rel=&quot;noreferrer&quot;&gt;http://www.willbell.com/tm/lens_design/index.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Practical Computer-Aided Lens Design, by G.H. Smith:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.willbell.com/tm/lensdesn.htm&quot; rel=&quot;noreferrer&quot;&gt;http://www.willbell.com/tm/lensdesn.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some good theoretical info, as always, can be found on telescope-optics.net&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.telescope-optics.net/chromatic.htm&quot; rel=&quot;noreferrer&quot;&gt;https://www.telescope-optics.net/chromatic.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2019-02-15T01:24:54.877" LastActivityDate="2019-02-15T01:24:54.877" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29607" PostTypeId="2" ParentId="29599" CreationDate="2019-02-14T22:37:22.720" Score="2" Body="&lt;p&gt;On short scales, atmospheric phenomena can cause variations in mass distribution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On relatively short sub-annual to annual timescales, variations in the length-of-day have been clearly shown to arise primarily as a consequence of the exchange of angular momentum between the solid Earth and its overlying atmosphere (e.g. Hide et al., 1980) and oceans. On the inter-annual timescale an important recent discovery concerning l.o.d. variability has been the documentation of a significant excitation associated with El Nino-Southern Oscillation events (Cox and Chao, 2002; Dickey et al., 2002). Concerning the sources of wobble excitation, it is clear that on the timescale of the seasonal cycle of climate change the excitation of the annual component of wobble variability is due to the inter-hemispheric exchange of atmospheric mass. The Chandler wobble, however, a free oscillation of Earth’s spin axis in a body fixed frame of reference with a period close to 14 months, is apparently significantly forced by the dynamical state of the oceans (e.g. Gross, 2000) as well as by the atmosphere. These most recent analyses of the problem of Chandler wobble excitation, to be reviewed in what follows, appear to have finally resolved what had remained an unresolved problem for decades.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;-- &lt;a href=&quot;ftp://ftp.gps.caltech.edu/pub/djs/peltier/History%20of%20Earth%20Rotation_nographics-1.doc&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;History of Earth Rotation&quot;, W. R. Peltier, Department of Physics,University of Toronto, Toronto, Ontario, Canada&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On longer time scales, glacier distribution, mantle convection, and effects from the larger planets on the orbit of Earth (which by altering the geometry of earth's orbit, would also alter solar tidal forces).&lt;/p&gt;&#xA;" OwnerUserId="22118" LastActivityDate="2019-02-14T22:37:22.720" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29608" PostTypeId="2" ParentId="29600" CreationDate="2019-02-15T00:02:33.310" Score="4" Body="&lt;p&gt;It's a good question!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went to the JPL Horizons &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;web page&lt;/a&gt; and looked up &lt;a href=&quot;https://en.wikipedia.org/wiki/3753_Cruithne&quot; rel=&quot;nofollow noreferrer&quot;&gt;3753 Cruithne&lt;/a&gt; and saw that the current orbital solution is based on observations from 1973 to 2018, about 45 years. That's a fraction of an orbit, but very careful measurements of it's position on the celestial sphere (RA, Dec) over time will allow astronomers to calculate its position and speed by fitting those observations to many possible trajectories and finding the one that's the best fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NVAuA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NVAuA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;JPL/HORIZONS               3753 Cruithne (1986 TO)         2019-Feb-14 23:48:43&#xA;Rec #:    3753 (+COV) Soln.date: 2018-Sep-27_06:26:23    # obs: 677 (1973-2018)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The orbit is elliptical but has a nearly-identical semimajor axis as Earth, so it's period &lt;em&gt;around the Sun&lt;/em&gt; is one year just like Earth. That means that it in 45 years it has been possible to observe about 45 orbits around the Sun, allowing astronomers to nail down its orbit with great precision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I plotted the distance from Earth to Cruithne from 1973 to 2018 below from the Horizons data. At a half of an AU and 0.1 arcsec astrometric position precision (averaging over many observation) that gives the position to tens of kilometers over a span of decades. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then using accurate orbital modeling using the gravity of all the solar system's bodies as &lt;a href=&quot;https://space.stackexchange.com/a/23409/12102&quot;&gt;described here&lt;/a&gt; it is possible to project the position with great accuracy, as long as non-gravitational forces like unpredictable outgassing aren't too strong, and &lt;em&gt;predictable&lt;/em&gt; non-gravitational forces like solar radiation pressure can be modeled as &lt;a href=&quot;https://space.stackexchange.com/q/18474/12102&quot;&gt;described here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DMsN1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DMsN1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Over time the orientation of the elliptical orbit slowly precesses around the Sun due to the gravitational effects of Earth. This is the ~770 year period you've mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the video &lt;a href=&quot;https://www.youtube.com/watch?v=dsHsYjuudVo&quot; rel=&quot;nofollow noreferrer&quot;&gt;Near-Earth Asteroid 3753 Cruithne&lt;/a&gt; they hold the position of the Earth fixed and you can see how Cruithne's orbit sloshes back and forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This GIF is too large to embed here, but it demonstrates the sloshing nicely: &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Animation_of_3753_Cruithne_orbit.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/File:Animation_of_3753_Cruithne_orbit.gif&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-02-15T00:21:09.277" LastActivityDate="2019-02-15T00:21:09.277" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29609" PostTypeId="1" AcceptedAnswerId="30024" CreationDate="2019-02-15T13:20:37.233" Score="6" ViewCount="1044" Body="&lt;p&gt;I noticed on many examples sun spots are near the equator of the Sun. Why are there no sun spots closer to the poles of the Sun?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastActivityDate="2019-03-17T22:13:53.913" Title="Why do sun spots appear near the equator of the Sun?" Tags="&lt;the-sun&gt;&lt;gravity&gt;&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29610" PostTypeId="1" CreationDate="2019-02-15T17:43:34.710" Score="3" ViewCount="856" Body="&lt;p&gt;I am a Computer Science Engineering student, in the 1st year of college. I'm insanely interested in making a project regarding astronomy/astrophysics but I don't know where to begin from. I know C, C++, Python(not Scipy,Numpy or Astropy, but the basics. But I'm willing to learn those too). Can someone please give me some insights on what do make/do which would satisfy my aptitude in this field? My C++ and C knowledge are almost up to date, and I know basics of python, with all the data structures and basic libraries of Python. I'm interested in simulation stuff, but I don't exactly have relevant resources to learn those.&lt;/p&gt;&#xA;" OwnerDisplayName="user25931" LastActivityDate="2019-02-15T17:43:34.710" Title="Astronomy/Astrophysics Project With C++/C/Python" Tags="&lt;data-analysis&gt;&lt;python&gt;&lt;n-body-simulations&gt;&lt;algorithm&gt;&lt;raw-data&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2019-02-15T18:42:31.527" ContentLicense="CC BY-SA 4.0" />
  <row Id="29612" PostTypeId="2" ParentId="29589" CreationDate="2019-02-16T04:27:23.157" Score="2" Body="&lt;p&gt;The interval between moonrise and moonset varies for the same reason that summer days are longer than winter days.&#xA;How long a celestial object appears above the horizon depends on the observer's latitude and the object's &lt;a href=&quot;https://en.wikipedia.org/wiki/Declination&quot; rel=&quot;nofollow noreferrer&quot;&gt;declination&lt;/a&gt; (angle north or south of the celestial equator).&#xA;At full moon on 2019-01-21, the Moon's declination was +20&lt;span class=&quot;math-container&quot;&gt;$^\circ$&lt;/span&gt;, and there were 15 hours between moonrise and moonset at 40&lt;span class=&quot;math-container&quot;&gt;$^\circ$&lt;/span&gt;N latitude.&#xA;At new moon on 2019-02-04, the Moon's declination was -18&lt;span class=&quot;math-container&quot;&gt;$^\circ$&lt;/span&gt;, and moonrise and moonset at 40&lt;span class=&quot;math-container&quot;&gt;$^\circ$&lt;/span&gt;N were only 10h10m apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to think of this is that when the Moon is in the first quarter, it's within about 5 degrees of where the Sun would be 3 months later; last quarter, the Moon is where the Sun would be 3 months earlier, etc.&#xA;A full Moon in June is up about as long as the Sun in December; a first quarter Moon in March is up about as long as the Sun in June.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2019-02-16T04:27:23.157" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29614" PostTypeId="1" AcceptedAnswerId="29638" CreationDate="2019-02-16T06:01:34.577" Score="2" ViewCount="624" Body="&lt;p&gt;From what I learnt, we can find what elements are in the sun by analysing spectrums since each elements has same emission spectrum, and when you look at the sun's spectrum, you can see absorption spectrum missing some lines which is the elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do we know how much are there? Are there other ways to find out how much of each elements contributes to the sun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And how can this become an evidence of Big Bang?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these even correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Can you explain it easily)&lt;/p&gt;&#xA;" OwnerUserId="25927" LastActivityDate="2019-02-18T12:29:41.243" Title="How do we know that the mass ratio of hydrogen and helium in the sun is approximately 3:1?" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29615" PostTypeId="1" AcceptedAnswerId="29619" CreationDate="2019-02-16T06:01:37.163" Score="0" ViewCount="991" Body="&lt;p&gt;Is there an official reference algorithm for calculating the Ekadashi days?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the textual description of &lt;a href=&quot;https://en.wikipedia.org/wiki/Ekadashi#Calculation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ekadashi calculation&lt;/a&gt; on wikipedia but can't really make sense of it in order to write the code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already have the (java) code for calculating the new/full moon phases, and originally assumed that Ekadashi would be as easy as appending 11 days to the time of new and full moon but apparently this approach is not providing correct results.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;&lt;em&gt;Update&lt;/em&gt;:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Based on James's answer the answer is &lt;a href=&quot;http://www.tifr.res.in/~vahia/tithi-calculations.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;11&lt;/a&gt; &lt;em&gt;lunar&lt;/em&gt; days and &lt;a href=&quot;http://www.tifr.res.in/~vahia/tithi-calculations.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;26&lt;/a&gt; &lt;em&gt;lunar&lt;/em&gt; days after new moon (&lt;em&gt;not&lt;/em&gt; gregorian days).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The required conditional adjustments are described in &lt;a href=&quot;http://gopal.home.sk/gcal/docs/GCalEventsDocumentation.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;chapter 1.4 of Vaisnava Events Calculation&lt;/a&gt; which don't match the ones described in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ekadashi#Calculation&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="25932" LastEditorUserId="25932" LastEditDate="2019-02-17T22:59:14.857" LastActivityDate="2019-02-17T22:59:14.857" Title="How to calculate the time of Ekadashi days?" Tags="&lt;moon-phases&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29616" PostTypeId="1" AcceptedAnswerId="29617" CreationDate="2019-02-16T12:05:50.593" Score="0" ViewCount="39" Body="&lt;p&gt;So I have this problem (which I  translated to English as accurately as could):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When studying a compact molecular cloud with radiotelescope, at a wavelength of 18 cm, hydroxyl (OH) maser(ic) radiation was registered. Radiation intensity changes at about 5 min cycle. Evaluate the maximum diameter of the cloud.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Actually we learned very little about molecular clouds at high school and I have no idea how to solve this problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the intensity change cycle is wave period, I can calculate wave velocity, which would be very low (as well as the frequency). I'm not asking for the answer, some hints would be enough. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="25939" LastEditorUserId="616" LastEditDate="2019-02-16T14:58:55.523" LastActivityDate="2019-02-16T14:58:55.523" Title="Finding the diameter of a molecular cloud using period of intensity variation of hydroxyl maser" Tags="&lt;astrophysics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29617" PostTypeId="2" ParentId="29616" CreationDate="2019-02-16T13:32:12.607" Score="1" Body="&lt;p&gt;The principle that you apply is that the properties of a body cannot change faster than it the time it takes for light to cross the body.  So pulsars can be very small, which we know from the high frequency of their variation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A body which varies with a period of 5 minutes can't be larger than about 5 light-minutes across.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-02-16T13:32:12.607" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29618" PostTypeId="1" CreationDate="2019-02-16T14:40:31.670" Score="1" ViewCount="125" Body="&lt;p&gt;I am wondering if this rock on picture below is meteor fragment? If so what composition you think it is? It is magnetic i tried with magnet and it sticks to it very strong, it does have a lot of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YiCN5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YiCN5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yyCo5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yyCo5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24173" LastActivityDate="2019-02-16T14:40:31.670" Title="Found green rock 20cm below surface, is this an meteor piece?" Tags="&lt;meteorite&gt;" AnswerCount="0" CommentCount="7" ClosedDate="2019-02-16T19:48:30.327" ContentLicense="CC BY-SA 4.0" />
  <row Id="29619" PostTypeId="2" ParentId="29615" CreationDate="2019-02-16T17:14:55.480" Score="1" Body="&lt;p&gt;It is not clear that it can be calculated in a simple way&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length of the lunar month varies between 29.3 and 29.8 days, due to perturbations from the sun. Since ekadashi is based on the actual astronomical moon, you would need to know the length of the current synodic month in order to work out when ekadashi is.  You can predict the motion of the moon very accurately, but the calculation is beyond this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However on average, each lunar month is 29.53 days, so ekadashi is the day which is 259.86 hours and 614.22 hours after each new moon. The calculation is&lt;br&gt;&#xA;24 hours/day * 11/30 * 29.53 days&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And New moon last occurred on 4th Feb at 21.03, you can add 29.53 days to find the next new moon etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The irregular motion of the moon is why the Christians choose to use an &quot;average moon&quot; and an &quot;average sun&quot; for their calculation of Easter.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-02-16T17:14:55.480" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29620" PostTypeId="1" AcceptedAnswerId="29777" CreationDate="2019-02-16T21:08:33.927" Score="4" ViewCount="666" Body="&lt;p&gt;Can a black hole free from an accretion disk collect photons in the photon sphere in a prolonged stable orbit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could enough photons be in orbit to shade the actual black hole with light?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would a rogue black hole appear to have an atmosphere made of photons from the light emitted from other distant stars entrapped and released?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2019-02-27T03:32:20.183" LastActivityDate="2019-02-27T10:42:42.380" Title="Can a photon have a stable orbit around a black hole?" Tags="&lt;orbit&gt;&lt;black-hole&gt;&lt;photons&gt;" AnswerCount="1" CommentCount="12" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29621" PostTypeId="1" AcceptedAnswerId="29622" CreationDate="2019-02-17T08:04:26.083" Score="31" ViewCount="4571" Body="&lt;p&gt;As far as I understand, a moon is an object in permanent orbit around a planet, dwarf-planet, asteroid, etc. If there was another object permanently orbiting this moon, would that be a moon-moon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any natural object of such kind that we know of?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no, why are such objects so unlikely?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: this is question is different from &lt;a href=&quot;https://astronomy.stackexchange.com/q/856/24789&quot;&gt;Do moons have moons?&lt;/a&gt; because I explicitly ask for a reason. Also, the terminology is included here.&lt;/p&gt;&#xA;" OwnerUserId="24789" LastEditorUserId="24789" LastEditDate="2019-02-18T21:32:37.400" LastActivityDate="2019-02-18T21:32:37.400" Title="Is there any known moon of a moon?" Tags="&lt;orbit&gt;&lt;natural-satellites&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="3" ClosedDate="2019-02-19T18:47:48.077" ContentLicense="CC BY-SA 4.0" />
  <row Id="29622" PostTypeId="2" ParentId="29621" CreationDate="2019-02-17T08:23:51.400" Score="25" Body="&lt;p&gt;No. &lt;a href=&quot;https://source.wustl.edu/2010/12/how-iapetus-saturns-outermost-moon-got-its-ridge/&quot; rel=&quot;nofollow noreferrer&quot;&gt;One explanation for the equatorial ridge on Iapetus&lt;/a&gt; involves a now-lost &lt;a href=&quot;https://en.wikipedia.org/wiki/Subsatellite&quot; rel=&quot;nofollow noreferrer&quot;&gt;subsatellite&lt;/a&gt;, so perhaps such objects existed in our Solar System's past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subsatellites tend to get rapidly removed by tidal forces causing changes to the orbit. Depending on the parameters of the system, the orbit may decay until the subsatellite collides with the moon or breaks up at its Roche limit. Alternatively it may migrate outwards until the subsatellite becomes unbound: at this point it would become a moon of the planet, though likely on an unstable orbit that would lead to scattering or a collision. The orbital evolution tends to be faster for more massive subsatellites. See &lt;a href=&quot;https://arxiv.org/abs/1810.03304&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kollmeier &amp;amp; Raymond (2018)&lt;/a&gt; for details, who note that any subsatellites orbiting the candidate Neptune-sized exomoon in the Kepler-1625 system would need to be smaller than Ceres to survive.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-02-18T19:56:55.530" LastActivityDate="2019-02-18T19:56:55.530" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29624" PostTypeId="1" AcceptedAnswerId="30054" CreationDate="2019-02-17T17:59:05.660" Score="5" ViewCount="220" Body="&lt;p&gt;Press-Schechter formalism allows one to universally quantify the number of massive dark matter halos in a given volume -- &lt;strong&gt;The Halo Mass Function&lt;/strong&gt;. The density contrast is given by the rms mass variance in Fourier space, which is the dependent on the linear power spectrum &lt;span class=&quot;math-container&quot;&gt;$P_{\rm lin}(k,z)$&lt;/span&gt; in a smooth window &lt;span class=&quot;math-container&quot;&gt;$\widetilde{W}(R;k)$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\sigma^{2}(R,z) &#xA;= \frac{1}{2\pi^{2}} &#xA;\int_{0}^{\infty} dk\ k^{2} P_{\rm lin}(k,z) \left| \widetilde{W}(k;R)\right|^{2}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The halo count can be scaled via observations in a defined window region of &lt;span class=&quot;math-container&quot;&gt;$R = 8\ h^{-1}\rm Mpc$&lt;/span&gt;. We can solve for this analytical/numerically and measure it from CMB fluctuations, given by the parameter &lt;span class=&quot;math-container&quot;&gt;$\sigma_{8}$&lt;/span&gt;. The power spectrum, as well as the halo mass function, is then scaled through the normalization of the analytical variance and the observed variance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is then, &lt;em&gt;why do we choose a smoothing radius of&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$R = 8\ h^{-1}\rm Mpc$&lt;/span&gt;? Is it arbitrarily chosen to compute the number variance in a cluster, or is their more too that? What prevents me from defining the window radius of &lt;span class=&quot;math-container&quot;&gt;$R = 6\ h^{-1}\rm Mpc$&lt;/span&gt; (just ignore the fact that the  WMAP and PLANCK results give &lt;span class=&quot;math-container&quot;&gt;$\sigma_{8}$&lt;/span&gt; in the radius of &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt;)?&lt;/p&gt;&#xA;" OwnerUserId="16260" LastEditorUserId="16260" LastEditDate="2019-02-17T18:35:49.953" LastActivityDate="2019-03-19T00:23:47.553" Title="Motivation for using $\sigma_{8}$?" Tags="&lt;cosmology&gt;&lt;cmb&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29625" PostTypeId="1" CreationDate="2019-02-17T18:22:49.030" Score="1" ViewCount="140" Body="&lt;p&gt;How do we calculate the mass of the universe taking into consideration dark matter? &#xA;Mass of visible matter can be computed by stellar method, but how do we calculate mass of dark matter which we don't see? Why do we assume that dark matter is more abundant than visible matter? &lt;/p&gt;&#xA;" OwnerUserId="25951" LastActivityDate="2019-07-18T03:03:30.327" Title="Mass of the universe" Tags="&lt;astrophysics&gt;&lt;cosmology&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29626" PostTypeId="1" AcceptedAnswerId="29647" CreationDate="2019-02-17T19:54:26.887" Score="0" ViewCount="230" Body="&lt;p&gt;How does Mars surface compare to Earth's surface? Mars is quite smaller than Earth, but most Earth's surface is covered with water. I thought making the math could be relatively easy but Earth isnt a sphere and I dont know about Mars? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the proportion of Mars' surface compared to Earth's surface uncovered of water?&lt;/p&gt;&#xA;" OwnerUserId="11562" LastActivityDate="2019-02-19T00:35:52.977" Title="What's the proportion of Mars' surface compared to Earth's surface uncovered of water?" Tags="&lt;earth&gt;&lt;mars&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29628" PostTypeId="2" ParentId="29625" CreationDate="2019-02-18T01:04:17.317" Score="-1" Body="&lt;p&gt;Check out the Wikipedia page on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Missing_baryon_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;missing baryon problem&lt;/a&gt;. It's how we compute the mass of ordinary matter. You can see there are methods of getting the mass of ordinary matter without adding them up. In other words, you can find ordinary matter even if they're invisible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is similar to how we find dark matter. Again, look at Wikipedia's article on dark matter, in particular the &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_matter#Observational_evidence&quot; rel=&quot;nofollow noreferrer&quot;&gt;observational evidence&lt;/a&gt; section. Ignore the more technical methods if you're new to the field and just look at the first few:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;With galaxy rotation curves, you know you need more matter to recover Kepler's laws. How much matter? You can tweak that number and therefore get an estimate for the dark matter needed. You can't prove that this dark matter isn't just invisible ordinary matter using this method, however.&lt;/li&gt;&#xA;&lt;li&gt;The same goes for velocity dispersions.&lt;/li&gt;&#xA;&lt;li&gt;Gravitational lensing can measure the total mass of an intervening cluster without knowing anything about the kinematics of the cluster.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you're wondering how we know dark matter can't just be invisible ordinary matter, again look at the Wikipedia article, in particular &lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_matter#Baryonic_matter&quot; rel=&quot;nofollow noreferrer&quot;&gt;this section&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Big bang nucleosynthesis constrains the amount of ordinary matter in the universe. If dark matter were just invisible ordinary matter, we also expect more helium, lithium, etc in the universe than we observe.&lt;/li&gt;&#xA;&lt;li&gt;Gravitational microlensing constrains the amount of MACHOs (massive compact halo objects, such as black holes and invisible planets) we can have in a large mass range, covering most of the plausible candidates.&lt;/li&gt;&#xA;&lt;li&gt;Data from the cosmic microwave background indicates a large fraction of matter does not react with photons.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="20348" LastActivityDate="2019-02-18T01:04:17.317" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29629" PostTypeId="1" AcceptedAnswerId="29636" CreationDate="2019-02-18T01:17:29.823" Score="2" ViewCount="129" Body="&lt;p&gt;I'm curious if there is a readily accessible tool available, or simply a reliable way to find out what the distance of the planets in the solar system were on a given date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I am trying to determine how far:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The Sun&lt;/li&gt;&#xA;&lt;li&gt;Mercury&lt;/li&gt;&#xA;&lt;li&gt;Venus&lt;/li&gt;&#xA;&lt;li&gt;Mars&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;were from Earth on May 3, 1954 ... how would I do that?&lt;/p&gt;&#xA;" OwnerUserId="25954" LastActivityDate="2019-02-18T11:13:25.180" Title="Historical Distance of Planets to Earth" Tags="&lt;the-sun&gt;&lt;distances&gt;&lt;mars&gt;&lt;venus&gt;&lt;mercury&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2019-02-20T05:57:12.807" ContentLicense="CC BY-SA 4.0" />
  <row Id="29630" PostTypeId="1" AcceptedAnswerId="29640" CreationDate="2019-02-18T03:59:46.080" Score="2" ViewCount="175" Body="&lt;p&gt;As the Moon loops around the Earth, on one side of the Earth it moves toward the Sun (let's call it downward side) and on the other side it moves away from the Sun (let's call it the upward side).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I imagine the Sun's gravity attraction accelerate the Moon on the downward side and slows it on the upward side. This should raise the Moon's apoapsis on the upward side and lowers its periapsis on the downward side, gradually putting the Moon on a collision course with Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet, this does not happen. I am curious to understand why. I suppose even a very very slim effect should build up upon billions of years, so is there a fault in my reasoning or some tidal effect correcting the Moon's trajectory?&lt;/p&gt;&#xA;" OwnerUserId="25955" LastActivityDate="2019-02-18T17:26:26.067" Title="How come the Sun does not make the Moon crash on the Earth?" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;satellite&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29631" PostTypeId="2" ParentId="856" CreationDate="2019-02-18T06:37:15.287" Score="13" Body="&lt;p&gt;It's &lt;strong&gt;theoretically&lt;/strong&gt; possible for moons of moons to exist in stable orbits, gravitationally speaking. However, as far as I'm aware, &lt;strong&gt;no natural moons of moons have ever been observed&lt;/strong&gt; as of yet. (Incidentally, there's a list of &lt;a href=&quot;https://qz.com/1419711/moonmoons-the-realish-term-for-a-moon-of-a-bigger-moon/&quot; rel=&quot;nofollow noreferrer&quot;&gt;proposed terms&lt;/a&gt; for moons of moons, the most popular of which seem to be &quot;submoon&quot; and &quot;moonmoon&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a &lt;a href=&quot;https://arxiv.org/pdf/1810.03304.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;2018 paper&lt;/a&gt; that explores this topic.  Their calculations show that several moons in the solar system are actually theoretically capable of hosting long-lived moons of moons, including Saturn's Titan and Iapetus, Jupiter's Callisto, and Earth's Moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.syfy.com/syfywire/can-moons-have-moons-yes-well-sometimes&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article&lt;/a&gt; has a good summary of the paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's true that the Hill sphere is an approximation, and that perturbations from other gravitational bodies and radiation can destabilize an orbit even within a body's Hill sphere. However, it's a decent estimate that if an orbit is within half of the Hill sphere's radius, then that orbit would be stable for on the order of billions of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper has some graphs showing that there are a few moons in the solar system that can have 10 km-scale submoons that are stable for at least the age of the solar system, under the influence of planet-moon-submoon tides:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://arxiv.org/pdf/1810.03304.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GMKyA.png&quot; alt=&quot;Figure 1. Moons of Moons – The parameter space in which the moon of a specified planet could host a long-lived submoon under the action of planet-moon-submoon tides&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors of the paper note however, that the above graphs do not take into account dynamic instabilities such as the Moon's unusual mass distributions, Sun-Earth perturbations, dynamical interactions between moons in multiple-moon systems, and dynamical scattering events between planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iapetus's equatorial ridge might hint at the existence of a past submoon, though. &lt;a href=&quot;http://dx.doi.org/10.1016/j.icarus.2011.05.031&quot; rel=&quot;nofollow noreferrer&quot;&gt;Levison et al. (2011)&lt;/a&gt; theorized that this ridge is from a submoon-generating collision, where the submoon was tidally pushed outwards and the debris belt was tidally pushed inward to create the ridge. Alternatively, &lt;a href=&quot;http://dx.doi.org/10.1029/2011JE004010&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dombard et al. (2012)&lt;/a&gt; theorized that the belt was caused by a submoon spirally inwards and being tidally shredded apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://apod.nasa.gov/apod/image/0502/iapetus_cassini_big.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jfzXt.png&quot; alt=&quot;Iapetus, with a ridge of mountains going around its equator&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why haven't we actually seen any submoons directly?&lt;/strong&gt; One reason might be that they're too small to be seen. It'd be pretty difficult to spot something 10 metres wide orbiting the Moon, let alone Titan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though the fact that we haven't seen any of these theoretically possible larger ones suggests that there may be some other reason why they're not commonplace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, it may be just too difficult for them to form in the first place, in the chaos of gas and dust circling a baby star. Tides also make the orbits of moons expand over time, so what is comfortable submoon real estate right now would not have been billions of years ago. For example, the Moon likely formed within several Earth radii from our planet, which would have been way too close to the planet for a submoon to be feasible. And perhaps it's just too rare and unlikely for a moon to capture an asteroid and have it become a submoon.&lt;/p&gt;&#xA;" OwnerUserId="13784" LastEditorUserId="13784" LastEditDate="2020-05-07T04:22:19.527" LastActivityDate="2020-05-07T04:22:19.527" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29632" PostTypeId="1" CreationDate="2019-02-18T07:13:36.077" Score="2" ViewCount="27" Body="&lt;p&gt;I have a fits image, and I need to identify an asteroid. What characteristics  do I look out for? &lt;/p&gt;&#xA;" OwnerUserId="25951" LastActivityDate="2019-02-18T07:13:36.077" Title="Asteroid identification in fits image" Tags="&lt;observational-astronomy&gt;&lt;asteroids&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29635" PostTypeId="2" ParentId="29609" CreationDate="2019-02-18T09:59:53.670" Score="6" Body="&lt;p&gt;Indeed the regions where sunspots  form tend to be equatorial, with sunspots concentrated near the regions at +/- 40° latitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sunspots are regions at temperature lower than that of the surrounding as for they are closing point of the sun magnetic field. At sunspots, the magnetic field is much higher than average and that hinders heat transport to the photosphere due to a reduced convection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the magnetic field of the sun is loosely speaking and as seen by far described as a dipole field (similar to that of Earth or of a magnetic bar), its local distribution is complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ecP9b.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ecP9b.jpg&quot; alt=&quot;Sun&amp;#39;s magnetic field&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture from &lt;a href=&quot;https://www.windows2universe.org/sun/sun_magnetic_field.html&quot; rel=&quot;noreferrer&quot;&gt;https://www.windows2universe.org/sun/sun_magnetic_field.html&lt;/a&gt; showing the magnetic field and spots forming at the loops closing areas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides the fact that the magnetic field flips every roughly 11 years, and keeps changing in meantime, sun, which is not rock solid, has a latitude depending rotational speed. This, superimposed to the dynamo mechanism for stars magnetic field generation, is responsible for &quot;carrying the local magnetic field  along with rotation&quot; and forces the field closing before the geometrical poles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the words of Wikipedia under the entry stellar magnetic field &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Stellar_magnetic_field&quot; rel=&quot;noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Stellar_magnetic_field&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;spots are regions of intense magnetic activity on the surface of a star. (On the Sun they are termed sunspots.) These form a visible component of magnetic flux tubes that are formed within a star's convection zone. Due to the differential rotation of the star, the tube becomes curled up and stretched, inhibiting convection and producing zones of lower than normal temperature&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't provide a clear picture of why the magnetic field behaves as such, but at least does tell that sunspots are a consequence of the magnetic field behaviour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Understanding the magnetic field of stars and even that of Earth is indeed ongoing research. A little more articulated answer is given by the above linked Wikipedia page, particularly by sections Field generation and Surface activity. &lt;/p&gt;&#xA;" OwnerUserId="17437" LastActivityDate="2019-02-18T09:59:53.670" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29636" PostTypeId="2" ParentId="29629" CreationDate="2019-02-18T11:01:16.947" Score="1" Body="&lt;h2&gt;Python&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I see that you are quite active in Stack Overflow but have no posts tagged with &lt;code&gt;python&lt;/code&gt; yet, so maybe it's time to give it a try!&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Run &lt;code&gt;pip install skyfield&lt;/code&gt; &lt;/li&gt;&#xA;&lt;li&gt;read the simple documentation: &lt;a href=&quot;https://rhodesmill.org/skyfield/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://rhodesmill.org/skyfield/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;calculate x, y, z positions in the solar system's coordinates with a single line of Python; e.g. `x, y, z = venus.at(time).position.km, or &lt;/li&gt;&#xA;&lt;li&gt;use &lt;code&gt;Topos&lt;/code&gt; to define a viewing location on Earth and the &lt;code&gt;.altaz()&lt;/code&gt; method to get the position of each in the sky.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;NASA JPL Horizions&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This is the same basic source of information that Skyfield uses.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Go to the Horizons website at: &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ssd.jpl.nasa.gov/horizons.cgi&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Follow the instructions I've written up in &lt;a href=&quot;https://space.stackexchange.com/a/25357/12102&quot;&gt;this answer&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;You can also read the documentation there to specify topographic coordinates and get positions in the sky, and you can do that &lt;em&gt;viewing from other planets and bodies as well!&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Have fun!&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-02-18T11:13:25.180" LastActivityDate="2019-02-18T11:13:25.180" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29637" PostTypeId="2" ParentId="29609" CreationDate="2019-02-18T11:23:58.083" Score="5" Body="&lt;p&gt;I would add some details to the answer above.   The interior of the sun is like a dipole or relatively consistent magnetic field that flips about once every 11 years, but it also does something interesting, it kind of goes from dipole to quadrupole before flipping.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See image and &lt;a href=&quot;https://astronomy.stackexchange.com/questions/973/how-can-only-1-magnetic-pole-of-the-sun-change-at-a-time&quot;&gt;answer to another stack exchange sunspot question.&lt;/a&gt;  Note, this image reflects the interior of the sun only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/70zxq.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/70zxq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That the Sun &quot;flips&quot; every 11 years suggests that it's much more dynamic that Earth's core which flips only after tens of thousands of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The entire sun is made up of plasma, so it's outer layers, not just it's inner layers can carry a charge and generate a magnetic field.   &lt;a href=&quot;https://www.windows2universe.org/?page=/sun/Solar_interior/Sun_layers/differential_rotation.html&quot; rel=&quot;noreferrer&quot;&gt;The Sun's outer layer flows faster at the equator&lt;/a&gt; (every 25 days) than it's poles (every 34 days) and that leads to rapid &lt;a href=&quot;https://www.windows2universe.org/sun/atmosphere/sunspot_form_jpg_image.html&quot; rel=&quot;noreferrer&quot;&gt;twisting of magnetic field lines&lt;/a&gt; while the Sun's interior rotates more as a single unit.   And we have a word for large twisting magnetic field lines on the Sun's surface.   We call them sunspots and they are visible dark spot on the sun, not visible to the naked eye, but observable with the right equipment.   Galileo is often credited with the first recorded observation of sunspots, but &lt;a href=&quot;http://hua.umf.maine.edu/China/astronomy/tianpage/0010H_sunspots6563w.html&quot; rel=&quot;noreferrer&quot;&gt;there are records of Chinese observation and recording&lt;/a&gt; of them going back to long before that, 1000 years before Christ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like the Sun, the Earth also undergoes a twisting of sorts, of it's magnetic field prior to a magnetic pole switch.   This generates something, not quite like sunspots, but there is a temporary complexity to Earth's magnetic field as well.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Geomagnetic_reversal#Causes&quot; rel=&quot;noreferrer&quot;&gt;See image here for Earth.  (having trouble embedding)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sunspots range from the Solar equator up to about 40 degrees North or South as noted in teh other answer.   During weak sunspot cycles, they mostly stay below 30 degrees, during strong ones, they mostly stay below 40 degrees however there are rare incidents of sunspots considerably above 40 degrees North or South, more common during strong sunspot periods, but still rare.  See image below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/My7ER.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/My7ER.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hudsonvalleygeologist.blogspot.com/2011/11/why-do-we-have-sunspots.html&quot; rel=&quot;noreferrer&quot;&gt;Source of image&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The reason for this&lt;/strong&gt; - finally getting to your question,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is basically that the twisting due to variations in rotation rate is more likely to happen in the 0-40 degree latitude range.  That said, actually understanding all the mechanisms of the Sun's magnetic field is a significant undertaking and not well understood.   This is perhaps an oversimplified answer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lucky for us, most sunspot driven coronal mass ejections or CMEs shoot above or below the solar equator which is near the ecliptic (tilted about 6 degrees off, so, OK, not all that near but near enough), but on average the majority of sunspots form with greater latitude than that, so many miss the Earth orbital region, shooting above or below the plane of the ecliptic or giving glancing blows.   A direct hit from a very large CME could cause havoc with our electrical grid and perhaps satellites as well.  Many of the larger sunspots during peak sunspot activity form further up in the 15-25 degree latitude range on the Sun.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2019-02-18T11:23:58.083" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29638" PostTypeId="2" ParentId="29614" CreationDate="2019-02-18T12:29:41.243" Score="1" Body="&lt;p&gt;You're asking two questions here but they are related, so that may be OK. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass ratio of the Sun is probably not 3:1 though, as the Sun has been turning hydrogen into Helium for nearly 5 billion years.   The interior of the Sun has a considerably higher helium ratio as a result, but the outer edges of the sun - which is what we see and read from spectroscopy, is close to the 3:1 mass ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is accurate &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/suncomp.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Sun overall is closer to about 2.6:1&lt;/a&gt;.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scientists can measure the ratio by observing spectral lines.  Stars make this easy because of how hot they are, but that information is limited to their surface.   We can't see the core of the sun.  Still more than half of the elements have been observed in the solar surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.atnf.csiro.au/outreach/education/senior/astrophysics/spectrographs.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;See on solar spectroscopy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or a &lt;a href=&quot;https://physics.stackexchange.com/questions/147513/how-do-scientists-estimate-elemental-abundance-in-the-universe&quot;&gt;related question from Physics Stack&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 25% helium by mass is confirmed both by spectroscopy and by quantum physics models based on the modeled expansion as the early Universe goes from hot to cold.   Helium is more stable than hydrogen, but, it requires an intermediate step in it's formation because deuterium has to form first.  Also, though I think this plays a smaller role, free neutrons decay and become protons which become Hydrogen.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More massive elements need hotter temperatures to form but the early universe went from hot (Quark soup), to less hot (formation of hadrons), to static - 92% Hydrogen, 8% helium by element or 75-25 by mass with tiny trace amounts of Lithium.   It stayed that way until the first stars and the formation of heavier elements.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a very simplified version.  You can visit the &lt;a href=&quot;https://physics.stackexchange.com/questions/199632/why-didnt-the-big-bang-create-heavy-elements?noredirect=1&amp;amp;lq=1&quot;&gt;stack exchange answer below for more details&lt;/a&gt; or &lt;a href=&quot;https://scienceblogs.com/startswithabang/2013/07/05/why-did-the-universe-start-off-with-hydrogen-helium-and-not-much-else&quot; rel=&quot;nofollow noreferrer&quot;&gt;this link here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both spectroscopy and quantum physics agree on the ratio, which is always a plus when the same answer is reached two different ways.   There is a lingering &lt;a href=&quot;https://phys.org/news/2017-02-universe-lithium-problem.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;lithium ratio problem&lt;/a&gt;, but even so, the 75%-25% ratio is pretty good.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2019-02-18T12:29:41.243" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29640" PostTypeId="2" ParentId="29630" CreationDate="2019-02-18T17:26:26.067" Score="5" Body="&lt;p&gt;The Sun's gravity does perturb the Moon's orbit but more subtly than you imagined.&#xA;The Moon's perigee and apogee migrate eastward in an 8.9 year cycle called &lt;a href=&quot;https://en.wikipedia.org/wiki/Apsidal_precession&quot; rel=&quot;noreferrer&quot;&gt;apsidal precession&lt;/a&gt;.&#xA;Also the plane of the lunar orbit, inclined about 5.1 degrees to Earth's orbit, shifts westward in an 18.6 year cycle called &lt;a href=&quot;https://en.wikipedia.org/wiki/Nodal_precession&quot; rel=&quot;noreferrer&quot;&gt;nodal precession&lt;/a&gt;.&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_theory&quot; rel=&quot;noreferrer&quot;&gt;Lunar theory&lt;/a&gt; also addresses various effects on shorter timescales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Moon were 4 times farther from the Earth (outside the Earth's &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_sphere&quot; rel=&quot;noreferrer&quot;&gt;Hill sphere&lt;/a&gt;), then the Sun would destabilize the lunar orbit.&#xA;Even so, the Moon would most likely drift into an independent orbit around the Sun, with little risk of collision with the Earth.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2019-02-18T17:26:26.067" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29641" PostTypeId="1" CreationDate="2019-02-18T19:12:41.777" Score="3" ViewCount="79" Body="&lt;p&gt;How does the orientation of stars axes of rotation and any planetary systems relate to the plane of the galaxy? Is there any relationship that has been detected, are they grouped from their parental nebulae. Do galactic magnetic fields have coherent orientations that would affect nebula growth and consequent stellar growth.&lt;/p&gt;&#xA;" OwnerUserId="7579" LastEditorUserId="7579" LastEditDate="2019-02-19T07:39:22.063" LastActivityDate="2019-06-26T14:50:30.700" Title="Star rotational axes and Solar system orientations" Tags="&lt;galaxy&gt;&lt;solar-system&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29642" PostTypeId="2" ParentId="29621" CreationDate="2019-02-18T19:13:41.533" Score="2" Body="&lt;p&gt;I would imagine that in most cases the sub satellite would be inside the instability limit of the planetary system in question and that such a scenario would be unstable. It would be interesting to model this with a simple 3-body system to see how the sub-satellite behaved over time (assuming a rigid body to start out). Someone else mentioned tidal perturbations and surely those would play a significant role too.&lt;/p&gt;&#xA;" OwnerUserId="25705" LastActivityDate="2019-02-18T19:13:41.533" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29643" PostTypeId="1" CreationDate="2019-02-18T19:30:07.010" Score="1" ViewCount="87" Body="&lt;p&gt;just want to try to clarify a few things about the nature of White Dwarfs and degenerate gases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if a white dwarf is help up by an electron degenerate gas, the electrons are unable to lose energy by emitting photons, correct? They basically are trapped at whatever energy level they occupy and that's that. The nuclei that make up the majority of the mass of the star are still constituents of a &quot;normal&quot; gas however, correct? This means that these nuclei and emit photons to lose kinetic energy? How does that work exactly? I understand that electron can emit photos inside an atom to drop to lower energy levels and that they can emit photons of any energy level if they are free (particularly if they're in a magnetic field, synchroton radiation, yeah?). To protons and neutrons do something similar? I ask because I never read anything about protons/nuclei emitting photons. If this basic picture is correct, is this the mechanism by which a white dwarf cools down over time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final bit, since most white dwarfs are quite hot, the peak wavelength is going to be in the UV. If I wanted to determine the percentage of such a star's luminosity that is given off in the optical, do I just need to integrate Planck's blackbody equation over the range of wavelengths I'm interested in and then compare that to the total luminosity given by &lt;span class=&quot;math-container&quot;&gt;$L = 4\pi R^2 \sigma T^4$&lt;/span&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and Cheers!&lt;/p&gt;&#xA;" OwnerUserId="25705" LastActivityDate="2019-02-18T21:32:32.287" Title="White Dwarf/Degenerate Gas Behaviour" Tags="&lt;white-dwarf&gt;&lt;degenerate-matter&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29644" PostTypeId="2" ParentId="29643" CreationDate="2019-02-18T21:32:32.287" Score="2" Body="&lt;p&gt;The main method by which heat is transferred in a white dwarf is thermal conduction. Degenerate electrons have very long mean free paths and can conduct heat excellently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;White dwarfs lose energy by emitting photons from their outer, non-degenerate surface layers. They may also emit neutrinos from their interiors when they are really young and hot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gamma rays and X-rays are of course emitted by nuclei. This works just the same as atomic emission - charged particles (protons) change energy levels within the nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most white dwarfs are &lt;em&gt;not&lt;/em&gt; hot, they are similar to the Sun or cooler. Hot ones are easier to see, but only young white dwarfs are hot. But yes, your scheme is approximately (in that white dwarfs are not blackbodies) correct.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-02-18T21:32:32.287" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29645" PostTypeId="1" AcceptedAnswerId="29648" CreationDate="2019-02-18T23:04:40.967" Score="0" ViewCount="597" Body="&lt;p&gt;All galaxies are moving away from each other (us) because of the expansion of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are they moving with same speed in each direction?&lt;/p&gt;&#xA;" OwnerUserId="25968" LastEditorDisplayName="user1569" LastEditDate="2019-02-19T10:16:25.717" LastActivityDate="2019-02-19T10:16:25.717" Title="Are galaxies moving in each direction with same speed?" Tags="&lt;observational-astronomy&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29647" PostTypeId="2" ParentId="29626" CreationDate="2019-02-19T00:35:52.977" Score="2" Body="&lt;p&gt;The surface area of Mars is 144798500 km&lt;sup&gt;2&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The land area of Earth is 148940000 km&lt;sup&gt;2&lt;/sup&gt; (figures from wikipedia)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the land area of Earth is slightly (about 3%) larger than the land area of Mars. However much of the Earth's surface is suitable for human habitation, but none of Mars is suitable for humans.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-02-19T00:35:52.977" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29648" PostTypeId="2" ParentId="29645" CreationDate="2019-02-19T01:11:54.663" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Are the galaxies moving towards each direction with same speed ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Short version : Locally, No.  On the cosmic scale, yes-ish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motions of galaxies are more complex than this questions suggests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start with the Milky Way.  From our point of view stationary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the Milky Way is part of a group of galaxies (&lt;a href=&quot;https://en.wikipedia.org/wiki/Local_Group&quot; rel=&quot;nofollow noreferrer&quot;&gt;the local group&lt;/a&gt;).  They essentially move in their mutual gravitational fields.  Even that field is complicated by the gravitational field of dark matter halos, which are invisible (unobservable directly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the local group is not the only group out there.  The local group is part of an even larger grouping known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Virgo_Supercluster&quot; rel=&quot;nofollow noreferrer&quot;&gt;Virgo Supercluster&lt;/a&gt;.  Again motions of these groups are complex and again dark matter plays a part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The local group contains 54 galaxies, and the super cluster contains something on the order of 100 such groups.  That's about 5400 galaxies in a fairly complex motion.  There's no one easy direction to spot here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the larger scale universe is thought governed by what is known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann%E2%80%93Lema%C3%AEtre%E2%80%93Robertson%E2%80%93Walker_metric&quot; rel=&quot;nofollow noreferrer&quot;&gt;FLRW metric&lt;/a&gt;, which (with dark matter, dark energy and some adjustments) fits the overall expansion of the universe we see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's only on that scale (essentially the entire universe excluding the nearest bits) that you see the kind of pattern you mean.  So the local bits are maybe 5400 galaxies, but the rest of the universe is very roughly two &lt;em&gt;trillion&lt;/em&gt; galaxies.  So our local part is on our scale vast and unimaginable, but on the scale that FLRW applies it's a microscopic dot lost in the averages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FLRW is a result of using General Relativity to model the expansion of the universe.  It does predict (when you match it to our universe's conditions) that from &lt;em&gt;any&lt;/em&gt; point everything else will be moving away from you (and we now think accelerating away).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an apparent contradiction between these two extreme (local group and universe away from that).  The resolution of this is that FLRW ignores local grouping and irregularities and assumes a nice continuous distribution (galaxies don't even feature on this grand scale).  On that local scale gravitational fields (currently) are large enough to overcome the expansion, hence our closest galaxy (Andromeda) is approaching us, not moving away.  But the further away you get, the larger the overall effects predicted by FLRW become and dominate &quot;normal&quot; gravitational effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FLRW does not say that they all move away at the same speed.  It's distance dependent - further away = faster moving.  So not the same speed, but from every viewpoint in the universe the same pattern of faster when further away.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2019-02-19T01:11:54.663" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29651" PostTypeId="1" AcceptedAnswerId="29657" CreationDate="2019-02-19T12:06:40.140" Score="3" ViewCount="235" Body="&lt;p&gt;I'm trying to simulate the Kepler-444 exoplanetary system using data from this &lt;a href=&quot;http://exoplanets.org/table&quot; rel=&quot;nofollow noreferrer&quot;&gt;website&lt;/a&gt;. The problem is that there is no argument of periapsis parameter for the Kepler-444 system, so while I think I'm getting the correct shape of the orbits, their periapsises and apoapsis are all aligned in the same direction, as you can see in the screen dump below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is how, or if, I can get the periapsis to be correctly aligned with the data that is available on this website for these planets (unfortunately I can't link to a query, but you can search Kepler-444 and the planets should pop up in the table)? The way I'm doing it now is that I take the semimajor axis and the eccentricity data from the website I linked to, and then I calculate the periapsis for each planet and then I set the x axis to be that value while I let y and z be zero. Once there, I use the vis-viva equation to calculate the velocity vectors for the planet, and then I iterate the simulation forwards in time and get the shapes of the orbits. With the solar system planets, I take the argument of periapsis and the inclination of the orbit and rotate the position and velocity vectors that I derived accordingly, and this way I end up with orbits that have the right shape and their periapsises are correctly aligned, too. Naturally I don't get the right position of a planet in its orbit at a given time using this approach, but for now that is not important; I just want the orbits to have the correct shape and to have their periapsises correctly aligned. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know very little about orbital mechanics, so any help would be much appreciated so that I can learn something and improve the quality of my simulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IwK4E.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IwK4E.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25978" LastEditorUserId="1891" LastEditDate="2019-02-19T13:07:33.320" LastActivityDate="2019-02-19T19:46:19.950" Title="How to calculate the argument of perihelion for exoplanets with incomplete orbital elements?" Tags="&lt;exoplanet&gt;&lt;orbital-mechanics&gt;&lt;orbital-elements&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29653" PostTypeId="1" CreationDate="2019-02-19T15:51:03.377" Score="1" ViewCount="97" Body="&lt;p&gt;I read that the New Horizon spacecraft passed by the Ultima Thule. I also got to know that it is a flat, snowman-shaped rock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is so special about it?&lt;/p&gt;&#xA;" OwnerUserId="25487" LastEditorUserId="1891" LastEditDate="2019-02-19T16:33:51.567" LastActivityDate="2019-02-20T10:11:30.190" Title="What is so special about the Ultima Thule passed by New Horizon spacecraft?" Tags="&lt;asteroids&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29654" PostTypeId="2" ParentId="29651" CreationDate="2019-02-19T16:06:57.320" Score="1" Body="&lt;h3&gt;Missing Orbital Parameters&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Six orbital parameters (seven if you include an epoch) are the minimum required to uniquely describe a simple two-body orbit like the type you are trying to simulate. The database you linked to contains entries for:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Semi-major Axis&lt;/li&gt;&#xA;&lt;li&gt;Eccentricity&lt;/li&gt;&#xA;&lt;li&gt;Argument of Periastron (aka argument of perigee, preiapsis)&lt;/li&gt;&#xA;&lt;li&gt;Inclination (missing some entries)&lt;/li&gt;&#xA;&lt;li&gt;Time of Periastron (also missing some entries)&lt;/li&gt;&#xA;&lt;li&gt;Right Ascension of the Ascending Node (aka big&lt;span class=&quot;math-container&quot;&gt;$\Omega$&lt;/span&gt; or just &lt;span class=&quot;math-container&quot;&gt;$\Omega$&lt;/span&gt; -- no entries)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It is the lack of values for item #6 that under-specifies the Kepler-144 orbits, as well as the missing values for the inclination and time of periastron. Without more information about the planet's orbits, there is no way to faithfully recreate their orbits without making up some values for the missing parameters. &lt;/p&gt;&#xA;" OwnerUserId="22897" LastActivityDate="2019-02-19T16:06:57.320" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29655" PostTypeId="1" AcceptedAnswerId="29658" CreationDate="2019-02-19T17:02:05.550" Score="3" ViewCount="162" Body="&lt;p&gt;According to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://space-facts.com/asteroid-belt/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Space facts&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Asteroid Belt is located in an area of space between the orbits of&#xA;  Mars and Jupiter. That places it between 2.2 and 3.2 astronomical&#xA;  units (AU) from the Sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But asteroids from the asteroid Velt like Pallas and Hygiea have an aphelion of&#xA;3.4119 AU and 3.5024 AU respectively. So my question is, which is the farthest distance a particular distant known-identified object from the asteroid belt can get from the sun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Updated since the question might seem unclear. I'm asking for a currently known object from the asteroid belt which gets closer to Jupiter.&lt;/p&gt;&#xA;" OwnerUserId="11562" LastEditorUserId="11562" LastEditDate="2019-02-19T17:52:05.503" LastActivityDate="2019-02-24T07:50:25.060" Title="Which is the farthest distance a particular distant known- identified object from the asteroid belt can get from the sun?" Tags="&lt;asteroid-belt&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29656" PostTypeId="1" AcceptedAnswerId="29659" CreationDate="2019-02-19T17:29:53.243" Score="3" ViewCount="185" Body="&lt;p&gt;Recently I bought a &lt;a href=&quot;https://www.meade.com/polaris-80mm-german-equatorial-refractor.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;telescope&lt;/a&gt; after a very good conversation with an astronomer friend of mine. The main goal is to see planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, Mars is so difficult and is far to find it in opposition. I really like the views of the Moon and there is where I spend most of my time observing. But I really want to go view the planets, but I seem a bit without an objective. I don't want only to see them at random; I want to do some kind of study using my observations. So, can you suggest something to do with my telescope to start some serious study?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a good base of Physics (I'm a computer engineer) and I really like Math, but I'm a complete amateur in astronomy.&lt;/p&gt;&#xA;" OwnerUserId="25980" LastEditorUserId="7982" LastEditDate="2019-02-20T00:08:05.360" LastActivityDate="2019-02-20T00:08:05.360" Title="What I can observe with a 80/900 refractor telescope? What can I study?" Tags="&lt;observational-astronomy&gt;&lt;telescope&gt;&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="29657" PostTypeId="2" ParentId="29651" CreationDate="2019-02-19T19:03:32.423" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/0801.2591&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ford et al. (2008)&lt;/a&gt; give the following relationship for eccentric transiting planets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{t_\mathrm{D}}{P} \simeq \frac{R_\ast}{\pi a \sqrt{1-e^2}}\sqrt{(1+r)^2 - b^2} \left(\frac{d_\mathrm{t}}{a}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$t_\mathrm{D}$&lt;/span&gt; is the total transit duration, &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; is the orbital period, &lt;span class=&quot;math-container&quot;&gt;$R_\ast$&lt;/span&gt; is the stellar radius, &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is the semimajor axis, &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt; is the eccentricity, &lt;span class=&quot;math-container&quot;&gt;$r \equiv R_\mathrm{p}/R_\ast$&lt;/span&gt; is the ratio of planetary radius &lt;span class=&quot;math-container&quot;&gt;$R_\mathrm{p}$&lt;/span&gt; to the stellar radius, &lt;span class=&quot;math-container&quot;&gt;$b \equiv d_\mathrm{t} \cos i / R_\ast$&lt;/span&gt; is the impact parameter, &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; is the inclination and &lt;span class=&quot;math-container&quot;&gt;$d_\mathrm{t} = a(1-e^2)/(1+e \cos \omega)$&lt;/span&gt; is the distance between the star and the planet during the transit. By inverting this you would be able to estimate &lt;span class=&quot;math-container&quot;&gt;$\cos \omega$&lt;/span&gt; from the parameters listed on the site, which unfortunately won't give you a unique value (and if you're unlucky, the various uncertainties may conspire to give you a value &lt;span class=&quot;math-container&quot;&gt;$\left|\cos \omega \right|&amp;gt;1$&lt;/span&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of Kepler-444, the website gives a link to the paper where the orbits were published, &lt;a href=&quot;https://iopscience.iop.org/article/10.1088/0004-637X/799/2/170/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;Campante et al. (2015)&lt;/a&gt;. Their table 4 tabulates both &lt;span class=&quot;math-container&quot;&gt;$e \cos \omega$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$e \sin \omega$&lt;/span&gt; for the planets. Unfortunately it is not possible to simply determine the value from these estimates without having access to the full distribution of samples, so this isn't entirely helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The good news is that the statistical significance of the non-zero eccentricities doesn't appear to be too high in this case, so you can probably treat the orbits as circular and not worry too much about it: tightly-packed systems like this usually have very low eccentricities otherwise they would become unstable.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-02-19T19:46:19.950" LastActivityDate="2019-02-19T19:46:19.950" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29658" PostTypeId="2" ParentId="29655" CreationDate="2019-02-19T23:34:39.527" Score="5" Body="&lt;p&gt;&lt;strong&gt;tl;dr:&lt;/strong&gt; The key part the question is the phrase&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...from the asteroid belt...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are lots of asteroids and only some of them are considered to be in &quot;the asteroid belt&quot; or to be &lt;em&gt;main-belt asteroids&lt;/em&gt;. As @zephyr's &lt;a href=&quot;https://astronomy.stackexchange.com/a/19973/7982&quot;&gt;answer&lt;/a&gt; to the question &lt;em&gt;Do astronomers generally agree that the distinction between comets and astroids is not so clear?&lt;/em&gt; points out, even the definition of what is or isn't an asteroid is under debate and review. To the question of which is the farthest a main belt asteroid can get from the Sun, &lt;em&gt;the answer depends on exactly how you draw your line.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also &lt;a href=&quot;https://en.wikipedia.org/wiki/Asteroid_belt#Families_and_groups&quot; rel=&quot;nofollow noreferrer&quot;&gt;families and groups&lt;/a&gt; of asteroids, and while these have visually clear clumps and clusters when plotted in certain ways, they may not have absolute universally accepted definitions of who's in and who's out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://en.wikipedia.org/wiki/File:AsteroidIncAu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Asteroid_osculating_vs_proper_elements.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; for some families within the main belt.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;While the OP&quot;s &lt;a href=&quot;https://space-facts.com/asteroid-belt/&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Asteroid Belt is located in an area of space between the orbits of Mars and Jupiter. That places it between 2.2 and 3.2 astronomical units (AU) from the Sun. The belt is about 1 AU thick. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;that wording seems awkward since &quot;between the orbits of between the orbits of Mars and Jupiter&quot; only really puts it between 1.4 and 5.5 AU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first sentence of Wikipedia's Asteroid belt says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The asteroid belt is the circumstellar disc in the Solar System &lt;strong&gt;located roughly&lt;/strong&gt; between the orbits of the planets Mars and Jupiter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;repeat, &quot;&lt;strong&gt;roughly&lt;/strong&gt;&quot; and it goes on to say&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The asteroid belt is also termed the main asteroid belt or main belt to distinguish it from other asteroid populations in the Solar System such as near-Earth asteroids and trojan asteroids.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The article also includes the figure shown below, showing that there are plenty of other groups of asteroids besides those designated as &quot;main belt&quot;. I am not sure if there is a 100% absolutely IAU-official definition that you can apply to any set of orbital parameters that will return Yes or No to the question &quot;is this one &lt;em&gt;main belt?&lt;/em&gt;&quot; but in general it looks like asteroids with a semimajor axis beyond 3.3 AU, or a high eccentricity, or a high inclination would not be called &quot;main belt&quot; by most astronomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @PM2Ring points out &lt;a href=&quot;https://en.wikipedia.org/wiki/Kirkwood_gap&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kirkwood Gaps&lt;/a&gt; delineate statistical sub-groupings of the main belt.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A Kirkwood gap is a gap or dip in the distribution of the semi-major axes (or equivalently of the orbital periods) of the orbits of main-belt asteroids. They correspond to the locations of orbital resonances with Jupiter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can see that while there is a minimum around 3.3 AU due to the 2:1 resonance, there are &quot;green&quot; stragglers extending to the right that continue on past the arbitrary 3.5 AU edge of this plot of &quot;main belt asteroids&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x1GKu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x1GKu.png&quot; alt=&quot;Kirkwood Gaps&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/File:Kirkwood_Gaps.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oGrWw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oGrWw.png&quot; alt=&quot;Kirkwood Gaps&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/e/ed/Main_belt_i_vs_a.png/983px-Main_belt_i_vs_a.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0Tma9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0Tma9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Screen shot from Scott Manley's video &lt;a href=&quot;https://www.youtube.com/watch?v=Jdb-DPjHEuw&quot; rel=&quot;nofollow noreferrer&quot;&gt;What Rock Star Brian May Discovered About Interplanetary Dust&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives a better idea that there are groups and stragglers than the image in the OP's link:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PeQQH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PeQQH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://space-facts.com/asteroid-belt/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-02-24T07:50:25.060" LastActivityDate="2019-02-24T07:50:25.060" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="29659" PostTypeId="2" ParentId="29656" CreationDate="2019-02-20T00:05:01.403" Score="6" Body="&lt;p&gt;There is a wonderful intersection of Physics, Mathematics, Applied Math (numerical analysis) and Observational Astronomy in a systematic analysis of the movement of four bright &lt;a href=&quot;https://en.wikipedia.org/wiki/Galilean_moons&quot; rel=&quot;noreferrer&quot;&gt;Galilean moons of Jupiter&lt;/a&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are easy to see in a small telescope, they have well defined orbits around Jupiter, and they have some trick up their sleeves that you can analyze.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an exercise you can do right now. When Jupiter is &lt;a href=&quot;https://en.wikipedia.org/wiki/Quadrature_(astronomy)&quot; rel=&quot;noreferrer&quot;&gt;anywhere near quadrature&lt;/a&gt;, it's shadow extends behind it &lt;em&gt;by an angle of almost 11 degrees&lt;/em&gt;, so the moons will often be eclipsed. That means instead of appearing or disappearing into Jupiter's overwhelming brightness, they can &quot;pop&quot; in or out of visibility as they pass out of or into Jupiter's shadow.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                a (km)&#xA;Io             421,800&#xA;Europa         671,100&#xA;Ganymede     1,070,400&#xA;Castillo     1,882,700&#xA;&#xA;Jupiter radius  71,492&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You can predict when these happen, and then watch to see them and confirm your calculations. You can start timing them to get more accurate information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a mathematical perspective you can check the periods of their orbits (from your measurements, or from a table) to see if there are any surprising numerical relationships!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can check the inclination of their orbital plane to see if the eclipses always happen or sometimes don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read the answers to &lt;a href=&quot;https://astronomy.stackexchange.com/q/23477/7982&quot;&gt;When will the next series of mutual eclipses of Jupiter's moons begin?&lt;/a&gt; and start planning for 2021 when these begin again!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can learn to do &lt;a href=&quot;https://space.stackexchange.com/a/23409/12102&quot;&gt;numerical integration of solar system bodies&lt;/a&gt; to simulate their orbits, or use the Python package &lt;a href=&quot;https://rhodesmill.org/skyfield/&quot; rel=&quot;noreferrer&quot;&gt;Skyfield&lt;/a&gt; or the &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;noreferrer&quot;&gt;JPL Horizons&lt;/a&gt; website for high accuracy positional information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These can also help you calculate the eclipse times by Jupiter, and the mutual eclipses of Jupiter's moons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you can try to measure the speed of light by seeing that the eclipses come sooner or later than average when Jupiter is closer or farther than average from the Earth. This is one of the early ways that the finite speed of light was verified.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-02-20T00:05:01.403" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29661" PostTypeId="1" AcceptedAnswerId="29662" CreationDate="2019-02-20T06:55:36.120" Score="4" ViewCount="477" Body="&lt;p&gt;I am an amateur in astronomy, but recently I wonder if it is possible to see (in a photo, maybe, if not in a telescope) Halley's Comet when its not close to earth. If there is such photo, how I can find it. Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="25983" LastEditorUserId="7982" LastEditDate="2019-02-20T07:32:45.577" LastActivityDate="2019-02-20T08:14:03.873" Title="Halley's Comet current location in sky and visibility?" Tags="&lt;comets&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29662" PostTypeId="2" ParentId="29661" CreationDate="2019-02-20T07:19:47.823" Score="4" Body="&lt;p&gt;In case you happen to have a &lt;a href=&quot;https://en.wikipedia.org/wiki/Very_Large_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;very large telescope&lt;/a&gt;, here's where to look for Halley's Comet in 2019. I got these from JPL's &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;Horizons&lt;/a&gt; ephemeris.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wZ4qJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wZ4qJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;date (UTC)    h  m  s        d  m  s        T-mag    N-mag&#xA;2019-Jan-01   08 26 12.95   +01 38 11.0     25.48    28.99&#xA;2019-Feb-01   08 22 37.55   +01 48 26.3     25.48    28.97&#xA;2019-Mar-01   08 19 35.31   +02 04 26.5     25.49    28.99&#xA;2019-Apr-01   08 17 21.46   +02 24 37.7     25.52    29.04&#xA;2019-May-01   08 16 49.80   +02 41 21.1     25.55    29.07&#xA;2019-Jun-01   08 17 59.21   +02 51 29.8     25.58    29.10&#xA;2019-Jul-01   08 20 21.84   +02 52 27.8     25.60    29.10&#xA;2019-Aug-01   08 23 27.41   +02 44 30.4     25.61    29.10&#xA;2019-Sep-01   08 26 24.57   +02 29 35.3     25.60    29.10&#xA;2019-Oct-01   08 28 23.88   +02 12 04.1     25.58    29.10&#xA;2019-Nov-01   08 28 56.98   +01 55 29.1     25.56    29.08&#xA;2019-Dec-01   08 27 45.99   +01 45 32.8     25.53    29.04&#xA;2020-Jan-01   08 24 59.69   +01 45 05.9     25.51    29.00&#xA;&#xA;T-mag N-mag =&#xA;   Comet's approximate apparent visual total magnitude (&quot;T-mag&quot;) and nuclear&#xA;   magnitude (&quot;N-mag&quot;) by following standard IAU definitions:&#xA;    T-mag =  M1 + 5*log10(delta) + k1*log10(r)&#xA;    N-mag =  M2 + 5*log10(delta) + k2*log10(r) + phcof*beta&#xA;   Units: MAGNITUDES&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Halley%27s_Comet#History&quot; rel=&quot;nofollow noreferrer&quot;&gt;Halley's Comet; History&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;After 1986&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Grainy, white-on-black image showing Halley as a barely distinguishable black dot&#xA;  Halley's Comet observed in 2003 at 28 AU from the Sun&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;On 12 February 1991, at a distance of 14.4 AU (2.15×109 km) from the Sun, Halley displayed an outburst that lasted for several months, releasing a cloud of dust 300,000 km across.&lt;sup&gt;49&lt;/sup&gt; The outburst likely started in December 1990, and then the comet brightened from magnitude 24.3 to magnitude 18.9.&lt;sup&gt;112&lt;/sup&gt; &lt;strong&gt;Halley was most recently observed in 2003 by three of the Very Large Telescopes at Paranal, Chile, when Halley's magnitude was 28.2. The telescopes observed Halley, at the faintest and farthest any comet has ever been imaged, in order to verify a method for finding very faint trans-Neptunian objects.&lt;sup&gt;9&lt;/sup&gt; Astronomers are now able to observe the comet at any point in its orbit.&lt;sup&gt;9&lt;/sup&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;sup&gt;9&lt;/sup&gt;&lt;a href=&quot;https://www.eso.org/public/news/eso0328&quot; rel=&quot;nofollow noreferrer&quot;&gt;New Image of Comet Halley in the Cold&quot;.&lt;/a&gt; European Southern Observatory. 1 September 2003. Retrieved 26 February 2018.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;sup&gt;49&lt;/sup&gt;J. C. Brandt. &quot;&lt;a href=&quot;http://www.mhest.com/spotlight/pluto/pdf/Sample_Halley&amp;#39;sComet.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;McGraw−Hill AccessScience: Halley's Comet&lt;/a&gt;&quot; (PDF). McGraw-Hill. Retrieved 27 November 2009.)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;sup&gt;112&lt;/sup&gt;D. Prialnik; A. Bar-Nun (1992). &quot;Crystallization of amorphous ice as the cause of Comet P/Halley's outburst at 14 AU&quot;. Astronomy and Astrophysics. 258 (2): L9–L12. &lt;a href=&quot;http://adsabs.harvard.edu/abs/1992A&amp;amp;A...258L...9P&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bibcode:1992A&amp;amp;A...258L...9P&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can view Halley's Comet's orbit in 3D here: &lt;a href=&quot;https://ssd.jpl.nasa.gov/sbdb.cgi?sstr=1P;orb=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ssd.jpl.nasa.gov/sbdb.cgi?sstr=1P;orb=1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KpBfi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KpBfi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;caption:&lt;/strong&gt; Seventeen years after the last passage of Comet Halley, the ESO Very Large Telescope at Paranal (Chile) has captured a unique image of this famous object as it cruises through the outer solar system. It is completely inactive in this cold environment. No other comet has ever been observed this far - 4200 million km from the Sun - or that faint - nearly 1000 million times fainter than what can be perceived with the unaided eye. This observation is a byproduct of a dedicated search &lt;a href=&quot;https://www.eso.org/public/news/eso0328/#1&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt; for small Trans-Neptunian Objects, a population of icy bodies of which more than 600 have been found during the past decade.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.eso.org/public/news/eso0328/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A record observation&#xA;  At 28.06 AU heliocentric distance (1 AU = 149,600,000 km - the mean distance between the Earth and the Sun), this is by far the most distant observation ever made of a comet &lt;a href=&quot;https://i.stack.imgur.com/KpBfi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;. It is also the faintest comet ever detected (by a factor of about 5); the previous record, magnitude 26.5, was co-held by comet Halley at 18.8 AU (with the ESO New Technology Telescope in 1994) and Comet Sanguin at 8.5 AU (with the Keck II telescope in 1997).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Interestingly, when Comet Halley reaches its largest distance from the Sun in December 2023, about 35 AU, it will only be 2.5 times fainter than it is now.&lt;/strong&gt; The comet would still have been detected within the present exposure time. &lt;strong&gt;This means that with the VLT, for the first time in the long history of this comet, the astronomers now possess the means to observe it at any point in its 76-year orbit!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-02-20T08:14:03.873" LastActivityDate="2019-02-20T08:14:03.873" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29663" PostTypeId="1" AcceptedAnswerId="29670" CreationDate="2019-02-20T07:27:53.393" Score="4" ViewCount="314" Body="&lt;p&gt;When researching for &lt;a href=&quot;https://astronomy.stackexchange.com/a/29662/7982&quot;&gt;this answer&lt;/a&gt; I was looking at Space.com's 2011 article &lt;a href=&quot;https://www.space.com/11552-photos-halleys-comet-images-astronomy.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Photos of Halley's Comet Through History&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw the image below, and read the caption.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h2&gt;Halley's Comet in 1910&lt;/h2&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9AR3o.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9AR3o.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;NASA/JPL&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Halley's Comet as photographed May 13, 1910, by a wide-angle camera at Lowell Observatory, Flagstaff, Ariz., during the comet's last appearance. A streak across the comet near the coma is a meteor trail, and not a scratch on the negative. Streaks at the bottom right are the city lights of Flagstaff Bright spot above the city lights is the planet Venus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that 109 years ago astrophotography wasn't what is is today, but the combination of the overexposed venus, the coincident meteor trails, the non-uniform star trails blending in with the comet's tail breakup along with the unlikely looking &quot;smoke trail&quot; from the upper side of the comet's tail make this photo look quite unusual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What's going on with that second tail thing above the main tail? The shape looks so unnatural, it seems to have a parallel extent rather than radiate from the coma. What caused this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gfjbb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gfjbb.jpg&quot; alt=&quot;between 3:27 and 4:31 a.m. M.S.T. May 13, 1910, Henry L. Giclas, Lowell Observatory&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;bonus points&quot; for identifying the photo in some catalog, or in a publication.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-02-27T06:31:51.717" LastActivityDate="2019-02-27T16:38:41.143" Title="What's going on in this photo of Halley's comet's tail?" Tags="&lt;photography&gt;&lt;comets&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29664" PostTypeId="2" ParentId="29653" CreationDate="2019-02-20T10:11:30.190" Score="2" Body="&lt;ul&gt;&#xA;&lt;li&gt;Ultima Thule is the first small Kuiper Belt Object (KBO) we have visited. For the first time, we can see a KBO as more than a tiny point of light in a telescope.&lt;/li&gt;&#xA;&lt;li&gt;it seems to be flat, rather than roughly round. This is really unusual (we've never seen that shape before in planets, moons, asteroids etc.).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="361" LastActivityDate="2019-02-20T10:11:30.190" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29665" PostTypeId="1" AcceptedAnswerId="29684" CreationDate="2019-02-20T10:15:12.330" Score="3" ViewCount="203" Body="&lt;p&gt;I do wonder, do interstellar asteroids eventually stop at one point in space after they gradually decelerate (or) even do they decelerate? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though there is no air like on earth and thus asteroids will not be affected by frictional forces, do they have friction with gravitational forces against their trajectories?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is so, they might eventually stop. Are there any known objects like that? &lt;/p&gt;&#xA;" OwnerUserId="23665" LastEditorUserId="23665" LastEditDate="2019-02-21T04:26:41.363" LastActivityDate="2019-02-25T07:57:23.497" Title="Do interstellar asteroids decelerate and eventually stop?" Tags="&lt;astrophysics&gt;&lt;asteroids&gt;&lt;fundamental-astronomy&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29667" PostTypeId="2" ParentId="29665" CreationDate="2019-02-20T14:10:10.897" Score="3" Body="&lt;p&gt;No. Gravity does not cause &quot;friction&quot;. A object will not stop unless it hits something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your comment clarifies that you mean &quot;interstellar bodies&quot; and not &quot;asteroids&quot;. Newton's first law of motion is &quot;A body will continue in a straight line and a constant speed, unless acted on by an external force&quot;. It won't decelerate in deep space where there is no friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When an interstellar body happens to approach a star it will start to fall towards it, and speed up. If it doesn't hit the star it will pass by and slow down as it moves up from the star. The velocity it loses as it moves away will be exactly the same as the velocity it gained while falling towards the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity is a conservative force, because the total energy (gravitational + Kinetic) remains constant. No energy is lost, and so there is no change in speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be aware that there is no such thing as &quot;not moving&quot; in the absolute sense.  The only words that you can use are &quot;not moving relative to something&quot;. When I say &quot;the car is not moving&quot;, what I mean is &quot;The car is not moving relative to the ground&quot;. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-02-20T14:10:10.897" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="29668" PostTypeId="2" ParentId="20122" CreationDate="2019-02-20T14:45:39.283" Score="6" Body="&lt;p&gt;The distance between Sirius A and B is between 8 and 31.5 AU and even when Sirius A becomes a red giant it  will be still above 6 AU. Such distance is too large and does not allow Sirius B to accrete significant mass, almost all mass lost by Sirius A as a red giant and later AGB will escape into space. Sirius B may become a recurrent nova due to some accretion, but will not gain enough mass to explode as supernova, it will hardly gain even 0.05-0.1 solar masses.&lt;/p&gt;&#xA;" OwnerUserId="25986" LastActivityDate="2019-02-20T14:45:39.283" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29669" PostTypeId="1" AcceptedAnswerId="29671" CreationDate="2019-02-20T14:57:45.923" Score="3" ViewCount="128" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Centaur_(minor_planet)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Centaur minor planet&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Centaurs are small Solar System bodies with a semi-major axis between those of the outer planets. They generally have unstable orbits because they cross or have crossed the orbits of one or more of the giant planets&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of all the known ones which I've found and I've seen data, all of them overlaps the orbit of Saturn, with the exception of  (514107) 2015 BZ509 which doesnt overlap the orbit of Saturn but the orbit of Jupiter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, are there known centaurs which dont overlap the orbits of Jupiter and Saturn but which orbit between those 2 orbits? Or does the definition of Centaur mean they have to cross orbits and an asteroid orbiting between those 2 orbits would be called another name if exists?&lt;/p&gt;&#xA;" OwnerUserId="11562" LastEditorDisplayName="user24157" LastEditDate="2019-02-20T18:05:22.647" LastActivityDate="2019-02-20T18:05:22.647" Title="Are there known centaurs which orbit between Jupiter and Saturn without crossing their orbits?" Tags="&lt;jupiter&gt;&lt;saturn&gt;&lt;centaur&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29670" PostTypeId="2" ParentId="29663" CreationDate="2019-02-20T15:33:48.310" Score="6" Body="&lt;p&gt;Negatives taken with various intstruments at Lowell Observatory on the morning of 1910-05-13 appear in a &lt;a href=&quot;https://ntrs.nasa.gov/search.jsp?R=19820006143&quot; rel=&quot;nofollow noreferrer&quot;&gt;1981 conference paper by Henry L Giclas&lt;/a&gt;.&#xA;The wide-angle images were taken with cameras piggyback on a telescope tracking the comet nucleus.&#xA;These exposures are naturally deeper in the center than at the edges; modern astrophotographers take flat fields to correct for this.&#xA;The stars in the upper right of the image in question were low enough to the horizon to be obscured for part of the exposure, making their trails shorter.&#xA;To make this more intuitive, rotate the image so that the tail points to the right and upward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Comet_tail&quot; rel=&quot;nofollow noreferrer&quot;&gt;Comet tails&lt;/a&gt; generally have two parts: a smooth, curved dust tail driven by steady radiation pressure; and a straight, blotchy plasma tail (also called an ion tail or gas tail) driven by the fluctuating solar wind.&#xA;A coronal mass ejection can severely disrupt the latter, as the &lt;a href=&quot;https://www.nasa.gov/mission_pages/stereo/news/encke.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;STEREO spacecraft observed with Comet Encke in 2007&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 1910 image in question may show a tail disconnection event in progress. If so, the highlighted feature is the old plasma tail drifting away from the nucleus, and the new plasma tail is not yet large enough to be distinct.&#xA;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1914ApJ....39..373B&quot; rel=&quot;nofollow noreferrer&quot;&gt;Barnard 1914&lt;/a&gt; includes &lt;a href=&quot;http://adsbit.harvard.edu//full/1914ApJ....39..373B/0000380P008.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;three photos&lt;/a&gt; of such an event on June 6-7, 1910.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-02-27T16:38:41.143" LastActivityDate="2019-02-27T16:38:41.143" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29671" PostTypeId="2" ParentId="29669" CreationDate="2019-02-20T17:22:55.837" Score="6" Body="&lt;p&gt;The &lt;a href=&quot;https://ssd.jpl.nasa.gov/sbdb_query.cgi&quot; rel=&quot;noreferrer&quot;&gt;JPL Small Body Database&lt;/a&gt; lists 5 asteroids and 5 comets with q &gt; 4.56 au and Q &amp;lt; 9.04 au.&#xA;Two of those asteroid orbits have large uncertainties due to few observations and may not actually meet your criteria. These are fairly well determined:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Asteroid &lt;a href=&quot;https://minorplanetcenter.net/db_search/show_object?object_id=494219&quot; rel=&quot;noreferrer&quot;&gt;494219&lt;/a&gt; has a highly inclined 14-year orbit which occasionally brings it within 2 au of Jupiter. It was discovered in 2012.&lt;/li&gt;&#xA;&lt;li&gt;Asteroid &lt;a href=&quot;https://minorplanetcenter.net/db_search/show_object?object_id=K00GD7M&quot; rel=&quot;noreferrer&quot;&gt;2000 GM137&lt;/a&gt; has a 22-year orbit which occasionally passes within 2 au of Saturn. It was discovered in 2000, unobserved for a few years, and recovered in 2014.&lt;/li&gt;&#xA;&lt;li&gt;Comet &lt;a href=&quot;https://en.wikipedia.org/wiki/29P/Schwassmann%E2%80%93Wachmann&quot; rel=&quot;noreferrer&quot;&gt;29P/Schwassmann-Wachmann 1&lt;/a&gt; was discovered in 1927, has a nearly circular 15-year orbit, and is known for frequent outbursts.&lt;/li&gt;&#xA;&lt;li&gt;Comet &lt;a href=&quot;https://en.wikipedia.org/wiki/39P/Oterma&quot; rel=&quot;noreferrer&quot;&gt;39P/Oterma&lt;/a&gt; was discovered in 1943 on an 8-year orbit. A 1963 encounter with Jupiter switched it to its present 20-year orbit, where it became dormant and was lost until 2001. Numerical simulations indicate a prior Jupiter encounter in 1937; it was likely also a Centaur before that.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="616" LastActivityDate="2019-02-20T17:22:55.837" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29672" PostTypeId="1" CreationDate="2019-02-20T19:20:37.390" Score="2" ViewCount="448" Body="&lt;p&gt;I was in Cuba from January 15th to the 22nd this year and there was a quarter moon. If I use a compass to explain the orientation of the quarter moon, it started in the north position around 4 pm. By 7 pm, it was in the east position, and by 11 pm, it was in the south position. Normally the quarter moon would start in the east and end in the east as it travels across the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please explain this phenomenon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="25987" LastEditorUserId="616" LastEditDate="2019-12-10T01:46:23.450" LastActivityDate="2019-12-10T01:46:23.450" Title="The movement of the moon" Tags="&lt;the-moon&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29673" PostTypeId="1" AcceptedAnswerId="29683" CreationDate="2019-02-21T02:37:19.293" Score="8" ViewCount="392" Body="&lt;p&gt;Suppose a very unusually small back hole with the mass of about, lets say, 1.5 solar masses(hawking radiation should destroy a lot of mass let's assume). It thus would have a Roche limit smaller than larger black holes, and the Roche limit should be outside of the black hole instead of inside it, like some supermassive black holes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, let's take a super massive neutron star, perhaps even with quark matter at its center. Even though this may not exist, let's say that is 3 solar masses by taking matter from a nearby red giant rather slowly. This would mean that it thus has a larger Roche limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, let's fit the two together in a binary system. Move the black hole inwards so that it is in the Roche limit of the neutron star, but its Roche limit is not affecting the neutron star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is that, since a black hole can't physically be ripped apart by gravity due to it being a hole in space time itself, can it's structure be altered in any way by the gravitational pull of the neutron star? Like from a sphere to an oval or something like that? Can it separate into two smaller ones due to the gravity of the neutron star? In the system, who would win the gravity war? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume that the black hole has angular momentum and change, giving it a magnetic field. Then make the neutron star a magnetar of the strongest kind, and place them in the same binary position as I described above. Would their magnetic fields have an effect on any of the bodies that are generating them? Would their magnetic fields have any effect on the other object's magnetic field?&lt;/p&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="5278" LastEditDate="2019-02-21T22:14:20.980" LastActivityDate="2019-02-21T22:14:20.980" Title="Roche Zone of a black hole vs Roche Zone of Neutron Star" Tags="&lt;black-hole&gt;&lt;neutron-star&gt;&lt;magnetic-field&gt;&lt;roche-limit&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29674" PostTypeId="5" CreationDate="2019-02-21T04:01:04.623" Score="0" Body="&lt;p&gt;The night sky is the visible realm during the night when a viewer from Earth can see various objects such as Planets, Stars, Galaxies, and many other objects.&lt;/p&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-02-21T10:56:01.673" LastActivityDate="2019-02-21T10:56:01.673" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29675" PostTypeId="4" CreationDate="2019-02-21T04:01:04.623" Score="0" Body="Questions regarding the visible night sky during nighttime from Earth." OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-02-21T08:14:41.107" LastActivityDate="2019-02-21T08:14:41.107" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29676" PostTypeId="5" CreationDate="2019-02-21T04:03:07.130" Score="0" Body="" OwnerUserId="25988" LastEditorUserId="6" LastEditDate="2019-02-21T14:43:17.023" LastActivityDate="2019-02-21T14:43:17.023" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29677" PostTypeId="4" CreationDate="2019-02-21T04:03:07.130" Score="0" Body="Questions on time, the indefinite progress of events in the Universe." OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-02-21T10:56:17.940" LastActivityDate="2019-02-21T10:56:17.940" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29678" PostTypeId="5" CreationDate="2019-02-21T04:05:06.363" Score="0" Body="" OwnerUserId="25988" LastEditorUserId="2153" LastEditDate="2019-02-21T14:44:08.917" LastActivityDate="2019-02-21T14:44:08.917" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29679" PostTypeId="4" CreationDate="2019-02-21T04:05:06.363" Score="0" Body="Questions about inducing chemistry, physics, and many more branches of science into astronomy." OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-02-21T10:55:45.180" LastActivityDate="2019-02-21T10:55:45.180" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29681" PostTypeId="2" ParentId="29672" CreationDate="2019-02-21T08:09:23.870" Score="5" Body="&lt;p&gt;I think there might have been a problem with your compass readings, or &quot;in the north&quot; really means &lt;em&gt;just a little north of east&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cuba is in the northern hemisphere at around 20 to 23°N (tropic of Cancer) and so the Sun can be directly over head and the Moon with an additional 5 degrees inclination with respect to the ecliptic can go several degrees North. (see &lt;a href=&quot;https://astronomy.stackexchange.com/a/27750/7982&quot;&gt;this answer&lt;/a&gt; for a bit about the Moon's motion).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;it started in the north position around 4 pm. &lt;/li&gt;&#xA;  &lt;li&gt;By 7pm, it was in the east position, &lt;/li&gt;&#xA;  &lt;li&gt;and by 11 pm, it was in the south position. &lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think there might have been something affecting your compass. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetic_declination&quot; rel=&quot;nofollow noreferrer&quot;&gt;Magnetic declination&lt;/a&gt; there is about 6-8 degrees, that might have had a tiny effect, but you might look for problems like &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;iron or other ferromagnetic material, &lt;/li&gt;&#xA;&lt;li&gt;mechanical compass from the &lt;a href=&quot;https://outdoors.stackexchange.com/q/21360/12619&quot;&gt;wrong hemisphere&lt;/a&gt; sticking due to dip angle&lt;/li&gt;&#xA;&lt;li&gt;or if it was an electronic compass, incorrect &lt;a href=&quot;https://electronics.stackexchange.com/q/249086/102305&quot;&gt;degaussing/calibration&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The other possibility is that by &quot;in the north&quot; you really mean only &quot;slightly north of east&quot; or &quot;east-north-east&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I wrote a short script in Python for your trip, and the three lines in the plot represent January 15 (lowest, blue), 17, and 19 from Santiago de Cuba, the southernmost area of Cuba. The moon does get to the zenith and perhaps a tiny bit past it (towards North) but I don't think that would be visually noticeable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Large black dots represent 7PM local time, and each dot to the west is one more hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RSVlk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RSVlk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some Python:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;from skyfield.api import Loader, Topos&#xA;&#xA;halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]&#xA;degs, rads        = 180/pi, pi/180&#xA;&#xA;load  = Loader('~/Documents/fishing/SkyData')  # single instance for big files&#xA;ts    = load.timescale()&#xA;de421 = load('de421.bsp')&#xA;earth = de421['earth']&#xA;moon  = de421['moon']&#xA;&#xA;Santiago_de_Cuba = earth + Topos(latitude_degrees  = 20.019833,&#xA;                                 longitude_degrees = -75.813917,&#xA;                                 elevation_m       = 10.)&#xA;&#xA;hours = np.arange(16, 23.1, 0.5) + 5&#xA;&#xA;days   = (15, 17, 19, 21)&#xA;&#xA;altazs, lines, linez= [], [], []&#xA;for day in days:&#xA;    times = ts.utc(2019, 1, day, hours)&#xA;    alt, az, d = (Santiago_de_Cuba).at(times).observe(moon).apparent().altaz()&#xA;    alt, az    = [thing.degrees for thing in (alt, az)]&#xA;    alt[alt&amp;lt;0] = np.nan&#xA;    altazs.append((alt, az))&#xA;    r = (1 - alt/90.)&#xA;    theta = rads * az&#xA;    lines.append((r, theta))&#xA;    x, y = [r*f(theta) for f in (np.sin, np.cos)]&#xA;    linez.append((x, y))&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;    plt.subplot(2, 1, 1)&#xA;    for alt, az in altazs:&#xA;        plt.plot(hours, alt)&#xA;        plt.plot(hours[::2], alt[::2], 'ok')&#xA;    plt.subplot(2, 1, 2)&#xA;    for alt, az in altazs:&#xA;        plt.plot(hours, az)&#xA;        plt.plot(hours[::2], az[::2], 'ok')&#xA;    plt.show()&#xA;&#xA;th = np.linspace(0, twopi, 201)&#xA;cth, sth = [f(th) for f in (np.cos, np.sin)]&#xA;if True:&#xA;    plt.figure()&#xA;    plt.plot(cth, sth, '-k', linewidth=1.5)&#xA;    plt.plot([0], [0], 'or', markersize=8)&#xA;    plt.plot([-0.1, 0.1], [ 0,   0  ], '-k')&#xA;    plt.plot([ 0,   0  ], [-0.1, 0.1], '-k')&#xA;&#xA;    plt.text(-0.05,  0.85, 'N', fontsize=16)&#xA;    plt.text( 0.9,  -0.05, 'E', fontsize=16)&#xA;    plt.text(-0.95, -0.05, 'W', fontsize=16)&#xA;    plt.text(-0.05, -0.90, 'S', fontsize=16)&#xA;&#xA;    for x, y in linez:&#xA;        linewidth= 1&#xA;        plt.plot(x, y)&#xA;        plt.plot(x[:-1:2], y[:-1:2], '.k')&#xA;        plt.plot(x[:1], y[:1], 'ok')&#xA;    plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-02-21T08:50:07.093" LastActivityDate="2019-02-21T08:50:07.093" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29683" PostTypeId="2" ParentId="29673" CreationDate="2019-02-21T11:59:40.683" Score="7" Body="&lt;p&gt;The Roche limit is defined in classical physics, dealing with materials that behave classically. That said, of course people have tried to estimate the counterpart in relativistic conditions. It turns out that &lt;a href=&quot;http://adsabs.harvard.edu/full/1973ApJ...185...43F&quot; rel=&quot;noreferrer&quot;&gt;it behaves roughly like the classical case&lt;/a&gt; when considering a liquid body orbiting a black hole. For the neutron star case and a more massive stellar mass black hole the limit is not too far away from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Innermost_stable_circular_orbit&quot; rel=&quot;noreferrer&quot;&gt;ISCO&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would a black hole be affected by tidal forces? A tidal force is after all just the effect of the radially varying curvature of spacetime, and a black hole embedded in such a curved spacetime does show distortion. Not only that, but &lt;a href=&quot;https://www.physics.uoguelph.ca/poisson/research/tidal.pdf&quot; rel=&quot;noreferrer&quot;&gt;one can apparently treat the black hole as a fluid sphere that can even dissipate&lt;/a&gt;! (&lt;a href=&quot;https://arxiv.org/abs/0907.0874&quot; rel=&quot;noreferrer&quot;&gt;paper&lt;/a&gt;, &lt;a href=&quot;https://www.physics.uoguelph.ca/poisson/research/BHtides.pdf&quot; rel=&quot;noreferrer&quot;&gt;presentation2&lt;/a&gt;) As the plot in the second presentation shows, yes, the horizon can get distorted by a nearby massive orbiting object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the laws of black hole thermodynamics do not allow the horizon area to go down. Splitting a mass 2M hole into two holes of mass M would change the area from &lt;span class=&quot;math-container&quot;&gt;$4 \times (4\pi G^2 M^2/c^4)$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$2\times (4\pi G^2 M^2/c^4)$&lt;/span&gt;. So that is not going to happen (unless there was a process that could release a black hole's worth of entropy in the outside universe - a very tall order).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, what would happen is that the deformations would dissipate orbital energy (by heating the neutron star and gravitational radiation) until the objects spiralled together, the neutron star would cross the relativistic Roche limit and shortly after the remnants would plunge inside the ISCO and be absorbed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnetic field case would complicate things in various ways, but remember that even magnetar magnetic fields only have the density of &lt;span class=&quot;math-container&quot;&gt;$10^4$&lt;/span&gt; times lead - compared to the neutron star density this is nothing. There would presumably be a lot of electromagnetic dissipation, jets, and complex torques, but I suspect the endgame is the same. &lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2019-02-21T11:59:40.683" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29684" PostTypeId="2" ParentId="29665" CreationDate="2019-02-21T12:31:36.717" Score="4" Body="&lt;p&gt;An object moving through the interstellar medium will experience a weak drag force. If the drag force is &lt;span class=&quot;math-container&quot;&gt;$f(v)$&lt;/span&gt; the velocity will decline as &lt;span class=&quot;math-container&quot;&gt;$dv/dt=-f(v)/M$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is the mass. In fluids &lt;span class=&quot;math-container&quot;&gt;$f(v)\propto v^2$&lt;/span&gt;; this appears to work even for low density gas. Note that this differential equation leads to a velocity that declines forever, slowly approaching zero but never reaching it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can make a simple ballpark estimate of the timescale of slowdown by estimating how long it takes for it to run into its own mass of interstellar material. If it has radius &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; it will encounter &lt;span class=&quot;math-container&quot;&gt;$\pi r^2 v \rho$&lt;/span&gt; kg of ISM per second. So for mass &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; the timescale of slowdown is &lt;span class=&quot;math-container&quot;&gt;$\tau \sim M/\pi r^2 v \rho$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take a &lt;span class=&quot;math-container&quot;&gt;$r=1$&lt;/span&gt; km spherical asteroid of mass &lt;span class=&quot;math-container&quot;&gt;$M= 8.8802\cdot 10^{12}$&lt;/span&gt; kg (assuming density 2.12 g/cm&lt;span class=&quot;math-container&quot;&gt;$^3$&lt;/span&gt;) moving at 26.33 km/s and use &lt;span class=&quot;math-container&quot;&gt;$\rho=10^{-20}$&lt;/span&gt; kg/m&lt;span class=&quot;math-container&quot;&gt;$^3$&lt;/span&gt; (this varies a lot) then the timescale is about &lt;span class=&quot;math-container&quot;&gt;$10^{22}$&lt;/span&gt; seconds, or 340 trillion years. So after maybe a quadrillion years the asteroid would be nearly at rest relative to the local gas if nothing disturbed its trajectory...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is fairly likely for the asteroid to encounter a star during this time. Stellar density is about &lt;span class=&quot;math-container&quot;&gt;$\rho_*=0.14$&lt;/span&gt; per cubic parsec, and the timescale for getting within 100 AU from a star is &lt;span class=&quot;math-container&quot;&gt;$\sim 1/\pi (100 AU)^2 v \rho_* \approx 10^{11}$&lt;/span&gt;years. This is likely to act as a gravity assist giving it some of the star's relative velocity (on the order of kilometers per second). So the true answer is that the asteroid will never settle down as long as there are stars in the galaxy. &lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2019-02-21T12:31:36.717" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29685" PostTypeId="1" CreationDate="2019-02-21T14:14:14.953" Score="1" ViewCount="57" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yOAAf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;Hi Everyone please check this pictures could you tell me this constellation name ?&lt;/p&gt;&#xA;" OwnerUserId="25993" LastActivityDate="2019-02-21T14:14:14.953" Title="Could you tell me this constellation name?" Tags="&lt;star&gt;&lt;constellations&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2019-02-21T22:14:48.273" ContentLicense="CC BY-SA 4.0" />
  <row Id="29686" PostTypeId="2" ParentId="29672" CreationDate="2019-02-21T15:01:42.110" Score="2" Body="&lt;p&gt;One picture tells more than thousands of words.&lt;br&gt;&lt;br&gt;&#xA;Check this out:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Moon rising (near East):&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nG1sf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nG1sf.jpg&quot; alt=&quot;Moon in the east&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Moon setting (near West):&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eMNoG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eMNoG.jpg&quot; alt=&quot;Moon in the west&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That picture can you show, that Moon rises in the East and sets in the West (in fact just little more to the North - 25°).&lt;br&gt;&#xA;It's probably just a technical problem with a compass because Moon cannot rise in the North. (The earth always revolves around its axis, which runs along the north-south direction.)&lt;br&gt;&lt;br&gt;&#xA;&lt;em&gt;(This picture is from the site &quot;theskylive.org&quot; for location Cuba and date 16/17th January 2019.)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="25994" LastEditorUserId="25994" LastEditDate="2019-02-22T05:45:38.113" LastActivityDate="2019-02-22T05:45:38.113" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29687" PostTypeId="1" AcceptedAnswerId="29692" CreationDate="2019-02-21T17:50:07.267" Score="10" ViewCount="233" Body="&lt;p&gt;Are other planets capable of producing rainbows? How would those rainbows appear? Can rain, clouds or ice from elements other than water produce rainbows?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related: &lt;a href=&quot;https://space.stackexchange.com/questions/34357/rainbow-space-probe&quot;&gt;https://space.stackexchange.com/questions/34357/rainbow-space-probe&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2019-02-21T21:14:29.017" LastActivityDate="2019-02-23T02:23:46.060" Title="How would rainbows appear on other planets?" Tags="&lt;planet&gt;&lt;atmospheric-effects&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29688" PostTypeId="2" ParentId="29687" CreationDate="2019-02-21T19:18:58.297" Score="9" Body="&lt;p&gt;Rainbows occur when sunlight shines through rain. This is rare in the solar system. Rain (of sulphuric acid) might be common enough under Venus's clouds, but there is no sun. Conversely, there is plenty of sun in Mars, but no rain, and only very rare clouds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It rains on Titan: methane rain. Methane has a lower refractive index than water (1.27 instead of 1.33), which would make the rainbows slightly larger (though not by much 42-&gt;52). However the atmosphere of Titan is hazy, and while there is some light on the surface, the sun's disc is not visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is rain in some layers of the gas giants, but again not on the outer layers where the sun is visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is likely that the Earth is the only place in the solar system where rainbows are a common phenomenon.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2019-02-22T07:29:06.827" LastActivityDate="2019-02-22T07:29:06.827" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29689" PostTypeId="1" AcceptedAnswerId="29690" CreationDate="2019-02-21T20:17:57.253" Score="1" ViewCount="66" Body="&lt;p&gt;As far as I know, our current knowledge regarding the exact shape and size of ʻOumuamua is based on the analysis of its light curve and the assumption that it has a uniform albedo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ʻOumuamua is very dark and too small to be resolved by even the largest telescopes as much more than a point of light. But so is Ultima Thule and we &lt;a href=&quot;https://www.nasaspaceflight.com/2019/01/2014-mu69-contact-binary-first-new-horizons-returns/&quot; rel=&quot;nofollow noreferrer&quot;&gt;did manage to get a really good idea of its size and shape by observing stellar occultations caused by it&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could the same technique be used to get a better idea of what shape ʻOumuamua really is? Or does its hyperbolic orbit render this kind of observation impractical? Have any attempts been made to perform such observations?&lt;/p&gt;&#xA;" OwnerUserId="211" LastEditorUserId="211" LastEditDate="2019-02-21T20:32:35.520" LastActivityDate="2019-02-21T21:06:44.193" Title="Determining the shape of ʻOumuamua via stellar occultation" Tags="&lt;occultation&gt;&lt;oumuamua&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29690" PostTypeId="2" ParentId="29689" CreationDate="2019-02-21T20:42:24.757" Score="3" Body="&lt;p&gt;Observing star occultations is dependent on the object passing in front of a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During its short time in the inner solar system, ʻOumuamua didn't pass in front of a star (from the point of view of Earth) It is now beyond the orbit of Saturn. It became too dim for Hubble in about January 2018. There are no forecast occultations of any star (and its orbit is probably not well enough determined to predict any, given the observation of non-gravitational and not well-understood accelleration). Any occultation that did occur would only happen for a very short time and only be visible in a very narrow strip of the Earth. We can't set up large telescopes every 100m on Earth, and constantly observe all the (very dim) stars that might possibly dim further for half a second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, Ultima Thule is a much larger object, with a well-known orbit, and the possibility of forecasting the exactly when and where an occultation of a reasonably bright star will occur, and set up observation stations on the narrow strip that is predicted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while an occultation could have told us about its shape, none occurred and none are expected.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2019-02-21T21:06:44.193" LastActivityDate="2019-02-21T21:06:44.193" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29691" PostTypeId="2" ParentId="29672" CreationDate="2019-02-21T21:35:28.727" Score="2" Body="&lt;p&gt;To avoid confusion with compass directions on the ground, let's express the orientation of the Moon by analogy to an imaginary clock dial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relative to the celestial equator and pole, the bright limb of a first quarter moon is always oriented between 2 and 4 o'clock depending on the time of year.&#xA;In astronomical terms, its &lt;a href=&quot;https://en.wikipedia.org/wiki/Position_angle&quot; rel=&quot;nofollow noreferrer&quot;&gt;position angle&lt;/a&gt; is between 240&amp;deg; and 300&amp;deg; counterclockwise from celestial north.&#xA;On the evening of 2019-01-15, JPL HORIZONS says the Moon's sub-Sun position angle was about 252&amp;deg;, or 3:36 on an equatorially mounted clock dial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relative to the horizon and zenith, an observer in the Arctic would see the Moon in a similar orientation, near 3 o'clock all evening.&#xA;In general, however, the celestial equator is inclined to the horizon at an angle of 90&amp;deg; minus the observer's geographic latitude.&#xA;As celestial objects cross the sky parallel to the equator, they roll through a range of &lt;a href=&quot;https://en.wikipedia.org/wiki/Parallactic_angle&quot; rel=&quot;nofollow noreferrer&quot;&gt;parallactic angles&lt;/a&gt;.&#xA;&lt;a href=&quot;https://astronomy.stackexchange.com/a/20625&quot;&gt;This answer&lt;/a&gt; illustrates it with the Sun, and &lt;a href=&quot;https://astronomy.stackexchange.com/a/26215&quot;&gt;this answer&lt;/a&gt; gives a standard formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YGAER.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YGAER.png&quot; alt=&quot;Moon at 13:40&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Q4jdJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q4jdJ.png&quot; alt=&quot;Moon at 19:50&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/hVBFa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hVBFa.png&quot; alt=&quot;Moon at 02:00&quot;&gt;&lt;/a&gt;&#xA;&lt;br/&gt;&lt;em&gt;Moon and &lt;a href=&quot;https://en.wikipedia.org/wiki/Equatorial_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;equatorial grid&lt;/a&gt; rendered by Stellarium&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observing from Havana (23&amp;deg;N) on January 15-16, the Moon's parallactic angle went from -67&amp;deg;  at 1:40pm, to 0&amp;deg; while crossing the meridian at 7:50pm, to +67&amp;deg; at 2:00am.&#xA;Subtracting this from the bright limb's position angle, we get angles of 319&amp;deg;, 252&amp;deg;, and 185&amp;deg; counterclockwise from the zenith, or 1:22, 3:36, and 5:50 on a vertical clock dial.&#xA;From a more northerly latitude, the rolling effect would be smaller; from the southern hemisphere, it would be reversed.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-02-25T03:00:28.900" LastActivityDate="2019-02-25T03:00:28.900" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29692" PostTypeId="2" ParentId="29687" CreationDate="2019-02-21T23:01:32.570" Score="8" Body="&lt;p&gt;&lt;strong&gt;note 1:&lt;/strong&gt; I've verified @JamesK's &lt;a href=&quot;https://astronomy.stackexchange.com/a/29688/7982&quot;&gt;answer&lt;/a&gt;'s index of refraction of 1.27 (since no source was cited), at least for a temperature of 111K, yay! On a colder day, say 90K, the index goes up and the rainbow will shrink a few degrees, close to the size of that on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source for methane:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://refractiveindex.info/?shelf=organic&amp;amp;book=methane&amp;amp;page=Martonchik-liquid-111K&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://refractiveindex.info/?shelf=organic&amp;amp;book=methane&amp;amp;page=Martonchik-liquid-111K&lt;/a&gt; which comes from &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.osapublishing.org/ao/abstract.cfm?uri=ao-33-36-8306&quot; rel=&quot;nofollow noreferrer&quot;&gt;Optical constants of liquid and solid methane&lt;/a&gt; John V. Martonchik and Glenn S. Orton, Applied Optics Vol. 33, Issue 36, pp. 8306-8317 (1994) &lt;a href=&quot;https://doi.org/10.1364/AO.33.008306&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://doi.org/10.1364/AO.33.008306&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;available at &lt;a href=&quot;https://www.researchgate.net/publication/47510751_Optical_constants_of_liquid_and_solid_methane&quot; rel=&quot;nofollow noreferrer&quot;&gt;ResearchGate&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;and &lt;a href=&quot;https://trs.jpl.nasa.gov/handle/2014/34442&quot; rel=&quot;nofollow noreferrer&quot;&gt;through NASA&lt;/a&gt; thanks to @mistertribs's &lt;a href=&quot;https://astronomy.stackexchange.com/questions/29687/how-would-rainbows-appear-on-other-planets/29692?noredirect=1#comment53778_29692&quot;&gt;comment&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source for water:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.osapublishing.org/ao/abstract.cfm?URI=ao-12-3-555&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.osapublishing.org/ao/abstract.cfm?URI=ao-12-3-555&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Optical Constants of Water in the 200-nm to 200-Mm Wavelength Region&lt;/li&gt;&#xA;&lt;li&gt;George M. Hale and Marvin R. Querry March 1973 / Vol. 12, No. 3 / APPLIED OPTICS 555&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now @CarlWitthoft &lt;a href=&quot;https://astronomy.stackexchange.com/a/29705/7982&quot;&gt;shows two unlabeled plots&lt;/a&gt; with no sources cited and very different values for &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;note 2:&lt;/strong&gt; @CarlWitthoft's &lt;a href=&quot;https://astronomy.stackexchange.com/a/29705/7982&quot;&gt;unsourced claim&lt;/a&gt; that methane has a significantly lower dispersion than water in visible light appears to be without merit. I've plotted both materials on the same axis and they are comparable. The rainbows will have slightly different spreading of colors, but I do not thing the rainbow will disappoint!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PXLoV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PXLoV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;@JamesK's &lt;a href=&quot;https://astronomy.stackexchange.com/a/29688/7982&quot;&gt;answer&lt;/a&gt; mentions that Titan could see rainbows from liquid methane rain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using math from &lt;a href=&quot;http://www.trishock.com/academic/rainbows.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;https://www.physics.harvard.edu/uploads/files/undergrad/probweek/sol81.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;, &lt;a href=&quot;https://www.stewartcalculus.com/data/ESSENTIAL%20CALCULUS%202e/upfiles/instructor/eclt_wp_0301_inst.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;3&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$k = \frac{n_{droplet}}{n_{atmosphere}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\alpha = \arcsin\left(\sqrt{ \frac{r-k^2}{3} } \right) $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\beta = \arcsin\left( \frac{\sin\alpha}{k} \right)  $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\theta = 2\phi = 4\beta - 2\arcsin(k \sin \beta)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RTpfT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RTpfT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/ivQre.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ivQre.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, lower index makes the rainbow &lt;em&gt;larger&lt;/em&gt;. Remember that red is on the &lt;em&gt;outside&lt;/em&gt;. With &lt;span class=&quot;math-container&quot;&gt;$k=4/3\approx1.33$&lt;/span&gt; the rainbow is at ~42°, for &lt;span class=&quot;math-container&quot;&gt;$k=1.27$&lt;/span&gt; it blows up to ~52°. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All else equal it would be a little brighter as well; with a larger incident angle at the back of the drop, the fresnel reflection will be a bit stronger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Fz1y2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fz1y2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.stewartcalculus.com/data/ESSENTIAL%20CALCULUS%202e/upfiles/instructor/eclt_wp_0301_inst.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# https://www.stewartcalculus.com/data/ESSENTIAL%20CALCULUS%202e/upfiles/instructor/eclt_wp_0301_inst.pdf&#xA;# https://www.physics.harvard.edu/uploads/files/undergrad/probweek/sol81.pdf&#xA;# nice math http://www.trishock.com/academic/rainbows.shtml&#xA;&#xA;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;&#xA;halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]&#xA;degs, rads = 180/pi, pi/180&#xA;&#xA;k = np.linspace(1.2, 1.5, 31)&#xA;&#xA;alpha = np.arcsin(np.sqrt((4.-k**2)/3.))&#xA;beta  = np.arcsin(np.sin(alpha)/k)&#xA;phi   = 2*beta - np.arcsin(k*np.sin(beta))&#xA;theta = 2 * phi&#xA;&#xA;things = (alpha, beta, theta)&#xA;names  = ('alpha', 'beta', 'theta = 2phi')&#xA;if True:&#xA;    plt.figure()&#xA;    for i, (thing, name) in enumerate(zip(things, names)):&#xA;        plt.subplot(3, 1, i+1)&#xA;        plt.plot(k, degs*thing)&#xA;        plt.title(name, fontsize=16)&#xA;        plt.plot(k[7],  degs*thing[7],  'ok')&#xA;        plt.plot(k[13], degs*thing[13], 'ok')&#xA;    plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-02-23T02:21:30.837" LastActivityDate="2019-02-23T02:21:30.837" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29693" PostTypeId="1" AcceptedAnswerId="29698" CreationDate="2019-02-21T23:44:10.443" Score="6" ViewCount="459" Body="&lt;p&gt;I have had one of these in my room for decades:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/rfsZp.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rfsZp.jpg&quot; alt=&quot;Map of The Universe&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have the booklet that seems to usually come with it, can someone explain to me what the months (with days below it) and hours mean? I imagine it has to do with the position of the earth and therefore the part of the universe that is possibly observable, but I get confused when I try to get the months in the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much&lt;/p&gt;&#xA;" OwnerUserId="25998" LastActivityDate="2020-03-24T16:43:41.890" Title="What do the months and hours mean in Tomas Filsinger's Map of The Universe" Tags="&lt;fundamental-astronomy&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29694" PostTypeId="2" ParentId="27522" CreationDate="2019-02-22T01:21:35.077" Score="2" Body="&lt;p&gt;&lt;strong&gt;Warning:&lt;/strong&gt; I am only rather mediocre on astronomy. It was a hobby of mine when I was in second grade to now, so I pretty much know a decent amount of info. But my speculation for this question may be some distance off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, let us look at the Sun and Earth. The Earth has a [25~65 microtesla] magnetic field due to the motion of convection currents its molten [mostly iron and nickel] outer core. These convection currents are caused by heat escaping from the solid inner core. Heat from below heats up the liquid, which rises, cools, sinks, heats up, rises... you get the point. These convection cells creates charge/electricity in the outer core. Now, when you hear electricity, and then, iron/metals, you will obviously know that iron would conduct electricity. Sort of like an electric magnet nowadays. That is exactly so in Earth's outer core. The electricity generated by the convection cells is [what I believe] conducted by the outer core and amplified/extended to form the magnetic field. This is the Dynamo theory, which contains three parts: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are three requisites for a dynamo to operate: &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1) An electrically conductive fluid medium&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2) Kinetic energy provided by planetary rotation&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;3) An internal energy source to drive convective motions within the fluid.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In the case of the Earth, the magnetic field is induced and constantly maintained by the convection of liquid iron in the outer core. A requirement for the induction of field is a rotating fluid. Rotation in the outer core is supplied by the Coriolis effect caused by the rotation of the Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The sun's [100 microtesla] magnetic field compared to Earth's is a really large and complicated. Since the sun is THAT LARGE :P, its magnetic field extends beyond the planets forming the heliosphere, a protection against cosmic rays from interstellar space. Unlike but similar to the Earth, the magnetic field of the sun is derived from the electricity conductive plasma inside the sun, but also from convection. Since the sun is not a stable ball, but rather a giant ball of fluid in a medium, it's rotation speeds on the equator and the poles are different, and this bends/stretches the magnetic field, creating sunspots, CMEs, Solar Flares, and many other &quot;space weather&quot; events.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To answer what you are asking, the magnetic attraction between Earth and the Sun should be rather strong, but not to an extreme extent. If you were to take 2 magnets of the same magnetic strength ratio, the stronger magnet the readily attract the weaker magnet. This same affect should be that of what the Sun is sort of doing. But even along with gravity, the Sun's pull on Earth cannot force it into a collision course due to Earth's sideways momentum. As commented on by uhoh, the interplanetary magnetic field is the magnetic fields of all objects in the Solar system. The sun, being the strongest one, takes dominance. But what is rather confusing is that there is a thing called Solar Wind. The Solar Wind is a rather &quot;pushing&quot; stream of charged that will &quot;push&quot; away particles if the objects doesn't have a magnetic field to repel that force. The Solar Wind carries the Sun's own magnetic fields away from it, but since magnetic lines cannot cross celestrial objects, they &quot;drape&quot; themselves onto the magnetic field-less object and form the &quot;magnetosphere&quot; of that object. I would imagine this would have little effect however on the attraction of the Sun and that object(earth, asteroid...) itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking for numbers, a hefty equation to calculate the electrical&#xA;magnetism oh an object is the Boit-Savart Law. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you have a&#xA;  current &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; flowing through an infinitesimal length&#xA;  &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}\vec{l}$&lt;/span&gt; then the field at a point &lt;span class=&quot;math-container&quot;&gt;$\vec{r}$&lt;/span&gt; is given by&#xA;  &lt;span class=&quot;math-container&quot;&gt;\begin{equation} \mathrm{d}\vec{B} = \frac{\mu_0}{4\pi} \frac{I&#xA;&amp;gt; \;\mathrm{d}\vec{l}\times\vec{r}}{|\vec{r}|^3} \end{equation}&lt;/span&gt; The&#xA;  field does still drop of with the inverse square of the distance, but&#xA;  the vector nature of the current makes the formula rather more&#xA;  complicated. Generally this expression has to be integrated over the&#xA;  length of a wire carrying the current. This can get quite complicated&#xA;  and so this method is generally only used for simple situations such&#xA;  as the field around a magnetic dipole.  Having found the field, the&#xA;  force on a infinitesimal length of current can be found from.&#xA;  &lt;span class=&quot;math-container&quot;&gt;\begin{equation} \vec{F} =&#xA;&amp;gt; I\;\mathrm{d}\vec{l}\times\vec{B}\end{equation}&lt;/span&gt; This is again&#xA;  analogous to the electrostatic case but with added complication due to&#xA;  current being a vector.&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/122258/what-is-equation-to-find-force-of-magnetic-attraction&quot;&gt;https://physics.stackexchange.com/questions/122258/what-is-equation-to-find-force-of-magnetic-attraction&lt;/a&gt;&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This can be used to calculate the force of Earth's magnetic field, and the Sun's magnetic field, since they both are electrically generated. The resulting numbers can be put into the magnetic attraction equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F=\frac{\mu q_{m1} q_{m2}}{4\pi r^4}$$&lt;/span&gt;&#xA;where&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; is force attraction in newtons.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$q_{m1}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$q_{m2}$&lt;/span&gt; is the magnitudes of magnetic poles in ampere-meters.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt; is the permeability of the intervening medium is tesla meter per ampere, henry per meter or newton per ampere squared.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; is the separation in meters.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus, the attraction between the Earth and the Sun can be calculated in the following means:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\implies \operatorname{B}(\text{Earth})=q_{m1}=7.84027\times 10^{22}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\implies \operatorname{B}(\text{Sun})=q_{m2}=3.48526\times 10^{29}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\implies \mu\approx 10^{-6}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\implies r=\operatorname{AU}=149597870700$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\implies F=\frac{10^{-6}\cdot (7.84427\times 10^{22})\cdot (3.48526\times 10^{29})}{4\pi\cdot 149597870700^4}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\implies F=\boxed{4.34165\text{ newtons}}$$&lt;/span&gt;&#xA;If the Earth doesn't have a liquid, conductive outer core, it wouldn't have a magnetic field. This would be similar to Venus, which has a super duper weak magnetic tail and magnetosphere from the Solar Wind(suprising is that it still retains an atmosphere. I will ask why...). However, we can use a real world example of such. Let's take Mars as the example. Mars used to have a magnetic field from its liquid outer core 4 bya, but this field was destroyed, either from repeated bombardment from large astroids which would have disrupted its internal structure, or from the cooling of its liquid core. Thru recent evidence from tests on Mars' crustal fields, it has shown that Mars' earlier magnetic field is more powerful that of Earth's. Now all that's left is some umbrella shaped fields concentrated on it's southern hemisphere. This offers little protection and would sufficiently answer your second question. You asked if not for the conductive, LIQUID, outer core, what would happen to the distance as possibly the attraction between the Sun diminishes and Earth wanders more outwards. I would suggest not such. Mars would fit your speculation, but its 4 bya orbit and current orbit has minute difference. I would postulate that the magnetic field degration would have some effect, but such effect would be rather unnoticeable, because even though electromagnetism is a very strong force, realize that gravity is the strongest force in the universe(black holes). The attraction between the celestrial bodies orbiting the Sun and the Sun itself is almost completely gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may be off, or way wrong, but to what I know, I believe this is more or less the right answer.&lt;/p&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="7982" LastEditDate="2019-03-13T05:17:08.137" LastActivityDate="2019-03-13T05:17:08.137" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29695" PostTypeId="2" ParentId="29363" CreationDate="2019-02-22T02:06:27.770" Score="1" Body="&lt;p&gt;[WARNING: I may be wrong. Astronomy is my hobby and I will answer to the extent that I know of. I may be a bit off.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars' atmosphere contains I guess what you can say 2 major parts. Gases, and debris picked up by winds on the lower levels of the atmosphere (like Earth). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Gases&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars' atmospheric gas composition mainly consists of CO&lt;span class=&quot;math-container&quot;&gt;$_2$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, in an article page by NASA, Mars' atmosphere contains&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Carbon dioxide: 95.32 percent&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Nitrogen: 2.7 percent&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Argon: 1.6 percent&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Oxygen: 0.13 percent&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Carbon monoxide: 0.08 percent&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Also, minor amounts of: water vapor, nitrogen oxide, neon, hydrogen-deuterium-oxygen, krypton and xenon.&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&lt;a href=&quot;http://www.answers.com/Q/What_are_the_contents_of_Mars%27_atmosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.answers.com/Q/What_are_the_contents_of_Mars%27_atmosphere&lt;/a&gt;&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dust/other stuff&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars' atmosphere, although 99~100 times thinner that that of Earth's, can support winds, weather/storms, and clouds. The winds are a major criterion to Mars' permanent atmospheric dust composition. Dust particles, airborne or not absorbing sunlight to some extent, which in turn warms up the atmosphere in and near their vicinity. Warm pockets of air flow toward colder regions, generating winds; the same effect here on Earth. Strong winds lift more dust off the ground, which in turn heats the atmosphere, raising more wind and kicking up more dust. This is amplified by Mars' axis tilt and momentum, which can contribute to Mars' massive dust storms encircling the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to the winds, the Martian atmosphere is dense enough to support vortexes. Dust devils routinely kick up the rusty dust that covers Mars' surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can see now the causes for such permanent content of dust in the atmosphere of Mars, with...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...higher amounts of dust in the northern fall and winter, and lower&#xA;  amounts in the northern spring and summer.&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&lt;a href=&quot;https://www.space.com/16903-mars-atmosphere-climate-weather.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.space.com/16903-mars-atmosphere-climate-weather.html&lt;/a&gt;&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's now take a look at Mars' surface dust composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its dust mainly is composed of Perchlorate salts; Chlorine; Iron(III) Sulfide; NiFe; Sulfur; Iron Sulfide; Nickel; Silicon; Oxygen; Iron; Magnesium; Aluminum; Calcium; Potassium; Pyroxene; Plagioclase; Glass; Olivine; and Feldspar. I would imagine that the dust in the air would contain many of these same components as the surface dust, although some heavier salts and chemicals would only last a brief time in the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The NASA Phoenix lander first detected chlorine-based compounds such as calcium perchlorate. The levels detected contain 0.6% of the salt perchlorate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The soil has a basic pH of 7.7&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Mars&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Mars&lt;/a&gt;&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At such high levels, there should be definitely perchlorate salt in the air, but not necessarily permanent though, as the atmosphere may not be able to support the density of the salt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a recent study of terrestrial dust devils has shown that around two thirds of the fine particles lifted by these vortices can remain suspended in the atmosphere and be transported around the globe. I would theorize that lighter salts would be able to remain suspended in the air, while heavier ones would slowly settle to the surface, given enough time. But hey, this is MARS! With those huge dust storms and continuous vortexes, I doubt there is enough time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this suggests that during the times when there are rather windy conditions, such as in a close proximity to a dust devil, or during one of the dust storms, the atmosphere should contain almost the same concentrations of salts as in the soil itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During calm periods, the air dust content should still remain the same.&lt;/p&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="14094" LastEditDate="2019-03-02T17:57:04.540" LastActivityDate="2019-03-02T17:57:04.540" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29696" PostTypeId="2" ParentId="19287" CreationDate="2019-02-22T02:28:17.460" Score="2" Body="&lt;p&gt;I wanted to make a comment on Sean's answer, but I do not have the reputation to do that. I guess I will take a stab at an answer with apologies to Sean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When converting from spherical coordinates to cartesian, you have to pay attention to the assumptions of the spherical coordinate system being used. In this case the assumption used in the equatorial coordinate system about the meaning of the Declination angle is different then the assumption used in the first equation in Sean's answer. We know this because the value for x is (cos δi * sin αi). This is correct for inclination, or the case where the angle is between the vector and the Z axis. For equatorial coordinates the angle is between the vector and the XY plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spherical coordinate systems using inclination are diagrammed here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Spherical_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia on spherical coordinate systems&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equatorial coordinate system is diagrammed here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Equatorial_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia on the equatorial coordinate system&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to calculate the Cartesian vectors for a given equatorial point from the usual astronomical notation to degrees or radians. Right Ascension is given in hours, minutes and seconds. An hour is 15 degrees. (The are 24 hours in a complete revolution and 360/24 is 15.) A minute is 1/60 of an hour. 15/60 gives 1/4 so every minute is 1/4 of a degree. Finally, a second is 1/60 of a minute. Multiplying the fraction of a minute in a second (1/60) by the number of degrees in a minute (1/4) gives the number degrees in a second. (1/(60*4)) = 1/240. Multiply the hours by 15, then the minutes by 1/4 and finally the seconds by 1/240, then add the three numbers together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More info here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Right_ascension&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia on Right Ascension&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a similar bit of work for Declination. Declination is in Degrees, minutes, seconds. We don't have to convert from hours to degrees since the declination does not use the idea of hours. Minutes and seconds convert as with Right Ascension. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Declination&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia on Declination&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this conversion you can transform the coordinates from spherical to Cartesian. In order to do this, we need to know the radius, or distance from the origin of the coordinate system and the object. Equatorial points don't usually come with this information. We can use 1 because the distance doesn't change the angle between the two points. It also makes the math easier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get X, we start by calculating the projection of the vector on the XY plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LenXY = 1*cos(Declination)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We next need to calculate the projection of the vector on the XY plane onto the X axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;X = LenXY*cos(Right Ascension) or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X = cos(Declination)*cos(Right Ascension)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process for Y is similar but of course we want the projection on the Y axis. This will be sin(Right Ascension) instead of cos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Y = LenXY*sin(Right Ascension) or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Y = cos(declination)*sin(Right Ascension)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally Z is the projection of the original vector on the Z axis. Since Declination is relative to the Z axis, although measured with respect to the XY plane, we can calculate the projection directly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Z = 1*sin(Declination) or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Z = sin(Declination)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the Haversine would work as it is meant for work with longitude and latitude. Latitude makes use of a declination angle the same way Declination in equatorial coordinates does. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the issue appears to be the limit of the amount of a precision typical floating point numbers in computers have, I would suggest looking at math libraries that support arbitrary precision. A library for Python is here:&#xA;&lt;a href=&quot;http://mpmath.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Link to Python Arbitrary Precision library&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, here is a link to a description on how to calculate the position of the moon and planets given their orbital elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://stjarnhimlen.se/comp/tutorial.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Computing Planetary Positions&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26000" LastActivityDate="2019-02-22T02:28:17.460" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29697" PostTypeId="1" CreationDate="2019-02-22T03:24:59.497" Score="-1" ViewCount="184" Body="&lt;p&gt;A black hole swallows a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The star remains intact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then goes supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would happen to the star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could we detect this event?&lt;/p&gt;&#xA;" OwnerUserId="26001" LastActivityDate="2019-02-25T14:19:40.163" Title="What would happen if a star went supernova inside a supermassive black hole?" Tags="&lt;supernova&gt;&lt;supermassive-black-hole&gt;" AnswerCount="1" CommentCount="13" ClosedDate="2019-02-24T10:16:56.927" ContentLicense="CC BY-SA 4.0" />
  <row Id="29698" PostTypeId="2" ParentId="29693" CreationDate="2019-02-22T03:26:01.597" Score="7" Body="&lt;p&gt;The hours show the lines of right ascension, where right ascension is one of two coordinates used to specify a point on the sky. (The other coordinate is declination, measured in degrees from the celestial equator.) Right ascension and declination are similar to longitude and latitude on the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The month and day show the right ascension of the Sun on a given date. The dotted circle passing through the zodiacal constellations (Pisces, Aries, etc.) is the actual path of the Sun (the ecliptic). Therefore, the position of the Sun on any day can be determined from the chart.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2019-02-22T03:26:01.597" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29700" PostTypeId="2" ParentId="29697" CreationDate="2019-02-22T08:46:49.233" Score="2" Body="&lt;p&gt;We could not-- whatever happens inside a black hole stays inside a black hole.  One way this is often explained is that the time coordinate inside the &quot;event horizon&quot; of a black hole looks to us on the outside like a radial coordinate-- events that occur at sequentially different times inside the event horizon, such as your supernova, seem to us on the outside to happen at sequentially smaller radii, until the last event for your dying star is to reach the center.  The same holds for all the ejecta and even the light and neutrinos emitted-- it all goes to the center in a finite time.  So nothing escapes from the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to say this is that &quot;black holes have no hair&quot;, meaning there is not information outside the event horizon stemming from anything happening on the inside.  All that happens to the outside stems from what falls into the hole-- how much mass, charge, and angular momentum crosses the event horizon.  What it does while it is in there has no effect on the outside, so could never be detected.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2019-02-25T14:19:40.163" LastActivityDate="2019-02-25T14:19:40.163" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29701" PostTypeId="2" ParentId="1203" CreationDate="2019-02-22T11:25:26.063" Score="2" Body="&lt;p&gt;By observing a source of light or radio waves that pass through the edge of a planet's atmosphere, the bending and the delay (both due to the lower speed of light in matter compared to vacuum) can be used to analyze that atmosphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;four answers&lt;/em&gt; to the question &lt;a href=&quot;https://space.stackexchange.com/q/30346/12102&quot;&gt;When did planetary scientists realize Venus' surface pressure was almost 100x that on Earth? How did they find out?&lt;/a&gt; will take a little time to browse through but they are worth it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By using a monochromatic carrier wave, one can detect integrated density by the total phase delays (number of cycles) that accumulate as the the beam passes through thicker and thicker layers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are from that answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ySRxe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ySRxe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hlPUC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hlPUC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19720013159.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J5jxF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J5jxF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; The Stanford 150 foot dish used to transmit the dual frequency signals to Mariner 5 for the radio occultation measurements of its atmospheric density. From instartupland.com &lt;a href=&quot;https://www.instartupland.com/?attachment_id=93&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;https://www.instartupland.com/?p=125&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;, &lt;a href=&quot;https://www.instartupland.com/?attachment_id=88&quot; rel=&quot;nofollow noreferrer&quot;&gt;3&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure radio refraction has been used to study the atmospheres of gas giants like Jupiter and Saturn, but I don't know yet if optical refraction has been modeled yet. This is from the (currently unanswered) question &lt;a href=&quot;https://astronomy.stackexchange.com/q/16703/7982&quot;&gt;Refraction by Saturn's atmosphere - how dense is it here?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1Q4rY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1Q4rY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; cropped portion of the NASA's Cassini image &lt;a href=&quot;http://photojournal.jpl.nasa.gov/catalog/PIA20491&quot; rel=&quot;nofollow noreferrer&quot;&gt;from here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-02-22T11:25:26.063" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29702" PostTypeId="2" ParentId="28513" CreationDate="2019-02-22T12:52:17.100" Score="3" Body="&lt;p&gt;Encounters between small planetesimals and the larger core of planets (especially Jupiter) transfer momentum between the two. But momentum is conserved. Momentum is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$p = m v$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; is momentum &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, smaller planetesimals gain more velocity than a larger planet core, for the same momentum exchange. Thus, smaller planetesimals are preferentially boosted to escape the planet, either to be scattered outward to the outer system, or inward to the sun. Some will have the right velocity and aggregate with to core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since smaller planetesimals get a larger boost than larger ones, say 1000x rather than 20x (just an example, not real numbers), then they are much more sensitive to initial velocities. Thus a much smaller range of initial velocities will have that exactly right velocity to be captured by the planet, as either a satellite or aggregating with the planet. In the case I just made up, 1000/20 i.e. 50 times narrower. Thus, for an equivalent distribution of velocities, the smaller planetesimals in this example will be 50 times less likely to be captured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for dust, beyond it will eventually clump together into larger rocks due to electrostatic attraction, a lot will be sucked into the growing core, or scattered outwards. But it will exert a net drag on the core (see the Grand Tack hypothesis) and gain momentum, and through velocity, energy (a.k.a. heat). And this energy will be averaged out through elastic collisions. Eventually, the dust will either aggregate into larger bodies, sucked into the planet, or scattered into the sun or outwards from the solar system.&lt;/p&gt;&#xA;" OwnerUserId="11625" LastEditorUserId="7472" LastEditDate="2019-02-22T13:21:44.050" LastActivityDate="2019-02-22T13:21:44.050" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29703" PostTypeId="1" AcceptedAnswerId="29847" CreationDate="2019-02-22T13:43:29.900" Score="7" ViewCount="346" Body="&lt;p&gt;Is it possible to verify negative mass with gravitational lensing?  The following video proposes an idea that galaxies are surrounded by negative mass: &lt;a href=&quot;https://youtu.be/MZtS7cBMIc4&quot; rel=&quot;noreferrer&quot;&gt;https://youtu.be/MZtS7cBMIc4&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could this be proven or disproven by looking at gravitational lensing of distant galaxies, by looking at Hubble Deep Field images?  I would expect if negative mass surrounded a galaxy, it should cause the opposite effect of the positive mass galaxy.  If there is surrounding negative mass, there should be an “anti” lensing effect just outside the normal lensing effect.  Could someone perform this analysis and confirm or deny that idea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; In the following image, and others, I have looked to see if I could see such an effect out further from the source visually, and it is hard to tell.  There are so many galaxies in front that aren't affected by the lensing.  It might take a computer analysis of images surrounding the &quot;ring&quot; effect to see if there is a &quot;radial effect&quot; around the ring.  Most images probably don't show enough of the surrounding area to where the radial effect as predicted in the paper &lt;a href=&quot;https://arxiv.org/abs/1305.5037&quot; rel=&quot;noreferrer&quot;&gt;Izumi et al. (2013)&lt;/a&gt; would be apparent.  If this is present, I would think it indicates negative mass surrounding the positive mass, and would be a huge discovery confirming negative mass!  If absent, it would likely show that there is at least not a significant amount of negative mass clustered around the galaxies (wouldn't be a total disproof of negative mass).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image from: &lt;a href=&quot;https://www.flickr.com/photos/mvannorden/8276107136&quot; rel=&quot;noreferrer&quot;&gt;Galaxies - Gravitational Lensing&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/CmxDg.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CmxDg.jpg&quot; alt=&quot;Galaxies - Gravitational Lensing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="1386" LastEditDate="2019-03-03T20:33:17.757" LastActivityDate="2020-10-22T20:23:57.547" Title="Is gravitational lensing a good way to search for negative mass?" Tags="&lt;galaxy&gt;&lt;gravity&gt;&lt;hubble-telescope&gt;&lt;gravitational-lensing&gt;" AnswerCount="3" CommentCount="11" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29705" PostTypeId="2" ParentId="29687" CreationDate="2019-02-22T16:49:32.253" Score="1" Body="&lt;p&gt;Take a look at these charts.  The methane one is the best I could find on a quick search, but it suggests the dispersion over the visible wavelength band is a fraction of the value for water.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6waQo.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6waQo.gif&quot; alt=&quot;refractive index&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ss4AL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ss4AL.jpg&quot; alt=&quot;refractive index&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the existence of a rainbow depends on the ability of the substance to 'bend' different wavelengths different amounts, you can see that methane, at least, would produce a rather unsatisfying rainbow. And even that assumes that you had an atmosphere which supported methane droplets of an appropriate size to achieve a prismatic effect.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly speaking, you would want the methane droplets to be larger than the water droplets which produce rainbows on Earth by the ratio of their dispersions.  This is because the angular output spread depends in part on the length of path thru the droplets. &lt;/p&gt;&#xA;" OwnerUserId="907" LastEditorUserId="7982" LastEditDate="2019-02-23T02:23:46.060" LastActivityDate="2019-02-23T02:23:46.060" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="29706" PostTypeId="1" CreationDate="2019-02-22T20:12:07.280" Score="1" ViewCount="108" Body="&lt;p&gt;Could water from Mars be fresh enough to drink? Could one run the water through a home water filter? I understand when water is frozen the salt is ejected as brine so would newly formed snow or ice be drinkable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://earthscience.stackexchange.com/questions/16076/how-salty-can-snow-sleet-be&quot;&gt;https://earthscience.stackexchange.com/questions/16076/how-salty-can-snow-sleet-be&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastActivityDate="2019-02-22T20:12:07.280" Title="Is the water on Mars drinkable?" Tags="&lt;mars&gt;&lt;water&gt;" AnswerCount="0" CommentCount="5" ClosedDate="2019-02-23T21:36:12.953" ContentLicense="CC BY-SA 4.0" />
  <row Id="29707" PostTypeId="1" CreationDate="2019-02-22T22:33:01.997" Score="2" ViewCount="101" Body="&lt;p&gt;Plotting the moons of Neptune in terms of semimajor axis and eccentricity using the values given on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Moons_of_Neptune&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page for the moons of Neptune&lt;/a&gt;, there appears to be a fairly large gap between the orbits of Larissa and the &lt;a href=&quot;http://hubblesite.org/news_release/news/2019-04&quot; rel=&quot;nofollow noreferrer&quot;&gt;newly-named satellite Hippocamp&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0Pqhp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0Pqhp.png&quot; alt=&quot;The inner moons of Neptune&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there stable orbits within the gap, and are there likely to be additional moons there?&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-02-22T23:54:35.370" LastActivityDate="2019-02-24T00:27:16.370" Title="Could there be additional Neptunian satellites between Larissa and Hippocamp?" Tags="&lt;orbit&gt;&lt;natural-satellites&gt;&lt;neptune&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29708" PostTypeId="1" AcceptedAnswerId="29723" CreationDate="2019-02-23T00:56:27.077" Score="4" ViewCount="815" Body="&lt;p&gt;Similar question &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10189/why-did-venus-not-lose-its-atmosphere-without-magnetic-field&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know as a fact that the magnetic field protects planets from Solar Wind, a damaging, continuous, atmosphere-stripping wind of charged ions. Thus, a magnetic field protected planet will have a stable atmosphere. Vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's take a look at some examples that logically fit this theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Earth&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth has a liquid iron outer core which conducts electricity generated by convection to form a huge electromagnet[technically].This magnetic field protects Earth from the Solar Wind stripping away its atmosphere. We can see its effect by looking at Earth currently, not nessesarily affected by the damaging effect of such wind on the atmosphere, only to have minor damage by direct hits of CMEs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mars&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars used to have a magnetic field. In fact, some analyzations of the crustal fields recently made by robots sent to Mars have shown that Mars may have even had a stronger magnetic field than Earth. Sadly, about 4 bya[according to Wikipedia], either repeated bombardment from large celestrial objects that disrupted the interior, or the solidification of its outer core have caused the degration of such. Consequently, Mars' atmosphere is being constantly stripped away by the Solar Wind, and its atmosphere is drastically thinner that what is started with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are obviously some pretty straightforward examples that fit very well with the first statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, there are two rather contradictory planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Venus&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus has no magnetic field. It has a very small and dilute magnetic tail according to Wikipedia, where reconnections and many different events happen. The Solar Wind carries the sun's magnetic field lines which wrap around Venus forming a not very protective induced magnetic field, and according to Wikipedia again the area magnetopause and ionopause is a strong[relatively saying] barrier which prevents the Solar Wind from penetrating deeper into the atmosphere. However, this offers in all little protection the Venus. Thus, naturally, we would expect Venus to act like Mars, with its atmosphere stripped away in a few million/billion years. Yet, this isn't the case. In face, what's more interesting is that Venus orbits closer the Sun than Mars, which means that it gets a higher &quot;dose&quot; of the Solar Wind, which should strip off the atmosphere rather quickly. Even more interesting is that Venus has a thicker atmosphere than Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other question I linked, MacUserT said that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...the ionosphere of Venus direct interaction with the solar winds causes an externally induced magnetic field...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even if this is the case, Venus lacks a intrinsic magnetic field[self generated], which is the &quot;main&quot; protection. Even with this induced field, this protection is substantially not enough, as the Solar Wind [according to space.com] penetrates relatively deep into the planetary exosphere and causes substantial atmosphere loss. This loss, like Earth thru its poles and the tail, would mainly occur on the tail of the induced magnetosphere. The main losses would be hydrogen, helium ions, and according to Wikipedia, also oxygen ions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is that how is Venus exactly protected by this &quot;induced field&quot; and what is the power of the field? How much atmosphere does Venus lose in one day, and how is this induced field generated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's look at another planet with opposite views.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mercury&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury has a large core Mercury also mas a liquid outer core. It is also spinning substantially fast enough so that it can generate a magnetic field stronger that what Mars and Venus have now[It is 300 nT]. However, due to its close orbit, it faces 3 times more pressure from the Solar Wind than that of Venus. This doesn't mean nessesarily that it can't support an atmosphere. My teacher said that Mercury has a very thin atmosphere constantly being stripped away by the Sun made of gasses such as H, He, O, Na, Ca, K, and many other elements from the planetary crust. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since it has a relatively speaking, strong magnetic field, doesn't is mean that it should at least have more or less the atmosphere density of Mars? What protection does the magnetic field offer of this isn't the case? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;You may be confused so I will I guess put my questions below&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Since Mercury it has a relatively speaking, strong magnetic field, doesn't is mean that it should at least have more or less the atmosphere density of Mars? What protection does the magnetic field offer of this isn't the case?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Answered&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;So my question is that how is Venus exactly protected by this &quot;induced field&quot; and what is the power of the field? How much atmosphere does Venus lose in one day, and how is this induced field generated?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Answered&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-02-24T01:04:23.353" LastActivityDate="2019-02-25T07:54:53.893" Title="What could possibly save an atmosphere other than a magnetic field? Why can't a magnetic field save the atmosphere in certain cases?" Tags="&lt;venus&gt;&lt;magnetic-field&gt;&lt;mercury&gt;&lt;core&gt;&lt;ions&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29709" PostTypeId="5" CreationDate="2019-02-23T00:57:30.200" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2019-02-23T00:57:30.200" LastActivityDate="2019-02-23T00:57:30.200" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29710" PostTypeId="4" CreationDate="2019-02-23T00:57:30.200" Score="0" Body="Questions regarding a molecule or molecules with a positive net electrical change, such as the Solar Wind." OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-02-24T22:30:24.430" LastActivityDate="2019-02-24T22:30:24.430" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29711" PostTypeId="1" AcceptedAnswerId="29715" CreationDate="2019-02-23T05:13:29.707" Score="2" ViewCount="493" Body="&lt;p&gt;Based on my extent of knowledge relating to astrophysics, I once thought space is a thing, which can be twisted a long side with Time regarding masses of objects(According to relativity theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I found out in this comment session of the following thread (I put the link following to this paragraph), I was pointed out as Space is not a thing, but space is the totality of geometric relations (distance, angle, etc) between things. And my confusion started. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/29665/do-interstellar-asteroids-decelerate-and-eventually-stop&quot;&gt;Do interstellar asteroids decelerate and eventually stop?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is it possible that the space can be twisted, if the space is not a thing? or is it? &lt;/p&gt;&#xA;" OwnerUserId="23665" LastActivityDate="2019-02-23T18:02:53.047" Title="Is space a thing?" Tags="&lt;astrophysics&gt;&lt;space-time&gt;&lt;space&gt;&lt;general-relativity&gt;" AnswerCount="3" CommentCount="3" ClosedDate="2019-02-27T01:43:05.060" ContentLicense="CC BY-SA 4.0" />
  <row Id="29714" PostTypeId="2" ParentId="29711" CreationDate="2019-02-23T12:50:58.707" Score="3" Body="&lt;p&gt;Einstein thought of space as a thing. See his 1920 &lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol7-trans/192?highlightText=%22neither%20homogeneous%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;Leyden Address&lt;/a&gt; where he said this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;“According to this theory the metrical qualities of the continuum of space-time differ in the environment of different points of&#xA;  space-time, and are partly conditioned by the matter existing outside&#xA;  of the territory under consideration. This space-time variability of&#xA;  the reciprocal relations of the standards of space and time, or,&#xA;  perhaps, the recognition of the fact that 'empty space' in its&#xA;  physical relation is neither homogeneous nor isotropic, compelling us&#xA;  to describe its state by ten functions (the gravitation potentials&#xA;  g&lt;sub&gt;μν&lt;/sub&gt;), has, I think, finally disposed of the view that space&#xA;  is physically empty”.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He ended by saying &lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol7-trans/197&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;&quot;Recapitulating, we may say that according to the general theory of relativity space is endowed with physical qualities; in this sense,&#xA;  therefore, there exists an ether&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also see the Wikipedia &lt;a href=&quot;https://en.wikipedia.org/w/index.php?title=Aether_theories&amp;amp;oldid=780870304#General_relativity&quot; rel=&quot;nofollow noreferrer&quot;&gt;aether theories&lt;/a&gt; article and note the quote by Robert B Laughlin: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;&quot;It is ironic that Einstein’s most creative work, the general theory of relativity, should boil down to conceptualizing space as a medium&#xA;  when his original premise [in special relativity] was that no such&#xA;  medium existed&quot;.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Laughlin also said space is more like a piece of window glass than ideal Newtonian emptiness. He finished up saying this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;&quot;the modern concept of the vacuum of space, confirmed every day by experiment, is a relativistic ether. But we do not call it this&#xA;  because it is taboo&quot;.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Einstein is supposed to have done away with the aether in 1905 but in the end, he didn’t. He thought of space as a something rather than a nothing. In his 1929 essay on &lt;a href=&quot;http://www.rain.org/~karpeles/einsteindis.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;the history of field theory&lt;/a&gt;, he described a field as a state of space. He was talking about gravitational fields and electromagnetic fields, and said this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;&quot;it can, however, scarcely be imagined that empty space has conditions or states of two essentially different kinds&quot;.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The point to appreciate is that according to Einstein a field isn’t something that exists in space, it’s a state of space. So yes, space is a thing.  &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2019-02-23T12:50:58.707" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="29715" PostTypeId="2" ParentId="29711" CreationDate="2019-02-23T13:15:30.070" Score="4" Body="&lt;p&gt;Much depends on what you mean by thing.  But first lets think about space, or space-time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a flat sheet of paper, and you draw a triangle on it, you will find that the angles add up to 180 degrees (I hope you are familiar with this fact). However, if you have curved surface and draw lines that are &quot;straight&quot; in the sense that they are the shortest distances between points, then you will find that the angles don't add up to 180 degrees. This is one of the rules of geometry on a curved surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you make a triangle in deep space the 180 degree rule works. But if you make a triangle in a gravitational field you will find that the angles don't add up to 180 degrees. (It is actually nearly impossible to measure this, but observations of gravity are exactly consistent with this).  We say that &quot;spacetime is curved&quot; to mean &quot;Geometry done in a gravitational field doesn't obey the rules for flat geometry, but for curved geometry.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do people mean by &quot;thing&quot; - If you mean &quot;an object made of some material&quot; then space is not a thing. If you mean &quot;the description of the location and geometric relationships between objects&quot; then space is a thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final insight is that it is not space that is curved, but four-dimensional &quot;spacetime&quot; and the geometric relationships between objects is described by a block of 16 numbers for each point in spacetime, that depend on the distribution and motion of mass, energy and charge. These 16 numbers can describe with incredible accuracy how objects in space will interact gravitationally, as they just follow straight lines in curved spacetime.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-02-23T13:15:30.070" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29716" PostTypeId="1" CreationDate="2019-02-23T14:33:28.697" Score="1" ViewCount="34" Body="&lt;p&gt;I have list of stellar pairs from The Washington Double Star Catalog (WDS), i.e. I have next parameters:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;angular separations&lt;/li&gt;&#xA;&lt;li&gt;position angles&lt;/li&gt;&#xA;&lt;li&gt;stellar magnitudes of both components&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Additionally I have GAIA DR2 parallaxes (and other parameters) from SIMBAD database. How to distinguish between binary and optical stellar pairs? &lt;/p&gt;&#xA;" OwnerUserId="761" LastEditorUserId="17355" LastEditDate="2019-02-23T15:24:29.653" LastActivityDate="2019-02-23T23:01:40.930" Title="How to separate binary and optical stellar pairs?" Tags="&lt;gravity&gt;&lt;binary-star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29717" PostTypeId="2" ParentId="29711" CreationDate="2019-02-23T18:02:53.047" Score="3" Body="&lt;p&gt;I think &quot;Twisted&quot; is a bad choice of description in your question. Suppose we used &quot;curved&quot; instead. Now you say you read that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;space is the totality of geometric relations (distance, angle, etc) between things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and it makes sense to say that collection of geometric relation is curved or flat. Consider three points in space and connected them by the &quot;straight&quot; (ie as short as possible) lines to make a triangle. Now measure the angles between those lines and add them up. If you always get 180 degrees then your space is flat -- it is &quot;the same shape as&quot; the standard space with real coordinates &lt;span class=&quot;math-container&quot;&gt;$(x,y,z)$&lt;/span&gt; and distance given be &lt;span class=&quot;math-container&quot;&gt;$\sqrt{x^2 + y^2 + z^2}$&lt;/span&gt;.  But that is not the only possibility. You might find that the angles always add up to &lt;em&gt;more than&lt;/em&gt; 180 degrees, with the &quot;excess&quot; angle proportional to the area of the triangle. This is called a positively curved space, and can be modelled by the surface of a four dimensional ball with lines being &quot;great circles&quot;. Or it might be &lt;em&gt;less than&lt;/em&gt; 180 degrees (again with the &quot;missing&quot; angle proportional to the area) which is called negatively curved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This all gets a little more complicated when time is also involved, but essentially Einstein's theory describes how the curvature of space-time (those geometrical relations again) have to match up with the amount of mass and energy in that space-time and how that mass and energy moves. &lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-02-23T18:02:53.047" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29718" PostTypeId="2" ParentId="29707" CreationDate="2019-02-23T20:12:18.633" Score="2" Body="&lt;p&gt;[My answer may be a bit off as astronomy is just my hobby]&#xA;&lt;img src=&quot;https://i.stack.imgur.com/6njj7.jpg&quot; alt=&quot;1&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;First off, I would assume eccentricity has little function into the investigation because why would it be a moon when it can escape Neptune's gravity?[assuming you mean Neptunian eccentricity] I mean, unless you are thinking about a moon escaping?&lt;/p&gt;&#xA;&lt;p&gt;Now, let's move onto the distances. You notice when you graph the moons that there is a considerable gap in distance between S/2004 N 1 and Larissa. I believe, after checking this on multiple sites, that there is indeed such a gap.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Neptune Moon Origin&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;First, let's take a look at one of the most accepted theories on Neptune and its moons. It is well known that Neptune is very close to the Kuiper Belt. In the Kuiper belt, there are objects like Pluto in orbit around the Sun. Well, Neptune having a very strong gravitational pull would actually yank objects out of orbit into its orbit. Most people believe that this is what happened with Triton. Making up more than 80% of the mass of all of Neptune's Moons,  if would have wreaked havoc on the then stable orbits of its original moons.&lt;/p&gt;&#xA;&lt;p&gt;This consequently could have cause many moons that were large[relatively speaking] to gain/lose eccentricity which would have cause some to spiral inside of the Roche Limit of Neptune ultimately causing the formation of several rings around Neptune. However, in the case of Saturn, we see moons inside of the Roche limit, so we can also deduce that the Roche limit of a planet varies on how massive/large an object is.&lt;/p&gt;&#xA;&lt;p&gt;We can now conclude that almost all of Neptune's current moons and their orbits are not its original ones. To answer your question, we look at the three body problem. I am not a physicist to I don't know much on the relative equations for the three body system to calculate the exact orbits of possible moons or calculate proof why there isn't a moon. However, I will give you a relative, not full equation blast, explanation.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Three body problem&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The three body system states that in long term, the three body problem may lead to either and for the most part, stabilization of the orbits of moons around a planet or, for some period ratios, be the cause of the chaotic motions.&lt;/p&gt;&#xA;&lt;p&gt;In Neptune's case, Triton introduction has cause for a short period, chaotic motions, which then stabilized into new stable orbits.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fvmw1.jpg&quot; alt=&quot;&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;In the image here, it shows the relative orbits of stable moons and Neptune's rings. We can conclude that the respective Roche limit of Neptune would cause any possibly try of ring matter collapsing into a moon to be foiled if the mass was too high. This means that only small moons form between rings Adams and Galle, with Galle marking the &amp;quot;will tear anything apart no matter the size&amp;quot; Roche limit.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Moons between the gaps&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;There can be many moons there in the gaps between the rings, as long as the moon is small enough to fit, and its gravity is not influencing the rings so much that it is noticeable, most likely very small to be picked up by our telescopes. It brings up another problem: How small must a moon be to be no longer a moon?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Moon definition&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;There isn't a restriction. However, the most widely accepted one is that a moon is an object with a relatively clear orbit[little debris]. We can say that there may be many moons between the rings, but are too small to be detected.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Moons in the large gap you notice&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now moving on, we have a gap between Larissa and Proteus / S/2004 N 1. Clearly, no rings call this area a home. So this gap is somehow mysteriously &amp;quot;uninhabited&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;I doubt that there isn't a &amp;quot;moon&amp;quot; there, even if it were really small. Hippocamp is a very small object. In fact, it is the smallest moon discovered around Neptune. The relative Roche limit is not here, so it probably is possible for ring material to recondense into a moon with a stable orbit, and many times, a moon might happen to be covered up when we search. As I said, there is no size limit to a moon, so, because this area is clear of debris, there might be quite a few that we can't find.&lt;/p&gt;&#xA;&lt;p&gt;A possible scenario [saying that there are a lot of moons] might be that the moons will gravitationally interact with each other, such as kick each other out, part of the three body problem. However, this can be easily detected, as rings would be seemingly affected. This obviously isn't so, due to everything being in a stable orbit, because the &amp;quot;chaotic&amp;quot; period is already passed, and debris settles down and recondense whenever possible.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Note&lt;/em&gt;: There can be moons also at the lagrange point if the point is near the &amp;quot;gap&amp;quot; you are taking about.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Another Note&lt;/em&gt;: Similar to lagrange points, uhoh mentioned another very good concept of co-orbital configurations, which is the interaction between two celetrial objects. Places with extremely possible stable orbits are called Trojan Points. One example of this is the Trojan asteroid groups which share the orbit with Jupiter. There are two Trojan Points near a given object so those points might harbor many stable moon orbits there too.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;So my guess is that yes, there are stable orbits within this gap, and there is a good chance there are additional moons out there.&lt;/p&gt;&#xA;&lt;p&gt;We are discovering more and more moons today. Jupiter is one good example.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;@uhoh I would think that a considerable gap must be necessary, because moons would kick each other out or collide if the space were too small, forming another chaotic period. Orbits must have settled for moons to reform and since there isn't another gravitational disturbance, there shouldn't be more chaos. Depending on the size of a given moon, the gap must be large enough so that the moon's gravity shouldn't disturb another moon's orbit. Such events would quickly settle if it would happen because objects tend to move to stable places.&lt;/p&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-02-24T00:27:16.370" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29719" PostTypeId="2" ParentId="29716" CreationDate="2019-02-23T23:01:40.930" Score="2" Body="&lt;p&gt;If they are both in the Gaia DR2 catalogue, then a physical binary should have components with similar parallax and proper motion.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-02-23T23:01:40.930" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29720" PostTypeId="1" AcceptedAnswerId="29722" CreationDate="2019-02-23T23:16:58.787" Score="25" ViewCount="4589" Body="&lt;p&gt;It always was interesting for me to understand the answer for the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What's the reason that we have a different number of days each month?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the month is fixed on the time that the moon circles the earth, then the time of the circling isn't equal each month? And if the month fixed by the time that the earth circles the sun and then divided by 12 (365:12=30.4), why one month has more days and the other month has less days?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before desktop computers, how did astronomers handle calculations of dates and times accurately, for example two plates taken decades apart, how would an astronomer calculate the time difference precisely?  &lt;/p&gt;&#xA;" OwnerUserId="26020" LastEditorUserId="7472" LastEditDate="2019-02-24T14:59:19.903" LastActivityDate="2020-02-29T00:13:14.903" Title="What's the reason that we have a different number of days each month?" Tags="&lt;history&gt;&lt;time&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="29721" PostTypeId="1" CreationDate="2019-02-24T00:08:45.670" Score="2" ViewCount="282" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Are there any kinds of binary pairings that haven't been discovered yet? Any that are particularly significant, or that might shed some light on binary stellar evolution or theories of capture processes, that are sought, but so far no examples have been found or at least suspected?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;pairing&quot; I mean a binary object where each is &lt;em&gt;a type of star&lt;/em&gt; or a &lt;em&gt;black hole&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars can be anything from brown dwarfs to neutron stars. Black holes should be very roughly comparable to stars in size. (I'm not asking about a star in orbit around a supermassive black hole in the center of a galaxy)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_gravitational_wave_observations&quot; rel=&quot;nofollow noreferrer&quot;&gt;recent observations of gravitational wave events&lt;/a&gt; have suggested a pair of black holes merging, and a pair of neutron stars merging.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="2153" LastEditDate="2019-02-25T14:33:56.663" LastActivityDate="2019-12-18T21:31:36.073" Title="Binary pairings that haven't been discovered yet?" Tags="&lt;black-hole&gt;&lt;stellar-evolution&gt;&lt;neutron-star&gt;&lt;binary-star&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="29722" PostTypeId="2" ParentId="29720" CreationDate="2019-02-24T00:19:07.167" Score="29" Body="&lt;p&gt;You make a great point. The reason behind the discrepancy between the dates is due to a complicated history behind it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calendar is based on the calendar created by ancient Romans, which is based on one Moon cycle. One lunar cycle is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;29.53 days.&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&lt;a href=&quot;http://www.universetoday.com/20620/lunar-year/&quot; rel=&quot;nofollow noreferrer&quot;&gt;www.universetoday.com/20620/lunar-year/&lt;/a&gt;&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which does not evenly divide into the 365.25 days of the year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note&lt;/em&gt;: The Romans borrowed the calendar from the ancient Greeks to develop the idea of a 10-month calendar that left approximately 60 days unaccounted for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, the earliest ancient Roman calendars had months that were either 29 or 30 days to account for this discrepancy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Romans started using the 10-month calendar in 738 B.C. Their months were initially called:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mensis Martius (March)&lt;/li&gt;&#xA;&lt;li&gt;Mensis Aprilis (April)&lt;/li&gt;&#xA;&lt;li&gt;Mensis Maius (May)&lt;/li&gt;&#xA;&lt;li&gt;Mensis Iunius (June)&lt;/li&gt;&#xA;&lt;li&gt;Mensis Quintilis (July, quin- meaning &quot;five&quot;)&lt;/li&gt;&#xA;&lt;li&gt;Mensis Sextilis (August, sex- meaning &quot;six&quot;)&lt;/li&gt;&#xA;&lt;li&gt;Mensis September (&quot;seven&quot;)&lt;/li&gt;&#xA;&lt;li&gt;Mensis October (&quot;eight&quot;)&lt;/li&gt;&#xA;&lt;li&gt;Mensis November (&quot;nine&quot;)&lt;/li&gt;&#xA;&lt;li&gt;Mensis December (&quot;ten&quot;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To account for the remaining &lt;span class=&quot;math-container&quot;&gt;$\sim60$&lt;/span&gt; days, Mensis Ianuarius (January) was added to the beginning of the year and Mensis Februarius (February) to the end of the year during Numa's reign around 700 B.C. with a leap year every 4 years. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Leap years were decreed by proclamation the pontifex maximus, in such years an extra month was added to the calendar.&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&lt;em&gt;Quoted from PM 2Ring&lt;/em&gt;&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The calendar stayed in that order until 452 B.C. when a small council of Romans, called the Decemvirs, moved February to follow January. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;in the initial 36 years of its adoption leap days were added every 3 years due to a misunderstanding (Romans used inclusive counting),&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&lt;em&gt;Quoted from PM 2Ring&lt;/em&gt;&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so that caused a big problem. In fact, by the start of the reign of Julius Caesar, the previous calendar was off by a whole week!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Julius Caesar noticed this and thus modified the Roman calendar in 46 B.C. to make each month have either 30 or 31 days, with the exception of Februarius[February], which had 29 days, with the three year leap year fixed into a four year leap year. Quintilis[July] was later renamed Julius[July] in his honor. Likewise, Sextilis[August] later became Augustus[August] to honor Augustus (Gaius Octavius, the first Roman Emperor). Augustus[August] was also given an extra day (taken away from Februarius[February]), so that Augustus and Julius would have an equal number of days, representing their equal power. This is the Julian Calendar.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This calendar was extremely accurate, but was still not accurate&#xA;  enough as it drifts by approximately 3 days every 400 years. So after&#xA;  a few hundred years, the drift becomes noticeable&#xA;  (i.e. the seasons don't properly match the calendar), which is&#xA;  important for farming, and therefore human survival. &lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&lt;em&gt;quoted from  CJ Dennis&lt;/em&gt;&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then a pope named Pope Gregory XIII modified the Julian calendar again in October 1582. This is the currently used &lt;strong&gt;Gregorian calendar&lt;/strong&gt;. He made it so that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Every year that is exactly divisible by four is a leap year, except for years that are exactly divisible by 100, but these centurial years are leap years if they are exactly divisible by 400.&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Gregorian_calendar&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Gregorian_calendar&lt;/a&gt;&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This calendar is the most accurate calendar currently used, and won't be a single day off until the year of &lt;span class=&quot;math-container&quot;&gt;$\pm 3200$&lt;/span&gt; AD. It isn't perfect though. Compared to the tropical year, it is 27 seconds too long. This is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;due to astronomical effects (the&#xA;  rotation of the Earth speeding up and slowing down, etc.) we can't be&#xA;  more precise than that. &lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&lt;em&gt;quoted from CJ Dennis&lt;/em&gt;&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As for your other question, unless the plates taken had accurate dates of when recorded, it would be rather impossible to calculate precisely.&lt;/p&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-10-22T22:11:05.927" LastActivityDate="2019-10-22T22:11:05.927" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="29723" PostTypeId="2" ParentId="29708" CreationDate="2019-02-24T00:35:55.283" Score="4" Body="&lt;p&gt;If I'm reading your question correctly, your suggestion is that Mercury, having a small magnetic field, should have some atmosphere, perhaps more atmosphere than Mars, with no magnetic field, though Mars does have some surface generated magnetic field regions, it just lacks a field generated by its core.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solar stripping due to no magnetic field isn't the only factor in maintaining a planet's atmosphere.   There's also &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_escape&quot; rel=&quot;nofollow noreferrer&quot;&gt;thermal escape or Jeans escape&lt;/a&gt;, which happens when the upper atmosphere's gas molecules move fast enough to escape the planet's gravity.  It's a byproduct of temperature and escape velocity.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional factors probably play a role, such as outgassing or how much atmosphere there was to begin with, or very large meteor impacts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jeans escape can happen whether there's a magnetic field or not.  Mercury has the hottest upper atmosphere temperature of the planets (Venus has hotter surface temperature, not upper atmosphere temperature), and it has the lowest escape velocity, so it's the worst of the 8 planets at maintaining an atmosphere.  Based on temperature and escape velocity, Mercury loses CO2, which is the heaviest common gas, so, even if Mercury had the strongest magnetic field imaginable, it would still lose its atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars is much colder with a slightly higher escape velocity so it maintains its atmosphere better, at least against thermal escape.   Mars should retain most of the CO2 and perhaps its N2 as well, but a strong solar wind can strip a planet of gas molecules that can't escape thermally, so both methods of atmospheric loss should be considered. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars very likely once had an atmosphere. Its polar ice and dried up river beds support that conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming it had them in the past, Mars may have lost its lighter gasses, such as Ammonia (NH3), Methane (CH4) and Water vapor (H20) as a result of atmospheric escape even before it lost its magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars may have also lost a share of its atmosphere &lt;a href=&quot;https://www.space.com/40344-crash-could-form-mars-moons.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;during a giant impact&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, while not directly related to your question.  Mercury doesn't really have what I'd call an atmosphere, &lt;a href=&quot;https://www.space.com/18644-mercury-atmosphere.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;it has an exosphere&lt;/a&gt;, which some people call an atmosphere, but I think that's poor terminology.   Mercury's exosphere is created by the same process that stripped Mars of it's atmosphere: the solar wind.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case Venus, while the &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Venus#Induced_magnetosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;induced magnetic field&lt;/a&gt; is considerably weaker and reaches less far from the planet than Earth's, it provides some protection of Venus's atmosphere.   &lt;a href=&quot;https://space.stackexchange.com/questions/13833/how-does-venus-thick-atmosphere-survive-against-the-solar-wind&quot;&gt;Stack exchange question on that here&lt;/a&gt;.   I think the answer could go into more detail, but it gets the gist of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus actually &lt;a href=&quot;https://blogs.ucl.ac.uk/science/2014/07/09/venus-is-losing-its-atmosphere/&quot; rel=&quot;nofollow noreferrer&quot;&gt;is losing atmosphere to the solar wind&lt;/a&gt;, but it has enough that the rate of loss hasn't stripped the planet, and may not happen for a long time, perhaps when the sun expands a bit . . . but I'm just throwing that out there).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how precise the data is on this, but Venus's magnetic field offers some protection, so does Venus's gravity relative to Mars's and the vast amount of atmosphere that Venus has also offers some protection.   Solar stripping is based on collisions, so a solar particle is likely to only remove one gas molecule. (kinda/sorta).   The solar wind is pretty dispersed, so it takes a long time for solar stripping to have an effect. I feel I should add that to my knowledge, that Mars lost its atmosphere to solar stripping is generally agreed with but not 100% certain. I personally like the big meteor impact as playing a role too, but I'm mostly guessing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a better model of atmospheric loss, we'd need to know what the planet's atmospheres were like billions of years ago, and we don't have that information, so there are some unknowns on the rates of loss.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="7411" LastEditDate="2019-02-24T18:08:01.503" LastActivityDate="2019-02-24T18:08:01.503" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29725" PostTypeId="2" ParentId="21948" CreationDate="2019-02-24T12:10:20.667" Score="2" Body="&lt;p&gt;Although userLTK's answer is right in addressing the when it was realised that the Earth should be an oblate and not an sphere - mostly by indirect means - the first direct measures of the departure from sphericity were by means of &lt;a href=&quot;https://en.wikipedia.org/wiki/Grade_measurement&quot; rel=&quot;nofollow noreferrer&quot;&gt;grade measurement&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, by measuring difference of astronomical latitude between two points (that is, elevation of the celestial North pole) along a given meridian and the topographical distance between those two points we can compute local radius of curvature of that meridian. By comparing local radius of curvature at different latitudes, we can adjust and ellipse or even test whether the Earth shape is actually an ellipsoid.&lt;/p&gt;&#xA;" OwnerUserId="14097" LastEditorUserId="14097" LastEditDate="2019-02-26T18:34:35.927" LastActivityDate="2019-02-26T18:34:35.927" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29727" PostTypeId="5" CreationDate="2019-02-24T16:43:28.733" Score="0" Body="&lt;p&gt;Celestial mechanics is a field regarding the motion and trajectory of certain objects in space. This field includes many sub-fields such as the launching and orbits artificial satellites, and lunar theory. Celestial mechanics currently consists of five main fields:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Orbital mechanics.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Planetary dynamics.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Astrodynamics and stellar dynamics.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Stellar dynamics and galactic dynamics.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Computational astrophysics.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All of the above fields obviously overlap, but mostly are treated as separate, especially the study of the motion of stars within galaxies and interactions between whole galaxies, which both tend to rely heavily on fluid mechanics.&lt;/p&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-02-24T18:27:11.823" LastActivityDate="2019-02-24T18:27:11.823" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29728" PostTypeId="4" CreationDate="2019-02-24T16:43:28.733" Score="0" Body="Questions regarding the motions of celestial objects." OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-02-24T18:27:19.687" LastActivityDate="2019-02-24T18:27:19.687" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29729" PostTypeId="2" ParentId="29708" CreationDate="2019-02-24T18:19:27.400" Score="5" Body="&lt;p&gt;usrLTK's answer provides a lot of good details and in particular explains why Mercury wouldn't have much of an atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me complicate the picture a little by pointed out that some recent research indicates that magnetic fields may not be the guaranteed, automatic atmosphere-protection devices that conventional wisdom suggests. In particular, &lt;a href=&quot;https://www.aanda.org/articles/aa/full_html/2018/06/aa32934-18/aa32934-18.html&quot; rel=&quot;noreferrer&quot;&gt;Gunell et al. (2018)&lt;/a&gt; point out that planetary magnetic fields may, in some circumstances, &lt;em&gt;increase&lt;/em&gt; atmospheric loss (basically, the cusps and polar caps of the magnetic field can accelerate atmospheric ions out into space). Their calculations suggest, for example, that if Venus had a magnetic field as strong as Earth's, it would lose atmosphere &lt;em&gt;faster&lt;/em&gt; than it currently does. Meanwhile, if Earth had only a Venus-style weak, induced magnetic field, it would lose hydrogen a little faster but oxygen &lt;em&gt;more slowly&lt;/em&gt; than it does now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intriguingly, Venus is currently losing atmosphere at a rate of about 0.5 kg/s, while Earth is losing atmosphere at 1.4 kg/s -- almost three times faster than Venus. (For Mars, the rate is somewhere between 0.7 and 2.1 kg/s.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2019-02-24T18:19:27.400" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29730" PostTypeId="1" AcceptedAnswerId="29758" CreationDate="2019-02-24T18:27:28.100" Score="2" ViewCount="150" Body="&lt;p&gt;Recently I bought a &lt;a href=&quot;https://www.meade.com/polaris-80mm-german-equatorial-refractor.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;telescope&lt;/a&gt; and I started to take pictures of my observations. The most basic solution, I'm using a &lt;a href=&quot;https://www.meade.com/smart-phone-adapter.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;smartphone adapter&lt;/a&gt; and my &lt;a href=&quot;https://www.mi.com/global/mi6/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Xiaomi Mi 6&lt;/a&gt; camera to do that but the images (except for the Moon sometimes) is too bad and Jupiter and its moons not even show in my screen. So I thought in use a more serious device to take the pictures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have in my personal finance control a monthly quota to &quot;astronomy&quot;, but to start I think in spend about 200€ (~$230,00).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I work with programming for 10 years (C#), so I can use something more &quot;crude&quot; if needed (like Raspberry Pi or Arduino). You guys can suggest something to start take decent shots of  Jupiter and its moons?&lt;/p&gt;&#xA;" OwnerUserId="25980" LastActivityDate="2019-02-27T20:19:58.230" Title="Options for first time astrophotography w/ 80/900 refractor?" Tags="&lt;amateur-observing&gt;&lt;photography&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29731" PostTypeId="2" ParentId="27522" CreationDate="2019-02-24T21:47:08.563" Score="-1" Body="&lt;p&gt;In SI units the &lt;a href=&quot;https://en.wikipedia.org/wiki/Force_between_magnets&quot; rel=&quot;nofollow noreferrer&quot;&gt;force between magnets&lt;/a&gt; is &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\approx 10^{-7}\frac{m_Sm_E}{r^4}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$=~10^{-7}\frac{(3.5 \times10^{29}) (7.84\times 10^{22})}{(1.5 \times 10^{11})^4} = 5.4 $$&lt;/span&gt; Newtons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnetic dipole moments &lt;span class=&quot;math-container&quot;&gt;$m_E$&lt;/span&gt; from &lt;a href=&quot;https://arxiv.org/pdf/1007.4497.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;span class=&quot;math-container&quot;&gt;$m_S$&lt;/span&gt; from &lt;a href=&quot;https://www.conservapedia.com/Sun&quot; rel=&quot;nofollow noreferrer&quot;&gt;this impeccable &quot;scientific&quot; source&lt;/a&gt;. From Figure 4 &lt;a href=&quot;https://ned.ipac.caltech.edu/level5/March03/Vallee2/Vallee2_3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, &lt;span class=&quot;math-container&quot;&gt;$m_S$&lt;/span&gt; is about &lt;span class=&quot;math-container&quot;&gt;$10^7 \times m_E$&lt;/span&gt; so the value for &lt;span class=&quot;math-container&quot;&gt;$m_S$&lt;/span&gt; looks about right.&lt;/p&gt;&#xA;" OwnerUserId="17743" LastEditorUserId="17743" LastEditDate="2019-03-13T18:25:00.193" LastActivityDate="2019-03-13T18:25:00.193" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29732" PostTypeId="5" CreationDate="2019-02-24T22:08:38.827" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2019-02-24T22:08:38.827" LastActivityDate="2019-02-24T22:08:38.827" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29733" PostTypeId="4" CreationDate="2019-02-24T22:08:38.827" Score="0" Body="Questions regarding the measurement of the amount of light coming from astronomical objects. " OwnerUserId="9527" LastEditorUserId="616" LastEditDate="2020-01-22T14:06:20.143" LastActivityDate="2020-01-22T14:06:20.143" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29734" PostTypeId="5" CreationDate="2019-02-24T22:14:17.293" Score="0" Body="&lt;p&gt;A meteorite is part of 3 main stages:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Meteoroid: A solid piece of debris in space.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Meteor: The light phenomenon when a meteoroid enters an atmosphere and burns due to atmospheric friction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Meteorite&lt;/em&gt;&lt;/strong&gt;: A solid piece of debris that originates from outer space and survives passage through the atmosphere and its impact with the Earth's surface. It can be in any shape, size, or color.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-02-25T08:06:34.283" LastActivityDate="2019-02-25T08:06:34.283" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29735" PostTypeId="4" CreationDate="2019-02-24T22:14:17.293" Score="0" Body="Questions regarding one of the three forms of space debris, this case a chunk of debris that survives its passage through the atmosphere and lands on a planet intact." OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-02-25T08:06:41.773" LastActivityDate="2019-02-25T08:06:41.773" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29736" PostTypeId="5" CreationDate="2019-02-24T22:18:36.847" Score="0" Body="&lt;p&gt;Declination is one of the two main angles that locate a point or an object on the celestial sphere, or the observable half-sphere for an observer on Earth, Part of the Equatorial Coordinate System. The angle of declination is measured north or south of the celestial equator, along the hour circle passing through the point in question.&lt;/p&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-02-25T08:06:51.527" LastActivityDate="2019-02-25T08:06:51.527" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29737" PostTypeId="4" CreationDate="2019-02-24T22:18:36.847" Score="0" Body="Questions regarding one of the two main angles(coordinates) of an object perceived by an observer on Earth." OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-02-25T08:06:44.547" LastActivityDate="2019-02-25T08:06:44.547" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29738" PostTypeId="5" CreationDate="2019-02-24T22:24:43.847" Score="0" Body="&lt;p&gt;A near-Earth object is any small or medium sized celestial body[mostly asteroids, but some are comets] whose orbit around the sun brings it into close proximity with Earth. However, other definition exist. Another one is that a celestial object whose perihelion is less than 1.3 AU is considered a near-Earth Object. Some near-Earth Objects have elliptical orbits that cross with the Earth's orbit. If such an object is larger than 140 meters across, it is then considered a potentially hazardous object. &lt;/p&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-02-25T08:06:38.707" LastActivityDate="2019-02-25T08:06:38.707" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29739" PostTypeId="4" CreationDate="2019-02-24T22:24:43.847" Score="0" Body="Questions regarding objects whose orbits bring them in close proximity to Earth, posing potential hazard." OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-02-25T08:06:31.310" LastActivityDate="2019-02-25T08:06:31.310" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29741" PostTypeId="1" AcceptedAnswerId="29817" CreationDate="2019-02-25T00:48:59.230" Score="2" ViewCount="140" Body="&lt;p&gt;Answers to the question &lt;a href=&quot;https://astronomy.stackexchange.com/q/4834/7982&quot;&gt;How did Kepler determine the orbital period of Mars?&lt;/a&gt; describing careful observations centuries ago got me thinking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What was the first analysis of observations that directly demonstrated that the movement of a planet &lt;em&gt;or asteroid&lt;/em&gt; in its orbit was affected by gravitational attraction by another planet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't mean inferences based on the spacing of orbits, I'm thinking of detailed positional measurements; something like &lt;em&gt;planet/asteroid A&lt;/em&gt; was X kilometers or Y arcseconds away from it's predicted position and it couldn't be accounted for if not for the gravitational effects of &lt;em&gt;planet B&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure if &lt;a href=&quot;https://astronomy.stackexchange.com/q/29742/7982&quot;&gt;First observation that the Sun and Jupiter (and friends) move around a common barycenter?&lt;/a&gt; is a completely separate question, or if this confirmation happened all at once, but I've currently asked it separately.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-02-28T05:20:12.440" LastActivityDate="2019-08-26T14:38:31.537" Title="First observation that the movement of a planet or asteroid in its orbit was affected by another planet?" Tags="&lt;gravity&gt;&lt;solar-system&gt;&lt;history&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29742" PostTypeId="1" CreationDate="2019-02-25T00:53:47.093" Score="7" ViewCount="303" Body="&lt;p&gt;Answers to the question &lt;a href=&quot;https://astronomy.stackexchange.com/q/4834/7982&quot;&gt;How did Kepler determine the orbital period of Mars?&lt;/a&gt; describing careful observations centuries ago got me thinking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was the first analysis of observations that directly demonstrated that the Sun and the outer giant planets were moving around a common barycenter rather than all planets rotating around a fixed Sun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure if &lt;a href=&quot;https://astronomy.stackexchange.com/q/29741/7982&quot;&gt;First observation that the movement of a planet or asteroid in its orbit was affected by another planet?&lt;/a&gt; is a completely separate question, or if this confirmation happened all at once, but I've currently asked it separately.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-02-28T05:20:29.840" LastActivityDate="2020-05-25T03:05:19.777" Title="First observation that the Sun and Jupiter (and friends) move around a common barycenter?" Tags="&lt;the-sun&gt;&lt;solar-system&gt;&lt;history&gt;" AnswerCount="1" CommentCount="19" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29743" PostTypeId="1" AcceptedAnswerId="29820" CreationDate="2019-02-25T01:23:01.083" Score="2" ViewCount="131" Body="&lt;p&gt;I'll still remember the handout my astronomy professor gave us more than several decades ago; &lt;em&gt;Our Friends the Polytropes&lt;/em&gt;. We spent a lot of time learning how &lt;a href=&quot;https://en.wikipedia.org/wiki/Polytrope&quot; rel=&quot;nofollow noreferrer&quot;&gt;polytropes&lt;/a&gt;, simple analytical power-law models could reproduce some basic thermodynamic properties of stars, and how those properties would tend to vary with mass (and perhaps other parameters, I don't remember now).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have polytrope models of stars (&lt;em&gt;or other simple analytical models of stars&lt;/em&gt;) ever been &quot;evolved&quot; analytically in time? In other words, in addition to variations on radius, have a set of equations ever been written which also included a &lt;span class=&quot;math-container&quot;&gt;$\frac{\partial}{\partial t}$&lt;/span&gt; in such a way that they were amenable to an analytic solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;note:&lt;/strong&gt; I've added the &lt;a href=&quot;/questions/tagged/history&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;history&amp;#39;&quot; rel=&quot;tag&quot;&gt;history&lt;/a&gt; tag since I'm asking about the &quot;gold 'ole days&quot; rather than about modern techniques.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-02-25T10:05:37.493" LastActivityDate="2019-03-02T09:28:08.533" Title="Has stellar evolution ever been modeled analytically?" Tags="&lt;stellar-evolution&gt;&lt;history&gt;&lt;modeling&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29744" PostTypeId="2" ParentId="29665" CreationDate="2019-02-25T02:45:22.573" Score="-1" Body="&lt;p&gt;The asteroids can be accelerated and de-accelerated depending on which side of a much larger rotating object the asteroid passes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If asteroid passes on the side in the direction of the rotation of the object, the asteroid experiences a gravity boost. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the asteroid passes on the side opposite to the direction of rotation, the asteroid experiences gravity breaking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the breaking slows down the asteroid sufficiently so it can be captured then it collides with object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of gravity assisted flight, the Mars probes used the gravity assist of the Earth to reach Venus, then the gravity assist of Venus to fling it towards Mar. Landing on Mars used gravity breaking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for the Mars lander to land on Mars required fuel and a parachute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity assisted acceleration and de-acceleration is accomplished by exchanging angular momentum with a larger rotating object and is based on conservation of angular momentum. &lt;/p&gt;&#xA;" OwnerUserId="26037" LastEditorUserId="26037" LastEditDate="2019-02-25T07:57:23.497" LastActivityDate="2019-02-25T07:57:23.497" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29749" PostTypeId="2" ParentId="29708" CreationDate="2019-02-25T07:54:53.893" Score="0" Body="&lt;p&gt;Nitrogen and oxygen have paramagnetic moments (strong) where as carbon dioxide has a diamagnetic moment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paramagnetic moments are typically much stronger than diamagnetic moments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since carbon dioxide is a heavy trace gas, it rarely leaves the troposphere.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, since Venus rotates in the direction opposite to all the other planets in the solar system, there was a conjecture that Venus might have had a collision with a carbon rich object in the distant past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And since Earth and Mars are down wind from Venus, most of the carbon on Earth and on Mars was a result of that collision. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury has a little carbon but the Sun is a heavy metal star (helium is considered to be a heavy metal) where 1% of the Sun is oxygen - all the elements found on the surface of Mercury can be produced by the Sun from that 1%.&lt;/p&gt;&#xA;" OwnerUserId="26037" LastActivityDate="2019-02-25T07:54:53.893" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29751" PostTypeId="1" CreationDate="2019-02-25T08:32:51.563" Score="1" ViewCount="125" Body="&lt;p&gt;When a rogue brown dwarf star hits a white dwarf star, will the collision at most cause a nova?&lt;br /&gt;&#xA;Or can the explosion be heavier?&lt;/p&gt;&#xA;" OwnerUserId="26045" LastEditorDisplayName="user1569" LastEditDate="2019-02-25T12:49:06.747" LastActivityDate="2019-02-25T21:29:11.610" Title="When a rogue brown dwarf star hits a white dwarf star, will the collision at most cause a nova?" Tags="&lt;star&gt;&lt;astrophysics&gt;&lt;deep-sky-observing&gt;&lt;collision&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2019-03-12T19:55:01.213" ContentLicense="CC BY-SA 4.0" />
  <row Id="29752" PostTypeId="2" ParentId="29751" CreationDate="2019-02-25T14:41:28.300" Score="5" Body="&lt;p&gt;Collisions between a &lt;strong&gt;rogue&lt;/strong&gt; brown dwarf and any other star would be very rare because the space between them is so vast.   I don't want to say it'd never happen, but it would be a rare event.   It's much more common for two stars that are already in the same system to collide by spiraling into each other, usually by tidal decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A collision with a brown dwarf and a white dwarf has been recognized.   The nova of 1670 didn't behave like an exploding star and recent research identified it as spiraling in collision &lt;a href=&quot;https://phys.org/news/2018-10-nova-white-dwarf-brown-collide.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;between a brown dwarf and a white dwarf&lt;/a&gt;.  &lt;a href=&quot;https://www.research.manchester.ac.uk/portal/files/77046206/CKVul_ALMA_RESUBMISSION_20180911.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Original article and abstract here&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for what kind of explosion is likely.  I would imagine something like a type-A supernova would be possible if there was sufficient mass to exceed the Chandrasekhar limit of about 1.4 solar masses.   Because brown dwarfs are much less mass than that, about .013 to about .08 solar masses, the white dwarf would need to be at least 1.32 solar masses prior to the collision.  Most white dwarfs are less than that.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/CK_Vulpeculae&quot; rel=&quot;nofollow noreferrer&quot;&gt;CK Vulpeculae&lt;/a&gt;, the 1670 nova wasn't a type 1a supernova.  If it had been, it would have been over 100 times brighter than Venus (ballpark estimate) based on it's distance 2280 light years (plus/minus 490).   It would have been many times brighter than it's visible but &lt;a href=&quot;http://www.astro.wisc.edu/~dolan/constellations/extra/brightest.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;not very bright peak of 2.6&lt;/a&gt;.   &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler%27s_Supernova&quot; rel=&quot;nofollow noreferrer&quot;&gt;SN 1604, for comparison&lt;/a&gt; was about 20,000 light years distant and it was brighter than any star in the sky and visible during the day at it's peak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the two colliding objects and energy of collision would have considerable effect on the brightness, but, unless the Chandrasekar limit is reached, and keeping in mind that direct collisions would be rare, and it wouldn't hit all at once, the result should be much less energetic than a supernova, but if it's close enough, it could still be a visible nova in the sky, like CK Vulpeculae in 1670.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2019-02-25T21:29:11.610" LastActivityDate="2019-02-25T21:29:11.610" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29753" PostTypeId="2" ParentId="29743" CreationDate="2019-02-25T15:01:40.743" Score="4" Body="&lt;p&gt;Analytic models have been applied to various phases of the evolution, though it would be impossible to apply a single model to all phases because the physics changes so much.   Also, a star will often have very different physics going on in various different parts of the star, so analytic models sometimes have to treat different parts separately and then unify the results across boundaries (just as numerical simulations do).  For example, I've seen different polytrope indices used in the convection, radiative, and core regions of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that your interest is when there is not a steady-state assumption, but rather time is a dynamical variable.  But it is not always necessary to use time as a dynamical variable to do evolution.  You can simply do a steady-state analytical model (like a polytrope, perhaps supported by a numerical piece that solves for some parameter like core pressure), and allow the parameters in your solution to be time varying in some simple analytical way.   For example, you could study how mass loss affects a star by doing steady-state analytical models and just let the mass change with time according to some analytical mass-loss prescription (such as &quot;Reimers' Law&quot;, something simple).  So steady-state analytical models can be elevated into evolutionary models if you simply have an analytical way to change the parameters with time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another simple example is the evolution of a fully convective protostar.  You simply fix the mass and initial radius, and you solve the interior by assuming it's all at the same entropy (a reasonable approximation for fully convective stars).  Then you fix the surface temperature to lie on the Hayashi track, which for a very simple model could mean just pegging the surface T to 3000 K.  That and the initial radius determines the luminosity, and the constant entropy assumption fixes the internal structure, so then you simply let it lose energy at the rate of the luminosity, and use the internal energy as the variable that changes with time, always updating the radius to be consistent with the new internal energy (a la the virial theorem), and that gives the new luminosity and so forth.  A simple fully analytic model until the star is no longer fully convective (and we then might call it a pre-main-sequence star).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also seen white dwarf cooling done analytically.  It's a similar idea-- take the surface temperature as an initial condition and just let it lose heat via its luminosity.  The radius doesn't really change, so all you need is to keep updating the surface temperature as a function of the internal energy, which can be handled via some analytic internal heat transport treatment.  If you are willing to make approximations that address the key physics, you can do almost anything analytically.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2019-02-25T15:01:40.743" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29754" PostTypeId="1" CreationDate="2019-02-25T15:14:36.557" Score="2" ViewCount="89" Body="&lt;p&gt;Most diagrams I see of sunlight look like something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vkTLX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vkTLX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They show that for basic geometric analyses, the sun's radiation can be thought of as 'thin lines' that strike earth. Further, at an equinox, those rays strike hit the horizon of every point on earth (except the poles) directly from the east. So, the diameter for the earth does not matter at all it terms of the angle at which the rays strike.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious - how far will this conceptual frame take me? What is a better way to conceptualize how the sun's light strikes the earth?&lt;/p&gt;&#xA;" OwnerUserId="26048" LastEditorUserId="7982" LastEditDate="2019-02-25T16:24:12.137" LastActivityDate="2019-07-25T21:01:01.163" Title="How should I think about sunlight for the purposes of analyzing the angles it makes w/ earth?" Tags="&lt;positional-astronomy&gt;&lt;sun-rays&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29755" PostTypeId="2" ParentId="29754" CreationDate="2019-02-25T15:29:58.203" Score="0" Body="&lt;p&gt;There isn't a better way to conceptualize this, because that's how it happens. The sun is very large and very far away, so the rays reaching us are effectively parallel to one another. This is already the best way to conceptualize how light strikes the earth - if there were a better one, you probably would have heard about it instead of this way.&lt;/p&gt;&#xA;" OwnerUserId="10488" LastActivityDate="2019-02-25T15:29:58.203" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29756" PostTypeId="2" ParentId="29754" CreationDate="2019-02-25T20:06:05.973" Score="1" Body="&lt;p&gt;This is a first-order approximation, and works well for simple models of irradiation of the atmosphere or the Earth's surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, a simple experiment with a pinhole camera or a single lens will show that this model doesn't work in imaging situations.  The sun is an &quot;extended object,&quot; meaning it is not a point source, so the image you get is not what you would expect from a point source.  Worse, even if the sun were a point source, the distance to the geometric focal point is different very slightly from the distance if all the sun's rays were perfectly parallel.  See the Lensmaker's Formula,  &lt;span class=&quot;math-container&quot;&gt;$ \frac{1}{f} = \frac{1}{p} + \frac{1}{q} $&lt;/span&gt; where p and q are the distances from the lens to the source and the image respectively.   (I'll warn that for a reasonable lens, say focal length &lt;span class=&quot;math-container&quot;&gt;$ f = 10 m$&lt;/span&gt;, the difference between focal location for rays that are parallel and rays from a point source &lt;span class=&quot;math-container&quot;&gt;$150\space million\space km$&lt;/span&gt; away is ridiculously small) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if you want to model the umbra and penumbra during a solar or lunar eclipse, you'll need to take the physical extent of the sun into consideration.  &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2019-02-25T20:06:05.973" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29758" PostTypeId="2" ParentId="29730" CreationDate="2019-02-25T22:37:28.803" Score="4" Body="&lt;p&gt;There is no &quot;best&quot; camera as it really depends on the subject you intend to image and the optical instrument you intend to use with it.  This could turn into a very long answer, so I'll try to offer a short response first and then offer a bit more to consider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For planetary imaging, use a camera capable of doing video frames and this does not need to be a camera with a large chip or high resolution.  Jupiter, for example, is just fractionally under 44 arc-&lt;em&gt;seconds&lt;/em&gt; diameter when viewed at opposition. (compare to the moon at roughly 30 arc-minutes or more than 40x the angular width).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically you'd capture perhap 30 seconds to a couple of minutes worth of frames and then feed those frames into stacking software.  Ideally the video frames use a &quot;lossless&quot; format.  Several image processing applications can stack the frames (e.g. AutoStakkert &amp;amp; Registax are popular and free).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A camera such as the ZWO ASI120MC-S is very popular and in your budget.  There's the ASI290MC for those with deeper pockets (and versions for those with much deeper pockets).    From QHYCCD there's a QHY5R-II ... more to the entry-range with lower resolution and a smaller chip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally your scope would be on a motorized equatorial mount so that it keeps the object in your field of view (otherwise the object will quickly drift out of the camera frame.  For this scope &amp;amp; field of view combination, that would happen in about 45 seconds for a planet in the &quot;center&quot; of the frame to drift out of frame completely).  If using a 2x barlow, it would happen in about half the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/01Y7Q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/01Y7Q.png&quot; alt=&quot;Simulated field of view using 80/900 refractor &amp;amp; ZWO ASI120MC-S camera&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A barlow is usually used for planetary imaging (the simulated field of view did not use a barlow).  To understand why, you'd want to read about Shannon-Nyquist sampling theory and how it applies to astrophotography.&lt;/p&gt;&#xA;" OwnerUserId="26051" LastActivityDate="2019-02-25T22:37:28.803" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29759" PostTypeId="1" AcceptedAnswerId="29760" CreationDate="2019-02-26T00:40:20.470" Score="2" ViewCount="262" Body="&lt;p&gt;As an exercise, I am trying to calculate the JD from a given date in the Gregorian Calendar at a time given in UT. Furthermore, I want to do so from first principles without relying on any formula. Let's say I choose UT 18:30 on 2/28/2010. I am not able to obtain a result consistent with the U.S. Naval Observatory results. I am in need of a calendar guru to show me where my calculation is going astray.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I find the number of years from 4713 BCE (Julian Date 0) and the target year making sure to add an extra one so as to include 0 CE. This is &lt;span class=&quot;math-container&quot;&gt;$4713 + 1 + 2010 = 6724$&lt;/span&gt; years. The number of these years which are leap is &lt;span class=&quot;math-container&quot;&gt;$\frac{6724}{4} - 3 = 1681$&lt;/span&gt;. This equation attempts to reflect that every fourth year is a leap year except for those years falling after 1582 (the year of the Gregorian Calendar reform) which are both divisible by &lt;span class=&quot;math-container&quot;&gt;$100$&lt;/span&gt; and &lt;em&gt;not&lt;/em&gt; by &lt;span class=&quot;math-container&quot;&gt;$400$&lt;/span&gt;. There are three such years which need to be deducted from the total-- 1700, 1800, 1900. This leaves &lt;span class=&quot;math-container&quot;&gt;$6724 - 1681 = 5043$&lt;/span&gt; non-leap years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I convert these to days and subtract the &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; days which were skipped as part of the reform from October 4 to October 15 in 1582. &lt;span class=&quot;math-container&quot;&gt;$5043*(365) + 1681*(366) - 10 = 2,455,928$&lt;/span&gt;. Next, I find the number of days from January 1st to the target day in the target year (February 28th). This is &lt;span class=&quot;math-container&quot;&gt;$59$&lt;/span&gt; days and so I have &lt;span class=&quot;math-container&quot;&gt;$2,455,987$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the time number of seconds afternoon on the target day is calculated as a fraction of the number of seconds in a 24 hour period. This works out to be &lt;span class=&quot;math-container&quot;&gt;$\frac{6(60)(60) + 30(60)}{24(60)(60)} \approx 0.27$&lt;/span&gt;. Adding this to the earlier result yields &lt;span class=&quot;math-container&quot;&gt;$JD = 2455987.27$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;actual&lt;/em&gt; answer is &lt;span class=&quot;math-container&quot;&gt;$2455255.77$&lt;/span&gt; which leaves a difference of &lt;span class=&quot;math-container&quot;&gt;$2455987.27 - 2455255.77 = 731.5$&lt;/span&gt; days. This is nearly two years. Obviously, I have either misunderstood the calculation or have performed it incorrectly. Any help would be much appreciated!&lt;/p&gt;&#xA;" OwnerUserId="26052" LastActivityDate="2019-02-26T22:22:47.193" Title="Determine Julian Date from Gregorian without formula" Tags="&lt;time&gt;&lt;celestial-mechanics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29760" PostTypeId="2" ParentId="29759" CreationDate="2019-02-26T04:48:17.587" Score="3" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Year_zero&quot; rel=&quot;nofollow noreferrer&quot;&gt;Year 0&lt;/a&gt; does not exist in BCE/CE and BC/AD notation. It jumps from 1 BCE to 1 CE directly. So if we look at the sequence of years in BCE/CE notation, we have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3 BCE&#xA;2 BCE&#xA;1 BCE&#xA;1 CE&#xA;2 CE&#xA;3 CE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you want to calculate the number of years between, say, January 1, 2 BCE and January 1, 2 CE, you can add the year numbers to get 4, but then you need to &lt;em&gt;subtract&lt;/em&gt; 1 to remove year 0 from the total, giving a total of 3 years. In your particular case, this leads to &lt;span class=&quot;math-container&quot;&gt;$4713 - 1 + 2010 = 6722$&lt;/span&gt; years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In astronomy, we often use another year numbering system called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_year_numbering&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomical year numbering&lt;/a&gt;, which includes a year 0 and makes things more convenient arithmetically when calculating time intervals between eras. It goes (-3, -2, -1, 0, 1, 2, etc.) To convert years BCE to the Astronomical year notation, we subtract 1 and add minus(-) in front of the year. For example, the year 4713 BCE is -4712 in Astronomical year notation. Your year interval is then calculated as &lt;span class=&quot;math-container&quot;&gt;$2010 - (\text{-}4712) = 2010 + 4712 = 6722$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the leap years calculation, since -4712 is a leap year, this simplifies matters, but don't forget to take the smallest integer greater than or equal to your result (the ceiling function). So in this case &lt;span class=&quot;math-container&quot;&gt;$\frac{6722}{4} - 3 = 1677.5$&lt;/span&gt;, and the ceiling is 1678, and that is the number of leap years between your years. Also, in your initial calculation, you had forgotten to subtract 3, because 6724 / 4 - 3 = 1678, not 1681. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the time interval between January 1 and February 28 is 58 days, not 59, because we do not count the last date as part of the interval. This gives a total of &lt;span class=&quot;math-container&quot;&gt;$5044 * 365 + 1678 * 366 - 10 + 58 + \frac{6 * 60 * 60 + 30 * 60}{24 * 60 * 60} = 2455256.27$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You state that the actual answer is 2455255.77, but in fact the answer really is 2455256.27. Perhaps you typed &quot;06:30&quot; instead of &quot;18:30&quot; in the converter. &lt;/p&gt;&#xA;" OwnerUserId="19814" LastEditorUserId="19814" LastEditDate="2019-02-26T22:22:47.193" LastActivityDate="2019-02-26T22:22:47.193" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29761" PostTypeId="1" AcceptedAnswerId="29766" CreationDate="2019-02-26T07:58:58.850" Score="2" ViewCount="132" Body="&lt;p&gt;I am pretty new to astronomy, and I have fallen down rabbit hole creating a new Metonic calendar in Javascript. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As part of this I would like to calculate when the sun is half way between a solstice and an equinox as viewed on a horison. Forgive me if my terminology is off but I believe this is known as the azimuth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My issue is that I do not know what the correct values are. I am already  calculating the solstice and equinoxes using the suncalc javascript library&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/mourner/suncalc&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/mourner/suncalc&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in this code are the following 4 lines that determine the solstices and equinoxes. 0 is the spring equinox and 3 is the winter solstice.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;switch( k ) {&#xA;     case 0: JDE0 = 2451623.80984 + 365242.37404*Y + 0.05169*POW2(Y) - 0.00411*POW3(Y) - 0.00057*POW4(Y); break;&#xA;     case 1: JDE0 = 2451716.56767 + 365241.62603*Y + 0.00325*POW2(Y) + 0.00888*POW3(Y) - 0.00030*POW4(Y); break;&#xA;     case 2: JDE0 = 2451810.21715 + 365242.01767*Y - 0.11575*POW2(Y) + 0.00337*POW3(Y) + 0.00078*POW4(Y); break;&#xA;     case 3: JDE0 = 2451900.05952 + 365242.74049*Y - 0.06223*POW2(Y) - 0.00823*POW3(Y) + 0.00032*POW4(Y); break;&#xA;   }&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;can anyone help me understand what the values might be to determine the halfway points between these solstice and equinox values?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thank you for your help. &lt;/p&gt;&#xA;" OwnerUserId="26058" LastEditorUserId="26058" LastEditDate="2019-02-26T09:12:48.380" LastActivityDate="2019-03-04T01:36:34.533" Title="Solar azimuth for celtic fire festivals" Tags="&lt;the-sun&gt;&lt;azimuth&gt;&lt;equinox&gt;&lt;sunrise&gt;&lt;solstice&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29763" PostTypeId="1" AcceptedAnswerId="29764" CreationDate="2019-02-26T10:53:38.883" Score="6" ViewCount="942" Body="&lt;p&gt;I am reading the book &quot;Fourier Analysis&quot; from T. W. Körner, and in chapter 95, he explains how you can compute the diameter of stars on a photographic plate. He says that &quot;Since observations of the nearest stars at six-monthly intervals (i.e. using a diameter of the earth's orbit as a surveyor's base line) enable astronomers to measure the distance of these stars, knowledge of the apparent diameter (i.e. the diameters of the discs on the photographic plate) would then enable us to calculate the true diameters of the nearest stars&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do you calculate the true diameter of a star given its apparent diameter on a photographic plate? Is it related to the &quot;zooming capacity&quot; of the telescope one uses?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if for instance, using a 100'' telescope, you measure that a star has an apparent diameter of 0.66mm, can you directly compute the true diameter of the star?&lt;/p&gt;&#xA;" OwnerUserId="26060" LastEditorUserId="22782" LastEditDate="2019-02-27T15:41:10.677" LastActivityDate="2019-02-27T15:41:10.677" Title="Calculate the true diameter of stars from photographic plate" Tags="&lt;observational-astronomy&gt;&lt;distances&gt;&lt;photography&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29764" PostTypeId="2" ParentId="29763" CreationDate="2019-02-26T12:03:55.703" Score="12" Body="&lt;p&gt;Let's pretend for the moment that this is possible (it's generally not, as I'll explain below) and see how we would go about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, this is basic trigonometry: you have a measured angle (&lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;, the diameter of the star), you know the distance to the star (&lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt;, from the six-month-separated parallax measurements), so to determine the linear diameter of the star (&lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt;, in the same units as the distance) it's a simple matter of &lt;span class=&quot;math-container&quot;&gt;$\sin \alpha = d/D$&lt;/span&gt;. In practice, the angles are small enough that you can use the small-angle approximation, so it's really just &lt;span class=&quot;math-container&quot;&gt;$\alpha \approx d/D \rightarrow d = \alpha D$&lt;/span&gt; (assuming &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; is in radians).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As your book notes, you need to know the distance to the star; for nearby stars, this can be obtained by &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_parallax&quot; rel=&quot;noreferrer&quot;&gt;stellar parallax&lt;/a&gt;, which involves the same basic trigonometric argument (only this time you know the &quot;diameter&quot;, which is the size of the Earth's orbit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why is this, as I said, not generally possible? In practice, &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_resolution&quot; rel=&quot;noreferrer&quot;&gt;the diameter of the &lt;em&gt;telescope&lt;/em&gt; sets a lower limit on &lt;em&gt;observed&lt;/em&gt; angular sizes&lt;/a&gt; -- anything with a true diameter smaller than the telescope's angular limit will appear to have an angular diameter &lt;span class=&quot;math-container&quot;&gt;$\approx$&lt;/span&gt; that limit. In addition, any telescope underneath the Earth's atmosphere will suffer the effects of blurring due to atmospheric turbulence, which sets an angular diameter limit of &lt;span class=&quot;math-container&quot;&gt;$\sim 0.5$&lt;/span&gt; arc seconds (full-width half-maximum) or worse at the best sites. Since the &lt;em&gt;largest&lt;/em&gt; stars in angular terms have &lt;a href=&quot;https://www.eso.org/public/news/eso9706/&quot; rel=&quot;noreferrer&quot;&gt;diameters of about 0.05 arc seconds&lt;/a&gt;, the result is that all stars in the image (photographic plate, CCD camera, etc.) will have the &lt;em&gt;same&lt;/em&gt; observed angular diameter (e.g., 0.5 arc seconds at a really good site), regardless of their true diameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It &lt;em&gt;is&lt;/em&gt; possible to do better for bright, nearby stars using large telescopes &lt;em&gt;in space&lt;/em&gt; or special interferometric techniques -- see the second link in the previous paragraph for an example -- but this is going beyond the &quot;stars on photographic plates&quot; scenario your book seems to envisage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And if for instance, using a 100'' telescope, you measure that a star has an apparent diameter of 0.66mm, can you directly compute the true diameter of the star?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leaving aside the problems mentioned above, you don't have enough information. You need to convert the linear diameter in mm to an &lt;em&gt;angular&lt;/em&gt; diameter, which means you need to know the plate scale (e.g., arcsec/mm) of the particular telescope + camera setup. You also need to know the distance to the star.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2019-02-26T12:03:55.703" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="29765" PostTypeId="1" CreationDate="2019-02-26T17:56:30.983" Score="1" ViewCount="24" Body="&lt;p&gt;I am studying astrophysics on the side and came across this problem in the textbook: &lt;a href=&quot;https://i.stack.imgur.com/IKfkw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IKfkw.png&quot; alt=&quot;Modeling Malmquist Bias Problem&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not quite sure how to approach it but I do know I need to account for the Malmquist bias to account for the correct &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt;. Is there any insight I can get here?&lt;/p&gt;&#xA;" OwnerUserId="26063" LastEditorUserId="7472" LastEditDate="2019-02-26T18:14:44.223" LastActivityDate="2019-02-26T18:14:44.223" Title="Modeling Malmquist Bias" Tags="&lt;galaxy&gt;&lt;astrophysics&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2019-02-27T23:34:47.103" ContentLicense="CC BY-SA 4.0" />
  <row Id="29766" PostTypeId="2" ParentId="29761" CreationDate="2019-02-26T18:03:52.693" Score="1" Body="&lt;p&gt;The azimuth of the sunrise (or sunset, or any object) is a function of the Sun's declination and observer's latitude. It can be calculated from the following forumla:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\cos(\theta_R)=-\frac{\sin(declination)}{\cos(latitude)}$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\theta_R$&lt;/span&gt; is measured from due south to the location where the object rises or sets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, at 55 degrees north latitude,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;declination = 0 at equinox (by definition), so &lt;span class=&quot;math-container&quot;&gt;$\theta_R$&lt;/span&gt; = 90 degrees from due south (90 degrees azimuth or due East when rising, 270 degrees or due west when setting).&lt;/li&gt;&#xA;&lt;li&gt;declination = 23.4 (approximately) at the summer solstice, so &lt;span class=&quot;math-container&quot;&gt;$\theta_R$&lt;/span&gt; = 133.8 degrees from due south (or an azimuth of 46.2 when rising, azimuth 313.8 when setting). [The declination at the solstice is plus or minus the obliquity of the ecliptic which is 23° 26′ 21″ = 23.4392° in the year 2000 and changes slowly.]&lt;/li&gt;&#xA;&lt;li&gt;Half way between those two points on the horizon is (90+133.8)/2 = 111.9 degrees from due south. Solving the equation for the declination gives a value of declination = 12.4 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Now find the date when the Sun's declination is +12.4 degrees (April 23 and August 24 according to &lt;a href=&quot;http://www.sci.ccny.cuny.edu/~luo/EAS488/Table_Declination_of_the_Sun.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Table of the Sun's Declination&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of course, the Sun will not be at the exact required declination at the time that it is rising (or setting), so you will want to select the date when the sunrise or sunset is closest. It will be accurate enough for most visual observations.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastEditorUserId="17355" LastEditDate="2019-03-04T01:36:34.533" LastActivityDate="2019-03-04T01:36:34.533" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29767" PostTypeId="1" CreationDate="2019-02-26T18:17:42.423" Score="2" ViewCount="343" Body="&lt;p&gt;AFAIK, Jupiter has the strongest magnetic field in the solar system, and it's so powerful that I've been told humans only could survive the radiation generated by it in its mooon Callisto, with proper protection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how about Saturn? Saturn has the second strongest magnetic field in the solar system. Which moons of Saturn could men survive from the radiation generated by the planet's magnetic field in?&lt;/p&gt;&#xA;" OwnerUserId="11562" LastEditorUserId="11562" LastEditDate="2019-02-27T00:29:57.370" LastActivityDate="2019-02-27T14:01:40.437" Title="How harmful is Saturn's magnetic field for human life?" Tags="&lt;jupiter&gt;&lt;magnetic-field&gt;&lt;saturn&gt;&lt;radiation&gt;&lt;astrobiology&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29769" PostTypeId="1" AcceptedAnswerId="29773" CreationDate="2019-02-26T22:26:33.107" Score="-1" ViewCount="99" Body="&lt;p&gt;Say, let's suppose the universe ends in an open universe, with the Universe expanding, but not going to such an extent that the big rip theory will take place. Obviously, the only things left would be stellar remnants and perhaps debris such as asteroids or comets in deep freeze. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then suppose that many civilizations have teamed together to survive this end, and they have complicated technology, such as, but not limited to faster than light travel, star or planet manipulation, and the ability to create artificial black holes, artificial wormholes(assuming they exist), and many other things if they have the energy and resources to do so. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance between galaxies are super large at this point, probably several trillion parsecs. Now, suppose that this time is where white dwarfs have all burned out, and neutron stars have either died due to neutron decay(assuming this is possible), or swallowed by black holes. Now, we see that that the universe is made of loose rocks and debris with black holes scattered around trillions of parsecs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, suppose a groups of living beings want to live on in such a universe. There are no other energy generation methods that will last, and last forever(ex. solar, wind, water, nuclear, putting asteroids in black holes and collecting the energy...). Thus, they use a simple energy/heat generation method, tidal forces. They make a binary planet system with the two planets close enough and rotating/orbiting around the barycenter fast enough with each other to generate heat via tidal forces. The tidal forces' generated heat power should power wind and water power via convection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we know that binary systems are not forever. They eventually spiral together. We can suppose that the living beings have thousands of engines that will activate and propel the planets away from each other so their previous orbit if the orbit spirals too in, powered by the heat energy from the tidal forces on each planet to avoid such thing from happening. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This model should be a feasible counter for such end of the universe. Now, is this model viable, or is there some kind of flaw in my thinking that will render such model null?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you need clarification, just ask and I will elaborate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max0815&lt;/p&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-02-27T21:15:49.357" LastActivityDate="2019-02-27T21:15:49.357" Title="Possible scenarios for life in the death of open universe possible?" Tags="&lt;fate-of-universe&gt;&lt;theories&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29770" PostTypeId="5" CreationDate="2019-02-26T22:27:03.030" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2019-02-26T22:27:03.030" LastActivityDate="2019-02-26T22:27:03.030" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29771" PostTypeId="4" CreationDate="2019-02-26T22:27:03.030" Score="0" Body="Questions regarding models proposed that can solve problems." OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-03-06T15:32:37.410" LastActivityDate="2019-03-06T15:32:37.410" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29772" PostTypeId="1" CreationDate="2019-02-27T02:22:49.390" Score="3" ViewCount="85" Body="&lt;p&gt;While the paradigm of cosmological inflation is well established, the nature of the inflation is not known and is tackled by introduction of a hypothetical &quot;inflaton&quot; field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if there are theories or research exploring the idea that after the Big Bang the Universe had just &lt;strong&gt;one spatial dimension&lt;/strong&gt; at first and therefore quite different expansion dynamics (compared to expansion along all dimensions at once), which may eliminate the need for the inflaton field. And as this simple Universe grew in size and complexity it gained the additional dimensions that slowed down the rate of expansion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of String theory this would probably mean that spatial dimensions were &quot;rolled-up&quot; initially and &quot;unfolded&quot; in stages with time after the Big Bang.&lt;/p&gt;&#xA;" OwnerUserId="26068" LastActivityDate="2019-02-27T02:22:49.390" Title="The nature of cosmological inflation — unfolding dimensions?" Tags="&lt;space-time&gt;&lt;cosmological-inflation&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29773" PostTypeId="2" ParentId="29769" CreationDate="2019-02-27T03:17:22.767" Score="3" Body="&lt;p&gt;Your scheme of creating energy from the tidal forces in a pair of orbiting planets cannot work because of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Conservation_of_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;law of conservation of energy&lt;/a&gt; (aka the 1st law of thermodynamics), which states that energy can neither be created nor destroyed; rather, it can only be transformed or transferred from one form to another. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It takes energy to set the 2 planets in orbit around each other, and any energy you obtain via tidal effects will reduce that store of orbital energy. In other words, the orbiting planets act as an energy storage system, they cannot create energy. &lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2019-02-27T03:17:22.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29774" PostTypeId="1" CreationDate="2019-02-27T04:36:52.137" Score="0" ViewCount="85" Body="&lt;p&gt;If one were to drop, say, a pebble, from a sufficient distance, into a black hole. The pebble would continue to accelerate until it were just short of the speed of light. In the next moment it's speed should exceed the speed of  light. Why not?&lt;/p&gt;&#xA;" OwnerUserId="26069" LastEditorUserId="1378" LastEditDate="2019-02-27T08:22:04.730" LastActivityDate="2019-02-27T08:22:04.730" Title="Dropping a pebble into a black hole" Tags="&lt;black-hole&gt;&lt;general-relativity&gt;&lt;special-relativity&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2019-02-27T10:39:09.323" ContentLicense="CC BY-SA 4.0" />
  <row Id="29775" PostTypeId="2" ParentId="29774" CreationDate="2019-02-27T06:44:58.590" Score="0" Body="&lt;p&gt;Gravity is no different from any other force. Doesn't matter it comes from a black hole, a star, a planet, or a lump of cheese floating in outer space. It's just a force, like any other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put a very powerful rocket engine on the pebble. In a short time it would accelerate the pebble near the speed of light. In the next moment it should exceed the speed of light. Why not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because that's not how it works. As you accelerate more and more, time dilation and space contraction also get stronger, so it gets harder and harder to get even closer to the speed of light, no matter what's pushing or pulling the pebble, be that a giant rocket engine or the enormous gravity of a large black hole. Same thing. The faster the pebble goes, the more warped space and time become, the harder it is to increase the speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It cannot even reach the speed of light because then time dilation, etc., would become infinite, effectively preventing it from reaching speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot cheat relativity.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note: This is not a rigorous explanation but it's close enough and should provide some intuitive understanding of what's going on.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2019-02-27T06:55:33.270" LastActivityDate="2019-02-27T06:55:33.270" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29777" PostTypeId="2" ParentId="29620" CreationDate="2019-02-27T09:46:08.047" Score="12" Body="&lt;p&gt;For any massive object the gravitational potential energy is given by Newton's law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ V(r) = -\frac{GMm}{r} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational potential energy is due to the attractive gravitational force, but for an orbiting object there is also a (fictitious) centrifugal force pushing it outwards. If we calculate the potential energy due to the centrifugal force and add it to the gravitational potential energy we get an effective potential energy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ V_\mathrm{eff}(r) = -\frac{GMm}{r} + \frac{L^2}{2mr^2} \tag{1} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; is the angular momentum, which is a constant for an orbiting object (because angular momentum is conserved in a central field). If we plot this graph for the Earth-Moon system we get a graph like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CLzP3.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CLzP3.gif&quot; alt=&quot;Effective potential&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(this comes from my answer to the question &lt;a href=&quot;https://physics.stackexchange.com/questions/170962/could-we-send-a-man-safely-to-the-moon-in-a-rocket-without-knowledge-of-general&quot;&gt;Could we send a man safely to the Moon in a rocket without knowledge of general relativity?&lt;/a&gt; on the Physics SE)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there is a minimum in the potential energy curve, and this minimum gives the radius of the stable orbit. Note also that because it's a minimum if we displace the object away from the minimum it will fall back towards the minimum again i.e. this is a stable orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for light we cannot simply use Newtonian mechanics because light is massless, but we can do the calculation using general relativity. The details are a bit intimidating, but we end up with an effective potential just as described above. For light the effective potential turns out to be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ V_\mathrm{eff}(r) = \sqrt{1 - \frac{2GM}{c^2r}}\frac{L}{r} \tag{2} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if we graph this it looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gPzqV.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gPzqV.gif&quot; alt=&quot;Light effective potential&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looks very different from our first graph, and it's different because light is massless and only ever travels at the same speed of &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;. The graph of &lt;span class=&quot;math-container&quot;&gt;$V_\mathrm{eff}$&lt;/span&gt; for light has a maximum not a minimum. The maximum corresponds to the position of a circular orbit, just like the minimum in the first graph, and we find the radius of this circular orbit is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{r}{r_\mathrm s} = 1.5 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$r_\mathrm s$&lt;/span&gt; is the Schwarzschild radius. This radius gives the position of the notorious &lt;a href=&quot;https://en.wikipedia.org/wiki/Photon_sphere&quot; rel=&quot;noreferrer&quot;&gt;photon sphere&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for light &lt;span class=&quot;math-container&quot;&gt;$V_\mathrm{eff}$&lt;/span&gt; has a maximum not a minimum. That means if we displace the light by even the tiniest distance from this maximum it will lower its potential energy by moving either inwards or outwards. The orbit at &lt;span class=&quot;math-container&quot;&gt;$1.5r_\mathrm s$&lt;/span&gt; is unstable and the tiniest perturbation will cause the light to spiral into the black hole or away from it. This means we cannot accumulate light in the photon sphere as the question asks. Any attempt to put light into this orbit is doomed to failure as even the tiniest perturbation (e.g. from other objects orbiting the black hole, or even from passing gravitational waves) will destabilise the orbit and the light will be lost.&lt;/p&gt;&#xA;" OwnerUserId="10687" LastEditorDisplayName="user8021" LastEditDate="2019-02-27T10:42:42.380" LastActivityDate="2019-02-27T10:42:42.380" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="29779" PostTypeId="2" ParentId="29767" CreationDate="2019-02-27T14:01:40.437" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetosphere_of_Saturn&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; says that &quot;The saturnian radiation belts are generally much weaker than those of Jupiter&quot; mostly due to the rings and other orbiting bodies absorbing the ions, I think. I have't yet found anything quantitative though&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-02-27T14:01:40.437" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29780" PostTypeId="1" AcceptedAnswerId="29795" CreationDate="2019-02-27T14:39:43.267" Score="1" ViewCount="85" Body="&lt;p&gt;The Local Interstellar Cloud is supposed to be around 30 light years large. Is it possible that in this region there are stars which are undetected? I would have thought that it wasn't possible, but there are stars which are very small or are very close to another star. BTW, with star I mean an astronomical object which is performing fusion, because I've seen lists of smallest or coldest stars which includes brown dwarfs which don't perform fusion or white dwarfs. &lt;/p&gt;&#xA;" OwnerUserId="11562" LastEditorUserId="7472" LastEditDate="2019-02-27T19:58:10.503" LastActivityDate="2019-02-28T15:40:15.413" Title="Is it possible that in the Local Interstellar Cloud there are undetected stars?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29782" PostTypeId="1" CreationDate="2019-02-27T20:18:07.200" Score="0" ViewCount="283" Body="&lt;p&gt;Wikipedia cites this formula for calculating solar azimuth:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JN6jQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JN6jQ.png&quot; alt=&quot;equation for solar azimuth&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\phi_s$&lt;/span&gt; is the solar azimuth angle&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\theta_s$&lt;/span&gt; is the solar zenith angle&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; is the hour angle, in the local solar time&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\delta$&lt;/span&gt; is the current sun declination&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\Phi$&lt;/span&gt; is the local latitude&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm implementing this equation in JavaScript, but it doesn't seem correct. Can someone here help me confirm whether or not this equation is correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my JS implementation. The only strange thing is at the end: limiting the acos domain to [-1, 1] to prevent NaN values.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;const getSolarAzimuth = function(&#xA;    zenithAngle, hourAngle, declination, latitude&#xA;) {&#xA;    const cos = Math.cos;&#xA;    const sin = Math.sin;&#xA;&#xA;    const cos_phi = (&#xA;        sin(declination) * cos(latitude) - cos(hourAngle) * cos(declination) *&#xA;            sin(latitude)) /&#xA;          sin(zenithAngle);&#xA;&#xA;    return Math.acos(Math.min(Math.max(cos_phi, -1), 1));&#xA;};&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="26075" LastEditorUserId="26075" LastEditDate="2019-02-27T20:22:34.960" LastActivityDate="2019-02-27T22:02:11.227" Title="How do I calculate the sun's azimuth based on zenith, hour angle, declination, and latitude?" Tags="&lt;azimuth&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29783" PostTypeId="2" ParentId="29730" CreationDate="2019-02-27T20:19:58.230" Score="1" Body="&lt;p&gt;A lot of people new to the hobby have no idea how difficult astrophotography can be.  Of course, a lot depends on your expectations.  If your expectations are for blurry, fuzzy images that might, or might not, be Jupiter, or M42, or any other astronomical body, then you have a lot of room.  But if the assumption is that you want to take reasonable quality pictures (I'm not talking great pictures or professional pictures, but ones where the object is readily identifiable and clear), things can get very difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first part of the problem is that eyes and cameras are not interchangeable.  They work very differently and have very different responses to light.  There's a lot of similarity and overlap, but a lot of difference as well.  As such, you can't just connect a camera to a telescope and hope to snap pictures as easy as you view objects.  We are talking about two very different activities here, with very different needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should also be stated that while some people have had some amount of success using cell phone cameras to capture images of astronomical objects, these cameras are fare from ideal for the purpose, and there's many, many more pictures people have taken this way that show nothing more than a blurry mess, if anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another important thing to understand is that Astrophotography is not a monolithic subject - it encompasses many sub-specialties, each with its own needs as far as technique and equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two simplest divisions in AP are solar-system imaging versus deep-sky imaging.  Solar-system imaging refers to anything in our own solar system, and for most amateurs really concentrates on four specific bodies: the moon, Mars, Jupiter, and Saturn.  Without going into details as to why, none of the other planets make good photographic targets without significant equipment upgrades.  But the moon, Mars, Jupiter and Saturn are fairly simple targets and make a good starting point.  These CAN be captured with a cell phone, but this is not always easy. In particular, focusing is a difficult task.  Even with a larger telescope on a mount intended for astrophotography, this can be difficult. With a manually-controlled telescope with fairly low aperture and magnification such as yours, keeping the object in the camera's field of view while you try to get the focus just right can be a major challenge.   The moon is probably your easiest bet, followed by Jupiter, Saturn, then Mars.  But you should also realize that Jupiter, Saturn, and Mars are only good targets when they're approaching opposition - the closest point to us in their orbit.  Even this varies.  At its largest size, when Jupiter is approaching its perihelion (closest point to the sun) and we are nearing aphelion (our furthest point from the sun) and those points happen to line up at the opposition point, Jupiter is only 51 arc-seconds in size.  Any other time, it will be smaller than that.  The best time to view or image any planet in its orbital cycle is when it is rising on the horizon at around midnight.  This means it's near opposition, and will present its best view to us.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another critical factor in planetary imaging is the atmosphere.  If you live in an area with a lot of air pollution, humidity, dust in the air, pollen, etc, this can all have an effect on the image quality by affecting the clarity and refractive properties of the air. Wind and turbulence in the air can do so as well, and it doesn't have to be at ground level.  Upper-atmospheric winds can be problems for imaging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For planetary imaging, most of the best images currently being produced by amateurs are actually done through video and make use of a process known as stacking.  In fact, nearly all AP these days uses stacking, which requires multiple sample points and, in essence, is a method of statistical analysis to determine the most likely correct value for each pixel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way this works starts with the image capture.  Let's say you record a 30-second video at 30 frames per second (fps).  This means that the end video has about 900 individual image frames.  When you load this into the stacking software, the first task is to align the frames.  The idea here is to make sure the target object in each frame is lined up, pixel-for-pixel, with the rest of the images in the &quot;stack&quot;.  Once this is accomplished, in theory each pixel location on each image in the stack should cover the same portion of the target object. At this point, then, the values of each of these pixels at the same location throughout the stack are analyzed.  If you had 900 frames, then you should have 900 individual samples of that value.  If you then average them out, you get the most likely &quot;true&quot; value for that pixel.  When you do this for every pixel location in the image, you can reduce the effect of individual bad frames.  You can also use the data to interpolate values for increasing resolution and detail.  The end result actually can produce a better image than the telescope itself is theoretically capable of providing, due to the processing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to try deep-sky imaging (anything outside of our solar system), then the process gets even harder - and the equipment gets even more expensive.  These objects (galaxies, nebulae, star clusters, etc...) tend to be very faint and require long-exposure images - shutter times measured in minutes or more.  This requires equipment that can compensate for the apparent motion of such objects across the sky caused by the rotation of the Earth.  In these cases, the telescope and camera take a back seat to the mount.  I know people who have spent tens of thousands of dollars on mounts alone for this purpose.  I myself have spent up to about $10,000 over 10 years to get to the point where I'm just starting to feel competent in this arena.  Unfortunately, that telescope you have is not suited for this at all, and simply will NOT provide good imaging prospects at all for this kind of imaging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the long run, if this is something you are interested in, then I strongly suggest you look for an astronomy club or society near you.  Most such clubs have at least a few people doing pretty good imaging.  Join, meet those people, and get to know them so you can learn from them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the long run, I would strongly recommend putting astrophotography aside for a bit, get used to using your scope visually and learning more about the sky above.  You might also consider learning to do astronomical sketching, which can help you better train your eyes and observing skills, which will help you enjoy the hobby more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck and clear skies!&lt;/p&gt;&#xA;" OwnerUserId="10546" LastActivityDate="2019-02-27T20:19:58.230" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29784" PostTypeId="2" ParentId="29782" CreationDate="2019-02-27T22:02:11.227" Score="1" Body="&lt;p&gt;I've confirmed that this formula is in fact correct. I did have the wrong units in the code, which @Glorfindel suggested: sun's hour angle was in hours instead of radians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did need to adjust the azimuth's angle offset based on the hour angle:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     // The angle offset needs to be adjusted based on whether the hour angle                                                    &#xA;     // is in the morning or the evening.                                                                                        &#xA;     //   https://en.wikipedia.org/wiki/Solar_azimuth_angle#Formulas                                                             &#xA;     if (hourAngle &amp;lt; 0 || hourAngle &amp;gt; Math.PI) {&#xA;         return az;&#xA;     }&#xA;&#xA;     return (Math.PI * 2) - az;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="26075" LastActivityDate="2019-02-27T22:02:11.227" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29785" PostTypeId="1" AcceptedAnswerId="29789" CreationDate="2019-02-27T23:55:43.367" Score="1" ViewCount="239" Body="&lt;p&gt;Similar question &lt;a href=&quot;https://astronomy.stackexchange.com/questions/29697/what-would-happen-if-a-star-went-supernova-inside-a-supermassive-black-hole&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question I linked got me thinking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that a black hole doesn't let anything escape its event horizon, not even light. Thus, naturally, anything that passes the event horizon would get &quot;eaten&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, assume a supermassive black hole(small Roche limit, large event horizon) was in the path of a high eccentricity, massive, dying star. Assume the star is travelling in a straight line at 5% the speed of light, and the black hole is stationary. The star would be swallowed by the black hole, and also swallowed intact. Now, the star goes supernova inside the black hole, but since its mass is very high, it too, creates a black hole. What would happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the supermassive black hole gain the mass of the star and things would go as normal for an outside viewer(everyday common things: black hole swallowing objects)? Would the black hole inside the larger one pull the larger one in? Or something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max0815&lt;/p&gt;&#xA;" OwnerUserId="25988" LastActivityDate="2019-02-28T01:19:40.433" Title="What happens if a black hole is produced inside another black hole?" Tags="&lt;black-hole&gt;&lt;supernova&gt;&lt;supermassive-black-hole&gt;&lt;theories&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29786" PostTypeId="1" AcceptedAnswerId="29810" CreationDate="2019-02-28T00:02:21.510" Score="1" ViewCount="133" Body="&lt;p&gt;I read on a Wikipedia article lately that tachyons are theoretical particles which always travel at or above the speed of light. That means that when one passes us, we see two images, a blue shifted one and a red shifted one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes on the other hand have gravity so strong that light gets pulled in. However, I haven't heard that black holes can exert a gravitational pull that can pull and object that is travelling greater than the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now, I am wondering that if it is possible to use/make tachyons and control those tachyons, could we send them into a black hole and use them like a &quot;echolocation&quot; thing, except the sensors would be on the other side of where the tachyons are first produced?&lt;/p&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-02-28T03:18:48.570" LastActivityDate="2019-03-01T20:09:27.433" Title="Possible to use tachyons to explore black hole?" Tags="&lt;black-hole&gt;&lt;light&gt;&lt;speed&gt;&lt;particles&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="29787" PostTypeId="1" AcceptedAnswerId="29809" CreationDate="2019-02-28T00:20:38.633" Score="2" ViewCount="304" Body="&lt;p&gt;I've hear of something like this: &quot;All planets lose their atmospheres eventually and they become a barren rock if given enough time.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three types of atmospheric loss ways:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Jeans Escape: Temperature and escape velocity factors determine the gasses and the amount that's lost. Jupiter cannot lose atmosphere via Jeans escape because Jupiter has too high escape velocity and too high of a temperature.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Charge exchange: Solar radiation creates electrons and positively charged ions in upper atmospheres by tearing electrons off atoms or molecules. Subsequently, charge attraction and repulsion in collisions accelerates ions. Jupiter most likely cannot lose its atmosphere via charge exchange because its magnetic field is strong and extensive while no(little?) material escapes via polar wind.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Vertical atmospheric escape/impact erosion: Energetic objects that strike a planet erodes its atmosphere by creating a plume of heated gas. Jupiter does not lose its atmosphere via impact erosion because it has very high gravity and escape velocity. This pulls particles ejected by impact back into its atmosphere.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since Jupiter doesn't lose atmosphere in the three types of atmospheric loss, how does it lose its atmosphere?&lt;/p&gt;&#xA;" OwnerUserId="25988" LastActivityDate="2019-03-01T19:48:20.123" Title="Does Jupiter lose some atmosphere at all?" Tags="&lt;planetary-atmosphere&gt;&lt;jupiter&gt;&lt;atmospheric-escape&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29788" PostTypeId="2" ParentId="29219" CreationDate="2019-02-28T01:09:50.167" Score="3" Body="&lt;p&gt;&lt;strong&gt;Yes ... and no.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Experimentally, there's no indication that the laws of physics are changing (c.f. conservation of energy: if the laws of physics change with time, energy would not be conserved via Noether's theorem). So if you're an empiricist, you'd conclude that the universe is stable (or at least its mean lifetime must be more than ~14 billion years, the age of the universe).&lt;/p&gt;&#xA;&lt;p&gt;However, if we believe string theory (which has no compelling experimental evidence), then we expect roughly &lt;span class=&quot;math-container&quot;&gt;$10^{500}$&lt;/span&gt; &amp;quot;vacuum states&amp;quot;, which correspond to &lt;span class=&quot;math-container&quot;&gt;$10^{500}$&lt;/span&gt; different possible universes, each with their own laws of physics. It's possible one of these other universes have a lower &amp;quot;total energy&amp;quot; than us (this word's meaning is vacuous since if one varies the laws of physics, the word 'energy' itself might lose meaning). This means it's in principle possible that our universe will quantum tunnel to a universe at lower energy, in which case our universe is metastable.&lt;/p&gt;&#xA;&lt;p&gt;Of course, this leads to other problems. How would one prove the multiverse exists? If a generally accepted way exists, I've not heard of it. But if you can't prove the multiverse exists, how would you prove our universe is metastable? For that matter, how do we know the electron is stable? Why can't it have a mean lifetime of, say, &lt;span class=&quot;math-container&quot;&gt;$10^{100}$&lt;/span&gt; years (so long that we don't observe it decaying)?&lt;/p&gt;&#xA;&lt;p&gt;At this point, we're purely in the realm of speculation. We're talking about things so far above the experimental threshold that some people will just dismiss it as pointless. Suffice to say that if the universe isn't stable and quantum tunnels to a lower energy, we won't be able to see the tunneling coming, and we're all instantly dead anyway. Whether the question is worth thinking about further depends on the individual.&lt;/p&gt;&#xA;&lt;p&gt;tl; dr:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Experimentally, universe is stable.&lt;/li&gt;&#xA;&lt;li&gt;Theoretically, maybe it isn't, depending on what your theory is because we don't have a widely-accepted, experimentally-backed theory.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="20348" LastEditorUserId="20348" LastEditDate="2021-05-25T02:43:56.093" LastActivityDate="2021-05-25T02:43:56.093" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29789" PostTypeId="2" ParentId="29785" CreationDate="2019-02-28T01:19:40.433" Score="3" Body="&lt;p&gt;From the no-hair theorem, black holes can have only three properties: mass, angular momentum, and electric charge. From this we'll see the original, supermassive black hole gain the mass, angular momentum and electric charge of the ingoing star. The black holes will merge (already merged actually, since one is inside the other). The singularities will merge too (all world lines, i.e. paths a particle can take, lead to the big singularity in the supermassive black hole). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the outside, we will see a more massive black hole with different angular momentum &amp;amp; electric charge, and that's it. We can't see the star go supernova because anything that happens inside the event horizon is invisible to us.&lt;/p&gt;&#xA;" OwnerUserId="20348" LastActivityDate="2019-02-28T01:19:40.433" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29790" PostTypeId="1" AcceptedAnswerId="29798" CreationDate="2019-02-28T06:06:21.843" Score="2" ViewCount="337" Body="&lt;p&gt;I found this image of what maybe a model of our solar system. Without a credible source I'm not sure I am looking at a star or black hole that comes close to our solar system and how it could throw Earth out of orbit from the sun and/or solar system?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found it while Googling for how a solar system travels through space and did a picture search with no luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this possible? And what am I looking at?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RzRXf.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RzRXf.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2019-02-28T06:27:45.683" LastActivityDate="2019-02-28T21:09:15.270" Title="Can Earth be ejected out of the solar system this way?" Tags="&lt;star&gt;&lt;black-hole&gt;&lt;solar-system&gt;&lt;earth&gt;&lt;identify-this-object&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29791" PostTypeId="2" ParentId="29790" CreationDate="2019-02-28T10:27:01.267" Score="3" Body="&lt;p&gt;It looks like the model is a simulation of what happens if a massive object flies by a very narrow ring system type of thing with a circular orbit.   The ring system setup seems improbable to me, and the massive object doesn't appear to fly in a straight line, so it's not very accurate for two reasons, but the gravitational effect seems accurate.    If a massive body was to fly into the solar system and pass near Earth it could send the Earth out of the solar system or in towards the sun.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately, bodies that massive almost never fly inside the solar system, so that scenario is unlikely to happen over the lifetime of the sun.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2019-02-28T10:27:01.267" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29792" PostTypeId="1" AcceptedAnswerId="40104" CreationDate="2019-02-28T10:46:51.593" Score="8" ViewCount="103" Body="&lt;p&gt;Reading a discussion on another website, it was proposed that if a large meteor struck the moon, that a few days later, Earth would receive a meteor shower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought about that and it seemed to me that one of the big unknowns would be what percentage of the impact would reflect back into space, away from the lunar gravity well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon, having no atmosphere, then velocity out, could be pretty close to the velocity in, minus whatever's lost in heat and pressure waves and perhaps phase transition, though that could increase the explosive rebound and maybe other energy changes I'm not considering.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is simply, is there a fairly simple formula or estimate of what percentage would blow back into space following a lunar impact.  I'd suspect that it would depend on both the velocity and mass of the meteor and perhaps, composition.  If it was large enough and a direct hit, it could blow the entire moon into space, for example, but I'm thinking more standard asteroid size impacts not giant impacts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say, for example, &lt;a href=&quot;https://en.wikipedia.org/wiki/99942_Apophis&quot; rel=&quot;noreferrer&quot;&gt;a fairly large impactor like 99942 Apophis&lt;/a&gt;, about 27 billion kg, 370 meters in diameter.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2020-11-30T05:28:31.873" Title="What percentage of a lunar meteor strike is blown back into space?" Tags="&lt;meteor&gt;&lt;impact&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29793" PostTypeId="2" ParentId="29790" CreationDate="2019-02-28T11:30:02.393" Score="2" Body="&lt;p&gt;That might be a simulation of Oort cloud. Loosely bound asteroids at the edge of sun gravity. Though it would be spherical instead of ring but it may be 2 d represention. Notice the fast moving inner planet. Since sun is stationary I think it is relative motion of stars  as sun revolves around milkyway. Possibly, how we get long distance comet. As suggested this type of encounter are rare in inner solar system but not impossible. Yes on close enough encounter, earth can be ejected outward. I guess that would be one way to travel inter stellar distances :)&lt;/p&gt;&#xA;" OwnerUserId="26086" LastEditorUserId="26086" LastEditDate="2019-02-28T11:40:09.067" LastActivityDate="2019-02-28T11:40:09.067" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29794" PostTypeId="1" CreationDate="2019-02-28T14:12:02.167" Score="3" ViewCount="39" Body="&lt;p&gt;When a spacecraft flys past a planet eg. Jupiter , its heliocentric energy increases dramatically allowing it to go further away. How is this possible without using any more rocket fuel?&lt;/p&gt;&#xA;" OwnerUserId="25925" LastActivityDate="2019-02-28T14:12:02.167" Title="Energy conservation in Interplanetary orbit transfer" Tags="&lt;orbital-mechanics&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2019-03-01T03:39:31.343" ContentLicense="CC BY-SA 4.0" />
  <row Id="29795" PostTypeId="2" ParentId="29780" CreationDate="2019-02-28T15:40:15.413" Score="4" Body="&lt;p&gt;It is unlikely, unless they are extremely faint. The reason is that the cloud obviously does not absorb enough light to keep us from detecting remote red and brown dwarfs, and hence a closer star would also be detectable. &lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2019-02-28T15:40:15.413" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29796" PostTypeId="1" CreationDate="2019-02-28T19:11:24.147" Score="5" ViewCount="67" Body="&lt;p&gt;Perusing the Planck 353 GHz polarisation Aitoff projection I noticed an apparent magnetically driven dust filament rising abruptly from the Galactic ß-field in the Chameleon-Mensa S polar region. It streams for ~25° before dwindling in Eridanus. It appears to be associated with Galactic cirrus clumps 80 PC below the disc plane. The &lt;a href=&quot;https://drive.google.com/open?id=1wqh7glVXZD7POVy5AEVAfDc2kZvu32F-&quot; rel=&quot;noreferrer&quot;&gt;Planck 353 GHz image is here&lt;/a&gt;, and a quickie &lt;a href=&quot;https://drive.google.com/open?id=1GcMCcmVR-xoJB828UpE_XEsCA9KAeVX4&quot; rel=&quot;noreferrer&quot;&gt;Chromoscope survey in optical, NIR, FIT mm/µm, and 21 cm is here&lt;/a&gt;. It is mentioned in only a few papers incl. Besla/Martinez-Delgado 2015, but there has been no specific study of its unusual morphology. It looks like a magnetically driven jet but originates from a region devoid of high-energy activity. If it is a Birkeland current it is a very odd one. I wonder if it is an old SN 1a remnant of the Hercules Loop and Aquila Rift type based on faint extinction filaments in Schlegel 1998 S Polar projection and 4 other Planck Aitoffs which seem to suggest traces of an old circular feature. Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="26088" LastActivityDate="2019-02-28T19:11:24.147" Title="Can anyone explain this odd Galactic dust filament near the LMC?" Tags="&lt;dust&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29797" PostTypeId="1" CreationDate="2019-02-28T19:35:43.677" Score="8" ViewCount="337" Body="&lt;p&gt;In browsing through Astronomy Stack Exchange, I came across this question about orbiting the moon (&lt;a href=&quot;https://astronomy.stackexchange.com/questions/20499/is-it-possible-to-achieve-a-stable-lunarstationary-orbit-around-the-moon&quot;&gt;Is it possible to achieve a stable &amp;quot;lunarstationary&amp;quot; orbit around the moon?&lt;/a&gt;).  In further looking, I came across (&lt;a href=&quot;https://astronomy.stackexchange.com/questions/11616/can-i-measure-the-moons-gravity&quot;&gt;Can I measure the moon&amp;#39;s gravity?&lt;/a&gt;) which seems to indicate that with sensitive enough equipment, the moon's gravitational pull &lt;em&gt;could&lt;/em&gt; be measured on earth.  This prompted me to wonder if the moon's gravitational pull can be observed on the Space Station's or geosynchronous satellite's orbital path?  And if so, if any compensation (either short-term or long-term) needs to be performed due to the moon's gravity.  I would think it to be relatively inconsequential, but am not sure, hence the question.&lt;/p&gt;&#xA;" OwnerUserId="19229" LastActivityDate="2019-11-26T02:04:29.977" Title="Can the moon's gravity influence be observed on the space station's orbit?" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;gravity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29798" PostTypeId="2" ParentId="29790" CreationDate="2019-02-28T21:09:15.270" Score="3" Body="&lt;p&gt;As far as I can tell, the origin of the image is &lt;a href=&quot;http://www.orbitsimulator.com/cgi-bin/yabb/YaBB.pl?num=1242850911/4&quot; rel=&quot;nofollow noreferrer&quot;&gt;this post on the Gravity Simulator forums&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From frankuitaalst's post there:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I started with the Onlyplantes.gsim , added about 200 bodies at ( I think ) 400 AU in circilar orbit , representing the Oort cloud and added 1 Solar mass originally in an excentric orbit with Sma 1000 AU .  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As regards the question, yes a sufficiently close stellar flyby could eject the planets (either directly or by triggering planet-planet scattering). Such flybys are unlikely to occur in our Solar System's current location, they would be more of an issue in crowded stellar clusters.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastActivityDate="2019-02-28T21:09:15.270" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29799" PostTypeId="2" ParentId="29721" CreationDate="2019-02-28T21:52:35.033" Score="1" Body="&lt;p&gt;&lt;strong&gt;tl;dr&lt;/strong&gt; Yes there have been theorized, and &lt;strong&gt;&lt;em&gt;extremely&lt;/em&gt;&lt;/strong&gt; possible binary systems that have not been observed. One such thing is a TZO, or a Thorne–Żytkow_object. This is a neutron star-red giant binary. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There are many kinds of binary pairings that have not been discovered yet, if they exist. A hypothetical star is the quasi-star, which you could say it is a binary between a black hole and an extremely massive star. You can also consider a Wolf–Rayet star, which can occur from a binary system. Other binary systems would include theoretical stars, also known as exotic stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I do know of a binary system that is composed of real observed types of stars or stellar remnants. This system is composed of a red giant and a neutron star, with the neutron star spiraling in towards the red giant's core. It starts off with a binary system with a star and a neutron star. When the star becomes a red giant, its atmospheric friction on the neutron star(the red giant envelops the neutron star) makes their orbit deteriorate, and the neutron star and the red giant's core will spiral into each other. After the neutron star meets the core, if the mass of the core does not overcome neutron degeneracy pressure, then the core is replaced with a neutron star. If it does, then a supernova resulting in a black hole occurs. This binary system is particularly significant because this can shed light onto how fusion works, and how binary systems develop, because in this model the neutron star does not absorb a lot of the red giant's mass. The surface of the neutron star would also be so hot that fusion can occur, demonstrating a new way of fusion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomers can observe these stars by their emitted wavelengths because the fusion creates heavy elements that may find their way to the surface of the giant:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To look for a TZO, the team searched the spectrums for some of the elements predicted to be produced in high quantities in these objects: lithium, rubidium, strontium, vanadium, zirconium, and molybdenum. Not all of these are easy to image, so they focused on lithium, rubidium, and molybdenum. They compared the levels of these elements with elements that create spectral features nearby but aren't expected to be enhanced in TZOs: potassium, calcium, iron, and nickel.&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&lt;a href=&quot;https://arstechnica.com/science/2014/06/red-supergiant-replaced-its-core-with-a-neutron-star/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arstechnica.com/science/2014/06/red-supergiant-replaced-its-core-with-a-neutron-star/&lt;/a&gt;&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would say that no true examples of this kind of binary pairing has been found. However, there are three possible candidates that are observed, though recently, these stars are under serious questioning on their validity(that may not answer part of your question on &quot;at least suspected :(&quot;), according to&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thorne%E2%80%93%C5%BBytkow_object&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thorne–Żytkow_object&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There is another one mentioned by mistertribs(all credits to him), that a black hole and neutron star binary is not observed to exist. These type of systems are possibly the most interesting from a gravitational point of view. They should produce gravitational waves, which could be detectable by LIGO. LIGO has yet to observe such a system and confirm such, which seems a bit puzzling, although something like the blue supergiant in Cygnus X-3 could evolve into a neutron star in the future. This is a case where more data and observations are needed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, scientists were only able to constrain the merger rate enough to eliminate the most optimistic models, and that it would take several more observing runs with no results before a continued failure became problematic. &lt;/p&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-03-05T23:36:17.983" LastActivityDate="2019-03-05T23:36:17.983" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="29800" PostTypeId="2" ParentId="29721" CreationDate="2019-02-28T22:21:03.330" Score="3" Body="&lt;p&gt;As far as I am aware, there are no known examples of black hole+neutron star (BHNS) binaries (e.g. &lt;a href=&quot;https://arxiv.org/abs/1806.00001&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper from 2018&lt;/a&gt; notes that &quot;No BHNS binaries (BHNSs) have been discovered so far by radio surveys&quot;). BHNS mergers should be observable with gravitational wave astronomy, so this situation may not last for too much longer. There is &lt;a href=&quot;https://www.ligo.caltech.edu/news/ligo20190502&quot; rel=&quot;nofollow noreferrer&quot;&gt;a report of a possible BHNS merger&lt;/a&gt; detected by LIGO and Virgo (event S190426c). This event has yet to be confirmed, follow-up searches by &lt;a href=&quot;https://arxiv.org/abs/1905.02186&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hosseinzadeh et al. (2019)&lt;/a&gt; and &lt;a href=&quot;https://arxiv.org/abs/1905.06980&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goldstein et al. (2019)&lt;/a&gt; for transients have failed to find a counterpart to the event.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Binaries containing an &lt;a href=&quot;https://en.wikipedia.org/wiki/Subdwarf_B_star&quot; rel=&quot;nofollow noreferrer&quot;&gt;sdB star&lt;/a&gt; and a neutron star have also not been found, though theoretical models predict they should exist in our galaxy (&lt;a href=&quot;https://arxiv.org/abs/1912.07705&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wu et al., 2019&lt;/a&gt;). Sufficiently close sdB+NS binaries should be strong gravitational wave sources, so &lt;a href=&quot;https://en.wikipedia.org/wiki/Laser_Interferometer_Space_Antenna&quot; rel=&quot;nofollow noreferrer&quot;&gt;LISA&lt;/a&gt; may discover them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on how you want to define various &quot;types&quot; of star, you can probably come up with more examples. For example, there is the concept of &lt;a href=&quot;http://adsabs.harvard.edu/abs/2008IAUS..246..273M&quot; rel=&quot;nofollow noreferrer&quot;&gt;singleton stars&lt;/a&gt; which have never been members of binary systems or had close encounters with other stars, which by definition have not been found in binary pairings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also some weird categories of objects with few known members. For example there are the &lt;a href=&quot;https://arxiv.org/abs/1902.05061&quot; rel=&quot;nofollow noreferrer&quot;&gt;LP&amp;nbsp;40-365 stars&lt;/a&gt;, which are runaway stars with atmospheres dominated by oxygen and neon, likely the partially-burnt remnants of peculiar type Iax supernovae. None of the four known examples of this class are binary systems, plus the circumstances that lead to their creation makes it seem fairly unlikely that we'll find one in a binary.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-12-18T21:31:36.073" LastActivityDate="2019-12-18T21:31:36.073" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29801" PostTypeId="1" AcceptedAnswerId="29802" CreationDate="2019-02-28T23:59:19.500" Score="2" ViewCount="271" Body="&lt;p&gt;Does our solar system have enough bodies in orbit to appear to twinkle to another distant solar system viewed from the side? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6nuHl.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6nuHl.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_asteroid-discovering_observatories&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/List_of_asteroid-discovering_observatories&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5zEAF.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5zEAF.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://phys.org/news/2018-11-nasa-lucy-sky-asteroids.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://phys.org/news/2018-11-nasa-lucy-sky-asteroids.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastActivityDate="2019-03-01T19:20:52.337" Title="Does the Sun twinkle?" Tags="&lt;the-sun&gt;&lt;orbit&gt;&lt;amateur-observing&gt;&lt;mass&gt;&lt;visualization&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29802" PostTypeId="2" ParentId="29801" CreationDate="2019-03-01T00:05:06.623" Score="5" Body="&lt;p&gt;While objects blocking the sun may have an effect on the twinkle if viewed on another planet, a twinkle is mainly caused by the effect the atmosphere of a planet has on the light, or something interfering with the light that is passing to your eyes.&lt;/p&gt;&#xA;&lt;p&gt;From Wikipedia's article on &lt;a href=&quot;https://en.wikipedia.org/wiki/Twinkling&quot; rel=&quot;nofollow noreferrer&quot;&gt;Twinkling&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Twinkling, or scintillation, is a generic term for variations in apparent brightness or position of a distant luminous object viewed through a medium. If the object lies outside the Earth's atmosphere, as in the case of stars and planets, the phenomenon is termed astronomical scintillation; within the atmosphere, the phenomenon is termed terrestrial scintillation.[2] As one of the three principal factors governing astronomical seeing (the others being light pollution and cloud cover), &lt;strong&gt;atmospheric twinkling is defined as variations in illuminance only.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Stars twinkle because they are so far from Earth that they appear as point sources of light easily disturbed by Earth's atmospheric turbulence&lt;/strong&gt;, which acts like lenses and prisms diverting the light's path. &lt;strong&gt;Large astronomical objects closer to Earth, like the Moon and other planets, encompass many points in space&lt;/strong&gt; and can be resolved as objects with observable diameters. With multiple observed points of light traversing the atmosphere, their light's deviations average out &lt;strong&gt;and the viewer perceives less variation in light coming from them.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-03-01T19:20:52.337" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29803" PostTypeId="2" ParentId="29797" CreationDate="2019-03-01T02:40:20.643" Score="7" Body="&lt;p&gt;According to NASA, satellites in Earth orbit don't experience strong enough interference from the Moon to &quot;whisk a spacecraft out of the game&quot;. To a satellite in Earth orbit, the gravitational pull of the Sun is actually ~160 times stronger than influence from the Moon. Source: &lt;a href=&quot;https://science.nasa.gov/science-news/science-at-nasa/2006/30nov_highorbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;A New Paradigm for Lunar Orbits&lt;/a&gt; (see EDIT note below for details).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, gravitational pull from the Sun, the Moon, planets such as Jupiter (as well as other factors such as solar radiation pressure, atmospheric drag, etc.) do cause enough interference to slightly alter the orbital path of any satellite orbiting Earth. To compensate for the deviations all satellites must periodically fire their thrusters for brief periods of time in order to maintain the correct desired orbit. If a satellite has no fuel to perform these &quot;orbital stationkeeping&quot; maneuvers it essentially becomes space junk. Source: &lt;a href=&quot;https://www.quora.com/Is-the-ISS-and-all-other-satellites-affected-by-the-gravitational-pull-of-the-moon&quot; rel=&quot;nofollow noreferrer&quot;&gt;Is the ISS, and all other satellites, affected by the gravitational pull of the moon?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Consider a satellite with mass of 3,000 kg rotating around Earth on geostationary (GEO) orbit of 42,164 km radius. Gravitational pull from the Earth is constant throughout the orbit and equals ~673 N (Newtons). As the satellite rotates around the Earth it gets closer or further away from the Moon as well as the Sun and, therefore, their gravitational pulls change. The pull from the Sun is about 17 Newtons but changes only by 0.02 N. The gravitational pull from the Moon varies between 0.12 N (when the satellite is between the Earth and the Moon) and 0.08 Newtons (when the satellite is furthest away from the Moon over the opposite side of the Earth). So even though the pull from the Moon is much smaller than from the Sun, the variation of the Moon's perturbation is about twice as large than from the Sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: the numbers above are computed under a simplified assumption that all orbits are on the same plane. In reality, due to the inclinations of orbits, the numbers can be somewhat lower, but the overall picture remains about the same. &lt;/p&gt;&#xA;" OwnerUserId="26068" LastEditorUserId="26068" LastEditDate="2019-11-26T02:04:29.977" LastActivityDate="2019-11-26T02:04:29.977" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29805" PostTypeId="1" CreationDate="2019-03-01T09:16:39.180" Score="17" ViewCount="444" Body="&lt;p&gt;The Tycho crater is thought to be the youngest major impact on the Moon, estimated at 108 million years ago. This puts it firmly in the reign of the dinosaurs, I find myself curious as to what our thunder-lizard brethren would have witnessed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8tMUS.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8tMUS.png&quot; alt=&quot;Tycho Crater by LRO&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly how bright would the initial flash have been? Would a glowing scar have been left on the Moon for days, weeks or months? Would the Earth have been subject to spectacular global meteor showers from ejected debris? Would any other effects have been felt on Earth? I'd guess that a simulation would be required to accurately answer these questions but I'm hoping that some reasonable estimates can be made by comparison of studies of other impacts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, was the Tycho impact of greater or smaller magnitude than the Shoemaker–Levy 9 impact in 1994?&lt;/p&gt;&#xA;" OwnerUserId="23640" LastActivityDate="2019-03-19T07:58:48.097" Title="How would the Tycho Impact have appeared from Earth?" Tags="&lt;the-moon&gt;&lt;impact&gt;&lt;crater&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29806" PostTypeId="1" AcceptedAnswerId="29816" CreationDate="2019-03-01T14:12:10.347" Score="6" ViewCount="187" Body="&lt;p&gt;Did any moon landing vehicle or orbiter took pictures of a solar eclipse happening on the earth, i.e. the moon's shadow on the earth? (Alternatively from other spacecraft showing the whole globe). A link to those pictures would be nice.&lt;/p&gt;&#xA;" OwnerUserId="26096" LastEditorUserId="26096" LastEditDate="2019-03-02T10:01:18.163" LastActivityDate="2021-05-26T10:49:15.177" Title="Are there pictures from a solar eclipse of the earth taken from a moon landing unit or orbiter?" Tags="&lt;the-moon&gt;&lt;earth&gt;&lt;solar-eclipse&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29807" PostTypeId="2" ParentId="2092" CreationDate="2019-03-01T15:21:43.037" Score="2" Body="&lt;p&gt;Beside of mass, the rotation also affects the shape of an object. The faster the object spins, the more oblate (like a flattened sphere) it is. This happens due to the centrifugal force at its equator. Examples are Haumea (a dwarf planet) and Regulus (a main sequence star).&lt;/p&gt;&#xA;" OwnerUserId="26097" LastEditorUserId="9527" LastEditDate="2020-03-25T14:14:29.537" LastActivityDate="2020-03-25T14:14:29.537" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29808" PostTypeId="2" ParentId="29806" CreationDate="2019-03-01T17:58:17.487" Score="0" Body="&lt;p&gt;Oddly enough,  yes-ish.  quoting from a &lt;a href=&quot;https://space.stackexchange.com/questions/12116/have-there-been-any-photos-taken-of-a-total-earth-sun-eclipse-from-the-moon-or&quot;&gt;space.SE&lt;/a&gt; post (pictures there),&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Yes, well, kind of; JAXA's Kaguya (SELENE) took images of the Earth&#xA;  during the February 10, 2009 penumbral lunar eclipse from lunar orbit&#xA;  of roughly 50 km altitude, using its HDTV camera&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There have been cases of Earth-orbiting satellites catching images too.  For example, &lt;a href=&quot;https://www.space.com/41489-double-solar-eclipse-satellite-photos-august-2018.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;space.com&lt;/a&gt; has an article about the Proba-2 satellite doing so. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2019-03-01T17:58:17.487" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29809" PostTypeId="2" ParentId="29787" CreationDate="2019-03-01T19:48:20.123" Score="4" Body="&lt;p&gt;The phrase you hear is correct! Atmosphere will always be lost, no matter if it is a star or planet. Stars lose their mass via solar wind. Planets lose theirs via the three types of atmospheric loss you mentioned. Frozen planets lost their atmosphere thru sublimation and other things such as ionization. So, yes Jupiter does indeed lose its atmosphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You compared the three types of atmospheric loss and how they relate to Jupiter. This is where you assumptions turn out to be wrong. The thermal loss aka Jean's escape affects Jupiter even though it may seem to be not affected in the chart. Periodical or random heat generated by things could excite particles enough to overcome the escape velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solar wind, even with Jupiter's extensive magnetic field, will reach the atmosphere and neutralize atoms, which in turn will then not be effected by the magnetic field and drift off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter is also prone to bombardment from asteroids and comets and other solar system junk. These can range from large sizes such as Comet Shoemaker-Levy 9, to minute small pieces that are barely larger than a pea. These impacts, though they add a considerable amount of stuff into Jupiter, eject mass into space, even with such strong gravitational pull via vertical atmospheric escape, creating a hot plume of material that rises and escapes from a planet. This effect nullifies most material Jupiter gains from the collisions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, will all things added together, we see that Jupiter is gaining mass only via collection of objects, which in turn removes mass from Jupiter via vertical atmospheric escape. This means only a small portion of all mass collected by Jupiter this way is a gain. Now, subtract off thermal and charge exchange escape from the gain, and you will see that Jupiter is not really gaining any material, but rather losing it. Over time, if enough is given, Jupiter will only be a barren core.&lt;/p&gt;&#xA;" OwnerUserId="25988" LastActivityDate="2019-03-01T19:48:20.123" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29810" PostTypeId="2" ParentId="29786" CreationDate="2019-03-01T20:09:27.433" Score="1" Body="&lt;p&gt;Let's look at the validity of tachyons first. Travelling faster than light, as said by userLTK, would create time travel, which would create the famous paradox, the tachyonic antitelephone. The tachyonic antitelephone is a violation of causality, and thus most likely, tachyons would not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, take a look at another assumption. Wikipedia is right almost all times, but sadly, it has some wrong information. Such is with tachyons. It gives us some contradictory statements of tachyons too. It is wrong to assume that tachyons always move faster than light. A tachyon, is a particle with an imaginary number as its rest mass. This, however, does not mean it can travel faster than the speed of light, nor that there's any conflict between their existence and the special theory of relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main idea here is that the typical intuition we have about particles -- them being billiard ball-like objects -- just fails in the quantum world. It turns out that the correct classical limit for quantum fields in many situations is classical fields rather than point particles, and so you must solve the field equations for a field with imaginary mass and see what happens rather than just naively assume the velocity will turn out to be faster than light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leandro M. wrote on physics stack exchange, that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a good way to think about tachyons is to imagine hanging several&#xA;  pendulums on a clothesline, one after the other. If you disturb one of&#xA;  them, some amount of force will be transmitted from one pendulum to&#xA;  the next and you'll see a traveling disturbance on the clothesline.&#xA;  You'll be able to identify a &quot;speed of light&quot; for this system (which&#xA;  will really be the speed of sound in the string). Now you can make a&#xA;  &quot;tachyon&quot; in this system by flipping all the pendulums upside down:&#xA;  they'll be in a very unstable position, but that's precisely what a&#xA;  tachyon represents. Nevertheless, there's absolutely no way that you&#xA;  could send a signal down the clothesline faster than the &quot;speed of&#xA;  light&quot; in the system, even with this instability.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This means that tachyons don't actually move faster than the speed of light, and your assumption is wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, StephenG wrote in a comment that tachyons essentially can't exist in a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;user106422 wrote in the linked question on physics stack exchange that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;in a nutshell, creating a tachyon itself would lead to some form of&#xA;  a spontaneous phase transition in which the tachyonic mode no longer&#xA;  exists. So from a rigourous QFT + GR setting, I don't think tachyons&#xA;  exist the way it is portrayed in simpler QM models. So no, tachyons&#xA;  will not escape the horizon simply because the theory will not allow&#xA;  for them to exist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, no, even if they exist, they don't really travel faster than the speed of light, and even if they do, that can't really &quot;travel&quot; through a black hole. &lt;strong&gt;So, it is not possible to use tachyons to explore a black hole.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="25988" LastActivityDate="2019-03-01T20:09:27.433" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29812" PostTypeId="2" ParentId="29703" CreationDate="2019-03-01T20:33:15.480" Score="3" Body="&lt;p&gt;One reference I found on this subject was &lt;a href=&quot;https://arxiv.org/abs/1305.5037&quot; rel=&quot;nofollow noreferrer&quot;&gt;Izumi et al. (2013)&lt;/a&gt;, who note that the images produced by lensing around a negative mass are radially distorted while lensing by a positive mass are tangentially distorted, and that this effect can be distinguished without prior knowledge of the lens position by measuring the alignment of lens images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not an expert in general relativity so I'm not able to evaluate their results, but assuming the paper is correct it does look like negative mass could indeed be detected by observations of gravitational lensing. The question of whether negative mass exists is another matter, so far it doesn't look like a likely prospect.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastActivityDate="2019-03-01T20:33:15.480" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29813" PostTypeId="1" AcceptedAnswerId="29815" CreationDate="2019-03-01T21:06:09.607" Score="2" ViewCount="191" Body="&lt;p&gt;It seems to me that I forgot the smallest mass of a star and its angular momentum in order to form a black hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I know that electron degeneracy pressure is overcome if the core is 1.4 solar masses, and neutron degeneracy pressure is overcome if the core is 3 solar masses, but if the neutron star is spinning fast enough, it can have a maximum mass of about 4 solar masses. But then, this is just the core of the star. How large does this star have to be in order to form a black hole, and how fast would its spin need to be in order to avoid forming a blitzar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've looked on google and bing but to no avail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help is appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max0815&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Please stop trying to make this a duplicate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why: &lt;strong&gt;How large does this star have to be in order to form a black hole, and &lt;em&gt;how fast would its spin need to be in order to avoid forming a blitzar?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-03-02T05:12:54.983" LastActivityDate="2019-03-04T17:17:41.210" Title="Smallest mass of star to be a black hole?" Tags="&lt;black-hole&gt;&lt;mass&gt;&lt;degenerate-matter&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2019-03-02T08:12:25.557" ContentLicense="CC BY-SA 4.0" />
  <row Id="29814" PostTypeId="2" ParentId="19097" CreationDate="2019-03-01T22:31:59.427" Score="1" Body="&lt;p&gt;We must consider angular velocities of the Earth rotation and the orbital Moon motion relative to the stars. Then, we can calculate the angular velocity of the Moon relative to the Earth:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$1-\frac{365}{27.32166\cdot 366}$&lt;/span&gt; in sidereal days. Thus, the length of the tidal day is - in hours:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\left(1-\frac{365}{27.32166\cdot 366}\right)^{-1}\cdot 24 \cdot \frac{ 365}{366} = 24.84631775 = \text{24h 50m 28.15s}$$&lt;/span&gt; &lt;/p&gt;&#xA;" OwnerUserId="26104" LastEditorDisplayName="user5434" LastEditDate="2019-03-02T08:11:55.957" LastActivityDate="2019-03-02T08:11:55.957" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29815" PostTypeId="2" ParentId="29813" CreationDate="2019-03-01T22:38:22.813" Score="0" Body="&lt;p&gt;Stars with 6 Jupiter Masses to 90 Jupiter Masses but are not able to heat up sufficiently to enable fusion become brown dwarfs, and die brown dwarfs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;White dwarfs form when stars have a mass range of about 85 ± 4 Jupiter Masses to 8~9 Solar Masses, or when a star sufficiently heats up to enable fusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutron stars generally form when the star is &gt;9 Solar Masses to about 24~25 Solar Masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as mentioned in the comments, neutron degeneracy pressure doesn't support the neutron star as a whole. Although it has some effects, a neutron star is supported by not only neutron degeneracy pressure but also by repulsive nuclear forces as a neutron star is comparable to that of an atomic nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your blitzar definition, I doubt 4 Solar Masses is the limit. Let's say that the rotation of a given neutron star is a little less than the speed of light. Supposing that if the neutron star doesn't fling itself apart, the mass limit would be much higher, because the centrifugal force forcing the neutron star is much, much more stronger. A blitzar's maximum mass is 18% more than the maximum mass for cold, nonrotating neutron stars.&#xA;For your question on blitzars, the star would need to spin rather slowly to avoid forming such, because a blitzar's spin is from the amplification of the original star's spin as is shrinks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes form when a star is greater than 25 Solar Masses to the point where stars are so massive that they form a black hole itself before even forming a stable star. Such is an example of a quasi-star and is theoretically real in the early universe.&lt;/p&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-03-04T17:17:41.210" LastActivityDate="2019-03-04T17:17:41.210" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29816" PostTypeId="2" ParentId="29806" CreationDate="2019-03-02T00:26:19.190" Score="7" Body="&lt;p&gt;Such images are most readily available from full-time Earth-observing satellites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://www.nesdis.noaa.gov/content/dscovr-deep-space-climate-observatory&quot; rel=&quot;noreferrer&quot;&gt;DSCOVR&lt;/a&gt; climate observatory sees the Earth from Lagrange point L1, 3.9 lunar distances toward the Sun.&#xA;Its EPIC camera captured sequences of images half an hour apart for the solar eclipses of &lt;a href=&quot;https://epic.gsfc.nasa.gov/?date=2016-03-09&quot; rel=&quot;noreferrer&quot;&gt;2016-03-09&lt;/a&gt; and &lt;a href=&quot;https://epic.gsfc.nasa.gov/?date=2017-08-21&quot; rel=&quot;noreferrer&quot;&gt;2017-08-21&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geostationary satellites orbit 0.11 lunar distance from the center of the Earth.&#xA;The CIMSS group at Wisconsin has a collection of various weather satellites' &lt;a href=&quot;http://cimss.ssec.wisc.edu/goes/blog/?s=eclipse&quot; rel=&quot;noreferrer&quot;&gt;images of several eclipses&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lunar point of view would fall between these two examples, qualitatively more like the former than the latter.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2019-03-02T00:26:19.190" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29817" PostTypeId="2" ParentId="29741" CreationDate="2019-03-02T01:21:09.920" Score="4" Body="&lt;p&gt;My best guess: &lt;strong&gt;The late periapsis of 1P Halley in 1759&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1705, with the mathematical assistance of Issac Newton, Edmond Halley published &lt;em&gt;Synopsis of the Astronomy of Comets&lt;/em&gt; which made the prediction that the comet observed in  1682 had previously been observed in 1531 and 1607 was the same object returning about every 76 years or so.  He included a rough estimation of the gravitational effects of Jupiter and Saturn, and predicted it would return in 1758.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://en.wikipedia.org/wiki/Halley%27s_Comet&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia: Halley's Comet&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Halley's prediction of the comet's return proved to be correct,&#xA;  although it was not seen until 25 December 1758, by Johann Georg&#xA;  Palitzsch, a German farmer and amateur astronomer. It did not pass&#xA;  through its perihelion until 13 March 1759, the attraction of Jupiter&#xA;  and Saturn having caused a retardation of 618 days. This effect was&#xA;  computed prior to its return (with a one-month error to 13 April) by a&#xA;  team of three French mathematicians, Alexis Clairaut, Joseph Lalande,&#xA;  and Nicole-Reine Lepaute. The confirmation of the comet's return was&#xA;  the first time anything other than planets had been shown to orbit the&#xA;  Sun. It was also one of the earliest successful tests of Newtonian&#xA;  physics, and a clear demonstration of its explanatory power.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This predates the discovery of Uranus(1781) and Neptune (1846).&lt;/p&gt;&#xA;" OwnerUserId="23813" LastActivityDate="2019-03-02T01:21:09.920" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29820" PostTypeId="2" ParentId="29743" CreationDate="2019-03-02T09:28:08.533" Score="1" Body="&lt;p&gt;Well yes, it is still a useful tool because it can give much more insight than the output from a black box computer code. However, you have to pick your problems or the complexity of the analysis can lose this advantage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A particular case is the evolution of low mass pre main sequence stars along the Hayashi track. Since these stars are fully convective they can be treated as single polytrope a (to first order). Myself and a colleague have used such analytical calculations to study the effects of spots and magnetic fields on the evolution of PMS stars and on the rate at which they &quot;burn&quot; lithium in their cores (e.g. &lt;a href=&quot;https://arxiv.org/abs/1404.0683&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jackson &amp;amp; Jeffries 2014a&lt;/a&gt;; &lt;a href=&quot;https://arxiv.org/abs/1410.0022&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jackson &amp;amp; Jeffries 2014b&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In turn, this work was inspired by a much earlier analytical treatment of the Li depletion problem by &lt;a href=&quot;https://arxiv.org/abs/astro-ph/9612155&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bildsten et al. (1997)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-03-02T09:28:08.533" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29821" PostTypeId="1" AcceptedAnswerId="29822" CreationDate="2019-03-02T10:25:27.340" Score="1" ViewCount="113" Body="&lt;p&gt;It seems to me every time I see stellar remnant binary systems where there is a white dwarf and another star or something, it appears that the white dwarf always pulls mass from the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if the white dwarf is with a neutron star? Will the white dwarf draw mass from the neutron star? What if the white dwarf was orbiting a supermassive star(not giant stage yet)?&lt;/p&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-03-02T10:30:51.733" LastActivityDate="2019-03-03T03:56:36.853" Title="Are white dwarfs the ones always absorbing mass from a companion?" Tags="&lt;star&gt;&lt;white-dwarf&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29822" PostTypeId="2" ParentId="29821" CreationDate="2019-03-02T10:30:22.697" Score="2" Body="&lt;p&gt;A white dwarf only draws mass because it is denser that it's companion star. If a nearby object was denser than the white dwarf, and exerts enough gravity on the white dwarf, then it will draw mass from the white dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the white dwarf is with a neutron star, and the neutron star is close enough to exert a gravitational pull that can distort a white dwarf, then the neutron star will draw mass from the white dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you see, stars are way less dense that stellar remnants. No matter how massive, it's the density that counts. So if the white dwarf orbits any star, it will draw mass from the star.&lt;/p&gt;&#xA;" OwnerUserId="25988" LastActivityDate="2019-03-02T10:30:22.697" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29823" PostTypeId="1" AcceptedAnswerId="29828" CreationDate="2019-03-02T10:56:30.850" Score="2" ViewCount="746" Body="&lt;p&gt;A black hole contains a singularity at its center. It is a zero-dimensional point, and  it's where all its mass is located. So, my first question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If a singularity contains all a black hole's mass, when a black hole absorbs matter, the matter should collect at the singularity. Then why does the black hole increase in size? The matter it absorbs is all in the singularity, not in the space around it...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, we know that spinning black holes can't have a singularity at the center as a point, because points can't spin, and when they do, they aren't a point. So there is such thing as a ringularity where it is a one-dimensional line bent in a circle that spins, and contains all the black hole's matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;However, is a ringularity really a bent one-dimensional line in a circle, or an infinite collection of points that form a circle-looking figure?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(My information on ringularities is from a paper published by Cornell University that compared the one-dimensional line to a bunch of zero-dimensional points, and videos by Kurgesagt.)&lt;/p&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="14094" LastEditDate="2019-03-02T11:45:36.917" LastActivityDate="2019-03-03T05:18:43.813" Title="Singularities/ringularities and their properties" Tags="&lt;black-hole&gt;&lt;singularity&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29824" PostTypeId="1" AcceptedAnswerId="29842" CreationDate="2019-03-02T11:52:26.220" Score="3" ViewCount="239" Body="&lt;p&gt;Spot checking Wikipedia, Google and JPL's &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;Horizions&lt;/a&gt; I found several musicians right away, including Lou Reed, Elvis, Freddy Mercury, and David Bowie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to find all of the asteroids that are named after musicians somehow? Or at least most of them? If there's not a compiled list, at least something for which an automated search for &quot;music&quot; or &quot;musician&quot; or related terms can be done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UVJLV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UVJLV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZWjP3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZWjP3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wf3hH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wf3hH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2lcWo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2lcWo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-09-10T09:46:12.603" Title="How many asteroids are named after musicians? Is there a searchable list for this?" Tags="&lt;asteroids&gt;&lt;naming&gt;" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="29825" PostTypeId="1" AcceptedAnswerId="31900" CreationDate="2019-03-02T13:51:17.060" Score="5" ViewCount="64" Body="&lt;p&gt;As mentioned &lt;a href=&quot;https://www.theguardian.com/science/2019/feb/22/japans-hayabusa-2-successfully-touches-down-on-ryugu-asteroid&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, that Ryugu is a part of the Solar System material that did not get swept into forming a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how can we be sure that it was not created later for example by an asteroid striking Mars or some other planet? Why do we think its &quot;primordial&quot;?&lt;/p&gt;&#xA;" OwnerUserId="26110" LastEditorUserId="1891" LastEditDate="2019-03-03T01:41:09.590" LastActivityDate="2019-05-16T20:12:45.053" Title="How can we be certain that Ryugu is a remnant of the material that did not form planets?" Tags="&lt;asteroids&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29826" PostTypeId="2" ParentId="29823" CreationDate="2019-03-02T15:44:34.647" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A black hole contains a singularity at its center. It is a zero-dimensional point, and it's where all its mass is located. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This isn't an established fact. A lot of people say they don't think there is a point-singularity at the centre of a black hole. Google on &lt;a href=&quot;https://www.google.com/search?ei=S556XNXWL5WE1fAP182hyAk&amp;amp;q=black+hole+%22no+point+singularity%22&amp;amp;oq=black+hole+%22no+point+singularity%22&amp;amp;gs_l=psy-ab.3..33i22i29i30.997570.1007608..1007961...0.0..0.90.2180.35......0....1..gws-wiz.......0i71j0i131j0i67j0j35i39j0i7i30j0i7i10i30j0i10j0i13j0i8i7i30j0i30j0i8i30j0i8i13i30j33i10j33i13i21j33i21.JQzCny23pcE&quot; rel=&quot;nofollow noreferrer&quot;&gt;black hole &quot;no point singularity&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So, my first question is: If a singularity contains all a black hole's mass, when a black hole absorbs matter, the matter should collect at the singularity. Then why does the black hole increase in size? The matter it absorbs is all in the singularity, not in the space around it...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;People typically talk about the extent of the event horizon when talking about the size of a black hole. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Second, we know that spinning black holes can't have a singularity at the center as a point, because points can't spin, and when they do, they aren't a point. So there is such thing as a ringularity where it is a one-dimensional line bent in a circle that spins, and contains all the black hole's matter. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That sounds like a ring singularity. See &lt;a href=&quot;https://en.wikipedia.org/wiki/Ring_singularity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; for something on that. Note however that a ring singularity is hypothetical, not proven fact.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, is a ringularity really a bent one-dimensional line in a circle, or an infinite collection of points that form a circle-looking figure?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ring_singularity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; article. It says this: &lt;em&gt;&quot;the minimal shape of the singularity that can support these properties is instead a ring with zero thickness but non-zero radius, and this is referred to as a ringularity or Kerr singularity&quot;&lt;/em&gt;. I don't endorse that by the way. I think the original &quot;frozen star&quot; description of a black hole was right. Google on &lt;a href=&quot;https://www.google.co.uk/search?q=%22frozen+star%22+oppenheimer+-farsight+-duffield&amp;amp;*&amp;amp;spf=380&amp;amp;cad=h&quot; rel=&quot;nofollow noreferrer&quot;&gt;frozen star and Robert Oppenheimer&lt;/a&gt; and you can find things like the 1971 Physics Today article &lt;a href=&quot;https://authors.library.caltech.edu/14972/1/Ruffini2009p1645Phys_Today.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;introducing the black hole&lt;/a&gt; by Remo Ruffini and John Wheeler. They said &lt;em&gt;“in this sense the system is a frozen star”&lt;/em&gt;.   &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2019-03-02T15:44:34.647" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29827" PostTypeId="2" ParentId="29824" CreationDate="2019-03-02T16:29:59.003" Score="3" Body="&lt;p&gt;I don't think you can do better than search the wikipedia pages, where there is a list &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_minor_planets_named_after_people&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/List_of_minor_planets_named_after_people&lt;/a&gt;. Or you can search yourself through &lt;a href=&quot;https://en.wikipedia.org/wiki/Meanings_of_minor_planet_names&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Meanings_of_minor_planet_names&lt;/a&gt;, if you don't trust wikipedian's curation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you may need to be quite subtle about how you search, since searching for &quot;music&quot; won't find those that are described as &quot;composers&quot;, nor those who are &quot;sopranos&quot;, or &quot;violinists&quot; and so forth.  There are also edge cases. William Herschel (who may have coined the name &quot;asteroid&quot;) was also a classical composer of note. Or &quot;644 Cosima&quot;, named after Wagner's second wife.  How about 1683 Castafiore, named for a fictional &quot;musician&quot;?&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2019-03-02T23:51:46.120" LastActivityDate="2019-03-02T23:51:46.120" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29828" PostTypeId="2" ParentId="29823" CreationDate="2019-03-02T17:24:07.577" Score="9" Body="&lt;p&gt;You need to be a bit cautious about statements like: &lt;em&gt;A black hole contains a singularity at its center. It is a zero-dimensional point, and it's where all its mass is located&lt;/em&gt;. What you are referring to is a mathematical structure called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_metric&quot; rel=&quot;noreferrer&quot;&gt;Schwarzschild metric&lt;/a&gt;. This is indeed a solution to the Einstein equations for a static black hole, but this particular solution takes an infinite time to form. So given that the universe is only a shade under 14 billion years old we can be confident that as I type this there are no Schwarzschild black holes in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, it is far from clear that a Schwarzschild black hole could ever form, even given infinite time, because black holes evaporate. This point is pursed in the question &lt;a href=&quot;https://physics.stackexchange.com/questions/95366/why-does-stephen-hawking-say-black-holes-dont-exist&quot;&gt;Why does Stephen Hawking say black holes don't exist?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, the Schwarzschild black hole doesn't have all its mass at the centre. In fact it is a vacuum solution and contains no matter at all. The &lt;em&gt;mass&lt;/em&gt; of the Schwarzschild black hole is actually a geometrical property called the &lt;a href=&quot;https://en.wikipedia.org/wiki/ADM_formalism#ADM_energy_and_mass&quot; rel=&quot;noreferrer&quot;&gt;ADM mass&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We run into the same problems with the spinning black hole. The structure with the ring singularity is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kerr_metric&quot; rel=&quot;noreferrer&quot;&gt;Kerr metric&lt;/a&gt;, and again while it's a perfectly valid solution to Einstein's equations, like the Schwarzschild metric it also takes an infinite time to form. Actually the Kerr metric is more troublesome still, because it is uncertain whether the structure is stable with respect to perturbations. It seems very likely that real collapsing stars would not form a structure resembling the Kerr metric. More precisely the region outside the event horizon would be well described by a Kerr metric, the region near the singularity would be different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we put these doubts aside and return to what you asked, the answer is that in the Kerr metric the ring singularity really is a ring and not a collection of points.&lt;/p&gt;&#xA;" OwnerUserId="10687" LastEditorUserId="10687" LastEditDate="2019-03-03T05:18:43.813" LastActivityDate="2019-03-03T05:18:43.813" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29829" PostTypeId="2" ParentId="29821" CreationDate="2019-03-02T20:43:50.800" Score="1" Body="&lt;p&gt;White dwarfs don't always draw mass from a companion, for example our Sun will one day be a white dwarf with no companion to draw mass from. But it sounds like you are wondering what controls which way the mass is transported.  That is normally controlled by whichever star is trying to grow in size.  The growth carries it across the &quot;Roche lobe&quot; and the mass is pulled to the other star.  White dwarfs don't have internal evolution that makes them try to grow, so that's why they don't normally give up mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility, however, is when two stars in a close binary have their orbits come together and the stars merge.  The merger of two white dwarfs is thought to be an important source of type Ia supernovae.  So we might not necessarily count a merger as a transfer of mass, but it shares some of the same attributes.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2019-03-02T20:43:50.800" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29830" PostTypeId="2" ParentId="29821" CreationDate="2019-03-03T03:56:36.853" Score="2" Body="&lt;p&gt;To address your other question&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What if the white dwarf was orbiting a supermassive star(not giant&#xA;  stage yet)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An interesting fact about main sequence stars is that the more massive the star, the lower the star's surface gravity, so the more massive you make the star, the more easily a smaller companion star, not necessarily a white dwarf but any smaller and close companion star, can draw matter away from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sirius&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sirius for example&lt;/a&gt;, is a little over twice the mass of our sun, but 1.7 times our sun's diameter.   That means it's surface gravity is 2/(1.7^2) or about 70% of our sun.   It's counterintuitive but largely true that the more massive a star during it's main sequence, the lower it's surface gravity.  It's also true that stars like our sun grow larger over time even as it loses mass, so the age of the star matters too, but generally, more massive stars have lower surface gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/TRAPPIST-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Trappist 1&lt;/a&gt;, a red dwarf star is about 1/11th the mass of the Sun and a bit under 1/8th the diameter of our Sun.  That gives it a surface gravity of about 6 times our Sun's surface gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why there's a theoretical limit of how large stars can get, somewhere around 150 times the mass of our Sun.  Around or a bit above that mass, the internal heat and outward pressure from fusion pushes the outer edges to the point where the outer layers of the star are hot enough and loosely held enough by gravity that they can escape the gravitational field of the star.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2019-03-03T03:56:36.853" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29832" PostTypeId="1" CreationDate="2019-03-03T05:42:16.233" Score="-2" ViewCount="1250" Body="&lt;p&gt;Firstly I’m genuinely interesting in a working explanation for this question. It is for this reason that I am editing the question to fine tune the question. In essense the question has remained the same.&lt;/p&gt;&#xA;&lt;p&gt;In order for me to check mark the best answer I’m going to ask for citations for the claims made in the answers because I’m getting lots of theory and responders are not in agreement, as far as I can tell. Also I cannot accept circular reasoning i.e. if we assume x then y, since y therefore x. That’s fallacious logical reasoning because x was never proven it was assumed and y might be independent of x.&lt;/p&gt;&#xA;&lt;p&gt;The vacuum of space is incredibly powerful, 1 x 10&lt;sup&gt;-17&lt;/sup&gt; torr, and the vacuum between the Earth and the Moon is 1 x 10&lt;sup&gt;-11&lt;/sup&gt; torr.&lt;/p&gt;&#xA;&lt;h3&gt;How can such a vacuum (very low pressure) in close proximity to the Earth’s atmosphere (high pressure) that goes to 8.5km elevation, coexist with the open system of the Earth’s atmosphere? How does this not defy the second law of thermodynamics and still remain true?&lt;/h3&gt;&#xA;&lt;p&gt;Space is low pressure and the earth’s atmosphere is high pressure. In order to have any pressure the gas requires (demands) that it press upon something.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Pressure is a force exerted by the substance per unit area on another substance. &lt;strong&gt;The pressure of a gas is the force that the gas exerts on the walls of its container&lt;/strong&gt;. When you blow air into a balloon, the balloon expands because the pressure of air molecules is greater on the inside of the balloon than the outside. Pressure is a property which determines the direction in which mass flows. If the balloon is released, the air moves from a region of high pressure to a region of low pressure and the balloon deflates. &lt;a href=&quot;https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Gas_Pressure&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Earth is an open system that presses against the vacuum of space, why therefore is the second law of thermodynamics suspended if indeed it is a law.&#xA;Space is low pressure, therefore the atmosphere which is not in a container should disperse into the low pressure space.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.raclub.org/Documents/Programs/VACUUM.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;VACUUM (There’s nothing to it…)&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;&lt;a href=&quot;https://web.pa.msu.edu/courses/2014fall/PHY451/Lectures/Vacuum.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vacuum Technology PHY451 October 22, 2014&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="26115" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-03-04T10:34:49.297" Title="Vacuum of space" Tags="&lt;cosmology&gt;&lt;planetary-atmosphere&gt;&lt;space&gt;&lt;vacuum&gt;" AnswerCount="5" CommentCount="17" ClosedDate="2019-03-04T14:10:04.953" ContentLicense="CC BY-SA 4.0" />
  <row Id="29833" PostTypeId="2" ParentId="29832" CreationDate="2019-03-03T06:12:40.083" Score="6" Body="&lt;p&gt;&lt;strong&gt;update:&lt;/strong&gt; This answer was written before the question was modified. I've tried to explain where a value like 10&lt;sup&gt;-17&lt;/sup&gt; Torr for deep space might come from, but it's since been dropped in lieu of 10&lt;sup&gt;-11&lt;/sup&gt; Torr at the Moon, &lt;em&gt;which is probably a better way to formulate the question.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the answer is the same, two points very far away can have very different pressures. They can coexist in the same solar system, just not right next to each other. I think &quot;Why doesn't the Moon have at least a small atmosphere?&quot; could also be an excellent, but very different question.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In a &lt;a href=&quot;https://astronomy.stackexchange.com/questions/29832/vacuum-of-space-question#comment54191_29832&quot;&gt;comment&lt;/a&gt; the OP links to the presentation &lt;a href=&quot;https://www.raclub.org/Documents/Programs/VACUUM.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;VACUUM (There’s nothing to it… )&lt;/a&gt; written from the perspective of an engineer in the semiconductor manufacturing industry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slide 6 gives examples of vacuum levels in different situations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going down:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Low vacuum: 760 Torr to 1 x 10&lt;sup&gt;-3&lt;/sup&gt; Torr&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Vacuum cleaner: to 600 Torr&lt;/li&gt;&#xA;&lt;li&gt;Thermos bottle 10&lt;sup&gt;-3&lt;/sup&gt; Torr&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;High vacuum: 10&lt;sup&gt;-3&lt;/sup&gt; to 10&lt;sup&gt;-9&lt;/sup&gt; Torr&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Electron microscope&lt;/li&gt;&#xA;&lt;li&gt;Ion Implanter&#xA;– Evaporator&#xA;– Sputterer&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Ultra high vacuum: 10&lt;sup&gt;-9&lt;/sup&gt; to 10&lt;sup&gt;-12&lt;/sup&gt; Torr&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;CERN LHC: 1 x 10&lt;sup&gt;-10&lt;/sup&gt; Torr&lt;/li&gt;&#xA;&lt;li&gt;Moon’s surface: 1 x 10&lt;sup&gt;-11&lt;/sup&gt; Torr&lt;/li&gt;&#xA;&lt;li&gt;Deep Space 1 x 10-&lt;sup&gt;17&lt;/sup&gt; Torr = 0.000,000,000,000,000,01 Torr&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So we can see that the value of 1 x 10-&lt;sup&gt;17&lt;/sup&gt; Torr is associated with a place in &quot;Deep Space&quot; which is (probably) beyond that of the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's see if we can figure out where the author is getting that number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the Wikipedia article on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Interstellar_medium&quot; rel=&quot;nofollow noreferrer&quot;&gt;interstellar medium&lt;/a&gt; (space between stars, far away from solar systems and other things):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In all phases, the interstellar medium is extremely tenuous by terrestrial standards. In cool, dense regions of the ISM, matter is primarily in molecular form, and reaches number densities of 106 molecules per cm&lt;sup&gt;3&lt;/sup&gt; (1 million molecules per cm&lt;sup&gt;3&lt;/sup&gt;). In hot, diffuse regions of the ISM, matter is primarily ionized, and the density may be as low as 10&lt;sup&gt;−4&lt;/sup&gt; ions per cm&lt;sup&gt;3&lt;/sup&gt;. Compare this with a number density of roughly 10&lt;sup&gt;19&lt;/sup&gt; molecules per cm&lt;sup&gt;3&lt;/sup&gt; for air at sea level, and 10&lt;sup&gt;10&lt;/sup&gt; molecules per cm&lt;sup&gt;3&lt;/sup&gt; (10 billion molecules per cm&lt;sup&gt;3&lt;/sup&gt;) for a laboratory high-vacuum chamber. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is harder to talk about pressure than density because pressure is related to both number density and temperature. The atmosphere is more than 10x hotter than interstellar medium, so let's look for a number density ratio of 1/10 of 1000 Torr versus 10&lt;sup&gt;-17&lt;/sup&gt; Torr, or a ratio of 10&lt;sup&gt;19&lt;/sup&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If (according to Wikipedia) Earth's atmosphere has a density of 10&lt;sup&gt;19&lt;/sup&gt; per cm&lt;sup&gt;3&lt;/sup&gt;, we're looking for a density of 1 per cm&lt;sup&gt;3&lt;/sup&gt;. Checking Wikipedia we can see that &lt;em&gt;there are components&lt;/em&gt; of the interstellar medium with number densities between 10&lt;sup&gt;6&lt;/sup&gt; and 10&lt;sup&gt;-4&lt;/sup&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like the value in the presentation a rough ballpark estimate, but isn't off by more than a handful of orders of magnitude ;-)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can such a vacuum coexist with the open system of earth’s atmosphere whereby debris from space can enter in?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While these two pressures can coexist in the same universe, they don't coexist in proximity at all. The interstellar medium is very, very far away from Earth's atmosphere, on the order of a lightyear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity keeps Earth's atmosphere nearby Earth, the solar system has gas produced by (and attracted by) the Sun's gravity. In interstellar space, there just isn't any source of gas, and what might have been there at one time has moved away, towards sources of gravity, over billions of years.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="11190" LastEditDate="2019-03-03T14:52:04.347" LastActivityDate="2019-03-03T14:52:04.347" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29836" PostTypeId="1" AcceptedAnswerId="29848" CreationDate="2019-03-03T13:16:55.130" Score="0" ViewCount="53" Body="&lt;p&gt;Phys.org's &lt;a href=&quot;https://phys.org/news/2019-03-black-hole.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hiding black hole found&lt;/a&gt; says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A research team led by Shunya Takekawa at the National Astronomical Observatory of Japan noticed HCN-0.009-0.044, &lt;strong&gt;a gas cloud moving strangely near the center of the galaxy 25,000 light-years away from Earth in the constellation Sagittarius.&lt;/strong&gt; They used ALMA (Atacama Large Millimeter/submillimeter Array) to perform high-resolution observations of the cloud and found that it is swirling around a massive invisible object.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Takekawa says, &amp;quot;Detailed kinematic analyses revealed that an enormous mass, 30,000 times that of the sun, was concentrated in a region much smaller than our solar system.&lt;/strong&gt; This and the lack of any observed object at that location strongly suggests an intermediate-mass black hole. By analyzing other anomalous clouds, we hope to expose other quiet black holes.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Tomoharu Oka, a professor at Keio University and coleader of the team, adds, &lt;strong&gt;&amp;quot;It is significant that this intermediate mass black hole was found only 20 light-years from the supermassive black hole at the galactic center.&lt;/strong&gt; In the future, it will fall into the supermassive black hole, much like gas is currently falling into it. This supports the merger model of black hole growth.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;These results were published as Takekawa et al. &amp;quot;Indication of Another Intermediate-mass Black Hole in the Galactic Center&amp;quot; in The Astrophysical Journal Letters on January 20, 2019.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The paper &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/aafb07/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;Indication of Another Intermediate-mass Black Hole in the Galactic Center&lt;/a&gt; (open access) is pretty hard to read as it details the careful analysis of ALMA data reduction and analysis.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How did the authors determine both the spatial size of gas cloud HCN-0.009-0.044 and the mass of the central object at the same time?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-03-04T10:17:34.740" Title="How did the authors determine both the spatial size of gas cloud HCN-0.009-0.044 and its central mass at the same time?" Tags="&lt;black-hole&gt;&lt;radio-astronomy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29837" PostTypeId="1" CreationDate="2019-03-03T14:00:56.083" Score="1" ViewCount="64" Body="&lt;p&gt;I'm part of a local astronomy club but the results from that have been vastly underwhelming with a membership count of around 50-200 active people at any given time. All that being said- I guess my question is for 2019 what are the biggest and most informative gatherings in terms of astronomy? Anything you know of from lectures to imaging to stargazing would be wonderful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;More specifically:&lt;/em&gt;&#xA;I was considering spending the 250 dollars to go to the Northeast Astronomy Forum this April 4th and 5th and the Northeast Astro Imaging Conference. It seems to be one of the larger scale events that happens anually. Is this a good event for beginners? Heres a link to their site:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.rocklandastronomy.com/neaf.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.rocklandastronomy.com/neaf.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.rocklandastronomy.com/neaic.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.rocklandastronomy.com/neaic.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If youve been Id love to hear from you. If this is off-topic I apologize, but the only other place I had to ask was our clubs forum and it has virtually no traffic.&lt;/p&gt;&#xA;" OwnerUserId="23729" LastEditorUserId="23729" LastEditDate="2019-03-03T14:21:56.617" LastActivityDate="2019-03-04T06:28:30.040" Title="What are some of the best astronomy gatherings in the USA?" Tags="&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29838" PostTypeId="2" ParentId="29832" CreationDate="2019-03-03T14:47:21.530" Score="2" Body="&lt;p&gt;I think uhoh covered proximity  but just to induce equilibrium to further elaborate:&lt;br&gt;&#xA;First of all, positive pressure and negative pressure are just terminology based on where we started i.e. 1 atm and above/below, we just went along. There is zero pressure and gradually moving up from that. Say, perfect vacuum starts at zero pressure  and move up as comes across high pressure open system. Things are always at equilibrium and if you want to move something from low pressure to high, you  do some work as you have mentioned second law of thermodynamics. Gravity does that work in this case till some point. Ignore everything,  lets say there is one good looking blue dot aka earth, and as you move close, gravity starts to get stronger. So, it will invite more molecule to be cuddly and at same time pressure gradient will move gas the other way. Eventually they will reach at equilibrium i.e. same transference. This is true for any height from earth. In absence of such equilibrium earth would loose its atmosphere or gain more (may be case at astronomical time scale). We start at one (gravity rules) to continuously (important) merge to space vacuum with equilibrium all along the line. Once gas truly escape earth gravity (i.e. temperature movement is way stronger than earth gravity), it has no reason to hang around. Same way we can gain by unsuspecting wandering gas molecules. Earth atmosphere looks at equilibrium at human timescale.  But earth loose and gain as these forces continuously change with distance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: let's try to come up based on fundamental property of universe:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take two points A and B sealed in perfect vacuum by a long tube. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 insert some gas at point A. You would expect, given enough time, gas will diffuse uniformly between point A and B. That is one property of universe: entropy always maximize. (Remember pressure is not a force). There is no answer why but it does and give rise to interesting open physics problem: arrow of time. Wikipedia has nice article on it. &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;Now second property: fundamental force gravity which attracts everything. Say gas is diffused and you switch on gravity at point A. You expect gas to move towards point A.&#xA;And it will.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now both at play. At any cross section , gas would cross to go to A due to gravity and B to maximize the entropy. Given enough time there will be equilibrium. Equal transference at any length between A and B. But you would expect A to be more concentrated since gravity wants all gas at A where entropy wants all uniformly distributed between A and B. So there will be gradient between A and B.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now think of tube between earth and space and make it disappear. Now B at low concentration, there can be escape but is it significant at human scale. In short atmosphere gradually thins out. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Answer based on escape velocity are fine but couple of things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Escape velocity is consequence of  fundamental force gravity.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You don't need escape velocity to leave earth. Only true for projectile thrown from ground.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Gas velocity is function of temperature. Earth atomposhere temperate decreases and then increases and then again decreases and guess  -increases again. (With height).See Wikipedia. So will Gas velocity. Where escape velocity will decrease with height.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Gas in space is hot because of high kinetic energy and lack of collision to transfer heat. Speed is not due to lack of gravity. You can argue lack of gravity explains lack of concentration but that feels like going in circles.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;All of this is in Wikipedia and basic science&#xA;Hope it is more clear..&lt;/p&gt;&#xA;" OwnerUserId="26086" LastEditorUserId="26086" LastEditDate="2019-03-04T08:47:46.667" LastActivityDate="2019-03-04T08:47:46.667" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29839" PostTypeId="2" ParentId="29832" CreationDate="2019-03-03T15:03:38.387" Score="7" Body="&lt;p&gt;The underlying reason that the molecules of Earth's atmosphere do not fly away into the surrounding vacuum is that they are slower than the escape velocity, which would be 11200 m/s. The typical molecule speed at ground level and room temperature appears to be 500 m/s. If it had a free path such a molecule could fly vertically for &lt;span class=&quot;math-container&quot;&gt;$t = v/a = \frac{500m/s}{9.81m/s^2} \approx 50s$&lt;/span&gt; before starting to fall back, with an average velocity of 250 m/s, thus reaching an altitude of 12 or 13 km. (In reality it would collide with other molecules on the way, transferring some kinetic energy to them, so that they could in turn rise higher. Obviously, molecules at the outer fringes of the atmosphere are the real escape candidates.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecules which are fast enough surely do escape Earth's gravity well. Some may have been accelerated by particles of the solar wind, some may just have been on the long tail of the standard distribution. The latter is more likely for light atoms and molecules which are faster, like Helium and Hydrogen. Hydrogen's average speed at room temperature is perhaps 2000 m/s. These gases have indeed mostly left Earth for good long ago. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By the way, the solar wind would likely &quot;blow away&quot; our atmosphere in the long run &amp;mdash; as it did on Mars &amp;mdash; if it weren't deflected by the Earth's magnetic field.)&lt;/p&gt;&#xA;" OwnerUserId="11190" LastEditorUserId="11190" LastEditDate="2019-03-04T10:34:49.297" LastActivityDate="2019-03-04T10:34:49.297" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29840" PostTypeId="1" AcceptedAnswerId="29943" CreationDate="2019-03-03T15:24:14.233" Score="6" ViewCount="276" Body="&lt;p&gt;I'm trying to understand how smooth the reflecting ocean surface would need to be to produce such a small bright spot as seen from the DSCOVR satellite at Sun-Earth L1. It appears to be only about 8E-05 rad, or about 0.3 arcminutes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is primarily about the geometrical optics involved in the reflection of light from the Sun off of the Earth. Interpretation of the apparent size and comparing to the image is secondary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a distance of 151.3 million km from Earth, the 1.391 million km diameter Sun has an angular width of 0.009196 radians. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Deep_Space_Climate_Observatory&quot; rel=&quot;noreferrer&quot;&gt;DSCOVR&lt;/a&gt; satellite is 1.586 million km from the Earth looking at the Sun's reflection from &lt;em&gt;somewhere around&lt;/em&gt; the Sun-Earth L1 point. It's close to the Earth-Sun axis (the Sun-Earth-DSCOVR angle is about 7.6 degrees) so let's simplify and assume it sits on-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth's radius is 6378 km; as a convex mirror its focal length is &lt;span class=&quot;math-container&quot;&gt;$f=-r/2$&lt;/span&gt; or -3189 km. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What would be the angular size of the Sun seen from DSCOVR reflecting in a spherical mirror model for the Earth in this configuration?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What I'm looking at (for context):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;https://epic.gsfc.nasa.gov/?date=2017-08-21&quot; rel=&quot;noreferrer&quot;&gt;sequence of images&lt;/a&gt; taken by the EPIC camera on the DSCOVR satellite near the Sun-Earth L1 point is notable for showing the progress of the Moon's shadow across the Earth, but that's not the topic of my question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first image shows the back-reflection of the Sun from the Earth. It is likely to be from a particularly calm spot on the Ocean, but that's not for sure, there are some images that show reflections from horizontally oriented ice crystals as well. See &lt;a href=&quot;https://earthscience.stackexchange.com/q/10436/6031&quot;&gt;Are there measurements or calculations that suggest atmospheric ice plates would be horizontal to within 0.1 degrees?&lt;/a&gt; and also &lt;a href=&quot;https://space.stackexchange.com/q/21587/12102&quot;&gt;How could the recently explained “glints” seen by DSCOVR appear so compact considering the finite size of the sun?&lt;/a&gt; for some interesting images and more on that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that doesn't matter for the purposes of my question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Width of Earth 6378 km x 2 / 1586000 km = 8.0428E-03 rad corresponds to ~1500 pixels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bright spot is about 15 pixels wide, or about 8E-05 rad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could such a small angular width be a geometrical reflection of the Sun from a very smooth ocean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://epic.gsfc.nasa.gov/archive/natural/2017/08/21/png/epic_1b_20170821151450.png&quot; rel=&quot;noreferrer&quot;&gt;Raw image at NASA&lt;/a&gt; seen &lt;a href=&quot;https://epic.gsfc.nasa.gov/?date=2017-08-21&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z9RPQ.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z9RPQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AwLKl.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AwLKl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-03-08T16:43:11.523" LastActivityDate="2019-03-14T01:31:54.783" Title="Angular diameter of the Sun's reflection from the ocean, seen from Sun-Earth L1?" Tags="&lt;telescope&gt;&lt;earth&gt;&lt;optics&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29841" PostTypeId="2" ParentId="29832" CreationDate="2019-03-03T21:00:54.310" Score="8" Body="&lt;p&gt;Your assertion that our atmosphere doesn't escape is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Helium and Hydrogen atoms have a low enough mass that they do have an escape velocity at the temperatures on the edge of our atmosphere.  This means that when those gasses are released, if they fail to react on their way out, then they will be lost forever to the planet.  This is why when you look at our atmosphere we just don't have any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also seem to assert that there's a &quot;line&quot; where it's the high pressure of our atmosphere on one side, and a low pressure of space on the other; that line doesn't exist.  The pressure is a gradient, in the same way as you swim down in a pool, at the top the pressure is low; at the bottom it's high; and it changes steadily as you traverse between the two.  Hence why climbers of Everest need to carry oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heavier gasses don't escape for the same reason that a rock you throw doesn't escape.  It requires energy to escape the gravitational well of Earth; and they don't have it; currently.  Of course, as the atmosphere heats up from global warming, heavier and heavier particles will gain the energy to leave our atmosphere... &lt;/p&gt;&#xA;" OwnerUserId="16969" LastActivityDate="2019-03-03T21:00:54.310" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29842" PostTypeId="2" ParentId="29824" CreationDate="2019-03-03T21:43:57.030" Score="3" Body="&lt;p&gt;There are many of them, e.g. Jimi Hendrix, Maria Callas, Frank Zappa, Dietrich Fischer-Dieskau, Mark Knopfler, Aretha Franklin, Ella Fitzgerald, Enrico Caruso, Paul McCartney, Jacques Brel, Bob Dylan, Niccolò Paganini.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Wikipedia has an entry for the asteroid you might get it by Wikidata (SPARQL) queries. E.g. &lt;a href=&quot;https://query.wikidata.org/#%0ASELECT%20%3Fitem%20%3FitemLabel%20%3FnamedAfter%20%3FnamedAfterLabel%20WHERE%20%7B%0A%20%20%3Fitem%20wdt%3AP31%20wd%3AQ3863%3B%0A%20%20%20%20%20%20%20%20wdt%3AP138%20%3FnamedAfter.%0A%20%20%3FnamedAfter%20wdt%3AP106%20wd%3AQ639669.%20%23%20musician%0A%20%20SERVICE%20wikibase%3Alabel%20%7B%20bd%3AserviceParam%20wikibase%3Alanguage%20%22en%22.%20%7D%0A%7D&quot; rel=&quot;nofollow noreferrer&quot;&gt;musicians - Q639669&lt;/a&gt;, &lt;a href=&quot;https://query.wikidata.org/#SELECT%20%3Fitem%20%3FitemLabel%20%3FnamedAfter%20%3FnamedAfterLabel%20WHERE%20%7B%0A%20%20%3Fitem%20wdt%3AP31%20wd%3AQ3863%3B%0A%20%20%20%20%20%20%20%20wdt%3AP138%20%3FnamedAfter.%0A%20%20%3FnamedAfter%20wdt%3AP106%20wd%3AQ177220.%0A%20%20SERVICE%20wikibase%3Alabel%20%7B%20bd%3AserviceParam%20wikibase%3Alanguage%20%22en%22.%20%7D%0A%7D%0A&quot; rel=&quot;nofollow noreferrer&quot;&gt;singers - Q177220&lt;/a&gt;, or &lt;a href=&quot;https://query.wikidata.org/#SELECT%20%3Fitem%20%3FitemLabel%20%3FnamedAfter%20%3FnamedAfterLabel%20WHERE%20%7B%0A%20%20%3Fitem%20wdt%3AP31%20wd%3AQ3863%3B%0A%20%20%20%20%20%20%20%20wdt%3AP138%20%3FnamedAfter.%0A%20%20%3FnamedAfter%20wdt%3AP106%20wd%3AQ855091.%20%23guitarist%0A%20%20SERVICE%20wikibase%3Alabel%20%7B%20bd%3AserviceParam%20wikibase%3Alanguage%20%22en%22.%20%7D%0A%7D&quot; rel=&quot;nofollow noreferrer&quot;&gt;guitarists - Q855091&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as not every asteroid has a complete Wikidata entry, these lists will not be exhaustive.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;&lt;a href=&quot;https://query.wikidata.org/#SELECT%20DISTINCT%20%3Fitem%20%3FitemLabel%20%3FnamedAfter%20%3FnamedAfterLabel%20WHERE%20%7B%0A%20%20%3Fitem%20wdt%3AP31%20wd%3AQ3863.%0A%20%20%3Fitem%20wdt%3AP138%20%3FnamedAfter.%0A%20%20%3FnamedAfter%20wdt%3AP106%20%3Foccupation.%0A%20%20%3Foccupation%20wdt%3AP279%20wd%3AQ639669.%0A%20%20SERVICE%20wikibase%3Alabel%20%7B%20bd%3AserviceParam%20wikibase%3Alanguage%20%22en%22.%20%7D%0A%7D&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is a query that selects all asteroids named after a person that is a subclass of musician (currently 288).&lt;/p&gt;&#xA;" OwnerUserId="10568" LastEditorUserId="10568" LastEditDate="2019-03-06T21:12:49.443" LastActivityDate="2019-03-06T21:12:49.443" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29843" PostTypeId="2" ParentId="29837" CreationDate="2019-03-04T01:30:32.570" Score="4" Body="&lt;p&gt;NEAF is pretty big (and probably the biggest).  There's also Southern California Astronomy Expo (SCAE).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For observing, there is the Texas Star Party in west Texas.  There is the Winter Star Party in the Florida Keys (that site for that event was severely damaged by a hurricane a couple of years ago, had to relocate up-state, but are now back in the keys.  This past year the had to limit attendees because the camp wasn't fully restored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A little closer to home for you (in Pennsylvania)... is the Cherry Springs Star Party (there is also the Black Forest Star Party at the same park ... different months).  &lt;/p&gt;&#xA;" OwnerUserId="26051" LastActivityDate="2019-03-04T01:30:32.570" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29844" PostTypeId="2" ParentId="29693" CreationDate="2019-03-04T03:40:51.620" Score="7" Body="&lt;p&gt;I am Tomás the guy that designed it following astrocartography traditions. The thing that makes it confusing is that the position of the little suns in the Ecliptic are plotted against the stars at 12 noon GMT each day (the ephemeris) meaning that the stars behind it are NOT visible because of the sunlight. The stars that are visible at 12 midnight (Obviously depending on your latitude) are the ones completely opposite. This confusion comes from the ancient times when it was believed that the Sun rotated around the earth. Hope it helps a little.&lt;/p&gt;&#xA;" OwnerUserId="26124" LastActivityDate="2019-03-04T03:40:51.620" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29845" PostTypeId="2" ParentId="29837" CreationDate="2019-03-04T06:23:07.190" Score="3" Body="&lt;p&gt;This one is quite famous; people travel from all over the world to attend.&lt;/p&gt;&#xA;&lt;p&gt;From Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellafane&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellafane&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;(The) &lt;strong&gt;Stellafane Convention&lt;/strong&gt; (is) a gathering of amateur telescope makers and amateur astronomers (star party) held every year at that location. The property is a &lt;strong&gt;National Historic Landmark&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/h0tEm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h0tEm.jpg&quot; alt=&quot;stellafane.org&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fO4s8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fO4s8.jpg&quot; alt=&quot;stellafane.org&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://stellafane.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://stellafane.org/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://stellafane.org/convention/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://stellafane.org/convention/index.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Current &amp;amp; Future Conventions&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;strong&gt;2019&lt;/strong&gt; Stellafane Convention will be held Thursday thru Sunday August 1-4. Registration is expected to open on or about May 1st.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;strong&gt;2020&lt;/strong&gt; Stellafane Convention will be held Thursday thru Sunday August 13-16.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;strong&gt;2021&lt;/strong&gt; Stellafane Convention will be held Thursday thru Sunday August 5-8.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Convention Reminders&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sign up for our &lt;a href=&quot;https://stellafane.org/about/announce.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellafane Announcements mail-list&lt;/a&gt; to receive an e-mail reminder when convention registration opens in the spring.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;About the Stellafane Convention&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Stellafane Convention is a gathering of amateur telescope makers. The Convention was started in 1926 to give amateur telescope makers an opportunity to gather to show off their creations and teach each other telescope making and mirror-grinding techniques. All telescopes, commercial and homemade, are welcome. If you have made your own telescope, we strongly encourage you to display it in the telescope fields near the Stellafane Clubhouse. If you wish, you can enter your homemade telescope in the mechanical and/or optical competition. There are also mirror-grinding and telescope-making demonstrations, technical lectures on telescope making and the presentation of awards for telescope design and craftsmanship. Vendor displays and the retail sale of commercial products are not permitted.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-03-04T06:28:30.040" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29846" PostTypeId="1" CreationDate="2019-03-04T08:27:07.700" Score="4" ViewCount="250" Body="&lt;p&gt;Is it also roughly the limit where gravity will be able to overcome space expansion? In other words, we can only hope to get to there, or at least see them, in the very far future?&lt;/p&gt;&#xA;" OwnerUserId="26128" LastActivityDate="2019-03-04T18:35:19.383" Title="What is the farthest blueshifted object to our galaxy?" Tags="&lt;galaxy&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="29847" PostTypeId="2" ParentId="29703" CreationDate="2019-03-04T08:49:45.673" Score="3" Body="&lt;p&gt;This was discussed in a recent paper by an Oxford cosmologist: &lt;a href=&quot;https://arxiv.org/abs/1712.07962&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/1712.07962&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It claims that there have already been some detections of negative mass using gravitational lensing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In  fact,  “puzzling  results”  in  clusters  such  as negative masses have been discussed in the literature (Andreon, Punzi, &amp;amp; Grado 2005). For example, Chandra observations of the merging cluster Abell 2034 found hints of a negative mass and  therefore  did  not  plot  those  data  (Kempner,  Sarazin,  &amp;amp; Markevitch  2003),  regions  of  the  mass  profile  in  the  galaxy NGC 4636 yielded “unphysical” negative masses (Johnson et al. 2009), in the NGC 3411 galaxy group the total mass was found  to  decline  with  increasing  radius  –  requiring  material with negative mass (O’Sullivan et al. 2007), measurements of galaxy clusters using the Sloan Digital Sky Survey yielded data that indicate a negative mass in poor clusters with fewer than five galaxies (Hansen et al. 2005), a CMB cluster lensing study found a cluster with “a fairly significant preference for negative mass” (Baxter et al. 2017), and a number of strong and weak gravitational lensing studies have discussed or found indications of negative masses in reconstructed mass distributions (Evans &amp;amp; Witt 2003; Clowe, Gonzalez, &amp;amp; Markevitch 2004; Clowe et al. 2006; Liesenborgs, De Rijcke, &amp;amp; Dejonghe 2006; Diego et al. 2007).&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="26129" LastActivityDate="2019-03-04T08:49:45.673" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29848" PostTypeId="2" ParentId="29836" CreationDate="2019-03-04T08:56:22.520" Score="1" Body="&lt;p&gt;The gas clouds are much larger than you think: From Fig. 1 of the paper (&lt;a href=&quot;https://arxiv.org/abs/1812.10733&quot; rel=&quot;nofollow noreferrer&quot;&gt;Takakawa 2019&lt;/a&gt;), the largest cloud (which they call the &quot;Balloon&quot;) is roughly 10 arcsec across. Now they can't actually measure its exact distance, but its projected distance from the Galactic center is only 7 pc, and the authors argue that it's at least 5 kpc away due to absorption from molecular gas which is known at this distance (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2006PASJ...58..335S&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sofue 2006&lt;/a&gt;), and since such hot and high density (&lt;span class=&quot;math-container&quot;&gt;$\sim 10^7\,\mathrm{cm}^{-3}$&lt;/span&gt;) molecular gas is abundant in the Galactic center, it is probably located there, i.e. at a distance of &lt;span class=&quot;math-container&quot;&gt;$R=8\,\mathrm{kpc}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cloud's absolute size is thus&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;D = 10'' \times 8\,\mathrm{kpc} = 80\,000\,\mathrm{AU} \simeq 0.4\,\mathrm{pc}.&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass is measured from dynamics, i.e. they measure the line-of-sight velocities of several clouds. These velocities depend both on the mass of the dynamical center (i.e. whatever object the clouds orbit), and on the total velocity of this object wrt. us. Since they only know the line-of-sight velocities and not the full, 3D motions, they cannot know with certainty the mass, but performing a chi-square fit of 3D models that fit the 2D observations, they determine the best-fit value of the mass to be &lt;span class=&quot;math-container&quot;&gt;$(3.2\pm0.6)\times10^4 M_\odot$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their Fig. 4 shows the model:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nUpeP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nUpeP.png&quot; alt=&quot;model&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;3D model of the motion of the clouds. The &lt;em&gt;red&lt;/em&gt; cloud is the Balloon. Other observed clouds are shown in &lt;em&gt;green&lt;/em&gt;, &lt;em&gt;blue&lt;/em&gt;, and &lt;em&gt;orange&lt;/em&gt; colors. The &lt;em&gt;black&lt;/em&gt; dot indicates the position of the black hole. The actual observation is the 2D projection, seen in the bottom plane.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model of the velocities also gives the actual trajectories of the clouds around the central mass. Because the radius of the Balloon's orbit is &lt;span class=&quot;math-container&quot;&gt;$\sim0.07\,\mathrm{pc}$&lt;/span&gt;, it means that this much mass must be concentrated within a region of maximum that size, and probably significantly smaller. This implies a density of &lt;span class=&quot;math-container&quot;&gt;$&amp;gt;2\times10^8 M_\odot\,\mathrm{pc}^{-3}$&lt;/span&gt;, at least an order of magnitude larger than even the most densely packed globular clusters, and since no luminous counterpart is observed at that location, the authors consider an intermediate-mass black hole as the most promising candidate for the gravitational source.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2019-03-04T10:17:34.740" LastActivityDate="2019-03-04T10:17:34.740" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="29849" PostTypeId="1" CreationDate="2019-03-04T09:12:07.133" Score="-4" ViewCount="143" Body="&lt;p&gt;My hypothesis is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;quantum tunneling in black hole&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matter which is consumed by a black holes is shifting to another part of the universe by quantum tunneling. Can it be possible?&lt;/p&gt;&#xA;" OwnerUserId="25466" LastActivityDate="2020-04-12T00:38:19.143" Title="Quantum tunneling in black holes" Tags="&lt;supermassive-black-hole&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" ClosedDate="2020-04-12T23:04:16.487" ContentLicense="CC BY-SA 4.0" />
  <row Id="29850" PostTypeId="2" ParentId="29832" CreationDate="2019-03-04T09:50:25.183" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Space is low pressure and the earth’s atmosphere is high pressure. In order to have any pressure the gas requires (demands) that it press upon something. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, and that 'something' is Earth's gravity. Gas molecules are pulled down toward the surface just like all other matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there's an equilibrium between gravity and gas pressure. At the upper end of the atmosphere, some gas escapes (molecules reach escape velocity), this is predominantly the light gases. &lt;/p&gt;&#xA;" OwnerUserId="361" LastActivityDate="2019-03-04T09:50:25.183" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="29851" PostTypeId="1" CreationDate="2019-03-04T11:20:13.870" Score="-4" ViewCount="92" Body="&lt;p&gt;As we all know that we cannot see black holes due to its strong gravitational force. But I think gravity is not the cause for invisibility of black holes. The temperature of a black hole is  1.4×10^-14 .At these low  temperature any body can act as super conductor due to Meissner effect.So that's why we are not seeing black holes due to this effect. Can it be possible?&lt;/p&gt;&#xA;" OwnerUserId="25466" LastActivityDate="2019-03-04T12:35:09.053" Title="Super conductivity of black holes" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29852" PostTypeId="2" ParentId="29851" CreationDate="2019-03-04T12:35:09.053" Score="2" Body="&lt;p&gt;Not really. Superconductors are not, in general, invisible (google images search for superconductor returns many results). Also, a black hole is more or less defined as a region of space from which light cannot escape due to gravity. &lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-03-04T12:35:09.053" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29853" PostTypeId="1" AcceptedAnswerId="29855" CreationDate="2019-03-04T13:32:47.040" Score="8" ViewCount="756" Body="&lt;p&gt;Phys.org's &lt;a href=&quot;https://phys.org/news/2019-03-black-hole.html&quot; rel=&quot;noreferrer&quot;&gt;Hiding black hole found&lt;/a&gt; says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A research team led by Shunya Takekawa at the National Astronomical Observatory of Japan noticed HCN-0.009-0.044, a gas cloud moving strangely near the center of the galaxy 25,000 light-years away from Earth in the constellation Sagittarius. They used ALMA (Atacama Large Millimeter/submillimeter Array) to perform high-resolution observations of the cloud and found that it is swirling around a massive invisible object.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Takekawa says, &amp;quot;Detailed kinematic analyses revealed that an enormous mass, 30,000 times that of the sun, was concentrated in a region much smaller than our solar system.&lt;/strong&gt; This and the lack of any observed object at that location strongly suggests an intermediate-mass black hole. By analyzing other anomalous clouds, we hope to expose other quiet black holes.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The paper is &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/aafb07/meta&quot; rel=&quot;noreferrer&quot;&gt;Indication of Another Intermediate-mass Black Hole in the Galactic Center&lt;/a&gt; and is &lt;em&gt;open access.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;The quote in the second paragraph is all over the internet. I have found that &lt;a href=&quot;https://www.sciencedaily.com/releases/2019/02/190228093614.htm&quot; rel=&quot;noreferrer&quot;&gt;sciencedaily&lt;/a&gt; and &lt;a href=&quot;https://www.iflscience.com/space/intermediatemass-black-hole-discovered-lurking-in-the-core-of-the-milky-way/&quot; rel=&quot;noreferrer&quot;&gt;iflscience&lt;/a&gt; both link back to the same quote on an &lt;a href=&quot;https://www.nao.ac.jp/en/news/science/2019/20190228-nro.html&quot; rel=&quot;noreferrer&quot;&gt;official NAO page here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Takekawa explains, “Detailed kinematic analyses revealed that an enormous mass, 30,000 times that of the Sun, was concentrated in a region much smaller than our Solar System.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt; How do they know that the central mass is &amp;quot;concentrated in a region much smaller than our Solar System&amp;quot;?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/29848/7982&quot;&gt;This answer&lt;/a&gt; about the paper and subsequent discussions below it don't seem to shed any light on this particular question. The NAO page does not indicate the origin of the quote, I'm wondering if it is a translation of a corresponding Japanese language page on the same site that explains further.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-03-05T11:04:57.017" Title="Why does the author believe that the central mass that gas cloud HCN-0.009-0.044 orbits is smaller than our solar system?" Tags="&lt;black-hole&gt;&lt;observational-astronomy&gt;&lt;radio-astronomy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29854" PostTypeId="1" CreationDate="2019-03-04T14:26:23.797" Score="2" ViewCount="45" Body="&lt;p&gt;My assignment is to identify how many planets are in this curve and what the orbital period is. I'm just confused on the way it looks, as most curves I've seen up to this date are more smooth and connected in appearance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bGIi8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bGIi8.png&quot; alt=&quot;Transit Method&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26133" LastEditorUserId="25994" LastEditDate="2019-03-06T15:22:04.743" LastActivityDate="2019-03-06T15:22:04.743" Title="Transit Method Curve" Tags="&lt;planetary-transits&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29855" PostTypeId="2" ParentId="29853" CreationDate="2019-03-04T15:51:28.297" Score="11" Body="&lt;p&gt;The authors were able to successfully model the motion of the gas streams as Keplerian orbits around an object of &lt;span class=&quot;math-container&quot;&gt;$\sim30000M_{\odot}$&lt;/span&gt;. In doing so, they derived some key quantities, such as the pericentric distances of these two gas components (the &quot;Balloon&quot; and the &quot;Stream&quot;). One has its closest approach at &lt;span class=&quot;math-container&quot;&gt;$\sim0.21$&lt;/span&gt; pc; the other has its closest approach at &lt;span class=&quot;math-container&quot;&gt;$\sim0.07$&lt;/span&gt; pc (&lt;span class=&quot;math-container&quot;&gt;$14000\text{ AU}$&lt;/span&gt;). This places the pericenter well inside the Oort Cloud, and therefore inside the Solar System.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, given that it's highly unlikely that the Balloon skims the surface of the central object, we can say that the object's radius is certainly less than that of the Oort Cloud, and by extension, the Solar System. This is a key part of the argument that the object is, in fact, an intermediate mass black hole, as they note on Page 5. It's basically the same logic used to constrain the size of Sagittarius A* &lt;a href=&quot;https://astronomy.stackexchange.com/q/28240/2153&quot;&gt;based on those infrared flares&lt;/a&gt; observed last year.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2019-03-04T16:02:09.397" LastActivityDate="2019-03-04T16:02:09.397" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29857" PostTypeId="1" CreationDate="2019-03-04T17:20:03.530" Score="4" ViewCount="851" Body="&lt;p&gt;Is a black hole a three dimensional or four-dimensional hole?&lt;/p&gt;&#xA;" OwnerUserId="26136" LastEditorUserId="2153" LastEditDate="2019-03-04T18:21:22.513" LastActivityDate="2019-03-05T22:23:10.853" Title="Is a black hole a three-dimensional or four-dimensional object?" Tags="&lt;black-hole&gt;" AnswerCount="2" CommentCount="6" ClosedDate="2019-03-05T21:40:27.450" ContentLicense="CC BY-SA 4.0" />
  <row Id="29859" PostTypeId="1" AcceptedAnswerId="29865" CreationDate="2019-03-04T18:28:39.803" Score="2" ViewCount="266" Body="&lt;p&gt;I saw the picture and seem incorrectly drawn. Its shows the the edge of the solar system leading facing the Heliosheath. It would not be like NASA to post something as incorrect as this. The solar system should at 60 degrees to the direction of travel?&#xA;I'm not sure maybe this picture is correct?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Qjyak.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qjyak.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://solarsystem.nasa.gov/news/784/nasas-voyager-2-probe-enters-interstellar-space/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://solarsystem.nasa.gov/news/784/nasas-voyager-2-probe-enters-interstellar-space/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2019-03-04T21:17:19.147" LastActivityDate="2019-03-10T18:30:20.520" Title="What is the solar system orientation to its direction of travel?" Tags="&lt;solar-system&gt;&lt;astrophysics&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2019-03-13T20:25:48.700" ContentLicense="CC BY-SA 4.0" />
  <row Id="29860" PostTypeId="2" ParentId="29846" CreationDate="2019-03-04T18:35:19.383" Score="3" Body="&lt;p&gt;My impression is that, aside from the very rare galaxy that is moving towards us, only stuff inside the local galactic gravitational field can be blueshifted.  Written perhaps more clearly,  from &lt;a href=&quot;https://www.spaceanswers.com/deep-space/where-did-the-first-particles-of-light-come-from/&quot; rel=&quot;nofollow noreferrer&quot;&gt;spaceanswers&lt;/a&gt; , &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Andromeda is not the only galaxy to be moving towards us. With the&#xA;  help of galaxy surveys, astronomers have found that around 100&#xA;  galaxies are moving towards us. Compared to the numbers of galaxies&#xA;  that we know of (hundreds of billions), blue-shifted galaxies are&#xA;  seemingly quite rare. Those that are moving towards us are either part&#xA;  of our Local Group, which means that we are gravitationally connected&#xA;  to each other, or they are found in the Virgo Cluster which everything&#xA;  in our Local Group is moving towards. The galaxies M90, M86 and M98&#xA;  are all in the Virgo Cluster and all show blue shifts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="907" LastActivityDate="2019-03-04T18:35:19.383" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29861" PostTypeId="1" AcceptedAnswerId="29877" CreationDate="2019-03-04T19:28:57.260" Score="3" ViewCount="135" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_refraction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia states&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Atmospheric refraction of the light from a star is zero in the zenith&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and other sources seem to agree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see why this should be so. Consider a situation in which air gets warmer from left to right. Then light should curve towards the warm air, and a star at the zenith would appear refracted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gZBGL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gZBGL.png&quot; alt=&quot;Cold air - Hot air&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why do so many sources claim there is no refraction at the zenith? What do they mean by this?&lt;/p&gt;&#xA;" OwnerUserId="1945" LastEditorUserId="25994" LastEditDate="2019-03-06T15:32:51.380" LastActivityDate="2019-03-06T15:32:51.380" Title="What is meant by &quot;there is no refraction at the zenith&quot;?" Tags="&lt;atmospheric-effects&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29862" PostTypeId="2" ParentId="29857" CreationDate="2019-03-04T20:09:34.213" Score="3" Body="&lt;p&gt;It would help to clarify your question on what exactly you mean by a 4th dimension.  Space-time for example is 4 dimensional everywhere, but it's 3 spatial dimensions and 1 time dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum theory allows for extra dimensions, but they're very small, smaller than an atom and impossible to measure by normal means.  CERN has looked for those folded dimensions but has found no evidence of them, and string theory defines 10 or 11 dimensions, most of them we can't see or measure, if they exist at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mean 4 spatial dimensions, then the answer is no, with a capital N.   There's no evidence that an extra spatial dimension would form in a black hole.  There's nothing in relativity that suggests that might happen.  An extra time dimension . . . probably not either, but it's fun to think about what 2 time dimensions would look like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens inside a black hole is that one of the dimensions becomes like time.  It only moves in 1 direction, but that doesn't make it a 4th dimension, it just changes the nature of 1 of the dimensions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you actually had a 4th dimension (and the book flatland discusses this in pretty good detail), but if you actually had that, very strange things can happen, for example, you can be flipped over, like flipping a card over, which can represent 2 dimensions, from face down to face up.  Your left side would become your right side.  Your heart would be on the other side of your chest and all the left handed molecules in your body would become right handed molecules, which would make your ability to absorb and use nutrients that you digest problematic.   It might even make things taste and smell different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't tie your shoes in the 4th dimension, but even if you could, your shoes could slide off your feet, no matter how well you tied them.   Food could fall out of your stomach, not through one of the two normal exits, but it could just slip out, like a ball rolling off a table.   Anything not attached could slip apart.   Sex would be much more difficult.  Bolts and nuts could just move sideways and detach, unless there was some electrostatic charge keeping them in place, in fact, the pressure that forms natural between a screw and a nut might cause all nuts and bolts to fly apart at considerable velocity once you entered 4th dimensional space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two people could also stand in exactly the same place, just in a different 4th dimensional space.   That brings a whole new aspect to the how many people can you fit in this Volkswagen game, as you could fit theoretically any number of people in the same Volkswagen.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 4th spatial dimension is probably impossible, because it breaks too many rules and laws as we know them, and it's never been observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes, by comparison, can do strange things like slow down time, and they aren't fully understood, but a 4th spatial dimension probably isn't possible, unless you mean at quantum distances, in which case . . . maybe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feel free to read this for a &lt;a href=&quot;https://astronomy.stackexchange.com/questions/9919/is-a-black-hole-a-5-dimensional-vortex?rq=1&quot;&gt;different approach to a somewhat similar question&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="9772" LastEditDate="2019-03-05T08:09:58.843" LastActivityDate="2019-03-05T08:09:58.843" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29863" PostTypeId="1" AcceptedAnswerId="29864" CreationDate="2019-03-04T22:16:26.023" Score="11" ViewCount="1721" Body="&lt;p&gt;I'm reading up on radio astronomy, and I came across &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1964Obs....84..191L&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;noreferrer&quot;&gt;this paper&lt;/a&gt; from 1964. At the bottom of page 193, the author uses a unit that I've not seen before in discussing radio power emission from stars:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now the outbursts on the Sun give an intensity on Earth of &lt;span class=&quot;math-container&quot;&gt;$10^{19}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$10^{20}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$wm^{-2}(c/s)^{-1}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm guessing it's &quot;Watts per square meter per [something] per second&quot;, but I'm not sure what the [something] is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar unit appears in &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1955ApJ...122..361H&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;noreferrer&quot;&gt;this paper&lt;/a&gt; on the first line on page 364:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The comparison band in the radiometer, being separated approximately 3.25 Mc from the signal band, never encounters the hydrogen range of frequencies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, this looks to me like mega[something]. Can anyone shed some light on this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On page 362 of the second paper, the unit appears as &lt;span class=&quot;math-container&quot;&gt;$(Watts/M^2&#xA;)/(C/S)$&lt;/span&gt; as a unit of flux. There, the &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; looks like coulombs, but that makes the &lt;span class=&quot;math-container&quot;&gt;$3.25 Mc$&lt;/span&gt; in the second quote seem weird.&lt;/p&gt;&#xA;" OwnerUserId="17535" LastEditorUserId="23881" LastEditDate="2019-03-05T08:57:55.763" LastActivityDate="2019-03-05T12:06:19.027" Title="A &quot;strange&quot; unit radio astronomy" Tags="&lt;radio-astronomy&gt;&lt;units&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29864" PostTypeId="2" ParentId="29863" CreationDate="2019-03-04T22:37:13.490" Score="23" Body="&lt;p&gt;I would expect the authors to be talking about the signal in terms of &lt;a href=&quot;https://en.wikipedia.org/wiki/Jansky&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;janskys&lt;/em&gt;&lt;/a&gt;, the now-commonly-used units of flux density. The typical definition is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$1\text{ Jansky}=10^{-26}\text{ Watts meters}^{-2}\text{ Hertz}^{-1}$$&lt;/span&gt;&#xA;One hertz is one cycle per second, which makes me suspect that the &quot;c&quot; stands for cycle. It might seem curious that the authors choose to use cycles/second instead of hertz, but as the papers were published in 1964 and 1955, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Hertz#History&quot; rel=&quot;noreferrer&quot;&gt;the hertz was only adopted on a large scale in 1964&lt;/a&gt;, the older term &lt;a href=&quot;https://en.wikipedia.org/wiki/Cycle_per_second&quot; rel=&quot;noreferrer&quot;&gt;&quot;cycles per second&quot;&lt;/a&gt; is more fitting, given the time period.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="18643" LastEditDate="2019-03-05T12:06:19.027" LastActivityDate="2019-03-05T12:06:19.027" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29865" PostTypeId="2" ParentId="29859" CreationDate="2019-03-04T23:12:33.270" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/316mj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/316mj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;  This is Venus, high in the morning sky, amid the faint pillar of light called the Zodiacal Light. The glow is sunlight reflected off cometary dust in the inner solar system.&lt;br&gt;&#xA;  Above is the centre of the Galaxy area of Sagittarius.  &lt;a href=&quot;https://amazingsky.net/2014/04/06/venus-in-the-zodiacal-light/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Alan Dyer&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since the  &lt;a href=&quot;https://www.skyandtelescope.com/astronomy-news/zodiacal-lights-mystery-solved/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zodiacal Light&lt;/a&gt; lies in the ecliptic plane, the orientation of the Solar system in the Galaxy is something like &lt;a href=&quot;http://www.xenology.info/Xeno/4.3.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bj6aX.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bj6aX.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun moves &lt;a href=&quot;http://espg.sr.unh.edu/ism/LISM.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;through the local interstellar gas cloud approximately into the direction of Scorpio with a speed of about 25 km/sec&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Scorpio is fairly close to the galactic center (Sagittarius), the Solar System is moving approximately edge-on through the gas cloud as in the NASA diagram.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Vcqba.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vcqba.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;(Image: NASA, ESA, Z. Levay (STScI) and A. Fujii)&lt;/p&gt;&#xA;" OwnerUserId="17743" LastEditorUserId="17743" LastEditDate="2019-03-10T18:30:20.520" LastActivityDate="2019-03-10T18:30:20.520" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29866" PostTypeId="2" ParentId="13178" CreationDate="2019-03-04T23:41:37.507" Score="2" Body="&lt;p&gt;Jupiter's &lt;a href=&quot;https://en.wikipedia.org/wiki/Jupiter&quot; rel=&quot;nofollow noreferrer&quot;&gt;orbital period is 11.862 years.&lt;/a&gt;The amount of variation will be miniscule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solar cycle, while the average cycle time is 11 years, it can vary from 9 to 14 years. From &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_cycle&quot; rel=&quot;nofollow noreferrer&quot;&gt;the wikipedia article&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since observations began cycles as short as 9 years and as long as 14 years have been observed, and in the double cycle of 1784–1799 one of the two component cycles had to be less than 8 years in length. Significant amplitude variations also occur.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17655" LastActivityDate="2019-03-04T23:41:37.507" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29867" PostTypeId="1" CreationDate="2019-03-05T00:11:07.297" Score="1" ViewCount="27" Body="&lt;p&gt;I'm creating a project for my astronomy class (introductory course) that requires knowledge of the galactocentric distances for many stars. I'm struggling to find a good source on this, and I'd rather not parse through dozens of papers to compile it myself if possible. Is there a good table of stars (preferably around 100 or so) that includes their galactocentric distances? (Obviously you could calculate it using galactic coordinates but that's too complicated for my students to do for several stars)&lt;/p&gt;&#xA;" OwnerUserId="26139" LastActivityDate="2019-03-05T00:11:07.297" Title="Data on galactocentric distances of various stars?" Tags="&lt;distances&gt;&lt;star-systems&gt;&lt;galaxy-center&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29868" PostTypeId="1" AcceptedAnswerId="29869" CreationDate="2019-03-05T02:50:05.327" Score="1" ViewCount="129" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If two stars have the same temperature but one is more luminous than&#xA;  the other, the more-luminous star must have a larger surface area, and&#xA;  therefore a larger radius than the dimmer star. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The light density of the two stars is obviously different. Must the brighter star have more lights?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Iuf3E.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Iuf3E.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25918" LastActivityDate="2019-03-05T03:55:26.650" Title="Why the more-luminous star has larger surface area when two stars have the same temperature?" Tags="&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29869" PostTypeId="2" ParentId="29868" CreationDate="2019-03-05T03:55:26.650" Score="7" Body="&lt;p&gt;I'd say your problem is that you are imagining the light rays emanating out from the center of the star.  Try it again with light emanating from each square area on the surface, and you will see that being at the same temperature means each square on the surface sends out the same number of rays, but there are just more such squares.  Then you will see a lot more light rays coming from the larger star, in proportion to its surface area.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2019-03-05T03:55:26.650" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29870" PostTypeId="2" ParentId="29849" CreationDate="2019-03-05T05:53:16.453" Score="2" Body="&lt;p&gt;A good hypothesis should be testable and falsifiable.  Can you think of consequences that could be verified if your hypothesis is correct? If not, then any other non-falsifiable hypothesis is equally valid which means we get nowhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To break it down... your hypotheses is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;(a) the black hole loses mass, and  &lt;/li&gt;&#xA;&lt;li&gt;(b) that mass is being transported to another part of the&#xA;universe.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I don't know how to test (b) but I can think of ways to test (a). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The consumed matter will increase the mass of the black hole.  If a black hole has  a mass of 15 Solar masses and we drop another star the size of our own Sun into it (1 Solar mass) it will have 16 Solar masses.  If matter is being transported out of the black hole, then the mass should decrease instead of increasing (or at least not increase as much as you expect if only a fraction of matter is transported).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to measure the mass of an object by observing the rate at which other (less massive) objects orbit the black hole.  Astronomers should be able to notice the gravitational pull of these objects getting weaker (observed by a change in the orbits of nearby stars.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes were all the rage in cosmology and heavily studied.  Stephen Hawking put forth a theory that black holes should be able to &quot;evaporate&quot; (and this helps get around the &quot;black hole information paradox&quot;) ... although based on his math, this should happen at an incredibly slow rate (not something you'd notice even if you could watch it for a few billion years.)  If Hawking is correct, then black holes can lose mass ... but very slowly.  (Also that mass is not transported across the universe... but really just beyond the event horizon.)  Read up on &quot;Hawking Radiation&quot; to learn more about his ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe is fairly young (it's less than 14 Billion years old) which isn't long enough for even modest sized black holes to have had time to evaporate based on Hawking radiation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should think that black hole studies would have noticed the loss of mass if matter were being transported out of the black hole ... and that would be &lt;em&gt;really&lt;/em&gt; big news if it were noticed (cosmologists would be &lt;em&gt;extremely&lt;/em&gt; interested).  As such, I don't think this has happened.&lt;/p&gt;&#xA;" OwnerUserId="26051" LastActivityDate="2019-03-05T05:53:16.453" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29871" PostTypeId="1" CreationDate="2019-03-05T09:20:19.317" Score="3" ViewCount="64" Body="&lt;p&gt;In a paper I'm writing, I've referenced an image I got from the NASA site at &lt;a href=&quot;https://fermi.gsfc.nasa.gov/science/eteu/agn/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://fermi.gsfc.nasa.gov/science/eteu/agn/&lt;/a&gt; regarding the Unified Model of AGN. I captioned it&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The AGN unified model, showing the BLR and NLR, as well as different types&#xA;based on the observational angle. Credit: Fermi and NASA: &#xA;https://fermi.gsfc.nasa.gov/science/eteu/agn/&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;but my supervisor said that I should use proper referencing (I see that the image seems to be derived from &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1995PASP..107..803U&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;). The image from the NASA site is much clearer and easier to understand, and I'd much rather use it for clarity's sake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the proper way to reference an image like this, which is derived from a paper?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe I should just use the original image?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or does anyone know of a better quality one from another paper, similar to the one from the NASA site?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I should mention that this is a follow-up question to one I posted &lt;a href=&quot;https://academia.stackexchange.com/questions/125372/properly-citing-an-image-from-a-website&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17535" LastEditorUserId="37191" LastEditDate="2021-04-16T08:35:12.500" LastActivityDate="2021-04-16T08:35:12.500" Title="An image for the unified model of AGN" Tags="&lt;active-galaxy&gt;&lt;resource-request&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29872" PostTypeId="1" CreationDate="2019-03-05T10:53:16.283" Score="3" ViewCount="43" Body="&lt;p&gt;I guess first off, do all/most planets with a tilted axis experience precession? And if so, is it usually in the same direction (clockwise or anticlockwise)?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm really curious about is if a planet is tidally locked to its star, could it still experience precession which would in effect slowly shift the day-night line on the planet?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware that most planets tidally locked to their stars are under more extreme gravitational forces whether it be a larger star than the sun or an orbital distance much smaller than 1AU. How would these gravitational forces affect the precession?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And lastly, is there any theoretical &quot;limit&quot; to the speed of precession? I know for the Earth it takes the axis ~26,000 years to complete a full circuit. Is it typical that this process takes so much longer than for the Earth to complete a rotation or revolution or could a planet theoretically have a much shorter precession cycle and what kind of effects would such movement have on the planet itself?  And similarly, if the planet's rotation speed is much slower than the Earth due to tidal locking (let's just say ~2-3 months) would the precession eventually stop all together or not until an object stops rotating completely? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone feels like being particularly helpful and doing some math that goes way over my head, perhaps you could give me a rough estimate of the speed of precession (i.e., length of time required for a complete circuit of the axis) for the following situations:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;An Earth-sized, moonless planet tidally locked in the goldilocks zone of an average red dwarf;&lt;/li&gt;&#xA;&lt;li&gt;A Mars-sized, moonless planet with the same conditions. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I've also read that other factors influence the precession such as the gravitational pull of large Jupiter-like planets. Is this minimal or considerable?&lt;/p&gt;&#xA;" OwnerUserId="26147" LastEditorUserId="1987" LastEditDate="2019-05-21T20:24:59.547" LastActivityDate="2019-05-21T20:24:59.547" Title="Does a planet's rotation speed and distance from its star influence the precession of its axis? And is the Earth's precession &quot;normal&quot;?" Tags="&lt;orbital-mechanics&gt;&lt;precession&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29873" PostTypeId="2" ParentId="29853" CreationDate="2019-03-05T11:04:57.017" Score="2" Body="&lt;p&gt;I just asked Shunya Takekawa, the first author of the paper, if they actually constrained the size to be &quot;smaller than the Solar System&quot;, or if it's just the media's take on &quot;&lt;span class=&quot;math-container&quot;&gt;$&amp;lt;0.07\,\mathrm{pc}$&lt;/span&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He answered:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Indeed, the latter is true. It might be an exaggerated expression.&lt;br&gt;&#xA;  Sorry for the confusion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2019-03-05T11:04:57.017" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29874" PostTypeId="2" ParentId="29871" CreationDate="2019-03-05T16:17:27.663" Score="3" Body="&lt;p&gt;I agree with your supervisor: If your paper is going to a peer-reviewed journal, you should cite peer-reviewed material. I also agree with you that the NASA version is clearer than the original figure, but if you're only going to &lt;em&gt;refer&lt;/em&gt; to the image — i.e. not show a reproduction — I'm sure the reader can figure it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you could create your own (color) version with the annotations that are relevant for your work, and then write something like &quot;…based on the schematic diagram of &lt;a href=&quot;http://adsabs.harvard.edu/abs/1995PASP..107..803U&quot; rel=&quot;nofollow noreferrer&quot;&gt;Urry &amp;amp; Padovani (1995)&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2019-03-05T16:17:27.663" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29875" PostTypeId="1" CreationDate="2019-03-05T18:35:31.503" Score="3" ViewCount="603" Body="&lt;p&gt;I wonder why saturn rings are so thin ? Rings A, B and C are approximately just 10 meters thick which is extremely thin compared to the size of Saturn. What are the explanations for that ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot in advance for your answers.&lt;/p&gt;&#xA;" OwnerUserId="23047" LastActivityDate="2019-03-05T18:47:31.747" Title="Why are saturn rings so thin?" Tags="&lt;planet&gt;&lt;saturn&gt;&lt;planetary-ring&gt;&lt;saturn-rings&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29876" PostTypeId="2" ParentId="29875" CreationDate="2019-03-05T18:47:31.747" Score="3" Body="&lt;p&gt;Not that &lt;a href=&quot;https://www.reddit.com/r/askscience/comments/2nwbst/how_thick_are_saturns_rings/&quot; rel=&quot;nofollow noreferrer&quot;&gt;reddit&lt;/a&gt; is the most reliable of sources, but here's a Q&amp;amp;A with a response that is at least consistent with orbital mechanics. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[Q} Oh. But why doesn't [the original particle set] just scatter&#xA;  around the planet, creating a big &quot;bubble&quot;?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;[A]Let's say there's a particle that moves around in an inclined&#xA;  orbit, so it has both a horizontal and vertical component to its&#xA;  velocity. If it collides with something, it'll either get more&#xA;  velocity, or it'll cancel out. If the vertical component is cancelled&#xA;  out, its orbit doesn't change, it just becomes less inclined. If the&#xA;  horizontal component is cancelled out, it'll fall to a different&#xA;  orbit. If you've got lots of particles going in different directions,&#xA;  their vertical components will cancel out, while those not going in&#xA;  the predominant horizontal direction will collide with others, lose&#xA;  their velocity, and fall inward. The net result will be a ring.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;[Q]That makes sense in two dimensions. But it doesn't explain why the&#xA;  rings are thin.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;[A]If the rings are thick, it means the particles still have a&#xA;  vertical component to the orbit, each being slightly inclined in a&#xA;  different way so as to make up thickness. They would have to move very&#xA;  precisely for them not to collide and cancel out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="907" LastActivityDate="2019-03-05T18:47:31.747" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29877" PostTypeId="2" ParentId="29861" CreationDate="2019-03-05T21:41:58.953" Score="4" Body="&lt;p&gt;To predict the position of a star one has to account for atmospheric refraction. In a simple model, the atmosphere is treated as being composed of layers of gas with variation in density and temperature due to height, but no variation with latitude or longitude and no turbulence. In such a model, one can predict the position of a star to a great deal of accuracy (so it is a useful model), and in this model there is no refraction at the zenith (as the light doesn't pass &lt;em&gt;at an angle&lt;/em&gt; through a medium of varying optical density).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In such a model there would be no twinkling, however modelling twinkling is not needed for positional astronomy. Large variation in optical density that could cause stars to appear significantly displaced at the zenith doesn't occur. The variation in optical density with height is much more significant for positional astronomy.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-03-05T21:41:58.953" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29878" PostTypeId="2" ParentId="29857" CreationDate="2019-03-05T21:55:29.993" Score="0" Body="&lt;p&gt;Given the current state of knowledge, the answer is yes: a black hole exists in three space dimensions plus one time dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spacetime around a black hole is often depicted using the rubber-sheet model, e.g. &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Flamm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Flamm's paraboloid&lt;/a&gt; for the &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_metric&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schwarzschild metric&lt;/a&gt;. This makes it look like spacetime is curving through a higher-dimensional space (i.e. the 3D space containing the sheet) but you have to bear in mind that this is just a visualisation. In order to make it comprehensible it needs to introduce an imaginary extra dimension outside of spacetime because humans are in general not equipped to visualise the geometry involved thanks to our evolution and experience in a region of the universe where space is near-Euclidean and time can be treated in everyday situations as being independent of space. There is at present no evidence that our 3+1-dimensional universe is embedded in a higher-dimensional space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is of course the possibility that whatever's actually going on (that general relativity and quantum field theory are merely very good approximations of) does involve extra dimensions. Extra dimensions have been invoked by several theoretical attempts to find a unified &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Theory_of_everything&quot; rel=&quot;nofollow noreferrer&quot;&gt;theory of everything&lt;/a&gt;&quot; but so far getting evidence for these theories being correct has proven extremely difficult. It doesn't help that the Large Hadron Collider served up the blandest Higgs boson on the menu, then proceeded to (so far) not find any evidence of things like supersymmetric particles that may have proven that these kind of theories were on the right track.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-03-05T22:23:10.853" LastActivityDate="2019-03-05T22:23:10.853" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29879" PostTypeId="1" AcceptedAnswerId="29889" CreationDate="2019-03-05T23:48:13.710" Score="5" ViewCount="129" Body="&lt;p&gt;I recently noticed that none of the planets in our solar system orbit the Sun in a retrograde direction. This is likely because they formed moving in the same direction as the planetary disk did in respect to the Sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there stars that move retrograde to the Milky Way galaxy's rotation? If so, where are they concentrated, and why?&lt;/p&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="7982" LastEditDate="2019-03-06T06:36:35.217" LastActivityDate="2019-03-07T03:26:28.450" Title="Do astronomers have any idea what percentage of our galaxy’s stars move in retrograde orbits?" Tags="&lt;star&gt;&lt;retrograde-orbit&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29880" PostTypeId="5" CreationDate="2019-03-05T23:48:57.240" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2019-03-05T23:48:57.240" LastActivityDate="2019-03-05T23:48:57.240" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29881" PostTypeId="4" CreationDate="2019-03-05T23:48:57.240" Score="0" Body="Questions regarding the movement of a celestial body that is in the opposite direction of another celestial body the original one is orbiting." OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-03-06T09:25:12.710" LastActivityDate="2019-03-06T09:25:12.710" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29882" PostTypeId="1" CreationDate="2019-03-05T23:51:54.213" Score="2" ViewCount="37" Body="&lt;p&gt;So the Moon is drifting away from Earth by 3.8 cm a year due to tidal interactions. With the Moon slowly receding from Earth, there will come a time when it no longer has sufficient angular size to cover the Sun completely. When will the last total solar eclipse occur?(when will the Moon's umbra no longer be able to project onto the Earth?)&lt;/p&gt;&#xA;" OwnerUserId="25988" LastActivityDate="2019-03-05T23:51:54.213" Title="Last possible total solar elipse in the Earth's history" Tags="&lt;solar-eclipse&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2019-03-06T09:25:47.530" ContentLicense="CC BY-SA 4.0" />
  <row Id="29883" PostTypeId="1" AcceptedAnswerId="29887" CreationDate="2019-03-06T00:09:01.157" Score="0" ViewCount="102" Body="&lt;p&gt;So a black hole is this large, sphere-blob shaped fluid thing that distorts space and time, making a weird hole in space(even after 5 years, I still don't exactly get what a black hole is exactly and where the mass goes when it eats xD). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's say, we make a huge antimatter(opposite charge, same mass as matter) star and send it into the black hole. What would happen? The antimatter would annihilate with the matter made black hole, or is it wrong to assume a black hole is made of matter?(I assume not much would happen if the antimatter can't react with the matter, because the black hole would probably get its original charge messed up.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about if we created a huge exotic matter(same charge, opposite mass) star and sent it into a black hole? Would the repulsive force successfully destroy a black hole? According to Kurzgesagt and several articles cited by them, exotic matter is able to keep a wormhole stable, so wouldn't it be able to annihilate the black hole or something? Or have I made a misunderstanding?&lt;/p&gt;&#xA;" OwnerUserId="25988" LastActivityDate="2019-03-06T00:40:04.760" Title="Scenarios: Abusing a black hole" Tags="&lt;matter&gt;&lt;antimatter&gt;&lt;exotic-matter&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29884" PostTypeId="5" CreationDate="2019-03-06T00:09:55.067" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2019-03-06T00:09:55.067" LastActivityDate="2019-03-06T00:09:55.067" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29885" PostTypeId="4" CreationDate="2019-03-06T00:09:55.067" Score="0" Body="Questions regarding non-baryonic matter, most notably, matter with negative mass or imaginary mass, such as tachyons." OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-03-08T08:11:30.103" LastActivityDate="2019-03-08T08:11:30.103" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29886" PostTypeId="1" AcceptedAnswerId="29888" CreationDate="2019-03-06T00:27:04.720" Score="1" ViewCount="211" Body="&lt;p&gt;The universe is expected to have a time when star formation slowly ceases. However, this seems bizarre as there is always an abundant amount of matter in the Universe. Massive amounts of hydrogen exists in several galaxies, but they are deemed dead. Any gravitational disturbance can send the gas into a protostar, and even if there weren't hydrogen left, enough helium or carbon can turn a star into a star that fuses heavier elements, until the universe is depleted of such and mostly elements heavier than iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why does star formation cease when these is still abundant hydrogen around? Is there some sort of interaction in these hydrogen supplies that prevents star formation? Is the gas kept at a temperature that is too hot for formation? Or have I made a misconception?&lt;/p&gt;&#xA;" OwnerUserId="25988" LastActivityDate="2019-03-08T22:39:40.687" Title="Why does star formation cease?" Tags="&lt;star&gt;&lt;star-formation&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29887" PostTypeId="2" ParentId="29883" CreationDate="2019-03-06T00:40:04.760" Score="7" Body="&lt;p&gt;Black holes are not &quot;made of matter&quot;. They are better described as structures of gravity/warped spacetime. However, they do grow when absorbing things with positive mass-energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Antimatter still has positive mass-energy, so the black hole cheerfully gobbles it up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, exotic matter with negative energy density might indeed decrease the size of the black hole, akin to some form of enforced evporation process. (&lt;a href=&quot;https://journals.aps.org/prd/abstract/10.1103/PhysRevD.81.124011&quot; rel=&quot;noreferrer&quot;&gt;paper1&lt;/a&gt;, &lt;a href=&quot;https://journals.aps.org/prd/abstract/10.1103/PhysRevD.79.121501&quot; rel=&quot;noreferrer&quot;&gt;paper2&lt;/a&gt;) This is highly speculative since there are various theoretical reasons why exotic matter does not exist, and many kinds of hypothetical exotic matter that might work differently.&lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2019-03-06T00:40:04.760" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29888" PostTypeId="2" ParentId="29886" CreationDate="2019-03-06T00:49:43.997" Score="5" Body="&lt;p&gt;Star formation in a galaxy can end for &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/0004-637X/825/2/113&quot; rel=&quot;nofollow noreferrer&quot;&gt;a number of reasons&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ram pressure affecting a galaxy falling into a cluster can strip away gas, which can end most star formation in the galaxy itself.&lt;/li&gt;&#xA;&lt;li&gt;The winds of massive stars (as well as supernovae shocks) can eject cold gas via radiation pressure.&lt;/li&gt;&#xA;&lt;li&gt;Mergers with other galaxies can also eject gas, although these interactions often lead to an increase in star formation, briefly.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As far as I'm aware, how strongly each of these phenomena contribute is still an open question, but we do know that they either eject gas or heat it up (and therefore make it unsuitable for star formation, which requires cold gas), thereby slowing or even effectively ending star formation in a given galaxy. A gas cloud is supported against gravity by thermal pressure (and occasionally magnetic fields). Clouds that are hot have higher thermal pressures and therefore can resist gravity; cold clouds cannot. Cold, dense clouds are therefore the best places for star formation (more gravity and less pressure).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As for the amount that can be ejected, the exact amount depends on a) the initial gas quantity, b) the mechanism, and c) likely a large number of other free parameters, so I don't think I can answer that for the general case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding if stars can form with little hydrogen: If hydrogen was ejected, so would other elements, so it's not as if all the hydrogen would be blown away and helium, lithium and others would remain. Hydrogen-poor stars exist, but they're older, evolved stars, not young ones.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="25988" LastEditDate="2019-03-07T23:34:06.770" LastActivityDate="2019-03-07T23:34:06.770" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29889" PostTypeId="2" ParentId="29879" CreationDate="2019-03-06T07:28:02.713" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;About two-thirds of the Milky Way’s estimated 100 billion stars reside in the disk, and these all move in direct orbits. Most of the remaining third are in the football-shaped central bulge, and while their orbits are more varied, on average they too are direct. &lt;strong&gt;A few percent of our galaxy’s stars&lt;/strong&gt; occupy a vast, spherical halo. These travel on randomly distributed orbits — accordingly, &lt;strong&gt;half are direct and half retrograde&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://www.skyandtelescope.com/astronomy-resources/astronomy-questions-answers/what-percentage-of-our-galaxys-stars-move-in-retrograde-orbits/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Do astronomers have any idea what percentage of our galaxy’s stars move in retrograde orbits?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And while &lt;a href=&quot;https://en.wikipedia.org/wiki/Galaxy_formation_and_evolution#Formation_of_disk_galaxies&quot; rel=&quot;nofollow noreferrer&quot;&gt;formation of disk galaxies&lt;/a&gt; is a complex process, which is not yet fully understood, the reason that almost all stars move in the same (direct) orbits is &lt;a href=&quot;https://www.quora.com/How-is-the-angular-momentum-of-a-galaxy-conserved&quot; rel=&quot;nofollow noreferrer&quot;&gt;the conservation of angular momentum when gravitational force pulls gas clouds to come together&lt;/a&gt;. It is the same mechanism that is responsible for &lt;a href=&quot;https://astronomy.stackexchange.com/questions/16379/why-do-the-planets-orbit-in-the-same-direction&quot;&gt;the planets to orbit the star in the same direction&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="26068" LastEditorUserId="26068" LastEditDate="2019-03-07T03:26:28.450" LastActivityDate="2019-03-07T03:26:28.450" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29890" PostTypeId="2" ParentId="29861" CreationDate="2019-03-06T07:29:54.150" Score="2" Body="&lt;p&gt;&lt;strong&gt;tl;dr:&lt;/strong&gt; Atmospheric refraction is different at radio frequencies than at optical frequencies due to water vapor and electrons, and needs to be though of in a different way. It can be much stronger than for optical observations, and &lt;strong&gt;it can indeed be non-zero at the zenith!&lt;/strong&gt; due to ionospheric effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quoted passage in the question should be only applied to optical observations where the refraction is very weak and doesn't vary much (on average) across the aperture of one optical telescope.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Refraction happens to all electromagnetic radiation, not only just visible light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At visible wavelengths the atmosphere is only very weakly absorbing/attenuating, so the refractive index is treated as a real number, with n ~ 1.0003 due to the dielectric properties of mostly N2 and O2 gas at the surface and dropping with decreasing density going up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However &lt;em&gt;things are very different at other wavelengths!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For radio astronomy there are (at least) two more sources of refraction:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;water vapor &lt;/li&gt;&#xA;&lt;li&gt;electrons &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;and the interaction of radio waves with these is more complex.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Water vapor&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Water vapor is fairly strongly absorbing in the microwave and infrared wavelengths, with a complicated distribution of vibrational and rotational states that strongly interact with electromagnetic radiation due to the molecule's strong polarization. And whenever you have absorption at one frequency (imaginary part of index of refraction), you have changes in the real index of refraction above and below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the distribution of water vapor with altitude differs from the distribution of the other gases, refraction can vary differently than a model for visible light refraction would predict. Since the refraction can be stronger than for visible light, accuracy become much more important as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For very high resolution inteferometric radio telescope arrays like ALMA for example, there are differences in refraction seen by each dish, and so every dish is equipped with its own &lt;em&gt;precipital water vapor monitor&lt;/em&gt; which I believe is some kind of broadband microwave monitor sensitive to the columnar density of water &lt;em&gt;in the direction that the dish is pointed&lt;/em&gt;, but I'm not sure. So I've asked &lt;a href=&quot;https://astronomy.stackexchange.com/q/18791/7982&quot;&gt;What is precipitable water vapor in millimeter-wave radioastronomy and how is it measured?&lt;/a&gt; and I'm still waiting for an authoritative reply.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Electrons&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8MO2b.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8MO2b.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;a href=&quot;http://www.arrl.org/files/file/Technology/pdf/119962.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt; from &lt;a href=&quot;https://astronomy.stackexchange.com/a/22402/7982&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth's atmosphere includes a layer called the ionosphere, and for every ion there's a free electron as well. The same way that the electron &quot;plasma&quot; in metals block and reflect light with frequency below the metal's plasma frequency, the ionosphere blocks and reflects radio waves with frequency below the ionosphere's plasma frequency. It's more complicated because the ionosphere is modeled in layers, but you can imagine a frequency somewhere between roughly 10 and 60 MHz depending on the time of day and the mood of the Sun below which the ionosphere is opaque. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At higher frequency and shorter wavelengths than this, the ionosphere will be mostly transparent, but the real part of the index of refraction will still be significant because it is close to an absorption edge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case refraction can be much larger than for optical telescopes. I'm still working on an answer to [How large does refraction become in radioastronomy?](How large does refraction become in radioastronomy?) While @RobJeffries &lt;a href=&quot;https://astronomy.stackexchange.com/a/22411/7982&quot;&gt;points out&lt;/a&gt; that for GHz type frequencies and above the refraction is not much different than for optical, it can get &lt;strong&gt;much larger&lt;/strong&gt; at much lower frequencies near the plasma frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slide #29 of &lt;a href=&quot;http://www.phys.unm.edu/~lwa/chicago08/talks/anderson_software.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;LOFAR Calibration of the Ionosphere and Other Fun Things&lt;/a&gt; shows that refraction can be several degrees at elevations that are still very useful for radio astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gnGDu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gnGDu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to address the OP's question directly, have a look at slide #21! A radio observation with an array of antennas has been fit to a simple model for the ionosphere with thin &quot;screen&quot; of refraction at 200 km. It shows a phase shift gradient of roughly 100 degrees over only 0.2 degrees in latitude, which is exactly the kind of refraction shown in the OP's question!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tS67Z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tS67Z.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-03-06T07:29:54.150" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29893" PostTypeId="1" AcceptedAnswerId="29896" CreationDate="2019-03-06T14:57:29.943" Score="5" ViewCount="547" Body="&lt;p&gt;Does it exist North of the Universe? (Which is probably used for determining coordinates of far galaxies.)&lt;br&gt;&#xA;If so, then where it is?&lt;br&gt;&#xA;On what is it aligned? (Is it orientated on Earth, Sun, Milky Way, Local group or on something more bigger?)&lt;/p&gt;&#xA;" OwnerUserId="25994" LastEditorUserId="25994" LastEditDate="2019-03-07T13:49:20.177" LastActivityDate="2019-03-07T13:49:20.177" Title="Where is the North of the Universe" Tags="&lt;coordinate&gt;&lt;observable-universe&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29895" PostTypeId="1" AcceptedAnswerId="29903" CreationDate="2019-03-06T17:30:33.010" Score="1" ViewCount="62" Body="&lt;p&gt;Is there any positive N on which it's agreed among astronomers that 10^-N is accepted as, or treated as 0 (zero)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think one of the sciences that most precision is required is astrophysics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware of my hardware capacity to lowest floating number, but that's not what I'm asking about.&lt;/p&gt;&#xA;" OwnerUserId="26161" LastEditorUserId="7982" LastEditDate="2019-03-06T23:20:09.227" LastActivityDate="2019-03-06T23:21:47.150" Title="Is there any positive N on which it's agreed among astronomers that 10^-N is accepted as 0 (zero)?" Tags="&lt;astrophysics&gt;&lt;mathematics&gt;" AnswerCount="1" CommentCount="11" ClosedDate="2019-03-06T23:18:15.463" ContentLicense="CC BY-SA 4.0" />
  <row Id="29896" PostTypeId="2" ParentId="29893" CreationDate="2019-03-06T17:56:19.133" Score="9" Body="&lt;p&gt;Sort of... There is a system called the International Celestial Reference System (ICRS) which has center at the Solar System Barycenter (normally inside the Sun but not the same as the Sun's center) and which has the x and y axis in the plane of the Earth's equator and the z-axis (&quot;North&quot; if you will) pointing towards the Celestial Pole. The x-axis points towards the equinox where the plane of the Earth's orbit and the plane of the Solar system (the ecliptic) cross (on average; this gets complicated quickly as everything related to the Earth is wobbling about on different timescales).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note although this reference system is close to the older mean equator and equinox J2000 or FK5 system it's not actually the same as it is now defined by a set of several hundred very distant extragalactic radio sources which are called the International Celestial Reference Frame (ICRF) and which we assume don't move on the sky. The latest version of this frame is called ICRF3 and was formally adopted at the IAU General Assembly last year in Vienna. More details on how the reference systems and frames work at &lt;a href=&quot;https://aa.usno.navy.mil/faq/docs/ICRS_doc.php&quot; rel=&quot;noreferrer&quot;&gt;the USNO page on the ICRS&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2019-03-06T17:56:19.133" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29897" PostTypeId="1" CreationDate="2019-03-06T18:39:35.153" Score="1" ViewCount="180" Body="&lt;p&gt;Is there a name for a point at the edge of the observable universe at which there is no visible matter, that appears completely dark?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking something like &quot;empty space&quot;, but that could equally apply to an empty void between galaxies, which isn't quite what I'm looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it would refer to a theoretical line in space from the observer to the edge of the universe which contains no visible matter. Is there a term for this?&lt;/p&gt;&#xA;" OwnerUserId="26160" LastEditorUserId="25994" LastEditDate="2019-03-06T19:17:03.827" LastActivityDate="2019-03-07T22:59:49.607" Title="Terminology for empty space at the edge of the observable universe?" Tags="&lt;observable-universe&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29898" PostTypeId="2" ParentId="29582" CreationDate="2019-03-06T19:24:18.703" Score="1" Body="&lt;p&gt;finally I think I grasped this.  As a novice, I did not understand some basic definitions that made that much harder for me to understand the paragraph. It might be a good idea to provide here an answer for reference to others that might encounter this issue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) parallax in this section refers to &lt;strong&gt;geocentric&lt;/strong&gt; parallax. i.e where one point of observation is the surface of the earth and the second is it's center. Evidently, in the Zenith the geocentric parallax vanishes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) the longitude and latitude in this section refer to the lines in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ecliptic_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ecliptic coordinate system&lt;/a&gt; - where the ecliptic is of latitude 0 of course. [I'm still not sure as to &lt;em&gt;why&lt;/em&gt; they used that system back then, instead of the more natural (at least to me ) Equatorial coordinate system. (maybe it was easier to calculate eclipses and plants transits using the ecliptic system)]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for the nonagesimal - it is simply the highest point (closest to the Zenith) in the ecliptic at any given moment. it might be helpful here to note that nonagesimal is &lt;em&gt;not&lt;/em&gt; the &lt;a href=&quot;https://en.wikipedia.org/wiki/Midheaven&quot; rel=&quot;nofollow noreferrer&quot;&gt;Midheaven&lt;/a&gt;. Simply because the highest point in the ecliptic does not necessarily intersect the meridian. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making those definitions clear, the rest in that paragraph follows: the great circle drawn from the Zenith through the nonagesimal is simply a longitude line in the ecliptic coordinate system since it is perpendicular to a latitude line (the ecliptic being latitude 0). so since the the plane being defined by the earth's center, the observer point on the surface and the nonagesimal, is a plane of one longitude, then the parallax could not affect the longitude at all.   &lt;/p&gt;&#xA;" OwnerUserId="25913" LastActivityDate="2019-03-06T19:24:18.703" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29900" PostTypeId="2" ParentId="29897" CreationDate="2019-03-06T21:01:26.493" Score="2" Body="&lt;p&gt;There is no name for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we look further in space, we also look further back in time. The &quot;edge of the observable universe&quot; is in the early universe. At an early time in the evolution of the universe, the whole of space was so hot and dense that it was opaque.  In whichever direction we look, we can't look back further than this. The is called the cosmic microwave background. There is no direction in which we can look in which we don't see the cosmic microwave background. There is no direction that contains no visible matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are directions that contain no galaxies or other more nearby objects, but there is no name for these directions.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-03-06T21:01:26.493" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29901" PostTypeId="2" ParentId="29693" CreationDate="2019-03-06T21:19:51.213" Score="3" Body="&lt;p&gt;Rafael and friends ;))  Here is the link to download the booklet that came with the original poster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.dropbox.com/sh/4ub9nnket6ki888/AADg_FCISkrbKdyNYPTLwfOLa?dl=0&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.dropbox.com/sh/4ub9nnket6ki888/AADg_FCISkrbKdyNYPTLwfOLa?dl=0&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have finished a new and improved version of the Map that I hope to get printed this year.  It now contains the calendar rounds of; 24 Hrs, the week (Visible Planets) and the 365 days within the months with their Sumerian-Babylonian-Greek-Roman patron Gods and Goddesses), the Octaederis (8 Earth- 5 Venus cycle), the 25,800 yr.12 &quot;ages&quot; Precession cycle and the 60 Galactic years since the big bang.&lt;/p&gt;&#xA;" OwnerUserId="26124" LastActivityDate="2019-03-06T21:19:51.213" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29902" PostTypeId="1" AcceptedAnswerId="29940" CreationDate="2019-03-06T22:42:56.543" Score="10" ViewCount="359" Body="&lt;p&gt;The mass region of objects between ~ 0.5 Jupiter masses and 80 Jupiter masses (gas giants through to brown dwarfs and red dwarfs) is typified by an almost flat relationship with object diameter. There are planets out there which are larger than some of the smallest stars.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smallest (currently fusing) star known, &lt;a href=&quot;https://en.wikipedia.org/wiki/EBLM_J0555-57&quot; rel=&quot;noreferrer&quot;&gt;EBLM-J0555-57&lt;/a&gt;, is estimated to be slightly larger than Saturn (at about 59000 km radius with 85 times Jupiter's mass).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the largest planets known that isn't a suspected brown dwarf, &lt;a href=&quot;https://en.wikipedia.org/wiki/WASP-79b&quot; rel=&quot;noreferrer&quot;&gt;WASP-79b&lt;/a&gt; is estimated to be twice Jupiter's diameter at 0.9 times Jupiter's mass. Many &lt;a href=&quot;https://en.wikipedia.org/wiki/Hot_Jupiter&quot; rel=&quot;noreferrer&quot;&gt;Hot jupiters and puffy planets&lt;/a&gt; with similar measurements are known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How likely are there to be systems where a planet is larger than its host star? Are there any examples known?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for currently fusing stars only, which rules out pulsar planets, etc.&lt;/p&gt;&#xA;" OwnerUserId="17655" LastActivityDate="2019-03-10T23:18:49.320" Title="Existence of planets larger than their host star?" Tags="&lt;planetary-formation&gt;&lt;size&gt;&lt;brown-dwarf&gt;&lt;red-dwarf&gt;&lt;hot-jupiter&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29903" PostTypeId="2" ParentId="29895" CreationDate="2019-03-06T23:04:55.510" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there any positive N on which it's agreed among astronomers that 10^-N is accepted as 0 (zero)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;No, there isn't.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Astronomy overlaps with cosmology and there can be some pretty big and small numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However most physical parameters also have &lt;em&gt;units&lt;/em&gt; and the units chosen affect the numerical values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A great example is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Jansky&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jansky&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The jansky (symbol Jy, plural janskys) is a non-SI unit of spectral flux density, or spectral irradiance, used especially in radio astronomy. It is equivalent to 10^−26 watts per square metre per hertz.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some cases one might even imagine going the other way and using L☉/ℓP² myr (solar luminosities per square planck-length per million years) which would require numerical values about 1x10^-110 times smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, astronomers and other scientists usually decide is something is small enough that it doesn't matter &lt;em&gt;within the context of the current discussion&lt;/em&gt;. THey use words like &lt;em&gt;negligible&lt;/em&gt; and &lt;em&gt;small&lt;/em&gt; to indicate they're going to ignore it.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-03-06T23:21:47.150" LastActivityDate="2019-03-06T23:21:47.150" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29904" PostTypeId="1" CreationDate="2019-03-06T23:53:59.023" Score="1" ViewCount="58" Body="&lt;p&gt;Recently I looked into the coldest(and hottest, but in another question) natural places. I discover this to be the Boomerang Nebula, with a freezing temperature of about −272.15 °C. Then I looked at the reason.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The outflowing gas is moving outwards at a speed of about 164 km/s and&#xA;expanding rapidly as it moves out into space; this gas expansion&#xA;results in the nebula's unusually low temperature.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Boomerang_Nebula&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Boomerang_Nebula&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This allows the gas to lose its heat. It works on the same principle&#xA;as that of refrigerators where expanding gas allows cooling.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://factslegend.org/20-interesting-boomerang-nebula-facts/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://factslegend.org/20-interesting-boomerang-nebula-facts/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now I am sort of confused. How exactly can expanding gas lose heat, and lose it to such extent? And gas should cool to the &amp;quot;universal surface temperature&amp;quot;, or the cosmic microwave background temperature if it stays there. However, it cools 2 °C lower. How does this happen? Would there be bose-einstein condensate matter there?&lt;/p&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-03-07T23:24:43.153" Title="What causes cold temperature extremes?" Tags="&lt;temperature&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2019-03-08T19:28:58.373" ContentLicense="CC BY-SA 4.0" />
  <row Id="29905" PostTypeId="1" AcceptedAnswerId="29918" CreationDate="2019-03-07T00:04:40.473" Score="1" ViewCount="61" Body="&lt;p&gt;Recently I looked into the hottest(and coldest, but in another question) natural places. I find this to be a &quot;cloud of searing gas that is surrounding a swarm of galaxies clustered together five billion light-years away in the constellation of Virgo&quot;. This place has a temperature of 300 million degrees Celsius, which is super hot even compared to the temperature at our Sun's core. A quick search why gives:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A group of galaxies collided violently with another swarm of galaxies at a speed of 2,500 miles per second.&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&lt;a href=&quot;https://www.skymania.com/wp/hottest-spot-in-universe-found/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.skymania.com/wp/hottest-spot-in-universe-found/&lt;/a&gt;&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, what mechanism exactly causes this temperature? Friction? Or something else? Would stars there be fusing stuff from the outside? &lt;/p&gt;&#xA;" OwnerUserId="25988" LastActivityDate="2019-03-07T23:29:41.447" Title="What causes hot temperature extremes?" Tags="&lt;temperature&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29906" PostTypeId="2" ParentId="29893" CreationDate="2019-03-07T00:08:08.640" Score="2" Body="&lt;p&gt;There are coordinate systems used on various scales, with &lt;a href=&quot;https://en.wikipedia.org/wiki/Ecliptic_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;the ecliptic system(s)&lt;/a&gt; useful for denoting where on Earth's sky things are, &lt;a href=&quot;https://en.wikipedia.org/wiki/Galactic_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;galactic coordinates&lt;/a&gt; for objects in and around the Milky Way, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Supergalactic_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;the supergalactic coordinate system&lt;/a&gt; for nearby galactic clusters. This is aligned along the supergalactic plane, a plane most local galaxy clusters are aligned with. The zero point is the intersection of this plane and the galactic plane. In the SGC the north supergalactic pole is approximately RA = 18.9h, Dec = +15.7° (epoch J2000), which seems to be somewhere in Aquila. The zero point (SGB = 0°, SGL = 0°) lies at 2.82h, +59.5°, between Perseus and Cassiopeia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/International_Celestial_Reference_System&quot; rel=&quot;nofollow noreferrer&quot;&gt;ICRS&lt;/a&gt; is basically the ecliptic system but centred on the solar system barycenter and oriented relative to a number of distant extragalactic radio sources unlikely to move much. So &quot;north&quot; in this system corresponds more or less to &quot;north&quot; on the usual sky sphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think one can make decent arguments that there are two possible &quot;north poles of the universe&quot; depending on which coordinate system one prefers.&lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2019-03-07T00:08:08.640" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29908" PostTypeId="1" CreationDate="2019-03-07T05:54:15.277" Score="13" ViewCount="320" Body="&lt;p&gt;The question &lt;a href=&quot;https://astronomy.stackexchange.com/q/29666/7982&quot;&gt;Who was the first to realize that the Earth is surrounded by vacuum?&lt;/a&gt; was closed because some users felt it was answered by answers to a different question in an different SE site: &lt;a href=&quot;https://hsm.stackexchange.com/q/2389/3869&quot;&gt;Who was the first to postulate that space was a vacuum?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me the distance between &lt;em&gt;realize&lt;/em&gt; and &lt;em&gt;postulate&lt;/em&gt; is astronomical!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That aside, I've asked a new question: &lt;em&gt;What was the first astronomical measurement which demonstrated that &quot;the Earth is surrounded by vacuum&quot;?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By demonstrated I'm talking about a measurement that shows it is, or at least is likely to be the case that Earth is surrounded by vacuum. It should be something such that it could be presented as scientifically convincing evidence to other scientists of the time, unencumbered by non-scientific predispositions (e.g. religion, rivalry, philosophy...) &lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-03-07T06:03:38.903" LastActivityDate="2019-03-14T15:04:26.677" Title="What was the first astronomical measurement which demonstrated that &quot;the Earth is surrounded by vacuum&quot;?" Tags="&lt;observational-astronomy&gt;&lt;planetary-atmosphere&gt;&lt;history&gt;&lt;atmospheric-effects&gt;" AnswerCount="2" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="29911" PostTypeId="2" ParentId="29859" CreationDate="2019-03-07T12:28:18.403" Score="4" Body="&lt;p&gt;The Sun is heading in the direction of the constellation Hercules: &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_apex&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Solar_apex&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's about 30 degrees above the Ecliptic, meaning the Sun's direction of motion is tilted by about 30 degrees from the plane of Earth's orbit.&lt;/p&gt;&#xA;" OwnerUserId="10892" LastActivityDate="2019-03-07T12:28:18.403" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29912" PostTypeId="1" CreationDate="2019-03-07T16:55:20.527" Score="1" ViewCount="182" Body="&lt;p&gt;I have a list of coordinates of black hole/neutron star systems in decimal degrees (ie. {(266.416,-29.009), (266.418,-29.009),...} from DS9 and would like to determine the distance from SgrA* in parsecs. To my knowledge, 25.8 parsecs ~ 1 degree, but the distance I calculate do not agree with values previously determined by a colleague. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I have a list of nine very faint X-ray transients (VFXTs) and their coordinates, one of which is (266.4191667, -29.0101111). Using the distance formula, I estimated the degree distance from SgrA* - from &lt;a href=&quot;https://en.wikipedia.org/wiki/Sagittarius_A*&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, in decimal degrees, (266.4168371,-29.0078106) using the Euclidean distance formula and multiplied my value to get&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(25.8 pc/deg)(0.00327404 deg) = 0.0844702 pc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;from the center; but this value &lt;strong&gt;does not agree with any&lt;/strong&gt; previously determined values of the VFXTs offsets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The list I received from my colleague for VFXT offsets is &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&quot;vfxt_d = [.7209, .8605, 1.0233, 3.1395, 3.1628, 10.4651, 14.8372, 25.6744, 29.2093] &quot; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I would appreciate some pointers to complete this project sometime today. Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="26174" LastEditorUserId="616" LastEditDate="2019-03-08T15:27:42.047" LastActivityDate="2019-04-08T13:52:37.960" Title="Determine distance from the galactic center in DS9" Tags="&lt;distances&gt;&lt;ds9&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29913" PostTypeId="2" ParentId="29902" CreationDate="2019-03-07T19:04:07.123" Score="6" Body="&lt;p&gt;Since the smallest stars are still the size of gas giant planets, the question ends up coming down to whether gas giants exist around stars at the bottom of the main sequence. Close-in gas giant planets are rare around low-mass stars, though there do seem to be long-period ones. This means the largest planetary radii for the systems in question are going to be similar to Jupiter, rather than inflated hot Jupiters. An exception would be the case of very young systems before the planets cool and shrink but in that case the star would also still be contracting so you probably don't win there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A problem is that these stars are extremely faint, so the radial velocity method is tricky - this may change a bit once more RV instruments that operate in the infrared (e.g. the &lt;a href=&quot;https://www.nist.gov/news-events/news/2019/02/nist-astrocomb-opens-new-horizons-planet-hunting-telescope&quot; rel=&quot;noreferrer&quot;&gt;Habitable Zone Planet Finder&lt;/a&gt;) come online. The long orbital periods for giant planets around these stars would also require longer observation times to make a detection. Unfortunately the long orbital periods would make transits unlikely, so most probably we wouldn't be able to determine the planet's radius and would not know for certain that the planet is larger than the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Direct imaging has spotted a few objects of a few Jupiter masses at fairly wide separations from objects near the hydrogen-burning limit, e.g. &lt;a href=&quot;https://exoplanetarchive.ipac.caltech.edu/cgi-bin/DisplayOverview/nph-DisplayOverview?objname=2MASS%20J02192210-3925225%20b&amp;amp;type=CONFIRMED_PLANET&quot; rel=&quot;noreferrer&quot;&gt;2MASS J02192210-3925225&lt;/a&gt; with an object at the deuterium-burning limit is located about 150 AU from a 0.1 solar mass star. It's not entirely clear what to call these objects and they may be very low-mass brown dwarfs rather than planets. Furthermore these systems are so young that the stars have not yet contracted to their main sequence radii. For low-mass stars this can take several billion years, by which time the planets will have cooled off and become much fainter (and less detectable). These kind of wide-separation systems may also end up being disrupted by stellar encounters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other approach that works for detecting these kind of systems is &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_microlensing&quot; rel=&quot;noreferrer&quot;&gt;gravitational microlensing&lt;/a&gt;, which tends to find objects near the system snowline, i.e. on scales more similar to our planetary system. An example of the kind of system that might have a planet larger than its star is &lt;a href=&quot;https://exoplanetarchive.ipac.caltech.edu/cgi-bin/DisplayOverview/nph-DisplayOverview?objname=KMT-2016-BLG-1107L%20b&amp;amp;type=CONFIRMED_PLANET&quot; rel=&quot;noreferrer&quot;&gt;KMT-2016-BLG-1107Lb&lt;/a&gt;, where the parameters suggest a ~3.3&amp;nbsp;Jupiter mass planet orbiting a ~0.087&amp;nbsp;solar mass star at ~0.34&amp;nbsp;AU. Unfortunately the uncertainties in the parameters are typically large because the lens systems are usually invisible. This means we also don't have radius information, so we can't say for sure that this system definitely has a planet larger than its star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it looks like systems do exist where a planet can be larger than the main-sequence star it orbits, though so far there is no confirmed case due to the difficulty of making the necessary observations.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-03-07T19:56:24.967" LastActivityDate="2019-03-07T19:56:24.967" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29915" PostTypeId="2" ParentId="29897" CreationDate="2019-03-07T22:59:49.607" Score="-1" Body="&lt;p&gt;The observable universe is merely how far we can see with current telescope technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space itself is endless. Oh, there's no end of people here who will aggressively argue this. But empty space, by definition, cannot have an end. It can only be infinite. Literally infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There could very well be an edge in the sense that there's no more matter, where stars, planets and galaxies etc simply tapers off. After that, it would be endless black space.&lt;/p&gt;&#xA;" OwnerUserId="22083" LastActivityDate="2019-03-07T22:59:49.607" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29916" PostTypeId="2" ParentId="29902" CreationDate="2019-03-07T23:11:04.433" Score="3" Body="&lt;p&gt;Beyond red dwarfs, another possibility is that of a planet orbiting a &lt;a href=&quot;https://en.wikipedia.org/wiki/Subdwarf_B_star&quot; rel=&quot;nofollow noreferrer&quot;&gt;type B subdwarf star&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some features of such stars:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Composed almost entirely of helium&lt;/li&gt;&#xA;&lt;li&gt;Thought to be formed through the merger of two white dwarfs or at a specific point in the evolution of some red giants&lt;/li&gt;&#xA;&lt;li&gt;Temperatures range from 20,000 K to 40,000 K&lt;/li&gt;&#xA;&lt;li&gt;Brightness is between 10 - 100 times the brightness of the Sun&lt;/li&gt;&#xA;&lt;li&gt;Mass is typically ~0.5 times the mass of the Sun&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Radius is around 0.15-0.25 times the radius of the sun&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This radius range puts it in overlap of the radius of the largest planets (~0.2 times solar radius). Since the progenitor star(s) is more massive, it leads to an increased likelihood of gas giants forming in the protoplanetary disk. &#xA;The question then becomes: &quot;Can a gas giant find its way to the inner star system so that it is able to puff up?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two type-B Subdwarf stars with planets are known. &#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/V391_Pegasi&quot; rel=&quot;nofollow noreferrer&quot;&gt;V391 Pegasi&lt;/a&gt; is perhaps the closest to fulfilling the planet-larger-than-star criterion. Orbiting the star at ~1.7 AU is a 2.5 - 3.99 &lt;span class=&quot;math-container&quot;&gt;$M_j$&lt;/span&gt; gas giant. This gas giant will receive more energy from its star per square metre than the Earth does from the sun, but this is not likely enough for the planet to heat up sufficiently to become sufficiently 'puffy' to overtake the star's size of 2.3 &lt;span class=&quot;math-container&quot;&gt;$R_j$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other known example is &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler-70&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kepler-70&lt;/a&gt;, a rather curious star that appears to be the remnant of a red giant. The Kepler 70 system is very compact, with the two small (sub-earth radius) planets orbiting with a blisteringly-fast period of 5 and 8 hours respectively. (Fascinatingly, these planets weren't detected by eclipsing their host star, but rather by the periodic &lt;em&gt;increase&lt;/em&gt; in luminosity as they begin to orbit behind the star. Both of these planets have surfaces hotter than the Sun, 7,600 K and 6,800 K respectively.) These planets are theorised to be the remnants of gas giants who were evaporated by being inside the star during its red giant phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From these tow examples, I conclude that there is no difficulty in having gas giants around small type-B subdwarf stars, although the mechanisms for bringing them close enough to become puffy planets is fraught with problems. You either have a red giant that boils all nearby gas giants before the subdwarf forms, or you have two white dwarfs that merge into a blue subdwarf, which requires a progenitor system of two close-binary stars that prohibit close-orbiting circumbinary planets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect for a planet-larger-than-host-star system to form, the gas giant has to migrate inwards somehow after the formation of the subdwarf star.&lt;/p&gt;&#xA;" OwnerUserId="17655" LastActivityDate="2019-03-07T23:11:04.433" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29918" PostTypeId="2" ParentId="29905" CreationDate="2019-03-07T23:29:41.447" Score="0" Body="&lt;p&gt;Within that galaxy cluster, things become chaotic as these galaxies are flung around but kept in orbit by their strong gravity. And as Steven Linton mentioned, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the galaxy's organized motion is turned primarily into disorganized&#xA;  motion within the material that is stuck together, which is what we&#xA;  know as heat. So the atoms of gas in these clouds collided and bounced&#xA;  off in all directions, where they hit other atoms, until the two&#xA;  clouds had formed one, with huge random internal motions (i.e. very&#xA;  hot).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="25988" LastActivityDate="2019-03-07T23:29:41.447" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29919" PostTypeId="1" CreationDate="2019-03-08T11:01:41.653" Score="2" ViewCount="139" Body="&lt;p&gt;I'm creating an application which will simulate the light created by sun or moon on a 3d topographic map of Earth. Something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9LLt1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9LLt1.jpg&quot; alt=&quot;map&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a word or definition for this light/light projection?&lt;/p&gt;&#xA;" OwnerUserId="26180" LastActivityDate="2019-03-08T11:01:41.653" Title="Definition of the light that is projected by a celestial body/luminary" Tags="&lt;the-sun&gt;&lt;the-moon&gt;&lt;light&gt;&lt;definition&gt;" AnswerCount="0" CommentCount="7" ClosedDate="2019-03-18T10:23:43.837" ContentLicense="CC BY-SA 4.0" />
  <row Id="29920" PostTypeId="2" ParentId="29908" CreationDate="2019-03-08T16:31:19.533" Score="5" Body="&lt;p&gt;&lt;strong&gt;I would argue the famous &lt;a href=&quot;https://en.wikipedia.org/wiki/Michelson%E2%80%93Morley_experiment&quot; rel=&quot;noreferrer&quot;&gt;Michelson-Morley experiment&lt;/a&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Luminiferous Aether&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A bit of background first before getting into the experiment itself. It is quite easy to surmise, via Newton's laws of motion, that Earth should be in a vacuum, otherwise the constant drag of traveling through some medium must eventually cause us to crash into the Sun. Despite this Isaac Newton himself proposed the concept of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Luminiferous_aether&quot; rel=&quot;noreferrer&quot;&gt;Luminiferous Aether&lt;/a&gt; which pervaded all of space and was the medium through which light propagated. Before this point there were various concepts of an &quot;Aether&quot; pervading space, but this, I believe, was the first truly scientific approach to the concept as a way of explaining physical phenomenon rather than a simple supposition of existence (such as the Greeks had done). The Luminiferous Aether was proposed as an almost magical concept to avoid numerous physical issues. I think Wikipedia describes it best.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The mechanical qualities of the aether had become more and more magical: it had to be a fluid in order to fill space, but one that was millions of times more rigid than steel in order to support the high frequencies of light waves. It also had to be massless and without viscosity, otherwise it would visibly affect the orbits of planets. Additionally it appeared it had to be completely transparent, non-dispersive, incompressible, and continuous at a very small scale.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The concept of the Luminiferous Aether was accepted on the authority of Newton and inability to explain the propogation of light otherwise.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Michelson-Morley Experiment (1887)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It wasn't until the Michelson-Morley Experiment that the Luminiferous Aether was seriously shot down. The goal of the Michelson-Morley Experiment was to find evidence of this Luminiferous Aether and the null result was very strong evidence that this Aether did not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The experiment itself was set up to measure the speed of light through this Aether. The idea was that as the Earth moved through the Aether, it would cause a sort of &quot;Aether wind&quot; that would slow down the speed of light. Thus if one measured the speed of light in the direction of the wind and perpendicular to it, one should get different speeds. The Michelson-Morley experiment set up exactly this scenario and used increasingly accurate measurements to try and find this difference in speed. Ultimate no difference was found and the Aether was ruled out as a material which existed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From that point onwards it was assumed that space was a vacuum, completely devoid of anything. Oddly enough, this assumption was so strong, that people didn't even really believe in the concept of the Solar Wind at first and were quite perplexed initially when they sent the first rockets into space with particle detectors that ended up detecting all sorts of charged particles in space. Regardless, I would say the Michelson-Morley experiment is the first time scientists had scientific evidence that space was a vacuum.&lt;/p&gt;&#xA;" OwnerUserId="11143" LastActivityDate="2019-03-08T16:31:19.533" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29921" PostTypeId="2" ParentId="29912" CreationDate="2019-03-08T19:03:08.080" Score="0" Body="&lt;p&gt;If r is the lateral (orthogonal to line of sight) distance between the VFXT and Sgr A*,&#xA;and R is the distance from the Sun to Sgr A*, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\theta = \sqrt{(\Delta\alpha \cos \delta)^2 + (\Delta\delta)^2 }$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then r = R tan &amp;theta; as suggested in the comments.&#xA;For the example in the question,&#xA;using R = 7860 &amp;plusmn; 140 pc&#xA;[&lt;a href=&quot;https://doi.org/10.3847/0004-637X/830/1/17&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boehle et al. 2016&lt;/a&gt; via &lt;a href=&quot;https://en.wikipedia.org/wiki/Sagittarius_A*&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;],&#xA;I get r = 0.4216 &amp;plusmn; 0.0075 pc, also not in your colleague's &lt;code&gt;vfxt_d&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where |&amp;theta;| &amp;lt; 1&amp;deg;, tan &amp;theta; &amp;approx; &amp;theta; in radians.&#xA;The approximation in the question used this but was off by a factor of 5.&#xA;A coefficient of 137.2 &amp;plusmn; 2.4 pc/deg should give a better result.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-03-08T19:16:40.610" LastActivityDate="2019-03-08T19:16:40.610" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29922" PostTypeId="5" CreationDate="2019-03-08T19:13:16.233" Score="0" Body="&lt;p&gt;Centaurs are a class of small solar system bodies located between the orbits of Jupiter and Neptune. The first identified member of the class was &lt;a href=&quot;https://en.wikipedia.org/wiki/2060_Chiron&quot; rel=&quot;nofollow noreferrer&quot;&gt;(2060) Chiron&lt;/a&gt;, also designated as a periodic comet under the designation 95P/Chiron. The object was named after Chiron, the wisest of the centaurs in Greek mythology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://minorplanetcenter.net//iau/Unusual.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IAU Minor Planet Center&lt;/a&gt; defines centaurs as having perihelia beyond the orbit of Jupiter and semimajor axes inside the orbit of Neptune. The &lt;a href=&quot;https://ssd.jpl.nasa.gov/sbdb_help.cgi?class=CEN&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL Small-Body Database&lt;/a&gt; defines centaurs as having semimajor axes between the orbits of Jupiter (5.5 AU) and Neptune (30.1 AU). Both of these definitions include the Uranus trojans &lt;a href=&quot;https://en.wikipedia.org/wiki/2011_QF99&quot; rel=&quot;nofollow noreferrer&quot;&gt;2011&amp;nbsp;QF&lt;sub&gt;99&lt;/sub&gt;&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/2014_YX49&quot; rel=&quot;nofollow noreferrer&quot;&gt;2014&amp;nbsp;YX&lt;sub&gt;49&lt;/sub&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-03-09T09:51:41.357" LastActivityDate="2019-03-09T09:51:41.357" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29923" PostTypeId="4" CreationDate="2019-03-08T19:13:16.233" Score="0" Body="Questions about centaurs, small Solar System bodies with orbits between Jupiter and Neptune." OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-03-09T09:51:28.140" LastActivityDate="2019-03-09T09:51:28.140" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29924" PostTypeId="1" CreationDate="2019-03-08T20:12:58.847" Score="1" ViewCount="31" Body="&lt;p&gt;What is the speed of progression of the sun toward and away from the galactic equator during its orbit?&lt;/p&gt;&#xA;" OwnerUserId="26182" LastEditorUserId="7472" LastEditDate="2019-03-08T20:32:22.313" LastActivityDate="2019-03-08T20:52:50.477" Title="Vertical speed vector at which the sun travels to and from the galactic equator" Tags="&lt;astrophysics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29925" PostTypeId="2" ParentId="29924" CreationDate="2019-03-08T20:52:50.477" Score="2" Body="&lt;p&gt;The Sun moves at about 15-20 km/s with respect to a local standard of rest defined by the general motion of stars in our vicinity around the Galaxy. In three-dimensions, this &quot;peculiar velocity&quot; is &lt;span class=&quot;math-container&quot;&gt;$U=10.00 \pm 0.36$&lt;/span&gt; km/s (radially inwards), &lt;span class=&quot;math-container&quot;&gt;$V=5.25 \pm 0.62$&lt;/span&gt; km/s (in the direction of Galactic rotation) and &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$W=7.17 \pm 0.38$&lt;/span&gt; km/s&lt;/strong&gt; (&lt;strong&gt;up and out of the plane&lt;/strong&gt;).  &lt;a href=&quot;http://adsabs.harvard.edu/abs/1998MNRAS.298..387D&quot; rel=&quot;nofollow noreferrer&quot;&gt;(Dehnen &amp;amp; Binney 1998)&lt;/a&gt;. Different authors arrive at velocities that differ by &lt;span class=&quot;math-container&quot;&gt;$\sim 1-2$&lt;/span&gt; km/s from these values and so this would probably be a more conservative estimate of the uncertainties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun executes oscillations around its mean orbit in the Galaxy, periodically crossing the Galactic plane. The Sun is currently above the plane and moving upwards, and each cycle takes about 70 million years with an amplitude of 100 pc; it will be roughly 30 million years before we cross the plane again.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-03-08T20:52:50.477" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29926" PostTypeId="2" ParentId="29886" CreationDate="2019-03-08T22:39:40.687" Score="0" Body="&lt;p&gt;Basically because this great amount of matter has many directions , you can see when many stars born in first time has same direction in a galaxy  (like to stars ursa mayor), when pass the time, this matter direction change, generating entropy, and is when stars formation cease.  &lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2019-03-08T22:39:40.687" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29927" PostTypeId="1" AcceptedAnswerId="29928" CreationDate="2019-03-08T23:12:22.583" Score="6" ViewCount="152" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Frost_line_(astrophysics)&quot; rel=&quot;noreferrer&quot;&gt;Frost line&lt;/a&gt; is the border that marks where volatiles such as water, methane, ammonia, etc. (commonly referred to as 'ices') are able to condense into solid grains in the protoplanetary disk around a forming protostar. Beyond this border, these ices form a significant part of any coalesced planetary body (like the many icy moons of the gas giants), whereas inside this border, the bodies that form are primarily rocky (the four terrestrial planets, Mercury, Venus, Earth, Mars).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can such a frost line exist in the accretion disk around a hot gas giant such as a proto-Jupiter?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some soft evidence for this: Jupiter's four Gallilean moons are of similar sizes and follow a clear trend from the innermost outwards:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Io: Mostly silicate rock and iron sulfide. Density is 3.5 &lt;span class=&quot;math-container&quot;&gt;$g/cm^3$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Europa: Mostly silicate rock with an estimated ~100 km layer of ice/water. 3 &lt;span class=&quot;math-container&quot;&gt;$g/cm^3$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Ganymede: Silicate with an estimated ~800 km layer of ice/water. 1.9 &lt;span class=&quot;math-container&quot;&gt;$g/cm^3$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Callisto: Approximately 50% ice. 1.83 &lt;span class=&quot;math-container&quot;&gt;$g/cm^3$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Source: Mainly articles in &lt;a href=&quot;https://en.wikipedia.org/wiki/Galilean_moons&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there are other mechanisms by which ices can be disproportionately lost from Io, mainly through its extreme volcanism. But this doesn't necessarily explain the trend seen in the other moons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;To be specific, I'm not asking if a protoplanetary frost line is responsible for the distribution of elements in Jupiter's moons, merely if such a frost line can exist and contribute to the observed element distribution&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17655" LastActivityDate="2019-03-09T08:59:09.913" Title="Can a frost line exist around a protoplanet?" Tags="&lt;planetary-formation&gt;&lt;proto-planetary-disk&gt;&lt;galilean-moons&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29928" PostTypeId="2" ParentId="29927" CreationDate="2019-03-09T08:59:09.913" Score="5" Body="&lt;p&gt;Yes they can exist. The effect of ice lines in circumplanetary discs has been studied, see for example &lt;a href=&quot;https://ui.adsabs.harvard.edu/#abs/2015ApJ...806..181H/abstract&quot; rel=&quot;noreferrer&quot;&gt;Heller &amp;amp; Pudritz (2015)&lt;/a&gt;, who propose that the ice lines may act as a barrier preventing moons from being lost as a result of inward type I migration.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastActivityDate="2019-03-09T08:59:09.913" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29930" PostTypeId="1" CreationDate="2019-03-09T18:58:02.160" Score="1" ViewCount="73" Body="&lt;p&gt;Does it remain void and pushes even more or will virtual particles eventually create normal matter?&lt;/p&gt;&#xA;" OwnerUserId="16136" LastEditorUserId="7472" LastEditDate="2019-03-09T19:16:43.457" LastActivityDate="2019-03-09T22:41:54.963" Title="What fills the new space created during expansion?" Tags="&lt;space&gt;&lt;expansion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29931" PostTypeId="1" AcceptedAnswerId="30459" CreationDate="2019-03-09T19:04:47.920" Score="2" ViewCount="672" Body="&lt;p&gt;What events could have to happen to Mars (which is thought to once look like Earth) that may have depleted the atmosphere and erase all proof of life ever existing at least on the surface? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know an asteroid could melt the planet but could there have been an event that could wipe the surface while leaving deposits under the ground intact?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no proof that Mars melted but there is proof that there at least was oceans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.yibada.com/articles/18182/20150309/red-planet-once-held-large-shallow-ocean-mean-positive-signs.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.yibada.com/articles/18182/20150309/red-planet-once-held-large-shallow-ocean-mean-positive-signs.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/1114/where-did-mars-water-go&quot;&gt;Where did Mars water go?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2019-03-18T03:13:58.967" LastActivityDate="2020-09-28T21:04:35.137" Title="Could Mars have oil?" Tags="&lt;planetary-atmosphere&gt;&lt;mars&gt;&lt;life&gt;&lt;geology&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29932" PostTypeId="1" AcceptedAnswerId="29939" CreationDate="2019-03-09T20:33:28.633" Score="1" ViewCount="1322" Body="&lt;p&gt;Is there any way to calculate declination of Moon for any day?&lt;br&gt;I'll subtract inclination of Earth on the ecliptic from the result. I'll get Moon's the shortest angular distance from Sun.&#xA;&lt;br /&gt;&lt;br /&gt;Thanks for all answers!&lt;/p&gt;&#xA;" OwnerUserId="25994" LastEditorUserId="25994" LastEditDate="2019-03-11T12:17:17.163" LastActivityDate="2019-03-11T12:17:17.163" Title="How to calculate declination of Moon?" Tags="&lt;the-moon&gt;&lt;fundamental-astronomy&gt;&lt;declination&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29933" PostTypeId="2" ParentId="29930" CreationDate="2019-03-09T22:41:54.963" Score="2" Body="&lt;p&gt;It's not the case that something needs to be put in there, it's just that space itself is expanding. So if the bit you are watching expand is void then it will still be void, only bigger - you don't get new space, you just get bigger space.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2019-03-09T22:41:54.963" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29934" PostTypeId="1" AcceptedAnswerId="29935" CreationDate="2019-03-09T23:00:22.020" Score="1" ViewCount="61" Body="&lt;p&gt;So I've been given the velocity curve, parallax and apparent magnitude of a star in a binary system with what is potentially a black hole. I've calculated from the apparent magnitude and parallax that the star is a type F5V, which puts the mass at about 1.4 Solar Masses. The velocity curve has an inclination of 90, and oscillates back and forth between +/- 75km/s. There is no data on the companion of this star, just the fact that it &lt;em&gt;could&lt;/em&gt; be a black hole. I'm supposed to estimate the mass by numerically approximating a polynomial.&#xA;So far I've used this equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{M^{3}}{(m+M)^{2}} = \frac{Pv^{2}}{2\pi G}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where M is the mass of the thing I don't know, m is the mass of the known companion (1.4 solar masses) P is the period (5.59 days) and v is of course the velocity (75km/s)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got lazy and wrote &lt;span class=&quot;math-container&quot;&gt;$\frac{Pv^{2}}{2\pi G}$&lt;/span&gt; as &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; and arrived at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$M^{3} - kM^{2} - km^{2} = 0$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using python's optimize library I found the mass of this unkown partner to be about 0.018 solar masses. My question here is where did I go wrong, and if I didn't go wrong anywhere is that a realistic mass for a super small black hole / other very small, dense and invisible object? &lt;/p&gt;&#xA;" OwnerUserId="26187" LastEditorUserId="26187" LastEditDate="2019-03-09T23:19:09.843" LastActivityDate="2019-03-10T11:42:21.417" Title="Mass of a potential black hole in a binary system" Tags="&lt;black-hole&gt;&lt;mass&gt;&lt;binary-star&gt;&lt;kepler&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29935" PostTypeId="2" ParentId="29934" CreationDate="2019-03-09T23:31:56.693" Score="2" Body="&lt;p&gt;Your first equation is incorrect. The left side has dimensions of mass, the right side has dimensions of mass&amp;nbsp;&amp;times;&amp;nbsp;time&amp;nbsp;&amp;times;&amp;nbsp;length&lt;sup&gt;-1&lt;/sup&gt;. The velocity semi-amplitude (usually denoted &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; rather than &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt;) should be raised to the third power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted by @PM-2Ring in the comments, your second equation is also incorrect as you didn't expand the &lt;span class=&quot;math-container&quot;&gt;$(m+M)^2$&lt;/span&gt; term correctly.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-03-10T11:42:21.417" LastActivityDate="2019-03-10T11:42:21.417" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29936" PostTypeId="1" CreationDate="2019-03-10T03:46:49.670" Score="9" ViewCount="187" Body="&lt;p&gt;I tried to research but what I found is pretty limited.    A very tiny but non zero percentage of primordial matter was Helium-3 or 3He.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars produce 3He as part of the proton-proton chain but they also consume 3He.   It has about a 400 year half life in our sun.   &lt;a href=&quot;https://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction#The_proton%E2%80%93proton_chain_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;From Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the Sun, each helium-3 nucleus produced in these reactions exists&#xA;  for only about 400 years before it is converted into helium-4.[6] Once&#xA;  the helium-3 has been produced, there are four possible paths to&#xA;  generate 4He&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question is 2-fold.   Is the primordial amounts of 3He significant or insignificant compared to what stars produce and eject by coronal mass ejections or blowing up into nebulas, and are there specific stars, due to the internal heat and rate of reaction that product and eject more 3He into their mass ejections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, would airless, rocky, magnetic field free worlds around red dwarfs be more saturated with 3He or would you find more around larger, hotter stars that undergo fusion faster.    If I wanted to go He3 mining, would I set my ship to a red dwarf system or a blue star system or a nebula?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not for a book or anything, I'm just personally curious, as 3He is potentially very useful stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware that primordial isn't particularly easy to collect, as anything gaseous and primordial would spread out and only collect in large gravity wells, gas giants or larger.   But primordial 3He could, for example, exist in Jupiter or Saturn from formation, though the formation of their magnetic fields probably prevents them from absorbing any ejected from stars.  Hence the primordial vs star ejected ratio and the type of star that produced the most question are somewhat related, so I thought one question rather than two, but I can split into two questions if desired.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2019-03-11T02:06:02.360" LastActivityDate="2019-09-10T20:46:19.087" Title="What percentage of Helium-3 is primordial vs produced in stars" Tags="&lt;nucleosynthesis&gt;&lt;helium&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29937" PostTypeId="1" AcceptedAnswerId="29938" CreationDate="2019-03-10T19:04:02.973" Score="7" ViewCount="1754" Body="&lt;p&gt;If we look back far enough we can see all the origins of the universe, so is it possible, even if not feasible, that we could trace the history of some matter as it moves through space-time? I want to understand how looking at different depths in space and time are correlated in regards to the matter being observed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example would it be possible to look deep into a certain part of space and time to find some galaxy that contributed to the matter that makes up the Milky Way today? Then somehow follow it through space-time by looking at different depths and locations in space, and see how it came to be part of the Milky Way?&lt;/p&gt;&#xA;" OwnerUserId="26195" LastActivityDate="2019-03-11T09:16:31.420" Title="Can we track matter through time by looking at different depths in space?" Tags="&lt;space-time&gt;&lt;cosmic-microwave-background&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29938" PostTypeId="2" ParentId="29937" CreationDate="2019-03-10T19:44:53.470" Score="14" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Would it be possible to look deep into a certain part of space and&#xA;  time to find some galaxy that contributed to the matter that makes up&#xA;  the Milky Way today?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, that's not possible. If we could do that, it'd mean that the matter traveled from there to here faster than its light got here, and matter can't travel faster through space than light does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All we &lt;em&gt;can&lt;/em&gt; do is look at similar galaxies to the Milky Way at earlier times. And because of the expansion of space those galaxies are now even further away from us than they were when they emitted the light that we're seeing now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxies develop (mostly) in isolation from one another, apart from the occasional merger or collision between neighbouring galaxies. Intergalactic distances are fairly huge, so it takes vast amounts of time for matter to travel from one galaxy to another, and matter is mostly bound by gravity to the galaxy it's in. Galactic escape speeds are pretty high, although the occasional star does get flung out of the galaxy by cataclysmic events like supernova explosions. But even then, such rogue bodies mostly end up in intergalactic space. The odds of them ending up in a different galaxy are pretty slim.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="16685" LastEditDate="2019-03-10T19:56:25.143" LastActivityDate="2019-03-10T19:56:25.143" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29939" PostTypeId="2" ParentId="29932" CreationDate="2019-03-10T22:23:13.757" Score="3" Body="&lt;p&gt;The complexity of the formula depends on the precision you need.&#xA;We can make a crude approximation with two sine waves: one for the Moon's travel eastward around the ecliptic over a&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_month#Tropical_month&quot; rel=&quot;nofollow noreferrer&quot;&gt;tropical month&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\delta_1 = 23.4^\circ \sin {2 \pi (u - 10.75) \over 27.32158}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and one for the Moon's travel north or south of the ecliptic over a&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_month#Draconic_month&quot; rel=&quot;nofollow noreferrer&quot;&gt;draconic month&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\delta_2 = 5.1^\circ \sin {2 \pi (u - 20.15) \over 27.21222}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;both modified as the Moon goes faster or slower over an &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_month#Anomalistic_month&quot; rel=&quot;nofollow noreferrer&quot;&gt;anomalistic month&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$u = t - 0.48 \sin {2 \pi (t - 3.45) \over 27.55455}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;em&gt;&lt;strong&gt;t&lt;/strong&gt;&lt;/em&gt; is the number of days since 2000-01-01 12:00 TT (J2000.0).&#xA;Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\delta = \delta_1 + \delta_2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is within 1&amp;deg; of the Moon's geocentric declination most of the time.&#xA;For a topocentric declination, subtract&#xA;1&amp;deg; sin (observer's geographic latitude).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need higher precision, see&#xA;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1979ApJS...41..391V&quot; rel=&quot;nofollow noreferrer&quot;&gt;van Flandern and Pulkkinen 1979&lt;/a&gt;&#xA;or &lt;a href=&quot;http://www.willbell.com/math/mc1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Meeus 1999&lt;/a&gt;&#xA;for more complicated formulas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: the &lt;strong&gt;u&lt;/strong&gt; and &lt;strong&gt;t&lt;/strong&gt; offsets above are near a 0&amp;deg; longitude crossing, an ascending node, and an apogee in January 2000 but are tweaked to minimize RMS differences from JPL HORIZONS lunar positions for 2000-2018.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2019-03-10T22:23:13.757" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29940" PostTypeId="2" ParentId="29902" CreationDate="2019-03-10T23:18:49.320" Score="4" Body="&lt;p&gt;The answer to the question depends on the exact definition of planet that is used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A possible example is the L dwarf 2M 0746+20 (2MASS J07464256+2000321) and its planet &lt;a href=&quot;http://exoplanet.eu/catalog/2m_0746+20_b/&quot; rel=&quot;nofollow noreferrer&quot;&gt;2M 0746+20 b&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radius of the planet is 12% greater than the radius of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{array}{lll}&#xA;\hline&#xA;\text{} &amp;amp; \text{Mass} &amp;amp; \text{Radius}\\&#xA;\hline&#xA;\text{Planet} &amp;amp; 12.21 \cdot M_J &amp;amp; 0.970 \cdot R_J\\&#xA;\text{Star} &amp;amp; 83.79 \cdot M_J &amp;amp; 0.089 \cdot R_{Sun} = 0.866 \cdot R_J\\&#xA;\hline&#xA;\end{array}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The reported planet mass of &lt;span class=&quot;math-container&quot;&gt;$12.21 (± 0.4) \cdot M_J$&lt;/span&gt; is slightly below the deuterium burning limit of 13 Jupiter masses.&lt;/p&gt;&#xA;" OwnerUserId="10568" LastActivityDate="2019-03-10T23:18:49.320" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29941" PostTypeId="2" ParentId="29937" CreationDate="2019-03-10T23:18:55.943" Score="5" Body="&lt;p&gt;You would have to catch up to the light that carries the information you seek. It's traveled for a few billion years at this point (Earth is ~4.3B). So, you could watch the formation of Earth (Milky Way, whatever), if you could instantly teleport billions of light years away from here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we watch distant galaxies and starts, what we're seeing is &quot;old&quot; light. The events that we see occurred many years ago. If we see a galaxy forming, and that galaxy is 10 billion light years away, then that galaxy has already formed. Its configuration at this time is very different from what we see. In fact, some of its stars have already burned out. Similarly, if you lived in (or could instantly teleport to) that galaxy, you would see the Milky Way as it appeared 10B year ago.&lt;/p&gt;&#xA;" OwnerUserId="26198" LastActivityDate="2019-03-10T23:18:55.943" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29942" PostTypeId="1" CreationDate="2019-03-11T04:59:31.737" Score="2" ViewCount="98" Body="&lt;p&gt;I live just south of Sydney Australia, and I went outside to watch the ISS pass overhead on Friday night. The ISS was following a smaller object easily visible to the naked eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sydney time 20.43 or UTC 9.43. (Friday March 8) The smaller object was ahead of the ISS by maybe 25 degrees. Appeared to be moving at exactly the same speed and trajectory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was this the the SpaceX Crew Dragon capsule? Sorry if this is the wrong place to ask.&lt;/p&gt;&#xA;" OwnerUserId="26202" LastEditorUserId="7982" LastEditDate="2019-03-11T11:44:36.467" LastActivityDate="2019-03-11T12:13:51.243" Title="Did I see the dragon capsule on Friday night?" Tags="&lt;identify-this-object&gt;&lt;iss&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29943" PostTypeId="2" ParentId="29840" CreationDate="2019-03-11T05:08:48.897" Score="3" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$d_o$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$h_o$&lt;/span&gt; are the distance and diameter of the Sun and &lt;span class=&quot;math-container&quot;&gt;$d_i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$h_i$&lt;/span&gt;  of the respective images. The Earth's radius is &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$f=\frac{r}{2}$&lt;/span&gt; is the focal length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mirror equation &lt;span class=&quot;math-container&quot;&gt;$$\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}$$&lt;/span&gt; in the approximation &lt;span class=&quot;math-container&quot;&gt;$d_o \gg f$&lt;/span&gt; gives &lt;span class=&quot;math-container&quot;&gt;$d_i \approx f$&lt;/span&gt;. Then, using the magnification equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$h_i = \frac {d_i}{d_o}h_o \approx \frac {f}{d_o}h_o.$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; is the distance to the satellite then the angular size is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{h_i}{S} \approx \frac{h_o}{d_o}\frac{f}{S} = \frac{h_o}{d_o}\frac{r}{2S} \approx \frac{1.4\times10^6}{150\times10^6}\frac{6.4\times10^3}{2\times 1.6\times 10^6}\approx 1.8 \times 10^{-5}$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where distances are in km. (We have also ignored that the image is &lt;span class=&quot;math-container&quot;&gt;$\frac{R}{2}$&lt;/span&gt; closer to the satellite than is the Earth's center.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a more descriptive note, the reflection on perfectly smooth water would look like a &lt;em&gt;very bright spot&lt;/em&gt; about 29 km in diameter. It is blurred out (due to sea waves) to about ten times that diameter on the image. It would be interesting to calculate the brightness (per area) of the (unblurred) spot compared to the brightness of the sunlit Earth's surface.&lt;/p&gt;&#xA;" OwnerUserId="17743" LastEditorUserId="7982" LastEditDate="2019-03-14T01:31:54.783" LastActivityDate="2019-03-14T01:31:54.783" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29945" PostTypeId="2" ParentId="29937" CreationDate="2019-03-11T09:16:31.420" Score="0" Body="&lt;p&gt;To see the Milky Way's formation, you'd need a way to observe the photons that are now ~5 billion light-years away from us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to accomplish this would be to find a mirror in the space far away, and look at our own reflection in it. The gravity of black holes can cause photons to make a full turn, and return to us: &lt;a href=&quot;https://physics.stackexchange.com/questions/225693/using-a-naked-black-hole-as-a-mirror&quot;&gt;https://physics.stackexchange.com/questions/225693/using-a-naked-black-hole-as-a-mirror&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, with our current technology that's still infeasible as the tiny amount of photons from our reflection would be lost among all the others, such as radiation from stars behind the black hole. It would require a huge telescope array to have enough resolution to collect enough and resolve their direction accurately enough. This is only a guess, but it would be something on a scale of  billions of telescopes accurately positioned over many light years of distance.&lt;/p&gt;&#xA;" OwnerUserId="14071" LastActivityDate="2019-03-11T09:16:31.420" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29946" PostTypeId="1" CreationDate="2019-03-11T10:29:44.357" Score="1" ViewCount="32" Body="&lt;p&gt;Is the Higgs field found everywhere? Is it responsible for our mass, and are we &lt;em&gt;in the Higgs field?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="25466" LastEditorUserId="7982" LastEditDate="2019-03-11T14:51:19.810" LastActivityDate="2019-03-11T14:51:19.810" Title="Higgs field interactions in universe" Tags="&lt;particles&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2019-03-11T15:13:04.063" ContentLicense="CC BY-SA 4.0" />
  <row Id="29947" PostTypeId="2" ParentId="29486" CreationDate="2019-03-11T10:45:51.363" Score="2" Body="&lt;p&gt;@Astrosnapper's &lt;a href=&quot;https://astronomy.stackexchange.com/a/29500/7982&quot;&gt;answer&lt;/a&gt; to &lt;em&gt;Helioseismology of the Sun, what is actually measured?&lt;/em&gt; begins with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The two main methods to detect the solar &lt;em&gt;p&lt;/em&gt;-mode oscillations, which have a period of about 5 minutes (so frequency of ~3 milliHertz) are line of sight Doppler shift and change in irradiance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3 mHz is the peak of the power spectrum, but a 2D histogram of shows significant power at frequencies beyond 8 mHz, corresponding to a period of only 128 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this mean? A dopplergram is collected over an extended period of time, maps of line-of-sight velocity determined via doppler shift of a given spectral line, and fit to a model based on spherical harmonics. Below is an example of a dopplergram and a histogram of power as a function of frequency and spherical harmonic. The shading left-to-right is the doppler shift associated with the rotation of the Sun, I'm not sure if the orbital motion of the Earth and other kinematic effects have been subtracted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went with 8 mHz and calculated the time and phase delay due to the different distances of parts of the 695,700 km radius Sun at a given instant of time. (script):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gwpl4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gwpl4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found some older data and a nice description &lt;a href=&quot;https://soho.nascom.nasa.gov/publications/CDROM1/papers/rhodes/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Measurements of Frequencies of Solar Oscillations from the MDI medium-l Program&lt;/a&gt; by E.J. Rhodes, Jr., A.G. Kosovichev, P.H. Scherrer, J. Schou &amp;amp; J. Reiter which comes from &lt;a href=&quot;http://soi.stanford.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Michelson Doppler Imager&lt;/a&gt; aboard &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_and_Heliospheric_Observatory&quot; rel=&quot;nofollow noreferrer&quot;&gt;SOHO&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zAjOf.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zAjOf.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Fig. 1c: Full disk Dopplergram. (full-scale version). Full-disk Dopplergram obtained with MDI on 9 July 1996. The large left-right variation is the signal from solar rotation. The &quot;texture&quot; is a combination of the motion from the sound waves that permeate the Sun and the large convective cells called &quot;super-granulation&quot;. The bright feature in the lower right quadrant is an active region. &lt;a href=&quot;https://soho.nascom.nasa.gov/publications/CDROM1/papers/rhodes/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/v9b6V.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/v9b6V.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Fig. 1b: Power spectrum obtained from 144 days the MDI Medium-l data for the modes averaged over the azimuthal order m. The power concentrates in ridges corresponding to solar acoustic (p) modes. The lowest weak ridge corresponds to the fundamental (f) mode. &lt;a href=&quot;https://soho.nascom.nasa.gov/publications/CDROM1/papers/rhodes/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-03-11T10:45:51.363" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29948" PostTypeId="2" ParentId="29942" CreationDate="2019-03-11T10:57:28.147" Score="2" Body="&lt;p&gt;I have certainly watched the ISS and an undocked Shuttle Orbiter chase each other across the sky. Since they appeared to be of equal brightness, I didn't know which was which. Their angular separation from my POV was approximately 20°, so their actual separation would have been about 150km. It was one of the last Shuttle flights, so I was very lucky to see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that you saw the Dragon capsule, and it would certainly have appeared to be be fainter than the ISS. The easiest way to establish the fact is by using a program like Stellarium. This will show you if the ISS was visible at your location, given the date and time of day.&lt;/p&gt;&#xA;" OwnerUserId="13887" LastActivityDate="2019-03-11T10:57:28.147" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29949" PostTypeId="2" ParentId="29942" CreationDate="2019-03-11T12:13:51.243" Score="1" Body="&lt;p&gt;The SpaceX YouTube video &lt;a href=&quot;https://www.youtube.com/watch?v=QVEBO6Zuppk&quot; rel=&quot;nofollow noreferrer&quot;&gt;SpaceX Crew Dragon Returns from Space Station&lt;/a&gt; shows the undocking and parts of the reentry and landing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbits work a little counterintuitively when it comes to velocity. To drop to a lower orbit the capsule will point it's engines forward so that they receive an impulse backward, but since they loose altitude this way the speed of the capsule actually &lt;em&gt;increases&lt;/em&gt;. As they loose altitude then they will continue to move forward of the ISS, until they start losing substantial amounts of velocity due to drag and begin ballistic reentry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes you will see the capsule &lt;em&gt;ahead of the ISS&lt;/em&gt; as it begins to loose altitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FYI you'll hear mention of the Keep-Out ellipse in the video, you can read more about that in answers to &lt;a href=&quot;https://space.stackexchange.com/q/20814/12102&quot;&gt;What is the “ISS's Keep Out Sphere” and what is its radius?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-03-11T12:13:51.243" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29950" PostTypeId="1" CreationDate="2019-03-11T12:16:20.217" Score="3" ViewCount="48" Body="&lt;p&gt;I am thinking, that if the Big Bang created time, it implies that before the Big Bang there was no time (correct me if I am wrong).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is that, if there was no time before the Big Bang, how did the Big Bang occured ?&lt;/p&gt;&#xA;" OwnerUserId="26148" LastActivityDate="2019-03-11T12:16:20.217" Title="How did the Big Bang occured if before that there was no time?" Tags="&lt;big-bang-theory&gt;&lt;time&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2019-03-11T12:52:49.910" ContentLicense="CC BY-SA 4.0" />
  <row Id="29951" PostTypeId="1" AcceptedAnswerId="29959" CreationDate="2019-03-11T15:36:07.120" Score="1" ViewCount="346" Body="&lt;p&gt;Recently, &lt;a href=&quot;https://agupubs.onlinelibrary.wiley.com/doi/abs/10.1029/2018JA026136&quot; rel=&quot;nofollow noreferrer&quot;&gt;a paper was published&lt;/a&gt; claiming that the Earth’s atmosphere extends far beyond the Moon. This has been reported on by a number of news sources and websites, including &lt;a href=&quot;https://m.esa.int/Our_Activities/Space_Science/Earth_s_atmosphere_stretches_out_to_the_Moon_and_beyond&quot; rel=&quot;nofollow noreferrer&quot;&gt;the ESA&lt;/a&gt;, &lt;a href=&quot;https://www.businessinsider.com/earth-atmosphere-extends-beyond-moon-hydrogen-gas-cloud-2019-2&quot; rel=&quot;nofollow noreferrer&quot;&gt;Business Insider&lt;/a&gt;, &lt;a href=&quot;https://www.space.com/earth-atmosphere-extends-beyond-moon.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Space.com&lt;/a&gt;, and &lt;a href=&quot;https://earthsky.org/earth/earth-atmosphere-geocorona-extends-beyond-moon&quot; rel=&quot;nofollow noreferrer&quot;&gt;EarthSky.org&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If these findings are true, how does this affect the current understanding related to the measured atmospheric pressure, the currently accepted pressure in space between the Earth and the Moon, the pressure on the Moon, the effect of Earth's gravity on the atmosphere, and Earth's ability to maintain this extensive atmosphere while traveling through the Solar System? Is the atmosphere gradually lost or replenished? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/N88aj.jpg&quot; alt=&quot;A diagram of the geocorona, the new part of Earth&amp;#39;s atmosphere&quot;&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;Image credit: European Space Agency&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="26115" LastEditorUserId="26115" LastEditDate="2019-03-11T17:10:09.563" LastActivityDate="2019-03-12T10:55:03.227" Title="Is the moon inside earth’s atmosphere? If so, what are the consequences?" Tags="&lt;the-moon&gt;&lt;gravity&gt;&lt;earth&gt;&lt;planetary-atmosphere&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29952" PostTypeId="1" CreationDate="2019-03-11T20:57:26.297" Score="1" ViewCount="204" Body="&lt;p&gt;I have a flux value in the infrared band of &lt;code&gt;8 mJy&lt;/code&gt; at a wavelength of &lt;code&gt;80&lt;/code&gt; microns that I want to translate into X-ray conventional flux of &lt;code&gt;keV/cm^2/s/keV&lt;/code&gt;.&#xA;I have tried many conversion factors but none of them is helpful (see e.g. &lt;a href=&quot;https://hea-www.harvard.edu/~pgreen/figs/Conversions.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://hea-www.harvard.edu/~pgreen/figs/Conversions.pdf&lt;/a&gt;). This table for example gives a conversion factor of &lt;span class=&quot;math-container&quot;&gt;$10^{-26}$&lt;/span&gt; from &lt;code&gt;mJy&lt;/code&gt; to &lt;code&gt;erg/cm^2/s/Hz&lt;/code&gt;, and this is correct. However, when I go from &lt;code&gt;erg/cm^2/s/Hz&lt;/code&gt; to &lt;code&gt;keV/cm^2/s/keV&lt;/code&gt;, an additional &lt;span class=&quot;math-container&quot;&gt;$10^{26}$&lt;/span&gt; factor comes out from the transformation of &lt;code&gt;ergs&lt;/code&gt; and &lt;code&gt;Hz&lt;/code&gt;. This cancels out with the previous factor and an unbelievable IR flux results, higher than the actually measured X-ray flux.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the IR flux is much less than the X-ray flux, but can not find an appropriate way to convert.&lt;/p&gt;&#xA;" OwnerUserId="26215" LastEditorUserId="26215" LastEditDate="2019-03-11T23:26:10.347" LastActivityDate="2019-03-11T23:26:10.347" Title="Conversion factors for flux" Tags="&lt;luminosity&gt;&lt;spectra&gt;&lt;units&gt;&lt;wavelength&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29954" PostTypeId="1" AcceptedAnswerId="30090" CreationDate="2019-03-11T22:46:34.640" Score="4" ViewCount="121" Body="&lt;p&gt;In &lt;a href=&quot;https://arxiv.org/abs/1706.06508&quot; rel=&quot;nofollow noreferrer&quot;&gt;a paper called &lt;em&gt;Further Observational Evidence for a Critical Ionising Luminosityin Active Galaxies&lt;/em&gt;&lt;/a&gt;, Section 2.1, the authors talk about using the ICRF2 to find a tuning frequency in search of the 21-cm absorption (of HI):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The &lt;em&gt;Second Realization of the International Celestial Reference Frame by Very Long Baseline Interferometry&lt;/em&gt; (ICRF2, Ma et al.2009), constitutes a sample of strong flat spectrum radio sources,of which 1682 now have known redshifts (Titov &amp;amp; Malkin 2009; Titov et al. 2013 and references therein), yielding a tuning frequency in the search for 21-cm absorption. Being VLBI sources, all have significant compact flux, thus maximising the chance of a high covering factor and thus optical depth (Curran et al. 2013a).The original aim of the survey was to form part of a large observing campaign to search and quantify the incidence of associated &lt;span class=&quot;math-container&quot;&gt;${\rm H\,{\small I}\,21}$&lt;/span&gt;-cm absorption over all redshifts, although observing time was only granted for the high redshift &lt;span class=&quot;math-container&quot;&gt;$(z \gtrsim 2.6)$&lt;/span&gt;  proposals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Since we already know the frequency of HI (1420.40575177 MHz), why is this so important? What am I missing about the functioning of radio telescopes?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/2295/what-determines-which-frequency-range-a-radio-telescope-has&quot;&gt;This question&lt;/a&gt; doesn't really answer my query.&lt;/p&gt;&#xA;" OwnerUserId="17535" LastEditorUserId="37191" LastEditDate="2021-05-12T10:23:52.480" LastActivityDate="2021-05-12T10:23:52.480" Title="Tuning radio telescopes" Tags="&lt;radio-astronomy&gt;&lt;redshift&gt;&lt;radio-telescope&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29956" PostTypeId="1" AcceptedAnswerId="29961" CreationDate="2019-03-12T01:51:55.310" Score="2" ViewCount="652" Body="&lt;p&gt;I am trying to use the periodogram to tell when a signal is periodic or not by following the tutorial for the astropy Lomb-scargle periodogram here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://docs.astropy.org/en/stable/stats/lombscargle.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://docs.astropy.org/en/stable/stats/lombscargle.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I simulated some data, one that is a sinusoid (period = 200) and one which is a skew gaussian (i.e, a single transient event). The hope was that periodogram would pick out the period for the periodic object and and give a  period for the transient that would imply a single event occured in the window. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the results don't make sense at all. I have attached code at the end and the figures generated below. Due to random noise in my simulation, each result is different and I provide two examples of the results below. I use the same method outlined in the link above where we use the function &lt;code&gt;LombScargle.model()&lt;/code&gt; on the best fit frequency from &lt;code&gt;frequency[argmax[power]]&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red line is the true function I simulated the data from. The green is the best fit from the periodogram. The right hand plots are the PSD from the periodogram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 1&#xA;&lt;a href=&quot;https://i.stack.imgur.com/06Bmn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/06Bmn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we can see the best fit frequency for the sinusoid (top right plot) picked out is 0.105 (i.e. a period of 9-10 days) which is not near a frequency of 0.05 I'd expect for something with a period of 200 days, yet when I feed this best fit frequency of 0.105 to the lomb-scargle model fitter, a nicely matching periodic curve comes out with a period of 200 days? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does not make sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DJbvQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DJbvQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I ran the code again and this time the results are switched around? It fit a very large period to the transient so that I can confidently say it is a single transient event yet the sinusoidal fit is terrible. The best frequency is still 0.105 (period=10) yet the lomb-scargle model fitter overlays something that appears to have a period of 60 days which is wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could I get clarification on if i'm doing something wrong? I've been told the periodogram is the de facto tool for unenvenly sampled data like this yet... the results seem awful half the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify my questions&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Can it be explained how in the first plot the best-fit frequncy of 0.105 that I feed into the astropy lomb-scargle model fitter somehow creates a correctly matching sinusoid with frequency 0.05? What is the explanation?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why are there 5 strong peaks in the top right plot for the periodogram in example 1 when I only expect 1? The middle two are close to the real value of 0.05 (at 0.045 and 0.055)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Here is the short code used to simulate and plot the data&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;import scipy.stats as sc&#xA;from astropy.stats import LombScargle&#xA;import math&#xA;&#xA;&#xA;#simulate parameters&#xA;data_range = [i for i in range(1,1001)]&#xA;number_of_samples = 50&#xA;gauss_skew = sc.skewnorm.pdf&#xA;skew = -10&#xA;period = 200&#xA;location = data_range[int(len(data_range)/2)]&#xA;&#xA;&#xA;y1= [(2 * (1. + np.sin(2. * np. pi * x/period)) + np.random.normal(loc =0.0, scale = 0.5)) for x in data_range]&#xA;errors1 = [np.random.normal(loc = 0.0, scale = 1) for x in data_range]&#xA;y2 = [(1000* (gauss_skew(x,skew,loc=location ,scale = 50)) + np.random.normal(loc =0.0, scale = 1)) for x in data_range]&#xA;errors2 = [np.random.normal(loc = 0.0, scale = 1) for x in data_range]&#xA;&#xA;sample_rate = int(len(data_range)/number_of_samples)# To thin the data a bit&#xA;y1 = y1[::sample_rate]&#xA;y2 = y2[::sample_rate]&#xA;errors1 = errors1[::sample_rate]&#xA;errors2 = errors2[::sample_rate]&#xA;x1= x2 =data_range[::sample_rate]&#xA;&#xA;truth1 = [2* (1. + np.sin(2. * np. pi * x/period)) for x in data_range]&#xA;truth2 = [1000 * (gauss_skew(x,-10,loc=location ,scale = 50)) for x in data_range]&#xA;&#xA;truths = [truth1,truth2]&#xA;x= [x1,x2]&#xA;y=[y1,y2]&#xA;errors = [errors1,errors2]&#xA;&#xA;fig,ax  = plt.subplots(nrows=2,ncols=2)&#xA;&#xA;ax[0][0].errorbar(x1,y1,yerr=errors1,fmt='o')&#xA;ax[0][0].set_xlabel('time')&#xA;ax[1][0].errorbar(x2,y2,yerr=errors2,fmt='o')&#xA;ax[1][0].set_xlabel('time')&#xA;ax[0][1].set_xlabel('frequency')&#xA;ax[1][1].set_xlabel('frequency')&#xA;&#xA;&#xA;for i in range(0,2):&#xA;    frequency, power = LombScargle(x[i],y[i],errors[i]).autopower()&#xA;    #Get the best fit frequency as in the tutorial&#xA;    best_frequency = frequency[np.argmax(power)]&#xA;    print('best frequency:',best_frequency)&#xA;    t_fit = np.linspace(x[i][0], math.floor(x[i][-1]),num =50)&#xA;&#xA;    #Fit the best fit frequency&#xA;    #plot the best best model based on the best fit&#xA;    y_fit = LombScargle(x[i], y[i], errors[i]).model(t_fit, best_frequency)&#xA;    ax[i][0].plot(t_fit,y_fit,'g')&#xA;    ax[i][0].plot(data_range,truths[i],'r')&#xA;&#xA;    #Plot the PSD&#xA;    ax[i][1].plot(frequency,power)&#xA;    ax[i][1].axvline(x=best_frequency,color='black', ls='--')&#xA;plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="26219" LastEditorUserId="26219" LastEditDate="2019-03-12T06:55:09.877" LastActivityDate="2019-03-12T11:28:52.130" Title="Making sense of the lomb-scargle periodogram" Tags="&lt;observational-astronomy&gt;&lt;fundamental-astronomy&gt;&lt;astropy&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29957" PostTypeId="1" CreationDate="2019-03-12T03:43:15.947" Score="3" ViewCount="151" Body="&lt;p&gt;A natural, tiny (meters-size, maybe 10.000kg mass) natural satellite could be trapped in a geostationary orbit. I wondered for quite some time:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;When and how are we able to detect these satellites?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I assume the distance of 35.700km of way too far for naked eye detection. So the next realistic opportunity would have been &lt;a href=&quot;https://en.wikipedia.org/wiki/Galileo_Galilei&quot; rel=&quot;nofollow noreferrer&quot;&gt;Galilei&lt;/a&gt; who first used telescopes for scientific observations of the sky. Could he have detected such an object?&lt;/p&gt;&#xA;&lt;p&gt;Now even if he had enough resolution, he certainly didn't systematically cover the whole &lt;strong&gt;4&lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; of the sky&lt;/strong&gt; (especially as the satellite could be syncronous with the other side of the earth, and he could never see it).&lt;/p&gt;&#xA;&lt;p&gt;Then, would it be at the times of &lt;a href=&quot;https://en.wikipedia.org/wiki/Edwin_Hubble&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble&lt;/a&gt; (because of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Mount_Wilson_Observatory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mount Wilson Observatory&lt;/a&gt; and other similarly powerful telescopes)? Would it at the times when quality equipment for &lt;a href=&quot;https://en.wikipedia.org/wiki/Amateur_astronomy&quot; rel=&quot;nofollow noreferrer&quot;&gt;hobby astronomers&lt;/a&gt; became cheap and therefore widespread enough (to cover huge areas of the sky)? Or would we - until today - not be able to detect such objects?&lt;/p&gt;&#xA;&lt;p&gt;To answer this question, one needs to consider both the technical ability as well as the area of the sky that is covered.&lt;/p&gt;&#xA;" OwnerUserId="2662" LastEditorUserId="7982" LastEditDate="2020-08-20T01:08:00.383" LastActivityDate="2020-08-20T01:08:00.383" Title="When would we detect a tiny meter size natural satellite in a geostationary orbits?" Tags="&lt;observational-astronomy&gt;&lt;amateur-observing&gt;&lt;natural-satellites&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29958" PostTypeId="1" AcceptedAnswerId="29966" CreationDate="2019-03-12T04:47:45.900" Score="6" ViewCount="265" Body="&lt;p&gt;My question is based in this Wikipedia article about the Jupiter magnetic field:&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetosphere_of_Jupiter&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Magnetosphere_of_Jupiter&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter magnetic field generates plasma and accelerates ionized molecules, eventually producing collisions. &lt;B&gt;Is anti matter generated naturally under this scenario?&lt;/B&gt;&lt;/p&gt;&#xA;" OwnerUserId="24506" LastActivityDate="2019-03-12T20:40:06.907" Title="Anti matter in Jupiter magnetic field?" Tags="&lt;jupiter&gt;&lt;magnetic-field&gt;&lt;antimatter&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29959" PostTypeId="2" ParentId="29951" CreationDate="2019-03-12T10:55:03.227" Score="8" Body="&lt;p&gt;It's long been known that Earth's atmosphere gradually reduces in density, and is detectable to very high altitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19770009539.pdf&quot; rel=&quot;noreferrer&quot;&gt;1976 atmospheric model&lt;/a&gt; has data up to 1000 km of altitude. At that altitude, atmospheric density is on the order of 10&lt;sup&gt;-15&lt;/sup&gt; kg/m&lt;sup&gt;3&lt;/sup&gt;. This is already better than the best vacuum we can create on Earth's surface. So we can call this an atmosphere (because it contains gases also found on Earth at ground level), but its consequences are limited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's sparse enough that at this altitude, atmospheric drag on satellites is negligible for the lifetime of a satellite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth's atmosphere loses some mass to space: at high altitude some atoms achieve escape velocity. This is replenished by evaporation and outgassing on Earth. For example, helium gas starts its life as alpha particles from uranium decay, the gas seeps through the ground, rises through the atmosphere (because it's much lighter than air) and eventually escapes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We already know with very high accuracy that the Moon slowly moves away from us as it's slowed down. Most of that slowdown is caused by tidal effects, but there will also be an atmospheric component. Due to the very low density, this component will be small. &lt;/p&gt;&#xA;" OwnerUserId="361" LastActivityDate="2019-03-12T10:55:03.227" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29961" PostTypeId="2" ParentId="29956" CreationDate="2019-03-12T11:28:52.130" Score="2" Body="&lt;p&gt;I think I see what is going on. Your best fit is actually a beat frequency between the true frequency (0.005) and double the sampling frequency (0.05). This does indeed produce a model that goes through your data points but because you have only plotted 50 points in the model you haven't been able to see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you change this line&#xA;t_fit = np.linspace(x[i][0], math.floor(x[i][-1]),num =1000)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so that it shows the model at more points you can see that it is indeed at a much higher frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick is to know (roughly) the answer before you start. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you limit your frequency range in the Lomb-Scargle command to below the Nyquist frequency:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;frequency, power = LombScargle(x[i],y[i],errors[i]).autopower(nyquist_factor=1.5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will find the behaviour you expected. e.g. (NB I reduced the error bar size to focus in on your problem, but also fixed the error bar size to a single &lt;strong&gt;positive&lt;/strong&gt; value. Your code was introducing very small and even negative errors and I do not know how the code would deal with those).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/k54Xn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k54Xn.png&quot; alt=&quot;Periodograms&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In unevenly spaced data, things are not so simple because the Nyquist frequency is not well defined.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-03-12T11:28:52.130" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29962" PostTypeId="1" AcceptedAnswerId="29964" CreationDate="2019-03-12T14:11:05.907" Score="3" ViewCount="86" Body="&lt;p&gt;Back in 2016, &lt;a href=&quot;http://adsabs.harvard.edu/abs/2016Natur.537..374M&quot; rel=&quot;nofollow noreferrer&quot;&gt;Marsh et al.&lt;/a&gt; reported that the binary system &lt;a href=&quot;https://en.wikipedia.org/wiki/AR_Scorpii&quot; rel=&quot;nofollow noreferrer&quot;&gt;AR Scorpii&lt;/a&gt; exhibits complex radio signals similar to those observed from traditional pulsars. In particular, pulsed synchrotron emission appears to be produced by interactions between the magnetospheres of a white dwarf and a red dwarf, powered in part by the spin-down of the white dwarf. The system has been dubbed the first &quot;white dwarf pulsar&quot; based on the signals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system is thought to be similar to, if not an extreme case of, &lt;a href=&quot;https://en.wikipedia.org/wiki/Intermediate_polar&quot; rel=&quot;nofollow noreferrer&quot;&gt;intermediate polars&lt;/a&gt;; we know of a number of these systems. Therefore, I'm wondering: Have there been any concerted searches for other intermediate polars exhibiting this pulsed emission, and if so, have any candidates been discovered (possibly serendipitously)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is partly belatedly in honor of &lt;a href=&quot;https://astronomy.meta.stackexchange.com/q/494/2153&quot;&gt;February's focus tag&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2019-03-12T19:10:24.003" Title="Have any more &quot;white dwarf pulsars&quot; been discovered or searched for?" Tags="&lt;radio-astronomy&gt;&lt;white-dwarf&gt;&lt;pulsar&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29963" PostTypeId="2" ParentId="29957" CreationDate="2019-03-12T18:30:45.300" Score="4" Body="&lt;p&gt;To first order:  the ratio of the Moon's radius to distance from Earth is&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \frac{1740e3}{380e6}  =  0.004578947 $&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;and the ratio of a 5-m radius satellite at geosync orbit is roughly&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{5}{36e6 } =  1.388889e-07 $&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;This means, for similar albedo, the light reaching your telescope (or eye) would be&#xA;&lt;span class=&quot;math-container&quot;&gt;$(\frac{1.388889e-07}{0.004578947})^2 =  9.200339e-10 $&lt;/span&gt; as much light as the full moon.  You're not going to see it even with a good telescope.&lt;/p&gt;&#xA;&lt;h2&gt;edit&lt;/h2&gt;&#xA;&lt;p&gt;As the comment points out, I was too glib there.  If you knew where to look, a decent 20 cm (aka 8-inch) telescope can easily show an object of that apparent magnitude.  The nice thing about geostationary is that you can spend plenty of nights sweeping the possible sky regions; the satellite won't move.&lt;/p&gt;&#xA;" OwnerUserId="907" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-03-12T23:59:31.430" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29964" PostTypeId="2" ParentId="29962" CreationDate="2019-03-12T19:10:24.003" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2017IJMPS..4560024O&quot; rel=&quot;nofollow noreferrer&quot;&gt;Edson et al. (2017)&lt;/a&gt; list two other candidates that may be white dwarf pulsars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/AE_Aquarii&quot; rel=&quot;nofollow noreferrer&quot;&gt;AE Aquarii&lt;/a&gt; has been described as pulsar-like (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/1998A&amp;amp;A...338..521I&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ikhsanov 1998&lt;/a&gt;) and was the best case for a white dwarf pulsar before AR Scorpii came along. From what I understand, the case for AE Aqr containing a white dwarf pulsar is not watertight. &lt;a href=&quot;https://arxiv.org/abs/1804.10105&quot; rel=&quot;nofollow noreferrer&quot;&gt;Blinova et al. (2018)&lt;/a&gt; explain the high spin-down luminosity as being the result of interactions with an accretion disc and the white dwarf having a magnetic field typical of that in &lt;a href=&quot;https://en.wikipedia.org/wiki/Intermediate_polar&quot; rel=&quot;nofollow noreferrer&quot;&gt;intermediate polars&lt;/a&gt;, rather than a pulsar-like mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ui.adsabs.harvard.edu/?#abs/2005ApJ...631L.143Z&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zhang &amp;amp; Gil (2005)&lt;/a&gt; proposed that the radio source &lt;a href=&quot;https://en.wikipedia.org/wiki/GCRT_J1745-3009&quot; rel=&quot;nofollow noreferrer&quot;&gt;GCRT J1745-3009&lt;/a&gt; is caused by a white dwarf pulsar. As far as I can tell, the case for this one isn't particularly secure either because there are other possible explanations. Follow-up observations have not yet confirmed the nature of the source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not aware of any specific surveys for such objects but so far it seems that AR Scorpii is the best example of a white dwarf pulsar.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastActivityDate="2019-03-12T19:10:24.003" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29965" PostTypeId="2" ParentId="25384" CreationDate="2019-03-12T20:19:37.360" Score="2" Body="&lt;p&gt;The Earth and Moon are in orbit about each other, which means that the centrifugal (outward inertial) force &lt;span class=&quot;math-container&quot;&gt;$M_iV_i^2/d_i$&lt;/span&gt; is balanced with the centripetal (real inward force), ie &lt;span class=&quot;math-container&quot;&gt;$GM_{\oplus}M_{\mathrm{Moon}}/(d_\oplus+d_{\mathrm{Moon}})^2$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$d_\oplus$&lt;/span&gt; is distance from Earth to Earth-Moon center of mass (c.m.) and &lt;span class=&quot;math-container&quot;&gt;$d_{\mathrm{Moon}}$&lt;/span&gt; is the Moon's c.m. distance.  Thus, most of the centrifugal force of the motion about the c.m. is canceled. Since we reside at one Earth radius from the center of the Earth, the cancellation is not exact, and the remainder is exactly what we call the Lunar Tide, i.e. the familiar tidal force of the Moon which causes most of the ocean tides (there is also a component from the orbit about the Sun).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tidal acceleration is described in &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_acceleration&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2019-03-12T20:19:37.360" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29966" PostTypeId="2" ParentId="29958" CreationDate="2019-03-12T20:40:06.907" Score="3" Body="&lt;p&gt;Yes it is. See this 2012 blog post by Paul Gilster:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.centauri-dreams.org/2012/05/17/powering-up-the-antimatter-engine/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.centauri-dreams.org/2012/05/17/powering-up-the-antimatter-engine/&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says things like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Antimatter in space is an idea that James Bickford (Draper Laboratory) analyzed in a Phase II study for NASA’s Institute for Advanced Concepts, for he had realized that high-energy galactic cosmic rays interacting with the interstellar medium (and also with the upper atmospheres of planets in the Solar System) produce antimatter. In fact, Bickford’s calculations showed that about a kilogram of antiprotons enter the Solar System every second, though little of this reaches the Earth. To harvest some of this incoming antimatter, you need a planet with a strong magnetic field, so Jupiter is a natural bet for Baxter’s scientists, who go there to forage. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But it also says Saturn's a better bet, and that Earth is handier, along with this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The problem with this — and this has been noted by The Physics &lt;a href=&quot;http://www.technologyreview.com/blog/arxiv/27847/&quot; rel=&quot;nofollow noreferrer&quot;&gt;arXiv Blog&lt;/a&gt; and Jennifer Ouellette in recent days — is that PAMELA could come up with only 28 antiprotons over the course of 850 days of data acquisition. There is no question that Bickford is right in seeing how antimatter can be produced locally. In fact, the paper on the PAMELA work says this: “The ﬂux exceeds the galactic CR antiproton ﬂux by three orders of magnitude at the current solar minimum, thereby constituting the most abundant antiproton source near the Earth.” But does the process produce enough antimatter to make local harvesting a serious possibility?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want some antimatter, buy a banana. I kid yet not. See the Symmetry article &lt;a href=&quot;https://www.symmetrymagazine.org/2009/07/23/antimatter-from-bananas&quot; rel=&quot;nofollow noreferrer&quot;&gt;antimatter from bananas&lt;/a&gt;. Flip Tanedo says this: &lt;em&gt;“While researching natural sources of antimatter, I discovered a curious article about a naturally occurring potassium isotope that, some fraction of the time, decays via positron emission. The conclusion was that: The average banana (rich in potassium) produces a positron roughly once every 75 minutes”&lt;/em&gt;. Better still, buy a whole bunch of bananas! &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2019-03-12T20:40:06.907" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29968" PostTypeId="2" ParentId="13841" CreationDate="2019-03-12T21:34:07.683" Score="2" Body="&lt;p&gt;The black holes will either merge or they won't. If they merge, they just become a bigger black hole, and we're back to square one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they don't merge, then...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When the black holes are closest to each other you should find&#xA;  yourself outside of the event horizon for a moment as the influence of&#xA;  one black hole should cancel out some part of the influence of the&#xA;  another black hole ( in similar fashion as the influences cancel out&#xA;  in the exact middle of a planet ); therefore, enabling you to escape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no similarity there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most people are fixated on a phrase, often repeated in popular science, that &quot;not even light can escape from a black hole&quot;, which leads people to believe that if you could somehow go faster than light, or if the bond with the black hole was a bit less strong, you could escape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, you can't. That bond is not a finite amount. It's effectively infinite (emphasis on &quot;effectively&quot;, because there's a trick here). Let me explain what I mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside the event horizon you're effectively as if in a different universe. That's a chunk of spacetime completely cut off from this universe. The reason why you can't escape is not because the pull is &quot;too strong&quot; or you can't move &quot;fast enough&quot;. The real reason is that &lt;strong&gt;there is no path from inside the event horizon to the outside&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick any point inside the event horizon. Draw a straight line from there in any direction. Keep drawing. Eventually the line will end at the center of the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, wrong direction. Pick another direction. Start drawing another straight line. Keep drawing. Ooops, hit the center again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to the initial point and pick a completely different direction. Draw the straight line. Damn, hit the center again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start over, but now scribble some randomly-shaped, winding line. Holy cow, I hit the center again!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's how it is. Spacetime inside the event horizon is closed into itself, will all possible trajectories ending at the center of the black hole. There is no possible escape route. No matter what, you will end up in the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bringing in another black hole makes no difference, because the two black holes will merge, first their event horizons, then they will complete the merger. At any stage in this process, the same rule applies to their common event horizon: there is no path from inside that can break through to the outer universe.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2019-03-12T21:34:07.683" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29970" PostTypeId="1" AcceptedAnswerId="29972" CreationDate="2019-03-12T22:16:17.323" Score="-2" ViewCount="100" Body="&lt;p&gt;Can our Sun's magnetic field strengthen in any way in a short period? Can the Sun pull anything trapped in close orbit around it? What would happen if an object in orbit passes over a sunspot for example?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="1891" LastEditDate="2019-03-18T07:58:35.943" LastActivityDate="2019-03-18T07:58:35.943" Title="Can the Sun pull an iron asteriods in it's orbit into it?" Tags="&lt;the-sun&gt;&lt;astrophysics&gt;&lt;asteroids&gt;&lt;magnetic-field&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2019-03-13T23:39:24.483" ContentLicense="CC BY-SA 4.0" />
  <row Id="29971" PostTypeId="1" AcceptedAnswerId="30466" CreationDate="2019-03-13T04:57:40.530" Score="5" ViewCount="393" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$v_\mathrm e=\sqrt{\frac{2GM}r}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we understand the escape velocity to be the speed needed to escape from the 'surface' of a gravitational object described by the above equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is equally true that the escape velocity is the speed an object pulled from rest will reach when it hits a certain point in the gravitational field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the event horizon is defined as when the escape velocity reaches the speed of light for a black hole, would matter accelerate near the speed of light as it approaches the singularity? Accretion disks are known to move at relativistic speeds near the event horizon, but perhaps oddities occur beyond this point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The event horizon can be derived from setting the escape velocity of an object to the speed of light, it is a point of no return. I understand there are better definitions, but this is not really germane to the question at hand. How about from the perspective of another in-falling observer?&lt;/p&gt;&#xA;" OwnerUserId="26226" LastEditorDisplayName="user8021" LastEditDate="2019-04-17T15:09:15.447" LastActivityDate="2019-04-17T15:09:15.447" Title="Does matter accelerate to the speed of light as it approaches the singularity?" Tags="&lt;black-hole&gt;&lt;general-relativity&gt;&lt;escape-velocity&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29972" PostTypeId="2" ParentId="29970" CreationDate="2019-03-13T06:00:53.357" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can our Sun's magnetic field be able to pull anything out of orbit&#xA;  closer than Mercury if it passes over a sunspot for example?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Almost nothing orbits as close as sunspots.    Sunspots are impressively large.  Large enough that the Earth can pass through a sunspot.  The larger ones approach the size of Jupiter, but that's still just 1/10th the diameter of the Sun.    When sunspots lead to coronal mass ejections, the ejected material and the height of the storm, prior to ejection can reach higher, but the majority of sunspots extend only a fraction of 1 solar radius from the Sun's surface.   That's insanely close for an orbit.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can our Sun's magnetic field be able to pull anything out of orbit&#xA;  closer than Mercury if it passes over a sunspot for example?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's three problems with this.   The first problem is that the Sun's magnetic field at any reasonable distance, becomes mostly insignificant.    Sunspots are local, twisted magnetic fields with little effect at high distance.   Their negative and positive poles are on the surface of the sun and from Mercury distance, very close to one another.  Perhaps 10 or 20 arc-minutes apart at maximum, a fraction of 1 degree.   Any magnetic force at that distance would be negligible.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem is that the Sun is a much more powerful gravity machine than it is a magnet.  By the time you get close enough for the Sun's magnetism to have significant effect, the Sun would fill half the sky.   It would be a bad day being that close to the sun.   Asteroids would be vaporizing under the immense radiation.  Free ions would be affected by magnetism, but the overall object would be more affected by gravity and heat from radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 3rd problem is that nearly everything in space is a weak magnet.   Even Jupiter, by mass, is a relatively weak magnet.   Asteroids and comets may have some natural magnetism to them, but not a lot.  Planets may induce a magnetic field, but relative to their mass, it's usually insignificant.    The magnetic interaction between stellar objects isn't very significant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for charged particles, it's very significant and many of the particles that fly through space are charged.   A charge particle is considerably affected by magnetic fields, but anything of any size, much less so.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, just for fun, if we were to send some neodymium magnets orbiting the sun inside the orbit of Mercury . . . . well, the heat might destroy the magnets, but lets ignore that for now.   Would those theoretical, artificial magnetic satellites indicate bumps in their orbit when they flew over a sunspot?    Perhaps.  I think at mercury distance the effects would be pretty small.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of the sunspot created orbital bumps would be dependent on the strength of the magnet relative to it's mass, so Neodymium magnet satellites would be a good choice to measure sunspot driven bumps in their orbits.   I can't run the math for you, but if you pick a magnet with a magnetism and a mass and an orbital distance and you define the size and power of the sunspot, maybe someone here could calculate the effect.   I still think it'd be pretty tiny.    When you talk about meteors, which would either be non-magnetic or weak magnets, the effect would probably be minuscule.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2019-03-13T06:00:53.357" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29973" PostTypeId="2" ParentId="29957" CreationDate="2019-03-13T06:10:59.123" Score="6" Body="&lt;p&gt;&lt;strong&gt;tl;dr:&lt;/strong&gt; At distances far enough from Earth that the motion with respect to the stars was slow, a serendipitous survey photographic plate from a large enough telescope might catch a trail, and &lt;em&gt;in a doubly serendipitous situation&lt;/em&gt; it might have been a short exposure, duplicated on the next night, an Earth orbit suspected, and a hunt for a second Earth satellite begun.&lt;/p&gt;&#xA;&lt;p&gt;However starting in the 1960's and 1970's radar and visual scans for artificial satellites in Earth orbit would have found this natural satellite in Earth orbit if it were low enough.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;I'll start with @CarlWitthoft's 5 meter asteroid refer to &lt;a href=&quot;https://space.stackexchange.com/a/25121/12102&quot;&gt;this answer&lt;/a&gt; and especially &lt;a href=&quot;https://astronomy.stackexchange.com/a/26021/7982&quot;&gt;this answer&lt;/a&gt;. Two equivalent equations to get the absolute magnitude of an asteroid are:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ H = C - 5 \log_{10} D - 2.5 \log_{10} p_V$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; is absolute magnitude, &lt;span class=&quot;math-container&quot;&gt;$p_V$&lt;/span&gt; is albedo, D is in km, and &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; = 15.618, and&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M_{Abs} = 5 \left(\log_{10}(1329) -\frac{1}{2}\log_{10}(\text{albedo}) -\log_{10}(D_{km})\right).$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;A 5 meter diameter asteroid with an albedo of 0.1 has an &lt;strong&gt;absolute magnitude of +29.6&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The apparent magnitude &lt;a href=&quot;https://space.stackexchange.com/a/25121/12102&quot;&gt;from this answer&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Knowing the absolute magnitude of an object, you calculate the apparent magnitude &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; using:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ m = M_{Abs} + 5 \log_{10}\left(\frac{d_{SR} \ d_{RE}}{1 \ \text{AU}^2 O(1)}\right), $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$d_{SR}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$d_{RE}$&lt;/span&gt; are the &lt;s&gt;Sun-Roadster and Roadster-Earth&lt;/s&gt; Sun-satellite and satellite-Earth distances, each normalized by 1 AU, and the factor &lt;span class=&quot;math-container&quot;&gt;$O(1)$&lt;/span&gt; is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Bond_albedo#Phase_integral&quot; rel=&quot;nofollow noreferrer&quot;&gt;phase integral&lt;/a&gt;, of order unity, taking into account the angular difference between the direction of illumination and the direction of viewing. In an order of magnitude calculation, this only becomes really significant when the body moves between the Sun and the viewer. See &lt;a href=&quot;https://en.wikipedia.org/wiki/Absolute_magnitude#Solar_System_bodies_(H)&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Absolute_magnitude#Solar_System_bodies_(H)&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Let's pick two distances. One is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Geostationary_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;geostationary distance&lt;/a&gt; where the satellite would appear to hover above the observer, probably drifting up and down in a roughly &lt;a href=&quot;https://en.wikipedia.org/wiki/Analemma&quot; rel=&quot;nofollow noreferrer&quot;&gt;annelema shape&lt;/a&gt; because the Earth's oblanteness will eventually tilt the orbit. See &lt;a href=&quot;https://en.wikipedia.org/wiki/Geostationary_orbit#Orbital_stability&quot; rel=&quot;nofollow noreferrer&quot;&gt;Geostationary orbit; Orbital stability&lt;/a&gt;. It will have a distance to Earth as close as &lt;strong&gt;36,000 km.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The other is a low Earth orbit, but high enough that it will not decay due to drag too soon. Call it &lt;strong&gt;1000 km altitude&lt;/strong&gt;, or a circular orbit with a 7378 km semi major axis.&lt;/p&gt;&#xA;&lt;p&gt;Plugging all of this into the equation above, I get:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;     orbit                   closest distance      visual magnitude&#xA;Geosynchronous altitude         36,000 km                +20.6&#xA;Low Earth Orbit                  1,000 km                +16.7 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;In low Earth orbit the apparent magnitude is almost as bright as Pluto, but it's going to be moving pretty fast. &lt;span class=&quot;math-container&quot;&gt;$\sqrt(GM/a)$&lt;/span&gt; gives 7350 m/s, at a distance of 1000 km that's about 0.4 degrees per second. Any large telescope that's being used in astronomy will be tracking the motion of stars, or close to that, so it will be fast +17 magnitude track rather than a dot, and last only a fraction of a second. That probably wouldn't expose a photographic plate or if it did it would be dismissed as an artifact, meteor, or scratch. Visually it wouldn't be noticed.&lt;/p&gt;&#xA;&lt;p&gt;At GEO type distances and +20.6 magnitude, the object would move about 0.25 degree in a minute, so it might also be captured on a photograph, but by the time the plate was developed it would be impossible to know when it appeared in a long exposure. However if the exposure (say at the Hale 200 inch telescope) were short, it really is possible that at a short-term trajectory on the celestial sphere could be considered. The problem is that nobody would suspect that it was in Earth orbit, and they would extrapolate to a heliocentric orbit and never find it again.&lt;/p&gt;&#xA;&lt;p&gt;If the plate happened to be a series and there was another exposure of the same patch of sky the next night, &lt;em&gt;then they would see it again and get pretty suspicious that it was in Earth orbit&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;However, in a post-Sputnik cold war era radar and optical searches of the sky for objects in Earth orbit became particularly interesting.&lt;/p&gt;&#xA;&lt;p&gt;So I would say that satellite surveys (both optical and radar) in the 1960's and 1970's would be the first &lt;em&gt;likely candidates&lt;/em&gt; to find this 5 meter, 0.1 albedo satellite.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;For some insight on optical tracking, see the two videos linked in &lt;a href=&quot;https://space.stackexchange.com/q/27436/12102&quot;&gt;Are commercial communications satellites in GEO being constantly monitored by telescopes?&lt;/a&gt;. Currently these links will take you to a new tab with the YouTube video:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=8ebIAUjFfZM&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=8ebIAUjFfZM&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=4FXX1kSNljU&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=4FXX1kSNljU&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;If you would like to see them here, then leave a comment or answer or vote at &lt;a href=&quot;https://astronomy.meta.stackexchange.com/q/482/7982&quot;&gt;Interest in looking into adding the YouTube viewer?&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-07-19T05:46:10.220" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29974" PostTypeId="1" AcceptedAnswerId="29975" CreationDate="2019-03-13T07:55:03.713" Score="2" ViewCount="46" Body="&lt;p&gt;What is meant by the &quot;mini disk&quot; of a blackhole? Is it the same as accretion disk? If not, whats the difference? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg: In the abstract of this paper &lt;a href=&quot;https://arxiv.org/abs/1806.05697&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/1806.05697&lt;/a&gt; it is mentioned &quot;When the accretion rate is relatively high, the circumbinary disk, accretion streams, and mini-disks combine to emit light in the UV/EUV bands&quot;&lt;/p&gt;&#xA;" OwnerUserId="26229" LastEditorUserId="26229" LastEditDate="2019-03-13T08:17:08.527" LastActivityDate="2019-03-13T10:40:30.857" Title="Mini disk of a blackhole meaning" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;accretion-discs&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29975" PostTypeId="2" ParentId="29974" CreationDate="2019-03-13T10:40:30.857" Score="3" Body="&lt;p&gt;This paper, &lt;a href=&quot;https://arxiv.org/abs/1712.05451&quot; rel=&quot;nofollow noreferrer&quot;&gt;Quasi-Periodic Behavior of Mini-Disks in Binary Black Holes Approaching Merger&lt;/a&gt;, which shares authors with the paper linked in the question, sheds some light on the terminology.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We present the first magnetohydrodynamic simulation in which a&#xA;  circumbinary disk around a relativistic binary black hole feeds mass&#xA;  to individual accretion disks (&quot;mini-disks&quot;) around each black hole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are talking about active binary black hole systems. The system as a whole has an accretion disk, the circumbinary disk, and each black hole has its own accretion disk, which they call the mini-disk.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2019-03-13T10:40:30.857" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29976" PostTypeId="1" CreationDate="2019-03-13T11:43:50.950" Score="2" ViewCount="126" Body="&lt;p&gt;Please could you let me know a (basic) formula for calculating at what time a star would become visible of an evening in clear weather conditions.  I presume the formula would involve the stars apparent magnitude, the location of the observer, and how far the sun has dipped below the horizon (so, the sky's limiting magnitude).  Working with these three parameters, I guess there must be a simple enough formula, however I cannot find it!  Could you help with this?  For example, assuming clear weather and good eyesight, how could I calculate at what time the star Deneb will become visible on the 1st May&#xA;2019 for a rural location?&lt;/p&gt;&#xA;&lt;p&gt;I am asking this as I want to create an art piece where something happens each time a new star (or stars) becomes visible around twilight time....&lt;/p&gt;&#xA;" OwnerUserId="26230" LastEditorUserId="37191" LastEditDate="2021-02-12T09:09:51.920" LastActivityDate="2021-02-12T09:09:51.920" Title="How to calculate the stellar visibility?" Tags="&lt;observational-astronomy&gt;&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29977" PostTypeId="1" AcceptedAnswerId="29983" CreationDate="2019-03-13T13:30:46.277" Score="6" ViewCount="306" Body="&lt;p&gt;I read somewhere that the reason why Venus has a runaway greenhouse effect and Earth doesn't is the former's inability to form separate tectonic plates due to slightly higher temperature softening the then-cooling tectonic plate, preventing its fracture and thus disabling the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{CO}_2$&lt;/span&gt; scrubbing function of tectonic activity, which absorbs &lt;span class=&quot;math-container&quot;&gt;$\mathrm{CO}_2$&lt;/span&gt; from air by rock weathering and then puts it back into the mantle. And once &lt;span class=&quot;math-container&quot;&gt;$\mathrm{CO}_2$&lt;/span&gt; is a runaway in the atmosphere &lt;span class=&quot;math-container&quot;&gt;$\text{H}_2\text{O}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\text{SO}_2$&lt;/span&gt; will join forces with it multiplying the greenhouse effect. Therefore, the surface temperature of Venus rose to astonishing values long before life formed on Earth. Is this theory at least possibly correct? &lt;/p&gt;&#xA;" OwnerUserId="26231" LastActivityDate="2019-03-14T10:14:59.730" Title="Why Venus evolved so differently from Earth?" Tags="&lt;venus&gt;&lt;earth-like-planet&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29978" PostTypeId="1" AcceptedAnswerId="32423" CreationDate="2019-03-13T16:46:27.947" Score="17" ViewCount="655" Body="&lt;p&gt;First, the planets. We have Mercury, which is rocky, no atmosphere. But then we have Venus, which is completely different: thick atmosphere, very hot, geologically active. Then Earth - blue, full of water. Mars, the opposite: red like nothing else. Jupiter and Saturn are fairly similar. Then Uranus and Neptune, fairly similar but still differ in color between each other and also totally different in color than the two gas giants.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand: satellites. Let's analyze satellites of Jupiter and Saturn.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ganymede and Callisto fairly similar, but then Europe, total opposite: completely icy. And then Io, again something completely different: strikingly yellow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn's moons: mostly rocky, but then, something completely different: Titan, with a thick atmosphere like no other satellite and oceans of liquid methane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If during the formation of the Solar system there was a protoplanetary disk of matter, wouldn't it be pretty homogenous and therefore give rise to similarly looking planets? I understand that gas giants can't look the same as rocky planets, but why are there differences even between similarly sized rocky planets? Granted, there are wildly different temperatures throughout the Solar system, depending on the distance from the Sun, which probably explain some of the differences. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then what I especially don't understand are the differences between the satellites. If say Jupiter had a disk of matter orbiting it, which eventually formed into satellites, wouldn't at least that &quot;local&quot; disk around a planet be fairly homogenous? But nevertheless it developed into wildly different satellites. For example, how did the &quot;yellowy&quot; thing get concentrated on Io, and not equally distributed on all the Jupiter's moons?&lt;/p&gt;&#xA;" OwnerUserId="24234" LastActivityDate="2021-01-12T08:33:44.760" Title="Why do planets and satellites in the Solar system look so wildly different if they came from more or less the same matter?" Tags="&lt;planet&gt;&lt;solar-system&gt;&lt;natural-satellites&gt;&lt;planetary-formation&gt;&lt;proto-planetary-disk&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29979" PostTypeId="2" ParentId="10433" CreationDate="2019-03-13T18:58:28.150" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://www.e-rara.ch/zut/content/titleinfo/162514&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.e-rara.ch/zut/content/titleinfo/162514&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Astronomia nova ..., seu physica coelestis, tradita commentariis de motibus stellae martis ... / Johann Kepler&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;E.g., the Excel line 60:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;1586   DIE 23 NOUEMBRIS A.M.   H.6 M.15    Declin. [MS] B. vno  7  19 3/4  11  65&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PDF page 112: Book Chapter 11, page 65.  The topic is Mars position, Brahe measurements.&lt;/p&gt;&#xA;" OwnerUserId="26235" LastEditorUserId="7472" LastEditDate="2019-03-13T19:52:31.980" LastActivityDate="2019-03-13T19:52:31.980" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29980" PostTypeId="1" AcceptedAnswerId="29984" CreationDate="2019-03-13T19:20:21.523" Score="1" ViewCount="90" Body="&lt;p&gt;What positions would the Earth, Sun and viewer need to be to see a glory (a rainbow-like effect) from space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What position would they need to be in to reproduce this picture?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hDbLO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hDbLO.jpg&quot; alt=&quot;glory seen by MODIS&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://www.gizmodo.com.au/2012/06/wow-its-a-double-rainbow-from-space/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gizmodo&lt;/a&gt; (&lt;a href=&quot;http://earthobservatory.nasa.gov/IOTD/view.php?id=78380&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA&lt;/a&gt; via &lt;a href=&quot;https://www.universetoday.com/95995/pacific-glory/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Universe Today&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://earthscience.stackexchange.com/questions/16308/rainbow-dynamics&quot;&gt;https://earthscience.stackexchange.com/questions/16308/rainbow-dynamics&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://space.stackexchange.com/questions/34357/rainbow-space-probe&quot;&gt;https://space.stackexchange.com/questions/34357/rainbow-space-probe&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="7982" LastEditDate="2019-03-14T01:18:47.010" LastActivityDate="2019-03-15T01:02:58.700" Title="Was is the angle needed to see a glory (a rainbow-like effect) from space off the Earth?" Tags="&lt;light&gt;&lt;atmospheric-effects&gt;&lt;spectra&gt;" AnswerCount="1" CommentCount="7" ClosedDate="2019-03-15T17:31:56.290" ContentLicense="CC BY-SA 4.0" />
  <row Id="29981" PostTypeId="1" AcceptedAnswerId="29982" CreationDate="2019-03-13T22:16:09.820" Score="2" ViewCount="85" Body="&lt;p&gt;Orbits can follow certain patterns, as I am aware.  Some include a circular orbit around the bulk of the planet, a polar orbit spinning about the pole...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'd like to know, for a planet with two moons (one large one in a circular orbit at the equator, more or less), could the smaller moon, at a larger range but higher velocity, be pulled into an orbit that looks like a polar orbit for part of it's cycle, then transition towards a circular one before &quot;flipping&quot; back towards a polar orbit on the opposing pole, dancing between the three states either in sequence or seemingly at random as the smaller moon approaches certain criteria, being tugged on by the planet, the larger moon, the sun, maybe other planets in the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this feasible, or does it break the laws of orbital dynamics to have such a situation occur?  This is for a story I am writing centered on a planet other than Earth, but is relatively Earth-like in it's properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Size and density the same as Earth.  Orbit a bit tighter and faster than Earth's in relation to it's sun.  The smaller moon is perhaps 200 square miles in surface area, tidally locked stone and ice.  The larger is perhaps 42 million square miles in surface area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thoughts on this?  The basic idea is that this smaller moon, Eamor, is going to be seen in the night's sky almost like a star to the naked eye, racing across the skies as if scared of it's own shadow, sometimes disappearing under the horizon only to dart back up again.  Hence the name, which means, in the language of one culture on the planet, &quot;one who is terrified&quot;.&lt;/p&gt;&#xA;" OwnerUserId="26238" LastActivityDate="2019-03-14T20:58:42.583" Title="Various moon orbits" Tags="&lt;orbit&gt;&lt;natural-satellites&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="29982" PostTypeId="2" ParentId="29981" CreationDate="2019-03-13T23:07:56.090" Score="1" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polar orbits aren't what you think they are. A polar orbit is a (more or less) circular orbit that travels over &lt;em&gt;both&lt;/em&gt; poles. A satellite in a polar orbit doesn't hover over one pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An orbit with greater &quot;range&quot; (ie further from the planet) will always be slower. This is Kepler's third law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't get exactly what you want in real physics.  Can I suggest a simpler alternative: A moon that is (for various unlikely reasons) in a highly elliptical orbit and tilted orbit. It would appear to grow, and move dramatically (for a moon) across the sky during perigee. Then shrink and fade and move much more slowly for most of the time of the orbit. It can't cross the orbit of the other moon (or they will eventually collide). The regular approach followed by becoming smaller justifies gives it the appearance of &quot;fleeing&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2019-03-14T20:58:42.583" LastActivityDate="2019-03-14T20:58:42.583" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="29983" PostTypeId="2" ParentId="29977" CreationDate="2019-03-14T10:14:59.730" Score="8" Body="&lt;p&gt;Venus orbits the Sun at 0.723332 AU. As solar radiation follows an inverse square law,  the intensity of the Sun at the top of Venus's atmosphere is almost twice that for the Earth (&lt;span class=&quot;math-container&quot;&gt;$\frac1{0.723332}^2\approx1.9113$&lt;/span&gt;). Even when the solar system was young and the Sun was considerably dimmer than it is now, Venus received 1/3 more solar radiation at the top of its atmosphere than the Earth does now. Unlike the Earth, Venus did not suffer a faint young Sun problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That faint young Sun was perhaps instead a blessing for Venus. Many have argued that shortly after cooling off, Venus started as wet planet, like the Earth. The intense and increasing solar radiation eventually resulted in Venus undergoing a wet greenhouse process. Instead of being a trace gas in the Earth's atmosphere, water vapor may well have been the dominant gas in Venus's atmosphere. Thanks to the potency of water vapor as a greenhouse gas, this resulting in a vicious positive feedback loop. The increased water vapor raised Venus's surface temperature, resulting in even more evaporation/boiling. Meanwhile, at the top of the atmosphere, solar radiation dissociated water vapor. The hydrogen slowly escaped, eventually leaving Venus more or less free of water in any form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water may be necessary as a lubricant for plate tectonics. Whether or not this is the case, Venus does not have plate tectonics. It instead has a stagnant lid geology, punctured occasionally by extreme vulcanism (Siberian traps level vulcanism, and then some). This eventually released some of Venus's deep carbon reserve into its atmosphere. The end result is a planet void of water with a very thick carbon dioxide atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2019-03-14T10:14:59.730" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29984" PostTypeId="2" ParentId="29980" CreationDate="2019-03-14T13:27:46.580" Score="2" Body="&lt;p&gt;A glory is a rainbow-colored bullseye appearing directly opposite the Sun due to diffraction by water droplets of a certain size.&#xA;From an airplane, you can sometimes see this effect surrounding the airplane's shadow.&#xA;As the &lt;a href=&quot;https://earthobservatory.nasa.gov/images/78380/a-glorious-view&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA article&lt;/a&gt; explains, the polar orbiting &lt;a href=&quot;https://terra.nasa.gov/about&quot; rel=&quot;nofollow noreferrer&quot;&gt;Terra&lt;/a&gt; satellite's scan pattern transformed the circular rings into elongated ovals.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-03-15T01:02:58.700" LastActivityDate="2019-03-15T01:02:58.700" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29985" PostTypeId="2" ParentId="29908" CreationDate="2019-03-14T15:04:26.677" Score="9" Body="&lt;p&gt;Torricelli, the inventor of the Mercury Barometer (~1644) argued that the height of the column of mercury was governed by atmospheric pressure (the &quot;weight of the atmosphere&quot; as he would have put it).   He asserted that the space above the mercury in his tube was a vacuum, a totally anti-Aristotlean concept at that time.&lt;br&gt;&#xA;To test this he enlisted the help of Blaise Pascal and Florin Perier in carrying one of two barometers up a mountain, the Puy de Dome in central France. His prediction was that the pressure at the top of the mountain would be less than at the base because of a lesser weight of air.&lt;br&gt;&#xA;The experiment showed a drop in height (ie of pressure) and Torriceli himself concluded that the height of the atmosphere would be about 20km, above which there would be vacuum - the same vacuum as in the top of the barometer tube. &#xA;The experiment was IMHO one of the greatest experiments in the  history of physics.  See for example  &lt;a href=&quot;https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3768090/&quot; rel=&quot;noreferrer&quot;&gt;https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3768090/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5301" LastActivityDate="2019-03-14T15:04:26.677" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29986" PostTypeId="1" AcceptedAnswerId="30001" CreationDate="2019-03-14T17:29:32.000" Score="17" ViewCount="4710" Body="&lt;p&gt;What is the official degree to one decimal point please, of the earth’s rotation in one single day. Can it be confirmed that it is exactly 360.0 degrees using official data? Thank you in advance. &lt;/p&gt;&#xA;" OwnerUserId="26115" LastActivityDate="2019-11-08T22:00:30.700" Title="Official degrees of earth’s rotation per day" Tags="&lt;earth&gt;&lt;rotation&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29987" PostTypeId="2" ParentId="29986" CreationDate="2019-03-14T18:29:19.750" Score="22" Body="&lt;p&gt;This is a bit more complicated than it seems. First off, the definition of a day that matters to us earthlings is the average amount of time from one solar noon to the next (or alternatively, the time it takes for the Sun to appear above the same meridian from day to day); it is called a solar day. The sidereal day, which is the time it takes for some given distant star to appear above the same meridian from day to day, is not the one that really matters to us; this is also the amount of time it takes for the Earth to rotate 360 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the Earth is rotating on its axis, it is also travelling along its orbit. In about the amount of time it takes to complete one revolution, it has also travelled about one degree along its orbital path so that in order for the Sun to appear above the same meridian, the Earth has to rotate about 361 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then near perihelion (its closest approach to the Sun, which is around January) it's travelling even faster, so it has to rotate more than 361 degrees. Near aphelion it's travelling slower, so the Earth has to rotate less than 361 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your actual question, given the complexity of Earth's orbital variations, I'm not sure it's answerable.&lt;/p&gt;&#xA;" OwnerUserId="7108" LastActivityDate="2019-03-14T18:29:19.750" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29989" PostTypeId="1" AcceptedAnswerId="30003" CreationDate="2019-03-14T22:01:52.620" Score="7" ViewCount="296" Body="&lt;p&gt;This &lt;a href=&quot;https://astronomy.stackexchange.com/a/29983/7982&quot;&gt;answer&lt;/a&gt; provides some insight into Venus' surface geology:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Water may be necessary as a lubricant for plate tectonics. Whether or not this is the case, Venus does not have plate tectonics. It instead has a stagnant lid geology, punctured occasionally by extreme vulcanism (Siberian traps level vulcanism, and then some).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What are the observations that lead to this conclusion? The existence of plate tectonics on Earth was first determined by putting a lot of pieces of the first-hand observational puzzle together. There's much less data available from Venus.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-03-15T12:12:14.083" Title="How can it be known that Venus does not have plate tectonics?" Tags="&lt;venus&gt;&lt;planetary-formation&gt;&lt;planetary-science&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29991" PostTypeId="2" ParentId="29986" CreationDate="2019-03-14T22:17:33.817" Score="14" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can it be confirmed that it is exactly 360.0 degrees using official data?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR:&lt;/strong&gt; No, it can not. Instead it can be confirmed to be 361.0 degrees.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;To my knowledge:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The Earth's rotation period is very close to 23 hours, 56 minutes, 4.1 seconds or 86164.1 sec. That's called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth&quot; rel=&quot;noreferrer&quot;&gt;sidereal day&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A day is defined as 24 hours, or 86400.0 sec.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in one day it turns &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$360° \times \frac{86400.0}{86164.1} \approx 360.986°$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rounded &quot;to one decimal point please&quot; that's 361.0°.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Using &quot;Official data&quot;:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;NASA Earth Fact Sheet: &lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html&quot; rel=&quot;noreferrer&quot;&gt;https://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Sidereal rotation period (hrs)   23.9345    &#xA;Length of day (hrs)              24.0000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$360° \times \frac{24.0000}{23.9345} \approx 360.985°$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rounded &quot;to one decimal point please&quot; again, that's still 361.0°.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-03-14T22:17:33.817" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29992" PostTypeId="2" ParentId="29989" CreationDate="2019-03-14T22:30:12.830" Score="5" Body="&lt;p&gt;Essentially, it boils down to the question of uniformity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Magellan_(spacecraft)&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Magellan&lt;/em&gt;&lt;/a&gt; reached Venus in the early 1990s, and was able to greatly improve on previous mapping attempts. The spacecraft was able to map essentially all of the surface, including the distribution of craters. &lt;a href=&quot;http://planetary.brown.edu/planetary/documents/2875.pdf&quot; rel=&quot;noreferrer&quot;&gt;By measuring the crater density&lt;/a&gt;, scientists found that the surface was fairly uniform; given our knowledge of cratering on other inner Solar System bodies (e.g. the Moon), they were able to say that the surface is roughly 500 million years old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's more, that's a uniform figure - while there are variations across the planet, it's believed that they're due to local resurfacing events from volcanism. One possible explanation for the planet's particular age is that it underwent a period of global volcanic activity about 500 million years ago, and the entire crust was essentially reformed.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2019-03-14T22:30:12.830" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="29993" PostTypeId="1" CreationDate="2019-03-15T00:15:23.783" Score="0" ViewCount="73" Body="&lt;p&gt;A friend and I are having a friendly chat about this. They are stars so I'm claiming they still move in orbit. &lt;/p&gt;&#xA;" OwnerUserId="26245" LastActivityDate="2019-03-15T01:30:42.437" Title="Are red giants stationary or do they move?" Tags="&lt;star&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2019-03-16T10:04:35.323" ContentLicense="CC BY-SA 4.0" />
  <row Id="29994" PostTypeId="1" AcceptedAnswerId="29998" CreationDate="2019-03-15T01:23:00.353" Score="1" ViewCount="209" Body="&lt;p&gt;I find it funny that the moon is the only object in the solar system to almost perfectly fit the sun, resulting in the spectacular solar eclipses. No other object except maybe a few irregular moons like Pandora from Saturn would do that, and for them, they aren't round, and thus can't fully block out all light. The moon also has a perfect rotation that is almost exactly the same as Earth's rotational period, resulting in a tidal lock, which causes tides. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a coincidence? Any speculation on why this happens? Is this the only observed place where such criteria play into such perfect coincidence?&lt;/p&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-03-15T02:48:19.013" LastActivityDate="2019-03-15T03:00:37.757" Title="Why is the moon that special? Is it really a well built coincidence?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2019-03-15T09:28:55.490" ContentLicense="CC BY-SA 4.0" />
  <row Id="29995" PostTypeId="2" ParentId="29993" CreationDate="2019-03-15T01:27:07.413" Score="0" Body="&lt;p&gt;Let's review some things. We know stars usually exist in galaxies. When they are, they orbit a central black hole in the center. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, there are rogue stars. They have been ejected out of a galaxy. They must be travelling faster than the galaxy's escape velocity to do that, so they are also moving, but very, very fast. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's review some basic logic. Are red giants stars? Yes they are. They are stars in the giant phase when the star ceases to fuse H into He and fuses heavier elements, to wit, not in the main sequence anymore. &lt;em&gt;But they are still stars&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, it is safe to say all red giants move. &lt;/p&gt;&#xA;" OwnerUserId="25988" LastActivityDate="2019-03-15T01:27:07.413" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29996" PostTypeId="2" ParentId="29993" CreationDate="2019-03-15T01:30:42.437" Score="4" Body="&lt;p&gt;Everything moves.  Entire galaxies move.  Our galaxy is moving towards Andromeda at over 100 km/s and our entire local group is moving towards the great attractor even faster than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a less massive object orbits a more massive one, both move, but the less massive one moves more.  Often the calculation is simplified to make the central body unmoving, but that's a simplification and not accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our Milkyway has a center of mass with a giant black hole in that region.   If anything in our galaxy didn't move, it would slowly fall into the center.   Stars don't move like that, they move around each other while orbiting within the Milky-Way at high speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Movement, I should add, is defined from a certain point of view, so it's always possible to pick one thing and say &quot;this isn't moving&quot; and that would be correct from that thing's point of view, but that object still sits in spacetime (relatively speaking), and space time defines a force, which (shockingly) moves things, so even if you say &quot;Lets say this object isn't moving&quot;, the unevenness of space time will move it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, everything moves.   That's a universal law.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2019-03-15T01:30:42.437" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29997" PostTypeId="1" AcceptedAnswerId="30002" CreationDate="2019-03-15T02:21:57.530" Score="5" ViewCount="234" Body="&lt;p&gt;In the Kurzgesagt video &lt;a href=&quot;https://youtu.be/ZW3aV7U-aik&quot; rel=&quot;nofollow noreferrer&quot;&gt;Atoms As Big As Mountains — Neutron Stars Explained&lt;/a&gt; they state that neutron stars have an atmosphere that reaches about 10 centimeters above the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this possible? Wouldn't the strong surface gravity pull all of the material onto the surface of the star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, Kurzgesagt also states that the neutron star has a surface variation of about 5mm. Still, wouldn't gravity pull it flat? Of is the gravity not strong enough to do so?&lt;/p&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="7982" LastEditDate="2021-02-19T23:23:01.300" LastActivityDate="2021-02-19T23:23:01.300" Title="A neutron star without a flat surface and an atmosphere?" Tags="&lt;gravity&gt;&lt;neutron-star&gt;&lt;stellar-atmospheres&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29998" PostTypeId="2" ParentId="29994" CreationDate="2019-03-15T03:00:37.757" Score="1" Body="&lt;p&gt;If the &quot;perfect rotation&quot; you describe is the fact that same face of the moon always faces the earth, this is explained by the 'Giant Impact Hypothesis' which also explains the formation of the moon. The molten earth was impacted by a large mass causing a part of the liquid earth to be flung out into space creating the moon. For a time, the two celestial bodies to be (the earth and moon) were connected to one another by some bridge of mass until eventually the centripetal force of their rotation finally pulled them apart. In fact, the two bodies are still moving apart today. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Giant-impact_hypothesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Giant-impact_hypothesis&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26246" LastActivityDate="2019-03-15T03:00:37.757" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29999" PostTypeId="1" CreationDate="2019-03-15T04:36:27.893" Score="2" ViewCount="50" Body="&lt;p&gt;I noted that the moon drifts away from Earth a few centimeters each year. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm thinking that the emission of gravitational waves would deter such thing and cause the exact opposite. Furthermore, the orbital decay due to tidal forces should also slow the mon down, which allows the Earth to pull and being the moon closer. What mechanism is working to drag the moon further away from Earth?&lt;/p&gt;&#xA;" OwnerUserId="25988" LastActivityDate="2019-03-15T05:09:20.083" Title="Why is the moon drifting? Shouldn't orbital decay bring it closer?" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;earth&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2019-03-15T08:47:46.287" ContentLicense="CC BY-SA 4.0" />
  <row Id="30001" PostTypeId="2" ParentId="29986" CreationDate="2019-03-15T05:46:37.227" Score="31" Body="&lt;p&gt;First, we need to decide which definition of &quot;day&quot; to employ. There are several types of days:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Apparent solar day&lt;/strong&gt;: the time between two successive culminations of the Sun (apparent Noon) from an fixed Earth-based observer;&lt;br&gt;&#xA;&lt;strong&gt;Mean solar day&lt;/strong&gt;: a more uniform, averaged solar day without seasonal variations;&lt;br&gt;&#xA;&lt;strong&gt;Stellar/Sidereal day&lt;/strong&gt;: the time needed for the Earth to rotate once relative to the stars;&lt;br&gt;&#xA;&lt;strong&gt;SI day&lt;/strong&gt;: a unit of time containing exactly 86,400 SI seconds defined by caesium atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we generally refer to the traditional day/night cycle when we say &quot;day&quot;, this means a form of solar time. To get a more averaged value, let's use the mean solar day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current formula linking the Earth Rotation Angle (ERA) to the modern approximation of mean solar time, UT1 (basically the Earth's clock following the mean day/night cycle), is by definition :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ERA = 2π(0.7790572732640 + 1.00273781191135448 T_u) \text{ radians}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$Tu$&lt;/span&gt; is the Julian UT1 Date - 2451545.0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So according to this formula, a (UT1) day is 1.00273781191135448 Earth rotations, which multiplied by 360° is about &lt;strong&gt;360.98561°&lt;/strong&gt;. However, the Earth's rotation and revolution are not constant, and are always changing at somewhat unpredictable rates, so the angle is not perfect, but the changes are very slow. So this is more a modern approximation rather than an exact value. Rounded to one decimal place, this gives you &lt;strong&gt;361.0°&lt;/strong&gt;, a figure that will likely remain true for at least several millenia. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to know the amount of Earth rotation for every SI day, you're in luck: it is possible to consult reports of the Earth's orientation (rotation and polar motion) thanks to the IERS. Values are tabulated for each 0h UTC every day in the IERS publications, allowing to derive the angle that Earth has rotated every 86,400 SI seconds, allowing to scientifically monitor variations in Earth's rotation compared to a very constant unit of time realized by atomic clocks. Nowadays, the Earth's rotation is measured and reported thanks to radio telescopes and &lt;a href=&quot;https://en.wikipedia.org/wiki/Very-long-baseline_interferometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;VLBI&lt;/a&gt; observing distant objects in the universe. Official reports of its orientation, the &lt;a href=&quot;https://www.iers.org/IERS/EN/Science/EarthRotation/EOP.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth Orientation Parameters&lt;/a&gt; are published on the &lt;a href=&quot;https://www.iers.org/IERS/EN/DataProducts/EarthOrientationData/eop.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IERS Website&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, here is a graph showing the amount of Earth rotation every 86,400 seconds constructed with IERS data of the last year:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ICTX1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ICTX1.png&quot; alt=&quot;Earth Rotation&quot;&gt;&lt;/a&gt;&#xA;As we can see, there are several seasonal, periodic and unpredictable variations in Earth's rotation.&lt;/p&gt;&#xA;" OwnerUserId="19814" LastEditorUserId="19814" LastEditDate="2019-07-16T01:55:36.623" LastActivityDate="2019-07-16T01:55:36.623" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30002" PostTypeId="2" ParentId="29997" CreationDate="2019-03-15T07:19:49.067" Score="12" Body="&lt;p&gt;There is a number of issues with the question, but let me sketch out some kind of answer, so you get something out of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atmosphere of a neutron star is a topic that's a bit speculative. Estimates vary a lot. Regardless, neutron stars can have an atmosphere - sure, gravity is huge, but they are also extremely hot. Some molecules are bound to jump up a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X-ray measurements have shown the atmosphere of a certain neutron star to be 10 cm thick. That's not 10 meters, but it's better than nothing. Other neutron stars surely have different kinds of atmospheres, some thicker, others thinner. It looks like this is a matter for further research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://spacemath.gsfc.nasa.gov/weekly/6Page78.pdf&quot; rel=&quot;noreferrer&quot;&gt;https://spacemath.gsfc.nasa.gov/weekly/6Page78.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say the relief on the surface could be up to 5 mm height differential. Earth has mountains 8.8 km tall, and ocean holes 11 km deep - that's a 20 km differential, more or less. That's 4 milion times greater than 5 mm. Keep in mind, the surface material on a neutron star also has different properties from the normal stuff on Earth, different composition, and it's under extreme compression and extreme magnetic fields, which give it different mechanical properties. So 5 mm is quite believable. However, that is an extreme, and most surface should be smooth as glass. Any surface irregularity should be unstable in the long run.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the model, in some cases, extremely hot neutron stars might have surfaces that are more like a fluid than a solid. Emphasis on &quot;might&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of the above is set in stone. Neutron star physics is an area of active and intense research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend the book called Dragon's Egg by physicist Robert L. Forward. It's sci-fi, but the author was a bona fide scientist with extensive work in the field of gravitational wave detection. In his own words, Dragon's Egg is &quot;a textbook on neutron star physics disguised as a novel&quot;. It's fun to read, too, one of my favorite hard sci-fi novels.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2019-03-15T07:34:58.320" LastActivityDate="2019-03-15T07:34:58.320" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30003" PostTypeId="2" ParentId="29989" CreationDate="2019-03-15T12:12:14.083" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;There's much less data available from Venus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some data exists. As mentioned in HDE 226868's answer, maps of Venus's surface exist. Like Earth's atmosphere, Venus's atmosphere is transparent to some low frequency electromagnetic radiation such as those used by radar. These observations are consistent with a planet that has stagnant lid tectonics and inconsistent with a planet that has active plate tectonics; more on this below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to these remote observations, the Soviet Union successfully sent several spacecraft into Venus's atmosphere, some of which landed and briefly operated on the surface of Venus. The initial attempts failed because nobody thought the surface conditions on Venus would be as brutal as they are. To make landing successful, the Soviet Union had to significantly downsize the parachutes and they had to use materials and avionics that could withstand very high temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every piece of evidence gathered to date regarding Venus is inconsistent with a planet with active plate tectonics:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Venus surface temperature and pressure are well above water's critical point. Venus cannot have any liquid water on its surface. Water is widely (but not universally) thought to be critical as a lubricant that enables plate tectonics to occur.&lt;/li&gt;&#xA;&lt;li&gt;The radar observations show a planet with a nearly universal surface age, about half a billion years old. This is inconsistent with a planet with active plate tectonics but consistent with a planet with stagnant lid tectonics.&lt;/li&gt;&#xA;&lt;li&gt;Venus's atmosphere is very thick, much thicker than the Earth's atmosphere, and is dominated by carbon dioxide. Plate tectonics recycles carbon dioxide at subduction zones. Stagnant lid tectonics does not. A planet with plate tectonics will see a gradual reduction in the amount of carbon dioxide in its atmosphere over geologically long periods of time. A planet with intermittently active stagnant lid tectonics will instead see a gradual increase in the amount of carbon dioxide in its atmosphere over geologically long periods of time.&lt;/li&gt;&#xA;&lt;li&gt;Multiple physics-based models of a planet with very high surface temperatures suggest that such planets will have rather thin and rather ductile crusts that can readily repair themselves against damage caused by subsurface tensions.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note very well: The stagnant lid tectonics of hot terrestrial planets such as Venus  and possibly Titan (Titan is &quot;hot&quot; because its geology is ice-based rather than rock-based) is rather different from the stagnant lid tectonics of cold terrestrial planets such as the Moon and Mars. The surfaces of the Moon and Mars are very old. The surface of Venus is much younger in comparison. Venus has undergone at least one somewhat recent nearly global resurfacing event. This has not happened on the Moon or Mars. Plate tectonics appears to require a planet whose surface is neither too cold nor too hot, and that has a good amount of liquid water on the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some references:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;David Bercovici and Yanick Ricard, &quot;Plate tectonics, damage and inheritance,&quot; &lt;em&gt;Nature&lt;/em&gt; 508.7497 (2014): 513.&lt;br&gt;&#xA;DOI: &lt;a href=&quot;http://dx.doi.org/10.1038/nature13072&quot; rel=&quot;noreferrer&quot;&gt;10.1038/nature13072&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;A. Davaille, S. E. Smrekar, and S. Tomlinson, &quot;Experimental and observational evidence for plume-induced subduction on Venus,&quot; &lt;em&gt;Nature Geoscience&lt;/em&gt; 10.5 (2017): 349.&lt;br&gt;&#xA;DOI: &lt;a href=&quot;http://dx.doi.org/10.1038/NGEO2928&quot; rel=&quot;noreferrer&quot;&gt;10.1038/NGEO2928&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;James F. Kasting and David Catling, &quot;Evolution of a habitable planet,&quot; &lt;em&gt;Annual Review of Astronomy and Astrophysics&lt;/em&gt; 41.1 (2003): 429-463.&lt;br&gt;&#xA;DOI: &lt;a href=&quot;https://doi.org/10.1146/annurev.astro.41.071601.170049&quot; rel=&quot;noreferrer&quot;&gt;10.1146/annurev.astro.41.071601.170049&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mikhail A. Kreslavsky, Mikhail A. Ivanov, and James W. Head, &quot;The resurfacing history of Venus: Constraints from buffered crater densities,&quot; &lt;em&gt;Icarus&lt;/em&gt; 250 (2015): 438-450.&lt;br&gt;&#xA;DOI: &lt;a href=&quot;https://doi.org/10.1016/j.icarus.2014.12.024&quot; rel=&quot;noreferrer&quot;&gt;10.1016/j.icarus.2014.12.024&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ignasi Ribas et al., &quot;Evolution of the solar activity over time and effects on planetary atmospheres. I. High-energy irradiances (1-1700 Å),&quot; &lt;em&gt;The Astrophysical Journal&lt;/em&gt; 622.1 (2005): 680.&lt;br&gt;&#xA;DOI: &lt;a href=&quot;https://doi.org/10.1017/S0074180900182427&quot; rel=&quot;noreferrer&quot;&gt;10.1017/S0074180900182427&lt;/a&gt;.&lt;br&gt;&#xA;Accessible pdf: &lt;a href=&quot;https://iopscience.iop.org/article/10.1086/427977/pdf&quot; rel=&quot;noreferrer&quot;&gt;https://iopscience.iop.org/article/10.1086/427977/pdf&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2019-03-15T12:12:14.083" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30004" PostTypeId="1" CreationDate="2019-03-15T17:20:58.330" Score="2" ViewCount="54" Body="&lt;p&gt;I have a project in which a Monte-Carlo method is used to find the optimal parameters for some model that is fit to a data set. I want to fit a model for the orbit of Mercury using this methods, but I can't find any data to fit it to. I have found a lot of papers on the orbit of Mercury that mention observational data, but no actual observational data. Where could I find data indicating the position of Mercury around the sun over a time period, or even a relatively accurate mathematical model from which I can generate my own data by defining it as a function that is stepped through some time interval?&lt;/p&gt;&#xA;" OwnerUserId="26253" LastEditorUserId="1891" LastEditDate="2019-03-16T09:48:35.207" LastActivityDate="2019-03-16T21:36:13.473" Title="Where can I find data on Mercury's orbit to fit a model to?" Tags="&lt;observational-astronomy&gt;&lt;data-analysis&gt;&lt;orbital-elements&gt;&lt;mercury&gt;&lt;raw-data&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2019-03-16T12:25:31.593" ContentLicense="CC BY-SA 4.0" />
  <row Id="30005" PostTypeId="1" AcceptedAnswerId="30006" CreationDate="2019-03-15T19:59:42.727" Score="4" ViewCount="430" Body="&lt;p&gt;There's no way around it: when I look at the full moon at night it looks like a mostly flat disc, with at most a bit convexity in the middle. Phases of moon look like a full moon seen through a cut-out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is that the case? I recall, when I was really young, looking at it through what at the time seemed to be a very strong telescope, and it appeared not only a lot more textured but also a lot rounder. A similar thing can be seen in a video I saw a couple months ago, at 2:05: &lt;a href=&quot;https://www.youtube.com/watch?v=XCrJ3NflOpE&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=XCrJ3NflOpE&lt;/a&gt; . So it's clearly a naked eye thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I presume it's related to the optics, which made me hesitant to post it on this stack exchange (thinking it might be more suited to the physics one), but chose against it presuming people here would have more experience with celestial bodies and &quot;large&quot; distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a hunch that the answer is very simple, but I really don't know what that might be, especially since it's done over what I view as large distances.&lt;/p&gt;&#xA;" OwnerUserId="26254" LastActivityDate="2019-03-16T15:52:55.510" Title="Why does the Moon appear to be flat?" Tags="&lt;the-moon&gt;&lt;telescope&gt;&lt;optics&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30006" PostTypeId="2" ParentId="30005" CreationDate="2019-03-15T20:59:18.247" Score="10" Body="&lt;p&gt;It is an optical illusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We perceive nearby objects in 3d because we have two eyes. As we see objects from two different viewpoints, our brain can put the images together to make a 3d image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objects that are more distant perceived as 3d if they are moving fast enough for us to see them change appearance. Again it is not in the eyes but in the brain that the perception of depth occurs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon is a long way away, far too far for any 3d from our two eyes, and doesn't change fast enough for to see any motion. It is too small for us to see shadows or any other clues that we could interpret as 3d structure, so it looks like a small  flat shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Telescopes can make shadows visible, and time-lapse can make the motion of the moon apparent, and so add to the perception of a solid.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-03-15T20:59:18.247" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30007" PostTypeId="2" ParentId="30004" CreationDate="2019-03-16T02:29:24.353" Score="0" Body="&lt;p&gt;The available astrometric data for Mercury come from meridian transit timings (right ascension and declination) and radar or spacecraft ranging (distance).&#xA;NASA JPL has some&#xA;&lt;a href=&quot;https://ssd.jpl.nasa.gov/?eph_data&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;,&#xA;and the French IMCCE has some&#xA;&lt;a href=&quot;http://www.geoazur.fr/astrogeo/?href=observations/base&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&#xA;To use those, you'll have to work out some geometry including the position of Earth in its own orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your purpose it may be easier to use numerical ephemerides via&#xA;JPL &lt;a href=&quot;https://ssd.jpl.nasa.gov/?ephemerides&quot; rel=&quot;nofollow noreferrer&quot;&gt;HORIZONS or SPICE&lt;/a&gt;,&#xA;or IMCCE &lt;a href=&quot;https://www.imcce.fr/services/ephemerides/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Miriade or Multisat&lt;/a&gt;,&#xA;which can output Cartesian coordinates if requested.&#xA;Of course these are fit to observational data themselves.&#xA;The underlying data files (e.g. DE430, INPOP17a) come with papers detailing sources, methods, uncertainties, etc.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-03-16T21:36:13.473" LastActivityDate="2019-03-16T21:36:13.473" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30008" PostTypeId="2" ParentId="30005" CreationDate="2019-03-16T04:41:43.073" Score="4" Body="&lt;p&gt;In addition to James K's perception methods, one can make a stereo pair of images by taking photos at different times in the Moon's libration as explained in &lt;a href=&quot;http://physics.kenyon.edu/EarlyApparatus/Astronomy/Moon_Stereo/Moon_Stereo.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stereoscopic Pictures of the Moon&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.stereocamera.ca/StereoMoon.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Newportts&lt;/a&gt; has a nice modern example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8yh3G.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8yh3G.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;If you can do the eye-crossing trick you can see a 3D Moon. It is more impressive if you open the &lt;a href=&quot;http://www.stereocamera.ca/Graphics/StereoFullmoonS.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;image&lt;/a&gt; in a browser window and fullscreen it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The pairs in &quot;Stereoscopic Pictures of the Moon&quot; are meant to be viewed left image with left eye and right image with right eye, so if you use the eye-crossing trick you will see a concave Moon.)&lt;/p&gt;&#xA;" OwnerUserId="17743" LastEditorUserId="17743" LastEditDate="2019-03-16T15:52:55.510" LastActivityDate="2019-03-16T15:52:55.510" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="30009" PostTypeId="2" ParentId="30005" CreationDate="2019-03-16T15:20:57.733" Score="2" Body="&lt;p&gt;Our brains rely on a variety of clues to determine depth. The most powerful clue is binocular vision but that only works at short distances, it's irrelevent here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other clues include the relative brightness of different parts of the image, the size and shape of shadows and the curving of linear features, the brain puts these clues together and sees a ball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you look at the moon with the naked eye at night it's difficult to see much detail because of it's small size and high brightness (compared to the rest of the night sky). So your brain can't see the clues that tell it the moon is not flat.&lt;/p&gt;&#xA;" OwnerUserId="11128" LastActivityDate="2019-03-16T15:20:57.733" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30013" PostTypeId="1" CreationDate="2019-03-17T01:46:57.750" Score="3" ViewCount="78" Body="&lt;p&gt;As observed from the surface of Earth:&#xA;Will the moon ever repeat the same path (trajectory)?&#xA;If yes, what is the period of repetition?&#xA;And also will the moon cover the night sky with its path completely over a period of time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More explanation:&#xA;I was observing the moon from my window and I was wondering will the moon follow the same path (trajectory) as it did before, and if I am taking photos all the different positions of the moon over a long period of time will it have covered the night sky?&lt;/p&gt;&#xA;" OwnerUserId="26262" LastActivityDate="2019-03-17T14:30:39.510" Title="Moon's path as seen from Earth" Tags="&lt;the-moon&gt;&lt;observational-astronomy&gt;&lt;amateur-observing&gt;&lt;moon-phases&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30014" PostTypeId="1" CreationDate="2019-03-17T05:53:54.933" Score="5" ViewCount="85" Body="&lt;p&gt;It occurs to me, that between the surface and interior of neutron stars, gravitational pressure might produce super-heavy elements and that signatures of such elements might be detectable in the star's spectrum.&lt;/p&gt;&#xA;" OwnerUserId="26263" LastActivityDate="2019-03-18T07:26:16.667" Title="Is there evidence of super-heavy elements in the x-ray and gamma-ray spectrums of neutron stars?" Tags="&lt;neutron-star&gt;&lt;x-ray&gt;&lt;spectra&gt;&lt;gamma-rays&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30015" PostTypeId="2" ParentId="30013" CreationDate="2019-03-17T14:30:39.510" Score="3" Body="&lt;p&gt;There is a beautiful composite picture demonstrating exactly this over at &lt;a href=&quot;http://www.twanight.org/newTWAN/guests_photos.asp?ID=5001716&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.twanight.org/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/By7K2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/By7K2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a composite of 90 photos across the four days, so you can easily see how far the moon moves each day (about 13 degrees)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://cseligman.com/text/sky/moonmotion.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://cseligman.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;on average the Moon crosses the sky once every 24 hours and 49 minutes (53 minutes longer than the stellar &quot;day&quot;). As a result, it rises (and sets) later and later every day, until after about 27 days, when it has gone once around the sky relative to the stars, it is back in its original position, rising and setting at its original time(s). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="43" LastActivityDate="2019-03-17T14:30:39.510" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30016" PostTypeId="2" ParentId="30014" CreationDate="2019-03-17T16:35:32.553" Score="7" Body="&lt;p&gt;The crusts of neutron stars will contain &quot;super-heavy&quot;, neutron-rich nuclei. This is an inevitable consequence of the high density material, the accompanying degenerate electrons (that block &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt;-decay) and what we know about nuclear physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the only things that contribute to a neutron star's observable spectrum are materials within a few cm of the neutron star surface. It is expected that this will be made of highly ionised &quot;iron-peak&quot; nuclei (Fe, Co, Mn etc.) and perhaps hydrogen and helium that has been accreted from the interstellar medium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;super-heavy elements&quot; won't be present until the densities are much higher than in the surface layers. In a low-density material, these exotic nuclei will simply undergo a series of &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; decays or even fission. In high density materials, the electron Fermi energy can exceed the maximum possible decay energy of any beta electrons and stymie the decay process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. For a surface density of order &lt;span class=&quot;math-container&quot;&gt;$10^{9}$&lt;/span&gt; kg/m&lt;span class=&quot;math-container&quot;&gt;$^{3}$&lt;/span&gt; (at which point nothing very exotic is created), the electron number density will be about &lt;span class=&quot;math-container&quot;&gt;$3\times 10^{35}$&lt;/span&gt; m&lt;span class=&quot;math-container&quot;&gt;$^{-3}$&lt;/span&gt;.&#xA;Soft X-rays emitted at &lt;span class=&quot;math-container&quot;&gt;$10^{6}$&lt;/span&gt;K, will undergo Thomson scattering with a mean free path of &lt;span class=&quot;math-container&quot;&gt;$1/n_e\sigma \sim 50$&lt;/span&gt; nm. No (detailed) information about the interior can emerge in such circumstances.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-03-18T07:26:16.667" LastActivityDate="2019-03-18T07:26:16.667" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30017" PostTypeId="2" ParentId="29997" CreationDate="2019-03-17T16:49:36.297" Score="4" Body="&lt;p&gt;The scale height of a neutron star atmosphere is indeed of cm scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An isothermal atmosphere will decay as&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \rho = \rho_0 \exp( -kT/mgh)\ ,$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; is the density, &lt;span class=&quot;math-container&quot;&gt;$\rho_0$&lt;/span&gt; is the density at &lt;span class=&quot;math-container&quot;&gt;$h=0$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; is the &quot;height&quot; above this datum point, &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; is the surface gravity of the neutron star, &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; is the mass of the particles making up the atmosphere and &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is the surface temperature. The exponential scale height of the atmosphere is therefore &lt;span class=&quot;math-container&quot;&gt;$kT/mg$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very young neutron star might have a surface temperature as high as &lt;span class=&quot;math-container&quot;&gt;$T \sim 10^6$&lt;/span&gt; K. The gravity &lt;span class=&quot;math-container&quot;&gt;$g = GM/R^2 \sim 2 \times 10^{12}$&lt;/span&gt; m/s&lt;span class=&quot;math-container&quot;&gt;$^{-2}$&lt;/span&gt; for a &lt;span class=&quot;math-container&quot;&gt;$1.5 M_{\odot}$&lt;/span&gt; neutron star of radius 10 km. In the best-case scenario we can assume there is a thin layer of hydrogen present (accreted from the ISM) so &lt;span class=&quot;math-container&quot;&gt;$m = 1.67\times 10^{-27}$&lt;/span&gt; kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The characteristic exponential scale height is therefore &lt;span class=&quot;math-container&quot;&gt;$kT/mg \sim 0.4$&lt;/span&gt; cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This scale height will be considerably less in older, colder neutron stars or where the surface composition is just heavier elements (which might be the case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scale height will be modified by strong magnetic fields (in an uncertain way).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-03-17T16:49:36.297" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30019" PostTypeId="1" CreationDate="2019-03-17T17:45:37.737" Score="-3" ViewCount="169" Body="&lt;p&gt;Oumuamua is apparently a cigar-shaped asteroid that recently passed through the solar system on an unusual orbit, exhibiting unexpected acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oumuamua is known to have fairly minimal powers of thrust other than gravitational free fall, insufficient in themselves to quickly turn around and accelerate the body, but they could be sufficient to enable it to influence itself to harness the gravitational effects of surrounding planetary bodies.  Assuming favourable thrust conditions, what is the soonest we could expect it to return?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*The &lt;a href=&quot;https://www.nature.com/articles/s41586-018-0254-4&quot; rel=&quot;nofollow noreferrer&quot;&gt;extragravitational thrust&lt;/a&gt; is of unknown cause.  The most popular explanation is that Oumuamua has some cometary properties and that it is caused by cometary outgassing in proximity to the Sun, although no cometary tail is apparent.&lt;/p&gt;&#xA;" OwnerUserId="12000" LastEditorUserId="12000" LastEditDate="2019-03-18T12:03:25.400" LastActivityDate="2019-03-18T12:03:25.400" Title="What's the soonest Oumuamua could return?" Tags="&lt;asteroids&gt;&lt;navigation&gt;&lt;oumuamua&gt;" AnswerCount="1" CommentCount="7" ClosedDate="2019-03-17T22:10:57.453" ContentLicense="CC BY-SA 4.0" />
  <row Id="30020" PostTypeId="1" CreationDate="2019-03-17T17:52:00.660" Score="2" ViewCount="106" Body="&lt;p&gt;This &lt;a href=&quot;https://astronomy.stackexchange.com/a/30015/7982&quot;&gt;answer&lt;/a&gt; includes a stunning photo of the Moon and Jupiter &quot;&lt;a href=&quot;http://www.twanight.org/newTWAN/guests_photos.asp?ID=5001716&quot; rel=&quot;nofollow noreferrer&quot;&gt;Trails of the Shifting Moon&lt;/a&gt;&quot; by photographer &lt;a href=&quot;http://www.twanight.org/newTWAN/guests_gallery_c.asp?Guest=Jin%20Lu&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jin Lu&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The caption reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From the photographer: &quot;I spent 4 days capturing the trail of the moon from the balcony of my house. This is a composite of 90 pictures, showing the shift of the path of the waxing Moon from crescent phase to near first quarter. Random floating clouds decorated the trails. The interval between each day's sequence of pictures ranges from 1.5 to 2 min. Besides the Moon, the most obvious celestial object in the composite is Jupiter. The trail of Jupiter is from pictures of Jan 9, the day the moon appeared closest to Jupiter in the evening sky. Although Jupiter looks closer to the Moon on Jan 8, that is not how it actually appeared then.&quot; - Jin Lu&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm confused by two things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The Moon is shown for January 7, 8, 9, 10 and the caption says &quot;The trail of Jupiter is from pictures of Jan 9&quot;. I'd expect Jupiter to have contributions from all four days, not just one day. How can that be?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Although Jupiter looks closer to the Moon on Jan 8, that is not how it actually appeared.&quot; If I measure the distance in pixels from the top-left Jupiter to the top-left Jan 8 and 9 Moons, the Jan 8 Moon is closer. How could that not be true in reality then?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/By7K2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/By7K2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-03-18T14:51:42.710" Title="How to understand this multi-day exposure photograph of Jupiter and the Moon" Tags="&lt;the-moon&gt;&lt;photography&gt;&lt;jupiter&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30022" PostTypeId="2" ParentId="30020" CreationDate="2019-03-17T20:00:54.413" Score="4" Body="&lt;p&gt;Looking toward the western horizon from 33&amp;deg;N latitude, hour angle increases in the 11-to-5 o'clock direction, and declination increases in the 8-to-2 o'clock direction.&#xA;The Moon was at -2.8&amp;deg; declination on Jan 8 and +2.2&amp;deg; on Jan 9, while Jupiter was near -2.0&amp;deg; on both nights.&#xA;Where they overlap in hour angle, the Jan 9 Jupiter images appear closer to the Jan 8 Moon images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Moon to appear at similar hour angles on consecutive nights, the photographer must have started taking images ~40 minutes later each night, increasing Jupiter's hour angle ~10&amp;deg; per night.&#xA;On Jan 7 and 8, Jupiter was probably still above the ~20&amp;deg; high frame at the ends of those short series of exposures.&#xA;When the Jan 10 series began, Jupiter was likely already behind the building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scenario is reproducible in Stellarium if the location is in the southwestern US and times are evenings in January 2011.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-03-18T14:51:42.710" LastActivityDate="2019-03-18T14:51:42.710" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30023" PostTypeId="2" ParentId="30019" CreationDate="2019-03-17T21:23:36.547" Score="3" Body="&lt;p&gt;Oumuamua as an object is remarkable, because it has a positive net energy, which means it is not bound to the gravitational well of our sun. Therefore it will never return.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2019-03-17T21:23:36.547" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="30024" PostTypeId="2" ParentId="29609" CreationDate="2019-03-17T22:03:00.257" Score="4" Body="&lt;p&gt;Here are some additional comments to the previous answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is observed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;At the &lt;strong&gt;beginning of a solar cycle&lt;/strong&gt; (that lasts on average 11 years), sunspots appear around &lt;strong&gt;30 degrees of latitude&lt;/strong&gt; and progressively they appear &lt;strong&gt;lower and lower in latitude&lt;/strong&gt;. At the end of the cycle they appear close to the equator. This is summarized in the butterfly diagram that shows the latitude in function of time:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f658C.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f658C.gif&quot; alt=&quot;butterfly diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to explain it:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Sunspots&lt;/strong&gt; are regions of &lt;strong&gt;strong magnetic fields&lt;/strong&gt;. The following images (from SDO satellite) represent the magnetic fields at the surface of the sun (left side) and a white light image taken at the same moment (right side). It is obvious that each sunspot correspond to a strong magnetic field (black/white patch).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fNrFg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fNrFg.jpg&quot; alt=&quot;magnetogram&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/G1uMV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G1uMV.jpg&quot; alt=&quot;white-light image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In order to produce sunspots, the magnetic field lines have to rise up and to come out of the sun, this configuration is called '&lt;strong&gt;toroidal magnetic field&lt;/strong&gt;'. On the other hand the simple configuration with only a north and a south pole is called a 'poloidal magnetic field'. Both configuration are illustrated below:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rvoqx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rvoqx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is thought that the sunspot cycle is produced by an oscillation between the poloidal field and the toroidal one. The &lt;strong&gt;toroidal field&lt;/strong&gt; is then created around &lt;strong&gt;30 degrees of latitude&lt;/strong&gt; and a mechanism of &lt;strong&gt;conveyor belt is responsible to make the latitude of appearance progressively lower and lower&lt;/strong&gt;. The cycle ends (after ~11 years) when sunspots reach the equator (that corresponds to a low toroidal field and strong poloidal field). Afterwards a new toroidal field is created and the sunspots of the new cycle appear at 30 degrees of latitude. This is illustrated on the figure below.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HOYSA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HOYSA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we are currently close to a solar minimum (few sunspots and high poloidal field). The first sunspots of the next cycle should appear at 'high' latitude (~30 degrees).&lt;/p&gt;&#xA;" OwnerUserId="17423" LastEditorUserId="17423" LastEditDate="2019-03-17T22:13:53.913" LastActivityDate="2019-03-17T22:13:53.913" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30025" PostTypeId="1" AcceptedAnswerId="30084" CreationDate="2019-03-17T22:32:57.310" Score="2" ViewCount="428" Body="&lt;p&gt;After reading about the Earth's orbit and how it moves faster in January than it does in July (based on its distance from the Sun), I started to wonder...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How far does the Earth travel around the Sun during each month of the year?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the math behind this calculation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I'm not asking about the average monthly distance, but the unique distance traveled for each month.  &lt;/p&gt;&#xA;" OwnerUserId="7895" LastEditorUserId="14094" LastEditDate="2019-03-21T11:15:53.140" LastActivityDate="2019-04-16T11:07:57.070" Title="How far does the Earth travel around the Sun during each month of the year?" Tags="&lt;orbit&gt;&lt;earth&gt;&lt;time&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30026" PostTypeId="1" AcceptedAnswerId="30036" CreationDate="2019-03-17T23:25:18.813" Score="14" ViewCount="4520" Body="&lt;p&gt;It’s very puzzling that the moon landing had no stars in the background, the ISS clips have no stars in the background. I listened to multiple astronaut interviews speak on what it looks like up in space and about half of them speak of the “darkest black space”. I’m sure there is a very good explanation for this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is star light only visible through the medium of earth atmosphere? But once in the vacuum of space where there is no medium they disappear? What’s the explanation? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pgojT.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pgojT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uMbTP.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uMbTP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XCoDo.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XCoDo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/17rRa.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/17rRa.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zwlVl.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zwlVl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://m.youtube.com/watch?v=BI_ZehPOMwIu&quot; rel=&quot;noreferrer&quot;&gt;Minute 47-49 stars, press conference all three Apollo 11 astronauts &lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://m.youtube.com/watch?v=PtdcdxvNI1o&quot; rel=&quot;noreferrer&quot;&gt;BBC interview with Neil Armstrong only&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26115" LastEditorUserId="26115" LastEditDate="2019-03-18T22:28:46.267" LastActivityDate="2019-03-19T02:50:55.970" Title="Why are there no stars visible in cislunar space?" Tags="&lt;the-moon&gt;&lt;earth&gt;&lt;space&gt;&lt;star-systems&gt;" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="30027" PostTypeId="1" CreationDate="2019-03-17T23:50:27.417" Score="1" ViewCount="141" Body="&lt;p&gt;I know that objects with lots of gravity are capable of warping light or slowing time (an example being a black hole). I also know this is because of the theory of relativity, but I need someone to expand on that and explain why/how that works. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, please write the answer in somewhat simple terms, as I am not (yet) an expert on this stuff :)&lt;/p&gt;&#xA;" OwnerUserId="20518" LastActivityDate="2020-06-18T14:48:41.013" Title="Why does gravity make time go slower and warp light?" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;space-time&gt;&lt;space&gt;&lt;time&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30028" PostTypeId="1" CreationDate="2019-03-18T00:44:52.947" Score="0" ViewCount="74" Body="&lt;p&gt;The March 15, 2019 Phys.org article &lt;a href=&quot;https://phys.org/news/2019-03-nasa-spacecraft-explored-edges-early.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;A NASA spacecraft may have explored the edges of an early Mars sea in 1997&lt;/a&gt; shows a stylized map of an area on Mars where the &lt;a href=&quot;https://en.wikipedia.org/wiki/Mars_Pathfinder&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mars Pathfinder&lt;/a&gt; landed and uses colors to indicate where bodies of surface water might have once existed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are labeled as&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Northern ocean&lt;/li&gt;&#xA;&lt;li&gt;Inland Sea&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and I noticed they are capitalized. Do these reflect proper nouns and names of possible bodies of water that are recognized by planetary scientists or Martian cartographers (or should I have said Mars cartographers?)? Do these possible bodies of water have other names as well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jAIW4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jAIW4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-03-18T18:55:11.813" Title="Does Mars' &quot;Northern ocean&quot; have another name? What about the &quot;Inland Sea&quot;?" Tags="&lt;mars&gt;&lt;water&gt;&lt;naming&gt;&lt;planetary-science&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30029" PostTypeId="2" ParentId="30026" CreationDate="2019-03-18T00:50:11.937" Score="37" Body="&lt;p&gt;It is a matter of exposure and dynamic range. A sensor like a camera can only handle inputs in a certain range of intensities, and much of photographic skill (or smart presets) is about mapping the outside light onto this range so the details you care about show up rather than turn into white or black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take a picture of a brightly lit scene, in order to make out the details of the bright parts (such as a lunar landscape, the Earth, the ISS etc) you will have to adjust the exposure making faint objects like the stars too dim to see against a dark sky background. You could try to set the exposure to show the stars instead, but now the landscape and Earth would be too bright (and likely also mess up the picture by causing flaring). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can try to work around it by taking several pictures at different exposure levels and later digitally compositing them together. But this requires a lot of extra work.&lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2019-03-18T00:50:11.937" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="30030" PostTypeId="1" CreationDate="2019-03-18T04:55:32.700" Score="1" ViewCount="22" Body="&lt;p&gt;In looking at the available &lt;a href=&quot;https://nssdc.gsfc.nasa.gov/nmc/spacecraft/displayDataset.action?spacecraftId=1978-079A&quot; rel=&quot;nofollow noreferrer&quot;&gt;datasets described by NASA for the International Space-Earth Explorer&lt;/a&gt;, I am only see gamma ray burst data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was the platform capable of recording cosmic rays other than gamma rays? If so, where is the data?&lt;/p&gt;&#xA;" OwnerUserId="2800" LastActivityDate="2019-03-18T04:55:32.700" Title="Did the International Space-Earth Explorer detect cosmic rays other than gamma rays?" Tags="&lt;space-probe&gt;&lt;cosmic-ray&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30032" PostTypeId="1" AcceptedAnswerId="30058" CreationDate="2019-03-18T07:21:18.287" Score="1" ViewCount="137" Body="&lt;p&gt;I find he following &lt;a href=&quot;https://astronomy.stackexchange.com/a/30023/7982&quot;&gt;answer&lt;/a&gt; to the question &lt;em&gt;What's the soonest Oumuamua could return?&lt;/em&gt; unsatisfying and oversimplifying and the last five words &quot;Therefore it will never return.&quot; unsupported at a minimum.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Oumuamua as an object is remarkable, because it has a positive net energy, which means it is not bound to the gravitational well of our sun. Therefore it will never return.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I'd like to ask: *What natural mechanisms could lead to the unlikely case of a rogue asteroid or planet passing through our solar system twice?&quot;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Stick to accepted scientific facts and theories&lt;/li&gt;&#xA;&lt;li&gt;Let's give an ample time limit of 10 billion years between visits.&lt;/li&gt;&#xA;&lt;li&gt;For the purposes of this question, the solar system includes the Kuiper belt (50AU) but not the Oort cloud&lt;/li&gt;&#xA;&lt;li&gt;Don't include intervention by artificial means or intervention (divine or otherwise)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-03-18T07:27:34.827" LastActivityDate="2020-05-10T02:28:49.920" Title="What natural mechanisms could lead to the unlikely case of the same rogue asteroid or planet passing through our solar system twice?" Tags="&lt;orbital-mechanics&gt;&lt;mathematics&gt;&lt;rogue-planet&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30033" PostTypeId="1" AcceptedAnswerId="30048" CreationDate="2019-03-18T07:38:54.973" Score="3" ViewCount="80" Body="&lt;p&gt;As I understand it, objects falling into an old/inactive/dead/non-pulsar neutron star would be ripped apart, form an accretion disc and eventually emit a burst of radiation as they fall onto the surface of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this happened to:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;a comet&lt;/li&gt;&#xA;&lt;li&gt;a dwarf planet&lt;/li&gt;&#xA;&lt;li&gt;an ice giant&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;...roughly how much energy/radiation would each produce? Also, if possible, over what kind of time frame? I'm only looking for ballpark figures in all cases.&lt;/p&gt;&#xA;" OwnerUserId="26258" LastActivityDate="2019-03-18T20:05:25.817" Title="Radiation from Neutron Star impact?" Tags="&lt;neutron-star&gt;&lt;radiation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30034" PostTypeId="1" CreationDate="2019-03-18T08:09:37.677" Score="1" ViewCount="34" Body="&lt;p&gt;I read that, unlike MOND theory, &lt;a href=&quot;https://en.wikipedia.org/wiki/Tully%E2%80%93Fisher_relation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tully-Fisher relation&lt;/a&gt; is believed to result from galaxy formation in dark matter theory. What is the exact current status regarding the explanation of Tully-Fisher relation from galaxy formation process?&lt;/p&gt;&#xA;" OwnerUserId="24850" LastEditorUserId="616" LastEditDate="2019-03-18T15:30:41.497" LastActivityDate="2019-03-18T15:30:41.497" Title="Can galaxy formation process explain Tully-Fisher relation in dark matter theory?" Tags="&lt;galaxy&gt;&lt;dark-matter&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30035" PostTypeId="1" CreationDate="2019-03-18T12:31:24.940" Score="9" ViewCount="3808" Body="&lt;p&gt;Am I correct in saying that the fusion process of the sun is constant, i.e. X amount of fusion happens per day, more or less?&#xA;Why does this not speed up, i.e. one fusion event creates energy for two fusion events, etc.?&#xA;Does every collision of atom cause a fusion event, or is the probability small for a fusion event to happen thus it's not a runaway reaction?&#xA;I have heard that the probability for fusion event to happen is only 1 in 10&lt;sup&gt;12&lt;/sup&gt; for every collision.&lt;/p&gt;&#xA;" OwnerUserId="26275" LastEditorUserId="7472" LastEditDate="2019-03-18T16:27:15.383" LastActivityDate="2019-03-19T16:31:21.803" Title="Why doesn't the fusion process of the sun speed up?" Tags="&lt;the-sun&gt;&lt;stellar-astrophysics&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30036" PostTypeId="2" ParentId="30026" CreationDate="2019-03-18T12:33:27.927" Score="14" Body="&lt;p&gt;Anders's answer is entirely fine, but I'd like to add some extra information. As evidenced by the transcripts, reflected Earth light is quite strong even at this distance:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The earthshine coming through the window is so bright you can read a book by it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is, even with the lights turned off, it would probably be tricky to see the stars unless you turned in a way that didn't allow the earthshine through the windows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as the capsule comes into the shadow cast by the Moon (a pure accident - they didn't plan for the approach to go this way), there comes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Houston, it's been a real change for us. Now we are able to see stars again and recognize constellations for the first time on the trip. It's - the sky is full of stars. Just like the nightside of Earth. But all the way here, we have only been able to see stars occasionally and perhaps through the monocular, but not recognize any star patterns. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So for a few minutes, they did see &quot;the sky full of stars&quot;. Other than that, they've seen a few stars once in a while, but only singular, bright stars (perhaps also when looking in a way that minimized the brightness from the Earth and Sun):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Houston, it's been a real change for us. Now we are able to see stars again and recognize constellations for the first time on the trip. It's - the sky is full of stars. Just like the nightside of Earth. But all the way here, we have only been able to see stars occasionally and perhaps through the monocular, but not recognize any star patterns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The core of Anders's answer is still true, though. Exposure is the main problem here - both cameras and human eyes have a certain dynamic range, and even the brightest stars are entirely too dim in comparison with both the Sun, the Earth (in distance comparable to the Moon's distance from the Earth) and the Lunar surface (if you're in sunshine, as most of the mission was). A modern camera &lt;em&gt;might&lt;/em&gt; be able to take a HDR picture that would allow the stars to be visible at the same time as the Earth or the Sun, and it'd be quite easy to do if you could occlude the main light sources (the same way we do it when photographing the Sun's corona etc.). But technically, that would be a &quot;doctored&quot; image - taken at two different exposures and combined in a way that uses different exposures for different parts of the image.&lt;/p&gt;&#xA;" OwnerUserId="10656" LastActivityDate="2019-03-18T12:33:27.927" CommentCount="15" ContentLicense="CC BY-SA 4.0" />
  <row Id="30037" PostTypeId="2" ParentId="29976" CreationDate="2019-03-18T13:36:18.297" Score="-1" Body="&lt;p&gt;Maybe that piece shouldn't consider necessarily individual stars rising from the East but entire constellations/parts of constellations. Trying to express the sky movement as a whole sounds really nice to me. There could be planets setting (like Mercury or Venus right after dusk in the West) or others rising (quite bright) Jupiter, Saturn Mars. Even large deep sky objects portrayed there could be amazing - like the Orion Nebula, Andromeda Galaxy, The Double Cluster, Pleiads, M44 Cluster, meteors, etc. I found a good expression of &lt;a href=&quot;https://www.astroapprentice.com/2019/03/07/learn-the-night-sky/&quot; rel=&quot;nofollow noreferrer&quot;&gt;items of the night sky&lt;/a&gt; here.&lt;/p&gt;&#xA;" OwnerUserId="23668" LastActivityDate="2019-03-18T13:36:18.297" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30038" PostTypeId="1" CreationDate="2019-03-18T13:40:10.127" Score="1" ViewCount="30" Body="&lt;p&gt;One hypothetical tenet of much astrophysics is that the Universe is uniform on large scales.  This actually hinges on many deep theoretical results from astrophysics, since what we &lt;em&gt;actually observe&lt;/em&gt; is that the Earth is at the center of the observable universe, and that stuff near the edges of the observable universe is quite different than stuff near the center.  However, these theoretical results could be verified by, e.g. traveling to a different point in the observable universe and observing the universe from that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, we are currently constrained to observation from Earth, so this concept of the &quot;constant observer&quot; is difficult to test directly.  At current technological levels, how far would we need to travel away from the Earth to perform an experiment directly verifying the &quot;constant observer&quot; effect?  &lt;/p&gt;&#xA;" OwnerUserId="26277" LastActivityDate="2019-03-18T13:40:10.127" Title="How far out would we need to travel to measure the &quot;constant observer&quot;?" Tags="&lt;observable-universe&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30039" PostTypeId="1" CreationDate="2019-03-18T15:34:40.030" Score="-1" ViewCount="143" Body="&lt;p&gt;For this question assume that.&lt;/p&gt;&#xA;&lt;p&gt;We found a way to go faster than light.&lt;/p&gt;&#xA;&lt;p&gt;I build a ship that could go faster than light and go to another planet with this ship in the span of mere hours. The ship was built in a year time.&lt;/p&gt;&#xA;&lt;p&gt;If I travel to another planet one light year from Earth, and I would look back.&lt;/p&gt;&#xA;&lt;p&gt;Would I see the ship being built?&lt;/p&gt;&#xA;&lt;p&gt;Since the light that comes from earth is 1 year old when it reaches me you would assume that I can see the ship being built 1 year back(with a powerful telescope)&lt;/p&gt;&#xA;&lt;p&gt;So my question is : &lt;strong&gt;If FTL travel is possible, can I look back in time to a younger earth?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="25592" LastEditorUserId="1891" LastEditDate="2020-12-23T22:54:44.580" LastActivityDate="2020-12-23T22:54:44.580" Title="If FTL travel is possible, can I look back in time to a younger earth?" Tags="&lt;light&gt;&lt;time&gt;&lt;interstellar-travel&gt;" AnswerCount="3" CommentCount="3" ClosedDate="2019-03-19T05:56:30.577" ContentLicense="CC BY-SA 4.0" />
  <row Id="30040" PostTypeId="2" ParentId="30035" CreationDate="2019-03-18T15:39:41.520" Score="22" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Am I correct in saying that the fusion process of the sun is constant, i.e. X amount of fusion happens per day, more or less? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, at least over human timescales. You could reasonably expect the fusion rate within the sun to be the same today as a few thousand years ago or into the future, give or take some small fraction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does this not speed up, i.e. one fusion event creates energy for two fusion events, etc.? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The energy released by fusion is quickly distributed as thermal energy in the centre of the sun, and the temperature difference between surface (around 6000K) and centre (estimated 15 million K) drives an energy flow from hot to cold.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does every collision of atom cause a fusion event, or is the probability small for a fusion event to happen thus it's not a runaway reaction? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fusion in the sun is not a runaway nuclear reaction (like a critical mass of uranium in a fission reaction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible in theory to have runaway fusion events, but the pressure and temperature for these to happen are not approached in the core of the sun. For stable stars like the sun, the forces and energy flows are in equilibrium - if the core grew slightly hotter, then the pressure would increase and the star expand slightly against the force of gravity to compensate. &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermal_runaway#Astrophysics&quot; rel=&quot;noreferrer&quot;&gt;Interesting things happen when stars fall out of equilibrium and runaway fusion ignition can happen in some scenarios&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, this equilibrium point moves during the lifetime of a star as its mix of elements changes due to fusion. This is predictable for many stars and forms the basis of the main sequence stars in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hertzsprung%E2%80%93Russell_diagram&quot; rel=&quot;noreferrer&quot;&gt;Hertzsprung-Russell diagram&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have heard that the probability for fusion event to happen is only 1 in 10^12 for every collision&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know the accuracy of that, but it seems reasonable. The definition of &quot;collision&quot; becomes somewhat arbitrary in such a hot dense environment. If you only include approaches close enough to make the strong nuclear force dominate the interaction, the ratio could be higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another fact that I found interesting in the same area is that the power density from fusion - i.e. the Watts per cubic metre of substance - in the sun is &lt;a href=&quot;https://en.wikipedia.org/wiki/Sun#Core&quot; rel=&quot;noreferrer&quot;&gt;roughly the same as that found in a typical compost heap&lt;/a&gt;. It is a very different environment to the inside of a fusion reactor experiment or a fusion bomb, which have much higher power densities.&lt;/p&gt;&#xA;" OwnerUserId="21532" LastEditorUserId="21532" LastEditDate="2019-03-18T22:06:18.123" LastActivityDate="2019-03-18T22:06:18.123" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="30041" PostTypeId="1" CreationDate="2019-03-18T16:07:35.553" Score="1" ViewCount="42" Body="&lt;p&gt;Nuclear astrophysics greatly interests me and i was wondering if there is something that could be observed with the LOFAR project in this wavelength that would benefit the field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also some general information about this wavelength in astronomy or pointers to information sources would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="22751" LastEditorUserId="7982" LastEditDate="2021-02-04T01:04:17.623" LastActivityDate="2021-02-04T01:04:17.623" Title="What could be observed with LOFAR in 230 MHz - 10.0 MHz wavelength?" Tags="&lt;observational-astronomy&gt;&lt;astrophysics&gt;&lt;radio-astronomy&gt;&lt;spectra&gt;&lt;wavelength&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30042" PostTypeId="2" ParentId="30039" CreationDate="2019-03-18T16:34:40.590" Score="5" Body="&lt;p&gt;&quot;Yes, but no.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could jump through space faster than light can travel, and you went 1 light year away to 'look back at earth', then you would observe the light emissions that began 1 year ago, and continue to observe them as they arrive. [And if you start moving towards or away from the emission source, then 'neat things' happen due to blue/red shift. Or red/blue shift, I'm waiting on coffee and forget which direction the shift is in.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Whether you could have the optical resolution to actually SEE something like a ship being built from that far away is another matter. At that distance the dominating signal would be the star's output as a whole, and digging other information out of it would get tricky. And if we handwave how exactly we punch a hole through space-time to actually make the FTL jump, then resolving details from that distance might actually be the &lt;em&gt;harder&lt;/em&gt; part of the question here...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is however not strictly speaking looking back 'through time', but merely observing a signal delay between the event's actual time and the point where you can detect the signal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you watch someone driving a fence post from a distance you'll be able to see the hammer strike, but the sound will be slightly delayed - We don't normally consider this as &quot;hearing through time&quot;, but it is ultimately the same effect: Observing a signal after it was emitted due to the time required for the signal to travel from its origin to the observer's location.&lt;/p&gt;&#xA;" OwnerUserId="26280" LastActivityDate="2019-03-18T16:34:40.590" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30043" PostTypeId="1" CreationDate="2019-03-18T16:52:57.567" Score="1" ViewCount="88" Body="&lt;p&gt;The basis and most observable metric for how the universe is understood to be expanding at an accelerating rate is the red shift in observable light.  I am assuming it has been thought about or maybe it is just obviously wrong to people with more knowledge, but why can we not say the universe is technically always collapsing on itself under gravity's pull when considering relativity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that space and time are affected by gravity so if we think in relative terms if everything is accelerating at an accelerating rate towards everything else this could explain the red shift (appearance of moving away). Things behind me would be accelerating slower towards the singularity things in front of us would be accelerating faster. Therefore relative to us it would actually look like it is all expanding at an accelerating rate, explaining the red shift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this very simplistic thinking, this would work for a universe that expands and contracts or is in a never ending state of contraction, which relative to us could go on infinitely since time and space is all intertwined with gravity.&lt;/p&gt;&#xA;" OwnerUserId="26279" LastEditorUserId="26279" LastEditDate="2019-03-21T13:23:04.067" LastActivityDate="2019-03-21T13:23:04.067" Title="Looking at the Expanding Universe through the lens of relativity" Tags="&lt;gravity&gt;&lt;universe&gt;&lt;space-time&gt;&lt;general-relativity&gt;&lt;expansion&gt;" AnswerCount="1" CommentCount="12" ClosedDate="2019-04-04T07:18:29.107" ContentLicense="CC BY-SA 4.0" />
  <row Id="30044" PostTypeId="2" ParentId="30039" CreationDate="2019-03-18T17:50:18.200" Score="3" Body="&lt;p&gt;You could do better than that.  You could walk around on a younger earth.  &lt;a href=&quot;https://physics.stackexchange.com/questions/52249/how-does-faster-than-light-travel-violate-causality&quot;&gt;This Physics.SE Q&amp;amp;A addresses why&lt;/a&gt;.  The short of the matter being that things are moving with respect to each other, and so have different inertial reference frames.  Your ability to travel faster than light lets you exploit the differing reference frames to conduct a round trip that arrives before it departs.  It doesn't matter how it is you manage to go faster than light.  Any FTL travel implies time travel.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2019-03-18T17:50:18.200" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30045" PostTypeId="2" ParentId="30039" CreationDate="2019-03-18T18:10:07.550" Score="0" Body="&lt;p&gt;If you have the technology to build a faster than light ship then you will probably have the technology to build equipment capable of receiving any signal ever transmitted on earth. Why stop one light year away. Go for a hundred or so. Get the first broadcast by Marconi, the first I love Lucy or the first text message ever sent. &lt;/p&gt;&#xA;" OwnerUserId="24928" LastActivityDate="2019-03-18T18:10:07.550" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30046" PostTypeId="2" ParentId="30035" CreationDate="2019-03-18T18:32:06.420" Score="13" Body="&lt;p&gt;No, the fusion rate of the Sun is not absolutely constant in time. The Sun is gradually becoming more luminous and that luminosity is provided for almost exclusively by fusion in the core. However, the rate of increase is not large, of order 10% per billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fusion process is extremely slow (and inefficient in terms of energy release per unit volume) - the Sun releases only 250 W/m&lt;span class=&quot;math-container&quot;&gt;$^3$&lt;/span&gt; in it's core. The reason for this is that fusion events are extremely unlikely, requiring two protons to overcome the Coulomb barrier between them &lt;em&gt;and&lt;/em&gt; for one of the protons to inverse beta-decay into a neutron so forming a deuterium nucleus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average lifetime of a proton against this process in the core is &lt;span class=&quot;math-container&quot;&gt;$10^{10}$&lt;/span&gt; years (the lifetime of the Sun), meaning the fusion rate per proton is about &lt;span class=&quot;math-container&quot;&gt;$3 \times 10^{-18}$&lt;/span&gt; s&lt;span class=&quot;math-container&quot;&gt;$^{-1}$&lt;/span&gt;. We can compare this to a collision rate between protons by assuming an average thermal speed of &lt;span class=&quot;math-container&quot;&gt;$v \simeq (3k_B T/m_p)^{1/2} = 600$&lt;/span&gt; km/s for a core temperature of &lt;span class=&quot;math-container&quot;&gt;$15\times 10^{6}$&lt;/span&gt; k, a proton number density of &lt;span class=&quot;math-container&quot;&gt;$n_p \sim 6 \times 10^{31}$&lt;/span&gt; m&lt;span class=&quot;math-container&quot;&gt;$^{-3}$&lt;/span&gt; in the core and a collisional cross-section of &lt;span class=&quot;math-container&quot;&gt;$\sigma \sim \pi (\hbar/mv)^2$&lt;/span&gt;, where the term in brackets is the reduced de Broglie wavelength. Putting these things together, the collision rate is &lt;span class=&quot;math-container&quot;&gt;$n_p \sigma v \sim 10^{12}$&lt;/span&gt; s&lt;span class=&quot;math-container&quot;&gt;$^{-1}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus comparing the two rates, we can conclude that only about 1 in &lt;span class=&quot;math-container&quot;&gt;$3\times 10^{29}$&lt;/span&gt; collisions ends up with fusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the fusion rate of the Sun did increase rapidly then what would happen is that the Sun would expand, the core would become less dense and the fusion rate would fall. This basically acts as a thermostat, keeping the Sun at exactly the right temperature to support its own weight and supply the luminosity emerging from its surface.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-03-19T13:03:31.050" LastActivityDate="2019-03-19T13:03:31.050" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30047" PostTypeId="2" ParentId="30028" CreationDate="2019-03-18T18:55:11.813" Score="1" Body="&lt;p&gt;According to the Wikipedia article on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Mars_ocean_hypothesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mars ocean hypothesis&lt;/a&gt;, the ocean is sometimes called the Paleo-Ocean, and sometimes called &quot;Oceanus Borealis&quot; (a.k.a. &quot;Northern Ocean&quot; for those who feel that &lt;em&gt;quidquid Latine dictum sit, altum videtur.&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inland sea in the article is located in the Simud Interior Basin (part of the Simud Valles). The &lt;em&gt;Nature&lt;/em&gt; article describing the findings (&lt;a href=&quot;https://www.nature.com/articles/s41598-019-39632-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rodriguez et al. 2019&lt;/a&gt;) doesn't appear to give the sea a name, although it does use the term &quot;SIB sea&quot; at one point.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastActivityDate="2019-03-18T18:55:11.813" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30048" PostTypeId="2" ParentId="30033" CreationDate="2019-03-18T20:05:25.817" Score="3" Body="&lt;p&gt;Back of the envelope? The energy released is about &lt;span class=&quot;math-container&quot;&gt;$GM/R$&lt;/span&gt; per unit mass. For a 1.5 solar mass neutron star of radius 10 km, this amounts to &lt;span class=&quot;math-container&quot;&gt;$2\times 10^{16}$&lt;/span&gt; J/kg, or approximately 2/9 of the rest mass energy of the accreted material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of order half of this would be radiated and half would go into heating the neutron star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Timescales are more difficult. Starting when? Once the material gets to the innermost stable circular orbit at around 15 km, then the final 5km to the surface takes a tiny fraction of a second. But from where an object is shredded (much further out), much would depend on its angular momentum.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-03-18T20:05:25.817" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30049" PostTypeId="2" ParentId="30027" CreationDate="2019-03-18T20:33:28.970" Score="-4" Body="&lt;p&gt;Sometimes things we think of as different are actually just different forms of the same thing. Electric fields and magnetic fields are forms of each other. That is why we now call it the electromagnetic force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, energy and mass are also just different forms of the same thing. We can even convert it too. &#xA;E = mc^2 ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And from what I understand, space and time are connected together as well. So it gets called spacetime. If you want to think about it that way, gravity works by warping space, like a heavy marble sitting on a fluffy blanket. The fluffy blanket is spacetime. The heavy marble is a black hole or other super massive object. So now light is still going &quot;straight&quot; along space (the fluffy blanket) but it is bent because the blanket is bent. (Caveat: this is where i am conceptualizing light as a wave, not photon packets. It is actually both.) And if gravity bends space, of course it bends time too, because space-time is connected somehow as the same substance.&lt;/p&gt;&#xA;" OwnerUserId="26282" LastActivityDate="2019-03-18T20:33:28.970" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30050" PostTypeId="2" ParentId="10978" CreationDate="2019-03-18T21:51:29.940" Score="3" Body="&lt;p&gt;According to &lt;a href=&quot;https://hal.archives-ouvertes.fr/hal-01221674v2/document&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt; I understood that scientists have found that redshift (z) of supernovae depends on relative distance (x) according to formula:&#xA;&lt;code&gt;z = 2 sinh(x)/(1 + x)&lt;/code&gt;. And IF nothing is pushing galaxies away with force then redshift should drop linearly with decrease of distance between galaxies. But it isn't so as relationship z(x) is a nonlinear one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets draw a graph:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/7aoGj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7aoGj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;In other words - when distance to a supernova drops in half, the redshift must drop in half too (i.e. linearly), if gravity is stopping expansion. But it isn't the case as can be seen in the graph. The only reasonable explanation: that there exists some force which doesn't let that happen and keeps increasing the expansion rate.&lt;/p&gt;&#xA;" OwnerUserId="26283" LastEditorUserId="14094" LastEditDate="2019-03-19T06:37:37.873" LastActivityDate="2019-03-19T06:37:37.873" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30051" PostTypeId="2" ParentId="30032" CreationDate="2019-03-18T22:12:11.697" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Space is big. Really big. You just won't believe how vastly, hugely, mind-bogglingly big it is (Douglas Adams)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The wikipedia page notes that 'Oumuamua &quot;has circulated the Milky Way several times&quot; so it was unlikely to have formed around any nearby star. While it would take 'Oumuamua  600,000 years to travel here from a local star, the time before it can be expected to have another stellar encounter would be much greater, hundreds of millions of years. And the next star that it meets is unlikely to be a currently local star, it could be almost any star in the galaxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once an interstellar object has left the solar system it will travel in a roughly straight line, (slowly curving with the general gravity of the galaxy). For it to return to the Earth it would have to have a gravitational encounter with another star (or more likely a series of encounters) that send it back towards the solar system. Even given a vast mind-boggling long period of time like 10 billion years the chance of this happening is minute. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such an object will probably orbit the galaxy several times between stellar encounters, perhaps it has an encounter every 250 million years. And it is equally likely to encounter any of the disc stars (conservatively 50 billion stars). This means that the mean time between encounters of the sun is on the order of ten quintillion years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It isn't impossible.  It is quite easy to design a free-return flight to the moon.   There is no reason why a free-return flight to another star can't be done. However, space is big &quot;therefore ʻOumuamua will never return&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2019-03-19T16:42:13.527" LastActivityDate="2019-03-19T16:42:13.527" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="30052" PostTypeId="1" AcceptedAnswerId="30056" CreationDate="2019-03-18T23:13:59.873" Score="8" ViewCount="509" Body="&lt;p&gt;The question is in the title. I'd like to find out the maximum angle the moon ever makes above the horizon at the North Pole. By &quot;the horizon at the North Pole,&quot; I mean the tangent plane to the earth at that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I'm not asking for the highest angle that the moon ever &lt;strong&gt;appears&lt;/strong&gt; to make above the horizon. Appearances are distorted by refraction, etc., although I'm not sure by how many degrees. Though I'd be interested in the exact distortion: is there a data repository where I can look up the highest observed angle? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would make the following prediction. The moon's orbital plane is inclined to the earth's equatorial plane at about &lt;span class=&quot;math-container&quot;&gt;$23.5+5\approx 28.5$&lt;/span&gt; degrees. Let &lt;span class=&quot;math-container&quot;&gt;$R_e$&lt;/span&gt; denote the radius of the earth and &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; denote the radius of the moon's orbit around the moon. A little geometry for the highest point suggests&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\theta_{\text{max}}=\tan^{-1}\left(\frac{R\sin(28.5^{\circ})-R_e}{R\cos(28.5^{\circ})}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But because &lt;span class=&quot;math-container&quot;&gt;$R&amp;gt;&amp;gt;R_e$&lt;/span&gt; (about 385,000 compared to 6300 km), &lt;span class=&quot;math-container&quot;&gt;$\theta_{\text{max}}$&lt;/span&gt; should be just a little less than &lt;span class=&quot;math-container&quot;&gt;$28.5^{\circ}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this right?&lt;/p&gt;&#xA;" OwnerUserId="26285" LastActivityDate="2020-01-18T22:53:55.090" Title="What is the highest angle the moon ever makes above the horizon at the North Pole?" Tags="&lt;the-moon&gt;&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30053" PostTypeId="2" ParentId="27597" CreationDate="2019-03-19T00:12:26.773" Score="-2" Body="&lt;p&gt;Thus far there has been no solid explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the person who discovered it, YouTube user Crrow777 has a video which explain his observations, and handily debunk Carl Witthoft's explanation of it being atmospheric turbulence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=xUGxysKSGEM&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=xUGxysKSGEM&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most obvious of which being that the distortion does not extent beyond the boundary presented by the structure of the moon in the image, even while the camera is moving around.&lt;/p&gt;&#xA;" OwnerUserId="26286" LastActivityDate="2019-03-19T00:12:26.773" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30054" PostTypeId="2" ParentId="29624" CreationDate="2019-03-19T00:23:47.553" Score="3" Body="&lt;p&gt;The specific radius of &lt;span class=&quot;math-container&quot;&gt;$8 h^{−1} \text{Mpc}$&lt;/span&gt; is used because the value of &lt;span class=&quot;math-container&quot;&gt;$\sigma_8$&lt;/span&gt; turned out to be close to unity. To quote e.g. from Amendola &amp;amp; Tsujikawa (2010), p. 39:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the cells have a radius of &lt;span class=&quot;math-container&quot;&gt;$8 h^{−1} \text{Mpc}$&lt;/span&gt;, it turns out that &lt;span class=&quot;math-container&quot;&gt;$\sigma_R$&lt;/span&gt; is close to unity. Conventionally the normalization of the power spectrum is therefore given by quoting &lt;span class=&quot;math-container&quot;&gt;$\sigma_8$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user26287" LastActivityDate="2019-03-19T00:23:47.553" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30055" PostTypeId="2" ParentId="30026" CreationDate="2019-03-19T02:50:55.970" Score="2" Body="&lt;p&gt;The reason is that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;To take a photograph under different lighting conditions, you need to use different camera settings to get a useful image.&lt;/li&gt;&#xA;&lt;li&gt;Cameras (and the human eye) do not have unlimited range in any given set of conditions, that is, they cannot represent objects of every brightness satisfactorily within one single image.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In particular, if one is photographing a subject that is brightly illuminated, one has to use camera settings that greatly limit the amount of light being recorded by the camera's sensor, otherwise it will be overwhelmed and fail to show useful detail. In the case of taking a subject that is only dimly illuminated (or, in this case, emitting only what amounts to a dim light), such as the stars, one needs to use settings which maximize the amount of light that the sensor absorbs to get useful detail in the image, or one will record nothing. These two types of settings are logically incompatible, and thus it is impossible (with existing camera technology) to capture simultaneously a very dim and very bright subject in a single (i.e. not a composite) photograph and have both of them look sensible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the Moon and stars are just such an incompatible pair. The Moon's surface is lit up effectively as brightly as the landscape of Earth in broad daylight. The stars are so dim they can only be seen at night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, you can demonstrate this right from Earth itself. Here are two photographs I took with my own camera about ten megaseconds or so ago, as of this posting. Both were shot at night, on the same night. The left hand photograph is shot with the camera set to daylight settings. Yes, these are the same settings you'd use to shoot a photograph in actual daylight, only being used at night, and the Moon registers loud and clear. That is how bright it is. Since surface brightness is not affected by distance, the Moon effectively amounts to a little piece of sunlit landscape in the sky, from our point of view, just like on a bright, sunny day on Earth. As you can see, the Moon's surface features are cleanly visible and, moreover, it is similar in coloration to your last photograph - as it should be, because that is its actual color. Note the complete absence of stars, exactly as in the NASA images. In the second photograph on the right, the camera was set to &quot;bulb&quot; mode to expose the sensor for a long time, and its sensitivity was greatly increased. You can now see the stars, but the Moon looks almost like a second Sun - its surface features completely obliterated as the sensor has been saturated with photons like a sponge that has already soaked up too much water and has now had enough, while bloom contaminates the rest of the image.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/MU8mW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MU8mW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;The reason you &quot;expect&quot; to see stars is likely because you have watched too many sci-fi movies. Movies depict stars for artistic effect. In reality, images capturing such, taken in a single bout, are not possible with today's tech, and the reason is that the factor between the two is on the order of a billion (90 dB) in brightness. (You could composite the above two images suitably to fake it, but it would be just that.)&lt;/p&gt;&#xA;" OwnerUserId="9807" LastActivityDate="2019-03-19T02:50:55.970" CommentCount="16" ContentLicense="CC BY-SA 4.0" />
  <row Id="30056" PostTypeId="2" ParentId="30052" CreationDate="2019-03-19T06:04:47.250" Score="10" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;${R_e}/{R}$&lt;/span&gt; is about 0.017 radian or 1&amp;deg;.&#xA;Using your trigonometry and &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbit_of_the_Moon&quot; rel=&quot;noreferrer&quot;&gt;these values&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Lunar orbit inclination = 5.15&amp;deg; (varies between 5.0&amp;deg; and 5.3&amp;deg;)&lt;/li&gt;&#xA;&lt;li&gt;Lunar perigee = 362600 km, apogee = 405400 km&lt;/li&gt;&#xA;&lt;li&gt;Earth polar radius = 6357 km&lt;/li&gt;&#xA;&lt;li&gt;Ecliptic obliquity = 23.44&amp;deg;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I get 27.7&amp;deg; at perigee and 27.8&amp;deg; at apogee.&#xA;At that altitude, &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_refraction&quot; rel=&quot;noreferrer&quot;&gt;atmospheric refraction&lt;/a&gt; is only about 0.03&amp;deg;.&#xA;The highest apparent altitude I found with &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;noreferrer&quot;&gt;JPL HORIZONS&lt;/a&gt; was 27.91&amp;deg; on 1987-09-15 around 17:15 UT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_precession&quot; rel=&quot;noreferrer&quot;&gt;nodal precession&lt;/a&gt;,&#xA;the Moon doesn't reach &amp;plusmn;28.6&amp;deg; geocentric declination every year, only near a major &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_standstill&quot; rel=&quot;noreferrer&quot;&gt;lunar standstill&lt;/a&gt; every 18 or 19 years, e.g. in 2006 or 2025.&#xA;Near a minor lunar standstill, e.g. in 2015, the Moon's declination is limited to &amp;plusmn;18.3&amp;deg;.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2019-03-19T06:04:47.250" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30057" PostTypeId="2" ParentId="29805" CreationDate="2019-03-19T07:58:48.097" Score="3" Body="&lt;p&gt;I found a partial answer in this article on the smaller and younger &lt;a href=&quot;https://en.wikipedia.org/wiki/Giordano_Bruno_(crater)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Giordano Bruno crater&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The impact creating the 22-km-wide crater would have kicked up 10 million tons of debris, triggering a week-long, blizzard-like meteor storm on Earth...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the secondary effects of the Tycho impact would have been even more spectacular. However, I don't know whether the fragments would have been large enough to devastate the Earth's surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experimenting with an online impact simulator it appears that the Tycho impact was somewhere in the order of 1.00x10^23 joules, judging by the size of the crater. This is about four times that of the G fragment impact of Shoemaker-Levy 9. When you consider all the fragments it would be fair to say humans have witnessed an impact of similar magnitude to Tycho. &lt;/p&gt;&#xA;" OwnerUserId="23640" LastActivityDate="2019-03-19T07:58:48.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30058" PostTypeId="2" ParentId="30032" CreationDate="2019-03-19T09:14:48.577" Score="-1" Body="&lt;p&gt;&lt;strong&gt;tl;dr:&lt;/strong&gt; It's a hard problem. While &lt;a href=&quot;https://astronomy.stackexchange.com/a/30051/7982&quot;&gt;this answer&lt;/a&gt; says &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, space is big &quot;therefore ʻOumuamua will never return&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I say &lt;em&gt;never say never!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What natural mechanisms could lead to the unlikely case of the same rogue asteroid or planet passing through our solar system twice?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can think of three mechanisms.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;1. Stellar and solar-system pinball&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Under normal situations an asteroid or planet could not swing around a massive object, make a U-turn, and come straight back nearly along it's same path after a 180° deflection. However deflections by two or more large bodies can do it. Those two bodies could be&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Two unrelated stars:&lt;/strong&gt; This would make a triangular path; Sun, star-1, star-2, Sun again.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;A double star:&lt;/strong&gt; Three-body unbound orbits are crazy and chaotic and just about anything can happen including a 180°U-turn. This is my favorite since it only requires one encounter and star systems of two or higher are pretty common.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;A star and its planet:&lt;/strong&gt; Stars with Jupiter-size (and larger) planets are pretty common, and a hyperbolic orbit around a star doing much/most of the bend with the planet doing the rest of the 180°U-turn is possible.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;other variations on this theme.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;2. General Relativity (or Neutron Stars)&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;A black hole:&lt;/strong&gt; As explained in answer(s) to &lt;a href=&quot;https://physics.stackexchange.com/q/467144/83380&quot;&gt;Could a trajectory around a large mass ever deflect by more than 180 degrees due to general relativistic effects?&lt;/a&gt; &quot;...there is no limit on how many &quot;laps&quot; a hyperbolic orbit can make before it returns to infinity.&quot; That's in the limiting case of a zero-mass object. For large mass ratios such as an asteroid or planet versus a black hole, the maxiumum number of turns an unbound orbit can make is about 0.41 times the mass ratio, which would be greater than a million for an Earth-sized planet and a vanilla stellar-collapse black hole.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;A neutron star:&lt;/strong&gt; I am not certain yet if the GR effects due to a neutron star are strong enough to produce a 180°U-turn without a destructively close approach; I'm currently working on this...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;3. Galactic roller-coaster&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The solar system orbits the galactic center at about 220 km/s with a period of about 250 million years. That speed is an &lt;em&gt;order of magnitude or possibly two faster&lt;/em&gt; than the orbital speeds of asteroids and planets within solar systems. Jupiter orbits at 13 km/s and (486958) 2014 MU69 (Ultima Thule) moves at 4.6 km/s. So no mater which direction our rogue friend leaves the solar system, it's going to be moving around the galactic center with a nearly parallel orbit to that of the solar system, and it's going to be doing a complicated dance above and below the galactic plane as our solar system is doing, due to the &lt;em&gt;restoring force&lt;/em&gt; of the galactic disks's gravity. This can bring us and our rogue friend back together in a way that would at first surprise us both until we remember how stars move in the disk.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Overall probability?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For all of these the probability is small, but the single collision options (double stars, solar systems, black holes, neutron stars) may make the probability better than the two unrelated star sequential scattering scenarios, and the galactic plane oscillations don't require a near approach to any other body at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think the statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...Therefore it will never return.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;could be considered cavalier and possibly scientifically unsupported.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a hard problem, so &lt;em&gt;never say never!&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2020-05-10T02:28:49.920" LastActivityDate="2020-05-10T02:28:49.920" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30059" PostTypeId="2" ParentId="29572" CreationDate="2019-03-19T13:23:43.470" Score="2" Body="&lt;p&gt;The geodesic equation below is a second-order partial differential equation for the spacetime coordinates &lt;span class=&quot;math-container&quot;&gt;$x^\mu$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\mu = 0 \ldots n$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$${d^{2}x^{\mu } \over ds^{2}}+\Gamma ^{\mu }{}_{\alpha \beta }{dx^{\alpha } \over ds}{dx^{\beta } \over ds}=0 \;,$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\Gamma ^{\mu }{}_{\alpha \beta }$&lt;/span&gt; are the Christoffel symbols, and the proper time &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; is the parameter of the curve (i.e. how ``far along'' you are on the path). You need to solve these &lt;span class=&quot;math-container&quot;&gt;$n+1$&lt;/span&gt; coupled equations as any such system, e.g. numerically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the parameter &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; is confusing, you can reparametrise the curve as a function of coordinate time &lt;span class=&quot;math-container&quot;&gt;$t = x^0$&lt;/span&gt;, as explained e.g. on Wikipedia's page on &lt;a href=&quot;https://en.wikipedia.org/wiki/Geodesics_in_general_relativity&quot; rel=&quot;nofollow noreferrer&quot;&gt;geodesics in general relativity&lt;/a&gt;, &lt;span class=&quot;math-container&quot;&gt;$${d^{2}x^{i} \over dt^{2}}=-\Gamma ^{i}{}_{{\alpha \beta }}{dx^{\alpha } \over dt}{dx^{\beta } \over dt} + \Gamma ^{0}{}_{{\alpha \beta }}{dx^{\alpha } \over dt}{dx^{\beta } \over dt}{dx^{i} \over dt} \;,$$&lt;/span&gt;&#xA;which gives you &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; partial differential equations for the spatial coordinates &lt;span class=&quot;math-container&quot;&gt;$i = 1 \ldots n$&lt;/span&gt;. (Note that there are remaining greek indices, for which &lt;span class=&quot;math-container&quot;&gt;$dx^0/dt = dt/dt = 1$&lt;/span&gt;.)&lt;/p&gt;&#xA;" OwnerDisplayName="user26287" LastActivityDate="2019-03-19T13:23:43.470" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30060" PostTypeId="2" ParentId="25850" CreationDate="2019-03-19T14:39:48.470" Score="3" Body="&lt;p&gt;While the Hubble constant describes the current expansion rate of the Universe, it should also be seen as a parameter (among others) of a given cosmological model (e.g., Lambda-CDM).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All methods to measure the Hubble constant are more or less indirect in some way, and they rely on very different assumptions. For example, the emission of the Cosmic Microwave Background (CMB) observed by Planck happened in the early Universe, when the expansion rate was very different from its current value. Therefore, a &quot;measurement&quot; of the Hubble constant from the Planck papers is only possible within a very specific cosmological model, that describes how this CMB would look today. If something is wrong with this model, the Hubble constant computed from Planck data would no longer correspond to the current expansion rate of the Universe, and so disagree with more local &quot;measurements&quot;. In this context, local means closer to the current moment in the expansion history.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why &quot;having trouble with measuring the Hubble constant&quot; is one symptom of &quot;learning about the cosmological model&quot;. The community broadly agrees that we don't know yet what causes the inconsistencies when Lambda-CDM is assumed. As the tension becomes more significant, it could be some form of unknown systematic error (observational or more theoretical), or it could be that Lambda-CDM is an inaccurate description of our Universe.&lt;/p&gt;&#xA;" OwnerUserId="9150" LastEditorUserId="9150" LastEditDate="2019-03-19T18:29:58.813" LastActivityDate="2019-03-19T18:29:58.813" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30061" PostTypeId="2" ParentId="30035" CreationDate="2019-03-19T15:50:05.610" Score="1" Body="&lt;p&gt;What's more, the usual explanation for why the fusion doesn't run away is incomplete. The simple story that can't be the full story is that if the fusion happens too fast, heat builds up and creates an overpressure.  That overpressure causes expansion, and expansion does work which lowers the temperature and dials back down the fusion until it matches the radiative escape rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this is incomplete is that expansion work &lt;em&gt;doesn't&lt;/em&gt; induce stability if it occurs only against a fixed external pressure, that amount of work is always insufficient to stabilize it (which leads to &quot;shell flashes&quot; later in the life of a star).  The only thing that is capable of stabilizing the fusion is the additional work against &lt;em&gt;gravity&lt;/em&gt;, as you can easily see from how gravity gets included in any such analysis.  So it must be important that a local runaway has the net result of lifting gas away from the solar center, thereby doing gravitational work-- an important detail normally left out of the explanations.  Indeed, it would be more fair to say that solar fusion is stabilized by a &lt;em&gt;combination&lt;/em&gt; of expansion work and gravitational lifting.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2019-03-19T16:31:21.803" LastActivityDate="2019-03-19T16:31:21.803" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30062" PostTypeId="1" CreationDate="2019-03-19T16:39:03.667" Score="4" ViewCount="123" Body="&lt;p&gt;After SOHO began watching the Sun from space with its coronagraphs 24/7 it was discovered that there are comets passing very close to the Sun quite regularly, and while some are destroyed, some are seen to emerge and continue on.&lt;/p&gt;&#xA;&lt;p&gt;From these observations and from all other comet and asteroid observations there should be quite a collection of objects whose orbits take them very close to the Sun.&lt;/p&gt;&#xA;&lt;p&gt;Is there any way to find out &lt;em&gt;What is the closest an asteroid or comet has passed the Sun and survived?&lt;/em&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Here is a GIF I made for &lt;a href=&quot;https://space.stackexchange.com/a/15429/12102&quot;&gt;this answer&lt;/a&gt; also used in &lt;a href=&quot;https://astronomy.stackexchange.com/a/28263/7982&quot;&gt;this answer&lt;/a&gt;. In addition to some comets you can see Venus on its approach to occultation by the Sun in 2016. Seeing the Pleiades so close to the Sun is exciting as well!&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;These &lt;a href=&quot;http://lasco-www.nrl.navy.mil/index.php?p=content/handbook/hndbk_3&quot; rel=&quot;nofollow noreferrer&quot;&gt;LASCO C3&lt;/a&gt; images from &lt;a href=&quot;http://sohowww.nascom.nasa.gov/home.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;SOHO&lt;/a&gt; were downloaded &lt;a href=&quot;http://sohodata.nascom.nasa.gov/cgi-bin/data_query&quot; rel=&quot;nofollow noreferrer&quot;&gt;sohodata.nascom.nasa.gov/cgi-bin/data_query&lt;/a&gt;. &lt;em&gt;The square frame is about 15.9 degrees wide&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LgrIs.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LgrIs.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-03-20T16:16:18.477" Title="What is the closest an asteroid or comet has passed the Sun and survived?" Tags="&lt;orbit&gt;&lt;asteroids&gt;&lt;comets&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30064" PostTypeId="1" CreationDate="2019-03-20T08:15:25.923" Score="2" ViewCount="565" Body="&lt;p&gt;I bought an amateur(and rather cheap) refracting telescope and while I can see moon quite clear and crispy I am having difficulties in observing some larger stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All stars appear as lines rather than dots. It is always one longer line. I read something about those lines but this stands only for reflecting telescopes that have some kind of structure inside that holds a mirror and this structure interfers thus producing these lines. But I don't have a reflecting telescope, mine is refracting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot seem to find anything on google regarding this. Maybe it is because I lack the vocabulary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT #1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.astroshop.eu/telescopes/ts-optics-telescope-ac-70-700-starscope-az-2/p,56947#tab_bar_1_select&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; is the telescope I have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit #2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my depiction of what I see. Forgive my MS pain skills. As I focus out(or in, not quite familliar with the naming) it goes more clearer and clearer but I never get to see a simple shiny dot. The closes to dot is the first image but this is so blurry that I barely see it. The red line is the one I perceive as the clearest. The orange lines is what I perceive as &quot;outline&quot; and the yellow inside is that shiny effect one usually gets when looking at objects that emmit light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Hd4Uy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hd4Uy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26304" LastEditorUserId="26304" LastEditDate="2019-03-20T20:06:25.823" LastActivityDate="2019-03-23T21:10:11.867" Title="Seeing a line rather than a dot when looking through refracting telescope" Tags="&lt;refractor-telescope&gt;" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="30065" PostTypeId="2" ParentId="30043" CreationDate="2019-03-20T11:02:51.140" Score="2" Body="&lt;p&gt;If you work out the equations of motion for the gravitational free-fall collapse of a clouds of particles (or galaxies), you will find that, from the point of view of &lt;em&gt;any&lt;/em&gt; of these particles, all other particles appear to be approaching. This is true in both Newtonian theory and general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculation is found in numerous text books and lecture notes, e.g. &lt;a href=&quot;http://www.astro.uu.se/~hoefner/astro/teach/apd_files/apd_collapse.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Susanne Höfner's&lt;/a&gt;, but a qualitative argument goes like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From your point of view, sitting on one of the particles, a particle farther from the center of the cloud does &quot;feel&quot; a smaller gravitational pull from the central particles. This is, I think, the reason that you think it should accelerate more slowly. However, any particle feels the gravity from &lt;em&gt;all&lt;/em&gt; particles that are closer to the center, so the farther you are from the center, the more particles will attract you. On the other hand, the gravity from all particles that are &lt;em&gt;farther&lt;/em&gt; away from the center cancels out, so you don't feel those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is, in essence, &lt;a href=&quot;https://en.wikipedia.org/wiki/Shell_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Newton's shell theorem&lt;/a&gt;, and it turns out to imply that &lt;em&gt;all particles reach the center at the same time&lt;/em&gt;. The distance between each particle is getting smaller and smaller, until the cloud collapses into an point after a a time inversely proportional to the square root of the density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is exactly opposite to what is observed.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2019-03-20T11:02:51.140" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30066" PostTypeId="2" ParentId="30062" CreationDate="2019-03-20T16:16:18.477" Score="3" Body="&lt;p&gt;Wikipedia offers some useful info on sungrazers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As of October 21, 2017, there are 1495 known comets that come within ~12 solar radii (~0.055 AU).[2] This accounts for nearly one third of all comets.[3] Most of these objects vaporize during their close approach, but a comet with a nucleus radius larger than 2–3 km is likely to survive the perihelion passage with a final radius of ~1 km.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sungrazing_comet&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Sungrazing_comet&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a bit about pyroxenes sublimating off at 7 solar radii, but no mention of whether any such object made it around the sun. Pyroxenes &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/Geophys/meltrock.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;generally melt&lt;/a&gt; at around 1000°C.&#xA;An Iron/Nickel body melting at 1538°C/1455°C might well make it through a 7 solar radii passage.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2019-03-20T16:16:18.477" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30067" PostTypeId="1" AcceptedAnswerId="30072" CreationDate="2019-03-21T01:26:33.633" Score="23" ViewCount="5594" Body="&lt;p&gt;Google News feed shows me the following. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does the term &quot;Super Worm Equinox Moon&quot; mean and has it ever been used before this 2019 clickbait instance?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oXhaV.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oXhaV.png&quot; alt=&quot;Super Worm Equinox Moon&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-03-22T15:52:30.343" Title="What (the heck) is a Super Worm Equinox Moon?" Tags="&lt;the-moon&gt;&lt;terminology&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30068" PostTypeId="2" ParentId="30067" CreationDate="2019-03-21T01:36:49.570" Score="16" Body="&lt;p&gt;All the monthly Full Moons are named e.g. &lt;a href=&quot;https://www.timeanddate.com/astronomy/moon/full-moon-names.html&quot; rel=&quot;noreferrer&quot;&gt;list at timeanddate.com&lt;/a&gt;, of which &quot;Harvest Moon&quot; is the one people are probably most familiar with. So the March Full Moon is indeed the &quot;Worm Moon&quot; although rarely referred to as such. The extra hyperbole (&quot;Super&quot;, &quot;Blood&quot; etc) seems to be a recent (within the last few years) media phenomenon for unknown reasons...&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2019-03-21T01:36:49.570" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="30069" PostTypeId="1" CreationDate="2019-03-21T01:37:31.393" Score="2" ViewCount="48" Body="&lt;p&gt;This &lt;a href=&quot;https://astronomy.stackexchange.com/a/30066/7982&quot;&gt;answer&lt;/a&gt; about Sun-grazing comets led me eventually to Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Helion_(meteoroid)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Helion (meteoroid)&lt;/a&gt;, which has a one-sentence long article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A Helion meteoroid is a meteoroid that arrives from the approximate direction of the Sun. They are thought to originate as debris from sun-grazing comets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It has one reference &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://science.nasa.gov/headlines/y2007/27aug_explodingeclipse.htm?list84849&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Exploding Lunar Eclipse&quot;, an article about an attempt to observe Helion meteoroids&lt;/a&gt;, Science@NASA, August 27, 2007.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but that link is now dead.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Was NASA's 2007 attempt to see meteoroids coming from a solar eclipse successful? What instruments were used?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Has it been attempted again since?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-03-21T03:56:33.277" LastActivityDate="2019-03-21T03:56:33.277" Title="Was NASA's 2007 attempt to see helions during a solar eclipse successful? What instruments were used?" Tags="&lt;solar-eclipse&gt;&lt;meteor&gt;&lt;nasa&gt;&lt;meteoroid&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30070" PostTypeId="2" ParentId="10978" CreationDate="2019-03-21T02:51:07.710" Score="1" Body="&lt;p&gt;When they look at the light from distant galaxies, they see peaks and troughs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The peaks and troughs are from the spectra of the atoms in the galaxies - for instance hydrogen, oxygen, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From frequency and the patterns of emission and absorption of the light - or the peaks and troughs, they can identify the element. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When they compare the spectra of an elements from the galaxies to the spectra of the elements in the lab, they find they're shifted to the red - or red shifted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is, they are not measuring velocity - they are inferring velocity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To infer velocity measurements from the spectra, one has to assume the processes which created the spectra in the star billions of years ago is identical to the processes which created the spectra in the lab. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And they have to assume there are no non-velocity red shifts.&lt;/p&gt;&#xA;" OwnerUserId="26037" LastEditorUserId="26037" LastEditDate="2019-03-21T02:58:23.093" LastActivityDate="2019-03-21T02:58:23.093" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30071" PostTypeId="1" AcceptedAnswerId="30121" CreationDate="2019-03-21T03:33:49.103" Score="2" ViewCount="142" Body="&lt;p&gt;Do the current models of the universe consider the energy release of gravitational waves?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. iirc, the last black hole merger detected had a gravitational wave energy equal to ~3 solar masses (hope I'm not mistaken in this fact).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same for other mergers, and supernovae: the energy related to their gravitational waves should sum up to something.&lt;/p&gt;&#xA;" OwnerUserId="9367" LastEditorUserId="14094" LastEditDate="2019-03-21T07:26:09.677" LastActivityDate="2019-03-25T13:58:26.030" Title="Are gravitational waves part of dark energy?" Tags="&lt;black-hole&gt;&lt;supermassive-black-hole&gt;&lt;gravitational-waves&gt;&lt;dark-energy&gt;" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="30072" PostTypeId="2" ParentId="30067" CreationDate="2019-03-21T04:47:55.037" Score="22" Body="&lt;p&gt;All those adjectives being smooshed together signify an uncommon event. That's why you've never seen them together like that before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All 3 conditions have to hold true:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It's a supermoon, which means the full moon coincides with the moons perigee or nearest approach. That can make it appear up to 30% brighter than one at apogee (farthest away). These happen about every 13 months and it doesn't have to be exact so usually, we get 2 of them in a row, like we did this year. February had a supermoon and so did March.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It's a worm moon, which means it is occurring in the month of March (see &lt;a href=&quot;https://astronomy.stackexchange.com/a/30068&quot;&gt;@astrosnapper's answer&lt;/a&gt; for a better explanation of that).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It's during an equinox, basically the first day of spring (or autumn).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If any one of those isn't happening then it can't be called a &lt;strong&gt;&lt;em&gt;Super Worm Equinox Moon&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, the term &lt;em&gt;supermoon&lt;/em&gt; (all one word, by the way) is a relatively recent thing. I tried to view it on Google N Gram viewer, &lt;a href=&quot;https://books.google.com/ngrams/graph?content=supermoon%2Csuper%20moon&amp;amp;year_start=1800&amp;amp;year_end=2019&amp;amp;corpus=15&amp;amp;smoothing=3&amp;amp;share=&amp;amp;direct_url=&quot; rel=&quot;nofollow noreferrer&quot;&gt;but...&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a particularly bright full moon and it does deserve to have its own terminology, IMO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: The rarity of the event is certainly relative. I saw a tweet from National Geographic that we also had a super worm equinox moon 19 years ago.&lt;/p&gt;&#xA;" OwnerUserId="10026" LastEditorUserId="-1" LastEditDate="2019-03-22T15:52:30.343" LastActivityDate="2019-03-22T15:52:30.343" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="30074" PostTypeId="1" AcceptedAnswerId="30075" CreationDate="2019-03-21T10:38:49.443" Score="17" ViewCount="6381" Body="&lt;p&gt;When astronauts are floating about in a spaceship or space station, they nearly always move very slowly. After doing a bit of research I can't see why being in zero gravity would restrict movement to such a degree. It's almost as if there's resistance to their movement, like they're moving through water.&lt;/p&gt;&#xA;" OwnerUserId="26312" LastEditorUserId="26312" LastEditDate="2019-03-22T09:40:43.647" LastActivityDate="2019-03-22T09:40:43.647" Title="Why do astronauts move so slowly in zero gravity?" Tags="&lt;gravity&gt;&lt;space&gt;" AnswerCount="3" CommentCount="4" ClosedDate="2019-03-21T21:05:04.090" ContentLicense="CC BY-SA 4.0" />
  <row Id="30075" PostTypeId="2" ParentId="30074" CreationDate="2019-03-21T10:52:54.113" Score="43" Body="&lt;p&gt;It's more for safety than anything else. Space is a very dangerous place for so many reasons. And making mistakes can very easily cause death.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being weightless does not mean you lose mass, so momentum is just as difficult as ever. But whereas on the ground you can easily use friction to stop, in space if you try to stop against the floor you will just move off it. You can only stop by holding something, or pressing against something close to perpendicular to your movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, imagine you jumped with all your force from one wall in the ISS. You will notice as you approach the other end that you are travelling at speed, head first, with no safe way to stop. Even reaching out to a handhold on a side wall will whip you round and into that wall, possibly injuring yourself or damaging instruments on the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at any video from the ISS to see how carefully they move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, outside the ISS, you want to do everything slowly so you don't damage your suit, miss a handhold or otherwise cause death.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2019-03-21T10:52:54.113" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30076" PostTypeId="2" ParentId="30067" CreationDate="2019-03-21T12:11:37.383" Score="8" Body="&lt;p&gt;A supermoon is when the full moon occurs at the time when the moon is at its closest point to us in its elliptical orbit.  This means that the moon is slightly bigger (honestly, you wouldn't notice it unless you made careful comparisons) and a chunk brighter than normal full moons.  Except that supermoons happen three or four times every year, so, actually, a supermoon is itself a pretty normal full moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the last few years, the media seems to have latched onto these as EXCITING! EVENTS!, despite the fact that they account for about a quarter of all new moons.  In the last year or so, perhaps they sensing that we have supermoon fatigue (&quot;What, another one? There was a supermoon only a couple of months ago...&quot;), they've started putting long chains of adjectives in front of the word &quot;supermoon&quot; to make it sound EVEN! MORE! EXCITING!!1! These adjectives mostly come from traditional names for each month's full moon, but they sound NEW! and EXCITING!! because we don't use them very often in the modern era.&lt;/p&gt;&#xA;" OwnerUserId="1501" LastEditorUserId="1501" LastEditDate="2019-03-22T09:45:52.607" LastActivityDate="2019-03-22T09:45:52.607" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30078" PostTypeId="2" ParentId="16687" CreationDate="2019-03-21T13:09:55.293" Score="5" Body="&lt;p&gt;I like @DrChuck's &lt;a href=&quot;https://astronomy.stackexchange.com/a/16688/7982&quot;&gt;answer&lt;/a&gt; and this Astronomy Picture of the Day shows how this has some plausibility:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...but you could get this sort of image without cheating, if it were taken a long way away... and the magnification cranked up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Screen shot from YouTube video &lt;a href=&quot;https://youtu.be/afHfMMC-MJE&quot; rel=&quot;nofollow noreferrer&quot;&gt;Moon Setting Behind Teide Volcano&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DZCW8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DZCW8.jpg&quot; alt=&quot;Astronomy Picture of the Day; 2018 June 4&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://apod.nasa.gov/apod/ap180604.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomy Picture of the Day; 2018 June 4&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Moon Setting Behind Teide Volcano &#xA;  Video Credit &amp;amp; Copyright: Daniel López (&lt;a href=&quot;http://www.elcielodecanarias.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;El Cielo de Canarias&lt;/a&gt;); Music: Piano della Moon (&lt;a href=&quot;https://vimeo.com/dansilvaworld&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dan Silva&lt;/a&gt;)&#xA;  Explanation: These people are not in danger. What is coming down from the left is just the Moon, far in the distance. Luna appears so large here because she is being photographed through a telescopic lens. What is moving is mostly the Earth, whose spin causes the Moon to slowly disappear behind Mount Teide, a volcano in the Canary Islands off the northwest coast of Africa. The people pictured are &lt;a href=&quot;https://apod.nasa.gov/apod/image/1806/teide_volcan_video.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;16 kilometers away&lt;/a&gt; and many are facing the camera because they are watching the Sun rise behind the photographer. It is not a coincidence that a full moon rises just when the Sun sets because the Sun is always on the opposite side of the sky from a full moon. The featured video was made last week during the full Milk Moon. The video is not time-lapse -- this was really how fast the Moon was setting.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zejDt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zejDt.jpg&quot; alt=&quot;Cropped Astronomy Picture of the Day; 2018 June 4 Mount Teide, a volcano in the Canary Islands&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xKU8n.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xKU8n.jpg&quot; alt=&quot;Astronomy Picture of the Day; 2018 June 4 Mount Teide, a volcano in the Canary Islands&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;update:&lt;/strong&gt; Following a bit of incredulity in comments, I'll mention that I don't think the image really is real, because it involved photography near the horizon where the air gets very thick, strongly attenuating of shorter wavelengths (blues) and wavy due to refraction and turbulence. The Image above shows the moon to be fuzzier than the people and the edges are a little ragged, *unlike the image in the question which looks quite sharp right down to the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a few more real images for comparison, from &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8KZ4Y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8KZ4Y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; The moon sets over the Dorchester neighborhood of Boston on the early morning of March 20, 2019. The third supermoon of 2019 on Wednesday isn't just special because it's the last supermoon of the year. It's also the first supermoon in nearly 20 years to fall near the spring equinox. The super worm equinox moon, as it's known, follows Januarys super blood wolf moon and Februarys super snow moon. This supermoon will hit its peak at 9:43 p.m. Wednesday, almost four hours after the spring equinox officially hits. (Craig F. Walker/The Boston Globe via Getty Images)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Bryz0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bryz0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; A lookout tower, left, and the broadcast tower of Antenna Hungaria at the top of Karancs mountain are backdropped by the rising moon as seen from the vicinity of Karancskeszi village, 80 miles northeast of Budapest, Hungary, Wednesday, March 20, 2019. (Peter Komka/MTI via AP)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-03-22T06:21:39.287" LastActivityDate="2019-03-22T06:21:39.287" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="30079" PostTypeId="1" CreationDate="2019-03-21T13:21:30.160" Score="2" ViewCount="210" Body="&lt;p&gt;I have imported my fits table, hdul0, into python. I'm using astropy.table. I opened it with hdul0 = fits.open('filename').  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;hdul0data = Table(hdul0[1].data)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I have then made a scatter plot of two columns in the table, HDA and OII. I have made a cut-out region on the plot (a short vertical line and then an upwards sloping line from there), and I would like to select the points under that line, like slice them from the original fits table and move the slice into another fits table that I then save. This is what I have done, but it is not working.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x = hdul0data['HDA']&#xA;y = hdul0data['OII']&#xA;x1 = [2.5, 2.5] #the x values of the vertical line cut &#xA;y1 = [0.0, 2.5] #the y values of the vertical line cut &#xA;x2 = [2.5,16.0] #the x values of the sloping cut &#xA;y2 = [2.5, 16.5] #the y values of the sloping cut  &#xA;&#xA;slopy = np.interp(x, x2, y2)&#xA;myslice = np.where((x &amp;gt; 2.5) &amp;amp; (y &amp;lt; slopy))[0]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The error message I get is &lt;em&gt;&quot;RuntimeWarning: invalid value encountered in greater&quot;&lt;/em&gt; (and the same for less). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help? How do I slice my table and get the selected points into a new fits file? I think maybe part of the issue is that some of the data points in the HDA and OII columns are NaN, but I don't know how to get rid off it. I read I could use masked, but then I read it might not work with numpy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sample values of x and y: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x = [1.0, 6.2, 10.4, NaN, 2.8, NaN, 7.0] &#xA;y = [10.3, NaN, 18.0, 22.9, 2.5, NaN, 31.8] &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="26315" LastEditorUserId="26315" LastEditDate="2019-03-21T15:07:01.637" LastActivityDate="2019-03-21T15:07:01.637" Title="How to &quot;slice&quot; a fits table in astropy?" Tags="&lt;python&gt;&lt;astropy&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="30080" PostTypeId="1" AcceptedAnswerId="30082" CreationDate="2019-03-21T13:34:13.713" Score="3" ViewCount="151" Body="&lt;p&gt;Media coverage of the &quot;super worm equinox moon&quot; (*eyeroll*) has stated that this is the last supermoon of 2019.  Since supermoons normally happen every three or four months, how can there be big gaps?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It feels like phenomena like supermoons, which are basically due to the relative phase of orbits, shouldn't have sharp transitions like this.&lt;/p&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2019-03-21T14:34:24.763" Title="How is this the last supermoon of 2019?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="30081" PostTypeId="2" ParentId="30074" CreationDate="2019-03-21T13:49:10.093" Score="-1" Body="&lt;p&gt;There is no air resistance in space. So if you increment your own speed you'll continue with that speed until you crash (considering the laws of thermodynamics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For safety, you should move slowly.&lt;/p&gt;&#xA;" OwnerUserId="26316" LastEditorUserId="-1" LastEditDate="2019-03-21T16:26:12.087" LastActivityDate="2019-03-21T16:26:12.087" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30082" PostTypeId="2" ParentId="30080" CreationDate="2019-03-21T14:34:24.763" Score="11" Body="&lt;p&gt;A supermoon does not occur every three or four months. There may be 2 or 3 consecutive supermoons (that is, separated by 1 month) that occur at about the same time each year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you ignore the precession of the Moon's orbit, then there is one time of the year when the Full Moon and perigee occur (point 1 in my diagram below). Two weeks later, the New Moon occurs near apogee (point 2). At any other time of the year, the Full Moon occurs at a different point in its orbit around the Earth, so it is farther from the Earth than at perigee. Six months later, the Full Moon is occurring at apogee (point 4), and no one cares about that!&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ie4dC.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ie4dC.png&quot; alt=&quot;Earth&amp;#39;s &amp;amp; Moon&amp;#39;s orbit&quot;&gt;&lt;/a&gt; (not to scale!)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Full Moon at perigee (closest to the Earth)&lt;/li&gt;&#xA;&lt;li&gt;New Moon at apogee (farthest from the Earth)&lt;/li&gt;&#xA;&lt;li&gt;New Moon at perigee&lt;/li&gt;&#xA;&lt;li&gt;Full Moon at apogee&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia article on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Supermoon&quot; rel=&quot;noreferrer&quot;&gt;Supermoon&lt;/a&gt; has a nice graphic showing the Full Moon and distance from the Earth. Depending on how close to the Earth the Moon needs to be to be &quot;super&quot;, you can see that there is a &quot;season&quot; when the supermoon occurs. (I have copied the image here, and added a dashed line at 360,000 km to show which moons might be &quot;super&quot; and which ones are not.)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/cH8np.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cH8np.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if only people would care about the Super First Quarter Moon. Then we would be celebrating on May 12, 2019! (Not really. That is farther than the 360,000 km criteria.)&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2019-03-21T14:34:24.763" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="30083" PostTypeId="2" ParentId="13716" CreationDate="2019-03-21T14:55:40.150" Score="5" Body="&lt;p&gt;Other answers are right at explaining why tidal forces move Earth and Moon apart but they don't move apart a pair of black holes. However, I think it is also needed to explain why the phenomenons making two black holes spiral inward don't make the Moon spiral inward to the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, every pair of rotating masses radiate gravitational waves. What makes the difference is that only very large masses rotating very close to each other produce gravitational waves large enough to meaningfully affect those masses orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_wave#Binaries&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Gravitational_wave#Binaries&lt;/a&gt; the time that takes a pair of masses to fall into each other due to radiated gravitational waves is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$t= \frac{5}{256}\, \frac{c^5}{G^3}\, \frac{r^4}{(m_1m_2)(m_1+m_2)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's plug the masses of Earth and Moon and its distance into that equation (all data taken from Wikipedia in SI units):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt; G &amp;lt;- 6.674e-11&#xA;&amp;gt; r &amp;lt;- 384e6&#xA;&amp;gt; mluna &amp;lt;- 7.342e22&#xA;&amp;gt; c &amp;lt;- 299792458&#xA;&amp;gt; mterra &amp;lt;- 5.97237e24&#xA;&amp;gt; (t &amp;lt;- 5/256*c^5/G^3*r^4/(mterra*mluna)/(mterra+mluna))&#xA;[1] 1.304925e+33&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;That is, left alone, radiating gravitational waves would make the Moon crash into the Earth in 1.3*10^33 seconds, that is 4.13*10^25 years or 3*10^15 times the current age of the universe. In other words, the effect of radiating gravitational waves in the motion of Earth and Moon is so tiny - specially compared with other forces like tide ones - that we can't absolutely forget about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for comparison, two one solar mass neutron stars orbiting each other at the same distance of the Earth and the Moon would fall into each other in:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt; msol &amp;lt;- 1.9885e30&#xA;&amp;gt; (t &amp;lt;- 5/256*c^5/G^3*r^4/(msol*msol)/(msol+msol))&#xA;[1] 2.19985e+14&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which is just about 7 million years, showing that changing masses have a large effect on the outcome. As stated in the beginning, gravitational waves make pairs of star sized objects to spiral inwards but they don't have noticeable effects on a satellite orbiting a planet.&lt;/p&gt;&#xA;" OwnerUserId="14097" LastEditorUserId="14097" LastEditDate="2019-03-22T19:13:50.313" LastActivityDate="2019-03-22T19:13:50.313" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30084" PostTypeId="2" ParentId="30025" CreationDate="2019-03-21T17:20:47.110" Score="4" Body="&lt;p&gt;By using the &lt;a href=&quot;https://aa.usno.navy.mil/faq/docs/SunApprox.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Approximate Solar Coordinates&lt;/a&gt; formulas on the U.S. Naval Observatory website, you can calculate the ecliptic longitude of the Sun (L) and the distance between the Earth and Sun (R) for any date. Then, The distance travelled between two dates is approximately s=R*(180/pi)*(L2-L1), where R is the average distance during the interval, and L2 and L1 are the ecliptic longitudes. Since the Earth's orbit is almost circular, this approach is reasonably accurate even when performing the calculation over an entire month. These are the results that I get, where the distance s is in Astronomical Units (AU, the average distance between the Earth and Sun).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Month     Distance Travelled,  days ave. daily  ave. change  accl.&#xA;Jan 2019      0.542 AU         31   .00175 AU       0 miles 0 f/ss&#xA;Feb           0.488            28   .00174 AU   - 475 miles 0.000672&#xA;Mar           0.536            31   .00173 AU  - 1300 miles 0.001840&#xA;Apr           0.514            30   .00171 AU  - 1490 miles 0.002104&#xA;May           0.527            31   .00170 AU  - 1210 miles 0.001709&#xA;Jun           0.508            30   .00169 AU   - 651 miles 0.000921&#xA;Jul           0.525            31   .00169 AU       0 miles 0&#xA;Aug           0.526            31   .00170 AU     279 miles 0.000395&#xA;Sep           0.513            30   .00171 AU    1300 miles 0.001840&#xA;Oct           0.535            31   .00173 AU    1490 miles 0.002104&#xA;Nov           0.522            30   .00174 AU    1300 miles 0.001840&#xA;Dec           0.542            31   .00175 AU     744 miles 0.001053&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Of course, the variation from month to month is heavily influenced by the number of days in each month, where as the variation from January to July is due to the distance from the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a check, I calculated the distance travelled for each day in January 2019 and totaled the distance. That result was s=0.5419 AU compared to s=0.5422 AU when using the calculation for the first and end of the month.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastEditorUserId="7895" LastEditDate="2019-04-16T11:07:57.070" LastActivityDate="2019-04-16T11:07:57.070" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30085" PostTypeId="2" ParentId="30074" CreationDate="2019-03-21T17:21:04.883" Score="2" Body="&lt;p&gt;Creating the illusion of weightlessness in movies actually requires a LOT of people operating cranes with harnesses and wires to hold the actors suspended in air. It's not easy moving (nor safe) the equipment holding the actors about as fast as real astronauts are able to move on the ISS. Also, it's important to not let the cranes and their crews fall into the view of the camera. As a result, there's quite a lot of choreography that goes into setting up weightless scenes, deciding the best camera angles, moving the actors, rehearsals, etc. All those factors go into slowing down the action we see in movies. Surprisingly, the wires are easier to hide either because their coloring and thickness are taken advantage of which causes them to be naturally hidden by the film's resolution in older movies, or they can be digitally masked out in post-production if needed in modern films.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mind you, astronauts don't fly about helterskelter on the ISS! They're being very careful, there, too. But moving about the station only needs a very slight push off a wall and the same force to stop them when they get to where they want to be in very short time. But while they're flying between, they'll frequently entertain themselves doing stunts in zeroG such as flips and twists.&lt;/p&gt;&#xA;" OwnerUserId="26323" LastActivityDate="2019-03-21T17:21:04.883" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30086" PostTypeId="1" CreationDate="2019-03-21T17:25:58.017" Score="8" ViewCount="203" Body="&lt;p&gt;Just what the question says. As the universe expands, and space expands, do objects like planets ever expand? Does the space INSIDE of those objects expand?   &lt;/p&gt;&#xA;" OwnerUserId="26324" LastEditorUserId="1891" LastEditDate="2019-03-21T17:39:23.227" LastActivityDate="2020-01-30T09:05:25.827" Title="As the Universe expands, do things like planets expand also, or just the space in between?" Tags="&lt;universe&gt;&lt;space&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="30087" PostTypeId="1" AcceptedAnswerId="30088" CreationDate="2019-03-22T02:38:32.203" Score="42" ViewCount="5311" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_rotation&quot; rel=&quot;noreferrer&quot;&gt;Sun's rotation period&lt;/a&gt; varies from about 25 days at the equator to about 38 days at the poles. As I understand it, this is because the Sun is not solid, and because of the way centripetal force works, the equator must move faster than the poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: if this works, why do Jupiter/Saturn/Uranus/Neptune have well-defined days? Why don't the equators of these planets rotate faster than the poles as well? For example, &lt;a href=&quot;https://en.wikipedia.org/wiki/Jupiter&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia's article on Jupiter&lt;/a&gt; gives the length of a Jovian day as 9h 55m 30s, which is so precise that it implies Jupiter does not have a rotational period which varies with latitude.&lt;/p&gt;&#xA;" OwnerUserId="20348" LastActivityDate="2019-03-22T19:55:59.160" Title="Why does the Sun have different day lengths, but not the gas giants?" Tags="&lt;the-sun&gt;&lt;rotation&gt;&lt;gas-giants&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30088" PostTypeId="2" ParentId="30087" CreationDate="2019-03-22T03:14:13.437" Score="52" Body="&lt;p&gt;It's a matter of how &quot;day&quot; is defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia's article on Jupiter cites &lt;a href=&quot;http://www.hnsky.org/iau-iag.htm&quot; rel=&quot;noreferrer&quot;&gt;this IAU/IAG paper&lt;/a&gt; for the length of a Jupiter day.  In it, footnote (e) of table I has the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The equations for W for Jupiter, Saturn, Uranus and Neptune refer to the rotation of their magnetic fields (System III)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The radio emissions of the gas giants have well-defined periodic variations.  These variations are caused by the rotation of the magnetic fields of those planets, and are evidence that they have a reasonably coherent core of some sort that's rotating at a uniform speed.  The periodic variations then represent the rotation speed of that object, which is taken as the rotation speed of the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're reasonably certain the Sun &lt;em&gt;doesn't&lt;/em&gt; have a coherent core.  Measuring the variation of the magnetic field doesn't show a well-defined period, and doesn't provide a useful definition of the Sun's rotation speed. &lt;/p&gt;&#xA;" OwnerUserId="1559" LastEditorUserId="1559" LastEditDate="2019-03-22T19:55:59.160" LastActivityDate="2019-03-22T19:55:59.160" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30089" PostTypeId="1" AcceptedAnswerId="30095" CreationDate="2019-03-22T03:16:39.103" Score="4" ViewCount="243" Body="&lt;p&gt;Days before, I saw this pic. I don't know what it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: This pic was taken by a Taoist at an ancient ruins in Mongolia on 2012-07-15 04:12 UTC+8 (day break). He said it looked strange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0Mjmd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0Mjmd.png&quot; alt=&quot;Lights in cloudy evening sky&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I installed Stellarium, changed the time to 2012-07-15 04:12 UTC+8, and faced east. It seems to be the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TywFQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TywFQ.png&quot; alt=&quot;It seems the Moon.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26329" LastEditorUserId="616" LastEditDate="2019-03-22T20:54:22.100" LastActivityDate="2019-03-23T05:09:01.217" Title="What's the object in the sky?" Tags="&lt;planet&gt;&lt;amateur-observing&gt;&lt;identify-this-object&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="30090" PostTypeId="2" ParentId="29954" CreationDate="2019-03-22T14:53:24.070" Score="2" Body="&lt;p&gt;The very first sentence of that paper makes it clear that it is concerned with observations of the 21cm line &lt;em&gt;in the redshifted universe&lt;/em&gt;, i.e. on objects so far away from us that redshift becomes significant. And just as any other electromagnetic emission, radio waves are affected by this. So, in section 2.1, the authors describe selecting a sample of radio sources with known redshifts so that they would know how much they had to tune their receivers downwards in frequency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is actually quite common and relevant in today's 21 cm radio observations. Redshifted signals are observed even down to 140 MHz and below as stated on &lt;a href=&quot;http://www.lofar.org/astronomy/eor-ksp/redshifted-21cm-hydrogen-line/redshifted-21cm-hydrogen-line&quot; rel=&quot;nofollow noreferrer&quot;&gt;LOFAR's Redshifted 21cm hydrogen&lt;/a&gt; Line page.&lt;/p&gt;&#xA;" OwnerUserId="20361" LastActivityDate="2019-03-22T14:53:24.070" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30091" PostTypeId="1" AcceptedAnswerId="39177" CreationDate="2019-03-22T18:16:01.427" Score="3" ViewCount="106" Body="&lt;p&gt;The phys.org article&lt;a href=&quot;https://phys.org/news/2018-10-chemical-complexity-saturn-planet-upper.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Surprising chemical complexity of Saturn's rings changing planet's upper atmosphere&lt;/a&gt; about the chemical complexity of Saturn's rings, quotes a member of the INMS (Thomas Cravens) as saying,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What the paper is describing is the environment in the gap between the inner ring and upper atmosphere, and some of the things found were expected, such as water,&quot; Cravens said. &quot;What was a surprise was the mass spectrometer saw methane—no one expected that. Also, it saw some carbon dioxide, which was unexpected. The rings were thought to be entirely water. But &lt;em&gt;the innermost rings are fairly contaminated, as it turns out, with &lt;strong&gt;organic&lt;/strong&gt; material caught up in ice&lt;/em&gt;. (italics and boldface are mine).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the nature of this &lt;em&gt;organic&lt;/em&gt; material?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does he simply mean a few carbon-based molecules like methane, propane, and butane, or maybe &lt;em&gt;only&lt;/em&gt; methane,  or is it a much more interesting cocktail of chemicals than that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also links to &lt;a href=&quot;http://science.sciencemag.org/content/362/6410/eaat2382&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chemical interactions between Saturn’s atmosphere and its rings&lt;/a&gt; published in Science, but the article is paywalled.&lt;/p&gt;&#xA;" OwnerUserId="10026" LastEditorUserId="7982" LastEditDate="2019-03-23T03:15:31.997" LastActivityDate="2020-10-03T08:22:53.767" Title="What is the organic material in Saturn's rings?" Tags="&lt;planetary-ring&gt;&lt;astrochemistry&gt;&lt;saturn-rings&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30095" PostTypeId="2" ParentId="30089" CreationDate="2019-03-23T05:09:01.217" Score="3" Body="&lt;p&gt;The photo shows a conjunction of the Moon and Jupiter.&#xA;A few hours later, observers in Europe&#xA;(&lt;a href=&quot;https://apod.nasa.gov/apod/ap120720.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;APOD&lt;/a&gt;,&#xA;&lt;a href=&quot;https://www.virtualtelescope.eu/2012/08/25/jupiter-occultation-by-the-moon-14-july-2012/&quot; rel=&quot;nofollow noreferrer&quot;&gt;virtualtelescope.eu&lt;/a&gt;)&#xA;saw the Moon pass directly in front of Jupiter, an event called an occultation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stellarium shows Venus and Aldebaran below them, but in the photo they are probably behind a cloud.&#xA;Capella is visible at far left, slightly higher than the Moon.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2019-03-23T05:09:01.217" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30096" PostTypeId="1" CreationDate="2019-03-23T06:48:38.990" Score="6" ViewCount="167" Body="&lt;p&gt;Suppose that we collected photons from a distant star, and they arrive regularly at 15 photons every microsecond onto a CCD, when a gravity wave bends space time, wouldn't the regular 15 photons graph wobble as a result of the gravitational wave?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the LISA observatory have to compare laser timings, if pulsars can be as precise as atomic clocks, can't it just compare a bunch of pulsar signals? A single satellite that records 50 pulsars would move compared the the pulsars and would have a frequency of 50Hz ticks with pulses which can be graphed at kHz frequencies.&lt;/p&gt;&#xA;" OwnerUserId="1528" LastEditorUserId="1528" LastEditDate="2019-03-23T08:52:05.947" LastActivityDate="2019-03-25T07:55:21.797" Title="Can't pulsars and stars be used for gravitational wave measurement?" Tags="&lt;gravity&gt;&lt;astrophysics&gt;&lt;gravitational-waves&gt;&lt;pulsar&gt;&lt;photons&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30098" PostTypeId="1" AcceptedAnswerId="30141" CreationDate="2019-03-23T13:27:40.707" Score="4" ViewCount="217" Body="&lt;p&gt;How do I get the days in this year when the sun will set exactly on top of &lt;a href=&quot;https://www.openstreetmap.org/#map=18/35.36304/138.73040&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fuji&lt;/a&gt; as viewed from &lt;a href=&quot;https://www.openstreetmap.org/#map=21/35.29875/139.47457&amp;amp;layers=C&quot; rel=&quot;nofollow noreferrer&quot;&gt;Enoshima&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I shamefully admit that I just want to take a nice picture of Fuji.&lt;/p&gt;&#xA;" OwnerUserId="26341" LastActivityDate="2019-03-27T12:39:14.353" Title="Sunset on [Mountain] from [Viewing Point]" Tags="&lt;the-sun&gt;&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30099" PostTypeId="2" ParentId="30098" CreationDate="2019-03-23T13:27:40.707" Score="3" Body="&lt;p&gt;My attempt at an answer of my own question. I have no confidence that it is correct, and I'd like to know whether there's an easier way to do this…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.movable-type.co.uk/scripts/latlong.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.movable-type.co.uk/scripts/latlong.html&lt;/a&gt; tells me that the bearing is 90.3°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick application of middle school trigonometry tells me that the visible elevation of Fuji is about 3413 m or 2.90°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, I went to Stellarium, fixed my position to Enoshima, and exported the Ephemeris (F10 menu) of the Sun in a 5 minute intervals with horizontal coordinates from February till May.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I used a quick python script to find the line in the Ephimeris export where the sun is closest to my desired coordinates (90.3°, 2.9°)&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import re&#xA;from pprint import pprint&#xA;&#xA;height = &quot;Height&quot;&#xA;azi = &quot;Azimut&quot;&#xA;&#xA;# Quick and dirty parser&#xA;pre = re.compile(&quot;([+-])(\\d+)°(\\d+)'([0-9.]+)\&quot;&quot;)&#xA;def ph(v):&#xA;    m = pre.match(v)&#xA;    if m:&#xA;        sign, *nums = m.groups()&#xA;        return sum([float(s) * 60 ** (-e) for e,s in enumerate(nums)]) * float(sign + &quot;1&quot;)&#xA;    try:&#xA;        return float(v)&#xA;    except:&#xA;        pass&#xA;    return v&#xA;&#xA;res = []&#xA;with open('ephemeris2.csv') as f:&#xA;    head = next(f).split(&quot;, &quot;)&#xA;    for line in f:&#xA;        php = [ph(v) for v in line.split(&quot;, &quot;)]&#xA;        res.append(dict(zip(head, php)))&#xA;&#xA;# This is where the magic happens!&#xA;filt = [k for k in res if k is not None and height in k and azi in k]&#xA;filt.sort(key = lambda v: (v[height] - 2.9) ** 2 + (v[azi] - 270.3) ** 2)&#xA;pprint(filt[:5])&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This yielded me&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{ 'Azimut': 270.13194444444446,&#xA;  'Datum and Time': '2019-03-25 17:39:00',&#xA;  'Height': 3.0792222222222225 }&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm unsure whether I can just convert the Bearing to Azimuth like that, but I suspect that a simple mathematical mistake is even more likely…&lt;/p&gt;&#xA;" OwnerUserId="26341" LastEditorUserId="26341" LastEditDate="2019-03-23T15:09:50.250" LastActivityDate="2019-03-23T15:09:50.250" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30100" PostTypeId="2" ParentId="30064" CreationDate="2019-03-23T21:10:11.867" Score="1" Body="&lt;p&gt;First off I want to say I am a complete amateur when it comes to stars and even bigger when it comes to optics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was looking at sky from indoors and my windows have double glass. It turns out that this was the problem or maybe the glass on the window is not some good quality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, all works well now and I am thrilled at what I'm seeing in the night sky.&lt;/p&gt;&#xA;" OwnerUserId="26304" LastActivityDate="2019-03-23T21:10:11.867" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30103" PostTypeId="1" AcceptedAnswerId="30106" CreationDate="2019-03-24T05:14:10.950" Score="0" ViewCount="70" Body="&lt;p&gt;If the heavier elements such as Uranium came from a supernova, then this supernova had to sit somewhere right in the middle of the original solar nebula, because although the ejected matter traveled extremely fast by star or planet standard, it was still not fast enough to truly get across the empty space between clusters of stars. Does this mean the supernova was actually birthed and died in the same nebula or at least in the immediate vicinity in a region that is rich in hydrogen? Could the solar system be still in this nebula? What could endow the solar system as a whole with so much momentum that it acquired a different trajectory around the galaxy than its mother nebula, if the solar nebula is in effect no where to be found?&lt;/p&gt;&#xA;" OwnerUserId="26231" LastActivityDate="2019-03-24T13:05:12.790" Title="How could a supernova seed solar nebula?" Tags="&lt;solar-system&gt;&lt;star-formation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30106" PostTypeId="2" ParentId="30103" CreationDate="2019-03-24T12:09:53.967" Score="3" Body="&lt;p&gt;The Uranium on Earth did not come from a &lt;em&gt;single&lt;/em&gt; supernova. The Earth formed from material that had been enriched by perhaps a billion stars that lived and died before the Sun and solar system were born. That material, enriched by heavier elements, is turbulently mixed within the galactic interstellar medium and is then gathered together to form new generations of stars. The mixing timescale is much shorter than the age of the Galaxy (see for example &lt;a href=&quot;https://astronomy.stackexchange.com/questions/16311/how-can-there-be-1-000-stellar-ancestors-before-our-sun?noredirect=1&amp;amp;lq=1&quot;&gt;How can there be 1,000 stellar ancestors before our Sun?&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is possible that one or two local supernovae did imprint a pattern of odd abundances for particularly rare or short-lived (radioactive) isotopes. It is widely believed that an explanation for the inferred high ratio of &lt;span class=&quot;math-container&quot;&gt;$^{26}$&lt;/span&gt;Al/&lt;span class=&quot;math-container&quot;&gt;$^{27}$&lt;/span&gt;Al (the former being radioactive with a half-life of less than a million years) in the early solar system, indicates that it was enriched by the ejecta from a nearby supernova. This may suggest that the Sun was born in a cluster of stars that was numerous enough to have contained very high mass stars capable of exploding as supernovae on short timescales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding the solar systems birth place or its siblings is proving to be a difficult (and perhaps impossible) task, given the time that has elapsed and the multitude of effects that act to disperse stars (and their natal gas) on much shorter timescales (see &lt;a href=&quot;https://astronomy.stackexchange.com/questions/22694/parent-stars-of-our-sun-where-are-its-remains?noredirect=1&amp;amp;lq=1&quot;&gt;Parent stars of our Sun - Where are its remains?&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="16685" LastEditDate="2019-03-24T13:05:12.790" LastActivityDate="2019-03-24T13:05:12.790" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30108" PostTypeId="2" ParentId="30096" CreationDate="2019-03-24T18:08:43.803" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can't pulsars and stars be used for gravitational wave measurement?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In theory, yes. See the 2017 NASA article &lt;a href=&quot;https://www.nasa.gov/feature/jpl/listening-for-gravitational-waves-using-pulsars&quot; rel=&quot;nofollow noreferrer&quot;&gt;Listening for Gravitational Waves Using Pulsars&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose that we collected photons from a distant star, and they arrive regularly at 15 photons every microsecond onto a CCD, when a gravity wave bends space time, wouldn't the regular 15 photons graph wobble as a result of the gravitational wave?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Kind of. When the gravitational wave is passing by, we'll be subject to a slight time dilation. So it would look as if the pulsar had speeded up. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does the LISA observatory have to compare laser timings, if pulsars can be as precise as atomic clocks, can't it just compare a bunch of pulsar signals? A single satellite that records 50 pulsars would move compared the pulsars and would have a frequency of 50Hz ticks with pulses which can be graphed at kHz frequencies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sorry, I don't know the answer to that. But see the &lt;a href=&quot;https://en.wikipedia.org/wiki/Laser_Interferometer_Space_Antenna&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; on LISA and note that it's scheduled for the 2030s. A lot can happen between now and then. Maybe it won't go ahead.   &lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2019-03-24T18:08:43.803" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30109" PostTypeId="2" ParentId="30096" CreationDate="2019-03-24T23:04:29.313" Score="4" Body="&lt;p&gt;&lt;strong&gt;How LIGO, LISA, etc. Detect Gravitational Waves&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point of instruments like LIGO and LISA is to measure time-varying changes in the distance within different arms of the instrument. In the case of an arm oriented in the direction of an incoming gravitational wave (GW), the length of the arm will increase and  decrease, while an arm oriented perpendicular will remain unchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/LIGO#Operation&quot; rel=&quot;nofollow noreferrer&quot;&gt;method for measuring this&lt;/a&gt; is to split a laser beam, send each half down the different arms, and then recombine them and look at the interference pattern. If the arms have identical lengths, then the beams will destructively interfere. If, however, one of the arms becomes longer (or shorter), the recombined beams will be out of phase, and you will no longer have perfect destructive interference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key to being able to measure this sort of thing accurately is two things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Arm length, so the beam traveling down the altered-length path has time&#xA;to accumulate a large enough phase shift for the interference pattern to&#xA;change enough to be measured. Longer arms lead to larger phase shifts, &#xA;which is why LISA aims to have arms that are 2.5 million km in length.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The wavelength of the light: the shorter the wavelength, the smaller the&#xA;actual shift in arm length needs to be to make the beam shift appreciably&#xA;in phase. If the recombined beams differ in phase by 0.001%, then the change&#xA;in the interference pattern will be very hard to detect and measure. But if they differ&#xA;in phase by 50%, then the difference will be obvious. Shorter wavelengths&#xA;make for larger phase shifts.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you want think of the interferometric measurement as a kind of timing measurement (that is, we are, in a sense, measuring the difference in time it takes light to travel down the two arms), then &lt;em&gt;optical light is vastly better than radio for &quot;timing&quot; measurements&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pulsar Timing&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you pointed out, there &lt;em&gt;are&lt;/em&gt; projects that use pulsars to try to detect gravitational waves (e.g., &lt;a href=&quot;https://en.wikipedia.org/wiki/Pulsar_timing_array&quot; rel=&quot;nofollow noreferrer&quot;&gt;pulsar timing arrays&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These work by measuring time delays in the arrival of radio pulses from millisecond pulsars. However, it's important to note the limitations: in order to measure the arrival time accurately, you have to observe a pulsar over a period of several minutes (or more) and &lt;em&gt;add up&lt;/em&gt; all the pulse arrivals to get a single, accurate value. (You are &lt;em&gt;not&lt;/em&gt; measuring changes between the arrival of one single pulse and the next.) You then repeat this measurement several weeks or months later to get another arrival time. Since you're taking measurements weeks or months apart, you can only detect variations on that same time scale, which is why pulsar timing arrays are hoping to detect GWs with periods of months to years (e.g., from binary supermassive black holes). The kinds of GWs that LIGO detects -- with periods of &lt;em&gt;fractions of a second&lt;/em&gt; -- are completely beyond what pulsar timing arrays can do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We cannot use optical light from stars, because we know of no star that produces the exquisitely regular periodicity of a millisecond pulsar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a final comment, your scenario of photons that &quot;arrive regularly at 15 photons every microsecond onto a CCD&quot; has the problem that Poisson noise (a.k.a. &lt;a href=&quot;https://en.wikipedia.org/wiki/Shot_noise&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;shot noise&quot;&lt;/a&gt;) means that you &lt;em&gt;wouldn't&lt;/em&gt; actually get 15 photons every microsecond; you'd get &lt;em&gt;around&lt;/em&gt; 15 every microsecond (a typical sequence might look like this: 16, 15, 18, 15, 14, 14, 15, 23, 16, 12, ...). With that much variation, it's really hard to detect subtle variations in intrinsic signal strength.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2019-03-24T23:21:29.317" LastActivityDate="2019-03-24T23:21:29.317" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30111" PostTypeId="1" CreationDate="2019-03-25T01:54:37.587" Score="0" ViewCount="62" Body="&lt;p&gt;The BBC News article &lt;a href=&quot;https://www.bbc.com/news/science-environment-43477352&quot; rel=&quot;nofollow noreferrer&quot;&gt;UK will lead European exoplanet mission&lt;/a&gt; says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Steady platform&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;And one aspect that would work in Ariel's favour is the absence of any moving parts in its build&lt;/strong&gt;, commented Plato team-member Dr Don Pollaco from Warwick University, UK.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;The issue with all of these planet experiments is that the signals you are looking for are so incredibly small that any systematics in the instrument itself will dominate the signal,&amp;quot; he explained.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;And the systematics are often associated with bits that move. So the great thing about Ariel is that &lt;strong&gt;it is fixed-format - nothing changes&lt;/strong&gt;,&amp;quot; he told BBC News.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What's fixed in Ariel that is moving in other planet-hunting space telescopes? Is it only aspects of the satellite's telescope and payload module, or does this difference include aspects of the satellite's attitude control system and other things in the service module as well? See ESA's &lt;a href=&quot;http://sci.esa.int/ariel/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ariel&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-03-29T03:29:54.177" Title="What's fixed in Ariel that is moving in other planet-hunting space telescopes?" Tags="&lt;exoplanet&gt;&lt;space-telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30112" PostTypeId="1" AcceptedAnswerId="30117" CreationDate="2019-03-25T03:41:49.540" Score="1" ViewCount="151" Body="&lt;p&gt;What orbit in our solar system would take the least time to perform 1 orbit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the fastest possible orbit to complete in the universe?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2019-03-25T03:58:50.400" LastActivityDate="2019-03-28T13:59:49.263" Title="What orbit is the smallest or takes the least time to complete?" Tags="&lt;orbit&gt;&lt;astrophysics&gt;&lt;natural-satellites&gt;&lt;orbital-mechanics&gt;&lt;time&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30113" PostTypeId="1" AcceptedAnswerId="30122" CreationDate="2019-03-25T04:22:18.720" Score="0" ViewCount="318" Body="&lt;p&gt;I've seen several papers that discuss the Lyman-Werner Background (LWB) when modeling star and galaxy formation in the early universe. One in particular (&lt;a href=&quot;https://arxiv.org/abs/1803.04527&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/1803.04527&lt;/a&gt;) even varies the strength (?) of the LWB using the units &lt;span class=&quot;math-container&quot;&gt;$J_{21}$&lt;/span&gt;. I assume the subscript is referencing the 21cm band, as opposed to a more generic use of &lt;span class=&quot;math-container&quot;&gt;$J_{c}$&lt;/span&gt;. But I am equally confused as to what the flux in terms of J is referring to. I've found and read on Lyman-Werner photons but going from these photons to a variable &quot;background&quot; is where I'm having a disconnect. &lt;/p&gt;&#xA;" OwnerUserId="10995" LastActivityDate="2019-03-25T14:27:55.053" Title="What is the lyman-werner background?" Tags="&lt;cosmology&gt;&lt;star-formation&gt;&lt;modeling&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30114" PostTypeId="2" ParentId="30096" CreationDate="2019-03-25T07:55:21.797" Score="0" Body="&lt;p&gt;The other point to raise, is the expected frequency of the gravitational waves you detect varies between Lisa and pulsar timing systems. This then alters the types of systems you will detect. Pulsar timing arrays should find massive black hole mergers, while Lisa should find; stellar mass black holes years before they merge, white dwarf binaries in the galaxy (also not merging), as well as intermediate mass black holes (merging). So the different detectors have different uses, even if one could be made more accurate than the other.&lt;/p&gt;&#xA;" OwnerUserId="16852" LastActivityDate="2019-03-25T07:55:21.797" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30115" PostTypeId="2" ParentId="30112" CreationDate="2019-03-25T09:33:05.083" Score="3" Body="&lt;p&gt;(This is a partial answer only addressing the second question; ideally this question should be split into at least two separate questions)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically, fastest possible orbits in terms of orbital velocity can be found around black holes. If we consider photons, the orbital speed would be equal to c in photon sphere. For massive particles, innermost stable circular orbit should be the fastest, both in terms of average speed as well as time it takes to complete one full orbit. For the latter, you’d want the smallest possible black hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have however no clear understanding what is the smallest black hole currently observed and whether we are able to detect distinct objects orbiting in order to establish a solid measurement.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One good reference point for this could be derived from LIGO observations of binary black hole mergers: I believe the last orbits take fractions of seconds as gravitational wave frequencies are peaking at around 700Hz which could mean 700 or 350 orbits per second if I understand the mechanics correctly. In this case, orbital radius being in the range of tens or hundreds km, the orbital speed is very close to the speed of light; can’t recall exact estimates right now though. &lt;/p&gt;&#xA;" OwnerUserId="10813" LastEditorUserId="10813" LastEditDate="2019-03-25T09:39:56.430" LastActivityDate="2019-03-25T09:39:56.430" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30117" PostTypeId="2" ParentId="30112" CreationDate="2019-03-25T11:28:21.197" Score="7" Body="&lt;p&gt;In general, the closer you are to the primary, the shorter the orbital period, except of course that you can't orbit &lt;em&gt;inside&lt;/em&gt; the primary. So, taking the primary to be a sphere of radius &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; and density &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; we find that the mass is &lt;span class=&quot;math-container&quot;&gt;$$\frac{4}{3}\pi R^3\rho$$&lt;/span&gt; so that the acceleration due to gravity at the surface is &lt;span class=&quot;math-container&quot;&gt;$$G\frac{4}{3}\pi R\rho$$&lt;/span&gt; For a surface-grazing orbit, this has to equal &lt;span class=&quot;math-container&quot;&gt;$R\omega^2$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; is the angular velocity, so we get &lt;span class=&quot;math-container&quot;&gt;$$\omega^2 = \frac{4}{3}\pi G\rho$$&lt;/span&gt; Now the orbital period &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$2\pi/\omega$&lt;/span&gt; so we get  &lt;span class=&quot;math-container&quot;&gt;$$P = \sqrt{\frac{3\pi}{G\rho}}$$&lt;/span&gt;&#xA;So the answer, at least for spherical primaries) is going to be a close orbit around the densest object in the solar system. The &lt;a href=&quot;https://starchild.gsfc.nasa.gov/docs/StarChild/teachers/densities.html&quot; rel=&quot;noreferrer&quot;&gt;densest planet&lt;/a&gt; is Earth! &#xA;There are denser objects, such as &lt;a href=&quot;https://en.wikipedia.org/wiki/LAGEOS&quot; rel=&quot;noreferrer&quot;&gt;the LAGEOS satellites&lt;/a&gt;, but I'm not sure if it is possible to orbit LAGEOS under its own gravity, because of the perturbing effects of Earth's gravity.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="21928" LastEditDate="2019-03-25T13:22:35.027" LastActivityDate="2019-03-25T13:22:35.027" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30121" PostTypeId="2" ParentId="30071" CreationDate="2019-03-25T13:43:45.600" Score="1" Body="&lt;p&gt;Gravitational waves cannot be part of dark energy, because -- just like electromagnetic waves -- their energy density dilutes as the universe expands: both due to the simple expansion of space &lt;em&gt;and&lt;/em&gt; due to the cosmological redshifting that the expansion produces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Dark_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dark energy&lt;/a&gt;, on the other hand, has an energy density that remains &lt;em&gt;constant&lt;/em&gt; as the universe expands (truly constant if it's the cosmological constant, very close to constant if it's something stranger like quintessence), so it can't be made up of GWs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GWs probably don't contribute much to the energy density we attribute to dark &lt;em&gt;matter&lt;/em&gt;, either, since dark matter is diluted by the cosmological expansion but &lt;em&gt;not&lt;/em&gt; by redshift. So if the energy density attributed to DM ten billion years ago were due to GWs, then it would be significantly lower than the DM energy density we measure today. (Also, current cosmological observations indicate significant dark matter was present very early in the universe, &lt;em&gt;before&lt;/em&gt; the star formation that was necessary to create binary black holes in the first place.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The GW energy from a binary-black-hole merger &lt;em&gt;is&lt;/em&gt; enormous, yes; but those are very rare events. The current estimate from LIGO observations is about 50 or 100 such mergers per year per cubic gigaparsec, which translates into a rate of less than one per million years per galaxy. &lt;a href=&quot;https://arxiv.org/abs/1710.05837&quot; rel=&quot;nofollow noreferrer&quot;&gt;This paper by the LIGO and Virgo Collaborations&lt;/a&gt; has estimates of the current GW energy density in units of the current energy density of the universe (their Figure 1): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xxpt1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xxpt1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The green curve shows the estimated energy density from binary-black-hole mergers as a function of the frequency of the GWs; the red curve shows the contribution from binary-neutron-star mergers. The combined peak is always less than &lt;span class=&quot;math-container&quot;&gt;$10^{-8}$&lt;/span&gt; of the critical energy density, so even summing up over frequencies, we have an energy density much smaller than that due to ordinary matter (&lt;span class=&quot;math-container&quot;&gt;$\Omega \sim 5$&lt;/span&gt;%), so it's probably safe to ignore its effects. &lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2019-03-25T13:58:26.030" LastActivityDate="2019-03-25T13:58:26.030" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30122" PostTypeId="2" ParentId="30113" CreationDate="2019-03-25T14:27:55.053" Score="3" Body="&lt;p&gt;The Lyman-Werner background (LWB) is the part of the &quot;meta-galactic&quot; background radiation consisting of photons that are able to photo-dissociate molecular hydrogen (&lt;span class=&quot;math-container&quot;&gt;$H_\mathrm{2}$&lt;/span&gt;), but not ionize atomic hydrogen (HI). These photons are in ultraviolet, with energies ranging from &lt;span class=&quot;math-container&quot;&gt;$11.2\,\mathrm{eV}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$13.6\,\mathrm{eV}$&lt;/span&gt;, originating mainly from hot, massive stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of the reason this radiation field is interesting particularly in the early Universe is that a sufficiently high intensity will prevent gas from cooling. In order for gas to form stars, it cannot be too hot, since a hot gas cannot collapse to form structure. A gas with a high metallicity is able to cool through the many different atomic transitions in the metals, but in the early Universe, where stars haven't yet produced the metals, the most important coolant at these temperatures is &lt;span class=&quot;math-container&quot;&gt;$H_\mathrm{2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also has implications for the formation of supermassive black holes in the early Universe, since once you form a large amount of stars, the feedback from the massive stars (both through radiation pressure and heating of the interstellar medium) makes it difficult to have gas collapse directly into the supermassive black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In simulations, you often have to &lt;em&gt;assume&lt;/em&gt; a value of the LWB, since performing the &quot;true&quot; radiative transfer is computationally very expensive. Hence, these authors apparently have experimented with two different fields of &lt;span class=&quot;math-container&quot;&gt;$100\,J_\mathrm{21}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$1000\,J_\mathrm{21}$&lt;/span&gt;, respectively. Here, the term &lt;span class=&quot;math-container&quot;&gt;$J_\mathrm{21}$&lt;/span&gt; doesn't have anything to do with &lt;span class=&quot;math-container&quot;&gt;$21\,\mathrm{cm}$&lt;/span&gt; radiation (which is in the other end of the spectrum), but is simply the intensity in units of &lt;span class=&quot;math-container&quot;&gt;$10^{-21}\,\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}\,\mathrm{Hz}^{-1}\,\mathrm{sr}^{-1}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2019-03-25T14:27:55.053" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30123" PostTypeId="1" CreationDate="2019-03-25T15:26:56.110" Score="1" ViewCount="78" Body="&lt;p&gt;Suppose 2 spaceships leave from the earth's north and South Pole going straight up which one will exit the galaxy first (assume they have the same speed and are going in a totally straight line) ? Edge of the galaxy means absence of stars, this question refers to the up-down edge of this spiral galaxy.&#xA;The question is about the current distance from the earths poles to the edge (top/down) not to the galactic plane, as far as I understand after crossing the plane the nearest pole to the edge will be reversed.&lt;/p&gt;&#xA;" OwnerUserId="17457" LastEditorUserId="17457" LastEditDate="2019-03-27T09:17:41.760" LastActivityDate="2019-03-27T09:17:41.760" Title="What pole is closer to the galactic edge?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;" AnswerCount="0" CommentCount="11" ClosedDate="2019-03-27T12:16:24.720" ContentLicense="CC BY-SA 4.0" />
  <row Id="30124" PostTypeId="1" AcceptedAnswerId="30129" CreationDate="2019-03-25T17:14:15.537" Score="4" ViewCount="193" Body="&lt;p&gt;When I try to find figures for the amount of helium today and just after the Big Bang, I get similar figures.  I was expecting that all of the stars since then would have created more helium.  Is the amount below our ability to estimate?  Is helium being consumed at a similar or greater rate to produce heavier elements?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, can we measure/ estimate a drop in the hydrogen?  I presume that little hydrogen is being created.  I am unaware of any common nuclear process which creates lone protons.  (Please correct me if appropriate.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarification: The Wikipedia Big Bang &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Big_Bang&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; mentions helium being created after a few minutes in the Big Bang nucleosynthesis.  So, it is helium created after this point that I am asking about.  I am presuming that negligible helium was created between this point and the first stars much later.  (Again, please correct me if I am wrong.)&lt;/p&gt;&#xA;" OwnerUserId="21104" LastEditorUserId="21104" LastEditDate="2019-03-26T08:22:59.487" LastActivityDate="2019-03-26T08:22:59.487" Title="How much helium has been created since the Big Bang?" Tags="&lt;big-bang-theory&gt;&lt;helium&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30125" PostTypeId="1" CreationDate="2019-03-25T21:04:48.233" Score="4" ViewCount="102" Body="&lt;p&gt;How do I calculate the thickness of the upper ice layer on cold ocean worlds like Europa, Enceladus, Ganymede, ...? I'm asking this for a programm I'm currently writing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given/Known is: mass, radius, average density of metal (assume Fe) core, rock (assume MgSiO3) layer, ice (assume H2O) layer; internal heat from tidal heating, radioactive decay and residual formation heat; external heating from solar radiation (assume equal constant heating distributed over the entire surface or ignore it if it is irrelevant/minuscule); pressure of the atmosphere above (assume pure N2 with variable pressure or no atmosphere)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wanted is: thickness frozen (crust) and molten layer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus stuff: If easy adaptation of the formula for the calculation of the rocky layers crust thickness is possible I would appreciate it. Should it be possible, an adaptation to find the rocky crust thickness and the ice crust thickness on wolds where both are needed would be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any answer and time spent of solving this.&lt;/p&gt;&#xA;" OwnerUserId="26374" LastActivityDate="2019-03-29T10:20:21.837" Title="How to calculate the thickness of the ice crust on a frozen ocean planet/moon?" Tags="&lt;planet&gt;&lt;water&gt;&lt;geology&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30126" PostTypeId="1" AcceptedAnswerId="30136" CreationDate="2019-03-25T22:50:35.543" Score="3" ViewCount="122" Body="&lt;p&gt;How strongly are objects bound to the Lagrange points of Earth that they inhabit in them in the same way Jupiter's Trojan meteors are entrapped in its Lagrange points?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the co-moving frame with the two main objects, how deep is the potential energy well of L4/L5?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="23748" LastEditDate="2019-03-26T15:28:33.870" LastActivityDate="2019-03-27T04:46:06.633" Title="How strong are the Lagrange Points of Earth?" Tags="&lt;magnitude&gt;&lt;lagrange-point&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="30127" PostTypeId="1" AcceptedAnswerId="30128" CreationDate="2019-03-26T01:05:03.223" Score="3" ViewCount="133" Body="&lt;p&gt;We know that presently Sun is 4.5 billion years into its main sequence. It has another 5 billion years before it enters the Red Giant phase. We also know that Sun's luminosity increases by 10% every billion years during the main sequence. I am interested in finding the temperature rise as we approach the end of main sequence. I got two different values for temperature on Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Future_of_Earth#Loss_of_oceans&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; entry says that temperature on Earth would be 422 k in 2.8 billion years. However, if we use the formula for effective temperature as discussed in this answer &#xA;&lt;a href=&quot;https://earthscience.stackexchange.com/a/4274/15299&quot;&gt;https://earthscience.stackexchange.com/a/4274/15299&lt;/a&gt; and L = 1.8 then, the temperature on Earth would be 330K. Also in &lt;a href=&quot;https://www.springer.com/us/book/9783319649603&quot; rel=&quot;nofollow noreferrer&quot;&gt;this book&lt;/a&gt;, the author does the same calculations on page 255. &lt;/p&gt;&#xA;" OwnerUserId="26375" LastActivityDate="2019-03-26T02:49:28.497" Title="What will be the temperature on Earth when Sun finishes its main sequence?" Tags="&lt;earth&gt;&lt;main-sequence&gt;&lt;red-giant&gt;&lt;hypothetical&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30128" PostTypeId="2" ParentId="30127" CreationDate="2019-03-26T02:49:28.497" Score="6" Body="&lt;p&gt;The difference is that your analysis is assuming that the albedo stays fixed, so the surface temperature simply scales like luminosity to the 1/4 power.  The Wiki entry is including feedback from the greenhouse effect, which will tend to further increase the surface temperature.  Note that an analysis that just looks at solar irradiation would get way too low a surface temperature for Venus, for example.  I can't speak to the accuracy of the feedback included-- as I understand it, that is far from a simple effect to include.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2019-03-26T02:49:28.497" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30129" PostTypeId="2" ParentId="30124" CreationDate="2019-03-26T03:22:47.723" Score="6" Body="&lt;p&gt;We can estimate the answer by considering the &quot;mass to light ratio&quot; for baryons, and assuming that ratio hasn't changed too much over the time that there has been galaxies, say for ten billion years.  The Wiki (&lt;a href=&quot;https://en.wikipedia.org/wiki/Mass-to-light_ratio&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Mass-to-light_ratio&lt;/a&gt;) on the mass to light ratio says this ratio for the universe as a whole is about 100 times what it is for the Sun right now, but that includes dark matter.  If we restrict to baryon mass-to-light, that would be more like 16 times the Sun right now.  The Sun's main-sequence lifetime is about ten billion years, conveniently, and in that time it will convert perhaps 20% of its hydrogen, so a mass-to-light ratio 16 times larger would convert only about 1% of its mass from hydrogen to helium.  So that says roughly 1% of all the hydrogen in the universe has already been converted to helium.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it started out 75% hydrogen and 25% helium by mass, now it should be more like 74% hydrogen and 26% helium by mass.  So although the estimate is crude, I think it's fair to say we are just now beginning to see a trace of an effect, which would be hard to see after only 1 billion years, and will be quite obvious after 100 billion years.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2019-03-26T03:22:47.723" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30131" PostTypeId="1" AcceptedAnswerId="30150" CreationDate="2019-03-26T11:07:14.613" Score="2" ViewCount="324" Body="&lt;p&gt;I know we have &quot;exoplanets&quot; etc. but those are nouns for specific objects outside our solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have extraterrestrial for objects outside Earth's atmosphere, but I don't know if we have a general adjective for objects outside our solar system.&lt;/p&gt;&#xA;" OwnerUserId="26379" LastEditorUserId="26379" LastEditDate="2019-03-26T11:35:40.963" LastActivityDate="2019-03-28T17:45:48.100" Title="Adjective for things outside our solar system" Tags="&lt;solar-system&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30132" PostTypeId="1" CreationDate="2019-03-26T15:49:11.890" Score="1" ViewCount="81" Body="&lt;p&gt;Given the orbit of the Earth around the Sun and the position of the Sun in the Galaxy, in which season are we (the Earth) closer to the galactic center?&lt;/p&gt;&#xA;" OwnerUserId="16793" LastActivityDate="2019-03-26T16:59:52.150" Title="Earth orbit around the Sun and its position in the Galaxy" Tags="&lt;orbit&gt;&lt;milky-way&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30133" PostTypeId="2" ParentId="30132" CreationDate="2019-03-26T16:59:52.150" Score="6" Body="&lt;p&gt;The radio source &lt;a href=&quot;https://en.wikipedia.org/wiki/Sagittarius_A*&quot; rel=&quot;noreferrer&quot;&gt;Sgr A*&lt;/a&gt; is at RA 17h46m Dec -29&amp;deg;, or about 6&amp;deg; south of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ecliptic&quot; rel=&quot;noreferrer&quot;&gt;ecliptic&lt;/a&gt; at longitude 267&amp;deg;.&#xA;The Sun appears most nearly opposite that position, ecliptic longitude 87&amp;deg;, about three days before the June solstice.&#xA;At that time of year, the brightest part of the Milky Way is highest in the sky around local solar midnight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the distance from here to Sgr A* is about 7.9 kpc or 1.6 billion &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_unit&quot; rel=&quot;noreferrer&quot;&gt;au&lt;/a&gt;.&#xA;The Earth is only infinitesimally closer to the galactic center in June than in December.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2019-03-26T16:59:52.150" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30135" PostTypeId="2" ParentId="11686" CreationDate="2019-03-27T00:29:37.160" Score="0" Body="&lt;h2&gt;No.&lt;/h2&gt;&#xA;&lt;p&gt;Per the beginning of @RobJeffries' &lt;a href=&quot;https://astronomy.stackexchange.com/a/23166/7982&quot;&gt;excellent answer&lt;/a&gt; to &lt;em&gt;Are binary neutron star mergers needed to explain the abundance of gold?&lt;/em&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The creation of some very heavy neutron-rich elements, like gold and platinum, requires the rapid capture of neutrons. This will only occur in dense, explosive conditions where the density of free neutrons is large. For a long time, the competing theories and sites for the r-process have been inside core-collapse supernovae and during the merger of neutron stars.&lt;/p&gt;&#xA;&lt;p&gt;My understanding is that it has become increasingly difficult for supernovae to produce (in theoretical models) sufficient r-process elements to match both the quantity and detailed abundance ratios of r-process elements in the solar system (see for example Wanajo et al. 2011; Arcones &amp;amp; Thielmann 2012). The conditions required, particularly a very neutron-rich environment in the neutrino-driven winds, are just not present without the fine tuning of parameters (see below).&lt;/p&gt;&#xA;&lt;p&gt;Instead, the models invoking neutron star mergers are much more robust to theoretical uncertainties and successfully produce r-process elements. The question mark appears to be only over their frequency at various times in the evolution of a galaxy and exactly how much enriched material is ejected.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This kind of environment has never existed and will not ever exist in the Sun's lifetime based on a pretty solid understanding of stellar evolution.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-03-27T00:29:37.160" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30136" PostTypeId="2" ParentId="30126" CreationDate="2019-03-27T04:46:06.633" Score="1" Body="&lt;p&gt;Because The triangular points (L4 and L5) are stable &lt;a href=&quot;https://en.wikipedia.org/wiki/Lagrangian_point&quot; rel=&quot;nofollow noreferrer&quot;&gt;equilibria&lt;/a&gt;, provided that the ratio of M1/M2, moon 0.166 g and earth 9.8 g  so 0.0169 g in L4 and L5&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2019-03-27T04:46:06.633" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30137" PostTypeId="1" CreationDate="2019-03-27T05:43:20.450" Score="2" ViewCount="133" Body="&lt;p&gt;The open access paper &lt;a href=&quot;https://agupubs.onlinelibrary.wiley.com/doi/full/10.1029/2019GL081961&quot; rel=&quot;nofollow noreferrer&quot;&gt;Formation of a New Great Dark Spot on Neptune in 2018&lt;/a&gt; (cited in Gizmodo's &lt;a href=&quot;https://gizmodo.com/formation-of-dark-vortex-on-neptune-captured-for-the-ve-1833555871&quot; rel=&quot;nofollow noreferrer&quot;&gt;Formation of Dark Vortex on Neptune Captured For the Very First Time&lt;/a&gt; and Phys.org's &lt;a href=&quot;https://phys.org/news/2019-03-hubble-captures-birth-giant-storm.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble captures birth of giant storm on Neptune&lt;/a&gt;) shows images of Neptune taken by the Hubble Space Telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The projections are called &quot;global cylindrical maps&quot; and those usually result in weirdly stretched features near the poles but do not result in the spiky hairdo or &lt;a href=&quot;https://en.wikipedia.org/wiki/Dragon%27s_Egg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cheela&lt;/a&gt;-like appearance in the Hubble  images below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TUV5G.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TUV5G.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://radiorobot-cnc.com/art&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What causes this effect seen in these projections?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vwI86.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vwI86.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Individual maps from Hubble images in 2017 through 2018, in the same filters and processing as Figure 1. These maps span ±90° of latitude and longitude, and tick marks indicate 30° latitude and longitude intervals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/30vBA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/30vBA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;False‐color global cylindrical maps from Hubble, spanning 180° of latitude and 360° of longitude, from images at 845 (red), 547 (green), and 467 nm (blue), and all use the same contrast scaling with a light unsharp mask applied. The maps are centered on 180°W longitude, except 2015, which was centered on 0° longitude. Dotted lines are shown at ~10° and 35°N to indicate the active region prior to the discovery of NDS‐2018.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Reference Cheela, from &lt;a href=&quot;https://www.amazon.fr/Starquake-Robert-L-Forward/dp/0345312333&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bC7Eh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bC7Eh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2021-03-18T23:55:15.773" LastActivityDate="2021-03-19T03:27:23.193" Title="Why do these Hubble images of Neptune look like Cheela?" Tags="&lt;photography&gt;&lt;data-analysis&gt;&lt;hubble-telescope&gt;&lt;neptune&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="30138" PostTypeId="1" CreationDate="2019-03-27T05:48:43.833" Score="3" ViewCount="140" Body="&lt;p&gt;We are used to viewing pictures of the Earth and the planets with their North Pole pointing towards the upper portion of the photo. But when a space telescope snaps photos of cosmic entities, do they rotate the photo around to get an attractive composition? Is it a subjective rotation, or are there rules? Which way is up?&lt;/p&gt;&#xA;" OwnerUserId="25289" LastEditorUserId="23881" LastEditDate="2019-03-27T13:33:18.137" LastActivityDate="2019-03-28T12:22:31.080" Title="Pictures of galaxies, nebulae: which way is up?" Tags="&lt;photography&gt;&lt;space-telescope&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30140" PostTypeId="2" ParentId="26909" CreationDate="2019-03-27T10:22:17.740" Score="1" Body="&lt;p&gt;With relatively simple equipment it is possible to observe the behaviour of the satellites of Jupiter. Assuming the hypothesis that Jupiter and all the planets rotate around the Earth, it should be expected that the occlusion of the satellites by Jupiter would happen on a highly regular basis. But what we see is the event happening at different times relative to Earth-bound clocks, even not very accurate ones, which proves that the orbit of Jupiter is not a simple epicycle around the Earth. Also the observation of any satellite not directly orbiting the Earth casts doubt on the Earth-centric view.&lt;/p&gt;&#xA;" OwnerUserId="26393" LastActivityDate="2019-03-27T10:22:17.740" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30141" PostTypeId="2" ParentId="30098" CreationDate="2019-03-27T12:20:16.060" Score="3" Body="&lt;h2&gt;You would like to do some &lt;a href=&quot;https://astronomy.stackexchange.com/q/23165/7982&quot;&gt;Viewing “Diamond Fuji”&lt;/a&gt;!&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Using Python and &lt;a href=&quot;https://rhodesmill.org/skyfield/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Skyfield&lt;/a&gt; and the GPS coordinates in your urls:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Mt_Fuji  = earth + Topos(latitude_degrees    =  +35.36304,&#xA;                         longitude_degrees   = +138.73040,&#xA;                         elevation_m         =  3776.0)&#xA;&#xA;Enoshima = earth + Topos(latitude_degrees    =  +35.29875,&#xA;                        longitude_degrees    = +139.47457,&#xA;                        elevation_m          =  10.0)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I get the following. &lt;strong&gt;I make no guarantees!&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Mt. Fuji peak from Enoshima&#xA;azimuth:     276.23 &#xA;altitude       2.86&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PTpEl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PTpEl.png&quot; alt=&quot;Sunset over Mt. Fuji from Enoshima&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/r3t8h.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/r3t8h.png&quot; alt=&quot;Sunset over Mt. Fuji from Enoshima&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;from skyfield.api import Loader, Topos&#xA;&#xA;load  = Loader('~/Documents/fishing/SkyData')  # single instance for big files&#xA;ts    = load.timescale()&#xA;de421 = load('de421.bsp')&#xA;sun   = de421['sun']&#xA;earth = de421['earth']&#xA;moon  = de421['moon']&#xA;&#xA;Mt_Fuji  = earth + Topos(latitude_degrees    =  +35.36304,&#xA;                         longitude_degrees   = +138.73040,&#xA;                         elevation_m         =  3776.0)&#xA;&#xA;Enoshima = earth + Topos(latitude_degrees    =  +35.29875,&#xA;                        longitude_degrees    = +139.47457,&#xA;                        elevation_m          =  10.0)&#xA;&#xA;hours  = 17 - 9 + np.arange(91)/60.&#xA;&#xA;days   = np.arange(1, 367)&#xA;&#xA;alt_Fuji, az_Fuji, d_Fuji = Enoshima.at(ts.now()).observe(Mt_Fuji).apparent().altaz()&#xA;alt_Fuji, az_Fuji         = [thing.degrees for thing in (alt_Fuji, az_Fuji)]&#xA;d_Fuji_km                  = d_Fuji.km&#xA;&#xA;Mt_Fuji_obs = Enoshima.at(ts.now()).observe(Mt_Fuji).apparent()&#xA;&#xA;if True:&#xA;    seps = []&#xA;    for day in days:&#xA;        times = ts.utc(2019, 1, day, hours)&#xA;        sunpos  = Enoshima.at(times).observe(sun).apparent()&#xA;        Fujipos = Enoshima.at(times).observe(Mt_Fuji)&#xA;        sep     = Fujipos.separation_from(sunpos)&#xA;        seps.append(sep)&#xA;&#xA;    sepz = [x.degrees for x in seps]&#xA;    SEP  = np.array(sepz)&#xA;&#xA;    if True:&#xA;        plt.figure()&#xA;        plt.imshow(SEP, vmin=0, vmax=5)&#xA;        plt.colorbar()&#xA;        plt.xlabel('minutes after 17:00 JST', fontsize=14)&#xA;        plt.ylabel('day number in 2019 JST', fontsize=14)&#xA;        plt.title('Sun sep (deg) from Mt. Fuji from Enoshima', fontsize=14)&#xA;        plt.show()&#xA;&#xA;# make a detailed plot&#xA;if True:&#xA;    days_1   = np.arange( 95, 100)   # april 5 thru 9&#xA;    days_2   = np.arange(246, 251)   # sept  3 thru 7&#xA;    both = []&#xA;    for days in (days_1, days_2):&#xA;        altazs = []&#xA;        for day in days:&#xA;            times = ts.utc(2019, 1, day, hours)&#xA;            alt, az, d = Enoshima.at(times).observe(sun).apparent().altaz()&#xA;            alt, az    = [thing.degrees for thing in (alt, az)]&#xA;            altazs.append((alt, az))&#xA;        both.append(altazs)&#xA;&#xA;    if True:&#xA;        hw_deg         = 5.0&#xA;        altmin, altmax = alt_Fuji - hw_deg, alt_Fuji + hw_deg&#xA;        azmin,  azmax  = az_Fuji  - hw_deg, az_Fuji  + hw_deg&#xA;        xFuji = [az_Fuji - 2*alt_Fuji, az_Fuji,  az_Fuji + 2*alt_Fuji, az_Fuji - 2*alt_Fuji]&#xA;        yFuji = [0,                    alt_Fuji, 0,                    0                   ]&#xA;        plt.figure()&#xA;        for i, altazs in enumerate(both):&#xA;            plt.subplot(2, 1, i+1)&#xA;            for (alt, az) in altazs:&#xA;                plt.plot(az, alt)&#xA;            plt.plot(xFuji, yFuji, '-k', linewidth=2)&#xA;            plt.plot([azmin, azmax], [0, 0], '-k')&#xA;            plt.xlim(azmin, azmax)&#xA;            plt.ylim(altmin, altmax)&#xA;            plt.ylabel('altitude(deg)', fontsize=14)&#xA;&#xA;        plt.xlabel('azimuth (deg)', fontsize=14)&#xA;        plt.suptitle('Sunset vs Mt. Fuji from Enoshima', fontsize=14)&#xA;        plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-03-27T12:39:14.353" LastActivityDate="2019-03-27T12:39:14.353" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30142" PostTypeId="1" AcceptedAnswerId="30143" CreationDate="2019-03-27T17:55:03.823" Score="0" ViewCount="95" Body="&lt;p&gt;Since iron is a stars waste could a star that goes Super nova eject pure iron to travel across the galaxy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://space.stackexchange.com/questions/35101/99-99-pure-iron-meteorite&quot;&gt;https://space.stackexchange.com/questions/35101/99-99-pure-iron-meteorite&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastActivityDate="2019-03-27T18:24:14.903" Title="Could pure iron from a star make it to Earth?" Tags="&lt;supernova&gt;&lt;meteor&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30143" PostTypeId="2" ParentId="30142" CreationDate="2019-03-27T18:24:14.903" Score="5" Body="&lt;p&gt;Sure. Iron is not made in significant quantities in the Big Bang, so 100% of the iron on Earth today was synthesized in a star. There's no other source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It escapes in many ways:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Supernovae&lt;/li&gt;&#xA;&lt;li&gt;Mergers between compact stars like white dwarfs and neutron stars&lt;/li&gt;&#xA;&lt;li&gt;Stellar winds from more ordinary stars&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Supernovae are the main source of modern-day iron, with r-process nucleosynthesis during the explosion producing a &lt;em&gt;lot&lt;/em&gt; of iron and elements near iron.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2019-03-27T18:24:14.903" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30144" PostTypeId="1" AcceptedAnswerId="30154" CreationDate="2019-03-27T20:17:46.483" Score="1" ViewCount="121" Body="&lt;p&gt;According to this video &lt;a href=&quot;https://youtu.be/-iWGtQ03OZM&quot; rel=&quot;nofollow noreferrer&quot;&gt;stars at end of time &lt;/a&gt; and many others on fate of universe they all agree on one  thing that stars will eventually convert  all the hydrogen into heavier elements and eventually there will be no light , i understand the documentaries are over simplified but is it really the case that universe had/has finite amount of hydrogen to start with and there is no known process that recycling hydrogen back into universe &lt;/p&gt;&#xA;" OwnerUserId="16136" LastActivityDate="2019-03-28T16:15:45.867" Title="Did universe have finite hydrogen" Tags="&lt;astrophysics&gt;&lt;universe&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30145" PostTypeId="2" ParentId="25061" CreationDate="2019-03-27T22:13:30.523" Score="2" Body="&lt;p&gt;The answer by David Hammen includes many of the interesting details of how Io's orbit evolves in time (and explains why Io can still be volcanic even though right now Io's orbit is &lt;em&gt;extremely&lt;/em&gt; circular).  It also explains that if Io was fully tidally locked, with no other moons, then it would not heat and its orbit would not change, which may be what the questioner was mostly wondering about.  Perhaps the only remaining question might then be, why does a moon that is in a circular orbit, but not rotating at the correct rate, find its orbit changing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this, there is an interesting result that if the moon is spinning faster than its orbit, the delay in the response of the moon's shape to its tidal equipotential means that the &quot;points&quot; of its bulges will get out ahead of alignment with the planet.  This produces a torque from gravity that tends to slow its spin.   The opposite holds if it is spinning slower than its orbit.  So that's how the spin gets tidally locked, and there is some heating associated with that.  But the planet-moon system (ignoring other moons) must conserve angular momentum, so if the spin slows, that angular momentum must show up somewhere else-- it shows up in the orbit.  So instead of thinking about the energy of the orbit (which is not conserved because heat is created and spins are changing), think about the angular momentum of spin plus orbit.  If the spin slows, the orbit must raise up to conserve angular momentum, and the opposite if the spin increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Io is not tidally locked, it isn't doing either, but in its history before it was locked, it would have done one or the other.  As for the Earth and Moon, the Moon is tidally locked but the Earth is spinning faster than the Moon's orbit, so the Earth's bulges get out ahead of the Moon and so the Moon is torquing down our spin.  That loss of angular momentum must go into the Moon's orbit, so that's why the Moon is getting farther away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think in terms of energy, then you see that the Earth is being heated by the Moon's gravity.  Also, the Moon's orbit is increasing in energy.  So there has to be a source for both of those, and it is the energy in the Earth's spin.  Here there is no question about how energy lost as heat radiated by the Earth can come out of the Moon's orbit, because in fact the orbital energy is increasing.  It's more clear how spin energy can go into both the heating and the orbit, because it is the spin that is creating the forces that are causing both the heating and the orbital effects.  Similarly, if a moon is spinning faster than its own orbit, then that spin creates forces on the moon that slow its spin, and some of that energy goes into heating the moon, and some goes into lifting its own orbit (to conserve angular momentum).  If the moon is spinning slower than its orbit, then energy is going into both heat, and its spin, and this will lower the orbit because of the forces that are doing both those things also must conserve angular momentum.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2019-03-27T22:13:30.523" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30146" PostTypeId="1" AcceptedAnswerId="30147" CreationDate="2019-03-27T22:17:04.953" Score="1" ViewCount="68" Body="&lt;p&gt;In &lt;a href=&quot;https://pdfs.semanticscholar.org/b678/e64654eb1a430c7373b2d04f0621019aa3e1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;, the authors describe the theoretical relationship between photometric colours, particularly in Figures 1 and 6:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KnaBE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KnaBE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/DOKWh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DOKWh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the colour is the difference between two magnitudes, and the redshift is defined as&#xA;&lt;span class=&quot;math-container&quot;&gt;$z=\frac{\lambda'-\lambda_0}{\lambda_0}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would have thought that the colour-z relationships would simply be a downward curve, since the light gets redshifted (goes down in frequency), and the ratio of two magnitudes is constant with redshift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do the colour-z relationships in this paper (and in others I've seen) show such a complex relationship?&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2019-03-27T23:44:24.307" Title="Relationship between photometric colour and redshift" Tags="&lt;redshift&gt;&lt;photometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30147" PostTypeId="2" ParentId="30146" CreationDate="2019-03-27T23:44:24.307" Score="4" Body="&lt;p&gt;The main reason is that the intrinsic spectra of galaxies are complex and therefore a redshift of their spectrum, whilst leading to a redder spectrum overall, does not necessarily lead to reddening in all colours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance if there is apeak in the intrinsic spectrum, then as that peak moves redward, then colours formed from bands on the same side or straddling the peak in wavelength would behave differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dramatic redward turn up at high redshifts (note, a large colour is redder) is caused by the &quot;Lyman break&quot; moving through the colour bands. Basically, the intrinsic spectrum is self-absorbed at wavelengths shortward of 91.2 nm by neutral hydrogen. This absorption edge is redshifts into the visible region at redshifts bigger than 3, causing the bluer bands to essentially disappear and any colour formed with them to become very red.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-03-27T23:44:24.307" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30148" PostTypeId="1" CreationDate="2019-03-28T00:04:59.557" Score="1" ViewCount="36" Body="&lt;p&gt;In the graphs in my previous question &lt;a href=&quot;https://astronomy.stackexchange.com/questions/30146/relationship-between-photometric-colour-and-redshift/30147#30147&quot;&gt;here&lt;/a&gt;, there are negative-sloped regions as the colour-z relationship oscillates. So if I look at how the I-W2 and W3-U colours change with redshift, as I've done here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3bcxP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3bcxP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with I-W2 and W3-U both getting lower (bluer, is that right?) does that mean that my graphs indicate one of the upward-sloping regions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What else does the redshift evolution of I-W2 against W3-U indicate?&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2019-03-28T00:04:59.557" Title="Relationship between photometric colour and redshift, part two" Tags="&lt;redshift&gt;&lt;photometry&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30150" PostTypeId="2" ParentId="30131" CreationDate="2019-03-28T06:06:10.213" Score="1" Body="&lt;p&gt;To cut it short, while there is no definitive term, things outside of our solar system is usually referred as Interstellar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we talk about things outside of our solar system, we either talk about other star systems (and their planets are called exoplanets as you mentioned) or we talk about interstellar medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interstellar term was not used for objects until 2017, when we discovered Oumuamua. Now, any object that doesn't belong to or that did not originate in our solar system is designated as Interstellar. For that matter Oumuamua's official designation is 1I/2017, for which the I stands for interstellar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I should note that planets that does not revolve around any star are called Rogue Planets. &lt;/p&gt;&#xA;" OwnerUserId="20818" LastActivityDate="2019-03-28T06:06:10.213" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30152" PostTypeId="2" ParentId="30138" CreationDate="2019-03-28T07:58:24.260" Score="3" Body="&lt;p&gt;Images are often presented with (Earth's) North at the top but there is no correct orientation. For example, in this image from the VLT north is 0.1 degrees from the vertical, &lt;a href=&quot;https://www.eso.org/public/unitedkingdom/images/eso1322a/&quot; rel=&quot;nofollow noreferrer&quot;&gt;the sidebar&lt;/a&gt; on the image page has details on the position, field of view and orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ui3OE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ui3OE.jpg&quot; alt=&quot;IC 2944&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, &lt;a href=&quot;https://www.eso.org/public/unitedkingdom/images/eso1238a/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this image&lt;/a&gt; from the same telescope has been rotated left 9 degrees, which is likely an artistic decision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/N1wwA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/N1wwA.jpg&quot; alt=&quot;Thor&amp;#39;s Helmet&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amateurs also regularly align the cameras on their scopes so that north is up as it can make it easier to diagnose tracking problems. However, it's fairly common to change the camera alignment on the telescope to frame particular targets or groups of targets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many classes as astronomical objects, such as elliptical galaxies, irregular galaxies or nebulae don't have a well defined or known axis of rotation, which is why Earth's poles are used as a reference.&lt;/p&gt;&#xA;" OwnerUserId="23640" LastEditorUserId="23640" LastEditDate="2019-03-28T12:22:31.080" LastActivityDate="2019-03-28T12:22:31.080" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30153" PostTypeId="2" ParentId="30112" CreationDate="2019-03-28T13:59:49.263" Score="1" Body="&lt;p&gt;I guess you only meant gravity based orbit but since you asked smallest, technically, other forces which are more stronger but act on small distances, would give much smaller orbit. For example electron orbit. But they are harder to think as circular and better explained by quantum number. (Sorry - maybe better as comment)&lt;/p&gt;&#xA;" OwnerUserId="26086" LastActivityDate="2019-03-28T13:59:49.263" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30154" PostTypeId="2" ParentId="30144" CreationDate="2019-03-28T16:15:45.867" Score="2" Body="&lt;p&gt;Are you sure it's true that all hydrogen will eventually be processed?   I can see that star formation rate should eventually slow to a point that there would be essentially no more light being created, but that's not the same thing.  Right now, most of the hydrogen in the universe is in the space within clusters of galaxies, and some of it falls into galaxies and some gets ejected, so you might imagine that all of that will eventually get processed.  But there is also hydrogen in between galaxy clusters, in the &quot;cosmic web&quot;, and since the universe is expanding quite rapidly, it seems quite possible to me that much or most of that gas will never fall into a galaxy, but instead just end up stranded in between the expanding clusters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's more, our models now include the &quot;cosmological principle&quot; that treats the universe as homogeneous, but that's just the universe we can see.   We have good evidence that it extends way beyond what we can see, so we have no way to know (or even to expect) that the homogeneity is a truly global property of the &lt;em&gt;entire&lt;/em&gt; universe (i.e., not just the &quot;observable&quot; universe).  On scales we cannot probe, the structure might be fractal, meaning that there could be pockets somewhere doing very different things, and there might always be stars forming somewhere.  Another possibility is that there could be a &quot;landscape&quot; of universes, which is hard to say if they count as the same universe or completely different ones, but it does mean there could be in existence hydrogen elsewhere in the &quot;landscape&quot; even after our universe has none.   So it's a tricky question to answer if &quot;all the hydrogen&quot; will eventually be processed, but I suppose you could look at all the hydrogen in the observable universe-- but you still have to wonder if the gas between the clusters ever gets into a galaxy.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2019-03-28T16:15:45.867" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30155" PostTypeId="2" ParentId="30131" CreationDate="2019-03-28T17:45:48.100" Score="3" Body="&lt;p&gt;As @AtmosphericPrisonEscape noted in comments, above, &quot;Extrasolar&quot; is the correct term.  E.g., we refer to &quot;extrasolar planets&quot;, not &quot;interstellar planets.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Interstellar&quot; refers to things among the stars (Merriam-Webster: &quot;located, taking place, or traveling among the stars especially of the Milky Way galaxy&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Extrasolar&quot; means &quot;outside the Solar System.&quot; (Merriam-Webster: &quot;originating or existing outside the solar system&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One distinction is that &quot;interstellar&quot; doesn't include all space outside the Solar System -- &quot;Interstellar space&quot; is usually spoken of as distinct from, and &lt;em&gt;much&lt;/em&gt; smaller than, &quot;Intergalactic space&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original question asked for &quot;a general adjective for objects outside our solar system&quot; and that would be &quot;extrasolar&quot;.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2019-03-28T17:45:48.100" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30156" PostTypeId="2" ParentId="20760" CreationDate="2019-03-28T20:35:04.037" Score="-8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can gravity cause a Big Crunch, if the cosmological principle holds?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. That's a myth that grew out of the Friedmann equations. Alexander Friedmann was a ballistics instructor on the Austrian front. He modelled the expanding universe as something like a cannonball fired aloft. See John Peacocks’s &lt;a href=&quot;https://ned.ipac.caltech.edu/level5/Peacock/Peacock3_2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;cosmological physics&lt;/a&gt; for an example: &lt;em&gt;“the dynamics of the entire universe are the same as those of a cannonball fired vertically against the Earth’s gravity. Just as the Earth’s gravity defines an escape velocity for projectiles, so a universe that expands sufficiently fast will continue to expand forever. Conversely, for a given rate of expansion there is a critical density that will bring the expansion asymptotically to a halt”&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MaGF4.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MaGF4.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;                                      &lt;sup&gt;                                                             Image from Norbert Rumiano’s article &lt;a href=&quot;http://nrumiano.free.fr/Ecosmo/cg_model.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;a cosmological model&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It’s totally wrong I'm afraid. Gravity doesn’t stop the universe expanding. A gravitational field is akin to a pressure-gradient in space. It makes matter fall down, but it doesn’t make space fall down. You can tell it’s wrong because Big Bang cosmology is said to be balanced on a knife edge, such that the density of the universe is incredibly close to the critical density. That’s just a re-heat of Einstein’s greatest blunder, wherein a slight increase in density would trigger contraction, and a slight decrease would trigger expansion. It’s also wrong because the 1998 supernova observations didn’t match any of the cannonball trajectories. They didn’t measure the expansion slowing down, they measured it speeding up. Like &lt;a href=&quot;http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/speed_of_light.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;the ascending light beam speeds up&lt;/a&gt;, because &lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol7-trans/156?highlightText=%22spatially%20variable%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;the speed of light is spatially variable&lt;/a&gt;. Unfortunately people like John Moffat and João Magueijo and other VSL guys said the speed of light in the universe is slowing down, not speeding up. Unfortunately the Friedmann equations feature not just a fixed c, but a curvature k. Even though a gravitational field is a place where the speed of light varies because space is &lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol7-trans/192?highlightText=%22neither%20homogeneous%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;neither homogeneous nor isotropic&lt;/a&gt;. Not a place where space is curved.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Say we have a closed universe that follows the cosmological principle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We have no evidence that the universe is closed, or that the cosmological principle is scientific. I think it's rather like saying &lt;em&gt;I live in a forest, so everybody must see trees&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As a result, mass is distributed equally and, if as far as I know, gravity should have no net effect on the movements of galaxies (as in, there are no peculiar velocities).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The issue isn't with galaxies, it's with space. Space is expanding like the raisin cake analogy, and gravity won't stop that. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, in the Big Crunch, the Universe stops expanding and gravity causes the scale factor to decrease. How is this possible? Shouldn't gravity have no effect on a homogenous and isotropic universe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't think it is possible. Note that Einstein described a gravitational field as a place where space was neither homogeneous nor isotropic. So a homogeneous isotropic universe is one where there is no gravity.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On that note, if gravity actually caused a collapse, would gravity cause the comoving coordinates of the galaxies to change?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would say no. I'd also say gravity doesn't cause a collapse.  &lt;/p&gt;&#xA;" OwnerUserId="9772" LastEditorUserId="9772" LastEditDate="2019-03-28T20:41:11.310" LastActivityDate="2019-03-28T20:41:11.310" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30157" PostTypeId="2" ParentId="30111" CreationDate="2019-03-29T03:29:54.177" Score="2" Body="&lt;p&gt;I think this is a case of selective quoting/editing in the media reporting otherwise something major has changed in the design and plan for the mission that hasn't been released or published officially. The large and comprehensive &lt;a href=&quot;http://sci.esa.int/cosmic-vision/59109-ariel-assessment-study-report-yellow-book/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ARIEL assessment report&lt;/a&gt; which was published less than 2 years ago discusses mounting all of the optical components on a large, thermally isolating optical bench to prevent movements of the optics introducing noise into the transit signals. However it also says (page 66) that secondary mirror (M2) will have the normal 3 axis position control to provide focus and tip-tilt. This is also mentioned in a &lt;a href=&quot;https://link.springer.com/article/10.1007/s10686-018-9604-3&quot; rel=&quot;nofollow noreferrer&quot;&gt;Phase A study report&lt;/a&gt; published in 2018. The stability will be improved by the launch of Ariel into the Earth-Sun L2 Lagrange point which is more thermally stable and doesn't suffer from Earth eclipses compared to Kepler or TESS which are closer to the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the normal and sensible accommodation incorporated into satellite optical systems in order to compensate for anything moving slightly during the pretty violent shaking that occurs during launch on a rocket. Locking everything down pre-launch with no adjustment would be a very risky undertaking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no more recent publications listed on the &lt;a href=&quot;https://ariel-spacemission.eu/ariel-publications/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ariel mission website&lt;/a&gt; that would suggest that the optical system has been redesigned to eliminate all moving parts although it is possible a study has been done and the result have not yet been communicated publicly.  &lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2019-03-29T03:29:54.177" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30158" PostTypeId="1" AcceptedAnswerId="30163" CreationDate="2019-03-29T03:54:51.263" Score="6" ViewCount="290" Body="&lt;p&gt;Come to think of it, the only things we hear about degeneracy is neutron and electron.&lt;br&gt;&#xA;I made a generalization:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;White Dwarf&lt;/strong&gt;: Supported by electron degeneracy.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Neutron Star&lt;/strong&gt;: Supported by neutron degeneracy and nuclear forces, and sometimes centrifugal force.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Quark Star&lt;/strong&gt;: Most likely supported by quark degeneracy.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, we can see that there is no mention of proton degeneracy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick search lead to the Wikipedia subsection &lt;a href=&quot;https://en.wikipedia.org/wiki/Degenerate_matter#Proton_degeneracy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Degenerate matter; Proton degeneracy&lt;/a&gt;, and the answer that yes, there is proton degeneracy, but is much less effective than other forms of degenerate matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question:&lt;/strong&gt; What is the force exerted by degenerate protons, and what equation is it modeled by?&lt;/p&gt;&#xA;" OwnerUserId="25988" LastEditorUserId="7982" LastEditDate="2019-03-29T22:14:27.183" LastActivityDate="2019-03-29T22:14:27.183" Title="Measures for the degeneracy from protons?" Tags="&lt;degenerate-matter&gt;&lt;proton&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="30159" PostTypeId="5" CreationDate="2019-03-29T03:56:26.237" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2019-03-29T03:56:26.237" LastActivityDate="2019-03-29T03:56:26.237" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30160" PostTypeId="4" CreationDate="2019-03-29T03:56:26.237" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2019-03-29T03:56:26.237" LastActivityDate="2019-03-29T03:56:26.237" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30162" PostTypeId="1" AcceptedAnswerId="30171" CreationDate="2019-03-29T06:50:07.933" Score="21" ViewCount="5758" Body="&lt;p&gt;The NY Times article &lt;a href=&quot;https://www.nytimes.com/2019/03/28/science/saturn-moons-rings.html&quot; rel=&quot;noreferrer&quot;&gt;Saturn’s Rings Are Sculpted by a Crew of Mini-Moons&lt;/a&gt; is really interesting and links to the recent paywalled paper in Science &lt;a href=&quot;http://science.sciencemag.org/content/early/2019/03/27/science.aat2349&quot; rel=&quot;noreferrer&quot;&gt;Close Cassini flybys of Saturn’s ring moons Pan, Daphnis, Atlas, Pandora, and Epimetheus&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I absolutely can't understand one of the photos in the NY Times article, shown below. Titan seems to be...&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;behind&lt;/em&gt; Saturn's rings, and yet it is&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;huge&lt;/em&gt; relative the spacing of the rings, and yet it&lt;/li&gt;&#xA;&lt;li&gt;appears to be &lt;em&gt;out of focus&lt;/em&gt; while the rings and Epimetheus are in focus.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Can someone help me understand how all of these can be true at the same time?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the foreground, the moon Epimetheus appearing to hover above Saturn's rings. Epimetheus is dwarfed by Titan in the background.CreditNASA/JPL/Space Science Institute&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SyQn5.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SyQn5.jpg&quot; alt=&quot;Epimetheus, Titan and Saturn&amp;#39;s Rings&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6UBhU.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6UBhU.png&quot; alt=&quot;Epimetheus, Titan and Saturn&amp;#39;s Rings cropped&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-03-29T09:05:40.397" LastActivityDate="2019-03-30T13:52:11.140" Title="Help understanding this unsettling image of Titan, Epimetheus, and Saturn's rings?" Tags="&lt;photography&gt;&lt;saturn&gt;&lt;cassini&gt;&lt;saturn-rings&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="30163" PostTypeId="2" ParentId="30158" CreationDate="2019-03-29T07:17:08.237" Score="10" Body="&lt;p&gt;Stars are electrically neutral. In the optimal case for proton degeneracy, a star is made of hydrogen, maximising the proton number density. The number density of electrons is exactly the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Fermi momentum &lt;span class=&quot;math-container&quot;&gt;$p_F$&lt;/span&gt; of a species is dependent only on number density&lt;span class=&quot;math-container&quot;&gt;$^1$&lt;/span&gt;, the Fermi momentum of the electrons and protons are the same. Pressure however&lt;span class=&quot;math-container&quot;&gt;$^1$&lt;/span&gt;, is an increasing function of &lt;span class=&quot;math-container&quot;&gt;$p_F/mc$&lt;/span&gt; and since the mass of an electron is 1800 times smaller, the electron degeneracy pressure is always much higher than the proton degeneracy pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Degenerate protons do exist. They form of order 1% of the density in the neutron fluid region (the bulk) of a neutron star. However, they contribute much less than a thousandth of the pressure and always lower than the degenerate electron pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, the form of the pressure equation (see below) does actually admit the possibility that the degeneracy pressures of electrons and protons become more similar as &lt;span class=&quot;math-container&quot;&gt;$p_F/mc \gg 1$&lt;/span&gt; (aka ultra-relativistic). But by that time, the energies of the electrons and protons are easily sufficient to create neutrons and an equilibrium is set up so that neutrons always outnumber the protons by at at least 8:1 and therefore their degeneracy pressure will dominate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) For those interested: The Fermi momentum is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$p_{F} = \left(\frac{3}{8\pi}\right)^{1/3} h n^{1/3}\ ,$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is the number density of fermions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whilst ideal fermion degeneracy pressure is given by&#xA;&lt;span class=&quot;math-container&quot;&gt;$$P = \frac{\pi m^4 c^5}{3 h^3} \left[x (2x^2 -3)(1 + x^2)^{1/2} + 3&#xA;  \sinh^{-1}(x) \right]\ , $$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$x = p_F/mc$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-03-29T14:19:05.537" LastActivityDate="2019-03-29T14:19:05.537" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30164" PostTypeId="2" ParentId="30162" CreationDate="2019-03-29T08:39:15.857" Score="30" Body="&lt;p&gt;&lt;a href=&quot;https://solarsystem.nasa.gov/resources/13794/titan-beyond-the-rings/&quot; rel=&quot;noreferrer&quot;&gt;This NASA page&lt;/a&gt; says this photo was taken on April 28 2006.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &lt;a href=&quot;https://celestia.space/&quot; rel=&quot;noreferrer&quot;&gt;Celestia&lt;/a&gt;, I managed to find the picture from Cassini that best lines up with the photo. It doesn't match up precisely, but that's to be expected as the calculated orbital elements of all these moons (and cassini) in the software won't necessarily match up to reality precisely.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/y3bvs.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y3bvs.jpg&quot; alt=&quot;cassini celestia shot of titan&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is the zoomed out version of this shot. You can see Titan in the centre and Epimetheus as a dot on top.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/CgA1d.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CgA1d.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;And here is the top down shot from Cassini to the moons. Circled are Epimetheus and Titan.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/gaaPZ.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gaaPZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer your question: Titan is really big compared to epimetheus (about 50x), Titan has an atmosphere and so appears fuzzy (It is actually in focus, everything in space is very far away and so is effectively at infinity for focusing purposes), and the rings are highly oblique, so you only see a small slice of them.&lt;/p&gt;&#xA;" OwnerUserId="17655" LastActivityDate="2019-03-29T08:39:15.857" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="30165" PostTypeId="1" AcceptedAnswerId="30170" CreationDate="2019-03-29T09:35:29.593" Score="3" ViewCount="103" Body="&lt;p&gt;I'm trying to create a geodesic dome with the stars of the northern night sky inscribed on the outside. The highest point of the dome would be the celestial north pole. How do I create a map that projects onto the &quot;net&quot; of the geodesic dome, such that when I assemble the dome, the stars are in the correct positions? I have an equirectangular map that I have rotated 90 degrees north so it is centered on the pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My existing map is a high-resolution jpeg file in an equirectangular projection. I have some experience creating and manipulating panoramic images, but much less experience converting between map projections! I would like to use a computer program to convert this projection into a printable template if possible. &lt;/p&gt;&#xA;" OwnerUserId="21997" LastEditorUserId="7472" LastEditDate="2019-04-06T08:42:11.607" LastActivityDate="2019-04-06T08:42:11.607" Title="What projection would I use to map the northern night sky onto a geodesic dome?" Tags="&lt;star-maps&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="30166" PostTypeId="2" ParentId="30162" CreationDate="2019-03-29T09:44:56.973" Score="6" Body="&lt;p&gt;&lt;strong&gt;note:&lt;/strong&gt; This is a supplemental answer adding some details to @ Ingolifs' &lt;a href=&quot;https://astronomy.stackexchange.com/a/30164/7982&quot;&gt;excellent answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;At &lt;em&gt;roughly&lt;/em&gt; &lt;code&gt;2006-Apr-28 08:30 UTC&lt;/code&gt; Cassini was both 1,800,000 km from Titan and 667,000 km from Epimetheus at the same time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used JPL's &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;noreferrer&quot;&gt;Horizons&lt;/a&gt; and saved the positions in the Saturn body centered coordinates every 5 minutes then ran the python script below to plot. I am not sure how to get the plane of the rings this way easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dCZzp.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dCZzp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pRU08.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pRU08.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;class Body(object):&#xA;    def __init__(self, name):&#xA;        self.name = name&#xA;&#xA;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;from mpl_toolkits.mplot3d import Axes3D&#xA;&#xA;fnames = ['Titan photo Cassini horizons_results.txt',&#xA;          'Titan photo Titan horizons_results.txt',&#xA;          'Titan photo Epimetheus horizons_results.txt' ]&#xA;&#xA;names  = ['Cassini', 'Titan', 'Epimetheus']&#xA;&#xA;bodies = []&#xA;&#xA;for name, fname in zip(names, fnames):&#xA;&#xA;    with open(fname, 'r') as infile:&#xA;&#xA;        lines = infile.read().splitlines()&#xA;&#xA;    iSOE = [i for i, line in enumerate(lines) if &quot;&lt;span class=&quot;math-container&quot;&gt;$$SOE&quot; in line][0]&#xA;    iEOE = [i for i, line in enumerate(lines) if &quot;$$&lt;/span&gt;EOE&quot; in line][0]&#xA;&#xA;    print iSOE, iEOE, lines[iSOE], lines[iEOE]&#xA;&#xA;    lines = zip(*[line.split(',') for line in lines[iSOE+1:iEOE]])&#xA;&#xA;    JD  = np.array([float(x) for x in lines[0]])&#xA;    pos = np.array([[float(x) for x in lines[i]] for i in 2, 3, 4])&#xA;    vel = np.array([[float(x) for x in lines[i]] for i in 5, 6, 7])&#xA;&#xA;    body = Body(name)&#xA;    bodies.append(body)&#xA;    body.JD  = JD&#xA;    body.pos = pos&#xA;    body.vel = vel&#xA;&#xA;Cassini, Titan, Epimetheus = bodies&#xA;&#xA;r_Titan      = np.sqrt(((Cassini.pos - Titan.pos     )**2).sum(axis=0))&#xA;r_Epimetheus = np.sqrt(((Cassini.pos - Epimetheus.pos)**2).sum(axis=0))&#xA;&#xA;hours = 24 * (JD - JD[0])&#xA;&#xA;r_Titan_target      = 1.8E+06 &#xA;r_Epimetheus_target = 6.67E+05&#xA;&#xA;hours_Titan      = hours[np.argmax(r_Titan &amp;lt; r_Titan_target)]&#xA;hours_Epimetheus = hours[np.argmax(r_Epimetheus[30:] &amp;gt; r_Epimetheus_target)+30]&#xA;&#xA;print hours_Titan, hours_Epimetheus&#xA;if True:    &#xA;    fig = plt.figure()&#xA;&#xA;    plt.subplot(2, 1, 1)&#xA;    plt.plot(hours, r_Titan)&#xA;    plt.plot(hours, 1.8E+06 * np.ones_like(r_Titan), '-k')&#xA;    plt.ylabel('Cassini-Titan distance (km)', fontsize=16)&#xA;&#xA;    plt.subplot(2, 1, 2)&#xA;    plt.plot(hours, r_Epimetheus)&#xA;    plt.plot(hours, 6.67E+05 * np.ones_like(r_Epimetheus), '-k')&#xA;    plt.ylabel('Cassini-Epimetheus distance (km)', fontsize=16)&#xA;    plt.xlabel('2006-Apr-28 hours', fontsize=16)&#xA;&#xA;    plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-03-29T09:50:35.217" LastActivityDate="2019-03-29T09:50:35.217" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30168" PostTypeId="2" ParentId="30125" CreationDate="2019-03-29T10:20:21.837" Score="1" Body="&lt;p&gt;Based on assumptions, if you are ok with approx solution then you can solve heat transfer equation in steady state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Heat_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heat transfer equation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring the time - long time has passed to attain equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat flux assuming outward and hence internal ocean would be sum of all internal heating mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Easiest would be if somehow you can come up with surface temperature value. Then you can solve conduction equation through ice for temperature and  weight for pressure and just see at what temperature and pressure ice will go into liquid. Simply asking what is ice layer thickness to maintain a internal heat flux and given surface temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need one boundary condition. Other way would be little harder, starting at centre with outward heatflux. You need to know both phase and composition since density and cp both changes with these conditions.&lt;/p&gt;&#xA;" OwnerUserId="26086" LastActivityDate="2019-03-29T10:20:21.837" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30169" PostTypeId="2" ParentId="30158" CreationDate="2019-03-29T16:38:46.600" Score="2" Body="&lt;p&gt;Another way to think of the answer is that the pressure of any nonrelativistic gas (degenerate or ideal) is always 2/3 the kinetic energy per unit volume.  So we can say that when the electrons go degenerate, and contribute almost all of the pressure, it must be because they also contain almost all of the kinetic energy.  This is true, the reason they contribute most of the pressure is because they hold most of the kinetic energy, but we can then ask, why do they contain most of the kinetic energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Framing the question this way focuses our attention on what really matters here: the thermodynamics of degeneracy.  A purely degenerate gas has zero temperature, that is what degeneracy means in the thermodynamic sense.  However, the kinetic energy is quite large.  So it is this &lt;em&gt;thermodynamic&lt;/em&gt; disconnect between the kinetic energy and the temperature that is the cause of the importance of the electron pressure.  Since we can assume the electrons and protons equilibrate to the same temperature (that assumption is always in place when you see these kinds of analyses), the protons must also have essentially zero temperature when the electrons are highly degenerate.  But protons are not degenerate, due to their higher mass (as explained above), so if they have low temperature, they also have low kinetic energy.  It's due to their different thermodynamic connection between kinetic energy and temperature (which is more like an ideal gas rather than a degenerate one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put this way, we see what degeneracy is doing-- it is causing the electrons to hog all the kinetic energy, because degenerate electrons have a very high ratio of kinetic energy to kT, whereas protons have a more classical such ratio.  That's why they produce the pressure, it's all about kinetic energy.  You don't need to think about any weird forces, you only need the quantum mechanical thermodynamics, which drives the temperature of electrons down, relative to their kinetic energy, owing to the fact that degenerate electrons are very stingy about giving up heat (and temperature is the statement of which objects will give up heat in equilibrium, as per the zeroth law of thermodynamics).&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2019-03-29T16:38:46.600" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30170" PostTypeId="2" ParentId="30165" CreationDate="2019-03-29T16:41:16.507" Score="3" Body="&lt;p&gt;The mathematically ideal scheme would be a separate gnomonic projection&#xA;(see &lt;a href=&quot;https://en.wikipedia.org/wiki/Gnomonic_projection&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; or&#xA;&lt;a href=&quot;https://www.imo.net/observations/methods/visual-observation/minor/gnomonic/&quot; rel=&quot;nofollow noreferrer&quot;&gt;IMO&lt;/a&gt;)&#xA;for each tile of the dome.&#xA;Many planetarium programs implement this projection as an option.&#xA;Fortunately, each tile boundary is a straight line on both adjoining maps.&#xA;Unfortunately, continuity across tile boundaries requires some care in choosing each tile's scale and center of projection.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2019-03-29T16:41:16.507" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30171" PostTypeId="2" ParentId="30162" CreationDate="2019-03-29T19:24:58.190" Score="13" Body="&lt;p&gt;The JPL &lt;a href=&quot;https://space.jpl.nasa.gov/&quot; rel=&quot;noreferrer&quot;&gt;Solar System Simulator&lt;/a&gt; doesn't show Epimetheus but does show Titan behind the Encke gap at 2006-04-28 08:12 UTC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/h0uBz.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h0uBz.jpg&quot; alt=&quot;Titan at 08:12, narrow field&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simulated surface texture is probably composed of&#xA;&lt;a href=&quot;https://photojournal.jpl.nasa.gov/catalog/PIA21923&quot; rel=&quot;noreferrer&quot;&gt;VIMS images&lt;/a&gt;&#xA;in infrared wavelengths where Titan's atmosphere is relatively transparent.&#xA;On the real Titan, haze scatters visible light so strongly that the surface is indistinct and the edge looks fuzzy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we zoom out, we see that we are looking near the outer edge of the rings at a very shallow angle.&#xA;This is why they cover less than half of Titan's 10-arcminute apparent diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Epimetheus appears above the rings while we are looking from below, it must be in front of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7ggTn.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7ggTn.jpg&quot; alt=&quot;Titan at 08:12, wide field&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Simulated images courtesy NASA/JPL-Caltech&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-03-30T13:52:11.140" LastActivityDate="2019-03-30T13:52:11.140" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30172" PostTypeId="1" CreationDate="2019-03-29T23:04:07.820" Score="1" ViewCount="51" Body="&lt;p&gt;I am trying to draw an orbit using the Transform3DGroup class in the C# .NET framework. I have orbital elements a, e, i, RAAN, ArgPer, trueAnom. The algorithm that spits these out for me also gives me x y z coordinates in ECI. For some reason, when I try to use Transform3DGroup to do my rotations in order to basically draw out where the satellite is supposed to go in ECI (using the orbit elements I gave above) my result is not matching up with the ECI coordinates that the original algorithm is spitting out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That original algorithm is my truth data. But I don't want to use those coordinates, I want to draw the orbit myself with the Keplarian elements using Transform3DGroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my code:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;public Point3D calculatePosition(double trueAnomaly)&#xA;{&#xA;    double angle = trueAnomaly;&#xA;    double a = semiMajorAxis;&#xA;    double e = eccentricity;&#xA;    double b = (a * (1 - e * e)) / (1 + e * Math.Cos(angle));&#xA;    double SA = Math.Sin(angle);&#xA;    double X = b * CA;//(CA * CR + SA * CI * SR); &#xA;    double Y = b * SA; //(CA * SR + SA * CI);&#xA;    double Z = 0; //b * SA * SI;&#xA;    Point3D p = new Point3D(X, Y, Z);//new Point3D(Math.Cos(angle) * b, &#xA;    Math.Sin(angle) * b, 0);&#xA;    return p;&#xA;}&#xA;public void orient2OrbitalElements(double _time)&#xA;{&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    orbitNorm = new Vector3D(0, 0, 1);&#xA;    positionPoint = calculatePosition(trueAnomaly);//Position of Orbiter&#xA;    Transform3DGroup tg = new Transform3DGroup();&#xA;    tg.Children.Add(new TranslateTransform3D(positionVec));//Translates to Orbit Position (relative to Parent)&#xA;    tg.Children.Add(new RotateTransform3D(new AxisAngleRotation3D(new Vector3D(1, 0, 0), inclinationDeg)));//Rotates Orbit to the set Inclination&#xA;    tg.Children.Add(new RotateTransform3D(new AxisAngleRotation3D(new Vector3D(0, 0, 1), RAAN)));//Rotates Orbiter's model RAAN&#xA;    orbitNorm = tgNoTranslation(tg).Transform(orbitNorm);&#xA;    tg.Children.Add(new RotateTransform3D(new AxisAngleRotation3D(orbitNorm, argOfPerigee)));//Rotates Orbiter's model to the Arg of Perigee&#xA;    Transform = tg;&#xA;    if (HasEphem)&#xA;    {&#xA;        Console.WriteLine(ObjectName + &quot;: &quot; + Transform.Value.OffsetX + &quot;, &quot; + Transform.Value.OffsetY + &quot;, &quot; + Transform.Value.OffsetZ);&#xA;        Console.WriteLine(ObjectName + &quot;: &quot; + cartesElems.xPos[ephemIndex] + &quot;, &quot; + cartesElems.yPos[ephemIndex] + &quot;, &quot; + cartesElems.zPos[ephemIndex]);&#xA;    }&#xA;}&#xA;public Transform3DGroup tgNoTranslation(Transform3DGroup tgWithTranslations)&#xA;{&#xA;    Transform3DGroup tempTg = tgWithTranslations.Clone();&#xA;    tempTg.Children.RemoveAt(0);&#xA;    return tempTg;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Basically I calculate an x,y position on the orbital plane w.r.t the Earth position using the basic eccentricity, semi-major axis, true anomaly formula as you'll see. Then I put the resulting coordinates into my Transform3DGroup as a TranslateTransform3D. This is the first object I put in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I perform the rotations:&#xA; --- a rotation is added for the inclination, rotating the position Vec about &#xA;     the unit x axis (1,0,0)&#xA; --- a rotation for the RAAN, about the unit z-axis(0,0,1)&#xA; --- then I create the orbit normal from that and perform a argPer rotation &#xA;     about that new axis&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding of how Transform3DGroup works is that it will apply the first one, then the second, and so on. Which means that the positionVec (which is just x, y coordinates in the reference frame of the 2d orbital plane) gets performed, then the rotations will all be performed as a rotation matrix on was is left after what was done before it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I performing the argPer rotation incorrectly? RAAN or inc? Perhaps there is a more exact way to calculate my position in my orbital plane or a particular order of operations I must adhere to? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more I think about it, the more I keep arriving at the thought that everything seems to be very simple if I use the Transform3DGroup and that I should be doing it right at this point, yet the ECI results don't line up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But lo and behold, the Transform.Value.OffsetX/Y/Z does want to come close to agreeing with the truth data. Perhaps Transform.Value.Offset is not the X,Y,Z as I am thinking of it?&lt;/p&gt;&#xA;" OwnerUserId="24452" LastActivityDate="2019-03-29T23:04:07.820" Title="How to draw an orbit from Orbital Elements using Transform3DGroup" Tags="&lt;orbital-mechanics&gt;&lt;orbital-elements&gt;&lt;frame-of-reference&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30176" PostTypeId="1" AcceptedAnswerId="30184" CreationDate="2019-03-30T09:53:03.970" Score="2" ViewCount="1126" Body="&lt;p&gt;Given our real solar system how would planet orbits plot look like if we fix earth in the middle (put earth in the middle and draw a real orbit of a given planet) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the shape of its orbit? Is there any software that allows you to do such plot ? Does anybody have an example of it ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried searching for such image but most results are connected with historical theory .&lt;/p&gt;&#xA;" OwnerUserId="26417" LastActivityDate="2019-05-11T23:32:40.537" Title="'Geocentric' view of orbits" Tags="&lt;orbit&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30179" PostTypeId="2" ParentId="30176" CreationDate="2019-03-30T21:35:29.327" Score="3" Body="&lt;p&gt;The shape of the orbit would be an &lt;a href=&quot;https://en.wikipedia.org/wiki/Epicycloid&quot; rel=&quot;nofollow noreferrer&quot;&gt;epicycloid&lt;/a&gt;.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/7QTwG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7QTwG.png&quot; alt=&quot;Wikimedia commons&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://en.wikipedia.org/wiki/Deferent_and_epicycle&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Deferent_and_epicycle&lt;/a&gt; for more information on the geocentric orbital view.&lt;/p&gt;&#xA;" OwnerUserId="17655" LastActivityDate="2019-03-30T21:35:29.327" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30180" PostTypeId="1" CreationDate="2019-03-31T04:26:30.737" Score="0" ViewCount="378" Body="&lt;p&gt;Wikipedia's Red Giant page says they range in mass from 0.3 to 0.8 solar masses.   Which corresponds to an initial mass of a bit less than 1 solar mass, about 0.8 or so, if I was to estimate and a few solar masses.   Given that over half the mass is lost prior to the Red Giant stage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any stars with less than 0.8 solar mass to begin with wouldn't have had enough time to go red giant, so it's possible that significantly smaller stars could eventually go red giant.   Is there an estimate for how light the main sequence mass can be where the star still becomes a red giant late in it's life?&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2019-03-31T05:14:14.053" LastActivityDate="2019-03-31T10:08:19.623" Title="Is there a theoretical minimum mass main sequence star that can become a red giant?" Tags="&lt;star&gt;&lt;red-giant&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30181" PostTypeId="2" ParentId="30180" CreationDate="2019-03-31T10:08:19.623" Score="2" Body="&lt;p&gt;Perhaps you are interested in the lower mass limit (given in the comments as 0.3 solar masses), and why that lower limit exists.  This is similar to the question of why stars become red giants in the first place.  You are right that given the age of the universe stars less massive than 0.8 solar masses won't have had time, so the lower mass limit is theoretical.  But what sets it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is, red giants happen because stars can exhaust all the hydrogen in their core, turning it all into helium.  Then there is no nuclear fuel to keep the star in equilibrium, so no balance between the light that leaks out (which sets the luminosity) and the fusion rate (which mostly just responds to that luminosity).  As a result, the core shrinks and heats, and fusion initiates in a shell around the core-- a shell that used to be too cool to fuse.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature in that shell, and key aspects of its fusion rate, are determined by how small and massive the core becomes (the core gets smaller with time because it is losing heat, and it gets more massive because helium &quot;ash&quot; keeps being added to it from the shell).  Eventually the free electrons in the core become what is called &quot;degenerate&quot;, meaning they approach their quantum mechanical ground state, and that implies the core gets about as small as the Earth. This also means the shell is extremely hot, and the fusion rate goes essentially berserk.  This requires the envelope to puff out into a red giant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, to have a red giant, you must have a degenerate core, and you must have shell fusion surrounding it.  But stars less massive than 0.3 solar masses have an internal structure that is &lt;em&gt;purely convective.&lt;/em&gt;  Thus they don't run out of hydrogen in their cores until they run out of it &lt;em&gt;everywhere&lt;/em&gt;  (and like you say, none have yet done this in our universe, it's a theoretical expectation).  That is why they cannot have shell fusion, and cannot become red giants.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2019-03-31T10:08:19.623" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30183" PostTypeId="2" ParentId="24363" CreationDate="2019-03-31T12:30:27.020" Score="1" Body="&lt;p&gt;Also maybe have a look at this: &lt;a href=&quot;https://www.astroapprentice.com/2019/03/30/what-should-i-expect-to-see-with-a-telescope/&quot; rel=&quot;nofollow noreferrer&quot;&gt;What should I expect to see with a telescope&lt;/a&gt;&#xA;Your dark site location will help you the most to observe planets and deep sky objects with amazing detail.&lt;/p&gt;&#xA;" OwnerUserId="23668" LastActivityDate="2019-03-31T12:30:27.020" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30184" PostTypeId="2" ParentId="30176" CreationDate="2019-03-31T14:40:52.943" Score="4" Body="&lt;p&gt;@Ingolifs' &lt;a href=&quot;https://astronomy.stackexchange.com/a/30179/7982&quot;&gt;answer&lt;/a&gt; is exactly right. I'll just add to it some more plotting using the Python package &lt;a href=&quot;https://rhodesmill.org/skyfield/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Skyfield&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;inner&quot; plotted bodies (from outer to inner) are Mars, (Earth), Moon, Venus, Mercury, The Sun. The data is for &lt;strong&gt;2000 days&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;outer&quot; plotted bodies are Uranus, Saturn, Jupiter, Earth, and The Sun. The data is for &lt;strong&gt;2000 months&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distances are in AU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click the plot to see larger, or just run the Python script and make your own plot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0bT3U.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0bT3U.png&quot; alt=&quot;inner and outer planets, heliocentric and geocentric&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;class Body(object):&#xA;    def __init__(self, name):&#xA;        self.name = name&#xA;&#xA;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;from skyfield.api import Topos, Loader, EarthSatellite&#xA;&#xA;halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]&#xA;degs, rads = 180/pi, pi/180&#xA;&#xA;AU = 149597870.700  # km&#xA;&#xA;load    = Loader('~/Documents/fishing/SkyData')  # avoids multiple copies of large files&#xA;data421    = load('de421.bsp')&#xA;data405    = load('de405.bsp')&#xA;data       = data405   # need the longer time range for the outer planets&#xA;ts      = load.timescale()&#xA;&#xA;things  = ('sun', 'mercury', 'venus', 'earth', 'moon', 'mars',&#xA;           'jupiter barycenter', 'saturn barycenter', 'uranus barycenter', )&#xA;&#xA;bodies = []&#xA;for thing in things:&#xA;    name = thing.split()[0]&#xA;    body = Body(name)&#xA;    bodies.append(body)&#xA;    body.obj = data[thing]&#xA;&#xA;sun, mercury, venus, earth, moon, mars, jupiter , saturn, uranus = bodies&#xA;&#xA;days_inner = np.arange(2000)&#xA;times_inner = ts.utc(2019, 1, days_inner)&#xA;&#xA;months_outer = np.arange(2000)&#xA;times_outer = ts.utc(2019, months_outer, 1)&#xA;&#xA;for body in bodies:&#xA;    body.posn_barycentric_inner = body.obj.at(times_inner).position.km&#xA;    body.posn_barycentric_outer = body.obj.at(times_outer).position.km&#xA;for body in bodies:&#xA;    body.posn_geocentric_inner = body.posn_barycentric_inner - earth.posn_barycentric_inner&#xA;    body.posn_geocentric_outer = body.posn_barycentric_outer - earth.posn_barycentric_outer&#xA;    body.posn_heliocentric_inner = body.posn_barycentric_inner - sun.posn_barycentric_inner&#xA;    body.posn_heliocentric_outer = body.posn_barycentric_outer - sun.posn_barycentric_outer&#xA;&#xA;print (jupiter.posn_geocentric_inner.shape)&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;    lw, fs = 0.7, 12&#xA;&#xA;    plt.subplot(2, 2, 1)&#xA;    plt.title('geocentric inner', fontsize=fs)&#xA;    for body in (sun, mercury, venus, earth, moon, mars):&#xA;        x, y, z = body.posn_geocentric_inner / AU&#xA;        plt.plot(x, y, linewidth=lw)&#xA;&#xA;    plt.subplot(2, 2, 3)&#xA;    plt.title('heliocentric inner', fontsize=fs)&#xA;    for body in (sun, mercury, venus, earth, moon, mars):&#xA;        x, y, z = body.posn_heliocentric_inner / AU&#xA;        plt.plot(x, y, linewidth=lw)&#xA;&#xA;    plt.subplot(2, 2, 2)&#xA;    plt.title('geocentric outer', fontsize=fs)&#xA;    for body in (earth, mars, saturn, uranus):&#xA;        x, y, z = body.posn_geocentric_outer / AU&#xA;        plt.plot(x, y, linewidth=lw)&#xA;&#xA;    plt.subplot(2, 2, 4)&#xA;    plt.title('heliocentric outer', fontsize=fs)&#xA;    for body in (earth, mars, saturn, uranus):&#xA;        x, y, z = body.posn_heliocentric_outer / AU&#xA;        plt.plot(x, y, linewidth=lw)&#xA;&#xA;    plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-05-11T23:32:40.537" LastActivityDate="2019-05-11T23:32:40.537" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30185" PostTypeId="1" AcceptedAnswerId="30187" CreationDate="2019-03-31T19:10:19.363" Score="1" ViewCount="114" Body="&lt;p&gt;I'm having a difficult time understanding certain behaviors of white dwarfs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand how mass is lost in the red giant to white dwarf transition process. I understand that white dwarfs can accumulate mass from a partner binary or some other source and I'm aware of the 1.4 M☉ Chandrasekhar limit. I get that they're not losing all their mass; in general they're just cooling and in the end there will be a lone black chunk of crystallized material floating forever in the void. I know they are really insanely hot when formed and very cold after an unfathomable amount of time. But is it that same, initial mass, or does it somehow diminish?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is that initial mass the mass the remnant has after its cooled to a black dwarf? I guess what I'm asking is that &lt;strong&gt;I'm trying to figure out if there is a correlation between temperature, mass and radius.&lt;/strong&gt; Some sort of equation?&lt;/p&gt;&#xA;" OwnerUserId="26425" LastEditorUserId="2531" LastEditDate="2019-03-31T21:49:54.533" LastActivityDate="2019-03-31T21:49:54.533" Title="Do white dwarfs lose mass as they fade to black dwarfs? Is there a correlation between temperature, mass, and radius?" Tags="&lt;white-dwarf&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30186" PostTypeId="2" ParentId="30185" CreationDate="2019-03-31T19:20:58.933" Score="0" Body="&lt;p&gt;The mass should be thought of as set by the history of the white dwarf, and it really doesn't change (unless there is mass transfer from a companion, but you know about that).  The radius is then set by the mass, and it really doesn't change either, because the pressure comes from the internal kinetic energy which is extremely large, and is set by the quantum mechanics of the degenerate electrons.  Also, the white dwarf is not really very luminous, such that it will never, in its entire lifetime, emit much energy compared to the vast amount that is already in there, and that means very little change in radius.  So given the mass, there is really only one radius that is appropriate for the white dwarf, which can be found from a formula called the &quot;mass-radius relation&quot; for a white dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the last thing you asked about is the temperature.  This is almost entirely separate from all the other things we've talked about.  It just depends on the age of the white dwarf, as the dwarf cools, so there is no formula that connects temperature to radius or mass (at least not in an important way, there is always   tiny connections because the white dwarf is not fully degenerate).&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2019-03-31T19:20:58.933" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30187" PostTypeId="2" ParentId="30185" CreationDate="2019-03-31T21:46:51.673" Score="2" Body="&lt;p&gt;The mass of a white dwarf continues to be that which it was &quot;born&quot; with. It will not change significantly unless it accreted material from a companion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radius of a white dwarf is, to first order, given by the &quot;mass-radius relationship&quot; and this relationship does not involve the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass-radius relationship is appropriate for a &quot;cold&quot; star. &quot;Cold&quot; in this context mean that the pressure that supports the white dwarf is only dependent on density, which is the case for the degenerate electrons in the interior, which have kinetic energies much greater than their thermal energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, degenerate electrons also have excellent thermal conductivity, so white dwarf interiors are isothermal. Yet, they have a density gradient - denser in the middle and less dense answer move outwards. At some point close to the surface the electrons are hot enough to stop being degenerate and the gas pressure becomes temperature sensitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means is that the radius of a hot white dwarf is bigger than that of a cold white dwarf of the same mass. The effect depends on both the mass of the white dwarf (bigger for lower mass white dwarfs with lower surface gravities) and its age (since white dwarfs cool as they get older).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This effect is not negligible and needs to be properly modelled in order to understand the luminosities of white dwarfs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plot below is from &lt;a href=&quot;https://arxiv.org/abs/1706.05016&quot; rel=&quot;nofollow noreferrer&quot;&gt;Parsons et al. (2017)&lt;/a&gt; and shows (as points) measurements of masses and radii of white dwarfs in eclipsing binary systems. The lines are model curves for white dwarfs with surface temperatures ranging from zero (dashed line) to 60,000 K (appropriate for a very young white dwarf) in steps of 10,000 K. Clearly, there &lt;em&gt;is not&lt;/em&gt; a unique mass radius relationship and the radii do apparently depend on temperature (and core composition), as the model curves suggest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/P6vOO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P6vOO.png&quot; alt=&quot;White dwarf mass-radius relation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-03-31T21:46:51.673" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30188" PostTypeId="1" AcceptedAnswerId="30203" CreationDate="2019-04-01T03:27:18.763" Score="3" ViewCount="324" Body="&lt;p&gt;I've been told white dwarfs don't produce stellar wind like the sun produces solar winds. I don't quite understand why that is and can't really find any evidence that confirms or denies that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that solar wind is the product of the sun's variable magnetic field and consists of charged particles. Does nuclear fusion have to be occurring at the star's core for these charged particles to exist?&lt;/p&gt;&#xA;" OwnerUserId="26425" LastEditorUserId="26425" LastEditDate="2019-04-02T01:29:43.490" LastActivityDate="2019-04-02T15:48:59.837" Title="Do white dwarfs produce stellar wind? If not, why is that?" Tags="&lt;white-dwarf&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30189" PostTypeId="1" CreationDate="2019-04-01T03:32:36.013" Score="-1" ViewCount="65" Body="&lt;p&gt;I'm given this set of data and I need to describe the habitability. Would this not be habitable since the temperature of the albedo and the equilibrium is similar to Earth's or does the Spectral Type M affect the planet that much?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Spectral Type: M&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Age of the Star (Gyr): 0.664&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Binary System: Single Star&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Planet Orbital Period (Years): 0.0387&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Planet Semi-Major Axis (AU): 0.0613&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Planet Mass (Earth Masses): 1.87&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Planet Radius (Earth Radii): 1.24&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Method of First Detection: Transit&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Eccentricity: 0.11 &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Atmospheric Composition (Percent by Volume): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;N2  77.8&lt;br /&gt;&#xA;CO2 0.00368&lt;br /&gt;&#xA;H2O 2.37&lt;br /&gt;&#xA;O2  19.8 &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Number of other (known) Planets in the System: 1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Location in the Galaxy: Open Cluster&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Distance from Earth (lightyears): 960&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Atmospheric Pressure (atm): 1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Stellar Mass (Solar Masses): 0.154&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Planetary Albedo: 0.271&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Planet Equilibrium Temperature (K): 243&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Stellar Radius (Solar Radii): 0.171&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Orbital Inclination (degrees): 5.7&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Planet Axial Tilt (degrees): 18.4&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Magnetic Field Strength (Units of Earth Magnetic Field): 0.109&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="26428" LastEditorDisplayName="user1569" LastEditDate="2019-04-01T07:20:30.803" LastActivityDate="2019-04-01T07:20:30.803" Title="Compare this planet to Earth and determined its habitability (can humans live in this environment?)" Tags="&lt;exoplanet&gt;&lt;spectral-type&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2019-04-02T09:43:10.277" ContentLicense="CC BY-SA 4.0" />
  <row Id="30190" PostTypeId="2" ParentId="700" CreationDate="2019-04-01T06:24:32.020" Score="3" Body="&lt;p&gt;Let's just go back to the time when a red supergiant goes supernova. When it goes supernova, its outer shells are blown off because of the explosion. What happens next depends on the mass of the remnant. If the mass is 1.4 to 3 times the mass of the sun, it becomes a neutron star. If it is 3 times the mass or greater it becomes a black hole. Neutron stars cannot have the event horizons of black holes because their supernova remnant was simply not massive enough.&lt;/p&gt;&#xA;" OwnerUserId="26430" LastActivityDate="2019-04-01T06:24:32.020" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30191" PostTypeId="2" ParentId="30189" CreationDate="2019-04-01T06:29:51.953" Score="2" Body="&lt;p&gt;A very low mass M-dwarf (what you have there is something like an M5-M6 dwarf) will remain a highly magnetically active star for several billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, the spectrum of light from such an object has a much higher proportion, by several orders of magnitude, of UV and X-ray emission than the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, if your planet is close enough to its star to receive a similar bolometric flux as the Earth does from the Sun, then it will get a very hefty dose of UV and X-rays. In addition to this, there will be flaring activity that raises these levels by more orders of magnitude for minutes or hours at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is not enough detail to know how well the planet is shielded from this, but it is unlikely to be habitable (by us).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-04-01T06:29:51.953" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30192" PostTypeId="2" ParentId="29525" CreationDate="2019-04-01T06:41:31.983" Score="3" Body="&lt;p&gt;Brown dwarfs are technically stars, but not in your typical sense, as they cannot start or sustain nuclear fusion. Think about it this way: stars are formed in clumps inside giant gas clouds called nebulae, along with planets, but what if one of these clumps is too big to become a planet and too small to become red dwarfs. These are brown dwarfs. Some consider them to be stars, others consider them to be planets, and some neither. So, opinions vary.&lt;/p&gt;&#xA;" OwnerUserId="26430" LastActivityDate="2019-04-01T06:41:31.983" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30193" PostTypeId="2" ParentId="26565" CreationDate="2019-04-01T08:00:01.000" Score="0" Body="&lt;p&gt;Not life as it has evolved on Earth, no. A Red Dwarf emits mostly in in the infrared so the rate of photosynthesis would be very slow, leading to the collapse of food chains even if the temperature was favourable. An old white dwarf that had cooled to Sun-like temperature might be able to support life on Earth for a period of time, but given how close the orbit would be I suspect the Earth would become tidally locked to the star, so that the same face always faced it. Eventually you'd have baking hot temperatures on the  day-side and a frigid wasteland on the night-side, with possibly a small habitable twilight zone.&lt;/p&gt;&#xA;" OwnerUserId="23640" LastActivityDate="2019-04-01T08:00:01.000" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30195" PostTypeId="1" CreationDate="2019-04-01T08:55:53.270" Score="3" ViewCount="47" Body="&lt;p&gt;I read that we can observe [OIII] from AGNs but what processes in AGNs produces these lines why we can use them to trace or tell if that galaxy is active or not?&lt;/p&gt;&#xA;" OwnerUserId="26434" LastActivityDate="2019-04-01T08:55:53.270" Title="What is the relationship between AGN and [OIII]" Tags="&lt;active-galaxy&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30198" PostTypeId="1" AcceptedAnswerId="30204" CreationDate="2019-04-01T15:34:39.703" Score="0" ViewCount="50" Body="&lt;p&gt;I just read the BBC News article &lt;a href=&quot;https://www.bbc.com/news/science-environment-47681239&quot; rel=&quot;nofollow noreferrer&quot;&gt;Exoplanet tally set to pass 4,000 mark&#xA;&lt;/a&gt;. It makes me wonder about the annual rate  that new exoplanets being recognized, and how it compares to the rate of new asteroids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also wondering if there is any chance there will be a crossover in the rates in the next few years. &lt;a href=&quot;https://space.stackexchange.com/a/26577/12102&quot;&gt;TESS&lt;/a&gt; has found it's first exoplanet, and &lt;a href=&quot;https://space.stackexchange.com/q/34762/12102&quot;&gt;hopefully WFIRST will be on schedule&lt;/a&gt; as well.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-04-01T22:35:30.337" Title="At what annual rate are new exoplanets being recognized? How does it compare to new asteroids?" Tags="&lt;exoplanet&gt;&lt;asteroids&gt;&lt;tess&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30199" PostTypeId="2" ParentId="28610" CreationDate="2019-04-01T15:53:26.193" Score="2" Body="&lt;p&gt;The SXS consortium has &lt;a href=&quot;https://www.black-holes.org/explore/movies&quot; rel=&quot;nofollow noreferrer&quot;&gt;a page&lt;/a&gt; linking  a number of their videos. From the text around the videos it seems they use colour to encode various different scalar quantities in different videos. For instance&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The colors encode the lapse function — the slowing of the rate of flow of time. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Colors indicate the vorticity of the apparent horizon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;with the color scale giving the density of matter within the star [this is a neutron star merger, not black holes]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the LIGO gallery has not preserved this information (nor  the links to published papers where more details seem to be given) so it is hard to say in this specific case. Also simply understanding what many of these quantities &lt;em&gt;are&lt;/em&gt; needs a fair level of differential geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You asked:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why can I see some evidence of rotation in the resulting merged object for a few rotations, then it disappears?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since we don't know what quantity is being depicted, it's a bit hard to be very precise, but there is an established concept of &quot;ringdown&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may have heard the phrase &quot;black holes have no hair&quot; meaning that from outside everything about a (non-quantum) black hole is determined by its mass, spin and charge, with no details (the hair) remaining. This is not exactly true. What is true is that any &quot;hair&quot; a black hole has is quickly radiated away as gravitational waves so that it rapidly approaches the ideal state given by its mass, spin and charge. This phenomenon is called &quot;ringdown&quot; and is what you are in seeing in the first moments after the collision. It can be detected in the gravitational waves.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="21928" LastEditDate="2019-04-02T06:24:28.343" LastActivityDate="2019-04-02T06:24:28.343" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30201" PostTypeId="1" AcceptedAnswerId="34096" CreationDate="2019-04-01T17:31:13.700" Score="5" ViewCount="701" Body="&lt;p&gt;I am looking for a list of all known identifiers used for stars. I'm not sure on the terminology. While I do want the &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_proper_names_of_stars&quot; rel=&quot;noreferrer&quot;&gt;proper names&lt;/a&gt;&quot;, I also want whatever identifier is used for each star detected, even those without proper names. I am assuming there are far more of the latter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If possible I would also like to have the coordinates and brightness (and any other interesting info) for each of these star. But just the list of identifiers would be great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does such data exist somewhere? Ideally I can just download a simple file (like a csv) and I can write my own software to process it without fiddling with some other software. However, it may be too large for that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what to expect... 1 million entries?&lt;/p&gt;&#xA;" OwnerUserId="26439" LastActivityDate="2019-11-26T11:59:37.713" Title="Is there a database of all known star names/identifiers?" Tags="&lt;star&gt;&lt;data-analysis&gt;&lt;naming&gt;&lt;raw-data&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="30203" PostTypeId="2" ParentId="30188" CreationDate="2019-04-01T18:59:36.590" Score="4" Body="&lt;p&gt;The strong gravity of the white dwarf star makes it difficult for stellar wind to escape. &lt;a href=&quot;https://www.aanda.org/component/article?access=bibcode&amp;amp;bibcode=&amp;amp;bibcode=2008A%2526A...486..923UFUL&quot; rel=&quot;nofollow noreferrer&quot;&gt;Unglaub (2008)&lt;/a&gt; shows that for surface gravity &lt;span class=&quot;math-container&quot;&gt;$\log g &amp;gt; 7.0$&lt;/span&gt; hydrogen-helium winds are not possible for solar or sub-solar metallicity white dwarfs and only a weak metallic wind would be possible (note that in astronomy, &quot;metal&quot; tends to refer to any element heavier than helium). There is a group of hot white dwarfs with temperatures above 60000&amp;nbsp;K called &quot;hot wind white dwarfs&quot; showing ultra-high excitation absorption lines. According to &lt;a href=&quot;https://ui.adsabs.harvard.edu/#abs/2019MNRAS.482L..93R/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reindl et al. (2019)&lt;/a&gt;, these features arise from wind material trapped in the magnetosphere of the white dwarfs.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-04-02T07:03:38.873" LastActivityDate="2019-04-02T07:03:38.873" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30204" PostTypeId="2" ParentId="30198" CreationDate="2019-04-01T21:39:21.033" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;At what annual rate are new exoplanets being recognized? How does it compare to new asteroids?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming exoplanets have been discovered at a constant rate since 1995 (not very true) then the rate is very approximately 117 exoplanets a year. Comparing to the &lt;a href=&quot;https://minorplanetcenter.net/iau/lists/ArchiveStatistics.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Minor Planet Center (MPC) archive stats&lt;/a&gt; which are plotted below: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/omCRY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/omCRY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which shows that approx. 761,000 asteroids of all types have been discovered since 1995 April or roughly 22,000 a year. (The lower black line is for the smaller number of asteroids that have been seen enough times and the orbit is secure enough that they can be numbered).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exoplanets suffer from magnitude and volume-limit problems in that to get a large enough number on the sky to discover large numbers of them, requires going faint. Since confirming an exoplanet and ruling out other explanations such as blended background stars and M dwarf or brown dwarf binaries (which give similar transit depths but different radial velocity amplitudes) requires a large additional effort and high precision spectroscopy to confirm them, this becomes the bottleneck requiring tens or more of hours of large (4m) telescope time to confirm each of the fainter exoplanet candidates. Additionally the most numerous exoplanets in the 2-15 times of the Earth range are very tough to confirm as the RV amplitude is very small (few m/s or smaller depending on the orbital period).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm also wondering if there is any chance there will be a crossover in the rates in the next few years. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because of the limitations discussed above, the chances of there being a cross-over in rates any time soon is slim to none.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastEditorUserId="7982" LastEditDate="2019-04-01T22:35:30.337" LastActivityDate="2019-04-01T22:35:30.337" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30205" PostTypeId="2" ParentId="30188" CreationDate="2019-04-01T22:13:11.310" Score="2" Body="&lt;p&gt;One thing to be aware of is that prior to becoming white dwarfs, low-mass stars shed their envelopes as planetary nebulae.  Then the star itself is called the central star of a planetary nebula (CSPN), which is a star on its way to becoming a white dwarf.  These stars are very luminous so they have strong winds, and indeed that's the tail end of the creation of the planetary nebula.  Also, the cores of these stars are degenerate, so they are basically white dwarfs that are wearing a &quot;coat&quot;, and they are in the process of shedding that coat.  The act of shedding that coat sounds a lot like a white dwarf core that has a very strong wind-- the wind is the envelope being shed.  So you could say that white dwarfs have a history of very strong winds, but if you don't count it as a white dwarf until it has finished shedding that envelope, then you are basically saying you are defining the white dwarf to be the star that no longer has much of a wind.  So that's why white dwarfs don't have much in the way of winds, they've already lost everything that isn't &quot;nailed down.&quot;  But since there is not necessarily clear demarcations between a CSPN, a hot white dwarf, and a cold white dwarf, whether you count it as a white dwarf with a wind, or a bare white dwarf that has already lost everything it will lose, kind of depends on where you draw those lines.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2019-04-01T22:13:11.310" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30206" PostTypeId="1" AcceptedAnswerId="30220" CreationDate="2019-04-02T05:26:56.930" Score="2" ViewCount="440" Body="&lt;p&gt;As we see every day in the morning the sun shows a red-orange colour but at midday has a blue-white like colour why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it concerned with the distance from the earth or about intensity? How do you see it?&lt;/p&gt;&#xA;" OwnerUserId="26443" LastEditorUserId="1891" LastEditDate="2019-04-02T20:06:13.380" LastActivityDate="2019-04-19T10:59:32.697" Title="Why does the Sun show different colour at midday and at morning?" Tags="&lt;astrophysics&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="30207" PostTypeId="2" ParentId="30188" CreationDate="2019-04-02T06:30:54.230" Score="3" Body="&lt;p&gt;Stellar winds tend to be either &quot;thermal&quot; in nature, driven by high temperatures in the gas surrounding the star (e.g. the solar wind); or they are &quot;radiatively driven&quot;, the wind is accelerated by radiation pressure from the stellar photosphere (e.g. the winds from most hot stars).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;White dwarfs are not surrounded by hot coronae - they have comparatively weak X-ray emission, suggesting very little hot plasma surrounding them. They are incapable of supporting thermally driven winds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;White dwarfs can be hot though, especially when young, and emit intense radiation from their surfaces. Whether a radiatively driven wind can be setup depends on a competition between the outward acceleration provided by radiation pressure and the inward acceleration due to gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outward radiative acceleration depends to first order on luminosity divided by distance from the star squared. But the luminosity depends on the radius of the star squared times its surface temperature to the fourth power. That means if we express distance as a multiple of the stellar radius, the radiative driving at a set number of stellar radii only depends on &lt;span class=&quot;math-container&quot;&gt;$T^4$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity on the other hand depends on the mass of the star divided by the square of the distance from it. So the gravitational force at a set number of stellar radii increases as the inverse square of the absolute stellar radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we can compare a hot white dwarf with a B star. The B star might have a mass of &lt;span class=&quot;math-container&quot;&gt;$\sim 10 M_{\odot}$&lt;/span&gt; and a radius of &lt;span class=&quot;math-container&quot;&gt;$\sim 10 R_{\odot}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$T\sim 20000$&lt;/span&gt;  K; it will have a strong, radiatively driven wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A young white dwarf at the same temperature has a mass of &lt;span class=&quot;math-container&quot;&gt;$\sim 1M_{\odot}$&lt;/span&gt; and radius &lt;span class=&quot;math-container&quot;&gt;$\sim 0.01R_{\odot}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider a point 1 stellar radius above the surface of each star, then material at this point will be outwardly accelerated by approximately &lt;em&gt;the same&lt;/em&gt; radiative flux. However, the inward gravitational force on that material will be a hundred thousand times higher for the white dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is why &lt;em&gt;most&lt;/em&gt; white dwarfs do not have strong winds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why &lt;em&gt;most&lt;/em&gt;? Well white dwarfs can be hotter than this. There are a fraction of much hotter white dwarfs (&lt;span class=&quot;math-container&quot;&gt;$T&amp;gt;60000$&lt;/span&gt; K) that do show high excitation UV emission lines with asymmetries suggesting weak mass loss via a wind. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2000A%26A...359.1042U&quot; rel=&quot;nofollow noreferrer&quot;&gt;Unglaub &amp;amp; Bues (2000)&lt;/a&gt; discuss theoretical calculations of mass loss rates for very young and hot white dwarfs. At &lt;span class=&quot;math-container&quot;&gt;$T\sim 80000$&lt;/span&gt; K, the radiative accelerations are increased by a factor of 250 over the case considered above. Simultaneously, hot white dwarfs are not completely degenerate and are larger (by factors of a few) than cool white dwarfs of the same mass (see &lt;a href=&quot;https://astronomy.stackexchange.com/questions/30185/do-white-dwarfs-lose-mass-as-they-fade-to-black-dwarfs-is-there-a-correlation-b&quot;&gt;Do white dwarfs lose mass as they fade to black dwarfs? Is there a correlation between temperature, mass, and radius?&lt;/a&gt;) and thus the gravitational accelerations are smaller by an order of magnitude. This allows the possibility (for white dwarfs with helium envelopes) of weak winds (of order &lt;span class=&quot;math-container&quot;&gt;$&amp;lt;10^{-11}\ M_{\odot}$&lt;/span&gt;/yr, which fade drastically on timescales of a few hundred thousand to a few million years as the white dwarf cools, the radiative flux decreases and the surface gravity increases. This level of mass loss is insufficient to affect the white dwarf's subsequent evolution.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-04-02T15:48:59.837" LastActivityDate="2019-04-02T15:48:59.837" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30209" PostTypeId="1" CreationDate="2019-04-02T10:04:13.637" Score="0" ViewCount="60" Body="&lt;p&gt;Assume that the expansion of the universe stops and it starts to contract again. What would happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would only distances between objects get smaller, or are there other effects we can expect?&lt;/p&gt;&#xA;" OwnerUserId="26447" LastEditorDisplayName="user1569" LastEditDate="2019-04-02T12:00:07.627" LastActivityDate="2019-04-02T12:09:43.233" Title="What would happen if space-time began to contract?" Tags="&lt;space-time&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2019-04-02T14:02:24.270" ContentLicense="CC BY-SA 4.0" />
  <row Id="30211" PostTypeId="1" AcceptedAnswerId="30214" CreationDate="2019-04-02T10:56:51.083" Score="0" ViewCount="54" Body="&lt;p&gt;I am a statistician with love and passion for astronomy. I try to explain some outliers I found in the values of quasars - color indexes. I found four (4) values close to twenty (20). Is that scientifically possible or there is an error in the CSV file? What values can a color index get?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ty in advance.&lt;/p&gt;&#xA;" OwnerUserId="22830" LastEditorUserId="25994" LastEditDate="2019-04-02T20:04:20.537" LastActivityDate="2019-04-03T11:20:01.700" Title="What values can a color index variable get?" Tags="&lt;observational-astronomy&gt;&lt;quasars&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30212" PostTypeId="2" ParentId="30209" CreationDate="2019-04-02T12:09:43.233" Score="1" Body="&lt;p&gt;Given that the expansion of the universe shows us that the distance between galaxies (or rather moreso between galaxy clusters) is increasing, without any change in the distances within or between atoms, or between stars within galaxies (apart from normal stellar motion), and that galaxies themselves are also not expanding, we could assume that a contraction would manifest as a decrease in the distances between galaxies/galaxy clusters. &lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2019-04-02T12:09:43.233" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30214" PostTypeId="2" ParentId="30211" CreationDate="2019-04-02T14:23:11.947" Score="3" Body="&lt;p&gt;The only physical mechanism I can think of for quasars to have such a large difference in two bands would be at very high redshift (&lt;span class=&quot;math-container&quot;&gt;$z\gtrsim6$&lt;/span&gt;) where you have the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gunn%E2%80%93Peterson_trough&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gunn-Peterson trough&lt;/a&gt; completely erasing everything blueward of 912 Å (restframe). But with typical magnitudes of &lt;span class=&quot;math-container&quot;&gt;$\sim20$&lt;/span&gt; or so at these redshifts, a color of &lt;span class=&quot;math-container&quot;&gt;$20$&lt;/span&gt; that would then mean that the magnitude in the bluest band would be of the order &lt;span class=&quot;math-container&quot;&gt;$\sim 40$&lt;/span&gt;, which no instrument is able to detect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So effectively, there would be no detection on the bluest band, in which case the color would be infinite, or rather undefined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, I think those values must be errors.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2019-04-03T11:20:01.700" LastActivityDate="2019-04-03T11:20:01.700" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30215" PostTypeId="2" ParentId="29090" CreationDate="2019-04-02T16:11:30.407" Score="1" Body="&lt;p&gt;As far as &lt;em&gt;microlensing&lt;/em&gt; is concerned:&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Not likely&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This is a bit of a dodge on the question, but it is hard to come up with a scenario where microlensing becomes the dominant PSF term. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Microlensing occurs during conjunction with a stellar or planetary object, and results in a momentary (seconds-to-days timescale) increase in the brightness of the observed star. It is an extremely rare event, &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0002510&quot; rel=&quot;nofollow noreferrer&quot;&gt;with rates around 100 in 17 million stars in the most recent surveys&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stellar densities (&lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt;) around our solar system are on the order of 0.14 stars per cubic parsec. With such a low number density, microlensing does not pose any practical limitation on the resolution of radio telescopes. Lets do some scratch-pad estimates to see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume a synthetic aperture (&lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt;) of 1AU (&lt;span class=&quot;math-container&quot;&gt;$4.84\times10^{-6}$&lt;/span&gt;parsecs) , a wavelength (&lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt;) of 1mm (&lt;a href=&quot;https://astronomynow.com/2017/06/30/see-the-sharpest-ever-view-of-giant-betelgeuse/&quot; rel=&quot;nofollow noreferrer&quot;&gt;similar to the view of Betelguese by the ALMA observatory&lt;/a&gt;) , and a desired spatial resolution (&lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt;) of 10,000 km at a distance &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{align}&#xA;\frac{s}{L}  &amp;amp;= 1.22 \frac{\lambda }{D} \\&#xA;L &amp;amp;= \frac{sD}{1.22 \lambda } = 40\;\text{kilo-parsec}&#xA;\end{align}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is roughly the diameter of the milky way galaxy. So an interferometer operating in the 1mm wavelength with a 1AU baseline has a 10,000 km resolution at a distance of 40 kpc.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many stars are between the 1AU-wide observatory and the target star? Using a cone of base diameter &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; and height &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; as a volume and the above stellar density, we get the average number of stars (&lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt;) between our aperture and the observed star:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ N = \rho \frac{\pi}{12}D^2 L = 3 \times 10^{-8} \; \text{stars}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would have to ramp &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; up by a factor of &lt;span class=&quot;math-container&quot;&gt;$10^4$&lt;/span&gt; before &lt;span class=&quot;math-container&quot;&gt;$N \approx 1$&lt;/span&gt; and 'microlensing aberrations' were more than just an extremely rare event.  &lt;/p&gt;&#xA;" OwnerUserId="22897" LastEditorUserId="7982" LastEditDate="2019-04-04T03:41:09.273" LastActivityDate="2019-04-04T03:41:09.273" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30216" PostTypeId="2" ParentId="26016" CreationDate="2019-04-02T19:46:22.200" Score="3" Body="&lt;p&gt;My understanding is that the stick-figure lines (in terms of applying them to every constellation) are a fairly modern invention - before that the individual stars were not (generally speaking) linked together directly but were instead highlighting important areas of each constellation figure. The constellation figures were fully 2D and potentially quite detailed, and there was not always a one to one correspondence between stars and details. In this sense one is free to modify the existing stick figure patterns to suit one's needs. Many of the stars visible to the naked eye have individual names (sometimes more than one name in fact) and these names usually mean/meant something about the constellation figure they are/were in. It's not always possible to follow these names when redrawing the figures (the constellations we use today are not all ancient and star names remain from figures we no longer use), but usually you can accommodate most of them. This is one of my main problems with H.A.Rey's versions - they often pay no attention at all to what the star names mean and you end up with a star named for a tail now marking the figure's mouth. :)&lt;/p&gt;&#xA;" OwnerUserId="26452" LastActivityDate="2019-04-02T19:46:22.200" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30217" PostTypeId="1" CreationDate="2019-04-02T21:03:28.527" Score="3" ViewCount="411" Body="&lt;p&gt;I am having a slight disagreement with my professor.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;We can measure the mass of a distant star cluster from:&lt;/p&gt;&#xA;&lt;p&gt;A.Its color&lt;/p&gt;&#xA;&lt;p&gt;B.Its radius&lt;/p&gt;&#xA;&lt;p&gt;C.Its metallicity&lt;/p&gt;&#xA;&lt;p&gt;D.Smearing of lines in its total spectrum&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;My thinking was:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;Law of Conservation of Energy&#xA;&#xA;Kinetic Energy = Gravitational force&#xA;&#xA;1/2*m*v^2 =  G* m*M/r&#xA;&#xA;v = sqrt(2GM/r)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;So it is dependant on both velocity and radius.&lt;/p&gt;&#xA;&lt;p&gt;Answer choice D would give you the velocity but I responded with answer choice B, radius.&lt;/p&gt;&#xA;&lt;p&gt;After talking it over with my professor, she responded with:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If we think about it conceptually, the dominant parameter is the velocity.  In other words, you can get a meaningful estimate of the cluster mass from the velocity dispersion without knowing the radius (because all clusters are the &amp;quot;same&amp;quot; size), but not vice versa.&lt;/p&gt;&#xA;&lt;p&gt;So I'd say it still seems pretty clear that the best answer is the velocities.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Is she right or is the question worded poorly?&lt;/p&gt;&#xA;" OwnerUserId="26453" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2021-04-22T12:09:50.897" Title="How to determine the mass of a galaxy?" Tags="&lt;galaxy&gt;&lt;gravity&gt;&lt;mass&gt;&lt;galaxy-cluster&gt;&lt;velocity&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30218" PostTypeId="1" AcceptedAnswerId="30225" CreationDate="2019-04-02T21:14:10.193" Score="1" ViewCount="468" Body="&lt;p&gt;I'm interested in the learning about the local densities around galaxies. I've found myself a bit confused on how to relate angular distances (arcseconds) with physical distances (parsecs) when thinking about distances between two distant galaxies. For example, let's say the average distance between galaxies within clusters is a few megaparsecs. If I had ra, dec coordinates of a particular galaxy and could hypothetically query an all-sky catalog, let's say I'd want to search within a few megaparsecs to get a feel for its density field. But generally when querying catalogs you specify a search radius in some arcseconds. How would these relate here? &lt;/p&gt;&#xA;" OwnerUserId="24076" LastActivityDate="2019-04-03T13:02:27.003" Title="How would I search within an angular distance of an object corresponding to a distance in parsecs?" Tags="&lt;distances&gt;&lt;parsec&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30219" PostTypeId="2" ParentId="30206" CreationDate="2019-04-02T23:47:58.600" Score="1" Body="&lt;p&gt;Neither. As @User123 &lt;a href=&quot;https://astronomy.stackexchange.com/questions/30206/why-does-the-sun-show-different-colour-at-midday-and-at-morning#comment55288_30206&quot;&gt;commented&lt;/a&gt;, it depends on the angle between where you are and the Sun. If the Sun is low in the sky, sunlight has to travel through much more of the atmosphere than at midday when the Sun is higher in the sky. The intensity of the Sun is basically constant, and the distance from the Sun does not vary between morning and midday.&lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="14094" LastEditDate="2019-04-19T10:59:32.697" LastActivityDate="2019-04-19T10:59:32.697" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30220" PostTypeId="2" ParentId="30206" CreationDate="2019-04-03T00:15:02.410" Score="3" Body="&lt;p&gt;Sunsets are red because the sky is blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air, while it appears colorless actually scatters light just a bit. This is not the same as absorbing the light, but more like how light is scattered by fog.  If you are driving through fog with your headlights on, part of the light from their beams bounces off the fog particles and lights up the fog well outside the beam and some of it even bounces multiple times coming back to your eyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air is much clearer than fog, but it does scatter light just a bit. What's more, it scatters blue light more than green, and green more than yellow, and yellow more than red.  So while the Sun's light in space is quite white, as it passes through the atmosphere on a sunny noonday a lot of the blue light gets scattered, while very little of the red does.  That blue light bounces many times and is the source of the blue sky light.  The loss of that blue light makes the sun appear redder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At noon, the light comes down from high overhead and passes through a minimal amount of atmosphere and only some of the blue light is scattered. At sunset (and sunrise) the light comes in at a long, low slant and passes through &lt;em&gt;much&lt;/em&gt; more of the atmosphere, and nearly all of the blue and even most of the green is scattered (and much light is absorbed, too), leaving mostly red and orange light left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the reddened sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://www.sciencemadesimple.com/sky_blue.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Why Is the Sky Blue?&lt;/a&gt; and &lt;a href=&quot;https://www.sciencemadesimple.com/space_black_sunset_red.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Why Is the Sunset Red?&lt;/a&gt; for a bit more including some diagrams.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2019-04-03T00:15:02.410" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30221" PostTypeId="2" ParentId="30218" CreationDate="2019-04-03T02:26:08.757" Score="1" Body="&lt;p&gt;Remember that an arcsecond is an angle, so imagine a pie wedge that extends out from your observatory, having that angle inside the point of the pie wedge.  That's all you can tell by looking at the sky, the angle in that pie wedge.    Also, notice that a field on the sky is not a line, it's more like a circle, so the pie wedge is really a cone with that angle in it.  But you don't know from just searching how far away anything is, so if you want to search on distance, then you have to go through and determine the distance to everything in that search angle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you center your search on a particular galaxy that you do know the distance to.  Now you have a search cone that has its point at your observatory, and the galaxy in question at its center some known distance away.  But within that angular search cone, you may see other objects at all kinds of distances.  There's no easy way to limit to things that have a similar distance as that galaxy, you get the whole cone unless you go through and find the distance to everything and throw out what you don't want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the galaxy density at nearly the same distance as that galaxy, then you will have to throw out things that are not nearly at the same distance.  Perhaps you have a lower and upper distance cutoff, so now you are searching a section from that cone, but it's not a sphere within some given distance from that galaxy, that shape would be much harder to figure out.  Still, the section I was talking about does have a volume, and you can count the galaxies within it, so that does give you a galaxy density around that galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To figure out that volume, you would need the linear distance across the cone at the distance to that galaxy.  That sounds like what you are asking about.  That is simple geometry, because the connection between angles and distances can be seen from the formula for the circumference of a circle.  The angle all the way around the center point of the circle is 2*Pi radians, and the circumference is 2*Pi*r, so that tells you that you take the angle in radians (you have to convert from arcsec) and multiply by the radius to get the length of the arc across from that angle.  That's not just true for 2*Pi radians, it's true for any smaller angle as well, as you know if you've ever sliced up a pizza.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for your application, the r is the distance to the galaxy, and the angle (converted into radians) is the angle of your search field, and when you multiply them you get the length of the arc that crosses the cone we are talking about, at the distance to that galaxy.  That length is the key to figuring out the volume you are counting galaxies in, as the rest of its dimensions depend on what distances you are using to limit what you are counting within that field.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2019-04-03T02:26:08.757" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30222" PostTypeId="1" AcceptedAnswerId="30224" CreationDate="2019-04-03T05:05:20.220" Score="6" ViewCount="250" Body="&lt;p&gt;Why is the outer section of the sun convective but the inner section stable and only radiative? They are both made up of the same kind of matter, both heated by the core.&lt;/p&gt;&#xA;" OwnerUserId="26231" LastActivityDate="2019-04-03T13:49:37.253" Title="Why does the sun have a convection layer and a radiation layer?" Tags="&lt;star&gt;&lt;the-sun&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30223" PostTypeId="2" ParentId="30217" CreationDate="2019-04-03T08:00:17.367" Score="0" Body="&lt;p&gt;I think she should have clarified the question by putting velocity and radius in the same choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By looking at her answer, it seems reasonable to think that velocity is more important. However  without knowing the radius we cannot determine the mass of the stellar system. She argued this idea by saying that the size is approximately known for stellar systems. But I don't think that is a solid argument. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoritically we should know both R and V to calculate M. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I did not understand why you wrote kinetic energy= gravitational force ? Thats not possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In stellar bounded system you should use virial theorem. By stating that &lt;span class=&quot;math-container&quot;&gt;$-2&amp;lt;E&amp;gt;=&amp;lt;U&amp;gt;$&lt;/span&gt;  and from there you get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M=v^2R/G$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="25424" LastEditorUserId="25424" LastEditDate="2019-04-03T08:06:56.123" LastActivityDate="2019-04-03T08:06:56.123" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30224" PostTypeId="2" ParentId="30222" CreationDate="2019-04-03T10:17:08.837" Score="5" Body="&lt;p&gt;The onset of convection is controlled by temperature gradients (see the famous experiment from &lt;a href=&quot;https://en.wikipedia.org/wiki/Rayleigh%E2%80%93B%C3%A9nard_convection&quot; rel=&quot;noreferrer&quot;&gt;Rayleigh &amp;amp; Bénard&lt;/a&gt;. When those become too strong, convection sets in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How strong a temperature gradient gets, depends on the local efficiency of energy transport inside the star. But for the sun (and other lower mass stars), the surface radiates energy away efficiently, dropping the temperature there compared to the rest of the star, making the outer layers convective. Lower temperatures (not gradients) compared to high-mass stars also play a role in controlling the energy transport for lower mass stars. Cool temperatures will allow the formation of &lt;span class=&quot;math-container&quot;&gt;$H^{-}$&lt;/span&gt; (negatively charged hydrogen ions) that give high opacities and thus prohibit energy transport.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For high-mass stars it is the enormously fast fusion processes in the core that 'overheat' this region, again letting convection take over.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2019-04-03T10:17:08.837" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30225" PostTypeId="2" ParentId="30218" CreationDate="2019-04-03T13:02:27.003" Score="2" Body="&lt;p&gt;Ken G's answer is essentially correct, but there is one important thing to keep in mind: The distance &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; between the galaxies that is interesting for you is the distance they had at the time they emitted the light you see (you don't care about what has happened to them since that time and how far apart they are today; you care about the physical properties &lt;em&gt;then&lt;/em&gt;). But the angle you observe them to span is not angle they spanned when they emitted the light (since they were closer to you in the past, so the light they emitted in your direction at that time would &quot;miss&quot; you by the time it reached that place).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that, in your calculation, the distance from you to the galaxies should not be the &quot;physical&quot; distance &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; (i.e. the actual distance right now to the galaxies, which also corresponds to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Comoving_and_proper_distances#Comoving_distance_and_proper_distance&quot; rel=&quot;nofollow noreferrer&quot;&gt;comoving distance&lt;/a&gt;), but the so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_diameter_distance&quot; rel=&quot;nofollow noreferrer&quot;&gt;angular diameter distance&lt;/a&gt;, &lt;span class=&quot;math-container&quot;&gt;$d_A$&lt;/span&gt;. These are related (in a flat universe) by&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;d_A = \frac{d}{1+z},&#xA;$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; is the redshift of the galaxies (I assume they have almost the same redshift; if they are too different, then the distance from each other is not simply given by the projected distance). Note that in a non-flat universe, the formula is slightly more complicated; see the link above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The comoving distance, in turn, is obtained by integration:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;d(z) = \frac{c}{H_0} \int_0^z \frac{dz'}{\left( \Omega_\mathrm{M}(1+z)^3 + \Omega_\Lambda \right)^{1/2}},&#xA;$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; is the speed of light, and &lt;span class=&quot;math-container&quot;&gt;$\Omega_\mathrm{M}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\Omega_\Lambda$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt; are the present matter and dark energy density parameters and Hubble constant, respectively. Here, I've again assumed a flat universe, and neglected radiation since when that was important, there were no galaxies anyway).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, this effect is not big for small redshifts, but &lt;span class=&quot;math-container&quot;&gt;$d_A$&lt;/span&gt; has the interesting property that it increases with physical distance only out to a certain point (roughly &lt;span class=&quot;math-container&quot;&gt;$z=1.6$&lt;/span&gt;), &lt;em&gt;after which it decreases&lt;/em&gt;, because the galaxies were closer to us in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the physical distance &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; between two galaxies spanning an angle &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;r = \frac{\theta}{d_A}&#xA;$$&lt;/span&gt;&#xA;or&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\boxed{&#xA;\frac{r}{\mathrm{kpc}} = \frac{\theta / \mathrm{arcsec}}{d / \mathrm{Mpc}} (1+z)&#xA; \, \frac{10^{-3} \times 180\times3600}{\pi},&#xA;}&#xA;$$&lt;/span&gt;&#xA;where the last factor accounts for converting Mpc to kpc and radians to arcseconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Python, arbitrarily for &lt;span class=&quot;math-container&quot;&gt;$z=2.2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\theta=12.3''$&lt;/span&gt;,  it's as easy as&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from astropy.cosmology import Planck15&#xA;from astropy import units as u&#xA;z     = 2.2                               # redshift of galaxies&#xA;theta = 12.3 * u.arcsec                   # angle&#xA;r_ang = Planck15.kpc_proper_per_arcmin(z) # phys. dist. per angle&#xA;r     = r_ang * theta                     # physical distance&#xA;print('Distance between galaxies: ', r.to(u.kpc))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which will print &lt;code&gt;Distance between galaxies:  104.22540165050975 kpc&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2019-04-03T13:02:27.003" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30226" PostTypeId="2" ParentId="30222" CreationDate="2019-04-03T13:49:37.253" Score="3" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Convection_zone&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia page&lt;/a&gt; linked by @pm-2ring in a comment contains a very clear summary of what decides if convection happens in a particular region:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The Schwarzschild criterion expresses the conditions under which a region of a star is unstable to convection. A parcel of gas that rises slightly will find itself in an environment of lower pressure than the one it came from. As a result, the parcel will expand and cool. If the rising parcel cools to a lower temperature than its new surroundings, so that it has a higher density than the surrounding gas, then its lack of buoyancy will cause it to sink back to where it came from. However, if the temperature gradient is steep enough (i. e. the temperature changes rapidly with distance from the center of the star), or if the gas has a very high heat capacity (i. e. its temperature changes relatively slowly as it expands) then the rising parcel of gas will remain warmer and less dense than its new surroundings even after expanding and cooling. Its buoyancy will then cause it to continue to rise. The region of the star in which this happens is the convection zone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;To work out the structure of a particular star (and in particular work out which parts of it are convective and which are not) you then have to solve a set of equations which relate the temperature and density at each radius to the flow of energy from the core to space and the hydrostatic balance within the star.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Star#Structure&quot; rel=&quot;nofollow noreferrer&quot;&gt;Another wikipedia page&lt;/a&gt; gives some more details and links to relevant books and papers. To quote&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The interior of a stable star is in a state of hydrostatic equilibrium: the forces on any small volume almost exactly counterbalance each other. The balanced forces are inward gravitational force and an outward force due to the pressure gradient within the star. The pressure gradient is established by the temperature gradient of the plasma; the outer part of the star is cooler than the core. The temperature at the core of a main sequence or giant star is at least on the order of &lt;span class=&quot;math-container&quot;&gt;$10^7$&lt;/span&gt; K. The resulting temperature and pressure at the hydrogen-burning core of a main sequence star are sufficient for nuclear fusion to occur and for sufficient energy to be produced to prevent further collapse of the star.&lt;/p&gt;&#xA;&lt;p&gt;As atomic nuclei are fused in the core, they emit energy in the form of gamma rays. These photons interact with the surrounding plasma, adding to the thermal energy at the core. Stars on the main sequence convert hydrogen into helium, creating a slowly but steadily increasing proportion of helium in the core.&lt;/p&gt;&#xA;&lt;p&gt;In addition to hydrostatic equilibrium, the interior of a stable star will also maintain an energy balance of thermal equilibrium. There is a radial temperature gradient throughout the interior that results in a flux of energy flowing toward the exterior. The outgoing flux of energy leaving any layer within the star will exactly match the incoming flux from below.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;These equations predict a variety of structures for different stars, depending on their mass and composition.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-04-03T13:49:37.253" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30227" PostTypeId="1" CreationDate="2019-04-03T15:38:40.143" Score="2" ViewCount="32" Body="&lt;p&gt;Having the solar flare fluxes on the two bands of GOES (0.05-0.4 nm and 0.1-0.8 nm) I need to calculate the flux between 0.05 and 0.2 nm.&lt;/p&gt;&#xA;" OwnerUserId="26460" LastEditorUserId="26460" LastEditDate="2019-04-06T19:51:38.757" LastActivityDate="2019-04-06T19:51:38.757" Title="Spectra from Temperature and Emission Measure" Tags="&lt;spectra&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2019-04-05T09:10:49.583" ContentLicense="CC BY-SA 4.0" />
  <row Id="30228" PostTypeId="1" AcceptedAnswerId="30230" CreationDate="2019-04-03T18:51:23.497" Score="3" ViewCount="114" Body="&lt;p&gt;In 2010 observations with the Fermi Gamma-ray Space Telescope discovered giant bubbles which have been named &lt;a href=&quot;https://fermi.gsfc.nasa.gov/science/constellations/pages/bubbles.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fermi Bubbles&lt;/a&gt;. Similar structures have also been discovered by &lt;a href=&quot;https://arxiv.org/abs/1603.07245&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pshirkov et al. (2016)&lt;/a&gt; surrounding M31 (the Andromeda Galaxy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our galaxy and M31 are barred spirals. Could Fermi Bubbles also form around other galaxy types, for example lenticular or elliptical galaxies?&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-04-03T21:51:53.483" LastActivityDate="2019-04-04T00:03:11.807" Title="Can Fermi Bubbles be produced in non-spiral galaxies?" Tags="&lt;galaxy&gt;&lt;gamma-rays&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30230" PostTypeId="2" ParentId="30228" CreationDate="2019-04-04T00:03:11.807" Score="3" Body="&lt;p&gt;The discovery paper (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2010ApJ...724.1044S&quot; rel=&quot;nofollow noreferrer&quot;&gt;Su et al. 2010&lt;/a&gt;) mentions that bipolar shells on the order of &lt;span class=&quot;math-container&quot;&gt;$0.5\text{ kpc}$&lt;/span&gt; were found on both sides of the nucleus of Centaurus A (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2006ApJ...641L..29Q&quot; rel=&quot;nofollow noreferrer&quot;&gt;Quillen et al. 2006&lt;/a&gt;). Centaurus A is a galaxy of contested morphology, likely either elliptical or lenticular - certainly not a spiral. This would seem to be a strong indicator that similar bubbles can indeed be found in non-spiral galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the nuclear shells Quillen et al. detected were seen at mid-infrared wavelengths (5.8, 8.0 and 24 &lt;span class=&quot;math-container&quot;&gt;$\mu\text{m}$&lt;/span&gt;), rather than via the gamma rays with which Su et al. discovered the Fermi bubbles. However, the authors believe they may have been caused by a burst of star formation a few million years ago, driven by winds. This is one of two main scenarios considered by Su et al. for the formation of the Milky Way's bubbles (the other being AGN activity), meaning that the same processes could indeed have been at work in both galaxies. I don't know if corresponding gamma-ray emission from the nuclear bubbles in Centaurus A has since been detected, but it certainly seems plausible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find it a little strange that Pshirkov et al. were so careful to focus on spiral galaxies. It's clear that starburst galaxies and the host galaxies of AGN can obviously have non-spiral morphologies. Perhaps the group was just interested in searching for other nuclear bubbles on galaxies similar to the Milky Way which might have similar conditions and evolutionary histories.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2019-04-04T00:03:11.807" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30231" PostTypeId="1" CreationDate="2019-04-04T01:30:08.193" Score="1" ViewCount="65" Body="&lt;p&gt;Given a right ascension in degrees and declination in degrees (say, 88 degrees RA, -63 degrees DEC) I want to find the angle between that location and the ecliptic pole (90 degrees RA, -66 degrees DEC... are these the right values? What are the decimals following the 66? Most of my values that I'm working with are similar to the one I give, so which ecliptic pole, if any, does this correspond to?) How would I find this angle, I need a convenient formula (such as a dot product to find this angle). Thank you!&lt;/p&gt;&#xA;" OwnerDisplayName="user23921" LastActivityDate="2019-04-04T20:58:08.590" Title="Finding angle between a given sky position and ecliptic pole" Tags="&lt;observational-astronomy&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30232" PostTypeId="1" AcceptedAnswerId="30233" CreationDate="2019-04-04T03:20:38.337" Score="0" ViewCount="816" Body="&lt;p&gt;Much of the Earth's atmosphere has a scale height &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; of roughly 7 to 8 kilometers such that the local density varies as &lt;span class=&quot;math-container&quot;&gt;$\exp(-(r-r_0)/h)$&lt;/span&gt; where r is the radius vector and &lt;span class=&quot;math-container&quot;&gt;$r_0$&lt;/span&gt; would be some reference radius above the surface of the Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works approximately, at least up to 50 km (six or seven scale heights).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivation of this approximation is based on a simple gas at constant temperature, and those probably don't apply to the Sun's atmosphere very well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, are there regions of the Sun's atmosphere where density varies approximately exponentially in such a way that a scale height characterization would work over the range of at least a few scale heights?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-04-04T11:25:21.833" Title="Does the Sun's atmosphere have a scale height?" Tags="&lt;the-sun&gt;&lt;stellar-atmospheres&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30233" PostTypeId="2" ParentId="30232" CreationDate="2019-04-04T06:43:03.493" Score="2" Body="&lt;p&gt;The exponential decrease in density comes out naturally whenever you have a gas in hydrostatic equilibrium. The scale height &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; is then given by the balance between the kinetic energy of the particles due to thermal motion, &lt;span class=&quot;math-container&quot;&gt;$kT$&lt;/span&gt;, and the gravitational energy of the particles, &lt;span class=&quot;math-container&quot;&gt;$mg$&lt;/span&gt;. This is often a good approximation, both in planetary and stellar atmospheres, and even in galaxies.&#xA;That is,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;H = \frac{kT}{mg}&#xA;$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; is Boltszmann's constant, &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is the temperature, &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; is the average mass of the particles, and&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;g = \frac{GM}{r^2}&#xA;$$&lt;/span&gt;&#xA;with &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; the gravitational constant, and &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; the mass inside the radius &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the surface of our Sun, &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; works out to &lt;span class=&quot;math-container&quot;&gt;$274\,\mathrm{m}\,\mathrm{s}^{-2}$&lt;/span&gt;, 27 times as high as on Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average particle mass depends weakly on metallicity, and mostly on the ionization state of the gas, since the small mass of free electrons compared to that of atoms pulls the average down. For a fully ionized gas, the &lt;em&gt;mean molecular mass&lt;/em&gt; — i.e. the mass in terms of the hydrogen mass — is &lt;span class=&quot;math-container&quot;&gt;$\mu \simeq 0.6$&lt;/span&gt;, whereas for a fully neutral gas it is (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/1996ima..book.....C&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carroll &amp;amp; Ostlie 1996&lt;/a&gt;)&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\mu \simeq \frac{1}{X + Y/4 + Z/15.5} \simeq 1.25,&#xA;$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$Y$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$Z$&lt;/span&gt; are the mass fractions of hydrogen, helium, and metals, respectively. I mistakenly wrote initially that the gas is fully ionized, but that's not true; it is only partially ionized, and the hydrogen is largely neutral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the average mass of a particle to be roughly equal to the proton mass &lt;span class=&quot;math-container&quot;&gt;$m_p$&lt;/span&gt; (i.e. setting &lt;span class=&quot;math-container&quot;&gt;$\mu=1$&lt;/span&gt;), and taking the temperature to be &lt;span class=&quot;math-container&quot;&gt;$T = 5770\,\mathrm{K}$&lt;/span&gt;, the scale height is thus&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;H_\odot = \frac{kTR_\odot}{GM_\odot m_p} \simeq 170\,\mathrm{km}.&#xA;$$&lt;/span&gt;&#xA;With &lt;span class=&quot;math-container&quot;&gt;$\mu=0.6$&lt;/span&gt; you'd get &lt;span class=&quot;math-container&quot;&gt;$H\simeq290\,\mathrm{km}$&lt;/span&gt;, while &lt;span class=&quot;math-container&quot;&gt;$\mu=1.25$&lt;/span&gt; yields &lt;span class=&quot;math-container&quot;&gt;$H = 140\,\mathrm{km}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Realistic density profiles&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The above calculation are quite basic, assuming a completely isotropic Sun. But observations and more realistic models, both 1D and 3D, do indeed predict exponential density profiles, although with quite large variations across the surface (according to a Solar physicist colleague down the hall). I found this model from &lt;a href=&quot;http://www2.gi.alaska.edu/~jvo/P472G/SolarAtmosphere.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;these lecture notes&lt;/a&gt; where the yellow curve shows the number density profile in the Sun's atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Jdj4h.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jdj4h.png&quot; alt=&quot;n_prof&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extracting the data and plotting on a log-linear scale shows reasonable agreement with an exponential decrease of scale height &lt;span class=&quot;math-container&quot;&gt;$H = 140\,\mathrm{km}\,\mathrm{s}^{-1}$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tcoja.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tcoja.png&quot; alt=&quot;n&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2019-04-04T11:25:21.833" LastActivityDate="2019-04-04T11:25:21.833" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="30234" PostTypeId="1" AcceptedAnswerId="30238" CreationDate="2019-04-04T15:05:41.187" Score="6" ViewCount="123" Body="&lt;p&gt;When we see multiple images of the same object because of the phenomenon of gravitational lensing, do all the images show the lensed object at the same point in time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it take the same amount of time for light from the lensed object to reach us no matter what image we look at or are we watching the lensed object at different points in time and if so, how big can the difference in time typically be?&lt;/p&gt;&#xA;" OwnerUserId="26403" LastActivityDate="2019-04-05T07:36:42.450" Title="Does gravitational lensing provide time evolution information?" Tags="&lt;gravitational-lensing&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30236" PostTypeId="2" ParentId="30231" CreationDate="2019-04-04T20:58:08.590" Score="1" Body="&lt;p&gt;The angle between an ecliptic pole and another sky position is complementary to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ecliptic_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;ecliptic latitude&lt;/a&gt; of that position.&#xA;To find ecliptic (&amp;lambda;, &amp;beta;) coordinates, one method is to convert the equatorial (&amp;alpha;, &amp;delta;) to Cartesian form:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \begin{align}&#xA;x_\mathrm{equ} &amp;amp;= \cos \alpha \cos \delta \\&#xA;y_\mathrm{equ} &amp;amp;= \sin \alpha \cos \delta \\&#xA;z_\mathrm{equ} &amp;amp;= \sin \delta&#xA;\end{align}$$&lt;/span&gt;&#xA;then rotate by &amp;epsilon; = 23.44&amp;deg; on the x axis:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \begin{align}&#xA;x_\mathrm{ecl} &amp;amp;= x_\mathrm{equ} \\&#xA;y_\mathrm{ecl} &amp;amp;= y_\mathrm{equ} \cos \varepsilon + z_\mathrm{equ} \sin \varepsilon \\&#xA;z_\mathrm{ecl} &amp;amp;= z_\mathrm{equ} \cos \varepsilon - y_\mathrm{equ} \sin \varepsilon&#xA;\end{align} $$&lt;/span&gt;&#xA;and convert to spherical form:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \begin{align}&#xA;\lambda &amp;amp;= \mathtt{atan2}(y_\mathrm{ecl}, x_\mathrm{ecl}) \\&#xA;\beta &amp;amp;= \arcsin z_\mathrm{ecl}&#xA;\end{align} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are not interested in the ecliptic longitude &amp;lambda;, you can express the ecliptic latitude as&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \beta = \arcsin (\sin \delta \cos \varepsilon - \sin \alpha \cos \delta \sin \varepsilon) $$&lt;/span&gt;&#xA;Then the angle off the north ecliptic pole (&amp;alpha;=18h0m, &amp;delta;=+66.56&amp;deg;) is 90&amp;deg; - &amp;beta;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working your example (&amp;alpha;=5h52m, &amp;delta;=-63&amp;deg;),&#xA;I get (&amp;lambda;=75.63&amp;deg;, &amp;beta;=-86.34&amp;deg;),&#xA;and the angle off the south ecliptic pole (&amp;alpha;=6h0m, &amp;delta;=-66.56&amp;deg;) is &amp;beta; + 90&amp;deg; = 3.66&amp;deg;.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2019-04-04T20:58:08.590" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30238" PostTypeId="2" ParentId="30234" CreationDate="2019-04-04T22:36:47.873" Score="3" Body="&lt;p&gt;Time delays between the light received from multiple lensed images is a well known and commonly studied phenomenon. A typical example can be found in &lt;a href=&quot;http://adsabs.harvard.edu/abs/2003ApJ...599...70K&quot; rel=&quot;nofollow noreferrer&quot;&gt;Koopmans et al. (2003)&lt;/a&gt; where they study a quadruple lens system formed by a forefround object at z=0.6 and a background source at z=1.4. The time delays measured are 31-77 days and this seems to be typical for the objects studied so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The delay depends on the difference in path length taken by the multiple images. This in turn depends on the distance to the source and the distance between the source and the lensing object and on the mass distribution around the lensing object. It also depends on the expansion rate of the universe, which gives a route to estimating the Hubble parameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rule of thumb appears to be that the delay is of order &lt;span class=&quot;math-container&quot;&gt;$r_s/c$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$r_s = 2GM/c^2$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is the mass of the lens interior to the path of the light ray. &#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \tau \sim 10^{-5} \left(\frac{M}{M_{\odot}}\right)\ {\rm sec}$$&lt;/span&gt;&#xA;So for a &lt;span class=&quot;math-container&quot;&gt;$10^{12}M_{\odot}$&lt;/span&gt; foreground lensing galaxy (a big spiral like the Milky Way), the delay is about 100 days.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-04-05T07:36:42.450" LastActivityDate="2019-04-05T07:36:42.450" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30239" PostTypeId="2" ParentId="11388" CreationDate="2019-04-04T22:44:28.600" Score="2" Body="&lt;p&gt;I believe his meaning is being misinterpreted as a claim that the universe actually is infinite.  If someone says &quot;in an infinite universe&quot;, they are not asserting that the universe is indeed infinite (as no scientist could ever claim to know), they are simply saying &quot;if the universe is infinite.&quot;  It is merely a case from which to begin the discussion.   For example, if you were having a discussion about whether to pass a seatbelt law, you might start with &quot;in a world where people always make the safer choice in everything they do, we would not need seatbelt laws.&quot;  That would not be a claim we live in such a world, it is merely setting a kind of baseline for going on to talk about what is the actual situation.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2019-04-04T22:44:28.600" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30240" PostTypeId="1" AcceptedAnswerId="30247" CreationDate="2019-04-05T03:55:15.520" Score="4" ViewCount="178" Body="&lt;p&gt;I found three NASA JPL images in the sequence PIA23133, PIA23134, and PIA23135. They were taken by the Curiosity Rover on Mars and show solar eclipses by Mars' two moons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Details of these images are given in &lt;a href=&quot;https://space.stackexchange.com/a/35306/12102&quot;&gt;this answer&lt;/a&gt; to &lt;em&gt;Puzzler: help understanding these amazing Curiosity eclipse GIFs&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Why is it that it's mostly a chunk of the right half of the sky that darkens during the eclipse in PIA23135? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this kind of effect visible in solar eclipses seen on Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enquiring minds want to know!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;fyi in case you are wondering, per @MarkAdler's &lt;a href=&quot;https://space.stackexchange.com/a/35078/12102&quot;&gt;ultra-concise answer&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They use neutral-density filters to look at the Sun, which reduce the light by a factor of 100,000. The two Pancam cameras each have one neutral-density filter, with the left one filtering blue and the right one filtering red.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://photojournal.jpl.nasa.gov/archive/PIA23133.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://photojournal.jpl.nasa.gov/archive/PIA23133.gif&lt;/a&gt; and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://photojournal.jpl.nasa.gov/archive/PIA23134.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://photojournal.jpl.nasa.gov/archive/PIA23134.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lm2Ci.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lm2Ci.gif&quot; alt=&quot;photojournal.jpl.nasa.gov/archive/PIA23133.gif&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/JyYOy.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JyYOy.gif&quot; alt=&quot;photojournal.jpl.nasa.gov/archive/PIA23134.gif&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://photojournal.jpl.nasa.gov/archive/PIA23135.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://photojournal.jpl.nasa.gov/archive/PIA23135.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a fairly slow) GIF:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hhpTR.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hhpTR.gif&quot; alt=&quot;photojournal.jpl.nasa.gov/archive/PIA23135.gif&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-07-25T02:26:40.160" LastActivityDate="2019-07-25T02:26:40.160" Title="Why is only a narrow chunk of sky darkened in this amazing Curiosity solar eclipse GIF from Mars?" Tags="&lt;mars&gt;&lt;solar-eclipse&gt;&lt;atmospheric-effects&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30241" PostTypeId="1" CreationDate="2019-04-05T10:03:21.617" Score="2" ViewCount="52" Body="&lt;p&gt;The question &lt;a href=&quot;https://astronomy.stackexchange.com/q/30234/7982&quot;&gt;Does gravitational lensing provide time evolution information?&lt;/a&gt; is excellent!&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;When we see multiple images of the same object because of the phenomenon of gravitational lensing, do all the images show the lensed object at the same point in time?&lt;/p&gt;&#xA;&lt;p&gt;Does it take the same amount of time for light from the lensed object to reach us no matter what image we look at or are we watching the lensed object at different points in time and if so, how big can the difference in time typically be?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The answer is generally no, different &amp;quot;images&amp;quot; are associated with substantially different time delays.&lt;/p&gt;&#xA;&lt;p&gt;But the lens and mirror-based optical systems we used to see those phenomenon pretty much strictly obey &lt;a href=&quot;https://en.wikipedia.org/wiki/Fermat%27s_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fermat's principle&lt;/a&gt; where every ray entering the telescope's aperture reaching given point on a detector takes the same amount of time through the telescope.&lt;/p&gt;&#xA;&lt;p&gt;Of course gravitational lenses are not manufactured imaging systems, but usually wherever you see that an image is produced in nature, the thing producing that image will obey Fermat's principle.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; So &lt;em&gt;at least within one given image&lt;/em&gt; of a multiple-impage producing gravitational lens, does Fermat's principle apply at least approximately?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-04-05T10:03:21.617" Title="Within a given image of a multiple-image producing gravitational lens, does Fermat's principle apply?" Tags="&lt;telescope&gt;&lt;optics&gt;&lt;gravitational-lensing&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30243" PostTypeId="1" AcceptedAnswerId="30244" CreationDate="2019-04-05T10:20:22.530" Score="12" ViewCount="1697" Body="&lt;p&gt;@RobJeffries' &lt;a href=&quot;https://astronomy.stackexchange.com/a/30238/7982&quot;&gt;answer&lt;/a&gt; to the question &lt;em&gt;Does gravitational lensing provide time evolution information?&lt;/em&gt; points out that there can be a substantial different in arrival times of light from a given source seen in different images from a gravitational lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The linked paper there shows &quot;&lt;span class=&quot;math-container&quot;&gt;$\Delta t$&lt;/span&gt;&quot; values of the order of 30 days, but it is hard for me to understand what the actual observable is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm asking for here is (ideally) if there is a well defined event that a lay person could understand, something blinking or disappearing or brightening substantially that has already been seen in one image produced by a gravitational lens that has &lt;em&gt;not yet been seen&lt;/em&gt; in one of the other images, and is expected to be seen in the (presumably near) future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If something like this does not exist, a substitute could be a case where this has happened, and the second sighting of the same event was predicted, waited for, and observed on time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea if this happens all the time, or has never happened yet. &lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-04-05T11:07:43.407" LastActivityDate="2019-04-05T11:54:18.737" Title="Are astronomers waiting to see something in an image from a gravitational lens that they've already seen in an adjacent image?" Tags="&lt;observational-astronomy&gt;&lt;gravitational-lensing&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30244" PostTypeId="2" ParentId="30243" CreationDate="2019-04-05T11:41:38.503" Score="15" Body="&lt;p&gt;What you do is cross-correlate the observational datasets for the multiple sources and look for the &quot;lag&quot; that maximises the cross-correlation function. Generally speaking, the &quot;events&quot; are not really individual flares or dips, but the summation of all the time variability that is seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variability in question usually comes about from the central portions of the &quot;central engine&quot; of a quasar or active galactic nucleus. For a supermassive black hole at the centre of a quasar, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Innermost_stable_circular_orbit&quot; rel=&quot;noreferrer&quot;&gt;innermost stable circular orbit&lt;/a&gt; is at 3 times the Schwarzschild radius (&lt;span class=&quot;math-container&quot;&gt;$= 6GM_{\rm BH}/c^2$&lt;/span&gt;). This basically defines the inner edge of any accretion disk and if we divide this by &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; then we get a a timescale for the most rapid variations in luminosity output. So this is very nearly the same formula as presented in the linked question&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\tau \sim 3\times 10^{-5} \left(\frac{M_{\rm BH}}{M_{\odot}}\right)\ {\rm sec}\, ,$$&lt;/span&gt;&#xA;except that the supermassive black holes are much less massive than entire foreground lensing galaxies (usually). This the timescale of variation is much shorter than the potential delay time due to gravitational lensing. It is this difference in timescales that means there is plenty of &quot;structure&quot; within the light curves that can be locked onto by the cross-correlation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a notable example however of a type Ia supernova being seen in a multiply lensed image (&lt;a href=&quot;https://arxiv.org/abs/1611.00014&quot; rel=&quot;noreferrer&quot;&gt;Goobar et al. (2017)&lt;/a&gt;, but the predicted delay in the light curves was &lt;span class=&quot;math-container&quot;&gt;$&amp;lt;35$&lt;/span&gt; hours and the light curves are not good enough to measure this. This technique is an active area of research and a major bit of science that is exprected to be achieved by the Large Synoptic Survey Telescope (&lt;a href=&quot;https://arxiv.org/abs/1903.00510&quot; rel=&quot;noreferrer&quot;&gt;Huber et al. 2019&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the thing you are really looking for &lt;em&gt;has&lt;/em&gt; happened in terms of &lt;strong&gt;SN &quot;Refsdal&quot;&lt;/strong&gt;. This was a type II supernova seen to &quot;go off&quot; in a multiply imaged galaxy, seen through/around a galaxy cluster. A prediction was made, based on a model for the cluster gravitational potential, that another image ought to appear within a year or two. This further image was then detected by &lt;a href=&quot;http://adsabs.harvard.edu/abs/2016ApJ...819L...8K&quot; rel=&quot;noreferrer&quot;&gt;Kelly et al. (2016)&lt;/a&gt; in a paper entitled &quot;Deja vu all over again&quot;. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://adsabs.harvard.edu/abs/2016ApJ...819L...8K&quot; rel=&quot;noreferrer&quot;&gt;Kelly et al. (2016)&lt;/a&gt; (&quot;Deja vu all over again&quot;). See &quot;SX&quot; in the third panel:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fYpHZ.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fYpHZ.jpg&quot; alt=&quot;MACS J1149.5+2223 galaxy-cluster field taken with HST&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Figure 1. Coadded WFC3-IR F125W and F160W exposures of the MACS J1149.5+2223 galaxy-cluster field taken with HST. The top panel shows images acquired in 2011 before the SN appeared in S1–S4 or SX. The middle panel displays images taken on 2015 April 20 when the four images forming the Einstein cross are close to maximum brightness, but no flux is evident at the position of SX. The bottom panel shows images taken on 2015 December 11 which reveal the new image SX of SN Refsdal. Images S1–S3 in the Einstein cross configuration remain visible in the 2015 December 11 coadded image (see Kelly et al. 2015a and Rodney et al. 2015b for analysis of the SN light curve).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Kelly, P. L., Brammer, G., Selsing, J., et al. &lt;strong&gt;2015a&lt;/strong&gt;, ApJ, submitted&#xA;  (arXiv:&lt;a href=&quot;http://arxiv.org/abs/1512.09093&quot; rel=&quot;noreferrer&quot;&gt;1512.09093&lt;/a&gt;)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Rodney, S. A., Strolger, L.-G., Kelly, P. L., et al. &lt;strong&gt;2015b&lt;/strong&gt;, ApJ, in press&#xA;  (arXiv:&lt;a href=&quot;http://arxiv.org/abs/1512.05734&quot; rel=&quot;noreferrer&quot;&gt;1512.05734&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="7982" LastEditDate="2019-04-05T11:54:18.737" LastActivityDate="2019-04-05T11:54:18.737" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30245" PostTypeId="2" ParentId="30240" CreationDate="2019-04-05T18:35:01.050" Score="2" Body="&lt;p&gt;The eclipse on sol 2359 occurred at&#xA;&lt;a href=&quot;https://mars.jpl.nasa.gov/msl/multimedia/raw/?s=2359&amp;amp;camera=MAST&quot; rel=&quot;nofollow noreferrer&quot;&gt;03:45 UTC on 2019-03-27&lt;/a&gt;,&#xA;and Curiosity saw Phobos pass wholly in front of the northern half of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The post-sunset eclipse on sol 2358 occurred 23 hours or 3 Phobos orbits earlier, around&#xA;&lt;a href=&quot;https://mars.jpl.nasa.gov/msl/multimedia/raw/?s=2358&amp;amp;camera=NAV_RIGHT&quot; rel=&quot;nofollow noreferrer&quot;&gt;04:49 UTC on 2019-03-26&lt;/a&gt;.&#xA;This time Phobos passed a little farther north, casting its shadow mostly to the north of Curiosity while points a few km south received full sunlight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Earth, a solar eclipse observer several km south of the centerline would see &#xA;the shadow sweep the northern horizon from west to east.&#xA;&lt;a href=&quot;https://vimeo.com/53641212&quot; rel=&quot;nofollow noreferrer&quot;&gt;Colin Legg's timelapse&lt;/a&gt; of the November 2012 solar eclipse faces eastward and shows the shadow moving from upper left to lower right.&#xA;Videos taken from outside the totality zone are harder to find.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-04-05T21:04:37.960" LastActivityDate="2019-04-05T21:04:37.960" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30246" PostTypeId="1" AcceptedAnswerId="30249" CreationDate="2019-04-05T18:40:09.313" Score="3" ViewCount="64" Body="&lt;p&gt;Based on what I know, the Earth accreted out of the solar nebula, had a period when it was entirely molten, and then slowly cooled. Could a planet form without ever passing through a molten phase? &lt;/p&gt;&#xA;" OwnerUserId="23788" LastActivityDate="2019-04-05T21:23:15.310" Title="Could a planet form without a molten state?" Tags="&lt;planetary-formation&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30247" PostTypeId="2" ParentId="30240" CreationDate="2019-04-05T18:52:40.217" Score="4" Body="&lt;p&gt;Phobos' shadow on Mars is very narrow -- only about 15 miles across -- and it didn't quite pass across the Sun as viewed from the lander on that particular orbit.  This image (taken by Mars Global Surveyor) gives you an excellent picture of just how small it is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JTvE4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JTvE4.jpg&quot; alt=&quot;Shadow of Phobos on the Martian surface&quot; /&gt;&lt;/a&gt; Credit: NASA/JPL/Malin Space Science Systems &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Phobos_shadow.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The penumbral shadow of Phobos is visible on the landscape of Mars, as photographed by the Mars Global Surveyor. The center of the shadow was at approximately 10.9°N 49.2°W (Map of area Map zoom) at 04:00:33.3 UTC Earth time.&lt;/p&gt;&#xA;&lt;p&gt;The image shows western Xanthe Terra on August 26 1999 at about 2:41 p.m. local solar time on Mars. The image covers an area about 250 kilometers across and is illuminated from the left. The dark spots on the crater floors are probably dark sand dunes. None of the craters pictured currently have names. Nanedi Valles is the meandering valley at the bottom right. The picture covers about 7°–15°N vertically and 52°–48°W horizontally. The vertical orientation of the image is 3.01° to the west of north; a north-pointing arrow superimposed on the image would point slightly to the right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It only can darken a portion of the sky.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-04-05T21:07:54.643" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30248" PostTypeId="1" AcceptedAnswerId="30253" CreationDate="2019-04-05T18:52:50.400" Score="3" ViewCount="408" Body="&lt;p&gt;What I understood about it till now is that space-time fabric is kind of broken there. Please tell me whether I am correct.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If it's true then does it mean that perception of time will stop there?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I really can't get what broken space fabric is like. If it's a hole then how is matter supposed to be present inside that?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is a black hole a discontinuity?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="26475" LastEditorUserId="14094" LastEditDate="2019-04-06T06:20:52.390" LastActivityDate="2019-04-06T06:40:46.643" Title="Discontinuity in space-time" Tags="&lt;black-hole&gt;&lt;space-time&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30249" PostTypeId="2" ParentId="30246" CreationDate="2019-04-05T21:23:15.310" Score="5" Body="&lt;p&gt;If you accrete a mass of &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; the total binding energy will be &lt;span class=&quot;math-container&quot;&gt;$3GM^2/5R$&lt;/span&gt;, which will at least partially turn into heat. If the melting temperature is &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; and specific heat capacity &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; we get the condition &lt;span class=&quot;math-container&quot;&gt;$$\frac{3GM^2}{5R}=CTM,$$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$M/R&amp;lt;5CT/3G$&lt;/span&gt;. So if we use a low silicate rock melting point of 600 C and basalt thermal capacity then &lt;span class=&quot;math-container&quot;&gt;$M/R&amp;lt;1.8313\cdot 10^{16}$&lt;/span&gt; kg/m, while the upper range 1200 C gives &lt;span class=&quot;math-container&quot;&gt;$3.0899\cdot 10^{16}$&lt;/span&gt; kg/m. So if we consider a basalt planetoid it will start to have some melt when cohering into one ball of radius larger than 1000 km. So, not much of a planet.&lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2019-04-05T21:23:15.310" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30250" PostTypeId="1" CreationDate="2019-04-05T21:32:24.873" Score="4" ViewCount="249" Body="&lt;p&gt;I found the picture below in an answer to the post &quot;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/1651/was-the-milky-way-ever-a-quasar&quot;&gt;Was the Milky Way ever a quasar&lt;/a&gt;&quot;. The picture is published on a website belonging to &quot;NASA/IPAC Extragalactic Database&quot;. In the answer that has 16 &quot;thumbs up&quot; it says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A quasar is simply an active galactic nucleus (AGN) that is viewed&#xA;  from a particular angle; see the picture below, in which quasars are&#xA;  labeled &quot;QSO&quot;. This is really a remarkable figure because historically&#xA;  all of the names in the figure were thought to correspond to different&#xA;  types of objects, when really they all refer to the same thing!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://ned.ipac.caltech.edu/level5/March04/Torres/Figures/figure5.jpg&quot; alt=&quot;AGN&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another answer to the question &lt;a href=&quot;https://astronomy.stackexchange.com/questions/6099/difference-between-quasar-and-active-galactic-nuclei&quot;&gt;&quot;Difference between quasar and Active Galactic Nuclei?&quot;&lt;/a&gt; where someone states that a quasars: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;main property is that it's galaxy axis points towards the earth and&#xA;  this way we receive its light and radio signals. That's why they are&#xA;  one of the most energetic AGN.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This answer has gotten six thumbs up. However, there is no reference to any source backing up that claim so it seems like private opinion.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Does the picture above represent the mainstream view among astronomers today or is it more like a private opinion of someone?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;According to Wikipedia, ten percent of all observed galaxies are classified as &lt;a href=&quot;https://en.wikipedia.org/wiki/Seyfert_galaxy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Seyfert galaxies&lt;/a&gt;. Would we perceive all of those as &quot;quasars&quot; if we happend to observe them from the right angle? (Maybe there is also a luminosity requirement)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;According to Wikipedia &lt;a href=&quot;https://en.wikipedia.org/wiki/Markarian_231&quot; rel=&quot;nofollow noreferrer&quot;&gt;Markarian 231&lt;/a&gt; is the closest quasar at a distance of 581 million light years. Also according to Wikipedia, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Circinus_Galaxy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Circinus Galaxy&lt;/a&gt; is the closest Seyfert galaxy, only 13 million light years away. Would we perceive there to be many quasars closer than 581 million light years away, if we just happend to view the active galaxy from the right angle?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Could someone please show me the scientific investigations proving that the galactical planes of galaxies perceived to host quasars always lies within a certain range of observational angles as viewed from the Earth? These kind of investigations should be available if the claim that whether or not an active galaxy is perceived to be a quasar is almost totally depending on observation angle has a real backing and is not just speculation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="26403" LastEditorUserId="26403" LastEditDate="2019-04-07T10:54:07.170" LastActivityDate="2020-08-19T16:32:13.687" Title="Are quasars simply AGNs that are viewed from a particular angle?" Tags="&lt;observational-astronomy&gt;&lt;quasars&gt;&lt;active-galaxy&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30251" PostTypeId="1" CreationDate="2019-04-06T02:46:44.127" Score="1" ViewCount="62" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Very-long-baseline_interferometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Very Long Baseline Interferometry&lt;/a&gt; or VLBI such as the Event Horizon Telescope or JPL's delta-DOR (see below) uses baselines of the order of the Earth's diameter. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://space.stackexchange.com/q/15347/12102&quot;&gt;Spacecraft position determination&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://space.stackexchange.com/q/21256/12102&quot;&gt;How do we track the exact location of the spacecraft which is millions or billions of miles away from us?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://space.stackexchange.com/q/31467/12102&quot;&gt;Help understanding BepiColombo's weak capture at Mercury's L1 and need for delta-differential one-way range measurements&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://space.stackexchange.com/q/6528/12102&quot;&gt;Tracking Arc and Time for Precise Orbit Determination&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The Earth's diameter is about 12,000 kilometers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Russian radio telescope satellite &lt;a href=&quot;https://en.wikipedia.org/wiki/Spektr-R&quot; rel=&quot;nofollow noreferrer&quot;&gt;Spektr-R&lt;/a&gt; allows for an even larger baseline, with an apoapsis that can reach to 300,000 kilometers!&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://space.stackexchange.com/q/27183/12102&quot;&gt;What artificial satellite has the farthest orbit around the Earth?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://space.stackexchange.com/q/33564/12102&quot;&gt;What are Spectr-R's major contributions to radio astronomy that could not have been done from Earth?&lt;/a&gt; (currently unanswered, &lt;em&gt;should I move it here?&lt;/em&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These result in maximum arrival times of about 42 milliseconds and 1 second, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to combine these signals, mathematical correlation is used, and the &quot;beam is steered&quot; by varying the time delay in the correlation. This can be done in real time, or during &quot;playback&quot;. In the case of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Event_Horizon_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Event Horizon Telescope&lt;/a&gt; (from which we are waiting for a press conference next Wednesday) suitcases full of hard drives are flown on airplanes from the various sites and collected before the correlation can even begin.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/q/20082/7982&quot;&gt;How does the Event Horizon Telescope implement the interferometry?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Have there been observations with much longer delays used in correlation of radio signals? Minutes? Months?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-04-06T02:52:11.890" LastActivityDate="2020-07-10T02:17:14.200" Title="Examples of radio correlations over times much longer than interferometric baselines?" Tags="&lt;observational-astronomy&gt;&lt;radio-astronomy&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="30252" PostTypeId="1" CreationDate="2019-04-06T06:22:19.337" Score="2" ViewCount="54" Body="&lt;p&gt;When observing 2 objects side by side close to the horizon with a very large array, one receding fast &amp;amp; red shifted, is the delay to correlate the signal from the closest dishes to the further ones longer for the red shifted object?&lt;/p&gt;&#xA;" OwnerUserId="26710" LastActivityDate="2019-04-08T14:44:58.410" Title="Is the delay to correlate the signal from one dish to another on a VLA radio telescope longer for red shifted objects?" Tags="&lt;radio-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30253" PostTypeId="2" ParentId="30248" CreationDate="2019-04-06T06:40:46.643" Score="2" Body="&lt;p&gt;A singularity in physics is basically a big signpost saying &quot;the theory you are using has probably ceased to be a good approximation of reality by the time you get here&quot;. An example is the infinite amount of short-wavelength radiation produced by a blackbody under the &lt;a href=&quot;https://en.wikipedia.org/wiki/Rayleigh%E2%80%93Jeans_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rayleigh-Jeans law&lt;/a&gt;. This is obviously an incorrect description of reality and the issue ended up being resolved by the development of quantum theory: taking into account the corrections gives &lt;a href=&quot;https://en.wikipedia.org/wiki/Planck&amp;#39;s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck's law&lt;/a&gt;, and the infinity goes away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside a black hole, &lt;a href=&quot;https://en.wikipedia.org/wiki/General_relativity&quot; rel=&quot;nofollow noreferrer&quot;&gt;general relativity&lt;/a&gt; does predict singularities, and it seems to be fairly common to talk of them as though they are physical objects. However we should be cautious: even though general relativity is our best description of gravity so far, we already know that it does not describe the actual universe. The issue here is that general relativity is incompatible with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Standard_Model&quot; rel=&quot;nofollow noreferrer&quot;&gt;Standard Model&lt;/a&gt; of quantum field theory, which is our best theory of how physics works excluding gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the time you get to the region near where general relativity predicts a singularity, you're working with a very small region with an intense gravitational field. Quantum corrections are almost certainly going to be relevant here, and unfortunately we do not yet have a self-consistent theory of &quot;quantum gravity&quot; to describe what should happen in such a region and so far experiments (such as the ones being performed at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Large_Hadron_Collider&quot; rel=&quot;nofollow noreferrer&quot;&gt;Large Hadron Collider&lt;/a&gt;) haven't given much of a clue as to how to proceed with developing one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a hypothetical universe where general relativity is correct then there would be a singularity (which is fortunately hidden from the rest of the universe by the event horizon), but we do not live in such a universe.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastActivityDate="2019-04-06T06:40:46.643" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30255" PostTypeId="2" ParentId="6643" CreationDate="2019-04-06T11:39:19.803" Score="1" Body="&lt;p&gt;I wondered the same thing until I focused my studies on the &quot;cosmological axis of evil&quot; regarding the CMB. It appears that some cosmologists believe that due to the fact the Cosmic Microwave Background appears to have an axis, this alludes to the expansion of the Universe having a rotation. Similar to how the earth rotates upon an axis. This, however, is not widely accepted. &lt;/p&gt;&#xA;" OwnerUserId="26485" LastEditorUserId="26485" LastEditDate="2019-04-06T12:02:36.033" LastActivityDate="2019-04-06T12:02:36.033" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30256" PostTypeId="1" CreationDate="2019-04-06T20:05:30.153" Score="2" ViewCount="145" Body="&lt;p&gt;I'm having a hard time understanding why there would be this ambiguity. Also, why would this be not the case if the planet was viewed pole on?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &quot;NOrth SOuth ambiguity&quot; is a term used in a paper that seemed most concise but I guess not. When using radar to doppler image a planetary surface which happens to be viewed equator on, we have no way of distinguishing whether the surface feature read in return signal belongs to the northern or southern hemisphere. WHat causes this?&lt;/p&gt;&#xA;" OwnerUserId="26488" LastEditorUserId="26488" LastEditDate="2019-04-07T23:03:54.000" LastActivityDate="2019-04-08T07:21:04.433" Title="What causes &quot;North-South ambiguity&quot; when doppler radar imaging a planet surface equator on?" Tags="&lt;planet&gt;&lt;radar&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="30257" PostTypeId="1" AcceptedAnswerId="30261" CreationDate="2019-04-06T21:37:41.733" Score="3" ViewCount="49" Body="&lt;p&gt;It is known that during Solar/Stellar Flare, plasma from the corona of the star is released into the space, and travels comparatively slowly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an hypothetical situation, does the coronal mass also consist of electromagnetic radiation within the optical wavelengths? And if so, is it possible to observe the incoming coronal mass ejection from the ground/surface of the Earth way before it approaches?&lt;/p&gt;&#xA;" OwnerUserId="26489" LastEditorUserId="43" LastEditDate="2019-04-06T22:20:34.393" LastActivityDate="2019-04-07T06:20:29.177" Title="Does coronal mass ejection include radiation at optical wavelengths?" Tags="&lt;astrophysics&gt;&lt;solar-flare&gt;&lt;coronal-mass-ejection&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30258" PostTypeId="2" ParentId="30257" CreationDate="2019-04-06T23:04:28.460" Score="0" Body="&lt;p&gt;We can see coronas normally by blocking out the main body of sun (think during eclipses) so I would assume yes? Except the sun is so bright in optical wavelength it would be hard to see&lt;/p&gt;&#xA;" OwnerUserId="26488" LastActivityDate="2019-04-06T23:04:28.460" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30259" PostTypeId="2" ParentId="30257" CreationDate="2019-04-07T01:37:25.173" Score="0" Body="&lt;p&gt;Yes.It is possible to see the ejection of mass into the corona not only during eclipses but also by means of an instrument named Coronograph which blinds the light of the photosphere.&lt;/p&gt;&#xA;" OwnerUserId="26460" LastActivityDate="2019-04-07T01:37:25.173" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30261" PostTypeId="2" ParentId="30257" CreationDate="2019-04-07T06:14:32.067" Score="5" Body="&lt;p&gt;Coronal mass ejections consist of a very hot, but thin, plasma. Their &lt;em&gt;very weak&lt;/em&gt; intrinsic emission would be dominated by ultraviolet and X-ray lines and bremsstrahlung continuum. There is very little optical radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, CMEs can and are monitored at optical wavelengths using the light that they &lt;em&gt;scatter&lt;/em&gt; from the Sun. The process is Thomson scattering from the free electrons in the plasma. This is done by spacecraft because the Earth's atmosphere makes it very difficult to achieve the contrast necessary to observe the scattered light close to the Sun (except during total solar eclipses).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A CME takes a couple of days to travel from the Sun to the Earth and such monitoring can give &quot;space weather forecasts&quot; associated with magnetospheric phenomena triggered by the impact of CMEs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example picture taken by the LASCO white light imager on board the SOHO spacecraft is shown below. Everything here is viewed in scattered light. The Sun is obscured by an occulting disk.I have picked an example showing the eruption of a CME.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QWJH0.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QWJH0.gif&quot; alt=&quot;LASCO image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-04-07T06:20:29.177" LastActivityDate="2019-04-07T06:20:29.177" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30262" PostTypeId="2" ParentId="1145" CreationDate="2019-04-07T15:36:31.890" Score="1" Body="&lt;p&gt;It's all about whether the gas (or whatever other dynamical object) can cool effectively while collapsing.  If it can, you get a disk oriented perpendicular to the average angular momentum, while if it cannot, you get a spherical object.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disks: spiral galaxies, black hole accretion disks, protostellar disks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;spheres: stars (dense enough to be opaque to cooling radiation), star clusters &amp;amp; elliptical galaxies (stars mostly don't collide, so cannot cool), gas in galaxy clusters (so low density that the atoms don't collide often enough to cool effectively)&lt;/p&gt;&#xA;" OwnerUserId="26497" LastActivityDate="2019-04-07T15:36:31.890" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30266" PostTypeId="1" CreationDate="2019-04-07T20:08:41.613" Score="5" ViewCount="137" Body="&lt;p&gt;I was wondering if any of the asteroids we have ever seen has a hot core like Earth's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking that Earth has a strong gravitational field and so formation would have involved a lot of gravitational energy being converted to heat, but not so much for the asteroids.&lt;/p&gt;&#xA;" OwnerUserId="26501" LastEditorUserId="7982" LastEditDate="2019-04-08T00:00:52.003" LastActivityDate="2019-04-09T18:48:24.810" Title="Does any asteroid we have ever seen have a hot core like earth.?" Tags="&lt;solar-system&gt;&lt;asteroids&gt;&lt;fundamental-astronomy&gt;&lt;planetary-formation&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30267" PostTypeId="2" ParentId="30256" CreationDate="2019-04-07T23:48:13.937" Score="4" Body="&lt;p&gt;It's a great question!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Delay-Doppler imaging is a technique to produce a kind of image that represents the topography of a rotating body, usually when it is so far away that there's no way to resolve the object with traditional imaging, radar or optical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A powerful radar signal is sent from Earth to an asteroid or a planet, and the reflected signal returning to Earth is recorded and analyzed. The &quot;special sauce&quot; that allows the technique to work is the pseudo-random sequence encoded into the modulation of the outgoing signal. Using correlation techniques, one can extract reflected components from a range of different delays due to different distances to each part of the planet, and different Doppler shifts due to different projected (radial) velocities of each part of the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The raw data is pretty messy, but after you record the body rotating a substantial fraction of a full turn, using modeling, you can recover a reconstructed image of the planet's topography.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The hard part&lt;/em&gt; is that while Doppler shift gives you &lt;em&gt;east-west&lt;/em&gt; discrimination since one side moves towards you while the other moves away from you (assuming the rotation axis is &quot;up-down&quot;), there's no easy way to tell the difference between the two hemispheres in the reflected signal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the &lt;em&gt;north-south&lt;/em&gt; ambiguity means that there is no temporal difference between a signal reflected from upper hemisphere and the lower. Say at 50 degrees North and 50 degrees South. The signals would have BOTH the same delay (since they are the same distance away) and the same doppler shift (since they are both rotating at the same speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that for a spherical-like object of radius R who's axis is perpendicular to us, the projected line-of-sight (radial) velocity will be: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$v_r = \frac{2\pi R}{T} \cos(\phi) \sin(\lambda)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is the period, &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt; is the latitude and &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; is the longitude. It's symmetrical about the equator (cosine behavior) but it varies monotonically east-west (sine).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hslw7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hslw7.png&quot; alt=&quot;delay-doppler imaging of rotating body&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fshk4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fshk4.png&quot; alt=&quot;delay-doppler imaging of rotating body&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There are solutions to this if the object is large enough though. If you have a huge dish and the object has an angular size on the same scale as the beam width of your dish, you can record data with the beam illuminating first one hemisphere more strongly, then the other. That provides some additional information, and with subtraction and sufficient modeling one can overcome the North-South ambiguity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if the body's rotation axis is oriented obliquely towards or away from us, then the symmetry is broken and the ambiguity can also be resolved in some cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related and possibly helpful: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://space.stackexchange.com/q/22493/12102&quot;&gt;Why do radar maps of the surface of Venus have missing slices?&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Emily Lakdawalla's &lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/2010/2462.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;How radio telescopes get &quot;images&quot; of asteroids&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Left:&lt;/strong&gt; A very slow GIF &lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/2010/2462.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Right:&lt;/strong&gt; A radar image shows asteroid 3122 Florence and tiny echoes from its two moons. &lt;a href=&quot;https://cneos.jpl.nasa.gov/images/news/florence.p5us.1Hz.s382.sep01.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is an animation&lt;/a&gt; that shows them more clearly. The direction of the radar illumination (and thus the direction toward Earth) is at the top.&quot; From &lt;a href=&quot;http://www.skyandtelescope.com/astronomy-news/asteroid-florence-has-two-moons/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. NASA / Jet Propulsion Laboratory. This is a small subset of the frames contained in the original 36 MB GIF, and the size has been decreased by a factor of 2 in order to fit in SE's 2 MB limit. From the question &lt;a href=&quot;https://astronomy.stackexchange.com/q/22480/7982&quot;&gt;What is the physical geometry of this apparent &quot;eclipse&quot; of a tiny moon of Asteroid Florence?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; These two images do not necessarily have the same orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sMeV5.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sMeV5.gif&quot; alt=&quot;Delay-Doppler Imaging Emily Lakdawalla&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/Pn291.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pn291.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-04-08T07:21:04.433" LastActivityDate="2019-04-08T07:21:04.433" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="30268" PostTypeId="2" ParentId="30266" CreationDate="2019-04-08T00:11:06.023" Score="-2" Body="&lt;p&gt;'ll explain, the asteroids are small rocky objects that orbit the sun. Although asteroids orbit the sun like planets, they are much smaller than planets. Remember that most of them live in the asteroid belt that is that region between the orbits of Mars and Jupiter. Additionally some asteroids are in the orbital trajectory of the planets. This means that the asteroid and the planet follow the same path around the sun. You should also consider that these asteroids were usually all formed at the same time and that most of them are located at the furthest distances from the Kuipier belt that is closer to the 100UA. The fact is that due to their size and their physical characteristics, these objects can not have active nuclei.&lt;/p&gt;&#xA;" OwnerUserId="13867" LastActivityDate="2019-04-08T00:11:06.023" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30270" PostTypeId="2" ParentId="30266" CreationDate="2019-04-08T06:53:25.897" Score="5" Body="&lt;p&gt;See &lt;a href=&quot;https://astronomy.stackexchange.com/questions/30246/could-a-planet-form-without-a-molten-state&quot;&gt;Could a planet form without a molten state?&lt;/a&gt;, where Anders Sandberg calculates that a body with radius ~ 1000 km would melt from the heat of formation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest asteroid, &lt;a href=&quot;https://en.wikipedia.org/wiki/Ceres_(dwarf_planet)&quot; rel=&quot;noreferrer&quot;&gt;Ceres&lt;/a&gt;, has a radius of 473 km, so it's a bit too small to have totally melted just from gravitational accretion, but it &lt;em&gt;may&lt;/em&gt; have picked up a little more energy from impacts that were faster than freefall. It could also generate some internal heat from radioactive decay, but it's not expected to contain a significant percentage of radioactive material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably, Ceres was plastic to some degree during formation, since it's (almost) spherical (and it's the only asteroid large enough for gravity to make it spherical). But that wouldn't have required a very high temperature, since Ceres contains a lot of water ice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even if it did totally melt when it formed, it's had over 4 billion years to cool down, so it's extremely unlikely to still have a hot core. However, it does emit more water vapour than expected, and it may be warm enough to contain some liquid water below the surface.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2019-04-08T06:53:25.897" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30271" PostTypeId="1" CreationDate="2019-04-08T07:13:19.110" Score="1" ViewCount="197" Body="&lt;p&gt;I am approaching this topic from a layman's perspective and view this as a logic puzzle, but when someone tries to get me to accept the logic, I'm never convinced by their arguments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Expansion_of_the_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;wiki page on the expanding universe&lt;/a&gt;, the universe is expanding (which is how we see redshifted light from distant galaxies), however this expansion does not result in the movement of objects across space away from each other. Instead, this expansion is the formation of new space such that all of the objects in space maintain a static relationship with space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should mean, that as light produced by a distant light source that is expanding away from us, it is not moving away from the photon. The photon then travels across space, and this trip takes longer as a result of the expansion, but once the photon is in space near to Earth, the earth is not moving away from the light. By the time the light is at the same distance from the telescope as a flashlight held in front of the telescope, the telescope is not moving away from the photon any more than it is moving away from the photon produced by the flashlight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, IF the expansion is not the result of motion across space, HOW do we show redshift?????? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another comment, laymans perspective&lt;br&gt;&#xA;The whole reason it can make sense that objects far enough away can expand away from us FASTER than the speed of light (which is the reason why telescopes can't see further) is that these objects are not moving across space to achieve this expansion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another comment in reference to the Raisin Cake. &#xA;understanding the expanding universe as a raisin cake swelling, therefore the actual cake that is in contact with any raisin stays in contact with that raisin,, the raisins do not move through the cake,,, &lt;em&gt;APART&lt;/em&gt;. BUT, in space, the raisins actually ARE moving through the cake. Also, the PHOTON that travels across space from the distant light source to the telescope, travels across space. additionally, it's NOT useful to think of this transaction of energy in a raisin cake, because the raisins DO travel across space away from each-other as the cake swells, but this is NOT how we can possibly understand objects in space expanding away from us on Earth faster than the speed of light. &lt;/p&gt;&#xA;" OwnerUserId="26506" LastEditorUserId="26506" LastEditDate="2019-04-09T03:40:21.453" LastActivityDate="2019-04-09T03:40:21.453" Title="If objects don't move when the universe expands, how can the expansion result in redshift?" Tags="&lt;universe&gt;&lt;redshift&gt;&lt;dark-energy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30273" PostTypeId="1" CreationDate="2019-04-08T09:47:29.667" Score="1" ViewCount="70" Body="&lt;p&gt;I've just been answering a question on this issue,in physics SE,and realised that my answer conceals a point I don't myself understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The expansion of metric isn't affected by the SR speed of light/signal propagation, and that this resulted in objects placed at separations (as measured within spacetime) where they can't be detected or signals received from them in the 13.7 bn years since the Big Bang.&lt;/li&gt;&#xA;&lt;li&gt;The rough scale of expansion discussed is ~ 10^26 linearly (roughly 1nm -&gt; 10 ly) and most of that took place in a tiny period of time, at about 10^-32 or so seconds, continuing much slower since then, and speeding up again in the last while.&lt;/li&gt;&#xA;&lt;li&gt;The expansion is theorised to be triggered by energies/effects of a phase change or decomposition of a unified field, meaning it didn't happen at all until it did. One implication seems to be that any &lt;s&gt;expansion&lt;/s&gt; increased distance between objects in space which might have arisen before expansion, as modelled by current theories, would have been a result of some physical movement affecting the distribution of energy/transient particles/whatever existed, within spacetime, and was not due to any earlier episode of metric expansion. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So far so good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't get is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we model the dominant modes of expansion of the universe and separation of objects within it, like this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;physical changes to distribution within spacetime (objects move apart within space) for first 10^-32 sec&lt;/li&gt;&#xA;&lt;li&gt;metric expansion instantaneously of the order 10^26, at 10^-32 sec or so.&lt;/li&gt;&#xA;&lt;li&gt;broadly, expansion in space since then - &lt;em&gt;I know its a mix of that and continuing but slower metric expansion, but not sure if that's affecting the line of logic&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Starting at a point at time zero, we find that two objects would be at most ~ 2 x 10^-32 light seconds apart at the start of expansion, or 2 x (3x10^8) x (10^-32) m apart = 6 x 10^-24 m. Therefore after expansion two objects could be &lt;em&gt;at most&lt;/em&gt; (6 x 10^-24) x (10^26) apart = 600m apart. That's still a pretty trivial distance for light or signals to propagate, though&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(I agree, if objects had initially been 1km or 1 ly apart immediately prior to the expansion epoch, they would now be billions of light years apart after expansion, and unobservable. But by definition, before expansion they just  didn't have time to become that far apart, they could at most become a tiny distance apart, because only ~ 10^-32 sec had elapsed)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So after initial expansion it seems that the universe could have been at most 600m diameter. Similar logic applies to the much slower expansion  since then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a non observable universe effect seems to imply that after initial expansion, the universe was light years across, and objects in it were able to be far enough apart to be unobservable now, meaning &gt; 13.7 ly apart. But that would imply a much larger expansion than 10^26, or a much larger size before expansion than there was time to achieve pre-expansion (which would have been predominantly SR limited and within spacetime, hence order of 10^-24m as above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the resolution of this paradox / point I'm missing?&lt;/p&gt;&#xA;" OwnerUserId="18601" LastEditorUserId="18601" LastEditDate="2019-04-08T21:24:13.953" LastActivityDate="2019-04-08T21:24:13.953" Title="Expansion of space - how do we end up with an unobservable region?" Tags="&lt;big-bang-theory&gt;&lt;expansion&gt;&lt;observable-universe&gt;&lt;size&gt;" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="30275" PostTypeId="2" ParentId="30271" CreationDate="2019-04-08T12:02:46.517" Score="4" Body="&lt;p&gt;The redshift is indeed not caused by the movement of the objects, but by the expansion itself. This is a theoretical result from considering the &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann%E2%80%93Lema%C3%AEtre%E2%80%93Robertson%E2%80%93Walker_metric&quot; rel=&quot;nofollow noreferrer&quot;&gt;FLRW metric&lt;/a&gt; for a light ray traveling on a null geodesic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A derivation can be found &lt;a href=&quot;https://ned.ipac.caltech.edu/level5/Watson/Watson2_2_3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cosmological redshift &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; is given by &lt;span class=&quot;math-container&quot;&gt;$z+1 = 1/a$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is the scale factor (the size of the Universe) at the time the light was emitted. Notice that this result doesn't care &lt;em&gt;how&lt;/em&gt; the Universe expanded. In principle, you could imagine a universe which was static when the light was emitted, and static when the light is observed, but expanded by the factor &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; at some point in between, and you'd observe the same redshift as if that universe had expanded gradually in the meantime. This is in sharp contrast to a normal Doppler shift, where in such a hypothetical case, the would be no observed redshift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that galaxies receding faster than the speed of light &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; is no hindrance for us to see them; all galaxies with a observed redshift larger than &lt;span class=&quot;math-container&quot;&gt;$z\simeq1.4$&lt;/span&gt; recede faster than &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;. The most distant observed galaxy GN-z11 has a redshift of &lt;span class=&quot;math-container&quot;&gt;$z=11.09$&lt;/span&gt;, and recedes at &lt;span class=&quot;math-container&quot;&gt;$v\simeq2.2c$&lt;/span&gt;. When it emitted the light we see, it receded even faster, namely at &lt;span class=&quot;math-container&quot;&gt;$v \simeq 4.3 c$&lt;/span&gt;. See the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ant_on_a_rubber_rope&quot; rel=&quot;nofollow noreferrer&quot;&gt;ant on a rubber rope &quot;paradox&quot;&lt;/a&gt; for an intuitive explanation.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2019-04-08T12:02:46.517" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30276" PostTypeId="2" ParentId="30271" CreationDate="2019-04-08T13:28:02.350" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If objects don't move when the universe expands, how can the expansion result in redshift?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because the objects &lt;em&gt;are&lt;/em&gt; moving in relation to one another. Neither is moving through space, but space is expanding. So two galaxies that used to be 1 billion light years apart are now 2 billion light years apart. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;According to the wiki page on the expanding universe, the universe is expanding (which is how we see redshifted light from distant galaxies), however this expansion does not result in the movement of objects across space away from each other. Instead, this expansion is the formation of new space such that all of the objects in space maintain a static relationship with space. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, they maintain a static relationship with their local space, but not with each other. See the &lt;a href=&quot;https://www.google.co.uk/search?q=raisin+cake+analogy&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ved=0ahUKEwikw_m6yMDhAhUEQxUIHbhYDGMQ_AUIDigB&amp;amp;biw=1280&amp;amp;bih=894&quot; rel=&quot;nofollow noreferrer&quot;&gt;raisin cake analogy&lt;/a&gt;. The galaxies are moving apart because they're embedded in space which is expanding. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This should mean, that as light produced by a distant light source that is expanding away from us, it is not moving away from the photon. The photon then travels across space, and this trip takes longer as a result of the expansion, but once the photon is in space near to Earth, the earth is not moving away from the light. By the time the light is at the same distance from the telescope as a flashlight held in front of the telescope, the telescope is not moving away from the photon any more than it is moving away from the photon produced by the flashlight. So, IF the expansion is not the result of motion across space, HOW do we show redshift?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because space expands, so two galaxies are moving apart, so they have a relative speed, so the photon looks redshifted. It hasn't actually been redshifted. It hasn't lost any energy. When you move away from a source of light it looks redshifted, but the light didn't change, instead you changed. Cosmologist Tamara Davis says the same thing in the 2010 Scientific American article &lt;a href=&quot;https://kaiserscience.files.wordpress.com/2015/08/is-the-universe-leaking-energy-law-conservation.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Is the Universe leaking energy?&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/To4tX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/To4tX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;em&gt;Image by Tamara Davis and Scientific American, see &lt;a href=&quot;https://kaiserscience.files.wordpress.com/2015/08/is-the-universe-leaking-energy-law-conservation.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Is the Universe leaking energy?&lt;/a&gt;&lt;/em&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The whole reason it can make sense that objects far enough away can expand away from us FASTER than the speed of light (which is the reason why telescopes can't see further) is that these objects are not moving across space to achieve this expansion. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can see galaxies that have always been receding from us faster than light. See &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0310808&quot; rel=&quot;nofollow noreferrer&quot;&gt;Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the Universe&lt;/a&gt; by Tamara Davis and Charles Lineweaver. Also see &lt;a href=&quot;https://en.wikipedia.org/wiki/Ant_on_a_rubber_rope&quot; rel=&quot;nofollow noreferrer&quot;&gt;the ant on the rubber rope&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9772" LastActivityDate="2019-04-08T13:28:02.350" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30277" PostTypeId="2" ParentId="29912" CreationDate="2019-04-08T13:52:37.960" Score="1" Body="&lt;p&gt;You cannot estimate the distance to anything or from anything in 3-dimensional space using a 2-dimensional coordinate alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may calculate an angular distance (as per Mike G's answer), then assume they are all at a common distance and translate that angular distance into a physical distance, but since they aren't all at the same distance, that won't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do what you want requires an estimate of the distance to these sources.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-04-08T13:52:37.960" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30278" PostTypeId="2" ParentId="30252" CreationDate="2019-04-08T14:35:28.230" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;...is the delay to correlate the signal from the closest dishes to the further ones longer for the red shifted object?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Good question!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short answer is no. The path difference between the closest and farthest dish in the array comes from local geometry, a big triangle in the air right above the VLA dishes. Here the speed of light is essentially constant (except for local temperature and humidity gradients in the air above the array, or maybe wind sheer) and so wouldn't vary between the to signals very much for reasons related to the sources' Doppler shifts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interference path length difference is divided by the local speed of light (in the air) to get the time difference needed for best correlation of those two dishes, and it doesn't make any difference what the source of the signal is or the speed that it was moving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doppler shifted photons don't look or act any different than non-shifted photons, there's no way to tell the difference in an experiment, unless you have some information about the original frequency before the shift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only teeny tiny difference might be if the air has a different index of refraction at the shifted frequency than the unshifted frequency. That's certainly taken into account during normal analysis, and I don't think it rises to the level and intent of your question.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-04-08T14:44:58.410" LastActivityDate="2019-04-08T14:44:58.410" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30279" PostTypeId="1" AcceptedAnswerId="30280" CreationDate="2019-04-08T14:48:07.770" Score="3" ViewCount="58" Body="&lt;p&gt;The classic paper &lt;a href=&quot;http://adsabs.harvard.edu/full/1965ApJ...142..419P&quot; rel=&quot;nofollow noreferrer&quot;&gt;A MEASUREMENT OF EXCESS ANTENNA TEMPERATURE AT 4080 Mc/s&lt;/a&gt; begins:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Measurements of the effective zenith noise temperature of the 20-foot horn-reflector antenna (Crawford, Hogg, and Hunt 1961) at the Crawford Hill Laboratory, Holmdel, New Jersey, at 4080 Mc/s have yielded a value about 3.5° K higher than expected. This excess temperature is, within the limits of our observations, isotropic, unpolarized, and free from seasonal variations (July, 1964-April, 1965). A possible explanation for the observed excess noise temperature is the one given by Dicke, Peebles, Roll, and Wilkinson (1965) in a companion letter in this issue.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;The total antenna temperature measured at the zenith is 6.7° K of which 2.3° K is due to atmospheric absorption.&lt;/strong&gt; The calculated contribution due to ohmic losses in the antenna and back-lobe response is 0.9° K.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So, why would it increase? And, what is the significance of zenith in this sentence?&lt;/p&gt;&#xA;" OwnerUserId="24850" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-04-09T06:16:18.810" Title="Why would atmospheric absorption increase rather than decrease the antenna temperature in Penzias and Wilson's famous CMB observation?" Tags="&lt;radio-astronomy&gt;&lt;cmb&gt;&lt;radio-telescope&gt;&lt;atmospheric-effects&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30280" PostTypeId="2" ParentId="30279" CreationDate="2019-04-08T15:35:13.210" Score="2" Body="&lt;p&gt;Short answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Great question! Ask yourself what absorbs better, something that is white, or black.Then ask yourself what radiates better. The answer is the same thing. It turns out that absorptivity and emissivity are linked pretty closely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kirchov's Law of Thermal Radiation&lt;/a&gt; can be stated as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to the absorptivity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you know that the atmospheric absorption at a given wavelength is 0.01, then you &lt;em&gt;could estimate&lt;/em&gt; the emissivity to be 0.01 as well. It doesn't mean it's exactly right, but it's a great start. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you then say the average temperature of the atmosphere in the region where absorption (and thermal radiation) at 4080 MHz is 230K, then the &lt;a href=&quot;https://en.wikipedia.org/wiki/Noise-equivalent_power&quot; rel=&quot;nofollow noreferrer&quot;&gt;Noise Equivalent Power&lt;/a&gt; (NEP) will be 0.01 time 230 K or 2.3 K.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;How can we express power in terms of a temperature?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Wait a minute, thermal radiated power varies as something like temperature to the &lt;em&gt;fourth power!&lt;/em&gt; How can we add two temperatures together, or multiply them by 0.01?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's because the energy of the photons we're talking about is way down on the left edge of Planck's blackbody distribution. &lt;strong&gt;See for example &lt;a href=&quot;https://space.stackexchange.com/a/25744/12102&quot;&gt;this answer&lt;/a&gt;&lt;/strong&gt; to the question &lt;em&gt;Why doesn't thermal radio emission from a DSN “hot dish” completely swamp the benefits of a cold LNA?&lt;/em&gt; Down at photon energies well below &lt;span class=&quot;math-container&quot;&gt;$k_B T$&lt;/span&gt; we can use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Rayleigh%E2%80%93Jeans_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rayleigh-Jeans law&lt;/a&gt; where the power at a given frequency depends linearly on temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plotted on a log-log graph, you can see that the peak heights increase three times faster than the straight-line low-frequency segments do, showing the &lt;span class=&quot;math-container&quot;&gt;$T^3$&lt;/span&gt; dependence of the spectral density peak height versus the linear behavior in the Rayleigh-Jeans regime. If you integrated the total area under the curve, you'd get the &lt;span class=&quot;math-container&quot;&gt;$T^4$&lt;/span&gt; dependence of the total power radiated by a blackbody.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; From New Jersey Institute of Technology's Dr. Dale Gary's nicely-written Physics 728 Radio Astronomy; &lt;a href=&quot;https://web.njit.edu/~gary/728/Lecture1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lecture #1&lt;/a&gt; notes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aZ7MW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aZ7MW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-04-09T06:16:18.810" LastActivityDate="2019-04-09T06:16:18.810" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30281" PostTypeId="1" CreationDate="2019-04-08T22:28:36.263" Score="1" ViewCount="26" Body="&lt;p&gt;In a paper titled &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Arp, H. (1984). Distribution of quasars on the sky. Journal of Astrophysics and Astronomy,&#xA;  5(1):31-41.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;at &lt;a href=&quot;https://link.springer.com/article/10.1007/BF02714970&quot; rel=&quot;nofollow noreferrer&quot;&gt;Journal of Astrophysics and Astronomy&lt;/a&gt;, Halton Arp wrote that there is a notable clustering of quasars at &lt;span class=&quot;math-container&quot;&gt;$21^h&amp;lt;RA&amp;lt;3^h$&lt;/span&gt;, and a marked lack of quasars at &lt;span class=&quot;math-container&quot;&gt;$9^h&amp;lt;RA&amp;lt;15^h$&lt;/span&gt;, but I haven't been able to find other references to the same distribution. The paper was published in 1984, so has anyone else published a similar result, or has this observation been&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;refuted, or&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;explained?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2019-04-08T22:28:36.263" Title="Anisotropy of quasar distribution" Tags="&lt;quasars&gt;&lt;resource-request&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30282" PostTypeId="1" CreationDate="2019-04-08T22:58:32.540" Score="2" ViewCount="76" Body="&lt;p&gt;&lt;a href=&quot;http://www.astro.utu.fi/%7Ecflynn/galdyn/intro_mw.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This page&lt;/a&gt; lists the total luminosity of the Milky Way (MW) Galaxy to be &lt;span class=&quot;math-container&quot;&gt;$4\times10^{10}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$L_{\rm sol}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$L_{\rm sol}=3.86 \cdot 10^{33} {\rm ergs.sec}^{-1}$&lt;/span&gt;; this gives a total luminosity for the MW of &lt;span class=&quot;math-container&quot;&gt;$1.54\cdot 10^{44}{\rm ergs.sec}^{-1}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;But &lt;a href=&quot;https://en.wikipedia.org/wiki/Seyfert_galaxy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Seyfert galaxies&lt;/a&gt; can have luminosities of &lt;span class=&quot;math-container&quot;&gt;$\sim 10^{40} - 10^{42}{\rm ergs.sec}^{-1}$&lt;/span&gt;, and this is due to the high luminosity of the galactic core of a Seyfert.&lt;/p&gt;&#xA;&lt;p&gt;Since the MW doesn't have an active SMBH, does this mean that the MW is particularly bright due to a high luminosity stellar population? Or am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="17535" LastEditorUserId="37191" LastEditDate="2021-04-15T10:53:08.040" LastActivityDate="2021-04-15T10:53:08.040" Title="Luminosity of the Milky Way compared to Seyfert galaxies" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;luminosity&gt;&lt;seyfert-galaxy&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30283" PostTypeId="1" CreationDate="2019-04-09T02:30:20.387" Score="1" ViewCount="33" Body="&lt;p&gt;While I was reading about &lt;a href=&quot;https://en.wikipedia.org/wiki/(486958)_2014_MU69&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ultima Thule&lt;/a&gt;, I was wondering if the discovery of contact binary objects had any effect on our current theories of how the Moon formed?&lt;/p&gt;&#xA;" OwnerUserId="883" LastEditorUserId="14094" LastEditDate="2019-04-09T04:13:48.493" LastActivityDate="2019-04-09T04:13:48.493" Title="Do contact binaries such as Ultima Thule affect our current theories on how the Moon formed?" Tags="&lt;the-moon&gt;&lt;contact-binaries&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30284" PostTypeId="1" AcceptedAnswerId="30299" CreationDate="2019-04-09T06:36:51.200" Score="4" ViewCount="204" Body="&lt;p&gt;In &lt;a href=&quot;https://astronomy.stackexchange.com/a/30280/7982&quot;&gt;this answer&lt;/a&gt; as well as &lt;a href=&quot;https://space.stackexchange.com/a/25744/12102&quot;&gt;this one&lt;/a&gt;, I've linked to &lt;a href=&quot;https://web.njit.edu/~gary/728/Lecture1.html&quot; rel=&quot;noreferrer&quot;&gt;New Jersey Institute of Technology's Dr. Dale Gary's nicely-written Physics 728 Radio Astronomy; Lecture #1 notes&lt;/a&gt; where I can see the image below. It links to Gizmodo.au's &lt;a href=&quot;https://www.gizmodo.com.au/2018/07/new-south-african-telescope-releases-epic-image-of-the-galactic-centre/&quot; rel=&quot;noreferrer&quot;&gt;New South African Telescope Releases Epic Image Of The Galactic Centre&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have heard of the &lt;a href=&quot;https://en.wikipedia.org/wiki/MeerKAT&quot; rel=&quot;noreferrer&quot;&gt;Meerkat&lt;/a&gt; array earlier (&lt;a href=&quot;https://astronomy.stackexchange.com/q/16621/7982&quot;&gt;Which 16 antenna locations are used for the MeerKAT radio image recently announced?&lt;/a&gt;), but I am wondering how to read further about this spectacular image. What is the scale for example? How would it look pasted into the arc of the Milky Way; just a tiny patch of a few arc-minutes, or several tens of degrees?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How will this compare to the radio image of Sag A* that we will see in the next day or two?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if anyone knows, what are those spiky arcs sticking up, and those puff-ball things on the left?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://www.gizmodo.com.au/2018/07/new-south-african-telescope-releases-epic-image-of-the-galactic-centre/&quot; rel=&quot;noreferrer&quot;&gt;Gizmodo&lt;/a&gt; &quot;The black hole at the centre of the Milky Way and filaments.&quot; Credit: SARAO, click for full size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HZYSK.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HZYSK.jpg&quot; alt=&quot;Meerkat Epic Image Of The Galactic Centre&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-04-09T21:24:16.723" Title="MeerKAT's Epic Image of the Galactic Centre, what are we seeing?" Tags="&lt;radio-astronomy&gt;&lt;identify-this-object&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30285" PostTypeId="1" CreationDate="2019-04-09T07:58:13.387" Score="9" ViewCount="208" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Gaia_(spacecraft)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gaia&lt;/a&gt; spacecraft is currently mapping the stars in the Milky Way by &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_parallax&quot; rel=&quot;nofollow noreferrer&quot;&gt;parallax&lt;/a&gt;. It orbits in the Earth-Sun L2 Lagrange point. The parallax observed from the spacecraft orbiting around the sun allows us to derive the distances to many stars in the milky way, as well as calculate their velocities.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Assuming no cost to placing such an observatory wherever we want around the sun, what would be the optimal orbit for optimal rate of science gains?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;I apologise if this seems a big vague, let me elaborate.&lt;/p&gt;&#xA;&lt;p&gt;If you imagine placing it in a smaller orbit, the parallax of distant stars (already very small) will be made even smaller. But information about the parallax will be returned quicker.&lt;/p&gt;&#xA;&lt;p&gt;If you place the observatory in the orbit of Uranus, you'll have a much greater parallax and will be able to measure the distances of distant stars with much greater precision. However, it will take 80 years to complete an orbit, and the rate of distance measuring would be intolerably slow.&lt;/p&gt;&#xA;&lt;p&gt;Intuitively, I feel like there is some obvious sweet spot that would produce measurements at an acceptable rate, though I also feel like my objective function here is ill-defined.&lt;/p&gt;&#xA;&lt;p&gt;Would a highly eccentric orbit be preferred to a circular one? Would a particular orbit optimise the 'science gains' for one particular aspect at the expense of others (such as accuracy of nearby stars traded for accuracy of distant stars, or accuracy of stellar proper motion)? If there are multiple objectives to optimise, what would  the tradeoffs be?&lt;/p&gt;&#xA;" OwnerUserId="17655" LastEditorUserId="2531" LastEditDate="2020-08-19T16:50:35.497" LastActivityDate="2020-08-19T16:50:35.497" Title="Is there an optimum orbit for a Hipparcos/Gaia-like parallax observatory" Tags="&lt;orbit&gt;&lt;parallax&gt;&lt;gaia&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="30287" PostTypeId="1" CreationDate="2019-04-09T09:32:47.353" Score="1" ViewCount="50" Body="&lt;p&gt;I think this picture is correct because for earth astronomical day is slightly shorter than solar day. (Is it a rule of thumb for planets? Where can I read such data?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Fpilq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fpilq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://de.wikipedia.org/wiki/Datei:Sidereal_day_(prograde).svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the angle between earth's angular momentum and earth's axis(from south pole to north pole) is 23.5° and not a (180°-23.5°).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is why observer is standing on the sun in the next picture. And the break in circle-shaped arrows is closer to the sun, then the rest of arrow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right? Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K3wnz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K3wnz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:%D0%97%D0%B5%D0%BC%D0%BB%D1%8F22.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26514" LastEditorUserId="26514" LastEditDate="2019-04-09T12:37:56.450" LastActivityDate="2019-04-09T12:37:56.450" Title="Angle between earth's angular momentum and earth's axis" Tags="&lt;the-sun&gt;&lt;earth&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="30288" PostTypeId="2" ParentId="30284" CreationDate="2019-04-09T11:13:46.750" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://www.syfy.com/syfywire/stunning-chaos-in-the-galactic-core&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article&lt;/a&gt; by the Bad Astronomer does a good job of explaining what is visible in the MeerKAT image.The field of view is one by two degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sa A* is the bright area right of centre, the arc to the left of that may be associated with the Arches cluster. The two bright joined blobs left of that are molecular clouds and further left are some supernova remnants. The filaments scattered over the image are somewhat mysterious but may be aligned with star-forming regions.&lt;/p&gt;&#xA;" OwnerUserId="23640" LastEditorUserId="23640" LastEditDate="2019-04-09T12:12:56.137" LastActivityDate="2019-04-09T12:12:56.137" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30289" PostTypeId="1" AcceptedAnswerId="30293" CreationDate="2019-04-09T12:06:28.173" Score="3" ViewCount="37" Body="&lt;p&gt;Exactly what is the non-thermal Sunyaev-Zel'dovich effect? From what I understand from reading several papers by Mark Birkinshaw and Sergio Colafrancesco, I get the rough idea that the non-thermal SZ effect has something to do with it being in the relativistic realm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Papers by Sergio Colafrancesco:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2016JCAP...07..031S/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Polarization of the Sunyaev-Zel'dovich effect: relativistic imprint of thermal and non-thermal plasma &lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2003A%26A...397...27C/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;The non-thermal Sunyaev-Zel'dovich effect in clusters of galaxies&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2001MmSAI..72..813C/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Non-thermal phenomena in galaxy clusters: cosmological relevance&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Review by Mark Birkinshaw:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/1999PhR...310...97B/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Sunyaev-Zel'dovich effect &lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify, I'm not referring to the kinematic SZ effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There appears to be a distinction by the authors between the SZ effect due to hot thermal electrons and relativistic electrons. I find the nomenclature ambiguous, as temperature is essentially kinetic energy on a smaller scale, so why would relativistic electrons that have more kinetic energy be 'non-thermal'? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, is the key distinction between them using a relativistic vs. a non-relativistic formulation?&lt;/p&gt;&#xA;" OwnerUserId="26516" LastEditorDisplayName="user1569" LastEditDate="2019-04-09T13:36:02.647" LastActivityDate="2019-04-09T16:50:12.880" Title="Non-thermal Sunyaev-Zel'dovich effect" Tags="&lt;astrophysics&gt;&lt;cosmology&gt;&lt;high-energy-astrophysics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30291" PostTypeId="1" AcceptedAnswerId="30297" CreationDate="2019-04-09T14:40:28.413" Score="4" ViewCount="112" Body="&lt;p&gt;The IAU recommends renaming the Hubble &lt;em&gt;law&lt;/em&gt; as the Hubble-Lemaître law, as discussed at its General Assembly in 2018 (see this &lt;a href=&quot;https://www.iau.org/news/pressreleases/detail/iau1812/&quot; rel=&quot;nofollow noreferrer&quot;&gt;press release&lt;/a&gt; and the links therein).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have now stumbled upon an occurrence of &quot;Hubble-Lemaître &lt;em&gt;constant&lt;/em&gt;&quot;. Are there any recommendations (or noteworthy statements, opinion papers) in favor or against this extended interpretation of the IAU resolution B4 to the terminology for &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="9150" LastEditorUserId="9150" LastEditDate="2019-04-09T15:18:49.273" LastActivityDate="2019-04-09T19:21:48.597" Title="Are there recommendations to use &quot;Hubble-Lemaître constant&quot; instead of &quot;Hubble constant&quot;?" Tags="&lt;history&gt;&lt;terminology&gt;&lt;hubble-constant&gt;&lt;iau&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30292" PostTypeId="1" CreationDate="2019-04-09T15:56:04.313" Score="4" ViewCount="36" Body="&lt;p&gt;I apologise if this is the wrong place to ask. I am trying to recreate part of the calculations by the JPL Horizons system using SPICE. Specifically the Observer Ecliptic Latitude and Longitudes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd here's the inputs.  (1 = RA, 3 = DEC, 31 = Observer Ecliptic Lat/Lon)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/G1UV2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G1UV2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the output...&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;*****************************************************************************************&#xA; Date__(UT)__HR:MN     R.A._(ICRF/J2000.0)_DEC dRA*cosD d(DEC)/dt    ObsEcLon    ObsEcLat&#xA;*****************************************************************************************&#xA;$$SOE&#xA; 2019-Apr-05 00:00     23 18 17.44 -05 37 22.3 94.28997  10.27670 348.4729140  -1.0458502&#xA; 2019-Apr-06 00:00     23 20 55.85 -05 32 12.3 102.7090  15.59295 349.1112323  -1.2238769&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I've been going through the documentation for SPICE and trying a few bits and pieces appear to be likely functions, such as SPKPOS_C and SPKCPO_C trying to work out how to recreate the above. I'm using de430.bsp to try and get positions to then work out the lat/lon, but I'm getting lost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if anyone familiar enough with the SPICE toolkit to point me at the functions/methods that will lead me to what I need? Thanks. &lt;/p&gt;&#xA;" OwnerUserId="26515" LastActivityDate="2019-04-09T15:56:04.313" Title="Recreating Observer Ecliptic Lat/Lon from JPL Horizons using SPICE toolkit" Tags="&lt;amateur-observing&gt;&lt;ecliptic&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30293" PostTypeId="2" ParentId="30289" CreationDate="2019-04-09T16:50:12.880" Score="4" Body="&lt;p&gt;The non-thermal S-Z effect is caused by inverse Compton scattering of the CMB photons from a non-thermal population of electrons - i.e. electrons that have high energies not because they are hot, but because they have been accelerated non-thermally. The usual mechanisms are accelerating by electromagnetic fields and the Lorentz force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest-mass energy of an electron is 0.511 keV. In order to attain &quot;relativistic energies&quot; then the electrons must have kinetic energies similar to this or higher. The average kinetic energy of a particle due to its temperature is just &lt;span class=&quot;math-container&quot;&gt;$3k_{B}T/2$&lt;/span&gt;. If we equate this to 0.511 keV then the temperature required to have thermal relativistic electrons is &lt;span class=&quot;math-container&quot;&gt;$T &amp;gt; 4 \times 10^{6}$&lt;/span&gt; K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gas at this temperature &lt;em&gt;does&lt;/em&gt; exist in the intracluster medium of massive galaxy clusters. But in addition there are populations of non-thermal electrons that are accelerated to relativistic speeds for instance in the jets and radio-emitting lobes of active galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distinction is important when studying the S-Z effect because the energy distributions of thermal and non-thermal populations can be quite different. Non-thermal populations are usually characterised by a power-law, whilst thermal electrons have a Maxwell-Boltzmann distribution. This results in a different inverse Comptonisation spectrum from electrons of the two populations that are often (usually) spatially unresolved. The non-thermal S-Z effect is essentially a contaminant that needs to be accounted for when using the thermal S-Z effect to investigate the structure and parameters of the intracluster gas and using the S-Z effect as a cosmological probe.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-04-09T16:50:12.880" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30294" PostTypeId="2" ParentId="1973" CreationDate="2019-04-09T18:27:27.300" Score="1" Body="&lt;p&gt;In Cosmology, the main contribution to redshift would be cosmological redshift. Namely the redshift that occurs due to the Hubble flow. According to Hubble's law,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{equation}&#xA;v = H_0 d&#xA;\end{equation}&lt;/span&gt;&#xA;The further two objects are away from each other, the faster they are moving. Of course, all this is under the assumption of an expanding universe, such as the widely accepted &lt;span class=&quot;math-container&quot;&gt;$\Lambda CDM$&lt;/span&gt; model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hubble flow is in a sense a type of &quot;speed&quot;, but this velocity can be faster than the speed of light. This is chiefly because the space between two points is expanding, causing the wavelength of the light to be stretched out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also gravitational redshift, something that occurs due to gravity's effect on spacetime. Although it might be better to think that &lt;em&gt;anything&lt;/em&gt; that can stretch the wavelength of light can cause redshift, suppose there exist some kind of &lt;em&gt;god hand&lt;/em&gt; that allows one to just pull and subsequently stretch the &lt;em&gt;light wave&lt;/em&gt; this would also cause redshift.&lt;/p&gt;&#xA;" OwnerUserId="26516" LastActivityDate="2019-04-09T18:27:27.300" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30295" PostTypeId="1" CreationDate="2019-04-09T18:46:12.937" Score="0" ViewCount="249" Body="&lt;p&gt;In the review paper on protoclusters, &lt;a href=&quot;https://link.springer.com/article/10.1007/s00159-016-0100-3&quot; rel=&quot;nofollow noreferrer&quot;&gt;Roderik Overzier (pg 25)&lt;/a&gt; mentions the concept of &lt;em&gt;virial shock heating&lt;/em&gt; which seems to be a process involved with galaxy cluster formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not fully certain what a virialized galaxy cluster is apart from the fact that it's in hydrostatic equilibrium, and is one of the key distinguishing features between a protocluster and a fully formed galaxy cluster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I vaguely understand that shock heating relates to the heating that occurs when something is travelling at supersonic speed within a medium and produces a shock.&lt;/p&gt;&#xA;" OwnerUserId="26516" LastActivityDate="2019-04-09T22:51:00.527" Title="Virial Shock Heating" Tags="&lt;cosmology&gt;&lt;galaxy-cluster&gt;&lt;hydrostatic-equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30296" PostTypeId="2" ParentId="30266" CreationDate="2019-04-09T18:48:24.810" Score="1" Body="&lt;p&gt;As far as is known no asteroid currently has a hot core. Things were probably a bit different in the past. The asteroid &lt;a href=&quot;https://en.wikipedia.org/wiki/4_Vesta&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vesta&lt;/a&gt; is differentiated: it has a silicate mantle surrounding an iron core. This implies that during its formation Vesta did get hot enough that the materials separated. &lt;a href=&quot;https://en.wikipedia.org/wiki/Ceres_(dwarf_planet)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ceres&lt;/a&gt; is also differentiated and may have a small metallic core, again implying an early hot interior state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also iron asteroids, the largest of which is &lt;a href=&quot;https://en.wikipedia.org/wiki/16_Psyche&quot; rel=&quot;nofollow noreferrer&quot;&gt;Psyche&lt;/a&gt;. These are probably remnants of differentiated objects (either fully or partially-differentiated) that got broken up: the iron asteroids are fragments of the metallic core.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastActivityDate="2019-04-09T18:48:24.810" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30297" PostTypeId="2" ParentId="30291" CreationDate="2019-04-09T19:21:48.597" Score="4" Body="&lt;p&gt;No, only the Hubble law was recommended to have its name changed (I'm a member of the IAU, so hopefully I'd have known if there were more votings).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, several astronomers (including myself) found the voting a bit… weird; while acknowledging the work of George Lemaître is admirable, many more people than him and Edwin Hubble contributed to the understanding that the Universe is expanding. So, &lt;a href=&quot;https://twitter.com/anisotropela/status/1057200200069140480&quot; rel=&quot;nofollow noreferrer&quot;&gt;some&lt;/a&gt; were joking that we should rename the Law to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Vesto_Slipher&quot; rel=&quot;nofollow noreferrer&quot;&gt;Slipher&lt;/a&gt;-&lt;a href=&quot;https://en.wikipedia.org/wiki/Alexander_Friedmann&quot; rel=&quot;nofollow noreferrer&quot;&gt;Friedmann&lt;/a&gt;-&lt;a href=&quot;https://en.wikipedia.org/wiki/Carl_Wilhelm_Wirtz&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wirtz&lt;/a&gt;-&lt;a href=&quot;https://sv.wikipedia.org/wiki/Knut_Lundmark&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lundmark&lt;/a&gt;-&lt;a href=&quot;https://en.wikipedia.org/wiki/Georges_Lema%C3%AEtre&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lemaître&lt;/a&gt;-&lt;a href=&quot;https://en.wikipedia.org/wiki/Howard_P._Robertson&quot; rel=&quot;nofollow noreferrer&quot;&gt;Robertson&lt;/a&gt;-&lt;a href=&quot;https://en.wikipedia.org/wiki/Edwin_Hubble&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble&lt;/a&gt;-&lt;a href=&quot;https://en.wikipedia.org/wiki/Willem_de_Sitter&quot; rel=&quot;nofollow noreferrer&quot;&gt;de Sitter&lt;/a&gt;-&lt;a href=&quot;https://en.wikipedia.org/wiki/Milton_L._Humason&quot; rel=&quot;nofollow noreferrer&quot;&gt;Humason&lt;/a&gt; law, or something like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Others joked that if the Hubble law should be renamed, we should also start referring to &quot;the Hubble-Lemaître Space Telescope&quot;, &quot;the Hubble-Lemaître Ultra Deep Field&quot;, &quot;the Hubble-Lemaître Bubble&quot;, and, as you have seen, &quot;the Hubble-Lemaître constant&quot;. Others probably also did this non-jokingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Btw., astronomer &lt;a href=&quot;https://en.wikipedia.org/wiki/Virginia_Louise_Trimble&quot; rel=&quot;nofollow noreferrer&quot;&gt;Virginia Trimble&lt;/a&gt; wrote a paper six years ago (&lt;a href=&quot;https://arxiv.org/abs/1307.2289&quot; rel=&quot;nofollow noreferrer&quot;&gt;Trimble 2013&lt;/a&gt;), arguing that we should name the Law after anybody but Hubble…&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2019-04-09T19:21:48.597" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30299" PostTypeId="2" ParentId="30284" CreationDate="2019-04-09T21:11:19.020" Score="4" Body="&lt;p&gt;&lt;a href=&quot;https://doi.org/10.1086/301168&quot; rel=&quot;nofollow noreferrer&quot;&gt;LaRosa et al 2000&lt;/a&gt; present a 90 cm VLA image and catalog of sources near the galactic center.&#xA;Here is their schematic diagram with the &lt;a href=&quot;https://www.skyandtelescope.com/astronomy-news/60-second-astronomy-news-must-see-images/&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&amp;deg;x1&amp;deg;, 18-33 cm band&lt;/a&gt; MeerKAT image region highlighted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G2Hz9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G2Hz9.png&quot; alt=&quot;Schematic of galactic center in 90 cm&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compact source Sgr A* is a small dot within the bright, pear-shaped blob &lt;a href=&quot;https://en.wikipedia.org/wiki/Sagittarius_A&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sgr A&lt;/a&gt; right of center.&#xA;Recent &lt;a href=&quot;https://eventhorizontelescope.org/news/lifting-veil-black-hole-heart-our-galaxy-results-gmvaalma-observations-sgr&quot; rel=&quot;nofollow noreferrer&quot;&gt;millimeter VLBI images&lt;/a&gt; (Issaoun et al 2019) indicate a width on the order of 200 &amp;mu;as, or 1/7000 of a pixel in the image you uploaded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bright pair of blobs left of center, Sgr B1 and &lt;a href=&quot;https://en.wikipedia.org/wiki/Sagittarius_B2&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sgr B2&lt;/a&gt;, are massive &lt;a href=&quot;https://en.wikipedia.org/wiki/Molecular_cloud&quot; rel=&quot;nofollow noreferrer&quot;&gt;molecular clouds&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the bubble shapes are either &lt;a href=&quot;https://en.wikipedia.org/wiki/Supernova_remnant&quot; rel=&quot;nofollow noreferrer&quot;&gt;supernova remnants&lt;/a&gt; (SNRs) or &lt;a href=&quot;https://en.wikipedia.org/wiki/H_II_region&quot; rel=&quot;nofollow noreferrer&quot;&gt;H II regions&lt;/a&gt; ionized by UV-bright young stars.&#xA;The far left pair Sgr D has one of each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The various non-thermal filaments (NTFs) align with the local magnetic field and emit &lt;a href=&quot;https://en.wikipedia.org/wiki/Synchrotron_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;synchrotron radiation&lt;/a&gt; but are not well understood otherwise.&#xA;The bundle of filaments near the center of the image are known as the galactic center radio arc (GCRA) or just the Arc.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-04-09T21:24:16.723" LastActivityDate="2019-04-09T21:24:16.723" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30302" PostTypeId="2" ParentId="30295" CreationDate="2019-04-09T22:51:00.527" Score="2" Body="&lt;p&gt;Virial shocks appear when gas falls in toward a galaxy, and the shocked gas stays hot so the pressure in the post-shock gas can support the shock and keep it away from the galactic disk. &#xA; The alternative is if the post-shock gas cools and loses pressure support, and the (non-virial) shock collapses right down to the galactic disk.  The significance is that if gas falls directly onto the high-density disk, it has different effects than if it meets a low-density shock well away from the galaxy.  There's a lot of information at &lt;a href=&quot;https://academic.oup.com/mnras/article/345/1/349/984798&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://academic.oup.com/mnras/article/345/1/349/984798&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2019-04-09T22:51:00.527" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30303" PostTypeId="1" CreationDate="2019-04-10T03:54:04.003" Score="1" ViewCount="68" Body="&lt;p&gt;I have seen it said on &lt;a href=&quot;https://www.quora.com/How-many-times-has-the-Earth-gone-around-the-Milky-Way-Galaxy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Quora&lt;/a&gt; and &lt;a href=&quot;https://www.reddit.com/r/askscience/comments/2xt7rd/how_many_rotations_has_our_solar_system_made/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reddit&lt;/a&gt; that it's about every 220-250 million years to go around the Milky Way from Earth. Wondering what the rough equation is for calculating it. That is, I guess, the angular momentum and distance from the center, so I could visualize it on a sped up animation and otherwise derive something like those numbers 220 to 250.&lt;/p&gt;&#xA;" OwnerUserId="7620" LastActivityDate="2019-04-10T04:28:11.537" Title="The equation for calculating how many times Earth (or our solar system) has revolved around the Milky Way" Tags="&lt;earth&gt;&lt;milky-way&gt;&lt;rotation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30304" PostTypeId="1" CreationDate="2019-04-10T04:11:14.670" Score="5" ViewCount="257" Body="&lt;p&gt;The Event Horizon Telescope is in the news now because they are about to release the first ever picture of a black hole. They have been working on this since 2006. Why did it take 13 years to produce the first picture? What was the critical mass that was recently hit that made the picture now complete?&lt;/p&gt;&#xA;" OwnerUserId="1549" LastActivityDate="2019-04-12T05:34:14.533" Title="Why did the Event Horizon Telescope take so long to take a photo of a black hole?" Tags="&lt;black-hole&gt;&lt;telescope&gt;&lt;radio-telescope&gt;&lt;event-horizon&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30305" PostTypeId="2" ParentId="30303" CreationDate="2019-04-10T04:28:11.537" Score="3" Body="&lt;p&gt;I'm not 100% sure what equation you would like, but I'll try at an answer to see how it goes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;observed speed&lt;/em&gt; of the Sun orbiting around the center of the Milky way is about 220 km/sec. The distance to the center is about 8.1 kiloparsec, which is 2.5E+17 kilometers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time it takes to travel around a circle is the circumference divided by the speed, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{2 \pi R}{v} \approx \frac{6.28 \times 2.5\times 10^{17}}{220} \approx 7.17 \times 10^{15} \text{seconds} \approx 227 \text{ million years}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed is a measured quantity. There isn't a simple way to calculate the speed with an equation, because the mass distribution in the galaxy is complicated and dominated by dark matter, whose presence we can only infer from the orbital motion of our own star as well as a large number of other stars in our galaxy.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/R1MKr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R1MKr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Rotation_curve_(Milky_Way).JPG&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Milky_Way&quot; rel=&quot;nofollow noreferrer&quot;&gt;Milky Way&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Galaxy rotation curve for the Milky Way. Vertical axis is speed of rotation about the Galactic Center. Horizontal axis is distance from the Galactic Center in kpcs. The Sun is marked with a yellow ball. The observed curve of speed of rotation is blue. The predicted curve based upon stellar mass and gas in the Milky Way is red. Scatter in observations roughly indicated by gray bars. The difference is due to dark matter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-04-10T04:28:11.537" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30306" PostTypeId="1" CreationDate="2019-04-10T06:10:13.457" Score="0" ViewCount="184" Body="&lt;p&gt;Does anyone know if Tycho Brahe was religious?&lt;/p&gt;&#xA;" OwnerUserId="26485" LastEditorUserId="7472" LastEditDate="2019-04-10T06:39:08.393" LastActivityDate="2019-04-12T05:23:41.113" Title="Tycho Brahe Danish Astronomer Religious?" Tags="&lt;history&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30308" PostTypeId="1" AcceptedAnswerId="30333" CreationDate="2019-04-10T06:55:34.087" Score="0" ViewCount="65" Body="&lt;p&gt;I have an access to a remote telescope at 05.15-05.45 am at Hawaii time. The sun rises at 06.15 am. I will use this telescope for educational purposes for high school students. I wonder 30 mins is enough to make a variable star observation or a spectroscopic measurement.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specs of the telescope could be found on &lt;a href=&quot;https://lco.global/observatory/2m/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://lco.global/observatory/2m/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26510" LastActivityDate="2019-04-10T19:11:07.120" Title="What to observe with a remote telescope close to sunrise time?" Tags="&lt;observational-astronomy&gt;&lt;telescope&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30309" PostTypeId="2" ParentId="30306" CreationDate="2019-04-10T07:08:09.427" Score="4" Body="&lt;p&gt;Like nearly everybody at the time, he &quot;believed in God&quot; and was a member of the Danish Lutheran church.  He from time to time used arguments based on bible verses to back up his solar system model (in which the sun orbits the Earth, and the planets orbit the Sun).  According to Tycho, the idea of a rotating and revolving Earth would be &quot;in violation not only of all physical truth but also of the authority of Holy Scripture, which ought to be paramount.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However he was not notably religious. He was suspected of heresy by some in the Lutheran church. He was not personally on good terms with the local bishops in Denmark. His interest in medicine and astrology were not sanctioned by the church and while he was a follower of Philip Melanchthon, he was not viciously sectarian, at a time when many others were.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He believed in God, but was not especially religious.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="7411" LastEditDate="2019-04-12T05:23:41.113" LastActivityDate="2019-04-12T05:23:41.113" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30310" PostTypeId="1" CreationDate="2019-04-10T07:10:54.027" Score="2" ViewCount="1538" Body="&lt;p&gt;What is the key difference between LST and ZT?&lt;br&gt;&#xA;Also, are UT, GMT and Greenwich Mean Solar Time the same?&lt;/p&gt;&#xA;" OwnerUserId="25925" LastEditorUserId="7472" LastEditDate="2019-04-10T11:25:14.507" LastActivityDate="2019-05-10T22:15:15.920" Title="Difference between zone time and local sidereal time?" Tags="&lt;time&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30311" PostTypeId="1" AcceptedAnswerId="30318" CreationDate="2019-04-10T08:13:22.297" Score="5" ViewCount="276" Body="&lt;p&gt;I went to &lt;a href=&quot;https://eventhorizontelescope.org/array&quot; rel=&quot;noreferrer&quot;&gt;https://eventhorizontelescope.org/array&lt;/a&gt; and read about the ten sites listed as part of the EHT, I have a mashed-up screen shot of them below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made a little script also shown below with approximate coordinates, and saw that Sgr A* is at a declination of about -29.5 degrees (South).  I spun it around the earth, calculated the dot product between each of the sites' position vectors and all normals pointing to Sgr A* as it rotates around the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then plotted the elevation angle above the horizon for all ten sites. I was quite surprised!&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Sgr A* is never above the horizon for the Greenland telescope!&lt;/li&gt;&#xA;&lt;li&gt;Sgr A* is never simultaneously visible even to all of the remaining nine telescopes&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Have I made a mistake? If not, how do these ten sites combine their data to image Sgr A*?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rbtYg.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rbtYg.png&quot; alt=&quot;Event Horizon Telescope visibility of Sgr A*&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GwjJR.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GwjJR.jpg&quot; alt=&quot;Event Horizon Telescope&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;class Site(object):&#xA;    def __init__(self, name, lat, lon, alt):&#xA;        self.name = str(name)&#xA;        self.lat  = rads * float(lat)&#xA;        self.lon  = rads * float(lon)&#xA;        self.alt  = km   * float(alt)&#xA;        (clat, slat), (clon, slon) = [[f(x) for f in (np.cos, np.sin)] for x in (self.lat, self.lon)]&#xA;        self.norm = np.array([clon*clat, slon*clat, slat])&#xA;&#xA;data = (('Northern Extended Millimeter Array', (44.634, 5.908, 2550.)),&#xA;        ('IRAM 30 meter telescope', (37.066, -3.393, 2850.)),&#xA;        ('The Greenland Telescope now near Thule Air Base', (76.531, -68.703, 10.)),&#xA;        ('Combined Array for Research in Millimeter-wave Astronomy CARMA', (37.280, -118.142, 2196.)),&#xA;        ('Kitt Peak National Observatory 12 meter Submillimeter Telescope (SMT)', (1.9583, -111.5967,  2096.)),&#xA;        ('Mt. Graham International Observatory 12 meter ALMA prototye', (32.701, -109.892, 3191.)),&#xA;        ('The Large Millimeter Telescope Alfonso Serrano', (18.985, -97.315, 4600.)),&#xA;        ('ALMA', (-22.971, -67.703, 4800.)),&#xA;        ('Caltech Submillimeter Observatory', (19.823, -155.476, 4140.)),&#xA;        ('South Pole Telescope', (-90.0, 0.0, 2800.)))&#xA;&#xA;# https://eventhorizontelescope.org/array&#xA;# https://astronomy.stackexchange.com/questions/26413/math-behind-a-uv-plot-in-interferometry&#xA;&#xA;datadict = dict(data)&#xA;&#xA;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;from skyfield.api import Topos, Loader, EarthSatellite&#xA;from mpl_toolkits.mplot3d import Axes3D&#xA;&#xA;halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]&#xA;degs, rads = 180/pi, pi/180&#xA;km         = 0.001&#xA;&#xA;sites = [Site(a, *b) for a, b in datadict.items()]&#xA;&#xA;Dec = rads * -29.5  # SgrA_star&#xA;cDec, sDec = [f(Dec) for f in (np.cos, np.sin)]&#xA;th         = twopi * np.linspace(0, 1, 100001)[:-1]&#xA;cth, sth   = [f(th) for f in (np.cos, np.sin)]&#xA;zth, oth   = np.zeros_like(th), np.ones_like(th)&#xA;&#xA;SgrA_star = np.stack([cDec*cth, cDec*sth, sDec*oth], axis=1)&#xA;&#xA;for site in sites:&#xA;    site.elev = np.arcsin(np.dot(SgrA_star, site.norm))&#xA;&#xA;if True:&#xA;    for site in sites:&#xA;        plt.plot(degs*site.elev)&#xA;    plt.plot(zth, '-k', linewidth=2)&#xA;    plt.ylim(-90, 90)&#xA;    plt.ylabel('elevation (deg)', fontsize=16)&#xA;    plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-04-10T15:25:54.757" Title="How can the Event Horizon Telescope image Sgr A* when it's not visible from all sites at one time?" Tags="&lt;radio-astronomy&gt;&lt;radio-telescope&gt;&lt;event-horizon-telescope&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30312" PostTypeId="2" ParentId="30311" CreationDate="2019-04-10T08:36:41.843" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Sgr A* is never above the horizon for the Greenland telescope!&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, Sgr A* is not the only focus of EHT; &lt;a href=&quot;https://eventhorizontelescope.org/about&quot; rel=&quot;nofollow noreferrer&quot;&gt;the website&lt;/a&gt; mentions it also studies other objects like &lt;a href=&quot;https://en.wikipedia.org/wiki/Messier_87&quot; rel=&quot;nofollow noreferrer&quot;&gt;the M87 galaxy in Virgo&lt;/a&gt; which, at +12&amp;deg; declination, &lt;em&gt;is&lt;/em&gt; visible from Greenland.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;Sgr A* is never simultaneously visible even to all of the remaining nine telescopes&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Very-long-baseline_interferometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Very Long Baseline Interferometry&lt;/a&gt; just combines the results of the telescopes which &lt;em&gt;can&lt;/em&gt; (and have been) pointed at a specific target at a specific time (plus or minus a few nanoseconds, because some telescopes might be further away from the target than others, and light takes slightly longer or shorter to reach them). So it will never use the full potential of all the telescopes in the array simultaneously.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At the location of the correlator the data are played back. The timing of the playback is adjusted according to the atomic clock signals on the (tapes/disk drives/fibre optic signal), and the estimated times of arrival of the radio signal at each of the telescopes. A range of playback timings over a range of nanoseconds are usually tested until the correct timing is found. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2019-04-10T08:36:41.843" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="30313" PostTypeId="1" AcceptedAnswerId="30315" CreationDate="2019-04-10T14:00:16.663" Score="26" ViewCount="7786" Body="&lt;p&gt;At the press conference this morning, the Event Horizon Telescope team didn't say much about Sagittarius A*, which was the target many of us have been waiting for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any explanation anywhere for this omission?&lt;/p&gt;&#xA;" OwnerUserId="22083" LastEditorUserId="22083" LastEditDate="2019-04-15T20:04:06.210" LastActivityDate="2019-04-15T20:04:06.210" Title="Why didn't the Event Horizon Telescope team mention Sagittarius A*?" Tags="&lt;black-hole&gt;&lt;supermassive-black-hole&gt;&lt;sagittarius-a&gt;&lt;event-horizon-telescope&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="30314" PostTypeId="2" ParentId="30313" CreationDate="2019-04-10T14:03:42.967" Score="21" Body="&lt;p&gt;I've found an explanation in Dutch &lt;a href=&quot;https://nos.nl/artikel/2279846-eerste-foto-van-zwart-gat-vergelijkbaar-met-columbus-of-maanlanding.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; by &lt;a href=&quot;https://en.wikipedia.org/wiki/Heino_Falcke&quot; rel=&quot;noreferrer&quot;&gt;Heino Falcke&lt;/a&gt;, one of the EHT founders. Translation:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;h3&gt;Hard to photograph&lt;/h3&gt;&#xA;&lt;p&gt;It was easiest to take a picture of M87. &amp;quot;It is very difficult to photograph the black hole in our Milky Way, because the material around it moves very fast: the vortex rotates around its axis in 20 minutes. Compare it to a toddler who has to sit still for hours to be photographed: that's not possible. With M87, the matter revolves around the hole in two days, so it's easier to photograph&amp;quot;, says Falcke.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;sub&gt;(The original text is as follows:)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;h3&gt;Lastig te fotograferen&lt;/h3&gt;&#xA;&lt;p&gt;Het lukte het beste om een foto te maken van M87. &amp;quot;Het is heel lastig om het zwarte gat in onze Melkweg op de foto te zetten, doordat de materie daaromheen heel snel beweegt: de draaikolk draait in 20 minuten om zijn as. Vergelijk het met een kleuter die urenlang stil moet zitten om op de foto te gaan: dat gaat niet. Bij M87 draait de materie in twee dagen om het gat heen, dus dat is makkelijker te fotograferen&amp;quot;, zegt Falcke.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7472" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-04-10T14:03:42.967" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30315" PostTypeId="2" ParentId="30313" CreationDate="2019-04-10T14:19:51.240" Score="41" Body="&lt;p&gt;There was a mention of Sagittarius A* during the Q+A portion of the press conference; the team indicated that they hope to produce an image sometime in the future (although they were careful to make no promises, and they're not assuming they'll be successful).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, I'm not wholly surprised that we ended up seeing M87, rather than Sgr A*, for a couple reasons which the team mentions in &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/ab0ec7#apjlab0ec7s8&quot; rel=&quot;noreferrer&quot;&gt;their first paper&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;As Glorfindel said, Sgr A*'s event horizon is much smaller, meaning matter orbiting the black hole has a shorter orbital period. This contributes to variability on the timescale of minutes. The observations of M87 took place over the course of a week - roughly the timescale over which that target varies, meaning the source should not change significantly over that time.&lt;/li&gt;&#xA;&lt;li&gt;Second - and this is the reason I've seen cited more often - Sgr A* lies in the center of our galaxy, and so thick clouds of gas and dust lie between it and us. That results in scattering, which is a problem. There are ways to mitigate this, of course, and the team has spent a long time on this, but it's simpler to just look at the black hole that doesn't have that problem in the first place. That's why M87's black hole is an attractive target.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Neither of these are impossible hurdles to overcome, but they're certainly very real difficulties that can't be ignored.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2019-04-10T20:03:26.047" LastActivityDate="2019-04-10T20:03:26.047" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30316" PostTypeId="1" AcceptedAnswerId="40652" CreationDate="2019-04-10T14:26:52.140" Score="9" ViewCount="152" Body="&lt;p&gt;Today, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Event_Horizon_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;EHT&lt;/a&gt; released (somewhat blurry as compared to simulations) &lt;a href=&quot;https://eventhorizontelescope.org/blog/wobbling-shadow-m87-black-hole&quot; rel=&quot;nofollow noreferrer&quot;&gt;images of the M87 black hole&lt;/a&gt;.&#xA;Did the image contain anything which might lead to a correction to General Relativity or were all the observations according to what had been predicted?&lt;/p&gt;&#xA;" OwnerUserId="22114" LastEditorUserId="37191" LastEditDate="2020-12-30T19:40:40.643" LastActivityDate="2020-12-30T19:40:40.643" Title="Is there any difference between M87's image and predictions?" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30317" PostTypeId="1" AcceptedAnswerId="30360" CreationDate="2019-04-10T14:35:24.393" Score="12" ViewCount="1620" Body="&lt;p&gt;So, as the title states, why are we able to actually see the 'blackness' of the black hole? I get that what we are actually &lt;em&gt;seeing&lt;/em&gt; is the event horizon, or accretion disk. But should this not extend all the way round? Surely the black hole is not a 2D thing so we can &quot;look into it from above&quot; (I use that term loosely as obviously there are no directions in space!), so why are we able to actually see the blackness?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing I can think of is that a black hole, much like our solar system has some sort of ecliptic, that the vast majority of matter is orbiting, and everywhere else just doesn't have enough matter for the light to be visible, kind of like why we are unable to see the Oort cloud. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that made sense, and I could be way off, but it is the only thing I could think of to explain it. If it is the case, then would we be able to get a similar image of Sagittarius A, seeing as we could be in this &quot;ecliptic&quot; of it, so surely we would only be able to see the heated matter around the event horizon and not the blackness?&lt;/p&gt;&#xA;" OwnerUserId="17096" LastActivityDate="2019-04-12T21:48:12.170" Title="M87 Black hole. Why can we see the blackness?" Tags="&lt;black-hole&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="30318" PostTypeId="2" ParentId="30311" CreationDate="2019-04-10T14:35:30.790" Score="2" Body="&lt;p&gt;I don't think it is necessary for all the telescopes to view the target simultaneously if you can make the assumption that the source you are observing is only varying on timescales longer than it takes the source to be viewable by all the telescopes. Sure, the maximum baseline you can get &lt;em&gt;will&lt;/em&gt; be defined by the most distant pair of simultaneously observing telescopes, but that doesn't stop you adding together images taken at other times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is very probably the reason why M87 is in the fact the first image presented by the EHT. The timescale for significant variability around this (big) black hole will be of order a few time &lt;span class=&quot;math-container&quot;&gt;$2GM/c^2$&lt;/span&gt;, which is a few days for the M87 black hole. Imaging Sgr A* is going to be more difficult (or at least more blurry) because the variability timescale is minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like (from the &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/ab0c57&quot; rel=&quot;nofollow noreferrer&quot;&gt;data reduction paper&lt;/a&gt;) that the first steps involve treating every baseline in a pair-wise way (as I suggested above) and then combining them across the network, with whatever baselines were observing at the time, to give data at the surprisingly (to me) high time-resolution of 10s.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-04-10T15:25:54.757" LastActivityDate="2019-04-10T15:25:54.757" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30319" PostTypeId="2" ParentId="30311" CreationDate="2019-04-10T14:42:07.417" Score="2" Body="&lt;p&gt;The recently announced M87 observations only used 8 of the 10 sites. This &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/ab0ec7&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper&lt;/a&gt; includes this diagram.Even the South Pole instrument seems to have been used only for calibratiom (the dashed lines).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xeAFk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xeAFk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-04-10T14:42:07.417" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30320" PostTypeId="2" ParentId="30317" CreationDate="2019-04-10T14:42:42.983" Score="-2" Body="&lt;p&gt;This black hole has an accretion disk, which is a disk of matter that rotates around the black hole at extreme speeds, causing it to heat up. The orange colour you see in the picture is that matter. The matter seems &quot;thicker&quot; on one side because the bottom of the disk is tilted slightly towards us. The &quot;blackness&quot; you see is simply the event-horizon stopping the light from that region from escaping.&lt;/p&gt;&#xA;" OwnerUserId="15468" LastActivityDate="2019-04-10T14:42:42.983" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="30321" PostTypeId="2" ParentId="30317" CreationDate="2019-04-10T14:45:11.657" Score="2" Body="&lt;p&gt;The paths light takes near a black hole are not anything like the ones it takes in empty space. Basically we're seeing the &quot;shadow&quot; of the black hole. Much of the light that we would expect to be coming towards us from that particular direction has been diverted elsewhere by the hole's gravity.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-04-10T14:45:11.657" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30322" PostTypeId="2" ParentId="30317" CreationDate="2019-04-10T14:48:22.387" Score="8" Body="&lt;p&gt;You have to think about how the light is going to get to you from where it is produced close to the black hole event horizon. Light produced between you and the black hole &lt;em&gt;can&lt;/em&gt; get to you. Light produced immediately behind the black hole cannot get to you (or at least it does not come to you from that direction). Light produced at other positions can get to you via various routes, one of which is to orbit the black hole and &lt;em&gt;then&lt;/em&gt; head in your direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result of this there is a concentration of the observed light into an apparent ring around the black hole and a dark(er) circle inside it which marks the region from which light cannot travel directly to you, but instead either falls into the black hole or loops around it. Asymmetries in the photon &quot;ring&quot; are going to be caused by the relativistic orbital motion of material that has the effect of boosting emission in the forward direction and also by the &quot;frame dragging&quot; caused by the blackhole rotation (which is why the shadow is &quot;off-centre&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rather academic description of the phenomenon is given by &lt;a href=&quot;https://iopscience.iop.org/article/10.1086/312423/meta&quot; rel=&quot;noreferrer&quot;&gt;Falcke et al. (2000)&lt;/a&gt; and &lt;a href=&quot;https://academic.oup.com/mnras/article/379/3/833/1036973&quot; rel=&quot;noreferrer&quot;&gt;Huang et al. (2007)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can observe the effects of &quot;shadowing&quot; for both Kerr and Schwarzschild blackholes at this &lt;a href=&quot;https://www.esa.int/gsp/ACT/phy/Projects/Blackholes/WebGL.html&quot; rel=&quot;noreferrer&quot;&gt;website&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-04-10T15:04:20.037" LastActivityDate="2019-04-10T15:04:20.037" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="30323" PostTypeId="2" ParentId="28929" CreationDate="2019-04-10T15:02:05.610" Score="4" Body="&lt;p&gt;&lt;a href=&quot;https://eventhorizontelescope.org/latest&quot; rel=&quot;nofollow noreferrer&quot;&gt;Today, April 10th 2019&lt;/a&gt;, there was a press conference where finally an image of M87 was released:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://static.projects.iq.harvard.edu/files/styles/os_files_xlarge/public/eht/files/20190410-78m-800x466.png?m=1554877319&amp;amp;itok=m0gPFSet&quot; alt=&quot;&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Scientists have obtained the first image of a black hole, using Event Horizon Telescope observations of the center of the galaxy M87. The image shows a bright ring formed as light bends in the intense gravity around a black hole that is 6.5 billion times more massive than the Sun. This long-sought image provides the strongest evidence to date for the existence of supermassive black holes and opens a new window onto the study of black holes, their event horizons, and gravity. Credit: Event Horizon Telescope Collaboration&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7472" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-04-10T15:02:05.610" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30325" PostTypeId="1" CreationDate="2019-04-10T16:33:32.950" Score="1" ViewCount="415" Body="&lt;p&gt;As far as I know, the plus-or-minus sign is used to indicate the &lt;a href=&quot;https://en.wikipedia.org/wiki/Plus-minus_sign#In_statistics&quot; rel=&quot;nofollow noreferrer&quot;&gt;statistical margin of error&lt;/a&gt; of a measurement. But what does it mean when there are two plus-or-minus signs in the same measurement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some examples from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The mass of &lt;a href=&quot;https://en.wikipedia.org/wiki/Messier_87#Supermassive_black_hole&quot; rel=&quot;nofollow noreferrer&quot;&gt;M87*&lt;/a&gt; is given as &lt;code&gt;(6.5 ± 0.2&lt;sub&gt;stat&lt;/sub&gt; ± 0.7&lt;sub&gt;sys&lt;/sub&gt;) × 10&lt;sup&gt;9&lt;/sup&gt; M☉&lt;/code&gt; (as measured by the Event Horizon Telescope).&lt;/li&gt;&#xA;&lt;li&gt;The measured distance to &lt;a href=&quot;https://en.wikipedia.org/wiki/Sagittarius_A*&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sagittarius A*&lt;/a&gt; is given as &lt;code&gt;7860 ± 140 ± 40 pc&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13784" LastActivityDate="2019-04-10T18:04:29.887" Title="What does it mean when there are two plus-or-minus signs for a measurement?" Tags="&lt;distances&gt;&lt;mass&gt;&lt;units&gt;&lt;event-horizon-telescope&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30326" PostTypeId="2" ParentId="30310" CreationDate="2019-04-10T16:42:15.600" Score="4" Body="&lt;p&gt;Universal Time (UT) and Greenwich Mean Time (GMT) are based on the average (mean) position of the sun. The mean position of the Sun crosses the meridian every 24 hours. There are subtle differences between UT and GMT, but they are the same for most practical purposes. (The Wikipedia article for &lt;a href=&quot;https://en.wikipedia.org/wiki/Greenwich_Mean_Time&quot; rel=&quot;nofollow noreferrer&quot;&gt;Greenwich Mean Time&lt;/a&gt; points out that the start of the day under GMT changed from noon to midnight sometime in the past. The Time and Date.com article &lt;a href=&quot;https://www.timeanddate.com/time/gmt-utc-time.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;GMT versus UTC&lt;/a&gt; describes GMT as a time zone and UTC as a time standard.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since our clocks are also based on the mean position of the Sun, the time in your time zone is a fixed number of hours before or after UT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Local sidereal time (LST) is the time according to the stars at your longitude. Let's assume that star X is on the meridian at 0 hours (midnight). The same star will return to the meridian after 23 hours 56 minutes 4 seconds (approximately) of clock time. In other words, the star time runs almost 4 minutes fast each day. Therefore, there are two key differences between sidereal time and the clock time:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Sidereal time is different at each longitude. Clock time is the same for all longitudes within a time zone.&lt;/li&gt;&#xA;&lt;li&gt;To convert from clock time to sidereal time, you need to know the time &lt;strong&gt;and&lt;/strong&gt; date.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2019-04-10T16:42:15.600" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30328" PostTypeId="2" ParentId="30325" CreationDate="2019-04-10T18:04:29.887" Score="3" Body="&lt;p&gt;The first error is due to the precision of the measurement, the second is an uncertainty associated with methodology used.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-04-10T18:04:29.887" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30330" PostTypeId="2" ParentId="30308" CreationDate="2019-04-10T18:39:48.697" Score="0" Body="&lt;p&gt;I have been an amateur astronomer for over 60 years. In all that time I have not had more that a passing interest in variable star observation or spectroscopic measurements. What makes you think a bunch of high school kids will?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I recall the best teachers I had were the ones that could make a subject interesting. Kid want candy. Give it to them. Point that monster at the most visually interesting objects. If you want to start a fire you use use a bit of kindling not a pail of cold water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put yourself in their position. A once in a lifetime opportunity and you want to turn it into a lecture on page 347 of the Science textbook. &lt;/p&gt;&#xA;" OwnerUserId="24928" LastActivityDate="2019-04-10T18:39:48.697" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30332" PostTypeId="1" AcceptedAnswerId="30338" CreationDate="2019-04-10T18:49:24.047" Score="3" ViewCount="529" Body="&lt;p&gt;On the image of the &lt;a href=&quot;https://eventhorizontelescope.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;M87 black hole released today by the Event Horizon Telescope team&lt;/a&gt;, which direction points towards the jet? Is it on the brighter side of the ring?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To resolve ambiguity, I'm talking about the main observed jet, not the &lt;a href=&quot;https://www.eso.org/public/news/eso9203/&quot; rel=&quot;nofollow noreferrer&quot;&gt;fainter counterjet&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastActivityDate="2019-04-12T12:34:30.457" Title="What is the orientation of the M87 black hole image relative to the jet?" Tags="&lt;supermassive-black-hole&gt;&lt;event-horizon-telescope&gt;&lt;astronomical-jet&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30333" PostTypeId="2" ParentId="30308" CreationDate="2019-04-10T19:03:10.373" Score="0" Body="&lt;p&gt;I would suggest a target which can only be observed before dawn with a large telescope: a faint comet just west of the Sun.&#xA;Two candidates listed on &lt;a href=&quot;https://theskylive.com/comets&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Sky Live&lt;/a&gt; are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;C/2018 R3 (&lt;a href=&quot;https://minorplanetcenter.net/db_search/show_object?utf8=%E2%9C%93&amp;amp;object_id=C%2F2018+R3&quot; rel=&quot;nofollow noreferrer&quot;&gt;MPC&lt;/a&gt;,&#xA;&lt;a href=&quot;https://ssd.jpl.nasa.gov/sbdb.cgi?spk=1003586&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;C/2019 D1 (&lt;a href=&quot;https://minorplanetcenter.net/db_search/show_object?utf8=%E2%9C%93&amp;amp;object_id=C%2F2019+D1&quot; rel=&quot;nofollow noreferrer&quot;&gt;MPC&lt;/a&gt;,&#xA;&lt;a href=&quot;https://ssd.jpl.nasa.gov/sbdb.cgi?spk=1003615&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They are moving targets, currently about 2 arcseconds per minute of time.&#xA;Pick one.&#xA;To predict its coordinates for your observation slot, you'll need to get an ephemeris.&#xA;Then if your images are successful, you can measure its actual position with astrometry software.&#xA;Good data could be useful in refining its orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're lucky, a stack of short exposures aligned on the comet nucleus might also make a pretty picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively you could try to get a light curve of an asteroid in the morning sky to estimate its rotation rate. &lt;a href=&quot;https://en.wikipedia.org/wiki/Northolt_Branch_Observatories&quot; rel=&quot;nofollow noreferrer&quot;&gt;Northolt Branch Observatories&lt;/a&gt; in the UK frequently post such results on social media.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-04-10T19:11:07.120" LastActivityDate="2019-04-10T19:11:07.120" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30334" PostTypeId="1" CreationDate="2019-04-10T19:19:51.990" Score="7" ViewCount="261" Body="&lt;p&gt;Interferometry is among the best ways (if not, the best way!) to have an image of a very distant object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently a picture of the black hole at the center of M87 was released. It is the result of several data collected by the Event Horizon Telescope, a series of arrays all across our world, working as one like a giant Earth sized telescope. The picture is not really cutting edge and high definition however it is still very surprising and in a way detailed enough, considering the fact that M87 is 53.49 million light years away... and this is where I arrive to Pluto which is just around 6 to 7 billion kilometers from us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we used an interferometer, perhaps the same size as the EHT (or just a smaller one, the size of an entire continent), and point all the arrays at Pluto, then we should have a picture with a resolution that is at least higher as the Hubble Space Telescope, but likely not as high as the pictures from the New Horizons spacecraft which directly made a fly by of Pluto... right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so then why don't we use interferometry to take pictures of Pluto from Earth?&lt;/p&gt;&#xA;" OwnerUserId="24220" LastActivityDate="2020-02-23T12:33:23.790" Title="Why not us interferometry to take a picture of Pluto?" Tags="&lt;planet&gt;&lt;telescope&gt;&lt;interferometry&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30335" PostTypeId="2" ParentId="30334" CreationDate="2019-04-10T19:54:41.230" Score="7" Body="&lt;p&gt;Radio interferometry can combine observations over very large baselines.  But optical interferometry cannot.  According to &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_astronomical_interferometers_at_visible_and_infrared_wavelengths&quot; rel=&quot;nofollow noreferrer&quot;&gt;a list&lt;/a&gt; of interferometry instruments on wikipedia, the largest baseline for optical measurements is less than a kilometer.  We can't take optical measurements with continent-sized instruments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then if you drop down to radio where the instruments do have that capability, I think you'll find Pluto is quite dim (it's not a radio source, and there's no strong radio emissions that it can reflect to us). There's no radio signal from Pluto that can be imaged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a page on &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_optical_interferometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;optical interferometry&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Interferometers are seen by most astronomers as very specialized&#xA;  instruments, as they are capable of a very limited range of&#xA;  observations. It is often said that an interferometer achieves the&#xA;  effect of a telescope the size of the distance between the apertures;&#xA;  this is only true in the limited sense of angular resolution. The&#xA;  combined effects of limited aperture area and atmospheric turbulence&#xA;  generally limit interferometers to observations of comparatively&#xA;  bright stars and active galactic nuclei.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2019-04-10T22:42:33.937" LastActivityDate="2019-04-10T22:42:33.937" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30336" PostTypeId="1" CreationDate="2019-04-10T20:59:12.723" Score="2" ViewCount="50" Body="&lt;p&gt;I have been given a distance, an average flux, and a time period. How do I find the light energy emitted of a supernova over that time period?&lt;/p&gt;&#xA;" OwnerUserId="26538" LastEditorUserId="37191" LastEditDate="2021-04-29T21:01:07.920" LastActivityDate="2021-04-29T21:01:07.920" Title="How to calculate the light energy of a supernova?" Tags="&lt;light&gt;&lt;supernova&gt;&lt;flux&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2021-05-01T17:41:37.090" ContentLicense="CC BY-SA 4.0" />
  <row Id="30338" PostTypeId="2" ParentId="30332" CreationDate="2019-04-10T22:01:49.560" Score="5" Body="&lt;p&gt;The jet is projected onto the image roughly in the E-W direction with the (main) jet coming towards us to the right of the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The brightening in the ring is perpendicular to this. It is probably caused by the spin of the black hole (aligned with the jet), which drags emitting material around and Doppler boosts the emission seen to the south.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This picture is from the fifth event horizon telescope paper published today. The jet is E-W (actually, I believe it is at PA=288 degrees, so rotated 18 degrees anticlockwise) and the black hole spin axis (indicated with the black arrows) is projected at the same PA and into the screen (the two left hand plots).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GFDgZ.jpg&quot; alt=&quot;Event Horizon black hole orientation&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-04-12T12:34:30.457" LastActivityDate="2019-04-12T12:34:30.457" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30339" PostTypeId="1" AcceptedAnswerId="30340" CreationDate="2019-04-10T23:44:51.460" Score="68" ViewCount="13315" Body="&lt;p&gt;There are closer galaxies than Messier 87 for sure, even ours! It sparked my curiosity that they went with one 53 million light years away. Is there a reason for this?&lt;/p&gt;&#xA;" OwnerUserId="26540" LastActivityDate="2019-04-13T22:46:34.420" Title="Why not take a picture of a closer black hole?" Tags="&lt;black-hole&gt;&lt;supermassive-black-hole&gt;&lt;event-horizon-telescope&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="30340" PostTypeId="2" ParentId="30339" CreationDate="2019-04-11T00:23:47.707" Score="83" Body="&lt;p&gt;I was surprised too when I first heard they were trying to image M87's black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short answer is because it's really, really big. It is 1500 times bigger (diameter) than our Sagittarius A*, and 2100 times farther away. This makes its apparent size about 70% of that of Sgr A*, which they are also attempting to image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cursory search of wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_most_massive_black_holes&quot; rel=&quot;noreferrer&quot;&gt;List of Largest black holes&lt;/a&gt; shows that there's no other black holes with a combination of size and closeness greater than these two. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of other candidates are not too far off. Andromeda's black hole is 50x the size of ours, and at 100x the distance, it would appear half the size of Sgr A*. The Sombrero galaxy is 380 times farther way than Sgr A*, and has a black hole estimated to be 1 billion solar masses, which is 232 times Sr A*, resulting in an angular diameter about 60% of Sgr A*.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There appear to be many other considerations to which black holes were chosen, as explained in &lt;a href=&quot;https://astronomy.stackexchange.com/questions/30313/why-didnt-the-event-horizon-telescope-team-mention-sagittarius-a&quot;&gt;this similar question.&lt;/a&gt; At a guess these would include how obscured each black hole is with foreground dust/stars etc, how active (and therefore bright) the nuclei are, and their inclination w.r.t earth affecting which observatories could observe them at which times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I've found another plausible candidate. &lt;a href=&quot;https://en.wikipedia.org/wiki/NGC_1600&quot; rel=&quot;noreferrer&quot;&gt;NGC_1600&lt;/a&gt; is 200 M light years away with a central black hole estimated to be 17 billion solar masses heavy. This would put it at about 40% the apparent diameter of Sgr A*. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cdsKB.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cdsKB.png&quot; alt=&quot;black hole size comparison, self made&quot;&gt;&lt;/a&gt;&#xA;&lt;strong&gt;Comparison of the apparent size of the largest nearby black holes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course &lt;a href=&quot;https://xkcd.com/1276/&quot; rel=&quot;noreferrer&quot;&gt;obligatory XKCD&lt;/a&gt; to remind us how small these objects really appear.&lt;/p&gt;&#xA;" OwnerUserId="17655" LastEditorUserId="17655" LastEditDate="2019-04-12T20:22:33.543" LastActivityDate="2019-04-12T20:22:33.543" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="30341" PostTypeId="2" ParentId="30339" CreationDate="2019-04-11T00:55:07.340" Score="34" Body="&lt;p&gt;There are a few criteria necessary to see a black hole with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Event_Horizon_Telescope&quot; rel=&quot;noreferrer&quot;&gt;Event Horizon Telescope&lt;/a&gt;. They are, in importance:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Active Feeding: you need a thick accretion disk with lots of matter accreting onto the black hole. M87 fits this criteria, and is a glut, &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0202238&quot; rel=&quot;noreferrer&quot;&gt;consuming about 90 Earth masses a day&lt;/a&gt;. &lt;/li&gt;&#xA;&lt;li&gt;Apparent size. Even though it is 53 million light-years away, M87 is 6.5 billion solar masses. Since the radius of the event horizon scales linearly with mass, its distance is made up for by sheer scale. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="22897" LastEditorUserId="25153" LastEditDate="2019-04-11T05:27:21.053" LastActivityDate="2019-04-11T05:27:21.053" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30342" PostTypeId="1" AcceptedAnswerId="30547" CreationDate="2019-04-11T01:20:03.077" Score="6" ViewCount="297" Body="&lt;p&gt;I was watching the video &lt;a href=&quot;https://www.youtube.com/watch?v=h3KOmt9onvc&quot; rel=&quot;nofollow noreferrer&quot;&gt;I Put a 1mm Size Black Hole Next to Earth and This Happened - Universe Sandbox 2&lt;/a&gt; discussing what would happen if a 1 millimeter black hole is orbiting the earth, and I began questioning myself, is it really possible to naturally have such a small black hole?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the smallest size that a black hole can form naturally in real life?&lt;/p&gt;&#xA;" OwnerUserId="26543" LastEditorUserId="7982" LastEditDate="2021-03-08T00:06:32.073" LastActivityDate="2021-03-09T02:09:38.597" Title="What is the smallest diameter black hole that can be formed naturally?" Tags="&lt;black-hole&gt;&lt;size&gt;&lt;universe-sandbox&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="30343" PostTypeId="1" AcceptedAnswerId="30348" CreationDate="2019-04-11T01:43:27.130" Score="39" ViewCount="7792" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tgIhr.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tgIhr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the recently released photo of a black hole shown above, which was created by using data from EHT, why is the lower region brighter than the one above? Is it because of the rotation of the accretion disk? Also what is the orientation of the accretion disk? Are we looking at it head on?&lt;/p&gt;&#xA;" OwnerUserId="9213" LastEditorUserId="-1" LastEditDate="2019-04-12T09:36:22.513" LastActivityDate="2019-04-29T06:38:14.643" Title="Why are there uneven bright areas in this photo of black hole?" Tags="&lt;black-hole&gt;&lt;event-horizon-telescope&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30344" PostTypeId="1" AcceptedAnswerId="30415" CreationDate="2019-04-11T03:18:34.010" Score="3" ViewCount="287" Body="&lt;p&gt;The BBC News article &lt;a href=&quot;https://www.bbc.com/news/science-environment-47873592&quot; rel=&quot;nofollow noreferrer&quot;&gt;First ever black hole image released&lt;/a&gt; shows the instantly-iconic radio image of the (almost certainly) supermassive black hole at the center of the M87 galaxy, just released by the Event Horizon Telescope collaboration, also shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though it calls it a &quot;first image&quot;, the article also shows a similarly edible-looking&lt;sup&gt;†&lt;/sup&gt; donut-shaped image taken earlier, presumably using a different technology. However the caption doesn't explain what that is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;†&lt;/sup&gt;&lt;a href=&quot;https://i.stack.imgur.com/J5TEe.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/J5TEe.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone recognize the earlier image and/or know what imaging technology and wavelengths were used to construct it? Is it roughly the same scale? Are these comparable donut-hole images?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1FZsz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1FZsz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Astronomers have suspected that the M87 galaxy has a supermassive black hole at its heart from false colour images such as this one. The dark centre is not a black hole but indicates that stars are densely packed and fast moving. Credit: DR JEAN LORRE/SCIENCE PHOTO LIBRARY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/h40s4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h40s4.jpg&quot; alt=&quot;mystery image of M87 (central black hole?)&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The first ever picture of a black hole: It's surrounded by a halo of bright gas. Credit: EHT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Overly entheusiastic &lt;a href=&quot;https://en.wikipedia.org/wiki/Messier_87&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia edit&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MbyHt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MbyHt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-04-12T23:18:53.390" Title="Is this the best non-radio image of whatever's at the center of M87? How was it taken?" Tags="&lt;black-hole&gt;&lt;photography&gt;&lt;identify-this-object&gt;&lt;event-horizon-telescope&gt;&lt;m87&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30345" PostTypeId="2" ParentId="483" CreationDate="2019-04-11T03:30:56.493" Score="0" Body="&lt;p&gt;There are posts mentioning it not being practical for amateurs because of the size of the antenna (generally) required combined with the need to &quot;point it&quot; in the right direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One answer to this is -- if you can just point a static antenna structure toward the sky (front poles supporting the forward-most element higher than the real-most support poles supporting it, thereby angling the beam toward space) then you simply let Earth be your antenna motor. The excitement comes from waiting for the precise time that Jupiter naturally moves through your window of opportunity. &#xA;This is a wonderfully feasible method of aiming huge wire antennas toward space for pretty low frequency signals of all kinds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as &quot;never knowing if it's just a terrestrial signal&quot;, for one thing, man's signals are not super-ultra-wide band like that. Receiver parameters should be set before being aimed at Jupiter, and results compared before and after. Plus there are all sorts of other tests and identification factors. With enough know-how, which is the free and laborless aspect of it, one can rule out man and extraneous natural interference.&lt;/p&gt;&#xA;" OwnerUserId="26545" LastActivityDate="2019-04-11T03:30:56.493" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30346" PostTypeId="2" ParentId="30343" CreationDate="2019-04-11T03:59:07.977" Score="9" Body="&lt;p&gt;I believe we are seeing one of the effects of the accretion disk rotating at very high speeds. This is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Relativistic_beaming&quot; rel=&quot;noreferrer&quot;&gt;relativistic beaming&lt;/a&gt;, and it occurs because particles (in this case matter in the accretion disk) that are travelling at relativistic speeds (say, upwards of .2c), tend to preferentially emit their radiation in a cone towards the direction of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This suggests that the matter at the bottom of the picture (the brightest blobs) are travelling towards us, and the darker parts are travelling away. Since the black hole tends to warp light around itself, I'm not sure from the photo of the orientation of the accretion disk.&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2019-04-11T03:59:07.977" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30347" PostTypeId="2" ParentId="30343" CreationDate="2019-04-11T04:37:37.417" Score="10" Body="&lt;p&gt;There's some recent information which is worthy of an update to the answer (despite the difficulty of typing MathJax on my phone). I've quoted minimally as I wouldn't have improved upon what these scientists have published. Previous edits remain beneath this addition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the paper &quot;&lt;a href=&quot;https://arxiv.org/abs/1904.07923&quot; rel=&quot;nofollow noreferrer&quot;&gt;Measurement of the spin of the M87 black hole from its observed twisted light&lt;/a&gt;&quot; (Apr 16 2019) by Fabrizio Tamburini, Bo Thidé, and Massimo Della Valle, they explain on page 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... The imaging techniques applied to this data set reveal the presence of an asymmetric ring with clockwise rotation and a “crescent” geometric structure that exhibits a clear central brightness depression. This indicates a source dominated by lensed emission surrounding the black hole shadow.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;From the analysis of the two data sets we obtain the asymmetry parameters &lt;span class=&quot;math-container&quot;&gt;$q_1$&lt;/span&gt; = 1.417 for epoch 1 and &lt;span class=&quot;math-container&quot;&gt;$q_2$&lt;/span&gt; = 1.369 for epoch 2. They give an averaged asymmetry in the spiral spectrum&#xA;  of &lt;span class=&quot;math-container&quot;&gt;$\bar{q}$&lt;/span&gt; = 1.393±0.024 in agreement with that of our numerical simulations, &lt;span class=&quot;math-container&quot;&gt;$q_{num}$&lt;/span&gt; = 1.375, of partially incoherent light emitted by the Einstein ring of a Kerr black hole with &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{a} \textbf{~}\!$&lt;/span&gt; &lt;strong&gt;0.9±0.1&lt;/strong&gt;, corresponding to a rotational energy &lt;span class=&quot;math-container&quot;&gt;$^{[10]}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\textbf{10}{^\textbf{64}}\!$&lt;/span&gt; &lt;strong&gt;erg&lt;/strong&gt;, which is &lt;strong&gt;comparable to the energy radiated by the &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_quasars#Most_powerful_quasars&quot; rel=&quot;nofollow noreferrer&quot;&gt;brightest quasars&lt;/a&gt; (~ 500 trillion &lt;span class=&quot;math-container&quot;&gt;$\odot$&lt;/span&gt;) over a Gyr (billion year) timescale&lt;/strong&gt;, and inclination &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; = 17° between the approaching jet and the line of sight, with the angular momenta of the accretion flow and of the black hole anti-aligned, showing clockwise rotation as described in Ref. 5.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This result is in good agreement with results from the analysis of the fiducial pipeline images of amplitude and phase&#xA;  plots for 11 April, 2017 of DIFMAP with &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; = 1.401, EHT&#xA;  &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; = 1.361 and SMILI, &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; = 1.319, &lt;span class=&quot;math-container&quot;&gt;$^{[6]}$&lt;/span&gt; giving for that day an averaged value &lt;span class=&quot;math-container&quot;&gt;$\bar{q}$&lt;/span&gt; = 1.360 that deviates of 0.09 from epoch 2 value estimated with TIE and &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; &gt; 0 confirm the clockwise rotation. The spiral spectra are reported in Fig. 2.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Then one determines the rotation parameter &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; by comparing&#xA;  those obtained by a linear interpolation with the asymmetry&#xA;  parameter &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; of various models, as reported in the numerical example of Table I for different values of inclination and rotation parameters &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt;. The results are depicted in Fig. 1.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;[1]&lt;/code&gt; Fabrizio Tamburini, Bo Thidé, Gabriel Molina-Terriza, and Gabriele Anzolin, “Twisting of light around rotating black holes,” Nature Phys. 7, 195–197 (2011).&lt;br&gt;&#xA;  &lt;code&gt;[4]&lt;/code&gt; EHT Collaboration et al., “Imaging the central supermassive black hole,” Astrophys. J. Lett. 875, L4(52) (2019), First M87 Event Horizon Telescope Results IV.&lt;br&gt;&#xA;  &lt;code&gt;[5]&lt;/code&gt; EHT Collaboration et al., “Physical origin of the asymmetric ring,” Astrophys. J. Lett. 875, L5(31) (2019), First M87 Event Horizon Telescope Results V.&lt;br&gt;&#xA;  &lt;code&gt;[6]&lt;/code&gt; EHT Collaboration et al., “The shadow and mass of the central black hole,” Astrophys. J. Lett. 875, L6(44) (2019), First M87 Event Horizon Telescope Results VI.&lt;br&gt;&#xA;  &lt;code&gt;[10]&lt;/code&gt; Demetrios Christodoulou and Remo Ruffini, “Reversible transformations of a charged black hole,” Phys. Rev. D 4, 3552–3555 (1971).&lt;br&gt;&#xA;  &lt;code&gt;[29]&lt;/code&gt; Bin Chen, Ronald Kantowski, Xinyu Dai, Eddie Baron, and Prasad Maddumage, “Algorithms and programs for strong gravitational lensing in Kerr space-time including polarization,” Astrophys. J. Suppl. Ser. 218, 4 (2015).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Figures:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KaUwv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KaUwv.jpg&quot; alt=&quot;Figure 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Figure 1. &lt;strong&gt;Experimental results&lt;/strong&gt;. Field components along the observer’s direction and spiral spectra obtained with the TIE method for epoch 1 and epoch 2. The asymmetry between the &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; = 1 and &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; = −1 components in both of the spiral spectra reveals the rotation of the black hole in M87. It also indicates that the electromagnetic vortex reconstructed from the TIE analysis of the EM field intensities extracted from the brightness temperature in a finite frequency bandwidth has components along the propagation direction to the observer that are compatible with twisted lensing of a black hole with &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; = 0.9±0.1 rotating clockwise with the spin pointing away from Earth and an Einstein ring with a gravitational radius &lt;span class=&quot;math-container&quot;&gt;$R_g$&lt;/span&gt; = 5, as indicated by an EHT analysis dominated by incoherent emission. For all days, the diameters of the ring features&#xA;  span the narrow range 38–44 µ-arcseconds and the observed peak brightness temperature of the ring is &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; ∼ 6×10&lt;span class=&quot;math-container&quot;&gt;$^9$&lt;/span&gt; K.&lt;span class=&quot;math-container&quot;&gt;$^{[6]}$&lt;/span&gt; The other components (&lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;) of the EM field derived from TIE equations do not show a predominant OAM component. This is expected &lt;span class=&quot;math-container&quot;&gt;$^{[1]}$&lt;/span&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZZ9ph.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZZ9ph.jpg&quot; alt=&quot;Figure 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Figure 2. &lt;strong&gt;Results from DIFMAP, EHT and SMILI data analyses and of numerical simulations from KERTAP&lt;/strong&gt;. The first three insets show the experimental spiral spectra obtained from the three fiducial pipeline images for 11 April 2017 from SMILI, EHT imaging, and DIFMAP &lt;span class=&quot;math-container&quot;&gt;$^{[4]}$&lt;/span&gt;. They represent the visibility amplitude and phase as a function of the vector baseline. In all the datasets the asymmetry parameter, the ratio between the &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; = 1 and &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; = −1 peaks in the spiral spectra, is &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; &gt; 1 indicating clockwise rotation: the black hole is found to have its spin pointing away from Earth and an inclination between the approaching jet and the line of sight of &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; = 17° (equivalent to a similar geometry with an inclination &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; = 163°, but where the angular momentum of the accretion flow and that of the BH are anti-aligned) (&lt;em&gt;left&lt;/em&gt;). &lt;em&gt;Fourth inset&lt;/em&gt;: spiral spectrum of the numerical simulations with KERTAP &lt;span class=&quot;math-container&quot;&gt;$^{[29]}$&lt;/span&gt; obtained from the normalized intensity and phase of the &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; component of the radiation field emitted from a spatially resolved image of the black hole accretion disc dominated by thermalized emission with Γ = 2. The coherence χ&#xA;  of the radiation emission is characterized by the ratio between the &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; = 0 and &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; = 1 peaks in the spiral spectra. The lower the χ, the higher the coherence in the emission. The experimental spiral spectra of SMILI, EHT imaging, and DIFMAP show higher coherence in the radiation emission (χ&lt;span class=&quot;math-container&quot;&gt;$_\text{SMILI}$&lt;/span&gt; = 1.198, χ&lt;span class=&quot;math-container&quot;&gt;$_\text{EHT}$&lt;/span&gt; = 1.798) and (χ&lt;span class=&quot;math-container&quot;&gt;$_\text{DIFMAP}$&lt;/span&gt; = 1.107) with respect to the simulated model of a simple thermalized accretion disk with power spectrum Γ = 2 (χ&lt;span class=&quot;math-container&quot;&gt;$_\text{KERTAP}$&lt;/span&gt; = 5.029) and with respect to that obtained in the TIE reconstruction of the wavefront (χ&lt;span class=&quot;math-container&quot;&gt;$_\text{ep1}$&lt;/span&gt; = 13.745 and χ&lt;span class=&quot;math-container&quot;&gt;$_\text{ep2}$&lt;/span&gt; = 14.649) in Fig.1. Even if the asymmetry &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; is well preserved, the TIE method can be improved by consecutive data acquisitions of the wavefront, separated by a much shorter time interval than one day and might therefore provide better information on the source emission.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That paper contains considerable additional information and illustrations well worth reviewing. &lt;strong&gt;Thank you &lt;a href=&quot;https://astronomy.stackexchange.com/users/7997/jack-r-woods&quot;&gt;Jack R. Woods&lt;/a&gt;&lt;/strong&gt; for the link which led me to the above information.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Previous edit&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the paper: &quot;&lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/ab0f43/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;First M87 Event Horizon Telescope Results. V. Physical Origin of the Asymmetric Ring&lt;/a&gt;&quot;, (Apr 10 2019), by The Event Horizon Telescope Collaboration, Kazunori Akiyama, Antxon Alberdi, Walter Alef, Keiichi Asada, Rebecca Azulay, Anne-Kathrin Baczko, David Ball, Mislav Baloković, John Barrett, et al., in one of several papers recently published they explain:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(4) The ring is brighter in the south than the north. This can be explained by a combination of motion in the source and Doppler beaming. As a simple example we consider a luminous, optically thin ring rotating with speed v and an angular momentum vector inclined at a viewing angle i &gt; 0° to the line of sight. Then the approaching side of the ring is Doppler boosted, and the receding side is Doppler dimmed, producing a surface brightness contrast of order unity if v is relativistic. The approaching side of the large-scale jet in M87 is oriented west–northwest (position angle &lt;span class=&quot;math-container&quot;&gt;$\mathrm{PA}\approx 288^\circ ;$&lt;/span&gt; in Paper VI this is called &lt;span class=&quot;math-container&quot;&gt;${\mathrm{PA}}_{\mathrm{FJ}}$&lt;/span&gt;), or to the right and slightly up in the image. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Figure 5 from that paper is included in Rob Jeffries answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conclusion that they reach, in part, is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;... The results of this comparison are consistent with the hypothesis that the compact 1.3 mm emission in M87 arises within a few &lt;span class=&quot;math-container&quot;&gt;${r}_{{\rm{g}}}$&lt;/span&gt; of a Kerr black hole, and that the ring-like structure of the image is generated by strong gravitational lensing and Doppler beaming. The models predict that the asymmetry of the image depends on the sense of black hole spin. If this interpretation is accurate, then the spin vector of the black hole in M87 points away from Earth (the black hole spins clockwise on the sky). The models also predict that there is a strong energy flux directed away from the poles of the black hole, and that this energy flux is electromagnetically dominated. If the models are correct, then the central engine for the M87 jet is powered by the electromagnetic extraction of free energy associated with black hole spin via the Blandford–Znajek process.&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First Draft&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article: &quot;&lt;a href=&quot;https://arxiv.org/abs/1807.08840v3&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ergoregion instability of exotic compact objects: electromagnetic and gravitational perturbations and the role of absorption&lt;/a&gt;&quot;, (Feb 15 2019), by Elisa Maggio, Vitor Cardoso, Sam R. Dolan, and Paolo Pani explains that this is due to &lt;a href=&quot;https://en.wikipedia.org/wiki/Superradiance#Rotational_superradiance&quot; rel=&quot;nofollow noreferrer&quot;&gt;rotational superradiance&lt;/a&gt; on page 10:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;... the instability can be understood in terms of waves trapped within the photon-sphere barrier and amplified by superradiant scattering&lt;span class=&quot;math-container&quot;&gt;$^{[43]}$&lt;/span&gt;&lt;br&gt;&#xA;  &lt;code&gt;[43]&lt;/code&gt; R. Brito, V. Cardoso, and P. Pani, Lect. Notes Phys. 906, pp.1 (2015), arXiv:&lt;a href=&quot;https://arxiv.org/abs/1501.06570&quot; rel=&quot;nofollow noreferrer&quot;&gt;1501.06570&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the article &quot;&lt;a href=&quot;https://arxiv.org/abs/1501.06570&quot; rel=&quot;nofollow noreferrer&quot;&gt;Superradiance&lt;/a&gt;&quot;, (above) while considerably longer, maybe much more approachable. On page 38 where they explain the &lt;a href=&quot;https://en.wikipedia.org/wiki/Penrose_process&quot; rel=&quot;nofollow noreferrer&quot;&gt;Penrose Process&lt;/a&gt; they offer a diagram which probably makes the understanding of this easier:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K7bmN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K7bmN.jpg&quot; alt=&quot;Penrose Process&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;Figure 7: Pictorial view of the original Penrose processes. A particle with energy E&lt;span class=&quot;math-container&quot;&gt;$_0$&lt;/span&gt; decays inside the ergosphere into two particles, one with negative energy E&lt;span class=&quot;math-container&quot;&gt;$_2$&lt;/span&gt; &amp;lt; 0 which falls into the BH, while the second particle escapes to infinity with an energy higher than the original particle, E&lt;span class=&quot;math-container&quot;&gt;$_1$&lt;/span&gt; &gt; E&lt;span class=&quot;math-container&quot;&gt;$_0$&lt;/span&gt;.&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From page 41:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2moLN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2moLN.jpg&quot; alt=&quot;Simplified Penrose Explanation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;Figure 8: The carousel analogy of the Penrose process. A body falls nearly from rest into a rotating cylinder, whose surface is sprayed with glue. At the surface the body is forced to co-rotate with the cylinder (analog therefore of the BH ergosphere, the surface beyond which no observer can remain stationary with respect to infinity). The negative energy states of the ergoregion are played by the potential energy associated with the sticky surface. If now half&#xA;  the object (in reddish) is detached from the first half (yellowish), it will reach infinity with more (kinetic) energy than it had initially, extracting rotational energy out of the system.&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A further more complicated model, believed to be beyond what was asked, from page 46:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uDjyO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uDjyO.jpg&quot; alt=&quot;Collisional Penrose Process&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;Figure 9: Pictorial view of the different collisional Penrose processes. Left: initial particleswith ingoing radial momentum (p&lt;span class=&quot;math-container&quot;&gt;$^r&#xA;_1$&lt;/span&gt; &amp;lt; 0 and p&lt;span class=&quot;math-container&quot;&gt;$^r_2$&lt;/span&gt; &amp;lt; 0). Particle 3 has initial ingoing radial&#xA;  momentum, but eventually finds a turning point and escapes to infinity. The maximum efficiency for this was shown to be quite modest η ∼ 1.5 &lt;span class=&quot;math-container&quot;&gt;$^{[168, 169, 170, 171]}$&lt;/span&gt;. Right: initial particles with p&lt;span class=&quot;math-container&quot;&gt;$^r_1$&lt;/span&gt; &gt; 0 and p&lt;span class=&quot;math-container&quot;&gt;$^r_2$&lt;/span&gt; &amp;lt; 0. In this case particle 1 must have p&lt;span class=&quot;math-container&quot;&gt;$^r_1$&lt;/span&gt; &gt; 0 inside the ergosphere. For this process the efficiency can be unbound for extremal BHs &lt;span class=&quot;math-container&quot;&gt;$^{[172, 173]}$&lt;/span&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;[168]&lt;/code&gt; T. Piran and J. Shaham, “Upper Bounds on Collisional Penrose Processes Near Rotating Black Hole Horizons,” Phys.Rev. D16 (1977) 1615–1635.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;[169]&lt;/code&gt; T. Harada, H. Nemoto, and U. Miyamoto, “Upper limits of particle emission from high-energy collision and reaction near a maximally rotating Kerr black hole,” Phys.Rev. D86 (2012) 024027, arXiv:1205.7088 [gr-qc].&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;[170]&lt;/code&gt; M. Bejger, T. Piran, M. Abramowicz, and F. Hakanson, “Collisional Penrose process near the horizon of extreme Kerr black holes,” Phys.Rev.Lett. 109 (2012) 121101, arXiv:1205.4350 [astro-ph.HE].&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;[171]&lt;/code&gt; O. Zaslavskii, “On energetics of particle collisions near black holes: BSW effect versus Penrose process,” Phys.Rev. D86 (2012) 084030, arXiv:1205.4410 [gr-qc].&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;[172]&lt;/code&gt; J. D. Schnittman, “A revised upper limit to energy extraction from a Kerr black hole,” arXiv:1410.6446 [astro-ph.HE].&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;[173]&lt;/code&gt; E. Berti, R. Brito, and V. Cardoso, “Ultra-high-energy debris from the collisional Penrose process,” arXiv:1410.8534 [gr-qc].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is a summary on page 170 (nowhere near the end of the paper) which explains:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;In gravitational theories, superradiance is intimately connected to tidal acceleration, even at Newtonian level. Relativistic gravitational theories predict the existence of BHs, gravitational vacuum solutions whose event horizon behaves as a one-way viscous membrane. This allows superradiance to occur in BH spacetimes, and to extract energy from vacuum even at the classical level. When semiclassical effects are taken into account, superradiance occurs&#xA;  also in static configurations, as in the case of Hawking radiation from a Schwarzschild BH.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The efficiency of superradiant scattering of GWs by a spinning (Kerr) BH can be larger than 100% and this phenomenon is deeply connected to other important mechanisms associated to spinning compact objects, such as the Penrose process, the ergoregion instability, the Blandford-Znajek effect, and the CFS instability. Rotational superradiance might be challenging to observe in the laboratory, but its BH counterpart is associated with a number&#xA;  of interesting effects and instabilities, which may leave an observational imprint. We have presented a unified treatment of BH superradiant phenomena including charged BHs, higher dimensions, nonasymptotically flat spacetimes, analog models of gravity and theories beyond GR.&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="18054" LastEditDate="2019-04-29T04:36:13.303" LastActivityDate="2019-04-29T04:36:13.303" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30348" PostTypeId="2" ParentId="30343" CreationDate="2019-04-11T06:43:27.250" Score="37" Body="&lt;p&gt;No, you aren't seeing the shape of the accretion disk. Although its plane is almost that of the picture it is far larger and fainter than the ring that is seen. The reason for this asymmetry is almost entirely due to Doppler beaming and boosting of radiation arising in matter travelling at relativistic speeds very close to the black hole. This in turn is almost entirely controlled by &lt;em&gt;the orientation of the black hole spin&lt;/em&gt;. The black hole sweeps up material and magnetic fields almost irrespective of the orientation of any accretion disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pictures below from &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/ab0f43&quot; rel=&quot;nofollow noreferrer&quot;&gt;the fifth event horizon telescope paper&lt;/a&gt; makes things clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HyyVY.jpg&quot; alt=&quot;Relative orientation of spin and accretion flow&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black arrow indicates the direction of black hole spin. The blue arrow indicates the initial rotation of the accretion flow. The jet of M87 is more or less East-West (projected onto the page), but the right hand side is pointing towards the Earth. It is assumed that the spin vector of the black hole is aligned (or anti-aligned) with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two left hand plots show agreement with the observations. What they have in common is that the black hole spin vector is mostly into the page (anti-aligned with the jet). Gas is forced to rotate in the same way and results in projected relativistic motion towards us south of the black hole and away from us north of the black hole. Doppler boosting and beaming does the rest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the paper says: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the location of the peak flux in the ring is controlled by the black hole spin: it always lies roughly 90 degrees counterclockwise from the projection of the spin vector on the sky.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-04-29T06:38:14.643" LastActivityDate="2019-04-29T06:38:14.643" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30349" PostTypeId="2" ParentId="22429" CreationDate="2019-04-11T07:42:45.680" Score="4" Body="&lt;p&gt;Smoke (fire) is not the only explication - Dust is also a good candidate. I've noticed a drop in &quot;seeing quality&quot; even when observing deep sky objects or planets trough a telescope due to large amounts of dust in the air.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The second reason for a red Moon is if there’s some kind of particle in the air. A forest fire or volcanic eruption can fill the air with tiny particles that partially obscure light from the Sun and Moon. Once again, these particles tend to scatter blue and green light away, while permitting red light to pass through more easily. When you see a red moon, high up in the sky, it’s probably because there’s a large amount of dust in the air.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(source: &lt;a href=&quot;https://www.universetoday.com/19969/red-moon/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Red Moon - Universe Today&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="23668" LastEditorUserId="7472" LastEditDate="2019-04-11T08:25:01.597" LastActivityDate="2019-04-11T08:25:01.597" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30350" PostTypeId="1" AcceptedAnswerId="30351" CreationDate="2019-04-11T08:06:00.110" Score="2" ViewCount="159" Body="&lt;p&gt;As for the title. The picture could have been plotted in grey scale, B&amp;amp;W, whatever couple of colours, or perhaps using a window of wavelength in order to show shift of the Doppler/relativistic Doppler kind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Discarding the obvious dark, shall the picture be watch at merely in term of brightness, or considering yellow and red as well? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And independently, why those colours have been choose?  Just an aesthetic and suggestive choice, or that should be the apparent image as seen by eyes if we could go there? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I resume: is the picture a radio waves intensity or brightness map, or conversely a kind of &quot;as it would be&quot; picture in the Vis (modelled from the behaviour of the collected radio waves and theories)?&lt;/p&gt;&#xA;" OwnerUserId="17437" LastEditorUserId="17437" LastEditDate="2019-04-11T08:19:41.300" LastActivityDate="2019-04-12T09:12:30.170" Title="Do the actual false colours in the M87 black hole picture convey information?" Tags="&lt;black-hole&gt;&lt;radio-astronomy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30351" PostTypeId="2" ParentId="30350" CreationDate="2019-04-11T08:37:08.643" Score="4" Body="&lt;p&gt;See &lt;a href=&quot;https://doi.org/10.3847%2F2041-8213%2Fab0ec7&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; recently published by the EHT collaboration describing how they generated the image. Specifically, chapter 5 of that article describes the origin of the image. To quote the article (specifically the caption of figure 3):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The image is shown in units of brightness temperature, &lt;span class=&quot;math-container&quot;&gt;${T}_{{\rm{b}}}=S{\lambda }^{2}/2{k}_{{\rm{B}}}{\rm{\Omega }}$&lt;/span&gt;, where S is the flux density, λ is the observing wavelength, &lt;span class=&quot;math-container&quot;&gt;$k_{B}$&lt;/span&gt; is the Boltzmann constant, and Ω is the solid angle of the resolution element.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The false colors in the image convey the surface T brightness (a sort of measurement of the intensity or flux from that area) of the material surrounding the black hole's event horizon. Brighter color means higher brightness temperature. There is not information about the physical T of the in falling materials as the collected radiation is of synchrotron type and not that of a black body. As you say, the image could have been shown in grayscale or any other color scale. The fact that it is shown in an orange-ish color is merely the convention chosen by the scientists who generated the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BUT there's a lot more information in the image than just how bright the material is. For example, the fact that the material is brighter on one side than on the other can tell us about the geometry of the material surrounding the black hole, or about how that material is rotating around it. If you have the time, check out &lt;a href=&quot;https://www.nature.com/articles/d41586-019-01155-0&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; and read up on the actual journal articles published by the &lt;a href=&quot;https://eventhorizontelescope.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Event Horizon Telescope collaboration&lt;/a&gt; linked at the bottom.&lt;/p&gt;&#xA;" OwnerUserId="14328" LastEditorUserId="17437" LastEditDate="2019-04-12T09:12:30.170" LastActivityDate="2019-04-12T09:12:30.170" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30352" PostTypeId="2" ParentId="30350" CreationDate="2019-04-11T08:54:24.557" Score="0" Body="&lt;p&gt;This comment is too long for a comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could not find a source (yet) stating explicitly that it is a scalar quantity that is plotted, so I thought I would investigate the colors themselves for any sign of additional information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I found was inconclusive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://eventhorizontelescope.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://eventhorizontelescope.org/&lt;/a&gt; I found the smaller size image &lt;code&gt;https://static.projects.iq.harvard.edu/files/styles/os_files_xlarge/public/eht/files/20190410-78m-800x466.png&lt;/code&gt; which I've imported below in png format (lossless-compression only) then broken into r, g, b components, flattened, and then plotted in 2D (and 3D) scatter plots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It certainly looks like some standard colormap was used to plot a image of a single scalar quantity,&lt;/strong&gt; but something has fuzzed up the color a bit. It could be the interpolation routings used to make the graphic for public release, or some &quot;internet color adjustment&quot; before publication. The world of publishing color is pretty confusing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, I've slimmed down the data before flattening and plotting:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;n = 8&#xA;img     = plt.imread(fname)[::n, ::n]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OtCXB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OtCXB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7ySFI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7ySFI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bBbcf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bBbcf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;from mpl_toolkits.mplot3d import Axes3D&#xA;&#xA;fname = '20190410-78m-800x466.png'&#xA;n = 8&#xA;img     = plt.imread(fname)[::n, ::n]&#xA;rgb     = np.array([x.flatten() for x in np.rollaxis(img, 2, 0)][:3])&#xA;r, g, b = rgb&#xA;rgbavg  = rgb.sum(axis=0)/3.&#xA;&#xA;if True:&#xA;    fig = plt.figure(figsize=[10, 8])  # [12, 10]&#xA;    ax  = fig.add_subplot(1, 1, 1, projection='3d')&#xA;    ax.plot(r, g, b, '.k')&#xA;    ax.set_xlim(0, 1)&#xA;    ax.set_ylim(0, 1)&#xA;    ax.set_zlim(0, 1)&#xA;    plt.show()&#xA;&#xA;titles = ('g vs r', 'b vs g', 'r vs b', 'r vs mean', 'g vs mean', 'b vs mean')&#xA;if True:&#xA;    plt.figure()&#xA;    for i in range(0, 3):&#xA;        plt.subplot(2, 3, i+1)&#xA;        plt.plot(rgb[i%3], rgb[(i+1)%3], '.k')&#xA;        plt.title(titles[i], fontsize=16)&#xA;        plt.xlim(0, 1)&#xA;        plt.ylim(0, 1)&#xA;    for i in range(3, 6):&#xA;        plt.subplot(2, 3, i+1)&#xA;        plt.plot(rgbavg, rgb[i%3], '.k')&#xA;        plt.title(titles[i], fontsize=16)&#xA;        plt.xlim(0, 1)&#xA;        plt.ylim(0, 1)&#xA;    plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-04-11T08:54:24.557" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="30353" PostTypeId="2" ParentId="30339" CreationDate="2019-04-11T09:10:20.737" Score="20" Body="&lt;p&gt;As Ingolifs says, Sgr A* and M87* are the obvious candidates. At the press conference, Heino Falcke explained why they got a picture of M87* first:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But it would take some more time because Sagittarius A Star is 1000 times faster and smaller. Its like a toddler who is moving constantly. In comparison, M87 is much slower, like a big bear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&amp;mdash; &lt;a href=&quot;https://www.deccanherald.com/science-and-environment/astronomers-release-first-photo-of-black-hole-727987.html&quot; rel=&quot;noreferrer&quot;&gt;The Deccan Herald&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26553" LastActivityDate="2019-04-11T09:10:20.737" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30354" PostTypeId="1" AcceptedAnswerId="30366" CreationDate="2019-04-11T10:34:54.640" Score="4" ViewCount="164" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0kabc.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0kabc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If black holes are spherical, so shouldn't it be absorbing light from all the directions? How would one explain the light ring. which seems more like a 2D ring than a sphere?&lt;/p&gt;&#xA;" OwnerUserId="26556" LastEditorUserId="1891" LastEditDate="2019-04-11T12:34:21.540" LastActivityDate="2019-04-11T14:21:53.040" Title="If blackholes are spherical, what is the prespective of the blackhole image?" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2019-04-27T08:39:25.243" ContentLicense="CC BY-SA 4.0" />
  <row Id="30355" PostTypeId="1" AcceptedAnswerId="30434" CreationDate="2019-04-11T10:59:26.250" Score="1" ViewCount="2693" Body="&lt;p&gt;I would like to project the Moon's position on to the surface of the Earth, starting with its position expressed as Right Ascension in degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the declination of the Moon corresponds to the terrestrial latitude but how can I &quot;convert&quot; the right ascension of the moon to find my longitude?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not an astronomer, so I'm hoping for a fairly simple answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My friend used this formula a long time ago:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the time at which I want to calculate the longitude,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;time_hrs = hours + minutes / 60  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;then using right ascension of Sun,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;delta_x = time_hrs * 15 + right_ascension_sun - 180;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The coordinates of the Moon's ground track at this time are then:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;longitude = Right_ascension_moon - delta_x;&#xA;latitude  = Declination_of_moon&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;but i don't know if is correct, or how precise it will be.&lt;/p&gt;&#xA;" OwnerUserId="26560" LastEditorUserId="7982" LastEditDate="2019-04-12T00:29:27.337" LastActivityDate="2019-04-13T21:46:21.180" Title="How to calculate the ground track of the Moon's position on the Earth's surface?" Tags="&lt;the-moon&gt;&lt;coordinate&gt;&lt;declination&gt;&lt;ephemeris&gt;&lt;right-ascension&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="30356" PostTypeId="1" AcceptedAnswerId="30372" CreationDate="2019-04-11T11:13:33.027" Score="0" ViewCount="46" Body="&lt;p&gt;I am stuck and a little embarrassed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to implement &lt;a href=&quot;https://astronomy.stackexchange.com/a/26467/7982&quot;&gt;this answer&lt;/a&gt;. I have a unit vector &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt; pointing in the direction of a radio source on the celestial sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; are the two other, mutually orthogonal unit vectors that point &lt;em&gt;east&lt;/em&gt; and &lt;em&gt;north&lt;/em&gt; from that point, tangent to the sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without back-converting to R.A. and Dec or using trigonometry, is there a simple way to generate &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; vectorially?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slide 38 &lt;a href=&quot;https://www.strw.leidenuniv.nl/radioastronomy/lib/exe/fetch.php?media=radio_astronomy_lec_4_ma_garrett.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; may or may not be helpful, I think my description is sufficient by itself.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-04-11T16:23:51.583" Title="Generate East and North unit vectors tangent to a point on the celestial sphere, using vectors?" Tags="&lt;mathematics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30357" PostTypeId="1" AcceptedAnswerId="30359" CreationDate="2019-04-11T11:17:53.207" Score="4" ViewCount="213" Body="&lt;p&gt;I'm still at my bachelor of electronic and communications engineering, so I'm not an expert of radiotelescopes; however, I surely know that, typically, radio communications are characterized by a RX that has at least one bandpass filter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, I am led to think that the &quot;photo&quot; of the black hole, recently released to the public, is a reconstruction picture taken from various radiofrequency signal samples that got mapped to the frequencies of visible light spectrum (since the Event Horizon Telescope project is based on radio telescopes), hence it doesn't depict what we would actually see (neglecting any impossibilities in doing so), but rather, it shows the radio emission from whatever is close to the event horizon of the black hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct, or is it actually a photo capturing visible light?&#xA;I mean, scientifically, this is a huge step; however, I have the feeling that the news got &quot;boosted&quot; a little too much, in medias, as they just talk about &quot;photos&quot; and &quot;light&quot; surrounding a black hole - not specifying anything else for a inexperienced audience.&lt;/p&gt;&#xA;" OwnerUserId="26559" LastEditorUserId="7982" LastEditDate="2019-04-11T11:54:23.433" LastActivityDate="2019-04-11T13:04:38.210" Title="Does the &quot;photo of a black hole&quot; taken by EHT actually depict a visible light-based scenario?" Tags="&lt;black-hole&gt;&lt;radio-astronomy&gt;&lt;supermassive-black-hole&gt;&lt;radio-telescope&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30359" PostTypeId="2" ParentId="30357" CreationDate="2019-04-11T12:01:53.437" Score="5" Body="&lt;p&gt;You are right that the image is not a visible light image. It was taken using very, very short wavelength microwaves -- the wavelength is &lt;em&gt;1.3 mm&lt;/em&gt; which is only a bit shorter than the Far IR!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you may be reading a bit too much into the word &quot;photo&quot; which, as you say, has the everyday meaning of &quot;taken with a camera using lenses in visible light.&quot;  I think most astronomers would use the term &quot;image&quot; rather than &quot;photo&quot; -- if you look at the &lt;a href=&quot;https://www.nsf.gov/news/news_summ.jsp?cntn_id=298276&amp;amp;org=NSF&amp;amp;from=news&quot; rel=&quot;noreferrer&quot;&gt;NSF's press release&lt;/a&gt;, for example, you'll see that it uses the term &quot;image&quot;.  (Links to all six of the papers the EHT team has published are in the NSF press release if you want further details.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'd bet that &quot;image&quot; was changed to &quot;photo&quot; somewhere in the journalistic process as being &quot;more meaningful&quot; to the average reader.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The appearance to the naked eye would probably not be terribly different, since the photon orbits don't depends on frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as whether it got &quot;boosted&quot; or not, you're right to be very skeptical -- university publicity offices are prone to exaggerate and sometimes pump relatively ordinary science into amazingly important-sounding &quot;breakthroughs.&quot;  In this case it all seems pretty straightforward, though. And there's nothing special about a visible-light image over a microwave image -- both are useful and neither tell us everything. (It is not uncommon for images in the visible to be among the least useful -- with X-ray or radio actually telling us much more.)&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2019-04-11T13:04:38.210" LastActivityDate="2019-04-11T13:04:38.210" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="30360" PostTypeId="2" ParentId="30317" CreationDate="2019-04-11T12:06:31.313" Score="13" Body="&lt;p&gt;Rob Jeffries' answer is excellent, I just wanted to add this picture trying to explain the geometry. Here, I assume a non-rotating black hole (BH); for a rotating BH the exact numbers are slightly different.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The photon sphere&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Photons move on straight lines, but in the heavily curved space around a BH these straight lines appear curved. Although the event horizon (EH) at a distance of &lt;span class=&quot;math-container&quot;&gt;$r = 2GM/c^2 \equiv r_\mathrm{S}$&lt;/span&gt; (the &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;noreferrer&quot;&gt;Schwarzschild radius&lt;/a&gt;) from the BH marks the region from which no photons may escape if emitted &lt;em&gt;radially&lt;/em&gt;, photons on partially tangential orbit will fall back out to a distance of &lt;span class=&quot;math-container&quot;&gt;$r = 1.5r_\mathrm{S}$&lt;/span&gt;, where photons traveling fully tangentially will stay on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Photon_sphere&quot; rel=&quot;noreferrer&quot;&gt;photon sphere&lt;/a&gt; (although this is an unstable orbit).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The innermost stable orbit and the accretion disk&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Ordinary matter will spiral inward out to twice this distance; hence, inside the &lt;a href=&quot;https://en.wikipedia.org/wiki/Innermost_stable_circular_orbit&quot; rel=&quot;noreferrer&quot;&gt;innermost stable circular orbit&lt;/a&gt; (ISCO) at &lt;span class=&quot;math-container&quot;&gt;$r=3r_\mathrm{S}$&lt;/span&gt;, matter is pretty much bound to be absorbed. Outside this region matter may orbit, forming the &lt;a href=&quot;https://en.wikipedia.org/wiki/Accretion_disk&quot; rel=&quot;noreferrer&quot;&gt;accretion disk&lt;/a&gt;, but since friction between the particles will cause them to lose energy, they will slowly approach the ISCO, after which they will rapidly fall into the BH. Note that the M87 BH doesn't have a thin accretion disk like the one &lt;a href=&quot;https://www.google.com/search?q=interstellar+black+hole&quot; rel=&quot;noreferrer&quot;&gt;depicted in the movie Interstellar&lt;/a&gt;; rather a thick &quot;cloud&quot; surrounding most of the BH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photons emitted tangentially &lt;em&gt;just&lt;/em&gt; outside the photon sphere will spiral around the BH many times, slowly increasing their distance, until eventually they escape at a projected distance of &lt;span class=&quot;math-container&quot;&gt;$\sqrt{27/4}r_\mathrm{S} \simeq 2.6r_\mathrm{S}$&lt;/span&gt; from the BH (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/1998bhp..book.....F&quot; rel=&quot;noreferrer&quot;&gt;Frolov &amp;amp; Novikov 1998&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The shadow&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Just as the path of light rays are curved around the BH, so are the sightlines from you toward the BH (you can think of sightlines as reversed photons). That means that all sightlines that are &lt;em&gt;closer&lt;/em&gt; than (a projected distance of) &lt;span class=&quot;math-container&quot;&gt;$2.6r_\mathrm{S}$&lt;/span&gt; to the BH will, eventually, end up on the EH, even if taking several orbits around the BH. These sightlines comprise the so-called &lt;em&gt;shadow&lt;/em&gt; (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2000ApJ...528L..13F&quot; rel=&quot;noreferrer&quot;&gt;Falcke et al.(2000)&lt;/a&gt;; &lt;a href=&quot;http://adsabs.harvard.edu/abs/2019ApJ...875L...1E&quot; rel=&quot;noreferrer&quot;&gt;Event Horizon Telescope Collaboration et al.(2019a)&lt;/a&gt;, ). On the other hand, along sightlines that a farther away, you see the radiation emitted from the matter falling into the BH, both in front of and behind the BH. And since the first sightlines that &lt;em&gt;don't&lt;/em&gt; terminate at the EH circle the photon sphere many times, those sightlines are actually very long paths through matter shining its last light before being engulfed, and hence they look exceptionally bright (e.g. &lt;a href=&quot;http://adsabs.harvard.edu/abs/2019ApJ...875L...5E&quot; rel=&quot;noreferrer&quot;&gt;Event Horizon Telescope Collaboration et al.(2019b)&lt;/a&gt;). This bright ring just outside the shadow is called the &lt;em&gt;photon ring&lt;/em&gt;, or the &lt;em&gt;emission ring&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The drawing&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The drawing below may help understand. All the red lines are sightlines toward the BH. Only the uppermost one just grazes the photon sphere (and the luminous matter behind). The rest terminate at the EH, and hence look black (except for luminous matter in front). Close to the center, you see the front of the EH; farther out you actually see the back of the EH; even farther out you &lt;em&gt;again&lt;/em&gt; see the front of the EH, and so on ad infinitum until you reach the photon ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2Kio5.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2Kio5.jpg&quot; alt=&quot;BHshadow&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The observation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Despite the observational resolution being an astonishing &lt;span class=&quot;math-container&quot;&gt;$\sim25$&lt;/span&gt; micro-arcseconds, the photon ring is smeared out over a larger region, resulting in the doughnut shape you've seen. That is, what you see in that image is not &quot;the EH in front of an accretion disk&quot;, but rather &quot;the EH seen from all sides at the same time and enlarged, with light emitted from the photons ring&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless the accretion disk is viewed exactly face-on, half of the accretion disk&lt;span class=&quot;math-container&quot;&gt;$^\dagger$&lt;/span&gt; has a velocity component toward you, making it brighter than the other half through a special relativistic effect called &lt;a href=&quot;https://en.wikipedia.org/wiki/Relativistic_beaming&quot; rel=&quot;noreferrer&quot;&gt;beaming&lt;/a&gt;. This is seen in the southern part of the M87 BH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure below (from &lt;a href=&quot;http://adsabs.harvard.edu/abs/2019ApJ...875L...5E&quot; rel=&quot;noreferrer&quot;&gt;Event Horizon Telescope Collaboration et al.(2019b)&lt;/a&gt;) shows, from left to right, the actual observation, a model where you see the rather sharp photon ring, and this model blurred to match the resolution of the observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KtLpj.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KtLpj.png&quot; alt=&quot;BHobs&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^\dagger$&lt;/span&gt;&lt;sub&gt;At least the material just before it plunges into the BH, which follows the rotation of the BH. Farther out, the rest of the accretion disk may in principle rotate the other way.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2019-04-12T21:02:12.563" LastActivityDate="2019-04-12T21:02:12.563" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="30361" PostTypeId="1" CreationDate="2019-04-11T12:24:30.973" Score="5" ViewCount="168" Body="&lt;p&gt;I was wondering how the researchers were able to minimize or neutralize confirmation bias regarding the data collected?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, the algorithm pruned much of the data, which would make sense, since it would likely be useless noise. See for example the video &lt;a href=&quot;https://youtu.be/XlrfBiuMzfE&quot; rel=&quot;nofollow noreferrer&quot;&gt;Event Horizon Telescope Press Conference. First Image Of a Black Hole.&lt;/a&gt; at about &lt;code&gt;38:40&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Five petabytes is a lot of data… The image you saw of course isn’t 5 petabytes in size, it’s a few hundred kilobytes, so our data analysis has to collapse this five petabytes of data into an imaget that’s more than a billion times smaller. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But apart from that, could certain features be misidentified as artifacts, and be thrown out in favour of preselecting what researchers expected to see?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only reason I ask is that the image looks almost identical to simulated predictions from two years ago - which may raise a slight cause for alarm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To follow up (based on answers section below):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first stage process - &quot;common features&quot; - are they like existing stamps of astronomical features (and image comparatives) that have been already observed in our universe (in both the visible and invisible spectrum and used for comparison)? And a second question - if we have never observed the earmarks of a black-hole before, how are we to use existing stamps of knowledge of astronomical features that we have seen to assess and validate an as of yet unknown (which would seem to directly impact what artifacts are preserved and what artifacts are thrown out and discarded)?&lt;/p&gt;&#xA;" OwnerUserId="26564" LastEditorUserId="26564" LastEditDate="2019-04-11T17:54:23.863" LastActivityDate="2019-04-11T17:54:23.863" Title="Event Horizon Telescope Data Cleanup" Tags="&lt;black-hole&gt;&lt;telescope&gt;&lt;radio-astronomy&gt;&lt;supermassive-black-hole&gt;&lt;radio-telescope&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30362" PostTypeId="1" AcceptedAnswerId="30364" CreationDate="2019-04-11T08:20:34.643" Score="7" ViewCount="496" Body="&lt;p&gt;The EHT has published the first image of a black hole.  It is the event horizon of the singularity in M87 against its accretion disk.  I've been through the reports in the popular press but none mention the light wavelengths this was taken at.  Is it an image in visible light, infra-red or longer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/j0EGm.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j0EGm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26593" OwnerDisplayName="Dave Gremlin" LastActivityDate="2019-04-12T23:03:23.887" Title="What part of the EM spectrum was used in the black hole image?" Tags="&lt;radio-telescope&gt;&lt;black-hole&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="30363" PostTypeId="2" ParentId="30362" CreationDate="2019-04-11T08:32:43.883" Score="5" Body="&lt;p&gt;According to the &lt;a href=&quot;https://eventhorizontelescope.org/&quot; rel=&quot;noreferrer&quot;&gt;EHT website&lt;/a&gt; The observations were done using radio telescopes observing at a wavelength of 1.3mm. Visible light or infrared from the accretion disk would be nearly impossible to detect at such large distances. The radio measurements from all the synchronized telescopes were then combined and converted into an image. &lt;/p&gt;&#xA;" OwnerDisplayName="Alexander Vandenberghe" LastActivityDate="2019-04-11T11:17:30.593" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30364" PostTypeId="2" ParentId="30362" CreationDate="2019-04-11T10:10:25.397" Score="5" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;What part of the EM spectrum was used in the black hole image?... Is it an image in visible light, infra-red or longer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;Microwaves, (&lt;a href=&quot;https://en.wikipedia.org/wiki/Extremely_high_frequency&quot; rel=&quot;nofollow noreferrer&quot;&gt;millimeter waves&lt;/a&gt; actually), and the hairy edge of far-infrared&lt;/h2&gt;&#xA;&lt;p&gt;at a nominal frequency of 230 GHz or 1.30 mm wavelength, the bandwidth is roughly 2 to 6 GHz wide depending on how many channels of data were used to produce the published image.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/ab0ec7&quot; rel=&quot;nofollow noreferrer&quot;&gt;First M87 Event Horizon Telescope Results. I. The Shadow of the Supermassive Black Hole&lt;/a&gt; the first of four articles published together:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;4. Observations, Correlation, and Calibration&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;We observed M87* on 2017 April 5, 6, 10, and 11 with the EHT. Weather was uniformly good to excellent with nightly median zenith atmospheric opacities at &lt;strong&gt;230 GHz&lt;/strong&gt; ranging from 0.03 to 0.28 over the different locations. The observations were scheduled as a series of scans of three to seven minutes in duration, with M87* scans interleaved with those on the quasar 3C 279. The number of scans obtained on M87* per night ranged from 7 (April 10) to 25 (April 6) as a result of different observing schedules. A description of the M87* observations, their correlation, calibration, and validated final data products is presented in Paper III and briefly summarized here.&lt;/p&gt;&#xA;&lt;p&gt;At each station, the astronomical signal in both polarizations &lt;strong&gt;and two adjacent 2 GHz wide frequency bands centered at 227.1 and 229.1 GHz&lt;/strong&gt; were converted to baseband using standard heterodyne techniques, then digitized and recorded at a total rate of 32 Gbps.[...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So if we use 230 GHz, the wavelength is given by &lt;span class=&quot;math-container&quot;&gt;$c/f$&lt;/span&gt; or 1.30 millimeters. It's hard for me to say right now if the image comes from only one 2GHz wide channel, or all three, which means that the bandwidth is either about 0.9% or  2.1%, but that's still pretty narrow compared to images taken at optical frequencies. That's (ultimately) because interferometry is done digitally these days and the computational size and time scales fairly fast with the size of the baseband.&lt;/p&gt;&#xA;&lt;p&gt;I should note that these days it's more and more common for astronomers to refer to all kinds of different wavelengths as &amp;quot;light&amp;quot; in a loose way. Circa 1.3 millimeter wavelength certainly could be thought of as far-infrared, though &lt;a href=&quot;https://en.wikipedia.org/wiki/Far_infrared&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; puts the cutoff at 1 mmm (300 GHz).&lt;/p&gt;&#xA;" OwnerUserId="7982" OwnerDisplayName="uhoh" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-04-12T23:03:23.887" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30365" PostTypeId="1" CreationDate="2019-04-11T14:14:39.747" Score="1" ViewCount="54" Body="&lt;p&gt;The EHT has published the first image of a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7qLeP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7qLeP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering what could be the size of the original image of the first black hole and the format of the image captured?&lt;/p&gt;&#xA;" OwnerUserId="26573" LastActivityDate="2019-04-11T14:14:39.747" Title="What description of the image can we expect from the image of first black hole?" Tags="&lt;black-hole&gt;&lt;telescope&gt;" AnswerCount="0" CommentCount="11" ClosedDate="2019-04-12T03:20:26.023" ContentLicense="CC BY-SA 4.0" />
  <row Id="30366" PostTypeId="2" ParentId="30354" CreationDate="2019-04-11T14:15:07.277" Score="2" Body="&lt;p&gt;One can never actually see the event horizon, one sees the gas that is about to enter the event horizon and gets hot along the way.  But this gas has a history, it doesn't just fall straight in.  The history imparts some angular momentum to it, like the water in a draining bathtub.  So it spirals in, not falling straight in, and this creates a disk shape called an accretion disk.  The disk does not have to be as thin as a pancake, but it's far from spherical, and that gets further distorted by the strong gravity (creating the brightness variations and the central dark sphere).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The disk does have a hole in the center because at some point the orbits go unstable and fall quickly below the event horizon.  It could be argued we are not actually seeing the event horizon where it goes black, we are seeing the drop in gas density where the orbits are no longer stable and the  slow spiral turns into a very rapid fall, but the way the strong gravity bends the light rays is always very important as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distortion by the strong gravity is what provides the test of general relativity that everyone is so excited about.  The strong gravity produces a &quot;shadow&quot; effect, so you don't have to see the accretion disk from above to get a spherical blackness at the center.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2019-04-11T14:21:53.040" LastActivityDate="2019-04-11T14:21:53.040" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30367" PostTypeId="2" ParentId="30339" CreationDate="2019-04-11T14:48:55.590" Score="15" Body="&lt;p&gt;Another quick note - They are trying to get a photo of Sag. A*:&lt;/p&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;https://www.space.com/first-black-hole-photo-by-event-horizon-telescope.html&quot; rel=&quot;noreferrer&quot;&gt;Space.com&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The project has been scrutinizing two black holes — the M87 behemoth, which harbors about 6.5 billion times the mass of Earth's sun, and our own Milky Way galaxy's central black hole, known as Sagittarius A*. This latter object, while still a supermassive black hole, is a runt compared to M87's beast, containing a mere 4.3 million solar masses.&lt;/p&gt;&#xA;&lt;p&gt;Both of these objects are tough targets because of their immense distance from Earth. Sagittarius A* lies about 26,000 light-years from us, and M87's black hole is a whopping 53.5 million light-years away.&lt;/p&gt;&#xA;&lt;p&gt;From our perspective, Sagittarius A*'s event horizon &amp;quot;is so small that it's the equivalent of seeing an orange on the moon or being able to read the newspaper in Los Angeles while you're sitting in New York City,&amp;quot; Doeleman said during the SXSW event last month.&lt;/p&gt;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&lt;p&gt;And in case you're wondering about Sagittarius A*: The EHT team hopes to get imagery of that supermassive black hole soon, Doeleman said today. The researchers looked at M87 first, and it's a bit easier to resolve than Sagittarius A* because it's less variable over short timescales, he explained.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10152" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-04-11T14:55:46.833" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30368" PostTypeId="1" CreationDate="2019-04-11T14:49:13.847" Score="0" ViewCount="163" Body="&lt;p&gt;In computer simulations of what a black hole might look like and many artist renderings of black holes, they all seem to show the brightly glowing gas BEHIND the black hole due to gravitational lensing.  In all such images (with a &quot;horizonal&quot; accretion disk) this light appears as a thin band on (the &quot;top&quot; or &quot;bottom&quot; of) the event horizon, that appears more and more intense as it gets closer to the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the recently released image of the black hole in M87, no such lensing effect is visible.  Rather, it looks more like a hazy version of plain ol'Saturnian rings.  Why is that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This link (&lt;a href=&quot;http://rantonels.github.io/starless/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://rantonels.github.io/starless/&lt;/a&gt;) contains several images, and some math on what I was expecting to see.  This animated gif of a black hole from various angles, I think, best shows what I was expecting:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/6jQw1.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6jQw1.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26576" LastEditorUserId="26576" LastEditDate="2019-04-11T15:05:10.050" LastActivityDate="2019-04-12T22:12:49.667" Title="Why don't we see the gas behind the black hole?" Tags="&lt;black-hole&gt;&lt;gravitational-lensing&gt;&lt;image-processing&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30369" PostTypeId="2" ParentId="30361" CreationDate="2019-04-11T15:39:39.530" Score="2" Body="&lt;p&gt;This process is described in great detail in &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/ab0e85&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;. The abstract says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To assess the reliability of these results, we implemented a two-stage&#xA;  imaging procedure. In the first stage, four teams, each blind to the&#xA;  others' work, produced images of M87 using both an established method&#xA;  (CLEAN) and a newer technique (regularized maximum likelihood). This&#xA;  stage allowed us to avoid shared human bias and to assess common&#xA;  features among independent reconstructions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They then generated a large number of &quot;plausible&quot; images, reconstructed the radio signals they would have given and ran them through the image analysis pipeline to see if it showed any biases.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the second stage, we reconstructed synthetic data from a large survey of imaging parameters and then compared the results with the corresponding ground truth images. This stage allowed us to select parameters objectively to use when reconstructing images of M87. Across all tests in both stages, the ring diameter and asymmetry remained stable, insensitive to the choice of imaging technique.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Incidentally, it's not really appropriate to think of the rejected data as image data in the sense you would get from a camera. It is (almost) recordings of the exact electromagnetic waveforms received by the telescopes. The image arises from subtle correlations between these datasets. So most of what is thrown away (apart from noise) is information about the exact phase of the radio signal on a nanosecond-by-nanosecond basis, which is not really interesting.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-04-11T15:39:39.530" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="30370" PostTypeId="1" AcceptedAnswerId="30378" CreationDate="2019-04-11T15:50:14.843" Score="5" ViewCount="478" Body="&lt;p&gt;I recently saw the first image of a black hole. As I understood, it is covered with bright, hot matter. In this case, how can we see the black disk (event horizon), instead of a bright disk due to the matter surrounding the black hole? Or is the matter around the black hole disposed in a disk shape and the astronomers got lucky that the disk was not oriented towards Earth?&lt;/p&gt;&#xA;" OwnerUserId="26578" LastActivityDate="2019-04-11T22:40:32.633" Title="Why isn't the black hole white?" Tags="&lt;black-hole&gt;&lt;matter&gt;&lt;disk&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30372" PostTypeId="2" ParentId="30356" CreationDate="2019-04-11T16:23:51.583" Score="1" Body="&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$\mathbf{\hat{n}}$&lt;/span&gt; points to the north celestial pole, then the eastward tangent vector is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathbf{\hat{u} = \frac{\hat{n} \times \hat{w}}{\|\hat{n} \times \hat{w}\|}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the northward tangent vector is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathbf{\hat{v} = \frac{\hat{w} \times \hat{u}}{\|\hat{w} \times \hat{u}\|}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\mathbf{\|\hat{w} \times \hat{u}\|}$&lt;/span&gt; analytically should be 1 but can be slightly different numerically.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2019-04-11T16:23:51.583" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30373" PostTypeId="1" AcceptedAnswerId="30376" CreationDate="2019-04-11T16:32:02.950" Score="4" ViewCount="500" Body="&lt;p&gt;What is the light source in the black hole images?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've &lt;a href=&quot;https://astronomy.stackexchange.com/search?q=%5Bblack-hole%5D%20light%20source%20is%3Aq&quot;&gt;searched the site&lt;/a&gt; before asking, as well as tried to google it. But perhaps the answer is buried due to the recent news.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding, is that the ring in the images is an illuminated accretion disc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I could not find explained is the light source, and by extension, why is a bright accretion disc – which is round by definition – a proof for light bending? (I'm not disputing the fact.)&lt;/p&gt;&#xA;" OwnerUserId="26579" LastActivityDate="2019-04-12T19:32:36.420" Title="What is the light source in the black hole images?" Tags="&lt;black-hole&gt;&lt;accretion-discs&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30374" PostTypeId="2" ParentId="30370" CreationDate="2019-04-11T16:56:31.027" Score="2" Body="&lt;p&gt;The black hole image that you saw is not a photograph in the traditional sense. A traditional photograph is created when visible light strikes a digital sensor or film in a camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image of M87 that you saw was created by an elaborate, complex, globe-spanning and labor-intensive operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To start with, the telescopes used were radio telescopes. These radio telescopes &quot;see&quot; radio waves with a 1.3mm wavelength like we see visible light, which has a wavelength from violet (380 nanometers) to red (700 nanometers). Think of it like infrared vision or night vision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eight radio telescopes all over the globe were pointed toward the same black hole at the same time over the course of four days. Each radio telescope collected an enormous amount of data by observing the black hole in the &lt;strong&gt;radio&lt;/strong&gt; spectrum, not the visible light spectrum. According to the &lt;a href=&quot;https://eventhorizontelescope.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Event Horizons Telescope website&lt;/a&gt;, each of the eight radio telescopes produced about 350 terabytes of data per day. This radio spectrum observation data was timestamped using an extremely accurate atomic clock at each location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data was then shipped to supercomputing centers to be processed by an algorithm. The computer algorithms work by lining up the data from the different telescopes using the accurate timestamps and by aligning the observed data spatially. This process is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Very-long-baseline_interferometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;very-long-baseline interferometry (VLBI)&lt;/a&gt;. The basic idea is that if you take two distinct observations of the same object from different locations and synchronize them. This helps to eliminate atmospheric noise by seeing what both different images have in common. While it may seem like a highly artificial process, it is very effective at eliminating noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image you see is therefore the result of a computer algorithm processing radio observations of the black hole from 8 different locations on earth. The choice of color was likely a choice made by the computer scientists who wrote the algorithm, but it appears to bear some relation to the intensity of the radio energies observed: more energy shows as brighter, while darker spots with less energy are deeper red or black. This choice of color might make sense because the strong gravity of a black hole will tend to &lt;a href=&quot;https://en.wikipedia.org/wiki/Redshift&quot; rel=&quot;nofollow noreferrer&quot;&gt;redshift&lt;/a&gt; any light emanating from nearby or passing close to the black hole. On the other hand, black holes are said to be enormously energetic in a broad spectrum of electromagnetic energy. First-hand accounts of nuclear explosions, which unleash X Rays and Gamma Rays and other far-ultraviolet energy have been described as having &lt;a href=&quot;https://www.newyorker.com/tech/annals-of-technology/the-first-light-of-the-trinity-atomic-test&quot; rel=&quot;nofollow noreferrer&quot;&gt;all kinds of exotic color&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this may seem a bit disappointing, it's worth noting that visible light does not propagate through space as well as radio. Radio penetrates dust and gas and all the intervening detritus in space much better. Visible light tends to be blocked and re-absorbed. Think of Wifi signals (radio) versus visible light. The Wifi penetrates walls, while the visible light gets blocked by walls, curtains, etc. The EHT website has some helpful &lt;a href=&quot;https://eventhorizontelescope.org/infographics&quot; rel=&quot;nofollow noreferrer&quot;&gt;infographics on VLBI&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16705" LastEditorUserId="16705" LastEditDate="2019-04-11T17:08:41.393" LastActivityDate="2019-04-11T17:08:41.393" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="30375" PostTypeId="2" ParentId="30373" CreationDate="2019-04-11T19:06:11.060" Score="2" Body="&lt;p&gt;Accretion disks are a complicated system which produces a wide variety of emission across the electromagnetic spectrum on a variety of timescales. The diagram below shows (some) of what is going on schematically, along with typical sizes scales (bars with arrows with sizes in parsecs (pc)). Details about how the &quot;central engine&quot; of supermassive black holes and AGN works and exactly how the energy is generated is an ongoing topic of research, which the Event Horizon Telescope images will very likely contribute to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HxlSd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HxlSd.png&quot; alt=&quot;Schematic diagram of an accretion disc&quot;&gt;&lt;/a&gt;&#xA;Friction and turbulence from the material orbiting the black hole causes the accretion disk to heat up and emit light. The inner parts of the accretion disk get the hottest and tend to emit at the shortest wavelengths (X-rays normally) with longer wavelength (optical, IR, radio) normally coming from the material further out and above the disk which is photo-ionized by the X-rays. Depending on the density of the material in and around the accretion disk, the angle we are viewing it from and whether the accretion disk flares up at larger radii, will control what the accretion disk/AGN will appear as which part of the spectrum will be seen most strongly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the sub-mm images produced by EHT, the emission is though to be coming from the ionized plasma closest to the inner edge of the accretion disk (the red region around the black hole in the bottom zoomed-in plot) via synchrotron emission as electrons are accelerated by the magnetic fields of the plasma. This is in contrast to the &lt;em&gt;jet&lt;/em&gt; radio emission which is on a larger scale and separation and often extends several galaxy radii which can be seen in e.g. &lt;a href=&quot;https://www.nrao.edu/pr/1999/m87big/layout.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this VLA/VLBI composite radio image&lt;/a&gt;.&#xA;There is some additional information on accretion disks and how they work at this &lt;a href=&quot;http://www.stronggravity.eu/public-outreach-tmp/accretion-disks/&quot; rel=&quot;nofollow noreferrer&quot;&gt;EU Strong Gravity research site&lt;/a&gt; and a more technical review article in &lt;a href=&quot;https://iopscience.iop.org/article/10.1088/1742-6596/548/1/012002/pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; conference proceedings.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastEditorUserId="23748" LastEditDate="2019-04-12T19:32:36.420" LastActivityDate="2019-04-12T19:32:36.420" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30376" PostTypeId="2" ParentId="30373" CreationDate="2019-04-11T19:45:26.127" Score="5" Body="&lt;p&gt;The radiation being detected in this case is mostly synchroton radiation, caused by energetic electrons spiralling around magnetic field line, rather than thermal radiation, but it comes from the same place (the disk). The actual evidence for light bending is, I believe the dark area in the middle, which is several times larger than the actual event horizon and is caused because the light that would have been coming to us from that direction has been bent away.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-04-11T19:45:26.127" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30377" PostTypeId="2" ParentId="30370" CreationDate="2019-04-11T19:48:55.520" Score="2" Body="&lt;p&gt;The black part in the centre of the image genuinely represents some directions from which less energy is arriving at the telescopes. I believe the intensity in the middle of it is about 10 times lower than the intensity in the bright ring around it.  So in that sense it really is a dark spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is described in the papers as the &quot;shadow&quot; of the black hole, although even that is stretching the word shadow a little. Basically the light that &quot;would have&quot; been coming to us from that direction has been bent away by gravity.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-04-11T19:48:55.520" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="30378" PostTypeId="2" ParentId="30370" CreationDate="2019-04-11T22:28:38.367" Score="5" Body="&lt;p&gt;What's going on here is that you have been misled into thinking the ring-like structure has anything to do with the accretion disk. It doesn't, or at least only indirectly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The disk is referred to as geometrically thick, but optically thin (see sections 1 and 2 of &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/ab0f43&quot; rel=&quot;noreferrer&quot;&gt;paper V&lt;/a&gt; issued by the Event Horizon Telescope collaboration on 10 April 2019). This is actually the &lt;em&gt;opposite&lt;/em&gt; of the disk visualised in the film &quot;Interstellar&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the disk is geometrically thick, it covers the whole picture. Because it is optically thin, we can see through it to the black hole. That is the basic answer to your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an optically thin plasma, the brightness you will see is proportional to the optical path length (physical length multiplied by an absorption coefficient) along that sightline. The photon ring marks radiation travelling towards us that has been bent around the black hole, or has even orbited it several times. Hence those sight lines have larger optical path lengths and that is why we see it as a bright ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radiation travelling towards us from plasma in front of the black hole, has a small optical path length and is not very bright. In addition, the sight lines to plasma behind the black hole cannot travel through the black hole or even close to the event horizon. Hence the circular &quot;shadow&quot; inside the photon ring.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-04-11T22:40:32.633" LastActivityDate="2019-04-11T22:40:32.633" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30379" PostTypeId="1" CreationDate="2019-04-12T00:54:47.057" Score="1" ViewCount="57" Body="&lt;p&gt;I have a spectrum of a galaxy from HST data and SDSS data (resulting in a nearly continuous SED from the FUV~1000 Angstroms through infrared). I have created a model SED with the galaxy simulation software &lt;code&gt;STARBURST99&lt;/code&gt; (for the correct star formation, age and type of evolution for the real galaxy) to extend from ~90 Angstroms through the infrared. I then reddened my model spectrum with the Cardelli et al. (1989) extinction law. I also have various photometric multi-band data from different surveys. I understand that I need to use this data to adjust my model spectrum so that it is a good fit of the observational SED and matches observations at each photometric point. But I'm confused on how I would go about this normalization/calibration process of my model spectrum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it as simple as linear offsets in flux at these various different points? I doubt that, however, because that would then seem difficult to keep my model spectrum smooth. I'm unsure if I really understand the process of proper SED fitting in the first place... Any help with this concept and how I could use photometric multi-band data to increase the goodness of fit of my model is greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="24076" LastActivityDate="2019-04-12T00:54:47.057" Title="How do you fit a model SED to an observed galaxy spectrum with photometric data?" Tags="&lt;galaxy&gt;&lt;spectroscopy&gt;&lt;photometry&gt;&lt;modeling&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30380" PostTypeId="1" AcceptedAnswerId="30399" CreationDate="2019-04-12T01:27:55.517" Score="5" ViewCount="245" Body="&lt;p&gt;This is the first of many similar questions I have to build up to a specific question or scenario that I want to explore and find an answer to.&lt;/p&gt;&#xA;" OwnerUserId="26596" LastEditorUserId="21928" LastEditDate="2019-04-12T12:19:44.587" LastActivityDate="2019-05-29T06:18:35.397" Title="Can a star fall in a super massive black hole without getting destroyed?" Tags="&lt;star&gt;&lt;supermassive-black-hole&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="30381" PostTypeId="2" ParentId="30380" CreationDate="2019-04-12T03:14:30.950" Score="-5" Body="&lt;p&gt;This is a great question! Simply put, we do not know. However, I would say that the star in question would likely be pulled apart therefore leaving the star structure destroyed but the energy of the star still intact. Imagine a mirror being shattered! The pieces (energy) of that mirror (STAR) are still existent, however the mirror is shattered (destroyed). I hope this makes sense? The energy of the star would likely continue to exist, but the formation of the star would likely be altered.&lt;/p&gt;&#xA;" OwnerUserId="26485" LastActivityDate="2019-04-12T03:14:30.950" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30383" PostTypeId="1" CreationDate="2019-04-12T05:01:33.367" Score="5" ViewCount="102" Body="&lt;p&gt;Event horizon achieved earth-size telescope by using  seperate telescopes around the world to interpolate an image of a black hole at resolution of 60 arcmicroseconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What resolution could we get if we built radio telescopes on the moon and used earth-moon orbit to interpolate an image? What could we see with it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if we used earth-moon-mars telescope?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it make sense to go even bigger?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all assuming we would have possibilities to build local telescopes like ones on earth in future.&lt;img src=&quot;https://i.stack.imgur.com/yyaO6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="26602" LastActivityDate="2019-04-12T08:23:22.380" Title="What resolution would earth-moon telescope achieve?" Tags="&lt;black-hole&gt;&lt;angular-resolution&gt;&lt;event-horizon&gt;&lt;very-large-telescope&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30384" PostTypeId="1" AcceptedAnswerId="30387" CreationDate="2019-04-12T05:03:56.647" Score="1" ViewCount="126" Body="&lt;p&gt;10 seconds before the M87 black hole was a black hole, what was it? A star 10 times the size of our solar system?&lt;/p&gt;&#xA;" OwnerUserId="26603" LastActivityDate="2019-04-12T05:33:46.533" Title="M87. What was the black hole before?" Tags="&lt;black-hole&gt;&lt;observational-astronomy&gt;&lt;astrophysics&gt;&lt;supermassive-black-hole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30385" PostTypeId="1" CreationDate="2019-04-12T05:15:22.907" Score="1" ViewCount="39" Body="&lt;p&gt;Could a binary star system have an earth-like planet orbiting star A, a second earth-like planet orbiting star B, AND a third earth-like planet in a stable circumbinary orbit? Or does the space that is required between the two stars to allow for stable habitable zone orbits for the first two planets eliminate the possibility of the third planet being both habitable and in a stable circumbinary orbit?&lt;/p&gt;&#xA;" OwnerUserId="26604" LastActivityDate="2019-04-12T05:15:22.907" Title="Are circumbinary and noncircumbinary habitable zones mutually exclusive?" Tags="&lt;orbit&gt;&lt;earth-like-planet&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30386" PostTypeId="2" ParentId="30304" CreationDate="2019-04-12T05:26:09.090" Score="-2" Body="&lt;p&gt;A single telescope does not have the right angle size to picture the black hole that far away. The scientists had to use 8 telescopes to take a picture of the same point in the sky and merge the data from all sites using super computer. The data retrieval must point to the same object at the same time. Given the diverse condition of the 8 locations, there are only 5 days in a year when all of them have clear enough skies to point to the same location in a year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Event_Horizon_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Event Horizon Telescope&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26605" LastEditorUserId="7472" LastEditDate="2019-04-12T05:34:14.533" LastActivityDate="2019-04-12T05:34:14.533" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="30387" PostTypeId="2" ParentId="30384" CreationDate="2019-04-12T05:33:46.533" Score="2" Body="&lt;p&gt;We don't know (yet). It could have started with the explosion of a very large star, or a primordial black hole formed very soon after the Big Bang. See &lt;a href=&quot;https://en.wikipedia.org/wiki/Supermassive_black_hole#Formation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The origin of supermassive black holes remains an open field of research. Astrophysicists agree that once a black hole is in place in the center of a galaxy, it can grow by accretion of matter and by merging with other black holes. There are, however, several hypotheses for the formation mechanisms and initial masses of the progenitors, or &amp;quot;seeds&amp;quot;, of supermassive black holes.&lt;/p&gt;&#xA;&lt;p&gt;One hypothesis is that the seeds are black holes of tens or perhaps hundreds of solar masses that are left behind by the &lt;strong&gt;explosions of massive stars&lt;/strong&gt; and grow by accretion of matter. Another model hypothesizes that before the first stars, large gas clouds could collapse into a &amp;quot;quasi-star&amp;quot;, which would in turn collapse into a black hole of around 20 M☉. These stars may have also been formed by dark matter halos drawing in enormous amounts of gas by gravity, which would then produce supermassive stars with tens of thousands of solar masses. The &amp;quot;quasi-star&amp;quot; becomes unstable to radial perturbations because of electron-positron pair production in its core and could collapse directly into a black hole without a supernova explosion (which would eject most of its mass, preventing the black hole from growing as fast). Given sufficient mass nearby, the black hole could accrete to become an intermediate-mass black hole and possibly a SMBH if the accretion rate persists.&lt;/p&gt;&#xA;&lt;p&gt;Another model involves a &lt;strong&gt;dense stellar cluster undergoing core-collapse&lt;/strong&gt; as the negative heat capacity of the system drives the velocity dispersion in the core to relativistic speeds. Finally, &lt;strong&gt;primordial black holes&lt;/strong&gt; could have been produced directly from external pressure in the first moments after the Big Bang. These primordial black holes would then have more time than any of the above models to accrete, allowing them sufficient time to reach supermassive sizes. Formation of black holes from the deaths of the first stars has been extensively studied and corroborated by observations. The other models for black hole formation listed above are theoretical.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;sub&gt;(emphasis mine)&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="7472" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-04-12T05:33:46.533" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30388" PostTypeId="2" ParentId="30383" CreationDate="2019-04-12T05:48:51.740" Score="2" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_resolution#Telescope_array&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, the angular resolution is inversely proportional to the baseline (the separation between the telescopes). The order of magnitude of the baseline of the Event Horizon Telescope is 10,000 km; the Earth-Moon distance is about 40 times larger, and Earth-Mars at its maximum 25,000 times larger. That means we could theoretically reach 1.5 arcmicroseconds (at this wavelength) with an additional telescope at the Moon, and 2.4 arcnanoseconds with Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, to obtain proper interferometry we need (regardless of the baseline size) to know the telescopes' precise location with the precision of a quarter of the wavelength. For 230 GHz radio waves, the wavelength is 1.3 mm, so the precision needs to be 0.3 mm. I suppose that this is incredibly difficult to reach for bodies moving through space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not well-versed enough in the matter to say what the effect of arranging the telescopes in a &lt;em&gt;line&lt;/em&gt; (on Earth-Moon or Earth-Mars scale, it's a line, unless you add space telescopes at e.g. L4 and L5 Lagrange points) vs. a &lt;em&gt;grid&lt;/em&gt; like the Event Horizon Telescope is. But interferometry does work for two telescopes already.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that angular resolution is not the only metric you should consider; objects become fainter when you enlarge them, so you'll need more telescopes (and/or a longer exposure time) to get the same level of brightness in the pictures.&lt;/p&gt;&#xA;" OwnerUserId="7472" LastEditorUserId="7472" LastEditDate="2019-04-12T06:07:37.087" LastActivityDate="2019-04-12T06:07:37.087" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30390" PostTypeId="1" AcceptedAnswerId="30394" CreationDate="2019-04-12T07:52:46.350" Score="3" ViewCount="106" Body="&lt;p&gt;By now we are all familiar with the official image of the black hole in M87:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0kabc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0kabc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in the universe, as far as I know, there is no up and down, nor left and right. All is relative. Hence my question: why the particular chosen angle of the image released? Why not instead release the image as, for example, below? Which is the point of reference chosen in the original image? Is it of any particular significance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z9IyN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z9IyN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26608" LastActivityDate="2019-04-12T08:38:33.577" Title="Why the chosen orientation for the M87 black hole image?" Tags="&lt;black-hole&gt;&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30392" PostTypeId="2" ParentId="30383" CreationDate="2019-04-12T08:23:22.380" Score="1" Body="&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Spektr-R&quot; rel=&quot;nofollow noreferrer&quot;&gt;space based radio telescope&lt;/a&gt; has been tried, although I can see little evidence of it being successful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get an image you do need baselines at multiple angles, not just a single line. This can be achieved on Earth with just two telescopes by combining measurements made at different times as the Earth rotates, and an Earth-Moon baseline could be treated similarly over a month, although this would require the image not to change over that time and be limited to objects in the Northern or Southern parts of the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could probably get direct line-of-sight high-bandwidth laser and microwave communications between the telescopes for a significant part of each day, so some of the synchronisation problems would actually be easier than for Earth-based VLBI. &lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-04-12T08:23:22.380" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30393" PostTypeId="1" CreationDate="2019-04-12T08:25:50.503" Score="4" ViewCount="140" Body="&lt;p&gt;I'm referring to the infamous relativistic jet, which consists of matter being accelerated to near the speed of light, originating at the poles of the black hole: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zKnOr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zKnOr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the image of the black hole in the radio spectrum nothing of the sort is visible:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MTf2C.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MTf2C.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would expect the relativistic jet to be one of the most prominent features since it's accelerated much more that other material orbiting the black hole.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because the jet is more prominently visible in the visible and infrared spectrum than in the radio spectrum?&lt;/p&gt;&#xA;" OwnerUserId="24234" LastActivityDate="2019-04-12T08:25:50.503" Title="Why isn't the relativistic jet visible in the image of the M87 black hole?" Tags="&lt;black-hole&gt;&lt;radio-astronomy&gt;&lt;m87&gt;&lt;astronomical-jet&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30394" PostTypeId="2" ParentId="30390" CreationDate="2019-04-12T08:38:33.577" Score="3" Body="&lt;p&gt;North is to the top and East is to the left. I believe it is a standard image in celestial coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to standardise so that the image can be compared with large scale features like the M87 jet.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-04-12T08:38:33.577" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30395" PostTypeId="1" CreationDate="2019-04-12T08:40:35.833" Score="3" ViewCount="85" Body="&lt;p&gt;For example, &lt;span class=&quot;math-container&quot;&gt;$0.1c &amp;lt; v &amp;lt; c$&lt;/span&gt;? Is this actually possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Background: is that true that the black hole in the center of the Milky Way rotates once per 20 minutes?&lt;/p&gt;&#xA;" OwnerUserId="22664" LastEditorUserId="22664" LastEditDate="2019-04-12T13:01:03.293" LastActivityDate="2019-04-12T13:01:03.293" Title="Are there known objects rotating at close-to-light speeds?" Tags="&lt;rotation&gt;&lt;speed&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30399" PostTypeId="2" ParentId="30380" CreationDate="2019-04-12T10:53:52.653" Score="8" Body="&lt;p&gt;Yes, it is easily &lt;em&gt;possible&lt;/em&gt;, but it depends critically on how massive the black hole is. A simple calculation will suffice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take a Newtonian approximation for the tidal acceleration across a star of radius &lt;span class=&quot;math-container&quot;&gt;$R_*$&lt;/span&gt; as it reaches the event horizon&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ a_{\rm tidal} \sim 2\frac{GM_{\rm BH}R_*}{r_s^3}\ ,$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$r_s = 2GM_{\rm BH}/c^2$&lt;/span&gt; is the Schwarzschild radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational acceleration at the surface of the star, due to its own mass is just&#xA;&lt;span class=&quot;math-container&quot;&gt;$$a_* = \frac{GM_*}{R_*^2}\ .$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take the ratio of the two, we can assume the star will survive being ripped up before it crosses the event horizon if this ratio is less than &lt;span class=&quot;math-container&quot;&gt;$\sim 1$&lt;/span&gt;. i.e. when&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{a_{\rm tidal}}{a_*}= \frac{c^6 R_*^3}{4G^3 M_{\rm BH}^2 M_*} &amp;lt;1$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tidal acceleration reduces markedly for high mass black holes. We can rearrange this inequality to give us the minimum black hole mass for survival:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ M_{\rm BH} &amp;gt; \left(\frac{c^6 R_*^3}{4G^3 M_*}\right)^{0.5} = 1.5\times 10^{8} \left(\frac{M_*}{M_{\odot}}\right)^{-1/2} \left(\frac{R_*}{R_{\odot}}\right)^{3/2}\ M_{\odot}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To first order, this is almost independent of the type of star swallowed. It depends on the inverse square root of the average stellar density, which doesn't vary a lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that a star could fall into a black hole like that in M87 (&lt;span class=&quot;math-container&quot;&gt;$M_{\rm BH}\simeq 6\times 10^9 M_{\odot}$&lt;/span&gt;), but  would not survive to the event horizon of Sgr A* (&lt;span class=&quot;math-container&quot;&gt;$M_{\rm BH}\simeq 4\times 10^6 M_{\odot}$&lt;/span&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think in practice this calculation can only be accurate to factors of a few. I have not taken into account shearing forces that would be present for a rotating black hole, or the compression due to tangential tidal forces, or any heating effects from hot gas near the black hole, or interaction with an accretion disk. A full hydrodynamic simulation is needed (and perhaps has been done?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note, this calculation is just for getting past the event horizon. The tidal forces grow as &lt;span class=&quot;math-container&quot;&gt;$r^{-3}$&lt;/span&gt;, so any normal star will be ripped up shortly after this, even for the most massive black holes we know of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note: This answer is written from the point of view of the falling star. A distant observer will not see the star (or anything else) cross the event horizon.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-05-29T06:18:35.397" LastActivityDate="2019-05-29T06:18:35.397" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="30400" PostTypeId="2" ParentId="30395" CreationDate="2019-04-12T10:54:52.920" Score="2" Body="&lt;p&gt;A particle travelling in a circle of radius &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; at velocity &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; has an acceleration of &lt;span class=&quot;math-container&quot;&gt;$v^2/R$&lt;/span&gt;. So either &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; is very large, or a very strong force is acting to stop the particles at the surface of the object flying away. We see no very large objects (galactic clusters, for instance) rotating this quickly, so we need to look for strong forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One option is gravity. We know of sub-millisecond pulsars, which we believe are neutron stars with a radius of perhaps 10 km, rotating 1000+ times per second. This gives a surface velocity at the equator of about &lt;span class=&quot;math-container&quot;&gt;$60\,000 km/s$&lt;/span&gt; which is &lt;span class=&quot;math-container&quot;&gt;$0.2c$&lt;/span&gt;. So those meet your conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Defining how fast the surface of a black hole is rotating is difficult on several levels, but black holes do have angular momentum, and there is a dimensionless number relating that to their mass (or equivalently radius) which can be loosely thought of as the fraction of light-speed at which the event horizon is moving. &lt;a href=&quot;https://academic.oup.com/mnrasl/article/403/1/L74/1187293&quot; rel=&quot;nofollow noreferrer&quot;&gt;This paper&lt;/a&gt; measures a value of about 0.44 for that parameter for the black hole in the centre of the milky way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the other extreme, &lt;a href=&quot;https://physics.stackexchange.com/questions/20187/how-fast-do-electrons-travel-in-an-atomic-orbital&quot;&gt;this stackexchange answer&lt;/a&gt; suggests that the electrons in a hydrogen atom rotate about the nucleus at something a bit less than &lt;span class=&quot;math-container&quot;&gt;$0.01c$&lt;/span&gt;. This scales up linearly with the charge on the nucleus, so the inner electrons in a lead or uranium atom are in some sense rotating at close to light-speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the particles in a large particle accelerator, rotate around the centre of the accelerator at velocities very close to that of light.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-04-12T10:54:52.920" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30403" PostTypeId="1" CreationDate="2019-04-12T13:58:13.407" Score="1" ViewCount="634" Body="&lt;p&gt;Today we know that a black hole represents a region of space where the gravitational forces are very intense, this consequence of a concentration of matter in a very small region of space. On April 10 of this year, scientists from the EHT project (event Horizont Telescope) presented to the world the image that I believe everyone has seen. The first image of a black hole in the galaxy M87. We know that what we are really seeing is the &quot;shadow&quot; of the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the &quot;photon sphere&quot;, where light can orbit a black hole, is actually at &lt;span class=&quot;math-container&quot;&gt;$1.5R_s$&lt;/span&gt;, (Schwarzschild radius), whereas the observed ring is a factor of 2 wider than this. Why we not see a ring of light at the radius of the photon sphere?&lt;/p&gt;&#xA;" OwnerUserId="13867" LastEditorUserId="2531" LastEditDate="2019-04-12T14:16:30.753" LastActivityDate="2019-04-12T21:38:02.720" Title="Why is the ring of light around the M87 black hole bigger than the photon sphere?" Tags="&lt;black-hole&gt;&lt;m87&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30404" PostTypeId="2" ParentId="30403" CreationDate="2019-04-12T14:44:29.943" Score="-1" Body="&lt;p&gt;If the light is orbiting the black hole, then it isn't escaping to our telescopes!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What matters is how close to the black hole the plasma of the accretion disk goes, since that's what produces the light. There are lots of papers on this, and the numbers are usually around 5 Schwarzchild radii, together with effects like gravitational redshift relativistic beaming and so on, and the geometry of the paths followed by light, this leads to the image we see.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-04-12T14:44:29.943" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30405" PostTypeId="2" ParentId="30403" CreationDate="2019-04-12T15:46:39.753" Score="-1" Body="&lt;p&gt;The innermost stable orbit (ISCO) of a massive object in circular orbit around a black hole or any other spherically symmetric nonspinning mass  is located at &lt;span class=&quot;math-container&quot;&gt;$3\text{R}_s$&lt;/span&gt;. From &lt;a href=&quot;https://en.wikipedia.org/wiki/Innermost_stable_circular_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The ISCO plays an important role in black hole accretion disks since it marks the inner edge of the disk.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the image in question depicts radiation emanating from matter in the accretion disk the &quot;ring of light&quot; that is the accretion disk is where it is supposed to be.&lt;/p&gt;&#xA;" OwnerUserId="26403" LastActivityDate="2019-04-12T15:46:39.753" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30406" PostTypeId="1" AcceptedAnswerId="30515" CreationDate="2019-04-12T16:39:15.970" Score="11" ViewCount="283" Body="&lt;p&gt;The Event Horizon Telescope was made possible in order to observe the details of supermassive black holes. This took a huge amount of work installing extra telescopes and developing the hardware and software needed to do VLBI at such short wavelengths. They have achieved spatial resolution around 25 micro arc seconds, comparable to the astrometric data from the Gaia space probe, but optimised for very different purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is what else could they usefully observe? What scientific targets have interesting detail at that angular scale and emit enough radiation at that wavelength to be observable?&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-04-16T22:40:46.317" Title="What else could the Event Horizon Telescope Observe?" Tags="&lt;radio-astronomy&gt;&lt;event-horizon-telescope&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30407" PostTypeId="1" CreationDate="2019-04-12T17:02:06.550" Score="1" ViewCount="48" Body="&lt;p&gt;In the following EHT and GMVA diagram, what do the yellow and blue lines represent? I don't think they're physical communication lines (that would be unfeasible). Do they depict anything specific? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8iijm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8iijm.jpg&quot; alt=&quot;the locations of the participating telescopes of the Event Horizon Telescope (EHT) and the Global mm-VLBI Array (GMVA)&quot;&gt;&lt;/a&gt;&#xA;Source: &lt;a href=&quot;https://www.eso.org/public/images/ann17015a/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ESO/O. Furtak&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13784" LastActivityDate="2019-04-13T15:55:33.250" Title="What do the lines in this Event Horizon Telescope and Global mm-VLBI Array infographic represent?" Tags="&lt;event-horizon-telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30408" PostTypeId="1" AcceptedAnswerId="30412" CreationDate="2019-04-12T17:05:14.820" Score="1" ViewCount="73" Body="&lt;p&gt;My understanding is that if the surface of a black hole is holographic, then information is potentially emitted from a black hole, say due to Hawking radiation or some other means. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can say about this information?  Is it physically possible and/or feasible that this type of emission would embed information about what went into it?  Also, how does this measure up against the idea that the world lines inside the event horizon are completely disconnected from the outside?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some knowledge of the information paradox, but I was really hoping someone could  explain it in simple terms or point me to a prior explanation along those lines.&lt;/p&gt;&#xA;" OwnerUserId="26629" LastEditorUserId="7472" LastEditDate="2019-04-12T17:10:04.807" LastActivityDate="2019-04-12T21:08:00.140" Title="What does it mean when we say that information is potentially recoverable from a black hole" Tags="&lt;black-hole&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30409" PostTypeId="2" ParentId="30403" CreationDate="2019-04-12T17:20:45.250" Score="3" Body="&lt;p&gt;The apparent radius of something residing in a Schwarzschild metric, when viewed from infinity is given by&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ R_{\rm obs} = R \left(1 - \frac{R_s}{R}\right)^{-1/2}\ ,$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$R_s$&lt;/span&gt; is the Schwarzschild radius &lt;span class=&quot;math-container&quot;&gt;$2GM/c^2$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;em&gt;enlargement&lt;/em&gt; is due to gravitational lensing and the formula is correct down to the &quot;photon sphere&quot; at &lt;span class=&quot;math-container&quot;&gt;$R =1.5 R_s$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the light in the EHT image comes from the photon sphere. It is therefore &lt;em&gt;observed&lt;/em&gt; to come from a radius&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ R_{\rm obs} =\frac{3R_s}{2}\left(1 - \frac{2}{3}\right)^{-1/2} = \frac{\sqrt{27}}{2}R_s\ .$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are small (&amp;lt;10%) corrections to this for a spinning black hole governed by the Kerr metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sketch which illustrates why gravitational lensing leads to an enlarged image can be found in Pela's &lt;a href=&quot;https://astronomy.stackexchange.com/a/30360/5264&quot;&gt;answer to a related question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-04-12T21:38:02.720" LastActivityDate="2019-04-12T21:38:02.720" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30411" PostTypeId="1" CreationDate="2019-04-12T17:36:46.900" Score="3" ViewCount="215" Body="&lt;p&gt;I am quite impressed with the recently published image of M87* black hole. It appears that this image achieves spatial resolution well above and beyond what was available until today. But how high this resolution is, exactly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article &quot;&lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/ab0ec7/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;First M87 Event Horizon Telescope Results. I. The Shadow of the Supermassive Black Hole&lt;/a&gt;&quot; states that Event Horizon Telescope has theoretical diffraction-limit resolution of ~25 μas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Messier_87#Supermassive_black_hole&quot; rel=&quot;nofollow noreferrer&quot;&gt;Another wiki article&lt;/a&gt; states that M87* has Swartzschild radius of 5.9×10&lt;sup&gt;-4&lt;/sup&gt; parsecs. Combined with the distance of 16.4 million parsecs, this gives me angular size of 14.8 microarceconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even with the shadow of the black hole being 2.5 bigger that its radius, the resolution of a widely distributed picture seems to be much higher than 25 microarcseconds stated above. Is the image actually blown up well beyond its original resolution (i.e. becoming more a less an &quot;artistic impression&quot;), or Event Horizon Telescope is actually capable of higher resolutions (10 microarcseconds or less)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YIdm6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YIdm6.png&quot; alt=&quot;M87* image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26628" LastEditorUserId="26628" LastEditDate="2019-04-12T17:50:02.873" LastActivityDate="2019-04-12T19:16:30.940" Title="What is the actual resolution of Event Horizon Telescope?" Tags="&lt;supermassive-black-hole&gt;&lt;angular-resolution&gt;&lt;event-horizon-telescope&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30412" PostTypeId="2" ParentId="30408" CreationDate="2019-04-12T17:44:34.227" Score="2" Body="&lt;p&gt;The first thing to understand is that we have &lt;em&gt;no actual knowledge of how information and black holes interact.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we do have are theories that are well supported by experiment (General Relativity, which predicts Black Holes but which says nothing about information) and Quantum Mechanics (which says a lot about information (in this technical sense) but nothing about BHs.)  We also have a variety of theories, none of which have any substantial experimental support, and all of which are incomplete (which means we don't know if there's actually a viable theory there -- all we have today is bits and pieces that look promising).  One of the goals of all of these theories is to -- somehow -- meld GR and QM into a single coherent theory with greater explanatory power than the two separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very fair to say that some of the work which has been done to date which tries to link information and BHs is plausible and (at least) does not contradict any experimental evidence we have.  And some of them (like what Hawking used in the prediction of Hawking radiation) are such a careful application of QM in the GR context that most physicists would be surprised if they turned out to be incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line is that GR and QM are both amazing theories which yet cannot be completely correct -- they must both be subsumed in some more encompassing theory. (Or at least everyone is convinced of that.  I am too, BTW.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the mathematical underpinnings of QM and GR, physicists vaguely can see how some generalization of QM might, possibly be able to include GR someday.  But virtually no one can see how GR could be extended to include QM.  So the smart money -- just about all the money, actually -- is on extending QM in some way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the most basic features of QM is that it is linear and preserves information (in a certain well-defined, but technical sense).  If GR could be derived from some extension of QM, we'd expect that that extension of QM preserves information, and that GR also must preserve information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the face of it, GR by itself says that information which falls into a BH is lost. Zap. Gone. Never to return.  Therefor finding ways to get GR to preserve information may be an important clue to the ultimate combined theory.  And extensions of GR which preserve information seem more likely to be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the paradox is that BHs, which are predicted by our marvelously correct theory of gravity, GR, do not conserve information.  But we have reasons to strongly suspect that BH built from the hypothetical gravitational extension of QM &lt;em&gt;would&lt;/em&gt; preserve information.  Something's wrong here!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the plausible theories that have been developed to resolve this says that Hawking radiation -- which seems very plausible, and which carries energy away from a BH in contradiction of vanilla GR -- might also carry the &quot;lost&quot; information away, eliminating the paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But remember that there isn't &lt;em&gt;any&lt;/em&gt; experimental evidence for any of this.  It's all based on smart hard-working, competent scientists trying to use their brains and their intuition to guess what the undiscovered theory which merges GR and QM will predict and to turn it into consistent math.  It's certainly better than a random-assed guess, but it is a far cry from established science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep watching -- I hope that all this work will bear fruit and a real theory with experimental evidence behind it might emerge in our lifetimes.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2019-04-12T17:44:34.227" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30413" PostTypeId="1" AcceptedAnswerId="34885" CreationDate="2019-04-12T17:55:44.573" Score="5" ViewCount="226" Body="&lt;p&gt;Is there a safe distance in which a black hole can be observed with the human eye? What is a safe distance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen many computer simulated and artist renditions, and I am not sure if they depicted the true colors and shape of BH. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below I have a few of examples of what may a BH may look like rendered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JCeS1.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JCeS1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;How light from the accretion disk maybe refracted by gravitational lensing.&#xA;&lt;a href=&quot;http://www.physics.utah.edu/~bromley/blackhole/index.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.physics.utah.edu/~bromley/blackhole/index.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AcIix.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AcIix.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/F0Wl2.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F0Wl2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.bbc.com/news/science-environment-38937141&quot; rel=&quot;noreferrer&quot;&gt;https://www.bbc.com/news/science-environment-38937141&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FxEHM.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FxEHM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.nasa.gov/connect/chat/black_hole_chat.html&quot; rel=&quot;noreferrer&quot;&gt;https://www.nasa.gov/connect/chat/black_hole_chat.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should these models be mixed to make a more complete rendition of a BH?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2019-04-12T18:11:52.800" LastActivityDate="2020-08-25T14:39:31.517" Title="What is the actual visual characteristics of a black hole?" Tags="&lt;black-hole&gt;&lt;light&gt;&lt;gravitational-lensing&gt;&lt;modeling&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="30414" PostTypeId="2" ParentId="30411" CreationDate="2019-04-12T19:16:30.940" Score="2" Body="&lt;p&gt;The resolution of an interferometer at a wavelength &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\theta\sim\lambda/b$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; is the longest baseline (distance between two telescopes) in the array. &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/ab0ec7#apjlab0ec7s3&quot; rel=&quot;nofollow noreferrer&quot;&gt;According to the first of this week's six papers&lt;/a&gt;, with observations at &lt;span class=&quot;math-container&quot;&gt;$\lambda=1.3\text{ mm}$&lt;/span&gt;, the longest baseline was &lt;span class=&quot;math-container&quot;&gt;$b\approx10700\text{ km}$&lt;/span&gt;, giving us &lt;span class=&quot;math-container&quot;&gt;$\theta\sim25.1\mu\text{as}$&lt;/span&gt;. &lt;a href=&quot;https://eventhorizontelescope.org/building-larger-array&quot; rel=&quot;nofollow noreferrer&quot;&gt;The team expects this limit to be lowered even further&lt;/a&gt; with the addition of more observatories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture, of course, seems much finer. The above limit can be lowered a bit by algorithms like &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/ab0e85&quot; rel=&quot;nofollow noreferrer&quot;&gt;CLEAN or regularized maximum likelihood&lt;/a&gt;; this brings the resolution down to about &lt;span class=&quot;math-container&quot;&gt;$20\mu\text{as}$&lt;/span&gt;. This still doesn't explain why the image seems even finer than that figure; that comes from image synthesis techniques that smooth out the data, rather than leaving it in a coarser form.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2019-04-12T19:16:30.940" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30415" PostTypeId="2" ParentId="30344" CreationDate="2019-04-12T19:50:45.847" Score="2" Body="&lt;p&gt;Your top image is &lt;a href=&quot;https://www.sciencephoto.com/media/333661/view&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; from Dr Jean Lorre stock science images.  Just a photo showing stars and the optical jet with a 22 MB CCD camera through a small telescope.  It is unfortunate that he chose a false color brightness scale with black at the top intensity.  But, that black region does not correspond to the true blackness in the interferometric image which has resolution 100,000 times higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparison, here's a &lt;a href=&quot;https://www.sciencephoto.com/media/333744/view&quot; rel=&quot;nofollow noreferrer&quot;&gt;False-col. optical image of spiral galaxy NGC 1097&lt;/a&gt; by the same &quot;artist&quot;, also with the dark center where the visible light intensity is actually a maximum.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="7982" LastEditDate="2019-04-12T23:18:53.390" LastActivityDate="2019-04-12T23:18:53.390" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30416" PostTypeId="1" CreationDate="2019-04-12T20:27:34.197" Score="3" ViewCount="280" Body="&lt;p&gt;Simulations suggest the early Moon formed about 20,000 to 30,000 km away from Earth which is less than the current geosynchronous orbit of 42,164 km.  The Moon is now 384,000 km away from Earth.  Does that mean the moon approximated a geosynchronous orbit for a time on its journey outward to its current orbit?  I recognize because of the Earth's tilt of 6.7 degrees the Moon does not orbit around the Earth's equator and therefore a geostationary orbit wasn't possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the early Earth was a water world covered almost entirely with water at the time it formed, what would happen when the Moon approached geosynchronous orbit and applied its much more amplified gravitational force upon the Earth?  Would there be tidal bulges of water at both ends exposing land at the polar regions or would the water eventually make its way toward the Moon side possibly exposing or making a bulge of land at the opposite end of the Earth?&lt;/p&gt;&#xA;" OwnerUserId="26635" LastEditorUserId="26635" LastEditDate="2019-04-12T23:27:28.063" LastActivityDate="2019-04-13T01:45:32.050" Title="Did the Moon ever approximate a geosynchronous orbit above the early Earth?" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30417" PostTypeId="1" CreationDate="2019-04-12T20:40:57.483" Score="0" ViewCount="142" Body="&lt;p&gt;I have an (observed) SED of a galaxy, and am attempting to match model SEDs to it. To do this I need to be able to see its shape better, which would involve some sort of smoothing of the spectrum. I would like to do this in Python. I've looked through scipy's various smoothing tools (like UnivariateSpline), but don't know if that's on the right track since that would only be some best fit to my spectral data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also used the Iris software to perform an interpolation of my spectrum with 1000 bins and the smoothing option turned on, but I do not understand the computational process behind this. How might I accomplish this in Python?&lt;/p&gt;&#xA;" OwnerUserId="24076" LastActivityDate="2019-04-12T20:40:57.483" Title="Smoothing an SED with binning" Tags="&lt;spectra&gt;&lt;python&gt;&lt;post-processing&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30418" PostTypeId="2" ParentId="30408" CreationDate="2019-04-12T21:08:00.140" Score="1" Body="&lt;p&gt;Just to add to Mark's excellent answer, experimental verification of any information carried by Hawking radiation is not easy, and it may remain so indefinitely, even if we could visit a black hole in a spaceship to make up-close measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hawking radiation (if it exists) is &lt;em&gt;extremely&lt;/em&gt; difficult to detect, even if you are near the black hole, which isn't a healthy place to be if the black hole is active. And of course the radiation from the matter in the accretion disk will totally swamp the Hawking radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the SMBH M87* emits Hawking radiation with a mean temperature of &lt;span class=&quot;math-container&quot;&gt;$9.5\times 10^{-18}$&lt;/span&gt; kelvin, at a luminosity of &lt;span class=&quot;math-container&quot;&gt;$2.13\times 10^{-48}$&lt;/span&gt; watts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A stellar mass BH is warmer &amp;amp; brighter, but still way too dim for us to detect. Eg, a 10 solar mass BH emits at a temperature of around a nanokelvin, with a luminosity of &lt;span class=&quot;math-container&quot;&gt;$9\times 10^{-31}$&lt;/span&gt; watts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used &lt;a href=&quot;https://www.vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator&quot; rel=&quot;nofollow noreferrer&quot;&gt;this online Hawking radiation calculator&lt;/a&gt; to obtain those figures.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2019-04-12T21:08:00.140" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30419" PostTypeId="1" CreationDate="2019-04-12T22:12:28.933" Score="2" ViewCount="206" Body="&lt;p&gt;The &quot;Schwarzschild solution&quot; in &quot;Schwarzschild metric&quot; looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text -c^2d\tau^2=-\left(1-\frac{2GM}{rc^2}\right)c^2dt^2+\left(1-\frac{2GM}{rc^2}\right)^{-1}dr^2+r^2(d\theta^2+\sin^2\theta\,d\phi^2)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this metric you get certain values for the radius of the &quot;Schwarzschild radius&quot;, the &quot;photon sphere&quot; and the &quot;least stable circular orbit&quot;, among other things. When the Jet Propulsion Laboratory or other institutions are dealing with orbital dynamics in the solar system, they instead use the &quot;isotropic metric&quot; or at least a first order expansion of the isotropic metric to calculate relativistic effects on the orbits. Now the Schwarzschild solution in isotropic coordinates looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ -c^2d\tau^2 = -\left(\frac{1-GM/2rc^2}{1+GM/2rc^2}\right)^2c^2dt^2 +(1+GM/2rc^2)^4[dr^2 +r^2( d\theta^2 +\sin^2\theta d\phi^2)]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in isotropic coordinates the &quot;Schwarzschild radius&quot;, the &quot;photon sphere&quot;, the &quot;least stable circular orbit&quot; among other things take other values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1&lt;/strong&gt;: Now that we have the measurements from the &quot;Event Horizon Telescope&quot; on the black hole, can something be said about what coordinates that fits the data better?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2&lt;/strong&gt;: It might be that both metrics fit the data equally well if we change the mass of the central black hole. Do we get different values on the mass of the central black hole if we fit the data using the isotropic coordinates instead of the Schwarzschild coordinates?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;According to the Schwarzschild solution in Schwarzschild coordinates the photon sphere is located at an &quot;r-parameter&quot; of &lt;span class=&quot;math-container&quot;&gt;$3GM/c^2$&lt;/span&gt;. According to &lt;a href=&quot;https://astronomy.stackexchange.com/a/30409/26403&quot;&gt;this post&lt;/a&gt; when using the Schwarzschild solution in Schwarzschild coordinates the apparent radius (&lt;span class=&quot;math-container&quot;&gt;$r_{obs}$&lt;/span&gt;) of something residing in a spherically symmetric gravitational potential when viewed from infinity is: &lt;span class=&quot;math-container&quot;&gt;$$r_{obs}=r(1-\frac{2GM}{rc^2})^{-0.5} $$&lt;/span&gt;&#xA;According to this formula we will observe the photon sphere at a radial distance of &lt;span class=&quot;math-container&quot;&gt;$r_{obs}=3\sqrt{3}GM/c^2$&lt;/span&gt;. Now I assume astronomers estimate the  &quot;observed radial distance&quot;, &quot;&lt;span class=&quot;math-container&quot;&gt;$r_{obs}$&lt;/span&gt;&quot; from the center of the M87 black hole to the photon sphere around the M87 black hole by measuring the angular size of the observed ring together with some estimate of the distance from earth to M87. If you use the formula for the photon sphere radius in Schwarzschild coordinates together with the formula for gravitational lensing enlargement you end up with a formula for the central mass of M87 as: &lt;span class=&quot;math-container&quot;&gt;$M=\frac{r_{obs}c^2}{{G3\sqrt{3}}}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if isotropic coordinates are used instead you will get another expression for the &quot;r-parameter&quot; of the photon sphere (I do not know what that expression looks like) and another expression for the &quot;gravitational lensing enlargement effect&quot; that makes objects in a spherically symmetric gravitational field look larger when observed from infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What i specifically would like to know in the second question above is how much the mass of the central body will &quot;change&quot; if you keep &lt;span class=&quot;math-container&quot;&gt;$r_{obs}$&lt;/span&gt; constant but you use isotropic coordinates instead of Schwarzschild coordinates and in both cases equate the &quot;r-coordinate&quot; with radial distance.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am a bit confused about the concept that you can do various coordinate transformations of the &quot;r-parameter&quot; in the Schwarzschild solution and still get a solution that fits reality.&lt;/p&gt;&#xA;" OwnerUserId="26403" LastEditorUserId="26403" LastEditDate="2019-04-13T17:54:02.167" LastActivityDate="2019-04-13T17:54:02.167" Title="Do the &quot;schwarzschild metric&quot; or the &quot;isotropic metric&quot; fit the Event Horizon telescope data on Messier 87 better?" Tags="&lt;black-hole&gt;&lt;observational-astronomy&gt;&lt;general-relativity&gt;&lt;event-horizon-telescope&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30420" PostTypeId="2" ParentId="30368" CreationDate="2019-04-12T22:12:49.667" Score="3" Body="&lt;p&gt;The reason is that your animated gif shows a geometrically thin, optically thick accretion disk, whereas the disk around the M87 black hole is a geometrically thick, but optically thin disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main difference is that you do not see any evidence of the structure of the disk in the picture - it is nearly transparent. You do however see plenty of gravitational lensing - that is what the bright ring is. Many of our sightlines to hot plasma run through a small annulus at the photon sphere (at 1.5 times the Schwarzschild radius) making it bright. Because of lensing, the apparent size of this ring, as viewed from infinity, is enlarged to a radius of 2.6 times the Schwarzschild radius. Contrary to your question, we see photons emitted from all around the black hole in that bright ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many GRMHD (general relativistic magneto-hydrodynamic) simulations that &lt;em&gt;do&lt;/em&gt; agree with what is observed.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-04-12T22:12:49.667" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30424" PostTypeId="1" CreationDate="2019-04-13T00:37:15.143" Score="2" ViewCount="229" Body="&lt;p&gt;There's some discussion that the image composed by the Event Horizons Telescope is &lt;em&gt;really&lt;/em&gt; just an accretion disk. To &quot;take an image&quot; of something, you need the light reflecting off of the surface of the object within frame. The &quot;light&quot; in the image is really just color-mapped gases within the accretion disc. Since black holes do not allow light to escape, is this actually an image of a black hole?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analogy I've heard is that it would be like taking a picture of an object in a completely dark room, showing someone the photograph taken, and saying, &quot;Hey look at this picture I took of this object.&quot; The other person could ask, &quot;Where is the object in this entirely black photo?&quot; to which the other person might respond, &quot;It's somewhere in the picture.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was a picture of the object really taken if you can't even see it? If you photograph the sky, have you &quot;captured an image of air molecules?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/inpbk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/inpbk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11851" LastEditorUserId="37191" LastEditDate="2021-02-18T13:37:42.360" LastActivityDate="2021-02-18T15:31:03.603" Title="If a black hole does not emit light, how can one take a picture of the black hole itself?" Tags="&lt;black-hole&gt;&lt;light&gt;&lt;photography&gt;&lt;accretion-discs&gt;&lt;event-horizon-telescope&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30425" PostTypeId="1" AcceptedAnswerId="30438" CreationDate="2019-04-13T01:19:45.793" Score="0" ViewCount="179" Body="&lt;p&gt;How close of an orbit to the Sun could an iron meteor get before it melts into a ball? &lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2019-04-13T22:50:44.757" LastActivityDate="2019-07-18T12:34:05.433" Title="Could the Sun have liquid iron in orbit?" Tags="&lt;the-sun&gt;&lt;orbit&gt;&lt;meteor&gt;&lt;roche-limit&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30426" PostTypeId="2" ParentId="30416" CreationDate="2019-04-13T01:45:32.050" Score="2" Body="&lt;p&gt;Maybe. The period of a 20,000 km orbit is about 7h 50m. &lt;a href=&quot;https://en.wikipedia.org/wiki/Giant-impact_hypothesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[The Earth] would have experienced a day some five hours long after&#xA;  the impact, and the Earth's equator and the Moon's orbit would have&#xA;  become coplanar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the present era, the Moon's orbital plane is tilted by &#xA;5.145° to the ecliptic (the Earth's orbital plane); the obliquity of the ecliptic (relative to Earth's equatorial plane) is about 23.4°. However, as &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbit_of_the_Moon&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Moon's orbit around Earth has many irregularities (perturbations),&#xA;  the study of which (lunar theory) has a long history.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The post-impact Earth was very hot, and had a magma ocean for a while. However, it soon had liquid water again because the high atmospheric pressure pushed the boiling point of water way up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Hadean#Atmosphere_and_oceans&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[From the Theia impact] a fair fraction of material should have been&#xA;  vaporized by this impact, creating a rock vapor atmosphere around the&#xA;  young planet. The rock vapor would have condensed within two thousand&#xA;  years, leaving behind hot volatiles which probably resulted in a&#xA;  heavy &lt;span class=&quot;math-container&quot;&gt;$\mathrm{CO_2}$&lt;/span&gt; atmosphere with hydrogen and water vapor. Liquid water&#xA;  oceans existed despite the surface temperature of 230 °C (446 °F)&#xA;  because at an atmospheric pressure of above 27 atmospheres, caused by&#xA;  the heavy &lt;span class=&quot;math-container&quot;&gt;$\mathrm{CO_2}$&lt;/span&gt; atmosphere, water is still liquid. As cooling&#xA;  continued, subduction and dissolving in ocean water removed most &lt;span class=&quot;math-container&quot;&gt;$\mathrm{CO_2}$&lt;/span&gt;&#xA;  from the atmosphere but levels oscillated wildly as new surface and&#xA;  mantle cycles appeared.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is difficult to estimate the effects of the interaction between the early Moon and that primordial ocean.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2019-04-13T01:45:32.050" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30428" PostTypeId="2" ParentId="30419" CreationDate="2019-04-13T07:36:58.183" Score="7" Body="&lt;p&gt;One of the things that can be confusing for beginners to GR is that there is no physical significance to the choice of coordinates we make. For example when studying static black holes we can use &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_metric&quot; rel=&quot;noreferrer&quot;&gt;Schwarzschild coordinates&lt;/a&gt;, &lt;a href=&quot;https://physics.stackexchange.com/questions/145342/schwarzschild-metric-in-isotropic-coordinates&quot;&gt;isotropic coordinates&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates&quot; rel=&quot;noreferrer&quot;&gt;Gullstrand-Painlevé coordinates&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Eddington%E2%80%93Finkelstein_coordinates&quot; rel=&quot;noreferrer&quot;&gt;Eddington-Finkelstein coordinates&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Kruskal%E2%80%93Szekeres_coordinates&quot; rel=&quot;noreferrer&quot;&gt;Kruskal-Szekeres coordinates&lt;/a&gt; and lots of others that I can't think of offhand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But regardless of what coordinates we choose it's the same metric, just written in different ways. You ask about the &lt;em&gt;Schwarzschild metric&lt;/em&gt; and the &lt;em&gt;isotropic metric&lt;/em&gt; as if they were different things, but it's the same metric just written in different coordinates. It would be better to call them the &lt;em&gt;static black hole metric written in Schwarzschild coordinates&lt;/em&gt; and the &lt;em&gt;static black hole metric written in isotropic coordinates&lt;/em&gt; but, well, life is too short and even experienced physicists will just say &lt;em&gt;Schwarzschild metric&lt;/em&gt; and &lt;em&gt;isotropic metric&lt;/em&gt; to save time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that the coordinates are not directly observable. We choose some coordinate system and then do a calculation to compute something that is observable e.g. the trajectories of photons. And the results will come out the same. After all the photons don't care what coordinate system a physicist chooses to compute their paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason we have all these different coordinates is that it's often the case that some calculations are easier in some coordinates than others. For example the Schwarzschild coordinates are the closest to our intuitive grasp of the geometry around the black hole. If we were considering simple questions like the time dilation for an observer hovering near the black hole I'd use Schwarzschild coordinates to do this because they make the calculation easy. But Schwarzschild coordinates don't work well at the event horizon. If I was explaining why light can't escape from a black hole &lt;a href=&quot;https://physics.stackexchange.com/questions/28297/why-is-a-black-hole-black&quot;&gt;I'd use Gullstrand-Painlevé coordinates&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To return to your question, I'm not familiar with the method the JPL team use for computing trajectories in the Solar System so I can't comment on why the calculations would be easier in the isotropic coordinates than in the Schwarzschild coordinates. Presumably the computational method they use is easier in isotropic coordinates, but the observables that they calculate will come out the same regardless of which coordinates are used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case neither coordinates are useful for the EHT calculations because Messier 87* is rotating not static and the spacetime geometry around it is described by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kerr_metric&quot; rel=&quot;noreferrer&quot;&gt;Kerr metric&lt;/a&gt; not the Schwarzschild metric. The Kerr metric can also be written in lots of different coordinate systems - I don't know which coordinate system the EHT team used.&lt;/p&gt;&#xA;" OwnerUserId="10687" LastEditorUserId="16685" LastEditDate="2019-04-13T07:53:17.997" LastActivityDate="2019-04-13T07:53:17.997" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30429" PostTypeId="2" ParentId="30424" CreationDate="2019-04-13T08:04:17.287" Score="6" Body="&lt;p&gt;The picture is of the central region of M87, taken at a wavelength where the gas is &quot;optically thin&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ring of bright light is pretty much exactly where it is expected to be for the synchrotron radiation emitted by the hot gas to have been gravitationally lensed by a black hole with the same mass as deduced previously from the motion of stars close to the centre of M87.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It &lt;em&gt;is&lt;/em&gt; a &quot;picture of a black hole&quot;, though all you see is the way it warps the light around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your analogies are poor. It isn't like taking a picture of someone in a dark room. The room here is illuminated and the effects of the black hole on that light have apparently been seen quite clearly. That, by the way is the only method by which you can see someone in an illuminated room, unless you have eyes that detect the mid-infrared light they emit. Of course that option is not available for a black hole, since they emit no light at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your other analogy is also curious. A cursory glance at the sky reveals it is full of air molecules. That is why it is blue. Your view though lacks the angular resolution to see individual air molecules, so you would not be able to see them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the resolution of the &quot;telescope&quot; is just good enough to reveal the essential structure and size of the photon ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, why isn't it a picture of an accretion disk? As I said, it is expected that the disk is optically thin and actually much bigger than the imaged area. Emission from the disk would not be expected to show anything but a haze with a vague central concentration. Given the spatial resolution of the image, what has actually been seen is a sharp, bright ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The accretion disk around a black hole should be terminated at or slightly beyond the innermost stable circular orbit. This is at 3 times the Schwarzschild radius, but gravitational lensing would make it &lt;em&gt;appear&lt;/em&gt; to have an inner radius of 3.7 times the Schwarzschild radius. The bright ring has a radius of 2.6 Schwarzschild radii; correct for a lensed photon sphere explanation, incorrect for the inner wall of the accretion disk (unless the black hole is 70% of the mass previously determined from stellar motions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orientation might be just about ok for an accretion disk explanation. If the rotation axis is aligned with the jet of M87, the disk should be about 17 degrees from face on. However, in order to see the inner parts having such a circular shape and well defined structure would require a geometrically thin and optically thick disk, the opposite of what is deduced from the physical conditions in the plasma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see more context, here is a wider view of M87 taken by Chandra, and from the &lt;a href=&quot;http://chandra.si.edu/blog/node/719&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chandra team's blog post&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Chandra was used to observe M87 and other targets during the EHT campaign. While Chandra can’t see the shadow itself, its field of view is much larger than the EHT’s, so Chandra can view the full length of the jet of high-energy particles launched by the intense gravitational and magnetic fields around the black hole. This jet extends more than 1,000 light years from the center of the galaxy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mgGVB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mgGVB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="43" LastEditDate="2019-04-13T09:18:39.080" LastActivityDate="2019-04-13T09:18:39.080" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="30431" PostTypeId="1" AcceptedAnswerId="30432" CreationDate="2019-04-13T12:44:05.183" Score="6" ViewCount="421" Body="&lt;p&gt;While reading publications in astrophysics I've encountered this kind of notation (fictitious example):&#xA;&lt;span class=&quot;math-container&quot;&gt;$$M = 6.1^{+0.6}_{-0.1} \times 10^{30} \textrm{ kg}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several examples are &lt;a href=&quot;https://doi.org/10.3847/2041-8213/ab0ec7&quot; rel=&quot;nofollow noreferrer&quot;&gt;here, p. 2&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;em&gt;assume&lt;/em&gt; that the superscript and subscript refer to the right and left uncertainty of the main magnitude. I have three questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Can someone explain to me what these three numbers (main magnitude, superscript, subscript) refer to, in terms of the probability distribution for the true value? For example, are they the mean and 10% and 90% quantiles? Or maybe the median and 25% and 75% quantiles? Or have I misunderstood them completely?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can someone provide early (or the first) examples of this particular &lt;em&gt;superscript-subscript&lt;/em&gt; notation? I only knew the &quot;&lt;span class=&quot;math-container&quot;&gt;$x \pm \triangle x$&lt;/span&gt;&quot; notation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can someone provide a written reference – textbook, manual, article, or similar – where this notation is explained?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I checked &lt;a href=&quot;https://astronomy.stackexchange.com/q/8841/26647&quot;&gt;this question&lt;/a&gt;, but it didn't really help me. I also checked the &lt;a href=&quot;https://www.bipm.org/en/publications/guides/gum.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Guide to the Expression of Uncertainty in Measurement&lt;/em&gt;&lt;/a&gt;, published by the &lt;em&gt;Joint Committee for Guides in Metrology&lt;/em&gt; (JCGM), formed by various organizations (BIPM, IEC, IFCC, ILAC, ISO, IUPAC, IUPAP, OIML). It is a standard for nomenclature and practice in reporting uncertainty, and makes important distinctions, for example between &quot;coverage intervals&quot; and &quot;confidence intervals&quot; (see &lt;a href=&quot;https://www.bipm.org/utils/common/documents/jcgm/JCGM_101_2008_E.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;JCGM 101:2008&lt;/a&gt; definitions 3.12–3.16). But I can't find anything about the superscript-subscript notation there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers!&lt;/p&gt;&#xA;" OwnerUserId="26647" LastEditorUserId="26647" LastEditDate="2019-04-14T12:47:53.333" LastActivityDate="2021-05-06T17:24:02.200" Title="Superscript-subscript uncertainty notation" Tags="&lt;astrophysics&gt;&lt;magnitude&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30432" PostTypeId="2" ParentId="30431" CreationDate="2019-04-13T13:43:26.913" Score="5" Body="&lt;p&gt;When quoted like this they would usually be the most likely value (peak of the probability distribution) or the median value (where half the distribution is above or below), and then the superscripts and subscripts would be increments that would contain &lt;span class=&quot;math-container&quot;&gt;$\sim 34$&lt;/span&gt;% of the probability distribution. In the M87 papers, the numbers are the 50th percentile (median) and lower (upper) limits corresponding to the 16th (84th) percentile. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total range corresponds to about a 68% confidence interval. The reason for this is just that it corresponds to &lt;span class=&quot;math-container&quot;&gt;$\pm$&lt;/span&gt; one sigma in a normal (Gaussian) distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is good practice to say what your error bars mean, but in the absence of that I would assume the above.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-04-13T14:21:07.633" LastActivityDate="2019-04-13T14:21:07.633" CommentCount="19" ContentLicense="CC BY-SA 4.0" />
  <row Id="30433" PostTypeId="2" ParentId="30407" CreationDate="2019-04-13T14:37:11.937" Score="4" Body="&lt;p&gt;I cant see a definitive statement, but I can see some infiormation and make some reasonable suggestions:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The two colours represent the two different collaborations. Blue for the EHT, yellow for the GMVA. We can see this by following the links on the source page of the image. &lt;a href=&quot;http://www3.mpifr-bonn.mpg.de/div/vlbi/globalmm/&quot; rel=&quot;nofollow noreferrer&quot;&gt;The GMVA page&lt;/a&gt; says:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Participating Stations:&lt;/p&gt;&#xA;&lt;p&gt;in Europe: Effelsberg (100m (Ef)), Onsala (20m (On)), Pico Veleta (30m (Pv)), Metsahovi (14m (Mh)), and Yebes (40m (Ys))&lt;/p&gt;&#xA;&lt;p&gt;in USA: VLBA (8x25m) - BR, NL, PT, LA, FD, KP, OV, MK (HN, SC are not equipped with 3mm receivers) and the GBT (100m)&lt;/p&gt;&#xA;&lt;p&gt;ALMA: for VLBI experiments involving phased ALMA the sensitivity depends on the number of antennas which are phased. For N=50 antennas and a typical phasing efficiency of 0.90 a gain of 1.3 K/Jy is obtained, and the effective diameter is ~ 80 m.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This matches the yellow lines.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://eventhorizontelescope.org/array&quot; rel=&quot;nofollow noreferrer&quot;&gt;This page&lt;/a&gt; describes the EHT array and more or less matches the blue lines.&lt;/p&gt;&#xA;&lt;p&gt;So now what are the lines? I'm pretty sure they are &amp;quot;baselines&amp;quot;. That is they connect pairs of telescopes whose signals are directly combined in great detail to produce information about the  image. The data from the various baselines is then combined in a later step. &lt;a href=&quot;https://en.wikipedia.org/wiki/Very-long-baseline_interferometry#How_VLBI_works&quot; rel=&quot;nofollow noreferrer&quot;&gt;This wikipedia article&lt;/a&gt; describes some of the physics. &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/ab0c57&quot; rel=&quot;nofollow noreferrer&quot;&gt;This paper&lt;/a&gt; refers to the individual baselines used for the EHT and their contribution to the image.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-04-13T15:55:33.250" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30434" PostTypeId="2" ParentId="30355" CreationDate="2019-04-13T21:46:21.180" Score="3" Body="&lt;p&gt;Given a date and time, the position of the Moon can be calculated to provide the declination and right ascension. The sub-point of the Moon (the point on the Earth at which the Moon is at the zenith) is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;latitude = declination of the Moon&lt;/li&gt;&#xA;&lt;li&gt;longitude can be found by calculating the local mean sidereal time (LMST) that equals the Moon's right ascension. (LMST dependents on the date, time, and longitude.) Everything is known except for the longitude.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Accurate Method (using GMST)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate the Local Mean Sidereal Time (LMST) from &lt;span class=&quot;math-container&quot;&gt;$LMST=GMST+long_{east}$&lt;/span&gt;, where GMST is the Greenwich Mean Sidereal Time and &lt;span class=&quot;math-container&quot;&gt;$long_{east}$&lt;/span&gt; is the longitude with positive values in the eastern hemisphere. From the post &lt;a href=&quot;https://astronomy.stackexchange.com/questions/24859/local-sidereal-time&quot;&gt;Local Sidereal Time&lt;/a&gt;, &#xA;&lt;span class=&quot;math-container&quot;&gt;$\mathit{GMST}_{\text{deg}} = 100.4606184 + 0.9856473662862 D + 15 H&#xA;$&lt;/span&gt; where D is the number of days (including the fraction of days) from J2000 (Jan 1, 2000 at 12 h UT = Julian Day 2,451,545.0) and H is the Universal Time (UT) in hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, at H=17 hour UT on Nov 1, 2000, I calculate the following values:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;RA_{Moon}=18h\;49m\;35s = 282.400° \\&#xA;Julian Day = 2,451,850.208 \\&#xA;\therefore D=2,451,850.208 - 2,451,545=305.209 \\&#xA;\therefore GMST = 296.288° \\&#xA;$$&lt;/span&gt;&#xA;Then&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;RA_{Moon}=LMST=GMST+long_{East} \\&#xA;282.400=296.288+long_{East} \\&#xA;\therefore long_{East}=-13.888°&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Approximate Method (using Sun's position)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your method of using the Sun's right ascension assumes that the Sun is over 0° longitude at 12 h UT. Written as a formula and allowing for different times and longitudes, it would be &lt;span class=&quot;math-container&quot;&gt;$LMST=RA_{sun}+15(Time \;in \;UT - 12h)+long_{east}$&lt;/span&gt;. Setting this LMST to the right ascension of the Moon gives the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;RA_{Moon}=LMST=RA_{sun}+15(Time \;in \;UT - 12h)+long_{east} \\&#xA;\therefore long_{East}=RA_{moon}-RA_{sun}-15(Time \;in \;UT - 12h)&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this is an approximation is because of the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The Sun is not at 0° longitude at 12:00:00 UT everyday because of the Equation of Time. The Sun can be off by up to 16 minutes of time (which is equivalent to 4° of longitude).&lt;/li&gt;&#xA;&lt;li&gt;The term 15(Time in UT - 12h) should be 15*1.002738*(Time in UT - 12 h), but this difference is small compared to the first approximation.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Continuing the example of Nov 1, 2000, I calculate the following:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;RA_{sun}=14h\;28m\;46s = 217.19° \\&#xA;long_{East}=RA_{moon}-RA_{sun}-15(Time \;in \;UT - 12h) \\&#xA;long_{East}=282.400-217.19-15(17-12) \\&#xA;\therefore long_{East}=-9.79°&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is different by 4° because I chose Nov 1 since it is close to the date of the largest Equation of Time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, if you wanted to calculate the sub-point of the Sun, it would be better to use the GMST and the Sun's right ascension by solving for the longitude from &lt;span class=&quot;math-container&quot;&gt;$RA_{sun}=LMST=GMST+long_{East}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2019-04-13T21:46:21.180" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30435" PostTypeId="2" ParentId="30339" CreationDate="2019-04-13T22:46:34.420" Score="0" Body="&lt;p&gt;The difficulty in imaging anything on the visible-light spectrum is dust.  Sagitarrius A is clouded by dust clouds which can be penetrated with infrared. The M87 fulfilled the criteria of being big and relatively close while at the same time enabling light to reflect off the event horizon and not being blocked by dust clouds.&lt;/p&gt;&#xA;" OwnerUserId="22963" LastActivityDate="2019-04-13T22:46:34.420" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30437" PostTypeId="1" AcceptedAnswerId="30450" CreationDate="2019-04-14T01:14:12.477" Score="4" ViewCount="123" Body="&lt;p&gt;I calculated the angular diameter of the Sun at the heliopause to be 0.004°.  If that is correct, would the Sun appear as a disk, or like the pinpoint of light from a star?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How bright would the light from the Sun appear?  I understand at Pluto the Sun appears as bright as the full moon at night on Earth.  Assuming the heliopause is 3x distant from the Sun as Pluto, does that mean there would be 1/9th as bright at the heliopause.  What would that brightness translate to an everyday measure?&lt;/p&gt;&#xA;" OwnerUserId="25088" LastActivityDate="2019-04-14T14:29:21.097" Title="What does the Sun look like from the heliopause?" Tags="&lt;the-sun&gt;&lt;angular-diameter&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30438" PostTypeId="2" ParentId="30425" CreationDate="2019-04-14T02:33:16.693" Score="2" Body="&lt;p&gt;As &lt;a href=&quot;https://books.google.de/books?id=c-v7OSaxGSsC&amp;amp;pg=PA112&amp;amp;lpg=PA112&amp;amp;dq=iron+albedo&amp;amp;source=bl&amp;amp;ots=XV_S83V-3f&amp;amp;sa=X#v=onepage&amp;amp;q=iron%20albedo&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; source stats, galvanized building iron has an albedo of 0.35&lt;sup&gt;1&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, we can assume, that a half of its total surface is shined by the Sun, with an average 45% angle. That means that it gets the radiation of the 6000 K Sun on &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{2} \cdot \frac{1}{\sqrt{2}} \approx 35\%$&lt;/span&gt;, as if it would be shined directly on its whole surface. It radiates back 35% of it, the remaining 65% heats it. That means, in its equilibrial temperature, the iron meteor radiates &lt;span class=&quot;math-container&quot;&gt;$\approx 23\%$&lt;/span&gt; in all direction due to its temperature, than it gets from the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, being &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; solar radii away from the Sun, it also gets the per-surface thermal energy of &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{n^2}$&lt;/span&gt;. Here I committed yet another &quot;approximation&quot;, because the Sun is not a pointlike radiator if &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is small, and the final result will be probably small, but it won't be too big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thermal radiation is proportional roughly with the 4th power of the temperature.&lt;sup&gt;2&lt;/sup&gt; Thus, &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; solar radii away, it gets &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{n^2}$&lt;/span&gt; per-surface solar heating, it radiates away &lt;span class=&quot;math-container&quot;&gt;$0.23\frac{1}{n^2}$&lt;/span&gt;, what makes its equilibrial temperature &lt;span class=&quot;math-container&quot;&gt;$\sqrt[4]{0.23 \cdot \frac{1}{n^2}} = 0.69\frac{1}{\sqrt n}$&lt;/span&gt; of the Sun's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average surface temperature of the Sun is 5778 K, the meltig point of iron is 1881K. Thus,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$5778\ \mathrm K \cdot 0.69\frac{1}{\sqrt n} = 1881\ \mathrm K$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{1}{\sqrt n}=\frac{1881\ \mathrm K}{5778\ \mathrm K\cdot 0.69}=0.47$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$n=\frac{1}{0.47^2} \approx \underline{\underline {4.5}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is: 4.5 solar radius from the solar center, which is &lt;span class=&quot;math-container&quot;&gt;$\approx$&lt;/span&gt; 1.8 solar diameter from the surface. It is far deeper than the orbit of the Mercury. As a comparison, the &lt;a href=&quot;http://parkersolarprobe.jhuapl.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Parker Solar Probe&lt;/a&gt; will near the Sun 8 solar radii away. A 2000 K maximal temperature is predicted on its heat shield (note, that probe won't have an equilibrial temperature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/suABms.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/suABms.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another remark: liquid things in vacuum evaporate. The evaporation for molten metals near their melting point is slow, but it exists. Such an object, if it does not have something to keep the iron vapor around it (f.e., gravitation), will slowly disappear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;&lt;sub&gt;Surprisingly, I did not found better data with google.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2&lt;/sup&gt;&lt;sub&gt;Also this is not exact, because it is true only for absolute black body radiators, but the spectrum of most ordinary materials does not cause a huge error in the calculation.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorDisplayName="user8021" LastEditDate="2019-07-18T12:34:05.433" LastActivityDate="2019-07-18T12:34:05.433" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30439" PostTypeId="1" AcceptedAnswerId="30447" CreationDate="2019-04-14T03:45:55.583" Score="1" ViewCount="71" Body="&lt;p&gt;The Forbes article &lt;a href=&quot;https://www.forbes.com/sites/startswithabang/2019/04/13/ask-ethan-how-does-very-long-baseline-interferometry-allow-us-to-image-a-black-hole/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ask Ethan: How Does Very-Long-Baseline Interferometry Allow Us To Image A Black Hole?&lt;/a&gt; includes an example of optical interferometry from the Large Binocular Telescope, shown below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The occultation of Jupiter's moon, Io, with its erupting volcanoes Loki and Pele, as occulted by Europa, which is invisible in this infrared image. The Large Binocular Telescope was able to do this owing to the technique of interferometry. Credit: LBTO&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;related: &lt;a href=&quot;https://astronomy.stackexchange.com/q/23477/7982&quot;&gt;When will the next series of mutual eclipses of Jupiter's moons begin?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The LBTO consists of only two telescopes &lt;a href=&quot;https://en.wikipedia.org/wiki/Large_Binocular_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;8.4 m (330 inch) wide mirrors, with centres 14.4 m apart&lt;/a&gt;, so I would expect to see higher resolution in one direction than the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in this GIF the resolution seems to be the same in both directions. While I see high spatial frequency ringing in the horizontal direction, I don't see any difference in resolution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the event is so fast (probably minutes or seconds) there is no chance that the natural rotation of the baseline during the course of a night could have helped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'd like to ask &lt;em&gt;How does the Large Binocular Telescope resolve so well in both orthogonal directions simultaneously?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tgr2W.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tgr2W.gif&quot; alt=&quot;Large Binocular Telescope&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Open image in new window for full size:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rxwF9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rxwF9.jpg&quot; alt=&quot;Large Binocular Telescope&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.lbto.org/leech-hr8799.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The LBTI (green structure in the center of the frame) between the two 8.4m mirrors of the LBT.  High resolution image Credit: LBTO - Enrico Sacchetti&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-04-14T12:03:08.487" Title="How does the Large Binocular Telescope resolve so well in both orthogonal directions simultaneously?" Tags="&lt;interferometry&gt;&lt;post-processing&gt;&lt;large-binocular-telescope&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30440" PostTypeId="1" CreationDate="2019-04-14T07:06:44.567" Score="0" ViewCount="44" Body="&lt;p&gt;We just pictured a medium distant black hole by combining data from various observatories around the world. That's when I wondered if interstellar cloud blocks the visibility at multiple position in the same direction in space such that we never accurately know the exact luminosity and distance between the black hole and Earth? I was not able to find much information about this particular subject so,  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Is it possible that the interstellar cloud reduces the visibility of an object behind it?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;And if so, is it possible that we are only able to observe interstellar cloud upto a certain distance in space? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If we observe a star that is behind a interstellar cloud, how do we determine the luminosity of the star? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can we accurately filter out interstellar cloud that are separated by x distance but at the same direction?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="26489" LastActivityDate="2019-04-14T16:57:22.983" Title="Does interstellar cloud obstruct or reduce the visibility?" Tags="&lt;observational-astronomy&gt;&lt;interstellar-medium&gt;&lt;interferometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30441" PostTypeId="1" AcceptedAnswerId="30485" CreationDate="2019-04-13T18:42:42.113" Score="7" ViewCount="773" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Event_Horizon_Telescope&quot; rel=&quot;noreferrer&quot;&gt;Event Horizon Telescope&lt;/a&gt; seems to include these radio telescopes currently:&#xA;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/e/eb/The_Event_Horizon_Telescope_and_Global_mm-VLBI_Array_on_the_Earth.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image by European Southern Observatory (ESO)/O. Furtak; &lt;a href=&quot;https://creativecommons.org/licenses/by/4.0/deed.en&quot; rel=&quot;noreferrer&quot;&gt;CC-BY 4.0&lt;/a&gt;-licensed, see &lt;a href=&quot;https://en.wikipedia.org/wiki/File:The_Event_Horizon_Telescope_and_Global_mm-VLBI_Array_on_the_Earth.jpg&quot; rel=&quot;noreferrer&quot;&gt;the source on Wikipedia Commons&lt;/a&gt; and &lt;a href=&quot;https://www.eso.org/public/images/ann17015a/&quot; rel=&quot;noreferrer&quot;&gt;the original image link&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also the map and list &lt;a href=&quot;https://eventhorizontelescope.org/array&quot; rel=&quot;noreferrer&quot;&gt;on their official website&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also said, that it has a “effective diameter of the entire planet”.&#xA;But it's striking, it does not include half-of the earth… It has &lt;strong&gt;no telescopes in Africa, Asia or Australia&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why were not they included?&lt;/li&gt;&#xA;&lt;li&gt;Would it increase the diameter if they would include some from there? If so, again why did not they? ()&lt;/li&gt;&#xA;&lt;li&gt;Or, similarly asked: How can it be the diameter of the earth if only half-the-earth is included in the network?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Please try to explain it in a simple way first. I am no astronomer. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What I've noticed so far when skimming their official pages is that &lt;a href=&quot;https://eventhorizontelescope.org/technology#boxes-box-1495646861&quot; rel=&quot;noreferrer&quot;&gt;they write here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As more telescopes are added to the EHT, we will be able to produce images of the emission around black holes. In general, the fidelity of images produced by an interferometric array increases as additional telescopes are added to the array. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This seems to support my question of why they did not use possibilities in Asia or so…&#xA;I guess, it would also make sense to have a more or less uniform distribution across the planet (or does not this matter?), that's why a &quot;dot&quot; in Africa might make sense too…&lt;/p&gt;&#xA;" OwnerUserId="26646" OwnerDisplayName="rugk" LastActivityDate="2019-04-20T13:38:00.540" Title="Why does the Event Horizon Telescope (EHT) not include telescopes from Africa, Asia or Australia?" Tags="&lt;radio-telescope&gt;&lt;black-hole&gt;&lt;radio-astronomy&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30442" PostTypeId="2" ParentId="30441" CreationDate="2019-04-13T18:50:35.110" Score="10" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Would it increase the diameter if they would include some from there?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. Not by much, at least. The telescopes are already ~20,000 km apart, so you can't create a longer baseline that still has a simultaneous view of the target. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't forget: Earth is a sphere. Only one half of that sphere can observe M87 at the same time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Telescopes in the Eastern hemisphere would enable more observations, but I don't know if that would bring any improvement over what they already had. &lt;/p&gt;&#xA;" OwnerUserId="361" OwnerDisplayName="Hobbes" LastActivityDate="2019-04-13T18:50:35.110" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30443" PostTypeId="2" ParentId="30441" CreationDate="2019-04-14T00:04:10.380" Score="5" Body="&lt;p&gt;&lt;strong&gt;tl;dr:&lt;/strong&gt; @Hobbes' &lt;a href=&quot;https://space.stackexchange.com/a/35504/12102&quot;&gt;answer&lt;/a&gt; is demonstrably wrong; the EHT takes a large fraction of its data when the target is not visible from one of the extreme sites. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there were sites distributed all the way around the earth, it would be tremendously advantageous to collect from all of them, and the image resolution would improve with the addition of longer duration data and more diversity in baseline vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &lt;a href=&quot;https://space.stackexchange.com/questions/35503/why-does-the-event-horizon-telescope-eht-not-include-telescopes-from-africa-a/35508#comment111508_35503&quot;&gt;pointed out&lt;/a&gt; by @AtmosphericPrisonEscape: &lt;strong&gt;The main requirements for a site to be used in the Event Horizon Telescope&lt;/strong&gt; include:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Low noise, stable receiving at 230 GHz (1.30 millimeters wavelength)&lt;/li&gt;&#xA;&lt;li&gt;High altitutde to minimize attenuation and dispersion at 230 GHz due to atmospheric water &lt;/li&gt;&#xA;&lt;li&gt;digitization of several 2GHz wide channels near 230 GHz for a total digitization rate of 32 Gigabits/second. Not too many sites are equipped to do all of this! See &lt;a href=&quot;https://astronomy.stackexchange.com/a/30364/7982&quot;&gt;this answer&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;infrastructure to calibrate, monitor atmospheric water, record and store huge amounts of data, monitor GPS time signals and support on-site atomic clock of sufficient precision for time stamping of these 1.3 millimeter wavelength signals.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Not so many sites currently can support all of these requirements simultaneously and sufficiently. But this may grow in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 1.30 millimeters wavelength (230 GHz), a huge dish antenna must maintain a parabolic shape to an accuracy of order of hundreds of microns while the dish tilts up and down during observations. Dishes with surface accuracy like that are &lt;em&gt;few and far between.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a further consideration. The EHT needs to be a &quot;movie camera&quot; because the swirling accretion disk around the black hole (the thing actually images) is constantly changing. For the M87 object things change more slowly than for the black hole at the center of our Milky Way (&lt;a href=&quot;https://astronomy.stackexchange.com/a/30318/7982&quot;&gt;days versus minutes&lt;/a&gt;) you would really want to have around-the-world telescope coverage to produce around-the-clock observations. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The results are described in four main papers (there are/will be more):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Paper I. &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/ab0ec7/meta&quot; rel=&quot;noreferrer&quot;&gt;First M87 Event Horizon Telescope Results. I. The Shadow of the Supermassive Black Hole&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Paper II. &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/ab0c96/meta&quot; rel=&quot;noreferrer&quot;&gt;First M87 Event Horizon Telescope Results. II. Array and Instrumentation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Paper III. &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/ab0c57/meta&quot; rel=&quot;noreferrer&quot;&gt;First M87 Event Horizon Telescope Results. III. Data Processing and Calibration&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Paper IV. &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/ab0e85/meta&quot; rel=&quot;noreferrer&quot;&gt;First M87 Event Horizon Telescope Results. IV. Imaging the Central Supermassive Black Hole&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;While the simple view is that the resolution is determined by the farthest baseline that the source is visible from simultaneously, the situation is more nuanced because with so few sites we really can't call this an aperture in a conventional sense. So you can still use a subset of all possible baselines to &quot;fill-in&quot; missing information and improve the image reconstruction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see between 04h and 06h UTC the eastern-most site (PV; 30m dish in (Spain)) cuts out and the western-most sites (JCMT (Hawaii), SMA (Hawaii)) come on line. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A brief reading suggests that they stop using the data from a given site when the object drops below 20 degrees elevation above the horizon at that site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second image below, you can see all sites used. The solid lines indicate baselines used to produce the image, the dashed lines indicate baselines used to collect data for calibration of the EHT network.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Iloao.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Iloao.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Figure 2. (of Paper &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/ab0c57/meta&quot; rel=&quot;noreferrer&quot;&gt;III&lt;/a&gt;). EHT 2017 observing schedules for M87 and 3C 279 covering the four days of observations. Empty rectangles represent scans that were scheduled, but were not observed successfully due to weather, insufficient sensitivity, or technical issues. The filled rectangles represent scans corresponding to detections available in the final data set. Scan duration varies between 3 and 7 minutes, as reflected by the width of each rectangle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tGGv1.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tGGv1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Figure 1. (of Paper &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/ab0ec7&quot; rel=&quot;noreferrer&quot;&gt;I&lt;/a&gt;). Eight stations of the EHT 2017 campaign over six geographic locations as viewed from the equatorial plane. Solid baselines represent mutual visibility on M87* (+12° declination). The dashed baselines were used for the calibration source 3C279 (see Papers &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/ab0c57/meta&quot; rel=&quot;noreferrer&quot;&gt;III&lt;/a&gt; and &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/ab0e85/meta&quot; rel=&quot;noreferrer&quot;&gt;IV&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" OwnerDisplayName="uhoh" LastActivityDate="2019-04-14T00:59:15.670" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30444" PostTypeId="1" CreationDate="2019-04-14T10:33:17.287" Score="5" ViewCount="332" Body="&lt;p&gt;The M87 black hole is at the centre, more or less, of an elliptical galaxy of 1 trillion or more stars. So certainly there are stars and other light emitting objects (emission nebula e.g.) in the intervening 53 million light years between earth and the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is it that the centre of the black hole image is perfectly dark, as if it is being observed directly, with not one point of light between the black hole and earth? I appreciate that space is mostly empty, but do we really have an unobscured line of sight to a black hole in the centre of an enormous elliptical galaxy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it have something to do with the type of light being observed?&lt;/p&gt;&#xA;" OwnerUserId="26657" LastEditorUserId="332" LastEditDate="2019-04-14T20:02:09.630" LastActivityDate="2019-04-14T20:02:09.630" Title="Where is the intervening light in the M87 black hole?" Tags="&lt;black-hole&gt;&lt;m87&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30445" PostTypeId="2" ParentId="27100" CreationDate="2019-04-14T11:24:53.713" Score="3" Body="&lt;p&gt;I found a convenient formula to determine Eclipse (satellite in shadow). From Ismail, et al., Equation 10 &lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/S2090997715000206&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image from same article:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x97EB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x97EB.jpg&quot; alt=&quot;Theta-E is the angle of eclipse&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So time in shadow = ((Theta-E)/360)*Orbital Period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do you calculate Theta-E? Well, for my part, I used the law of similar triangles. You will have two similar isosceles triangles - one from the Earth to the point of the umbra cone and one from the Sun to the point of the umbra cone. Cut the isosceles triangles in half and use the right triangles. The sides of similar triangles are proportional and using the distance of 1 AU and the radii of the Sun and Earth, you can find the distance to the point of the umbra cone with a little algebra. Then use the arc-tangent to find the half-angle to the distance at GEO, then doubled it to find Theta-E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be aware, however, that you are dealing with (1) circular orbits with (2) 0 inclination. The time of year also can change things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My result was very close to that of James K. with only a few seconds difference.&lt;/p&gt;&#xA;" OwnerUserId="26658" LastActivityDate="2019-04-14T11:24:53.713" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30446" PostTypeId="2" ParentId="30444" CreationDate="2019-04-14T11:29:58.190" Score="7" Body="&lt;p&gt;Based on the comment from @SteveLinton&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radius of M87 is given to be 60,000 light years, which is a volume of 900 trillion cubic light years.  Given that it contains roughly one trillion stars, there is an average of 900 cubic light years for each star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several resources referred to the diameter of the black hole in the range of 1.5 light days (38 billion km).  That's 0.004 light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider the cylinder&lt;sup&gt;1&lt;/sup&gt; extending from the black hole to the edge of M87, the cylinder is 0.004 light years in diameter with a height of 60,000 light years.  The volume of that cylinder is just over 3 cubic light years.  Since there is one star for every 900 cubic light years, it follows that the cylinder encompassing our line of sight between the black hole and the edge of M87 is empty of any stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether or not any star would even be bright enough to be visible is another matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; &lt;sub&gt;Technically a conic frustum but I don't think the difference in volume is significant.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="26657" LastEditorUserId="26657" LastEditDate="2019-04-14T11:37:30.437" LastActivityDate="2019-04-14T11:37:30.437" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30447" PostTypeId="2" ParentId="30439" CreationDate="2019-04-14T11:39:46.230" Score="2" Body="&lt;p&gt;The simple answer is that it &lt;em&gt;isn't&lt;/em&gt; resolving in both orthogonal directions equally well. The horizontal dimension is the binocular dimension, and from looking at your animation, it looks to have about ~3 times the resolution. The horizontal banding, I'm pretty sure is &lt;em&gt;not ringing&lt;/em&gt;, and is in fact, representative of additional information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomynow.com/2015/04/30/lava-lake-of-ios-volcano-loki-viewed-with-large-binocular-telescope/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article&lt;/a&gt; does a nice job of explaining what is being seen, and it has a picture, both of a simulated depiction of Io from one of the two telescope mirrors, and the combined image.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/CICeG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CICeG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;The thing you should notice from the above image is that the 8.4 m telescope picture is at the same resolution as the vertical axis of your animation.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;What is going on with the interferometer image?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;By increasing the separation between telescopes, we increase the angular resolution. Just about everyone with a casual interest in astronomy will have learned that fact &lt;a href=&quot;https://www.jpl.nasa.gov/news/news.php?feature=7372&quot; rel=&quot;nofollow noreferrer&quot;&gt;in this past week&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the other thing you do, is introduce interference patterns. The two telescopes essentially act like a double slit experiment.&#xA;In the vertical (non binocular) dimension, the diffraction can be treated like a single-slit experiment. The angular intensity formula for a &lt;em&gt;single slit&lt;/em&gt; is as follows:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ I(\theta) = I_0 sinc^2(\frac{\pi b sin(\theta)}{\lambda})$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; is the diameter of each mirror, and &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; is the wavelength of the light. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/5XqPP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5XqPP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horizontal (binocular) dimension acts as a double slit. The double slit formula is as follows:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ I(\theta) = I_0 cos^2(\frac{\pi d sin(\theta)}{\lambda})sinc^2(\frac{\pi b sin(\theta)}{\lambda})$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; is &lt;em&gt;the distance between the centres of the two telescopes&lt;/em&gt;.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/IwASL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IwASL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you combine these two together, using the telescopes dimensions that you quotes (b=8.4, d=14.4), you come up with a pattern remarkably close to what you actually see from the telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fyUNM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fyUNM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;On the left, screencapture from above animation, on the right, the predicted double slit intensity.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Fringe removal&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It seems that the animation you saw is based off unprocessed images from the LBT. Obviously, they have methods for removing the fringe bands. As to how, I have no idea. I &lt;a href=&quot;http://www.jb.man.ac.uk/~njj/int.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;found a paper&lt;/a&gt; that discusses interferometry in depth and they mention:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What we therefore have is a series of fringes, whose amplitude is given by the Fourier transform of the source intensity distribution.&#xA;  In practice, steps are usually taken to get rid of the fringes using a&#xA;  phase rotation whose rate is known (as both B and s are known). This&#xA;  is done in optical interferometers by use of accurate delay lines to&#xA;  compensate for the path difference, and in radio interferometers by&#xA;  the insertion of electronic delays. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I have no idea what that means. Perhaps some boffin from Physics.SE would be nice enough to answer.&lt;/p&gt;&#xA;" OwnerUserId="17655" LastEditorUserId="17655" LastEditDate="2019-04-14T12:03:08.487" LastActivityDate="2019-04-14T12:03:08.487" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30449" PostTypeId="1" CreationDate="2019-04-14T14:24:01.507" Score="4" ViewCount="128" Body="&lt;p&gt;We have been repeatedly assured that their ain't enough brown dwarfs to account for all the dark matter in our galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is have we been looking in the right places?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or in more detail. Where exactly does computer modelling put the missing mass? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would expect the bulk of it to be on the galaxy's plain and on the edge. And if I am not mistaken there is too much junk in the way to look.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further I can't see any reason why the dark matter won't be unevenly distributed in clumps just like everything else. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, given the above an extrapolation from what we have detected will be way low if it assumes an even distribution in all directions, and horribly low if the LMG and GMG happen to be between clumps. &lt;/p&gt;&#xA;" OwnerUserId="26660" LastActivityDate="2019-04-15T12:50:31.803" Title="Are we looking for brown dwarfs where the dark matter has to be?" Tags="&lt;dark-matter&gt;&lt;brown-dwarf&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30450" PostTypeId="2" ParentId="30437" CreationDate="2019-04-14T14:29:21.097" Score="5" Body="&lt;p&gt;0.004° &lt;a href=&quot;http://www.kylesconverter.com/angle/arcseconds-to-degrees&quot; rel=&quot;noreferrer&quot;&gt;converts to&lt;/a&gt; about 14.4 arc-seconds.&#xA;That's within the range typical for &lt;a href=&quot;https://www.cloudynights.com/topic/619945-mars-angular-size-some-plots/&quot; rel=&quot;noreferrer&quot;&gt;Mars; 5-25 arc-seconds&lt;/a&gt;. &#xA;You'll see the sun as a point with the naked eye.&#xA;Heliopause is about 14 billion miles out. Absolute magnitude of sun is 4.83.&#xA;14 billion miles = 0.0007297pc.&#xA;&lt;a href=&quot;http://www.calctool.org/CALC/phys/astronomy/star_magnitude&quot; rel=&quot;noreferrer&quot;&gt;Apparent magnitude&lt;/a&gt; will be -15.8543. &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_magnitude&quot; rel=&quot;noreferrer&quot;&gt;Magnitude of the full moon&lt;/a&gt;, from earth, is about −12.74 &#xA;So you'll see a point source of light, brighter than the full moon.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2019-04-14T14:29:21.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30452" PostTypeId="2" ParentId="30449" CreationDate="2019-04-14T14:47:39.857" Score="5" Body="&lt;p&gt;The dark matter has to have a &lt;strong&gt;roughly&lt;/strong&gt; spherical distribution and a smooth radial distribution if it is to account for the dynamics of gas and stars in our and other Galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact shape of any required dark matter halo is still the subject of debate and research. Whilst computer simulations of galaxy formation predict mildly non-spherical or triaxial profiles (e.g. &lt;a href=&quot;https://arxiv.org/abs/1809.07255&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kunting et al. 2018&lt;/a&gt;) and some stellar kinematic data appear to confirm sphericity (e.g. &lt;a href=&quot;https://arxiv.org/abs/0902.2709&quot; rel=&quot;nofollow noreferrer&quot;&gt;Smith et al. 2009&lt;/a&gt; ); others disagree and suggest there is some significant prolateness ( &lt;a href=&quot;https://arxiv.org/abs/1604.06885&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bowden et al. 2016&lt;/a&gt;) or oblateness suggested by large scale surveys (e.g. &lt;a href=&quot;https://arxiv.org/abs/1408.5388&quot; rel=&quot;nofollow noreferrer&quot;&gt;Loebman et al. 2014&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any concentration of dark matter into a disk is &lt;strong&gt;totally ruled out&lt;/strong&gt; by observations of the dynamics of stars above and below the plane.&#xA;e.g. The latest word, using Gaia DR2 data, is that there can only be about 10% of the &lt;em&gt;local&lt;/em&gt; mass density in the form of dark matter of any kind (&lt;a href=&quot;https://arxiv.org/abs/1808.05603&quot; rel=&quot;nofollow noreferrer&quot;&gt;Buch et al. 2019&lt;/a&gt;, agreeing with lots of previous work. For instance this was established fact at least 30 years ago (&lt;a href=&quot;http://adsabs.harvard.edu/abs/1989MNRAS.239..651K&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kuijken &amp;amp; Gilmore 1989&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of dark matter required at different radii (assuming Newtonian gravity is the correct formulation)  is deduced from the rotation curve. The image below, courtesy of Nick Strobel at www.astronomynotes.com, shows a sketch of the current thinking on how much dark matter needs to be at what radius (labelled as &quot;Corona&quot;) and takes the dark matter to be distributed spherically symmetrically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark matter already &quot;outweighs&quot; luminous matter by 2:1 at the solar radius, yet there is room for only 10% of the local density to be in a &quot;dark&quot; form of any kind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UCrsA.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UCrsA.gif&quot; alt=&quot;Distribution of dark matter&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brown dwarfs are not actually &quot;dark objects&quot; that cannot be found. They emit infrared light and surveys near the Sun reveal that stars outnumber them by 4:1 and that they contribute less than a few percent the local mass density.&#xA;They can also be found via gravitational microlensing, in surveys that have been performed towards the Magellanic clouds (sampling the halo) and towards the Galactic bulge (sampling the inner disk). In neither case has evidence been found for huge numbers of brown dwarfs (or cold white dwarfs or black holes) and these have been ruled out as significant contributors to the Galactic dark matter problem (e.g. &lt;a href=&quot;https://www.nature.com/articles/d41586-018-07006-8&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pietrzynski 2018&lt;/a&gt;). In particular, the numbers of microlensing events towards the Galactic bulge is perfectly consistent with the brown dwarf/star fraction in the local neighborhood (&lt;a href=&quot;https://arxiv.org/abs/1901.07120&quot; rel=&quot;nofollow noreferrer&quot;&gt;Niikura et al. (2019)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a final remark, dark matter (whether in the form of brown dwarfs or non-baryonic particles) doesn't clump very easily because it cannot readily dissipate it's kinetic energy. There is little evidence for any clumping of dark matter on scales smaller than dwarf galaxies. In particular no evidence for dark matter is found in globular or open clusters of stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stellar disk forms because the stars formed more-or-less in a disk that had already formed from dissipative baryonic gas that had already collapsed into a disk.  As brown dwarfs must also form from baryonic gas then there are really only two alternatives for their distribution. Either they follow the disk and should be associated with disk stars or they should be very old and associated with halo stars that formed in a spherical gas cloud before it collapsed to form a disk and therefore still have a spherical distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dark matter needs to be much more extended than the luminous matter of the Galaxy. In neither scenario described above do we end up with tens of trillions of brown dwarfs (for that is what would be required) distributed a long way outside the solar circle and unassociated with any luminous matter.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-04-15T12:50:31.803" LastActivityDate="2019-04-15T12:50:31.803" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="30453" PostTypeId="2" ParentId="30440" CreationDate="2019-04-14T16:57:22.983" Score="3" Body="&lt;p&gt;Yes it is possible that a cloud reduce the visibility because the matter in it will absorb some of the light which is going through. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, we cannot see all the universe and so we can see interstellar clouds up to a certain distance. Indeed, light takes time to reach us. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a star behind a cloud you will have difficulties to determine its true luminosity. In general, we define the luminosity of stars thanks to a reference star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure to really understand the fourth question ... &lt;/p&gt;&#xA;" OwnerUserId="25214" LastActivityDate="2019-04-14T16:57:22.983" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30454" PostTypeId="1" AcceptedAnswerId="30456" CreationDate="2019-04-14T18:01:24.733" Score="3" ViewCount="107" Body="&lt;p&gt;I understand that the so called observable universe has a limit we cannot see through. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is whether we might get information through an intermediate hypothetical actor located halfway to the limit. This actor would have, I assume, a limit of the observable universe further away than ours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We, in turn, might interact with this actor in order to get information about a nova we cannot see from the Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Information about such a star might perhaps have reached the Earth earlier at a time before it changed into a nova. The information about the nova would therefore be new. &lt;/p&gt;&#xA;" OwnerUserId="15135" LastActivityDate="2019-04-15T01:48:41.300" Title="Can the range of the observable universe be extended through an intermediary?" Tags="&lt;general-relativity&gt;&lt;observable-universe&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30455" PostTypeId="2" ParentId="30454" CreationDate="2019-04-14T18:05:32.783" Score="2" Body="&lt;p&gt;This can't happen. One of the fundamental principles of relativity is &quot;there is no overtaking of light by light&quot;. So the light carrying the message from the intermediary cannot overtake the light from the nova, which never gets to Earth thanks to the accelerating expansion of the universe. &lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-04-14T18:05:32.783" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30456" PostTypeId="2" ParentId="30454" CreationDate="2019-04-14T22:58:09.023" Score="4" Body="&lt;p&gt;The simple answer is &lt;em&gt;no&lt;/em&gt;, but it might be easier to understand if we look at the difference between the different &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmological_horizon&quot; rel=&quot;nofollow noreferrer&quot;&gt;cosmological horizons&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since no information can travel faster than light, and nothing can have travelled for longer than the age of the Universe, then the limit of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Observable_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;observable universe&lt;/a&gt; is calculated by multiplying the speed of light by the conformal time since the Big Bang (the age of the Universe, adjusted for the &lt;a href=&quot;https://en.wikipedia.org/wiki/Scale_factor_(cosmology)&quot; rel=&quot;nofollow noreferrer&quot;&gt;scale factor&lt;/a&gt; representing the expansion of the Universe). This limit is also known as the &lt;em&gt;particle horizon&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current limit is a sphere of ~46.5 billion light-years in radius. If we could detect any signals emitted from matter that far away, we'd be seeing the Big Bang itself! However, the most distant &lt;em&gt;light&lt;/em&gt; (EMR) signal we will ever be able to detect is the cosmic microwave background radiation (CMBR), which represents the earliest light after photon decoupling, emitted from a shell ~45.7 billion light-years away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CMBR was emitted about 380,000 years after the Big Bang, while the earliest &lt;em&gt;stars&lt;/em&gt; didn't form for at least another 100 million years, and the earliest galaxy we've observed so far is &lt;a href=&quot;https://en.wikipedia.org/wiki/GN-z11&quot; rel=&quot;nofollow noreferrer&quot;&gt;GN-z11&lt;/a&gt; which dates back to about 400 million years after the Big Bang. Nonetheless, there's been enough time since then for the light from those earliest structures – albeit highly redshifted and extremely weak – to reach us directly, so there's no need for an &quot;intermediary&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the particle horizon defines the &lt;em&gt;current&lt;/em&gt; limit of the observable universe, what about future observations? In a Universe with accelerating expansion, the outer parts of the observable universe are now receding from us faster than the speed of light, and information from those areas may never reach us at all. The point at which light emitted now can never reach us is called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmological_horizon#Event_horizon&quot; rel=&quot;nofollow noreferrer&quot;&gt;cosmic event horizon&lt;/a&gt;. The &lt;em&gt;current&lt;/em&gt; cosmic event horizon is around 16 billion light-years away, which is a lot less than the ~46b ly to the edge of the observable universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make this a bit clearer, let's look at the path the information would take via the intermediary. To avoid confusion, let's call your distant object a &lt;em&gt;supernova&lt;/em&gt; rather than just a nova, and let's assume the intermediary is at the midpoint of a straight line between us and the supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The information – whether light waves or gravitational waves – leaves the supernova and travels at the speed of light to the intermediary, where (let's say) the information is immediately relayed in a boosted signal – also travelling at the speed of light – towards Earth. Apart from the momentary delay in relaying the signal at the midpoint, the information will be following exactly the same path as if it had left the supernova and travelled directly to Earth. In other words, if we can receive the information about the distant supernova from the intermediary, we must also be able to receive the information straight from the supernova itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in the case of a supernova that's just beyond our cosmic event horizon, the intermediary can receive the light from the supernova, but the space between us and the intermediary will have expanded in the billions of years it takes for the supernova's light to get to them. By that time, the Earth is receding from the intermediary so fast that their relayed signal can never overcome the accelerating expansion of space on the second leg of its journey. We never receive their signal, as they too have receded over the cosmic event horizon.&lt;/p&gt;&#xA;" OwnerUserId="14094" LastEditorUserId="14094" LastEditDate="2019-04-15T01:48:41.300" LastActivityDate="2019-04-15T01:48:41.300" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30457" PostTypeId="1" AcceptedAnswerId="30492" CreationDate="2019-04-14T23:00:09.340" Score="0" ViewCount="513" Body="&lt;p&gt;Standing in Melbourne at 4 am in the morning on 14th April 2019,  Sirus RA = 06h 45m 08.9s, Dec = -16° 42' 58&quot;..How do I Locate this. Standing facing east I turn myself towards south approx 90 Deg and then look up about 16 Deg, I do not find any star.&lt;/p&gt;&#xA;" OwnerUserId="26665" LastEditorUserId="1891" LastEditDate="2019-07-14T21:42:03.787" LastActivityDate="2019-07-14T21:42:03.787" Title="Knowing the RA/DEC of a star how do I locate a star from ground?" Tags="&lt;amateur-observing&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30458" PostTypeId="2" ParentId="30457" CreationDate="2019-04-14T23:24:53.340" Score="1" Body="&lt;p&gt;You know that stars seem to move in the sky because the Earth is rotating. The RA and Dec tell you the position of the star relative to other stars. But because the Earth is rotating, the RA and Dec don't tell you where to look in the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, you need to know two more pieces of information: where you are standing on Earth and the time. It is then possible to calculate the position in the sky. This position is sometimes called the Alt(itude) (angle above horizon) and Az(imuth) (bearing from North). See for example &lt;a href=&quot;https://astronomy.stackexchange.com/questions/12677/software-to-convert-ra-and-dec-into-alt-and-az&quot;&gt;Software to convert RA and DEC into ALT and AZ&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculations basically involve spherical trigonometry, so are too difficult for most people to do in their head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But rather than try to do this yourself, it is better to get some planetarium software, such as Stellarium. This shows you what the sky looks like at any given time and where stars and planets can be seen.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you prefer a coding solution, you can use &lt;code&gt;astropy&lt;/code&gt;, or &lt;code&gt;pyephem&lt;/code&gt;. Libraries also exist for other languages. Search for &quot;Ephemeris&quot; and the name of your favourite language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even better is to learn your way around the sky, by recognising bright constellations and asterisms.  You find Sirus by first finding Orion, and then extending the line of his belt to the west, and there you will see a surprisingly bright star, Sirius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you won't see Sirus at 4 am in Melbourne since it will have already set.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2019-07-14T16:50:05.537" LastActivityDate="2019-07-14T16:50:05.537" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30459" PostTypeId="2" ParentId="29931" CreationDate="2019-04-15T00:53:00.053" Score="5" Body="&lt;p&gt;Planet solidifying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No molten core iron -&gt; no magnetosphere -&gt; solar wind strips atmosphere.&lt;br&gt;&#xA;Without atmosphere to cycle in exposed water also leaves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why did Mars freeze solid &amp;amp; Earth has not (yet).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;is further from sun&lt;/li&gt;&#xA;&lt;li&gt;smaller means less volume to hold heat&lt;/li&gt;&#xA;&lt;li&gt;Earth may have the iron cores from two planets (moon forming impact)&lt;/li&gt;&#xA;&lt;li&gt;Radioactive decay?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;On oil, possibly.&lt;br&gt;&#xA;If there was life there and it colonized the land and it found it needed to create a scaffolding structure to compete for photons such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Lignin&quot; rel=&quot;noreferrer&quot;&gt;ligand&lt;/a&gt; as happened here, and there were several hundred millions years until other life there figured out how to digest those ligands then there would be a layer there like our &lt;a href=&quot;https://en.wikipedia.org/wiki/Carboniferous&quot; rel=&quot;noreferrer&quot;&gt;Carboniferous&lt;/a&gt; era in which a large amount of carbon is sequestered underground. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But chances are vanishingly slim.&lt;br&gt;&#xA;If there was life there it most likely   &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;recycled carbon just as we have here for all time except that little &lt;em&gt;blip*&lt;/em&gt; after trees showed up and before bacteria figured out how to digest them.  &lt;/li&gt;&#xA;&lt;li&gt;did not survive long enough to need to compete for photons by growing very tall&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(*)  300,000,000 / 4,300,000,000  is about 1/14th of life on earth's history.&lt;/p&gt;&#xA;" OwnerUserId="24480" LastActivityDate="2019-04-15T00:53:00.053" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30460" PostTypeId="5" CreationDate="2019-04-15T03:07:15.927" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2019-04-15T03:07:15.927" LastActivityDate="2019-04-15T03:07:15.927" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30461" PostTypeId="4" CreationDate="2019-04-15T03:07:15.927" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2019-04-15T03:07:15.927" LastActivityDate="2019-04-15T03:07:15.927" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30462" PostTypeId="1" AcceptedAnswerId="30503" CreationDate="2019-04-15T03:09:25.027" Score="1" ViewCount="133" Body="&lt;p&gt;Is there a limit in which the mass in space obscure the ability to detect anything farther even as technology progresses?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/30454/can-the-range-of-the-observable-universe-be-extended-through-an-intermediary&quot;&gt;Can the range of the observable universe be extended through an intermediary?&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/496/which-galaxy-is-receding-from-the-milky-way-the-fastest-what-is-known-of-the-me&quot;&gt;Which galaxy is receding from the Milky Way the fastest? What is known of the mechanism behind its recession?&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="10687" LastEditDate="2019-04-15T18:38:19.600" LastActivityDate="2019-04-16T13:59:45.270" Title="What is the farthest we can see into the universe?" Tags="&lt;astrophysics&gt;&lt;distances&gt;&lt;radio-astronomy&gt;&lt;identify-this-object&gt;&lt;weak-lensing&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="30466" PostTypeId="2" ParentId="29971" CreationDate="2019-04-15T07:35:24.360" Score="7" Body="&lt;p&gt;There is indeed a sense that infalling objects accelerate to the speed of light as it reaches the event horizon, but you need to be careful what you mean by the speed of the object because the speed is observer dependent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is explained in detail on the Physics site in my answer to &lt;a href=&quot;https://physics.stackexchange.com/questions/170502/will-an-object-always-fall-at-an-infinite-speed-in-a-black-hole&quot;&gt;Will an object always fall at an infinite speed in a black hole?&lt;/a&gt; An observer watching from far from the black hole would see the falling object initially accelerate towards the black hole but then decelerate to a halt at the event horizon. However an observer hovering a distance &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; above the event horizon would see the falling object pass them at a speed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ v = c \sqrt{\frac{r_s}{r_s + d}} \tag{1} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$r_s$&lt;/span&gt; is the radius of the event horizon. As the distance above the horizon &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; goes to zero the speed calculated from equation (1) goes to the speed of light &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10687" LastActivityDate="2019-04-15T07:35:24.360" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30467" PostTypeId="1" AcceptedAnswerId="30470" CreationDate="2019-04-15T09:25:05.587" Score="0" ViewCount="117" Body="&lt;p&gt;Is it possible to get some gases like hydrogen from gas planets (Saturn, Jupiter,  Uranus or Neptune)?&lt;/p&gt;&#xA;" OwnerUserId="26672" LastEditorUserId="26672" LastEditDate="2019-04-16T01:47:12.730" LastActivityDate="2019-04-16T01:47:12.730" Title="Can we get the gas compositions from a gas planets?" Tags="&lt;gas-giants&gt;&lt;gas&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2019-04-16T01:28:46.193" ContentLicense="CC BY-SA 4.0" />
  <row Id="30470" PostTypeId="2" ParentId="30467" CreationDate="2019-04-15T11:52:34.913" Score="2" Body="&lt;p&gt;Yes.  It's certainly feasible to obtain a sample from Jupiter or Saturn, effectively getting gas from the planet, but it's not easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Massive planets, as a byproduct of their high mass, create large gravity wells which require a lot of fuel to fly close enough to get a sample from and return with it to where it's needed.  It would take perhaps a ton of conventional fuel to return from Jupiter with a few pounds of gas, which isn't a good return on investment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Collecting hydrogen from Jupiter would be expensive almost certainly not worth the cost.   It would be much easier to obtain hydrogen in space from frozen water ice, frozen methane or frozen ammonia, where hydrogen is abundant.     This can be obtained in icy comets and there's even some water-ice on the moon in craters, which has a much smaller gravity well.   Also on Mars' ice caps and inside Ceres, it's believed there's about an entire oceans worth of water, perhaps more water inside Ceres than on Earth.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a lot of hydrogen that would be far easier to harvest than hydrogen from Jupiter.   And while it would need to be separated from Oxygen, chemical separation would require many times less energy than collecting hydrogen from a gas planet's gravity well.   You'd also have the advantage of not just harvesting hydrogen but oxygen as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not impossible to harvest hydrogen from Jupiter or Saturn, it's just difficult to imagine it being worth the trouble.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2019-04-15T20:44:43.557" LastActivityDate="2019-04-15T20:44:43.557" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30471" PostTypeId="1" CreationDate="2019-04-15T11:59:27.167" Score="4" ViewCount="117" Body="&lt;p&gt;I tried looking this up without success and I imagine there's a formula for it, with some ridiculous numbers like it drops off to the 5th power of the distance or maybe even more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a planet like Jupiter, large equatorial bulge and imagine an artificial satellite orbiting in a polar orbit, not an equatorial orbit.    Jupiter's enormous equatorial bulge should disrupt the satellite orbit and in time (perhaps a very long time), turn it into an equatorial orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that correct or incorrect thinking on my part and is there a formula that applies?    I imagine that some kind of measure for equatorial bulge would be needed as well and I'm not sure how that would be measured in a gravitational field from orbit.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="22897" LastEditDate="2019-04-18T18:05:48.940" LastActivityDate="2019-04-18T20:08:01.997" Title="What is the force that changes a satellite’s inclination when orbiting a planet with an equatorial bulge" Tags="&lt;orbital-mechanics&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="30472" PostTypeId="2" ParentId="30285" CreationDate="2019-04-15T12:15:24.007" Score="2" Body="&lt;p&gt;There is several reasons why the L2 point is an optimum point for the satellite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First you have a stable orbit with low gravitational perturbation and which is not too far away. So, you won't have a huge travel time. And the parallax is good for what the satellite is supposed to do.&#xA;Secondly in a Lagrange point you can easily point out the same point. In orbit, you will have some grey area. &#xA;Eventually, I think you are doing a mistake when you say : &quot;Assuming no cost&quot;. In my opinion, if we are not sending a satellite like Gaia around another planet, it is because we want to avoid risks and sending the satellite, around Mars for example, adds a ton of risk. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why we use the L2 point to place a satellite in regards to all the parameters. But, you can choose another location if you want to maximize something in particular.&lt;/p&gt;&#xA;" OwnerUserId="25214" LastActivityDate="2019-04-15T12:15:24.007" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30473" PostTypeId="1" AcceptedAnswerId="30475" CreationDate="2019-04-15T14:49:20.877" Score="2" ViewCount="73" Body="&lt;p&gt;Is the first image of a black hole that was captured by the Event Horizon Telescope what would be seen with a large enough optical telescope (or with the naked eye from a proper distance of the black hole)? If so, how do radio telescopes gather information about visible light?  &lt;/p&gt;&#xA;" OwnerUserId="26675" LastEditorUserId="7472" LastEditDate="2019-04-15T14:50:55.660" LastActivityDate="2019-04-15T14:58:51.920" Title="How do radio telescopes gather information about visible light?" Tags="&lt;light&gt;&lt;radio-telescope&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30475" PostTypeId="2" ParentId="30473" CreationDate="2019-04-15T14:58:51.920" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How do radio telescopes gather information about visible light?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They don't; the picture you saw all over the news was a &lt;a href=&quot;https://en.wikipedia.org/wiki/False_color#False_color&quot; rel=&quot;nofollow noreferrer&quot;&gt;false color image&lt;/a&gt;, where frequencies are shifted from the radio spectrum to the visible spectrum (blue meaning shorter radio wavelengths and red meaning longer radio wavelengths) and/or the intensities mapped to different colors (instead of just affecting the brightness as with true color images).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, to quote an old &lt;a href=&quot;https://starchild.gsfc.nasa.gov/docs/StarChild/questions/question20.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;StarChild question&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The term &quot;false color&quot; is used to describe what astronomers (and others) often do to images to make them more comprehensible. Long ago, when radio astronomers first started generating images of sources, they wound up with essentially images that were just shades of gray - ranging from pure black to pure white. Each shade represented the intensity of the radio emission from a particular part of the object. Radio astronomers took their shades-of-gray images and converted them to color ones by assigning red to the most intense radio emission and blue to the least intense emission recorded on the image. Intermediate colors (orange, yellow, and green) were assigned to the intermediate levels of radio intensity. Black was assigned to places in the image in which there appeared to be no radio emission.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2019-04-15T14:58:51.920" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30476" PostTypeId="1" CreationDate="2019-04-15T17:52:31.133" Score="6" ViewCount="220" Body="&lt;p&gt;I learned &lt;a href=&quot;https://astronomy.stackexchange.com/questions/30473/how-do-radio-telescopes-gather-information-about-visible-light/30475#30475&quot;&gt;here&lt;/a&gt; that the picture of the black hole we saw all over the news was a false color image, based on the radio spectrum emissions. How would the black hole at the center of M87 look like when seen by the naked eye from a close enough distance, or by a large enough optical telescope? (please tell me if the two would be different)&lt;/p&gt;&#xA;" OwnerUserId="26675" LastActivityDate="2020-01-13T10:15:56.587" Title="How does the black hole at the center of M87 look like in the visible spectrum?" Tags="&lt;black-hole&gt;&lt;telescope&gt;&lt;spectra&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30478" PostTypeId="1" CreationDate="2019-04-15T22:11:16.280" Score="3" ViewCount="826" Body="&lt;p&gt;Say I have a spectrum of a galaxy at a redshift &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;, in flux density units of erg/s/cm^2/Angstrom. I'd like to recover the spectrum (in the same flux density units) at z=0, i.e. at its rest wavelengths. How do I do this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw this paper (&lt;a href=&quot;https://arxiv.org/pdf/astro-ph/9905116.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Distance measures in cosmology, Hogg (2000)&lt;/a&gt;) but am still confused, because of their use of luminosity and luminosity densities, and in my class I only have fluxes.&lt;/p&gt;&#xA;" OwnerUserId="24076" LastActivityDate="2020-12-10T15:04:35.570" Title="How to correct observed flux densities for redshift" Tags="&lt;galaxy&gt;&lt;cosmology&gt;&lt;redshift&gt;&lt;spectra&gt;&lt;luminosity&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30479" PostTypeId="1" CreationDate="2019-04-16T00:13:22.000" Score="1" ViewCount="34" Body="&lt;p&gt;I have created a model galaxy spectrum using &lt;code&gt;STARBURST99&lt;/code&gt; and want to compare it to an observed spectrum of a particular galaxy. I understand that I need to redden my model spectrum to do this, so I am following the Cardelli et al. (1989) parameterization, which returns A(lambda)/A(V) for different wavelengths. I understand that the extinction correction in magnitudes is A_V, and I know that R_V = A_V/E(B-V). However I don't know how these terms relate to each other to apply them to my model spectrum (flux density).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read that the commonly adopted R_V value for the Milky Way is 3.1, but what is the E(B-V) value? If I want to apply extinction to a model galaxy for both the Milky Way AND the other galaxy's internal extinction, how would I go about this? For the external galaxy, I have a measured E(B-V) value to use (or does this only apply to the MW extinction?), but in this case I don't know what to use for R_V.&lt;/p&gt;&#xA;" OwnerUserId="24076" LastActivityDate="2019-04-16T00:13:22.000" Title="Clarity on dust extinction terms and implementation: R_V, A_V, and E(B-V)" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;dust&gt;&lt;extinction&gt;&lt;modeling&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30480" PostTypeId="1" AcceptedAnswerId="30482" CreationDate="2019-04-16T00:48:28.117" Score="1" ViewCount="1801" Body="&lt;p&gt;What is the Difference of or the Structure of a GAS PLANET AND AN ICE GIANT Why they called ICE GIANTS like Neptune and Uranus. Do they have a similarity ? &lt;/p&gt;&#xA;" OwnerUserId="26672" LastActivityDate="2019-04-16T20:46:10.620" Title="Difference of Gas Giant Planet and Ice Giant?" Tags="&lt;planet&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30482" PostTypeId="2" ParentId="30480" CreationDate="2019-04-16T01:14:41.043" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Gas_giant&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gas Giants&lt;/a&gt; consist primarily of the gases hydrogen and helium, with a rocky/iron core. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interiors of &lt;a href=&quot;https://en.wikipedia.org/wiki/Ice_giant&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ice Giants&lt;/a&gt; consist primarily of what astronomers call 'Ices'. These are compounds like water, methane and ammonia, and are so-named presumably because they are solid at cold (-200 C) temperatures, unlike hydrogen and helium, which remain gaseous up until very close to Absolute Zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen and helium are very light, and it only takes a small amount of heat energy to make them fast enough to escape the gravitational field of most planets. Jupiter and Saturn are the only planets heavy enough to retain a significant fraction of hydrogen and helium (though the atmospheres of ice giants Uranus and Neptune consist mainly of hydrogen and helium, it's mainly ices further down).&lt;/p&gt;&#xA;" OwnerUserId="17655" LastEditorUserId="17655" LastEditDate="2019-04-16T20:46:10.620" LastActivityDate="2019-04-16T20:46:10.620" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30484" PostTypeId="1" CreationDate="2019-04-16T04:16:06.377" Score="1" ViewCount="1713" Body="&lt;p&gt;I'm Being confused what is that light in the middle of milky way galaxy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image is from &lt;a href=&quot;https://www.nasa.gov/jpl/charting-the-milky-way-from-the-inside-out&quot; rel=&quot;nofollow noreferrer&quot;&gt;Charting the Milky Way From the Inside Out&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The text with the image explains a lot about the spiral arms and the location of the Earth, but the central bright area isn't explained in great depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DEC0t.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DEC0t.jpg&quot; alt=&quot;NASA: Charting the Milky Way From the Inside Out&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26672" LastEditorUserId="7982" LastEditDate="2019-04-16T10:38:57.000" LastActivityDate="2019-04-16T10:47:07.293" Title="What is that light of the middle of the milky way galaxy?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;galaxy-center&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30485" PostTypeId="2" ParentId="30441" CreationDate="2019-04-16T04:42:16.687" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does the Event Horizon Telescope (EHT) not include telescopes from Africa, Asia or Australia? Why were not they included?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Africa &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_radio_telescopes#Africa&quot; rel=&quot;noreferrer&quot;&gt;doesn't have a radio telescope&lt;/a&gt; in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Event_Horizon_Telescope#Telescope_array&quot; rel=&quot;noreferrer&quot;&gt;frequency range necessary&lt;/a&gt; (230-450GHz) to participate in the EHT array. For Asia Wikipedia lists the &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Yevpatoria_RT-70_radio_telescope&quot; rel=&quot;noreferrer&quot;&gt;Yevpatoria RT-70 radio telescope&lt;/a&gt;&quot; as capable up to 300 GHz and located in &lt;a href=&quot;https://en.wikipedia.org/wiki/Yevpatoria&quot; rel=&quot;noreferrer&quot;&gt;western Crimea&lt;/a&gt;. For Australia Wikipedia lists the &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Parkes_Observatory&quot; rel=&quot;noreferrer&quot;&gt;Parkes Observatory&lt;/a&gt;&quot; whose frequency is also &lt;a href=&quot;https://www.parkes.atnf.csiro.au/observing/documentation/user_guide/pks_ug_3.html#Receivers-and-Correlators&quot; rel=&quot;noreferrer&quot;&gt;too low&lt;/a&gt;, with a maximum of 26 GHz. In Japan there is the &quot;&lt;a href=&quot;http://www.nro.nao.ac.jp/~nma/index.html&quot; rel=&quot;noreferrer&quot;&gt;Nobeyama Millimeter Array&lt;/a&gt;&quot; but it only goes up to 230 GHz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only radio telescopes that go up to the required frequency range (according to the &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_radio_telescopes#Africa&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia webpage&lt;/a&gt;, which may not be a complete list) are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Galenki RT-70 radio telescope (Russia) (Incomplete &lt;a href=&quot;https://en.wikipedia.org/wiki/Galenki_RT-70_radio_telescope&quot; rel=&quot;noreferrer&quot;&gt;English Wikipedia&lt;/a&gt; webpage) (More complete &lt;a href=&quot;https://translate.googleusercontent.com/translate_c?depth=1&amp;amp;nv=1&amp;amp;rurl=translate.google.com&amp;amp;sl=auto&amp;amp;sp=nmt4&amp;amp;tl=en&amp;amp;u=https://ru.wikipedia.org/wiki/%25D0%2592%25D0%25BE%25D1%2581%25D1%2582%25D0%25BE%25D1%2587%25D0%25BD%25D1%258B%25D0%25B9_%25D1%2586%25D0%25B5%25D0%25BD%25D1%2582%25D1%2580_%25D0%25B4%25D0%25B0%25D0%25BB%25D1%258C%25D0%25BD%25D0%25B5%25D0%25B9_%25D0%25BA%25D0%25BE%25D1%2581%25D0%25BC%25D0%25B8%25D1%2587%25D0%25B5%25D1%2581%25D0%25BA%25D0%25BE%25D0%25B9_%25D1%2581%25D0%25B2%25D1%258F%25D0%25B7%25D0%25B8&amp;amp;xid=17259,15700022,15700186,15700191,15700253,15700256,15700259&amp;amp;usg=ALkJrhgmRNbUCp1750VZKODxG5v5YSM11g&quot; rel=&quot;noreferrer&quot;&gt;Russian Wikipedia&lt;/a&gt; webpage, translated) (&lt;a href=&quot;https://www.google.com/maps/place/44%C2%B000&amp;#39;58.0%22N+131%C2%B045&amp;#39;25.2%22E/@44.0161,131.757,17z/data=!3m1!1e3?hl=en&quot; rel=&quot;noreferrer&quot;&gt;Google Maps&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Large_Latin_American_Millimeter_Array&quot; rel=&quot;noreferrer&quot;&gt;Large Latin American Millimeter Array&lt;/a&gt; (LLAMA) is under construction and is intended to operate between 40-900 GHz when completed. (&lt;a href=&quot;https://www.google.com/maps/place/Observatorio+Los+Chorrillos,+Susques,+Jujuy,+Argentina/@-24.1912761,-66.4756052,19z/data=!3m1!1e3!4m5!1m2!3m1!2zMjTCsDExJzMxLjQiUyA2NsKwMjgnMjkuNCJX!3m1!1s0x941ce6f7cfeccacb:0x72a8d2af88a740a6?hl=en&quot; rel=&quot;noreferrer&quot;&gt;Google Maps&lt;/a&gt; shows some digging, and the &quot;&lt;a href=&quot;https://translate.googleusercontent.com/translate_c?depth=1&amp;amp;nv=1&amp;amp;rurl=translate.google.com&amp;amp;sl=auto&amp;amp;sp=nmt4&amp;amp;tl=en&amp;amp;u=https://laplata.conicet.gov.ar/el-megatelescopio-llama-ya-viaja-camino-a-la-puna-saltena/&amp;amp;xid=17259,15700022,15700186,15700191,15700253,15700256,15700259&amp;amp;usg=ALkJrhiqxkdVII-znPmMKZOcW7pl7COP3Q&quot; rel=&quot;noreferrer&quot;&gt;Technological Scientific Center CONICET La Plata&lt;/a&gt;&quot; shows the trucks being loaded (on Jan 19 2018), but it doesn't appear that it will be finished this year.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/South_Pole_Telescope&quot; rel=&quot;noreferrer&quot;&gt;South Pole Telescope&lt;/a&gt; (Google Maps and others I checked don't show that far south in any detail, but they have a &lt;a href=&quot;https://maps.app.goo.gl/UyHX8&quot; rel=&quot;noreferrer&quot;&gt;Google Maps photo webpage&lt;/a&gt; and &lt;a href=&quot;https://pole.uchicago.edu/public/receiver.html&quot; rel=&quot;noreferrer&quot;&gt;website&lt;/a&gt;). It's &lt;strong&gt;&lt;em&gt;not&lt;/em&gt;&lt;/strong&gt; clear that this radio telescope reaches the required frequency but it does have continuous view of the southern sky due to its unique location.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_Submillimeter_Telescope&quot; rel=&quot;noreferrer&quot;&gt;Solar Submillimeter Telescope&lt;/a&gt; (SST) (San Juan Province, Argentina) (&lt;a href=&quot;https://www.google.com/maps/place/Astronomical+Observatory+El+Leoncito,+Ingreso+A+Casleo+-+Pnel,+San+Juan,+Argentina/@-31.7984947,-69.2956224,20z/data=!3m1!1e3!4m5!1m2!3m1!2zMzHCsDQ3JzU2LjMiUyA2OcKwMTcnNDguMSJX!3m1!1s0x9686e9fba74f7b57:0xd5f9bbc6d84e548a?hl=en&quot; rel=&quot;noreferrer&quot;&gt;Google Maps&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As you can see the list is certainly incomplete. Another list of the world's radio telescopes is the website &lt;a href=&quot;http://theskyisnotthelimit.org/telescopes&quot; rel=&quot;noreferrer&quot;&gt;TheSkyIsNotTheLimit.org&lt;/a&gt; which offers this graphic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OmXdk.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OmXdk.jpg&quot; alt=&quot;Countries with Radio Telescopes&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you center Google Earth on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Atacama_Pathfinder_Experiment&quot; rel=&quot;noreferrer&quot;&gt;Atacama Pathfinder Experiment&lt;/a&gt; (APEX), the center of the EHT array, you'll notice that those other radio telescopes aren't included in the view:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/D1Mix.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D1Mix.png&quot; alt=&quot;Earth - Centered on Atacama Pathfinder Experiment&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Would it increase the diameter if they would include some from there? If so, again why did not they?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The west coast of Africa could assist if they had a modern radio telescope on a mountain top, but they don't. Even so, such a radio telescope would be pointing sideways, through the thickness of the atmosphere over the ocean; radio telescopes work best pointing upwards and you want the ability to pan, not to be constricted to a limited range of movement. Such distant points in the array will be simultaneously active only for a short period of time, but they serve as a hand-off as one rotates into position and the other rotates out of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you rotate the globe a bit you lose Hawaii but but gain western Europe (including Crimea, if they upgrade) and the west coast of Africa, including Gamsberg Mountain (in close proximity to the &lt;a href=&quot;https://en.wikipedia.org/wiki/High_Energy_Stereoscopic_System&quot; rel=&quot;noreferrer&quot;&gt;High Energy Stereoscopic System&lt;/a&gt; (H.E.S.S.)) in Namibia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fadxN.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fadxN.jpg&quot; alt=&quot;Earth - Rotated to include Gamsberg Mountain, Namibia Africa&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reason to &lt;a href=&quot;https://www.ias-observatory.org/index.php/en/observatories/gamsberg-2&quot; rel=&quot;noreferrer&quot;&gt;be interested in Gamsberg&lt;/a&gt; is because the mountain is owned by the &quot;&lt;a href=&quot;https://www.ias-observatory.org/index.php/en/about-the-ias&quot; rel=&quot;noreferrer&quot;&gt;Internationale Amateursternwarte&lt;/a&gt;&quot; referred to as the IAS. It was established in Germany March 1999 by a group of dedicated amateur astronomers and supported by the Max Planck Institute for Astronomy. This image tells the tale:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MY9yU.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MY9yU.jpg&quot; alt=&quot;Gamsberg Mountain Namibia Africa&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are discussions underway with respect to Gamsberg, see : &quot;&lt;a href=&quot;https://www.researchgate.net/publication/323925242_The_African_Millimetre_Telescope&quot; rel=&quot;noreferrer&quot;&gt;The African Millimetre Telescope&lt;/a&gt;&quot; (June 2017), by Michael Backes, Cornelia Müller, John E. Conway and Roger Deane, on page 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Although the EHT already constitutes an impressive VLBI network of mm-wave radio telescopes, their spacial distribution is clustered around the Americas, thus limiting its capabilities. Significant improvement can be achieved by adding a single mm-wave radio telescope in Africa, nominally called the Africa Millimetre Telescope.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;3. The Africa Millimetre Telescope&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Adding the Africa Millimetre Telescope (AMT)&lt;span class=&quot;math-container&quot;&gt;$^{12}$&lt;/span&gt;, a single mm-wave radio telescope on the African continent, to the EHT network will significantly increase the coverage in the &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt;-&lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt;-plane (see fig.2). This will greatly improve the imaging capabilities of the EHT and, hence, advance its capabilities to directly image the ‘shadow’ of Sgr A∗. In particular, it will enable the ‘Eastern sub-array’, including the IRAM 30 m telescope, NOEMA, SPT, ALMA, and AMT to perform imaging observations, thus &lt;strong&gt;adding a significant amount of daily observing time of Sgr A∗ to the EHT&lt;/strong&gt; (see fig. 1 on the right). As shown in fig. 1, the AMT will have common baselines for observations of Sgr A∗ with all the high-sensitivity telescopes (the IRAM 30 m telescope on Pico Veleta, NOEMA, ALMA, and the LMT) as well as with all the other EHT telescopes. The &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt;−&lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt;-plane coverage of the current EHT configuration is presented in figure 2; included is the improvement in coverage by adding the AMT.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6jyPH.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6jyPH.jpg&quot; alt=&quot;Gamsberg Mountain - Namibia Africa&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Figure 1&lt;/strong&gt;: Baselines of the current EHT VLBI network (in yellow) and additional baselines provided by the AMT (in red). Note: the Combined Array for Research in Millimeter-wave Astronomy (CARMA) ceased observations in 2015, IRAM PV denotes the IRAM 30 m telescope on Pico del Veleta, Plateau de Bure denotes NOEMA, and Gamsberg denotes the AMT.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Potential sites on the African continent where Sgr A∗ can be observed at elevation angles ≥ 40° and that are at sufficiently high altitude to ensure an average precipitable water vapour column of less than 6 mm include sites on &lt;strong&gt;Mt. Kilimanjaro&lt;/strong&gt; (≥ 4,300 m a.s.l.) in Tanzania and the &lt;strong&gt;Sani Pass&lt;/strong&gt; in the Drakensberg mountains in Lesotho (≥ 3,050 m a.s.l.). However, primarily because&#xA;  of temporal overlap in visibility of Sgr A∗ with the mm-wave telescopes in the Americas, the western-most site, &lt;strong&gt;Mt. Gamsberg&lt;/strong&gt; (2,347 m a.s.l.) in Namibia has been chosen as the primary site&#xA;  for further investigations. Additional supporting arguments for this site are that the land is &lt;strong&gt;owned by the Max-Planck Society&lt;/strong&gt;, and that the &lt;strong&gt;government of Namibia is very encouraging&lt;/strong&gt; of astronomy development.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The addition of Gamsberg will significantly improve the coverage, but construction hasn't been approved yet.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;In general&lt;/strong&gt;, the fidelity of images produced by an interferometric array increases as additional telescopes are added to the array. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, but while adding one &lt;em&gt;anywhere&lt;/em&gt; does &lt;em&gt;something&lt;/em&gt; adding one or more in &lt;strong&gt;&lt;em&gt;perfect&lt;/em&gt;&lt;/strong&gt; locations maximizes the return on the enormous investment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &quot;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1999ASPC..180.....T&quot; rel=&quot;noreferrer&quot;&gt;Synthesis Imaging in Radio Astronomy II&lt;/a&gt;&quot;, A Collection of Lectures from the Sixth NRAO/NMIMT Synthesis Imaging Summer School. Edited by G. B. Taylor, C. L. Carilli, and R. A. Perley. ASP Conference Series, Vol. 180, 1999 (&lt;a href=&quot;http://www.phys.unm.edu/~gbtaylor/astr423/s98book.pdf&quot; rel=&quot;noreferrer&quot;&gt;.PDF&lt;/a&gt; - Warning: &lt;strong&gt;43 MB&lt;/strong&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a collection of papers, on page 537:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;27. Interferometric Array Design&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;M.A. Holdaway &amp;amp; Tamara T. Helfer&lt;br&gt;&#xA;  National Radio Astronomy Observatory, Tucson, AZ 85721, U.S.A.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Abstract. We investigate some of the principles which lead to the design of radio interferometric arrays and array configurations, including both abstract issues such as sensitivity and &lt;strong&gt;Fourier plane coverage&lt;/strong&gt;, and practical issues such as moving antennas and site topographical constraints. We draw on the design and history of existing arrays and also give a glimpse of what ideas and algorithms are helping design new instruments such as the Submillimeter Array &#xA;  (SMA) and the Millimeter Array (MMA).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;1. Introduction&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Array design can include a variety of topics: how many antennas should the telescope have, and how big should they be? Are there &lt;strong&gt;astronomical requirements which dictate an aspect of the array layout&lt;/strong&gt;? How many &lt;strong&gt;antenna configurations&lt;/strong&gt; will there be, and how will the different configurations work together? How should we design each individual configuration? But &lt;strong&gt;the central topic of array &#xA;  design deals with how to efficiently sample the Fourier plane&lt;/strong&gt;. Each interferometer, or pair of antennas, at a given moment in time samples a single point in the Fourier plane, and &lt;strong&gt;we need to arrange the antennas in such a way that the set of sampled points permits us to make high quality, high sensitivity images&lt;/strong&gt;. Since most antennas require a fair amount of infrastructure with not insubstantial capital costs on the ground beneath their bases (called antenna pads), it is important to design a good set of antenna configurations which adequately sample the Fourier plane before the array is built.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The above applies to compact arrays (all the antennae involved are interconnected and local to each other) and VLBI where data is recorded and combined at a later date. When the wavelength being measured is less than a millimeter any unwanted movement (or movement not accounted for) of even a millimeter introduces serious errors, each of which needs to be calculated and removed; leaving multiple sources of very tiny errors (which randomly add and subtract from each other, introducing noise).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read on to page 547:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;4.3. VLA-Y and GMRT-Y&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The main benefit of the VLA's &quot;&lt;strong&gt;Y&lt;/strong&gt;&quot; configuration is that it is a convenient 2-D arrangement of antennas which gives &lt;strong&gt;reasonable 2-D snapshot Fourier plane coverage&lt;/strong&gt;. The BIMA and OVRO &quot;&lt;strong&gt;T&lt;/strong&gt;&quot; arrays are similar in concept. The bad things about a &quot;&lt;strong&gt;Y&lt;/strong&gt;&quot; or a &quot;&lt;strong&gt;T&lt;/strong&gt;&quot; are that the regularity in the antenna directions along the arms will lead to a sort of grating response in the point spread function, and that it will take several hours of earth rotation synthesis for the Fourier samples to overcome this deficit.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The &quot;&lt;strong&gt;Y&lt;/strong&gt;&quot; and &quot;&lt;strong&gt;T&lt;/strong&gt;&quot; configurations are &lt;strong&gt;compromises&lt;/strong&gt;: they seek to maintain the convenience of a 1-D array in reconfiguring the antennas, but would also like to get good Fourier plane coverage. As such, they are like arrays of fractal dimension 1.5: better than 1-D, not so good as fully 2-D arrays. Examples for the VLA's coverage and beams for a snapshot and a full track are shown in Figures 27-3 and 27-4.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The Giant Metrewave Radio Telescope (GMRT) in India has 14 inner antennas in a 1 km configuration and 16 in an outer configuration in the shape of an irregular &quot;&lt;strong&gt;Y&lt;/strong&gt;&quot; (see Figure 27-5). The two configurations will often be used separately (the 30 m antennas were not designed for reconfiguration). The irregularity of the &quot;&lt;strong&gt;Y&lt;/strong&gt;&quot; mainly stems from where land could be obtained, but it also produces a snapshot beam with 16 antennas which has much lower sidelobes than the VLA snapshot beam has with 27 antennas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The wye (Y-shaped) configuration uses the least antenna groups to obtain the best result by filling in enough points that it's convenient to fill in the Fourier plane using an optimized algorithm, it is also convenient to position such a shape on existing continents. A five armed star provides better results, again with fewer antennas, but is obviously more expensive than the wye configuration. The &lt;em&gt;best&lt;/em&gt; configurations, such as the randomly filled circle, don't fit multiple continents and their intervening oceans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice the &lt;strong&gt;Y&lt;/strong&gt;-shaped configuration of the current EHT array sites in user rugk's question. More sites in a narrow slice of the Earth's circumference allow increased capabilities but if a particular sector is competently (capability) covered it's a duplication of resources and an increase of processing time compared to a perfect location that fills in a &lt;em&gt;hole&lt;/em&gt;. A site that is near the edge is useful to increase the observation window, but one too far away to be able to simultaneously observe with a sufficient number of other site isn't anywhere near as useful. Wherever the site is it must be capable of operating at the necessary frequencies and have enough sensitivity to gather useful data at that distance. It's a tall order.&lt;/p&gt;&#xA;" OwnerUserId="18054" LastActivityDate="2019-04-16T04:42:16.687" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30487" PostTypeId="1" CreationDate="2019-04-16T05:23:03.510" Score="1" ViewCount="135" Body="&lt;p&gt;Our &lt;a href=&quot;https://clbin.com/C7Jcz&quot; rel=&quot;nofollow noreferrer&quot;&gt;data set&lt;/a&gt; has &lt;span class=&quot;math-container&quot;&gt;$10^4$&lt;/span&gt; data points, but has a long baseline and many gaps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we bin the data, there would be &lt;span class=&quot;math-container&quot;&gt;$10^8$&lt;/span&gt; data points (&lt;span class=&quot;math-container&quot;&gt;$[t,\rm {value}]$&lt;/span&gt;), but only about 1% are non-zero values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to improve the detection efficiency? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is a multi-threading way possible (especially for Lomb-Scargle)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I use LombScargle in &lt;em&gt;astropy.stats&lt;/em&gt;,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;freq, power = LombScargle.autopower(minimum_frequency=0.5, maximum_frequency=1.5, normalization='standard')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;there are two problems:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It is slow and easy to get &lt;code&gt;memoryError&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;It can't utilize the computing power as much as possible.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So my question is about the efficiency. For the data set above, what is the best method?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="1891" LastEditDate="2019-04-16T19:17:12.377" LastActivityDate="2020-01-18T21:03:15.503" Title="How to improve efficiency in periodicity detection?" Tags="&lt;binary-star&gt;&lt;pulsar&gt;&lt;algorithm&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30490" PostTypeId="2" ParentId="30484" CreationDate="2019-04-16T06:03:34.460" Score="3" Body="&lt;p&gt;The galactic bulge is formed by tightly packed stars and interstellar dust&#xA;Also most stars are in the direction towards the galactic center&#xA;Thus it would appear brightest if it were viewed from outside the plane of the galaxy as this image is intended to show.&lt;/p&gt;&#xA;&lt;p&gt;However, we don't see this from Earth, and the reason why is explained in  Phys.org's &lt;a href=&quot;https://phys.org/news/2015-07-center-milky.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Why can't we see the center of the Milky Way?&lt;/a&gt; explains why we don't see this at night.&lt;/p&gt;&#xA;&lt;p&gt;A brief outline of what is described there is as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;When it is dark enough, and conditions are clear, the dusty ring of the Milky Way can certainly be discerned in the night sky. However, we can still only see about 6,000 light years into the disk with the naked eye, and relying on the visible spectrum. Here's a rundown on why that is.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Size and Structure&lt;/li&gt;&#xA;&lt;li&gt;Low Surface Brightness&lt;/li&gt;&#xA;&lt;li&gt;Dust and Gas&lt;/li&gt;&#xA;&lt;li&gt;Limited Instrumentation&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;However we could see a bright spot if we were outside the Earth's atmosphere and could see in certain wavelengths of infrared. The article shows the following image from COBE&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iklDG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iklDG.jpg&quot; alt=&quot;Milky Way in infrared. Credit: COBE&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Milky Way in infrared. Credit: COBE &lt;a href=&quot;https://phys.org/news/2015-07-center-milky.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;False-color image of the near-infrared sky as seen by the DIRBE. Data at 1.25, 2.2, and 3.5 µm wavelengths are represented respectively as blue, green and red colors. The image is presented in Galactic coordinates, with the plane of the Milky Way Galaxy horizontal across the middle and the Galactic center at the center. The dominant sources of light at these wavelengths are stars within our Galaxy. The image shows both the thin disk and central bulge populations of stars in our spiral galaxy. Our Sun, much closer to us than any other star, lies in the disk (which is why the disk appears edge-on to us) at a distance of about 28,000 light years from the center. The image is redder in directions where there is more dust between the stars absorbing starlight from distant stars. This absorption is so strong at visible wavelengths that the central part of the Milky Way cannot be seen. DIRBE data will facilitate studies of the content, energetics and large scale structure of the Galaxy, as well as the nature and distribution of dust within the Solar System. The data also will be studied for evidence of a faint, uniform infrared background, the residual radiation from the first stars and galaxies formed following the Big Bang.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="25925" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-04-16T10:47:07.293" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30492" PostTypeId="2" ParentId="30457" CreationDate="2019-04-16T07:37:18.493" Score="2" Body="&lt;p&gt;Calculating local sidereal time exactly is not trivial as shown by the calculation for local sidereal time given at &lt;a href=&quot;https://aa.usno.navy.mil/faq/docs/GAST.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Approximate Sidereal Time&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Greenwich mean sidereal time in hours is: &lt;span class=&quot;math-container&quot;&gt;$$GMST = 6.697374558 + 0.06570982441908 D_0 + 1.00273790935 H + 0.000026 T^2$$&lt;/span&gt; &#xA;  where &lt;span class=&quot;math-container&quot;&gt;$T = D/36525$&lt;/span&gt; is the number of centuries since the year 2000.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fortunately thereare a number of online calculators that will do that for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can, however ballpark it knowing that at the equinox in March (~21st, depending on the year, because you also need to take leap years into account), local sidereal time is 12 hours different than local time (without daylight saving). i.e. 6 pm (18:00) is 06:00 LST. This means that RA 06h00m is overhead at 18:00 local time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then add 1 hour for each &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{2}$&lt;/span&gt; month or 2 hours for each month since then. So, mid-April is about 1 month after the March equinox which means that RA 08h00m is overhead at 18:00 local time, and mid-May will be 2 months so RA 10h00m is overhead at 18:00.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there, it is just like a clock - midnight is 6 hours later than 18:00 so you add 6 hours to the local sidereal time calculated earlier meaning that in mid-April RA 14h00m is overhead at midnight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this tells us is that around the March equinox (~21st), Sirius (RA 06h45m) is overhead at about 18:45, which we can round off to around 19:00 (7 pm) given the roughness of the calculation so far. We can guess then that it will set then around 6 hours later, at about 1 am.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In mid-April, RA 21h00m will be overhead at around 19:00 (7 pm), which means Sirius will have been overhead about 2 hours earlier - around 17:00 (5 pm), which means it will set around 11 pm, and in mid-May will set around 9 pm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Melbourne is located at ~145° E, ~38° S. From this you can work out the relative angle from the declination (Sirius dec. is -16°43'). Melbourne is at 38° south, so declination -38 is directly overhead. Sirius is at dec. -16°43' (so about -17°) so it will be about 20° north of directly overhead. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you would face towards north and look upwards almost to directly overhead. Knowing the LST, you know the RA that is directly overhead and can count towards the west (or east, depending on the month), and knowing that in mid-April Sirius was overhead at around 5 pm and will set around 11 pm, you can find it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB. James K's answer is really worth reading.&lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2019-04-16T07:37:18.493" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30494" PostTypeId="1" CreationDate="2019-04-16T08:49:11.300" Score="2" ViewCount="66" Body="&lt;p&gt;I've been wanting to make a mass-volume (or mass-radius) chart for astronomical objects ranging from asteroids to supermassive black holes on a log-log scale. A sort of 'Hertzsprung-Russel' type diagram but with mass and volume as the axes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to check first to see if anyone else has done this before. Google doesn't seem to turn up anything of the sort, and so I'm wondering if anybody in the esteemed audience of Astronomy.SE would know of the existence of such a diagram. &lt;/p&gt;&#xA;" OwnerUserId="17655" LastActivityDate="2019-09-13T12:26:10.293" Title="Existence of a mass-volume astronomical chart?" Tags="&lt;mass&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30495" PostTypeId="2" ParentId="30406" CreationDate="2019-04-16T09:07:20.360" Score="5" Body="&lt;p&gt;How about &lt;a href=&quot;https://en.wikipedia.org/wiki/Betelgeuse&quot; rel=&quot;noreferrer&quot;&gt;Betelgeuse&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Betelgeuse is about 640 light years away compared to 54 million light years for M87.  It has an angular diameter of 0.042 to 0.056 arc seconds while the quoted resolution of the EHT was 0.000025 arc seconds so you'd expect some detail on its surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Betelgeuse seems to be undergoing a number of rapid changes right now.  It is a young but very massive star and is racing through its evolution.  It is only ten million years old but is expected to explode as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Type_II_supernova&quot; rel=&quot;noreferrer&quot;&gt;type II supernova&lt;/a&gt; within the next million years.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best image of Betelgeuse is this from &lt;a href=&quot;https://en.wikipedia.org/wiki/Atacama_Large_Millimeter_Array&quot; rel=&quot;noreferrer&quot;&gt;ALMA&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nJbOj.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nJbOj.jpg&quot; alt=&quot;Image of Betegeuse by ALMA telescope&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the problem with this question is that there are many, many researchers and teams who would like EHT time to investigate their stuff, but those individual telescopes are already committed to other projects.  The &lt;a href=&quot;https://en.wikipedia.org/wiki/Event_Horizon_Telescope#Collaboration&quot; rel=&quot;noreferrer&quot;&gt;EHT people&lt;/a&gt; managed to pull this off because of the sensational nature of the target, a black hole!  No other subject in astronomy would get so much attention, so many institutes working together, and bring in the funds and time&lt;/p&gt;&#xA;" OwnerUserId="26593" LastActivityDate="2019-04-16T09:07:20.360" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30497" PostTypeId="2" ParentId="30494" CreationDate="2019-04-16T10:40:27.653" Score="1" Body="&lt;p&gt;Something like this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/astro-ph/9608168&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/astro-ph/9608168&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you can do literature search about the topic using NASA ADS for example, or astro-ph arxiv. They are pretty comprehensive databases of papers from astronomy/astrophysics field. If you dont find anything there you are pretty sure that no one did it. If you do find some results there (and I know for a fact that mass-size diagrams are explored/created for galaxies, clusters, etc) it might shed some light on your problem as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NASA ADS:&#xA;&lt;a href=&quot;http://adsabs.harvard.edu/abstract_service.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abstract_service.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;astro-ph arXiv:&#xA;&lt;a href=&quot;https://arxiv.org/archive/astro-ph&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/archive/astro-ph&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7094" LastEditorUserId="616" LastEditDate="2019-09-13T12:26:10.293" LastActivityDate="2019-09-13T12:26:10.293" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30498" PostTypeId="2" ParentId="1269" CreationDate="2019-04-16T11:26:54.080" Score="2" Body="&lt;p&gt;We can only see edge-on, but that gives us enough information to establish the shape of our galaxy. For many stars, we don't just have their position in our sky, we also know their distance and relative motion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This branch of astronomy is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Astrometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;astrometry&lt;/a&gt;. This is done from Earth, but in recent years we've had 2 satellite missions, &lt;a href=&quot;https://en.wikipedia.org/wiki/Astrometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hipparcos&lt;/a&gt; and &lt;a href=&quot;http://sci.esa.int/gaia/28820-summary/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gaia&lt;/a&gt;, which have given us unprecedented levels of precision for more than a billion stars, which is plenty to give us a good idea of what our galaxy looks like. &lt;/p&gt;&#xA;" OwnerUserId="361" LastActivityDate="2019-04-16T11:26:54.080" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30499" PostTypeId="1" AcceptedAnswerId="30505" CreationDate="2019-04-16T12:13:12.480" Score="3" ViewCount="849" Body="&lt;p&gt;How does more compression relate to a stronger gravitational pull. Like, when we say that a black hole is a tiny space that has 20-30 suns compressed in it, how does this increase its density and gravitational pulling power (I'm open to mathematical answers but I prefer a layman type answer for better understanding)&lt;/p&gt;&#xA;" OwnerUserId="26686" LastEditorUserId="2365" LastEditDate="2019-04-16T14:00:18.613" LastActivityDate="2019-04-20T16:16:33.730" Title="Black hole density" Tags="&lt;black-hole&gt;&lt;supermassive-black-hole&gt;&lt;density&gt;" AnswerCount="4" CommentCount="16" ContentLicense="CC BY-SA 4.0" />
  <row Id="30502" PostTypeId="2" ParentId="30487" CreationDate="2019-04-16T13:23:14.327" Score="1" Body="&lt;h2&gt;AstroPy&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Jake Vanderplas (Director of Open Software at the University of Washington eScience Institute) has written a Pythonic Preambulations post &lt;a href=&quot;https://jakevdp.github.io/blog/2015/06/13/lomb-scargle-in-python/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fast Lomb-Scargle Periodograms in Python&lt;/a&gt; which &lt;em&gt;by definition appears to be&lt;/em&gt; a good place to start to learn how to do Lomb-Scargle Periodograms... Fast... in Python. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not read it yet in full, but I've found other posts of his to be extremely useful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He starts out by explaining the implementation in &lt;a href=&quot;https://www.astropy.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;AstroPy&lt;/a&gt; which if you are going to be doing Astronomy, is probably a good thing to become familliar with.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;SciPy&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Alternatively, there is &lt;a href=&quot;https://docs.scipy.org/doc/scipy/reference/generated/scipy.signal.lombscargle.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;code&gt;scipy.signal.lombscargle&lt;/code&gt;&lt;/a&gt; which you can try assuming you have a current version of SciPy installed but want to try something different than AstroPy or want to compare performance or results. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;StackOverflow&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I found &lt;a href=&quot;https://stackoverflow.com/search?q=Lomb-Scargle&quot;&gt;over 40 posts containing 'Lomb-Scargle' in Stack Overflow&lt;/a&gt; which is a very good sign!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I notice that you've been active there and know how to ask well-received questions. Since your asking about speed and errors, StackOverflow is your go-to source! However, don't forget to be as clear as possible and quote all error messages and symptoms, and see if you can somehow add a short Minimum Complete and Verifiable Example or &lt;a href=&quot;https://meta.stackoverflow.com/q/366988&quot;&gt;MCVE&lt;/a&gt;. That's hard when your problem results from a very large dataset I know.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-04-16T13:29:24.877" LastActivityDate="2019-04-16T13:29:24.877" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30503" PostTypeId="2" ParentId="30462" CreationDate="2019-04-16T13:23:45.363" Score="3" Body="&lt;p&gt;It depends. In some areas, far-away stuff is obscured by obstacles in the foreground. In other places, we can see almost all the way to the earliest (and furthest-away) galaxies, like in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble_Deep_Field&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble Deep Field&lt;/a&gt; image: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NNPEv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NNPEv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The very earliest objects have so much redshift they disappear out of view for visible-light telescopes. That's where the &lt;a href=&quot;https://www.jwst.nasa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;JWST&lt;/a&gt; comes in: because it's an infrared telescope, it can capture objects that have too much redshift to be visible in a visible-light telescope. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Advances help us see more in other ways too. Gaia data &lt;a href=&quot;http://www.astronomy.com/news/2018/11/gaia-ghost&quot; rel=&quot;nofollow noreferrer&quot;&gt;revealed some small nearby galaxies&lt;/a&gt; which are mostly obscured by our own galaxy. &lt;/p&gt;&#xA;" OwnerUserId="361" LastEditorUserId="361" LastEditDate="2019-04-16T13:59:45.270" LastActivityDate="2019-04-16T13:59:45.270" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30505" PostTypeId="2" ParentId="30499" CreationDate="2019-04-16T14:27:46.367" Score="5" Body="&lt;p&gt;We can understand gravity as following a set of mathematical equations called &quot;General Relativity&quot; which were discovered by Einstein (and others) around the start of the 20th century.  The same gravitational equations apply to black holes, stars, planets, people, apples etc. These equations are very hard to solve. Fortunately there is a very good approximation, that was discovered by Isacc Newton about 350 years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says that there is a force between any two objects that is proportional to the mass of the each object and inversely proportional to the square of the distance between the objects. The closer the objects are to each other, the stronger is the gravitational force.  For normal objects (like you, and an apple) the size of this force is so small that it is almost undetectable.  But if one of the objects is very big (like a planet) then it becomes a very strong force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you get a couple of balls of styrofoam. They have some mass and so there is a force of gravity between them. But because they are not very dense they cannot come very close together.  If you crush the styrofoam, you make it more dense. This would let you get the balls closer together, and so the force of gravity on the surface would be larger.  If you don't push the balls closer together then the force between the balls would stay the same.  It is the distance between the masses that is important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a ball of any matter, there will be some gravitational force pulling it together. If the object is very big, its only gravity will pressurise the centre of the object. For example, the centre of the Earth is under very high pressure, due to the Earth's own gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it weren't for gravity, the gas that makes a star would fly out into space. What keeps it in one place is gravity - its own gravity. The star is literally compressed by its own gravity, and the pressure is huge. This is due to every atom being gravitationally attracted to all other atoms in the star, they are all pulled together like that - this is what we mean by &quot;the star's own gravity&quot;. If the star is massive enough, its own gravity will crush it until it collapses into a neutron star, or even into a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A star is very massive, and its own gravity would be enough to crush it, if it didn't have a nuclear furnace inside which provides the energy to stop this.  But when a star runs out of fuel, its own gravity is enough to crush the core of the star.  Since you now have the same amount of mass in a smaller ball, the gravity on the surface is greater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a black hole this process runs away (in a way that can only be described accurately by General Relativity). The gravity gets so strong that nothing can prevent the star's complete collapse to a single point (it is a lot weirder than this, because space and time are bent by the mass). Around this is a region of space from which even light can't escape, which is why black holes look black. Furtherout from the black hole, gravity is normal. Black holes don't &quot;suck&quot; they just have strong gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A black hole is not a &quot;hole&quot; in anything. Nor is it a solid object. It is a region of spacetime that is curved so much that nothing can escape this region.  All of the actual black holes that we have observed seem to have formed from collapsed stars (there are other ways to make a black hole in theory, but in practice, only stars are massive enough).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of this answers the question &quot;why does gravity get weaker as distance increases.  Perhaps that is due to how gravity spreads out from a mass. It gets weaker in a way that is analogous to how light gets weaker as you get further from a lamp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nor does this explain why gravity is proportional to mass. There doesn't seem to be an answer to this (except that in a universe with no gravity, it seems likely that no structures with living creatures could form, so we wouldn't be here to ask the question)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2019-04-20T16:16:33.730" LastActivityDate="2019-04-20T16:16:33.730" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="30507" PostTypeId="1" AcceptedAnswerId="30555" CreationDate="2019-04-16T16:10:50.740" Score="1" ViewCount="103" Body="&lt;p&gt;Measuring the total width of the Moon's umbra and penumbra at EM-L2, &lt;a href=&quot;http://www.oasis-nss.org/articles/2001/08/lagrange.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;64,700 km&lt;/a&gt;, how wide would the shadow be?  &lt;/p&gt;&#xA;" OwnerUserId="25088" LastEditorUserId="7982" LastEditDate="2019-04-20T01:57:18.110" LastActivityDate="2019-04-20T01:57:18.110" Title="How wide is the Moon's umbra and penumbra at EM-L2?" Tags="&lt;the-moon&gt;&lt;lagrange-point&gt;&lt;eclipse-non-earth-related&gt;&lt;umbra&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="30508" PostTypeId="2" ParentId="30406" CreationDate="2019-04-16T16:38:14.747" Score="0" Body="&lt;p&gt;Maybe you could take a look at some of the &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_the_most_distant_astronomical_objects&quot; rel=&quot;nofollow noreferrer&quot;&gt;most distant objects&lt;/a&gt; known? There is at least &lt;a href=&quot;https://en.wikipedia.org/wiki/GN-z11&quot; rel=&quot;nofollow noreferrer&quot;&gt;one known galaxy&lt;/a&gt; at Z=11. It will be far infrared at the source so I guess you might see a bit of it even if it does not emit much at radio frequencies. You also have the &lt;a href=&quot;https://en.wikipedia.org/wiki/ULAS_J1342%2B0928&quot; rel=&quot;nofollow noreferrer&quot;&gt;most distant quasar&lt;/a&gt;, the &lt;a href=&quot;https://en.wikipedia.org/wiki/3C_273&quot; rel=&quot;nofollow noreferrer&quot;&gt;optically brightest quasar&lt;/a&gt; and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Markarian_231&quot; rel=&quot;nofollow noreferrer&quot;&gt;nearest quasar&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/1603.04221&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cygnus A&lt;/a&gt; seems to already have been investigated using VLBI a few years ago.&lt;/p&gt;&#xA;" OwnerUserId="26403" LastActivityDate="2019-04-16T16:38:14.747" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30509" PostTypeId="1" AcceptedAnswerId="30511" CreationDate="2019-04-16T16:43:14.933" Score="2" ViewCount="257" Body="&lt;p&gt;I just can't understand why light and other massless particles are not traveling at different speeds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity effects anything with a mass but light has no mass but a black hole, and weak lensing changes the trajectory of light to create more red shifted light more than blue shifted light. That tells me light has mass. Can someone clarify this to me in Laymons terminology? &#xA;Pictures would be helpful.&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2019-04-18T19:11:53.130" LastActivityDate="2019-04-22T05:59:59.830" Title="Is blue light not traveling faster then red light in space?" Tags="&lt;redshift&gt;&lt;mass&gt;&lt;velocity&gt;&lt;quantum-mechanics&gt;&lt;blueshift&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30510" PostTypeId="2" ParentId="29405" CreationDate="2019-04-16T17:29:16.563" Score="1" Body="&lt;p&gt;Well it looks like it has finally showed up! The Institute for Astronomy in Hawaii has put up the following press release: &lt;a href=&quot;http://www.ifa.hawaii.edu/info/press-releases/Kepler47d/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Scientists Fill Out A Circumbinary Planetary System&lt;/a&gt;. Kepler-47d turns out to be the largest of the planets, with a radius 7 times that of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relevant paper is Orosz et al. (2019) &quot;&lt;a href=&quot;https://arxiv.org/abs/1904.07255&quot; rel=&quot;nofollow noreferrer&quot;&gt;Discovery of a Third Transiting Planet in the Kepler-47 Circumbinary System&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-04-17T17:41:53.417" LastActivityDate="2019-04-17T17:41:53.417" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30511" PostTypeId="2" ParentId="30509" CreationDate="2019-04-16T17:34:25.343" Score="7" Body="&lt;p&gt;Photons are massless. This doesn't depend on their energy, so doesn't depend on their frequency or wavelength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Massless particles travel at the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if we abandon particles and look at classical electrodynamics, we find that the speed of an electromagnetic wave (in vacuum) has a fixed value. It doesn't depend on wavelength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational lensing affects light of all wavelengths equally. Lensing is caused by light travelling on null geodesics - basically the &quot;straightest line&quot; it can follow in space-time that has been curved by the presence of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT (Not required for any understanding of my answer, which stands as is, but to address some comments below).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational bending of light &lt;em&gt;must&lt;/em&gt; take into account the curvature of space-time. If that is not done, then one gets the &quot;Newtonian&quot; formula which is half the deflection of a correct GR treatment. For particles with mass (which isn't what this question is about), the appropriate formula for their deflection will depend on their speed. At low speeds, the Newtonian formula will work just fine. At high speeds (approaching &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;) the deflection asymptotically approaches that for a light beam (i.e. twice the Newtonian value and the same as light). A mathematical proof of the latter in the Schwarzschild and Kerr metrics is provided by &lt;a href=&quot;https://arxiv.org/abs/gr-qc/0311010&quot; rel=&quot;nofollow noreferrer&quot;&gt;Barrabes &amp;amp; Hogan 2004&lt;/a&gt; (see their eqn 4.5). The reason, as discussed by &lt;a href=&quot;https://press.princeton.edu/titles/11169.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Misner, Thorne &amp;amp; Wheeler&lt;/a&gt; (in &quot;Gravitation&quot;, 2017 Princeton University Press, chapter 1, section 1.6 - freely available from the linked site) is that gravitation is not just due to the curvature of space, but the curvature of space-time. The tracks of light and (non-relativistic) particles follow distinctly different paths, with different curvatures, through &lt;em&gt;space&lt;/em&gt;; but their tracks through space-time have the same curvature. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-04-18T12:34:43.977" LastActivityDate="2019-04-18T12:34:43.977" CommentCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="30512" PostTypeId="1" CreationDate="2019-04-16T18:55:02.193" Score="3" ViewCount="145" Body="&lt;p&gt;We analyze the light spectrum of stars and distant planets to know characteristics like: chemical composition, temperature, mass, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting from the same principle, a probe at the border of the solar system could analyze the spectrum of the light reflected by the Earth. In theory, comparing the spectra captured by the probe with what was happening here on Earth at a time, the light was reflected to the space we could discover new spectrographic patterns that would help us look for planets similar to Earth and maybe even life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if any study like this has been done, where a whole-Earth spectrum from a distant spacecraft has been captured and analyzed, with the intent of working the problem as if it were a spectrum from an exoplanet, in order to correlate with known Earth conditions.&lt;/p&gt;&#xA;" OwnerUserId="26690" LastEditorUserId="7982" LastEditDate="2019-04-16T23:25:16.870" LastActivityDate="2019-04-19T16:37:26.750" Title="Is there a far away probe that has recorded the light spectrum from Earth as if it were an exoplanet?" Tags="&lt;earth-like-planet&gt;&lt;spectra&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30513" PostTypeId="1" AcceptedAnswerId="30531" CreationDate="2019-04-16T21:10:12.220" Score="3" ViewCount="798" Body="&lt;p&gt;How to determine the orbital eccentricity of a planet, knowing the instantaneous orbital speed of it and the distance from the Sun to the planet at that moment?&lt;/p&gt;&#xA;" OwnerUserId="26691" LastEditorUserId="7982" LastEditDate="2019-04-17T23:30:14.360" LastActivityDate="2019-04-18T01:25:06.970" Title="How to determine the eccentricity, knowing the orbital speed and distance" Tags="&lt;fundamental-astronomy&gt;&lt;orbital-mechanics&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30514" PostTypeId="1" CreationDate="2019-04-16T22:31:32.170" Score="4" ViewCount="242" Body="&lt;p&gt;So according to Relativity, time moves at different rates for different observers, and distant observers will actually disagree on the order of events.  Now from what I gather this gets sorted out for objects with relative speeds less than &lt;em&gt;c&lt;/em&gt;, but would ultimately cause a paradox for anything moving faster because it would time travel.  However, I don't really see how.  On a spacetime diagram light always moves at a 45° angle, or at least some angle between 0° and 90°, and the only way I see anything effecting the past is if something moves at an angle greater than 90°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can someone please give a visual representation that explains the issue?&lt;/p&gt;&#xA;" OwnerUserId="26693" LastEditorUserId="6" LastEditDate="2020-09-18T12:40:12.990" LastActivityDate="2020-09-18T12:40:12.990" Title="Can anyone one show how speeds greater than c cause a paradox on a space time diagram?" Tags="&lt;special-relativity&gt;&lt;time-dilation&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30515" PostTypeId="2" ParentId="30406" CreationDate="2019-04-16T22:40:46.317" Score="4" Body="&lt;p&gt;&lt;strong&gt;If&lt;/strong&gt; we could assume that most of the mm-wave emission from an ordinary star is photospheric, then the EHT could make a massive contribution to measuring the radii of stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment, this fundamental property can only be measured for stars in short-period eclipsing binaries or for a small set of nearby stars and more distant giant stars using infrared interferometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The state of the art for the latter is the &lt;a href=&quot;http://www.chara.gsu.edu/public/tour-overview&quot; rel=&quot;nofollow noreferrer&quot;&gt;CHARA array&lt;/a&gt;, with an angular resolution of 200 microarcsec. The EHT can do 10 times better, opening up a thousand times more targets for angular radius measurements, that can now be combined with Gaia parallaxes to yield physical radii.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would mean that we could properly investigate the mass-radius relationship in low-mass stars, establishing whether rapid rotation and/or magnetic fields make them bigger. This would also lead to better determinations of the properties of transiting exoplanets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That much I know about, but I suspect there are other, rarer types of star that might be brought into reach and others could be studied with more precision. I imagine that following the radius evolution of pulsating variables like Mira would be &lt;em&gt;easy&lt;/em&gt; - they have angular dianeters of &lt;span class=&quot;math-container&quot;&gt;$\sim 10$&lt;/span&gt; milliarcsec. But the nearest Cepheids have radii of around 40 times the Sun at about 400 light years (e g. Polaris). This would have an angular diameter of 1 milliarcsec, so significant progress might be made here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another place where super resolution at mm wavelengths would be highly advantageous, is in the study of protoplanetary disks. The mm-wave observatory ALMA has already yielded some exquisite images of disks around nearby young stars at angular resolutions of tens of milliarcsec. These reveal the possible traces of rings and gaps marking the onset of planetary formation. Presumably, observations at a much finer scale could be used to test detailed hydrodynamic models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, I have no idea if any of the above is feasible in terms of source surface brightness!&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-04-16T22:40:46.317" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30516" PostTypeId="2" ParentId="30513" CreationDate="2019-04-16T23:14:33.630" Score="-1" Body="&lt;p&gt;You can't do that. The object has 6 degrees of freedom and hence needs 6 variables to describe it. A common method is to use the orbital elements given &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_elements&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&#xA;A description is given by parameters such as eccentricity, semi-major axis and inclination.  &lt;/p&gt;&#xA;" OwnerUserId="16926" LastActivityDate="2019-04-16T23:14:33.630" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30517" PostTypeId="2" ParentId="30512" CreationDate="2019-04-16T23:43:02.430" Score="5" Body="&lt;p&gt;&lt;strong&gt;TL;DR: (1) we don't need to go very far to measure the spectrum of Earth's reflected light: a satellite in orbit around the Earth could easily do that; however, (2) detecting the reflected light of an Earth-size planet is &lt;em&gt;extremely&lt;/em&gt; difficult: we currently use other techniques with much greater success.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia article on &lt;a href=&quot;https://en.wikipedia.org/wiki/Methods_of_detecting_exoplanets&quot; rel=&quot;noreferrer&quot;&gt;how scientists detect exoplanets&lt;/a&gt; provides a good introduction and overview - well worth a quick read. In summary:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the thousands of exoplanets detected so far have been found using the &lt;em&gt;transit photometry&lt;/em&gt; method, where the measured brightness of the star temporarily dips because a planet is blocking some of the light as it transits across the face of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next most successful technique is to measure changes in the &lt;em&gt;radial velocity&lt;/em&gt; of the star to deduce the presence of a planet (by the &quot;wobble&quot; its gravity causes). This was the most productive method until 2012, after which transit photometry became the dominant detection method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;However, neither of these methods are looking for reflected light from the planet.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to deduce the presence of a planet from periodic changes to a star's brightness that would correspond with a nearby planet going through its phases (like how the Moon is bright when full and &quot;dark&quot; when new), but we're typically talking about a Jupiter-sized exoplanet in a close orbit - definitely not a relatively &quot;tiny&quot; Earth-like planet further out from the star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also possible to &lt;em&gt;directly detect&lt;/em&gt; the reflected light, or the infrared light emitted by a &quot;hot&quot; planet, or light that has been &lt;em&gt;polarised&lt;/em&gt; by the planet's atmosphere – but all these methods rely on a much-bigger-than-Jupiter planet orbiting a relatively small/dim star, and preferably in our local stellar neighbourhood, since the light we're trying to detect is exceedingly faint. Note that currently the &lt;em&gt;smallest&lt;/em&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_directly_imaged_exoplanets&quot; rel=&quot;noreferrer&quot;&gt;directly-imaged exoplanet&lt;/a&gt; is still more massive than Jupiter, and Jupiter itself is over 300 times more massive than the Earth (and ten times the Earth's radius, which is even more relevant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that &lt;a href=&quot;https://en.wikipedia.org/wiki/Proxima_Centauri_b&quot; rel=&quot;noreferrer&quot;&gt;Proxima b&lt;/a&gt; (an Earth-size planet orbiting our very closest neighbour, the red dwarf Proxima Centauri) was discovered using the radial velocity method, and while efforts are under way to observe it directly, &lt;em&gt;no direct image has yet been obtained&lt;/em&gt;. This might give you an indication of just how hard it is to directly image such a (relatively) tiny planet!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, we don't need to send a probe to the edge of the solar system to work out what we're looking for, as we already know exactly what the reflected light from Earth looks like – the iconic &lt;a href=&quot;https://en.wikipedia.org/wiki/Earthrise&quot; rel=&quot;noreferrer&quot;&gt;Earthrise&lt;/a&gt; being the most famous example. And in 1990 &lt;a href=&quot;https://en.wikipedia.org/wiki/Voyager_1&quot; rel=&quot;noreferrer&quot;&gt;Voyager 1&lt;/a&gt; took a famous photo of the Earth from 6 billion kilometres (41 AU) away: see the &lt;a href=&quot;https://en.wikipedia.org/wiki/Pale_Blue_Dot&quot; rel=&quot;noreferrer&quot;&gt;Pale Blue Dot&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Sept 2017, Voyager 1 was 21 billion km (140 AU) from the Sun - by far the most distant probe we've ever sent. Nonetheless, that's a long way from the &quot;border of the solar system&quot;. Voyager has entered the interstellar medium but it hasn't yet left the Kuiper Belt, and it will be another 300 years before it reaches the Oort Cloud, and &lt;em&gt;tens of thousands of years&lt;/em&gt; before it reaches the real edge of the Solar System.&lt;/p&gt;&#xA;" OwnerUserId="14094" LastEditorUserId="14094" LastEditDate="2019-04-17T07:42:03.253" LastActivityDate="2019-04-17T07:42:03.253" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30518" PostTypeId="1" CreationDate="2019-04-16T23:46:16.820" Score="1" ViewCount="1076" Body="&lt;p&gt;Why is Earth called a terrestrial planet even though it is 70% covered by water?&lt;/p&gt;&#xA;" OwnerUserId="26672" LastEditorUserId="16685" LastEditDate="2019-04-17T06:36:31.007" LastActivityDate="2019-04-17T19:41:36.570" Title="Why is the Earth called terrestrial? Can we call the Earth an ocean planet?" Tags="&lt;galaxy&gt;&lt;earth&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30519" PostTypeId="2" ParentId="30499" CreationDate="2019-04-17T04:19:00.460" Score="5" Body="&lt;p&gt;It's simply because you can &quot;get closer&quot; to it, that's all. No special sauce.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know how gravity is pretty weak far away, and gets stronger close by? The closer you get to the Sun - more specifically to the center of the Sun, because that's how you measure the distance - the greater the pull.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, once you reach the Sun's surface, there's a problem. As you dive under the surface, the stuff above you, the layers of the Sun you're leaving behind as you keep diving in more deeply, are pulling you in the opposite direction. Part of the Sun is pulling you towards the center as before, but the other part is pulling you up. They fight each other. So gravity actually gets weaker when you dive under the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if the Sun was smaller? Same mass, just more compressed. Then you could get closer to it (measured to the center) without diving under the surface (which would weaken the pull). Gravity could get stronger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now make it even smaller. You could get pretty freakin' close to it without touching the surface. The pull of gravity could be enormous, and from the same mass, too. It's just that the distance to the center could get smaller before you even touch it, that's what makes the difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes are just an extreme example of this. There's no magic, they're just super-tiny for their mass. So you can get very close to them, and gravity just keeps increasing A LOT as you do so. Eventually you reach the event horizon where gravity is so strong that space itself gets weird and you can't get out anymore.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2019-04-17T07:52:33.553" LastActivityDate="2019-04-17T07:52:33.553" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="30520" PostTypeId="2" ParentId="30518" CreationDate="2019-04-17T06:07:56.820" Score="10" Body="&lt;p&gt;There really isn't a lot of water on the Earth. The image below, from the &lt;a href=&quot;https://www.usgs.gov/special-topic/water-science-school/science/where-earths-water?qt-science_center_objects=0#qt-science_center_objects&quot; rel=&quot;noreferrer&quot;&gt;USGS science school&lt;/a&gt;, graphically illustrates the volume of the water on (and in) the crust of the Earth, relative to the volume of everything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SQqmi.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SQqmi.jpg&quot; alt=&quot;water on Earth&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bigger blue ball is all the salt water, from all the oceans &amp;amp; seas, the small one is all the fresh water, from the rivers, lakes and aquifers. You may need to zoom in to see the fresh water ball...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is more water dissolved in the mantle, but we can't get at it (with current technology), apart from what's released in volcanic eruptions. &lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2019-04-17T06:07:56.820" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30522" PostTypeId="2" ParentId="30499" CreationDate="2019-04-17T09:50:57.667" Score="0" Body="&lt;p&gt;Black hole has very high gravitational pull not because it is small but because it has huge mass. Gravitational pull does not depend on SIZE but only MASS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example is EARTH is compressed to a pebble size( 10 mm in diameter) it will become a tiny black hole. However it will not affect the moon!. Moon will not understand where the earth has gone!. It will remain in the same orbit as before. &lt;/p&gt;&#xA;" OwnerUserId="26665" LastActivityDate="2019-04-17T09:50:57.667" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30523" PostTypeId="2" ParentId="29971" CreationDate="2019-04-17T13:13:44.543" Score="-6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does matter accelerate to the speed of light as it approaches the singularity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In a way. But let's get rid of that word &quot;singularity&quot;, and replace it with &quot;black hole&quot;. Yes, falling bodies fall faster and faster. As you know, falling bodies do not slow down. Taken to the limit, the falling body would be falling at the speed of light. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since the event horizon is defined as when the escape velocity reaches the speed of light for a black hole&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where did you get that definition from? I don't recognise it. Let's just agree that the event horizon is the location from which the upward light beam cannot escape. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Would matter accelerate near the speed of light as it approaches the singularity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, but again let's replace singularity with black hole. As the falling body approaches the black hole, it falls faster and faster. Eventually it approaches the speed of light. But there's a catch, and it is a &lt;em&gt;whopper&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Accretion disks are known to move t relativistic speeds near the event horizon, but perhaps oddities occur beyond this point?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Perhaps. However we don't know what happens on the other side of the event horizon. But we do know that oddities happen on this side of the event horizon. Such as gamma ray bursts. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The event horizon can be derived from setting the escape velocity of an object to the speed of light, it is a point of no return...I understand there are better definitions, but this is not really germane to the question at hand. How about from the perspective of another in-falling observer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See what Einstein said in &lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol7-trans/156?highlightText=%22spatially%20variable%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;1920&lt;/a&gt;: &lt;em&gt;“Second, this consequence shows that the law of the constancy of the speed of light no longer holds, according to the general theory of relativity, in spaces that have gravitational fields. As a simple geometric consideration shows, the curvature of light rays occurs only in spaces where the speed of light is spatially variable”&lt;/em&gt;. The falling body is falling into the black hole because the speed of light is reducing. It falls faster and faster, until at some point it's falling at the local speed of light. Then it erupts into a gamma ray burst.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some strange reason not many people know about this. But see the 2013 AMPS paper &lt;a href=&quot;https://arxiv.org/abs/1304.6483&quot; rel=&quot;nofollow noreferrer&quot;&gt;an apologia for firewalls&lt;/a&gt;. Tucked away in the conclusion is footnote 31, containing a reference 87 to Friedwardt Winterberg’s 2001 paper &lt;a href=&quot;https://www.degruyter.com/downloadpdf/j/zna.2001.56.issue-12/zna-2001-1216/zna-2001-1216.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;gamma ray bursters and Lorentzian relativity&lt;/a&gt;. Winterberg talks about the direct conversion of an entire stellar rest mass into gamma ray energy. See the Wikipedia &lt;a href=&quot;https://en.wikipedia.org/w/index.php?title=Gamma-ray_burst&amp;amp;oldid=871852709&quot; rel=&quot;nofollow noreferrer&quot;&gt;gamma ray burst article&lt;/a&gt; and note that &lt;em&gt;“a typical burst releases as much energy in a few seconds as the Sun will in its entire 10-billion-year lifetime”&lt;/em&gt;. I think this is the hard scientific evidence that Winterburg is correct, it ties in with what Einstein said about the speed of light, and it clears up the issue wherein falling bodies are said to accelerate to the speed of light &lt;em&gt;and&lt;/em&gt; decelerate to a halt.   &lt;/p&gt;&#xA;" OwnerUserId="9772" LastEditorUserId="9772" LastEditDate="2019-04-17T13:20:14.283" LastActivityDate="2019-04-17T13:20:14.283" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30524" PostTypeId="2" ParentId="30509" CreationDate="2019-04-17T13:51:20.553" Score="-7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is blue light not travelling faster then red light in space?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No it isn't. Oddly enough, it isn't travelling faster than red light in a prism either. Here's an analogy: A little old lady scurries along the pavement at 4mph. A tall young man strides along the pavement at 4mph. When the pavement is clear, that's like space. When the pavement is crowded, that's like a prism. Then the little old lady and the tall young man are both moving at 4mph, but the tall young man has big strides so he spends longer getting round the people in his way. So the little old lady makes better progress. So it looks like she's moving faster.      &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I just can't understand why light and other massless particles are not traveling at different speeds?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As far as we can tell photons and neutrinos travel at the same speed in space regardless of their energy. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Gravity effects anything with a mass but light has no mass but a black hole, and weak lensing changes the trajectory of light to create more red shifted light more than blue shifted light. That tells me light has mass. Can someone clarify this to me in Laymans terminology? Pictures would be helpful.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gravity doesn't just affect anything with mass. Light is massless, but gravity affects it. As to why, see what Einstein said in &lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol7-trans/156?highlightText=%22spatially%20variable%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;1920&lt;/a&gt;: &lt;em&gt;“Second, this consequence shows that the law of the constancy of the speed of light no longer holds, according to the general theory of relativity, in spaces that have gravitational fields. As a simple geometric consideration shows, the curvature of light rays occurs only in spaces where the speed of light is spatially variable”&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did you see what he said about light curving because the speed of light is spatially variable? It's rather like the way sonar waves curve in the ocean:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/r6dpT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/r6dpT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Image from FAS and the US Navy, see &lt;a href=&quot;https://fas.org/man/dod-101/navy/docs/es310/syllabus.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;course ES310 chapter 20&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also see Professor Ned Wright’s &lt;a href=&quot;http://www.astro.ucla.edu/~wright/deflection-delay.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Deflection and Delay of Light&lt;/a&gt; article. He says this: &lt;em&gt;“In a very real sense, the delay experienced by light passing a massive object is responsible for the deflection of the light. The figure below shows a bundle of rays passing the Sun at various distances”&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DU50e.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DU50e.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Gif from Ned Wright’s Deflection and Delay of Light&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That delay ties in what Einstein said. Which ties in with the Shapiro delay. See the &lt;a href=&quot;https://en.wikipedia.org/wiki/Shapiro_time_delay&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; on that, and note this: &lt;em&gt;&quot;according to the general theory, the speed of a light wave depends on the strength of the gravitational potential along its path&quot;&lt;/em&gt;. It doesn't matter what the wavelength is, the light curves the same regardless of wavelength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Ned Wright points out that the deflection of light is twice the Newtonian deflection of light. The latter is based on the deflection of matter, see for example &lt;a href=&quot;https://arxiv.org/abs/physics/0508030&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/physics/0508030&lt;/a&gt;. That ought to tell you that what you usually hear about light and matter  &lt;em&gt;&quot;following the curvature of spacetime&quot;&lt;/em&gt; is just &lt;a href=&quot;https://en.wikipedia.org/wiki/Lie-to-children&quot; rel=&quot;nofollow noreferrer&quot;&gt;lies to children&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="9772" LastEditorUserId="9772" LastEditDate="2019-04-18T10:48:56.760" LastActivityDate="2019-04-18T10:48:56.760" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30525" PostTypeId="1" AcceptedAnswerId="30532" CreationDate="2019-04-17T14:18:09.893" Score="2" ViewCount="47" Body="&lt;p&gt;Disclaimer: I have no formal scientific training.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understand, surface temperature effects on Earth extend only so far into the ground so that at a certain depth the temperature remains constant around the globe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has any attempt been made to identify/predict subsurface temperature fluctuations and/or constant temperature depth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The reason I am interested in looking under that particular rock, so to speak is that given the dark side/light side temperature difference on the moon I am wondering if we can predict the ideal location and depth on the moon to find a shallow surface layer that leads to a constant temperature in a range suitable for human/plant life without requiring machinery to preserve the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.physicsforums.com/threads/temperature-for-3-or-4ft-underground.826193/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.physicsforums.com/threads/temperature-for-3-or-4ft-underground.826193/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26701" LastActivityDate="2019-04-18T00:21:49.487" Title="Do we know the fluctuations and constant temperature depth of the moon?" Tags="&lt;the-moon&gt;&lt;temperature&gt;&lt;astrobiology&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30526" PostTypeId="1" AcceptedAnswerId="30537" CreationDate="2019-04-17T14:58:19.153" Score="4" ViewCount="68" Body="&lt;p&gt;Disclaimer: I have no formal training in any scientific field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have we located any material suitable for solar panels already on the moon? Or does it appear that they would have be imported?&lt;/p&gt;&#xA;" OwnerUserId="26701" LastActivityDate="2019-04-18T04:52:32.000" Title="Do we know the availability of any material suitable for solar cells on the moon?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30527" PostTypeId="1" AcceptedAnswerId="33580" CreationDate="2019-04-17T15:40:29.883" Score="1" ViewCount="116" Body="&lt;p&gt;In the event that we want to dig into the surface of either face of the moon do we know yet if the surfaces of the moon are particularly difficult or particularly easy to dig in?&lt;/p&gt;&#xA;" OwnerUserId="26701" LastActivityDate="2019-10-04T02:04:53.873" Title="How impermeable are the surfaces of the moon?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="30528" PostTypeId="2" ParentId="30518" CreationDate="2019-04-17T17:58:16.683" Score="3" Body="&lt;p&gt;From John Dvorak's comment, the word &quot;&lt;a href=&quot;https://en.wiktionary.org/wiki/terrestrial#Etymology&quot; rel=&quot;nofollow noreferrer&quot;&gt;terrestrial&lt;/a&gt;&quot; derives from the Latin &quot;&lt;a href=&quot;https://en.wiktionary.org/wiki/terrestris#Latin&quot; rel=&quot;nofollow noreferrer&quot;&gt;terrestris&lt;/a&gt;&quot;, meaning &quot;of or pertaining to the Earth&quot;, which itself is related to the Latin word &quot;&lt;a href=&quot;https://en.wiktionary.org/wiki/terra#Latin&quot; rel=&quot;nofollow noreferrer&quot;&gt;terra&lt;/a&gt;&quot; meaning &quot;Earth&quot;. So Earth is &lt;em&gt;by definition&lt;/em&gt; a terrestrial planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &lt;a href=&quot;https://astronomy.stackexchange.com/a/30520/24157&quot;&gt;PM 2Ring notes&lt;/a&gt;, the quantity of water in the Earth is relatively small, corresponding to roughly 0.02% of the mass of the planet. There may be a similar amount in the mantle that's not accounted for in the estimate linked there, but it is still a very small proportion of the mass. The majority of the Earth is in the form of rock, with the second component being the iron/nickel core accounting for 33% of the mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term &quot;ocean planet&quot; is often used to refer to planets that are scaled-up versions of Ganymede or Pluto that have migrated into the habitable zone, see for example Léger et al. (2004) '&lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2004Icar..169..499L/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;A new family of planets? &quot;Ocean-Planets&quot;&lt;/a&gt;'. These planets contain far greater quantities of water: the case considered in that paper has 50% water by mass, similar to Ganymede. In particular the base of the ocean on such planets is not the water/rock boundary as it is on Earth: instead the ocean is separated from the rocky core by a layer of high-pressure ice.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-04-17T19:41:36.570" LastActivityDate="2019-04-17T19:41:36.570" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30529" PostTypeId="2" ParentId="30514" CreationDate="2019-04-17T20:09:42.257" Score="1" Body="&lt;p&gt;Sometimes spacetime diagrams can be misleading about paradoxes, though, because they might tend to suggest to you that information traveling from your present here, to what you regard (in your coordinates) as the past &lt;em&gt;somewhere else&lt;/em&gt; is &#xA; paradoxical, but it's actually not obvious if and when such a situation should be regarded as any kind of problem.  That's because what you regard as the past is purely a coordinate issue, it's just the language you have chosen to describe when things happen.  Thus if you use a strange coordinate system, you can send information into what you call the past (in the sense of an earlier time coordinate) without any problems at all.  So there has to be some kind of limitation on the type of coordinate being used before this should be regarded as a paradox.  As such, we want a way to understand what is the paradox in a way that is not married to the coordinates used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it's true that if we stick to coordinates motivated by the Einstein simultaneity convention, it turns out that we cannot send information anywhere that we regard as the past in those coordinates.  But it's not obvious why that would lead to a paradox if we could, because the only true paradox is if we can send information into our &lt;em&gt;own&lt;/em&gt; past &lt;em&gt;here&lt;/em&gt; where we are.  If we could do that, it would be a paradox in any coordinate system we are using, because we could tell ourselves not to do something that we've already done.  So your question could be reframed, if it were possible to send information faster than light in Einstein-type coordinates (the coordinates that Lorentz transform into each other), how would that allow us to send information to ourselves in our own past?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, you don't need a spacetime diagram, you only need to understand the relativity of simultaneity.  It's also easiest to use the limit of superfast information, i.e., infinite speed.  If you use Einstein coordinates to match up &quot;today&quot; with somewhere that is stationary with respect to you and is the distance of the Andromeda galaxy, you get the &quot;Andromeda paradox,&quot; which says that if you simply break into a sprint toward that place, and reassess &quot;what day is it there&quot;, you will find your Einstein coordinates tell you it is now &lt;em&gt;tomorrow&lt;/em&gt; there, not today any more.  If you then turn around and sprint away from that place, and reasses, you will conclude it is now &lt;em&gt;yesterday&lt;/em&gt; there.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, by repeating that sprint back and forth, you can make that distant place relive today over and over again.   Of course it would be silly to think beings in that place really would relive that day, but that's what the coordinates give you.  That's not a paradox, that's just exactly what does happen in Einstein coordinates under the Lorentz transformation.  But if you couple this with two-way superluminal signals, taken to the extreme of infinite speed to make it really clear, what could happen is that if when you are standing still, you and someone at this Andromeda distance agree the date is April 17 in both places, when you sprint toward Andromeda you will think it is April 18 there and April 17 here, such that when you send an infinitely fast signal (in that reference frame) you will know they will not receive it until April 18.  But if you send an infinitely fast signal when you are sprinting away (infinitely fast in that reference frame), you will know they will receive it on April 16.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's only problematic if people at the Andromeda distance can also send an infinitely fast signal, in their reference frame, back to you.  Then if you say &quot;hello&quot; while sprinting away, they will receive that on April 16, and can send back a &quot;same to you&quot; signal that you will receive on April 16!  That's yesterday for you, and you will be puzzled what they were talking about, since you don't even know you will be doing some sprints the next day, let alone that you will be sending messages to Andromeda.  So that's what we said is the real paradox-- the ability to send information back into our &lt;em&gt;own&lt;/em&gt; past right where we are.  That requires two-way superluminal communication-- one-way superluminal speed could just be an artifact of weird coordinates chosen.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2019-04-17T20:16:24.640" LastActivityDate="2019-04-17T20:16:24.640" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30530" PostTypeId="2" ParentId="30514" CreationDate="2019-04-17T23:16:28.773" Score="8" Body="&lt;p&gt;Suppose there was a magic gun that fired a bullet at ten times the speed of light relative to the firer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have the only such gun, and I don't move then there is no paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now suppose these guns are generally available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At time zero, an observer flies past me at velocity 0.5 c and I give them one of these guns. 7 years later, I fire the gun after them, so in my frame of reference the bullet will catch them at about time  7.36 years when they are about 3.68 ly away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get these numbers I just need to solve &lt;span class=&quot;math-container&quot;&gt;$10*(t-7) = 0.5*t$&lt;/span&gt; to calculate when the bullet overtakes the spaceship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In their frame of reference though, that occurs about &#xA;time 6.38 years after they pass me, at which point, I am about 3.19 ly away from them (we both agree on our relative velocity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is calculated using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lorentz_transformation&quot; rel=&quot;noreferrer&quot;&gt;Lorentz transformations&lt;/a&gt;. &lt;span class=&quot;math-container&quot;&gt;$$t' = \gamma(t - vx)$$&lt;/span&gt; taking &lt;span class=&quot;math-container&quot;&gt;$c = 1$&lt;/span&gt;. Here &lt;span class=&quot;math-container&quot;&gt;$t = 7.38$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\gamma = \frac{1}{\sqrt{1-v^2}} \sim 1.15$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$v = 0.5$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$x = 3.68$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They now fire back and we can do a similar calculation, but in their frame of reference. So in their frame the bullet will pass me at time &lt;span class=&quot;math-container&quot;&gt;$t_1$&lt;/span&gt; when &#xA;&lt;span class=&quot;math-container&quot;&gt;$(t_1 - 6.38)*10 = 0.5 t_1$&lt;/span&gt; which tells us that &lt;span class=&quot;math-container&quot;&gt;$t_1 = 6.71$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we use the Lorentz transformation again to determine what time that is in our frame. We find that it is 5.81 years. A bit more than a year before we fired our bullet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key thing is that &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; times the speed of light in one frame will be backwards in time in another frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diagrams below illustrates this scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IjiEk.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IjiEk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both diagrams show exactly the same events. The one on the left uses the time and space coordinates of the original observer (&quot;you&quot;) the one on the right uses the coordinates of the other observer. In bith cases time is on the y axis, space on the x axis. Units are years and light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black diagonal lines are some possible world lines of light rays.&#xA;The red lines are lines of constant x (solid) or t (dashed) as determined in &quot;your&quot; frame of reference. The thick red arrow (at x = 0) is your world-line. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blue lines are lines of constant x (solid) or t (dashed) as determined by the ship. The thick blue arrow is the ship's world-line. As can be seen you meet at the event that you both agree is x = 0, t= 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The green line is the bullet that &quot;you&quot; fire. It crosses three solid red lines and goes a bit less than half way from one dashed red line to the next (travels 3+ ly in one 0.36 y). The cyan line is the return bullet. It crosses 3 solid blue lines and goes rather less than half way from one dashed blue line to the next (travels 3+ ly in about 0.3y). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming I've got this right, all the intersections and crossing patterns of red, blue and black lines, and the four world-lines should agree between the two diagrams, since they really do depict the same events.  In particular both diagrams show a causal loop, in which the cyan bullet arrives back at the red observer before the green bullet is fired. They disagree about which bullet travels back in time, but that's relativity for you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this tells us practically is that any physics which allows FTL travel but not time travel must &lt;strong&gt;not&lt;/strong&gt; be Lorentz invariant, so your gun will work differently in different frames of reference. We know of no such physics at present, and see no clues which suggest it might exist.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="21928" LastEditDate="2019-04-23T09:31:24.803" LastActivityDate="2019-04-23T09:31:24.803" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="30531" PostTypeId="2" ParentId="30513" CreationDate="2019-04-17T23:59:23.927" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How to determine the orbital eccentricity of a planet, knowing the instantaneous orbital speed of it and the distance from the Sun to the planet at that moment?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are given three things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; &quot;instantaneous orbital speed&quot; (a scalar)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; &quot;distance from the Sun to the planet at that moment&quot; (a scalar)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$GM$&lt;/span&gt; since the OP specifies &quot;the Sun&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Let's see how far we can get with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Vis-viva_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;vis-viva equation&lt;/a&gt; and look at the problem &lt;em&gt;purely from an energy standpoint&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$v^2 = GM(2/r-1/a)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$a = 1/(2/r-v^2/GM).$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at a numerical example. Say &lt;span class=&quot;math-container&quot;&gt;$r = 1AU \approx 1.5 \times 10^{11}$&lt;/span&gt; meters, and &lt;span class=&quot;math-container&quot;&gt;$v = 25,000$&lt;/span&gt; m/s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know then that the semimajor axis of the orbit must be 0.773 AU. If the orbit reaches 1AU, the velocity will be 25,000 m/s. If it reaches 1.1 AU (which it might or might not, depending on eccentricity) it's velocity will be 21,500 m/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're almost there but not quite. We need one more piece of information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The periapsis and apoapsis &lt;span class=&quot;math-container&quot;&gt;$r_p$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$r_a$&lt;/span&gt; are related to the semimajor axis and eccentricity by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$r_p = (1-e)a$$&lt;/span&gt; &#xA;&lt;span class=&quot;math-container&quot;&gt;$$r_a = (1+e)a$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for example if you had mentioned that the apoapsis was 1.1 AU, then we could calculate eccentricity from&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$e = r_a/a -1$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and get &lt;span class=&quot;math-container&quot;&gt;$e = 0.4228$&lt;/span&gt; AU and a periapsis &lt;span class=&quot;math-container&quot;&gt;$r_p = 0.387$&lt;/span&gt; AU.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-04-18T01:25:06.970" LastActivityDate="2019-04-18T01:25:06.970" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30532" PostTypeId="2" ParentId="30525" CreationDate="2019-04-18T00:21:49.487" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Do we know the fluctuations and constant temperature depth of the moon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, at some Apollo landing sites, it's been carefully measured! The astronauts make deep holes in the Moon's regolith and put in thermal probes that precisely measured temperature. These signals were radioed back to Earth by a solar-power transmitter, in some cases for years. These experimental packages were called ALSEP and included seismometers and passive retroreflectors and other &quot;goodies&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to quote verbatim &lt;a href=&quot;https://space.stackexchange.com/a/26503/12102&quot;&gt;this answer&lt;/a&gt; from Space Exploration SE since it fits so nicely here.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;tl;dr:&lt;/strong&gt; The number is often given as 252 +/- 3 K (or about -21°C or -6°F), so @BowlOfRed pretty much &lt;a href=&quot;https://space.stackexchange.com/questions/19906/constant-lunar-sub-surface-temperature/26503#comment53095_19906&quot;&gt;nailed it&lt;/a&gt; using first principles! &lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;Without atmosphere, equilibrium temperature at 1AU is about -17C. I would expect that to be the temp &quot;near&quot; the surface around the equator. Colder as the latitude increases.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;p&gt;Reading the references, it seems that the &lt;a href=&quot;https://en.wikipedia.org/wiki/Diurnal_cycle&quot; rel=&quot;nofollow noreferrer&quot;&gt;diurnal oscillation&lt;/a&gt; (what Earthies might call &quot;monthly&quot; or Lunies might call &quot;daily&quot;) in temperature decays with a 1/e depth of about 30 centimeters, so by the time one reaches 1 meter, the temperature &quot;stabilizes&quot; &lt;em&gt;temporally&lt;/em&gt;, but of course will increase with depth since the Moon's interior is hot. You can see this in Figure 2 of the last paper (&quot;Subsurface temperature histories, covering 3.5 years...&quot;) where the oscillations in temperature at different depths along the probe are plotted.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Further, some of this data can already be found in @KimHolder's question &lt;a href=&quot;https://space.stackexchange.com/q/14422/12102&quot;&gt;Does the temperature near the surface of the Moon rise rather quickly with depth?&lt;/a&gt; who quotes as NASA News article &lt;a href=&quot;https://science.nasa.gov/science-news/science-at-nasa/2010/12jul_rabbithole/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Down the Lunar Rabbit-hole&lt;/a&gt; which then quotes Mark Robinson of Arizona State University, principal investigator for the LRO camera.&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&quot;The tunnels offer a perfect radiation shield and a very benign thermal environment,&quot; says Robinson. &quot;Once you get down to 2 meters under the surface of the Moon, the temperature remains fairly constant, probably around -30 to -40 degrees C.&quot;&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;p&gt;Why the difference between the &quot;ground truth&quot; actual temperature measurements discussed below and the quote of a quote?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SsGMn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SsGMnm.jpg&quot; alt=&quot;astronaut measures lunar temperature&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(image &lt;a href=&quot;https://ntrs.nasa.gov/search.jsp?R=19750006612&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt; also &lt;a href=&quot;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19780005001.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;) My guess is that the -30 to -40 number is a mid-latitude number, while the following concentrates on nearer to equator values. There's no physical way the temperature can drop by 10 or 20C in 1 meter of increasing depth, so the difference is more context and location.&lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;I'm not an expert on planetary science, but the oft-cited number for an average sub-surface temperature about 1 meter below the surface seems to come from the following papers:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;a href=&quot;https://link.springer.com/article/10.1007/BF00562006&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Apollo 15 lunar heat-flow measurement&lt;/a&gt;, Langseth, M.G., Clark, S.P., Chute, J.L. et al. The Moon (1972) 4: 390. &lt;a href=&quot;https://doi.org/10.1007/BF00562006&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://doi.org/10.1007/BF00562006&lt;/a&gt; &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&lt;strong&gt;Abstract:&lt;/strong&gt;&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;The heat-flow experiment is one of the Apollo Lunar Surface Experiment Package (ALSEP) instruments that was emplaced on the lunar surface on Apollo 15. This experiment is designed to make temperature and thermal property measurements in the lunar subsurface so as to determine the rate of heat loss from the lunar interior through the surface. About 45 days (1 1/2 lunations) of data has been analyzed in a preliminary way. This analysis indicates that the vertical heat flow through the regolith at one probe site is 3.3 × 10−6 W/cm2 (±15%). This value is approximately one-half the Earth's average heat flow. Further analysis of data over several lunations is required to demonstrate that this value is representative of the heat flow at the Hadley Rille site. &lt;strong&gt;The mean subsurface temperature at a depth of 1 m is approximately 252.4K at one probe site and 250.7K at the other&lt;/strong&gt;. These temperatures are approximately 35K above the mean surface temperature and indicate that conductivity in the surficial layer of the Moon is highly temperature dependent. Between 1 and 1.5m, the rate of temperature increase as a function of depth is 1.75K/m (±2%) at the probe 1 site. In situ measurements indicate that the thermal conductivity of the regolith increases with depth. Thermal-conductivity values between 1.4 × 10−4 and 2.5 × 10−4 W/cm K were determined; these values are a factor of 7 to 10 greater than the values of the surface conductivity. If the observed heat flow at Hadley Base is representative of the moonwide rate of heat loss (an assumption which is not fully justified at this time), it would imply that overall radioactive heat production in the Moon is greater than in classes of meteorites that have formed the basis of Earth and Moon bulk composition models in the past.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;&lt;a href=&quot;https://ntrs.nasa.gov/search.jsp?R=19750006612&quot; rel=&quot;nofollow noreferrer&quot;&gt;In-situ measurements of lunar heat flow&lt;/a&gt; (also &lt;a href=&quot;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19780005001.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;) in Soviet-American Conference on Geochemistry of the Moon and Planets, Langseth, M. G., and S. J. Keihm, NASA Spec. Publ., NASA SP-30, 283 – 293, 1977.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/p9zpF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p9zpF.png&quot; alt=&quot;In-situ measurements of lunar heat flow&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8aZ27.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8aZ27.png&quot; alt=&quot;In-situ measurements of lunar heat flow&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;The number is often given as 252 +/- 3 K (or about -21°C or -6°F)&lt;/strong&gt; as for example in &lt;a href=&quot;https://agupubs.onlinelibrary.wiley.com/doi/abs/10.1029/2001JE001497&quot; rel=&quot;nofollow noreferrer&quot;&gt;Latitude variation of the subsurface lunar temperature: Lunar Prospector thermal neutrons&lt;/a&gt; R. C. Little W. C. Feldman S. Maurice I. Genetay D. J. Lawrence S. L. Lawson O. Gasnault B. L. Barraclough R. C. Elphic T. H. Prettyman A. B. Binder&#xA;  First published: 29 May 2003 &lt;a href=&quot;https://doi.org/10.1029/2001JE001497&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://doi.org/10.1029/2001JE001497&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;Some of the experimental and theoretical details behind the data and its analysis can be read in &lt;a href=&quot;http://adsabs.harvard.edu/full/1976LPSC....7.3143L&quot; rel=&quot;nofollow noreferrer&quot;&gt;Revised lunar heat-flow values&lt;/a&gt;, Langseth, M. G., Keihm, S. J., &amp;amp; Peters, K., Lunar Science Conference, 7th, Houston, Tex., March 15-19, 1976, Proceedings. Volume 3. (A77-34651 15-91) New York, Pergamon Press, Inc., 1976, p. 3143-3171.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jOtPe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jOtPe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zSMAa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zSMAa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SjuET.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SjuET.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-04-18T00:21:49.487" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30533" PostTypeId="1" AcceptedAnswerId="39351" CreationDate="2019-04-18T01:19:24.497" Score="9" ViewCount="409" Body="&lt;p&gt;The abstract of the new Nature paper &lt;a href=&quot;https://www.nature.com/articles/s41586-019-1090-x&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astrophysical detection of the helium hydride ion HeH+&lt;/a&gt; tell us that infrared spectroscopy from &lt;a href=&quot;https://en.wikipedia.org/wiki/Stratospheric_Observatory_for_Infrared_Astronomy&quot; rel=&quot;nofollow noreferrer&quot;&gt;SOFIA&lt;/a&gt; detected&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;...rotational ground-state transition of HeH+ at a wavelength of 149.1 micrometres in the planetary nebula NGC 7027.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It also says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In this metal-free and low-density environment, &lt;strong&gt;neutral helium atoms formed the Universe’s first molecular bond&lt;/strong&gt; in the helium hydride ion HeH+ through radiative association with protons. As recombination progressed, the destruction of HeH+ created a path to the formation of molecular hydrogen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Why was helium hydride (HeH+) the universe's first molecule? Why was it the &lt;a href=&quot;https://en.wikipedia.org/wiki/Helium_hydride_ion&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrogen helium cation&lt;/a&gt; rather than the &lt;a href=&quot;https://en.wikipedia.org/wiki/Dihydrogen_cation&quot; rel=&quot;nofollow noreferrer&quot;&gt;dihydrogen cation&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2021-05-14T05:06:01.567" LastActivityDate="2021-05-14T05:06:01.567" Title="Why was helium hydride (HeH+) the universe's first molecule?" Tags="&lt;astrochemistry&gt;&lt;infrared&gt;&lt;sofia&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="30534" PostTypeId="2" ParentId="25384" CreationDate="2019-04-18T02:47:58.257" Score="1" Body="&lt;p&gt;Something interesting from &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbit_of_the_Moon#Path_of_Earth_and_Moon_around_Sun&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1haFN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1haFN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;  Scale model of the Earth–Moon system: Sizes and distances are to&#xA;  scale. It represents the mean distance of the orbit and the mean radii&#xA;  of both bodies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Earth orbits the common barycenter of the Earth–Moon system in about &lt;strong&gt;28 days&lt;/strong&gt;, the same time the Moon orbits Earth (or more accurately, the common barycenter of the Earth–Moon system which is below Earth surface). It's estimated that the centrifugal force caused by this should be minimal on Earth surface, and only observable by a very large scale (e.g., the tides).&lt;/p&gt;&#xA;" OwnerUserId="26711" LastActivityDate="2019-04-18T02:47:58.257" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30535" PostTypeId="1" CreationDate="2019-04-18T02:54:36.873" Score="1" ViewCount="30" Body="&lt;p&gt;I'm working with &lt;strong&gt;astroML&lt;/strong&gt; in Python and I need to get the fluxes from the magnitudes in r band of &lt;strong&gt;fetch_imaging_sample()&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from astroML.datasets import fetch_imaging_sample&#xA;data = fetch_imaging_sample()&#xA;mag = data['rRaw'] # these are the magnitudes in r band&#xA;err = data['rErr'] # these are the errors in r band&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How can I get the fluxes with their errors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explain of fetch_imaging_sample: &lt;a href=&quot;http://www.astroml.org/modules/generated/astroML.datasets.fetch_imaging_sample.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.astroml.org/modules/generated/astroML.datasets.fetch_imaging_sample.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26709" LastActivityDate="2019-04-18T02:54:36.873" Title="How can I get the fluxes of fetch_imaging_sample? - astroML" Tags="&lt;python&gt;&lt;magnitude&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30536" PostTypeId="2" ParentId="16583" CreationDate="2019-04-18T03:18:30.107" Score="2" Body="&lt;p&gt;I apologize for the length of this answer, but this is a complicated subject to address.  Thank you all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I owned a camera store for many years, and have 60+ years in photo.  I got my start in photo via Astronomy in high school.  Went to the University of Iowa, started in Physics and Astronomy, but my math skills were not up to the task, so went to Radio-TV-Film.  The physics was a great basis for photography as you might well imagine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I knew just enough Physics and science to be a danger to myself and those around me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several years ago, a product called ROR (Residual Oil Remover) came to market, invented by a man from Illinois whom I knew casually.  His premise was that in today's world, much of the 'dirt and dust' we perceive on optics of all kinds, is suspended in microscopic droplets of oil in the atmosphere, and as we wear glasses and walk in or move about, those microscopic droplets, also containing dirt/dust of all sorts, coat optical surfaces and everything else, and we breath the stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would guess that this oily deposit may also be a cause for some of the 'rapid' deterioration of optical coatings on telescope mirrors, it builds up over time, dries, which forms a glue or sticky layer which also causes dust and dirt to build up more rapidly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can find ROR in a local camera store (or similar), get a bottle, I used a Kleenex tissue to clean my glasses.  Spray the ROR directly on the glass, both sides, one lens only, gently rub it until it dries, then put on the glasses and compare the views, you should be amazed.  The other effect of ROR was that as you cleaned the glass once, then a second time, the glass simply becomes slippery.  The ROR, somehow, cleans the micro pores of the glass and removes the deep down sticky, oily material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it can be used on any optical glass surface as far as I know&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was endorsed, as I remember, by NASA, FBI, Nat Geo, etc. so good credentials.&lt;/p&gt;&#xA;" OwnerUserId="26712" LastActivityDate="2019-04-18T03:18:30.107" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30537" PostTypeId="2" ParentId="30526" CreationDate="2019-04-18T04:10:54.610" Score="6" Body="&lt;p&gt;Welcome to Astronomy.SE! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon, like the earth, consists of &lt;a href=&quot;https://en.wikipedia.org/wiki/Geology_of_the_Moon&quot; rel=&quot;noreferrer&quot;&gt;25% silicon&lt;/a&gt; by mass. Silicon is one of the main elements used in producing solar cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So sourcing the material is not the problem. &lt;strong&gt;The manufacturing is.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume your interest in this matter doesn't just stop at finding the materials, but also includes &lt;em&gt;in-situ&lt;/em&gt; manufacture. I won't go for a detailed answer, but I will outline a few considerations.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;You need your own silicon industry.&lt;/strong&gt; It's the sort of thing that requires a large infrastructure behind it. For instance, monocrystalline silicon solar cells require &lt;a href=&quot;https://en.wikipedia.org/wiki/Monocrystalline_silicon&quot; rel=&quot;noreferrer&quot;&gt;high purity silicon&lt;/a&gt; (on the order of a few ppm) and pristine growing conditions for the crystal to grow. That means you need to have the infrastructure in place to mine moon rock, reduce the silica to silicon, and separate out as much impurity as you can. With the sorts of purity levels we're talking about, you'd probably need at the very least a medium-sized moon base already in place.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Making &lt;em&gt;any&lt;/em&gt; sort of solar panel is hard.&lt;/strong&gt; You need an industry to make the silicon. You need another industry to make the solar panels. This sort of stuff requires precision engineering, and complex (and heavy) tooling which you'll have to import from earth. You need a 'pyramid' of support structures and expertise to maintain such expensive equipment.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Amorphous SI is probably your best bet.&lt;/strong&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Amorphous_silicon&quot; rel=&quot;noreferrer&quot;&gt;Amorphous silicon&lt;/a&gt; seems to be your best bet. It can be applied through some form of vapour deposition, and from a brief bit of reading around, appears to be one of the easiest ways of making solar cells, though the efficiencies tend to be low, with the best reported efficiency being 14% and realistically achievable efficiencies being likely much lower than 10%. (see the bottom green line of &lt;a href=&quot;https://en.wikipedia.org/wiki/Thin-film_solar_cell#/media/File:Best_Research-Cell_Efficiencies.png&quot; rel=&quot;noreferrer&quot;&gt;this diagram&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17655" LastEditorUserId="17655" LastEditDate="2019-04-18T04:52:32.000" LastActivityDate="2019-04-18T04:52:32.000" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30538" PostTypeId="2" ParentId="30471" CreationDate="2019-04-18T08:53:43.890" Score="2" Body="&lt;p&gt;I can explain it for rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phenomenon you're looking for is called nodal precession. Meaning, when the planet has an equatorial bulge, all non-equatorial orbits will precess, will wobble around the spin axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Nodal_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Nodal_precession&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because of the bulge the gravitational force on the satellite is not&#xA;  directly toward the center of the central body, but is offset toward&#xA;  the equator. Whichever hemisphere the satellite is in it is&#xA;  preferentially pulled slightly toward the equator. This creates a&#xA;  torque on the orbit. This torque does not reduce the inclination;&#xA;  rather, it causes a torque-induced gyroscopic precession, which causes&#xA;  the orbital nodes to drift with time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For rings, any such precessing orbits would be unstable - ring particles would collide with each other. The only non-precessing, non-colliding orbits are the equatorial ones. So that's where the rings persist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for satellites, I would venture a guess that their orbits actually do not migrate towards the equatorial plane, they stay where they are, just slowly precessing. Someone please correct me if your orbital mechanics is in better shape than mine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: There's a ton of perturbations from an oblate planet, and some do apply to the inclination of the orbit. So it's not just precession. Some terms are periodic. Google &quot;inclination of orbit oblate&quot; without the quotes. It's... complex. I'll edit this answer tomorrow if I find something relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT2: This is far more complex than I thought. A thorough treatment is given in Kozai, Y. The motion of a close earth satellite:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1959AJ.....64..367K&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/1959AJ.....64..367K&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chapter 11.15 in J.M.A. Danby, Fundamentals Of Celestial Mechanics, is entirely dedicated to this topic. He talks about critical values for initial parameters where his results are not applicable, so obviously there isn't a simple conclusion that applies in all cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not going to interpret the math, sorry. That's all I had to say on this topic.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2019-04-18T20:08:01.997" LastActivityDate="2019-04-18T20:08:01.997" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30539" PostTypeId="2" ParentId="30512" CreationDate="2019-04-18T09:46:00.793" Score="4" Body="&lt;p&gt;As an addendum to Chappo's very good answer, I'll note that people have indeed used satellite and spacecraft observations of the Earth to estimate what a similar exoplanet might look like (assuming we had better telescopes and instrumentation than we currently do). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the NASA &lt;a href=&quot;https://en.wikipedia.org/wiki/Deep_Impact_(spacecraft)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Deep Impact&lt;/a&gt; spacecraft (which had visited the comet Tempel 1 in 2005) was re-purposed as &lt;a href=&quot;https://en.wikipedia.org/wiki/EPOXI&quot; rel=&quot;nofollow noreferrer&quot;&gt;EPOXI&lt;/a&gt; for more comet and asteroid flybys, searches for extrasolar planet occultations of other stars -- and, along the way, observations of the Earth in 2008. A brief discussion of one of the papers analyzing the data can be found &lt;a href=&quot;https://astrobites.org/2011/02/21/the-colors-of-a-second-earth/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And in the early 1990s, spectroscopic observations made by the Galileo spacecraft during one of its gravitational-slingshot flybys of Earth &lt;a href=&quot;http://adsabs.harvard.edu/abs/1993Natur.365..715S&quot; rel=&quot;nofollow noreferrer&quot;&gt;were analyzed (Sagan et al. 1993)&lt;/a&gt; to see what indications of life might be detectable from space, such as atmospheric oxygen and methane.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important to understand that there's no need to put something far away from the Earth (e.g., &quot;at the border of the solar system&quot;): you can take a nearby observation of the Earth and blur it and add noise to get the same effect. The important thing is to be able to look at the Earth from outside its atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternate approach is to look at the effect of a planet's atmosphere on starlight passing through it. This has been done for the Earth &lt;a href=&quot;http://www.ing.iac.es/PR/press/eclipse.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;by observing light reflected off the Moon during a lunar eclipse&lt;/a&gt;, since this sunlight passing through the Earth's atmosphere on its way to the Moon.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2019-04-18T15:28:16.810" LastActivityDate="2019-04-18T15:28:16.810" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30541" PostTypeId="2" ParentId="21943" CreationDate="2019-04-18T18:42:30.553" Score="-4" Body="&lt;p&gt;First off, Jesus death was on a Wednesday and not a Friday. Why, Creation of the Sun, moon, and stars was on the 4th day. Also, anything to do with the life of Yeshua, has to line up with Leviticus 23 which is the word Season's in Genesis 1:14. Years is the Shemitah and Jubilee Years.&#xA;    The Menorah has 7 candles. The fourth candle is called the Christ candle or servant Candle and also you have to line up His death to Daniel 9:27 in the MIDST of/Middle of so the year would have been on Passover (not a Sabbath) on Wednesday, 4000 years from Creation (Aleph-Tav in Genesis 1:1) in 30 CE (Wednesday) between 27(Sunday) to 33 CE (Saturday).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2). NASA (NOT a straight aswer) is trying to hide Trappist, Nemesis, El Shaddai, the Destroyer who brings in the Trumpet and Bowl Judgments a of Revelation in 2019 between Passover (April 20th) and Day of Atonement on October 10th. The orbital path of Nibiru( the Crossing) is an elliptical orbit and comes around every 300-350 years. Gill Broussard's work will help you out on this one or Jardalkalataol.blogspot.com.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, there was an asteroid that did occur at Yeshua's death just like the one's at  His second Coming during this week.&#xA;Asteroid 2016 JP&#xA;21.1000 H&#xA;7,292,673 km&#xA;19 April 2019&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asteroid 2017 UT2&#xA;24.6000 H&#xA;19,652,952 km&#xA;21 April 2019&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asteroid 2016 WQ3&#xA;28.8000 H&#xA;10,862,251 km&#xA;21 April 2019&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asteroid 2019 FY&#xA;22.9030 H&#xA;9,793,095 km&#xA;26 April 2019&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asteroid 2013 KJ6&#xA;19.9000 H&#xA;17,478,956 km&#xA;26 April 2019&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asteroid 137805 1999 YK5&#xA;16.6000 H&#xA;7,923,219 km&#xA;4 May 2019&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.asteroidsnear.com/year?year=2019&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.asteroidsnear.com/year?year=2019&lt;/a&gt; &#xA;As you can tell we are in the baseball park and parking lot for an exciting two weeks of hitting balls (asteroids)  as they come by the earth. At the time of this reporting, there will not be any home runs until May to October time frame with 2 hits (Asteroid by Sea and asteroid by land (wormwood) of Revelation chapter 8. &lt;/p&gt;&#xA;" OwnerUserId="26721" LastActivityDate="2019-04-18T18:42:30.553" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30542" PostTypeId="2" ParentId="27597" CreationDate="2019-04-18T19:09:46.610" Score="4" Body="&lt;p&gt;Video astronomy requires much more equipment than traditional eyepiece viewing. It helps to understand what is going on between the camera lens and the video screen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most important is the conversion of the cameras analog signal to the computer digital format. This is usually done with a frame grabber and its associated drivers. The cameras typically have 700 ish lines of resolution which are usually translated into 640 by 480 digital signal. The camera analog signal is 60 cycle interlaced. The computer display frequency is whatever the PC is set to. I'll let the Math people tell us how the drivers fit all those numbers together and make a nice picture. I think the answer is PI or at least within a decimal place or two. The numbers get rounded and every once in a while the computer screen hiccups. National Enquirer sort of news. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The frame grabbers come in a wide range of cost and reliability. There's no rhyme or reason to their reliability. They depend on the computers operating system which is not identical on every computer. The frame grabber and driver will work well on one PC but not another similarly equipped PC. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically the astro video cameras two output connectors, a coax and an S-Video. The person making the video could have proven the &quot;Waves&quot; were and artifact of the PC by simply connecting one of the cameras outputs to an analog video screen, a television with a composite input for example and the other output to the computers frame grabber. Provided the camera and analog video are functioning properly he could have shown us a rock steady image on the analog screen and the waves on the Computer screen. With a little skill he could have shown us an alien or two surfing on the waves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smoke, mirrors and snake oil. Suckers are said to be born at the rate of one per minute. I have a feeling the con artist birth rate is up there as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a bit of a shame, Video or near live viewing as many of us call it, is gaining in popularity. Having someone use what is not unusual for a Video astronomer to see just for a bit of sensationalism serves no useful purpose.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point to edit in. The uTube is a video of a video. The image we see is taken by a second video cameras pointed at a video image being displayed on a computer screen. Crunch the computer screen frequency and resolution with the frequency and resolution of the second camera. You end up with images of airplane propellers that appear to stand still or move very slowly. I wonder how much adverting revenue that sort of video would bring into uTube. &lt;/p&gt;&#xA;" OwnerUserId="24928" LastEditorUserId="24928" LastEditDate="2019-04-18T20:31:09.150" LastActivityDate="2019-04-18T20:31:09.150" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30543" PostTypeId="1" CreationDate="2019-04-18T20:13:38.013" Score="2" ViewCount="165" Body="&lt;p&gt;So the Earth's been orbiting the sun in a relatively stable orbit for billions of years now. during that time, it's been plowing through dust, and hit by comets and asteroids, presumably mostly from the outer reaches more than the inner parts of the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if the drag and increased mass should amount to a tiny, tiny fraction of Earth's mass, shouldn't that have been enough to cause our orbit to spiral inwards? Both due to slowing of the Earth's motion, and an increase in mass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="23390" LastActivityDate="2019-04-19T10:21:29.807" Title="Why hasn't the Earth fallen into the sun yet?" Tags="&lt;orbit&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30544" PostTypeId="1" CreationDate="2019-04-18T20:29:59.307" Score="4" ViewCount="216" Body="&lt;p&gt;The Big Bang is a proposed answer to the beginning of the universe, and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cosmic Microwave Background&lt;/a&gt; or CMB is a central part of the evidence for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose the measurement returned a different value than 2.7K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if it were 0.27K or 27K?  Would the Big Bang theory still work with a few minor adjustments or is the specific temperature of 2.7K critical evidence that the theory works?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I have removed a paragraph which was not mine asking about small changes in temperature. It is a fundamentally different question. I am more interested in how firm the evidence for the BB theory is, not its details. At least for this question. &#xA;However if uhoh who seems to have done the editing thinks it a fair question, I would certainly welcome it - in another post. &lt;/p&gt;&#xA;" OwnerUserId="26660" LastEditorUserId="26660" LastEditDate="2019-04-19T13:18:54.717" LastActivityDate="2019-04-20T04:03:24.790" Title="cosmic-microwave-background - 2.7 degrees but does the Big Bang care?" Tags="&lt;cosmology&gt;&lt;cosmic-microwave-background&gt;" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="30545" PostTypeId="2" ParentId="30543" CreationDate="2019-04-18T23:05:16.117" Score="6" Body="&lt;p&gt;No not nearly enough drag.  There just isn't much dust in space, and the Earth has so much momentum that the tiny amount of interplanetary gas has a tiny effect on the Earth's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's only about 5 particles of mostly hydrogen gas per cubic cm in the region around the Earth. It has a drag on the earth of a few thousand Newtons. But the mass of the Earth is &lt;span class=&quot;math-container&quot;&gt;$5\times10^{24}$&lt;/span&gt; kg, so the deceleration is of the order &lt;span class=&quot;math-container&quot;&gt;$10^{-20}$&lt;/span&gt; m/s2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this rate it would take longer than the age of the universe for drag to make a significant change to the orbit of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things were different when the planets orbited in an accretion disc. At this time, large planets could make excursions from the outer part of a stellar system to the inner, due to friction from the dust that they encounter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth does gain a little mass from dust, rocks and asteroids that continuously hit it, but it also loses a little mass to atmospheric lose. The total is slightly negative (overall Earth loses a little mass) but this doesn't affect the orbit, as orbital velocity is independent of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Large bodies do change the orbit very very slightly, but comet and asteroid impacts happen from all directions, and there is little cumulative effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravity of other planets does perturb the orbit of the Earth, but not to make it spiral into the sun, but to cause the elliptical orbit to become first more, then less eccentric, and change the orientation of the ellipse.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2019-04-19T05:38:23.287" LastActivityDate="2019-04-19T05:38:23.287" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30547" PostTypeId="2" ParentId="30342" CreationDate="2019-04-19T03:13:02.027" Score="3" Body="&lt;p&gt;The mass at which a star is able to condense into a black hole upon its death is called the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Tolman%E2%80%93Oppenheimer%E2%80%93Volkoff_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tolman–Oppenheimer–Volkoff Limit&lt;/a&gt;, which is roughly 2.17 solar masses. At this mass, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schwarzschild radius&lt;/a&gt; of the black hole (the size of its event horizon) is about 6410 meters by the equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$r_s = \frac{2GM}{c^2}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I'd also like to add that the theoretical minimum mass for a black hole is the Planck mass, &lt;span class=&quot;math-container&quot;&gt;$2.2\times 10^{−8}$&lt;/span&gt; kg, which would yield a radius of &lt;span class=&quot;math-container&quot;&gt;$3.267\times 10^{-35}$&lt;/span&gt; meters, though these would not be formed by stellar collapse, though they could have formed as primordial black holes in the very early universe.&lt;/p&gt;&#xA;" OwnerUserId="26727" LastEditorUserId="616" LastEditDate="2019-04-19T17:56:40.837" LastActivityDate="2019-04-19T17:56:40.837" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="30548" PostTypeId="2" ParentId="30544" CreationDate="2019-04-19T09:15:37.617" Score="4" Body="&lt;p&gt;The expansion of the universe is described by a &lt;a href=&quot;https://en.wikipedia.org/wiki/Scale_factor_(cosmology)&quot; rel=&quot;nofollow noreferrer&quot;&gt;scale factor&lt;/a&gt; that we normally call &lt;span class=&quot;math-container&quot;&gt;$a(t)$&lt;/span&gt;. We take the scale factor to be equal to one right now, so if &lt;span class=&quot;math-container&quot;&gt;$a(t)$&lt;/span&gt; is less than one that means the universe has contracted while if &lt;span class=&quot;math-container&quot;&gt;$a(t)$&lt;/span&gt; is greater than one the universe has expanded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General relativity allows us to calculate how the scale factor depends on time. If you're interested I do this in the question &lt;a href=&quot;https://physics.stackexchange.com/questions/136056/how-does-the-hubble-parameter-change-with-the-age-of-the-universe&quot;&gt;How does the Hubble parameter change with the age of the universe?&lt;/a&gt; on the Physics SE. This is what the results look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lWUhz.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lWUhz.gif&quot; alt=&quot;Scale factor&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point of this is that the CMB temperature is directly related to the scale factor so if we can calculate how the scale factor has changed with time then we can calculate the current CMB temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the way the scale factor changes with time depends on the density of matter and energy in the universe. Specifically it depends on how much light, matter (normal and dark) and dark energy is present. The matter/energy density could in principle have any values, that is there is no fundamental principle in physics that specifies what the densities of the of the various types of matter and energy have to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our current best measurements of the matter/energy density come from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Planck_(spacecraft)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planck experiment&lt;/a&gt;, and they suggest the total density has a value close to the &lt;a href=&quot;http://astronomy.swin.edu.au/cosmos/C/Critical+Density&quot; rel=&quot;nofollow noreferrer&quot;&gt;critical density&lt;/a&gt;. This density gives us the form of the scale factor that I drew above, and it gives us the CMB temperature of 2.7K. If the density were lower than critical we would get a lower CMB temperature and if the density were higher than critical we get a higher CMB temperature. So the answer to your question is that the CMB temperature is not uniquely determined and it could be higher or lower. Ultimately the CMB temperature is an experimental parameter that we have to determine by making experimental measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we should note that there are excellent reasons why we would expect the matter/energy density to be equal to the critical density, so we would expect the CMB temperature to be what we observe. If the CMB temperature were different by a factor of ten that would be hard to explain theoretically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally we should note that there are other independent constraints of the form of the scale factor function. Rob Jeffries describes some of these &lt;a href=&quot;https://astronomy.stackexchange.com/a/30550/10687&quot;&gt;in his answer&lt;/a&gt; so I won't go through them again here, but for example the ratio of hydrogen to helium depends on how quickly &lt;span class=&quot;math-container&quot;&gt;$a(t)$&lt;/span&gt; increased after the Big Bang. If the scale factor had increased significantly faster or more slowly it would have changed the hydrogen to helium ratio and we'd be able to measure this change.&lt;/p&gt;&#xA;" OwnerUserId="10687" LastEditorUserId="10687" LastEditDate="2019-04-20T04:03:24.790" LastActivityDate="2019-04-20T04:03:24.790" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30549" PostTypeId="2" ParentId="30544" CreationDate="2019-04-19T09:26:41.320" Score="0" Body="&lt;p&gt;I'd say it doesn't care all that much,  but that 27 degrees Kelvin is a step too far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the 1965 NASA csomic times article &lt;a href=&quot;https://imagine.gsfc.nasa.gov/educators/programs/cosmictimes/online_edition/1965/hiss.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;big hiss missed by others&lt;/a&gt;. You can read how Andrei Doroshkevic and Igor Novikov published a Big Bang study in 1964 saying the remaining heat would now be between 1 and 10 degrees Kelvin. They proposed the use of the Holmdel horn measurements made in 1961 by Edward Ohm, who &lt;em&gt;“had already identified in his data what seemed to be a 3.3 degrees Kelvin background radiation”&lt;/em&gt;. The article also says &lt;em&gt;“Ten years ago Émile Le Roux reported a microwave background radiation of 3 degrees Kelvin, plus or minus 2 degrees at the Nançay Radio Observatory”&lt;/em&gt;. And this: &lt;em&gt;“In 1957 it was Tigran Shmaonov who nearly made the discovery. He reported measuring background temperature of 4 degrees Kelvin, give or take 3 degrees”&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, everybody knows about the famous back-to-back papers to the Astrophysical Journal in July 1965. They were &lt;a href=&quot;http://adsabs.harvard.edu/abs/1965ApJ...142..419P&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Measurement of Excess Antenna Temperature at 4080 Mc/s&lt;/a&gt; by Penzias and Wilson, and &lt;a href=&quot;http://adsabs.harvard.edu/abs/1965ApJ...142..414D&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cosmic Black-Body Radiation&lt;/a&gt; by Dicke, Peebles, Roll, and Wilkinson. What people don't generally know is that Dicke et al were talking about &lt;em&gt;&quot;a closed universe, oscillating all the time&quot;&lt;/em&gt;. So according to them, the evidence for the &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Bounce&quot; rel=&quot;nofollow noreferrer&quot;&gt;Big Bounce&lt;/a&gt; had been found! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally I don't see any way by which the universe can ever contract. So I think the Big Bounce is a non-starter. However even though I'm 110% sure that the universe is expanding, I'm not happy with inflation or creation ex nihilo. So I think you're right to question the &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Bang&quot; rel=&quot;nofollow noreferrer&quot;&gt;Big Bang&lt;/a&gt;. The $64,0000 question is this: &lt;em&gt;what banged?&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="9772" LastEditorUserId="9772" LastEditDate="2019-04-19T09:31:59.420" LastActivityDate="2019-04-19T09:31:59.420" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30550" PostTypeId="2" ParentId="30544" CreationDate="2019-04-19T09:31:05.527" Score="4" Body="&lt;p&gt;Yes, I think it is a constraint, but don't think I can tell you how different it could be from 2.73 K to cause a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CMB tells us that the universe was once much hotter (&lt;span class=&quot;math-container&quot;&gt;$&amp;gt;3000$&lt;/span&gt; K), so that hydrogen nuclei and electrons had sufficient kinetic energies to remain uncombined. In the big-bang model, the universe expands and cools, then at temperatures just below this the protons&#xA;and electrons combine to form atoms that are almost transparent to the blackbody radiation that filled the universe before that. This radiation is then redshifts by the cosmological expansion and we still see it today as microwaves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the big-bang model, the temperature of the universe scales as &lt;span class=&quot;math-container&quot;&gt;$(1+z)^{-1}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; is the redshift corresponding to a particular look back time. That we see the CMB at 2.73 K now, means it occurred at &lt;span class=&quot;math-container&quot;&gt;$z=1100$&lt;/span&gt;. Your question amounts to - would it matter if this were &lt;span class=&quot;math-container&quot;&gt;$z=110$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$z=11000$&lt;/span&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's less complicated to deal with the lower of these possibilities. At &lt;span class=&quot;math-container&quot;&gt;$z&amp;lt;3400$&lt;/span&gt;, the energy density of the universe is matter dominated, and the scale factor &lt;span class=&quot;math-container&quot;&gt;$a \simeq (t/t_0)^{2/3}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$t_0$&lt;/span&gt; is the age of the universe and &lt;span class=&quot;math-container&quot;&gt;$a(t_0)=1$&lt;/span&gt;. Since &lt;span class=&quot;math-container&quot;&gt;$1+z = a^{-1}$&lt;/span&gt;, this means the connection between age and redshift is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ t(z) \simeq t_0 (1+z)^{-3/2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the CMB was formed at &lt;span class=&quot;math-container&quot;&gt;$z=1100$&lt;/span&gt; this would be 370,000 years after the big bang (using the usual accepted cosmological parameters and an age of the universe of 13.7 billion years&lt;span class=&quot;math-container&quot;&gt;$^1$&lt;/span&gt;). If instead we use &lt;span class=&quot;math-container&quot;&gt;$z=110$&lt;/span&gt;, the CMB formed about 14 &lt;em&gt;million&lt;/em&gt; years after the big bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That in itself is not a problem. We are just changing the expansion rate of the universe to match the redshift of the CMB and demanding this occurs just below 3000 K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this alters the temperature versus time relationship at all later and earlier times. How might this affect some other cosmological observations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think one area where the change in timescales would be hugely important is in primordial nucleosynthesis. By slowing down (or speeding up) the expansion rate, one allows an order of magnitude more (or less) time for nuclear reactions to take place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example: The neutron to proton ratio in the early universe freezes out at 1/6, but neutron decay with a half-life of 15 minutes, reduces the final ratio to 1/7 during the window of time available to nuclear reactions. If this window is made much bigger, the ratio will decrease further, if it is made much smaller the ratio will remain at 1/6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost all of those neutrons end up in helium nuclei, so the primordial He abundance would vary between 10 H atoms per He atom through to something considerably bigger. The primordial He abundance is known to better than 1%, and it is around 12 H atoms per He atom. i.e. There was just enough time when primordial nuclear reactions were taking place for &lt;em&gt;some&lt;/em&gt; of the free neutrons to decay, but not all. &lt;strong&gt;This would certainly seem to rule out order of magnitude changes in the CMB temperature.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may add to this as other things occur to me. There is likely another constraint based on the observed formation redshift of the first stars and the epoch of reionization.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-04-19T13:47:37.067" LastActivityDate="2019-04-19T13:47:37.067" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30554" PostTypeId="2" ParentId="30512" CreationDate="2019-04-19T16:37:26.750" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I wonder if any study like this has been done, where a whole-Earth spectrum from a distant spacecraft has been captured and analyzed, with the intent of working the problem as if it were a spectrum from an exoplanet, in order to correlate with known Earth conditions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Looking at your question from a different viewpoint, note that what you are suggesting is studying &lt;em&gt;just one&lt;/em&gt; example of a planet with life to make general predictions about exoplanets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is fundamentally flawed and the way I would explain it to you is this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have never seen some particular animal before and I give you one sample of that animal, does this tell you anything about other members of the same species ?  No.  You don't know if you have an extreme statistical outlier or even a mutant of that species.  You certainly cannot extrapolate from that that all animals will be similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot take a single example of anything and reliably or even usefully extrapolate anything from that about a larger population you know nothing about.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In theory, comparing the spectra captured by the probe with what was happening here on Earth at a time, the light was reflected to the space we could discover new spectrographic patterns that would help us look for planets similar to Earth and maybe even life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is not correct.  Earth, as far as we can tell, has had life of some form on it for &lt;em&gt;at least&lt;/em&gt; 3.5 billion years of it's 4.5 billion year history.  We &lt;em&gt;suspect&lt;/em&gt; life may have existed earlier, but there is no direct evidence for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During that long period conditions on Earth have radically altered.  Atmospheric and surface chemistry have changed completely from what we have now.  The spectrum of light from early Earth (with life) would be completely different because of this.  Ocean temperatures on Earth are estimated to have been as much as &lt;span class=&quot;math-container&quot;&gt;$55-85^\circ C$&lt;/span&gt; between 2 billion and 3.5 billion years ago (when life existed).  Up until about 2.5 billion years ago Earth had no free atmospheric Oxygen to speak of, something that a spectrum would indicate.  Now if we see an exoplanet with no apparent free oxygen can we say it has life or not ?  Well it seems not from our own planet's history.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the converse cannot be assumed either : just because we found free oxygen in the spectrum of an exoplanet it does not tell us anything about life on that planet, not even it's presence or absence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So extrapolating from Earth's current spectrum is next to useless in terms of finding life on exoplanets.&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2019-04-19T16:37:26.750" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30555" PostTypeId="2" ParentId="30507" CreationDate="2019-04-19T17:28:16.520" Score="2" Body="&lt;p&gt;The angle between the inner and outer edges of the penumbra is the same as the Sun's apparent angular diameter.&#xA;At 1.0 au from the Sun, that's 32 arcmin or 0.0093 radian.&#xA;At the distance from the Moon to EM-L2, this angle spans 0.0093 * 64700 km = 602 km.&#xA;The Moon's diameter is 3474 km, so the umbra diameter would be 2872 km and the penumbra diameter would be 4076 km.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-04-20T01:29:53.400" LastActivityDate="2019-04-20T01:29:53.400" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30556" PostTypeId="1" CreationDate="2019-04-20T01:54:50.807" Score="2" ViewCount="32" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/30555/7982&quot;&gt;This answer&lt;/a&gt; to the question &lt;em&gt;How wide is the Moon's umbra and penumbra at EM-L2?&lt;/em&gt; calculates the widths of the umbra and penumbra &lt;em&gt;at a distance from the Moon equal to the distance from EM-L2 from lunacenter&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, EM-L2 spends most of its time in direct sunlight because the Earth-Moon line rarely coincides with the Sun-Earth line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking into account the inclination difference between the Earth's orbital plane and that of the Moon as well as the size of the umbra, roughly what fraction of the time is EM-L2 in total solar eclipse — within the Moon's umbra?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/X3Ve8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X3Ve8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Earth-Moon.PNG&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-04-20T01:54:50.807" Title="What fraction of the time is EM-L2 in darkness on average?" Tags="&lt;lagrange-point&gt;&lt;eclipse-non-earth-related&gt;&lt;umbra&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30557" PostTypeId="2" ParentId="30499" CreationDate="2019-04-20T03:25:58.457" Score="0" Body="&lt;p&gt;&lt;a href=&quot;http://blogs.discovermagazine.com/badastronomy/2008/10/30/ten-things-you-dont-know-about-black-holes/#.XLqLvd-YVrl&quot; rel=&quot;nofollow noreferrer&quot;&gt;Phil Plait&lt;/a&gt;:&lt;br&gt;&#xA;&lt;strong&gt;Black holes can be low density&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... the event horizon grows linearly with the mass. In other words, if you double the black hole’s mass, the event horizon radius doubles as well. ...&lt;br&gt;&#xA;  Density is how much mass is packed into a given volume. Keep the size the same and add mass, and the density goes up. Increase the volume, but keep the mass the same, and the density goes down. ...&lt;br&gt;&#xA;  A regular black hole — that is, one with three times the Sun’s mass — with have an event horizon radius of about &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt; km. That means it has a huge density, about two quadrillion grams per cubic cm &lt;span class=&quot;math-container&quot;&gt;$(2 \times 10^{15})$&lt;/span&gt;. But double the mass, and the density drops by a factor of four. Put in &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; times the mass and the density drops by a factor of &lt;span class=&quot;math-container&quot;&gt;$100$&lt;/span&gt;. A billion solar mass black hole (big, but we see them this big in galaxy centers) would drop that density by a factor of &lt;span class=&quot;math-container&quot;&gt;$10^{18}$&lt;/span&gt;. That would give it a density of roughly &lt;span class=&quot;math-container&quot;&gt;$1/1000$&lt;/span&gt; of a gram per cc… and that’s the density of air!&lt;br&gt;&#xA;  A billion solar mass black hole would have an event horizon &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; billion km in radius — roughly the distance of Neptune to the Sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17743" LastActivityDate="2019-04-20T03:25:58.457" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30558" PostTypeId="1" AcceptedAnswerId="30562" CreationDate="2019-04-20T04:16:25.080" Score="4" ViewCount="361" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Ulysses_(spacecraft)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ulysses mission&lt;/a&gt; has a compelling story. It was sent to Jupiter to perform a gravitational assist shooting it out of the plane of the ecliptic in order to fly over the Sun's north and south poles to perform &amp;quot;fast latitude scans&amp;quot;. Because of its design it was used for several important lines of scientific study.&lt;/p&gt;&#xA;&lt;p&gt;Ulysses contained a pair of coherent transponders which received signals from Earth, shifted them in frequency in a coherent way using phase-locked loops and beamed them immediately back to Earth at two different frequencies.&lt;/p&gt;&#xA;&lt;p&gt;From ESA's write up of the &lt;a href=&quot;http://sci.esa.int/ulysses/40406-instruments/?fbodylongid=1940&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ulysses Gravitational Wave Experiment&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In the spacecraft Doppler tracking method, the Earth and spacecraft constitute the two objects whose time-varying separation is monitored to detect a passing gravitational wave. The monitoring is accomplished with high-precision Doppler tracking in which a constant frequency microwave radio signal (S-band) is transmitted from the Earth to the spacecraft (uplink); the signal is transponded (received and coherently amplified) at the spacecraft; and then transmitted back to Earth (downlink) in both S- and X-band signals. This Dual frequency downlink is required in order to calibrate the interplanetary media which affects the two frequency bands differently. The downlink signal is recorded at Earth and its frequency is compared to the constant uplink frequency f0 to extract the Doppler signal, δf / f0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The article goes on to say:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Since the optimum size of a gravitational wave detector is the wave length, interplanetary dimensions are needed for detecting gravitational waves in the mHz range. Doppler tracking of Ulysses provides sensitive detections of gravitational waves in this low frequency band. The driving noise source is the fluctuations in the refractive index of interplanetary plasma. This dictates the timing of the experiment to be near solar opposition and sets the target accuracy for the fractional frequency change at 3.0 × 10-14 for integration times of the order of 1000 seconds.&lt;/p&gt;&#xA;&lt;p&gt;SUMMARY OF OBJECTIVES&lt;/p&gt;&#xA;&lt;p&gt;The objective of the gravitational wave investigation on Ulysses is to search for low frequency gravitational waves crossing the Solar System. Because of the great distance to the spacecraft, &lt;strong&gt;this method is most sensitive to wave periods between about 100 - 8000 seconds, a band which is not accessible to ground-based experiments which are superior for periods below 1 second.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You can read more about Ulysses in &lt;a href=&quot;https://earth.esa.int/web/eoportal/satellite-missions/u/ulysses#foot46%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;eoPortal's Ulysses&lt;/a&gt; where I found both the link above and the following:&lt;/p&gt;&#xA;&lt;p&gt;B. Bertotti, R. Ambrosini, S. W. Asmar, J. P. Brenkle, G. Comoretto, G. Giampieri, L. Iess, A. Messeri, H. D. Wahlquist, “The gravitational wave experiment,” Astronomy and Astrophysics Supplement Series, Ulysses Instruments Special Issue, Vol. 92, No. 2, pp. 431-440, Jan. 1992&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What produces gravitational waves with &amp;quot;periods between about 100 - 8000 seconds&amp;quot;?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2021-04-23T05:40:02.813" LastActivityDate="2021-04-23T06:05:38.153" Title="What produces gravitational waves with &quot;periods between about 100 - 8000 seconds&quot;?" Tags="&lt;gravitational-waves&gt;&lt;artificial-satellite&gt;&lt;nasa&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30562" PostTypeId="2" ParentId="30558" CreationDate="2019-04-20T07:56:37.187" Score="5" Body="&lt;p&gt;Any binary system produces gravitational waves at twice it's orbital frequency, i.e. with periods of half it's orbital period. So binary systems with periods between 200s and 16000s will produce such waves.&lt;/p&gt;&#xA;&lt;p&gt;We can use Kepler's third law to say something about these:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ a = \left(\frac{GM}{4\pi}\right)^{1/3} P^{2/3},$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; is the orbital period, &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is the total mass of the binary system and &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is the orbital separation.&lt;/p&gt;&#xA;&lt;p&gt;For a binary with &lt;span class=&quot;math-container&quot;&gt;$M\sim 1M_{\odot}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$200&amp;lt;P&amp;lt;16000$&lt;/span&gt;s, then &lt;span class=&quot;math-container&quot;&gt;$0.11 &amp;lt; a &amp;lt; 2.00\ R_{\odot}$&lt;/span&gt;. Since normal stars of mass &lt;span class=&quot;math-container&quot;&gt;$\sim 0.5M_{\odot}$&lt;/span&gt; have radii that are similar to this, then the stars would probably need to be stellar remnants (white dwarfs, neutron stars or black holes) except right at the longest period end, where it might be possible to observe W Uma binaries. More massive binaries have separations that increase as &lt;span class=&quot;math-container&quot;&gt;$M^{1/3}$&lt;/span&gt;, but the radii of normal stars increases more like &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt;, so this conclusion is even firmer at larger masses.&lt;/p&gt;&#xA;&lt;p&gt;It could be possible to have a compact binary involving a low mass star plus a compact object - perhaps a Roche lobe filling one, so as well as &amp;quot;double degenerates&amp;quot;, the long period end of this range would include Cataclysmic Variables and Low Mass X-ray binary counterparts, with orbital periods of a few hours. Here is a prime example &lt;a href=&quot;https://astronomy.stackexchange.com/questions/43435/time-domain-astronomy-and-fastest-eclipsing-binary-ztf-j15395027-20-mag-6-91&quot;&gt;Time domain astronomy and fastest eclipsing binary ZTF J1539+5027 (+20 mag, 6.91 minutes): How to measure its minimum brightness?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Of course gravitational wave strain goes as something like &lt;span class=&quot;math-container&quot;&gt;$M P^{-4/3} d^{-1}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; is the distance. These binaries are much longer period than the (presumably rare) massive, merging black holes seen so far and so probably need to be close, in our own Galaxy, to be detected.&lt;/p&gt;&#xA;&lt;p&gt;e.g. LIGO was capable of detecting &lt;span class=&quot;math-container&quot;&gt;$M \sim 30 M_{\odot}$&lt;/span&gt; merging black holes, with &lt;span class=&quot;math-container&quot;&gt;$P \sim 0.02$&lt;/span&gt; s at distances of a billion light years. A similar strain amplitude would be produced by a &lt;span class=&quot;math-container&quot;&gt;$M\sim 2M_{\odot}$&lt;/span&gt; binary with &lt;span class=&quot;math-container&quot;&gt;$P= 200$&lt;/span&gt; s at a distance of 300 light years.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2021-04-23T06:05:38.153" LastActivityDate="2021-04-23T06:05:38.153" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30563" PostTypeId="1" AcceptedAnswerId="30574" CreationDate="2019-04-20T08:15:18.600" Score="2" ViewCount="368" Body="&lt;p&gt;First of all I know there is a similar question &lt;a href=&quot;https://astronomy.stackexchange.com/questions/20032/difference-between-luminosity-luminous-flux-and-radiant-flux&quot;&gt;here&lt;/a&gt;, but I feel is wasn't answered at all.&lt;/p&gt;&#xA;&lt;p&gt;My question is that I have seen many texts which says that Radiative Flux is equal to the Irradiance of a star. In Carroll &amp;amp; Ostlie book &amp;quot;An Introduction to Modern Astrophysics&amp;quot; even say that:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/s6AjB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/s6AjB.png&quot; alt=&quot;Screenshot from the book&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_irradiance#Irradiance_on_Earth%27s_surface&quot; rel=&quot;nofollow noreferrer&quot;&gt;Irradiance on Earth's surface&lt;/a&gt; says this:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Average annual solar radiation arriving at the top of the Earth's atmosphere is roughly 1361 W/m².&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Following this I first I assume that Irradiance and Radiative Flux are the same thing, but when searching for &lt;a href=&quot;https://en.wikipedia.org/wiki/Irradiance&quot; rel=&quot;nofollow noreferrer&quot;&gt;Irradiance&lt;/a&gt; on Wikipedia says that:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In radiometry, irradiance is the radiant flux (power) received by a surface per unit area.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Which means that Irradiance is Flux divided by Area, so Radiant Flux has units of J/s and Irradiance has units of W m^-2 which I remember from my Optics class, then again Carroll &amp;amp; Ostlie say that:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yV6GG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yV6GG.png&quot; alt=&quot;S and O screenshot&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;That doesn't mean if Irradiance is equal to Radiant Flux divided by Area, so F = I*A that would make Irradiance equal Luminosity / Area^2 ?&#xA;In that case the bolometric magnitude should be calculated with the integral of Irradiance, not Luminosity nor Flux right?&lt;/p&gt;&#xA;&lt;p&gt;Probably I'm making it complicated for myself and it's not that hard to understand. I'm just sharing my confusion with you guys, please help me with that question.&lt;/p&gt;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="26737" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-04-28T10:36:52.417" Title="Radiative Flux, Irradiance and Luminosity confusion" Tags="&lt;light&gt;&lt;luminosity&gt;&lt;radiation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30566" PostTypeId="1" CreationDate="2019-04-20T13:12:11.877" Score="3" ViewCount="658" Body="&lt;p&gt;I came across a question in which moon's highest altitude had to be calculated as seen from a certain latitude, also the month and the hour and the constellation in which moon would be in had to be calculated......&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could not understand why or how would moons path change during the year&lt;/p&gt;&#xA;" OwnerUserId="25925" LastActivityDate="2019-04-20T22:13:31.780" Title="Why does moon's highest altitude change during the year" Tags="&lt;the-moon&gt;&lt;observational-astronomy&gt;&lt;ecliptic&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30568" PostTypeId="1" CreationDate="2019-04-20T15:41:43.407" Score="2" ViewCount="65" Body="&lt;p&gt;As an amateur astronomer for 60 years and always as interested in the science as in the observing, I thought I had a pretty good understanding of the fundamentals. But I'm distance-auditing a third-year course in the fundamentals of applied astronomy, and I found that I had a misconception about the color index. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had assumed the CI was derived by measuring the star's own magnitude in each of, say, the &lt;em&gt;B&lt;/em&gt; and &lt;em&gt;V&lt;/em&gt; passbands, then subtracting to get the CI (e.g., by taking the ratio of flux intensities, expressed as a magnitude).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not so! It's derived by measuring the star's magnitude &lt;em&gt;relative to standard zero points&lt;/em&gt; (the magnitude of Vega at each of those wavelengths) that are &lt;em&gt;defined to be&lt;/em&gt; magnitude 0 &lt;em&gt;for those wavelength&lt;/em&gt;, then subtracting to get the CI.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Why measure two fluxes relative to a standard and then compare those two values, rather than simply measuring them relative to each other? Why &lt;em&gt;not&lt;/em&gt; just measure the flux at each wavelength, take the ratio, express this as a magnitude, and use that as the CI? The relationships are the same either way, just with a different zero point (think 0K and -273°C). It seems like finding how far apart Cleveland and Chicago are by measuring the distance of each from NYC and subtracting, rather than just measuring the distance between them. I don't see that the convention accomplishes anything other than assigning a CI of 0 to &quot;neutral&quot; white stars. Or am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="26742" LastEditorUserId="26742" LastEditDate="2019-04-20T15:49:32.837" LastActivityDate="2019-04-20T20:37:28.257" Title="Basis of color index" Tags="&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30569" PostTypeId="2" ParentId="30566" CreationDate="2019-04-20T17:32:28.597" Score="5" Body="&lt;p&gt;The moon's orbit around the Earth is inclined at about 5° to the ecliptic &#xA;So as the Earth progresses in its own orbit around the sun, the moon's altitude also changes &lt;/p&gt;&#xA;&#xA;&lt;p&gt;During summers the part of the orbit on the other side of the Earth(night side) is below the ecliptic. While during winters it is the upper part which is on the night side&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Lyp9Y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lyp9Y.png&quot; alt=&quot;Moon&amp;#39;s orbit is inclined to the ecliptic&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Earth-Moon.PNG&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25925" LastEditorUserId="7982" LastEditDate="2019-04-20T22:13:31.780" LastActivityDate="2019-04-20T22:13:31.780" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30570" PostTypeId="2" ParentId="30568" CreationDate="2019-04-20T20:37:28.257" Score="2" Body="&lt;p&gt;What you may be missing is that no two telescope, filter and detector combinations are the same. A colour index defined by your method would be measured differently by different observers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is to define standards that  have defined magnitudes and colours that all observers can use to calibrate their observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In actual fact, the basic measurement is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ B-V = \alpha (b-v) + z + \beta X,$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; are native measured magnitudes (-2.5 log the detected adu with a correction for exposure time) and &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; is the airmass. The constants &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;, zeropoint &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; and extinction coefficient &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; are determined by observing a network of secondary standards with a range of colours at a range of airmasses.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-04-20T20:37:28.257" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30571" PostTypeId="2" ParentId="30563" CreationDate="2019-04-20T22:31:35.147" Score="0" Body="&lt;p&gt;It looks to me like irradiant flux is the same as radiant flux is the same as irradiance, both are energy per second divided by area.  That interpretation would seem to be consistent with everything you quoted.  In more precise terms, these should all be &quot;flux density&quot;, not &quot;flux&quot;, because flux just means the amount of something per time, but that convention is never followed much any more.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2019-04-20T22:31:35.147" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30573" PostTypeId="1" CreationDate="2019-04-21T02:44:37.100" Score="3" ViewCount="125" Body="&lt;p&gt;I am aware that a &quot;naked singularity&quot; is not possible. However, what would happen if I added angular momentum to a black hole near the maximal angular momentum limit? Does it asymptote?&lt;/p&gt;&#xA;" OwnerUserId="26747" LastActivityDate="2019-04-21T10:55:22.573" Title="What happens if I were to add angular momentum to a maximally spinning black hole?" Tags="&lt;black-hole&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2019-04-22T03:33:00.650" ContentLicense="CC BY-SA 4.0" />
  <row Id="30574" PostTypeId="2" ParentId="30563" CreationDate="2019-04-21T08:02:40.510" Score="0" Body="&lt;p&gt;I think the confusion you're having stems from the fact that there are different terms for the same phenomena being used in different fields, and some of these terms use the same underlying words, but in different, inconsistent ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, in astrophysics we most commonly use the following terms&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;luminosity&lt;/strong&gt; -- total power emitted by an object, with units of energy per time (e.g. J s&lt;span class=&quot;math-container&quot;&gt;$^{-1}$&lt;/span&gt; = W);&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;flux&lt;/strong&gt; -- power crossing a specific area, with units of energy per time per area (e.g. W m&lt;span class=&quot;math-container&quot;&gt;$^{-2}$&lt;/span&gt;);&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;intensity&lt;/strong&gt; -- flux per solid angle (e.g. W m&lt;span class=&quot;math-container&quot;&gt;$^{-2}$&lt;/span&gt; ster&lt;span class=&quot;math-container&quot;&gt;$^{-1}$&lt;/span&gt;).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But in &lt;a href=&quot;https://en.wikipedia.org/wiki/Radiometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;radiometry&lt;/a&gt;, the corresponding terms (with the same units) are &lt;strong&gt;radiant flux&lt;/strong&gt;, &lt;strong&gt;irradiance&lt;/strong&gt; (also called &lt;strong&gt;flux density&lt;/strong&gt;), and &lt;strong&gt;radiance&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Carroll &amp;amp; Ostlie book you're referencing follows the astrophysical convention, but unintentionally confuses things by (sometimes) referring to flux as &quot;radiant flux&quot;, probably unaware that this term means something different (i.e., luminosity) in radiometry. (And thus you get confused when you look thngs up on Wikipedia and encounter the &lt;em&gt;radiometric&lt;/em&gt; terms.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing Carroll &amp;amp; Ostlie do this to remind the reader that this is flux of &lt;em&gt;light&lt;/em&gt; as opposed to something else (Google search within the book shows a brief discussion of &quot;convective flux&quot; later on). In practice, astrophysicists almost always use just &quot;flux&quot; for light, and will use extra words to indicate if they're talking about flux of something else (e.g. dust grains, cosmic rays, neutrinos, or whatever).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For what it's worth, the astrophysical use of &quot;flux&quot; is consistent with the general &lt;a href=&quot;https://en.wikipedia.org/wiki/Flux&quot; rel=&quot;nofollow noreferrer&quot;&gt;mathematics/physics use of the term&lt;/a&gt; (e.g., &quot;flux is defined as the rate of flow of a property per unit area, which has the dimensions [quantity] &lt;span class=&quot;math-container&quot;&gt;$\cdot$&lt;/span&gt; [time]&lt;span class=&quot;math-container&quot;&gt;$^{−1}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\cdot$&lt;/span&gt; [area]&lt;span class=&quot;math-container&quot;&gt;$^{-1}$&lt;/span&gt;. The area is of the surface the property is flowing &quot;through&quot; or &quot;across&quot;. For example, the magnitude of a river's current, i.e. the amount of water that flows through a cross-section of the river each second, or the amount of sunlight energy that lands on a patch of ground each second, are kinds of flux.&quot;). So the radiometric term &quot;radiant flux&quot; is arguably a misuse of the word &quot;flux&quot;, though I doubt radiometrists are losing sleep over this.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2019-04-28T10:36:52.417" LastActivityDate="2019-04-28T10:36:52.417" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30575" PostTypeId="2" ParentId="20032" CreationDate="2019-04-21T08:40:46.007" Score="0" Body="&lt;p&gt;In astronomy, &lt;strong&gt;luminosity&lt;/strong&gt; is exactly as you've defined it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;em&gt;radiometry&lt;/em&gt;, the usual term for this is &lt;strong&gt;radiant flux&lt;/strong&gt;. So, yes, they are the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Luminous flux&lt;/strong&gt;, however, is different. It is a term from &lt;a href=&quot;https://en.wikipedia.org/wiki/Photometry_(optics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;photometry&quot;&lt;/a&gt;, which is the measurement of light *as perceived by the human eye&quot; (I put it in scare quotes because &lt;em&gt;in astronomy&lt;/em&gt;, the word &quot;photometry&quot; usually means measurement of light without any reference to human perceptions, and is sometimes casually used to refer to imaging or analysis of images). It is &lt;em&gt;analogous&lt;/em&gt; to luminosity/radiant flux, but not the same thing, because it involves weighting the energy of the light by how well or poorly the human eye can sense it. (See the linked Wikipedia article for more details.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a radio source that doesn't emit any visible light. An astronomer would say it has a luminosity of (say) &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; watts. From a radiometric perspective, it has a radiant flux of exactly the same amount. But it would have a &lt;em&gt;luminous flux&lt;/em&gt; of zero, since the human eye can't detect radio photons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;this website ... says lumens are a unit of luminosity, but how if lumens are candela times steradians?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically, that website isn't really correct, since it's casually equating luminosity (or radiant flux) with luminous flux.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2019-04-21T15:17:16.463" LastActivityDate="2019-04-21T15:17:16.463" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30577" PostTypeId="2" ParentId="30573" CreationDate="2019-04-21T10:55:22.573" Score="2" Body="&lt;p&gt;How would you add angular momentum? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way to do that is to drop something &lt;em&gt;with&lt;/em&gt; angular momentum (relative to the centre of mass of the BH and the thing, if you want to get picky) into the BH. Could be a spinning object, or an object fired in at a oblique angle, or circularly polarised light or even suitably polarised gravity waves.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatever you do, you increase both the mass and the angular momentum of the BH. Now the limit on the spin of a black hole is expressed in terms of a relation between these two quantities and the Schwarzschild radius of the black hole (which is determined by the mass) and it turns out that whatever you add, and however you add it (if you're firing something in obliquely, it has to actually &lt;em&gt;hit&lt;/em&gt; the BH so it can't be too ar from the centre of mass) you cannot increase the angular momentum enough relative to the increase in mass to break that limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations you have to look at are on &lt;a href=&quot;https://en.wikipedia.org/wiki/Kerr_metric&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt; see especially &lt;a href=&quot;https://en.wikipedia.org/wiki/Kerr_metric#Overextreme_Kerr_solutions&quot; rel=&quot;nofollow noreferrer&quot;&gt;this section&lt;/a&gt;. One way of expressing the limit is given there as &#xA;&lt;span class=&quot;math-container&quot;&gt;$GM^2 &amp;lt; Jc$&lt;/span&gt;  where &lt;span class=&quot;math-container&quot;&gt;$J$&lt;/span&gt; is the angular momentum. Suppose you have a black hole just under this limit and add an object of mass &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; angular momentum just under &lt;span class=&quot;math-container&quot;&gt;$Gm^2/c$&lt;/span&gt; velocoty just under &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; and on a trajectory that just grazes the Schwarzchild radius of the hole at &lt;span class=&quot;math-container&quot;&gt;$\frac{2GM}{c^2}$&lt;/span&gt;. The angular moment of the result is just under &lt;span class=&quot;math-container&quot;&gt;$$GM^2/c + 2GMmc/c^2 + Gm^2/c$$&lt;/span&gt; where the terms are respectively the original angular moment of the hole, the angular moment from the grazing impact and the angular momentum of the added body. This works out to be (just under) &lt;span class=&quot;math-container&quot;&gt;$$G(M+m)^2/c$$&lt;/span&gt;, the limit for the (now-enlarged) black hole.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-04-21T10:55:22.573" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30579" PostTypeId="1" AcceptedAnswerId="30580" CreationDate="2019-04-21T17:21:58.067" Score="67" ViewCount="8775" Body="&lt;p&gt;As the title says, what happens when a gravitational wave approaches a black hole?  I would presume that something interesting happens because of the way spacetime works near black holes but I have no knowledge to back it up.&lt;/p&gt;&#xA;" OwnerUserId="21184" LastEditorUserId="10687" LastEditDate="2019-04-21T18:50:02.060" LastActivityDate="2019-04-24T11:00:48.233" Title="Can gravitational waves pass through a black hole?" Tags="&lt;black-hole&gt;&lt;general-relativity&gt;&lt;gravitational-waves&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="30580" PostTypeId="2" ParentId="30579" CreationDate="2019-04-21T18:26:06.677" Score="62" Body="&lt;p&gt;No, gravitational waves cannot pass through a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A gravitational wave follows a path through spacetime called a null geodesic. This is the same path that would be followed by a light ray travelling in the same direction, and gravitational waves are affected by black holes in the same way that light rays are. So for example gravitational waves can be refracted by gravitational lenses just as light waves are. And just like light waves, if a gravitational wave crosses the event horizon surrounding a black hole it is then doomed to travel inwards to the singularity and can never escape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one caveat to this. When we talk about a gravitational wave we generally mean a ripple in spacetime that is relatively small. Specifically it is small enough that the energy of the gravitational wave does not significantly affect the spacetime curvature. So when we calculate the trajectory of a gravitational wave near a black hole we take the black hole geometry as fixed, i.e. unaffected by the wave, and we compute the trajectory of the wave in this fixed background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is exactly the same approach as we use for calculating the trajectories of light rays. Since light rays carry energy and momentum then, at least in principle, they have their own gravitational fields. But for both the light rays and gravitational waves likely to exist in the universe the energy carried is too small to make a significant contribution to the spacetime curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you say in your question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I would presume that something interesting happens because of the way spacetime works near black holes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would guess you are thinking that the gravitational wave could change the geometry near a black hole, but as described above typical gravitational waves don't have enough energy to do this. It would be reasonable to ask what happens if we give the wave enough energy, but the answer turns out to be that it no longer behaves like a simple wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational waves exist in a regime called &lt;a href=&quot;https://en.wikipedia.org/wiki/Linearized_gravity&quot; rel=&quot;noreferrer&quot;&gt;linearised gravity&lt;/a&gt; where they obey a wave equation that is basically similar to the wave equation light obeys. If we increase the energy so much that gravity becomes non-linear (as if the case for black holes) then the oscillations in the spacetime curvature no longer obey a wave equation and need to be described by the full Einstein equations. For example it has been suggested, but not proven, that really high energy gravitational (or light) waves could interact with each other to form a bound state called &lt;a href=&quot;https://en.wikipedia.org/wiki/Geon_(physics)&quot; rel=&quot;noreferrer&quot;&gt;a geon&lt;/a&gt;. I confess that I'm unsure how much work has been done studying oscillations in this regime.&lt;/p&gt;&#xA;" OwnerUserId="10687" LastActivityDate="2019-04-21T18:26:06.677" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="30581" PostTypeId="1" CreationDate="2019-04-21T21:19:45.573" Score="1" ViewCount="130" Body="&lt;p&gt;So I know that the Eddington luminosity is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$L_{Eddington} = 4\pi GMc/\kappa$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to calculate this luminosity for a plasma of purely ionized helium, as well as for an electron-positron plasma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I find &lt;span class=&quot;math-container&quot;&gt;$\kappa$&lt;/span&gt; in each of these cases? I know sometimes this is approximated as &lt;span class=&quot;math-container&quot;&gt;$m/\sigma_T$&lt;/span&gt;, but I'm not sure when to invoke this approximation, and what the respective mass would be.&lt;/p&gt;&#xA;" OwnerDisplayName="user23921" LastActivityDate="2019-07-23T17:02:30.253" Title="Eddington luminosity in various plasmas" Tags="&lt;star&gt;&lt;luminosity&gt;&lt;high-energy-astrophysics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30584" PostTypeId="2" ParentId="16510" CreationDate="2019-04-21T22:56:25.177" Score="0" Body="&lt;p&gt;There is an AstroImageJ. Available on windows, mac, and linux&#xA;&lt;a href=&quot;https://www.astro.louisville.edu/software/astroimagej/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.astro.louisville.edu/software/astroimagej/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26751" LastActivityDate="2019-04-21T22:56:25.177" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30585" PostTypeId="1" CreationDate="2019-04-22T00:42:43.070" Score="9" ViewCount="252" Body="&lt;p&gt;I recently read something about a major earthquake in Japan having possibly shortened Earth's day due to an increase in rotation. Here's an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.nasa.gov/topics/earth/features/japanquake/earth20110314.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.nasa.gov/topics/earth/features/japanquake/earth20110314.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I read some more about Earth's day length, I learned that the length of Earth's day is believed to have been shorter in the past and getting longer over time. Here's an example from this site:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/27966/what-was-the-length-of-the-solar-day-73-million-years-ago&quot;&gt;What was the length of the solar day 73 million years ago?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As some may know from an introductory psychology course, humans tend to settle into a day that is longer than 24 hours when placed in an environment with no external time or light cues. For most people, their circadian system is able to adjust their day to the 24-hour Earth day. Some blind people lose this ability, as seen from commercials about medication to help them gain a 24-hour rhythm. Here is some background about that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Non-24-hour_sleep%E2%80%93wake_disorder&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Non-24-hour_sleep%E2%80%93wake_disorder&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the risk of asking an unacceptable &quot;opinion&quot; question, how convincing is the evidence of historical Earth day lengths? Is the evidence at a point where shorter Earth days in the past are an accepted certainty? Or are things still at a point of being a best guess without really knowing? It would seem rather odd for our nervous systems to have evolved with a preference for days longer than 24 hours. It is certainly possible, but I would appreciate some input here before spending time contemplating why evolution might have worked that way. I'm only a hobbyist, so there would be no need to go into too much depth for me specifically. Thanks in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The current answers and comments are interesting and appreciated, but to clarify, the intended focus is the extent and quality of the astronomical evidence that could be used in a discussion (elsewhere) about circadian issues.&lt;/p&gt;&#xA;" OwnerUserId="26752" LastEditorUserId="26752" LastEditDate="2019-04-23T03:23:58.683" LastActivityDate="2020-02-14T20:38:23.200" Title="Historical Day Lengths on Earth" Tags="&lt;earth&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30586" PostTypeId="1" AcceptedAnswerId="30590" CreationDate="2019-04-22T01:20:11.933" Score="0" ViewCount="110" Body="&lt;p&gt;I am always hearing that after travelling inside a given BH's event horizon, the escape velocity is greater than &lt;em&gt;c&lt;/em&gt;, which makes it impossible to escape. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose an ship is inside a BH. Assume it is possible to travel faster than &lt;em&gt;c&lt;/em&gt;. If the ship applied the faster-than-light travel, would it be able to escape a black hole?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, it seems to me that above event horizon, the escape velocity is less than &lt;em&gt;c&lt;/em&gt;, while inside event horizon, the escape velocity is greater than &lt;em&gt;c&lt;/em&gt;. This leads to a further question: Does the escape velocity increase the closer you get to a BH's singularity?&lt;/p&gt;&#xA;" OwnerUserId="25988" LastActivityDate="2019-04-22T08:31:25.923" Title="Could an object, assuming possible, escape black hole by travelling above constant speed?" Tags="&lt;black-hole&gt;&lt;light&gt;&lt;escape-velocity&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2019-04-23T19:26:00.497" ContentLicense="CC BY-SA 4.0" />
  <row Id="30588" PostTypeId="2" ParentId="30585" CreationDate="2019-04-22T06:34:23.897" Score="9" Body="&lt;p&gt;From a physics point of view, there's no obvious mechanism that could provide significant, consistent increases in the earth's rotation speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon's gravitation appears to be the main driver of changes to the earth's rotation.  And since it revolves more slowly than the earth's rotation, it tends to pull energy from the rotation, lengthening the day.  This is completely consistent with the evidence that points to ancient years on earth having more (and therefore shorter) days than is true in the modern era.   This is also consistent with the observed gradual recess of the moon away from the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also possible to directly measure the length of a day and see that days on earth are significantly longer than they were in 1900.  Again, consistent with a slowing rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The implications of this for the evolution of sleep cycles would indeed be better in another location (perhaps Biology SE).  But I'd suggest that this is a very slow process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that this is something that increases the length of the &lt;em&gt;average&lt;/em&gt; day.  But there are lots of other variations between days during a year.  An organism might for instance synch to something like sunrise.  But in spring, sunrises come more than a minute earlier than 24 hours apart each day, and in fall they come more than a minute later than average each day.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, it was probably more than a million years ago that the average day was shorter by a full minute.  So the increase in day length over time is definitely happening, but at a rate that is almost irrelevant to biology.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2019-04-22T07:06:05.017" LastActivityDate="2019-04-22T07:06:05.017" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30589" PostTypeId="2" ParentId="30585" CreationDate="2019-04-22T07:02:12.737" Score="1" Body="&lt;p&gt;Our biological clocks are ancient. One theory about the &gt;24 hour period is that the clocks evolved in the sea, where the tidal period can be more important than the solar day length. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tides are primarily caused by the Moon, and in the modern era the Moon rises roughly 50 minutes later each day, plus or minus 20 minutes, due to the eccentricity, inclination, and general complexity of the lunar orbit. The actual tidal changes at a given location are quite complex, but that lunar period gives a reasonable first approximation. So it's actually a good strategy to have a clock with a period of 24h 50m but which can easily be adjusted to be in synch with sensory input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That 24h 50m lunar period would have been different hundreds of millions of years ago, when our ancestors were in the sea. But I'm not sure how different, since although the solar day length was shorter, so was the lunar orbital period, and so the Moon' daily motion relative to the Sun was greater.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="16685" LastEditDate="2019-04-22T07:52:09.883" LastActivityDate="2019-04-22T07:52:09.883" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30590" PostTypeId="2" ParentId="30586" CreationDate="2019-04-22T08:20:13.230" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Assume it is possible to travel faster than &lt;em&gt;c&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We normally cannot use physics to answer questions about scenarios that break the laws of physics. But for this scenario, we can say that if the ship can travel backwards through time, then it can escape from the event horizon. Just about every FTL scheme can be used for backwards time travel, and as far as we know, backwards time travel is physically impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather than thinking about this problem in terms of escape velocity, its better to think in terms of worldlines. The velocity of a body depends on the observer measuring it, and which coordinates they're using, and dealing with that stuff near a black hole gets tricky. The usual Schwarzschild coordinates break down at the EH, and different observers measure quite different things, depending on their location and motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside the event horizon of a Schwarzschild black hole, &lt;em&gt;all&lt;/em&gt; worldlines travel towards the singularity, without exception. As you move towards the future, you move towards the singularity. So the reason you can't escape from inside the EH is that there are no paths that go outside the EH. Unless you can magically travel backwards in time, so you can follow an in-falling worldline in reverse.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="16685" LastEditDate="2019-04-22T08:31:25.923" LastActivityDate="2019-04-22T08:31:25.923" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30593" PostTypeId="1" AcceptedAnswerId="30623" CreationDate="2019-04-22T11:19:15.737" Score="2" ViewCount="142" Body="&lt;p&gt;Imagine you have 3 star points in 3D space A,B and C.&#xA;From your unknown position P in space you can measure the angles APC, APB, BPC and you know coordinates of A,B and C and therefore the distances between AB,BC,CA. Is there a way to calculate your own position P? Imagine you are floating in space and you're not on a sphere like earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps this is a frequent problem in astronomy and am wondering if there are any techniques to solve it. If you think there is, please point me in the right direction!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Greetings,&#xA;Matlab M.&lt;/p&gt;&#xA;" OwnerUserId="26760" LastActivityDate="2019-04-23T23:38:06.357" Title="Can you find your own location in space by measuring angles between stars?" Tags="&lt;distances&gt;&lt;coordinate&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30594" PostTypeId="1" AcceptedAnswerId="30605" CreationDate="2019-04-22T11:33:23.790" Score="3" ViewCount="125" Body="&lt;p&gt;The short, interesting paper &lt;a href=&quot;https://arxiv.org/abs/1904.07224&quot; rel=&quot;nofollow noreferrer&quot;&gt;Discovery of a Meteor of Interstellar Origin&lt;/a&gt; describes a proposed discovery of a second interstellar object. This one hit the Earth, and was discovered not by telescope, but by &quot;data mining&quot; the CNEOS fireball catalog &lt;a href=&quot;https://cneos.jpl.nasa.gov/fireballs/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://cneos.jpl.nasa.gov/fireballs/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not hard to find in the table based on the date given in the paper, but the information listed is sparse, there's no ID number or information how this was obtained:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;peak brightness UTC   lat     lon    alt(km)  v(km/s)  vx     vy     vz &#xA;2014-01-08 17:05:34   1.3S   147.6E   18.7     44.8   -3.4  -43.5  -10.3 &#xA;&#xA;total radiated kJ   calc total imact kt&#xA;     3.1e10                0.11&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; With what instrument was this particular fireball observed and recorded, and how was it analyzed and processed into a trajectory and state vector to be entered into this table? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HrN5T.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HrN5T.png&quot; alt=&quot;Fig. 1, Discovery of a Meteor of Interstellar Origin&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Figure 1. Trajectory of the January 8, 2014 meteor (red), shown intersecting with that of Earth (blue) at the time of impact, ti = 2014-01-08 17:05:34.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aGMIB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aGMIB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Figure 2. Size distribution of interstellar objects based on the detection of ‘Oumuamua and of the meteor detected at 2014-01-08 17:05:34 UTC. Red lines indicate the envelope for possible power-law fits (slopes of -1.9 to -3.8), given 95% Poisson distribution confidence intervals on each number density (based on a single detection for each). The range of possible power-law slopes is consistent with that inferred for small bodies in the Kuiper belt, -2.5 to -3 (Kenyon &amp;amp; Bromley 2004).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7411" LastEditDate="2019-04-28T10:41:08.910" LastActivityDate="2019-04-28T10:41:08.910" Title="How was this interstellar fireball observed and its trajectory analyzed?" Tags="&lt;observational-astronomy&gt;&lt;orbital-mechanics&gt;&lt;orbital-elements&gt;&lt;meteor&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30595" PostTypeId="2" ParentId="30581" CreationDate="2019-04-22T13:48:37.827" Score="1" Body="&lt;p&gt;I think you mean that kappa = sigma / m.   The sigma is the &quot;cross section,&quot; like the area you expose when you slice a watermelon in half.  The m is the mass of the particle that has the cross section in question (it's not literally the cross section of the physical particle, but an effective cross section for interaction with light), or more often, the mass of all the particles that come along with the cross section in question (often there is one particle contributing most of the cross section, but several particles that come with it which provide the mass).  For example, if you mix protons and electrons, the electrons have a larger cross section because their low mass makes them interact more easily with light, but that lower mass also means it is not the electrons, but the protons, that provide the &quot;m&quot; in the above formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So these quantities depend on the composition of the gas you have in mind.  It's easy for an electron-proton [edited] plasma, because if you don't worry about resonant interactions in the plasma, you can just consider the cross sections of the individual particles, and you get &quot;Thomson opacity&quot; of 0.4 square centimeters per gram (that's sigma/m for the sigma of an electron and the m of a proton, rounded to one significant figure because nothing is ever really a pure electron-proton plasma).  Purely ionized helium is also easy, because if you again ignore any resonant interactions between the electrons and the nuclei, it's just the same Thomson opacity except reduced by the extra neutrons [edited] in the heavy helium nuclei.  Since there is one neutron per proton, that reduces 0.4 by factor of 2, yielding 0.2.  [edited for forgetting that the 0.4 already includes the protons.]&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2019-06-23T16:43:26.150" LastActivityDate="2019-06-23T16:43:26.150" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30597" PostTypeId="2" ParentId="30579" CreationDate="2019-04-22T18:55:25.170" Score="9" Body="&lt;p&gt;Gravitational waves should be lensed by massive objects in a very similar way to light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light rays (and by extension, gravitational waves) from a distant object, that pass within 1.5 times the Schwarzschild radius (for a non-spinning black hole) have trajectories that take then towards the event horizon. Waves on such trajectories cannot escape from the black hole, so the basic answer is no, gravitational waves cannot &quot;pass through a black hole&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, far from &quot;hiding&quot; a source of gravitational waves, an intervening black hole would cause the presence of lensed and magnified images. For perfect alignment of source, black hole and observer, there would be an intense &quot;Einstein ring&quot; at an angular radius that depends on the relative distances of the source and the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course gravitational waves cannot be imaged at present, so what would be detected is an abnormally strengthened gravitational wave signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the above is in the &lt;strong&gt;geometric optics&lt;/strong&gt; limit that the wavelength is small compared with the lens. If the black hole is small enough (which depends on its mass), or the gravitational wave wavelength is large enough, then the behaviour ought to be analogous to a plane wave encountering a small, opaque disc (&lt;a href=&quot;https://arxiv.org/abs/astro-ph/0305055&quot; rel=&quot;nofollow noreferrer&quot;&gt;Takahashi &amp;amp; Nakamura 2003&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In which case we would get a &lt;strong&gt;diffraction pattern&lt;/strong&gt; and perhaps a &quot;bright&quot; &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Arago_spot&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arago spot&lt;/a&gt; in the centre, though I am not aware of any such calculations in the literature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not an unlikely scenario. For example, the gravitational waves detected by LIGO have relatively high frequencies of 10-1000 Hz and therefore wavelengths of 30,000-300 km, which are as big as the Schwarzschild radii of 10,000 - 100 solar mass black holes and certainly bigger than black hole remnants of stellar evolution.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-04-23T21:17:45.867" LastActivityDate="2019-04-23T21:17:45.867" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30598" PostTypeId="1" CreationDate="2019-04-22T23:36:52.743" Score="3" ViewCount="2388" Body="&lt;p&gt;In my class we are studying objects in our solar system and this question seemed to just pop up. And since I cannot answer this, I've been really frustrated for quite a while now and would like some help on understanding this.&lt;/p&gt;&#xA;" OwnerUserId="26769" LastActivityDate="2019-04-23T00:03:43.907" Title="If gravity precedes the formation of a solar system, where did the mass come from that caused the gravity?" Tags="&lt;solar-system&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30599" PostTypeId="2" ParentId="30598" CreationDate="2019-04-23T00:03:43.907" Score="15" Body="&lt;p&gt;The start of it all is usually something called a Giant Molecular Cloud, a particular kind of nebula which is denser than usual and cooler than usual.  A GMC will typically be 10x to 1000x the mass of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That mass is enough to cause the outer layers to fall in on the inner parts and the GMC starts to collapse and shrink. (Diffuse though it is, the mass exerts the same force on the outer edges as it would if the entire mass was concentrated at the center.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cloud collapses and frequently fragments into a bunch of smaller collapsing blobs centered on especially dense sections of the cloud. At this point it looks a lot like the Oriion Nebula. The young stars light up and blow away the remaining bits of the GMC, and in the end, you have a cluster of young stars with planetary disks looking something like the Pleiades. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There never was a central mass -- none is needed.  The GMC collapses under its own self-attraction.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2019-04-23T00:03:43.907" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="30601" PostTypeId="1" CreationDate="2019-04-23T02:17:49.960" Score="2" ViewCount="83" Body="&lt;p&gt;Gravitational waves detected are from merging of black holes few millions of light years away. While waves have reached us after travelling so far, they must have encountered many black-holes and have reached us after refracting/gravitational lens effects. Then how do one decide the correct direction of those merging black holes?&lt;/p&gt;&#xA;" OwnerUserId="26665" LastEditorUserId="25988" LastEditDate="2019-04-23T07:29:43.940" LastActivityDate="2019-04-23T11:34:23.300" Title="Gravitational wave distortion" Tags="&lt;gravitational-waves&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30602" PostTypeId="2" ParentId="30601" CreationDate="2019-04-23T04:12:19.487" Score="1" Body="&lt;p&gt;Gravitational waves detected by the LIGO that originate a few million light years away is not far away, relatively speaking, at all. As user Chappo has noted, space is mostly empty. Although gravitational waves can be scattered or distorted by an sufficiently massive and dense celestial body, it has a very low probability of happening. Although the technology today has no way of determining the original position of where the waves originate if the waves are disturbed, such disturbance is very unlikely. So it is safe to say that almost all waves detected by the LIGO can be traced back to their original positions.&lt;/p&gt;&#xA;" OwnerUserId="25988" LastActivityDate="2019-04-23T04:12:19.487" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30604" PostTypeId="2" ParentId="30601" CreationDate="2019-04-23T08:06:21.283" Score="3" Body="&lt;p&gt;Gravitational waves are affected by gravitational lensing in the same way as light. Most lensing events are &lt;em&gt;not&lt;/em&gt; caused by black holes, but by more common objects with similar mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There largest deflection examples of the light from distant galaxies, quasars etc being diverted by foreground sources (clusters of galaxies) is about an arcminute (e.g. &lt;a href=&quot;https://arxiv.org/abs/1407.3769&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zitrin et al. 2014&lt;/a&gt;). Current gravitational wave detectors are not capable of locating sources even to a few degrees. So source location isn't an issue at present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source strength might be. A lensed source can be magnified or even split into multiple images that arrive at different times. This will lead to a luminosity distance underestimate from a gravitational wave source and an overestimate of its mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those that are interested, &lt;a href=&quot;https://arxiv.org/abs/1807.02584&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oguri et al. (2018)&lt;/a&gt; have done an extensive set of simulations, showing the effects of gravitational lensing on estimates of mass and redshift distributions for black hole binary gravitational wave sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conclusion is that for detectors like LIGO there should be a (very) small tail of highly magnified events with underestimated redshifts and overestimated masses. These are predominantly at (chirp) masses &lt;span class=&quot;math-container&quot;&gt;$&amp;gt;40 M_{\odot}$&lt;/span&gt; (which is also roughly the minimum mass of the primary component in a binary) and more massive than any of the LIGO events seen so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;optical depth&quot; to lensing as a function of source redshift is shown in Fig. 3 of that paper. What this shows is that the chance of any individual source being strongly (&lt;span class=&quot;math-container&quot;&gt;$&amp;gt;$&lt;/span&gt; factor of 10 or split into multiple images) lensed is less than one in a million at typical LIGO source distances of a billion light years (redshifts of less than 0.1). But this probability can increase by several orders of magnitude at the high redshifts that might be probed by detectors in the future.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-04-23T11:34:23.300" LastActivityDate="2019-04-23T11:34:23.300" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30605" PostTypeId="2" ParentId="30594" CreationDate="2019-04-23T09:34:59.737" Score="4" Body="&lt;p&gt;&quot;US Government sensors is a &lt;a href=&quot;https://www.universetoday.com/116355/we-are-not-alone-government-sensors-shed-new-light-on-asteroid-hazards/&quot; rel=&quot;nofollow noreferrer&quot;&gt;euphemism&lt;/a&gt; for &quot;nuclear weapon warning early warning satellites&quot;, which is why we won't know details of the actual sensor: It is a military satellite and secret.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This particular event occurred in the Bismark sea, North of Papua New Guinea, it probably could have been seen from Papua or parts of Indonesia, if the sky was clear (which it rarely is, in the middle of the tropical rain band), but if it was seen, it wasn't reported to the American meteor society (&lt;a href=&quot;https://fireball.amsmeteors.org/members/imo_view/browse_reports?country=&amp;amp;start_date=2014-01-07%2000:00:00&amp;amp;end_date=2014-01-09%2023:59:59&amp;amp;event=&amp;amp;event_id=&amp;amp;event_year=&amp;amp;page=1)&quot; rel=&quot;nofollow noreferrer&quot;&gt;reports from that time&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The satellites are designed to look for rocket launches, and to track the missiles as the climb through the atmosphere to determine the likely target. Thus they are able to get accurate position and velocity vectors of meteors. If you know the position and velocity vector of the meteoroid just prior to impact, it is an exercise in Newtonian gravity to work out the orbit of the meteoroid around the sun. (It is a three body problem as the meteoroid is moving in both the Earth's and the Sun's gravitational field)&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-04-23T09:34:59.737" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30607" PostTypeId="1" CreationDate="2019-04-23T14:29:31.327" Score="-1" ViewCount="47" Body="&lt;p&gt;The gravity caused by a black hole is said to be so strong such that even light(photons) can not escape from it. since gravity exist only in between objects with mass, How does a black hole attract a mass less photon?&lt;/p&gt;&#xA;" OwnerUserId="26528" LastActivityDate="2019-04-23T18:25:07.960" Title="Mass of Photon and Black holes" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;supermassive-black-hole&gt;&lt;mass&gt;&lt;photons&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2019-04-24T00:09:25.877" ContentLicense="CC BY-SA 4.0" />
  <row Id="30609" PostTypeId="1" CreationDate="2019-04-23T16:11:31.850" Score="0" ViewCount="75" Body="&lt;p&gt;It was found from the observations of the radiopulsars that the intervals between their pulses are changing &#xA;periodically, besides, for all pulsars this period is the same and equal to one stellar (sidereal) year. Explain &#xA;this phenomenon&lt;/p&gt;&#xA;" OwnerUserId="25925" LastActivityDate="2019-05-23T23:03:15.403" Title="Constant interval between pulses from all pulsars" Tags="&lt;radio-astronomy&gt;&lt;pulsar&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30610" PostTypeId="2" ParentId="30609" CreationDate="2019-04-23T16:33:06.830" Score="1" Body="&lt;p&gt;The variation comes from the Earth's movement around the Sun. When the Earth is moving away from the pulsar in question, each subsequent pulse has slightly further to travel (up to 30km for a 1 second pulsar on the ecliptic down to zero for a pulsar located due North or South) and so is slightly delayed compared to the one before. Six months later, each subsequent pulse has slightly less far to travel, and so arrives a little earlier compared to the one before. The variation ranges from zero at the North or South poles to 1 part in 10000 on the ecliptic (this just comes from the ration between the Earth's orbital velocity and the speed of light).&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="7982" LastEditDate="2019-04-23T22:59:37.993" LastActivityDate="2019-04-23T22:59:37.993" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30611" PostTypeId="1" AcceptedAnswerId="30620" CreationDate="2019-04-23T16:36:38.563" Score="1" ViewCount="356" Body="&lt;p&gt;I understand that one could theoretically replace the sun with a blackhole without affecting the orbits of the planets. This leads me to believe that a blackhole isn't a &quot;cosmic vacuum cleaner&quot; so much as it is a concentration of mass that has the strongest gravitational field known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, this must mean that there is a threshold beyond which escaping the blackhole's gravity becomes impossible even at light speed, which is the event horizon. Since a Planck length is the smallest unit known at which quantum gravitational effects become apparent, would it be reasonable to assume that a photon that is one Planck length away from the event horizon of a blackhole can escape its gravitational pull?&lt;/p&gt;&#xA;" OwnerUserId="26788" LastEditorUserId="26788" LastEditDate="2019-04-23T18:26:39.723" LastActivityDate="2019-04-24T23:16:28.543" Title="If a photon were 1 Planck length away from the event horizon of a blackhole, would it be able to escape the blackhole's gravitational pull?" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;general-relativity&gt;&lt;event-horizon&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30612" PostTypeId="2" ParentId="30611" CreationDate="2019-04-23T16:53:25.160" Score="0" Body="&lt;p&gt;Yes it should be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every body has a Schwarzschild radius which defines the radius at which light would not be able to escape:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$r_s= \frac{2GM}{c^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By this formula, if a black hole had the mass of the Earth its radius would be about 9mm, but as the Earth has a bigger radius than this we do not see light being captured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A black hole is an object smaller than its Schwarzschild radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I found an article on Wikipedia which says that photons will orbit the black hole at &lt;span class=&quot;math-container&quot;&gt;$\frac{3}{2}r_s$&lt;/span&gt;. Within that &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Photon_sphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;photon sphere&lt;/a&gt; no stable orbit is possible:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Any free fall orbit that crosses it from the outside spirals into the black hole. Any orbit that crosses it from the inside escapes to infinity.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="25925" LastEditorUserId="7982" LastEditDate="2019-04-24T23:16:28.543" LastActivityDate="2019-04-24T23:16:28.543" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30613" PostTypeId="2" ParentId="30607" CreationDate="2019-04-23T18:08:23.277" Score="0" Body="&lt;p&gt;I will try to give a simplified answer here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We used to describe gravity as a force starting back with Newton. But as more and more unexpected effects of gravity came into play we had to create a better description.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best description we have today, General Relativity or short GR, is now over a hundred years old (published by Einstein in 1915). It has been proven with lots of experiments and with extremely high precision.  I do find it slightly annoying that it is not taught in schools. Well, next year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GR gets rid of the concept that gravity is a force. There is no hidden rubberband force pulling masses to each other. Instead mass modifies the shape of space (and time, but let us leave that aside now). A photon moves with the speed of light and follows a &quot;straight&quot; line in space. Because any mass modifies space, the line will no longer look straight in the normal sense. It will follow the modified &quot;straight&quot; line in space as the mass creates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A photon that passes by a mass will change direction. More for a large mass and more when close. This was first demonstrated clearly by looking at stars passing behind the sun (it did take a solar eclipse to be able to measure it). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I already said, GR is an extremely good description. It is not the &quot;full truth&quot; as there are things that it cannot describe. Even better descriptions will be needed, but when they will around is hard to say.&lt;/p&gt;&#xA;" OwnerUserId="26789" LastEditorUserId="7472" LastEditDate="2019-04-23T18:25:07.960" LastActivityDate="2019-04-23T18:25:07.960" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30614" PostTypeId="1" CreationDate="2019-04-23T18:47:18.640" Score="1" ViewCount="22" Body="&lt;p&gt;Say I have a flux density spectrum of a galaxy, as well as polarization spectra for the object in U, Q, and P. I want to find the polarization of some emission line in the spectrum. First, I can fit a line to my flux spectrum which will give me the continuum (i.e. without the peak(s)). I can then subtract the continuum from the whole spectrum to get just the profile of the emission line. But I'm unclear on how to relate these results with the polarization spectrum (right now I'm just working with the Q). I'm guessing I need to subtract the continuum polarization from my polarization spectrum, which would also involve converting my flux continuum to some polarization units, but how do I do this? Do I need the U or P (total) portion as well to do this?&lt;/p&gt;&#xA;" OwnerUserId="24076" LastActivityDate="2019-04-23T18:47:18.640" Title="Polarization of an emission line" Tags="&lt;galaxy&gt;&lt;light&gt;&lt;spectra&gt;&lt;polarization&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30615" PostTypeId="2" ParentId="30611" CreationDate="2019-04-23T19:43:19.840" Score="0" Body="&lt;p&gt;Probably no. It all depends on definition of photon location as photons have both wave and particle properties. In terms of macrocosmos location of photons are well (enough) defines, but as we speak lengths closer to its wavelength (Planck length being in the very extreme of this), it is simply difficult to pinpoint whether the photon is on this or that side. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would assume that it needs to be at least one wavelength away from the EH in order to be able to escape, but then again - this is highly theoretical as we also need something that’ll emit that photon. I would be very interested in hearing some real professional opinions on this one!&lt;/p&gt;&#xA;" OwnerUserId="10813" LastActivityDate="2019-04-23T19:43:19.840" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30616" PostTypeId="2" ParentId="22766" CreationDate="2019-04-23T20:43:29.417" Score="-1" Body="&lt;p&gt;Someone else asked a similar question that was linked here as a duplicate. It seems that question isn't quite the same, but similar enough that it makes sense to put it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That person asked:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;I know that the Sun is made up of hydrogen and helium. &#xA;Also, hydrogen explodes when it touches a flame. &#xA;But why does the Sun not explode?&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The kind of explosion that happens when hydrogen touches flame is an exothermic chemical reaction in which hydrogen is rapidly combined with oxygen (oxidation).  When heat is applied (flame) to hydrogen (H2) in the presence of oxygen (O2), it causes a molecule of O2 to separate and each oxygen atom to bind with a molecule of H2.  When this happens, it releases a large amount of energy ( believe this is caused by the breaking of the O2 bonds), and you end up with water (probably steam) and heat (fire).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kind of &quot;buring&quot; that happens in stars is thermonuclear fission. When two atoms of hydrogen are brought together by immense heat and pressure, they fuse into an atom of helium.  In this process, enormous amounts of energy are released.  This is the same nuclear reaction that takes place in a thermonuclear fusion bomb.  It does not involve a chemical change, but a nuclear one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me take a half-step back on that.  A chemical change or reaction is when the atoms rearrange themselves in molecules.  In most such reactions, energy is either absorbed (endothermic) or released (exothermic).   Sometimes fairly large amounts are released.  Explosives such as gunpowder, TNT, C4, and the like are all chemical explosives and mostly their reactions involve very rapid oxidation of the substance being burned. The burning of a match, candle or campfire is the same type of chemical oxidation reaction, but at a much slower pace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nuclear reactions come in two main types: fission and fusion.  Fission is the splitting of larger atoms into smaller ones.  The isotope of uranium with an atomic mass of 235 (92 protons and 143 neutrons) has a half-life of about 700 million years (which is far shorter than the 4.5 billion year half-life of the more common U238).  Atoms naturally decay over time, and in a span of 700 million years, a mass of U235 will naturally split on its own and the number of atoms will reduce by half (hence the term half-life).  When an atom of U235 splits, it releases a bunch of energy, produces an atom of krypton and one of barium, and releases a few extra neutrons. When one of these neutrons meets another atom of U235, it joins that atom, which destabilizes it, and causes the reaction to happen again.  Note: because we are talking about reactions in the nucleus of the atom and resulting changes to the atoms themselves, we are talking about nuclear reactions, not chemical ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction I mentioned above takes place in U235 naturally. If you don't have much of it and it's not in a dense enough state, it just sits there like a lump of rock and seems to do nothing.  Most of the neutrons it releases don't find another atom to play with, and end up just flying off into space.  But if you squeeze it down enough or get enough of it together, more neutrons are meeting up with other atoms than are flying off into nowhere.  This creates a chain reaction, which is the basis for nuclear power and nuclear weapons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reactor works by controlling the rate of this reaction, keeping it from going too fast.  Lumps of uranium are held just close enough together to react,but not rapidly, and rods of neutron-absorbing material (usually graphite) are slid between the lumps to help absorb the neutrons and prevent them from causing more reactions. A nuclear meltdown occurs when the reaction gets out of control and heats up so much that it literally melts the framework holding it all together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't enough, however, to create a nuclear explosion.  That occurs when you apply pressure and squeeze the material down, increasing its density.   U235 isn't actually suited for this, but element 94, plutonium, is. When you squeeze plutonium down to a high enough density, this reaction happens extremely rapidly.  First one atom of Pu239 splits, releasing its energy in the form of light and heat along with additional neutrons.  Then those extra neutrons find other atoms of Pu239 and cause them to split.  Every split creates at least twice as many extra neutrons as splitting atoms.  The first atom splits to cause 2 to split, which split and cause 2 each for four, then eight, sixteen, 32, 64, 128, 256, 512, 1024, 2048, 4096... each &quot;generation&quot; of these splits is double the previous. This all happens at a time scale measured in tiny fractions of a second, the time it takes light to go only a few millimeters.  This all builds up tremendous amounts of heat and pressure, which, unconstrained, erupt out of the core of the bomb.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it turns out, only a small percentage of the original mass is actually lost to fission. The heat and pressure build up so great so fast that it forces the atoms away from each other in the explosion, thus ending the reaction.  In fact, in a fission bomb, before anything other than high-energy x- and gamma rays leaves the bomb case, the reaction is complete, all that's left is for it to expand outward due to the immense pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other kind of nuclear reaction, fusion, is also used in bombs.  In fusion, as I mentioned above, huge amounts of pressure are used to force atoms of lighter elements, typically hydrogen, together.  Every time two atoms of hydrogen fuse into one, a huge amount of heat and energy are released (to be honest, I've never understood why energy is released here and not absorbed, but I'm not a nuclear physicist, just a guy who likes to read a lot).  In bombs, the 'trigger' for the fusion is actually a fission bomb.  The fission bomb creates the heat and pressure necessary, and when hydrogen atoms are added to the mix (usually in the form of tritium or deuterium), the fusion reaction begins and starts adding its heat and pressure to the pot, which helps sustain the reaction long enough to put out an amount of energy several times that which the fusion bomb started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you remember back to your high school chemistry lessons, Boyle's Law states &lt;span class=&quot;math-container&quot;&gt;$P_1V_1 = P_2V_2$&lt;/span&gt;. P is pressure, V is volume.  This formula shows that pressure and volume are inextricably interconnected.  Charles' law states that &lt;span class=&quot;math-container&quot;&gt;$V_1T_2 = V_2T_1$&lt;/span&gt;, showing the interrelation between volume and temperature.  Gay-Lussac's law, &lt;span class=&quot;math-container&quot;&gt;$P_1T_2 = P_2T_1$&lt;/span&gt; shows the relation between pressure and temperature. These all come together into the combined gas law and Avogadro's law.  They show that if you increase the pressure, volume, or temperature of a mass, the others are affected as well. An increase in pressure results in an increase of temperature.  A decrease of volume creates an increase in pressure which create an increase in temperature.  If you increase the volume, the temperature and pressure decrease.  Etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, so, now here's where it all applies to stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the result of the Big Bang, most of the matter created was hydrogen. There was some helium and a little lithium, but the overwhelming majority was hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the hydrogen molecules floating through space started to clump together through gravity, they started to condense down into smaller and smaller volumes.  Separately, the gravity of a single atom of hydrogen is minuscule.  But as more and more of them interacted, the combined gravity began to build up.  The clouds became more and more compact - denser, and their gravity brought in more and more mass, which added their gravity to the total.  Things began to heat up, due to the pressure caused by the volume.  The gravity started to get to be enough to continue to pull things toward the center of mass, into the &quot;gravity well&quot; caused by the cloud.  As the heat increased due the the pressure caused by gravity, it caused outward pressure.  The forces continued to balance each other.  As more mass fell in, gravity increased, causing more heat, which caused more pressure, which pushed outward... and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a certain point, the gravity gets so great, that it exerts enough pressure on the center of the mass to cause nuclear fusion.  This, in turn, causes more pressure outward, but the force of gravity of the entire mass is great enough to keep it all together.  This is when a star is born.  The pressure of gravity has caused a gravitational collapse to ignite into a star.  The balance of the forces of gravity and heat and pressure are what keep it going.  If there wasn't enough pressure outward, it would collapse further into a singularity (a black hole), if there's too much pressure outward, it would either explode (type II supernova) or at least expand outward until balance, what we call hydrostatic equilibrium, is achieved again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a star ages and burns through its supply of hydrogen, the equilibrium changes and at some point it begins to burn helium. After that, carbon. Depending on the original mass, it may fuse other materials up to iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some stars, the mass is enough that when the fusion can no longer be supported, there is no longer enough outward pressure, and the core collapses in on itself.  The gravity increases as it shrinks, and there's not enough heat or pressure to overcome it.  It collapses to a point of infinite density and becomes a gravitational singularity and a black hole is created.  In other stars, the mass isn't quite enough to keep collapsing, and the pressure overcomes it in a massive explosion (type II supernova).  In other stars, the star belches out wave after wave of gas as it expands and contracts, finally belching enough away to no longer have enough reactive mass.  It shrinks down into a white dwarf - a hot ball of atomic particles that slowly cools over billions and billions of years until they are no longer luminous, becoming a black dwarf (the time for this to occur is believed to be so long that there are, as yet, no black dwarfs in the universe... but there someday should be, long, long in the future).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to circle back to the original questions... why doesn't the sun explode, and why doesn't the hydrogen in the sun explode... the main reason is gravity. It reaches a balance.  The heat and pressure pushing  are balanced against the gravity pulling in.  There is no chemical oxidation going on, because the atoms of hydrogen are being fused into helium and there's little oxygen to be consumed in the oxidation process to begin with.  Someday we expect the sun will run out of hydrogen to burn.  As it starts burning helium, it will expand out past the orbit of Mars, then at some point it will shed its cooler, outer layers and contract into a white dwarf which will slowly cool down to a dying ember in the ever darkening universe.&lt;/p&gt;&#xA;" OwnerUserId="10546" LastEditorUserId="14094" LastEditDate="2019-06-01T00:24:57.653" LastActivityDate="2019-06-01T00:24:57.653" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30617" PostTypeId="1" AcceptedAnswerId="30619" CreationDate="2019-04-23T21:01:33.133" Score="1" ViewCount="287" Body="&lt;p&gt;If I understand correctly, in countries to the north of the tropics the sun's noon shadow points north.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens south of the tropics? Is it pointing south?  &lt;/p&gt;&#xA;" OwnerUserId="25200" LastEditorUserId="14094" LastEditDate="2019-04-24T13:46:08.943" LastActivityDate="2019-04-24T13:46:08.943" Title="South of the tropics, where does the sun's noon shadow point?" Tags="&lt;the-sun&gt;&lt;solar-system&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30618" PostTypeId="1" CreationDate="2019-04-23T21:35:31.060" Score="-5" ViewCount="493" Body="&lt;p&gt;I need to know this for a test. Please help. The question states &quot;If a spacecraft was parked on Venus and needed to make a flight to Jupiter, how far would it need to travel? (Assume both planets are aligned with the sun and are on the same side of the sun.)&quot;&lt;/p&gt;&#xA;" OwnerUserId="26795" LastEditorUserId="616" LastEditDate="2019-04-24T16:24:21.867" LastActivityDate="2019-04-24T19:25:00.107" Title="What is the distance from Venus to Jupiter in km?" Tags="&lt;planet&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30619" PostTypeId="2" ParentId="30617" CreationDate="2019-04-23T21:59:38.390" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it pointing south?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, the shadow points south when you are south of the tropics (south of the Tropic of Capricorn).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the tropics (between the Tropic of Cancer and the Tropic of Capricorn), the direction of the Sun's shadow depends on the location and the date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the angle between the sun and the earth's axis changes during the year, the apparent path of the sun in the sky changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Outside the tropics, this change is insufficient to ever bring the sun directly overhead.  Regardless of the date, the sun will be seen to be toward the equator at local noon.  (Same as saying the shadow points away from the equator).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Within the tropics, the changing path sometimes brings the sun directly overhead.  The position at local noon depends on the specific location and the date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the December solstice, the sun will be overhead for locations on the Tropic of Capricorn and will cross to the south at local noon for all locations in the tropics (and those further north).  During the June solstice, the situation is reversed for the tropics.  All locations south of the Tropic of Cancer would see the sun pass to the north.  &lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2019-04-23T23:26:18.237" LastActivityDate="2019-04-23T23:26:18.237" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="30620" PostTypeId="2" ParentId="30611" CreationDate="2019-04-23T22:47:30.937" Score="3" Body="&lt;p&gt;It depends on the direction the photon travels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the photon is directed straight away from the black hole, it will escape (but will be &lt;em&gt;very&lt;/em&gt; redshifted). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously all directions towards the event horizon will lead to capture. But since the spacetime is so curved, this also includes many directions that point away from the hole: the photon trajectory will curve around and hit the event horizon anyway. Closer to the event horizon the cone of directions where photons can escape gets smaller and smaller, until it is essentially just the direction straight away. (&lt;a href=&quot;https://www.zarm.uni-bremen.de/fileadmin/user_upload/space_science/gravitational_theory/tarusa.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;See slide 6 in this presentation&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also  &lt;a href=&quot;http://spiro.fisica.unipd.it/~antonell/schwarzschild/live/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this interactive web page&lt;/a&gt;; try it for a fixed position looking along the &quot;orbit&quot;. The higher up in the picture the closer you get to the horizon. I have found that one can get past the &lt;span class=&quot;math-container&quot;&gt;$1.5R_s$&lt;/span&gt; limit (the photon sphere, where exactly half of the sky is directions leading into the hole) by moving to a different browser tab, scrolling, and returning; the effect between 1 and 1.5 &lt;span class=&quot;math-container&quot;&gt;$R_s$&lt;/span&gt; is to make the sky look like an ever smaller ball - photons emitted in any other direction will be captured.&lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2019-04-23T22:47:30.937" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30621" PostTypeId="1" CreationDate="2019-04-23T22:54:29.477" Score="4" ViewCount="76" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/30602/7982&quot;&gt;This answer&lt;/a&gt; and comments got me thinking. Astrometry 101 tells us that while we can use &lt;span class=&quot;math-container&quot;&gt;$\lambda/D$&lt;/span&gt; as an estimator of resolution, &lt;em&gt;if we can assume a point source&lt;/em&gt; we can determine the centroid or position to far higher precision. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;https://en.wikipedia.org/wiki/Gaia_(spacecraft)#Objectives&quot; rel=&quot;nofollow noreferrer&quot;&gt;GAIA's design targets for precision were roughly 7, 20, and 200 micro-arcseconds for visual magnitudes 10, 15, and 20, respectively&lt;/a&gt; whereas &lt;span class=&quot;math-container&quot;&gt;$\lambda/D$&lt;/span&gt;  gives 70,000 x 200,000 micro-arcseconds for its rectangular mirror.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limitations are both &lt;em&gt;instrumental&lt;/em&gt; beyond just the aperture or baseline, and connected to the nature of the signal, and that's were determining the direction of a gravitational wave source is very different than determining the direction of a star's light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; For gravitational waves then, what actually determines the angular uncertainty of the source direction? Does it turn out to be baseline-limited (e.g. &lt;span class=&quot;math-container&quot;&gt;$\lambda/D$&lt;/span&gt;) or instrument-limited, or limited by the very short-lived and chirped nature of the signal? Or is it something else, perhaps in the modeling and reconstruction of the event itself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From ligo.caltech.edu's &lt;a href=&quot;https://www.ligo.caltech.edu/image/ligo20171016e&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gravitational-Wave Observatories Across the Globe&lt;/a&gt; just for background:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AtUy6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AtUy6.jpg&quot; alt=&quot;Gravitational-Wave Observatories Across the Globe&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2020-09-02T15:12:18.310" LastActivityDate="2020-09-02T15:12:18.310" Title="What actually determines the angular uncertainty of the source of a detected gravitational wave?" Tags="&lt;observational-astronomy&gt;&lt;gravitational-waves&gt;&lt;angular-resolution&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30622" PostTypeId="1" AcceptedAnswerId="31991" CreationDate="2019-04-23T23:14:01.633" Score="2" ViewCount="56" Body="&lt;p&gt;I ran across the image below in the August 2000 ESA Bulletin # 103 &lt;a href=&quot;http://www.esa.int/esapub/bulletin/bullet103/perryman103.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;GAIA – Unravelling the Origin and&#xA;Evolution of Our Galaxy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Figure 3. This simulation shows how a galaxy halo, like our own, may have been&#xA;  built up by the accretion of 50 dwarf galaxies colliding with the galaxy at various&#xA;  times during the last 10 billion years. The unit of distance is the kpc, or kiloparsec, where 1 kpc is a little more than 3000 lightyears. GAIA would be able&#xA;  to detect the fossil streams of these ancient merging events (Courtesy of Paul Harding)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Twenty years later, have results from GAIA been used to generate any data of this nature, identifying groups of stars that are likely to have come from nearby dwarf galaxies merging with our own?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rHsdy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rHsdy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-05-01T14:43:08.060" LastActivityDate="2019-05-22T07:44:33.687" Title="What has GAIA found out about the dwarf galaxy heritage of stars in the Milky Way, if anything?" Tags="&lt;galaxy&gt;&lt;galactic-dynamics&gt;&lt;gaia&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30623" PostTypeId="2" ParentId="30593" CreationDate="2019-04-23T23:38:06.357" Score="1" Body="&lt;p&gt;The related observation in astronomy is &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_parallax&quot; rel=&quot;nofollow noreferrer&quot;&gt;stellar parallax&lt;/a&gt;:&#xA;the observer knows the Earth's position relative to the Sun, measures tiny annual oscillations in the direction to another star, and estimates the distance to the star.&#xA;A star at a distance of 1 &lt;a href=&quot;https://en.wikipedia.org/wiki/Parsec&quot; rel=&quot;nofollow noreferrer&quot;&gt;parsec&lt;/a&gt; (pc) appears to shift by 1 arcsecond (4.85 &amp;mu;rad) per &lt;a href=&quot;https://en.wikipedia.org/wiki/Astronomical_unit&quot; rel=&quot;nofollow noreferrer&quot;&gt;astronomical unit&lt;/a&gt; (au) of lateral displacement.&#xA;Since parallax is inversely related to distance, the relative error of a parallax-based distance estimate increases with distance.&#xA;For any given star, the position uncertainty ellipsoid is shaped like a needle pointing toward the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aboard an interstellar spacecraft with limited observational facilities, it may not be practical to measure angles between nearby stars directly.&#xA;Instead you might individually measure their positions relative to several distant &lt;a href=&quot;https://en.wikipedia.org/wiki/Catalogues_of_Fundamental_Stars&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference stars&lt;/a&gt; near them in the sky.&#xA;For best results, you would need to account for the stars' &lt;a href=&quot;https://en.wikipedia.org/wiki/Proper_motion&quot; rel=&quot;nofollow noreferrer&quot;&gt;proper motion&lt;/a&gt; over the duration of your journey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After doing that data reduction, you can compute angles APB, BPC, APC fairly precisely.&#xA;However, the Sun distances SA, SB, SC have significant uncertainties which propagate to other quantities computed from them, e.g. distances AB, BC, AC.&#xA;You can estimate distances PA, PB, PC by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Law_of_sines&quot; rel=&quot;nofollow noreferrer&quot;&gt;law of sines&lt;/a&gt;, but these are uncertain by similar amounts.&#xA;Alternatively you can estimate PA, PB, PC by comparing &lt;a href=&quot;https://en.wikipedia.org/wiki/Apparent_magnitude&quot; rel=&quot;nofollow noreferrer&quot;&gt;apparent magnitudes&lt;/a&gt; to &lt;a href=&quot;https://en.wikipedia.org/wiki/Absolute_magnitude&quot; rel=&quot;nofollow noreferrer&quot;&gt;absolute magnitudes&lt;/a&gt;, which are subject to the same uncertainties.&#xA;Then you can find the intersection of spheres around A, B, C to estimate position P, but still no more precisely than distances SA, SB, SC.&#xA;P is not so much a point as a probability distribution cloud which expands as you move away from the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the real world, our most distant spacecraft &lt;a href=&quot;https://voyager.jpl.nasa.gov/mission/status/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Voyager 1&lt;/a&gt; is only 145 au (0.0007 pc) away.&#xA;Within about 400 times that distance, or 1/5 the distance to &amp;alpha; Cen, the Sun is the brightest star in the sky by far and would remain your primary point of reference.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2019-04-23T23:38:06.357" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30624" PostTypeId="5" CreationDate="2019-04-23T23:58:27.027" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2019-04-23T23:58:27.027" LastActivityDate="2019-04-23T23:58:27.027" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30625" PostTypeId="4" CreationDate="2019-04-23T23:58:27.027" Score="0" Body="Questions regarding a substance that takes up space, has volume, and has mass. Matter usually occurs in 5 forms, namely plasma, gas, liquid, solid, and bose-einstein condensate." OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-04-24T14:59:26.957" LastActivityDate="2019-04-24T14:59:26.957" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30626" PostTypeId="5" CreationDate="2019-04-24T00:00:00.397" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2019-04-24T00:00:00.397" LastActivityDate="2019-04-24T00:00:00.397" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30627" PostTypeId="4" CreationDate="2019-04-24T00:00:00.397" Score="0" Body="Questions regarding a sudden brightness increase due to the release of magnetic energy from the sun." OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-04-24T14:32:48.180" LastActivityDate="2019-04-24T14:32:48.180" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30628" PostTypeId="5" CreationDate="2019-04-24T00:01:54.640" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2019-04-24T00:01:54.640" LastActivityDate="2019-04-24T00:01:54.640" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30629" PostTypeId="4" CreationDate="2019-04-24T00:01:54.640" Score="0" Body="Questions regarding a type of elementary particle with 0 mass and is a quantum of the electromagnetic field. Photons always travel at constant speed in a vacuum." OwnerUserId="25988" LastEditorUserId="25988" LastEditDate="2019-04-24T14:59:30.303" LastActivityDate="2019-04-24T14:59:30.303" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30630" PostTypeId="5" CreationDate="2019-04-24T00:05:12.447" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2019-04-24T00:05:12.447" LastActivityDate="2019-04-24T00:05:12.447" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30631" PostTypeId="4" CreationDate="2019-04-24T00:05:12.447" Score="0" Body="Questions regarding a phenomenon when an object has an orbital period that is equal to its rotational period due to gravitational tidal forces." OwnerUserId="25988" LastEditorUserId="7982" LastEditDate="2019-04-24T02:12:01.220" LastActivityDate="2019-04-24T02:12:01.220" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31622" PostTypeId="2" ParentId="30618" CreationDate="2019-04-24T14:34:49.037" Score="2" Body="&lt;p&gt;If you want the distance between the planets, then you can easily look up the orbital radius of both planets, subtract the orbital distance of Venus from that of Jupiter, and there you have the distance between Venus and Jupiter.... but this is problematic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, both planets orbit in an elliptical path.  Their distances from the sun vary.  For example, Venus's Perihelion, the closest it gets to the sun, is about 107,447,000 km.  It's Aphelion (furthest from the sun) is 108,939,000.  Jupiter's Perihelion is 740,520,000 km and its Aphelion is 816,650,000.   If Venus is at Aphelion and Jupiter is at Perihelion when Jupiter is in opposition to Venus (meaning directly opposite from the sun), then the distance is about 631,581,000 km.  If Venus is at Perihelion while Jupiter is at Aphelion while at opposition, then the distance is 709,173,000 km.  If they're not both at opposition, the distance can be as far as 925,559,000 km if they're both at Aphelion and on opposite sides of the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But these are straight line distances.  If you're flying between the two, as you leave one and head to the other, you won't travel in a straight line.  You'll follow a curved path to get from one to the other.  The distance traveled will then be highly variable depending on the velocity.  The higher the Delta-V, the &quot;straighter&quot; the path between them and, thus, the shorter the distance.  But a smaller Delta-V, the longer the trip, and the wider the radius of the arc, and, thus, the longer the distance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, there's not enough information given to provide an accurate answer.&lt;/p&gt;&#xA;" OwnerUserId="10546" LastActivityDate="2019-04-24T14:34:49.037" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31624" PostTypeId="1" CreationDate="2019-04-24T16:31:48.860" Score="1" ViewCount="43" Body="&lt;p&gt;How do astronomers distinguish hot stars, reddened by the interstellar light absorption, from the actual cold &#xA;red stars?&lt;/p&gt;&#xA;" OwnerUserId="25925" LastActivityDate="2019-04-24T21:55:32.697" Title="Difference in observing hot stars that have been reddened and stars that are cool and red" Tags="&lt;observational-astronomy&gt;&lt;extinction&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31625" PostTypeId="1" CreationDate="2019-04-24T16:38:31.800" Score="1" ViewCount="1034" Body="&lt;p&gt;Suppose that iron and calcium have a spectral line with the same wavelength. How would you determine which element is present in the atmosphere of a star if you found this line in its spectrum?&lt;/p&gt;&#xA;" OwnerUserId="25925" LastActivityDate="2019-04-25T21:05:55.183" Title="How to differentiate elements that have same spectral lines in a star?" Tags="&lt;star&gt;&lt;spectra&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31626" PostTypeId="2" ParentId="31625" CreationDate="2019-04-24T17:14:06.410" Score="5" Body="&lt;p&gt;If you only had that single line, you would be unable to differentiate them.  This is especially true of redshifted objects where we couldn't tell the actual wavelength of an isolated peak.  But a good spectrograph will give you information over a range of wavelengths.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each element has multiple lines.  It is the overall pattern of relative strengths that is definitive.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, the stronger lines for a few elements are shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NlFYn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NlFYn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;(From &lt;a href=&quot;https://cnx.org/contents/LnN76Opl@10.1:H5KhIDcK@3/Spectroscopy-in-Astronomy&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://cnx.org/contents/LnN76Opl@10.1:H5KhIDcK@3/Spectroscopy-in-Astronomy&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see there's a cyan line that is similar in both hydrogen and calcium.  But the other lines are so different, you could tell if one or both were present, even if they were overlaid on a single chart.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2019-04-25T21:05:55.183" LastActivityDate="2019-04-25T21:05:55.183" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31628" PostTypeId="2" ParentId="30618" CreationDate="2019-04-24T19:13:27.937" Score="5" Body="&lt;p&gt;To save energy, let's consider a &lt;a href=&quot;https://en.wikipedia.org/wiki/Hohmann_transfer_orbit&quot; rel=&quot;noreferrer&quot;&gt;Hohmann transfer orbit&lt;/a&gt;, an ellipse tangent to opposite sides of Venus's and Jupiter's orbits.&#xA;The spacecraft would follow the outbound half of this orbit, departing ahead of Venus and arriving as Jupiter overtakes it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on where in Jupiter's orbit the spacecraft meets it, this orbit has semi-major axis a = (r&lt;sub&gt;V&lt;/sub&gt; + r&lt;sub&gt;J&lt;/sub&gt;) / 2 = 2.96&amp;plusmn;0.03 au and &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law_of_Kepler&quot; rel=&quot;noreferrer&quot;&gt;period&lt;/a&gt; of a&lt;sup&gt;3/2&lt;/sup&gt; = 5.10&amp;plusmn;0.08 years.&#xA;The outbound half has a &lt;a href=&quot;https://www.mathsisfun.com/geometry/ellipse-perimeter.html&quot; rel=&quot;noreferrer&quot;&gt;path length&lt;/a&gt; of 7.78&amp;plusmn;0.07 au and takes 2.55&amp;plusmn;0.04 years.&#xA;As Jupiter &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Second_law_of_Kepler&quot; rel=&quot;noreferrer&quot;&gt;moves&lt;/a&gt; 72&amp;deg; to 84&amp;deg; around the Sun in that time, the launch window is when Venus's heliocentric ecliptic longitude is 96&amp;deg; to 108&amp;deg; behind Jupiter's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To convert au to km, multiply by 1.496 &amp;times; 10&lt;sup&gt;8&lt;/sup&gt; km / au.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-04-24T19:25:00.107" LastActivityDate="2019-04-24T19:25:00.107" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31631" PostTypeId="2" ParentId="31624" CreationDate="2019-04-24T21:55:32.697" Score="3" Body="&lt;p&gt;Reddening does not change the details of the spectrum of a star. The differences in absorption lines that are present in a hot star vs a cool star are still there after reddening. So if you have a spectrum then distinguishing the two scenarios is easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't have a spectrum but have two colours then it is still possible to distinguish the scenarios by tracing the star back along an extinction vector in a colour 1 vs colour 2 diagram until it intercepts an intrinsic locus for (e.g.) main sequence stars. This can be definitive, but can be ambiguous depending on the types of star and particular colours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just have a colour, you can't tell.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-04-24T21:55:32.697" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31632" PostTypeId="1" AcceptedAnswerId="31638" CreationDate="2019-04-24T22:15:51.763" Score="1" ViewCount="75" Body="&lt;p&gt;I was looking to find some example carbon star spectra and came across &lt;a href=&quot;https://www.cfa.harvard.edu/~pberlind/atlas/htmls/cstars.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page on the Harvard website&lt;/a&gt; which shows several plots of carbon star spectra. The horizontal axis is labelled as wavelength in ångströms, but the vertical axis appears to be labelled with &quot;g-98&quot; which I am not familiar with - any idea what quantity this is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nLess.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nLess.gif&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorUserId="7982" LastEditDate="2019-04-25T22:52:05.583" LastActivityDate="2019-04-25T22:52:05.583" Title="What is &quot;g-98&quot; in plots of stellar spectra?" Tags="&lt;spectra&gt;&lt;units&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31633" PostTypeId="1" CreationDate="2019-04-25T04:34:01.747" Score="2" ViewCount="50" Body="&lt;p&gt;I've read many interesting threads concerning the composition of different planet cores in the solar system while trying to see if this question has already been asked. It appears as if Earth is unique regarding magnetic field strength and interior activity for our size compared with our rocky neighbors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if the general consensus was that most rocky-type planets our size would share the same characteristics (such as magnetic field strength and convection flow) as us? Or if we are the outliers and most extra solar planets similar to us are expected to be more like Mars or Venus?&lt;/p&gt;&#xA;" OwnerUserId="27808" LastActivityDate="2019-04-25T04:34:01.747" Title="Frequency of Earth type cores?" Tags="&lt;planet&gt;&lt;exoplanet&gt;&lt;core&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31635" PostTypeId="1" CreationDate="2019-04-25T09:15:47.300" Score="35" ViewCount="5547" Body="&lt;p&gt;From what I understand so far, when one of virtual particles crosses the event horizon and the other does not, they can not annihilate each other. The latter wanders off into the universe (btw, is it still virtual at this point, and what does 'virtual' mean at this point, if so?), while the other gets consumed by the black hole. I don't see how this event contributes to evaporation of the black hole (since the particles do not originate from the black hole). Shouldn't the consumed particle actually add-up to the black hole mass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest question to mine is &lt;a href=&quot;https://astronomy.stackexchange.com/questions/25501/does-hawking-radiation-in-fact-bring-mass-into-the-universe&quot;&gt;Does Hawking radiation in fact bring mass into the universe?&lt;/a&gt;, but I don't find the answers satisfactory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. &quot;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/20496/is-there-a-better-explanation-of-hawking-radiation&quot;&gt;the escaped virtual particle is 'boosted' by black hole's gravitational field into becoming a real particle&lt;/a&gt;&quot;, rather adds to the question then answer it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I am humbled by the knowledge presented in the answers and feel incompetent to mark any as the most suitable. I hope this is okay.&lt;/p&gt;&#xA;" OwnerUserId="27810" LastEditorUserId="28132" LastEditDate="2019-07-15T18:00:46.917" LastActivityDate="2019-07-15T21:23:06.213" Title="How exactly does Hawking radiation decrease the mass of black holes?" Tags="&lt;black-hole&gt;&lt;hawking-radiation&gt;&lt;particles&gt;" AnswerCount="5" CommentCount="7" FavoriteCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="31636" PostTypeId="2" ParentId="31635" CreationDate="2019-04-25T10:36:22.583" Score="28" Body="&lt;p&gt;I'm going to give you an intuitive answer. Keep in mind, this is not the &quot;actual&quot; answer, as the Hawking radiation is quite a bit more complex than the typical pop-sci explanation with virtual particles. But some intuitive justification is possible nevertheless.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I don't see how this event contributes to evaporation of the black&#xA;  hole (, since the particles do not originate from the black hole).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're missing a key point here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the pair was generated, those were virtual particles. After one side of the pair was absorbed by the black hole, and the other side was released, the released part is a real particle. Huge difference there - virtual vs real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Virtual particles don't really exist the same way that you and me exist. They seem to exist for a very short time; the more energetic they are, the shorter the interval of their virtual &quot;existence&quot;, per the Heisenberg equation. In many ways they are just a mathematical trick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of the vacuum, where no real particles exist. Before, it's just vacuum. Right now, a virtual pair flickers briefly, then it's gone. In the future, it's vacuum again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was the energy before? Zero. What is the energy in the future? Zero. What's the energy during the flicker? Well, it basically zero, within the limits permitted by Heisenberg's equations. Bottom line is, virtual particles come and go, and they do not contribute to the energy balance of some empty chunk of space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am ignoring here the concept of vacuum energy, for the sake of an intuitive explanation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let's say one of the virtual particles gets trapped by the black hole, so it cannot annihilate with its counterpart. The other particle flies off in the opposite direction and escapes the black hole. What's worse, this is now a real particle - we've exceeded the duration permitted by the Heisenberg equations, so the one that escapes is not virtual anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did that particle become real?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a big issue, because virtual particles don't require an energy budget to briefly exist, while real particles do carry energy forever. Something prevented the virtual pair from annihilating itself, and boosted one of the components to the status of real particle. The virtual pair has zero energy. The real particle that gets away has non-zero energy. That energy has to come from somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It comes from the black hole. The black hole gives up some of its mass / energy (same thing) to boost one particle from virtual to real. The other particle is captured - but being virtual anyway, it doesn't really matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this intuitive explanation doesn't say is how the boost actually happens. I dunno, magic. Somehow one of the virtual particles gets a chunk of energy from the black hole and becomes real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, this is not the actual process. The actual process &lt;a href=&quot;http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html&quot; rel=&quot;noreferrer&quot;&gt;is more complex&lt;/a&gt;. This is just a pop-sci fairy tale.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: To hit closer to home, Hawking radiation is more like a close relative to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Unruh_effect&quot; rel=&quot;noreferrer&quot;&gt;Unruh effect&lt;/a&gt;. Say an inertial observer sees empty space here in this chunk of volume. An accelerating observer would not see empty space in the same volume, but instead would see blackbody radiation. That's the Unruh effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, gravity and acceleration are the same thing, per general relativity. So the strong gravity near a black hole is equivalent to strong acceleration. Something similar to the Unruh effect must happen there. That's the Hawking radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://backreaction.blogspot.com/2015/12/hawking-radiation-is-not-produced-at.html&quot; rel=&quot;noreferrer&quot;&gt;http://backreaction.blogspot.com/2015/12/hawking-radiation-is-not-produced-at.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT2: The other answers currently on this page provide useful alternative points, so check them out too.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2019-04-26T05:40:33.677" LastActivityDate="2019-04-26T05:40:33.677" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="31637" PostTypeId="1" CreationDate="2019-04-25T11:26:09.727" Score="1" ViewCount="71" Body="&lt;p&gt;.. and if so, wouldn't this also mean that during this event, time would be created as a collateral, as a canvas for this property to manifest?&lt;/p&gt;&#xA;" OwnerUserId="27810" LastActivityDate="2019-04-25T11:26:09.727" Title="Is it possible for 'pre-big bang' empty space to have the property of being able to spawn matter and/or energy into existence?" Tags="&lt;big-bang-theory&gt;" AnswerCount="0" CommentCount="11" ClosedDate="2019-04-25T13:06:25.120" ContentLicense="CC BY-SA 4.0" />
  <row Id="31638" PostTypeId="2" ParentId="31632" CreationDate="2019-04-25T14:05:16.093" Score="3" Body="&lt;p&gt;It looks like the floating &quot;g-98&quot; is just an extension of the header of the plot (made by &lt;a href=&quot;http://ast.noao.edu/data/software&quot; rel=&quot;nofollow noreferrer&quot;&gt;IRAF&lt;/a&gt;), which has wrapped around the image for lack of room on the right-hand side. Looking at &lt;a href=&quot;https://www.cfa.harvard.edu/~pberlind/atlas/images/HD144578.N0.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;the N0 plot&lt;/a&gt;, we see that the header should read&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;NOAO/IRAF V2.10.4EXPORT perry@sparky.sao.arizona.edu Mon 06:04:51 03-Aug-98&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which just contains information about the software, user and creation date behind the plot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Hat tip to Pela for confirming my guess!)&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2019-04-25T14:05:16.093" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31639" PostTypeId="1" AcceptedAnswerId="31645" CreationDate="2019-04-25T14:36:27.060" Score="3" ViewCount="252" Body="&lt;p&gt;According to the formula for gravitational acceleration &lt;span class=&quot;math-container&quot;&gt;$a = GM/r^2$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$G=6.674 \times 10^{-11}$&lt;/span&gt;, Jupiter with mass 1.89813 × 10^27 kg and radius 69,911,000 m gives a gravity around 25m/s^2. The Sun with mass 1.989 × 10^30 kg and radius 695,510,000m will have a gravity around 270m/s^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But using the data for Kepler 39B, mass 18 times that of Jupiter 3.416634 × 10^28, and radius about 5/4 of Jupiter's 87,220,000m, I get almost 300m/s^2, higher than the Sun's. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did and redid it, and still concerned that I am doing something wrong.&lt;/p&gt;&#xA;" OwnerUserId="1707" LastEditorUserId="7982" LastEditDate="2019-04-25T22:48:01.710" LastActivityDate="2019-04-26T02:08:31.787" Title="Is the gravity on Kepler 39b higher than on the Sun?" Tags="&lt;gravity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31640" PostTypeId="2" ParentId="31635" CreationDate="2019-04-25T15:24:45.993" Score="7" Body="&lt;p&gt;Heisenberg's principle allows you to temporarily violate energy conservation laws (e.g. create pairs of particles out of nothing) as long as you repay everything in time. The larger the particle-antiparticle pair, the quicker it has to be repaid. Converting a virtual pair to a real pair can be seen as generating a bit of negative-energy &quot;exotic matter&quot; (whatever that is) to represent the unpaid debt. Its energy is equal in size to the pair with the opposite sign. This then falls into the black hole along with one of the particles, decreasing the mass of the black hole overall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horizon of the black hole gets in the way of recombining some virtual pairs, so these conversions virtual-&gt;real will happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this lecture with the same idea (more detailed and less butchered):&#xA;&lt;a href=&quot;http://teacher.pas.rochester.edu/Ast102/LectureNotes/Lecture19/Lecture19.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://teacher.pas.rochester.edu/Ast102/LectureNotes/Lecture19/Lecture19.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user27815" LastEditorDisplayName="user27815" LastEditDate="2019-04-25T15:42:19.173" LastActivityDate="2019-04-25T15:42:19.173" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31641" PostTypeId="2" ParentId="18791" CreationDate="2019-04-25T16:28:25.900" Score="3" Body="&lt;p&gt;The reason it is called &quot;precipitable&quot; water vapor (PWV) is that the water vapor is in the form of gaseous vapor which may further condense into cloud form and then into actual precipitation i.e. rain (which often overwhelm the sensors). It is often also called &lt;em&gt;total column water vapor&lt;/em&gt; which more accurately reflects what is being measured i.e. a height of the equivalent column of liquid water, hence the measurement in mm. The American Meteorological Society &lt;a href=&quot;http://glossary.ametsoc.org/wiki/Precipitable_water&quot; rel=&quot;nofollow noreferrer&quot;&gt;glossary entry on precipitable water (or precipitable water vapor)&lt;/a&gt; similarly defines it as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The total atmospheric water vapor contained in a vertical column of unit cross-sectional area extending between any two specified levels, commonly expressed in terms of the height to which that water substance would stand if completely condensed and collected in a vessel of the same unit cross section.&#xA;  The total precipitable water is that contained in a column of unit cross section extending all of the way from the earth's surface to the &quot;top&quot; of the atmosphere. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For very high and dry sites e.g. the Atacama, Antarctica will have less than 5mm PWV, sea level sites will be 50+mm of PWV. ESO's Cerro Paranel observing site (home of the VLT), has median PWV of 2.5mm (see e.g. the histograms in this &lt;a href=&quot;http://www.eso.org/gen-fac/pubs/astclim/paranal/seeing/ASM_STATS%20at%20Paranal-2016-2018.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;ASM 2016-2018 report&lt;/a&gt;). The Atacama Pathfinder EXperiment (APEX) sub-mm telescope on Chajnantor (also home to the ALMA array) has had a weather station for several years and for example the &lt;a href=&quot;http://www.apex-telescope.org/weather/year_weather/index.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;APEX 2017 weather statistics&lt;/a&gt; show that the APEX site (~5100m altitude) had a PWV &amp;lt;1.5mm approx. 67% of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several ways to measure the PWV. The Earth orbiting satellites such as the AIRS and MODIS instruments on NASA's &lt;em&gt;Aqua&lt;/em&gt; and &lt;em&gt;Terra&lt;/em&gt; satellites measure the radiance in several IR wavebands (typically from approx. 0.5 to 15 um). Some of these bands are essentially dominated by water absorption bands so performing differences between bands containing water and without water will give a measure of the water column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Microwave radiometers at 210 or 225 GHz are traditionally what have been used to measure PWV, with more modern radiometers shifting to the 350um (856 GHz) window/band. An example is the one at the Caltech Submillimeter Observatory on Mauna Kea described in &lt;a href=&quot;http://www.gemini.edu/sciops/telescopes-and-sites/observing-condition-constraints/mk-water-vapour-statistics&quot; rel=&quot;nofollow noreferrer&quot;&gt;this link at Gemini Observatory&lt;/a&gt;. They work similarly by measuring the depth of the absorption by the water vapor in the molecular lines/bands. More water vapor produces deeper lines with more optical depth. More details are in this (freely available) &lt;a href=&quot;https://arxiv.org/pdf/1107.5633.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper by Radford&lt;/a&gt; which shows the effect of increasing PWV on the absorption in the sub-mm, the amount of PWV at various sites and more info on the measurement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally this can also be done by dual frequency GPS receivers at the L1 (1575 MHz) and L2 (1224 MHz) and measuring the path length excess through the atmosphere above what is caused by the receiver-satellite distance alone. By using two frequencies, you can remove the effect of the variable ionosphere on the path length. The remaining zenith path distance breaks down into two parts, a hydrostatic or &quot;dry&quot; component which can be easily estimated using e.g. the &lt;a href=&quot;https://gssc.esa.int/navipedia/index.php/Galileo_Tropospheric_Correction_Model&quot; rel=&quot;nofollow noreferrer&quot;&gt;Saastamoinen&lt;/a&gt; model, leaving the &quot;wet&quot; component caused by water vapour. Once this is corrected to the zenith, this gives a measure of the PWV.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastEditorUserId="23748" LastEditDate="2019-04-26T15:44:18.703" LastActivityDate="2019-04-26T15:44:18.703" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31642" PostTypeId="2" ParentId="31635" CreationDate="2019-04-25T21:04:22.083" Score="9" Body="&lt;p&gt;&lt;a href=&quot;http://teacher.pas.rochester.edu/Ast102/LectureNotes/Lecture19/Lecture19.pdf&quot; rel=&quot;noreferrer&quot;&gt;These lecture notes&lt;/a&gt; address the issues to some degree, especially on slides 33-35.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because in the strongly warped spacetime near the&#xA;   horizon, virtual particles made from vacuum fluctuations&#xA;   turn out to have negative energy density.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Energy density = energy per unit volume.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;These particles indeed have positive mass -- look at the&#xA;   one that escaped! -- but their mass is distributed very&#xA;   strangely over spacetime. (Quantum-mechanically&#xA;   speaking, particles have nonzero volume; this is an aspect&#xA;   of the wave-particle duality.)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Matter with negative energy density is generally called&#xA;  &lt;em&gt;exotic matter&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and, a bit later:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Quantum mechanical vacuum fluctuations in flat&#xA;   spacetime - far from any strong gravitational field -&#xA;   always have zero net energy density; they can never be&#xA;   exotic.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;However, in warped spacetime, vacuum fluctuations are&#xA;   in general exotic: their net energy density is negative,&#xA;   according to a distant observer measuring the energy&#xA;   density by observation of the deflection of light by the&#xA;   ensemble of fluctuations. The stronger the curvature, the&#xA;   more negative the energy density looks. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the best explanation I have seen so far. &lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-04-25T21:04:22.083" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="31643" PostTypeId="1" AcceptedAnswerId="31674" CreationDate="2019-04-25T23:50:05.427" Score="2" ViewCount="140" Body="&lt;p&gt;The Phys.org article &lt;a href=&quot;https://phys.org/news/2019-04-hubble-universe-faster.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mystery of the universe's expansion rate widens with new Hubble data&lt;/a&gt; says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Astronomers have been using Cepheid variables as cosmic yardsticks to gauge nearby intergalactic distances for more than a century. But trying to harvest a bunch of these stars was so time-consuming as to be nearly unachievable. &lt;strong&gt;So, the team employed a clever new method, called DASH (Drift And Shift), using Hubble as a &amp;quot;point-and-shoot&amp;quot; camera to snap quick images of the extremely bright pulsating stars, which eliminates the time-consuming need for precise pointing.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;When Hubble uses precise pointing by locking onto guide stars, it can only observe one Cepheid per each 90-minute Hubble orbit around Earth. So, it would be very costly for the telescope to observe each Cepheid,&amp;quot; explained team member Stefano Casertano, also of STScI and Johns Hopkins. &lt;strong&gt;&amp;quot;Instead, we searched for groups of Cepheids close enough to each other that we could move between them without recalibrating the telescope pointing.&lt;/strong&gt; These Cepheids are so bright, we only need to observe them for two seconds. This technique is allowing us to observe a dozen Cepheids for the duration of one orbit. So, we stay on gyroscope control and keep 'DASHing' around very fast.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I understand that the technique skips the use of guide stars, but I'm wondering exactly what does happen. What does the word &lt;em&gt;recalibrating&lt;/em&gt; mean in &amp;quot;..move between them without recalibrating the telescope pointing...&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Do they just execute a &lt;em&gt;blind change&lt;/em&gt; in direction using Hubble's attitude control system of gyros and reaction wheels, expose, and move on to the next field, and hope the target object was imaged?&lt;/p&gt;&#xA;&lt;p&gt;For a little background see &lt;a href=&quot;https://space.stackexchange.com/q/31180/12102&quot;&gt;Hubble's one-gyro mode; how does it work for attitude control, stabilization, and slewing?&lt;/a&gt; and answer(s).&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-04-28T12:11:37.280" Title="How does the Hubble Space Telescope &quot;Drift and Shift&quot;?" Tags="&lt;hubble-telescope&gt;&lt;cepheids&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31645" PostTypeId="2" ParentId="31639" CreationDate="2019-04-26T02:08:31.787" Score="6" Body="&lt;p&gt;You are probably overlooking &lt;em&gt;why&lt;/em&gt; the radius is what it is, or equivalently why the planet is denser than the Sun.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun is undergoing fusion in its core, which releases energy, which causes it to expand (though it achieves an equilibrium state, as expansion decreases rate of fusion which leads to contraction, which leads to accelerated fusion which leads to expansion...).  Jupiter-like bodies, however, don't produce heat from fusion (except maybe some deuterium fusion very early in their lives); they slowly cool and contract over time.  So such a body can compress into a denser state, allowing it achieve higher surface gravity (give or take difficulties in defining &quot;surface&quot; for a gaseous body).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eyeballing the calculation, you can see that the sun is 10 times the radius of Jupiter and 1000 times the mass, so it's surface gravity should be about &lt;span class=&quot;math-container&quot;&gt;$1000/10^2=10$&lt;/span&gt; times greater than Jupiter's, which is what you find.  But Kepler 39b is 18 times more massive than Jupiter and only marginally larger. That's a big uptick in density. So that it's surface gravity would be about &lt;span class=&quot;math-container&quot;&gt;$18/(1.25)^2=11.52$&lt;/span&gt; times stronger than Jupiter's (also what you find).  And &lt;span class=&quot;math-container&quot;&gt;$11.52&amp;gt;10$&lt;/span&gt;, so Kepler 39b has higher surface gravity.&lt;/p&gt;&#xA;" OwnerUserId="2188" LastActivityDate="2019-04-26T02:08:31.787" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31647" PostTypeId="1" CreationDate="2019-04-26T12:50:49.397" Score="2" ViewCount="45" Body="&lt;p&gt;The transition between metal dish and mirror happens around hundreds of microns. For example, the shortest wavelength the &lt;a href=&quot;https://en.wikipedia.org/wiki/Atacama_Large_Millimeter_Array&quot; rel=&quot;nofollow noreferrer&quot;&gt;ALMA&lt;/a&gt;(metal dish) can detect is about 0.3mm, while the longest wavelength the &lt;a href=&quot;https://en.wikipedia.org/wiki/Spitzer_Space_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Spitzer&lt;/a&gt;(optical mirror) can detect is about 160 micrometer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions is what determines the surface material of the ALMA and Spitzer telescope? Could a metal dish observe shorter wavelength than a mirror telescope?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="7982" LastEditDate="2019-04-26T15:21:36.257" LastActivityDate="2019-04-26T22:48:19.590" Title="What determines the surface material of the ALMA and Spitzer telescope?" Tags="&lt;telescope&gt;&lt;radio-telescope&gt;&lt;infrared&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31648" PostTypeId="2" ParentId="31647" CreationDate="2019-04-26T21:34:44.147" Score="3" Body="&lt;p&gt;They're all mirrors, they just operate in different wavelength ranges. A wire fence could be a mirror if the wavelength is big enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface is an easy problem to solve. Pick any material and do vacuum deposition with it on the substrate. It's the substrate, the base material that matters a lot - the thick part that keeps the shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A metal mirror is easier to make. Standard engineering principles apply. That's why this technology is used for ALMA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there's a catch - precision. This is dictated by wavelength. Imagine all the radiation captured by the mirror, focused in one point. If the mirror is perfect, it's all good. If the mirror has deviations from perfection, parts of the wavefront arrive too soon or too late. The wavefront begins to interfere with itself, reducing the amplitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interference is completely destructive (the signal is completely lost) when the difference between the paths is exactly half the wavelength: λ/2. Then the peak wave (maximum) from one side falls into the trough wave (minimum) from the other side, and they cancel each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you get a λ/2 difference between two parts of the wavefront? If the surface error on the mirror is λ/4. One part of the wavefront begins to be reflected immediately, while the other keeps traveling λ/4 to get to the mirror, and then wastes λ/4 again on its way back up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mirrors with λ/4 errors are bad - since the error is never exactly half-and-half, parts of the wavefront are not cancelled so the mirror keeps working somewhat, but poorly. Around λ/8 the mirrors become decent. λ/16 are very good, and beyond λ/20 are essentially indistinguishable from perfection. This is for optical mirrors (visible spectrum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why optical telescope mirrors are made of glass. It takes such a smooth, rigid material to reach the precision necessary for optical observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Spitzer mirror is actually made of beryllium, cooled to 5.5 K. This is fine. Precision requirements are a bit more relaxed, compared to visible light. In fact, solid metal mirrors have been made for visible light telescopes even by amateurs - but in these conditions glass is preferred.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ALMA would have to be manufactured to a precision better than 75 micron, in reality much better. That's doable with regular metal parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other materials used for visible light telescopes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Granite. It takes an excellent polish, just like glass, it's very rigid, and the coefficient of expansion is similar to glass. As long as it doesn't have voids, it's a decent material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Marble - same as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speculum metal - back in the days of Herschel they were casting mirrors out of this bronze-like alloy. Easy to work with but it tarnishes very quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Special ceramics such as Zerodur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quartz - like glass, but better (low thermal expansion). Takes more work to grind and polish because it's harder than glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TLDR: It's mostly about precision and ease of manufacturing.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Once the mirror is made, it could be coated with anything you like, so it could reflect any wavelength up to hard UV. Like I said, the surface, the reflective layer, is the easy part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply make a list of all materials that provide good reflectivity in the intended range of wavelengths. Most of these materials turn out to be metals, unsurprisingly. Now compare them based on such criteria as: durability, price, ease of coating. Pick one and drop it in the tungsten &quot;boat&quot; in the vacuum chamber. Put the mirror in the vacuum chamber and run some amperage through the tungsten boat until the metal layer on the mirror is thick enough so it's not transparent anymore - but not thicker than that (for precision). Done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative that is sometimes used in infrared and visible light is &lt;a href=&quot;https://en.wikipedia.org/wiki/Dielectric_mirror&quot; rel=&quot;nofollow noreferrer&quot;&gt;dielectric coating&lt;/a&gt;. It's a stack of fairly transparent layers, and there are multiple reflections from all the surfaces between layers. The thickness is such that the path difference between reflections is exactly one wavelength, 1 λ. That strengthens all reflections (constructive interference).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dielectric mirrors have extremely high reflection coefficients, and are used for lasers, etc. Also tend to have preferred wavelengths where they work well (though broadband dielectric mirrors also exist). They're quite expensive and are not commonly used for telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, after all that, telescope mirrors sometimes get a protective coating to shield the reflective metal layer from corrosion. A commonly used material is silicon oxide (SiOx, not SiO2 which is quartz and it's different). If the mirror operates in a vacuum or it can be recoated often then this is less important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vast majority of amateur telescopes nowadays use:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;borosilicate glass (e.g. Schott Supremax) for the substrate; it's a low expansion glass, cheap enough. The original Pyrex glass (not made anymore) was almost the exact same thing.&lt;/li&gt;&#xA;&lt;li&gt;aluminum for the reflective layer; sometimes silver is used, which is slightly more reflective initially, but decays more quickly&lt;/li&gt;&#xA;&lt;li&gt;SiOx top coat for durability&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some mirrors have &quot;enhanced coating&quot; which is conceptually similar to dielectric mirrors, in addition to the metal layer.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2019-04-26T22:48:19.590" LastActivityDate="2019-04-26T22:48:19.590" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31649" PostTypeId="1" AcceptedAnswerId="31667" CreationDate="2019-04-27T00:16:24.873" Score="1" ViewCount="86" Body="&lt;p&gt;astrosnapper@'s &lt;a href=&quot;https://astronomy.stackexchange.com/a/31641/7982&quot;&gt;thorough answer&lt;/a&gt; to the question &lt;em&gt;What is precipitable water vapor in millimeter-wave radioastronomy and how is it measured?&lt;/em&gt; links to the paper &lt;a href=&quot;https://arxiv.org/abs/1107.5633&quot; rel=&quot;nofollow noreferrer&quot;&gt;Observing Conditions for Submillimeter Astronomy&lt;/a&gt; which discusses spectral windows of atmospheric transmission in the 100 ~ 1500 GHz range. These windows are &quot;delineated&quot; or limited by strong absorption lines of molecular water vapor in the atmosphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure 2 is shown below. The caption reads&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Fig. 2. Energy levels of the lowest rotational states of water with transitions labeled by the frequency in GHz. The lines at 557, 752, 988, 1113, and 1229 GHz delineate the principal submillimeter windows.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and displays &lt;em&gt;ortho-&lt;/em&gt; and &lt;em&gt;para-&lt;/em&gt; water separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Curious to find out what those were, I went to &lt;a href=&quot;http://www1.lsbu.ac.uk/water/martin_chaplin.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Martin Chapin&lt;/a&gt;'s (also &lt;a href=&quot;http://www.lsbu.ac.uk/about-us/people/people-finder/prof-martin-chaplin&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;) site &lt;a href=&quot;http://www1.lsbu.ac.uk/water/water_structure_science.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Water Structure and Science&lt;/a&gt; which discusses the physics of &lt;em&gt;ortho-&lt;/em&gt; and &lt;em&gt;para-&lt;/em&gt; water, and describes experiments on single water molecules trapped and isolated in C60 fullerene &quot;cages&quot; (&lt;a href=&quot;https://en.wikipedia.org/wiki/Endohedral_fullerene&quot; rel=&quot;nofollow noreferrer&quot;&gt;Endohedral fullerenes&lt;/a&gt;): &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www1.lsbu.ac.uk/water/ortho_para_water.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ortho-Water and para-Water&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Have caged molecules ever been observed in space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These could be trapped in fullerenes (which I believe have been observed) or otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ijapj.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ijapj.gif&quot; alt=&quot;ortho- or para- water trapped in a fullerene cage&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www1.lsbu.ac.uk/water/ortho_para_water.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-04-27T23:53:39.090" LastActivityDate="2019-04-27T23:53:39.090" Title="Have caged molecules ever been observed in space?" Tags="&lt;radio-astronomy&gt;&lt;astrochemistry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31650" PostTypeId="2" ParentId="30594" CreationDate="2019-04-27T00:49:04.497" Score="0" Body="&lt;p&gt;@JamesK's answer is spot-on. I'll add some further background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out I should have known the answer to this question because about six weeks ago I asked &lt;a href=&quot;https://space.stackexchange.com/q/34896/12102&quot;&gt;How was the 18 December 2018 Bering Sea Fireball detected and characterized? Was it a serendipity or a weather satellite?&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll draw upon &lt;a href=&quot;https://space.stackexchange.com/a/34947/12102&quot;&gt;my answer&lt;/a&gt; to that question here.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Scott Manley just addressed this in his new video &lt;a href=&quot;https://www.youtube.com/watch?v=fpaxvjFh-qA&quot; rel=&quot;nofollow noreferrer&quot;&gt;173 Kiloton Explosion Over Bering Sea Was Asteroid Breaking Up&lt;/a&gt; a bit after &lt;code&gt;01:00&lt;/code&gt;:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But there are sensors all over the world &lt;strong&gt;looking for these big blasts because they look a lot like nuclear blasts; they have the same kind of energy&lt;/strong&gt;, and there are many people who want to know if someone is testing nuclear weapons, so these things get picked up by that same network. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But the nature of this network means that we don’t really get the results from it very quickly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-04-27T00:49:04.497" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31651" PostTypeId="1" CreationDate="2019-04-27T04:14:46.430" Score="3" ViewCount="171" Body="&lt;p&gt;I'm very new to radio astronomy and I am curious how an the signals received by a radio antenna are converted to data on a computer and then images? Is there a popular software to use for this, or do astronomers typically write their own code?&#xA;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="27831" LastEditorUserId="20361" LastEditDate="2019-08-13T14:01:41.510" LastActivityDate="2019-08-13T14:01:41.510" Title="How do astronomers convert radio signals received by their antenna to images?" Tags="&lt;radio-astronomy&gt;&lt;data-analysis&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31652" PostTypeId="2" ParentId="31635" CreationDate="2019-04-27T04:55:15.930" Score="2" Body="&lt;p&gt;I don't know if the experts will agree with this description, but here is how I understand it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both space and the event horizon are in constant quantum fluctuation. Essentially, the event horizon has tiny ripples. At points where the event horizon ripples up (above the average radius of the black hole), it has an above average amount of local energy. The intense gravity rapidly pulls that local bump back down, the falling bump sends that local energy concentration back across the rest of the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's consider possible virtual particle-pairs near the hole. If a stationary virtual particle-pair appears just above the event horizon, it will either recombine and vanish or the entire thing gets pulled into the hole and vanishes in zero. We need a virtual particle-pair that has an apparent motion away from the black hole, at nearly the speed of light. If that that virtual particle-pair is going fast enough to completely escape, they recombine and vanish. Zero net effect. We need a virtual particle-pair that is moving away from the black hole at nearly the speed of light, and we need a ripple in horizon which only catches one virtual particle. I believe the ripple must be under extreme downwards acceleration for it to pull away from the second virtual particle, to avoid catching both. And here's the key part: The energy-debt between the particle pair intensely pulls them towards each other. The trapped particle is being pulled upwards, effectively pulling upwards on the horizon that trapped it. This slows the fall of horizon-ripple, diminishing the energy that the falling ripple returns to the rest of the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy required to pull the two virtual particles apart equals the combined energy of the two non-virtual particles. So the falling ripple loses energy equal to two particles, and the hole eats one particle. Everything balances out with the one escaped particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe it works the same, regardless of whether the virtual particles are photons or a matter-antimatter pair.&lt;/p&gt;&#xA;" OwnerUserId="27832" LastEditorUserId="27832" LastEditDate="2019-04-27T17:15:56.997" LastActivityDate="2019-04-27T17:15:56.997" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31654" PostTypeId="1" AcceptedAnswerId="31668" CreationDate="2019-04-27T07:13:58.773" Score="0" ViewCount="72" Body="&lt;p&gt;These days with high speed internet, observational astronomy can be done from remote locations. Still I believe from time to time an experimenter might travel to work on-site for particularly challenging or unusual/non-standard observations, where working closely with the instrument support team is necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When astronomers travel to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Atacama_Large_Millimeter_Array&quot; rel=&quot;nofollow noreferrer&quot;&gt;Atacama Large Millimeter Array&lt;/a&gt; or ALMA:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Where do they live?&lt;/li&gt;&#xA;&lt;li&gt;Where do they work during observing sessions?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Representative photos would be &lt;em&gt;greatly appreciated!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing it looks a little different than this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ehMXJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ehMXJ.jpg&quot; alt=&quot;The WSRT telescope RT5 equipped with the APERTIF prototype - cropped&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: Cropped from &lt;a href=&quot;https://arxiv.org/abs/0912.0093&quot; rel=&quot;nofollow noreferrer&quot;&gt;Apertif - the focal-plane array system for the WSRT&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Figure 2. The WSRT telescope RT5 equipped with the APERTIF prototype. Shown in the inserts is the Vivaldi array in the lab (top right) and in the focal plane of the telescope (bottom right).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-04-28T02:44:21.103" LastActivityDate="2019-04-28T02:44:21.103" Title="Where do local users of ALMA live and work?" Tags="&lt;observational-astronomy&gt;&lt;radio-astronomy&gt;&lt;observatory&gt;" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="31655" PostTypeId="1" AcceptedAnswerId="31656" CreationDate="2019-04-27T07:35:21.323" Score="3" ViewCount="525" Body="&lt;p&gt;I realized that after answering the question &lt;a href=&quot;https://astronomy.stackexchange.com/q/31651/7982&quot;&gt;How do astronomers convert radio signals received by their antenna to images?&lt;/a&gt; I had missed the point that the OP was only asking about software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've written this question about experimental techniques as a companion question in order to move my answer from there to here.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-04-27T07:48:36.073" Title="Which techniques are used to convert radio signals received by antenna to images?" Tags="&lt;observational-astronomy&gt;&lt;radio-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31656" PostTypeId="2" ParentId="31655" CreationDate="2019-04-27T07:35:52.003" Score="3" Body="&lt;p&gt;I can think of at least four very different ways that &quot;astronomers convert radio signals received by their antenna to images.&quot;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;1. Scanning&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This is the &lt;em&gt;oldest&lt;/em&gt; and easiest to understand way to image. &lt;a href=&quot;https://astronomy.stackexchange.com/a/16242/7982&quot;&gt;The answer&lt;/a&gt; to the question &lt;a href=&quot;https://astronomy.stackexchange.com/q/16238/7982&quot;&gt;How did single dish (or single receiver) radio telescopes originally generate images?&lt;/a&gt; sums it up concisely:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They scan the object, if you point the dish a a point in the sky as the Earth rotates the dish scans across astronomical objects, then move the dish to point at a slightly different position and let it scan across the object again, and again. After a while you can re-construct an image from the scan lines in a similar way to analogue TV.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is an early 1D projection of scanning the plane of our Milky Way galaxy from &lt;a href=&quot;http://adsabs.harvard.edu/abs/1958MNRAS.118..379O&quot; rel=&quot;nofollow noreferrer&quot;&gt;The galactic system as a spiral nebula&lt;/a&gt; Oort, J. H.; Kerr, F. J.; Westerhout, G.  MNRAS 118, (1958) p. 379. This image is reversed tone for better clarity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information see answer(s) to &lt;a href=&quot;https://astronomy.stackexchange.com/q/28083/7982&quot;&gt;Why the blank wedges in this very early 21 cm map of the Milky Way? (Oort et al. 1958)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bjDoE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bjDoE.jpg&quot; alt=&quot;Oort 1958 Milky Way radio image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;2. Interferometry (most common technique today)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If you receive an incoming wavefront from two or more different locations, which in this case means two or more antennas, you can &lt;em&gt;use math&lt;/em&gt; to reconstruct an image. This is called interferometric imaging and these days it's done by feeding the signals from each antenna to a large, fast computer known as a correlator. However in the early days of interferometric imaging the interference was actually done electronically, with amplified signals from each antenna combined in a mixer and other electronics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, one of the earliest techniques of radio interferometry used a single antenna which collected both a direct signal from the source and a signal taking a longer path by reflecting off of water! This is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Sea_interferometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sea interferometry&lt;/a&gt; and in this case the water acts as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Lloyd%27s_mirror&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lloyd's Mirror&lt;/a&gt;. This is one dimensional imaging only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read more about this in &lt;a href=&quot;http://www.atnf.csiro.au/outreach/about/history/doverheights/index.html#DoverInterferometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Radio Astronomy at Dover Heights&lt;/a&gt; as well as in the paper &lt;a href=&quot;http://adsabs.harvard.edu/abs/1953AuJPh...6..420B&quot; rel=&quot;nofollow noreferrer&quot;&gt;Galactic Radiation at Radio Requencies V. The Sea Interferometer&lt;/a&gt; J. G. Bolton and O. B. Slee, Australian Journal of Physics, vol. 6, p.420, December 1953.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zYzxy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zYzxy.jpg&quot; alt=&quot;Radio Astronomy at Dover Heights&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DM7V1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DM7V1.jpg&quot; alt=&quot;Radio Astronomy at Dover Heights&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qkO5q.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qkO5q.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Very_Large_Array&quot; rel=&quot;nofollow noreferrer&quot;&gt;Very Large Array&lt;/a&gt; (&lt;a href=&quot;https://en.wikipedia.org/wiki/File:USA.NM.VeryLargeArray.02.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may have seen images of (material around) a black hole recently. In this case the signals from each antenna were recorded on to hard drives, which were then shipped to a central location before being put into a computer correlator, but mathematically it is the same process. See answer(s) to &lt;a href=&quot;https://astronomy.stackexchange.com/q/20082/7982&quot;&gt;How does the Event Horizon Telescope implement the interferometry?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;3. Direct imaging using focal plane arrays&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The question &lt;a href=&quot;https://astronomy.stackexchange.com/q/16237/7982&quot;&gt;What is the highest granularity focal-plane array on a dish radio telescope? Or is this the ONLY ONE?&lt;/a&gt; shows the first image below, of the &lt;a href=&quot;http://www.atnf.csiro.au/research/multibeam/.overview.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Parkes 21cm Multibeam Receiver&lt;/a&gt; and seems to be dated 1997. In &lt;a href=&quot;https://astronomy.stackexchange.com/a/24045/7982&quot;&gt;this answer&lt;/a&gt; links to &lt;a href=&quot;https://www.rug.nl/fse/news/first-image-with-apertif_-a-new-life-for-the-westerbork-radio-telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;First image with Apertif: a new life for the Westerbork radio telescope&lt;/a&gt; which is shown in the second image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lE5ke.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lE5ke.png&quot; alt=&quot;Parkes 21cm Multibeam Receiver&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YIp9W.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YIp9W.jpg&quot; alt=&quot;Westerbork radio telescope&amp;#39;s Apertif&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;4. Delay-Doppler radar&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This is the coolest method! It is based on radar, which means that from one dish astronomers beam a carefully modulated radio signal towards a planet or asteroid, and receive the reflected beam in a separate dish. This kind of radar is called delay-doppler because it extracts two kinds of information from the returned signal, the total delay time giving distances, and the doppler shift caused by the velocity of the surface doing the reflecting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the body is rotating, different areas will be moving towards or away from us depending on their location, and different parts will be closer or farther away from us, also depending on location. If they receive enough signals over a long enough period of time, they can reconstruct a hypothetical image of the shape of the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See The Planetary Society's Senior Editor and Planetary Evangelist Emily Lakdawalla's &lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/2010/2462.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;How radio telescopes get &quot;images&quot; of asteroids&lt;/a&gt; and &lt;a href=&quot;https://astronomy.stackexchange.com/a/30267/7982&quot;&gt;this answer&lt;/a&gt; for more on this technique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Left:&lt;/strong&gt; A very slow GIF &lt;a href=&quot;http://www.planetary.org/blogs/emily-lakdawalla/2010/2462.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Right:&lt;/strong&gt; A radar image shows asteroid 3122 Florence and tiny echoes from its two moons. &lt;a href=&quot;https://i.stack.imgur.com/YIp9W.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is an animation&lt;/a&gt; that shows them more clearly. The direction of the radar illumination (and thus the direction toward Earth) is at the top.&quot; From &lt;a href=&quot;https://i.stack.imgur.com/YIp9W.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. NASA / Jet Propulsion Laboratory. This is a small subset of the frames contained in the original 36 MB GIF, and the size has been decreased by a factor of 2 in order to fit in SE's 2 MB limit. From the question &lt;a href=&quot;https://astronomy.stackexchange.com/q/22480/7982&quot;&gt;What is the physical geometry of this apparent &quot;eclipse&quot; of a tiny moon of Asteroid Florence?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; These two images do not necessarily have the same orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sMeV5.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sMeV5.gif&quot; alt=&quot;Delay-Doppler Imaging Emily Lakdawalla&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/Pn291.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pn291.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-04-27T07:48:36.073" LastActivityDate="2019-04-27T07:48:36.073" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31659" PostTypeId="1" AcceptedAnswerId="31660" CreationDate="2019-04-27T10:47:52.990" Score="2" ViewCount="159" Body="&lt;p&gt;I have heard the term &quot;plane of the spirit level&quot; e.g. the sun's shadows is X degrees to the plane of the spirit level to signify the surface of the earth and I see it is a common term used in surveying and as an instrument to show if there is a curvature.&lt;br&gt;&#xA;My question is: where is this term coming from?&lt;/p&gt;&#xA;" OwnerUserId="25200" LastEditorUserId="25200" LastEditDate="2019-04-27T13:38:55.650" LastActivityDate="2019-04-27T13:38:55.650" Title="What is the origin of the term &quot;plane of the spirit level&quot;?" Tags="&lt;earth&gt;&lt;history&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31660" PostTypeId="2" ParentId="31659" CreationDate="2019-04-27T11:15:26.263" Score="3" Body="&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Spirit_level&quot; rel=&quot;nofollow noreferrer&quot;&gt;spirit level&lt;/a&gt; is an instrument used to find a horizontal or vertical plane. In surveying, the spirit level is used as a reference. Surveying instruments like the &lt;a href=&quot;https://en.wikipedia.org/wiki/Theodolite&quot; rel=&quot;nofollow noreferrer&quot;&gt;theodolite&lt;/a&gt; have a spirit level built in. This is used to set up the instrument, after which you can measure angles relative to that reference plane.&lt;/p&gt;&#xA;&lt;p&gt;This makes sure you have a repeatable reference, and not one that depends on the incline of the local surface.&lt;/p&gt;&#xA;&lt;p&gt;As to the origin of the name:&lt;/p&gt;&#xA;&lt;p&gt;it's called a &lt;em&gt;level&lt;/em&gt; because it is used to find a level surface (i.e. one that's horizontal). It's called a &lt;em&gt;spirit&lt;/em&gt; level because the liquid inside is usually a spirit (i.e. a mix of alcohol and water).&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Early tubular spirit levels had very slightly curved glass vials with constant inner diameter at each viewing point. These vials are incompletely filled with a liquid, usually a colored spirit or alcohol, leaving a bubble in the tube. They have a slight upward curve, so that the bubble naturally rests in the center, the highest point. At slight inclinations the bubble travels away from the marked center position. Where a spirit level must also be usable upside-down or on its side, the curved constant-diameter tube is replaced by an uncurved barrel-shaped tube with a slightly larger diameter in its middle.&lt;/p&gt;&#xA;&lt;p&gt;Alcohols such as ethanol are often used rather than water. Alcohols have low viscosity and surface tension, which allows the bubble to travel the tube quickly and settle accurately with minimal interference from the glass surface. Alcohols also have a much wider liquid temperature range, and won't break the vial as water could due to ice expansion. A colorant such as fluorescein, typically yellow or green, may be added to increase the visibility of the bubble.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;why &amp;quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Liquor#Etymology&quot; rel=&quot;nofollow noreferrer&quot;&gt;spirit&lt;/a&gt;&amp;quot;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The term &amp;quot;spirit&amp;quot; in reference to alcohol stems from Middle Eastern alchemy. These alchemists were more concerned with medical elixirs than with transmuting lead into gold. The vapor given off and collected during an alchemical process (as with distillation of alcohol) was called a spirit of the original material&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="361" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-04-27T13:25:00.677" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="31662" PostTypeId="1" CreationDate="2019-04-27T16:03:44.827" Score="2" ViewCount="59" Body="&lt;p&gt;I have fit images of binary systems and want a program that will help determine which ones are highest quality, the seeing. The simplest way is apparently to equate it to the FWHM of the Gaussian of the point spread function of one of the stars. But I'm at a loss as to how to do this. I've searched for any helpful tutorials online but can't find an understandable method.&lt;/p&gt;&#xA;" OwnerUserId="27840" LastEditorUserId="7472" LastEditDate="2019-04-27T16:15:59.947" LastActivityDate="2019-04-27T16:15:59.947" Title="How would I code to find the seeing of a binary star system in python?" Tags="&lt;python&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31664" PostTypeId="1" AcceptedAnswerId="31665" CreationDate="2019-04-27T19:06:37.663" Score="1" ViewCount="130" Body="&lt;p&gt;If the sun never rises above the tropic of cancer and never goes bellow (southern) of the tropic of capricorn then it seems that it should appear to the observer lower and lower in the horizon for as we move north. But from what I have seen this is not what I have observed traveling in northern and southern countries.&lt;br&gt;&#xA;Am I not remembering right or is my expectation wrong?&lt;/p&gt;&#xA;" OwnerUserId="25200" LastActivityDate="2019-04-29T16:20:03.910" Title="Should the sun appear very low in the horizon if its path is always between the two tropics?" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;horizon&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31665" PostTypeId="2" ParentId="31664" CreationDate="2019-04-27T19:16:10.377" Score="3" Body="&lt;p&gt;Your explanation is correct, so you are not remembering correctly. The farther north you go, the lower the sun is to the horizon at noon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course it depends on how far north you go. If you only travel 5° farther north in latitude, then the sun is only 5° lower in the sky. 5° is a &quot;long&quot; ways to travel on the earth, but in the sky 5° lower is not that noticeable.&lt;/p&gt;&#xA;" OwnerUserId="17355" LastActivityDate="2019-04-27T19:16:10.377" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="31666" PostTypeId="1" AcceptedAnswerId="31671" CreationDate="2019-04-27T19:23:04.467" Score="1" ViewCount="400" Body="&lt;p&gt;Here is my problem statement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a temperature of &lt;span class=&quot;math-container&quot;&gt;$7.3 \times 10^5 \ \text{K}$&lt;/span&gt;, make a graph of the Plank function (Eq. 3.24), plotting &lt;span class=&quot;math-container&quot;&gt;$\log_{10}{\nu} B_\nu(T)$&lt;/span&gt; vs &lt;span class=&quot;math-container&quot;&gt;$\log_{10} \nu$&lt;/span&gt; for &lt;span class=&quot;math-container&quot;&gt;$\log_{10} \nu$&lt;/span&gt; between 15.5 and 17.5. How does the behaviour of your graph of a blackbody compare with that of Fig. 28.14 for the continuous spectrum of the quasar 3C 273?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the equation 3.24&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{equation}&#xA;    B_\nu (T) = \frac{2h \nu^3/c^2}{e^{h\nu/kT}-1}.&#xA;\end{equation}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have put together an excel spreadsheet and created a graph of &lt;span class=&quot;math-container&quot;&gt;$\log_{10}({\nu} B_\nu(T))$&lt;/span&gt; vs &lt;span class=&quot;math-container&quot;&gt;$\log_{10} (\nu)$&lt;/span&gt;, however, it is just a straight line (which is what I would have expected although the graph in the book is significantly different). I am wondering if I am misunderstanding the problem statement. What I did to get my data points for &lt;span class=&quot;math-container&quot;&gt;$\log_{10}{\nu} B_\nu(T)$&lt;/span&gt; is I first computed &lt;span class=&quot;math-container&quot;&gt;$\nu$&lt;/span&gt; values from the given range of &lt;span class=&quot;math-container&quot;&gt;$\log_{10} \nu$&lt;/span&gt; values, then I computed B(T) values with the above-given equation. Then I substituted the product into logarithm : &lt;span class=&quot;math-container&quot;&gt;$\log_{10}({\nu} B_\nu(T))$&lt;/span&gt;, thus obtaining my &lt;span class=&quot;math-container&quot;&gt;$y = f(x)$&lt;/span&gt; values. and then I plotted it against  &lt;span class=&quot;math-container&quot;&gt;$x = \log_{10} \nu$&lt;/span&gt;. Am I doing things wrong or is this fine?&lt;/p&gt;&#xA;" OwnerUserId="26479" LastActivityDate="2019-04-28T08:41:32.163" Title="Trouble with plotting a graph for the Planck function" Tags="&lt;black-body&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31667" PostTypeId="2" ParentId="31649" CreationDate="2019-04-27T20:44:04.983" Score="1" Body="&lt;p&gt;Fullerenes have been detected in space, including C&lt;sub&gt;60&lt;/sub&gt;, C&lt;sub&gt;60&lt;/sub&gt;&lt;sup&gt;+&lt;/sup&gt; and C&lt;sub&gt;70&lt;/sub&gt; and appear to be responsible for the &lt;a href=&quot;https://en.wikipedia.org/wiki/Diffuse_interstellar_bands&quot; rel=&quot;nofollow noreferrer&quot;&gt;diffuse interstellar bands&lt;/a&gt; (DIBs), which are absorption features caused by the interstellar medium. The question of endohedral fullerenes is an interesting one, it is discussed in the paper &quot;&lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2016A%26A...590A..52O/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Interstellar fullerene compounds and diffuse interstellar bands&lt;/a&gt;&quot; by Omont (2016).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In section 3.1.2, the situation for metallofullerenes M@C&lt;sub&gt;60&lt;/sub&gt; is discussed:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Analysis of the optical spectra of metallofullerenes thus&#xA;  presents the same difficulties as do C&lt;sub&gt;60&lt;/sub&gt; or C&lt;sub&gt;60&lt;/sub&gt;&lt;sup&gt;-&lt;/sup&gt;. It is therefore not surprising that even theoretical studies of their&#xA;  energy levels and excitation are lacking, incomplete or ambiguous, and we are still waiting for laboratory spectra.&#xA;  The accuracies of wavelength determinations are still far&#xA;  from being useful in constraining possible associations of&#xA;  visible DIB features with metallofullerenes. Estimates of&#xA;  line strengths are practically always lacking.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From this it would appear that more lab work is necessary before we are able to identify metallofullerenes from astronomical spectra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In section 7, Omont describes the situation for hydrogen molecules in fullerenes (H&lt;sub&gt;2&lt;/sub&gt;@C&lt;sub&gt;60&lt;/sub&gt;), with section 7.2 discussing the prospects for finding interstellar H&lt;sub&gt;2&lt;/sub&gt;@C&lt;sub&gt;60&lt;/sub&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It seems that the presence of endohedral H&lt;sub&gt;2&lt;/sub&gt; would not&#xA;  change the basic properties of interstellar fullerenes much,&#xA;  making it difficult to reveal its presence, except thanks to&#xA;  specific spectroscopic features. However, even if it were&#xA;  significant, the presence of endohedral H&lt;sub&gt;2&lt;/sub&gt; should be difficult&#xA;  to confirm because the abundance of H&lt;sub&gt;2&lt;/sub&gt;@C&lt;sub&gt;60&lt;/sub&gt;&lt;sup&gt;+&lt;/sup&gt;, for instance,&#xA;  could hardly be expected to exceed or even reach 10&lt;sup&gt;−9&lt;/sup&gt;.&#xA;  There is thus practically no chance to ever detect its near-IR H&lt;sub&gt;2&lt;/sub&gt; 2 µm absorption. UV spectroscopy is more sensitive&#xA;  by orders of magnitude than absorption of weak IR lines.&#xA;  It can detect H&lt;sub&gt;2&lt;/sub&gt; column densities lower than ∼10&lt;sup&gt;14&lt;/sup&gt; cm&lt;sup&gt;−2&lt;/sup&gt;&#xA;  (e.g. Gillmon et al. 2006). But even so, the UV detection&#xA;  of H&lt;sub&gt;2&lt;/sub&gt;@C&lt;sub&gt;60&lt;/sub&gt; seems out of reach.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A quick check of the papers citing Omont (2016) doesn't turn up any detections of endohedral fullerenes, so it looks like the situation remains as described.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastActivityDate="2019-04-27T20:44:04.983" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31668" PostTypeId="2" ParentId="31654" CreationDate="2019-04-27T20:59:09.503" Score="1" Body="&lt;p&gt;No one travels to ALMA to observe; the observations are all carried out by staff working at the Operations Support Facility. (This is at an altitude of 2900 meters; the actual telescopes are 28 km away at the Array Operations Site, which is at an altitude of 5000 meters and is not permanently staffed.) Any &quot;unusual/non-standard observations&quot; would have to be specified in advance and encoded into the Scheduling Blocks (prepared files specifying the observations you want).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://almascience.nrao.edu/about-alma/alma-site&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://almascience.nrao.edu/about-alma/alma-site&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the staff at the Operations Support Facility stays in the &lt;a href=&quot;https://www.eso.org/public/teles-instr/alma/residencia/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Residencia&lt;/a&gt;. (More pictures &lt;a href=&quot;https://www.eso.org/public/images/archive/search/?ranking=0&amp;amp;fov=0&amp;amp;release_id=&amp;amp;minimum_size=0&amp;amp;description=&amp;amp;published_until_year=0&amp;amp;published_until_month=0&amp;amp;title=&amp;amp;subject_name=ALMA+Residencia&amp;amp;credit=&amp;amp;published_until_day=0&amp;amp;published_since_day=0&amp;amp;published_since_month=0&amp;amp;id=&amp;amp;published_since_year=0&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.kouvo-partanen.fi/alma_e.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;; pictures of the OSF, including the telescope control room, can be found &lt;a href=&quot;https://www.almaobservatory.org/en/images-tags/osf-en/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I wonder if some future movie will film scenes there, as the James Bond movie &lt;em&gt;Quantum of Solace&lt;/em&gt; did at the Paranal Observatory's Residencia in 2008...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Visitors (who cannot stay overnight) are permitted on Saturdays and Sundays, though you have &lt;a href=&quot;https://www.almaobservatory.org/en/outreach/alma-observatory-public-visits/&quot; rel=&quot;nofollow noreferrer&quot;&gt;reserve your visit in advance&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2019-04-27T20:59:09.503" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31670" PostTypeId="1" AcceptedAnswerId="31673" CreationDate="2019-04-27T23:55:36.527" Score="2" ViewCount="113" Body="&lt;p&gt;I'm interested in finding the combined mass of the Centaurs and the scattered disk. I looked around but I wasn't able to find any estimates. This seems to suggest that either the combined masses of the Centaurs or the scattered disk are negligible in comparison to that of the asteroid belt, or nobody ever thought of making an estimate. All I could find was that the asteroid belt is approximately 0.04 moon masses and that the Jupiter Trojans are approximately 1/5 of the mass of the asteroid belt. My feeling is that these combined masses should be way more than that of the Jupiter Trojans.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance, and yes, I tried Wiki.&lt;/p&gt;&#xA;" OwnerUserId="27844" LastEditorDisplayName="user24157" LastEditDate="2019-04-28T11:54:16.080" LastActivityDate="2019-04-28T20:53:27.477" Title="How would the combined masses of the Centaurs or the scattered disk compare to the combined mass of the asteroid belt?" Tags="&lt;solar-system&gt;&lt;trans-neptunian-object&gt;&lt;centaur&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31671" PostTypeId="2" ParentId="31666" CreationDate="2019-04-28T03:16:05.243" Score="3" Body="&lt;p&gt;Here's what I got for &lt;span class=&quot;math-container&quot;&gt;$\nu B_{\nu}(T)$&lt;/span&gt; using Python, which is easy enough to read that maybe you can check against your Excel calculation. I also plotted just &lt;span class=&quot;math-container&quot;&gt;$B_{\nu}(T)$&lt;/span&gt; for comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither looks like a straight line and they shouldn't. If you print out &lt;span class=&quot;math-container&quot;&gt;$\log_{10}(k_B T / c)$&lt;/span&gt; you get about 16.2 which is right in the middle of your range. They've specified a range of frequencies centered roughly on the peak of the Planck distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U4PNJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U4PNJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def Bnu(nu, T):&#xA;    top = 2 * h * nu**3 / c**2&#xA;    bot = np.exp(h * nu / (kB * T)) - 1.&#xA;    return top/bot&#xA;&#xA;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;&#xA;h  = 6.6261E-34 # m^2 kg / s&#xA;c  = 2.9979E+08 # m / s&#xA;kB = 1.3806E-23 # m^2 kg / s^2 K&#xA;&#xA;T   = 7.3E+05  # K&#xA;&#xA;print &quot;log10(kB T / h): &quot;, np.log10(kB * T / h)&#xA;&#xA;nu  = np.logspace(15.5, 17.5, 201) # Hz&#xA;&#xA;lam = c / nu&#xA;&#xA;print nu.min(), nu.max(), 'Hz'&#xA;print 1E+09*lam.min(), 1E+09*lam.max(), 'nm'&#xA;&#xA;B = Bnu(nu, T)&#xA;&#xA;plt.figure()&#xA;&#xA;plt.subplot(1, 2, 1)&#xA;plt.title('B(nu, T)', fontsize=16)&#xA;plt.plot(nu, B)&#xA;plt.yscale('log')&#xA;plt.xscale('log')&#xA;&#xA;plt.subplot(1, 2, 2)&#xA;plt.title('nu x B(nu, T)', fontsize=16)&#xA;plt.plot(nu, nu * B)&#xA;plt.yscale('log')&#xA;plt.xscale('log')&#xA;&#xA;plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-04-28T08:41:32.163" LastActivityDate="2019-04-28T08:41:32.163" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31673" PostTypeId="2" ParentId="31670" CreationDate="2019-04-28T11:34:52.953" Score="2" Body="&lt;p&gt;According to Sheppard et al. (2000) &quot;&lt;a href=&quot;https://iopscience.iop.org/article/10.1086/316805&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Wide-Field CCD Survey for Centaurs and Kuiper Belt Objects&lt;/a&gt;&quot;, the total mass of the centaurs is likely around 10&lt;sup&gt;-4&lt;/sup&gt; Earth masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gomes et al. (2008) &quot;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2008ssbn.book..259G&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Scattered Disk: Origins, Dynamics, and End States&lt;/a&gt;&quot; estimate the mass of the scattered disc as between 0.01 and 0.1 Earth masses. (Going by the abstract, I don't have a copy of the book to check the details.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you noted in your question, the asteroid belt is around 4% of the mass of the Moon, or about 0.0005 Earth masses. So the centaurs have a smaller or similar mass to the asteroid belt (bearing in mind the uncertainties on the values), while the scattered disc is considerably more massive.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-04-28T20:53:27.477" LastActivityDate="2019-04-28T20:53:27.477" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31674" PostTypeId="2" ParentId="31643" CreationDate="2019-04-28T11:42:04.873" Score="2" Body="&lt;p&gt;The &quot;drift-and-shift&quot; (DASH) mode was described in a paper by &lt;a href=&quot;https://iopscience.iop.org/article/10.1088/1538-3873/129/971/015004/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;Momcheva et al. (2016)&lt;/a&gt;, and works the way you supposed: a sequence of nearby images is taken using blind offsets without guide-star recalibration: move to the first field and take an image (possibly first acquiring a guide star), move to a nearby target field and take another image (no guide star acquisition), repeat...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Guide-star &quot;calibration&quot; means correcting the internal model of &quot;where in the sky &lt;em&gt;HST&lt;/em&gt; is pointing&quot; by locating a bright star with well-known coordinates, tweaking the pointing of &lt;em&gt;HST&lt;/em&gt; to put the guide star in the right position within the telescope's focal plane (so that the center of the focal plane is now pointing where the observing proposal specifies), and setting up a monitoring process where the position of the guide star is continually tracked during an observation to make sure the telescope pointing is not drifting (and correcting it when it is). The whole acquisition-and-setup process introduces an overhead of about six minutes; since a typical &lt;em&gt;HST&lt;/em&gt; observing window is only about 50 minutes (before the spacecraft's orbit carries it around to the other side of the Earth and it can't see the target any more), trying to change target fields multiple times would rapidly eat up your actual observation time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In some [most?] cases, it might make sense to do guide-star acquisition for the very first field, to which &lt;em&gt;HST&lt;/em&gt; has likely just made a very large slew, to ensure the initial pointing is accurate and that the (smaller) offsets to new fields will be reasonably accurate.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not having a guide star &lt;em&gt;does&lt;/em&gt; impact the stability of the telescope pointing &lt;em&gt;during&lt;/em&gt; an observation: guiding using just the gyros means a drift of about 0.001 to 0.002 arc seconds per second of time. So long exposures are not a good idea, because they'll be smeared by the telescope drift: an exposure of 100 seconds means a typical drift of several pixels during that time. But since Riess et al. were taking very short exposures (just a few seconds), this wasn't a problem for them.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2019-04-28T12:11:37.280" LastActivityDate="2019-04-28T12:11:37.280" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31675" PostTypeId="2" ParentId="27673" CreationDate="2019-04-28T11:59:21.760" Score="1" Body="&lt;p&gt;&lt;strong&gt;Indeed&lt;/strong&gt;, the IAU's definition of a planet has been determined by vote (of actually a very small part of the IAU as the majority didn't come to the notorious meeting in 2006) and bares any scientific footing. If Pluto (and Eris) were no planets due to not having &quot;cleared their neighbourhood&quot; then many other planets including Earth and Jupiter wouldn't be planets either but so-called &quot;dwarf planets&quot;. So all in all the IAU decision has no logic and shouldn't be taken seriously by anybody. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A scientific definition of &quot;planet&quot; has been proposed by Kirby Runyon, PhD student from the Department of Earth and Planetary Sciences at Johns Hopkins University, which is known as the &lt;em&gt;geophysical definition of a planet&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A planet is a sub-stellar mass body that has never undergone nuclear fusion and that has sufficient self-gravitation to assume a spheroidal shape adequately described by a triaxial ellipsoid regardless of its orbital parameters.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Runyon's definition any spherical celestial body which isn't a star is a planet, including moons. One doesn't have to agree with it (I don't stand up for that definition) but it is a logical scientific and consequent possible definition of a planet and it is better to stand up for this one than to hold to a definition that was voted in by astronomers who obviously have not much clue of science. Sorry to say it so harshly but the IAU definition really lacks any science or logic, but it would be too much to describe it all here. But one more point: planets don't have to orbit stars so that part of their definition is wrong too. &lt;/p&gt;&#xA;" OwnerUserId="27845" LastEditorUserId="27845" LastEditDate="2019-04-28T12:21:58.507" LastActivityDate="2019-04-28T12:21:58.507" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31677" PostTypeId="2" ParentId="25937" CreationDate="2019-04-28T16:29:58.027" Score="0" Body="&lt;p&gt;Here is a comment on the physics behind planetary diameters which are derived from the nebular hypothesis proposed by the rshis of long ago (and reproposed in the 18th century). The planets condensed from a solar nebula!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The planets are too far away to estimate diameters without powerful telescopes.Diameters can be estimated only by scaling rules based on physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“The Surya Siddhanta also estimates the diameters of the planets. The estimate for the diameter of Mercury is 3,008 miles, an error of less than 1% from the currently accepted diameter of 3,032 miles. It also estimates the diameter of Saturn as 73,882 miles, which again has an error of less than 1% from the currently accepted diameter of 74,580. Its estimate for the diameter of Mars is 3,772 miles, which has an error within 11% of the currently accepted diameter of 4,218 miles. It also estimated the diameter of Venus as 4,011 miles and Jupiter as 41,624 miles, which are roughly half the currently accepted values, 7,523 miles and 88,748 miles, respectively.” from the wiki at&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Surya_Siddhanta&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Surya_Siddhanta&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[The Surya Siddhanta is the name of a Sanskrit treatise in Indian astronomy from late … It calculates the earth’s diameter to be 8,000 miles (modern: 7,928 miles), diameter of moon as 2,400 miles (actual ~2,160) and the distance between moon …&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How were the planetary sizes determined? &lt;strong&gt;What are the possible scaling rules?&lt;/strong&gt; Let us define&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D = Planetary diameter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R= Orbit radius&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M = planetary mass ~ D^3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T = Orbit time = 2 Pi/ Omega&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I = angular momentum = Integral of R^2 dm .Omega ~ R^2 D^3/ T&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Possible relations:&lt;/strong&gt; (This law of gravitation is most unlikely to be known so early in History)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If force is G M_sun M_planet /R^2 = Mplanet. Omega^2 R&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R^3. Omega^2 = constant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R^3 ~T^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Snow plow theory:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Larger radii sweep more particles leading to bigger planets&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D^3 ~ 2Pi R no pf particles ~ 2Pi R n. volume ~ 2 pi R R. thickness&lt;/p&gt;&#xA;&#xA;&lt;p&gt;~ R^2 if thickness is fixed, n being number density&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D^3 ~ R^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thin disk:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case the planet grows to bigger than the thickness and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D^3 ~ R^3 or D ~ R&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears that ancients assumed the last possibility. The planetary diameter scales with the orbital size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears that ancients assumed the last possibility. The planetary diameter scales with the orbital size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Body mile(D) “Relative Size `Relative size (modern)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury 3008. &lt;code&gt;0.38&lt;/code&gt;3.87E-01&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus 4011.00 &lt;code&gt;&lt;/code&gt;0.50 `7.20E-01&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth 8.00E+03 &lt;code&gt;1.00&lt;/code&gt;1.00E+00&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars “3.77E+03 “0.47 `1.52E+00&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter 4.16E+04 &lt;code&gt;5.20&lt;/code&gt;5.19E+00&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn 7.39E+04 “9.24E+00 9.24 9.53E+00&lt;/p&gt;&#xA;&#xA;&lt;p&gt;( Ignore the`. I had to use it to line numbers up.) Relative size is from Surya Siddhanta and is compared to modern measurements of orbits compared to Earth’s orbit. Venus and Mars have the most disagreement . They are the rocky planets. The agreement is very good in general. Especially good for gas giants!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;: &lt;em&gt;The vedas predicted planetary sizes using the acretion model from the initial solar nebula. So it automatically means they were the first to hit upon the idea of sun-centered&lt;/em&gt; planetary system!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also means Indian Astronomy was developed by Indians with no input from the Greeks. No others have planetary diameters! The kalachakra was a giant astronomical clock and was used to calculate orbits of planets visible to the naked eye from which planet sizes were determined, a far cry from the flat earth theory of Christianity! See&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.quora.com/How-was-kalachakra-used-in-Indian-Asronomy&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.quora.com/How-was-kalachakra-used-in-Indian-Asronomy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This uses four dials, one of which is the Zodiac, the same as the Greek one, most likely copied by the Greeks, an exact translation from Sanskrit! For fixed stars, the kalachakra uses bright stars with Sanskrit names. The Indians have been observing stars from long enough to know the period of nutation of Earth is 27,000 years. At least 5000 years of data is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rg Veda is variously dated from 1500 BC to 8000 BC (from internal evidence on the order of time when River Saraswati was still flowing). (The internal reference also mentions a big earthquake that made River Saraswati stop flowing because of tectonic shifts.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No luck was involved! Just thinking smart!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me add background information on astronomy of ancient India.&#xA;There were many before Aryabhata.&#xA;Box 3. Indian Astronomers before Aryabhata &lt;/p&gt;&#xA;&#xA;&lt;p&gt;“As mentioned in the Chandogya Upanishad (VII.1.2, 4), nakshatra-vidya (science of asterisms) was among the core disciplines of study in the Vedic era. Astronomers were called nakshtra-darsa (star-observers) or gal}aka. The sage Atri (who was among the originators of the oldest Vedic hymns) and his descendants were distinguished for expertise in accnrate edipse prediction and planetary astronomy. The Rg-Veda describes a solar eclipse observed by Atri (dated &lt;strong&gt;3928 BCE&lt;/strong&gt; in [5], p.1l6; [6], pp.173-l74). The Taittiriya Brahmana (Ill.lO.9) eulogises Ahina, Devabhaga and for attaining bliss due to their absorption in the science of the Sun, i.e., astronomy ([7], pp.20-2l); sage Matsya is also mentioned (1.5.2, 1) in the context of astronomy. &#xA;Garga is the most ancient astronomer referred to in post-Vedic treatises. The Mahahharata (XII.59.ll) refers to him as the court-astronomer of the great King Prthvi. TIte cpic (IX.37.14-17) mentions that a holy tirtha on the Sarasvati was named after the &#xA;as Garga-srota (&quot;stream of Garga&quot;). This was the sacred place where Vrddha Garga performed ascetic penance for self-purification and attained mastery over astonomy. Rishis of high merit and rigorous discipline l1sed to assemble here to acqllire thc profollnd knowledge of astronomy from the venerated Rshi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ancient Indian traditions mention a list of 18 astronomy texts called siddhiintm.; (&quot;established theories&quot;) named after SfIrya, Pitamaha, Vyasa, Vasistha, Atri, Parasara, Kasyapa, Narada, Garga, Marici, Manu, Ailgira, Lomasa, Plllisa, Cyavana, Yavana,Brughu and Saunaka. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Varahamihira (505-587 CE), himself a prominent astronomer, also mentions Asita-Devala, Maya, Badarayalla and Nagllajit. Many of the treatises by the above astronomers got lost even by the time of Varahamihira; None are available in their original forms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While considerable astronomical knowledge is embedded in early Vedic literatures, the oldest available treatise devoted exclusively to astronomy is the Vedanta Jyothisha (G. &lt;strong&gt;1300 BCE&lt;/strong&gt; vide [6]) composed by sage Lagadha. The Vedanga era represents a transitional pcriod in Indian civilisation when the Vedic culture was on the wane and there was a consequent attempt to organise and formulate the extant knowledge and systematise them into various branches called shastras .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aryahhatlya (499 CE) is the earliest extant astronomy treatise after the Vedanga Jyotisha. It was composed during the &quot;Classical Age&quot; of post-Vedic India. Towards the beginning (Ganita 1) and the end (Gola 48-50) of his text, Aryabhata had made a general acknowledgement of his predecessors.&quot; from&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.ias.ac.in/describe/article/reso/011/05/0058-0072&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.ias.ac.in/describe/article/reso/011/05/0058-0072&lt;/a&gt;&#xA;by Amar Kumar Dutta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;More info on why I am doing this&lt;/strong&gt;:&#xA;Surya Siddhanta is dated by wiki which says : Surya Siddhanta is a Hindu text on astronomy from late 4th-century or early 5th-century CE[1] Above is verse 1.1, which pays homage to Brahma.[2 Some scholars refer to Panca siddhantika as the old Surya Siddhanta and date it to 505 CE.[17] &lt;a href=&quot;https://en.wikipedia.org/wiki/Surya_Siddhanta&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Surya_Siddhanta&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is pretty old and gives a list of planetary diameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have to explain how the data was known for orbit radii and how planetary diameters were calculated. They didn’t have telescopes and such in those days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Data must have been available for whatever method was used before Surya Siddhanta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I show the data was available .  ( Kak says “ My discovery that the Rgveda is an altar of mantras came rather suddenly. Although I had been studying the altars described in the Shatapatha Brahmana for several months, I had never thought of any connection of these with the Rgveda. It was while reading some unrelated matter the idea flashed that the Rgveda itself is a symbolic altar. “)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worth reading Subash Kak’s  “The Astronomical Code of the Ṛgveda (Third Edition) “ for a fascinating insight into Hindu Civilization and Astronomical knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.695.536&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.695.536&amp;amp;rep=rep1&amp;amp;type=pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My contribution is to show that only scaling offers a way to calculate planetary sizes and the simplest is the snow plow model of accretion where larger orbits gather more particles.&lt;/p&gt;&#xA;" OwnerUserId="27847" LastEditorUserId="27847" LastEditDate="2019-05-16T15:55:36.733" LastActivityDate="2019-05-16T15:55:36.733" CommentCount="17" ContentLicense="CC BY-SA 4.0" />
  <row Id="31678" PostTypeId="1" AcceptedAnswerId="31680" CreationDate="2019-04-28T16:46:19.977" Score="46" ViewCount="11711" Body="&lt;p&gt;Why is Polaris, the North Star, always above (or near) the North Pole? If Earth is tilted, Polaris' path should be in winter 23 degrees away from its path in summer, or not?&lt;/p&gt;&#xA;" OwnerUserId="27848" LastEditorUserId="616" LastEditDate="2021-05-29T22:42:46.647" LastActivityDate="2021-05-29T22:42:46.647" Title="If Earth is tilted, why is Polaris always above the same spot?" Tags="&lt;observational-astronomy&gt;&lt;earth&gt;&lt;amateur-observing&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31680" PostTypeId="2" ParentId="31678" CreationDate="2019-04-28T17:46:14.227" Score="106" Body="&lt;p&gt;You are correct that the axis of the Earth's rotation is tilted with respect to the plane of its orbit by 23 degrees. But it is incorrect that the direction that the axis points changes by a large amount (it should be 2*23 degrees) over a 6 month time span.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uvB4q.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uvB4q.png&quot; alt=&quot;Incorrect orientation of axis&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Your assumption&lt;/em&gt;: If axis it pointed at Polaris at (1), then it should be pointed at a different star at (2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The axis remains pointed in the same direction throughout the entire year because the laws of physics are that the axis of a spinning object remains pointed in the same direction unless a torque acts on the body to change its orientation. There are torques acting on the Earth (namely the Sun and Moon), but it takes 13000 years to change the direction from &quot;one way to the opposite way&quot;, not 6 months.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/prbAG.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/prbAG.png&quot; alt=&quot;Correct orientation of axis&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Correct&lt;/em&gt;: The orientation of the Earth's spin axis remains pointed at the same star throughout the year (ignoring the slow, 26000 year precession cycle).&lt;/p&gt;&#xA;" OwnerUserId="17355" LastEditorUserId="17355" LastEditDate="2019-05-01T00:51:36.337" LastActivityDate="2019-05-01T00:51:36.337" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31681" PostTypeId="1" CreationDate="2019-04-28T18:19:31.517" Score="3" ViewCount="357" Body="&lt;p&gt;During the year the Sun moves along the ecliptic passing all the signs of the zodiac belt, in my understanding. This means that it takes ~1 month to go from one zodiac sign to the next.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equinoxes are approximately on 20th of March and 23rd of September. This means the vernal equinox Sun is in Pisces (or Aries) and the autumnal equinox Sun is in Libra (or Virgo).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But also I found that there is the observation of &lt;a href=&quot;https://www.hq.nasa.gov/office/hqlibrary/aerospacedictionary/aerodictall/p.html#precession%20of%20the%20equinoxes&quot; rel=&quot;nofollow noreferrer&quot;&gt;precession of the equinoxes&lt;/a&gt;, where the equinox Sun passes through all zodiac signs in a calendar round that lasts ~25,000 years.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this possible?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my understanding about the Sun's yearly round across the zodiac is correct, the equinox Sun should always be at the same zodiac signs. How can this precession happen?&lt;/p&gt;&#xA;" OwnerUserId="25200" LastEditorUserId="14094" LastEditDate="2019-04-30T08:47:39.483" LastActivityDate="2019-05-02T22:56:52.513" Title="Is the equinox Sun always at the same zodiac sign in the ecliptic belt?" Tags="&lt;the-sun&gt;&lt;ecliptic&gt;&lt;precession&gt;&lt;equinox&gt;&lt;zodiac&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31682" PostTypeId="2" ParentId="27673" CreationDate="2019-04-28T18:47:59.477" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;We suggest attempts to build consensus around planetary taxonomy not rely on the non-scientific process of voting, but rather through precedent set in scientific literature and discourse ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Good luck with that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are (at least) two distinct camps in the planet versus not a planet debate. There is a third camp as well: What should those things that orbit stars other than the Sun be called? Per the current definition, there are exactly eight planets in the entire universe. I'll ignore that issue for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One side of the planet versus not-a-planet debate looks at things from the perspective of roundness. That Titan and Pluto are not called planets, despite having active geologies, apparently much more active geologies than Mercury or Mars, is a bit befuddling to this group. The other side of the debate looks at things from the perspective of dynamics. That Titan and Pluto are not called planets while Mercury and Mars are is crystal clear. There's a three to six order of magnitude difference (depending on the metric) between Mars and Pluto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, while the planetologist viewpoint does have some credence, it has a boundary problem. There is no clearcut boundary between planet and not-a-planet from this perspective. Any definition would be highly arbitrary, and there will be boundary cases on both sides that go against the grain of the definition. On the other hand, the three to six order of magnitude difference between the dynamicists' distinction of planet versus not-a-planet means that boundary is crystal clear, at least in our solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is not a problem given that there are only eight planets in the entire universe. The thousands of objects discovered in the last few decades to be orbiting other stars are not planets. They're exoplanets. Eventually the exoplanet problem will have to be solved. (Given the amount of rancor that occurred, and continues to occur, over the demotion of Pluto to not-a-planet, that eventuality is probably well into the future.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someday in the not too distant future, a group of astronomers will have discovered a 20 Jupiter mass object that they will have deduced to have formed from planetary rather than stellar formation processes, making it an exoplanet, with a Neptune sized object orbiting this exoplanet. At the same time, another group of astronomers will have discovered a 10 Jupiter mass object that they will have deduced to have formed from stellar rather than planetary formation processes, making it a brown dwarf, with a Mars-sized object orbiting this brown dwarf. The Neptune-sized object will not be deemed to be a planet (better: an exoplanet) while the Mars-sized object will.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2019-04-28T18:47:59.477" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31683" PostTypeId="2" ParentId="31681" CreationDate="2019-04-28T19:19:09.740" Score="7" Body="&lt;p&gt;As you said, the Sun takes a year to cycle the Zodiac, and it takes a year between two spring or autumn equinoxes. Both statements are true, but &quot;year&quot; in them means &lt;a href=&quot;https://en.wikipedia.org/wiki/Year#Sidereal,_tropical,_and_anomalistic_years&quot; rel=&quot;noreferrer&quot;&gt;two different things&lt;/a&gt;. It takes a sidereal year to cycle the Zodiac, but it takes a tropic year to cycle between equinoxes. Since the tropic year is about 20 minutes shorter, the equinoctial points move across the Zodiac cycling it in about 25,000 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calendar year we use in the Gregorian calendar is designed to approximate the tropic year to have it synchronised with seasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for all it is that the Earth's axis is not fixed but its orientation &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_precession&quot; rel=&quot;noreferrer&quot;&gt;slowly rotates&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="14097" LastEditorUserId="14097" LastEditDate="2019-04-29T09:41:54.803" LastActivityDate="2019-04-29T09:41:54.803" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31684" PostTypeId="2" ParentId="10213" CreationDate="2019-04-29T00:54:42.070" Score="-2" Body="&lt;p&gt;The names are in Sanskrit and mean something in Sanskrit.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Chandra is the Moon&lt;/li&gt;&#xA;&lt;li&gt;Surya is the Sun&lt;/li&gt;&#xA;&lt;li&gt;Mangala is Mars&lt;/li&gt;&#xA;&lt;li&gt;Bhuda is Mercury&lt;/li&gt;&#xA;&lt;li&gt;Jupiter is Guru (weighty in Sanskrit as its size was known from Surya Siddhanta)&lt;/li&gt;&#xA;&lt;li&gt;Shukra is Venus&lt;/li&gt;&#xA;&lt;li&gt;Saturn is Shanischara (slow-moving in Sanskrit)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The week days are ordered in that order of increasing distances except for Venus (I do not know why.) All planets are considered gods!&lt;/p&gt;&#xA;" OwnerUserId="27847" LastEditorUserId="27850" LastEditDate="2019-04-29T04:07:09.923" LastActivityDate="2019-04-29T04:07:09.923" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31685" PostTypeId="1" CreationDate="2019-04-29T10:28:34.937" Score="2" ViewCount="57" Body="&lt;p&gt;Is there any benefit to view the 3D (real) universe as 3D projective space?&lt;/p&gt;&#xA;" OwnerUserId="27855" LastActivityDate="2019-04-29T10:28:34.937" Title="Universe as 3D projective space?" Tags="&lt;galaxy&gt;&lt;universe&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2019-04-29T22:12:58.050" ContentLicense="CC BY-SA 4.0" />
  <row Id="31687" PostTypeId="1" AcceptedAnswerId="31691" CreationDate="2019-04-29T12:36:17.583" Score="5" ViewCount="198" Body="&lt;p&gt;I'm currently reading &lt;a href=&quot;https://en.wikipedia.org/wiki/Atom_(Krauss_book)&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Atom&lt;/em&gt; (Krauss book)&lt;/a&gt;, and in chapter 11 there is this text:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This collision can explain many things, including the high rate of spin of the Earth (the grazing collision would have twisted it around like two football players who collide while running in opposite directions).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is in reference to a proto-planet hitting the early Earth, which led to the formation of the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is Earth's spin considered high rate, given that Mars' spin is pretty similar without having a big moon? Or have I misunderstood what is being referred to?&lt;/p&gt;&#xA;" OwnerUserId="26579" LastActivityDate="2019-04-30T06:49:23.407" Title="Why is Earth's rotation considered fast?" Tags="&lt;earth&gt;&lt;rotation&gt;&lt;proto-planetary-disk&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31688" PostTypeId="2" ParentId="31664" CreationDate="2019-04-29T16:20:03.910" Score="1" Body="&lt;p&gt;I am not sure how your observations are so mistaken, but this is exactly what happens. In fact at high latitudes, there will be parts of the year when the sun never rises above the horizon. Try going north of the Arctic Circle in winter and you'll see what I mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I can think is that either you have not travelled very far north or south, or you have travelled to northern or southern countries only in their summer and not paid attention to the sun's apparent movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I grew up at 59 degrees North, and one of the very obvious signs of the seasons is how low the sun is to the horizon for the winter months. This makes for long sunsets while the sun moves along the horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, at the equator, the sun goes overhead, and sunsets are rapid, as it comes down near perpendicularly to the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at &lt;a href=&quot;http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit2/motions.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this ohio-state.edu&lt;/a&gt; lecture page for some useful info on apparent motion of celestial bodies, like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/V0UNT.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V0UNT.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2019-04-29T16:20:03.910" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31690" PostTypeId="2" ParentId="31678" CreationDate="2019-04-29T21:00:03.740" Score="6" Body="&lt;p&gt;You may hear about something called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Celestial_sphere&quot; rel=&quot;noreferrer&quot;&gt;celestial sphere&lt;/a&gt;. While we now know the Earth is rotating on its axis, to ancient people it appeared as though the stars were rotating around us (and they are according to our own frame of reference). This rotating globe of stars can itself be mapped on a globe in the same fashion that we can map the Earth's land and water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/98BHs.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/98BHs.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few reasons why Polaris is always seen in the same position on this celestial sphere.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;As the globe of the Earth and the celestial sphere rotate relative to each other, they share an axis. This center of rotation does not move.&lt;/li&gt;&#xA;&lt;li&gt;While some planets (like Mars) rotate on an axis that is itself wobbling relative to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ecliptic&quot; rel=&quot;noreferrer&quot;&gt;ecliptic&lt;/a&gt;, Earth has a stable axis &lt;a href=&quot;https://astronomy.stackexchange.com/questions/476/why-does-the-moon-stablize-earths-axis&quot;&gt;thanks to our large moon&lt;/a&gt;. This axis always points in the same direction and so it's always aligned with the same stars in our galaxy.&lt;/li&gt;&#xA;&lt;li&gt;The stars in our galaxy are orbiting the galactic center very slowly from our perspective, so human civilization would not have observed any significant change in the stars' positions across its lifetime.&lt;/li&gt;&#xA;&lt;li&gt;During the Earth's orbit around the Sun, its own change in position is so slight relative to the galactic scale that no significant &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_parallax&quot; rel=&quot;noreferrer&quot;&gt;stellar parallax&lt;/a&gt; can be expected.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="18119" LastEditorUserId="18119" LastEditDate="2019-04-30T16:25:25.390" LastActivityDate="2019-04-30T16:25:25.390" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31691" PostTypeId="2" ParentId="31687" CreationDate="2019-04-29T21:36:56.697" Score="4" Body="&lt;p&gt;The interactions of the Earth and its moon have slowed the Earth's rotation significantly over the billions of years since their formation.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth%27s_rotation#Origin&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; cites a paper that gives a figure of approximately five hours for rotational period in the past.  We don't know of a similar method to slow down Mars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that the Earth must have had a much faster rotation than Mars in the past.  &lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="14094" LastEditDate="2019-04-30T06:49:23.407" LastActivityDate="2019-04-30T06:49:23.407" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31692" PostTypeId="2" ParentId="13509" CreationDate="2019-04-29T23:20:37.497" Score="1" Body="&lt;p&gt;Conservation of angular momentum largely preserves angular momentum when gaseous planetary nebulas condense to form planets in spite of friction and collisions.This is illustrated below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angular momentum of bodies in our solar system are given in &lt;a href=&quot;http://www.zipcon.net/~swhite/docs/astronomy/Angular_Momentum.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.zipcon.net/~swhite/docs/astronomy/Angular_Momentum.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are not constant but the gaseous planets are of the same order of magnitude. Orbital Angular Momentum Body orbital radius(km) orbital period(days) mass(kg) L&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury 58.e6 87.97 3.30e23 9.1e38&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Venus 108.e6 224.70 4.87e24 1.8e40&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth 150.e6 365.26 5.97e24 2.7e40&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mars 228.e6 686.98 6.42e23 3.5e39&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter 778.e6 4332.71 1.90e27 1.9e43&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn 1429.e6 10759.50 5.68e26 7.8e42&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uranus 2871.e6 30685.00 8.68e25 1.7e42&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neptune 4504.e6 60190.00 1.02e26 2.5e42&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are of order e^43. (Mars has less angular momentum. Some may have been distributed to the asteroid belt.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Each outer planet seems to carry the same angular momentum!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I originally thought that Surya Siddhanta used constancy of angular momentum but it is even simpler. It is simply a snow plough theory which makes larger orbits collect more particles. See &quot;How did the authors of Surya Siddhanta find the diameters of other planets in the solar system?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am giving this table to illustrate the constancy of angular momentum even in our solar system presumed to have condensed out of the primordial solar nebula, a fact ancients could have used to determine planetary diameters. Constancy of angular momentum requires planets to spin and orbit around the Sun (or the center of mass).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there was an angular momentum to begin with is understood. Any large mass of gas or nebula will form eddies eventually by turbulence with rotations in opposite directions as rotations arise naturally (by fluid instability). If each part condenses to a star (and solar system) planetary systems will occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our solar system may have been formed with another mechanism which is a passing star that imparted angular momentum to the original solar nebula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very large scale bodies also condense to galaxies (say) and must have black holes at their centers to trap angular momentum. Angular momentum can't be destroyed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish to add this, the rotational angular momentum of all bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rotational Angular Momentum, L&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Body/ mass kg/  radius(km)  rotational period (days)/ L &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sun  /695000    /24.6   /1.99e30    /1.1e42&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth/6378  /0.99   /5.97e24    /7.1e33&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter /71492  /0.41   /1.90e27    /6.9e38&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the Sun's rotational angular momentum is is also e^42. The spin angular momenta of all planets are small compared to orbital angular momenta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The outer planets and the Sun have the same angular momenta!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some kind of equipartition of angular momenta at work?&lt;/p&gt;&#xA;" OwnerUserId="27847" LastEditorUserId="27847" LastEditDate="2019-05-01T01:04:18.793" LastActivityDate="2019-05-01T01:04:18.793" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31693" PostTypeId="2" ParentId="31678" CreationDate="2019-04-30T01:09:10.833" Score="4" Body="&lt;p&gt;Kick (or toss) an AFL/rugby football (a prolate spheroid) so that it spins rapidly (around its long axis) at the same as it traverses an arc (falling under gravity). You should be able to observe that the spin axis is &lt;em&gt;parallel transported&lt;/em&gt; (at least if the perturbation by air resistance is weak enough to neglect).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The earth is spinning. To change the direction it spins in (with respect to the inertial frame of distant stars) it would be necessary to apply some torque to the earth. A sun's gravity does not apply any torque to a spherical planet (due to the symmetries of the Newtonian forces applied).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the earth is not quite a perfect sphere (e.g. is slightly oblate) there is a small torque applied by the sun's gravitational field gradient, so over millennia the planet's north axis gradually precesses to point toward different constellations.&lt;/p&gt;&#xA;" OwnerUserId="27869" LastActivityDate="2019-04-30T01:09:10.833" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31694" PostTypeId="1" AcceptedAnswerId="31696" CreationDate="2019-04-30T01:37:22.263" Score="2" ViewCount="515" Body="&lt;p&gt;I am using a code on EUCLID future mission.&#xA;The original author of this code has set a value for the density of galaxy equal to : &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ng = 354543085.80106884&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I think this is expressed in inverse steradian. I think that EUCLID mission has a &lt;span class=&quot;math-container&quot;&gt;$30\text{ arcmin}^{-2}$&lt;/span&gt; value for density of galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the conversion correct betweeen &lt;span class=&quot;math-container&quot;&gt;$30\text{ arcmin}^{-2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$354543085.8010688 \text{ sr}^{-1}$&lt;/span&gt; ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and how to do this conversion ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, I would like to calculate ng with a density of &lt;span class=&quot;math-container&quot;&gt;$48 \text{ arcmin}^{-2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards&lt;/p&gt;&#xA;" OwnerUserId="16492" LastEditorUserId="18643" LastEditDate="2019-04-30T07:50:18.743" LastActivityDate="2019-04-30T08:27:09.677" Title="How to find inverse steradian from $\text{arcmin}^{-2}$ for density of galaxy" Tags="&lt;galaxy&gt;&lt;angular-resolution&gt;&lt;density&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31696" PostTypeId="2" ParentId="31694" CreationDate="2019-04-30T08:14:14.427" Score="2" Body="&lt;p&gt;There are 60 arc minutes in a degree and &lt;span class=&quot;math-container&quot;&gt;$180/ \pi$&lt;/span&gt; degrees in a radian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 1 radian is 57.2957795 degrees or 3437.746771 arc minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as you are talking about units, you can square any of these angular units to obtain solid angle units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 1 steradian is 3282.80635 square degrees, or 11818102.860 square arc minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's try your theory:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/354543085.801 steradians is 9.259259E-06 square degrees, or 0.0333333 square arc minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that it is 1/30th of a square arc minute, so &lt;strong&gt;YES it is 30 inverse square arc minutes!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then 48 per square arcminute is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$48 \times 60^2 \times \frac{180^2}{\pi^2} \approx 567268937.282 \ \ \text{sr}^{-1}.$$&lt;/span&gt; &lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-04-30T08:27:09.677" LastActivityDate="2019-04-30T08:27:09.677" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31699" PostTypeId="2" ParentId="31681" CreationDate="2019-04-30T16:24:45.397" Score="2" Body="&lt;p&gt;The equinoxes are where the ecliptic crosses the equator.&#xA;As the Earth's axis and equatorial plane pivot under &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;precession&lt;/a&gt;,&#xA;the equinox points migrate westward along the ecliptic, 360&amp;deg; in 25800 years or 1.4&amp;deg; per century.&#xA;This animation shows the vernal equinox drifting 28&amp;deg; in 20 centuries of precession:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/baDyp.gif&quot; alt=&quot;Vernal equinox movement 100-2100 CE&quot;&gt;&#xA;&lt;em&gt;Images generated by &lt;a href=&quot;https://stellarium.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellarium&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun moves eastward (right to left) 360&amp;deg; per &lt;a href=&quot;https://en.wikipedia.org/wiki/Sidereal_year&quot; rel=&quot;nofollow noreferrer&quot;&gt;sidereal year&lt;/a&gt; relative to the stars along the ecliptic.&#xA;Meanwhile the vernal point moves 0.014&amp;deg; per year the other way, so the Sun moves that much less between vernal equinoxes - a &lt;a href=&quot;https://en.wikipedia.org/wiki/Tropical_year&quot; rel=&quot;nofollow noreferrer&quot;&gt;tropical year&lt;/a&gt;.&#xA;This makes a tropical year 20.4 minutes (1/25800 year) shorter than a sidereal year.&#xA;The modern calendar approximates the tropical year to keep the average equinox and solstice dates from drifting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While some traditions define a sidereal zodiac anchored to the stars, the Western zodiac is tropical, linked to the equinoxes and solstices and migrating along with them.&#xA;The sign of Aries is defined as the first 30&amp;deg; along the ecliptic east of the vernal equinox, even as that point moves through the constellations Pisces and Aquarius.&#xA;Similarly, the sign of Libra remains the first 30&amp;deg; east of the autumnal equinox despite its present alignment with the constellation Virgo.&#xA;We can blame the resulting confusion on Ptolemy, who understood precession yet named the 12 equal signs after the nearest constellations in his own lifetime anyway.&#xA;For more about the distinction between constellations and signs, see &lt;a href=&quot;https://astronomy.stackexchange.com/a/26369&quot;&gt;this answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a tropical year and a tropical zodiac, the Sun is always in the same &lt;em&gt;sign&lt;/em&gt; at the same time of year.&#xA;However, as the equinoxes and solstices drift westward over the centuries, so does the Sun's position among the &lt;em&gt;constellations&lt;/em&gt; at a given time of year.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-05-02T22:56:52.513" LastActivityDate="2019-05-02T22:56:52.513" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31700" PostTypeId="1" CreationDate="2019-04-30T22:01:36.887" Score="0" ViewCount="69" Body="&lt;p&gt;I know that Earth rotates around the Sun in an oval route:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/liDSg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/liDSg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I know the seasons happen, because the axis of the Earth is not vertical. I think it is 23 degree slope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How come the distance from the sun when the earth at the closest point to the Sun has no effect and if it has effect what is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see that when it is summer on the northern hemisphere it is winter in the southern and vise versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shouldn't be it is hotter time for both sides when earth is closest to sun and colder when it is furthest?&lt;/p&gt;&#xA;" OwnerUserId="17046" LastEditorUserId="1891" LastEditDate="2019-04-30T22:32:40.700" LastActivityDate="2019-05-01T01:46:44.310" Title="Seasonal effect based on Earth position" Tags="&lt;the-sun&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2019-05-01T00:37:21.487" ContentLicense="CC BY-SA 4.0" />
  <row Id="31701" PostTypeId="2" ParentId="31700" CreationDate="2019-05-01T01:46:44.310" Score="1" Body="&lt;p&gt;The amount of insolation (the &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_irradiance&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar irradiation&lt;/a&gt; a planet receives) follows the inverse-square law: half the distance to the Sun results in four times the insolation for a given surface area. However, Earth's &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_eccentricity&quot; rel=&quot;nofollow noreferrer&quot;&gt;orbital eccentricity&lt;/a&gt; is very low – currently just 0.0167 – which means our orbit is nearly circular, so the change in insolation is never so dramatic as a quadrupling: our perihelion is 147.1 million km and our aphelion 152.1 million km – a difference of just 5 million km – which means the Earth receives &lt;span class=&quot;math-container&quot;&gt;$\left(\frac {152.1}{147.1}\right)^2$&lt;/span&gt; = 1.069 times (i.e. 6.9%) more insolation at perihelion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another effect of orbital eccentricity is that under Kepler's second law, Earth's orbital velocity is greater at perihelion than at aphelion: Earth spends about 4.7 fewer days in the seasonal quadrant at perihelion than in the opposite quadrant at aphelion. In other words, the northern winter and southern summer are slightly shorter, and the northern summer and southern winter slightly longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These variations do indeed affect the seasonal weather: the southern hemisphere currently has slightly hotter, shorter summers and colder, longer winters, and the northern hemisphere has slightly milder, longer summers and milder, shorter winters. 13,000 years ago, when the Earth's axis was pointing in the opposite direction, it was the reverse: slightly hotter but shorter northern summer, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are many other things that affect insolation and its impact on Earth, including variations in the Sun's radiation (for just one example, see the &lt;a href=&quot;https://en.wikipedia.org/wiki/Maunder_Minimum&quot; rel=&quot;nofollow noreferrer&quot;&gt;Maunder Minimum&lt;/a&gt; in sunspot activity), variations in albedo, scattering and absorption (e.g. greater northern hemisphere land mass means more heat absorbed; loss of Arctic ice in summer reduces surface albedo so more heat absorbed), variations in Earth's axial tilt, etc. It's &lt;em&gt;very&lt;/em&gt; complicated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of these variations are cyclic in nature. The combined impact on Earth's insolation from cyclic variations in orbital eccentricity, axial tilt (obliquity of the ecliptic) and axial direction (precession of the equinoxes) are described by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Milankovitch_cycles&quot; rel=&quot;nofollow noreferrer&quot;&gt;Milankovitch cycles&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="14094" LastActivityDate="2019-05-01T01:46:44.310" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31702" PostTypeId="1" CreationDate="2019-05-01T02:16:21.853" Score="1" ViewCount="79" Body="&lt;p&gt;How can I convert the observed magnitude to Bolometric Luminosity? I know the relation for the star but should it be same for a Quasar? Basically I want to convert the X-axis of this plot to luminosity (or mass). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rckSN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rckSN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20177" LastActivityDate="2019-05-01T11:26:56.017" Title="How to convert the observed magnitude of a Quasar to it's bolometric luminosity?" Tags="&lt;black-hole&gt;&lt;observational-astronomy&gt;&lt;astrophysics&gt;&lt;supermassive-black-hole&gt;&lt;quasars&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31704" PostTypeId="2" ParentId="6183" CreationDate="2019-05-01T02:56:00.480" Score="-3" Body="&lt;p&gt;The outer planets of our Solar system are not only moving the same way but have almost equal angular momenta. This includes the Sun spinning on its axis. Why it is so is not clear.&lt;/p&gt;&#xA;" OwnerUserId="27847" LastActivityDate="2019-05-01T02:56:00.480" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31705" PostTypeId="1" CreationDate="2019-05-01T06:30:46.610" Score="1" ViewCount="37" Body="&lt;p&gt;Could anyone direct me to papers with info on minimal flux that can be registered with radiotelescopes?&lt;/p&gt;&#xA;" OwnerUserId="22098" LastActivityDate="2019-05-31T10:04:00.523" Title="Need some sources on radiotelescope sensitivity" Tags="&lt;radio-astronomy&gt;&lt;radio-telescope&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31706" PostTypeId="2" ParentId="31705" CreationDate="2019-05-01T09:05:21.997" Score="2" Body="&lt;p&gt;I'll try a quick estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strength of a radio source &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; is often measured in Jansky's and 1 Jy is &lt;span class=&quot;math-container&quot;&gt;$1 \times 10^{-26}$&lt;/span&gt; W/m^2/Hz. That means every square meter of collection area is exposed to &lt;span class=&quot;math-container&quot;&gt;$1 \times 10^{-26}$&lt;/span&gt; per Hz of bandwidth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total power received would then be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P_S = S A \Delta \nu$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is the receiving area of the dish and &lt;span class=&quot;math-container&quot;&gt;$\Delta \nu$&lt;/span&gt; is the bandwidth in Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The noise limit of an RF receiving system is often expressed as an equivalent temperature. A typical number might be 50 Kelvin, which includes a receiver temperature, sky temperature, spillover (thermal radio emission radiated from the ground) etc. Though it seems counterintuitive, we can add temperatures as a proxy for adding powers because we are in the Rayleigh-Jeans regime where &lt;span class=&quot;math-container&quot;&gt;$\nu &amp;lt;&amp;lt; k_B T/h$&lt;/span&gt; as &lt;a href=&quot;https://space.stackexchange.com/a/25744/12102&quot;&gt;explained in this answer&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You convert the noise temperature to a power in Watts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P_N = k_B T \Delta \nu$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$k_B$&lt;/span&gt; is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Boltzmann_constant&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boltzmann constant&lt;/a&gt; which is about &lt;span class=&quot;math-container&quot;&gt;$1.381 \times 10^{-23}$&lt;/span&gt; Joules/K, and &lt;span class=&quot;math-container&quot;&gt;$\Delta \nu$&lt;/span&gt; is again bandwidth in Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's set the minimum strength easily detected to be equal to the noise, and equate the two powers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P_S = P_T $$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$S A \Delta \nu = k_B T \Delta \nu$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$S A = k_B T$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$S = k_B T / A$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the bandwidth drops out since both the thermal noise and most radio sources are fairly broadband. Even if the original line is fairly narrow, you'll tend set your minimum bandwidth of a specific measurement to cover the spectral region of interest.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Ballpark Estimate&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Arecibo_Observatory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arecibo&lt;/a&gt; dish has a radius of about 150 meters, so the area &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is about 70,700 square meters. A typical radio astronomy system temperature is 50 Kelvin (it can be significantly higher or lower depending on many factors, NASA deep space 70 meter dish liquid helium cooled front-end is 20K (see &lt;a href=&quot;https://space.stackexchange.com/q/21269/12102&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://space.stackexchange.com/a/24343/12102&quot;&gt;here&lt;/a&gt;), so plugging that into the equation above, we get &lt;span class=&quot;math-container&quot;&gt;$ = 9.8 \times 10^{-27}$&lt;/span&gt; watts, &lt;strong&gt;which is&lt;/strong&gt; (probably not coincidentally) &lt;strong&gt;almost exactly 1 Jansky!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Confirm that&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;According to Joeri van Leeuwen's &lt;a href=&quot;https://www.astron.nl/astrowiki/lib/exe/fetch.php?media=ra_uva:ra_uva_lecture5.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Radio Astronomy lecture 5, The Radio Telescope&lt;/a&gt; at the bottom of page 79, the &lt;em&gt;System equivalent Flux Density&lt;/em&gt; or &lt;strong&gt;SEFD of Arecibo is 3 Jansky, so a basic back of the envelope calculation gets us quite close!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other radio telescopes will very roughly scale based on area, a smaller area dish will have a higher SEFD or signal level equivalent to the system noise.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Another example, VLA&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A VLA dish is 25 meters in diameter, which is about 150 times less area. My back-of-the-envelope calculation &lt;strong&gt;would then be 150 Jansky.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Table 3.2.1 on the NRAO page &lt;a href=&quot;https://science.nrao.edu/facilities/vla/docs/manuals/oss/performance/sensitivity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Performance of the VLA during the Next Semester; Sensitivity&lt;/a&gt; at its lowest, a VLA dish's &lt;strong&gt;best SEFD is 250 Jansky.&lt;/strong&gt; I've plotted the table below. Still checks out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sxSYY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sxSYY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-05-01T09:18:01.550" LastActivityDate="2019-05-01T09:18:01.550" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31709" PostTypeId="2" ParentId="25781" CreationDate="2019-05-01T10:23:35.223" Score="1" Body="&lt;p&gt;Complementing @KenG's &lt;a href=&quot;https://astronomy.stackexchange.com/a/26173/7982&quot;&gt;answer&lt;/a&gt;, Here's an actual datapoint.&lt;/p&gt;&#xA;&lt;p&gt;The new paper in Nature &lt;a href=&quot;https://www.nature.com/articles/s41550-019-0738-7&quot; rel=&quot;nofollow noreferrer&quot;&gt;Polarized reflected light from the Spica binary system&lt;/a&gt; (downloadable &lt;a href=&quot;https://www.researchgate.net/publication/332133160_Polarized_reflected_light_from_the_Spica_binary_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;) is notable in that the measurement of the &lt;em&gt;polarized component of the reflected light speaks to it being an actual reflection&lt;/em&gt;, rather than one star heating the other producing a more brightly radiating area.&lt;/p&gt;&#xA;&lt;p&gt;Their model of the Spica system uses mostly adopted parameters, but of the derived parameters, the &lt;em&gt;geometric albedo&lt;/em&gt; of the A and B components are about 3.6% and 1.4% respectively.&lt;/p&gt;&#xA;&lt;p&gt;You can read further about Bond albedo and geometric albedo in @zephyr's &lt;a href=&quot;https://astronomy.stackexchange.com/a/20803&quot;&gt;excellent answer&lt;/a&gt; to the question &lt;em&gt;Why is Enceladus's albedo greater than 1?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2020-07-09T08:41:10.030" LastActivityDate="2020-07-09T08:41:10.030" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31711" PostTypeId="1" CreationDate="2019-05-01T10:31:17.767" Score="3" ViewCount="993" Body="&lt;p&gt;What is the furthest star or celestial object whose distance has been calculated with parallax and how does it compare to the theoretical limit using today's telescopes? And how exactly does telescope aperture relate to the maximum distance measurable (other than the bigger the aperture the bigger the distance)?&lt;/p&gt;&#xA;" OwnerUserId="23042" LastEditorUserId="23042" LastEditDate="2019-05-01T15:40:17.900" LastActivityDate="2020-05-26T14:03:20.323" Title="What is the maximum distance measurable with parallax?" Tags="&lt;telescope&gt;&lt;distances&gt;&lt;parallax&gt;" AnswerCount="2" CommentCount="9" ClosedDate="2020-05-27T03:28:04.767" ContentLicense="CC BY-SA 4.0" />
  <row Id="31712" PostTypeId="2" ParentId="31702" CreationDate="2019-05-01T11:26:56.017" Score="1" Body="&lt;p&gt;The bolometric magnitude of a quasar can be determined by the same method as a star (using total energy radiated)......but for the fact that quasars do not radiate isotropically and this distribution is dependant on wavelength&#xA;A further complication comes as many photons are absorbed and re-emitted , etc&#xA;Some standard methods of calculation can be found here......&#xA;&lt;a href=&quot;https://academic.oup.com/mnras/article/408/3/1598/1074070&quot; rel=&quot;nofollow noreferrer&quot;&gt;Quasar bolometric corrections&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25925" LastActivityDate="2019-05-01T11:26:56.017" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31713" PostTypeId="1" AcceptedAnswerId="31755" CreationDate="2019-05-01T12:50:15.170" Score="2" ViewCount="149" Body="&lt;p&gt;A trans-Neptunian object, 1 million km from the observer, with an angular diameter of 0.126° occults the Sun (angular diameter of 0.004°) and the TNO and the observer are moving in the same direction, in the same plane.  The TNO is moving 8m/sec with respect to the observer, how long would the occultation last?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the observer is 18 billion km from the Sun.&lt;/p&gt;&#xA;" OwnerUserId="25088" LastEditorUserId="25088" LastEditDate="2019-05-03T02:03:53.023" LastActivityDate="2019-05-05T01:57:05.540" Title="How long would an occultation by a TNO last?" Tags="&lt;the-sun&gt;&lt;occultation&gt;&lt;trans-neptunian-object&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31714" PostTypeId="2" ParentId="31711" CreationDate="2019-05-01T13:47:01.640" Score="1" Body="&lt;p&gt;Quick Google reveals a couple simple analyses.  For &lt;a href=&quot;https://socratic.org/questions/why-can-t-astronomers-use-parallax-to-measure-distances-to-other-galaxies&quot; rel=&quot;nofollow noreferrer&quot;&gt;example&lt;/a&gt;, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Andromeda Galaxy, M31, is the nearest major galaxy to the Milky&#xA;  Way. The distance to M31 has been measured using other techniques to&#xA;  be &#xA;  2.5⋅10^6 light years , or 7.6⋅10^5 parsecs. Using the slightly modified parallax formula, we can find the necessary parallax angle to&#xA;  measure the distance to Andromeda. &lt;span class=&quot;math-container&quot;&gt;$ p = \frac{1}{d} =&#xA;&amp;gt; \frac{1}{7.6*10^5} parsec = 1.3 *10^{-6} arc-seconds $&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This is an incredibly small angle. For comparison, the resolution of&#xA;  the Hubble Space Telescope is 0.05  arc-seconds, so even Hubble would&#xA;  not be able to detect the necessary angular shift of the nearest&#xA;  galaxy to effectively use parallax as a measure of its distance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="907" LastActivityDate="2019-05-01T13:47:01.640" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31715" PostTypeId="1" CreationDate="2019-05-01T14:20:16.600" Score="1" ViewCount="34" Body="&lt;p&gt;A number of pictures and some animations similar to the ones below have had much press recently. Clearly the colours are artificial, but I wondered how accurately the represent the geometry of the system -- the relative sizes of the event horizon and the thin and thick accretion disks, the distance from the companion star, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/a6J7z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a6J7z.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XrHME.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XrHME.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The images are taken from popular press sites, but seem to originate from &lt;a href=&quot;https://www.icrar.org/Cygni&quot; rel=&quot;nofollow noreferrer&quot;&gt;ICRAR&lt;/a&gt;, where they are described as &quot;artists impressions&quot;. I would nevertheless be interested to know how accurate a brief their artists were working from.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="21928" LastEditDate="2019-05-01T14:45:07.797" LastActivityDate="2019-05-01T14:45:07.797" Title="How accurate are the new visualisations of V404 Cygni" Tags="&lt;black-hole&gt;&lt;binary-star&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31716" PostTypeId="2" ParentId="31711" CreationDate="2019-05-01T15:14:14.940" Score="1" Body="&lt;p&gt;Per Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Gaia_(spacecraft)#Objectives&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gaia (spacecraft); Objectives&lt;/a&gt; which I linked to in the question &lt;a href=&quot;https://astronomy.stackexchange.com/q/30621/7982&quot;&gt;What actually determines the angular uncertainty of the source of a detected gravitational wave?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Determine the position, parallax, and annual proper motion of 1 billion stars with an accuracy of about 20 microarcseconds (µas) at 15 mag, and 200 µas at 20 mag.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;20 (µas) is about &lt;span class=&quot;math-container&quot;&gt;$1 \times 10^{-10}$&lt;/span&gt; radians. If the Earth's amplitude is 2 AU, then the farthest distance that could be &lt;em&gt;detected&lt;/em&gt; is &lt;span class=&quot;math-container&quot;&gt;$2 \times 10^{10}$&lt;/span&gt; AU. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to measure to about 10% accuracy, then that distance is  &lt;span class=&quot;math-container&quot;&gt;$2 \times 10^{9}$&lt;/span&gt; AU or about &lt;s&gt;3,000&lt;/s&gt; 30,000 light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That sounds surprisingly far away!&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-05-02T10:16:28.210" LastActivityDate="2019-05-02T10:16:28.210" CommentCount="17" ContentLicense="CC BY-SA 4.0" />
  <row Id="31717" PostTypeId="1" CreationDate="2019-05-01T16:14:51.727" Score="2" ViewCount="586" Body="&lt;p&gt;Is it only possible to measure objects that form around the same time? Is it possible to measure clusters from distant galaxies other than our own?&lt;/p&gt;&#xA;" OwnerUserId="27895" LastActivityDate="2020-01-18T23:47:03.587" Title="What limits the use of the H-R diagram to measure distance (main sequence fitting), what distances is it useful for?" Tags="&lt;star&gt;&lt;distances&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31718" PostTypeId="2" ParentId="31717" CreationDate="2019-05-01T16:37:17.277" Score="4" Body="&lt;p&gt;It depends how precise you need to be. Main sequence fitting assumes that the star(s) in question is(are) on the main sequence. If you have a cluster of stars (at the same age) then defining what is on the main sequence and what isn't becomes much, much easier and of course you have lots of stars with which to beat down the statistical uncertainty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, for coeval groups of stars what you term &quot;main sequence fitting&quot; is rarely done. The process is the fitting of an &quot;isochrone&quot; (a line that links points in the HR diagram at a single age), so that both the distance, age (also extinction and metallicity) are possible free parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The subtle distinction here is that what is termed &quot;the main sequence&quot; does not really exist in practice, or at least not as a uniquely defined locus in the HR diagram. Star begin their main sequence lives on the &quot;zero age main sequence&quot; (ZAMS) and end their main sequence careers at the &quot;terminal age main sequence&quot; (TAMS), gradually changing their luminosity and temperature as they do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a diagram from &lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/S1384107613000791&quot; rel=&quot;nofollow noreferrer&quot;&gt;Martignoni et al. (2014)&lt;/a&gt; showing the ZAMS and TAMS for stars of different masses. there is typically a factor 2-3 in luminosity between them (a larger difference at larger masses). That means whether you use the ZAMS, the TAMS, or something in between to determine the distance from a vertical displacement in the HR diagram, you could vary your answer for the distance by &lt;span class=&quot;math-container&quot;&gt;$\sqrt{2}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\sqrt{3}$&lt;/span&gt;. In other words you &lt;strong&gt;need to know the age of a main sequence star before &quot;main-sequence fitting&quot; can give you an accurate distance.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, lower mass stars are longer lived. Anything of say 0.7 solar masses or below has hardly moved from the ZAMS in the age of the Galaxy, so there would be little error in assuming a ZAMS locus. Conversely, the effects of age are far more rapid and therefore far more important on the main sequence at higher masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to try to use main-sequence fitting to estimate the distance to &lt;em&gt;individual&lt;/em&gt; stars then there are several hazards. For one, it can be nearly impossible to estimate the age of an individual star. Therefore if it has a mass greater than 0.7 solar masses then there will be an uncertainty in its position in the absolute HR diagram that leads to an inevitable uncertainty in estimated distance. Further, the intrinsic position in the HR diagram depends on the star's chemical composition. Such ancillary information might be available, but it might not, in which case that is another source of error. A further source of systematic uncertainty is stellar rotation. Fast rotating stars have extended lifetimes on the main sequence and somewhat different intrinsic positions on the HR diagram; again a source of systematic uncertainty that is especially problematic for high mass stars. Finally, it can be that what you think is an isolated main sequence star is in fact a binary system. A companion can increase the luminosity of the system and make a star appear closer than it actually is (by up to a factor &lt;span class=&quot;math-container&quot;&gt;$\sqrt{2}$&lt;/span&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/W2v1S.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W2v1S.jpg&quot; alt=&quot;A HR diagram showing ZAMS and TAMS&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2020-01-18T23:47:03.587" LastActivityDate="2020-01-18T23:47:03.587" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31719" PostTypeId="1" AcceptedAnswerId="31720" CreationDate="2019-05-01T17:23:28.797" Score="5" ViewCount="1004" Body="&lt;p&gt;My question is related to black hole image released in April. As far as I understand idea of EHT, it joins observations from multiple locations to work like one telescope with radius that is equal to the distance between the farthest telescopes in array. Moreover Earth atmosphere affects observation, e.g. by introducing delays. Would it beneficial to add Hubble Space Telescope to EHT and if yes, why HST was not included in observation array?&lt;/p&gt;&#xA;" OwnerUserId="27905" LastActivityDate="2020-12-16T15:44:07.017" Title="Would Hubble Space Telescope improve black hole image observed by EHT if it joined array of telesopes?" Tags="&lt;black-hole&gt;&lt;supermassive-black-hole&gt;&lt;event-horizon-telescope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31720" PostTypeId="2" ParentId="31719" CreationDate="2019-05-01T17:40:26.070" Score="16" Body="&lt;p&gt;No, it would not, because it operates in the visible spectrum and the EHT is an array of radio telescopes. For the &amp;quot;very long baseline interferometry&amp;quot; technique to work, all the telescopes have to be operating at the same wavelength, because combining the signals involves measuring exactly how well the peaks and troughs of the radio waves from the different telescopes line up.&lt;/p&gt;&#xA;&lt;p&gt;You can do VLBI in the visible spectrum, but you have to match up the waves even more precisely since light has shorter wavelengths than radio waves. The EHT collected all the data first, and then spent a lot of time combining it by computer, but that required using very precise clocks to sync the data. We don't have clocks precise enough to do that for light, so a direct optical connection is required between the multiple telescopes. So there isn't a good way to do a planetwide VLBI optical telescope yet.&lt;/p&gt;&#xA;&lt;p&gt;Addendum inspired by the comments, especially from TazAstroSpacial: To get the same resolution with optical light, you can use a much smaller telescope array. I didn't mention that fact because I was thinking about keeping the size of the array the same and getting better resolution. But in any case the problem is that the challenges of doing the interferometry at smaller wavelengths more than make up for the advantages of needing a smaller aperture. At least at the current state of the art.&lt;/p&gt;&#xA;" OwnerUserId="7752" LastEditorUserId="7752" LastEditDate="2020-12-16T15:44:07.017" LastActivityDate="2020-12-16T15:44:07.017" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="31721" PostTypeId="1" AcceptedAnswerId="31723" CreationDate="2019-05-01T18:15:08.477" Score="13" ViewCount="3028" Body="&lt;p&gt;If so, how do we know it speeds up?  Doesn't time slow down as gravity increases?  If time slows down around a black hole, is it possible matter doesn't actually speed up?&lt;/p&gt;&#xA;" OwnerUserId="27906" LastActivityDate="2019-05-02T13:34:16.300" Title="As matter approaches a black hole, does it speed up?" Tags="&lt;black-hole&gt;&lt;gravity&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31722" PostTypeId="2" ParentId="31721" CreationDate="2019-05-01T18:48:51.130" Score="5" Body="&lt;p&gt;The dilation of time is only relevant from the perspective of someone far away from the black hole. Close to the black hole time is still progressing forward at what would appear to be a normal rate to someone who is close to the black hole.  The movie &lt;em&gt;Interstellar&lt;/em&gt; had a great depiction this phenomenon, with the astronauts Copper and Brand on Miller's planet, near the black hole, spending only a few hours, but the astronaut Romilly aging decades as he remained far from the planet.  Copper and Brand didn't experience any change in the passage of time, from their perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matter falling into a black hole would not experience any change in its perspective of time, so would not appear to change speed, other than what would be expected by the gravitational attraction.&lt;/p&gt;&#xA;" OwnerUserId="25088" LastActivityDate="2019-05-01T18:48:51.130" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31723" PostTypeId="2" ParentId="31721" CreationDate="2019-05-01T19:17:28.393" Score="20" Body="&lt;p&gt;The answer is neither yes or no or possibly both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a simple example. If something falls freely towards a black hole along a radial path, and is observed by someone who is far from the black hole, its velocity (according to the distant observer) is given by&#xA;&lt;span class=&quot;math-container&quot;&gt;$$v = -\left(1 - \frac{r_s}{r}\right)\left(\frac{r_s}{r}\right)^{1/2}c\, ,$$&lt;/span&gt;&#xA;(e.g. see chapter 6 of &lt;a href=&quot;http://www.eftaylor.com/exploringblackholes/&quot; rel=&quot;noreferrer&quot;&gt;Exploring Black Holes&lt;/a&gt; by Taylor, Wheeler &amp;amp; Bertschinger - freely available) where &lt;span class=&quot;math-container&quot;&gt;$r_s$&lt;/span&gt; is the Schwarzschild radius and the negative sign just indicates an inward velocity with &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; decreasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you plot this function (see Fig.2 in Ch.6 of Taylor et al. - freely available) you will see that initially the magnitude of the velocity increases as &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; decreases, but as &lt;span class=&quot;math-container&quot;&gt;$r\rightarrow r_s$&lt;/span&gt; then &lt;span class=&quot;math-container&quot;&gt;$v \rightarrow 0$&lt;/span&gt; and the falling object appears to come to a standstill (actually, because the light from the object is gravitationally redshifted, this may not actually be observed). However, if the velocity first increases and then slows to a standstill, then it must go through a maximum!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum &lt;em&gt;observed&lt;/em&gt; speed in this scenario is achieved at &lt;span class=&quot;math-container&quot;&gt;$r=3r_s$&lt;/span&gt; and is &lt;span class=&quot;math-container&quot;&gt;$0.384c$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this story is different for different observers. If you are the falling object then your speed just keeps increasing through the event horizon and towards the singularity. On the other hand, an observer who was somehow able to hover just above the event horizon would measure the falling object's speed as just below &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; as it passed. &lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-05-02T13:34:16.300" LastActivityDate="2019-05-02T13:34:16.300" CommentCount="22" ContentLicense="CC BY-SA 4.0" />
  <row Id="31725" PostTypeId="1" CreationDate="2019-05-01T21:49:14.567" Score="4" ViewCount="140" Body="&lt;p&gt;I live in South America and I'm willing to travel to Chile in July 2019 to observe (in a curious way, not professional) the Solar Total Eclipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to my research, one of the best cities to observe the eclipse is in the coastal city of La Serena at 3:20 PM Local Time. However, I don't know if it can be seen from &quot;anywhere&quot;. I was thinking of observing it from the beach. Does that make sense?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know that I need some kind of special sunglasses for safety. I found some on Ebay but I don't know if they are authentic. I would appreciate if someone could point me to a website that sells authentic eclipse glasses for a reasonable price and international shipping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, my question is, will I be able to amazingly see the eclipse by just traveling to La Serena and go to the beach on the specific day and time? Is there a better place to see it? Is it better to see it from a high ground? Any tips are welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="27912" LastEditorUserId="1891" LastEditDate="2019-05-01T22:01:39.420" LastActivityDate="2019-05-07T05:00:07.780" Title="How to properly observe Solar Total Eclipse in 2019?" Tags="&lt;observational-astronomy&gt;&lt;solar-eclipse&gt;" AnswerCount="3" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="31726" PostTypeId="2" ParentId="31681" CreationDate="2019-05-01T21:51:15.750" Score="1" Body="&lt;p&gt;Terminology is critical. As mentioned in other answers, precession causes the equinoxes (the points where the ecliptic intersects the celestial equator) to move from one &lt;strong&gt;constellation&lt;/strong&gt; to another. For example, the March equinox was in the &lt;strong&gt;constellation&lt;/strong&gt; Aries in the time of the ancient Greeks, is in the &lt;strong&gt;constellation&lt;/strong&gt; Pisces today, and will be in the &lt;strong&gt;constellation&lt;/strong&gt; Aquarius around the year 2600.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A zodiac &lt;strong&gt;sign&lt;/strong&gt;, also called an astrological &lt;strong&gt;sign&lt;/strong&gt; in Western astronomy, is defined as starting at the March equinox with Aries, then Taurus, and so on until the 12th sign, Pisces. Each sign occupies 30 degrees, even though the actual constellations are a variety of different angular widths.&lt;/p&gt;&#xA;" OwnerUserId="7385" LastActivityDate="2019-05-01T21:51:15.750" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31728" PostTypeId="1" CreationDate="2019-05-01T23:05:53.027" Score="2" ViewCount="42" Body="&lt;p&gt;In other words, is the assumption, that redshift is caused by universe expansion, definitive? Or are other theories still being considered? Could it be, that light is being slowed down by dark energy, dark matter, massive objects, etc., as it travels through space? &lt;/p&gt;&#xA;" OwnerUserId="27810" LastActivityDate="2019-05-01T23:05:53.027" Title="Is it possible for redshift to be caused by light being somehow slowed down during travel?" Tags="&lt;expansion&gt;&lt;redshift&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" ClosedDate="2019-05-02T02:08:15.343" ContentLicense="CC BY-SA 4.0" />
  <row Id="31729" PostTypeId="2" ParentId="31681" CreationDate="2019-05-02T07:49:36.567" Score="0" Body="&lt;p&gt;&quot;We say equinox is at same place&quot; - This is an approximation. Due to precession,equinox  keeps on shifting every year by a fraction of 360 Deg/26000 year=0.0138 Deg/year.&#xA;In fact based on certain references in Indian scripts the Year of particular incidence say- &quot;Epic Mahabharat&quot; is estimated 5000 years back. If interested I can provide more details &lt;/p&gt;&#xA;" OwnerUserId="26665" LastActivityDate="2019-05-02T07:49:36.567" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31733" PostTypeId="1" CreationDate="2019-05-02T10:25:24.347" Score="3" ViewCount="124" Body="&lt;p&gt;In the comments of another question on this Stack Exchange site, I talk with another user about the possibility of building an optical space telescope with a 6.5 km aperture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, this is much larger than any rocket we have constructed, so it would need to be launched from Earth in a number of components and assembled in orbit in a similarly massive number of spacewalks, and the sheer scale of it would likely require active heat management systems to prevent heat expansion of the components from sunlight warping the mirrors or stretching the distance between the mirrors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other obstacles that would prevent this project from being possible?&lt;/p&gt;&#xA;" OwnerUserId="27914" LastEditorUserId="27914" LastEditDate="2019-05-02T12:12:55.647" LastActivityDate="2019-05-02T15:40:00.353" Title="What would be the obstacles to building a 6.5km space telescope besides cost?" Tags="&lt;optics&gt;&lt;space-telescope&gt;&lt;very-large-telescope&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31735" PostTypeId="1" CreationDate="2019-05-02T11:49:40.493" Score="1" ViewCount="59" Body="&lt;p&gt;If we plot graph between accumulate of forming mass (y-axis) and mass (x-axis), where is the maximum ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a story how I get this question. It start when I think about question &quot;Is our sun is big or small ?&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.quora.com/What-is-the-average-mass-of-a-star&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.quora.com/What-is-the-average-mass-of-a-star&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average mass ends up being around 0.5 solar masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But very high mass stars have very short life span so there are more O-type main sequence stars ever form more than O-type main sequence stars we see today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it would be better to measure our sun by using average mass of forming stars than average mass of stars we see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what is average mass of forming stars ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I think a litle bit more. There is no exact boundry between large brown dwarf and small red dwarf stars. So, if we include them too, what is the most mass that matter likely to form ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is inverse-square law, so graph between accumulate mass (y-axis) and mass (x-axis) will be y=c/x where c is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And make the maximum is zero or mean that most of matter aren't form star. At this point, I think I should limit the minimum mass of forming object at some point. But is it inverse-square law for all type of forming object?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If most of matter aren't form star, will be gas cloud. So we don't count them (both molecular gas and mono-atomic gas but dust is still count because it is solid not molecular or mono-atomic object), what is the most mass that matter likely to form?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me or tell me what should read to answer or give keyword to me.&lt;/p&gt;&#xA;" OwnerUserId="24925" LastEditorUserId="24925" LastEditDate="2019-05-02T12:36:52.150" LastActivityDate="2019-05-02T12:36:52.150" Title="What is the most mass that matter likely to form?" Tags="&lt;star&gt;&lt;planet&gt;&lt;star-formation&gt;&lt;planetary-formation&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31736" PostTypeId="1" CreationDate="2019-05-02T14:06:35.750" Score="1" ViewCount="94" Body="&lt;p&gt;I have ephemeris data for state vectors of a satellite in a J2000.0 coordinate system. This data is already filtered and cleaned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data sample rate ranges between 1 second and 60 seconds, and I want to resample the data down to &amp;lt;10 milliseconds with a constant, regular time step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have already managed to interpolate using methods from &lt;a href=&quot;https://docs.scipy.org/doc/scipy/reference/interpolate.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;scipy.interpolate&lt;/a&gt; with cubic interpolation; &lt;a href=&quot;https://docs.scipy.org/doc/scipy/reference/generated/scipy.interpolate.interp1d.html#scipy.interpolate.interp1d&quot; rel=&quot;nofollow noreferrer&quot;&gt;interp1d&lt;/a&gt; and &lt;a href=&quot;https://docs.scipy.org/doc/scipy/reference/generated/scipy.interpolate.griddata.html#scipy.interpolate.griddata&quot; rel=&quot;nofollow noreferrer&quot;&gt;griddata&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads me to the question of how to choose between various methods to resample/interpolate ephemeris state vectors. Certainly the interpolation of tabular ephemerides is something that astronomers and orbital dynamicists need to do and so there must exist some guidelines or best practices or recognized-better or -worse ways to so this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How to choose between methods to resample/interpolate ephemeris state vectors? Are there methods recognized as superior for orbital data?&lt;/p&gt;&#xA;" OwnerUserId="27924" LastEditorUserId="7982" LastEditDate="2019-05-04T00:45:38.340" LastActivityDate="2019-05-04T16:20:09.123" Title="How to choose between methods to resample/interpolate ephemeris state vectors? Are there recognized techniques for orbital data?" Tags="&lt;orbital-mechanics&gt;&lt;artificial-satellite&gt;&lt;ephemeris&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31737" PostTypeId="2" ParentId="31733" CreationDate="2019-05-02T14:32:38.180" Score="2" Body="&lt;p&gt;Large-aperture space-based telescopes are all based on a segmented primary design. Not only is it impractical to launch &lt;em&gt;any&lt;/em&gt; rocket with mega-diameter cross-section, but thermal and gravitational stress relaxations would cause the mirror to go ugly after deployment.&lt;br&gt;&#xA;These segmented primaries are surrounded by heat shields and other goodies specifically to avoid thermal effects.  We already have the technology and computer power to align and phase a zillion (that's the technical term :-) ) segments into the desired base curve for this telescope.&lt;br&gt;&#xA;OK, so aside from cost, why not do this?  For one thing, it's far easier to improve the Quantum Efficiency (QE) and the dark noise(signal-independent electronic noise) in the sensors than to struggle to increase the total light collected.   That takes care of concerns about seeing dim objects.&lt;br&gt;&#xA;Next, it's also much easier to deploy very long baseline interferometric (VLBI) systems, essentially a widely spaced array of small telescopes, to achieve the diffraction-limited resolving power of an equivalent giant single telescope.   &quot;Easier&quot; here is still very difficult, since you need to know and maintain the telescopes' spacing to a fraction of a wavelength.  &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2019-05-02T14:32:38.180" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31738" PostTypeId="2" ParentId="31733" CreationDate="2019-05-02T14:37:20.643" Score="4" Body="&lt;p&gt;The basic issue with a large optical telescope is to keep all of the mirror in an extremely stable position relative to the rest of the mirror. Depending on the wavelength of your light you probably want to limit deviations from your perfect shape to less than about 100nm, so, you need to account for and deal with somehow any force which might act on one part of your mirror differently from another with enough force to bend the whole mirror by more than this tiny amount. At a glance, therefore you will need to deal with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;tidal forces from Earth, Moon and Sun (I think you can ignore tides from Venus and Jupiter, but I'm not certain)&lt;/li&gt;&#xA;&lt;li&gt;Orbital perturbations arising because Earth is not spherical&lt;/li&gt;&#xA;&lt;li&gt;temperature variations from moving in and out of Earth's shadow at different times and from sunlight and Earthlight (and maybe moonlight) hitting different parts of the structure differently&lt;/li&gt;&#xA;&lt;li&gt;interactions of any conducting components of the mirror with Earth's magnetic field&lt;/li&gt;&#xA;&lt;li&gt;flexing and oscillations of the structure arising from thrusts used to point the mirror&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and probably a lot more that I haven't thought of.  These problems would become somewhat simpler if you put the telescope much further from the Earth, or even better much further from the Sun (say a very distant orbit around Neptune) but that would make assembling it by spacewalking impossible.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="21928" LastEditDate="2019-05-02T15:40:00.353" LastActivityDate="2019-05-02T15:40:00.353" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31739" PostTypeId="1" CreationDate="2019-05-02T15:39:33.203" Score="2" ViewCount="35" Body="&lt;p&gt;As I understand it there can be said to be two &quot;kinds&quot; of electromagnetic radiation filling up the universe. First you have the primordial radiation from a time soon after Big Bang. Then you have &quot;newer&quot; radiation from the fact that you start up with very diffuse clouds of mainly hydrogen that falls together and forms planets, stars, black holes etc. basically converting potential energy to heat radiation and in the stars you of course also have fusion producing heat that radiates away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you say something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X percent of the energy of the electromagnetic radiation in the universe is primordial and (100 -x) percent of the energy of the electromagnetic radiation in the universe is of newer origin, and if so, what is X?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean if you &quot;magically removed&quot; all &quot;primordial radiation&quot; some 13 billion years ago we would still have a lot of radiation in the universe today.&lt;/p&gt;&#xA;" OwnerUserId="26403" LastActivityDate="2019-05-02T15:39:33.203" Title="How large part of the energy of the electromagnetic radiation in the universe is primordial" Tags="&lt;cmb&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31740" PostTypeId="1" CreationDate="2019-05-02T16:02:22.740" Score="0" ViewCount="159" Body="&lt;p&gt;Could the known universe be inside a black hole and it's curvature really be the event horizon of the black hole?&lt;/p&gt;&#xA;" OwnerUserId="27925" LastEditorUserId="1891" LastEditDate="2019-05-04T08:14:30.747" LastActivityDate="2019-05-04T08:14:30.747" Title="Is our Universe inside a black hole?" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2019-05-02T21:09:53.053" ContentLicense="CC BY-SA 4.0" />
  <row Id="31741" PostTypeId="2" ParentId="31740" CreationDate="2019-05-02T18:03:07.170" Score="0" Body="&lt;p&gt;No. The metrics look very differently.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The metrics of the the Universe is determined by the FLRW metrics. Its spacelike sections are planar.&lt;/li&gt;&#xA;&lt;li&gt;The spacelike sections of the spacetime around a Schwarzschild (non-rotating) BH is spherical.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;More elementarily formulated: the black hole is like a static sphere. The Universe is like an expanding sheet. However, this is what we currently see, matched by our current models. These are not so strong facts, like the density of the iron, what anybody could measure on his desk. I think the correct explanation is roughly this: &lt;em&gt;&quot;it doesn't really looks so, but no one knows&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="16685" LastEditDate="2019-05-02T21:14:02.473" LastActivityDate="2019-05-02T21:14:02.473" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31742" PostTypeId="1" CreationDate="2019-05-02T20:11:32.737" Score="0" ViewCount="73" Body="&lt;p&gt;Based on reading the &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Bang&quot; rel=&quot;nofollow noreferrer&quot;&gt;wiki page&lt;/a&gt; of the big bang, a simple question with a probably not a simple answer came to mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement on the wikipedia page was.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since Georges Lemaître first noted in 1927 that an expanding universe could be traced back in time to an originating single point, scientists have built on his idea of cosmic expansion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question is:  &lt;strong&gt;Have we measured or found evidence of a detectable origin point of the big bang?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="25592" LastActivityDate="2019-05-02T20:23:02.540" Title="Have we measured or found evidence of a detectable origin point of the big bang?" Tags="&lt;universe&gt;&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ClosedDate="2019-05-03T05:31:43.633" ContentLicense="CC BY-SA 4.0" />
  <row Id="31743" PostTypeId="2" ParentId="31742" CreationDate="2019-05-02T20:19:16.530" Score="0" Body="&lt;p&gt;&lt;strong&gt;There is no single origin point.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before the big bang, the entire universe was contained in an extremely dense singularity - there was no &quot;space&quot; for the universe to expand into, the very fabric of spacetime was contained entirely in the singularity. There is no single &quot;origin point&quot; in the universe where the big bang took place where we'd see the entire universe receding from us. Rather, from &lt;em&gt;any point in the universe&lt;/em&gt; we can see everything else receding. The big bang happened &lt;strong&gt;everywhere&lt;/strong&gt;, since the whole universe, including space itself, came from that singularity.&lt;/p&gt;&#xA;" OwnerUserId="10488" LastEditorUserId="10488" LastEditDate="2019-05-02T20:23:02.540" LastActivityDate="2019-05-02T20:23:02.540" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31744" PostTypeId="1" AcceptedAnswerId="31746" CreationDate="2019-05-02T21:20:44.740" Score="2" ViewCount="301" Body="&lt;p&gt;So, if a black hole is so dense, that light cannot escape, what happens to the photons as the approach and go past the event horizon (or some other horizon, about which I am ignorant)?  Do photons slow down at some point, or is it somehow binary, where at one moment they are moving at the speed of light and the next they aren't moving at all?  Or, do they continue to move at the speed of light, just never moving beyond the black hole?&lt;/p&gt;&#xA;" OwnerUserId="27906" LastActivityDate="2019-05-14T20:56:28.833" Title="Do photons slow down as they approach the event horizon of a black hole?" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;light&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31745" PostTypeId="2" ParentId="31744" CreationDate="2019-05-02T21:46:38.913" Score="1" Body="&lt;p&gt;The last option &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;do they continue to move at the speed of light, just never moving beyond the black hole?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is the most nearly correct. But it is &lt;strong&gt;essential&lt;/strong&gt; to realise that in GR questions like &quot;how fast is something moving&quot; cannot be answered until you know where the observer is and how they are moving. There simply is no universal answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to think of it is that space-time is constantly rushing into the black hole, dragged in by its gravity. &lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-05-02T21:46:38.913" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31746" PostTypeId="2" ParentId="31744" CreationDate="2019-05-02T22:57:14.867" Score="5" Body="&lt;p&gt;Photons travel on null geodesics through space time. In the curved space-time near a non-spinning black hole, a distant observer could infer that the speed of light (travelling on a radial path) is given by&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ v_{\rm light} = c \left(1 - \frac{r_s}{r}\right),$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$r_s$&lt;/span&gt; is the Schwarzschild radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &lt;span class=&quot;math-container&quot;&gt;$r \rightarrow r_s$&lt;/span&gt; then we see that &lt;span class=&quot;math-container&quot;&gt;$v_{\rm light} \rightarrow 0$&lt;/span&gt;. So yes, according to an observer far from a black hole, the speed of light approaching a black hole appears to slow down and asymptotically approaches zero at the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the locally measured speed of light is always &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-05-05T15:16:47.660" LastActivityDate="2019-05-05T15:16:47.660" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31747" PostTypeId="1" AcceptedAnswerId="32134" CreationDate="2019-05-02T22:59:41.703" Score="1" ViewCount="184" Body="&lt;p&gt;Per Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Gaia_(spacecraft)#Objectives&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gaia (spacecraft); Objectives&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Determine the position, parallax, and annual proper motion of 1 billion stars with an accuracy of about 20 microarcseconds (µas) at 15 mag, and 200 µas at 20 mag.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As @RobJeffries painstakingly points out to us in comments below &lt;a href=&quot;https://astronomy.stackexchange.com/a/31716/7982&quot;&gt;this answer&lt;/a&gt; when we talk about &lt;em&gt;parallax as a unit&lt;/em&gt;, it is referenced to a baseline of 1 AU. It doesn't refer to the actual cyclic motion seen in the sky due to the orbital motion of whatever is moving around the Sun at its own particular distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the accuracy of both &quot;position&quot; and &quot;annual proper motion&quot; have nothing to do with 1 AU and are measures of positional accuracy of the telescope and how its astrometric data is analyzed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can all three of these accuracies simultaneously be 20 µas for 15 magnitude stars?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-05-02T23:05:35.020" LastActivityDate="2019-06-03T08:06:27.223" Title="How can GAIA's accuracy for position, parallax, and annual proper motion for a given star be the same number?" Tags="&lt;astrometry&gt;&lt;parallax&gt;&lt;gaia&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31748" PostTypeId="1" CreationDate="2019-05-03T02:53:48.040" Score="1" ViewCount="121" Body="&lt;p&gt;I often see questions on this site that only apply to non-spinning black holes and often wonder how the answer would be different for a spinning black hole.  As the title says, what are the major differences between the behavior of spinning and non-spinning black holes, especially when it comes to the effect on space-time?&lt;/p&gt;&#xA;" OwnerUserId="21184" LastActivityDate="2019-05-03T02:53:48.040" Title="How does a spinning black hole affect space-time compared to a non-spinning black hole?" Tags="&lt;black-hole&gt;&lt;space-time&gt;&lt;rotation&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31749" PostTypeId="2" ParentId="21487" CreationDate="2019-05-03T04:49:26.150" Score="-1" Body="&lt;p&gt;This article describes it.&#xA;&lt;a href=&quot;https://www.ias.ac.in/article/fulltext/reso/011/03/0051-0068&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.ias.ac.in/article/fulltext/reso/011/03/0051-0068&lt;/a&gt;&#xA;Aryabhata described it and the Greeks most likely had a similar method.&#xA;The title of this article is Aryabhata and Axial Rotation of Earth - Khagola (The Celestial Sphere)&quot; by &#xA;Amartya Kumar Dutta. There are a lot of diagrams here.&lt;/p&gt;&#xA;" OwnerUserId="27847" LastActivityDate="2019-05-03T04:49:26.150" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31750" PostTypeId="2" ParentId="30585" CreationDate="2019-05-03T05:36:20.533" Score="0" Body="&lt;p&gt;It can be observed. The astronomical constants at 2000AD_500AD _ 1604AD are given below. The Àryabhatíya of Àryabhata, the oldest precise astronomical constant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some tables.This table is from Jacobs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Table 1. Comparison of The Àryabhatiya of Àryabhata and Astronomic values.(orbit times are slowing down ever so slightly)&#xA;Astronomy Constants AD2000 _ AD 500 _ 1604 BC&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotations per solar orbit 366.25636031_ 366.2563589_ 366.25635656&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Days per solar orbit 365.25636031_ 365.2563589_ 365.25635656&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Days per lunar orbit 27.32166120_ 27.3216638_ 27.32166801&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotations per lunar orbit 27.39646289_ 27.39646514_ 27.39646936&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the number of decimal places. It goes on like this to the Rg veda. India had a passion for astronomy and was very good at it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other links.&#xA;Amartya Kumar Dutta — Aryabhata and Axial Rotation of Earth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbits decay ever so slightly in 1000 year time scales!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.ias.ac.in/article/fulltext/reso/011/03/0002-0003&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.ias.ac.in/article/fulltext/reso/011/03/0002-0003&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.ias.ac.in/article/fulltext/reso/011/03/0051-0068&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.ias.ac.in/article/fulltext/reso/011/03/0051-0068&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jqjacobs.net/astro/aryabhata.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.jqjacobs.net/astro/aryabhata.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="27847" LastEditorUserId="27847" LastEditDate="2019-05-03T16:02:04.720" LastActivityDate="2019-05-03T16:02:04.720" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31751" PostTypeId="1" CreationDate="2019-05-03T06:53:28.720" Score="5" ViewCount="229" Body="&lt;p&gt;In this publication from 1981, was shown that &lt;a href=&quot;https://www.nature.com/articles/292435a0&quot; rel=&quot;nofollow noreferrer&quot;&gt;Neptune is raining diamonds&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Many of the current models of Uranus and Neptune postulate a three-layer structure, consisting of an inner rocky core, a middle ‘ice’ layer of fluid, &lt;span class=&quot;math-container&quot;&gt;$\rm H_2\rm O$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\rm{CH}_4$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\rm{NH}_3$&lt;/span&gt; and an outer hydrogen–helium layer of solar composition. The estimated pressures and temperatures of the ice layer range from about 6Mbar and 7,000K at the inner core-ice boundary to &lt;span class=&quot;math-container&quot;&gt;$\approx$&lt;/span&gt; 0.2 Mbar and 2,200K at the outer ice/hydrogen–helium boundary. I point out here, that shockwave experiments on these liquids, as well as theoretical studies, imply that the &lt;span class=&quot;math-container&quot;&gt;$\rm H_2\rm O$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\rm{NH}_3$&lt;/span&gt; in the ice layer are almost totally ionized, and the &lt;span class=&quot;math-container&quot;&gt;$\rm{CH}_4$&lt;/span&gt; has been pyrolysed to carbon, possibly in the metallic or diamond form.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; In the decades since this work was published in 1981 have these predictions been addressed further, or even proven or disproven?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If this turns out to be true&lt;/em&gt;, would these diamonds have different physical properties than those on Earth because of the different formation mechanism? &lt;/p&gt;&#xA;" OwnerUserId="26672" LastEditorUserId="7982" LastEditDate="2019-05-04T00:11:09.180" LastActivityDate="2019-05-06T02:03:02.217" Title="Diamonds in Neptune?" Tags="&lt;earth&gt;&lt;neptune&gt;&lt;astrochemistry&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31752" PostTypeId="1" CreationDate="2019-05-03T11:19:08.947" Score="2" ViewCount="97" Body="&lt;p&gt;.. because, how would all the current matter and energy manifest then? Isn't everything coming to complete stop, minimal temperature, zero interaction state impossible? Wouldn't there still be some rare minor events occuring? And if so, given infinite amount of time couldn't these result in events resulting in some kind of chain-reaction?&lt;/p&gt;&#xA;" OwnerUserId="27810" LastActivityDate="2019-05-03T21:59:39.350" Title="Doesn't the energy conservation law imply, that the universe can't come to a Big Freeze?" Tags="&lt;fate-of-universe&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="31753" PostTypeId="1" CreationDate="2019-05-03T14:41:32.710" Score="1" ViewCount="81" Body="&lt;p&gt;In the movie, &lt;a href=&quot;https://en.m.wikipedia.org/wiki/First_Man_%28film%29?wprov=sfla1&quot; rel=&quot;nofollow noreferrer&quot;&gt;First Man&lt;/a&gt; Neil Armstrong and his colleague David Scott, as part of the Gemini 8 program, attempt to dock with the Agena. But it turns out to be disastrous as they go into a continuous &lt;em&gt;roll over&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the need for docking with the Agena and then later undocking so as to land on the moon in a smaller vessel? Why can't there be one rocket which takes the passengers, lands as a single unit and then comes back (much like how an airplane is) without docking and undocking? Wouldn't that be easier?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What caused the continuous roll over and how did the &lt;em&gt;RCS Brakes&lt;/em&gt; help in the stabilization of the vessel?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="26686" LastEditorUserId="1891" LastEditDate="2019-05-03T18:50:03.860" LastActivityDate="2019-05-03T18:50:03.860" Title="Agena docking in First Man" Tags="&lt;the-moon&gt;&lt;space&gt;&lt;nasa&gt;&lt;moonlanding&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2019-05-03T18:48:55.637" ContentLicense="CC BY-SA 4.0" />
  <row Id="31755" PostTypeId="2" ParentId="31713" CreationDate="2019-05-03T17:12:39.137" Score="6" Body="&lt;p&gt;From the observer's point of view 1 million km away, the TNO's apparent angular motion is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{\frac{8~m/s}{10^9~m} = 8 \times 10^{-9}~rad/s = 0.00165~^\circ/h}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the observer at 18 billion km = 120 au is in a circular orbit around the Sun, the orbital period is&#xA;120&lt;sup&gt;3/2&lt;/sup&gt; = 1320 years, making the Sun appear to move the other way at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{\frac{360~^\circ}{1320~y} = 0.27~^\circ/y = 3.1 \times 10^{-5~\circ}/h},$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the TNO's apparent motion relative to the Sun is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{0.00165~^\circ/h + 3.1 \times 10^{-5~\circ}/h = 0.00168~^\circ/h}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun would be totally occulted for&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{\frac{0.126~^\circ - 0.004~^\circ}{0.00168~^\circ/h} = 72.6~h}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and partially occulted for&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{\frac{0.004~^\circ}{0.00168~^\circ/h} = 2.4~h}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at each end.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-05-05T01:57:05.540" LastActivityDate="2019-05-05T01:57:05.540" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31756" PostTypeId="1" AcceptedAnswerId="31766" CreationDate="2019-05-03T21:18:39.443" Score="7" ViewCount="2153" Body="&lt;p&gt;In the news recently it's been said that in 2029 the asteroid called Apophis, aka 'The God of Chaos', will fly by very close to earth. To quote &lt;a href=&quot;https://www.geek.com/news/nasa-giant-god-of-chaos-asteroid-will-fly-past-earth-in-2029-1785300/&quot; rel=&quot;noreferrer&quot;&gt;Geek.com&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 2029, a massive asteroid called 99942 Apophis will fly past our&#xA;  planet at an approximate distance of 19,000 miles within the distance&#xA;  of some orbiting spacecraft.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's said the chances of a collision are about 1 in 100,000, which is small but not not completely infeasible either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The asteroid has a diameter of around 350 meters with a mass of 4×10^10 kg and an average orbital speed of around 30km/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of damage could we expect on earth if it were to collide with us?&lt;/p&gt;&#xA;" OwnerUserId="27933" LastActivityDate="2019-05-04T14:10:37.103" Title="What would happen if asteroid 99942 Apophis collides with the earth?" Tags="&lt;asteroids&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31757" PostTypeId="2" ParentId="31752" CreationDate="2019-05-03T21:59:39.350" Score="2" Body="&lt;p&gt;The current model (&lt;span class=&quot;math-container&quot;&gt;$\Lambda$&lt;/span&gt;CDM) predicts that space will expand exponentially, and keep on doing it indefinitely far into the future. The end result, after the stars burn out, galaxies disperse, protons decay and black holes evaporate is a very thin soup of stable particles (see &lt;a href=&quot;https://arxiv.org/abs/astro-ph/9701131&quot; rel=&quot;nofollow noreferrer&quot;&gt;Adams &amp;amp; Laughlin 1997&lt;/a&gt; or their popular book &lt;em&gt;The Five Ages of the Universe&lt;/em&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since each stable particle finds itself separated by exponentially growing distances due to the expanding space they will not be able to interact. Their density scales as &lt;span class=&quot;math-container&quot;&gt;$\rho(t)\sim \rho_0/a(t)^3 = \rho_0 e^{-3t/H_0}$&lt;/span&gt;. The probability of a collision over time &lt;span class=&quot;math-container&quot;&gt;$dt$&lt;/span&gt; for a particle is roughly &lt;span class=&quot;math-container&quot;&gt;$\sigma v \rho(t) dt$&lt;/span&gt;, which if we integrate to infinity becomes &lt;span class=&quot;math-container&quot;&gt;$\Pr[\text{any  collision}]=\sigma v \rho_0 \int_0^\infty e^{-3t/H_0} dt = \sigma v \rho_0 H_0/3 $&lt;/span&gt; which is finite - there will only be a finite number of interactions for each particle, ever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point temperatures are extremely low since all background radiation has been redshifted to undetectable levels, and only the very cold horizon radiation remains (&lt;span class=&quot;math-container&quot;&gt;$\approx 10^{-30}$&lt;/span&gt; K). Particles end up in a equilibrium state with this heat bath, and this is the real heat death of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This low temperature is still enough to rarely excite the particles, or generate new particles... &lt;a href=&quot;https://physics.stackexchange.com/questions/434877/can-the-universe-build-itself-back-together-after-the-heat-death/434886#434886&quot;&gt;or Boltzmann brains or entire worlds&lt;/a&gt;. Maybe. Eternity is a very long time, so low-probability processes will happen if they are allowed. They will however be so rare that the effects of one will typically dissipate &lt;em&gt;long&lt;/em&gt; before anything else happens.&lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2019-05-03T21:59:39.350" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31758" PostTypeId="2" ParentId="20560" CreationDate="2019-05-03T23:26:38.543" Score="1" Body="&lt;p&gt;I've implemented what &lt;a href=&quot;https://astronomy.stackexchange.com/questions/20560/how-to-calculate-the-position-of-the-sun-in-long-lat/20585#20585&quot;&gt;Thomas suggested&lt;/a&gt; as an approximation. It seems to do its job for a small rendered globe toy in JavaScript.&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function getSunEuler(date) {&#xA;        const now = date || new Date();&#xA;&#xA;        // The boilerplate: fiddling with dates&#xA;        const soy = (new Date(now.getFullYear(), 0, 0)).getTime();&#xA;        const eoy = (new Date(now.getFullYear() + 1, 0, 0)).getTime();&#xA;        const nows = now.getTime();&#xA;        const poy = (nows - soy) / (eoy - soy);&#xA;&#xA;        const secs = now.getUTCMilliseconds() / 1e3&#xA;                + now.getUTCSeconds()&#xA;                + 60 * (now.getUTCMinutes() + 60 * now.getUTCHours());&#xA;        const pod = secs / 86400; // leap secs? nah.&#xA;&#xA;        // The actual magic&#xA;        const lat = (-pod + 0.5) * Math.PI * 2;&#xA;        const lon = Math.sin((poy - .22) * Math.PI * 2) * .41;&#xA;&#xA;        return new THREE.Euler(0, lat, lon, 'YZX');&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The returned Euler angle can be used to transform an X unit vector to find a point on a sphere where &lt;em&gt;0 N 0 E&lt;/em&gt; is at &lt;em&gt;(1, 0, 0)&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got curious how accurate the solution is and fiddled with &lt;a href=&quot;https://rhodesmill.org/skyfield/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Skyfield&lt;/a&gt; a bit to get closer to the truth:&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from skyfield.api import Topos, load&#xA;ts = load.timescale()&#xA;planets = load('de421.bsp')&#xA;sun = planets['sun']&#xA;earth = planets['earth']&#xA;t = ts.utc(2019, 6, 21, 12, 0, 0)&#xA;&#xA;# A rather correct answer can probably be obtained from&#xA;earth.at(t).observe(sun).apparent().position.au&#xA;# but I ended up being too lazy to calculate the angles myself&#xA;&#xA;# This serves as a convenient workaround for manual verification&#xA;np = earth + Topos('90 N', '180 W')&#xA;np.at(t).observe(sun).apparent().altaz()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The approximation seems to be indeed accurate within a few degrees, but is in an different coordinate system. I'd love it if someone could provide a clean solution using skyfield.&lt;/p&gt;&#xA;" OwnerUserId="26341" LastActivityDate="2019-05-03T23:26:38.543" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31759" PostTypeId="1" AcceptedAnswerId="31762" CreationDate="2019-05-04T02:07:57.360" Score="1" ViewCount="115" Body="&lt;p&gt;C|Net's &lt;a href=&quot;https://www.cnet.com/news/hubble-spots-drop-dead-gorgeous-spiral-galaxy-tucked-into-leo/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble spots drop-dead gorgeous spiral galaxy tucked into Leo&lt;/a&gt; links to NASA's &lt;a href=&quot;https://www.nasa.gov/image-feature/goddard/2019/hubble-spots-stunning-spiral-galaxy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble Spots Stunning Spiral Galaxy&lt;/a&gt; which shows the image below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The caption on the NASA page doesn't mention the color coding, nor does it mention a reference to a technical page for the image's construction or history.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is this close to a &quot;straight RGB&quot; image of NGC 2903, or were filters used at certain wavelengths, possibly including IR or UV, in order to highlight certain regions and emissions, then assigned false colors. If the latter, is it a standard color coding?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The NASA page says: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;NGC 2903 is located about 30 million light-years away in the constellation of Leo (the Lion), &lt;strong&gt;and was studied as part of a Hubble survey of the central regions of roughly 145 nearby disk galaxies.&lt;/strong&gt; This study aimed to help astronomers better understand the relationship between the black holes that lurk at the cores of galaxies like these, and the rugby-ball-shaped bulge of stars, gas and dust at the galaxy’s center — such as that seen in this image. Text credit: ESA (European Space Agency)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Click to view full size!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MRzzb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MRzzb.jpg&quot; alt=&quot;New NGC 2903 Hubble image, wow!&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-05-04T02:22:28.633" LastActivityDate="2019-05-04T15:10:13.800" Title="False-color wavelength assignments in this &quot;drop-dead gorgeous&quot; image of NGC 2903?" Tags="&lt;photography&gt;&lt;hubble-telescope&gt;&lt;image-processing&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31760" PostTypeId="1" AcceptedAnswerId="31770" CreationDate="2019-05-04T03:58:47.117" Score="2" ViewCount="63" Body="&lt;p&gt;I want to start with the fact that I am by no means an astronomer, or even a hobbyist in the field. I am attempting to build a fantasy world which still reflects accepted physical laws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem comes from the fact that I have no idea how to go about figuring out the period of axial precession for a fantasy earth-like planet. For the Earth, I believe the period is roughly ~26,000 years. I've spent a few days trying to find resources, but in  all honesty I can't wrap my head around half of the equations I've seen presented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go about doing this calculation for a given planet, and which of its properties would I have to specify in order to do this calculation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit &lt;em&gt;Important? Details&lt;/em&gt;:&#xA;My Earthlike planet has a mass 1.2597x the mass of our earth, and two fairly large moons (actually, one of them may be large enough to classify it as a binary planet but I'm not sure).&lt;/p&gt;&#xA;" OwnerUserId="27936" LastEditorUserId="27936" LastEditDate="2019-05-04T18:21:44.670" LastActivityDate="2019-05-04T23:09:55.403" Title="How would one figure out the rate of axial precession of a planet?" Tags="&lt;earth-like-planet&gt;&lt;precession&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31761" PostTypeId="1" AcceptedAnswerId="31810" CreationDate="2019-05-04T05:31:48.880" Score="0" ViewCount="94" Body="&lt;p&gt;A &lt;em&gt;really enjoyable interview&lt;/em&gt; with Mansi Kasliwal &amp;amp; Samaya Nissanke (&quot;From discovery to making precision measurements&quot;) is available as a short audio podcast in the BBC News article &lt;a href=&quot;https://www.bbc.com/news/science-environment-48137011&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gravitational waves hunt now in overdrive&lt;/a&gt; describes the very recent gravitational wave events and the way optical searches for the source were initiated by pressing a &quot;go button&quot; on a cell phone!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a excerpt from the BBC's Science in Action episode &lt;a href=&quot;https://www.bbc.co.uk/programmes/w3csym1m&quot; rel=&quot;nofollow noreferrer&quot;&gt;What is behind the Indian Ocean Cyclones?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://www.bbc.com/news/science-environment-48137011&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; is quite a good read as well, and talks about the &lt;a href=&quot;https://www.ztf.caltech.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zwicky Transient Facility (ZTF)&lt;/a&gt;, based on the 48-inch Samuel Oschin Schmidt telescope with several upgrades including a state of the art silicon CCD array replacing the 14 inch square photographic plates it originally used, faster slewing and AI-based recognition of candidate objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The huge aperture allows for almost 50 square degree field to be checked down to +20 magnitude every five minutes. That's about one steradian in 5 hours. According to the podcast, it can run as fast as 4000 square degrees per hour, about 1.2 sr/h presumably with a brighter limiting magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Does the ZTF only start surveying when triggered by a cellphone or other alert, or does it continue to search for new objects and events on some nominal routine whenever there is good visibility?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-05-04T05:59:17.460" LastActivityDate="2019-05-08T22:52:37.293" Title="Does the Zwicky Transient Facility only run when triggered by a cell phone?" Tags="&lt;observational-astronomy&gt;&lt;observatory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31762" PostTypeId="2" ParentId="31759" CreationDate="2019-05-04T06:02:37.570" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://www.spacetelescope.org/images/potw1917a/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Spacetelescope.org&lt;/a&gt;&#xA;indicates under &quot;Colours &amp;amp; filters&quot; that the blue channel of the image is from a 658 nm filter (red)&#xA;and the red channel is from an 814 nm filter (near infrared).&#xA;The cyan and orange rows of the table also list these filters; presumably the green channel is a blend of the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.stsci.edu/hst/acs/documents/handbooks/current/c05_imaging2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ACS instrument handbook&lt;/a&gt; says filter F658N is narrow, passing only NII or slightly redshifted H&amp;alpha; emission (Table 5.1, Figure 5.4).&#xA;Filter F814W is wide, passing wavelengths between 710 nm and 960 nm (Table 5.2, Figure 5.1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NGC 2903 is one of the brighter galaxies not in the Messier catalog.&#xA;Here is a wider angle, more natural RGB image by &lt;a href=&quot;http://bf-astro.com/ngc2903/ngc2903.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bob Franke&lt;/a&gt;, rotated for easier comparison.&#xA;The reddish emission nebulae here appear blue in the HST image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JHDtw.jpg&quot; alt=&quot;NGC 2903 by Bob Franke&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-05-04T15:10:13.800" LastActivityDate="2019-05-04T15:10:13.800" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31763" PostTypeId="2" ParentId="31751" CreationDate="2019-05-04T12:34:43.367" Score="5" Body="&lt;p&gt;People have &lt;a href=&quot;https://escholarship.org/uc/item/1zd805nx&quot; rel=&quot;noreferrer&quot;&gt;demonstrated high-pressure diamond formation using laser-driven shocks&lt;/a&gt;, much in line with the original idea. The diamond rain idea seems to be doing well (&lt;a href=&quot;https://www.americanscientist.org/article/on-neptune-its-raining-diamonds&quot; rel=&quot;noreferrer&quot;&gt;popular science article about it&lt;/a&gt;, with references). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is quite likely that the diamonds could have different properties because of formation mechanisms. There are &lt;a href=&quot;https://en.wikipedia.org/wiki/Allotropes_of_carbon#Other_possible_forms&quot; rel=&quot;noreferrer&quot;&gt;a lot of unusual allotropes of carbon&lt;/a&gt;, and local conditions of formation (not to mention impurities) may cause the carbon to turn into these forms rather than the standard diamond lattice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.pnas.org/content/103/5/1204.long&quot; rel=&quot;noreferrer&quot;&gt;Simulations show that beyond 1000 GPa diamond has a BC8 structure&lt;/a&gt;. However, this is a pressure unlikely to occur inside Neptune's atmosphere: it occurs deep in the mantle, so sinking diamonds may shift crystal form. &lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2019-05-04T12:34:43.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31764" PostTypeId="1" AcceptedAnswerId="31765" CreationDate="2019-05-04T13:37:49.180" Score="2" ViewCount="1884" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Two-body_problem_in_general_relativity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;According to Einstein's theory of general relativity, particles of&#xA;  &lt;strong&gt;negligible mass&lt;/strong&gt; travel along geodesics in the space-time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does that &quot;negligible mass&quot; only refer to particles or objects like a spacecraft, whose mass are literally negligible to that of a black hole, are also taken into account?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the mass of a spacecraft is not negligible, how has one to modify the geodesics equation corresponding to that mass?&lt;/p&gt;&#xA;" OwnerUserId="7843" LastActivityDate="2019-05-05T23:56:26.770" Title="What does &quot;negligible mass&quot; mean in the formulation of geodesics equation?" Tags="&lt;general-relativity&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31765" PostTypeId="2" ParentId="31764" CreationDate="2019-05-04T14:02:37.353" Score="5" Body="&lt;p&gt;Yes its means that you ignore the effect that an object such as a spacecraft has on the central mass that it is orbiting, that the orbiting body has infinitely less mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I think there is no exact solution found of the geodesics equations when that approximation is no longer valid or when you try to devolop geodesics equation for a spacecraft under the influence of two (or more) spherically symmetric mass distributions such as the combination of planets and the Sun in our solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the solar system JPL who calculates orbits of planets and other celestial objects use what is called the &quot;post Newtonian expansion&quot;. You can look at their official documentation, &lt;a href=&quot;http://Formulation%20forObserved%20and%20Computed%20Values%20ofDeep%20Space%20Network%20Data%20Typesfor%20Navigation&quot; rel=&quot;noreferrer&quot;&gt;Formulation for Observed and Computed Values of Deep Space Network Data Types for Navigation&lt;/a&gt;, expression 4-26 on page 4-19. Some of the the terms may be zero if you only have two non-neglibly massive objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this is not really a &quot;geodesics equation&quot; per se. Nasa/JPL uses what I believe is a first order expansion of the Schwarzshild solution in isotropic coordinate and then uses a scheme I do not quite follow to get to the relativistic acceleration terms of the post-Newtonian that they add to the classical Newtonian gravitational acceleration term to calculate the orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know if there is a simpler way to do it.&lt;/p&gt;&#xA;" OwnerUserId="26403" LastActivityDate="2019-05-04T14:02:37.353" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31766" PostTypeId="2" ParentId="31756" CreationDate="2019-05-04T14:10:37.103" Score="3" Body="&lt;p&gt;If it hits the deep ocean, not much relatively speaking. In the recent past, people wildly exaggerated the capability of an asteroid (at least ones in the size range of Apophis) to cause massive tsunamis; in reality, the ocean efficiently and quickly absorbs the impact. That said, it'd still be terrifying to see such an impact in deep water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it hits a city, it would cause violent devastation, immediately killing 10's or 100's of thousands with its initial impact, and millions with the resulting shock wave and debris.&lt;/p&gt;&#xA;" OwnerUserId="22083" LastActivityDate="2019-05-04T14:10:37.103" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31767" PostTypeId="1" CreationDate="2019-05-04T14:43:42.423" Score="2" ViewCount="426" Body="&lt;p&gt;The earth is tilted at 23.5 degrees which means that in the solstices the north/south hemisphere is more radiated by the sun and the tilt to the sun is maximum.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the equinoxes, the tilt is not towards the sun but to the &quot;side&quot; so the equator has direct exposure to the sun. As a result, we have 12 hours on day and night on the equinoxes.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand is why is it only for these 2 days (equinoxes) that this happens. In the next day(s) I think the equator should have the same exposure to the sun as the equinoxes. Has the tilt changed so much?&lt;br&gt;&#xA;I couldn't find any animation that shows the exact path of the tilt throughout the year.&lt;br&gt;&#xA;Can someone please help me understand this?&lt;/p&gt;&#xA;" OwnerUserId="25200" LastEditorUserId="25994" LastEditDate="2019-05-04T15:33:30.440" LastActivityDate="2020-06-26T02:59:27.990" Title="Why is the equinoxes only 2 days and not more?" Tags="&lt;the-sun&gt;&lt;earth&gt;&lt;ecliptic&gt;&lt;equinox&gt;&lt;solstice&gt;" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31768" PostTypeId="2" ParentId="31767" CreationDate="2019-05-04T15:29:38.980" Score="8" Body="&lt;p&gt;First, don't think that equinox is when the day and night are the same. It is the moment, when the sun's declination is zero, or when the sun is directly over the equator. Let me explain ...&lt;/p&gt;&#xA;&lt;p&gt;Earth is always rotating around the Sun, so &lt;a href=&quot;https://en.wikipedia.org/wiki/Declination&quot; rel=&quot;nofollow noreferrer&quot;&gt;declination&lt;/a&gt; of the Sun is always changing (except at the solstices when it stops for a minute and goes in the other way). As @JohnHolst has pointed out, Sun's declination is changing for 1 angular minute per hour (mark 1 mm on the 3 m of distance – this is nearly 1 angular minute). An equinox doesn't last for a day. It lasts for a moment when the Earth is totally aligned and straight with the sun:&#xA;&lt;em&gt;(this figure shows the geocentric model, which is easier to represent and imagine)&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OZffQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OZffQ.png&quot; alt=&quot;declination = 0&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Also, Wikipedia says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;An equinox is commonly regarded as the &lt;strong&gt;instant of time&lt;/strong&gt; when the plane (extended indefinitely in all directions) of Earth's equator passes through the center of the Sun. This occurs twice each year, around 20 March and 23 September. &lt;strong&gt;In other words, it is the &lt;em&gt;moment&lt;/em&gt; at which the center of the visible Sun is directly above the equator.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;P. S.&lt;/p&gt;&#xA;&lt;p&gt;You can look &lt;a href=&quot;https://astro.unl.edu/classaction/animations/coordsmotion/eclipticsimulator.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this simulation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="25994" LastEditorUserId="25994" LastEditDate="2020-06-25T14:22:55.433" LastActivityDate="2020-06-25T14:22:55.433" CommentCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="31769" PostTypeId="2" ParentId="31767" CreationDate="2019-05-04T16:38:41.827" Score="5" Body="&lt;p&gt;The easiest graph to consider is the graph of day-length. If the day-length varies from 8 to 16 hours, there can only be two equinoxes:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/XmFEw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XmFEw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the time of equinox, the day length changes at the fastest rate, by a few minutes in temperate regions, compared to the solstices, which change by a few seconds every day. So, the equinox doesn't exist perfectly, the night time will be 11.57, 12.01, 12.04... It can be off by a minute, but the next day, it is changed by a few minutes...&#xA;march 20th: &#xA;&lt;a href=&quot;https://www.timeanddate.com/sun/france/paris?month=3&amp;amp;year=2019&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.timeanddate.com/sun/france/paris?month=3&amp;amp;year=2019&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't try to understand with words, try to understand with images and objects in your hands: This image shows you. On the equinox, does the north pole and the south pole face away from the sun? &lt;a href=&quot;https://youtu.be/tX3Y5bzNDiU?t=35&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/tX3Y5bzNDiU?t=35&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't over think it, rotations are confusing, just have a basic idea that the north pole becomes parallel with the sun on the equinox, and points away from the sun in winter, and towards the sun in summer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you can pick up an apple or a fruit and put it near a lightbulb / lamp and do the movement mechanics for yourself. try youtube for &quot;seasons earth animation&quot; &lt;a href=&quot;https://youtu.be/WLRA87TKXLM?t=5&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/WLRA87TKXLM?t=5&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="27938" LastEditorUserId="27938" LastEditDate="2019-05-10T08:05:48.880" LastActivityDate="2019-05-10T08:05:48.880" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="31770" PostTypeId="2" ParentId="31760" CreationDate="2019-05-04T23:09:55.403" Score="1" Body="&lt;p&gt;Wikipedia's page on &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;Axial Precession&lt;/a&gt; has a good deal of mathematics on it and, unless you have absolutely got to have precise numbers for some obscure story reason (which is probably what I'd call the tail wagging the dog - change the story to avoid that problem) then making up the numbers or just forgetting all about Axial Precession is the way to go.    The periods involved will be very long by the standards of any story you're likely to write, so why burden yourself with something you probably don;t need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maths that follows only gives a rough approximation anyway, and you don't want to even contemplate the kind of things you need to do to get a better one : it's not worth it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said let's have a look at the very basic theory result that Wikipedia gives :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two component to axial precession that matter (for Earth) : the one due to the Moon and the one due to the Sun.  The Moon's effect is actually larger, but these numbers are very sensitive to the values you use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Solar Contribution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{d\psi}{dt}=\left[ \frac {GM_s}{a_s^3\left(1-e_s^2\right)^\frac 3 2} \right] \left[ \frac {C-A}C \frac {cos\epsilon}\omega \right]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lots of symbols so what do they mean ?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; - the Universal Gravitational Constant also famous from &lt;span class=&quot;math-container&quot;&gt;$F = \frac {GM_1M_2} {r^2}$&lt;/span&gt; Newtons law for gravitation.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$M_s$&lt;/span&gt; - The Sun's mass - in your case you need the mass of your planet's star, of course.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$a_s$&lt;/span&gt; - The semi-major axis of the orbit of the planet around it's star&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$e_s$&lt;/span&gt; - The eccentricity of the planet's orbit around it's star.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now that second term in square bracket, which is also in the expression for the Lunar contribution.  This one is trickier.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; - moment of inertia (of Earth) around the axis of rotation&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; - moment of inertia around the equator&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\epsilon$&lt;/span&gt; - the angle between the equatorial plane and the ecliptic plane (see below)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; - Earth's angular velocity (due to it's rotation, not it's orbit)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now this expression is really poorly dealt with in Wikipedia because of two problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$epsilon$&lt;/span&gt; in Wikipedia is assumed to be the same for both the Solar and Lunar contributions.  This is not (AFAIK) correct.  The angle should be the angle between axis of rotation of the body and the plane of the orbit of the other body (which means it's different for the Sun and Moon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;span class=&quot;math-container&quot;&gt;$C-A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; terms are really hard to deal with for mere mortals (and frankly just messy for anyone else).  For your purposes I would propose the following compromise term instead.  It's based on modeling the planetary bulge as an ellipsoid of constant density compared with the &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; value for an ideal sphere - both objects have the same mass and density, which I'm taking as constant .  I'll spare you the derivation :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac {C-A} C \approx 1 - \frac{R^2} {a^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where in this case &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; is the &lt;em&gt;average&lt;/em&gt; radius of the planet and &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is the equatorial radius of the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Lunar Contribution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{d\psi}{dt}=\left[ \frac {GM_l}{a_l^3\left(1-e_l^2\right)^\frac 3 2} \left(1-\frac 3 2 sin^2i\right) \right] \left[ \frac {C-A}C \frac {cos\epsilon}\omega \right]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not much change here expect that the masses and so on refer to the Moon and not the Sun (hence the different subscripts).  There is one additional term which is the factor :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\left(1-\frac 3 2 sin^2i\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This corrects for the effect that the angle of inclination of Moon's orbit to the ecliptic is not zero.  The ecliptic being the plane with the Sun and the Earth's orbit in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to decide these numbers for yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The total effect :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total effect is simply the sum of the two other effects so :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac {d\psi}{dt} = \frac {d\psi_l}{dt} + \frac {d\psi_s}{dt}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had multiple moons you would need multiple lunar correction terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for clarification that &lt;span class=&quot;math-container&quot;&gt;$\frac {d\psi}{dt}$&lt;/span&gt; means the rate of change of the angle &lt;span class=&quot;math-container&quot;&gt;$\psi$&lt;/span&gt; with respect to time &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt;.  To get how big an angle you'd move through in a century you do this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta \psi \approx \frac {d\psi}{dt} \Delta t$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="10196" LastActivityDate="2019-05-04T23:09:55.403" CommentCount="1" CommunityOwnedDate="2019-05-04T23:09:55.403" ContentLicense="CC BY-SA 4.0" />
  <row Id="31772" PostTypeId="1" CreationDate="2019-05-05T10:45:42.127" Score="2" ViewCount="51" Body="&lt;p&gt;How to get ellipticities of a list of random galaxies. I saw few programs written in SQL and they were only giving stokes parameters and isophotes values for ONE galaxy(for which we input the ObjID). I am not familiar with syntax of SQL. Is there any method to put the objid in a loop so that the program generates stoke parameters values for a list of galaxies? &lt;/p&gt;&#xA;" OwnerUserId="27937" LastActivityDate="2019-05-17T12:04:16.083" Title="How to get ellipticities of galaxies from SDSS using casjobs?" Tags="&lt;galaxy&gt;&lt;galaxy-cluster&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31773" PostTypeId="2" ParentId="31725" CreationDate="2019-05-05T12:45:27.997" Score="3" Body="&lt;p&gt;You can get some equipment especially designed for the eclipse.&lt;/p&gt;&#xA;&lt;p&gt;But don't forget about a simple &lt;a href=&quot;https://www.timeanddate.com/eclipse/make-pinhole-projector.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;pinhole camera&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Important to note that you shouldn't look through the pinhole. Instead, you are using the pinhole to project an image of the sun on the ground behind you.&lt;/p&gt;&#xA;&lt;p&gt;The gaps between leaves in a tree act as natural pinholes so if there are any trees around you, you should be able to see lots of &lt;a href=&quot;https://whyy.org/articles/see-the-solar-eclipse-using-natures-pinhole-camera-tree-leaves/&quot; rel=&quot;nofollow noreferrer&quot;&gt;eclipse images&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;One of the easiest ways to safely watch a solar eclipse is to use 2 sheets of cardboard and make your own simple pinhole projector.&lt;/p&gt;&#xA;&lt;p&gt;A partial solar eclipse projected on cardboard. Solar eclipses can look spectacular, but the Sun’s UV radiation can cause permanent eye damage or even blindness. Make your own pinhole projector to view a solar eclipse.&lt;/p&gt;&#xA;&lt;p&gt;Projecting the Sun through a box projector, &lt;strong&gt;or projecting using binoculars or telescope&lt;/strong&gt; (edit: shown below), or simply 2 pieces of card is a safe and easy way to view a solar eclipse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.timeanddate.com/eclipse/make-pinhole-projector.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fHyWQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fHyWQ.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Light filtering through the leaves of trees during a solar eclipse creates images of the crescent sun, like the image created by a pinhole camera. (Bigstock)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://whyy.org/articles/see-the-solar-eclipse-using-natures-pinhole-camera-tree-leaves/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bEkle.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bEkle.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="27904" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-05-07T05:00:07.780" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31774" PostTypeId="2" ParentId="31725" CreationDate="2019-05-05T15:08:39.060" Score="3" Body="&lt;p&gt;The Eclipse will be visible from the La Serena Beach (weather permitting). The eclipse will start at 1522, but totality won't be reached until 1638. Totality will last for about 2 minutes. However, there is a higher risk of clouds and/or fog on the coast. Places in the interior, like Vicuna, have better weather prospects, but the eclipse is low so the positioning is more critical if you don't want the sun to be hidden behind mountains. The Argentina side is also an option, and has relatively dry regions near Bella Vista / San Juan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total Solar Eclipses attract lots of visitors, so finding accommodation or transport can be difficult. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted in comments &lt;strong&gt;viewing the partial eclipse without eye protection can cause permanent damage to the retina, you should ensure that you have proper eye protection.&lt;/strong&gt; You should organise this before you leave. You can have a look at NASA's website, concerning &lt;a href=&quot;https://eclipse2017.nasa.gov/safety&quot; rel=&quot;nofollow noreferrer&quot;&gt;Safety&lt;/a&gt; when viewing eclipses. Also, they published a list of &lt;a href=&quot;https://eclipse.aas.org/resources/solar-filters&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reputable Vendors&lt;/a&gt; of eclipse glasses.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="7982" LastEditDate="2019-05-07T01:00:47.080" LastActivityDate="2019-05-07T01:00:47.080" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31775" PostTypeId="1" AcceptedAnswerId="31778" CreationDate="2019-05-05T19:59:09.737" Score="24" ViewCount="5599" Body="&lt;p&gt;I'm researching a scene for a sci-fi novel in which the near-future protagonists observe earth through a station-mounted telescope in Mars orbit. My goal is to understand how much detail they reasonably could discern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hubble telescope is probably a reasonable comparison for my purposes. I found images Hubble took of Mars during a close approach, but I don't know if those represent the best resolution possible or simply the resolution that was selected or available at the time, or indeed, if the whole planet was imaged at a higher resolution than the photo published in popular media outlets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can a Hubble-like telescope observe significantly greater detail than displayed in the article below? If so, what might reasonably be resolved? Large cities? Individual buildings?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Vox.com's &lt;a href=&quot;https://www.vox.com/science-and-health/2018/7/27/17621488/hubble-telescope-photo-mars-opposition-saturn-rings&quot; rel=&quot;noreferrer&quot;&gt;Hubble can see galaxies impossibly far away. Here’s what happens when it looks at Mars and Saturn.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/omKtD.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/omKtD.jpg&quot; alt=&quot;Mars, ASA, ESA, and STScI&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Cropped from &lt;a href=&quot;https://www.vox.com/science-and-health/2018/7/27/17621488/hubble-telescope-photo-mars-opposition-saturn-rings&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt; NASA, ESA, and STScI&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K49Ev.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K49Ev.jpg&quot; alt=&quot;Mars, NASA/Hubble&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Cropped from &lt;a href=&quot;https://www.vox.com/science-and-health/2018/7/27/17621488/hubble-telescope-photo-mars-opposition-saturn-rings&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt; NASA/Hubble&lt;/p&gt;&#xA;" OwnerUserId="1315" LastEditorUserId="7982" LastEditDate="2019-05-05T23:45:16.297" LastActivityDate="2019-05-07T03:31:31.683" Title="What detail can Hubble see on Mars?" Tags="&lt;telescope&gt;&lt;mars&gt;&lt;hubble-telescope&gt;&lt;angular-resolution&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31776" PostTypeId="2" ParentId="31775" CreationDate="2019-05-05T20:29:24.217" Score="10" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble_Space_Telescope&quot; rel=&quot;noreferrer&quot;&gt;Hubble space telescope&lt;/a&gt; has a 2.4m mirror and is pretty much diffraction limited, so at near-UV wavelengths of say 240nm it has an angular resolution of about &lt;span class=&quot;math-container&quot;&gt;$10^{-7}$&lt;/span&gt; radians. Mars' &lt;a href=&quot;https://mars.nasa.gov/allaboutmars/nightsky/mars-close-approach/&quot; rel=&quot;noreferrer&quot;&gt;closest distance to Earth&lt;/a&gt; is about 54.6 million km, so the theoretical minimal resolution is between 5 ad 6 km. So large cities might be visible if they have lots of contrast.  Since at closest approach an observer on Mars is looking straight at the night side of the Earth, a well lit city might be easy to spot, on the other hand, they are also looking more or less straight at the Sun, which might give some problems. Still there will be times when the distance is only 70 or 80 million km and the angle from the Sun is more manageable, so a resolution of 10km in the UV, 20 in visible light is credible.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-05-05T20:29:24.217" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31777" PostTypeId="1" CreationDate="2019-05-05T23:00:18.227" Score="3" ViewCount="306" Body="&lt;p&gt;Since dimensions usually refer to space and we naturally acknowledge three of them, we also perceive time and we separate space from time (&quot;space AND time&quot;), why is it, that time has ended up in the same shelf as spatial dimensions. Is time really a dimension &quot;by nature&quot;? Or is it a concesus, shifting our natural understanding what dimensions are, for practical reasons, similarly to defining particles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: by &lt;code&gt;Is time really a dimension &quot;by nature&quot;?&lt;/code&gt; I asked if there are good intrinsic reasons, given by to label time as a dimension, as opposed to us labeling it as such, to make it more tangible, explainable. &lt;/p&gt;&#xA;" OwnerUserId="27810" LastEditorUserId="27810" LastEditDate="2019-05-06T19:43:19.733" LastActivityDate="2019-05-10T06:44:09.520" Title="Why is time considered a (fourth) dimension?" Tags="&lt;space-time&gt;" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="31778" PostTypeId="2" ParentId="31775" CreationDate="2019-05-05T23:07:55.510" Score="34" Body="&lt;p&gt;Forget about magnification. People who know telescopes don't think in terms of magnification. What matters is the angular resolution, or the resolving power: the angular size of the smallest details that you could see in an instrument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rule of thumb: the resolving power of a telescope with a diameter of 10 cm is 1 arcsecond when using visible light. The numbers are inversely proportional. A 20 cm telescope resolves details 0.5 arcsec in size. A 1 meter telescope resolves 0.1 arcsec.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hubble has an aperture (diameter) of 2.4 m, so its resolving power is 0.04 arcsec.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimum distance between Earth and Mars is about 55 million km and it only happens very rarely. The maximum distance is 400 mil km. The &quot;average&quot; distance is 225 mil km (but actual distance varies all the time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's apply the tangent of 0.04 arcsec at 55 mil km:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.wolframalpha.com/input/?i=tan(0.04%20arcseconds)%20*%2055000000&quot; rel=&quot;noreferrer&quot;&gt;https://www.wolframalpha.com/input/?i=tan(0.04+arcseconds)+*+55000000&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's 10 km. It would only be able to see the major geographic features.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see buildings (down to the scale of 10 m), it would need a 1000x increase in resolution. That means an aperture of 2.4 km. None of the classic telescope designs can provide that. It would have to be some kind of interferometric design - a large, flat field where several mirrors are placed several km apart and are coupled optically to function as a single huge mirror (well, sort of - this is more of an intuitive explanation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be similar to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Navy_Precision_Optical_Interferometer&quot; rel=&quot;noreferrer&quot;&gt;Navy Precision Optical Interferometer&lt;/a&gt; near Flagstaff, Arizona.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mBCVJ.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mBCVJ.jpg&quot; alt=&quot;NPOI&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the wide, flat parts of Valles Marineris might provide a good location for the interferometer. Acidalia Planitia would provide even more space for building huge interferometers, and should be a good place to build structures in general - flat to beyond horizon; it's the place where much of the book/movie The Martian set their story. But any big, reasonably flat field would work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the above assumes the distance of closest approach between Earth and Mars. In practice, the distance is greater than that, so aperture must increase. You're contemplating an interferometer with a base of dozens of km if you want to distinguish structures such as buildings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conceivably, the interferometer could be built in orbit, but you must ensure that the distance between mirrors is maintained with extraordinary precision. On the planetary surface, the ground provides the required rigidity. In space you'd have to... I dunno, use space magic.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2019-05-07T03:31:31.683" LastActivityDate="2019-05-07T03:31:31.683" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="31779" PostTypeId="2" ParentId="31764" CreationDate="2019-05-05T23:56:26.770" Score="0" Body="&lt;p&gt;Something of a tautology. Objects of negligible mass have a mass that can be neglected. By definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it another way, if you have decided to not bother calculating in the mass of some object, because you have decided for whatever reason that it's small enough that it won't significantly affect the result of some calculation, then you would compute that that object travels along a geodesic.&lt;/p&gt;&#xA;" OwnerUserId="27947" LastActivityDate="2019-05-05T23:56:26.770" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31780" PostTypeId="1" CreationDate="2019-05-06T00:45:26.953" Score="2" ViewCount="219" Body="&lt;p&gt;If there are supposedly galaxies, or whole other parts of the universe, that we can't ever see due to them receding away from us faster than the speed of light, then why isn't or hasn't this occurred with the cosmic microwave background?&lt;/p&gt;&#xA;" OwnerUserId="27948" LastActivityDate="2019-05-08T00:59:25.030" Title="Why hasn't the CMB already receded away faster than the speed of light?" Tags="&lt;expansion&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="31781" PostTypeId="2" ParentId="31751" CreationDate="2019-05-06T02:03:02.217" Score="1" Body="&lt;p&gt;There are probably a variety of factors.   How fast they form, how fast they sink through the atmosphere.   Like rain, there's likely a part in the atmosphere where they form and grow, and a part, deeper in the atmosphere where the temperature gets too high and the diamonds dissolve again into plasma, perhaps releasing some free/charged carbon to return upwards and be part of the cycle all over again.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I was to guess, I would guess that the diamonds formed in gas giant atmospheres would be much more impure.   The interior of the Earth is relatively slow moving and consistent in temperature, often cooling very slowly.   It's the same process which helps form veins of metals form, which are followed in mining.   This very gradual cooling in Earth's core probably plays a role in the purity of diamonds.    Earth's core also has very little nitrogen,    Diamonds formed in Earth are about 99.95% carbon atoms on average, with nitrogen, often the primary impurity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.gia.edu/diamond-description&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.gia.edu/diamond-description&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.dianerdiamonds.com/diamond_info/nitrogen_in_diamonds&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.dianerdiamonds.com/diamond_info/nitrogen_in_diamonds&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relative abundance of nitrogen available in gas giant planets suggests to me that on average, gas giant diamonds would be more impure.   Less consistency in temperature during formation and an availability of nitrogen suggests to me that 99.95% pure carbon diamonds are unlikely.   There's also the problem of how large they'd be.   If diamonds form slowly, which seems likely and as they grow larger, they would have a density problem, likely slowly falling deeper into the gas giant's atmosphere, where they would get too hot and likely become plasma and dissipate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I was to guess, I'd guess that the diamonds that form inside the 4 gas giant planets in our solar-system are more along the grain of sand to perhaps raindrop size and probably quite impure.   But I'm mostly guessing.   There might be a planet out there somewhere, (perhaps a mini Neptune) with just the right mix of lower gravity and possible suspension and temperature where it forms grapefruit or basketball sized diamonds of high purity.   It's certainly possible, but my guess is that diamonds inside our solar-systems 4 gas giants would be pretty uninteresting, though the chemical and gas-giant geological processes that form them would still be interesting, but I don't think the diamonds would stand out as gem quality or worth harvesting.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2019-05-06T02:03:02.217" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31782" PostTypeId="1" AcceptedAnswerId="31786" CreationDate="2019-05-06T03:01:24.767" Score="25" ViewCount="2549" Body="&lt;p&gt;There are very roughly a &quot;mole&quot; of stars in the universe. &lt;a href=&quot;https://en.wikipedia.org/wiki/Star&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; quotes an estimate of &lt;span class=&quot;math-container&quot;&gt;$3 \times 10^{23}$&lt;/span&gt; though the number is associated with some debate and uncertainly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know if there are estimates of when the number of stars in the universe will maximize. Is it expected to increase asymptotically to some maximum, or will it peak and then decrease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose this could depend on what the definition of &quot;star&quot; is taken to be, if brown or black dwarf objects are counted or not. I don't want to pre-specify, it's more likely that a good, well-informed answer will include this information.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-05-07T00:43:36.997" LastActivityDate="2019-05-07T00:43:36.997" Title="When will the number of stars be a maximum?" Tags="&lt;cosmology&gt;&lt;stellar-evolution&gt;&lt;galactic-dynamics&gt;&lt;fate-of-universe&gt;&lt;modeling&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31783" PostTypeId="1" CreationDate="2019-05-06T05:54:37.267" Score="2" ViewCount="226" Body="&lt;p&gt;I read about Hawking and Leonard Susskind and their debates about whether information is destroyed or not if it falls into a black hole.  It was resolved in favor of information always being conserved, preserved, whatever (i.e. not destroyed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see how this could be consistent with determinism.  However, how is this in keeping with quantum physics, when there is so much randomness.  It seems that at any moment, all matter, energy, even every atom in our bodies, is indeterminate and &quot;scrambled&quot; somewhat.  Even if you could pinpoint or describe every atom and its position perfectly in your body at one moment, by the next moment there is uncertainty.  So how is this compatible with conservation of information?&lt;/p&gt;&#xA;" OwnerUserId="27948" LastEditorUserId="37191" LastEditDate="2021-02-12T20:50:22.927" LastActivityDate="2021-02-12T20:50:22.927" Title="Law of Conservation of Information?" Tags="&lt;black-hole&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31786" PostTypeId="2" ParentId="31782" CreationDate="2019-05-06T08:11:19.717" Score="20" Body="&lt;p&gt;&lt;strong&gt;TL; DR Somewhere between now and a few hundred billion years time. (For a co-moving volume)&lt;/strong&gt; Now read on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If stellar remnants are included, then the answer is very far in the future indeed, if and when the constituents of baryons begin to decay. So let's assume  that &quot;stars&quot; means those things that are undergoing nuclear fusion reactions to power their luminosity. Let's further assume that the stellar mass function, (&lt;span class=&quot;math-container&quot;&gt;$N(m)$&lt;/span&gt; is the number of stars per unit mass) we see in the neighbourhood of the Sun is representative of populations in all galaxies at all times (difficult to make a start, without assuming this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of stars that have been born is equal to the sum over time (the integral) and over mass of &lt;span class=&quot;math-container&quot;&gt;$N(m)$&lt;/span&gt; multiplied by the rate at which mass is turned into stars in a comoving volume of the universe &lt;span class=&quot;math-container&quot;&gt;$\Phi(t)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then need to subtract a sum over time and mass of the rate of stellar death in the same comoving volume. The rate of stellar death is the rate of stellar birth at a time &lt;span class=&quot;math-container&quot;&gt;$t-\tau(m)$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\tau(m)$&lt;/span&gt; is the mass-dependent stellar lifetime. We ignore mass transfer in binary systems and assume that multiples can be treated as independent stellar components. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the number of stars at time &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; is approximately&#xA;&lt;span class=&quot;math-container&quot;&gt;$$N_*(t) = \int_0^{t} \int_m N(m) \Phi(t') - N(m)\Phi(t'-\tau(m))\ dm\ dt'\ .$$&lt;/span&gt;&#xA;To find where this is a maximum, we differentiate with respect to time and then equate to zero. i.e. We look for the time when the stellar birth and death rates are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was going to (and possibly still will) attempt some sort of analytic approximation, but &lt;a href=&quot;https://arxiv.org/abs/1403.0007&quot; rel=&quot;noreferrer&quot;&gt;Madau &amp;amp; Dickinson (2014)&lt;/a&gt; have done it better and taken into account the metallicity dependence of stellar lifetimes and the chemical evolution of galaxies. The star formation rate peaked about 10 billion years ago, is more than an order of magnitude lower now and is exponentially decreasing with a time constant of 3.9 billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integrated stellar mass is shown in their Fig 11 (shown below). It is still increasing today, but at a very low rate and has &lt;em&gt;not&lt;/em&gt; passed through a maximum. The reason for this is that most stars have masses of 0.2-0.3 solar masses and lifetimes much longer than the age of the universe. Even if these stars are added at a very slow rate, their death rate is zero at present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vHOvT.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vHOvT.png&quot; alt=&quot;Integrated stellar mass density (from Madau &amp;amp; Dickinson 2014)&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If star formation did continue at a low-level then the number of stars would only begin to significantly diminish once the stars near the peak of the stellar mass function, that were born at the earliest times, start to die. The lifetime of a 0.25 solar mass star is around a trillion years (&lt;a href=&quot;https://iopscience.iop.org/article/10.1086/304125&quot; rel=&quot;noreferrer&quot;&gt;Laughlin et al. 1997&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand if star formation ceased &lt;em&gt;now&lt;/em&gt; then the number of stars would immediately begin to diminish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps we could argue that the current exponential decline will continue and the peak &#xA;will come in another few billion years when stars of 0.8-0.9 solar masses begin dying off. However, that is futurology given that we have no first principles theory that explains the time-dependence of star formation, so I believe the best answer that can be given is somewhere between now and a few hundred billion years time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this answer assumes a co-moving volume. If the question asked is phrased in terms of the observable universe then because the number of stars has nearly reached a plateau, then the answer becomes close to whatever age the volume of the observable universe is maximised. I say &quot;close to&quot; because you have to factor in that the observable universe includes stars in distance slices at all cosmic epochs. I am unwilling to undertake this horrendous calculation, but note that the current concordance cosmological model has our observable universe slowly increasing from around a radius of 45 billion light years now, to about 60 billion light years in the far future &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0310808&quot; rel=&quot;noreferrer&quot;&gt;Davis &amp;amp; Lineweaver 2005&lt;/a&gt;, and this may compensate for a slow decline in the number of stars in a co-moving volume.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-05-06T17:56:41.453" LastActivityDate="2019-05-06T17:56:41.453" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31787" PostTypeId="1" CreationDate="2019-05-06T08:43:09.703" Score="0" ViewCount="54" Body="&lt;p&gt;Is it proven that the frequency of EMR does not decay (at all, ever), and if not, could the observations that led to Hubble's Law simply be that the frequency of light decays over very large durations/distances?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it does not decay innately simply due to duration, what about the various interactions it would have with particles / fields / &quot;empty space&quot; over very large distances; could these have a (reducing) effect on the frequency?&lt;/p&gt;&#xA;" OwnerUserId="27952" LastActivityDate="2019-05-06T08:43:09.703" Title="Could Hubble's Law simply be light's frequency decaying over time?" Tags="&lt;light&gt;&lt;hubble-constant&gt;" AnswerCount="0" CommentCount="5" ClosedDate="2019-05-06T13:58:08.443" ContentLicense="CC BY-SA 4.0" />
  <row Id="31788" PostTypeId="1" CreationDate="2019-05-06T11:58:35.387" Score="2" ViewCount="64" Body="&lt;p&gt;Asteroids and smaller celestial objects tend to be odd-shaped, whereas planets are always spherical (or slightly oblate due to their spin). The reason obviously is that for sufficiently large masses/sizes the gravitational force overwhelms any resistance of even rocky materials to maintain a shape other than a sphere (which minimises the gravitational energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The characteristic scale at which this transition occurs depends on the material and is different for ice and rock as mentioned in &lt;a href=&quot;https://astronomy.stackexchange.com/a/11924/633&quot;&gt;this answer&lt;/a&gt; (to a &lt;a href=&quot;https://astronomy.stackexchange.com/questions/2092/what-is-the-minimum-mass-required-so-that-objects-become-spherical-due-to-its-ow&quot;&gt;related question&lt;/a&gt;) which also links &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrostatic_equilibrium#Planetary_geology&quot; rel=&quot;nofollow noreferrer&quot;&gt;this useful Wikipedia article&lt;/a&gt;. I wonder whether this limit can be estimated from simple physics (involving material constants for ice/rock).&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2019-05-06T13:19:45.283" LastActivityDate="2019-05-08T18:02:27.840" Title="A quantitive estimate of the size above which celestial bodies must be near-spherical" Tags="&lt;planet&gt;&lt;gravity&gt;&lt;asteroids&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2019-05-06T14:50:52.543" ContentLicense="CC BY-SA 4.0" />
  <row Id="31789" PostTypeId="2" ParentId="31777" CreationDate="2019-05-06T22:04:18.013" Score="2" Body="&lt;p&gt;(The speed of light) times (time) has the properties of a linear dimension.  Just like length, width, or height, it can be used to measure how far you can go.  The Lorentz contraction is consistent with applying the Pythagorean theorem to it and the other dimensions.&lt;/p&gt;&#xA;" OwnerUserId="10491" LastActivityDate="2019-05-06T22:04:18.013" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31790" PostTypeId="1" AcceptedAnswerId="31792" CreationDate="2019-05-06T22:24:31.147" Score="3" ViewCount="168" Body="&lt;p&gt;I assume that for space telescopes to get good images, they must be put in perfect standstill, as even the slightest deviation in the viewing angle is amplified by the distance of observed objects, results in observing a different area of space. How is this feat accomplished? How can you stabilize an instrument in space, so that it can take images of galaxies gazillion light years away? And how do you know you have succeeded, apart from getting quality imagery?&lt;/p&gt;&#xA;" OwnerUserId="27810" LastActivityDate="2019-05-07T00:10:42.897" Title="How are space telescopes stabilised to a perfect standstill?" Tags="&lt;photography&gt;&lt;space-telescope&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="31791" PostTypeId="2" ParentId="31725" CreationDate="2019-05-06T23:19:03.593" Score="0" Body="&lt;h2&gt;The wrong way&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vFDu8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vFDu8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Possibly the right way (adult supervision required)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9zdIh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9zdIh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-05-06T23:19:03.593" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31792" PostTypeId="2" ParentId="31790" CreationDate="2019-05-06T23:51:33.813" Score="3" Body="&lt;h2&gt;Pointing&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In some ways it's easier to keep the direction that a telescope is pointing stable in orbit in space than it is on Earth. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There's no wind &lt;a href=&quot;https://astronomy.stackexchange.com/q/29116/7982&quot;&gt;Why is the opening in the Anglo-Australian Telescope's dome so small?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;You aren't on a spinning planet requiring you to constantly move the telescope with electric motors and gears&lt;/li&gt;&#xA;&lt;li&gt;there's no sagging due to gravity&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are tiny torques due to differential solar photon pressure on different parts of a space telescope, but these are &lt;em&gt;small, very reproducible and can be learned, predicted, and compensated.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The James Webb Space Telescope has a giant sunshield that keeps the infrared telescope cold, and this can produce substantial torque. So it has a second reflector used to passively null out most of the torque. You can read more about it in answers to &lt;a href=&quot;https://space.stackexchange.com/q/35399/12102&quot;&gt;How will JWST manage solar pressure effects to maintain attitude and station keep it's unstable orbit?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In low Earth orbit (LEO) telescopes like Hubble will also experience tiny torques due to tidal forces; for example, one end of Hubble is in a higher or lower orbit than the other end. Without damping, this could cause it to slowly oscillate &lt;a href=&quot;https://space.stackexchange.com/a/33781/12102&quot;&gt;see this answer&lt;/a&gt;. Again, this will be &lt;em&gt;small, very reproducible and can be learned, predicted, and compensated.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can somehow stop something in space from rotating, it will simply stay stopped and point in a stable direction, at least for a short time. If you can predict the torques due to solar pressure and gravitational tidal forces, you can apply tiny controlling torques to minimize those effects and it will stay pointed longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Sensitive gyroscopes to measure tiny rates of rotation (inertial stabilization)&lt;/li&gt;&#xA;&lt;li&gt;Watch the stars at the edge of the telescope's field of view, using them as guide stars&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Both are shown in the image below, taken from &lt;a href=&quot;https://www.nasa.gov/content/goddard/hubble-space-telescope-pointing-control-system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble Space Telescope Pointing Control System&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objects labeled &quot;Gyros&quot; are gyroscopes used to measure tiny rates of rotation of the telescope and send those measurement to the attitude control system for the telescope. Systems can use real spinning wheels, or fiber optic rings with light going in both directions, or other techniques, but the result is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently Hubble is working without all three gyroscopic sensors on-line because it has redundant systems: You can read more about that in &lt;a href=&quot;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20080023344.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble Space Telescope Reduced-gyro control law design, implementation, and on-orbit performance&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objects labeled Fine Guidance Sensors use CCDs at the edge of the Hubble's field of view to track stars:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The FGSs use starlight captured by the telescope’s mirrors to find and maintain a lock on guide stars to ensure that the spacecraft’s attitude does not change. One FGS can also be used as a scientific instrument to determine a star’s position with high accuracy. The level of stability and precision that the FGSs provide gives Hubble the ability to remain pointed at a target with no more than 0.007 arcsecond of deviation over extended periods of time. This is the same as holding a laser beam focused on Franklin D. Roosevelt’s head on a dime roughly 200 miles away — which is about the distance from Washington, D.C., to the Empire State Building in New York City — for 24 hours.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Feed this and other bits of information from other star cameras into the computer, combine it with the predicted torques from tidal forces and photon pressure, and the computer decides on a tiny correction torque necessary to keep it pointed stably. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That message then goes to the objects labeled as Reaction Wheels. These are four heavy, spinning wheels whose angular momentum is balanced such that it normally applies no torque to the telescope. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the computer senses that a little torque is needed to keep the telescope pointed in the right direction, or to move it to a new field, motors on the spinning reaction wheels will speed them up or slow them down a little bit, and that will apply a compensating torque to the telescope.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Vibrations&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Vibrations and oscillations of a space telescope's structure can also be a problem. These are addressed through very careful engineering and special materials. Structures are design to minimize vibration and oscillation by damping, isolation of moving parts (pumps, solar panels, etc.) and by use of light weight and stiff materials like silicon carbide to move resonant frequencies as high as possible where they are most easily damped and isolated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silicon carbide is a very popular material in newer space telescopes and is found in the optical system of too man of them for me to remember, but here is mention of it in GAIA's optical bench &lt;a href=&quot;https://space.stackexchange.com/a/18237/12102&quot;&gt;answer&lt;/a&gt; and in New Horizion's LORRI telescope &lt;a href=&quot;https://astronomy.stackexchange.com/a/25276/7982&quot;&gt;answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MFmOF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MFmOF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Gaya's Silicon Carbide Optical Bench, with the two objective mirrors of it's twin telescopes pointing 106.5° apart. From &lt;a href=&quot;http://spaceflight101.com/gaia/gaia-instrument-information/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Spaceflight 101&lt;/a&gt;, image credit: ESA/Astrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PKujo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PKujo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; The LORRI telescope is described for example in the ArXiv preprint &lt;a href=&quot;https://arxiv.org/abs/0709.4278&quot; rel=&quot;nofollow noreferrer&quot;&gt;Long-Range Reconnaissance Imager on New Horizons&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; From &lt;a href=&quot;https://www.nasa.gov/content/goddard/hubble-space-telescope-pointing-control-system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble Space Telescope Pointing Control System&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/72Las.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/72Las.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-05-07T00:10:42.897" LastActivityDate="2019-05-07T00:10:42.897" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31793" PostTypeId="1" AcceptedAnswerId="31795" CreationDate="2019-05-07T00:42:19.573" Score="2" ViewCount="447" Body="&lt;p&gt;Words escape me, but by &quot;whole universe&quot; (I think) I mean everything that's spatially connected to the observable universe in a conventional sense. If there is a better term for it, please let me know!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was surprised to find out that theories and measurements about the big bang seem to mostly relate to the size of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Observable_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;observable universe&lt;/a&gt; via expansion, and if I understand correctly don't say much about the size of the whole universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have I got this right? Are there any ways at all to try to estimate size of the &lt;em&gt;whole&lt;/em&gt; universe? Does the term even mean anything?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See @RobJeffries' &lt;a href=&quot;https://astronomy.stackexchange.com/a/31786/7982&quot;&gt;excellent answer&lt;/a&gt; to &lt;em&gt;When will the number of stars be a maximum?&lt;/em&gt; and @Acccumulation's &lt;a href=&quot;https://astronomy.stackexchange.com/questions/31782/when-will-the-number-of-stars-be-a-maximum#comment57862_31782&quot;&gt;comment&lt;/a&gt; for background on this question.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-05-10T05:31:00.533" Title="Are there ways to estimate size of the &quot;whole universe&quot;?" Tags="&lt;universe&gt;&lt;cosmology&gt;&lt;cosmological-inflation&gt;&lt;cosmological-horizon&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31794" PostTypeId="1" AcceptedAnswerId="31797" CreationDate="2019-05-07T00:58:07.467" Score="2" ViewCount="230" Body="&lt;p&gt;The Wikipedia subsection &lt;a href=&quot;https://en.wikipedia.org/wiki/Observable_universe#Misconceptions_on_its_size&quot; rel=&quot;nofollow noreferrer&quot;&gt;Observable Universe; Misconceptions on its size&lt;/a&gt; shows this image of a plaque, with the caption:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An example of the misconception that the radius of the observable universe is 13 billion light-years. This plaque appears at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Rose_Center_for_Earth_and_Space&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rose Center for Earth and Space&lt;/a&gt; in New York City.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Rose_Center_for_Earth_and_Space&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Rose Center for Earth and Space is a part of the American Museum of Natural History in New York City. The Center's complete name is The Frederick Phineas and Sandra Priest Rose Center for Earth and Space. The main entrance is located on the northern side of the museum on 81st Street near Central Park West in Manhattan's Upper West Side. Completed in 2000, it includes the new Hayden Planetarium, the original of which was opened in 1935 and closed in 1997. &lt;strong&gt;Neil deGrasse Tyson is its first and, to date, only director.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That last line is the operative part. &lt;strong&gt;Dr. Tyson is a world-renowned stickler for truth and accuracy, and so a problem in is own back yard needs to be investigated.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What &lt;em&gt;exactly&lt;/em&gt; is wrong with this plaque photographed in 2011? &lt;/li&gt;&#xA;&lt;li&gt;What should it have said?&lt;/li&gt;&#xA;&lt;li&gt;If something is indeed wrong, has it been fixed?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wj6Uz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wj6Uz.jpg&quot; alt=&quot;Incorrect plaque at the Rose Center for Earth and Space, April 2011&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Incorrect_plaque_at_the_Rose_Center_for_Earth_and_Space,_April_2011.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Incorrect plaque at the Rose Center for Earth and Space, April 2011&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; Rose Center for Earth and Space &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Rose_Center_for_Earth_and_Space.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt; which &lt;em&gt;literally&lt;/em&gt; contains the famous Hayden Planetarium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/peN0V.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/peN0V.jpg&quot; alt=&quot;Rose Center for Earth and Space which literally contains the Hayden Planetarium.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; I just noticed that even in 2011 the plaque had already been amended (at least) once!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nEMyA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nEMyA.jpg&quot; alt=&quot;Correction to Rose Center for Earth and Space which literally contains the Hayden Planetarium.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-05-07T10:10:59.200" LastActivityDate="2019-05-07T10:32:18.653" Title="What exactly is wrong with this plaque in Neil deGrasse Tyson's &quot;back yard&quot;? Has it been fixed?" Tags="&lt;cosmology&gt;&lt;observable-universe&gt;&lt;cosmological-horizon&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="31795" PostTypeId="2" ParentId="31793" CreationDate="2019-05-07T01:51:04.633" Score="5" Body="&lt;p&gt;&lt;strong&gt;tl; dr&lt;/strong&gt; The universe is probably infinite, but if that's the case it's impossible to verify. If the universe is finite, and small enough, and the global curvature is equal to the curvature of our observable universe, then we will be able to estimate its size.&lt;/p&gt;&#xA;&lt;hr&gt;&#xA;&lt;p&gt;If the global curvature of the universe isn't positive, then the size of the universe is infinite, (and it's always been infinite since the dawn of time, at the Big Bang), assuming the topology of the universe is trivial.&lt;/p&gt;&#xA;&lt;p&gt;Measurements of the observable universe indicate that the curvature may be zero (giving a flat universe), or even negative; if the curvature is positive, then its value is &lt;em&gt;very&lt;/em&gt; small. We assume that the observable universe is typical of the whole universe, but of course that's impossible to verify.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Shape_of_the_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia says&lt;/a&gt; that&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;experimental data from various independent sources (WMAP, BOOMERanG,&#xA;and Planck for example) confirm that the observable universe is flat&#xA;with only a 0.4% margin of error.&lt;/p&gt;&#xA;&lt;p&gt;[...]&lt;/p&gt;&#xA;&lt;p&gt;The latest research shows that even the most powerful future&#xA;experiments (like SKA, Planck..) will not be able to distinguish&#xA;between flat, open and closed universe if the true value of&#xA;cosmological curvature parameter is smaller than &lt;span class=&quot;math-container&quot;&gt;$10^{-4}$&lt;/span&gt;. If the true&#xA;value of the cosmological curvature parameter is larger than &lt;span class=&quot;math-container&quot;&gt;$10^{-3}$&lt;/span&gt; we&#xA;will be able to distinguish between these three models even now.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So even if the total universe &lt;em&gt;does&lt;/em&gt; have positive curvature, its size is much larger than the observable universe, assuming that our observable universe isn't a patch that's abnormally flat.&lt;/p&gt;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://scienceblogs.com/startswithabang/2012/07/18/how-big-is-the-entire-universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;How Big is the Entire Universe?&lt;/a&gt; by Ethan Siegel, if the curvature is positive, then the diameter of the total universe is at least 14 trillion lightyears, but that article is from 2012, more recent calculations may give a larger value.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-05-07T06:49:17.800" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31797" PostTypeId="2" ParentId="31794" CreationDate="2019-05-07T10:32:18.653" Score="5" Body="&lt;p&gt;I see 2 candidates: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the universe is &lt;a href=&quot;https://www.space.com/24054-how-old-is-the-universe.html&quot; rel=&quot;noreferrer&quot;&gt;13.8 billion years&lt;/a&gt; old, so should be rounded off to 14&lt;/li&gt;&#xA;&lt;li&gt;the cosmic horizon is &lt;a href=&quot;https://en.wikipedia.org/wiki/Particle_horizon&quot; rel=&quot;noreferrer&quot;&gt;46 billion ly&lt;/a&gt; away&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="361" LastActivityDate="2019-05-07T10:32:18.653" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31798" PostTypeId="1" CreationDate="2019-05-07T15:15:32.223" Score="14" ViewCount="2591" Body="&lt;p&gt;It is Ramzan and the usual heated disagreements between clerics here in Pakistan as to the sighting of the crescent. This time the minister of Science and Technology (who is not a scientist himself btw) &lt;a href=&quot;https://www.dawn.com/news/1480849/fawad-chaudhry-forms-committee-to-resolve-moonsighting-controversy&quot; rel=&quot;noreferrer&quot;&gt;has intervened and formed a committee&lt;/a&gt; comprising of meteorologists and our space agency. This committee is tasked to form a 5 year lunar calendar, specially one that will fix the issue of moon sighting of major religious months and events such a Ramzan, Hajj (pilgrimage), Eids etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is that: Is it scientifically possible to predict (simulate...) in future with 100% accuracy as to the location of the moon? My surface reading of &lt;a href=&quot;https://en.wikipedia.org/wiki/Lyapunov_time&quot; rel=&quot;noreferrer&quot;&gt;Lyapunov time&lt;/a&gt; suggests that the solar system can be predicted with accuracy for 5 million years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know your understand on this matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also some commenters mentioned that this question might be a duplicate of an other one. I would like to add that there are school of thoughts and in one the condition is that the crescent is to be observed from the surface of the Earth. So I think this question is unique in that sense that it includes that component too.&lt;/p&gt;&#xA;" OwnerUserId="27974" LastEditorUserId="27974" LastEditDate="2019-05-07T16:32:06.893" LastActivityDate="2019-05-08T06:38:43.213" Title="Is it possible to do moon sighting in advance for 5 years with 100% accuracy?" Tags="&lt;orbit&gt;&lt;the-moon&gt;&lt;amateur-observing&gt;&lt;moon-phases&gt;&lt;star-gazing&gt;" AnswerCount="2" CommentCount="14" FavoriteCount="3" ClosedDate="2019-05-08T06:45:25.817" ContentLicense="CC BY-SA 4.0" />
  <row Id="31799" PostTypeId="2" ParentId="31783" CreationDate="2019-05-07T16:32:59.397" Score="1" Body="&lt;p&gt;The real reason of conservation of information in quantum mechanics is that the von Neumann entropy &lt;span class=&quot;math-container&quot;&gt;$S(\rho) = -Tr(\rho \log_2 \rho)$&lt;/span&gt; of an isolated quantum system with density operator &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; does not change in time. This happens because &lt;span class=&quot;math-container&quot;&gt;$S(\rho)$&lt;/span&gt; only depends on the spectrum of &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt;, and the unitarity of time evolution preserves the spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you start out with a perfectly certain state it will evolve unitarily, and remain perfectly certain - but realistically, there are going to be subspaces you did not measure and their uncertainty are going to &quot;leak out&quot; and get mixed with the state. This is basically &lt;a href=&quot;https://en.wikipedia.org/wiki/No-hiding_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;the no hiding theorem&lt;/a&gt; which states that when your system interacts with the outside world and decoheres the information about the original state ends up on the outside and cannot be recovered by operating on just the internal state (but it is indeed conserved for the whole system). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that none of this depends on uncertainty per se nor the issue of what measurement does, but rather that the deterministic effects of the Schrödinger equation/unitary evolution makes things dependent on degrees of freedom you have little control over.&lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2019-05-07T16:32:59.397" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31800" PostTypeId="2" ParentId="31798" CreationDate="2019-05-07T17:01:52.160" Score="24" Body="&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; this answer was posted under duress; though I mentioned in a comment under the question that I was composing an answer, several users have decided to close the question out from under me. Therefore I've put this together a little hastily. It's late here and I'll come back in the morning to address any questions or requests for clarification, and probably add some more information.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it scientifically possible to predict (simulate...) in future with 100% accuracy as to the location of the moon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nothing is 100% accurate or possible. But it is highly likely that the apparent position of the Moon from a given geographic position on Earth can be predicted to very high precision over the next five years. Each aspect:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Geometry&lt;/li&gt;&#xA;&lt;li&gt;Geography&lt;/li&gt;&#xA;&lt;li&gt;Time&lt;/li&gt;&#xA;&lt;li&gt;Meteorology/atmospherics&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;is predictable based on well-understood scientific principles which have been tested against a substantial extent of data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without a specific specification in the question, I can present these aspects of the problem and give an example of the precision to which they can be predicted.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Geometry&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We have records of bits and pieces of the Moon's motion for several thousand years from accounts of eclipses, and several hundred years with much more precision with respect to time and to other stars from records of lunar occultations of stars as seen through telescopes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combine this with laser ranging measurements of the moon which fit theoretical predictions to &lt;strong&gt;centimeter accuracy over a period longer than forty years&lt;/strong&gt;, and it seems that the relative positions of the Earth and Moon are probably going to continue to be within a few centimeters for the next five years as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read further about that in &#xA; - &lt;a href=&quot;https://arxiv.org/abs/gr-qc/0411113&quot; rel=&quot;noreferrer&quot;&gt;Progress in Lunar Laser Ranging Tests of Relativistic Gravity&lt;/a&gt;&#xA; - &lt;a href=&quot;http://tmurphy.physics.ucsd.edu/papers/rop-llr.pdf&quot; rel=&quot;noreferrer&quot;&gt;Lunar laser ranging: the millimeter challenge&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the position of the Earth relative to the Sun is a different matter. If you want to read about how these calculations are done you can have a look &lt;a href=&quot;https://space.stackexchange.com/a/23409/12102&quot;&gt;at this answer&lt;/a&gt; to &lt;em&gt;How to calculate the planets and moons beyond Newtons's gravitational force?&lt;/em&gt; These days the movements of the Sun, Earth, planets, and other bodies with enough mass to affect their motions substantially are know extremely well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at one of the major &lt;a href=&quot;https://en.wikipedia.org/wiki/Jet_Propulsion_Laboratory_Development_Ephemeris&quot; rel=&quot;noreferrer&quot;&gt;JPL Development Ephemerides&lt;/a&gt;. In the abstract of &lt;a href=&quot;https://naif.jpl.nasa.gov/pub/naif/generic_kernels/spk/planets/de430_and_de431.pdf&quot; rel=&quot;noreferrer&quot;&gt;The Planetary and Lunar Ephemerides&#xA;DE430 and DE431&lt;/a&gt; there are a few sentences that give us some idea of what's discussed in this pretty in-depth report:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The planetary and lunar ephemerides DE430 and DE431 are generated by fitting&#xA;  numerically integrated orbits of the Moon and planets to observations. &lt;strong&gt;The present-day&#xA;  lunar orbit is known to submeter&lt;/strong&gt; accuracy through fitting lunar laser ranging data with&#xA;  an updated lunar gravity field from the Gravity Recovery and Interior Laboratory (GRAIL)&#xA;  mission. &lt;strong&gt;The orbits of the inner planets are known to subkilometer accuracy&lt;/strong&gt; through fitting radio tracking measurements of spacecraft in orbit about them. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Geography&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The visibility of a crescent moon from a given spot on Earth needs more than the positions of the center of masses of the Moon, Earth, and Sun. We need the position of the viewer. These days we can get GPS coordinates of the site, but those coordinates are fixed to the Earth, and the Earth rotates around its axis and its axis nutates and wobbles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That motion is pretty small on the scale of five years, you can see the image below showing the motion of the Tropic of Cancer each year.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Time&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The rotational speed of the Earth increases and decreases in a somewhat unsteady way, and this is why the placement of leap seconds looks a bit uneven. However, the fact that we are talking about one second makes it clear that the uncertainty over five years is likely to be a handful of seconds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, &lt;strong&gt;this may be the single biggest risk in the accuracy of the prediction five years into the future!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There could be a major seismic event, magma movement, or big chunk of ice that falls into the ocean. Specific details about how these can affect the rotational speed of the Earth (and the exact direction of its axes) need to be addressed in detail in separate questions, but this could be an upset. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course we would know about it, and a request for an updated five year report could be issued. The time error wouldn't suddenly jump to some big number, it would slowly start accumulating, and there would be time to recalculate before it became significant.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Atmospherics and Meteorology&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This is the hardest one. &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_refraction&quot; rel=&quot;noreferrer&quot;&gt;Refraction from the atmosphere&lt;/a&gt; can move the apparent position of objects significantly, up to a fraction of a degree. But the amount of refraction is fairly well predictable. It depends a little bit on the Weather because index of refraction's deviation from 1.0 depends almost linearly on the density of the air, and slightly on the humidity. But this is a very small correction on top of a small correction.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Libration&quot; rel=&quot;noreferrer&quot;&gt;Lunar Libration&lt;/a&gt; image from &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Lunar_libration_with_phase2.gif&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lLTA5.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lunar Laser Ranging images &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment&quot; rel=&quot;noreferrer&quot;&gt;from here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/j7j38.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j7j38.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/DGsBA.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DGsBA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iqHwP.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iqHwP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Tr%C3%B3pico_de_C%C3%A1ncer_en_M%C3%A9xico_-_Carretera_83_(V%C3%ADa_Corta)_Zaragoza-Victoria,_Km_27%2B800.jpg&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Mrpe3.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mrpe3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:BennettAtmRefractVsAlt.png&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-05-07T17:16:11.033" LastActivityDate="2019-05-07T17:16:11.033" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31801" PostTypeId="1" AcceptedAnswerId="31949" CreationDate="2019-05-07T17:19:07.853" Score="2" ViewCount="321" Body="&lt;p&gt;As seen in &lt;a href=&quot;https://astronomy.stackexchange.com/a/31723/26403&quot;&gt;this answer&lt;/a&gt; and also in &lt;a href=&quot;http://www.eftaylor.com/exploringblackholes/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this book&lt;/a&gt; chapter 6, page 6-10 expression 22,  when you are falling in radially from being at rest at infinity your velocity in coordinate time, as seen from a distant observer, can be described by the expression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{dr}{dt} = -\left(1 - \frac{r_s}{r}\right)\left(\frac{r_s}{r}\right)^{1/2}c $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume ”&lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt;” to be the real radial distance and not just the ”r-parameter of Schwarzschild coordinates” this is the velocity &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt;. I want to know the expression for the acceleration as a function of ”&lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt;” and ”&lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt;” for a small object of mass &lt;span class=&quot;math-container&quot;&gt;$m&amp;lt;&amp;lt;M$&lt;/span&gt; moving purely radially inwards or outwards from a compact spherically symmetric mass distribution in the general case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By taking the derivative with respect to &quot;r&quot; of the expression above I get:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{dv}{dr}=\frac{0.5}{r}{\sqrt{\frac{r_s}{r}}(1-\frac{3r_s}{r})}c$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From these two expressions we see that the maximum velocity is &lt;span class=&quot;math-container&quot;&gt;$v=\frac{2}{3\sqrt{3}}c$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$r=3r_s$&lt;/span&gt;. From this finding, &lt;span class=&quot;math-container&quot;&gt;$\frac{dv}{dt}$&lt;/span&gt; should not be so hard but this will not be a general expression but only the acceleration of something dropped from rest at infinity. In Schwarzschild coordinates the radial velocity of light goes as &lt;span class=&quot;math-container&quot;&gt;$v_{light}=c(1-r_s/r)$&lt;/span&gt;. At &lt;span class=&quot;math-container&quot;&gt;$r=3r_s$&lt;/span&gt; this becomes &lt;span class=&quot;math-container&quot;&gt;$v_{light}=2c/3$&lt;/span&gt; so when falling in from being at rest at infinity you reach maximal velocity when &lt;span class=&quot;math-container&quot;&gt;$v/v_{light}=1/\sqrt{3}$&lt;/span&gt;. Maybe you can say that gravitation becomes repulsive whenever you are travelling towards the center of the gravitational field faster than &lt;span class=&quot;math-container&quot;&gt;$1/\sqrt{3}$&lt;/span&gt; times the local speed of light?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: What does the general expression for the acceleration of an object in coordinate time look like for an object moving purely radially inwards/outwards from the center of the gravitational field assuming a spherically symmetric non-spinning mass distribution and that the r-parameter of the Schwarzschild coordinates is the real physical distance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for an expression for the instantaneous acceleration as a function of &quot;r&quot; and &quot;v&quot;. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I know that in the weak fields of our solar system Nasa/JPL are using this expression to calculate Newtonian plus relativistic acceleration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\left(1-\frac{4GM}{rc^2}+\frac{v^2}{c^2}\right)\hat{r} +\frac{4GM}{r^2}\left(\hat{r}\cdot \hat{v}\right)\frac{v^2}{c^2}\hat{v}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is based upon a first order expansion of the Schwarzshild solution in isotropic coordinates. For pure radial infall you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\left(1-\frac{4GM}{rc^2}-3\frac{v^2}{c^2}\right)\hat{r} $$&lt;/span&gt; which becomes repulsive whenever you are approaching the central mass faster than: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$v=\frac{c}{\sqrt{3}}\sqrt{1-\frac{4GM}{rc^2}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am hoping for an exact solution for pure radial acceleration in &quot;Schwarzschild coordinate space&quot; and coordinate time similar to the approximate solution for pure radial acceleration you can get from the Nasa equation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As per the answer by amateurAstro down below the expression should be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{dv}{dt}=-\frac{GM}{r^2}\left(1-2\frac{v^2}{c^2(1-\frac{2GM}{rc^2})}-\frac{v^2}{c^2(1-\frac{2GM}{rc^2})^2}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is consistent with stuffing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\gamma=\frac{1}{\sqrt{1-\frac{2GM}{rc^2}-\frac{v^2}{c^2\left((1-\frac{2GM}{rc^2})(\hat{r}\cdot\hat{v})^2+|\hat{r}\times\hat{v}|^2\right)}}}\frac{1}{\sqrt{1-\frac{2GM}{rc^2}}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;into&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{d(m\gamma\bar{v})}{dt}=-\frac{GMm\gamma}{r^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to get (at least this is true for pure radial and pure non-radial motion):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\left(\hat{r}-2\frac{v^2(\hat{r}\cdot\hat{v})\hat{v}}{c^2(1-\frac{2GM}{rc^2})} -\frac{v^2(\hat{r}\cdot\hat{v})\hat{v}}{c^2(1-\frac{2GM}{rc^2})^2}\left((\hat{v}\cdot\hat{r})^2+(1-\frac{2GM}{rc^2}-\frac{v^2}{c^2})|\hat{v}\times\hat{r}|^2\right)\right)$$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am trying to find a velocity-dependent term that together with the classical Newtonian acceleration should reproduce Schwarzshild orbits exactly)&lt;/p&gt;&#xA;" OwnerUserId="26403" LastEditorUserId="26403" LastEditDate="2019-05-24T08:51:14.957" LastActivityDate="2019-05-24T08:51:14.957" Title="What is the coordinate acceleration for pure radial motion?" Tags="&lt;black-hole&gt;&lt;orbit&gt;&lt;general-relativity&gt;&lt;nasa&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31802" PostTypeId="2" ParentId="31780" CreationDate="2019-05-07T17:51:02.267" Score="4" Body="&lt;p&gt;Let me try this from a building blocks point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine the Universe is basically uniform, that is, it's pretty much the same everywhere at any given time, or, at any specific age.  We think this is essentially true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, imagine that the Universe is ridiculously, impossibly large, much larger than we can see, because when we look at things very far away, we also look back in time and we can only see as far as the universe is old.   We can't see most of the Universe because most of it is much too far away.   Given the observed expansion, we'll never see most of the universe.  The light from objects too distant will never reach us.   We assume there are countless galaxies out there, similar to the galaxies we can see, but we'll never see them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets skip the very early parts when the observable universe was the size of a baseball, or the size of the solar system, because that happened long before the cosmic background radiation and it's not relevant to your question, and we can begin in the, lets say, around 370,000 years of age, when the Universe was much less spread out than it is today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At that age, about 370,000 years after the big bang, the Universe was still hot everywhere, kind of like fire thinly spread out everywhere.   There were no galaxies or stars as they hadn't coalesced yet.  Just spread out clouds of fiery gas still spread out everywhere, but beginning to clump together under gravity.  Basically just hydrogen and helium nuclei and free electrons, hot enough to be in a plasma state, not too different than flame, and flame is opaque.   It's not fully opaque like a solid object, but flame diffuses light, &lt;a href=&quot;https://physics.stackexchange.com/questions/372117/shadow-of-fire-doesnt-exist&quot;&gt;see more details here&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so, 370,000 years of age, the entire universe was kind of a glowing fog.  You couldn't see very far in any direction.  You could maybe see an AU or a few hundred AU before the opaque plasma blocked your view if you happened to be around at that time, but basically everywhere you looked, you'd just see the heat-glow of the hot universe.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This all changed around the time when the universe was 379,000 years old, &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background#Features&quot; rel=&quot;nofollow noreferrer&quot;&gt;during a time called recombination&lt;/a&gt;.   The universe expanded enough and became cold enough that most of the free electrons rejoined with the atomic nuclei and the plasma became atoms.   As a byproduct of this, the universe went from a glowing fog to transparent for the first time.   This didn't happen all at once, it was more of a transition into transparency, but for the purposes of this question we can think of this as happening at 379,000 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, because when we look in any direction, the farther away things are the older they are, so at the edge of this transparency, there's always going to be the cosmic background radiation, that was true shortly after the 379,000 year recombination and it's been true ever since.   What has changed, is the expansion of the universe keeps stretching the wavelength of the CMB and it keeps moving further away as we move forward in time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What a person would actually see, when the Universe was 380,000 years old, is the cosmic background radiation in a sphere around them, about 1000 light years distant and in visible light, because the expansion wouldn't have affected it much over just 1000 years.   The CMB is kind of like a sphere of time in addition to a sphere of background radiation.   It's the furthest thing we can ever see in the EM spectrum, about 379,000 years younger than the age of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing to keep in mind about this sphere is the universe is much bigger than we can see.   The glow was everywhere, but the glow we see at any given time is the age of the universe minus 379,000 years old, it's a kind of observable universe bubble, past which we can't see.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first stars didn't show up &lt;a href=&quot;https://news.nationalgeographic.com/2018/02/first-stars-universe-big-bang-edges-space-science/&quot; rel=&quot;nofollow noreferrer&quot;&gt;until about 180 million years later&lt;/a&gt; and the first galaxies, &lt;a href=&quot;https://www.space.com/11386-galaxies-formation-big-bang-hubble-telescope.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;another 20 million years after that&lt;/a&gt;, so it was a long time between recombination and the first stars.  If a person was around 180 million years after the big bang, they could observe the first stars forming, but only stars that were close to them.  Because looking further out looks back in time, so the first stars anyone could observe would be in their neighborhood, but over time, we saw stars form further and further away, but never as far as the CMB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what we can see doesn't change what the largely uniform universe did.  Stars and Galaxies began to form everywhere, both inside and outside our visible edge of the CMB.   Every star and every galaxy is younger than the CMB, but we can infer, even though we'll never see them, that a great many stars and galaxies formed outside or our visible edge of the universe because the universe is much bigger than we can see.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's kind of a step by step overview, perhaps more wordy than it needs to be.   Let me know if that helps.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="6841" LastEditDate="2019-05-08T00:59:25.030" LastActivityDate="2019-05-08T00:59:25.030" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31803" PostTypeId="2" ParentId="31801" CreationDate="2019-05-07T18:53:25.293" Score="2" Body="&lt;p&gt;Edit: This is only the freefall from infinity case; not the general case. More needs to be added.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acceleration can be written as&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{d^2 r}{dt^2} =  v \frac{dv}{dr} = -\left(1 - \frac{r_s}{r}\right)\left(\frac{r_s}{2r^2}\right){(1-\frac{3r_s}{r})}c^2  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acceleration is negative (i.e. &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; becomes more negative) whilst &lt;span class=&quot;math-container&quot;&gt;$r&amp;gt;3r_s$&lt;/span&gt;, changes sign to a &quot;deceleration&quot; when &lt;span class=&quot;math-container&quot;&gt;$r&amp;lt;3r_s$&lt;/span&gt;, reaches a minimum and then increases to zero at &lt;span class=&quot;math-container&quot;&gt;$r = r_s$&lt;/span&gt; (but never gets there since &lt;span class=&quot;math-container&quot;&gt;$v \rightarrow 0$&lt;/span&gt;).&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-05-07T21:29:27.480" LastActivityDate="2019-05-07T21:29:27.480" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31804" PostTypeId="2" ParentId="31798" CreationDate="2019-05-08T06:38:43.213" Score="1" Body="&lt;p&gt;The location is fairly easily possible and even ancient civilizations had managed to do such predictions. In fact for a large part of history large parts of clergy we're committed to just keeping track of time, predicting moon phases, winter/summer solstices and such. Some could even predict eclipses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if it's interpreted that a specific person has to see it and that determines the time something starts, then that is impossible - and it becomes a problem of determining authority rather than anything else. A member of clergy is unlikely to give up such a position of authority as to be the one who has to spot the crescent. However you can still use a calendar to check about what time the specific person should be able to spot it, barring a storm or whatever. The persons importance is greatly diminished though if you don't need him to do the observation anymore and is unlikely to agree to that the sun, earth and moon go on a predictable path.&lt;/p&gt;&#xA;" OwnerUserId="27980" LastActivityDate="2019-05-08T06:38:43.213" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31805" PostTypeId="2" ParentId="14769" CreationDate="2019-05-08T11:33:43.760" Score="2" Body="&lt;p&gt;If you know the angular size of your image (e.g., 10&quot;, 1' etc), then pixel size  (in &quot;/pixel units) is just image size / number of pixels.&#xA;For beam size, depends a lot on the wavelength-range of your observations (for instance sub-mm/radio observations are conceptually very different from UV/optical), but to have an idea of the beam/psf size, you can fit a 2D gaussian/moffat profile to the emission of a point-like object (i.e., a star) in the field of your image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that most of the time these info are contained in the header.&lt;/p&gt;&#xA;" OwnerDisplayName="user19333" LastActivityDate="2019-05-08T11:33:43.760" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31806" PostTypeId="1" CreationDate="2019-05-08T13:11:22.027" Score="2" ViewCount="69" Body="&lt;p&gt;Looking at the big moons of the outer solar system a clear pattern can be seen. Thick icey surface layers covering rocky mantles and deep down metallic cores. The only outlier is Io as it has no icey, but a rocky surface. &lt;strong&gt;Did Io form this way or has Jupiter's tidal heating now powering it's intense vulcanic activity cooked away its oceans?&lt;/strong&gt; After all water vapor could not be retained on Io so if the surface ice where to melt it would escape Io for good. &lt;/p&gt;&#xA;" OwnerUserId="26374" LastActivityDate="2019-05-08T13:11:22.027" Title="Did Io start out as an icemoon?" Tags="&lt;planetary-formation&gt;&lt;io&gt;&lt;atmospheric-escape&gt;&lt;galilean-moons&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31807" PostTypeId="1" CreationDate="2019-05-08T20:24:53.187" Score="-1" ViewCount="282" Body="&lt;p&gt;Is the Universe a sphere with water or ice surrounding its' confinement? Are there any theories that imply the Universe being confined within a sphere of water/ice material?&lt;/p&gt;&#xA;" OwnerUserId="26485" LastActivityDate="2019-05-10T06:48:06.630" Title="What covers the outside of the Universe?" Tags="&lt;universe&gt;&lt;cosmology&gt;&lt;theories&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ClosedDate="2019-05-10T15:36:23.853" ContentLicense="CC BY-SA 4.0" />
  <row Id="31808" PostTypeId="2" ParentId="31807" CreationDate="2019-05-08T20:33:13.163" Score="1" Body="&lt;p&gt;It's very difficult to answer a question &quot;Are there any theories ....&quot; in the negative, since if nothing else, the questioner has probably just formulated such a theory. I can say with reasonable confidence that no such theory has any traction or credibility in any of the communities of research and scholarship that consider such things (astronomy, cosmology, etc.). Most of the theories I am aware of have the universe unconfined and without boundaries, although possibly of finite extent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some consider the possibility of &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Domain_wall_(string_theory)&quot; rel=&quot;nofollow noreferrer&quot;&gt;domain walls&lt;/a&gt;&quot; -- two dimensional surfaces that are vaguely analagous to the boundaries between crystals in a solid -- the fundamental structure of the universe has &quot;crystallised&quot; differently on the two sides.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-05-08T20:33:13.163" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="31809" PostTypeId="2" ParentId="31807" CreationDate="2019-05-08T20:37:01.143" Score="6" Body="&lt;p&gt;No.  The universe is not a sphere with water ice surrounding it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asking about &quot;outside the universe&quot; is akin to asking about what a candle flame is made of &lt;em&gt;before it is lit&lt;/em&gt;.  Just because you can put the words together doesn't mean the question makes sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe is &quot;all that exists&quot;.  Nothing can exist outside the universe, because then it would exist, and if it exists then it is part of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is probably best to imagine the universe as stretching infinitely in all directions. This is not the only possible topology of the universe, but it does help to avoid visualisation of the universe as being like a &quot;sphere&quot; with something outside it. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-05-08T20:37:01.143" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31810" PostTypeId="2" ParentId="31761" CreationDate="2019-05-08T22:52:37.293" Score="2" Body="&lt;p&gt;The ZTF is operating on most clear nights on an automated queue of pointings. When they discover transients, they inform &lt;a href=&quot;https://wis-tns.weizmann.ac.il/search?&amp;amp;name=&amp;amp;name_like=0&amp;amp;isTNS_AT=all&amp;amp;public=all&amp;amp;unclassified_at=0&amp;amp;classified_sne=0&amp;amp;ra=&amp;amp;decl=&amp;amp;radius=&amp;amp;coords_unit=arcsec&amp;amp;groupid%5B%5D=48&amp;amp;classifier_groupid%5B%5D=null&quot; rel=&quot;nofollow noreferrer&quot;&gt;TNS&lt;/a&gt; which may trigger other observatories to point at the same object, just as ZTF may respond to transient events from other observatories. &lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2019-05-08T22:52:37.293" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31811" PostTypeId="1" CreationDate="2019-05-09T05:20:55.527" Score="0" ViewCount="175" Body="&lt;p&gt;How does a gas giant planet form like Saturn and Jupiter, and why just gas? Why didn't they form any solid surface? &lt;/p&gt;&#xA;" OwnerUserId="26672" LastEditorUserId="7411" LastEditDate="2019-06-10T13:17:07.907" LastActivityDate="2019-06-10T13:17:07.907" Title="Birth of the Gas Giant Planets?" Tags="&lt;space&gt;&lt;jupiter&gt;&lt;gas-giants&gt;&lt;saturn&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31813" PostTypeId="1" AcceptedAnswerId="31814" CreationDate="2019-05-09T06:07:55.423" Score="21" ViewCount="2944" Body="&lt;p&gt;A meteorite according to the Oxford English Dictionary is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A piece of rock or metal that has fallen to the earth's surface from outer space as a meteor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But, meteors impact other planets.  Are they still meteorites?  Or is there another word for them.  And, if they are still called meteorites, what is the threshold for an object to be impacted upon by a meteor or meteoroid in this case, and still have the meteoroid be called a meteorite.  Are meteoroids impacting Pluto or Pallas meteorites?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We expect the OED to be authoritative so I am just curious if they have let us down or if the clever space brains have a taxonomy to differentiate them.&lt;/p&gt;&#xA;" OwnerUserId="27983" LastEditorUserId="7472" LastEditDate="2019-05-09T06:24:13.627" LastActivityDate="2020-03-25T23:46:10.207" Title="What is the name of meteoroids which hit Moon, Mars, or pretty much anything that isn’t the Earth?" Tags="&lt;terminology&gt;&lt;meteorite&gt;&lt;meteoroid&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31814" PostTypeId="2" ParentId="31813" CreationDate="2019-05-09T06:23:44.273" Score="18" Body="&lt;p&gt;Oxford English Dictionary is an authoritative source, but it's aimed at the &lt;em&gt;general public&lt;/em&gt;. I think no astronomer will frown upon you when you're using the terms meteoroid and meteorite to describe a small body impacting another body in the solar system. As an example, &lt;a href=&quot;https://en.wikipedia.org/wiki/Meteorite&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; doesn't impose the limitation that it must hit Earth:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A meteorite is a solid piece of debris from an object, such as a comet, asteroid, or meteoroid, that originates in outer space and survives its passage through the atmosphere to reach the surface of a planet or moon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2019-05-09T06:23:44.273" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31815" PostTypeId="2" ParentId="31813" CreationDate="2019-05-09T08:10:24.310" Score="14" Body="&lt;p&gt;Wikipedia's entry for &lt;a href=&quot;https://en.wikipedia.org/wiki/Heat_Shield_Rock&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heat Shield Rock&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Heat Shield Rock is a basketball-sized iron-nickel &lt;strong&gt;meteorite found on Mars&lt;/strong&gt; by the Mars rover Opportunity in January 2005. The meteorite was formally named Meridiani Planum meteorite by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Meteoritical_Society&quot; rel=&quot;nofollow noreferrer&quot;&gt;Meteoritical Society&lt;/a&gt; in October, 2005 (meteorites are always named after the place where they were found)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Meteorical Society publishes the journal &lt;a href=&quot;https://en.wikipedia.org/wiki/Meteoritics_%26_Planetary_Science&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Meteoritics &amp;amp; Planetary&lt;/strong&gt; Science&lt;/a&gt;. &quot;The journal's broad focus is planetary science.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it seems the words meteoritics and meteorite are in fact used without restriction to Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Literature surrounding the discovery of Egg Rock meteorite on Mars calls it a meteorite:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Arizona State University's Red Planet Report website: &lt;a href=&quot;http://redplanet.asu.edu/?p=20977&quot; rel=&quot;nofollow noreferrer&quot;&gt;Curiosity: Egg Rock, a small &lt;strong&gt;meteorite&lt;/strong&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Debora Bird in Earthsky.org: &lt;a href=&quot;http://earthsky.org/space/curiosity-meteorite-mars-october-30-2016&quot; rel=&quot;nofollow noreferrer&quot;&gt;Curiosity Finds Small &lt;strong&gt;Meteorite on Mars&lt;/strong&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Gizmodo: &lt;a href=&quot;https://www.gizmodo.com.au/2016/11/curiosity-just-found-a-freaky-metal-meteorite-on-mars/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Curiosity Just Found&lt;/strong&gt; A Freaky Metal &lt;strong&gt;Meteorite On Mars&lt;/strong&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Images from &lt;a href=&quot;http://earthsky.org/space/curiosity-meteorite-mars-october-30-2016&quot; rel=&quot;nofollow noreferrer&quot;&gt;EarthSky&lt;/a&gt;, credit NASA/JPL/&lt;a href=&quot;http://redplanet.asu.edu/?p=20977&quot; rel=&quot;nofollow noreferrer&quot;&gt;ASU&lt;/a&gt;, captions &quot;October 30, 2016 image via Curiosity rover on Mars&quot; (click for larger size):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7Nuup.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Nuupm.jpg&quot; alt=&quot;&amp;quot;Egg Rock&amp;quot; meteorite on Mars&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/QshE4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QshE4m.jpg&quot; alt=&quot;&amp;quot;Egg Rock&amp;quot; meteorite on Mars&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/gZS9p.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gZS9pm.png&quot; alt=&quot;&amp;quot;Egg Rock&amp;quot; meteorite on Mars&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there is the meteorite named &lt;a href=&quot;https://en.wikipedia.org/wiki/Heat_Shield_Rock&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heat Shield Rock&lt;/a&gt;, also on Mars:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Heat Shield Rock is a basketball-sized iron-nickel &lt;strong&gt;meteorite found on Mars&lt;/strong&gt; by the Mars rover Opportunity in January 2005. The meteorite was formally named Meridiani Planum meteorite by the Meteoritical Society in October, 2005 (meteorites are always named after the place where they were found)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/p1How.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p1How.jpg&quot; alt=&quot;Heat Shield Rock&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/File:PIA07269-Mars_Rover_Opportunity-Iron_Meteorite.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It's not limited to Mars either&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Earth and Space Science News: &lt;a href=&quot;https://eos.org/articles/apollo-may-have-found-an-earth-meteorite-on-the-moon&quot; rel=&quot;nofollow noreferrer&quot;&gt;Apollo May Have Found an Earth &lt;strong&gt;Meteorite on the Moon&lt;/strong&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Science Focus: &lt;a href=&quot;https://www.sciencefocus.com/space/how-often-do-large-meteorites-hit-the-moon/&quot; rel=&quot;nofollow noreferrer&quot;&gt;How often do large &lt;strong&gt;meteorites hit the Moon&lt;/strong&gt;?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Space.com &lt;a href=&quot;https://www.space.com/41356-watch-two-meteorites-hit-the-moon.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Watch Two &lt;strong&gt;Meteorites Hit the Moon&lt;/strong&gt;!&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2020-03-25T23:46:10.207" LastActivityDate="2020-03-25T23:46:10.207" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31817" PostTypeId="1" AcceptedAnswerId="31821" CreationDate="2019-05-09T13:04:54.547" Score="2" ViewCount="510" Body="&lt;p&gt;Sometimes I observed there is moon on sky at noon and sun is also shining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also observed dull moon is still on sky at 7:00 AM to 8:00 AM while sun is rising. My local time is UTC+5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If possible, please explain it in simple words without using advanced astronomy terminology if possible, thanks!&lt;/p&gt;&#xA;" OwnerUserId="27987" LastEditorUserId="7982" LastEditDate="2019-05-09T13:23:08.077" LastActivityDate="2019-05-09T17:32:53.457" Title="How moon is visible during sun is still shining, simple explanation please!" Tags="&lt;the-sun&gt;&lt;the-moon&gt;" AnswerCount="3" CommentCount="9" ClosedDate="2019-05-09T19:03:50.823" ContentLicense="CC BY-SA 4.0" />
  <row Id="31818" PostTypeId="2" ParentId="31817" CreationDate="2019-05-09T13:14:33.300" Score="5" Body="&lt;p&gt;Let's look at what the difference is between day and night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the day time, the air that's between you and the Moon is in the sunlight, but at night, that air is in the dark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Day or night though, the air is still transparent (you can see the Moon through it), and the light from the moon is mostly unaffected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it's in sunlight, the air scatters some of the sunlight in your direction. For the parts of the Moon that are bright, they are so bright that they look white. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the dark parts of the moon don't look black to you on Earth because there's still the fainter blue-sky light scattered by the air.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Moon during the day with bright blue sky light&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eGbmu.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eGbmu.jpg&quot; alt=&quot;Moon during the day&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Moon_day_(32148170671).jpg&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Moon at sunset with faint blue sky light mixed with Earthshine&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4AHKh.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4AHKh.jpg&quot; alt=&quot;Crescent Moon with blue sky&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ast.wikipedia.org/wiki/Ficheru:Crescent_Moon_(2558144570).jpg&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-05-09T13:27:27.427" LastActivityDate="2019-05-09T13:27:27.427" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31819" PostTypeId="2" ParentId="31777" CreationDate="2019-05-09T14:16:42.133" Score="0" Body="&lt;p&gt;Dimensions has been abstract concept in math for long but we usually associate dimensions as a mean to locate object or measure its property like movement. You need to specify spatial plus time to convey any moving object location. Here, all X,y,z and t are independent and convey some point along that axis. But time has special property in sense that only movement in one direction is allowed. Don't need anything fancy till now but time is dimension in simple sense. However it becomes coupled with space since later we observed that time move differently for different observers. Basically different time passage at different X,y,z. So in a sense it is coupled by nature with space. However time still retains special status by moving in one direction so it is still different from space. I guess it would boil down to how strictly you define your dimension to be alike or not..but they are dependent on each other for practical purposes.&lt;/p&gt;&#xA;" OwnerUserId="26086" LastActivityDate="2019-05-09T14:16:42.133" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31820" PostTypeId="2" ParentId="31777" CreationDate="2019-05-09T15:51:46.243" Score="2" Body="&lt;p&gt;The equations of general relativity especially the fundamental equation relating the shape of space to its contents:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathbf {G} ={\frac {8\pi G}{c^{4}}}\mathbf {T}$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can be expressed and manipulated in a very nice and simple way, that allows us to perform useful calculations by talking in terms of a four-dimensional &quot;space-time&quot; with certain properties.  So basically we treat time as a dimension because it makes the equations nicer and the calculations easier. &lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-05-09T15:51:46.243" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31821" PostTypeId="2" ParentId="31817" CreationDate="2019-05-09T16:51:25.423" Score="4" Body="&lt;p&gt;If you turn your back to the sun and look at a building, the sun shines on the building and you can see it.  If the building is very far away but very big, you can still see it because the sun is still shining on it during the day.  If you think of the moon as a very large object very far away in the same direction as that building, you can see that the sun lights it up.  The only time the sun doesn't light it up is when the moon is on the side of the earth towards the sun.  Otherwise, the sun will always light up the moon.  You can see it because it's bright enough that the sun doesn't wash it out like it does the stars.&lt;/p&gt;&#xA;" OwnerUserId="15161" LastActivityDate="2019-05-09T16:51:25.423" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31822" PostTypeId="2" ParentId="31817" CreationDate="2019-05-09T17:21:28.453" Score="4" Body="&lt;p&gt;Let's look at our solar system &lt;a href=&quot;http://esc11.weebly.com/final-exam-review.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;from above&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PVrMN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PVrMN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The night side of Earth is the hemisphere facing away from the Sun. The Moon orbits around the Earth, on a path that takes 28 days. So for half of that orbit (14 days, from Last Quarter to New Moon to First Quarter), the Moon is visible from the day side of the Earth. &lt;/p&gt;&#xA;" OwnerUserId="361" LastEditorUserId="361" LastEditDate="2019-05-09T17:32:53.457" LastActivityDate="2019-05-09T17:32:53.457" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31823" PostTypeId="1" CreationDate="2019-05-09T23:25:02.257" Score="3" ViewCount="166" Body="&lt;p&gt;I just built a DIY spectroscope using a CD and after making some tests, I noticed a really strong emission line located in the orange-yellowish zone of the fire spectrum (specifically, burning wood).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that the strong line could be actually sodium, but wouldn't that mean that wood contains a lot of sodium in order to see such a strong emission line?&lt;/p&gt;&#xA;" OwnerUserId="26737" LastEditorUserId="14094" LastEditDate="2019-05-11T12:45:25.140" LastActivityDate="2019-05-11T12:45:25.140" Title="Strong orange line in wood fire spectrum?" Tags="&lt;spectroscopy&gt;&lt;spectra&gt;" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="31824" PostTypeId="1" AcceptedAnswerId="31825" CreationDate="2019-05-10T01:52:26.650" Score="3" ViewCount="355" Body="&lt;p&gt;As we know, an object that is beyond the Roche limit doesn't disintegrate (obviously) because the tidal forces upon the object are weaker than the gravitational pull of the object towards its centre, so the planet remains intact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: if that's so, why does the body deform even before crossing the Roche limit? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the gravitational pull of the particles in the object outweighs the tidal forces, wouldn't the object have to stay totally round? There couldn't have been any deformation because the tidal forces are totally cancelled out.&lt;/p&gt;&#xA;" OwnerUserId="9949" LastEditorUserId="14094" LastEditDate="2019-05-11T12:45:08.967" LastActivityDate="2019-05-16T13:37:06.623" Title="Question about tidal forces and the Roche limit" Tags="&lt;tidal-forces&gt;&lt;roche-limit&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31825" PostTypeId="2" ParentId="31824" CreationDate="2019-05-10T03:17:24.750" Score="6" Body="&lt;p&gt;When we do mathematical calculations, sometimes terms cancel and we get to draw a line through them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the real world cancellations and &quot;total cancellations&quot; don't really happen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forces just add. Often there's a big force and one or more little forces, and as long as the big force substantially &lt;em&gt;dominates&lt;/em&gt; the little forces, we tend to notice only the big force. But it doesn't cancel, it just covers up the effects of the others to the casual observer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's think about gravity. We are so used to moving around in 1 g of gravity that we don't notice it. You could say that our brain &quot;cancels&quot; out 1 g from our conscious awareness and we compensate for it automatically. We don't feel and notice the 1 g when we stand or sit or lay down, even though we make use of it to stay put.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, get in an elevator or an amusement ride and add a little acceleration, and we notice it right away! That sinking feeling when a fast elevator starts accelerating downward for example watch (but don't listen to) the video &lt;a href=&quot;https://www.youtube.com/watch?v=m8iP5WsiSbc&quot; rel=&quot;noreferrer&quot;&gt;The Equivalence Principle&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xLmbF.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xLmbF.jpg&quot; alt=&quot;The Equivalence Principle screenshots youtube.com/watch?v=m8iP5WsiSbc&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of the above discussion is to discount the idea of &quot;total cancellations&quot;. A bit of rock or soil on the surface of a body feels the gravitation of the rest of the body pulling down, the &quot;centrifugal force pulling&quot; it up if the body is rotating, and the gravitation of nearby bodies (e.g. Sun or Moon or planet) pulling up or down or sideways depending on orientation. It feels all of these all the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the &lt;em&gt;net force&lt;/em&gt;, the sum of all the forces, pulls up, then the body starts to disintegrate. If the net force is still down, it doesn't. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of it as a threshold, rather than a cancellation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...if that's so, ¿why does the body deforms even before crossing the Roche limit?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As long as the body is made of material capable of deforming or flowing (which rock certainly is, especially when hot) then it will continue to move or flow until the sum of all the forces reaches zero. Gravity usually dominates, but it's the force of the compressed layer below pushing up that counters the sum of all the forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the body is rotating, then the net force is weaker on the equator than at the poles, and so less pressure is needed to compensate. That means the higher pressure at the poles will cause some of the material to move to the equator.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To read more about several of the competing forces pulling on us all the time as we stand on the Earth, see:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://physics.stackexchange.com/a/141981/83380&quot;&gt;@DavidHammen's cool table&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://space.stackexchange.com/a/34181/12102&quot;&gt;my (sadly down-voted) calculation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;all of the answers to &lt;a href=&quot;https://space.stackexchange.com/q/14947/12102&quot;&gt;Did the Mars rovers actually confirm the gravity of Mars?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Once you are comfortable with the idea of equilibrium, multiple forces adding and pressure, then have a look at all the interesting answers to these particular Roche limit-related questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/q/26955/7982&quot;&gt;Is there any orbit at which the Roche limit can be “felt”?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/q/25535/7982&quot;&gt;Roche limit - calculation with centrifugal forces&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/q/19926/7982&quot;&gt;How Wide is Earth's Roche Limit?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/q/25874/7982&quot;&gt;Filling the Roche Lobe&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-05-10T03:37:04.897" LastActivityDate="2019-05-10T03:37:04.897" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31826" PostTypeId="2" ParentId="31793" CreationDate="2019-05-10T05:31:00.533" Score="1" Body="&lt;p&gt;So far our estimates of the size a the Universe is from what it is expected to be (i.e. calculations) rather than what we see. But there are several problems.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Age of the Universe&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;We are pretty sure of the age of the Universe, 13.8B years old, and the time when the first light was emitted. This gives us a relatively good idea of the size of the Universe since we consider that the furthest &lt;em&gt;edge&lt;/em&gt; of &lt;em&gt;our&lt;/em&gt; Universe is the light that it has emitted so far which has gone outward and never hit anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a perfect sphere, a.k.a. a perfect Euclidean Universe (which it isn't,) then you would think the Universe is a sphere with a radius of about 13.8 billion light years and so its size would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$size_{universe} = {\frac {4 \pi \times r^3} {3}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With r = 13.8 billion light years.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Hubble Discovery: Expansion of the Universe&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Only we know that our Universe is in expansion and there is nothing that says it's going to stop expanding. This affects the size of the Universe. Unfortunately, when we look around the cosmos, we can't see the expansion happening. We inferred it because of what we call the redshift. Most of the far away galaxies are moving away from us, the further they are, the longer the wave length of their light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed at which the expansion happens is not yet set in stone, though. Therefore, we have many arguing about the &lt;em&gt;current&lt;/em&gt; size of the Universe.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Is there an Horizon?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;One problem we currently have is the weight of the observable Universe. We calculated a weight which is much higher than what we see... so we added Black Matter. Matter which we can't see because it doesn't emit any kind of radiation and it's not a small weight: 85% of the calculated weight is missing!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another theory in that regard is the possibility that the Expansion of the Universe has already moved most of the &lt;em&gt;would be visible Universe&lt;/em&gt; over the Horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you go to the Ocean and look at the water, at some point you see the edge of the Earth. This is called the Horizon. Whatever is behind that line, you can't see it unless you go toward the horizon (when we observe the Universe, though, we stay in our solar system...) At such a size, the Universe is expanding faster than the speed of light. So light over the Horizon will never reach us because it can't go faster than the Expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more theory which is going to be hard to prove...&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Limited Size?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;There is a funny word circulating these days: Circumnavigation. This is like Pacman going out on the left side of the screen to re-appear on the right side. There are theories, which so far have been disproved, that the Universe would be finite and that things going in one direction reappear in on the other side. (If I'm correct, this is directly linked to the string theory.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess some people think that this would have been great since that would mean our Universe has a specific size and things will always be around. They just float and go around, come back where they were... in an infinite loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, we have not been able to see light of galaxies coming from the opposite direction. This is also quite contradictory with the possibility of a Horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also personally think this is contradictory with the Expansion theory. If the Universe had a limited size, how could it also expand?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;So what's the size?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;We actually calculate a size from what we can observe using a radius of 46.5 billion light years. This gives us an estimation of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$3.57×10^{80} m^3$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This size takes our best estimate of the redshift in account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For additional details about how the size is calculated, I would suggest reading the &lt;a href=&quot;https://en.wikipedia.org/wiki/Observable_universe#Size&quot; rel=&quot;nofollow noreferrer&quot;&gt;Size of the Observable Universe&lt;/a&gt; section on Wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="27850" LastActivityDate="2019-05-10T05:31:00.533" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31827" PostTypeId="2" ParentId="31824" CreationDate="2019-05-10T05:33:55.317" Score="4" Body="&lt;p&gt;Slightly long for a comment, so I'll put it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A comparison could be drawn to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Equatorial_bulge&quot; rel=&quot;nofollow noreferrer&quot;&gt;equatorial bulge&lt;/a&gt;.   A planet that rotates is fatter at the equator than the poles.   Nothing &quot;lifts&quot; off the surface because the outward force from the rotation is generally much weaker than the gravity, but the planet still bulges because the forces get added together and the planet effectively has less gravitation on it's axis of rotation.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as Earth's rotation, which on a person's weight might only be a fraction of a lb between equator and pole, that variation is still enough to give the Earth a 42 km tidal bulge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Equatorial_bulge#Other_celestial_bodies&quot; rel=&quot;nofollow noreferrer&quot;&gt;Equatorial bulge of the planets&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.sciencefocus.com/planet-earth/why-dont-we-weigh-more-at-the-poles-than-we-do-at-the-equator/&quot; rel=&quot;nofollow noreferrer&quot;&gt;0.5% weight loss at the Equator vs North Pole.&lt;/a&gt;  (note 80% of your equatorial weight loss is due to rotation, 20% is due to further distance from the center).    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tidal force on Earth is quite a bit weaker than that and the bulge is measured in feet, not 21 km radius, but Earth being as large as it is, even a small force over the entire surface creates a bulge.   Here's a fun answer on the &lt;a href=&quot;https://space.stackexchange.com/questions/3977/would-it-be-possible-to-fall-off-phobos&quot;&gt;tidal force calculated on the surface of Phobos&lt;/a&gt;.   &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2019-05-10T05:33:55.317" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31831" PostTypeId="2" ParentId="30310" CreationDate="2019-05-10T22:15:15.920" Score="1" Body="&lt;p&gt;Solar time is the ordinary 24 hour day.  Local 'noon' is the time when the Sun is located at the central meridian on the sky (the mid-point separating the eastern and western sides of the sky).  With timezones, we cheat a bit by grouping everyone in the same timezone to agree to use the same time on the clock ... whether or not the Sun is actually at the central meridian at noon (for most observers it will be near the meridian ... but not &lt;em&gt;at&lt;/em&gt; the meridian).  There are other nuances since the Earth does not travel at a constant speed in its orbit around the Sun.  The elliptical shape of our orbit causes the planet to slightly speed up or slow down at parts of the year.  We don't adjust clocks for this ... we just ignore it.  That means even if you live at a position in the timezone where the sun actually is at the meridian at noon on one day ... several months later it is not likely to be at the meridian at noon.  If you were to photograph the Sun at noon each day of the year for an entire year (assuming you never have cloudy skies), and combined the images, the shape you would see in the final merged image would show a path in a figure 8 on the sky.  The altitude of the Sun above the horizon changes based on season due to Earth's tilt.  But the east/west variation is based on the changes in Earth's velocity through space as we orbit the Sun.  This shape is called the Sun Analemma. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sidereal time is based on the amount of time it takes Earth to complete one revolution on its axis.  In other words if you select a star &lt;em&gt;other&lt;/em&gt; than the Sun and determine how long it takes for that star to move from the mid-point in the sky one day until mid-point in the sky the following day ... that length of time is a sidereal day.  In liberally rounded values... it's roughly 4 minutes short of a 24 hour solar day.  (approximately 23 hours 56 minutes 4 seconds)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference occurs because in the time that it takes the  Earth to complete one rotation on its axis, the planet &lt;em&gt;also&lt;/em&gt; moves forward in orbit around the Sun.  There are roughly 365 days per year and 360 degrees in a circle.  This means the Earth will have progressed just shy of 1 degree in it's orbit (our orbit is not a perfect circle so there are times of the year when the planet is moving slightly faster than other times of the year ... but on average it's just a tiny bit short of a degree.)  Even though we've completed an entire revolution on our axis, the Sun wont return to the same position because the Earth has progressed in its orbit.  We need to let the Earth keep spinning for nearly 4 more minutes to get the Sun to return to the same position in the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This fact causes the local sidereal time to diverge from solar time by roughly 4 minutes per day.  In 15 days your solar vs. sidereal clocks will have diverged by about an hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 00 hours point is also called the &quot;First point of Aries&quot; (&lt;a href=&quot;https://en.wikipedia.org/wiki/First_Point_of_Aries&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/First_Point_of_Aries&lt;/a&gt;).  To make things more confusing ... the point is actually located in Pisces ... thanks to the precession cycle of the Earth's axial wobble (when the position was first established ... it actually was in Aries)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The First Point of Aries is defined as the position of the Sun in the sky when the Sun passes from the Southern Hemisphere to the Northern Hemisphere at the Vernal Equinox (the start of Northern-Hemisphere Spring / Southern Hemisphere Fall) in March.  That point becomes the &quot;00&quot; hour point.  The right-ascension coordinate of all objects in the sky is based off that point.&lt;/p&gt;&#xA;" OwnerUserId="26051" LastActivityDate="2019-05-10T22:15:15.920" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31832" PostTypeId="1" AcceptedAnswerId="31834" CreationDate="2019-05-11T09:47:54.697" Score="2" ViewCount="4500" Body="&lt;p&gt;Few days a go, I came across these two pictures of paths of planets as observed from Earth. They have a perfect geometric shapes.  Actually, to me they look so beautiful and perfect, that they appear fake or made by a pseudo-scientist or some propaganda society (e.g. flat Earth like community people) who talks about divine geometry etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Auxiliary questions: Is this the reason, why Earth was consider to be in the center of the solar system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zWljc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zWljc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/z2ae0.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z2ae0.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23759" LastEditorUserId="23759" LastEditDate="2019-05-11T13:37:53.857" LastActivityDate="2019-05-11T23:48:30.290" Title="Are these real paths of planets as traced from Earth?" Tags="&lt;planet&gt;&lt;solar-system&gt;&lt;earth&gt;&lt;rotation&gt;&lt;planetary-transits&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31833" PostTypeId="2" ParentId="31811" CreationDate="2019-05-11T11:15:37.927" Score="1" Body="&lt;p&gt;As @RobJeffries says, this information is mostly in &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Formation_and_evolution_of_the_Solar_System&quot; rel=&quot;nofollow noreferrer&quot;&gt;the relevant wikipedia article&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is really in two parts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, they don't have solid surfaces because the elements they are made of don't have solid forms at the temperatures and pressures that arise inside the planets. In particular, Jupiter and Saturn are made mostly of hydrogen and helium which only become solid at rather low temperatures and the planets are still hot due to the heat created by their formation, heating from the Sun and some heat released by internal processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, of course, begs the question of why those two planets are made mostly of hydrogen and helium, when the others are not. I should start with the caveat that a lot of aspects of this are still areas of active research and debate among scientists, and could not be considered settled  or universally accepted.  As more evidence is gathered and computer models (used for making checkable predictions from underlying theories) are refined,views of this may change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the solar system started out as a cloud of gas and dust. While it was mostly hydrogen and helium (still around since the big bang) it had significant amounts of other elements present in it which had been produced in older generations of stars. Some chance concentration of mass, or external shock (eg from a supernova) caused it to start to collapse and, under the influence of its own gravity, fluid dynamics and the conservation of angular momentum, a large part of it formed a ball (called a protostar) at the centre which was heated up by the collapse process itself. Around that, gas and dust formed a flat disk (oriented according to the original net angular momentum of the clump). Within that disk, tiny grains of solid matter formed. Far from the protostar, those grains were mostly made of compounds relatively light elements (carbon, oxygen, nitrogen, hydrogen) which we collectively call &quot;ices&quot;. Closer to the protostar, its heat stopped ices from solidifying, but still allowed grains of iron, rock-like materials and so on.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When these grains collided, they stuck together, gradually forming bigger and bigger lumps. In the outer system they were lumps of ice, a lot like comets, in the inner system, something more like lumps of rock (like some asteroids). These then started holding together and pulling in more material by their own gravity and eventually grew into planets. At some point during this process fusion started in the Sun, and drove awat most of the remaining gas and dust. After that the planets grew by collisions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the inner system, you got rocky planets like Mercury, Venus, Earth and Mars. A bit further out you got icy planets like Uranus, Neptune and Pluto. A couple of those icy planets  happened to get big enough, quickly enough, to attract a lot of hydrogen and helium gas before the Sun lit up. They are Jupiter and Saturn.   &lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-05-11T11:15:37.927" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31834" PostTypeId="2" ParentId="31832" CreationDate="2019-05-11T13:56:41.463" Score="4" Body="&lt;p&gt;From &lt;a href=&quot;https://astronomy.stackexchange.com/a/30184/7982&quot;&gt;this answer&lt;/a&gt; here are views of the inner and outer planets. I used the python package Skyfield which uses the NASA JPL ephemerides to generate positions of all the planets, then I subtracted the Earth's position to generate the geocentric positions.&lt;/p&gt;&#xA;&lt;p&gt;I agree with your guess that the first image is too perfect.&lt;/p&gt;&#xA;&lt;p&gt;The paths will not be closed, because there is not a rational number relationship (e.g. 5:9, 2:3, 1:1) between the periods of the planets.&lt;/p&gt;&#xA;&lt;p&gt;The ones in the image are &lt;em&gt;reminiscent of&lt;/em&gt; the &lt;a href=&quot;https://en.wikipedia.org/wiki/Spirograph&quot; rel=&quot;nofollow noreferrer&quot;&gt;Spirograph&lt;/a&gt; drawings I used to make a very long time ago.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;...plotting using the Python package &lt;a href=&quot;https://rhodesmill.org/skyfield/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Skyfield&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The &amp;quot;inner&amp;quot; plotted bodies (from outer to inner) are Mars, (Earth), Moon, Venus, Mercury, The Sun. The data is for &lt;strong&gt;2000 days&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The &amp;quot;outer&amp;quot; plotted bodies are Uranus, Saturn, Jupiter, Earth, and The Sun. The data is for &lt;strong&gt;2000 months&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Distances are in AU.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0bT3U.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0bT3U.png&quot; alt=&quot;inner and outer planets, heliocentric and geocentric&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-05-11T23:48:30.290" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31835" PostTypeId="2" ParentId="29487" CreationDate="2019-05-12T00:34:08.617" Score="0" Body="&lt;p&gt;Astronomer Emily Petroff (&lt;a href=&quot;http://www.ebpetroff.com/my-research&quot; rel=&quot;nofollow noreferrer&quot;&gt;informative website&lt;/a&gt;) studies FRB time structure and has recently commented on the rapidly increasing number and diversity of observations of FRBs &lt;a href=&quot;https://twitter.com/ebpetroff/status/1118815497066303493&quot; rel=&quot;nofollow noreferrer&quot;&gt;in twitter&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These show that in addition to the overall widths several milliseconds for the dispersion &quot;corrected&quot; (or compensated) spectral histograms, &lt;em&gt;there is finer structure at order 100 microseconds&lt;/em&gt; and possibly below!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When you correct for this DM and extract the intrinsic pulse, we see that some FRBs have structure in the pulse, but some don't&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Y3XzT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y3XzT.jpg&quot; alt=&quot;https://twitter.com/ebpetroff/status/1118815497066303493&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-05-12T00:44:18.433" LastActivityDate="2019-05-12T00:44:18.433" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31837" PostTypeId="1" CreationDate="2019-05-12T06:52:19.223" Score="0" ViewCount="81" Body="&lt;p&gt;since we observe galaxies and other steller objects in very tiny tiny scale, and how we can understand that the speed of expansion on universe is higher than the speed of light ?&#xA;even using the powerful telescope we just see the lights from a galaxy not the object itself in real time, also the lights are red shifted as well. So what's the basis of proving univers's expansion rate is higher than of light speed?&lt;/p&gt;&#xA;" OwnerUserId="28002" LastActivityDate="2019-05-13T15:23:50.077" Title="how to define the universe itself is expanding greater than than the speed of light?" Tags="&lt;universe&gt;&lt;light&gt;&lt;expansion&gt;&lt;speed&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31838" PostTypeId="1" CreationDate="2019-05-12T07:51:25.577" Score="4" ViewCount="62" Body="&lt;p&gt;Is stellar ignition an instantaneous thing - before, dark, after, shining brightly - or more of a process over time? In fusion research 'ignition' appears to be an event, but maybe in a protostar it starts locally and then spreads?&lt;/p&gt;&#xA;" OwnerUserId="28003" LastActivityDate="2019-05-12T12:17:27.640" Title="Do stars come on like LEDs or wick lamps?" Tags="&lt;star-formation&gt;" AnswerCount="0" CommentCount="5" ClosedDate="2019-05-12T15:53:32.590" ContentLicense="CC BY-SA 4.0" />
  <row Id="31839" PostTypeId="1" AcceptedAnswerId="31847" CreationDate="2019-05-12T08:49:06.410" Score="1" ViewCount="468" Body="&lt;p&gt;I have a picture of a power spectrum of matter in the Universe. What is the connection to Lyman alpha? Please describe it for me (horizontal and vertical and shape).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8PF1r.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8PF1r.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="28004" LastEditorUserId="5264" LastEditDate="2019-05-14T17:32:21.020" LastActivityDate="2019-05-15T11:30:21.850" Title="Power spectrum of matter in the Universe" Tags="&lt;cosmology&gt;&lt;identify-this-object&gt;" AnswerCount="1" CommentCount="13" FavoriteCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31841" PostTypeId="1" CreationDate="2019-05-12T12:41:04.643" Score="2" ViewCount="189" Body="&lt;p&gt;According to this January 2005 NASA JPL News item &lt;a href=&quot;https://www.jpl.nasa.gov/news/news.php?feature=716&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA Details Earthquake Effects on the Earth&lt;/a&gt;, the Three Gorges Dam has slowed down Earth's rotation by &lt;strong&gt;0.06 microseconds per day&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having that weight of Earth is an estimate, and so is the volume / weight of water held by the dam and also that these two resulting figures are apart by several orders, how can the resulting deceleration be not only precise, but also calculated at all? The error margin must be much larger, then the result. Or not? How did NASA get the numbers so accurate?&lt;/p&gt;&#xA;" OwnerUserId="27810" LastEditorUserId="7982" LastEditDate="2019-05-12T14:11:01.170" LastActivityDate="2019-05-12T14:11:01.170" Title="How to caclulate the chinese Three Gorges Dam impact on Earth rotation?" Tags="&lt;earth&gt;&lt;rotation&gt;&lt;nasa&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2019-05-13T04:59:56.100" ContentLicense="CC BY-SA 4.0" />
  <row Id="31843" PostTypeId="1" AcceptedAnswerId="31844" CreationDate="2019-05-12T16:48:33.667" Score="7" ViewCount="986" Body="&lt;p&gt;I assume they should be in the Milky Way galaxy. I have read about it a little but did not find information on where they are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which galaxy do the Pleiades belong to?&lt;/p&gt;&#xA;" OwnerUserId="28007" LastEditorUserId="23881" LastEditDate="2019-05-13T14:29:05.907" LastActivityDate="2019-05-18T19:48:08.760" Title="Whereabouts of the Pleiades" Tags="&lt;galaxy&gt;&lt;milky-way&gt;&lt;star-cluster&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31844" PostTypeId="2" ParentId="31843" CreationDate="2019-05-12T17:45:47.007" Score="17" Body="&lt;p&gt;With the exception of the Andromeda galaxy and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Magellanic_Clouds&quot; rel=&quot;noreferrer&quot;&gt;Magellanic clouds&lt;/a&gt;&lt;sup&gt;1&lt;/sup&gt;, every star, star cluster and nebula that is visible to the naked eye is part of the Milky Way. The Pleiades is a star cluster in the Milky Way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objects in other galaxies are too far and too dim to be visible with the naked eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;&lt;sub&gt;The Triangulum galaxy (M33) may also be visible to the naked eye from a dark site to trained observers with excellent vision&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="7982" LastEditDate="2019-05-18T19:48:08.760" LastActivityDate="2019-05-18T19:48:08.760" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31845" PostTypeId="2" ParentId="31843" CreationDate="2019-05-12T21:32:42.293" Score="13" Body="&lt;p&gt;To give you a perspective the Milky Way Galaxy is between 150,000 and 200,000 light years in across.  The Pleiades is less than 450 light years from Earth.  In a galactic perspective the Pleiades is incredibly close to us and so clearly within the Milky Way Galaxy.&lt;/p&gt;&#xA;" OwnerUserId="25088" LastEditorUserId="25088" LastEditDate="2019-05-12T22:10:43.203" LastActivityDate="2019-05-12T22:10:43.203" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31847" PostTypeId="2" ParentId="31839" CreationDate="2019-05-13T12:55:54.590" Score="5" Body="&lt;h3&gt;Power spectrum&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The distribution of matter in the Universe is not uniform&lt;span class=&quot;math-container&quot;&gt;$^\dagger$&lt;/span&gt;. Rather, it has clumped together by gravity to form the structure we see, from planets and stars, to galaxies, to groups and clusters of galaxies. Exactly &lt;em&gt;how&lt;/em&gt; much the Universe clumps on the various scales can be described statistically by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Correlation_function_(astronomy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;two-point correlation function&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$\xi(r)$&lt;/span&gt;, which gives the excess probability of finding a clump of matter at a certain distance &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; from another clump, relative to a random, Poisson-distributed matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Fourier transform of &lt;span class=&quot;math-container&quot;&gt;$\xi(\mathbf{r})$&lt;/span&gt; is called the matter power spectrum &lt;span class=&quot;math-container&quot;&gt;$P(k)$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$k = 2\pi/r$&lt;/span&gt; is the wavenumber. The power spectrum thus shows readily how much structure there is on various physical scales, i.e. a high value means much structure. It evolves (increases in amplitude) with time or, correspondingly, redshift &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;, and so may be written &lt;span class=&quot;math-container&quot;&gt;$P(k,z)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power spectrum can be written as the product of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Primordial_fluctuations&quot; rel=&quot;nofollow noreferrer&quot;&gt;primordial power spectrum&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$P_*(k) \propto k^{n_s-1}$&lt;/span&gt; originating from &lt;a href=&quot;https://en.wikipedia.org/wiki/Inflation_(cosmology)&quot; rel=&quot;nofollow noreferrer&quot;&gt;cosmic inflation&lt;/a&gt;, and a transfer function &lt;span class=&quot;math-container&quot;&gt;$T(k,z)$&lt;/span&gt; (squared) describing what happens at a later epoch. A good description is given by &lt;a href=&quot;http://adsabs.harvard.edu/abs/1980lssu.book.....P&quot; rel=&quot;nofollow noreferrer&quot;&gt;Peebles (1980)&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The figure&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The figure you show is taken from &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0207047v3&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tegmark &amp;amp; Zaldarriaga (2002)&lt;/a&gt;. It shows the power spectrum as a function of wavenumber, but on the top &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; axis you see the corresponding spatial scale. This scale is expressed in &lt;em&gt;comoving megaparsecs&lt;/em&gt; (Mpc) (where 1 parsec equals 3.26 light-years). This factors out the size of the expanding Universe — or, equivalently, the epoch at which an observation is made — such that observations at different epochs in the history of the Universe can be readily compared. A distance &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; at redshift &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; will expand to &lt;span class=&quot;math-container&quot;&gt;$r(1+z)$&lt;/span&gt; at redshift &lt;span class=&quot;math-container&quot;&gt;$z=0$&lt;/span&gt;. Using comoving coordinates, no matter at what value of &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; we observe &lt;span class=&quot;math-container&quot;&gt;$P(k,z)$&lt;/span&gt;, we &quot;extrapolate&quot; to the current power spectrum &lt;span class=&quot;math-container&quot;&gt;$P(k)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, the distance is divided by the dimensionless Hubble constant, &lt;span class=&quot;math-container&quot;&gt;$h \equiv H_0\,/\,100\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1} \simeq 0.7$&lt;/span&gt;. The reason for this is that different observations may have made different assumptions about the exact value of &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt;; in general inferred distances will scale as &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt;, so measuring distances in &lt;span class=&quot;math-container&quot;&gt;$h^{-1}\mathrm{Mpc}$&lt;/span&gt; allows for direct comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, &lt;span class=&quot;math-container&quot;&gt;$1 h^{-1}\mathrm{Mpc} \simeq 4.7\times10^6$&lt;/span&gt; light-years at redshift &lt;span class=&quot;math-container&quot;&gt;$z=0$&lt;/span&gt;, but at redshift, say, &lt;span class=&quot;math-container&quot;&gt;$z=4$&lt;/span&gt; — when the Universe was five times smaller in all directions — we had that &lt;span class=&quot;math-container&quot;&gt;$1 h^{-1}\mathrm{Mpc} \simeq 9.3\times10^5$&lt;/span&gt; light-years.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Observational probes&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The power spectrum depends on various cosmological parameters, mostly on the matter density parameter &lt;span class=&quot;math-container&quot;&gt;$\Omega_\mathrm{m}$&lt;/span&gt; and the expension rate &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt;, but also the baryons (&lt;span class=&quot;math-container&quot;&gt;$\Omega_\mathrm{b}$&lt;/span&gt;) and dark energy (&lt;span class=&quot;math-container&quot;&gt;$\Omega_\Lambda$&lt;/span&gt;).&#xA;Given these cosmological parameters, the transfer function and the resulting matter power spectrum can be calculated. In the plot, the red line shows &lt;span class=&quot;math-container&quot;&gt;$P(k)$&lt;/span&gt; as calculated theoretically from &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmological_perturbation_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;linear perturbation theory&lt;/a&gt;. For details on the calculations, see &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0207047v3&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tegmark &amp;amp; Zaldarriaga (2002)&lt;/a&gt;. It is compared to constraints from various observational probes of the cosmological parameters:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Black&lt;/strong&gt;: Anisotropies in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot; rel=&quot;nofollow noreferrer&quot;&gt;CMB&lt;/a&gt; shows how matter clumped in the very early Universe, and its polarization pattern, caused by subsequent scattering, gives constraints on matter distribution in later epochs, on the largest scales.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Green&lt;/strong&gt;: Clustering of galaxies (here from the &lt;a href=&quot;https://en.wikipedia.org/wiki/2dF_Galaxy_Redshift_Survey&quot; rel=&quot;nofollow noreferrer&quot;&gt;2dF survey&lt;/a&gt;) shows how matter has formed structure on (comoving) scales roughly from a few tens of Mpc (a bit larger than the typical distance between galaxies) to several hundred Mpc (above which the Universe is homogeneous).&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Blue&lt;/strong&gt;: Abundances of galaxy clusters. Here, the error bars reflect the spread in the literature.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Magenta&lt;/strong&gt;: &lt;a href=&quot;https://en.wikipedia.org/wiki/Weak_gravitational_lensing&quot; rel=&quot;nofollow noreferrer&quot;&gt;Weak lensing&lt;/a&gt; measures photons that have traveled through fluctuations in the metric caused by intervening matter; this technique probes the power spectrum on &lt;span class=&quot;math-container&quot;&gt;$\sim10\,\mathrm{Mpc}$&lt;/span&gt; scales. These data are from &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0204295&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hoekstra et al. (2002)&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Red&lt;/strong&gt;: The smallest scales (largest wavenumbers) are probed by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lyman-alpha_forest&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lyman &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; forest&lt;/a&gt; (LAF). Whereas all the previous techniques measure light, the LAF is the &lt;em&gt;absence&lt;/em&gt; of light, namely absorption lines in a bright background source, caused by the neutral hydrogen in all the diffuse clouds and filaments of gas lying between the source and us. The background source is usually a &lt;a href=&quot;https://en.wikipedia.org/wiki/Quasar&quot; rel=&quot;nofollow noreferrer&quot;&gt;quasar&lt;/a&gt;, but may also be, e.g., a &lt;a href=&quot;https://en.wikipedia.org/wiki/Gamma-ray_burst&quot; rel=&quot;nofollow noreferrer&quot;&gt;gamma-ray burst&lt;/a&gt;. As light blueward of the quasar's own Ly&lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; emission line travels through the Universe, it is gradually redshifted toward the rest wavelength of Ly&lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; (1216 Å), so whenever a parcel of hydrogen gas is present along the line of sight, it will cause an absorption line at the corresponding wavelength, creating a &quot;forest&quot; of absorption lines in the blue part of the quasar's spectrum. These data are (primarily) from &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0012324&quot; rel=&quot;nofollow noreferrer&quot;&gt;Croft et al. (2002)&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^\dagger$&lt;/span&gt;&lt;sub&gt;This is fortunate, since otherwise you wouldn't be here to ask the question.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2019-05-15T11:30:21.850" LastActivityDate="2019-05-15T11:30:21.850" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31848" PostTypeId="2" ParentId="31837" CreationDate="2019-05-13T15:23:50.077" Score="1" Body="&lt;p&gt;You don't see &lt;em&gt;anything&lt;/em&gt; in real time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you check your phone, you look a nanosecond back in time. When you look at the Sun,&lt;span class=&quot;math-container&quot;&gt;$\!^\dagger$&lt;/span&gt; you look 8 minutes back in time. How do you know your phone and the Sun are still there? Because you have a model of how the world works, and that model tells you that, to the best of your knowledge, they won't just disappear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, we have a &lt;a href=&quot;https://en.wikipedia.org/wiki/Lambda-CDM_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;model&lt;/a&gt; of how the Universe works, based on theories and observational evidence for these theories. This model predicts that the Universe expands in such a way that two objects recede from one another at a rate proportional to their distance (the so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Hubble%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble-Lemaître law&lt;/a&gt;). Hence, for large enough distances, this rate exceed the speed of light. This is described by general relativity, and is not in conflict with the speed of light being a fundamental &quot;speed limit&quot;, since that is a prediction of &lt;em&gt;special&lt;/em&gt; relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In physics, we can never be 100% sure of anything, but we can have more or less faith in our models. If you're asking more specifically how can we trust the models, then that would become a longer description of the assumptions entering the hypothesis, the theoretical framework, the equations, the observational evidence, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, the fact that we look farther back in time, the more distant an object is, has &lt;em&gt;helped&lt;/em&gt; us developing our models, since this allows us to observe the Universe not only &lt;em&gt;today&lt;/em&gt;, but throughout its entire history. This is one of the coolest differences between astronomy and other sciences, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The redshifting of light is also sometimes helpful. The earliest galaxies (i.e. the most distant ones) are often only visible due to a certain type of light called &lt;a href=&quot;https://en.wikipedia.org/wiki/Lyman-alpha_line&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lyman &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;&lt;/a&gt;. This light is ultraviolet, and hence doesn't easily penetrate our atmosphere. But when observing distant galaxies, the light has been redshifted into the optical or even infrared, and so makes it possible to observe from ground, rather than form space-based telescopes (which is much more expensive).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^\dagger$&lt;/span&gt;&lt;sub&gt;Don't.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2019-05-13T15:23:50.077" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31850" PostTypeId="1" AcceptedAnswerId="31851" CreationDate="2019-05-13T18:25:15.120" Score="2" ViewCount="240" Body="&lt;p&gt;Can the Sun go through seasons or cycles to cause a temperature change? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could the Sun have to go through any kind of thermal cycle to change the climate on Earth and by how much?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorDisplayName="user1569" LastEditDate="2019-05-16T06:53:34.933" LastActivityDate="2019-05-16T06:56:07.467" Title="Does the sun have cycles causing temperature changes on Earth?" Tags="&lt;the-sun&gt;&lt;orbit&gt;&lt;earth&gt;&lt;magnetic-field&gt;&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31851" PostTypeId="2" ParentId="31850" CreationDate="2019-05-13T20:11:25.403" Score="7" Body="&lt;p&gt;The Sun's magnetic field is way too weak to have any measurable effect on the Earth's orbit. The sunspot cycle does produce a small but detectable signature in the global average temperature (about &lt;span class=&quot;math-container&quot;&gt;$0.2\sideset{^{\circ}}{}{\mathrm{C}}$&lt;/span&gt;). Over a billion-year timescale the Sun is getting hotter as helium builds up in its core. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="15986" LastEditDate="2019-05-16T00:33:33.090" LastActivityDate="2019-05-16T00:33:33.090" CommentCount="1" CommunityOwnedDate="2019-05-13T20:11:25.403" ContentLicense="CC BY-SA 4.0" />
  <row Id="31853" PostTypeId="1" CreationDate="2019-05-14T04:32:02.787" Score="25" ViewCount="7566" Body="&lt;p&gt;If the universe is expanding outward, what is the processes for one galaxy to get off track enough to collide with another? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Andromeda_Galaxy&quot; rel=&quot;noreferrer&quot;&gt;Andromeda Galaxy&lt;/a&gt; and the Milky Way.&lt;/p&gt;&#xA;" OwnerUserId="28014" LastEditorUserId="332" LastEditDate="2019-05-15T01:52:28.660" LastActivityDate="2021-04-20T13:45:48.250" Title="Why do galaxies collide?" Tags="&lt;big-bang-theory&gt;" AnswerCount="6" CommentCount="7" FavoriteCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31854" PostTypeId="2" ParentId="31853" CreationDate="2019-05-14T05:43:22.870" Score="38" Body="&lt;p&gt;The universe is expanding on a large scale. But locally things are always messy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Locally, galaxies are not set in stone, they move relative to each other, and the directions are random. If they're moving towards each other fast enough, then they will collide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there's gravity. Some galaxies are bound to each other by gravity, and that will tend to pull them together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to why galaxies move at all, relative to each other - well, things in this universe have kinetic energy, and it's distributed randomly. Being distributed randomly, all kinds of scenarios are possible - things running away from each other, zooming past each other, bumping into each other, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a messy and random universe, and the order of expansion becomes apparent only on the largest scale.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2019-05-14T05:43:22.870" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="31855" PostTypeId="2" ParentId="31853" CreationDate="2019-05-14T08:16:05.947" Score="9" Body="&lt;p&gt;The galaxies don't really get &quot;off track&quot; - it's not impossible, but that kind of thing probably doesn't happen anymore (as space continues to expand). What actually happens is that galaxies form gravitationally bound clusters - within the cluster, the acceleration due to gravity is larger than the equivalent expansion of space between the galaxies, so rather than growing more distant, the galaxies in question actually get closer together over time. Eventually, this results in a collision and a merger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the expansion remains roughly constant, there will come a point where we'll no longer be able to see any galaxies outside of our own cluster. But for those close enough, this has little effect - just like the expansion of space doesn't cause atoms, planets, solar systems or galaxies to get bigger.&lt;/p&gt;&#xA;" OwnerUserId="10656" LastActivityDate="2019-05-14T08:16:05.947" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31856" PostTypeId="1" CreationDate="2019-05-14T11:27:00.093" Score="2" ViewCount="105" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Mars water  ice clouds are often associated with regional- &#xA;  to-global circulation and saturation conditions, as opposed to local convective overturning, and so typically exhibit cirrus rather than cumulus forms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://www.cambridge.org/core/books/atmosphere-and-climate-of-mars/CA5316BF24A89891563B2E17EF185E19&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Atmosphere and Climate of Mars&lt;/a&gt; Edited by  Robert M. Haberle et al., 2017 - chapter 5 - subchapter 5.2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are cirrus clouds more common than cumulus clouds on mars?&lt;/p&gt;&#xA;" OwnerUserId="28018" LastEditorUserId="7982" LastEditDate="2019-05-14T13:07:58.283" LastActivityDate="2019-05-15T03:18:46.030" Title="Why are cirrus clouds more common on Mars?" Tags="&lt;planet&gt;&lt;planetary-atmosphere&gt;&lt;mars&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31857" PostTypeId="2" ParentId="31843" CreationDate="2019-05-14T12:52:55.007" Score="7" Body="&lt;p&gt;This screen grab from the program &lt;em&gt;&lt;a href=&quot;http://www.thinkastronomy.com/M13/&quot; rel=&quot;noreferrer&quot;&gt;Where is M13?&lt;/a&gt;&lt;/em&gt; shows the location of the Sun and Pleiades in our galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KQnQd.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KQnQd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun is the orange dot, the Pleiades the yellow dot.&lt;/p&gt;&#xA;" OwnerUserId="23640" LastActivityDate="2019-05-14T12:52:55.007" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31859" PostTypeId="2" ParentId="29064" CreationDate="2019-05-14T15:51:02.387" Score="6" Body="&lt;p&gt;The aurora depicted in the artist's conception is of ultraviolet light.  It would not be seen with the naked eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;NASA's Hubble Space Telescope observed a pair of auroral belts encircling the Jovian moon Ganymede. &lt;a href=&quot;http://hubblesite.org/image/3505/news_release/2015-09&quot; rel=&quot;noreferrer&quot;&gt;The belts were observed in ultraviolet light&lt;/a&gt; by the Space Telescope Imaging Spectrograph and are colored blue in this illustration. They are overlaid on a visible-light image of Ganymede taken by NASA's Galileo orbiter. The locations of the glowing aurorae are determined by the moon's magnetic field, and therefore provide a probe of the moon's interior, where the magnetic field is generated. The amount of rocking of the magnetic field, caused by its interaction with Jupiter's own immense magnetosphere, provides evidence that the moon has a subsurface ocean of saline water.&quot;&lt;/p&gt;&#xA;" OwnerUserId="25088" LastActivityDate="2019-05-14T15:51:02.387" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31860" PostTypeId="2" ParentId="31853" CreationDate="2019-05-14T17:02:04.203" Score="9" Body="&lt;p&gt;I'm not sure that anyone has answered the question asked. The root cause is indeed that gravitationally bound structures with freefall timescales that are much shorter than the age of the universe are not greatly affected by the general expansion of the universe (NB: Structures with freefall timescales longer than this are &lt;em&gt;not&lt;/em&gt; going to be the source of many galaxy collisions). That is, locally, the expansion within such structures is negligible. However, that does not necessarily lead to collisions on a a timescale shorter than the age of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first reason for galaxy collisions is that galaxy clusters have a very large number density - that is, the spacing between galaxies is not vastly larger than the &quot;size&quot; of a galaxy, where here, &quot;size&quot; means the effective interaction cross-sectional radius. As a result of these high densities, the freefall dynamical timescales in rich clusters (and even smaller groups of galaxies) are of order billions of years and so there is plenty of time for the galaxies to interact. As a contrast, think about how you might construct a &lt;strong&gt;scale&lt;/strong&gt; model of stars in the local neighbourhood and compare the sizes of stars with their separations. It would in fact be difficult to make such a scale model with any meaningfully sized stars. On the other hand, you &lt;em&gt;can&lt;/em&gt; make a scale model of say the local group of galaxies because their separations are only &lt;span class=&quot;math-container&quot;&gt;$\sim 10$&lt;/span&gt; times their sizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second reason is that many galaxies contain gas and that gas can easily dissipate kinetic energy and also transfer angular momentum. Another factor is that massive clusters of galaxies contain intracluster gas that can also serve to dissipate kinetic energy. In a gravitationally bound system, then objects that are in orbit around each other or around a common centre of mass need ways in which kinetic energy and angular momentum can be lost in order for a collision to take place. Even without gas, the fact that galaxies exist in groups and clusters means that n-body interactions can serve to dissipate energy and angular momentum to make a collision happen.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-05-15T15:51:18.727" LastActivityDate="2019-05-15T15:51:18.727" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31861" PostTypeId="2" ParentId="31853" CreationDate="2019-05-14T19:34:40.090" Score="1" Body="&lt;p&gt;I think it's also worth noting that cosmological expansion is measurable only on the very largest of scales. Hubble's law tells us that the farther away an object is, the faster it's being pulled away by the expansion. That rate is approximately &lt;span class=&quot;math-container&quot;&gt;$70 km.s^{-1}Mpc^{-1}$&lt;/span&gt; -- in other words, for every megaparsec (Mpc -- about 3 million light years) that you look farther out in space, an additional &lt;span class=&quot;math-container&quot;&gt;$70 km.s^{-1}$&lt;/span&gt; is tacked onto its recessional speed. For a relatively close galaxy, its actual velocity through space can be much larger than this, and in the specific case of the Andromeda Galaxy, which is only about 2.5 million light years (0.77 MPc) away, it is actually approaching us at around &lt;span class=&quot;math-container&quot;&gt;$110 km.s^{-1}$&lt;/span&gt;. For much, much farther galaxies, billions of light years away, all of those &lt;span class=&quot;math-container&quot;&gt;$70 km.s^{-1}$&lt;/span&gt; add-ons stack up to a much, much higher recessional speed.&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2019-05-14T19:34:40.090" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31862" PostTypeId="2" ParentId="31744" CreationDate="2019-05-14T20:56:28.833" Score="1" Body="&lt;p&gt;I believe it just appears that way, the photons racing away reach a point of equilibrium where their speed getting away from the black hole matches the speed of the space following into it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the light speed limit only applies to the speed which everything inside the universe can move within it. It doesn’t apply to the universe or space itself.  That can move much quicker than the speed of light.  At some it all flows back into the singularity faster than anything can move away from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could still get away from it if you could somehow go faster than light but since you can’t you’re screwed along with the photons.&lt;/p&gt;&#xA;" OwnerUserId="28024" LastActivityDate="2019-05-14T20:56:28.833" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31863" PostTypeId="1" CreationDate="2019-05-14T21:34:09.680" Score="1" ViewCount="65" Body="&lt;p&gt;In Steven Weinberg's 'Cosmology' Chapter 2.3 (pg 113), he begins with a number density for some particle via the Maxwell-Boltzmann Formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RS550.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RS550.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I have some familiarity with statistical mechanics, I can't figure out how this equation was derived. The spin states and fugacity are obvious enough for any quantum microcanonical ensemble. The &lt;span class=&quot;math-container&quot;&gt;$(2\pi\hbar)^{-3}$&lt;/span&gt; and the integral however I am not sure where they come from. I am used to working with canonical ensembles in terms of partition functions and that doesn't seem to be the case here unless I am missing something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I have also posted &lt;a href=&quot;https://physics.stackexchange.com/q/480096&quot;&gt;this question in physics.stackexchange&lt;/a&gt; because it is an application of physics (statistical mechanics). &lt;/p&gt;&#xA;" OwnerUserId="10995" LastEditorUserId="10995" LastEditDate="2019-05-15T06:06:35.840" LastActivityDate="2019-05-15T06:06:35.840" Title="Understanding the statistical mechanics of Recombination Epoch (&quot;Cosmology&quot; Weinberg)" Tags="&lt;cosmology&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2021-04-30T01:19:43.070" ContentLicense="CC BY-SA 4.0" />
  <row Id="31864" PostTypeId="2" ParentId="31853" CreationDate="2019-05-14T22:31:22.567" Score="2" Body="&lt;p&gt;Galaxies don't get &quot;off track&quot; - to see how collisions happen, we need to go right back to galaxy formation early on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, Big Bang happens. Space starts to expand - dramatically and to a huge extent. That's space itself expanding, not galaxies moving within space, by the way - distances themselves change. (Which is why it's called a &quot;metric&quot; expansion, metric being a term for distance-measures, and also why cosmologists say the Big Bang happened &quot;everywhere&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some tiny fraction of a second, the massive expansion winds down. Space continues to expand, but at a much slower rate. The last of the fundamental forces breaks away, and the cosmos is left as an insanely hot dense mix, so hot that even basic particles like protons, neutrons and electrons can't yet exist - although quarks can. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are some very subtle things going on. Even though expansion left us with an incredibly uniform, homogenous universe, the density does sliiightly vary between places. As things cool down, and particles start to condense (and annihilate, and other things), the universe is left with what cosmologists call acoustic waves - basically standing waves. And if you've ever seen &lt;a href=&quot;https://youtube.com/watch?v=wvJAgrUBF4w&quot; rel=&quot;nofollow noreferrer&quot;&gt;videos of a tray of sand being vibrated&lt;/a&gt;, you'll know that one effect is that it leaves some places with more sand, some with less, due to interference patterns. So our universe ends up, as it expands, with some areas denser, some less dense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second effect comes into play. You'll know (or have heard of) dark matter. We don't know what it is made of, but we know it exists (galaxies couldn't form without it, they'd fly apart or take longer than the age of the universe to form), and we know a lot about how it behaves - what forces it responds to, and what forces it doesn't. Interact via gravity - yes, very weakly. Interact via electromagnetic force - no, not at all. That latter bit is crucial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When &quot;ordinary&quot; matter collapses, it heats up. That's how we get stars, for example. The radiation released during collapse also acts as a kind of pressure, opposing collapse, slowing it down. That's why stars like our sun are stable for so long. Dark matter doesn't interact electromagnetically (as far as we know) so it can't experience or create electromagnetic radiation. So when it collapses, it doesn't get hot, it doesn't release radiation... I think you can see where this is going. &lt;em&gt;There's no radiation released during collapse to resist further collapse, so it can collapse much faster than ordinary matter&lt;/em&gt;. As an aside, &lt;strong&gt;because&lt;/strong&gt; it can't release radiation, it also can't jettison the energy that must be got rid of to allow dense objects to form. So it ends up quickly collapsing to a hazy diffuse &quot;halo&quot;, but then can't collapse much more. And no surprise, it collapses in those places where the universe was fractionally denser. So you get what cosmologists call &quot;filaments&quot; and &quot;halos&quot; of dark matter, a bit like a sponge or a swiss cheese, with comparative &quot;voids&quot; separating them. Ordinary matter is more strongly attracted to these already-existing dark matter filaments and halos. It collapses towards them. The self-gravity of the ordinary matter is enhanced by the gravity due to the concentrations of dark matter there - and ordinary matter &lt;strong&gt;can&lt;/strong&gt; lose energy by radiation, so it collapses more than dark matter does, to form the galaxies and their contents which we can see today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity can do this, because the expansion of the universe has by now slowed down so much from its &quot;heyday&quot;, that gravity can pull some of the matter together &lt;strong&gt;within&lt;/strong&gt; space faster than the expansion can &lt;strong&gt;add space between them&lt;/strong&gt;. Over cosmic distances, gravity is much much weaker, and expansion dominates, so clusters and superclusters still move apart, but within clusters, the galaxies and groups of galaxies are accelerated by gravity enough that they mostly stay in their groups and clusters, and move around or orbit within them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we end up with a universe that, on a cosmic scale, we see expansion &quot;winning&quot; as gravity is weak, so we see superclusters moving apart. But within clusters and galaxy groups, we see gravity &quot;winning&quot; because it's stronger over smaller distances, so clusters and gravitationally bound entities like galaxies stay together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this in turn means, is that galaxies and galaxy groups are bound by gravity more than they are separated by expansion. So they remain moving within their clusters and groups, despite universal expansion. And, occasionally, because motion of 3 or more separate bodies under gravity is chaotic (and clusters can contain billions or trillions of galaxies),  entire galaxies will get ejected, or collide, or do whatever galaxies do. And that's how it happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Although you didn't ask, it's a natural question to wonder what happens next. We believe that the rate of expansion has slowly sped up. That means that in the far future (tens and hundreds of billions of years), that galaxies will have to be even closer together, for gravity to dominate expansion. So clusters that are stable now, might break apart in the far future. If expansion accelerates enough then even smaller bodies could ultimately break up, perhaps galaxies themselves, or even stars and atoms. But that's something nobody knows.)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="18601" LastEditorUserId="18601" LastEditDate="2019-05-15T20:54:05.507" LastActivityDate="2019-05-15T20:54:05.507" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31865" PostTypeId="2" ParentId="31856" CreationDate="2019-05-15T03:18:46.030" Score="2" Body="&lt;p&gt;Hmm... here's my guess.&#xA;Cumulus clouds form due to warm, moist air rising. With the atmosphere of Mars being so cold, there's little heat to facilitate the formation of cumulus clouds. Cirrus clouds, on the other hand, are composed of sparse ice crystals, which makes them probably more common on Mars than cumulus clouds. After all, supercooled water does become ice.&lt;/p&gt;&#xA;" OwnerUserId="28028" LastActivityDate="2019-05-15T03:18:46.030" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31867" PostTypeId="2" ParentId="31853" CreationDate="2019-05-15T12:24:11.483" Score="0" Body="&lt;p&gt;Although the Universe is expanding and, in general, the further away a galaxy is from us the faster it appears to be moving away from us. This does not apply to the galaxies in the Local Group. which is a gravitationally bound structure. The Andromeda galaxy is moving towards the Milky Way at about 400,000 km/h and the Milky Way and Andromeda are expected to collide in about 4 billion years time. When this happens, a large new single galaxy will be formed.  The new galaxy which will be formed by the merger is sometimes given the name Milkomeda&#xA;. For more details see &lt;a href=&quot;https://thesciencegeek.org/2015/05/08/ultimate-fate-of-the-universe/&quot; rel=&quot;nofollow noreferrer&quot;&gt;my recent blog post&lt;/a&gt; on this topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over billions of years, Milkomeda will gradually absorb the other Local Group members.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general any gravitationally bound structure such as: stellar systems (e.g. solar system ) our galaxy and groups and clusters of galaxies will not get larger as the Universe expands)&lt;/p&gt;&#xA;" OwnerUserId="28038" LastEditorUserId="28038" LastEditDate="2019-05-16T09:14:09.013" LastActivityDate="2019-05-16T09:14:09.013" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31868" PostTypeId="2" ParentId="31850" CreationDate="2019-05-15T12:30:08.790" Score="1" Body="&lt;p&gt;The Sun is gradually increasing in brightness by 1% every hundred million years. This increase in solar output will, over the long term, cause a gradual warming of the Earth’s surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In roughly 1 billion years, which is long before it becomes a red giant, the combination of the Sun’s gradual increase in output and the &quot;moist greenhouse effect&quot; will make the Earth’s surface too hot for liquid water to exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details see my blog post &lt;a href=&quot;https://thesciencegeek.org/2015/04/26/the-future-of-humanity/&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Future of humanity&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="28038" LastEditorDisplayName="user1569" LastEditDate="2019-05-16T06:56:07.467" LastActivityDate="2019-05-16T06:56:07.467" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31869" PostTypeId="2" ParentId="31767" CreationDate="2019-05-15T12:38:48.480" Score="5" Body="&lt;p&gt;Two further points to make are that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) because of the way we define sunrise and sunset as being when the entire Sun is below the horizon not just the centre of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2)  When the Sun is close to the horizon the Sun’s ray are refracted (bent) by the Earth’s atmosphere so that the Sun appears to be slightly higher in the sky that it would be if the Earth had no atmosphere. For more details see &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_refraction&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Atmospheric_refraction&lt;/a&gt;. This means that the Sun &lt;strong&gt;&lt;em&gt;appears&lt;/em&gt;&lt;/strong&gt; to be above the horizon when it is actually just below it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of these facts there are always more than 12 hours of daylight and thus slightly less than 12 hours of night on the dates of the equinox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dates when there are almost exactly 12 hours between sunrise and sunset are know as the equiluxes and vary according to latitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details see my recent blog post of this topic. &lt;em&gt;(Apologies if this seems like self promotion, it just pulls it all the information together about equinoxes and equiluxes into one single place.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://thesciencegeek.org/2018/03/04/20-march-2018-the-equinox/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://thesciencegeek.org/2018/03/04/20-march-2018-the-equinox/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="28038" LastEditorUserId="28038" LastEditDate="2019-05-17T11:43:07.783" LastActivityDate="2019-05-17T11:43:07.783" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31870" PostTypeId="1" CreationDate="2019-05-15T13:56:39.070" Score="3" ViewCount="441" Body="&lt;p&gt;If I stare at the sun, it's bright and blinding. However, if I look at a star in the sky, it doesn't hurt. &lt;strong&gt;What causes this loss of energy that would otherwise damage one's eyes?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="24434" LastEditorUserId="24434" LastEditDate="2019-05-15T14:10:16.080" LastActivityDate="2019-05-20T06:20:07.973" Title="Why doesn't it hurt to look at stars?" Tags="&lt;star&gt;&lt;light&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31871" PostTypeId="2" ParentId="31870" CreationDate="2019-05-15T15:22:47.553" Score="2" Body="&lt;p&gt;The sun is one &quot;astronomic unit&quot; away. The unit is defined that way. The closest other star is &lt;a href=&quot;https://imagine.gsfc.nasa.gov/features/cosmic/nearest_star_info.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Proxima Centauri&lt;/a&gt; at a distance of 268,770 astronomic units away. The intensity of the light reduces at distance squared; if Proxima Centauri sent out just as much light as the sun we would see it as (268,770)&lt;sup&gt;2&lt;/sup&gt; times fainter. That is approximately 72 billion times fainter. Different stars vary in their luminosity but they are all a long way away, except for the Sun. That is why you can look at them without hurting your eyes.&lt;/p&gt;&#xA;" OwnerUserId="26403" LastEditorUserId="24434" LastEditDate="2019-05-15T18:22:22.720" LastActivityDate="2019-05-15T18:22:22.720" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31873" PostTypeId="2" ParentId="31870" CreationDate="2019-05-15T15:42:05.400" Score="4" Body="&lt;p&gt;What's more, stars are so small to our eyes that they look like &quot;point sources,&quot; meaning that each resolution element in our eyes gathers light from a much larger region of the sky than a distant star can fill.  If you were to look at the Sun from, say, Uranus, and you have good eyes, it would still hurt almost as much because you could still resolve the Sun at that distance (sharp eyes can resolve objects 1/25 the width of the Sun).  That means whatever spot on your optic nerve the sunlight is focused to would be getting the same heat that it gets now-- there would be fewer such spots if you were at Uranus, but a burning spot still hurts.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But any distance larger than 25 astronomical units or so, you would start to not be able to resolve the Sun so even that single spot on the optic nerve would be mixing in light from the dark region around the Sun.  At &lt;span class=&quot;math-container&quot;&gt;$10^4$&lt;/span&gt; times farther away than Uranus, the distance mentioned in the other answer, it means the sunlight would get diluted by the darkness around the Sun &lt;span class=&quot;math-container&quot;&gt;$10^8$&lt;/span&gt; times larger than the Sun, accordingly reducing the heat that spot on the optic nerve receives by &lt;span class=&quot;math-container&quot;&gt;$10^{-8}$&lt;/span&gt;.  Far enough away, you can't even see there is a star there.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="25994" LastEditDate="2019-05-15T18:22:11.950" LastActivityDate="2019-05-15T18:22:11.950" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31874" PostTypeId="1" AcceptedAnswerId="31885" CreationDate="2019-05-15T15:46:55.670" Score="1" ViewCount="179" Body="&lt;p&gt;I’ve written a script based on Meeus’s Astronomical Algorithms to calculate the position of the sun, moon etc. up to about 4000 BC.&#xA;Everything looks good, except when I go to 4000 BC the solstice are about a month after expected dates (they fall mid-July and mid-January.&#xA;I checked through this forum and came across this: &lt;a href=&quot;https://astronomy.stackexchange.com/a/13009&quot;&gt;https://astronomy.stackexchange.com/a/13009&lt;/a&gt; . I looked at the results and saw that we have very similar dates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the precession mean the solstices and equinoxes can shift to this extent? Or have I forgotten a step?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="28041" LastEditorUserId="7472" LastEditDate="2019-05-15T15:57:31.320" LastActivityDate="2019-05-16T15:24:38.247" Title="Date of solstices pre-0 AD" Tags="&lt;the-sun&gt;&lt;solstice&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31875" PostTypeId="1" CreationDate="2019-05-15T16:37:22.270" Score="3" ViewCount="63" Body="&lt;p&gt;In several books I find reported that the first scientist to perform spectrographic observations of celestial objects was Fraunhofer with the spectrographs he himself produced. I do not find though a reference, a paper, in which he presents this work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the NASA astrophysics data systems I find only &lt;a href=&quot;https://ui.adsabs.harvard.edu/search/q=%20%20author%3A%22%5EFraunhofer%2C%20J%22&amp;amp;sort=date%20desc%2C%20bibcode%20desc&quot; rel=&quot;nofollow noreferrer&quot;&gt;the papers concerning his technical work with glasses&lt;/a&gt; but not the observations of Venus, Mars, and other heavenly bodies, he performed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find books on Google Books reporting &lt;a href=&quot;https://books.google.de/books?id=iCs4AAAAIAAJ&amp;amp;pg=PA28&amp;amp;lpg=PA28&amp;amp;dq=fraunhofer%20spectroscopic%20observations%20venus&amp;amp;source=bl&amp;amp;ots=tyiTvtTfXu&amp;amp;sig=ACfU3U3so31AyGlfPSrx5FfR5KwtRCOZZw&amp;amp;hl=it&amp;amp;sa=X&amp;amp;ved=2ahUKEwiy14nz-p3iAhWJYlAKHUjGDhkQ6AEwC3oECAcQAQ#v=onepage&amp;amp;q=fraunhofer%20spectroscopic%20observations%20venus&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;such observations&lt;/a&gt;&#xA;but unfortunately I can view limited parts of these books and not their bibliographies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anybody can find a reference for any of these observations?&lt;/p&gt;&#xA;" OwnerUserId="28042" LastEditorUserId="28042" LastEditDate="2019-05-15T19:27:23.010" LastActivityDate="2019-05-16T07:28:35.913" Title="Reference: First spectrographic observations by Fraunhofer" Tags="&lt;history&gt;&lt;spectroscopy&gt;&lt;resource-request&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31876" PostTypeId="2" ParentId="19085" CreationDate="2019-05-15T16:42:08.037" Score="0" Body="&lt;p&gt;As an aside, on its descent Cassini was supposed to receive Huygens' signal over two channels, but because of an operational commanding error, only one channel was used. Unfortunately only 350 pictures were received on the way down instead of 700 that were expected. More details &lt;a href=&quot;https://www.astrobio.net/titan/what-happened-when-huygens-landed-on-titan-eight-years-ago/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="28038" LastEditorUserId="7472" LastEditDate="2019-05-15T17:00:48.507" LastActivityDate="2019-05-15T17:00:48.507" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31877" PostTypeId="1" CreationDate="2019-05-15T18:17:50.567" Score="1" ViewCount="929" Body="&lt;p&gt;As the title says I am trying to calculate solar coordinates of the sun for a given location and time. I have spent many 10's of hours on this without success so I will sincerely appreciate help on this.  My RA value seems to be right on but the Dec, Alt and Az are off. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alt value is within 6 degrees but I cannot figure out why it is that close because the latitude is in radians instead of degrees like the other values. If I change latitude to degrees the values don't even make sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have included all the values of the intermediate variables to help see at a glance what might be off. I hard coded the JD and the Local Hour Angle just to take those out of the equation for debugging.  Links are shown as to where I got those values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The C# code is below although the programming language doesn't matter. The formulas are the same in any language.&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  //for formulas&#xA;            double D = 2458599.125000 - 2451545.0; //the  first  term is the jd of interest; 2nd term is jd of 1-1-2000 at 12:00:00.0&#xA;            double Latitude = 41.369341;&#xA;            double Longitude = 81.885493;&#xA;&#xA;            double gTmp = 357.529 + 0.98560028 * D;  //mean anomaly of the sun  = 7310.0765751550007&#xA;            double gTmp1 = (gTmp % 360);     //subtract multiples of 360 (returns remainder in degrees)  =  110.07657515500068&#xA;            double g = gTmp1 * Math.PI / 180;   //Convert to radians = 1.9211986657737494&#xA;            double qTmp = 280.459 + 0.98564736 * D; //mean longitude of the sun = 7233.33868336&#xA;            double qTmp1 = (qTmp % 360); //subtract multiples of 360 (returns remainder in degrees) =  33.338683359999777&#xA;            double q = qTmp1;   //don't convert to radians here; convert after terms in degrees are added in the computation of L; =33.338683359999777&#xA;            double L_tmp = q + 1.915 * Math.Sin(g) + 0.020 * Math.Sin(2 * g);//geocentric apparent ecliptic longitude of the sun(adjusted for aberration) = 35.124421105295383&#xA;            var L = (L_tmp % 360) * Math.PI / 180;   //Convert to radians; //subtract multiples of 360 = 0.61303679614439033&#xA;            double eTmp = 23.4393 - 0.00000036 * D;  // = 23.436760515&#xA;            double e = eTmp * Math.PI / 180;   //Convert to radians;   = 0.40904863698815186&#xA;&#xA;            //http://www.neoprogrammics.com/de405_usno_ae98/DE405_Sun.php&#xA;            // formulas per https://aa.usno.navy.mil/faq/docs/SunApprox.php&#xA;            double RA = (Math.Atan2(Math.Cos(e) * Math.Sin(L), Math.Cos(L))) * 180 / Math.PI; //right ascension converted from radians to degrees = 32.838762087228496&#xA;            double dec = (Math.Asin(Math.Sin(e) * Math.Sin(L))) * 180 / Math.PI;  //declination converted from radians to degrees =  3.228747909189217         &#xA;            string[] sunCoordinates = new string[2];   &#xA;            sunCoordinates[0] = ToHMSStr(RA);    // = &quot;2h 11m 21.3s&quot;&#xA;            sunCoordinates[1] = ToDMSstr(dec);  //  = &quot;13d 13m 43.49s&quot;&#xA;&#xA;            //compute alt and az per https://aa.usno.navy.mil/faq/docs/Alt_Az.php and https://aa.usno.navy.mil/faq/docs/GAST.php&#xA;            //LHA for testing is from https://ssd.jpl.nasa.gov/horizons.cgi#results and https://www.vercalendario.info/en/how/convert-ra-degrees-hours.html&#xA;            //from http://www.neoprogrammics.com/de405_usno_ae98/DE405_Sun.php  &#xA;            double LHA = 307.39;   //Local Hour Angle        &#xA;            double lat = Latitude * Math.PI / 180; //Convert to radians; //Latitude is defined above; We get sign (below) after converted to Rad????  = 0.72203120983028346&#xA;            double alt = (Math.Cos(LHA) * Math.Cos(dec) * Math.Cos(lat) + Math.Sin(dec) * Math.Sin(lat)) * 180 / Math.PI; //convert from radians  = 53.266598643452909&#xA;            double az =Math.Atan2(-Math.Sin(LHA) * Math.PI / 180, (Math.Tan(dec)*Math.Cos(lat) * Math.PI / 180 - Math.Sin(lat) * Math.Cos(LHA)) * Math.PI / 180); //convert each term to radians   = 2.4585704411338072&#xA;            az = az * 180 / Math.PI;   //convert from radians   = 140.86570991258415&#xA;&#xA;&lt;span class=&quot;math-container&quot;&gt;```&lt;/span&gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="28044" LastActivityDate="2019-06-15T07:57:36.583" Title="How to calculate solar coordinates. (Right Ascension, Declination, and ultimately the Altitude and Azimuth of the Sun)" Tags="&lt;the-sun&gt;&lt;declination&gt;&lt;azimuth&gt;&lt;right-ascension&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="31882" PostTypeId="2" ParentId="21039" CreationDate="2019-05-15T19:28:26.320" Score="9" Body="&lt;p&gt;The electromagnetic energy density is dominated by the cosmic microwave background and the optical/IR background. &lt;a href=&quot;https://physics.stackexchange.com/questions/196366/does-the-number-density-of-photons-n-gamma-approx-108-mathrm-m-3-refer/196373#196373]&quot;&gt;This Physics SE answer&lt;/a&gt;, contains the plot below, showing the contribution of energy density at different frequencies. You can integrate under this &quot;by eye&quot; to see that the CMB contribution is the largest followed closely by an optical/IR bump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A plot of &lt;span class=&quot;math-container&quot;&gt;$\nu I_\nu$&lt;/span&gt; (proportional to energy density per logarithmic frequency interval), taken from &lt;a href=&quot;https://arxiv.org/abs/1802.03694&quot; rel=&quot;noreferrer&quot;&gt;Hill et al. (2018)&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G8uCd.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G8uCd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CMB is approximately isotropic blackbody radiation with a temperature of 2.73 K and therefore a specific intensity of&#xA;$I_{\nu} = \frac{2h\nu^3}{c^2}\frac{1}{\exp[h \nu/k_B T] -1} $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral of this over all frequencies is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ u = \frac{8\pi^5 (k_B T)^4}{15 (hc)^3} = 4.2 \times 10^{-14}\ \ {\rm J\ m}^{-3} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of power, we just multiply by &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; to get &lt;span class=&quot;math-container&quot;&gt;$1.3\times 10^{-5}$&lt;/span&gt; Wm&lt;span class=&quot;math-container&quot;&gt;$^{-2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you can't harness this unless you have a receiver that is colder than the CMB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do a rough calculation for the contribution of optical background light by noting that the dark sky at a good, dark astronomical site is around 22 mag per square arcsecond in the V band. Only about 0.1-0.2 mag of this could be attributed to light pollution at the best astronomical sites (see for example &lt;a href=&quot;http://www.ing.iac.es/Astronomy/observing/conditions/skybr/skybr.html&quot; rel=&quot;noreferrer&quot;&gt;Benn &amp;amp; Ellison 2007&lt;/a&gt;). Noting that the zero point for the V-band magnitude scale is &lt;span class=&quot;math-container&quot;&gt;$3.6 \times 10^{-12}$&lt;/span&gt; Wm&lt;span class=&quot;math-container&quot;&gt;$^{-2}$&lt;/span&gt; per angstrom for &lt;span class=&quot;math-container&quot;&gt;$V=0$&lt;/span&gt; and taking the optical band as 3000 angstroms, and noting there are &lt;span class=&quot;math-container&quot;&gt;$5.34\times 10^{11}$&lt;/span&gt; square arcseconds in the sky, then a point in space will receive &lt;span class=&quot;math-container&quot;&gt;$10^{-5}$&lt;/span&gt; Wm&lt;span class=&quot;math-container&quot;&gt;$^{-2}$&lt;/span&gt; (or perhaps about half this, because a significant fraction of the dark sky brightness is caused by airglow).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This roughly agrees with what is shown in the plot. Contributions at mid-infrared and at shorter wavelengths than optical do not contribute appreciably.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be thought that discrete sources or the Milky Way might contribute more, but I think that isn't the case. The Milky Way surface brightness is only about a factor of three above the dark sky brightness, but of course occupies a tiny fraction of the total sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A way to see that discrete stars don't contribute much is to note that the optical flux from the entire dark sky is equivalent to about 1000 zeroth magnitude stars or &lt;span class=&quot;math-container&quot;&gt;$10^7$&lt;/span&gt; stars at 10th magnitude. Both numbers are higher by orders of magnitude than the actual numbers of Galactic stars at these brightnesses.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-05-16T13:57:08.183" LastActivityDate="2019-05-16T13:57:08.183" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="31883" PostTypeId="2" ParentId="31877" CreationDate="2019-05-15T21:06:10.467" Score="0" Body="&lt;p&gt;I recommend you to use astropy, it's a python package with plenty of astronomical libraries.&lt;br&gt;&#xA;Check this tutorial on &lt;a href=&quot;http://docs.astropy.org/en/stable/generated/examples/coordinates/plot_obs-planning.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Determining and plotting the altitude/azimuth of a celestial object&lt;/a&gt;.&lt;br&gt;&#xA;You have the specific function&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from astropy.coordinates import get_sun&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hope it helps.&lt;/p&gt;&#xA;" OwnerUserId="28042" LastActivityDate="2019-05-15T21:06:10.467" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31884" PostTypeId="1" CreationDate="2019-05-15T22:33:07.360" Score="3" ViewCount="163" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Himawari_8&quot; rel=&quot;nofollow noreferrer&quot;&gt;Himawari-8&lt;/a&gt; geostationary weather satellite recently had a lunar intrusion into it's primary sensor (the Advance Himawari Imager, or AHI):&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/CYUlg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CYUlg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zooming in on the moon:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/6UxSh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6UxSh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;I see a strange cupping (or 'peeling') of the moon away from the horizon: &#xA;&lt;a href=&quot;https://i.stack.imgur.com/PnItr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PnItr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: what is the explanation for this 'peeling' aberration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: if your answer involves the lunar terminator, please account for the following concerns:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;At the time of day of the photograph, we see the Earth with ~5 degrees of longitude in night. The Earth presents approximately 17 degrees across from GEO, so the moon is &lt;span class=&quot;math-container&quot;&gt;$17^{\circ}/2=8.5^{\circ}$&lt;/span&gt; degrees away from nadir, and so I expect the moon to have &lt;span class=&quot;math-container&quot;&gt;$5^{\circ}+8.5^{\circ}= 13.5^{\circ}$&lt;/span&gt; of longitude in night. With barely half the moon peaking out above the Earth's horizon, the terminator is well behind the Earth.&lt;/li&gt;&#xA;&lt;li&gt;The cupping aberration appears equal on both edges. The Sun is in the top right corner of the picture, and so I would expect a terminator related effect &#xA;to be more pronounced on the bottom of the moon than the top.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="22897" LastActivityDate="2019-05-16T22:36:22.317" Title="Why does the moon 'peel' away from the horizon in this photograph?" Tags="&lt;the-moon&gt;&lt;photography&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31885" PostTypeId="2" ParentId="31874" CreationDate="2019-05-15T22:40:22.547" Score="4" Body="&lt;p&gt;Sounds like Julian Calendar slippage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gregorian Calendar, the one we use now, was created to fix a problem with the Julian Calendar: The fact that the Solar Year wasn't exactly 365.25 days. As a result, compared to the calendar year, the date of the Vernal Equinox (and more importantly at the time, Easter) was slipping forward. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To remedy this, ten calendar days in 1582 (or up to thirteen calendar days in 1918, depending on how long a particular country waited to adopt the Gregorian Calendar) were omitted, and the Gregorian Calendar added a modification to the Julian calendar's rule of One leap year every 4 years: No leap year on century-years that aren't divisible by 400.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your code automatically switches from Gregorian to Julian Calendar at the appropriate year, the adoption of the Julian Calendar probably winds up at about the expected days when the Julian calendar was adopted mid-first-century BCE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But every 400 years further back, your calendar is slipping about three extra days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And at 4000 BCE, that's thirty days.&lt;/p&gt;&#xA;" OwnerUserId="23813" LastEditorUserId="23813" LastEditDate="2019-05-16T15:24:38.247" LastActivityDate="2019-05-16T15:24:38.247" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31889" PostTypeId="2" ParentId="31875" CreationDate="2019-05-16T07:28:35.913" Score="2" Body="&lt;p&gt;With further research I found an Answer in &quot;The Cosmic Century&quot; by Longair (Cambridge).&lt;br&gt;&#xA;When Longair says &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 1823, Fraunhofer made further observations of the spectra of the planets and the brightest stars, anticipating by about 40 years the next serious attempts to measure the spectra of the stars (Fraunhofer, 1823)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the article referenced is &lt;em&gt;Kurzer Bericht von den Resultaten neuerer Versuche über die Gesetze des Lichtes, und die Theorie derselben&lt;/em&gt; (A short account of the results of recent experiments upone the laws of light and its theory.) which was actually fetched by my research in the NASA ADS (&lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/1823AnP....74..337F/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ui.adsabs.harvard.edu/abs/1823AnP....74..337F/abstract&lt;/a&gt;). But the title did not mention any planet or star and I could not access the full text, so I thought it was not the right reference.&lt;/p&gt;&#xA;" OwnerUserId="28042" LastActivityDate="2019-05-16T07:28:35.913" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31890" PostTypeId="1" CreationDate="2019-05-16T11:47:37.007" Score="1" ViewCount="168" Body="&lt;p&gt;Does Moon rotate around itself while it makes one revolution around Earth or not?&lt;/p&gt;&#xA;" OwnerUserId="28048" LastActivityDate="2019-05-16T13:22:46.050" Title="Moon's revolution around the Earth and it's rotation around it's axis?" Tags="&lt;the-moon&gt;&lt;rotation&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2019-05-17T05:04:58.687" ContentLicense="CC BY-SA 4.0" />
  <row Id="31891" PostTypeId="2" ParentId="31890" CreationDate="2019-05-16T12:14:05.583" Score="-2" Body="&lt;p&gt;Yes. Exactly once. This is how we get the light and dark side of the moon. The reason for it being exactly once is due to something called tidal forces. Wikipedia explains this quite well: &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_locking&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Tidal_locking&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="28055" LastActivityDate="2019-05-16T12:14:05.583" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31892" PostTypeId="2" ParentId="31884" CreationDate="2019-05-16T12:22:59.853" Score="1" Body="&lt;p&gt;Looks like an inferior mirage to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://aty.sdsu.edu/explain/simulations/inf-mir/Kaplan_photos.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://aty.sdsu.edu/explain/simulations/inf-mir/Kaplan_photos.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has a work through for the disk of the sun but I think the same general principles apply here as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific effect will be somewhat different since in the image from space rays are moving from space down into and through the atmosphere, and then back up again, whereas in the image below the rays terminate at the Earth's surface.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Simulation and photograph from Andrew T. Young's page &lt;a href=&quot;https://aty.sdsu.edu/explain/simulations/inf-mir/Kaplan_photos.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Photographs of an Inferior-Mirage Sunset&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The inferior mirage is very clearly shown in a series of sunset images taken by George Kaplan, of the U. S. Naval Observatory. Here, I describe what's seen in these images, which are actually on the &lt;a href=&quot;https://aa.usno.navy.mil/index.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Naval Observatory's website&lt;/a&gt;. They are such a “textbook example” of the inferior-mirage sunset that I begged for some details, and permission to show the images here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wce4V.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wce4V.gif&quot; alt=&quot; from Andrew T. Young&amp;#39;s page Photographs of an Inferior-Mirage Sunset&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jpDKn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jpDKn.jpg&quot; alt=&quot;sunset images taken by George Kaplan, of the U. S. Naval Observatory&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="28055" LastEditorUserId="7982" LastEditDate="2019-05-16T22:36:22.317" LastActivityDate="2019-05-16T22:36:22.317" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31894" PostTypeId="2" ParentId="31824" CreationDate="2019-05-16T13:37:06.623" Score="3" Body="&lt;p&gt;It may also help your mental picture to consider a phenomenon many people see every day. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth orbits the Moon quite a distance outside the Moon’s Roche limit. Nevertheless, it is distorted by this. The distortion is a bit less than a metre: that is, a perfectly fluid frictionless inertialess material would settle into an ellipsoidal shape pointing directly towards (and away from) the Moon. Water is not frictionless and inertialess, and in addition the oceans are shallow containers rather like a soup plate, and the water in them sloshes like soup which is why the tides are often much higher (and sometimes non-existent). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aim of this picture is to show that distortion outside the Roche limit is not hard to believe but, on the contrary, an everyday experience. &lt;/p&gt;&#xA;" OwnerUserId="12016" LastActivityDate="2019-05-16T13:37:06.623" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31896" PostTypeId="1" CreationDate="2019-05-16T15:05:20.793" Score="0" ViewCount="43" Body="&lt;p&gt;I have data that contains galactic latitude and longitude of stars. I have to determine all the stars that are within our milky way. How do I find it?&lt;/p&gt;&#xA;" OwnerUserId="28058" LastActivityDate="2019-05-17T05:09:02.090" Title="How do I find stars within milky way using galactic latitude and longitude" Tags="&lt;star&gt;&lt;milky-way&gt;&lt;galactic-center&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31897" PostTypeId="2" ParentId="10295" CreationDate="2019-05-16T16:30:18.203" Score="1" Body="&lt;p&gt;Lookback time, T, and distance, D, can be calculated from the object’s redshift, z, and cosmological parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lookback time, or (less appropriately) light travel time, answers the following similar questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-How old are the photons presently recorded in the object's spectrum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-How long did it take the light to traverse the distance between the object and the observer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-What distance did the light have to travel through an expanding universe in order to arrive at the present time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-When we observe a remote object, how far back in time do we see?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relation between redshift and lookback time is quite complex, and is presented here with simple equations derived by regression analysis from actual results. Correlation coefficients are better than 0.999. The equations are based on the following cosmological parameters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Omega Matter = 0.272&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Omega Radiation = 8.12E-5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Omega Lambda = 0.728&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hubble Constant now = 70.4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is given in billions of years (By)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T = ( 19292 x Z ) / ( 1.3878 + Z ), for Z between 0 and 0.2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T = ( 0.589 + 13.87 x Z^1.25 ) / ( 0.852 + Z^1.25), for Z between 0.2 and 20&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving the second equation for quasar APM 8279, with Z = 3.911, T = 12.10 By &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lookback time also describes the distance the quasar's light travelled through an expanding universe before reaching the observer. This lookback time distance, D, determines by how much the object's light intensity falls off due to distance and due to intergalactic medium extinction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D = C x T,  where C = 1, describes D in light years,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D= ( C x T ) / 3.26, where C = 1, describes D in parsecs&lt;/p&gt;&#xA;" OwnerUserId="28051" LastEditorUserId="28051" LastEditDate="2019-05-18T02:21:32.957" LastActivityDate="2019-05-18T02:21:32.957" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31898" PostTypeId="1" AcceptedAnswerId="31899" CreationDate="2019-05-16T19:34:24.183" Score="5" ViewCount="71" Body="&lt;p&gt;The wobble of a star around its barycenter is interesting (the wobble caused by the pull of the rest of the solar-system on the sun).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question explains why we don't see this wobble from within our own solar-system (because we are mostly orbiting the sun, and not our solar-system's barycenter):&#xA;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/18169/what-point-does-earth-actually-orbit&quot;&gt;What point does Earth actually orbit?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is it possible to detect from here on Earth, the wobble of neighbouring solar-systems relative to the wobble of our solar-system?&lt;/p&gt;&#xA;" OwnerUserId="28060" LastActivityDate="2019-05-16T19:57:22.230" Title="Can we detect the wobble of neighbouring stars around their barycenters?" Tags="&lt;solar-system&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31899" PostTypeId="2" ParentId="31898" CreationDate="2019-05-16T19:57:22.230" Score="6" Body="&lt;p&gt;Yes it is. The Gaia astrometric satellite is and will be doing exactly that. The subtle movement of the photocentre of a target star due to any orbiting planets can be separated from the proper motion and the annual parallax due to our motion around the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The technique is most sensitive to massive planets that are widely separated from  parent stars that are close to the Sun and have low mass themselves, since these provide the biggest angular shift in the star's position with respect to the barycentre. However, the separations cannot be too wide or the orbital period of the star/planet system will be too long for the motion to be detectable on the timescales of the mission. In practice therefore, Gaia will be most sensitive to planets with orbital periods of 1-10 years (or separations of 1-5 au). Simulations suggest that tens of thousands of exoplanets may be revealed in this way (&lt;a href=&quot;https://iopscience.iop.org/article/10.1088/0004-637X/797/1/14/meta&quot; rel=&quot;noreferrer&quot;&gt;Perryman et al. 2014&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ESA have even provided a little &lt;a href=&quot;http://sci.esa.int/gaia/58788-detecting-exoplanets-with-astrometry/&quot; rel=&quot;noreferrer&quot;&gt;animation&lt;/a&gt; to show how it works.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-05-16T19:57:22.230" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31900" PostTypeId="2" ParentId="29825" CreationDate="2019-05-16T20:01:57.983" Score="1" Body="&lt;p&gt;Most of my information is from &lt;a href=&quot;https://science.sciencemag.org/content/357/6355/1026&quot; rel=&quot;nofollow noreferrer&quot;&gt;this link&lt;/a&gt;. So here's a watered-down version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the Solar System first formed, numerous concentrations of mass, called 'planetesimals', formed over time from slowly accreting material. Some of these would go on to combine with other planetesimals and form planets in their own right, but some would simply float in space. Over time, these planetesimals would shatter apart into primordial asteroids from impacts with smaller bodies. However, these remnant asteroids would originate from a single body, so they would have about the same physical properties. In addition with our modern orbital calculation techniques*, we would able to identify groups of these primordial asteroids, and flag them as such. However, an asteroid created from a planetary impact would likely be ruled out as non-primordial, since there would be no &quot;brother asteroids&quot; to compare it with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*To clarify, we can trace orbits of asteroids back over billions of years to see where the bodies might have been in the distant past. Asteroids that originated from a single parent body would be observed to converge to a single point.&lt;/p&gt;&#xA;" OwnerUserId="28028" LastEditorUserId="28028" LastEditDate="2019-05-16T20:12:45.053" LastActivityDate="2019-05-16T20:12:45.053" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31901" PostTypeId="1" AcceptedAnswerId="31902" CreationDate="2019-05-16T20:31:53.103" Score="6" ViewCount="292" Body="&lt;p&gt;When I look at a &lt;a href=&quot;https://en.wikipedia.org/wiki/Hertzsprung%E2%80%93Russell_diagram&quot; rel=&quot;noreferrer&quot;&gt;Hertzsprung-Russell diagram&lt;/a&gt;, I'm able to see a main-sequence line from the most luminous main-sequence stars at the top left to the dimmest main-sequence stars at the bottom right. However, this line isn't straight - it's quite squiggly. Why is the line squiggly, and not defined by a polynomial or rational function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JlYXq.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JlYXq.png&quot; alt=&quot;The Hertzsprung-Russell diagram, courtesy of Wikipedia. Shows the main-sequence line as &amp;quot;V: Main Sequence&amp;quot;.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="28028" LastEditorUserId="16685" LastEditDate="2019-05-17T05:27:59.427" LastActivityDate="2019-05-18T20:27:17.987" Title="On a Hertzsprung-Russell diagram, why is the main-sequence line squiggly?" Tags="&lt;star&gt;&lt;main-sequence&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31902" PostTypeId="2" ParentId="31901" CreationDate="2019-05-16T21:22:48.597" Score="7" Body="&lt;p&gt;Here is another plot of a Hertzsprung Russell diagram (luminosity versus temperature), but this time based on theoretical models. (The plot is from D. Prialnik 2000, An Introduction to the theory of stellar structure and evolution). Note that the zero age main sequence &lt;em&gt;is&lt;/em&gt; well behaved in this plot. Luminosity and temperature are related by smoothly changing power laws that are largely due to the varying temperature dependence of energy generation and modes of energy transport within stars of different mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wvqeW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wvqeW.png&quot; alt=&quot;A Hertzsprung Russell diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your diagram is actually an absolute magnitude versus colour diagram. This is not the same thing. The transformation between absolute magnitude and luminosity and especially between colour and temperature are distinctly non-linear and dependent on what particular spectral features happen to be present in the photometric bands that are used to construct the plot. For example, there is a knee for cool M-dwarfs because the B-V colour &quot;saturates&quot; at a value of about 1.6 because there is hardly any flux in the B and V bands and their ratio doesn't change very much with temperature because of the characteristics of the opacity sources in the atmospheres of these stars at those wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To demonstrate that the plot is not necessarily &quot;wiggly&quot;, the plot below shows the colour-magnitude for the Hyades and Praesepe clusters (basically at the ZAMS) in the Gaia filter system (from &lt;a href=&quot;https://arxiv.org/abs/1804.09378&quot; rel=&quot;nofollow noreferrer&quot;&gt;Babusiaux et al. 2018&lt;/a&gt;). Nice and smooth - though it doesn't cover as big a luminosity range as the Hipparcos plot in your question; the main wiggle at B-V~0.3 is still in the Gaia data (off the top of the plot), but the M-dwarf &quot;knee&quot; is smoothed out by the redder filter response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iOCH9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iOCH9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-05-18T20:27:17.987" LastActivityDate="2019-05-18T20:27:17.987" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31903" PostTypeId="1" AcceptedAnswerId="31926" CreationDate="2019-05-16T22:33:09.737" Score="2" ViewCount="93" Body="&lt;p&gt;This is closely related, but distinct question to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/22545/what-are-the-chances-of-a-star-colliding-with-another-during-a-galactic-collision&quot;&gt;What are the chances of a star colliding with another during a galactic collision?&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The accepted answer to that question places the odds of the Sun colliding with another star at about &lt;span class=&quot;math-container&quot;&gt;$10^{-12}$&lt;/span&gt;, what I want to ask is how near a near-miss is expected? In other words, what is the expected closest encounter between the Sun and another star during the collision? A more sophisticated answer would give the expected distribution of close encounters by distance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more general version of the question is -- what disruptive events would a typical star such as the Sun expect to experience during the merger?&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-05-18T18:07:03.297" Title="What is the likely closest encounter for a typical star during the Milky Way Andromeda Merger?" Tags="&lt;galaxy&gt;&lt;milky-way&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31904" PostTypeId="2" ParentId="31870" CreationDate="2019-05-16T23:31:45.960" Score="2" Body="&lt;p&gt;Distance &lt;em&gt;is&lt;/em&gt; important, as is surface brightness, but there's also one more thing that has not been let onto here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, a recap. For one, the fact that while &lt;em&gt;total&lt;/em&gt; brightness falls with distance, &lt;em&gt;surface&lt;/em&gt; brightness does not is indeed that - an unassailable fact, as simply observing whether or not the surface brightness of a wall changes as you move closer to, or further from, it. (It doesn't.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you get twice as far from the Sun (e.g. 300 Gm versus 150 Gm in the case of &quot;as far&quot; being Earth's distance), then while its angular size halves and area quarters, as you might expect, that quarter-sized area has just as much brightness - just as much &quot;punch&quot;, if you will - as the full-size one at our distance and thus just as much is capable of burning (that much smaller) a hole in your retina.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you get to the distance of the nearest stars at Proxima and Alpha Centauri, that is 40 Pm distant, or &lt;span class=&quot;math-container&quot;&gt;$40 \times 10^6\ \mathrm{Gm}$&lt;/span&gt;. Centauri provides a reasonable test case since it is a binary system of two stars that are roughly similar to the Sun. Thus, we can roughstimate it as being like a Sun some &lt;span class=&quot;math-container&quot;&gt;$\frac{40\ 000\ 000}{150}$&lt;/span&gt; or about 270,000 times, further away. That makes its angular size on the sky, and thus actual size on our retina - about 1/270,000 that of the Sun. If the spot on the retina is on the order of 1 mm from the ordinary Sun (say 0.1-1 - don't know offhand), then this will be in the range of &lt;span class=&quot;math-container&quot;&gt;$10^{-6}\ \mathrm{mm}$&lt;/span&gt;, i.e. the single-digit &lt;em&gt;nanometer&lt;/em&gt; range, at best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory, such a spot could still perhaps get a few atoms plenty hot depending on just where in that range it falls and could, say, perhaps puncture a cell membrane, but the trick is, in fact, you never actually &lt;em&gt;can&lt;/em&gt; get a spot in single-digit nanometers with visible light, much less near-infrared, because of &lt;em&gt;diffraction&lt;/em&gt;: the wavelengths of visible light are at about 400-750 nm, and that sets the scale of the largest &quot;spot&quot; possible since a full wave cycle could not fit in any smaller an area. Effectively, the distant star is &lt;em&gt;not&lt;/em&gt; a tiny spot, even with perfect focus, but actually a tiny &lt;em&gt;blur&lt;/em&gt; on your retina and even then, I believe many, if not all, human eyes don't even reach the diffraction limit to begin with (&lt;em&gt;esp.&lt;/em&gt; if you're myopic!), making it even &lt;em&gt;more&lt;/em&gt; blurred. At even the low end of this range, 400 nm, the surface brightness now &lt;em&gt;is&lt;/em&gt; diluted because it's the same total light - that entered the eyeball - spread over a larger area, on the order of &lt;span class=&quot;math-container&quot;&gt;$400^2$&lt;/span&gt; or &lt;em&gt;16,000&lt;/em&gt; times fainter &lt;em&gt;per unit area&lt;/em&gt;. Clearly, that is not going to injure anything &lt;em&gt;at all&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9807" LastEditorUserId="9807" LastEditDate="2019-05-17T12:17:56.900" LastActivityDate="2019-05-17T12:17:56.900" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31906" PostTypeId="2" ParentId="31896" CreationDate="2019-05-17T05:09:02.090" Score="1" Body="&lt;p&gt;Stars beyond our own galaxy and its immediate neighbors are indistinct without a large telescope, so I assume the question is about the band across the sky, rich with faint stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Milky Way's boundary is fuzzy, so its extent is expressed in terms of isophotes: lines of constant brightness per unit solid angle (e.g. square degree).&#xA;It is wider near the central bulge (galactic longitudes around 0&amp;deg; or 360&amp;deg;) than away from it (around 180&amp;deg;).&#xA;Most stars within galactic latitudes &amp;plusmn;10&amp;deg; are in the Milky Way; most outside &amp;plusmn;15&amp;deg; are not.&#xA;At galactic latitudes in between, galactic longitude and the choice of isophote determine whether a star is in or out.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2019-05-17T05:09:02.090" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31908" PostTypeId="1" CreationDate="2019-05-17T08:49:14.150" Score="1" ViewCount="75" Body="&lt;p&gt;I believe there is a difference between the mass calculated using the orbital speed and the stellar mass of the galaxy. So how is the stellar mass calculated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you :)&lt;/p&gt;&#xA;" OwnerUserId="28063" LastEditorUserId="28063" LastEditDate="2019-05-17T13:18:46.603" LastActivityDate="2019-05-17T21:32:21.430" Title="How is stellar mass of a galaxy obtained?" Tags="&lt;star&gt;&lt;mass&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31910" PostTypeId="1" AcceptedAnswerId="31915" CreationDate="2019-05-17T10:44:21.907" Score="1" ViewCount="49" Body="&lt;p&gt;If a Planet have 2 Moons (so suppose system of 3 bodies) and we say &quot;One Moon is tidally locked to Planet&quot; how can we say that it is tidally locked when Baricenter of 3 bodies has its trajectory?  &lt;/p&gt;&#xA;" OwnerUserId="28048" LastActivityDate="2019-05-17T17:44:55.720" Title="tidaly locked celestial body in three planet system" Tags="&lt;tidal-locking&gt;&lt;planetary-systems&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31911" PostTypeId="2" ParentId="31772" CreationDate="2019-05-17T12:04:16.083" Score="2" Body="&lt;p&gt;Just found out this method in casjobs.Previously I came across one SQL code in Stackexchange in which you have to input one ObjID to get its stokes parameters and isoA and isoB values.In sample codes in DR7, you can edit the line which gives Feild values and coordinates to get Stokes parameters(type q_r,u_r in their place). &lt;a href=&quot;https://i.stack.imgur.com/VxDjj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VxDjj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From stokes parametres you get the ellipiticity of the galaxy from this equation (which can be done in python.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$e  = 1 - \frac{b}{a} = 1 - \frac{1 - \sqrt{Q^2 + U^2}}{1 + \sqrt{Q^2 + U^2}}.&#xA;$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if you are downloading the Casjobs query data in CSV or excel format,You need to write python code to read the file and apply the formula&#xA;&lt;a href=&quot;https://dzone.com/articles/python-101-reading-and-writing&quot; rel=&quot;nofollow noreferrer&quot;&gt;(to learn how to read csv files in python click here)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="27937" LastActivityDate="2019-05-17T12:04:16.083" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31914" PostTypeId="2" ParentId="18154" CreationDate="2019-05-17T16:39:10.477" Score="2" Body="&lt;p&gt;Two questions that don't seem to have been answered yet:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why had people previously thought that the Sun is fixed?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fact, since the time of Newton in the late 1600's people have known that the Sun is in motion, wobbling around the center of mass of the solar system. And, before Copernicus, people thought the Earth was fixed, and the Sun moved around it. So the idea that the Sun was perfectly fixed really only lasted about 130 years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the reason Copernicus assumed it was fixed is that, at the time, that was the simplest model that would match the motions of the planets that he could observe.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this correct planets in helix path or spring like path?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, sort of. That picture looks like a frame from a video that's been circulating on the internet, and the video overall is hugely wrong in a lot of ways. But it is true that the whole solar system is moving around the center of the galaxy, and the plane of the solar system is tilted about 60&lt;span class=&quot;math-container&quot;&gt;$^\circ$&lt;/span&gt; compared to the plane of the galaxy. So, if you combine the motion of the planets around the sun with the motion of the whole solar system around the center of the galaxy, you do get a kind of corkscrew pattern. But if you hear the term &quot;vortex&quot; for the solar system, that's a sign that you're in nonsense territory.&lt;/p&gt;&#xA;" OwnerUserId="7752" LastActivityDate="2019-05-17T16:39:10.477" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31915" PostTypeId="2" ParentId="31910" CreationDate="2019-05-17T17:44:55.720" Score="3" Body="&lt;p&gt;Tidal locking just means that one side of the Moon will face the planet all the time. The period of rotation around its own axis is the same as the period of rotation around the planet. It is not absolutely &quot;locked&quot; exactly in that configuration, but on average the two periods will be the same. Another moon or two could possibly also be in the same condition, and there wouldn't be strong interference between them. There could be some slow processes that could work against both remaining tidally locked, or some slow processes that kept them that way, but it's a complicated problem&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-05-17T17:44:55.720" CommentCount="0" CommunityOwnedDate="2019-05-17T17:44:55.720" ContentLicense="CC BY-SA 4.0" />
  <row Id="31917" PostTypeId="1" AcceptedAnswerId="31924" CreationDate="2019-05-17T20:43:20.333" Score="5" ViewCount="498" Body="&lt;p&gt;The TESS input catalog (TIC) &quot;uses the GAIA DR2 catalog as a base&quot; according to the &lt;a href=&quot;https://tess.mit.edu/science/tess-input-catalogue/&quot; rel=&quot;noreferrer&quot;&gt;catalog's website&lt;/a&gt;. The Gaia DR2 database provides &lt;code&gt;source_id&lt;/code&gt;s for each observation of a target. I want to get the TIC number for a target based on the Gaia &lt;code&gt;source_id&lt;/code&gt;. Is there an accepted way of doing this that I'm missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more specific, I can get a list of Cepheids from Gaia DR2 using &lt;a href=&quot;https://gea.esac.esa.int/archive/&quot; rel=&quot;noreferrer&quot;&gt;their archive&lt;/a&gt; and the ADQL command &lt;code&gt;select * from gaiadr2.vari_cepheid&lt;/code&gt;. The first result has the Gaia &lt;code&gt;source_id&lt;/code&gt; as &lt;code&gt;4052922352453886976&lt;/code&gt;. How can I find the TIC number for this target star? Is the best way of doing this using the RA and DEC of the Gaia observation and finding the closest match in the TIC? Since the TIC is based on the Gaia DR2, I would assume there would be a direct corresponding target and finding the closest RA and DEC match would be unnecessary.&lt;/p&gt;&#xA;" OwnerUserId="28071" LastActivityDate="2019-05-18T17:19:17.270" Title="Obtaining TESS input catalog (TIC) IDs for Gaia `source_ids`?" Tags="&lt;gaia&gt;&lt;tess&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31918" PostTypeId="2" ParentId="31908" CreationDate="2019-05-17T21:32:21.430" Score="1" Body="&lt;p&gt;In short, you count stars. Of course you can't count or even see all the stars, so you have to count certain types of (bright) stars and then assume that the ratio of these to all the stars in their neighbourhood is similar to the ratio in a local part of the Galaxy where we can do a more complete census.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we have an idea of a number for the stars in any region, we can use an assumed mass distribution for the stars to calculate the total stellar mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are details and nuances (e.g. the form of the stellar mass distribution depends on the age of the population), but that's the basic process in our own Galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other galaxies, where you don't resolve individual stars, you work on the basis of mass to luminosity ratios for typical populations to convert a luminosity to a stellar mass.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-05-17T21:32:21.430" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31920" PostTypeId="2" ParentId="539" CreationDate="2019-05-18T05:00:20.870" Score="1" Body="&lt;p&gt;Well, it depends on how you define a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://astronomiac.com/how-many-planets-are-there-in-the-solar-system/&quot; rel=&quot;nofollow noreferrer&quot;&gt;IAU&lt;/a&gt; (InternationalAstronomical Union), an object should satisfy three conditions for it to be called a planet&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It should revolve around the sun.&lt;/li&gt;&#xA;&lt;li&gt;It should have enough gravity to form itself into a sphere.&lt;/li&gt;&#xA;&lt;li&gt;It should have cleared space around itself (it should be a dominant object in that area)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Pluto was demoted as a planet because it failed the &lt;strong&gt;third criteria&lt;/strong&gt;. &#xA;Currently, &lt;a href=&quot;https://astronomiac.com/how-many-planets-are-there-in-the-solar-system/&quot; rel=&quot;nofollow noreferrer&quot;&gt;there are only Eight planets&lt;/a&gt; that satisfy the three conditions&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Mercury&lt;/li&gt;&#xA;&lt;li&gt;Venus&lt;/li&gt;&#xA;&lt;li&gt;Earth&lt;/li&gt;&#xA;&lt;li&gt;Mars&lt;/li&gt;&#xA;&lt;li&gt;Jupiter&lt;/li&gt;&#xA;&lt;li&gt;Saturn&lt;/li&gt;&#xA;&lt;li&gt;Uranus&lt;/li&gt;&#xA;&lt;li&gt;Neptune&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are five dwarf planets officially recognised by IAU &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Pluto&lt;/li&gt;&#xA;&lt;li&gt;Ceres&lt;/li&gt;&#xA;&lt;li&gt;Eris&lt;/li&gt;&#xA;&lt;li&gt;Haumea&lt;/li&gt;&#xA;&lt;li&gt;Makemake&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28039" LastActivityDate="2019-05-18T05:00:20.870" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31923" PostTypeId="1" CreationDate="2019-05-18T16:19:56.243" Score="2" ViewCount="70" Body="&lt;p&gt;I need this for worldbuilding, but it is a physics question. I want to be able calculate the energy over distance from a hypernova of a variable star like Eta Carinae. Understanding what energy levels would be dangerous to a planet, and dangerous on what level, would also be necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scientists have numbers saying that a normal supernova would damage Earth from within 50 to 100 light years. I'm more interested in hypernovae, because Eta Carinae is 7500 light years from Earth, and some scientists are saying it could strip our ozone layer literally any minute, day, year, or century now. I have to wonder: if it might be that bad here, how bad is it 5000 light years or 1000 light years away?&lt;a href=&quot;https://en.wikipedia.org/wiki/Eta_Carinae#Possible_effects_on_Earth&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Eta_Carinae#Possible_effects_on_Earth&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know we have models to predict these, but after repeated attempts at searching I cannot find them, and maybe I just don't know the correct terms to search. I like this answer to a similar question: &lt;a href=&quot;https://worldbuilding.stackexchange.com/a/19002&quot;&gt;https://worldbuilding.stackexchange.com/a/19002&lt;/a&gt; but it appears to be about planets orbiting the dying star having a normal supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to be able to take the characteristics of the star and create a distance spectrum that goes from planetary evaporation to an aurora light show for a wide range of planets: rocky, Neptunian, or Jovian . Being able to estimate that of any star would be a bonus, but I mostly just want to be able to do massive variable stars.&lt;/p&gt;&#xA;" OwnerUserId="28076" LastActivityDate="2019-05-18T17:35:41.107" Title="Astrophysics of Supernovae - Energy over Distance?" Tags="&lt;astrophysics&gt;&lt;supernova&gt;&lt;high-energy-astrophysics&gt;&lt;variable-star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31924" PostTypeId="2" ParentId="31917" CreationDate="2019-05-18T17:19:17.270" Score="3" Body="&lt;p&gt;You can query TIC by Gaia ID. Using the &lt;a href=&quot;https://mast.stsci.edu/portal/Mashup/Clients/Mast/Portal.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;MAST Portal&lt;/a&gt;, MAST Catalogs, TESS Input v7, Advanced Search form, I specified:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;GAIA ID: 4052922352453886976&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and got a single result:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;TIC ID (ID):            50559830&#xA;RA (ra):                275.945292 (18:23:46.870)&#xA;Dec (dec):              -25.672353 (-25:40:20.47)&#xA;TESS Mag. (Tmag):       13.579&#xA;Version (version):      20180610&#xA;UCAC ID (UCAC):         322-168973&#xA;2MASS ID (TWOMASS):     18234686-2540204&#xA;GAIA ID (GAIA):         4052922352453886976&#xA;GAIA G Mag. (GAIAmag):  14.5068&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since the MAST portal currently provides TICv7, the Gaia ID is probably from Gaia DR1.&#xA;The &lt;a href=&quot;https://outerspace.stsci.edu/display/TESS/TIC+and+CTL+Data+Release+Notes+Home+Page&quot; rel=&quot;nofollow noreferrer&quot;&gt;TIC release notes&lt;/a&gt; say Gaia DR2 data are new in TICv8.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;9&amp;nbsp;&amp;nbsp;GAIA&amp;nbsp;&amp;nbsp;The matched Gaia DR2 identifier has been updated in TICv8.&#xA;  Gaia IDs in TICv8 are included for stars that are found in the Gaia-provided Gaia-2MASS look-up table.&#xA;  These identifiers have been updated by the Gaia team from DR1 to DR2, and may not be consistent between TICv7 and TICv8.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="616" LastActivityDate="2019-05-18T17:19:17.270" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31925" PostTypeId="2" ParentId="31923" CreationDate="2019-05-18T17:35:41.107" Score="1" Body="&lt;p&gt;In short, there are no nice standard formulas for this. One can make some order-of-magnitude calculations, though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key formula you need is the inverse-square law: the intensity of a spherical energy source falls off with the inverse square of distance. &lt;span class=&quot;math-container&quot;&gt;$$I(r)=\frac{I}{r^2}.$$&lt;/span&gt; The useful thing is that if you know that some source with intensity &lt;span class=&quot;math-container&quot;&gt;$I_1$&lt;/span&gt; has certain effects at distance &lt;span class=&quot;math-container&quot;&gt;$r_1$&lt;/span&gt; (say, a supernova being dangerous at 50 light years) then you get the same effects from a source with intensity &lt;span class=&quot;math-container&quot;&gt;$I_2$&lt;/span&gt; at a distance &lt;span class=&quot;math-container&quot;&gt;$$r_2=\sqrt{\frac{I_2}{I_1}}r_1.$$&lt;/span&gt; So a hypernova 10 times more luminious than a supernova would be dangerous out to about 158 lightyears (assuming the initial 50 lightyear range - this is where different papers and models might quibble a fair bit). Notice that you need a fairly large luminosity to get a vast range in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hypernovas are likely progenitors of gamma-ray bursts that release most of the energy along fairly narrow jets, 2-20 degrees across. That would amp up the range significantly. If the jet covers a fraction &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt; of the sky, that means the range is now &lt;span class=&quot;math-container&quot;&gt;$$r_{2 GRB}=\sqrt{\frac{I_2}{f I_1}}r_1$$&lt;/span&gt; where the intensities are the true energy releases. For an opening-half angle &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$f=1-\cos(\theta)$&lt;/span&gt; so we should expect &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt; to be between 0.00015 and 0.015, basically increasing the range by a factor 81 to 8.1 if one is unlucky enough to be in the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact effect on different planets depends on many complex factors. Planetary evaporation is not a simple process unless the star delivers far more energy than the gravitational binding energy. The absorbed energy &lt;span class=&quot;math-container&quot;&gt;$\pi \epsilon R^2 I(r)$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$0&amp;lt;\epsilon&amp;lt;1$&lt;/span&gt; is the absorption efficiency must be much larger than &lt;span class=&quot;math-container&quot;&gt;$3GM^2/5R$&lt;/span&gt;, or &lt;span class=&quot;math-container&quot;&gt;$$ I(r) \gg \frac{3GM^2}{5\pi\epsilon R^3} $$&lt;/span&gt; This scales proportional to mass and density, so to evaporate a ten times more massive planet you need ten times more radiation, assuming the same density. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Merely melting the surface or blowing off the atmosphere would depend on a lot of geophysical details. &lt;span class=&quot;math-container&quot;&gt;$\epsilon$&lt;/span&gt; depends a lot on wavelengths, whether there is a plasma layer forming and similar messy issues. The light curve timecourse also matters, since some of the processes will be hydrodynamic flows rather than instant shockwaves or long-term heating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A crude model might be that the absorbed energy heats up the atmosphere initially as &lt;span class=&quot;math-container&quot;&gt;$$\Delta T = \frac{\epsilon I(r)}{ C_P M_a}$$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$C_P$&lt;/span&gt; is the specific heat and &lt;span class=&quot;math-container&quot;&gt;$M_a$&lt;/span&gt; the mass of a one-square meter column of atmosphere. For Earth, &lt;span class=&quot;math-container&quot;&gt;$M_a=10300$&lt;/span&gt; kg and &lt;span class=&quot;math-container&quot;&gt;$C_P=1.00$&lt;/span&gt; kJ/kg.K. A &lt;span class=&quot;math-container&quot;&gt;$10^{47}$&lt;/span&gt; J hypernova at a distance of 50 ly radiating evenly would deposit &lt;span class=&quot;math-container&quot;&gt;$10^{47}/4\pi (50 \text{ly})^2\approx 35\cdot 10^9$&lt;/span&gt; W/m&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt;, heating atmosphere if we assume &lt;span class=&quot;math-container&quot;&gt;$\epsilon=0.1$&lt;/span&gt; by 345 K - enough to broil us, but not enough to evaporate the oceans (there &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is 1000 kg per &lt;em&gt;meter&lt;/em&gt; depth, and &lt;span class=&quot;math-container&quot;&gt;$C_P=4.2$&lt;/span&gt; kJ/kg.K). Were it just a supernova 100 times less intensive the effect is a balmy 3.45 K. On the other hand, with a focused GRB beam we can get a hundred to 10,000 times more heating (at which point the above formula stops being valid since the air ionises and becomes plasma). &lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2019-05-18T17:35:41.107" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31926" PostTypeId="2" ParentId="31903" CreationDate="2019-05-18T18:07:03.297" Score="2" Body="&lt;p&gt;Suppose the sun moves through a column of Andromeda stars &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; parsecs long with density &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; and random positions. The probability of encountering at least one star within &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; of the sun along this column would be &lt;span class=&quot;math-container&quot;&gt;$$F(r)=1-e^{-\pi r^2 L \rho}$$&lt;/span&gt; (basically, what is the probability of no &quot;success&quot; when doing a Poisson sample of the column volume times density). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting &lt;span class=&quot;math-container&quot;&gt;$F(r_{1/2})=1/2$&lt;/span&gt; to find the distance where we should expect a 50% chance of a star more nearby we get &lt;span class=&quot;math-container&quot;&gt;$r_{1/2}=\sqrt{\ln(2)/\pi L \rho}$&lt;/span&gt;. If we plug in &lt;span class=&quot;math-container&quot;&gt;$\rho=0.14 $&lt;/span&gt; per cubic parsec and &lt;span class=&quot;math-container&quot;&gt;$L=1000$&lt;/span&gt; parsec (assuming a flat disk-disk collision) I get &lt;span class=&quot;math-container&quot;&gt;$r_{1/2}=0.02$&lt;/span&gt; parsec. That is 4125.296 AU, reassuringly far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the mean distance we can use &lt;a href=&quot;https://stats.stackexchange.com/questions/10159/find-expected-value-using-cdf&quot;&gt;the not very well known but useful formula&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$E[r]=\int_0^\infty (1-F(r)) dr =\int_0^\infty e^{-\pi r^2 L \rho} dr = 1/2\sqrt{L \rho}\approx 0.042$&lt;/span&gt; parsec.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on your views of disk thickness and angle on merger &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; may vary a fair bit. A sideways collision might have &lt;span class=&quot;math-container&quot;&gt;$L=67$&lt;/span&gt; kpc reducing the distances by a factor of 8.1. Direct collisions still remain very unlikely, but messed up Oort clouds seem very likely. &lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2019-05-18T18:07:03.297" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31927" PostTypeId="1" CreationDate="2019-05-18T18:42:22.090" Score="6" ViewCount="169" Body="&lt;p&gt;I was puzzled by this press release by the Royal Astronomical Society: &lt;a href=&quot;https://phys.org/news/2019-05-small-hardy-planets-survive-death.html&quot; rel=&quot;noreferrer&quot;&gt;&quot;Small, hardy planets most likely to survive death of their stars&quot; - phys.org, May 14, 2019&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Astrophysicists from the Warwick Astronomy and Astrophysics Group modeled the chances of different planets being &lt;strong&gt;destroyed by tidal forces&lt;/strong&gt; when their host stars become white dwarfs and have determined the most significant factors that decide whether they avoid destruction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to the shell theorem, the gravitational forces of a spherically symmetrical body's regions, in terms of how they act on external objects, sum up to the functional equivalent of a point mass. A star therefore acts like a point gravity source both before and after becoming a white dwarf (albeit a lower-mass one after, due to casting off matter that becomes a planetary nebula — a drop in attraction that I'd expect to cause any remaining planets, those not close enough to have been slowed and swallowed by the red giant phase, to migrate outward).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why would a star, after turning into a white dwarf, exert &lt;em&gt;stronger&lt;/em&gt; tidal forces on its planets?&lt;/p&gt;&#xA;" OwnerUserId="22118" LastEditorUserId="7982" LastEditDate="2019-05-18T19:34:43.900" LastActivityDate="2019-05-19T14:35:57.483" Title="Why would tidal forces on planets become more intense when a star becomes a white dwarf?" Tags="&lt;planet&gt;&lt;tidal-forces&gt;&lt;white-dwarf&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31928" PostTypeId="1" CreationDate="2019-05-19T00:26:08.577" Score="1" ViewCount="178" Body="&lt;p&gt;I’m studying specific intensity for a course on astronomical instrumentation. I can’t quite get my head around the proper definition of the term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One definition I’ve found in various lecture notes / books / powerpoint slides online is that specific intensity is the power per unit frequency INCIDENT upon a unit area INCIDENT from a unit solid angle. An often given example, for instance in the book of Lena, is that if a telescope records a power P from a star, incident from a solid angle S onto a pixel of area A, then the specific intensity of the source would be P/SA per unit frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the second definition I keep finding is that the specific intensity is the power per unit frequency, this time passing through a unit area falling into a unit solid angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, are these two equivalent? From the 2D geometry equivalent , I know that an incident pencil of rays through a point causes the same pencil to come out again on the other side. But this feels wrong to appeal to, since the physical interpretations feel so different: incident light is very different from emergent light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my above interpretation is true, shouldnt there at least be a sign difference between the emergent and incident definitions such that the integral of the specific intensity over all solid angles yields 0 if no absorption/emission takes place?&lt;/p&gt;&#xA;" OwnerUserId="18857" LastActivityDate="2019-05-19T00:26:08.577" Title="A good definition of specific intensity?" Tags="&lt;radiative-transfer&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31929" PostTypeId="2" ParentId="31927" CreationDate="2019-05-19T02:15:45.033" Score="2" Body="&lt;p&gt;From &lt;a href=&quot;https://ras.ac.uk/sites/default/files/2019-05/1904.03195.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;the paper, p.2&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The star eventually transitions into a white dwarf, which is comparable in size to the Earth but has a Roche radius which extends outward to about one Solar radius ...How objects can be emplaced at one Solar radius from several au – and then circularised – is still subject to debate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are considering planets which have been observed in close orbits (orbital period as short as 2 hours) around white dwarfs.&lt;/p&gt;&#xA;" OwnerUserId="17743" LastActivityDate="2019-05-19T02:15:45.033" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31930" PostTypeId="1" CreationDate="2019-05-19T03:01:24.843" Score="6" ViewCount="98" Body="&lt;p&gt;I tried looking around but couldn't find any sources attempting to answer this particular question.  It's obviously impossible to know the exact answer, but all I'm looking for is an estimate.  My best guess would be that it's about a tenth (or less) of the mass of the asteroid belt, with a composition roughly 10-15% metallic and 85-90% carbonaceous or silicate.  Is anyone aware of any sources that attempt to give an estimate?  If not, what would you say is a reasonable estimate? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.  &lt;/p&gt;&#xA;" OwnerUserId="27844" LastActivityDate="2021-01-05T06:20:23.843" Title="Are there any estimates for the combined mass of all near earth objects (NEOs)?" Tags="&lt;asteroids&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31931" PostTypeId="1" CreationDate="2019-05-19T07:32:37.310" Score="3" ViewCount="63" Body="&lt;p&gt;I have some data that I collected from observations at a specific RA, DEC and I am trying to figure out what telescope surveys were looking at that portion of the sky at that particular time? I am currently trying to use NAVO's python notebooks to query databases but I'm struggling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated! &lt;/p&gt;&#xA;" OwnerUserId="27993" LastActivityDate="2019-05-19T18:52:43.547" Title="How to check what telescopes were looking at a certain portion of sky at a particular time?" Tags="&lt;telescope&gt;&lt;asteroids&gt;&lt;sky-survey&gt;&lt;meteoroid&gt;&lt;raw-data&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31932" PostTypeId="1" AcceptedAnswerId="31940" CreationDate="2019-05-19T11:52:54.063" Score="1" ViewCount="115" Body="&lt;p&gt;What is the relationship between Periapsis and Apoapsis longitude ? Are they 180° apart ? (ie) Periapsis + 180° = Apoapsis ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related question: if apoapsis is not known, what orbital elements are required to compute it ?&lt;/p&gt;&#xA;" OwnerUserId="28080" LastEditorUserId="28080" LastEditDate="2019-05-19T12:13:40.383" LastActivityDate="2020-12-02T01:36:41.043" Title="Relationship between Periapsis and apoapsis longitude" Tags="&lt;orbital-elements&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31933" PostTypeId="1" AcceptedAnswerId="31948" CreationDate="2019-05-19T12:54:41.070" Score="3" ViewCount="93" Body="&lt;p&gt;I have recently read a bit a out Bok globules. One thing, that is repeatedly emphasised is that they are among the coldest objects in space, with a temperature of ~3 K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is that so? Naively, I would expect the coldest objects to be far out in the super voids. Also, I would expect these globules to have retained heat from the gravitational collapse. After all, the kinetic energy of their free fall should be dissipated, shouldn't it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe someone can elaborate. &lt;/p&gt;&#xA;" OwnerUserId="28082" LastActivityDate="2019-05-19T23:43:03.860" Title="Why are Bok globules so cold?" Tags="&lt;astrophysics&gt;&lt;gas&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31934" PostTypeId="2" ParentId="11255" CreationDate="2019-05-19T13:06:34.963" Score="-2" Body="&lt;p&gt;Jupiter is the most radioactive planet. It has the greatest Electromagnetic field of all planets in Solar system according to Juno space mission ( &lt;a href=&quot;https://en.wikipedia.org/wiki/Juno_(spacecraft)&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Juno_(spacecraft)&lt;/a&gt; ). When we know the size of Jupiter it is obviously logical that it must contain farther most mass of radioactive elements like uranium.&lt;/p&gt;&#xA;" OwnerUserId="28048" LastEditorUserId="28048" LastEditDate="2019-05-20T12:44:22.773" LastActivityDate="2019-05-20T12:44:22.773" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31936" PostTypeId="2" ParentId="18417" CreationDate="2019-05-19T14:02:25.083" Score="-2" Body="&lt;p&gt;There couldn't be a universe where no matter &amp;amp; energy exists;the universe IS matter &amp;amp; energy,so without it,no universe.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-05-19T14:02:25.083" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31938" PostTypeId="2" ParentId="31927" CreationDate="2019-05-19T14:35:57.483" Score="1" Body="&lt;p&gt;They wouldn't become more intense when a star becomes a white dwarf provided the distance from the centre of the star to the planet doesn't decrease. On the contrary,tidal forces would diminish because in order to become a white dwarf the star would have to expel a certain amount of mass.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-05-19T14:35:57.483" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31939" PostTypeId="1" CreationDate="2019-05-19T15:54:28.020" Score="1" ViewCount="56" Body="&lt;p&gt;I am looking at your &lt;a href=&quot;https://astronomy.stackexchange.com/a/1104&quot;&gt;response&lt;/a&gt; to the above question.  You stated:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For Northern Hemisphere, a star that is crossing the North local meridian (that is, between the Zenith and the North Horizon) the latitude is 90−alt=dec−lat&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For Northern Hemisphere, a star that is crossing the South local meridian (that is, between the Zenith and the South Horizon) the latitude is 90−alt=lat−dec.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am simply not seeing this when I looked at image for the culimination of X in : &lt;a href=&quot;http://star-www.st-and.ac.uk/~fv/webnotes/chapter4.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://star-www.st-and.ac.uk/~fv/webnotes/chapter4.htm&lt;/a&gt;&quot; ???&lt;/p&gt;&#xA;" OwnerDisplayName="user28084" LastEditorUserId="616" LastEditDate="2019-05-19T16:08:03.593" LastActivityDate="2019-05-20T05:59:50.000" Title="Question on dec 2013 answer to&quot;How to measure the altitude and azimuth of a star?&quot;" Tags="&lt;fundamental-astronomy&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31940" PostTypeId="2" ParentId="31932" CreationDate="2019-05-19T17:17:53.040" Score="2" Body="&lt;p&gt;The periapsis and apoapsis are 180&amp;deg; apart.&#xA;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kepler's first law&lt;/a&gt;,&#xA;an orbit is an ellipse with the central body at one focus.&#xA;The periapsis (P), central body (F), and apoapsis (A) all lie on the major axis of that ellipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/p1Yq8.png&quot; alt=&quot;Elliptical orbit&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2019-05-19T17:17:53.040" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31941" PostTypeId="1" AcceptedAnswerId="31943" CreationDate="2019-05-19T17:17:53.687" Score="1" ViewCount="104" Body="&lt;p&gt;May anyone tell me does it known if Enceladus has water and is that water salty?&lt;/p&gt;&#xA;" OwnerUserId="28048" LastEditorUserId="2365" LastEditDate="2019-05-19T18:33:05.647" LastActivityDate="2019-05-19T21:05:35.733" Title="Is Enceladus the moon with salty water?" Tags="&lt;saturn&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31942" PostTypeId="1" CreationDate="2019-05-19T18:23:38.857" Score="4" ViewCount="101" Body="&lt;p&gt;There are at least three catalogues that are counting up-to-date discovered exoplanets. What criteria do they use to count the number of planets? Every catalogue has a different count value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at this moment various catalogues as:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;exoplanet.eu&lt;/li&gt;&#xA;&lt;li&gt;Open exoplanet catalogue&lt;/li&gt;&#xA;&lt;li&gt;NASA exoplanet archive&lt;/li&gt;&#xA;&lt;li&gt;etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;show values of confirmed number of exoplanets between approx. 3500 and 4050.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do they disagree with each other?&lt;/p&gt;&#xA;" OwnerUserId="28048" LastEditorUserId="9527" LastEditDate="2020-04-23T12:30:52.503" LastActivityDate="2020-04-23T12:30:52.503" Title="Why does the number of extrasolar planets vary in different sources?" Tags="&lt;exoplanet&gt;&lt;star-catalogues&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31943" PostTypeId="2" ParentId="31941" CreationDate="2019-05-19T18:42:12.637" Score="5" Body="&lt;p&gt;Like many bodies in the outer solar system, Enceladus is covered in water ice. It is also strongly believed to have liquid water underneath the ice, as we have observed (with the Cassini probe) plumes of water from its polar regions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water is salty and alkali due to the dissolved salts and the reactions of water with the underlying rocks. The plumes contain sodium chloride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is a salty ocean under the ice of Enceladus.  There is also likely to be water under the ice of other moons, such as Europa.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2019-05-19T21:05:35.733" LastActivityDate="2019-05-19T21:05:35.733" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31944" PostTypeId="2" ParentId="31931" CreationDate="2019-05-19T18:52:43.547" Score="3" Body="&lt;p&gt;The areas of the sky covered by the major Near Earth Object (NEO) surveys are reported to the Minor Planet Center. You can plot visualizations of that sky coverage data using the &lt;a href=&quot;https://www.minorplanetcenter.net/iau/SkyCoverage.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;sky coverage form&lt;/a&gt; where you can filter by depth, date and survey. The raw data is available, after the surveys give permission for it to be released, from &lt;a href=&quot;https://www.minorplanetcenter.net/iau/info/PointingData.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;the raw data page&lt;/a&gt;. This provides a download link to a tarball of all the pointing data so far, broken down by survey site and then a file per night. The description of the pointing fields is also shown on that page; hopefully it shouldn't be too hard to use a combination of e.g. astroPy's &lt;code&gt;Table&lt;/code&gt; and 'SkyCoord` to read and parse the data.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2019-05-19T18:52:43.547" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31947" PostTypeId="1" AcceptedAnswerId="31976" CreationDate="2019-05-19T23:07:30.807" Score="3" ViewCount="86" Body="&lt;p&gt;If I'm given a star's starting mass, starting composition (metallicity), and its current age, How do I calculate the star's luminosity, current mass, and radius?&#xA;I want to know how to do this for any star, from the minimum size of 0.08 solar masses all the way up to stars 100-150 solar masses for a personal project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that there are equations that relate these things, but my desperate googling has only found bits and pieces of the answer I'm seeking. Also, many of the equations I've found only apply to stars within a certain mass range (usually near the sun.)&#xA;I would seriously appreciate some guidance from an expert in the field.&lt;/p&gt;&#xA;" OwnerUserId="28089" LastActivityDate="2019-05-21T06:10:48.683" Title="Given a star's mass, age, and composition, how do I calculate the rest of its properties?" Tags="&lt;star&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31948" PostTypeId="2" ParentId="31933" CreationDate="2019-05-19T23:43:03.860" Score="3" Body="&lt;p&gt;Most of my information is sourced from here:&#xA;&lt;a href=&quot;https://apod.nasa.gov/apod/ap990511.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://apod.nasa.gov/apod/ap990511.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize the source, Bok globules are extremely cold because of their composition -- they're generally just clouds of pre-stellar material which block out all light. This makes the interior of Bok globules &quot;shielded&quot; from interstellar radiation that would otherwise heat them up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you claim that the globules should have retained some heat from gravitational collapse. However, since the globules don't encounter much friction in space, and there aren't many major collisions that produce changes in kinetic energy, there isn't much energy to dissipate. After all, even Bok globules aren't dense objects by any stretch of the imagination.&lt;/p&gt;&#xA;" OwnerUserId="28028" LastActivityDate="2019-05-19T23:43:03.860" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31949" PostTypeId="2" ParentId="31801" CreationDate="2019-05-20T01:21:42.817" Score="5" Body="&lt;p&gt;As Rob Jeffries suggested in a comment, formulating a solution for arbitrary dr/dt requires another parameter, either the energy per unit mass, E/m, or the radius of the shell where the object was dropped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A shell observer at &lt;span class=&quot;math-container&quot;&gt;$r_0$&lt;/span&gt;, sees an object passing by at proper radial velocity &lt;span class=&quot;math-container&quot;&gt;$v_0$&lt;/span&gt;, with corresponding energy/momentum as measured in the local Minkowski frame (using units with c = G = 1)&#xA;&lt;span class=&quot;math-container&quot;&gt;$$m^2=E_0^2-p_0^2=E_0^2-\gamma _0^2 m^2 v_0^2$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{E_0^2}{m^2}=\gamma _0^2 v_0^2+1=\frac{v_0^2}{1-v_0^2}+1=\frac{1}{1-v_0^2}=\gamma _0^2$$&lt;/span&gt;&#xA;where the subscripts indicate local shell coordinates. In Schwarzschild coordinates,  energy per unit mass is constant and defined as&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{E}{m}={ \left(1-\frac{2 M}{r}\right)}\frac{dt}{d\tau}$$&lt;/span&gt;&#xA;From equations 11 and 12 in section 6.3 of the Taylor, Wheeler, Bertschinger book (TWB) linked in the original question&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{E_0}{m}=\left(1-\frac{2M}{r_0}\right)^{-1/2}\frac{dt}{d\tau}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{E}{m}=\sqrt{1-\frac{2M}{r_0}}\frac{E_0}{m}=\gamma _0 \sqrt{1-\frac{2 M}{r_0}}=\sqrt{\frac{1-\frac{2M}{r_0}}{1-v_0^2}}$$&lt;/span&gt;&#xA;The equation for direct radial inward motion is obtained from TWB section 8.4, equation 17, by setting the angular momentum to zero and taking the negative root .&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \left(\frac{dr}{d\tau}\right)^2=(E/m)^2-\left(1-\frac{2M}{r}\right)$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{dr}{dt}=\frac{dr}{d\tau}\frac{d\tau}{dt}= &#xA;\frac{1-\frac{2 M}{r}}{E/m}\frac{dr}{d\tau}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{dr}{dt}=-\left(1-\frac{2M}{r}\right)\sqrt{1-\frac{1-\frac{2M}{r}}{(E/m)^2}}$$&lt;/span&gt;&#xA;Differentiating, and a bit of algebra yields the following result for the acceleration.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{d^2 r}{dt^2}=-\frac{2M}{r^2}\left(1-\frac{2 M}{r}\right)&#xA;\left(1-\frac{3}{2}\frac{1-\frac{2 M}{r}}{(E/m)^2}\right)$$&lt;/span&gt;&#xA;These can be rewritten in terms of the Schwarzschild radius &lt;span class=&quot;math-container&quot;&gt;$r_s=2M$&lt;/span&gt; and proper velocity, &lt;span class=&quot;math-container&quot;&gt;$v_0$&lt;/span&gt;, at &lt;span class=&quot;math-container&quot;&gt;$r_0$&lt;/span&gt;.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{dr}{dt}=-\left(1-\frac{r_s}{r}\right)\sqrt{1-\frac{1-\frac{r_s}{r}}{(E/m)^2}}=-\left(1-\frac{r_s}{r}\right)\sqrt{1-(1-v_0^2)\frac{1-\frac{r_s}{r}}{1-\frac{r_s}{r_0}}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{d^2 r}{dt^2}=-\frac{r_s}{r^2}\left(1-\frac{r_s}{r}\right)&#xA;\left(1-\frac{3}{2}\frac{1-\frac{r_s}{r}}{(E/m)^2}\right)&#xA;=-\frac{r_s}{r^2}\left(1-\frac{r_s}{r}\right)&#xA;\left(1-\frac{3}{2}(1-v_0^2)\frac{1-\frac{r_s}{r}}{1-\frac{r_s}{r_0}}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For known &lt;span class=&quot;math-container&quot;&gt;$dr/dt$&lt;/span&gt; at a radius &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$E/m$&lt;/span&gt; can be determined by inverting the previous expression for &lt;span class=&quot;math-container&quot;&gt;$dr/dt$&lt;/span&gt;. Then substitute &lt;span class=&quot;math-container&quot;&gt;$E/m$&lt;/span&gt; in the equation for acceleration above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$(E/m)^2=\frac{(1-r_s/r)^3}{(1-r_s/r)^2-\left(\frac{dr}{dt}\right)^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alternate approach, dropping an object from rest at radius &lt;span class=&quot;math-container&quot;&gt;$r_0$&lt;/span&gt;, uses equations 37 and 38 from section 6.7 of TWB, resulting in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{dr}{dt}=-\left(1-\frac{r_s}{r}\right)&#xA;\sqrt{\frac{\frac{r_s}{r}-\frac{r_s}{r_0}}{1-\frac{r_s}{r_0}}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{d^2 r}{dt^2}=\frac{r_s\left(1-\frac{r_s}{r}\right)}&#xA;{2 r^2\left(1-\frac{r_s}{r_0}\right)}&#xA;\left(1+\frac{2r_s}{r_0}-\frac{3r_s}{r}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shell radius &lt;span class=&quot;math-container&quot;&gt;$r_0$&lt;/span&gt; necessary to obtain a desired velocity at shell &lt;span class=&quot;math-container&quot;&gt;$r_1$&lt;/span&gt; can be found using&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{E_0}{E_1}=\sqrt{\frac{1-\frac{r_s}{r_1}}{1-\frac{r_s}{r_0}}}&#xA;=\frac{\gamma_0}{\gamma_1}$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\gamma_0=1$&lt;/span&gt; for an object dropped from rest at &lt;span class=&quot;math-container&quot;&gt;$r_0$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, a plot is shown below using &lt;span class=&quot;math-container&quot;&gt;$v_1=1/2$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$r_1=2r_s$&lt;/span&gt;. For these values &lt;span class=&quot;math-container&quot;&gt;$\gamma_1=2/\sqrt{3}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$E/m=\sqrt{2/3}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$r_0=3r_s$&lt;/span&gt;. For reference, the green and red curves are Schwarzchild coordinate velocity and acceleration for an object dropped from rest at infinity, &lt;span class=&quot;math-container&quot;&gt;$E/m=1$&lt;/span&gt;. The dashed blue and brown curves are for an object dropped from rest at &lt;span class=&quot;math-container&quot;&gt;$r_0$&lt;/span&gt;, and the solid blue and brown curves for an object launched from &lt;span class=&quot;math-container&quot;&gt;$r_1$&lt;/span&gt; with a Schwarzschild velocity of&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\left(\frac{dr}{dt}\right)_{r_1}&#xA;=-\left(1-\frac{r_s}{r_1}\right)v_1&#xA;=\frac{1}{4}$$&lt;/span&gt;&#xA;Note that the dashed curves are overlaid by the solid ones from &lt;span class=&quot;math-container&quot;&gt;$r_1$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$r_s$&lt;/span&gt;, since the initial conditions were chosen to result in the same velocity at &lt;span class=&quot;math-container&quot;&gt;$r_1$&lt;/span&gt;.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/eYE39.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eYE39.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="28086" LastEditorUserId="28086" LastEditDate="2019-05-20T17:06:15.680" LastActivityDate="2019-05-20T17:06:15.680" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31951" PostTypeId="1" AcceptedAnswerId="31960" CreationDate="2019-05-20T04:34:34.113" Score="-2" ViewCount="279" Body="&lt;p&gt;Mizar and Alcor, each about 83 light-years away (in fact Mizar presents system of four stars and Alcor is binary system. It is not yet known if those 6 stars presents one or two star systems. In further years, data provided by GAIA space telescope will show us if there is one or two systems), have a fainter line-of-sight companion known as&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sidus_Ludoviciana&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ludwig's star&lt;/a&gt;, which was mistaken for a planet in 1722.&#xA;The Wikipedia page only says it is &quot;roughly five times more distant,&quot; citing &lt;a href=&quot;http://www.leosondra.cz/en/mizar/&quot; rel=&quot;nofollow noreferrer&quot;&gt;A New View of Mizar&lt;/a&gt; by Leos Ondra.&#xA;Is a better distance estimate available?&lt;/p&gt;&#xA;" OwnerUserId="28048" LastEditorUserId="28048" LastEditDate="2019-05-27T15:10:12.950" LastActivityDate="2019-05-27T15:10:12.950" Title="Ludwig's star distance" Tags="&lt;star&gt;&lt;distances&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31952" PostTypeId="2" ParentId="31951" CreationDate="2019-05-20T04:34:34.113" Score="-1" Body="&lt;p&gt;Ludwig's star, or &lt;a href=&quot;https://en.wikipedia.org/wiki/Sidus_Ludoviciana&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sidus Ludoviciana&lt;/a&gt; is an apporoximately +8 magnitude star between &lt;a href=&quot;https://en.wikipedia.org/wiki/Mizar_and_Alcor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mizar and Alcor&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first Wikipedia article states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Sidus Ludoviciana&lt;/strong&gt; /ˈsaɪdəs ˌljuːdoʊˌvɪsiˈeɪnə/ is an 8th-magnitude star in the asterism of the Big Dipper in the constellation Ursa Major, halfway between Mizar and Alcor. It was discovered on 2 December 1722 by Johann Georg Liebknecht, who mistook it for a planet and named it after Louis V, Landgrave of Hesse-Darmstadt. A line-of-sight companion with Mizar and Alcor (with a spectral type similar to the latter), &lt;strong&gt;it is roughly five times more distant.&lt;/strong&gt; It has the spectral type A8/F0 III.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since Mizar and Alcor are 83 light-years away from the Sun, that puts Sidus Ludoviciana at &lt;strong&gt;a little more than 400 light-years from the Sun.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="28048" LastEditorUserId="14094" LastEditDate="2019-05-22T00:45:37.347" LastActivityDate="2019-05-22T00:45:37.347" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31960" PostTypeId="2" ParentId="31951" CreationDate="2019-05-20T13:23:15.503" Score="5" Body="&lt;p&gt;&lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=HD+116798&quot; rel=&quot;nofollow noreferrer&quot;&gt;SIMBAD&lt;/a&gt; lists a Gaia DR2 parallax of 10.85&amp;plusmn;0.03 milli-arcseconds for that star.&#xA;Without going into the biases and uncertainties in current Gaia parallax data,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$r \approx \frac{1}{\varpi}$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\varpi$&lt;/span&gt; is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stellar_parallax&quot; rel=&quot;nofollow noreferrer&quot;&gt;parallax&lt;/a&gt; in arcseconds and &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; is the distance in &lt;a href=&quot;https://en.wikipedia.org/wiki/Parsec&quot; rel=&quot;nofollow noreferrer&quot;&gt;parsecs&lt;/a&gt;.&#xA;This works out to a distance of roughly 92 parsecs or 300 light-years.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-05-22T16:10:02.687" LastActivityDate="2019-05-22T16:10:02.687" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31962" PostTypeId="1" AcceptedAnswerId="31979" CreationDate="2019-05-20T07:01:20.120" Score="2" ViewCount="140" Body="&lt;p&gt;I am big fan of Asimov's Foundation and I am wondering about the plausibility of some of Foundation's imagined words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://asimov.fandom.com/wiki/Terminus&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; Terminus was a special world with little metals:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It was the sole planet orbiting its isolated star and had &lt;strong&gt;almost no&#xA;  metals.&lt;/strong&gt; The nearest planet was Anacreon, 8 parsecs (26 light years)&#xA;  away. Being on the fringe of the galaxy, there are almost no stars in&#xA;  the sky.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since the planet hosted the first Foundation it is clearly habitable (i.e. atmosphere, no extreme temperatures). However, I am wondering how could it have so little metals if it seems so similar to Earth (habitability-wise) which has lots of metals. Is this possible in our Universe? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in answers that use current discovered exoplanets information or some articles that deal with planets come to be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is it plausible for a planet that is positioned in the habitable area of a solar system to have little extractable metals?&lt;/p&gt;&#xA;" OwnerUserId="28093" OwnerDisplayName="Alexei" LastEditorUserId="28093" LastEditDate="2019-05-20T16:18:20.560" LastActivityDate="2019-05-22T13:15:19.750" Title="Is it plausible for a planet that is positioned in the habitable area of a solar system to have little extractable metals?" Tags="&lt;planetary-science&gt;&lt;exoplanet&gt;" AnswerCount="4" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="31963" PostTypeId="2" ParentId="31962" CreationDate="2019-05-20T10:37:33.023" Score="4" Body="&lt;p&gt;Interesting question, though probably a better fit for  world-building. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly &quot;metals&quot; is often used by astronomers to refer to any element other than hydrogen or helium. That is clearly not the intended use here. Terminus has a solid surface at temperatures that humans can stand, and a breathable atmosphere (perhaps after terraforming). So, what can it be made of?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most obvious option is silicate rock (mainly aluminium and magnesium silicates) similar to what most of the outer layers of the Earth are made of. It is certainly imaginable that a planet might be &lt;strong&gt;more&lt;/strong&gt; thoroughly differentiated than Earth, so that all the iron etc. ended up in the core, and there were no, or almost no, useful ore bodies in the crust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the one hand, this clearly fails, since magnesium and aluminium are metals. On the other hand, it is quite possible in context that what is meant is &quot;easily extractable metals&quot; and/or &quot;the metals required for a specific range of technologies&quot;, which might include iron, copper, etc. (given that Asimov was writing in the '30s that is probably what he was thinking of). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, with enough energy you can extract aluminium or magnesium from these rocks. We don't do this, because it's cheaper to find rare compounds of those elements from which the metal is more easily extracted, but the Foundation could certainly have had aluminium and magnesium in this scenario assuming they had enough power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A stranger possibility would be carbon. Several gas giants are believed to have carbon-rich cores. Suppose one of those lost its outer layers in some cataclysm, the resulting smaller planet could differentiate to have a thick diamondoid crust over a rocky mantle and iron core. Although if diamonds were literally as common as dirt on Terminus, I can't help feeling someone would have remarked on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar but slightly less exotic option would be silica -- so a planet with a mainly or entirely quartz crust. I can't see a scenario for that to form -- one where you'd have silicon and oxygen and not much else around when a planet was forming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we're definitely into worldbuilding territory, but I wonder is ice is a contender? If you found a world with an icy crust a few hundred km thick, is there a way to terraform it &lt;strong&gt;without&lt;/strong&gt; melting all (or even most) of the ice? YOu could imagine covering the ice in a layer of vacuum-filled diamond honeycomb and then dumping a few meters of imported or manufactured soil over it, creating an atmosphere somehow and then putting a heavy greenhouse layer (CO2 and SF6 and so on) to warm the surface. Could the first-empire scientists have done something like that?   Taking this back to the original question -- an ice world could exist at the outer edge of the habitable zone (or just beyond it, if you like) and then be made habitable by adding a greenhouse layer and changing its albedo. If you can avoid melting the ice, that would meet the requirements.&lt;/p&gt;&#xA;" OwnerUserId="21928" OwnerDisplayName="Steve Linton" LastEditorUserId="21928" LastEditDate="2019-05-20T14:54:23.490" LastActivityDate="2019-05-20T14:54:23.490" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31964" PostTypeId="2" ParentId="31962" CreationDate="2019-05-20T14:57:58.010" Score="3" Body="&lt;p&gt;Confirming the absence of an element on an exoplanet is not within our current means. Given the absence of any context for the lacks metals (does it have to be present in the crust? elemental form? How scare counts as absent?) its hard to conjecture what level of oddity would be require to achieve &quot;absence of metal&quot;. However, I think its safe to say its not super likely that rocky planet was completely devoid of metal.&lt;/p&gt;&#xA;" OwnerUserId="28055" LastActivityDate="2019-05-20T14:57:58.010" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31965" PostTypeId="1" AcceptedAnswerId="31968" CreationDate="2019-05-20T16:41:50.593" Score="2" ViewCount="116" Body="&lt;p&gt;For the past few dates, I've been observing the rotation of the earth around the sun and noticed that the azimuth of the sun doesn't vary uniformly throughout the day. The azimuth varies faster during some parts of the day, depending on the position &#xA;&lt;a href=&quot;https://planetcalc.com/4270/&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this happen? I've been thinking about the cause, but can't come up with a proper explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/r2PVh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/r2PVh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26691" LastEditorUserId="26691" LastEditDate="2019-05-20T18:01:08.563" LastActivityDate="2019-05-20T20:15:43.413" Title="Why doesn't the solar azimuth vary uniformly" Tags="&lt;the-sun&gt;&lt;rotation&gt;&lt;azimuth&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31966" PostTypeId="1" CreationDate="2019-05-20T16:44:23.200" Score="0" ViewCount="45" Body="&lt;p&gt;I don't know ANYTHING about astronomy but I know what I saw and 2 of my siblings saw it too. I initially thought it was a shooting star, but a Google search quickly made it clear that it wasn't.&#xA;it was of similar brightness to the surrounding stars and weirdly wasn't moving in a single direction. it was moving erratically in no particular path, was not constrained to a specific area and almost mimicked what a laser point would look like if a human was moving it randomly. I do not believe satellites move at the speed at which it was moving and as randomly.. &#xA;what could this possibly be? &lt;/p&gt;&#xA;" OwnerUserId="28094" LastEditorUserId="28094" LastEditDate="2019-05-20T20:19:07.797" LastActivityDate="2019-05-20T20:19:07.797" Title="really fast moving star-like light" Tags="&lt;star&gt;&lt;light&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2019-05-20T23:20:22.187" ContentLicense="CC BY-SA 4.0" />
  <row Id="31968" PostTypeId="2" ParentId="31965" CreationDate="2019-05-20T20:15:43.413" Score="4" Body="&lt;p&gt;The Sun's &lt;a href=&quot;https://en.wikipedia.org/wiki/Hour_angle&quot; rel=&quot;nofollow noreferrer&quot;&gt;hour angle&lt;/a&gt;, measured around the celestial equator, changes at a nearly constant rate.&#xA;The gnomon of a sundial is aligned with the polar axis to take advantage of this.&#xA;The celestial equator is oblique to the horizon except at the Earth's equator and poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lines of azimuth converge at the observer's zenith and nadir, so the Sun's azimuth changes more rapidly near one of those points (e.g. summer midday, winter midnight) and more slowly near the horizon.&#xA;The amplitude of this fluctuation depends on the Sun's &lt;a href=&quot;https://en.wikipedia.org/wiki/Declination&quot; rel=&quot;nofollow noreferrer&quot;&gt;declination&lt;/a&gt; and the observer's geographic latitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an observer in the tropics, on a day when the Sun passes directly overhead, its azimuth swings abruptly from 90&amp;deg; to 270&amp;deg; at noon.&#xA;For an observer near the north or south pole, the Sun remains near the same altitude all day, and its azimuth rate stays near the average 15&amp;deg;/hour.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2019-05-20T20:15:43.413" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31970" PostTypeId="1" AcceptedAnswerId="31977" CreationDate="2019-05-20T22:25:28.447" Score="1" ViewCount="82" Body="&lt;p&gt;Giant planets such as Jupiter have oscillations which enable analyses using the techniques of asteroseismology, for example &lt;a href=&quot;https://arxiv.org/abs/1106.3714&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gaulme et al. (2011)&lt;/a&gt; detected global modes on Jupiter via radial velocity measurements. Is there a specific term for this technique as applied to giant planets instead of stars?&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastActivityDate="2019-10-26T00:56:34.810" Title="Is there a term for asteroseismology as applied to giant planets?" Tags="&lt;gas-giants&gt;&lt;terminology&gt;&lt;asteroseismology&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31972" PostTypeId="2" ParentId="31970" CreationDate="2019-05-20T23:28:17.960" Score="0" Body="&lt;p&gt;Yes. Although there is not much point looking for them on the surfaces of the giant planets, as those are too dynamic, unfortunately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the rings of Saturn can give us some insight into internal oscillation modes, this area of research is then dubbed &lt;a href=&quot;https://en.wiktionary.org/wiki/kronoseismology&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kronoseismology&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to my (very limited) view of the developments in this subfield of planetary science, results were coming in only slowly and with low accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since Cassinis grand finale hit us, we have close up observations of moving density waves in the rings, together with high-precision orbital data, so it seems &lt;a href=&quot;https://www.sciencedirect.com/science/article/abs/pii/S0019103518303397&quot; rel=&quot;nofollow noreferrer&quot;&gt;research is picking up speed&lt;/a&gt; in recent years.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2019-05-20T23:28:17.960" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31973" PostTypeId="1" AcceptedAnswerId="31980" CreationDate="2019-05-21T01:43:11.333" Score="3" ViewCount="99" Body="&lt;p&gt;I am trying to understand how to extract stellar velocity and velocity dispersion maps from spectral cubes, but with no success so far. To be more precise, I do not understand why people use stellar spectra templates to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For gas kinematics (e.g., Ha emission), the usual approach is to fit a Gaussian function to each spectrum. Then, the means and sigmas correspond to the line-of-sight velocities and velocity dispersions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For stellar kinematics, most people use &lt;code&gt;pPXF&lt;/code&gt;. To my understanding, in the simplest case, &lt;code&gt;pPFX&lt;/code&gt; fits a parametric function (e.g., Gaussian) convolved with a template stellar spectrum to the data. Why is that? I tried to read the &lt;code&gt;pPFX&lt;/code&gt; papers but there is no explanation for this. It assumes that the use of a template stellar spectrum is obvious and there is no justification for this. Actually the majority of papers that use &lt;code&gt;pPFX&lt;/code&gt; assume that the use of a template is a de facto thing and provide no explanation.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is the rationale behind using a template stellar spectrum?&lt;/li&gt;&#xA;&lt;li&gt;Why can't we just use an inverted Gaussian and fit that to the absorption line of each spectrum?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;After some more reading, some papers imply (but do not explicitly say it) that the absorption lines of stellar spectra have complex shapes that can't be modeled with a simple Gaussian. But that is also the case for emission lines! Emission lines can be asymmetric due to multiple kinematic components or beam smearing. Yet, nobody uses templates when extracting kinematic maps from them. I do understand that fitting a single Gaussian to an asymmetric emission line can result in loss of information. But why is this acceptable for emission lines and not acceptable for absorption lines?&lt;/p&gt;&#xA;" OwnerUserId="1986" LastActivityDate="2019-05-21T07:40:37.463" Title="Extracting kinematics from stellar spectra. Why are template stellar spectra necessary?" Tags="&lt;spectroscopy&gt;&lt;kinematics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31974" PostTypeId="1" CreationDate="2019-05-21T02:49:42.583" Score="-6" ViewCount="243" Body="&lt;p&gt;Rg Veda of ancient India contains a code that gives all planetary orbital information visible in the night sky. It is worth knowing the origins of our solar system and its development through the ages. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good calendar and star atlas for navigation were important for India and both were developed early on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is its description? Information is available in &quot;The Astronomical code of Rg Veda&quot;  by Subash Kak and also in  &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.695.536&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.695.536&amp;amp;rep=rep1&amp;amp;type=pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="27847" LastEditorUserId="27847" LastEditDate="2019-05-22T16:03:40.430" LastActivityDate="2019-05-22T19:15:05.457" Title="How were the orbital sizes determined in ancient India?" Tags="&lt;observational-astronomy&gt;" AnswerCount="1" CommentCount="7" ClosedDate="2019-05-21T14:03:16.977" ContentLicense="CC BY-SA 4.0" />
  <row Id="31975" PostTypeId="1" CreationDate="2019-05-21T03:04:24.377" Score="2" ViewCount="86" Body="&lt;p&gt;I have been watching too much TV related to black holes.  Most black holes are depicted with a disk of matter that spirals into the black hole. Does the mass and diameter of the disk have a bearing on the rotational speed, maximum rotational speed of a black hole? Also, do black holes have gravity? It looks like they must in order to collect matter. Does the gravity of a black hole create gravitational waves much as the electromagnetic waves created around the earth?  Do these waves have any influence on the speed of the black hole's rotation. What about if the rotation if the black hole is not symmetric? &lt;/p&gt;&#xA;" OwnerUserId="28099" LastEditorUserId="2365" LastEditDate="2019-05-21T06:28:54.163" LastActivityDate="2019-05-22T09:53:56.563" Title="Question from a non-astronomer /non-mathematician regarding rotation speeds" Tags="&lt;black-hole&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31976" PostTypeId="2" ParentId="31947" CreationDate="2019-05-21T06:10:48.683" Score="3" Body="&lt;p&gt;In astronomy research, models of stars are computed using large software packages that solve a set of differential equations that describe the structure of a star as it evolves, mostly because of nuclear fusion transforming elements, under a number of assumptions.  (I once wrote a &lt;a href=&quot;https://physics.stackexchange.com/questions/95128/what-is-a-simple-mathematical-model-of-a-star/95165#95165&quot;&gt;more detailed answer about this on Physics.SE&lt;/a&gt;.) So I'd say the first option is to find and operate one of these programs. One of the easiest to access and use is &lt;a href=&quot;http://mesa.sourceforge.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;MESA&lt;/a&gt; but there are many others that exist, although they often aren't freely available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option is to use (probably by interpolating in) large tables of models that have been produced using those codes.  MESA is used to produce the &lt;a href=&quot;http://waps.cfa.harvard.edu/MIST/&quot; rel=&quot;nofollow noreferrer&quot;&gt;MESA Isochrones and Stellar Tracks&lt;/a&gt; for which the &lt;a href=&quot;https://isochrones.readthedocs.io/en/latest/&quot; rel=&quot;nofollow noreferrer&quot;&gt;isochrones Python package&lt;/a&gt; provides an interface, but there's also the &lt;a href=&quot;http://stellar.dartmouth.edu/models/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dartmouth Stellar Evolution Database&lt;/a&gt; and the &lt;a href=&quot;http://stev.oapd.inaf.it/cgi-bin/cmd&quot; rel=&quot;nofollow noreferrer&quot;&gt;PARSEC&lt;/a&gt; grid, among others.  (Apologies to all the isochrone grids I've missed!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also try your hand with the program &lt;a href=&quot;http://astronomy.swin.edu.au/~jhurley/&quot; rel=&quot;nofollow noreferrer&quot;&gt;SSE&lt;/a&gt;, which uses analytic fits to a grid of models from the late 1990s. While SSE implements them, the analytic fits themselves are described in &lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2000MNRAS.315..543H/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hurley, Pols &amp;amp; Tout (2000)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1577" LastActivityDate="2019-05-21T06:10:48.683" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31977" PostTypeId="2" ParentId="31970" CreationDate="2019-05-21T06:32:23.013" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a specific term for this technique (asteroseismology) as applied to giant planets instead of stars?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No it seems there is not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, people just use normal literal constructions.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://arxiv.org/abs/1411.1740&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Seismology of Giant Planets&quot;&lt;/a&gt;&quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://arxiv.org/abs/1106.3714&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Jovian Seismology&quot; and &quot;giant planet seismology&quot;&lt;/a&gt; or (&lt;em&gt;your planet here&lt;/em&gt;) seismology in general&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://space.stackexchange.com/a/28047/12102&quot;&gt;&quot;Planetary normal-mode seismology&quot;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A good example of the Fourier imaging technique is SYMPA:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.aanda.org/articles/aa/pdf/2007/42/aa7019-06.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;SYMPA, a dedicated instrument for Jovian seismology I. Principle and performance&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.aanda.org/articles/aa/pdf/2008/41/aa09512-08.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;SYMPA, a dedicated instrument for Jovian seismology&#xA;II. Real performance and first results&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;JOVIAL is next-gen implementation, also from Earth: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ajax.ehu.es/Juno_amateur_workshop/talks/01_06_Jupiter_oscillations_Schmider&quot; rel=&quot;nofollow noreferrer&quot;&gt;JOVIAL; &lt;strong&gt;J&lt;/strong&gt;ovian &lt;strong&gt;O&lt;/strong&gt;scillations through radial &lt;strong&gt;V&lt;/strong&gt;elocimetry &lt;strong&gt;I&lt;/strong&gt;m&lt;strong&gt;A&lt;/strong&gt;ging observations at several &lt;strong&gt;L&lt;/strong&gt;ongitudes&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www-n.oca.eu/jovial2016/Kick-off_Avril2016/Monday/JOVIAL_Guillot.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jupiter’s interior: from gravimetry to seismology&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;note:&lt;/strong&gt; after posting I realized that one of the linked papers is the same as in the OP&quot;s question. I'll leave it here for completeness.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-10-26T00:56:34.810" LastActivityDate="2019-10-26T00:56:34.810" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31979" PostTypeId="2" ParentId="31962" CreationDate="2019-05-21T06:48:22.720" Score="4" Body="&lt;p&gt;I'm going to approach this question in two steps: what metals are you talking about, and could you have a planet where those metals are not easily extractable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What metals?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get the sense that you're specifically referring to the &lt;em&gt;non-&lt;a href=&quot;https://en.wikipedia.org/wiki/Goldschmidt_classification#Lithophile_elements&quot; rel=&quot;nofollow noreferrer&quot;&gt;lithophile&lt;/a&gt;&lt;/em&gt; metals, which include the d-block &lt;a href=&quot;https://en.wikipedia.org/wiki/Transition_metal&quot; rel=&quot;nofollow noreferrer&quot;&gt;transition metals&lt;/a&gt; iron, nickel, copper and gold, and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Goldschmidt_classification#Chalcophile_elements&quot; rel=&quot;nofollow noreferrer&quot;&gt;chalcophile&lt;/a&gt; metals zinc, tin, lead, arsenic, mercury and silver. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an important distinction, since the advent of metallurgy was a critical step in human development in taking us out of the Stone Age, and it might be reasonable to posit that any advanced civilisation would follow a similar trajectory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final phase of the Stone Age, as we transitioned from the Neolithic to the Bronze Age, is called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Chalcolithic&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chalcolithic&lt;/a&gt; because it's marked by the first use of &lt;em&gt;copper&lt;/em&gt; (although lead-smelting may have slightly preceded copper-working in some places). The Bronze Age initially alloyed copper with &lt;em&gt;arsenic&lt;/em&gt;, but bronze made from copper and &lt;em&gt;tin&lt;/em&gt; turned out to be both less toxic and more durable; bronze, in turn, was replaced by superior steel tools. Steel requires a sophisticated ferrous metallurgy involving &lt;em&gt;iron&lt;/em&gt; alloys with a carbon content, and knowing how to make it is the mark of a culture's entry into the Iron Age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[NB: There are plenty of iron relics that predate the Iron Age, but these were made from &lt;em&gt;meteoric&lt;/em&gt; iron, an iron-nickel alloy that requires no prior smelting of ores; &lt;em&gt;terrestrial&lt;/em&gt; iron's high melting point is well beyond the temperatures that could be achieved in Stone Age pottery kilns.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Easily extractable?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/History_of_Earth#Solar_System_formation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; describes the early stage of the Earth's formation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The proto-Earth grew by accretion until its interior was hot enough to melt the heavy, siderophile metals. Having higher densities than the silicates, these metals sank. This so-called iron catastrophe resulted in the separation of a primitive mantle and a (metallic) core only 10 million years after the Earth began to form, producing the layered structure of Earth and setting up the formation of Earth's magnetic field. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we assume that rocky exoplanets would follow a similar trajectory, we can certainly postulate that the non-lithophile metals (see my first section) would be depleted in the lithophile mantle. However, I'd expect there would still be ore bodies of even the most heavily depleted metals (e.g. iron), due to volcanism or as a relic of the earliest crust before the &quot;iron catastrophe&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Giant-impact_hypothesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;giant impact hypothesis&lt;/a&gt; provides one mechanism for an exoplanet to form with a much higher concentration of lithophile elements. It's proposed that a solid object (Theia) crashed into the early Earth, ejecting a significant proportion of the two mantles (mostly lighter lithophile elements) into space. Much of this ejecta ended up forming the Moon. A similar scenario but with one or both bodies more massive could certainly result in an even bigger body than the Moon forming out of this lighter ejecta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it's also suggested that the giant impact not only re-liquified the Earth's mantle but caused significant remixing of the mantle with the outer core. It's not clear to me whether, in the absence of a giant impact, there would have been an even stronger differentiation of elements: in other words, even less residual iron in the mantle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all supposition, of course, and there are many questions still to be answered in the Theia model. &lt;strong&gt;Nonetheless, it seems reasonable to say that it is indeed &lt;em&gt;plausible&lt;/em&gt; for an exoplanet in the habitable zone to have a significant deficiency of non-lithophile elements in the planet's crust.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="14094" LastActivityDate="2019-05-21T06:48:22.720" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31980" PostTypeId="2" ParentId="31973" CreationDate="2019-05-21T07:40:37.463" Score="2" Body="&lt;p&gt;(Optically thin) Emission lines in nebulae are generally unresolved and have simple shapes that might be approximated by Gaussians. Stellar spectra are complicated - there are blends, there is rotational broadening, the lines are not Gaussian, there are molecular bands, there may be two stars!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By cross-correlating (or fitting) a template you are effectively modelling your spectrum with what you already think is the best representation of the spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If emission lines are not unresolved and show velocity structure, as you say, then fitting a simple Gaussian will give an incorrect result. A more complex model would be required, but as far as I know there are no template spectra that can help you out here. The situation might be thought analogous to a stellar spectrum with rotational broadening, but there the template can simply be convolved with the appropriate rotational broadening kernel.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-05-21T07:40:37.463" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31981" PostTypeId="1" CreationDate="2019-05-21T11:55:02.227" Score="3" ViewCount="105" Body="&lt;p&gt;&lt;strong&gt;update:&lt;/strong&gt; I still haven't been able to get my hands on the &lt;em&gt;Icarus&lt;/em&gt; paper linked below (I'll try other libraries) but these are newer and &lt;em&gt;quite interesting!&lt;/em&gt;:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.planetary.brown.edu/pdfs/5256.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Constraining the thickness of polar ice deposits on Mercury using the&#xA;Mercury Laser Altimeter and small craters in permanently shadowed&#xA;regions&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://agupubs.onlinelibrary.wiley.com/doi/epdf/10.1002/2017JE005500&quot; rel=&quot;nofollow noreferrer&quot;&gt;nvestigating Mercury’s South Polar Deposits: Arecibo RadarObservations and High-Resolution Deter minationof Illumination Conditions&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;How did Arecibo make radar images of ice on Mercury's poles?&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How were observations of the poles of Mercury geometrically possible from Earth?&lt;/li&gt;&#xA;&lt;li&gt;How was &amp;quot;&lt;a href=&quot;https://www.sciencedirect.com/science/article/abs/pii/S0019103510003131&quot; rel=&quot;nofollow noreferrer&quot;&gt;high-resolution (1.5-km) imaging&lt;/a&gt;&amp;quot; possible from Earth?&lt;/li&gt;&#xA;&lt;li&gt;How was ice identified and imaged?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;https://www.nap.edu/read/21729/chapter/8#136&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chapter 6; Planetary Radar Astronomy&lt;/a&gt; in &lt;em&gt;A Strategy for Active Remote Sensing Amid Increased Demand for Radio Spectrum (2015):&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Radar observations from Earth continue to provide new information about the terrestrial planets and the Moon despite the many spacecraft flybys, orbiters, and landers. The unique radar back-scattering properties of water ice led to the discovery by ground-based radar of ice deposits in the permanently shadowed portions of impact craters at Mercury’s poles. High-resolution (1.5 km) radar imaging of these deposits has mapped out their distribution with high precision&lt;sup&gt;9&lt;/sup&gt; (Figure 6.9).&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;9&lt;/sup&gt;&lt;a href=&quot;https://www.sciencedirect.com/science/article/abs/pii/S0019103510003131&quot; rel=&quot;nofollow noreferrer&quot;&gt;Radar imagery of Mercury’s putative polar ice: 1999–2005 Arecibo results&lt;/a&gt; Icarus, 211, (1) January 2011, pp 37-50&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DzReN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DzReN.jpg&quot; alt=&quot;FIGURE 6.9 Arecibo radar image of the distribution of ice deposits at the north pole of Mercury...&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;FIGURE 6.9 Arecibo radar image of the distribution of ice deposits at the north pole of Mercury (yellow), superimposed on a mosaic of Messenger orbiter images showing the coincidence of the ice deposits with the shadowed portions of impact craters. SOURCE: NASA/Johns Hopkins University Applied Physics Laboratory/Carnegie Institution of Washington.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2020-07-13T10:45:01.433" LastActivityDate="2020-07-13T10:45:01.433" Title="How did Arecibo make radar images of ice on Mercury's poles?" Tags="&lt;radio-astronomy&gt;&lt;radio-telescope&gt;&lt;mercury&gt;&lt;water&gt;" AnswerCount="0" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31982" PostTypeId="2" ParentId="31974" CreationDate="2019-05-21T12:11:00.853" Score="5" Body="&lt;p&gt;The calculation starts with a quote from Pañcavimśa Brahmana, which postdates the Rgveda&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The world of heaven is as far removed from this world,&#xA;  they say, as a thousand earths (literally &quot;cows&quot;) stacked one above the&#xA;  other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then we suppose that the sun is halfway between heaven and earth, or 500 earth-diameters (this is supposed as there are three &quot;planets&quot; (Moon, Mercury and Venus) that have shorter periods and so are supposed to be closer, and three (Mars, Jupiter, Saturn) that are further, so the sun must be about half way) However, you will note that this is pure supposition, and is not accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now suppose that both Moon and sun are moving around the Earth at the same speed. This combined with the known length of the year and month imply that the moon is 40 Earth-diameters, and that the sun is about 4.6 times bigger than the Earth, while the moon is much smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth was taken to be about 7500 miles in diameter, probably from measurements of shadows at the solstice. And so the distances to other bodies can be calculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A careful reader will note that while this is an impressive attempt at calculation, the actual values obtained are very very far from the accepted modern values. 500×7500 gives 3.75million miles, or 6million km, compared to a modern value of a shade under 150 million km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Rgveda may contain hidden information. But there is no evidence presented in your source that the composers of the Rgveda, or later texts Pañcavimśa Brahmana had any actual ability to calculate accurate values for the distance to the sun. Indeed, this was not done until the calculations of Edmund Halley and the observations of the transit of Venus.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2019-05-22T19:15:05.457" LastActivityDate="2019-05-22T19:15:05.457" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31983" PostTypeId="1" AcceptedAnswerId="31986" CreationDate="2019-05-21T12:33:52.140" Score="2" ViewCount="143" Body="&lt;p&gt;I took a look at &lt;a href=&quot;https://en.wikipedia.org/wiki/Proxima_Centauri_b#Habitability&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia's page for Proxima Centauri b&lt;/a&gt;, the exoplanet that orbits &lt;a href=&quot;https://en.wikipedia.org/wiki/Proxima_Centauri&quot; rel=&quot;nofollow noreferrer&quot;&gt;Proxima Centauri&lt;/a&gt;, where it is said it is expected to receive 2000 times of stellar wind pressures than on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not say anything about Proxima b having a magnetic field, but I found &lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/S0032063317303380&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt; where a magnetic field for the planet is predicted:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In our simulations, most Proxima b analogues develop intrinsic dynamos&#xA;  that last for 4 Gyr (the estimated age of the host star). If alive,&#xA;  the dynamo of Proxima b have a dipole moment . These results are not&#xA;  restricted to Proxima b but they also apply to earth-like planets &#xA;  having similar observed properties.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question is if an Earth-like magnetic field, if Proxima b has one, could protect life as it does on Earth, or, conversely, habitability could not be possible due to stellar wind.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/S0032063317303380&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Jorge I.Zuluaga, Sebastian Bustamante (2018):&lt;/strong&gt; &quot;&lt;em&gt;Magnetic properties of Proxima Centauri b analogues&lt;/em&gt;&quot;. &lt;strong&gt;Planetary and Space Science&lt;/strong&gt;.&#xA;Volume 152, March 2018, Pages 55-67&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user28103" LastEditorUserId="616" LastEditDate="2019-05-21T15:42:00.993" LastActivityDate="2019-05-21T20:11:43.800" Title="Could an Earth-like magnetic field protect life at exoplanet Proxima Centauri b?" Tags="&lt;exoplanet&gt;&lt;life&gt;&lt;magnetic-field&gt;&lt;earth-like-planet&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31984" PostTypeId="1" AcceptedAnswerId="31985" CreationDate="2019-05-21T13:14:25.743" Score="7" ViewCount="1600" Body="&lt;p&gt;Given that the moon has long been tidally locked with the Earth, why isn't Earth (or any of our other solar system's planets) tidally locked to the sun?&lt;/p&gt;&#xA;" OwnerUserId="1987" LastEditorUserId="1987" LastEditDate="2019-05-21T15:43:17.797" LastActivityDate="2020-12-12T22:03:41.500" Title="Why isn't Earth tidally locked to the sun?" Tags="&lt;planet&gt;&lt;orbital-mechanics&gt;&lt;tidal-locking&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31985" PostTypeId="2" ParentId="31984" CreationDate="2019-05-21T15:07:29.367" Score="6" Body="&lt;p&gt;Really, it's just because the &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_locking#Timescale&quot; rel=&quot;nofollow noreferrer&quot;&gt;tidal locking timescale&lt;/a&gt; is so long for Earth:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$t\propto\frac{a^6m_{s}}{m_{p}^2R_s^3}$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is semi-major axis, &lt;span class=&quot;math-container&quot;&gt;$m_s$&lt;/span&gt; is the mass of the secondary object, &lt;span class=&quot;math-container&quot;&gt;$m_p$&lt;/span&gt; is the mass of the primary, and &lt;span class=&quot;math-container&quot;&gt;$R_s$&lt;/span&gt; is the radius of the secondary. If we compare the Sun-Earth system to the Earth-Moon system, we see&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{a_1}{a_2}\approx380,\quad \frac{m_{s_1}}{m_{s_2}}\approx80,\quad \frac{m_{p_1}}{m_{p_2}}\approx333000,\quad \frac{R_{s_1}}{R_{s_2}}\approx3.67$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$_1$&lt;/span&gt; denotes the Sun-Earth system and &lt;span class=&quot;math-container&quot;&gt;$_2$&lt;/span&gt; denotes the Earth-Moon system. To make up for these differences (and to have Earth be tidally locked by now), assuming similar Love numbers and dissipation functions for Earth and the Moon, we would need Earth's initial spin to be substantially smaller than the Moon's initial spin by many, many orders of magnitude, and this just wasn't the case.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="31410" LastEditDate="2020-12-12T22:03:41.500" LastActivityDate="2020-12-12T22:03:41.500" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31986" PostTypeId="2" ParentId="31983" CreationDate="2019-05-21T20:11:43.800" Score="3" Body="&lt;p&gt;According to Garcia-Sage et al. (2017) &quot;&lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/aa7eca/pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;On the Magnetic Protection of the Atmosphere of Proxima Centauri b&lt;/a&gt;&quot;, who assume that Proxima b has a magnetic field strength identical to the Earth, the situation does not look promising:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;With an O&lt;sup&gt;+&lt;/sup&gt; and H&lt;sup&gt;+&lt;/sup&gt; escape at 1600&amp;nbsp;km altitude of 4.2×10&lt;sup&gt;9&lt;/sup&gt;&amp;nbsp;cm&lt;sup&gt;−2&lt;/sup&gt;&amp;nbsp;s&lt;sup&gt;−1&lt;/sup&gt; as found for thermospheric densities of 3×T,&#xA;  the total mass-loss rate is 4.3×10&lt;sup&gt;5&lt;/sup&gt;&amp;nbsp;g&amp;nbsp;s&lt;sup&gt;−1&lt;/sup&gt; over the dayside&#xA;  hemisphere. Using a total atmospheric mass of 5×10&lt;sup&gt;18&lt;/sup&gt;&amp;nbsp;kg&#xA;  (Trenberth &amp;amp; Smith 2005) results in a loss of the entire mass of&#xA;  the atmosphere over 365 million years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(T is the thermosphere temperature of the Earth)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They go on to note that Proxima would have had higher EUV and X-ray emission in the past, which would have made things worse. Ongoing outgassing of volatiles might be possible to create habitable conditions but the resultant situation would be very different from the Earth.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastActivityDate="2019-05-21T20:11:43.800" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31987" PostTypeId="1" CreationDate="2019-05-21T20:27:30.173" Score="1" ViewCount="482" Body="&lt;p&gt;Given a specific date and time, and coordination of a location on earth, how can I calculate (myself, or using a python package) range of visible RA and Dec? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example (but I'm looking for a general solution and formula):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;input:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;date and time: 2019-05-01 23:00:00&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;earth coordinates: 32.0853° N, 34.7818° E&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;output:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;RA min: ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;RA max: ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Dec min: ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Dec max: ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28105" LastEditorUserId="28105" LastEditDate="2019-05-22T20:26:10.473" LastActivityDate="2019-05-22T20:26:10.473" Title="Calculating the range of visible Right Ascension and Declination from specific location + time" Tags="&lt;coordinate&gt;&lt;star-maps&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31989" PostTypeId="1" AcceptedAnswerId="31992" CreationDate="2019-05-22T04:07:57.133" Score="31" ViewCount="6073" Body="&lt;p&gt;I've been thinking about black holes, specifically during the final moments before two merge.  I'm wondering if black holes, or I guess more specifically their event horizons, are always spherical.  It seems to me that in the moments before two merge, their respective event horizons will be stretched, somewhat like how the Moon causes our ocean's tides. I have drawn a (poor) diagram of what I think they may look like.  Notice how the event horizons are closer to the singularity on the inner side, this is because the gravity from each black hole is in opposition.  The event horizons are further from the singularity on the outer side because the gravity from each black hole adds up.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/MiUsn.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MiUsn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="28106" LastEditorUserId="1378" LastEditDate="2019-05-22T07:51:28.450" LastActivityDate="2019-05-24T01:11:43.170" Title="Are black holes spherical during merger?" Tags="&lt;black-hole&gt;&lt;gravity&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31990" PostTypeId="1" CreationDate="2019-05-22T07:35:34.873" Score="4" ViewCount="1113" Body="&lt;p&gt;I want to learn astrophysics and astronomy, I'm very interested in it because there are many questions that popped up in my head, and really curious about things, but I'm afraid of a little bit on math, but I'm a fast learner, I can understand things quickly and easily especially if it is translated on my native language, I have learned basic trigonometric functions, special angles, trigonometry, algebra, quadratic equations, and some basic math but not calculus, will proceed on that later, and I have learned basic physics, until vertical dimension, but still afraid a little on math and equations, how can I counteract that fear? I mean I really want to learn astrophysics but I'm afraid of numbers,  can you give me tips? And can you suggest me some book? Or can I learn astrophysics without interacting too much with math? Btw, I'm just taking advancement because I'm really curious and interested in astrophysics, I'm still at high school, hope you can help me.&lt;/p&gt;&#xA;" OwnerUserId="28108" LastEditorUserId="616" LastEditDate="2020-04-29T16:56:38.923" LastActivityDate="2020-05-05T08:32:06.677" Title="Can I learn astrophysics and astronomy with my basic math?" Tags="&lt;astrophysics&gt;&lt;resource-request&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31991" PostTypeId="2" ParentId="30622" CreationDate="2019-05-22T07:44:33.687" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/1804.11339&quot; rel=&quot;nofollow noreferrer&quot;&gt;Malhan et al. (2018)&lt;/a&gt; summarise progress so far (the Gaia astrometry is improving all the time as more data are collected) using Gaia DR2 data. In particular they produce &quot;panoramic maps&quot; of streams in the Galaxy halo, which I believe is exactly what you are looking for (see below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qVBVv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qVBVv.jpg&quot; alt=&quot;Streams in the southern Galactic sky&quot;&gt;&lt;/a&gt;&#xA;&lt;em&gt;Coherent tangential velocity structures, a.k.a. &quot;streams&quot;, seen in the southern Galactic hemisphere by Gaia DR2 - from Malhan et al. (2018).&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-05-22T07:44:33.687" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31992" PostTypeId="2" ParentId="31989" CreationDate="2019-05-22T07:48:01.530" Score="45" Body="&lt;p&gt;No need to guess. There's solid research done in this field. Even Wikipedia has some info:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As two black holes approach each other, a ‘duckbill’ shape protrudes&#xA;  from each of the two event horizons towards the other one. This&#xA;  protrusion extends longer and narrower until it meets the protrusion&#xA;  from the other black hole. At this point in time the event horizon has&#xA;  a very narrow X-shape at the meeting point. The protrusions are drawn&#xA;  out into a thin thread. The meeting point expands to a roughly&#xA;  cylindrical connection called a bridge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Binary_black_hole#Shape&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Binary_black_hole#Shape&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are research papers with images showing the results of calculations of the shape of the event horizons during merger. Here's an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x9wFR.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x9wFR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image above is taken from this paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/1110.1668&quot; rel=&quot;noreferrer&quot;&gt;On Toroidal Horizons in Binary Black Hole Inspirals&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We examine the structure of the event horizon for numerical&#xA;  simulations of two black holes that begin in a quasicircular orbit,&#xA;  inspiral, and finally merge. We find that the spatial cross section of&#xA;  the merged event horizon has spherical topology (to the limit of our&#xA;  resolution), despite the expectation that generic binary black hole&#xA;  mergers in the absence of symmetries should result in an event horizon&#xA;  that briefly has a toroidal cross section.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2019-05-22T07:54:24.457" LastActivityDate="2019-05-22T07:54:24.457" CommentCount="16" ContentLicense="CC BY-SA 4.0" />
  <row Id="31993" PostTypeId="2" ParentId="31990" CreationDate="2019-05-22T08:00:03.703" Score="3" Body="&lt;p&gt;Generally, astrophysics (and astronomy) is math-heavy. That should not discourage you, but rather act as a way of learning math: it is often easier to learn topics that you have a use for and are part of some personal project than just getting lectured about them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astrophysics is based on math because it is based on many forms of physics - mechanics, thermodynamics, nuclear physics, relativity theory and so on - plus chemistry. Mathematics is a common language that binds these fields together and allows you to combine them to build models to understand what is going on. Some parts just use some basic math, but in general calculus really is needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematics in the more common &quot;calculations with numbers&quot; form also matters because it all starts with measuring things - where things are, how much they change, brightness and so on - and then we build and test our models and understanding from this. That also means that statistics is important for handling measurement noise and finding patterns in data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One cannot really do astronomy and astrophysics without math. It would be just looking at things and making up explanations without actually checking if and how they hold true.&lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2019-05-22T08:00:03.703" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31995" PostTypeId="2" ParentId="31975" CreationDate="2019-05-22T09:53:56.563" Score="2" Body="&lt;p&gt;This is a bunch of questions, many of which have answers already, but they are linked.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Most black holes are depicted with a disk of matter that spirals into the black hole. Does the mass and diameter of the disk have a bearing on the rotational speed, maximum rotational speed of a black hole?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's much more the other way round. The size of the accretion disk is determined mainly by the mass of the black hole, and by the supply of new material passing close enough to the black hole to possibly get trapped in the disk. The rotation of the black hole only plays a role in the innermost part of the disk, where the strange effects of General Relativity cause the material of the disk to be dragged around.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, do black holes have gravity? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. Indeed they don't really have much else.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the gravity of a black hole create gravitational waves much as the electromagnetic waves created around the earth?  Do these waves have any influence on the speed of the black hole's rotation. What about if the rotation if the black hole is not symmetric?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A black hole is only asymmetric for very brief periods after large objects fall into it (for example when two black holes merge). In this case it produces very large amounts of gravitational wave energy, rapidly shedding the asymmetry. This process is called &quot;ringdown&quot;. A symmetrical spinning object does not radiate gravitational waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very good source for seeing these kinds of things are the videos produced by the &lt;a href=&quot;https://www.black-holes.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;SXS consortium&lt;/a&gt; with the caveat that they are visualisations of aspects of what is happening, not images. There are far too many dimensions of data to show the whole story.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-05-22T09:53:56.563" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31996" PostTypeId="1" CreationDate="2019-05-22T10:55:15.700" Score="1" ViewCount="39" Body="&lt;p&gt;Is it possible to calculate the day lengths of moons or moons of moons? If so, does the day length depend on the longitude as well as the latitude? Also, if we calculate the day length, is it possible to take into account the duration the sun is eclipsed by the parent body?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit of a background. I am playing a game, Elite Dangerous, and the game provides basic orbital information about planetary bodies like orbital period, rotation period, axial tilt and some more. I was planning to build a tool around this but I have no idea how to calculate such a thing.&lt;/p&gt;&#xA;" OwnerUserId="28109" LastEditorUserId="23748" LastEditDate="2019-05-22T14:41:46.553" LastActivityDate="2019-05-22T14:41:46.553" Title="Calculation of day lengths of moons or moons of moons" Tags="&lt;orbital-mechanics&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31997" PostTypeId="1" CreationDate="2019-05-22T13:04:24.550" Score="3" ViewCount="116" Body="&lt;p&gt;I was wondering this for a school paper I'm writing.&lt;/p&gt;&#xA;" OwnerUserId="28111" LastEditorUserId="7472" LastEditDate="2019-05-22T17:32:56.903" LastActivityDate="2019-05-22T19:15:54.817" Title="What is the temperature about 10 or 15 feet below Mars' surface?" Tags="&lt;mars&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31998" PostTypeId="2" ParentId="31962" CreationDate="2019-05-22T13:15:19.750" Score="4" Body="&lt;p&gt;It depends a bit on whether you are talking about &lt;strong&gt;extractable&lt;/strong&gt; metals or about extractable &lt;strong&gt;metals&lt;/strong&gt;. Which is to say:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Iron is pretty universal, so if you are thinking about iron and its cousins then your concern will be whether it has all sunk down into the core, whether there is a thick layer of ice between you and the ground – in a word, extractability.&lt;/li&gt;&#xA;&lt;li&gt;If you mean more precious metals such as gold and uranium, the question is not extractability, it is about the things being there in the first place. It is no longer thought that most of these &lt;em&gt;r&lt;/em&gt;-process elements are created in an exploding supernova, because a supernova explosion is too slow and too gentle. The current trend is to see them as made of &quot;splashed neutronium&quot;: either the débris of colliding neutron stars or in the wreck of a neutron star which has been disrupted by a black hole. The reason the question then becomes more interesting is that these things do not happen very often.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The key paper to read is &lt;a href=&quot;https://www.nature.com/articles/s41586-019-1113-7&quot; rel=&quot;nofollow noreferrer&quot;&gt;A nearby neutron-star merger explains the actinide abundances in the early Solar System&lt;/a&gt; by Imre Bartos &amp;amp; Szabolcs Marka in the May 1 issue of &lt;em&gt;Nature&lt;/em&gt;: the link has an abstract. The key points are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Neutron star collisions are rare. They mention something like dozens of collisions per 100 million years (as I understand it, this rate is for our galaxy taken as a whole).&lt;/li&gt;&#xA;&lt;li&gt;Their paper specifically identifies, by isotopic analysis, that our solar system was within a reasonable radius of such a collision shortly before it came into being, where &quot;reasonable radius&quot; is something quite small, of the order of 1,000 light years. (Very small, in the context of the size of the galaxy).&lt;/li&gt;&#xA;&lt;li&gt;It follows that where freshly created &lt;em&gt;r&lt;/em&gt;-process matter is concerned, our galaxy is pretty spotty. When your solar system is born, you are either near a recent neutron star merger or you are not; but you are much more likely not to be.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The question that needs looking into further is how spotty the galaxy is if you are willing to accept stale &lt;em&gt;r&lt;/em&gt;-process matter as well as fresh. The staler matter will be depleted in radioactive elements (for instance, there will be less uranium and thorium and more lead) but of course the abundance of stable elements such as gold will be unchanged. If you include the staler matter, is the galaxy still blotchy, at least? If it is, you can easily have exoplanets which were born outside a blotch and so have no heavy elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, since a bit under half of the Earth's internal heat (no reference, so please check!) comes from uranium and thorium, a planet which was born far away from a neutron star merger would not only lack uranium mines but might also have a much colder core and a different system of tectonics to deal with the smaller heat flow.  &lt;/p&gt;&#xA;" OwnerUserId="12016" LastActivityDate="2019-05-22T13:15:19.750" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31999" PostTypeId="2" ParentId="31997" CreationDate="2019-05-22T14:48:16.880" Score="3" Body="&lt;p&gt;This is one of key questions that NASA's Mars InSight probe is trying to answer but it has only recently arrived and starting taking data and it will be a while before the data is analyzed and published in peer reviewed journals. The &quot;mole&quot; which would burrow below the surface and make the temperature measurements encountered some initial issues and NASA and the instrument's builders were running further tests and looking at those data before proceeding according to &lt;a href=&quot;https://mars.nasa.gov/news/8419/more-testing-for-mars-insights-mole/?site=insight&quot; rel=&quot;nofollow noreferrer&quot;&gt;this mission press release&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2019-05-22T14:48:16.880" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32001" PostTypeId="1" AcceptedAnswerId="32047" CreationDate="2019-05-22T17:21:40.157" Score="0" ViewCount="164" Body="&lt;p&gt;I was perusing the exoplanet database and looking for some of the more extreme orbits, when I happened upon &lt;strong&gt;HD 20782 B&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://exoplanets.org/detail/HD_20782_b&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://exoplanets.org/detail/HD_20782_b&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is also listed (in two sentences) &lt;a href=&quot;https://en.wikipedia.org/wiki/HD_20782_b&quot; rel=&quot;nofollow noreferrer&quot;&gt;on wikipedia&lt;/a&gt;, so there's not a whole lot of information. However, the first thing I noticed was that it was almost at escape velocity!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tjV9u.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tjV9u.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess not much is known, but my overall question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Would an exoplanet on an escape trajectory offer us any opportunities for research that have yet to be performed? What would those opportunities be? The answer can be a justified no (E.G. once the planet leaves the host star, we can't observe the doppler shift anymore).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Imaginary Bonus Points: have there ever been any observed as escaping/becoming rogue planets?&lt;/p&gt;&#xA;" OwnerUserId="23729" LastActivityDate="2019-05-25T22:14:55.703" Title="Are exoplanets which are above escape velocity of particular interest?" Tags="&lt;exoplanet&gt;&lt;escape-velocity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32002" PostTypeId="2" ParentId="31990" CreationDate="2019-05-22T19:04:00.593" Score="3" Body="&lt;p&gt;For most parts of astronomy and astrophysics the level of mathematics required is not very high, compared to what a &lt;em&gt;mathematician&lt;/em&gt; would call mathematics. (You are not going to spend much time on proving theorems, for example). So that is the first reason for telling you not to be discouraged by feeling nervous about maths. Much of the time, you only need to understand it well enough to be comfortable with using it as a tool. (It depends on the field, obviously: the shape of black hole event horizons is going to be a lot more advanced than the orbits of exoplanets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second reason not to be discouraged is that not only is most of the mathematics easier, it also comes with a built-in relevance to real life. What I mean is this. When you are studying maths &lt;em&gt;as maths&lt;/em&gt;, your teacher (or your book) gives you a problem and you have to find the answer, and the answer could be anything at all. When you are using maths as a tool in astrophysics, the answer can't be &quot;anything at all&quot;. It has to make physical sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a recent paper I was reading was trying to answer the question &quot;When did the violent neutron-star merger occur in which our heavy elements (such as gold and uranium) were created?&quot;. They measured the relative abundance, in cosmic dust grains, of certain pairs of elements which would have been the ultimate decay products of the original radioactive elements which were created in the merger. If you did such a calculation, and came up with the answer &quot;the neutron stars merged 400 years ago&quot;, you would know the answer was nonsense. If you did it, and came up with the answer &quot;the neutron stars merged 4 trillion years ago&quot;, you would also know the answer was nonsense. This built-in sanity check is actually reassuring, when you compare it with theoretical, &quot;mathematical&quot; mathematics. You know in advance what sort of answer you ought to be getting.&lt;/p&gt;&#xA;" OwnerUserId="12016" LastActivityDate="2019-05-22T19:04:00.593" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32003" PostTypeId="2" ParentId="31997" CreationDate="2019-05-22T19:15:54.817" Score="1" Body="&lt;p&gt;For a term-paper kind of answer you don't actually need to wait for Mars InSight to tell you. 15 feet is deep enough for you to able to assume that the temperature will be the &lt;em&gt;average&lt;/em&gt; of temperatures on the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So first find a graph of the temperature on a typical Martian spring or autumn day, then work out the average temperature over the 25 hours. That is a good first guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that 15 feet is deep enough for variations &lt;em&gt;across the year&lt;/em&gt;, between summer and winter, to be averaged out fairly effectively as well, though that is something you may want to discuss a little in your paper. &lt;a href=&quot;https://en.wikipedia.org/wiki/Permafrost&quot; rel=&quot;nofollow noreferrer&quot;&gt;This Wikipedia article about permafrost&lt;/a&gt; is worth reading to see how the Earth averages its own temperatures at depth over the course of a year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a planetary scale 15 feet is very shallow indeed, so the amount of heat flowing from the inside of Mars towards the surface will be tiny (this is what that InSight probe is measuring) and I would suggest mentioning it only in order to say that you are ignoring it.&lt;/p&gt;&#xA;" OwnerUserId="12016" LastActivityDate="2019-05-22T19:15:54.817" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32004" PostTypeId="1" AcceptedAnswerId="32012" CreationDate="2019-05-22T23:28:29.923" Score="1" ViewCount="110" Body="&lt;p&gt;In some images with coloring like in this one: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K4jTp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K4jTp.jpg&quot; alt=&quot;NGC 4302&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a noticeably bluer region in the &quot;bottom left&quot; of the image, and the bottom part of the galaxy on the left. Being an inexperienced galaxy viewer, I can't figure out what to make of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the spectrum of that region more complex than it seems? What is it?&lt;/p&gt;&#xA;" OwnerUserId="24727" LastActivityDate="2019-05-23T19:33:56.987" Title="Why a blue spot in some hubble images of NGC 4302?" Tags="&lt;galaxy&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32005" PostTypeId="2" ParentId="32001" CreationDate="2019-05-22T23:59:45.677" Score="0" Body="&lt;p&gt;Source: &lt;a href=&quot;https://exoplanets.nasa.gov/5-ways-to-find-a-planet/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://exoplanets.nasa.gov/5-ways-to-find-a-planet/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, we have five known ways of detecting exoplanets. Your question asked about observing planets that are escaping stellar systems. These are detecting shifts in stellar velocity, detecting shifts in stellar position, searching for transits, direct imaging, and detecting the gravitational lensing of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A planet that's being ejected from the orbit of a star would have increasingly small gravitational effects on its parent star as it flies further away through space. It would also no longer have a regular orbit and transits and would be poorly lit in the interstellar medium, making our detection efforts largely in vain. So I would suggest that the data we could glean from a planet going rogue is about the same data we can glean from any regular exoplanet.&lt;/p&gt;&#xA;" OwnerUserId="28028" LastActivityDate="2019-05-22T23:59:45.677" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32007" PostTypeId="1" AcceptedAnswerId="32008" CreationDate="2019-05-22T14:38:22.800" Score="3" ViewCount="838" Body="&lt;p&gt;From certain positions in outer space the Earth will appear to have a blue ring around it when the viewer is in the shadow of the Earth and sunlight is passing through the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a simulation from Google Earth:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/94WHd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/94WHd.png&quot; alt=&quot;image of the earth in shadow with blue ring around it&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sample sentence: The astronaut looked down at the &lt;em&gt;blue ring formed by the sunlight passing through the atmosphere&lt;/em&gt; around the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a word that describes this ring?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had assumed the ring would be blue, based on this photograph from the Tumblr site &lt;a href=&quot;https://colchrishadfield.tumblr.com/post/50605588853/spaceflight-finale-to-some-this-may-look-like-a&quot; rel=&quot;nofollow noreferrer&quot;&gt;Col. Chris Hadfield&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wVwpF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wVwpF.png&quot; alt=&quot;tumblr Col. Chris Hadfield&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25088" OwnerDisplayName="Bob516" LastEditorUserId="7982" LastEditDate="2019-05-23T13:57:52.627" LastActivityDate="2019-05-23T13:57:52.627" Title="Word for blue ring around the Earth" Tags="&lt;earth&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="32008" PostTypeId="2" ParentId="32007" CreationDate="2019-05-22T23:06:53.683" Score="4" Body="&lt;p&gt;There are actually a number of words / expressions related to this. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Aureole&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[Bright edge of object - sometimes with colour of light source vs object] This is when an object is silhouetted against a bright light source and it gives a ring of colour. This typically occurs when the light source is smaller than the sillhoutte. I think this is the closest meaning because its meaning implies the colouration caused by diffusion (or even refraction - in this case) of light &lt;strong&gt;around&lt;/strong&gt; the sillhoutte. For example, a girls head in front of a candle: the fine granularity of the hair diffuses the light, bending it and causes a bright circumference of light around the extremities of her hair. This word is associated in optical physics with a number of other diffusion phenomina too, such as the aureole effect.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Corona&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[Colour of the light source, shape of light source] This is a ring of light when the sillhoutte is much smaller than the light source (or even no sillhoutte at all!), and there is practically no interaction between the light and the extremities of the sillhoutte, as a result most of the colour is from the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noteable mentions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A diamond ring&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[Angle of light source to object] is where the eclipse has passed either side of total and there's a flare on one side.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bailey's beads&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[Shape of object to shape of light source] are when there is a total eclipse and there are flares all around the circumference, caused by imperfections in the planet's surface and / or the irregularity of the star's corona - entirely diffraction based.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Glory&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[Entirely colour of diffraction / refraction of light source around object] This is an optical phenomina which is less common, and is almost entirely based on refraction, giving colour bands in the vicinity of the object, typically in concentric colour bands. A bit like a rainbow. This only occurs in the context of where the observer and object are in an atmosphere so is least applicable to this context - although technically the whole solar system is within the atmosphere (or heliosphere) of the sun, right up to the helioshock, where the heliosphere interacts with either other stars or forms a boundary with outter space, but it's not inconceivable that this could occur in a nebula, or even give rise to a&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Brocken spectre&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Incidently: the blue aura, in this case, is actually caused by a physics phenomina called Rayleigh Scattering which is where the atmosphere has the effect of appearing to change the wavelength of light as it passes through. Often this can be blue, however in some situations it might be red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a whole family of words for these kinds of meanings, and a many of them are poetic. Some of the words relate to colour, some to shape, some to diffraction, some to refraction. This is one of those situations where knowing a little science could help you chose the right word.&lt;/p&gt;&#xA;" OwnerUserId="28160" OwnerDisplayName="Owl" LastActivityDate="2019-05-23T00:11:52.980" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32009" PostTypeId="2" ParentId="13551" CreationDate="2019-05-23T08:17:00.360" Score="2" Body="&lt;p&gt;If the two black holes miss are more than a few Schwarzschild radii apart at closest approach then what will happen will look very much like the pure Newtonian solution -- similar to a close encounter between two planets or stars -- but with some small perturbations to the orbits due to GR and some energy lost to gravitational radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they go much closer than that, the GR effects become much larger. Firstly the trajectories change, so they may &quot;bend&quot; or &quot;spiral&quot; into one another (quotes because saying anything geometric about what happens near a black hole where space and time are so curved risks being misleading). As they do so, when they get close enough the event horizons will merge rapidly deforming to enclose all the space that was previously inside either of them, and some more. The resulting event horizon will then quickly settle down to one of the standard symmetrical forms, radiating a lot of gravitational waves in the process. The energy and angular momentum of the initial situation will end up either in the spin and mass of the final black hole, or in those gravitational waves.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-05-23T08:17:00.360" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32010" PostTypeId="1" AcceptedAnswerId="32011" CreationDate="2019-05-23T08:31:18.247" Score="4" ViewCount="53" Body="&lt;p&gt;In one of the answers to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/31962/is-it-plausible-for-a-planet-that-is-positioned-in-the-habitable-area-of-a-solar&quot;&gt;this question&lt;/a&gt; @MartinKochanski made the interesting point that the abundance of elements heavier than iron (r-process elements) in the solar system is probably due to a fairly recent and close (at the time) neutron star merger -- a relatively rare event on a galactic scale.  This means the early solar system was enriched with &quot;fresh&quot; r-process elements, as compared to the &quot;stale&quot; r-process elements that would be typical. For instance the early solar system is conjectured to have been rich in uranium and thorium compared to a &quot;typical&quot; system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is what differences might this have made to the formation and evolution of the solar system? Would it have had any bearing on the origins of life on Earth?&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-05-23T12:53:16.290" Title="How much difference did the supply of &quot;fresh&quot; r-process elements make to the Solar System.?" Tags="&lt;solar-system&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32011" PostTypeId="2" ParentId="32010" CreationDate="2019-05-23T11:44:57.927" Score="4" Body="&lt;p&gt;The discussion in that answer is slightly misleading. Here's the current picture, taking into account the &lt;a href=&quot;http://adsabs.harvard.edu/abs/2019Natur.569...85B&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Nature&lt;/em&gt; paper by Bartos &amp;amp; Marka referenced there&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Elements heavier than iron up through rubidium (atomic number 37) are produced by the &lt;a href=&quot;https://en.wikipedia.org/wiki/R-process&quot; rel=&quot;nofollow noreferrer&quot;&gt;r-process&lt;/a&gt; in supernovae.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Elements heaver than rubidium are produced by a combination of the &lt;a href=&quot;https://en.wikipedia.org/wiki/S-process&quot; rel=&quot;nofollow noreferrer&quot;&gt;s-process&lt;/a&gt; in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Asymptotic_giant_branch&quot; rel=&quot;nofollow noreferrer&quot;&gt;asymptotic giant branch stage&lt;/a&gt; of intermediate-mass stars (i.e., masses of 2-10 solar masses) &lt;em&gt;and&lt;/em&gt; the r-process in neutron-star mergers.[1] &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The argument in the &lt;em&gt;Nature&lt;/em&gt; paper is that when the solar system formed, a significant fraction of actinide metals &lt;em&gt;with short half-lives&lt;/em&gt; (e.g., isotopes of curium and plutonium, with half lives &lt;span class=&quot;math-container&quot;&gt;$&amp;lt; 100$&lt;/span&gt; million years) came from one recent, nearby neutron-star merger. (These are the &quot;fresh&quot; r-process elements.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Stable heavy elements produced by neutron-star mergers would have come from the cumulative contributions of hundreds (possibly a few thousand) NS mergers over the history of the Milky Way, mixed over hundreds of millions or billions of years into the Galaxy's interstellar gas (these are the &quot;stale&quot; r-process elements).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Unstable heavy elements with &lt;em&gt;long&lt;/em&gt; half-lives -- e.g., U-238 (4.5 billion years) and Th-232 (14 billion years) -- would be mostly in the &quot;stale&quot; category, and so would &lt;em&gt;not&lt;/em&gt; be significantly affected by the presence of said recent, nearby NS merger. In fact, the &lt;em&gt;Nature&lt;/em&gt; paper estimates that their hypothesized nearby NS merger would account for only 0.3% of the solar system's initial Th-232. Even U-235, with a half-life of 700 million years, would probably only be moderately affected by that one NS merger.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In general, then, the presence or absence of metals from a single nearby NS merger would have made very little difference to the formation of the solar system or to the evolution of life on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] See &lt;a href=&quot;https://en.wikipedia.org/wiki/S-process#/media/File:Nucleosynthesis_periodic_table.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this figure&lt;/a&gt; for an approximate breakdown by element -- note that the &quot;dying low-mass stars&quot; label in the figure refers to the AGB stage of what I call intermediate-mass stars, which are &quot;low mass&quot; only in comparison to stars with masses &lt;span class=&quot;math-container&quot;&gt;$&amp;gt; 10$&lt;/span&gt; times that of the Sun.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="16685" LastEditDate="2019-05-23T12:53:16.290" LastActivityDate="2019-05-23T12:53:16.290" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32012" PostTypeId="2" ParentId="32004" CreationDate="2019-05-23T12:15:58.247" Score="5" Body="&lt;p&gt;That's a region of recent star formation in the disk of NGC 4302. Regions of recent star formation include short-lived, high-mass O and B stars, which are blue (because they're hot) and luminous enough to outshine the (redder) lower-mass stars that are also present. If you look carefully at the image, you can see hints of blue light from other parts of the disk (e.g., in the upper right and upper left sides of the disk). You can also see lots of recent star formation in the disk of the neighboring galaxy, NGC 4298.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the full-resolution (84 MB) version of the image &lt;a href=&quot;http://hubblesite.org/image/4019/gallery&quot; rel=&quot;noreferrer&quot;&gt;available here&lt;/a&gt;, you can see that the blue region resolves into many blue pinpricks in the midst of a fuzzier blue haze, a clear indication that it's made up of lots of bright blue stars (or small clusters of bright, blue stars) and a larger number of somewhat fainter blue stars. You can see the same thing going on in the blue regions of NGC 4298.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are probably other, similar regions in the disk of NGC 4302; the reason this particular region shows up so well is probably because it's located in the outer part of the disk on the side closest to us, so there's relatively little dust blocking its light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Although there is undoubtedly ionized gas in the vicinity of those stars, the gas is not producing enough light to show up in the image -- you would in fact see a &lt;em&gt;reddish/pinkish&lt;/em&gt; region if it was mostly light from ionized gas.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2019-05-23T12:15:58.247" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32014" PostTypeId="1" AcceptedAnswerId="32042" CreationDate="2019-05-23T16:22:09.403" Score="6" ViewCount="869" Body="&lt;p&gt;I recently came across a short &lt;a href=&quot;https://en.wikipedia.org/wiki/Fast_Fourier_Transform_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; mentioning the idea for a telescope built without mirrors or lenses that instead uses an array of antennas and massive computing power to resolve an image. Such a thing will apparently become possible if &quot;Moore's law continues to hold&quot;, but doesn't explain how it will work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a link to the original paper: &lt;a href=&quot;https://arxiv.org/abs/0805.4414&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tegmark and Zaldarriaga 2009&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We will show that it acts much like a single dish telescope with a&#xA;  dramatically larger field of view, yet is potentially much cheaper&#xA;  than a standard interferometer with comparable area. If a modern&#xA;  all-digital interferometer such as the MWA &lt;a href=&quot;https://www.youtube.com/watch?v=spUNpyF58BY&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt; is scaled up to a very&#xA;  large number of antennas N , its price becomes completely dominated by&#xA;  the computing hardware cost for performing of order N&lt;sup&gt;2&lt;/sup&gt; correlations&#xA;  between all its antenna pairs. The key idea behind the FFT Telescope&#xA;  is that, if the antennas are arranged on a rectangular grid, this cost&#xA;  can be cut to scale merely as N log&lt;sub&gt;2&lt;/sub&gt; N using Fast Fourier Transforms.&#xA;  As we will see, this design also eliminates the need for individual&#xA;  antennas that are pointable (mechanically or electronically),&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unlike a traditional interferometer, which consists of individual telescopes that focus light (or radio waves) to a point before ultimately combining their images, the antennas don't have a dish that focuses the radio waves from a certain point in the sky. Instead it's just an array of antennas that images the entire sky at once:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The results above also show that the FFTT can be thought of as a cheap&#xA;  maximally compact interferometer with a full-sky primary beam.  To&#xA;  convert a state-of- the-art interferometers such as MWA &lt;a href=&quot;https://www.youtube.com/watch?v=spUNpyF58BY&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;, LOFAR&#xA;  [2], 12 PAPER[4], 21CMA [3] into an FFTT, one would need to do three&#xA;  things:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Move all antenna tiles together so that they nearly touch.&lt;/li&gt;&#xA;  &lt;li&gt;Get rid of any beamformer that “points” tiles to- wards a specific sky direction by adding relative phases to its component antennas, and&#xA;  treat each antenna as independent instead, thus allowing the array to&#xA;  image all sky directions simultaneously.&lt;/li&gt;&#xA;  &lt;li&gt;Move the antennas onto a rectangular grid to cut the correlator cost from N&lt;sup&gt;2&lt;/sup&gt; to N log&lt;sub&gt;2&lt;/sub&gt; N .&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can someone explain how a computer can resolve an image without a lens, dish or mirror to make parallel rays of light converge on a sensor? I thought that a sensor without a lens, dish or mirror will just produce a blurry mess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I think I understand it now. Watching &lt;a href=&quot;https://www.youtube.com/watch?v=spUNpyF58BY&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video&lt;/a&gt; was massively helpful for me in understanding how Fourier transformations allow you to isolate individual waveforms from a signal containing many different waveforms.&lt;/p&gt;&#xA;" OwnerUserId="28132" LastEditorUserId="37191" LastEditDate="2021-02-12T20:52:48.613" LastActivityDate="2021-02-12T20:52:48.613" Title="How would a Fast Fourier Transform Telescope work without a mirror, dish or lenses?" Tags="&lt;telescope&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32015" PostTypeId="1" CreationDate="2019-05-23T17:43:35.733" Score="0" ViewCount="110" Body="&lt;p&gt;Prior to the era of radar and other forms of radio/RF/EM ranging, what approaches, methods, and techniques did early astronomers (&lt;em&gt;e.g.,&lt;/em&gt; Kepler, Cassini, Copernicus) use to measure the distance(s) from Earth to various objects (such as: Mars, Venus, Alpha Centauri, etc.)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an especially informative article (say, &lt;em&gt;Scientific American, Sky&lt;/em&gt;) or book that explains their efforts and results (&lt;em&gt;e.g.,&lt;/em&gt; how accurate were their computations?)?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;References and citations are warmly welcomed!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user28134" LastEditorUserId="2365" LastEditDate="2019-05-23T22:20:18.220" LastActivityDate="2019-05-23T22:24:11.797" Title="How Did Early Astronomers Measure Distances?" Tags="&lt;observational-astronomy&gt;&lt;distances&gt;&lt;fundamental-astronomy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32016" PostTypeId="2" ParentId="253" CreationDate="2019-05-23T18:32:50.050" Score="3" Body="&lt;p&gt;Thermonuclear fusion has nothing to do with the central temperature of the Sun. You can get a rough estimate of the temperature (with some necessary simplification) following this line of reasoning:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The material of the Sun is an ideal, completely ionized, gas (all electrons are separated from nuclei);&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This means that pressure of the gas is proportional to its temperature and to number of gas particles in unit volume;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Pressure in the center (innermost part) of the Sun must be large enough to support the weight of all the layers above;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you suppose that the Sun is made only from hydrogen you get a central temperature of some 23 million degrees.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="25871" LastEditorUserId="14094" LastEditDate="2019-05-24T14:39:37.923" LastActivityDate="2019-05-24T14:39:37.923" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32017" PostTypeId="1" AcceptedAnswerId="32128" CreationDate="2019-05-23T19:43:17.377" Score="4" ViewCount="138" Body="&lt;p&gt;Recently there has been &lt;a href=&quot;https://www.quantamagazine.org/black-hot-superionic-ice-may-be-natures-most-common-form-of-water-20190508/&quot; rel=&quot;nofollow noreferrer&quot;&gt;experimental verification of superionic water ice&lt;/a&gt; (suggested to be called ice&amp;nbsp;XVIII, not sure if this nomenclature is official yet), which turns out to be a solid (or at least, solid-ish). This has been proposed to comprise a substantial fraction of the volume of ice giant planets like Uranus and Neptune, which is something of a departure from the idea that the bulk of these planets is fluid, though potentially the presence of substances other than water may modify things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is whether there are any observational constraints on the presence of solid material in Uranus and Neptune, e.g. from the gravitational moments, tidal parameters, or global oscillations (&quot;seismology&quot;) - I'm aware that &lt;a href=&quot;https://arxiv.org/abs/1702.02943v1&quot; rel=&quot;nofollow noreferrer&quot;&gt;the K2 mission failed to detect any global oscillations on Neptune&lt;/a&gt; so I'd guess this latter method may not help much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There is a &lt;a href=&quot;https://astronomy.stackexchange.com/q/28895/24157&quot;&gt;previous question about solids in gas giants&lt;/a&gt;, but that one seems to be more focused on rock rather than ice and has ended up addressing Jupiter rather than the ice giant planets.)&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-05-23T19:58:37.920" LastActivityDate="2019-06-01T19:02:52.167" Title="What constraints are there on solid material in the ice giants?" Tags="&lt;water&gt;&lt;neptune&gt;&lt;uranus&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32018" PostTypeId="1" CreationDate="2019-05-23T19:52:46.770" Score="3" ViewCount="66" Body="&lt;p&gt;Over the last decade we've discovered that &lt;em&gt;many&lt;/em&gt; exoplanets not only have highly eccentric orbits, but also orbit very close to their sun. This is true of many classes of stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I definitely have a gap in my understanding, because I'm perplexed as to why the orbits don't circularize when the planets are so close to their star. The tidal forces can be immense, yet the proportionally large number of eccentric planets indicates that many are likely not circularizing (at least over time periods of a handful of giga years). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand the physics of this well enough to say. Could it have something to do with with spin-orbital resonance of the planet? A paper I found from Anthony R. Dobrovolskis titled &lt;em&gt;Spin states and climates of eccentric exoplanets&lt;/em&gt; indicates that highly eccentric planets in the tidal-lock zone would likely lock into spin states other than 1:1. This could be 2:1, 3:2, etc. Much like mercury. Does the achievement of this spin-orbital resonance do something to preserve the eccentricity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd appreciate any help that could be provided! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because it was asked for, I've included a list of some exoplanets that have both short orbital periods as well as highly eccentric orbits (varying from ≈0.15 all the way up to around 0.5). Ages of systems vary widely or is not known, from possibly as low as about 2gy up to over 10gy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.openexoplanetcatalogue.com/planet/CoRoT-16%20b/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.openexoplanetcatalogue.com/planet/CoRoT-16%20b/&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.openexoplanetcatalogue.com/planet/HD%2011964%20A%20c/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.openexoplanetcatalogue.com/planet/HD%2011964%20A%20c/&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.openexoplanetcatalogue.com/planet/HD%20109271%20c/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.openexoplanetcatalogue.com/planet/HD%20109271%20c/&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.openexoplanetcatalogue.com/planet/HD%2017156%20b/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.openexoplanetcatalogue.com/planet/HD%2017156%20b/&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.openexoplanetcatalogue.com/planet/HD%2017156%20b/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.openexoplanetcatalogue.com/planet/HD%2017156%20b/&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.openexoplanetcatalogue.com/planet/KOI-1299%20b/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.openexoplanetcatalogue.com/planet/KOI-1299%20b/&lt;/a&gt; - VERY eccentric&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possibly highly eccentric, but measurement delta is too large to confirm eccentricity above ≈0.11:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.openexoplanetcatalogue.com/planet/Kapteyn%20b/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.openexoplanetcatalogue.com/planet/Kapteyn%20b/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update 2:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More sources showing the existence of near-orbiting highly eccentric planets. While higher eccentricities are common farther out from a star, it looks like moderately high eccentricities (0.2, 0.3, 0.4) are not infrequent close-in to a star, particularly if the system lacks many other planets. Very high eccentricities are also possible close-in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ctP4i.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ctP4i.png&quot; alt=&quot;Chart&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PLqgL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PLqgL.png&quot; alt=&quot;Chart&quot;&gt;&lt;/a&gt;&#xA;From: &lt;a href=&quot;https://academic.oup.com/mnras/article/425/1/757/1004680&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://academic.oup.com/mnras/article/425/1/757/1004680&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8R4LU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8R4LU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&quot;Eccentricity versus semimajor axis going from low (left) to high multiplicity (right). A red dot is shown for where Jupiter would appear on the one- and two-planet distributions, and for Saturn on the two-planet distribution.&quot;&lt;/p&gt;&#xA;" OwnerUserId="20372" LastEditorUserId="20372" LastEditDate="2019-05-24T16:29:55.647" LastActivityDate="2019-05-24T16:29:55.647" Title="What maintains the high orbital eccentricity of many exo-planets near their sun?" Tags="&lt;orbit&gt;&lt;exoplanet&gt;&lt;rotation&gt;&lt;eccentric-orbit&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32019" PostTypeId="2" ParentId="24038" CreationDate="2019-05-23T21:01:34.573" Score="3" Body="&lt;p&gt;You should use 3CRR. It contains a few sources of large angular diameter that were missed in the original surveys.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Third_Cambridge_Catalogue_of_Radio_Sources#3CRR&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Third_Cambridge_Catalogue_of_Radio_Sources#3CRR&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://3crr.extragalactic.info/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://3crr.extragalactic.info/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="28137" LastEditorUserId="7982" LastEditDate="2019-05-24T01:15:09.013" LastActivityDate="2019-05-24T01:15:09.013" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32020" PostTypeId="2" ParentId="32015" CreationDate="2019-05-23T22:24:11.797" Score="2" Body="&lt;p&gt;They were not very successful, but not for a want of trying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ancient greeks tried to measure the angle between the sun and the moon when the moon is half full. They got a value of 87 degrees, (when really it is 89.85). From this they could estimate the relative distances of the Moon and the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further observations of the moon from different locations on Earth could, in principle find the relative distance to the moon in terms of the radius of the Earth. And the radius of the Earth could be determined by noting the different lengths of shadows at the same time in different locations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several Greeks attempted to put these together to find the distance of the sun, and while their methods were geometrically correct, their observations were not accurate enough to achieve an accurate value for the distance of the sun. Typically they estimates were about 5% of the true distance. Attempts based on similar methods were made in other centres of astronomy, in China, India and the Middle East. They all tended to dramatically underestimate the distance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even after the invention of the telescope allowed the observations to be done more accurately the distance was usually under-estimated. only when careful measurements of the transit of Venus were made, was Jerome Lalande able to calculate the distance of Venus, and use Kepler's laws to derive the distance of the sun and other planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kepler had created a model of the solar system that predicted the relative distances of each planet very successfully, but not the absolute distance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nearly all other distances are based on the Earth-sun distance. The distances of the planets can be found using Kepler's laws. The actual distance of stars was unknown we were able to measure the slight change in position of a star as the Earth orbits the sun. If you know the distance to the sun you can calculate the distance to a star. Prior to the observation of this &quot;stellar parallax&quot; estimates of the distanct to stars was essentially guesses, and usually significant underestimates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So prior to radar ranging, the methods were geometric, and based on a ladder of measurements. The size of the Earth was used to determine the distance to the moon and sun. Then the distance to the sun was used to find the distance to the stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more distant stars there can be considerable uncertainty about the distance. If a star is too far to have parallax, then it's true distance will not be known to accuracy.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-05-23T22:24:11.797" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32021" PostTypeId="2" ParentId="31989" CreationDate="2019-05-24T01:11:43.170" Score="2" Body="&lt;p&gt;From the question &lt;a href=&quot;https://astronomy.stackexchange.com/q/28610/7982&quot;&gt;What can be learned from, or noted in this LIGO Orrery video?&lt;/a&gt; we can watch the video &lt;a href=&quot;https://youtu.be/gmmD72cFOU4&quot; rel=&quot;nofollow noreferrer&quot;&gt;LIGO Orrery&lt;/a&gt; (which was inspired by the mind-blowing &lt;a href=&quot;https://youtu.be/_DnDeBa0KFc&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kepler Orrery IV&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've made a small, low quality GIF from screen shots here, the video is much more interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's been difficult to be sure what exactly is depicted as mentioned in &lt;a href=&quot;https://astronomy.stackexchange.com/a/30199/7982&quot;&gt;this answer&lt;/a&gt; but it's likely the surfaces depict something at least a bit like an event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VGVRR.gif&quot; alt=&quot;LIGO Orrery&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-05-24T01:11:43.170" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32022" PostTypeId="1" CreationDate="2019-05-24T03:30:30.743" Score="1" ViewCount="107" Body="&lt;p&gt;I have read various discussions of the Julian day, including that on the Wikipedia, but none of these appear to define the maximum Julian day or proleptic Gregorian year in which that maximum day would occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a maximum Julian day? If so, what is it?&lt;/p&gt;&#xA;" OwnerUserId="2800" LastActivityDate="2019-05-24T13:55:20.403" Title="Is there a maximum Julian day?" Tags="&lt;time&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32024" PostTypeId="2" ParentId="32022" CreationDate="2019-05-24T13:55:20.403" Score="7" Body="&lt;p&gt;Since the Julian day is defined as a continuous count of seqential days starting at Jan 1 4013 BC, and there is no limit to the number of integers available for  the count, there is no specified maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are asking when the Julian Day Number will no longer be useful, possible answers are when there is no one left to count it, or when it is replaced by another concept or definition. Note that there are already several shortened versions, such as Modified JD and Truncated JD, used when JDN became too large to fit in smaller computer word sizes with the desired precision.&lt;/p&gt;&#xA;" OwnerUserId="28086" LastActivityDate="2019-05-24T13:55:20.403" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32025" PostTypeId="2" ParentId="32014" CreationDate="2019-05-24T15:09:31.187" Score="2" Body="&lt;p&gt;Like all science-fiction, it is a mixture of real technology and projections. Obtaining images from lensless telescopes is mainstream radioastronomy (like the &lt;a href=&quot;https://eventhorizontelescope.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;recently famous picture of a black hole&lt;/a&gt;). We don't have technology to do it for visible light, but it's all electromagnetic radiation following the same rules so one might imagine building the necessary receivers and processing the data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, exposing your typical image sensor to light without a lens is not going to produce anything interesting (&lt;a href=&quot;https://arxiv.org/pdf/1702.06619.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;unless you are these guys&lt;/a&gt;), but radioastronomy has one advantage over visible light observations: they see the shape of the incoming waves, not just a photon count. Mathematically combining the waves in a correct way reveals what would happen if the lens were really there, creating the image similarly to what the real lens does.&lt;/p&gt;&#xA;" OwnerUserId="7654" LastActivityDate="2019-05-24T15:09:31.187" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32026" PostTypeId="1" CreationDate="2019-05-24T16:46:03.497" Score="-1" ViewCount="91" Body="&lt;p&gt;We know that when two huge black holes collide they produces gravitational waves in space time curvature. So if I am not wrong another object in space if goes in that direction during production of waves then that huge black holes can push that object like star, sun, planet into opposite direction. So, indirectly, two objects in space can repel each other in a very few cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it also the case that two planets can repel each other in a few cases?&lt;/p&gt;&#xA;" OwnerUserId="28146" LastEditorUserId="2365" LastEditDate="2019-05-25T06:14:55.540" LastActivityDate="2019-05-25T06:24:52.707" Title="Planets may repel each other in few cases" Tags="&lt;gravity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32027" PostTypeId="2" ParentId="32026" CreationDate="2019-05-24T16:56:24.677" Score="2" Body="&lt;p&gt;Gravitational waves mostly just shake things, there's not much push-- especially in comparison to the gravity of the object that created the waves.  It seems unlikely there could ever be a net push.  Still, it is thought that empty space already has a tendency to &quot;push&quot; objects apart, based on the idea of &quot;dark energy.&quot;  So two objects in an otherwise empty universe, if they are far enough apart, would be seen to push each other apart based on a gravitational effect-- but it would not be viewed as the gravity of the objects, but rather the antigravity, in effect, of all the empty space between them.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2019-05-24T16:56:24.677" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32028" PostTypeId="1" AcceptedAnswerId="32033" CreationDate="2019-05-24T17:43:18.007" Score="5" ViewCount="344" Body="&lt;p&gt;In layman terms:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;nothing ever escapes the pull of a black hole, not even light&lt;/li&gt;&#xA;&lt;li&gt;when a super massive star reaches the end of it's life you get a supernova&lt;/li&gt;&#xA;&lt;li&gt;sometimes the &lt;em&gt;&quot;remains&quot;&lt;/em&gt; of these stars can turn turn into black holes&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My question is: can a black hole &lt;em&gt;&quot;supernova&quot;&lt;/em&gt; the same way a large star does?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm saying is, could this process occur in reverse? &lt;strong&gt;I.E:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;black hole&lt;/code&gt; -&gt; &lt;code&gt;supernova&lt;/code&gt; -&gt; &lt;code&gt;particles form dust cloud&lt;/code&gt; -&gt; &lt;code&gt;new star&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Apologies if any of this sounds stupid (or vague), I'm just curious.&lt;/p&gt;&#xA;" OwnerUserId="28147" LastEditorUserId="28147" LastEditDate="2019-07-23T14:13:06.767" LastActivityDate="2019-08-03T09:26:44.623" Title="Can a black hole &quot;supernova&quot;?" Tags="&lt;star&gt;&lt;black-hole&gt;&lt;gravity&gt;&lt;supernova&gt;&lt;explosion&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32029" PostTypeId="2" ParentId="7700" CreationDate="2019-05-24T18:30:10.690" Score="1" Body="&lt;p&gt;The expression that is used by NASA/JPL to approximate the relativistic effects on the orbit of a single planet plus the Sun in our solar system is known as the &quot;&lt;a href=&quot;https://space.stackexchange.com/a/23409/29864&quot;&gt;post-Newtonian expansion&lt;/a&gt;&quot; and looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\left(1-\frac{4GM}{rc^2}+\frac{v^2}{c^2}\right)\hat{r} +\frac{4GM}{r^2}\left(\hat{r}\cdot \hat{v}\right)\frac{v^2}{c^2}\hat{v}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are many planets the expression becomes more complex. You can compare this to the classical Newtonian acceleration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\hat{r}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not a big fan of this approximation but it is what is mostly used.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For a pure circular orbit in Schwarzschild coordinates you get the same orbital velocity in GR (in coordinate time) as classically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I you are dropping an object from rest the initial acceleration in GR (in coordinate time) is the same as classically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally if you want to know if GR or classical Newtonian gravitation results in more acceleration you have to decide if you are interested in the result in &quot;coordinate time&quot; or in &quot;proper time&quot; and the fraction will also vary depending on in what direction the planet is moving as compared to the Sun.&lt;/p&gt;&#xA;" OwnerUserId="26403" LastActivityDate="2019-05-24T18:30:10.690" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32030" PostTypeId="1" AcceptedAnswerId="32031" CreationDate="2019-05-24T18:49:08.157" Score="3" ViewCount="96" Body="&lt;p&gt;&lt;a href=&quot;https://arxiv.org/abs/1801.04318&quot; rel=&quot;nofollow noreferrer&quot;&gt;This paper on Arxiv&lt;/a&gt; indicates that the signal period of known pulsars is 1 ms-15 s (0.07-1,000 Hz since F=1/T). I have an EE background and it's a long story how I ended up here, but suffice it to say I'm weak on astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Radio_telescope#/media/File:Atmospheric_electromagnetic_opacity.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;This chart on Wikipedia&lt;/a&gt; implies that the entire range of pulsar frequencies should be absorbed by the atmosphere, but &lt;a href=&quot;http://pulsarsearchcollaboratory.com/wp-content/uploads/2016/01/PSC_search_guide.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this guide&lt;/a&gt; and even &lt;a href=&quot;https://www.rtl-sdr.com/amateur-pulsar-observations-rtl-sdr/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this budget RF telescope writeup&lt;/a&gt; clearly state that pulsars &lt;em&gt;can&lt;/em&gt; be detected from Earth's surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions is: how are pulsars detected from Earth--what am I missing here? Are RF telescopes simply that receptive? Any insights appreciated.&lt;/p&gt;&#xA;" OwnerUserId="28148" LastActivityDate="2019-05-24T19:58:51.083" Title="How are pulsars detected from Earth?" Tags="&lt;radio-astronomy&gt;&lt;pulsar&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32031" PostTypeId="2" ParentId="32030" CreationDate="2019-05-24T19:58:51.083" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;the signal period of known pulsars is 1 ms-15 s (0.07-1,000 Hz since F=1/T).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's the frequency of how often the signals occur, &lt;em&gt;not&lt;/em&gt; the frequency of the waves they emit. Just compare it to a lighthouse, which has a signal period of a few seconds but a wave frequency of &lt;span class=&quot;math-container&quot;&gt;$10^{15}$&lt;/span&gt; Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first known pulsars, for example &quot;Little Green Man&quot; or &lt;a href=&quot;https://en.wikipedia.org/wiki/PSR_B1919%2B21&quot; rel=&quot;nofollow noreferrer&quot;&gt;PSR B1919+21&lt;/a&gt;, were radio pulsars; your graph shows those waves &lt;em&gt;are&lt;/em&gt; detectable from Earth's surface. I stumbled upon &lt;a href=&quot;https://academic.oup.com/mnras/article/431/4/3624/1161659&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; which lists the frequency of some of the pulsars:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;After the discovery of pulsars at 81.5 MHz (Hewish et al. 1968), the first pulsar detected at 40 MHz was PSR B1133+16 (Arecibo Observatory, 1968, unpublished). In 1969–1970 the pulsars PSR B1133+16 and B0809+74 were detected at 25 MHz (Bruck 1970).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Below is the full detection range for the telescopes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Therefore joint observations (simultaneous or not) will be highly beneficial, providing a continuous frequency coverage from 8–240 MHz.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is roughly equivalent to wavelengths of 40m to 1m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X-ray pulsars like &lt;a href=&quot;https://en.wikipedia.org/wiki/Centaurus_X-3&quot; rel=&quot;nofollow noreferrer&quot;&gt;Centaurus X-3&lt;/a&gt; were discovered later, and not by telescopes on the surface:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Centaurus X-3 was first observed during experiments of cosmic X-ray sources made on May 18, 1967. These initial X-ray spectrum and location measurements were performed using a sounding rocket.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2019-05-24T19:58:51.083" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32032" PostTypeId="1" CreationDate="2019-05-24T22:18:28.410" Score="4" ViewCount="72" Body="&lt;p&gt;Reading &lt;a href=&quot;https://astronomy.stackexchange.com/questions/32014/how-would-a-fast-fourier-transform-telescope-work-without-a-mirror-dish-or-lens&quot;&gt;this question&lt;/a&gt; I was struck by the fact that the &quot;pointing&quot; of an interferometer is effectively done in the correlator, at least within the common field of view of all the telescopes, which is many thousands of times wider than the imaged area. Could the same data be reprocessed to reveal images of other &quot;tiles&quot; from this common field of view? Could it image the accretion disk, for example, or the base of the jets? Are there any targets that are close enough and bright enough to be worth doing this?&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="7982" LastEditDate="2021-03-26T23:28:06.177" LastActivityDate="2021-03-26T23:28:06.177" Title="Could images of (very) nearby targets be recovered from Event Horizon Telescope data?" Tags="&lt;radio-astronomy&gt;&lt;event-horizon-telescope&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32033" PostTypeId="2" ParentId="32028" CreationDate="2019-05-24T22:28:32.317" Score="12" Body="&lt;p&gt;No, it cannot. A black hole is something vastly different from a star. It's vastly different from anything else. It's not a thing, really, but more like a portion of very distorted spacetime. Nothing escapes from it simply because there is no way out - spacetime is distorted in such a way that all trajectories lead to the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there is a mechanism where radiation is generated just outside the black hole, sucking energy from its gravitational field and therefore from its mass. This is called Hawking radiation. It does not come from inside the BH, but the way it's created it leeches the BH's energy / mass. In time, this would diminish the mass of the BH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out, the smaller the BH, the stronger the Hawking radiation. This in turn makes the BH lose weight even faster, so the HR gets even stronger, and so on. It's a vicious cycle. If the black hole is small enough, the process turns into an explosion - HR gets so intense, the whole mass of the black hole is converted into radiation at once. It's not a supernova, not as big as that, but it is a powerful explosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only issue here is - HR is very weak when BHs are big. So regular BHs will take an incredibly long time to slim down, if ever, to the tiny size where HR spirals out of control. Basically this never happens in reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TLDR: Not a supernova, just a huge bomb, but not very likely.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2019-05-24T22:28:32.317" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32034" PostTypeId="1" CreationDate="2019-05-24T23:11:32.033" Score="13" ViewCount="82637" Body="&lt;p&gt;My husband and I have just seen a line of about 15 lights moving in a straight line, quite high above the cloud level with a gap, and then 5 lights, a gap, another few, a gap, 1 light, a gap, and a few more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They were moving at the same sort of speed as I would expect a plane to move at - almost like a train in the sky with a banner flying behind. They were in the sky for at least a minute. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had seen a shooting star a few minutes before and it wasn’t one of them... what could it be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it be Elon Musk’s satellites?&lt;/p&gt;&#xA;" OwnerUserId="28149" LastEditorUserId="43" LastEditDate="2019-06-10T09:46:54.533" LastActivityDate="2019-06-10T10:20:06.707" Title="Line of lights moving in a straight line, with a few following" Tags="&lt;satellite&gt;&lt;identify-this-object&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32039" PostTypeId="2" ParentId="32034" CreationDate="2019-05-25T06:03:53.513" Score="22" Body="&lt;p&gt;Given the date and timing, this could be most like &lt;a href=&quot;https://en.wikipedia.org/wiki/Starlink_(satellite_constellation)&quot; rel=&quot;noreferrer&quot;&gt;the Starlink satellites&lt;/a&gt; in their &quot;stacked&quot; configuration. They are currently in a line, but they will later move to separate orbits. Dr Marco Langbroek has &lt;a href=&quot;https://twitter.com/Marco_Langbroek/status/1132058893897621504&quot; rel=&quot;noreferrer&quot;&gt;a video of a line of 60 satellites&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They won't stay in a line; they have ion engines that they will use to spread out into separate orbits.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="332" LastEditDate="2019-06-10T10:20:06.707" LastActivityDate="2019-06-10T10:20:06.707" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32040" PostTypeId="2" ParentId="32026" CreationDate="2019-05-25T06:24:52.707" Score="3" Body="&lt;p&gt;No, gravity always pulls things together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational waves can cause a stretching of an object as they pass, but not much &quot;push&quot;. After the waves pass, the object will return to its unstretched shape and it will not have moved. There is an effect in which an object can gain momentum (ie be pulled) when it enters the region around a rotating black hole. However it is dragged sideways, it isn't pushed away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &quot;frame dragging effect&quot; also happens around other rotating bodies, and has been measured around the Earth. It is a very subtle effect that required some very sensitive measurements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no situations in which two planets could actually &quot;repel&quot; each other. The dominant term in the gravity between two planets will always pull them together.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-05-25T06:24:52.707" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32042" PostTypeId="2" ParentId="32014" CreationDate="2019-05-25T07:55:13.563" Score="5" Body="&lt;p&gt;From one perspective, the purpose of a telescope lens, or mirror is to delay the light arriving in the centre of the instrument a bit relative to the light arriving at the edges. A lens is thicker in the middle, and light travels more slowly in glass, the path to the centre of the mirror and back to the detector is longer, because of the curve of the mirror than the one to the edge, and so on. The result is that all the waves which hit any part of the lens or mirror from a point source end up arriving at the detector &lt;em&gt;in phase&lt;/em&gt; so they add up, while waves from anywhere else cancel out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A phased array antenna (common for short wavelength radio operations) achieves the same delay by capturing the signal at lots of points on the (flat) antenna and then artificially delaying it before combining it with the signals from the other points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interferometer (such as the Event Horizon telescope) does the same thing digitally. The full waveform of the signal (more or less) is recorded in multiple places and then they are combined in a large computer with various slightly different delays to separate waves coming from slightly different directions (I am over-simplifying greatly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this article proposes is mainly a way to do that for more antennae at less computing cost by placing the antennae in a specific layout and using a bit of clever maths (the FFT) to do all the combining in one go.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-05-25T07:55:13.563" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32046" PostTypeId="1" CreationDate="2019-05-25T20:39:36.190" Score="1" ViewCount="30" Body="&lt;p&gt;I customised my own landscape with stellarium, created my own horizon with the polyline setting like horizon_Geneve.txt file.&#xA;Manually, I can increase/decrease time to see when the sun's upper limb  touches the horizon, so my question is:&#xA;Is there a possibility or a script, to make the mission automatically, for many days instead to test each day then note it?&lt;/p&gt;&#xA;" OwnerUserId="16985" LastActivityDate="2019-05-25T20:39:36.190" Title="Stellarium script for sunrise times affected by topography" Tags="&lt;software&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32047" PostTypeId="2" ParentId="32001" CreationDate="2019-05-25T22:08:37.337" Score="3" Body="&lt;p&gt;The problem is that planet ejection is something that takes place over a very short timescale compared to the ages of the systems in question, so we're unlikely to see it actually take place. The main prospect of detection would be imaging of the ejected planets, for which you'd want a young giant planet as these retain enough heat from their formation to be sufficiently luminous in the infrared for detection even when not illuminated by a nearby star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There has actually been a claim for the detection of a (proto)planet in the process of being ejected from a binary star system: the case of TMR-1C as reported by Terebey et al. (1998) &quot;&lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/1998ApJ...507L..71T/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Candidate Protoplanet in the Taurus Star-forming Region&lt;/a&gt;&quot;. Unfortunately the nature of the object is not confirmed, see Terebey et al. (2000) &quot;&lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2000AJ....119.2341T/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;The spectrum of TMR-1C is consistent with a background star&lt;/a&gt;&quot;. There has been some back-and-forth on this: Riaz &amp;amp; Martín (2011) &quot;&lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2011A%26A...525A..10R/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Large-amplitude photometric variability of the candidate protoplanet TMR-1C&#xA;&lt;/a&gt;&quot; reject the background star hypothesis, while Riaz et al. (2013) &quot;&lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2013A%26A...559A.109R/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;A near-infrared variability campaign of TMR-1: New light on the nature of the candidate protoplanet TMR-1C&lt;/a&gt;&quot; regard it as a strong candidate for being a young stellar object (YSO).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another relevant case might be the circumbinary planet HD 106906 (AB) b, which may have been saved from ejection by a stellar flyby, see De Rosa &amp;amp; Kalas (2019) &quot;&lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2019AJ....157..125D/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Near-coplanar Stellar Flyby of the Planet Host Star HD 106906&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ejected planets are of interest due to their contribution to the population of low-mass objects in star clusters and the field (this population may well contain both ejected planets and low-mass brown dwarfs), and the impacts on the orbital architecture of the surviving planets after episodes of planet-planet scattering. One example of the latter I found is Carrera et al. (2019) &quot;&lt;a href=&quot;https://arxiv.org/abs/1903.02564&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planet-planet scattering as the source of the highest eccentricity exoplanets&lt;/a&gt;&quot;, there are many other papers on this topic.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-05-25T22:14:55.703" LastActivityDate="2019-05-25T22:14:55.703" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32048" PostTypeId="2" ParentId="32014" CreationDate="2019-05-26T00:22:14.657" Score="3" Body="&lt;p&gt;I'll add some perspective in addition to the other existing excellent answers.&lt;/p&gt;&#xA;&lt;p&gt;In a sense, single imaging optical telescopes or even single radio telescope dishes with focal plane arrays &lt;em&gt;can be thought of&lt;/em&gt; as Fourier transform telescopes.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Fourier_optics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fourier optics&lt;/a&gt; tells us that the relationship between the electromagnetic field at the aperture or pupil of an imaging system and the field at the image plane is Fourier in nature. A star viewed on-axis produces a flat plane wave at the pupil, and the FT of that is a delta function at zero. If you treat the finite diameter of the pupil, then the FT of the circular area is an &lt;a href=&quot;https://en.wikipedia.org/wiki/Airy_disk&quot; rel=&quot;nofollow noreferrer&quot;&gt;Airy disk&lt;/a&gt;, which we say is &amp;quot;due to diffraction&amp;quot;. If you have more stuff in the aperture (circular obstructions for secondaries, etc) you can take the FT of that and generate a point-spread function-like plot.&lt;/p&gt;&#xA;&lt;p&gt;If the star is off-axis, you still have the plane wave with a flat amplitude, but the phase is now ramped across the pupil. The FT of that is the same delta function or Airy disk, but &lt;em&gt;offset in position&lt;/em&gt;. Your second star is off to one side on the image plane.&lt;/p&gt;&#xA;&lt;p&gt;The principle works for microscopes, cameras and projectors as well, the last one being vice-versa; the inverse FT of the projected object represents the field at the output of the projector.&lt;/p&gt;&#xA;&lt;h2&gt;Conventional Radiotelescope Arrays and correlators&lt;/h2&gt;&#xA;&lt;p&gt;With limited resources, an array might have tens or hundreds of dishes or receivers, and to get resolution at small angular distances these are spread out quite sparsely over a very large area. The patterns look like Y's or spirals or random jumbles, usually with most of the dishes near the center but a fraction at very large distances.&lt;/p&gt;&#xA;&lt;p&gt;The computers that take in all this data are called &amp;quot;correlators&amp;quot; because they perform pair-by-pair correlations of the incoming digitized signals. The results of all of these correlations are then fed into an algorithm that generates an output image, but this requires some serious amount of modeling and careful processing to avoid wrong images or spurious artifacts appearing. An illustration of that is that the results of the Event Horizon Telescope were independently generated by different groups of researchers just to make sure that they came up with the same image after all that processing. There were so few telescopes used in the analysis that it would have been easy to end up with a crazy image.&lt;/p&gt;&#xA;&lt;h2&gt;Fourier Transform Telescope&lt;/h2&gt;&#xA;&lt;p&gt;A Fourier transform telescope would absolutely probably be built with a complete 2^M x 2^N evenly-spaced grid of receiving antennas or telescopes. Now instead of performing pair-wise correlations, you can just dump the whole thing into a computer architecture that has been optimized for performing a parallelized Fast Fourier transform on the data. It's going to be a big computer, but it can take advantage of hardware optimized for FFTs.&lt;/p&gt;&#xA;&lt;p&gt;This analysis &lt;em&gt;solves the same mathematical problem that a focusing lens or mirror solves&lt;/em&gt;, reproducing the Fourier transform of the electromagnetic field at the aperture.&lt;/p&gt;&#xA;&lt;p&gt;You could build a giant dish the same size as this array, and put a giant 2^M x 2^N array of receiving feed horns at the focal plane of this dish and recover an image that way, but performing the mathematical FT on the output of the flat aperture is thought to be more practical at some point in the near future, when the hardware is fast enough.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;below x2:&lt;/strong&gt; Example of a small dish with a focal plane array, cropped from &lt;a href=&quot;https://en.wikipedia.org/wiki/File:CSIRO_ScienceImage_2161_Close_up_of_a_radio_astronomy_telescope_with_several_more_in_the_background.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;CSIRO ScienceImage 2161 Close up of a radio astronomy telescope with several more in the background&lt;/a&gt;. Click for full size.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Rqq4X.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rqq4Xm.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/03IcN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/03IcNm.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;There are more radio telescope focal plane array images in &lt;a href=&quot;https://youtu.be/D2XPjOs5qYQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;this answer&lt;/a&gt; and a lonely one in &lt;a href=&quot;https://astronomy.stackexchange.com/q/31654/7982&quot;&gt;this answer&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;That the two different processes can both be used to produce wide-field images when the antennas are on a regular array is shown in Figure 1 of your linked paper, Tegmark &amp;amp; Zaldarriaga 2009 &lt;a href=&quot;https://arxiv.org/abs/0805.4414&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Fast Fourier Transform Telescope&lt;/a&gt;, and some differences are nicely summarized in the caption, along with the unhappy face on the correlator.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0SH3Q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0SH3Q.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;FIG. 1: When the antennas are arranged in a rectangular grid as in the FFT Telescope, &lt;strong&gt;the signal processing pipeline can be dramatically accelerated by eliminating the correlation step (indicated by a sad face)&lt;/strong&gt;: its computational cost scales as N2 a, because it must be performed for all pairs of antennas, whereas all other steps shown scale linearly with Na. The left and right branches recover the same images on average, but with slightly different noise. Alternatively, if desired, the FFT Telescope can produce images that are mathematically identical to those of the right branch (while retaining the speed advantage) by replacing the correlation step marked by the sad face by a spatial FFT, “squaring,” and an inverse spatial FFT.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;Interferometry?&lt;/h2&gt;&#xA;&lt;p&gt;When we have two telescopes, optical or radio, and we combine their signals together in hardware (using a beam splitter or electrical equivalent), we can call it an interferometer. If you have more antennas and you combine the signals with incremented phase offsets for each signal, you can also call that an interferometer. But by the time you have an ALMA or VLA we really don't call the combination of all signals as interferometry, because it's done in a more complicated way to recover images, rather than a single output like a stellar interferometer.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-05-26T00:44:41.747" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32049" PostTypeId="1" CreationDate="2019-05-26T04:06:38.203" Score="3" ViewCount="2294" Body="&lt;p&gt;I bought my first telescope, a 3-inch refractor a few weeks ago. It has been a wonderful experience seeing the Moon, Jupiter, Saturn magnified and Andromeda, M33, M81 and much more for the first time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I wish to get my second telescope. I was considering : &#xA;Explore Scientific ED APO 102mm F/7&#xA;Bresser Maksutov Cassegrain 152mm F/12.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I live in a high altitude region with Bortle 3 skies. I hope to see basic details on Jupiter, Saturn and Mars at high contrast and clarity and perhaps the disc of Neptune. Also occasionally deep sky objects. Portability and fast setup time are also a consideration. The MC is about 60% of the price of the APO. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to my 26 &amp;amp; 12 mm Plossl eyepieces I plan on getting a 50mm Erfle (Field Stop 40mm) and a 6mm Plossl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For mount, I was contemplating between Bresser Exos 2 and iOptron CEM25P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish to know your experience with refractors and MCs, in general, or with these specific models. Since the MC is cheaper and has a bigger aperture, will the APO have any advantages eg.contrast? Also, what are your thoughts on the mounts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your answer!&lt;/p&gt;&#xA;" OwnerUserId="25234" LastEditorUserId="25994" LastEditDate="2019-12-09T20:05:32.733" LastActivityDate="2019-12-09T20:05:32.733" Title="Second Telescope: Maksutov Cassegrain vs Refractor" Tags="&lt;observational-astronomy&gt;&lt;telescope&gt;&lt;amateur-observing&gt;&lt;refractor-telescope&gt;&lt;cassegrain-telescope&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32050" PostTypeId="1" AcceptedAnswerId="32053" CreationDate="2019-05-26T13:53:59.010" Score="5" ViewCount="418" Body="&lt;p&gt;There's a strikingly visible boulder sitting right at the top of the peak complex in the center of Tycho Crater on the Moon.&lt;/p&gt;&#xA;&lt;p&gt;Is it just a coincidence that this gigantic, bright rock is sitting dead center of the peak complex or is it a known piece of the original impactor that caused the wider crater? Does this remarkable rock have a name?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:LRO_Tycho_Central_Peak_0.25.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;On 10 June 2011 the LRO spacecraft slewed 65° to the west, allowing the LROC NACs to capture this dramatic sunrise view of Tycho crater. A very popular target with amateur astronomers, Tycho is located at 43.37°S, 348.68°E, and is ~82 km (51 miles) in diameter. The summit of the central peak is 2 km (6562 ft) above the crater floor, and the crater floor is about 4700 m (15,420 ft) below the rim. Many &amp;quot;clasts&amp;quot; ranging in size from 10 meters to 100s of meters are exposed in the central peak slopes. Were these distinctive outcrops formed as a result of crushing and deformation of the target rock as the peak grew? Or do they represent preexisting rock layers that were brought intact to the surface? Imagine future geologists carefully making their way across these steep slopes, sampling a diversity of rocks brought up from depth. Tycho's features are so steep and sharp because the crater is young by lunar standards, only about 110 million years old. Over time, micrometeorites, and not so micro meteorites, will grind and erode these steep slopes into smooth mountains. For a preview of what Tycho's central peak may look like in a few billion years, visit Bhabha crater.&lt;/p&gt;&#xA;&lt;p&gt;This is a version with about one fourth the linear pixel density of the original NASA full size image.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Original sources &lt;a href=&quot;http://lroc.sese.asu.edu/news/?archives/411-Tycho-Central-Peak-Spectacular!.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://wms.lroc.asu.edu/lroc_browse/view/M162350671&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;  Credit: NASA / GSFC / Arizona State University&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AECkH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AECkH.jpg&quot; alt=&quot;Tycho crater central peak&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13928" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2020-08-22T08:04:40.520" Title="Is the boulder on the peak of Tycho Crater the core of the impactor, or is it a random rock?" Tags="&lt;the-moon&gt;&lt;crater&gt;&lt;planetary-science&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32051" PostTypeId="2" ParentId="32032" CreationDate="2019-05-26T13:59:22.317" Score="3" Body="&lt;p&gt;The question asks about &quot;targets&quot; but doesn't specify black holes. However the other &quot;targets&quot; are related to infall of black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, I think the answer is a solid &quot;yes&quot; either way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Data is recorded separately at each site. Hours and hours of digitization of one or more 2 GHz wide data channels around 230 GHz, down converted to baseband and synched using a local atomic clock. See &lt;a href=&quot;https://astronomy.stackexchange.com/a/30443/7982&quot;&gt;this answer&lt;/a&gt; and &lt;a href=&quot;https://astronomy.stackexchange.com/a/20086/7982&quot;&gt;this answer&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Any signal reaching the feed horn of each radio telescope is going to be recorded in its data stream.&lt;/strong&gt; So if there is another source nearby that is covered by the radiation pattern of all of the dishes in the EHT collection, then it certainly could be imaged to some (indeterminate) level of resolution, &lt;strong&gt;with (at least) the following caveats:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You have to run the numerical correlations again, with a new range of time-offsets for each telescope. If the source is much brighter, you wouldn't have to use the whole dataset. But if it's just as dim or dimmer, you will have to correlate lots of data.&lt;/li&gt;&#xA;&lt;li&gt;If the source is far away from the imaged event horizon, there may be &lt;em&gt;several different types of calibrations&lt;/em&gt; for each dish and site that have been optimized, and may not work as well for a substantially removed object. The existing data may result in slightly lower resolution far from the target, depending on how calibrations were done. In addition to signals of the black hole, signals from other sources were recorded to assist in calibration of the whole array.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I need to read further on that, I'll update the part about calibration tomorrow.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-05-26T14:07:45.370" LastActivityDate="2019-05-26T14:07:45.370" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32053" PostTypeId="2" ParentId="32050" CreationDate="2019-05-26T17:07:08.833" Score="7" Body="&lt;p&gt;It is just a rock.&lt;/p&gt;&#xA;&lt;p&gt;A complex crater like Tycho is formed in several stages as the rock behaves like a fluid. The initial impact completely destroys the impactor and excavates a large cavity in the moon, and ejecta is shot out to all sides.&lt;/p&gt;&#xA;&lt;p&gt;Powerful shock waves push the rock to the side building large walls. As these push down on the edges of the crater, the centre is pushed up forming a central peak. Further faulting and cracking will continue to modify the crater and the peak.&lt;/p&gt;&#xA;&lt;p&gt;It is possible that this is part of the breccia that falls back to the surface having been shot upwards in the initial impact, but that seems very unlikely. It is more probable that this is part of the ridges of the central peak that cracked and fell into the shallow valley on the top of the central peak, which formed during the faulting as the peak was pushed up from the crater floor. In all likelyhood, this is a piece of moon rock. As it hasn't been exposed to solar radiation for billions of years, it is comparatively light. It can't be a piece of the impactor, as that would have been completely destroyed&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2020-08-22T08:04:40.520" LastActivityDate="2020-08-22T08:04:40.520" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32054" PostTypeId="1" AcceptedAnswerId="32055" CreationDate="2019-05-26T18:04:32.487" Score="19" ViewCount="3450" Body="&lt;p&gt;I saw this picture on the &lt;a href=&quot;https://www.bbc.com/news/world-us-canada-48407023&quot; rel=&quot;noreferrer&quot;&gt;BBC news website&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UZEeR.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UZEeR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There appears to be a ring of craters in the center of the picture, aligned in an approximate sub vertical orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I just seeing things or does this apparent ring of craters a feature and does it extend to the far side of the moon? If it does exist, what form of bombardment or gravitational attraction would have create such a ring?&lt;/p&gt;&#xA;" OwnerUserId="28167" LastActivityDate="2019-05-27T22:06:11.083" Title="Apparent Ring of Craters on the Moon" Tags="&lt;the-moon&gt;&lt;crater&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32055" PostTypeId="2" ParentId="32054" CreationDate="2019-05-26T18:32:12.623" Score="44" Body="&lt;p&gt;Actually, most of the surface of the Moon is covered with craters like that. The exception are the large dark 'seas', which are in fact &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_mare&quot; rel=&quot;noreferrer&quot;&gt;basaltic plains&lt;/a&gt;. The seas are mostly present on this side of the moon; the far side of the moon is &lt;a href=&quot;https://en.wikipedia.org/wiki/Moon#/media/File:Moon_Farside_LRO.jpg&quot; rel=&quot;noreferrer&quot;&gt;almost entirely composed of craters&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason you see a ring near the &lt;a href=&quot;https://en.wikipedia.org/wiki/Terminator_(solar)#Lunar_terminator&quot; rel=&quot;noreferrer&quot;&gt;terminator&lt;/a&gt; (the boundary between the illuminated and dark part) is that the shadows make the craters much more visible.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Due to the angle at which sunlight strikes this portion of the Moon, shadows cast by craters and other geological features are elongated, thereby making such features more apparent to the observer. This phenomenon is similar to the lengthening of shadows on Earth when the Sun is low in the sky.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7472" LastActivityDate="2019-05-26T18:32:12.623" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32056" PostTypeId="2" ParentId="32049" CreationDate="2019-05-26T19:01:13.800" Score="1" Body="&lt;p&gt;I have viewed through many telescopes over the years and in doing so have met many people like yourself. To suggest a F/12.5 Maksutov will satisfy you would be the equivalent of telling someone to dive into the deep end of a swimming pool without knowing if that person can swim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's bang for buck that motivated your question I think you would get the most if you just explore the capabilities of the 3 inch refractor for a while. There is a lot to be seen with that telescope and you have barely scratched the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some prefer electronic eyepieces. Have you seen what they are capable of doing? I viewed with eyepieces for fifty years before trying a Mallincam ten years ago. I haven't used an eyepieces for serious viewing since. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at the NightSkiesNetwork. It is a group that regularly broadcasts their observing sessions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.nightskiesnetwork.ca/about&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.nightskiesnetwork.ca/about&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot think of any better advice for someone like yourself than to stick your toe in the water for a few seconds before jumping in.&lt;/p&gt;&#xA;" OwnerUserId="24928" LastActivityDate="2019-05-26T19:01:13.800" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32059" PostTypeId="1" AcceptedAnswerId="32061" CreationDate="2019-05-27T00:27:18.087" Score="61" ViewCount="17178" Body="&lt;p&gt;I understand this is a silly hypothetical but I'm asking for a 7 year old so please bear with me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine an interstellar stray gas giant comes flying through our solar system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were not concerned that it would also steal our atmosphere and create tidal forces that destroyed everything... How close would it need to come to us to exert enough gravity to lift people off the ground and pull them into its own orbit?&lt;/p&gt;&#xA;" OwnerUserId="5405" LastEditorUserId="-1" LastEditDate="2019-05-28T10:08:04.947" LastActivityDate="2019-05-30T06:11:28.157" Title="If a massive object like Jupiter flew past the Earth how close would it need to come to pull people off of the surface?" Tags="&lt;gravity&gt;" AnswerCount="3" CommentCount="14" FavoriteCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32060" PostTypeId="2" ParentId="32028" CreationDate="2019-05-27T00:58:42.350" Score="4" Body="&lt;p&gt;One other supernovae-like but not a supernovae is a tidal disruption event. If a star passes close enough to a black hole it can be fully disrupted into a stream of gas. As the material passes its closest approach to the black hole it can be compressed and ignited. The resulting explosion will be very bright, like a supernovae, but looks different to a normal supernovae.&lt;/p&gt;&#xA;" OwnerUserId="16852" LastEditorUserId="16852" LastEditDate="2019-08-03T09:26:44.623" LastActivityDate="2019-08-03T09:26:44.623" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32061" PostTypeId="2" ParentId="32059" CreationDate="2019-05-27T01:06:21.783" Score="87" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;TL:DR Jupiter isn't dense enough for its gravity gradient over Earth's radius to produce a 1g tidal acceleration, even right at Jupiter's surface.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/32059/if-a-massive-object-like-jupiter-flew-past-the-earth-how-close-would-it-need-to/32061?noredirect=1#comment58605_32061&quot;&gt;thanks to PeterCordes&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Jupiter's gravity will pull on the Earth itself, as well as everything on it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not like a vacuum cleaner that selectively lifts small and light objects, the gravitational force will scale with the mass of each object; if the Earth is a zillion times more massive than we are, then Jupiter's gravitational force will also be about a zillion times larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What that means is that Earth will accelerate towards Jupiter, and we will accelerate along with it, and so we won't &quot;feel the tug&quot; anywhere near as strongly as one might suspect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, let's think about the size of the Earth, and the fact that people on the near side will be closer to Jupiter than the center of mass of the Earth, and people on the far side will be farther away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since people nearer to Jupiter will feel a &lt;em&gt;slightly stronger&lt;/em&gt; acceleration than the center of mass of the Earth, they will feel a quite gentle tug. We'll calculate that in a minute. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But believe it or not, people on the far side of the Earth, feeling less of a tug than the Earth's center of mass, will believe they are being pulled in the opposite direction! They won't really be pulled away from Jupiter, but they will not accelerate towards Jupiter as fast as the Earth, and so it will feel like they are being repelled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This kind of force is called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Tidal_force&quot; rel=&quot;noreferrer&quot;&gt;tidal force&lt;/a&gt; and this is the picture that's often used with the concept:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hr0OE.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Field_tidal.svg&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt; Replace &quot;Satellite&quot; with &quot;Jupiter&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The acceleration we feel due to gravity is expressed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$a_G = \frac{GM}{r^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; is the gravitational constant and equal to about &lt;span class=&quot;math-container&quot;&gt;$6.674 \times 10^{-11}$&lt;/span&gt; m^3/kg s^2 and M is each mass that's pulling on you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you put in 6378137 meters and the mass of the Earth (&lt;span class=&quot;math-container&quot;&gt;$5.972 \times 10^{+24}$&lt;/span&gt; kg) you get the familiar 9.8 m/s^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Jupiter were 114,000,000 meters or 114,000 kilometers away, the Earth would accelerate at 1 g towards it, but people on the close and far side would accelerate very differently. On the close side, being 6,378 kilometers closer, would feel an acceleration 1.2 m/s^2 greater, so they would feel that they weighed 12% less. And people on the far side would also feel about the same amount lighter because they felt less acceleration than the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Jupiter were so close that it were practically touching the Earth, it still wouldn't pull is off of Earth, assuming that Earth remained intact. &lt;strong&gt;But that wouldn't last very long!!!&lt;/strong&gt; Earth would be accelerating towards Jupiter at about 20.9 m/s^2, and people on the near side would feel acceleration of 24.8 towards Jupiter, but relative to Earth that's only 3.9 m/s^2, so not enough to overcome Earth's gravity of -9.8 m/s^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the far side of Earth it's similar; the acceleration towards Jupiter would be 17.8 m/s^2 but minus Earth's acceleration of - 20.9 it's -3.0 m/s^2 away, but that's also not enough to overcome the attraction to Earth of in this case +9.8 m/s^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;When Earth touches Jupiter, we will feel about 40% lighter on the near side and 31% lighter on the far side of Earth, but we would not leave the surface.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in just minutes we'd be pulled so deep into Jupiter that we would be crushed by Jupiter's internal atmospheric pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would certainly be fun, but it wouldn't last long!&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-05-30T06:11:28.157" LastActivityDate="2019-05-30T06:11:28.157" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="32062" PostTypeId="2" ParentId="32028" CreationDate="2019-05-27T03:25:53.097" Score="0" Body="&lt;p&gt;I'll try to explain as easiest as possible. It's okay to ask this type of question. It's natural! Okay, when the small star(smaller than our sun) dies, it just becomes a white dwarf. Not even supernova. When the big star dies (bigger than our sun), a supernova occurs. Your question, black hole, is the process AFTER the supernova. If a star is pretty big but not quite that big, it explodes and becomes a neutron star. If a star is GIGANTIC, it becomes a black hole after a supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll answer your second question. First, you have to know why a large star becomes a black hole. Do you know why stars can make light itself, like our sun? Because a nuclear fusion is occurring inside the star. If you don't know what nuclear fusion is, you might want to search in youtube (Hydrogen bomb is nuclear fusion bomb). This power is a force that is trying to make stars bigger or pushing forces outward. While the gravity of a star is pushing inward. The shape and size of the stars are stable because these two forces are equal. Stars &quot;die&quot; mean they are out of their fuel to make a nuclear fusion. If the fusion stops, the gravity will overwhelm, which will keep pushes the star inward. If the star is big, the supernova occurs after the shrinkage. Just imagine, if a massive star became infinitely small, its gravity will become infinitely larger. I hope I answered your second question because this process cannot occur in reverse. &lt;/p&gt;&#xA;" OwnerUserId="24317" LastActivityDate="2019-05-27T03:25:53.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32063" PostTypeId="1" CreationDate="2019-05-27T03:31:57.550" Score="-1" ViewCount="89" Body="&lt;p&gt;How does a rocket propel its fuel in a space, where no air exists? Newton's Third Law cannot apply to this because, in Earth, rockets can push the air to propel, but in space, there is nothing to push.&lt;/p&gt;&#xA;" OwnerUserId="24317" LastActivityDate="2019-05-27T04:52:14.637" Title="How does a rocket propel in a space?" Tags="&lt;space-probe&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2019-05-27T06:59:42.730" ContentLicense="CC BY-SA 4.0" />
  <row Id="32064" PostTypeId="2" ParentId="32063" CreationDate="2019-05-27T04:44:48.697" Score="5" Body="&lt;p&gt;This is not a question for this stack. It's purely a physics question. It may get closed for that reason, but I don't want you to leave here empty-handed, so I am going to stick a reply to it anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is actually one of the oldest misconceptions about rockets, the belief that they are &quot;pushing against air&quot;. In reality, that is not what happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sit in a boat, on a lake. Have someone load a bunch of boulders onto the boat - not too many, so you don't sink. Paddle away to the middle of the lake. Stop the boat. Now start throwing boulders overboard, vigorously pushing them one by one in the same direction. What happens then? The boat will start moving, in the direction opposite to the boulders.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same effect could be achieved by throwing anything overboard, either something heavy like boulders, or something you could throw very fast, like bullets. There are videos on Youtube showing how you could &quot;paddle&quot; a boat by just shooting a gun in the same direction repeatedly - the boat will start moving in the opposite direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=U_zaa_uqBSE&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=U_zaa_uqBSE&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how rockets work. They are pushing, alright, you got that part straight. But they are not pushing against air - or against anything around them. They are pushing against the fuel. By throwing the fuel out at high velocity (like  boulders or bullets from a boat), the rocket gets a kick in the opposite direction. That's what Newtons's third law says.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When one body exerts a force on a second body, the second body&#xA;  simultaneously exerts a force equal in magnitude and opposite in&#xA;  direction on the first body.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;One body&quot; here is the rocket. The &quot;second body&quot; here is the fuel; it's not the air. Fuel is also a body. Really, any body would do, just throw it fast enough. It doesn't have to be rocket fuel. It could be fluffy bunnies for all we care.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rocket exerts a force on the burning fuel by throwing it out via the nozzle. Have you noticed how fast the fuel comes out of there? Yeah, there's a lot of force pushing it out of the nozzle, otherwise it would not come out like that. But then what does Newton's 3rd law say? Since the rocket is exerting force on the fuel, then the fuel, like for like, exerts force on the rocket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's how the rocket gets pushed up. It pushes the fuel really hard, down. So the fuel pushes the rocket up. One body pushes the other, and viceversa. No air needed at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps. Physics is fun!&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2019-05-27T04:52:14.637" LastActivityDate="2019-05-27T04:52:14.637" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32065" PostTypeId="1" CreationDate="2019-05-27T04:51:14.763" Score="0" ViewCount="55" Body="&lt;p&gt;My fiancé and I took our youngest sons out to sky watch at the river near by. The older of the 2 was looking up and said - &amp;quot;look it's Santa's reindeers!&amp;quot; All of us looked up and saw a literal straight line of what looked like hundreds of stars in a straight line, like in a formation, moving due north. They didn't come from the south, it's more like they just appeared and disappeared. Does anyone have any idea what we may have seen?&lt;/p&gt;&#xA;" OwnerUserId="28171" LastEditorUserId="35106" LastEditDate="2020-10-21T14:05:28.813" LastActivityDate="2020-10-21T14:05:28.813" Title="Strange lights moving North" Tags="&lt;identify-this-object&gt;" AnswerCount="0" CommentCount="5" ClosedDate="2019-05-27T07:06:13.130" ContentLicense="CC BY-SA 4.0" />
  <row Id="32066" PostTypeId="1" CreationDate="2019-05-27T07:36:24.583" Score="1" ViewCount="54" Body="&lt;p&gt;About 20 minutes ago, we saw four lights lined up and moving relatively quickly across the sky. These lights looked exactly like the stars around them - not red or blue or flashing or anything, just four stars at first glance. The spacing between them remained constant as they traveled in a straight line together (moving from West to East). The two in the middle were slightly closer to one another than the two in the front and back of the line, but the spacing was otherwise equal and constant - as if they'd been strung together on a line that was begging pulled forward. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They moved as quickly as a plane might when making a final approach to land, but there were no other lights visible around them and no sound. After watching for a minute, I called a friend out to reassure me I wasn't just seeing things, but they saw the same and described them as resembling stars more than anything else as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do satalites travel in groups? Would they form a consistent line like this? That's the next closest thing I can think of, but I've never seen satalites quite that bright/large or traveling in locked step with others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could these have been military planes flying high enough to only appear as one whitish, twinkling light but still as fast as a lower flying plane? There is a naval Base not to far from us; that's the only other thing I can think of. &lt;/p&gt;&#xA;" OwnerUserId="28173" LastActivityDate="2019-05-27T07:36:24.583" Title="Four Lights (star like) Moving in a Line Together" Tags="&lt;artificial-satellite&gt;&lt;identify-this-object&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2019-05-27T09:55:40.703" ContentLicense="CC BY-SA 4.0" />
  <row Id="32067" PostTypeId="1" CreationDate="2019-05-27T08:48:01.887" Score="1" ViewCount="277" Body="&lt;p&gt;I'm trying to make it more general code so I can trace even a parabola or hyperbola for &lt;span class=&quot;math-container&quot;&gt;$e=1, E=0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$e&amp;gt;1, E&amp;gt;0$&lt;/span&gt; respectively. And after achieving that general code I want to make the same code working for perihelion precession motion of Mercury just by adding an additional term (&lt;span class=&quot;math-container&quot;&gt;$-A/r^3$&lt;/span&gt;) in the energy term. &lt;em&gt;Feel free to change my entire code and energy term.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from numpy import *&#xA;import matplotlib.pyplot as plt&#xA;&#xA;&#xA;L = 9.11*10**38     #L = angular momentum&#xA;m = 3.28*10**23     #m = mass of mercury&#xA;M = 1.99*10**30     #M = mass of sun&#xA;a = 5.8*10**10       #a = semi-major axis&#xA;G = 6.674*10**-11   #G = Gravitationl constant&#xA;k = G*M*m&#xA;E = -k/(2*a)        #E = energy&#xA;##E = 0&#xA;p = L**2/(m*k)   &#xA;c = 1 + (2*E*L**2)/(m*k**2)&#xA;e = sqrt(c)       #e = eccentricity&#xA;print(e,c)&#xA;def fx(x):&#xA;    r = p/(1 + e*cos(x))&#xA;    return float(r)&#xA;&#xA;n = 1000&#xA;phi =linspace(0,2*pi,n)&#xA;radius = zeros([n])&#xA;theta = zeros([n])&#xA;x = zeros([n])&#xA;y = zeros([n])&#xA;&#xA;for i in range(0,n):&#xA;    radius[i] = fx(phi[i])&#xA;    theta[i] = 180*phi[i]/pi&#xA;&#xA;for i in range(0,n):&#xA;    x[i] = radius[i]*cos(phi[i])&#xA;&#xA;for i in range(0,n):&#xA;    y[i] = radius[i]*sin(phi[i])&#xA;&#xA;print('r =',radius)&#xA;print('x =',x)&#xA;print('y =',y)&#xA;plt.plot(x,y)&#xA;plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="28172" LastEditorUserId="7472" LastEditDate="2019-05-27T09:27:13.177" LastActivityDate="2019-05-27T09:27:13.177" Title="To show perihelion precession motion of Mercury in Python with matplotlib" Tags="&lt;orbit&gt;&lt;planet&gt;&lt;astrophysics&gt;&lt;python&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32068" PostTypeId="1" CreationDate="2019-05-27T08:59:37.953" Score="2" ViewCount="219" Body="&lt;p&gt;If a satellite loses energy it starts to circle lower and finally hits the atmosphere. The moon is moving like an inch a year away from the earth even though it is losing energy pulling the oceans around.&lt;/p&gt;&#xA;" OwnerUserId="28175" LastEditorUserId="2618" LastEditDate="2019-05-27T12:24:25.417" LastActivityDate="2019-05-27T12:24:25.417" Title="If the moon is losing energy dragging the oceans around, why is it moving away from earth instead of falling like any other satelite would?" Tags="&lt;the-moon&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2019-05-27T12:28:37.023" ContentLicense="CC BY-SA 4.0" />
  <row Id="32069" PostTypeId="2" ParentId="32059" CreationDate="2019-05-27T09:16:05.037" Score="18" Body="&lt;p&gt;About 70,000 km. If the Earth orbited Jupiter (or flew by) at a closer distance, not just we would leave the surface, but the whole Earth would disintegrate since all its mass will be leaving too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;70,000 km is Jupiter's &lt;a href=&quot;https://en.wikipedia.org/wiki/Roche_limit&quot; rel=&quot;noreferrer&quot;&gt;Roche limit&lt;/a&gt; (although its actual value varies a lot depending on the other involved body), the radius where tidal forces (already explained in uhoh's answer) overwhelm gravitational forces and any orbiting body can't hold itself by its own weight. In that context, people on the surface don't behave differently than rocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Btw., this scenario is also explored &lt;a href=&quot;https://www.youtube.com/watch?v=ym7bDBg9jQ4&quot; rel=&quot;noreferrer&quot;&gt;in a youtube video&lt;/a&gt;. I wouldn't say it's very good but it may be helpful to explain the Roche limit to a 7 year old.&lt;/p&gt;&#xA;" OwnerUserId="14097" LastEditorUserId="14097" LastEditDate="2019-05-27T12:53:06.653" LastActivityDate="2019-05-27T12:53:06.653" CommentCount="16" ContentLicense="CC BY-SA 4.0" />
  <row Id="32070" PostTypeId="2" ParentId="32068" CreationDate="2019-05-27T09:18:39.567" Score="5" Body="&lt;p&gt;No. The energy budget of the Moon is actually increasing. The Earth's spin is slowing down, and that energy gets transferred to the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/420/why-is-the-moon-receding-from-the-earth-due-to-tides-is-this-typical-for-other&quot;&gt;Why is the Moon receding from the Earth due to tides? Is this typical for other moons?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatever else may be happening (like the losses you mention), this transfer from Earth's spin to Moon's orbital energy is the main process responsible for raising the Moon's orbit.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2019-05-27T09:18:39.567" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32072" PostTypeId="1" CreationDate="2019-05-27T17:19:33.803" Score="1" ViewCount="71" Body="&lt;p&gt;According to one theory, the region between Mars and Jupiter was once populated by several primordial planets with a combined mass of roughly the earth.  These planets provided the material that made up the original asteroid belt.  Since then, 99.9% of this material was lost.  The current composition of the asteroid belt is about 10% metallic and 90% silicate/carbonaceous.  Could we assume that the original asteroid belt had the same composition?  Or, at least a very similar composition?  &lt;/p&gt;&#xA;" OwnerUserId="27844" LastActivityDate="2019-05-30T20:41:57.000" Title="What was the composition of the primordial asteroid belt?" Tags="&lt;asteroid-belt&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32074" PostTypeId="1" CreationDate="2019-05-27T22:15:34.070" Score="1" ViewCount="91" Body="&lt;p&gt;What appears to be a string of satellites moving nearly overhead towards NW at 10:30pm to 10:50pm Near Eagle River WI on Sun May 26, 2019.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was lead by 3 equally spaced (approx 1/2 in total length of three at arms length), appro magnitude 4-3.  A couple of min later approx 10 other objects followed in the same line but much dimmer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clear night. No Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was this phenomenon? &lt;/p&gt;&#xA;" OwnerUserId="28185" LastActivityDate="2019-05-27T22:15:34.070" Title="String of satellites" Tags="&lt;identify-this-object&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2019-05-28T00:56:58.390" ContentLicense="CC BY-SA 4.0" />
  <row Id="32075" PostTypeId="1" CreationDate="2019-05-28T10:14:33.187" Score="2" ViewCount="101" Body="&lt;p&gt;If we consider orbital decay of the Earth's orbit, then can we counter argue with the following reasons?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Due to tidal effects transferring small amounts of angular momentum from the Sun's spin to Earth's orbit, can we conclude that Earth's orbit is actually moving slightly further away from the Sun?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can we say that the orbit is increasing in size because the sun loses millions of tons of mass each day, mostly to the ejection of particles called solar winds?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="26758" LastEditorUserId="26758" LastEditDate="2019-05-28T10:20:55.090" LastActivityDate="2019-05-28T10:20:55.090" Title="Orbital decay of the Earth's orbit and Earth's moving closer to the Sun" Tags="&lt;the-sun&gt;&lt;orbit&gt;&lt;earth&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32078" PostTypeId="2" ParentId="79" CreationDate="2019-05-28T19:26:02.177" Score="1" Body="&lt;p&gt;The hottest stars that are still fusing in their cores are Wolf-Rayet stars that are on the extreme end of the WC sequence, appropriately classified as WO stars, which display prominent Oxygen emission lines. The hottest known star is WR 102, which has a spectral type of WO2 and a surface temperature of 210,000 Kelvin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WR 102 is thought to have a mass of ~16.7 solar masses. Since this is a highly evolved Wolf-Rayet star, the majority of this mass is composed of the fusing core with a very thin radiative layer surrounding it. For reference, the threshold for being an O-type star is about 16 solar masses, with just a fraction of that mass the fusing core. This means WR 102 probably started with around 50-60 solar masses at ZAMS. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point it is unknown what exactly produces a WO star, whether it is an evolutionary stage following being a WC star or if it takes an extraordinary massive star that goes straight to WO after shifting through a WN stage. The number of WO stars known presently is in the single digits so there is still a lot to learn about these sort of stars. &lt;/p&gt;&#xA;" OwnerUserId="23212" LastEditorUserId="23212" LastEditDate="2019-05-28T19:47:17.963" LastActivityDate="2019-05-28T19:47:17.963" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32079" PostTypeId="2" ParentId="28721" CreationDate="2019-05-28T22:08:20.500" Score="1" Body="&lt;p&gt;TL;DR: by flapping wings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have the book at hand, but as far as I can remember Isaac Asimov argued in &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/039531545X&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Change! : 71 Glimpses of the Future&lt;/a&gt; that humans could fly in the Moon (assuming a closed pressurised venue) just by attaching wings to their arms and flapping - as the mythological &lt;a href=&quot;https://en.wikipedia.org/wiki/Icarus&quot; rel=&quot;nofollow noreferrer&quot;&gt;Icarus&lt;/a&gt; allegedly did, but benefiting from the much lower gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surface gravity in Titan is lower than in the Moon, and atmospheric pressure is larger than in Earth. Therefore, flying in Titan should be even easier than flying in the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, walking in low gravity is not practical in the Moon, and it will be even less practical against a tick atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, flapping wings with our arms would be a convenient way of moving around in Titan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a more moderate view, a combination of hoping and paddling with small hand-held wings (probably similar to holding light shields) could be more practical for most people. &lt;/p&gt;&#xA;" OwnerUserId="14097" LastEditorUserId="14097" LastEditDate="2019-05-30T08:58:29.270" LastActivityDate="2019-05-30T08:58:29.270" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32080" PostTypeId="2" ParentId="32059" CreationDate="2019-05-28T23:07:46.570" Score="15" Body="&lt;p&gt;As other answers point out, Jupiter is not quite massive enough to pull a planet of Earth's density apart. But we can use a slightly heavier object instead -- say, a small cold brown dwarf massing 13 Jupiters, or about 4000 Earths. According to the rigid-body Roche formula, its Roche limit is then &lt;span class=&quot;math-container&quot;&gt;$\sqrt[3]{2\cdot 4000}=20$&lt;/span&gt; earth radii, or 130,000 km. The radius of the brown dwarf is not much larger than Jupiter's (since they're both made of compressible gas), so smaller than 100,000 km, and there's some room for Earth to be destroyed without actually colliding with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our brown dwarf, wandering sedentarily around in the galaxy, spots our sun, and decides to take a closer look. It comes screaming through the inner solar system on a hyperbolic orbit, which means that it will be moving at somewhat more than solar escape velocity when it narrowly misses Earth -- call it 100 km/s. Switching perspective, we can say that &lt;em&gt;Earth&lt;/em&gt; comes at the brown dwarf at 100 km/s and just about misses. This speed allows us to spend a almost half an hour inside the Roche limit, if we almost touch the brown dwarf at the time of closest approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that sounds like it will be unnecessarily dramatic, so it's instead have the brown dwarf give is a slightly wider berth such that at the time of closest approach, the acceleration of gravity at ground zero will be a more pedestrian &lt;span class=&quot;math-container&quot;&gt;$-0.1\;\rm m/s^2$&lt;/span&gt;. That will be the case when our distance to the brown dwarf is &lt;span class=&quot;math-container&quot;&gt;$\sqrt[3]{\frac{9.82}{0.1+9.82}} = 0.997$&lt;/span&gt; of the Roche limit, or 128,000 km. The length of our path through the Roche zone is then about 20,000 km, which means the encounter takes 200 seconds. Call it three minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The sharp-eyed reader will notice that these numbers mean that the far side of the earth is actually never inside the 130,000-km limit, but what really counts is the first derivative of the brown dwarf's gravitational field, so if you're standing on the antipodal point you'll still have the Earth's center-of-mass pulled away under you even if you yourself is outside the limit. The numbers are all approximate anyway).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(On the other hand, a few minutes clearly not enough time for the molten inside of the Earth to flow into a hydrostatic equilibrium in the new situation, so using the rigid-body formula is appropriate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What happens then?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, of course, it is an &lt;strong&gt;awesome sight&lt;/strong&gt;. The brown dwarf dominates the sky with an angular diameter of somewhere between 60° and 100°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, it may be getting &lt;strong&gt;uncomfortably hot&lt;/strong&gt;. Not necessarily &quot;the mountains are melting&quot; hot or even &quot;the seas boil away&quot; hot. But &lt;a href=&quot;https://www.space.com/22659-brown-dwarf-stars-hot-as-oven.html&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; says that the coldest brown dwarves have about the temperature of a baking oven, and having a significant part of the sky at 150 °C can make anyone sweat. No worries, though -- it will all be over in a few hours, so just go inside and crank up the AC; that'll deal with it fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right at ground zero &lt;strong&gt;gravity decreases smoothly&lt;/strong&gt; while we approach Roche. When it passes zero G you're in free fall and start floating gently upwards. Except that everything around you -- cars, houses, trees, the soil itself -- is also in free fall since the only thing that kept them down was gravity. So to a first approximation your &lt;em&gt;local&lt;/em&gt; experience is not about bring ripped off Earth, but just of weightlessness. (&lt;em&gt;Or is it?&lt;/em&gt; See below.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ditto at the antipodal point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One problem that shows up here is that &lt;strong&gt;the atmosphere is escaping&lt;/strong&gt; into space. Since there is no gravity to keep it down, it escapes rather faster than the gentle floating of cars, trees, and people, propelled by its own pressure. Even before we reach Roche, the air may have become too thin to breathe. On the other hand fresh air will rush in from the surrounding areas to fill the void, creating the great-great-grandmother of all hurricanes. (And a great-great-grandfather around the antipode, of course).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a great circle 90° from ground zero, gravity increases to about 1.7 G. You feel heavy. Ho hum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Between&lt;/em&gt; these areas dramatic things happen. At about 45° from ground zero (or the antipode) the tidal force is at right angles to vertical, so the &lt;em&gt;strength&lt;/em&gt; of gravity is about what we're used to -- but its direction is different. It's as if &lt;strong&gt;the world is tilted by tens of degrees&lt;/strong&gt;, rather like how bad sci-fi movies pretend &quot;entering a gravitational field&quot; works. Tall buildings tip over; many not-so-tall ones just collapse. Lakes and seas do things that make the word &quot;tsunami&quot; pack up and go home, hopelessly outclassed. What the water doesn't get, unstoppable rockslides will. And don't forget the hypercane-force gales as the &lt;em&gt;atmosphere&lt;/em&gt; slides &quot;downwards&quot; almost unimpeded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This assumes that the ground below is rigid, of course. It isn't quite, though it probably has enough structural integrity that the preceding paragraph is still true. In any case, the entire crust of Earth starts sliding &quot;down&quot; towards ground zero (or, as always, the antipode). Different parts of the crust slide at different velocities, though. Near the &quot;ho hum&quot; zone the crust is stretched; at ground zero or antipode, crust piles up. Nothing actually has time to move more than (very roughly) some tens of kilometers from its starting position at best, but that is quite sufficient to get &lt;strong&gt;cataclysmic hyper-earthquakes&lt;/strong&gt; at every tectonically active zone on earth. Where there is no active zone to take up the stress, new ones open up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not entirely sure what the mantle is doing, but it probably isn't something nice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing the mantle &lt;em&gt;is&lt;/em&gt; doing happens around ground zero. Without any net gravity to keep the crust down, hydrostatic pressure in the lower lithosphere drops towards zero. Dissolved &lt;strong&gt;volatiles in magmas everywhere&lt;/strong&gt; attempt to outgas, forming bubbles and expanding the magma until the sheer inertia of the overlying rocks resists it. The effect is to push the crust upwards &lt;em&gt;faster&lt;/em&gt; than it is being pulled by the tides. So standing at ground zero you may not get to experience weightlessness after all. Instead you get to stand right on top of the greatest volcano eruption in the history of the planet. Truly the experience of a lifetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the three minutes are up and the brown dwarf recedes again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At ground zero you're now at least a kilometer higher than you started out, together with everything around you, and still moving upwards at tens of kilometers an hour. That's far less than escape velocity, so what goes up must come down again. Except &quot;down&quot; is now most likely a boiling volcanic inferno. Aren't you glad you didn't get roasted by the brown dwarf to start with?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's still time for the sliding tectonic plates to slide to a halt, and for the new rifts in the &quot;ho hum&quot; zone to start rivaling the ground zero volcano. Unless the interior of the earth deformed &lt;em&gt;elastically&lt;/em&gt; so everything now tries to slide &lt;em&gt;back&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;planet&lt;/em&gt; still exists, though. No mass was actually lost. On the other hand, the encounter did change our &lt;em&gt;collective&lt;/em&gt; velocity by several tens of kilometers per second, which is broadly comparable to our usual orbital motion. That's going to wreak total havoc on the seasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh well. It's not as if any of us would be around to complain about that.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(The sharp-eyed reader from before will note that most of these calamities would happen even without getting all the way to the Roche limit. So if the world has to end, Jupiter's gravity field &lt;em&gt;might&lt;/em&gt; be capable enough, after all.)&lt;/p&gt;&#xA;" OwnerUserId="5350" LastActivityDate="2019-05-28T23:07:46.570" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32081" PostTypeId="1" AcceptedAnswerId="40127" CreationDate="2019-05-29T00:32:06.663" Score="3" ViewCount="152" Body="&lt;p&gt;Is there any reason to think a Trans-Neptunian dwarf planet, the size of Pluto, could not exist in a Keplerian orbit 120AU from the Sun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If such an object existed, and came within 0.1AU of the heliopause, is there anything about the environment near the heliopause that would prevent a stable orbit.&lt;/p&gt;&#xA;" OwnerUserId="25088" LastEditorUserId="25088" LastEditDate="2019-06-02T15:15:54.270" LastActivityDate="2020-12-01T06:46:38.790" Title="Trans-Neptunian Dwarf Planet near Heliopause" Tags="&lt;dwarf-planets&gt;&lt;trans-neptunian-object&gt;" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="32083" PostTypeId="1" CreationDate="2019-05-29T01:50:44.833" Score="2" ViewCount="119" Body="&lt;p&gt;The new New York Times article &lt;a href=&quot;https://www.nytimes.com/2019/05/28/science/solar-eclipse-einstein-physics.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Eclipse That Made Einstein Famous&lt;/a&gt; describes several events and discussions on the hundred year anniversary of the the 1919 solar eclipse and verification of the deflection of light by the Sun as predicted by General Relativity. The roles of Albert Einstein, Arthur Eddington, and others are discussed in the article, the play performed at the Royal Astronomical Society in London, in the Nature Physics article &lt;a href=&quot;https://www.nature.com/articles/s41567-019-0519-3&quot; rel=&quot;nofollow noreferrer&quot;&gt;A hundred years of the first experimental test of general relativity&lt;/a&gt; and the book &lt;a href=&quot;https://press.princeton.edu/titles/17193.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;No Shadow of a Doubt; The 1919 Eclipse That Confirmed Einstein's Theory of Relativity&lt;/a&gt; by Daniel Kennefick.&lt;/p&gt;&#xA;&lt;p&gt;The instruments used for the measurement are shown in the illustration and photograph below.&lt;/p&gt;&#xA;&lt;p&gt;My question is about the astrometric technique used, and the calibration and verification.&lt;/p&gt;&#xA;&lt;p&gt;The instruments are long and laying on their sides, viewing the eclipse and the stars near the Sun by reflection in coelostats (mirrors in front of telescopes, moved by clockwork mechanisms).&lt;/p&gt;&#xA;&lt;p&gt;I can imagine several sources of instrumental and experimental errors due to variations in &lt;em&gt;field distortion&lt;/em&gt; and &lt;em&gt;magnification&lt;/em&gt;, presumably the measurement was done by comparing the eclipse images with similar images taken by the same instruments of the same field of stars in the same position in the field of view a few months later, at night, in the dark, in order to show that the tiny deflections were not due to variation from one instrument to another.&lt;/p&gt;&#xA;&lt;p&gt;Is that actually how it was done? Was the experimental set-up left in-place between day and night measurements, or was it packed up and moved?&lt;/p&gt;&#xA;&lt;p&gt;Was the flatness of the coelostat mirrors independently confirmed somehow?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Images from the new New York Times article &lt;a href=&quot;https://www.nytimes.com/2019/05/28/science/solar-eclipse-einstein-physics.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Eclipse That Made Einstein Famous&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tZbV7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tZbV7.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Einstein’s theory suggested that starlight would bend near a massive body like the sun.CreditChronicle/Alamy&lt;/p&gt;&#xA;&lt;p&gt;In Sobral, Brazil, two coelostats with movable mirrors were used to direct images of the eclipsed sun into a pair of horizontal telescopes.&#xA;Credit Royal Observatory, Greenwich/Science Museum Group&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/E8f92.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E8f92.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-05-30T05:50:23.490" Title="How was the astrometry of the famous 1919 solar eclipse confirmation of General Relativity calibrated and verified?" Tags="&lt;observational-astronomy&gt;&lt;telescope&gt;&lt;history&gt;&lt;astrometry&gt;&lt;instruments&gt;" AnswerCount="0" CommentCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="32085" PostTypeId="2" ParentId="9924" CreationDate="2019-05-29T10:03:49.177" Score="-2" Body="&lt;p&gt;Unsure where the original poster obtained an idea that objects inside the Earth’s orbit, as seen from Earth, “move &lt;strong&gt;ordinarily&lt;/strong&gt; with respect to the stars” from west to east.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, of which “west” and “east” do we speak? To be specific, let’s understand these for the region of sky not far from the meridian. That is, when the object &lt;em&gt;X&lt;/em&gt; culminates, then nearby objects having the right ascension greater than RA(&lt;em&gt;X&lt;/em&gt;) are on the east from the meridian (not yet culminated), whereas those having RA &amp;lt; RA(&lt;em&gt;X&lt;/em&gt;) are on the west from the meridian (already culminated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Sun itself, every educated person can remember that its diurnal motion is slightly &lt;em&gt;slower&lt;/em&gt; than diurnal motion of the inertial celestial sphere (fixed stars) – that is, with respect to those stars Sun goes from west to east. Its right ascension increases monotonically. For other objects inside the Earth’s orbit, their right ascension is within 90° of the Sun’s, hence &lt;em&gt;in average&lt;/em&gt; they move “from west to east” as well, and quicker do it when are behind the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But definitely the terrestrial picture is opposite for e.g. Venus when it overtakes Earth in their orbital motions near the inferior conjunction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/b/b7/Venusorbitsolarsystem.gif&quot; alt=&quot;Venus and Earth orbit the Sun&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During this part of the orbit the right ascension of Venus decreases – seen on this animation near the end of cycle. “Experts” asserting that it’s impossible to see mistake – Venus is not very close to Sun during it, and anyway perfectly observable (albeit rare) during Venus’s transits before the Sun.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastActivityDate="2019-05-29T10:03:49.177" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32086" PostTypeId="1" CreationDate="2019-05-29T11:37:29.060" Score="3" ViewCount="1189" Body="&lt;p&gt;Earth is part of Solar system, which is said to have originated from condensation and mutual attraction in dense clouds of gasses and dust. If Sun is basically more than ninty percent hydrogen and rest is helium. Nuclear fusion inside sun is said to pause near iron(56/26) and also several other planets like Jupiter and Saturn are gas giants. Where did heavy metals and other heavy elements on earth come from if second most nearest star(/star-system) is some four light years away (and mass would take long to come from those sources) ?&lt;/p&gt;&#xA;" OwnerUserId="28207" LastActivityDate="2020-02-13T00:59:01.050" Title="Origin of heavier elements on earth?" Tags="&lt;observational-astronomy&gt;&lt;solar-system&gt;&lt;earth&gt;" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32087" PostTypeId="1" CreationDate="2019-05-29T07:15:55.883" Score="5" ViewCount="388" Body="&lt;p&gt;This question is inspired by the one asked here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://space.stackexchange.com/questions/1911/is-there-any-stable-orbit-around-a-black-hole&quot;&gt;Is there any stable orbit around a black hole?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I note that (though I've heard of these before) there are, apparently, 3 important &lt;em&gt;fundamental&lt;/em&gt; &quot;zones of inescapability&quot; with regard to a black hole that can be identified on the basis of physical principles and which can be defined by the Schwarzschild &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; coordinate and its critical radius &lt;span class=&quot;math-container&quot;&gt;$r_S$&lt;/span&gt;: one of these is, of course, the event horizon, at &lt;span class=&quot;math-container&quot;&gt;$1\ r_S$&lt;/span&gt;, from which no escape &lt;em&gt;at all&lt;/em&gt; is possible. The second one is the &quot;photon sphere&quot;, here referred to as the zone at the &quot;innermost bound circular orbit&quot;, at &lt;span class=&quot;math-container&quot;&gt;$1.5\ r_S$&lt;/span&gt;, at which a photon (particle of light) traveling at right angles (i.e. constant &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt;-coordinate) can complete one circuit around the hole and within which only &lt;em&gt;impossibly&lt;/em&gt; powerful rockets could get you out. Orbits above this point but below the &quot;innermost &lt;em&gt;stable&lt;/em&gt; circular orbit&quot;, out to &lt;span class=&quot;math-container&quot;&gt;$3\ r_S$&lt;/span&gt;, are possible but require a continuous application of (likely ungodly large) engine propulsion. Above this, more-or-less normal orbits are possible, though they precess more and more rapidly as the general-relativistic effect that is incipient with the orbits of the planets becomes fully mature in these environs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the trick, of course, is the qualifiers regarding propulsion: &quot;impossibly powerful&quot;, &quot;ungodly large&quot; which basically are likely words for &quot;it can't be done with realistic technology&quot;. Which, then, of course, is what begs this question - what &lt;em&gt;can&lt;/em&gt; be done with such?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For my &quot;realistic&quot; scenario, a generous interpretation will be used which is &quot;a matter-antimatter fueled spacecraft with a wet mass 100x its dry mass&quot;. The goal is for the craft to execute one of two manoeuvres:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;lower itself into an orbit around the hole, orbit a few times, and then take off again (so it must have fuel for both trips) and return to where it launched from, and&lt;/li&gt;&#xA;&lt;li&gt;establish a &quot;slingshot&quot; trajectory around the hole that takes it as close as possible before shooting it back out into space (this may be easier?).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I consider this &quot;reasonable&quot; because any travel to a black hole will already be an extreme problem in interstellar travel, much less the kinds of targets that will already be in use for this kind of proposal which will almost surely have to be supermassive black holes to be interesting and thus requiring travel at least to the center of the Galaxy (~200 000 Pm) if not intergalactically (distances of many Zm), with tera- or petaseconds of travel time (thousands to millions of years) at least, or something like wormholes if they are possible. Hence I presume the technology to make such a craft will be readily available to any civilization that is capable of pulling this feat off to begin with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason we need supermassive black holes is tidal forces and ideally we want to get as close to the horizon as possible. Possible targets may be Sagittarius A* and M87* (the black hole that the picture was taken of), both supermassive black holes, the latter much more so still, at the centers of galaxies. However, M87* presents the additional challenge of an extremely active matter environment around it, which may foreclose that possibility even further, though even Sgr A* might be too much. Hence, failing that, we can just consider a hypothetical black hole of mass equal to either, that is suitably &quot;quiet&quot;, if nothing else is possible. (or perhaps imagine that with M87*, it has entered a quiescent period by the time the ships arrive as compared to the time when it sent the images we saw)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the closest approach possible, in both scenarios, while still having enough fuel to get away (or not being destroyed by other effects)? We should also assume that the craft starts out at great distance where gravity is hopefully unremarkable, but do not budget for more than 250 Ms shipboard total mission time (a bit less than 2893 days of 86.4 ks each), measured from time of release from mission departure station to time of return.&lt;/p&gt;&#xA;" OwnerUserId="9807" OwnerDisplayName="The_Sympathizer" LastEditorUserId="31410" LastEditDate="2021-05-10T16:34:43.923" LastActivityDate="2021-05-10T16:34:43.923" Title="What is the closest one could get to a black hole and still stand a realistic chance of escape?" Tags="&lt;black-hole&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32088" PostTypeId="2" ParentId="32086" CreationDate="2019-05-29T14:40:33.070" Score="4" Body="&lt;p&gt;A very quick answer -- someone else may have a better and more referenced one. The heat of the newly formed sun drove the light elements away from the inner system (mostly). The rocky planets formed from what was left, the relatively small proportion of heavier elements in the dust and gas. Those elements were there at all because previous generations of stars, some much larger than the Sun, as well as violent events like neutron star collisions, had released them into space, contaminating the primordial clouds of hydrogen and helium.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-05-29T14:40:33.070" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32089" PostTypeId="2" ParentId="32087" CreationDate="2019-05-29T14:42:40.480" Score="0" Body="&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;We can imagine that the Sun was a black hole with the same mass it has now. Then if you accelerate a rocket 30/km per second from the Earth in the direction opposite to  the direction of the Earth around the Sun, then that rocket would have no angular momenta with respect to the Sun and fall right into it. This is because the Earth travels at 30 km/s in a rouhly spherical orbit. In order to establish a really close encounter, and ignoring tidal effects, you thus only have to accelerate the velocity of the rocket with less than 30 km/s.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Since the solar system travels Sagittarius A with a velocity of roughly 230 km/s you would have to accelerate the rocket with less then 230 km/s to get a really close encounter with Sagitarrius A. If this is possible, you have to ask a rocket engineer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In general relativty, the energy of a body in circular motion around a spherically symmetric gravitational field can be written as (I belive this is correct):&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$E=mc^2\left(\frac{{1-\frac{2GM}{rc^2}}}{\sqrt{1-\frac{3GM}{rc^2}}}\right)$$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;setting &lt;span class=&quot;math-container&quot;&gt;$E=mc^2$&lt;/span&gt; at rest at infinity. In case of Sagitarrius A you could use this expression for, depending on how good your rocket engines are, how close to Sagittarius A you can accelerate/brake your self into a circular orbit and then get back up again.&lt;/p&gt;&#xA;" OwnerUserId="26403" LastEditorUserId="26403" LastEditDate="2019-05-29T15:01:59.770" LastActivityDate="2019-05-29T15:01:59.770" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32093" PostTypeId="2" ParentId="32086" CreationDate="2019-05-29T18:29:24.097" Score="4" Body="&lt;p&gt;Supernova ejecta expand at velocities between 5,000 km/s and 30,000 km/s.  Lets take 10,000 km/s as typical.  The speed of light, c, is 300,000 km/s, so 10,000 km/s = c/30. So, it would take 120 years for metals from the nearest stars to get to the Sun.  How much time was available?  Star formation in galaxies was well along 1 billion years after the Big Bang, 12.7 Gyr ago, and the Sun is 4.7 Gyr old.  So that gives 8 Gyr of travel time for debris from the first massive stars which become supernova in less then 100 million years.  Thus, the traveled length of the debris is 8 Gyr * c/30  is about 270 million light years.  In other words, there was sufficient time for supernova debris from our galaxy and hundreds of other galaxies to reach the solar protoplanetary nebula.  The story of course is different for the metals that come from progenitors that are not massive stars because of the longer delay for them to form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should add that the debris is likely to encounter intergalactic material along the way which will eventually halt its motion.  So realistically material will rarely move quite that far.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastEditorUserId="2424" LastEditDate="2020-02-12T23:23:21.863" LastActivityDate="2020-02-12T23:23:21.863" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32094" PostTypeId="2" ParentId="32086" CreationDate="2019-05-29T19:44:35.090" Score="8" Body="&lt;p&gt;Different types of stars are responsible for creating different groups of elements via &lt;a href=&quot;https://en.wikipedia.org/wiki/Nucleosynthesis&quot; rel=&quot;noreferrer&quot;&gt;nucleosynthesis&lt;/a&gt;. Most of the &quot;lighter&quot; metals (Na, Al, Ga, Ge, Se), were formed by exploding massive stars, and most of the &quot;heavier&quot; metals (I, Ir, Ag, Au, Pt, U, Pu) were created by merging neutron stars. Exploding white dwarfs are partially responsible for elements like Ti, V, Cr, Fe, Co, Ni. Dying low mass stars were involved with the creation of some of the elements Sr, Y, Zr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following periodic table of elements (from Wikipedia), shows the elements color coded according to their source of creation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WGdiN.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WGdiN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the elements on Earth were created by various stars that seeded our region of the universe prior to the formation of our solar system.&lt;/p&gt;&#xA;" OwnerUserId="28167" LastActivityDate="2019-05-29T19:44:35.090" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32095" PostTypeId="1" CreationDate="2019-05-29T21:58:30.743" Score="3" ViewCount="139" Body="&lt;p&gt;Wondering how to calculate the following in a general way:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sunrise/sunset&lt;/li&gt;&#xA;&lt;li&gt;Equinoxes/solstices&lt;/li&gt;&#xA;&lt;li&gt;Moon phases&lt;/li&gt;&#xA;&lt;li&gt;Planetary positions&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For hundreds to thousands of years ago. Wondering what the calculation is, or where I go about figuring it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to be able to say, in 1541 on Jan 1 the sun rose at x time, Jupiter was in y position, etc.&lt;/p&gt;&#xA;" OwnerUserId="28221" LastActivityDate="2019-05-29T21:58:30.743" Title="How to calculate planetary positions hundreds or thousands of years ago" Tags="&lt;amateur-observing&gt;&lt;positional-astronomy&gt;&lt;equinox&gt;&lt;solstice&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32096" PostTypeId="1" AcceptedAnswerId="32101" CreationDate="2019-05-30T00:11:52.900" Score="20" ViewCount="6135" Body="&lt;p&gt;Can a planet, star or otherwise have a magnetic field that is stronger or have more range than its gravity?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2019-05-31T18:54:12.160" LastActivityDate="2019-06-01T00:27:58.783" Title="Can a magnetic field of an object be stronger than its gravity?" Tags="&lt;gravity&gt;&lt;fundamental-astronomy&gt;&lt;magnetic-field&gt;&lt;stellar-astrophysics&gt;&lt;stellar-dynamics&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32097" PostTypeId="2" ParentId="32096" CreationDate="2019-05-30T00:49:32.330" Score="37" Body="&lt;p&gt;It depends on what object it's acting on.  There are many objects, including stars, that have magnetic fields where &lt;a href=&quot;https://en.wikipedia.org/wiki/Lorentz_force&quot; rel=&quot;noreferrer&quot;&gt;Lorentz forces&lt;/a&gt; on charged particles like electrons and protons are stronger than the gravitational force on them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also remember that the strength of the Lorentz force depends on the speed of the particle moving through it, so a fast enough moving electron even here on Earth will receive a larger magnetic force than a force of gravity. This is how the Earth's magnetic field is able to contain charged particles in the Van Allen belts that its gravity could not contain.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="7982" LastEditDate="2019-05-30T04:44:18.087" LastActivityDate="2019-05-30T04:44:18.087" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="32099" PostTypeId="1" CreationDate="2019-05-30T02:24:50.907" Score="2" ViewCount="78" Body="&lt;p&gt;I understand that we can measure the sound horizon from the anisotropies in the CMB. Large galaxy surveys today can construct two-point correlation functions, which reveal peaks at distances equal to today's sound horizon, I think because the BAOs at decoupling essentially created dark matter gravitational wells for structure to begin forming in (correct me if I'm wrong). I'm under the impression that the sound horizon is a fixed physical distance, so its measured comoving distance at any redshift should always be the same. Am I right or wrong in this assumption?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read some information from a few different sources on this topic which confused me. On &lt;a href=&quot;https://en.wikipedia.org/wiki/Accelerating_expansion_of_the_universe#Supernova_observation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;: &quot;Peaks have been found in the correlation function (the probability that two galaxies will be a certain distance apart) at 100 h−1 Mpc, indicating that this is the size of the sound horizon today, and by comparing this to the sound horizon at the time of decoupling (using the CMB), we can confirm the accelerated expansion of the universe.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if the sound horizon is always at the same physical scale, how can we use measurements today to determine the rate of expansion? Do the correlation functions give you &lt;em&gt;proper&lt;/em&gt; separations, and from that with different cosmological models we can determine what &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$H(z)$&lt;/span&gt; value gives a &lt;em&gt;comoving&lt;/em&gt; separation equal to the sound horizon? And is this the same as the BAO signal?&lt;/p&gt;&#xA;" OwnerUserId="24076" LastActivityDate="2019-05-31T08:20:09.223" Title="How is the expansion rate of the universe obtained from Baryon Acoustic Oscillations?" Tags="&lt;galaxy&gt;&lt;cosmology&gt;&lt;expansion&gt;&lt;cmb&gt;&lt;hubble-constant&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32100" PostTypeId="1" CreationDate="2019-05-30T02:32:06.050" Score="1" ViewCount="44" Body="&lt;p&gt;Current estimates for the number of stars in the universe are about 10^22. However, that number has changed several times as new observations have come forth. How has the estimate of the number of stars in the universe changed over time? What was the estimate after Hubble's findings? What was the estimate in the 70s, 90s, etc., and what new discoveries forced those estimates to change? &lt;/p&gt;&#xA;" OwnerUserId="28224" LastEditorUserId="28224" LastEditDate="2019-05-30T16:29:57.533" LastActivityDate="2019-05-30T16:29:57.533" Title="How has the estimate of the number of stars in the universe changed over time?" Tags="&lt;observable-universe&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32101" PostTypeId="2" ParentId="32096" CreationDate="2019-05-30T04:40:37.280" Score="17" Body="&lt;p&gt;Let's look at the proper magnetic force (as opposed to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lorentz_force&quot; rel=&quot;noreferrer&quot;&gt;Lorentz&lt;/a&gt; force on a &lt;em&gt;moving, charged object&lt;/em&gt; described in &lt;a href=&quot;https://astronomy.stackexchange.com/a/32097/7982&quot;&gt;@KenG's answer&lt;/a&gt;) on a specimen &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; of magnetized material with mass &lt;span class=&quot;math-container&quot;&gt;$M_S$&lt;/span&gt; as a way to try to compare. Let's arbitrarily assume it has a fixed, permanent magnetic moment &lt;span class=&quot;math-container&quot;&gt;$m_S$&lt;/span&gt;. We can't use iron because it will saturate too easily. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then let's look at how the forces scale differently with distance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathbf{F_G} = -\frac{G M_S M}{r^2}\mathbf{\hat{r}} \tag{1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathbf{F_B} =\nabla (\mathbf{m_S} \cdot \mathbf{B}(\mathbf{r})) \tag{2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we reduce these to scalar equations at a radius &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; (assume &lt;span class=&quot;math-container&quot;&gt;$\mathbf{m_S}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathbf{B}$&lt;/span&gt; are parallel) assume all forces are attractive, and evaluate the potentials and their gradients on the equator of the body &lt;em&gt;at it's physical radius&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt;. Since the magnetic force on our dipole specimen drops off faster than the gravitational force, we have to evaluate the two at the closest physically possible distance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F_G = \frac{G M_S M}{R^2} \tag{3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F_B = \frac{3 m_S B_{r=R}}{R} \tag{4}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where our specimen is a distance &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; from our field source, and it's moment &lt;span class=&quot;math-container&quot;&gt;$m_S$&lt;/span&gt; is a magnetization of 1 Tesla times the volume of a 1 kg rare earth magnet, about 0.000125 cubic meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All MKS units, all rough, ballpark numbers with emphasis on strongest magnetic fields&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Body             R (m)      M (kg)    B(r=R) (T)    F_G  (N)    F_B (N)    F_B/F_G&#xA;Earth            6.4E+06    6.0E+24   5.0E-05       9.8E+00     2.9E-15    3.0E-16&#xA;Jupiter          7.1E+07    1.9E+27   4.2E-04       2.5E+01     2.2E-15    8.8E-17&#xA;Neutron Star     1.0E+04    4.0E+30   5.0E+10       2.7E+12     1.9E+03    7.0E-10&#xA;Magnetar         1.0E+04    4.0E+30   2.0E+11       2.7E+12     7.6E+03    2.8E-09&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So even for a &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetar&quot; rel=&quot;noreferrer&quot;&gt;Magnetar&lt;/a&gt; (see also &lt;a href=&quot;https://arxiv.org/abs/1703.00068&quot; rel=&quot;noreferrer&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://www.physics.mcgill.ca/~pulsar/magnetar/main.html&quot; rel=&quot;noreferrer&quot;&gt;2&lt;/a&gt;) a kind of neutron star with a very strong magnetic field), the magnetic force on our 1kg specimen of permanent magnet is only 3 parts per billion as strong as the gravitational force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might see a much more favorable ratio if you compared two subatomic particles at short ranges (e.g. 1E-15 meters) but for astronomical objects, gravity seems to win smartly.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-05-31T07:51:21.737" LastActivityDate="2019-05-31T07:51:21.737" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32102" PostTypeId="1" CreationDate="2019-05-30T07:57:53.423" Score="10" ViewCount="131" Body="&lt;p&gt;Papers, like this one from &lt;a href=&quot;https://ai.google/research/pubs/pub46789/&quot; rel=&quot;noreferrer&quot;&gt;Google&lt;/a&gt;, show that Deep Learning and Machine Learning are effective in detecting exoplanets from luminosity data provided by telescopes like TESS or Kepler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These models seem extremely quick and lightweight. Has it become somewhat of a standard practice to use these on freshly obtained TESS data? If not, why, and what is done with the data instead? Are there drawbacks to using these seemingly accurate models?&lt;/p&gt;&#xA;" OwnerUserId="21492" LastActivityDate="2020-04-14T01:28:13.170" Title="How actively is Deep Learning being used in exoplanet identification?" Tags="&lt;exoplanet&gt;&lt;luminosity&gt;&lt;kepler&gt;&lt;algorithm&gt;&lt;tess&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32103" PostTypeId="1" CreationDate="2019-05-30T08:05:12.033" Score="1" ViewCount="33" Body="&lt;p&gt;While reading the research paper &lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/1973A%26A....24..181F/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Excitation of the Fe XIII Spectrum in the Solar Corona&quot; by D.R. Flower and G. Pineau des Forets&lt;/a&gt; in the summary I found the concept of statistical equilibrium equation hard to understand. &#xA;Please explain me the ways in which I can obtain these equations and what is the physical meaning of these? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;Solution of the Statistical Equilibrium Equation&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;In coronal conditions, the full statistical equilibrium equations need be solved for only the first few (&lt;span class=&quot;math-container&quot;&gt;$i=1 \text{ to } m$&lt;/span&gt;) energy levels. More highly excited states (&lt;span class=&quot;math-container&quot;&gt;$k=m+1 \text{ to } n$&lt;/span&gt;) contribute only through the indirect process of electron collisional excitation followed by radiative cascade. In the case of Fe XIII, we solve the statistical equilibrium equations for the states of the ground &lt;span class=&quot;math-container&quot;&gt;$3s^2 3p^2$&lt;/span&gt; configuration (&lt;span class=&quot;math-container&quot;&gt;$m=5$&lt;/span&gt;) and include the indirect contribution of the states of the excited &lt;span class=&quot;math-container&quot;&gt;$3s 3p^3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$3s^2 3p 3d$&lt;/span&gt; configuration (&lt;span class=&quot;math-container&quot;&gt;$k=6 \text{ to } 27$&lt;/span&gt;). The error committed in neglecting all other bound states of the ion and the continuum is considered separately later in this Section. In this approximation, the statistical equilibrium equations of levels &lt;span class=&quot;math-container&quot;&gt;$1 \text{to} m$&lt;/span&gt; may be written simply as &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\sum_{j=1}^{m} d_{ij} N_j=0 \tag{6}$$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;with &lt;span class=&quot;math-container&quot;&gt;$d_{ji}$&lt;/span&gt; defined by…(several long expressions)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://articles.adsabs.harvard.edu/full/1973A%26A....24..181F&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://articles.adsabs.harvard.edu/full/1973A%26A....24..181F&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://articles.adsabs.harvard.edu/pdf/1973A%26A....24..181F&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://articles.adsabs.harvard.edu/pdf/1973A%26A....24..181F&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Bitcode: 1973A&amp;amp;A....24..181F &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="28229" LastEditorUserId="7982" LastEditDate="2019-05-30T22:51:50.267" LastActivityDate="2019-05-30T22:51:50.267" Title="what/how do we mean/derive statistical equilibrium equation of certain energy levels?" Tags="&lt;the-sun&gt;&lt;spectroscopy&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32105" PostTypeId="1" CreationDate="2019-05-30T12:27:59.547" Score="1" ViewCount="82" Body="&lt;p&gt;Are there enough neutrinos &amp;amp; other invisible particles in a cubic light year of space to significantly affect the expansion of the universe,&amp;amp; at a very rough estimate without taking into account WIMPS &amp;amp; other exotic particles for which there is no firm evidence,what would be the total mass of this quantity of matter? Only a tiny proportion of all the neutrinos emitted since the Big Bang have been absorbed,the rest are still flying around in search of a home.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastEditorUserId="28083" LastEditDate="2019-05-30T21:47:14.510" LastActivityDate="2019-05-31T11:37:04.833" Title="Invisible Mass Affecting the Expansion of the Universe" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32107" PostTypeId="1" AcceptedAnswerId="32116" CreationDate="2019-05-30T17:52:04.370" Score="4" ViewCount="250" Body="&lt;p&gt;If the dark energy equation of state &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; evolves with time, and becomes larger, this would result in decelerated expansion and eventually a Big Crunch. If &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; decreases with time, becoming more negative, then a Big Rip would occur. I understand that the Big Rip has not yet been ruled out by current observations, but I often hear that the Big Crunch has been ruled out since we know the universe is expanding (and accelerating).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Big Rip can't currently be ruled out, then how can the Big Crunch be ruled out? Shouldn't they both still be on the table, unless we can somehow infer which direction &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; evolves in under a quintessence model?&lt;/p&gt;&#xA;" OwnerUserId="24076" LastActivityDate="2019-06-02T12:45:49.473" Title="Has the Big Crunch been ruled out?" Tags="&lt;cosmology&gt;&lt;expansion&gt;&lt;dark-energy&gt;&lt;fate-of-universe&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32108" PostTypeId="1" CreationDate="2019-05-30T19:15:23.410" Score="1" ViewCount="74" Body="&lt;p&gt;I have two short telescopes. The tubes that the eyepieces fit in are different; one is short and one is long, such as the tubes on the Celestron first scope and the Jason Comet chaser. Why is this so and what is the advantages of one over the other? They both take 1.25 eyepieces. &lt;/p&gt;&#xA;" OwnerUserId="28238" LastEditorUserId="16462" LastEditDate="2019-06-01T22:54:45.350" LastActivityDate="2019-06-01T22:54:45.350" Title="Long tube for eyepiece" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32110" PostTypeId="2" ParentId="32072" CreationDate="2019-05-30T20:41:57.000" Score="-2" Body="&lt;p&gt;We can be pretty sure that the asteroid belt once held one or more sizable planetesimals,as there is evidence that it or they were large enough for gravity &amp;amp; internal heating to drag the heavy metals (mainly iron &amp;amp; nickel) to the centre,forming a metallic core. Bodes Law decrees that there should be a planet where the asteroid belt is now,but it seems that as this planet was forming it was shattered by an impact &amp;amp; the remnants scattered. Some reached Earth,which is how we know that their materials had differentiated into metallic core &amp;amp; rocky mantle. It is likely that some of the huge asteroids which have struck the Earth in the distant past were fragments thrown out of the asteroid belt &amp;amp; that some hit Mars &amp;amp; Venus as well. There may be more material in the asteroid belt than we think,because some fragments are so small &amp;amp; widely dispersed that they don't register in our telescopes. I think we can assume that the primordial planetesimals had a similar though not identical composition to the asteroid belt of today.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-05-30T20:41:57.000" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32111" PostTypeId="2" ParentId="32096" CreationDate="2019-05-30T20:43:37.203" Score="9" Body="&lt;p&gt;It isn't impossible, but the short answer is &quot;no&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A gravitational field will accelerate all matter and energy equally while a magnetic field will only accelerate moving electric charges (other magnets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force due to gravity is proportional to the &lt;em&gt;inverse square&lt;/em&gt; of the distance, and the force due to magnetism asymptotically approaches the &lt;em&gt;inverse cube&lt;/em&gt; of the distance. At some critical distance the gravitational force will become stronger than the magnetic force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless most of the large body was magnetic, even over the magnetic poles the magnetic field would probably be too low to levitate a typical magnet in the large body's gravitational field.&lt;/p&gt;&#xA;" OwnerUserId="28240" LastActivityDate="2019-05-30T20:43:37.203" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32112" PostTypeId="1" AcceptedAnswerId="32113" CreationDate="2019-05-30T22:41:57.720" Score="4" ViewCount="179" Body="&lt;p&gt;This image was created using the NICER instrument on the ISS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.nasa.gov/feature/goddard/2019/nicer-s-night-moves-trace-the-x-ray-sky&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.nasa.gov/feature/goddard/2019/nicer-s-night-moves-trace-the-x-ray-sky&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7xXyZ.jpg&quot; alt=&quot;a map of X-ray emissions in the sky, with some pulsars and other X-ray sources labelled, and curved lines between them&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the big curves between the X-ray sources?&lt;/p&gt;&#xA;" OwnerUserId="16160" LastActivityDate="2019-05-31T01:10:11.303" Title="What are the curved lines in this X-ray image of the sky?" Tags="&lt;x-ray&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32113" PostTypeId="2" ParentId="32112" CreationDate="2019-05-31T00:33:15.993" Score="6" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dO0sK.gif&quot; alt=&quot;NICER X-ray telescope on the International Space Station&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GIFs too big to post here directly: &lt;a href=&quot;https://media.giphy.com/media/Yq2QfBg1WZeUu9ETOc/giphy.gif&quot; rel=&quot;noreferrer&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;https://media.giphy.com/media/fsVr6XOUkugVA9mqjT/giphy.gif&quot; rel=&quot;noreferrer&quot;&gt;2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NICER is all about time resolution and energy resolution and is used to collect the time and energy-resolved X-ray spectra from pulsars. It isn't an imaging telescope and so can't record &quot;patches of the sky&quot; like other devices can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The curved lines are the paths that the narrow field of view (FOV) of the NICER telescope has traced out across the celestial sphere during this 22 month (intermittent) observation period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://heasarc.gsfc.nasa.gov/docs/nicer/mission_guide/&quot; rel=&quot;noreferrer&quot;&gt;this page&lt;/a&gt; the FOV is only about 30 square arc minutes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Large effective area: ~1900 cm2 at 1.5 keV&lt;/li&gt;&#xA;&lt;li&gt;Broad Bandpass: 0.2 &amp;lt; E &amp;lt; 12.0 keV&lt;/li&gt;&#xA;&lt;li&gt;Absolute timing precision of &amp;lt; 300 ns&lt;/li&gt;&#xA;&lt;li&gt;Moderate spectral Resolution: 6 &amp;lt; E/ΔE &amp;lt; 80 from 0.5 keV to 8 keV&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Restricted field of view: 30 arcmin²&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With limited observing time and effectively a narrow single pixel field, it can only trace out thread-like arcs across the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NICER is mounted on the constantly moving International Space Station and has its own pointing mechanisms on top of that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you see several arcs meet at a certain point, this is an interesting X-ray source, usually a neutron star (the primary purpose of the device &lt;a href=&quot;https://en.wikipedia.org/wiki/Neutron_Star_Interior_Composition_Explorer&quot; rel=&quot;noreferrer&quot;&gt;and the &quot;N&quot; in NICER&lt;/a&gt;), and the telescope passes over it several times in order to collect more data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Watch the COOL YouTube video &lt;a href=&quot;https://youtu.be/kk0ry3_R2pE&quot; rel=&quot;noreferrer&quot;&gt;NASA'S NICER Does the Space Station Twist&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Watch the COOL YouTube video &lt;a href=&quot;https://www.youtube.com/watch?v=sDd2PjbXEg8&quot; rel=&quot;noreferrer&quot;&gt;ISS NICER - First Deployment &amp;amp; Range of Motion Checks&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Watch more NICER videos here: &lt;a href=&quot;https://heasarc.gsfc.nasa.gov/docs/nicer/nicer_gallery.html&quot; rel=&quot;noreferrer&quot;&gt;https://heasarc.gsfc.nasa.gov/docs/nicer/nicer_gallery.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Watch the YouTube video &lt;a href=&quot;https://youtu.be/IOEPDf2DYNM&quot; rel=&quot;noreferrer&quot;&gt;Unlocking Secrets of Neutron Stars with NICER&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read the presentation &lt;a href=&quot;https://heasarc.gsfc.nasa.gov/docs/nicer/slides/NICER_Science_Overview.pdf&quot; rel=&quot;noreferrer&quot;&gt;NICER Science Overview&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that it is a number of grazing-incidence curved crystals used as concentrators, with &quot;single-pixel&quot; X-ray detectors behind them in the photo below (from &lt;a href=&quot;https://space.stackexchange.com/q/19882/12102&quot;&gt;How will NavCube (actually) be important for the XCOM testing and demonstration?&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; NICER X-ray timing telescope/concentrator array, from &lt;a href=&quot;https://www.nasa.gov/content/about-nicer&quot; rel=&quot;noreferrer&quot;&gt;About NICER&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/V64w2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-05-31T01:10:11.303" LastActivityDate="2019-05-31T01:10:11.303" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32115" PostTypeId="2" ParentId="32099" CreationDate="2019-05-31T08:20:09.223" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm under the impression that the sound horizon is a fixed physical&#xA;  distance, so its measured comoving distance at any redshift should&#xA;  always be the same. Am I right or wrong in this assumption?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are both right and wrong at the same time. Right in the sense that, yes, in comoving coordinates, the distance is the same at all redshifts. But when we use galaxy catalogues to measure the BAO distance scale, aka the BAO peak, it usually varies with redshift. This variation is very tiny, of the order of a 1-2 h&lt;span class=&quot;math-container&quot;&gt;$^{-1}$&lt;/span&gt;Mpc. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But if the sound horizon is always at the same physical scale, how can&#xA;  we use measurements today to determine the rate of expansion? Do the&#xA;  correlation functions give you proper separations, and from that with&#xA;  different cosmological models we can determine what 0 or ()&#xA;  value gives a comoving separation equal to the sound horizon? And is&#xA;  this the same as the BAO signal?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Before redshift of galaxies were measured, people were just using the galaxy positions on the sky (Right ascension RA, Declination DEC) to calculate the angular correlation function. You can also see the BAO peak in the angular correlation function. Nowadays, since we have accurate redshift measurements, people calculate the 3d correlation function &lt;span class=&quot;math-container&quot;&gt;$\xi(r)$&lt;/span&gt;. To get constraints from BAO peaks in a very simple way, we can in brief, do the following. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do so, we first need to convert RA, DEC and redshift to X,Y,Z cartesian coordinates. It is at this step, we assume a fiducial cosmological model, as in &lt;span class=&quot;math-container&quot;&gt;$\Omega_{M}=0.27, \Omega_{\Lambda}=0.7, h=0.67$&lt;/span&gt; etc. Now, we have 3D information of our galaxies and we can count pairs and get the &lt;span class=&quot;math-container&quot;&gt;$\xi(r)$&lt;/span&gt;, from which we can calculate the BAO peak. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;People can then use theoretical model predictions for &lt;span class=&quot;math-container&quot;&gt;$\xi(r)$&lt;/span&gt;, given a specific set of cosmological parameters. It is here that cosmologists vary the set of cosmological parameters, which will give us a specific &lt;span class=&quot;math-container&quot;&gt;$\xi(r)$&lt;/span&gt;. &#xA;Lets say, we have 10 variations of &lt;span class=&quot;math-container&quot;&gt;$\Omega_{M}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\Omega_{\Lambda}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt;. This would give us 30 theoretical &lt;span class=&quot;math-container&quot;&gt;$\xi(r)$&lt;/span&gt;'s, and we can compute the BAO peak for all of them using an empirical fit model. These peaks can now be compared with the observed peak from the data and we can calculate the &lt;span class=&quot;math-container&quot;&gt;$\chi^{2} = \frac{(peak_{obs} - peak_{theory})^{2}}{err(peak_{obs})^{2}}$&lt;/span&gt;. This can give us constraints on the different cosmological parameters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is just a basic answer to your question and there are a lot of other processes involved! You can check out &lt;a href=&quot;https://arxiv.org/pdf/1903.09651.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper which was recently published in arXiv&lt;/a&gt; which uses only the BAO peak information to get constraints on cosmological distance measures. &lt;/p&gt;&#xA;" OwnerUserId="2655" LastActivityDate="2019-05-31T08:20:09.223" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32116" PostTypeId="2" ParentId="32107" CreationDate="2019-05-31T09:30:57.200" Score="4" Body="&lt;p&gt;The Big Rip happens if the equation of state for the dark energy has &lt;span class=&quot;math-container&quot;&gt;$p/\rho = w&amp;lt;-1$&lt;/span&gt;, and all empirical data give us &lt;span class=&quot;math-container&quot;&gt;$w\approx -1$&lt;/span&gt;. A Big Crunch requires a pretty high value of &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt; (it must go above -1/3 to just stop the acceleration), but of course if dark energy is changing over time it might do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So tentatively ruling out Big Crunches is the reasonable thing to do given observational evidence and the assumption that the future will not be wildly different. If one thinks we should be more open to dark energy changing (which is going to be rather unconstrained by observation for the foreseeable future) then we should also be more open to &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt; moving across the Big Rip boundary. &lt;/p&gt;&#xA;" OwnerUserId="19816" LastEditorUserId="19816" LastEditDate="2019-06-02T12:45:49.473" LastActivityDate="2019-06-02T12:45:49.473" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32117" PostTypeId="2" ParentId="32105" CreationDate="2019-05-31T11:37:04.833" Score="-2" Body="&lt;p&gt;Yes, there is enough of these things, especially dark matter which is accelerating the expansion of the universe.&lt;/p&gt;&#xA;" OwnerUserId="27997" LastActivityDate="2019-05-31T11:37:04.833" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32119" PostTypeId="1" AcceptedAnswerId="32122" CreationDate="2019-05-31T13:46:29.963" Score="4" ViewCount="245" Body="&lt;p&gt;I know that you can calculate the distance to another star for example, by using the phenomenon redshift to determine how much it has redshifted because of the expansion of the universe. But how do you know what the original frequency/wavelength of the light was? (And if all of my above statements are wrong, how do you measure the distance?)&lt;/p&gt;&#xA;" OwnerUserId="28254" LastEditorUserId="2365" LastEditDate="2019-06-01T07:01:56.887" LastActivityDate="2019-06-03T12:47:40.367" Title="How do we measure distance in space?" Tags="&lt;star&gt;&lt;light&gt;&lt;distances&gt;&lt;redshift&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32120" PostTypeId="1" CreationDate="2019-05-31T13:59:47.773" Score="0" ViewCount="178" Body="&lt;p&gt;It is said that neutrinos can travel for billons of years at almost the speed of light without interacting with anything. That being the case,what do they do - is it possible they could orbit the universe indefinitely? They certainly can't leave it.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-05-31T16:57:24.117" Title="Where do Neutrinos Go if They Don't Interact?" Tags="&lt;cosmology&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="32121" PostTypeId="2" ParentId="32119" CreationDate="2019-05-31T14:34:33.033" Score="1" Body="&lt;p&gt;Apparent brightness is one way of measuring distance; the further away a certain type of star is, the fainter it will be. We have Cepheid variables whose intrinsic luminosity is indicated by the period of their variability, so we know with certainty how bright they really are. Most stars (as opposed to galaxies) are too close for cosmic redshift to be a factor, so redshift or blueshift indicates their proper motion away or towards us.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastEditorUserId="14094" LastEditDate="2019-06-01T23:49:53.267" LastActivityDate="2019-06-01T23:49:53.267" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32122" PostTypeId="2" ParentId="32119" CreationDate="2019-05-31T16:39:21.270" Score="8" Body="&lt;p&gt;Red shift is used for measuring the distance to very distant stars (galaxies mostly, in fact). The secret is to use spectral lines. Specific elements when very hot emit light at very specific colours and you can spot the pattern of those colours when the whole pattern has been shifted towards the red and see how far it has shifted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For closer stars, there are other techniques, using types of star whose brightness can be worked out in some way, or, for the closest, by observing how the position of the star compared to more distant ones changes as the Earth moves from one side of the Sun to the other.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-05-31T16:39:21.270" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32123" PostTypeId="2" ParentId="32120" CreationDate="2019-05-31T16:57:24.117" Score="3" Body="&lt;p&gt;It depends on the ultimate fate of universe, as with everything else in the universe. Fundamentally, neutrinos are not much different from “ordinary matter”. Yes, they do interact only via weak force and gravity, but this only quantitatively changes the frequency of interaction, but does not change the game qualitatively. As long as there’s non-zero chance for them to interact, and as long as there is non-zero density, there is non-zero probability they will eventually interact with another particle (or black hole), and given infinite timeframe, it will happen for sure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accelerating expansion of universe does however complicate things a bit, but again, eventually there is bound to be some random fluctuation which will knock the neutrino off from its track. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually neutrinos would meet anti-neutrinos and annihilate each other. Since the chances for this are really low, I would suspect neutrino-antineutrino collisions being the very last blink of this universe - given heat death is the ultimate fate of this existence.&lt;/p&gt;&#xA;" OwnerUserId="10813" LastActivityDate="2019-05-31T16:57:24.117" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="32125" PostTypeId="2" ParentId="32108" CreationDate="2019-06-01T13:37:57.320" Score="3" Body="&lt;p&gt;The reason that some telescopes have long eye-piece tubes is that the focal plane of the objective has been placed far enough outside the main tube assembly to allow erecting prisms or binocular adapters to be fitted, if desired.&lt;/p&gt;&#xA;" OwnerUserId="13887" LastActivityDate="2019-06-01T13:37:57.320" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32126" PostTypeId="1" CreationDate="2019-06-01T14:59:39.093" Score="3" ViewCount="53" Body="&lt;p&gt;The right ascension and declination of stars change on a day to day basis. Some of the changes are quite large. The right ascension of Polaris, for example, changes by more than a degree over the course of some years. The excursions are irregular. Their time and extent differ from year to year. I know that precession will change coordinates of all the stars over tens of thousands of years, but I can't find an explanation for these short term changes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polaris started 2017 with a sidereal hour angle of 316 deg 23.9 min. The SHA increased monotonically to 317  3.6 by the end of April, then decreased in small steps to 315  58.3 by November 9, then increased again, ending the year at 316 &#xA; 6.2.&lt;/p&gt;&#xA;" OwnerUserId="28264" LastEditorUserId="28264" LastEditDate="2019-06-01T18:07:26.033" LastActivityDate="2019-06-01T18:07:26.033" Title="Short term changes in celestial coordinates of stars" Tags="&lt;observational-astronomy&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32127" PostTypeId="1" AcceptedAnswerId="32133" CreationDate="2019-06-01T16:49:09.750" Score="5" ViewCount="98" Body="&lt;p&gt;My understanding is that much of the visible light that we see during the day is given off from the Sun, and that means that most of the colors that we see are dictated by (with the exception of atmospheric effects) the particular wavelengths of color that the Sun emits, which does not include the entire continuous color spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.noao.edu/image_gallery/images/d5/sun.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is an image of the wavelengths emitted by the Sun&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is also that, at least during a particular part of the life cycle of a star, all stars are burning balls of hydrogen and helium. If that is the case, then do all stars at that point in their life give off emission spectra of visible light that is similar to our own? Is it possible that an observer on a planet with an atmosphere similar to that of Earth might see different ranges of color than we do, simply due to, perhaps, the composition of their star? Could there be planets similar to Earth where the sky would appear purple rather than blue?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if any of my assumptions are incorrect.&lt;/p&gt;&#xA;" OwnerUserId="28265" LastEditorUserId="14094" LastEditDate="2019-06-01T23:23:36.087" LastActivityDate="2019-06-02T04:12:03.653" Title="Do all stars of a given age and mass have relatively the same emission spectrum?" Tags="&lt;light&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32128" PostTypeId="2" ParentId="32017" CreationDate="2019-06-01T19:02:52.167" Score="2" Body="&lt;p&gt;NASA conducted a study on the Ice Giants in 2016 after recognition as a high-priority flagship mission target in the 2013-2022 Planetary Decadal Survey (&lt;a href=&quot;http://www.planetary.org/get-involved/be-a-space-advocate/become-an-expert/what-is-the-decadal-survey.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;overview of the Decadal Survey from The Planetary Society&lt;/a&gt;), a community document which helps set priorities and goals. The &lt;a href=&quot;https://www.lpi.usra.edu/icegiants/mission_study/Full-Report.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;full report&lt;/a&gt; and summary are available from &lt;a href=&quot;https://www.lpi.usra.edu/icegiants/&quot; rel=&quot;nofollow noreferrer&quot;&gt;LPI's Ice Giants Mission Planning&lt;/a&gt; page and is intended to review the science and mission options leading into planning for the next decadal survey report. In the Science section starting on page 29, the introduction has a table of the major components of the different planet types (terrestrial, gas giant and ice giant) where ice giants are listed as ~10% gas, ~65% ice, ~25% rock with a total mass of ~15 times that of the Earth. This is in comparison to terrestrial planets which are almost 100% rock and gas giants which are ~85% gas. This is illustrated in the figure below taken from the report (their Figure 3-1): &lt;img src=&quot;https://i.stack.imgur.com/wUKTk.png&quot; alt=&quot;Illustration of compositional differences among the giant planets and their relative sizes.&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the lead-in to the subsections on the various science questions and subsections, the report goes on to say: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We  therefore  consider  the  objectives  of determining  the  interior  structure  of an ice  giant  (Section  3.1.1)  and  the  bulk  composition (including noble gases and key isotopic ratios, Section 3.1.2) as the highest-priority science to be done. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Section 3.1.1 on interior structure states that there is a presumption that these &quot;ice giants&quot; are dominated by water but there is very little direct evidence. It then explains that ice giants are normally modelled with three layers: an outer Hydrogen/Helium envelope, below that a thick fluid ionic water layer (or water mixed with other ices like ammonia and methane), and at the center, a small rocky core. However the limited amount of gravity field data from the Voyager 2 flybys showed that both planets are not as centrally condensed (less mass in the center) as expected in the simple 3 layer structure and also:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;there is no strong motivation, beyond computational convenience, for a 3-layer structure (Helled et al. 2010).&quot; &lt;a href=&quot;https://iopscience.iop.org/article/10.1088/0004-637X/726/1/15/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;Link to Helled et al. 2010 journal article (free access)&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Both Uranus and Neptune have bulk densities a little under compressed water, implying that there has to be Hydrogen/Helium (H/He) envelope and which is indeed seen in spectra of the visible surface. Below this envelope though little is known apparently. The low density and the limited gravity data would imply a mostly water interior with other gasses such as ammonia or more H/He to keep the density lower than water and a very small amount of rock in the core &lt;a href=&quot;https://arxiv.org/pdf/1207.2309.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nettelmann et al. 2013 (arXiv paper link)&lt;/a&gt;. This model however predicts an ice:rock ratio of more than 10:1 which is not seen in any of the other icy moons and the model does not explain well the heat flux coming from the interior. This &lt;a href=&quot;https://www.hou.usra.edu/meetings/icegiants2014/presentations/Day_2/Nettleman_2034.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;presentation by Nettelmann&lt;/a&gt; at a 2014 ice giants meetings talks more about these models, the constraints from the gravity data from Voyager 2 and the heat flux problems and shows possible internal structure diagrams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interior structure section of the Ice Giants report ends by calling for measurements of the higher order gravitational moments &lt;em&gt;J2, J4&lt;/em&gt; &amp;amp; &lt;em&gt;J6&lt;/em&gt; to higher precision, particularly for Neptune. However this is limited to the bulk structure and doesn't constrain the layering which would need a Doppler imager on a spacecraft to measure planetary oscillations analogous to solar helioseismology and/or an entry probe. The Doppler imager is discussed more in Sections 3.3.1 and 3.3.2 of the report which note that these new instruments have the potential to provide great information on the interior structure, but that Doppler imagers are technologically immature and that even with the largest telescopes on Earth or near-Earth spacecraft it has not been possible to see if the oscillations exist and the oscillation amplitude may be too small to detect. The report also notes such an instrument is also large, power hungry and produces a large data volume, all risks on a constrained mission to the outer planets but the benefits outweigh the risks. (Using up most of your mass and power budget on an instrument that may fail, or not produce a measurable result, makes people very nervous...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, there are some limited constraints on the bulk components of the ice giants but this is still a poorly constrained and understood problem which is a subject of ongoing research but is need of additional constraints and data from a dedicated mission. Such a mission would need to be flagship class which makes it expensive and would need broad support from the majority of the planetary community and would likely need to be the top-ranked mission from the next decadal survey covering the 2023-2032 timeframe to stand a chance of being funded and flying.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2019-06-01T19:02:52.167" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32129" PostTypeId="1" AcceptedAnswerId="32130" CreationDate="2019-06-01T19:33:13.307" Score="3" ViewCount="177" Body="&lt;p&gt;I read something in a book the other day which has gotten me really confused. They were talking about the expansion of the universe, and if I understood it all, they said that the expansion of the universe increases with distance. I’m wondering if I understood this correctly, because I was reading about Hubble’s law and it said that the redshift if galaxies far away (caused by the expansion of the universe) is proportional to the distance (up to a few hundred megaparsecs away). Doesn’t this mean that the expansion is the same but the redshift increases? (I don’t see why the redshift would be proportional and a linear graph if the universe expanded from us proportionally to the distance, or at least more further away). I hope you understand what I mean by all this. &lt;/p&gt;&#xA;" OwnerUserId="28254" LastEditorUserId="28254" LastEditDate="2019-06-01T21:08:29.503" LastActivityDate="2019-06-03T12:06:16.940" Title="Does the expansion of the universe increase with distance?" Tags="&lt;universe&gt;&lt;expansion&gt;&lt;redshift&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32130" PostTypeId="2" ParentId="32129" CreationDate="2019-06-01T21:59:03.890" Score="6" Body="&lt;p&gt;I'm not sure why you say that &quot;&lt;em&gt;expansion is the same&lt;/em&gt;&quot; when you also say that &quot;&lt;em&gt;expansion increases with distance&lt;/em&gt;&quot;, but I wouldn't say the &quot;expansion increases&quot;, but rather that &quot;the recession velocity increases&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The recession velocity &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; of galaxies increases linearly with distance &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt;. The constant of proportionality is called the Hubble constant &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt;. Thus, Hubble's law states that&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;v = H_0 d,&#xA;$$&lt;/span&gt;&#xA;and currently, &lt;span class=&quot;math-container&quot;&gt;$H_0 \simeq 70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$&lt;/span&gt;, but in the past it was higher. That means that, for every Mpc (mega-parsec, &lt;span class=&quot;math-container&quot;&gt;$\simeq 3.26\times10^6$&lt;/span&gt; light-years) a galaxy is from us, its speed away from us is roughly &lt;span class=&quot;math-container&quot;&gt;$70\,\mathrm{km}\,\mathrm{s}^{-1}$&lt;/span&gt;. This is the speed it has &lt;em&gt;right now&lt;/em&gt;, but not the speed we see it have, since we see it in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, we don't really see a speed at all. We measure a galaxy's redshift &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; also increases with distance, but not linearly. Thats's because it depends on how much the Universe has expanded since the light was emitted, and the expansion rate changes with time. But for small enough distances, i.e. for galaxies that are not seen that far into the past, the relation is close to linear, so that &lt;span class=&quot;math-container&quot;&gt;$v = cz$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; is the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is indeed approximately true out to a few hundred Mpc. For larger distances, the linear relation becomes inaccurate, and one must instead apply a cosmological model of the expansion history of the Universe, which depends non-trivially on the densities of its various constituents.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Acceleration of the Universe&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The term &quot;accelerated expansion rate&quot; refers not to the value of &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;, but to the rate at which the scale factor &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; — the &quot;size&quot; of the Universe — increases (the Universe may or may not be infinitely large; &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is normalized such that, today, &lt;span class=&quot;math-container&quot;&gt;$a=1$&lt;/span&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given value of &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;, we know the recession velocity of any two points separated by some distance &lt;em&gt;at a given time&lt;/em&gt;. At a later time, these two points will have increased their distance, and hence — by Hubble's law — their recession velocity will have increased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was higher in the past is the Hubble parameter &lt;span class=&quot;math-container&quot;&gt;$H(t)$&lt;/span&gt;. That number will, in the future, asymptotically decrease toward a slightly smaller value of&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;H(t\!\!\rightarrow\!\!\infty) = H_0\sqrt{\Omega_\Lambda} \simeq 56\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}.&#xA;$$&lt;/span&gt;&#xA;The Hubble parameter is defined as &lt;span class=&quot;math-container&quot;&gt;$H\equiv\dot{a}/a$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\dot{a}\equiv da/dt$&lt;/span&gt; is the Universe's change in size per unit time. If &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; is constant, and &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; increases, then &lt;span class=&quot;math-container&quot;&gt;$\dot{a}$&lt;/span&gt; also increases. That is, the increase in velocity of any two points accelerates. In fact, we can solve the differential equation and get that, in the future,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;a(t\!\!\rightarrow\!\!\infty) \propto e^{Ht},&#xA;$$&lt;/span&gt;&#xA;i.e. the scale factor increases exponentially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure below shows the evolution of &lt;span class=&quot;math-container&quot;&gt;$a(t)$&lt;/span&gt; until &lt;span class=&quot;math-container&quot;&gt;$t\sim130\,\mathrm{Gyr}$&lt;/span&gt;. On the left is a zoom-in of the first &lt;span class=&quot;math-container&quot;&gt;$20\,\mathrm{Gyr}$&lt;/span&gt; where you can see how the Universe &lt;em&gt;de&lt;/em&gt;celerated the first &lt;span class=&quot;math-container&quot;&gt;$\sim10\,\mathrm{Gyr}$&lt;/span&gt;, until dark energy started taking over the dynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mYza6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mYza6.png&quot; alt=&quot;a(t)&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment the acceleration is actually quite modest, but it will increase without bounds in the future; for instance, in 50 Gyr from now, &lt;span class=&quot;math-container&quot;&gt;$a\sim10$&lt;/span&gt;, i.e. galaxies will be &lt;span class=&quot;math-container&quot;&gt;$10\times$&lt;/span&gt; farther from each other, but in 100 Gyr we have &lt;span class=&quot;math-container&quot;&gt;$a\sim150$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2019-06-03T07:56:30.727" LastActivityDate="2019-06-03T07:56:30.727" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32131" PostTypeId="1" CreationDate="2019-06-01T22:09:39.550" Score="1" ViewCount="55" Body="&lt;p&gt;Today we have various data available from independent probes to constrain dark energy, i.e. its equation of state &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt;. These include Type 1a supernovae, BAO, CMB, large-scale galaxy surveys, etc. Are any of these probes better/more powerful at constraining dark energy than the others? Why or why not? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I assume that if a galaxy survey includes 1000 galaxies in a redshift range &lt;span class=&quot;math-container&quot;&gt;$0.1&amp;lt;z&amp;lt;1.2$&lt;/span&gt;, that would result in higher constraints on &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt; than just using 500 Type 1a supernovae in a range &lt;span class=&quot;math-container&quot;&gt;$0.1&amp;lt;z&amp;lt;0.8$&lt;/span&gt;. But how much of a difference does this, among other factors I'm unaware of, make? Quantitatively, how do the errors/uncertainties scale with larger samples and larger redshift intervals (I think for a larger sample the uncertainty decreases as &lt;span class=&quot;math-container&quot;&gt;$1/\sqrt N$&lt;/span&gt;)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also makes sense that dark energy models with more free parameters would come with a larger uncertainity (for example, &lt;span class=&quot;math-container&quot;&gt;$wCDM$&lt;/span&gt; has 1 more free parameter, &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt;, than &lt;span class=&quot;math-container&quot;&gt;$\Lambda CDM$&lt;/span&gt;). But how much is the constraining power diminished for each new degree of freedom? See plots (Figure 9, &lt;a href=&quot;https://arxiv.org/pdf/1709.01091.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/pdf/1709.01091.pdf&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oFqxg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oFqxg.png&quot; alt=&quot;Figure 9, https://arxiv.org/pdf/1709.01091.pdf&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the plot on the right, clearly the constraints are much looser because more free parameters were introduced in fitting the data to this time-dependent dark energy model. We can also see that the SN1a data, for example, seems to constrain dark energy less than CMB or BAO data. Why is this? Is it because SN1a have only been observable at lower redshifts so far?&lt;/p&gt;&#xA;" OwnerUserId="24076" LastActivityDate="2019-06-01T22:09:39.550" Title="How are constraints on dark energy improved the most?" Tags="&lt;cosmology&gt;&lt;supernova&gt;&lt;expansion&gt;&lt;dark-energy&gt;&lt;cmb&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32133" PostTypeId="2" ParentId="32127" CreationDate="2019-06-02T04:00:03.980" Score="4" Body="&lt;p&gt;The spectrum of light emitted by a star depends mainly on its surface temperature, with smaller variations associated with composition, surface gravity and magnetic activity&lt;span class=&quot;math-container&quot;&gt;$^{1}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface temperature of a star is determined by its mass and evolutionary phase and therefore by its age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars spend most of their lives on the core hydrogen burning main sequence. During this phase, a simple rule is that more massive stars have hotter surface temperatures and their spectra are more heavily weighted to shorter (bluer) wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The appearance of the sky is caused both by the spectrum of our Sun and the physiology of the eye. Light is scattered in our atmosphere according to the inverse fourth power of wavelength. At the same time, there is comparatively less violet light in the spectrum of sunlight at the top of the atmosphere &lt;em&gt;and&lt;/em&gt; the eye is less sensitive to that light (and some of it is absorbed, most of its ultraviolet wavelengths). It is the case that a hotter star than the Sun would have a spectrum richer in violet light and this would increase the amount of violet scattered light from the sky. Whether that would be sufficient for us to view the sky as purple, with our eye physiology, I am unsure; but probably not (see &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10510/why-are-there-no-green-stars?rq=1&quot;&gt;Why are there no green stars?&lt;/a&gt; ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More massive stars also &quot;evolve&quot; faster, because they burn their available fuel much faster. An &quot;evolved&quot; star is one which has finished fusing hydrogen in its core and moved on to &quot;shell&quot; burning around the core or burning heavier fuels in its core. Shell burning stars tend to be bigger and cooler than main sequence stars of the same mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^{1}$&lt;/span&gt; Increased magnetic activity can drastically increase the amount of light from a star at UV and X-ray wavelengths, by orders of magnitude. This is primarily a feature of comparatively low-mass stars (like the Sun or lower) and is driven by rapid rotation. Rapidly rotating (single) stars are young. This extra short wavelength light is not energetically important to the star, in that the vast majority of its radiation is still at visible and infrared wavelengths.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-06-02T04:12:03.653" LastActivityDate="2019-06-02T04:12:03.653" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32134" PostTypeId="2" ParentId="31747" CreationDate="2019-06-02T04:49:31.667" Score="4" Body="&lt;p&gt;Parallax and proper motion are determined from a &lt;em&gt;series&lt;/em&gt; of position measurements taken over the course of (for Gaia DR2) 22 months. A &quot;5-parameter&quot; astrometric model is fitted to these position measurements, consisting of a sky position at some epoch, a parallax and a proper motion in each of the celestial coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The precisions of each of these parameters improves as the number of measurements increases, but parallax and mean position precision are basically unaffected by the timespan of the observations (as long as they are taken over more than a year), whereas proper motion precision is improved by a longer timespan given the same number of observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a star had no proper motion, then a basic parallax measurement would be the difference between two (extreme) positions on the sky, and the parallax precision would be around &lt;span class=&quot;math-container&quot;&gt;$\sqrt{2}$&lt;/span&gt; of an individual position precision. The mean position of the star is something like the average of the two measurements, so might have a precision that is a factor 2 better. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, you have many more than 2 observations and measuring the parallax and mean position is like trying to fit a sine wave (with a DC offset), of known period (1 year), where the mean position is the DC offset and the parallax is the amplitude.  In such cases, the mean position has a variance of &lt;span class=&quot;math-container&quot;&gt;$\sigma^2/n$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; is the uncertainty in a single position measurement and &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is the number of observations. It took me a while to track it down, but it can be shown that the variance of your best estimate of the amplitude is &lt;em&gt;also&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$\sigma^2/n$&lt;/span&gt;, &lt;em&gt;when the frequency is known and &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is large&lt;/em&gt; (e.g. &lt;a href=&quot;https://pdfs.semanticscholar.org/df9b/0c876b488be088fd416387866cf8499fa438.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://pdfs.semanticscholar.org/df9b/0c876b488be088fd416387866cf8499fa438.pdf&lt;/a&gt;). &lt;strong&gt;Thus we expect the mean position and parallax to have similar uncertainties and for these to get smaller as &lt;span class=&quot;math-container&quot;&gt;$\sqrt{n}$&lt;/span&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proper motion precision is different. If we imagine a set of measurements where parallax can be ignored (e.g. at the same time every year), the proper motion precision is akin to the precision in the gradient of a line fitted to two or more points (with time as the x-axis). The precision improves with both the number and precision of the position points and with the timespan of observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to your question is therefore coincidence. You have picked on a magnitude of star for which the typical individual position measurement precision, typical number of measurements and typical timespan lead to a rough coincidence in the precision of the 5 parameters when proper motion is expressed in units of per year. For other magnitudes or for stars observed for an atypical number of times or duration, that isn't the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quoting the abstract of &lt;a href=&quot;https://www.aanda.org/articles/aa/abs/2018/08/aa32727-18/aa32727-18.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lindegren et al. (2018)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For the sources with five-parameter astrometric solutions, the median uncertainty in parallax and position at the reference epoch J2015.5 is about 0.04 mas for bright (G &amp;lt; 14 mag) sources, 0.1 mas at G = 17 mag, and 0.7 masat G = 20 mag. In the proper motion components the corresponding uncertainties are 0.05, 0.2, and 1.2 mas yr−1, respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-06-03T08:06:27.223" LastActivityDate="2019-06-03T08:06:27.223" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32135" PostTypeId="2" ParentId="32119" CreationDate="2019-06-02T05:12:30.047" Score="2" Body="&lt;p&gt;There are various ways of measuring distances in the universe which depend on the range in which we are measuring.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Parallax method: Up to 100 - 1000 ly&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Exploiting period luminosity relation of Cephied variables: Up to 3 - 10 Mpc&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Tully Fisher relationship: Intergalactic distances&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Hubble parameter: Large scale structures in the universe&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Magnitude measurement of type 1a supernova&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="25925" LastEditorUserId="14094" LastEditDate="2019-06-03T12:47:40.367" LastActivityDate="2019-06-03T12:47:40.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32136" PostTypeId="1" CreationDate="2019-06-02T10:02:18.203" Score="0" ViewCount="48" Body="&lt;p&gt;Where to find please theoretical values of radius and mass of O9.5 II star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean something like in &lt;a href=&quot;http://www.mpe.mpg.de/~martins/pub/calib_Ostar.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt;. However, there are only O I, O III and O V stars in this article.&lt;/p&gt;&#xA;" OwnerUserId="28268" LastEditorUserId="7472" LastEditDate="2019-06-02T10:32:16.373" LastActivityDate="2019-06-02T10:32:16.373" Title="Theoretical values of radius and mass of O9.5 II star" Tags="&lt;star&gt;&lt;stellar-astrophysics&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32138" PostTypeId="1" AcceptedAnswerId="32140" CreationDate="2019-06-02T19:48:46.323" Score="1" ViewCount="80" Body="&lt;p&gt;The only galaxies I can think of (not being an astronomer) are Andromeda and Milky Way. There are &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_nearest_galaxies&quot; rel=&quot;nofollow noreferrer&quot;&gt;51&lt;/a&gt; near galaxies, but they all pretty much say &quot;satellite of Milky way&quot; or &quot;satellite of Andromeda&quot;. There are &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_galaxies&quot; rel=&quot;nofollow noreferrer&quot;&gt;100k+&lt;/a&gt; galaxies in the local supercluster, and that page seems to have a better list:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Corvus    &lt;/li&gt;&#xA;&lt;li&gt;Coma Berenices&lt;/li&gt;&#xA;&lt;li&gt;Ursa Major    &lt;/li&gt;&#xA;&lt;li&gt;Virgo &lt;/li&gt;&#xA;&lt;li&gt;Sculptor  &lt;/li&gt;&#xA;&lt;li&gt;etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you had to rank them in order of prominence in the scientific community or in popular science, wondering what the top 10 or 20 galaxies would be (for which we have photos).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to come up with a list of images for educational purposes that are potentially somewhat familiar to laymen audiences, or which would be useful to introduce to laymen audiences.&lt;/p&gt;&#xA;" OwnerUserId="7620" LastEditorUserId="1891" LastEditDate="2019-06-03T13:11:49.343" LastActivityDate="2019-06-03T13:11:49.343" Title="What are the most popular galaxies for which we have images?" Tags="&lt;galaxy&gt;&lt;photography&gt;&lt;nasa&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2019-06-03T15:09:50.750" ContentLicense="CC BY-SA 4.0" />
  <row Id="32139" PostTypeId="1" CreationDate="2019-06-02T22:04:47.047" Score="2" ViewCount="198" Body="&lt;p&gt;It is unlikely that the largest supermassive black hole in the universe has already been discovered, there may be much larger ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there, or could there be, supermassive black holes large enough to swallow whole galaxies as opposed to the occasional star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there &lt;em&gt;other factors beyond size&lt;/em&gt; that would also affect a SMBH's ability to swallow it's own galaxy?&lt;/p&gt;&#xA;" OwnerUserId="28083" LastEditorUserId="7982" LastEditDate="2019-06-03T15:10:20.147" LastActivityDate="2019-06-04T07:53:38.493" Title="Could the largest SMBH swallow an entire galaxy?" Tags="&lt;supermassive-black-hole&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32140" PostTypeId="2" ParentId="32138" CreationDate="2019-06-02T23:22:49.183" Score="4" Body="&lt;p&gt;Any such list is going to be terribly subjective. Since I'm an astronomer who studies galaxies, I'll go ahead and throw out a subjective list of the more famous, photogenic, and/or scientifically well-studied galaxies. The first six are in the Local Group (LMC and SMC are satellites of the Milky Way, M32 is a satellite of Andromeda).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Milky Way&lt;/li&gt;&#xA;&lt;li&gt;Andromeda (M31)&lt;/li&gt;&#xA;&lt;li&gt;Large Magellanic Cloud&lt;/li&gt;&#xA;&lt;li&gt;Small Magellanic Cloud&lt;/li&gt;&#xA;&lt;li&gt;Triangulum (M33)&lt;/li&gt;&#xA;&lt;li&gt;M32&lt;/li&gt;&#xA;&lt;li&gt;Sombrero (M104)&lt;/li&gt;&#xA;&lt;li&gt;Pinwheel (M101)&lt;/li&gt;&#xA;&lt;li&gt;Whirlpool (M51a)&lt;/li&gt;&#xA;&lt;li&gt;M64 (Black Eye)&lt;/li&gt;&#xA;&lt;li&gt;M74 (NGC 628)&lt;/li&gt;&#xA;&lt;li&gt;M81&lt;/li&gt;&#xA;&lt;li&gt;M82 (Cigar)&lt;/li&gt;&#xA;&lt;li&gt;M87&lt;/li&gt;&#xA;&lt;li&gt;M100&lt;/li&gt;&#xA;&lt;li&gt;NGC 891&lt;/li&gt;&#xA;&lt;li&gt;NGC 1068 (M77)&lt;/li&gt;&#xA;&lt;li&gt;NGC 1300&lt;/li&gt;&#xA;&lt;li&gt;NGC 1365&lt;/li&gt;&#xA;&lt;li&gt;Centaurus A&lt;/li&gt;&#xA;&lt;li&gt;Cygnus A&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2019-06-02T23:22:49.183" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32141" PostTypeId="1" CreationDate="2019-06-03T00:40:08.710" Score="2" ViewCount="236" Body="&lt;p&gt;I have been looking for images of the planets and have found pretty decent ones of Earth and &lt;a href=&quot;https://solarsystem.nasa.gov/resources/688/computer-simulated-global-view-of-venus/?category=planets_venus&quot; rel=&quot;nofollow noreferrer&quot;&gt;Venus&lt;/a&gt; from NASA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/b4yO6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/b4yO6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That detail is opposed to this one of Uranus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yRZFy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yRZFy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And these of &lt;a href=&quot;https://www.spacetelescope.org/images/opo0124a/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mars&lt;/a&gt;, (and &lt;a href=&quot;https://www.spacetelescope.org/images/opo0745a/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;https://www.spacetelescope.org/images/opo0534a/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3PA64.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3PA64.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am disappointed in the quality of the public domain images from NASA I have found for these planets:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mars&lt;/li&gt;&#xA;&lt;li&gt;Jupiter&lt;/li&gt;&#xA;&lt;li&gt;Saturn&lt;/li&gt;&#xA;&lt;li&gt;Uranus&lt;/li&gt;&#xA;&lt;li&gt;Neptune&lt;/li&gt;&#xA;&lt;li&gt;Pluto&lt;/li&gt;&#xA;&lt;li&gt;Earth's Moon&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am happy with the ones found for:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mercury&lt;/li&gt;&#xA;&lt;li&gt;Venus&lt;/li&gt;&#xA;&lt;li&gt;Earth&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They seem to be high quality in terms of detail and pixel size (in the MBs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wondering if one could explain how I go about finding these public domain planet images from NASA (or elsewhere, but I imagine they are all from the Hubble telescope, I don't know too much) that are high quality in terms of &lt;strong&gt;detail&lt;/strong&gt; and &lt;strong&gt;pixels/size&lt;/strong&gt;. If nothing else it would be helpful to simply link to them if there is no pattern to where they are located.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a pattern to where the high quality images are located, I would be interested too to find ones of comets/asteroids/stars. I have found many high quality images of galaxies which is neat, but not of the planets yet.&lt;/p&gt;&#xA;" OwnerUserId="7620" LastActivityDate="2019-06-03T22:19:55.510" Title="High quality, detailed images of planets on web" Tags="&lt;planet&gt;&lt;photography&gt;&lt;nasa&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32142" PostTypeId="1" CreationDate="2019-06-03T01:04:35.417" Score="2" ViewCount="237" Body="&lt;p&gt;Hi I am a novice in the field of astronomy. I want to find out the dates for lunar and solar eclipse 9000-11000 yrs ago. I found out that Stellarium is based on VSOP87 which is recommended for 4000 yrs ago. However, the software allows one to use DE431 which can provide planet ephemerides upto 13000 yrs ago. I am wondering if I can calculate eclipses observed in India accurately using DE431.&lt;/p&gt;&#xA;" OwnerUserId="26375" LastActivityDate="2019-06-03T01:04:35.417" Title="Stellarium software with DE431 for eclipse 10000 yrs ago" Tags="&lt;solar-eclipse&gt;&lt;lunar-eclipse&gt;" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="32143" PostTypeId="1" CreationDate="2019-06-03T04:40:52.930" Score="4" ViewCount="306" Body="&lt;p&gt;In &lt;a href=&quot;https://en.wikipedia.org/wiki/Eternal_inflation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eternal Inflation Theory&lt;/a&gt;, there exists the idea of multiple &quot;bubble&quot; Universes that grow and according to this &lt;a href=&quot;https://iopscience.iop.org/article/10.1088/1475-7516/2009/04/025/pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;theory&lt;/a&gt;, can &quot;collide&quot; with each other. The paper suggests that you can search for evidence that our Universe collided with other &quot;bubble&quot; Universes, by looking for an imprint on the CMB radiation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the concept of a Universe existing as a bubble that moves relative to other universes makes no sense:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How can you have a space &quot;outside&quot; of ours that allows the Universes to&#xA;move relative to each other?&lt;/li&gt;&#xA;&lt;li&gt;In what dimension are the other &quot;Universes&quot; moving relative to ours?&lt;/li&gt;&#xA;&lt;li&gt;If time started in our Universe at the Big Bang, then in what time dimension (outside of ours) do collisions between Universes occur?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;PBS Spacetime looks at this idea here: &lt;a href=&quot;https://www.youtube.com/watch?v=lrS0rxX-UhA&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=lrS0rxX-UhA&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: A winning answer will answer all three points above.&lt;/p&gt;&#xA;" OwnerUserId="28132" LastEditorUserId="28132" LastEditDate="2019-08-03T19:06:46.937" LastActivityDate="2020-12-07T09:28:22.310" Title="How can cosmological bubble universes &quot;collide&quot; with each other?" Tags="&lt;universe&gt;&lt;cmb&gt;&lt;multiverse&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32144" PostTypeId="2" ParentId="32129" CreationDate="2019-06-03T10:50:05.517" Score="-1" Body="&lt;p&gt;To put the matter at its simplest, imagine an inflated balloon with a number of dots scattered on its surface. The balloon represents the universe and the dots are the galaxies. Inflate the balloon further and you will see that an ant placed on one of the dots would see himself as being at rest with all the other dots moving away, and with the furthest dots moving away fastest of all. The ant, of course represents you, and the expansion of the ballon represents the expansion of the universe. An ant placed on any other dot would also see himself as being at rest, with your dot moving away from him with a speed proportional to its distance. The universe deviates from this analogy inasmuch as the galaxies which are very close to us, Andromeda for example, have a proper motion which masks cosmic expansion and will sometimes produce a blueshift instead of a redshift. &lt;/p&gt;&#xA;" OwnerUserId="28083" LastEditorUserId="14094" LastEditDate="2019-06-03T12:06:16.940" LastActivityDate="2019-06-03T12:06:16.940" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32145" PostTypeId="1" CreationDate="2019-06-03T12:29:14.820" Score="1" ViewCount="115" Body="&lt;p&gt;So we have data for a globular cluster M13. we have plotted the radius in pixels against the surface brightness in magnitudes per square arcsec. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to find the king model for this. We dont exactly know how to do that. We think we just have to set the units to solar luminosities and then plot it again but if that is the case we dont really know what our graph tells us that the first graph did not tell us. So we think we are doing that wrong. If we are acually doing it right we dont know what this tells us. Can somebody help?&lt;/p&gt;&#xA;" OwnerUserId="28274" LastEditorUserId="1891" LastEditDate="2019-06-03T13:09:15.997" LastActivityDate="2019-06-03T13:09:15.997" Title="How do we find the king model of a globular cluster?" Tags="&lt;globular-clusters&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32146" PostTypeId="1" AcceptedAnswerId="32151" CreationDate="2019-06-03T13:05:42.260" Score="2" ViewCount="327" Body="&lt;p&gt;I guess this is more a question about stellar models than anything else. I was wondering what is predicted to be the hottest possible stars that would still be hydrogen-burning. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What complicates this more is that the most massive stars will probably only sit close to the main sequence (as very early O-type) for several thousand years before gaining a WNh spectrum despite still being relatively early in the lifetime. So I'll be somewhat more precise and ask what is the hottest possible star predicted at ZAMS that won't just blow itself apart by radiation pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also tangential to the topic but have there been any stars found with an O1 spectrum, and is this even a spectral type that has models made for it?  &lt;/p&gt;&#xA;" OwnerUserId="23212" LastActivityDate="2019-06-06T16:38:20.107" Title="Hottest Possible Hydrogen-Fusing Stars" Tags="&lt;star&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32147" PostTypeId="1" CreationDate="2019-06-03T14:19:36.377" Score="4" ViewCount="112" Body="&lt;p&gt;Why is it that sunrise and sunset occur at different times when compared between the vernal and autumnal equinoxes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider Pontianak, Indonesia, 0 deg N, for simplicity. Timeanddate.com gives sunrise and sunset times 5:46 AM, 5:53 PM at 3/21 (vernal equinox) and 5:31 AM, 5:38 PM at 9/23 (autumnal equinox). Given the symmetry of the problem, I would expect that sunrise would occur at the exact same time during the two events. Same for sunset. What is the reason for the 15 minute difference between the time sunrise occurs on the vernal equinox and the time sunrise occurs on the autumnal equinox?   &lt;/p&gt;&#xA;" OwnerUserId="28275" LastActivityDate="2019-06-09T10:22:06.527" Title="Sunrise and sunset times when compared between the vernal and autumnal equinoxes" Tags="&lt;equinox&gt;&lt;sunrise&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32148" PostTypeId="2" ParentId="32147" CreationDate="2019-06-03T14:33:38.460" Score="4" Body="&lt;p&gt;Instead of looking at sunset/sunrise times, it's easier to look at noon, which is right in the middle between sunset and sunrise: 11:50am in March and 11:35am in September.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that noon doesn't always happen at the same time of the day (when measured with steady clocks) is expressed by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Equation_of_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;Equation of Time&lt;/a&gt;. Two factors play a role here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The plane in which Earth rotates around the Sun (the ecliptic) is tilted with respect to the one it rotates around its axis (the equator)&lt;/li&gt;&#xA;&lt;li&gt;The Earth's orbit around the Sun is an ellipse, not a circle, and it moves faster than average in its orbit from (roughly) October up to and including March, and slower in the other months&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7472" LastEditorUserId="7982" LastEditDate="2019-06-03T14:52:00.503" LastActivityDate="2019-06-03T14:52:00.503" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32149" PostTypeId="2" ParentId="32147" CreationDate="2019-06-03T15:01:14.867" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Given the symmetry of the problem...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's the key to the answer. While we like to imagine motion in the system as having some symmetries, it's all a bit wobbly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @Glorfindel &lt;a href=&quot;https://astronomy.stackexchange.com/a/32148/7982&quot;&gt;points out&lt;/a&gt; the particular asymmetry is connected to the direction that Earth's tilted axis points relative to the shape of Earth's elliptical orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If the Earth's orbit&lt;/strong&gt; were a circle and you were using local sidereal time, then maybe you would have your symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But it's not&lt;/strong&gt;, it's an ellipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But if&lt;/strong&gt; Earth reached it's perihelion on January 1 or thereabouts every year, you might still have most of the symmetry you're expecting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But it doesn't&lt;/strong&gt;. &lt;a href=&quot;http://www.astropixels.com/ephemeris/perap2001.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;It tends to wobble&lt;/a&gt; around in the first week of January by several days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So since these two directions don't quite match up (and there's no reason to expect them to either) the imperfect symmetry yields times that are close, but don't quite match. &lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-06-09T10:22:06.527" LastActivityDate="2019-06-09T10:22:06.527" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32151" PostTypeId="2" ParentId="32146" CreationDate="2019-06-03T16:25:52.947" Score="8" Body="&lt;p&gt;&lt;em&gt;An&lt;/em&gt; answer to your question is contained within &lt;a href=&quot;https://astronomy.stackexchange.com/questions/11594/what-is-the-largest-hydrogen-burning-star?rq=1&quot;&gt;What is the largest hydrogen-burning star?&lt;/a&gt; The hottest observed main sequence stars are of type O3V, with photospheric temperatures of about 50,000 K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is indeed possible that hotter main sequence stars may exist in the present-day universe, but have simply evolved into Wolf-Rayet stars (and lost a lot of mass). Indeed, &lt;a href=&quot;https://arxiv.org/abs/1007.3284&quot; rel=&quot;noreferrer&quot;&gt;Crowther et al. (2010)&lt;/a&gt; claim to see evidence for such objects in the cluster R136 in the Large Magellanic Cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How hot could such objects be? A theoretical study by &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0007248&quot; rel=&quot;noreferrer&quot;&gt;Bromm et al. (2001)&lt;/a&gt; suggests that very massive, metal-rich present-day stars might reach 65,000 K at a few hundred solar masses (see their Fig.1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hottest zero age main sequence stars ever would probably be very massive (&lt;span class=&quot;math-container&quot;&gt;$\sim 1000M_{\odot}$&lt;/span&gt;) ultra-metal poor, or even metal-free population III stars. These actually begin life as He burning stars until they produce enough carbon to commence the CNO H-burning cycle. Equation 6 and Fig.1 in &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0007248&quot; rel=&quot;noreferrer&quot;&gt;Bromm et al. (2001)&lt;/a&gt;, who provide a theoretical study of such objects, suggests effective temperatures of about &lt;span class=&quot;math-container&quot;&gt;$1.1\times 10^{5}$&lt;/span&gt; K for such stars, with very little detailed dependence on mass and metallicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess such a star might be classified as O1, but according to Bromm et al. they are basically blackbodies once the mass exceeds about 300 solar masses.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-06-03T16:25:52.947" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32152" PostTypeId="2" ParentId="32139" CreationDate="2019-06-03T21:17:09.797" Score="9" Body="&lt;p&gt;This question might be related to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/24196/is-there-a-maximum-size-for-a-black-hole&quot;&gt;Is there a maximum size for a black hole?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Typical SMBHs do not contribute much to the overall mass of a galaxy, much less than the sun is of the total mass of our solar system. And as we can see, even such mass centered quite nearby does not swallow all objects in the solar system very easily.&lt;/p&gt;&#xA;&lt;p&gt;While there is no theoretical maximum, in practice - based on simulations - maximum mass for a black hole seems to be around 50 billion solar masses&lt;sup&gt;†&lt;/sup&gt;, which will give a radius of 0.01 light years. This is very small compared to the size of the galaxy, so it will not affect the situation much. Thus the main effect of huge black hole vs galaxy would be via plain old gravitation.&lt;/p&gt;&#xA;&lt;p&gt;The main impediment for a black hole to swallow anything is angular momentum. Any object a black hole is going to consume must shed its angular momentum in order to get closer to event horizon. This is not very easy thing to do, and the main reason black holes are surrounded by discs of extremely hot gas - friction will turn angular momentum into heat, allowing matter to descend towards black hole.&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;†&lt;/sup&gt;practical maximum of 50 billion suns is due to fact that accretion disc becomes unstable due to new stars forming from it. See for example Andrew King's 2015 paper &lt;a href=&quot;https://arxiv.org/abs/1511.08502&quot; rel=&quot;nofollow noreferrer&quot;&gt;How Big Can a Black Hole Grow?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Abstract:&lt;/p&gt;&#xA;&lt;p&gt;I show that there is a physical limit to the mass of a black hole, above which it cannot grow through luminous accretion of gas,  and so cannot appear as a quasar or active galactic nucleus. The limit is &lt;span class=&quot;math-container&quot;&gt;$M_{max} ≃ 5 \times 10^{10}$&lt;/span&gt; M⊙ for typical parameters, but can reach &lt;span class=&quot;math-container&quot;&gt;$M_{max} ≃ 2.7 \times 10^{11}$&lt;/span&gt; M⊙ in extreme cases (e.g. maximal prograde spin). The largest black hole masses so far found are close to but below the limit. The Eddington luminosity &lt;span class=&quot;math-container&quot;&gt;$≃ 6.5 \times 10^{48} \text{erg s}^{-1}$&lt;/span&gt; corresponding to &lt;span class=&quot;math-container&quot;&gt;$M_{max}$&lt;/span&gt; is remarkably close to the largest AGN bolometric luminosity so far observed. The mass and luminosity limits both rely on a reasonable but currently untestable hypothesis about AGN disc formation, so future observations of extreme SMBH masses can therefore probe fundamental disc physics. Black holes can in principle grow their masses above &lt;span class=&quot;math-container&quot;&gt;$M_{max}$&lt;/span&gt;  by non–luminous means such as mergers with other holes, but cannot become luminous accretors again. They might nevertheless be detectable in other ways, for example through gravitational lensing. I show further that black holes with masses ∼ &lt;span class=&quot;math-container&quot;&gt;$M_{max}$&lt;/span&gt;  can probably grow above the values specified by the black–hole – host–galaxy scaling relations, in agreement with observation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10813" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-06-04T07:53:38.493" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32154" PostTypeId="2" ParentId="32141" CreationDate="2019-06-03T22:19:55.510" Score="3" Body="&lt;p&gt;The best NASA images of planets are not from the &lt;em&gt;Hubble Space Telescope&lt;/em&gt;; they are from the spacecraft that flew by or went into orbit around them. For Uranus and Neptune, this means Voyager 2; for Pluto, this means New Horizons. For Jupiter, this means primarily Voyagers 1 and 2, and especially Galileo and Juno; for Saturn, Voyagers 1 and 2 and (especially) Cassini.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you google for &quot;[planet-name] planet gallery&quot;, you should get some pretty good images that way; also try adding the name of a relevant spacecraft (if you're stuck, check the Wikipedia page for each planet to learn which spacecraft visited them). A few NASA web pages:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://photojournal.jpl.nasa.gov/mission/Viking&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://photojournal.jpl.nasa.gov/mission/Viking&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://solarsystem.nasa.gov/planets/saturn/galleries&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://solarsystem.nasa.gov/planets/saturn/galleries&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://solarsystem.nasa.gov/planets/neptune/galleries&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://solarsystem.nasa.gov/planets/neptune/galleries&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://solarsystem.nasa.gov/planets/dwarf-planets/pluto/galleries&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://solarsystem.nasa.gov/planets/dwarf-planets/pluto/galleries&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and so forth. (Note that Uranus, even when seen by Voyager 2, is pretty featureless and boring, at least at visible wavelengths.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also try &lt;a href=&quot;https://apod.nasa.gov/cgi-bin/apod/apod_search&quot; rel=&quot;nofollow noreferrer&quot;&gt;searching at the Astronomy Picture of the Day site&lt;/a&gt;. (Not all of those are necessarily public domain, thoug.)&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2019-06-03T22:19:55.510" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32155" PostTypeId="1" CreationDate="2019-06-03T23:50:01.270" Score="2" ViewCount="82" Body="&lt;p&gt;While link-clicking for &lt;a href=&quot;https://space.stackexchange.com/q/36553/12102&quot;&gt;Where did Herschel Space Telescope go in 2013?&lt;/a&gt; I ran across the 2009 Time Magazine article &lt;a href=&quot;http://content.time.com/time/health/article/0,8599,1898174,00.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Two Telescopes to Measure the Big Bang&lt;/a&gt; which shows the file photo below with the caption:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Herschel's 3.5-m diameter mirror is checked with ultraviolet light at Europeís Spaceport in Kourou, French Guiana T VALLEE/AFP/Getty&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Herschel_Space_Observatory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Herschel Space Observatory&lt;/a&gt; the telescope covers a wavelength range from far-infrared to sub millimeter waves; 55–672 µm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why then check the mirror's figue with a wavelength 200 times smaller in the ultraviolet? Even visible light would have been more convenient, and there should be several IR bands where coherent sources and imaging detectors work even in the humid French Guiana climate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LCqM4.jpg&quot; alt=&quot;Herschel&amp;#39;s 3.5-m diameter mirror is checked with ultraviolet light at Europeís Spaceport in Kourou, French Guiana T VALLEE/AFP/Getty&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-06-03T23:50:01.270" Title="Why look at an infrared telescopes's mirror with ultraviolet light? (Herschel Space Observatory)" Tags="&lt;telescope&gt;&lt;optics&gt;&lt;space-telescope&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32157" PostTypeId="1" CreationDate="2019-06-04T10:57:21.500" Score="0" ViewCount="798" Body="&lt;p&gt;Unlike its sister planet Uranus,Neptune has a substantial internal heat source which gives rise to a dynamic atmosphere &amp;amp; weather. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Neptune#Internal_heating&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia says&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;As with Uranus, the source of this heating is unknown&lt;/strong&gt;, but the discrepancy is larger: Uranus only radiates 1.1 times as much energy as it receives from the Sun; whereas Neptune radiates about 2.61 times as much energy as it receives from the Sun. Neptune is the farthest planet from the Sun, yet its internal energy is sufficient to drive the fastest planetary winds seen in the Solar System. Depending on the thermal properties of its interior, the heat left over from Neptune's formation may be sufficient to explain its current heat flow, though it is more difficult to simultaneously explain Uranus's lack of internal heat while preserving the apparent similarity between the two planets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Have plausible theories been proposed for the source of the heat necessary to keep Neptune's ocean liquid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could there be sufficient methane in the atmosphere to produce a significant greenhouse effect?&lt;/p&gt;&#xA;" OwnerUserId="28083" LastEditorUserId="7982" LastEditDate="2019-06-04T23:54:06.400" LastActivityDate="2019-06-06T12:17:45.110" Title="Neptune's Internal Heat Source" Tags="&lt;neptune&gt;&lt;planetary-science&gt;&lt;outer-solar-system&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32158" PostTypeId="1" CreationDate="2019-06-04T11:10:37.530" Score="2" ViewCount="368" Body="&lt;p&gt;If a binary star system has a non-zero inclination, how would that show up in a graph of radial velocity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could I then extract the inclination from the graph?&lt;/p&gt;&#xA;" OwnerUserId="16506" LastEditorUserId="7982" LastEditDate="2019-06-04T23:45:21.963" LastActivityDate="2019-06-06T10:47:15.387" Title="How does one calculate inclination from a graph of radial velocity?" Tags="&lt;binary-star&gt;&lt;inclination&gt;&lt;radial-velocity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32159" PostTypeId="2" ParentId="32158" CreationDate="2019-06-04T16:08:04.047" Score="6" Body="&lt;p&gt;From its RV graph alone, with no other information, you cannot calculate inclination.  This is why RV velocity measurements are typically reported as &quot;&lt;span class=&quot;math-container&quot;&gt;$v \sin{i}$&lt;/span&gt;&quot;, because what you're actually measuring is the orbital velocity projected along the line of sight.  Without other information, you cannot disentangle the orbital velocity from the viewing angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What additional information could be had that would allow you to disentangle?  The following list is probably not exhaustive.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you can observe the two stars as separate objects, then you can track their relative positions through an orbit and can measure inclination that way.&lt;/li&gt;&#xA;&lt;li&gt;If you have &quot;light curves&quot; (brightness measurements over time) for the star(s) and you see periodic eclipses corresponding to the period and correct epoch from the radial velocity data, then you know the stars are aligned such that they periodically have their light blocked by one, the other, or both.  In such a case, you know that the inclination must be close to 90 degrees else the eclipses could not occur.&lt;/li&gt;&#xA;&lt;li&gt;If you have reliable distance measurements to the stars and determine somehow that they are main sequence stars, you can determine the star's masses via theoretical stellar evolution models and calculations given a measurement of absolute luminosity (from a measurement of the apparent luminosity coupled with the distance measure).  Once you know the star's masses, you can calculate the expected orbital velocity, compare with your &lt;span class=&quot;math-container&quot;&gt;$v\sin{i}$&lt;/span&gt; measure, and determine the inclination.&lt;/li&gt;&#xA;&lt;li&gt;If the stars are not main sequence stars in the above scenario, then you'll need some additional information (their ages should be sufficient) to correctly extract a mass from their apparent luminosity &lt;em&gt;and&lt;/em&gt; color.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="128" LastActivityDate="2019-06-04T16:08:04.047" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="32160" PostTypeId="1" CreationDate="2019-06-04T17:05:10.523" Score="3" ViewCount="794" Body="&lt;p&gt;I will be going to the desert this month and it's my understanding that the night sky will be very dark there. I've got some practice locating the planets but I'd really like to get a look at a nebula or galaxy. I understand from &lt;a href=&quot;https://astronomy.stackexchange.com/questions/27780/is-the-celestron-astromaster-130-eq-model-able-to-see-nebulas-and-galaxies&quot;&gt;Is the Celestron astromaster 130 EQ Model able to see nebulas and galaxies?&lt;/a&gt; that nebulas and galaxies are visible with modest magnification. I'd really appreciate some tips on locating stuff that's cool to look at other than planets and the moon. I've got an &lt;a href=&quot;https://www.celestron.com/products/astromaster-114eq-telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;AstroMaster 114&lt;/a&gt; (114mm, f/8.77) with a couple of eyepieces and a Barlow lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've searched a bit and these targets seem like a good short list.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Albireo&lt;/li&gt;&#xA;&lt;li&gt;Orion Nebula, M42&lt;/li&gt;&#xA;&lt;li&gt;Andromeda Galaxy, M31&lt;/li&gt;&#xA;&lt;li&gt;Hercules Cluster, M13&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I understand that my telescope has an equatorial mount and I do take care to orient it northwards and adjust it to match my latitude, but I've never truly bothered to center it on the north star. I find locating the north star a bit daunting. All the instructions I've encountered say to locate the big dipper (aka the plow) and follow the line of two if its stars. Finding constellations is not my strong suit -- especially with the haze and obstructions here in Los Angeles. I can usually find Orion pretty well, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an app on my iphone called &lt;a href=&quot;https://itunes.apple.com/us/app/planets/id305793334&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planets&lt;/a&gt; which I find quite helpful for locating the planets, but I'm not sure if it will operate properly out there where there's no telephone service. It also doesn't provide any tools to identify or locate deep space objects. It's just for planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My telescope does have some &quot;setting circles&quot; (little dials demarcated in degrees or hours, see photo) but my understanding of ascension and declination is sketchy at best and I also understand that these are not especially accurate. I'm wondering if they might have some use to get me in the ball park of the objects I'm seeking -- maybe help me at least find the constellation from which I can locate the deep space objects of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to summarize, I'm hoping for tips on locating the north star and establishing a proper baseline position for my telescope and I'm also hoping for good, basic instructions for locating some interesting, non-planetary objects to view in my scope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4Rwfz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4Rwfz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16705" LastEditorUserId="7982" LastEditDate="2019-06-04T23:18:44.430" LastActivityDate="2019-06-09T03:48:27.810" Title="How to locate nebulas visible through AstroMaster 114 telescope without device assistance?" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;nebula&gt;&lt;deep-sky-observing&gt;&lt;orion&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32161" PostTypeId="1" CreationDate="2019-06-04T17:54:01.040" Score="2" ViewCount="49" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\delta^2 u}{\delta \theta^2} + u = \frac{\mu}{h^2} + 3 \mu u^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is &lt;a href=&quot;https://en.wikipedia.org/wiki/Binet_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Binet's relativistic equation&lt;/a&gt;. Here &lt;span class=&quot;math-container&quot;&gt;$u = 1/r$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; is the radial distance, and &lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt; is the reduced mass of the Sun and Mercury.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to calculate the perihelion precession path of Mercury &lt;a href=&quot;https://astronomy.stackexchange.com/q/32067/7982&quot;&gt;numerically in Python&lt;/a&gt; and for doing so I would like to use a dimensionless form of this expression.&lt;/p&gt;&#xA;" OwnerUserId="28172" LastEditorUserId="28172" LastEditDate="2019-06-05T06:33:49.123" LastActivityDate="2019-06-05T06:33:49.123" Title="How to write a normalized, dimensionless form of Binet's relativistic equation?" Tags="&lt;orbit&gt;&lt;astrophysics&gt;&lt;mercury&gt;&lt;precession&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32163" PostTypeId="2" ParentId="4755" CreationDate="2019-06-04T18:55:55.207" Score="0" Body="&lt;p&gt;According to this &lt;a href=&quot;https://en.wikipedia.org/wiki/Lunar_phase&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article on Lunar phase&lt;/a&gt;, 0.1% is the minimum percentage. However, at the very best I've observed 0.17% with my naked eye.&lt;/p&gt;&#xA;" OwnerUserId="28283" LastEditorUserId="7472" LastEditDate="2019-06-04T19:16:18.963" LastActivityDate="2019-06-04T19:16:18.963" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32164" PostTypeId="2" ParentId="32160" CreationDate="2019-06-04T19:04:19.247" Score="3" Body="&lt;p&gt;If you have a Laptop or Notebook you can download one of several planetarium programs like Stellarium or Cartes du Ciel. They will display large areas of the sky that will help orient you initially. Zoom the computer view to match the view in your finder then zoom in again and get a view on the laptop that is close to the view in your eyepiece. Compare the star field views between the laptop and telescope then move the telescope accordingly. Many years ago there was a method of finding objects called star hopping. This is basically the same procedure but with a bit of assistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Patience, practice and persistence. Get hold of as much as you can before heading out to the desert. Nothing else will help you as much as those.&lt;/p&gt;&#xA;" OwnerUserId="24928" LastActivityDate="2019-06-04T19:04:19.247" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32165" PostTypeId="2" ParentId="32160" CreationDate="2019-06-04T20:59:22.177" Score="4" Body="&lt;p&gt;Those targets are good choices for beginners at the right times of year.&#xA;However, in June M42 is a daytime object, and M31 rises in the wee hours, leaving only M13 and Albireo in the evening.&#xA;Other good summer targets include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;M8, the Lagoon Nebula&lt;/li&gt;&#xA;&lt;li&gt;M11, the Wild Duck Cluster&lt;/li&gt;&#xA;&lt;li&gt;M57, the Ring Nebula&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Precise polar alignment is only required if you plan to take long exposure photographs.&#xA;Setting circles help when you can't see much of the sky at once, e.g. under an observatory dome or partial cloud cover.&#xA;In order to be useful, the right ascension circle must be reset each observing session, or more frequently if your mount isn't motorized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://skymaps.com/downloads.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Skymaps.com Evening Sky Map&lt;/a&gt; can help you find your way around the night sky.&#xA;It shows the whole sky as it appears 2-3 hours after sunset, including several deep-sky objects visible with a 100mm or larger telescope.&#xA;A multi-page atlas such as the &lt;a href=&quot;http://www.shopatsky.com/pocket-sky-atlas&quot; rel=&quot;nofollow noreferrer&quot;&gt;S&amp;amp;T Pocket Sky Atlas&lt;/a&gt; provides enough detail to actually get those objects in the telescope's field of view.&#xA;A little red light for the maps minimizes the impact on your night vision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With practice, you can learn to locate any target of interest by triangulation, first among the bright stars, then among the faint stars in between.&#xA;For example, M13 is about 60% of the way from Arcturus to Vega and 30% of the way from &amp;eta; Her to &amp;zeta; Her.&#xA;A detailed atlas also shows two 7th magnitude stars each &amp;frac14; degree away from M13 in an obtuse triangle.&#xA;If the object does not at first appear in the telescope field, carefully sweep the surrounding area.&#xA;Some mounts provide fine motion controls to help with this.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-06-09T03:48:27.810" LastActivityDate="2019-06-09T03:48:27.810" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32167" PostTypeId="1" CreationDate="2019-06-05T08:45:18.930" Score="3" ViewCount="173" Body="&lt;p&gt;Is &lt;a href=&quot;https://en.wikipedia.org/wiki/Titius%E2%80%93Bode_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bode's Law&lt;/a&gt; just a series of remarkable coincidences,or would it hold good for other solar systems where the star is similar to our sun?&lt;/p&gt;&#xA;" OwnerUserId="28083" LastEditorUserId="37191" LastEditDate="2020-12-26T22:59:37.767" LastActivityDate="2021-02-18T19:55:45.420" Title="How Reliable is Bode's Law?" Tags="&lt;exoplanet&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32168" PostTypeId="2" ParentId="32157" CreationDate="2019-06-05T12:53:58.063" Score="4" Body="&lt;p&gt;Neptune and Uranus have estimated core temperatures of 7273 K and 5255 K respectively. Uranus and Neptune however likely do not have liquid water oceans. The combination of pressure and temperature on both planet results in unsuitable conditions for liquid water oceans. In fact for liquid water oceans to form on Neptune, it's predicted it would have to cool down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paper on this at: &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0609723&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cornell University&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Irrelevant, but fun) Diamonds oceans might exist. Uranus and Neptune are made up of 10% carbon some of which is methane. Methane when exposed to the conditions available in Uranus and Neptune can form diamond. Diamond can melt and form liquid diamond under again, the conditions on Neptune and Uranus. Therefore there could potentially be diamond oceans on Neptune and Uranus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paper on this at; &lt;a href=&quot;https://www.nature.com/articles/nmat1863&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nature Materials&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two plausible theories considered for Neptune source of heat is that as you said, the methane concentration in the atmosphere is high enough to retain the formation heat after the formation of the planet. The other one is that the tidal forces generated by Triton's retrograde degrading elliptical orbit, but the paper on it concludes that it doesn't explain the difference between Neptune and Uranus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to consider is that Uranus might be cooler than it should be rather than Neptune being warmer. This could be a result of an Earth sized &#xA;indirect collision earlier in its lifetime. The collision causes this decrease in emitted temperature because it deposited a significant amount of material around the core which made heat transfer from the core to the outer parts of the planet much slower. Note that if this collision were to occur, it is what would've caused the unusual tilt of 97.7°. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paper on this collision at: &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/1538-4357/aac725/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;The American Astronomical Society&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the second question, the methane in the atmosphere of Neptune, as covered before, can produce a significant greenhouse effect.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Neptune emits 2.7 times more energy than it receives from the Sun. This access energy powers the atmosphere to produce the storms that are not seen on its twin planet Uranus. The source of internal energy cannot be due solely to leftover energy from formation (i.e. Jupiter) since Neptune is smaller and would have radiated away the energy long ago. Nor is it due to an unusual chemical change, such as the helium rain for Saturn. Rather, it appears that Neptune is more efficient at trapping leftover formation heat due to the fact that methane is highly abundant in Neptune's atmosphere, and methane is an excellent insulator of heat (i.e. the greenhouse effect). Neptune has a sub-zero type greenhouse effect that is trapping formation heat that should have been radiated billions of years ago like Uranus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This extract is from: &#xA;&lt;a href=&quot;http://abyss.uoregon.edu/~js/ast121/lectures/lec20.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;University of Oregon&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In regards to why Uranus atmospheric methane doesn't appear to have as much of an effect as on Neptune, it could be that either Neptune simply has more methane in it's atmosphere or that a combination of factors, some of which mentioned above causes the difference. Beyond that I do not know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paper on Tritons Tidal Heating:&#xA;&lt;a href=&quot;http://articles.adsabs.harvard.edu//full/1974ApJ...193..477T/0000480.000.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;University of Texas&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="28286" LastEditorUserId="28286" LastEditDate="2019-06-06T12:17:45.110" LastActivityDate="2019-06-06T12:17:45.110" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32169" PostTypeId="1" CreationDate="2019-06-05T13:06:08.600" Score="1" ViewCount="40" Body="&lt;p&gt;Is there an online source where historic observational data regarding the position of the planets can be found? I was always intrigued by the fact that Neptune was mathematically predicted before it was actually observed (&lt;a href=&quot;https://en.wikipedia.org/wiki/Discovery_of_Neptune&quot; rel=&quot;nofollow noreferrer&quot;&gt;wiki article)&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curios in finding more about this mathematical prediction regarding the existence of Neptune. Here are a few questions related to this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Are there any detailed references on the subject? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is it possible to find the data which lead to this discovery online somewhere? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Are there online resources regarding the position of the planets (now or at various points in the past)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="28288" LastActivityDate="2019-06-05T13:06:08.600" Title="Historic observational data - position of the planets" Tags="&lt;planet&gt;&lt;data-analysis&gt;&lt;raw-data&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32170" PostTypeId="1" CreationDate="2019-06-05T13:36:22.733" Score="1" ViewCount="160" Body="&lt;p&gt;I watched an &lt;a href=&quot;https://youtu.be/TgA2y-Bgi3c?t=435&quot; rel=&quot;nofollow noreferrer&quot;&gt;interview with Neil deGrasse Tyson on The Late Show with Stephen Colbert&lt;/a&gt; where he briefly mentioned a disturbing thought. It is something that I have never thought of before, and the idea borders on the eerie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is this (and here comes a mouthful): If all our knowledge of the Universe is based on observations of distant galaxies (and Milkyway stars), and we accept that the Universe is expanding at such an accelerating rate that at some point in the future only our nearest stars will be visible to whoever inhabits Earth, and that then those future inhabitants will need to explain the Universe based only on what they can observe, then it is possible that we are today trying to explain the Universe based on largely incomplete data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like, imagine that there is much more stuff out there, stuff that could help us explain some of the scientific &quot;mysteries&quot; such as dark energy/matter, but it is beyond our horizon. We don't know what is missing from our view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so I guess the question is this: &lt;strong&gt;How does this affect the validity of our current models of the Universe?&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="28287" LastEditorUserId="1891" LastEditDate="2019-06-05T13:54:06.263" LastActivityDate="2019-06-05T21:27:30.273" Title="How much of the Universe is invisible to us, and how does it affect our theories?" Tags="&lt;observable-universe&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32171" PostTypeId="1" CreationDate="2019-06-05T16:41:47.610" Score="3" ViewCount="102" Body="&lt;p&gt;I've seen multiple sources repeat the information that the &lt;a href=&quot;https://en.wikipedia.org/wiki/Deep_Space_Atomic_Clock&quot; rel=&quot;nofollow noreferrer&quot;&gt;Deep Space Atomic Clock&lt;/a&gt; will 'at first' orbit Earth, before moving on out of Earth orbit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Deep Space Atomic Clock (DSAC) is a miniaturized, ultra-precise&#xA;  mercury-ion atomic clock for precise radio navigation in deep space.&#xA;  It is orders of magnitude more stable than existing navigation clocks,&#xA;  and has been refined to limit drift of no more than 1 nanosecond in 10&#xA;  days.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Deep_Space_Atomic_Clock&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;: &quot;Its development will include a test flight in low-Earth orbit&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.jpl.nasa.gov/news/news.php?feature=7415&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL&lt;/a&gt;: &quot;the instrument will be tested in Earth orbit for one year, with the goal of being ready for future missions to other worlds&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would it then orbit? The Sun? Or some L point? Other planets? Or will multiple ones orbit all of the above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: for those wanting to close based on &quot;not astronomy&quot;, have you considered astronomy does not have to be done from earth and indeed much of the data astronomers use often comes from probes and satellites in space.  Therefore 'how/where will this space beacon be used?' has strong impact on future of astronomy.&lt;/p&gt;&#xA;" OwnerUserId="28290" LastEditorUserId="1891" LastEditDate="2019-06-06T21:21:28.147" LastActivityDate="2019-06-06T21:21:28.147" Title="Where will the planned Deep Space Atomic Clock orbit &quot;after&quot; Earth?" Tags="&lt;orbit&gt;&lt;artificial-satellite&gt;&lt;navigation&gt;" AnswerCount="2" CommentCount="5" ClosedDate="2019-06-06T10:38:10.620" ContentLicense="CC BY-SA 4.0" />
  <row Id="32172" PostTypeId="2" ParentId="32170" CreationDate="2019-06-05T16:53:47.250" Score="3" Body="&lt;p&gt;It well might be the case that universe beyond our observable universe is filled with pink flying elephants, but this just seems highly unlikely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our current models are usually based on the fact that universe is isotropic, ie there’s nothing special about our observable universe vs everything else there is. Obviously there are smaller and larger deviations, but  on larger scales these are thought to cancel out. This means the simplest position here is to assume all of universe looks more or less the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While we can’t currently / no longer observe what is going on on the other side of horizon, there is some evidence supporting above view. Firstly, cosmic microwave background measurements seems to support isotropic universe. Secondly, inflation models seem to provide explanation for isotropic&#xA;universe via early causal connection beyond our current observable universe, and as our early universe model is able to explain many other things to great detail, it gives some confidence to this claim as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since early universe models do support the view universe outside our observable universe is pretty similar to what we can see now, this helps us to rely on observable universe for overall model building. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there does not seem to be any large mass concentrations outside the observable universe, since these would be seen as bias for movement of galaxies and galaxy groups, which also supports that the rest of universe is at least very close to what we can observe directly. This gives further confidence that our current models are probably not affected much by things beyond our currently observable universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since everything outside our observable universe is now causally not connected to us, anything that might be going on over there does and will not affect us - thus it is not necessary nor valid to include any of that in our models either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize my answer: no, there’s probably no effect on universe beyond the observable horizon that would need to affect our models. &lt;/p&gt;&#xA;" OwnerUserId="10813" LastEditorUserId="7411" LastEditDate="2019-06-05T19:32:39.320" LastActivityDate="2019-06-05T19:32:39.320" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32173" PostTypeId="2" ParentId="4755" CreationDate="2019-06-05T19:17:41.560" Score="0" Body="&lt;p&gt;As already mentioned naked eye sightings are dependent on location and conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some have made a point of photographing the earliest new Moon with interesting results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a site that might add some additional information for those interested in the topic.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://earthsky.org/astronomy-essentials/young-moon-visibility&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://earthsky.org/astronomy-essentials/young-moon-visibility&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Please keep in mind viewing or photographing anywhere near the sun is not safe for the viewers eyes or equipment. Be aware of the dangers and protect yourself appropriately.  &lt;/p&gt;&#xA;" OwnerUserId="24928" LastEditorUserId="24928" LastEditDate="2019-06-05T19:44:18.947" LastActivityDate="2019-06-05T19:44:18.947" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32174" PostTypeId="2" ParentId="32171" CreationDate="2019-06-05T20:13:07.227" Score="4" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;The instrument will be tested in Earth orbit for one year, ...&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;What would it then orbit?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It will continue to orbit the Earth, at least for a while. It will eventually end up not orbiting the Earth because it will be placed in low Earth orbit, where orbits decay due to atmospheric drag.&lt;/p&gt;&#xA;&lt;p&gt;The project is a technology demonstrator, which means a limited amount of funding.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-06-05T20:13:07.227" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32175" PostTypeId="2" ParentId="32171" CreationDate="2019-06-05T20:46:01.733" Score="3" Body="&lt;p&gt;The DSAC to be launched in June 2019 is a &lt;a href=&quot;https://www.nasa.gov/mission_pages/tdm/main/overview.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Technology Demonstration Mission&lt;/a&gt;,&#xA;in which they only test whether it works as well in space as it does on the ground.&#xA;If it passes the tests, similar devices could simplify navigation for future missions elsewhere in the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is one of many payloads sharing space on an &lt;a href=&quot;http://www.ga.com/general-atomics-orbital-test-bed-satellite-arrives-at-cape-canaveral&quot; rel=&quot;nofollow noreferrer&quot;&gt;Orbital Test Bed&lt;/a&gt; satellite, which will probably remain in Earth orbit until its orbit decays.&#xA;The OTB is itself one of several payloads on the same &lt;a href=&quot;https://www.spacex.com/stp-2&quot; rel=&quot;nofollow noreferrer&quot;&gt;launch vehicle&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2019-06-05T20:46:01.733" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32177" PostTypeId="1" AcceptedAnswerId="32180" CreationDate="2019-06-05T21:03:48.267" Score="3" ViewCount="367" Body="&lt;p&gt;How is an object's size determined from other solar systems than our own? How are new large bodies being found in our solar system currently?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="7411" LastEditDate="2019-06-06T07:02:23.880" LastActivityDate="2019-06-06T21:23:13.997" Title="How is the size of a distant planet determined?" Tags="&lt;planet&gt;&lt;distances&gt;&lt;space-geometry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32178" PostTypeId="2" ParentId="19129" CreationDate="2019-06-05T21:17:58.933" Score="10" Body="&lt;p&gt;It’s been a while since I studied this, and meteorites are very much not my field, so I’ll share my current understanding but would welcome corrections from those more informed :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These meteorites are mostly iron with a small amount of nickel. For a given composition of iron and nickel, there are different crystal structures that can be formed. Of relevance here are two crystal structures called “taenite” and “kamacite” (in the iron-carbon system where very similar patterns can be formed, these would be called “austenite” and “ferrite”, but they refer to the same crystal structures).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether taenite or kamacite is more stable depends both on the temperature and the amount of nickel present. At high temperatures, when the meteorite is young, it will be all taenite. However, as it cools, we eventually reach a temperature where, energetically speaking, the universe would like a mix of both taenite and kamacite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, just because it’s now energetically favorable for kamacite to form, there still needs to be a mechanism by which some of the taenite crystal can &lt;em&gt;transform&lt;/em&gt; into kamacite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Widmanstätten patterns only form at very low undercoolings, in other words when the temperature is only &lt;em&gt;just&lt;/em&gt; in this range where kamacite is starting to become stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you’re actually seeing in this pattern are the plates of kamacite that have formed. The exact pattern you see will depend which slice of the meteorite you decide to polish, and is related to the underlying symmetry of the crystal structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason you don’t see these patterns be so large on earth is for two reasons: one is simply that the grain sizes themselves are much smaller, and so the size of the microstructure is limited by that. The other reason is that the cooling rate is much faster, so the patterns can’t grow as large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When these plates nucleate, the extent (length/width) of the Widmanstätten plates can grow rapidly, but they will only get thicker at a rate that's limited by the diffusion of nickel atoms between the iron atoms. This is because the sides of the plates have a coherent interface with the parent taenite phase and diffusion is slower, whereas the tips/edges of the plate have an incoherent interface where diffusion is much more rapid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the rate of the plate growing longer is related to the degree of undercooling, so at very slow cooling rates like those in space, the plates grow longer much more slowly. This then gives time for atoms diffuse on the plate sides for the plate to grow thicker. And therefore, eventually, we have this beautiful microstructure large enough to see with the human eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That’s my understanding, however I think there isn’t yet consensus on exactly how Widmanstätten patterns form, the role that diffusion plays, and also the role of other additions in mediating this (e.g. the presence of phosphorous in the case of iron-nickel meteorites).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="28292" LastEditorUserId="28292" LastEditDate="2019-06-09T04:37:09.687" LastActivityDate="2019-06-09T04:37:09.687" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32180" PostTypeId="2" ParentId="32177" CreationDate="2019-06-05T22:48:34.313" Score="7" Body="&lt;p&gt;One method of finding planets orbiting stars is to analyze the transit effect. This is when a planet crosses the face of a star between the star and observer.  This will block some of the light reaching the observer. By measuring this decrease in light from the star we can determine how much of the surface area of star is occluded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed of a planet across the face of a star will let us calculate the orbital radius considering a circular orbit. The temperature of the star can be determined from its brightness and location on HR diagram and spectroscopic data. Combining this information lets us determine the size and of the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using space-based or ground-based telescopes and automated data collection and analysis methods large regions of the sky can be viewed for these transits. This method is highly accurate and will probably surpass the radial velocity method as most common method to discover new planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method for finding planets is known as the radial velocity method. When a planet orbits a star it is really the planet and star rotating about the center-of-mass of the system. Hence the star wobbles due to the bodies orbiting the star. It wobbles because usually the center-of-mass is inside the star and this results in the star rotating about an axis running thru the star but offset from the center. By doing a doppler analysis it can be determined whether the star is revolving or not and the period of revolution. The radial velocity method is the most effective method. The transit method is very limited by the viewing angle, the planet must pass between observer and star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good source to better learn about these methods is &lt;a href=&quot;http://www.planetary.org/explore/space-topics/exoplanets/how-to-search-for-exoplanets.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="16926" LastEditDate="2019-06-06T21:23:13.997" LastActivityDate="2019-06-06T21:23:13.997" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32181" PostTypeId="2" ParentId="32158" CreationDate="2019-06-06T00:00:41.983" Score="3" Body="&lt;p&gt;@NeutronStar's &lt;a href=&quot;https://astronomy.stackexchange.com/a/32159/7982&quot;&gt;excellent answer&lt;/a&gt; sums up the situation nicely:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From its RV (radial velocity) graph alone, with no other information, you cannot calculate inclination. This is why RV velocity measurements are typically reported as &quot;sin&quot;, because what you're actually measuring is the orbital velocity projected along the line of sight. Without other information, you cannot disentangle the orbital velocity from the viewing angle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After being confused, I finally &lt;a href=&quot;https://astronomy.stackexchange.com/questions/32158/how-does-one-calculate-inclination-from-a-graph-of-radial-velocity/32159?noredirect=1#comment58904_32159&quot;&gt;convinced myself&lt;/a&gt; that this is the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a plot of radial velocity versus time, assuming the period is fairly short, we have a cyclic plot which gives us a specific shape and amplitude as well as a period. The period gives us the combination of the semi major axis associated with the pair's orbit around their center of mass, and mass of the system. From &lt;a href=&quot;https://space.stackexchange.com/a/27101/12102&quot;&gt;this answer&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T = 2 \pi \sqrt{\frac{a^3}{m_1 + m_2}},$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could have a heavy system or a light system with the same period, just different size orbits. The &lt;em&gt;shapes&lt;/em&gt; of the two orbit could still be the same, and this means the &lt;em&gt;shape&lt;/em&gt; of the velocity profiles could be the same as well, just with different scale factors. For velocity, the vis-viva equation tells us that that for smaller orbits the velocity profile will scale as &lt;span class=&quot;math-container&quot;&gt;$1/\sqrt{a}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @NeutronStar's answer explains, the inclination of the orbit relative to our line of sight will reduce the radial velocity by a geometrical multiplicative factor &lt;span class=&quot;math-container&quot;&gt;$\cos(\theta_{inc})$&lt;/span&gt;. This will have no effect on the shape, it's strictly a scaling factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is no way to tell the difference between the geometrical reduction of a radial velocity curve and a physical reduction of the velocity due to lower reduced mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In curve-fitting parlance two parameters are 100% correlated and so they can never be independently determined from radial velocity alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By invitation I have included a simple Python script that builds a library of &lt;em&gt;unique&lt;/em&gt; radial velocity profiles from elliptical orbits of varying eccentricity, inclination, and orientation. It doesn't span all possible Keplerian orbits, but it covers all possible shapes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first plot shows a sparse sampling, for each combination of eccentricity and rotation is plotted 0 and 60 degree inclinations. They have the same shape, just different amplitude. This is illustrated by the second plot which is the 0 degree versus 60 degree velocity profile, showing slopes of 2.0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last plot the &quot;library&quot; of shapes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When &lt;em&gt;curve fitting via library search&lt;/em&gt;, one would compare a velocity profile shape to all of these and select the best fit, then either interpolate between nearest neighbor curves or do a quick iterative fit with an orbit generator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QiSxB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mpxsf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bre1J.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def deriv(X, t):&#xA;    x, v = X.reshape(2, -1)&#xA;    acc  = -x * ((x**2).sum())**-1.5&#xA;    return np.hstack((v, acc))&#xA;&#xA;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;from scipy.integrate import odeint as ODEint&#xA;&#xA;halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]&#xA;degs, rads        = 180/pi, pi/180&#xA;&#xA;eccs   = np.linspace(0, 0.7,    30)&#xA;incs   = np.linspace(0, halfpi, 19)   # not used&#xA;thetas = np.linspace(0, pi,  41)[:-1]&#xA;times  = np.linspace(0, twopi, 257)[:-1]&#xA;&#xA;A = np.array([f(thetas) for f in (np.cos, np.sin, np.ones_like)]).T[:, None, :, None]&#xA;B = np.array([f(incs) for f in (np.cos, np.ones_like, np.sin)]).T[:, None, None, :, None]   # not used&#xA;&#xA;# Set semimajor axis a to 1.0, period to twopi&#xA;&#xA;# build small library of orbits&#xA;velocities = [] &#xA;for ecc in eccs:&#xA;    peri   = 1. - ecc&#xA;    v0     = np.sqrt(2./peri - 1.)&#xA;    X0     = np.array([peri, 0, 0] +  [0, v0, 0])&#xA;    answer, info = ODEint(deriv, X0, times, full_output=True)&#xA;    xx, vv = answer.T.reshape(2, 3, -1)&#xA;    velocities.append(vv)&#xA;&#xA;velocities  = np.array(velocities)&#xA;&#xA;big_library = B * A * velocities    # not used  # build larger library of rotations&#xA;med_library = A * velocities # build medium library of rotations&#xA;&#xA;# library     = big_library[...,0, :].reshape(-1, big_library.shape[-1])   # not used&#xA;library     = med_library[...,0, :].reshape(-1, med_library.shape[-1])&#xA;noinc_library     = med_library[..., 0, :].reshape(-1, med_library.shape[-1])&#xA;&#xA;print library.shape&#xA;&#xA;a, b = big_library[[0, 12], ::10, ::10, 0].reshape(2, 12, -1) # small sample&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;    for i, (c, d) in enumerate(zip(a, b)):&#xA;        plt.subplot(4, 3, i+1)&#xA;        plt.plot(c)&#xA;        plt.plot(d)&#xA;        #plt.plot(c/d)&#xA;    plt.show()&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;    for i, (c, d) in enumerate(zip(a, b)):&#xA;        plt.subplot(4, 3, i+1)&#xA;        plt.plot(d, c)&#xA;    plt.show()&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;    for v in noinc_library:&#xA;        plt.plot(times/twopi, v)&#xA;    plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="16506" LastEditDate="2019-06-06T10:47:15.387" LastActivityDate="2019-06-06T10:47:15.387" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32182" PostTypeId="1" CreationDate="2019-06-06T05:10:42.917" Score="0" ViewCount="154" Body="&lt;p&gt;As the Sun moves along with the planets does the sun exhibit vertical movement? That is, does it traverse in different planes?&lt;/p&gt;&#xA;" OwnerUserId="28296" LastActivityDate="2019-06-07T06:45:37.833" Title="Does the sun have a vertical movment?" Tags="&lt;the-sun&gt;&lt;movement&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32184" PostTypeId="2" ParentId="32177" CreationDate="2019-06-06T06:50:36.087" Score="-2" Body="&lt;p&gt;The size of a planet is determined by&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The amount of gas &amp;amp; dust available to build it.&lt;/li&gt;&#xA;&lt;li&gt;The type of star it orbits.&lt;/li&gt;&#xA;&lt;li&gt;Its distance from the star.&lt;/li&gt;&#xA;&lt;li&gt;Whether or not the sort of resonances we see in Bode's Law come into play. We don't know for sure whether Bode's Law operates for all stars, some stars, or only our star.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28083" LastEditorUserId="7472" LastEditDate="2019-06-06T08:12:28.083" LastActivityDate="2019-06-06T08:12:28.083" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32185" PostTypeId="2" ParentId="925" CreationDate="2019-06-06T11:30:03.393" Score="0" Body="&lt;p&gt;The largest &quot;complex, less cohesively bound structure, which may consist of multiple bodies or even other objects with substructures&quot; known object is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Shapley_Supercluster&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shapley Supercluster&lt;/a&gt;, which comprises 76000 galaxies. It is supossed that &lt;a href=&quot;https://es.wikipedia.org/wiki/Laniakea&quot; rel=&quot;nofollow noreferrer&quot;&gt;Laniakea&lt;/a&gt; (our own supercluster) moves towards it and both may be part of a bigger structure, yet to be confirmed.&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2019-06-06T11:30:03.393" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32186" PostTypeId="2" ParentId="32182" CreationDate="2019-06-06T12:13:52.907" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the sun have a vertical movment?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Sure it does!&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;People who follow solar system bodies with the highest accuracy, including NASA's JPL, use the solar system barycenter (center of mass) for the origin of the plane of the ecliptic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the big planets (Jupiter et al.) have some inclination, they pull the Sun up and down out of the plane of the Ecliptic by tens of thousands of kilometers. It's a small fraction of it's 600,000 kilometer radius, but it certainly happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see the data from the &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL Horizons&lt;/a&gt; website, or if you like Python, you can get virtually the same numbers by running the &lt;a href=&quot;https://rhodesmill.org/skyfield/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Skyfield&lt;/a&gt; package.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Horizions I downloaded the position of the Sun and the four major planets of the solar system in 10 day steps from the years 1900 to 2100. Below I plot the three components X, Y and Z in the J2000.0 Solar system barycentric coordinates using the ecliptic as the reference frame. The black line is the motion of the Sun, and the color lines are for the four planets, but I have multiplied each by the the mass of the planet divided by the Mass of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second plot, I added the four together, and you can see that together they mirror the motion of the Sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last two show the movement of the Sun in 3D (two views). Note that the vertical scale is much smaller than the horizontal scale. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Horizons setup:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CJHyh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ysp17.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; Black thick line: motion of the Sun in barycentric coordinates in the J2000.0 X, Y and Z directions, relative to the ecliptic. Four thinner color lines: motions of the four largest planets, each weighted by its mass relative to the Sun's mass. Short period lines are Jupiter and Saturn with Jupiter the largest amplitude, long periods are Uranus and Neptune, with Neptune the largest amplitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Syq7T.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; Black thick line: motion of the Sun in barycentric coordinates in the J2000.0 X, Y and Z directions, relative to the ecliptic. Thin red line: sum of the motions of the four largest planets, each weighted by its mass relative to the Sun's mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nC5R0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; Two views  of the 3D motion of the Sun in barycentric coordinates between the years 1900 and 2100. Units are kilometers, and note that the scales are different for each axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lvBX2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OS7ju.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;class Body(object):&#xA;    def __init__(self, name):&#xA;        self.name = name&#xA;&#xA;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;from mpl_toolkits.mplot3d import Axes3D&#xA;&#xA;names = ('Sun', 'Jupiter', 'Saturn', 'Uranus', 'Neptune')&#xA;&#xA;bodies = []&#xA;for name in names:&#xA;    fname = 'sun motion ' + name + ' horizons_results.txt'&#xA;    body  = Body(name)&#xA;    bodies.append(body)&#xA;&#xA;    with open(fname, &quot;r&quot;) as infile:&#xA;        lines = infile.read().splitlines() &#xA;&#xA;    iSOE = [i for i, line in enumerate(lines) if &quot;&lt;span class=&quot;math-container&quot;&gt;$$SOE&quot; in line][0]&#xA;    iEOE = [i for i, line in enumerate(lines) if &quot;$$&lt;/span&gt;EOE&quot; in line][0]&#xA;&#xA;    print (name, iSOE, iEOE, lines[iSOE], lines[iEOE])&#xA;&#xA;    lines = [line.split(',') for line in lines[iSOE+1:iEOE]]&#xA;    JD  = np.array([float(line[0]) for line in lines])&#xA;    pos = np.array([[float(item) for item in line[2:5]] for line in lines])&#xA;    body.lines = lines&#xA;    body.JD    = JD&#xA;    body.pos   = pos.T.copy()&#xA;&#xA;years = 2000 + (JD - JD[3652])/365.2564&#xA;Sun, Jupiter, Saturn, Uranus, Neptune = bodies&#xA;&#xA;for body in bodies:&#xA;    body.rsun = np.sqrt(((body.pos - Sun.pos)**2).sum(axis=0))&#xA;&#xA;planets = [b for b in bodies if 'sun' not in b.name.lower()]&#xA;&#xA;massdict = {'Jupiter':1.898E+27, 'Saturn':5.683E+26, 'Uranus':8.681E+25,&#xA;            'Neptune':1.024E+26, 'Sun':1.989E+30}&#xA;&#xA;for b in bodies:&#xA;    b.mass = massdict[b.name]&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;    for i, thing in enumerate(Sun.pos):&#xA;        plt.subplot(3, 1, i+1)&#xA;        plt.plot(years, thing, '-k', linewidth=1.5)&#xA;    for p in planets:&#xA;        for i, thing in enumerate(p.pos):&#xA;            plt.subplot(3, 1, i+1)&#xA;            thing = thing * (p.mass/Sun.mass)&#xA;            plt.plot(years, thing)&#xA;    plt.show()&#xA;&#xA;all_four = Body('all_four')&#xA;all_four.pos = np.array([p.pos*(p.mass/Sun.mass) for p in planets]).sum(axis=0)&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;    for i, thing in enumerate(Sun.pos):&#xA;        plt.subplot(3, 1, i+1)&#xA;        plt.plot(years, thing, '-k', linewidth=1.5)&#xA;    for i, thing in enumerate(all_four.pos):&#xA;        plt.subplot(3, 1, i+1)&#xA;        plt.plot(years, thing, '-r')&#xA;    plt.show()&#xA;&#xA;if True:    &#xA;    fig = plt.figure(figsize=[12, 10])  # [12, 10]&#xA;    ax  = fig.add_subplot(1, 1, 1, projection='3d')&#xA;&#xA;    x, y, z = Sun.pos&#xA;    ax.plot(x, y, z)&#xA;&#xA;    plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-06-07T06:45:37.833" LastActivityDate="2019-06-07T06:45:37.833" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="32187" PostTypeId="1" CreationDate="2019-06-06T15:28:19.593" Score="0" ViewCount="78" Body="&lt;p&gt;I'm doing research for a story that I'm writing and I'd like some physics advice on whether a hypothetical system of celestial bodies is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A system of planets orbiting a star, similar in size to our own.&lt;/li&gt;&#xA;&lt;li&gt;A cloud of debris (Asteroids, technological refuse, and dust) surrounds the entire system, similar to the description of a Dyson swarm.&lt;/li&gt;&#xA;&lt;li&gt;The outermost planet orbiting on a path that intersects the cloud during a small portion of it's year, but is within the cloud for the majority of it's year.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is this configuration possible within our understanding of physics, or would gravity cause the objects of the cloud and the outermost planet to 'snap' to an orbit that does not intersect? Would the planet 'capture' objects within the swarm in it's own orbit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My story takes place on the inside of a Dyson swarm or bubble, where interplanetary travel has developed to a reasonable degree, but cannot escape the swarm. An area of intrigue might be the outermost planet. If it can be landed upon while it's orbit is within the swarm, people could 'ride' the planet through to the other side if they stayed there for a number of years or went into stasis there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thoughts or opinions are greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="28309" LastActivityDate="2019-07-06T20:01:00.597" Title="If a solar system were surrounded by a cloud of debris, is it possible for a planet's orbit to intersect it?" Tags="&lt;orbit&gt;&lt;planet&gt;&lt;gravity&gt;&lt;asteroids&gt;&lt;space-debris&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32188" PostTypeId="2" ParentId="32146" CreationDate="2019-06-06T15:55:44.807" Score="2" Body="&lt;p&gt;Another key point to bear in mind is that if one is asking for the &quot;hottest&quot; star, then presumably it is &quot;effective temperature&quot; that is referenced.  Effective temperature is related to the luminosity L and the radius R from which the light emerges (not the static surface of the star, R can be well out in a dense stellar wind) by the Stefan-Boltzmann law, which says T_eff ~ L^(1/4) / R^(1/2).   Normally for main-sequence stars, whose R rises much more slowly than L, that means the T_eff is higher at higher L.  But really high L stars have dense radiatively-driven winds, and eventually the wind could get so dense that the mass-loss rate from the star might be proportional to L.  If so, the radius R from which the light emerges (where the optical depth is near unity) is also proportional to L, so then T_eff ~ L^(1/4) / L^(1/2) ~ L^(-1/4).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This says we should expect T_eff to reach a peak and then drop if L is further increased.  The peak T_eff should be when the winds are getting so dense that the mass-loss rate becomes proportional to L (when a significant amount of L is being used to lift the mass), and that is more or less also where the WNh spectrum replaces the O spectrum (something like what is trying to be the O1 spectrum).  So the kinds of T_eff already referred to, above 50,000 K or so, is going to be the highest T_eff possible for a core hydrogen burning star.  After that, even higher L will simply mean that you see a much larger star with a lower T_eff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, this probably points to a flaw in the Bromm et al. paper cited above.  That paper is really about Pop III stars, so zero metallicity and not much in the way of stellar winds also.  So when they talk about T_eff, they appear to use an expression that involves L and the static radius, but that's not the T_eff that matters for observing stars, or for their UV flux.  The latter should use the R from which the light emerges, which can be in the wind.  The distinction is not important for the pop III conclusions of their paper, but it does matter for the Pop I results shown in Figure 1.  That is probably not the correct T_eff, so I doubt T_eff really can get up to 65,000 K for Pop I stars.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2019-06-06T16:38:20.107" LastActivityDate="2019-06-06T16:38:20.107" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32189" PostTypeId="1" CreationDate="2019-06-06T16:43:01.587" Score="1" ViewCount="116" Body="&lt;p&gt;I tried to solve the three body problem numerically.&#xA;While after 1000 time steps the system looks stable (See Fig.1)&#xA;it is completely unstable after 10000 time steps (See Fig.2).&#xA;Is that an intrinsic property of the three body problem or a mistake in my code or initial conditions?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/l59Lk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l59Lk.png&quot; alt=&quot;N=1000&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/UgqJa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UgqJa.png&quot; alt=&quot;N=10000&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="28310" LastActivityDate="2019-06-06T16:43:01.587" Title="Three Body Problem" Tags="&lt;astrophysics&gt;&lt;python&gt;&lt;n-body-simulations&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32190" PostTypeId="1" CreationDate="2019-06-06T16:50:25.073" Score="7" ViewCount="113" Body="&lt;p&gt;This question is inspired by, but different from, &lt;a href=&quot;https://astronomy.stackexchange.com/questions/2357/what-would-happen-if-an-ice-cube-is-left-in-space&quot;&gt;What happens if an ice cube is left in space?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mark's accepted answer says &quot;…if you stick your ice cube out in the Oort Cloud, it'll grow: the mean surface temperature is 40K or below,…&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To which the obvious response is &lt;em&gt;Everything has a vapour pressure, so everything in the vacuum of space will evaporate eventually&lt;/em&gt;. Accordingly I thought it might be worth investigating why ice grains &lt;em&gt;do&lt;/em&gt; exist. For which we need some numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Vapour pressure of water at 40K&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most charts do not go that low. The best figure I have come up with was by looking at the pictures in &lt;a href=&quot;https://trace.tennessee.edu/cgi/viewcontent.cgi?referer=https://www.google.com/&amp;amp;httpsredir=1&amp;amp;article=2364&amp;amp;context=utk_graddiss&quot; rel=&quot;noreferrer&quot;&gt;Water Ice Films in Cryogenic Vacuum Chambers&lt;/a&gt;, a doctoral thesis by Jesse Michael Labello. Figure 4 on page 4 has graphs which, extrapolated by guesswork, suggest that the vapour pressure of water at 40K is less than &lt;span class=&quot;math-container&quot;&gt;$10^{-26}$&lt;/span&gt; torr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Atmospheric pressure in space&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon's atmospheric pressure is said to be &lt;span class=&quot;math-container&quot;&gt;$10^{-11}$&lt;/span&gt; torr, and the New World Encyclopedia says interstellar space is &lt;span class=&quot;math-container&quot;&gt;$10^{-16}$&lt;/span&gt; torr, and some answers on Physics Stack Exchange suggest about &lt;span class=&quot;math-container&quot;&gt;$10^{-25}$&lt;/span&gt; for intergalactic space.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It appears, therefore, that ice grains at 40K do not evaporate because their vapour pressure is not high enough. Does it follow, therefore:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;That ice grains would evaporate, even at 40K, if they were in a vacuum?&lt;/li&gt;&#xA;&lt;li&gt;That, if the ice is at equilibrium with its vapour, the relative humidity of interstellar space is 100%? Or is the process of evaporation/condensation &lt;em&gt;kinetically&lt;/em&gt; so slow that the universe is too young for it to have reached equilibrium?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="12016" LastActivityDate="2021-02-01T21:27:03.830" Title="Water vapour in space" Tags="&lt;space&gt;&lt;water&gt;&lt;vacuum&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32191" PostTypeId="2" ParentId="29026" CreationDate="2019-06-06T17:48:36.667" Score="2" Body="&lt;p&gt;There are more or less two possibility to get debris in a sufficient amount: collisions and breakup/ejection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Collisions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is mostly the case for asteroids. At the point of impact you get a cloud of debris which basically moves as if the colliding bodies combined into a bigger one. However, for small objects, the fragments in the cloud are not gravitationally bound to each other and thus can separate over time. If larger objects survive the initial collision, they can undergo further impacts at different orbital positions leading to the seasonal meteor showers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Breakup/Ejection&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is mostly the case for comets. Particles get ejected from the original body by heating of the surface, or other effects. This is only possible for small grains, which are massively influenced by stellar radiation pressure. The result is an effectively smaller gravity of the star and thus a different orbit. Being on different orbits, theses particles naturally have to separate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases you get a physical separation over time. Being that all of the quantities involved are on a continuum, you effectly have a smeared distribution of fragments over an orbit. Depending on the special case, you can get a distribution over the whole ellipse, but a part of it seems more likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: As you asked about catalogues, I would recommend you looking at &lt;a href=&quot;https://www.ta3.sk/IAUC22DB/MDC2007/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.ta3.sk/IAUC22DB/MDC2007/&lt;/a&gt; plus at &lt;a href=&quot;http://adsabs.harvard.edu/abs/2006mspc.book.....J&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/abs/2006mspc.book.....J&lt;/a&gt; for a scientific read.&lt;/p&gt;&#xA;" OwnerUserId="28279" LastEditorUserId="28279" LastEditDate="2019-06-06T17:59:46.313" LastActivityDate="2019-06-06T17:59:46.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32192" PostTypeId="2" ParentId="1961" CreationDate="2019-06-06T18:06:50.757" Score="2" Body="&lt;p&gt;For a more fundamental understanding, it is helpful to realize the difficulties of fusing He-4 into C-12. This is called the Triple-Alpha process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When two He-4 nuclei (alpha particles) have sufficient energy to overcome to the Coulomb barrier and have their cross sections align, it produces Be-8. The Be-8 nucleus is so unstable (due to it being energetically favourable for the subject nucleons to be arranged in two alpha particles) that it has a half-life of around 10^-17 seconds which is amazingly brief. Therefore, to produce C-12 three alpha particles have to come together nearly instantaneously, two produce Be-8 and in that half-life threshold, a third interacts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a moment to think about how extreme the conditions of the core must be to allow the probability of three alpha particles to come together and successfully interact nearly instantaneously and for it to happen enough times to produce the energy needed to bring the core out of degeneracy. Helium fusion takes about 100 million K to start as opposed to the 15 million K of the sun's core (undergoing proton-proton chain for about 99% of reactions) at present time. This temperature is provided both by the incredible pressure of the degenerate core and by additional energy supplied by the shell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shell fusion starts before the triple-alpha process because as the core contracts and becomes degenerate, there is so much energy being radiated from the core that it heats the immediate surrounding layers to the point where it can start to fuse H-to-He, in fact it is so hot that shell fusion is by the CNO cycle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outer layers of the star expand rapidly as there is an enormous amount of energy being radiated away from this shell, which is fusing at a temperature much hotter than the core is today. &lt;/p&gt;&#xA;" OwnerUserId="23212" LastActivityDate="2019-06-06T18:06:50.757" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32193" PostTypeId="2" ParentId="32187" CreationDate="2019-06-06T19:32:13.493" Score="1" Body="&lt;p&gt;The solar system IS surrounded by a cloud of debris,left over from its formation. This debris is called the Oort Cloud,&amp;amp; consists of thousands of comet-like bodies,sometimes described as dirty snowballs. None of the known major planets passes through them,but a few years ago some astronomers seemed certain that  they had detected signs of a 9th major planet much further out than the others. They predicted it was only a matter of time before this planet would be discovered,but so far it hasn't been. If there is indeed such a planet &amp;amp; part of its orbit passes through the Oort Cloud,it would probably sweep a lane through it eventually. The planet would not be a nice place to live,as the likelihood is that it has a temperature close to that of liquid nitrogen,&amp;amp; is frequently struck by comets &amp;amp; other debris.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-06-06T19:32:13.493" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32195" PostTypeId="1" CreationDate="2019-06-07T00:03:32.690" Score="3" ViewCount="45" Body="&lt;p&gt;As the comet orbits around the Sun, I gather that its tail loses mass due to solar wind and tidal waves. Approximately, from what distance (with respect to Sun) will the mass loss begin?&lt;/p&gt;&#xA;" OwnerUserId="28296" LastActivityDate="2019-06-07T02:03:34.563" Title="Comet mass loss" Tags="&lt;comets&gt;&lt;mass&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32196" PostTypeId="1" AcceptedAnswerId="32201" CreationDate="2019-06-07T00:48:22.700" Score="3" ViewCount="107" Body="&lt;p&gt;Why is a true and average anomaly of the planet important? What useful information do they give us?&lt;/p&gt;&#xA;" OwnerUserId="28313" LastEditorUserId="28286" LastEditDate="2019-06-07T01:28:11.760" LastActivityDate="2019-06-07T10:58:17.557" Title="Why is true and average anomaly of planet important?" Tags="&lt;orbit&gt;&lt;planet&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32197" PostTypeId="2" ParentId="32195" CreationDate="2019-06-07T02:03:34.563" Score="2" Body="&lt;p&gt;The mass loss begins at approximately 450-600 million kilometers from the sun. This begins roughly in an area between the end of the asteroid belt and 1 AU from Jupiters orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extract from wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The coma is generally made of H&#xA;  2O and dust, with water making up to 90% of the volatiles that outflow from the nucleus when the comet is within 3 to 4 astronomical units (450,000,000 to 600,000,000 km; 280,000,000 to 370,000,000 mi) of the Sun. &lt;a href=&quot;https://en.wikipedia.org/wiki/Comet&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="28286" LastActivityDate="2019-06-07T02:03:34.563" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32198" PostTypeId="1" CreationDate="2019-06-07T02:03:56.750" Score="0" ViewCount="158" Body="&lt;p&gt;If you had the distance from a celestial object, in parsecs, how would you calculate the velocity of the said object?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I'm sorry I don't have more context, the specific questions is: &quot;A galaxy that is 10 million parsecs away will be moving at what velocity?&quot;&lt;/p&gt;&#xA;" OwnerUserId="28314" LastEditorUserId="28314" LastEditDate="2019-06-07T13:52:04.040" LastActivityDate="2019-09-06T04:47:23.943" Title="How can we calculate velocity of an object given the distance from an object in parsecs?" Tags="&lt;galaxy&gt;&lt;velocity&gt;&lt;parsec&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32199" PostTypeId="1" AcceptedAnswerId="32500" CreationDate="2019-06-07T03:05:13.903" Score="4" ViewCount="108" Body="&lt;p&gt;Gizmodo's &lt;a href=&quot;https://gizmodo.com/astronomers-spot-mysterious-10-million-light-year-long-1835301269&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astronomers Spot Mysterious, 10-Million-Light-Year-Long Magnetic Field Connecting Two Galaxy Clusters&lt;/a&gt; shows the image below, and Space.com's &lt;a href=&quot;https://www.space.com/radio-ridge-between-galaxy-clusters.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Weird 'Radio Bridge' 10 Million-Light Years Long Links Colliding Galaxy Clusters&lt;/a&gt; discusses it as well.&lt;/p&gt;&#xA;&lt;p&gt;They both link to the new paper in Science &lt;a href=&quot;https://science.sciencemag.org/content/364/6444/981/tab-pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;A radio ridge connecting two galaxy clusters in a filament of the cosmic web&lt;/a&gt; (also &lt;a href=&quot;https://arxiv.org/abs/1906.07584&quot; rel=&quot;nofollow noreferrer&quot;&gt;arXiv&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;p&gt;What I've gathered so far is that seen between two known and characterized clusters of galaxies, there is a region with some excess continuum radio power at 140 MHz and some excess continuum X-ray power.&lt;/p&gt;&#xA;&lt;p&gt;Neither of these are spectral features so their distances can't be associated with the galactic clusters by doppler shift.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How do they know that the X-ray and Radio continuum are both really coming from the space between these two clusters of galaxies, and not in front of or behind it?&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;bonus points&amp;quot; how does Planck's &amp;quot;y parameter&amp;quot; figure in to this mix?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;“A radio ridge connecting two galaxy clusters in a filament of the cosmic web”, F. Govoni et al. 2019, Science. Optical: DSS and Pan STAARS (insets) — Red, X-rays: XMM-Newton — Yellow, y-parameter: Planck satellite — Blue, radio 140 MHz: LOFAR. Image credits: M. Murgia - INAF&lt;/p&gt;&#xA;&lt;p&gt;The two galactic clusters and the intervening radio ridge. Image: DSS and Pan-STARRS1 (Optical), XMM-Newton (X-rays) PLANCK Satellite (y paramter), F. Govoni, M. Murgia, INAF&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DEtRX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DEtRX.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="37191" LastEditDate="2021-03-19T08:38:19.950" LastActivityDate="2021-03-19T08:38:19.950" Title="How is it determined that the X-ray and radio intensity come from a magnetic field bridge between two clusters of galaxies?" Tags="&lt;radio-astronomy&gt;&lt;magnetic-field&gt;&lt;cmb&gt;&lt;x-ray&gt;&lt;xmm-newton&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32200" PostTypeId="2" ParentId="32198" CreationDate="2019-06-07T10:36:00.857" Score="-1" Body="&lt;p&gt;If the object is moving directly towards or away from us &amp;amp; we can get a spectrum,the Doppler shift will give us its velocity. Blue shift means it's approaching us,reshift means it's receding. If the object is orbiting some larger body with a known mass,the period of its orbit will indicate velocity (if the orbit is elliptical,its velocity will vary,speeding up as it gets nearer to the larger body). If its distance is neither increasing nor decreasing,we can track its motion within a certain timescale to calculate its velocity. Sometimes more than one method may need to be used.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-06-07T10:36:00.857" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32201" PostTypeId="2" ParentId="32196" CreationDate="2019-06-07T10:58:17.557" Score="6" Body="&lt;p&gt;Given one body orbiting one another, the radial distance between the two and the true anomaly are the polar coordinates of the orbiting body. This wouldn't be all that useful if there was no way to predict where the orbiting body will be at some point in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mean anomaly along with Kepler's laws of motion do just that. Kepler's first law says that the planets orbits about the Sun are ellipses. His second law says that the rate at which a planet sweeps out area (with the Sun as the central point) is constant. His third law says that the period of a planet's orbit depends on the semi-major axis length but not eccentricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that a planet in an eccentric orbit and a planet in a circular orbit, both with the same semi-major axis length, will, on average, exhibit the same motion. In particular, their periods will be exactly the same. It's easy to predict where the planet in a circular orbit will be at some point in the future because that planet's true anomaly is a linear function of time. Calling this the mean anomaly, this linear relationship means that&#xA;&lt;span class=&quot;math-container&quot;&gt;$$M(t) = M(t_0) + (t-t_0)n$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Mean_motion&quot; rel=&quot;noreferrer&quot;&gt;mean motion&lt;/a&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$n = \frac{2\pi}{T} = \sqrt{\frac{\mu}{a^3}}$$&lt;/span&gt;&#xA;Here, &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is the period, &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is the semi-major axis length, and &lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt; is the constant of proportionality in Kepler's third law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's needed is a bridge between the mean anomaly and the true anomaly that makes the true anomaly satisfy Kepler's second law. This bridge is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Eccentric_anomaly&quot; rel=&quot;noreferrer&quot;&gt;eccentric anomaly&lt;/a&gt;. Mean anomaly and eccentric anomaly are related via &lt;a href=&quot;https://en.wikipedia.org/wiki/Kepler%27s_equation&quot; rel=&quot;noreferrer&quot;&gt;Kepler's equation&lt;/a&gt;,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$M = E - e \sin E$$&lt;/span&gt;&#xA;Here, &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; is the eccentric anomaly and &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt; is the eccentricity. Note that by this equation, it's trivial to calculate the mean anomaly given the eccentric anomaly and the eccentricity. Simply plug the values in the right hand side and evaluate. What's wanted is a means to calculate the eccentric anomaly given the mean anomaly and the eccentricity. Unfortunately, there is no close-formed inverse of Kepler's equation in the elementary functions. Fortunately, there are many ways to solve Kepler's equation for the eccentric anomaly. One easy way is to use Newton-Raphson iteration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's still needed is a relationship between eccentric anomaly and true anomaly. This relationship can be represented in closed form:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\tan\frac\theta2 = \sqrt{\frac{1+e}{1-e}}\tan\frac E2$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2019-06-07T10:58:17.557" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32202" PostTypeId="1" CreationDate="2019-06-07T13:26:57.723" Score="7" ViewCount="133" Body="&lt;p&gt;Are any of the images of bow shocks created from visible light, or are they all from infrared light?  While I'm including two of the better known images, my question is about any known bow shock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NXw5F.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NXw5F.png&quot; alt=&quot;LL Orionis&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;LL Orionis&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9SScT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9SScT.jpg&quot; alt=&quot;Zeta Ophiuchi&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zeta Ophiuchi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Bow_shocks_in_astrophysics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source of image-Wilipedia&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25088" LastEditorUserId="25088" LastEditDate="2019-06-08T01:47:41.910" LastActivityDate="2019-07-24T21:45:19.577" Title="Are any bow shocks seen in the visible spectrum?" Tags="&lt;bow-shock&gt;&lt;visible-light&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32203" PostTypeId="1" AcceptedAnswerId="32210" CreationDate="2019-06-07T18:37:38.147" Score="4" ViewCount="308" Body="&lt;p&gt;I'm aware that voids are relatively empty regions of the universe, but just how empty can they be? Wikipedia states that voids have 'few or no' galaxies, but I can't find much else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make my question specific, suppose I were in the centre of a large void, way outside of the few galaxies that may still exist in this void. Would I occasionally encounter asteroids (if I may call them such), rogue planets, or even some occasional intergalactic star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: I'm aware that this question is a bit of a one-up of &lt;a href=&quot;https://astronomy.stackexchange.com/q/307/7982&quot;&gt;What is there in the intergalactic space?&lt;/a&gt;. If this is too much of a duplicate, let me know and I remove the question.&lt;/p&gt;&#xA;" OwnerUserId="28319" LastEditorUserId="7982" LastEditDate="2019-06-07T21:53:21.457" LastActivityDate="2019-06-12T09:42:25.110" Title="What would you find within a void?" Tags="&lt;asteroids&gt;&lt;rogue-planet&gt;&lt;intergalactic-space&gt;&lt;interstellar&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32204" PostTypeId="2" ParentId="28339" CreationDate="2019-06-07T19:06:49.407" Score="1" Body="&lt;p&gt;Islam has a word, (Asr). The name of their prayer which takes place exactly between solar noon (Dhuhr) and sunset (Maghrib). I was trying to figure out the name for the time exactly between sunrise and solarnoon, as Islam does not call to prayer then. Islam also has names of prayer opposite solar noon (mid night Isha) and again at sunrise (Fair).&lt;/p&gt;&#xA;" OwnerUserId="28320" LastActivityDate="2019-06-07T19:06:49.407" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32205" PostTypeId="1" CreationDate="2019-06-07T20:14:13.737" Score="5" ViewCount="64" Body="&lt;p&gt;The IAU Minor Planet Center lists &lt;span class=&quot;math-container&quot;&gt;$(486958)\space 2014 \space MU_{69}$&lt;/span&gt; &quot;Ultima Thule&quot; as having an uncertainty parameter of 2, based on the observation arc of 851 days, from its discovery in 2014, to the last observation submitted from Hubble in 2016.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/(486958)_2014_MU69#2017_occultations&quot; rel=&quot;noreferrer&quot;&gt;Other&lt;/a&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/(486958)_2014_MU69#2018_occultations&quot; rel=&quot;noreferrer&quot;&gt;sources&lt;/a&gt; indicate that there have been a number of ground-based observations of Ultima Thule occulting various background stars. Are these &quot;observations&quot; purposely not included/submitted as further sightings to increase the observation arc?&lt;/p&gt;&#xA;" OwnerUserId="9760" LastEditorUserId="7982" LastEditDate="2020-12-03T13:11:46.803" LastActivityDate="2020-12-03T13:11:46.803" Title="Are occultation observations used/useful for orbit determination?" Tags="&lt;observational-astronomy&gt;&lt;orbital-elements&gt;&lt;dwarf-planets&gt;&lt;occultation&gt;&lt;trans-neptunian-object&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32206" PostTypeId="2" ParentId="32203" CreationDate="2019-06-07T21:46:50.867" Score="-4" Body="&lt;p&gt;Apart from what you read in Wikipedia, nobody knows. All the things you mention could be in there, but these voids are just too far away to detect them. Commonsense dictates that like the rest of space, they must be seething with neutrinos, photons &amp;amp; other particles. QM mechanics will also tell you that space,including the voids,is seething with virtual particles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the risk of being condemned for heresy, I have to admit that so far as virtual particles are concerned, I'm an atheist. I also have a suspicion that in some of these voids; the &lt;a href=&quot;https://en.wikipedia.org/wiki/Bo%C3%B6tes_void&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boötes void&lt;/a&gt; for example, there is something mysterious in there or has passed through there to create the void, but that is speculation.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastEditorUserId="7982" LastEditDate="2019-06-07T21:51:47.683" LastActivityDate="2019-06-07T21:51:47.683" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32207" PostTypeId="1" AcceptedAnswerId="32213" CreationDate="2019-06-07T21:57:59.113" Score="0" ViewCount="70" Body="&lt;p&gt;I am trying to plot superwasp lightcurves in python from the fits files, using TMID and FLUX2, getting this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OmFZH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OmFZH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a way to transform the information so I can plot the lightcurves with the kepler format, a example from kepler is something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/koS3w.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/koS3w.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is useful.&lt;/p&gt;&#xA;" OwnerUserId="28321" LastEditorUserId="14094" LastEditDate="2019-06-10T15:23:02.393" LastActivityDate="2019-06-10T15:23:02.393" Title="Plot SuperWasp Lightcurves" Tags="&lt;amateur-observing&gt;&lt;exoplanet&gt;&lt;python&gt;&lt;light-curve&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32209" PostTypeId="1" CreationDate="2019-06-08T14:47:32.983" Score="2" ViewCount="38" Body="&lt;p&gt;I know that RR Lyrae stars are convective (this is suspected to be part of the cause of the Blazhko effect), but I would like to know more generally:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Are RR Lyrae fully convective?&lt;/li&gt;&#xA;&lt;li&gt;Are all HB stars convective, and fully convective?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The detailed articles I've found so far to try to answer this question are all quite focused on RR Lyrae variables and so don't readily provide the answers to these questions as far as I've seen.&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2019-06-08T14:47:32.983" Title="Are horizontal branch stars fully convective?" Tags="&lt;star&gt;&lt;stellar-evolution&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32210" PostTypeId="2" ParentId="32203" CreationDate="2019-06-08T16:00:54.970" Score="4" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Void_(astronomy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article on voids&lt;/a&gt; is pretty good (though IMO unusually awkwardly written.) The key thing is that voids are not &lt;em&gt;empty&lt;/em&gt;, they are just large volumes which have a lower density (typically around 10% of average) compared with the rest of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These low density areas still contain stars and galaxies, just fewer of them and the galaxies they contain tend to be smaller. The Lambda-Cold-Dark-Matter (LCDM) model, which is currently favored as our best description of the universe, predicts that the voids should not have many large galaxies in them, but should have a fair scattering of small, diffuse galaxies (SDGs).  (A small diffuse galaxy is, well, small compared with the Milky Way -- &#xA;sized more like the Magellanic Clouds or one of the small galaxies which are being devoured by the Milky Way today such as the Sagittarius dwarf. They are also more spread out than the Magellanic Clouds.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an &lt;a href=&quot;http://file:///C:/Users/mlo/Downloads/Rochelle_Steele-Senior%20Thesis.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;interesting senior thesis&lt;/a&gt; reporting on a search for small diffuse galaxies in voids. (They think they found some, but need confirmation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You ask what the night sky would look like from within a void. First, we'll assume you're in one of the SDGs, though this wouldn't make a &lt;em&gt;huge&lt;/em&gt; difference.  What you'd see mostly is whatever planets share your solar system -- they'd be comparable in brightness to our own planets from Earth. You'd also see a few dim stars.  SDGs tend not to have much star formation, so there would be very few young stars and thus very few intrinsically bright stars. The visible stars would all be ones which happened to be nearby. And because SDGs are diffuse, there'd be many fewer nearby stars. At a &lt;em&gt;guess&lt;/em&gt; there would be 20-200 naked eye stars, with only a handful of magnitude 3 or brighter. (Compare that with 6000 naked eye stars visible from Earth.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There would be no naked eye galaxies. The SDG itself would not appear as a milky way, but more like the Zodiacal Light -- a vague glow dimly visible under exceptional conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be a world whose astronomy would be heavily focused on the planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Planetary formation might be quite different in SDGs, but I don't know even to try to speculate on how it would be different. So I assume that the planetary system would be similar to ours.  YMMV.)&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2019-06-08T21:06:22.827" LastActivityDate="2019-06-08T21:06:22.827" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32211" PostTypeId="1" CreationDate="2019-06-08T20:56:03.400" Score="0" ViewCount="29" Body="&lt;p&gt;How far a spaceship should travel to take a full picture of our own galaxy from the outside?&lt;/p&gt;&#xA;" OwnerUserId="28325" LastEditorUserId="28325" LastEditDate="2019-06-08T22:42:28.723" LastActivityDate="2019-06-08T22:42:28.723" Title="Full picture of our galaxy from the outside" Tags="&lt;milky-way&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2019-06-09T06:37:17.790" ContentLicense="CC BY-SA 4.0" />
  <row Id="32212" PostTypeId="1" AcceptedAnswerId="32225" CreationDate="2019-06-08T22:11:11.690" Score="4" ViewCount="400" Body="&lt;p&gt;In &lt;a href=&quot;https://space.stackexchange.com/a/36633/12102&quot;&gt;this answer&lt;/a&gt; I mention &lt;em&gt;day number&lt;/em&gt; which is 1 on the first day of each calendar year (January 1) and increments to 365 or 366 on December 31 of that year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was an edit proposed, which included a sentence explaining that this could be called the Julian Day. I left a message (elsewhere) to the author of the edit I rejected saying:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hi, I rejected your edit on &lt;a href=&quot;https://space.stackexchange.com/a/36633/12102&quot;&gt;this answer&lt;/a&gt; because &lt;a href=&quot;https://en.wikipedia.org/wiki/Julian_day#Julian_date_calculation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Julian day&lt;/a&gt; is not correct. JD is monotonic and is currently 2,458,643.06&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to which the author replied&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;@uhoh many thanks for the comment! However, there is such a thing as Julian Day of the year. See this NASA web page: &lt;a href=&quot;https://landweb.modaps.eosdis.nasa.gov/browse/calendar.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://landweb.modaps.eosdis.nasa.gov/browse/calendar.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Looking further I saw several similar pages: &lt;a href=&quot;https://landweb.modaps.eosdis.nasa.gov/browse/calendar.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;https://ltdr.modaps.eosdis.nasa.gov/browse/calendar.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;, &lt;a href=&quot;http://uop.whoi.edu/UOPinstruments/frodo/aer/julian-day-table.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;3&lt;/a&gt;, &lt;a href=&quot;https://pubs.usgs.gov/of/2013/1063/images2/Julian_Day_Calendar.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;4&lt;/a&gt;, &lt;a href=&quot;http://people.biology.ucsd.edu/patrick/julian_cal.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;5&lt;/a&gt; Some are called &lt;strong&gt;Julian Day Calendar&lt;/strong&gt; and the rest are called &lt;strong&gt;Julian Day Table&lt;/strong&gt;, but each is a table of what I have defined as &lt;em&gt;day numbers&lt;/em&gt;, 1 on January 1, and 365 or 366 on December 31.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone help me sort out all of the possible, acceptable terms for the following numbers corresponding to yesterday which include some form of &lt;code&gt;Julian&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please be generous; include usages that are less common or have fallen out of use but still have some historical use.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;2,458,643&lt;/code&gt; (integer)&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;2,458,643.06&lt;/code&gt;  (float)&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;160&lt;/code&gt;  (day number for June 8 on non-leap years)&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;June 8, 2019&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2020-11-04T15:08:27.010" Title="Sorting out Julian Day, Julian Date, Julian Day number, Julian Day Calendar, and Julian Day Table" Tags="&lt;history&gt;&lt;time&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32213" PostTypeId="2" ParentId="32207" CreationDate="2019-06-09T00:46:45.523" Score="3" Body="&lt;p&gt;Not sure what you are expecting to see from the two datasets. Both datasets are examples of light curves, flux against time with different arbitrary origins for the time axis; SuperWASP &lt;code&gt;TMID&lt;/code&gt; is integer seconds from Julian Date 2453005.5, Kepler uses BKJD = Barycentric Kepler Julian Date, but offset by 2454833.0. i.e., BKJD = BJD - 2454833.0. The SuperWASP datasets typically span 4 years (it's a little difficult to read the plot scales of the example) and are from a ground-based set of small (11.1 cm diameter) lenses with a CCD and so will have day gaps and weather. Kepler/K2 data are from a space-based 95cm telescope which was able to stare at the same patch of sky continuously for about 90 days before a break (and of course had no weather). Consequently there will be a considerable difference in appearance and photometric precision of the lightcurves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://exoplanetarchive.ipac.caltech.edu/docs/API_superwasptimeseries_columns.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This page&lt;/a&gt; at the NASA Exoplanet Archive gives details on the SuperWASP public data and &lt;a href=&quot;https://github.com/spacetelescope/notebooks/blob/master/notebooks/MAST/K2/K2_Lightcurve/k2_lightcurve.ipynb&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Python Jupyter notebook&lt;/a&gt; has info on plotting K2 light curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Disclaimer: I worked on the SuperWASP project for several years)&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2019-06-09T00:46:45.523" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32214" PostTypeId="1" AcceptedAnswerId="32226" CreationDate="2019-06-09T01:35:19.093" Score="3" ViewCount="279" Body="&lt;p&gt;I understand that it isn't &quot;fire&quot; in the combustion of oxygen sense, but I can't find an answer online about what we're witnessing happening at the visible surface of the sun if all actual fusion explosive stuff is occurring near the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would love a primer or a link to an explanation.&lt;/p&gt;&#xA;" OwnerUserId="5405" LastActivityDate="2019-06-10T17:20:38.853" Title="If fusion only happens near the center of the sun, what's with all the &quot;fire&quot; on the surface?" Tags="&lt;the-sun&gt;" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32215" PostTypeId="1" AcceptedAnswerId="32248" CreationDate="2019-06-09T03:49:36.900" Score="6" ViewCount="265" Body="&lt;p&gt;The speed of EM radiation is &lt;em&gt;very slightly&lt;/em&gt; less than &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;, because space is not quite a vacuum.  Say EM travels at &lt;span class=&quot;math-container&quot;&gt;$(1-\varepsilon)c$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, this results in a slight delay between receiving a gravitational wave and detecting an associated EM emission from the body that caused it.  For an object about 130 million light years away (GW170817) I've seen this delay quoted as everything from &lt;span class=&quot;math-container&quot;&gt;$1.74$&lt;/span&gt; seconds (so &lt;span class=&quot;math-container&quot;&gt;$\varepsilon\approx 4\times 10^{-16}$&lt;/span&gt;) to 27 minutes (so &lt;span class=&quot;math-container&quot;&gt;$\varepsilon\approx 4\times 10^{-13}$&lt;/span&gt;).  I'm sure these figures depend on all sorts of interesting astronomy to do with black hole mergers and gamma ray bursts, but the actual value of &lt;span class=&quot;math-container&quot;&gt;$\varepsilon$&lt;/span&gt; doesn't need to know any of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm asking here because anyone working on LIGO must surely know our best estimate for the value of &lt;span class=&quot;math-container&quot;&gt;$\varepsilon$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="28326" LastActivityDate="2019-06-11T01:45:17.580" Title="What's the actual speed of electromagnetic radiation in space?" Tags="&lt;light&gt;&lt;gravitational-waves&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32216" PostTypeId="2" ParentId="32214" CreationDate="2019-06-09T09:39:39.540" Score="-1" Body="&lt;p&gt;The fusion reactions only happen in the central regions, because immense heat and pressure are essential for these reactions to take place. The mutual repulsion of &#xA;like charges has to be overcome before the various nuclei can fuse. Then the heat and energy generated has to find a way out, otherwise the star would explode, so this heat and energy gradually makes its way to the surface, taking some time to do it. Fortunately some energy is instantly carried away by neutrinos, which reach the surface at the speed of light. However, scientists are baffled by the surprisingly high temperature of the corona, which mysteriously is much higher than at the sun's surface. So far, no one has come up with a mechanism which explains this.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastEditorUserId="14094" LastEditDate="2019-06-10T01:50:54.807" LastActivityDate="2019-06-10T01:50:54.807" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32217" PostTypeId="1" CreationDate="2019-06-09T12:09:39.177" Score="-2" ViewCount="61" Body="&lt;p&gt;Quantum mechanics requires a vacuum energy 40 orders of magnitude higher than that indicated by cosmological observations. Which side is most likely to be right,&amp;amp; when the truth is finally established beyond doubt,as one day it will be,won't this be extremely embarrassing for whichever group turns out to be wrong? Forty orders of magnitude is not a trivial error.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-06-09T16:10:44.247" Title="Discrepancy Between Results of Cosmological Observations &amp; the Assertions of Quantum Mechanics" Tags="&lt;cosmology&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2019-06-09T23:20:49.827" ContentLicense="CC BY-SA 4.0" />
  <row Id="32218" PostTypeId="2" ParentId="32214" CreationDate="2019-06-09T16:03:49.493" Score="6" Body="&lt;p&gt;99% of nuclear fusion of hydrogen nuclei into heavy nuclei happens at the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Solar_core&quot; rel=&quot;nofollow noreferrer&quot;&gt;core&lt;/a&gt; of the Sun. This is because only at the core is the temperature high enough to overcome the &lt;a href=&quot;https://en.wikipedia.org/wiki/Coulomb_barrier&quot; rel=&quot;nofollow noreferrer&quot;&gt;coulomb barrier&lt;/a&gt; caused by the electrostatic repulsion between the nuclei. The solar core produces a lot of energy &lt;span class=&quot;math-container&quot;&gt;$(3.828×10^{26} W)$&lt;/span&gt; by fusion and since all that energy has to go somewhere, it eventually reaches the surface making the surface of the Sun hot (6000K).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most objects like a red hot iron or tungsten filament of a bulb glow when heated. An idealized version of this is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Black-body_radiation?wprov=sfla1&quot; rel=&quot;nofollow noreferrer&quot;&gt;black body radiation&lt;/a&gt;. The colour at which the body glows or the frequency of the electromagnetic radiation depends on the temperature. Since the surface of the Sun is hot, it glows. This is not combustion but natural heat glow. Moreover glowing hot plasma rises up in columns while cooler denser gases settle at the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Photosphere?wprov=sfla1&quot; rel=&quot;nofollow noreferrer&quot;&gt;surface&lt;/a&gt; of the Sun at speeds upto 7km/s in a process called &lt;a href=&quot;https://en.wikipedia.org/wiki/Granule_(solar_physics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar granulation&lt;/a&gt; causing the Sun to look like a fierce fire ball.&lt;/p&gt;&#xA;" OwnerUserId="16469" LastEditorUserId="16469" LastEditDate="2019-06-10T17:20:38.853" LastActivityDate="2019-06-10T17:20:38.853" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32220" PostTypeId="2" ParentId="32217" CreationDate="2019-06-09T16:10:44.247" Score="2" Body="&lt;p&gt;I'm not sure the language of &quot;sides&quot; really applies here. There is a model, one of the predictions seems to be wrong, so we will try to find a better model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By way of analogy, look at the model of black-body radiation developed in the 19th century, one of its predictions was that an infinite amount of energy should be radiated at very short wavelengths. That prediction was ridiculous, so over the early part of the 20th century, people developed a better model, that &quot;better model&quot; became known as quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's nothing embarrassing about a model being wrong. All models are wrong, buts some are more useful than others. QM is a very useful model of reality in many many situations but seems to give some surprising values. So physicists try to find something better. There aren't &quot;groups&quot;. This isn't a school playground. Being wrong isn't embarrassing, it just means the theories (or the observations) are incomplete.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-06-09T16:10:44.247" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32221" PostTypeId="1" AcceptedAnswerId="32236" CreationDate="2019-06-09T17:09:59.290" Score="3" ViewCount="101" Body="&lt;p&gt;I am currently trying to make a Keplerian telescope from &quot;scratch&quot; with plano-convex lenses bought online. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yN81W.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yN81W.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found my objective and eyepiece lenses. I'm just not sure if they will suffice to observe Mars with detail, I'm especially doubting my eyepiece choice because of the lens' diameter. Here are my lenses:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Objective: &lt;a href=&quot;https://i.stack.imgur.com/Jom4r.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jom4r.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Eyepiece: &lt;a href=&quot;https://i.stack.imgur.com/pcmWW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pcmWW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if there's any other lenses combo that you could recommend that aren't too pricey (~300$CAD) that could make observing Mars possible, that would be wonderful!!&lt;/p&gt;&#xA;" OwnerUserId="28328" LastEditorUserId="28328" LastEditDate="2019-06-09T17:19:27.147" LastActivityDate="2019-06-10T17:34:07.487" Title="Custom telescope" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;mars&gt;&lt;newtonian-telescope&gt;&lt;telescope-making&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32222" PostTypeId="1" CreationDate="2019-06-09T19:31:13.410" Score="1" ViewCount="467" Body="&lt;p&gt;I am trying to save a NumPy RGB array (a 3d array with dimension 512x512x3), to a .fits file using &lt;code&gt;astropy.io&lt;/code&gt;.&#xA;Saving this array to a .png file works fine, but when I'm trying to save it to a .fits file using &lt;code&gt;HDUList.writeto&lt;/code&gt;, it produces just garbage colors. &#xA;Also, I have to reshape the image like so:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;img = img.reshape(img.shape[2], img.shape[0], img.shape[1])&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or it will try to store just 3 column of the image instead of 512.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each element in the 3d array contains a value between 0 and 255.&lt;/p&gt;&#xA;" OwnerUserId="28329" LastEditorUserId="28329" LastEditDate="2019-06-10T17:08:31.130" LastActivityDate="2019-06-10T17:08:31.130" Title="Converting an RGB image to fits (astropy)" Tags="&lt;python&gt;&lt;astropy&gt;&lt;image-processing&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32223" PostTypeId="2" ParentId="32214" CreationDate="2019-06-10T00:41:32.650" Score="3" Body="&lt;p&gt;Probably the &quot;explosive&quot; events at the surface that you are asking about are solar flares.  The cause of solar flares is thought to be release of magnetic energy, because it is known that there are strong magnetic fields threading the surface of the Sun, and they get tangled up by the convection of the Sun as hot gas rises to the surface and cool gas drops back down.   The gas density above the surface of the Sun drops very fast, so it is easy to heat that gas up to very high temperatures, though the details of how this hot &quot;corona&quot; is created are not entirely certain (it appears to be a combination of heating by pressure waves and various other plasma and magnetic waves).  Usually that heating proceeds in small spurts, but occasionally a region of very strong and very twisted fields is created, and it can release energy more explosively, which produces a strong solar flare and often comes with a &quot;coronal mass ejection.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the bottom line is, the physics of explosions at and above the surface are magnetic in nature, and are very different from nuclear fusion. These &quot;explosions&quot; are only possible because the density is so low, which is also why fusion is not important, even though the temperatures are comparable.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2019-06-10T00:41:32.650" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32224" PostTypeId="2" ParentId="32222" CreationDate="2019-06-10T04:00:41.467" Score="4" Body="&lt;p&gt;I haven't done much astronomical image processing before, but as this question is unanswered I'll give it a shot - hopefully to some avail. If the problem is more specific, a code sample/image sample would probably be useful for further diagnosis, but otherwise &lt;a href=&quot;http://docs.astropy.org/en/stable/generated/examples/io/split-jpeg-to-fits.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this example&lt;/a&gt; may help. It discusses the process of writing a 3-channel image to separate FITS images. I would try this first, and check if this output is also garbage - it might just be an issue with how you're using the HDUList.writeto() function. The relevant code is pasted below (with a couple of edits, as the example uses the Pillow Image class at first):&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Split the three channels (RGB):&lt;/h3&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;r, g, b = img[:, :, 0], img[:, :, 1], img[:, :, 2]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h3&gt;Write out the channels as separate FITS images&lt;/h3&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;red = fits.PrimaryHDU(data=r)&#xA;red.writeto('red.fits')&#xA;&#xA;green = fits.PrimaryHDU(data=g)&#xA;green.writeto('green.fits')&#xA;&#xA;blue = fits.PrimaryHDU(data=b)&#xA;blue.writeto('blue.fits')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Also out of curiosity - why is your &quot;numpy RGB array&quot; storing values between 0 and 512? The standard for images is usually integers between 0-255 or floats between 0 and 1; but my guess is that astronomy requires a larger bit depth? In any case, you should also make sure the dtype of your numpy array is large enough for the desired bit depth - often images are stored with dtype=&quot;uint8&quot; which ranges from 0 to 255.&lt;/p&gt;&#xA;" OwnerUserId="28330" LastEditorUserId="28330" LastEditDate="2019-06-10T04:18:05.363" LastActivityDate="2019-06-10T04:18:05.363" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32225" PostTypeId="2" ParentId="32212" CreationDate="2019-06-10T06:30:21.970" Score="7" Body="&lt;p&gt;The &lt;a href=&quot;https://web.archive.org/web/20190503053806/https://aa.usno.navy.mil/data/docs/JulianDate.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Julian Date&lt;/a&gt; (JD) is the real number of days since 12:00 UT, January 1, 4713 BCE (proleptic Julian calendar).&#xA;The Julian Day Number (JDN) is floor(JD), the nearest integer ≤ JD.&lt;/p&gt;&#xA;&lt;p&gt;The &lt;a href=&quot;https://web.archive.org/web/20190503041626/https://tycho.usno.navy.mil/mjd.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Modified Julian Date&lt;/a&gt; (MJD) is JD - 2400000.5, or the real number of days since 1858-11-17 0:00 UT.&#xA;The IAU officially approves either JD or MJD as long it's clear which quantity is given.&lt;/p&gt;&#xA;&lt;p&gt;A number of 366 or less in a bureaucratic or informatics context is a day number or day of the year.&#xA;In his MJD &lt;a href=&quot;https://web.archive.org/web/20190503041626/https://tycho.usno.navy.mil/mjd.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;explainer&lt;/a&gt;, the late USNO Time Service director Gernot Winkler strongly opposed calling this a &amp;quot;Julian&amp;quot; date:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The MJD (and even more so the JD) has to be well distinguished from this day of the year (DOY).&#xA;This is also often but erroneously called Julian Date, when in fact it is a Gregorian Date expressed as number of days in the year.&#xA;This is a grossly misleading practice that was introduced by some who were simply ignorant and too careless to learn the proper terminology.&#xA;It creates a confusion which should not be taken lightly.&#xA;Moreover, a continuation of the use of expressions &amp;quot;Julian&amp;quot; or &amp;quot;J&amp;quot; day in the sense of a Gregorian Date will make matters even worse.&#xA;It will inevitably lead to dangerous mistakes, increased confusion, and it will eventually destroy whatever standard practices exist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Expressing the example date of 2019-06-08 13:26 UT in various forms:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Julian day number: 2458643&lt;/li&gt;&#xA;&lt;li&gt;Julian date: 2458643.06&lt;/li&gt;&#xA;&lt;li&gt;Modified Julian date: 58642.56&lt;/li&gt;&#xA;&lt;li&gt;Day of year: 159&lt;/li&gt;&#xA;&lt;li&gt;Ordinal date: 2019-159&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2020-11-04T15:08:27.010" LastActivityDate="2020-11-04T15:08:27.010" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32226" PostTypeId="2" ParentId="32214" CreationDate="2019-06-10T06:47:13.810" Score="6" Body="&lt;p&gt;The &quot;fire&quot; on the surface arises because the core is really, really hot. If something has a hot interior, then its exterior will also get hot, even if not &lt;em&gt;as&lt;/em&gt; hot as that interior is. You can see this in an everyday situation by thinking about an oven. If the oven has been running for a while, you will probably not want to touch the outside surface of the door, even if that is not the inside surface and not as hot. Heat transfers from the hot center to the exterior through the intervening matter in a number of ways, and the same is the case with a star like the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And at 15 million kelvins, that is a &lt;em&gt;lot&lt;/em&gt; of heat transfer, so much so that the outside surface, even with on the order of &lt;span class=&quot;math-container&quot;&gt;$10^9\ \mathrm{Yg}$&lt;/span&gt; of stellar matter piled on top, has to then be at 5800 K - over twice the heat of a tungsten bulb filament - in order to balance it. The reason that surface is cooler than the core is in part because it is a lot bigger than the core surface and also because the matter immediately surrounding the core transfers a sizeable amount of the heat right back in, thus slowing down the rate at which it goes out. Heat &quot;percolates&quot; from core to surface, instead of just rushing out all at once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Insofar as the relation to fusion is concerned, the answer is that fusion &lt;em&gt;keeps&lt;/em&gt; the Sun's core hot. It did not &lt;em&gt;make&lt;/em&gt; it hot to begin with: that heat was originally provided by the gravitational contraction of the gas cloud that formed the star in the first place. That's how the core first got to 15 MK, at which point the rate of fusion became enough to sustain it there in an equilibrium by replenishing the heat being lost. (Actually, it was likely cooler. As the Sun has aged, it has steadily grown hotter, for reasons that are a bit complicated to get into in full detail but relate to the depletion of the hydrogen fusion fuel over time causing the star to contract and thus release more gravitational energy.) The surface would still be just as hot and interesting were there no fusion - although if that were the case, the Sun would gradually contract further and eventually become a dense white dwarf, and would have done so long before the present day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you are asking about the various complicated phenomena on the surface, i.e. things like the &quot;bubbly&quot; appearance it has, the flares and prominences, sunspots, etc. - these things are all basically the result of the fact that all that hot gas also moves and churns thanks to the temperature gradient and, moreover, due to its heat, it is actually not just &quot;gas&quot; but a plasma, which then adds complicated electromagnetic effects into the mix as it generates its own magnetic fields with which it also interacts.&lt;/p&gt;&#xA;" OwnerUserId="9807" LastEditorUserId="9807" LastEditDate="2019-06-10T06:54:33.993" LastActivityDate="2019-06-10T06:54:33.993" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32227" PostTypeId="1" AcceptedAnswerId="32252" CreationDate="2019-06-10T08:43:01.807" Score="1" ViewCount="160" Body="&lt;p&gt;I made a scatter plot in Python that looks like this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7STO3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7STO3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I want galactic longitude, l, to be plotted from 180 to -180 like these graphs from Iorio and Belokurov (2019): &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1Ni5G.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1Ni5G.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is the code for my plot: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;sns.set_style(&quot;ticks&quot;)&#xA;sns.set_context(&quot;poster&quot;)&#xA;%matplotlib inline&#xA;from matplotlib import pyplot as plt&#xA;&#xA;fig = plt.figure(figsize=(8,5))&#xA;&#xA;ax  = fig.add_subplot(1,1,1, aspect='equal')&#xA;&#xA;ax.scatter(rrl_pm.l, rrl_pm.b, s=1, color='black', alpha=0.1)&#xA;&#xA;ax.set_xlim(0, 360)&#xA;ax.set_ylim(-90., 90.)&#xA;&#xA;ax.set_xlabel(&quot;l [deg]&quot;)&#xA;ax.set_ylabel(&quot;b [deg]&quot;)&#xA;textstr = 'N = ' + str(len(rrl_pm.l))&#xA;props = dict(boxstyle='round', facecolor='white', alpha=0.5)&#xA;ax.text(0.05, 0.95, textstr, transform=ax.transAxes, fontsize=16,&#xA;        verticalalignment='top', bbox=props)&#xA;plt.title(&quot;Scatter Plot of Cleaned Sample of RRLs in Sgr dSph with Spatial AND PM Cuts&quot;, fontsize = 16)&#xA;plt.savefig(&quot;SP-Clean-Spatial-PM.png&quot;, dpi = 1000)&#xA;plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If I simply change the xlim to run from 180 to -180 degrees the following graph is produced:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RrfE1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RrfE1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So essentially it only covers half the sky. Why exactly is this the case and what transformations do I need to apply to the galactic longitude for the 2nd plot to &quot;wrap around&quot; and display both sides? &lt;/p&gt;&#xA;" OwnerUserId="28331" LastEditorUserId="7982" LastEditDate="2019-06-11T12:38:44.387" LastActivityDate="2019-06-11T13:01:09.690" Title="Plotting Galactic Longitude from 180 to -180" Tags="&lt;coordinate&gt;&lt;python&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32229" PostTypeId="1" AcceptedAnswerId="32244" CreationDate="2019-06-10T10:58:56.927" Score="2" ViewCount="205" Body="&lt;p&gt;I am attempting this past paper question and am not sure how to tackle this, this is not homework! :)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The spectrum of a distant quasar shows two broad emission lines with observed&#xA;  wavelengths of 317.7nm and 404.7nm. The strongest lines are most likely to be either Lyman-α with a rest wavelength 121.6nm, CIV with a rest wavelength of 154.9nm or MgII with a rest wavelength of 280.0nm.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(a) What is the redshift of the quasar? Please show your working.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(b) What two lines have been detected?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(c) The two lines have widths of 10nm and 20nm respectively. If I have measured&#xA;  the mass of the black hole in the quasar to be 10^9Msun, how far from the centre of the quasar are the regions of ionised gas responsible for the emission lines?&#xA;  Please specify any assumptions that you make to obtain your result.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am assuming I must find z to be the ratio of the change in wavelength to the rest wavelength. However I'm unsure of how to identify these in the question.&lt;/p&gt;&#xA;" OwnerUserId="28333" LastActivityDate="2019-06-10T22:55:54.480" Title="Using emission lines to determine redshift of a quasar" Tags="&lt;redshift&gt;&lt;wavelength&gt;&lt;spectra&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32230" PostTypeId="2" ParentId="32229" CreationDate="2019-06-10T11:08:31.623" Score="4" Body="&lt;p&gt;So the two observed lines must be two of the three suggested lines, red-shifted by &lt;strong&gt;the same amount&lt;/strong&gt;. That means the &lt;em&gt;ratio&lt;/em&gt; of their wavelengths will be unchanged, so we need two of the suggested lines whose wavelength is close to the ratio &lt;span class=&quot;math-container&quot;&gt;$404.7/317.6$&lt;/span&gt;.  It's easy to check that only the first two are close to that ratio, so they must be those two. So the redshift should be equal to &lt;span class=&quot;math-container&quot;&gt;$317.7/121.6 = 2.61$&lt;/span&gt;  and to &lt;span class=&quot;math-container&quot;&gt;$404.7/154.9$&lt;/span&gt; which is also &lt;span class=&quot;math-container&quot;&gt;$2.61$&lt;/span&gt; (such close agreement in real data would be a bit unlikely, and lead one to check the authenticity of the data.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-06-10T11:08:31.623" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32231" PostTypeId="1" CreationDate="2019-06-10T12:05:13.950" Score="-3" ViewCount="116" Body="&lt;p&gt;Bearing in mind the provisions of the 1st Law of Thermodynamics, which decrees that energy can neither be created nor destroyed, all the energy in the Universe must have been there at the time of the Big Bang. Dark energy began to accelerate the expansion about 5 billion years ago, but in what form was it stored before it mysteriously began to manifest itself?&lt;/p&gt;&#xA;" OwnerUserId="28083" LastEditorUserId="14094" LastEditDate="2019-06-10T15:32:44.437" LastActivityDate="2019-06-10T15:32:44.437" Title="In what form was dark energy stored 6 billion years ago?" Tags="&lt;dark-energy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32232" PostTypeId="2" ParentId="32231" CreationDate="2019-06-10T12:12:14.173" Score="0" Body="&lt;p&gt;We don't really know enough about Dark Energy now let alone back then. We're not even entirely sure why there's so much conventional matter versus antimatter, even though logically there should be the same amount of each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 1st Law of Thermodynamics is seemingly true for Generalistic observations, but when we look at Quantum levels, even a perfect vacuum has particles and antiparticles (and therefore energy) being created from nothing, before recombining into nothing. So the 1st Law of Thermodynamics isn't explicitly accurate for that scale&lt;/p&gt;&#xA;" OwnerUserId="11525" LastActivityDate="2019-06-10T12:12:14.173" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32233" PostTypeId="2" ParentId="32215" CreationDate="2019-06-10T12:36:14.683" Score="1" Body="&lt;p&gt;Given that different frequencies of light, or energies of photons, take your pick, travel through different the same medium at different rates, e.g. red light is slowed less by a glass prism and blue is slowed more, it would seemingly take higher energy signals longer to reach the observer, be it cosmic dust, interplanetary/galactic gas, or whatever else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way to calculate would be by observing a singular pulse over multiple synchronised frequencies simultaneously. Apparently with GW170817 as you mentioned there was an observed Gamma Ray Burst by the Fermi and INTEGRAL spacecraft 1.7 seconds after the LIGO detected event, but as far as I'm aware that was the only other observation instantaneously. Most other observations were telescopes directed there after being alerted by the GRB&lt;/p&gt;&#xA;" OwnerUserId="11525" LastActivityDate="2019-06-10T12:36:14.683" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32234" PostTypeId="1" CreationDate="2019-06-10T14:41:32.470" Score="1" ViewCount="311" Body="&lt;p&gt;I found this NASA website which says that to predict eclipse, we should solve the Besselian elements ..what does this mean, how can I solve for Besselian elements? Is there a numerical method? I need to find the circumstance (length of eclipse etc) at a given coordinate..Sorry I am  very new to astronomy so my question might be basic..&#xA;&lt;a href=&quot;https://i.stack.imgur.com/qXUJN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qXUJN.png&quot; alt=&quot;see the picture below&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="28335" LastActivityDate="2019-06-10T14:41:32.470" Title="evaluate solar eclipse Besselian elements?" Tags="&lt;fundamental-astronomy&gt;&lt;solar-eclipse&gt;&lt;positional-astronomy&gt;&lt;moon-phases&gt;&lt;mathematics&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2019-06-11T13:37:14.160" ContentLicense="CC BY-SA 4.0" />
  <row Id="32235" PostTypeId="1" AcceptedAnswerId="32249" CreationDate="2019-06-10T17:17:02.660" Score="5" ViewCount="294" Body="&lt;p&gt;I'm having trouble understanding how to solve a certain problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is from the book &lt;em&gt;Astronomy: Principles and Practice&lt;/em&gt; by Roy and Clarke. It is actually a solved example, but I am unable to understand parts of the solution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Draw the celestial sphere for latitude 30&amp;deg; N, showing the star Sirius (right ascension&#xA;  6h 40m, declination 17&amp;deg; S) at rising and draw the ecliptic. Estimate from your diagram the approximate&#xA;  date when Sirius rises with the Sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Vo2DP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vo2DP.png&quot; alt=&quot;This is the image from the solution&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the image from the solution. I'll copy the entire solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/33RRa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/33RRa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DZDWx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DZDWx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here's what I don't understand. If Sirius is at X, and if X is on a parallel of declination, why isn't this parallel parallel to the equatorial plane? Does the same go for the Sun? And how do we get the RA of the Sun?&lt;/p&gt;&#xA;" OwnerUserId="28337" LastEditorUserId="616" LastEditDate="2019-06-10T21:10:22.930" LastActivityDate="2019-06-11T05:02:31.993" Title="Draw the celestial sphere for latitude, showing the star Sirius, Estimate the approximate date when Sirius rises with the Sun" Tags="&lt;observational-astronomy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32236" PostTypeId="2" ParentId="32221" CreationDate="2019-06-10T17:34:07.487" Score="1" Body="&lt;p&gt;There are a lot of potential problems here. First,  double-convex lenses are not necessarily the best so far as chromatic distortion, or, more important, astigmatism &amp;amp; coma go. Most likely plano-convex will work better.&lt;br&gt;&#xA;As you suggest, in general the ocular (eyepiece) lens can be much smaller in diameter and still cover the stop-limited field of view of the objective.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend you do some searching for &quot;make your own Keplerian&quot; to see what sort of lenses are recommended vs. telescope size, power, cost.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also strongly recommend you get or borrow a copy of Smith's &quot;Modern Optical Engineering,&quot; which will give you a great introduction to simple optical systems and the difference between axial ray-trace and full-aperture performance.  Once you're comfortable with the concepts of Field Stop and Aperture Stop, and lower-order aberrations, you may be better set to design your own 'scope. &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2019-06-10T17:34:07.487" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32237" PostTypeId="1" CreationDate="2019-06-10T18:49:53.053" Score="2" ViewCount="422" Body="&lt;p&gt;How can the gravitational redshift of a very distant quasar be distinguished from its cosmological redshift? Quasars are very massive objects,thought to be supermassive black holes,so therefore must have substantial gravitational red shift,but they are so far away they must also have a cosmological redshift. Is it possible to distinguish one from the other?&lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-06-10T20:03:38.510" Title="Quasar Redshifts" Tags="&lt;redshift&gt;&lt;quasars&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32238" PostTypeId="1" AcceptedAnswerId="32242" CreationDate="2019-06-10T19:23:55.347" Score="11" ViewCount="901" Body="&lt;p&gt;I was wondering if it would be possible to record the emission/absorption spectra of stars without expensive spectroscopy equipment. Would it be possible to somehow utilize diffraction grating? I own a Newtonian with a 130mm aperture if that helps.&lt;/p&gt;&#xA;" OwnerUserId="28338" LastActivityDate="2019-06-10T20:11:48.497" Title="Recording Spectral Lines at Home" Tags="&lt;telescope&gt;&lt;spectroscopy&gt;&lt;spectra&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32239" PostTypeId="2" ParentId="32235" CreationDate="2019-06-10T19:36:30.027" Score="2" Body="&lt;p&gt;The the image does not display the parallel of declination that X is on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only declination line shown in the image is the equator. The two arcs shown intersecting at X are the observer's horizon (NWSXE) and the Right Ascension/Hour Angle arc passing through Sirius (PAXQ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm afraid I'm not sufficiently familiar with the methods in the mentioned book to work out how the Sun's  Right ascension was determined from the answer snippet.&lt;/p&gt;&#xA;" OwnerUserId="23813" LastEditorUserId="23813" LastEditDate="2019-06-11T00:36:10.780" LastActivityDate="2019-06-11T00:36:10.780" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32240" PostTypeId="2" ParentId="32237" CreationDate="2019-06-10T19:46:59.243" Score="3" Body="&lt;p&gt;Both cosmological redshifts, and gravitational redshifts, can be thought of as coming from the same source-- the equations of general relativity.  In that sense, the distinction between them is somewhat artificial, though useful, but they are redshifts that affect all the lines in the same way.  So there is no way to disentangle their effects, other than to model both of their sources.  This is not so unusual-- if you have normal Doppler shifts in a reference frame where you don't know the motion of either the source or the detector, then the Doppler shift registered will represent a combination of those motions that you cannot disentangle simply from what you observe-- you will have to model the effects of all contributing effects.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That generally means you need to bring in more independent information, such that you can test your models.  In the case of quasars, you could look for quasars that appear to be similar except for their distance from you, and then you can assume their gravitational redshift is similar but their cosmological redshift is very different.  In particular, you can have relatively nearby quasars that do not have a lot of cosmological redshift, perhaps even comparable to their gravitational redshift.   And you can have very distant ones that are predominantly cosmological redshifts.  You can also use the fact that the cosmological redshift is the same for all the lines of a given quasar, but gravitational redshifts will depend on how deep in the gravity well each line forms.  There is also a connection between where a line forms and what its Doppler shift is, because the gas is generally moving, so we even have a third source of frequency shift, but it connects with gravitational redshift in ways that alter the line shape.  So we have new information from line shape, that allows the models to be tested, even though we must combine no less than three separate sources of frequency shift (special relativistic, general relativistic due to local gravity, general relativistic due to global effects of the expanding universe). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have other information also, such as temperatures and Doppler shifts, and time dependence, so you put all that together and you try to model the quasar.  That's the only way to disentangle the redshifts, you need self-consistent models that succeed at producing a wide range of observables.  Remember that we had quasar spectra for many years before we were even able to figure out what they were, and debate still continues about the sources of the various spectral components.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2019-06-10T20:01:30.300" LastActivityDate="2019-06-10T20:01:30.300" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32241" PostTypeId="2" ParentId="32237" CreationDate="2019-06-10T20:03:38.510" Score="7" Body="&lt;p&gt;The optical emission lines of quasars do not come from sufficiently close the the central supermassive black hole to be appreciably gravitationally redshifted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they did arise from gas near the &quot;innermost stable circular orbit&quot;, then the maximum gravitational redshift would be about 0.2. In addition, the lines would have a characteristic profile caused by a combination of gravitational redshift and the blue/redshift and Doppler boosting caused by the fast-moving gas (such line profiles are sometimes seen at &lt;strong&gt;X-ray&lt;/strong&gt; wavelengths). Of course many if not most quasars have redshifts that far exceed 0.2 and can only be explained by cosmic expansion.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-06-10T20:03:38.510" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32242" PostTypeId="2" ParentId="32238" CreationDate="2019-06-10T20:06:18.677" Score="11" Body="&lt;p&gt;Here is a link to a &lt;a href=&quot;https://www.rspec-astro.com/star-analyser/&quot; rel=&quot;noreferrer&quot;&gt;diffraction grating&lt;/a&gt; that can do what you want. It is mounted in a 1.25&quot; filter ring that attaches to an eyepiece, or to most astro cameras. I believe they also sell adapters for other cameras and software to extract spectra from the photos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link to a spectra of &lt;a href=&quot;https://gvk.smugmug.com/Astrophotography/Spectroscopy/i-jk4zSHs/A&quot; rel=&quot;noreferrer&quot;&gt;Vega&lt;/a&gt; that I took with a 120 mm refractor using  the Star Analyser grating.&lt;/p&gt;&#xA;" OwnerUserId="28086" LastEditorUserId="28086" LastEditDate="2019-06-10T20:11:48.497" LastActivityDate="2019-06-10T20:11:48.497" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32243" PostTypeId="1" CreationDate="2019-06-10T22:34:03.243" Score="-1" ViewCount="210" Body="&lt;p&gt;In this YT video Michelle Thaller says The Big Bang wasn’t an explosion but an expansion and as such there's no empty center where the explosion would've been.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain expansion she uses the analogies of surface of a balloon being blown up and stretching a rubber sheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But isn't it true that any shape, be it having an edge (the sheet) or a balloon/sphere, would have a point that isn't expanding? ... the center of an expanding 2D rectangle, a sphere or cube, etc?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it more accurate to say that it's not like a sheet or balloon but that either every point in space is expanding three dimensionally, or, that space is actually not expanding, because it doesn't exist, and that what we call 'space expanding' is actually cosmic entities traveling in uncoordinated directions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=veQtF3_2tOE&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=veQtF3_2tOE&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="28342" LastActivityDate="2019-06-12T17:04:38.787" Title="What does it mean for space to expand?" Tags="&lt;big-bang-theory&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32244" PostTypeId="2" ParentId="32229" CreationDate="2019-06-10T22:55:54.480" Score="2" Body="&lt;p&gt;Steve Linton answers correctly that the line ratios must be (nearly) identical and hence identifies the two lines as Lyman &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; and C IV, but it's not true that the two obtained redshifts are suspiciously alike. In fact, when you do the calculation, you get (from &lt;span class=&quot;math-container&quot;&gt;$z \equiv \lambda_\mathrm{obs}/\lambda_\mathrm{rest} - 1$&lt;/span&gt;),&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{array}{rcl}&#xA;z_{\mathrm{Ly}\alpha} &amp;amp; = &amp;amp; \frac{317.7\,\mathrm{nm}}{125.67\,\mathrm{nm}} &amp;amp; \simeq &amp;amp; 1.6134 \\&#xA;z_\mathrm{C\,IV}      &amp;amp; = &amp;amp; \frac{404.7\,\mathrm{nm}}{154.9\,\mathrm{nm}} &amp;amp; \simeq &amp;amp; 1.6127.&#xA;\end{array}&#xA;$$&lt;/span&gt;&#xA;That is, Lyman &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; is slightly more redshifted than C IV. This is often seen, and is due to Lyman &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; scattering resonantly on the neutral hydrogen enshrouding the quasar / host galaxy. This effect is more pronounced the higher the redshift, and by &lt;span class=&quot;math-container&quot;&gt;$z\simeq6$&lt;/span&gt;, the blue part of the spectrum is completely gone (this is the so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Gunn%E2%80%93Peterson_trough&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gunn-Peterson trough&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relating the widths &lt;span class=&quot;math-container&quot;&gt;$\Delta\lambda$&lt;/span&gt; of the lines to the motion around the black hole, the velocity of the gas responsible for emitting the lines is given by &lt;span class=&quot;math-container&quot;&gt;$v/c = \Delta\lambda / \lambda$&lt;/span&gt;, so the velocities are&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{array}{rcl}&#xA;v_{\mathrm{Ly}\alpha} &amp;amp; = &amp;amp; c \frac{10\,\mathrm{nm}}{317.7\,\mathrm{nm}} &amp;amp; \simeq &amp;amp; 9\,400\,\mathrm{km}\,\mathrm{s}^{-1} \\&#xA;v_\mathrm{C\,IV}      &amp;amp; = &amp;amp; c \frac{20\,\mathrm{nm}}{404.7\,\mathrm{nm}} &amp;amp; \simeq &amp;amp; 14\,800\,\mathrm{km}\,\mathrm{s}^{-1}.&#xA;\end{array}&#xA;$$&lt;/span&gt;&#xA;From simple dynamics, this can be converted to a distance &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; from the black hole, knowing its mass &lt;span class=&quot;math-container&quot;&gt;$M_\bullet$&lt;/span&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{array}{rcl}&#xA;r &amp;amp; = &amp;amp; \frac{G M_\bullet}{v^2} &amp;amp; \Rightarrow \\&#xA;r_{\mathrm{Ly}\alpha} &amp;amp; \simeq &amp;amp; 0.05\,\mathrm{pc} &amp;amp; \simeq &amp;amp; 58\,\mathrm{light\text{-}days} \\&#xA;r_{\mathrm{C\,IV}} &amp;amp; \simeq &amp;amp; 0.02\,\mathrm{pc} &amp;amp; \simeq &amp;amp; 23\,\mathrm{light\text{-}days},&#xA;\end{array}&#xA;$$&lt;/span&gt;&#xA;which is consistent with the fact that these lines are formed in the so-called &lt;em&gt;broad line region&lt;/em&gt; which has dimension from tens of light-days to several parsec (in contrast, the narrow line region has dimensions of several 100s to ~1000 light-years).&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2019-06-10T22:55:54.480" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32245" PostTypeId="2" ParentId="32202" CreationDate="2019-06-10T23:26:41.407" Score="3" Body="&lt;p&gt;I'm far from an expert in this topic, but I was curious about this question and looked into it myself. Sadly, I was only able to come up with a pretty weak answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The very &lt;a href=&quot;https://en.wikipedia.org/wiki/Bow_shocks_in_astrophysics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page you got the images from&lt;/a&gt; contains this line,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bow shocks are also a common feature in Herbig Haro objects, in which a much stronger collimated outflow of gas and dust from the star interacts with the interstellar medium, producing bright bow shocks that are visible at optical wavelengths.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But it is not cited. Perhaps it should have a [citation needed]. That sentence does link to &lt;a href=&quot;https://en.wikipedia.org/wiki/Herbig%E2%80%93Haro_object&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page about the Herbig Haro Object&lt;/a&gt;, which does contain a cited line,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;HH objects associated with very young stars or very massive protostars are often hidden from view at optical wavelengths by the cloud of gas and dust from which they form. The intervening material can diminish the visual magnitude by factors of tens or even hundreds at optical wavelengths. Such deeply embedded objects can only be observed at infrared or radio wavelengths.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So it seems to me the answer to your question is that no, images of bow shocks in visible light are not produced because the visible light from the bow shock cannot reach us due to there being stuff in the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that an answer based on 2 Wikipedia pages as sources is pretty weak by Stack Exchange standards, and that these Herbig Haro Objects are not the only thing in the universe that create bow shocks, and therefore the possibility exists that visible bow shocks around other objects may exist. I hope this answer inspires an expert to give an answer deserving of that green check mark. &lt;/p&gt;&#xA;" OwnerUserId="14255" LastActivityDate="2019-06-10T23:26:41.407" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32248" PostTypeId="2" ParentId="32215" CreationDate="2019-06-11T01:45:17.580" Score="4" Body="&lt;p&gt;I believe this question was addressed in the &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/aa920c/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper&lt;/a&gt; , Abbot, et. al., Gravitational Waves and Gamma-Rays from a Binary Neutron Star Merger: GW170817 and GRB 170817A , ApJ. Lett., 848:L13, 2017 October 16.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;We use the observed time delay of (&lt;span class=&quot;math-container&quot;&gt;$+1.74 \pm 0.05$&lt;/span&gt;) s between GRB 170817A and GW170817 to: (i) constrain the difference between the speed of gravity and the speed of light to be between &lt;span class=&quot;math-container&quot;&gt;$-3 \times  10^{-15}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$+7 \times 10^{-16}$&lt;/span&gt; times the speed of light, ...&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In Section 2.2 of the paper they discuss the uncertainty of delay between GW and GRB emission, predicted to be a few seconds. In Section 4 of the paper they compute the range of speed uncertainty quoted above. They conservatively use 26 MPc, their lower bound on the 90% credible distance interval, and for the upper bound on speed difference they assume the GW and GRB were emitted simultanously, and use the 1.74 s &lt;span class=&quot;math-container&quot;&gt;$\Delta t$&lt;/span&gt;. For a lower bound on speed of gravity, they assume the GRB were emitted 10 s after the GW, and a faster EM signal made up some of the difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that X-rays were also detected several (~8-9) days later. This delay was attributed to time needed for the shock wave from the merger to interact with surrounding matter. I was unable to find any reference to signal arrival 27 minutes after GW detection in the LIGO papers.&lt;/p&gt;&#xA;" OwnerUserId="28086" LastActivityDate="2019-06-11T01:45:17.580" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32249" PostTypeId="2" ParentId="32235" CreationDate="2019-06-11T01:48:48.117" Score="5" Body="&lt;p&gt;The figure doesn't show the -17&amp;deg; parallel of declination.&#xA;If it were shown, it would be parallel to the equator as you expected, as would the parallel of the Sun's declination.&#xA;PAQ and PBQ are semicircles of right ascension, intersecting at the celestial poles P and Q.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an interactive aid, try &lt;a href=&quot;https://stellarium.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stellarium&lt;/a&gt;.&#xA;You can highlight the meridian, equator, ecliptic, etc. in the View &gt; Markings dialog.&#xA;Under View &gt; Sky, orthographic projection renders the sky as a celestial sphere, mirror-reversed from the book figure.&#xA;To reproduce the problem scenario, set a location at 30&amp;deg; N and a date of...July 24.&#xA;The Sun's right ascension on that date is closer to 8h10m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors don't explain how they determined an RA of 8h40m for the Sun, and they're off by half an hour; maybe they eyeballed it as 30&amp;deg; east of Sirius in the diagram.&#xA;The Sun's RA is exactly 0h at the March equinox, 6h at the June solstice, 12h at the September equinox, and 18h at the December solstice.&#xA;Between those milestones, the Earth's orbital motion makes the Sun's RA increase about 4m per day or 1h in 15 days.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-06-11T05:02:31.993" LastActivityDate="2019-06-11T05:02:31.993" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32250" PostTypeId="1" AcceptedAnswerId="32255" CreationDate="2019-06-11T04:00:30.500" Score="2" ViewCount="91" Body="&lt;p&gt;I've read that you can change the longitude of ascending node of a satellite (i.e. changing the orbital plane without changing the inclination) to change its orbital plane by using the torque imparted by the oblateness of Earth. How does this happen exactly? I mean how does the oblateness of Earth cause a &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Precession#Classical_(Newtonian)&quot; rel=&quot;nofollow noreferrer&quot;&gt;torque induced precession&lt;/a&gt; of a satellite?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. I think I don't understand how torque induced precession works and that's why I am not able to understand this. So if any of you kind gentlefolk are willing to explain, I thank you in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I just came across &lt;a href=&quot;https://en.wikipedia.org/wiki/Nodal_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nodal Precession&lt;/a&gt;. And from what I understood that due to Earth's equatorial bulge, the gravitational force vector is offset from its center, thus producing torque. I can visualize how the longitude of ascending node will change, but can't visualise how will the change in the torque caused by slightly offset gravitational force vector will cause it.&lt;/p&gt;&#xA;" OwnerUserId="28345" LastEditorUserId="28345" LastEditDate="2019-08-10T03:34:01.510" LastActivityDate="2019-08-10T03:34:01.510" Title="How can the oblateness of Earth impart torque to change the longitude of ascending node of a satellite?" Tags="&lt;orbit&gt;&lt;orbital-mechanics&gt;&lt;precession&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32252" PostTypeId="2" ParentId="32227" CreationDate="2019-06-11T13:01:09.690" Score="3" Body="&lt;p&gt;Assuming&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x, y = rrl_pm.l, rrl_pm.b&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The transformation you want is &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;xx = [(q+180)%360 - 180 for q in x]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Add 180, do modulo 360, then subtract 180.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then set your limits&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ax.set_xlim(180, -180)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/k3wEG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6OfYr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import matplotlib.pyplot as plt&#xA;&#xA;x = (0, 10, 20, 40, 80, 160, 200, 280, 320, 340, 350)&#xA;y = (0, 10, 20, 30, 40, 50,  -50,  -40, -30, -20, -10)&#xA;&#xA;fig = plt.figure(figsize=(8,5))&#xA;ax  = fig.add_subplot(1,1,1, aspect='equal')&#xA;&#xA;ax.scatter(x, y)&#xA;&#xA;ax.set_xlim(0, 360)&#xA;ax.set_ylim(-90., 90.)&#xA;&#xA;plt.title('Old', fontsize=18)&#xA;&#xA;plt.show()&#xA;&#xA;xx = [(q+180)%360 - 180 for q in x]  # DO THIS!&#xA;&#xA;fig = plt.figure(figsize=(8,5))&#xA;ax  = fig.add_subplot(1,1,1, aspect='equal')&#xA;&#xA;ax.scatter(xx, y)&#xA;&#xA;ax.set_xlim(180, -180)&#xA;ax.set_ylim(-90., 90.)&#xA;&#xA;plt.title('New', fontsize=18)&#xA;&#xA;plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-06-11T13:01:09.690" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32253" PostTypeId="1" AcceptedAnswerId="32254" CreationDate="2019-06-11T14:18:52.213" Score="8" ViewCount="1259" Body="&lt;p&gt;I guess that's pretty much it. I just heard that that a meteor fragment was found to be solid iron and it just seems inexplicable that it was floating around.&lt;/p&gt;&#xA;" OwnerUserId="28346" LastActivityDate="2019-06-11T15:10:16.027" Title="Why didn't all the iron and heavier elements find their way to the center of the accretion disc in the early solar system?" Tags="&lt;solar-system&gt;&lt;meteorite&gt;&lt;accretion-discs&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32254" PostTypeId="2" ParentId="32253" CreationDate="2019-06-11T15:10:16.027" Score="13" Body="&lt;p&gt;The process of &lt;a href=&quot;https://en.wikipedia.org/wiki/Planetary_differentiation&quot; rel=&quot;noreferrer&quot;&gt;differentiation&lt;/a&gt; is how metals like iron and nickel can become separated from less dense substances. It generally occurs in large, partly solid bodies in the early solar system, whose interiors are melted either via accretion energy, radioactivity or some combination of the two. A solid iron meteor is likely a fragment of a differentiated larger body that has been disrupted by a collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think what you are asking is why such a differentiation process does not occur in the primordial disc from which the solar system formed. The answer lies in turbulence and other mixing mechanisms that act against differentiation. If the mixing timescale in the disc is shorter than the differentiation timescale then the differentiation just won't occur. This is for example the same reason that stars like the Sun do not have all their heavy chemical elements at the core.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-06-11T15:10:16.027" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32255" PostTypeId="2" ParentId="32250" CreationDate="2019-06-11T17:55:50.690" Score="1" Body="&lt;p&gt;I'd like to start by quoting from the Wiki page on Nodal Precession you mentioned: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A non-rotating body of planetary scale or larger would be pulled by&#xA;  gravity into a sphere. Virtually all bodies rotate, however. The&#xA;  centrifugal force deforms the body so that it has an equatorial bulge.&#xA;  Because of the bulge the &lt;strong&gt;gravitational force on the satellite is not&#xA;  directly toward the center of the central body, but is offset toward&#xA;  the equator&lt;/strong&gt;. Whichever hemisphere the satellite is in it is&#xA;  preferentially pulled slightly toward the equator. This creates a&#xA;  torque on the orbit. This torque does not reduce the inclination;&#xA;  rather, it causes a torque-induced gyroscopic precession, which causes&#xA;  the orbital nodes to drift with time.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(emphasis mine).  Note that this is a small force and the precession is slow compared with orbital speeds.  Precession towards (or coasting away from) the equator thus is a change in latitude compared with what would be expected, and this &quot;twist&quot; changes the expected longitude at a given point in time because the satellite's vector velocity changes (I believe in both direction and magnitude, but I'm not certain about the magnitude). &lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2019-06-11T17:55:50.690" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32256" PostTypeId="2" ParentId="32203" CreationDate="2019-06-11T23:58:07.660" Score="3" Body="&lt;p&gt;I can address the part of the question that states &quot;have 'few or no' galaxies, but I can't find much else&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In recent years there has been a lot of work published on void galaxies.  Identifying void galaxies is not easy: you first have to find voids and then you have to be able to establish whether a galaxy is or is not located in one of those voids.  One way of doing this is to use a geometric definition of a void as described in this paper:&#xA;&lt;a href=&quot;https://www.astro.rug.nl/~weygaert/vgs_jenam_weygaert.col.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.astro.rug.nl/~weygaert/vgs_jenam_weygaert.col.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am a member of the Groningen Group, led by Rien van de Weygaert, who are responsible for that work, but I am not a co-author).  The &quot;void galaxies&quot; found in this way are truly isolated and not selected by virtue of brightness or surface brightness, so it's a fairly objective sample.  This sample of void galaxies was studied in the following papers (more technical):&#xA;&lt;a href=&quot;https://arxiv.org/abs/1204.5185&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/1204.5185&lt;/a&gt;&#xA;&lt;a href=&quot;https://arxiv.org/abs/1410.6597&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/1410.6597&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the difficulty of finding even these galaxies, it is unlikely that we could find isolated stars or planets, even in our &quot;Local Void&quot; - we lie on the border of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This picture is a slice through a 3D reconstruction of the density field in a relatively small volume surrounding our galaxy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iHPfk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iHPfk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slice cuts through a large number of voids, some of which we can identify from ealrier discoveries.  The voids are indeed almost empty of galaxies.  The image is the work of Johan Hidding, a member of the Groningen Group. See the paper:&#xA;&lt;a href=&quot;https://arxiv.org/pdf/1611.01222.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/pdf/1611.01222.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5301" LastEditorUserId="5301" LastEditDate="2019-06-12T09:42:25.110" LastActivityDate="2019-06-12T09:42:25.110" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32258" PostTypeId="1" CreationDate="2019-06-12T03:03:18.143" Score="1" ViewCount="40" Body="&lt;p&gt;As far as I can see, there are two main sources for the emission of energy from an accretion disk:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;release of gravitational potential energy of the infalling matter;&lt;/li&gt;&#xA;&lt;li&gt;friction from differential rotation within the disk.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How much of the emitted energy is due to each process? For that matter, are there any other processes that contribute to the spectrum?&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2019-06-12T03:03:18.143" Title="Causes of emission of accretion disk" Tags="&lt;black-hole&gt;&lt;quasars&gt;&lt;accretion-discs&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32259" PostTypeId="1" AcceptedAnswerId="32261" CreationDate="2019-06-12T03:26:24.983" Score="1" ViewCount="195" Body="&lt;p&gt;For a report I'm writing, I'm interested in comparing the exposure time required for spectroscopy versus photometry for a given magnitude. I found the figures on pages 7 and 8 of &lt;a href=&quot;http://www.astro.wisc.edu/~sheinis/ast500/AY500_lect5.ppt.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this lecture&lt;/a&gt; particularly informative. I tried emailing the author, but his email address is inactive. Where can I get similar information? &lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2019-06-13T06:45:39.850" Title="Exposure time for spectroscopy compared to photometry" Tags="&lt;observational-astronomy&gt;&lt;spectroscopy&gt;&lt;photometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32261" PostTypeId="2" ParentId="32259" CreationDate="2019-06-12T07:01:39.020" Score="3" Body="&lt;p&gt;Your question is a bit unclear. Compare in what sense?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem boils down to that you have a fixed number of photons in a wavelength band and you can choose, with the appropriate filters or dispersion elements, how many &quot;bins&quot; to divide these photons into. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To accumulate the same total number of photons with the same telescope will take the same amount of time, except putting more optical elements in the way will reduce the efficiency of the instrument. Filters are more efficient than spectrographs by factors of a few to an order of magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However a further factor to consider, is that although in principle we could divide the light up as finely as we want, there may be static sources of noise that are independent of the signal. In the case of a CCD detector, this means the read noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want read noise &lt;em&gt;not&lt;/em&gt; to limit the observations, then the product of the bin size and exposure time needs to be large enough that the square root of the accumulated counts (including any sky background) in a bin exceeds the read noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example might help. Suppose I have a star that I observe for 10s through a V-band filter with a width of 100 nm and it produces 1000 detected photons on a CCD, where the star image effectively illuminates 10 pixels, each of which has a read noise of 10 counts. The total read noise is &lt;span class=&quot;math-container&quot;&gt;$10\sqrt{10} = \sqrt{1000}$&lt;/span&gt; counts (the square root of the total variance) and equals the Poisson noise from the source plus sky. Ignoring the sky, the signal to noise ratio is &lt;span class=&quot;math-container&quot;&gt;$1000/\sqrt{1000+1000}=22$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose I do slitless spectroscopy, passing the starlight through a diffraction grating, splitting the light into 1 nm bins. My signal strength in each bin is divided by 100, but the read noise is approximately the same. To obtain a spectrum with a similar signal to noise ratio you would need to expose for 100 times longer, i.e.1000 s (probably longer because a diffraction grating is not as efficient as a filter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is no general answer to your question, it depends on the spectrograph efficiency, the &quot;resolution&quot; of the spectrograph, the relative contributions of source, sky and read noise and the signal to noise ratio desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A simple rule of thumb to get an exposure that is just limited by read noise for an imaging filter of wavelength width &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt; and a spectrograph with a resolution element &lt;span class=&quot;math-container&quot;&gt;$\Delta \lambda$&lt;/span&gt;, would be to observe for a factor of &lt;span class=&quot;math-container&quot;&gt;$w/\Delta \lambda$&lt;/span&gt; longer, with an extra factor of a few to account for poorer spectrograph efficiency.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;To test my ideas I go to the Isaac Newton group of telescopes' signal to noise calculator.&lt;a href=&quot;http://catserver.ing.iac.es/signal/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://catserver.ing.iac.es/signal/&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I try observing a 20th magnitude star for 10s with a V filter on the WHT prime focus, assuming 1 arcsec seeing and no moon. This gives me a SNR of 46.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the same SNR in a low resolution spectrum withe ISIS spectrograph and 158R grating (lowest resolution available) takes 15,000 s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this stack up with what I said above? Each pixel in the spectrum is 0.16 nm. If the V band filter is about 100nm wide, my crude formula above suggested an increase in exposure time of 100/0.16 = 625. But the instrument throughput is only 33% (compared with essentially 100% for the filter) so we have to add another factor 3, to get a required exposure that is 1875 times as long as for the photometry. Pretty close!&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-06-13T06:45:39.850" LastActivityDate="2019-06-13T06:45:39.850" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32262" PostTypeId="1" AcceptedAnswerId="32263" CreationDate="2019-06-12T08:28:50.760" Score="23" ViewCount="11590" Body="&lt;p&gt;I googled it and checked a few Q&amp;amp;A and there's only things about &quot;Earth's rotation&quot;. But why can't we feel the revolution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They say we can't feel the rotation because the Earth spins at a constant speed. Okay, I get what happens for the rotation, but isn't it different when it comes to the revolution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xiWmA.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xiWmA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At night we feel the sum of speed of green and blue, whereas at day we should feel the sum of speed of green and &lt;strong&gt;&lt;em&gt;minus&lt;/em&gt;&lt;/strong&gt; blue, shouldn't we? In other words, shouldn't we feel the changes in velocity by times?&lt;/p&gt;&#xA;" OwnerUserId="28352" LastActivityDate="2019-06-13T16:44:54.887" Title="Why can't we feel the Earth's revolution?" Tags="&lt;gravity&gt;&lt;astrophysics&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32263" PostTypeId="2" ParentId="32262" CreationDate="2019-06-12T10:13:23.900" Score="24" Body="&lt;p&gt;Firstly the speeds are massively different (about 1000 mph (1610 kph) on the equator for Earth's rotation and 70,000 mph (112,654 kph) for the revolution), so the change is not large. Secondly, the green line is far straighter than it appears in your picture (because the orbit is so large) so Earth's motion around the Sun is pretty close to motion at constant velocity, which Einstein tells us cannot change the outcome of any experiment.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="26040" LastEditDate="2019-06-13T10:10:38.993" LastActivityDate="2019-06-13T10:10:38.993" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32264" PostTypeId="1" CreationDate="2019-06-12T11:32:16.187" Score="1" ViewCount="135" Body="&lt;p&gt;Radius of the observable universe is 47.5 billion light years,which means we are seeing things which are about 48 billion light years away. At a rough estimate,at what distance would our finest instruments and telescopes be unable to see or detect anything at all,&amp;amp; is there anything in the idea that if we were able to see far enough,we might see our own galaxy as it appears from an immense distance,its light having circled the universe? And as space is not a perfect vacuum,wouldn't the light be significantly retarded by the end of such a long journey?&lt;/p&gt;&#xA;" OwnerUserId="28083" LastEditorUserId="7982" LastEditDate="2019-06-13T01:24:24.757" LastActivityDate="2019-06-13T06:23:40.677" Title="What Is The Maximum Distance Our Finest Instruments Could See When They're Perfected?" Tags="&lt;radio-astronomy&gt;&lt;observable-universe&gt;&lt;cmb&gt;" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="32265" PostTypeId="1" CreationDate="2019-06-12T11:43:00.693" Score="1" ViewCount="48" Body="&lt;p&gt;Most reasonings about hypothetical wormholes say, that you need an extreme amount of energy to keep them open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that, isn’t the worm hole too close to the black hole gargantua? The gravity between the two would make the spin and eventually collapse together?&lt;/p&gt;&#xA;" OwnerUserId="28354" LastActivityDate="2019-06-12T11:43:00.693" Title="Hypothetical: Is the interstellar wormhole too close to the black hole?" Tags="&lt;interstellar&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2019-06-12T15:34:20.547" ContentLicense="CC BY-SA 4.0" />
  <row Id="32266" PostTypeId="2" ParentId="32243" CreationDate="2019-06-12T17:04:38.787" Score="2" Body="&lt;p&gt;When we talk about the expansion of the Universe, we're really saying that space is being created between all matter. Let me explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine setting up a grid that keeps track of all points in space. &quot;Expansion&quot; just means that the distances between objects on the grid are getting larger. In essence, more space is being created between the objects. Below is a gif I've made to demonstrate this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/amQeZ.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/amQeZ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more useful way to describe this is to say the grid is expanding — that space itself, as a coordinate system, is growing. As an analogy, imagine are walking your dog. Suddenly, the ground begins expanding between you. You and your dog will separated and continue receding away from each other. So the same thing is happening with our universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The grid is in fact growing, and objects are being swept away with it. A consequence of this is that they can recede away from each other faster than light; while objects are indeed limited in how fast they can move &lt;em&gt;through&lt;/em&gt; space, there is no limit as to how fast space can be created between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we've gotten the core concepts down, I'll introduce one more bit of terminology. The &quot;scale factor of the Universe&quot; refers to how much the Universe has expanded, compared to now. For example, if in a billion years the scale factor is 3, that means that every object in the Universe is 3 times farther from each other compared to now. If the scale factor 700 million years ago was 0.8, then everything was closer by a factor of 0.8 at that time. By definition, the scale factor is 1 right now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if the Universe is expanding now, we'd expect it to be smaller as we look further back in time — i.e. the scale factor would be less. General relativity predicts the scale factor to be zero at 13.8 billion years ago. This would mean that every object would be zero times its current distance from us — in other words, there would be no space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think a Universe without space is impossible, you're correct. We apparently have a contradiction. In GR, you can't have a spacetime with zero space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our modern physical theories work fine up a few fractions of a second after the moment of contradiction, and our observations do agree with the idea of an extremely dense early universe. However, our theories break down as we try to model the Universe at earlier and earlier times, until they no longer prove accurate, preventing us from explaining the most interesting moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why the moment of the Big Bang is one of the biggest mysteries in cosmology. Theories like quantum gravity have arisen to try to explain the conditions near the Big Bang, but none are sufficient as of now.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2019-06-12T17:04:38.787" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="32267" PostTypeId="2" ParentId="32264" CreationDate="2019-06-12T18:54:11.047" Score="5" Body="&lt;p&gt;The furthest we can &quot;see&quot; is the cosmic microwave background at a redshift of about 1100.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proper distance of the CMB-emitting gas that we see today is about 46 billion light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are talking about galaxies, then the first are thought to have formed at redshifts of about 20 (current distance 36 billion light years) and beyond that are the universe's &quot;dark ages&quot; where no stars or galaxies should exist; the most distant galaxies observed at the moment are at redshifts of about 10 (current distance 31 billion light years).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;All current distances depend on the adopted cosmological model. The numbers are for a flat &lt;span class=&quot;math-container&quot;&gt;$\Lambda$&lt;/span&gt;CDM universe with the Planck cosmological parameters.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There is no chance of seeing our own Galaxy, since it only formed about 12 billion years ago and the universe is a lot bigger than 12 billion light years.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-06-13T06:23:40.677" LastActivityDate="2019-06-13T06:23:40.677" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32268" PostTypeId="1" CreationDate="2019-06-12T19:45:13.897" Score="4" ViewCount="108" Body="&lt;p&gt;We know the CNO cycle produces the majority of energy production in a &quot;higher mass star&quot; of approx. over 2 solar masses and the core is convective due to the large temperature gradient. My professor has noted that radius, luminosity, T_eff, and T_center all increase but rho_center (density) does not. I cannot find anything in my text that explains this, but I assume it is because convection works against gravity. If someone could clear this concept up (as to why rho_center) stays generally constant, I'd appreciate it. Conceptual explanations are helpful!&lt;/p&gt;&#xA;" OwnerUserId="28360" LastActivityDate="2019-07-24T04:05:50.573" Title="Does the density in the core of a &quot;higher mass&quot; star stay generally constant while it evolves on the MS?" Tags="&lt;star&gt;&lt;stellar-evolution&gt;&lt;main-sequence&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32269" PostTypeId="2" ParentId="32262" CreationDate="2019-06-12T21:25:54.420" Score="11" Body="&lt;p&gt;The problem with how you're looking at it is that &lt;em&gt;velocities&lt;/em&gt; don't cause or result from forces, but &lt;em&gt;accelerations&lt;/em&gt; do. Think of Newton's 2nd law, &lt;span class=&quot;math-container&quot;&gt;$F = m a$&lt;/span&gt;. Circular motion is motion at constant speed but changing direction, this changing direction is a type of acceleration because velocity is a vector (has direction) and acceleration is change in velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do actually feel a difference due to this acceleration - if you think about a sphere rotating around an axis, the points on near the intersection of that axis with the surface are rotating slower than anywhere else, and so if you measure the local gravitational acceleration &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; at the poles you get a slightly larger value than you do at the equator (about 0.3%). This is because the rotation acts in opposition to the force of gravity due to the earth's mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a very small effect from the earth's orbit around the sun. In this case, the force due to acceleration is exactly the same as the force that holds the earth in orbit, so you can just look at the gravitational effects as done in &lt;a href=&quot;https://physics.stackexchange.com/a/121785/11645&quot;&gt;this answer on physics.se&lt;/a&gt;, which came up with a value of around 26 parts per billion, or 0.0000026%. The interesting thing is that you become lighter both when the sun is overhead, and when it's directly on the opposite side of the earth to you.&lt;/p&gt;&#xA;" OwnerUserId="25532" LastActivityDate="2019-06-12T21:25:54.420" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32271" PostTypeId="2" ParentId="32262" CreationDate="2019-06-13T03:04:29.703" Score="58" Body="&lt;p&gt;&lt;strong&gt;You do, but it's too small to really notice&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, it's not correct to say that we don't feel Earth's rotation because it's rotating at a constant speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about driving a car, or riding in an airplane.  Whether you're cruising down the road at 90 kph, or soaring through the air at 900 kph, you don't really &quot;feel the speed&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, When you take a sharp turn, or take off from the runway, you definitely feel &lt;em&gt;something&lt;/em&gt;.  That's the &lt;strong&gt;acceleration&lt;/strong&gt;.  It doesn't matter if your speedometer stays steady - if you take a sudden 90 degree turn, you're going to feel it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More relaxed turns, such as going through a roundabout, or when the airplane circles the airport before landing, are much less likely to spill your drink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if Earth is spinning at a constant speed, the &lt;strong&gt;spin is a change in direction, which requires acceleration&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acceleration is quite noticeable, &lt;strong&gt;depending on its magnitude&lt;/strong&gt;.  Even just sitting down, you can feel the pull of Earth's 9.8 m/s² gravity - your body's &quot;weight&quot;, as it were.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how large is the acceleration keeping Earth in orbit?  &lt;a href=&quot;https://van.physics.illinois.edu/qa/listing.php?id=1063&amp;amp;t=gravitational-pull-of-the-sun&quot; rel=&quot;noreferrer&quot;&gt;About 0.0059 m/s².&lt;/a&gt;&#xA;What about the acceleration of Earth's rotation?  &lt;a href=&quot;https://van.physics.illinois.edu/qa/listing.php?id=186&quot; rel=&quot;noreferrer&quot;&gt;A ever so slightly larger 0.0339 m/s².&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small wonder that it seems like you can't feel these forces!&lt;/p&gt;&#xA;" OwnerUserId="28363" LastActivityDate="2019-06-13T03:04:29.703" CommentCount="20" ContentLicense="CC BY-SA 4.0" />
  <row Id="32272" PostTypeId="1" CreationDate="2019-06-13T04:06:41.863" Score="1" ViewCount="686" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iRrAG.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;a href=&quot;http://www.moonconnection.com/moon_phases.phtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically this diagram is how the phases of the moon are explained but the issue is this, namely, the phases seen on the right half are not possible. All the phases should appear on the left half during the night when the moon is typically most visible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how on earth are the phases even generated? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most remarkable phases are waning moon phases because the moon is lit up on the face away from the sun light source. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’ve edited the question by adding two additional photos showing the progression of the moon in the night sky over the course of a given time interval and several days, as well as a synodic month which would hopefully indicate based on the position of the moon the sequence of days and based on the height the time of day/night. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/By7K2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/By7K2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.twanight.org/newTWAN/guests_photos.asp?ID=5001716&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GuXKd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GuXKd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.skyandtelescope.com/online-gallery/lunar-phases/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link to the explanation of how she generated her photo and over what time span but you must have a Facebook account to read it. It is easier to follow the Italian than the English but both are available&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://m.facebook.com/901571186524532/posts/2329156960432607/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Explanation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26115" LastEditorUserId="26115" LastEditDate="2019-06-13T18:34:24.437" LastActivityDate="2019-11-17T23:24:56.460" Title="Struggling to understand the phases of the moon" Tags="&lt;the-moon&gt;&lt;moon-phases&gt;" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32273" PostTypeId="2" ParentId="32272" CreationDate="2019-06-13T05:14:45.647" Score="4" Body="&lt;p&gt;As noted in a comment, you should imagine the picture is looking down on the Earth, and remember that the actual moon is much further from the Earth, relative to their sizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a person standing on the left side of the Earth (in the diagram). The time would be midnight for them. They could see the moon when it is full, and they could see the moon when it is a gibbous moon (but the gibbous moon would be lower in the sky)  They could even see the moon at first quarter, but it would be on the horizon.  You can't see a crescent moon at midnight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine a person standing in the top left on the Earth (in the diagram). The time for them would be about 9pm. (and night time) That person could see the half moon and even a crescent moon (but the crescent would be on the horizon)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is only possible to see a crescent moon at dawn or dusk, you never see a crescent moon at midnight. At 6pm (at dusk) you can see the crescent moon. If you look to the moon, you see mostly the side in shadow, but a little of the light shining around the side, so it looks like a crescent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you are also confused about the illumination side shown in the waning phases, This is because the picture shows the view of the moon that a person would see if they were standing on the Earth. Because that person is upside down (from our point of view) they see the moon reversed. In other words, from our position above the North Pole, we see the right side of the moon illuminated.  From their position, it is the left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the diagram still confuses, discard it and think in terms of angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the perspective of the Earth, the sun and moon go around in circles at different rates, so the angle made between the sun and the moon changes. When the angle between the sun and the moon is about 90 degrees, then half of the near side of the moon is lit by then sun, and a half moon is visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the angle is close to 180 degrees, then nearly all the moon is lit by the sun, and the moon is full.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the angle is close to 0 degrees, then none of the near side will be lit, and the moon won't be visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the angle is small, ie the moon is close to the sun, then a crescent moon will be visible. A crescent moon can only be seen if it is near to the sun, so the sun cannot be far below the horizon. This means that the a crescent moon cannot be seen late at night when the sun is far below the horizon, but at some times and locations the sun might not set until late in the evening, so the setting time of the moon might be quite late.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless it is the angle between the sun and the moon that determines the phase of the moon, and nothing else.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2019-06-13T20:49:32.423" LastActivityDate="2019-06-13T20:49:32.423" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32275" PostTypeId="2" ParentId="10597" CreationDate="2019-06-13T08:48:17.497" Score="1" Body="&lt;p&gt;You can use &lt;em&gt;Sky and Telescope's interactive tool for observing Jupiter's moons&lt;/em&gt;. &lt;a href=&quot;https://www.skyandtelescope.com/observing/celestial-objects-to-watch/jupiters-moons-javascript-utility/#&quot; rel=&quot;nofollow noreferrer&quot;&gt;This website&lt;/a&gt; describes how to use it to determine the positions of the Galilean moons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It works in the past or future.&lt;/p&gt;&#xA;" OwnerUserId="28369" LastEditorDisplayName="user1569" LastEditDate="2019-06-13T09:33:29.907" LastActivityDate="2019-06-13T09:33:29.907" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32278" PostTypeId="2" ParentId="32272" CreationDate="2019-06-13T12:24:12.463" Score="3" Body="&lt;p&gt;The image is mostly correct, but also a bit misleading. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the Earth in the middle, the 8 small moons around it and 'sunlight' are seen from above. &lt;/li&gt;&#xA;&lt;li&gt;the 8 larger Moon faces are what you see from Earth.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="361" LastActivityDate="2019-06-13T12:24:12.463" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32279" PostTypeId="2" ParentId="32272" CreationDate="2019-06-13T15:08:05.540" Score="4" Body="&lt;p&gt;The text directly underneath the diagram from the &lt;a href=&quot;http://www.moonconnection.com/moon_phases.phtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;original source&lt;/a&gt; says&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Diagram Explanation&lt;/p&gt;&#xA;&lt;p&gt;The illustration may look a little complex at first, but it's easy to explain.&lt;/p&gt;&#xA;&lt;p&gt;Sunlight is shown coming in from the right. The earth, of course, is at the center of the diagram. The moon is shown at 8 key stages during its revolution around the earth. The moon phase name is shown alongside the image. The dotted line from the earth to the moon represents your line of sight when looking at the moon. The large moon image shows what you would see at that point in the cycle. For the waning gibbous, third quarter, and waning crescent phases you have to mentally turn yourself upside down when imagining the line of sight. When you do this, you'll &amp;quot;see&amp;quot; that the illuminated portion is on your left, just as you see in the large image.&lt;/p&gt;&#xA;&lt;p&gt;One important thing to notice is that exactly one half of the moon is always illuminated by the sun. Of course that is perfectly logical, but you need to visualize it in order to understand the phases. At certain times we see both the sunlit portion and the shadowed portion -- and that creates the various moon phase shapes we are all familiar with. Also note that the shadowed part of the moon is invisible to the naked eye; in the diagram above, it is only shown for clarification purposes. Finally, please realize this diagram is only meant to demonstrate how the phases work; the small inner moons in the diagram do not show the fact that the same side of the moon always faces Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I do not see what is problematic with this.  The diagram is drawn such that the observer is always looking straight up at the Moon. In such situations a New Moon and crescent moons can only be seen during the day (and of course the Moon is plainly visible during the day). In practice we can observe the Moon at different altitudes as the Earth spins so crescent moons can indeed be viewed, either rising before dawn or setting after dusk at night. e.g. Consider the waxing crescent position. You could view this from a position on the Earth that you could describe as the &amp;quot;11 o'clock&amp;quot; position that would be in darkness, but not from the &amp;quot;9 o'clock&amp;quot; position in the middle of the night.&lt;/p&gt;&#xA;&lt;p&gt;Of course the diagram cannot be used for quantitative calculation because the Earth-Moon separation on this diagram is too small and the Earth-Sun-Moon do not move in the same plane.&lt;/p&gt;&#xA;&lt;p&gt;Your comment about the waxing moon being lit up from the wrong side is incorrect. As the explanation above says, the large Moon images around the outside of the diagram represent what you would see from a point on Earth. The inner, smaller moon images show how it is (consistently) illuminated by the Sun.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-06-13T15:19:51.457" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32280" PostTypeId="2" ParentId="32262" CreationDate="2019-06-13T16:44:54.887" Score="9" Body="&lt;p&gt;I will give a biologically motivated answer to this question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feeling the Earth rotation carries no meaning to us. It is always approximately the same and we will blend out such background information and concentrate on news that are really important to us: Is there a danger approaching? Is the some food to gain? What are our peers doing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because there was no evolutionary pressure to feel the Earth rotation, we did not develop a sense for it. The fact that it is weak and really difficult to measure adds to this. We need an apparatus like Foucault's pendulum to see its effect.&lt;/p&gt;&#xA;" OwnerUserId="11617" LastActivityDate="2019-06-13T16:44:54.887" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32281" PostTypeId="1" CreationDate="2019-06-13T18:31:53.783" Score="0" ViewCount="51" Body="&lt;p&gt;Black holes are things with immense gravity.Inside black holes gravity is much stronger than any other fundamental forces.But is strong enough to overcome Pauli repulsion?&lt;/p&gt;&#xA;" OwnerUserId="28383" LastActivityDate="2019-06-13T18:31:53.783" Title="Black hole properties" Tags="&lt;black-hole&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2019-06-14T02:03:30.390" ContentLicense="CC BY-SA 4.0" />
  <row Id="32282" PostTypeId="1" CreationDate="2019-06-13T21:37:06.087" Score="1" ViewCount="311" Body="&lt;p&gt;Hubble deep field image at maximum resolution shows a few of the furthest galaxies very small &amp;amp; faint, only just visible, but suggests that there is more to see beyond if only the resolution were high enough. I know the age of the universe is 'only' about 14 billion years, but if Hubble can only show galaxies 13.2 billion light years away, how do we know the radius of the universe is 47 billion light years?&lt;/p&gt;&#xA;" OwnerUserId="28083" LastEditorUserId="9301" LastEditDate="2019-06-14T09:39:46.460" LastActivityDate="2019-06-14T09:39:46.460" Title="How do we know the radius of the universe is 47 billion light years?" Tags="&lt;observable-universe&gt;&lt;hubble-telescope&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32284" PostTypeId="1" AcceptedAnswerId="32287" CreationDate="2019-06-14T01:14:41.747" Score="6" ViewCount="894" Body="&lt;p&gt;The Space.com article &lt;a href=&quot;https://www.space.com/4271-huge-hole-universe.html&quot; rel=&quot;noreferrer&quot;&gt;Huge Hole Found in the Universe&lt;/a&gt; says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The gargantuan hole was found by examining observations made using the Very Large Array (VLA) radio telescope, funded by the National Science Foundation.&lt;/p&gt;&#xA;&lt;p&gt;There is a &amp;quot;remarkable drop in the number of galaxies&amp;quot; in a region of sky in the constellation Eridanus, Rudnick said.&lt;/p&gt;&#xA;&lt;p&gt;The region had been previously been dubbed the &amp;quot;WMAP Cold Spot,&amp;quot; because it stood out in a map of the Cosmic Microwave Background (CMB) radiation made by NASA's Wilkinson Microwave Anisotopy Probe (WMAP) satellite. The CMB is an imprint of radiation left from the Big Bang, the theoretical beginning of the universe.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Although our surprising results need independent confirmation, the slightly colder temperature of the CMB in this region appears to be caused by a huge hole devoid of nearly all matter roughly 6 to 10 billion light-years from Earth,&amp;quot; Rudnick said.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Photons of the CMB gain a small amount of energy when they pass through normal regions of space with matter, the researchers explained. But when the CMB passes through a void, the photons lose energy&lt;/strong&gt;, making the CMB from that part of the sky appear cooler. (emphasis added)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How do CMB photons gain energy when they pass through &amp;quot;normal regions of space with matter&amp;quot; and lose energy when they pass through voids?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-06-14T06:57:41.170" Title="How do CMB photons 'gain energy when they pass through normal regions of space with matter' and 'lose energy when they pass through voids'?" Tags="&lt;cmb&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32285" PostTypeId="2" ParentId="32282" CreationDate="2019-06-14T01:24:02.143" Score="6" Body="&lt;p&gt;The observable universe's edge, by definition, is the farthest part of the Universe that light could have reached us from. Let's say light from the observable universe's edge has just reached us. The longest it could have possibly travelled for would be 13.8 billion years, since the Universe is only 13.8 billion years old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, one might assume that the light came from 13.8 billion light years away, and hence the observable universe's radius should be 13.8 billion light years. The problem is that this overlooks a key phenomenon: the expansion of the universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The location from where the light had been emitted has grown increasingly distant from us. In other words, by the time the photon has reached us and we are able to see the object emitting it, the object has moved much farther than when it emitted the photon. This is why we are able to see objects of distances far beyond 13.8 billion light years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to work out the actual numbers, let's suppose an object emitted a photon at the beginning of the universe &lt;span class=&quot;math-container&quot;&gt;$t=0$&lt;/span&gt;, and it just reached us today (&lt;span class=&quot;math-container&quot;&gt;$t_0 = 13.8 \ Gyr$&lt;/span&gt;). Evidently, this object lies at the edge of the observable universe, so finding its distance allows us to find the observable universe's radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider a given time &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; in the past. Suppose our photon travelled for a small interval of time &lt;span class=&quot;math-container&quot;&gt;$dt$&lt;/span&gt;; it follows that the distance it just travelled away from its source would be &lt;span class=&quot;math-container&quot;&gt;$dx = c \ dt$&lt;/span&gt;. Due to the expansion of the universe, that &lt;span class=&quot;math-container&quot;&gt;$dx$&lt;/span&gt; will expand over time; compared to today, it was a factor &lt;span class=&quot;math-container&quot;&gt;$a(t) = \frac{dx(t)}{dx(t_0)}$&lt;/span&gt; smaller (this is called the scale factor at time &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt;). Hence, the distance it gained from its source grew to be &lt;span class=&quot;math-container&quot;&gt;$dx(t_0) = \frac{dx(t)}{a(t)} = \frac{c \ dt}{a(t)}$&lt;/span&gt; by today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At every time &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt;, it increased by an amount &lt;span class=&quot;math-container&quot;&gt;$dx$&lt;/span&gt;. So if the photon reached us at time &lt;span class=&quot;math-container&quot;&gt;$t_0 = 13.8 Gyr$&lt;/span&gt; after it was emitted, then the total distance it gained from its source is the sum of these changes:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$r=\int_0^{t_0} dx(t_0) = \int_0^{t_0} \frac{c}{a(t)} dt \approx 46 \ Gly$$&lt;/span&gt;&#xA;Which is therefore the distance of the object from us.&lt;/p&gt;&#xA;" OwnerUserId="9301" LastActivityDate="2019-06-14T01:24:02.143" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32286" PostTypeId="1" CreationDate="2019-06-14T03:04:04.110" Score="2" ViewCount="336" Body="&lt;p&gt;This is a question that has been bugging me for some time. Let's see if I am able to make me understood!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you are travelling in a spaceship at 99.9999999999% the speed of light and you are somehow within a spaceship without windows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, is there ANY kind of experiment that would show you any time/space dilation effect as compared to being stationary? Is there ANY lab or thought experiment you could make or think that would tell you there are weird things happening with time and space outside your spaceship? I mean, would physics work differently than they would at home? Would radioactive atoms take longer to decay in a way you could measure and compare with your earth-based decay tables? Anything?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or the only way to find out would be simply to step out of the spaceship and find out that life on Earth has gone extinct in the last couple of hours?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And another question related to the first. If you were travelling in the spaceship at that speed in the X-axis direction and you handthrew a baseball in that same direction, would that ball's mass or inertia explode asyntotically to practically infinity since you are increasing a fantastically high speed a tiny winy bit more? Or perhaps you could not threw the ball at all because of the huge mass increase being that close already to the speed of light? Or you wouldn't notice nothing special and a regular sized with normal weight baseball would impact the other side of the room at a relative speed to you of 60 km/h?&lt;/p&gt;&#xA;" OwnerUserId="28388" LastActivityDate="2019-06-14T21:29:18.190" Title="Is it possible to know if you are travelling close to the speed of light if you have nothing to compare your speed with?" Tags="&lt;special-relativity&gt;" AnswerCount="3" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="32287" PostTypeId="2" ParentId="32284" CreationDate="2019-06-14T06:57:41.170" Score="18" Body="&lt;p&gt;It is the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Sachs%E2%80%93Wolfe_effect&quot; rel=&quot;noreferrer&quot;&gt;late time integrated Sachs Wolfe effect&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As they travel towards us, apart from the general expansion, photons from the CMB &lt;em&gt;gain&lt;/em&gt; energy when they fall into potential wells (where matter is). Of course, they lose it again as they emerge on the other side of the well, but the cosmic expansion means that the well isn't quite as deep by the time that happens. As a result, photons gain a little energy &lt;em&gt;compared to the average CMB sightline&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The opposite happens for voids. Energy is required to enter a void (the potential is higher than average), but all this energy is not retrieved because the potential flattens due to expansion as the photon traverses it. Hence the directions of voids would be a bit cooler.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-06-14T06:57:41.170" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32288" PostTypeId="2" ParentId="32286" CreationDate="2019-06-14T10:21:58.573" Score="-4" Body="&lt;p&gt;First of all, common sense should tell you you’re not travelling at that speed; only cosmic rays and accelerated particles can reach such speeds. The incredible amount of energy needed to boost a spaceship to such a speed, far more than is produced by the largest multi-megaton hydrogen bomb, is one of the reasons it couldn’t happen. But let’s assume for the sake of argument that your spaceship has managed to reach this speed, there are ways and means to see what is ahead without having windows. All the photons coming toward you would be dramatically blue-shifted, while those coming from behind would be dramatically redshifted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space is not a perfect vacuum, even in the most transparent regions there is a proton every few metres. At speeds close to the speed of light your spaceship would be subject to a headwind of fast protons, and you would feel the walls of the spaceship heating up. If you had the misfortune to run into a cloud of gas or dust, your spaceship would rapidly burn up, and your problems would be over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming there was no gas or dust in your path and space was apparently perfectly transparent, I still wouldn’t advise stepping out of your spaceship; the proton storm and other radiations would kill you. Only objects which didn’t share your motion would have relativistic mass increase, so the ball you threw would behave as though you had thrown it in the international space station. Inside the spaceship, you would not notice any change in the decay rates of radioactive elements. The spaceship itself, when viewed by an outside observer who saw himself as at rest, would have acquired a colossal amount of relativistic mass increase because both you and the spaceship are now made of relativistic particles. This mass increase couldn’t come from nowhere, it would have to come from the energy provided by the fuel used, and no such fuel exists.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastEditorDisplayName="user8021" LastEditDate="2019-06-14T21:29:18.190" LastActivityDate="2019-06-14T21:29:18.190" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32289" PostTypeId="2" ParentId="32286" CreationDate="2019-06-14T12:43:43.270" Score="3" Body="&lt;p&gt;Velocity is relative. From your perspective, your velocity is zero. Without windows (or sensors) on your spaceship, you cannot know the difference. To measure your velocity, you have to define a reference point, eg a star, the average velocity of surounding stars or the galaxy center. If you have such a huge velocity relative to them, from you perspective, they are this fast and not you. So you can observe the time and space dilation of these objects, while your time appears to be normal. You ship is an inertial reference frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for throwing the baseball: From your perspective, you have the speed zero, and the baseball eg. the speed 0.5c. A &quot;fixed&quot; observer, which observes you at 0.99c will measure the balls velocity at 0.997c. With those relative effects, speeds don't add up, only kinetic energy does. &lt;a href=&quot;http://curious.astro.cornell.edu/about-us/139-physics/the-theory-of-relativity/special-relativity/1016-why-can-t-relative-velocities-add-up-to-more-than-the-speed-of-light-intermediate&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://curious.astro.cornell.edu/about-us/139-physics/the-theory-of-relativity/special-relativity/1016-why-can-t-relative-velocities-add-up-to-more-than-the-speed-of-light-intermediate&lt;/a&gt; The increased mass is only a relativistic observation, resulting from the kinetic energy (relative to the observer).&lt;/p&gt;&#xA;" OwnerUserId="28395" LastActivityDate="2019-06-14T12:43:43.270" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32291" PostTypeId="1" CreationDate="2019-06-14T13:11:18.693" Score="0" ViewCount="95" Body="&lt;p&gt;I had a meeting in the evening and thought of grabbing a snack before that. I stepped out of the office along with a colleague. He was on his phone and I was simply staring at the sky wishing it would rain. Although there was no trace of a cloud, however, what I saw was beyond what I expected and could ever imagine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw three stars aligned in an inverted triangle, moving at a decent speed away from each other. For a moment I didn't believe what I was seeing. I quickly called my colleague and asked him if he could see something moving in the sky. Well, we both were astonished as we saw the stars moving away from each other and changing there position in the sky. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that evening I went back home a was awake till 4 am sitting at the terrace staring at those three stars, wishing them to move again. It's been three days and the stars are still in the same position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, I can detect those three stars as they were right above the north star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In these three days, I have done a bunch of research on satellite, shifting stars, supernova and so on but nothing could explain what we saw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would love to know if anyone had the same experience as us.&lt;/p&gt;&#xA;" OwnerUserId="28396" LastActivityDate="2019-06-14T13:43:52.757" Title="Stars readjusting themselves in the sky" Tags="&lt;star&gt;&lt;star-formation&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2019-06-15T13:17:53.413" ContentLicense="CC BY-SA 4.0" />
  <row Id="32293" PostTypeId="1" CreationDate="2019-06-14T15:07:47.823" Score="15" ViewCount="2325" Body="&lt;p&gt;I've seen quite a few news about astronomy but I'm not deep into knowing how it works or how people observe it. It got me thinking: if I want to watch a supernova for example, how long will I have to look for it to watch the whole &quot;event&quot;? Or events like NASA &quot;&lt;a href=&quot;https://interestingengineering.com/nasa-sees-for-the-first-time-ever-something-coming-out-of-a-black-hole-while-it-feeds&quot; rel=&quot;noreferrer&quot;&gt;seeing something coming out of a black hole&lt;/a&gt;&quot;, were they &quot;looking&quot; at the blackhole when something came out really quickly or did they observe it for months? Or maybe the birth of a blackhole?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I do know black holes that quite a few trillion of years to vanish, I'm just curious how much time does it take for astronomers to observe such events.&lt;/p&gt;&#xA;" OwnerUserId="28398" LastEditorUserId="-1" LastEditDate="2019-06-15T13:17:14.667" LastActivityDate="2019-06-16T04:55:27.337" Title="How &quot;fast&quot; do astronomical events occur?" Tags="&lt;observational-astronomy&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32294" PostTypeId="2" ParentId="32293" CreationDate="2019-06-14T15:53:22.217" Score="10" Body="&lt;p&gt;A supernova (and many other interesting events) starts abruptly -- a flash of neutrinos lasting just a fraction of a second, but then different aspects of the event play out over minutes, days, hours or years.  The astronomy community has for some years been developing increasingly sophisticates ways to deal with such events early on. Instruments with a very wide field of view (gravitational observatories, neutrino observatories, gamma ray detectors on satellites) detect the events initially and bring first networks of robotic telescopes and then humans and still larger telescopes into the loop to try and get as much information as possible about the violent and fast developing early stages. See for example &lt;a href=&quot;http://growth.caltech.edu/&quot; rel=&quot;noreferrer&quot;&gt;http://growth.caltech.edu/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other things are slower. The &quot;something coming out of a black hole&quot; (not really that, just something coming from very close to a black hole) event takes a long time, this was just the first occasion when they'd observed that black hole with the right instrument.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-06-14T15:53:22.217" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32295" PostTypeId="2" ParentId="32293" CreationDate="2019-06-14T16:28:57.543" Score="3" Body="&lt;p&gt;In our galaxy, supernovae visible from Earth are very rare events, so if you wait for astronomers to raise the alarm when they see the next one you might have to wait hundreds of years. The prospects of seeing one in faraway galaxies are much better, but even so they are not everyday events. The chances of seeing one from the very beginning when the star collapses and explodes are almost non-existent because when they occur, they have to reach considerable brightness to become noticeable from Earth. However, searching for supernovae is something a well-equipped amateur can do, but it requires a lot of patience. As for the energetic particle beams that some black holes send out, they don't arise from within the black hole itself but from the matter in the accretion disc which is accelerated by the poles of the black hole's magnetic field. As this process continues for many years, you don't need to react so rapidly as a supernova discoverer, but you would need some very high tech equipment to observe and photograph it (the particle beams eject large amounts of matter which form large swirls, sometimes light years in length, from both poles of the black hole, and need a radio telescope to image them).&lt;/p&gt;&#xA;" OwnerUserId="28083" LastEditorDisplayName="user8021" LastEditDate="2019-06-15T11:40:24.947" LastActivityDate="2019-06-15T11:40:24.947" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32297" PostTypeId="1" CreationDate="2019-06-14T18:07:32.603" Score="0" ViewCount="517" Body="&lt;p&gt;I was reading a book by Stephen Hawking, it was written that universe is made up of 2 ingredients,i.e., space and energy and space was the negative energy, so they add up to 0. But since space is energy does that mean it is finite?&lt;/p&gt;&#xA;" OwnerUserId="28399" LastActivityDate="2020-04-18T10:10:20.127" Title="If space is negative energy and matter is positive energy then does that mean the universe is finite?" Tags="&lt;astrophysics&gt;&lt;big-bang-theory&gt;&lt;general-relativity&gt;&lt;singularity&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32298" PostTypeId="2" ParentId="32286" CreationDate="2019-06-14T18:39:00.423" Score="2" Body="&lt;p&gt;If you have nothing to compare yourself with then there is no speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speed is like marriage - you can't be &quot;married&quot; if you're alone. To define speed, you need an external reference. Then you say &quot;my speed is XYZ km/s relative to object ABC&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what most people get wrong about speed. They think it's something you have in yourself, like the number of atoms, or the net electric charge. That is not true. Speed is always relative - you always, ALWAYS, measure your speed relative to an external object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a completely empty universe, where only you would exist, speed would be meaningless. You could not define your speed in any way, because how would you measure it? You need an external reference, always.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And to prevent another question that people usually ask: you cannot measure your speed relative to &quot;space&quot;. Speed can only be measured relative to other things, and space is not a thing. Space is just the background where the relationship called &quot;distance&quot; takes place. You cannot grab the blue marker pen and put a big X on &quot;space&quot;. But you can grab the blue marker pen and put a big blue X on an asteroid, and then say &quot;measure speed relative to this&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you move at 0.999c relative to object A, then space contraction and time dilation apply to you (and to A), as calculated from relativity. But if at the very same time your speed is only 0.5c relative to object B, then space contraction / time dilation are different as seen by B (or seen by you with regards to B), again as calculated from relativity. Object A will see a certain amount of space contraction being applied to you; object B will see a different amount of space contraction applied to you. &lt;strong&gt;Both are right&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why it's called &quot;relativity&quot; - because nothing is absolute, everything is relative, and it all depends on the relative speeds between objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't &quot;contract&quot; in an absolute way when you move - because motion (speed) is always relative. The contraction is just something that happens between you and the external object you use to measure your speed. Again, see the comparison with marriage - it's something between you and the other person, and applies only to the two of you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, that doesn't mean that space contraction is an &quot;illusion&quot;. It is very much real. If you move at 0.999c relative to object A, you're shrinking length-wise from the p.o.v. of object A. But if at the very same time you're not moving at all relative to object B, then B will say your length remains the same. &lt;strong&gt;Both are right&lt;/strong&gt;. Both are for real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We grow up learning that length and duration are absolute and fixed, but that's just an illusion - &lt;strong&gt;that&lt;/strong&gt;, actually, is &lt;strong&gt;the&lt;/strong&gt; illusion. They're not fixed, they're not absolute. They're just relative attributes, that depend on your motion relative to other things. Relativity gives you the exact math to calculate the amount of length/time change, depending on relative speed (well, &quot;relative speed&quot; is like saying &quot;wet water&quot; - speed is always relative, by definition).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing absolute in this universe is the speed of light - it's always c in your local frame of reference, no matter what. Everything else sort of shifts around and gets adjusted as needed.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2019-06-14T18:52:54.340" LastActivityDate="2019-06-14T18:52:54.340" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32299" PostTypeId="2" ParentId="32293" CreationDate="2019-06-14T20:45:24.010" Score="10" Body="&lt;p&gt;As described in other answers, most astronomical events vary over rather long time scales. Even a supernova, which starts suddenly, has a subsequent light curve that peaks, then decays usually over many months. However, there are some shorter events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the recent LIGO gravitational wave observations, the final stage inspiral and merger of binary black holes or neutron stars generates a detectable GW signal for less than a second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other shorter term astronomical observations occur within the Solar system. Of course, the Sun and planets move across the sky over the year, and the Moon changes its phase continuously, repeating about every month. Occasional Solar and Lunar eclipses occur over a time scale of hours, with totality lasting only a few minutes for a Solar eclipse. Similiar time frames apply to other Solar system transits. Occultations, where a solar system object passes in front of a star, are nearly instantaneous for a visual observer. Meteor trails in the atmosphere are common, and are rarely visible for more than a few seconds.&lt;/p&gt;&#xA;" OwnerUserId="28086" LastActivityDate="2019-06-14T20:45:24.010" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32301" PostTypeId="2" ParentId="32293" CreationDate="2019-06-15T00:14:37.040" Score="1" Body="&lt;p&gt;The universe is rather robust in that if physics allows a particular situation, there will usually be some place or time in the universe where that situation will occur.  Thus any timescales that are allowed by physics, the universe will have astrophysical phenomena that occur on those timescales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a long way of providing the answer &quot;as fast as can be and as slow as can be, and everything in between.&quot;&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2019-06-15T00:14:37.040" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32302" PostTypeId="1" CreationDate="2019-06-15T00:46:43.307" Score="1" ViewCount="97" Body="&lt;p&gt;Space with distance of 1megaparsec is expanding with rate of 70km/s, and if space with a distance of 4296 megaparsec is expanding at rate of 300,720km/s, then why Andromeda is coming to us? Even there has a dark energy that accelerate the space expansion? Why?&lt;/p&gt;&#xA;" OwnerUserId="28108" LastActivityDate="2019-06-15T01:28:47.843" Title="If the space is expanding near the speed of light or the speed of light, why Andromeda is approaching or coming near to our galaxy (Milky way)?" Tags="&lt;space&gt;&lt;expansion&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2019-06-15T13:02:51.887" ContentLicense="CC BY-SA 4.0" />
  <row Id="32304" PostTypeId="2" ParentId="32302" CreationDate="2019-06-15T01:28:47.843" Score="6" Body="&lt;p&gt;The Andromeda galaxy is about 2.5 Mly away, less than 1 MPc, so if the Hubble flow were relevant, it would have Andromeda moving at about 55 km/s away from us. Andromeda is actually approaching at about 300 km/s, or almost 6X that calculated due to the expansion of space. However, Andromeda and the Milky Way are graivitationally bound along with the rest of the galaxies in our local cluster. The individual (so called, peculiar) velocities of these local galaxies are the result of orbital motion within groups and around the local cluster barycenter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expansion dominates on larger spatial scales, such as distances between clusters and superclusters of galaxies.&lt;/p&gt;&#xA;" OwnerUserId="28086" LastActivityDate="2019-06-15T01:28:47.843" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32307" PostTypeId="2" ParentId="31877" CreationDate="2019-06-15T07:57:36.583" Score="1" Body="&lt;p&gt;For the Sun at JD 2458599.125, RA = 32.8&amp;deg; and dec = 13.2&amp;deg; are correct regardless of geographic location.&#xA;In the &lt;a href=&quot;https://aa.usno.navy.mil/faq/docs/Alt_Az.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;alt/az calculation&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;LHA&lt;/code&gt; should be 323.6&amp;deg; and converted to radians, and&lt;/li&gt;&#xA;&lt;li&gt;the expression for &lt;code&gt;alt&lt;/code&gt; needs a call to &lt;code&gt;Math.Asin&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Trigonometric expressions are easier to verify if they are free of inline unit conversions.&#xA;Store angles in radians where possible; otherwise &lt;a href=&quot;https://www.joelonsoftware.com/2005/05/11/making-wrong-code-look-wrong/&quot; rel=&quot;nofollow noreferrer&quot;&gt;make it obvious&lt;/a&gt; with a name suffix such as &lt;code&gt;_deg&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When computing the local hour angle, you can approximate &lt;a href=&quot;https://aa.usno.navy.mil/faq/docs/GAST.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;GAST&lt;/a&gt; with GMST.&#xA;The difference eqeq is smaller than the approximation error of the &lt;a href=&quot;https://aa.usno.navy.mil/faq/docs/SunApprox.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sun position algorithm&lt;/a&gt; you're using, and the alternative GMST formula is valid for D within a few centuries of J2000.0.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2019-06-15T07:57:36.583" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32309" PostTypeId="1" CreationDate="2019-06-15T14:12:09.623" Score="1" ViewCount="138" Body="&lt;p&gt;There are millions of galaxies out there in the entire universe. When a telescope is set to a direction to to identify the stellar objects, it can capture all lights from all the objects. How do we identity that the particular light is emitted from one star/galaxy and not the other one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if we precisely point to one star, there are chances that other luminous objects interfere with the light ray.&lt;/p&gt;&#xA;" OwnerUserId="28002" LastEditorUserId="1891" LastEditDate="2019-06-16T19:31:02.613" LastActivityDate="2019-06-16T19:31:02.613" Title="How to find the source of the light?" Tags="&lt;telescope&gt;&lt;light&gt;&lt;optics&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ClosedDate="2019-06-21T07:29:33.233" ContentLicense="CC BY-SA 4.0" />
  <row Id="32311" PostTypeId="1" CreationDate="2019-06-15T18:50:22.573" Score="7" ViewCount="7024" Body="&lt;p&gt;In the solar system, things float around. We are so certain that the gravity of our sun exists. Still, why does that gravity not influence satellites and other objects there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our solar system is also orbiting the center of our galaxy, and the gravity of our black hole, but still many objects float around. I wonder why.&lt;/p&gt;&#xA;" OwnerUserId="28002" LastEditorUserId="-1" LastEditDate="2019-06-17T17:31:43.183" LastActivityDate="2019-06-17T17:31:43.183" Title="Why do things float in space, though the gravity of our star is always present?" Tags="&lt;gravity&gt;&lt;solar-system&gt;&lt;supermassive-black-hole&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="3" ClosedDate="2019-06-17T18:13:48.890" ContentLicense="CC BY-SA 4.0" />
  <row Id="32312" PostTypeId="2" ParentId="32293" CreationDate="2019-06-15T19:14:20.673" Score="2" Body="&lt;p&gt;The answer is that some phenomenas are incredibly fast. Others take billions or trillions of years (or more). But, most observed events don't just occur one time. So we can watch for longer events many times, and observe &quot;snapshots&quot; of many different times during the process, and interpolate to get then process as a whole.&lt;/p&gt;&#xA;" OwnerUserId="18601" LastActivityDate="2019-06-15T19:14:20.673" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32313" PostTypeId="2" ParentId="32311" CreationDate="2019-06-15T19:26:15.123" Score="42" Body="&lt;p&gt;It is not true that &quot;objects float around&quot; in the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you have seen video from the space station, and you can see things floating. This is not because there is no gravity, but because everything in the space station going at the same speed in the same direction. This makes it look as if things are floating. In fact the space station and everything in it, is moving at about 7km per second around the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth and everything on it is is also moving at 30 km/s around the sun. The satellites around the Earth are also affected by the sun's gravity, but because they are affect the same as the Earth, they are also moving at about 30km/s around the sun. If two things are moving at the same speed and in the same direction, it will look as if they are floating. In fact they are &quot;whizzing&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun, Earth and satellites are also whizzing around in the gravity of the galaxy.  The black hole is only a very small part of this; most of the gravity of the galaxy is in the dark matter that we can't see.  But we don't feel this pull because the Sun, Earth, the satellites and you are all being pulled at the same time. &lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="2365" LastEditDate="2019-06-15T19:32:21.080" LastActivityDate="2019-06-15T19:32:21.080" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32316" PostTypeId="2" ParentId="32293" CreationDate="2019-06-16T04:55:27.337" Score="2" Body="&lt;p&gt;Your question strikes me as being more about humanity than astronomy. We define what is interesting to us: there is no physical law as to what &quot;interesting&quot; is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;interesting&quot; part of a supernova might be a couple of months (when it's at its brightest), however, if we could predict when a star was within ten years of becoming a supernova, that would increase the window of &quot;interesting&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun has been observed by humans for as long as humans have existed. We still find it interesting. We now know much more about it than when humans first appeared on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, the length of an interesting event is entirely up to us, and each type of interesting event will have its own typical length. It could be milliseconds to millennia.&lt;/p&gt;&#xA;" OwnerUserId="1435" LastActivityDate="2019-06-16T04:55:27.337" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32317" PostTypeId="2" ParentId="32311" CreationDate="2019-06-16T05:23:45.563" Score="33" Body="&lt;p&gt;To help with James K's excellent answer, a visual representation might help. Let's look at a thought experiment - &lt;a href=&quot;https://en.wikipedia.org/wiki/Newton%27s_cannonball&quot; rel=&quot;noreferrer&quot;&gt;Newton's Cannonball&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you have a cannon, high enough that it's being held above Earth's atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You fire it, and it falls to Earth a little ways away (&quot;D&quot; in the below diagram).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You fire another one with more power so it's moving quicker, so that it falls to Earth farther away. (&quot;E&quot; in the below diagram)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CGw7Y.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CGw7Y.jpg&quot; alt=&quot;Newton&amp;#39;s Cannonball&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, you fire a cannonball with such extreme velocity, that it's &quot;falling&quot; around the Earth fast enough that it never reaches the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is orbit. Orbit isn't necessarily being really high up and moving slowly. More often than not, orbit is going sideways fast enough that you're falling without losing height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why on places like the ISS, does it look like things are just floating around? Let's quickly go back to the cannonball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine the cannon fires &lt;em&gt;two&lt;/em&gt; cannonballs at the same time, both going fast enough to be in orbit. These cannonballs are going blisteringly fast... but they were fired at the same time, going the same speed, so they stay together. If you can imagine being one of the cannonballs, the other cannonball would look like it's just floating next to you as you fly around the Earth. This is because, &lt;em&gt;relative to each other&lt;/em&gt;, the cannonballs have next to no &lt;em&gt;relative&lt;/em&gt; velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ISS, similarly, is traveling about 7.66 km/s, or around &lt;strong&gt;27,600 km/h&lt;/strong&gt; (about 17,150 miles per hour for those using imperial measurements). It's going &lt;em&gt;fast&lt;/em&gt;. But when you're on there, everything is going the same speed, because you're all on the ISS together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you let go of a pen on the ISS, it's still traveling the same speed as you - around 27,600 km/h. But &lt;em&gt;because&lt;/em&gt; it's going the same speed as you, relative to you it just looks like it's floating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth isn't just floating in orbit around the Sun, it's in orbit at (on average) 107,000 km/h (or 67,000 miles per hour).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our Solar System isn't just floating around the center of our galaxy, it's in orbit at around 828,000 km/h (or around 514,500 miles per hour).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are all hard to comprehend speeds - we can all agree though, that this isn't just floating around. Things are moving fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things can &lt;em&gt;appear&lt;/em&gt; to be just &quot;floating&quot; because their relative velocity to the observer is small. But I hope this all gives an explanation of how just because something appears to be slowly floating around, that doesn't mean it's not still moving fast from a different point of view.&lt;/p&gt;&#xA;" OwnerUserId="28415" LastActivityDate="2019-06-16T05:23:45.563" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32318" PostTypeId="2" ParentId="32309" CreationDate="2019-06-16T10:03:59.933" Score="-5" Body="&lt;p&gt;The positions of many galaxies are recorded, so astronomers can point their telescopes to the exact point in the sky that they wish to observe. In addition,the relatively nearby galaxies can be recognised by their appearance,&amp;amp; these are the galaxies most intensively studied. Professional astronomers have memorised most of the stars you will see in the night sky, and the constellations (Sagittarius, Ursa Major, Orion etc) guide them to their target. Those galaxies near the limit of observation are of lttle interest except when something unusual &amp;amp; very luminous happens within them,such as a supernova or gamma ray burst.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-06-16T10:03:59.933" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32319" PostTypeId="1" AcceptedAnswerId="32321" CreationDate="2019-06-16T10:07:55.893" Score="1" ViewCount="90" Body="&lt;p&gt;I am currently trying to understand dynamical modelling using Schwarzschild's method. I have been doing some reading the past days, but I suspect that I am getting something really wrong. According to my current understanding, the Schwarzschild method for dynamical modelling using spatially-resolved kinematic data works like this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Fit one or more 2D profiles (e.g., Gaussian) to the image or zeroth moment of the observation.&lt;/li&gt;&#xA;&lt;li&gt;De-project the above fitted model to 3D space and create a 3D luminosity distribution.&lt;/li&gt;&#xA;&lt;li&gt;Convert the 3D luminosity distribution to mass using a bunch of mass particles.&lt;/li&gt;&#xA;&lt;li&gt;Create a series of possible orbits for the above mass particles and calculate the resulting velocity vectors.&lt;/li&gt;&#xA;&lt;li&gt;From the above velocity vectors create velocity and velocity dispersion maps and calculate the chi-squared between them and the observed velocity and velocity dispersion maps. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The last step is what troubles me. To my understanding, the orbits of the particles in a Schwarzschild model can be counter-rotating. Right? In a galaxy observation, when we have two collisionless components (e.g., stellar components) that counter-rotate, it is possible that we will observe double-peaked spectral lines. Right? This information is impossible to be stored using a velocity and a velocity dispersion map, but it can be stored in a spectral cube. Yet, all papers on dynamical modelling using the Schwarzschild method fit models to velocity and velocity dispersion maps instead of a spectral cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, from the above I have concluded that my understanding of the Schwarzschild method is not correct. I would really appreciate if someone could help me to understand what I am getting wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="1986" LastEditorUserId="1986" LastEditDate="2019-06-16T13:31:29.203" LastActivityDate="2019-06-17T10:51:34.743" Title="Dynamical modelling of galaxies with counter-rotating components using the Schwarzschild method" Tags="&lt;galactic-dynamics&gt;&lt;stellar-dynamics&gt;&lt;modeling&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32321" PostTypeId="2" ParentId="32319" CreationDate="2019-06-16T14:56:06.727" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Convert the 3D luminosity distribution to mass using a bunch of mass particles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, you probably could do that, but I suspect it would be inaccurate and extremely time-consuming. Most Schwarzschild modeling approaches parameterize the 3D luminosity-density distribution using functions from which the gravitational potential can be relatively easily calculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, one popular approach is to approximate the observed 2D light distribution using the sum of multiple Gaussians (as you mention), which can be deprojected under certain assumptions into the sum of multiple 3D Gaussians to get the 3D luminosity-density distribution. Then you multiply this distribution by a mass-to-light ratio (usually one of the parameters of the model you're ultimately trying to fit to the data) and derive the gravitational potential from that. No need to approximate things with a bunch of particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also allows you to easily add extra hypothetical &quot;dark&quot; components, like dark-matter halos or central supermassive black holes, which have simple analytical descriptions and don't need to modeled with particles.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To my understanding, the orbits of the particles in a Schwarzschild model can be counter-rotating.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Correct. In practice, if the system is (assumed to be) axisymmetric, you can calculate a set of orbits, and then easily generate their counter-rotating counterparts by just reversing the velocity vectors.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Yet, all papers on dynamical modelling using the Schwarzschild method fit models to velocity and velocity dispersion maps instead of a spectral cube.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're clearly not reading the right papers ;-). &#xA;Let me divide this into two parts:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;1. Converting spectroscopic data to stellar-kinematic data&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Before any modeling takes place, the observed spectroscopic data (long-slit and/or integral-field-unit &quot;spectral cubes&quot;) must be analyzed to extract the stellar-kinematic data. This is done by dividing up the data into different spatial bins on the sky (bins along the slit, or 2D bins for a spectral cube) and summing the spectra within each bin, to improve the S/N. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you fit each spectrum with a weighted combination of stellar spectra (observed or from models), broadened by a model&lt;a href=&quot;https://i.stack.imgur.com/lGeit.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt; line-of-sight velocity distribution (LOSVD). In very simple cases, the LOSVD is assumed to be just a Gaussian with a specific central velocity &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; and velocity dispersion &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt;. The more common present-day approach is to model the LOSVD with a Gaussian which is multiplied by third- and fourth-order Gauss-Hermite polynomials. The strength of the 3rd-order polynomial is parameterized by &lt;span class=&quot;math-container&quot;&gt;$h_{3}$&lt;/span&gt; and measures skew; the 4th-order polynomial is parameterized by &lt;span class=&quot;math-container&quot;&gt;$h_{4}$&lt;/span&gt; and measures kurtosis. For galaxies with counter-rotating components, &lt;span class=&quot;math-container&quot;&gt;$h_{3}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$h_{4}$&lt;/span&gt; provide &lt;em&gt;some&lt;/em&gt; information about the non-single-Gaussian nature of the LOSVD, although it's rather fuzzy and poorly constrained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more sophisticated approach (requiring high-S/N data) is to &quot;non-parametrically&quot; model the LOSVD with a histogram having twenty, thirty, or more velocity bins. To show how the non-parametric approach works, consider the following stellar-kinematic data for the S0 galaxy NGC 4191:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first plot, from Fig. B1 of &lt;a href=&quot;http://cdsads.u-strasbg.fr/abs/2011MNRAS.414.2923K&quot; rel=&quot;nofollow noreferrer&quot;&gt;Krajnovic et al. 2011 (&lt;em&gt;MNRAS&lt;/em&gt;, 414, 2923)&lt;/a&gt;, shows &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; maps (from the SAURON IFU). In the velocity field, you can see that the velocity reverses between &lt;span class=&quot;math-container&quot;&gt;$r \sim 5$&lt;/span&gt; and 10 arc seconds; at roughly the same radius, &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; peaks. The paired off-center peaks in &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; are suggestive of a region where the two counter-rotating components are about equal (as argued by Krajnovic et al. in their paper).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lGeit.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lGeit.png&quot; alt=&quot;Based on Fig.B1 of Krajnovic et al. (2011)&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second plot, based on Fig. 2 of &lt;a href=&quot;http://cdsads.u-strasbg.fr/abs/2015A%26A...581A..65C&quot; rel=&quot;nofollow noreferrer&quot;&gt;Coccato et al. 2015 (&lt;em&gt;A&amp;amp;A&lt;/em&gt;, 581, A65)&lt;/a&gt;, shows the &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; map (colored panel, rotated 90 degrees with respect to the figure above) and non-parametric LOSVDs (31 velocity bins) for three of the bins (histograms) from VIRUS-W IFU observations of NGC 4191. You can see how, in the second histogram, the double peaks due to the overlapping, counter-rotating stellar components are clearly represented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Lmbew.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lmbew.png&quot; alt=&quot;Based on Fig.2 of Coccato et al. (2015)&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;2. Comparing the Schwarzschild model with the kinematic data&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;After computing the library of stellar orbits for a given galaxy potential and assigning a set of weights to the individual orbits, you &quot;observe&quot; the model by orienting the galaxy to match the observed orientation, dividing it up into projected 2D spatial bins that match the 2D binning used for determining real galaxy's stellar kinematics. Then (for each bin) you compute the weighted LOSVD by integrating along the line of sight through that bin, adding up the radial-velocity contributions from each orbit that intersects the line of sight. If your data consists of &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$h_3$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$h_4$&lt;/span&gt;, then you fit the resulting model LOSVD for that bin with the Gauss-Hermite function to get the model bin's &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$h_3$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$h_4$&lt;/span&gt;. If your data are non-parametric LOSVDs, then you can simply bin the model LOSVD using the same velocity-binning scheme you used for the data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This set of predicted LOSVD values (either &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$h_3$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$h_4$&lt;/span&gt; or the non-parametric binned velocity histogram) is what you compare with data's LOSVD values for that spatial bin, repeating for all the spatial bins with kinematic data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lGeit.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt; &quot;Model&quot; in this case having nothing to do with the Schwarzschild modeling.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2019-06-17T10:51:34.743" LastActivityDate="2019-06-17T10:51:34.743" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="32322" PostTypeId="2" ParentId="32311" CreationDate="2019-06-17T00:00:53.557" Score="20" Body="&lt;p&gt;Ok, gotta &lt;a href=&quot;https://what-if.xkcd.com/58/&quot; rel=&quot;noreferrer&quot;&gt;quote XKCD&lt;/a&gt; on this.&lt;/p&gt;&#xA;&lt;p&gt;This is not how space works:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/h23k2.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h23k2.png&quot; alt=&quot;Not space&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;This is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kB4GI.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kB4GI.png&quot; alt=&quot;Space&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Gravity in low Earth orbit is almost as strong as gravity on the surface.&lt;/strong&gt; The Space Station hasn't escaped Earth's gravity at all; it's experiencing about 90% the pull that we feel on the surface.&lt;/p&gt;&#xA;&lt;p&gt;To avoid falling back into the atmosphere, you have to go sideways &lt;strong&gt;really, really fast.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The reason things in space tend to stay in space, or why things &amp;quot;float&amp;quot; in the international space station is because they're going so fast gravity doesn't do anything more than keep those objects moving around the planet: the parabolic trajectory that all falling objects have is so wide that it misses the horizon and comes back around again. Aka, orbit. As the space station and all of its contents are moving at the same speed, and are in constant in free fall, things appear to float: there's no object at rest for things to fall &lt;em&gt;towards;&lt;/em&gt; everything is falling and everything is falling at the same speed, all the time.&lt;/p&gt;&#xA;&lt;p&gt;If the ISS had an arm long enough&lt;sup&gt;1&lt;/sup&gt; and someone put something perfectly still in the center of the space at the far end, that object wouldn't stay there. It would be ever so slightly off the orbital path that the station would either catch up to it, or fall away from it. But it would be very, very slow. And it would happen because the ISS as a whole and the &amp;quot;floating object&amp;quot; would be in slightly different, intersecting, orbits.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Still, why does that gravity not influence satellites and other objects there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It does. It just isn't very strong. Gravity falls off at an inverse-square relationship with distance, just like light does. Except that gravity is &lt;em&gt;way way&lt;/em&gt; weaker of a force than electromagnetism is (&amp;quot;no its not, I can feel the Earth right now!&amp;quot; Yes, and the Earth is a few trillion-trillion times heavier than you are, how about a table? Can you feel the table's gravity?). The gravitational acceleration towards the sun is on the order of &lt;a href=&quot;https://van.physics.illinois.edu/qa/listing.php?id=1063&amp;amp;t=gravitational-pull-of-the-sun&quot; rel=&quot;noreferrer&quot;&gt;about 0.005m/s&lt;sup&gt;2&lt;/sup&gt;&lt;/a&gt; (and that's still 50 million times greater than the gravitational pull from &lt;a href=&quot;https://what-if.xkcd.com/136/&quot; rel=&quot;noreferrer&quot;&gt;all the spiders everywhere&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;sub&gt;Technically this is already true, just that the effect is so subtle I'm not sure you'd see it in timespans less than &amp;quot;days.&amp;quot; In either case, my google fu isn't strong enough to find any videos of such an experiment.&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="25370" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-06-17T00:00:53.557" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32323" PostTypeId="2" ParentId="32311" CreationDate="2019-06-17T08:43:06.867" Score="2" Body="&lt;p&gt;The real answer is that, locally (strictly speaking, in an infinitesimal region), gravity is not a thing. Whatever your movement is relatively to earth or the sun (even without going &quot;fast to the side&quot; as previously stated), you won't feel any gravity in a local experiment. Even when forces enter into account (like air friction or the reaction of the ground), you can't say without looking at the window if you're act on by gravity or inertia coming from, say, thrusters from your rocket. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the equivalence principle, the one responsible for the miracle that astronauts in the ISS and the ISS fall at the same rate, leaving astronauts weightless, in Newton's theory, through the dubious force = mass x acceleration and gravitational force = mass x gravity field, hence m a = m g, so a = g for any mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton himself knew that the strict equality between inertial and gravitational masses for any body, whatever its mass, composition or whatever, could not come from something random. This laid the groung for general relativity.&lt;/p&gt;&#xA;" OwnerUserId="20131" LastActivityDate="2019-06-17T08:43:06.867" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32325" PostTypeId="1" AcceptedAnswerId="32328" CreationDate="2019-06-17T13:13:03.080" Score="1" ViewCount="187" Body="&lt;p&gt;For an application I am working on I am creating a simulation of the solar system. In this simulation, a user needs to be able to 'move' to any position on any planet, and have the sky be accurately simulated around him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the examples &amp;amp; calculations provided at &lt;a href=&quot;http://stjarnhimlen.se/comp/ppcomp.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://stjarnhimlen.se/comp/ppcomp.html&lt;/a&gt; and the data provided at &lt;a href=&quot;https://ssd.jpl.nasa.gov/txt/aprx_pos_planets.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ssd.jpl.nasa.gov/txt/aprx_pos_planets.pdf&lt;/a&gt; I am able to compute the POSITIONS of the planets at any date, however, what I am still looking for is the ORIENTATION of these planets at that position (i.e. which way is Long 0 Lat 0 pointing, and which way is Long 0 Lat 90 pointing (with respect to the vernal equinox)). I need this orientation in order to figure out what the rotation (quaternion) of this planet needs to be..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been looking around, but have so far not been able to find a source for any kind of information like this. The closest I've found was &lt;a href=&quot;https://en.wikipedia.org/wiki/Poles_of_astronomical_bodies&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Poles_of_astronomical_bodies&lt;/a&gt; , but that only gives me an up-vector, not a 'forward' (and is for J2000, not the current Epoch).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone be of assistance?&lt;/p&gt;&#xA;" OwnerUserId="28433" LastActivityDate="2021-05-10T14:56:59.773" Title="How to calculate the orientation of planets at current epoch" Tags="&lt;planet&gt;&lt;solar-system&gt;&lt;rotation&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32326" PostTypeId="1" CreationDate="2019-06-18T05:36:21.763" Score="1" ViewCount="124" Body="&lt;p&gt;We have a Celestron 70AZ &lt;code&gt;70mm f/10 refractor&lt;/code&gt; that has been working perfectly. Our children knocked it over the other day, and now when we try to view the moon, it fills the lens as a giant, blurry ball of light. Have they knocked something loose? We tried adjusting the different focus knobs but I am worried it's messed up and I have zero clue how to fix this, diagnose it or even where to go to get it looked at. &lt;/p&gt;&#xA;" OwnerUserId="28443" LastEditorUserId="7982" LastEditDate="2019-06-18T17:24:48.470" LastActivityDate="2019-06-18T17:24:48.470" Title="Telescope was working and now it's not" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32328" PostTypeId="2" ParentId="32325" CreationDate="2019-06-18T07:36:50.837" Score="2" Body="&lt;p&gt;The &lt;a href=&quot;https://www.usgs.gov/centers/astrogeology-science-center/science/iau-wgccre&quot; rel=&quot;nofollow noreferrer&quot;&gt;IAU WGCCRE&lt;/a&gt; (Working Group on Cartographic Coordinates and Rotational Elements) seem to be the authorities on this.&#xA;Table 1 in their &lt;a href=&quot;https://astrogeology.usgs.gov/search/map/Docs/WGCCRE/WGCCRE2015reprint&quot; rel=&quot;nofollow noreferrer&quot;&gt;2015 report&lt;/a&gt; gives the following information for the major planets:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;α&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt;, &lt;em&gt;δ&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt; are the ICRF right ascension and declination of the planet's north pole. If the planet's axial precession is known, these are expressed in terms of &lt;em&gt;T&lt;/em&gt;, the number of Julian centuries since epoch J2000.0.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;W&lt;/em&gt; is the rotation angle eastward from reference point &lt;em&gt;Q&lt;/em&gt; to the planet's prime meridian at time &lt;em&gt;d&lt;/em&gt;, where &lt;em&gt;Q&lt;/em&gt; is the intersection of the planet's equator and the ICRF equator at RA &lt;em&gt;α&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt; + 90°, Dec 0°, and &lt;em&gt;d&lt;/em&gt; is the number of days since epoch J2000.0.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;They define a planet's north pole as the one pointing north of the solar system's &lt;a href=&quot;https://en.wikipedia.org/wiki/Invariable_plane&quot; rel=&quot;nofollow noreferrer&quot;&gt;invariable plane&lt;/a&gt;.&#xA;This means that &lt;em&gt;W&lt;/em&gt; decreases with time for Venus and Uranus.&lt;/p&gt;&#xA;&lt;p&gt;Their figure 1 appears to indicate &lt;em&gt;Q&lt;/em&gt;'s RA as 90° - &lt;em&gt;α&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt;, in conflict with the text. I believe the text is correct.&lt;/p&gt;&#xA;&lt;p&gt;When you are ready to validate your calculations, the &lt;a href=&quot;https://www.skyandtelescope.com/observing/interactive-sky-watching-tools/mars-which-side-is-visible/&quot; rel=&quot;nofollow noreferrer&quot;&gt;S&amp;amp;T Mars Profiler&lt;/a&gt; shows which side of Mars faces Earth at a given time.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2021-05-10T14:56:59.773" LastActivityDate="2021-05-10T14:56:59.773" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32329" PostTypeId="1" CreationDate="2019-06-18T11:20:05.720" Score="4" ViewCount="153" Body="&lt;p&gt;I am a complete beginner in astronomy. While learning DS9, I came up with the following question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I want to exclude a boxed region from an image FITS. The corresponding parameters according to &lt;a href=&quot;http://cxc.harvard.edu/ciao/ahelp/dmregions.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chandra&lt;/a&gt; is&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Box = (xcenter,ycenter,width,height)&lt;/li&gt;&#xA;&lt;li&gt;Box = (xcenter,ycenter,width,height,angle)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Although the name is self-explanatory yet I am confused about their respective units. What unit is used to label these parameters?  &lt;/p&gt;&#xA;" OwnerUserId="28447" LastEditorUserId="28447" LastEditDate="2019-06-18T20:32:21.570" LastActivityDate="2019-06-18T20:32:21.570" Title="Excluding a rectangular region from an image in FITS" Tags="&lt;ds9&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32330" PostTypeId="2" ParentId="32329" CreationDate="2019-06-18T12:47:38.377" Score="3" Body="&lt;p&gt;Usually pixel coordinates.&#xA;What I would do is to let DS9 make a region file, as your documentation site says it supports them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do so:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Open file in ds9&lt;/li&gt;&#xA;&lt;li&gt;From region-&gt;shape select the kind of region you want.&lt;/li&gt;&#xA;&lt;li&gt;Click on edit-&gt;region&lt;/li&gt;&#xA;&lt;li&gt;Draw the shapes you want (just click on the img)&lt;/li&gt;&#xA;&lt;li&gt;Click region-&gt;save. In the coordinates box choose physical pixel coordinates.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Probably that should work... Good luck!&lt;/p&gt;&#xA;" OwnerUserId="28449" LastActivityDate="2019-06-18T12:47:38.377" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32331" PostTypeId="2" ParentId="32329" CreationDate="2019-06-18T13:02:22.453" Score="1" Body="&lt;p&gt;Several different units are possible. If you double click on the box a dialogue window will open where you can adjust the size/location. In the drop-down menus on the right hand site, you can choose which unit and coordinate system you want.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ct5dP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ct5dP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="950" LastActivityDate="2019-06-18T13:02:22.453" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32332" PostTypeId="2" ParentId="32297" CreationDate="2019-06-18T13:32:32.480" Score="3" Body="&lt;p&gt;As everyone is saying, this is a very loose &quot;plain English&quot; version of some complex ideas, but if space was infinite, you could view this suggestion (that the overall total energy of the universe is zero) as applying to any sufficiently large region of space. So if you pick any 10 billion lightyear diameter sphere you would expect to find the energy of the matter in that sphere more or less cancelled out by the gravitational potential energy arising from the separation of that matter.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-06-18T13:32:32.480" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32333" PostTypeId="1" CreationDate="2019-06-18T14:00:12.213" Score="-2" ViewCount="99" Body="&lt;p&gt;How essential to modern cosmology is the vacuum energy concept, and what would happen to our model of the expanding universe if it were proved to be non-existent? &lt;/p&gt;&#xA;" OwnerUserId="28083" LastEditorUserId="21928" LastEditDate="2019-06-18T14:03:56.730" LastActivityDate="2019-06-19T09:39:38.693" Title="How Essential is the Vacuum Energy to Our Present Model of the Expanding Universe?" Tags="&lt;cosmology&gt;&lt;vacuum&gt;" AnswerCount="1" CommentCount="7" ClosedDate="2019-06-19T12:23:14.403" ContentLicense="CC BY-SA 4.0" />
  <row Id="32336" PostTypeId="1" CreationDate="2019-06-18T22:13:45.793" Score="8" ViewCount="469" Body="&lt;p&gt;Reading this response &lt;a href=&quot;https://astronomy.stackexchange.com/questions/10538/when-the-universe-expands-does-it-create-new-space-matter-or-something-else&quot;&gt;When the universe expands does it create new space, matter, or something else?&lt;/a&gt; I understand Universe is not expanding like if it is &lt;em&gt;stretching its space&lt;/em&gt;, but it is actually adding space that matter can travel through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if the Universe can &lt;em&gt;create&lt;/em&gt; more space, and space and time are related, shouldn't that mean Universe is &lt;em&gt;creating&lt;/em&gt; also more &lt;em&gt;time&lt;/em&gt;? And if it is the case, shouldn't this effect alter results of our measurements about how much time light spends to reach us from distant stars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that photons have no mass, travel at speed of light, and time has no sense for them. But at the same time they are influenced in their trajectories by folds in the space-time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If gravity deforms space-time, shouldn't an expanding Universe &lt;em&gt;adding new space&lt;/em&gt; doing the same?&lt;/p&gt;&#xA;" OwnerUserId="28452" LastEditorUserId="28452" LastEditDate="2019-06-18T23:44:04.467" LastActivityDate="2019-06-19T17:17:23.497" Title="Universe is expanding, why doesn't time expand with it?" Tags="&lt;space-time&gt;&lt;time&gt;&lt;time-dilation&gt;&lt;photons&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32337" PostTypeId="2" ParentId="32336" CreationDate="2019-06-19T04:48:07.120" Score="2" Body="&lt;p&gt;The short answer is, &quot;gravity&quot;.  The time of &quot;co-moving observers&quot; (observers moving with the expansion of the universe&quot;) is expanding, but our &quot;proper time&quot; is not, because we're within a local group of galaxies that are bound together gravitationally.  The principles involved are detailed in the Wikipedia article &quot;Co-moving and proper distances&quot;, and its links.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in any reading about General Relativity--the currently-accepted theory of gravity-- it may be helpful to understand that &quot;observers&quot; are really any sort of hypothesized material or energetic object, however small:  I believe they were supposed to be hypothetical but possible beings, when use of that terminology began--back around the time when GR's predecessor, Special Relativity, was written--but are now involved in so many inaccessible and inhospitable situations that &quot;they&quot; are not necessarily considered to be sentient, or even alive.  It may also be useful to understand that, although gravity is realistically considered to be the curvature of spacetime, &quot;time&quot; itself is best viewed as comparable to the three well-known spatial dimensions, or at least as something comparably definite, like &quot;duration&quot;:  The temporal aspect of spacetime actually includes most of its curvature, as described in a website about General Relativity by the University of Pittsburgh's Dr. Norton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your secondary question, the delay in the arrival of particles (photons) from distant stars is actually affected by several factors, generally based on inflationary cosmology, that are in a well-known diagram shown in the first Wikipedia item I suggested, but visible in larger versions with explanatory text:  Searches by the names of its originators, &quot;Lineweaver and Davis&quot;, will reach references on various levels of sophistication.      &lt;/p&gt;&#xA;" OwnerUserId="22115" LastEditorUserId="22115" LastEditDate="2019-06-19T17:17:23.497" LastActivityDate="2019-06-19T17:17:23.497" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32339" PostTypeId="2" ParentId="32333" CreationDate="2019-06-19T09:39:38.693" Score="2" Body="&lt;p&gt;I'm neither a quantum physicist, nor a cosmologist, so I'm going only from my laymans understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The   framework for constructing theories to explain everything in the universe &lt;em&gt;except gravity&lt;/em&gt; which has proved most popular and useful for the last 50 or so years, is &lt;a href=&quot;https://en.wikipedia.org/wiki/Quantum_field_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;quantum field theory&lt;/a&gt;. In this theory, there are multiple fields (conceptually similar to the electric and magnetic fields) suffusing every point in space and time. From the values of these fields, we can,  in principle, compute the probabilities of the outcomes of any experiment, including the probability of detecting a particular kind of particle at a particular place and time, and from the rules of these computations we can derive all the normal &quot;laws of physics&quot; such as the conservation of energy, the types of particles which exist and so on, as &quot;emergent&quot; properties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I described QFT as a &quot;framework&quot; because there are many possible sets of fields and rules for their interactions, each of which gives different physics. From the experiments we can do, we can develop some constraints on the overall structure, but we also know that the fields and interactions that we know about cannot be the whole story. Over short distance and time scales and large energy scales they would give rise to non-sensical predictions, so they must be wrong, but we cannot yet do experiments to find out what really happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the vacuum is the configuration of those fields which has the lowest possible energy, but that does not have all the fields being zero, and when we calculate that minimal energy (as far we can) using only the known particles and fields we get a ridiculously high value (that's one of the non-sensical predictions I mentioned). This is not taken as a sign that the vacuum energy &lt;em&gt;is&lt;/em&gt; that high, but that the unknown high-energy fields must somehow contribute to lowering it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of this is very relevant to cosmology, which, except for the very earliest moments of the universe, does not really depend on the details of QFT, but only  on the &quot;emergent&quot; properties such as the behaviour of gasses, stars, light waves, neutrinos, etc. and on the one thing that (so far) does not seem to fit into QFT, namely gravity (or general relativity, if you prefer). For, sure, if the vacuum energy actually were the silly value computing from known forces using QFT, the gravitational effects of that energy would make anything like the universe we see impossible, but that is just taken as confirming that there is as yet unexplored physics on small scales. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the summary is &quot;not essential at all&quot;. Cosmology and the expanding universe depends on GR and properties of matter and radiation on large scales and at relatively low energies. The Vacuum energy relates to the properties of the fields that underly that matter and energy on very small scales.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-06-19T09:39:38.693" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32340" PostTypeId="1" CreationDate="2019-06-19T15:57:12.530" Score="4" ViewCount="68" Body="&lt;p&gt;The &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt;-process in supernovae happens when a high flux of neutrons irradiates the heaviest abundant nuclei (in the region of iron), resulting in endothermic nuclear reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the extent that at least some of the highest &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt;-process elements are now thought to be synthesised in neutron star mergers, is the mechanism the same: that is, is it a question of  the (hypothesised) iron crust of the neutron stars being irradiated by a sudden intense flux of neutrons at the moment of the merger? Or is it that the &quot;splashed&quot; neutronium resolves itself into a range of atomic nuclei, especially the heavier &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt;-process ones?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I appreciate that this is a question about hypothesis built upon hypothesis, but it would be interesting to get some idea of current thinking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT (from a comment):&lt;/strong&gt; &lt;a href=&quot;https://physics.stackexchange.com/a/379404/97235&quot;&gt;This Stack Exchange answer&lt;/a&gt; suggests that splashed neutronium will end up as the lightest elements, but tantalisingly raises another interesting question: whether the &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt;-process elements are created in the merger or whether they are constantly present at the bottom of the crust, in some sort of equilibrium between the infinite supply of neutrons within the star and the iron in the crust itself. In which case the merger would create nothing, merely liberate what is already there.&lt;/p&gt;&#xA;" OwnerUserId="12016" LastEditorUserId="12016" LastEditDate="2019-06-19T17:25:09.667" LastActivityDate="2019-06-19T17:25:09.667" Title="Neutron star r-process nucleosynthesis" Tags="&lt;neutron-star&gt;&lt;nucleosynthesis&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32341" PostTypeId="1" CreationDate="2019-06-19T15:57:14.700" Score="2" ViewCount="68" Body="&lt;p&gt;How can I explain approximately the same depth of primary and secondary minimum in light curve? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I observe star HD 36486. I tried filters MOST and BRITE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it related with photometric filter? When I change the filter in program for analysis the result is the same.&lt;/p&gt;&#xA;" OwnerUserId="28268" LastEditorUserId="2424" LastEditDate="2019-06-20T22:17:47.010" LastActivityDate="2019-06-20T22:17:47.010" Title="Depth of primary and secondary minimum in a light curve" Tags="&lt;data-analysis&gt;&lt;binary-star&gt;&lt;light-curve&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32342" PostTypeId="1" AcceptedAnswerId="32347" CreationDate="2019-06-19T17:51:53.663" Score="3" ViewCount="71" Body="&lt;p&gt;Was the expansion of space time during big bang uniform if so how come the stars and planets be formed because and matter as well as gravity would have been same everywhere&lt;/p&gt;&#xA;" OwnerUserId="28461" LastEditorUserId="616" LastEditDate="2019-06-19T21:46:08.920" LastActivityDate="2019-06-19T23:25:10.370" Title="Question about expansion of space time" Tags="&lt;early-universe&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32343" PostTypeId="1" CreationDate="2019-06-19T18:51:59.233" Score="1" ViewCount="419" Body="&lt;p&gt;A new exoplanet has been just discovered around the star Teegarden, at only 12 light years away.&lt;/p&gt;&#xA;" OwnerUserId="18421" LastActivityDate="2019-06-21T18:33:44.863" Title="Is the newly discovered exoplanet 'Teegarden b' habitable?" Tags="&lt;space&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32344" PostTypeId="2" ParentId="32343" CreationDate="2019-06-19T18:51:59.233" Score="-1" Body="&lt;p&gt;It is potentially habitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Teegarden b has a minimum mass almost identical to Earth and it orbits within the star's habitable zone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exoplanet has a 60% chance of having a temperate surface environment where&#xA;temperature should be closer to 28°C assuming a similar terrestrial atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, it is the exoplanet with the highest Earth Similarity Index discovered so far: 95%. This means that it has the closest mass and insolation to terrestrial values.&lt;/p&gt;&#xA;" OwnerUserId="18421" LastEditorUserId="2153" LastEditDate="2019-06-19T18:57:10.897" LastActivityDate="2019-06-19T18:57:10.897" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32347" PostTypeId="2" ParentId="32342" CreationDate="2019-06-19T23:19:49.557" Score="3" Body="&lt;p&gt;Your question is one that bothered cosmologists for a long time, how do you get clumps of matter out of a smooth initial condition, in the time allowed?  It was long understood that a perfectly smooth matter distribution could not be maintained forever, because gravity creates instabilities-- the smallest overdensity in one region would tend, over time, to pull in more matter and increase the overdensity.  But this takes time, and the CMB measurements were coming in surprisingly smooth-- so how did galaxies and stars form in less than a billion years, given that very uniform initial density?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer seems to be dark matter.  Dark matter increases the mass and therefore the gravity, so it reduces the time it takes for instabilities to develop and overdensities to grow.  So the simple answer to your question is that even though cosmology assumes a smooth initial density, it does not say the density is &lt;em&gt;exactly&lt;/em&gt; smooth, and indeed the CMB tells us there are tiny density variations (you can look at the tiny variations seen by the PLANCK satellite at &lt;a href=&quot;https://en.wikipedia.org/wiki/Planck_(spacecraft)#/media/File:Ilc_9yr_moll4096.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Planck_(spacecraft)#/media/File:Ilc_9yr_moll4096.png&lt;/a&gt;, for example-- the scale shows variations of 1 part in ten thousand).  Then gravitational instability increases the overdensity, but needs stronger gravity to do this in the time allowed, so that appears to be provided by dark matter.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2019-06-19T23:25:10.370" LastActivityDate="2019-06-19T23:25:10.370" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32350" PostTypeId="1" AcceptedAnswerId="32467" CreationDate="2019-06-20T02:51:24.177" Score="6" ViewCount="289" Body="&lt;p&gt;I'm looking for a modern astrophysics text. I've had Barbara Ryden's book &lt;em&gt;Foundations of Astrophysics&lt;/em&gt; recommended to me and it sounds okay, but the only review of it I've seen slagged off the problem sets as &quot;more complex than they would initially seem, and often don't actually derive from the chapter.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other opinions on this or recommendations for other texts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an okay grasp of calculus and I'm currently teaching myself introductory mechanics, so I'm not a complete novice (close, but not quite). I'm definitely looking for a college text, not popsci. I'm looking for an introductory astronomy text that assumes 2 semesters of calculus and 2 semesters of introductory physics - basic mechanics and basic electricity and magnetism, essentially. So nothing too high-powered, but also not too watered down.&lt;/p&gt;&#xA;" OwnerUserId="25705" LastEditorUserId="2153" LastEditDate="2019-06-29T03:24:11.233" LastActivityDate="2019-06-29T03:42:41.340" Title="Book recommendation" Tags="&lt;astrophysics&gt;&lt;resource-request&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32354" PostTypeId="1" CreationDate="2019-06-20T13:25:52.630" Score="0" ViewCount="151" Body="&lt;p&gt;For what I know, this torque exists because of the misalignment of the tidal bulge with the apsidal line Earth-Moon a certain angle &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;, which makes the earth rotation to slow down a bit. But, how that pulls away the moon? I thought this &quot;loss&quot; of rotational momentum transferred into tidal heating. I fail to see how this energy is transferred into more orbital momentum, because I thought the conservation of angular momentum didn't have to do with the rotation of the objects. I mean, isn't true that &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; remains constant in the two-body problem (being &lt;span class=&quot;math-container&quot;&gt;$c = \vec{r}\times \dot{\vec{r}}$&lt;/span&gt;)? How the rotation plays into that?&lt;/p&gt;&#xA;" OwnerUserId="9949" LastEditorUserId="9949" LastEditDate="2019-06-20T14:07:56.200" LastActivityDate="2019-06-20T20:17:16.047" Title="Why the torque exerted by the moon upon the Earth makes the Moon to increase his orbit?" Tags="&lt;tidal-forces&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2019-06-21T06:48:01.397" ContentLicense="CC BY-SA 4.0" />
  <row Id="32355" PostTypeId="1" AcceptedAnswerId="32365" CreationDate="2019-06-20T13:46:21.640" Score="4" ViewCount="121" Body="&lt;p&gt;Sunspot maps are created from a projection of the sun, so the coordinates of sunspots (or active regions in general) are assigned according to it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do these coordinates correspond to actual spherical positions, or are they altered as a result of the projection? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; After seeing the answer, I realized that a part of what I am asking about was not clear in the initial question. So here it is: Are the values of the sunspot coordinates &lt;em&gt;as provided officially&lt;/em&gt; in spherical positions or in projection? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A commonly used instance of &quot;official&quot; records is X-ray measurements from the NOAA GOES. The text files are found &lt;a href=&quot;https://www.ngdc.noaa.gov/stp/space-weather/solar-data/solar-features/solar-flares/x-rays/goes/xrs/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, see file for &lt;a href=&quot;https://www.ngdc.noaa.gov/stp/space-weather/solar-data/solar-features/solar-flares/x-rays/goes/xrs/goes-xrs-report_2000.txt&quot; rel=&quot;nofollow noreferrer&quot;&gt;2000&lt;/a&gt;. The output is documented &lt;a href=&quot;https://www.ngdc.noaa.gov/stp/space-weather/solar-data/solar-features/solar-flares/x-rays/goes/xrs/documentation/miscellaneous/software/xraydatareports.pro&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, under &quot;Output File Specification&quot;. &lt;/p&gt;&#xA;" OwnerUserId="16462" LastEditorUserId="16462" LastEditDate="2019-06-26T19:41:17.963" LastActivityDate="2019-06-27T00:33:58.403" Title="Is the longitude assigned to sunspots spherical or projected?" Tags="&lt;the-sun&gt;&lt;coordinate&gt;&lt;positional-astronomy&gt;&lt;sunspots&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32356" PostTypeId="1" AcceptedAnswerId="32357" CreationDate="2019-06-20T14:16:04.573" Score="6" ViewCount="463" Body="&lt;p&gt;I was wondering where spectra of galaxies at various redshifts are to be found. I'm looking for ones that can be used to find the recessional velocity of the galaxy and eventually the Hubble Constant. Here's one I found:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pL3Os.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pL3Os.jpg&quot; alt=&quot;Sample galaxy spectrum from 4000 to 9000 Angstroms, showing hydrogen alpha and oxygen lines&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="28470" LastEditorUserId="2153" LastEditDate="2019-06-20T22:59:42.030" LastActivityDate="2019-06-20T22:59:42.030" Title="Where can I find a database of galactic spectra?" Tags="&lt;galaxy&gt;&lt;redshift&gt;&lt;spectra&gt;&lt;hubble-constant&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32357" PostTypeId="2" ParentId="32356" CreationDate="2019-06-20T14:54:12.733" Score="10" Body="&lt;p&gt;The &lt;a href=&quot;https://www.sdss.org/dr15/spectro/&quot; rel=&quot;noreferrer&quot;&gt;Sloan Digital Sky Survey Data Release 15&lt;/a&gt; contains over 4 million spectra of both galactic and extra-galactic origin from the multi-fiber spectrographs. Of these spectra, 0.7 million came from the original spectrographs during the SDSS-I/II Legacy Survey and the remainder from the upgraded spectrographs as part of the &lt;a href=&quot;https://www.sdss.org/surveys/boss/&quot; rel=&quot;noreferrer&quot;&gt;BOSS survey&lt;/a&gt; during SDSS-III (see &lt;a href=&quot;https://www.sdss.org/surveys/&quot; rel=&quot;noreferrer&quot;&gt;SDSS surveys details pages&lt;/a&gt;). These data went through an automatic pipeline that determined the redshift automatically (where the data quality was good enough) and the details of the process are given &lt;a href=&quot;https://www.sdss.org/dr15/spectro/galaxy/&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spectroscopy data are available through the links to the &lt;a href=&quot;https://skyserver.sdss.org/public/en/tools/quicklook/quickobj.aspx&quot; rel=&quot;noreferrer&quot;&gt;SkyServer Quicklook&lt;/a&gt; which lets you visualize and download the spectra and through the &lt;a href=&quot;http://skyserver.sdss.org/dr15/en/tools/search/SQS.aspx&quot; rel=&quot;noreferrer&quot;&gt;SkyServer Spectroscopic Query Form&lt;/a&gt; which lets you search for galaxies. (Try setting the 'Parameters to return' to 'typical' under the 'Spectroscopy' section' and setting the 'Classification=GALAXY' under the 'Spectroscopy constraints' section to only return objects that have been classified as galaxies). &lt;/p&gt;&#xA;" OwnerUserId="23748" LastEditorUserId="23748" LastEditDate="2019-06-20T20:41:44.527" LastActivityDate="2019-06-20T20:41:44.527" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32358" PostTypeId="1" CreationDate="2019-06-19T21:05:45.250" Score="1" ViewCount="55" Body="&lt;p&gt;How do planets still orbit around the empty space where a black hole was after a black hole absorbs the milky way's black hole and keeps moving through space without disturbing surrounding planets that were orbiting that black hole?&lt;/p&gt;&#xA;" OwnerDisplayName="jehovahsays" LastActivityDate="2019-06-20T16:07:42.687" Title="Undisturbed planetary alignment after galactic collision" Tags="&lt;planet&gt;&lt;celestial-mechanics&gt;&lt;black-hole&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32359" PostTypeId="2" ParentId="32358" CreationDate="2019-06-20T16:07:42.687" Score="4" Body="&lt;p&gt;The short answer is that they don't. If an even larger SMBH passed through the centre of the Milky Way and &quot;swept up&quot; the existing SMBH, all the stars and planets that are close to it (within a few tens of light-years, maybe more) would experience huge disruption and might end up orbiting the new black hole and being carried away, or flying free, or whatever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stars and planets much further away, which make up the bulk of the Milky Way wouldn't be much disturbed, because they are held in their orbits by their gravitational attraction for each other, and for the dark matter halo which suffuses the galaxy and only a small contribution comes from the gravity of the SMBH.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-06-20T16:07:42.687" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32360" PostTypeId="1" CreationDate="2019-06-20T16:40:07.473" Score="0" ViewCount="41" Body="&lt;p&gt;So I was out with my friends at a bonfire and we decided to sit down and star gaze, my other friend pointed out a star that looked to be moving and we all followed it with our eyes and t was smaller than the other stars and would disappear for a second and then come back but not blinking. And once it went out of view another one would appear again and just go opposite directions in kind of a circular path not a straight one and they were really fast too. This was like by the Big Dipper btw super confused is there like something that moved in our space that does that, a satellite would’ve gone in a non crazy direction right? And not so many of them..&lt;/p&gt;&#xA;" OwnerUserId="28472" LastEditorUserId="28472" LastEditDate="2019-06-20T22:30:25.223" LastActivityDate="2019-06-20T22:30:25.223" Title="Saw moving stars in the sky" Tags="&lt;star&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2019-06-20T19:02:25.513" ContentLicense="CC BY-SA 4.0" />
  <row Id="32362" PostTypeId="2" ParentId="32354" CreationDate="2019-06-20T20:07:59.303" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;because I thought the conservation of angular momentum didn't have to do with the rotation of the objects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The angular momentum of the system is conserved.  As rotation can contribute to angular momentum, it does affect the equation.  You can ignore it for point-like objects.  In the case of the earth-moon system, the earth's rotation is a significant portion of the total momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A not-too-wrong version would be to assume an axis centered on the earth.  Then the total angular momentum of the system could be found by summing the individual momenta for:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the earth's rotation&lt;/li&gt;&#xA;&lt;li&gt;the moon's rotation&lt;/li&gt;&#xA;&lt;li&gt;the moon's orbit&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Assuming the moon is tidally locked to the earth and has a nearly circular orbit, this gives you only two variables.  You can then calculate that, if you slow the earth's rotation, the only way to keep angular momentum of the system constant is to increase the angular momentum of the moon by an equal amount.  This would correspond to a larger orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further if you calculate the KE from both configurations, you find the second one has less energy.  So there must be an energy loss to transition between them.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2019-06-20T20:17:16.047" LastActivityDate="2019-06-20T20:17:16.047" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32363" PostTypeId="2" ParentId="32343" CreationDate="2019-06-20T23:01:32.440" Score="6" Body="&lt;p&gt;TLDR version: at this point there are too many unknowns to determine whether this planet candidate is habitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discovery paper for this system, Zechmeister et al. (2019) &quot;&lt;a href=&quot;https://arxiv.org/abs/1906.07196&quot; rel=&quot;nofollow noreferrer&quot;&gt;The CARMENES search for exoplanets around M dwarfs. Two temperate Earth-mass planet candidates around Teegarden's Star&lt;/a&gt;&quot; is available on the arXiv. Table 1 lists the properties of the star and table 2 lists the properties of the planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The planetary quantities we have reliable estimates for are the minimum masses, the orbital periods and the distances. The eccentricities are consistent with zero so I'm going to assume for the moment that the orbits are circular. The reason that minimum mass is relevant is that this is an example of a radial velocity detection, which results in a degeneracy between the inclination of the planet's orbit to the line-of-sight and the mass of the planet.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Habitable zone&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;According to figure 13 in the discovery paper, Teegarden b is too close to the star to be in the conservative estimate of the habitable zone, though it is in the optimistic habitable zone. The Earth Similarity Index (ESI) appears to be a fairly misleading metric (I am of the opinion that ESI is a bad metric that should be abandoned), from the discovery paper:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In addition, planet b receives nearly the same stellar flux&#xA;  as the Earth and therefore has a nearly identical equilibrium temperature, but is outside the conservative HZ. As discussed by&#xA;  &lt;a href=&quot;https://arxiv.org/abs/1404.5292&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kopparapu et al. (2014)&lt;/a&gt;, this is caused by a runaway greenhouse&#xA;  effect that is due to water vapour starting at lower insolation for&#xA;  low-mass stars. This effect is neglected in the ESI calculation&#xA;  and leads to the curious fact that planet b has a higher ESI value&#xA;  than planets inside the conservative HZ.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The position of Teegarden b therefore suggests it may have undergone a runaway greenhouse effect, though this might be avoided if it can build up enough reflective cloud cover on the daylight hemisphere, or if it is a desert planet without enough water to destabilise the climate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if a planet is located in the habitable zone, it is not necessarily habitable. An obvious counterexample would be the Moon.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Terrestrial or not?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The next question: are these planets terrestrial? Unfortunately we don't know: Teegarden b does not transit the star, and there has so far been insufficient coverage of Teegarden c to say whether it transits. As a result, we do not know the radii of the planets so we cannot tell whether or not the planets have densities compatible with being terrestrial. The minimum masses are low enough that these planets do likely fall below the transition region between rocky and non-rocky planets suggested by Kepler data, see for example Rogers (2015) &quot;&lt;a href=&quot;https://arxiv.org/abs/1407.4457&quot; rel=&quot;nofollow noreferrer&quot;&gt;Most 1.6 Earth-Radius Planets are not Rocky&lt;/a&gt;&quot; - using the mass-radius relationships of Zeng et al. (2016) &quot;&lt;a href=&quot;https://arxiv.org/abs/1512.08827&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mass-Radius Relation for Rocky Planets based on PREM&lt;/a&gt;&quot; the transition is somewhere around 3–5 Earth masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point of comparison here is the other known system of low-mass planets around an ultracool dwarf star: TRAPPIST-1. In this case, despite the small size of the planets, the inferred masses indicate the presence of substantial amounts of volatiles, e.g. Dorn et al. (2018) &quot;&lt;a href=&quot;https://arxiv.org/abs/1808.01803&quot; rel=&quot;nofollow noreferrer&quot;&gt;Interior characterization in multiplanetary systems: TRAPPIST-1&lt;/a&gt;&quot; where the volatile contents of the planets are estimated to correspond to a few percent by mass, compared with 0.02% by mass for the Earth. A planet like this would be very different to the Earth, in an optimistic scenario it might result in an ocean planet. The habitable zone for ocean planets would be different to that for Earthlike planets, since the negative feedback of continental weathering would not occur.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Red dwarf stars&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Red dwarf stars have a variety of problems when it comes to habitability. One issue is that they have an extended high-luminosity pre-main sequence phase which might result in runaway greenhouse effects and/or severe water-loss on close-in planets, see Ramirez &amp;amp; Kaltenegger (2014) &quot;&lt;a href=&quot;https://arxiv.org/abs/1412.1764&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Habitable Zones of Pre-Main-Sequence Stars&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An additional problem is that red dwarf stars have severe stellar activity that would erode an Earth-like atmosphere in a very short amount of time, as an example Garcia-Sage et al. (2017) &quot;&lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/aa7eca/pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;On the Magnetic Protection of the Atmosphere of Proxima Centauri b&lt;/a&gt;&quot; estimate an Earthlike atmosphere at Proxima Centauri would be lost in less than 400 million years. Teegarden's Star is fairly quiet for its spectral type, but it is an old star and was likely far more active in its youth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These issues are not necessarily fatal, outgassing from volcanism might be able to replenish the atmosphere to some extent after the stellar activity dies down, or if the planet was born with a large amount of volatiles it might be able to emerge as a habitable planet. Unfortunately we don't have sufficient data on habitable zone planets around red dwarf stars to know for sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A further issue is that the atmospheric chemistry on planets around stars like Teegarden's Star would likely result in substantial quantities of carbon monoxide accumulating in the atmosphere, see Schwieterman et al. (2019) &quot;&lt;a href=&quot;https://arxiv.org/abs/1902.04720&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Limited Habitable Zone for Complex Life&lt;/a&gt;&quot;. This would be lethally unpleasant for any human visitors (or other organisms that use haemoglobin as their oxygen carrier) caught outside without an air supply. Note that haemoglobin is not the only possible oxygen carrier: some groups of life on Earth use alternatives (haemerythrin, haemocyanin) that are less vulnerable to carbon monoxide poisoning, as noted by Howell (2019) &quot;&lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2515-5172/ab2a76&quot; rel=&quot;nofollow noreferrer&quot;&gt;Comment on &quot;A Limited Habitable Zone for Complex Life&quot; (Schwartzman et al. 2019 ApJ 878, 19)&lt;/a&gt;&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In short, there is a notable diversity of extant invertebrates on this planet that would be quite comfortable outside of the range of CO concentrations proposed by Schwieterman et al. to constrain the habitable zone. There is no a priori reason to rule out the possibility that these lineages could have given rise to intelligent life during the history of life on this planet. A significantly higher atmospheric CO concentration would have prevented the evolution of hemoglobin, but not necessarily prevented astronomically observable biosignatures from Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Summary&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;It is too early to be able to say whether the planets around Teegarden's Star are habitable. There are plenty of reasons to be sceptical but worlds like Teegarden b are still promising enough to be worth investigating in more detail. It is unfortunate that it is not a transiting planet, but similar systems like TRAPPIST-1 should be helpful in constraining what the possibilities are.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-06-21T18:33:44.863" LastActivityDate="2019-06-21T18:33:44.863" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32365" PostTypeId="2" ParentId="32355" CreationDate="2019-06-21T00:03:21.563" Score="4" Body="&lt;p&gt;The transformation from projected geometry to heliographic coordinates is straightforward.&#xA;As &lt;a href=&quot;http://www.petermeadows.com/html/location.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Peter Meadows explains&lt;/a&gt;, this can be done either mathematically, or graphically using a latitude/longitude grid template.&#xA;Since the solar equator is inclined about 7&amp;deg; to the ecliptic, different &lt;a href=&quot;http://solar-center.stanford.edu/solar-images/latlong.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stonyhurst disks&lt;/a&gt; are used at different times of year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://doi.org/10.1051/0004-6361:20054262&quot; rel=&quot;nofollow noreferrer&quot;&gt;W T Thompson 2006&lt;/a&gt; section 2 describes the two main heliographic coordinate systems.&#xA;Latitude in both systems is measured from the solar equator.&#xA;The central meridian of longitude in Stonyhurst coordinates is fixed facing the Earth.&#xA;The Carrington coordinate system rotates rigidly with a sidereal period of 25.38 days, approximating &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_rotation&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar rotation&lt;/a&gt; at a typical sunspot latitude of &amp;plusmn;26&amp;deg;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The NOAA Space Weather Prediction Center's daily &lt;a href=&quot;https://www.swpc.noaa.gov/products/solar-region-summary&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solar Region Summary&lt;/a&gt; provides spherical coordinates in both forms.&#xA;Here are excerpts of two reports four days apart:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;SRS Number 129 Issued at 0030Z on 09 May 2019&#xA;Report compiled from data received at SWO on 08 May&#xA;I.  Regions with Sunspots.  Locations Valid at 08/2400Z &#xA;Nmbr Location  Lo  Area  Z   LL   NN Mag Type&#xA;2740 N08E14   307  0160 Cao  04   03 Beta&#xA;2741 N05E52   269  0160 Hsx  03   01 Alpha&#xA;&#xA;SRS Number 133 Issued at 0030Z on 13 May 2019&#xA;Report compiled from data received at SWO on 12 May&#xA;I.  Regions with Sunspots.  Locations Valid at 12/2400Z &#xA;Nmbr Location  Lo  Area  Z   LL   NN Mag Type&#xA;2740 N08W40   308  0070 Hax  03   02 Alpha&#xA;2741 N06W02   270  0210 Hsx  03   02 Alpha&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;According to their &lt;a href=&quot;ftp://ftp.swpc.noaa.gov/pub/forecasts/SRS/README&quot; rel=&quot;nofollow noreferrer&quot;&gt;README&lt;/a&gt;,&#xA;&lt;code&gt;Location&lt;/code&gt; is a latitude and longitude relative to the central meridian, and &lt;code&gt;Lo&lt;/code&gt; is a Carrington longitude.&#xA;A typical sunspot region moves about 13&amp;deg;/day from east to west in Stonyhurst longitude and less than &amp;plusmn;1&amp;deg;/day in Carrington longitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GOES X-ray event report columns 29-34 also appear to be Stonyhurst spherical coordinates.&#xA;For example, sunspot region 9042 was associated with several X-ray events in June 2000.&#xA;If you filter the report by that region number, you can see that its progress from E90 to W90 longitude is roughly linear in time.&#xA;In projection, the plot of east-west position vs. time would be sinusoidal, flattening at the limbs.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-06-27T00:33:58.403" LastActivityDate="2019-06-27T00:33:58.403" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32367" PostTypeId="1" CreationDate="2019-06-21T03:09:21.203" Score="8" ViewCount="218" Body="&lt;p&gt;In the past, the asteroid belt was thought to be the remains of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Phaeton_(hypothetical_planet)&quot; rel=&quot;noreferrer&quot;&gt;shattered planet&lt;/a&gt;. Of course, that was all pseudoscience, the mass of the asteroid belt being &lt;a href=&quot;https://en.wikipedia.org/wiki/Asteroid_belt&quot; rel=&quot;noreferrer&quot;&gt;4% of the moon&lt;/a&gt;. I've been wondering, what would happen if this did occur. What would an asteroid belt with around 50% of the Earth's mass be like? How often would asteroids hit earth? Would it influence Mars?&lt;/p&gt;&#xA;" OwnerUserId="19304" LastActivityDate="2019-06-21T10:33:00.173" Title="What if there was an asteroid belt with the mass of a planet?" Tags="&lt;asteroids&gt;&lt;asteroid-belt&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32368" PostTypeId="2" ParentId="32367" CreationDate="2019-06-21T10:33:00.173" Score="-6" Body="&lt;p&gt;The mass of a planet? There probably was, billions of years ago, and perhaps still is, but the debris of this planet-in-the-making has been shattered and dispersed by impacts. We know from fragments which have struck the Earth that the planet which was forming there had grown large enough to have a molten interior where the heavy metals, mostly iron and nickel, had sunk to the centre and differentiated from the rocky mantle. A massive impact must have shattered this forming planet about 4 or 5 billion years ago, and the debris has become so dispersed that it is difficult to assess how much there originally was. Some was kicked out of the asteroid belt completely, the rest is still there but much of it so finely shattered and dispersed that we can't see it (the asteroid belt covers an enormous area, much larger than the orbit of the Earth-moon system). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting absence of a body with substantial gravitational field to attract comets &amp;amp; other objects has enabled them to pass on by, perhaps to impact the Earth and inner planets. Bodes Law says here should be a planet there, but whether it would have been a Mars-sized planet or not, nobody knows. The differentiation of nickel-iron from the rocky material becomes hard to explain without a fairly substantial planetesimal there to separate them, and in our solar system at least, Bodes Law seems fairly reliable.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-06-21T10:33:00.173" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32369" PostTypeId="1" CreationDate="2019-06-21T16:09:29.847" Score="5" ViewCount="78" Body="&lt;p&gt;Recently, the fly-through the Solar system of the inter-stellar object (ISO) `Oumuamua caused quite a stir in the astronomy community. I wonder whether there are reliable estimates for the (local) number density of such objects. Such estimates could be based on the `Oumuamua event (when they are necessarily rather crude owing to very low-number statistics) or on any other constraints.&#xA;A related question is: how close to Earth must an ISO pass to be detected with today's instruments, or 10, 25, 50 years ago?&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="16685" LastEditDate="2019-06-21T17:07:00.270" LastActivityDate="2019-06-21T17:07:00.270" Title="What's the number density of `Oumuamua like objects?" Tags="&lt;asteroids&gt;&lt;oumuamua&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32370" PostTypeId="1" CreationDate="2019-06-21T17:47:49.720" Score="1" ViewCount="76" Body="&lt;p&gt;Given the phase of a planet or satellite, I can find the area illuminated visible to us, &lt;strong&gt;if&lt;/strong&gt; we consider a 2D surface. But how do I find the percentage of illuminated area visible considering it a 3D surface and thus finding the magnitude of the partially illuminated planet/satellite.&lt;/p&gt;&#xA;" OwnerUserId="16506" LastActivityDate="2019-06-21T20:21:44.653" Title="Relation between phase and magnitude" Tags="&lt;moon-phases&gt;&lt;apparent-magnitude&gt;&lt;magnitude&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32371" PostTypeId="2" ParentId="32370" CreationDate="2019-06-21T20:21:44.653" Score="-1" Body="&lt;p&gt;In general, you'd have to project the light from the sun (Assuming we're talking about objects in our solar system) onto the 3D surface and project it onto a 2D surface. This can be accomplished via a ray tracing algorithm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This procedure will give you the fraction of the surface illuminated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you are concerned about the fraction of the illuminated surface that is reflected in a specific directly -- so that it is &quot;viewable&quot; from that direction -- you have to do the ray tracing in 3D from the source (Sun) to the object and on to the viewer. There packages you can google that will do this type of ray tracing if you can model the objects in 3D... &lt;/p&gt;&#xA;" OwnerUserId="9648" LastActivityDate="2019-06-21T20:21:44.653" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32372" PostTypeId="1" AcceptedAnswerId="32380" CreationDate="2019-06-22T03:23:22.050" Score="3" ViewCount="79" Body="&lt;p&gt;The recent paper &lt;a href=&quot;https://arxiv.org/abs/1905.12566&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thermal Emission from the Uranian Ring System&lt;/a&gt; has &lt;a href=&quot;https://news.berkeley.edu/2019/06/20/astronomers-see-warm-glow-of-uranuss-rings/&quot; rel=&quot;nofollow noreferrer&quot;&gt;been in the news&lt;/a&gt; recently. The introduction mentions:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The ε ring, the brightest and most massive of the narrow rings, &lt;strong&gt;was shown to maintain an appreciable eccentricity (e = 0.00794)&lt;/strong&gt; and an azimuthally-varying width; the ring is five times wider and ∼2.5 times brighter in reflected sunlight at apoapsis than at periapsis (French et al. 1988&lt;sup&gt;1&lt;/sup&gt;; Karkoschka 2001a&lt;sup&gt;2&lt;/sup&gt;). However, many fundamental parameters about the ring system remain unknown, including the filling factor, composition, thickness,&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;French, R. G., Elliot, J. L., French, L. M., et al. 1988, Icarus, 73, 349, doi: 10.1016/0019-1035(88)90104-2&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;2&lt;/sup&gt;Karkoschka, E. 2001a, Icarus, 151, 51, doi: 10.1006/icar.2001.6596&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I've read in Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Rings_of_Uranus#%CE%B5_ring&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rings of Uranus&lt;/a&gt; that several rings are expected to be fairly young, and there may be interaction with some small moons, but I am surprised that the rings could maintain a significant eccentricity, and do not circularize or smear out due to collisions.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What kind of dynamical effects can maintain Uranus' rings' eccentricities?&lt;/p&gt;&#xA;&lt;p&gt;Uranus itself has a &lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/uranusfact.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;huge J2&lt;/a&gt; and so any ring that isn't perfectly equatorial should have an eccentricity-dependent precession. The paper states that the apoapsis of the rings is five times wider than the periapsis, so there's definitely a population of different eccentricities and semi-major axes within the ε ring.&lt;/p&gt;&#xA;&lt;p&gt;List of &lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/uraniansatfact.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;rings and their orbital parameters&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-06-22T17:18:51.977" Title="What kind of effects can maintain Uranus' rings' eccentricities?" Tags="&lt;orbital-mechanics&gt;&lt;planetary-ring&gt;&lt;eccentric-orbit&gt;&lt;uranus&gt;&lt;eccentricity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32373" PostTypeId="1" AcceptedAnswerId="32390" CreationDate="2019-06-22T04:04:37.203" Score="-1" ViewCount="132" Body="&lt;p&gt;The Minkowski diagrams that are employed to illustrate Special Relativity depict changes in the motion of objects through the dimensions of space as occurring within a field (that's otherwise divided only by imaginary lines at right angles to each other), in relation to changes of their motion through that field within the dimension of time.  Since it accounts for gravity, General Relativity is more complex than Special Relativity, but I've also seen references to time being &quot;orthogonal to&quot; (meaning that it's &quot;at right angles to&quot;) space, in discussions of GR.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because &quot;spacetime&quot; is a relativistic concept implying that space and time are integral to each other, they can't be separated, under the assumption that GR is as correct as experiments and observations have shown it to be.  A surface or volume of material will eventually tear apart if it's pulled, over a certain interval and with equal force, in orthogonal directions that might otherwise have allowed its lengthening as well as its widening, and, since a &quot;tearing apart&quot; of space from time is impossible in GR, I'm consequently assuming that those orthogonal tensions are factors in the development of the curvature that defines gravity in that theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a &quot;local universe&quot; within a larger &quot;multiverse&quot;, or in any temporal iteration of any universe comprising a sequence of such iterations (such as a &quot;bouncing&quot; cosmology), it would, consequently, seem that the continuity of spacetime would require any expansion of the spatial volume of a universe or other causally-separated region to be accompanied by a contraction of its temporal duration, and vice-versa. (By &quot;vice-versa&quot;, I mean that an expansion of its temporal duration would be accompanied by a contraction of its spatial volume, that a contraction of its temporal duration would result in an expansion of its spatial volume, and that a contraction of its spatial volume would result in an expansion of its temporal duration.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know whether this inference is correct or mistaken.  (I'm thinking that it might be clearer in the non-Euclidean geometry employed in GR, with which I'm almost completely unfamiliar, and, as well, that the inference's idea of a balancing of spatial against temporal components of reality might have some relation to that &quot;compactification&quot; of extra dimensions which is posited by string theory.)&lt;/p&gt;&#xA;" OwnerUserId="22115" LastEditorUserId="22115" LastEditDate="2019-06-23T01:32:25.017" LastActivityDate="2019-06-23T03:07:45.443" Title="Does space being at right angles to time imply that the expansion or contraction of each occurs in a direction opposite the other's?" Tags="&lt;cosmology&gt;&lt;general-relativity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32374" PostTypeId="1" CreationDate="2019-06-22T07:29:33.337" Score="2" ViewCount="270" Body="&lt;p&gt;I came to know that the cosmic horizon is the part of universe beyond which we cannot see. So if the light from 13.7 billion years ago is what we can observe, does that mean the universe is older than that and we can't see it?&lt;/p&gt;&#xA;" OwnerUserId="28399" LastActivityDate="2019-06-25T15:13:50.873" Title="Is the universe older than 13.7 Billion years?" Tags="&lt;cosmology&gt;&lt;cosmological-horizon&gt;&lt;cosmic-microwave-background&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32375" PostTypeId="2" ParentId="32374" CreationDate="2019-06-22T07:52:09.887" Score="5" Body="&lt;p&gt;No, the oldest light we can see is as old as the Universe&lt;span class=&quot;math-container&quot;&gt;$^\dagger$&lt;/span&gt;. This is simply because light started traveling at that time, so after one year you'd se light that had been traveling for one light-year, after 1 million years you'd see light that had been traveling for a million light-years, and so on. There  is (most likely) more Universe beyond the horizon, but light from those regions just haven't had the time to reach us yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might think that the &lt;em&gt;distance&lt;/em&gt; to the cosmic horizon today is 13.7 billion light-years, but because the Universe is expanding, the distance is much larger (in fact 43.6 billion light-years).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^\dagger$&lt;/span&gt;&lt;sub&gt;Because the Universe was opaque the first 380,000 years, we can't see &lt;em&gt;all&lt;/em&gt; the way, but that's a technical detail.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2019-06-22T07:52:09.887" CommentCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="32376" PostTypeId="1" CreationDate="2019-06-22T09:24:23.900" Score="2" ViewCount="39" Body="&lt;p&gt;Can anyone point me to the reference where I can find the density of Jupiter or Saturn as a function of the distance from its center?&lt;/p&gt;&#xA;" OwnerUserId="24850" LastActivityDate="2019-06-22T09:24:23.900" Title="What is the radial dependence of the density of Jupiter or Saturn?" Tags="&lt;planet&gt;&lt;jupiter&gt;&lt;saturn&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32378" PostTypeId="1" CreationDate="2019-06-22T13:31:30.083" Score="1" ViewCount="124" Body="&lt;p&gt;Between years 2009-2011, I was located in Greece, watching the perfectly dark night sky around 21:00-23:00 during summer, when I suddenly noticed a very dull dark-brown object that looked like a solid rock/potato having a perceived size around half the diameter of the moon in the sky, moving in a perfect straight line looking up &#xA; and perceived as fast (it took 3-4 seconds to disappear in the horizon) - so total exposure if I had seen it from start to end would have been less than 10-12 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could barely notice it, it had some texture like a rock but blurry. It had almost no light reflection and it wasn't rotating or anything. It wasn't spherical and it was somewhat elongated along the axis of movement. I was barely able to detect it above me with naked eyes, as it was passing from left to right. I only did when I saw it was hiding a small region of the night sky stars. This way I perceived its irregular shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I cannot tell it's actual size or how fast it was going. Is it an object of a kind known to astronomers? I couldn't find any similar video footage or anyone seeing something similar talking in the internet. Could it be a space rock (asteroid?) that missed the earth atmosphere and at the same time be seen with naked eyes? Any estimates on space rock type, size and speed based on the time of sky exposure? Any astronomical id from a database or some guidance on how I could search it? &#xA;Could it be an asteroid occultation like this one: &lt;a href=&quot;http://call4obs.iota-es.de/wp-content/uploads/2011/08/palma_south.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://call4obs.iota-es.de/wp-content/uploads/2011/08/palma_south.gif&lt;/a&gt;&#xA;Thanks &lt;/p&gt;&#xA;" OwnerUserId="28483" LastEditorUserId="28483" LastEditDate="2019-06-22T19:31:41.320" LastActivityDate="2019-06-22T19:31:41.320" Title="dull rock/potato-shaped object seen with naked eye, moving in straight line during night. What could it be and is size and speed estimation possible?" Tags="&lt;identify-this-object&gt;&lt;near-earth-object&gt;" AnswerCount="1" CommentCount="9" ClosedDate="2019-06-23T08:07:09.233" ContentLicense="CC BY-SA 4.0" />
  <row Id="32380" PostTypeId="2" ParentId="32372" CreationDate="2019-06-22T17:18:51.977" Score="3" Body="&lt;p&gt;A relevant paper is Papaloizou &amp;amp; Melita (2004) &quot;&lt;a href=&quot;https://arxiv.org/abs/astro-ph/0404579&quot; rel=&quot;nofollow noreferrer&quot;&gt;Structuring eccentric-narrow planetary rings&lt;/a&gt;&quot; which starts off promisingly with the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The nature of the dynamical mechanism that maintains the apse alignment of narrow-eccentric planetary rings is one of the most interesting and challenging problems of Celestial&#xA;  Mechanics.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;According to the leading model (Goldreich and Tremaine 1979) the self-gravity of the&#xA;  ring counter-acts the differential precession induced by the oblateness of the central planet.&#xA;  Using this hypothesis, a prediction of the total mass of the ring can be made, which, in&#xA;  general, is not in good agreement with the inferred mass of the observed eccentric rings in&#xA;  the Uranus system&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It then goes on to describe subsequent refinements taking account of additional effects including particle interactions and perturbations that bring the predictions more in line. The approach described is given in the introduction:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In this work we build, from first principles, a simple general continuum or fluid like&#xA;  model of a narrow-eccentric ring. The eccentric pattern in the ring can be described as&#xA;  being generated by a normal mode of oscillation of wave-number &lt;em&gt;m&lt;/em&gt; = 1 which may be&#xA;  considered to be a standing wave. Dissipation can be allowed to occur due to inter-particle&#xA;  collisions leading to a viscosity which would lead to damping of the mode. However, this&#xA;  global &lt;em&gt;m&lt;/em&gt; = 1 mode can also be perturbed by neighboring-shepherd satellites which can&#xA;  inject energy and angular momentum through resonances. In this way losses due to particle&#xA;  collisions may be balanced. It is that process that is the focus of this paper. Other possible&#xA;  mechanisms, such as mode excitation through self-excitation through viscous overstability,&#xA;  that could arise with an appropriate dependence of viscosity on physical state variables (see&#xA;  Papaloizou and Lin 1988, Longaretti &amp;amp; Rappaport 1995), are beyond the scope of this paper&#xA;  and accordingly not investigated here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The specific case of the ε-ring of Uranus is described in section 10.2, where the main satellite forcing considered is due to the 47:49 second-order mean-motion resonance between the ring and Cordelia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper goes into quite a bit of mathematical detail on the various processes involved.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastActivityDate="2019-06-22T17:18:51.977" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32381" PostTypeId="5" CreationDate="2019-06-22T18:07:57.230" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2019-06-22T18:07:57.230" LastActivityDate="2019-06-22T18:07:57.230" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32382" PostTypeId="4" CreationDate="2019-06-22T18:07:57.230" Score="0" Body="Questions about 1I/2017 U1 (ʻOumuamua), an interstellar asteroid or comet that passed through the Solar System in 2017." OwnerDisplayName="user24157" LastEditorDisplayName="user24157" LastEditDate="2019-06-23T04:28:11.367" LastActivityDate="2019-06-23T04:28:11.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32383" PostTypeId="2" ParentId="32374" CreationDate="2019-06-22T18:11:53.343" Score="1" Body="&lt;p&gt;The universe might be eternal to the past, as well as to the future.  The simplest theory for that is described in &quot;Steady state eternal inflation&quot;, formulated by Anthony Aguirre and Steve Gratton, and available free online:  It is compatible with observations of the Cosmic Microwave Background, and is centered on a Cauchy surface from which time extends in opposite spatial directions.  A similar but somewhat more elaborate cosmology, by Carroll and Chen, is analyzed, together with the Aguirre-Gratton one, by Vilenkin, in his online paper &quot;Arrows of time and the beginning of the universe&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Borde-Guth-Vilenkin Theorem is usually construed as preventing the &quot;past eternality&quot; of inflation by requiring a balancing of universal expansion by universal contraction, but the references in the last (2003) formulation of that theorem by its three authors include a footnote exempting the Aguirre &amp;amp; Gratton theory from its strictures.  Also, the BGV Theorem does not rule out a start of the universe at arbitrarily long times before the start of inflation (quasi-exponential expansion), as hypothesized in Martin Bojowald's 2010 book titled &quot;Once Before Time:  A Whole Story of the Universe&quot;, which utilizes Loop Quantum Cosmology. Some of the &quot;bouncing&quot; cosmologies, of which the Wiki &quot;Big Bounce&quot; provides an overview, also imply past eternality, and substitute a repetition of &quot;bounces&quot; for the &quot;Big Bang&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What these cosmologists, as well as the three mentioned in the last of my comments below, are trying to do is resolve that contradiction of infinite density in the zero volume of the singularity at which General Relativity has been considered to break down:  Because densities in &quot;negative time&quot; might cancel densities in&quot;positive time&quot;, the densities of the objects duplicated in either version of the temporal dimension might  cancel each other, thereby replacing that contradiction with zero density in zero volume, which makes sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that a widespread misconception leading to the belief that the universe, or its massive and energetic contents, are finite is the fact that their limitation to some finite amount would provide the simplest explanation for the fact that the sky is dark at night:  However, because of the fact that expansion is different from speed or velocity, and is (consequently) NOT limited to a rate no higher than the speed of light, such accelerations as were witnessed by the Supernova 1A observations in the late 1990's are entirely permissible in General Relativity. The simplest explanations do not necessarily provide the best explanation of scientific phenomena.&lt;/p&gt;&#xA;" OwnerUserId="22115" LastEditorUserId="22115" LastEditDate="2019-06-25T15:13:50.873" LastActivityDate="2019-06-25T15:13:50.873" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32384" PostTypeId="2" ParentId="32378" CreationDate="2019-06-22T19:00:04.103" Score="2" Body="&lt;p&gt;Rock-shaped doesn't mean very much, rocks come in all shapes and sizes. It couldn't possibly have been an asteroid. It must have been quite large for you to see it at night, but it is surprising you could see the colour of it at night, when colour vision fades. Large, fast, elongated in the direction of motion - it sounds as though it could have been a dirigible, aided by a tail wind. Dirigibles are airships vaguely similar to a zeppelin. The ones with no rigid skeleton are called blimps. They have come back into fashion for certain tasks in the last 20 years or so; there are some in Europe and in USA and probably other countries too, Russia for example. Whether there are any in Greece, I couldn't say, but one thing I can say is that it is far more likely to have been an airship than an asteroid. One possible reason for its rapid disappearance is that it could have disappeared into or behind a cloud.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-06-22T19:00:04.103" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32385" PostTypeId="1" CreationDate="2019-06-22T22:33:45.367" Score="2" ViewCount="109" Body="&lt;p&gt;We look at galaxies that are billions of light years distant and draw conclusions based on the red shift of the light. How is the age of the light accounted for. We see the reality of a galaxy perhaps a billion years ago. Why not conclude that redshift is related to time instead of distance? &lt;/p&gt;&#xA;" OwnerUserId="28490" LastActivityDate="2019-06-23T08:32:36.767" Title="How to deal with data that's billions of years old?" Tags="&lt;astrophysics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32388" PostTypeId="1" AcceptedAnswerId="32417" CreationDate="2019-06-23T02:17:08.457" Score="4" ViewCount="195" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/14676/7982&quot;&gt;This answer&lt;/a&gt; points out that the near Earth asteroid Apophis will likely be close to 2 arcseconds in diameter as seen from Earth during its close approach in 2029. I speculate that if the Hubble Space Telescope were still operational then, it could potentially image the asteroid in visible light at a few dozen pixels in diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads me to wonder if the Hubble has ever been used to image&lt;sup&gt;†&lt;/sup&gt; or at least spatially resolve in some way an asteroid during a close pass to the Earth before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;†&lt;/sup&gt;Here the verb &quot;image&quot; should be taken to mean the act of producing a resolved image of an object so that different pixels correspond to intensity from different parts of the body being imaged. For the purposes of this question please don't consider telescope images in which an asteroid happens to appear but is too far away to be resolved. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-06-25T10:12:52.997" LastActivityDate="2019-06-30T13:38:32.980" Title="Has Hubble ever been used to try to image a near Earth asteroid?" Tags="&lt;asteroids&gt;&lt;hubble-telescope&gt;&lt;near-earth-object&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32389" PostTypeId="1" CreationDate="2019-06-23T02:42:33.437" Score="0" ViewCount="41" Body="&lt;p&gt;I was working on reshift details for various galaxies but do not know any database to find the z values given to redshift galaxies. Can anyone help?&lt;/p&gt;&#xA;" OwnerUserId="28470" LastActivityDate="2019-07-23T15:03:10.407" Title="Where a database of different z values of redshift can be found?" Tags="&lt;redshift&gt;&lt;velocity&gt;&lt;hubble-constant&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32390" PostTypeId="2" ParentId="32373" CreationDate="2019-06-23T03:07:45.443" Score="1" Body="&lt;p&gt;I found the answer as given by Ben Crowell on the Physics Stack Exchange (with an upvote of 3, including my own), in response to a question titled &quot;Special and temporal dimensions orthogonality&quot; and tagged &quot;General-Relativity&quot; (but not &quot;Special-Relativity&quot;). It lacked any expressed relation to the speculative issues I'd imagined, and included the following information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When two spacelike vectors are orthogonal, it means what we have in mind in Euclidean geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a timelike vector is orthogonal to a spacelike vector, it means that to an observer who is moving along the timelike vector, the spacelike vector is purely spatial, i.e., it connects events that are simultaneous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lightlike vector is orthogonal to itself.&lt;/p&gt;&#xA;" OwnerUserId="22115" LastActivityDate="2019-06-23T03:07:45.443" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32391" PostTypeId="2" ParentId="32385" CreationDate="2019-06-23T08:32:36.767" Score="5" Body="&lt;p&gt;The redshift you're referring to is a &lt;em&gt;cosmological&lt;/em&gt; redshift, which is fundamentally different from a Doppler redshift. At the heart of it, it has nothing to do with neither time, nor distance. It is a consequence of light traveling through an expanding space, no more and no less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because space expands, the emitting source is carried away to a certain distance from us, so we can &lt;em&gt;relate&lt;/em&gt; the redshift to a distance. And because it takes time for the light to reach us, we can &lt;em&gt;relate&lt;/em&gt; the redshift to a time. But this requires a model for &lt;em&gt;how&lt;/em&gt; the Universe has expanded through time, and how light travels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model that best describes our reality is the so-called &lt;em&gt;standard cosmological model&lt;/em&gt;, or &lt;a href=&quot;https://en.wikipedia.org/wiki/Lambda-CDM_model&quot; rel=&quot;noreferrer&quot;&gt;ΛCDM&lt;/a&gt; (&quot;Lambda-CDM&quot;) model, which is a mathematically based, physically motivated, and observationally constrained model. It uses the so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann%E2%80%93Lema%C3%AEtre%E2%80%93Robertson%E2%80%93Walker_metric&quot; rel=&quot;noreferrer&quot;&gt;Friedmann–Lemaître–Robertson–Walker metric&lt;/a&gt;, which is a description of space itself, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;noreferrer&quot;&gt;Friedmann equations&lt;/a&gt;, which describes how space expands, and an &lt;a href=&quot;https://en.wikipedia.org/wiki/Equation_of_state_(cosmology)&quot; rel=&quot;noreferrer&quot;&gt;equation of state&lt;/a&gt;, which describes the relation between pressure and density. The various parameters that enter the equations are then observed and constrained in a number of different ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this model, we can relate the distance to an object, and its age, to the redshift we observe.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2019-06-23T08:32:36.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32392" PostTypeId="1" AcceptedAnswerId="32393" CreationDate="2019-06-23T13:32:03.157" Score="2" ViewCount="112" Body="&lt;p&gt;Take DS Geminorum for example, it has a cz of -2.33 [2.96]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is what does cz mean, what does is represent? and what does the value in the brackets mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=ds%20gem&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://simbad.u-strasbg.fr/simbad/sim-id?Ident=ds%20gem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5401" LastActivityDate="2019-06-23T13:40:10.940" Title="What does cz mean?" Tags="&lt;star&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32393" PostTypeId="2" ParentId="32392" CreationDate="2019-06-23T13:40:10.940" Score="4" Body="&lt;p&gt;It stands for the speed of light (&lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;) times the redshift (&lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;) as measured from the redshift of the spectrum.&#xA;It is equal to the line-of-sight velocity that the star has. Positive means going away from us.&#xA;It is given in a heliocentric reference frame, that is corrected for the rotation of the Earth around its axis/the Sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The -2.33 is the velocity (km/s) and the [2.96] is its uncertainty.&lt;/p&gt;&#xA;" OwnerUserId="28449" LastActivityDate="2019-06-23T13:40:10.940" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32394" PostTypeId="2" ParentId="32389" CreationDate="2019-06-23T14:29:23.030" Score="1" Body="&lt;p&gt;There are some great databases with a lot of information in them, when you found the galaxy you looked for, somewhere on the page there should be redshift information if it is has been measured at some point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The NED/ipac database is quite nice for galaxies (&lt;a href=&quot;https://ned.ipac.caltech.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ned.ipac.caltech.edu/&lt;/a&gt;).&#xA;Or SIMBAD is also great (&lt;a href=&quot;http://simbad.u-strasbg.fr/simbad/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://simbad.u-strasbg.fr/simbad/&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="28449" LastActivityDate="2019-06-23T14:29:23.030" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32395" PostTypeId="2" ParentId="32268" CreationDate="2019-06-23T16:55:40.853" Score="0" Body="&lt;p&gt;I was able to get some insight from my professor, so I figured I'd answer this question. Basically, convection works AGAINST gravity and keeps the density constant. &lt;/p&gt;&#xA;" OwnerUserId="28360" LastActivityDate="2019-06-23T16:55:40.853" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32396" PostTypeId="1" CreationDate="2019-06-23T20:05:16.673" Score="3" ViewCount="29" Body="&lt;p&gt;What technology is required to detect a coconut-sized sphere in Earth orbit, at average altitude of communications satellites? &lt;/p&gt;&#xA;" OwnerUserId="28493" LastEditorUserId="7982" LastEditDate="2019-06-24T00:31:57.337" LastActivityDate="2019-06-24T00:31:57.337" Title="Small orbital body detection" Tags="&lt;orbit&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2019-06-24T12:55:46.750" ContentLicense="CC BY-SA 4.0" />
  <row Id="32397" PostTypeId="1" CreationDate="2019-06-23T21:37:16.017" Score="20" ViewCount="2472" Body="&lt;p&gt;A meteor &lt;a href=&quot;https://www.theguardian.com/australia-news/2019/jun/23/queensland-sky-lit-up-by-plummeting-meteor&quot; rel=&quot;noreferrer&quot;&gt;recently flashed over Queenland.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The meteor has been described by various people as appearing cyan or aqua, so the blueness of the image can't be due to a camera issue. The colour doesn't appear to be anything found along the black body spectrum either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What (chemical/physical process) would cause this meteor to appear cyan?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G8trh.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G8trhm.jpg&quot; alt=&quot;The Guardian: Queensland sky lit up by plummeting meteor&quot;&gt;&lt;/a&gt; click for full size, &lt;a href=&quot;https://www.theguardian.com/australia-news/2019/jun/23/queensland-sky-lit-up-by-plummeting-meteor&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;‘I saw it come right down in the sky’: the meteor over south-east Queensland on Saturday night. Photograph: Craig Turton/AAP&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17655" LastEditorUserId="7982" LastEditDate="2019-06-24T00:25:08.503" LastActivityDate="2019-06-24T03:24:17.053" Title="Why did this meteor appear cyan?" Tags="&lt;meteor&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32399" PostTypeId="2" ParentId="32268" CreationDate="2019-06-24T03:14:30.637" Score="1" Body="&lt;p&gt;The most important thing that happens is hydrogen is turned into helium, which essentially means that electrons and protons are getting merged into neutrons (the neutrons inside the helium). That reduces the number of particles.  For example, if you turn a pure hydrogen gas into a pure helium gas, you now have half the electrons you had before.  Also, you have 3 particles (a He nucleus and 2 electrons) where before you would have had 8 (in the case of pure H initially, just as an example).  Losing particles means you lose pressure at a given temperature and volume (it would be proportional to number of particles), so the core must contract.  The temperature does go up a little, but not much because nuclear burning is very temperature sensitive.  But the particle density drops a great deal, so for the density to stay the same, the core pressure must drop quite a bit, perhaps by almost a factor of 1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this might not require the core density to change is that the core pressure scales roughly like the radius to the -4 power (because of the requirements of force balance, which balances the pressure times the surface area, which scales like P*R^2, with the gravity, which scales like 1/R^2).   So that's quite a steep drop in P as R rises a little, and apparently the small rise in R is enough to compensate for the large drop in core P, to keep the density the same (based on your professor's statement that it does stay nearly the same).  It must be essentially a coincidence of how these things work out, it's not obvious to show why the density should be exactly constant but at least we can see that a large drop in P is compensated well by a small rise in R.  The physics of convection is pretty tricky, and there is only an approximate treatment called &quot;mixing length theory&quot; but it's never really been clear how to do this in a formally correct way.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastEditorUserId="14229" LastEditDate="2019-06-24T03:29:44.860" LastActivityDate="2019-06-24T03:29:44.860" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32400" PostTypeId="2" ParentId="32397" CreationDate="2019-06-24T03:24:17.053" Score="19" Body="&lt;p&gt;The colour of a meteor depends mostly on its chemical composition emission spectrum. To simplify magnesium meteors are noted to have an emission spectrum which results in blue to green colour. Changing factors like adding additional metals and the speed of the meteor could results in a cyan meteor compared to a blue one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &#xA;&lt;a href=&quot;https://leonid.arc.nasa.gov/meteor.html&quot; rel=&quot;noreferrer&quot;&gt;NASA on meteors&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="28286" LastActivityDate="2019-06-24T03:24:17.053" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32402" PostTypeId="2" ParentId="32374" CreationDate="2019-06-24T07:47:54.263" Score="1" Body="&lt;p&gt;We didn't determine the age of the universe by looking at what the farthest object is we can see. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, we looked at closer objects and &lt;a href=&quot;https://en.wikipedia.org/wiki/Age_of_the_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;determined their distance, speed and rate of acceleration&lt;/a&gt;. When you know those, you can work backwards and find a time when all those galaxies would have been in the same place, which is 13.7 billion years ago. &lt;/p&gt;&#xA;" OwnerUserId="361" LastActivityDate="2019-06-24T07:47:54.263" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32403" PostTypeId="1" AcceptedAnswerId="32404" CreationDate="2019-06-24T08:27:22.777" Score="4" ViewCount="98" Body="&lt;p&gt;There has been research that suggests that the changing brightness of Brown Dwarves are due to bands of Iron/Silicate clouds. Several models have been made that supports that hypothesis. My question is: Which part of the model are the actual clouds? Are the clouds the bright orange bits? Or are those bright areas “breaks” in the clouds and a view of the Brown Dwarf below.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GWRQO.jpg&quot; alt=&quot;Artist Rendition&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;5 MB animated GIF: &lt;a href=&quot;https://photojournal.jpl.nasa.gov/figures/PIA21752_200.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brown Dwarf Cloud Model&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="28500" LastEditorUserId="31410" LastEditDate="2021-05-27T02:20:16.587" LastActivityDate="2021-05-27T02:20:16.587" Title="Brown Dwarf Cloud Models" Tags="&lt;brown-dwarf&gt;&lt;modeling&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32404" PostTypeId="2" ParentId="32403" CreationDate="2019-06-24T11:05:49.153" Score="2" Body="&lt;p&gt;Initial research proved to be wrong as intuition does not match up with expectations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the paper referenced by the OP seems to sum up the answer to the question pretty well:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In  this  explanation,  the  dark  areas  of  our  map represent  thicker  clouds  that  obscure  deeper,  hotter  parts  of  the  atmosphere  and present a  higher-altitude (and  thus  colder) emissive  surface, whereas bright  regions correspond to holes in the upper cloud layers that provide a view of the hotter, deeper interior. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Extract from &lt;a href=&quot;https://www.researchgate.net/publication/259984905_A_global_cloud_map_of_the_nearest_known_brown_dwarf&quot; rel=&quot;nofollow noreferrer&quot;&gt;A global cloud map of the nearest known brown dwarf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially I assumed that the metallic clouds would reflect radiation from the nearby star, however as it turns out the hot atmosphere is emitting infrared radiation which affects the brightness much more significantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More on Brown Dwarfs at &lt;a href=&quot;http://www.scholarpedia.org/article/Brown_dwarfs&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="28286" LastEditorUserId="28286" LastEditDate="2019-06-24T14:49:30.617" LastActivityDate="2019-06-24T14:49:30.617" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32405" PostTypeId="1" CreationDate="2019-06-24T04:45:40.953" Score="1" ViewCount="16" Body="&lt;p&gt;There has been research that suggests that the changing brightness of Brown Dwarves are due to bands of Iron/Silicate clouds. Several models have been made that supports that hypothesis. My question is: Which part of the model are the actual clouds? Are the clouds the bright orange bits? Or are those bright areas “breaks” in the clouds and a view of the Brown Dwarf below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://www.jpl.nasa.gov/images/spitzer/20170817/PIA21752-label.gif&quot; alt=&quot;Brown Dwarf Cloud Model&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GWRQO.jpg&quot; alt=&quot;Artist Rendition&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="28500" OwnerDisplayName="Hippeus_Lancer" LastActivityDate="2019-06-24T13:00:26.207" Title="Brown dwarf cloud models" Tags="&lt;modeling&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2019-06-24T13:00:44.907" ContentLicense="CC BY-SA 4.0" />
  <row Id="32406" PostTypeId="2" ParentId="20276" CreationDate="2019-06-24T13:10:15.027" Score="1" Body="&lt;p&gt;I'm sure I worked this out in detail in another answer, but I can't find it now. Just to add a point addressing some comments above. The limiting angular momentum of a black hole is (in suitable units) the square of its mass, while the Schwarzschild radius grows as the mass. So consider a large (near) maximally spinning black holes of mass &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; which will have Schwarzchild radius &lt;span class=&quot;math-container&quot;&gt;$2M$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum orbital angular momentum you can add to it by firing a particle of mass &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt;  just inside the event horizon and velocity almost &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; (which is 1 in these units) is therefore &lt;span class=&quot;math-container&quot;&gt;$2Mm$&lt;/span&gt;. If that particle is a smaller maximally spinning black hole of mass &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; and angular momentum &lt;span class=&quot;math-container&quot;&gt;$m^2$&lt;/span&gt; then the total angular momentum of the coalesced holes is &lt;span class=&quot;math-container&quot;&gt;$M^2+2Mm+m^2$&lt;/span&gt; which is exactly &lt;span class=&quot;math-container&quot;&gt;$(M+m)^2$&lt;/span&gt;, so the new black hole is still just maximally spinning.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-06-24T13:10:15.027" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32407" PostTypeId="1" AcceptedAnswerId="32408" CreationDate="2019-06-24T15:14:23.373" Score="-1" ViewCount="77" Body="&lt;p&gt;White dwarfs rank among the hottest things in the universe.In the early years of the life of the universe ,universe was very hot and the Higgs field was deactivated causing all the particles generated by quantum flunctuations to travel at the speed  of light(being massless).Is the Higgs field inside a white dwarf also deactivated so any pair of particles-antiparticles will travel at the speed of light?&lt;/p&gt;&#xA;" OwnerUserId="28503" LastActivityDate="2019-06-24T17:44:55.613" Title="Higgs Field inside a White Dwarf" Tags="&lt;white-dwarf&gt;&lt;high-energy-astrophysics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32408" PostTypeId="2" ParentId="32407" CreationDate="2019-06-24T17:44:55.613" Score="3" Body="&lt;p&gt;In a word, no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe was certainly quite hot for the first few hundred thousand years, but most of the important events of the Big Bang happened in the first few &lt;em&gt;seconds&lt;/em&gt;, when the universe was &lt;em&gt;extremely&lt;/em&gt; hot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the time the universe was about 10 seconds old it was cool enough for protons and neutrons to combine into deuterium and thence helium, as well as a little bit of lithium-7. (Small amounts of tritium and beryllium-7 were also produced, but they are both radioactive with short halflives, and they soon decayed to helium &amp;amp; lithium, respectively).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That phase, known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Big_Bang_nucleosynthesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Big Bang nucleosynthesis&lt;/a&gt;, lasted until the universe was about 20 minutes old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Particles acquired mass via the Higgs mechanism when the universe was only about 1 picosecond (&lt;span class=&quot;math-container&quot;&gt;$10^{-12}$&lt;/span&gt; seconds) old, if not earlier. That phase is known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Chronology_of_the_universe#Electroweak_epoch&quot; rel=&quot;nofollow noreferrer&quot;&gt;electroweak epoch&lt;/a&gt;. At the start of that epoch, the temperature of the universe was around &lt;span class=&quot;math-container&quot;&gt;$10^{28}$&lt;/span&gt;K. The temperature in the core of a white dwarf is in the order of &lt;span class=&quot;math-container&quot;&gt;$10^7$&lt;/span&gt;K, which is  far too cool to disturb the Higgs field.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2019-06-24T17:44:55.613" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32409" PostTypeId="1" AcceptedAnswerId="32419" CreationDate="2019-06-24T19:18:10.100" Score="3" ViewCount="146" Body="&lt;p&gt;Since other planets have their own gravitational pull, I'm wondering if their effect on Earth's orbit has ever been measured? I know that in comparison with the Sun's pull the effect might be rather small, but I'm wondering if it's measurable...&lt;/p&gt;&#xA;" OwnerUserId="28506" LastActivityDate="2019-06-25T13:30:58.483" Title="Has the effect of other planets on Earth's orbit been ever measured?" Tags="&lt;orbit&gt;&lt;planet&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32410" PostTypeId="1" CreationDate="2019-06-24T21:26:14.230" Score="55" ViewCount="12249" Body="&lt;p&gt;Hypothetical question based on my understanding that two event horizons that overlap (touch) can't ever separate again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a 1 billion solar mass black hole (so the event horizon is massive and very gravitationally weak) is travelling at a velocity of 0.9c through empty flat intergalactic space; now imagine an identical 1 billion solar mass black hole travelling at 0.9c but in exactly the opposite direction so the two are heading roughly towards each other. The black holes' paths, once all the space time warping is taken into account, aren't on a direct collision but the outermost edges of the event horizons will just 'clip' each other, ordinarily only overlap for a fraction of a nanosecond as these two bodies are travelling at such incredibly fast velocities and in opposite directions to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So firstly, am I right in thinking that if two event horizons overlap they can never 'unlap'?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, what would happen to this incredible amount of momentum of each other the black holes? Would it just get instantly turned into gravitational energy? Bearing in mind when black holes normally merge, it happens very slowly as black holes slowly move closer and closer together over millions of years giving off gravitational energy as that happens, so not in a fraction of a nanosecond as in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And thirdly, what would this look like? Would the event horizons remain fairly spherical and the radiated energy just insane or would they stretch and warp into a kind of long thin elastic event horizon as they shoot past each other and then over time slow down and snap back to each other?&lt;/p&gt;&#xA;" OwnerUserId="28507" LastEditorUserId="14019" LastEditDate="2019-06-26T15:35:28.090" LastActivityDate="2019-06-26T15:35:28.090" Title="If two black hole event horizons overlap (touch) can they ever separate again?" Tags="&lt;black-hole&gt;&lt;supermassive-black-hole&gt;&lt;event-horizon&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="32411" PostTypeId="2" ParentId="32410" CreationDate="2019-06-25T00:03:54.850" Score="53" Body="&lt;p&gt;If the event horizons ever touch and become one continuous surface, their fate is sealed - the two black holes will merge all the way in. They can never separate again, no matter what.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several possible ways to explain it, with varying degrees of rigorousness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An intuitive explanation is that escape velocity at the event horizon equals the speed of light. But nothing can move as fast as light, not even a black hole. In order for the two black holes to separate, parts of one would have to &quot;escape&quot; the other, or move faster than light, which is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Another intuitive &quot;explanation&quot; (a.k.a. lots of handwaving) - inside the event horizon, all trajectories lead to the center. There is no possible path from any place within the horizon to the outside. Whichever way you turn, you're looking at the center. Whichever way you move, you move towards the center. If the event horizons have merged, for the black holes to split up again, parts of them would have to move &quot;away from center&quot; (or away from one of the centers), which is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the above is about as &quot;rigorous&quot; as &quot;explaining&quot; general relativity with steel balls on a rubber sheet. It's just metaphor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More rigorously, see this paper by Stephen Hawking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://projecteuclid.org/euclid.cmp/1103857884&quot; rel=&quot;noreferrer&quot;&gt;Black holes in general relativity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As time increases, black holes may merge together and new black holes&#xA;  may be created by further bodies collapsing but a black hole can never&#xA;  bifurcate. (page 156)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Event horizons don't really &quot;just clip each other&quot;. Perfectly spherical event horizons are a theoretical abstraction (a non-rotating black hole in an otherwise empty universe). In reality, anything near a BH will deform the event horizon, which will &quot;reach out&quot; towards that mass. If it's a small mass, the effect is negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if two black holes get close to each other, the EHs become egg-shaped, as if trying to touch each other. If they're close enough, then eventually a very narrow bridge will form in between, and the EHs will merge. At that moment, the full merger is decreed and will procede with absolute certainty until it's complete. &lt;strong&gt;Nothing&lt;/strong&gt; can stop it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See this answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/31989/are-black-holes-spherical-during-merger/31992#31992&quot;&gt;Are black holes spherical during merger?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;what would happen to this incredible amount of momentum of each other&#xA;  the black holes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The resulting black hole after the merger is going to have a heck of a lot of spin, if the collision is not perfectly frontal. Whatever energy cannot be stuffed into spin, is probably going to be radiated away as gravitational waves (as others have indicated already in comments to your question).&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2019-06-25T19:05:43.203" LastActivityDate="2019-06-25T19:05:43.203" CommentCount="23" ContentLicense="CC BY-SA 4.0" />
  <row Id="32412" PostTypeId="1" AcceptedAnswerId="32418" CreationDate="2019-06-25T00:45:35.007" Score="3" ViewCount="254" Body="&lt;p&gt;I'm having a hard time understanding this image in the CNN News item &lt;a href=&quot;https://edition.cnn.com/2019/06/24/world/muse-glowing-ring-of-light-scn-trnd/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This glowing ring of light is from a distant galaxy&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The caption (&lt;a href=&quot;https://www.abc17news.com/news/national-world/this-glowing-ring-of-light-is-from-a-distant-galaxy/1088915478&quot; rel=&quot;nofollow noreferrer&quot;&gt;from this version&lt;/a&gt; of the article) says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;This  image shows observations taken with the NASA/ ESA Hubble Space Telescope of two magnified galaxies behind massive galaxy clusters: the pink glowing haloes reveal the gas surrounding the distant galaxies and its structure. The gravitational lensing effect from the clusters multiplies the images of the galaxies, producing up to 4 images of the same source.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Trying to parse the caption, I think that the two objects in the center are magnified images of the galaxies in the background, and so should be related to the source of the light in the Einstein ring. But they are yellow and the ring is pink.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; So the Einstein ring is picking up light not from the background galaxies, but from a halo of gas around them? Is there a better source to read about this image and everything that's going on here?&lt;/p&gt;&#xA;&lt;p&gt;The image contains the following label at the top right corner:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/F5sjC.png&quot; alt=&quot;gravitationally lensed galaxies and Einstein ring - label&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;click for full size:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/exZqT.jpg&quot; alt=&quot;gravitationally lensed galaxies and Einstein ring&quot; /&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The article then says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The MUSE instrument on the European Southern Observatory's Very Large Telescope has identified halos around distant galaxies. Previously, images of these glowing rings lacked detail, but astronomers were able to use a gravitational lensing effect of galaxy clusters to see detail and structure.&lt;/p&gt;&#xA;&lt;p&gt;Gravitational lensing occurs as an effect of Albert Einstein's theory of general relativity: mass bends light. So the gravitational fields of galaxy clusters bend light, acting like a magnifier of the galaxies behind them.&lt;/p&gt;&#xA;&lt;p&gt;The best observations of the halos by MUSE were released Monday, presented by Adélaïde Claeyssens during the annual meeting of the European Astronomical Society in Lyon, France. Claeyssens is a Ph.D. student at the Centre de Recherche Astrophysique de Lyon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Which further confuses me because the caption mentions Hubble but the article mentions data from MUSE at the VLT. I wonder if the image is a file photo and only peripherally related to the content of the news article?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-06-25T09:48:47.123" Title="Why is this Einstein ring pink?" Tags="&lt;galaxy-cluster&gt;&lt;gravitational-lensing&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32414" PostTypeId="2" ParentId="32409" CreationDate="2019-06-25T05:13:40.847" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Has the effect of other planets on Earth's orbit been ever measured?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...but I'm wondering if it's measurable...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;set up the question in an interesting way because it calls into question (intentionally or not) what kind of &lt;em&gt;measurement&lt;/em&gt; would satisfy the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motion of the Earth and other planets can be &lt;em&gt;numerically reproduced&lt;/em&gt; with incredible accuracy, and this can only be done when taking into account a large number of planet-planet gravitational effects. Here's an example of a &lt;a href=&quot;https://space.stackexchange.com/a/23409/12102&quot;&gt;python script&lt;/a&gt; doing an approximate calculation and being compared to data from the &lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;nofollow noreferrer&quot;&gt;Horizons web interface&lt;/a&gt; which uses data from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Jet_Propulsion_Laboratory_Development_Ephemeris&quot; rel=&quot;nofollow noreferrer&quot;&gt;JPL Development ephemerides&lt;/a&gt; or DE's which can reproduce the motion of the inner planets to kilometers and outer planets to tens of kilometers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the Sun, Jupiter's effect is huge, but interaction between the inner planets is quite big as well, with Venus's and Earth's effects upon each other being quite important. Venus also has a big effect on Mercury as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to affecting the inner planets directly, the massive outer planets Jupiter through Neptune move the Sun around by a million kilometers as well, and the inner planets will track the Sun's motion slow movements on the scale of decades or centuries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But besides showing that simulations match measurements, I don't think there has been any way to perform an &lt;em&gt;actual measurement&lt;/em&gt; of the effect of one Planet on another. So the answer to your question could be yes or no depending on how you would like to think of a measurement.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-06-25T05:26:00.390" LastActivityDate="2019-06-25T05:26:00.390" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32417" PostTypeId="2" ParentId="32388" CreationDate="2019-06-25T08:58:13.640" Score="6" Body="&lt;p&gt;[rewritten to address the revised question]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe, depending on how fussy you want to be about &quot;resolved&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/1995Icar..113..353N/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; is a study from 1995, using observations of asteroid 4179 Toutatis made in 1992 with &lt;em&gt;HST&lt;/em&gt;. They reported marginal resolution of the asteroid, as suggested by this figure comparing a deconvolved image of a star (observed with the same filter and imager location) and a similarly deconvolved image of the asteroid itself (each pixel corresponds to about 450 m at the distance of the asteroid):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Mt32U.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mt32U.jpg&quot; alt=&quot;Figure 4 from Noll et al. 1995&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The appearance of the asteroid is pretty clearly &lt;em&gt;not&lt;/em&gt; a point source, but it's also fair to say it's only partly resolved, and mostly just in one direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(My admittedly vague impression is that this is one of the best, if not &lt;em&gt;the&lt;/em&gt; best, case of &lt;em&gt;HST&lt;/em&gt; &quot;resolving&quot; a near-Earth asteroid.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most observations of near-Earth objects with &lt;em&gt;HST&lt;/em&gt; are, I think, aimed at getting optical information on compositions not possible from other wavelengths, and sometimes refining estimates of rotation rates, as was done this year (using data from 2012) for the asteroid &lt;a href=&quot;https://arxiv.org/abs/1903.00096&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bennu&lt;/a&gt;, currently being visited by OSIRIS-REx.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, you get &lt;em&gt;much&lt;/em&gt; better spatial resolution &lt;a href=&quot;https://echo.jpl.nasa.gov&quot; rel=&quot;nofollow noreferrer&quot;&gt;using radar&lt;/a&gt; (including line-of-sight distance variations due to the structure of the asteroids from time-of-return measurements, allowing you to construct 3D models of them), so there's not much point in trying to resolve them with &lt;em&gt;HST&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastEditorUserId="7411" LastEditDate="2019-06-30T13:38:32.980" LastActivityDate="2019-06-30T13:38:32.980" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32418" PostTypeId="2" ParentId="32412" CreationDate="2019-06-25T09:48:47.123" Score="6" Body="&lt;p&gt;OK, guessing rather wildly in the absence of solid information....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The press-release page is &lt;a href=&quot;http://ewass.kuoni-congress.info/press/2019/06/22/muse-reveals-a-glowing-ring-of-light-in-the-distant-universe/&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;; that provides a bit more information&#xA;than the news article, and includes &lt;em&gt;two&lt;/em&gt; images of (different regions of) the cluster and a hint that the cluster is named MACS0940.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the main image is a color composite from &lt;em&gt;HST&lt;/em&gt; imaging. It shows a galaxy cluster (the yellow galaxies). Also visible is light from an Einstein ring, the gravitational lensing by the cluster of a (single) distant background galaxy. The thin, white, ringlike streaks within the pink arcs are the stellar light from this galaxy (probably UV light from massive, recently formed stars); the broad pink arcs themselves are (I'm guessing) the Lyman-alpha emission as seen by MUSE. (So the image is a composite of both HST data &lt;em&gt;and&lt;/em&gt; MUSE data.) Thus it is plausible to say that the Einstein ring is made up of stellar light from the background galaxy and (in pink) emission from gas in and around that same galaxy; the assumption is that said gas is mostly in a halo around the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second &quot;magnified&quot; galaxy referred to in the news article is probably in a different part of the cluster, and doesn't form a nice Einstein ring; it is presumably the four pink blobs in the second image on the press-release page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.stsci.edu/hst/phase2-public/15696.pro&quot; rel=&quot;noreferrer&quot;&gt;This&lt;/a&gt; is the abstract for an &lt;em&gt;HST&lt;/em&gt; proposal on which Claeyssens was a co-I; the data were taken on May 1, 2019, so it's plausible the image is from those observations:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Recent deep integral-field spectroscopy observations using VLT/MUSE&#xA;  revealed two     high-redshift (z&gt;4) lensed arcs in the galaxy cluster&#xA;  MACS0940. Notably, both lensed galaxies show highly extended&#xA;  Lyman-alpha emission (&gt; 5 arcsec)... one of these galaxies forms an&#xA;  almost complete Einstein ring, with the Lyman-alpha arc reaching an&#xA;  unrivaled extent of 30 arcsec.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2019-06-25T09:48:47.123" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32419" PostTypeId="2" ParentId="32409" CreationDate="2019-06-25T13:30:58.483" Score="7" Body="&lt;p&gt;For decades, geologists have understood that climate changes are due in part to gradual shifts in the Earth’s orbit, which are caused primarily by Venus for its proximity and Jupiter for its mass, and repeat regularly every 405,000 years. In 2018 a team of geologists and Earth scientists unearthed the first evidence of these changes – sediments and rock core samples that provide a geological record of how and when these changes took place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.pnas.org/content/115/24/6153&quot; rel=&quot;noreferrer&quot;&gt;https://www.pnas.org/content/115/24/6153&lt;/a&gt;&#xA;&quot;Empirical evidence for stability of the 405-kiloyear Jupiter–Venus eccentricity cycle over hundreds of millions of years&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So given the timescales involved, this is as close to measuring this effect as one could expect to get.&lt;/p&gt;&#xA;" OwnerUserId="22580" LastActivityDate="2019-06-25T13:30:58.483" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32420" PostTypeId="2" ParentId="32410" CreationDate="2019-06-25T15:28:39.457" Score="55" Body="&lt;p&gt;You have already got some good answers, but I'll just try to provide one more intuitive solution on why the event horizons will never separate again if overlapping each other:&lt;br&gt;&#xA;First, imagine a speck of dust that comes inside the EH of a black hole. I believe we'll agree this speck can never escape the black hole, because nothing can come back from behind the event horizon.&lt;br&gt;&#xA;Now, imagine the same speck of dust, but inside the overlapping parts of the EH of two black holes passing each other. This speck of dust will never escape any of those two black holes, because it is inside the EH of them both. If these black holes would be able to separate again, the speck caught between them would obviously escape at least one of the black holes, after being behind it's event horizon.&lt;br&gt;&#xA;Since this can not happen, the two black holes will be united from the point their event horizons are overlapping, no matter their speed.&lt;/p&gt;&#xA;" OwnerUserId="28521" LastActivityDate="2019-06-25T15:28:39.457" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="32421" PostTypeId="5" CreationDate="2019-06-25T19:50:06.747" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2019-06-25T19:50:06.747" LastActivityDate="2019-06-25T19:50:06.747" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32422" PostTypeId="4" CreationDate="2019-06-25T19:50:06.747" Score="0" Body="Questions regarding the darker areas on the solar disk formed by emerging magnetic fields." OwnerUserId="16462" LastEditorUserId="16462" LastEditDate="2019-06-26T06:43:49.020" LastActivityDate="2019-06-26T06:43:49.020" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32423" PostTypeId="2" ParentId="29978" CreationDate="2019-06-25T20:31:34.717" Score="13" Body="&lt;p&gt;This questions can be split in two; for planets and satellites.&lt;/p&gt;&#xA;&lt;p&gt;The diversity of planets reflects in part the diversity in terms of chemical composition of the protoplanetary disk. We know that UV radiation from the sun can dissociate complex molecules or even very simple ones; for example, when UV rays split water molecules the result is free hydrogen and oxygen atoms. Since hydrogen is extremely lightweight they can be transported easily on the flux of stellar winds. So water, to keep with that example, if close to the sun could end up been dissociated and depleted from the region of the disk, but above the so called &lt;a href=&quot;https://en.wikipedia.org/wiki/Frost_line_(astrophysics)&quot; rel=&quot;noreferrer&quot;&gt;&amp;quot;snow line&amp;quot;&lt;/a&gt; UV radiation from the Sun was so weak that this couldn't happen as often and thus water molecules (which are very heavy in comparison to single hydrogen atoms), stayed there. That only explains the dichotomy between interior and exterior planets in terms of water content, and even then, some processes (like the &lt;a href=&quot;https://en.wikipedia.org/wiki/Late_Heavy_Bombardment&quot; rel=&quot;noreferrer&quot;&gt;late heavy bombardment&lt;/a&gt;) might add some water to the interior (as it happened on Earth). But this reasoning is not only for water, carbon dioxide, ammonia, methane and hundredths of different molecules have their own &amp;quot;frost lines&amp;quot;. Closer to the sun the carbon can't be methane is a volatile gas that gets quickly pushed outwards, but at some tenths of AUs methane can remain in stable conditions and can even condensate into liquid droplets.&lt;/p&gt;&#xA;&lt;p&gt;All of this just to say that the protoplanetary disk was NOT homogenous in terms of chemical composition, and was not homogeneous in terms of density or pressure. The thermal and chemical gradient across the nebula ensures some diversity and complexity for the entire planetary system.&lt;/p&gt;&#xA;&lt;p&gt;Here you have a beautiful diagram showing how different chemical compounds could condensate in different temperatures and pressures on the protoplanetary disk.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CExeK.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CExeK.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Also accretion of planetesimals is more energetic closer to the Sun (meaning that breakups can occur more frequently and is difficult for a planet to grow large), while in the outer regions planets can increase in mass with regularity since collisions with other planetesimals are performed at lower relative speeds (because of how two similar orbits have a difference in periods that gets larger when you get closer to the Sun and thus larger relative velocities). This coupled with the gravitational interactions of the protoplanets and the early disk (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Planetary_migration&quot; rel=&quot;noreferrer&quot;&gt;planetary migration&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Nice_model&quot; rel=&quot;noreferrer&quot;&gt;nice model&lt;/a&gt; etc...) allow for different accretion rates and the accretion of materials of different composition of what was found in the original place of formation of a particular planetesimal. This also helps to keep a wide variety on planetary masses.&lt;/p&gt;&#xA;&lt;p&gt;A wide variety in planetary masses is the starting point for a larger variation as the planets evolve in time and diverge from their initial conditions. A rocky small planet (Mercury) might have less heat trapped inside than a larger one (Earth), due to the smaller energy released by smaller accretion rates. Thus it might get cold rapidly and a magnetosphere due to a melted interior can't happen. The absence of a magnetosphere allows for solar wind charged particles to erode your atmosphere by &lt;a href=&quot;https://en.wikipedia.org/wiki/Sputtering&quot; rel=&quot;noreferrer&quot;&gt;sputtering&lt;/a&gt;. Instead on a planet like Earth, the larger mass has led to a melted interior that in turn generated a magnetosphere that lasted for billions of years, in Mars it lasted some time but now is almost gone so the atmosphere has also been almost destroyed. On Earth the presence of an atmosphere leds to all sorts of chemical erosions, and phenomena. Also, it's melted interior coupled with the specifics of its chemical composition and thickness of the crust allow for a mechanism called plate tectonics. Tectonics can't happen on Venus because the crust is not as thick (due to different composition) and thus it doesn't break apart in plates but just &lt;a href=&quot;https://en.wikipedia.org/wiki/Tessera_(Venus)&quot; rel=&quot;noreferrer&quot;&gt;deforms and folds&lt;/a&gt; in a complex behaviour that is unique to Venus.&lt;/p&gt;&#xA;&lt;p&gt;Also collisions with planetesimals can alter the future evolution of similar planets. Venus was probably very similar to Earth (similar mass, very similar composition and not so dissimilar temperatures as one might think) but their paths diverged completely as tectonics on Earth recycled the lithosphere and on Venus the carbon dioxide got more trapped in a greenhouse effect, and because Earth had a collision with another planet that have us our Moon, which is a mechanical stabilizer while a random collision with Venus (with different impact parameters) led to an extremely slow rotation and long days (but no moons). Longer days mean different insulation, and that changes drastically the climate of a planet. On Mars days are similar to Earth's but since it is smaller and the atmosphere has gone many things are very dissimilar from Earth. Also, Mars has no tectonics (the crust is thicker than Earth's and moves like a monolithic object) so there are few volcanoes and they grow huge (while on earth a single hot-spot generated by a mantle plume drills several holes on the crust as the plate moves on what is called a volcano chain) and there are stress faults like Valles Marineris (unique in the Solar System) that on Earth would have been relaxed by tectonic movements.&lt;/p&gt;&#xA;&lt;p&gt;To see how different the evolution of two planetary objects can be just by making them different mass take a look at our Moon. It has the same chemical composition (it is a chunk from Earth in fact), it is basically at the same distance to the Sun as Earth, it lives in the same interplanetary environment (same solar radiation, solar wind, impact rates etc...), and still it is completely different. This is all due to mass! The moon can't retain a large atmosphere as Earth because it has less gravitational pull. The same temperature for our atmosphere there means that the particles reach escape velocity easily and start to escape from the gravity well. With no atmosphere, not internal heat, the moon lacks almost any type of erosion in across billions of years of evolution. Processes of erosion on Earth have made the diversity of geological formations to explode in comparison to those found on the Moon. Even then the moon has its own peculiarities and dynamical features unique to it.&lt;/p&gt;&#xA;&lt;p&gt;Now we are getting closer to the satellites question. They in fact should look almost the same, since they are formed from very very similar material in extremely similar conditions. And indeed we believe that moons originally were very similar (for example the 4 Galilean moons). But Io is to close to Jupiter and the other moons interact with it in such a way that the geological processes are completely different. Water and volatiles evaporated quickly as it heated up by tidal forces from Jupiter. These tidal forces were not as strong in Europa since it is farther away, thus it only melted part of the icy crust creating an ice analog of plate tectonics which generated a plethora of diverse formations. Satellites evolve. Enceladus shoots jets because of tidal interactions and orbital resonances with other moons. Some moons like Japeto have a dual colored surface because of the material sprayed by Enceladus landing on on one of its sides. Some moons like Triton have nothing to do with the other because they formed in another region of the solar System and later got trapped by the gravitational pull of a planet (Neptune in this case).&lt;/p&gt;&#xA;&lt;p&gt;As I mentioned before. Atmospheres (density, composition and pressure) are largely dependent on the mass of the planet or moon. Look at this graph:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/c5LPJ.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c5LPJ.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;It shows the velocity of gas molecules in relation to temperature of the gas. For larger temperatures gas molecules move faster. In a planet with low mass, the escape velocity is lower than one with a larger mass. Thus a planet closer to the sun (at higher temperature) needs to be larger in size if it wants to preserve the same gas molecules in its atmosphere as a planet that is farther away (colder). You can see why Earth's atmosphere could trap and retain water, oxygen, carbon dioxide, ammonia, methane nitrogen and other gases while it is unable to trap hydrogen and helium (because they are lighter and thus for the same temperature they can move as fast as needed to escape Earth). Meanwhile, the Moon, which has the same heat coming from the Sun as Earth, since it is less massive it can't retain almost any gases (maybe a bit of Xenon). Titan, is a huge moon thus it can retain many gaseous molecules like Nitrogen and Oxygen (those in turn make the pressure high enough to retain also volatiles like methane in a liquid form on the surface). But why doesn't Ganymede has the same atmosphere as Titan if they are basically the same size? Because Ganymede is closer to the Sun, larger temperature means that the molecules move faster and thus they escape its attraction easily.&lt;/p&gt;&#xA;&lt;p&gt;As you can see the complex processes of atmospheres of a moon or a planet change everything (erosion, recycling processes, chemical corrosion, etc...) and in turn that diversity in atmospheres comes from a diversity of masses and distances to the Sun.&lt;/p&gt;&#xA;&lt;p&gt;I think the Solar System is a chaotic system, dynamically, geologically, chemically etc... Chaos means that for an small difference in initial conditions the system will evolve in exponentially diverging different states. Planets and moons might have started as similar objects but history and the chaotic dynamics of the system have evolved into completely different environments. Not only that, but the truth is that planets didn't started as equals but were very different from the beginning, so imagine how far away is Venus to become a Titan, or an Io to become an Earth.&lt;/p&gt;&#xA;&lt;p&gt;Also there are processes and conditions that are specially well suited for divergence. For example: Earth is very dynamic while Mars, Venus, Mercury, the Moon and others are totally not. Why? because on Earth water can exist in 3 different states of matter. We can find liquid water, water vapor and ice in different regions and seasons. And that is because Earth is at an average temperature and it's atmosphere has just the right pressure to allow this. Earth's conditions are very close to the triple point of water (where all three states of matter coexist), that's why we have a water cycle on Earth, with rivers and glaciers eroding the landscape and clouds regulating the climate.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5cr0j.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5cr0j.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Mars, Venus, Mercury, all have temperatures and pressures were this can't happen, not only on water but on many compounds present there. You know where this can happen? On Pluto! This was very surprising, Pluto shows a variety of terrains and geological features that exceeds all expectations. Now we know this is because Pluto is extremely dynamic (as Earth) and a lot of erosion and geo-chemical processes can occur, but this is not because of water (since Pluto has low pressure and low temperatures), but because of Nitrogen and Neon! Both elements have their triple point inside to Pluto's range of conditions and thus Neon rivers, Nitrogen glaciers and hazes are expected on this dwarf planet.&lt;/p&gt;&#xA;&lt;p&gt;It is indeed an interesting question. How incredible are the laws of nature that allow for extreme variety even between brothers. I wonder how a planet around any other star might be, our simplistic categories of hot Jupiters, mini-Neptunes, super-Terras, etc... are just so primitive and restrictive. What wonders await us in this complex and diverse cosmos is beyond our comprehension.&lt;/p&gt;&#xA;" OwnerUserId="28496" LastEditorUserId="33547" LastEditDate="2021-01-11T21:20:40.627" LastActivityDate="2021-01-11T21:20:40.627" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32424" PostTypeId="1" CreationDate="2019-06-26T02:57:59.540" Score="3" ViewCount="455" Body="&lt;p&gt;How long does it take an asteroid to get from the asteroid belt, to Earth? I know the asteroids don't go directly from the asteroid belt to Earth, but I know it's not quick, either. The time probably varies, but what's the average time?&lt;/p&gt;&#xA;" OwnerUserId="19304" LastActivityDate="2019-06-26T15:59:06.977" Title="How long does it take for an asteroid to get to Earth?" Tags="&lt;asteroids&gt;&lt;asteroid-belt&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32425" PostTypeId="1" CreationDate="2019-06-26T06:34:20.730" Score="4" ViewCount="84" Body="&lt;p&gt;After recording the ripples of  passing gravity waves I would imagine that some high-power math would be required to derive information about the sizes, spin, and distance of the merging black holes. Is it possible to get some insight into how they figured out one was 31 solar masses and the other was 20? How do you &quot;see&quot; spin from a wave?&lt;/p&gt;&#xA;" OwnerUserId="28175" LastEditorUserId="1203" LastEditDate="2019-06-26T12:27:09.387" LastActivityDate="2019-06-26T12:27:09.387" Title="Gravitational waves: how can scientists derrive the sizes and spin and distance of two merging black holes?" Tags="&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32426" PostTypeId="1" AcceptedAnswerId="32463" CreationDate="2019-06-26T07:11:44.023" Score="6" ViewCount="874" Body="&lt;p&gt;First of all, for the purposes of this question Pluto is a planet. So the eight planets being considered are Mercury, Venus, Mars, Jupter, Saturn, Uranus, Neptune and Pluto. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: can all eight of these planets &lt;em&gt;go retrograde&lt;sup&gt;†&lt;/sup&gt;&lt;/em&gt; at the same time? I know that seven can do so on rare instances; there were 12 days of that in the 20th century all in the 1940's and the 1980's. But to show how rare that is there was a block of 400 years in the middle ages with no days of seven retrogrades. Has eight ever happened? Or is it impossible? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;†&lt;/sup&gt;Retrograde motion as seen from Earth, not retrograde orbits around the Sun.&lt;/p&gt;&#xA;" OwnerUserId="28528" LastEditorUserId="10196" LastEditDate="2019-06-26T10:42:23.883" LastActivityDate="2020-09-09T04:53:33.290" Title="Can Eight Retrograde (seen from Earth) planets occur?" Tags="&lt;orbital-mechanics&gt;&lt;positional-astronomy&gt;&lt;ecliptic&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32427" PostTypeId="2" ParentId="32425" CreationDate="2019-06-26T08:41:22.977" Score="5" Body="&lt;p&gt;One element of the answer is that they do it in the other direction. They simulate &lt;a href=&quot;https://arxiv.org/abs/1904.04831&quot; rel=&quot;nofollow noreferrer&quot;&gt;many different possible black hole mergers&lt;/a&gt; and then look for the nearest matches to what they observe. I imagine there is some scope for interpolating between the simulations as well as running &lt;a href=&quot;https://www.ligo.caltech.edu/video/ligo20181203v1&quot; rel=&quot;nofollow noreferrer&quot;&gt;bespoke simulations&lt;/a&gt; to confirm the match. Spin affects the wave because the shape of space around a spinning black hole is different from that around a non-spinning one, which affects both the motion of the black holes and the processes by which the gravitational waves are radiated.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="21928" LastEditDate="2019-06-26T09:13:14.297" LastActivityDate="2019-06-26T09:13:14.297" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32428" PostTypeId="1" AcceptedAnswerId="32432" CreationDate="2019-06-26T08:58:47.930" Score="11" ViewCount="253" Body="&lt;p&gt;The leading hypothesis for how the Moon was formed was that another object collided with the Earth, throwing off a large lump of matter which formed into the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this thought to be the case with all satellites? e.g. Phobos, Deimos, Charon, Titan, Europa, etc. ? If not, then why do we need such a dramatic &quot;origin story&quot; for the Moon but not the others? Why do we not just assume the Moon was formed the same way as all satellites?&lt;/p&gt;&#xA;" OwnerUserId="24696" LastActivityDate="2019-06-27T00:00:46.780" Title="Are all satellites thought to be the result of collisions with the planets?" Tags="&lt;the-moon&gt;&lt;natural-satellites&gt;&lt;formation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32430" PostTypeId="2" ParentId="29641" CreationDate="2019-06-26T13:52:10.893" Score="2" Body="&lt;p&gt;No. there's in no relation as far as we know. Ecliptic plane orientations of other planetary systems with respect to the galactic plane are completely random as far as we know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know this also because we can find exoplanets by the transit method nearly in any direction of the sky; if we look at &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; stars located at galactic latitude of (let's say) &lt;span class=&quot;math-container&quot;&gt;$20^\circ$&lt;/span&gt; (thus their ecliptic plane has relative inclinations of &lt;span class=&quot;math-container&quot;&gt;$20^\circ$&lt;/span&gt; with respect to the galactic plane) we find roughly the same number of transit events as if we look to &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; stars at &lt;span class=&quot;math-container&quot;&gt;$0^\circ$&lt;/span&gt; latitude (thus their ecliptic planes have relative inclinations of &lt;span class=&quot;math-container&quot;&gt;$0^\circ$&lt;/span&gt; with respect to the galactic plane). Since we are probably not the center of a planetary conspiracy on which everything is aligned so that we see the same distribution in all the sky we can assume that inclinations are random and we just happen to detect a similar fraction of transits everywhere because we are looking at at the same fraction of the same random population in each direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Humanity has just started to observe other planetary systems so this might change a little but in principle if there is any &quot;preferential orientation&quot;, for now, with our current knowledge it is completely camouflaged by the main random component of this distribution.&lt;/p&gt;&#xA;" OwnerUserId="28496" LastEditorUserId="28496" LastEditDate="2019-06-26T14:50:30.700" LastActivityDate="2019-06-26T14:50:30.700" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32431" PostTypeId="2" ParentId="32424" CreationDate="2019-06-26T15:59:06.977" Score="4" Body="&lt;p&gt;The full orbital period of an object orbiting the sun is &lt;span class=&quot;math-container&quot;&gt;$$T=2\pi \sqrt{\frac{a^3}{GM_\odot}},$$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is the semi-major axis of the orbit and &lt;span class=&quot;math-container&quot;&gt;$M_\odot$&lt;/span&gt; the mass of the sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the asteroid has apoapsis in the asteroid belt (&lt;span class=&quot;math-container&quot;&gt;$r_{max}= 3$&lt;/span&gt; AU) and periapsis near Earth (&lt;span class=&quot;math-container&quot;&gt;$r_{min}=1$&lt;/span&gt; AU) then the semimajor axis will be &lt;span class=&quot;math-container&quot;&gt;$a=(r_{max}+r_{min})/2=2$&lt;/span&gt; AU and going in from the belt takes half a period, so the answer is &lt;span class=&quot;math-container&quot;&gt;$\pi\sqrt{(2 AU)^3/GM_\odot}=1.4140$&lt;/span&gt; years. An asteroid coming from further out, say 4 AU, would take 1.9762 years. From 2 AU, 0.9184 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, were one to just give an asteroid a big push it could move nearly arbitrarily fast (up to the lightspeed limit if we want to be absurd) along an Earth-intersecting orbit, but I suspect the question was about normal asteroids. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is worth noting that long-periodic comets can move fast (tens of km/s). Such a comet (or an interstellar asteroid) would have a velocity on the order of &lt;span class=&quot;math-container&quot;&gt;$\sqrt{2GM_\odot/(1AU)}\approx 42.1285$&lt;/span&gt; km/s if it came in from the outer edge of the solar system. That would cross the 2 AU from the asteroid belt to Earth in about 82 days. &lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2019-06-26T15:59:06.977" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32432" PostTypeId="2" ParentId="32428" CreationDate="2019-06-26T17:00:53.980" Score="14" Body="&lt;p&gt;There are three main formation scenarios for planetary moons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The giant impact hypothesis:&lt;/strong&gt; The satellite forms as a consecuence of an &lt;a href=&quot;https://en.wikipedia.org/wiki/Giant-impact_hypothesis&quot; rel=&quot;noreferrer&quot;&gt;impact between the planet an a large planetesimal&lt;/a&gt;. The Moon is an example, and one of the arguments is that the chemical composition of the Moon matches that of Earth with a significant accuracy suggesting that it is a in part our planet and in part the original impactor (&lt;a href=&quot;https://en.wikipedia.org/wiki/Theia_(planet)&quot; rel=&quot;noreferrer&quot;&gt;Theia&lt;/a&gt;). We also know that the Moon has been getting farther away from Earth because we have evidence that it gained orbital potential energy by absorbing it from the rotational energy of Earth. We know this because days were not 24 hours long a few million years ago and we can keep track of those changes in the rotational period of Earth using rings in fossillized corals (those have rigns like the tree rings but ones that generate on a daily basis). We then can see that the moon was extremely close to the Earth some billion years ago (we have more evidence of this from the fact that tides were huge in those times and led geological evidence of daily floods across the recently formed planet). If you keep going back in time you see that the Moon was basically emerging from Earth. There are many more evidences of this scenario for our moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The accretion scenario:&lt;/strong&gt; The satellite coalesced from a disc of material around the new-born planet (just as the planet accreted from the protoplanetary disc), the so called circumplanetary disc. As an example we have the &lt;a href=&quot;https://en.wikipedia.org/wiki/Galilean_moons&quot; rel=&quot;noreferrer&quot;&gt;four galilean moons&lt;/a&gt; around Jupiter (Io, Europa, Ganymede and Callisto). Since the disk was relatively flat the moons formed in the same orbital plane, also they move in the same direction as the planet rotates (which makes sense as both are generated from the same material spining with a certain angular momentum). This is the most frequent scenario for large moons. Our moon couldn't form like this because the expected size of the circumplanetary disc was in no way as massive as our moon is today (Earth is a tiny planet and it has a huge moon in relative terms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Capture scenario:&lt;/strong&gt; The satellite formed elsewhere in the Solar System as an independent minor body. With time some dynamical interaction might have led the object close to a planet and both got gravitationally binded. An example of this is Triton, Neptune's largest moon. The retrograde orbit is inexplicable in terms of the accretion scenario and the energy needed for a gian-impact scenario to work on Neptune is too large. &lt;a href=&quot;https://en.wikipedia.org/wiki/Triton_(moon)#Capture&quot; rel=&quot;noreferrer&quot;&gt;Triton was captured&lt;/a&gt; (we think it formed as another planetesimal in the Kuiper Belt as it shares &lt;a href=&quot;https://www.nature.com/articles/311355a0&quot; rel=&quot;noreferrer&quot;&gt;many chemical features of Pluto&lt;/a&gt; and other objects of the region). There are not so many moons on Neptune probably because they dissapeared (crashing into the planet or been ejected) just as Triton arrived to the system and dynamically destabilized their orbits. Another clear example are the tiny &lt;a href=&quot;https://en.wikipedia.org/wiki/Moons_of_Jupiter#Irregular_satellites&quot; rel=&quot;noreferrer&quot;&gt;irregular satellites of Jupiter&lt;/a&gt;. This scenario is very difficult to immagine for Earth since capturing a massive moon like ours and making the orbit circular would have been a feat in terms of how precisely tunned would the orbit insertion parameters had to be. The giant-impact scenario leads to the current situation in simulations for a larger range of impact parameters, thus is much more probable in statistical terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some less frequent and some speculative scenarios also:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Eyecta fragments from other moons:&lt;/strong&gt; Some satellites might have their origin on other satellites. A large impact might eject material into orbit. An example would be &lt;a href=&quot;https://en.wikipedia.org/wiki/Hippocamp_(moon)&quot; rel=&quot;noreferrer&quot;&gt;Hippocamp&lt;/a&gt; (a Neptunian moon) which is now regarded to be a fragment stripped out from &lt;a href=&quot;https://en.wikipedia.org/wiki/Proteus_(moon)&quot; rel=&quot;noreferrer&quot;&gt;Proteus&lt;/a&gt; (a larger moon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lagrangian/&lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_objects_at_Lagrangian_points#Saturn%E2%80%93Tethys_Lagrangian_points&quot; rel=&quot;noreferrer&quot;&gt;Trojan&lt;/a&gt; moons:&lt;/strong&gt; This is similar to the circumplanetary disc scenario but here the accretion in the disc of the planet is further stimulated in certain regions because of a moon that formed a bit earlier. An orbiting body can generate five equilibrium points (&lt;a href=&quot;http://forum.spaceengine.org/viewtopic.php?t=446&amp;amp;start=45#p21605&quot; rel=&quot;noreferrer&quot;&gt;Lagrange points&lt;/a&gt;) by sculpting the gravitational landscape. Two of those equilibrium points (L4 and L5) are stable equilibrium points; thus they are like gravitational traps where matter can accumulate until a new moon is formed. As a potential example we have &lt;a href=&quot;https://en.wikipedia.org/wiki/Telesto_(moon)&quot; rel=&quot;noreferrer&quot;&gt;Telesto&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Calypso_(moon)&quot; rel=&quot;noreferrer&quot;&gt;Calypso&lt;/a&gt; in the Saturnian system. They both lie on the L4 and L5 Lagrange points of &lt;a href=&quot;https://en.wikipedia.org/wiki/Tethys_(moon)&quot; rel=&quot;noreferrer&quot;&gt;Tethys&lt;/a&gt; (a much larger moon with great gravitational influence). They might have formed as regular objects and then got trapped on the equilibrium points or they might have actually formed there as matter coalesced on those gravitational traps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sprayed into existence by &lt;a href=&quot;https://en.wikipedia.org/wiki/Cryovolcano&quot; rel=&quot;noreferrer&quot;&gt;cryovolcanism&lt;/a&gt; from another moon&lt;/strong&gt;: It sounds ridiculous because it is an hypothetical scenario, I just think might happen somewhere in the universe. Take a look to &lt;a href=&quot;https://en.wikipedia.org/wiki/Enceladus&quot; rel=&quot;noreferrer&quot;&gt;Enceladus&lt;/a&gt; (a large active moon of Saturn). Enceladus has &lt;a href=&quot;https://en.wikipedia.org/wiki/Tiger_stripes_(Enceladus)&quot; rel=&quot;noreferrer&quot;&gt;water plumes and jets that shoot material from its interior into space through craks in the icy crust&lt;/a&gt; (because tidal stresses heat the interior like a pressure cooker and the pressure gets liberated that way). The entire &lt;a href=&quot;https://en.wikipedia.org/wiki/Rings_of_Saturn#E_Ring&quot; rel=&quot;noreferrer&quot;&gt;E-ring on Saturn&lt;/a&gt; was created by orbiting ice and dust grains sprayed by Enceladus. We know that the E-ring has a mass of nearly &lt;a href=&quot;https://www.sciencedirect.com/science/article/abs/pii/S0019103511003708&quot; rel=&quot;noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$12\cdot10^8\; kg$&lt;/span&gt;&lt;/a&gt; and we know that it is so far from Saturn as to make possible the coalescence of this matter (tidal forces wouldn't disrupt it: see &lt;a href=&quot;https://en.wikipedia.org/wiki/Roche_limit&quot; rel=&quot;noreferrer&quot;&gt;Roche limit&lt;/a&gt;). Thus it is possible that the material from the ring could make a moon with the density of &lt;a href=&quot;https://en.wikipedia.org/wiki/Aegaeon_(moon)&quot; rel=&quot;noreferrer&quot;&gt;Aegaeon&lt;/a&gt; (another saturnian moon) and with a diameter of just &lt;span class=&quot;math-container&quot;&gt;$162\; m$&lt;/span&gt; (a third of the size of Aegaeon). Enceladus prohibits this because of the gravitational influence it excerts on the E-ring, it also prohibits it to grow (the ring would be more massive if the material was not continuosly reabsorbed by Enceladus). But if Enceladus migrated to another orbit in relatively short timescales then at least I think is possible for this to happen. The new moon greated from the material of the ring could then end up coliding with Enceladus since both would probably still interact chaotically. The material once sprayed out of its interior would had returned to home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Centrifugal breakup scenario:&lt;/strong&gt; This is also hypothetical but we think this happens a lot on asteroids. Lightweight moons could be ruble piles just as many comets and asteroids are. Losely bounded material with little cohesion. If the moon starts to spin faster and faster (because of some mechanism like the &lt;a href=&quot;https://en.wikipedia.org/wiki/Yarkovsky%E2%80%93O%27Keefe%E2%80%93Radzievskii%E2%80%93Paddack_effect&quot; rel=&quot;noreferrer&quot;&gt;Yarkovsky-O'Keefe–Radzievskii–Paddack effect&lt;/a&gt;), it &lt;a href=&quot;https://www.researchgate.net/publication/5237852_Rotational_breakup_as_the_origin_of_small_binary_asteroids&quot; rel=&quot;noreferrer&quot;&gt;could eventually break into two pieces&lt;/a&gt; due to the extreme centrifugal forces (described as seen from the co-rotating reference frame of the original moon). It is believed that asteroids like &lt;a href=&quot;https://en.wikipedia.org/wiki/(66391)_1999_KW4&quot; rel=&quot;noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$1999 \;KW_4$&lt;/span&gt;&lt;/a&gt; splitted in two because of this effect. I don't see any reason why a moon couldn't do the same, forming a new independent moon in the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Moon made of chunks of other disrupted moons:&lt;/strong&gt; As crazy as it seems, it is one of the formation hypothesis regarding the formation of &lt;a href=&quot;https://en.wikipedia.org/wiki/Miranda_(moon)&quot; rel=&quot;noreferrer&quot;&gt;Miranda&lt;/a&gt; (one of Uranus satellites). Miranda's surface is so complex and varied that some speculate it might have formed as several pieces that orbited Uranus came toghether gently. These pieces might have been chunks from other moons or might have been chunks from an earlier iteration of Miranda itself, fragmented after a disruptive event. The geology on each chunk would have evolved independently until they re-assembled together. But this is also quite speculative.&lt;/p&gt;&#xA;" OwnerUserId="28496" LastEditorUserId="28496" LastEditDate="2019-06-27T00:00:46.780" LastActivityDate="2019-06-27T00:00:46.780" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32433" PostTypeId="1" CreationDate="2019-06-26T21:03:43.900" Score="0" ViewCount="47" Body="&lt;p&gt;I was wondering how the distance in Mpc could be calculated between Earth and a galaxy using the luminosity provided? Do not be shy to use all math necessary :)&lt;/p&gt;&#xA;" OwnerUserId="28470" LastActivityDate="2019-06-26T21:03:43.900" Title="How to calculate the distance to galaxies with standard candles?" Tags="&lt;galaxy&gt;&lt;distances&gt;&lt;luminosity&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32434" PostTypeId="2" ParentId="32350" CreationDate="2019-06-26T21:45:17.580" Score="4" Body="&lt;p&gt;What about Astronomical Algorithms by Jean Meeus. It's a mathematical discussion of the formulae behind astronomical events such as solar eclipses, occultations and transits. I have read it. It is a wonderful book. The only aspect that might not be satisfactory is the modern aspect. It's somewhat old. &lt;/p&gt;&#xA;" OwnerUserId="28528" LastActivityDate="2019-06-26T21:45:17.580" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32435" PostTypeId="1" CreationDate="2019-06-26T22:54:36.183" Score="2" ViewCount="171" Body="&lt;p&gt;When calculating luminosity/radius/temperature for several stars, I've noticed that while the majority of stars conform to the luminosity relationship &lt;span class=&quot;math-container&quot;&gt;$L=4\pi R^{2}\sigma T^{4}$&lt;/span&gt;, there are a few that don't. Take 8 Ursae Minoris. As far as I can tell it has a luminosity 56 times that of the sun, a radius of 8.5 solar radii and an effective temperature of about 5000 K. When I substitute those numbers in they don't match. The math isn't wrong so I can only assume either the data isn't reliable, or maybe there is a certain class of stars that don't conform to that relation. I don't know enough astronomy to determine which is correct, but I suspect obtaining a reliable luminosity value for some stars could be the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know the answer to this riddle? &lt;/p&gt;&#xA;" OwnerUserId="28542" LastEditorUserId="28542" LastEditDate="2019-06-26T23:08:14.253" LastActivityDate="2019-06-26T23:08:14.253" Title="Stars that don't fit $L=4\pi R^{2}\sigma T^{4}$?" Tags="&lt;star&gt;&lt;temperature&gt;&lt;luminosity&gt;" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32436" PostTypeId="1" CreationDate="2019-06-26T23:37:22.573" Score="1" ViewCount="102" Body="&lt;p&gt;The Phys.org news item &lt;a href=&quot;https://phys.org/news/2019-06-hubble-ultimate-multitasker-asteroids.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble is the ultimate multitasker: Discovering asteroids while it's doing other observations&lt;/a&gt; mentions the Hubble Space Telescope's use of &lt;em&gt;parallel fields&lt;/em&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A couple years ago, the NASA/ESA Hubble Space Telescope was observing very distant galaxies, some of them billions of light years away. It was part of the Frontier Fields Project (FFP,) an ambitious observing program which aimed to observe six massive galaxy clusters. The FFP pushed Hubble to its limits.&lt;/p&gt;&#xA;&lt;p&gt;[...]&lt;/p&gt;&#xA;&lt;p&gt;This image is a couple years old now, and it shows what's called the parallel field for the Abell 370 galaxy cluster, about 4 billion light years away. It's called a parallel field because of the technique used to capture it.&lt;/p&gt;&#xA;&lt;p&gt;A press release explains parallel fields this way:&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;While observing each cluster with one of the cameras on Hubble, the team also used a different camera, pointing in a slightly different direction, to photograph six so-called 'parallel fields.&amp;quot; This maximised Hubble's observational efficiency in doing deep space exposures, imaging a myriad of far away galaxies.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How are &amp;quot;parallel fields&amp;quot; implemented on the Hubble Space Telescope? How are two cameras able to take different exposures in different directions at the same time? What optical components points them in different directions?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-06-30T08:43:12.310" Title="How are &quot;parallel fields&quot; implemented on the Hubble Space Telescope?" Tags="&lt;observational-astronomy&gt;&lt;hubble-telescope&gt;&lt;optics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32437" PostTypeId="1" AcceptedAnswerId="32441" CreationDate="2019-06-26T23:47:07.200" Score="3" ViewCount="146" Body="&lt;p&gt;The Phys.org news item &lt;a href=&quot;https://phys.org/news/2019-06-hubble-ultimate-multitasker-asteroids.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hubble is the ultimate multitasker: Discovering asteroids while it's doing other observations&lt;/a&gt; includes the image below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What causes these asteroid trails to be so strangely shaped and repeated? Usually asteroids appear as short straight segments in long exposures from terrestrial telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RZOqe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some asteroids from within our Solar System have photobombed deep images of the Universe taken by the NASA/ESA Hubble Space Telescope. The asteroidal streaks in this image are created by our virtual neighbours; asteroids in the main asteroid belt between Mars and Jupiter. In the background are thousands of colourful galaxies, some of them billions of light years away. Credit: NASA, ESA, and B. Sunnquist and J. Mack (STScI); CC BY 4.0; Acknowledgment: NASA, ESA, and J. Lotz (STScI) and the HFF Team&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-06-30T13:47:15.963" Title="What causes these asteroid trails to be so strangely shaped and repeated?" Tags="&lt;asteroids&gt;&lt;hubble-telescope&gt;&lt;deep-sky-observing&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32438" PostTypeId="2" ParentId="32437" CreationDate="2019-06-27T04:36:48.980" Score="3" Body="&lt;p&gt;The asteroids leave &lt;strong&gt;corkscrew&lt;/strong&gt; shaped trails which are interrupted by not recording when the Earth passes through the field of view.  The corkscrew shape echoes the path of the telescope moving around the Earth as the Earth moves around the Sun.  The &lt;strong&gt;straighter&lt;/strong&gt; trails you are familiar with are merely zoomed-out versions of the same -- asteroid hunting needs long exposures but not high magnification.&lt;/p&gt;&#xA;" OwnerUserId="10007" LastActivityDate="2019-06-27T04:36:48.980" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32439" PostTypeId="2" ParentId="32436" CreationDate="2019-06-27T05:43:27.180" Score="6" Body="&lt;p&gt;As the question &lt;a href=&quot;https://astronomy.stackexchange.com/questions/2509/instrument-aperture-sizes-on-hubble-telescope?rq=1&quot;&gt;Instrument aperture sizes on Hubble Telescope&lt;/a&gt; shows, the focal plane area is large enough to focus on several instruments at the same time (but with each capturing a different area).&lt;br&gt;&#xA;If two objects of interest are separated by a certain angle (the instruments are fixed within the focal plane), the telescope can be rotated so that two instruments are exposed, one to each object.  If desired, the telescope can later be rotated to swap the instruments on each object, so that both objects get a wider spectral coverage, for example.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No optical components point the instruments (cameras/spectrometers) at different objects; rather, the whole telescope orients on one direction in space, and by rotating around the main mirror axis to a particular angle (matching the positions of the fixed instruments) can choose to expose multiple instruments to multiple objects in that Field-Of-View.  All instruments (except the WFC UV/IR sensors) are always exposed to something, but data collection/transmission stresses the systems, so 'parallel' viewing is used selectively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only the Wide Field Camera uses a pick-off mirror (steerable for UV or IR sensors). The other instruments are aligned with their patches of the focal plane and used COSTAR lenses for correction until Service Mission 4 removed COSTAR and replaced the actual instruments (which now have built in corrections). The Wide Field Camera was replaced several times and had its own corrective mirrors instead of a lens via COSTAR. &lt;/p&gt;&#xA;" OwnerUserId="10007" LastEditorUserId="10007" LastEditDate="2019-06-30T08:43:12.310" LastActivityDate="2019-06-30T08:43:12.310" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32440" PostTypeId="1" AcceptedAnswerId="32451" CreationDate="2019-06-27T09:42:27.657" Score="4" ViewCount="128" Body="&lt;p&gt;Gravitational time dilation like the one shown in the movie Interstellar, causes time to move slowly for someone within the high gravity reference frame. Now let's say that from such a place, I was able to view through a telescope somewhere far away which is not gravitationally dilated. And assume that I am able to view the near live video stream of it. Would those people look fast moving? &lt;/p&gt;&#xA;" OwnerUserId="28550" LastActivityDate="2019-06-27T23:28:14.080" Title="If I could see from a Gravitational-Time dilated place to somewhere far which is not dilated, would I see fast moving people?" Tags="&lt;gravity&gt;&lt;time-dilation&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32441" PostTypeId="2" ParentId="32437" CreationDate="2019-06-27T11:17:05.233" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What causes these asteroid trails to be so strangely shaped and repeated? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The linked image is a composite of several long duration exposures captured by the Hubble Advanced Camera for Surveys of the same region of the distant sky. Each individual exposure was about ~2200 seconds long. Nearby objects such as asteroids would have been subject to the proper motion of the asteroid with respect to the Earth and parallax due to the Hubble being in orbit about the Earth during those ~2200 second long exposures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each individual streak comprises both proper motion and parallax effects. The parallax effects are what make the streaks have a nonlinear shape. One orbit later (~5725 seconds), another ~2200 second long exposure showed a similar path for the same asteroid, but shifted a bit because of the asteroid's proper motion. This is why the composite image shows twenty streaks (per the linked article; I see sixteen) for only seven asteroids.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2019-06-27T11:17:05.233" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32442" PostTypeId="2" ParentId="20122" CreationDate="2019-06-27T12:55:29.577" Score="1" Body="&lt;p&gt;Sirius b is 1.02 solar masses and is a carbon-oxygen white dwarf. The critical mass depends on its composition, an iron composition white dwarf may have a critical mass of as low as 1.0667 solar masses, see column seven of  the last row of table 3 in &lt;a href=&quot;https://iopscience.iop.org/article/10.1088/1674-4527/17/6/61/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;. They get lower than the usual values through considering inverse beta decay (proton plus electron combine to make a neutron) also called neutronization. It’s a difficult calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also stack exchange comment here&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/345282/why-dont-electrons-and-protons-in-a-white-dwarf-combine-before-the-electrons-be&quot;&gt;Why don't electrons and protons in a white dwarf combine before the electrons become ultrarelativistic?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Anyway with that paper’s suggested correction, again using their table 3, it’s 1.3846 for oxygen and 1.3916 for carbon white dwarfs. Sirius b is still well below critical mass.&#xA;When that does happen, Sirius b is probably too far away to accumulated enough gas from it to go supernova and anyway all the stars are moving around the galaxy, chances are that Sirius is a vast distance from us before this can happen, see &lt;a href=&quot;https://astronomy.stackexchange.com/a/20124/11432&quot;&gt;Rob Jeffries' answer&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another possibility, a sub-Chandrasekar white dwarf. This is a recent idea that it’s possible for a lighter white dwarf to go supernova without reaching the Chandrasekhar limit, making a less energetic supernova.&#xA;According to &lt;a href=&quot;http://lclavelli.people.ua.edu/uploads/9/5/0/7/95076160/1609.02742r.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt; there is some statistical evidence that there may be fewer white dwarfs of mass of 1.1 solar masses and higher than you’d expect from the predicted range of birth masses, suggesting that maybe some of them go supernova as they cool down.&#xA;It's unlikely to apply to Sirius b though.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Sirius b, the right-most magenta diamond, is unlikely to produce a supernova unless the critical density is significantly lower than the value taken in fig. 3. In this improbable case many of the high mass dwarfs from ref. [16] would have to be undergoing significant accretion in order to survive to the measured ages.&quot; &lt;a href=&quot;http://lclavelli.people.ua.edu/uploads/9/5/0/7/95076160/1609.02742r.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Degeneracy Breakdown as a Source of Supernovae Ia&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11432" LastActivityDate="2019-06-27T12:55:29.577" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32443" PostTypeId="1" AcceptedAnswerId="32455" CreationDate="2019-06-27T13:20:11.667" Score="7" ViewCount="819" Body="&lt;p&gt;The estimated &lt;a href=&quot;https://en.wikipedia.org/wiki/Age_of_the_universe&quot; rel=&quot;noreferrer&quot;&gt;age of the universe&lt;/a&gt; is 14 billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The estimated &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_radius#Parameters&quot; rel=&quot;noreferrer&quot;&gt;Schwarzschild radius&lt;/a&gt; (event horizon) of the observable universe is 14 billion light-years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the ramifications?&lt;/p&gt;&#xA;" OwnerUserId="28551" LastActivityDate="2021-02-05T19:17:10.607" Title="Observable universe equals its Schwarzschild radius (event horizon)?" Tags="&lt;black-hole&gt;&lt;universe&gt;&lt;cosmology&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32445" PostTypeId="1" AcceptedAnswerId="32448" CreationDate="2019-06-27T14:42:28.437" Score="6" ViewCount="473" Body="&lt;p&gt;Inspired by &lt;a href=&quot;https://astronomy.stackexchange.com/q/32410&quot;&gt;If two black hole event horizons overlap (touch) can they ever separate again?&lt;/a&gt;, I was wondering what would happen if two black holes collided into each other in a head-on collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this model there's two black holes on a 3d Cartesian plane. Each black hole has a mass of 1 billion solar masses, and they're both going 0.9c. They're traveling on the z axis in opposite directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If both black holes are not spinning, one traveling positive z, the other in negative z, and collide at the origin in a perfect head-on collision what would be the result?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How long would a full merger take, and what would the resulting spin be? Would there be a large wave of gravitational energy, or some other emission?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't believe this to be a duplicate of &lt;a href=&quot;https://astronomy.stackexchange.com/questions/26465/would-ligo-detect-head-on-collision&quot;&gt;Would LIGO Detect Head-On Collision?&lt;/a&gt; as suggested by &lt;a href=&quot;https://astronomy.stackexchange.com/questions/32445/head-on-collision-of-two-black-holes#comment59714_32445&quot;&gt;@John Rennie&lt;/a&gt; or &lt;a href=&quot;https://astronomy.stackexchange.com/questions/32445/head-on-collision-of-two-black-holes#comment59717_32445&quot;&gt;@uhoh&lt;/a&gt; beucase I'm more asking what's the effect on the BH themselves and what type of emissions are involved. Not necessarily ones detectable by LIGO.&lt;/p&gt;&#xA;" OwnerUserId="28554" LastEditorUserId="28554" LastEditDate="2019-06-28T14:21:39.153" LastActivityDate="2019-06-28T14:21:39.153" Title="Head on collision of two black holes" Tags="&lt;black-hole&gt;&lt;supermassive-black-hole&gt;&lt;collision&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2019-06-28T21:32:56.960" ContentLicense="CC BY-SA 4.0" />
  <row Id="32448" PostTypeId="2" ParentId="32445" CreationDate="2019-06-27T15:05:44.463" Score="9" Body="&lt;p&gt;Well the spin is easy. Your system has zero angular momentum, so the spin will be zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there will also be no gravitational radiation. for the slightly technical reason that gravitational waves only come from changes to the quadrupole moment of a system and in your system the quadrupole moment is always zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So everything ends up in the final black hole which will have mass about 4.6 billion solar masses, coming from the mass of the two colliding holes, and their kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure exactly how long the merger would last, but it will be comparable to the Schwarzschild radii involved, divided by &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;, so around three hours (as seen by a distant observer).&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-06-27T15:05:44.463" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32449" PostTypeId="2" ParentId="32440" CreationDate="2019-06-27T17:13:03.667" Score="1" Body="&lt;p&gt;Yes they would.  In fact, you can interpret cosmological observations we now do as precisely this phenomenon, only backward-- we are looking from a less dilated place to more dilated places, and they appear to be moving slowly as a result.  For example, cosmologically distant supernovae seem to play out in slow motion as we watch them.  So to have a situation as in your question, we only would have needed a universe that is contracting instead of expanding, and we'd be seeing things playing out in fast motion at large distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note this general relativistic effect, which appears at order (v/c)^2 where v is a measure of the strength of gravity in velocity units (similar to an escape speed, for example), is not to be confused with a time-of-flight effect, which appears to order (v/c) for some actual v, and also can make things appear like they are moving fast or slow depending on the sign of v.  For example, if you drive toward a drive-in movie screen, the movie will appear to play out slightly faster, but this is just a time-of-flight effect and would happen for a sound concert as well.  It is also not to be confused with the special relativistic effect of time dilation, which always makes everything else outside of you appear to play out slower than it would have otherwise, regardless of whether it is moving toward you or away from you.  The general relativistic effect is a gravity effect, and cosmological effects can be interpreted similarly to gravity effects as the gravity of the universe as a while evolves.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2019-06-27T17:13:03.667" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32451" PostTypeId="2" ParentId="32440" CreationDate="2019-06-27T19:04:45.850" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;a Gravitational-Time dilated place to somewhere far which is not dilated&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Dilated&quot; or &quot;not dilated&quot; are not absolutes. It's always a relative thing. &quot;Dilated&quot; relative to me, but perhaps &quot;not dilated&quot; relative to you. There is no dilation that happens &quot;in general&quot;; it's always relative to another place/observer/reference.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Would those people look fast moving?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, everything coming out of that place would seem like it's &quot;speeding up time&quot; relative to you. If you're looking at them through a telescope, all light would be blue-shifted. If you're receiving a TV transmission, the frequency of the carrier would also be higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind, this time dilation is not an illusion. It's not just &quot;how you see things&quot;. Their flow of time, as measured by you, is indeed faster. This affects everything in that place, as measured by you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's also relative. Some other observer may be in a place where time flows differently from yours - they would look through the telescope and would see a different rate of change. And it's not an illusion for them either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time dilation is real. What's not real is a unique time, ticking the same for everyone - that's the actual illusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time is a local thing. (more precisely, the rate of ticking for a clock is a local thing)&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2019-06-27T23:28:14.080" LastActivityDate="2019-06-27T23:28:14.080" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32455" PostTypeId="2" ParentId="32443" CreationDate="2019-06-28T11:42:07.180" Score="10" Body="&lt;p&gt;There's an error in your source, but even if there weren't, it wouldn't mean that the Universe is a black hole (see below): The Schwarzschild radius of the observable Universe is &lt;em&gt;not&lt;/em&gt; equal to 13.7 Glyr. Wikipedia cites some random, non-refereed paper that uses the Hubble radius as the radius of the observable Universe, which is too small by a factor of &lt;span class=&quot;math-container&quot;&gt;$\gtrsim3$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The Schwarzschild radius of the Universe&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Although the age of the Universe is indeed ~13.8 Gyr, its radius is much larger than 13.8 Glyr, because it expands. In fact, the radius is &lt;span class=&quot;math-container&quot;&gt;$R \simeq 46.3\,\mathrm{Glyr}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mean density of the Universe is very close to the &lt;em&gt;critical&lt;/em&gt;&lt;span class=&quot;math-container&quot;&gt;$^\dagger$&lt;/span&gt; density &lt;span class=&quot;math-container&quot;&gt;$\rho_\mathrm{c} \simeq 8.6\times10^{-30}\,\mathrm{g}\,\mathrm{cm}^{-3}$&lt;/span&gt;.&#xA;Hence, the total mass (including &quot;normal&quot;, baryonic matter, dark matter, and dark energy) is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;M = \rho_\mathrm{c}V = \rho_\mathrm{c}\frac{4\pi}{3}R^3 \simeq 3.0\times10^{57}\,\mathrm{g},&#xA;$$&lt;/span&gt;&#xA;and the corresponding Schwarzschild radius is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;R_\mathrm{S} \equiv \frac{2GM}{c^2} \simeq 475\,\mathrm{Glyr}.&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The Universe is not a black hole&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Even worse! you might say. If our Universe is much smaller than its Schwarzschild radius, does that mean we live in a black hole?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, it doesn't. A black hole is a region in space where some mass is squeezed inside its Schwarzschild radius, but the Universe is not &quot;a region in space&quot;. There's no &quot;outside the Universe&quot;. If anything, you might call it a &lt;em&gt;white hole&lt;/em&gt;, the time-reversal of a black hole, in which case you could say that the singularity is not something that everything will fall into in the future, but rather something that everything came from in the past. You may call that singularity Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The error in the source&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;You will see questions like this many places on the internet. As I said above, they all assume the Hubble radius, &lt;span class=&quot;math-container&quot;&gt;$R_\mathrm{H} \equiv c/H_0$&lt;/span&gt; for the radius. But this radius is well within our observable Universe, and doesn't really bear any physical significance. In this case, the age in Gyr works out to be exactly equal to the radius in Glyr, by definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what does that tell us?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing, really. Except that our Universe is flat, i.e. has &lt;span class=&quot;math-container&quot;&gt;$\rho\simeq\rho_\mathrm{c}$&lt;/span&gt;, which we already knew and used in the calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, setting &lt;span class=&quot;math-container&quot;&gt;$R = R_\mathrm{H} \equiv c/H_0$&lt;/span&gt;,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{array}{rcl}&#xA;R_\mathrm{S} &amp;amp; \equiv &amp;amp; \frac{2GM}{c^2}\\&#xA;             &amp;amp; =      &amp;amp; \frac{2G}{c^2} \rho V \\&#xA;             &amp;amp; =      &amp;amp; \frac{2G}{c^2} \rho \frac{4\pi}{3}R^3 \\&#xA;             &amp;amp; =      &amp;amp; \frac{2G}{c^2} \rho \frac{4\pi}{3} \left(\!\frac{c}{H_0}\!\right)^3 \\&#xA;             &amp;amp; =      &amp;amp; \frac{8\pi G}{3 H_0^2} \rho \frac{c}{H_0} \\&#xA;             &amp;amp; =      &amp;amp; \frac{8\pi G}{3 H_0^2} \rho  R,&#xA;\end{array}&#xA;$$&lt;/span&gt;&#xA;so if &lt;span class=&quot;math-container&quot;&gt;$R=R_\mathrm{S}$&lt;/span&gt;, we have&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\rho = \frac{3H_0^2}{8\pi G},&#xA;$$&lt;/span&gt;&#xA;which is exactly the expression for the critical density you get from the Friedmann equation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^\dagger$&lt;/span&gt;&lt;sub&gt;The density that determines the global geometry of the Universe.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2019-06-28T12:58:22.027" LastActivityDate="2019-06-28T12:58:22.027" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32456" PostTypeId="1" CreationDate="2019-06-28T13:36:00.523" Score="4" ViewCount="333" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Shape_of_the_universe#Infinite_or_finite&quot; rel=&quot;nofollow noreferrer&quot;&gt;shape of the universe&lt;/a&gt;  is the question of whether the universe is flat, has a positive curvature or a negative one. More recently astronomers have compared radiation coming from extremely distant points of the observable universe and have concluded that it is &lt;a href=&quot;https://www.space.com/19027-universe-baby-picture-wmap.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;flat with a 0.4% margin of error,&lt;/a&gt; Based on articles I've read elsewhere this seems to have created the consensus that the universe is indeed flat. But isn't the universe also believed to be infinite in size? If it is really infinite then shouldn't such measurements be inconsequential as an infinite curved universe would still appear perfectly flat to a local observer?&lt;/p&gt;&#xA;" OwnerUserId="24410" LastEditorUserId="16685" LastEditDate="2019-06-28T18:42:34.200" LastActivityDate="2019-06-29T20:52:07.173" Title="Why do scientists assume they can measure the shape of the universe if it is also widely believed to be infinite?" Tags="&lt;cosmic-microwave-background&gt;&lt;infinite&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32457" PostTypeId="1" AcceptedAnswerId="32514" CreationDate="2019-06-28T15:20:03.217" Score="2" ViewCount="1145" Body="&lt;p&gt;I am interested in two aspects of this question. Imagine that Venus had a similar rotation speed and the same direction of rotation that we have here on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would change for Venus itself? I'm interested about changes in the core and mantle behaviour, a potential magnetosphere, changes in the atmosphere and tectonic plates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be different for the other planets near the Sun? I'm interested about changes in gravitational resonance and potential orbit changes.&lt;/p&gt;&#xA;" OwnerUserId="28544" LastEditorUserId="14094" LastEditDate="2019-07-02T06:17:29.473" LastActivityDate="2019-07-03T00:10:02.007" Title="What would change if Venus rotated at a speed and direction similar to Earth?" Tags="&lt;orbit&gt;&lt;rotation&gt;&lt;venus&gt;&lt;speed&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32458" PostTypeId="2" ParentId="32456" CreationDate="2019-06-28T18:39:10.377" Score="9" Body="&lt;p&gt;Yes, the universe is believed to be infinite in size. That's what you get if the curvature is zero or negative, assuming a simple topology. The curvature has to be positive for a finite universe, once again, assuming a simple topology, and no weird stuff like edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it's &lt;em&gt;possible&lt;/em&gt; that the universe has a very small positive curvature, so that it's finite, but it's so large that it looks flat to us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it's reasonable to assume that the observable universe is representative of the whole thing, and not just coincidentally a region of anomalous curvature. Of course, that's impossible to verify, but if the curvature were significantly greater outside our observable patch we'd expect the curvature to be a bit higher near the edges than it is in the middle of the patch, and we don't see that in the data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;em&gt;if&lt;/em&gt; the global curvature equals the maximum positive curvature consistent with the 0.4% margin of error of the WMAP, BOOMERanG, and Planck data, then the radius of curvature of the whole universe is (currently) around 150 times larger than the radius of the observable universe. And of course, in the future it will continue to expand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That figure comes from &lt;a href=&quot;https://medium.com/starts-with-a-bang/how-big-is-the-entire-universe-f3fdd468d3db&quot; rel=&quot;noreferrer&quot;&gt;How Big is the Entire Universe?&lt;/a&gt; by astrophysicist &lt;a href=&quot;https://en.wikipedia.org/wiki/Ethan_Siegel&quot; rel=&quot;noreferrer&quot;&gt;Ethan Siegel&lt;/a&gt;. That article has a great explanation of curvature, with lots of helpful diagrams.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="16685" LastEditDate="2019-06-28T18:45:56.720" LastActivityDate="2019-06-28T18:45:56.720" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32461" PostTypeId="1" AcceptedAnswerId="32468" CreationDate="2019-06-28T19:35:07.763" Score="-4" ViewCount="216" Body="&lt;p&gt;Does time dilation drop off like a gravity or does resemble a magnetic field model?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could a smaller body with a magnetic field have more time dilation then a larger body without and to what degree?&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2019-06-30T19:55:14.573" LastActivityDate="2019-07-02T06:14:27.347" Title="What is the time dilation less a magnetic field?" Tags="&lt;astrophysics&gt;&lt;space-time&gt;&lt;redshift&gt;&lt;magnetic-field&gt;" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="32462" PostTypeId="1" AcceptedAnswerId="32466" CreationDate="2019-06-28T20:35:07.017" Score="3" ViewCount="81" Body="&lt;p&gt;I am studying &lt;a href=&quot;https://en.wikipedia.org/wiki/Circumbinary_planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;circumbinary planets&lt;/a&gt; and I made a python model to calculate certain properties of the orbits, and when I calculate the inclination I get a sine wave. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not see a way to predict the shape of the curve in advance, despite it being a rather simple curve. Furthermore, when I plot the eccentricity of the orbit the frequencies of the curves do not align.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would anyone know why it oscillates at this rate and amplitude?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And a side question, would anyone know some literature on the subject of circumbinary planets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yZota.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yZota.png&quot; alt=&quot;The inclination curve&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/q3Ij8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q3Ij8.png&quot; alt=&quot;The eccentricity curve&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="28382" LastEditorUserId="7982" LastEditDate="2019-06-29T00:04:39.383" LastActivityDate="2019-06-29T02:38:24.437" Title="periodic inclination change circumbinary planet" Tags="&lt;exoplanet&gt;&lt;orbital-mechanics&gt;&lt;orbital-elements&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32463" PostTypeId="2" ParentId="32426" CreationDate="2019-06-28T21:44:05.417" Score="8" Body="&lt;p&gt;&lt;strong&gt;EDIT: I used the DE431 ephemeris to get the results below. However, if you visit HORIZONS (&lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;noreferrer&quot;&gt;https://ssd.jpl.nasa.gov/horizons.cgi&lt;/a&gt;) and create an ephemeris for Pluto (actually Pluto's barycenter, object 999), the results include this warning:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Horizons output log may report using DE431 as the source of target&#xA;  or center data, but it is nevertheless using DE433 as an override if&#xA;  Pluto is involved. This is to maintain ephemeris consistency for the&#xA;  New Horizons  spacecraft encounter at Pluto.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Therefore, my calculations may be inaccurate, above and beyond the normal inaccuracy of predicting planetary data in the past and future.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 2 times in the past 15,000 years and 5 times in the next 15,000 years where all 8 planets are in retrograde as observed from Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The table below lists the times when this happens:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The last column is the ephemeris time, the number of seconds since '2000-JAN-01 12:00:00 UTC', which NASA refers to as &quot;the epoch&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The first line in each pair shows the start time of when all 8 planets are in retrograde, and the final planet to enter retrograde to make this true.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The second line in each pair shows the end time when at least one planet is no longer in retrograde, and the planet whose retrograde period has ended.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&#xA;B.C. 12435-AUG-10 07:52 JUPITER STARTS RETROGRADE -455482152466.038696&#xA;B.C. 12435-AUG-24 17:12 MERCURY ENDS RETROGRADE -455480909251.797852&#xA;&#xA;B.C.  4139-MAR-23 15:02 MARS STARTS RETROGRADE -193692373029.630005&#xA;B.C.  4139-MAR-28 21:57 MERCURY ENDS RETROGRADE -193691916114.978455&#xA;&#xA;A.D.  7662-APR-27 19:33 MARS STARTS RETROGRADE 178685595220.195740&#xA;A.D.  7662-MAY-12 05:19 NEPTUNE ENDS RETROGRADE 178686840014.478882&#xA;&#xA;A.D.  8807-JAN-07 02:02 MERCURY STARTS RETROGRADE 214808680981.726929&#xA;A.D.  8807-JAN-20 01:05 JUPITER ENDS RETROGRADE 214809800771.851685&#xA;&#xA;A.D. 10575-FEB-14 21:34 VENUS STARTS RETROGRADE 270604748086.593933&#xA;A.D. 10575-FEB-24 18:53 NEPTUNE ENDS RETROGRADE 270605602468.037781&#xA;&#xA;A.D. 13679-NOV-06 16:03 VENUS STARTS RETROGRADE 368580427477.379639&#xA;A.D. 13679-NOV-19 01:05 MERCURY ENDS RETROGRADE 368581496795.683594&#xA;&#xA;A.D. 16364-MAR-11 19:25 MERCURY STARTS RETROGRADE 453290109945.561157&#xA;A.D. 16364-MAR-13 08:53 MARS ENDS RETROGRADE 453290244824.895996&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Notes and caveats:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I wrote &lt;a href=&quot;https://github.com/barrycarter/bcapps/tree/master/ASTRO/bc-retrograde.c&quot; rel=&quot;noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/tree/master/ASTRO/bc-retrograde.c&lt;/a&gt; to print out retrograde times for all planets.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The file &lt;a href=&quot;https://github.com/barrycarter/bcapps/tree/master/ASTRO/all-retrogrades.txt.bz2&quot; rel=&quot;noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/tree/master/ASTRO/all-retrogrades.txt.bz2&lt;/a&gt; shows start and end retrograde times for all 8 planets. NOTE: the lines indicating that planets leave retrograde at &quot;17191-MAR-01 00:00&quot; is incorrect-- that's simply the maximum time for which I could make calculations.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I wrote &lt;a href=&quot;https://github.com/barrycarter/bcapps/tree/master/ASTRO/bc-moving-backwards.pl&quot; rel=&quot;noreferrer&quot;&gt;https://github.com/barrycarter/bcapps/tree/master/ASTRO/bc-moving-backwards.pl&lt;/a&gt; to determine which planets were in retrograde at any given time, and used the results to find the times above where all 8 planets were in retrograde.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I spot checked my results against &lt;a href=&quot;https://www.calsky.com/&quot; rel=&quot;noreferrer&quot;&gt;https://www.calsky.com/&lt;/a&gt; -- note that I am computing &quot;ecliptic retrograde&quot;, when a planet's ecliptic longitude is decreasing, not &quot;equatorial retrograde&quot;, when a planet's right ascension is decreasing.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;As always, I invite people to double check my results, and remind everyone not to use my results for anything critical.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user21" LastEditorDisplayName="user21" LastEditDate="2019-06-29T18:35:10.907" LastActivityDate="2019-06-29T18:35:10.907" CommentCount="16" ContentLicense="CC BY-SA 4.0" />
  <row Id="32464" PostTypeId="2" ParentId="32456" CreationDate="2019-06-28T22:41:49.223" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If it is really infinite then shouldn't such measurements be inconsequential as an infinite curved universe would still appear perfectly flat to a local observer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, it wouldn't necessarily. It could, but it's not mandatory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine an infinite line that just takes a sharp turn at some point. It would definitely be &quot;curvy&quot; at that point, even though it's infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a parabola. It is infinite, but has a definite curvature in every point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same goes for our 3D universe.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2019-06-28T22:41:49.223" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32466" PostTypeId="2" ParentId="32462" CreationDate="2019-06-29T02:38:24.437" Score="1" Body="&lt;p&gt;The period of the stars' orbits is likely so much shorter than the planet's orbit that from the planet's perspective the pair looks like two donuts, or toroidal-shaped masses, possibly overlapping depending on the masses and sizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In orbital mechanics you can think of this roughly as a single star with a really huge J2 term (gravitational quadrupole moment).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to &lt;a href=&quot;https://en.wikipedia.org/wiki/Apsidal_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;apsidal&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Nodal_precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;nodal precession&lt;/a&gt; (which you would have seen) J2 is well known to induce oscillations in inclination. For example see &lt;a href=&quot;https://space.stackexchange.com/q/22266/12102&quot;&gt;J2 long-period perturbations in the inclination&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also from &lt;a href=&quot;http://help.agi.com/stk/index.htm#stk/vehSat_orbitProp_2bodyJ2J4.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Two-Body, J2 Perturbation &amp;amp; J4 Perturbation Propagators&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The solutions produced by the J2 Perturbation and J4 Perturbation propagators are approximate, based upon Keplerian mean elements. In general, forces on a satellite cause the Keplerian mean elements to drift over time (secular changes) and oscillate (usually with small amplitude). &lt;strong&gt;In particular, the J2 and J4 terms cause only periodic oscillations to semi-major axis, eccentricity and inclination, while producing drift in argument of perigee, right ascension and mean anomaly.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-06-29T02:38:24.437" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32467" PostTypeId="2" ParentId="32350" CreationDate="2019-06-29T03:42:41.340" Score="3" Body="&lt;p&gt;I used &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/0321595580&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Foundations of Astrophysics&lt;/em&gt;&lt;/a&gt; by Ryden and Peterson in my first college astronomy course, which focused on stars, the ISM, and galaxies. The textbook itself goes far beyond those topics, delving into cosmology, planetary science, and more, and we actually use it for another intro-level course for majors covering those topics. I think it could be used well as the text for any such course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I liked a couple of things about Ryden and Peterson:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The style is casual and accessible to a beginning astrophysics student. There's also humor tucked away in footnotes here and there; the book certainly doesn't try to be imposing - and it's not. I think that's important to have from time to time in an introductory text.&lt;/li&gt;&#xA;&lt;li&gt;It covers &lt;em&gt;everything&lt;/em&gt;, as I alluded to before. I do wish that the authors provided some resources for going further into particular subtopics, if the reader so desired, but then again, most students would have access to a professor/TA to go to for that.&lt;/li&gt;&#xA;&lt;li&gt;I actually agree with the point that the problems go &lt;em&gt;slightly&lt;/em&gt; beyond the scope of each chapter in some cases, but I also don't think that's a bad thing. The extra critical thinking helped me. I don't remember them making me want to bang my head against the wall as much as some other physics textbooks I could mention - maybe because Ryden and Peterson had a habit of breaking each problem into more manageable chunks, if I remember correctly.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I do think Ryden and Peterson satisfies the other requirements quite well; it doesn't use more than single-variable calculus (and even that only sparingly, I think), and it actually doesn't require much mechanics or electricity and magnetism, so in turns of prerequisites, you should be fine. One thing to note is that Ryden and Peterson doesn't include example problems done out in the same way many physics textbooks do - I don't know if that's a turn-off for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should add a disclaimer that my opinion might be slightly skewed because of the textbook used in my seminar this past semester, LeBlanc's &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/B005UQD0H4&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;An Introduction to Stellar Astrophysics&lt;/em&gt;&lt;/a&gt;. LeBlanc is comparatively drier, denser, and more specialized (on, of course, stellar astrophysics). I found it substantially less accessible, even though I had much more experience in the subject at that point. This, then, may have affected my hindsight view of Ryden and Peterson. Ironically, we did use Carroll &amp;amp; Ostlie as supplementary reading (as well as Ryden and Peterson) on certain sections, which was much more helpful, and I daresay a refreshing change of pace.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2019-06-29T03:42:41.340" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32468" PostTypeId="2" ParentId="32461" CreationDate="2019-06-29T08:19:53.900" Score="2" Body="&lt;p&gt;There is a magnetic field whenever there is a moving charge, and since &quot;moving&quot; is a relative term, there is at least the possibility of some observer seeing a magnetic field whenever you have any charged object at all. However, there are thought experiments which show time dilation without involving any charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example in special relativity would be a beam of Z&lt;sup&gt;0&lt;/sup&gt; bosons (which are, as far as we know uncharged fundamental particles). If the beam was moving fast relative to some observer, then they would appear to decay less quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A GR example is an uncharged black hole (possible in theory, but unlikely in reality – any black hole will have a bit of charge, just by chance). A clock (our bunch of Z&lt;sup&gt;0&lt;/sup&gt; bosons for example) close to the black hole will seem to run slower according to a distant observer at rest relative to the black hole.  The exact degree of dilation will be affected by the charge on the black hole, but it would have to be very highly charged for the difference to be much.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="14094" LastEditDate="2019-07-02T06:14:27.347" LastActivityDate="2019-07-02T06:14:27.347" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32469" PostTypeId="2" ParentId="32456" CreationDate="2019-06-29T10:29:16.453" Score="-4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why do scientists assume they can measure the shape of the universe if it is also widely believed to be infinite?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because not everybody thinks it's infinite. Take a look at &lt;a href=&quot;http://www.physicsoftheuniverse.com/topics_bigbang_timeline.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;the timeline of the Big Bang&lt;/a&gt; written by Luke Mastin in 2009. He said &lt;em&gt;“The linear dimensions of the early universe increases during this period of a tiny fraction of a second by a factor of at least 10&lt;span class=&quot;math-container&quot;&gt;$^{26}$&lt;/span&gt; to around 10 centimetres (about the size of a grapefruit)”&lt;/em&gt;. You can find John Gribbin talking about a universe &lt;a href=&quot;https://books.google.co.uk/books?id=metLkRzY618C&amp;amp;pg=PT82&amp;amp;lpg=PT82&amp;amp;dq=gribbin+universe+biography+grapefruit&amp;amp;source=bl&amp;amp;ots=hm55-Kce2P&amp;amp;sig=lyLObWshya0PwbWUoxr1TxnZt54&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwiDl_vx95fVAhWMblAKHexFCogQ6AEINTAD#v=onepage&amp;amp;q=grapefruit&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;the size of a grapefruit&lt;/a&gt; in his 2008 book &lt;em&gt;The Universe: A Biography&lt;/em&gt;. You can find Marcus Chown talking about a universe the size of a grapefruit in &lt;a href=&quot;http://www.physicsoftheuniverse.com/topics_bigbang_timeline.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;his review&lt;/a&gt; of Alan Guth’s 1997 book &lt;em&gt;The Inflationary Universe&lt;/em&gt;. You can also find Jeremiah Ostriker and Paul Steinhardt talking about a universe the size of a grapefruit in &lt;a href=&quot;http://sci-hub.tw/10.1038/scientificamerican0101-46&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Quintessential Universe&lt;/a&gt; in SciAm in 2002. They weren't talking about the observable universe. They were talking about the whole universe. And if it was the size of a grapefruit 13.8 billion years ago, it can't be infinite now. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The shape of the universe is the question of whether the universe is flat, has a positive curvature or a negative one. More recently astronomers have compared radiation coming from extremely distant points of the observable universe and have concluded that it is flat with a 0.4% margin of error.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's right. Two out of three options were always going to be wrong. IMHO if you've read the Einstein digital papers it's obvious that the universe was going to be flat.  Yes, Einstein came up the idea of a closed curved universe, see his 2nd February &lt;a href=&quot;https://einsteinpapers.press.princeton.edu/vol8-trans/309?highlightText=outlandish&quot; rel=&quot;nofollow noreferrer&quot;&gt;postcard&lt;/a&gt; to Willem de Sitter. But by 1932 he and de Sitter had dropped the idea in favour of flat space in their &lt;a href=&quot;https://en.wikipedia.org/w/index.php?title=Einstein%E2%80%93de_Sitter_universe&amp;amp;oldid=892851453&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein-de Sitter universe&lt;/a&gt;. See John Baez and Emory Bunn’s &lt;a href=&quot;http://math.ucr.edu/home/baez/einstein/node2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;preliminaries&lt;/a&gt; article dating from 2006: &lt;em&gt;“Similarly, in general relativity gravity is not really a ‘force’, but just a manifestation of the curvature of spacetime. Note: not the curvature of space, but of spacetime. The distinction is crucial”.&lt;/em&gt; Curved space is not the same thing as curved spacetime. Unfortunately &lt;a href=&quot;https://medium.com/starts-with-a-bang/how-big-is-the-entire-universe-f3fdd468d3db&quot; rel=&quot;nofollow noreferrer&quot;&gt;some people&lt;/a&gt; confuse the two. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Based on articles I've read elsewhere this seems to have created the consensus that the universe is indeed flat. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think that consensus is correct. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But isn't the universe also believed to be infinite in size? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A lot of people claim that a flat universe must be an infinite universe. But it's a non-sequitur. Going back to Einstein again, he thought of space as a thing. That’s why in his 1929 essay on the &lt;a href=&quot;http://www.rain.org/~karpeles/einsteindis.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;history of field theory&lt;/a&gt;, he described a field as a state of space. In his &lt;a href=&quot;http://sci-hub.tw/10.1126/science.71.1850.608&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nottingham lecture&lt;/a&gt; in 1930 he said space &lt;em&gt;“remains the sole medium of reality”&lt;/em&gt;. So I think it's reasonable to envisage the universe as a sphere of space, with no space beyond it. Maybe light rays would undergo something akin to total internal reflection. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If it is really infinite then shouldn't such measurements be inconsequential as an infinite curved universe would still appear perfectly flat to a local observer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's what some people say. But they also say things like &lt;em&gt;&quot;the universe was infinite at the time of the Big Bang and it's even more infinite now&quot;&lt;/em&gt;. I'm not fond of that because it doesn't seem to fit in with Big Bang cosmology, and I just don't see how an infinite universe can expand. &lt;/p&gt;&#xA;" OwnerUserId="9772" LastEditorUserId="9772" LastEditDate="2019-06-29T10:41:25.347" LastActivityDate="2019-06-29T10:41:25.347" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32470" PostTypeId="1" AcceptedAnswerId="32475" CreationDate="2019-06-29T10:58:46.357" Score="4" ViewCount="282" Body="&lt;p&gt;We have just passed one of the solstices and are approaching aphelion.  The two events are close but not simultaneous.  There is no very obvious (to me) reason why the two should coincide or be close.  Is there a less obvious reason why they are close?  If so, is the gap closing, growing, or remaining approximately constant?&lt;/p&gt;&#xA;" OwnerUserId="21104" LastActivityDate="2019-06-30T20:40:49.040" Title="Aphelion and the solstice" Tags="&lt;orbit&gt;&lt;solstice&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32471" PostTypeId="2" ParentId="32437" CreationDate="2019-06-29T11:23:58.307" Score="0" Body="&lt;p&gt;I've accepted the other answer. Then I ran across this article so I'm posting this as a supplemental answer.&lt;/p&gt;&#xA;&lt;p&gt;It turns out that at the bottom of the Phys.org article there was a link to another, older article &lt;a href=&quot;https://phys.org/news/2018-06-image-asteroids.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image: Dance of the asteroids&lt;/a&gt; which did contain an explanation, as well as another image:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The asteroid trails look curved due to an observational effect called parallax. As Hubble orbits around Earth, an asteroid will appear to move along an arc with respect to the vastly more distant background stars and galaxies. The motion of Earth around the Sun, and the motion of the asteroids along their orbits, are other contributing factors to the apparent skewing of asteroid paths.&lt;/p&gt;&#xA;&lt;p&gt;All the asteroids were found manually, the majority by &amp;quot;blinking&amp;quot; consecutive exposures to capture apparent asteroid motion. Astronomers found a unique asteroid for every 10 to 20 hours of exposure time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Note also:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;These asteroid trails should not be confused with the mysterious-looking arcs of blue light that are actually distorted images of distant galaxies behind the cluster.&lt;/strong&gt; Many of these far-flung galaxies are too faint for Hubble to see directly. Instead, in a dramatic example of &amp;quot;gravitational lensing,&amp;quot; the cluster functions as a natural telescope, warping space and affecting light traveling through the cluster toward Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Qyzw9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qyzw9.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Credit: NASA, ESA, and B. Sunnquist and J. Mack (STScI) Acknowledgment: NASA, ESA, and J. Lotz (STScI) and the HFF Team&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-06-30T13:47:15.963" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32472" PostTypeId="2" ParentId="32457" CreationDate="2019-06-29T12:24:27.900" Score="-1" Body="&lt;p&gt;I don't think anything substantial would have changed. Four billion years ago, Venus was similar to Earth but hotter. It had seas, and like Earth, a very substantial atmosphere composed mainly of CO2. The rotation rate was probably not quite the same as it is today, but for your scenario let's suppose it was once every 24 hours. Owing to its greater proximity to the sun and the greenhouse effect of a CO2 atmosphere with a strong flavouring of methane, regardless of a 24 hour rotation rate the seas would have eventually evaporated away, increasing the greenhouse effect. Meanwhile, outgassing from volcanoes would have added more CO2 and sulphur dioxide to the toxic mix, forming sulphurous acid  in combination with the water vapour. At a guess, I would think this took about a billion years, perhaps slightly faster than an earth-like rotation would have done. Venus's 8-month rotation rate would have meant that one side of the planet was exposed far longer to the sun's rays than was the case on Earth, so it also had very long nights. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation today is that despite the very long days and very long nights, the atmosphere is so thick that there is no significant difference in temperature between the night side of the planet and the day side. Had it been rotating every 24 hours, there might have been a temperature difference between night and day for the first few million years, but this would have decreased as the greenhouse atmosphere built up to become a more effective blanket. In other words, had Earth been 30 million miles closer to the sun, the same sort of thing would have happened here, so the answer to your question is that a more rapid rate of rotation wouldn't have made much difference. &lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-06-29T12:24:27.900" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32473" PostTypeId="2" ParentId="32087" CreationDate="2019-06-29T15:38:32.343" Score="3" Body="&lt;p&gt;The simplest case is for a rocket that does not fire its engine when it approaches the black hole from afar, and hence follows a &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_geodesics&quot; rel=&quot;nofollow noreferrer&quot;&gt;geodesic trajectory&lt;/a&gt;. In this case the closest you can get is &lt;span class=&quot;math-container&quot;&gt;$3r_s$&lt;/span&gt;, but you can make an arbitrary number of turns around the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.physics.ucc.ie/apeer/PY4112/Sch.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;The reason for this&lt;/a&gt; is that the counterpart of a hyperbolic orbit corresponds to moving in the effective potential &lt;span class=&quot;math-container&quot;&gt;$V(r)=-GMm/r+L^2/2\mu r^2 - G(M+m)L^2/c^2\mu r^3$&lt;/span&gt; (this is the Wikipedia version of the formula; different authors use different measures). For a given energy and angular momentum this means that either you can come in from infinity and fall into the black hole (for low energy and angular momentum) or the potential has a maximum greater than your energy and you will turn back. The closest you can get the maximum to go to the hole is &lt;span class=&quot;math-container&quot;&gt;$r=3r_s$&lt;/span&gt; (as your angular momentum increases). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the angular velocity behaves as &lt;span class=&quot;math-container&quot;&gt;$d\theta / d\tau = (L/\mu)/r^2$&lt;/span&gt;: you will swirl around a lot as you get close to the black hole. Since a carefully selected trajectory will have energy and angular momentum so that you turn around almost exactly at the potential maximum where the derivative &lt;span class=&quot;math-container&quot;&gt;$V'(r)\approx 0$&lt;/span&gt; the proper time spent there will be long and you can get as many turns as you want. The orbital frequency as seen from the outside is 2200 Hz &lt;span class=&quot;math-container&quot;&gt;$(M_\odot/M)$&lt;/span&gt; (!) (&lt;a href=&quot;https://www.astro.umd.edu/~miller/teaching/Brazil/lecture02.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that all of this is freefall: the only thing to worry about is tidal forces, but that is not much of an issue near a SMBH. The velocity needed is of course in principle zero at infinity if you come in along a parabolic orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The second question is if one can &quot;slingshot&quot;&lt;/strong&gt;. I assume this refers to an &lt;a href=&quot;https://en.wikipedia.org/wiki/Oberth_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oberth manoeuvre&lt;/a&gt; rather than a gravitational slingshot, which is unpowered. Gravitational slingshots from black holes can give you any change in angle, but the velocity change is just like the classical case. Rotation does not affect this, except by making the angular shift weirder. Note that for &lt;a href=&quot;https://en.wikipedia.org/wiki/Innermost_stable_circular_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;prograde orbits around an extremal spinning black hole the ISCO approaches the horizon radius&lt;/a&gt;, so the above unpowered flybys can in principle get right up  to the horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the powered manoeuvre I don't have a full answer; I would actually like to see it answered (see &lt;a href=&quot;https://physics.stackexchange.com/questions/321835/limit-to-the-oberth-effect-in-a-supermassive-black-hole-flyby&quot;&gt;this question and my half-answer to it&lt;/a&gt;). The basic story looks like that for velocity-at-infinity increasing manoeuvres you should not get closer than &lt;span class=&quot;math-container&quot;&gt;$3r_s$&lt;/span&gt;. But for a sightseeing trip close to the horizon spending some of your fuel (and mass) ought to work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you need to do is to fire the engines powerfully enough to change the sign of your radial velocity &lt;em&gt;and&lt;/em&gt; retain more total kinetic energy than the potential maximum. The potential maximum is due to angular momentum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the freefall radial trajectory with no angular momentum you will have &lt;span class=&quot;math-container&quot;&gt;$(dr/d\tau)^2=(E/mc^2)^2-c^2(1-r_s/r)$&lt;/span&gt;, so if you start with zero velocity at &lt;span class=&quot;math-container&quot;&gt;$r=\infty$&lt;/span&gt; (hence &lt;span class=&quot;math-container&quot;&gt;$E=mc^2$&lt;/span&gt;) the inward radial velocity will be &lt;span class=&quot;math-container&quot;&gt;$dr/d\tau = c\sqrt{r_s/r}$&lt;/span&gt;. So you need &lt;span class=&quot;math-container&quot;&gt;$\Delta v =2c\sqrt{r_s/r}$&lt;/span&gt; to escape again this way. We can rearrange the formula into &lt;span class=&quot;math-container&quot;&gt;$$r_{escape}= \frac{r_s}{\left(\Delta v / 2c\right)^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case &lt;span class=&quot;math-container&quot;&gt;$m_0/m_1=100$&lt;/span&gt; and I assume &lt;span class=&quot;math-container&quot;&gt;$I_{sp}\approx 0.6c$&lt;/span&gt;,  &lt;span class=&quot;math-container&quot;&gt;$\Delta v = c \tanh \left( \frac{I_{sp}}{c}\log\left(\frac{m_0}{m_1}\right)\right) = 0.9921c$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$r_{escape}=4.0642 r_s $&lt;/span&gt;. That demonstrates that the purely radial approach is not a good choice; adding a bit of lateral thrust would allow a closer approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have the full answer for the lateral case. If you come in along a flyby orbit that approaches the ISCO you can slow down a bit so you now are on an inspiralling trajectory. To escape you need to pay roughly twice the potential difference between your final radial distance orbit and the maxmimum of the new potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, rotating black holes allow you to use the &lt;a href=&quot;https://www.nature.com/articles/physci229177a0.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Penrose process&lt;/a&gt; to gain a lot of angular momentum by dumping reaction mass in the right direction. However, the efficiency is proportional to the total momentum of the reaction mass and is rather negligible for non-relativistic rockets. Hence rockets not powered by antimatter will likely not benefit much from this possibility.&lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2019-06-29T15:38:32.343" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32475" PostTypeId="2" ParentId="32470" CreationDate="2019-06-29T17:19:57.513" Score="5" Body="&lt;p&gt;The two coincided about 800 years ago. The December solstice and perihelion date &lt;a href=&quot;https://www.timeanddate.com/astronomy/perihelion-aphelion-solstice.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;coincided in 1246&lt;/a&gt;, 773 years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many different concepts of what qualifies as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Year#Astronomical_years&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;year&quot;&lt;/a&gt;. Three of them are the sidereal, tropical, and anomalistic years. The sidereal year measures how long it takes for the Earth to complete one orbit about the Sun with respect to the remote stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tropical year measures how long it takes for the Earth to pass from one March equinox to the next. The tropical year is about 20 minutes shorter than the sidereal year due to the slow changes in the Earth's rotation axis caused by the Moon and the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The anomalistic year measures how long it takes for the Earth to pass from one perihelion to the next (or one aphelion to the next). The anomalistic year is about 7.8 minutes longer than the sidereal year due to gravitational perturbations from the other planets, with Jupiter and Venus dominating over all the others. The tropical year is thus shorter than the anomalistic year by about 28 minutes. That the tropical year and anomalistic year have slightly different frequencies means that there is a very slow &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/Sound/beat.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;beat frequency&lt;/a&gt; with a beat period of about 26000 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Human civilization arose about half of this period ago (about 12000 years ago) when the last glaciation (&quot;ice age&quot;) ended. At that time, the anomalistic and tropical years were almost 180 degrees out of phase with respect to where they are now. The end of a glaciation typically occurs when northern hemisphere summers coincide with perihelion. That makes summers in far northern latitudes short in duration but rather hot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the key proxies regarding whether glaciations start or end is the mean July temperature at far northern latitudes. When perihelion occurs a few weeks after the June solstice, the short but hot summers that result makes snow from the previous winter melt rather than build up. The melting that results from such short but hot summers triggers a positive feedback that makes the end of a glaciation very rapid, in a geological context of &quot;rapid&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;To clarify things,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So, there is no particular reason in orbital mechanics; it just happens that we live in an era when they are close together?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Exactly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or maybe not. The end of the last glaciation, which occurred about 12000 years ago, massively spurred human development. The cycle then was almost exactly opposite what it is now. It took about 10000 years for humanity to develop to a somewhat advanced level (e.g., Roman Empire level), and then only another couple of thousand to develop to the stage where we are now. Whether humanity can survive for another 13000 years without wiping itself out will determine whether this was happenstance.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2019-06-30T20:40:49.040" LastActivityDate="2019-06-30T20:40:49.040" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="32476" PostTypeId="1" AcceptedAnswerId="32477" CreationDate="2019-06-29T17:34:14.777" Score="-5" ViewCount="66" Body="&lt;p&gt;A magnetic fields generally fall off as 1 / r^3 rather than 1 / r^2 for gravity.&#xA;How does time dilation fall off from a large body?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where would the gravity and magnetic field line be in this chart provided from comments?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pnL6c.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pnL6c.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15415" LastEditorUserId="15415" LastEditDate="2019-06-29T18:57:53.643" LastActivityDate="2019-06-29T18:57:53.643" Title="How does time dilation from a large body fall off?" Tags="&lt;black-hole&gt;&lt;planet&gt;&lt;distances&gt;&lt;time-dilation&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2019-07-01T07:47:19.917" ContentLicense="CC BY-SA 4.0" />
  <row Id="32477" PostTypeId="2" ParentId="32476" CreationDate="2019-06-29T17:43:56.103" Score="3" Body="&lt;p&gt;The formula is given on &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_time_dilation#Outside_a_non-rotating_sphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$t_{0}=t_{f}{\sqrt {1-{\frac {2GM}{rc^{2}}}}}=t_{f}{\sqrt {1-{\frac {r_{s}}{r}}}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here &lt;span class=&quot;math-container&quot;&gt;$t_0$&lt;/span&gt; is the time measured by clock distance &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; from an object whose Schwarzschild radius is &lt;span class=&quot;math-container&quot;&gt;$r_s$&lt;/span&gt; as calculated by a distant observer at rest whose clock measures &lt;span class=&quot;math-container&quot;&gt;$t_f$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is however, dependent on how the distant observer decides which ticks of your clock occur at the same time as which ticks of theirs, for which there are a number of perfectly reasonably choices, each of which gives a distinct answer. This formula is for what are called Schwarzschild coordinates, but there are other coordinate frames. Each one forces the distant observer to do a different calculation based on when they receive light signals from the close-in clocks ticks, in order to decide  at what time the tick happened.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="21928" LastEditDate="2019-06-29T17:55:32.350" LastActivityDate="2019-06-29T17:55:32.350" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32479" PostTypeId="1" CreationDate="2019-06-29T19:49:26.170" Score="3" ViewCount="95" Body="&lt;p&gt;Can the eccentricity of a planet's orbit be found just from the analemma of its star? We can draw an analemma given the axial tilt and eccentricity, but how do we find the reverse?&lt;br&gt;&#xA;Note: I already know that the highest and lowest points indicate the amount of axial tilt&lt;/p&gt;&#xA;" OwnerUserId="16506" LastActivityDate="2019-06-29T19:49:26.170" Title="Eccentricity from Analemma" Tags="&lt;coordinate&gt;&lt;data-analysis&gt;&lt;celestial-mechanics&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32484" PostTypeId="1" CreationDate="2019-06-29T22:02:36.110" Score="6" ViewCount="427" Body="&lt;p&gt;If the Martian polar ice caps melted, given a surface temperature and pressure that allow for liquid water, how large would the oceans be? Would they match their &lt;a href=&quot;https://en.wikipedia.org/wiki/Mars_ocean_hypothesis&quot; rel=&quot;noreferrer&quot;&gt;ancient glory&lt;/a&gt;, or would they be smaller?&lt;/p&gt;&#xA;" OwnerUserId="19304" LastActivityDate="2019-06-30T20:56:51.327" Title="If the Mars polar caps were melted, how deep would the oceans be?" Tags="&lt;mars&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32485" PostTypeId="2" ParentId="32484" CreationDate="2019-06-29T23:58:48.190" Score="9" Body="&lt;p&gt;The oceans would be smaller because Mars has lost most of its water. From Wikipedia's article on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Martian_polar_ice_caps&quot; rel=&quot;noreferrer&quot;&gt;Martian polar ice caps&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Evidence that Mars once had enough water to create a global ocean at&#xA;  least 137 m deep has been obtained from measurement of the &lt;span class=&quot;math-container&quot;&gt;$HDO$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$H_2O$&lt;/span&gt;&#xA;  ratio over the north polar cap. In March 2015, a team of scientists&#xA;  published results showing that the polar cap ice is about eight times&#xA;  as enriched with deuterium, heavy hydrogen, as water in Earth's&#xA;  oceans. This means that Mars has lost a volume of water 6.5 times as&#xA;  large as that stored in today's polar caps. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The water for a time may have formed an ocean in the&#xA;  low-lying Vastitas Borealis and adjacent lowlands&#xA;  (Acidalia, Arcadia and Utopiaplanitiae). Had the water ever all been&#xA;  liquid and on the surface, it would have covered 20% of the planet and&#xA;  in places would have been almost a mile deep.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So if all the water melted (and the atmospheric pressure allowed liquid water) the ocean would cover roughly 6% of the planet, and be roughly half as deep as the original ocean. This crude estimate totally ignores the topography of the Martian surface. In the comments, David Hammen writes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given the marked unevenness of Mars's surface, if both of its ice cap melted completely the resulting water would cover a fraction of its northern hemisphere, plus a few depressions such as the Hellas Planitia (8 kilometers deep!), but to a depth much greater than 137 meters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Heavy water is slightly heavier than normal water, so it evaporates slower and has a higher boiling point. So in a mixture of normal and heavy water the normal water evaporates at a slightly faster rate, which increases the relative concentration in the unevaporated water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That same article mentions that the total volume of both Martian polar caps is around 3.2 million cubic kilometres. Most of that ice is water ice, but some of it is frozen carbon dioxide (aka &quot;Dry ice&quot;). The article doesn't state the proportions, but it says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;During each year on Mars as much as a third of Mars' thin carbon&#xA;  dioxide (&lt;span class=&quot;math-container&quot;&gt;$CO_2$&lt;/span&gt;) atmosphere &quot;freezes out&quot; during the winter in the&#xA;  northern and southern hemispheres.&lt;br&gt;&#xA;  [...]&lt;br&gt;&#xA;  Analysis of data showed that if these deposits were all changed&#xA;  into gas, the atmospheric pressure on Mars would double.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that not all of Mars's water is in its polar caps, some may be frozen underground, especially near the poles, and some underground water may even be in a liquid state, due to a high salt concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further details, please see the linked article.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="16685" LastEditDate="2019-06-30T20:56:51.327" LastActivityDate="2019-06-30T20:56:51.327" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32487" PostTypeId="1" CreationDate="2019-06-30T10:39:02.787" Score="-2" ViewCount="149" Body="&lt;p&gt;Can we bend two points of spacetime in a way  we would create a wormhole ?Why havent we tried it yet?How much energy would we need to do it?&lt;/p&gt;&#xA;" OwnerUserId="28503" LastActivityDate="2019-07-02T13:34:59.577" Title="Wormholes man made are real?" Tags="&lt;gravity&gt;&lt;wormhole&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2019-07-02T06:20:52.987" ContentLicense="CC BY-SA 4.0" />
  <row Id="32488" PostTypeId="2" ParentId="32487" CreationDate="2019-06-30T13:37:33.197" Score="5" Body="&lt;p&gt;Short answer: We haven't because there is no imaginable way to do it with current technology nor with any vaguely plausible extension of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two fundamental problems and some technical ones:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space is stiff. You can measure the stiffness of a material by measuring the energy that it takes to bend it.  Applying that same approach to measuring the &quot;stiffness&quot; of space, you find that it is far, far stiffer than the strongest steel. The energies that must be manipulated to bend space to produce a macroscopic wormhole are colossal. They far exceed what we can produce or might plausibly produce.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, wormholes are unstable. &lt;em&gt;Very&lt;/em&gt; unstable. Even if a wormhole could somehow be created, it would snap shut faster than light could traverse it. There are theoretical ways to stabilize a wormhole, but they require the use of &quot;exotic matter&quot; -- matter with a very high negative mass -- and exotic matter has never been observed and we have no slightest clue how it might be produced. (The only reason that it's even considered is that our current theories don't actually say that exotic matter is impossible.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technical issues include the problem that the energies involved in warping space enough to produce a macroscopic wormhole would themselves produce gravitational effect which would probably destroy Earth, so it would likely need to be done in deep space. There is also theoretical work which suggests that either (a) wormholes can produce a quantum feedback that would destroy the universe or (b) wormholes are always a slower route to the other end than moving through empty space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: If we ever develop a good theory of quantum gravity, this may all change. But today that's speculation piled on speculation.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastEditorUserId="23115" LastEditDate="2019-07-02T13:34:59.577" LastActivityDate="2019-07-02T13:34:59.577" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32493" PostTypeId="1" CreationDate="2019-06-30T23:20:47.767" Score="3" ViewCount="85" Body="&lt;p&gt;Urry and Padovani &lt;a href=&quot;http://adsabs.harvard.edu/abs/1995PASP..107..803U&quot; rel=&quot;nofollow noreferrer&quot;&gt;1995&lt;/a&gt; proposed a unified model for Active Galactic Nuclei, which has since been recognised as a good model for AGN. If AGN are randomly oriented, then I'd expect there to be more obscured AGN than unobscured, because there would be more orientations that would make us look through the sides, than along the axis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NXFXL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NXFXL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the case? In other words, how many obscured-type AGNs (Seyfert 1 or 2 Galaxies, radio galaxies, radio-quiet quasars, NLRs) are there, compared to unobscured AGN (BLRs, blazars, radio-loud quasars)?&lt;/p&gt;&#xA;" OwnerUserId="17535" LastEditorUserId="37191" LastEditDate="2021-04-15T14:17:14.820" LastActivityDate="2021-04-15T14:17:14.820" Title="Number of obscured AGN compared to unobscured AGN" Tags="&lt;galaxy&gt;&lt;quasars&gt;&lt;active-galaxy&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32494" PostTypeId="1" AcceptedAnswerId="32517" CreationDate="2019-07-01T00:42:40.470" Score="2" ViewCount="76" Body="&lt;p&gt;Where might I find a detailed table or figure containing the absorption lines and their corresponding wavelengths for each type of main sequence star? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to get the chemical abundance of some elements on a G0V spectrum but I don't know which lines I'm looking at.&lt;/p&gt;&#xA;" OwnerUserId="28572" LastEditorUserId="7982" LastEditDate="2019-07-01T05:23:24.547" LastActivityDate="2019-07-03T01:26:59.887" Title="Examples of absorption lines for main sequence stars" Tags="&lt;spectroscopy&gt;&lt;spectra&gt;&lt;spectral-type&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32496" PostTypeId="2" ParentId="10450" CreationDate="2019-07-01T05:48:48.893" Score="2" Body="&lt;p&gt;The very short answer is &quot;no one knows&quot;. The available evidence very strongly suggests that the Universe 13.8 billion years ago was hot, dense and expanding and the further back we look, the more indirect the clues we get are, but the hotter and denser it seems to have been.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are multiple theories for what happens as we follow this process back and back, but theories is all they are. Scientists work to make testable predictions that would follow from some of them but not others, and allow us to whittle down the pool, but it's hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that said, at the broadest level there seem to be two common current views as to how to address the question &quot;what happened before the big bang&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is to argue that it is simply a bad question: time, like space, is &quot;curved&quot; by the presence of a lot of mass or energy and the question may be analagous to asking &quot;what is North of the North pole&quot;. There simnply is no &quot;before&quot; direction from &quot;the Big Bang&quot;, or to put it another way, it is the nature of this universe to have a time dimension which only goes a finite distance in the &quot;past&quot; direction. This, very loosely, is the &quot;no boundary&quot; approach of Hawking and others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other view, which has names like &quot;eternal inflation&quot; or &quot;cosmic inflation&quot; is that the original state of the universe was a smooth and fairly featureless, but extremely rapidly expanding (exponentially rapidly) one, which was only meta-stable. Part or all of this universe, spontaneously &quot;changed phase&quot; to produce normal space and time, and as a result of the energy released by the phase change, was filled with very hot, very dense, matter and energy, still expanding, but no longer exponentially. Indeed, gravity then started to slow the expansion.  In some theories just a small region would undergo this phase change, in others it would be the whole universe. Either way, all of the universe we see comes from a small region of the inflating universe.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastEditorUserId="21928" LastEditDate="2019-07-01T07:28:30.947" LastActivityDate="2019-07-01T07:28:30.947" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32500" PostTypeId="2" ParentId="32199" CreationDate="2019-07-01T12:01:24.697" Score="3" Body="&lt;p&gt;I think the answer is that it's primarily a matter of association on the sky, though there are some (weak) additional constraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing to note is that there are &lt;em&gt;three&lt;/em&gt; observations that are associated on the sky: the X-ray emission (which has been known about since the mid 1990s), the newly detected 140 MHz LOFAR emission, and evidence for a &lt;a href=&quot;https://en.wikipedia.org/wiki/Sunyaev%E2%80%93Zeldovich_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sunyaev-Zel'dovich effect&lt;/a&gt; in the same region, from Planck data (e.g., &lt;a href=&quot;https://www.aanda.org/articles/aa/abs/2018/01/aa31699-17/aa31699-17.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bonjean et al. 2018&lt;/a&gt;). From the caption to Figure 3 in the new paper: &quot;The Planck data show a bridge of gas between the pair of galaxy clusters Abell 0399 – Abell 0401, in the same location as the LOFAR ridge.&quot; and from the main text: &quot;The radio ridge is located along the filament of gas connecting Abell 0399 and Abell 0401 detected with Planck (9, 10).&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The &quot;Planck &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;-parameter&quot;, also known as the &quot;Compton parameter&quot;, refers to the measurement of the Sunyaev-Zel'dovich effect due to inverse Compton scattering of cosmic background radiation photons, derived from Planck satellite data. I believe it's proportional to the line-of-sight integral of the gas pressure.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Suzaku X-ray spectrum of this region (e.g., Figure 2 of &lt;a href=&quot;https://academic.oup.com/pasj/article/60/sp1/S343/1461794&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fujita et al. 2008&lt;/a&gt;; see below) does have some spectral features, including the Fe K emission line. Due to the low spectral resolution, these features don't constrain the redshift of the X-ray-emitting gas very strongly (and I think Fujita et al. fixed the redshift at the mean of the two clusters in their analysis), but it would probably be noticeable if it differed by more than &lt;span class=&quot;math-container&quot;&gt;$\sim 0.1$&lt;/span&gt; in redshift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BR3m7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BR3m7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Figure 2 of Fujita et al. (2008): Suzaku X-ray spectrum of part of the &quot;bridge&quot; region between the two clusters. Black is observed data, red is the combined model, including Galactic and extragalactic background (dashed blue), and emission from the inter-cluster gas (green).&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2019-07-01T12:01:24.697" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32502" PostTypeId="2" ParentId="2302" CreationDate="2019-07-01T12:47:15.343" Score="1" Body="&lt;p&gt;The gravitational pull of the Sun is so big, that it fuses hydrogen to helium-2. Helium-2 is unstable and decays to H-2 through the weak interaction. This releases energy in form of heat. Now again the gravitational pull of the sun is so big that it fuses H-2 to He-4 through the strong interaction. This releases energy in form of heat. &lt;/p&gt;&#xA;" OwnerUserId="28503" LastEditorUserId="1891" LastEditDate="2019-07-01T13:16:46.033" LastActivityDate="2019-07-01T13:16:46.033" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32503" PostTypeId="1" CreationDate="2019-07-01T14:28:49.860" Score="0" ViewCount="906" Body="&lt;p&gt;I don't really understand how it's drawn on sketches of the celestial sphere. At first I thought it goes through east and west, but it does not. How do you know where to draw it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example from Roy and Clarke, &lt;em&gt;Astronomy: Principles and Practice&lt;/em&gt;:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Vo2DP.png&quot; alt=&quot;celestial sphere diagram&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="28337" LastEditorUserId="616" LastEditDate="2019-08-02T15:05:06.890" LastActivityDate="2019-08-02T18:05:55.480" Title="How do you draw the ecliptic on a celestial sphere sketch?" Tags="&lt;declination&gt;&lt;ecliptic&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32504" PostTypeId="2" ParentId="32503" CreationDate="2019-07-01T16:26:02.987" Score="0" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Ecliptic&quot; rel=&quot;nofollow noreferrer&quot;&gt;ecliptic&lt;/a&gt; is the mean apparent path of the Sun throughout the year. Thus it is the plane of the Earth's orbit.  The ecliptic crosses the celestial equator from south to north at the vernal equinox. &lt;/p&gt;&#xA;" OwnerUserId="28086" LastActivityDate="2019-07-01T16:26:02.987" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32507" PostTypeId="1" CreationDate="2019-07-02T04:55:42.477" Score="1" ViewCount="175" Body="&lt;p&gt;I am currently working on a project on which I am calculating the luminosity of galaxies by using the SDSS data. They use a system of asinh magnitudes, and according to their website they use a system of softening parameters as it follows in &lt;a href=&quot;https://www.sdss.org/dr12/algorithms/magnitudes/#asinh&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.sdss.org/dr12/algorithms/magnitudes/#asinh&lt;/a&gt;:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/toOQi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/toOQi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/T61XM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/T61XM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used their cmodel magnitude that they advise for galaxies' photometry. For example, I have the r magnitude of this galaxy that is 14.68, and I am trying to find its luminosity. They say that to convert to flux density, one must follow the following equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;S = 3631 Jy * f/f0, where for the r band the AB conversion and shift is minimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I plug the numbers into the equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://chart.googleapis.com/chart?cht=tx&amp;amp;chl=L%20%3D%204%5Cpi%20d%5E%7B2%7Df&quot; alt=&quot;formula&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance of the galaxy is 63.3 Mpc&#xA;But it seems that when I plug the numbers, the final value does not match, how is that? And yes, I converted to the units including the Jansky to 10^-26 W m^-2 Hz^-1 and the Mpc to m. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish to find the luminosity of the galaxy in terms of solar luminosity and I know that one solar luminosity unit is the same as 3.827e+26 W. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was only able to find the correct answer when I did the following equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://chart.googleapis.com/chart?cht=tx&amp;amp;chl=%5Cfrac%7BL_%7Bgal%7D%7D%7BL_%7Bsun%7D%7D%3D%20%5Cleft%20(%20%5Cfrac%7Bd_%7Bgal%7D%7D%7Bd_%7Bsun%7D%7D%20%5Cright%20)%5E%7B2%7D%5Ctimes%20%5Cleft%20(%20%5Cfrac%7Bf_%7Bgal%7D%7D%7Bf_%7Bsun%7D%7D%20%5Cright%20)&quot; alt=&quot;formula&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://latex.codecogs.com/gif.latex?%5Cleft%20%28%20%5Cfrac%7Bf_%7Bgal%7D%7D%7Bf_%7Bsun%7D%7D%20%5Cright%20%29%3D10%5E%7B-0.4%5Cleft%20%28mag_%7Bgal%7D%20-%2026.7%20%5Cright%20%29%7D&quot; alt=&quot;formula&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone knows what I am doing wrong point out, please. Is it because flux intensity is different from flux? &lt;/p&gt;&#xA;" OwnerUserId="28470" LastActivityDate="2019-07-02T04:55:42.477" Title="Q: Calculating the luminosity of a galaxy" Tags="&lt;galaxy&gt;&lt;redshift&gt;&lt;luminosity&gt;&lt;sloan-dss&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32508" PostTypeId="1" AcceptedAnswerId="32509" CreationDate="2019-07-02T07:13:26.847" Score="3" ViewCount="307" Body="&lt;p&gt;I have recently purchased a &lt;a href=&quot;https://www.ontariotelescope.com/CG-4-POLAR-AXIS-FINDER_p_766.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;CG-4 Polar Axis Finder&lt;/a&gt; for my CG-4 mount. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This PAF has markings on the lens for aligning with Big Dipper or Cassiopeia, which looks like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/G4IMw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G4IMw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have aligned the PAF to the mount during the day, and I am now attempting to use it at night. My understanding is that I need to rotate the mount in RA to match the image in the PAF to the actual stars in either Big Dipper or Cassiopeia. This will give me the correct position to then re-target at Polaris to find NCP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been trying for 2 nights, and still no luck! The Big Dipper and Cassiopeia are in front of me - I can see them very clearly with the naked eye. Yet when I look through the PAF, I cannot see any star patterns resembling either of my targets. There are stars, but it's like the magnification is too high, maybe? No stars forms the pattern I am looking for, in any orientation. I cannot derive the connection between what I see with the naked eye right in front of me, and what I see through the PAF. I understand that the image in the PAF will be mirrored vertically and horizontally - still no luck. I can easily find my target through the Finder Scope on my main tube, but not through the PAF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have devised an experiment. I have positioned my tripod under a power line running high above me. I have aligned my view at Alkaid (western tip of the Big Dipper handle) with the wire, so I can see it in the PAF. Good news - I can see it in the PAF, and I can see Alkaid (it's the brightest in the area, so I am fairly sure that I can use it as a reference point between what I see with the naked eye and what I see through the PAF). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bad news - no matter how I rotate my mount in RA while keeping the handle tip aligned with Alkaid, there's nothing that matches the rest of Big Dipper...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is going on? I can't even explain this with a &#xA;&quot;Defective PAF Theory&quot;...&lt;/p&gt;&#xA;" OwnerUserId="28582" LastEditorUserId="28582" LastEditDate="2019-07-02T07:34:26.947" LastActivityDate="2019-07-02T07:53:15.133" Title="Unable to locate Big Dipper through Polar Axis Finder" Tags="&lt;telescope&gt;&lt;alignment&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32509" PostTypeId="2" ParentId="32508" CreationDate="2019-07-02T07:53:15.133" Score="5" Body="&lt;p&gt;My guess is that the drawing is a &quot;cartoon&quot; and not to scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The declination of &lt;a href=&quot;https://en.wikipedia.org/wiki/Polaris&quot; rel=&quot;noreferrer&quot;&gt;Polaris&lt;/a&gt; is &lt;code&gt;+89° 15′ 50.8″&lt;/code&gt; so it is only 0.746° away from the pole, (where the &quot;+&quot; is in the center of your drawing). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the inner circle's 143 pixel diameter corresponds to 1.472°, the outer circle's 824 pixel diameter is only 8.48°, or 85.76 North declination. The big dipper and Caseopia are at 60° to 50° North declination, way way outside the view of your scope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;They are shown only to provide a general direction to help decide how to rotate your finder.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/a3dyn.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Octans_IAU.svg&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-07-02T07:53:15.133" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32510" PostTypeId="1" AcceptedAnswerId="32513" CreationDate="2019-07-02T11:53:19.607" Score="46" ViewCount="21419" Body="&lt;p&gt;We know most of the objects in the Universe have a spherical or elliptical shape. The object which has less mass and gravitational pull orbits around the nearest object with more mass and gravitational pull. For example:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Moon orbits around Earth&lt;/li&gt;&#xA;&lt;li&gt;Earth orbits around Sun&lt;/li&gt;&#xA;&lt;li&gt;Sun orbits around Sagittarius A* which is the center of Milky Way.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thus, is the Milky Way orbiting around some object or perhaps Black Hole?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the Milky Way is going towards Andromeda as they are &lt;a href=&quot;https://qr.ae/TWtMSA&quot; rel=&quot;noreferrer&quot;&gt;attracting each other&lt;/a&gt; and they will collide with each other after 3 billion years to 6 billion years. But it is possible that the Milky way is orbiting around some object at the same time? Perhaps both galaxies are present in a group of galaxies which is orbiting around some object. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Milky Way is not orbiting around some object then is there any proof found by the scientists for that? &lt;/p&gt;&#xA;" OwnerUserId="28584" LastEditorUserId="14094" LastEditDate="2019-07-03T04:13:15.610" LastActivityDate="2020-10-03T09:00:59.303" Title="Does the Milky Way orbit around anything?" Tags="&lt;orbit&gt;&lt;galaxy&gt;&lt;milky-way&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32511" PostTypeId="1" AcceptedAnswerId="32512" CreationDate="2019-07-02T12:56:13.817" Score="3" ViewCount="128" Body="&lt;p&gt;Scientists claim that objects left on the Moon can lie hundreds of years in the same condition like astronaut Charles Duke family photo. Why we don't see remains of asteroids inside or around craters? Are asteroids going destroyed on impact or go under the surface even without any residue? Why do we not see asteroids that hit  Moon with a small force and remained on the surface for hundreds of years like astronaut's photo? Can craters form for other reasons? For example, from underground collapse?&lt;/p&gt;&#xA;" OwnerUserId="28585" LastActivityDate="2019-07-02T18:14:26.053" Title="Where asteroids disappear" Tags="&lt;the-moon&gt;&lt;asteroids&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32512" PostTypeId="2" ParentId="32511" CreationDate="2019-07-02T18:14:26.053" Score="6" Body="&lt;p&gt;Asteroids hit the surface of the moon at speeds of about 25 km per sec, so they are mostly vaporised on impact. If a geologist inspected a crater on foot and looked hard enough, he probably would find a few relatively small fragments, particularly if the meteorite was of the iron/nickel variety. A few such fragments were recovered from the Barringer Crater in the United States. It is unlikely an impact would leave no residue at all. but the residue might be difficult to identify. When trying to visualise the impact of, say, a 100-ton meteorite, it is helpful to imagine that it turned into 100 tons of high explosive on impact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, craters can form for other reasons. Most craters on the moon are impact craters, but a small number are volcanic. Underground collapse sometimes occurs in the vicinity of volcanoes; the resulting holes are usually elongated, and are called graben. Lava tubes sometimes collapse too, leaving a  long, winding  valley like a river bed. These indentations are called rills. Objects left on the moon tend to survive for centuries in good condition, because there is no weather worth talking about, just solar wind (charged particles), cosmic rays and electromagnetic radiation.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-07-02T18:14:26.053" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32513" PostTypeId="2" ParentId="32510" CreationDate="2019-07-02T18:47:33.163" Score="61" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The object which has less mass and gravitational pull orbits around the nearest object with more mass and gravitational pull.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Actually, both the heavier and the lighter object orbit around their common center of mass. It's just that the heavier object doesn't move much (has a tiny orbit), while the lighter object moves a lot (has a wide orbit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. our Sun actually orbits the center of mass of the whole solar system, but that motion is tiny, it barely budges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a double star, where both partners have about the same mass, you can clearly see how both are making similar orbits around their common mass center.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sun Orbits Around Sagittarius A* which us center of Milky Way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With galaxies, including ours, it's a little different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no super-heavy thing at the center, around which everything else is orbiting. Not even the very large black hole at the center of our galaxy is heavy enough for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather, galaxies are clumps of matter that create a common gravitational field. Stars, and everything else, are trapped in this common field and orbit around the common center of mass.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So the question is that is that is Milky Way is orbiting around some object or perhaps Black Hole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Same idea. There is no single point-object nearby massive enough for our galaxy to &quot;orbit&quot; around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our galaxy, along with Andromeda, and a handful of other galaxies, are bound together in what is known as the Local Group. Each galaxy is moving within the common gravitational field of the whole group. The Local Group has a diameter of about 10 million light-years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Local Group is part of a larger structure, the Virgo Supercluster, which is about 100 million light-years in diameter and has at least 100 galaxies. However, the Virgo Supercluster is more &quot;loose&quot; - it is not gravitationally bound together.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2019-07-02T18:55:04.390" LastActivityDate="2019-07-02T18:55:04.390" CommentCount="16" ContentLicense="CC BY-SA 4.0" />
  <row Id="32514" PostTypeId="2" ParentId="32457" CreationDate="2019-07-03T00:10:02.007" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm interested about changes in the core and mantle behavior, a&#xA;  potential magnetosphere, changes in the atmosphere and tectonic&#xA;  plates.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A lot of this is very theoretical.   Venus' atmosphere is nearly 100 times as massive Earth's atmosphere.  It's also nearly all CO2, a greenhouse gas, which means it's good at trapping heat.   That wouldn't change if Venus' rotation rate was faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Core and mantle behavior&quot; is a bit of an unusual statement, cores rotate with the planet and they slowly cool, there's some convection vs conduction going on, but it's difficult to study the cores of other planets.  It's hard enough studying the core of Earth.   But what we know of the core of Earth, or think we know is that Earth's core and Earth's crust rotate at slightly different rates.  This may be due to Earth's crust slowing down over time and the core, not slowing down as fast.   There's a slight variation in rotation rate.   Also, Earth's inner core is solid, Earth's outer core is somewhat liquid (though extremely viscus), and as it cools, the inner core slowly grows, the outer core shrinks and in the process heat is released.   This process creates a flow, which may be partially influenced by the Moons orbit, and together the circulations are thought to create Earth's magnetic field.   This isn't certain, but the flow of the outer core is thought to be the cause and Earth's rotation may play a role in that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if Venus rotated much faster, in combination with it's proximity to the Sun and the solar tidal forces, it's possible a faster rotating Venus would have a permanent magnetic field.   It's far from certain, but that's a possible outcome.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Plate tectonics is more tricky.  Various studies suggest that planetary size has a lot to do with plate tectonics, and abundant water plays a role too, oceans making plate tectonics easier.   Venus is on the small size and it has almost no water, so it's not a good candidate for plate tectonics.   Venus also has very little granite.    This is geology more than astronomy, but abundant granite (IMHO) plays a role in plate tectonics.   The rocky bodies in our solar-system.   Mercury, Venus, Mars, the Moon, are all, basically basalt like on their surfaces.   Earth formed large granite land masses that float above the more dense basalt like layers under the oceans.   Without granite land masses and without oceans, I don't think Venus is a good candidate for plate tectonics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did Venus ever have plate tectonics, like billions of years ago?   Maybe.   I have no idea, but maybe.   Would it have plate tectonics if it rotated faster?   I doubt it.  A magnetic field - maybe.  Plate tectonics, I don't think so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is probably too general for stack exchange and it may be closed, but I don't believe that just spinning Venus faster would change all that much apart from maybe giving it a magnetic field, and that wouldn't change all that much on Venus. It has an induced magnetic field anyway.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could give Venus an ocean, maybe make it more massive and move it further away from the sun, then some interesting things might happen like plate tectonics, which could have lead the planet down a different path, perhaps more similar to Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth was able to absorb CO2 into it's vast oceans and take in CO2 by photosynthesis and rocks were able to absorb CO2 by plate tectonics, and Earth was far enough from the sun to not have it's surface water evaporated and lost like what probably happened to Venus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think planetary formation, based on size, type of star, temperature, initial materials, etc, will be a very interesting field of study when/(if) big enough telescopes are set up to actually get a look at distant planets.   It's very difficult to even observe, much less, study a planet that's light-years away.   The so far twice or three ties delayed James Webb Space Telescope might get the ball rolling on this subject with better observations of exo-planet atmospheres, but it's a long ways away from being well understood.  The James Webb ST is an enormously difficult undertaking, so we should be patient.  With luck, it'll be up and taking in images within a few years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, would Venus' rotation, if it was faster, affect other planets?    Basically not at all.   Jupiter's rotation doesn't affect Earth.   Rotation rate doesn't affect other objects at great distances.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2019-07-03T00:10:02.007" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32515" PostTypeId="1" AcceptedAnswerId="32554" CreationDate="2019-07-03T00:16:07.913" Score="0" ViewCount="109" Body="&lt;p&gt;As we know magnetars have strong magnetic field but what if we swap earth with a magnetar and the sun will released a solar wind. Is the magnetar affected by the solar wind or not ? &lt;/p&gt;&#xA;" OwnerUserId="28511" LastActivityDate="2019-07-06T07:38:58.830" Title="The sun's SOLAR WIND VS MAGNETAR" Tags="&lt;the-sun&gt;&lt;magnetic-field&gt;&lt;solar-wind&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32517" PostTypeId="2" ParentId="32494" CreationDate="2019-07-03T01:26:59.887" Score="1" Body="&lt;p&gt;There are several online sources of spectra (eg SDSS), but they usually just provide a FITS file with data, or a plot vs wavelength without any annotation. Of course, you could use the  &lt;a href=&quot;https://www.nist.gov/pml/atomic-spectra-database&quot; rel=&quot;nofollow noreferrer&quot;&gt;NIST atomic spectra database&lt;/a&gt; to identify lines, but that quickly becomes tedious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many figures containing spectra for various classes of stars in the book by Gray and Corbally, &lt;strong&gt;Stellar Spectral Classification&lt;/strong&gt;. Many of these figures were previously available on Gray's web site, but the links no longer work. I scanned one that may be of use, &lt;a href=&quot;https://photos.smugmug.com/Photography/Miscellaneous/i-HP6R5tb/0/77e58ecc/O/Fig7.1GtoK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Figure 7.1&lt;/a&gt; containing spectra of G through K stars. In particular, one of these spectra is for a reference standard G0V star, &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; CVn, with the stronger Ca, Fe and H-Balmer lines identified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another good resource for annotated spectra of many star classes is Richard Walker's, &lt;strong&gt;Spectroscopic Atlas for Amateur Astronomers&lt;/strong&gt;. I have the printed version, but pdf files of &lt;a href=&quot;https://www.unitronitalia.com/prodotti/spectroscopic-atlas-3_0-english.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;older versions&lt;/a&gt; are available online. This atlas contains spectra of a G0IV star, and the Sun, G2V, that have similar absorption lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also collections of standard spectra, such as the Miles library. These are often included in software used for processing spectra, along with tables of common lines for various elements that can be shown on the plot. Programs I have used for amateur spectroscopy that display these include Rspec (commercial), Vspec and ISIS (Christian Buil).&lt;/p&gt;&#xA;" OwnerUserId="28086" LastActivityDate="2019-07-03T01:26:59.887" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32519" PostTypeId="1" AcceptedAnswerId="32522" CreationDate="2019-07-03T05:25:22.983" Score="1" ViewCount="121" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/E40Dt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E40Dt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;India is going to launch a spacecraft to the moon on 16 July which will consist of a lander, an orbiter and a rover. I found an article in today's newspaper regarding   this. If you zoom in a bit, you will be able to see the list of other lunar missions that have taken place over the years. A common difference that I found in lunar missions of 20th century and lunar missions of 21st century was the time. Earlier, the spacecrafts reached moon in a day or two but recent missions take over 2 months. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With developed technology, it should take even less time, but it's the other way round. Can anyone explain the reason behind this.&lt;/p&gt;&#xA;" OwnerUserId="28209" LastActivityDate="2019-07-03T08:37:17.327" Title="Time taken for journey to the moon" Tags="&lt;moonlanding&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2019-07-04T17:36:00.440" ContentLicense="CC BY-SA 4.0" />
  <row Id="32520" PostTypeId="2" ParentId="20122" CreationDate="2019-07-03T05:33:03.950" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IPtSz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IPtSz.jpg&quot; alt=&quot;accretion Sirius A - B&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is certainly possible, see the picture.&lt;/p&gt;&#xA;" OwnerUserId="28592" LastEditorUserId="7472" LastEditDate="2019-07-03T05:54:50.310" LastActivityDate="2019-07-03T05:54:50.310" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32521" PostTypeId="1" CreationDate="2019-07-03T06:24:00.510" Score="7" ViewCount="279" Body="&lt;p&gt;Adaptive optics generally apply slight changes to mirrors to account for atmospheric turbulence. These generally require either a mirror to be broken up into smaller mirrors, each with their own actuator, or for a single large mirror to be slightly deformed by many different actuators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to do much of this digitally? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Couldn't software slightly attenuate or amplify each pixel at each time-slice collected from a passive mirror? These changes would depend on the distortions measured by a guide star or laser (this digital system would still require a guiding light)? I suspect this would probably be less accurate than true adaptive optics, but it would also be far less expensive (as you wouldn't need so many actuators).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has this approach been studied and dropped? Is it ridiculous and not even theoretically possible? I'm curious if anything has researched this idea.&lt;/p&gt;&#xA;" OwnerUserId="28593" LastEditorUserId="9582" LastEditDate="2021-05-16T21:20:04.600" LastActivityDate="2021-05-16T21:20:04.600" Title="Are digital adaptive optics possible?" Tags="&lt;adaptive-optics&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32522" PostTypeId="2" ParentId="32519" CreationDate="2019-07-03T08:37:17.327" Score="4" Body="&lt;p&gt;Some trajectories need less energy than others, so to save on fuel, scientists often choose the lengthy, fuel efficient journey rather than a shorter one. In the case of the Apollo astronauts, it was important to get them to the moon in the shortest possible time, so a short trajectory was chosen.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-07-03T08:37:17.327" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32525" PostTypeId="2" ParentId="32510" CreationDate="2019-07-03T14:43:04.633" Score="12" Body="&lt;p&gt;The Milky Way would likely orbit the Great Attractor, the gravitational center of our local supergroup, but the metric expansion of space overwhelms gravitational attraction at that scale. The metric expansion of space only produces significant effects at the intergalactic scale so anything inside a galaxy is not affected much. In between galaxies however, and ever moreso with increasing scale, the metric expansion of space rips apart gravitational attraction so we see a foamy appearance rather than orbits.&lt;/p&gt;&#xA;" OwnerUserId="28600" LastEditorUserId="28600" LastEditDate="2019-07-25T18:15:32.900" LastActivityDate="2019-07-25T18:15:32.900" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32527" PostTypeId="2" ParentId="32521" CreationDate="2019-07-03T15:29:58.423" Score="5" Body="&lt;p&gt;The problem with your approach is that the deformable mirror changes the &lt;strong&gt;phase&lt;/strong&gt; of the light across the mirror, where the light is not focussed.  The light at the sensor array is focussed, and what you get is the &lt;strong&gt;intensity&lt;/strong&gt; which is, roughly, the Fourier Transform of the phase front at the mirror.   By the time you have intensity, the phase information has been lost.  &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;edit for clarity:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The sensor array &lt;em&gt;measures&lt;/em&gt; the intensity, at which point the phase info is irretrievable.  If you remove the sensor array and re-measure past the focal plane, yes, you can get phase info -- see &lt;a href=&quot;https://en.wikipedia.org/wiki/Light-field_camera&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Plenoptic cameras.&quot;&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there are other techniques -- you might be interested in searching for papers on &lt;a href=&quot;https://en.wikipedia.org/wiki/Lucky_imaging&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Lucky Imaging&quot;&lt;/a&gt; , which basically takes as many images as possible and throws away the distorted ones. &lt;/p&gt;&#xA;" OwnerUserId="907" LastEditorUserId="907" LastEditDate="2019-07-04T13:01:02.333" LastActivityDate="2019-07-04T13:01:02.333" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32528" PostTypeId="2" ParentId="32521" CreationDate="2019-07-03T18:45:28.570" Score="-2" Body="&lt;p&gt;With enough computational power every pixel of a camera sensor could be processed individually. Outputs amplified or reduced to compensate for variations of light levels across the sensor. Stacking algorithms can discard those pixels that do not receive a constant input. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photo shopping to the max. Make the Astro Photographers life easier. The problem for everyone else would be in deciding where science becomes art.&lt;/p&gt;&#xA;" OwnerUserId="24928" LastActivityDate="2019-07-03T18:45:28.570" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32529" PostTypeId="2" ParentId="32521" CreationDate="2019-07-03T19:16:39.447" Score="2" Body="&lt;p&gt;Each part of the mirror contributes to every pixel in the image, and the pockets of atmospheric distortion may be only 10-20 cm wide.&#xA;If you look at a bright planet through a 30+ cm telescope in poor seeing conditions, the image looks like a stack of several sub-images shifting in and out of alignment, each from a different part of the mirror.&#xA;To oversimplify, adaptive optics systems continually tweak the mirror segments to keep those sub-images aligned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can accept a much higher detector cost in exchange for eliminating the actuator cost, it's possible to make an array of small telescopes, each with its own video camera, and align the shifting sub-images in software.&#xA;But then diffraction by the smaller apertures would limit the system resolution unless you built an optical interferometer, which has its own category of difficulties to overcome.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2019-07-03T19:16:39.447" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32530" PostTypeId="2" ParentId="32521" CreationDate="2019-07-04T07:57:56.907" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/32527/7982&quot;&gt;@CarlWitthoft's answer&lt;/a&gt; is misleading if not wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Fourier transform of a field does not lose phase information. If you let the light drift another focal length and then use an identical mirror, you completely recover the initial electric field distribution incident at the telescopes entrance aperture. Information is not lost here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is not related to the optics at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that the detector (silicon CCD or photographic plate or whatever) measures time-averaged intensity of the light, which is the &lt;em&gt;absolute value squared&lt;/em&gt; of the field at the detector's surface. It's this &lt;strong&gt;squaring&lt;/strong&gt; and &lt;strong&gt;averaging&lt;/strong&gt; in conventional detection schemes that makes phase information unrecoverable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a whole field of research into trying to make phase-sensitive pixels for cameras but it's pretty academic, and you lose significantly in resolution and other performance metrics when you try to do this.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;But what about longer wavelengths?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Your idea &lt;strong&gt;can and indeed does work in radio astronomy&lt;/strong&gt;, even for microwaves and millimeterwaves. This is because the electric field received by each pixel (which is one dish antenna of an array of them) can indeed be absolutely digitized. They down-convert frequencies as high as a THz to one or two GHz and then amplify and digitize it with extremely fast ADCs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once that is done, you can correct for distortions in the wavefront arriving at your array of antennas in software. This process is explained further in answers to &lt;a href=&quot;https://astronomy.stackexchange.com/q/13930/7982&quot;&gt;Would Adaptive Optics be Useful in Radio Astronomy?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-07-04T07:57:56.907" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32531" PostTypeId="1" CreationDate="2019-07-04T13:14:42.420" Score="8" ViewCount="1869" Body="&lt;p&gt;Visible light imaging of the heavens is limited on Earth primarily due to the atmosphere. I know the ISS is moving at 17000mph, but given the delays to the James Webb telescope, is there any merit in attaching a space-rugged visible light telescope to the outside of the ISS? Could it ever approach the visible light performance of Hubble?&lt;/p&gt;&#xA;" OwnerUserId="22580" LastActivityDate="2019-07-05T23:13:25.103" Title="Attach a visible light telescope to the outside of the ISS" Tags="&lt;photography&gt;&lt;space&gt;&lt;iss&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32532" PostTypeId="2" ParentId="32531" CreationDate="2019-07-04T13:51:24.287" Score="-1" Body="&lt;p&gt;I think there is nothing wrong with your idea in principle, there is already a modified Boeing jumbo jet specifically for the purpose of carrying a large telescope above the densest layers of the atmosphere in order to minimise atmospheric interference and absorption by atmospheric water vapor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the telescope and the personnel who man it use up a considerable amount of space, though  on the aircraft there are no competing activities going on beyond those necessary to fly the aircraft. On the ISS the main problems would be in finding the space to accommodate the telescope and its crew and making the necessary modifications to the fabric of the space station. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think such a telescope could compete with Hubble, but that doesn't mean it couldn't do anything useful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;One advantage&lt;/em&gt; would be that if anything should go wrong,it could be somewhat easier to repair than it is with Hubble. Now that the Space Shuttle is no longer available, a crewed mission to Hubble would be extremely challenging and a robotic mission would require technology not yet available.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastEditorUserId="7982" LastEditDate="2019-07-05T23:13:25.103" LastActivityDate="2019-07-05T23:13:25.103" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32533" PostTypeId="2" ParentId="32531" CreationDate="2019-07-04T14:44:27.673" Score="16" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A self-sufficient orbiting telescope is basically Hubble mkII and would never get off the ground, literally and metaphorically&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hubble was expensive because it was state-of-the-art, requiring development of many new systems. The systems it needed to function as a standalone satellite (compared to being attached to the ISS) were cheap by comparison (reaction control, power, communications) because they mostly didn't require new development. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a Hubble successor attached to the ISS would not be much cheaper than a free-flying one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any telescope attached to the ISS has to deal with: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;high vibration levels which degrade performance&lt;/li&gt;&#xA;&lt;li&gt;outgassing which degrades performance&lt;/li&gt;&#xA;&lt;li&gt;severe pointing constraints because it's surrounded by a large structure&lt;/li&gt;&#xA;&lt;li&gt;bandwidth constraints because it has to share a transmitter with ISS operations&lt;/li&gt;&#xA;&lt;li&gt;end of mission in 5 years when the station reaches the end of its life.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The advantages of attaching to the ISS: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;easier servicing,&lt;/li&gt;&#xA;&lt;li&gt;you get the option to store lots of data on the ISS and bring it down on a capsule. Downside is a latency of several months.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;IMO you'd be better off building a standalone space telescope.&lt;/p&gt;&#xA;" OwnerUserId="361" LastActivityDate="2019-07-04T14:44:27.673" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32534" PostTypeId="2" ParentId="32531" CreationDate="2019-07-04T15:24:46.547" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Attach a visible light telescope to the outside of the ISS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a reasonable idea and it has been thought of before, but usually for other-than-visible light. Other answers do a good job of explaining why the disadvantages substantially outweigh the advantage. The cost to put something on the ISS large enough to outperform the top few dozen telescopes on Earth or approach the performance of the Hubble would be... &lt;em&gt;Astronomical!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is a moveable telescope mounted on the outside of the ISS &lt;em&gt;right now&lt;/em&gt; but read further to see why.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Had the Apollo missions gone on longer there might have been an Apollo space telescope. See answers to &lt;a href=&quot;https://space.stackexchange.com/q/31190/12102&quot;&gt;How would the Apollo telescope have worked in the Apollo command module? Where would it be located and how would it be operated?&lt;/a&gt; and the YouTube video &lt;a href=&quot;https://youtu.be/QUHyDnMS5oo?t=277&quot; rel=&quot;noreferrer&quot;&gt;Missions we Lost When Apollo was Cancelled&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Skylab&quot; rel=&quot;noreferrer&quot;&gt;Skylab&lt;/a&gt; space station did have a space telescope!&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;See for example &lt;a href=&quot;https://space.stackexchange.com/q/32497/12102&quot;&gt;How did Skylab's electrographic camera work?&lt;/a&gt;. This telescope took advantage of being above Earth's atmosphere in order to photograph using &lt;em&gt;ultraviolet light&lt;/em&gt;, something that you simply can not do on the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Apollo astronauts also had an ultraviolet telescope on the Moon (see answers to &lt;a href=&quot;https://space.stackexchange.com/q/33504/12102&quot;&gt;How was the Moon's first telescope used? (Apollo 16)&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Chang%27e_3#Lunar-based_ultraviolet_telescope_(LUT)&quot; rel=&quot;noreferrer&quot;&gt;Chang'e-3 lander also has an ultraviolet telescope&lt;/a&gt;! (see also GBTimes' &lt;a href=&quot;https://gbtimes.com/chinas-telescope-moon-still-working-and-could-do-30-years&quot; rel=&quot;noreferrer&quot;&gt;China’s telescope on the Moon is still working, and could do for 30 years&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/International_Space_Station&quot; rel=&quot;noreferrer&quot;&gt;ISS&lt;/a&gt; also has a space telescope!&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Finally, NASA has indeed attached a robotic, point-able telescope to the outside of the ISS. It is called NICER or &lt;a href=&quot;https://en.wikipedia.org/wiki/Neutron_Star_Interior_Composition_Explorer&quot; rel=&quot;noreferrer&quot;&gt;Neutron Star Interior Composition Explorer&lt;/a&gt; and it is an X-ray telescope, also something that you can't see from he ground and need to put your telescope in space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://astronomy.stackexchange.com/a/32113/7982&quot;&gt;this answer&lt;/a&gt; and links therein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dO0sK.gif&quot; alt=&quot;NICER X-ray telescope on the ISS GIF&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-07-04T23:24:48.593" LastActivityDate="2019-07-04T23:24:48.593" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32537" PostTypeId="1" CreationDate="2019-07-04T21:51:38.023" Score="3" ViewCount="132" Body="&lt;p&gt;In the wikipedia article for ephemeris time as first adopted in 1952, it does not explain clearly how or why it was superseded. For example, it reads:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In 1976 the IAU resolved that the theoretical basis for its current&#xA;(1952) standard of Ephemeris Time was non-relativistic, and that&#xA;therefore, beginning in 1984, Ephemeris Time would be replaced by two&#xA;relativistic timescales intended to constitute dynamical timescales:&#xA;Terrestrial Dynamical Time (TDT) and Barycentric Dynamical Time&#xA;(TDB).[28] Difficulties were recognized, which led to these being in&#xA;turn superseded in the 1990s by time scales Terrestrial Time (TT),&#xA;Geocentric Coordinate Time GCT(TCG) and Barycentric Coordinate Time&#xA;BCT(TCB). [16]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So, from this it appears that ephemeris time was never superseded and it continues to be evolved with a different theoretical basis?&lt;/p&gt;&#xA;&lt;p&gt;From reading Robert Newton's books such as &amp;quot;Ancient planetary observations and the validity of ephemeris time&amp;quot;, it would seem that ephemeris time is invalid because of non-secular accelerations in the motion of Mercury and Venus.&lt;/p&gt;&#xA;&lt;p&gt;So, I guess my question is: does current astronomical theory consider ephemeris time to be invalid or valid, and if it considers it to be valid, then what is current evaluation of Newton's calculations and their implications for the validity of ephemeris time?&lt;/p&gt;&#xA;" OwnerUserId="2800" LastEditorUserId="16685" LastEditDate="2021-01-28T16:49:21.787" LastActivityDate="2021-01-28T16:49:21.787" Title="Validity of ephemeris time" Tags="&lt;time&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32541" PostTypeId="1" CreationDate="2019-07-05T06:53:50.660" Score="1" ViewCount="266" Body="&lt;p&gt;Models of the solar system show each planet moving around on its own orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is is really because of the Sun that every planets remains in its own fixed orbit?&lt;/p&gt;&#xA;" OwnerUserId="28624" LastEditorUserId="7982" LastEditDate="2019-07-05T07:23:48.950" LastActivityDate="2019-07-05T11:14:18.103" Title="What keeps the planets of the solar system moving in their fixed orbits?" Tags="&lt;the-sun&gt;&lt;planet&gt;&lt;solar-system&gt;&lt;earth&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32542" PostTypeId="1" CreationDate="2019-07-05T08:44:35.253" Score="1" ViewCount="102" Body="&lt;p&gt;I was using SkySafari with my phone, but the compass didn't quite work.&#xA;So as I knew where the cardinal points were, I think I have shot the Sagittarius star cloud, at least a part of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you guys tell me if I am wrong? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CNBpR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CNBpR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was the 5th of July at 00:10 local time. I was at N 45.61&amp;deg;, E 7.35&amp;deg;.&lt;/p&gt;&#xA;" OwnerUserId="28625" LastEditorUserId="616" LastEditDate="2019-07-07T01:55:33.637" LastActivityDate="2019-07-07T01:55:33.637" Title="Sagittarius star cloud?" Tags="&lt;galaxy&gt;&lt;photography&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32543" PostTypeId="2" ParentId="32541" CreationDate="2019-07-05T11:14:18.103" Score="1" Body="&lt;p&gt;It is ,of course, gravity which keeps the planets circling the sun. They are actually falling toward the sun, but as the sun is a sphere, its surface keeps curving away, so they never get any closer to it. Because they are in a state of free fall, they are in effect weightless with respect to the sun, just as the space station is weightless with respect to the Earth, and as they use no energy in falling they can do it for ever. If you are wondering why you don't feel weightless, that's because of the gravitational field of the Earth. If the space station was a massive body like the moon.the astronauts wouldn't be weightless,but although the space station has a gravitational field of its own it is so trivial that it has no measurable effect on the astronauts.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-07-05T11:14:18.103" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32544" PostTypeId="2" ParentId="32542" CreationDate="2019-07-05T13:11:50.097" Score="-2" Body="&lt;p&gt;You've definitely got the Milky Way here. The galactic centre in Sagittarius is slightly thickened and noticeably brighter than the rest of the Milky way, so I think you might have got it in the bottom right hand corner of your photograph.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-07-05T13:11:50.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32546" PostTypeId="2" ParentId="32542" CreationDate="2019-07-05T15:49:09.590" Score="7" Body="&lt;p&gt;The image scale is a bit small, but I can see the constellation of Lyra just above center. The bright star Deneb (in the constellation of Cygnus) is left of center, and the star Altair (in the constellation of Aquila) is at the bottom (just right of center). Thus, your photo does not include Sagittarius. Sagittarius is off to the bottom right. The Scutum star cloud is the bright patch toward the right corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IGcNr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IGcNr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be helpful if you include the time of the photograph and location (latitude and longitude).&lt;/p&gt;&#xA;" OwnerUserId="17355" LastEditorUserId="2365" LastEditDate="2019-07-06T22:18:37.810" LastActivityDate="2019-07-06T22:18:37.810" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32547" PostTypeId="1" CreationDate="2019-07-05T15:58:46.053" Score="1" ViewCount="43" Body="&lt;p&gt;Anyone know if we have any chemical reactions occuring in this quasar?&lt;/p&gt;&#xA;" OwnerUserId="28633" LastEditorUserId="28633" LastEditDate="2019-07-07T01:58:54.363" LastActivityDate="2019-07-07T01:58:54.363" Title="What compounds are present in 3C273?" Tags="&lt;quasars&gt;&lt;astrochemistry&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32548" PostTypeId="1" CreationDate="2019-07-05T18:24:39.503" Score="1" ViewCount="295" Body="&lt;p&gt;This is a random question that popped into my mind. I'm not a physicist or astronomist, however, I am genuinely curious. Also, this is not intended as a science fiction question, but an actual science question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had doubts about where this question should be posted - the Astronomy or the Physics SE page, in the end, I decided to post it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sub-questions that I have, all being closely related to the one in the title:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Do the liftoffs of spaceships have any impact on Earth's orbit? As negligible as they might be?&lt;/li&gt;&#xA;&lt;li&gt;Every bit of mass lost (sent into space) decreases the gravity of our planet, as tiny as the change might be. Is that correct?&lt;/li&gt;&#xA;&lt;li&gt;Could the loss of Earth's mass over time (Hypothetically - centuries in the future, with tens of thousands of spaceships having left Earth.) change the orbit of Earth or produce any side-effects?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28634" LastActivityDate="2019-07-06T07:24:59.913" Title="What effect does Earth losing a part of its mass have, is there an impact on the Earth's orbit around the Sun?" Tags="&lt;orbit&gt;&lt;gravity&gt;&lt;earth&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32549" PostTypeId="2" ParentId="32548" CreationDate="2019-07-05T23:22:57.073" Score="0" Body="&lt;p&gt;The thing that is unaffected by liftoffs is the center of mass of all the material that goes into the rockets and the Earth.  Since it is likely the rockets will launch in all directions, if their mass was ever anything to account for (quite unlikely), it would still not affect the orbit of the Earth.  Also, the (tiny) reduction in Earth mass would also have little impact, because the Earth mass does not affect its own orbit, and the Earth has very little effect on the orbit of anything else other than the Moon.  So there could be a tiny affect on the Moon's orbit, which in turn could affect Earth tides, but we are talking completely unmeasurably small here.  Also, launching rockets in the direction of Earth's spin might ultimately produce a (minutely small) reduction in the Earth's spin, but again, hardly measurable even if you launched a million rockets.&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2019-07-05T23:22:57.073" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32550" PostTypeId="2" ParentId="32548" CreationDate="2019-07-05T23:25:30.500" Score="-2" Body="&lt;p&gt;Yes, the blast off of a rocket from Earth obeys Newton's 3rd Law: for every action there is an equal but opposite reaction. Think of it in terms of a man firing a gun. The bullet goes one way, the man's shoulder, absorbing the recoil, goes the other way, though only a very short distance because the man is heavier than the bullet. The same principle applies to the rocket fired into space: the rocket goes one way, the Earth goes the other. That is the theory, but because the mass of the Earth is trillions of times greater than that of the rocket, the movement of the Earth in the opposite direction to the rocket is so tiny as to be undetectable and unmeasurable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the decrease in Earth's mass. Again you must bear in mind that the Earth has trillions of times more mass than a large rocket, so not only is the mass loss trivial, but most of the objects sent into space eventually come back down to Earth. Also. there is a constant rain of space debris into Earth's atmosphere: space dust, meteors, meteorites and occasionally asteroids, so the Earth gains more mass than it loses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could loss or gain of mass eventually change the Earth's orbit? No, not to any measurable extent. Just as the mass of the Earth far exceeds any mass that leaves us, it also far exceeds the amount which arrives, so it doesn't have any detectable effect on our orbit.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-07-05T23:25:30.500" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32551" PostTypeId="1" AcceptedAnswerId="32557" CreationDate="2019-07-06T04:47:07.303" Score="1" ViewCount="55" Body="&lt;p&gt;The third paragraph of the introduction to the ArXiv preprint &lt;a href=&quot;https://arxiv.org/abs/1901.06007&quot; rel=&quot;nofollow noreferrer&quot;&gt;Constraining Quasar Relativistic Reflection Regions and Spins with Microlensing&lt;/a&gt; says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Quasar microlensing has significantly improved our understanding of the accretion disks (e.g., [five refs]) and non-thermal emission regions (e.g., [seven refs]) of quasars, and the demographics of microlenses in the lens galaxy (e.g., [three refs]) . Since the magnification diverges on the caustics produced by the lensing stars, quasar microlensing can constrain arbitrarily small emission regions if they can be isolated from other emission, in position, velocity, or energy. &lt;strong&gt;In particular, microlensing can be used to constrain the spin of black holes by measuring the ISCO size.&lt;/strong&gt; (emphasis added)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;ISCO was previously identified as &lt;em&gt;innermost stable circular orbit&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;From what I understand from Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_microlensing#How_it_works&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gravitational microlensing; How it works&lt;/a&gt;, a microlensing observation records a short term brightening of an object due to focusing by an intervening gravitational body, without useful spatial resolution.&lt;/p&gt;&#xA;&lt;p&gt;See also the Planetary Society's explanation &lt;a href=&quot;http://www.planetary.org/explore/space-topics/exoplanets/microlensing.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Microlensing; Beyond our Cosmic Neighborhood&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How are microlensing events used to constrain the size of innermost stable circular orbits around spinning black holes? What instruments are used and what kind of analysis is done to obtain a ISCO constraint?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eb3ep.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eb3ep.png&quot; alt=&quot;Example light curve of Gravitational Microlensing event - OGLE-2005-BLG-006 &quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Example light curve of Gravitational Microlensing event - OGLE-2005-BLG-006 &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Gravitational.Microlensing.Light.Curve.OGLE-2005-BLG-006.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2021-04-01T05:03:25.073" LastActivityDate="2021-04-01T05:03:25.073" Title="How are microlensing events used to constrain the size of innermost stable circular orbits around spinning black holes?" Tags="&lt;black-hole&gt;&lt;gravitational-lensing&gt;&lt;accretion-discs&gt;&lt;microlensing&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32553" PostTypeId="2" ParentId="32548" CreationDate="2019-07-06T07:24:59.913" Score="3" Body="&lt;p&gt;According to &lt;a href=&quot;https://earthscience.stackexchange.com/a/16725/6031&quot;&gt;this answer&lt;/a&gt; to &lt;em&gt;Is Earth getting heavier or lighter?:&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;tl;dr:&lt;/strong&gt; The Earth receives 40,000 tons of dust from space every year, but looses 95,000 tons of Hydrogen and 1,600 tons of Helium every year as well. After all additional effects are balanced, &lt;strong&gt;the Earth looses about 50,000 tons a year&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://space.stackexchange.com/a/26735/12102&quot;&gt;this answer&lt;/a&gt; to &lt;em&gt;Does launching a device into orbit change earth's orbit?&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;As long as &amp;quot;what happens in GEO stays in GEO&amp;quot; and there's no mass lost to deep space, the Shuttle and the Earth will mostly orbit around their common center of mass.&lt;/p&gt;&#xA;&lt;p&gt;When the Shuttle is moving (relative to Earth) in the prograde direction (Earth's motion around the Sun), the Earth will be moving slightly slower than it normally would, but a half-orbit later when the Shuttle is moving retrograde, the Earth will be moving slightly faster than it normally would.&lt;/p&gt;&#xA;&lt;p&gt;When the Shuttle returns, to 1st order (and maybe 2nd order as well) the Earth will be in the same place it would be if the Shuttle had never taken off.&lt;/p&gt;&#xA;&lt;p&gt;It's quite a bit like the Earth and Moon orbiting around their common barycenter, and it being pretty much &lt;em&gt;the EM barycenter that moves in a nice elliptical orbit around the Sun&lt;/em&gt;, except that as @RussellBorogove &lt;a href=&quot;https://space.stackexchange.com/a/26734/12102&quot;&gt;points out&lt;/a&gt; very effectively, it's a quite a bit smaller effect than that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;So as long as the launched mass stays in orbit around the Earth, the Earth's orbit is essentially unaffected. If the spacecraft goes off into deep space, then there will be an extremely tiny change in the Earth's velocity (rather than due to its mass change) by conservation of momentum.&lt;/p&gt;&#xA;&lt;p&gt;If the Earth loses mass by natural process isotropically, the orbit will be essentially unaffected. Of course the Earth and Sun orbit around their common center of mass, so if the Earth gets lighter it will have an extremely tiny second-order effect on the orbit, but that's nothing compared to the effect that the outer planets have on the Sun.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-07-06T07:24:59.913" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32554" PostTypeId="2" ParentId="32515" CreationDate="2019-07-06T07:38:58.830" Score="0" Body="&lt;p&gt;The solar wind is just a stream of charged particles, mostly protons and electrons, some alpha particles and a tiny trace amount of other stuff.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the solar wind approached an uncharged Neutron star, then the particles would simply fall onto the neutron star or circle around it in a hyperbolic orbit.  The Neutron star's gravity would add considerable velocity to the particles due to it's strong gravity.   Any particle that falls within about 2 radii of the neutron star would inevitably fall towards it due to the space-time properties of relativistic orbits, so there would be some spiraling in and probably some kind of light-show in the process.  The impacts of particles onto the surface would likely be in the x-ray or gamma ray spectrums and might not be that visible.   Ultimately it wouldn't be that interesting, but perhaps there would be some kind of light-show.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we give the Neutron star a strong magnetic field, then the solar wind would simply get redirected, similar to how it gets redirected by Earth's magnetic field, except in this case, the magnetic field is much stronger and the central object, much smaller.  Probably very few of the solar wind particles would hit the Neutron star and most would just circle around it's magnetic field, which would produce (I think) less of a light show.   There wouldn't be much effect that I can think of.  The Magnetar star is very massive with a very strong magnetic field.  The solar wind is comparatively, especially after spreading out for 93 million miles, quite diffuse and weak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a person got hit directly by the solar wind, it might be deadly.   We're not very tough, but a magnetar is one of the most indestructible objects in the universe.   The effect would be negligible.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be more interesting if you moved the Magnetar much closer to the star.   Magnetic field lines can't cross each other, but they can push each other and flatten out, and in the case of solar flares they can get very twisted (but still never cross).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A magnetar is several orders of magnitude smaller than a sunspot.   The interaction between the two magnetic fields, if the magnetar passed close to a sunspot would be interesting to see. (though the gravitational effect would still probably be the more impressive of the two)  But if you move a Magnetar as distant as the Earth, the solar wind would have lost much of it's strength and it would just be ejected material from a solar magnetic storm.  It wouldn't have much effect on the magnetar.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's my hobbyist understanding anyway.   If I missed any key points, please clarify your question or ask a new question.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2019-07-06T07:38:58.830" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32555" PostTypeId="1" CreationDate="2019-07-06T09:04:19.023" Score="1" ViewCount="214" Body="&lt;p&gt;I'm wondering - how much delta-v do typical geostationary satellites carry with them once they get into the orbit for things like orbital adjustment or movement to the graveyard orbit?&lt;/p&gt;&#xA;" OwnerUserId="28506" LastActivityDate="2019-07-06T14:43:35.447" Title="How much delta-v does a typical geostationary satellite carry?" Tags="&lt;artificial-satellite&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2019-07-08T00:46:04.150" ContentLicense="CC BY-SA 4.0" />
  <row Id="32556" PostTypeId="2" ParentId="32555" CreationDate="2019-07-06T14:43:35.447" Score="0" Body="&lt;p&gt;There's a mountain of readily available information, of which the following is one example.  From   &lt;a href=&quot;http://spaceflight101.com/goes-r/goes-r-spacecraft/&quot; rel=&quot;nofollow noreferrer&quot;&gt;GOES page&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h2&gt;edit&lt;/h2&gt;&#xA;&lt;p&gt;There is more information there about the thrust produced, so I suspect you can calculate the total delta-kinetic energy available.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A2100 has been used for over 40 missions and is available in different versions catered to specific mission needs in terms of&#xA;dimensions and power availability with satellite mass varying between&#xA;2,000 and 6,740 Kilograms.&lt;/p&gt;&#xA;&lt;p&gt;Each GOES-R-class satellite is 6.1 by 5.6 by 3.9 meters in size and has a dry mass of 2,857 Kilograms. At liftoff, the satellite&#xA;weighs 5,192 Kilograms, loaded with propellants to support a mission&#xA;of at least 14 years.&lt;br /&gt;&#xA;GOES-R, S, T, U launches with a propellant and pressurant load of&#xA;2,335 Kilograms with Mixed Oxides of Nitrogen and Hydrazine acting as&#xA;main propellant components and Helium gas used for the pressurization&#xA;of tanks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="907" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-07-06T14:43:35.447" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32557" PostTypeId="2" ParentId="32551" CreationDate="2019-07-06T17:46:53.287" Score="1" Body="&lt;p&gt;I think most of the paper is devoted to describing how they do it.  (This may be more of a comment than an answer, because I'm an amateur and not prepared to go through the paper in detatil).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that microlensing here is a little different than microlensing as described in the Wikipedia.  In the Wikipedia page and the Planetary Society page you reference, microlensing refers to observing the change in brightness over time of a small, relatively nearby object when a closer but non-luminous object passes in front of it from our line-of-sight.  In this paper, microlensing refers to the extra magnification that can be provided to tiny sub-sections of a &quot;macrolensed&quot; (gravitationally-lensed cosmologically distant object) by the chance placement of a star at the right position in the lensing geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors of the paper figure out how &lt;em&gt;this&lt;/em&gt; sort of microlensing might apply to the innermost parts of a quasar, where it might highlight small parts of the accretion disk, and so the contribution of those parts to the spectrum (in this case, one particular X-ray emission line characteristic of active galactic nuclei accretion disks) of the &quot;macrolensed&quot; object as seen from Earth.  They apply a statistical model of the likelihood of such microlensing and how it would affect the measured spectrum of a range of possible active-galactic nuclei (resolving a particular part of the accretion disk with microlensing would enhance that particular parts contribution to the whole, and each part of the accretion disk would have a different relativistic shift of the emission line as it whips around the central black hole), and then see how their model matches the actual measured spectrum.  It sounds like they are operating at the limits of statistical significance-- at least for most of the objects in their study-- but got more robust data for one particular object.&lt;/p&gt;&#xA;" OwnerUserId="14434" LastActivityDate="2019-07-06T17:46:53.287" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32558" PostTypeId="1" AcceptedAnswerId="32560" CreationDate="2019-07-06T20:43:41.533" Score="2" ViewCount="60" Body="&lt;p&gt;I'm trying to understand the results of the ExoMars Trace Gas Orbiter vs the Mars Express in terms of their abilities to find methane on the surface of Mars. The solution that I'm coming to is that it might be a result of the Sun's spectra masking what is actually there. Is there a place that has a complete infrared spectra of the Sun, focusing in on the region around 3-4 um? Specifically, I'm interested in wave numbers (&lt;span class=&quot;math-container&quot;&gt;$cm^{-1}$&lt;/span&gt;) 3000-3100 or so, something with a resolution of 1 wave number. Some things that would help would be anything along the following lines, so long as they apply to the true spectra, and not the one as seen from Earth.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A generally accepted solar spectra (NASA, ESA, NIST, or some other body)&lt;/li&gt;&#xA;&lt;li&gt;A time varying solar spectra.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="182" LastActivityDate="2019-07-07T04:31:10.153" Title="Where can I find a good solar irradiance chart for the infrared?" Tags="&lt;the-sun&gt;&lt;spectroscopy&gt;&lt;spectra&gt;&lt;infrared&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32559" PostTypeId="1" CreationDate="2019-07-06T23:16:59.090" Score="3" ViewCount="75" Body="&lt;p&gt;first I have posted originally on the physics stackexchange but I have not had answers about my question, so I try my luck on this specific forum (if forums leaders want to delete the original post on physics exchange, they can do it without problems, I don't want to make duplicates).&lt;/p&gt;&#xA;&lt;p&gt;From this article : &lt;a href=&quot;https://arxiv.org/pdf/1611.09787.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/pdf/1611.09787.pdf&lt;/a&gt; , I try to deduce the equation that my teacher told me which links 2 quantities :&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;the global number density of galaxies&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the local number density of galaxies&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the contrast of Dark matter density&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;The relation that I would like to find (the relation given by my teacher) is very simple :&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$N_{1} = n_{1} b_{1}\,\delta_{\text{DM}}\quad\quad(1)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$$N_{1}$$&lt;/span&gt; is the local number density of galaxies in Universe, &lt;span class=&quot;math-container&quot;&gt;$$n_{1}$$&lt;/span&gt; is the global number density, &lt;span class=&quot;math-container&quot;&gt;$$b_{1}$$&lt;/span&gt; is the bias (cosmological bias of galaxies) and &lt;span class=&quot;math-container&quot;&gt;$$\delta_{\text{DM}}$$&lt;/span&gt; the contrast in dark matter density. When I say &amp;quot;local&amp;quot;, I mean in the volume of scale that I consider (in a cluster of galaxies for example, doesn't it ?)&lt;/p&gt;&#xA;&lt;p&gt;for the moment, I can't find this equation.&lt;/p&gt;&#xA;&lt;p&gt;Into the article above, they define the bias by doing the relation &lt;span class=&quot;math-container&quot;&gt;$(1.1)$&lt;/span&gt; (equation reference on the article) :&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\delta_{g}(\vec{x}) = \dfrac{n_{g(\vec{x})}}{\overline{n_{g}}}-1 = b_{1}\,\delta_{\text{DM}}(\vec{x}) = b_{1}\big(\dfrac{\rho_{m}(\vec{x})}{\overline{\rho_{m}}}-1\big)\quad\quad(2)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;with &lt;span class=&quot;math-container&quot;&gt;$b_{1}$&lt;/span&gt; the bias.&lt;/p&gt;&#xA;&lt;p&gt;As you can see, in this article, authors are reasoning with the contrast of density number of galaxies (&lt;span class=&quot;math-container&quot;&gt;$\delta_{g}(\vec{x}))$&lt;/span&gt; and the contrast of matter density of Dark matter (&lt;span class=&quot;math-container&quot;&gt;$\delta_{\text{DM}}(\vec{x})$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;p&gt;I tried to modify this equation &lt;span class=&quot;math-container&quot;&gt;$(2)$&lt;/span&gt; to get &lt;span class=&quot;math-container&quot;&gt;$(1)$&lt;/span&gt; but I am stuck by the following difference : on one side, one takes number densities and on the other one, they take contrasts of density (with contrast density number and Dark matter contrast).&lt;/p&gt;&#xA;&lt;p&gt;Multiplying the both by the volume &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; is not enough since there is the value &amp;quot;-1&amp;quot; in the definition of contrast : I don't know if I have to write :&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text{Global Number of galaxies} = \overline{n_{g}}\quad V$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text{Local Number of galaxies} = \overline{n_{g}}\quad V$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;???&lt;/p&gt;&#xA;&lt;p&gt;I think that I have to use the following relations : &lt;span class=&quot;math-container&quot;&gt;$N_{g}\equiv N_{1}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\overline{n_{g}}=n_{1}$&lt;/span&gt; in the relation of my teacher but I am not sure.&lt;/p&gt;&#xA;&lt;p&gt;Anyone could help me to find the equation (1) from the equation (2) of article cited ?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;If I take the relation eq&lt;span class=&quot;math-container&quot;&gt;$(2)$&lt;/span&gt;, I can write :&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$n_{g(\vec{x})} = \overline{n_{g}}\,b_{1}\,\delta_{\text{DM}}+\overline{n_{g}}\quad\quad(3)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;As you can see, &lt;span class=&quot;math-container&quot;&gt;$(3)$&lt;/span&gt; is not equal to the equation &lt;span class=&quot;math-container&quot;&gt;$(1)$&lt;/span&gt; that I would like to get (since a second term &lt;span class=&quot;math-container&quot;&gt;$\overline{n_{g}}$&lt;/span&gt;)&lt;/p&gt;&#xA;&lt;p&gt;With the notations of the equation&lt;span class=&quot;math-container&quot;&gt;$(1)$&lt;/span&gt;, in order to be coherent, I think that I have to assimilate &lt;span class=&quot;math-container&quot;&gt;$N_{1}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$n_{g}(\vec{x})$&lt;/span&gt; (local density) and &lt;span class=&quot;math-container&quot;&gt;$n_{1}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\overline{n_{g}}$&lt;/span&gt; (global or mean density).&lt;/p&gt;&#xA;&lt;p&gt;How can I circumvent this issue about the presence of this second term into eq&lt;span class=&quot;math-container&quot;&gt;$(3)$&lt;/span&gt; compared to eq&lt;span class=&quot;math-container&quot;&gt;$(1)$&lt;/span&gt; ?&lt;/p&gt;&#xA;&lt;p&gt;I am near from the equality between both, this is frustrating . Maybe it is a problem of convention about the factor &lt;span class=&quot;math-container&quot;&gt;$b_{1}$&lt;/span&gt; called &amp;quot;bias&amp;quot; ?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: This is also cross-posted at &lt;a href=&quot;https://physics.stackexchange.com/questions/489437/demonstration-about-large-scale-bias-of-galaxies&quot;&gt;https://physics.stackexchange.com/questions/489437/demonstration-about-large-scale-bias-of-galaxies&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16492" LastEditorUserId="37191" LastEditDate="2021-03-11T13:16:19.547" LastActivityDate="2021-03-11T13:16:19.547" Title="Cosmology : Formula for the bias of galaxies" Tags="&lt;galaxy&gt;&lt;astrophysics&gt;&lt;cosmology&gt;&lt;dark-matter&gt;&lt;density&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32560" PostTypeId="2" ParentId="32558" CreationDate="2019-07-07T04:20:21.243" Score="1" Body="&lt;p&gt;Thanks to @PearsonArtPhoto's &lt;a href=&quot;https://astronomy.stackexchange.com/questions/32558/where-can-i-find-a-good-solar-irradiance-chart-for-the-infrared?noredirect=1#comment60008_32558&quot;&gt;comment&lt;/a&gt; data with 1 cm^-1 wavenumber resolution is available in the link labeled &lt;code&gt;MODTRAN data&lt;/code&gt; here: &lt;a href=&quot;https://www.nrel.gov/grid/solar-resource/spectra.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.nrel.gov/grid/solar-resource/spectra.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a quickie plot and script for the downloaded ASCII data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are six columns of spectral intensity data (Watts/m^2/nm) with labels 'MCebKur', 'MChKur', 'MNewKur', 'MthKur', 'MoldKur', 'MODWherli_WMO'. The plot shows them to be very similar in the thermal IR (not surprising) and differ substantially in the visible and UV where differences in the spectrometers will show up in the region rich with spectral emission and absorption lines. This is illustrated in the second plot, showing a randomly chosen zoom in on a few nanometers of visible light, and the Fraunhofer &quot;A&quot; line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0haxz.png&quot; alt=&quot;MODTRAN data&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LVUBE.png&quot; alt=&quot;MODTRAN data&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;&#xA;fname = 'AllMODEtr MODTRAN.txt' # MODTRAN data https://www.nrel.gov/grid/solar-resource/spectra.html&#xA;&#xA;with open (fname, 'r') as infile:&#xA;    lines = infile.readlines()&#xA;&#xA;lines = [line.split() for line in lines[19:-1]] # skip a few lines with missing data&#xA;&#xA;print('line lengths: ', set([len(line) for line in lines]))&#xA;&#xA;data = [[float(x) for x in line] for line in lines]&#xA;&#xA;data = np.array(zip(*data))&#xA;&#xA;print('data.shape: ', data.shape)&#xA;&#xA;# data[0]: wavenumber (cm^-1)&#xA;# data[1]: wavelength (nm)&#xA;# data[2:8] MCebKur, MChKur, MNewKur, MthKur, MoldKur, MODWherli_WMO        &#xA;&#xA;labels = 'MCebKur', 'MChKur', 'MNewKur', 'MthKur', 'MoldKur', 'MODWherli_WMO'&#xA;wavelength = data[1]&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;&#xA;    plt.subplot(2, 1, 1)&#xA;&#xA;    for (thing, label) in zip(data[2:], labels):&#xA;        plt.plot(wavelength, thing)&#xA;    plt.xlabel('wavelength (nm)', fontsize=16)&#xA;    plt.ylabel('Watts/m^2/nm', fontsize=16)&#xA;    plt.yscale('log')&#xA;    plt.xscale('log')&#xA;    plt.xlim(190, 2E+05)&#xA;&#xA;    plt.subplot(2, 1, 2)&#xA;&#xA;    n700 = np.argmax(wavelength &amp;gt; 700)&#xA;    for (thing, label) in zip(data[2:], labels):&#xA;        plt.plot(wavelength[:n700], thing[:n700])&#xA;    plt.xlabel('wavelength (nm)', fontsize=16)&#xA;    plt.ylabel('Watts/m^2/nm', fontsize=16)&#xA;    plt.yscale('log')&#xA;    # plt.xscale('log')&#xA;    plt.xlim(190, wavelength[n700])&#xA;&#xA;    plt.show()&#xA;&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;&#xA;    plt.subplot(2, 1, 1)&#xA;&#xA;    n1 = np.argmax(wavelength &amp;gt; 516.5)&#xA;    n2 = np.argmax(wavelength &amp;gt; 519.5)&#xA;    for (thing, label) in zip(data[2:], labels):&#xA;        plt.plot(wavelength[n1:n2], thing[n1:n2], label=label)&#xA;    plt.xlabel('wavelength (nm)', fontsize=16)&#xA;    plt.ylabel('Watts/m^2/nm', fontsize=16)&#xA;    # plt.yscale('log')&#xA;    # plt.xscale('log')&#xA;    plt.xlim(wavelength[n1], wavelength[n2])&#xA;    # plt.legend()&#xA;&#xA;    plt.subplot(2, 1, 2)&#xA;&#xA;    n1 = np.argmax(wavelength &amp;gt; 757)&#xA;    n2 = np.argmax(wavelength &amp;gt; 760)&#xA;    for (thing, label) in zip(data[2:], labels):&#xA;        plt.plot(wavelength[n1:n2], thing[n1:n2], label=label)&#xA;    plt.xlabel('wavelength (nm)', fontsize=16)&#xA;    plt.ylabel('Watts/m^2/nm', fontsize=16)&#xA;    # plt.yscale('log')&#xA;    # plt.xscale('log')&#xA;    plt.xlim(wavelength[n1], wavelength[n2])&#xA;    plt.title('Fraunhoffer &quot;A&quot; line')&#xA;    # plt.legend()&#xA;&#xA;    plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-07-07T04:31:10.153" LastActivityDate="2019-07-07T04:31:10.153" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32561" PostTypeId="1" AcceptedAnswerId="32786" CreationDate="2019-07-07T04:42:44.173" Score="1" ViewCount="103" Body="&lt;p&gt;Ralf Vandebergh's &lt;a href=&quot;https://twitter.com/ralfvandebergh/status/1116090016873238529&quot; rel=&quot;nofollow noreferrer&quot;&gt;tweet&lt;/a&gt; shows his own reconstruction of an image of the accretion disk around a black hole, generated presumably from Event Horizon Telescope data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming that the data came from a publicly available source, but I don't know for sure. That would make sense, they used two completely independent groups to reconstruct the images to cross-check each other before they were published, because with so few of the telescopes in the array recording data at any one time, the reconstruction took a bit of artistic license or &quot;filling in&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the &lt;em&gt;raw data&lt;/em&gt; is huge, boxes and boxes of hard drives with high speed digital recording of GHz rate ADC conversions, I don't think there's an ftp site for the actual raw data. And since calibrations and baselining is pretty complicated, it would be hard to reconstruct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm wondering what kind of data is available that would allow someone to reconstruct their own EHT image?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FYI Ralf Vandebergh's astrophotography can be found in links within &lt;a href=&quot;https://photo.stackexchange.com/q/109370/50110&quot;&gt;Why exactly do “CMOS astrophotographers” prefer CMOS sensors?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://twitter.com/ralfvandebergh/status/1116090016873238529&quot; rel=&quot;nofollow noreferrer&quot;&gt;tweet&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This is one of my processings of the #blackhole #EHTBlackHole first ever image. @RHdijkgraaf @govertschilling @NASAblueshift @ehtelescope @jordydavelaar @UofA @perimeter @smithsonian @TheNRAO @PhysicsWorld&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gkJFE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gkJFEm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; click for larger&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-07-25T07:39:43.643" LastActivityDate="2019-07-25T08:39:59.570" Title="From whence is the Event Horizon Telescope black hole data available for amateur reconstruction?" Tags="&lt;radio-astronomy&gt;&lt;data-analysis&gt;&lt;event-horizon-telescope&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32562" PostTypeId="1" AcceptedAnswerId="32566" CreationDate="2019-07-07T11:06:25.273" Score="16" ViewCount="2381" Body="&lt;p&gt;Sorry for the noob question but I don't seem to be able to find the answer on the internet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been looking through some telescopes and binoculars and noticed that shops typically give different specifications for the two groups. For example, for telescopes I'll often see focal length, aperture ratio, or limit value, but I haven't seen any of those for binoculars; for binoculars, on the other hand, they'll mention exit pupil, field of view, or glass material, none of which I've seen given for a telescope. Why is that? Are the two devices so different as to be incomparable?&lt;/p&gt;&#xA;" OwnerUserId="28647" LastActivityDate="2019-07-08T09:08:43.113" Title="Why different specifications for telescopes and binoculars?" Tags="&lt;telescope&gt;&lt;optics&gt;&lt;binoculars&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32563" PostTypeId="2" ParentId="32562" CreationDate="2019-07-07T11:43:31.017" Score="11" Body="&lt;p&gt;Binoculars tend to be mostly used for daytime observing (of birds, ships etc) Telescopes are mostly used for astronomical observing. The users of the two types of equipment want different information.  You may find large, mounted, astronomical binoculars described more like a telescope. Similarly, small &quot;mononcular&quot; may be described in the same terms as binoculars.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-07-07T11:43:31.017" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32564" PostTypeId="1" CreationDate="2019-07-07T17:57:43.337" Score="1" ViewCount="211" Body="&lt;p&gt;I am having trouble finding the quality of old telescopes. My question is how does a decent modern telescope that cost about a couple hundred USD compare to the top-notch telescopes in the nineteenth century. Can one make the same observations now with such a modern telescope as one could do in the nineteenth century with a top-notch one?&lt;/p&gt;&#xA;" OwnerUserId="28648" LastEditorUserId="28648" LastEditDate="2019-07-07T18:09:15.477" LastActivityDate="2019-07-08T01:09:09.227" Title="Old Telescopes versus New" Tags="&lt;telescope&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32565" PostTypeId="2" ParentId="32537" CreationDate="2019-07-07T19:44:52.210" Score="3" Body="&lt;p&gt;It depends on what one means by &quot;Ephemeris Time.&quot; Officially, the Newtonian-based Ephemeris Time has been superseded by the relativistically-correct Barycentric Dynamic Time (TDB) and Barycentric Coordinate Time (TCB). However, the concept of what was meant by Ephemeris Time carries on to this day. The intent of Ephemeris Time was to use the solar system as a timekeeping device. This is what TDB and TCB do, but in a relativistic rather than Newtonian setting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TCB is a theoretical time scale whose clocks are infinitely removed from the solar system, are co-moving with the solar system barycenter, and that tick at the rate of one tick per 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom (at a temperature of 0 K). (Note well: This is the same definition currently used for an SI second.) That TCB clocks are well removed from the solar system means that a second measured by an atomic clock on the surface of the Earth ticks a tiny bit slower than does a TCB clock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These slightly different tick rates was perceived as a problem by some. The solution was eventually to redefine TDB as a scaled version of TCB such that, on average, TCB ticks at the same rate as atomic clocks on the surface of the Earth. Atomic clocks on the surface of the Earth tick faster / slower than would TCB clocks based on whether the Earth is further from / closer to the Sun than average. But averaged over the course of years, TCB and TAI remain more or less in sync.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the more or less: TDB is updated every year. As technology improves, these subtle updates change the timing of past events. TAI does not do this. As technology improves, older atomic clocks are replaced with newer ones. Atomic clocks have a shelf life of a decade or so. This can result in discontinuities in atomic time. This is not a problem with TDB as it is re-issued annually. This also means that TDB is independent of atomic time.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2019-07-07T19:44:52.210" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32566" PostTypeId="2" ParentId="32562" CreationDate="2019-07-07T20:41:36.647" Score="24" Body="&lt;p&gt;With a binocular, all its optical components are fixed -  the user can't change them.  What's important for the user to know is the size of the front lens, which determines the brightness (and in theory sharpness) of the image, the magnification, and the field of view.  These are all useful things to know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A telescope has an interchangable component, namely the eyepiece.  The choice of eyepiece determines:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The magnification (= focal length of telescope divided by focal length of eyepiece)&lt;/li&gt;&#xA;&lt;li&gt;The field of view, which depends on magification and design of the eyepiece&lt;/li&gt;&#xA;&lt;li&gt;The exit pupil size (diameter of objective divided by magnification)&lt;/li&gt;&#xA;&lt;li&gt;and a few other things&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hence, based on a particular eyepiece type you can work out the same information as you get with binoculars.&lt;/p&gt;&#xA;" OwnerUserId="11822" LastEditorUserId="11822" LastEditDate="2019-07-08T07:07:18.537" LastActivityDate="2019-07-08T07:07:18.537" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32568" PostTypeId="2" ParentId="32564" CreationDate="2019-07-07T21:20:55.797" Score="1" Body="&lt;p&gt;Modern telescopes, even those that can be purchased from stores for, say, under $10000 (which is already &lt;em&gt;a lot&lt;/em&gt;), were produced with computer-controlled equipment to polish the mirror and make the lenses and mirrors, and the lenses usually have nano-coating to reduce reflections and increase light transmission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall you have a century of progress &amp;amp; experience building them since then. Mounts are better as well, mechanically and electronically (you have auto-guiding whereas it was manual a century ago, with an astronomer watching a star through a reticle). In other words, modern telescopes, even amateur ones, are vastly improved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can compare photos (from various books such as &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/0307276600&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amazon.com/Day-We-Found-Universe/dp/0307276600&lt;/a&gt;) to modern astrophotography done by amateurs (eg &lt;a href=&quot;https://www.astrobin.com/explore/top-picks/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.astrobin.com/explore/top-picks/&lt;/a&gt;). The modern ones (also benefiting from sensors more sensitive than film) are much better.&lt;/p&gt;&#xA;" OwnerUserId="28651" LastEditorUserId="7982" LastEditDate="2019-07-08T01:09:09.227" LastActivityDate="2019-07-08T01:09:09.227" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32569" PostTypeId="1" CreationDate="2019-07-07T22:31:40.720" Score="2" ViewCount="60" Body="&lt;p&gt;Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Coronagraph#Invention&quot; rel=&quot;nofollow noreferrer&quot;&gt;Coronagraph; Invention&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;High Altitude Observatory's Mark IV Coronagraph on top of Mauna Loa, uses polarization to distinguish sky brightness from the image of the corona: both coronal light and sky brightness are scattered sunlight and have similar spectral properties, but the coronal light is Thomson-scattered at nearly a right angle and therefore undergoes scattering polarization, while the superimposed light from the sky near the Sun is scattered at only a glancing angle and hence remains nearly unpolarized.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How does this work? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I looked at the daytime sky through a polarizing filter and I didn't notice substantial sky darkening? How could a polarizer have darkened the sky sufficiently to make the solar corona more visible?&lt;/p&gt;&#xA;" OwnerUserId="28650" LastEditorUserId="7982" LastEditDate="2019-07-08T01:05:59.090" LastActivityDate="2019-07-08T08:09:47.937" Title="How could the polarizer in early coronagraphs make the Sun's corona more visible against the daytime sky?" Tags="&lt;the-sun&gt;&lt;atmospheric-effects&gt;&lt;corona&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32570" PostTypeId="1" AcceptedAnswerId="32571" CreationDate="2019-07-08T01:03:13.977" Score="1" ViewCount="56" Body="&lt;p&gt;In this diagram of the regions and approximate scale of an AGN:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dYrG2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dYrG2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is from &lt;a href=&quot;https://arxiv.org/pdf/1709.00019.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;, what do the x and y scales &lt;code&gt;log(z/pc)&lt;/code&gt; and &lt;code&gt;log(r/pc)&lt;/code&gt; mean? I'm guessing &lt;code&gt;z&lt;/code&gt; is the z-direction ('upwards') and &lt;code&gt;r&lt;/code&gt; is radius, but what is the meaning of the pc bit?&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2019-07-08T02:31:31.980" Title="Regions and size of an AGN" Tags="&lt;quasars&gt;&lt;active-galaxy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32571" PostTypeId="2" ParentId="32570" CreationDate="2019-07-08T02:31:31.980" Score="1" Body="&lt;p&gt;&lt;code&gt;pc&lt;/code&gt; is the abbreviation for &lt;a href=&quot;https://en.wikipedia.org/wiki/Parsec&quot; rel=&quot;nofollow noreferrer&quot;&gt;parsec&lt;/a&gt;, a common unit of distance objects outside the Solar system. In order to compress the scales, they have also taken the logarithm (to base 10) of the distances in parsecs. So the scale in the figure on the x-axis in the radial (&lt;code&gt;r&lt;/code&gt;) direction from the central black hole spans from 1e-5 to 100 parsecs. The z-axis, which is perpendicular or normal to the plane of the galaxy and the accretion disk, spans from 1e-5 to 1000 parsecs.&lt;/p&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2019-07-08T02:31:31.980" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32572" PostTypeId="2" ParentId="32569" CreationDate="2019-07-08T07:27:46.530" Score="2" Body="&lt;p&gt;When you look at the Sun in white (visible) light, what you see is actually sunlight that has scattered from free electrons in the solar corona. This is known as &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thomson_scattering&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thomson scattering&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we view the corona around the Sun (from our perspective), then the scattered light that reaches us has scattered through a large angle (roughly 90 degrees). The sunlight incident upon the electrons in the corona is unpolarised (that is, the electric field vector of the light is randomly oriented but at right angles to the direction of travel of the light). Such light accelerates the free electrons which re-emit light (this is what scattering is). But the scattered light &lt;em&gt;is&lt;/em&gt; polarised when it is scattered through a large angle, because we effectively can only &quot;see&quot; the electrons accelerating along a line that forms a tangent to the limb of the Sun. The component of the acceleration along the line of sight does not cause any radiation to be emitted towards us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The light we see from the daylight sky is also produced by scattering. In this case it is &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Rayleigh_scattering&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rayleigh scattering&lt;/a&gt;&quot; from electrons that are bound into molecules and atoms in the Earth's atmosphere. The polarisation effects of this scattering are almost identical in nature to that caused by Thomson scattering. However, when we look at the sky right next to the Sun, then this will have been scattered through a very small angle. In these circumstances the scattered light will be &lt;em&gt;unpolarised&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have a means of seeing a contrast between (polarised) light coming from the corona and (unpolarised) light coming from the sky in the same line of sight. By viewing the coronal area using a linear polariser, and then presumably rotating the plane of polarisation and taking a series of pictures, and comparing this with images taken without a polarising filter, we can isolate the small fraction of light from around the Sun that is due to Thomson scattering in its corona.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You made the comment that you don't see the sky darken when looking through a polarising filter. Well, either your filter isn't working or you were looking at the sky roughly 90 degrees away from the Sun and had the polarising filter in just the right orientation, or your sky is very polluted (the scattering caused by particulates does not cause the same degree of polarisation). Daytime sky is most definitely polarised depending on angular separation from the Sun. &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/6/6e/Mapdeg-ortho.gif/330px-Mapdeg-ortho.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;This map&lt;/a&gt; shows how the degree of linear polarisation changes with position as the Sun moves across the sky. The Sun is where the black circle is (black indicating unpolarised light). See also &lt;a href=&quot;https://physics.stackexchange.com/a/196057/83380&quot;&gt;this answer&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="7982" LastEditDate="2019-07-08T08:09:47.937" LastActivityDate="2019-07-08T08:09:47.937" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32574" PostTypeId="2" ParentId="32562" CreationDate="2019-07-08T09:08:43.113" Score="2" Body="&lt;p&gt;You've been given good answers already. Let me just drive this point home:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you buy a truck, you might be interested in things such as how much stuff it can carry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you buy a sedan for your daily commute, you might be interested in the gas mileage, and the comfort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you buy a race car, all you care about is top speed and acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same with binoculars and telescopes. Manufacturers emphasize the things that are important for each category.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastActivityDate="2019-07-08T09:08:43.113" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32575" PostTypeId="1" CreationDate="2019-07-08T11:18:16.163" Score="3" ViewCount="77" Body="&lt;p&gt;This &lt;a href=&quot;https://astronomy.stackexchange.com/a/32565/7982&quot;&gt;in-depth answer&lt;/a&gt; mentions that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...TDB is updated every year. As technology improves, these subtle updates change the timing of past events. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This makes me wonder...&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Does this updating result in yearly discontinuities in TDB relative to a smoothly-running (non-discontinuous) atomic clock?&lt;/li&gt;&#xA;&lt;li&gt;If not, does it result in discontinuities in the first derivative of the difference?&lt;/li&gt;&#xA;&lt;li&gt;Is TDB ever non-monotonic?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="616" LastEditDate="2019-07-09T20:12:37.157" LastActivityDate="2019-07-09T20:12:37.157" Title="Is TDB monotonic? Does it or any of its derivatives have yearly discontinuities with respect to an atomic clock?" Tags="&lt;fundamental-astronomy&gt;&lt;general-relativity&gt;&lt;time&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32576" PostTypeId="1" AcceptedAnswerId="43589" CreationDate="2019-07-08T14:05:57.373" Score="4" ViewCount="312" Body="&lt;p&gt;Recently I am searching for a Shack-Hartmann wavefront sensor for university laboratory usage. I was expecting my target sensor to be cheap, which is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Low spatial wavefront resolution (50 x 50) for beam measurement;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Low frame rate;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Portable design, and easy to use.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Most quotations are beyond my expectation. On ThorLabs it costs 4k USD (which is, &quot;cheaper&quot; compared to most competitors):&#xA;&lt;a href=&quot;https://www.thorlabs.com/thorproduct.cfm?partnumber=WFS30-5C&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.thorlabs.com/thorproduct.cfm?partnumber=WFS30-5C&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following reasons may to blame for this high price:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Expensive industrial sensors. But for &lt;a href=&quot;https://www.flir.com/products/grasshopper3-usb3/?model=GS3-U3-14S5M-C&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.flir.com/products/grasshopper3-usb3/?model=GS3-U3-14S5M-C&lt;/a&gt; it takes only 1.7k USD for a high performance sensor.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Microlens array (MLA) fabrication?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;MLA-sensor alignment?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Software development? (SDK &amp;amp; GUI)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But I am not convinced by myself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas why a Shack-Hartmann sensor is so expensive?&lt;/p&gt;&#xA;" OwnerUserId="16553" LastActivityDate="2021-04-28T23:20:48.683" Title="Why are Shack-Hartmann sensors so expensive (4k+ USD)?" Tags="&lt;optics&gt;&lt;adaptive-optics&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32577" PostTypeId="2" ParentId="20928" CreationDate="2019-07-08T18:26:46.637" Score="5" Body="&lt;p&gt;As the older answers stated, Earthshine is what you're seeing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compared to the Moon as seen from Earth, Earth as seen from the Moon has almost four times the apparent diameter, and close to 14 times the apparent surface area. If Earth's reflectivity were the same as the Moon's, that would make Earthlight on the Moon 14 times brighter than moonlight on the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But Earth is &lt;em&gt;shinier&lt;/em&gt; than the Moon. The Moon's &quot;albedo&quot; (the proportion of light it reflects instead of absorbing) averages 0.12, about the same as a well-traveled asphalt road; that means it reflects back only 1/8 of the light that hits it. Earth's albedo averages 0.37, three times greater -- and if there's a lot of cloud cover, it can be even higher (since clouds are white).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, think about how well you can see the Earth's landscape under a nearly-full Moon. If you were standing on the Moon's night side under a nearly-full Earth, there would be &lt;em&gt;nearly 50 times&lt;/em&gt; as much light. No wonder we can see it from Earth!&lt;/p&gt;&#xA;" OwnerUserId="26791" LastActivityDate="2019-07-08T18:26:46.637" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32578" PostTypeId="1" CreationDate="2019-07-08T19:36:05.927" Score="6" ViewCount="426" Body="&lt;p&gt;Imagine you have a particle accelerator that you can crank up to arbitrarily high energies.  Because of General Relativity, the particles get heavier and heavier as you dump more energy into them.  Will these particles ever become a black hole from this relativistic mass?  If so, what will an observer particle moving parallel to the black hole, just below that speed, see?  What I'm getting at is, kinetic energy is relative.  So therefore relativistic mass is as well.  So would these particles be black holes to some observers but not others?&lt;/p&gt;&#xA;" OwnerUserId="28106" LastActivityDate="2020-05-05T19:05:54.757" Title="Can you create a black hole using kinetic energy?" Tags="&lt;black-hole&gt;&lt;general-relativity&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32579" PostTypeId="1" AcceptedAnswerId="32960" CreationDate="2019-07-08T20:54:26.587" Score="1" ViewCount="72" Body="&lt;p&gt;What are some examples of timekeeping that are based on the height/angle/position of the sun. Two examples that come to mind are the times of the Islamic prayers times and the golden hour in photography. &lt;/p&gt;&#xA;" OwnerUserId="28659" LastEditorUserId="28659" LastEditDate="2019-07-08T21:24:05.030" LastActivityDate="2019-08-07T22:14:01.240" Title="What are some examples of routine human activity times for certain cultures/peopes that are based on the position of the sun?" Tags="&lt;the-sun&gt;&lt;time&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32580" PostTypeId="2" ParentId="32579" CreationDate="2019-07-08T21:15:43.830" Score="1" Body="&lt;p&gt;We are now dependent for timekeeping on clocks, most clocks are synced (directly or indirectly) to atomic clocks that keep time without direct reference to the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the principal definition of clock time, UTC, is coordinated to the position of the sun, through the insertion of leap seconds every 18 months or so. So anything that happens at a particular time is coordinated to the position of the sun, though not directly.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-07-08T21:15:43.830" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32581" PostTypeId="2" ParentId="32578" CreationDate="2019-07-08T21:50:18.357" Score="-4" Body="&lt;p&gt;Relativistic mass increase is inertial, not gravitational, so no, you couldn't make a black hole by accelerating a particle. What happens is that the more energy you use to accelerate the particle, the more its inertial mass increases, but the more its inertial mass increases, the harder it is to accelerate. There comes a point near the speed of light when virtually all the energy you pump into the system is manifested as mass increase rather than velocity, and you can't make the particle go any faster. The best idea, if you want a really fast particle with huge amounts of energy, much more than could be produced by an accelerator, is to use a cosmic ray. Cosmic rays are fast protons or atomic nuclei which have been accelerated by natural events such as supernova explosions to much higher energies than you could get in an accelerator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cosmic rays have been used for particle experiments, but the snag is that they are random events, so you don't know exactly where to put your particle detectors to intercept the best ones. To give you an idea of how great their energies can sometimes be, the most powerful cosmic ray particles have the kinetic energy of a rifle bullet, yet the atomic nucleus which carries this energy is so minute that it cannot be seen in even the most powerful electron microscope!&lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-07-08T21:50:18.357" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32582" PostTypeId="1" AcceptedAnswerId="32584" CreationDate="2019-07-09T05:39:32.633" Score="9" ViewCount="3983" Body="&lt;p&gt;I know that we have measured the rate a lot of stars and galaxies move away from us using the doppler shift, and I know that the further a star / galaxy is the faster they accelerate away from us due to space inflation. I'm wondering, did anyone categorise which star / galaxy is moving towards us / away from us the fastest? How fast are they going in relation to our solar system?&lt;/p&gt;&#xA;" OwnerUserId="28506" LastEditorUserId="7982" LastEditDate="2019-07-09T07:03:22.340" LastActivityDate="2021-03-31T10:52:11.633" Title="Which star / galaxy is moving away from us the fastest?" Tags="&lt;star&gt;&lt;galaxy&gt;&lt;redshift&gt;&lt;speed&gt;&lt;movement&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32583" PostTypeId="2" ParentId="32576" CreationDate="2019-07-09T06:47:36.740" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why are Shack-Hartmann sensors so expensive (4k+ USD)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I had the same feeling the first time I saw their prices. In principle they are just an array of microlenses one focal length in front of a CCD or CMOS imaging chip. It can even be a pinhole array in some specific cases, depending on what you need to resolve and the nature of your incoming wavefront.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some examples of  being used to analyze the optical performance of astronomical telescopes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In preparation for the Nasmyth focus of one of the VLT unit telescopes: &lt;a href=&quot;https://www.eso.org/sci/libraries/SPIE2008/7012-124.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Design and pe rformances of the Shac k-Ha rtmann sensor within&#xA;the A ctive Phasing E xpe riment&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;ResearchGate: &lt;a href=&quot;https://www.researchgate.net/publication/227199650_Shack-Hartmann_wavefront_sensor_for_testing_the_quality_of_the_optics_of_the_25-m_SAI_telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shack-Hartmann wavefront sensor for testing the quality of the optics of the 2.5-m SAI telescope&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Alcor System: &lt;a href=&quot;http://www.alcor-system.com/new/SH/Shackscope40x40.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shack-Hartman&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Spot Optics: &lt;a href=&quot;http://www.spot-optics.com/downloads/pdf/telsc_cassegrainfocus.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shack-Hartmann wavefront sensor test of a telescope at Cassegrain&#xA;focus&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If the 4k+ is well outside the range you'd like to spend on this application, then consider building your own for some simple tests. After trying that a while you may discover that your results are off, or non-repeatable, or getting the software to reliably extract a reasonable wavefront shape is harder than you thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But yes, &lt;em&gt;in principle&lt;/em&gt; it's just a dark box with an array of lenses or holes and a CCD or CMOS imaging array, and &lt;em&gt;a bunch of software&lt;/em&gt; that you'd have to write from scratch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll have to mount the array and the sensor rigidly, take care to manage every bit of stray light (especially any with spatial structure!) think about thermal expansion, write the capture, processing, and wavefront-extraction software, etc yourself. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There are several things that drive the price up higher than you would initially expect. There's quality in workmanship, a warrantee, a guaranteed spec that it will meet, software, support, accessible repair, a large installed base and track record, etc. In the end these add up to the costs that you encounter. Add to that the idea that if they made a low-end sensor, people wouldn't be forced to buy the expensive one! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ihieV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ihieVm.jpg&quot; alt=&quot;Shack-Hartmann sensor Alcor-Systems&quot;&gt;&lt;/a&gt; Source: &lt;a href=&quot;http://www.alcor-system.com/new/SH/Shackscope40x40.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Alcor-Systems&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-07-09T06:58:30.647" LastActivityDate="2019-07-09T06:58:30.647" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32584" PostTypeId="2" ParentId="32582" CreationDate="2019-07-09T07:33:13.520" Score="20" Body="&lt;p&gt;When a galaxy recedes from us, the light we see from it is redshifted. For galaxies at cosmological distances, this redshift is fundamentally different from a Doppler shift; whereas the latter is due to a velocity difference between the emitter and the receiver, a cosmological redshift is due to photons traveling through an expanding space&lt;span class=&quot;math-container&quot;&gt;$^\dagger$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Hence, as @uhoh comments, your question is equivalent to asking &amp;quot;Which galaxy has the highest measured redshift?&amp;quot;. Redshift is arguably the single most important concept in astronomy, and indeed we do catalogue the redshift of all galaxies, if possible. For our adopted cosmological model, the cosmological redshift can be translated to both a recession velocity, a distance, and an age of the Universe when the light was emitted.&lt;/p&gt;&#xA;&lt;p&gt;The answer is &lt;a href=&quot;https://en.wikipedia.org/wiki/GN-z11&quot; rel=&quot;nofollow noreferrer&quot;&gt;GN-z11&lt;/a&gt; (&lt;a href=&quot;https://arxiv.org/abs/1603.00461&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oesch et al. 2016&lt;/a&gt;) which has a redshift of &lt;span class=&quot;math-container&quot;&gt;$z=11.09$&lt;/span&gt;. This corresponds to a distance of &lt;span class=&quot;math-container&quot;&gt;$d = 32.2\,\mathrm{Glyr}$&lt;/span&gt; (i.e. billion light-years), and hence, by Hubble's law, to a recession velocity of&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;v = H_0\,d = 670\,000\,\mathrm{km}\,\mathrm{s}^{-1},&#xA;$$&lt;/span&gt;&#xA;or more than twice the speed of light. Furthermore, the light we see today was emitted when the Universe was only 410 Myr (i.e. million years), or 3% of its current age.&lt;/p&gt;&#xA;&lt;p&gt;You may think that &amp;quot;twice the speed of light&amp;quot; violates the theory of relativity, but this velocity is not a velocity &lt;em&gt;through&lt;/em&gt; space. Both our galaxy (the Milky Way) and GN-z11 move &lt;em&gt;through&lt;/em&gt; space at modest velocities of a few 100 km/s. The recession is merely due to space expanding, and space is allowed to expand at whatever rate is wishes.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^\dagger$&lt;/span&gt;&lt;sub&gt;&amp;quot;Fundamentally different&amp;quot; might be a too strong statement, since the cosmological redshift can be interpreted as infinitely many infinitesimally small Doppler shifts. However, a hypothetical scenario that emphasizes the difference between the two types of redshifts is the following:&lt;br&gt;&#xA;If an emitter and an observer are stationary wrt. each other when the emitter emits a photon, then start moving away from each other while the photon is traveling, then stop again before the observer receives the photon, the observer would measure zero redshift.&lt;br&gt;&#xA;On the other hand, if space is static when the emitter emits the photon, then while the photon is traveling suddenly expands by a factor of, say, four, and then is static again when the observer receives the photon, the observer would measure a redshift of &lt;span class=&quot;math-container&quot;&gt;$z+1 = 4$&lt;/span&gt;.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5264" LastEditorUserId="5264" LastEditDate="2021-03-31T10:52:11.633" LastActivityDate="2021-03-31T10:52:11.633" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32585" PostTypeId="2" ParentId="32582" CreationDate="2019-07-09T07:37:29.763" Score="4" Body="&lt;p&gt;From what I know I don't think they have categorized that this specific galaxy, for example, is the fastest one moving away from the Milky Way Galaxy. There are two reasons for this:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First of all, this isn't really that important. Measuring which galaxy is moving that fastest away from us can have some implications but not a whole lot. Scientists are however interested in the fastest speed any galaxy can move away from us. That does not, of course, violate the Theory of Relativity. And that speed is faster than the speed of light. Here is some extra info:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The Hubble constant is the measure of how fast the Universe is expanding today and its value has been measured to be 70 km/s per Megaparsec (a parsec is just a unit of distance equal to about 3.26 light-years, and a Megaparsec is a million parsecs). This means that on average, for every Megaparsec two galaxies are separated by, they are moving away from each other by 70 km/s. Therefore, to be moving away from each other at the speed of light, two galaxies would need to be separated by a distance of about 4,300 million parsecs. This is smaller than the radius of the observable Universe, therefore not only are there galaxies in the Universe that are moving away from us faster than light, but we can still see them!&lt;/p&gt;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://curious.astro.cornell.edu/legal-information/104-the-universe/cosmology-and-the-big-bang/expansion-of-the-universe/1066-can-two-galaxies-move-away-from-each-other-faster-than-light-intermediate&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://curious.astro.cornell.edu/legal-information/104-the-universe/cosmology-and-the-big-bang/expansion-of-the-universe/1066-can-two-galaxies-move-away-from-each-other-faster-than-light-intermediate&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Second, we can't really be sure even if we research really hard, because every day we detect new objects circulating in our solar system, and for galaxies, we are talking about mega-big distances.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion:&lt;/strong&gt; So it is possible, but it is really time-consuming and abstract in a way, so it isn't something scientists are looking into today. But if you want some galaxies which are moving really fast away from us you can search the internet or read some books and you will find a lot. Of course, there is this certain galaxy which at the moment is categorized as the fastest or most-redshifted (if you understand what I'm saying) that we have detected (remark: detected) and it is &lt;a href=&quot;https://en.wikipedia.org/wiki/GN-z11&quot; rel=&quot;nofollow noreferrer&quot;&gt;GN-Z11&lt;/a&gt;. It is the most distant found galaxy in the whole observable universe, so the fact that it is the fastest moving away from us makes sense.&lt;/p&gt;&#xA;" OwnerUserId="28662" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-08-06T09:35:38.803" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="32587" PostTypeId="1" CreationDate="2019-07-09T09:02:00.147" Score="4" ViewCount="78" Body="&lt;p&gt;I have two methods which allow for a detection of a signal that is at most of magnitude x. If method 2 can detect signal that are half a magnitude larger, i.e. fainter, let's say x.5 mag, is it correct to say that method 2 can detect signals that are about 60% fainter than what method 1 can detect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the 60% correct? I got this number by converting the magnitudes &lt;span class=&quot;math-container&quot;&gt;$m_i$&lt;/span&gt; to flux &lt;span class=&quot;math-container&quot;&gt;$f_i$&lt;/span&gt;, i.e.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$100^{(m_1-m_2)/2.5}=\frac{f_2}{f_1}\approx 0.40$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\Rightarrow f_2=0.40 f_1\sim 40\% f_1$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, &lt;span class=&quot;math-container&quot;&gt;$f_2$&lt;/span&gt; is about &lt;span class=&quot;math-container&quot;&gt;$40\% f_1$&lt;/span&gt; which means about &lt;span class=&quot;math-container&quot;&gt;$60\%$&lt;/span&gt; fainter. I have never worked with magnitudes before, so is this a valid statement?&lt;/p&gt;&#xA;" OwnerUserId="28663" LastEditorUserId="28663" LastEditDate="2019-07-09T10:59:30.113" LastActivityDate="2019-07-09T20:31:19.437" Title="Comparing different magnitudes. Is this statement correct?" Tags="&lt;apparent-magnitude&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32589" PostTypeId="2" ParentId="32587" CreationDate="2019-07-09T18:54:39.063" Score="5" Body="&lt;p&gt;Textbooks typically express the flux vs. magnitude relation something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$m_2 - m_1 = -2.5 \log_{10} \frac{f_2}{f_1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which we can transform into this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{f_2}{f_1} = 10^{(m_1 - m_2) / 2.5}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the original version of your formula was correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If m&lt;sub&gt;2&lt;/sub&gt; - m&lt;sub&gt;1&lt;/sub&gt; = 0.5, then f&lt;sub&gt;2&lt;/sub&gt; / f&lt;sub&gt;1&lt;/sub&gt; = 0.63.&#xA;Some readers find comparisons like &quot;x% fainter&quot; confusing, so I would say method 2 is more sensitive than method 1 by a factor of 1.58.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-07-09T20:31:19.437" LastActivityDate="2019-07-09T20:31:19.437" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32590" PostTypeId="1" CreationDate="2019-07-09T19:53:28.630" Score="2" ViewCount="61" Body="&lt;p&gt;Let's say we have a gas giant composed primarily of hydrogen and hydrogen-rich molecules that is ejected from its star system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this planet embarks on its potentially infinite journey out into the cold darkness of space, away from the light and warmth of its host star, what would happen to its atmosphere? Would the gasses freeze and fall, piling up as winds cease? Would mass be lost? Or would the planet just remain in a relatively similar state? I know Jupiter radiates more heat than it receives from the Sun.&lt;/p&gt;&#xA;" OwnerUserId="26425" LastActivityDate="2019-07-09T19:53:28.630" Title="What would happen to the atmosphere of a rogue jovian planet?" Tags="&lt;planetary-atmosphere&gt;&lt;gas-giants&gt;&lt;rogue-planet&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32592" PostTypeId="1" AcceptedAnswerId="32602" CreationDate="2019-07-09T22:26:11.797" Score="3" ViewCount="296" Body="&lt;p&gt;If the solar system were suddenly placed in the middle of a large void... would everything still function as normal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would we even notice any difference here on Earth apart from the night sky?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...and speaking of the night sky, would that even look much different to how it is already? I imagine just less colourful and and more sparse, but seeing whole galaxies instead of individual stars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or would moving the solar system into a void just mean the end of the world as we know it?&lt;/p&gt;&#xA;" OwnerUserId="7845" LastActivityDate="2019-07-17T12:37:27.173" Title="Could the solar system survive in a void?" Tags="&lt;galaxy&gt;&lt;solar-system&gt;&lt;earth&gt;&lt;life&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32593" PostTypeId="1" AcceptedAnswerId="32599" CreationDate="2019-07-09T22:50:55.667" Score="0" ViewCount="253" Body="&lt;p&gt;There are some stars in the NASA Exoplanet Archive that I am interested in but I am confused by the physical meaning of the terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, there's a star BD+15 2940 with a log10 of Solar Luminosity of 2.01 +/-0.75 where the variation is also a log.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is the linear equivalents 18.2 to 575.44 with most likely value 102.33?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went to the citing paper &lt;a href=&quot;https://arxiv.org/pdf/1304.6755.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nowak et al. 2013&lt;/a&gt;and they quote it as log L*/LSun 2.01 +/- 0.75 as per the NASA archive.&lt;/p&gt;&#xA;" OwnerUserId="25513" LastEditorUserId="25513" LastEditDate="2019-07-09T23:32:53.090" LastActivityDate="2019-07-10T07:47:47.240" Title="Converting the log (Stellar Luminosity/Solar Luminosity) and the log of the plus and minus variation" Tags="&lt;luminosity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32594" PostTypeId="2" ParentId="32592" CreationDate="2019-07-10T00:12:34.137" Score="1" Body="&lt;p&gt;As far as we can tell, the solar system is in such a void already. The region of space outside of the solar system is thought to be pretty sparse but that is hard to tell. We know the nearest star is Alpha Centauri which is over 4 light-years away. If the solar system was moved to an even less dense area if would not be noticeable. There may be planetary bodies nearby to our solar system but we currently have no way of finding these. &lt;/p&gt;&#xA;" OwnerUserId="16926" LastActivityDate="2019-07-10T00:12:34.137" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32595" PostTypeId="1" AcceptedAnswerId="32600" CreationDate="2019-07-10T00:22:53.827" Score="4" ViewCount="161" Body="&lt;p&gt;The Phys.org article &lt;a href=&quot;https://phys.org/news/2019-07-interstellar-iron-isnt-plain-sight.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Interstellar iron isn't missing, it's just hiding in plain sight&lt;/a&gt; links to the paywalled article &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/1538-4357/ab22b7&quot; rel=&quot;nofollow noreferrer&quot;&gt;On the Structure, Magnetic Properties, and Infrared Spectra of Iron Pseudocarbynes in the Interstellar Medium&lt;/a&gt; (Tarakeshwar, Buseck and Timmes, Arizona State University), and says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The new work may also solve another longstanding puzzle. &lt;strong&gt;Carbon chains with more than nine atoms are unstable, the team explains.&lt;/strong&gt; Yet observations have detected more complex carbon molecules in interstellar space. How nature builds these complex carbon molecules from simpler carbon molecules has been a mystery for many years.&lt;/p&gt;&#xA;&lt;p&gt;Buseck explained, &amp;quot;&lt;strong&gt;Longer carbon chains are stablized by the addition of iron clusters.&lt;/strong&gt;&amp;quot; This opens a new pathway for building more complex molecules in space, such as polyaromatic hydrocarbons, of which naphthalene is a familiar example, being the main ingredient in mothballs.&lt;/p&gt;&#xA;&lt;p&gt;Said Timmes, &amp;quot;&lt;strong&gt;Our work provides new insights into bridging the yawning gap between molecules containing nine or fewer carbon atoms and complex molecules&lt;/strong&gt; such as C60 buckminsterfullerene, better known as 'buckyballs.'&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; On Earth we don't see a big difference between hydrocarbon chains with lengths below and above 9 (think kerosene, wax...), why is there such a cutoff in stability in interstellar space? What is it about hydrocarbon chains longer than 9 atoms that makes them unstable there but not here?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-07-15T17:26:08.203" Title="Why do these astronomers say that &quot;Carbon chains with more than nine atoms are unstable&quot;?" Tags="&lt;interstellar-medium&gt;&lt;astrochemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32596" PostTypeId="1" AcceptedAnswerId="32598" CreationDate="2019-07-10T03:35:17.613" Score="1" ViewCount="47" Body="&lt;p&gt;When a telescope observes a distant star, all it sees is the light being emitted by the star. Observing the wavelength of this light, we determine the velocity of the star. If the wavelength is red-shifted (longer) then the star is moving away from us, and if the wavelength is blue-shifted (shorter) then the star is moving towards us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do we know that the wavelength that we see is not the red shifted wavelength but the original wavelength emitted by the star? Do we already know what wavelength of light the star is supposed to emit (if yes, how?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, using this method we should be able to identify only the radial component of velocity. Because the tangential component of velocity of the star would have no effect on the wavelength that we observe? So how do we determine the actual velocity (including the proper direction) of the star?&lt;/p&gt;&#xA;" OwnerUserId="28672" LastActivityDate="2019-07-10T06:09:22.410" Title="How to identify component of velocity of a star from its red shift?" Tags="&lt;redshift&gt;&lt;velocity&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2019-07-10T19:01:32.440" ContentLicense="CC BY-SA 4.0" />
  <row Id="32597" PostTypeId="2" ParentId="32592" CreationDate="2019-07-10T04:38:34.887" Score="3" Body="&lt;p&gt;If you took away the Milky Way except for the Solar system (but kept the MW's satellite galaxies), the  &lt;a href=&quot;http://curious.astro.cornell.edu/our-solar-system/98-the-universe/galaxies/observing-galaxies/556-can-any-galaxies-be-seen-with-the-naked-eye-beginner&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;naked eye&quot; night sky&lt;/a&gt; would feature the Andromeda Galaxy, 2.5  million light-years away, the Large Magellanic Cloud (LMC), 160,000 light-years away and the&#xA;Small Magellanic Cloud (SMC), 200,000 light years away.&lt;/p&gt;&#xA;" OwnerUserId="17743" LastActivityDate="2019-07-10T04:38:34.887" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32598" PostTypeId="2" ParentId="32596" CreationDate="2019-07-10T06:09:22.410" Score="4" Body="&lt;p&gt;The first part of this question is already answered at &lt;a href=&quot;https://astronomy.stackexchange.com/questions/11874/how-do-we-know-about-redshift&quot;&gt;How Do we know about redshift?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the second part, measuring tangential velocity is &lt;strong&gt;much&lt;/strong&gt; harder, and indeed impossible for stars that are far away. Basically the method is to observe the position of the star very accurately compared to very distant stars which are assumed not to be moving and then observe it again a few years later. After adjusting for known effects like aberration and parallax, what remains is &quot;proper motion&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most sensitive current telescope for this kind of measurement is the Gaia mission, which is able to measure the movemnt of stars up to a few thousand light years away with reasonable accuracy.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-07-10T06:09:22.410" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32599" PostTypeId="2" ParentId="32593" CreationDate="2019-07-10T07:47:47.240" Score="1" Body="&lt;p&gt;You're completely right, except that the authors, as far as I can see, don't specify whether the central value (i.e. &lt;span class=&quot;math-container&quot;&gt;$\log L = 2.01$&lt;/span&gt;, or &lt;span class=&quot;math-container&quot;&gt;$L = 102$&lt;/span&gt;) is the &lt;em&gt;most likely value&lt;/em&gt; (i.e. the &lt;em&gt;mode&lt;/em&gt;), or the &lt;em&gt;average&lt;/em&gt;, or the &lt;em&gt;median&lt;/em&gt;. My guess is it's the average, but especially for logarithmically distributed values, my opinion is that the median would makes more sense to quote, since in that case you'll get the same result whether you take the median of &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; or of &lt;span class=&quot;math-container&quot;&gt;$\log L$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2019-07-10T07:47:47.240" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32600" PostTypeId="2" ParentId="32595" CreationDate="2019-07-10T08:48:00.153" Score="6" Body="&lt;p&gt;Unlike the saturated hydrocarbons in kerosene, &lt;a href=&quot;https://en.wikipedia.org/wiki/Linear_acetylenic_carbon&quot; rel=&quot;nofollow noreferrer&quot;&gt;carbynes&lt;/a&gt; are unsaturated carbon chains with alternating single and triple bonds.&#xA;Molecules containing such chains are called &lt;a href=&quot;https://en.wikipedia.org/wiki/Polyyne&quot; rel=&quot;nofollow noreferrer&quot;&gt;polyynes&lt;/a&gt;, e.g. the short &lt;a href=&quot;https://en.wikipedia.org/wiki/Cyanopolyyne&quot; rel=&quot;nofollow noreferrer&quot;&gt;cyanopolyyne&lt;/a&gt; HC&lt;sub&gt;5&lt;/sub&gt;N:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H&amp;minus;C&amp;equiv;C&amp;minus;C&amp;equiv;C&amp;minus;C&amp;equiv;N&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those carbon atoms readily interact, and long chains (if they form; see comments) are more likely to crosslink or form cycles than to remain linear.&#xA;Tarakeshwar et al. suggest that iron clusters in &quot;pseudocarbynes&quot; inhibit this by bonding to some of the carbons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2016MNRAS.463.4175L/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Loomis et al. 2016&lt;/a&gt; looked for cyanopolyynes in radio spectra of &lt;a href=&quot;https://en.wikipedia.org/wiki/Taurus_Molecular_Cloud&quot; rel=&quot;nofollow noreferrer&quot;&gt;Taurus molecular cloud 1&lt;/a&gt;.&#xA;They got a good HC&lt;sub&gt;9&lt;/sub&gt;N signal but did not detect the HC&lt;sub&gt;11&lt;/sub&gt;N lines &lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/1996ApJ...469L..65T/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Travers et al. 1996&lt;/a&gt; observed in the lab.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-07-15T17:26:08.203" LastActivityDate="2019-07-15T17:26:08.203" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32601" PostTypeId="1" AcceptedAnswerId="32615" CreationDate="2019-07-10T17:49:43.593" Score="8" ViewCount="4396" Body="&lt;p&gt;I want to make a projection of millisecond pulsars on the galactic plane, much like this one from &lt;a href=&quot;https://www.esa.int/gsp/ACT/projects/pulsar-navigation.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sala et al. 2004&lt;/a&gt;: &#xA;&lt;a href=&quot;https://i.stack.imgur.com/8sUOm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8sUOm.png&quot; alt=&quot;From Josep Sala, &amp;quot;Feasibility Study for a Spacecraft Navigation System...&amp;quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried several methods, and have not gotten anywhere. This is my current code, along with what it is producing:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import math as m&#xA;import numpy as np&#xA;import csv&#xA;import matplotlib.pyplot as plt&#xA;import pandas as pd&#xA;import astropy.coordinates as coord&#xA;import astropy.units as u&#xA;from astropy.io import ascii&#xA;from astropy.coordinates import SkyCoord&#xA;&#xA;data = pd.read_csv('galacticwperiod.csv')&#xA;&#xA;xarr = np.array(data.iloc[:,0])&#xA;yarr = np.array(data.iloc[:,1])&#xA;eq = SkyCoord(xarr[:], yarr[:], unit=u.deg)&#xA;gal = eq.galactic&#xA;&#xA;#print(xarr)&#xA;#xarr = np.array(df.iloc[:,0])&#xA;#yarr = np.array(df.iloc[:,1])&#xA;#zarr = np.array(df.iloc[:,2])&#xA;&#xA;#ra = coord.Angle(xarr[:], unit=u.hour)&#xA;#ra.degree&#xA;#ra = ra.wrap_at(180*u.degree)&#xA;#dec = coord.Angle(yarr[:], unit=u.deg)&#xA;#print(ra)&#xA;&#xA;plt.figure(figsize=(6,5))&#xA;fig = plt.figure()&#xA;ax = fig.add_subplot(111, projection=&quot;aitoff&quot;)&#xA;plt.plot(gal.l.wrap_at(180*u.deg), gal.b.wrap_at(180*u.deg), linestyle='None')&#xA;ax.scatter(gal.l, gal.b, linestyle='None')&#xA;&#xA;&#xA;#ax.set_facecolor('xkcd:battleship grey')&#xA;#fig.patch.set_facecolor('white')&#xA;#ax.tick_params(axis='both', which='major', labelsize=10)&#xA;#ax.grid(color='b', linestyle='solid')&#xA;fig.show()&#xA;#plt.savefig('millisecondcoloraitoff.png', dpi=600)&#xA;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here are a few lines of the input file 'galacticwperiod.csv':&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Gl,Gb&#xA;111.383,-54.849&#xA;305.898,-44.888&#xA;305.913,-44.877&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is the image it produces:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/FowhM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FowhM.png&quot; alt=&quot;My plot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am almost certain this is wrong because they are not distributed along the galactic plane, which they should be. The data I am using is from the &lt;a href=&quot;http://www.atnf.csiro.au/research/pulsar/psrcat/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ATNF Pulsar Catalog&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the sites I have already looked at, for reference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://stackoverflow.com/questions/33105898/astropy-matplotlib-and-plot-galactic-coordinates&quot;&gt;https://stackoverflow.com/questions/33105898/astropy-matplotlib-and-plot-galactic-coordinates&lt;/a&gt;&#xA;&lt;a href=&quot;https://astropy4scipy2014.readthedocs.io/_static/notebooks/06_Celestial_Coordinates_solutions.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://astropy4scipy2014.readthedocs.io/_static/notebooks/06_Celestial_Coordinates_solutions.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help with this would be greatly appreciated. &lt;/p&gt;&#xA;" OwnerUserId="28671" LastEditorUserId="616" LastEditDate="2019-07-17T17:45:29.477" LastActivityDate="2019-07-17T17:45:29.477" Title="How do I plot galactic coordinates using matplotlib and astropy in python?" Tags="&lt;coordinate&gt;&lt;python&gt;&lt;astropy&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32602" PostTypeId="2" ParentId="32592" CreationDate="2019-07-10T19:12:05.813" Score="2" Body="&lt;p&gt;The solar system could continue to exist outside a galaxy, and apart from a lack of stars in the night sky, I see no reason why anything much would change. It would be very unlikely to form there, even if some chance provided a sufficiently dense gas cloud. The cloud would be made up of almost nothing except hydrogen and helium (since there would not have been previous generations of stars to enrich it) so (a) it would be hard for a star as small as the Sun to form and (b) there would certainly be no solid planets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A void is just a region with a relatively low density of galaxies. Relatively low does not mean zero, so the solar system could continue to exist, or even form, in a galaxy in a void with no problems, but in most parts of most voids it could not form.&lt;/p&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-07-10T19:12:05.813" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32603" PostTypeId="2" ParentId="32592" CreationDate="2019-07-10T20:12:32.747" Score="2" Body="&lt;p&gt;Actually the further away the Solar System is from anything else the better for life. There would be less radiation and fewer mass extinction events from supernovas. We would not see galaxies instead of stars; their brightness would not increase. In fact if we were located in one of the great intergalactic voids the night sky would be completely dark to the naked eye.&lt;/p&gt;&#xA;" OwnerUserId="28600" LastActivityDate="2019-07-10T20:12:32.747" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32604" PostTypeId="1" CreationDate="2019-07-10T21:52:06.777" Score="4" ViewCount="129" Body="&lt;p&gt;I'm thinking about the recent famous photo of a black hole, some months ago. It got me wondering if there isn't a way to more easily make a direct observation of these. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We detect exoplanets in part by phenomenae that occur when they transit their star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also have excellent time lapse photography of numerous stars orbiting the Milky Way's central Black Hole (Sgr A*), with timescales quite amenable to human timescales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could we observe Sgr A* directly by observing one of these stars passing behind, or almost behind it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diameter of the BH is estimated at 60m km, more than 40x the diameter of our sun (1.4m km). At that distance, parallax related effects and relative sizes wouldn't be an issue, as the star and BH are virtually at the same distance both very far from earth. A star also wouldn't have to be occluded, just pass close enough behind, for gravitational lensing to be detected on that fraction of its orbit (but not other places).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, a star could also be occluded by an accretion disk (if any), and its light affected in a way that would presumably show something of the nature of what it was passing behind. Admittedly Sgr A* may have little or no disk, but we might detect a disk of low enough density/size/activity not to be active, if a residual disk did in fact exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are we likely to be able to see in time lapse photography, stars vanishing and reappearing behind an apparently empty patch of space, in this way? Has it been done?&lt;/p&gt;&#xA;" OwnerUserId="18601" LastEditorUserId="7982" LastEditDate="2019-07-11T01:17:59.833" LastActivityDate="2019-07-11T01:28:18.067" Title="How feasible is it that we could see the Central Black Hole Sgr A* occlude one of it's closely orbiting stars?" Tags="&lt;black-hole&gt;&lt;observational-astronomy&gt;&lt;supermassive-black-hole&gt;&lt;sagittarius-a&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32605" PostTypeId="2" ParentId="32604" CreationDate="2019-07-11T01:16:33.923" Score="3" Body="&lt;p&gt;I'm sure someone will write a longer/better answer, so I'll keep this short and just mention that when a black hole passes between the observer an a star, we usually say that it becomes &lt;em&gt;gravitationally lensed&lt;/em&gt; rather than simply occulted. This can potentially provide much richer data than an occultation but requires more complex analysis. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Optical/Infrared&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It also takes heroic efforts to try to image near Sgr A* optically. You need to use infrared wavelengths because of dust and an array of telescopes using &lt;a href=&quot;https://en.wikipedia.org/wiki/Adaptive_optics&quot; rel=&quot;nofollow noreferrer&quot;&gt;adaptive optics&lt;/a&gt;, combined interferometrically. See &lt;a href=&quot;https://astronomy.stackexchange.com/q/28940/7982&quot;&gt;How did they make a video of the center of the galaxy, and what is it exactly that's flashing there?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there was a &lt;em&gt;dedicated, second&lt;/em&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Very_Large_Telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Very Large Telescope&lt;/a&gt; array pointed at SgrA* all night every night, there is certainly the possibility that a &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_microlensing&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravitational microlensing event&lt;/a&gt; could occur, and an otherwise invisible star could suddenly brighten substantially as it passed directly behind Sgr A*. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Are there candidates?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As far as I know there are no currently known stars closely orbiting Sgr A* that are likely to pass directly behind it. However, I suppose that there could be one so dim that it has not yet been observed, and a lensing event would be the first time it gets noticed, so this is certainly a possibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But lensing can brighten any background star, it would not have to be in a close orbit around Sgr A* at all. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Radio&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Gravitational microlensing works essentially the same at all electromagnetic wavelengths, so it can be observed in radio as well as it can optically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, individual stars tend to radiate most of their energy optically and are not so bright at radio wavelengths. It would be hard to pick out the radio radiation of a background star above the strong emission of heated plasma in SgrA*'s accretion disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To this end I have just asked &lt;a href=&quot;https://astronomy.stackexchange.com/q/32606/7982&quot;&gt;How far have individual stars been seen by radio telescopes?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think (but I'm not sure) that you'd have to pick a time of extreme quiet (little/no infall) to even try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eXQ0F.gif&quot; alt=&quot;homemade GIF from ESA video ESOcast 173: First Successful Test of Einstein’s General Relativity Near Supermassive Black Hole&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; homemade GIF from ESA video &lt;a href=&quot;https://www.youtube.com/watch?v=if2opecmev8&quot; rel=&quot;nofollow noreferrer&quot;&gt;ESOcast 173: First Successful Test of Einstein’s General Relativity Near Supermassive Black Hole&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-07-11T01:28:18.067" LastActivityDate="2019-07-11T01:28:18.067" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32606" PostTypeId="1" AcceptedAnswerId="32787" CreationDate="2019-07-11T01:27:36.800" Score="2" ViewCount="305" Body="&lt;p&gt;Writing &lt;a href=&quot;https://astronomy.stackexchange.com/a/32605/7982&quot;&gt;this answer&lt;/a&gt; got me thinking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understand it, &lt;em&gt;stars themselves&lt;/em&gt; radiate most of their energy at optical wavelengths (near-IR through near-UV).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various types of stellar objects may flare in various ways when in-falling matter from an accretion disk or companion is heated in magnetic fields or undergoes transient nucleosynthesis, &lt;strong&gt;but I'm excluding these types of processes&lt;/strong&gt; and asking only about radiation from the stellar object itself, in steady-state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How far have individual stars been seen by radio telescopes?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2020-02-01T22:05:56.793" Title="How far have individual stars been seen by radio telescopes?" Tags="&lt;radio-astronomy&gt;" AnswerCount="3" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="32607" PostTypeId="1" AcceptedAnswerId="32609" CreationDate="2019-07-11T02:49:56.127" Score="0" ViewCount="90" Body="&lt;p&gt;The Fox News article &lt;a href=&quot;https://www.foxnews.com/science/nasa-moon-landing-flagstaff-lunar-history&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arizona city played critical role in moon exploration history&lt;/a&gt; covers several interesting activities that took place in Arizona in preparation for the Apollo Moon landings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It shows the telescope (shown below) and mentions that it was used to draw lunar maps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is it possible to identify which telescope this is exactly? Is there some place where one can read more about it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;click images for full size viewing&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/smZyN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Lowell Observatory telescope used by scientists who collaborated with artists to map out the moon for Apollo astronauts (Fox News)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; Screenshot from July 9, 2019 Fox News video &lt;a href=&quot;https://video.foxnews.com/v/6057102900001/#sp=show-clips&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA lunar legacy in the Arizona desert&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q3VOj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-07-12T09:43:33.497" Title="Which Lowell Observatory telescope is this? (used to draw Moon maps for the Apollo program)" Tags="&lt;telescope&gt;&lt;identify-this-object&gt;&lt;observatory&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32608" PostTypeId="1" AcceptedAnswerId="32616" CreationDate="2019-07-11T03:01:05.763" Score="1" ViewCount="221" Body="&lt;p&gt;The Fox News article &lt;a href=&quot;https://www.foxnews.com/science/nasa-moon-landing-flagstaff-lunar-history&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arizona city played critical role in moon exploration history&lt;/a&gt; covers several interesting activities that took place in Arizona in preparation for the Apollo Moon landings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It shows the telescope (shown below) and mentions that it was used to draw lunar maps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is more information about this telescope &lt;a href=&quot;http://scopeviews.co.uk/LowellClarkRefractor.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://lowell.edu/history/the-clark-refractor/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; as mentioned in &lt;a href=&quot;https://astronomy.stackexchange.com/a/32609/7982&quot;&gt;this answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Why does this Lowell Observatory telescope have so many knobs? What do they all do? Is it possible to identify the function of all of the annotated controls and devices on this complicated-looking telescope?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've added some numbers to facilitate discussion. There is also a chain going between the locations of #4 and #5, though I don't know exactly what it couples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;click images for full size viewing&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2YnGs.jpg&quot; alt=&quot;a Lowell Observatory telescope, annotated&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; Lowell Observatory telescope used by scientists who collaborated with artists to map out the moon for Apollo astronauts (Fox News) &lt;a href=&quot;https://www.foxnews.com/science/nasa-moon-landing-flagstaff-lunar-history&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; Screenshot from July 9, 2019 Fox News video &lt;a href=&quot;https://video.foxnews.com/v/6057102900001/#sp=show-clips&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA lunar legacy in the Arizona desert&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q3VOj.jpg&quot; alt=&quot;a Lowell Observatory telescope&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-07-12T06:09:38.417" LastActivityDate="2019-07-12T09:50:30.090" Title="Why does this Lowell Observatory telescope have so many knobs? What do they all do?" Tags="&lt;telescope&gt;&lt;identify-this-object&gt;&lt;observatory&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32609" PostTypeId="2" ParentId="32607" CreationDate="2019-07-11T06:06:09.997" Score="4" Body="&lt;p&gt;That's the historic &lt;a href=&quot;https://lowell.edu/history/the-clark-refractor/&quot; rel=&quot;nofollow noreferrer&quot;&gt;24-inch Alvan Clark refractor&lt;/a&gt;.&#xA;It was installed in 1896 and restored in 2014-15.&#xA;Besides the moon mapping project, Percival Lowell used it to observe Mars, and Vesto Slipher used it for some of the first measurements of galaxy redshifts.&#xA;Its current mission is education; in mild weather the observatory lets visitors look through it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roger Vine's &lt;a href=&quot;http://scopeviews.co.uk/LowellClarkRefractor.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Scope Views&lt;/a&gt; has more photos and a detailed description.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2019-07-11T06:06:09.997" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32610" PostTypeId="1" AcceptedAnswerId="32648" CreationDate="2019-07-11T08:39:22.003" Score="1" ViewCount="27" Body="&lt;p&gt;I'm trying to compile a table showing the wavelengths and flux at m=0 for WISE, Spitzer and GALEX, but I can't find the right data for GALEX. Here's what I have so far:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        Band    Wavelength      F_0&#xA;                micrometers    Jansky&#xA;        ------------------------------&#xA;        FUV     0.152   &#xA;        NUV     0.227   &#xA;        U       0.365           1810&#xA;        B       0.445           4260&#xA;        V       0.551           3640&#xA;        R       0.658           3080&#xA;        I       0.806           2550&#xA;        J       1.220           1600&#xA;        H       1.630           1080&#xA;        K       2.190           670&#xA;        W1      3.4             309.54&#xA;        W2      4.6             171.787&#xA;        SPIT8   8               64.13&#xA;        W3      12              37.674&#xA;        W4      22              8.363&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I got most of the info from &lt;a href=&quot;https://www.cfa.harvard.edu/~dfabricant/huchra/ay145/mags.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. The information I found for GALEX on page 687 of &lt;a href=&quot;https://pdfs.semanticscholar.org/5321/ac845601d0882b5e7d32f6dfdd4d60acb1f0.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt; gives m_0 for NUV and FUV, but no F_0. How can I change m_0 to F_0?&lt;/p&gt;&#xA;" OwnerUserId="17535" LastActivityDate="2019-07-15T09:02:54.880" Title="Comparing info for WISE and GALEX" Tags="&lt;telescope&gt;&lt;space-telescope&gt;&lt;magnitude&gt;&lt;wavelength&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32612" PostTypeId="2" ParentId="32601" CreationDate="2019-07-11T16:01:02.827" Score="-1" Body="&lt;pre&gt;&lt;code&gt;from astropy.io import fits    &#xA;from astropy import wcs    &#xA;import matplotlib.pyplot as plt&#xA;&#xA;x = fits.open('f160w_noopt_drz.fits')['SCI']&#xA;y = wcs.WCS(x.header)&#xA;&#xA;plt.subplots(figsize=(8, 8), subplot_kw={'projection':y})&#xA;plt.imshow(x.data, cmap='viridis', origin='lower', vmin=0, vmax=10)&#xA;plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="22468" LastEditorUserId="2153" LastEditDate="2019-07-11T17:15:10.823" LastActivityDate="2019-07-11T17:15:10.823" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32613" PostTypeId="1" CreationDate="2019-07-11T18:22:47.643" Score="2" ViewCount="48" Body="&lt;p&gt;We are looking into the possibility of building a low budget horn antenna for detecting 21 cm hydrogen line radiation (~1420 MHz) and and possibly measuring its spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the types of parameters and specifications we should look for in an amplifier and filter that would be connected to the input dipole at the end of the horn?&lt;/p&gt;&#xA;" OwnerUserId="28688" LastEditorUserId="7982" LastEditDate="2019-07-12T13:39:11.103" LastActivityDate="2019-07-12T13:39:11.103" Title="Requirements for an amplifier and filter for a horn antenna-based 21 cm radio telescope?" Tags="&lt;observational-astronomy&gt;&lt;amateur-observing&gt;&lt;radio-astronomy&gt;&lt;instruments&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32614" PostTypeId="1" CreationDate="2019-07-11T18:54:46.353" Score="2" ViewCount="50" Body="&lt;p&gt;One of the controversies surrounding the detection of exoplanets around post-common envelope binaries is the possibility that the orbital period variations observed might be the result of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Applegate_mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;Applegate mechanism&lt;/a&gt; as opposed to reflex motion from an orbiting planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would these concerns also apply to &lt;a href=&quot;https://en.wikipedia.org/wiki/AM_Canum_Venaticorum_star&quot; rel=&quot;nofollow noreferrer&quot;&gt;AM CVn systems&lt;/a&gt; or other close white dwarf+white dwarf binaries? The Applegate mechanism itself is explained in terms of the activity cycle of the main sequence component of a cataclysmic variable, could a similar mechanism operate or would the changes in the quadrupole moment of the star be too small to produce observable effects?&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastEditorUserId="6" LastEditDate="2019-08-27T18:37:22.450" LastActivityDate="2019-08-27T18:37:22.450" Title="Could an Applegate-like mechanism operate in AM CVn systems?" Tags="&lt;exoplanet&gt;&lt;binary-star&gt;&lt;white-dwarf&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32615" PostTypeId="2" ParentId="32601" CreationDate="2019-07-12T01:11:48.583" Score="6" Body="&lt;p&gt;This code reads coordinates as equatorial (ra, dec) and transforms them to galactic (l, b):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;eq = SkyCoord(xarr[:], yarr[:], unit=u.deg)&#xA;gal = eq.galactic&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The contents of 'galacticwperiod.csv' are already in galactic coordinates and should not be transformed.&#xA;Something like this may give better results:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;gal = SkyCoord(xarr[:], yarr[:], frame='galactic', unit=u.deg)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The other issue is that pyplot's geographic projections seem to expect angles in radians.&#xA;This code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;plt.subplot(111, projection='aitoff')&#xA;plt.grid(True)&#xA;plt.scatter(gal.l.wrap_at('180d').radian, gal.b.radian)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;produces this plot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2qQiC.png&quot; alt=&quot;Successful plot in Aitoff projection&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-07-17T07:39:57.533" LastActivityDate="2019-07-17T07:39:57.533" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32616" PostTypeId="2" ParentId="32608" CreationDate="2019-07-12T05:01:11.037" Score="3" Body="&lt;p&gt;Here is part of a drawing from the Lowell Observatory booklet &quot;100 Years of Good Seeing: The History of the 24-Inch Clark Telescope&quot; (22 MB (i.e. large file warning) &lt;a href=&quot;https://lowell.edu/wp-content/uploads/2014/08/100-Years-of-Good-Seeing-Clark-Telescope.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;PDF&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bHUs6.png&quot; alt=&quot;Lower half of 24-inch refractor&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dial 6 is the right ascension clock.&#xA;Without closely watching observatory staff operate the telescope, &lt;em&gt;the rest is speculation:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being near dial 6 and linked by chains, knobs 4, 5, and 7 are probably for adjustments in right ascension.&#xA;Knob 5 could be a clutch to disengage for slewing and engage for sidereal tracking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of knobs 1-3 is for an iris near the objective, to limit chromatic aberration by reducing the aperture.&#xA;The other two knobs are probably for adjustments in declination; again one may be a clutch.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="7982" LastEditDate="2019-07-12T05:04:06.080" LastActivityDate="2019-07-12T05:04:06.080" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32617" PostTypeId="1" CreationDate="2019-07-12T05:22:37.203" Score="1" ViewCount="53" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/32616/7982&quot;&gt;@MikeG's answer&lt;/a&gt; to &lt;em&gt;Why does this Lowell Observatory telescope have so many knobs? What do they all do?&lt;/em&gt; explains that item #6 labeled in the image there (and cropped version here) is likely to be a &quot;Right Ascension clock&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is not specifically about the item in the photo, but instead is more general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are operating a large research telescope in the past, when things are all done manually (clock drives, pen and paper), and there was such an object on your telescope, how would you answer the following?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What exactly is a Right Ascension clock and how is one (mounted on a telescope) used in practice?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm asking for more than a one-liner comment. If I had a RA value that I wanted to point the telescope at, what steps would I have to execute to get it pointed there? Do I use a second real-time clock as well, or is the RA clock the only timepiece I need? Is it a timepiece, or is it an indicator of the telescope's position? Do I do simple subtraction? Do I also need to read something else out from the telescope separately?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How exactly would this have worked in practice back when the &quot;clock&quot; was first installed on this historic telescope?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J0TSg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J0TSg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; cropped and annotated, from &lt;a href=&quot;https://www.foxnews.com/science/nasa-moon-landing-flagstaff-lunar-history&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; Credit: Fox News&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; screen shot from Restoring Lowell Observatory's Clark Refractor (mentioned &lt;a href=&quot;https://astronomy.stackexchange.com/questions/32608/why-does-this-lowell-observatory-telescope-have-so-many-knobs-what-do-they-all/32616#comment60115_32616&quot;&gt;here&lt;/a&gt;) cropped, rotated, enlarged, sharpened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SIA4A.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SIA4A.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-07-12T06:10:02.190" LastActivityDate="2019-07-12T06:10:02.190" Title="What exactly is a Right Ascension clock and how was one (mounted on a telescope) used in practice, historically?" Tags="&lt;observational-astronomy&gt;&lt;coordinate&gt;&lt;time&gt;&lt;astrometry&gt;&lt;right-ascension&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32620" PostTypeId="2" ParentId="32607" CreationDate="2019-07-12T09:43:33.497" Score="1" Body="&lt;p&gt;I contacted Dr. Danielle Adams, Deputy Director for Marketing and Communications at the Lowell Observatory. She was kind enough to reply, and generously provided the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...this is the 24&quot; Alvan Clark &amp;amp; Sons refractor, commissioned by observatory founder Percival Lowell in 1895 and completed in 1896. It was fully restored in 2015, a process that took 18 months, since we restored or replaced everything, down to the screw. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The Clark Refractor, as we typically call it, was first used to study Mars and other planets by Percival Lowell, but it also has a storied history in studying deep-space objects. In 1912-1914, VM Slipher used a spectrograph attached to the Clark Refractor to study the redshift of M31, the Andromeda Galaxy, and in doing so he found the first evidence that the universe was expanding. (Edwin Hubble later used this data and his own research to refine this.) &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In the 1960's ('61-'69), the Clark Refractor was used by the USAF's Aeronautical Chart and Information Center to develop highly-detailed maps of the moon, using a combination of photography and professional illustration at the telescope during exquisite moments of seeing. Today, the Clark Refractor is no longer used for research, which allows us to dedicate it fully to public observing for guests who visit the observatory. We also recently started after hours sessions for guests who want to spend more time with the Clark after the observatory has officially closed.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Lowell Observatory's Historian, Kevin Schindler, literally wrote the book(s) on the Clark Refractor. There's a link to his latest book (post-restoration) and a mini-documentary about the process here: &lt;a href=&quot;https://lowell.edu/history/the-clark-refractor&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://lowell.edu/history/the-clark-refractor&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In response to &lt;a href=&quot;https://astronomy.stackexchange.com/questions/32607/which-lowell-observatory-telescope-is-this-used-to-draw-moon-maps-for-the-apol#comment60105_32607&quot;&gt;my comment&lt;/a&gt; above. she continues:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, note that in response to the connected question about which wood was used to line observatory copulas, in the case of both the Clark Refractor and the Pluto Astropgraph at Lowell Observatory, local Ponderosa Pine is the sole that was used for the domes. Flagstaff is home to the largest contiguous grove of Ponderosa Pine in the world, a tree that grows tall and straight, making it perfect for building.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-07-12T09:43:33.497" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32621" PostTypeId="2" ParentId="32608" CreationDate="2019-07-12T09:50:30.090" Score="9" Body="&lt;p&gt;I contacted Dr. Danielle Adams, Deputy Director for Marketing and Communications at the Lowell Observatory. She was kind enough to reply, and generously provided the following (lightly edited for formatting):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I spoke with one of our senior educators about the Clark knobs. The numbers below correspond to the numbers in the image posted in &lt;a href=&quot;https://astronomy.stackexchange.com/questions/32608/why-does-this-lowell-observatory-telescope-have-so-many-knobs-what-do-they-all&quot;&gt;Why does this Lowell Observatory telescope have so many knobs? What do they all do?&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Slow motion control for adjusting Right Ascension (east-west movements) while the Right Ascension clutch (3) is locked down.&lt;/li&gt;&#xA;  &lt;li&gt;Iris control that allows us to control the effective aperture of the telescope. There is an adjustable aperture that can be adjusted from 24&quot; (wide open) to 6&quot;.&lt;/li&gt;&#xA;  &lt;li&gt;Right Ascension clutch that when tightened prevents the telescope from being moved by pushing, allowing the telescope drive to operate.&lt;/li&gt;&#xA;  &lt;li&gt;and 7. Slow motion controls for adjusting declination (north-south movements) while the Declination clutch (5) is locked down. The two controls are linked with bicycle chain so when one moves, so does the other. This allows operation of the control from either side of the viewing position.&lt;/li&gt;&#xA;  &lt;li&gt;Declination clutch that when tightened prevents the telescope from being moved by pushing, allowing the telescope drive to operate.&lt;/li&gt;&#xA;  &lt;li&gt;Sidereal clock that permits astronomers to point their telescopes to a given celestial object more easily.The sidereal day is about 4 minutes shorter than the 24 hour solar day. This is due to the fact that the Earth revolves around the Sun while it rotates about its axis.&lt;/li&gt;&#xA;  &lt;li&gt;see (4)&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UhwUB.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UhwUB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-07-12T09:50:30.090" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32622" PostTypeId="1" AcceptedAnswerId="32623" CreationDate="2019-07-12T10:55:50.317" Score="18" ViewCount="2132" Body="&lt;p&gt;The CNet article &lt;a href=&quot;https://www.cnet.com/news/astronomers-discover-two-supermassive-black-holes-in-a-death-spiral/&quot; rel=&quot;noreferrer&quot;&gt;Astronomers discover two supermassive black holes in a death spiral&lt;/a&gt; links to &lt;a href=&quot;https://iopscience.iop.org/article/10.3847/2041-8213/ab2a14&quot; rel=&quot;noreferrer&quot;&gt;Discovery of a Close-separation Binary Quasar at the Heart of a z ~ 0.2 Merging Galaxy and Its Implications for Low-frequency Gravitational Waves&lt;/a&gt; (available in &lt;a href=&quot;https://arxiv.org/abs/1907.03757&quot; rel=&quot;noreferrer&quot;&gt;ArXiv&lt;/a&gt;) and says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Supermassive black holes are usually found at the center of galaxies, including our own, and during a galaxy merger they end up beginning a dance of death, spinning around each other in a near-endless waltz, until finally merging. However, researchers are currently unclear as to the time it takes for black holes to merge -- or indeed, if they merge at all.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;It's a major embarrassment for astronomy that we don't know if supermassive black holes merge,&amp;quot; said Jenny Greene, a professor of astrophysical sciences at Princeton and co-author of the study. &amp;quot;For everyone in black hole physics, observationally this is a long-standing puzzle that we need to solve.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;This puzzle is dubbed the &amp;quot;final-parsec problem.&amp;quot; Some astronomers believes that once two supermassive black holes get close enough together, reducing their distance to 1 parsec (3.2 light years), they may dance for an eternity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; If it turns out that supermassive black holes can't merge, or have a tough time doing so, what might the reasons be?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-07-12T23:56:02.983" Title="Why can't supermassive black holes merge? (or can they?)" Tags="&lt;black-hole&gt;&lt;gravity&gt;&lt;supermassive-black-hole&gt;&lt;gravitational-waves&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32623" PostTypeId="2" ParentId="32622" CreationDate="2019-07-12T12:42:42.650" Score="22" Body="&lt;p&gt;The main problem is angular momentum. In order for two gravitationally bound objects to merge (whether black holes, supermassive black holes, planets, stars, etc.), they must shed enough angular momentum for their orbital separation to become small enough.  Average orbital separation (semi-major axis) is determined entirely from the angular momentum of the orbit (at least in classical mechanics; I don't know if that is true for relativistic situations such as merging black holes when they get close together).  Removal of angular momentum requires interactions with other objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When two galaxies merge, their supermassive black holes both have angular momentum.  Through a phenomenon known as &quot;dynamical friction,&quot; gravitational interactions with other stars sap the black holes of much of their angular momentum, until they are brought within a few parsecs or so of each other.  At this point, the black holes have flung out all of the stars that were in the region and there is (presumably) nothing left for dynamical friction to sap their angular momentum.  Once the black holes are close enough (I don't know off the top of my head how close), emission of gravitational waves will sap the orbiting pair of their remaining angular momentum and a merger becomes inevitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to &lt;strong&gt;answer your question&lt;/strong&gt;, the reason for supermassive black holes to be unable to merge is that they are too close together for there to be any remaining material (stars, gas, etc.) in the center of the galaxy to remove angular momentum from the orbiting pair, having already removed the material themselves, but are not close enough for emission of gravitational waves to remove angular momentum fast enough for their merger to occur any time soon (in an astronomical sense).&lt;/p&gt;&#xA;" OwnerUserId="128" LastEditorUserId="10675" LastEditDate="2019-07-12T21:02:13.040" LastActivityDate="2019-07-12T21:02:13.040" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="32624" PostTypeId="1" CreationDate="2019-07-12T13:50:56.710" Score="25" ViewCount="3459" Body="&lt;p&gt;If I had a nice amateur telescope&lt;sup&gt;†&lt;/sup&gt;, are there any multiple star systems that I could observe over a few years or a few decades and actually see the movement of one or both of them over time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My short human lifespan and limited telescope put heavy constraints on the orbital distance, brightness, and distance from the Sun, so I am guessing that if there are any at all, the number is probably small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;†&lt;/sup&gt;arbitrarily defined as say 8-inch (20 cm) aperture, with a good set of eyepieces, and a sketch pad.&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-07-13T00:19:52.773" LastActivityDate="2019-07-14T14:47:56.700" Title="Are there any double stars that I can actually see orbit each other?" Tags="&lt;observational-astronomy&gt;&lt;amateur-observing&gt;&lt;binary-star&gt;" AnswerCount="4" CommentCount="9" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32625" PostTypeId="1" CreationDate="2019-07-12T14:25:33.193" Score="1" ViewCount="32" Body="&lt;p&gt;In the Gaia catalogue I have the magnitude of several stars in the G band. I also have the distances to these stars (parallax is known). Additionally, I oftentimes have the effective temperature in Kelvins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to convert this magnitude to an approximate spectral luminosity (units of W/Hz) at some frequency band? You may assume that we are in the Rayleigh Jean's regime (Intensity proportional to frequency&lt;sup&gt;2&lt;/sup&gt;).&lt;/p&gt;&#xA;" OwnerDisplayName="user28693" LastEditorUserId="7472" LastEditDate="2019-07-12T14:50:35.857" LastActivityDate="2019-07-12T14:50:35.857" Title="How to convert magnitude to spectral luminosity?" Tags="&lt;observational-astronomy&gt;&lt;astrophysics&gt;&lt;luminosity&gt;&lt;magnitude&gt;" AnswerCount="0" CommentCount="5" ClosedDate="2019-07-13T15:14:09.893" ContentLicense="CC BY-SA 4.0" />
  <row Id="32626" PostTypeId="2" ParentId="32624" CreationDate="2019-07-12T16:00:40.213" Score="28" Body="&lt;h2&gt; Vir (12h 42m, –01° 27′)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Probably &lt;a href=&quot;http://stars.astro.illinois.edu/sow/porrima.html&quot; rel=&quot;noreferrer&quot;&gt;Porrima&lt;/a&gt;, &lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt; Vir, is the best candidate for most observers in the Northern Hemisphere to see changes in a binary orbit, particularly using a small telescope. It is a pair of stars with similar size and visual magnitude, of about 3.6. Their orbital period is about 169 years, but the orbit is eccentric, e = 0.88. They are also relatively close at about 40 ly. Periapsis was in 2005, so the stars are now moving away from each other, but their rate of separation is decreasing. Separation at periapsis was about 0.4 arcsec, so would not have been resolved using a small telescope in 2005. By 2015 their separation was ~2.5 arcsec, and will increase to ~3 arcsec by 2020. I estimated a position angle change between 2015-2020 of ~7 degrees. These changes should be detectable with a 100-200 mm (4-8 inch) telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since most short period binaries are close together, with nearly circular orbits, and often more distant, they are very difficult or impossible to resolve with a small telescope.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Sirius B (06h 45m, −16° 43′)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As mentioned in &lt;a href=&quot;https://astronomy.stackexchange.com/a/32628/7982&quot;&gt;@MichaelWalsby's answer&lt;/a&gt;, it is also possible to observe the orbit of the white dwarf binary companion to Sirius, the brightest star visible in the night sky. Sirius is only 8.6 ly away, and their orbit has a semi-major axis of about 7.5 arcsec, an eccectricity of e = 0.59, and a period of about 50 years. If this pair were similar in brightness, they would be an easy answer to this question. Unfortunately, Sirius B, or the &lt;a href=&quot;https://www.skyandtelescope.com/observing/sirius-b-a-new-pup-in-my-life/&quot; rel=&quot;noreferrer&quot;&gt;Pup&lt;/a&gt; (as the companion is known to amateur astronomers), is ~10 magnitudes dimmer than Sirius, and usually lost in its glare. It takes a night with excellent seeing, i.e. a stable, non-turbulent atmosphere, especially since Sirius never gets much above 30 degrees elevation at the mid-northern latitude where I live. I have only seen the Pup 4 or 5 times (one view was probable but not certain) over almost 6 decades of observing, and I have never seen it in a telescope with aperture under 300 mm. I know other amateurs that have seen Sirius B in 150-200 mm telescopes, but mostly at lower latitudes. However, by seeing Sirius B at intervals separated by decades, I have observed its polar angle change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the separation of Sirius B is now over 10 arcsec, and still increasing slightly. So for the next couple of decades, observing it might be a bit easier. In recent winters I have  tried with telescopes from a 120 mm refractor to a 250 mm Dobsonian, and occasionally larger, but still have not seen it for several years. This Hubble photo of &lt;a href=&quot;https://www.spacetelescope.org/images/heic0516a/&quot; rel=&quot;noreferrer&quot;&gt;Sirius&lt;/a&gt; gives some idea why Sirius B is hard to observe in small telescopes.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Indirect methods&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Also, many eclipsing binaries are often observed, and their light curves measured. By &lt;a href=&quot;http://www.physics.sfasu.edu/astro/ebstar/ebstar.html&quot; rel=&quot;noreferrer&quot;&gt;analyzing light curves&lt;/a&gt;, orbital elements can be estimated. However, these indirect orbital observations are  probably stretching the intent of the original question.&lt;/p&gt;&#xA;" OwnerUserId="28086" LastEditorUserId="28086" LastEditDate="2019-07-14T14:47:56.700" LastActivityDate="2019-07-14T14:47:56.700" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32627" PostTypeId="1" CreationDate="2019-07-12T16:55:29.843" Score="1" ViewCount="42" Body="&lt;p&gt;I'm analyzing the spectra of M dwarfs. Some authors characterize the lines with the oscillator strengths &lt;span class=&quot;math-container&quot;&gt;$\log{gf}$&lt;/span&gt; while others follow different paths with the Einstein &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;-coefficient. I would like to know if there is  relation between the oscillator strengths &lt;span class=&quot;math-container&quot;&gt;$\log{gf}$&lt;/span&gt;, and the Einstein &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;-coefficient, rotational quantum number J', vibrational levels with the Hönl-London, factors or similar parameters.&lt;/p&gt;&#xA;" OwnerUserId="28419" LastActivityDate="2019-07-12T16:55:29.843" Title="Relation between oscillator strengths $\log{gf}$ and the Einstein $A$-coefficient, rotational quantum number," Tags="&lt;spectroscopy&gt;&lt;spectra&gt;&lt;stellar-astrophysics&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32628" PostTypeId="2" ParentId="32624" CreationDate="2019-07-12T18:39:21.200" Score="6" Body="&lt;h2&gt;Sirius (α Canis Majoris, 06h 45m, −16° 43′)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You could try Sirius B, which is a white dwarf orbiting Sirius A around their common centre of mass. Sirius is one of the closest stars to Earth, and Sirius B has a 50 year orbital period. At maximum separation the white dwarf is about 30 AU from Sirius A, which should allow you to separate them with a good amateur telescope, but I'm not sure where Sirius B is right now. There should be a detectable change in position within 5 years or less. Needless to say, Sirius is a very btight star, and its companion rather faint, so you won't be able to separate them when they are close together.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastEditorUserId="7982" LastEditDate="2019-07-13T11:25:30.627" LastActivityDate="2019-07-13T11:25:30.627" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32629" PostTypeId="1" CreationDate="2019-07-12T19:47:18.277" Score="2" ViewCount="29" Body="&lt;p&gt;I was supposed to find the precession of a certain frequency in a time series. I tried breaking down the time series into smaller intervals and folded them over the average frequency I'm interested in. Then I plotted all of the folded light curves on top of each other.&lt;/p&gt;&#xA;&lt;p&gt;I notice a slight oscillation in the shape of the light curve as I iterate through all of these light curves. I'm not sure how do I pin down the exact frequency of oscillation. The light curve doesn't resemble a pure sine wave so I don't think I can just fit a sinusoid and find the phase variation.&lt;/p&gt;&#xA;&lt;p&gt;What all can I do to find the period of oscillation? Thanks a ton!&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yLJux.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yLJux.jpg&quot; alt=&quot;T&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10731" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-07-12T19:47:18.277" Title="Finding the precession period of folded light curves" Tags="&lt;light-curve&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32630" PostTypeId="2" ParentId="32624" CreationDate="2019-07-13T04:11:48.503" Score="1" Body="&lt;h2&gt;Mizar and Alcor (ζ Ursae Majoris, 13h 23m, +54° 55′)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Zeta_Ursae_Majoris&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zeta Ursae Majoris&lt;/a&gt; is a binary star system in the Big Dipper constellation that is visible to naked eye. In fact, the ability to resolve Mizar and Alcor is taken as a test of good eye sight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, in the Hindu marriage tradition, immediately after the marriage the couple are asked to look at this binary star system (Alcor is called Arundhathi) as a sign of fidelity.&lt;/p&gt;&#xA;" OwnerUserId="28672" LastEditorUserId="7982" LastEditDate="2019-07-13T11:23:47.913" LastActivityDate="2019-07-13T11:23:47.913" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32631" PostTypeId="2" ParentId="32624" CreationDate="2019-07-13T08:11:21.697" Score="16" Body="&lt;h2&gt;α Centauri (14h 40m, –60° 50′)&lt;/h2&gt;&#xA;&lt;p&gt;The most obvious visual multiple system, where orbital changes can be observed is Alpha Cen A+B, (together with Proxima Centauri). The A/B system has an orbital period of 80 years, but because it is so close (1.34 parsec), the semi-major axis is a whopping 17.5 arcsecond.&lt;/p&gt;&#xA;&lt;p&gt;The two stars are currently separated by 5 arcseconds on the sky. This will increase to around 10 arcseconds by 2030, decrease to about 2.5 arcseconds in 2037 and then increase again to about 22 arcseconds over the course of the next 15 years, with a changing relative angle all the time. The progress of this could easily be followed with binoculars or a small telescope.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Orbit Alpha Centauri AB &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Orbit_Alpha_Centauri_AB_arcsec.png&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0eHu7.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Trajectory of Alpha Centauri B relative to A (fixed to the coordinate origin) as seen from the Earth (inclined ellipse) and face-on (horizontal ellipse). The orbit parameters are taken from Pourbaix et al. (2002). The graph has been created with gnuplot based on data generated by solving Kepler's equation.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Note 1: The time stamps refer to the motion as seen from Earth, i.e. delayed with respect to the true motion by the travel of light by about 4.4 years. North is down, as usual in many astronomical charts.&lt;/li&gt;&#xA;&lt;li&gt;Note 2: For simplicity, the points refer to steps of 1/80 of an orbit rather than to exact 1-year steps (which would be 1/79.91 of an orbit).&lt;/li&gt;&#xA;&lt;li&gt;Note 3: The ascending node is the point where B (with respect to A) penetrates the celestial plane and becomes more distant to Earth than A.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-07-13T11:40:49.243" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32632" PostTypeId="1" CreationDate="2019-07-13T18:18:30.290" Score="4" ViewCount="160" Body="&lt;p&gt;I am making a video game which features a planet and its moon. And I thought I would add rings too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that got me wondering... how do a planet's moon and a planet's ring influence each other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does a moon clear out all the debris in its orbit? In other words: does it make a gap in the ring? If so, how large is the gap in relation to the moon's radius?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, does the moon travel along as all the other matter in the ring? Would it not make the debris fall to the moon, thereby growing the moon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, are both the rings and moons typically in the same plane?&lt;/p&gt;&#xA;" OwnerUserId="28698" LastEditorUserId="7982" LastEditDate="2019-07-13T22:55:13.340" LastActivityDate="2019-07-14T13:41:32.163" Title="How do a planet's moons and a planet's rings interact?" Tags="&lt;natural-satellites&gt;&lt;orbital-mechanics&gt;&lt;planetary-ring&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32633" PostTypeId="2" ParentId="32632" CreationDate="2019-07-13T19:43:40.457" Score="6" Body="&lt;p&gt;First, there are different types of rings. Using saturn as an example: there are icy particles, dust bands and more. These interact differently if they were to hit a moon, for example enceladus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every little particle in a ring is a „mini-moon“ and obeys the same laws (see kepler) as the big ones. &#xA;They technically all have their own orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main rings of saturn have a high enough particle density to exert gravitational influence on each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are hypothesis that suggest that rings are made of shattered moons (like through collision) or another one that says that the rings are made of particles that „failed“ to make a moon in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now, what happens when rings collide with moons is dependent on the type of ring.&#xA;Enceladus is a small icy moon in the so called E-Ring of saturn. This ring is small and made of water/ice, just like the moon. Because the ring is so very small and such a water/icy „cloud“ would  not really stay together, it is believed that it is „refilled“ by enceladus constantly.&#xA;The surface of enceladus was observed to be smooth, not really having craters one might expect, but this could also be due to geological activity. Enceladus also „geysirs“ out water all the time, so this might be how the ring is kept. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I recommend checking out this link: &lt;a href=&quot;https://courses.lumenlearning.com/suny-astronomy/chapter/planetary-rings/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://courses.lumenlearning.com/suny-astronomy/chapter/planetary-rings/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="28687" LastEditorUserId="28687" LastEditDate="2019-07-14T13:41:32.163" LastActivityDate="2019-07-14T13:41:32.163" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32634" PostTypeId="1" CreationDate="2019-07-13T22:50:07.373" Score="3" ViewCount="138" Body="&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/32623/7982&quot;&gt;This answer&lt;/a&gt; to &lt;em&gt;Why can't supermassive black holes merge? (or can they?)&lt;/em&gt; describes the barrier to merging that two supermassive black holes face when two galaxies are in the process of merging or have &quot;successfully&quot; merged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a bit of the answer, but its worth taking a moment to &lt;a href=&quot;https://astronomy.stackexchange.com/a/32623/7982&quot;&gt;go there&lt;/a&gt; and read the whole thing:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When two galaxies merge, their supermassive black holes both have angular momentum. Through a phenomenon known as &quot;dynamical friction,&quot; gravitational interactions with other stars sap the black holes of much of their angular momentum, until they are brought within a few parsecs or so of each other. At this point, the black holes have flung out all of the stars that were in the region and there is (presumably) nothing left for dynamical friction to sap their angular momentum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What kind of timescale are we talking about to reach &quot;the last (few) parsec(s)? Is it similar to the time it takes for the merged galaxy to reach some kind of centralized shape (things rotating around a common center rather than having two identifiable centers of rotation)? Much longer or shorter?&lt;/li&gt;&#xA;&lt;li&gt;What kinds of processes are involved that allow the two black holes to reach the last few parsecs of separation? What is &quot;dynamical friction&quot; and &quot;star-flinging&quot;?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-07-13T22:50:07.373" Title="How do two supermassive black holes reach &quot;the last parsec&quot; in merging galaxies?" Tags="&lt;supermassive-black-hole&gt;&lt;galactic-dynamics&gt;&lt;galactic-center&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32635" PostTypeId="1" CreationDate="2019-07-14T04:47:09.110" Score="0" ViewCount="75" Body="&lt;p&gt;What resources do &lt;a href=&quot;https://en.wikipedia.org/wiki/C-type_asteroid&quot; rel=&quot;nofollow noreferrer&quot;&gt;C-Type&lt;/a&gt; asteroids contain? The research I've done has come up with carbon, but what form is the carbon in? Graphite, diamonds, coal, or some other material?&lt;/p&gt;&#xA;" OwnerUserId="19304" LastActivityDate="2019-07-14T11:57:08.543" Title="What resources do C-Type(Carbonaceous) Asteroids contain?" Tags="&lt;asteroids&gt;&lt;asteroid-belt&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32637" PostTypeId="1" CreationDate="2019-07-14T11:16:15.887" Score="2" ViewCount="520" Body="&lt;p&gt;Is it a coincidence that Earth's axial tilt and maximum declination are both approximately 23 degrees? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mechanically speaking, do planets physically cross the Solar equator, or is declination due to the axial tilt angle? The later is difficult to visualize given variety of planetary axial tilts i.e Mercury 0, Jupiter 4, Earth and Saturn 23 degrees approximately. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll try and give a concise example. At JPL's Horizons you can download data of RA and DEC for all planets and objects placing either Earth or the Sun at the center. This will calculate the Declination of a planet geocentrically or heliocentrically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculations suggest all planets will obtain a maximum declination North and South between 23&amp;deg; and 28&amp;deg; (i.e Mercury reaches 28&amp;deg; when viewed from the Earth). Earth's axial tilt (obliquity) at 23.45&amp;deg; matches its maximum declination N/S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, Mercury's Axial tilt is 0&amp;deg; and Jupiter 3.13&amp;deg; so if declination is the angle between the ecliptic plane and the celestial equator how do Mercury and Jupiter obtain maximum declination of 23&amp;deg; and 28&amp;deg; N/S?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll appreciate any help clarifying this.&lt;/p&gt;&#xA;" OwnerUserId="28705" LastEditorUserId="616" LastEditDate="2019-07-19T20:36:03.063" LastActivityDate="2019-08-19T05:02:35.353" Title="Planetary Declination and Axial Tilt" Tags="&lt;celestial-mechanics&gt;&lt;declination&gt;&lt;inclination&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32638" PostTypeId="1" AcceptedAnswerId="32643" CreationDate="2019-07-14T11:28:34.553" Score="0" ViewCount="158" Body="&lt;p&gt;I am looking at this video about the timeline of the future.&#xA;It says that the universe will stop expanding in about 1 googol years and become a cold bubble but I am not sure that will be true. There are several theories that can be true, so I am not sure which one. Anyone have some proof that one will be true? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="28702" LastActivityDate="2019-07-14T16:42:43.510" Title="What will be the end of the universe?" Tags="&lt;fate-of-universe&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32639" PostTypeId="2" ParentId="32635" CreationDate="2019-07-14T11:57:08.543" Score="1" Body="&lt;p&gt;Carbonaceous asteroids contain carbon in the form of chondrites (small marble-sized nodules),tiny crystals and other sorts of carbon dust. This is mixed in a sort of conglomerate with dust and nodules if silicate material, sometimes even nickel-iron nodules. What resources do they contain? They don't contain diamonds, because diamonds require enormous heat and pressure to form, so sometimes micro-diamonds are found in the craters of large meteorite impacts on Earth. Some of them probably contain valuable materials, but they would have to be very valuable indeed to be worth the trouble and expense of retrieving them from the asteroid belt. Prospectors there wouldn't restrict their attentions solely to carbonaceous types, and would probably explore large asteroids like Ceres. Asteroids and meteorites couldn't possibly contain coal, which is a biological product formed from thick layers of plant debris transformed by heat and pressure over millions of years in the bowels of the Earth.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-07-14T11:57:08.543" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32640" PostTypeId="1" AcceptedAnswerId="32644" CreationDate="2019-07-14T12:30:06.467" Score="19" ViewCount="3067" Body="&lt;p&gt;When I go to the NASA HORIZONS webpage&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ssd.jpl.nasa.gov/horizons.cgi&quot; rel=&quot;noreferrer&quot;&gt;https://ssd.jpl.nasa.gov/horizons.cgi&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and ask for the VECTOR coordinates of the Earth on 1 January 2000, it tells me that Earth has the following coordinates in the J2000 epoch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X =-1.756637922977122E-01 Y = 9.659912850526895E-01 Z = 2.020629118443605E-04&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are measured in astronomical units. Quoting from the webpage:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Reference epoch: J2000.0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;XY-plane: plane of the Earth's orbit at the reference epoch&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X-axis  : out along ascending node of instantaneous plane of the Earth's             orbit and the Earth's mean equator at the reference epoch&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Z-axis  : perpendicular to the xy-plane in the directional (+ or -) sense              of Earth's north pole at the reference epoch.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;But these figures don't make sense! The z value should be EXACTLY zero, since we are measuring at 1 January 2000. Also the x should be exactly 1, and the y should be zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is going on?&lt;/p&gt;&#xA;" OwnerUserId="28707" LastActivityDate="2019-07-15T15:55:23.127" Title="Why does the Earth have a z-component at the start of the J2000 epoch?" Tags="&lt;orbit&gt;&lt;nasa&gt;&lt;ephemeris&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32641" PostTypeId="1" AcceptedAnswerId="32646" CreationDate="2019-07-14T14:06:22.243" Score="18" ViewCount="2370" Body="&lt;p&gt;I thought that solar systems and galaxies are disk-shaped because that is the most stable shape under gravitation. Globular clusters are very old, often times older than their host galaxies, so why haven't they flattened out?&lt;/p&gt;&#xA;" OwnerUserId="28708" LastActivityDate="2019-07-15T07:06:42.467" Title="Why aren't globular clusters disk shaped" Tags="&lt;globular-clusters&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32642" PostTypeId="1" CreationDate="2019-07-14T15:32:57.427" Score="1" ViewCount="248" Body="&lt;p&gt;If we have the RA/DEC of an object that we want to observe, how can we tell if that object is visible from our place on earth (at a particular time of the year)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the part of the sky we are observing is changing throughout the year, it seems fairly complicated.&lt;/p&gt;&#xA;" OwnerUserId="14536" LastActivityDate="2019-07-16T00:41:50.837" Title="How can we estimate if a certain astronomical object will be visible from a place on earth?" Tags="&lt;observational-astronomy&gt;&lt;amateur-observing&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32643" PostTypeId="2" ParentId="32638" CreationDate="2019-07-14T16:42:43.510" Score="-4" Body="&lt;p&gt;There are several theories about the future of the universe, but I don't think we can be completely certain about any of them. The favourite at the moment, and probably the one with the most evidence to support it, is the dark energy universe which will go on expanding for ever, with the galaxies getting further and further apart and more isolated until they burn themselves out into total darkness. This raises problems about a unique, one-off beginning, with no cause, and vast quantities of matter and energy created magically out of nothing, thus violating the First Law of Thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My preferred option is the Big Bang-Big Crunch-Big Bounce universe in which the expanding universe is brought to a halt by gravity and collapses on itself to form a blue-shifted, shrinking universe which ends in a fireball and singularity similar to the one it was created from. Except it doesn't end. It erupts again (the Big Bounce) in a gigantic white hole which we call the Big Bang, and this cycle is repeated endlessly. It had no beginning, it will never end, and it doesn't violate the Fist Law of Thermodynamics.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-07-14T16:42:43.510" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32644" PostTypeId="2" ParentId="32640" CreationDate="2019-07-14T16:47:01.940" Score="28" Body="&lt;p&gt;You made the same fundamental mistake that Anton Gromov made in &lt;a href=&quot;https://space.stackexchange.com/questions/18645&quot;&gt;his question&lt;/a&gt; on the sister &lt;a href=&quot;https://space.stackexchange.com&quot;&gt;Space Exploration&lt;/a&gt; StackExchange network site: You used the solar system barycenter rather than the Sun as the frame origin. Had you used the Sun, the apparent discrepancy would have dropped by almost two orders of magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A much lesser flaw is that you apparently used midnight rather than noon. The J2000.0 epoch is 12 noon Terrestrial Time on 1 January 2000. This doesn't change things by much, but it is nonetheless important. That half day offset has bitten me more than once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The z coordinate of the Earth's position would not have dropped to zero had you made the above corrections (Sun-centered rather than solar system barycenter, and 12 noon rather than midnight). It would not have dropped to zero even if you had used the Earth-Moon barycenter rather than the center of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One reason it would not have dropped to zero is that the Earth's orbit about the Sun is not an ellipse. The Earth's orbit doesn't even lie on a plane! The key cause of this non-planar motion is that the Moon's orbit about the Earth is inclined by about 5.15° with respect to the ecliptic, making the Earth bob up and down with respect to the ecliptic&lt;sup&gt;1&lt;/sup&gt;. Because the Earth's orbit is not truly elliptical or even planar, the ecliptic plane at some epoch is a time-averaged plane centered about the epoch time that makes the Earth's position with respect to the Sun (or the Sun's position with respect to the Earth) average to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lesser reason is that the J2000.0 mean ecliptic and mean equinox frame was defined over 36 years ago. HORIZONS uses DE431, which was released 6 years ago. JPL's ability to model the solar system improved significantly in the intervening three decades.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;sup&gt;1&lt;/sup&gt; Strictly speaking, it makes the Sun appear to bob up and down with respect to the ecliptic. The concept of the ecliptic plane dates back to the ancient Greeks, and it retains a vestige of this definition to this day.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2019-07-14T16:47:01.940" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32645" PostTypeId="2" ParentId="32641" CreationDate="2019-07-15T02:56:29.387" Score="10" Body="&lt;p&gt;While some galaxies are disk-shaped, others (elliptical galaxies) are more spherical. Disk-shaped systems are those in which conservation of angular momentum of the star-forming gas played a central role; for more spherical systems, this was not the case. Globular clusters are thought to form in the densest portions of gas in galaxies with the highest external pressure, possibly resulting from collisions of gas clouds. In this case, the initial angular momentum of the larger-scale gas distribution is irrelevant, and the dynamics of the gas on much smaller scales is more important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it another way: it is not the case that disks are the most stable shape under gravitation; spheroidal or elliptical systems are also stable equilibria. The physics of collapse and formation determine the final shape. For globular clusters, the initial angular momentum of the gas is likely irrelevant because they condense out of much smaller pockets of gas.&lt;/p&gt;&#xA;" OwnerUserId="28713" LastActivityDate="2019-07-15T02:56:29.387" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32646" PostTypeId="2" ParentId="32641" CreationDate="2019-07-15T07:06:42.467" Score="24" Body="&lt;p&gt;In order to result in a disk-like stellar system there are two conditions that need to be satisfied. (a) The initial gas from which the stars form must have a significant ratio of rotational to gravitational energy. (b) The star formation would have to occur slowly enough, that the gas collapses to a disk &lt;em&gt;before&lt;/em&gt; star formation is complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formation of a disk from a spherically symmetric state requires a reduction (more negative) gravitational potential energy, whilst conserving angular momentum. This can only happen if there is some means to lose energy from the system in the form of dissipative interactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pure stellar system is almost collisionless and there is no means to dissipate energy. That means once a spherical system of stars has formed, there is no way for it to become more disk-like. Thus if star formation takes place in a spherical way, then those stars will end up in a spherical system. Only in cases where the gas collapses to a disk before star formation occurs do we end up with disk-like systems. This is true of the disk of our Galaxy (and of the solar system).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of globular clusters (and other types of star cluster in the Galaxy) it appears the star formation is rapid enough that the gas cannot have collapsed to a disk before most star formation is complete. NB. Many globular clusters &lt;em&gt;do&lt;/em&gt; have measurable rotation.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-07-15T07:06:42.467" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32647" PostTypeId="2" ParentId="19476" CreationDate="2019-07-15T07:21:45.547" Score="1" Body="&lt;p&gt;I won't go through your equations carefully nor your code, but since you haven't mentioned the exact starting positions that you used, my guess is that you haven't first &lt;em&gt;solved for a stable horseshoe orbit&lt;/em&gt; state vector first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The positions for the Sun and Earth are fixed in the synodic (rotating) frame. But where did you start your astroid? For a given starting position, you have to also give the asteroid the correct starting velocity (speed and direction) otherwise it may not be stable and just wander off or spiral around as you mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at &lt;a href=&quot;https://astronomy.stackexchange.com/a/29596/7982&quot;&gt;this answer&lt;/a&gt; where I show the correct equations and then solve for stable horseshoe orbits before plotting them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I start with a position on the x-axis, opposite the Earth, and a little farther or closer from the Sun than the Earth is. Then I launch it with a low velocity (in the rotating frame) exactly in the direction of the y-axis. I watch it drift towards Earth and then boomerang back to the region of the starting area. When it crosses the x-axis again, I check to see how close the velocity is to y. I use an optimization zero solver &lt;code&gt;brentq&lt;/code&gt; to adjust the initial velocity until the return velocity is in exactly the opposite direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's true, then the orbit will be periodic and repeating, and we can call it a stale orbit under the constraints of the CR3BP.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://astronomy.stackexchange.com/a/29596/7982&quot;&gt;this answer&lt;/a&gt; (there's a lot more to read there, including several references!):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ddot{x} = x + 2\dot{y} - \frac{(1-\mu)(x+\mu)}{r_1^3} - \frac{\mu(x-1+\mu)}{r_2^3}$$&lt;/span&gt; &#xA;  &lt;span class=&quot;math-container&quot;&gt;$$\ddot{y} = y - 2\dot{x} - \frac{(1-\mu)y}{r_1^3} - \frac{\mu y}{r_2^3}$$&lt;/span&gt;&#xA;  &lt;span class=&quot;math-container&quot;&gt;$$\ddot{z} =  -\frac{(1-\mu) z}{r_1^3} - \frac{\mu z}{r_2^3} $$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OkuSe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; half-cycles of some wobbly horseshoe orbits&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rWKMP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; times to first x-axis crossings of the same wobbly horseshoe orbits, used to calculate half-cycle times.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cHG8s.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;above:&lt;/strong&gt; cycle times from this calculation (black dots) versus from the synodic period estimation method (red dots). Good qualitative agreement. Also the starting y velocities at each starting point in x.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;below:&lt;/strong&gt; Python script for these plots.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def x_acc(x, ydot):&#xA;    r1    = np.abs(x-x1)&#xA;    r2    = np.abs(x-x2)&#xA;    xddot = x + 2*ydot  -  ((1-mu)/r1**3)*(x+mu) - (mu/r2**3)*(x-(1-mu))&#xA;    return xddot&#xA;&#xA;def C_calc(x, y, z, xdot, ydot, zdot):&#xA;    r1 = np.sqrt((x-x1)**2 + y**2 + z**2)&#xA;    r2 = np.sqrt((x-x2)**2 + y**2 + z**2)&#xA;    C = (x**2 + y**2 + 2.*(1-mu)/r1 + 2.*mu/r2 - (xdot**2 + ydot**2 + zdot**2))&#xA;    return C&#xA;&#xA;def deriv(X, t): &#xA;    x, y, z, xdot, ydot, zdot = X&#xA;    r1 = np.sqrt((x-x1)**2 + y**2 + z**2)&#xA;    r2 = np.sqrt((x-x2)**2 + y**2 + z**2)&#xA;    xddot = x + 2*ydot  -  ((1-mu)/r1**3)*(x+mu) - (mu/r2**3)*(x-(1-mu))&#xA;    yddot = y - 2*xdot  -  ((1-mu)/r1**3)*y      - (mu/r2**3)*y&#xA;    zddot =             -  ((1-mu)/r1**3)*z      - (mu/r2**3)*z&#xA;    return np.hstack((xdot, ydot, zdot, xddot, yddot, zddot))&#xA;&#xA;# http://cosweb1.fau.edu/~jmirelesjames/hw4Notes.pdf&#xA;&#xA;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;from scipy.integrate import odeint as ODEint&#xA;from scipy.optimize import brentq&#xA;&#xA;halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]&#xA;&#xA;mu = 0.001&#xA;&#xA;x1 = -mu&#xA;x2 = 1. - mu&#xA;&#xA;x = np.linspace(-1.4, 1.4, 1201)&#xA;y = np.linspace(-1.4, 1.4, 1201)&#xA;&#xA;Y, X = np.meshgrid(y, x, indexing='ij')&#xA;Z    = np.zeros_like(X)&#xA;&#xA;xdot, ydot, zdot = [np.zeros_like(X) for i in range(3)]&#xA;&#xA;C = C_calc(X, Y, Z, xdot, ydot, zdot)&#xA;C[C&amp;gt;8] = np.nan&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;    plt.imshow(C)&#xA;    plt.colorbar()&#xA;    levels = np.arange(2.9, 3.2, 0.04) &#xA;    CS = plt.contour(C, levels,&#xA;                 origin='lower',&#xA;                 linewidths=2) &#xA;    plt.show()&#xA;&#xA;ydot0s   = np.linspace(-0.08, 0.08, 20)&#xA;x0ydot0s = []&#xA;for ydot0 in ydot0s:&#xA;    x0, infob =  brentq(x_acc, -1.5, -0.5, args=(ydot0), xtol=1E-11, rtol=1E-11,&#xA;                           maxiter=100, full_output=True, disp=True)&#xA;    x0ydot0s.append((x0, ydot0))&#xA;&#xA;states = [np.array([x0, 0, 0, 0, ydot0, 0]) for (x0, ydot0) in x0ydot0s]&#xA;&#xA;times  = np.arange(0, 150, 0.01)&#xA;&#xA;results = []&#xA;for X0 in states:&#xA;    answer, info = ODEint(deriv, X0, times, atol = 1E-11, full_output=True)&#xA;    results.append(answer.T.copy())&#xA;&#xA;resultz = []&#xA;for x0ydot0, thing in zip(x0ydot0s, results):&#xA;    y     = thing[1]&#xA;    check = y[2:]*y[1:-1] &amp;lt; 0&#xA;    zc    = np.argmax(y[2:]*y[1:-1] &amp;lt; 0) + 1&#xA;    if zc &amp;gt; 10:&#xA;        resultz.append((thing, zc, x0ydot0))&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;    hw = 1.6&#xA;    for j, (thing, zc, x0ydot0) in enumerate(resultz):&#xA;        x, y = thing[:2,:zc]&#xA;        plt.plot(x, y)&#xA;    plt.xlim(-hw, hw)&#xA;    plt.ylim(-hw, hw)&#xA;    plt.plot([x1], [0], 'ok')&#xA;    plt.plot([x2], [0], 'ok')&#xA;    plt.show()&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;    for j, (thing, zc, x0ydot0) in enumerate(resultz):&#xA;        x, y = thing[:2]&#xA;        plt.plot(times[:zc], y[:zc])&#xA;    plt.show()&#xA;&#xA;if True:&#xA;    plt.figure()&#xA;    for j, (thing, zc, x0ydot0) in enumerate(resultz):&#xA;        x0, ydot0 = x0ydot0&#xA;        cycle_time = 2. * times[zc] / twopi&#xA;        ratio = abs(x0/x2)&#xA;        T_simple_model = twopi * abs(x0/x2)**1.5&#xA;        T_synodic_simple_model = 1. / (1. - twopi/T_simple_model) # https://astronomy.stackexchange.com/a/25002/7982&#xA;        plt.subplot(2, 1, 1)&#xA;        plt.plot(x0, cycle_time, 'ok')&#xA;        plt.plot(x0, abs(T_synodic_simple_model), 'or')&#xA;        plt.subplot(2, 1, 2)&#xA;        plt.plot(x0, ydot0, 'ok')&#xA;    plt.subplot(2, 1, 1)&#xA;    plt.xlabel('x0', fontsize=16)&#xA;    plt.ylabel('cycle times (periods)', fontsize=16)&#xA;    plt.subplot(2, 1, 2)&#xA;    plt.xlabel('x0', fontsize=16)&#xA;    plt.ylabel('ydot0', fontsize=16)&#xA;    plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-07-15T07:31:23.930" LastActivityDate="2019-07-15T07:31:23.930" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32648" PostTypeId="2" ParentId="32610" CreationDate="2019-07-15T09:02:54.880" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://galex.stsci.edu/GR6/?page=faq#ANSWER2&quot; rel=&quot;nofollow noreferrer&quot;&gt;GALEX magnitudes&lt;/a&gt; use the &lt;a href=&quot;https://en.wikipedia.org/wiki/AB_magnitude&quot; rel=&quot;nofollow noreferrer&quot;&gt;AB system&lt;/a&gt;, which means that &lt;span class=&quot;math-container&quot;&gt;$F_{0} \approx 3631$&lt;/span&gt; janskys for both NUV and FUV, &lt;em&gt;by definition&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7411" LastActivityDate="2019-07-15T09:02:54.880" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32649" PostTypeId="1" CreationDate="2019-07-15T09:47:22.120" Score="3" ViewCount="471" Body="&lt;p&gt;Lunar and solar eclipses go in tandem so the two often happens two weeks apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I noticed if there is a solar eclipse in my area then the preceding/following lunar eclipse is not visible from my area because the Moon is set when it happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly when lunar eclipse is visible in my area, the corresponding solar eclipse is not visible at my area because the Sun is set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently the same thing happens in July 2 when there were a total solar eclipse on the other side of Earth, and soon we will see a partial lunar eclipse from my area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only exception was when the eclipse happens near the sunrise or sunset.&#xA;For example on May 31 in 2003 a partial eclipse could be seen from Hungary at the morning when the Sun rose already partially eclipsed just few minutes before the maximum occultation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two weeks before that there was a total lunar eclipse where we could only see the beginning of the eclipse as the Moon was already set when totality occurred.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These eclipse pairs seem to occur almost exactly at the same time of day. &#xA;Why?&lt;/p&gt;&#xA;" OwnerUserId="5269" LastEditorUserId="5269" LastEditDate="2019-07-15T11:14:59.993" LastActivityDate="2019-07-16T01:29:57.203" Title="Why can't we observe a solar eclipse and the preceding/following lunar eclipse from the same place?" Tags="&lt;solar-eclipse&gt;&lt;lunar-eclipse&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32650" PostTypeId="2" ParentId="32649" CreationDate="2019-07-15T11:06:56.400" Score="2" Body="&lt;p&gt;For a solar eclipse, the Moon has to be between the Earth and Sun. For a lunar eclipse, the Earth has to be between the Moon and Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it takes the Moon about 4 weeks to orbit the Earth, it will take it about 2 weeks to go from a lineup of Moon-Earth-Sun to a lineup of Earth-Moon-Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make this more complicated, the Earth is rotating so that it takes about 24 hours to go from midday to midday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, say we have a solar eclipse at midday - there's a line between the centre of the Earth, through you, the Moon and the Sun. Two weeks later, when the Moon has moved around to the other side of the Earth this means that when you go out at midday, the line is now Moon-Earth-you-Sun. So you can't see the lunar eclipse, because the Earth is now in the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see both eclipses two weeks apart when they are around sunrise/sunset because you're no longer on the line through the centre of the Earth between the Sun and Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was going to do the math - but I think that this &lt;a href=&quot;https://www.timeanddate.com/eclipse/list.html?starty=2010&quot; rel=&quot;nofollow noreferrer&quot;&gt;map of eclipses in 2019&lt;/a&gt; demonstrates it better. If you compare the solar eclipses with their paired lunar eclipses, we see that there is a bit of overlap (since eclipses are not instantaneous) - but since these are at the edges of the eclipse, it's practically impossible to see a full solar eclipse two weeks after seeing a full lunar eclipse from the exact same location.&lt;/p&gt;&#xA;" OwnerUserId="14176" LastEditorUserId="14176" LastEditDate="2019-07-16T01:29:57.203" LastActivityDate="2019-07-16T01:29:57.203" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32651" PostTypeId="1" CreationDate="2019-07-14T08:36:14.367" Score="0" ViewCount="27" Body="&lt;p&gt;I thought that solar systems and galaxies are disk-shaped because that is the most stable shape under gravitation. Globular clusters are very old, often times older than their host galaxies, so why haven't they flattened out?&lt;/p&gt;&#xA;" OwnerUserId="28708" OwnerDisplayName="G griffo" LastActivityDate="2019-07-15T12:41:32.030" Title="Why aren't globular clusters disk-shaped?" Tags="&lt;gravity&gt;&lt;solar-system&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" ClosedDate="2019-07-15T12:41:46.633" ContentLicense="CC BY-SA 4.0" />
  <row Id="32652" PostTypeId="1" CreationDate="2019-07-15T13:46:22.617" Score="9" ViewCount="244" Body="&lt;p&gt;If extraterrestrial civilizations were sending probes by Earth similar to our plans for Breakthrough Starshot (where the probes are only a few meters in diameter and will be moving at up to 0.2c when they fly by the planet), would we be able to detect them, technologically speaking? And if so, what are the odds we would catch a glimpse of one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT/ADDITION: An additional question I thought of is whether astronauts would potentially be able to perceive such objects with the naked eye, or would the probes be moving far too fast to register? &lt;/p&gt;&#xA;" OwnerUserId="22038" LastEditorUserId="22038" LastEditDate="2019-07-17T15:56:21.980" LastActivityDate="2020-01-14T12:03:05.447" Title="Breakthrough Starshot; could we detect similar probes flying by Earth?" Tags="&lt;space-probe&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32653" PostTypeId="1" AcceptedAnswerId="32662" CreationDate="2019-07-15T14:52:49.090" Score="8" ViewCount="295" Body="&lt;p&gt;A few hundreds of millions of years ago, the moon was closer to the Earth than it is today and hence was of a bigger apparent size. This made every solar eclipse either total or partial. Over the course of time, due to tidal forces between the moon and the Earth, the moon slowly drifted far from Earth decreasing its apparent size when viewed from Earth and hence causing annular eclipses to happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That would mean there must have been some day in history when Earth witnessed its first ever annular eclipse. Is there an estimate on when this happened? &lt;/p&gt;&#xA;" OwnerUserId="16469" LastActivityDate="2019-07-16T00:25:05.817" Title="When did the first annular eclipse happen?" Tags="&lt;earth&gt;&lt;solar-eclipse&gt;&lt;history&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32654" PostTypeId="1" AcceptedAnswerId="32659" CreationDate="2019-07-15T15:03:05.093" Score="4" ViewCount="612" Body="&lt;p&gt;A substantial number of discovered exoplanet systems are likely contain more than one gas giant with distances far closer to the star than in our solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that in solar system evolution gravitational effects between objects can lead to some of them migrating outwards while others migrate inwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has there been any study of what happens when a gas giant approaches its star close enough to start losing mass to it? In this case could the planet end up completely in the star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the purposes of this question let's choose a solar-mass G-type star and a gas giant such as Jupiter, since we have so much first-hand knowledge of these objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter has plenty of hydrogen and this would represent the addition of more nucleosynthetic fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this be enough to cause the Sun to become the red giant in this case?&lt;/p&gt;&#xA;" OwnerUserId="28721" LastEditorUserId="7982" LastEditDate="2019-07-16T23:39:17.197" LastActivityDate="2019-07-17T19:06:47.960" Title="What would happen if a Sun-like star were to consume a Jupiter-like planet?" Tags="&lt;star&gt;&lt;the-sun&gt;&lt;solar-system&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32655" PostTypeId="2" ParentId="22822" CreationDate="2019-07-15T15:51:57.333" Score="5" Body="&lt;p&gt;I can't speak for the JWST, but I do work at a company that designs telescopes for spacecraft and I can tell you that a lot of the structure of the telescope is to protect the instruments during the initial ascent to space where the launch vibrations and g-loading will be the worst. Plus mirrors for telescopes that big have really precise alignment tolerances and if the secondary mirror isn't very rigidly attached to the spacecraft even simple microstrains from heat loads or spacecraft pointing can move the mirror out of alignment. Even the fact that the telescope is going from a 1g environment to a 0g environment can move the mirror because it was aligned on Earth when the beam was slightly bent and then it goes back to being straight in microgravity and the mirror is misaligned! It stinks that there is stuff in the way of some beautiful science that you're trying to take a picture of, but it's the only way to make sure the telescope survives up there!&lt;/p&gt;&#xA;" OwnerUserId="28501" LastActivityDate="2019-07-15T15:51:57.333" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32656" PostTypeId="2" ParentId="32640" CreationDate="2019-07-15T15:55:23.127" Score="2" Body="&lt;p&gt;In addition to the small z-coordinate discrepancy explained in the previous answer and links, the assertion that the x-coordinate should be 1 AU at JD 2451545, and the y-axis zero, is incorrect. In the question you included the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;X-axis : out along ascending node of instantaneous plane of the&#xA;  Earth's orbit and the Earth's mean equator at the reference epoch&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, the x-axis points outward from the Vernal equinox at the J2000.0 epoch along the line of nodes, which is the intersection of the ecliptic and equatorial planes. The Sun is about 2 2/3 months away from the ascending node on Jan 1, so the Earth-Sun radius is not directly on the x-axis.&lt;/p&gt;&#xA;" OwnerUserId="28086" LastActivityDate="2019-07-15T15:55:23.127" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32659" PostTypeId="2" ParentId="32654" CreationDate="2019-07-15T17:25:22.957" Score="10" Body="&lt;p&gt;It won't make much difference. The sun has a mass that is 1000 times more than Jupiter. Adding Jupiter to the sun will change the mass from &lt;span class=&quot;math-container&quot;&gt;$1.989\times10^{30}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$1.991\times10^{30}$&lt;/span&gt;kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding more mass to the sun will slightly reduce the life of the sun. The greater mass will cause the core to compress and heat up, increasing the rate of nuclear fusion. The sun will use up its fuel faster. Bigger stars live for shorter time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a relatively small addition of mass would not change much.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-07-15T17:25:22.957" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32660" PostTypeId="2" ParentId="22822" CreationDate="2019-07-15T19:05:58.377" Score="5" Body="&lt;p&gt;I'm an amateur telescope maker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Single vane designs do exist. Their main problem is lack of stability. You would have to use a very thick vane to hold the secondary in place in a stable way. That would block off a significant amount of light and may impact the performance of the instrument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Active systems, while theoretically possible, would be very expensive. Many vibrations are quite fast, you'd need the system to track alignment, detect deviations, make decisions very quickly and apply corrections. Meanwhile, for large telescopes, the mass of the secondary mirror is not insignificant. The actuators themselves would flex and tilt the vane while they're steering the mirror around. This is not a trivial job at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is much simpler to just use multiple vanes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One lesser-known fact: each straight vane actually makes 2 spikes, diametrically opposed. So even the single vane system would still make 2 spikes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 3 vane system makes 6 spikes, and a 4 vane system makes 8 spikes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why the 4 vane systems appear to only make 4 spikes? Because they coincide two by two. Within each pair of opposed vanes, each vane is carefully aligned with its opposed partner. This will merge the spikes two by two and will make the overall figure simpler.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You can actually make the spikes disappear. Each spike is perpendicular to the vane causing it. It's light being diffracted at the edge, going sideways from the edge of the vane. Each point on the edge diffracts light - the spike is the result of all points on the edge each diffracting a little bit of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what happens if you curve the vane? The spike &quot;spreads out&quot;, like when you open a fan. It's not a sharp line anymore, it's a whole area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the vane is curved into exactly half a circle, then the spike is maximally washed out - it becomes a full 360 degree figure. In practice, when it's spread out that much, the spike becomes invisible to the eye of the observer. Telescopes with curved vanes are said to have no spikes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there's no free lunch. Now that you've spread out the spikes, that diffracted light is still there. What happens is that the field of view becomes overall slightly more &quot;foggy&quot;, as in a fog of light. The effect is very slight, but it matters for critical observations.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Finally, you can always install a flat glass plate at the top of the instrument and glue the secondary mirror support to it. This will definitely eliminate all diffraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's hard to manufacture a large, optically flat plate like that. It will be thick and heavy. This will not scale beyond the size of small reflector telescopes. There are also issues with cooling, which becomes slower and more complex, because the telescope is now closed at the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SCT telescopes that use a corrector plate (not flat) do this anyway. Same with Maksutov-Cassegrain systems (MCT). The plate is there to correct the instrument, you might as well glue the secondary to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Actually, I lied. The SCT and MCT designs are done on purpose so that the location of the corrector plate, and the location of the secondary mirror, coincide. It's just easier to build it that way in practice. But in theory the plate and the mirror do not have to coincide.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So the main choices are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Three or four straight vanes. Maximum stability. You get 6 or 4 (nominally 6 or 8) spikes, but the image in between spikes is clean.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;One curved, half-circle vane. Sometimes two opposing curved vanes, like an X with arms curved up/down into two half-circles, or like the number 8 with the top and bottom caps lopped off (this is more stable but generates more diffraction). No visible spikes. Somewhat reduced contrast because the spikes have been smeared all over the image.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;One single, thick vane. Either not very stable, or reduces the amount of light captured by the instrument.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No vanes, just a glass plate. Expensive, heavy, doesn't scale. But if the plate is not flat you could use it to apply corrections to the instrument, so then its existence is justified.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Choice #1 is the most popular for professional and amateur reflector telescopes. #2 is used sometimes by people who object to spikes on esthetic grounds. #3 is rarely seen in practice. #4 is used with corrected SCT systems, MCT, and the like, because the plate is already there by design.&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditDate="2019-07-15T21:13:22.147" LastActivityDate="2019-07-15T21:13:22.147" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32661" PostTypeId="2" ParentId="31635" CreationDate="2019-07-15T21:23:06.213" Score="1" Body="&lt;p&gt;Here's an analogy to Quantum Mechanics. A particle in QM can tunnel through an impossible barrier, which is how elements heavier than lead can have some of their neutrons &quot;tunnel&quot; out of the nucleus escaping the bonds of the Strong Force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ymt1c.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ymt1c.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small Black Hole is like a quantum barrier that a particle can be tunnel through to escape. The smaller the barrier (Event Horizon), the more likely it is to be able to tunnel out. So a micro black hole with a mass of &lt;strong&gt;228 tons&lt;/strong&gt; and a Event Horizon of &lt;strong&gt;3.4 x 10^-7 femtometers&lt;/strong&gt; (literally less than 1 millionth the size of a proton) won't hold on to it's particles for very long and at all. In fact it will explode in a burst of Hawking Radiation after exactly &lt;strong&gt;1 second&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A larger &lt;strong&gt;Earth Mass&lt;/strong&gt; black hole with a radius of a whole &lt;strong&gt;centimeter&lt;/strong&gt;, will last much longer: &lt;strong&gt;8 x 10^50 years&lt;/strong&gt; because it's much less likely for a particle to tunnel through a whole centimeter to get to freedom.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Source: Quantum tunneling from three-dimensional black holes: &lt;a href=&quot;https://arxiv.org/abs/1306.6380&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/1306.6380&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: Hawking Radiation modelled as a Quantum Effect: &lt;a href=&quot;http://cscanada.net/index.php/ans/article/view/j.ans.1715787020120502.1817&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://cscanada.net/index.php/ans/article/view/j.ans.1715787020120502.1817&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="28132" LastActivityDate="2019-07-15T21:23:06.213" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32662" PostTypeId="2" ParentId="32653" CreationDate="2019-07-16T00:12:51.190" Score="3" Body="&lt;p&gt;A quick and dirty estimate is as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An angular eclipse occurs first if the Moon is at apogee and Earth at perihelion. Assuming that these two events coincide (which happens every ~8.85 years owing to lunar apsidal precession), an angular eclipse requires that (&lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; denoting orbital semi-major axis and &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; radius)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;      \frac{R_\supset}{(1+e_\supset)a_\supset} &amp;lt; \frac{R_\odot}{(1-e_\oplus)a_\oplus}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;      a_\supset &amp;gt; \frac{R_\supset}{R_\odot} \frac{1-e_\oplus}{1+e_\supset} a_\oplus\approx 348300\,\mathsf{km}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where I assumed that the orbital eccentricities, Lunar and Solar radii, and Earth's semi-major axis didn't change (over the period considered here) and are given by&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Orbit_of_the_Moon&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$e_\supset=0.0549$&lt;/span&gt;&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth%27s_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$e_\oplus=0.0167$&lt;/span&gt;&lt;/a&gt;, &lt;span class=&quot;math-container&quot;&gt;$R_\supset=1737.1\,$&lt;/span&gt;km, &lt;span class=&quot;math-container&quot;&gt;$R_\odot=695510\,$&lt;/span&gt;km, and &lt;span class=&quot;math-container&quot;&gt;$a_\oplus=149.6\times10^6\,$&lt;/span&gt;km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the present we have &lt;span class=&quot;math-container&quot;&gt;$a_\supset=384400\,$&lt;/span&gt;km. If we assume an &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbit_of_the_Moon#Tidal_evolution&quot; rel=&quot;nofollow noreferrer&quot;&gt;average annual change by 22 mm&lt;/a&gt;, we find a time of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$1.6\times10^9\,\mathsf{years},$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;though it could have been more recent if the annual change of &lt;span class=&quot;math-container&quot;&gt;$a_\supset$&lt;/span&gt; was larger some hundred million years ago (which may well have been the case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a similar way, one may also estimate when the &lt;strong&gt;last full Solar eclipse&lt;/strong&gt; will occur. This time, the Moon is at perigee and Earth at apohelion, when&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;      a_\supset &amp;lt; \frac{R_\supset}{R_\odot} \frac{1+e_\supset}{1-e_\oplus} a_\oplus\approx 400800\,\mathsf{km}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which last occurs in &lt;span class=&quot;math-container&quot;&gt;$430\times10^6$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$745\times10^6$&lt;/span&gt; years, depending on whether the current lunar recession rate of 38mm/yr or the long-term average of 22mm/yr is used.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditDate="2019-07-16T00:25:05.817" LastActivityDate="2019-07-16T00:25:05.817" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32663" PostTypeId="2" ParentId="32642" CreationDate="2019-07-16T00:41:50.837" Score="1" Body="&lt;p&gt;A simple estimate can give you a rough idea what area of the sky will be visible at different times of the year, using only a bit of mental arithmetic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just remember one number, the right ascension (RA) at (or nearest to) the meridian (between due S and the zenith) at midnight on Jan 1st. This is (roughly) equivalent to local sidereal time. Actually, I just remember the hour, 06h, at my longitude of just under 72 degrees E. This changes very slowly from year to year due to precession (I may have to use 05h by around 2050 or so, if I am still alive!). Since the difference between sidereal time and UTC is about 4 minutes per day, RA on the meridian at midnight shifts about 2 hours later per month. Now around Jul 15th, about 19h RA is on the meridian at midnight here. Latitude determines what range of declination can be seen. At ~42 degrees N, I usually cannot see much below -30 degrees declination, depending on the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing what RA is near the meridian (I can easily subtract or add hours corresponding to time before or after midnight), by looking at a sky chart I can easily estimate what objects should be visible, or well placed for observing. I used this simple technique for many years, before digital everything! Now I can just use SkySafari on my iPad. :-)&lt;/p&gt;&#xA;" OwnerUserId="28086" LastActivityDate="2019-07-16T00:41:50.837" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32664" PostTypeId="2" ParentId="22822" CreationDate="2019-07-16T00:56:44.207" Score="2" Body="&lt;p&gt;&lt;strong&gt;tl;dr:&lt;/strong&gt; Using one vane may not provide any optical advantage over three vanes since the primary mirror is already so highly segmented and &amp;quot;edgy&amp;quot;, and it certainly provides a lot of mechanical disadvantages in terms of stiffness against lateral vibration translation and breaking of symmetry during thermal expansion!&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;note:&lt;/strong&gt; I'll be updating this answer soon, once I get access to some SPIE papers.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Stiffness&lt;/h2&gt;&#xA;&lt;p&gt;Spacecraft, even space telescopes contain many active sources of vibration. Pumps, valves, refrigerators, thrusters, reaction wheels, actuators (JWST has a movable solar reflector for trimming radiation pressure) movable antennas and solar panels (Hubble at least, not sure about JWST, will check...) and even the science payloads may have movable mirrors or focusing mechanisms.&lt;/p&gt;&#xA;&lt;p&gt;A three-vane mount provides &lt;em&gt;substantially more stiffness&lt;/em&gt; against in-plane movement, since flexure of one vane would result in tension and compression in the other two vanes.&lt;/p&gt;&#xA;&lt;p&gt;For a single vane; you have a mass on the end of a stick. My first telescope was an Edmund 4¼ inch Newtonian with a big glass right-angle prism on the end of a metal rod. You could tap the tube near the top and that thing would ring, making the image a blur.&lt;/p&gt;&#xA;&lt;p&gt;Three vanes vs one will also help in axial vibrations (toward/away from the primary) but not as much as transverse, but &lt;em&gt;small transverse oscillations are much worse than small axial vibrations&lt;/em&gt; because the secondary is curved. Move it sideways and the image will slide back and forth across the image sensors, blurring across the pixels. Tens of microns of lateral motion of the image will kill the resolution.&lt;/p&gt;&#xA;&lt;p&gt;However, tens of microns of axial motion would hardly make a difference in focus in a long focal length and high f/no. system.&lt;/p&gt;&#xA;&lt;h2&gt;Symmetry under thermal expansion&lt;/h2&gt;&#xA;&lt;p&gt;With the huge multi-layer Sun shield JWST shows a major effort to keep the temperature of all of the optics and detector system very low so that it does not contribute thermal infrared radiation and so the sensors will have low thermal noise.&lt;/p&gt;&#xA;&lt;p&gt;However you would also like to keep the large structural elements as thermally stable as possible to minimize drift in focus and position of the image. A three-vane spider offers some amount of lateral symmetry, and if there are temperature changes it might reduce the amount of lateral drift, compared to a single vane where expansion-contraction obviously would just cause the secondary to move back and forth along the vane's axis.&lt;/p&gt;&#xA;&lt;h2&gt;Diffraction, PSF and Hubble&lt;/h2&gt;&#xA;&lt;p&gt;The Hubble Space Telescope does suffer from diffraction effects from its four vanes &lt;em&gt;as well as from its giant secondary and three mirror through-holes for mounting&lt;/em&gt;, but a major effort in management of the HST's point spread function (e.g. &amp;quot;optical modeling using Tiny Tim&amp;quot;) seems to handle this well.&lt;/p&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;https://photo.stackexchange.com/a/89217/50110&quot;&gt;this answer&lt;/a&gt; in Photography SE:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;(see also) @scottbb's &lt;a href=&quot;https://photo.stackexchange.com/a/89186/50110&quot;&gt;(excellent) answer&lt;/a&gt;... For scientific analysis of images from Hubble, one has to account for all aberrations, diffraction, and pixilation. There is a nice writeup on this: &lt;a href=&quot;http://www.stsci.edu/institute/org/telescopes/Reports/krist_tinytim_spie.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;20 years of Hubble Space Telescope optical modeling using Tiny Tim&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vLXtn.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tHbOo.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kBdYe.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Vane number&lt;/h2&gt;&#xA;&lt;p&gt;It's important to note that WFIRST went the other way, and had &lt;em&gt;six vanes!&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.stsci.edu/wfirst/science-planning-toolbox/psf-simulation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Introduction to WebbPSF for WFIRST&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ilPmF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ilPmF.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h2&gt;JWST's Point Spread Function - don't forget the segmented primary!&lt;/h2&gt;&#xA;&lt;p&gt;The following image is taken out of context (at the moment, I'll retrieve it later today) from the paywalled 2014 SPIE paper &lt;a href=&quot;https://www.spiedigitallibrary.org/conference-proceedings-of-spie/9143/91433X/Updated-point-spread-function-simulations-for-JWST-with-WebbPSF/10.1117/12.2056689.short&quot; rel=&quot;nofollow noreferrer&quot;&gt;Updated point spread function simulations for JWST with WebbPSF&lt;/a&gt;. Considering that the primary mirror presents much more total &amp;quot;diffracting edge&amp;quot; transverse to the incoming wavefront than the secondary's three vanes, it's possible or even likely that the spider's contribution even with three vanes is small compared to that from the segmenting of the primary.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/d5r5J.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;https://buildmedia.readthedocs.org/media/pdf/webbpsf/latest/webbpsf.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;webbpsf Documentation Release 0.8.0&lt;/a&gt; (click for larger size)&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HXFB2.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Fig. 1: Sample PSFs for JWST’s instrument suite, all on the same angular scale and display stretch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;Conclusion&lt;/h2&gt;&#xA;&lt;p&gt;In other words, using one vane may not provide any advantage over three vanes since the primary mirror is already so highly segmented and &amp;quot;edgy&amp;quot;, and it certainly provides a lot of mechanical disadvantages in terms of stiffness against lateral vibration translation and breaking of symmetry during thermal expansion!&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-07-16T06:25:37.863" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32665" PostTypeId="1" AcceptedAnswerId="32671" CreationDate="2019-07-16T02:42:27.007" Score="2" ViewCount="142" Body="&lt;p&gt;The NPR.org article &lt;a href=&quot;https://www.npr.org/2019/07/15/741990200/hawaii-protesters-block-access-road-to-stop-construction-of-massive-telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hawaii Protesters Block Access Road To Stop Construction Of Massive Telescope&lt;/a&gt; briefly summarizes the situation, one paragraph says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The TMT, as it is called, has been the object of intense opposition for nearly a decade and the building project has faced numerous delays, protests and lawsuits. But a decision by the state's Supreme Court in October cleared the way for construction to begin after reinstating a building permit that was revoked years earlier. And last week, Gov. David Ige announced several roads would be closed beginning Monday to allow for the movement of large equipment onto the mountain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and another says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Hawaii is a unique place,&quot; Meisenzahl said, noting that protesters gathered on the mountain were engaging in a peaceful, nonviolent form of protest called Kapu Aloha. &quot;It's a small island and everybody knows each other so law enforcement and opponents are working together. Despite what is happening today it's a positive environment,&quot; he added. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What would be the scientific impact of locating the Thirty Meter Telescope at its alternate site? Is it a loss of resolution, or infrared spectrum, or both, or something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the Canary Island site would be the alternate to Mauna Kea, since the E-ELT would be located in Chile and putting both (extremely) large telescopes in the same hemisphere is undesirable. &lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-08-26T10:55:43.850" Title="What would be the scientific impact of locating the Thirty Meter Telescope at its alternate site?" Tags="&lt;observational-astronomy&gt;&lt;observatory&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32666" PostTypeId="1" CreationDate="2019-07-16T04:15:55.127" Score="2" ViewCount="137" Body="&lt;p&gt;When looking at a diagram of the optical path through a newtonian reflector, the lines always cross before the eyepiece. I understand that the eyepiece is used to focus this diverging light cone so that your eye can observe an image. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use a camera in &quot;prime focus&quot; i.e. Without an eyepiece to rectify this diverging light cone, it can discern an image anyway? Would this mean that the focus adjustment (distance from the focal point) merely determines the size of the image on the sensor? Or is there still a point where the image is in focus? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we ensure that this occurs while the light cone is still smaller than the sensor? Or is the whole image always varied in focus across the cone, just that the depth of field allows all of it to be in focus enough for a good image?&lt;/p&gt;&#xA;" OwnerUserId="28726" LastEditorUserId="7982" LastEditDate="2019-07-16T06:15:06.547" LastActivityDate="2019-07-16T06:15:06.547" Title="Telescope focal point and inversion" Tags="&lt;telescope&gt;&lt;photography&gt;&lt;optics&gt;&lt;telescope-lens&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32667" PostTypeId="2" ParentId="32666" CreationDate="2019-07-16T05:20:32.130" Score="4" Body="&lt;p&gt;The main mirror of a telescope forms an image at the prime focus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see the lines being brought to a single point, but that doesn't mean all the light entering the telescope is focused to a single point. It means that all the light from a single point in space is focussed at a single point. Light from different sources is focussed at different points, these points form a plane in the body of the telescope, the &quot;focal plane&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image can be seen. You can project an image using a single lens, or if you hold up a lens at arm's length and look through it, you will see a small inverted image. The purpose of the eyepiece is to act as a magnifying glass and make this small inverted image appear bigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image is in focus only at the focal plane. If a sensor (without any further lens) is placed at any other point, no image will be seen, or at best a blurry image. We can see the image when we look through the eyepiece only because we have another lens (the cornea of our eye) bringing the light to a second focus on our retina.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-07-16T05:20:32.130" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32668" PostTypeId="1" CreationDate="2019-07-16T05:58:48.343" Score="1" ViewCount="310" Body="&lt;p&gt;I found that Earth's perihelion occurs roughly two weeks after winter solstice similar thing happens for summer solstice and aphelion. I understand what those terms mean but I don't understand the reason for the two-week gap. I actually thought they occur on the same day. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I not understanding here? Can anyone explain the reason for the two weeks? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that the earth's aphelion and perihelion occur on the major axis of earth's elliptical orbit. Why are they not the extremities of &lt;a href=&quot;https://en.wikipedia.org/wiki/Analemma&quot; rel=&quot;nofollow noreferrer&quot;&gt;analemma&lt;/a&gt; which correspond to solstices? Where is my thinking wrong?&lt;/p&gt;&#xA;" OwnerUserId="28708" LastEditorUserId="7982" LastEditDate="2019-07-16T06:08:14.557" LastActivityDate="2019-07-16T06:11:43.163" Title="Earth's perihelion vs winter solstice" Tags="&lt;the-sun&gt;&lt;orbit&gt;&lt;earth&gt;&lt;solstice&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2019-07-16T12:39:44.343" ContentLicense="CC BY-SA 4.0" />
  <row Id="32669" PostTypeId="2" ParentId="32668" CreationDate="2019-07-16T06:11:43.163" Score="4" Body="&lt;p&gt;There is no connection between the date of the solstice and the perihelion. It is merely a minor coincidence that perihelion occurs close to the solstice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship isn't fixed. Precession in the Earth's orbit (caused by gravitational perturbations of Jupiter and other planets) will change the relative time of perihelion and solstice over a period of thousands of years.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-07-16T06:11:43.163" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32670" PostTypeId="1" AcceptedAnswerId="32674" CreationDate="2019-07-16T06:27:59.483" Score="1" ViewCount="134" Body="&lt;p&gt;I mentioned my old telescope's secondary mirror in &lt;a href=&quot;https://astronomy.stackexchange.com/a/32664/7982&quot;&gt;this answer (stiffness)&lt;/a&gt; but I'm not certain I am remembering correctly. My first telescope was a Edmund Scientific 4¼ inch Newtonian reflector. This was in the 1970's. I seem to remember that the secondary mirror was a right angle prism rather than an elliptical front-surface mirror.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I dreaming or mis-remembering, or did Edmund use prisms on at least some 4¼ inch Newtonians?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, any discussion on the tradeoffs between front-surface elliptical mirrors and 90° prism secondaries would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UaO29.gif&quot; alt=&quot;example of Newtonian with prism secondary&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: John M. Pierce's HobbyGraph Articles &lt;a href=&quot;http://bobmay.astronomy.net/johnpierce/hg15.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;HobbyGraph #15; Diagonals and Diagonal Supports.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-07-16T06:33:23.433" LastActivityDate="2019-07-19T01:47:24.743" Title="Did Edmund Scientific 4¼ inch Newtonians have 90° prisms as secondary mirrors in the 1970's?" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;history&gt;&lt;newtonian-telescope&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32671" PostTypeId="2" ParentId="32665" CreationDate="2019-07-16T07:04:18.743" Score="3" Body="&lt;p&gt;The Canary Islands have slightly poorer seeing on average, but the main difference would be the higher column density of water vapour. Mauna Kea is a  higher and &quot;drier&quot; site, which means you can observe into the mid-IR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the &lt;a href=&quot;https://www.tmt.org/page/science-instruments&quot; rel=&quot;nofollow noreferrer&quot;&gt;first light instrument package&lt;/a&gt; I would guess that WFOS (wide field imaging at 0.3-1 micron) would not be badly affected. But there would be some impact on IRIS (near-IR spectroscopy) towards the long-wavelength end of its range. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The near-IR adaptive optics would also be somewhat affected by the slightly poorer natural seeing, but I'm not sure I can quantify this.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-07-16T12:45:32.083" LastActivityDate="2019-07-16T12:45:32.083" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32672" PostTypeId="1" AcceptedAnswerId="32678" CreationDate="2019-07-16T12:54:22.090" Score="17" ViewCount="2678" Body="&lt;p&gt;The Kepler space telescope detects planets based on the dip in brightness caused by planets moving past the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wouldn’t that mean that there are an unknown amount of planets that have an orbit that wouldn’t be detected because their orbits don’t cross that path between the star and the telescope?&lt;/p&gt;&#xA;" OwnerUserId="28708" LastEditorUserId="18643" LastEditDate="2019-07-17T12:43:17.677" LastActivityDate="2019-07-18T01:05:10.040" Title="Kepler space telescope undetected planets" Tags="&lt;planet&gt;&lt;kepler&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32673" PostTypeId="2" ParentId="32672" CreationDate="2019-07-16T14:29:36.147" Score="8" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of a transit taking place is something like &lt;span class=&quot;math-container&quot;&gt;$r/a$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; is the stellar radius and &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is the radius of the planetary orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you assume that planet orbits are randomly inclined to our line of sight, then each detected planet corresponds to &lt;span class=&quot;math-container&quot;&gt;$a/r$&lt;/span&gt; planets in actual fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this approximation is ok for circular orbits where &lt;span class=&quot;math-container&quot;&gt;$r \ll a$&lt;/span&gt; and where the planetary radius is much smaller than the stellar radius. Much more complicated expressions are required (and of course used) in planet population analyses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://iopscience.iop.org/article/10.1086/522039&quot; rel=&quot;nofollow noreferrer&quot;&gt;Burke et al. (2008)&lt;/a&gt;, the correction for eccentricity is just to &lt;em&gt;divide&lt;/em&gt; the probability by &lt;span class=&quot;math-container&quot;&gt;$(1 - e^2)$&lt;/span&gt; if &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is now the semi-major axis of the orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correction for the size of the planet is simply to add the planet radius to the stellar radius. Thus:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ p \simeq \frac{r_p + r}{a (1 - e^2)},$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$r_p$&lt;/span&gt; is the planet radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final bit of detail that can't be captured by a simple equation are the chances of capturing the transit because of limited cadence or duty cycle of the observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even for a mission like Kepler there comes a limit when the duration of the transit may only cover one or two observation points, and it becomes difficult to discern a transit. Ditto of course if the mission duration only covers a single transit so that the planetary nature cannot be confirmed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These kinds of effects are important at larger &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt;, where planets have longer orbital periods and shorter transits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you have to take into account the signal to noise ratio of the observations. Smaller planets around fainter stars produce transit signals that are harder to detect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These issues can (and are) only be dealt with by making simulations of the observational data.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="2531" LastEditDate="2019-07-17T08:08:40.737" LastActivityDate="2019-07-17T08:08:40.737" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32674" PostTypeId="2" ParentId="32670" CreationDate="2019-07-16T15:35:58.503" Score="3" Body="&lt;p&gt;My Edmund F/10 4 1/4&quot; was bought in 1975, and had a rectangular front-surface mirror as the diagonal, with a single arm support attached to the tube near the eyepiece holder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the &lt;a href=&quot;https://en.wikipedia.org/wiki/Astroscan&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astroscan 2001&lt;/a&gt; (or at least some models of it) used a prism for the diagonal, supported by the flat glass plate at the front of the tube; haven't found a reference though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantage of a prism is that it is more maintenance-free then a mirror which will eventually need to be re-aluminized.  The disadvantages are that there are two optical surfaces instead of one in the light path that could reflect or scatter light, you can get color dispersion, and a rectangular prism will block more light than an elliptical mirror (especially important in a fast scope like the AstroScan).  The &lt;a href=&quot;https://en.wikipedia.org/wiki/Star_diagonal&quot; rel=&quot;nofollow noreferrer&quot;&gt;star diagonal Wikipedia page&lt;/a&gt; has a pretty good discussion of the trade-offs. &lt;/p&gt;&#xA;" OwnerUserId="14434" LastActivityDate="2019-07-16T15:35:58.503" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32675" PostTypeId="1" AcceptedAnswerId="32677" CreationDate="2019-07-16T15:37:11.873" Score="4" ViewCount="259" Body="&lt;p&gt;If there is no matter inside any of the black holes and all of them have a singularity in the center of them, which is infinitely small and infinitely dense, then how can they differ in size and weight? &lt;/p&gt;&#xA;" OwnerUserId="28732" LastEditorUserId="16685" LastEditDate="2019-07-19T05:27:53.950" LastActivityDate="2019-07-19T05:27:53.950" Title="Size of a black hole" Tags="&lt;black-hole&gt;&lt;supermassive-black-hole&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32676" PostTypeId="1" AcceptedAnswerId="32748" CreationDate="2019-07-16T16:14:27.533" Score="9" ViewCount="803" Body="&lt;p&gt;Scientists strongly suspect that several moons in our solar system have frozen-over oceans of water-ammonia mixture. I've also read speculations on the possibility of surface water-ammonia oceans on exoplanets and exomoons. What I haven't seen a lot of is a discussion on how those oceans might behave differently than the water oceans of Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, &lt;strong&gt;how would the surface water-ammonia ocean behavior differ from our water ocean behavior&lt;/strong&gt; in terms of: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Ice(s) formation&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Evaporation&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Surface and underwater currents&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Varying ratios of water-ammonia based on temperature or other&#xA;factors&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I'd like to know if this has been studied and if these behaviors have been described in papers, books, or in-depth articles on this topic.&lt;/strong&gt; Especially anyting that goes into the specifics of the water/ammonia ratios.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'll add some personal speculation here. &lt;strong&gt;While it is not part of the primary question stated above&lt;/strong&gt;, it helps to explain the type of problem I'm trying to understand, so may be helpful to narrow down which resources may be most useful.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When subject to increasing levels of heat would the ammonia molecules&#xA;in the ocean evaporate before the water molecules, causing warmer&#xA;areas of the ocean to have an increased amount of H2O relative to&#xA;NH3? (And the cooler areas like the poles having an increasing ratio&#xA;of NH3 to H2O).&lt;/li&gt;&#xA;&lt;li&gt;If the above was true, might the equatorial waters that have less NH3&#xA;facilitate the formation of water-ice? If so I could imagine&#xA;exoplanets having water-ice-covered equators &lt;em&gt;because&lt;/em&gt; they were&#xA;warmer, and the cooler poles might be ice-free thanks to the greater&#xA;amount of NH3 in the ocean.&lt;/li&gt;&#xA;&lt;li&gt;Would the ice formed on the ocean be composed of a molecular mix&#xA;matching that of the ocean (with H2O and NH3 in the same ratio as the&#xA;ocean), or would some molecules freeze-out first, possibly causing&#xA;different layers or areas of varying types of ices?&lt;/li&gt;&#xA;&lt;li&gt;Might the poles of some water-ammonia planets have &quot;continents&quot; of&#xA;water-ammonia ices that are frozen from ocean bed up to the surface&#xA;(thanks to the fact that frozen ammonia is more dense than liquid,&#xA;allowing for entire sections of ocean to freeze solid)?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If my speculation is correct (and I could be wildly wrong since I don't yet understand how a mix of water-ammonia actually behaves) then an exoplanet with an equatorial band of water-ice, mid-latitudes of liquid ocean, and solid polar-caps/continents of water/ammonia ices could be possible. The ammonia in the ocean would evaporate more quickly near the warmer equator, causing that area of a global ocean to have a greater ratio of H2O, and the ammonia precipitation and ocean might freeze at the poles, causing the build up of vast polar continents of ammonia/water ices. I would imagine the currents would function very differently than on earth, with surface and deep-level currents of varying ammonia-water ratios and temperatures. I I also see the possibility of frozen ocean beds in the more ammonia-heavy parts of an ocean, with the surface unfrozen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the ratio of ammonia to water will be very important. I'd love to see specific information on that!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any informed clarification appreciated!&lt;/p&gt;&#xA;" OwnerUserId="20372" LastEditorUserId="1891" LastEditDate="2019-07-20T22:54:12.887" LastActivityDate="2019-07-21T17:17:25.070" Title="How would water-ammonia oceans behave?" Tags="&lt;exoplanet&gt;&lt;water&gt;&lt;surface&gt;&lt;terrestrial-planets&gt;" AnswerCount="1" CommentCount="14" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32677" PostTypeId="2" ParentId="32675" CreationDate="2019-07-16T18:22:07.660" Score="5" Body="&lt;p&gt;The size of a black hole is determined by its energy and angular momentum. For a non-spinning black hole, there is a simple linear relationship between the black hole's mass and its &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;noreferrer&quot;&gt;Schwarzschild radius&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$r_s=\frac{2GM}{c^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$r_s$&lt;/span&gt; is the Schwarzschild radius, &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; is the universal gravitational constant, &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; is the speed of light, and &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is the black hole's mass. The mass of the black hole comes from all the matter and energy that went into creating the black hole, as well as any additional matter &amp;amp; energy that's fallen into it afterwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strictly speaking, the Schwarzschild solution is for a universe containing nothing but a single eternal black hole. But although it's unphysical it's still a useful solution when we want to model spacetime with spherical spatial symmetry (including planets &amp;amp; stars, not just black holes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a rotating black hole, the situation is more complex, and there are several horizons. Please see Wikipedia's article on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kerr_metric&quot; rel=&quot;noreferrer&quot;&gt;Kerr metric&lt;/a&gt; for details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We cannot say exactly what happens at the core of a black hole. As discussed in &lt;a href=&quot;https://astronomy.stackexchange.com/a/2254/16685&quot;&gt;this answer by Florin&lt;/a&gt;, we need a theory that unites General Relativity and Quantum mechanics to answer such questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pure GR black hole has a mathematical singularity at its core, but most astrophysicists believe that's unphysical, and that a proper quantum gravity theory will eliminate that singularity. However, it's likely that the core of a black hole is still very small, since the quantum gravity corrections probably don't kick in until the size gets smaller than an atom. And of course even with a quantum gravity we still won't ever be able to observe black hole cores to validate the theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we cannot currently say exactly what happens to energy and matter once it reaches the core of a black hole. But we do know that once anything crosses the event horizon of a black hole it can no longer affect the universe outside the event horizon, and that anything inside the event horizon must rapidly (in its own proper time) fall towards the core of the black hole, and that no known force can prevent that process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational field of a black hole is sometimes described as a &lt;a href=&quot;http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_gravity.html&quot; rel=&quot;noreferrer&quot;&gt;&quot;fossil field&quot;&lt;/a&gt;. All matter &amp;amp; energy falling into the black hole modifies the spacetime curvature as it approaches the event horizon. And once it crosses the event horizon it can no longer change the spacetime curvature outside the horizon, so those curvature changes are preserved (until something else comes along to add its own curvature changes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it doesn't really matter what happens at the core of the black hole, since the exact nature of the black hole core doesn't affect what's happening outside the event horizon. &lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2019-07-16T18:22:07.660" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32678" PostTypeId="2" ParentId="32672" CreationDate="2019-07-16T18:47:08.027" Score="37" Body="&lt;p&gt;That's right. The inclination of the orbital plane around stars is considered to be random throughout the galaxy, thus the planets we can detect by the transit method is just a tiny fraction of the planets that we should expect in our stellar neighbourhood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transit method allows for planetary detection only when the line of sight from Earth to the system is contained, or &lt;strong&gt;almost&lt;/strong&gt; contained, in the orbital plane of the planet. This means that only a tiny range of orbital inclinations on each star are good for detection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why did I say almost? Because there is some range of inclinations that still would yield a transit. This range is not fixed, and it depends on the distance of the planet to its host star. As you can see in this diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zrXgc.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zrXgc.png&quot; alt=&quot;Enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planet A is closer to the star and thus creates a wider shadow. If an observer is located in that shadowed region far away it can detect planet A. Planet B instead is farther from the star and thus its shadow is narrower. It is interesting to note, that even if both planets here share the exact same orbital plane there are places from where you would only detect planet A and never detect planet B (see the green arrows). This is the reason we have a bias towards planets orbiting closer to their star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This effect is in fact quite strong: consider our Solar System from an exoplanetary perspective. If you were located in a random star in the sky, what are the chances you would spot an Earth transit? Well, it turns out that it is way more probable to detect a Mercury transit, even if Mercury is the smallest planet, just because of its vicinity to the Sun. &lt;a href=&quot;https://arxiv.org/pdf/1709.02211.pdf&quot; rel=&quot;noreferrer&quot;&gt;A recent paper&lt;/a&gt; showed this diagram of the regions of the sky where some alien inhabitants would spot a transit for each of our planets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5lDMq.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5lDMq.png&quot; alt=&quot;Enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see Mercury has the wider strip. Also it's interesting to note that due to these differences in the size of the orbits (let's use the semi-major axis, &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt;, as a reference) and due to small differences in orbital inclinations there is no place in the entire sky from which an alien could detect simultaneously more than four of our planets by the transit method. No place in the universe where all the Solar System's planets would be detectable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The detection method also depends on the relative sizes of the star, &lt;span class=&quot;math-container&quot;&gt;$R_s$&lt;/span&gt;, and the planet &lt;span class=&quot;math-container&quot;&gt;$R_p$&lt;/span&gt;: A larger star has a larger disk (as viewed from Earth) that can be easily &lt;a href=&quot;https://en.wikipedia.org/wiki/Photobombing&quot; rel=&quot;noreferrer&quot;&gt;photobombed&lt;/a&gt; by a planet and a larger planet can photobomb more easily if it is larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YAdd1.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YAdd1.png&quot; alt=&quot;Enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is that the probability of detecting a planet increases as we increase both/either &lt;span class=&quot;math-container&quot;&gt;$R_p$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$R_s$&lt;/span&gt; and increases as we decrease the distance to the host star &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt;. The relation is then of this form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P \sim (R_s+R_p)/a$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This relation imposes several observational biases. We can see exoplanets that are large and closer to their star, but we can't see planets that are small and farther. That is the reason the first detected exoplanets are the so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Hot_Jupiter&quot; rel=&quot;noreferrer&quot;&gt;hot Jupiters&lt;/a&gt;: giant planets much closer to their stars than Mercury is to the Sun. This diagram shows all the exoplanet detections plotted on size vs. orbital distance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1F3zx.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1F3zx.jpg&quot; alt=&quot;Enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, small planets are only detectable if they have very small orbits around their stars. We have yet to find a planet the size of Earth (quite small) and with a 365 day orbital period (1 AU distance) using the transit method. There is no reason to think that this is representative of the overall population of planets. The black region of the plot is probably filled with dots, but our instruments can't scout that region yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Kepler telescope had a camera with a field of view on which it could detect more than half a million stars, but the actual number of stars monitored during the mission was around &lt;a href=&quot;https://arxiv.org/pdf/1001.0349.pdf&quot; rel=&quot;noreferrer&quot;&gt;150,000 stars (these stars had good signals and were perfect targets for the mission)&lt;/a&gt;. For these 150,000 stars &lt;a href=&quot;https://exoplanetarchive.ipac.caltech.edu/docs/counts_detail.html&quot; rel=&quot;noreferrer&quot;&gt;Kepler found 2,345 exoplanets&lt;/a&gt; distributed in 1,205 stars. So we can say that for each star targeted by Kepler, the average probability of finding some planets there is around &lt;span class=&quot;math-container&quot;&gt;$0.8\;\%$&lt;/span&gt;. That should give you an estimate of the occurrence of orbital inclinations that result in transits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The truth is that this number is too small, because Kepler has several more biases. For example, Kepler only confirmed planets after three transits were detected. Since the Kepler mission lasted for four years and four months we can say that in the best case scenario Kepler was able to detect a planet with an orbital period as long as two years and two months, but this is not even the case since for that to happen a transit should have been detected just at the beginning of the mission, halfway, and at the exact end of it, and this coincidence didn't happen. Thus Kepler had no chance to discover any planet with periods longer than two years (enough for Earth, but not enough for our Jupiter for example), &lt;strong&gt;even if the orbital inclination matched perfectly for the transit&lt;/strong&gt;. So you might expect more possible transits than those actually portrayed by the Kepler telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, for planets close to their stars, &lt;a href=&quot;https://en.wikipedia.org/wiki/Methods_of_detecting_exoplanets#Transit_photometry&quot; rel=&quot;noreferrer&quot;&gt;it has been estimated&lt;/a&gt; that the probability of a random alignment to allow a transit gets up to &lt;span class=&quot;math-container&quot;&gt;$10\;\%$&lt;/span&gt;. For the case of stars as big as our Sun and planets at the same distance as Earth, the probability of this random occurrence drops to &lt;span class=&quot;math-container&quot;&gt;$0.47 \;\%$&lt;/span&gt;. So with all the diversity of planets (in terms of sizes and distances to their host star) it is reasonable to expect a &lt;span class=&quot;math-container&quot;&gt;$0.8\;\%$&lt;/span&gt; detection rate for Kepler (if we also add the time restriction to observe three transits).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;span class=&quot;math-container&quot;&gt;$0.47\;\%$&lt;/span&gt; is an amazing number! It means that for every Earth-like planet we detect by the transit method we should expect another 213 Earth-like planets orbiting other stars that are undetectable by the transit method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This kind of reasoning has been expanded. We have many difficulties to detect them, but if you mathematically model that difficulty and the corresponding biases associated with the known instruments and you assume random configurations, you can see that each discovery yields statistical significance to the amount of possible planets that are really out there. There are so many detections now that we can finally establish with statistical confidence that there are more planets than stars in our galaxy (even if we have probed an infinitesimal fraction of the entire population), even if this was something that could be expected we have now strong evidence for that thanks to Kepler. This means that there &lt;a href=&quot;https://www.nasa.gov/mission_pages/kepler/news/kepler20130103.html&quot; rel=&quot;noreferrer&quot;&gt;could be around a trillion or more plants just in the Milky Way&lt;/a&gt;. Now we are also able to establish some statistical constraints on the occurrence of Earth-like planets (orbiting in the habitable zone of their sun-like star) thanks to Kepler. &lt;a href=&quot;https://arxiv.org/pdf/1311.6806.pdf&quot; rel=&quot;noreferrer&quot;&gt;There are probably around 11 billion planets in our galaxy with these specifications&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many more planets than the ones we can detect by the transit method, between 10 and 100 times more depending on the size and orbital period of the planet you are searching for.&lt;/p&gt;&#xA;" OwnerUserId="28496" LastEditorUserId="1435" LastEditDate="2019-07-18T01:05:10.040" LastActivityDate="2019-07-18T01:05:10.040" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="32679" PostTypeId="1" CreationDate="2019-07-16T19:53:00.180" Score="4" ViewCount="92" Body="&lt;p&gt;Has there been any theoretical or observational study of the possibility of detecting &quot;local dark matter&quot;, either within the solar system or near the Sun in interstellar space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, what methods could possibly be used to detect it?&lt;/p&gt;&#xA;" OwnerUserId="775" LastEditorUserId="7982" LastEditDate="2019-07-16T23:25:42.257" LastActivityDate="2019-07-16T23:25:42.257" Title="Any studies of the detection of dark matter in or near the solar system?" Tags="&lt;observational-astronomy&gt;&lt;solar-system&gt;&lt;dark-matter&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32680" PostTypeId="2" ParentId="32679" CreationDate="2019-07-16T20:55:22.887" Score="1" Body="&lt;p&gt;There is dark matter in the solar system in the sense of ordinary matter which is too little, too dispersed, and reflects too little light to be seen. Occasionally one of the larger fragments will fly close enough to be noticed by our radars and telescopes, but there is not enough of it to affect the dynamics of the solar system. If you mean exotic dark matter in the form of WIMPs, we can't say with 100 percent certainty that there is none at all, but we can say there are not massive amounts of it, otherwise its gravity would betray its presence. Personally, I don't believe there is any at all, but I couldn't prove it.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-07-16T20:55:22.887" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32681" PostTypeId="2" ParentId="32679" CreationDate="2019-07-16T21:10:11.957" Score="5" Body="&lt;p&gt;We don't know what dark matter is, and we have never directly detected the exotic dark matter that is believed to make up much of the mass of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We expect dark matter to be in the form of a particle that interacts only with gravity and the weak nuclear interaction. Such a particle would be invisible (as light wouldn't interact with it) and it could pass through normal matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since such dark matter doesn't interact much, except gravitationally, it would be spread out in a large elliptical cloud around the galaxy. There would be dark matter everywhere in the galaxy, including in the solar system, but there are no clumps of dark matter. There are no dark matter planets or stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, we would expect about &lt;a href=&quot;http://cdms.berkeley.edu/Education/DMpages/FAQ/question36.html&quot; rel=&quot;noreferrer&quot;&gt;1 proton mass of dark matter in every 3 cubic cm&lt;/a&gt;. That is a very small amount of mass. In the solar system, there is a billion-billion times more normal matter than exotic dark matter. And as I've noted, nobody has ever directly observed a dark matter particle, we don't know what they are.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-07-16T21:10:11.957" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32682" PostTypeId="1" AcceptedAnswerId="32689" CreationDate="2019-07-16T22:43:45.630" Score="1" ViewCount="334" Body="&lt;p&gt;If a hypothetical black hole with the mass of our sun would appear in place of the sun. &#xA;How long would it take before our entire solar system (everything inside our heliosphere) to be swallowed by this black hole?&lt;/p&gt;&#xA;" OwnerUserId="25592" LastEditorUserId="25592" LastEditDate="2019-07-17T08:08:00.230" LastActivityDate="2019-07-17T12:25:09.080" Title="How long would it take for a black hole the mass of our sun to swallow our solar system?" Tags="&lt;black-hole&gt;&lt;solar-system&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32683" PostTypeId="1" AcceptedAnswerId="32684" CreationDate="2019-07-16T23:20:52.163" Score="1" ViewCount="41" Body="&lt;p&gt;I know there's a difference between the ugriz and u'g'r'i'z' systems of photometric magnitude, but I also sometimes see UGRIZ written. Is there a difference between a UGRIZ magnitude and a ugriz magnitude?&lt;/p&gt;&#xA;&lt;p&gt;I'm trying to apply the method from the paper &lt;a href=&quot;https://iopscience.iop.org/article/10.1086/322132/pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Photometric Redshifts of Quasars&lt;/a&gt; (Richards et al. 2001, ApJ 122, 1151-1162, Equation 1) to my data, which contain data in U, B, V, R, etc. Do I need to do any transformations to my data in order to apply their equation?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;First, we construct an empirical color-redshift &lt;span class=&quot;math-container&quot;&gt;$\chi^2$&lt;/span&gt; (for each redshiftÈ indicated by the subscript z) is computed as&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\chi_z^2 = {\left((u'-g')-(u'-g')_z \right)^2 \over \sigma^2_{u'-g'} - \sigma^2_{(u'-g')_z}} + C_{gr} + C_{ri} + C_{iz}  \tag{1}\label{eq1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$(u'-g')$&lt;/span&gt; is the measured &lt;span class=&quot;math-container&quot;&gt;$u'-g'$&lt;/span&gt; color of the object, &lt;span class=&quot;math-container&quot;&gt;$(u'-g')_z$&lt;/span&gt; is the color from the median color-redshift relation at a given redshift, &lt;span class=&quot;math-container&quot;&gt;$\sigma_{u'-g'}$&lt;/span&gt; is the photometric error in the &lt;span class=&quot;math-container&quot;&gt;$u'-g'$&lt;/span&gt; color, which is given by &lt;span class=&quot;math-container&quot;&gt;$(\sigma_{u'}^2+\sigma_{g'}^2)^{1/2}$&lt;/span&gt;, and  &lt;span class=&quot;math-container&quot;&gt;$\sigma_{(u'-g')_z}$&lt;/span&gt;  is the 1 &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; error width of the median color-redshift relation as a function of redshift.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17535" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-07-18T02:31:50.097" Title="Photometry systems" Tags="&lt;photometry&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32684" PostTypeId="2" ParentId="32683" CreationDate="2019-07-17T01:08:55.173" Score="1" Body="&lt;p&gt;Doing photometric reduction (from image to brightness) includes the following steps:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;image cleaning (bias + dark + flat + etc.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;identify/measure brightness of the object of interests from the cleaned image (using e.g. sextractor). Let's say the image was taken in U filter with the brightness (mA_U_image) in any arbitrary magnitude given any arbitrary zeropoint (ZP).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;identify/measure brightness of standard objects (which are objects that we know their brightness) using U filter and the same ZP. Let's say the brightness is mB_U_image.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Find transformation: mB_U_real(AB) = mB_U_image(arbitrary with zeropoint = ZP) + correction where mB_U_real(AB) = magnitude of B in U filter in AB system (or in other magnitude system as identified by the database of B given U filter).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Apply the transformation: mA_U_real(AB) = mA_U_image(arbitrary with zeropoint = ZP) + correction&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note: this method is simplified because both science objects and standards were assumed to be taken from the same U filter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to answer your question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Typically, there are two frequently used magnitude systems: Vega and AB. You might need to get the Vega to AB correction if the transformation involve changing from one system to another.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;UGRIZ and ugriz: I found that sometimes UGRIZ is used to notify that the magnitude is expressed in Vega system and ugriz is in AB system, or vice versa. For an example, &lt;a href=&quot;https://arxiv.org/abs/1605.01054&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://arxiv.org/abs/1605.01054&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;U and u and u' are slightly different (check their response curves from the telescope handbooks). So, transforming from one filter to another might need to include more terms in the transformation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If standards and science objects were taken from different filter/telescope/setting (even in the case of U and u), you might need to also include a color correction term in the transformation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Eq. 1 is for calculating chi-sq to be used in the optimization problem. I believe that the equation is expressed in u-g color index as an example. You should be able to use any color index pair. If you really want to use u-g index, but you have UBVR, here is what I think you should do: find u = U + correction, and g = B + correction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="22468" LastActivityDate="2019-07-17T01:08:55.173" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32687" PostTypeId="1" CreationDate="2019-07-17T07:42:06.857" Score="2" ViewCount="109" Body="&lt;p&gt;As an outsider to the field, I am surprised to learn that astronomers commonly use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Equatorial_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;equatorial coordinate system&lt;/a&gt;, in which fixed stars are not actually fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't that making life harder than it needs to be? Why would one not define a coordinate system in terms of fixed stars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example like this: Fixed star A is defined as (0°, 0°), fixed star B is defined as (0°, X). That would constrain the 3 degrees of freedom of a spherical coordinate system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to be clear, by fixed stars I mean stars that do not change their apparent relative position over long time scales, say a million years. So stars in the Milky Way would likely be excluded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When making a new observation, these two reference stars may not be visible, which would make calibration difficult. But there could be a reference catalog of fixed stars with precisely known coordinates. Any new observation could be made relative to a visible subset of that catalog. This would be similar in principle to the International Temperature Scale of 1990, which defines temperature relative to a set of reference materials and states.&lt;/p&gt;&#xA;" OwnerUserId="28708" LastEditorUserId="7982" LastEditDate="2019-07-18T02:07:55.143" LastActivityDate="2019-07-18T02:07:55.143" Title="Why do astronomers use the equatorial coordinate system that moves, instead of one based on fixed, distant stars?" Tags="&lt;star&gt;&lt;fundamental-astronomy&gt;&lt;coordinate&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32688" PostTypeId="2" ParentId="32601" CreationDate="2019-07-17T07:44:54.650" Score="0" Body="&lt;p&gt;Looking at the plot you have, I notice that there is a concentration of dots at the &quot;poles&quot;.  The concentration should be along the equator, and particularly towards the centre of the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you have your axes switched, so points along the galactic equator are being drawn along the vertical &quot;prime meridian&quot;.  &lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-07-17T07:44:54.650" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32689" PostTypeId="2" ParentId="32682" CreationDate="2019-07-17T12:25:09.080" Score="6" Body="&lt;p&gt;Is a common missconception to think black holes suck everything around them by definition as if the fact they are black holes gave them some kind of different gravitational influence than the one excerted by any other object with a similar mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main difference between the Sun and a solar-mass black hole its the density. The Sun would contain a solar mass in a sphere with 1.3 million km in diameter while the &lt;a href=&quot;https://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;noreferrer&quot;&gt;Schwarzschild diameter&lt;/a&gt; of a solar-mass black hole would be below 6 km. But the mass is still the same. Take a look to Newton's Universal Law of Gravity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F = -G\frac{m_1m_2}{r^2}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see the gravitational force depends on the masses of the two bodies interacting, &lt;span class=&quot;math-container&quot;&gt;$m_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$m_2$&lt;/span&gt; (for example, the Sun and the Earth), but does not depend on the density of the bodies in any way, thus the interaction has the exact same behaviour in both the Sun and the solar-mass black hole scenarios. The planets would continue to circle around the barycenter of the Solar System as if the Sun was there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, besides the gravitational influence, a lot of horrible consequences would take place. The absence of the solar light flux would freeze the entire system pretty fast. The first few days of darkness would be fine for Earth (even if air currents, migration patterns and circadian rythms would be totally disrupted). In a matter of months the Earth's surface would start to be sterilized for complex organisms unless moved to geothermal facilities. By the first year life on Earth would have been eradicated as the world oceans become a thick ice crust. After 10 years or so even Earth's atmosphere would start to collapse as air turns liquid and freeze upon arrival to the ground. With some luck the ice shell would still insulate the water below and microbial life could still be there, thriving with the geothermal energy of Earth's interior. The best option for us would be to transfer to Venus since atmospheric greenhouse gases would probably allow for larger thermal inertia and thus the surface would freeze slower than on Earth. In the end, after a few decades, even Venus would be sterilized. The entire Solar System would be a dark cold planetary ghost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a mechanism to allow for the black hole to suck the planets eventually and that is gravitational radiation. We know that orbiting bodies emmit gravitational waves. These waves carry angular momentum away from the system shrinking the sizes of the orbits. We in fact have detected strong gravitational waves coming from inspiraling black holes. Therefore, with enought time, gravitational waves radiated by the planets would put them so close to the black hole as to be &quot;swallowed&quot;, but this makes no difference with the Sun scenario we live in, orbits are decaying right now by the same process. But we need to have some orders of magnitude in mind, this process is crazily insignificant, more for small masses like the ones of planets, and even more for the huge orbital periods our planets have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.journalijar.com/uploads/687_IJAR-14086.pdf&quot; rel=&quot;noreferrer&quot;&gt;How long you would have to wait for this to happen?&lt;/a&gt; In the case of Earth, it would be swallowed in &lt;span class=&quot;math-container&quot;&gt;$10^{13}$&lt;/span&gt; times the history of the universe. Yep, those are not 13 zeros of years but 13 zeros of times the universe age. Mercury, is not swallowed much earlier, in fact only half that time (which is still trillions of times the age of the universe). And for the entire Solar System to be swallowed (considering Neptune the last object of it), you would have to wait 145.000 times that. Not to mention the exponentially increasing waits for Kuiper belt objects and Oort cloud comets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these crazy spans of time we also have to think about other strange effects. &lt;a href=&quot;https://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;noreferrer&quot;&gt;Hawking radiation&lt;/a&gt; would evaporate the solar-mass black hole slowly, meaning that the central mass of the system would start to decrease. This means that planetary orbits would start to get larger (since a planet orbiting a less massive body with the same orbital speed would climb the gravitational dwell by some amount). The solar black hole would evaporate in &lt;span class=&quot;math-container&quot;&gt;$10^{58}$&lt;/span&gt; times the age of the universe. Until the planets get free from the influence of an ancient mass that no longer exists in the Solar System. This is a way larger characteristic time period than the one required for orbital decay by gravitational radiation. So the gravitational decaying of the orbits would be the predominant effect even if the wait would be a bit larger (since the mass of the black hole decreases slowly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yeah, you can wait &lt;span class=&quot;math-container&quot;&gt;$10^{29}$&lt;/span&gt; years until all the planets get swallowed by your black hole. The Sun would swallow them in the same way only that they would touch the Sun's surface even before than the event horizon of your black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only that, the Earth is probably going to be swallowed by the Sun way before than in the solar-mass black hole scenario. We know because of stellar evolution that the Sun is going to get larger to the point of reaching our current position in just 6 billion years at max (even if there is some discussion about the solar mass loss in this era enlarging Earth's orbit and making it &quot;safe&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to keep your solar-mass black hole scenario think about those crazy trillions of times the age of the universe wait. Many things are going to happen way before that would disrupt the entire Solar System. &lt;a href=&quot;https://perso.imcce.fr/jacques-laskar/papers/2009/nature08096&amp;amp;s1.pdf&quot; rel=&quot;noreferrer&quot;&gt;It is expected that the orbital perturbations between the planets will eject several planets in a matter of a few billion years&lt;/a&gt;. That is a blink of an eye in the wait for them to decay. At these extreme timescales even close flybys of other stars will certainly dirsupt the system if planets don't do it by themselves in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my answer is that in real life, turning the Sun into a black hole, would never end in the Solar System been swallowed by it, more like everything disspersed in interstellar space in some hundred of billion years at most.&lt;/p&gt;&#xA;" OwnerUserId="28496" LastActivityDate="2019-07-17T12:25:09.080" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32692" PostTypeId="1" CreationDate="2019-07-17T12:40:05.080" Score="1" ViewCount="79" Body="&lt;p&gt;Currently, I am a Ph.D. student(mathematics) but always loved the field of astronomy hence I would like to read something on the &quot;side&quot; in this field along with my work. So is there any good introductory book(s) for me to read? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, any help/guidance will be greatly appreciated&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I, personally, wouldnt call it 'too broad' as the question is asking for some introductory books but if not then, its fine :)&lt;/p&gt;&#xA;" OwnerUserId="28744" LastEditorUserId="28744" LastEditDate="2019-07-18T12:58:33.207" LastActivityDate="2019-07-18T12:58:33.207" Title="Any good books to read for complete beginners?" Tags="&lt;astrophysics&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2019-07-17T15:45:36.473" ContentLicense="CC BY-SA 4.0" />
  <row Id="32693" PostTypeId="1" CreationDate="2019-07-17T15:39:00.377" Score="2" ViewCount="102" Body="&lt;p&gt;If not, is there any reason such a planet couldn't exist? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask only because that planet would literally have a twilight zone and I want to know that that's a thing somewhere. &lt;/p&gt;&#xA;" OwnerUserId="28747" LastActivityDate="2019-07-18T17:09:26.320" Title="Do we know of any tidally-locked planets with atmospheres?" Tags="&lt;rotation&gt;&lt;planetary-atmosphere&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32694" PostTypeId="2" ParentId="32652" CreationDate="2019-07-17T17:45:18.293" Score="1" Body="&lt;p&gt;I'm not in a position to toss real numbers around at the moment, but here's a thought experiment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iridium satellites reflect sunlight toward Earth's surface, producing bright flares, up to magnitude -8. That's about 1 million times brighter than the faintest light source a fully dark-adapted eye can perceive -- a five-magnitude difference in brightness is a factor of 100, and under &lt;em&gt;really&lt;/em&gt; dark skies humans with excellent vision can see down to magnitude 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think these reflections are coming off a panel of about 3 x 1.5 meters, so it's comparable in size to your probe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These satellites orbit about 800km above Earth's surface. How far away could one be, and still produce a visible reflection? Brightness falls off as the inverse square of distance, so the greatest distance at which you could &lt;em&gt;barely&lt;/em&gt; see a reflection would be 1,000 times greater, or 800,000km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light travels at 300,000km/s. A probe traveling at 0.2c would cover 60,000km/s. Traveling 800,000km would take about 13 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that something the size of an Iridium satellite, reflecting sunlight directly toward an observer during its entire passage, could be visible for up to almost half a minute -- blue-shifted for the first part of that, and red-shifted for the last part. (Unless it's quite far away at closest approach, which would make it dimmer, the transition from &quot;basically heading straight for you&quot; to &quot;basically heading straight away from you&quot; would take place in the literal blink of an eye, a matter of milliseconds.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this is probably an upper bound -- probes won't have the same shape and reflectivity as an Iridium satellite, and won't be oriented to reflect directly toward an observer during their entire transit. But, somewhat to my surprise, it looks like you &lt;em&gt;might&lt;/em&gt; be able to see one approaching and receding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it's also possible that an object making 0.2c might have a visible interaction with the solar wind. (&lt;em&gt;That&lt;/em&gt; requires more numbers than I'm able to run at the moment, but having thought on it a little bit, it seems unlikely.)&lt;/p&gt;&#xA;" OwnerUserId="26791" LastEditorUserId="16685" LastEditDate="2019-08-17T10:07:13.170" LastActivityDate="2019-08-17T10:07:13.170" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32695" PostTypeId="2" ParentId="32654" CreationDate="2019-07-17T19:06:47.960" Score="16" Body="&lt;p&gt;There are scenarios where the effects of planet accretion on a sun-like star would be very significant indeed, at least in the short term. Whilst the amount of mass accreted by the star would be a tiny perturbation, the amount of accreted energy and/or angular momentum may not be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scenario 1:&lt;/strong&gt; The scenario where a hot Jupiter just drops into the star from a distant radius would certainly have drastic short-term effects. But short-term here means compared with the lifetime of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinetic energy of say Jupiter falling to the star's surface from a distant (more than a few solar radii) would be of order &lt;span class=&quot;math-container&quot;&gt;$GM_{\odot}M_\mathrm{Jup}/R_{\odot} \sim 4\times 10^{38}$&lt;/span&gt; joules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparison, the solar luminosity is &lt;span class=&quot;math-container&quot;&gt;$3.83 \times 10^{26}\ \mathrm{J/s}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The addition of this much energy (if it is allowed to thermalise) would &lt;em&gt;potentially&lt;/em&gt; affect the luminosity of the Sun for timescales of tens of thousands of years. The exact effects will depend on where the energy is deposited. Compared with the binding energy of the star, the additional energy is negligible, but if the energy is dissipated in the convection zone then the kinetic energy would do work and lift the convective envelope. In other words, the star would both increase in luminosity and in radius. If the effects were just limited to the convective envelope, then this has a mass of around &lt;span class=&quot;math-container&quot;&gt;$0.02 M_{\odot}$&lt;/span&gt; and so could be &quot;lifted&quot; by &lt;span class=&quot;math-container&quot;&gt;$\sim 4\times 10^{38} R_{\odot}^2/GM_{\odot}M_{\rm conv} \sim 0.05 R_{\odot}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this scenario, the star would both expand and become more luminous. The relevant timescale is the Kelvin-Helmholtz timescale &lt;em&gt;of the convective envelope&lt;/em&gt;, which is of order &lt;span class=&quot;math-container&quot;&gt;$GM_{\odot}M_{\rm conv}/R_{\odot} L_{\odot} \sim $&lt;/span&gt;few &lt;span class=&quot;math-container&quot;&gt;$10^5$&lt;/span&gt; years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the planet somehow survived and punched its way to the centre of the star, then much less energy would be deposited in the convection zone and the effects would be lessened. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On longer timescales the star would settle back down to the main sequence, with a radius and luminosity only slightly bigger than it was before, commensurate with its tiny (0.1%) increase in mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This all assumes that the planet can remain intact as it falls. It certainly wouldn't &quot;evaporate&quot; in this direct infall scenario, but would it get tidally shredded before it can disappear below the surface? The Roche limit is of order &lt;span class=&quot;math-container&quot;&gt;$R_{\odot} (\rho_{\odot}/\rho_{\rm Jup})^{1/3}$&lt;/span&gt;. But the average densities of the star and (gas giant) planet are almost identical. So it seems likely that the planet would be &lt;em&gt;starting&lt;/em&gt; to be tidally ripped apart, but as it is travelling towards the star at a few hundred km/s at this point, tidal breakup could not be achieved before it had disappeared below the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my conclusion is that dropping a Jupiter into a Sun-like star &lt;em&gt;in this scenario&lt;/em&gt; would be like dropping a depth charge, with a lag of order &lt;span class=&quot;math-container&quot;&gt;$10^{5}$&lt;/span&gt; years before the full effects became apparent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scenario 2:&lt;/strong&gt; A hot Jupiter arrives at the Roche limit (just above the stellar surface) having lost a large amount of angular momentum. In this case the effects may be experienced on human timescales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case what will happen is the planet will be (quickly) shredded by the tidal field, possibly leaving a substantial core. At an orbital radius of &lt;span class=&quot;math-container&quot;&gt;$2 R_{\odot}$&lt;/span&gt;, the orbital period will be about 8 hours, the orbital speed about  &lt;span class=&quot;math-container&quot;&gt;$300\ \mathrm{km/s}$&lt;/span&gt; and the orbital angular momentum about &lt;span class=&quot;math-container&quot;&gt;$10^{42}\ \mathrm{kg\ m^2\ s^{-1}}$&lt;/span&gt;. Assuming total destruction, much of the material will form an accretion disc around the star, since it must lose some of its angular momentum before it can be accreted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much of the star's light is blocked is uncertain. It mainly depends on how the material is distributed in the disk, especially the disk scale height. This in turn depends on the balance of the heating and cooling mechanisms and hence the temperature of the disk. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some sort of minimal estimate could be to assume the disk is planar and spread evenly between the solar surface and &lt;span class=&quot;math-container&quot;&gt;$2R_{\odot}$&lt;/span&gt; and that it gets close to the solar photospheric temperature at &lt;span class=&quot;math-container&quot;&gt;$\sim 5000\ \mathrm K$&lt;/span&gt;. In which case the disk area is &lt;span class=&quot;math-container&quot;&gt;$3 \pi R_{\odot}^2$&lt;/span&gt;, with an &quot;areal density&quot; of &lt;span class=&quot;math-container&quot;&gt;$\sigma \sim M_{\rm Jup}/3\pi R_{\odot}^2$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In hydrostatic equilibrium, the scale height will be &lt;span class=&quot;math-container&quot;&gt;$\sim kT/g m_\mathrm H$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; is the gravitational field and &lt;span class=&quot;math-container&quot;&gt;$m_\mathrm H$&lt;/span&gt; the mass of a hydrogen atom. The gravity (of a plane) will be &lt;span class=&quot;math-container&quot;&gt;$g \sim 4\pi G \sigma$&lt;/span&gt;. Putting in &lt;span class=&quot;math-container&quot;&gt;$T \sim 5000\ \mathrm K$&lt;/span&gt;, we get a scale height of &lt;span class=&quot;math-container&quot;&gt;$\sim 0.1 R_{\odot}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How long the accretion disk would remain, I am unsure how to calculate. It depends on the assumed viscosity and temperature structure and how much mass is lost through evaporation/winds. The accreted material will have radiated away a large fraction of its gravitational potential energy, so the energetic effects will be much less severe than Scenario 1. However, the star will accrete &lt;span class=&quot;math-container&quot;&gt;$\sim 10^{42}\ \mathrm{kg\ m^2\ s^{-1}}$&lt;/span&gt; of angular momentum, which is comparable to its current angular momentum of the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The accretion of an exoplanet in this way is therefore &lt;strong&gt;sufficient to increase the angular momentum of a star like the Sun by a significant amount&lt;/strong&gt;. In the long term this will have a drastic effect on the magnetic activity of a Sun-like star – increasing it by a factor of a few to an order of magnitude.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2019-07-17T19:06:47.960" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32696" PostTypeId="1" CreationDate="2019-07-18T02:19:34.427" Score="2" ViewCount="46" Body="&lt;p&gt;Due to the curvature of the earth, and the way the moon interacts with the atmosphere (especially close to the horizon), is there a known function that describes the speed of the moon relative to its angle in the sky?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For bonus points, imagine that you are at altitude in a plane, and for some reason want to descend at a rate that would keep the total % of the moon illuminated constant. Is there a function for that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance, this is not for an assignment, just for my own curiosity as I watched moonrise tonight. I have tried googling these questions, to no avail. &lt;/p&gt;&#xA;" OwnerUserId="28753" LastActivityDate="2019-07-18T02:19:34.427" Title="Speed of Moonrise relative to fixed point on ground" Tags="&lt;the-moon&gt;&lt;natural-satellites&gt;&lt;orbital-mechanics&gt;&lt;mathematics&gt;" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32697" PostTypeId="1" CreationDate="2019-07-18T04:26:45.023" Score="3" ViewCount="101" Body="&lt;p&gt;I see the term &quot;assembly bias&quot; being used quite a bit in cosmology literature? I have not been able to find a good primer on what it means, and why it is important. &lt;/p&gt;&#xA;" OwnerUserId="28756" LastEditorUserId="1891" LastEditDate="2019-07-18T06:05:03.030" LastActivityDate="2019-07-18T11:13:16.623" Title="What is assembly bias in context of cosmology?" Tags="&lt;cosmology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32698" PostTypeId="2" ParentId="32693" CreationDate="2019-07-18T09:06:56.070" Score="3" Body="&lt;p&gt;As we haven't measured the atmosphere of many rocky planets yet, there aren't many examples. But I found 55 Cancri e:&#xA;It is a super-earth with very close tidally locked orbit and an atmosphere that has been measured by Hubble. A year on the exoplanet lasts for only 18 hours and temperatures on the surface are thought to reach around 2000 degrees Celsius. The detected atmosphere seems to consists mainly of hydrogen and helium. So you do have a constant Twilight Zone, just a rather hot one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zSQTP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zSQTP.jpg&quot; alt=&quot;Poster:&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources: &#xA;- &lt;a href=&quot;https://asd.gsfc.nasa.gov/blueshift/index.php/2013/07/12/jasons-blog-next-stop-diamond-planets/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Beeing tidally locked&lt;/a&gt;&#xA;- &lt;a href=&quot;https://phys.org/news/2016-02-super-earth-atmosphere.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Detection of the Atmosphere&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="28760" LastActivityDate="2019-07-18T09:06:56.070" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32700" PostTypeId="5" CreationDate="2019-07-18T10:51:59.673" Score="0" Body="&lt;p&gt;The Spitzer Space Telescope (SST), formerly the Space Infrared Telescope Facility (SIRTF), is an infrared space telescope launched in 2003 and is planned to be retired in January 2020.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The planned mission period was to be 2.5 years with a pre-launch expectation that the mission could extend to five or slightly more years until the onboard liquid helium supply was exhausted. This occurred on 15 May 2009. Without liquid helium to cool the telescope to the very low temperatures needed to operate, most of the instruments are no longer usable. However, the two shortest-wavelength modules of the IRAC camera are still operable with the same sensitivity as before the cryogen was exhausted, and have continued to be used to the present in the Spitzer Warm Mission. All Spitzer data, from both the primary and warm phases, are archived at the Infrared Science Archive (IRSA).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In keeping with NASA tradition, the telescope was renamed after its successful demonstration of operation, on 18 December 2003. Unlike most telescopes that are named after famous deceased astronomers by a board of scientists, the new name for SIRTF was obtained from a contest open to the general public.&lt;/p&gt;&#xA;" OwnerUserId="2894" LastEditorUserId="2894" LastEditDate="2019-07-18T11:12:43.767" LastActivityDate="2019-07-18T11:12:43.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32701" PostTypeId="4" CreationDate="2019-07-18T10:51:59.673" Score="0" Body="The Spitzer Space Telescope (SST), formerly the Space Infrared Telescope Facility (SIRTF), is an infrared space telescope launched in 2003 and is planned to be retired in January 2020." OwnerUserId="2894" LastEditorUserId="2894" LastEditDate="2019-07-18T12:34:08.920" LastActivityDate="2019-07-18T12:34:08.920" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32702" PostTypeId="2" ParentId="32697" CreationDate="2019-07-18T11:13:16.623" Score="3" Body="&lt;p&gt;Early analytic calculations of the distribution of the masses of dark matter halos (in which galaxies reside) by &lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/1974ApJ...187..425P/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Press &amp;amp; Schechter (1974)&lt;/a&gt;, &lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/1991ApJ...379..440B/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bond et al. (1991)&lt;/a&gt;, &lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/1996MNRAS.282..347M/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mo &amp;amp; White (1996)&lt;/a&gt; and several others showed that the clustering of these halos depended only the redshift and the mass of the halos. However, later work showed that clustering depends on several other parameters, such as the assembly history of the halos (&lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2004MNRAS.350.1385S/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sheth &amp;amp; Tormen 2004&lt;/a&gt;; &lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2005MNRAS.363L..66G/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gao et al. 2005&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This fact is known as the &lt;em&gt;assembly bias&lt;/em&gt;. I think the first to use the term was &lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2007MNRAS.374.1303C/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Croton et al. (2007)&lt;/a&gt;. Later, the term was &quot;expanded&quot; to also refer to the dependence of the clustering on halo spin, concentration, and shape (ellipticity) of the halos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find some more recent good discussions in &lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2017JCAP...03..059L/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lazeyras et al. (2017)&lt;/a&gt; and &lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/2019MNRAS.486..582P/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Padilla et al. (2019)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5264" LastActivityDate="2019-07-18T11:13:16.623" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32703" PostTypeId="1" AcceptedAnswerId="32704" CreationDate="2019-07-18T12:12:51.660" Score="1" ViewCount="69" Body="&lt;p&gt;I have a set of data that is absolute magnitude of a star and its redshift. Is there any way to compute the value of the hubble parameter given that magnitude and redshift?&lt;/p&gt;&#xA;" OwnerUserId="28761" LastActivityDate="2019-07-18T13:15:15.710" Title="Relationship between Hubble paramater and absolute magnitude" Tags="&lt;hubble-constant&gt;&lt;absolute-magnitude&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32704" PostTypeId="2" ParentId="32703" CreationDate="2019-07-18T13:15:15.710" Score="0" Body="&lt;p&gt;Maybe. To calculate the Hubble constant (H=v/d), you need the know the velocity (v≈c*z￼￼) and distance of a distant object (or rather as many objects as possible). The velocity of these objects must be dominated by the expansion of the Universe. So stars in our local group that are still bound to us via gravity wont work. So as long as the stars you use are not really bright or even supernovae in distant galaxies there is no way to get the Hubble constant.&#xA;If your stars are far enough away, however, you still have get their distance. This you can do with the distance modulus, i.e., difference between absolute and apparent magnitude: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;m-M = 5 log(d)-5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So assuming the stars are distant stars for which you also know their apparent magnitude (e.g. Cepheid variables), than yes you can calculate H. &#xA;If they are just random stars, than not. &lt;/p&gt;&#xA;" OwnerUserId="28760" LastActivityDate="2019-07-18T13:15:15.710" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32705" PostTypeId="1" CreationDate="2019-07-18T13:17:43.300" Score="2" ViewCount="108" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/szZYn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/szZYn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am using Astropy's basic &lt;strong&gt;models.Gaussian1D&lt;/strong&gt; model and a &lt;strong&gt;fitter = fitting.LevMarLSQfitter()&lt;/strong&gt; to fit 3 Gaussian pulses with some added random noise. The fit is good, but when I type&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;fitter.fit_info&lt;/strong&gt;, according to  &lt;a href=&quot;https://docs.astropy.org/en/stable/api/astropy.modeling.fitting.LevMarLSQFitter.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;documentation&lt;/a&gt; here, it should automatically calculate covariance matrix and add it to the fit_info, but it doesn't, &lt;strong&gt;fit_info['param_cov']&lt;/strong&gt; is simply &lt;strong&gt;None&lt;/strong&gt;. What is going on?&lt;/p&gt;&#xA;" OwnerUserId="26488" LastEditorUserId="7982" LastEditDate="2019-07-18T15:29:58.283" LastActivityDate="2019-07-18T15:29:58.283" Title="Covariance matrix not calculated when fitting gaussians?" Tags="&lt;python&gt;&lt;software&gt;&lt;astropy&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32706" PostTypeId="1" AcceptedAnswerId="32709" CreationDate="2019-07-18T13:32:40.860" Score="10" ViewCount="2068" Body="&lt;p&gt;If a white dwarf compresses to the limit of electron degeneracy, and a neutron star compresses to the limit of neutron degeneracy, what does a black hole compress to the limit of?&lt;/p&gt;&#xA;" OwnerUserId="14085" LastEditorUserId="16685" LastEditDate="2019-07-19T05:31:21.640" LastActivityDate="2019-07-19T05:31:21.640" Title="Pauli exclusion principle - black holes" Tags="&lt;black-hole&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32708" PostTypeId="2" ParentId="32706" CreationDate="2019-07-18T14:19:39.360" Score="7" Body="&lt;p&gt;As far as current physics knows, nothing. This is the reason why it is commonly thought that a singularity  exists in the middle of a black hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, singularities are also thought to be non-physical, so there is most likely something else inside a black hole — we just don’t have the science to describe it just yet. &lt;/p&gt;&#xA;" OwnerUserId="10813" LastActivityDate="2019-07-18T14:19:39.360" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32709" PostTypeId="2" ParentId="32706" CreationDate="2019-07-18T14:28:21.000" Score="20" Body="&lt;p&gt;In classical General Relativity, there &lt;em&gt;is&lt;/em&gt; no limit to the compression in a black hole, hence you get a singularity. However, many astrophysicists feel that's unphysical, and that a theory which unites General Relativity and Quantum Mechanics will impose some kind of limit, perhaps something connected with the quantization of spacetime itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't have a working theory of quantum gravity, so at this stage we don't exactly know what happens in the core of a black hole. OTOH, we're fairly confident that the core has to be very small, since quantum gravity effects probably don't kick in until a scale much smaller than the size of an atom, and probably smaller than a proton, somewhere around the scale of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Planck_length&quot; rel=&quot;noreferrer&quot;&gt;Planck length&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2019-07-18T14:28:21.000" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32710" PostTypeId="1" CreationDate="2019-07-18T15:37:07.960" Score="2" ViewCount="207" Body="&lt;p&gt;The current mean temperature in Titan is &amp;minus;180 ºC (93 K), the same as in Saturn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I was wondering what would be the case if the atmosphere of Titan would be like the Venus atmosphere, whose composition is mainly CO&lt;sub&gt;2&lt;/sub&gt; and has a pressure of 90 bar, instead of approximately 1.5 bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What temperature could Titan reach in that case? Would the greenhouse effect allow the moon to be heated enough to maintain liquid water or is it too far from the Sun to be affected by any kind of greenhouse effect? Maybe the latest question allows to ask another one: How far from the Sun could Venus be so it can have temperatures above 0 ºC?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question may be answered given two possible alternatives:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Same composition as Venus, but same pressure as in Titan.&lt;/li&gt;&#xA;&lt;li&gt;Same composition and pressure as in Venus.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="957" LastEditorUserId="7982" LastEditDate="2019-07-19T14:18:37.303" LastActivityDate="2019-07-19T23:06:14.420" Title="What if the atmosphere of Titan were like the one on Venus?" Tags="&lt;planetary-atmosphere&gt;&lt;temperature&gt;&lt;venus&gt;&lt;titan&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32711" PostTypeId="2" ParentId="32710" CreationDate="2019-07-18T16:05:06.697" Score="2" Body="&lt;p&gt;If the pressure were the same as the current pressure on Titan, the planet would be slightly warmer for a while, but not for long. The CO2 would soon be frozen out. It is probable that 4 billion years ago the atmospheres of Venus,Earth,Masrs and Titan were much more similar. A hotter Saturn may have warmed Titan in those days. It probably,like the others, had a lot of CO2 in its atmosphere,but that has long ago been frozen out. There is plenty of frozen water on Titan, and if gelogists ever dig down far enough,they are likely to find a layer of frozen CO2 there as well.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-07-18T16:05:06.697" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32712" PostTypeId="2" ParentId="32693" CreationDate="2019-07-18T17:09:26.320" Score="2" Body="&lt;p&gt;You asked if we know of any such planets, this raises the question of how sure do you want to be. There are definitely a variety of planets located close to their stars which have atmospheres (hot Jupiters, hot Neptunes, various volatile-rich super-Earths), that have short tidal spindown timescales. These planets are generally assumed to be in a 1:1 spin-orbit state for this reason. As far as I'm aware none of these planets has had their rotation measured, so while it may be reasonable to assume they are tidally locked, we don't know for sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This then brings up a substantial caveat: atmospheres can act against tidal locking. The usual example given here is Venus, where the atmospheric &quot;thermal tides&quot; make the current rotation state more favourable than the tidally-locked one. A hypothetical airless Venus might well have ended up tidally locked. You don't necessarily need an atmosphere as massive as the one Venus has: according to Leconte et al. (2015) &quot;&lt;a href=&quot;https://arxiv.org/abs/1502.01952&quot; rel=&quot;nofollow noreferrer&quot;&gt;Asynchronous rotation of Earth-mass planets in the habitable zone of lower-mass stars&lt;/a&gt;&quot; an Earth-like atmosphere would be sufficient to prevent an Earth-mass planet from being tidally locked around stars down to about 0.5–0.7 solar masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even below this limit, planet-planet interactions may act to keep planets out of the synchronous state. According to Vinson et al. (2019) &quot;&lt;a href=&quot;https://arxiv.org/abs/1905.11419&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Chaotic Nature of TRAPPIST-1 Planetary Spin States&lt;/a&gt;&quot; several of the TRAPPIST-1 planets may be in non-synchronous rotation states as a result of this, despite being subject to strong tides from the host star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermal tides are also relevant for hot Jupiters and hot Neptunes: again these planets are usually considered to be in synchronous rotation but this is not necessarily the case. For example, see Auclair-Desrotour &amp;amp; Leconte (2018) &quot;&lt;a href=&quot;https://arxiv.org/abs/1801.07519&quot; rel=&quot;nofollow noreferrer&quot;&gt;Semidiurnal thermal tides in asynchronously rotating hot Jupiters&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'd say the answer is a definite maybe. It is likely that the list of known exoplanets does contain some examples of synchronously-rotating planets with atmospheres, unfortunately at the current time we cannot say for sure whether an individual planet is actually in the synchronous rotation state.&lt;/p&gt;&#xA;" OwnerDisplayName="user24157" LastActivityDate="2019-07-18T17:09:26.320" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32714" PostTypeId="2" ParentId="14762" CreationDate="2019-07-19T01:35:40.667" Score="1" Body="&lt;p&gt;I believe the earliest known depiciton of the Moon is that on the 'Nebra Sky Disc' which is said to be 4000 years old. Also, it could be (but not confirmed) depicted on the Tal-Quadi slab which is dated over 5000 years old.&lt;/p&gt;&#xA;" OwnerUserId="28769" LastActivityDate="2019-07-19T01:35:40.667" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32715" PostTypeId="2" ParentId="32670" CreationDate="2019-07-19T01:47:24.743" Score="0" Body="&lt;p&gt;Supplemental to &lt;a href=&quot;https://astronomy.stackexchange.com/a/32674/7982&quot;&gt;@antlersoft's answer&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was poking around a bit further after I posted and ran across this &quot;1960's Edmund Scientific 'Space Conqueror', 3&quot; reflector&quot; with a prism diagonal at &lt;a href=&quot;http://www.retrotechnology.com/glass/edmund.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.retrotechnology.com/glass/edmund.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not exactly 1970's but at least they were using them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WCTti.jpg&quot; alt=&quot;&amp;quot;Edmund Prism&amp;quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-07-19T01:47:24.743" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32716" PostTypeId="1" AcceptedAnswerId="32721" CreationDate="2019-07-19T01:51:14.693" Score="0" ViewCount="106" Body="&lt;p&gt;Supplemental to &lt;a href=&quot;https://astronomy.stackexchange.com/a/32674/7982&quot;&gt;this answer&lt;/a&gt; to &lt;em&gt;Did Edmund Scientific 4¼ inch Newtonians have 90° prisms as secondary mirrors in the 1970's?&lt;/em&gt; I was poking around a bit further and ran across this &quot;1960's Edmund Scientific 'Space Conqueror', 3&quot; reflector&quot; with a prism diagonal at &lt;a href=&quot;http://www.retrotechnology.com/glass/edmund.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.retrotechnology.com/glass/edmund.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Does anybody recognize what kind of prism this is? It doesn't look like a simple right-angle prism, but possibly something that has additional reflections and does some additional flipping or rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any chance that it isn't positioned correctly, it doesn't look like it is centered along the eyepiece holder's axis, but the image has a strong perspective component and we can't see the business end of the prism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WCTti.jpg&quot; alt=&quot;&amp;quot;Edmund Prism&amp;quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-07-19T15:05:43.030" Title="Does anybody recognize this kind of prism?" Tags="&lt;telescope&gt;&lt;amateur-observing&gt;&lt;optics&gt;&lt;identify-this-object&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32718" PostTypeId="2" ParentId="14762" CreationDate="2019-07-19T10:34:33.443" Score="1" Body="&lt;p&gt;In the geocentric system, the Moon is still, in a sense, a satalite of the Earth, as is the sun, the planets and the stars as they all orbit the Earth.  Aristotle is often credited for the geocetric model, but there are writings that preceed him that describe it.  The idea that the Moon orbits the Earth may be even older than that, but I couldn't find anything on that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think one of the interesting things the ancient Greeks came up with, was they may have been the first society for whom the Moon was no longer a god.  Their gods were human like, not the Sun, the Moon, Nature, etc.  The Sun and Moon became objects in the sky instead of part of their mythology.   But I digress.      &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first written and worked out calculation that the Moon was smaller than the Earth and orbited the Earth as a &quot;moon&quot;, and that the Earth was smaller than the Sun and orbited the Sun came from &lt;a href=&quot;https://www.britannica.com/biography/Aristarchus-of-Samos&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aristarchus of Samos&lt;/a&gt;, who came not long after Aristotle in the 3rd century BC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Aristarchus had been a few years earlier, perhaps his analysis would have swayed Aristotle, but that's just speculation on my part.  Aristotle's ideas came first and he was well respected and well read.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ptolomy of Alexandria who'd writings follow about 3 centuries later, knew of both Aristotle's and Aristarchus' models, but he found Aristarchus' model troubling because he thought that the Earth's rotation speed would create high speed winds not seen on Earth, so he leaned towards Aristotle's model and, in addition to that, using epicycles, the Ptolemaic model was highly predictable.  It explained planetary motion and people like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Predictions, a model that fits what we see as well as the respect for Aristotle's teachings and the Church saying &quot;this is so&quot; may have all played a role in the 15 centuries that the Ptolemaic model endured.  We shouldn't judge that as stupid, because there was a logic to it and a lack of evidence to say otherwise.  If a system works and can't be shown to be false, it's likely to stick around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may have been the occasional scholar who read Aristarchus during the middle ages and agreed with him, and I've read that Copernicus credits Aristarchus, but that wouldn't be enough to sway the accepted theory.   Tycho Brahe, for example, who came after Copernicus but preceded and later coincided with Galileo, went back to a revised geocentric model.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Tychonic_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;See Tychonic system&lt;/a&gt; where the Moon, Sun and Stars orbit the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galileo's telescope, however, provided proof to the contrary of the Ptolemaic system.  He discovered Jupiter's 4 large moons, later named the Galilean moons and he observed the phases of Venus and later, he timed the tides to the Moon's and Sun's orbit.   Galileo was the first to provide proof that the Moon was a satellite of Earth and that other planets could have moons.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2019-07-19T10:34:33.443" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32720" PostTypeId="2" ParentId="30478" CreationDate="2019-07-19T11:07:22.163" Score="1" Body="&lt;p&gt;You are familiar with the relationship between flux and luminosity, right? This is usually taken as the definition of luminosity distance in extragalactic astronomy,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ F = \frac{L}{4\pi D_L^2}.$$&lt;/span&gt;&#xA;The quickest way to convert this into a relationship with spectral variables is to use &lt;A HREF=&quot;https://en.wikipedia.org/wiki/Differential_form&quot; rel=&quot;nofollow noreferrer&quot;&gt;differential 1-forms&lt;/A&gt;. So, &lt;span class=&quot;math-container&quot;&gt;$L \rightarrow L_\lambda \,\mathrm{d}\lambda_e$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$F \rightarrow F_\lambda \,\mathrm{d}\lambda_o$&lt;/span&gt;, with &lt;span class=&quot;math-container&quot;&gt;$\lambda_e$&lt;/span&gt; the photon wavelength at emission (also called the &lt;em&gt;rest frame&lt;/em&gt; wavelength) and &lt;span class=&quot;math-container&quot;&gt;$\lambda_o$&lt;/span&gt; the observed wavelength at &lt;span class=&quot;math-container&quot;&gt;$z=0$&lt;/span&gt; (also called the &lt;em&gt;comoving&lt;/em&gt; wavelength). Thus&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  F_\lambda \,\mathrm{d}\lambda_o &amp;amp; = \frac{L_\lambda \,\mathrm{d}\lambda_e}{4\pi D_L^2} \Rightarrow\\&#xA;  F_\lambda &amp;amp; = \frac{L_\lambda}{4\pi D_L^2} \times \frac{\mathrm{d}\lambda_e}{\mathrm{d}\lambda_o}. &#xA;\end{align}&lt;/span&gt;&#xA;Now you just need to take your derivative using the definition of redshift, &lt;span class=&quot;math-container&quot;&gt;$\lambda_o = (1+z)\lambda_e$&lt;/span&gt;, and use some way of calculating the luminosity distance in your cosmology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally, a lot of people like to use &lt;span class=&quot;math-container&quot;&gt;$\nu F_\nu = \lambda F_\lambda$&lt;/span&gt; as though it were a total flux, but it's not. The derivation will work, but only because redshifting scales all frequencies/wavelengths by a single constant. &lt;span class=&quot;math-container&quot;&gt;$\nu F_\nu$&lt;/span&gt; isn't a measure of the flux under any part of the spectrum, it's a spectral flux using a different coordinate system. Consider the following&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;   F &amp;amp; = F_\nu \,\mathrm{d}\nu \\&#xA;    &amp;amp;= \nu F_\nu \frac{\mathrm{d}\nu}{\nu} \\&#xA;    &amp;amp;= F_{\ln \nu} \,\mathrm{d}\ln \nu.&#xA;\end{align}&lt;/span&gt;&#xA;In other words, &lt;span class=&quot;math-container&quot;&gt;$\nu F_\nu$&lt;/span&gt; is still a spectral flux. If &lt;span class=&quot;math-container&quot;&gt;$F_\nu$&lt;/span&gt; is in &lt;span class=&quot;math-container&quot;&gt;$\mathrm{ergs}\,\mathrm{cm}^{-2}\,\mathrm{Hz}^{-1}$&lt;/span&gt; then &lt;span class=&quot;math-container&quot;&gt;$\nu F_\nu$&lt;/span&gt; will be in &lt;span class=&quot;math-container&quot;&gt;$\mathrm{ergs}\,\mathrm{cm}^{-2}\,(e\text{-fold})^{-1}$&lt;/span&gt;. The &lt;span class=&quot;math-container&quot;&gt;$e\text{-fold}$&lt;/span&gt; is a pseudo-unit, like radians, dex, octave, decibel, or magnitudes. In principle it can be omitted, just like how angular frequencies can be in radians per second, or just &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}^{-1}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fun math exercise: show that &lt;span class=&quot;math-container&quot;&gt;$\ln(10) \nu F_\nu$&lt;/span&gt; has units of &lt;span class=&quot;math-container&quot;&gt;$\mathrm{ergs}\,\mathrm{cm}^{-2}\,\mathrm{dex}^{-1}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="14244" LastActivityDate="2019-07-19T11:07:22.163" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32721" PostTypeId="2" ParentId="32716" CreationDate="2019-07-19T15:05:43.030" Score="1" Body="&lt;p&gt;My guess is that it's not a prism (or at least doesn't function as one optically).  It looks like it is a right triangle in one cross section.  The diagonal face would be silvered to act as a star diagonal. If aligned correctly, the black plate it's glued to that attaches to the support arm would be perpendicular to the tube axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like it's rough beveled on other sides, maybe to get an ellipsoidal cross section?&lt;/p&gt;&#xA;" OwnerUserId="14434" LastActivityDate="2019-07-19T15:05:43.030" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32722" PostTypeId="1" CreationDate="2019-07-19T15:10:42.930" Score="2" ViewCount="28" Body="&lt;p&gt;I am extracting gamma ray spectrum from Fermi-LAT data. My spectrum is a lot higher than the ones on previous studies of the same object. What could cause a spectrum to go too high?&lt;/p&gt;&#xA;" OwnerUserId="28780" LastActivityDate="2019-07-19T15:10:42.930" Title="Fermi-LAT gamma ray spectrum too high" Tags="&lt;observational-astronomy&gt;&lt;astrophysics&gt;&lt;gamma-rays&gt;" AnswerCount="0" CommentCount="5" ClosedDate="2019-07-29T02:55:42.390" ContentLicense="CC BY-SA 4.0" />
  <row Id="32723" PostTypeId="1" AcceptedAnswerId="32788" CreationDate="2019-07-19T18:39:18.943" Score="2" ViewCount="116" Body="&lt;p&gt;During its lifetime, a star can synthesize elements to iron because of the pressure and temperature. A neutron star is composed of iron nuclei 56 and up to krypton 118.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to that, a black hole (which concentrates matter to an increasingly tiny space during infall) must also have an unimaginable pressure/temperature during its formation and as additional matter falls into it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can it therefore synthesize new elements? &lt;/p&gt;&#xA;" OwnerUserId="28781" LastEditorUserId="7982" LastEditDate="2019-07-19T21:02:22.460" LastActivityDate="2019-07-25T17:31:40.383" Title="Does nucleosynthesis happen near a black hole or during its formation?" Tags="&lt;black-hole&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="13" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32724" PostTypeId="1" CreationDate="2019-07-19T19:32:05.597" Score="2" ViewCount="64" Body="&lt;p&gt;If there was a radio telescope, as sophisticated as our best radio telescopes within a 100 light year radius from earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AND if its sole purpose was search for signs of artificial/intelligent radio transmissions like our SETI program. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What chance would such radio telescope have of detecting our radio emissions regardless of its three dimensional location relative to earth and how would that chance change the further it is from us?&lt;/p&gt;&#xA;" OwnerUserId="28782" LastActivityDate="2019-07-19T19:32:05.597" Title="Would radio telescope located anywhere within a 100 light year radius detect our radio emissions?" Tags="&lt;radio-astronomy&gt;&lt;radio-telescope&gt;" AnswerCount="0" CommentCount="5" ClosedDate="2019-07-21T14:52:27.343" ContentLicense="CC BY-SA 4.0" />
  <row Id="32725" PostTypeId="2" ParentId="32143" CreationDate="2019-07-19T20:34:56.860" Score="0" Body="&lt;p&gt;There is a bit of a semantic mess in multiverse talk.  The eternal inflation theory claims that, on the largest scales, inflation (exceptionally rapid expansion) continues for a long time, perhaps forever.   Which would mean, that what we have been calling &quot;The Universe&quot; would be a subregion of space-time in which inflation ended earlier, in the first few seconds, and is now just an insignificantly small piece of the whole.  The whole of space-time would then be a multiverse because it contains many other non-inflating sections like ours.  Perhaps, if this theory turns out to be correct, we should call these slower growing sections &quot;island universes&quot; to be clearer.  The problem though is that the literature has historically called this one &quot;The Universe&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2019-07-19T20:34:56.860" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32726" PostTypeId="1" CreationDate="2019-07-19T21:41:05.640" Score="0" ViewCount="59" Body="&lt;p&gt;Would a lunar eclipse seen from low (or even high) Earth orbit look different to the naked eye than it does from the surface of the Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only image I could find taken from orbit was the photo by Alexander Gerst from the ISS.  &lt;a href=&quot;https://www.flickr.com/photos/astro_alex/42962319354/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;The slight hue of blue is actually the Earth's atmosphere, just before the Moon is diving into it.&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KE7Kg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KE7Kg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25088" LastEditorUserId="25088" LastEditDate="2019-07-20T14:34:21.850" LastActivityDate="2019-07-20T19:19:22.383" Title="Lunar eclipse above the atmosphere" Tags="&lt;orbit&gt;&lt;lunar-eclipse&gt;&lt;naked-eye&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32727" PostTypeId="1" AcceptedAnswerId="32737" CreationDate="2019-07-19T22:14:39.160" Score="0" ViewCount="533" Body="&lt;p&gt;The Sun's photosphere contains the Sun's surface as defined by opacity = 2/3 point. I'd like to see the profile of mass density from bottom to top of the photosphere. I did a quick search and got confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image below is found on the Wikipedia &lt;a href=&quot;https://en.wikipedia.org/wiki/Photosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;Photosphere&lt;/a&gt; page. If I trace the dotted line labeled &quot;Density&quot; to the photosphere layer and read the density axis at the top, I read something like &lt;span class=&quot;math-container&quot;&gt;$8 \text{ to } 3 \times 10^{-7} \text{ g/cm}^{-3}$&lt;/span&gt;, which you could call &lt;span class=&quot;math-container&quot;&gt;$1 \times 10^{-6} \text{ g/cm}^{-3}$&lt;/span&gt;. However, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Photosphere#Sun&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sun section&lt;/a&gt; of Wikipedia page where this image is shown says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Sun's photosphere has a temperature between 4,500 and 6,000 K (4,230 and 5,730 °C) (with an effective temperature of 5,777 K (5,504 °C)) &lt;strong&gt;and a density of about 1×10−6 kg/m3;&lt;/strong&gt; increasing with depth into the sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and links to the solar-center.stanford.edu page &lt;a href=&quot;http://solar-center.stanford.edu/vitalstats.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Sun's Vital Statistics&lt;/a&gt; for the &lt;span class=&quot;math-container&quot;&gt;$1 \times 10^{-6} \text{ kg/m}^{-3}.$&lt;/span&gt; Converting the units, this is only &lt;span class=&quot;math-container&quot;&gt;$1 \times 10^{-9} \text{ g/cm}^{-3}.$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is it possible to clear up this disparity, and to see a plot of the density versus depth from the bottom to the top of the Sun's photosphere, which would likely contain both positive and negative heights above the Sun's surface?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Source: &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Sun_Atmosphere_Temperature_and_Density_SkyLab.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;File:Sun Atmosphere Temperature and Density SkyLab.jpg&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Original source: &lt;a href=&quot;https://history.nasa.gov/SP-402/p2.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;SP-402 A New Sun: The Solar Results From Skylab&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/S24zQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S24zQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-07-19T23:37:09.707" LastActivityDate="2019-11-14T05:13:54.977" Title="What is the density profile within the Sun's photosphere? Which one of these is wrong?" Tags="&lt;the-sun&gt;&lt;stellar-atmospheres&gt;&lt;photosphere&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32728" PostTypeId="2" ParentId="32710" CreationDate="2019-07-19T23:06:14.420" Score="2" Body="&lt;p&gt;One thing to keep in mind is that planet temperature doesn't follow a neat equation when an atmosphere is involved.   Without an atmosphere, it's pretty straight forward and there's even an equation for it.  &lt;a href=&quot;https://berketurkay.wordpress.com/2014/07/23/effective-temperature-of-planets/&quot; rel=&quot;nofollow noreferrer&quot;&gt;See here&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Effective_temperature#Surface_temperature_of_a_planet&quot; rel=&quot;nofollow noreferrer&quot;&gt;or here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Q54M9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q54M9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, the first link above has an error in it.  It says Venus' clouds block 10% of the sun.   I think it's closer to 80% of the sun, but apart from that error it's a nice article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Titan's effective temperature, no atmosphere, albedo of 0.22 works out to about 84 kelvin, 11 degrees less than it's surface temperature.  Earth's atmosphere, by comparison &lt;a href=&quot;https://en.wikipedia.org/wiki/Effective_temperature#Earth_Effective_Temperature&quot; rel=&quot;nofollow noreferrer&quot;&gt;traps about 31 degrees&lt;/a&gt; and Venus' atmosphere, about 500 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Titan, being an icy-moon, likely has more CO2 than Venus.  It certainly has much more water and much more methane than Venus, but only it's Methane is currently able to exist in the gaseous state.   Even so, with about 5% methane in it's atmosphere, Titan has a strong greenhouse gas atmosphere already.   Considerably stronger than Earths.   The reason it traps less heat (about 11 degrees to 31 is because there's less heat to trap).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atmospheric equilibrium temperatures get hugely complicated and way beyond my paygrade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take Earth, for example.   If you move Earth 1/2% further from the Sun, so it gets about a 1% drop in solar energy due to the inerse square law, and using the effective temperature formula,the drop in temperature would be the 4th root of that, or about 0.25%.   But we know studying Earth's history that ice ages can be triggered by roughly the equivalent of a 1% drop in energy, leading to as much as 4 degrees globally, about 1.25% of Earth's actual temperature and that's because of various feedback mechanisms like greater oceanic absorption of CO2 and greater ice cover, lowering Earth's albedo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transition of gases from ice in the upper atmosphere to liquid would affect the equilibrium temperature.  It's no longer a neat equation, but a very complicated one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Titan gets very little heat from the sun.  At it's distance, it gets about 1/90th the heat that Earth receives per square meter, so if it did have a thick enough greenhouse gas atmosphere that trapped heat very effectively, it would still take a very long time for Titan to heat up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not doing any math, I'd say offhand that Titan would need a Venus pressure atmosphere to begin holding significantly more heat, because it has a pretty strong greenhouse gas atmosphere already, &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Titan&quot; rel=&quot;nofollow noreferrer&quot;&gt;being 5.6% methane&lt;/a&gt;.   You wouldn't need to bring CO2 to Titan though, you could simply heat up Titan's surface and initially some CH4 would boil off and if you got the atmosphere hot enough, some CO2 would begin to bubble up from it's frozen ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not sure how effectively, even a very dense greenhouse gas atmosphere would keep Titan warm, because it only receives 1/90th the sunlight that Earth receives per square meter, and that's a very small amount of heat going in.   I'm not sure where the warm point would be, but it would take a considerable amount of greenhouse atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe there's someone here who can do a calculation about surface temperature equilibrium under a thick atmosphere.  Mine is just a partial answer.&lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2019-07-19T23:06:14.420" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32729" PostTypeId="1" AcceptedAnswerId="32734" CreationDate="2019-07-19T23:16:42.093" Score="4" ViewCount="364" Body="&lt;p&gt;I'd like to write a short supplementary answer to &lt;a href=&quot;https://space.stackexchange.com/q/12293/12102&quot;&gt;A probe floating in Sun's surface&lt;/a&gt; but in order to do so I need to know at &lt;em&gt;roughly&lt;/em&gt; what depth relative to the Sun's surface the mass density reaches 1 g/cm^3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the bottom of the photosphere the density seems to be either 1E-06 or 1E-09 g/cm^3 as discussed in the question &lt;a href=&quot;https://astronomy.stackexchange.com/q/32727/7982&quot;&gt;What is the density profile within the Sun's photosphere?&lt;/a&gt; but how deep do you have to go below that to reach 1 g/cm^3?&lt;/p&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="7982" LastEditDate="2019-07-19T23:40:23.973" LastActivityDate="2019-07-20T01:52:31.157" Title="At what depth below the Sun's surface does the density reach that of water?" Tags="&lt;the-sun&gt;&lt;stellar-atmospheres&gt;&lt;photosphere&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32730" PostTypeId="1" CreationDate="2019-07-20T00:18:44.640" Score="1" ViewCount="109" Body="&lt;p&gt;I heard a theory (possibly from a crackpot) that there was an event in our solar system where two objects collided and created the asteroid belt, the orbits of some of the planets were altered, and that this event took place late enough for homo sapiens to have referenced observations made during the previous orbit in their oral traditions. (Mars parsing close enough to Earth for people to observe its moons)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is any of this possible given readily observable evidence?&#xA;Are asteroid belts ever created in this way?&lt;/p&gt;&#xA;" OwnerUserId="15605" LastEditorUserId="23748" LastEditDate="2019-07-20T00:41:42.470" LastActivityDate="2019-07-21T22:34:33.290" Title="How young could the asteroid belt be?" Tags="&lt;solar-system&gt;&lt;orbital-migration&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32731" PostTypeId="2" ParentId="20082" CreationDate="2019-07-20T00:32:30.537" Score="0" Body="&lt;p&gt;Supplementary graphical representation for the &lt;a href=&quot;https://astronomy.stackexchange.com/a/20086/7982&quot;&gt;accepted answer&lt;/a&gt;, from &lt;a href=&quot;https://twitter.com/MIT_CSAIL/status/1116007460039483392/photo/1&quot; rel=&quot;nofollow noreferrer&quot;&gt;tweet&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4ugoo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7982" LastActivityDate="2019-07-20T00:32:30.537" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32732" PostTypeId="2" ParentId="32727" CreationDate="2019-07-20T01:12:42.390" Score="0" Body="&lt;p&gt;The density is what you read from the graph, correctly.  Don't worry about what it says in that quote, it's all just a matter of what is meant by the &quot;photosphere&quot;, a term that is rather vaguely defined and used to mean different things in different places.  You can see the problem in the temperatures used in that quote-- they correspond to what the graph considers to be entirely above the photosphere.  The quote seems to think of the photosphere as the region from the tau ~ 2/3 point to the minimum in the temperature, whereas the graph seems to think of the photosphere as something quite noticeably hotter.  Other places regard the photosphere as a shell of zero width, right where tau ~ 2/3.  It's all just the different ways the word is used, there's nothing to worry about.  The graph matches up density with height and temperature, so you can just use that-- and note that even that is a kind of average situation, the reality is much more complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for positive and negative heights, why would you care what point is getting called x=0?  It's completely arbitrary where the zero height is set, every different source could likely use a different meaning for &quot;the top of the photosphere.&quot;&lt;/p&gt;&#xA;" OwnerUserId="14229" LastActivityDate="2019-07-20T01:12:42.390" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32734" PostTypeId="2" ParentId="32729" CreationDate="2019-07-20T01:22:59.387" Score="5" Body="&lt;p&gt;The sun's density is 1 gm/cc at approximately 50% of the way down towards the core. if radius of Sun is R then at R/2 the density will be that of water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/questions/32729/at-what-depth-below-the-suns-surface-does-the-density-reach-that-of-water/32733#comment60429_32729&quot;&gt;@astrosnapper's comment&lt;/a&gt; links to &lt;a href=&quot;https://physics.stackexchange.com/a/201239/83380&quot;&gt;this answer&lt;/a&gt; in Physics SE. A reverse image search of that unsourced graphic leads to &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Thesis: &lt;a href=&quot;https://www.mps.mpg.de/phd/theses/multi-spacecraft-analysis-of-the-solar-coronal-plasma&quot; rel=&quot;nofollow noreferrer&quot;&gt;Multi-spacecraft analysisof the solar coronal plasma&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;NASA: &lt;a href=&quot;https://solarscience.msfc.nasa.gov/interior.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Solar Interior&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://cellcode.us/quotes/density-graph-sun-profile.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Whatever this is&lt;/a&gt; shown below&lt;/li&gt;&#xA;&lt;li&gt;the link in that image &lt;a href=&quot;http://www.sns.ias.edu/~jnb/SNdata/Export/BP2004/bp2004stdmodel.dat&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.sns.ias.edu/~jnb/SNdata/Export/BP2004/bp2004stdmodel.dat&lt;/a&gt; &lt;strong&gt;provides some raw data&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;the reference in that image is standard solar model BP2004 J. N. Bahcall and M. H. Pinsonneault, Phys. Rev. Lett. 92 (2004) 121301 &lt;a href=&quot;https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.92.121301&quot; rel=&quot;nofollow noreferrer&quot;&gt;What do we (not) know theoretically about solar neutrino fluxes?&lt;/a&gt; (also in &lt;a href=&quot;https://arxiv.org/abs/astro-ph/0402114&quot; rel=&quot;nofollow noreferrer&quot;&gt;ArXiv&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NRIm8.png&quot; alt=&quot;standard solar model BP2004 J. N. Bahcall and M. H. Pinsonneault, Phys. Rev. Lett. 92 (2004) 121301&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16926" LastEditorUserId="7982" LastEditDate="2019-07-20T01:52:31.157" LastActivityDate="2019-07-20T01:52:31.157" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32735" PostTypeId="2" ParentId="32637" CreationDate="2019-07-20T02:28:16.813" Score="1" Body="&lt;p&gt;A planet's apparent position in the sky is determined by orbital geometry alone.&#xA;The tilt of its axis only affects its orientation as we see it in a telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easier to think of the solar system in terms of &lt;a href=&quot;https://en.wikipedia.org/wiki/Ecliptic_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;ecliptic coordinates&lt;/a&gt;.&#xA;The Sun's geocentric latitude is always near 0&amp;deg;, as is the Earth's heliocentric latitude.&#xA;A planet's orbital inclination is measured relative to the ecliptic, and its maximum excursion in latitude is related to that inclination.&#xA;Each planet crosses the ecliptic northward at its ascending node and southward at its descending node.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/saturnfact.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Saturn&lt;/a&gt;'s orbit has a semimajor axis of 9.6 au and is inclined 2.5&amp;deg; to the ecliptic.&#xA;Its heliocentric latitude runs that far north or south of the ecliptic over its 29.5 year period.&#xA;Earth's orbit brings us 1 au closer at opposition, so Saturn's geocentric ecliptic latitude can reach &amp;plusmn;2.8&amp;deg;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/mercuryfact.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mercury&lt;/a&gt;'s orbit has a semimajor axis of 0.4 au and is inclined 7&amp;deg;, so its heliocentric ecliptic latitude range is &amp;plusmn;7&amp;deg;.&#xA;At inferior conjunction ~0.6 au away, its geocentric ecliptic latitude range is about &amp;plusmn;5&amp;deg;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transformation from ecliptic to equatorial coordinates is a 3D rotation by the 23.4&amp;deg; obliquity, not a simple sum.&#xA;Also a planet's orbit and the Earth's axis may be tilted in different directions.&#xA;Saturn's ascending node is 114&amp;deg; longitude east of the J2000 vernal equinox; its orbital inclination and our axial tilt are that far out of phase.&#xA;A JPL HORIZONS ephemeris for 29 years of Saturn oppositions and conjunctions gives geocentric declinations in a range of &amp;plusmn;22.6&amp;deg;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturn's 26.7&amp;deg; axial tilt relative to its orbital plane does not affect its declination at all.&#xA;The effect we do see is that at different points in its orbit, its rings appear open to the north, open the south, or edge-on and slanted east or west, as &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Saturnoppositions.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;shown here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you mentioned the solar equator, it's inclined 7.2&amp;deg; to the ecliptic.&#xA;The planets do cross it, but their declinations are not related to it either.&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="616" LastEditDate="2019-07-20T04:05:03.633" LastActivityDate="2019-07-20T04:05:03.633" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32737" PostTypeId="2" ParentId="32727" CreationDate="2019-07-20T05:10:53.120" Score="1" Body="&lt;p&gt;I usually don't answer my own questions, but sometimes when the question itself is &lt;a href=&quot;https://astronomy.stackexchange.com/a/32732/7982&quot;&gt;called into question&lt;/a&gt; I make an exception.&lt;/p&gt;&#xA;&lt;p&gt;The density of the photosphere at &lt;span class=&quot;math-container&quot;&gt;$\tau_{5000}=1$&lt;/span&gt; is predicted to be &lt;span class=&quot;math-container&quot;&gt;$3 \times 10^{-7} \text{g/cm}^3$&lt;/span&gt; in the Holweger-Müller Model Atmosphere&lt;sup&gt;7&lt;/sup&gt;.&lt;/p&gt;&#xA;&lt;p&gt;As pointed out &lt;a href=&quot;https://astronomy.stackexchange.com/questions/32727/what-is-the-density-profile-within-the-suns-photosphere-which-one-of-these-is/32737?noredirect=1#comment60449_32737&quot;&gt;in comments&lt;/a&gt; there is a spread in values here.&lt;/p&gt;&#xA;&lt;p&gt;The 1E-06 g/cm^3 density value (plot) is more consistent with  = 1 or &amp;quot;bottom&amp;quot; of the photosphere, while the density in the quote is more consistent with the cooler &amp;quot;top&amp;quot; of the photosphere (circa 4300 K).&lt;/p&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;https://people.smp.uq.edu.au/TimoNieminen/papers/thesis/chapter2_photosphere.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chapter 2: The Photosphere&lt;/a&gt; of Timo Nieminen's thesis &lt;a href=&quot;https://people.smp.uq.edu.au/TimoNieminen/thesis.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solar Line Asymmetries: Modelling the Effect of Granulation on the Solar Spectrum&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JS8JR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JS8JR.png&quot; alt=&quot;density versus hight above the Sun's surface&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/s6zLZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/s6zLZ.png&quot; alt=&quot;density and other parameters versus hight above the Sun's surface&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Figure 2-3: The Holweger-Müller Model Atmosphere&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;7&lt;/sup&gt; Holweger, H. and Müller, E. A. “The Photospheric Barium Spectrum: Solar Abundance and Collision Broadening of Ba II Lines by Hydrogen”, Solar Physics 39, pg 19-30 (1974). Extra points have been cubic spline interpolated by J. E. Ross. The optical properties (such as the optical depth and the opacity) of a model atmosphere are, obviously, very important, and will be considered later. See table C-4 for complete details of the Holweger-Müller model atmosphere including all depth points used.&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;8&lt;/sup&gt;&lt;strong&gt;The height scale is not arbitrary.&lt;/strong&gt; The base of the photosphere (height = 0 km) is chosen to be at standard optical depth of one (i.e.  &lt;sub&gt;5000Å&lt;/sub&gt; = 1 ).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7982" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-11-14T05:13:54.977" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32738" PostTypeId="2" ParentId="17956" CreationDate="2019-07-20T08:08:47.160" Score="0" Body="&lt;p&gt;Logically moon has partially completed tidal locking and is  on its way to complete its tidal locking in future..ie will eventually have zero angular spin with respect to earth. In past, after moon just stopped spinning, it must have started to go back in reverse directiion and continued back and forth ever since, just like a pendulam having harmonic motion and continued till today. The amplitude of the harmonic motion must be reducing with time due to kinetic losses. When the amplitude finally becomes zero in distant future, at that time it will be in absolute tidal locking with earth.&lt;/p&gt;&#xA;" OwnerUserId="28788" LastActivityDate="2019-07-20T08:08:47.160" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32739" PostTypeId="2" ParentId="32730" CreationDate="2019-07-20T08:28:29.520" Score="-1" Body="&lt;p&gt;Some of this, as you say, is crackpot stuff, but there have, long ago, been some major collisions in the asteroid belt. The asteroid belt was formed about the same time as the Earth, and these collisions took place quite early on, billions of years before there were any humans. Mars never passed close enough for humans to observe its moons until telescopes were invented, but it does come within 35 million miles on closest approach. Only Venus comes closer at a distance of 26 million miles. The only asteroid belt we can study is our own, but common sense tells us that there must be other solar systems with asteroid belts, created in much the same way as ours. Some of the fragments from these asteroid belt collisions have been propelled Earthwards and left many huge craters, but only the biggest from the last couple of billion years have left identifiable traces. Those from earlier times, especially those that fell in deep ocean, have been eroded and left no visible trace.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-07-20T08:28:29.520" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32741" PostTypeId="2" ParentId="32726" CreationDate="2019-07-20T19:19:22.383" Score="1" Body="&lt;p&gt;It wouldn't look much different. The atmosphere absorbs some light and &quot;seeing&quot; reduces the effective resolution, but these have little effect on the visual impact of a lunar eclipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a sufficiently high orbit, you might be able to see the terminator between the partially lit &quot;red&quot; side of the moon and the (nearly) completely dark side on the far side from the Earth (and the sun).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lunar eclipse looks pretty cool from the Earth, and it would look equally cool from space.&lt;/p&gt;&#xA;" OwnerUserId="2365" LastActivityDate="2019-07-20T19:19:22.383" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32743" PostTypeId="1" AcceptedAnswerId="32744" CreationDate="2019-07-20T21:58:58.820" Score="9" ViewCount="1068" Body="&lt;p&gt;Would it be possible to put a satellite into the same orbit as the Moon, but far enough ahead or behind the Moon to remain in place?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has this ever been done in practice?&lt;/p&gt;&#xA;" OwnerUserId="28793" LastEditorUserId="1891" LastEditDate="2019-07-20T23:15:44.783" LastActivityDate="2019-07-21T13:54:34.147" Title="Satellite in orbit in front of and behind the Moon" Tags="&lt;the-moon&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32744" PostTypeId="2" ParentId="32743" CreationDate="2019-07-20T23:20:00.623" Score="11" Body="&lt;p&gt;Yes. &lt;a href=&quot;https://en.wikipedia.org/wiki/Trojan_(celestial_body)&quot; rel=&quot;noreferrer&quot;&gt;Trojan orbits&lt;/a&gt; (60° before or after the Moon) are even stable in most cases. These points are called the L₄ and L₅ points, which are two of the five &lt;a href=&quot;https://en.wikipedia.org/wiki/Lagrangian_point&quot; rel=&quot;noreferrer&quot;&gt;Lagrangian points&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These orbits are mostly impractical (nothing is there).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today the most known is the Chinese Queqiao satellite, which is a communication relay to their probe on the far side of the Moon. Instead of using L₄ or L₅, Queqiao is in a &lt;a href=&quot;https://en.wikipedia.org/wiki/Halo_orbit&quot; rel=&quot;noreferrer&quot;&gt;halo orbit&lt;/a&gt; around another Lagrangian point; Earth-Moon L₂.&lt;/p&gt;&#xA;" OwnerUserId="1891" LastEditorUserId="16685" LastEditDate="2019-07-21T13:54:34.147" LastActivityDate="2019-07-21T13:54:34.147" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32747" PostTypeId="2" ParentId="32743" CreationDate="2019-07-21T12:05:04.003" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Has this ever been done in practice?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not intentionally, and not by humankind. But by nature, the answer is yes. Thousands of objects are known to co-orbit with Jupiter about the Sun-Jupiter L4 and L5 points, millions are thought to exist. &lt;a href=&quot;https://en.wikipedia.org/wiki/Trojan_(celestial_body)&quot; rel=&quot;noreferrer&quot;&gt;Trojans&lt;/a&gt; (the generic name for small objects that co-orbit with a somewhat large object about an even larger object, with the small objects in pseudo orbits about the triangular Lagrange points) are known to exist for Neptune, Mars, Uranus, and Earth. Two of Saturn's moons, Tethys and Dione, have trojans in pseudo orbits about their triangular Lagrange points.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2019-07-21T12:05:04.003" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32748" PostTypeId="2" ParentId="32676" CreationDate="2019-07-21T17:17:25.070" Score="5" Body="&lt;p&gt;It's a big question, but kind of a favorite subject of mine, thinking about exoplanets, so I can give a ballpark answer, and I invite anyone to give correction or give a more technical answer if they like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ice(s) formation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An ammonia-water ocean wouldn't be friendly towards ice formation because water ice would sink in the ammonia-water solution and Ammonia ice would sink in liquid ammonia.  There's no range where the ice floats unless you remove nearly all of the ammonia and had it a high percentage of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, oceans are likely to have dissolved salts in them, especially on rocky worlds that have a very high rock to water ratio.    A water world might have much less salty oceans, but, lets not get too sidetracked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there's enough dissolved salts or iron in the water-ammonia mix, then the density might be sufficient that water-ice could remain on top.  Ice tends to form out of nearly pure water, with very little ocean salt, which increases it's buoyancy.   It's possible, with enough saltiness, that ice could form in an ammonia-water ocean.  It's also possible that ice would sink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Density&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to point this out now, because it's important.    Using our oceans as a model, higher salt concentration sinks.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.windows2universe.org/earth/Water/density.html&quot; rel=&quot;noreferrer&quot;&gt;Salt concentration by density&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xpMcp.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xpMcp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's kind of common sense and is likely to happen with a water-ammonia ocean as well.   Higher ammonia concentration near the surface, higher water (and higher salt) concentration as you go lower.  That's speculation on my part, but it seems reasonable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Varying ratios of water-ammonia based on temperature or other factors&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Higher water concentrations would sink and higher ammonia concentrations would rise and flow towards the lower.  That's simple physics.   Like on Earth, the variation of salinity of surface water isn't all that much.  The same thing would likely happen with an ammonia-water ocean, the surface currents would try to equalize the different ratios.  There would be some variation, but likely not as pronounced as the variation with depth.   You're not likely to get a 90% water ocean in one region and a 40-60 ratio in another region because gravity abhors an imbalance like that.   The 90% water region would be heavier and as a result, lower.  You can have that kind of density imbalance with a solid crust, but not a flowing ocean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The varying density of the ocean is important for another reason.   On Earth, the salinity being relatively consistent across the ocean surface, the cold polar ocean is more dense and it sinks, and this is what's thought to drive &lt;a href=&quot;https://oceanservice.noaa.gov/education/tutorial_currents/05conveyor2.html&quot; rel=&quot;noreferrer&quot;&gt;the oceanic conveyer.&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a range of uncertainty on whether an ammonia-water ocean would conveyer or not.   You'd need a sufficient temperature and/or density variation, keeping in mind that if you begin with a ratio imbalance the ocean would work to correct it, so the planet needs an engine to maintain the conveyer.   On Earth, that engine is the warm equator and cold poles.  Different planets would have different temperature variations.  Venus, for example, has practically no temperature variation at it's surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked, but couldn't find a table on ammonia-water solution by temperature.  That would be a factor if anyone can find it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Evaporation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ammonia in water has a &lt;a href=&quot;https://en.wikipedia.org/wiki/Ammonia_solution#Saturated_solutions&quot; rel=&quot;noreferrer&quot;&gt;saturation point&lt;/a&gt;, based on temperature and pressure.   If the ammonia-water solution exceeds that saturation point, say, on a hot day, then ammonia would bubble out of the ocean fairly quickly.  Perhaps quickly enough to generate a weather system and trigger a density change, and in this case, warmer water sinking and a kind of reverse conveyer, driven by sinking water at the planet's warmest regions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phase transition, liquid to gas would remove heat from the water as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A planet with an ammonia-water ocean and a surface temperature that exceeds the saturation point could be quite dynamic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the planet is slightly colder and/or the ammonia percentage stays below the saturation point, then it's more of an equilibrium system.   Water is a good solvent and it takes in many gases, such as CO2, in small amounts.    Ammonia and Water, unlike CO2 and Water, mix very happily, so evaporation of an ammonia-water ocean below saturation wouldn't be that different to water on Earth, where evaporation is driven primarily by wind and sunlight.  (contrary to logic temperature isn't as big factor in evaporation as direct sunlight).  Wind is also a big player and on Earth, running water from rain or snow melt and transpiration from plants all return water to the atmosphere, which quickly rains or snows back to the surface.   At any given time, Earth's percentage of water vapor in the atmosphere is about 1%, but it varies locally based on air temperature and relative humidity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ammonia would enter the atmosphere similar to water, but how long it would stay in the atmosphere is a factor.    Air temperature and 100% relative humidity are the upper limits for how much water vapor can exist in the atmosphere.  There are no such upper limits for Ammonia at Earthlike temperature and pressure because Ammonia, unlike water, is a gas at standard temperature and pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, a theoretical Earth-like planet with an ammonia-water ocean is also, very likely, going to have a lot of ammonia in the atmosphere, not the trace 1% of water that's in our atmosphere, but a lot more.   You can't have an ammonia ocean without ammonia in the atmosphere at well, in high concentrations.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, to have ammonia raindrops at 1 ATM, the air temperature needs to be colder than -33 C because that's ammonia's boiling/liquification temperature.   Unless the planet was really cold, a lot of the ammonia would just stay in the atmosphere, though, presumably the planet would have water-rain and that water rain would pull some of the ammonia back out and return it to the oceans.   That's the equilibrium, and it's a complicated thing to try to calculate, but the question isn't just how fast would it evaporate, but also, how fast would it return and the return rate would depend on the method of return.   Ammonia, being a gas at what we think of as normal planetary temperatures, means it would likely remain in the atmosphere as well, it wouldn't all go into the ocean.  It would be some of both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps over time, the ocean ratio and the rain-water/ammonia ratio would equalize, but again, that's just speculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the planet has very cold poles, where the ammonia could rain out of the atmosphere, or perhaps tidal locking, and a colder night-side of the planet, then the ammonia could return from the atmosphere more quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I'd like to know if this has been studied and if these behaviors have been described in papers, books, or in-depth articles on this topic.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know of any papers.  Theoretical planetary formation is fun to think about, but to do it right, requires some pretty serious computer modeling.   What I've read on the subject is much less ambitious than your question.   Models are done on water worlds and on tidally locked worlds and on plate tectonics for different mass planets.   I've never seen a study on ammonia-water oceans, but I think it's a fun subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earth had to have fairly abundant ammonia at some point because it's one of the primary &quot;ices&quot; in comets.   Comets are mostly ammonia, water, CO2 and CH4, with smaller amounts of other elements.  It's worth asking where Earth's ammonia went.  It may be more chemically reactive than the other elements.   If Earth lost it's ammonia, other planets may lose their ammonia as well in the process of planet formation and the chemistry that goes with it.  It's also possible that ammonia-water oceans are common, perhaps more common than the pure water oceans like the ones we have on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps time and better telescopes will tell us more about this subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take my answer with a grain of salt as I'm a hobbyist.  &lt;/p&gt;&#xA;" OwnerUserId="6841" LastActivityDate="2019-07-21T17:17:25.070" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32749" PostTypeId="2" ParentId="32730" CreationDate="2019-07-21T19:31:49.877" Score="3" Body="&lt;p&gt;Dating the age of the asteroid belt is a problem that has to date no general solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rocks are usually dated in the laboratory via &lt;a href=&quot;https://en.wikipedia.org/wiki/Mass_spectrometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;isotopical analysis&lt;/a&gt;. This is possible for meteorites, but what to do about those rocks that remain in space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several methods to date the asteroid belt are known to me, apart from ad-hoc assumptions that it has the same age as the planets:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Creating links between dated meteorites and asteroid families: The meteorite family of &lt;a href=&quot;https://en.wikipedia.org/wiki/L_chondrite&quot; rel=&quot;nofollow noreferrer&quot;&gt;L chondrites&lt;/a&gt; have possible origin families in the asteroid belt. Those 'families' are small &lt;a href=&quot;https://en.wikipedia.org/wiki/Asteroid_family&quot; rel=&quot;nofollow noreferrer&quot;&gt;blobs in the eccentricity-inclination&lt;/a&gt; diagrams, that probably originated in one common parent-body breakup event. Dating meteorites originating from those families, gives minimum ages for the parent bodies of i.e. 500 Myrs for the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ordovician_meteor_event&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ordovician meteor event.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Dynamical age modeling: After a breakup of a parent body, the cloud of fragments experience different strengths of &lt;a href=&quot;https://en.wikipedia.org/wiki/Yarkovsky%E2%80%93O%27Keefe%E2%80%93Radzievskii%E2%80%93Paddack_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;YORP&lt;/a&gt; and tidal forces, which causes their semimajor axis distribution to spread. Taking a given distribution, one can &lt;a href=&quot;https://www.sciencedirect.com/science/article/abs/pii/S0019103502968302&quot; rel=&quot;nofollow noreferrer&quot;&gt;calculate back&lt;/a&gt; to an age of when the cloud was one particle, i.e. the moment of breakup. This again is a minimum age and found for the Flora family to be more than &lt;span class=&quot;math-container&quot;&gt;$10^8$&lt;/span&gt;, but less than &lt;span class=&quot;math-container&quot;&gt;$10^9$&lt;/span&gt; years. This does exclude a very young belt, but does not say much about the upper age limit.&lt;/li&gt;&#xA;&lt;li&gt;Crater ages: For the largest bodies, i.e. Ceres, Vesta, etc. we can assign &lt;a href=&quot;https://en.wikipedia.org/wiki/Crater_counting&quot; rel=&quot;nofollow noreferrer&quot;&gt;relative ages&lt;/a&gt; to the lunar cataclysm. Those usually give ages in excess than 3-4 Gyrs. For bodies like Vesta, where the surface is saturated with craters, this age is a lower limit.&lt;/li&gt;&#xA;&lt;li&gt;Dating of crustal ejecta: There is another family of meteorites that can be linked specifically to Vesta via spectroscopy, called the &lt;a href=&quot;https://en.wikipedia.org/wiki/HED_meteorite&quot; rel=&quot;nofollow noreferrer&quot;&gt;HED meteorites&lt;/a&gt;. Those meteorites look like they're coming from a differentiated parent body (melted, low siderophile element content), and they containt &lt;a href=&quot;https://en.wikipedia.org/wiki/Calcium%E2%80%93aluminium-rich_inclusion&quot; rel=&quot;nofollow noreferrer&quot;&gt;CAIs&lt;/a&gt;. Thus, we can date Vesta at least back to the origin of the solar system, as CAIs give us the effective &lt;span class=&quot;math-container&quot;&gt;$t0$&lt;/span&gt; for the formation of the solar system.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditDate="2019-07-21T22:34:33.290" LastActivityDate="2019-07-21T22:34:33.290" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32750" PostTypeId="1" CreationDate="2019-07-21T23:55:25.867" Score="1" ViewCount="66" Body="&lt;p&gt;Does a nova, not a supernova, get hot enough that nickel is produced despite its formation by fusion being an endothermic reaction? If nickel is made, does it make it out of the star into the nebula?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I'm finding is this: &lt;a href=&quot;https://arxiv.org/pdf/0709.0041.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Spectral Analysis of Central Stars of Planetary Nebulae&lt;/a&gt; which is not a very satisfying read.&#xA;I'd expect some nickel synthesis even in the core of non-nova stars. there's some equilibrium constant for the reaction after all, but it does not appear to be much talked about.&lt;/p&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="2031" LastEditDate="2019-07-22T06:17:53.987" LastActivityDate="2019-07-22T06:17:53.987" Title="Does Nickel show up in spectra of Nova?" Tags="&lt;spectroscopy&gt;&lt;nebula&gt;&lt;nova&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32753" PostTypeId="1" CreationDate="2019-07-22T09:17:06.360" Score="4" ViewCount="218" Body="&lt;p&gt;Narrators of astrophysics documentaries constantly state that matter spirals into black holes as if it were intuitive that this would happen because of gravity. But the moon isn't spiralling into the earth. Newtonian physics suggest that an object would either fall past the event horizon in less than half a revolution, or it would orbit indefinitely until acted upon by a third party.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why would an orbit gradually degrade?&lt;/p&gt;&#xA;" OwnerUserId="15605" LastEditorUserId="15605" LastEditDate="2019-07-22T12:00:35.563" LastActivityDate="2019-07-24T03:40:03.883" Title="Why does matter spiral into blackholes?" Tags="&lt;black-hole&gt;&lt;orbit&gt;" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32754" PostTypeId="2" ParentId="32753" CreationDate="2019-07-22T09:38:27.683" Score="3" Body="&lt;p&gt;The moon isn't losing energy by its interaction with the rest of the solar system. On the contrary, it is gaining energy from its gravitational interaction with the Earth, while the Earth itself is gradually losing rotational energy and transferring it to the moon. The moon therefore is slowly moving away from us at the rate of several centimetres per year. The accretion disc orbiting a black hole, on the other hand, is losing energy through radiation caused by friction and compression, so the particles in the disc are gradually falling into the event horizon.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-07-22T09:38:27.683" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32755" PostTypeId="2" ParentId="32753" CreationDate="2019-07-22T10:56:58.393" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why would an orbit gradually degrade?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With planetary orbits being stable around a star for long times, one would think that the change of angular momentum (its usually angular momentum, not energy that you want to talk about for orbital decay) for celestial objects must be about zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact it is more the rule than the exception, that an objects orbital angular momentum changes with time. Early in their history, the giant planets in our solar system migrated quite significantly (&lt;a href=&quot;http://adsabs.harvard.edu/abs/2019A%26A...623A.169P&quot; rel=&quot;nofollow noreferrer&quot;&gt;for which there is more and more evidence&lt;/a&gt;), and as you stated yourself, the Moon also recedes from us. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a fundamental reason why this can happen, and it is the same reason why things must inspiral into a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First things first, I have to define a bit of math. A potential &lt;span class=&quot;math-container&quot;&gt;$\Phi(x,y,z)$&lt;/span&gt; is a scalar function, whose 3D-derivative gives a force vector &lt;span class=&quot;math-container&quot;&gt;$\vec F(x,y,z) = \nabla \Phi(x,y,z)$&lt;/span&gt;.&lt;br&gt;&#xA;If you're unfamiliar with this notation, then the key message here is: A potential contains essentially all information that a conservative (!) force does, and physicists use it all the time because handling one scalar is easier than handling three scalars (which is the force vector).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, without proof, I state a theorem from classical mechanics, namely that the only force laws that can produce &lt;strong&gt;closed orbits&lt;/strong&gt; are potentials of the form &lt;span class=&quot;math-container&quot;&gt;$\Phi\sim r^2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\Phi \sim 1/r$&lt;/span&gt;. The former is a Hooke spring-like law, the latter is a perfect two-body gravitational potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tides and perturbations from the planets are relatively weak contributions compared to central gravity. In the case of a black hole however, you pick up an additional strong-gravity potential term &lt;span class=&quot;math-container&quot;&gt;$\Phi_{\rm BH}\sim 1/r^3$&lt;/span&gt; which has a much stronger radial dependency than your regular gravity term, and is purely attractive.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of this, black holes posses a so called &lt;a href=&quot;https://en.wikipedia.org/wiki/Innermost_stable_circular_orbit&quot; rel=&quot;nofollow noreferrer&quot;&gt;innermost stable circular orbit (ISCO)&lt;/a&gt;, beyond which nothing can orbit the black hole without drifting radially inward. Because one can show that ISCO and the black hole's Schwarzschild radius are related via &lt;span class=&quot;math-container&quot;&gt;$r_{\rm ISCO} = 3 r_{\rm s}$&lt;/span&gt;, one will never encounter the ISCO in regular solar system dynamics.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastActivityDate="2019-07-22T10:56:58.393" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32757" PostTypeId="2" ParentId="32753" CreationDate="2019-07-23T08:07:51.597" Score="9" Body="&lt;p&gt;There are probably four different processes, any or all of which may be described as &quot;spiralling into a black hole&quot;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;The effects of general relativity.&lt;/strong&gt; as @AtmosphericPrisonEscape has indicated, once you get very close to a Black hole Newtonian physics ceases to be a good approximation and must be replaced by general relativity (GR). When you do the calculations, you find that all circular orbits within 3 Schwarzschild radii of the black hole are unstable. Any object orbiting closer than that will definitely spiral in.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Gravitational radiation&lt;/strong&gt; GR tells us that any two orbiting objects will emit gravitational waves, which carry away energy and angular momentum, slowly reducing the separation between the objects. This effect is, however, incredibly small, unless the objects are very massive and orbiting very close to one another (which means they must also be very dense). Thus this is mostly relevant for black holes and neutron stars in close orbits.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Turbulence in an accretion disk&lt;/strong&gt; When a lot of gas or dust (which may be the remnant of larger objects pulled apart by tides) is orbiting a black hole, it tends to form a disk, called an &quot;accretion disk&quot;. Basically the disk is oriented according to the average angular momentum of the stuff falling in, and anything orbiting in a different plane will collide with lots of the material in the disk and quickly join the disk. This is the same mechanism that caused the early proto solar system to form a disk, and indeed Saturn's rings (I know I'm oversimplifying somewhat).  Anyway, to remain in orbit, the inner parts of such a disk must be rotating faster than the outer parts, and this &quot;shear&quot; in the flow of gas tends to cause turbulence, which, it turns out, tends to transfer angular momentum outwards through the disk. This results in material at the inner edge of the disk getting slowly closer to the Black Hole until it reaches the Innermost Stable Circular orbit, at which point effect number 1 takes over.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Dynamical &quot;friction&quot;&lt;/strong&gt;  If there are a lot of larger bodies (stars, asteroids, etc.) orbiting a large central mass, they will, from time to time, pass near each other, and  change each others orbits, transferring angular momentum between them. It turns out that averaged over many encounters, the effect of this is to drive larger bodies inwards and smaller ones outwards (possibly ejecting the smaller bodies from the system completely) so that it acts as something like friction on the larger bodies. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="21928" LastActivityDate="2019-07-23T08:07:51.597" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32759" PostTypeId="1" AcceptedAnswerId="32770" CreationDate="2019-07-23T15:12:28.420" Score="4" ViewCount="137" Body="&lt;p&gt;We know that our universe is expanding i.e.  volume of the universe is getting larger with time with total energy being conserved. My question is whether the energy density is getting smaller with time.&#xA;If this is true then what is the effect of this phenomenon on all the celestial bodies?&lt;/p&gt;&#xA;" OwnerDisplayName="user28809" LastEditorDisplayName="user28809" LastEditDate="2019-07-23T15:23:15.357" LastActivityDate="2019-07-24T15:33:28.090" Title="What is the effect of this expanding universe on the energy density of it?" Tags="&lt;astrophysics&gt;&lt;universe&gt;&lt;cosmology&gt;&lt;general-relativity&gt;&lt;dark-matter&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32760" PostTypeId="1" AcceptedAnswerId="32761" CreationDate="2019-07-23T16:44:54.250" Score="2" ViewCount="168" Body="&lt;p&gt;I am fairly new to pulsar astronomy, and I need to download PSRCHIVE and then Tempo2. I have followed the instructions online here: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.pulsarastronomy.net/wiki/Software/PSRSoft#toc5&quot; rel=&quot;nofollow noreferrer&quot;&gt;PulsarAstronomy.net&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also pretty new to linux commands/Mac terminal in general, so I think I might be missing something obvious. &#xA;So far I have untarred the tar file, which gave me a directory 'psrsoft' in Downloads. &#xA;I also attempted to open a psrsoft bash profile, but I don't think it worked because none of the psrsoft commands worked. &#xA;The instructions mention '$PSRSOFT_DIR' a lot, and I am not sure where that is supposed to be or what it represents.&#xA;Right now, my goal is to get the package installation options listed and then from there install Tempo2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what psrsoft contains:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Downloads/psrsoft&gt;bin: pkg_files, psrsoft, psrsoft-selfupdate, psrsoft-shell, psrsoft-update-index, psrsoft.py, shell_files, usr&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Downloads/psrsoft&gt;config: profile.example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="28671" LastActivityDate="2019-07-23T17:08:39.740" Title="How do I download PSRCHIVE and Tempo2 for pulsar timing on Mac Mojave?" Tags="&lt;radio-astronomy&gt;&lt;software&gt;&lt;pulsar&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32761" PostTypeId="2" ParentId="32760" CreationDate="2019-07-23T17:08:39.740" Score="3" Body="&lt;p&gt;The documentation is somewhat confusingly written. It refers to a &lt;code&gt;$PSRSOFT_DIR&lt;/code&gt; as a shortcut to where you unpacked the tarball, which everything is done relative to but doesn't require this shell variable to be set. So there is 1 thing to do and then you can do 1 of 2 things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mandatory thing (as it says in the docs) is to copy the &lt;code&gt;psrsoft/config/profile.example&lt;/code&gt; to &lt;code&gt;psrsoft/config/profile&lt;/code&gt; and edit if necessary (I didn't need any changes but I'm running on pretty vanilla CentOS/RHEL 7; MacOS X may well be different). Then your two options are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You can then manually &quot;substitute in&quot; for the unset &lt;code&gt;$PSRSOFT_DIR&lt;/code&gt; variable e.g.  run &lt;code&gt;/path/to/Downloads/psrsoft/bin/psrsoft psrchive&lt;/code&gt; when the docs say to run &lt;code&gt;$PSRSOFT_DIR/bin/psrsoft psrchive&lt;/code&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can set &lt;code&gt;PSRSOFT_DIR&lt;/code&gt; in your user's shell startup files such as &lt;code&gt;~/.bashrc&lt;/code&gt; or &lt;code&gt;~/.bash_profile&lt;/code&gt; (&lt;code&gt;~&lt;/code&gt; is a shortcut to your home directory which works on Unix/Linux and I assume MacOS X) by adding &lt;code&gt;export PSRSOFT_DIR=/path/to/Downloads/psrsoft/&lt;/code&gt;. You should then be able to run the commands as shown in the documentation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="23748" LastActivityDate="2019-07-23T17:08:39.740" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32762" PostTypeId="2" ParentId="32759" CreationDate="2019-07-23T18:26:28.630" Score="-6" Body="&lt;p&gt;The experts say that as fast as the universe expands the increased space fills instantly with countless trillions of virtual particles which arrive from nowhere. Not only does this contravene the 1st Law of Thermodynamics (mass/energy can neither be created nor destroyed),but there is a discrepancy of 40 orders of magnitude (yes,40 orders of magnitude!) between the vacuum energy density required by Quantum Mechanics and cosmological observations. It has no effect on celestial bodies, though it should have, which is where the conflict with cosmological observations comes in. Supposedly, the 1st Law of Thermodynamics doesn't apply on a cosmological scale.&lt;/p&gt;&#xA;" OwnerUserId="28083" LastActivityDate="2019-07-23T18:26:28.630" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32764" PostTypeId="1" CreationDate="2019-07-23T22:59:38.787" Score="3" ViewCount="104" Body="&lt;p&gt;What's the difference between Supergalactic and Equatorial coordinates?&lt;/p&gt;&#xA;" OwnerUserId="28814" LastActivityDate="2019-07-24T08:51:07.400" Title="Supergalactic and Equatorial coordinates" Tags="&lt;milky-way&gt;&lt;galactic-dynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32765" PostTypeId="2" ParentId="28814" CreationDate="2019-07-24T01:17:31.210" Score="1" Body="&lt;h3&gt;Blazar sequence&lt;/h3&gt;&#xA;&lt;p&gt;Let's first look at which kind of sources are studied.&lt;/p&gt;&#xA;&lt;p&gt;Blazars are special type of radio-loud AGN. One of their special properties is that their spectral energy distributions (SED) form the so-called Blazar sequence. They all have a characteristic double peak SED and the position of the peaks mainly depend on their luminosity. The higher the luminosity the lower the peak frequencies (see Fossati et al., 1998 for details)&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MJuOV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MJuOV.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;(image source: Donato et al., 2001)&lt;/p&gt;&#xA;&lt;p&gt;Leaving details like sub-classes aside, this means that SED of Blazars can be modelled very well, especially if looking at a huge ensemble of Blazars.&lt;/p&gt;&#xA;&lt;h3&gt;Extra-galactic background light (EBL) absorption&lt;/h3&gt;&#xA;&lt;p&gt;Now the high energetic light has to travel from the Blazar to us. The Blazars being far away, much can happen during this travel, notably it can interact with the extra-galactic background light (EBL absorption). If a high energetic photon from the Blazar encounters an EBL photon in the optical or ultraviolet range they can pair produce an electron/positron pair (this is the dominant process for the involved energies).&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/k8mP6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k8mP6.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;These charged particles will most likely be subject to magnetic fields or other interactions diverting them from their direct way towards earth.&lt;/p&gt;&#xA;&lt;p&gt;This means not all of the high energy photons from the Blazar reach Earth and how large the absorption is depends on:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the distance between the Blazar and Earth&lt;/li&gt;&#xA;&lt;li&gt;the energy of the high energy photons from the Blazar&lt;/li&gt;&#xA;&lt;li&gt;the energy of the interaction partners from the EBL&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h3&gt;Bringing both together&lt;/h3&gt;&#xA;&lt;p&gt;Fermi looked at a large sample of SEDs from Blazars of various luminosities and grouped them based on their place in the Blazar sequence. For each group they could compare the expected SED from the Blazar model with their measured SED and thus determine how large the EBL absorption is depending on the energy. This allows to calculate the density and energy distribution of the EBL.&lt;/p&gt;&#xA;&lt;p&gt;Knowing the spectral energy distribution and density of the EBL allows to go even one step further and make assumptions on where the photos in the EBL came from. How many stars it took, how long they had to emit photons etc.&lt;/p&gt;&#xA;&lt;p&gt;As if this alone would not already be interesting enough, Fermi even repeated this for samples of Blazars at different distances, thus probing the EBL at different ages of the universe (remember that we observe the light that reaches the Earth now, thus the light from Blazars further away was emitted earlier than the light from &amp;quot;nearby&amp;quot; Blazars)&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f0Wx9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f0Wx9.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;(image source: &lt;a href=&quot;https://fermi.gsfc.nasa.gov/science/eteu/ebl/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://fermi.gsfc.nasa.gov/science/eteu/ebl/&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="950" LastEditorUserId="-1" LastEditDate="2020-06-17T09:47:40.653" LastActivityDate="2019-07-24T01:17:31.210" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32766" PostTypeId="1" AcceptedAnswerId="32774" CreationDate="2019-07-24T02:35:31.883" Score="3" ViewCount="205" Body="&lt;p&gt;As I was reading about the heliocentric model, a question came up: How was Nicolaus Copernicus able to figure out that the sun is at the center of the solar system, and that all planets orbit around it, even though the telescope had not been invented yet? During his time the geocentric model was the standard.&lt;/p&gt;&#xA;" OwnerUserId="28791" LastEditorUserId="16685" LastEditDate="2021-01-30T01:17:58.373" LastActivityDate="2021-01-30T22:51:25.013" Title="Nicolas Copernicus discovery" Tags="&lt;observational-astronomy&gt;&lt;history&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32767" PostTypeId="2" ParentId="32753" CreationDate="2019-07-24T03:40:03.883" Score="0" Body="&lt;p&gt;Any orbit requires velocity to maintain. If the moon's orbital velocity slowed its orbit it would come closer to Earth. If it were to speed up it would move further away. If the moon were to stop in it's orbit it would immediately fall like a giant stone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside a black hole a velocity greater than light is necessary just to maintain a stable orbit. Since such velocity is impossible then the only possibility is a decaying orbit (&quot;spiraling in&quot;) that inevitably leads to collision with the singularity.&lt;/p&gt;&#xA;" OwnerUserId="28600" LastActivityDate="2019-07-24T03:40:03.883" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32768" PostTypeId="2" ParentId="32766" CreationDate="2019-07-24T08:33:56.250" Score="4" Body="&lt;p&gt;Basically, because it was a more elegant solution to describe the motion of the planets. This could be see easily without any telescopes. The big difference was that it needed way less so called epicycles (he still needed those as all his orbits were circular not elliptic): &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One of the glaring mathematical problems with this model was that the planets, on occasion, would travel backward across the sky over several nights of observation. Astronomers called this retrograde motion. To account for it, the current model, based on the Greek astronomer and mathematician Ptolemy's view, incorporated a number of circles within circles — epicycles — inside of a planet's path. Some planets required as many as seven circles, creating a cumbersome model many felt was too complicated to have naturally occurred.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://www.space.com/15684-nicolaus-copernicus.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;-Space.com&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Besides that, the idea wasn't entirely new. Aristarchus proposed the idea of a moving earth in 250 BC and Copernicus was probably familiar with that. &lt;a href=&quot;https://www.jstor.org/stable/228080&quot; rel=&quot;nofollow noreferrer&quot;&gt;(Source)&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="28760" LastActivityDate="2019-07-24T08:33:56.250" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32769" PostTypeId="2" ParentId="32764" CreationDate="2019-07-24T08:51:07.400" Score="5" Body="&lt;p&gt;Equatorial coordinates have their equator and poles at the equator of the earth and the poles at the earth each projected onto the sky. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J0O1Z.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J0O1Z.gif&quot; alt=&quot;Equatorial Coordinates:&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supergalactic coordinates on the other hand have their equator in the same plane as the &quot;Supergalactic plane&quot;, which comes from the distribution of nearby galaxy clusters. And the poles just perpendicular to the the plane.You can kind of see this preferred plane in the image below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZZ1Q9.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZZ1Q9.gif&quot; alt=&quot;Credit: R. Powell&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is therefore just a rotated or tilted version of the other. To be precise the north poles of Supergalactic coordinates lies at approximately RA = 18.9h, Dec = +15.7° in Equatorial coordinates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also here &lt;a href=&quot;https://en.wikipedia.org/wiki/Celestial_coordinate_system&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; for a comparison of common celestial coordinate systems. &lt;/p&gt;&#xA;" OwnerUserId="28760" LastActivityDate="2019-07-24T08:51:07.400" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32770" PostTypeId="2" ParentId="32759" CreationDate="2019-07-24T09:21:43.750" Score="3" Body="&lt;p&gt;I am no expert and I am sure there will be a more rigorous answer later, but here my 2 cents:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears to me that the term &quot;energy&quot; and energy conservation becomes a bit more complicated in general relativity and cosmology, taking into account more abstract forms as the energy of the gravitational field or space curvature. As you seem to be more interested in the implications I just want to focus on the common forms of energy, i.e., radiation, matter and dark energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given expanding volume of space only the energy from matter (dark and normal) is dilutes as expected. You could say that with respect to expansion its energy is conserved. This means its energy density decreases with &lt;span class=&quot;math-container&quot;&gt;$\sim a(t)^{-3}$&lt;/span&gt;, where a is the scale factor (or just the diameter) of the universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radiation additionally looses energy via cosmological red-shift, so its energy density decreases even faster with  &lt;span class=&quot;math-container&quot;&gt;$\sim a(t)^{-4}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally you have the dark energy density. What makes it special is, that its density does not depend on &lt;span class=&quot;math-container&quot;&gt;$a(t)$&lt;/span&gt;. One can think of it as the energy associated with space itself. So if the universe expands new space and dark energy &quot;appears&quot; and its density stays constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So over time the average (matter and radiation) energy density decrease and approaches the energy of empty space itself (or vacuum energy). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, however, that this is only true for the average densities. In regions of space, that are held together by gravity, as our local group, the matter density will not decrease. On the contrary, these region are still contraction due to gravity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer your question on the effect on all the celestial bodies:&#xA;Those which are gravitationally bound, won't notice much but others will drift apart never to see each other again. &lt;/p&gt;&#xA;" OwnerUserId="28760" LastEditorUserId="28760" LastEditDate="2019-07-24T15:33:28.090" LastActivityDate="2019-07-24T15:33:28.090" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32771" PostTypeId="1" AcceptedAnswerId="32772" CreationDate="2019-07-24T12:31:11.540" Score="6" ViewCount="418" Body="&lt;p&gt;I'm wondering if there's a way to formulate the luminosity of a star not only as a function of mass, but of age, too, and if so, what the formula for luminosity would look like. In the case of the Sun I'm aware it gets 10% brighter every billion or so years, but I doubt this relationship holds true for all mass ranges of stars, just like you have to adjust the formula for the luminosity mass relationship for stars of different masses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone could &quot;shed some light&quot; on this matter, it would be very much appreciated. &lt;/p&gt;&#xA;" OwnerUserId="25978" LastEditorUserId="7982" LastEditDate="2019-07-25T00:00:11.887" LastActivityDate="2019-07-25T00:00:11.887" Title="Is there a way to factor age into the mass-luminosity relationship for stars?" Tags="&lt;star&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32772" PostTypeId="2" ParentId="32771" CreationDate="2019-07-24T15:32:29.533" Score="5" Body="&lt;p&gt;What you are looking seems to be the vertical path of a star in the Hertzsprung-Russel diagram (HRD). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only problem is that stellar evolution is pretty complicated. See here a few simulated trajectories for different masses and chemical compositions from &lt;a href=&quot;https://www.aanda.org/articles/aa/pdf/2008/24/aa9165-07.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt; Bertelli et al. 2008 &lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vp4NL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vp4NL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Especially before and after the main sequence (i.e., regular boring hydrogen burning) the luminosity evolution becomes extremely variable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the simulations above one can also derive respective age-luminosity relations (taken from  &lt;a href=&quot;http://%20file:///home/thalhammer/Downloads/Effect_of_Metallicity_on_the_Evolution_of_the_Habi.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Danchi 2013&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kjtsF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kjtsF.png&quot; alt=&quot;age-luminosity&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, you can see that the relation is very complicate for young or old stars and is fairly constant during the main sequence. The problem is, that it changes too little during most of the main sequence. So for most stars even with good mass and metallic measurement you won't be able to accurately estimate age just from luminosity. In addition to that, there are still uncertainties in our models. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing one could do is trying to improve the mass-luminosity-relation by including age. I think this is also what the title of your question implies. The problem here it simply that we generally don't know the age of a star. But if we do, as Rob Jeffries explained in his answer, it is in principle possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One think that is done to get age from luminosity is to go all in and calculate so called isochrones in the HRD. These are lines with stars of same age but varying masses and metallicities and can be derived from simulations. If one then measures the luminosity, temperature and metallic one can look on which isochrone the star falls in the HRD (and  therefore what age it has). This however is still pretty inaccurate, especially on the main sequence, and is mostly done with whole star cluster, where statistics makes things easier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is however not my field of expertise so I would be happy it an actual expert could chime in. :)&lt;/p&gt;&#xA;" OwnerUserId="28760" LastEditorUserId="28760" LastEditDate="2019-07-24T19:18:38.427" LastActivityDate="2019-07-24T19:18:38.427" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32773" PostTypeId="2" ParentId="32606" CreationDate="2019-07-24T15:51:22.490" Score="1" Body="&lt;p&gt;From &lt;a href=&quot;https://academic.oup.com/mnras/article/348/4/1282/973630&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vega stars types, observations studies&lt;/a&gt;, include star HD 23362 (Table I.) with a distance 1004 light years&lt;/p&gt;&#xA;" OwnerUserId="775" LastEditorUserId="7982" LastEditDate="2019-07-24T16:14:27.227" LastActivityDate="2019-07-24T16:14:27.227" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32774" PostTypeId="2" ParentId="32766" CreationDate="2019-07-24T18:45:33.753" Score="5" Body="&lt;p&gt;Copernicus did not figure out that the Sun &lt;em&gt;was&lt;/em&gt; at the centre of the solar system. He merely proposed that a heliocentric model &lt;em&gt;simplified the calculation of planetary orbits&lt;/em&gt;. Epicycles (circles within circles) were still needed in the Copernican model, for instance for the orbit of Mars, but there were fewer of them than in the Ptolemaic model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Copernican model received a boost when Kepler, discovering that planetary orbits were elliptical, removed the need for epicycles completely, thus increasing Copernicus' advantage over Ptolemy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first evidence that the Copernican model was a &lt;em&gt;true representation&lt;/em&gt; of actual reality came with the observation of stellar parallax in 1806. Until then, it was preferred only because it was a &lt;em&gt;simpler description&lt;/em&gt; - a much simpler one, after Kepler.&lt;/p&gt;&#xA;" OwnerUserId="12016" LastActivityDate="2019-07-24T18:45:33.753" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32775" PostTypeId="2" ParentId="32771" CreationDate="2019-07-24T19:04:15.850" Score="4" Body="&lt;p&gt;I basically agree with spacebread that it is complicated, but then so is the basic luminosity-mass relation if you start including stars that aren't on the main sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we do restrict ourselves to the main sequence, then you can see that mass is the dominant variable, with the luminosity perhaps increasing by about a factor of two 